\documentclass[10pt,leqno,draft]{article}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsmath}

\setcounter{MaxMatrixCols}{10}
%TCIDATA{OutputFilter=Latex.dll}
%TCIDATA{Version=5.50.0.2890}
%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
%TCIDATA{BibliographyScheme=Manual}
%TCIDATA{LastRevised=Saturday, February 16, 2019 21:28:45}
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
%TCIDATA{Language=American English}

\pagestyle{plain}
\frenchspacing
\sloppy
\parskip=0cm
\renewcommand{\baselinestretch}{1.3}
\setlength{\rightmargin}{0 cm}
\setlength{\textwidth}{17.0cm}
\setlength{\textheight}{22.0cm}
\setlength{\oddsidemargin}{1.0cm}
\setlength{\evensidemargin}{1.0cm}
\setlength{\topmargin}{-1cm}
\setlength{\footskip}{1cm}
\newcommand{\be}{\begin{equation}}
\newcommand{\ee}{\end{equation}}
\newcommand{\bea}{\begin{eqnarray*}}
\newcommand{\eea}{\end{eqnarray*}}
\newcommand{\ba}{\begin{array}}
\newcommand{\ea}{\end{array}}
\newcommand{\bi}{\begin{itemize}}
\newcommand{\ei}{\end{itemize}}
\newcommand{\bc}{\begin{center}}
\newcommand{\ec}{\end{center}}
\newcommand{\bfr}{\begin{flushright}}
\newcommand{\efr}{\end{flushright}}
\newcommand{\f}{\frac}
\newcommand{\ov}{\overline}
\newcommand{\un}{\underline}
\newcommand{\vs}{\vspace}
\newcommand{\tg}{{\rm tg\ }}
\newcommand{\arctg}{{\rm arctg\ }}
\newcommand{\ctg}{{\rm ctg\ }}
\newcommand{\R}{{\rm Re \ }}
\newcommand{\I}{{\rm Im \ }}
\newcommand{\la}{\langle}
\newcommand{\ra}{\rangle}
\newcommand{\p}{\partial}
\newcommand{\eps}{\varepsilon}
\newcommand{\Ri}{\ \Rightarrow\ }
\newcommand{\Le}{\ \Leftarrow\ }
\newcommand{\Lr}{\ \Leftrightarrow\ }
\newcommand{\ds}{\displaystyle}
\newcommand{\ung}{\widehat}
\newcommand{\tr}{\triangle}
\newcommand{\q}{\quad}
\newcommand{\cc}{^\circ}
\newcommand{\va}{\varphi}
\newcommand{\bb}{\mathbb}
\renewcommand{\vec}{\overrightarrow}
\newcommand{\til}{\widetilde}
\numberwithin{equation}{section}
\renewcommand{\theequation}{\thesection.\arabic{equation}}
\newtheorem{thm}{Theorem}[section]
\newtheorem{cor}{Corollary}[section]
\newtheorem{definition}{Definition}[section]
\newenvironment{defi}{\begin{definition}\rm}{\end{definition}}
\newtheorem{example}{Example}[section]
\newenvironment{ex}{\begin{example}\rm}{\end{example}}
\newtheorem{lem}{Lemma}[section]
\newtheorem{prop}{Proposition}[section]
\newtheorem{remark}{Remark}[section]
\newenvironment{rem}{\begin{remark}\rm}{\end{remark}}
\newenvironment{dem}[1][Proof]{\noindent\textbf{#1.} }{\ $\square$}
%\input{tcilatex}
\begin{document}

\title{ Differential sandwich theorem for certain class of analytic functions associated with an integral operator}
\author{Lumini\c{t}a-Ioana Cot\^{\i}rl\u{a} and Adriana C\u{a}ta\c{s}}
\date{}
\maketitle

\begin{abstract}
In this paper we obtain some applications of first order differential subordination and superordination result involving an
integral operator for certain normalized analytic function.
\end{abstract}

\textbf{2014 Mathematics Subject Classification:} 30C45.

\textbf{Keywords: } integral operator, subordination and superordination,
analytic functions, sandwich theorem.

\section{Introduction and preliminaries}

Let $\mathcal{A}$ denote the class of functions of the form
\begin{equation}
f(z)=z+\displaystyle\sum_{k=2}^{\infty }a_{k}z^{k},\text{ }a_{k}\geq 0,
\label{1.1}
\end{equation}%
which are analytic and univalent in the open unit disk $U=\{z:|z|<1\}$.

