Geometric properties of certain analytic functions associated with generalized fractional integral operators

Abstract

Let be the class of normalized analytic functions in the unit disk and define the class $\mathcal{P}(\beta )=\{f\in \mathcal{A}:\mathrm{\exists }\phi \in \mathbb{R}\text{ such that }Re[e^{i\phi }(f^{\prime }(z)-\beta )]>0,z\in \mathcal{U}\}$. In this paper we find conditions on the number β and the non-negative weight function $\lambda (t)$ such that the integral transform $V_{\lambda }(f)(z)={\int }_0^1\lambda (t)\frac{f(tz)}{t}dt$ is convex of order γ ($0\le \gamma \le 1/2$) when $f\in \mathcal{P}(\beta )$. Some interesting further consequences are also considered.

MSC:30C45, 33C50.

1 Introduction and definitions

Let denote the class of functions of the form

$$f(z)=z+\sum _{n=2}^{\mathrm{\infty }}{a}_n{z}^n$$

(1.1)

which are analytic in the open unit disk $\mathcal{U}=\{z\in \mathbb{C}:|z|<1\}$. Also let , ${\mathcal{S}}^{\ast }(\gamma )$ and $\mathcal{K}(\gamma )$ denote the subclasses of consisting of functions which are univalent, starlike of order γ and convex of order γ in , respectively. In particular, the classes ${\mathcal{S}}^{\ast }(0)={\mathcal{S}}^{\ast }$ and $\mathcal{K}(0)=\mathcal{K}$ are the familiar ones of starlike and convex functions in , respectively.

We note that

$$f(z)\in \mathcal{K}(\gamma )⟺z{f}^{\prime }(z)\in {\mathcal{S}}^{\ast }(\gamma )$$

(1.2)

for $0\le \gamma <1$.

Let a, b, and c be complex numbers with $c\ne 0,-1,-2,\dots$ . Then the Gaussian hypergeometric function ${}_{2}F_1$ is defined by

$${}_{2}F_1(a,b;c;z)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(a)}_n{(b)}_n}{{(c)}_n}\frac{z^n}{n!},$$

(1.3)

where ${(\lambda )}_n$ is the Pochhammer symbol defined, in terms of the Gamma function, by

$${(\lambda )}_n=\frac{\mathrm{\Gamma }(\lambda +n)}{\mathrm{\Gamma }(\lambda )}=\{\begin{array}{ll}1& (n=0),\\ \lambda (\lambda +1)\cdots (\lambda +n-1)& (n\in \mathbb{N}).\end{array}$$

For functions $f_j(z)$ ($j=1,2$) of the forms

$$f_j(z):=\sum _{n=0}^{\mathrm{\infty }}{a}_{j,n+1}{z}^{n+1}(a_{j,1}:=1;j=1,2),$$

let $(f_1\ast f_2)(z)$ denote the Hadamard product or convolution of $f_1(z)$ and $f_2(z)$, defined by

$$(f_1\ast f_2)(z):=\sum _{n=0}^{\mathrm{\infty }}{a}_{1,n+1}{a}_{2,n+1}{z}^{n+1}(a_{j,1}:=1;j=1,2).$$

By using (1.3), Hohlov [1] introduced the convolution operator $H_{a,b,c}$ by

$$H_{a,b,c}(f)(z):=z_2{F}_1(a,b;c;z)\ast f(z)$$

(1.4)

for $f\in \mathcal{A}$. The three-parameter family of operators given by (1.4) contains as special cases several of the known linear integral or differential operators studied by a number of authors. This operator has been studied extensively by Ponnusamy [2], Kim and Rønning [3] and many others [4, 5]. In particular, if $a=1$ in (1.4), then $H_{1,b,c}$ is the operator $\mathcal{L}(b,c)$ due to Carlson and Shaffer [6] which was defined by

$$\mathcal{L}(b,c)f(z)=z_2{F}_1(1,b;c;z)\ast f(z).$$

Clearly, $\mathcal{L}(b,c)$ maps onto itself, and $\mathcal{L}(c,b)$ is the inverse of $\mathcal{L}(b,c)$, provided that $b\ne 0,-1,-2,\dots$ . Furthermore, $\mathcal{L}(b,b)$ is the unit operator and

$$\mathcal{L}(b,c)=\mathcal{L}(b,e)\mathcal{L}(e,c)=\mathcal{L}(e,c)\mathcal{L}(b,e)(c,e\ne 0,-1,-2,\dots ).$$

(1.5)

Also, we note that

$$\mathcal{K}(\gamma )=\mathcal{L}(1,2){\mathcal{S}}^{\ast }(\gamma )(0\le \gamma <1)$$

and

$${\mathcal{S}}^{\ast }(\gamma )=\mathcal{L}(2,1)\mathcal{K}(\gamma )(0\le \gamma <1).$$

(1.6)

Various definitions of fractional calculus operators are given by many authors. We use here the following definition due to Saigo [7] (see also [5, 8]).

