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\title{Differential subordinations obtained by using a fractional operator}
\author{Eszter Szatmari}
\address{``Babe\c{s}-Bolyai'' University, \\ Faculty of Mathematics and Computer Sciences\\
1, Kog\u{a}lniceanu Street,\\
400084 Cluj-Napoca,\\
Romania}
\email{szatmari.eszter@math.ubbcluj.ro}
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\subjclass{30C45}
\keywords{ differential subordination, analytic function, fractional operator, convex function.}
\begin{abstract}
We investigate several differential subordinations using the fractional operator $\mathbb{D}_\lambda^{\nu, n}:\mathcal{A}\to\mathcal{A}$, for $-\infty<\lambda<2, \nu>-1, n\in\mathbb{N}_0=\{0,1,2,...\}$, introduced in \cite{7}.
\end{abstract}
\maketitle

\section{Introduction}
$\; \; \; $ Let $\mathcal{H}(U)$ denote the class of functions which are analytic in the open unit disk $U=\{z\in\mathbb{C}:|z|<1\}$.

For $a\in\mathbb{C}$ and $k\in\mathbb{N}=\{1, 2,...\}$, let
$$\mathcal{H}[a,k]=\{f\in\mathcal{H}(U):f(z)=a+a_kz^k+a_{k+1}z^{k+1}+...\},$$
and
$$\mathcal{A}=\{f\in\mathcal{H}(U):f(z)=z+a_2z^2+a_3z^3+..., z\in{U}\}.$$

In \cite{3}, the fractional integral operator $D_z^{-\mu}$ of order $\mu, \mu>0$, for the function $f\in\mathcal{A}$, is defined by
$$ D_z^{-\mu}f(z)=\frac{1}{\Gamma(\mu)}\int\limits _0^z\frac{f(t)}{(z-t)^{1-\mu}}dt , z\in{U},$$
where the multiplicity of $(z-t)^{\mu-1}$ is removed by requiring $\log(z-t)$ to be real when $z-t>0$.

Also, the fractional derivative operator $D_z^\lambda$ of order $\lambda, \lambda\geq0$, for the function $f\in\mathcal{A}$, is defined by
 $$D_z^{\lambda}f(z)=\begin{cases} \frac{1}{\Gamma(1-\lambda)} \frac{d}{dz}\int\limits_0^z \frac{f(t)}{(z-t)^\lambda}dt, & 0\leq\lambda<1 \\ \frac{d^n}{dz^n} D_z^{\lambda-n}f(z), & n\leq\lambda<n+1\end{cases},
n\in\mathbb{N}_0,$$

where the multiplicity of $(z-t)^{-\lambda}$ is understood in a similar way.

In \cite{4} is defined the fractional differintegral operator $\Omega_z^\lambda: \mathcal{A}\to\mathcal{A}, -\infty<\lambda<2$, by
$$ \Omega_z^\lambda f(z)=\Gamma(2-\lambda)z^\lambda D_z^\lambda f(z) , z\in{U},$$
where $D_z^\lambda f(z)$ is the fractional integral of order$\lambda, -\infty<\lambda<0$, and a fractional derivative of order $\lambda, 0\leq\lambda<2$.

In \cite{6}, the S\v{a}l\v{a}gean operator $\mathcal{D}^n$ of order $n$, $n\in\mathbb{N}_0$, for $f\in\mathcal{A}$, is defined by
$$\mathcal{D}^0f(z)=f(z)$$
$$\mathcal{D}^1f(z)=\mathcal{D}f(z)=zf'(z)$$
$$\mathcal{D}^nf(z)=\mathcal{D}(\mathcal{D}^{n-1}f(z)), n\in\mathbb{N}.$$

In \cite{5}, the Ruscheweyh operator $\mathcal{R}^\lambda: \mathcal{A}\to\mathcal{A}$ for $\lambda\geq-1$ is defined by
$$\mathcal{R}^\lambda f(z)=\frac{z}{(1-z)^{1+\lambda}}*f(z), z\in U,$$
where $"*"$ is the Hadamard product or convolution.

