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\title[Application of Ruscheweyh $q$-differential operator]{Application of
Ruscheweyh $q$-differential operator to analytic functions of reciprocal
order}
\author[S. Mahmood]{ Shahid Mahmood$^{A,\ast }$}
\address{ $^{A}$Department of Mechanical Engineering, Sarhad University of
Science \& I. T Landi Akhun Ahmad, Hayatabad Link. Ring Road, Peshawar,
Pakistan}
\email{shahidmahmood757@gmail.com (S. Mahmood)}
\author[S. Mustafa]{Saima Mustafa$^{B}$}
\address{$^{B}$Deptt. of Statistics \& Mathematics, PMAS-Arid Agriculture
University, Rawalpindi}
\email{saimamustafa28@gmail.com (S. Mustafa)}
\author[I. Khan]{Imran Khan$^{A}$}
\email{ikhanqau1@gmail.com (I. Khan)}
\keywords{Analytic functions, Subordination, Functions with positive real
part, Ruscheweyh $q$-differential operator, reciprocal order.}
\date{Received: November 30, 2018\\
\indent$^{\ast }$ Corresponding author\\
2010\textit{\ Mathematics Subject Classification. }30C45, 30C50.}

\begin{abstract}
The core object of this paper is to define and study new class of analytic
function using Ruscheweyh $q$-differential operator. We also investigate a
number of useful properties such as inclusion relation, coefficient
estimates, subordination result,for this newly subclass of analytic
functions.
\end{abstract}
\maketitle
\section{Introduction}

\noindent Quantum calculus ($q$-calculus) is simply the study of classical
calculus without the notion of limits. The study of $q$-calculus attracted
the researcher due to its applications in various branches of mathematics
and physics, see detail \cite{Gupta12}. Jackson \cite{Jackson1,Jackson2} was
the first to give some application of $q$-calculus and introduced the $q$%
-analogue of derivative and integral. Later on Aral and Gupta \cite%
{Gupta1,Gupta2,Gupta3} defined the $q$-Baskakov Durrmeyer operator by using $%
q$-beta function while the author's in \cite{6,7,8} discussed the $q$%
-generalization of complex operators known as $q$-Picard and $q$%
-Gauss-Weierstrass singular integral operators. Recently, Kanas and R\u{a}%
ducanu \cite{Kanasq} defined $q$-analogue of Ruscheweyh differential
operator using the concepts of convolution and then studied some of its
properties. The application of this differential operator was further
studied by Mohammed and Darus \cite{Maslina2} and Mahmood and Sok\'{o}{\l }
\cite{Shahid}. The aim of the current paper is to define a new class of
analytic functions of reciprocal order involving $q$-differetial operator.

\noindent Let $\mathcal{A}$ be the class of functions having the form%
\begin{equation}
f(z)=z+\sum\limits_{n=2}^{\infty }a_{n}z^{n},  \label{taylor series}
\end{equation}%
which are analytic in the open unit disk $\mathbb{U}=\left\{ z\in \mathbb{C}%
:\left\vert z\right\vert <1\right\} $. Let $\mathcal{M}(\alpha )$ denote a
subclass of $\mathcal{A}$ consisting of functions which satisfy the
inequality%
\begin{equation*}
\mathfrak{Re}\frac{zf^{\prime }(z)}{f(z)}<\alpha \ \ \left( z\in \mathbb{U}%
\right) ,
\end{equation*}%
for some $\alpha \left( \alpha >1\right) $. And let $\mathcal{N}(\alpha )$
be the subclass of $\mathcal{A}$ consisting of functions $f$ which satisfy
the inequality:
\begin{equation*}
\mathfrak{Re}\frac{\left( zf^{\prime }(z)\right) ^{\prime }}{f^{\prime }(z)}%
<\alpha \ \ \ (z\in \mathbb{U}),
\end{equation*}%
for some $\alpha \left( \alpha >1\right) $. These classes were studied by
Owa et al. \cite{owa1,owa2}. Shams et al. \cite{Shams} have introduced the $%
k $-uniformly starlike $\mathcal{SD}\left( k,\alpha \right) $ and $k$%
-uniformly convex $\mathcal{CD}\left( k,\alpha \right) $ of order $\alpha ,$
for some $k\left( k\geq 0\right) $ and $\alpha \left( 0\leq \alpha <1\right)
$. Using these ideas in above defined classes, Junichi et al. \cite{junchi}
introduced the following classes.

\begin{definition}
Let $f\in \mathcal{A}$. Then $f$ is said to be in class $\mathcal{MD}\left(
k,\alpha \right) $ if it satisfies
\begin{equation*}
\mathfrak{Re}\frac{zf^{\prime }(z)}{f(z)}<k\left \vert \frac{zf^{\prime }(z)%
}{f(z)}-1\right \vert +\alpha \ \ \left( z\in \mathbb{U}\right) ,
\end{equation*}%
for some $\alpha \left( \alpha >1\right) $ and $k\left( k\leq 0\right) $.
\end{definition}

\begin{definition}
An analytic function $f$ of the form \eqref{taylor series} belongs to the
class $\mathcal{ND}\left( k,\alpha \right) $, if and only if
\begin{equation*}
\mathfrak{Re}\frac{\left( zf^{\prime }(z)\right) ^{\prime }}{f^{\prime }(z)}%
<k\left \vert \frac{\left( zf^{\prime }(z)\right) ^{\prime }}{f^{\prime }(z)}%
-1\right \vert +\alpha \ \ \left( z\in \mathbb{U}\right) ,
\end{equation*}%
for some $\alpha \left( \alpha >1\right) $ and $k\left( k\leq 0\right) $.
\end{definition}

