1 Introduction
Let
${\mathcal H}$
be the class of analytic functions in
${\mathbb D}:= \{z \in {\mathbb C} : |z|<1 \}$
of the form
$$ \begin{align} f(z) = \sum_{n=1}^{\infty}a_{n}z^{n},\quad z\in{\mathbb D}, \end{align} $$
and
${\mathcal A}$
be the subclass of f normalised by
$f'(0)=1.$
Let
${\mathcal S}$
denote the subclass of
${\mathcal A}$
of univalent functions.
Let
$$ \begin{align} F_f(z):=\log\frac{f(z)}{z}=2\sum_{n=1}^{\infty}\gamma_n(f) z^n,\quad z\in{{\Bbb D}},\ \log 1:=0, \end{align} $$
be a logarithmic function associated with
$f\in {\mathcal S}.$
The numbers
$\gamma _n:=\gamma _n(f)$
are called the logarithmic coefficients of
$f.$
As is well known, the logarithmic coefficients play a crucial role in Milin’s conjecture (see [Reference Milin17]; see also [Reference Duren9, page 155]). It is surprising that for the class
${\mathcal S}$
sharp estimates of single logarithmic coefficients are known only for
$\gamma _1$
and
$\gamma _2,$
namely,
$$ \begin{align*} |\gamma_1|\le 1,\quad |\gamma_2|\le \frac12+\frac1e=0.635\dots \end{align*} $$
and are unknown for
$n\ge 3.$
Logarithmic coefficients have recently attracted considerable interest (see [Reference Ali and Vasudevarao2, Reference Ali, Vasudevarao, Thomas and Lecko3, Reference Cho, Kowalczyk, Kwon, Lecko and Sim6, Reference Girela10, Reference Kumar and Vasudevarao13, Reference Thomas20]).
For
$q,n \in {\mathbb N},$
the Hankel determinant
$H_{q,n}(f)$
of
$f\in {\mathcal A}$
of the form (1.1) is defined as
$$ \begin{align*} H_{q,n}(f) := \begin{vmatrix} a_{n} & a_{n+1} & \cdots & a_{n+q-1} \\ a_{n+1} & a_{n+2} & \cdots & a_{n+q} \\ \vdots & \vdots & \vdots & \vdots \\ a_{n+q-1} & a_{n+q} &\cdots & a_{n+2(q-1)} \end{vmatrix}. \end{align*} $$
Recently the second and the third Hankel determinants
$H_{2,2}(f)$
and
$H_{3,1}(f)$
have been examined for selected subclasses of
${\mathcal A}$
and particularly of
${\mathcal S}$
(see [Reference Cho, Kowalczyk, Kwon, Lecko and Sim5, Reference Kowalczyk, Lecko and Sim12] for further references).
Based on these ideas, we propose the study of the Hankel determinant
$H_{q,n}(F_f/2)$
whose entries are logarithmic coefficients of
$f\in {\mathcal S},$
that is,
$$ \begin{align*} H_{q,n}(F_f/2) = \begin{vmatrix} \gamma_{n} & \gamma_{n+1} & \cdots & \gamma_{n+q-1} \\ \gamma_{n+1} & \gamma_{n+2} & \cdots & \gamma_{n+q} \\ \vdots & \vdots & \vdots & \vdots \\ \gamma_{n+q-1} & \gamma_{n+q} &\cdots & \gamma_{n+2(q-1)} \end{vmatrix}. \end{align*} $$
The authors do not know of previous work on this topic. Due to the importance of logarithmic coefficients, the proposed problem seems reasonable and interesting. The results broaden the knowledge of logarithmic coefficients. In this paper we deal with
$H_{2,1}(F_f/2)=\gamma _1\gamma _3-\gamma _2^2$
when f is a convex or starlike univalent function. Note that
$H_{2,1}(F_f/2)$
corresponds to the well-known functional
$H_{2,1}(f)=a_3-a_2^2$
over the class
${\mathcal S}$
or its subclasses and is associated with the coefficient
$b_1$
in the class
$\Sigma $
and the area theorem (see [Reference Pommerenke18, pages 18–19]). For the class
$\cal S$
the functional
$H_{2,1}(f)$
was estimated in 1916 by Bieberbach (see [Reference Goodman11, Volume I, page 35]).
