Certain Properties for a Class of Analytic Functions Associated with Hypergeometric Functions

Khadeejah Rasheed Alhindi$^\ast $ Maslina Darus$^\S $

February 24, 2014.
Published online: January 23, 2015.

$^\ast $School of Mathematical Sciences, Faculty of Science and Technology, Universiti Kebangsaan Malaysia, 43600 Bangi, Selangor, Malaysia, e-mail: m.kh83@yahoo.co.uk. The work of this author has been supported by Zamalah and GUP-2013-004

$^\S $School of Mathematical Sciences, Faculty of Science and Technology, Universiti Kebangsaan Malaysia, 43600 Bangi, Selangor, Malaysia, e-mail: maslina@ukm.edu.my (Corresponding author). The work of this author has been supported by GUP-2013-004

In this particular paper, we investigate coefficient inequalities, closure theorems, convolution properties for the functions belonging to the class $ \mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta )$. Further, integral transforms of functions in the same class are also discussed.

MSC. 30C45, 33D15, 05A30

Keywords. Analytic function, coefficient inequalities, Schwarz inequality, closure theorems, Hadamard product, convolution properties, integral operator.

1 Introduction

Let $\mathcal{A}$ be the class of analytic functions of the form

\begin{equation} \label{one} f(z)=z+\sum _{k=2}^{\infty } a_k z^k;\mathrm\qquad z\in (\mathbb {U}=\{ z\in {C}:|z|<1\} ) \end{equation}

and $\mathcal{S}$ be the subclass of $\mathcal{A}$ consisting of univalent functions, and $\mathcal{S(\alpha )}$, $\mathcal{C(\alpha )}$ $(0 {\lt} \alpha \leq 1)$ denote the subclasses of $\mathcal{A}$ consisting of functions that are starlike of order $\alpha $ and convex of order $\alpha $ in $\mathbb {U}$, respectively.

For two analytic functions $ f(z)=z+\sum _{k=2}^{\infty } a_k z^k $ and $ g(z)=z+\sum _{k=2}^{\infty } b_k z^k$ in the open unit disc $\mathbb {U}=\{ z\in {C}:|z|{\lt}1\} $, the Hadamard product (or convolution) $f\ast g$ of $f$ and $g$ is defined by

\begin{equation} \label{hadamard} f(z)\ast g(z) = (f*g)(z)=z+\sum _{k=2}^{\infty } a_k b_k z^k. \end{equation}

For complex parameters $\alpha _1,...\alpha _r$ and $\beta _1,...\beta _s$ $(\beta _j\neq 0,-1,-2,...;j=1...s)$, Dziok and Srivastava [ 1 ] defined the generalized hypergeometric function

\[ _rF_s(\alpha _1,...,\alpha _r;\beta _1,...,\beta _s;z) \]

\begin{equation} \label{F} _rF_s(\alpha _1,...,\alpha _r;\beta _1,...,\beta _s;z)=\sum _{k=0}^{\infty }\tfrac {(\alpha _1)_k...(\alpha _r)_k}{(\beta _1)_k,...,(\beta _s)_k }\tfrac {z^k}{k!}; \end{equation}

\begin{equation} (r\leq s+1; r,s \in \mathbb {N}_0= \mathbb {N} \cup {0}; z \in \mathcal{U}), \end{equation}

where $(x)_k$ is the Pochhammer symbol defined, in terms of Gamma function $\Gamma $, by

\begin{equation*} (x)_k =\tfrac {\Gamma (x+k)}{\Gamma (x)}= \begin{cases} 1, & \text{if } k=0,\\ x(x+1)...(x+k-1), & \text{if } k\in \mathbb {N}. \end{cases}\end{equation*}

Dziok and Srivastava [ 1 ] defined also the linear operator

\begin{equation} \label{H} H(\alpha _1,...,\alpha _r;\beta _1,...,\beta _s)f(z)= z + \sum _{k=2}^{\infty } \Gamma _k a_k z^k \end{equation}

where

\begin{equation} \label{g} \Gamma _k = \frac{(\alpha _1)_{k-1}...(\alpha _r)_{k-1}}{(\beta _1)_{k-1},...,(\beta _s)_{k-1}(k-1)!}. \end{equation}