\bigskip\ If $f$ \ and \ $g$ are analytic functions in $U$, we say that $f$
is subordinate to $g$ in $U$, written symbolically as $f\prec g$ or $%
f(z)\prec g(z)$ if there exists a Schwarz function $w(z)$ analytic in $U,$
with $w(0)=0$ and $\left\vert w(z)\right\vert <1,$ such that $f(z)=g(w(z))$,
$z\in U$. In particular, if the function $g$ is univalent in $U$, the
subordination $f\prec g$ \ is equivalent to $f(0)=g(0)\quad $and$\quad
f(U)\subset g(U)$, (see $\cite{2}$, $\cite{3}$).

For the function $f$ given by $\left( \ref{1.1}\right) $ and $g$ $\in
\mathcal{A}$ given by $g(z)=z+\displaystyle\sum_{k=2}^{\infty }b_{k}z^{k},$
the Hadamard product (or convolution) of $f$ and $g$ is defined by
\begin{equation*}
\left( f\ast g\right) \left( z\right) =z+\displaystyle\sum_{k=2}^{\infty
}a_{k}b_{k}z^{k}=\left( g\ast f\right) \left( z\right) .
\end{equation*}

The set of all functions $f$ that are analytic and injective on $\overline{U}%
-E(f)$,\bigskip\ denote by $Q$ where%
\begin{equation*}
E(f)=\{\zeta \in \partial U:\ \lim\limits_{z\rightarrow \zeta }f(z)=\infty \}
\end{equation*}%
and are such that $f^{\prime }(\zeta )\neq 0$ for $\zeta \in \partial
U\backslash E(f)$, (see {\rm \cite{4}}).

If $\psi :\mathbb{C}^{3}\times U\rightarrow \mathbb{C}$ and $h$ is univalent
in $U$ with $q\in Q$. In {\rm \cite{3}} Miller and Mocanu consider the problem
of determining conditions on admissible functions $\psi $ such that
\begin{equation}
\psi (p(z),zp^{\prime }(z),z^{2}p^{\prime \prime }(z);z)\prec h(z)
\label{1.2}
\end{equation}%
implies that $\ p(z)\prec q(z)$ for all functions $p\in {\mathcal{H}}[a,n]$
that satisfy the differential subordination $\left( \ref{1.2}\right) $.

Let $\phi :\mathbb{C}^{3}\times U\rightarrow \mathbb{C}$ and $h\in \mathcal{H%
}$ with $q\in $ ${\mathcal{H}}[a,n].$ In $\cite{4}$ and {\rm \cite{5}} is
studied the dual problem and determined conditions on $\phi $ such that%
\begin{equation}
h(z)\prec \phi (p(z),zp^{\prime }(z),z^{2}p^{\prime \prime }(z);z)
\label{1.3}
\end{equation}%
implies $q(z)\prec p(z)$ for all functions $p\in Q$ that satisfy the above
subordination. They also found conditions so that the functions $q$ is the
largest function with this property, called the best subordinant of the
subordination $\left( \ref{1.3}\right) .$

Let ${\mathcal{H}}\left( U\right) $ be the class of analytic functions in
the open unit disc.

For $n$ a positive integer and $a\in
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
$ let
\begin{equation*}
{\mathcal{H}}[a,n]=\left\{ f\in {\mathcal{H}}:f(z)=a+a_{n}z^{n}+\ldots
\right\} .
\end{equation*}

The integral operator $I^{m}$ of a function $f$ is defined in {\rm \cite{6}} by%
\begin{equation*}
I^{0}f(z)=f(z),
\end{equation*}%
\begin{equation*}
I^{1}f(z)=If(z)=\int_{0}^{z}f(t)t^{-1}dt,
\end{equation*}%
\begin{equation*}
\ldots
\end{equation*}%
\begin{equation*}
I^{m}f(z)=I\left( I^{m-1}f(z)\right) ,\text{ }z\in U.
\end{equation*}

\begin{lem}
\bigskip \label{Lema1.1} {\rm \cite{3}} Let $q$ be univalent in $U,$  $\zeta \in
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
^{\ast }$ and suppose that
\begin{equation}
\mathrm{Re\ }\left\{ 1+\displaystyle\frac{zq^{\prime \prime }(z)}{q^{\prime
}(z)}\right\} >\max \left\{ 0,-\R\left( \frac{1}{\zeta }\right)
\right\} .  \label{1.4}
\end{equation}%
If $p$ is analytic in $U$ with $p(0)=q(0)$ and
\begin{equation}
p(z)+\zeta zp^{\prime }(z)\prec q(z)+\zeta zq^{\prime }(z)  \label{1.5}
\end{equation}%
then $p(z)\prec q(z)$ and $q$ is the best dominant.
\end{lem}

