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\begin{document}
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\title[Hankel and Symmetric Toeplitz determinants]{Hankel and Symmetric Toeplitz determinants for Sakaguchi starlike functions}
\author[S. Kumar]{Sushil Kumar}
\address{Department of Applied Science\\Bharati Vidyapeeth's college of Engineering\\
Delhi-110063\\
India}
\email{sushilkumar16n@gmail.com}
%
\author[S. Anand]{Swati Anand}
\address{Department of Mathematics\\ University of Delhi\\
Delhi-110007\\
India}
\email{swati\_anand01@yahoo.com}
%
\author[N. K. Jain]{Naveen Kumar Jain}
\address{Department of Mathematics \\ Aryabhatta College\\
Benito Juarez Road\\
New Delhi-110021\\
India}
\email{naveenjain05@gmail.com}
%
\subjclass{30C45, 30C80}
\keywords{Starlike function, Sakaguchi starlike functions,  Zalcman conjecture, third and forth order Hankel determinants, second, third and fourth order symmetric Toeplitz determinants.}
\begin{abstract}
In this paper, we consider the class of starlike functions with respect to symmetric points which are also known as Sakaguchi starlike functions. We determine best possible bounds on Zalcman conjecture
$\vert a_n^2-a_{2n-1}\vert$ and generalized Zalcman conjecture $\vert a_ma_n-a_{m+n-1}\vert$ for $n=2$ and $n=4$, $m=2$, respectively for such functions. Further, we compute estimate on third order and fourth order  Hankel determinants. As well, we also obtain estimates on third and fourth symmetric Toeplitz determinants.
\end{abstract}
\maketitle

\section{Introductory text}
Let $\mathcal{A}$ be the family of all normalized analytic functions $f$  defined on $\D=\{z \in \C : \vert z\vert  < 1\}$ with   series expansion $f(z) = z +a_2 z^2+a_3 z^3 + \cdots$.
The subfamily $\mathcal{S}\subset \mathcal{A}$ contains  univalent  functions. Let $\mathcal{S}^*$ and $\mathcal{K}$ represent the subfamily of $\mathcal{S}$ containing starlike and convex functions, respectively. Analytically,
$\mathcal{S}^*= \{f \in \mathcal{S}: \RE\left({zf'(z)}/{f(z)}\right)>0, \, z\in \D\}$
and
$ \mathcal{K}= \{f \in \mathcal{S}: 1+\RE\left({zf''(z)}/{f'(z)}\right)>0, \, z\in \D\}$\cite{goodman}. The  class $\mathcal{P}$ consists of all analytic functions $p:\mathbb{D}\to\mathbb{C}$ satisfying conditions $p(0)=1$ and $\RE p(z)>0$. Recent results for a more general  class of $\mathcal{P}$  can be found in   \cite{ali1}.  In 1959, Sakaguchi \cite{saka} studied the subclass $\mathcal{S_{S}^*}$ of $\mathcal{S}$ consisting of starlike functions with respect to the symmetric points. The analytical description of these functions is
\[\mathcal{S_{S}^*}=\left\{f \in \mathcal{S}:\frac{2zf'(z)}{f(z)-f(-z)}\in \mathcal{P}, \, z\in \D\right\}.\]
The functions $f\in\mathcal{S_{S}^*}$ are also called  Sakaguchi starlike functions. The coefficient estimates related literature gives the geometric properties of univalent functions. The bound on the initial coefficient $a_2$ contribute in  growth, distortion and covering theorems.
%In 1916, Bieberbach \cite{bieberbach} attained the bound on the second coefficient of univalent functions. Furthermore, it was conjectured that $|a_n| \le n$ which was later proved by De-Branges \cite{branges}.
 Zalcman conjecture and Hankel determinants are two of the coefficient problems that have been discussed by several authors. In recent years, many authors have studied the Toeplitz determinant  $T_q(n)$ for various values of $q$ and $n$ for several subclasses of analytic functions.
A significant problem concerning the coefficients in the series expansion of the the function $f \in \mathcal{A}$ is the Zalcman conjecture which is defined as
\[\vert a_n^2-a_{2n-1}\vert  \le (n-1)^2, \, n\ge 2.\] From \cite{Brown86}, we observe that the Zalcman conjecture implies the Bieberbach conjecture. Ma \cite{ma88} verified Zalcman conjecture $(n\geq 4)$ for close-to-convex functions.
Further, Ma \cite{ma} explored  the generalized Zalcman conjecture which is defined as
\[\vert a_ma_n-a_{m+n-1}\vert  \le (m-1)(n-1); \quad\,\,m \ge 2,\, n \ge 2\]
for the starlike functions and the univalent functions with real coefficients.
In~\cite{ravi1}, Ravichandran and Verma established the generalized Zalcman conjecture for certain starlike and convex functions.  In~\cite{Vasu20}, the Zalcman conjecture and the generalized
Zalcman conjecture for the locally univalent functions were discussed  using extreme point theory.
Recently, in~\cite{obradovic}  the Zalcman conjecture and the generalized Zalcman conjecture were shown  for the class $\mathcal{U}$ defined as
$\mathcal{U}= \{z \in \mathcal{A}: \left\vert\left({z}/{f(z)}\right)^2f'(z)-1\right\vert<1, \, z\in \D\}.$
For $q \ge 1$ and $n \ge 1$, the $q^{th}$ Hankel determinant $H_q(n)(f)$ for a function $f \in \mathcal{S}$ is given by $
H_q(n)(f):=\det \{a_{n+i+j-2}\}^q_{i,j}$, $ 1 \leq i,\,j \leq q,$
%\begin{equation} \label{eqdet}
%H_q(n)(f) = \begin{array}{|cccc|}
%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+2q-2}
%\end{array}\,
%\end{equation}
where $a_1=1$. For $q=2$ and $n=1$, the Hankel determinant $H_2(1)= a_3-a_2^2$ is the Fekete Szeg\"{o} functional.
The study of Hankel determinant was initiated by Pommerenke \cite{pom1,pom2} for the starlike functions.
Since then the growth of $H_q(n)(f)$ has been studied for different subclasses of univalent functions. One of the notable results in this direction is by Hayman \cite{hayman} giving the best possible upper bound as $Mn^{1/2}$ on $H_2(n)(f)$, where $M$ is an absolute constant.
For $q=2$ and $n=2$, Janteng \emph{et al.}\cite{janteng} obtained the sharp estimates on second order Hankel determinant  $H_2(2)(f)= a_2a_4-a_3^2$ for the classes of starlike and convex functions. However, the sharp bound for the whole class $\mathcal{S}$ is not known till now. For the class of Bazilevic functions, Krishna and RamReddy \cite{krish} determined $H_2(2)(f)$. Recently, Anand \emph{et al.} \cite{sharma} studied the second order Hankel determinant for a class of normalized analytic functions.

