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# Properties of multivalent functions associated with the integral operator defined by the hypergeometric function

Journal of Inequalities and Applications20132013:458

https://doi.org/10.1186/1029-242X-2013-458

• Received: 15 February 2013
• Accepted: 6 September 2013
• Published:

## Abstract

In this paper, we introduce a new class of multivalent functions by using a generalized integral operator defined by the hypergeometric function. Some properties such as inclusion, radius problem and integral preserving are considered.

MSC:30C45, 30C50.

## Keywords

• Bazilevic functions
• multivalent functions
• integral operator
• hypergeometric functions

## 1 Introduction and preliminaries

Let ${A}_{p}$ denote the class of functions $f\left(z\right)$ of the form
$f\left(z\right)={z}^{p}+\sum _{n=p+1}^{\mathrm{\infty }}{a}_{n}{z}^{n}\phantom{\rule{1em}{0ex}}\left(p\in \mathbb{N}=\left\{1,2,3,\dots \right\}\right),$
(1.1)
which are analytic in the open unit disc E. Also ${A}_{1}=A$, the usual class of analytic functions defined in the open unit disc $E=\left\{z:|z|<1\right\}$. A function $f\in {A}_{p}$ is a p-valent starlike function of order ρ if and only if
$Re\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}>\rho ,\phantom{\rule{1em}{0ex}}0\le \rho
This class of functions is denoted by ${S}_{p}^{\ast }\left(\rho \right)$. It is noted that ${S}_{p}^{\ast }\left(0\right)={S}_{p}^{\ast }$. Let $f\left(z\right)$ and $g\left(z\right)$ be analytic in E, we say $f\left(z\right)$ is subordinate to $g\left(z\right)$, written $f\prec g$ or $f\left(z\right)\prec g\left(z\right)$ if there exists a Schwarz function $w\left(z\right)$, $w\left(0\right)=0$ and $|w\left(z\right)|<1$ in E, then $f\left(z\right)=g\left(w\left(z\right)\right)$. In particular, if g is univalent in E, then we have the following equivalence
$f\left(z\right)\prec g\left(z\right)\phantom{\rule{1em}{0ex}}⟺\phantom{\rule{1em}{0ex}}f\left(0\right)=g\left(0\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}f\left(E\right)\subset g\left(E\right).$
For any two analytic functions $f\left(z\right)$ and $g\left(z\right)$ with
$f\left(z\right)=\sum _{n=0}^{\mathrm{\infty }}{b}_{n}{z}^{n+1}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}g\left(z\right)=\sum _{n=0}^{\mathrm{\infty }}{c}_{n}{z}^{n+1},\phantom{\rule{1em}{0ex}}z\in E,$
the convolution (the Hadamard product) is given by
$\left(f\ast g\right)\left(z\right)=\sum _{n=0}^{\mathrm{\infty }}{b}_{n}{c}_{n}{z}^{n+1},\phantom{\rule{1em}{0ex}}z\in E.$
A function $f\in A$ is said to be in the class, denoted by $\mathit{SD}\left(k,\delta \right)$ ($0\le \delta <1$), if and only if
$Re\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}>k|\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-1|+\delta ,\phantom{\rule{1em}{0ex}}k\ge 0,z\in E.$
(1.2)