Definition 1 For $\lambda >0$, $\mu ,\nu \in \mathbb{R}$, the fractional integral operator ${\mathcal{I}}_{0,z}^{\lambda ,\mu ,\nu }$ is defined by

$${\mathcal{I}}_{0,z}^{\lambda ,\mu ,\nu }f(z)=\frac{z^{-\lambda -\mu }}{\mathrm{\Gamma }(\lambda )}{\int }_0^z(z-\zeta )^{\lambda -1}{}_2{F}_1(\lambda +\mu ,-\nu ;\lambda ;1-\frac{\zeta }{z})f(\zeta )d\zeta ,$$

where $f(z)$ is taken to be an analytic function in a simply connected region of the z-plane containing the origin with the order

$$f(z)=\mathcal{O}\left({|z|}^{ϵ}\right)(z\to 0)$$

for $ϵ>max\{0,\mu -\nu \}-1$, and the multiplicity of ${(z-\zeta )}^{\lambda -1}$ is removed by requiring $log(z-\zeta )$ to be real when $z-\zeta >0$. With the aid of the above definition, Owa et al. [9] defined a modification of the fractional integral operator ${\mathcal{J}}_{0,z}^{\lambda ,\mu ,\nu }$ by

$${\mathcal{J}}_{0,z}^{\lambda ,\mu ,\nu }f(z)=\frac{\mathrm{\Gamma }(2-\mu )\mathrm{\Gamma }(2+\lambda +\nu )}{\mathrm{\Gamma }(2-\mu +\nu )}{z}^{\mu }{\mathcal{I}}_{0,z}^{\lambda ,\mu ,\nu }f(z)$$

for $f(z)\in \mathcal{A}$ and $min\{\lambda +\nu ,-\mu +\nu ,-\mu \}>-2$. Then it is observed that ${\mathcal{J}}_{0,z}^{\lambda ,\mu ,\nu }$ also maps onto itself and

$${\mathcal{J}}_{0,z}^{\lambda ,\mu ,\nu }f(z)=\mathcal{L}(2,2-\mu )\mathcal{L}(2-\mu +\nu ,2+\lambda +\nu )f(z).$$

(1.7)

The function

$$s_{\alpha }(z)=\frac{z}{{(1-z)}^{2(1-\alpha )}}(0\le \alpha <1)$$

is the well-known extremal function for the class ${\mathcal{S}}^{\ast }(\alpha )$. A function $f(z)\in \mathcal{A}$ is said to be in the class $\mathcal{R}(\alpha ,\gamma )$ if

$$(f\ast s_{\alpha })(z)\in {\mathcal{S}}^{\ast }(\gamma )(0\le \alpha <1;0\le \gamma <1).$$

Note that

$$\mathcal{R}(\alpha ,\gamma )=\mathcal{L}(1,2-2\alpha ){\mathcal{S}}^{\ast }(\gamma )$$

(1.8)

and $\mathcal{R}(\alpha ,\alpha )\equiv \mathcal{R}(\alpha )$ is the subclass of consisting of prestarlike functions of order α which was introduced by Suffridge [10]. In [11], it is shown that $\mathcal{R}(\alpha )\subset \mathcal{S}$ if and only if $\alpha \le 1/2$. For $\beta <1$ we denote the class

$$\mathcal{P}(\beta )=\{f\in \mathcal{A}:\mathrm{\exists }\phi \in \mathbb{R}\text{ such that }Re\left[e^{i\phi }(f^{\prime }(z)-\beta )\right]>0,z\in \mathcal{U}\}.$$

Throughout this paper we let $\lambda :[0,1]\to \mathbb{R}$ be a non-negative function with

$${\int }_0^1\lambda (t)dt=1.$$

(1.9)

For certain specific subclasses of $f\in \mathcal{A}$, many authors considered the geometric properties of the integral transform of the form

$$V_{\lambda }(f)(z)={\int }_0^1\lambda (t)\frac{f(tz)}{t}dt.$$

(1.10)

More recently, starlikeness of this general operator $V_{\lambda }(f)$ was discussed by Fournier and Ruscheweyh [12] by assuming that $f\in \mathcal{P}(\beta )$. The method of proof is the duality principle developed mainly by Ruscheweyh [13]. This result was later extended by Ponnusamy and Rønning [14] by means of finding conditions such that $V_{\lambda }(f)$ carries $\mathcal{P}(\beta )$ into starlike functions of order γ, $0\le \gamma \le 1/2$.

