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\title{{\bf {Inclusion properties of
hypergeometric type functions and related integral transforms
}}}

\author{
Lateef Ahmad Wani
}
\address{
Department of  Mathematics  \\
Indian Institute of Technology, Roorkee-247 667,
Uttarkhand,  India
}
\email{lateef17304@gmail.com}


\author{
A. Swaminathan
}
\address{
Department of  Mathematics  \\
Indian Institute of Technology, Roorkee-247 667,
Uttarkhand,  India
}
\email{swamifma@iitr.ac.in, mathswami@gmail.com}


\bigskip
\begin{abstract}
In this work,
conditions on the parameters $a, b$ and $c$ are given so that
the normalized Gaussian hypergeometric function $zF(a,b;c;z)$, where
\begin{align*}
F(a,b;c;z)=\sum_{n=0}^{\infty} \frac{(a)_n(b)_n}{(c)_n(1)_n}z^n,
\quad |z|<1,
\end{align*}
is in certain class of analytic functions. Using Taylor coefficients of functions in certain classes,
inclusion properties of the Hohlov integral transform involving $zF(a,b;c;z)$ are obtained. Similar
inclusion results of the Komatu integral operator related to the generalized polylogarithm are also obtained. Various results for the particular values of these parameters are deduced and compared with the existing literature.
\end{abstract}



\subjclass[2000]{30C45, 33C45, 33A30}

\keywords{Univalent; Convex; Starlike; Close-to-convex functions, Gaussian hypergeometric functions,
Incomplete beta functions, Komatu integral operator, Polylogarithm}


\maketitle

\pagestyle{myheadings}
\markboth{
Lateef Ahmad Wani and A. Swaminathan
}{
Inclusion properties of hypergeometric type transforms
}

\section{Introduction }\label{sec1}
Let ${\mathcal A}$ denote the class of functions of the form
\begin{align}\label{eq:series}
f(z) = z + \sum \sb{k=2}^{\infty} a\sb{k} z^{k},
\end{align}
analytic in the open unit disk $\mathbb{D}=\{ z: |z|<1\}$, and ${\mathcal
S}$ denote the subclass of $\mathcal{A}$ that contains functions univalent in
$\mathbb{D}$.
A function $f\in{\mathcal A}$ is called starlike, denoted by $f\in{\mathcal S}^{\ast}$,
if $tw\in f(\mathbb{D})$ whenever $w\in f(\mathbb{D})$ and $t\in[0,1]$.
The class of all convex functions, denoted by ${\mathcal{C}}$, consists of the functions $f\in {\mathcal{A}}$
such that $zf'$ is starlike.
%A function $f\in {\mathcal A}$ that maps the unit disc $\mathbb{D}$ onto a convex domain
%is called a convex function.
%Let ${\mathcal K}$ denote the class of all functions $f\in {\mathcal A}$ that
%are convex.
%A well-known fact is that ${\mathcal K}\subsetneq {\mathcal S}^{\ast}
%\subsetneq {\mathcal S}$. Further $f$ is convex if, and only if, $zf'$ is starlike.
A function $f\in\mathcal{A}$ is
said to be {\em{close-to-convex}} with respect to a fixed starlike function $g\in S^{\ast}$ if and only if
$\displaystyle
{\rm Re \,} \left( e^{i\lambda} \frac{z f'(z)}{g(z)}\right)>0$ for $z\in \mathbb{D}$ and $\lambda \in {\mathbb{R}}$. .
Let ${\mathcal K}$ denote the subclass of all such close-to-convex functions, where $\lambda=0$.
Various generalization of these classes and various other subclasses of $S$ exist in the literature. For example the class
of starlike functions of order $\sigma$, denoted by $S^*(\sigma)$, $0\leq \sigma<1$,
which has the analytic characterization ${\rm Re \,} \dfrac{zf'(z)}{f(z)}>\sigma$, is the generalization of the class $S^*(0)=S^*$.
Note that $C(\sigma)$, the class of convex functions of order $\sigma$ contains all functions $f\in S$ for which $zf' \in S^*(\sigma)$.

We introduce the class $R_{\gamma,\alpha}^{\tau}(\beta)$, with
$0\leq \gamma<1$, $0\leq \alpha\leq 1$, $\tau\in {\mathbb{C}}\backslash\{0\}$ and $\beta <1$ as
{\small{
\begin{align}\label{defn-class}
R_{\gamma,\alpha}^{\tau}(\beta):=
\left\{f\in{\mathcal A}:\!\!
\left|
\frac{(1-\alpha+2\gamma)\frac{f}{z}+(\alpha-2\gamma)f'+\gamma zf'' -1}
{2\tau(1-\beta)+(1-\alpha+2\gamma)\frac{f}{z}+(\alpha-2\gamma)f'+\gamma zf''-1}
\right|<1, \, z\in\mathbb{D} \right\}.
\end{align}
}}
Note that few particular cases of this class discussed in the literature.
\begin{enumerate}[1.]
\item  The class $R_{\gamma,\alpha}^{\tau}(\beta)$ for $\alpha=2\gamma+1$, was considered in \cite{Sw-camwa}, where references about other particular cases in this direction are provided.
\item
The class $R_{\gamma,\, \alpha}^{\tau}(\beta)$ for $\tau=e^{i\eta}\cos\eta$,
where $-\pi/2<\eta <\pi/2$ is considered in \cite{ali11} (see also \cite{ali10, ali12}), and
the properties of certain integral transforms of the type
\begin{align}\label{eq-vlambda}
V_{\lambda}(f)=\int_0^1\lambda(t)\frac{f(tz)}{t}dt, \quad
f\in R_{0,\gamma}^{(e^{i\eta}\cos\eta)}(\beta)
\end{align}
with $\beta<1$, $\gamma<1$ and $|\eta|<\pi/2$,
under suitable restriction on $\lambda(t)$ was discussed using duality techniques for various
values of $\gamma$ in \cite{ali11}.
For other interesting cases, we refer to \cite{ali12, Sw-camwa} and references therein.
\item The class $R_{0, 1}^{\tau}(0)$ with $\tau=e^{i\eta}\cos\eta$ was considered in \cite{Lin} with reference
to the univalency of partial sums.
%\item $f \in R_{\gamma}^{(e^{i\eta}\cos\eta)}(\beta)$ whenever $zf'\in P_{\gamma}^{\tau}(\beta)$, the class considered in \cite{S2}.
\end{enumerate}

%The main interest of this class is due to the fact that,
It is clear that the geometric properties of certain integral transforms under duality techniques, which is one of recent research
interest (for example, see \cite{ali11, ali12} and references therein),
cannot be proved easily as the results involve certain multiple integrals and it is difficult to check the conditions given
for the existence of the inclusion results for these integral transforms.
For this purpose, the inclusion properties of certain special functions to be in the analytic subclasses like
$R_{\gamma, \, \alpha}^{(e^{i\eta}\cos\eta)}(\beta)$ are studied using techniques other than duality methods which motivates this work.

Among various results related to the integral operator $\eqref{eq-vlambda}$ available in the literature,
an important and interesting result is application of the operator
$\eqref{eq-vlambda}$ when $\lambda(t)$ is related to the function $zF(a,b;c;z)$.
Here by $F(a,b;c;z)$ we mean the well-known Gaussian hypergeometric function
\begin{align}\label{eqn-Gauss-series}
F(a,b;c;z)=\sum\sb{n=0}^{\infty}\frac{(a)_n(b)_n}{(c)_n(1)_n}
z^n
\end{align}
$z\in\mathbb{D}$, with $(\lambda)_n$ being the Pochhammer symbol given by $(\lambda)_n=\lambda(\lambda+1)_{n-1}$, $(\lambda)_0=1$.
Also, there has been considerable interest to find conditions on the
parameters $a,b,$ and $c$ such that the normalized hypergeometric functions
$(c/ab)\left(F(a,b;c;z)-1\right)$ or $zF(a,b;c;z)$ belong
to one of the known subclasses of ${\mathcal S}$.
For more details on the basic ideas of Gaussian hypergeometric functions, we refer to \cite{Rv} and on the applications related
to geometric function theory, we refer to \cite{ali11, Sw-JCAM, Sw-itsf, Sw-camwa} and references therein.