Al-Abbadi and Darus [ 2 ] defined the analytic function

\begin{equation} \Phi ^m_{\lambda _1,\lambda _2}=z+\sum _{k=2}^{\infty }\tfrac {(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}z^k, \end{equation}

where $m\in \mathbb {N}_0 = \{ 0,1,2,....\} $ and $ \lambda _2 \geq \lambda _1 \geq 0$.

Using the Hadamard product (2), we can derive the generalized derivative operator $\mathcal{K}^{m,r,s}_{\lambda _1,\lambda _2} $ as follows

\begin{equation} \mathcal{K}^{m,r,s}_{\lambda _1,\lambda _2}f(z)=z+\sum _{k=2}^{\infty }\tfrac {(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m} {\Gamma _k} a_k z^k \end{equation}

where $ \Gamma _k$ is as given in (8).

Remark 1

When ($\lambda _1=\lambda _2=0$), ($\lambda _1=m=0$) or ($\lambda _2=0$ and $m=1$) we get Dziok-Srivastava operator [ 1 ] .

Also there are three cases to get the Hohlov operator [ 3 ] , by giving ($\lambda _1=\lambda _2=0,\alpha _i=0,\beta _j=0 $), ($\lambda _1=m=0,\alpha _i=0,\beta _j=0$) or ($\lambda _2=0, m=1, \alpha _i=0, \beta _j=0$) where ($i=1...r $ and $j=1...s$).

Putting ($\lambda _1=\lambda _2=0, \alpha _2=1,\alpha _3=...=\alpha _r=0,\beta _2=...=\beta _s=0$), ($\lambda _1=m=0, \alpha _2=1,\alpha _3=...=\alpha _r=0,\beta _2=...=\beta _s=0$) or $(\lambda _2=0, m=1, \alpha _2=1,\alpha _3=...=\alpha _r=0,\beta _2=...=\beta _s=0)$, we obtain the Carlson-Shaffer operator [ 4 ] .

There are six cases to get the Ruscheweyh operator [ 5 ] as follows: ($\lambda _1=\lambda _2=0,\alpha _2=\alpha _3=...=\alpha _r=0,\beta _1=\beta _2=...=\beta _s=0$), ($\lambda _1=m=0,\alpha _2=\alpha _3=...=\alpha _r=0,\beta _1=\beta _2=...=\beta _s=0$), ($\lambda _2=0, m=1,\alpha _2=\alpha _3=...=\alpha _r=0,\beta _1=\beta _2=...=\beta _s=0$),($\lambda _1=\lambda _2=0, \alpha _2=\alpha _3=...=\alpha _r=0,\beta _2=...=\beta _s=0$), ($\lambda _1=m=0, \alpha _2=\alpha _3=...=\alpha _r=0,\beta _2=...=\beta _s=0$) or $(\lambda _2=0, m=1, \alpha _2=\alpha _3=...=\alpha _r=0,\beta _2=...=\beta _s=0)$.

If($\lambda _2=0,m=2,\alpha _2=\alpha _3=...=\alpha _r=0,\beta _1=\beta _2=...=\beta _s=0$), we get the generalized Ruscheweyh derivative operator as well [ 6 ] .

Moreover, if we put ($\alpha _2=\alpha _3=...=\alpha _r=0,\beta _1=\beta _2=...=\beta _s=0$) or ($\alpha _2=\alpha _3=...=\alpha _r=0,\beta _1=\beta _2=...=\beta _s=0$), we can get operator given by Al-Abbadi and Darus [ 2 ] .