In this paper, we find conditions on β and the function $\lambda (t)$ such that $V_{\lambda }(f)$ carries $\mathcal{P}(\beta )$ into $\mathcal{K}(\gamma )$. As a consequence of this investigation, a number of new results are established.

2 Preliminaries

We begin by recalling the following results.

Lemma 1 ([15]; see also [5])

If $f\in \mathcal{A}$ and $c-a+1>b>0$, then

$$H_{a,b,c}(f)(z)=\frac{\mathrm{\Gamma }(c)}{\mathrm{\Gamma }(a)\mathrm{\Gamma }(b)}{\int }_0^1\frac{{(1-t)}^{c-a-b}}{\mathrm{\Gamma }(c-a-b+1)}t^{b-2}{}_2{F}_1(c-a,1-a;c-a-b+1;1-t)f(tz)dt.$$

Remark 1 In view of Lemma 1, we see that the convolution operator (1.4) is an integral operator of the form (1.10) with

$$\lambda (t)={\frac{\mathrm{\Gamma }(c)t^{b-1}{(1-t)}^{c-a-b}}{\mathrm{\Gamma }(a)\mathrm{\Gamma }(b)\mathrm{\Gamma }(c-a-b+1)}}_2{F}_1(c-a,1-a;c-a-b+1;1-t).$$

For $\mathrm{\Lambda }:[0,1]\to \mathbb{R}$ being integrable and positive on $(0,1)$, we define

$$L_{\mathrm{\Lambda }}(h_{\gamma })=\underset{z\in \mathcal{U}}{inf}{\int }_0^1\mathrm{\Lambda }(t)[Re\frac{h_{\gamma }(zt)}{zt}-\frac{1-\gamma (1+t)}{(1-\gamma ){(1+t)}^2}]dt$$

and

$$M_{\mathrm{\Lambda }}(h_{\gamma })=\underset{z\in \mathcal{U}}{inf}{\int }_0^1\mathrm{\Lambda }(t)[Re{h}_{\gamma }^{\prime }(zt)-\frac{1-t-\gamma (1+t)}{(1-\gamma ){(1+t)}^3}]dt,$$

where $0\le \gamma <1$ and

$$h_{\gamma }(z)=\frac{z(1+\frac{ϵ+2\gamma -1}{2-2\gamma }z)}{{(1-z)}^2},|ϵ|=1.$$

(2.1)

In [16], Ponnusamy and Rønning proved the following lemmas.

Lemma 2 Let $\mathrm{\Lambda }(t)$ be integrable on $[0,1]$ and positive on $(0,1)$. If $\mathrm{\Lambda }(t)/(1+t){(1-t)}^{1+2\gamma }$ is decreasing on $(0,1)$, then for $0\le \gamma \le 1/2$ we have $L_{\mathrm{\Lambda }}(h_{\gamma })\ge 0$.

Lemma 3 Let $0\le \gamma <1$ and let $\lambda (t)$ be given by (1.9). Define $\beta <1$ by

$$\frac{\beta }{1-\beta }=-{\int }_0^1\lambda (t)[\frac{1+\gamma -(1-\gamma )t}{(1-\gamma )(1+t)}-\frac{2\gamma }{1-\gamma }\frac{log(1+t)}{t}]dt.$$

Assume that ${lim}_{t\to 0+}t\mathrm{\Lambda }(t)=0$, where

$$\mathrm{\Lambda }(t)={\int }_t^1\lambda (s)ds/s.$$

Then $V_{\lambda }(\mathcal{P}(\beta ))\subset {\mathcal{S}}^{\ast }(\gamma )$ if and only if $L_{\mathrm{\Lambda }}(h_{\gamma })\ge 0$.