Related to $F(a,b;c;z)$ is the Hohlov operator $H_{\,a,\, b,\, c\, }(f)(z)= zF(a,b;c;z)\ast f(z)$, where $\ast$ denotes the well-known Hadamard product
or convolution.
This operator is particular case of a general integral transform studied in \cite{FR1}.
To be more specific, the properties of certain
integral transforms of
the type
\begin{equation}\label{eq50}
V_{\lambda}(f)=\int_0^1\lambda(t)\frac{f(tz)}{t}dt, \quad
f\in R_{\gamma, \, \alpha}^{(e^{i\eta}\cos\eta)}(\beta)
\end{equation}
under suitable restriction on $\lambda(t)$ was discussed by many authors
\cite{ali11, ali12, FR1}. In particular, if
$$
\lambda(t)= \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}t^{b-1}(1-t)^{c-b-1}
$$
then $V_{\lambda}(f)={\mathcal L}(b,c)(f)(z)$ which is the well-known Carlson-Schaffer operator.
Note that $H_{\,1,\, b,\, c\, }(f)(z)={\mathcal L}(b,c)(f)(z)$.
The following lemma exhibits the relation between the integral operator
in discussion with the Hohlov operator.
\begin{lemma}\label{lem41}%\cite{ali11}
If $f \in {\mathcal A}$ and
$c-a+1>b>0$, then
$$
V_{\lambda}(f)(z)=H_{a,b,c}(f)(z)
$$
where
$$
H_{\,a,\, b,\, c\, }(f)(z)=
\frac{\Gamma(c)}{\Gamma(a)\Gamma(b)} \int_{0}^{1} \frac{
(1-t)^{c-a-b}}{\Gamma(c-a-b+1)} t^{b-2} F(c-a, 1-a;c-a-b+1;1-t) f(t z) d t.
$$
\end{lemma}
The Komatu operator $K_a^p:\mathcal{A}\rightarrow\mathcal{A}$ \cite{Komatu} is defined as
\begin{align*}
K_a^p[f](z)=\dfrac{(1+a)^p}{\Gamma(p)}\int_0^1\big{(}\log(\dfrac{1}{t})\big{)}^{p-1}t^{a-1}f(tz)dt,
\end{align*}
where $a>-1$ and $p\geq0$. It has a series representation as
\begin{align*}
K_a^p[f](z)=z+\sum_{n=2}^\infty\dfrac{(1+a)^p}{(n+a)^p}a_nz^n
\end{align*}
and in terms of convolution, we can write
\begin{align*}
K_a^p[f](z)=\mathcal{K}_a^p(z)\ast f(z),
\end{align*}
where $\mathcal{K}_a^p(z)=z+\sum_{n=2}^\infty\dfrac{(1+a)^p}{(n+a)^p}z^n$.


In this paper we study the operators $H_{\,a, \, b,\, c \,}(f)(z)$  and $K_a^p[f](z)$ for various choices
of the function $f$.

The paper is organized as follows. In Section $\ref{sec-coeff-cond}$, some preliminary results about the Gaussian hypergeometric function
$F(a,b;c;z)$ and conditions
on the Taylor coefficients of $f \in R_{\gamma,\alpha}^{\tau}(\beta)$ are given which are used in the subsequent sections.
Conditions on the triplets $a,b,c$ are obtained so that in Section $\ref{sec-incl-zF}$ inclusion properties of $F(a,b;c;z)$
and its normalized case to be in the class $R_{\gamma,\alpha}^{\tau}(\beta)$ are discussed and in Section $\ref{sec-incl-H-f}$, inclusion properties
of $zF(a,b;c;z)\ast f(z)$ for $f$ in various subclasses of $S$ are discussed. Similar type of inclusion results for the Komatu operator is discussed in Section $\ref{sec-incl-K-f}$. In the last section, certain remarks are given to provide
motivation for further research in this direction.

\section{Preliminary results}\label{sec-coeff-cond}
The following result is available in \cite{Sw-camwa}, which can also be easily verified by simple computation.
\begin{lemma}\label{lem-hyper-sum-condition}
Let $F(a,b;c;z)$ be the Gaussian hypergeometric function as given in \eqref{eqn-Gauss-series}. Then we
have the following
\begin{enumerate}[{\rm(i)}]
\item
For ${\rm Re \,}(c-a-b)>0$ and $c\neq 0,-1, -2, \ldots$,
\begin{align}\label{eq-gauss-sum}
F(a,b;c;1)=\dfrac{\Gamma(c-a-b)\Gamma(c)}{\Gamma(c-a)\Gamma(c-b)}
\end{align}
\item
For $a,b>0$, $c>a+b+1$,
\begin{align}\label{eqlm331}
\sum_{n=0}^\infty \frac{(n+1)(a)_n(b)_n}{(c)_n(1)_n}=F(a,b;c;1)\left[ \frac{ab}{c-1-a-b}%
+1\right] .
\end{align}
\item
For $a\neq 1$, $b\neq 1$ and $c\neq 1$ with $c>\max \{0,a+b-1\}$,
\begin{align}\label{eqlm333}
\sum_{n=0}^{\infty}\frac{(a)_n(b)_n}{(c)_n(1)_{n+1}} = \frac{(c-1)}{(a-1)(b-1)}
\bigg[F(a-1,b-1;c-1;1)- 1\bigg] .
\end{align}
\item
For $a\neq 1$ and $c\neq 1$ with $c>\max \{0,2\mathrm{Re}\,a -1\}$,
\begin{align}\label{eqlm334}
\sum_{n=0}^{\infty} \frac{|(a)_n|^2}{(c)_n(1)_{n+1}} = \frac{(c-1)}{|a-1|^2}
\bigg[ F(a-1,\bar{a}-1;c-1;1)-1 \bigg ].
\end{align}
\end{enumerate}
\end{lemma}
\begin{proof}
Part (i) is the well-known Gauss summation formula.
Part (ii) follows from splitting the left hand side into two parts and applying \eqref{eq-gauss-sum}.
For part (iii), using the fact that $\lambda(\lambda+1)_{m}=(\lambda)_{m+1}$, in place of $a$, $b$ and $c$,
the required result follows. Part (iv) is nothing but Part (iii) with $b=\bar{a}$.
\end{proof}

In order to obtain the objective, we need conditions on the Taylor coefficients of $R_{\gamma,\alpha}^{\tau}(\beta)$
which is given in the following results.

