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\begin{document}
\title[Subordinations
for multivalent analytic functions]{\bf Subordinations for
multivalent analytic functions associated with Wright generalized
hypergeometric function}

%--------------------------------------------------------------------
\author{N. Sarkar$^A$, P. Goswami$^B$, J. Dziok$^C$ and J. Sok\'{o}{\l}$^D$}
  %--------------------------------------------------------------------
\address{$^A$Department of Mathematics, Jadavpur University,
         Kolkata-700032, India}
         \email{nantu.math@gmail.com}
%--------------------------------------------------------------------
\address{$^B$Department of Mathematics, AMITY University Rajasthan,
          Jaipur-302002, India}
          \email{pranaygoswami83@gmail.com}
%--------------------------------------------------------------------
\address{$^C$Institute of Mathematics, University of Rzesz\'{o}w,
         Rejtana 16A, 35-310 Rzesz\'{o}w, Poland}
         \email{jdziok@univ.rzeszow.pl}
%--------------------------------------------------------------------
\address{$^D$Department of Mathematics, Rzesz\'{o}w University of Technology,
         W. Pola 2, 35-959 Rzesz\'{o}w, Poland}
         \email{jsokol@prz.edu.pl}
%--------------------------------------------------------------------
\date{}

\begin{abstract}
Recently  M. K. Aouf and T. M. Seoudy, (2011, {\it Integral Trans.
Spec. Func.} {\bf 22}(6) (2011), 423--430) have introduced families
of analytic functions associated with the Dziok--Srivastava
operator. In this paper we use the Dziok--Raina operator to consider
classes of multivalent analytic functions. It is connected with
Wright generalized hypergeometric function, see J. Dziok and R. K.
Raina (2004, {\it Demonstratio Math.}, {\bf 37}(3) 533–-542).  In
this paper we present a new result which extends  some of the
earlier results and give other properties of these classes. We have
made use of differential subordinations and properties of
convolution in geometric function theory.
\end{abstract}

%---------------------------------------------------------------------
\keywords {Dziok--Srivastava operator; Dziok--Raina operator;
Hadamard product; univalent functions; $p$--valent functions;
convolution; differential subordination.}

\subjclass[2000]{Primary 30C45, secondary 30C80}
%---------------------------------------------------------------------
\maketitle
%---------------------------------------------------------------------




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

\section{Introduction}
\indent Let $\Lambda_p$ denote the class of functions $f(z)$ of the form
\begin{eqnarray}\label{1.1}
f(z)=z^p+\sum_{k=p+1}^\infty c_kz^k,~~~~~~(p\in
\mathbb{N}:=\{1,2,3,\ldots\}),
\end{eqnarray}
which are analytic in the open unit disk
$$\mathbb{U}:=\{ z:z\in \mathbb{C}\ \ {\rm and}\ \ |z|<1 \}.$$
\indent For analytic functions $f\in\Lambda_p$, given by (1.1) and $g\in\Lambda_p$ given by
\begin{eqnarray}\label{1.2}
g(z)=z^p+\sum_{k=p+1}^\infty b_kz^k, (p\in \mathbb{N})
\end{eqnarray}
the Hadamard product (or convolution) of $f$ and $g$ is defined by
\begin{eqnarray*}
(f\ast g)(z)=z^p+\sum_{k=p+1}^\infty c_kb_kz^k=(g\ast f)(z).
\end{eqnarray*}
\indent Let $a_1,A_1,\ldots,a_q,A_q$ and $b_1,B_1,\ldots,b_s,B_s$
$(q,s\in \mathbb{N})$ be positive real parameters such that
\begin{eqnarray*}
1+\sum_{k=1}^s B_k\geq \sum_{k=1}^q A_k.
\end{eqnarray*}
\indent The Wright generalized hypergeometric function \cite{wright}
\begin{eqnarray*}
_q\Psi_s[(a_1,A_1),\ldots,(a_q,A_q);(b_1,B_1),\ldots,(b_s,B_s);z]:={_q\Psi_s}[(a_m,A_m)_{1,q};(b_m,B_m)_{1,s};z]
\end{eqnarray*}
is defined by
\begin{eqnarray}\label{1.4}
_q\Psi_s[(a_k,A_k)_{1,q};(b_k,B_k)_{1,s};z] :=\sum_{m=0}^\infty
\left\{\frac{\Pi _{m=1}^q\Gamma(a_m+nA_m)}{\Pi
_{m=1}^s\Gamma(b_m+nB_m)}\right\} \frac{z^m}{m!},~(z\in \mathbb{U}).
\end{eqnarray}
If $A_m=1~(m=1,\ldots,q)$ and $B_m=1~(m=1,\ldots,s)$, we have the
following relationship
\begin{eqnarray*}
\omega{_q\Psi_s}[(a_m,1)_{1,q};(b_m,1)_{1,s};z]:={_qF_s}(a_1,\ldots,a_q;b_1,\ldots,b_s;z),
\end{eqnarray*}
where $_qF_s(a_1,\ldots,a_q;b_1,\ldots,b_s;z)$ is the generalized
hypergeometric function (see for details \cite{aouf3,dziok,dziok3}
and any many others) and
\begin{eqnarray}\label{1.7}
\omega=\left\{\Pi_{m=1}^q\Gamma(a_m)\right\}^{-1}\left\{\Pi_{m=1}^s\Gamma(b_m)\right\}.
\end{eqnarray}

