# 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 . In this paper we find conditions on the number β and the non-negative weight function $\lambda \left(t\right)$ such that the integral transform ${V}_{\lambda }\left(f\right)\left(z\right)={\int }_{0}^{1}\lambda \left(t\right)\frac{f\left(tz\right)}{t}\phantom{\rule{0.2em}{0ex}}dt$ is convex of order γ ($0\le \gamma \le 1/2$) when $f\in \mathcal{P}\left(\beta \right)$. Some interesting further consequences are also considered.

MSC:30C45, 33C50.

## 1 Introduction and definitions

Let denote the class of functions of the form

$f\left(z\right)=z+\sum _{n=2}^{\mathrm{\infty }}{a}_{n}{z}^{n}$
(1.1)

which are analytic in the open unit disk $\mathcal{U}=\left\{z\in \mathbb{C}:|z|<1\right\}$. Also let , ${\mathcal{S}}^{\ast }\left(\gamma \right)$ and $\mathcal{K}\left(\gamma \right)$ 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 ${\mathcal{S}}^{\ast }\left(0\right)={\mathcal{S}}^{\ast }$ and $\mathcal{K}\left(0\right)=\mathcal{K}$ are the familiar ones of starlike and convex functions in , respectively.

We note that

$f\left(z\right)\in \mathcal{K}\left(\gamma \right)\phantom{\rule{1em}{0ex}}⟺\phantom{\rule{1em}{0ex}}z{f}^{\prime }\left(z\right)\in {\mathcal{S}}^{\ast }\left(\gamma \right)$
(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

${}_{2}F_{1}\left(a,b;c;z\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(a\right)}_{n}{\left(b\right)}_{n}}{{\left(c\right)}_{n}}\frac{{z}^{n}}{n!},$
(1.3)

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

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

For functions ${f}_{j}\left(z\right)$ ($j=1,2$) of the forms

${f}_{j}\left(z\right):=\sum _{n=0}^{\mathrm{\infty }}{a}_{j,n+1}{z}^{n+1}\phantom{\rule{1em}{0ex}}\left({a}_{j,1}:=1;j=1,2\right),$

let $\left({f}_{1}\ast {f}_{2}\right)\left(z\right)$ denote the Hadamard product or convolution of ${f}_{1}\left(z\right)$ and ${f}_{2}\left(z\right)$, defined by

$\left({f}_{1}\ast {f}_{2}\right)\left(z\right):=\sum _{n=0}^{\mathrm{\infty }}{a}_{1,n+1}{a}_{2,n+1}{z}^{n+1}\phantom{\rule{1em}{0ex}}\left({a}_{j,1}:=1;j=1,2\right).$

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

${H}_{a,b,c}\left(f\right)\left(z\right):={z}_{2}{F}_{1}\left(a,b;c;z\right)\ast f\left(z\right)$
(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 $\mathcal{L}\left(b,c\right)$ due to Carlson and Shaffer [6] which was defined by

$\mathcal{L}\left(b,c\right)f\left(z\right)={z}_{2}{F}_{1}\left(1,b;c;z\right)\ast f\left(z\right).$

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

$\mathcal{L}\left(b,c\right)=\mathcal{L}\left(b,e\right)\mathcal{L}\left(e,c\right)=\mathcal{L}\left(e,c\right)\mathcal{L}\left(b,e\right)\phantom{\rule{1em}{0ex}}\left(c,e\ne 0,-1,-2,\dots \right).$
(1.5)

Also, we note that

$\mathcal{K}\left(\gamma \right)=\mathcal{L}\left(1,2\right){\mathcal{S}}^{\ast }\left(\gamma \right)\phantom{\rule{1em}{0ex}}\left(0\le \gamma <1\right)$

and

${\mathcal{S}}^{\ast }\left(\gamma \right)=\mathcal{L}\left(2,1\right)\mathcal{K}\left(\gamma \right)\phantom{\rule{1em}{0ex}}\left(0\le \gamma <1\right).$
(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

