\documentclass[10pt]{studiamnew} \usepackage{graphicx} \usepackage{amsmath} \usepackage{amssymb} \sloppy \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \theoremstyle{definition} \newtheorem{remark}[theorem]{Remark} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{problem}[theorem]{Problem} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{solution}[theorem]{Solution} \newtheorem{notation}[theorem]{Notation} \renewcommand{\theequation}{\thesection.\arabic{equation}} \numberwithin{equation}{section} \begin{document} % \setcounter{page}{1} \setcounter{firstpage}{1} \setcounter{lastpage}{11} \renewcommand{\currentvolume}{56} \renewcommand{\currentyear}{2011} \renewcommand{\currentissue}{2} \newcommand{\sumful}[1]{\displaystyle\sum^\infty_{#1}} \title{Some Properties of a new subclass of analytic univalent functions defined by Multiplier Transformation} \author{Saurabh Porwal} \address{Department of Mathematics\\ UIET, CSJM University, Kanpur-208024\\ (U.P.), India\\} \email{saurabhjcb@rediffmail.com} % \author{Surya Pratap Singh} \address{Department of Mathematics\\ UIET, CSJM University, Kanpur-208024\\ (U.P.), India\\} %\email{Your e-mail address} % % \subjclass{ 30C45, 30C50, 30C55.} \keywords{Analytic, Univalent functions, Multiplier Transformation.} \begin{abstract} The purpose of the present paper is to study the integral operator of the form \[ \int_{0}^{z}\left\{\frac{I^n_{\mu}f(t)}{t}\right\}^{\delta}dt\] where $f$ belongs to the subclass $C(n,\alpha,\beta, \mu)$ and $\delta$ is a real number. We obtain integral characterization for the subclass $C(n,\alpha,\beta, \mu)$ and also prove distortion, rotation and radii theorem for this class. 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(z)=z+\sumful{k=2}a_kz^k, \end{equation} which are analytic in the open unit disk $U=\{z:z\in \mathbb{C}$ and $|z|<1\}$ and satisfy the normalization condition $f(0)=f'(0)-1=0$. Let $S$ be the subclass of $A$ consisting of functions of the form \eqref{eq11} which are also univalent in $U$. A function $f$ of $S$ is said to be starlike of order $\alpha (0\le \alpha <1),$ denoted by $f\in S^{*}(\alpha)$, if and only if \[ \Re\left\{\frac{zf'(z)}{f(z)}\right\}>\alpha,~~~z\in U,\] and is said to be convex of order $\alpha (0\le \alpha <1),$ denoted by $f\in K(\alpha)$, if and only if \[ \Re\left\{1+\frac{zf''(z)}{f'(z)}\right\}>\alpha,~~~z\in U.\] The classes $S^{*}$ and $K$ of starlike and convex functions, respectively, are identified by $S^{*}(0)\equiv S^{*}$ and $K(0)\equiv K$. These classes were first studied by Robertson \cite{14}. In 2003 Cho and Srivastava \cite{3a}, (see also \cite{4a}) introduced the multiplier transformation for functions $f$ of the form \eqref{eq11} as follows \[ I^n_{\mu}f(z)=z+\sum _{k=2}^{\infty}\left(\frac{k+\mu}{1+\mu}\right)^n a_{k} z^{k}.