\begin{lem}
\label{Lema1.2} {\rm \cite{3}} \ Let the function $q$ be univalent in the unit
disk and let $\theta ,\varphi $ be analytic in domain $D$ containing  $q(U)$
with $\varphi (w)\neq 0$, where $w\in q(U).$ Set
\begin{equation*}
Q(z)=zq^{\prime }(z)\varphi (q(z))\text{ \ and \ }h(z)=\theta (q(z))+Q(z).
\end{equation*}%
Suppose that
\begin{equation*}
\text{\ }Q\text{ is starlike univalent in }U;
\end{equation*}%
\begin{equation*}
\text{\ \ }\mathrm{Re\ }\left\{ \displaystyle\frac{zh^{\prime }(z)}{Q(z)}%
\right\} >0,\text{ for }z\in U.
\end{equation*}%
\ \ \ \ \ If $p$ is analytic with $p(0)=q(0),$ $p(U)\subseteq D$ and
\begin{equation}
\theta (p(z))+zp^{\prime }(z)\varphi (p(z))\prec \theta (q(z))+zq^{\prime
}(z)\varphi (q(z))  \label{1.6}
\end{equation}%
then $p(z)\prec q(z)$ and $q$ is the best dominant.
\end{lem}

\begin{lem}
\label{Lema1.3} {\rm \cite{1}} Let $q$ be convex in the~unit disc $U,$ $q(0)=a$
and $\zeta \in
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
$, $\mathrm{Re}\left( \zeta \right) >0.$ If $p\in {\mathcal{H}}[a,1]\cap Q$ and  $%
p(z)$ $+\zeta zp^{\prime }(z)$ is univalent in $U$ then
\begin{equation}
q(z)+\zeta zq^{\prime }(z)\prec p(z)+\zeta zp^{\prime }(z)  \label{1.7}
\end{equation}%
implies $q(z)\prec p(z)$ and $q$ is the best subordinant.
\end{lem}

\begin{lem}
\label{Lema1.4}  {\rm \cite{2}} Let the function $q$ be convex and univalent in the unit
disc $U$ and $\theta $ and $\varphi $ be analytic in a domain $D$ containing
$q(U)$. Suppose that

\noindent 1. $\mathrm{Re\ }\left\{ \displaystyle\frac{\theta ^{\prime }(q(z))%
}{\varphi (q(z))}\right\} >0$ for $z\in U$ and

\noindent 2. $Q(z)=zq^{\prime }(z)\varphi (q(z))$ is starlike univalent in $U
$.

If $p\in {\mathcal{H}}[q(0),1]\cap Q$ with $p(U)\subseteq D$  and $\theta
(p(z))+zp^{\prime }(z)\varphi (p(z))$ is univalent in $U$ and
\begin{equation}
\theta (q(z))+zq^{\prime }(z)\varphi (q(z))\prec \theta (p(z))+zp^{\prime
}(z)\varphi (p(z))  \label{1.8}
\end{equation}%
then $q(z)\prec p(z)$ and $q$ is the best subordinant.
\end{lem}

\bigskip

\section{Main results}

\begin{thm}
\label{T2.1} \textit{Let }$q$ be univalent in $U,$ with $q(0)=1$ and $q(z)
$ $\neq 0$ for all  $z\in U,$ and let $\sigma \in
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
^{\ast },$ $f\in \mathcal{A}$ and suppose that $f$ and $g$ satisfy the next
conditions:%
\begin{equation}
\frac{I^{m+1}\left( f(z)\right) }{z}\neq 0, z\in U  \label{2.1}
\end{equation}%
and%
\begin{equation}
\mathrm{Re\ }\left\{ 1+\displaystyle\frac{zq^{\prime \prime }(z)}{q^{\prime
}(z)}-\displaystyle\frac{zq^{\prime }(z)}{q(z)}\right\} >0\text{ for }z\text{
}\in U.  \label{2.2}
\end{equation}%
If
\begin{equation}
\frac{I^{m}\left( f(z)\right) }{I^{m+1}\left( f(z)\right) }\prec 1+\frac{%
zq^{\prime }(z)}{\sigma q(z)},  \label{2.3}
\end{equation}%
then%
\begin{equation*}
\left( \frac{I^{m+1}\left( f(z)\right) }{z}\right) ^{\sigma }\prec q(z)
\end{equation*}%
and $q$ is the best dominant of  (\ref{2.3}).
\end{thm}