Similarly, a function $f\in A$ is said to be in the class, denoted by $\mathit{CD}\left(k,\delta \right)$ of k-uniformly convex of order δ ($0\le \delta <1$), if
$Re\left\{1+\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\right\}>k|\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}|+\delta ,\phantom{\rule{1em}{0ex}}k\ge 0,z\in E.$
(1.3)
Geometric interpretation The functions $f\in \mathit{SD}\left(k,\delta \right)$ and $f\in \mathit{CD}\left(k,\delta \right)$ if and only if $\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}$ and $\frac{z{f}^{″}\left(z\right)}{f\left(z\right)}+1$, respectively, take all the values in the conic domain ${\mathrm{\Omega }}_{k,\delta }$ defined by
${\mathrm{\Omega }}_{k,\delta }=\left\{u+iv:u>k\sqrt{{\left(u-1\right)}^{2}+{v}^{2}}+\delta \right\}$
with $p\left(z\right)=\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}$ or $p\left(z\right)=\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}+1$ and considering the functions which map E onto the conic domain ${\mathrm{\Omega }}_{k,\delta }$ such that $1\in {\mathrm{\Omega }}_{k,\delta }$. One may rewrite the conditions (1.2) or (1.3) in the form
$p\left(z\right)\prec {q}_{k,\delta }\left(z\right).$
The function ${q}_{k,\delta }\left(z\right)$ plays the role of extremal for these classes and is given by
${q}_{k,\delta }\left(z\right)=\left\{\begin{array}{cc}\frac{1+\left(1-2\delta \right)z}{1-z},\hfill & k=0,\hfill \\ 1+\frac{2\delta \gamma }{{\pi }^{2}}{\left(log\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{2},\hfill & k=1,\hfill \\ 1+\frac{2\delta }{1-{k}^{2}}{sinh}^{2}\left[\left(\frac{2}{\pi }arccosk\right)arctanh\sqrt{z}\right],\hfill & 01.\hfill \end{array}$
(1.4)
For $f\left(z\right)$ in ${A}_{p}$, the operator ${D}^{\mu +p-1}:{A}_{p}⟶{A}_{p}$ is defined by
${D}^{\mu +p-1}f\left(z\right)=\frac{{z}^{p}}{{\left(1-z\right)}^{\mu +p}}\ast f\left(z\right)\phantom{\rule{1em}{0ex}}\left(\mu >-p\right),$
or equivalently
${D}^{\mu +p-1}f\left(z\right)=\frac{{z}^{p}{\left({z}^{\mu -1}f\left(z\right)\right)}^{\mu +p-1}}{\left(\mu +p-1\right)!},$
(1.5)
where μ is any integer greater than −p. If $f\left(z\right)$ is given by (1.1), then it follows that
${D}^{\mu +p-1}f\left(z\right)={z}^{p}+\sum _{n=p+1}^{\mathrm{\infty }}\frac{\left(\mu +n-1\right)!}{\left(n-p\right)!\left(\mu +p-1\right)!}{a}_{n}{z}^{n}.$
The symbol ${D}^{\mu +p-1}$ when $p=1$, was introduced by Ruscheweyh  and ${D}^{\mu +p-1}$ is called the $\left(\mu +p-1\right)$th order Ruscheweyh derivative. We now introduce a function ${\left({z}_{2}^{p}{F}_{1}\left(a,b,c;z\right)\right)}^{-1}$ given by
$\left(z^{p}{}_{2}{F}_{1}\left(a,b,c;z\right)\right)\ast {\left(z^{p}{}_{2}{F}_{1}\left(a,b,c;z\right)\right)}^{-1}=\frac{{z}^{p}}{{\left(1-z\right)}^{\mu +p}}\phantom{\rule{1em}{0ex}}\left(\mu >-p\right),$
and the following linear operator
${I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)={\left(z^{p}{}_{2}{F}_{1}\left(a,b,c;z\right)\right)}^{-1}\ast f\left(z\right),$
(1.6)