After that, if($\lambda _2=0, m=m+1, \alpha _2=\alpha _3=...=\alpha _r=0,\beta _1=\beta _2=...=\beta _s=0$), we get the generalized derivative operator by Al-Shaqsi and Darus [ 7 ] .
â–¡

Definition 2

Let $f\in \mathcal{A}$. Then $f(z)\in \mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta )$ if and only if

\begin{equation} \Re \bigg\{ \frac{z[\mathcal{K}^{m,r,s}_{\lambda _1,\lambda _2}f(z)]'}{\mathcal{K}^{m,r,s}_{\lambda _1,\lambda _2}f(z)}\bigg\} >\eta ,\quad 0\leq \eta <1, z\in \mathcal{U}. \end{equation}

In this present paper, we obtain the coefficient inequalities, closure theorems, convolution properties for the functions belonging to the class $ \mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta )$. Finally, the preserving integral operators of the form

\begin{equation} G_c(z)=c\int _0^1u^{c-2}f(uz)du; \quad \quad (c>0) \end{equation}

for the class $\mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta )$ is considered. We employ techniques similar to those used earlier by [ 8 ] .

2 Coefficient Estimate for the Class $\mathcal{S}^{\lowercase {m,r,s}}_{\lambda _1,\lambda _2}(\eta )$

Theorem 3

Let $f(z)\in \mathcal{A}$. If

\begin{equation} \label{co} \sum _{k=2}^\infty (k-\eta )\Big[\tfrac {(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\Big] {\Gamma _k} |a_k|\leq 1-\eta ,\qquad 0\leq \eta \leq 1 \end{equation}

then $f(z)\in \mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta )$. The result (13) is sharp.

Proof â–¼

Suppose that (13) holds. Since

\begin{eqnarray*} 1-\eta & \geq & \sum _{k=2}^\infty (k-\eta ) \Big[\tfrac {(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\Big] {\Gamma _k} |a_k|\nonumber \\ & \geq & \sum _{k=2}^\infty \eta \Big[\tfrac {(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\Big] {\Gamma _k} |a_k|- \sum _{k=2}^\infty k \Big[\tfrac {(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\Big] {\Gamma _k} |a_k|\nonumber \\ \end{eqnarray*}

we deduce that

\begin{equation*} \tfrac {1+\sum \limits _{k=2}^\infty k \big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k} |a_k|}{1+\sum \limits _{k=2}^\infty \big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k} |a_k|} {\gt} \eta . \end{equation*}

Thus

\[ \Re \Bigg\{ \frac{z\big[\mathcal{K}^{m,r,s}_{\lambda _1,\lambda _2}f(z)\big]'}{\mathcal{K}^{m,r,s}_{\lambda _1,\lambda _2}f(z)}\Bigg\} {\gt}\eta , \qquad 0\leq \eta {\lt}1, z\in \mathcal{U}. \]

We note that the assertion (13) is sharp, moreover, the extremal function can be given by

\begin{equation*} f(z)=z+\sum _{k=2}^\infty \tfrac {(1-\eta )}{(k-\eta )\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}}z^k. \end{equation*}

Corollary 4

If the hypotheses of Theorem 3 is satisfied, then

\begin{equation} |a_k| \leq \tfrac {(1-\eta )}{(k-\eta )\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}},\qquad \forall k\geq 2. \end{equation}

3 Closure Theorems

Let the functions $f_j(z)$ be defined by

\begin{equation} \label{clo} f_j(z)=z+\sum _{k=2}^\infty a_{k,j}z^{k};\quad \quad \quad (a_{k,j}\geq 0 , z \in \mathcal{U}). \end{equation}

Theorem 5

Let the functions $f_j(z)$ defined by (13) be in the class $\mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta )$ for every $j=1,2,...,l$. Then the function $G(z)$ defined by

\begin{equation*} G(z)= z+\sum _{k=2}^\infty b_{k}z^{k};\quad \quad \quad (b_{k}\geq 0 , z \in \mathcal{U}) \end{equation*}

is a member of the class $\mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta )$, where

\begin{equation*} b_k=\tfrac {1}{l}\sum _{j=1}^l a_{k,j};\quad \quad \quad (k\geq 2). \end{equation*}