Differentiating (1.2) and using (1.1),
$$ \begin{align*} \gamma_{1}=\tfrac{1}{2}a_{2},\quad \gamma_{2}=\tfrac{1}{2}(a_{3}-\tfrac{1}{2}a_{2}^{2}),\quad \gamma_{3}=\tfrac{1}{2}(a_{4}-a_{2}a_{3}+\tfrac{1}{3}a_{2}^{3}). \end{align*} $$
Therefore,
$$ \begin{align} H_{2,1}(F_f/2)=\tfrac14(a_2a_4-a_3^2+\tfrac{1}{12}a_2^4). \end{align} $$
Note that when
$f\in {\mathcal S},$
then, for
$f_\theta (z):=e^{-i\theta }f(e^{\mathrm {i} \theta }z),\ \theta \in \Bbb R,$
$$ \begin{align} H_{2,1}(F_{f_\theta}/2)=\frac{e^{4i\theta}}{4}\bigg(a_2a_4-a_3^2+\frac{1}{12}a_2^4\bigg)=e^{4i\theta}H_{2,1}(F_f/2). \end{align} $$
The main goal of this paper is to find sharp upper bounds for
$H_{2,1}(F_f/2)$
when f belongs to the class of convex or starlike functions. Recall that a function
$f\in {\mathcal A}$
is called starlike if
$f\in \cal S$
and
$f({\Bbb D})$
is a starlike domain in
${\Bbb C}$
, that is,
$[0,w]\subset f({\Bbb D})$
for every
$w\in f({\Bbb D}).$
The class of starlike functions, usually denoted by
${\mathcal S}^*$
, was introduced by Alexander in [Reference Alexander1]. It is well known that starlike functions are characterised by the condition
$$ \begin{align} \operatorname{\mathrm{Re}}\frac{zf'(z)}{f(z)}>0,\quad z\in{\Bbb D} \end{align} $$
(see [Reference Alexander1], [Reference Duren9, page 41] and [Reference Lecko, Partyka, Krzyż, Lewandowski and Szapiels14]). Further, a function
$f\in {\mathcal A}$
is called convex if
$f\in \cal S$
and
$f({\Bbb D})$
is a convex domain in
${\Bbb C}$
, that is,
$[w_1,w_2]\subset f({\Bbb D})$
for every
$w_1,w_2\in f({\Bbb D}).$
The class of convex functions, denoted here by
${\mathcal S}^c$
, was introduced by Study in [Reference Study19], where the well-known analytic characterisation of
$f\in \cal S^c$
was given as
$$ \begin{align} \operatorname{\mathrm{Re}}\bigg\{1+\frac{zf''(z)}{f'(z)}\bigg\}>0,\quad z\in{\Bbb D}. \end{align} $$
By (1.5) and (1.6), both classes
$\cal S^c$
and
$\cal S^*$
have a representation using the Carathéodory class
${\cal P}$
, the class of analytic functions p in
${\Bbb D}$
of the form
$$ \begin{align} p(z) = 1 + \sum_{n=1}^{\infty}c_{n}z^{n}, \quad z\in{\mathbb D}, \end{align} $$
having a positive real part in
${\mathbb D},$
and so the coefficients of functions in
$\cal S^c$
and
$\cal S^*$
can be expressed by coefficients of functions in
${\cal P}.$
Therefore, to get the upper bound of
$H_{2,1}(F_f/2),$
we base our calculations on the well-known formula for the coefficient
$c_2$
(see [Reference Pommerenke18, page 166]) and the formula for
$c_3$
due to Libera and Zlotkiewicz [Reference Libera and Zlotkiewicz15, Reference Libera and Zlotkiewicz16]. For some further remarks related to extremal functions, see [Reference Cho, Kowalczyk and Lecko7].
