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\title{Fekete-Szeg\"{o} Problem For A Class of Analytic Functions Defined By Carlson-Shaffer Operator}
\author{Saurabh Porwal}
\address{Department of Mathematics\\
UIET, CSJM University, Kanpur-208024\\
(U.P.), India\\}
\email{saurabhjcb@rediffmail.com}
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\author{Kaushal Kumar}
\address{Department of Mathematics\\
UIET, CSJM University, Kanpur-208024\\
(U.P.), India\\}
%\email{Your e-mail address}
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\subjclass{30C45.}
\keywords{Analytic Function, Fekete-Szeg\"{o} Problem, Carlson-Shaffer operator.}
\begin{abstract}
In the present paper, authors study a Fekete-Szeg\"{o} problem for a class of analytic functions defined by Carlson-Shaffer operator. Relevant connections of the results presented here with various known results are briefly indicated.\\
\end{abstract}
\maketitle

\section{Introduction}
Let $A$ denote the class of functions $f$  of the form
\begin{equation}\label{eq11}
f\left( z \right)=z+\sum\limits_{k=2}^{\infty }{{{a}_{k}}{{z}^{k}}},
\end{equation}
which are analytic in the open unit disk $U = \left\{z : z\in  C~ \text{and}~ |z| < 1\right\}$ and $S$ denote the subclass of $A$ that are univalent in $U$.
	Fekete and Szeg\"{o} \cite{10} proved a interesting result that the estimate
	\begin{equation}\label{eq12}
\left| {{a}_{3}}-\lambda a_{2}^{2} \right|\leq 1+2\exp \left( \frac{-2\lambda }{1-\lambda } \right)
\end{equation}
holds for any normalized univalent function $f(z)$ of the form \eqref{eq11} in the open unit disk $U$ for
$0\le \lambda \le 1$. This inequality is sharp for each $\lambda$.

The coefficient functional
\begin{equation}\label{eq13}
{{\phi }_{\lambda }}\left( f \right)={{a}_{3}}-\lambda a_{2}^{2}=\frac{1}{6}\left( f'''\left( 0 \right)-\frac{3\lambda }{2}{{\left[ f''\left( 0 \right) \right]}^{2}} \right),
\end{equation}
on normalized analytic functions $f$ in the unit disk represents various geometric quantities, for example, when $\lambda = 1$, ${{\phi }_{\lambda }}\left( f \right)={{a}_{3}}-a_{2}^{2}$, becomes $\frac{S_f(0)}{6}$, where $S_f$ denote the Schwarzian derivative $\left( f\,'''/f' \right)'-{{\left( f''/f' \right)}^{2}}/2$ of locally univalent functions $f$ in $U$.  The problem of maximising the absolute value of the functional  $\phi_{\lambda}(f)$ is called the Fekete-Szeg\"{o} problem.

	The Fekete-Szeg\"{o} problem is one of the interesting problems in Geometric Function Theory. This attracts many researchers (see the work of \cite{1}-\cite{4}, \cite{6}-\cite{8}, \cite{12}, \cite{13}, \cite{16}, \cite{18}, \cite{21} and \cite{22}) to study the Fekete-Szeg\"{o} problem for the various classes of analytic univalent functions . Very recently, Bansal \cite{3} introduced the class $R^{\tau}_{\gamma}(\phi)$ of functions in $f\in S$ for which
\[1+\frac{1}{\tau }\left( f'(z)+\gamma zf''(z)-1 \right)\ \prec \phi (z),~~~z\in U\] 	
where $0\le \gamma <1$,  $\tau \in C\setminus \{0\}$, $\phi(z)$ is an analytic function with positive real part on $U$ with $\phi(0)=0,~\phi^{'}(0)>0$   which maps the unit like disk $U$ onto a starlike region with respect to 1 which is symmetric with respect to the real axis and $\prec$ denotes the subordination between analytic functions and studied the Fekete-Szeg\"{o} problem for this class.