Proof â–¼

Since $f_j(z)\in \mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta )$, from Theorem 3 we can write

\begin{equation*} \sum _{k=2}^\infty (k-\eta )\Big[\tfrac {(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\Big] {\Gamma _k} |a_{k,j}|\leq 1-\eta , \qquad 0\leq \eta \leq 1 \end{equation*}

for every $j=1,2,...,l$. Thus

\begin{equation*} \begin{split} & \sum _{k=2}^\infty (k-\eta )\Big[\tfrac {(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\Big] {\Gamma _k} |b_{k}|=\ \\ & = \sum _{k=2}^\infty (k-\eta )\Big[\tfrac {(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\Big] {\Gamma _k}\bigg|\tfrac {1}{l}\sum _{j=1}^l a_{k,j}\bigg| \\ & \leq \tfrac {1}{l}\sum _{j=1}^l\Bigg(\sum _{k=n}{\infty }(k-\eta )\Big[\tfrac {(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\Big] {\Gamma _k} |a_{k,j}|\Bigg) \\ & = \tfrac {1}{l}\sum _{j=1}^l( 1-\eta )=( 1-\eta ). \end{split}\end{equation*}

In view of Theorem 3, we conclude that $G(z)\in \mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta )$.

Theorem 6

The class $ \mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta )$ is closed under convex linear combination.

Proof â–¼

Let $f_j(z)$ defined by (13) be belonged to $\mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta )$ for every $j=1,2,...,l$, it is sufficient to prove that the function

\begin{equation*} h(z)=\mu f_1(z)+(1-\mu )f_2(z) \end{equation*}

is also in the class $\mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta )$.

Let us write, for $0 \leq \mu \leq 1$,

\begin{equation*} h(z)=z+\sum _{k=n}^\infty \{ \mu a_{k,1}+(1-\mu )a_{k,2}\} z^k, \end{equation*}

we note that

\begin{align*} & \sum _{k=n}^{\infty }(k-\eta )\Big[\tfrac {(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\Big] {\Gamma _k}\Big|\mu a_{k,1}+(1-\mu )a_{k,2} \Big|=\ \\ & \leq \sum _{k=n}^{\infty }(k-\eta )\Big[\tfrac {(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\Big] {\Gamma _k}|\mu a_{k,1}|+\\ & \quad +\sum _{k=n}^{\infty }(k-\eta )\Big[\tfrac {(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\Big] {\Gamma _k}|(1-\mu )a_{k,2}| \\ & =\mu \sum _{k=n}^{\infty }(k-\eta )\Big[\tfrac {(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\Big] {\Gamma _k}|a_{k,1}|\\ & \quad +(1-\mu )\sum _{k=n}^{\infty }(k-\eta )\Big[\tfrac {(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\Big] {\Gamma _k}|a_{k,2}|\\ & \leq \mu (1-\eta )+(1-\mu )(1-\eta )= (1-\eta ). \end{align*}

It follows from Theorem 3 that $ h(z)\in \mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta )$, which completes the proof.

Theorem 7

Let

\begin{equation*} f_0(z)=z \end{equation*}

and

\begin{equation*} f_k(z)=z+\tfrac {(1-\eta )}{(k-\eta )\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}}z^k. \end{equation*}

Then $ f(z)\in \mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta )$ if and only if it can be expressed in the form

\begin{equation} f(z)=\sum _{k=0}^\infty \rho _kf_k(z) \end{equation}

where $\rho _k\geq 0$ and $\sum _{k=0}^\infty \rho _k =1$.

