1. Introduction
Let
$\mathcal {H}$
denote the class of analytic functions in the unit disk
$\mathbb {D}:=\{z\in \mathbb {C}:\, |z|<1\}$
. Here
$\mathcal {H}$
is a locally convex topological vector space endowed with the topology of uniform convergence over compact subsets of
$\mathbb {D}$
. Let
$\mathcal {A}$
denote the class of functions
$f\in \mathcal {H}$
such that
$f(0)=0$
and
$f'(0)=1$
. Let
$\mathcal {S}$
denote the subclass of
$\mathcal {A}$
consisting of functions which are univalent (that is, one-to-one) in
$\mathbb {D}$
. If
$f\in \mathcal {A}$
, then it has the series representation
$$ \begin{align} f(z)= z+\sum_{n=2}^{\infty}a_n z^n, \quad z\in \mathbb{D}. \end{align} $$
For
$q,n \in \mathbb {N}$
, the Hankel determinant
$H_{q,n}(f)$
of the Taylor coefficients of the function
$f \in \mathcal {A}$
of the form (1.1) is
$$ \begin{align*} H_{q,n}(f) = \begin{vmatrix} a_n & a_{n+1} & \cdots & a_{n+q-1} \\ a_{n+1} & a_{n+2} & \cdots & a_{n+q} \\ \vdots & \vdots & \ddots &\vdots \\ a_{n+q-1} & a_{n+q} & \cdots & a_{n+2(q-1)} \end{vmatrix}. \end{align*} $$
Hankel determinants of various orders have been studied in many contexts (see for instance [Reference Allu, Lecko and Thomas5]). The Fekete–Szegö functional is the second Hankel determinant
$H_{2,1}(f)$
. Fekete–Szegö obtained estimates for
$|a_3 - \mu a_2 ^2|$
with
$\mu $
real (see [Reference Duren10, Theorem 3.8]).
Let g be the inverse function of
$f\in \mathcal {S}$
defined in a neighbourhood of the origin with the Taylor series expansion
$$ \begin{align} g(w)=f^{-1}(w)=w+\sum_{n=2}^{\infty}A_n w^n, \end{align} $$
where we may choose
$|w| < 1/4$
from Koebe’s
$1/4$
-theorem. Using variational methods, Löwner [Reference Löwner16] obtained the sharp estimate
$$ \begin{align*} |A_n| \leq K_n \quad \text{for each}\ n \in \mathbb{N},\end{align*} $$
where
$K_n = (2n)!/(n!(n + 1)!)$
and
$K(w) = w + K_2w^2 + K_3w^3 + \cdots $
is the inverse of the Koebe function. Let
$f(z)= z+\sum _{n=2}^{\infty }a_n z^n $
be a function in class
$\mathcal {S}$
. Since
$f(f^{-1})(w)=w$
, it follows from (1.2) that
$$ \begin{align*} & A_{2}=-a_2,\\ & A_{3}=-a_3+2a_2^2,\\ & A_{4}=-a_4+5a_2a_3-5a_2^3. \end{align*} $$
The logarithmic coefficients
$\gamma _{n}$
of
$f\in \mathcal {S}$
are defined by
$$ \begin{align} F_{f}(z):= \log\frac{f(z)}{z}=2\sum\limits_{n=1}^{\infty}\gamma_{n}z^{n}, \quad z \in \mathbb{D}. \end{align} $$
Few exact upper bounds for
$\gamma _{n}$
have been established. The significance of this problem in the context of the Bieberbach conjecture was pointed out by Milin [Reference Milin17]. Milin’s conjecture that for
$f\in \mathcal {S}$
and
$n\ge 2$
,
$$ \begin{align*}\sum\limits_{m=1}^{n}\sum\limits_{k=1}^{m}\bigg(k|\gamma_{k}|^{2}-\frac{1}{k}\bigg)\le 0,\end{align*} $$
led De Branges, by proving this conjecture, to the proof of the Bieberbach conjecture [Reference de Branges9]. For the Koebe function
$k(z)=z/(1-z)^{2}$
, the logarithmic coefficients are
$\gamma _{n}=1/n$
. Since the Koebe function k plays the role of extremal function for most of the extremal problems in the class
$\mathcal {S}$
, it might be expected that
$|\gamma _{n}|\le 1/n$
holds for functions in