\begin{lemma}\label{th31}
Let $f(z)\in{\mathcal S}$ and is of the form $\eqref{eq:series}$. If $f(z)$ is in
$R_{\gamma,\alpha}^{\tau}(\beta)$, then
\begin{align}\label{eqhyp321}
|a_n|\leq \frac{2|\tau|(1-\beta)}{1+(n-1)(\alpha-2\gamma+\gamma n)}, \quad n=2,3, \ldots.
\end{align}
Equality holds for the function
\begin{align}\label{eqn-coeff-sharp}
f(z)=\dfrac{1}{z^{(1/\nu)-1}}\dfrac{1}{\mu\nu}
\int_0^z \dfrac{1}{t^{\frac{1}{\mu}-\frac{1}{\nu}+1}}\int_0^t w^{\dfrac{1}{1-\frac{1}{\mu}}}
\left(1+\dfrac{2(1-\beta)\tau w^{n-1}}{1-w^{n-1}}\right)dw,
\end{align}
where $\mu+\nu =  \alpha-\gamma$ and $\mu\nu=\gamma$.
\end{lemma}
\begin{proof}
Clearly $f\in R_{\gamma,\alpha}^{\tau}(\beta)$ is equivalent to
%
%Since $f\in P_{\gamma}^{\tau}(\beta)$, we have
\begin{align*}
1+\frac{1}{{\tau}}\left((1-\alpha+2\gamma)\frac{f}{z}+(\alpha-2\gamma)f'+\gamma zf''-1\right)
=\frac{1+(1-2\beta)w(z)}{1-w(z)},
\end{align*}
where $w(z)$ is analytic in $\mathbb{D}$ and satisfies the condition $w(0)=0$,
$|w(z)|<1$ for $z\in\mathbb{D}$. Hence we have
\newline
$\displaystyle
\frac{1}{{\tau}}
\left((1-\alpha+2\gamma)\frac{f}{z}+(\alpha-2\gamma)f'+\gamma zf''-1\right)
$
\begin{align*}
=w(z)\left(2(1-\beta)+
\frac{1}{{\tau}}\left((1-\alpha+2\gamma)\frac{f}{z}+(\alpha-2\gamma)f'+\gamma zf''-1\right)\right).
\end{align*}
Using ($\ref{eq:series}$) and $w(z)=\sum_{n=1}^{\infty}b_nz^n$ we have
\newline
$\displaystyle
\left[2(1-\beta)+\frac{1}{{\tau}}
\left(\sum_{n=2}^{\infty}[1+(\alpha-2\gamma+\gamma n)(n-1)]a_nz^{n-1}\right)\right]
\left[\sum_{n=1}^{\infty}b_nz^n\right]
$
\begin{align*}
=\frac{1}{{\tau}}\sum_{n=2}^{\infty}[1+(\alpha-2\gamma+\gamma n)(n-1)]a_nz^{n-1}.
\end{align*}
Equating the coefficients of the powers of $z^{n-1}$ on both sides of the above equation, it is easy to observe that
the coefficient $a_n$ in right hand side of the above expression depends
only on $a_2,\ldots, a_{n-1}$ and the left hand side of the above expression.
Hence, for $n\geq 2$ this gives
\newline $\displaystyle
\left[2(1-\beta)+\frac{1}{{\tau}}
\left(\sum_{n=2}^{k-1}[1+(\alpha-2\gamma+\gamma n)(n-1)]a_nz^{n-1}\right)\right]w(z)
$
$$
=\frac{1}{{\tau}}\sum_{n=2}^k[1+(\alpha-2\gamma+\gamma n)(n-1)]a_nz^{n-1}
+\sum_{n=k+1}^{\infty}d_nz^{n-1}.
$$
Using $|w(z)|<1$, this reduces to the inequality
\newline $\displaystyle
\left|2(1-\beta)+\frac{1}{{\tau}}
\left(\sum_{n=2}^{k-1}[1+(\alpha-2\gamma+\gamma n)(n-1)]a_nz^{n-1}\right)\right|
$
$$
>\left|\frac{1}{{\tau}}\sum_{n=2}^k[1+(\alpha-2\gamma+\gamma n)(n-1)]a_nz^{n-1}
+\sum_{n=k+1}^{\infty}d_nz^{n-1}\right|.
$$
Squaring the above inequality and integrating around $|z|=r$,
$0<r<1$,
we get
\newline $\displaystyle
4(1-\beta)^2+\frac{1}{{|\tau|^2}}
\left(\sum_{n=2}^{k-1}[1+(\alpha-2\gamma+\gamma n)(n-1)]^2|a_n|^2r^{2(n-1)}\right)
$
\begin{align*}
>\frac{1}{{|\tau|^2}}\sum_{n=2}^k[1+(\alpha-2\gamma+\gamma n)(n-1)]^2|a_n|^2 r^{2(n-1)}
+\sum_{n=k+1}^{\infty}|d_n|^2 r^{2(n-1)}.
\end{align*}
and letting $r\rightarrow 1$
we obtain
$$
4(1-\beta)^2\geq \frac{1}{|{\tau}|^2}[1+(\alpha-2\gamma+\gamma n)(n-1)]^2|a_n|^2
$$
which
gives the desired result.
For sharpness, consider the function
\begin{align*}
(1-\alpha+2\gamma)\frac{f}{z}+(\alpha-2\gamma)f'+\gamma zf''=1+\dfrac{2(1-\beta)\tau z^{n-1}}{1-z^{n-1}}:=p(z).
\end{align*}
Simplifying and using the fact
$\mu+\nu =  \alpha-\gamma$ and $\mu\nu=\gamma$ gives \eqref{eqn-coeff-sharp}.
\end{proof}

\begin{remark}
The condition given in \eqref{eqhyp321} is equivalent to the condition
\begin{align}\label{eqhyp321-equivalent}
|a_n|\leq \frac{2|\tau|(1-\beta)}{1+\alpha(n-1)+\gamma(n-1)(n-2)}, \quad n=2,3, \ldots,
\end{align}
which will be used in the sequel.
\end{remark}
\begin{lemma}\label{th32}
Let $f(z)$ be of the of the form $\eqref{eq:series}$. Then a sufficient
condition for $f(z)$ to be in $R_{\gamma,\alpha}^{\tau}(\beta)$ is
\begin{align}\label{eqn-suff-condition}%{eqhyp322}
\sum_{n=2}^{\infty}[1+(n-1)(\alpha-2\gamma+\gamma n)]|a_n|\leq |\tau|(1-\beta).
\end{align}
This condition is also necessary if $\eta=0$ in \eqref{defn-class} and $a_n<0$ in \eqref{eq:series}.
\end{lemma}
\begin{proof}
Using \eqref{eq:series} it is easy to see that
\newline
$\displaystyle
{\rm Re\,}e^{i\eta}
\left((1-\alpha+2\gamma)\frac{f}{z}+(\alpha-2\gamma)f'+\gamma zf'' -\beta\right)
$
\begin{align*}
=& \quad (1-\beta)\cos\eta +{\rm Re \,} e^{i\eta}\sum_{n=2}^{\infty} \left(1+(\alpha-2\gamma+\gamma n)(n-1)\dfrac{}{}\right)a_nz^{n-1} \\
%\geq & \quad (1-\beta)\cos\eta - \sum_{n=2}^{\infty} \left|\left(1+(\alpha-2\gamma+\gamma n)(n-1)\dfrac{}{}\right)\right| |a_n|\, |z^{n-1}| \\
\geq & \quad (1-\beta)\cos\eta -\sum_{n=2}^{\infty} \left|\left(1+(\alpha-2\gamma+\gamma n)(n-1)\dfrac{}{}\right)\right| |a_n| \geq 0,
\end{align*}
using \eqref{eqn-suff-condition}. The resultant obtained above is
equivalent to the analytic characterization of $f \in R_{\gamma,\alpha}^{\tau}(\beta)$
and the proof is complete.
\end{proof}



\section{Inclusion results for $zF(a,b;c;z)$}\label{sec-incl-zF}
\begin{theorem}\label{th33}
Let $a,b,c$ and $\gamma$ satisfy any one of the following conditions
such that $T_i(a,b,c,\gamma)\leq |\tau|(1-\beta)$
for $i=1,2,3$.
\begin{itemize}
\item[{\rm(i)}] $a,b>0$, $c>a+b+2$ and
{\Small{
\begin{align*}
T_1(a,b,c,\gamma)=
F(a,b;c;1)+\alpha\dfrac{a\,b}{c}
F(a+1,b+1;c+1;1)
+\gamma \dfrac{(a)_2\, (b)_2}{(c)_2}F(a+2,b+2;c+2;1)-1.
\end{align*}
}}
\item[{\rm(ii)}]
$a,b\in{\mathbb{C}}\backslash\{0\}, \, |a|\neq 1, |b|\neq 1, \, c>|a|+|b|+2$
and
\newline
$\displaystyle
T_2(a,b,c,\gamma)
$
{\small{
\begin{align*}
=F(|a|+1,|b|+1;c+1;1)
\left(\alpha\dfrac{|ab|}{c}+\gamma\dfrac{(|a|)_2 (|b|)_2 }{(c)(c-|a|-|b|-2)}+\dfrac{c-|a|-|b|-1}{c}\right)-1.
\end{align*}
}}
\item[{\rm(iii)}]
$-1<a<0$, $-1<b<0$, $c>0$ and
\newline
$\displaystyle
T_3(a,b,c,\gamma)
$
{\small{
\begin{align*}
=F(a+1,b+1;c+1;1)
\left(\alpha\dfrac{ab}{c}+\gamma\dfrac{(a)_2 (b)_2 }{(c)(c-a-b-2)}+\dfrac{c-a-b-1}{c}\right)-1.
\end{align*}
}}
\end{itemize}
Then $zF(a,b;c;z)$ is in $R_{\gamma,\alpha}^{\tau}(\beta)$.
\end{theorem}

\begin{proof}
Clearly $zF(a,b;c;z)$ has the series representation
of the form $\eqref{eq:series}$ where
\begin{align}\label{eq-an}
a_n=\frac{(a)_{n-1}(b)_{n-1}}{(c)_{n-1}(1)_{n-1}}.
\end{align}
Using Lemma $\ref{th32}$, it suffices to prove that
\begin{align*}
\sum_{n=2}^{\infty}[1+(n-1)(\alpha-2\gamma+\gamma n)]|a_n|\leq |\tau|(1-\beta),
\end{align*}
which is equivalent in writing
\begin{align}\label{eqhyp321-equivalent2}
\sum_{n=2}^{\infty}[1+\alpha(n-1)+\gamma (n-1)(n-2)]|a_n|\leq |\tau|(1-\beta) \Longrightarrow f\in R_{\gamma,\alpha}^{\tau}(\beta).
\end{align}
{\underline{Case (i)}}: Let $a,b>0$ and $c>a+b+2$. Then the series in the left hand side of \eqref{eqhyp321-equivalent2} can be written
as
{\Small{
\begin{align*}
S:&
=\sum_{n=2}^{\infty}\bigg(1+\alpha(n-1)+\gamma(n-1)(n-2)\bigg)
\frac{(a)_{n-1}(b)_{n-1}}{(c)_{n-1}(1)_{n-1}}\\ \nonumber
&=\sum_{n=1}^{\infty}\frac{(a)_n(b)_n}{(c)_n(1)_n}
+\alpha\frac{ab}{c}\sum_{n=2}^{\infty}\frac{(a+1)_{n-2}(b+1)_{n-2}}{(c+1)_{n-2}(1)_{n-2}}
+\gamma\dfrac{(a)_2(b)_2}{(c)_2}\sum_{n=3}^{\infty}
\frac{(a+2)_{n-3}(b+2)_{n-3}}{(c+2)_{n-3}(1)_{n-3}}.
\end{align*}
}}
An easy computation by using the hypothesis of the theorem and applying $\eqref{eq-gauss-sum}$,
we get the required result.