\begin{dem}
Let
\begin{equation}
p(z)=\left( \frac{I^{m+1}\left( f(z)\right) }{z}\right) ^{\sigma },\text{ }%
z\in U.  \label{2.4}
\end{equation}%
Because the integral operator $I^{m}$ satisfies the identity%
\begin{equation*}
z\left[ I^{m+1}\left( f(z)\right) \right] ^{^{\prime }}=I^{m}\left(
f(z)\right)
\end{equation*}%
and the function $p(z)$ is analytic in $U$, by differentiating $\left( \ref%
{2.4}\right) $ logarithmically with respect to $z$, we obtain%
\begin{equation}
\frac{zp^{\prime }(z)}{p(z)}=\sigma \left( \frac{I^{m}\left( f(z)\right) }{%
I^{m+1}\left( f(z)\right) }-1\right) .  \label{2.5}
\end{equation}%
In order to prove our result we will use Lemma \ref{Lema1.2}. In this lemma
we consider%
\begin{equation*}
\theta (w)=1\text{ and }\varphi (w)=\frac{1}{\sigma w},
\end{equation*}%
then $\theta $ is analytic in $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
$ and $\varphi (w)\neq 0$ is analytic in $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
^{\ast }.$ Also, if we let%
\begin{equation*}
Q\left( z\right) =zq^{\prime }(z)\varphi (q(z))=\frac{zq^{\prime }(z)}{%
\sigma q(z)}
\end{equation*}%
and%
\begin{equation*}
h(z)=\theta \left( q(z)\right) +Q\left( z\right) =1+\frac{zq^{\prime }(z)}{%
\gamma \sigma q(z)}
\end{equation*}%
from $\left( \ref{2.2}\right) $ we see that $Q\left( z\right) $ is a
starlike function in $U.$ We also have%
\begin{equation*}
\mathrm{Re\ }\left\{ \displaystyle\frac{zh^{\prime }(z)}{Q(z)}\right\} =%
\mathrm{Re\ }\left\{ 1+\displaystyle\frac{zq^{\prime \prime }(z)}{q^{\prime
}(z)}-\displaystyle\frac{zq^{\prime }(z)}{q(z)}\right\} >0\text{ for }z\text{
}\in U
\end{equation*}%
and then, by using Lemma  \ref{Lema1.2} we deduce that subordination (\ref{2.3}) implies $p(z)\prec q(z)$ and the function $q$ is the
best dominant of $\left( \ref{2.3}\right) .$
\end{dem}

Taking $q(z)=\frac{1+Az}{1+Bz},$ $\left( -1\leq B<A\leq 1\right) $ in
Theorem \ref{T2.1}, it easy to check that the assumption
\begin{equation*}
p(z)+\frac{1}{\sigma }zp^{\prime }(z)\prec q(z)+\frac{\alpha }{\sigma }%
zq^{\prime }(z)
\end{equation*}%
holds, hence we obtain the next result.

\begin{cor}
\label{C2.2} \textit{Let }$\sigma \in
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
^{\ast }$ and  $f\in \mathcal{A}$ . Suppose
\begin{equation*}
\frac{I^{m+1}\left( f(z)\right) }{z}\neq 0, z\in U.
\end{equation*}%
If
\begin{equation*}
\frac{I^{m}\left( f(z)\right) }{I^{m+1}\left( f(z)\right) }\prec 1+\frac{%
z\left( A-B\right) }{\sigma \left( 1+Az\right) \left( 1+Bz\right) },
\end{equation*}%
then%
\begin{equation*}
\left( \frac{I^{m+1}\left( f(z)\right) }{z}\right) ^{\sigma }\prec \frac{1+Az%
}{1+Bz}\text{ }
\end{equation*}%
and $q(z)=\frac{1+Az}{1+Bz}$ is the best dominant$.$
\end{cor}

Taking $q(z)=\frac{1+z}{1-z}$ in Theorem \ref{T2.1}, it easy to check that
the assumption
\begin{equation*}
p(z)+\frac{1}{\sigma }zp^{\prime }(z)\prec q(z)+\frac{\alpha }{\sigma }%
zq^{\prime }(z)
\end{equation*}%
holds, hence we obtain the next result.