where a, b, c are real or complex numbers other than $0,-1,-2,\dots$ , $\mu >-p$, $z\in E$ and $f\left(z\right)\in {A}_{p}$. This operator was recently introduced in . In particular, for $p=1$, this operator is studied by Noor . For $b=1$, this operator reduces to the well-known Cho-Kwon-Srivastava operator ${I}_{\mu ,p}\left(a,c\right)$, which was studied by Cho et al. , and for $\mu =1$, $b=c$, $a=n+p$, see . For $a=n+p$, $b=c=1$, this operator was investigated by Liu  and Liu and Noor .

Simple computations yield
${I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)={z}^{p}+\sum _{n=p+1}^{\mathrm{\infty }}\frac{{\left(c\right)}_{n}{\left(\mu +p\right)}_{n}}{{\left(a\right)}_{n}{\left(b\right)}_{n}}{a}_{n}{z}^{n}.$
From (1.6), we note that
$\begin{array}{r}{I}_{\mu ,1}\left(a,b,c\right)f\left(z\right)={I}_{\mu }\left(a,b,c\right)f\left(z\right)\phantom{\rule{1em}{0ex}}\text{(see )},\\ {I}_{0,p}\left(a,p,a\right)f\left(z\right)=f\left(z\right),{I}_{1,p}\left(a,p,a\right)f\left(z\right)=\frac{z{f}^{\prime }\left(z\right)}{p}.\end{array}$
Also, it can be easily seen that
$z{\left({I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)\right)}^{\prime }=\left(\mu +p\right){I}_{\mu +1,p}\left(a,b,c\right)f\left(z\right)-\mu {I}_{\mu ,p}\left(a,b,c\right)f\left(z\right),$
(1.7)
and
$z{\left({I}_{\mu ,p}\left(a+1,b,c\right)f\left(z\right)\right)}^{\prime }=a{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)-\left(a-p\right){I}_{\mu ,p}\left(a,b,c\right)f\left(z\right).$

We define the following class of multivalent analytic functions by using the operator ${I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)$ above.

Definition 1.1 Let $f\in {A}_{p}$ for $p\in \mathbb{N}$. Then $f\in U{B}_{\mu ,p}^{\alpha }\left(a,b,c;\gamma ,k,\delta \right)$ for $a,b,c\in \mathbb{R}\mathrm{\setminus }{\mathbb{Z}}_{0}^{-}$, $\mu >-p$, $\alpha >0$, $k\ge 0$, $0\le \delta <1$ and $\gamma >0$ if and only if
$\begin{array}{r}\left(1-\gamma \right){\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)}^{\alpha }+\gamma \frac{{I}_{\mu +1,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu +1,p}\left(a,b,c\right)g\left(z\right)}{\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)}^{\alpha -1}\\ \phantom{\rule{1em}{0ex}}\prec {q}_{k,\delta }\left(z\right),\end{array}$
(1.8)
where $g\in {A}_{p}$ is such that
$q\left(z\right)=\frac{{I}_{\mu +1,p}\left(a,b,c\right)g\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\in P\left(\rho \right),\phantom{\rule{1em}{0ex}}\rho =\frac{k+\delta }{k+1},z\in E.$
(1.9)
Furthermore, for different choices of parameters being involved, we obtain many other well-known subclasses of the class ${A}_{p}$ and A as special cases.
1. (i)

$a=c$, $b=1$, $k=0$, $\mu =m\in {\mathbb{N}}_{0}$, we have ${B}_{m,p}^{\alpha }\left(\gamma ,\delta \right)$ studied in .

2. (ii)
$a=c=b=p=\gamma =1$, $k=\mu =0$, $g\left(z\right)=z$, the class $U{B}_{\mu ,p}^{\alpha }\left(a,b,c;\gamma ,k,\delta \right)$ reduces to the class
${B}^{\alpha }\left(\delta \right)=\left\{f\in A\left(1\right):\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}{\left(\frac{f\left(z\right)}{z}\right)}^{\alpha }\in P\left(\delta \right)\right\}$

studied in .

3. (iii)

$a=c=b=p=\gamma =1$, $k=\mu =0$, $g\left(z\right)=z$, the $U{B}_{\mu ,p}^{\alpha }\left(a,b,c;\gamma ,k,\delta \right)$ reduces to $B\left(\alpha \right)$ is the class of Bazilevich functions investigated by Singh .

4. (iv)
$a=c=b=p=\alpha =1$, $\gamma =0$, $k=\mu =0$, $g\left(z\right)=z$, the class $U{B}_{\mu ,p}^{\alpha }\left(a,b,c;\gamma ,k,\delta \right)$ reduces to the class
${B}_{\delta }=\left\{f\in A\left(1\right):\frac{f\left(z\right)}{z}\in P\left(\delta \right)\right\},$

the class studied by Chen .