\noindent If $f$ and $g$ are analytic in $\mathbb{U},$ we say that $f$ is
subordinate to $g,$ written as $f\prec g$ or $f(z)\prec g(z),$ if there
exists a Schwarz function $w,$ which is analytic in $\mathbb{U}$ with $%
w\left( 0\right) =0$ and $\left\vert w(z)\right\vert <1$ such that $\
f(z)=g(w(z))$. Furthermore, if the function $g(z)$ is univalent in $\mathbb{U%
},$ then we have the following equivalence holds, see \cite{Goodman,milr}.
\begin{equation*}
\begin{tabular}{llllll}
$f(z)\prec g(z)$ & $(z\in \mathbb{U})$ & $\Longleftrightarrow $ & $f(0)=g(0)$
& $\mathrm{and}$ & $\ f(\mathbb{U})\subset g(\mathbb{U}\mathcal{)}.$%
\end{tabular}%
\end{equation*}%
For two analytic functions
\begin{equation*}
\begin{tabular}{lll}
$f(z)=\sum\limits_{n=1}^{\infty }a_{n}z^{n}$ & $g(z)=\sum\limits_{n=1}^{%
\infty }b_{n}z^{n}$ & $\left( z\in \mathbb{U}\right) ,$%
\end{tabular}%
\end{equation*}%
For $t\in \mathbb{R}$ and $q>0$, $q\neq 1$, the number $\left[ t,q\right] $
is defined in \cite{Shahid} as
\begin{equation*}
\left[ t,q\right] =\frac{1-q^{t}}{1-q},\ \ \left[ 0,q\right] =0.
\end{equation*}%
For any non-negative integer $n$ the $q$-number shift factorial is defined
by
\begin{equation*}
\left[ n,q\right] !=\left[ 1,q\right] \left[ 2,q\right] \left[ 3,q\right]
\cdots \left[ n,q\right] ,\text{ \ }\left( \left[ 0,q\right] !=1\right) .
\end{equation*}%
We have $\lim\limits_{q\rightarrow 1}\left[ n,q\right] =n$. Throughout in
this paper we will assume $q$ to be fixed number between $0$ and $1$.

\noindent The $q$-derivative operator or $q$-difference operator for $f\in
\mathcal{A}$ is defined as
\begin{equation*}
\partial _{q}f(z)=\frac{f\left( qz\right) -f(z)}{z\left( q-1\right) },\text{
}z\in \mathbb{U}.
\end{equation*}%
It can easily be seen that for $n\in \mathbb{N}:=\left\{ 1,2,3,\ldots
\right\} $ and $z\in \mathbb{U}$
\begin{equation*}
\partial _{q}z^{n}=\left[ n,q\right] z^{n-1},\ \ \partial _{q}\left\{
\sum_{n=1}^{\infty }a_{n}z^{n}\right\} =\sum_{n=1}^{\infty }\left[ n,q\right]
a_{n}z^{n-1}.
\end{equation*}%
The $q$-generalized Pochhammer symbol for $t\in \mathbb{R}$ and $n\in
\mathbb{N}$ is defined as
\begin{equation*}
\left[ t,q\right] _{n}=\left[ t,q\right] \left[ t+1,q\right] \left[ t+2,q%
\right] \cdots \left[ t+n-1,q\right] ,
\end{equation*}%
and for $t>0$, let $q$-gamma function is defined as
\begin{equation*}
\Gamma _{q}\left( t+1\right) =\left[ t,q\right] \Gamma _{q}\left( t\right)
\text{ and }\Gamma _{q}\left( 1\right) =1.
\end{equation*}

%DEFINITIONN ==========================================================

\begin{definition}
\cite{Shahid} For a function $f(z)\in \mathcal{A},$ the Ruscheweyh $q$%
-differential operator is defined as
\begin{equation}
\mathfrak{D}_{q}^{\mu }f(z)=\phi \left( q,\mu +1;z\right) \ast
f(z)=z+\sum\limits_{n=2}^{\infty }\Phi _{n-1}a_{n}z^{n},~\text{\ \ }\left(
z\in \mathbb{U}\text{ and }\mu >-1\right) ,  \label{L-series}
\end{equation}%
where
\begin{equation}
\phi \left( q,\mu +1;z\right) =z+\sum\limits_{n=2}^{\infty }\Phi _{n-1}z^{n},
\label{fi(a,c)}
\end{equation}%
and
\begin{equation}
\Phi _{n-1}=\frac{\Gamma _{q}\left( \mu +n\right) }{\left[ n-1,q\right]
!\Gamma _{q}\left( \mu +1\right) }=\frac{\left[ \mu +1,q\right] _{n-1}}{%
\left[ n-1,q\right] !}.  \label{psi}
\end{equation}
\end{definition}