${\mathcal{I}}_{0,z}^{\lambda ,\mu ,\nu }f\left(z\right)=\frac{{z}^{-\lambda -\mu }}{\mathrm{\Gamma }\left(\lambda \right)}{\int }_{0}^{z}\left(z-\zeta \right)^{\lambda -1}{}_{2}{F}_{1}\left(\lambda +\mu ,-\nu ;\lambda ;1-\frac{\zeta }{z}\right)f\left(\zeta \right)\phantom{\rule{0.2em}{0ex}}d\zeta ,$

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

$f\left(z\right)=\mathcal{O}\left({|z|}^{ϵ}\right)\phantom{\rule{1em}{0ex}}\left(z\to 0\right)$

for $ϵ>max\left\{0,\mu -\nu \right\}-1$, and the multiplicity of ${\left(z-\zeta \right)}^{\lambda -1}$ is removed by requiring $log\left(z-\zeta \right)$ 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

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

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

${\mathcal{J}}_{0,z}^{\lambda ,\mu ,\nu }f\left(z\right)=\mathcal{L}\left(2,2-\mu \right)\mathcal{L}\left(2-\mu +\nu ,2+\lambda +\nu \right)f\left(z\right).$
(1.7)

The function

${s}_{\alpha }\left(z\right)=\frac{z}{{\left(1-z\right)}^{2\left(1-\alpha \right)}}\phantom{\rule{1em}{0ex}}\left(0\le \alpha <1\right)$

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

$\left(f\ast {s}_{\alpha }\right)\left(z\right)\in {\mathcal{S}}^{\ast }\left(\gamma \right)\phantom{\rule{1em}{0ex}}\left(0\le \alpha <1;0\le \gamma <1\right).$

Note that

$\mathcal{R}\left(\alpha ,\gamma \right)=\mathcal{L}\left(1,2-2\alpha \right){\mathcal{S}}^{\ast }\left(\gamma \right)$
(1.8)

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

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

${\int }_{0}^{1}\lambda \left(t\right)\phantom{\rule{0.2em}{0ex}}dt=1.$
(1.9)

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

${V}_{\lambda }\left(f\right)\left(z\right)={\int }_{0}^{1}\lambda \left(t\right)\frac{f\left(tz\right)}{t}\phantom{\rule{0.2em}{0ex}}dt.$
(1.10)

More recently, starlikeness of this general operator ${V}_{\lambda }\left(f\right)$ was discussed by Fournier and Ruscheweyh [12] by assuming that $f\in \mathcal{P}\left(\beta \right)$. 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 ${V}_{\lambda }\left(f\right)$ carries $\mathcal{P}\left(\beta \right)$ into starlike functions of order γ, $0\le \gamma \le 1/2$.

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

## 2 Preliminaries

We begin by recalling the following results.

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

${H}_{a,b,c}\left(f\right)\left(z\right)=\frac{\mathrm{\Gamma }\left(c\right)}{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{\int }_{0}^{1}\frac{{\left(1-t\right)}^{c-a-b}}{\mathrm{\Gamma }\left(c-a-b+1\right)}t^{b-2}{}_{2}{F}_{1}\left(c-a,1-a;c-a-b+1;1-t\right)f\left(tz\right)\phantom{\rule{0.2em}{0ex}}dt.$

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

$\lambda \left(t\right)={\frac{\mathrm{\Gamma }\left(c\right){t}^{b-1}{\left(1-t\right)}^{c-a-b}}{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)\mathrm{\Gamma }\left(c-a-b+1\right)}}_{2}{F}_{1}\left(c-a,1-a;c-a-b+1;1-t\right).$