Lemma 1.1. If
$p \in {{\cal P}}$
is of the form (1.5) with
$c_1\ge 0,$
then
$$ \begin{align} c_1=2\zeta_1, \end{align} $$
$$ \begin{align} c_2=2\zeta_1^2+2(1-\zeta_1^2)\zeta_2 \end{align} $$
and
$$ \begin{align} c_3=2\zeta_1^3+4(1-\zeta_1^2)\zeta_1\zeta_2-2(1-\zeta_1^2)\zeta_1\zeta_2^2+2(1-\zeta_1^2)(1-|\zeta_2|^2)\zeta_3 \end{align} $$
for some
$\zeta _1\in [0,1]$
and
$\zeta _2,\zeta _3\in \overline {{\Bbb D}}.$
For
$\zeta _1\in \Bbb T,$
there is a unique function
$p\in {\cal P}$
with
$c_1$
as in (1.8), namely,
$$ \begin{align*} p(z)=\frac{1+\zeta_1z}{1-\zeta_1z},\quad z\in{\Bbb D}. \end{align*} $$
For
$\zeta _1\in {\Bbb D}$
and
$\zeta _2\in \Bbb T,$
there is a unique function
$p\in {\cal P}$
with
$c_1$
and
$c_2$
as in (1.8)–(1.9), namely,
$$ \begin{align*} p(z)=\frac{1+(\overline{\zeta_1}\zeta_2+\zeta_1)z+\zeta_2z^2}{1+(\overline{\zeta_1}\zeta_2-\zeta_1)z-\zeta_2z^2},\quad z\in{\Bbb D}. \end{align*} $$
For
$\zeta _1,\zeta _2\in {\Bbb D}$
and
$\zeta _3\in \Bbb T,$
there is a unique function
$p\in {\cal P}$
with
$c_1$
,
$c_2$
and
$c_3$
as in (1.8)–(1.10), namely,
$$ \begin{align} p(z)=\frac{1+(\overline{\zeta_2}\zeta_3+\overline{\zeta_1}\zeta_2+\zeta_1)z+(\overline{\zeta_1}\zeta_3+\zeta_1\overline{\zeta_2}\zeta_3+\zeta_2)z^2+\zeta_3z^3} {1+(\overline{\zeta_2}\zeta_3+\overline{\zeta_1}\zeta_2-\zeta_1)z+(\overline{\zeta_1}\zeta_3-\zeta_1\overline{\zeta_2}\zeta_3-\zeta_2)z^2-\zeta_3z^3},\quad z\in{\Bbb D}. \end{align} $$
We will also apply the following lemma. Here
$\overline {{\mathbb D}}:=\{z\in {\mathbb C}:|z|<1\}.$
Lemma 1.2. [Reference Choi, Kim and Sugawa8]
Given real numbers
$A,\ B, C,$
let
$$ \begin{align*} Y(A,B,C) := \max\{|A+Bz+Cz^2|+1-|z|^2: z\in\overline{\mathbb{D}}\}. \end{align*} $$
I. If
$AC\ge 0,$
then
$$ \begin{align*} Y(A,B,C)=\left\{ \begin{array}{@{}ll} |A|+|B|+|C|, & |B|\ge 2(1-|C|),\\ 1+|A|+\dfrac{B^2}{4(1-|C|)}, & |B|<2(1-|C|). \end{array} \right.\end{align*} $$
II. If
$AC<0,$
then
$$ \begin{align*} Y(A,B,C) =\left\{\begin{array}{@{}lll} 1-|A|+\dfrac{B^2}{4(1-|C|)}, &\quad -4AC(C^{-2}-1)\le B^2 \wedge |B|<2(1-|C|), \\ 1+|A|+\dfrac{B^2}{4(1+|C|)}, &\quad B^2<\min\{4(1+|C|)^2,-4AC(C^{-2}-1)\}, \\ R(A,B,C), &\quad {otherwise}, \end{array} \right. \end{align*} $$
where
$$ \begin{align*} R(A,B,C):=\left\{ \begin{array}{@{}lll} |A|+|B|-|C|, &\quad |C|(|B|+4|A|)\le |AB|, \\ -|A|+|B|+|C|, &\quad |AB|\le |C|(|B|-4|A|), \\ (|C|+|A|)\sqrt{1-\dfrac{B^2}{4AC}}, &\quad {otherwise}. \end{array} \right. \end{align*} $$
2 Convex functions
Note that using (1.6) it is easy to show that
$|\gamma _n(f)|\le 1/(2n)$
for
$n\in {\Bbb N}$
, when
$f\in {\cal S}^c,$
and the inequality is sharp. Now we deal with
$H_{2,1}(F_f/2)$
in the class
$\cal S^c.$
Theorem 2.1. If
$f\in {\mathcal S}^c,$
then
$$ \begin{align} |\gamma_{1} \gamma_{3} - \gamma_{2}^{2} | \le \tfrac{1}{33}. \end{align} $$
The inequality is sharp.