%==========================================================================

\noindent From (\ref{L-series}), it can be seen that
\begin{equation*}
L_{q}^{0}f(z)=f(z)\text{ and }L_{q}^{1}f(z)=z\partial _{q}f(z),
\end{equation*}%
and
\begin{equation*}
L_{q}^{m}f(z)=\frac{z\partial _{q}^{m}\left( z^{m-1}f(z)\right) }{\left[ m,q%
\right] !},~\text{\ \ }\left( m\in \mathbb{N}\right) .
\end{equation*}%
\begin{equation*}
\lim\limits_{q\rightarrow 1^{-}}\phi \left( q,\mu +1;z\right) =\frac{z}{%
\left( 1-z\right) ^{\mu +1}},
\end{equation*}%
and
\begin{equation*}
\lim\limits_{q\rightarrow 1^{-}}\mathfrak{D}_{q}^{\mu }f(z)=f(z)\ast \frac{z%
}{\left( 1-z\right) ^{\mu +1}}.
\end{equation*}%
This shows that in case of $q\rightarrow 1^{-},$ the Ruscheweyh $q$%
-differential operator reduces to the Ruscheweyh differential operator $%
D^{\delta }\left( f(z)\right) $ (see \cite{Ruscheweyh}). From (\ref{L-series}%
) the following identity can easily be derived.
\begin{equation}
z\partial \mathfrak{D}_{q}^{\mu }f(z)=\left( 1+\frac{\left[ \mu ,q\right] }{%
q^{\mu }}\right) \mathfrak{D}_{q}^{\mu }f(z)-\frac{\left[ \mu ,q\right] }{%
q^{\mu }}\mathfrak{D}_{q}^{\mu }f(z).  \label{id}
\end{equation}%
If $q\rightarrow 1^{-}$, then
\begin{equation*}
z\left( \mathfrak{D}_{q}^{\mu }f(z)\right) ^{\prime }=\left( 1+\mu \right)
\mathfrak{D}_{q}^{\mu }f(z)-\mu \mathfrak{D}_{q}^{\mu }f(z).
\end{equation*}

\noindent Now using the Ruscheweyh $q$-differential operator, we define the
following class.

\begin{definition}
Let $f\in \mathcal{A}$. Then $f$ is in the class $\mathcal{KD}_{q}\left(
k,\alpha ,\gamma \right) $ if
\begin{equation*}
\mathfrak{Re}\left\{ 1+\frac{1}{\gamma }\left( \frac{z\partial _{q}\mathfrak{%
D}_{q}^{\mu }f(z)}{\mathfrak{D}_{q}^{\mu }f(z)}-1\right) \right\}
<k\left\vert \frac{1}{\gamma }\left( \frac{z\partial _{q}\mathfrak{D}%
_{q}^{\mu }f(z)}{\mathfrak{D}_{q}^{\mu }f(z)}-1\right) \right\vert +\alpha ,
\end{equation*}%
for some $k\left( k\leq 0\right) $, $\alpha \left( \alpha >1\right) $ and
for some $\gamma \in \mathbb{C}\setminus \{0\}$.
\end{definition}

\noindent We note that $\mathcal{LD}_{2}^{0}\left( 1,1,\alpha \right) =%
\mathcal{M}(\alpha )$ and $\mathcal{LD}_{1}^{0}\left( 1,1,\alpha \right) =%
\mathcal{N}(\alpha ),$ the classes introduced by Owa et al. \cite{owa1,owa2}%
. When we take $\gamma =1,2$, $c=1$, and $a=1$ the class $\mathcal{KD}%
_{q}\left( k,\alpha ,\gamma \right) $ reduces to the classes $\mathcal{MD}%
\left( k,\alpha \right) $ and $\mathcal{ND}\left( k,\alpha \right) $ (see
\cite{junchi}). For $1<\alpha <4/3$ the classes $\mathcal{M}(\alpha )$ and $%
\mathcal{N}(\alpha )$ were investigated by Uralegaddi et al. \cite%
{Uralegaddi}.


\section{Preliminary Results}

\begin{lemma}
\cite{shahid1}For a positive integer $t$, we have
\begin{equation}
\sigma \sum\limits_{j=1}^{t}\frac{(\sigma )_{j-1}}{(j-1)!}=\frac{(\sigma
)_{t}}{(t-1)!}.  \label{kkk}
\end{equation}
\end{lemma}

%----------------------------------------------------------------------------
%----------------------------------------------------------------------------