Now, by using the Carlson-Shaffer operator we introduce a new subclass $R^{\tau}_{\gamma}(\phi,a,c)$ for functions $f\in A$ and  $0\le \gamma <1$,  $\tau \in C\setminus \left\{0\right\}$, $a,c\in C, \{c\ne 0,-1,-2,....\}$ satisfying the condition
\begin{equation}\label{eq14}
1+\frac{1}{\tau }\left({\left( L(a,c)f(z) \right)}^{'}+\gamma z{\left( L(a,c)f(z) \right)}^{''}-1 \right)\prec \phi (z)~~\left( z\in U \right)
\end{equation}		
where $\phi(z)$ is defined the same as above and $L(a,c)$ denotes the Carlson-Shaffer operator introduced in \cite{5} and defined in the following way:
\[L(a,c)f(z)=\quad f(z)*z h(a,c;z),\]
where 				\[h (a,c;z)=\quad 1+\sum\limits_{n=1}^{\infty }{\frac{{{\left( a \right)}_{n}}}{{{\left( c \right)}_{n}}}{{z}^{n}}}\]
				 $L(a,c)$ maps $A$ into itself. $L(c,c)$is the identity and if $a\ne 0,-1,-2...$, then $L(a,c)$  has a continuous inverse $L(c,a)$  and is an one-to-one mapping of $A$   onto itself.  $L(a,c)$ provides a convenient representation of differentiation and integration. If $g(z)=z{{f}^{/}}(z)$, then $g=L(2,1)f$ and $f=L(1,2)g$.


If we set
\[\phi (z)=\frac{1+Az}{1+Bz},\quad \quad \quad (-1\le B<A\le 1;z\in U),\]				

in \eqref{eq14},we obtain
\begin{footnotesize}\[R_{\gamma }^{\tau }\left( \frac{1+Az}{1+Bz},a,c \right)=R_{\gamma }^{\tau }(A,B,a,c)=\left\{ f\in A :\left| \frac{{{(L(a,c)f(z))}^{/}}+\gamma z{{\left( L(a,c)f(z) \right)}^{//}}-1}{\tau (A-B)-B\left( {{\left( L(a,c)f(z) \right)}^{/}}+\gamma z{{\left( L(a,c)f(z) \right)}^{//}}-1 \right)} \right|<1 \right\}\]
\end{footnotesize} which is again a new class.

By specializing parameters in the subclass $R_{\gamma }^{\tau }(A,B,a,c)$   we obtain the following known subclasses studied earlier by various authors.
\begin{enumerate}
	\item $R_{\gamma }^{\tau }(A,B,a,a)\equiv  R_{\gamma }^{\tau }(A,B)$  studied by Bansal \cite{3}.
	\item $R_{\gamma }^{\tau }(1-2\beta,-1,a,a)\equiv R_{\gamma }^{\tau }(\beta)$ for $0\le \beta<1$,  studied by Swaminathan \cite{23}.
    \item $R_{\gamma }^{\tau }(1-2\beta,-1,a,a)\equiv R_{\gamma }^{\tau }(\beta)$ for $\tau=e^{i \eta}cos \eta, 0\le \beta<1$, where $-\pi/2<\eta<\pi/2$ introduced by Ponnusamy and R{\o}nning  \cite{20}, (see also \cite{19}).
        \item $R_{1}^{\tau }(0,-1,a,a)\equiv  R^{\tau }(\beta)$ for $\tau=e^{i \eta}cos \eta$ was considered in \cite{14}.
        \end{enumerate}



To prove our main result, we shall require the following lemma.
\begin{lemma} (see \cite{11}, \cite{15}).
If $p\left( z \right)=1+{{c}_{1}}z+{{c}_{2}}{{z}^{2}}+{{c}_{3}}{{z}^{3}}+...\,\,\,\left( z\in U \right)$ is a function with positive real part, then for any complex number $\mu$ ,
\begin{equation}\label{eq17}
\left| {{c}_{2}}-\mu c_{1}^{2} \right| \le \,\,\,2\,\,\,\max \left\{ 1,\,\,\left| 2\mu -1 \right| \right\}
\end{equation}
and the result is sharp for the functions given by
			
\begin{equation}\label{eq18}
p\left( z \right)=\frac{1+{{z}^{2}}}{1-{{z}^{2}}},\,\,\,\,\,p\left( z \right)=\frac{1+z}{1-z}\,\,\,\,\,\left( z\in U \right).
\end{equation}
\end{lemma}