$\mathcal {S}$
. However, this is not true in general, even in order of magnitude. Indeed, there exists a bounded function f in the class
$\mathcal {S}$
with logarithmic coefficients
$\gamma _{n}\ne O(n^{-0.83})$
(see [Reference Duren10, Theorem 8.4]). By differentiating (1.3) and equating coefficients,
$$ \begin{align} \nonumber & \gamma_{1}=\tfrac{1}{2}a_{2}, \\ & \gamma_{2}=\tfrac{1}{2}(a_{3}-\tfrac{1}{2}a_{2}^{2}),\\ \nonumber & \gamma_{3}=\tfrac{1}{2}(a_{4}-a_{2}a_{3}+\tfrac{1}{3}a_{2}^{3}). \end{align} $$
If
$f\in \mathcal {S}$
, it is easy to see that
$|\gamma _{1}|\le 1$
, because
$|a_2| \leq 2$
. Using the Fekete–Szegö inequality [Reference Duren10, Theorem 3.8] for functions in
$\mathcal {S}$
in (1.3), we obtain the sharp estimate
$$ \begin{align*}|\gamma_{2}|\le\tfrac{1}{2}(1+2e^{-2})=0.635\ldots.\end{align*} $$
For
$n\ge 3$
, the problem seems much harder and no significant bound for
$|\gamma _{n}|$
when
$f\in \mathcal {S}$
appears to be known. In 2017, Ali and Allu [Reference Ali and Allu1] obtained initial logarithmic coefficient bounds for close-to-convex functions. For recent results on several subclasses of close-to-convex functions, see [Reference Ali and Allu2, Reference Cho, Kowalczyk, Kwon, Lecko and Sim6, Reference Ponnusamy, Sharma and Wirths21].
The notion of logarithmic inverse coefficients, that is, logarithmic coefficients of the inverse of f, was proposed by Ponnusamy et al. [Reference Ponnusamy, Sharma and Wirths20]. The logarithmic inverse coefficients
$\Gamma _n$
,
$n \in \mathbb {N}$
, of f are defined by the equation
$$ \begin{align*} F_{f^{-1}}(w):= \log\frac{f^{-1}(w)}{w}=2\sum\limits_{n=1}^{\infty}\Gamma_{n}w^{n}, \quad |w|<1/4. \end{align*} $$
In [Reference Ponnusamy, Sharma and Wirths20], Ponnusamy et al. found sharp upper bounds for the logarithmic inverse coefficients for the class
$\mathcal {S}$
, namely
$$ \begin{align*}|\Gamma_n| \leq \frac{1}{2n} \bigg( \begin{matrix} 2n \\ n \end{matrix} \bigg), \quad n \in \mathbb{N}, \end{align*} $$
with equality only for the Koebe function or one of its rotations. Ponnusamy et al. [Reference Ponnusamy, Sharma and Wirths20] also obtained sharp bounds for the initial logarithmic inverse coefficients for some of the important geometric subclasses of
$ \mathcal {S}$
.
Recently, Kowalczyk and Lecko [Reference Kowalczyk and Lecko12] proposed the study of the Hankel determinant whose entries are logarithmic coefficients of
$ f \in \mathcal {S}$
, given by
$$ \begin{align*} H_{q,n}(F_f/2) = \begin{vmatrix} \gamma_n & \gamma_{n+1} & \cdots & \gamma_{n+q-1} \\ \gamma_{n+1} & \gamma_{n+2} & \cdots & \gamma_{n+q} \\ \vdots & \vdots & \ddots &\vdots \\ \gamma_{n+q-1} & \gamma_{n+q} & \cdots & \gamma_{n+2(q-1)} \end{vmatrix}. \end{align*} $$
Kowalczyk and Lecko [Reference Kowalczyk and Lecko12] obtained a sharp bound for the second Hankel determinant
$H_{2,1}(F_f/2)$
for starlike and convex functions. Sharp bounds for
$H_{2,1}(F_f/2)$
for various subclasses of
$\mathcal {S}$
are considered in [Reference Allu and Arora3, Reference Allu, Arora and Shaji4, Reference Kowalczyk and Lecko11, Reference Kowalczyk and Lecko13, Reference Mundalia and Kumar18]).