For $q=3$ and $n=1,2,3$, the third Hankel determinants are given as
\begin{align}
H_3(1)(f)  &= a_3(a_2a_4-a_3^2) + a_4(a_2a_3-a_4)+a_5(a_3-a_2^2) \label{h31}\\
H_3(2)(f) &= a_2(a_4a_6-a_5^2) - a_3(a_3a_6-a_4a_5)+a_4(a_3 a_5-a_4^2)\label{h32}\\
H_3(3)(f) &=a_3(a_5a_7-a_6^2) - a_4(a_4a_7-a_5a_6)+a_5(a_4 a_6-a_5^2)\label{h33}.
\end{align}
The study of the third order Hankel determinant $H_3(1)(f)$ for the classes $\mathcal{S}^*$ and $\mathcal{K}$ was initiated by Babalola~$(2010)$ \cite{babalola}  which was later improved by Zaprawa \cite{zap1}. However, the bounds obtained in \cite{zap1} were not sharp. The best possible bound on third order Hankel determinant $H_3(1)(f)$ for the class of convex functions was computed by Kowalczyk \emph{et al.}~\cite{kowa2}. Also, Lecko \emph{et al.} \cite{lecko} computed the best possible upper bound on $H_3(1)(f)$ for the starlike functions of order $1/2$. Krishna \emph{et al.} \cite{krishna} obtained the bound on  $H_3(1)(f)$ for the class $\mathcal{S}_S^*$. Recently, Kumar  \emph{et al.}~\cite{vsv} improved the existing bound for the class $\mathcal{S}_S^*$.
For more recent developments on coefficient estimates and third order Hankel determinant, see \cite{kowa1, krishna,prajapat,zap2, SA1, SR1, ven}.
For $q=4$ and $n=1$, the fourth order Hankel determinant is given by
\begin{equation}\label{eq11}
H_4(1)(f)=a_7H_3(1)(f)-a_6\Delta_1+a_5\Delta_2-a_4\Delta_3
\end{equation}
where
%$\Delta_1= a_3(a_2a_4-a_3^2) - a_4(a_4-a_2a_3)+a_5(a_3-a_2^2)$,
$\Delta_1=(a_3a_6-a_4a_5)-a_2(a_2a_6-a_3a_5)+a_4(a_2a_4-a_3^2)$, $\Delta_2=(a_4a_6-a_5^2) -a_2(a_3a_6-a_4a_5)+a_3(a_3a_5-a_4^2)$ and $\Delta_3 =a_2(a_4a_6-a_5^2)-a_3(a_3a_6-a_4a_5)+a_4(a_3a_5-a_4^2)$.
Arif  \emph{et al.}~\cite{arif} obtained the bound on $H_4(1)(f)$ for the functions with bounded turning. Cho and Kumar \cite{cho} computed the bound on $H_4(1)(f)$ for starlike functions associated with a lune-shaped region. For recent results on fourth order Hankel determinant,  see~\cite{sushil21,cho21}.
For $q \ge 1$ and $n \ge 1$, the symmetric Toeplitz determinant $T_q(n)$ for a function $f \in \mathcal{S}$ is defined as
 \begin{equation*}
 T_q(n) = \begin{array}{|cccc|}
 a_n & a_{n+1} &\cdots & a_{n+q-1} \\
 a_{n+1} & a_n & \cdots & a_{n+q} \\
 \vdots & \vdots & \vdots &\vdots\\
 a_{n+q-1} &a_{n+q} & \cdots & a_{n}
 \end{array}\,
 \end{equation*}
 where $a_1=1$.
 In particular, for $q=2$ and $n=2,3$ the second Toeplitz deteminants are given by
  $   T_2(2) =  a^2_3-a^2_2$ and $T_2(3) =   a_4^2-a_3^2$.

 For $q=3$ and $n=1,2$ the third Toeplitz determinants are as follows
      \begin{equation}\label{eq20}
         T_3(1) = 1+2a_2^2(a_3-1)-a_3^2 \,\, \text{and} \,\,
        T_3(2) =  (a_2-a_4)(a_2^2-2a_2^3+a_2a_4).
            \end{equation}
For $q=4$ and $n=2$ the fourth Toeplitz determinant is given by
 \begin{align}\label{eq16}
  T_4(2)=  (a^2_2-a^2_3)^2&+2(a_3^2-a_2a_4)(a_2a_4-a_3a_5)-(a_2a_3-a_3a_4)^2\\
  & +(a_4^2-a_3a_5)^2-(a_3a_4-a_2a_5)^2.\notag
  \end{align}
 In 2019, Zhang \emph{et al.} \cite{zhang1} computed the upper bound on the Toeplitz determinant $T_3(2)$ for the  starlike functions associated with the sine function. Ahuja \emph{et al.} \cite{ahuja} studied the Toeplitz determinants $T_2(2)$ and $T_3(1)$ for unified  class  of starlike and convex functions. Recently, in ~\cite{zhang2}, Zhang and  Tang obtained the upper bound on fourth Toeplitz determinant $T_4(2)$ for the  starlike functions associated with the sine function. For more recent details, see~\cite{ahuja1,SA}

In this manuscript,  we prove Zalcman Conjecture  $\vert a_n^2-a_{2n-1}\vert  \le (n-1)^2$ for $n=2$ and generalised Zalcman Conjecture $\vert a_m a_n-a_{m+n-1}\vert  \le (m-1)(n-1)$ for $ m=2, \, n=4$. Further, we obtain the estimates on  the third order Hankel determinant $H_3(1)(f)$ for such functions which is an improvement to the existing estimate computed in \cite{vsv}. In addition, we compute the bounds on third order Hankel determinants $H_3(2)(f)$, $H_3(3)(f)$ and the fourth order Hankel determinant $H_4(1)(f)$. Moreover, bounds on the symmetric Toeplitz determinants $T_2(2), T_2(3), T_3(1)$,$T_3(2)$ and $T_4(2)$ are also determined.

\section{Inductive Lemmas}
In order to establish the main results, we need  following lemmas related to coefficient estimates.
\begin{lemma}\label{lem1}\cite{prokhorov}
Let $w(z) = c_1 z+c_2 z^2+ \cdots$ be a Schwarz function. Then $\vert c_3+\mu c_1c_2+ \nu c_1^3\vert  \le 1$,
 where ${1}/{2} \le \vert \mu\vert  \le 2,\,\, {4}(\vert \mu\vert +1)^3/{27}-(\vert \mu\vert +1)\le \nu \le 1$.
\end{lemma}
Let $\mathcal{B}$ be the class of functions $f\in \mathcal{A}$  satisfying $\vert f(z)\vert <1$ for all $z \in \D$.
\begin{lemma} \cite{carlson}\label{lem2}
Let $f(z) = a_0+\sum\limits_{n=1}^{\infty}a_nz^n$ be in $\mathcal{B}$. Then
\begin{equation} \label{eq2}
\vert a_{2n+1}\vert  \le 1-\vert a_0\vert ^2-\vert a_1\vert ^2- \cdots -\vert a_n\vert ^2,\,\, n= 0,1, \cdots
\end{equation}
and
\begin{equation}\label{eq3}
\vert a_{2n}\vert  \le 1-\vert a_0\vert ^2-\vert a_1\vert ^2- \cdots -\vert a_{n-1}\vert ^2-\frac{\vert a_n\vert ^2}{1+\vert a_0\vert },\,\, n= 1,2, \cdots .
\end{equation}
Equality in (\ref{eq2}) holds for
\[f(z) = \frac{a_0+a_1z+ \cdots+a_nz^n+\varepsilon z^{2n+1}}{1+(\overline{a_n}z^{n+1}+\overline{a_{n-1}}z^{n+2}+\cdots + \overline{a_0}z^{2n+1}) \varepsilon}, \,\, \vert \varepsilon\vert  =1\]
and in (\ref{eq3}) for
\[f(z) = \frac{a_0+a_1z+ \cdots+a_{n-1}z^{n-1}+\frac{a_n}{1+\vert a_0\vert }+\varepsilon z^{2n}}{1+(\frac{\overline{a_n}}{1+\vert a_0\vert }z^n+\overline{a_{n-1}}z^{n+1}+\cdots + \overline{a_0}z^{2n}) \varepsilon}, \,\, \vert \varepsilon\vert  =1\]
where $a_0 \overline{a_n}^2 \varepsilon$ is non-positive real.
\end{lemma}
%If $f\in \mathcal{B}$, then $w(z) = zf(z)$ is a Schwarz function.
In view of Lemma \ref{lem2},  for a Schwarz function $w(z)=c_1z+c_2z^2+ \cdots$, we have
\begin{equation}\label{eq4}
\vert c_2\vert  \le 1-\vert c_1\vert ^2, \,\, \vert c_3\vert  \le 1-\vert c_1\vert ^2-\frac{\vert c_2\vert ^2}{1+\vert c_1\vert } \,\,\text{and}\,\, \vert c_4\vert  \le 1-\vert c_1\vert ^2-\vert c_2\vert ^2.
\end{equation}