\begin{proof}
Consider
\begin{eqnarray*}
&&\sigma \sum \limits_{j=1}^{t}\frac{(\sigma )_{j-1}}{(j-1)!} \\
&=&\sigma \left( 1+\frac{\sigma }{1}+\frac{(\sigma )_{2}}{2!}+\frac{(\sigma
)_{3}}{3!}+\frac{(\sigma )_{4}}{4!}+\cdots +\frac{(\sigma )_{t-1}}{(t-1)!}
\right) \\
&=&\sigma (1+\sigma )\left( 1+\frac{\sigma }{2}+\frac{\sigma (\sigma +2)}{
2\times 3}+\cdots +\frac{\sigma (\sigma +2)\cdots (\sigma +t-2)}{2\times
\cdots \times (t-1)}\right) \\
&=&\sigma (1+\sigma )\frac{(\sigma +2)}{2}\left( 1+\frac{\sigma }{3}+\cdots
+ \frac{\sigma (\sigma +3)\cdots (\sigma +t-2)}{3\times 4\times \cdots
\times (t-1)}\right) \\
&=&\sigma (1+\sigma )\frac{(\sigma +2)}{2}\frac{(\sigma +3)}{3}\left( 1+
\frac{\sigma }{4}+\cdots +\frac{\sigma (\sigma +4)\cdots (\sigma +t-2)}{
4\times \cdots \times (t-1)}\right) \\
&=&\sigma (1+\sigma )\frac{(\sigma +2)}{2}\frac{(\sigma +3)}{3}\frac{(\sigma
+4)}{4}\left( 1+\frac{\sigma }{5}+\cdots +\frac{\sigma \cdots (\sigma +t-2)}{
5\times 6\times \cdots \times (t-1)}\right) \\
&=&\sigma (1+\sigma )\frac{(\sigma +2)}{2}\frac{(\sigma +3)}{3}\frac{(\sigma
+4)}{4}\cdots \left( 1+\frac{\sigma }{t-1}\right) \\
&=&\sigma (1+\sigma )\frac{(\sigma +2)}{2}\frac{(\sigma +3)}{3}\frac{(\sigma
+4)}{4}\cdots \left( \frac{\sigma +(t-1)}{t-1}\right) \\
&=&\frac{(\sigma )_{t}}{(t-1)!}.
\end{eqnarray*}
\end{proof}

%----------------------------------------------------------------------------
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\section{Main Results}

%----------------------------------------------------------------------------
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With the help of the definition of $\mathcal{KD}_{q}\left( k,\alpha ,\gamma
\right) ,$ we prove the following results.%
%----------------------------------------------------------------------------
%--3.1--------------------------------------------------------------------------

\begin{theorem}
\label{t3.1} If $f(z)\in \mathcal{KD}_{q}\left( k,\alpha ,\gamma \right) ,$
then
\begin{equation*}
f(z)\in \mathcal{KD}_{q}\left( 0,\frac{\alpha -k}{1-k},\gamma \right) .
\end{equation*}%
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\begin{proof}
Because $k\leq 0$, we have
\begin{eqnarray*}
\mathfrak{Re}\left\{ 1+\frac{1}{\gamma }\left( \frac{z\partial _{q}\mathfrak{%
D}_{q}^{\mu }f(z)}{\mathfrak{D}_{q}^{\mu }f(z)}-1\right) \right\}
&<&k\left\vert \frac{1}{\gamma }\left( \frac{z\partial _{q}\mathfrak{D}%
_{q}^{\mu }f(z)}{\mathfrak{D}_{q}^{\mu }f(z)}-1\right) \right\vert +\alpha ,
\\
&\leq &k\mathfrak{Re}\left( \frac{1}{\gamma }\left( \frac{z\partial _{q}%
\mathfrak{D}_{q}^{\mu }f(z)}{\mathfrak{D}_{q}^{\mu }f(z)}-1\right) \right)
+\alpha -k,
\end{eqnarray*}%
which implies that
\begin{equation*}
\left( 1-k\right) \mathfrak{Re}\frac{1}{\gamma }\left( \frac{z\partial _{q}%
\mathfrak{D}_{q}^{\mu }f(z)}{\mathfrak{D}_{q}^{\mu }f(z)}-1\right) <\alpha
-k.
\end{equation*}%
After simplification, we obtain
\begin{equation}
\mathfrak{Re}\left[ 1+\frac{1}{\gamma }\left( \frac{z\partial _{q}\mathfrak{D%
}_{q}^{\mu }f(z)}{\mathfrak{D}_{q}^{\mu }f(z)}-1\right) \right] <\frac{%
\alpha -k}{1-k},\left( k\leq 0,\text{ }\alpha >1\text{ and }\right) .
\label{1t3.1}
\end{equation}%
This completes the proof.
\end{proof}
\end{theorem}

%----------------------------------------------------------------------------
%---3.2-------------------------------------------------------------------------

\begin{theorem}
\label{coeff bound} If $f(z)\in \mathcal{KD}_{q}\left( k,\alpha ,\gamma
\right) $ and if $f(z)$ has the form \eqref{taylor series}, then
\begin{equation}
\left\vert a_{n}\right\vert \leq \frac{(\sigma )_{n-1}}{(n-1)!\Phi _{n-1}},
\end{equation}%
where
\begin{equation}
\sigma =\frac{2|\gamma |(\alpha -1)}{q(1-k)}.  \label{sigma}
\end{equation}
\end{theorem}

%----------------------------------------------------------------------------
%----------------------------------------------------------------------------