 The Wright generalized hypergeometric functions \eqref{1.4} have been recently
involved in the geometric function theory, see , as well as:
\cite{bansal,deziok3,dziok3,dziok1,dziok2,raina,raina1}.  Using the
Wright generalized hypergeometric functions we introduce the linear
operator
\begin{eqnarray}
\Theta_p[(a_m,A_m)_{1,q};(b_m,B_m)_{1,s}]:\Lambda_p\mapsto\Lambda_p
\nonumber
\end{eqnarray}
defined by the convolution
\begin{eqnarray*}
\Theta_p[(a_m,A_m)_{1,q};(b_m,B_m)_{1,s}]f(z) =\omega\left\{
z^p{_q\Psi_s}[(a_m,A_m)_{1,q};(b_m,B_m)_{1,s};z]\right\} \ast f(z).
\end{eqnarray*}
In particular, the operator
\begin{eqnarray}\label{1.8}
\Theta[(a_m,A_m)_{1,q};(b_m,B_m)_{1,s}]=\Theta_1[(a_m,A_m)_{1,q};(b_m,B_m)_{1,s}]
\end{eqnarray}
was investigated by Dziok and Raina \cite{dziok1}. We observe that
for a function $f$ defined by \eqref{1.1} we have
\begin{eqnarray}\label{1.9}
   \Theta_p[(a_m,A_m)_{1,q};(b_m,B_m)_{1,s}]f(z)
   =z^p+\sum_{k=p+1}^\infty
   \Omega_k[a_i;b_i]c_kz^k,
\end{eqnarray}
where
\begin{eqnarray}\label{1.10}
   \Omega_k[a_i;b_i]
   =\omega\frac{\Gamma(a_1+A_1(k-p))\ldots\Gamma(a_q+A_q(k-p))}{\Gamma(b_1+B_1(k-p))\ldots\Gamma(b_s+B_s(k-p))
(k-p)!}
\end{eqnarray}
and $\omega$ is given by \eqref{1.7}. For our convenience, we write
\begin{eqnarray*}
   \Theta_p[a_i]f:=\Theta_p[(a_m,A_m)_{1,q};(b_m,B_m)_{1,s}]f.
\end{eqnarray*}

 Let $f$ and $g$ be analytic in $\mathbb{U}$. Then we say that the function $g$ is subordinate to $f$
if there exists an analytic function $w$ in $\mathbb{U}$ such that
\begin{equation*}
    w(0)=0,\ \ \ |w(z)|<1\ \ \ (z\in \mathbb{U})\ \ \ {\rm and}\ \ \
    g(z)=f\left(w(z)\right).
\end{equation*}

 For this subordination, the symbol $g(z)\prec f(z)$ is used. In
case $f$ is univalent in $\mathbb{U}$, the subordination $g\prec f$
is equivalent to
\begin{equation*}
    g(0)=f(0)\ \ \ {\rm and}\ \ \ g(\mathbb{U})\subset
    f(\mathbb{U}).
\end{equation*}