\] For $\mu=1$, the operator $I^n_{\mu}\equiv I^n$ was studied by Uralegaddi and Somanatha \cite{12} and for $\mu=0$ the operator $I^n_{\mu}$ reduce to well-known S\u{a}l\u{a}gean operator introduced by S\u{a}l\u{a}gean \cite{15}. \newcommand{\cb}[1]{\left\{{#1}\right\}} Using the multiplier transformation we introduce the class $S(n, \alpha, \mu)$ of functions of the form \eqref{eq11} satisfying the following condition \begin{equation} \label{eq12} \Re\cb{\frac{z\left(I^{n}_{\mu}f(z)\right)'}{I^n_{\mu}f(z)}}>\alpha,~~z\in U. \end{equation} It is worthy to note that for $\mu=0$ the class $S(n, \alpha, \mu)$ reduce to the class $S(n, \alpha)$ was first introduced by S\u{a}l\u{a}gean \cite{15} and further studied by Kadio\v{g}lu \cite{1a}. It should be worthy to note that $S(0,\alpha, 0)=S^{\ast}(\alpha)$ and $S(1,\alpha,0)=K(\alpha)$. A function $f$ of $A$ belongs to the class $C(n,\alpha,\beta, \mu)$ if there exists a function $F\in S^{*}(\alpha)$ such that \[ \left|\arg \frac{I^n_{\mu}f(z)}{F(z)}\right|<\frac{\beta \pi}{2},~~~z\in U,\] where $n\in N_0,~0\leq \alpha<1,~0<\beta\le 1$, $\mu>-1$. By specializing the parameters in $C(n,\alpha,\beta, \mu)$ we obtain the following known subclasses of $A$ studied earlier by various researchers. \begin{itemize} \item[(1)] $C(0,\alpha,\beta,0)\equiv CS^{*}(\alpha,\beta)$ studied by Mishra \cite{6}. \item[(2)] $C(1,\alpha,\beta,0)\equiv C(\alpha,\beta)$ studied by Mishra \cite{6}. \item[(3)] $C(0,0,\beta,0)\equiv CS^{*}(\beta)$ studied by Reade \cite{13}. \item[(4)] $C(1,0,\beta,0)\equiv C(\beta)$ studied by Kaplan \cite{2}. \item[(5)] $C(0,0,1,0)\equiv S^{*}$ studied by Roberston \cite{14}, (see also \cite{1}, \cite{17}). \item[(6)] $C(1,0,1,0)\equiv K$ studied by Roberston \cite{14}, (see also \cite{1}, \cite{17}). \end{itemize} In the present paper, we study the integral operator \begin{equation} \label{eq12} h(z)=\int^{z}_{0}\left\{\frac{I^n_{\mu}f(t)}{t}\right\}^{\delta}dt \end{equation} where $n\in N_{0}$ and $\delta$ is a real number. For $n=0$ and $n=1$ this integral operator was studied by Kim \cite{3}, Merkes and Wright \cite{5}, Mishra \cite{6}, Nunokawa(\cite{7}, \cite{8}), Pfaltzgraff \cite{10}, Royster \cite{14a}, Patil and Tahakare \cite{9} and Shukla and Kumar \cite{16}, (see also \cite{16a}). To prove our main results, we shall require the following definition and lemmas. \begin{definition} Let $P(\alpha)$ denote the class of functions of the form $P(z)=1+\sum_{k=1}^{\infty}p_kz^k$ which are regular in $U$ and satisfy $\Re\left\{P(z)\right\}>\alpha,~~z\in U$. \end{definition} \begin{lemma}\label{l11} Let $P(z)=1+\sumful{k=1}p_kz^k$ be analytic in $U$. If $\Re\left\{P(z)\right\}>\alpha$ in $U$, then \begin{equation} \label{eq13} \alpha(\theta_2-\theta_1)<\int^{\theta_2}_{\theta_1}\Re\left\{P(re^{i\theta})\right\}d\theta<2\pi(1-\alpha)+\alpha(\theta_2-\theta_1), \end{equation} where $0\leq \theta_1<\theta_2\leq 2\pi$, $z=re^{i\theta}$ and $0\le r<1$. \end{lemma} \begin{proof} Since \[\Re\left\{P(z)\right\}>\alpha.