\begin{cor}
\label{C2.3} \textit{Let }$\sigma \in
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
^{\ast }$ and $f\in \mathcal{A}$ . Suppose
\begin{equation*}
\frac{I^{m+1}\left( f(z)\right) }{z}\neq 0,z\in U.
\end{equation*}%
If
\begin{equation*}
\frac{I^{m}\left( f(z)\right) }{I^{m+1}\left( f(z)\right) }\prec 1+\frac{2z}{%
\sigma \left( 1-z\right) \left( 1+z\right) },
\end{equation*}%
then%
\begin{equation*}
\left( \frac{I^{m+1}\left( f(z)\right) }{z}\right) ^{\sigma }\prec \frac{1+z%
}{1-z}\text{ }
\end{equation*}%
and $q(z)=\frac{1+z}{1-z}$  is the best dominant$.$
\end{cor}

\begin{thm}
\label{T2.4} Let $q$ be univalent in $U$, with $q(0)=1.$ Let $\sigma \in
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
^{\ast }$ and $t,\nu ,\eta \in
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
$ with $\nu +\eta \neq 0.$ Let $f\in \mathcal{A}$ and suppose that $f$ and $g
$ satisfy the next conditions%
\begin{equation}
\frac{\upsilon I^{m+1}\left( f(z)\right) +\eta I^{m+2}\left( f(z)\right) }{%
\left( \upsilon +\eta \right) z}\neq 0,z\in U  \label{2.6}
\end{equation}%
and
\begin{equation}
\R\left\{ 1+\displaystyle\frac{zq^{\prime \prime }(z)}{q^{\prime
}(z)}\right\} >\max \left\{ 0,-\R{t}\right\} \text{,  }z\text{
}\in U.  \label{2.7}
\end{equation}%
If
\begin{equation}
\psi (z)=t\left[ \frac{\upsilon I^{m+1}\left( f(z)\right) +\eta
I^{m+2}\left( f(z)\right) }{\left( \upsilon +\eta \right) z}\right] ^{\sigma
}+\sigma \left[ \frac{\upsilon z\left( I^{m+1}\left( f(z)\right) \right)
^{\prime }+z\eta \left( I^{m+2}\left( f(z)\right) \right) ^{\prime }}{%
\upsilon I^{m+1}\left( f(z)\right) +\eta I^{m+2}\left( f(z)\right) }-1\right]
\label{2.8}
\end{equation}%
and%
\begin{equation}
\psi (z)\prec tq(z)+\frac{zq^{\prime }(z)}{q(z)}  \label{2.9}
\end{equation}%
then%
\begin{equation*}
\left[ \frac{\upsilon I^{m+1}\left( f(z)\right) +\eta I^{m+2}\left(
f(z)\right) }{\left( \upsilon +\eta \right) z}\right] ^{\sigma }\prec q(z)
\end{equation*}%
and $q$ is the best dominant.
\end{thm}