In this paper, we consider the second Hankel determinant for logarithmic inverse coefficients. From (1.4), for
$ f \in \mathcal {S}$
given by (1.1), the second Hankel determinant of
$F_{f^{-1}}/2$
is given by
$$ \begin{align}H_{2,1}(F_{f^{-1}}/2) =\Gamma_1\Gamma_3-\Gamma_{2}^2 &=\tfrac{1}{4}(A_2A_4-A_3^2+\tfrac{1}{4}A_2^4)\nonumber\\ &=\tfrac{1}{48}(13a_2^4-12a_2^2a3-12a_3^2+12a_2a_4).\end{align} $$
We note that
|$H_{2,1}(F_{f^{-1}}/2)|$
is invariant under rotation, since for
$f_{\theta }(z):=e^{-i \theta } f(e^{i \theta } z)$
,
$\theta \in \mathbb {R}$
and
$f \in \mathcal {S}$
,
$$ \begin{align*} H_{2,1}(F_{f_{\theta}^{-1}}/2)=\frac{e^{4 i \theta}}{48}(13a_2^4-12a_2^2a3-12a_3^2+12a_2a_4)=e^{4 i \theta} H_{2,1}(F_{f^{-1}}/2). \end{align*} $$
The main aim of this paper is to find a sharp upper bound for
$|H_{2,1}(F_{f^{-1}}/2)| $
when f belongs to the class of convex or starlike functions. A domain
$\Omega \subseteq \mathbb {C}$
is said to be starlike with respect to a point
$z_{0}\in \Omega $
if the line segment joining
$z_{0}$
to any point in
$\Omega $
lies entirely in
$\Omega $
. If
$z_0$
is the origin, then we say that
$\Omega $
is a starlike domain. A function
$f \in \mathcal {A}$
is said to be starlike if
$f(\mathbb {D})$
is a starlike domain. We denote by
$\mathcal {S}^*$
the class of starlike functions f in
$\mathcal {S}$
. It is well known that a function
$f \in \mathcal {A}$
is in
$\mathcal {S}^*$
if and only if
$$ \begin{align} \mathrm{Re\,}\bigg( \frac{zf'(z)}{f(z)} \bigg)> 0 \quad \text{for}\ z \in \mathbb{D}. \end{align} $$
Further, a domain
$\Omega \subseteq \mathbb {C}$
is called convex if the line segment joining any two points of
$\Omega $
lies entirely in
$\Omega $
. A function
$f\in \mathcal {A}$
is called convex if
$f(\mathbb {D})$
is a convex domain. We denote by
$\mathcal {C}$
the class of convex functions in
$\mathcal {S}$
. A function
$f \in \mathcal {A}$
is in
$\mathcal {C}$
if and only if
$$ \begin{align} \mathrm{Re\,}\bigg( 1+\frac{zf"(z)}{f'(z)} \bigg)> 0 \quad \text{for}\ z \in \mathbb{D}. \end{align} $$
2. Preliminary results
In this section, we present the key lemmas which will be used to prove the main results of this paper. Let
$\mathcal {P}$
denote the class of all analytic functions p having positive real part in
$\mathbb {D}$
, with the form
$$ \begin{align} p(z)=1+c_{1} z+c_{2} z^{2}+c_{3} z^{3}+ \cdots. \end{align} $$
A member of
$\mathcal {P}$
is called a Carathéodory function. It is known that
$|c_{n}| \leq 2, n \geq 1$
, for
$p \in \mathcal {P}$
. By using (1.6) and (1.7), functions in the classes
$\mathcal {S}^*$
and
$\mathcal {C}$
can be represented in terms of functions in the Carathéodory class
$\mathcal {P}$
.
Parametric representations of the coefficients are often useful. In Lemma 2.1, (2.2) is due to Carathéodory [Reference Duren10]. Equation (2.3) can be found in [Reference Pommerenke19]. In 1982, Libera and Zlotkiewicz [Reference Libera and Zlotkiewicz14, Reference Libera and Zlotkiewicz15] derived (2.4) with the assumption that
$c_1 \geq 0$
. Later, Cho et al. [Reference Cho, Kowalczyk and Lecko7] derived (2.4) in the general case and gave the explicit form of the extremal function.