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

${L}_{\mathrm{\Lambda }}\left({h}_{\gamma }\right)=\underset{z\in \mathcal{U}}{inf}{\int }_{0}^{1}\mathrm{\Lambda }\left(t\right)\left[Re\frac{{h}_{\gamma }\left(zt\right)}{zt}-\frac{1-\gamma \left(1+t\right)}{\left(1-\gamma \right){\left(1+t\right)}^{2}}\right]\phantom{\rule{0.2em}{0ex}}dt$

and

${M}_{\mathrm{\Lambda }}\left({h}_{\gamma }\right)=\underset{z\in \mathcal{U}}{inf}{\int }_{0}^{1}\mathrm{\Lambda }\left(t\right)\left[Re{h}_{\gamma }^{\prime }\left(zt\right)-\frac{1-t-\gamma \left(1+t\right)}{\left(1-\gamma \right){\left(1+t\right)}^{3}}\right]\phantom{\rule{0.2em}{0ex}}dt,$

where $0\le \gamma <1$ and

${h}_{\gamma }\left(z\right)=\frac{z\left(1+\frac{ϵ+2\gamma -1}{2-2\gamma }z\right)}{{\left(1-z\right)}^{2}},\phantom{\rule{1em}{0ex}}|ϵ|=1.$
(2.1)

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

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

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

$\frac{\beta }{1-\beta }=-{\int }_{0}^{1}\lambda \left(t\right)\left[\frac{1+\gamma -\left(1-\gamma \right)t}{\left(1-\gamma \right)\left(1+t\right)}-\frac{2\gamma }{1-\gamma }\frac{log\left(1+t\right)}{t}\right]\phantom{\rule{0.2em}{0ex}}dt.$

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

$\mathrm{\Lambda }\left(t\right)={\int }_{t}^{1}\lambda \left(s\right)\phantom{\rule{0.2em}{0ex}}ds/s.$

Then ${V}_{\lambda }\left(\mathcal{P}\left(\beta \right)\right)\subset {\mathcal{S}}^{\ast }\left(\gamma \right)$ if and only if ${L}_{\mathrm{\Lambda }}\left({h}_{\gamma }\right)\ge 0$.

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

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

1. (ii)

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

$\frac{\beta -\frac{1}{2}}{1-\beta }=-{\int }_{0}^{1}\lambda \left(t\right)\frac{1-\gamma \left(1+t\right)}{\left(1-\gamma \right){\left(1+t\right)}^{2}}\phantom{\rule{0.2em}{0ex}}dt.$

Then ${V}_{\lambda }\left(\mathcal{P}\left(\beta \right)\right)\subset \mathcal{K}\left(\gamma \right)$ if and only if ${M}_{\mathrm{\Lambda }}\left({h}_{\gamma }\right)\ge 0$.

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

$\begin{array}{rcl}{I}_{\gamma }& =& {\int }_{0}^{1}\mathrm{\Lambda }\left(t\right)\left[Re{h}_{\gamma }^{\prime }\left(zt\right)-\frac{1-t-\gamma \left(1+t\right)}{\left(1-\gamma \right){\left(1+t\right)}^{3}}\right]\phantom{\rule{0.2em}{0ex}}dt\\ =& {\int }_{0}^{1}\mathrm{\Lambda }\left(t\right)\frac{d}{dt}\left[Re\frac{{h}_{\gamma }\left(zt\right)}{z}-\frac{t\left(1-\gamma \left(1+t\right)\right)}{\left(1-\gamma \right){\left(1+t\right)}^{2}}\right]\phantom{\rule{0.2em}{0ex}}dt\\ =& -{\int }_{0}^{1}t{\mathrm{\Lambda }}^{\prime }\left(t\right)\left[Re\frac{{h}_{\gamma }\left(zt\right)}{zt}-\frac{1-\gamma \left(1+t\right)}{\left(1-\gamma \right){\left(1+t\right)}^{2}}\right]\phantom{\rule{0.2em}{0ex}}dt.\end{array}$

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

1. (ii)

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

$F\left(z\right)\in \mathcal{K}\left(\gamma \right)\phantom{\rule{1em}{0ex}}⟺\phantom{\rule{1em}{0ex}}\frac{1}{z}\left(z{F}^{\prime }\left(z\right)\ast {h}_{\gamma }\left(z\right)\right)\ne 0,$
(2.2)

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

${f}^{\prime }\left(z\right)=\left(1-\beta \right)\frac{1-xz}{1-yz}+\beta \phantom{\rule{1em}{0ex}}\left(|x|=|y|=1\right).$