\begin{lemma} \cite{saka}\label{lem3}
Let $f(z) = z+\sum\limits_{n=2}^{\infty}a_nz^n$ be univalent and starlike with respect to symmetric points in $\mathbb{D}$. Then
\[\vert a_n\vert  \le 1, \, n \ge 2\]
equality being attained by the function $z/(1+\varepsilon z)$, $\vert \varepsilon\vert <1$.
\end{lemma}

\begin{lemma} \cite%[Lemma 2.3, p.507]
{ ravi2} \label{lem4}
If $p(z) = 1+p_1z+p_2 z^2+p_3z^3+ \cdots \in \mathcal{P}$ then for all $n,m \in \mathbb{N}$
\begin{equation*}
\vert \mu p_np_m-p_{m+n}\vert  \le \left\{ \begin{array}{ll}
                2,& 0 \le \mu \le 1 \\
                  2\vert 2\mu-1\vert ,& \mbox{elsewhere}.

                             \end{array}
                       \right.
                       \end{equation*}
 If $0< \mu <1$, the inequality is sharp for the function $p(z) = (1+z^{n+m})/(1-z^{n+m})$. In other cases, the inequality is sharp for the function $p(z)= (1+z)/(1-z)$.
\end{lemma}

%\begin{lemma}
%\cite%[Lemma 2.1, p.1055]
%{vkumar} \label{lem5}
%Let $p(z) = 1+p_1z+p_2 z^2+p_3z^3+ \cdots \in \mathcal{P}$. Then for any number $\mu$,
%\begin{equation}
%|\mu p_3-p_1^3| \le \left\{ \begin{array}{ll}
%                2|\mu-4|,&  \mu \le \dfrac{4}{3} \\
%                  2\mu \sqrt{\dfrac{\mu}{\mu-1}},&\dfrac{4}{3}<\mu .
%
%                             \end{array}
%                       \right.
%                       \end{equation}
% The above estimate is sharp. If $\mu\leq\frac{4}{3}$, equality holds for the function $p_0(z):=(1+z)/(1-z)$, and if $\mu >\frac{4}{3}$, then equality holds for the function
%$p_1(z):={(1-z^2)}/{(z^2-2 \sqrt{{\mu }/{(\mu -1)}}\;z+1)}.$
%\end{lemma}

%\begin{lemma} \cite  %[Lemma 2, p.254]
% {LIB1}\label{lem6}
%If $p(z) = 1+ \sum\limits_{n=1}^{\infty}p_n z^n \in \mathcal{P}$, then
%\begin{align*}
%2p_2 &= p^2_1+x(4-p^2_1)\\
%4p_3 &=p^3_1+2p_1(4-p^2_1)x-p_1(4-p^2_1)x^2 +2(4-p^2_1)(1-|x|^2)z
%\end{align*}
%for some $x$ and $z$ such that $|x| \le 1$ and $|z|\le1$.
%\end{lemma}

\section{Zalcman Conjecture}
In this section, we first prove Zalcman conjecture $(n=2)$ for starlike functions with respect to the symmetric points .
\begin{theorem}
If the function $f \in \mathcal{S_{S}^*}$ is of the form $f(z) = z+a_2z^2+a_3z^3+ \cdots$. Then
 \[\vert a_2^2-a_3\vert  \le 1.\]
The inequality is sharp.
\end{theorem}

\begin{proof}
Let $f \in \mathcal{S_{S}^*}$. Then we have
\[\frac{2 zf'(z)}{f(z) -f(-z)} \prec \frac{1+z}{1-z}\] for all $z\in\mathbb{D}$ so that
\[\frac{2zf'(z)}{f(z) -f(-z)}=p(z)\] where $\prec$ denotes subordination and $p(z) = 1+p_1z+p_2z^2+\cdots \in \mathcal{P}$.
 On comparing the coefficients of like power terms on both sides, we get
 \begin{align}
 a_2 &= \frac{p_1}{2}; \label{a2}  \\
 a_3 &= \frac{p_2}{2}; \label{a3} \\
 a_4 &= \frac{1}{8} (p_1p_2+2p_3); \label{a4} \\
 a_5 &= \frac{1}{8} (p_2^2+2p_4); \label{a5}\\
 a_6 &= \frac{1}{48}(4p_2p_3+p_1(p_2^2+2p_4)+8p_5);\label{a6}\\
 a_7 &= \frac{1}{48} (p_2^3+6p_2p_4+8p_6); \label{a7}
 %a_8 &= \frac{1}{384} (p_1(p_2^3+6p_2p_4+8p_6)+6(p_2^2p_3+2p_3p_4+4p_2p_5+8p_7)); \label{a8}\\
% a_9 &= \frac{1}{384} (p_2^4+12p_2^2p_4+32p_2p_6+12(p_4^2+4p_8)). \label{a9}
 \end{align}
 It follows from (\ref{a2}) and (\ref{a3}) that
\[a_2^2-a_3= \frac{p_1^2}{4}-\frac{p_2}{2}.\]
By using Lemma \ref{lem4}, we get
\begin{align*}
\vert a_2^2-a_3\vert  &= \frac{1}{2}\left\vert \frac{1}{2}p_1^2-p_2\right\vert \leq1.
\end{align*}
The inequality is sharp for the function
\begin{equation}\label{th1}
\frac{2zf'(z)}{f(z) -f(-z)}= \dfrac{1+z^2}{1-z^2}= 1+2 z^2+2z^4+ \cdots
\end{equation}
by noting  the fact $a_2 =0, a_3=1$ implies $\vert a_2^2-a_3\vert =1$.
\end{proof}


Next we prove the generalized Zalcman conjecture for $m=2$ and $n=4$.
\begin{theorem}
Let the function $ f(z)=z+a_2z^2+a_3z^3+ \cdots\in \mathcal{S_{S}^*}$. Then
 \[\vert a_2a_4-a_5\vert  \leq 1.\]
The inequality is sharp.
\end{theorem}

\begin{proof}
If the function $f  \in \mathcal{S_{S}^*}$,  then using (\ref{a2}),(\ref{a4}) and (\ref{a5}), we get
\begin{align*}
a_2a_4-a_5 &=\frac{1}{16}p_1(p_1p_2+2p_3)-\frac{1}{8}(p_2^2+2p_4)\\
& =\frac{1}{8}p_2\left(\frac{1}{2}p_1^2-p_2\right)+\frac{1}{4}\left(\frac{1}{2}p_1p_3-p_4\right).
\end{align*}
Using triangle inequality
\[
\vert a_2a_4-a_5\vert \le \frac{1}{8}\vert p_2\vert \left\vert \frac{1}{2}p_1^2-p_2\right\vert +\frac{1}{4}\left\vert \frac{1}{2}p_1p_3-p_4\right\vert .
\]
Applying Lemma \ref{lem4} and the fact $\vert p_n\vert  \le 2$, we get
\[\vert a_2a_4-a_5\vert  \le 1.\]
The inequality is sharp for the function $f$ defined by \eqref{th1}.\qedhere
%\[\frac{2zf'(z)}{f(z) -f(-z)}= \dfrac{1+z^2}{1-z^2}= 1+2 z^2+2z^4+ \cdots.\]
%Here we have $a_2 =0, a_4=0, a_5=1$ and hence $|a_2a_4-a _5|=1$.
\end{proof}