Proof. Let
$f \in \cal S^c$
be of the form (1.1). Then, by (1.6),
$$ \begin{align} f'(z)+zf''(z)=p(z)f'(z),\quad z\in{\Bbb D}, \end{align} $$
for some
$p \in {\cal P}$
of the form (1.7). Since the class
${\cal P}$
is invariant under rotation and (1.4) holds, we may assume that
$c_1 \in [0,2]$
(see [Reference Carathéodory4]; see also [Reference Goodman11, Volume I, page 80, Theorem 3]), that is, in view of (1.8), that
$\zeta _1\in [0,1].$
Substituting the series (1.1) and (1.7) into (2.2) and equating the corresponding coefficients,
$$ \begin{align*} a_2= \tfrac12c_1, \quad a_3=\tfrac16(c_2+c_1^2),\quad a_4 = \tfrac{1}{24}(2c_3+3c_1c_2+c_1^3). \end{align*} $$
Hence, by (1.3),
$$ \begin{align*} H_{2,1}(F_f/2)=\gamma_1\gamma_3-\gamma_2^2=\tfrac{1}{2304}(24c_1c_3+4c_1^2c_2-16c_2^2-c_1^4). \end{align*} $$
Now, using (1.8)–(1.10), by straightforward algebraic computation,
$$ \begin{align} \gamma_1\gamma_3-\gamma_2^2=\tfrac{1}{144}[3\zeta_1^4+6(1-\zeta_1^2)\zeta_1^2\zeta_2-2(1-\zeta_1^2)(2+\zeta_1^2)\zeta_2^2 +6(1-\zeta_1^2)(1-|\zeta_2|^2)\zeta_1\zeta_3]. \end{align} $$
A. Suppose that
$\zeta _1=1.$
Then, by (2.3),
$$ \begin{align*} |\gamma_1\gamma_3-\gamma_2^2|=\tfrac{1}{48}. \end{align*} $$
B. Suppose that
$\zeta _1=0.$
Then, by (2.3),
$$ \begin{align*} |\gamma_1\gamma_3-\gamma_2^2|=\tfrac{1}{36}|\zeta_2|^2\le\tfrac{1}{36}. \end{align*} $$
C. Suppose that
$\zeta _1\in (0,1).$
Since
$|\zeta _3|\le 1$
, from (2.3),
$$ \begin{align} |\gamma_1\gamma_3-\gamma_2^2|&\le \tfrac{1}{144}[|3\zeta_1^4+6(1-\zeta_1^2)\zeta_1^2\zeta_2-2(1-\zeta_1^2)(2+\zeta_1^2)\zeta_2^2| +6(1-\zeta_1^2)(1-|\zeta_2|^2)\zeta_1] \notag\\ &=\tfrac{1}{24}\zeta_1(1-\zeta_1^2)[|A+B\zeta_2+C\zeta_2^2|+1-|\zeta_2|^2], \end{align} $$
where
$$ \begin{align*} A:=\frac{\zeta_1^3}{2(1-\zeta_1^2)},\quad B:=\zeta_1,\quad C:=-\frac{2+\zeta_1^2}{3\zeta_1}. \end{align*} $$
Since
$AC<0,$
we apply Lemma 1.2 only for the case II.