\begin{dem}
Let
\begin{equation}
p(z)=\left[ \frac{\upsilon I^{m+1}\left( f(z)\right) +\eta I^{m+2}\left(
f(z)\right) }{\left( \upsilon +\eta \right) z}\right] ^{\sigma },z\in U.
\label{2.10}
\end{equation}
According to $\left( \ref{2.3}\right) $ the function $p(z)$ is analytic in $U
$ and differentiating $\left( \ref{2.10}\right) $ logarithmically with
respect to $z$, we obtain%
\begin{equation}
\frac{zp^{\prime }(z)}{p(z)}=\sigma \left[ \frac{\upsilon z\left(
I^{m+1}\left( f(z)\right) \right) ^{\prime }+z\eta \left( I^{m+2}\left(
f(z)\right) \right) ^{\prime }}{\upsilon I^{m+1}\left( f(z)\right) +\eta
I^{m+2}\left( f(z)\right) }-1\right]   \label{2.11}
\end{equation}%
and hence%
\begin{equation*}
zp^{\prime }(z)=\sigma \left[ \frac{\upsilon I^{m+1}\left( f(z)\right) +\eta
I^{m+2}\left( f(z)\right) }{\left( \upsilon +\eta \right) z}\right] ^{\sigma
}\cdot \left[ \frac{\upsilon z\left( I^{m+1}\left( f(z)\right) \right)
^{\prime }+z\eta \left( I^{m+2}\left( f(z)\right) \right) ^{\prime }}{%
\upsilon I^{m+1}\left( f(z)\right) +\eta I^{m+2}\left( f(z)\right) }-1\right]
.
\end{equation*}%
In order to prove our result we will use Lemma \ref{Lema1.2}. In this lemma
we consider%
\begin{equation*}
\theta \left( w\right) =tw\text{ \ and \ }\varphi \left( w\right) =\frac{1}{w%
}
\end{equation*}%
then $\theta $ is analytic in $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
$ and $\varphi \left( w\right) \neq 0$ is analytic in $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
^{\ast }.$ Also if we let%
\begin{equation*}
Q\left( z\right) =zq^{\prime }(z)\varphi (q(z))=\sigma \left[ \frac{\upsilon
z\left( I^{m+1}\left( f(z)\right) \right) ^{\prime }+z\eta \left(
I^{m+2}\left( f(z)\right) \right) ^{\prime }}{\upsilon I^{m+1}\left(
f(z)\right) +\eta I^{m+2}\left( f(z)\right) }-1\right]
\end{equation*}%
and%
\begin{equation*}
h(z)=\theta (q(z))+Q\left( z\right) =
\end{equation*}%
\begin{equation*}
=t\left[ \frac{\upsilon I^{m+1}\left( f(z)\right) +\eta I^{m+2}\left(
f(z)\right) }{\left( \upsilon +\eta \right) z}\right] ^{\sigma }+\sigma %
\left[ \frac{\upsilon z\left( I^{m+1}\left( f(z)\right) \right) ^{\prime
}+z\eta \left( I^{m+2}\left( f(z)\right) \right) ^{\prime }}{\upsilon
I^{m+1}\left( f(z)\right) +\eta I^{m+2}\left( f(z)\right) }-1\right]
\end{equation*}%
from (\ref{2.6}) we see that $Q\left( z\right) $ is a starlike
function in $U$. We also have%
\begin{equation*}
\mathrm{Re\ }\left\{ \displaystyle\frac{zh^{\prime }(z)}{Q(z)}\right\} =%
\mathrm{Re\ }\left\{ t+1+\displaystyle\frac{zq^{\prime \prime }(z)}{%
q^{\prime }(z)}\right\} >0\text{ for }z\text{ }\in U
\end{equation*}%
and then, by using Lemma \ref{Lema1.2} we deduce that the subordination $%
\left( \ref{2.9}\right) $ implies $p(z)\prec q(z)$.
\end{dem}

Taking $q(z)=\frac{1+Az}{1+Bz},$ $\left( -1\leq B<A\leq 1\right) $ in
Theorem \ref{T2.4} and according to
\begin{equation*}
\frac{zp^{\prime }(z)}{p(z)}=\sigma \left( \frac{I^{m+1}\left( f(z)\right) }{%
I^{m+2}\left( f(z)\right) }-1\right)
\end{equation*}%
the condition $\left( \ref{2.7}\right) $ becomes $\max \left\{ 0,-\R %
\left( t\right) \right\} \leq \frac{1-\left\vert B\right\vert }{1+\left\vert
B\right\vert }.$ Hence, for the special case $\upsilon =1$ and $\eta =0$ we
obtain the following result.

\begin{cor}
\bigskip \label{C2.5} Let $t\in
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
$ with $\max \left\{ 0,-\R\left( t\right) \right\} \leq \frac{%
1-\left\vert B\right\vert }{1+\left\vert B\right\vert }.$ Let $f\in \mathcal{%
A}$ and suppose that%
\begin{equation*}
\frac{I^{m+1}\left( f(z)\right) }{z}\neq 0,z\in U.
\end{equation*}%
If
\begin{equation*}
t\left[ \frac{I^{m+1}\left( f(z)\right) }{z}\right] ^{\sigma }+\sigma \left[
\frac{z\left( I^{m+1}\left( f(z)\right) \right) ^{\prime }}{I^{m+1}\left(
f(z)\right) }-1\right] \prec t\frac{1+Az}{1+Bz}+\frac{\left( 1-B\right) z}{%
\left( 1+Az\right) \left( 1+Bz\right) }
\end{equation*}%
then%
\begin{equation*}
\left( \frac{I^{m+1}\left( f(z)\right) }{z}\right) ^{\sigma }\prec \frac{1+Az%
}{1+Bz}
\end{equation*}%
and $\ q(z)=\frac{1+Az}{1+Bz}$ is the best dominant.
\end{cor}

Taking $\upsilon =m=1,\ \eta =0$ and  $q(z)=\frac{1+z}{1-z}$ in Theorem \ref%
{T2.1}, we obtain the next result.