Now we define two subclasses $\mathcal{S}_p[A,B]$ and
$\mathcal{K}_p[A,B]$ of the class $\Lambda_p$, for \\$-1\leq B<A\leq
1$, $p\in \mathbb{N}$, as follows:
\begin{eqnarray}\label{S_p[A,B]}
   \mathcal{S}_p[A,B]=\left\{f\in \Lambda_p: \frac{zf'(z)}{f(z)}\prec
   p\frac{1+Az}{1+Bz},~(z\in \mathbb{U})\right\},
\end{eqnarray}
\begin{eqnarray}\label{K_p[A,B]}
   \mathcal{K}_p[A,B]=\left\{f\in \Lambda_p: 1+\frac{z f''(z)}{f'(z)}\prec
   p\frac{1+Az}{1+Bz}, ~(z\in \mathbb{U})\right\}.
\end{eqnarray}
Clearly
\begin{eqnarray}
   zf'\in \mathcal{S}_p[A,B]\Leftrightarrow f\in \mathcal{K}_p[A,B].
\end{eqnarray}
We note that for special choices of the parameters we obtain the
classes  of $p$-valent starlike of order $\alpha$ and of $p$-valent
convex of order $\alpha$
\begin{equation*}
    \mathcal{S}_1[1-2\alpha,-1]=\mathcal{S}^*(\alpha)~,~~\mathcal{K}_1[1-2\alpha,-1]=\mathcal{K}(\alpha),
\end{equation*}
\begin{equation*}
    \mathcal{S}_p[1-2\alpha,-1]=\mathcal{S}^*_p(\alpha)~,~~\mathcal{K}_p[1-2\alpha,-1]=\mathcal{K}_p(\alpha),
\end{equation*}
where $0\leq \alpha<1$. Next, using the  operator $\Theta_p[a_i]$,
we introduce the following classes of analytic functions for $q,s\in
\mathbb{N}$ and $-1\leq B<A\leq 1$
\begin{eqnarray}\label{S^*_{p,q,s}[a_i;A,B]}
   \mathcal{S}^{*}_{p,q,s}[a_1;A,B]=\left\{f\in \Lambda_p:
   \Theta_p[a_i]f\in \mathcal{S}_p[A,B]\right\}
\end{eqnarray}
and
\begin{eqnarray}\label{K_{p,q,s}[a_i;A,B]}
   \mathcal{K}_{p,q,s}[a_i;A,B]=\left\{f\in \Lambda_p:
   \Theta_p[a_i]f\in \mathcal{K}_p[A,B]\right\}.
\end{eqnarray}
\indent We also note that
\begin{equation*}
    zf'\in \mathcal{S}^{*}_{p,q,s}[a_i;A,B]\Leftrightarrow f\in
    \mathcal{K}_{p,q,s}[a_i;A,B].
\end{equation*}
If $A_m=1~(m=1,\ldots,q)$ and $B_m=1~(m=1,\ldots,s)$, then the
operator $\Theta_p[a_i]f$ becomes the operator $\mathcal{H}_p$:
\begin{eqnarray*}
    \mathcal{H}_p(\alpha_1,\ldots,\alpha_q;\beta_1,\ldots,\beta_s)f
    =z^p{_qF_s}(a_1,\ldots,a_q;b_1,\ldots,b_s;z)*f,
\end{eqnarray*}
and the class $\mathcal{S}^{*}_{p,q,s}[a_1;A,B]$ becomes the class
$V_p(\alpha_i;A,B)$ considered in \cite{1AMC08}, see also
\cite{KPJS08,AMC08,CAMWA10}.

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

\section{Main results}
\indent We assume throughout this section that $0\leq\theta<2\pi$,
$-1\leq B<A\leq 1$ and $\Omega_k[a_i]$ is defined by \eqref{1.10}.

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

\begin{theorem}\label{th1}
Suppose that the function $f$ is defined by \eqref{1.1}. Then $f$ is
in the class $\mathcal{S}_p[A,B]$ if and only if
\begin{eqnarray}\label{1th1}
\frac{1}{z^p}\left[f(z)\ast \frac{z^p-Dz^{p+1}}{(1-z)^2}\right]\neq
0
\end{eqnarray}
for all $z\in \mathbb{U}$ and $0\leq\theta<2\pi$, where
\begin{equation}\label{D}
    D=1+\frac{e ^{-i\theta}+B}{p(A-B)}.
\end{equation}
\end{theorem}