\section{Hankel Determinants}
Using the technique discussed in \cite{zap2}, the following theorem gives an improved estimate on $H_3(1)$ for the functions $f$ in the class $\mathcal{S}_S^*$.
\begin{theorem}\label{thmh31}
Let the function $ f \in \mathcal{S_{S}^*}$ be of the form $f(z) = z+a_2z^2+a_3z^3+ \cdots$. Then
\[\vert H_3(1)(f)\vert  \le \frac{329}{400} \simeq 0.8225.\]
\end{theorem}

\begin{proof}
Let $f \in \mathcal{S_{S}^*}$. Then we have
\[\frac{2 zf'(z)}{f(z) -f(-z)} \prec \frac{1+z}{1-z}\]
so that
\[\frac{2 zf'(z)}{f(z) -f(-z)}=\frac{1+w(z)}{1-w(z)},\qquad z\in \D\]
where $\prec$ denotes subordination and $w(z) = c_1z+c_2 z^2+ \cdots$ is a Schwarz function.
%\[2zf'(z)(1-w(z))= (f(z)-f(-z)) (1+w(z)).\]
On comparing the coefficients of like powers of $z$, we get
\begin{align}\label{eq5}
a_2 &= c_1,\, a_3 = c_1^2+c_2,\, a_4 = \frac{1}{2}(c_3+3 c_1 c_2+2c_1^3)\\
a_5 &= \frac{1}{2}(c_4+2c_1c_2+5c_1^2c_2+2c_1^4+2c_2^2)\label{eqq1}.
\end{align}
Therefore, in view of (\ref{h31}), \eqref{eq5} and \eqref{eqq1} the third order Hankel determinant $H_3(1)$ becomes
\begin{align*}
H_3(1) (f) &= \frac{1}{4}(c_1^2c_2^2+2c_1c_2c_3-c_3^2+2c_2c_4)\\
&= \frac{1}{4}(-2c_3(c_3-c_1c_2)+c_3^2+c_1^2c_2^2+2c_2c_4).
\end{align*}
Hence, applying Lemma \ref{lem1} $(\mu=-1, \nu =0)$ and inequalities given in (\ref{eq4}), we get
\begin{align*}
\vert H_3(1)(f)\vert  &\leq \frac{1}{4}(2\left(1-\vert c_1\vert ^2-\frac{\vert c_2\vert ^2}{1+\vert c_1\vert }\right)+\left(1-\vert c_1\vert ^2-\frac{\vert c_2\vert ^2}{1+\vert c_1\vert }\right)^2\\
&\qquad\qquad\qquad\qquad+\vert c_1\vert ^2\vert c_2\vert ^2+2\vert c_2\vert \left(1-\vert c_1\vert ^2-\vert c_2\vert ^2\right))\\
&=\frac{1}{4} G(\vert c_1\vert ,\vert c_2\vert ).
\end{align*}
The function $G(x,y)$ is given by
\[G(x,y) = g_1(x,y)+g_2(x,y)+g_3(x,y),\]
where
\begin{align*}
g_1(x,y) &= \left(\frac{y}{(1+x)^2}-1\right)y^3\\
g_2(x,y) &= 2(1-x^2)y- \frac{4-3x^2-x^3}{1+x} y^2\\
g_3(x,y) &= -y^3+ x^4-4 x^2+3
\end{align*}
where $x=\vert c_1\vert $ and $y=\vert c_2\vert $. In view of $\vert c_2\vert  \le 1-\vert c_1\vert ^2$, we maximize the function $G(x,y)$ in the region
\[\Omega= \{(x,y) : x\ge 0,y\ge 0, y\le 1-x^2\}.\]
It is noted that
\begin{equation} \label{eq8}
g_1(x,y) \le 0.
\end{equation}
Since $g_2(x,y)$ is a quadratic expression in $y$, so it attains its maximum value at \[y_0= \dfrac{(1-x^2) (1+x)}{4-3x^2-x^3}.\] Also $y_0 < 1-x^2$ for all $x \in [0,1]$ and thus we have
\[g_2(x,y) \le g_2(x,y_0) = \frac{(1-x)(1+x)^3}{(2+x)^2}=:f(x).\]
A simple calcultaion shows that $x_2 = 0.3$ is a critical point of the function $f$ in $(0,1)$. Hence,
\begin{equation} \label{eq9}
g_2(x,y) \le f(x_2) = \frac{29}{100}.
\end{equation}
For the function $g_3(x,y)$, it is evident that
\[g_3(x,y) \le g_3(x,0) = x^4-4x^2+3 =:h(x).\]
Now $h'(x) = 4x(x^2-2)$, so $h(x) \le h(0)$. This gives
\begin{equation} \label{eq10}
g_3(x,y) \le h(0) = 3.
\end{equation}
Using (\ref{eq8}),(\ref{eq9}) and (\ref{eq10}), we get $G(x,y) \le {329}/{100}.$
Therefore, we have $\vert H_3(1)(f)\vert  \le  {329}/{400} \simeq 0.8225. $
\end{proof}

\begin{remark}
The obtained upper bound $\frac{329}{400} \simeq 0.8225$ on $H_3(1)(f)$~\eqref{thmh31} improves the existing bound $\frac{5}{4}\simeq 1.25$ \cite[Theorem 2.3, p.227]{vsv}  for the functions $f \in \mathcal{S_{S}^*}$.
\end{remark}