\begin{cor}
\label{C2.6} Let $f\in \mathcal{A}$ and suppose that $\frac{I^{2}\left(
f(z)\right) }{z}\neq 0,z\in U,$ $\sigma \in
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
^{\ast }.$ If
\begin{equation*}
t\left[ \frac{I^{2}\left( f(z)\right) }{z}\right] ^{\sigma }+\sigma \left[
\frac{z\left( I^{2}\left( f(z)\right) \right) ^{\prime }}{I^{2}\left(
f(z)\right) }-1\right] \prec t\frac{1+z}{1-z}+\frac{2z}{\left( 1+z\right)
\left( 1-z\right) }
\end{equation*}%
then%
\begin{equation*}
\left[ \frac{I^{2}\left( f(z)\right) }{z}\right] ^{\sigma }\prec \frac{1+z}{%
1-z}
\end{equation*}%
and $q(z)=\frac{1+z}{1-z}$ is the best dominant.
\end{cor}

\begin{thm}
\label{T2.7} Let $q$ be convex in $U$, with $q(0)=1$. Let $\sigma \in
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
^{\ast }$ and $\ t,\nu ,\eta \in
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}$
%EndExpansion
 with $\nu +\eta \neq 0$ and  $\mathrm{\R}t>0$. Let $f\in \mathcal{%
A}$ and suppose that $f$ satisfies the next conditions:%
\begin{equation}
\frac{\upsilon I^{m+1}\left( f(z)\right) +\eta I^{m+2}\left( f(z)\right) }{%
\left( \upsilon +\eta \right) z}\neq 0,z\in U  \label{2.12}
\end{equation}%
and
\begin{equation}
\left[ \frac{\upsilon I^{m+1}\left( f(z)\right) +\eta I^{m+2}\left(
f(z)\right) }{\left( \upsilon +\eta \right) z}\right] ^{\sigma }\in \mathcal{%
H}\left[ q(0),1\right] \cap Q.  \label{2.13}
\end{equation}%
If the function $\psi $ given by $\left( \ref{2.8}\right) $ is univalent in $%
U$ and%
\begin{equation}
tq(z)+\frac{zq^{\prime }(z)}{q(z)}\prec \psi (z),  \label{2.14}
\end{equation}%
then%
\begin{equation*}
q(z)\prec \left[ \frac{\upsilon I^{m+1}\left( f(z)\right) +\eta
I^{m+2}\left( f(z)\right) }{\left( \upsilon +\eta \right) z}\right] ^{\sigma
}
\end{equation*}
and  $q(z)$ is the best subordinant of $\left( \ref{2.14}\right) .$
\end{thm}

\begin{dem}
Let%
\begin{equation}
p(z)=\left[ \frac{\upsilon I^{m+1}\left( f(z)\right) +\eta I^{m+2}\left(
f(z)\right) }{\left( \upsilon +\eta \right) z}\right] ^{\sigma },\text{ }%
z\in U.  \label{2.15}
\end{equation}%
According to $\left( \ref{2.12}\right) $ the function $p(z)$ is analytic in $%
U$ and differentiating $\left( \ref{2.15}\right) $ logarithmically with
respect to $z$, we obtain
\begin{equation}
\frac{zp^{\prime }(z)}{p(z)}=\sigma \left[ \frac{\upsilon z\left(
I^{m+1}\left( f(z)\right) \right) ^{\prime }+z\eta \left( I^{m+2}\left(
f(z)\right) \right) ^{\prime }}{\upsilon I^{m+1}\left( f(z)\right) +\eta
I^{m+2}\left( f(z)\right) }-1\right] .  \label{2.16}
\end{equation}%
In order to prove our result we will use Lemma $\ref{Lema1.4}.$ In this
lemma we consider

\begin{equation*}
Q\left( z\right) =zq^{\prime }(z)\varphi (q(z))=\sigma \left[ \frac{\upsilon
z\left( I^{m+1}\left( f(z)\right) \right) ^{\prime }+z\eta \left(
I^{m+2}\left( f(z)\right) \right) ^{\prime }}{\upsilon I^{m+1}\left(
f(z)\right) +\eta I^{m+2}\left( f(z)\right) }-1\right]
\end{equation*}%
and%
\begin{equation*}
h(z)=\theta (q(z))+Q\left( z\right) =
\end{equation*}%
\begin{equation*}
=t\left[ \frac{\upsilon I^{m+1}\left( f(z)\right) +\eta I^{m+2}\left(
f(z)\right) }{\left( \upsilon +\eta \right) z}\right] ^{\sigma }+\sigma %
\left[ \frac{\upsilon z\left( I^{m+1}\left( f(z)\right) \right) ^{\prime
}+z\eta \left( I^{m+2}\left( f(z)\right) \right) ^{\prime }}{\upsilon
I^{m+1}\left( f(z)\right) +\eta I^{m+2}\left( f(z)\right) }-1\right]
\end{equation*}%
from (\ref{2.12}) we see that $Q\left( z\right) $ is a starlike
function in $U$. We also have%
\begin{equation*}
\mathrm{Re\ }\left\{ \displaystyle\frac{zh^{\prime }(z)}{Q(z)}\right\} =%
\mathrm{Re\ }\left\{ t+1+\displaystyle\frac{zq^{\prime \prime }(z)}{%
q^{\prime }(z)}\right\} >0\text{ for }z\text{ }\in U
\end{equation*}%
and then, by using Lemma \ref{Lema1.4} we deduce that the subordination $%
\left( \ref{2.14}\right) $ implies $q(z)\prec p(z)$ and the proof is
completed.
\end{dem}