For $\lambda\in\mathbb{N}_0$ this operator is defined by
$$\mathcal{R}^\lambda f(z)=\frac{z(z^{\lambda-1}f(z))^\lambda}{\lambda !}, z\in U.$$

In \cite{7}, the fractional operator $\mathbb{D}_\lambda^{\nu, n}:\mathcal{A}\to\mathcal{A}$ for $-\infty<\lambda<2, \nu>-1, n\in\mathbb{N}_0$, is defined as a composition of fractional differintegral operator, the S\v{a}l\v{a}gean operator and the Ruscheweyh operator:
$$\mathbb{D}_\lambda^{\nu, n} f(z)=\mathcal{R}^\nu \mathcal{D}^n \Omega_z^\lambda f(z).$$

The series expression of $\mathbb{D}_\lambda^{\nu, n}f(z)$ for $f\in\mathcal{A}$ of the form $f(z)=z+\sum\limits_{k=1}^{\infty}a_{k+1}z^{k+1}$ is given by
$$\mathbb{D}_\lambda^{\nu, n}f(z)=z+\sum_{k=1}^{\infty} \frac{(\nu+1)_k}{(2-\lambda)_k}(k+1)^{n+1}a_{k+1}z^{k+1},$$
$-\infty<\lambda<2, \nu>-1, n\in\mathbb{N}_0, z\in U$,
where the symbol $(\gamma)_k$ denotes the usual Pochhammer symbol, for $\gamma\in\mathbb{C}$, defined by
$$(\gamma)_k=\begin{cases} 1, k=0 \\ \gamma(\gamma+1)...(\gamma+k-1), k\in\mathbb{N} \end{cases}= \frac{\Gamma(\gamma+k)}{\Gamma(\gamma)}, \gamma\in\mathbb{C}\setminus\mathbb{Z}_0^-.$$

\begin{remark}\cite{7}
The fractional operator $\mathbb{D}_0^{\nu, 0}$ is precisely the Ruscheweyh derivative operator $\mathcal{R}^\nu$ of order $\nu, \nu>-1$, and $\mathbb{D}_\lambda^{0,0}$ is the fractional differintegral operator $\Omega_z^\lambda$ of order $\lambda, -\infty<\lambda<2$, while $\mathbb{D}_0^{0, n}=\mathcal{D}^n$ and $\mathbb{D}_\lambda^{1-\lambda, n}=\mathcal{D}^{n+1}$ are the S\v{a}l\v{a}gean operators, respectively, of order n and n+1, $n\in\mathbb{N}_0$.
\end{remark}

\begin{remark}\cite{7}
The operator $\mathbb{D}_\lambda^{\nu, n}$ satisfies the following identity:\begin{equation}\label{eq:1}
\mathbb{D}_\lambda^{\nu+1, n}f(z)=\frac{\nu}{\nu+1}\mathbb{D}_\lambda^{\nu, n}f(z)+\frac{1}{\nu+1}z(\mathbb{D}_\lambda^{\nu, n}f(z))', \end{equation}
where $-\infty<\lambda<2, \nu>-1, n\in\mathbb{N}_0.$
\end{remark}

\begin{remark}\cite{8}
The operator $\mathbb{D}_\lambda^{\nu, n}$ satisfies the following identities:\begin{equation}\label{eq:2}
\mathbb{D}_\lambda^{\nu, n+1}f(z)=z(\mathbb{D}_\lambda^{\nu, n}f(z))',\end{equation}
where $-\infty<\lambda<2, \nu>-1, n\in\mathbb{N}_0,$

and\begin{equation}\label{eq:3}
\mathbb{D}_{\lambda+1}^{\nu, n}f(z)=-\frac{\lambda}{1-\lambda}\mathbb{D}_\lambda^{\nu, n}f(z)+\frac{1}{1-\lambda}z(\mathbb{D}_\lambda^{\nu, n}f(z))',\end{equation}
where $-\infty<\lambda<1, \nu>-1, n\in\mathbb{N}_0.$
\end{remark}

\begin{definition} \cite [p. 4] {1}
Let $f, F \in\mathcal{H}(U)$. The function $f$ is said to be subordinate to $F$, written $f \prec F$, or $f(z) \prec F(z)$, if there exists a function $w\in\mathcal{H}(U)$, with $w(0)=0$ and $\big|w(z)\big|<1, z\in U$, such that $f(z)=F\big[w(z)\big], z\in U.$
\end{definition}

In order to prove our results we shall need the following lemma.