\section{Main Results}
Our main result is contained in the following theorem.
\begin{theorem}\label{t21}
Let $\phi (z)=1+{{B}_{1}}z+{{B}_{2}}{{z}^{2}}+{{B}_{3}}{{z}^{3}}+...$, where $\phi (z)\in A$ with $\phi^{'}(0)>0$. If $f(z)$ given by \eqref{eq11} belongs to $R_{\gamma }^{\tau }(\phi ,a,c)(0\le \gamma \le 1,\tau \in C\setminus \{0\}, a,c\in C, \{c\ne 0,-1,-2,....\}, z\in U),$ then for any complex number $\mu$
\begin{equation}\label{eq21}
\left| {{a}_{3}}-\mu a_{2}^{2} \right|\le \frac{{{B}_{1}}\left| \tau  \right|c(c+1)}{3a(a+1)(1+2\gamma )}\max \left\{ 1,\left| \frac{{{B}_{2}}}{{{B}_{1}}}-\frac{3\mu \tau {{B}_{1}}c(a+1)(1+2\gamma )}{4a(c+1){{(1+\gamma )}^{2}}} \right| \right\}.
\end{equation}			
 This result is sharp.

\end{theorem}
\begin{proof}
 If $f(z)\in R_{\gamma }^{\tau }(\phi ,a,c)$, then there exists a Schwarz function $w(z)$ analytic in $U$ with $w(0) = 0$ and $|w(z)|<1$  in $U$ such that
\begin{equation}\label{eq22}
1+\frac{1}{\tau }\left( {{\left( L(a,c)f(z) \right)}^{/}}+\gamma z{{\left( L(a,c)f(z) \right)}^{//}}-1 \right)=\phi \left( w\left( z \right) \right),\quad \quad \quad \quad \left( z\in U \right).
\end{equation}
Define the function $p_1(z)$ by
	\begin{equation}\label{eq23}
p_{1}\left( z \right)=\frac{1+w\left( z \right)}{1-w\left( z \right)}=1+c_{1}z+c_{2}{z}^{2}+......
\end{equation}
Since $w(z)$ is a Schwarz function, we see that  $Re\left\{p_1(z)\right\}> 0$ and $p_1(0) = 1$.

 Define the function $p(z)$ by,
	 	\begin{equation}\label{eq24}
p(z)=1+\frac{1}{\tau }\left( {{\left( L(a,c)f(z) \right)}^{/}}+\gamma z{{\left( L(a,c)f(z) \right)}^{//}}-1 \right)=1+{{b}_{1}}z+{{b}_{2}}{{z}^{2}}+....
\end{equation}
In view of \eqref{eq22}, \eqref{eq23}, \eqref{eq24}, we have
				\[p\left( z \right)=\phi \left( \frac{{{p}_{1}}\left( z \right)-1}{{{p}_{1}}\left( z \right)+1} \right)=\phi \left( \frac{{{c}_{1}}z+{{c}_{2}}{{z}^{2}}+.....}{2+{{c}_{1}}z+{{c}_{2}}{{z}^{2}}+.....} \right)\]	
	\begin{equation}\label{eq25}=\phi \left( \frac{1}{2}{{c}_{1}}z+\frac{1}{2}\left( {{c}_{2}}-\frac{c_{1}^{2}}{2} \right){{z}^{2}}+..... \right)
	\end{equation}
\[=1+{{B}_{1}}\frac{1}{2}{{c}_{1}}z+{{B}_{1}}\frac{1}{2}\left( {{c}_{2}}-\frac{c_{1}^{2}}{2} \right){{z}^{2}}+{{B}_{2}}\frac{1}{4}c_{1}^{2}{{z}^{2}}+....\]	
Thus,
				\begin{equation}\label{eq26}
{{b}_{1}}=\frac{1}{2}{{B}_{1}}{{c}_{1}};\,\,\,\,\,\,{{b}_{2}}=\frac{1}{2}{{B}_{1}}\left( {{c}_{2}}-\frac{c_{1}^{2}}{2} \right)+\frac{1}{4}{{B}_{2}}c_{1}^{2}.
\end{equation}
From \eqref{eq24}, we obtain 			
\begin{equation}\label{eq27}
{{a}_{2}}=\frac{\tau \,{{B}_{1}}{{c}_{1}}c}{4a\left( 1+\gamma  \right)};\,\,\,\quad \quad \quad \quad \,\,{{a}_{3}}=\frac{\tau \,c(c+1)}{6a(a+1)\left( 1+2\gamma  \right)}\left[ {{B}_{1}}\left( {{c}_{2}}-\frac{c_{1}^{2}}{2} \right)+\frac{1}{2}{{B}_{2}}c_{1}^{2} \right].
\end{equation}