\begin{cor}
\label{C2.8} Let $q_{1},q_{2}$ are two convex functions in $U$, with $%
q_{1}(0)=q_{2}(0)=1,$ $\sigma \in
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
^{\ast }$ , $\ t,\nu ,\eta \in
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C} $
%EndExpansion
 with $\nu +\eta \neq 0$ and $\mathrm{\R}t>0$. Let $f\in \mathcal{A%
}$ and suppose that $f$ satisfies the next conditions:%
\begin{equation*}
\frac{\upsilon I^{m+1}\left( f(z)\right) +\eta I^{m+2}\left( f(z)\right) }{%
\left( \upsilon +\eta \right) z}\neq 0,z\in U
\end{equation*}%
and%
\begin{equation*}
\left( \frac{\upsilon I^{m+1}\left( f(z)\right) +\eta I^{m+2}\left(
f(z)\right) }{\left( \upsilon +\eta \right) z}\right) ^{\sigma }\in \mathcal{%
H}\left[ q(0),1\right] \cap Q.
\end{equation*}%
If the function $\psi (z)$ given by $\left( \ref{2.8}\right) $ is univalent
in $U$ and%
\begin{equation*}
tq_{1}(z)+\frac{zq_{1}^{\prime }(z)}{q_{1}(z)}\prec \psi (z)\prec tq_{2}(z)+%
\frac{zq_{2}^{\prime }(z)}{q_{2}(z)}
\end{equation*}%
then%
\begin{equation}
q_{1}(z)\prec \left( \frac{\upsilon I^{m+1}\left( f(z)\right) +\eta
I^{m+2}\left( f(z)\right) }{\left( \upsilon +\eta \right) z}\right) ^{\sigma
}\prec q_{2}(z)  \label{2.17}
\end{equation}%
and $q_{1},q_{2}$ are respectively, the best subordinant and the best
dominant of $\left( \ref{2.17}\right) .$
\end{cor}

\begin{thebibliography}{9}
\bibitem{1} T. Bulboac\u{a}, \textit{Classes of first-order differential superordinations}%
, Demonstr. Math., 35, (2)(2002), 287-292.

\bibitem{2} T. Bulboac\u{a}, \textit{Differential Subordinations and
Superordinations, }Recent Results, \textit{\ }House of\textit{\ S}cientific
Book Publ., Cluj-Napoca, 2005.

\bibitem{3} S.S. Miller and P.T. Mocanu, \textit{Differential
Subordinations: Theory and Applications}, in Series on Monographs and
Textbooks in  Pure and Applied Mathematics, No. 225, Marcel Dekker, New
York, (2000).

\bibitem{4} S.S. Miller and P.T. Mocanu, \textit{Subordinants of
differential superordinations}, Compelx Variables, 48(10)(2003), 815-826.

\bibitem{5} S.S. Miller and P.T. Mocanu, \textit{Briot-Bouquet differential
superordinations and sandwich theorems,} J. Math., Anal. Appl., 329(1),
(2007), 327-335.

\bibitem{6} G. \c{S}t. S\u{a}l\u{a}gean, \textit{Subclasses of univalent
functions}, Complex Analysis, Fifth Romanian-Finnish Seminar, Part 1
(Bucharest, 1981), Lecture Notes in Math., vol. 1013, Springer, Berlin,
1983, 362-372.
\end{thebibliography}

\noindent Technical University of Cluj-Napoca\newline
Department of Mathematics \newline
Cluj-Napoca, Romania\newline
E-mail: \newline
luminita.cotirla@yahoo.com\newline
Luminita.Cotirla@math.utcluj.ro\newline

\noindent University of Oradea, Romania \newline
Department of Mathematics and Computer Science \newline
e-mail:\newline
acatas@gmail.com

\end{document}