Proof â–¼

Firstly, suppose that

\begin{equation} f(z)=\sum _{k=0}^\infty \rho _k f_k(z) \end{equation}

where $\rho _k\geq 0$ and $\sum _{k=0}^\infty \rho _k =1$. Then

\begin{eqnarray*} f(z) & =& \sum _{k=0}^\infty \rho _k f_k(z) = \rho _0f_0(z)+\sum _{k=1}^\infty \rho _k f_k(z)\\ & =& z+ \tfrac {(1-\eta )}{(k-\eta )\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}}\rho _{k} z^k. \end{eqnarray*}

We observe that

\begin{equation*} \begin{split} & \sum _{k=n}^{\infty }(k-\eta )\Big[\tfrac {(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\Big] {\Gamma _k} \cdot \bigg[\tfrac {(1-\eta )}{(k-\eta )\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}}\bigg]\rho _{k}=\ \\ & = (1-\eta )\sum _{k=1}^\infty \rho _k = (1-\eta )(1-\rho _0)\leq (1-\eta ). \end{split}\end{equation*}

In view of Theorem 3, we conclude that $f(z)\in \mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta )$.

Conversely, let us suppose that $f(z)\in \mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta )$. Since

\begin{equation} a_k \leq \tfrac {(1-\eta )}{(k-\eta )\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}}, \qquad \forall k\geq 2. \end{equation}

Then by Corollary 4, we set

\begin{equation*} \rho _k = \tfrac {(k-\eta )\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}}{(1-\eta )}a_k, \end{equation*}

and

\begin{equation*} \rho _0 = 1- \sum _{k=1}^{\infty }\rho _k. \end{equation*}

We thus conclude that $ f(z)=\sum _{k=0}^\infty \rho _k f_k(z)$. This completes the proof of the theorem.

4 Convolution Properties

For functions $f_j(z)\in \mathcal{A};(j=1,2,...,m)$ given by

\begin{equation*} f_j(z)= z+ \sum _{k=2}^{\infty }a_{k,j}z^k;\quad \quad \quad (z\in \mathcal{U}), \end{equation*}

the Hadamard product (or convolution) of $f_1(z),f_2(z),...,f_m(z)$ is defined by

\begin{equation*} G_m(z)=(f_1\ast f_2\ast ...\ast f_m)(z)=z+\sum _{k=2}^\infty \Big( \textstyle \prod \limits _{j=1}^m a_{k,j}z^k\Big). \end{equation*}

Theorem 8

If $f_j(z)\in \mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta )$ for each $(j=1,2,...,m)$, then $G_m(z)\in \mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta )$ with

\begin{equation*} \eta ^*=\tfrac {\prod \limits _{j=1}^m(1-\eta _j)}{\prod \limits _{j=1}^m(2-\eta _j)\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}-\prod \limits _{j=1}^m(1-\eta _j)}. \end{equation*}

Proof â–¼

We use the mathematical induction to get to the required result. Firstly, we have to show that $G_2(z)\in \mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta )$ for $f_1(z)$ and $f_2(z)$ belonging to $\mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta _1)$ , $\mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta _1)$ respectively. We can write

\begin{equation*} \sum _{k=2}^\infty \tfrac {(k-\eta _j)\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}}{(1-\eta _j)} |a_{k,j}|\leq 1;\quad \quad \quad (j=1,2). \end{equation*}

Applying the Schwarz inequality, we have the following inequality

\begin{equation*} \sum _{k=2}^\infty \sqrt{\tfrac {(k-\eta _1)\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}}{(1-\eta _1)}\tfrac {(k-\eta _2)\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}}{(1-\eta _2)}} \sqrt{|a_{k,1}|\cdot |a_{k,2}|}\leq 1. \end{equation*}

Then, we will determine the largest $\eta ^*$ such that

\begin{equation*} \sum _{k=2}^\infty \tfrac {(k-\eta ^*)\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}}{(1-\eta ^*)} |a_{k,1}|\cdot |a_{k,2}|\leq 1. \end{equation*}

That is

\begin{equation*} \begin{split} & \sum _{k=2}^\infty \tfrac {(k-\eta ^*)\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}}{(1-\eta ^*)} |a_{k,1}|\cdot |a_{k,2}|\leq \ \\ & \leq \sum _{k=2}^\infty \sqrt{\tfrac {(k-\eta _1)\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}}{(1-\eta _1)}\tfrac {(k-\eta _2)\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}}{(1-\eta _2)}} \sqrt{|a_{k,1}||a_{k,2}|}. \end{split}\end{equation*}