\begin{proof}
First, suppose $f$ is in the class $\mathcal{S}_p[A,B]$. Then by
definition of subordination it is equivalent to
\begin{eqnarray}\label{2th1}
\frac{z f'(z)}{f(z)}\neq
p\frac{1+Ae^{i\theta}}{1+Be^{i\theta}}\nonumber
\end{eqnarray}
for all $z\in \mathbb{U}$ and $0\leq\theta<2\pi$. Since,
\begin{eqnarray}\label{3th1}
\frac{z^p}{(1-z)^2}=z^p+\sum_{k=p+1}^\infty(k-p+1)z^k,~~~\mbox{and~~~}
\frac{z^{p+1}}{(1-z)^2}=\sum_{k=p+1}^\infty(k-p)z^k
\end{eqnarray}
 we have
\begin{eqnarray*}
&~~&\frac{1}{z^p}\left[f(z)\ast \frac{z^p-Dz^{p+1}}{(1-z)^2}\right]\nonumber\\
&=&\frac{1}{z^p}\left[f(z)\ast\{z^p+\sum_ {k=p+1}^\infty z^k+(1-D)
\sum_ {k=p+1}^\infty kz^k+p(D-1)\sum_ {k=p+1}^\infty z^k\}\right]\nonumber\\
&=&\frac{1}{z^p}\left[f(z)+(1-D)\sum_{k=p+1}^\infty k c_k z^k+p(D-1)\sum_{k=p+1}^\infty c_k z^k\right]\nonumber\\
&=&\frac{1}{z^p}\left[f(z)+(1-D)[zf'(z)-pz^p]+p(D-1)[f(z)-z^p\right]\nonumber\\
&=&\frac{1}{z^p}\left[f(z)\left\{\frac{zf'(z)}{f(z)}(1-D)+(1-p+pD)\right\}\right]\\
&\neq& \frac{1}{z^p}f(z)\left[-\frac{e^{-i\theta}+A}{A-B}+\frac{e^{-i\theta}+A}{A-B}\right]\nonumber\\
&=& 0\nonumber
\end{eqnarray*}
which proves the necessary part.\\
Again, if the condition \eqref{1th1} hold, then because
\begin{eqnarray}\label{4th1}
\frac{1}{z^p}\left[f(z)\ast \frac{z^p-Dz^{p+1}}{(1-z)^2}\right]
=\frac{1}{z^p}\left[f(z)\left\{\frac{zf'(z)}{f(z)}(1-D)+(1-p+pD)\right\}\right],
\end{eqnarray}
we can write
\begin{eqnarray}
\frac{1}{z^p}\left[f(z)\left\{\frac{zf'(z)}{f(z)}(1-D)+(1-p+pD)\right\}\right]\neq
0\nonumber
\end{eqnarray}
for all $z\in \mathbb{U}$ and $0\leq\theta<2\pi$. Then  we easily
obtain the required result as
\begin{eqnarray}
\frac{z f'(z)}{f(z)}\neq
p\frac{1+Ae^{i\theta}}{1+Be^{i\theta}}\nonumber
\end{eqnarray}
for all $z\in \mathbb{U}$ and $0\leq\theta<2\pi$, which proves that
$f\in \mathcal{S}_p[A,B]$.
\end{proof}

%=======================================================================
%=======================================================================
\begin{theorem}\label{th2}
Suppose that the function $f$ is defined by \eqref{1.1}. Then $f$ is
in the class $\mathcal{K}_p[A,B]$ if and only if
\begin{eqnarray}\label{1th2}
  \frac{1}{z^p}\left[f(z)\ast\frac{pz^p-\{p+D(p+1)-2)\}z^{p+1}+D(p-1)z^{p+2}}{(1-z)^3}\right]
  \neq 0
\end{eqnarray}
for all $z\in \mathbb{U}$ and $0\leq\theta<2\pi$, where $D$ is given
by \eqref{D}.
\end{theorem}

\begin{proof}
 Choose
\begin{equation*}
    g(z)=\frac{z^p-Dz^{p+1}}{(1-z)^2}
\end{equation*}
and we note that
\begin{eqnarray}\label{2th2}
   zg'(z)=\left[\frac{pz^p-\{p+D(p+1)-2)\}z^{p+1}+D(p-1)z^{p+2}}{(1-z)^3}\right].
\end{eqnarray}
From the identity $zf'(z)\ast g(z)=f(z)\ast zg'(z)$ and the fact that
\begin{eqnarray*}
   f\in \mathcal{K}_p[A,B]\Leftrightarrow zf'(z)\in \mathcal{S}_p[A,B]
\end{eqnarray*}
we can say that $f\in \mathcal{K}_p[A,B]$ if and only if
\begin{eqnarray*}
    &\,&\frac{1}{z^p}\left[zf'(z)\ast g(z)\right]\neq 0,~~~~\mbox{(By Theorem \ref{th1})}\\
    &\Leftrightarrow& \frac{1}{z^p}\left[f(z)\ast zg'(z)\right]\neq 0
\end{eqnarray*}
which, on using \eqref{2th2} gives the required  result
\eqref{1th2}.
\end{proof}

%================================================================================================
\begin{theorem}\label{th3}
Suppose that the function $f$ is defined by \eqref{1.1}. Then a
necessary and sufficient condition for the function $f$  to be in
the class $\mathcal{S}^{*}_{p,q,s}[a_i;A,B]$ is that
\begin{eqnarray}\label{1th3}
1-\sum_{k=p+1}^\infty
\left[\frac{(k-p)e^{-i\theta}-Ap+B(k-p+1)}{p(A-B)}\right]
\Omega_k[a_i;b_i]c_kz^{k-p}\neq 0
\end{eqnarray}
for all $z\in \mathbb{U}$ and $0\leq\theta<2\pi$.
\end{theorem}