Let $f\in {A}_{p\cdot }$ and ${F}_{\eta ,p}:{A}_{p\cdot }\to {A}_{p\cdot }$ be defined by
${F}_{\eta ,p}\left(z\right)=\frac{\left(\eta +p\right)}{{z}^{\eta }}{\int }_{0}^{z}{t}^{\eta -1}f\left(t\right)\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}\eta >-p.$
(1.10)

We need the following lemmas which will be used in our main results.

Lemma 1.2 

Let $u={u}_{1}+i{u}_{2}$ and $v={v}_{1}+i{v}_{2}$, and let $\psi :D\subset {\mathbb{C}}^{2}\to \mathbb{C}$ be a complex-valued function satisfying the conditions:
1. (i)

$\psi \left(u,v\right)$ is continuous in a domain $D\subset {\mathbb{C}}^{2}$,

2. (ii)

$\left(1,0\right)\in D$ and $\psi \left(1,0\right)>0$,

3. (iii)

$Re\psi \left(i{u}_{2},{v}_{1}\right)\le 0$, whenever $\left(i{u}_{2},{v}_{1}\right)\in D$ and ${v}_{1}\le -\frac{1}{2}\left(1+{u}_{2}^{2}\right)$.

If $h\left(z\right)=1+{c}_{1}z+{c}_{2}{z}^{2}+\cdots$ is analytic in E such that $\left(h,z{h}^{\prime }\right)\in D$ and $Re\psi \left(h\left(z\right),z{h}^{\prime }\left(z\right)\right)>0$ for $z\in E$, then $Reh\left(z\right)>0$.

Lemma 1.3 

Let h be convex in the unit disc E, and let $A\ge 0$. Suppose that $B\left(z\right)$ is analytic in E with $ReB\left(z\right)\ge A$. If g is analytic in E and $g\left(0\right)=h\left(0\right)$. Then
$A{z}^{2}{g}^{″}\left(z\right)+B\left(z\right)z{g}^{\prime }\left(z\right)+g\left(z\right)\prec h\left(z\right)\phantom{\rule{1em}{0ex}}\mathit{\text{implies that}}\phantom{\rule{1em}{0ex}}g\left(z\right)\prec h\left(z\right).$

Lemma 1.4 

Let F be analytic and convex in E. If $f,g\in {A}_{p}$ and $f,g\prec F$. Then
$\sigma f+\left(1-\sigma \right)g\prec F,\phantom{\rule{1em}{0ex}}0\le \sigma \le 1.$

Lemma 1.5 

Let h be convex in E with $h\left(0\right)=a$ and $\beta \in \mathbb{C}$ such that $Re\beta \ge 0$. If $p\in H\left[a,n\right]$ and
$p\left(z\right)+\frac{z{p}^{\prime }\left(z\right)}{\beta }\prec h\left(z\right),$
then $p\left(z\right)\prec q\left(z\right)\prec h\left(z\right)$, where
$q\left(z\right)=\frac{\beta }{n{z}^{\beta /n}}{\int }_{0}^{z}h\left(t\right){t}^{\beta /n-1}\phantom{\rule{0.2em}{0ex}}dt$

and $q\left(z\right)$ is the best dominant.

## 2 Main results

Theorem 2.1 Let $f\in U{B}_{\mu ,p}^{\alpha }\left(a,b,c;\gamma ,k,\delta \right)$ for $a,b,c\in \mathbb{R}\mathrm{\setminus }{\mathbb{Z}}_{0}^{-}$, $\mu >-p$, $p\in \mathbb{N}$, $\alpha >0$, $k\ge 0$, $0\le \delta <1$ and $\gamma >0$. Then $f\in U{B}_{\mu ,p}^{\alpha }\left(a,b,c;0,k,\delta \right)$.