\begin{lemma}\cite{2}
Let $q$ be a convex function in $U$ and let 
$$h(z)=q(z)+n\alpha zq'(z),$$
where $\alpha>0$ and $n$ is a positive integer. If
$$p(z)=q(0)+p_nz^n+...\in\mathcal{H}[q(0), n]$$
and
$$p(z)+\alpha zp'(z)\prec h(z)$$
then
$$p(z)\prec q(z),$$
and this result is sharp.
\end{lemma}

\section{Main results}
\begin{theorem}
Let $g$ be a convex function, $g(0)=1$ and let $h$ be a function such that 
$$h(z)=g(z)+\frac{1}{\nu+1}zg'(z), \nu>-1.$$
If $f\in\mathcal{A}$ verifies the differential subordination\begin{equation}\label{eq:4}
\big(\mathbb{D}_\lambda^{\nu+1, n}f(z)\big)'\prec h(z)\end{equation}
then
$$\big(\mathbb{D}_\lambda^{\nu, n}f(z)\big)'\prec g(z).$$
The result is sharp.
\end{theorem}

\begin{proof}
If we denote by
$$p(z)=\big(\mathbb{D}_\lambda^{\nu, n}f(z)\big)',$$
where $p(z)\in\mathcal{H}[1,1]$, then, by $\left(\ref{eq:1}\right)$, we get\begin{equation}\label{eq:5}
\big(\mathbb{D}_\lambda^{\nu+1, n}f(z)\big)'=p(z)+\frac{1}{\nu+1}zp'(z), z\in U.\end{equation}
From $\left(\ref{eq:4}\right)$ and $\left(\ref{eq:5}\right)$ we obtain
$$p(z)+\frac{1}{\nu+1}zp'(z)\prec g(z)+\frac{1}{\nu+1}zg'(z)\equiv h(z).$$
Applying Lemma 1, we get
$$p(z)\prec g(z)$$
or
$$\big(\mathbb{D}_\lambda^{\nu, n}f(z)\big)'\prec g(z).$$
This result is sharp.
\end{proof}

\begin{theorem}
Let $g$ be a convex function, $g(0)=1$ and let $h$ be a function such that 
$$h(z)=g(z)+\frac{1}{1-\lambda}zg'(z), -\infty<\lambda<1.$$
If $f\in\mathcal{A}$ verifies the differential subordination\begin{equation}\label{eq:6}
\big(\mathbb{D}_{\lambda+1}^{\nu, n}f(z)\big)'\prec h(z)\end{equation}
then
$$\big(\mathbb{D}_\lambda^{\nu, n}f(z)\big)'\prec g(z).$$
The result is sharp.
\end{theorem}

\begin{proof}
If we denote by
$$p(z)=\big(\mathbb{D}_\lambda^{\nu, n}f(z)\big)',$$
where $p(z)\in\mathcal{H}[1,1]$, then, by $\left(\ref{eq:3}\right)$, we get\begin{equation}\label{eq:7}
\big(\mathbb{D}_{\lambda+1}^{\nu, n}f(z)\big)'=p(z)+\frac{1}{1-\lambda}zp'(z), z\in U.\end{equation}
From $\left(\ref{eq:6}\right)$ and $\left(\ref{eq:7}\right)$ we obtain
$$p(z)+\frac{1}{1-\lambda}zp'(z)\prec g(z)+\frac{1}{1-\lambda}zg'(z)\equiv h(z).$$
Applying Lemma 1, we get
$$p(z)\prec g(z)$$
or
$$\big(\mathbb{D}_\lambda^{\nu, n}f(z)\big)'\prec g(z).$$
This result is sharp.
\end{proof}

\begin{theorem}
Let $g$ be a convex function, $g(0)=1$ and let $h$ be a function such that 
$$h(z)=g(z)+zg'(z), n\in\mathbb{N}_0.$$
If $f\in\mathcal{A}$ verifies the differential subordination\begin{equation}\label{eq:8}
\big(\mathbb{D}_{\lambda}^{\nu, n+1}f(z)\big)'\prec h(z)\end{equation}
then
$$\big(\mathbb{D}_\lambda^{\nu, n}f(z)\big)'\prec g(z).$$
The result is sharp.
\end{theorem}