Lemma 2.1. If
$p \in \mathcal {P}$
is of the form (2.1), then
$$ \begin{align} c_{1}=2 p_{1}, \end{align} $$
$$ \begin{align} c_{2}=2 p_{1}^{2}+2(1-p_{1}^{2}) p_{2} \end{align} $$
and
$$ \begin{align} c_{3}=2 p_{1}^{3}+4\big(1-p_{1}^{2}\big) p_{1} p_{2}-2\big(1-p_{1}^{2}\big) p_{1} p_{2}^{2}+2\big(1-p_{1}^{2}\big)\big(1-|p_{2}|^{2}\big) p_{3} \end{align} $$
for some
$p_{1}, p_{2}, p_{3} \in \overline {\mathbb {D}}:=\{z \in \mathbb {C}:|z| \leq 1\}$
.
For
$p_{1} \in \mathbb {T}:=\{z \in \mathbb {C}:|z|=1\}$
, there is a unique function
$p \in \mathcal {P}$
with
$c_{1}$
as in (2.2), namely
$$ \begin{align*} p(z)=\frac{1+p_{1} z}{1-p_{1} z}, \quad z \in \mathbb{D}. \end{align*} $$
For
$p_{1} \in \mathbb {D}$
and
$p_{2} \in \mathbb {T}$
, there is a unique function
$p \in \mathcal {P}$
with
$c_{1}$
and
$c_{2}$
as in (2.2) and (2.3), namely
$$ \begin{align} p(z)=\frac{1+(p_{1}+\overline{p_{1}} p_{2}) z+p_{2} z^{2}}{1-(p_{1}-\overline{p_{1}} p_{2}) z-p_{2} z^{2}}. \end{align} $$
For
$p_{1}, p_{2} \in \mathbb {D}$
and
$p_{3} \in \mathbb {T}$
, there is unique function
$p \in \mathcal {P}$
with
$c_{1}, c_{2}$
and
$c_{3}$
as in (2.2)–(2.4), namely
$$ \begin{align*} p(z)=\frac{1+(\overline{p_{2}} p_{3}+\overline{p_{1}} p_{2}+p_{1}) z+(\overline{p_{1}} p_{3}+p_{1} \overline{p_{2}} p_{3}+p_{2}) z^{2}+p_{3} z^{3}}{1+(\overline{p_{2}} p_{3}+\overline{p_{1}} p_{2}-p_{1}) z+(\overline{p_{1}} p_{3}-p_{1} \overline{p_{2}} p_{3}-p_{2}) z^{2}-p_{3} z^{3}}, \quad z \in \mathbb{D}. \end{align*} $$
Next we recall the following well-known result due to Choi et al. [Reference Choi, Kim and Sugawa8].
Lemma 2.2. Let
$A, B, C$
be real numbers and
$$ \begin{align*} Y(A, B, C):=\max _{z \in \overline{\mathbb{D}}}(|A+B z+C z^{2}|+1-|z|^{2}). \end{align*} $$
-
(i) If
$A C \geq 0$
, then
$$ \begin{align*} Y(A, B, C)= \begin{cases}|A|+|B|+|C|, & |B| \geq 2(1-|C|), \\ 1+|A|+\displaystyle\frac{B^{2}}{4(1-|C|)}, & |B|<2(1-|C|) .\end{cases} \end{align*} $$
-
(ii) If
$A C<0$
, then
$$ \begin{align*} Y(A, B, C)= \begin{cases}1-|A|+\displaystyle\frac{B^{2}}{4(1-|C|)}, & -4 A C(C^{-2}-1) \leq B^{2} \wedge|B|<2(1-|C|), \\ 1+|A|+\displaystyle\frac{B^{2}}{4(1+|C|)}, & B^{2}<\min \{4(1+|C|)^{2},-4 A C(C^{-2}-1)\}, \\ R(A, B, C), & \text {otherwise, }\end{cases} \end{align*} $$
where
$$ \begin{align*} R(A, B, C)= \begin{cases}|A|+|B|+|C|, & |C|(|B|+4|A|) \leq|A B|, \\ -|A|+|B|+|C|, & |A B| \leq|C|(|B|-4|A|), \\ (|A|+|C|) \sqrt{1-\displaystyle\frac{B^{2}}{4 A C}}, & \! \ \text{otherwise.}\end{cases} \end{align*} $$
3. Main results
Now we will prove the first main result of this paper. We obtain the following sharp bound for
$H_{2,1}(F_{f^{-1}}/2)$
for functions in the class
$\mathcal {C}$
.