\begin{proof}
Let us define a function
\begin{equation}
p(z)=\frac{\left( \alpha -k\right) -\left( 1-k\right) \left[ 1+\frac{1}{%
\gamma }\left( \frac{z\partial _{q}\mathfrak{D}_{q}^{\mu }f(z)}{\mathfrak{D}%
_{q}^{\mu }f(z)}-1\right) \right] }{\alpha -1}.  \label{p(z)}
\end{equation}%
Then $p(z)$ is analytic in $\mathbb{U},$ $p(0)=1$ and $\mathfrak{Re}\left\{
p(z)\right\} >0$ for $z\in \mathbb{U}$. We can write
\begin{equation}
\left[ 1+\frac{1}{\gamma }\left( \frac{z\partial _{q}\mathfrak{D}_{q}^{\mu
}f(z)}{\mathfrak{D}_{q}^{\mu }f(z)}-1\right) \right] =\frac{(\alpha
-k)-(\alpha -1)p(z)}{1-k}  \label{zf}
\end{equation}%
If we take $p(z)=1+\sum\limits_{n=1}^{\infty }p_{n}z^{n}$, then \eqref{zf}
can be written as
\begin{equation*}
z\partial _{q}\mathfrak{D}_{q}^{\mu }f(z)-\mathfrak{D}_{q}^{\mu }f(z)=-\frac{%
\gamma \left( \alpha -1\right) }{1-k}\left( \mathfrak{D}_{q}^{\mu
}f(z)\right) \left( \sum_{n=1}^{\infty }p_{n}z^{n}\right) .
\end{equation*}%
this implies that
\begin{equation*}
\left[ \sum\limits_{n=2}^{\infty }q\left[ n-1\right] \Phi _{n-1}a_{n}z^{n}%
\right] =-\frac{\gamma (\alpha -1)}{1-k}\left( \sum\limits_{n=1}^{\infty
}\Phi _{n-1}a_{n}z^{n}\right) \left( \sum_{n=1}^{\infty }p_{n}z^{n}\right) .
\end{equation*}%
Using Cauchy product $\left( \sum\limits_{n=1}^{\infty }x_{n}\right) \cdot
\left( \sum\limits_{n=1}^{\infty }y_{n}\right) =\sum\limits_{j=1}^{\infty
}\sum\limits_{k=1}^{j}x_{k}y_{k-j}$, we obtain
\begin{equation*}
q\left[ n-1\right] \Phi _{n-1}a_{n}z^{n}=-\frac{\gamma (\alpha -1)}{1-k}%
\sum\limits_{n=2}^{\infty }\left( \sum\limits_{j=1}^{n-1}\Phi
_{j-1}a_{j}p_{n-j}\right) z^{n}.
\end{equation*}%
Comparing the coefficients of $nth$ term on both sides, we obtain
\begin{equation*}
a_{n}=\frac{-\gamma (\alpha -1)}{q\left[ n-1\right] \Phi _{n-1}\left(
1-k\right) }\sum\limits_{j=1}^{n-1}\Phi _{j-1}a_{j}p_{n-j}.
\end{equation*}%
By taking absolute value and applying triangle inequality, we get
\begin{equation*}
\left\vert a_{n}\right\vert \leq \frac{\left\vert \gamma \right\vert (\alpha
-1)}{q\left[ n-1\right] \Phi _{n-1}\left( 1-k\right) }\sum%
\limits_{j=1}^{n-1}\Phi _{j-1}\left\vert a_{j}\right\vert \left\vert
p_{n-j}\right\vert .
\end{equation*}%
Applying the coefficient estimates $\left\vert p_{n}\right\vert \leq 2$ $%
(n\geq 1)$ for Caratheodory functions \cite{Goodman}, we obtain
\begin{eqnarray}
|a_{n}| &\leq &\frac{2\left\vert \gamma \right\vert (\alpha -1)}{q\left[ n-1%
\right] \Phi _{n-1}(1-k)}\sum\limits_{j=1}^{n-1}\Phi _{j-1}\left\vert
a_{j}\right\vert  \notag  \label{coeff} \\
&=&\frac{\sigma }{\left[ n-1\right] \Phi _{n-1}}\sum\limits_{j=1}^{n-1}\psi
_{j-1}\left\vert a_{j}\right\vert ,
\end{eqnarray}%
where $\sigma =2|\gamma |(\alpha -1)/q(1-k)$. To prove \eqref{coeff bound}
we apply mathematical induction. So for $n=2$, we have from \eqref{coeff}
\begin{equation}
\left\vert a_{2}\right\vert \leq \frac{\sigma }{\Phi _{1}}=\frac{\left(
\sigma \right) _{2-1}}{\left[ 2-1\right] !\Phi _{2-1}},  \label{a2}
\end{equation}%
which shows that \eqref{coeff bound} holds for $n=2$. For $n=3$, we have
from \eqref{coeff}
\begin{equation*}
\left\vert a_{3}\right\vert \leq \frac{\sigma }{\left[ 3-1\right] \Phi _{3-1}%
}\left\{ 1+\Phi _{1}\left\vert a_{2}\right\vert \right\} ,
\end{equation*}%
using \eqref{a2}, we have
\begin{equation*}
\left\vert a_{3}\right\vert \leq \frac{\sigma }{\left[ 2\right] \Phi _{2}}%
\left( 1+\sigma \right) =\frac{\left( \sigma \right) _{3-1}}{\left[ 3-1%
\right] \Phi _{3-1}},
\end{equation*}%
which shows that \eqref{coeff bound} holds for $n=3$. Let us assume that %
\eqref{coeff bound} is true for $n\leq t,$ that is,
\begin{equation}
\left\vert a_{t}\right\vert \leq \frac{\left( \sigma \right) _{t-1}}{\left[
t-1\right] !\Phi _{t-1}}\ \ \ j=1,2,\ldots ,t.  \label{coeff 3}
\end{equation}%
Using \eqref{coeff} and \eqref{coeff 3}, we have
\begin{eqnarray*}
\left\vert a_{t+1}\right\vert &\leq &\frac{\sigma }{t\Phi _{t}}%
\sum\limits_{j=1}^{t}\Phi _{j-1}\left\vert a_{j}\right\vert \\
&\leq &\frac{\sigma }{t\Phi _{t}}\sum\limits_{j=1}^{t}\psi _{j-1}\frac{%
\left( \sigma \right) _{j-1}}{\left[ j-1\right] !\Phi _{j-1}} \\
&=&\frac{\sigma }{t\Phi _{t}}\sum\limits_{j=1}^{t}\frac{\left( \sigma
\right) _{j-1}}{\left[ j-1\right] !}.
\end{eqnarray*}%
Applying \eqref{kkk}, we have
\begin{eqnarray*}
\left\vert a_{t+1}\right\vert &\leq &\frac{1}{t\Phi _{t}}\frac{(\sigma )_{t}%
}{\left[ t-1\right] !} \\
&=&\frac{1}{\Phi _{t}}\frac{(\sigma )_{t}}{\left[ t\right] !}.
\end{eqnarray*}%
Consequently, using mathematical induction, we have proved that
\eqref{coeff
bound} holds true for all $n$, $n\geq 2$. This completes the proof.
\end{proof}