Therefore, we need to find the largest $\eta ^*$ such that

\begin{equation*} \begin{split} & \tfrac {(k-\eta ^*)\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}}{(1-\eta ^*)} \sqrt{|a_{k,1}|\cdot |a_{k,2}|}\leq \ \\ & \leq \sqrt{\tfrac {(k-\eta _1)\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}}{(1-\eta _1)}\tfrac {(k-\eta _2)\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}}{(1-\eta _2)}} \end{split}\end{equation*}

for all $k\geq 2$. Thus we can write

\begin{align*} & \tfrac {(k-\eta ^*)\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}}{(1-\eta ^*)}\leq \\ & \leq \bigg\{ \tfrac {(k-\eta _1)\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}}{(1-\eta _1)}\bigg\} \bigg\{ \tfrac {(k-\eta _2)\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}}{(1-\eta _2)}\bigg\} . \end{align*}

After some calculations, we get

\begin{equation*} \eta ^* \leq 1- \tfrac {(k-1)(1-\eta _1)(1-\eta _2)} {(k-\eta _1)(k-\eta _2)\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big]\Gamma _k-(1-\eta _1)(1-\eta _2)}. \end{equation*}

We note that the right hand side of the above inequality is an increasing function for all $k\geq 2$. This implies that

\begin{align} \label{*} \eta ^* =& \min _{k\geq 2}\bigg\{ \tfrac {(k-1)(1-\eta _1)(1-\eta _2)} {(k-\eta _1)(k-\eta _2)\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big]\Gamma _k-(1-\eta _1)(1-\eta _2)}\bigg\} \nonumber \\ =& \tfrac {(1-\eta _1)(1-\eta _2)} {(2-\eta _1)(2-\eta _2)\big[\frac{(1+\lambda _1)^{m-1}}{(1+\lambda _2)^m}\big]\Gamma _2-(1-\eta _1)(1-\eta _2)}. \end{align}

Thus $G_2(z)\in \mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta )$. Therefore the theorem is true for $m=2$. Now, we suppose that $G_{m-1}(z)\in \mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta _0)$ and $f_m(z)\in \mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta _m)$, where

\begin{equation*} \eta _0 = \tfrac {\prod \limits _{j=1}^{m-1}(1-\eta _j)}{\prod \limits _{j=1}^{m-1}(2-\eta _j)\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^{m-1}}\big] {\Gamma _k}-\prod \limits _{j=1}^{m-1}(1-\eta _j)}. \end{equation*}

Replacing $\eta _1$ by $\eta _0$, and $\eta _2$ by $\eta _m$ in the inequality (27), we get

\begin{align*} \label{*} \eta ^* =& \tfrac {(1-\eta _0)(1-\eta _m)} {(2-\eta _0)(2-\eta _m)\big[\frac{(1+\lambda _1)^{m-1}}{(1+\lambda _2)^m}\big]\Gamma _2-(1-\eta _0)(1-\eta _m)}\ \\ =& \tfrac {\prod \limits _{j=1}^m(1-\eta _j)}{\prod \limits _{j=1}^m(2-\eta _j)\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}-\prod \limits _{j=1}^m(1-\eta _j)}. \end{align*}

For the integer $m$ the theorem ia also true. By the mathematical induction, the proof of the theorem is complete.

5 Integral Operator

In this section we consider integral transforms of functions in the class $ \mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta )$.