Theorem 3.1. For
$f\in \mathcal {C}$
given by (1.1),
$$ \begin{align} |H_{2,1}(F_{f^{-1}}/2)| \leq \tfrac{1}{33}. \end{align} $$
The inequality is sharp.
Proof. Let
$f\in \mathcal {C}$
be of the form (1.1). Then by (1.7),
$$ \begin{align} 1+\frac{zf"(z)}{f'(z)}=p(z) \end{align} $$
for some
$p \in \mathcal {P}$
of the form (2.1). Since the class
$\mathcal {C}$
is invariant under rotation and the function is also rotationally invariant, we can assume that
$c_1 \in [0,2]$
. Comparing the coefficients on both sides of (3.2) yields
$$ \begin{align*} & a_2=\tfrac{1}{2}c_1, \\ & a_3=\tfrac{1}{6}(c_2+c_1^2), \\ & a_4=\tfrac{1}{24}(2c_3+3c_1c_2+c_1^3 ). \end{align*} $$
Hence, by (1.5),
$$ \begin{align*} H_{2,1}(F_{f^{-1}}/2)=\tfrac{1}{2304}(11c_1^4-20c_1^2c_2-16c_2^2+24c_1c_3). \end{align*} $$
By (2.2)–(2.4), after simplification,
$$ \begin{align} \nonumber H_{2,1}(F_{f^{-1}}/2) &=\frac{p_1^4}{48} -\frac{1}{24}(1-p_1^2)p_1^2p_2-\frac{1}{72}(1-p_1^2)(2+p_1^2)p_2^2 \\ &\quad+\frac{1}{24}(1-p_1^2)(1-|p_1^2|)p_1p_3. \end{align} $$
We consider three cases according to the value of
$p_1$
.
Case 1:
$p_1=1$
. By (3.3),
$$ \begin{align*}|H_{2,1}(F_{f^{-1}}/2)|=\tfrac{1}{48}.\end{align*} $$
Case 2:
$p_1=0$
. By (3.3),
$$ \begin{align*}|H_{2,1}(F_{f^{-1}}/2)|=\tfrac{1}{36}|p_2^2|\leq\tfrac{1}{36}.\end{align*} $$
Case 3:
$p_1 \in (0,1)$
. Since
$|p_3| \leq 1$
, applying the triangle inequality in (3.3) gives
$$ \begin{align} \nonumber |H_{2,1}(F_{f^{-1}}/2)| &=\frac{1}{24}p_1(1-p_1^2)\bigg(\bigg| \frac{p_1^3}{2(1-p_1^2)}-p_1 p_2-\frac{2+p_1^2}{3p_1}p_2^2 \bigg|+1-|p_2^2|\bigg) \notag \\ &\leq \frac{1}{24}p_1(1-p_1^2)(| A+B p_2+C p_2^2 |+1-|p_2^2|), \end{align} $$
where
$$ \begin{align*} A:=\frac{p_1^3}{2(1-p_1^2)}, \quad B:=-p_1,\quad C:=-\frac{2+p_1^2}{3p_1}. \end{align*} $$
Since
$AC < 0$
, we can apply Lemma 2.2(ii). The argument now divides into five parts.
3(a). For
$p_{1} \in (0,1)$
,
$$ \begin{align*} -4 A C\bigg(\frac{1}{C^{2}}-1\bigg)-B^{2}=-\frac{p_{1}^{2}(14+p_1^2)}{3(2+p_{1}^{2})} \leq 0. \end{align*} $$
The inequality
$|B|<2(1-|C|)$
is equivalent to
$p_1(4 - 6 p_1 + 5 p_1^2) < 0$
which is not true for
$p_{1} \in (0,1)$
.