%----------------------------------------------------------------------------
%---3.3-------------------------------------------------------------------------

\begin{theorem}
\label{subordinate}If a function $f\in \mathcal{KD}_{q}\left( k,\alpha
,\gamma \right) $, then
\begin{equation}
\frac{z\partial _{q}\mathfrak{D}_{q}^{\mu }f(z)}{\mathfrak{D}_{q}^{\mu }f(z)}%
\prec 1+2\left( \alpha _{1}-1\right) -\frac{2\left( \alpha _{1}-1\right) }{%
1-z}\ \ (z\in \mathbb{U}),  \label{sub inq}
\end{equation}%
\begin{equation}
\alpha _{1}=\frac{\alpha -k}{1-k}.  \label{B1}
\end{equation}
\end{theorem}

%----------------------------------------------------------------------------
%----------------------------------------------------------------------------

\begin{proof}
If $f(z)\in \mathcal{KD}_{q}\left( k,\alpha ,\gamma \right) $, then by %
\eqref{1t3.1}
\begin{equation}
\mathfrak{Re}\left\{ 1+\frac{1}{\gamma }\left( \frac{z\partial _{q}\mathfrak{%
D}_{q}^{\mu }f(z)}{\mathfrak{D}_{q}^{\mu }f(z)}-1\right) \right\} <\alpha
_{1}.  \label{1t3.7}
\end{equation}%
Then there exists a Schwarz function $w(z)$ such that
\begin{equation}
\frac{\alpha _{1}-\left\{ 1+\frac{1}{\gamma }\left( \frac{z\partial _{q}%
\mathfrak{D}_{q}^{\mu }f(z)}{\mathfrak{D}_{q}^{\mu }f(z)}-1\right) \right\}
}{\alpha _{1}-1}=\frac{1+w(z)}{1-w(z)},  \label{sub inq 2}
\end{equation}%
and
\begin{equation*}
\mathfrak{Re}\left\{ \frac{1+w(z)}{1-w(z)}\right\} >0,\ \ (z\in \mathbb{U}).
\end{equation*}%
Therefore, from \eqref{sub inq 2}, we obtain%
\begin{equation*}
\frac{z\partial _{q}\mathfrak{D}_{q}^{\mu }f(z)}{\mathfrak{D}_{q}^{\mu }f(z)}%
=1+\gamma \left( \alpha _{1}-1\right) \left( 1-\frac{1+w(z)}{1-w(z)}\right) .
\end{equation*}%
This gives
\begin{equation*}
\frac{z\partial _{q}\mathfrak{D}_{q}^{\mu }f(z)}{\mathfrak{D}_{q}^{\mu }f(z)}%
=1+2\gamma \left( \alpha _{1}-1\right) -\frac{2\gamma \left( \alpha
_{1}-1\right) }{1-w(z)}
\end{equation*}%
and hence
\begin{equation*}
\frac{z\partial _{q}\mathfrak{D}_{q}^{\mu }f(z)}{\mathfrak{D}_{q}^{\mu }f(z)}%
\prec 1+2\gamma \left( \alpha _{1}-1\right) -\frac{2\gamma \left( \alpha
_{1}-1\right) }{1-z}\ \ (z\in \mathbb{U}).
\end{equation*}%
which was required in \eqref{sub inq}.
\end{proof}

%----------------------------------------------------------------------------
%----3.8------------------------------------------------------------------------

\begin{theorem}
If function $f\in \mathcal{KD}_{q}\left( k,\alpha ,\gamma \right) $, then we
have
\begin{equation}
\frac{1-\left[ 1+2\gamma (\alpha _{1}-1)\right] r}{1-r}\leq \mathfrak{Re}%
\left\{ \frac{z\partial _{q}\mathfrak{D}_{q}^{\mu }f(z)}{\mathfrak{D}%
_{q}^{\mu }f(z)}\right\} \leq \frac{1+\left[ 1+2\gamma (\alpha _{1}-1)\right]
r}{1+r},  \label{inequality}
\end{equation}%
for $\left\vert z\right\vert =r<1$ and $\alpha _{1}$ is defined by \eqref{B1}%
.
\end{theorem}