Proof Consider
$h\left(z\right)={\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)}^{\alpha },$
(2.1)
where h is analytic in E with $h\left(0\right)=1$, and $g\in {A}_{p}$ satisfies condition (1.9). Differentiating (2.1) logarithmically and using (1.7), we have
$\frac{z{h}^{\prime }\left(z\right)}{h\left(z\right)}=\alpha \left(\mu +p\right)\left\{\left(\frac{{I}_{\mu +1,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}-\frac{{I}_{\mu +1,p}\left(a,b,c\right)g\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)\right\}.$
Using (2.1) and simplifying, we obtain
$\begin{array}{rcl}h\left(z\right)+\frac{\gamma z{h}^{\prime }\left(z\right)}{\alpha \left(\mu +p\right)q\left(z\right)}& =& \left(1-\gamma \right){\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)}^{\alpha }\\ +\gamma \frac{{I}_{\mu +1,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu +1,p}\left(a,b,c\right)g\left(z\right)}{\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)}^{\alpha -1}.\end{array}$
Since $f\in U{B}_{\mu ,p}^{\alpha }\left(a,b,c;\gamma ,k,\delta \right)$, therefore, we can write
$h\left(z\right)+\frac{\gamma z{h}^{\prime }\left(z\right)}{\alpha \left(\mu +p\right)q\left(z\right)}\prec {q}_{k,\delta }\left(z\right),\phantom{\rule{1em}{0ex}}z\in E.$

Now using Lemma 1.3 for $A=0$ and $B\left(z\right)=\frac{\gamma }{\alpha \left(\mu +p\right)q\left(z\right)}$ with $Req\left(z\right)>0$, we have $ReB\left(z\right)\ge 0$, therefore, $h\left(z\right)\prec {q}_{k,\delta }\left(z\right)$. Hence $f\in U{B}_{\mu ,p}^{\alpha }\left(a,b,c;0,k,\delta \right)$. □

Theorem 2.2 Let $f\in U{B}_{\mu ,p}^{\alpha }\left(a,b,c;\gamma ,0,\delta \right)$ for $a,b,c\in \mathbb{R}\mathrm{\setminus }{\mathbb{Z}}_{0}^{-}$, $\mu >-p$, $p\in \mathbb{N}$. Then $f\in U{B}_{\mu ,p}^{\alpha }\left(a,b,c;0,0,{\delta }_{1}\right)$, where
${\delta }_{1}=\frac{2\alpha \delta \left(p+\mu \right)|q\left(z\right){|}^{2}+\gamma \rho }{2\alpha \left(p+\mu \right)|q\left(z\right){|}^{2}+\gamma \rho }.$
Proof Consider
$h\left(z\right)=\frac{1}{\left(1-{\delta }_{1}\right)}\left\{{\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)}^{\alpha }-{\delta }_{1}\right\},$
(2.2)
where h is analytic in E with $h\left(0\right)=1$, and $g\in {A}_{p}$ satisfies condition (1.9). Differentiating (2.2), we have
$\begin{array}{rcl}\frac{\left(1-{\delta }_{1}\right)}{\alpha }{h}^{\prime }\left(z\right)& =& {\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)}^{\alpha -1}\\ ×\left\{\frac{{\left({I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)\right)}^{\prime }}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}-\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\frac{{\left({I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)\right)}^{\prime }}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right\}.\end{array}$