\] It is easy to see that \[ \left.\left(\Re\left\{P(z)\right\}-\alpha\right)\right|_{z=0}=1-\alpha.\] Then by mean value theorem, we have \[ 0\le \int_{\theta_1}^{\theta_2}\left(\Re\left\{P(re^{i\theta})\right\}-\alpha\right)d\theta \le \int_{0}^{2\pi}\left(\Re\left\{P(re^{i\theta})\right\}-\alpha\right)d\theta=2\pi\left(1-\alpha\right).\] or, equivalently \[ 0\le \int_{\theta_1}^{\theta_2}\left(\Re\left\{P(re^{i\theta})\right\}\right)d\theta-\alpha(\theta_2-\theta_1)\le 2\pi\left(1-\alpha\right),\] or \[\alpha(\theta_2-\theta_1)<\int^{\theta_2}_{\theta_1}\Re\left\{P(re^{i\theta})\right\}d\theta<2\pi(1-\alpha)+\alpha(\theta_2-\theta_1). \] \end{proof} The following lemma is a direct consequence of Lemma \ref{l11}, and improves a result of Patil and Thakare (\cite{9}, Lemma 2.2). \begin{lemma}\label{l12} If $f\in S^{*}(\alpha)$, then \begin{equation} \label{eq14} \alpha(\theta_2-\theta_1)<\int^{\theta_2}_{\theta_1}\Re\left\{\frac{zf'(z)}{f(z)}\right\}d\theta<2\pi(1-\alpha)+\alpha(\theta_2-\theta_1), \end{equation} where $0\leq \theta_1<\theta_2\leq 2\pi$, $z=re^{i\theta}$ and $0\le r<1$. \end{lemma} In the following lemma, we obtain integral characterization for the class $C(n,\alpha,\beta, \mu)$. \begin{lemma}\label{l13} If $f\in C(n,\alpha,\beta, \mu)$, then \begin{equation} \label{eq15} -\beta\pi+\alpha(\theta_2-\theta_1)<\int^{\theta_2}_{\theta_1}\Re\left\{\frac{z\left(I^{n}_{\mu}f(z)\right)'}{I^n_{\mu}f(z)}\right\}d\theta<\beta\pi +2\pi(1-\alpha)+\alpha(\theta_2-\theta_1), \end{equation} where $0\leq \theta_1<\theta_2\leq 2\pi$, $z=re^{i\theta}$ and $0\le r<1$. Conversely, let $f$ be analytic and satisfying $I^n_{\mu}f(z)\neq 0$ in $U$, if \[ \int^{\theta_2}_{\theta_1} \Re\left\{\frac{z\left(I^{n}_{\mu}f(z)\right)'}{I^n_{\mu}f(z)}\right\}d\theta>-\beta\pi+\alpha(\theta_2-\theta_1)\] then $f\in C(n,\alpha,\beta, \mu)$. \end{lemma} \begin{proof} $f\in C(n,\alpha,\beta,\mu)$ implies that there exists a function $F\in S^{*}(\alpha)$ such that \[ \left|\arg \frac{I^n_{\mu}f(z)}{F(z)}\right|<\frac{\beta \pi}{2},~~~z\in U.\] Therefore \[ -\frac{1}{2}\beta \pi< \arg I^n_{\mu}f(z)-\arg F(z)<\frac{1}{2}\beta \pi.\] Let $0\le \theta_1<\theta_2\le 2\pi$. Then with $z=re^{i\theta_2}$, we have \begin{equation} \label{eq16} -\frac{1}{2}\beta \pi< \arg I^n_{\mu}f(re^{i\theta_2})-\arg F(re^{i\theta_2})<\frac{1}{2}\beta \pi. \end{equation} and with $z=re^{i\theta_1}$, we have \begin{equation} \label{eq17} -\frac{1}{2}\beta \pi< -\arg I^n_{\mu}f(re^{i\theta_1})+\arg F(re^{i\theta_1})<\frac{1}{2}\beta \pi. \end{equation} Combining \eqref{eq16} and \eqref{eq17}, we obtain \begin{small}\[ -\beta \pi+\arg F(re^{i\theta_2})-\arg F(re^{i\theta_1})<\arg I^n_{\mu}f(re^{i\theta_2})-\arg I^n_{\mu}f(re^{i\theta_1})<\beta \pi+\arg F(re^{i\theta_2})-\arg F(re^{i\theta_1}),\] \end{small} or \[-\beta \pi+\int^{\theta_2}_{\theta_1}d \arg F(re^{i\theta})<\int^{\theta_2}_{\theta_1}d \arg I^n_{\mu} f(re^{i\theta})<\beta \pi+\int^{\theta_2}_{\theta_1}d \arg F(re^{i\theta}),\] or \begin{equation} \label{eq18} -\beta \pi+\int^{\theta_2}_{\theta_1}\Re\left\{\frac{zF'(z)}{F(z)}\right\}d\theta<\int^{\theta_2}_{\theta_1}\Re\left\{\frac{z\left(I^{n}_{\mu}f(z)\right)'}{I^n_{\mu}f(z)}\right\}d\theta<\beta \pi+\int^{\theta_2}_{\theta_1}\Re\left\{\frac{zF'(z)}{F(z)}\right\}d\theta. \end{equation} But $F\in S^{*}(\alpha)$, then using Lemma \ref{l12} in \eqref{eq18}, we have \[ -\beta\pi+\alpha(\theta_2-\theta_1)<\int^{\theta_2}_{\theta_1}\Re\left\{\frac{z\left(I^{n}_{\mu}f(z)\right)'}{I^n_{\mu}f(z)}\right\}d\theta<\beta\pi +2\pi(1-\alpha)+\alpha(\theta_2-\theta_1)\] and this completes the proof of direct part of the lemma. To prove the converse part, we follow the techniques of Kaplan \cite{2} and Patil and Thakare \cite{9} and can obtain the desired result. \end{proof} \begin{remark} If we put $n=1$, $\mu=0$ in Lemma \ref{l13}, we obtain the following result If $f\in C(\alpha,\beta)$, then \begin{equation} \label{eq19} -\beta\pi+\alpha(\theta_2-\theta_1)<\int^{\theta_2}_{\theta_1}\Re\left\{1+\frac{zf''(z)}{f'(z)}\right\}d\theta<\beta\pi +2\pi(1-\alpha)+\alpha(\theta_2-\theta_1), \end{equation} where $0\leq \theta_1<\theta_2\leq 2\pi$, $z=re^{i\theta}$ and $0\le r<1$. Conversely, let $f$ be analytic and satisfying $f'(z)\neq 0$ in $U$, if \begin{equation} \label{eq110} \int^{\theta_2}_{\theta_1} \Re\left\{1+\frac{zf''(z)}{f'(z)}\right\}d\theta>-\beta\pi+\alpha(\theta_2-\theta_1) \end{equation} then $f\in C(\alpha,\beta)$. \end{remark} \section{\bf Main Results}\label{sec2} \begin{theorem}\label{t21} If $f\in C(n,\alpha,\beta, \mu)$, then $h\in C(\eta, \gamma)$, provided \begin{equation} \label{eq21} \frac{-\gamma}{\beta+2(1-\alpha)}\le \delta \le \frac{\gamma+2(1-\eta)}{\beta+2(1-\alpha)}. \end{equation} The result is sharp when (i) $\gamma=0$ (ii) $\eta=0, \gamma=1$. \end{theorem} \begin{proof} From relation \eqref{eq12} we have \[ h'(z)=\left\{\frac{I^n_{\mu}f(z)}{z}\right\}^{\delta}.\] Applying logarithmic differentiation and then taking real parts of both sides, we obtain \[ Re\left\{1+\frac{zh''(z)}{h'(z)}\right\}=\delta Re\left\{\frac{z\left(I^{n}_{\mu}f(z)\right)'}{I^n_{\mu}f(z)}\right\}+(1-\delta).\] For $\delta>0$, using Lemma \ref{l13}, we get \[ \int^{\theta_2}_{\theta_1}Re\left\{1+\frac{zh''(z)}{h'(z)}\right\}d\theta=\delta\int^{\theta_2}_{\theta_1}Re\left\{\frac{z\left(I^{n}_{\mu}f(z)\right)'}{I^n_{\mu}f(z)}\right\}d\theta+(1-\delta)(\theta_2-\theta_1)\] \[ >\delta[-\beta \pi+\alpha(\theta_2-\theta_1)]+(1-\delta)(\theta_2-\theta_1)\] \[ =-\beta \delta \pi+[1-(1-\alpha)\delta](\theta_2-\theta_1).