We now find conditions on β and the non-negative weight function $\lambda (t)$ such that $V_{\lambda }(\mathcal{P}(\beta ))\subset \mathcal{K}(\gamma )$.

Lemma 4 (i) Let $\mathrm{\Lambda }(t)$ be monotone decreasing on $[0,1]$ satisfying $\mathrm{\Lambda }(1)=0$ and ${lim}_{t\to 0+}t\mathrm{\Lambda }(t)=0$. For $0\le \gamma \le 1/2$ if $t{\mathrm{\Lambda }}^{\prime }(t)/(1+t){(1-t)}^{1+2\gamma }$ is increasing on $(0,1)$, then $M_{\mathrm{\Lambda }}(h_{\gamma })\ge 0$.

• (ii) Let $0\le \gamma \le 1/2$ and let $\lambda (t)$ and $\mathrm{\Lambda }(t)$ be as in Lemma 3. Define $\beta <1$ by

  $$\frac{\beta -\frac{1}{2}}{1-\beta }=-{\int }_0^1\lambda (t)\frac{1-\gamma (1+t)}{(1-\gamma ){(1+t)}^2}dt.$$

Then $V_{\lambda }(\mathcal{P}(\beta ))\subset \mathcal{K}(\gamma )$ if and only if $M_{\mathrm{\Lambda }}(h_{\gamma })\ge 0$.

Proof (i) Let $M_{\mathrm{\Lambda }}(h_{\gamma })={inf}_{z\in \mathcal{U}}{I}_{\gamma }$. Then, by using the conditions $\mathrm{\Lambda }(1)=0$ and ${lim}_{t\to 0+}t\mathrm{\Lambda }(t)=0$, an integration by parts yields

$$\begin{array}{rcl}{I}_{\gamma }& =& {\int }_0^1\mathrm{\Lambda }(t)[Re{h}_{\gamma }^{\prime }(zt)-\frac{1-t-\gamma (1+t)}{(1-\gamma ){(1+t)}^3}]dt\\ =& {\int }_0^1\mathrm{\Lambda }(t)\frac{d}{dt}[Re\frac{h_{\gamma }(zt)}{z}-\frac{t(1-\gamma (1+t))}{(1-\gamma ){(1+t)}^2}]dt\\ =& -{\int }_0^{1}t{\mathrm{\Lambda }}^{\prime }(t)[Re\frac{h_{\gamma }(zt)}{zt}-\frac{1-\gamma (1+t)}{(1-\gamma ){(1+t)}^2}]dt.\end{array}$$

Since $t{\mathrm{\Lambda }}^{\prime }(t)/(1+t){(1-t)}^{1+2\gamma }$ is increasing on $(0,1)$, by Lemma 2, ${inf}_{z\in \mathcal{U}}{I}_{\gamma }\ge 0$, which evidently completes the proof of (i).

• (ii) We state this proof only in outline here because the proof is similar to that of [[3], Theorem 2.1]. Let $F(z)=V_{\lambda }(f)(z)$. Then, by convolution theory [[13], p.94] and (1.2), we have

  $$F(z)\in \mathcal{K}(\gamma )⟺\frac{1}{z}(z{F}^{\prime }(z)\ast h_{\gamma }(z))\ne 0,$$

  (2.2)

where $h_{\gamma }(z)$ is given by (2.1). Since $f\in \mathcal{P}(\beta )$, by the duality principle [[13], p.23], it is enough to verify this with f given by

$$f^{\prime }(z)=(1-\beta )\frac{1-xz}{1-yz}+\beta (|x|=|y|=1).$$

In the same way as in [[3], Theorem 2.1], we conclude that (2.2) holds if and only if

$$Re{\int }_0^1\lambda (t)[\frac{h_{\gamma }(zt)}{zt}-\frac{1-\gamma (1+t)}{(1-\gamma ){(1+t)}^2}]dt>0.$$

(2.3)

Integrating by parts, we find that the inequality (2.3) is equivalent to

$$Re{\int }_0^1\mathrm{\Lambda }(t)[h_{\gamma }^{\prime }(zt)-\frac{1-t-\gamma (1+t)}{(1-\gamma ){(1+t)}^3}]dt\ge 0,$$

which again is equivalent to $M_{\mathrm{\Lambda }}(h_{\gamma })\ge 0$. □

Remark 2 In particular, taking $\gamma =0$ in Lemma 4, we obtain the result due to Ali and Singh [[17], Theorem 1].