\begin{proof}
From Theorem \ref{th1}, we find that $f\in \mathcal{S}_p^*[a_i;A,B]$
if and only if
\begin{eqnarray}\label{2th3}
\frac{1}{z^p}\left[\Theta_p[a_i]f(z)\ast\frac{z^p-Dz^{p+1}}{(1-z)^2}\right]\neq
0
\end{eqnarray}
for all $z\in \mathbb{U}$ and $0\leq\theta<2\pi$. Using relations
\eqref{3th1} and after a long calculations with the help of
\eqref{1.9}, we can write from \eqref{1th3} that
\begin{eqnarray}\label{3th3}
&\,&1+\sum_{k=p+1}^\infty [1+(p-k)(D-1)]\Omega_k[a_i;b_i]c_kz^{k-p}\neq 0\nonumber\\
&\Leftrightarrow& 1-\sum_{k=p+1}^\infty
\left[\frac{(k-p)e^{-i\theta}-Ap+B(k-p+1)}{p(A-B)}\right]\Omega_k[a_i;b_i]c_kz^{k-p}\neq
0\nonumber
\end{eqnarray}
This proves Theorem \ref{th3}.
\end{proof}

%================================================================================================
\begin{theorem}\label{th4}
Suppose that the function $f$ is defined by \eqref{1.1}. Then a
necessary and sufficient condition for the function $f$  to be in
the class $\mathcal{K}_{p,q,s}[a_i;A,B]$ is that
\begin{eqnarray}\label{1th4}
1-\sum_{k=p+1}^\infty
k\frac{(k-p)e^{-i\theta}-Ap+Bk}{p(A-B)}\Omega_k[a_i;b_i]c_kz^{k-p}\neq
0
\end{eqnarray}
for all $z\in \mathbb{U}$ and $0\leq\theta<2\pi$.
\end{theorem}

\begin{proof} From Theorem \ref{th2}, we find that $f\in \mathcal{K}_{p,q,s}[a_i;A,B]$ if and only if
\begin{eqnarray}\label{2th4}
\frac{1}{z^p}\left[\Theta_p[a_i]f(z)\ast
\frac{pz^p-\{p+D(p+1)-2)\}z^{p+1}+D(p-1)z^{p+2}}{(1-z)^3}\right]\neq
0
\end{eqnarray}
for all $z\in \mathbb{U}$ and $0\leq\theta<2\pi$. Now, it can be
easily shown that
\begin{eqnarray}
\frac{z^p}{(1-z)^3}&=&z^p+\sum_ {k=p+1}^\infty \frac{(k-p+1)(k-p+2)}{2}z^k,\label{4th4}\\
\frac{z^{p+1}}{(1-z)^3}&=&\sum_ {k=p+1}^\infty \frac{(k-p)(k-p+1)}{2}z^k,\label{5th4}\\
\frac{z^{p+2}}{(1-z)^3}&=&\sum_ {k=p+1}^\infty
\frac{(k-p-1)(k-p)}{2}z^k.\label{6th4}
\end{eqnarray}
Using \eqref{4th4}--\eqref{6th4} in  \eqref{2th4} and noting that
$\Theta_p[a_i]f(z)\ast z^p=z^p$, we can say that \eqref{2th4} is
equivalent to
\begin{eqnarray*}
p&+&\frac{1}{2}\sum_ {k=p+1}^\infty (k-p+1)[p(k-p+2)-(k-p)\{(p-2)+D(p+1)\}\\
&+&D((k-p)(p-1)]\Omega_k[a_i;b_i]c_kz^{k-p}\neq 0\\
&\Leftrightarrow& p-\sum_ {k=p+1}^\infty
k\frac{(k-p)e^{-i\theta}-Ap+Bk}{p(A-B)}\Omega_k[a_i;b_i]c_kz^{k-p}\neq
0\\
&\Leftrightarrow& 1-\sum_{k=p+1}^\infty
k\frac{(k-p)e^{-i\theta}-Ap+Bk}{p(A-B)}\Omega_k[a_i;b_i]c_kz^{k-p}\neq
0
\end{eqnarray*}
Thus, the proof of Theorem \ref{th4} is completed.
\end{proof}