\begin{proof}
If we denote by
$$p(z)=\big(\mathbb{D}_\lambda^{\nu, n}f(z)\big)',$$
where $p(z)\in\mathcal{H}[1,1]$, then, by $\left(\ref{eq:2}\right)$, we get\begin{equation}\label{eq:9}
\big(\mathbb{D}_{\lambda}^{\nu, n+1}f(z)\big)'=p(z)+zp'(z), z\in U.\end{equation}
From $\left(\ref{eq:8}\right)$ and $\left(\ref{eq:9}\right)$ we obtain
$$p(z)+zp'(z)\prec g(z)+zg'(z)\equiv h(z).$$
Applying Lemma 1, we get
$$p(z)\prec g(z)$$
or
$$\big(\mathbb{D}_\lambda^{\nu, n}f(z)\big)'\prec g(z).$$
This result is sharp.
\end{proof}

\begin{theorem}
Let $g$ be a convex function, $g(0)=1$ and let $h$ be a function such that 
$$h(z)=g(z)+zg'(z), z\in U.$$
If $f\in\mathcal{A}$ verifies the differential subordination\begin{equation}\label{eq:10}
\big(\mathbb{D}_\lambda^{\nu, n}f(z)\big)'\prec h(z), z\in U\end{equation}
then
$$\frac{\mathbb{D}_\lambda^{\nu, n}f(z)}{z}\prec g(z).$$
The result is sharp.
\end{theorem}

\begin{proof}
Let
$$p(z)=\frac{\mathbb{D}_\lambda^{\nu, n}f(z)}{z}, z\in U.$$
Differentiating we obtain
$$p'(z)=\frac{\big(\mathbb{D}_\lambda^{\nu, n}f(z)\big)'}{z}-\frac{p(z)}{z}.$$
We get
$$\big(\mathbb{D}_\lambda^{\nu, n}f(z)\big)'=p(z)+zp'(z).$$
The subordination $\left(\ref{eq:10}\right)$ becomes
$$p(z)+zp'(z)\prec g(z)+zg'(z).$$
Applying Lemma 1, we get
$$p(z)\prec g(z)$$
or
$$\frac{\mathbb{D}_\lambda^{\nu, n}f(z)}{z}\prec g(z).$$
This result is sharp.
\end{proof}

\begin{theorem}
Let $g$ be a convex function, $g(0)=1$ and let $h$ be a function such that 
$$h(z)=g(z)+zg'(z), z\in U.$$
If $f\in\mathcal{A}$ verifies the differential subordination\begin{equation}\label{eq:11}
\bigg(\frac{z\mathbb{D}_\lambda^{\nu+1, n}f(z)}{\mathbb{D}_\lambda^{\nu, n}f(z)}\bigg)'\prec h(z), z\in U,\end{equation}
then
$$\frac{\mathbb{D}_\lambda^{\nu+1, n}f(z)}{\mathbb{D}_\lambda^{\nu, n}f(z)}\prec g(z), z\in U.$$
The result is sharp.
\end{theorem}

\begin{proof}
Let
$$p(z)=\frac{\mathbb{D}_\lambda^{\nu+1, n}f(z)}{\mathbb{D}_\lambda^{\nu, n}f(z)}.$$
We obtain
$$\bigg(\frac{z\mathbb{D}_\lambda^{\nu+1, n}f(z)}{\mathbb{D}_\lambda^{\nu, n}f(z)}\bigg)'=p(z)+zp'(z).$$
The subordination $\left(\ref{eq:11}\right)$ becomes
$$p(z)+zp'(z)\prec g(z)+zg'(z).$$
Applying Lemma 1, we get
$$p(z)\prec g(z)$$
or
$$\frac{\mathbb{D}_\lambda^{\nu+1, n}f(z)}{\mathbb{D}_\lambda^{\nu, n}f(z)}\prec g(z), z\in U.$$
This result is sharp.
\end{proof}

\begin{theorem}
Let $g$ be a convex function, $g(0)=1$ and let $h$ be a function such that 
$$h(z)=g(z)+zg'(z), z\in U.$$
If $f\in\mathcal{A}$ verifies the differential subordination\begin{equation}\label{eq:12}
\bigg(\frac{z\mathbb{D}_{\lambda+1}^{\nu, n}f(z)}{\mathbb{D}_\lambda^{\nu, n}f(z)}\bigg)'\prec h(z), z\in U, -\infty<\lambda<1, \end{equation}
then
$$\frac{\mathbb{D}_{\lambda+1}^{\nu, n}f(z)}{\mathbb{D}_\lambda^{\nu, n}f(z)}\prec g(z), z\in U.$$
The result is sharp.
\end{theorem}