C1. Note that the inequality
$$ \begin{align*} -4AC\bigg(\frac{1}{C^2}-1\bigg)-B^2=\frac{2(2+\zeta_1^2)\zeta_1^2}{3(1-\zeta_1^2)}\bigg(\frac{9\zeta_1^2}{(2+\zeta_1^2)^2}-1\bigg)-\zeta_1^2\le 0 \end{align*} $$
is equivalent to
$-(1-\zeta _1^2)(14+\zeta _1^2)\le 0,$
which is true for
$\zeta _1\in (0,1).$
Moreover, the inequality
$|B|<2(1-|C|)$
is equivalent to
$5\zeta _1^2-6\zeta _1+4<0,$
which is false for
$\zeta _1\in (0,1).$
C2. Since
$$ \begin{align*}4(1+|C|)^2=4\frac{(\zeta_1^2+3\zeta_1+2)^2}{9\zeta_1^2}>0,\quad -4AC\bigg(\frac{1}{C^2}-1\bigg)=-\frac{2\zeta_1^2(4-\zeta_1^2)}{3(2+\zeta_1^2)}<0,\end{align*} $$
we see that the inequality
$$ \begin{align*}\zeta_1^2=B^2<\min\bigg\{4(1+|C|)^2,-4AC\bigg(\frac{1}{C^2}-1\bigg)\bigg\}=-\frac{2\zeta_1^2(4-\zeta_1^2)}{3(2+\zeta_1^2)}\end{align*} $$
is false for
$\zeta _1\in (0,1).$
C3. Observe that the inequality
$$ \begin{align*} |C|(|B|+4|A|)-|AB|=\frac{2+\zeta_1^2}{3\zeta_1}\bigg(\zeta_1+\frac{2\zeta_1^3}{1-\zeta_1^2}\bigg)-\frac{\zeta_1^4}{2(1-\zeta_1^2)}\le 0\end{align*} $$
is equivalent to
$\zeta _1^4-6\zeta _1^2-4\ge 0,$
which is false for
$\zeta _1\in (0,1).$
C4. Note that the inequality
$$ \begin{align*} |AB|-|C|(|B|-4|A|)=\frac{\zeta_1^4}{2(1-\zeta_1^2)}-\frac{2+\zeta_1^2}{3\zeta_1}\bigg(\zeta_1-\frac{2\zeta_1^3}{1-\zeta_1^2}\bigg)\le 0\end{align*} $$
is equivalent to
$9\zeta _1^4+10\zeta _1^2-4\le 0,$
which is true for
$$ \begin{align*} 0< \zeta_1\le \zeta':= \tfrac{1}{3}{\textstyle \sqrt{\sqrt{61}-5}}\approx 0.5588. \end{align*} $$
$$ \begin{align*} |\gamma_1\gamma_3-\gamma_2^2|\le \tfrac{1}{24}\zeta_1(1-\zeta_1^2)(-|A|+|B|+|C|)=\varphi(\zeta_1)\le \varphi\Big(\sqrt{{2}/{11}}\Big)=\tfrac{1}{33}, \end{align*} $$
where
$$ \begin{align*} \varphi(t):=\tfrac{1}{144}(-11t^4+4t^2+4),\quad 0\le t\le\zeta_1'. \end{align*} $$
Note that
$\sqrt {2/11}\approx 0.4264<\zeta '.$
C5. It remains to consider the last case in Lemma 1.2, which, taking into account C4, holds for
$\zeta '<\zeta _1<1.$
Then, by (2.4),
$$ \begin{align} |\gamma_1\gamma_3-\gamma_2^2|&\le \frac{1}{24}\zeta_1(1-\zeta_1^2)(|C|+|A|) {\textstyle \sqrt{1-\dfrac{B^2}{4AC}}}=\psi(\zeta_1) \notag\\ &\le \psi(\zeta')=\frac{125-7\sqrt{61}}{5832} \sqrt{\frac{35-3\sqrt{61}}{2}}\approx 0.029, \end{align} $$
where
$$ \begin{align*} \psi(t):=\frac{1}{144}(t^4-2t^2+4) \sqrt{\frac{7-t^2}{2(2+t^2)}},\quad \zeta'\le t\le 1. \end{align*} $$
Indeed, to see that the last inequality in (2.5) is true, observe that
$$ \begin{align*} \psi'(t)=\frac{t}{288}\sqrt{\frac{2(2+t^2)}{7-t^2}}\cdot\frac{-4t^6+15t^4+54t^2-92}{(2+t^2)^2}=0 \end{align*} $$
if and only if
$-4t^6+15t^4+54t^2-92=0.$
However, all real roots of the last equation are
$t_{1,2}\approx \pm 1.1714$
and
$t_{3,4}\approx \pm 2.335.$
Therefore,
$\psi '(t)<0$
for
$\zeta _1'<t<1,$
so
$\psi $
is decreasing and hence
$\psi (t)\le \psi (\zeta ')$
for
$\zeta '\le t\le 1.$
D. Summarising, the inequality (2.1) follows from parts A–C. Equality holds for the function
$f\in {\mathcal A}$
given by (2.2), where the function
$p\in {\cal P}$
is of the form (1.11) with
$\zeta _1=\sqrt {2/11}$
,
$\zeta _2=-1$
and
$\zeta _3=1,$
that is,
$$ \begin{align*} p(z)=\frac{1-z^2}{1-2\sqrt{{2}/{11}}z+z^2},\quad z\in{\Bbb D}.\\[-42pt] \end{align*} $$
3 Starlike functions
Using (1.5), it is easy to show that
$|\gamma _n(f)|\le 1/n$
for
$n\in \Bbb N,$
when
$f\in {\mathcal S}^*,$
and that the inequality is sharp. We discuss now
$H_{2,1}(F_f/2)$
for the class
${\mathcal S}^*.$
Theorem 3.1. If
$f\in {\mathcal S}^*,$
then
$$ \begin{align} |\gamma_1\gamma_3-\gamma_2^2|\le \tfrac{1}{4}. \end{align} $$
The inequality is sharp.