\vspace*{6pt}

\noindent
{\underline{Case (ii)}}:
Let $a,b\in{\mathbb{C}}\backslash\{0\}, \, c>|a|+|b|+2$. Since $|(a)_{n}|\leq (|a|)_n$, we have from $\eqref{eqhyp321-equivalent2}$,
\newline
$\displaystyle
S :=\sum_{n=2}^{\infty}\bigg(1+\alpha(n-1)+\gamma(n-1)(n-2)\bigg)|a_n|
$
\begin{align}\label{eqth333}
\leq \frac{|ab|}{c}\sum_{n=0}^{\infty}\frac{(|a|+1)_n(|b|+1)_n}{(c+1)_n(1)_{n+1}}
+\alpha\sum_{n=0}^{\infty}\frac{(|a|)_{n+1}(|b|)_{n+1}}{(c)_{n+1}(1)_{n}}
+\gamma\sum_{n=1}^{\infty} (n-1) \frac{(|a|)_{n}(|b|)_{n}}{(c)_{n}(1)_{n-1}}.
\end{align}
Note that the third sum in the right hand side of \eqref{eqth333} is equivalent to
\newline
$\displaystyle
\sum_{n=0}^{\infty}n\frac{(|a|)_{n+1}(|b|)_{n+1}}{(c)_{n+1}(1)_{n}}
$
\begin{align*}
&=\sum_{n=0}^{\infty}(n+1)\frac{(|a|)_{n+1}(|b|)_{n+1}}{(c)_{n+1}(1)_{n}}-\sum_{n=0}^{\infty}\frac{(|a|)_{n+1}(|b|)_{n+1}}{(c)_{n+1}(1)_{n}}\\
&=\dfrac{|ab|}{c}\sum_{n=0}^{\infty}(n+1)\frac{(|a|+1)_{n}(|b|+1)_{n}}{(c+1)_{n}(1)_{n}}
        -\dfrac{|ab|}{c}\sum_{n=0}^{\infty}\frac{(|a|+1)_{n}(|b|+1)_{n}}{(c+1)_{n}(1)_{n}}.
\end{align*}
Using the above value in \eqref{eqth333} we get that the inequality \eqref{eqth333} is equivalent to
\newline
$\displaystyle
S\leq \frac{|ab|}{c}\sum_{n=0}^{\infty}\frac{(|a|+1)_n(|b|+1)_n}{(c+1)_n(1)_{n+1}}
$
\begin{align}\label{eqth31-split}
+(\alpha-\gamma)\dfrac{|ab|}{c}\sum_{n=0}^{\infty}\frac{(|a|+1)_n(|b|+1)_n}{(c+1)_n(1)_{n}}
+\gamma\dfrac{|ab|}{c}\sum_{n=0}^{\infty}(n+1)\frac{(|a|+1)_{n}(|b|+1)_{n}}{(c+1)_{n}(1)_{n}}.
\end{align}
Now applying \eqref{eqlm333} and the hypothesis of the theorem in the first sum of \eqref{eqth31-split} gives
\begin{align}\label{eqth31-split-1}
\left(\dfrac{c-|a|-|b|-1}{c}F(|a|+1,|b|+1;c+1;1)-1\right).
\end{align}
Similarly applying \eqref{eqlm331} and the hypothesis of the theorem in the third sum of \eqref{eqth31-split} gives
\begin{align}\label{eqth31-split-3}
\dfrac{|ab|}{c}\left(F(|a|+1,|b|+1;c+1,1)\left(\dfrac{(|a|+1)(|b|+1)}{c-|a|-|b|-2}+1\right)\right).
\end{align}
Clearly the second sum of \eqref{eqth31-split} is related to \eqref{eq-gauss-sum} which gives
$\displaystyle \dfrac{|ab|}{c}F(|a|+1,|b|+1;c+1;1)$. Now substituting this resultant and \eqref{eqth31-split-1} and \eqref{eqth31-split-3}
in \eqref{eqth31-split} gives the required result.


\noindent
{\underline{Case (iii)}}:
Let $-1<a<0$, $-1<b<0$ and $c>0$. The result follows by  proceeding in a similar way to the previous case.
\end{proof}

Since the substitution $a={\overline{b}}$ in Theorem $\ref{th33}$ is useful in characterizing polynomials
with positive coefficients when $b$ is some negative integer, we give the corresponding result independently, wherein only the second case can be
applied.

\begin{corollary}\label{cor1}
Let $c>2\,{\rm Re\,} b+2$
and $T_4(b,c,\gamma)\leq |\tau|(1-\beta)$ where
\begin{align*}
T_4(b,c,\gamma)=
F({\overline{b+1}},b+1;c+1;1)
\left(\alpha\dfrac{|b|^2}{c}+\gamma\dfrac{{(|b|)_2}^2}{c(c-2{\rm Re \,} b -2)} + \dfrac{c-2{\rm Re \,} b -1}{c}\right)-1
\end{align*}
Then $zF({\overline{b}},b;c;z)$ is in $R_{\gamma,\alpha}^{\tau}(\beta)$.
\end{corollary}

Note that the results in Corollary $\ref{cor1}$ can also be obtained directly by using $\eqref{eqlm334}$ instead of $\eqref{eqlm333}$,
as used in Theorem \ref{th33}.

Further, if we set $\alpha=1$ and $\gamma =0$, then by choosing $\beta=0$ and $\tau=e^{i\eta}\cos\eta$ with $-\pi/2<\eta<\pi/2$,
we get the functions in the class $R_{\gamma,\alpha}^{\tau}(\beta)$ satisfying the analytic criterion ${\rm Re \,}f'>0$ which implies that
$f(z)$ is close-to-convex with respect to the starlike function $g(z)=z$. Hence the following result is immediate.

\begin{corollary}\label{cor2}
Let $c>2|b-1|+3$ and
\begin{align}\label{eqcor21}
F({\overline{b}}, b;c;1)\leq \dfrac{2(c-1)}{|b-2|^2+c-3},
\end{align}
then $zF({\overline{b}},b;c;z)$ is close-to-convex with respect to the starlike function $g(z)=z$.
\end{corollary}

\begin{remark}\label{rem1}
Corollary $\ref{cor2}$, with the absence of $\alpha$, $\beta$, $\gamma$ and $\tau$, is much
useful, in particular, for extracting polynomials with positive coefficients, which is the main idea behind choosing $a={\overline{b}}$.
Moreover, if we take $b=-m$, then $\eqref{eqcor21}$ gives
\begin{align*}
 F(-m,-m;c;1)\left(\dfrac{m^2+4m+c+1}{2(c-1)}\right)\leq 1 .
\end{align*}
But, when $m$ is sufficiently large, $c$ has to be chosen so large to have the value in the left side bounded by $1$. This
is given by the condition that $c>2m+5$. In the case of $m=2$, $c$ need to be larger than $9$ and
should satisfy $c^3-18c^2-75 c -104 \geq 0$ so that the corresponding polynomial
$ 1+\dfrac{4}{c}z+\dfrac{2}{c(c+1)}z^2$
is close-to-convex. It is easy to see that the condition is satisfied for $c$ more than $21.68057259\ldots$,
which is obtained using mathematical software. Hence if $m$ is chosen as a larger negative
integer then this result is true for polynomials having their coefficients very small, which is not interesting.