\begin{proof}
Let
$$p(z)=\frac{\mathbb{D}_{\lambda+1}^{\nu, n}f(z)}{\mathbb{D}_\lambda^{\nu, n}f(z)}.$$
We obtain
$$\bigg(\frac{z\mathbb{D}_{\lambda+1}^{\nu, n}f(z)}{\mathbb{D}_\lambda^{\nu, n}f(z)}\bigg)'=p(z)+zp'(z).$$
The subordination $\left(\ref{eq:12}\right)$ becomes
$$p(z)+zp'(z)\prec g(z)+zg'(z).$$
Applying Lemma 1, we get
$$p(z)\prec g(z)$$
or
$$\frac{\mathbb{D}_{\lambda+1}^{\nu, n}f(z)}{\mathbb{D}_\lambda^{\nu, n}f(z)}\prec g(z), z\in U.$$
This result is sharp.
\end{proof}

\begin{theorem}
Let $g$ be a convex function, $g(0)=1$ and let $h$ be a function such that
$$h(z)=g(z)+zg'(z), z\in U.$$
If $f\in\mathcal{A}$ verifies the differential subordination\begin{equation}\label{eq:13}
\bigg(\frac{z\mathbb{D}_\lambda^{\nu, n+1}f(z)}{\mathbb{D}_\lambda^{\nu, n}f(z)}\bigg)'\prec h(z), z\in U,
\end{equation}
then
$$\frac{\mathbb{D}_\lambda^{\nu, n+1}f(z)}{\mathbb{D}_\lambda^{\nu, n}f(z)}\prec g(z), z\in U.$$
The result is sharp.
\end{theorem}

\begin{proof}
Let
$$p(z)=\frac{\mathbb{D}_\lambda^{\nu, n+1}f(z)}{\mathbb{D}_\lambda^{\nu, n}f(z)}.$$
We obtain
$$\bigg(\frac{z\mathbb{D}_\lambda^{\nu, n+1}f(z)}{\mathbb{D}_\lambda^{\nu, n}f(z)}\bigg)'=p(z)+zp'(z).$$
The subordination $\left(\ref{eq:13}\right)$ becomes
$$p(z)+zp'(z)\prec g(z)+zg'(z).$$
Applying Lemma 1, we get
$$p(z)\prec g(z)$$
or
$$\frac{\mathbb{D}_\lambda^{\nu, n+1}f(z)}{\mathbb{D}_\lambda^{\nu, n}f(z)}\prec g(z), z\in U.$$
This result is sharp.
\end{proof}

%\subsection{First subsection}

\renewcommand{\abstractname}{Acknowledgements}
\begin{abstract}
The author wish to express her sincere thanks to the reviewer of this paper for several useful comments and suggestions.
\end{abstract}

\begin{thebibliography}{99}
 \bibitem{1}  S. S. Miller, P. T. Mocanu, 
\emph{Differential Subordinations: Theory and Applications,} Pure and Applied Mathematics No. 225, Marcel Dekker, New York, 2000.

 \bibitem{2} S. S. Miller, P. T. Mocanu,
\emph{ On some classes of first order differential subordinations,} Michigan Math. J., \textbf{32}(1985), 185-195.

 \bibitem{3}   S. Owa,
\emph{On the distortion theorems I,} Kyungpook Math. J., \textbf{18}(1978), 53-59.

\bibitem{4}   S. Owa, H. M. Srivastava,
\emph{Univalent and starlike generalized hypergeometric functions,} Canad. J. Math., \textbf{39}(1987), 1057-1077.

\bibitem{5}   St. Ruscheweyh,
\emph{New criteria for univalent functions,} Proc. Amer. Math. Soc., \textbf{49}(1975), 109-115.

\bibitem{6}  G.  \c{S}. S\v{a}l\v{a}gean,
\emph{Subclasses of univalent functions,} Lecture Notes in Math. (Springer Verlag), \textbf{1013}(1983), 362-372.

\bibitem{7} P. Sharma, R. K. Raina, G. \c{S}. S\v{a}l\v{a}gean, 
\emph{Some Geometric Properties of Analytic Functions Involving a new Fractional Operator, } Mediterr. J. Math., \textbf{13}(2016), 4591-4605.

\bibitem{8} E. Szatmari,
\emph{On a class of analytic functions defined by a fractional operator,} submitted

\end{thebibliography}

\end{document}