Proof. Let
$f \in \cal S^*$
be of the form (1.1). Then, by (1.5),
$$ \begin{align} zf'(z)=p(z)f(z),\quad z\in{\Bbb D}, \end{align} $$
for some
$p \in {\cal P}$
of the form (1.7). As in the proof of Theorem 2.1, we may assume that
$c_1 \in [0,2],$
that is, in view of (1.8), that
$\zeta _1\in [0,1].$
Substituting the series (1.1) and (1.7) into (3.2) and equating the corresponding coefficients,
$$ \begin{align*} a_2=c_1, \quad a_3=\tfrac12(c_2+c_1^2),\quad a_4 = \tfrac{1}{6}(2c_3+3c_1c_2+c_1^3). \end{align*} $$
Hence, by (1.3),
$$ \begin{align*} H_{2,1}(F_f/2)=\gamma_1\gamma_3-\gamma_2^2=\tfrac{1}{48}(4c_1c_3-3c_2^2). \end{align*} $$
$$ \begin{align} \gamma_1\gamma_3-\gamma_2^2 =\tfrac{1}{12}[\zeta_1^4+2(1-\zeta_1^2)\zeta_1^2\zeta_2-(1-\zeta_1^2)(3+\zeta_1^2)\zeta_2^2+4(1-\zeta_1^2)(1-|\zeta_2|^2)\zeta_1\zeta_3]. \end{align} $$
A. Suppose that
$\zeta _1=1.$
Then, by (3.3),
$$ \begin{align*} |\gamma_1\gamma_3-\gamma_2^2|=\tfrac{1}{12}. \end{align*} $$
B. Suppose that
$\zeta _1=0.$
Then, by (3.3),
$$ \begin{align*} |\gamma_1\gamma_3-\gamma_2^2|=\tfrac{1}{4}|\zeta_2|^2\le\tfrac{1}{4}. \end{align*} $$
C. Suppose that
$\zeta _1\in (0,1).$
Since
$|\zeta _3|\le 1$
, from (3.3),
$$ \begin{align*} |\gamma_1\gamma_3-\gamma_2^2| &\le \tfrac{1}{12}[|\zeta_1^4+2(1-\zeta_1^2)\zeta_1^2\zeta_2-(1-\zeta_1^2)(3+\zeta_1^2)\zeta_2^2|+4(1-\zeta_1^2)(1-|\zeta_2|^2)\zeta_1]\\ &=\tfrac{1}{3}\zeta_1(1-\zeta_1^2)[|A+B\zeta_2+C\zeta_2^2|+1-|\zeta_2|^2], \end{align*} $$
where
$$ \begin{align*} A:=\frac{\zeta_1^3}{4(1-\zeta_1^2)},\quad B:=\frac12\zeta_1,\quad C:=-\frac{3+\zeta_1^2}{4\zeta_1}. \end{align*} $$
Since
$AC<0,$
we apply Lemma 1.2 only for the case II.