%================================================================================================
\begin{theorem}\label{th5}
If the function $f$ is defined by \eqref{1.1} and it  belongs to
$\mathcal{S}^{*}_{p,q,s}[a_i;A,B]$, then
\begin{eqnarray}
\sum_{k=p+1}^\infty [(k-p)+|Ap-B(k-p+1)|]\Omega_k[a_i;b_i]|c_k|\leq
p(A-B).
\end{eqnarray}
\end{theorem}

\begin{proof}
Since
\begin{eqnarray*}
&\,&\left|1-\sum_{k=p+1}^\infty \left[\frac{(k-p)e^{-i\theta}-Ap+B(k-p+1)}{p(A-B)}\right]
\Omega_k[a_i;b_i]c_kz^{k-p}\right|\\
&>& 1-\sum_{k=p+1}^\infty
\left|\frac{(k-p)e^{-i\theta}-Ap+B(k-p+1)}{p(A-B)}\right|\Omega_k[a_i;b_i]\left|c_k\right|
\end{eqnarray*}
and
\begin{eqnarray*}
&\,&\left|\frac{(k-p)e^{-i\theta}-Ap+B(k-p+1)}{p(A-B)}\right|\\
&=&\frac{\left|(k-p)e^{-i\theta}-Ap+B(k-p+1)\right|}{p(A-B)}\\
&\leq& \frac{(k-p)+\left|Ap-B(k-p+1)\right|}{p(A-B)}
\end{eqnarray*}
the results follows from Theorem \ref{th3}.
\end{proof}

 In the same way, we can also prove the following theorem.

%================================================================================================
\begin{theorem}\label{th6}
If the function $f$ is defined by \eqref{1.1}  and it  belongs to
$\mathcal{K}_{p,q,s}[a_i;A,B]$, then
\begin{eqnarray}
\sum_{k=p+1}^\infty k\{(k-p)+\left|Ap-Bk\right|\}\Omega_k[a_i;b_i]|
c_k|\leq p(A-B).
\end{eqnarray}
\end{theorem}



%================================================================================================
We will need the following Lemma on Briot--Bouquet differential
subordination.

%LEMMA 1-----------------------------------------------------------------

\begin{lemma}\label{lem1} \cite{EMMR} Let $h$ be
convex univalent in $\mathbb{U}$, with $\mathfrak{Re}[(\beta
h(z)+\gamma )]\geq 0$. If $q$ is analytic in $\mathbb{U}$, with
$q(0)=h(0)$, then
\begin{equation*}
    q(z)+\frac{zq^{\prime }(z)}{\beta q(z)+\gamma }\prec h(z)\Rightarrow
    q(z)\prec h(z).
\end{equation*}
\end{lemma}


%COROLLARY 1-------------------------------------------------------------
\begin{corollary}\label{cor1} If $q$ is an
analytic function in $\mathbb{U}$, $q(0)=p$ and
\begin{equation*}
    q(z)+\frac{zq^{\prime }(z)}{q(z)+\gamma }\prec p\frac{1+Az}{1+Bz}\ ,
    \ \ \ \ \mathfrak{Re}\left[ \gamma +p\frac{1+A}{1+B}\right]\geq 0,
\end{equation*}
then
\begin{equation*}
    q(z)\prec p\frac{1+Az}{1+Bz}.
\end{equation*}
\end{corollary}

%================================================================================================
\begin{theorem}\label{th10}
Suppose that $i\in\{1,\ldots,q\}$ and
\begin{equation*}
    \frac{a_i}{A_i}\geq p\frac{A-B}{1+B}.
\end{equation*}
Then for $m\in\mathbb{N}$ we have
\begin{equation*}
    \mathcal{K}_p[a_i+m;A,B]\subset \mathcal{K}_p[a_i;A,B].
\end{equation*}
\end{theorem}
%PROOF--------------------------------------------------------------------------