3(b). It is easy to check that
$$ \begin{align*} \min \bigg\{4(1+|C|)^{2},-4 A C\bigg(\frac{1}{C^{2}}-1\bigg)\bigg\}=-4 A C\bigg(\frac{1}{C^{2}}-1\bigg), \end{align*} $$
and from 3(a),
$$ \begin{align*} -4 A C\bigg(\frac{1}{C^{2}}-1\bigg) \leq B^{2}. \end{align*} $$
Therefore, the inequality
$B^{2} < \min \{4(1+|C|)^{2},-4 A C({1}/{C^{2}}-1)\}$
does not hold for
$0<p_1<1$
.
3(c). The inequality
$|C|(|B|\hspace{-0.5pt}+\hspace{-0.5pt}4|A|)\hspace{-0.5pt}-\hspace{-0.5pt}|A B|\hspace{-0.5pt} \leq\hspace{-0.5pt} 0$
is equivalent to
${4\hspace{-0.5pt}+\hspace{-0.5pt}6p_1^2\hspace{-0.5pt}-\hspace{-0.5pt}p_1^4\hspace{-0.5pt}\leq\hspace{-0.5pt} 0}$
, which is false for
$p_{1} \in (0,1)$
.
3(d). The inequality
$$ \begin{align*}|A B|-|C|(|B|-4|A|)=\frac{9 p_1^4+10 p_1^2-4}{1 - p_1^2} \leq 0 \end{align*} $$
is equivalent to
$9 p_1^4+10 p_1^2-4\leq 0$
, which is true for
$$ \begin{align*} 0<p_1 \leq p_1'=\tfrac{1}{3}\sqrt{\sqrt{61}-5}\approx 0.5588. \end{align*} $$
It follows from Lemma 2.2 and (3.4) that
$$ \begin{align*} |H_{2,1}(F_{f^{-1}}/2)| \leq \tfrac{1}{24}p_1(1-p_1^2)(-|A|+|B|+|C|) = \tfrac{1}{144}(4+4p_1^2-11p_1^4)=h(p_1), \end{align*} $$
where
$h(x)=4+4x^2-11x^4$
. By a simple calculation, the maximum of the function
$h(x)$
for
$0<x\leq p_1'$
occurs at the point
$x_0=\sqrt {2/11}$
. We conclude that
$$ \begin{align*} |H_{2,1}(F_{f^{-1}}/2)|\leq h\Big(\sqrt{\tfrac{2}{11}}\Big)=\tfrac{1}{33}. \end{align*} $$
3(e). For
$p_1'<p_1<1$
, we use the last case of Lemma 2.2 together with (3.4) to obtain
$$ \begin{align*} |H_{2,1}(F_{f^{-1}}/2)| &\leq \frac{1}{24}p_1(1-p_1^2)(|C|+|A|) \sqrt{1-\frac{B^{2}}{4 A C}} \notag \\ &= \frac{1}{144}(p_1^4-2p_1^2+4)\sqrt{\frac{7-p_1^2}{4+2p_1^2}}=k(p_1), \end{align*} $$
where
$$ \begin{align*}k(x)=\frac{1}{144}(x^4-2x^2+4)\sqrt{\frac{7-x^2}{4+2x^2}}. \end{align*} $$
We want to find the maximum of
$k(x)$
over the interval
$p_1'<x<1$
. Observe that
$$ \begin{align*} k'(x)=\frac{x}{144} \sqrt{\frac{7 - x^2}{ 4 + 2 x^2}}\bigg( \frac{92 - 54x^2 - 15 x^4 + 4 x^6}{(-7 + x^2) (2 + x^2)}\bigg)=0 \end{align*} $$
if and only if
$92 - 54x^2 - 15 x^4 + 4 x^6=0$
. However, all the real roots of this equation lie outside the interval
$p_1'<x<1$
and
$k'(x)<0$
for
$p_1'<x<1$
. So k is decreasing and hence
$k(x)\leq k(p_1')$
for
$p_1'<x<1$
. We conclude that, for
$p_1'<x<1,$
$$ \begin{align*} |H_{2,1}(F_{f^{-1}}/2)|\leq k(p_1') \approx 0.0290035. \end{align*} $$
The desired inequality (3.1) follows from Cases 1–3. By tracking back in the proof, we see that equality in (3.1) holds when
$$ \begin{align*} p_1=\sqrt{\tfrac{2}{11}},\quad p_3=1, \end{align*} $$
and
$$ \begin{align} |A+Bp_2+Cp_2^2|+1-|p_2^2|=-|A|+|B|+|C|, \end{align} $$
where
$$ \begin{align*} A=\tfrac{1}{9}\sqrt{\tfrac{2}{11}},\quad B=-\sqrt{\tfrac{2}{11}},\quad C=4\sqrt{\tfrac{2}{11}}. \end{align*} $$
Indeed, we can easily verify that one of the solutions of (3.5) is
$p_2=1.$
In view of Lemma 2.2, we conclude that equality holds for the function
$f\in \mathcal {A}$
given by (1.7), corresponding to the function
$p \in \mathcal {P}$
of the form (2.5) with
$p_1=\sqrt {2/11},p_2=1$
and
$p_3=1$
, that is,
$$ \begin{align*} p(z)=\frac{1+2\sqrt{2/11}z+z^2}{1-z^2}. \end{align*} $$
This complete the proof.