Instead, if we consider, Theorem $\ref{th33}$, with either $a=-m$ or $b=-m$ we can still extract polynomials that can have smaller
values of $c$, with coefficients having alternate signs, that satisfy the hypothesis given in Theorem $\ref{th33}$.
\end{remark}

In Theorem $\ref{th33}$, if we take $a=1$, we get the result for
the incomplete beta function $zF(1,b;c;z)$. Since the incomplete beta function plays an important role
in geometric function theory (for example, see \cite{Sw-itsf}),  we give the result for the incomplete beta function independently as
\begin{theorem}\label{th34}
Let $b,c$ and $\gamma$ satisfy any one of the following conditions \newline
such that $T_i(b,c,\gamma) \leq |\tau|(1-\beta)$
for $i=1,2$.
\begin{itemize}
\item[{\rm(i)}] $b>0$, $c>b+3$ and
{\Small{
\begin{align*}
T_1(b,c,\gamma)=
F(1,b;c;1)+\alpha\dfrac{b}{c}
F(2,b+1;c+1;1)
+\gamma \dfrac{2\, (b)_2}{(c)_2}F(3,b+2;c+2;1)-1.
\end{align*}
}}
\item[{\rm(ii)}]
$b\in{\mathbb{C}}\backslash\{0\}, \, c>|b|+3$
and
{\small{
\begin{align*}
T_2(b,c,\gamma)=
F(2,|b|+1;c+1;1)
\left(\alpha\dfrac{|b|}{c}+\gamma\dfrac{2 (|b|)_2 }{(c)(c-|b|-3)}+\dfrac{c-|b|-2}{c}\right)-1.
\end{align*}
}}
\end{itemize}


Then the incomplete beta function $\phi(b;c;z): = zF(1,b;c;z)$
is in $R_{\gamma,\alpha}^{\tau}(\beta)$.
\end{theorem}

\begin{remark}\label{rem2}
Note that at $\alpha=1$, $\gamma=0$, $\beta=0$ and $\tau=e^{i\eta}\cos\eta$ with $-\pi/2<\eta<\pi/2$ the above result reduces to
$c>b+3$, $b>0$. Under these conditions, the normalized incomplete beta function $zF(1,b;c;z)$ is close-to-convex
with respect to the starlike function $g(z)=z$.
\end{remark}

%\section{An Integral operator}
Consider the operator of the form
$\displaystyle G(a,b;c;z):=\int_0^z F(a,b;c;t)dt $.
Then we have
\begin{align*}
G(a,b;c;z)
:= z+\sum\sb{n=2}^{\infty}\frac{(a)_{n-1}(b)_{n-1}}{(c)_{n-1}(1)_{n}}z^{n}
=  z+\sum_{n=2}^{\infty}\dfrac{a_n}{n} z^n,
\end{align*}
where $a_n$ is given as in $\eqref{eq-an}$.
This is the normalized form of the hypergeometric function $F(a,b;c;z)$ which has many interesting properties. Note that
a function may fail to inherit its geometric properties under such normalization. For example, $1+z$ is convex univalent in ${\mathbb{D}}$, whereas
its normalized form $z(1+z)$ is not even univalent.

\begin{theorem}\label{th35}
Let $a,b \in {\mathbb{C}}\backslash\{0\}$ with $|a|\neq 1$, $|b|\neq 1$ and $|c|>|a|+|b|+1$ such that $T(a,b,c,\gamma)\le|\tau|(1-\beta)$ where
\newline
$\displaystyle
T(a,b,c,\gamma)
$
\begin{align*}
=F(a,b;c;1)\bigg(\dfrac{\gamma\,ab}{c-a-b-1}+ \alpha +\dfrac{(1-\alpha+2\gamma)(c-a-b)}{(a-1)(b-1)}\bigg)
-\dfrac{(1-\alpha+2\gamma)(c-1)}{(a-1)(b-1)}.
\end{align*}
Then $G(a,b;c;z)$ is in $R_{\gamma,\alpha}^{\tau}(\beta)$.
\end{theorem}
\begin{proof}
We have
$\displaystyle
G(a,b;c;z)=z+\displaystyle\sum_{n=2}^{\infty}\dfrac{(a)_{n-1}(b)_{n-1}}{(c)_{n-1}(1)_{n}}z^n.
$
So it is sufficient to prove that
\begin{align*}
\sum_{n=2}^{\infty}\left[ 1+\alpha(n-1)+\gamma(n-1)(n-2)\right]|a_n|\le|\tau|(1-\beta).
\end{align*}
The left hand side of the above inequality can be expressed as
%{\small{
\begin{align}\label{eq-int-1}
\sum_{n=1}^{\infty}\dfrac{(a)_n(b)_n}{(c)_n(1)_{n+1}}+\alpha \sum_{n=1}^{\infty} n \dfrac{(a)_{n}(b)_{n}}{(c)_{n}(1)_{n+1}}
+\gamma\sum_{n=1}^{\infty}n(n-1)\dfrac{(a)_{n}(b)_{n}}{(c)_{n}(1)_{n+1}}.
\end{align}
%}}
For the third part \eqref{eq-int-1}, writing $n(n-1)=n(n+1)-2(n+1)+2$ and adding with the second part of \eqref{eq-int-1} gives
\begin{align}\label{eq-int-split}
(1-\alpha+2\gamma)\sum_{n=1}^{\infty}\dfrac{(a)_n(b)_n}{(c)_n(1)_{n+1}}+(\alpha-2\gamma) \sum_{n=1}^{\infty}\dfrac{(a)_{n}(b)_{n}}{(c)_{n}(1)_{n}}
+\dfrac{\gamma a b}{c}\sum_{n=0}^{\infty}\dfrac{(a+1)_{n}(b+1)_{n}}{(c+1)_{n}(1)_{n}}.
\end{align}
Now, using the hypothesis and comparing the first part of \eqref{eq-int-split} with \eqref{eqlm333},
second and third part of \eqref{eq-int-split} with \eqref{eq-gauss-sum} gives the required result upon simplification.
\end{proof}

Check, if at $a={\overline{b}}$ in the above result gives the following Corollary.

\begin{corollary}
Let $a=\overline{b}$, $0<b\neq\,1$ and $c>2\rm{Re}\,b+1$ such that $T(\overline{b},b,c,\gamma)\le|\tau|(1-\beta)$ where
\newline
$
\displaystyle
T(\overline{b},b,c,\gamma)
$
\begin{align*}
=
F(b, {\overline{b}}; c;1)
%\dfrac{\Gamma(c-2\rm{Re}\,b)\Gamma(c)}{\Gamma(c-\overline{b})\Gamma(c-b)}
\bigg(\dfrac{\gamma\,|b|^2(\alpha-2\gamma)}{c-2\rm{Re}\,b-1}
+\dfrac{(1-\alpha+2\gamma)(c-2\rm{Re}b)}{|b-1|^2}\bigg)
-\bigg(\dfrac{(1-\alpha+2\gamma)(c-1)}{|b-1|^2}\bigg)
\end{align*}
Then $G(\overline{b},b;c;z)$ is in $R_{\gamma,\alpha}^{\tau}(\beta)$.
\end{corollary}

\section{Inclusion properties of $H_{\,a,\,b \,c}(f)(z)$}\label{sec-incl-H-f}

Our next interest is to find the inclusion properties of $H_{\,a,\,b \,c}(f)(z) = zF(a,b;c;z)\ast f(z)$, where $f(z)$ is in certain
subclass of $S$. For this, we recall certain subclasses that are necessary for further discussion.
We begin with the following definition.

\begin{definition}\cite{BPS2}
Let $f\in {\mathcal A}$, $0\leq k<\infty$, and $0\leq \sigma<1$.
Then $f\in k-UCV(\sigma) $ if and only if
\begin{align}\label{eqkucv}
{\rm{Re}} \left(1+\frac{zf''(z)}{f'(z)}\right) \geq k
\left|\frac{zf''(z)}{f'(z)}\right|+\sigma .
\end{align}
\end{definition}
This class generalizes various other classes which are worthy to mention here.
The class $k-UCV(0)$, called as $k$-uniformly convex is due to \cite{Kan1},
and has the geometric characterization that for $0\leq k < \infty$, the function $f\in {\mathcal A}$ is
said to be $k$-uniformly convex in ${\mathbb{D}}$, if $f$ is convex in ${\mathbb{D}}$,
and the image of every circular arc $\gamma$ contained in
${\mathbb{D}}$, with center $\zeta$, where $|\zeta|\leq k$, is convex.