\begin{proof} It is clear that it is sufficient to prove this theorem
for $m=1$. Let a function $f$ belong to the class $\mathcal{K}
_{p}[a_i+1;A,B]$, then from  \eqref{K_{p,q,s}[a_i;A,B]}, we can
write
\begin{equation}\label{df2}
1+\frac{z[\Theta_p[a_i+1]f(z)]''}{ [\Theta_p[a_i+1]f(z)]'}\prec
p\frac{1+Az }{1+Bz}\ \ \left( z\in\mathbb{U}\right).
\end{equation}
From \eqref{1.10} we get
\begin{equation*}
    \Omega_k[a_i+1;b_i]=\frac{a_i+A_i(k-p)}{a_i}\Omega_k[a_i;b_i],
\end{equation*}
thus through \eqref{1.9} we obtain
\begin{eqnarray}\label{Theta[ai+1]}
   \Theta_p[a_i+1]f(z)
   &=& z^p+\sum_{k=p+1}^\infty\Omega_k[a_i+1;b_i]c_kz^k \nonumber\\
   &=& z^p+\sum_{k=p+1}^\infty\frac{a_i+A_i(k-p)}{a_i}\Omega_k[a_i;b_i]c_kz^k \nonumber\\
   &=& \left(1-\frac{pA_i}{a_i}\right)\left[z^p+\sum_{k=p+1}^\infty\Omega_k[a_i;b_i]c_kz^k
   \right]\nonumber\\
   &\,&\ +\ \
   \frac{A_i}{a_i}\left[pz^p+\sum_{k=p+1}^\infty\Omega_k[a_i;b_i]kc_kz^k\right]\nonumber\\
   &=&\left(1-\frac{pA_i}{a_i}\right)\Theta_p[a_i]f(z)+\frac{A_i}{a_i}z\left[\Theta_p[a_i]f(z)\right]'.
\end{eqnarray}
Therefore, after some calculations we obtain from
\eqref{Theta[ai+1]}
\begin{equation}\label{3th9}
   1+\frac{z\left[{\Theta}_p[a_i+1]f(z)\right]''}{[{\Theta}_p[a_i+1]f(z)]'}
   =F(z)+
   \frac{zF'(z)}
   {F(z)+\frac{a_i}{A_i}-p},
\end{equation}
where
\begin{equation*}
    F(z)=1+\frac{z\left[{\Theta}_p[a_i]f(z)\right]''}{[{\Theta}_p[a_i]f(z)]'}
\end{equation*}
Thus from \eqref{df2} the right side of \eqref{3th9} is subordinated
to the function $h(z)=p\frac{ 1+Az}{1+Bz}$ which is convex and
univalent in $\mathbb{U}$. Therefore, by Corollary \ref{cor1}
\begin{equation*}
    1+\frac{z\left[{\Theta}_p(a_i)f(z)\right]''}{[{\Theta}_p(a_i)f(z)]'}\prec
    p\frac{ 1+Az}{1+Bz}\ \ \left( z\in\mathbb{U}\right),
\end{equation*}
whenever
$\mathfrak{Re}\left[p\frac{1+Az}{1+Bz}+\frac{a_i}{A_i}-p\right]>0$
for $z\in\mathbb{U}$, which follows from the assumption that
$\frac{a_i}{A_i}\geq p\frac{A-B}{1+B}$ and $-1\leq B<A\leq 1$.
\end{proof}


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




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

{\bf Remarks :}\\
(i) On substituting $p = 1$ and $A_i=B_j=1$, where $i = 1,\ldots,q$
and $j=1,\ldots,s$ in Theorems \ref{th1}-\ref{th6}, we obtain known
results given by Aouf and Seoudy [2, Theorems 1-6]. \\
(ii) On substituting $p =m = 1$ and $A_i =  B_j = 1$, where $i =
1,\ldots,q$ and $j = 1,\ldots,s$ in Theorems \ref{th10}, we obtain
known results given by Aouf and Seoudy [2, Theorem 8] but the proof
is not convincing there.

Using Lemma \ref{lem1} one can obtain  a sufficient conditions for
functions to be in the class $\mathcal{S}^*_{p}[a_i;A,B]$.  This
problem was considered earlier in  \cite{DziokRaina} in a more
general situation. This result is presented below,  in current
notation.{\medskip}

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

\begin{theorem}\label{th11} \cite{DziokRaina}
Let $p\in\mathbb{N}$, $i\in\{1,\ldots,q\}$, $0\leq B<A\leq 1$ and
$\frac{a_i}{A_i}\geq p\frac{A-B}{1+B}$. If a function
$f\in\Lambda_{p}$ satisfies the following inequality
\begin{eqnarray}
   \left|\frac{{\Theta}_p[{a_i}+2]f(z)}{{\Theta}_p[{a_i}+1]f(z)}-1\right|
   <\frac{p(A-B)(1+A_i)}{(1+B)(1+{a_i})}\;\;\;\;\left( z\in \mathbb{U}
   \right),
\label{1th11}
\end{eqnarray}
then $f$  belongs to the class $\mathcal{S}^*_p[a_i;A,B]$.
\end{theorem}



To obtain next theorem which is in a way the sharp version of
Theorem \ref{th11} we shall recall an another basic lemma in the
theory of Briot--Bouquet differential subordinations.

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



\begin{lemma}\label{lem2} \cite{MM}
Let $\beta,\gamma,\delta\in\mathbb{C}$ with $\mathfrak{
Re}[\beta+\gamma]>|\beta\delta|$. If $s(z)=1+a_1z+a_2z^2+\cdots$
satisfies
\begin{equation*}
   s(z)+\frac{zs'(z)}{\beta s(z)+\gamma}\prec1+\delta z\ \ \left( z\in
   \mathbb{U}\right),
\end{equation*}
then
\begin{equation*}
   s(z)\prec q(z)\prec 1+\delta z\ \ \left( z\in \mathbb{U}\right),
\end{equation*}
where
\begin{equation*}
   q(z)=\frac{1}{\beta g(z)}-\frac{\gamma}{\beta},
\end{equation*}
and
\begin{equation*}
   g(z)=\int\limits_0^1e^{\beta\delta(t-1)z}
   t^{\beta+\gamma-1}{\rm{d}}t= {_1F_1}(1,\beta+\gamma+1;-\beta\delta
   z)(\beta+\gamma)^{-1}.
\end{equation*}
Moreover, the function $q$ is the best dominant in the sense that if
$s\prec q_1,$ then $q\prec q_1$.
\end{lemma}

 Lemma \ref{lem2} in a more general case one can be found in
\cite[p.~109]{MM}.