C1. Note that the inequality
$$ \begin{align*} -4AC\bigg(\frac{1}{C^2}-1\bigg)-B^2=\frac{(3+\zeta_1^2)\zeta_1^2}{4(1-\zeta_1^2)}\bigg(\frac{16\zeta_1^2}{(3+\zeta_1^2)^2}-1\bigg) -\frac14\zeta_1^2\le 0 \end{align*} $$
is equivalent to
$-(9-\zeta _1^2)\le 3+\zeta _1^2$
, which evidently holds for
$\zeta _1\in (0,1).$
Moreover, the inequality
$|B|<2(1-|C|)$
is equivalent to
$2\zeta _1^2-4\zeta _1+3<0,$
which is false for
$\zeta _1\in (0,1).$
C2. Since
$$ \begin{align*} 4(1+|C|)^2=4\frac{(\zeta_1^2+4\zeta_1+3)^2}{16\zeta_1^2}>0,\quad -4AC\bigg(\frac{1}{C^2}-1\bigg)=-\frac{\zeta_1^2(9-\zeta_1^2)}{4(3+\zeta_1^2)}<0, \end{align*} $$
we see that the inequality
$$ \begin{align*} \frac14\zeta_1^2=B^2<\min\bigg\{4(1+|C|)^2,-4AC\bigg(\frac{1}{C^2}-1\bigg)\bigg\}=-\frac{\zeta_1^2(9-\zeta_1^2)}{4(3+\zeta_1^2)} \end{align*} $$
is false for
$\zeta _1\in (0,1).$
C3. Observe that the inequality
$$ \begin{align*}|C|(|B|+4|A|)-|AB|=\frac{3+\zeta_1^2}{4\zeta_1}\bigg(\frac12\zeta_1+\frac{\zeta_1^3}{1-\zeta_1^2}\bigg)-\frac{\zeta_1^4}{8(1-\zeta_1^2)}\le 0\end{align*} $$
is equivalent to
$3+4\zeta _1^2\le 0,$
which is false for
$\zeta _1\in (0,1).$
C4. Note that the inequality
$$ \begin{align*} |AB|-|C|(|B|-4|A|)=\frac{\zeta_1^4}{8(1-\zeta_1^2)}-\frac{3+\zeta_1^2}{4\zeta_1}\bigg(\frac12\zeta_1- \frac{\zeta_1^3}{1-\zeta_1^2}\bigg)\le 0\end{align*} $$
is equivalent to
$4\zeta _1^4+8\zeta _1^2-3\le 0,$
which is true for
$$ \begin{align*}0< \zeta_1\le \zeta'':={\textstyle \sqrt{\sqrt{7}/2-1}}\approx 0.5682.\end{align*} $$
$$ \begin{align*} |\gamma_1\gamma_3-\gamma_2^2|\le \tfrac{1}{3}\zeta_1(1-\zeta_1^2))(-|A|+|B|+|C|)=\tfrac{1}{12}(-4\zeta_1^4+3)\le \tfrac14 \end{align*} $$
for
$0<\zeta _1\le \zeta ''.$
C5. It remains to consider the last case in Lemma 1.2, which, taking into account C4, holds for
$\zeta ''<\zeta _1<1.$
Then, by (3.3),
$$ \begin{align} |\gamma_1\gamma_3-\gamma_2^2|&\le \frac{1}{3}\zeta_1(1-\zeta_1^2)(|C|+|A|){\textstyle\sqrt{1-\dfrac{B^2}{4AC}}}=\psi(\zeta_1) \notag\\ &\le \psi(\zeta'')=\frac{5-\sqrt{7}}{3{\textstyle\sqrt{8+2\sqrt{7}}}}\approx 0.21525, \end{align} $$
where
$$ \begin{align*} \psi(t):=\frac{3-2t^2}{6\sqrt{3+t^2}},\quad \zeta''\le t\le 1. \end{align*} $$
Indeed, to see that the last inequality in (3.4) is true, observe that since
$$ \begin{align*} \psi'(t)=-\frac{15t+2t^3}{6(3+t^2)^{3/2}}<0,\quad \zeta''< t<1, \end{align*} $$
the function
$\psi $
is decreasing and therefore
$\psi (t)\le \psi (\zeta '')$
for
$\zeta ''\le t\le 1.$
Summarising, the inequality (3.1) follows from parts A–C. Equality holds for the function
$f\in {\mathcal A}$
given by
$$ \begin{align*} \frac{zf'(z)}{f(z)}=\frac{1+z^2}{1-z^2},\quad z\in{\Bbb D}, \end{align*} $$
for which
$a_2=a_4=0$
and
$a_3=1.$