The class $1-UCV(0)=UCV$ \cite{Good-ucv} (see also \cite{Ronn2}) describes geometrically the domain of
values of the expression $\displaystyle p(z)=1+\frac{z f''  (z)}{f'(z)}, z\in {\mathbb{D}},$
as $f\in UCV$ if and only if $p$ is in the conic region
$$
\Omega = \{ \omega \in \mathbb C: ({\rm Im }\omega)^2 < 2~{\rm Re\,
}\omega -1 \}.
$$
Using Alexander transform a related class $k-{\mathcal S}_p(\sigma)$ is obtained as
$f\in k-UCV(\sigma)
\Longleftrightarrow zf' \in k-{\mathcal S}_p(\sigma)$.
%The classes $UCV$ and $S_p$ :($1-{\mathcal S}_p(0)$),
%are unified and studied using certain fractional
%calculus operator in \cite{hmsri2}. We refer to \cite{hmsri3, Kan1, Kan2, Ronn2}
%and references therein for some interesting results in these directions.
Results for the condition on the Taylor coefficients of functions in these classes are available in the literature. Among them, we
mention the results that serve our purpose.

\begin{lemma}\label{lm:ucvab}\cite{BPS2}
A function $f\in{\mathcal A}$ is in $k-UCV(\sigma)$ if it satisfies the
condition
\begin{align}\label{eq:ucvab}
\sum\sb{n=2}^{\infty} n \left[
n(1+k)-(k+\sigma)\right]|a_{n}| \leq 1-\sigma.
\end{align}
\end{lemma}
It was also found that the condition ($\ref{eq:ucvab}$) is
necessary if $f\in {\mathcal{A}}$ given by $\eqref{eq:series}$ has $a_n<0$.
Further that the condition
\begin{align}\label{eq:spab}
\sum\sb{n=2}^{\infty} \left[
n(1+k)-(k+\sigma)\right]|a_{n}| \leq 1-\sigma.
\end{align}
is sufficient for $f$ to be in $k-{\mathcal S}\sb{p}(\sigma)$ and
turns out to be also necessary if $f\in{\mathcal A}$ given by $\eqref{eq:series}$ has $a_n<0$.



\begin{theorem}\label{th51}
Let $f\in {\mathcal A}$ be defined as in $(\ref{eq:series})$.
Suppose that $a,b\in\mathbb C\backslash\{0\}$, $c>|a|+|b|+1$ be such that,
for $k\geq 0$, $0\leq \sigma <1$,
\newline
$\displaystyle
F(|a|+1, |b|+1;c+1;1)
$
\begin{align}\label{eq51}
%\nonumber
%\frac{\Gamma(c-|a|-|b|-1)\Gamma(c)}{\Gamma(c-|a|)\Gamma(c-|b|)}
(|ab|(1+k)+(1-\sigma)(c-|a|-|b|-1)
\leq c (1-\sigma)\left(1+\frac{\alpha-3 \gamma}{2|\tau|(1-\beta)}\right).
\end{align}
Then, for $f \in R_{\gamma,\alpha}^\tau (\beta)$, $0\leq \gamma\leq 1$, $0\leq \alpha \leq 1$ and $0\leq \beta<1$,
$H_{\,a,\,b \,c}(f)(z)\in k-UCV(\sigma)$.
\end{theorem}

\begin{proof}
Let $f\in{\mathcal A}$ be defined as in Theorem \ref{th51}.
%From Lemma $\ref{th31}$ we have that
%\begin{align}\label{eq52}
%f(z)=z+\sum_{n=2}^{\infty}a_nz^n \in R_{\gamma,\alpha}^\tau (\beta) \Longrightarrow
%|a_n|\leq \frac{2|\tau|(1-\beta)}{1+\alpha(n-1)+\gamma(n-1)(n-2)}.
%\end{align}
Considering ($\ref{eq:ucvab}$), from Lemma $\ref{th31}$, we need to prove that
if $f\in {\mathcal A}$ satisfies ($\ref{eqhyp321}$), then
\begin{align}\label{eq53}
\sum_{n=2}^{\infty}n\left(\frac{}{}n(1+k)-(k+\sigma)\right)|A_n|\leq 1-\sigma,
\end{align}
where
$$
A_n=\frac{(a,n-1)(b,n-1)}{(c,n-1)(1,n-1)}  a_n, \quad \quad n\geq 2.
$$
Since $1+\alpha(n-1)+\gamma(n-1)(n-2) \geq n(\alpha -3\gamma)$ for $0\leq\gamma\leq 1$ and $n\geq 2$,
using $|(a,n)|\leq (|a|,n)$ it is enough if we prove that
\begin{align*}
%\quad \quad \quad \quad
T:=\sum_{n=2}^{\infty} n\frac{(n)(1+k)-(k+\sigma)}{n}
\frac{(|a|,n-1)(|b|,n-1)}{(|c|,n-1)(1,n-1)}
\leq \frac{(1-\sigma)(\alpha - 3\gamma)}{2|\tau|(1-\beta)}.
\end{align*}
Using $(n+2)(1+k)-(k+\sigma)=(n+1)(1+k)+(1-\sigma)$ and
$$
F(a,b;c;1) =\sum_{n=0}^{\infty}\frac{(a,n)(b,n)}{(c,n)(1,n)}
=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)},
\quad \quad
{\rm Re \,}(c-a-b)>0,
$$
we get,
\begin{align*}
T &= & (1+k)\sum_{n=0}^{\infty}(n+1)\frac{(|a|,n+1)(|b|,n+1)}{(c,n+1)(1,n+1)}
        + (1-\sigma)\sum_{n=0}^{\infty}\frac{(|a|,n+1)(|b|,n+1)}{(c,n+1)(1,n+1)}\\
%  &= & (1+k)\frac{ab}{c}\sum_{n=0}^{\infty}\frac{(|a|+1,n)(|b|+1,n)}{(c+1,n)(1,n)}
%        + (1-\sigma)\sum_{n=1}^{\infty}\frac{(|a|,n)(|b|,n)}{(c,n)(1,n)}\\
  &= & (1+k)\frac{ab}{c}\left(\frac{\Gamma(c-|a|-|b|-1)
        \Gamma(c+1)}{\Gamma(c-|a|)\Gamma(c-|b|)}\right)
        + (1-\sigma)\left(\frac{\Gamma(c-|a|-|b|)\Gamma(c)}{\Gamma(c-|a|)\Gamma(c-|b|)}-1
        \right)\\
  &= & \left(\frac{\Gamma(c-|a|-|b|-1)\Gamma(c)}{\Gamma(c-|a|)\Gamma(c-|b|)}\right)
        \left(\frac{}{}|ab|(1+k)+(1-\sigma)(c-|a|-|b|-1)\right)-(1-\sigma),
\end{align*}
which by using the hypothesis, gives the required result.
\end{proof}

Another sufficient condition for the class $k-UCV$ is also given
in \cite{Kan1} by the following result.
\begin{lemma}\cite{Kan1}\label{lem52}
Let $f\in{\mathcal S}$ and has the form $(\ref{eq:series})$.
If for some $k$, $0\leq k <\infty$, the inequality
\begin{align}\label{eq:kan}
\sum_{n=2}^{\infty}n(n-1)|a_n|\leq \frac{1}{k+2},
\end{align}
holds, then $f\in k-UCV$. The number $1/(k+2)$ cannot be increased.
\end{lemma}

It is interesting to observe that, even though $\sigma$ is not involved in
this sufficient condition, this condition holds for $f\in k-UCV(\sigma)$, by
the method of proof given for Lemma \ref{lem52} in \cite{Kan1}.
Also that,  using the Alexander transform,
a result for $f\in k-S_p(\sigma)$ analogous to ($\ref{eq:kan}$) cannot be obtained by
replacing $a_n$ by $a_n/n$ as in many other situations.