Next theorem gives bound on $H_3(2)$ for the functions $f \in \mathcal{S_{S}^*}$.
\begin{theorem}
If $ f \in \mathcal{S_{S}^*}$ is of the form $f(z) = z+a_2z^2+a_3z^3+ \cdots$. Then
\[\vert H_3(2)(f)\vert < \frac{83}{24} \simeq3.45 .\]
\end{theorem}
\begin{proof}
On substituting the values of $a_4$, $a_5$ and $a_6$ from \eqref{a4}, \eqref{a5} and \eqref{a6}, respectively in the expression $a_4a_6-a_5^2$, we have
\begin{align*}
&a_4a_6-a_5^2  \\
& =\frac{1}{384}(p_1p_2+2p_3)(4p_2p_3+p_1p_2^2+2p_1p_4+8p_5)-\frac{1}{64}(p_2^4+4p_4^2+4p_2^2p_4)\\
&= \frac{1}{384}(4p_1p_2^2p_3+p_1^2p_2^3+2p_1^2p_2p_4+8p_1p_2p_5+8p_2p_3^2+2p_1p_2^2p_3+4p_1p_3p_4\\
&\quad \quad\qquad+16p_3p_5)-\frac{1}{64}p_2^4-\frac{1}{16}p_4^2-\frac{1}{16}p_2^2p_4\\
&= \frac{1}{384}p_1^2p_2^3+\frac{1}{64}p_1p_2^2p_3+\frac{1}{48}p_2p_3^2+\frac{1}{192}p_1^2p_2p_4+\frac{1}{96}p_1p_4p_3+\frac{1}{48}p_1p_2p_5\\
&\quad \quad\qquad+\frac{1}{24}p_3p_5-\frac{1}{64}p_2^4-\frac{1}{16}p_4^2-\frac{1}{16}p_2^2p_4\\
&= \frac{1}{64}p_2^3\left(\dfrac{1}{6}p_1^2-p_2\right)+\frac{1}{16}p_2^2\left(\dfrac{1}{4}p_1p_3-p_4\right)+\frac{1}{24}p_3\left(\dfrac{1}{2}p_2p_3+p_5\right)\\
& \quad \quad+\frac{1}{16}p_4\left(\dfrac{1}{6}p_1p_3-p_4\right)+\frac{1}{48}p_1p_2\left(\dfrac{1}{4}p_1p_4+p_5\right).
\end{align*}
By triangle inequality, we get
\begin{align*}
\vert a_4a_6-a_5^2\vert  &\le \frac{1}{64}\vert p_2^3\vert \left\vert \dfrac{1}{6}p_1^2-p_2\right\vert +\frac{1}{16}\vert p_2^2\vert \left\vert \dfrac{1}{4}p_1p_3-p_4\right\vert +\frac{1}{24}\vert p_3\vert \left\vert \dfrac{1}{2}p_2p_3+p_5\right\vert \\
& \quad \quad+\frac{1}{16}\vert p_4\vert \left\vert \dfrac{1}{6}p_1p_3-p_4\right\vert +\frac{1}{48}\vert p_1\vert \vert p_2\vert \left\vert \dfrac{1}{4}p_1p_4+p_5\right\vert .
\end{align*}
Using Lemma \ref{lem4} and the inequality $\vert p_n\vert  \le 2$, we get
\begin{align}\label{eqa10}
\vert a_4a_6-a_5^2\vert  &\le
%\frac{1}{64}(8)(2)+\frac{1}{16}(4)(2)+\frac{1}{24}(2)(4)\notag \\
%&\quad+\frac{1}{16}(2)(2)+\frac{1}{48}(4)(3) \notag\\
 \frac{19}{12}.
\end{align}
Again, on substituting the values of $a_3$, $a_4$, $a_5$ and $a_6$ from (\ref{a3}),  (\ref{a4}), (\ref{a5}) and (\ref{a6}), respectively in the expression $a_3a_6-a_4a_5$, we have
\begin{align*}
a_3a_6-a_4a_5
&= \frac{1}{96}(4p_2^2p_3+p_1p_2^3+2p_1p_2p_4+8p_2p_5)\\
& \quad\quad-\frac{1}{64}(p_1p_2^3+2p_1p_2p_4+2p_2^2p_4+2p_2^2p_3+4p_3p_4)\\
&= \frac{1}{96}p_2^2p_3-\frac{1}{192}p_1p_2^3-\frac{1}{96}p_1p_2p_4+\frac{1}{12}p_2p_5-\frac{1}{16}p_3p_4.
\end{align*}
So that
\[
\vert a_3a_6-a_4a_5 \vert   \le\frac{1}{16}\vert p_3\vert \left\vert \frac{1}{12}p_2^2-p_4\right\vert + \frac{1}{192}\vert p_2^2\vert \vert p_1p_2-p_3\vert + \frac{1}{12}\vert p_2\vert \left\vert \frac{1}{8}p_1p_4-p_5\right\vert
\]
By Lemma \ref{lem4} and the fact $\vert p_n\vert  \le 2$, we have
\begin{align}\label{eqa9}
\vert a_3a_6-a_4a_5\vert  &\le
% \frac{1}{16}(2)(2)+\frac{1}{192}(4)(2)+\frac{1}{12}(2)(2)\notag \\
%& =
\frac{5}{8}.
\end{align}
On substituting the values of $a_3$, $a_4$ and $a_5$ from (\ref{a3}), (\ref{a4}) and (\ref{a5}), respectively in the expression $a_3a_5-a_4^2$, we have
\begin{align*}
a_3a_5-a_4^2
&= -\frac{1}{16}p_2^2\left(\frac{1}{4}p_1^2-p_2\right)- \frac{1}{8}p_2\left(\frac{1}{2}p_1p_3-p_4\right)- \frac{1}{16}p_3^2
\end{align*}
so that
\[\vert a_3a_5-a_4^2 \vert  \le \frac{1}{16}\vert p_2\vert ^2\left\vert \frac{1}{4}p_1^2-p_2\right\vert +\frac{1}{8}\vert p_2\vert \left\vert \frac{1}{2}p_1p_3-p_4\right\vert + \frac{1}{16}\vert p_3\vert ^2. \]
Using Lemma \ref{lem4} and the inequality $\vert p_n\vert  \le 2$, we have
\begin{align}\label{eqa8}
\vert a_3a_5-a_4^2\vert  &\le
%\frac{1}{16}(4)(2)+\frac{1}{8}(2)(2)+\frac{1}{16}(4) =
\frac{5}{4}.
\end{align}
It follows from (\ref{h32}) that
\[\vert H_3(2)(f)\vert  \le \vert  a_2\vert \vert a_4a_6-a_5^2\vert  + \vert a_3\vert \vert a_3a_6-a_4a_5\vert +\vert a_4\vert \vert a_3 a_5-a_4^2\vert .\]
Using inequality (\ref{eqa10}), (\ref{eqa9}) and (\ref{eqa8}) and Lemma \ref{lem3}, we have $\vert H_3(2)(f)\vert  \le {83}/{24} \simeq3.45.$
\end{proof}
In the next theorem we estimate third order Hankel determinant $H_3(3)$ for $f \in \mathcal{S_{S}^*}$.
\begin{theorem}
If $f(z) = z+ \sum\limits_{n=2}^{\infty}a_nz^n \in \mathcal{S_{S}^*}$, then
\[\vert H_3(3)(f)\vert \leq \frac{89}{24}\simeq 3.7.\]
\end{theorem}
\begin{proof}
On substituting (\ref{a5}),(\ref{a6}) and (\ref{a7}), we have
\begin{align*}
&a_5a_7-a_6^2\\
&=\frac{1}{384}(p_2^5+6p_2^3p_4+8p_2^2p_6+2p_2^3p_4+12p_2p_4^2+16p_4p_6)-\frac{1}{2304}(16p_2^2p_3^2\\
& \quad +p_1^2p_2^4+4p_1^2p_4^2+64p_5^2+8p_1p_2^3p_3+16p_1p_2p_3p_4+64p_2p_3p_5+4p_1^2p_2^2p_4\\
& \quad +16p_1p_2^2p_5+32p_1p_4p_5)\\
&= \frac{1}{96}p_2^2\left(p_6-\dfrac{2}{3}p_3^2\right)+\frac{1}{384}p_2^4\left(p_2-\dfrac{1}{6}p_1^2\right)+\frac{1}{192}p_4^2\left(p_2-\dfrac{1}{3}p_1^2\right)\\
& \quad+\frac{1}{64}p_2^3\left(p_4-\dfrac{2}{9}p_1p_3\right)+\frac{5}{192}p_2p_4\left(p_4-\dfrac{4}{15}p_1p_3\right)+\frac{1}{192}p_2^2p_4\\
&\quad\left(p_2-\dfrac{1}{3}p_1^2\right)+\frac{1}{24}p_4\left(p_6-\dfrac{1}{3}p_1p_5\right)+\frac{1}{96}p_2^2\left(p_6-\dfrac{2}{3}p_1p_5\right)-\frac{1}{36}p_5^2\\
&\quad-\frac{1}{36}p_2p_3p_5.
\end{align*}
Using triangle inequality, Lemma \ref{lem4} and the fact $\vert p_n\vert  \le 2$, we get
\begin{align}\label{eqa12}
\vert a_5a_7-a_6^2\vert  &\le
%\frac{1}{96}(4)(2)+\frac{1}{384}(16)(2)+\frac{1}{192}(4)(2)  +\frac{1}{64}(8)(2)+\frac{5}{192}(4)(2)+\frac{1}{192}(8)(2)\notag\\
%&\quad+\frac{1}{24}(2)(2)+\frac{1}{96}(4)(2)+\frac{1}{36}(4)+\frac{1}{36}(8) \notag\\
%& =
 \frac{4}{3}.
\end{align}
Again in view of (\ref{a4}),(\ref{a5}),(\ref{a6}) and (\ref{a7}), we have