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

$Re{\int }_{0}^{1}\lambda \left(t\right)\left[\frac{{h}_{\gamma }\left(zt\right)}{zt}-\frac{1-\gamma \left(1+t\right)}{\left(1-\gamma \right){\left(1+t\right)}^{2}}\right]\phantom{\rule{0.2em}{0ex}}dt>0.$
(2.3)

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

$Re{\int }_{0}^{1}\mathrm{\Lambda }\left(t\right)\left[{h}_{\gamma }^{\prime }\left(zt\right)-\frac{1-t-\gamma \left(1+t\right)}{\left(1-\gamma \right){\left(1+t\right)}^{3}}\right]\phantom{\rule{0.2em}{0ex}}dt\ge 0,$

which again is equivalent to ${M}_{\mathrm{\Lambda }}\left({h}_{\gamma }\right)\ge 0$. □

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

$\phi \left(1-t\right)=1+\sum _{n=1}^{\mathrm{\infty }}{b}_{n}{\left(1-t\right)}^{n}\phantom{\rule{1em}{0ex}}\left({b}_{n}\ge 0\right)$
(3.1)

and

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

${P}_{a,b,c}\left(f\right)\left(z\right)={\int }_{0}^{1}\lambda \left(t\right)\frac{f\left(tz\right)}{t}\phantom{\rule{0.2em}{0ex}}dt,$

where $\lambda \left(t\right)$ is given by (3.2). Special choices of $\phi \left(1-t\right)$ and C led to various interesting geometric properties concerning certain linear operators. For example, if we take $\phi \left(1-t\right){=}_{2}{F}_{1}\left(c-a,1-a;c-a-b+1;1-t\right)$ and

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

by virtue of Remark 1,

${P}_{a,b,c}\left(f\right)\left(z\right)={H}_{a,b,c}\left(f\right)\left(z\right).$
(3.3)

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

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

$\frac{\beta -\frac{1}{2}}{1-\beta }=-{\int }_{0}^{1}\lambda \left(t\right)\frac{1-\gamma \left(1+t\right)}{\left(1-\gamma \right){\left(1+t\right)}^{2}}\phantom{\rule{0.2em}{0ex}}dt.$

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

Proof Let $C>0$ and

$\mathrm{\Lambda }\left(t\right)={\int }_{t}^{1}\frac{\lambda \left(s\right)}{s}\phantom{\rule{0.2em}{0ex}}ds,$

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

$u\left(t\right)=\frac{\lambda \left(t\right)}{\left(1+t\right){\left(1-t\right)}^{1+2\gamma }}$
(3.4)

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

$\frac{{u}^{\prime }\left(t\right)}{u\left(t\right)}=\frac{{\lambda }^{\prime }\left(t\right)}{\lambda \left(t\right)}+\frac{2\left(\gamma +\left(1+\gamma \right)t\right)}{1-{t}^{2}}.$
(3.5)

Since

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

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

$\left(c-a-3-2\gamma \right){t}^{2}+\left(c-a-b-2\gamma \right)t+1-b\ge -t\left(1-{t}^{2}\right)\frac{{\phi }^{\prime }\left(1-t\right)}{\phi \left(1-t\right)}\phantom{\rule{1em}{0ex}}\left(0
(3.6)

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

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, $0, and $c\ge a+b+2\gamma +1$. Define $\beta =\beta \left(a,b,c,\gamma \right)$ by

$\begin{array}{rl}\frac{\beta -\frac{1}{2}}{1-\beta }=& -\frac{\mathrm{\Gamma }\left(c\right)}{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{\int }_{0}^{1}\frac{{\left(1-t\right)}^{c-a-b}{t}^{b-1}}{\mathrm{\Gamma }\left(c-a-b+1\right)}\frac{1-\gamma \left(1+t\right)}{\left(1-\gamma \right){\left(1+t\right)}^{2}}\\ {×}_{2}{F}_{1}\left(c-a,1-a;c-a-b+1;1-t\right)\phantom{\rule{0.2em}{0ex}}dt.\end{array}$