To compare the results we are interested in giving a theorem equivalent to
Theorem \ref{th51}, by using ($\ref{eq:kan}$) instead of ($\ref{eq:ucvab}$).
Since $\sigma$ is not involved in ($\ref{eq:kan}$), we present this result
for the case $\sigma=0$ only. The proof of this theorem is similar to
Theorem \ref{th51} and we omit details.

\begin{theorem}\label{th52}
Let $f\in {\mathcal A}$ be defined as in $(\ref{eq:series})$.
Suppose that $a,b\in\mathbb C\backslash\{0\}$, $c>|a|+|b|+1$ be such that,
for $k\geq 0$, $0\leq \alpha <1$,
\begin{align}\label{eq55}
F(|a|+1, |b|+1;c+1;1)\dfrac{|ab|}{c}
%\frac{\Gamma(c-|a|-|b|-1)\Gamma(c)}{\Gamma(c-|a|)\Gamma(c-|b|)} ~~|ab|
\leq \frac{(\alpha - 3 \gamma)}{2|\tau|(1-\beta)(k+2)}.
\end{align}
Then, for $f \in R_{\gamma , \alpha}^ \tau (\beta)$, $0\leq \gamma\leq 1$,$0\leq \alpha \leq 1$  and $0\leq \beta<1$,
$H_{\,a,\,b, \,c\,}(f)(z)\in k-UCV$.
\end{theorem}

If we let  $a={\overline{b}}$ in $F(a,b;c;z)$ we get polynomials
with positive coefficients when $b$ is some negative integer. Hence the
above Theorems are useful in characterizing convex polynomials and we give
the corresponding results independently.

\begin{corollary}\label{cl53}
Let $f\in {\mathcal A}$ be defined as in $(\ref{eq:series})$.
Suppose that $b>0$, $c>2 \rm{Re}b+1$ and $b, c$ satisfy
\begin{align}\label{eq53-cor}
\nonumber
F(b+1, {\overline{b}}+1;c+1;1)
%\frac{\Gamma(c-2 \rm{Re}b-1)\Gamma(c)}{\Gamma(c-b)\Gamma(c-\overline {b})}
(|b|^2 (1+k)+(1-\sigma)(c-2\rm{Re}b-1)\\
\leq c(1-\sigma)\left(1+\frac{\alpha-3 \gamma}{2|\tau|(1-\beta)}\right).
\end{align}
Then, for $f \in R_{\gamma,\alpha}^\tau (\beta)$, $0\leq \gamma\leq 1$, $0\leq \alpha \leq 1$ and $0\leq \beta<1$,
$H_{\,\overline {b},\,b \,c}(f)(z)\in k-UCV(\sigma)$.
\end{corollary}

\begin{corollary}\label{cl54}
Let $f\in {\mathcal A}$ be defined as in $(\ref{eq:series})$.
Suppose that $b>0$, $c>2\rm{Re}{b}+1$ be such that
\begin{align}\label{eq55-cor}
F(b+1, {\overline{b}}+1;c+1;1)
%\frac{\Gamma(c-2\rm{Re}{b}-1)\Gamma(c)}{\Gamma(c-\overline{b})\Gamma(c-b)}
\dfrac{|b|^2}{c}
\leq \frac{(\alpha - 3 \gamma)}{2|\tau|(1-\beta)(k+2)}.
\end{align}
Then, for $f \in R_{\gamma , \alpha}^ \tau (\beta)$, $0\leq \gamma\leq 1$,$0\leq \alpha \leq 1$  and $0\leq \beta<1$,
$H_{\,\overline {b},\,b, \,c\,}(f)(z)\in k-UCV$ where $k\geq 0$.
\end{corollary}

The Hohlov operator $H_{\,a,\,b,\, c}(f)(z)$ reduces to the Carlson-Shaffer operator
${\mathcal L}(b,c)(f)(z)$ if $a=1$. Hence we give the statement of the following results.

\begin{corollary}
Let $f\in {\mathcal A}$ be defined as in $(\ref{eq:series})$.
Suppose that $b>0$, $c>b+2$ are such that, for $k\geq 0$, $0\leq \sigma <1$ and
\begin{align}\label{eq53-cor2}
%\nonumber
\frac{(c-1)}{(c-b-1)(c-b-2)}
(|b| (1+k)+(1-\sigma)(c-b-2)
\leq (1-\sigma)\left(1+\frac{\alpha-3 \gamma}{2|\tau|(1-\beta)}\right).
\end{align}
Then, for $f \in R_{\gamma,\alpha}^\tau (\beta)$, $0\leq \gamma\leq 1$, $0\leq \alpha \leq 1$ and $0\leq \beta<1$,
${\mathcal L}(b,c)(f)(z)\in k-UCV(\sigma)$.
\end{corollary}

\begin{corollary}
Let $f\in {\mathcal A}$ be defined as in $(\ref{eq:series})$.
Suppose that $b>0$, $c>b+2$ are such that, for $k\geq 0$, $0\leq \sigma <1$ and
\begin{align}\label{eq53-cor3}
\dfrac {(\alpha-3\gamma)}{2|{\tau}|(1-\beta)(k+2)}\left(\dfrac{}{}(c-1)^2+(2b+1)(c-1)+b(b+1)\right)-b(c-1)\geq 0.
\end{align}
Then, for $f \in R_{\gamma,\alpha}^\tau (\beta)$, $0\leq \gamma\leq 1$, $0\leq \alpha \leq 1$ and $0\leq \beta<1$,
${\mathcal L}(b,c)(f)(z)\in k-UCV$.
\end{corollary}


Let ${\mathcal S}^{\ast}_{\lambda}$ ($\lambda>0$),
denotes the class of functions in ${\mathcal S}$ such that
$\displaystyle \left|\frac{zf'(z)}{f(z)}-1 \right|<\lambda$.

A sufficient condition
for $f\in{\mathcal A}$ of the form ($\ref{eq:series}$) to be in
${\mathcal S}^{\ast}_1\subset {\mathcal S}^{\ast}$, is given by
$\displaystyle \sum_{n=2}^{\infty}n|a_n|\leq 1$,
and is proved by many authors.
For example, see \cite{G1}. A particular
extension of this, due to \cite{Sil1}, is
\begin{align}\label{eq513}
\sum_{n=2}^{\infty}(n+\lambda-1)|a_n|\leq \lambda
\Longrightarrow f\in {\mathcal S}^{\ast}_{\lambda}.
\end{align}
%We further note that when $f(z)$ is of the form ($\ref{eqnegseries}$),
%the condition ($\ref{eq513}$) is both necessary and sufficient
%for $f\in{\mathcal S}^{\ast}_{\lambda}$.

\begin{theorem}\label{th53}
Let $a,b>0$ or $a\in{\mathbb C}\backslash\{0\}$ with $a={\overline{b}}$.
Further, let $|a|\neq 1$, $|b|\neq 1$,
and $0\neq c>a+b$ be such that
\begin{align}\label{eq514}
%\nonumber
F(a,b;c;1)
\left(1+\dfrac{(\lambda -1)(c-|a|-|b|)}{(|a|-1)(|b|-1)}\right)
 \leq \frac{(\lambda-1)(c-1)}{(|a|-1)(|b|-1)}
   +\lambda\left(1+\frac{(\alpha -3\gamma)}{2|\tau|(1-\beta)}\right).
\end{align}
Suppose that $f\in {\mathcal A}$ be defined as in $(\ref{eq:series})$.
Then, for $f \in R_{\gamma, \alpha}^{\tau}(\beta)$, $0\leq \gamma\leq 1$, $0\leq \alpha \leq 1$, $0\leq \beta<1$,
and $\lambda>0$,
$H_{\,a,\,b, \,c}(f)(z)\in {\mathcal S}^{\ast}_{\lambda}$.
\end{theorem}

\begin{proof}
Let $f(z)$ be of the form ($\ref{eq:series}$).
In view of ($\ref{eq513}$), it suffices to prove that
\begin{align}\label{eq515}
\sum_{n=2}^{\infty}(n+\lambda-1)|A_n|\leq \lambda,
\end{align}
where
$$
A_n=\frac{(a,n-1)(b,n-1)}{(c,n-1)(1,n-1)}\,\,a_n, \quad \quad \quad n\geq 2.
$$
Since $f\in R_{\gamma, \alpha}^{\tau}(\beta)$, using ($\ref{eqhyp321}$) and
$1+\alpha(n-1)+\gamma(n-1)(n-2) \geq n(\alpha -3\gamma)$, we need only to show that
\begin{align*}
T:  =  \sum_{n=2}^{\infty}\frac{(|a|,n-1)(|b|,n-1)}{(c,n-1)(1,n-1)}
    +(\lambda-1)\sum_{n=2}^{\infty}\frac{(|a|,n-1)(|b|,n-1)}{(c,n-1)(1,n)}
    \leq \frac{\lambda\,(\alpha-3 \gamma )}{2|{\tau}|(1-{\beta})}.
\end{align*}
But this last inequality is true by the hypothesis of the theorem and
($\ref{eqlm333}$).
\end{proof}