Next, we obtain the sharp bound for
$H_{2,1}(F_{f^{-1}}/2)$
for functions in the class
$\mathcal {S}^*$
.
Theorem 3.2. For
$f\in \mathcal {S}^*$
given by (1.1),
$$ \begin{align} |H_{2,1}(F_{f^{-1}}/2)| \leq \tfrac{13}{12}. \end{align} $$
The inequality is sharp.
Proof. Let
$f\in \mathcal {S}^*$
be of the form (1.1). By (1.6),
$$ \begin{align} \frac{zf'(z)}{f(z)}=p(z) \end{align} $$
for some
$p \in \mathcal {P}$
of the form (2.1). By comparing the coefficients on both sides of (3.7),
$$ \begin{align*} & a_2=c_1, \\ & a_3=\tfrac{1}{2}(c_2+c_1^2), \\ & a_4=\tfrac{1}{6}(2c_3+3c_1c_2+c_1^3). \end{align*} $$
Hence, by (1.5),
$$ \begin{align*} H_{2,1}(F_{f^{-1}}/2)=\tfrac{1}{48}\big(6c_1^4-6c_1^2c_2-3c_2^2+4 c_1c_3\big). \end{align*} $$
From (2.2)–(2.4), by straightforward computation,
$$ \begin{align} \notag H_{2,1}(F_{f^{-1}}/2)& = \tfrac{13}{12}p_1^4-\tfrac{5}{2}(1-p_1^2)p_2-\tfrac{1}{12}(1-p_1^2)(3+p_1^2)p_2^2 \\[5pt] & \quad +\tfrac{1}{3}(1-p_1^2)(1-|p_1^2|)p_1p_3. \end{align} $$
Now we consider three cases according to the value of
$p_1$
.
Case 1:
$p_1=1$
. By (3.8),
$$ \begin{align*}|H_{2,1}(F_{f^{-1}}/2)|=\tfrac{13}{12}.\end{align*} $$
Case 2:
$p_1=0$
. By (3.8),
$$ \begin{align*}|H_{2,1}(F_{f^{-1}}/2)|=\tfrac{1}{4}|p_2^2|\leq\tfrac{1}{4}.\end{align*} $$
Case 3:
$p_1 \in (0,1)$
. Applying the triangle inequality in (3.8) and using the fact that
$|p_3| \leq 1$
,
$$ \begin{align*} |H_{2,1}(F_{f^{-1}}/2)| &=\frac{1}{3}p_1(1-p_1^2)\bigg(\bigg| \frac{13p_1^3}{4(1-p_1^2)}-\frac{5}{2}p_1 p_2-\frac{3+p_1^2}{4p_1}p_2^2 \bigg|+1-|p_2^2|\bigg) \notag \\[5pt] &\leq \frac{1}{24}p_1(1-p_1^2)(| A+B p_2+C p_2^2 |+1-|p_2^2|), \end{align*} $$
where
$$ \begin{align*} A:=\frac{13p_1^3}{4(1-p_1^2)}, \quad B:=-\frac{5}{2}p_1,\quad C:=-\frac{3+p_1^2}{4p_1}. \end{align*} $$
Since
$AC < 0$
, we can apply Lemma 2.2(ii).