\begin{align*}
&a_4a_7-a_6a_5\\
&=\frac{1}{384}(p_1p_2^4+6p_1p_2^2p_4+8p_1p_2p_6+2p_2^3p_3+12p_2p_3p_4+16p_3p_6)\\
&\quad-\frac{1}{384}(4p_2^3p_3+8p_2p_3p_4+p_1p_2^4+4p_1p_2^2p_4+4p_1p_4^2+8p_2^2p_5+16p_4p_5)\\
& =\frac{1}{48}p_2^2\left(\dfrac{1}{4}p_1p_4-p_5\right)+\frac{1}{24}p_4\left(\dfrac{1}{8}p_2p_3-p_5\right)+\frac{1}{192}p_2p_3(p_4-p_2^2)\\
& \quad +\frac{1}{24}p_6\left(\dfrac{1}{2}p_1p_2+p_3\right)+\frac{1}{96}p_1p_4^2.
\end{align*}
Using triangle inequality, by Lemma  \ref{lem4} and $\vert p_n\vert  \le 2$, we have
\begin{align}\label{eqa11}
\vert a_4a_7-a_6a_5\vert  &\le
%\frac{1}{48}(4)(2)+\frac{1}{24}(2)(2)+\frac{1}{192}(4)(2)+\frac{1}{24}(2)(4)+\frac{1}{96}(2)(4) \notag\\
%& =
\frac{19}{24}.
\end{align}
It follows from (\ref{h33}) that
\[\vert H_3(3)\vert  \le \vert  a_3\vert \vert a_5a_7-a_6^2\vert  + \vert a_4\vert \vert a_4a_7-a_6a_5\vert +\vert a_5\vert \vert a_4 a_6-a_5^2\vert .\]
Using  (\ref{eqa10}),(\ref{eqa12}), and (\ref{eqa11})  and Lemma \ref{lem3}, we have
$\vert H_3(3)(f)\vert  \le {89}/{24}\simeq 3.7.$
\end{proof}
Next we compute an estimate on the fourth Hankel determinant $H_4(1)$.
\begin{theorem}
Let $ f \in \mathcal{S_{S}^*}$ be of the form $f(z) = z+a_2z^2+a_3z^3+ \cdots$. Then
\begin{equation*}
\vert H_4(1)(f)\vert \leq 1.84.
\end{equation*}
\end{theorem}

\begin{proof}
Since $ f \in \mathcal{S_{S}^*}$, then in view of (\ref{eq11}), (\ref{a2}), (\ref{a3}), (\ref{a4}), (\ref{a5}),(\ref{a6}) and (\ref{a7}), we get
\begin{align*}
&36864 H_4(1)(f)=p_1^4(p_2^2-4p_4)^2+8p_1^3(p_2^2-4p_4)(p_2p_3-4p_5)\\
&-32p_1(2p_2^4p_3-2p_2^2p_3p_4-2p_2^3p_5+12p_3(-2p_4^2+p_3p_5)\\
&+p_2(-3p_3^3+20p_4p_5-12p_3p_6))+8(3p_2^6-6p_2^4p_4+4p_2^2(9p_4^2+20p_3p_5)\\
&-4p_2^3(p_3^2+12p_6)+6(3p_3^4-12p_4^3+16p_3p_4p_5-8p_3^2p_6)\\
&-32p_2(3p_3^2p_4+2p_5^2-3p_4p_6))-8p_1^2(p_2^5-8p_2^3p_4\\
&+16p_2(p_4^2+p_3p_5)-p_2^2(5p_3^2+12p_6)+4(3p_3^2p_4-8p_5^2+12p_4p_6)).
\end{align*}

\begin{align*}
&36864 H_4(1)(f)=p_1^4p_2^4-8p_1^2p_2^5+24p_2^6+8p_1^3p_2^3p_3-64p_1p_2^4p_3+40p_1^2p_2^2p_3^2\\
&\quad-32p_2^3p_3^2+96p_1p_2p_3^3+144p_3^4-8p_1^4p_2^2p_4+64p_1^2p_2^3p_4-48p_2^4p_4\\
&\quad-32p_1^3p_2p_3p_4+64p_1p_2^2p_3p_4-96p_1^2p_3^2p_4-768p_2p_3^2p_4+16p_1^4p_4^2\\
&\quad-128p_1^2p_2p_4^2+288p_2^2p_4^2+768p_1p_3p_4^2-576p_4^3-32p_1^3p_2^2p_5+64p_1p_2^3p_5\\
&\quad-128p_1^2p_2p_3p_5+640p_2^2p_3p_5-384p_1p_3^2p_5+128p_1^3p_4p_5\\
&\quad-640p_1p_2p_4p_5+768p_3p_4p_5+256p_1^2p_5^2-512p_2p_5^2+96p_1^2p_2^2p_6\\
&\quad-384p_2^3p_6+384p_1p_2p_3p_6-384p_3^2p_6-384p_1^2p_4p_6+768p_2p_4p_6.
\end{align*}
A simple calculation gives
\begin{align*}
&36864 H_4(1)(f)= 8p_1^4p_2^2\left(\dfrac{1}{8}p_2^2-p_4\right)+\frac{1}{2}p_1^3 \left(\dfrac{1}{4}p_2^2-p_4\right)\left(\dfrac{1}{4}p_2p_3-p_5\right)\\
&\quad-64p_1^2p_2^3\left(\dfrac{1}{8}p_2^2-p_4\right)-64p_1p_2^2p_3(p_2^2-p_4)+32p_2^2p_3^2\left(\dfrac{5}{4}p_1^2-p_2\right)\\
&\quad+48p_2^4\left(\dfrac{1}{2}p_2^2-p_4\right) -96p_1p_3p_4(p_1p_3-p_4)+576p_4^2(p_1p_3-p_4)\\
&\quad -768p_3p_4(p_2p_3-p_5)-96p_1p_4^2(p_1p_2-p_3)-640p_2p_3p_5\left(\dfrac{1}{5}p_1^2-p_2\right)\\
&\quad-640p_2p_4(p_1p_5-p_6)+384p_3p_6(p_1p_2-p_3)-128p_4p_6(3p_1^2-p_2)\\
&\quad+512p_5^2\left(\dfrac{1}{2}p_1^2-p_2\right) +192p_2^2p_6\left(\dfrac{1}{2}p_1^2-p_2\right)+192p_2^3\left(\dfrac{1}{3}p_1p_5-p_6\right)\\
&\quad-384p_1p_3^2\left(\dfrac{1}{8}p_2p_3-p_5\right)-288p_2p_4^2\left(\dfrac{1}{9}p_1^2-p_2\right)\\
&\quad-144p_3^3\left(-\dfrac{1}{3}p_1p_2-p_3\right)
\end{align*}
which implies
\begin{align*}
&36864 \vert H_4(1)(f)\vert  \le 8p_1^4p_2^2\left\vert \dfrac{1}{8}p_2^2-p_4\right\vert +\frac{1}{2}p_1^3 \left\vert \dfrac{1}{4}p_2^2-p_4\right\vert \left\vert \dfrac{1}{4}p_2p_3-p_5\right\vert\\
&\quad +64p_1^2p_2^3\left\vert \dfrac{1}{8}p_2^2-p_4\right\vert+64p_1p_2^2p_3\vert p_2^2-p_4\vert +32p_2^2p_3^2\left\vert \dfrac{5}{4}p_1^2-p_2\right\vert  \\
&\quad +48p_2^4\left\vert \dfrac{1}{2}p_2^2-p_4\right\vert +96p_1p_3p_4\vert p_1p_3-p_4\vert +576p_4^2\vert p_1p_3-p_4\vert\\
&\quad  +768p_3p_4\vert p_2p_3-p_5\vert +96p_1p_4^2\vert p_1p_2-p_3\vert +640p_2p_3p_5\left\vert \dfrac{1}{5}p_1^2-p_2\right\vert\\
&\quad  +640p_2p_4\vert p_1p_5-p_6\vert +384p_3p_6\vert p_1p_2-p_3\vert +128p_4p_6\vert 3p_1^2-p_2\vert \\
&\quad +512p_5^2\left\vert \dfrac{1}{2}p_1^2-p_2\right\vert +192p_2^2p_6\left\vert \dfrac{1}{2}p_1^2-p_2\right\vert +192p_2^3\left\vert \dfrac{1}{3}p_1p_5-p_6\right\vert\\
&\quad  +384p_1p_3^2\left\vert \dfrac{1}{8}p_2p_3-p_5\right\vert +288p_2p_4^2\left\vert \dfrac{1}{9}p_1^2-p_2\right\vert +144p_3^3\left\vert -\dfrac{1}{3}p_1p_2-p_3\right\vert .
\end{align*}
Using Lemma \ref{lem4} and the fact $\vert p_n\vert  \le 2$, we get
%\begin{align*}
%36864|H_4(1)(f)| &\le
% 8(2)^4(2)^2(2)+\frac{1}{2}(2)^3(2)(2)+64(2)^2(2)^3(2)+64(2)(2)^2(2)(2)+32(2)^2(2)^22\\
%&+48(2)^4(2)+96(2)(2)(2)(2)+576(2)^2(2)+768(2)(2)(2)+96(2)(2)^2(2)\\
%&+640(2)(2)(2)(2)+640(2)(2)(2)+384(2)(2)(2)+128(2)(2)(2)+512(2)^2(2)\\
%&+192(2)(2)(2)(2)+192(2)^3(2)+384(2)(2)^2(2)+288(2)(2)^2(2)+144(2)^3(\frac{10}{3}).
%\end{align*}
%so that
\begin{align*}
\vert H_4(1)(f)\vert  &\le \frac{4241}{2304} \simeq1.84.
\end{align*}
Thus, we have the required bound for $\vert H_4(1)(f)\vert$.
\end{proof}