%----------------------------------------------------------------------------
%----------------------------------------------------------------------------

\begin{proof}
By the virtue of Theorem \eqref{subordinate}, let us take the function $\phi
(z)$ defined by
\begin{equation*}
\phi (z)=1+2\gamma \left( \alpha _{1}-1\right) -\frac{2\gamma (\alpha _{1}-1)%
}{1-z}\ \ \left( z\in \mathbb{U}\right) .
\end{equation*}%
Letting $z=re^{i\theta }(0\leq r<1),$ we see that
\begin{equation*}
\mathfrak{Re}\phi (z)=1+2\gamma \left( \alpha _{1}-1\right) +\frac{2\gamma
\left( 1-\alpha _{1}\right) \left( 1-r\cos \theta \right) }{1+r^{2}-2r\cos
\theta }.
\end{equation*}%
Let us define
\begin{equation*}
\psi (t)=\frac{1-rt}{1+r^{2}-2rt}\ \ \left( t=\cos \theta \right) .
\end{equation*}%
Since $\psi ^{\prime }(t)=\frac{r\left( 1-r^{2}\right) }{\left(
1+r^{2}-2rt\right) ^{2}}\geq 0,$ because $r<1$. Therefore we get
\begin{equation*}
1+2\gamma \left( \alpha _{1}-1\right) -\frac{2\gamma \left( \alpha
_{1}-1\right) }{1-r}\leq \mathfrak{Re}\phi (z)\leq 1+2\gamma \left( \alpha
_{1}-1\right) -\frac{2\gamma \left( \alpha _{1}-1\right) }{1+r}.
\end{equation*}%
After simplification, we have
\begin{equation*}
\frac{1-\left[ 1+2\gamma \left( \alpha _{1}-1\right) \right] r}{1-r}\leq
\mathfrak{Re}\phi (z)\leq \frac{1+\left[ 1+2\gamma \left( \alpha
_{1}-1\right) )\right] r}{1+r}.
\end{equation*}%
Since we note that $\frac{z\partial _{q}\mathfrak{D}_{q}^{\mu }f(z)}{%
\mathfrak{D}_{q}^{\mu }f(z)}\prec \phi (z),\left( z\in \mathbb{U}\right) $
by Theorem \ref{subordinate} and $\phi (z)$ is analytic in $\mathbb{U},$ we
proved the inequality (\ref{inequality}).
\end{proof}

%----------------------------------------------------------------------------
%----------------------------------------------------------------------------
%----3.14------------------------------------------------------------------------

\begin{theorem}
\label{t14} If $f\in \mathcal{A}$ satisfies
\begin{equation}
\left\vert \frac{z\partial _{q}\mathfrak{D}_{q}^{\mu }f(z)}{\mathfrak{D}%
_{q}^{\mu }f(z)}-1\right\vert <\frac{(\alpha -1)|\gamma |}{(1-k)}\ \ \ z\in
\mathbb{U},  \label{1t14}
\end{equation}%
for some $k\left( k\leq 0\right) $, $\alpha \left( \alpha >1\right) $ and $%
\gamma \in \mathbb{C}\setminus \{0\}$. Then $f\in \mathcal{KD}_{q}(k,\alpha
,\gamma )$.
\end{theorem}

%----------------------------------------------------------------------------
%----------------------------------------------------------------------------

\begin{proof}
\begin{eqnarray*}
&~&\left\vert \frac{z\partial _{q}\mathfrak{D}_{q}^{\mu }f(z)}{\mathfrak{D}%
_{q}^{\mu }f(z)}-1\right\vert <\frac{(\alpha -1)|\gamma |}{(1-k)} \\
&\Rightarrow &\left\vert \frac{1}{\gamma }\left( \frac{z\partial _{q}%
\mathfrak{D}_{q}^{\mu }f(z)}{\mathfrak{D}_{q}^{\mu }f(z)}-1\right)
\right\vert <\frac{\alpha -1}{1-k} \\
&\Rightarrow &(1-k)\left\vert \frac{1}{\gamma }\left( \frac{z\partial _{q}%
\mathfrak{D}_{q}^{\mu }f(z)}{\mathfrak{D}_{q}^{\mu }f(z)}-1\right)
\right\vert +1<\alpha \\
&\Rightarrow &\left\vert \frac{1}{\gamma }\left( \frac{z\partial _{q}%
\mathfrak{D}_{q}^{\mu }f(z)}{\mathfrak{D}_{q}^{\mu }f(z)}-1\right)
\right\vert +1<k\left\vert \frac{1}{\gamma }\left( \frac{z\partial _{q}%
\mathfrak{D}_{q}^{\mu }f(z)}{\mathfrak{D}_{q}^{\mu }f(z)}-1\right)
\right\vert +\alpha \\
&\Rightarrow &\mathfrak{Re}\left\{ 1+\frac{1}{\gamma }\left( \frac{z\partial
_{q}\mathfrak{D}_{q}^{\mu }f(z)}{\mathfrak{D}_{q}^{\mu }f(z)}-1\right)
\right\} +1<k\left\vert \frac{1}{\gamma }\left( \frac{z\partial _{q}%
\mathfrak{D}_{q}^{\mu }f(z)}{\mathfrak{D}_{q}^{\mu }f(z)}-1\right)
\right\vert +\alpha \\
&\Rightarrow &f\in \mathcal{LD}_{b}^{k}(a,c,\beta )
\end{eqnarray*}
\end{proof}