3 Main results

We define

$$\phi (1-t)=1+\sum _{n=1}^{\mathrm{\infty }}{b}_n{(1-t)}^n(b_n\ge 0)$$

(3.1)

and

$$\lambda (t)=C{t}^{b-1}{(1-t)}^{c-a-b}\phi (1-t),$$

(3.2)

where C is a constant satisfying the condition (1.9). For $f\in \mathcal{A}$ Balasubramanian et al. [4] defined the operator $P_{a,b,c}$ by

$$P_{a,b,c}(f)(z)={\int }_0^1\lambda (t)\frac{f(tz)}{t}dt,$$

where $\lambda (t)$ is given by (3.2). Special choices of $\phi (1-t)$ and C led to various interesting geometric properties concerning certain linear operators. For example, if we take $\phi (1-t){=}_2{F}_1(c-a,1-a;c-a-b+1;1-t)$ and

$$C=\frac{\mathrm{\Gamma }(c)}{\mathrm{\Gamma }(a)\mathrm{\Gamma }(b)\mathrm{\Gamma }(c-a-b+1)},$$

by virtue of Remark 1,

$$P_{a,b,c}(f)(z)=H_{a,b,c}(f)(z).$$

(3.3)

First, by applying Lemma 4, we prove the following.

Theorem 1 Let $0\le \gamma \le 1/2$, $a>0$, $0<b\le 1$, and $c\ge a+b+2\gamma +1$, and let $\lambda (t)$ be given by (3.2). Define $\beta =\beta (a,b,c,\gamma )$ by

$$\frac{\beta -\frac{1}{2}}{1-\beta }=-{\int }_0^1\lambda (t)\frac{1-\gamma (1+t)}{(1-\gamma ){(1+t)}^2}dt.$$

If $f(z)\in \mathcal{P}(\beta )$, then $P_{a,b,c}(f)(z)\in \mathcal{K}(\gamma )$. The value of β is sharp.

Proof Let $C>0$ and

$$\mathrm{\Lambda }(t)={\int }_t^1\frac{\lambda (s)}{s}ds,$$

where $\lambda (t)$ is given by (3.2). Then it is easily seen that $\mathrm{\Lambda }(t)$ is monotone decreasing on $[0,1]$ and ${lim}_{t\to 0+}t\mathrm{\Lambda }(t)=0$. In order to apply Lemma 4, we want to prove that the function

$$u(t)=\frac{\lambda (t)}{(1+t){(1-t)}^{1+2\gamma }}$$

(3.4)

is decreasing on $(0,1)$, where $\lambda (t)$ is given by (3.2). Making use of the logarithmic differentiation of both sides in (3.4), we have

$$\frac{u^{\prime }(t)}{u(t)}=\frac{{\lambda }^{\prime }(t)}{\lambda (t)}+\frac{2(\gamma +(1+\gamma )t)}{1-t^2}.$$

(3.5)

Since

$${\lambda }^{\prime }(t)=C{t}^{b-2}{(1-t)}^{c-a-b-1}[\phi (1-t)\{(b-1)(1-t)-t(c-a-b)\}-t(1-t){\phi }^{\prime }(1-t)],$$

from (3.4) and (3.5) we find that $u^{\prime }(t)\le 0$ on $(0,1)$ is equivalent to

$$(c-a-3-2\gamma )t^2+(c-a-b-2\gamma )t+1-b\ge -t(1-t^2)\frac{{\phi }^{\prime }(1-t)}{\phi (1-t)}(0<t<1).$$

(3.6)

In view of (3.1), $\phi (1-t)>0$ and ${\phi }^{\prime }(1-t)\ge 0$ on $(0,1)$, so that the right hand side of the inequality (3.6) is non-positive for all $t\in (0,1)$. If we assume that $0\le \gamma \le 1/2$, $a>0$, $0<b\le 1$, and $c\ge a+b+2\gamma +1$, then $(c-a-3-2\gamma )t^2+(c-a-b-2\gamma )t+1-b\ge 0$ for $t\in (0,1)$. Thus, the inequality (3.6) holds for all $t\in (0,1)$. Hence, from Lemma 4 we obtain $P_{a,b,c}(f)(z)\in \mathcal{K}(\gamma )$. □

The same techniques as in the proof of [[5], Theorem 1] show that the value β is sharp.

By using (3.3) and Theorem 1, we have the following.