Using (1.7) and simplifying, we obtain
$\begin{array}{r}\left(1-\gamma \right){\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)}^{\alpha }+\gamma \frac{{I}_{\mu +1,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu +1,p}\left(a,b,c\right)g\left(z\right)}{\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)}^{\alpha -1}\\ \phantom{\rule{1em}{0ex}}=\left(1-{\delta }_{1}\right)h\left(z\right)+{\delta }_{1}+\frac{\gamma \left(1-{\delta }_{1}\right)z{h}^{\prime }\left(z\right)}{\alpha \left(p+\mu \right)q\left(z\right)}.\end{array}$
Since $f\in U{B}_{\mu ,p}^{\alpha }\left(a,b,c;\gamma ,0,\delta \right)$, therefore we have
$\left(1-{\delta }_{1}\right)h\left(z\right)+{\delta }_{1}+\frac{\gamma \left(1-{\delta }_{1}\right)z{h}^{\prime }\left(z\right)}{\alpha \left(p+\mu \right)q\left(z\right)}\prec \frac{1+\left(1-2\delta \right)z}{1-z},\phantom{\rule{1em}{0ex}}0\le \delta <1,z\in E.$
This implies that
$\frac{1}{1-\delta }\left\{\left(1-{\delta }_{1}\right)h\left(z\right)+{\delta }_{1}-\delta +\frac{\gamma \left(1-{\delta }_{1}\right)z{h}^{\prime }\left(z\right)}{\alpha \left(p+\mu \right)q\left(z\right)}\right\}\in {q}_{0,0}\left(E\right)=P.$
To obtain our desired result, we show that $h\in P$, for $z\in E$. Let $u={u}_{1}+i{u}_{2}$, $v={v}_{1}+i{v}_{2}$, and let $\mathrm{\Psi }:D\subset {\mathbb{C}}^{2}\to \mathbb{C}$ be a complex-valued function such that $u=h\left(z\right)$, $v=z{h}^{\prime }\left(z\right)$. Then
$\mathrm{\Psi }\left(u,v\right)=\left(1-{\delta }_{1}\right)u+{\delta }_{1}-\delta +\frac{\gamma \left(1-{\delta }_{1}\right)v}{\alpha \left(p+\mu \right)q\left(z\right)}.$
The first two conditions of Lemma 1.2 are easily verified. To verify the third condition, we consider
$\begin{array}{r}Re\mathrm{\Psi }\left(i{u}_{2},{v}_{1}\right)\\ \phantom{\rule{1em}{0ex}}=Re\left\{\left(1-{\delta }_{1}\right)i{u}_{2}+{\delta }_{1}-\delta +\frac{\gamma \left(1-{\delta }_{1}\right){v}_{1}}{\alpha \left(p+\mu \right)q\left(z\right)}\right\}\\ \phantom{\rule{1em}{0ex}}={\delta }_{1}-\delta +Re\frac{\gamma \left(1-{\delta }_{1}\right){v}_{1}}{\alpha \left(p+\mu \right)q\left(z\right)}\\ \phantom{\rule{1em}{0ex}}\le {\delta }_{1}-\delta -Re\frac{\gamma \left(1-{\delta }_{1}\right)\left(1+{u}_{2}^{2}\right)\overline{q\left(z\right)}}{2\alpha \left(p+\mu \right)|q\left(z\right){|}^{2}}\\ \phantom{\rule{1em}{0ex}}\le {\delta }_{1}-\delta -\frac{\gamma \left(1-{\delta }_{1}\right)\left(1+{u}_{2}^{2}\right)\rho }{2\alpha \left(p+\mu \right)|q\left(z\right){|}^{2}}=\frac{A+B{u}^{2}}{C},\end{array}$