\section{Toeplitz Determinants}
In this section, we first compute the bound on second Toeplitz determinant $T_2(2)$.
\begin{theorem} \label{thmt22}
If $ f \in \mathcal{S_{S}^*}$ be of the form $f(z) = z+a_2z^2+a_3z^3+ \cdots$. Then
\[\vert  T_2(2)(f)\vert  \le 2.\]
The inequality is sharp.
\end{theorem}

\begin{proof}
Since $ f \in \mathcal{S_{S}^*}$, then on putting the values of $a_2$ and $a_3$ from (\ref{a2}) and (\ref{a3}) in expression $T_2(2)=a_3^2-a_2^2$, we get
\[\vert  T_2(2)\vert  =\vert  a^2_3-a^2_2\vert   = \left\vert  \frac{p^2_2}{4}-\frac{p^2_1}{4}\right\vert  .\]
Applying triangle inequality and using the fact $\vert  p_n\vert   \le 2$, we get
\[\vert  T_2(2)\vert  \le 2.\]
The inequality is sharp for the function $f: \D \to \C$ defined as
\[f(z) = \frac{z}{1-i z}.\] It is noted that $a_2 = i, a_3 =-1$ and thus $\vert  a^2_3-a^2_2\vert  =2$.\qedhere
\end{proof}
Next, we obtain an estimate for second Toeplitz determinant $T_2(3)$.
\begin{theorem}
Let $ f \in \mathcal{S_{S}^*}$ be of the form $f(z) = z+a_2z^2+a_3z^3+ \cdots$. Then
\[\vert  T_2(3)(f)\vert   \le 2.\]
The inequality is sharp.
\end{theorem}

\begin{proof}
For $ f \in \mathcal{S_{S}^*}$, then on putting the values of $a_3$ and $a_4$ from  (\ref{a3}) and (\ref{a4}) in expression $T_2(3)=a^2_4-a^2_3$, we get
\begin{align*}
\vert  T_2(3)\vert  =\vert  a^2_4-a^2_3\vert   &= \left\vert  \frac{1}{64}(p_1p_2+2p_3)^2-\frac{p^2_2}{4}\right\vert  \\
&=\left\vert  \frac{1}{16}\left(-\dfrac{1}{2}p_1p_2-p_3\right)^2-\frac{p^2_2}{4}\right\vert  .
\end{align*}
Applying triangle inequality, Lemma \ref{lem4} and the fact $\vert  p_n\vert   \le 2$, we get

\[\vert  T_2(3)\vert   \le 2.\]
To prove the sharpness, consider the function $f: \D \to \C$ defined as
\[f(z) = \frac{z}{1-i z}.\] Here $a_3 = -1$ and $ a_4 =- i$ and thus $\vert  a^2_4-a^2_3\vert  =2$. \qedhere
\end{proof}
In the next theorem we obtain an estimate for the bound on third Toeplitz determinant $T_3(1)$.
\begin{theorem}
If $ f \in \mathcal{S_{S}^*}$ be of the form $f(z) = z+a_2z^2+a_3z^3+ \cdots$. Then
\[\vert  T_3(1)(f)\vert  \le 4.\]
The inequality is sharp.
\end{theorem}

\begin{proof}
Let $ f \in \mathcal{S_{S}^*}$. Then in view of \eqref{eq20}, (\ref{a2}) and (\ref{a3}), we get
\begin{align*}
\vert  T_3(1)\vert  =\vert  1+2a^2_2(a_3-1)-a^2_3\vert   &= \left\vert  1+2 \frac{p_1^2}{4}\left(\frac{p_2}{2}-1\right)-\left(\frac{p_2^2}{4}\right)\right\vert  \\
&= \frac{1}{4}\left\vert  4+ p_1^2 p_2-2 p_1^2-p_2^2\right\vert  \\
&=\frac{1}{4}\left\vert  4+ p_2(p_1^2-p_2)-2 p_1^2\right\vert  .
\end{align*}
Using triangle inequality, we obtain
\[\vert  T_3(1)\vert   \le \frac{1}{4}(4+ \vert  p_2\vert  \vert  p_1^2-p_2\vert  +2 \vert  p_1^2\vert  ). \]
Applying Lemma \ref{lem4} and using the fact that $\vert  p_n\vert   \le 2$, we get
\[\vert  T_3(1)\vert   \le 4.\]
For the function
$f(z) = \dfrac{z}{1-i z}$, we have $a_2 =i$ and  $a_3 =-1$. Thus, we get
 \[\vert  1+2a^2_2(a_3-1)-a^2_3\vert  =4.\]
 This proves the sharpness of the result.
 \end{proof}