Corollary 1 Let $0\le \gamma \le 1/2$, $0<a\le 1$, $0<b\le 1$, and $c\ge a+b+2\gamma +1$. Define $\beta =\beta (a,b,c,\gamma )$ by

$$\begin{array}{rl}\frac{\beta -\frac{1}{2}}{1-\beta }=& -\frac{\mathrm{\Gamma }(c)}{\mathrm{\Gamma }(a)\mathrm{\Gamma }(b)}{\int }_0^1\frac{{(1-t)}^{c-a-b}{t}^{b-1}}{\mathrm{\Gamma }(c-a-b+1)}\frac{1-\gamma (1+t)}{(1-\gamma ){(1+t)}^2}\\ {\times }_2{F}_1(c-a,1-a;c-a-b+1;1-t)dt.\end{array}$$

If $f(z)\in \mathcal{P}(\beta )$, then $H_{a,b,c}(f)(z)\in \mathcal{K}(\gamma )$. The value of β is sharp.

Proof If we put

$$\lambda (t)={\frac{\mathrm{\Gamma }(c)t^{b-1}{(1-t)}^{c-a-b}}{\mathrm{\Gamma }(a)\mathrm{\Gamma }(b)\mathrm{\Gamma }(c-a-b+1)}}_2{F}_1(c-a,1-a;c-a-b+1;1-t),$$

then, by applying (3.3) and Theorem 1, we obtain the desired result. □

Setting $a=1$ in Corollary 1, we obtain the following.

Corollary 2 Let $0\le \gamma \le 1/2$, $0<b\le 1$, and $c\ge b+2\gamma +2$. Also let

$$\beta (1,b,c,\gamma )=1-\frac{1-\gamma }{2[1{-}_2{F}_1(2,b;c;-1)-\gamma (1{-}_2{F}_1(1,b;c;-1))]}.$$

If $\beta (1,b,c,\gamma )<\beta <1$ and $f(z)\in \mathcal{P}(\beta )$, then $\mathcal{L}(b,c)f(z)\in \mathcal{K}(\gamma )$.

Next we find a univalence criterion for the operator ${\mathcal{J}}_{0,z}^{\lambda ,\mu ,\nu }$.

Theorem 2 Let $0\le \gamma \le 1/2$, $0\le \mu <2$, $\lambda \ge 2(1+\gamma )-\mu$, and $\mu -2<\nu \le \mu -1$. Define $\beta =\beta (\lambda ,\mu ,\nu ,\gamma )$ by

$$\beta =1-\frac{1-\gamma }{2[1{-}_2{F}_1(2,2-\mu +\nu ;2+\lambda +\nu ;-1)-\gamma (1{-}_2{F}_1(1,2-\mu +\nu ;2+\lambda +\nu ;-1))]}.$$

If $f(z)\in \mathcal{P}(\beta )$, then ${\mathcal{J}}_{0,z}^{\lambda ,\mu ,\nu }f(z)\in \mathcal{R}(\mu /2,\gamma )$.

Proof Making use of (1.5) and (1.7), we note that

$$\begin{array}{rcl}{\mathcal{J}}_{0,z}^{\lambda ,\mu ,\nu }f(z)& =& \mathcal{L}(2,2-\mu )\mathcal{L}(2-\mu +\nu ,2+\lambda +\nu )f(z)\\ =& \mathcal{L}(1,2-\mu )\mathcal{L}(2,1)\mathcal{L}(2-\mu +\nu ,2+\lambda +\nu )f(z).\end{array}$$

(3.7)

By using Corollary 2, we obtain

$$\mathcal{L}(2-\mu +\nu ,2+\lambda +\nu )f(z)\in \mathcal{K}(\gamma ).$$

Since $0\le \mu <2$, from (1.6), (1.8) and (3.7) we have ${\mathcal{J}}_{0,z}^{\lambda ,\mu ,\nu }f(z)\in \mathcal{R}(\mu /2,\gamma )$, which completes the proof of Theorem 2. □

Taking $\mu =2\gamma$ in Theorem 2, we get the following.