where $A=2\alpha \left(p+\mu \right)\left({\delta }_{1}-\delta \right)|q\left(z\right){|}^{2}-\gamma \rho \left(1-{\delta }_{1}\right)$, $B=-\gamma \rho \left(1-{\delta }_{1}\right)\le 0$ if $0\le {\delta }_{1}<1$ and $C=2\alpha \left(p+\mu \right)|q\left(z\right){|}^{2}>0$. From the relation ${\delta }_{1}=\frac{2\alpha \delta \left(p+\mu \right)|q\left(z\right){|}^{2}+\gamma \rho }{2\alpha \left(p+\mu \right)|q\left(z\right){|}^{2}+\gamma \rho }$, we have $A\le 0$. This implies that $Re\mathrm{\Psi }\left(i{u}_{2},{v}_{1}\right)\le 0$. Using Lemma 1.2, we have $h\in P$ for $z\in E$. This completes the proof. □

Theorem 2.3 Let $a,b,c\in \mathbb{R}\mathrm{\setminus }{\mathbb{Z}}_{0}^{-}$, $\mu >-p$, $p\in \mathbb{N}$, $\alpha >0$, $k\ge 0$ and $0\le \delta <1$. Then
$U{B}_{\mu ,p}^{\alpha }\left(a,b,c;{\gamma }_{2},k,\delta \right)\subset U{B}_{\mu ,p}^{\alpha }\left(a,b,c;{\gamma }_{1},k,\delta \right),\phantom{\rule{1em}{0ex}}0\le {\gamma }_{1}<{\gamma }_{2},z\in E.$
Proof Since $f\in U{B}_{\mu ,p}^{\alpha }\left(a,b,c;{\gamma }_{2},k,\delta \right)$, therefore, we have
$\begin{array}{r}\left(1-{\gamma }_{2}\right){\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)}^{\alpha }+{\gamma }_{2}\frac{{I}_{\mu +1,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu +1,p}\left(a,b,c\right)g\left(z\right)}{\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)}^{\alpha -1}\\ \phantom{\rule{1em}{0ex}}={h}_{1}\left(z\right)\prec {q}_{k,\delta }\left(z\right),\end{array}$
(2.3)
where $g\in {A}_{p}$ satisfies condition (1.9). From Theorem 2.1, we write
${\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)}^{\alpha }={h}_{2}\left(z\right)\prec {q}_{k,\delta }\left(z\right),\phantom{\rule{1em}{0ex}}z\in E.$
(2.4)
Now, for ${\gamma }_{1}\ge 0$, we obtain
$\begin{array}{r}\left(1-{\gamma }_{1}\right){\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)}^{\alpha }+{\gamma }_{1}\frac{{I}_{\mu +1,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu +1,p}\left(a,b,c\right)g\left(z\right)}{\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)}^{\alpha -1}\\ \phantom{\rule{1em}{0ex}}=\left(1-\frac{{\gamma }_{1}}{{\gamma }_{2}}\right){\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)}^{\alpha }\\ \phantom{\rule{2em}{0ex}}+\frac{{\gamma }_{1}}{{\gamma }_{2}}\left\{\left(1-{\gamma }_{2}\right){\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)}^{\alpha }+{\gamma }_{2}\frac{{I}_{\mu +1,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu +1,p}\left(a,b,c\right)g\left(z\right)}{\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)}^{\alpha -1}\right\}\\ \phantom{\rule{1em}{0ex}}=\frac{{\gamma }_{1}}{{\gamma }_{2}}{h}_{1}\left(z\right)+\left(1-\frac{{\gamma }_{1}}{{\gamma }_{2}}\right){h}_{2}\left(z\right).\end{array}$
Using the convexity of the class of the function ${q}_{k,\delta }\left(z\right)$ and Lemma 1.4, we write
$\frac{{\gamma }_{1}}{{\gamma }_{2}}{h}_{1}\left(z\right)+\left(1-\frac{{\gamma }_{1}}{{\gamma }_{2}}\right){h}_{2}\left(z\right)\prec {q}_{k,\delta }\left(z\right),\phantom{\rule{1em}{0ex}}z\in E,$

where ${h}_{1}$ and ${h}_{2}$ are given by (2.3) and (2.4), respectively. This implies that $f\in U{B}_{\mu ,p}^{\alpha }\left(a,b,c;{\gamma }_{1},k,\delta \right)$. Hence the proof of the theorem is completed. □

Theorem 2.4 Let $f\in U{B}_{\mu ,p}^{1}\left(a,b,c;\gamma ,k,\delta \right)$, $a,b,c\in \mathbb{R}\mathrm{\setminus }{\mathbb{Z}}_{0}^{-}$, $\mu >-p$, $p\in \mathbb{N}$, $\alpha >0$, $k\ge 0$, $\gamma \ge 1$ and $0\le \delta <1$. Then $f\in U{B}_{\mu +1,p}^{1}\left(a,b,c;1,k,\delta \right)$.

Proof Since $f\in U{B}_{\mu ,p}^{1}\left(a,b,c;\gamma ,k,\delta \right)$, therefore, we have
$\left(1-\gamma \right)\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)+\gamma \frac{{I}_{\mu +1,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu +1,p}\left(a,b,c\right)g\left(z\right)}\prec {q}_{k,\delta }\left(z\right).$
Now, consider
$\begin{array}{rcl}\gamma \frac{{I}_{\mu +1,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu +1,p}\left(a,b,c\right)g\left(z\right)}& =& \left(1-\gamma \right)\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)+\gamma \frac{{I}_{\mu +1,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu +1,p}\left(a,b,c\right)g\left(z\right)}\\ +\left(\gamma -1\right)\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right).\end{array}$
This implies that
$\begin{array}{rcl}\frac{{I}_{\mu +1,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu +1,p}\left(a,b,c\right)g\left(z\right)}& =& \frac{1}{\gamma }\left\{\left(1-\gamma \right)\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)+\gamma \frac{{I}_{\mu +1,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu +1,p}\left(a,b,c\right)g\left(z\right)}\right\}\\ +\left(1-\frac{1}{\gamma }\right)\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right).\end{array}$