Therefore, we have
\begin{equation}\label{eq28}			
{{a}_{3}}-\mu a_{2}^{2}=\frac{{{B}_{1}}\tau \,c(c+1)}{6a(a+1)\left( 1+2\gamma  \right)}\left( {{c}_{2}}-\nu c_{1}^{2} \right)
\end{equation}
where
\begin{equation}\label{eq29}				
\nu =\frac{1}{2}\left( 1-\frac{{{B}_{2}}}{{{B}_{1}}}+\frac{3\tau \mu {{B}_{1}}c(a+1)\left( 1+2\gamma  \right)}{4a(c+1){{\left( 1+\gamma  \right)}^{2}}} \right).
\end{equation}
Our result now is followed by an application of Lemma 1.1. Also, by the application of Lemma 1.1 equality in (2.1) is obtained when
\begin{equation}\label{eq210}				
{{p}_{1}}\left( z \right)=\frac{1+{{z}^{2}}}{1-{{z}^{2}}}~~\text{or}~~	{{p}_{1}}\left( z \right)=\frac{1+z}{1-z}
\end{equation}
but
\begin{equation}\label{eq211}		
p\left( z \right)=1+\frac{1}{\tau }\left( {{\left( L(a,c)f(z) \right)}^{/}}+\gamma z{{\left( L(a,c)f(z) \right)}^{//}}-1 \right)=\phi \left( \frac{{{p}_{1}}\left( z \right)-1}{{{p}_{1}}\left( z \right)+1} \right).
\end{equation}
Putting value of $p_1(z)$ we get the desired results.							
Thus the proof of Theorem \ref{t21} is established.
\end{proof}
For the class $R_{\gamma }^{\tau }\left( A,\,\,B,a,c \right)$,
\begin{equation}\label{eq212}
\phi \left( z \right)=\frac{1+Az}{1+Bz}=\left( 1+Az \right){{\left( 1+Bz \right)}^{-1}}=1+\left( A-B \right)z-\left( AB-{{B}^{2}} \right){{z}^{2}}+...
\end{equation}
Thus, putting ${{B}_{1}}=A-B$ and ${{B}_{2}}=-B\left( A-B \right)$ in Theorem \ref{t21}, we get the following corollary.


\begin{corollary}\label{c21}
 If $f(z)$  given by \eqref{eq11} belongs to $R_{\gamma }^{\tau }\left( A,\,\,B,a,c \right)$  , then
	\begin{equation}\label{eq213}
\left| {{a}_{3}}-\mu a_{2}^{2} \right| \le \,\,\frac{\left( A-B \right)\,\,\left| \tau  \right|c(c+1)}{3a(a+1)\left( 1+2\gamma  \right)}\max \left\{ 1,\,\,\left| B+\frac{3\tau \mu c(a+1)\left( A-B \right)\left( 1+2\gamma  \right)}{4a(c+1){{\left( 1+\gamma  \right)}^{2}}} \right| \right\}.
\end{equation}
\end{corollary}


If we put $a = c$	in Theorem \ref{t21}, then we obtain the following result of Bansal \cite{3}.
\begin{corollary}\label{c22}
Let $\phi (z)=1+{{B}_{1}}z+{{B}_{2}}{{z}^{2}}+{{B}_{3}}{{z}^{3}}+...$ , where $\phi(z)\in A$ with $\phi^{'}(0)>0$ . If  $f(z)$ given by \eqref{eq11} belongs to $R_{\gamma }^{\tau }(\phi )(0\le \gamma \le 1,\tau \in C\backslash \{0\},z\in U)$  then for any complex number $\mu$
\[\left| {{a}_{3}}-\mu a_{2}^{2} \right|\le \frac{{{B}_{1}}\left| \tau  \right|}{3(1+2\gamma )}\max \left\{ 1,\left| \frac{{{B}_{2}}}{{{B}_{1}}}-\frac{3\mu \tau {{B}_{1}}(1+2\gamma )}{4{{(1+\gamma )}^{2}}} \right| \right\}.\]			
 This result is sharp.

\end{corollary}


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\end{document}