\] To prove that $h\in C(\eta, \gamma)$, we have to show that the right hand side of the above inequality is not less than $-\gamma \pi+\eta(\theta_2-\theta_1),$ provided \begin{equation}\label{eq22} 0\le \delta \le \frac{\gamma+2(1-\eta)}{\beta+2(1-\alpha)}. \end{equation} Similarly, for $\delta<0$, using Lemma \ref{l13}, we get \[ \int^{\theta_2}_{\theta_1}\Re\left\{1+\frac{zh''(z)}{h'(z)}\right\}d\theta>\delta\left[\beta \pi+2(1-\alpha)+\alpha(\theta_2-\theta_1)\right]+(1-\delta)(\theta_2-\theta_1).\] To show that $h\in C(\eta, \gamma)$, we have to prove that the right-hand side of the above inequality is not less than $-\gamma \pi+\eta(\theta_2-\theta_1)$, provided \begin{equation}\label{eq23} \frac{-\gamma}{\beta+2(1-\alpha)}\le \delta \le 0. \end{equation} Combining \eqref{eq22} and \eqref{eq23}, we obtain \eqref{eq21}. Thus the proof of Theorem \ref{t21} is established. To show the sharpness, let us take the function $f(z)$ defined by the relation \begin{equation}\label{eq24} I^n_{\mu}f(z)=\frac{z}{(1-z)^{2(1-\alpha)+\beta}}, \end{equation} then it is easy to see that this function belongs to $C(n,\alpha,\beta, \mu)$ with respect to the function $\frac{z}{(1-z)^{2(1-\alpha)}}$ belonging to $S^{\ast}(\alpha)$. Then \begin{equation}\label{eq25} h(z)=\int^{z}_{0}\frac{dt}{(1-t)^{[2(1-\alpha)+\beta]\delta}} \end{equation} and from condition \eqref{eq110} this functions belongs to $C(0,1)$ if and only if \[ \frac{-1}{2(1-\alpha)+\beta}\le \delta \le \frac{3}{2(1-\alpha)+\beta}.\] Again for $\gamma=0$, from \eqref{eq25} we have \[1+\frac{zh''(z)}{h'(z)}=\frac{1+\left[1-2\left(1-\frac{\left\{2(1-\alpha)+\beta\right\}\delta}{2}\right)\right]z}{1-z}\] and $\Re\left\{1+\frac{zh''(z)}{h'(z)}\right\}>\eta$ if and only if \[ 1-\frac{\left\{2(1-\alpha)+\beta\right\}\delta}{2}\ge \eta\] \[ \Rightarrow 0\le \delta \le \frac{2(1-\eta)}{\beta+2(1-\alpha)}.\] \end{proof} \begin{remark} The undermentioned results are particular cases of Theorem \ref{t21}. \begin{itemize} \item[(i)] If we put $n=0$ and $n=1$ with $\mu=0$ in Theorem \ref{t21} we obtain the corresponding results of Mishra \cite{6}. \item[(ii)] If we put $n=1, \beta=0, \gamma=0$ with $\mu=0$ we obtain a result of Patil and Thakare \cite{9}. \item[(iii)] If we put $n=1, \beta=0, \eta=0$ with $\mu=0$ we obtain a result of Patil and Thakare \cite{9}. \item[(iv)] If we put $n=1, \alpha=0, \eta=0$ with $\mu=0$ we obtain a result of Patil and Thakare \cite{9}. \item[(v)] If we put $n=0, \beta=0, \eta=0$ we obtain a result of Patil and Thakare \cite{9}. \item[(vi)] If we put $n=1, \alpha=0, \beta=0, \eta=0$ and $\gamma=1$ with $\mu=0$ we obtain a result of Nunokawa \cite{8} as well as that of Merkes and Wright \cite{5}. \item[(vii)] If we put $n=0, \alpha=0, \beta=0, \eta=0$ and $\gamma=1$ with $\mu=0$ we obtain a result of Nunokawa \cite{8} as well as that of Merkes and Wright \cite{5}. \item[(viii)] If we put $n=1, \alpha=0, \beta=1, \eta=0$ and $\gamma=1$ with $\mu=0$ we obtain a result of Nunokawa \cite{8} as well as that of Merkes and Wright \cite{5}. \item[(ix)] If we put $n=0, \alpha=0, \eta=0$ with $\mu=0$ we obtain a result of Shukla and Kumar \cite{16}. \item[(x)] If we put $n=0, \alpha=0, \beta=1, \eta=0$ and $\gamma=1$ with $\mu=0$ we obtain a result of Kim \cite{3}. \item[(xi)] If we put $n=0, \alpha=1/2, \beta=0, \eta=0$ and $\gamma=1$ with $\mu=0$ we obtain a result of Nunokawa \cite{8} as well as that of Merkes and Wright \cite{5}. \end{itemize} \end{remark} \begin{theorem}\label{t22} Let $f\in C(n,\alpha,\beta, \mu)$. Then for $|z|=r$ \[ \frac{r(1-r)^{\beta}}{(1+r)^{\beta+2(1-\alpha)}}\le |I^n_{\mu}f(z)|\le \frac{r(1+r)^{\beta}}{(1-r)^{\beta+2(1-\alpha)}}\] The result is sharp. \end{theorem} \begin{proof} By definition $f\in C(n,\alpha,\beta, \mu)$ if and only if there exists a function $P\in P(0)$ and $F(z) \in S^{*}(\alpha)$ such that \[ \frac{I^n_{\mu}f(z)}{F(z)}=[P(z)]^{\beta}.\] Therefore \[\left|I^n_{\mu}f(z)\right|=|P(z)|^{\beta}|F(z)|.\] Now using the well-known inequalities (see \cite{1}) \[ \frac{1-r}{1+r}\le |P(z)| \le \frac{1+r}{1-r}\] and \[ \frac{r}{(1+r)^{2(1-\alpha)}}\le |F(z)| \le \frac{r}{(1-r)^{2(1-\alpha)}},\] we obtain the required inequalities. Sharpness follows if we take $f(z)$ connected by the relation \[ I^n_{\mu}f(z)=\frac{z(1+z)^{\beta}}{(1-z)^{\beta+2(1-\alpha)}}\] and \[ F(z)=\frac{z}{(1-z)^{2(1-\alpha)}}.\] \end{proof} \begin{theorem}\label{t23} If $f\in C(n,\alpha,\beta, \mu)$, then \[ \left|\arg \frac{I^n_{\mu}f(z)}{z}\right| \le \beta \sin^{-1} \frac{2r}{1+r^2}+2(1-\alpha)\sin^{-1}r. \] The result is sharp. \end{theorem} \begin{proof} If $f\in C(n,\alpha,\beta, \mu)$, then \[\frac{I^n_{\mu}f(z)}{F(z)}=[P(z)]^{\beta},\] for some $P(z)\in P(0)$ and $F(z)\in S^{*}(\alpha)$. Thus \begin{equation}\label{eq26} \left|\arg \frac{I^n_{\mu}f(z)}{z}\right|\le \alpha\left|\arg P(z)\right|+\left|\arg \frac{F(z)}{z}\right|. \end{equation} Now using the well-known results \begin{equation}\label{eq27} \left|\arg P(z)\right|\le \sin^{-1} \frac{2r}{1+r^{2}} \end{equation} and a result of Pinchuk \cite{11} \begin{equation}\label{eq28} \left|\arg \frac{F(z)}{z}\right|\le 2(1-\alpha)\sin^{-1} r, \end{equation} Using \eqref{eq27} and \eqref{eq28} in \eqref{eq26} we get the required result. Sharpness follows if we take $f(z)$ to be the same as in Theorem \ref{t22}. \end{proof} \begin{theorem}\label{t24} If $f\in C(n,\alpha,\beta, \mu)$, then $f\in S(n)$ for $|z|