\section{Inclusion properties of $K_a^p[f](z)$}\label{sec-incl-K-f}
\begin{theorem}
Let $f\in\mathcal{A}$ be as in \eqref{eq:series}. Suppose $a>-1,p\geq0$ and
\begin{equation} \label{j}
\sum_{n=2}^\infty[n(1+k)-(k+\sigma)]B_n(a,p)\leq\frac{(1-\sigma)(\alpha-3\gamma)}{2|\tau|(1-\beta)},
\end{equation}
where $B_n(a,p)=\frac{(1+a)^p}{(n+a)^p}$. Then for $f\in\mathcal{R}_{\gamma,\alpha}^\tau(\beta), 0\leq\gamma\leq1, 0\leq\alpha\leq1$ and $0\leq\beta<1$, we have $K_a^p[f](z)\in k-UCV(\sigma)$.
\end{theorem}
\begin{proof}
Since $f\in\mathcal{R}_{\gamma,\alpha}^\tau(\beta)$, we have from Lemma $\ref{th31}$ and the fact $1+\alpha(n-1)+\gamma(n-1)(n-2)\geq n(\alpha-3\gamma),n\geq2$ that
\begin{align*} \label{k}
|a_n|\leq\frac{2|\tau|(1-\beta)}{n(\alpha-3\gamma)}.
\end{align*}
Now using Lemma $\ref{lm:ucvab}$, it is enough to show that
\begin{align*}
\sum_{n=2}^{\infty} n\left[\dfrac{}{}n(1+k)-(k+\sigma)\right]|A_n|\leq 1-\sigma,
\end{align*}
where $A_n=B_n(a,p)a_n$. Clearly, the above inequality is true if (\ref{j}) holds.
\end{proof}
\noindent
It is easy to see that, for all $n\geq 2$,
\begin{align*}
B_n(a,p)=\frac{(1+a)^p}{(n+a)^p}<1, \qquad a>-1, \quad p\geq0
\end{align*}
which leads to
\begin{corollary}
Let $f\in\mathcal{A}$ be as in \eqref{eq:series}. Suppose $a>-1,p\geq0$ and
$$\sum_{n=2}^\infty[n(1+k)-(k+\sigma)]\leq\frac{(1-\sigma)(\alpha-3\gamma)}{2|\tau|(1-\beta)}.$$
%where $B_n(a,p)=\frac{(1+a)^p}{(n+a)^p}$.
Then for $f\in\mathcal{R}_{\gamma,\alpha}^\tau(\beta), 0\leq\gamma\leq1, 0\leq\alpha\leq1$ and $0\leq\beta<1$, we have $K_a^p[f](z)\in k-UCV(\sigma)$.
\end{corollary}

\begin{theorem}
Let $p\geq0$, $a>-1$ and $f\in\mathcal{A}$ be as in \eqref{eq:series}. Suppose that
\begin{equation} \label{m}
\sum_{n=2}^\infty[n+\lambda-1]\frac{B_n(a,p)}{n}\leq\frac{\lambda(\alpha-3\gamma)}{2|\tau|(1-\beta)},
\end{equation}
where $B_n(a,p)=\dfrac{(1+a)^p}{(n+a)^p}$. Then for $f\in\mathcal{R}_{\gamma,\alpha}^\tau(\beta), 0\leq\gamma\leq1, 0\leq\alpha\leq1$ and $0\leq\beta<1$, we have $K_a^p[f](z)\in\mathcal{S}_\lambda^*$.
\end{theorem}
\begin{proof}
Since $f\in\mathcal{R}_{\gamma,\alpha}^\tau(\beta)$, Lemma $\ref{th31}$ gives
\begin{align*}
|a_n|\leq\frac{2|\tau|(1-\beta)}{1+\alpha(n-1)+\gamma(n-1)(n-2)}.
\end{align*}
Using the fact that $1+\alpha(n-1)+\gamma(n-1)(n-2)\geq n(\alpha-3\gamma),n\geq2$, we obtain
\begin{equation} \label{n}
|a_n|\leq\frac{2|\tau|(1-\beta)}{n(\alpha-3\gamma)}.
\end{equation}
Now $K_a^p[f](z)\in\mathcal{S}_\lambda^*$ if
\begin{align*}
                  &\sum_{n=2}^\infty[n+\lambda-1]\Big{|}\frac{(1+a)^p}{(n+a)^p}a_n\Big{|}\leq\lambda\\
\Longrightarrow\quad    &\sum_{n=2}^\infty[n+\lambda-1]\frac{(1+a)^p}{(n+a)^p}\big{|}a_n\big{|}\leq\lambda\\
\Longrightarrow\quad &\sum_{n=2}^\infty[n+\lambda-1]\frac{(1+a)^p}{(n+a)^p}\frac{2|\tau|(1-\beta)}{n(\alpha-3\gamma)}\leq\lambda,\quad \quad \mbox{using }\eqref{n}\\
\Longrightarrow\quad &\sum_{n=2}^\infty[n+\lambda-1]\frac{(1+a)^p}{(n+a)^p}\frac{1}{n}\leq\frac{\lambda(\alpha-3\gamma)}{2|\tau|(1-\beta)},
\end{align*}
which is the hypothesis and the proof is complete.
\end{proof}

\section{Concluding remarks}

\begin{remark}
If $k=0$ then it is clear from the analytic characterization that $k-UCV(\sigma)$
reduces to the class of Convex functions of order $\sigma$, denoted by
${\mathcal C}(\sigma)$. Similarly, (using  Alexander transform),
$k-{\mathcal S}_p(\sigma)$ reduces to the class of Starlike
functions of order $\sigma$, (${\mathcal S}^{\ast}(\sigma)$). For results regarding
to these classes we refer to \cite{G1}. Further results on the restriction
$k=0$ can be found in the literature, e.g. see
\cite{Kan1}.
\end{remark}

\begin{remark}
We note that Theorem \ref{th51} and Theorem \ref{th52} are not sharp.
In particular, for $a, b$ real with $\eta=0$, $k=0$ and $\sigma=0$, we get from ($\ref{eqhyp321}$),
\begin{align}\label{eq56}
F(|a|+1, |b|+1; c+1;1)\dfrac{|ab|}{c}+(c-|a|-|b|-1) \leq 1+ \dfrac{\alpha-3\gamma}{2(1-\beta)}.
\end{align}
This inequality for $\alpha =1$ and $\gamma=0$ further reduces to
\begin{align}\label{eq57}
F(|a|+1, |b|+1; c+1;1)\dfrac{|ab|}{c}+(c-|a|-|b|-1) \leq 1+ \dfrac{1}{2(1-\beta)}.
\end{align}
Similarly,  $\eqref{eq55}$ reduces to
\begin{align}\label{eq58}
F(|a|+1, |b|+1; c+1;1)\dfrac{|ab|}{c} \leq \dfrac{1}{4(1-\beta)}.
\end{align}
From $\eqref{eq57}$ and $\eqref{eq58}$, it is easy to see that
Theorem \ref{th52} is better for all $c$ lying between $|a|+|b|+1$ and $|a|+|b|+\dfrac{3}{2}$ and for all
other values of $c$ satisfying $c> |a|+|b|+\dfrac{3}{2}$, Theorem \ref{th51} is better.
\end{remark}

Note that, in Theorem \ref{th53}, $|a|\neq 1$ and $|b|\neq 1$. Hence Theorem \ref{th53} cannot be reduced to the important transforms
such as Carlson--Schaffer integral operator, which leads to the following.

\begin{problem}\label{prob:CarlsonOperator}
To find conditions on $b$ and  $c$ such that the Carlson--Schaffer operator ${\mathcal{L}}(b,c)(f)(z)$ maps the class
$R_{\gamma,\alpha}^{\tau}(\beta)$ onto $S_{\lambda}^*$.
\end{problem}

Note that, for $p=1$, the results given in Section $\ref{sec-incl-K-f}$ for the Komatu operator $K_a^p[f](z)$ reduce to the results for the Bernardi integral operator and coincide with the results of Section $\ref{sec-incl-H-f}$ for particular values of $a$, $b$ and $c$. However, for no values of $p$ or $a$, the Komatu operator $K_a^p[f](z)$ can be reduced to the Carlson--Schaffer operator ${\mathcal{L}}(b,c)(f)(z)$. Hence Problem $\ref{prob:CarlsonOperator}$ gains further significance.

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