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



\begin{theorem}\label{th13}
Let $p\in\mathbb{N}$, $i\in\{1,\ldots,q\}$, $0\leq B<A\leq 1$ and
$\frac{a_i}{A_i}\geq p\frac{A-B}{1+B}$. If a function
$f\in\Lambda_{p}$ satisfies the following inequality
\begin{eqnarray}
   \left|\frac{{\Theta}_p[{a_i}+2]f(z)}{{\Theta}_p[{a_i}+1]f(z)}-1\right|
   <r=\frac{p(A-B)(1+A_i)}{(1+B)(1+{a_i})}\;\;\;\;\left( z\in \mathbb{U}
   \right),
\label{1th13}
\end{eqnarray}
then
\begin{equation}\label{2th13}
   \frac{[{\Theta}_p[{a_i}]f(z)]'}{{\Theta}_p[{a_i}]f(z)}\prec
   q(z)\prec p+\frac{(a_i+1)r}{A_i+1}z \;\;\;\;\left( z\in \mathbb{U} \right),
\end{equation}
where the function $q$ is given by
\begin{equation}\label{3th13}
   q(z)=\left[\int\limits_0^1e^{\frac{rz(a_i+1)(t-1)}{1+A_i}}
   t^{\frac{a_i}{A_i}-1}{\rm{d}}t\right]^{-1}-\left(\frac{a_i}{A_i}-p\right)
\end{equation}
and it is the best dominant.
\end{theorem}
\medskip



%PROOF----------------------------------------------------------------------------------------
\begin{proof}
By using \eqref{Theta[ai+1]} the inequality \eqref{1th11} becomes
\begin{equation}\label{4th13}
    \frac{z\left[{\Theta}_p[a_i+1]f(z)\right]'}{{\Theta}_p[a_i+1]f(z)}
    \prec p+\frac{a_i+1}{A_i+1}rz\;\;\;\;\left( z\in\mathbb{U}\right),
\end{equation}
where
\begin{equation*}
r=\frac{p(A-B)(1+A_i)}{(1+B)(1+{a_i})}.
\end{equation*}
From  \eqref{4th13} trough \eqref{Theta[ai+1]} we have
\begin{eqnarray}
        \frac{z\left[\Theta_p[a_i+1]f(z)\right]'}{p{\Theta}_p[a_i+1]f(z)}
    &=& \frac{z\left[\Theta_p[a_i]f(z)\right]'}{p\Theta_p[a_i]f(z)}+
        \frac{z\left[\frac{z\left[{\Theta}_p[a_i]f(z)\right]'}{p\Theta_p[a_i]f(z)}\right]'}
        {p\frac{z\left[\Theta_p[a_i]f(z)\right]'}{p\Theta_p[a_i]f(z)}+\frac{a_i}{A_i}-p}\nonumber  \\
    &\prec & 1+\frac{r(1+a_i)}{p(1+A_i)}z.\label{5th13}
\end{eqnarray}

If we apply Lemma \ref{lem2} to \eqref{5th13} with $\delta
=\frac{r(1+a_i)}{p(1+A_i)}$, $\beta=p$, $\gamma=\frac{a_i}{A_i}-p$
and
\begin{equation*}
    s(z)=\frac{z\left[\Theta_p[a_i]f(z)\right]'}{p\Theta_p[a_i]f(z)}
\end{equation*}
we obtain
\begin{equation*}
    s(z)=\frac{z\left[\Theta_p[a_i]f(z)\right]'}{p\Theta_p[a_i]f(z)}
    \prec\frac{1}{p\int\limits_0^1e^{\frac{rz(a_i+1)(t-1)}{1+A_i}}
   t^{\frac{a_i}{A_i}-1}{\rm{d}}t}-\left(\frac{a_i}{pA_i}-1\right),
\end{equation*}
what gives \eqref{2th13} and the best dominant \eqref{3th13}.
\end{proof}
%-----------------------------------------------------------------
\noindent{\bf Acknowledgment}

The authors would like to express their sincerest  thanks to the
referees for a careful reading and various suggestions made for the
improvement of the paper.

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

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