Next we compute the bound on third Toeplitz determinant $T_3(2)$.
\begin{theorem}
Let $ f \in \mathcal{S_{S}^*}$ be of the form $f(z) = z+a_2z^2+a_3z^3+ \cdots$. Then
\[ \vert T_3(2)(f)\vert  =\vert  (a_2-a_4)(a_2^2-2 a_2^3+a_2 a_4)\vert   \le 6.\]
\end{theorem}
\begin{proof}
In view of (\ref{a2}) and (\ref{a4}), we get
\begin{align*}
&\vert  T_3(2)(f)\vert \\
&=\vert  (a_2-a_4)(a_2^2-2 a_2^3+a_2 a_4)\vert   \\
&= \left\vert  \left(\frac{p_1}{2}-\frac{1}{8}(p_1p_2+2p_3)\right)\left(\frac{p_1^2}{4}-\frac{p_1^3}{4}+\frac{p_1}{16}(p_1p_2+2p_3)\right)\right\vert  \\
& =\left\vert  \frac{p_1^3}{8}- \frac{p_1^4}{8}+\frac{1}{32}p_1^4p_2-\frac{1}{128}p_1^3p_2^2+\frac{1}{16}p_1^3p_3-\frac{1}{32}p_1^2p_2p_3-\frac{1}{32}p_1p_3^2\right\vert   \\
&=\left\vert  \frac{p_1^3}{8}- \frac{p_1^4}{8}-\frac{1}{16}p_1^3\left(-\frac{1}{2}p_1p_2-p_3\right)+\frac{1}{32}p_1^2p_2\left(-\frac{1}{4}p_1p_2-p_3\right)-\frac{1}{32}p_1p_3^2\right\vert  .
\end{align*}
Using triangle inequality, we obtain
 \begin{align*}
 \vert  T_3(2)(f)\vert   &\le \frac{1}{8}\vert  p_1\vert  ^3+ \frac{1}{8}\vert  p_1\vert  ^4+\frac{1}{16}\vert  p_1\vert  ^3\left\vert  -\frac{1}{2}p_1p_2-p_3\right\vert  +\frac{1}{32}\vert  p_1\vert  \vert  p_3\vert  ^2\\
 &\quad+\frac{1}{32}\vert  p_1\vert  ^2\vert  p_2\vert  \left\vert  -\frac{1}{4}p_1p_2-p_3\right\vert.
 \end{align*}
By using Lemma \ref{lem4} and the inequality $\vert  p_n\vert   \le 2$, we get $\vert  T_3(2)(f)\vert   \le 6.$
\end{proof}


The following theorem gives an estimate on fourth Toeplitz deteminant $T_4(2)$.
\begin{theorem}
Let $ f \in \mathcal{S_{S}^*}$ be of the form $f(z) = z+a_2z^2+a_3z^3+ \cdots$. Then
\begin{align*}
\vert  T_4(2)(f)\vert  =&\vert  (a^2_2-a^2_3)^2+2(a_3^2-a_2a_4)(a_2a_4-a_3a_5)-(a_2a_3-a_3a_4)^2\\
   &
   +(a_4^2-a_3a_5)^2-(a_3a_4-a_2a_5)^2\vert  \le 15.12.
\end{align*}
\end{theorem}

\begin{proof}
In view of (\ref{eq16}), (\ref{a2}), (\ref{a3}), (\ref{a4}) and (\ref{a5}) and on rearranging the terms,  we have
\begin{align*}
&4096 T_4(2)(f)\\
&= 256(p_1^2-p_2^2)^2-16p_2^2(p_1(-4+p_2)+2p_3)^2-64(p_2p_3-p_1p_4)^2\\
&\quad+((p_1p_2+2p_3)^2-4p_2(p_2^2+2p_4))^2-32(p_1^2p_2-4p_2^2+2p_1p_3)\\
&\quad(p_1^2p_2+2p_1p_3-p_2(p_2^2+2p_4))\\
& = 256p_1^4-768p_1^2p_2^2-32p_1^4p_2^2+256p_1^2p_2^3+256p_2^4+16p_1^2p_2^4+p_1^4p_2^4\\
&\quad -128p_2^5-8p_1^2p_2^5+16p_2^6-128p_1^3p_2p_3+512p_1p_2^2p_3+8p_1^3p_2^3p_3\\
&\quad-32p_1p_2^4p_3-128p_1^2p_3^2-128p_2^2p_3^2+24p_1^2p_2^2p_3^2-32p_2^3p_3^2+32p_1p_2p_3^3\\
&\quad+16p_3^4+64p_1^2p_2^2p_4-256p_2^3p_4-16p_1^2p_2^3p_4+64p_2^4p_4+256p_1p_2p_3p_4\\
&\quad-64p_1p_2^2p_3p_4-64p_2p_3^2p_4-64p_1^2p_4^2+64p_2^2p_4^2\\
&=-256p_1^2p_2^2\left(\frac{1}{8}p_1^2-p_2\right)+128p_2^4\left(\frac{1}{8}p_1^2-p_2\right)-16p_2^5\left(\frac{1}{2}p_1^2-p_2\right)\\
&\quad-512p_1p_2p_3\left(\frac{1}{4}p_1^2-p_2\right)-64p_2^4\left(\frac{1}{2}p_1p_3-p_4\right)+32p_2^2p_3^2\left(\frac{3}{4}p_1^2-p_2\right)\\
&\quad+16p_1^2p_2^3\left(\frac{1}{2}p_1p_3-p_4\right)+64p_2p_3^2\left(\frac{1}{2}p_1p_3-p_4\right)+64p_1^2p_4(p_2^2-p_4)\\
&\quad-64p_2^2p_4(p_1p_3-p_4)+256p_1^4-768p_1^2p_2^2+256p_2^4+p_1^4p_2^4-128p_1^2p_3^2\\
&\quad-128p_2^2p_3^2+16p_3^4-256p_2^3p_4+256p_1p_2p_3p_4.
\end{align*}
Using triangle inequality, we get
\begin{align*}
& 4096\vert  T_4(2)(f)\vert  \le 256\vert  p_1\vert  ^2\vert  p_2\vert  ^2\left\vert  \frac{1}{8}p_1^2-p_2\right\vert  +128\vert  p_2\vert  ^4\left\vert  \frac{1}{8}p_1^2-p_2\right\vert \\
 &\quad+16\vert  p_2\vert  ^5\left\vert  \frac{1}{2}p_1^2-p_2\right\vert +512\vert  p_1\vert  \vert  p_2\vert  \vert  p_3\vert  \left\vert  \frac{1}{4}p_1^2-p_2\right\vert  +64\vert  p_2\vert  ^4\left\vert  \frac{1}{2}p_1p_3-p_4\right\vert    \\
&\quad+32\vert  p_2\vert  ^2\vert  p_3\vert  ^2\left\vert  \frac{3}{4}p_1^2-p_2\right\vert+16\vert  p_1\vert  ^2\vert  p_2\vert  ^3\left\vert  \frac{1}{2}p_1p_3-p_4\right\vert     \\
&\quad+64\vert  p_2\vert  \vert  p_3\vert  ^2\left\vert  \frac{1}{2}p_1p_3-p_4\right\vert +64\vert  p_1\vert  ^2\vert  p_4\vert  \vert  p_2^2-p_4\vert  \\
&\quad+64\vert  p_2\vert  ^2\vert  p_4\vert  \vert  p_1p_3-p_4\vert+256\vert  p_1\vert  ^4+768\vert  p_1\vert  ^2\vert  p_2\vert  ^2+256\vert  p_2\vert  ^4 \\
&\quad+\vert  p_1\vert  ^4\vert  p_2\vert  ^4 +128\vert  p_1\vert  ^2\vert  p_3\vert  ^2+128\vert  p_2\vert  ^2\vert  p_3\vert  ^2+16\vert  p_3\vert  ^4+256\vert  p_2\vert  ^3\vert  p_4\vert\\
&\quad +256\vert p_1\vert  \vert  p_2\vert  \vert  p_3\vert  \vert  p_4\vert  .
\end{align*}
Applying Lemma \ref{lem4} and the fact that $\vert  p_n\vert  \le 2$, we get $\vert  T_4(2)(f)\vert   \le 15.12.$
\end{proof}


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\end{document}