Corollary 3 Let $0\le \gamma \le 1/2$, $\lambda \ge 2$, and $2(\gamma -1)<\nu \le 2\gamma -1$. Define $\beta =\beta (\lambda ,\nu ,\gamma )$ by

$$\beta =1-\frac{1-\gamma }{2[1{-}_2{F}_1(2,2(1-\gamma )+\nu ;2+\lambda +\nu ;-1)-\gamma (1{-}_2{F}_1(1,2(1-\gamma )+\nu ;2+\lambda +\nu ;-1))]}.$$

If $f(z)\in \mathcal{P}(\beta )$, then ${\mathcal{J}}_{0,z}^{\lambda ,2\gamma ,\nu }f(z)\in \mathcal{R}(\gamma )\subset \mathcal{S}$.

Proof If we put $\mu =2\gamma$ in Theorem 2, then

$${\mathcal{J}}_{0,z}^{\lambda ,2\gamma ,\nu }f(z)\in \mathcal{R}(\gamma ,\gamma )=\mathcal{R}(\gamma ).$$

Since $\gamma \le 1/2$, $\mathcal{R}(\gamma )\subset \mathcal{S}$, so that the proof is completed. □

Remark 3 In [4], Balasubramanian et al. found the conditions on the number β and the function $\lambda (t)$ such that $P_{a,b,c}(f)(z)\in {\mathcal{S}}^{\ast }(\gamma )$ ($0\le \gamma \le 1/2$). Since ${\mathcal{J}}_{0,z}^{\lambda ,\mu ,\nu }f(z)=P_{1-\nu ,2,\lambda -\nu +2}(f)(z)$ with $\phi (1-t){=}_2{F}_1(\lambda +\mu ,-\nu ;\lambda ;1-t)$ and

$$C=\frac{\mathrm{\Gamma }(2-\mu )\mathrm{\Gamma }(2+\lambda +\nu )}{\mathrm{\Gamma }(\lambda )\mathrm{\Gamma }(2-\mu +\nu )},$$

the condition on β and $\lambda (t)$ is easily found such that ${\mathcal{J}}_{0,z}^{\lambda ,\mu ,\nu }f(z)\in {\mathcal{S}}^{\ast }(\gamma )$.

Finally, by using Lemma 4 again, we investigate convexity of the operator ${\mathcal{J}}_{0,z}^{\lambda ,\mu ,\nu }$.

Theorem 3 Let $0\le \gamma \le 1/2$, $0<\lambda \le 1+2\gamma$, $2<\mu <3$, $0<\nu \le 1$, and $\nu >\mu -2$. Define $\beta =\beta (\lambda ,\mu ,\nu ,\gamma )$ by

$$\frac{\beta -\frac{1}{2}}{1-\beta }=-\frac{\mathrm{\Gamma }(2-\mu )\mathrm{\Gamma }(2+\lambda +\nu )}{\mathrm{\Gamma }(\lambda )\mathrm{\Gamma }(2-\mu +\nu )}{\int }_0^1{\frac{t{(1-t)}^{\lambda -1}(1-\gamma (1+t))}{(1-\gamma ){(1+t)}^2}}_2{F}_1(\lambda +\mu ,-\nu ;\lambda ;1-t)dt.$$

If $f(z)\in \mathcal{P}(\beta )$, then ${\mathcal{J}}_{0,z}^{\lambda ,\mu ,\nu }f(z)\in \mathcal{K}(\gamma )$. The value of β is sharp.

Proof Let $0\le \gamma \le 1/2$, $0<\lambda \le 1+2\gamma$, $2<\mu <3$, and $\nu >\mu -2$, and let

$$\lambda (t)=\frac{\mathrm{\Gamma }(2-\mu )\mathrm{\Gamma }(2+\lambda +\nu )}{\mathrm{\Gamma }(\lambda )\mathrm{\Gamma }(2-\mu +\nu )}t(1-t)^{\lambda -1}{}_2{F}_1(\lambda +\mu ,-\nu ;\lambda ;1-t).$$

(3.8)

Then we can easily see that ${\int }_0^1\lambda (t)dt=1$, $\mathrm{\Lambda }(t)={\int }_t^1\lambda (s)ds/s$ is monotone decreasing on $[0,1]$ and ${lim}_{t\to 0+}t\mathrm{\Lambda }(t)=0$. Also we find that the function $u(t)=\lambda (t)/(1+t){(1-t)}^{1+2\gamma }$ is decreasing on $(0,1)$, where $\lambda (t)$ is given by (3.8). Hence, $t{\mathrm{\Lambda }}^{\prime }(t)/(1+t){(1-t)}^{1+2\gamma }=-u(t)$ is increasing on $(0,1)$. From Lemma 4, we obtain the desired result. □