3(a). For
$p_{1} \in (0,1)$
,
$$ \begin{align*} -4 A C\bigg(\frac{1}{C^{2}}-1\bigg)-B^{2}=-\frac{3p_{1}^{2}(16+p_1^2)}{(3+p_{1}^{2})} \leq 0. \end{align*} $$
The inequality
$|B|<2(1-|C|)$
is equivalent to
$3 - 4 p_1 + 2 p_1^2 < 0$
which is not true for
$p_{1} \in (0,1)$
.
3(b). It is easy to see that
$$ \begin{align*} \min \bigg\{4(1+|C|)^{2},-4 A C\bigg(\frac{1}{C^{2}}-1\bigg)\bigg\}=-4 A C\bigg(\frac{1}{C^{2}}-1\bigg), \end{align*} $$
and from
$3(a),$
$$ \begin{align*} -4 A C\bigg(\frac{1}{C^{2}}-1\bigg) \leq B^{2}. \end{align*} $$
Therefore, the inequality
$B^{2} < \min \{4(1+|C|)^{2},-4 A C({1}/{C^{2}}-1)\}$
does not hold for
$0<p_1<1$
.
3(c). The inequality
$|C|(|B|+4|A|)-|A B| \leq 0$
is equivalent to the inequality
$44p_1^4-68p_1^2 -16-p_1^4\geq 0$
, which is false for
$p_{1} \in (0,1)$
.
3(d). The inequality
$$ \begin{align*}|A B|-|C|(|B|-4|A|)=\frac{96 p_1^4+88 p_1^2-15}{1 - p1^2} \leq 0 \end{align*} $$
is equivalent to
$96 p_1^4+88 p_1^2-15\leq 0$
, which is true for
$$ \begin{align*} 0<p_1\leq p_1"=\frac{1}{2}\sqrt{\frac{\sqrt{211}-11}{6}} \approx 0.38328. \end{align*} $$
$$ \begin{align}| H_{2,1}(F_{f^{-1}}/2)| \leq \tfrac{1}{3}p_1(1-p_1^2)(-|A|+|B|+|C|) = \tfrac{1}{12}(3+8p_1^2-24p_1^4)=h(p_1), \end{align} $$
where
$h(x)=3+8x^2-24x^4$
. Since
$h'(x)>0$
in
$ 0< x\leq p_1"$
, we have
$h(x)\leq h(p_1")$
for
$ 0< x\leq p_1"$
. Therefore,
$$ \begin{align*} |H_{2,1}(F_{f^{-1}}/2)|\leq \tfrac{1}{48}(-58+5\sqrt{211}) \approx 0.304775. \end{align*} $$
3(e). Furthermore, for
$p_1"<p_1<1$
, from (3.8) and Lemma 2.2,
$$ \begin{align*} |H_{2,1}(F_{f^{-1}}/2)| &\leq \frac{1}{24}p_1(1-p_1^2)(|C|+|A|) \sqrt{1-\frac{B^{2}}{4 A C}} \notag \\ &= \frac{1}{6}(12p_1^4 -2p_1^2+3)\sqrt{\frac{16-3p_1^2}{39+13p_1^2}}=k(p_1), \end{align*} $$
where
$$ \begin{align*}k(x)=\frac{1}{6}\sqrt{\frac{16-3x^2}{39+13x^2}}(12x^4 -2x^2+3). \end{align*} $$
As
$k'(x)=0$
has no solution in
$(p_1",1)$
and
$k'(x)>0$
, the maximum occurs at
$x=1$
and we conclude that
$$ \begin{align*} |H_{2,1}(F_{f^{-1}}/2)| \leq k(1)=\tfrac{13}{12} \quad \text{for} \ p_1"<x<1. \end{align*} $$
The desired inequality (3.6) follows from Cases 1–3. For the equality, consider the Koebe function
$$ \begin{align*} k(z)=\frac{z}{(1-z)^2}. \end{align*} $$
Clearly,
$k \in \mathcal {S}^*$
and it is easy to show that
$$ \begin{align*} |H_{2,1}(F_{k^{-1}}/2)|=\tfrac{13}{12}. \end{align*} $$
This completes the proof.