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

Proof If we put

$\lambda \left(t\right)={\frac{\mathrm{\Gamma }\left(c\right){t}^{b-1}{\left(1-t\right)}^{c-a-b}}{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)\mathrm{\Gamma }\left(c-a-b+1\right)}}_{2}{F}_{1}\left(c-a,1-a;c-a-b+1;1-t\right),$

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, and $c\ge b+2\gamma +2$. Also let

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

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

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

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

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

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

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

$\begin{array}{rcl}{\mathcal{J}}_{0,z}^{\lambda ,\mu ,\nu }f\left(z\right)& =& \mathcal{L}\left(2,2-\mu \right)\mathcal{L}\left(2-\mu +\nu ,2+\lambda +\nu \right)f\left(z\right)\\ =& \mathcal{L}\left(1,2-\mu \right)\mathcal{L}\left(2,1\right)\mathcal{L}\left(2-\mu +\nu ,2+\lambda +\nu \right)f\left(z\right).\end{array}$
(3.7)

By using Corollary 2, we obtain

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

Since $0\le \mu <2$, from (1.6), (1.8) and (3.7) we have ${\mathcal{J}}_{0,z}^{\lambda ,\mu ,\nu }f\left(z\right)\in \mathcal{R}\left(\mu /2,\gamma \right)$, 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 $2\left(\gamma -1\right)<\nu \le 2\gamma -1$. Define $\beta =\beta \left(\lambda ,\nu ,\gamma \right)$ by

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

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

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

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

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

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

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

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

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 $\beta =\beta \left(\lambda ,\mu ,\nu ,\gamma \right)$ by

$\frac{\beta -\frac{1}{2}}{1-\beta }=-\frac{\mathrm{\Gamma }\left(2-\mu \right)\mathrm{\Gamma }\left(2+\lambda +\nu \right)}{\mathrm{\Gamma }\left(\lambda \right)\mathrm{\Gamma }\left(2-\mu +\nu \right)}{\int }_{0}^{1}{\frac{t{\left(1-t\right)}^{\lambda -1}\left(1-\gamma \left(1+t\right)\right)}{\left(1-\gamma \right){\left(1+t\right)}^{2}}}_{2}{F}_{1}\left(\lambda +\mu ,-\nu ;\lambda ;1-t\right)\phantom{\rule{0.2em}{0ex}}dt.$

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

$\lambda \left(t\right)=\frac{\mathrm{\Gamma }\left(2-\mu \right)\mathrm{\Gamma }\left(2+\lambda +\nu \right)}{\mathrm{\Gamma }\left(\lambda \right)\mathrm{\Gamma }\left(2-\mu +\nu \right)}t\left(1-t\right)^{\lambda -1}{}_{2}{F}_{1}\left(\lambda +\mu ,-\nu ;\lambda ;1-t\right).$
(3.8)

Then we can easily see that ${\int }_{0}^{1}\lambda \left(t\right)\phantom{\rule{0.2em}{0ex}}dt=1$, $\mathrm{\Lambda }\left(t\right)={\int }_{t}^{1}\lambda \left(s\right)\phantom{\rule{0.2em}{0ex}}ds/s$ is monotone decreasing on $\left[0,1\right]$ and ${lim}_{t\to 0+}t\mathrm{\Lambda }\left(t\right)=0$. Also we find that the function $u\left(t\right)=\lambda \left(t\right)/\left(1+t\right){\left(1-t\right)}^{1+2\gamma }$ is decreasing on $\left(0,1\right)$, where $\lambda \left(t\right)$ is given by (3.8). Hence, $t{\mathrm{\Lambda }}^{\prime }\left(t\right)/\left(1+t\right){\left(1-t\right)}^{1+2\gamma }=-u\left(t\right)$ is increasing on $\left(0,1\right)$. From Lemma 4, we obtain the desired result. □

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## Acknowledgements

The first author was supported by Yeungnam University (2013).

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Correspondence to Jae Ho Choi.

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Kim, Y.C., Choi, J.H. Geometric properties of certain analytic functions associated with generalized fractional integral operators. J Inequal Appl 2014, 177 (2014). https://doi.org/10.1186/1029-242X-2014-177