Theorem 9

Let the function $f(z)$ defined by (1) be in the class $ \mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta )$. Then the integral transforms

\begin{equation} G_c(z)=c\int _0^1u^{c-2}f(uz)du; \quad \quad (c>0) \end{equation}

are in the class $ \mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\gamma )$, where

\begin{equation} \gamma =1-\tfrac {c(1-\eta )}{(2-\eta )(c+1)-c(1-\eta )}. \end{equation}

Proof â–¼

Let $f(z)=z+\sum _{k=2}^{\infty } a_k z^k$. Then we have

\begin{equation} G_c(z)=c\int _0^1u^{c-2}f(uz)du = z+\sum _{k=2}^{\infty }\Big( \tfrac {c}{c+k-1}\Big) a_k z^k. \end{equation}

Since $f(z)\in \mathcal{S}^{m,r,s}_{\lambda _1,\lambda _2}(\eta )$, we have

\begin{equation} \sum _{k=2}^\infty \tfrac {(k-\eta )\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}}{(1-\eta )} |a_{k}|\leq 1. \end{equation}

In view of Theorem 3, we shall find the largest $\gamma $ for which

\begin{equation} \sum _{k=2}^\infty \tfrac {(k-\gamma )\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}}{(1-\gamma ) \big( \frac{c}{c+k-1}\big)}|a_{k}|\leq 1. \end{equation}

Let us find the range of values of $\gamma $ for which

\begin{equation*} \tfrac {(k-\gamma )\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}}{(1-\gamma )\big( \frac{c}{c+k-1}\big)} \leq \tfrac {(k-\eta )\big[\frac{(1+\lambda _1(k-1))^{m-1}}{(1+\lambda _2(k-1))^m}\big] {\Gamma _k}}{(1-\eta )}, \qquad (k\geq 2). \end{equation*}

After some calculations, we obtain from the above inequality that

\begin{equation*} \gamma \leq 1-\tfrac {c(k-1)(1-\eta )}{(k-\eta )(c+k-1)-c(1-\eta )}. \end{equation*}

We note that the right hand side of the above inequality is an increasing function for all $k\geq 2$. This implies that

\begin{eqnarray*} \gamma & =& \min _{k\geq 2}\bigg\{ 1-\tfrac {c(k-1)(1-\eta )}{(k-\eta )(c+k-1)-c(1-\eta )}\bigg\} \nonumber \\ & =& 1-\tfrac {c(1-\eta )}{(2-\eta )(c+1)-c(1-\eta )}. \end{eqnarray*}

The proof is complete.

Acknowledgements

The authors wish to thank the referee for the comments to improve the contents of the manuscript.

Bibliography

1: J. Dziok and H. M. Srivastava, Classes of analytic functions associated with the generalized hypergeometric function, Appl. Math. Comput., 103(1) (1999), pp. 1–13. $\includegraphics[scale=0.1]{ext-link.png}$
2: M. H. Al-Abbadi and M. Darus, Differential subordination for new generalized derivative operator, Acta Univ. Apulensis, 20 (2009), pp. 265–280.
3: Yu. E. Hohlov, Operators and operations in the class of univalent functions, Izv. Vyssh. Uchebn. Zaved. Mat., 10(10) (1978), pp. 83–89.
4: B. C. Carlson and D. B. Shaffer, Starlike and prestarlike hypergeometric function, SIAM J. Math. Anal., 15(4) (1984), pp. 737–745, 1984. $\includegraphics[scale=0.1]{ext-link.png}$
5: S. Ruscheweyh, New criteria for univalent functions, Proc. Amer. Math. Soc., 49 (1975), pp. 109–115. $\includegraphics[scale=0.1]{ext-link.png}$
6: K. Al-Shaqsi and M. Darus, On univalent functions with respect to the k-symmetric points defined by a generalization Ruscheweyh derivative operators, Jour. Anal. Appl., 7(1) (2009), pp. 53–61.
7: K. Al-Shaqsi and M. Darus, Differential Subordination with generalized derivative operator, Int. J. Comp. Math. Sci., 2(2) (2008), pp. 75–78.
8: J. Nishiwaki and S. Owa, Convolutions for certain analytic functions, Gen. Math., 15 (2007), no. 2–3, pp. 38–51.

Certain Properties for a Class of Analytic Functions Associated with Hypergeometric Functions

1 Introduction

2 Coefficient Estimate for the Class \(\mathcal{S}^{\lowercase {m,r,s}}_{\lambda _1,\lambda _2}(\eta )\)

3 Closure Theorems

4 Convolution Properties

5 Integral Operator

Bibliography