\begin{corollary}
\label{c15} Let $f\in \mathcal{A}$ be of the form \eqref{taylor
series} and satisfies
\begin{equation}
\left\vert \frac{\sum_{n=2}^{\infty }\left[ n-1\right] \Phi
_{n-1}a_{n}z^{n-1}}{1+\sum_{n=2}^{\infty }\Phi _{n-1}a_{n}z^{n-1}}%
\right\vert <\frac{(\alpha -1)|\gamma |}{q(1-k)}\ \ \ z\in \mathbb{U},
\label{1c15}
\end{equation}%
for some $k\left( k\leq 0\right) $, $\beta \left( \beta >1\right) $ and for
some $b\in \mathbb{C}\setminus \{0\}$. Then $f\in \mathcal{KD}_{q}(k,\alpha
,\gamma )$..
\end{corollary}

%----------------------------------------------------------------------------
%----------------------------------------------------------------------------

\begin{proof}
We have
\begin{equation*}
\mathfrak{D}_{q}^{\mu }f(z)=z+\sum\limits_{n=2}^{\infty }\Phi
_{n-1}a_{n}z^{n}
\end{equation*}%
and by \eqref{id}
\begin{equation*}
z\partial \mathfrak{D}_{q}^{\mu }f(z)=z+\sum\limits_{n=2}^{\infty }\left[ n%
\right] \Phi _{n-1}a_{n}z^{n}.
\end{equation*}%
Therefore, \eqref{1t14} follows immediately \eqref{1c15}.
\end{proof}

%----t16------------------------------------------------------------------------

\begin{theorem}
\label{t16} Let $f\in \mathcal{A}$ be of the form \eqref{taylor
series} and satisfies
\begin{equation}
\sum_{n=2}^{\infty }\left( \left[ n-1\right] +y\right) |\Phi
_{n-1}||a_{n}|<y\ \ \ z\in \mathbb{U},  \label{1t16}
\end{equation}%
\newline
for some $k\left( k\leq 0\right) $, $\beta \left( \beta >1\right) $ and for
some $b\in \mathbb{C}\setminus \{0\}$ and where
\begin{equation*}
y=\frac{(\alpha -1)|\gamma |}{q(1-k)}>0.
\end{equation*}%
Then $f\in \mathcal{KD}_{q}(k,\alpha ,\gamma )$.
\end{theorem}

%----------------------------------------------------------------------------
%----------------------------------------------------------------------------

\begin{proof}
We have
\begin{eqnarray}
&~&\sum_{n=2}^{\infty }\left( \left[ n-1\right] +y\right) |\Phi
_{n-1}||a_{n}|<y  \notag  \label{2t16} \\
&\Rightarrow &\sum_{n=2}^{\infty }\left( \left[ n-1\right] +y\right) |\Phi
_{n-1}||a_{n}|<y-y\sum_{n=2}^{\infty }|\Phi _{n-1}||a_{n}|  \notag \\
&\Rightarrow &0<y-y\sum_{n=2}^{\infty }|\Phi _{n-1}||a_{n}|  \notag \\
&\Rightarrow &0<y-y\sum_{n=2}^{\infty }|\Phi _{n-1}||a_{n}||z^{n-1}|  \notag
\\
&\Rightarrow &0<y\left\vert 1+\sum_{n=2}^{\infty }\Phi
_{n-1}a_{n}z^{n-1}\right\vert
\end{eqnarray}%
We have
\begin{eqnarray*}
&~&\sum_{n=2}^{\infty }\left( \left[ n-1\right] +y\right) |\Phi
_{n-1}||a_{n}|<y \\
&\Rightarrow &\sum_{n=2}^{\infty }\left( \left[ n-1\right] +y\right) |\Phi
_{n-1}||a_{n}||z^{n-1}|<y \\
&\Rightarrow &\sum_{n=2}^{\infty }\left[ n-1\right] |\Phi
_{n-1}||a_{n}||z^{n-1}|<y-y\sum_{n=2}^{\infty }|\Phi _{n-1}||a_{n}||z^{n-1}|
\\
&\Rightarrow &\left\vert \sum_{n=2}^{\infty }\left[ n-1\right] \Phi
_{n-1}a_{n}z^{n-1}\right\vert <y\left\vert 1+\sum_{n=2}^{\infty }\Phi
_{n-1}a_{n}z^{n-1}\right\vert \\
&\Rightarrow &\left\vert \frac{\sum_{n=2}^{\infty }\left[ n-1\right] \Phi
_{n-1}a_{n}z^{n-1}}{1+\sum_{n=2}^{\infty }\Phi _{n-1}a_{n}z^{n-1}}%
\right\vert <y,
\end{eqnarray*}%
because of \eqref{2t16}. By \eqref{1c15} it follows $f\in \mathcal{LD}%
_{b}^{k}(a,c,\beta )$.
\end{proof}

\noindent \textbf{Competing interests}

\noindent The authors declare that they have no competing interests.

%----------------------------------------------------------------------------
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