Using Theorem 2.1, Lemma 1.4 and the convexity of ${q}_{k,\delta }\left(z\right)$, we have the required result. □

Now, using the operators ${I}_{\mu ,p}\left(a,b,c\right)$ and ${F}_{\eta ,p}$ defined by (1.7) and (1.10), respectively, we have
$z{\left({I}_{p}^{\mu }\left(a,b,c\right){F}_{\eta ,p}\left(f\right)\left(z\right)\right)}^{\prime }=\left(p+\eta \right){I}_{p}^{\mu }\left(a,b,c\right)f\left(z\right)-\eta {I}_{p}^{\mu }\left(a,b,c\right){F}_{\eta ,p}\left(f\right)\left(z\right),\phantom{\rule{1em}{0ex}}\eta >-p.$
(2.5)
Theorem 2.5 Let $f\in {A}_{p}$ and ${F}_{\eta ,p}$ be given by (1.10). If
$\left(1-\gamma \right)\frac{{I}_{\mu ,p}\left(a,b,c\right){F}_{\eta ,p}\left(f\right)\left(z\right)}{{z}^{p}}+\gamma \frac{{I}_{\mu ,p}\left(a,b,c\right)\left(f\left(z\right)\right)}{{z}^{p}}\prec {q}_{k,\delta }\left(z\right),\phantom{\rule{1em}{0ex}}z\in E,$
(2.6)
with $a,b,c\in \mathbb{R}\mathrm{\setminus }{\mathbb{Z}}_{0}^{-}$, $\mu ,\eta >-p$, $p\in \mathbb{N}$, $\gamma >0$, then
$\frac{{I}_{\mu ,p}\left(a,b,c\right){F}_{\eta ,p}\left(f\left(z\right)\right)}{{z}^{p}}\prec h\left(z\right)\prec {q}_{k,\delta }\left(z\right),\phantom{\rule{1em}{0ex}}z\in E,$
where
$h\left(z\right)=\frac{p+\eta }{\gamma {z}^{\left(p+\eta \right)/\gamma }}{\int }_{0}^{z}{q}_{k,\delta }\left(z\right){t}^{{}^{\left(p+\eta \right)/\gamma }-1}\phantom{\rule{0.2em}{0ex}}dt.$
Proof Let
$\frac{{I}_{\mu ,p}\left(a,b,c\right){F}_{\eta ,p}\left(f\left(z\right)\right)}{{z}^{p}}={h}_{1}\left(z\right),\phantom{\rule{1em}{0ex}}z\in E,$
where ${h}_{1}$ is analytic in E with ${h}_{1}\left(0\right)=1$. Then
$z{\left({I}_{\mu ,p}\left(a,b,c\right){F}_{\eta ,p}\left(f\left(z\right)\right)\right)}^{\prime }=p{z}^{p}{h}_{1}\left(z\right)+{z}^{p+1}{h}_{1}^{\prime }\left(z\right).$
Using (2.5), we have
$\gamma \left(p+\eta \right)\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{z}^{p}}-\gamma \eta \frac{{I}_{\mu ,p}\left(a,b,c\right){F}_{\eta ,p}\left(f\right)\left(z\right)}{{z}^{p}}=p\gamma {h}_{1}\left(z\right)+\gamma z{h}_{1}^{\prime }\left(z\right).$
Thus,
$\left(1-\gamma \right)\frac{{I}_{\mu ,p}\left(a,b,c\right){F}_{\eta ,p}\left(f\left(z\right)\right)}{{z}^{p}}+\gamma \frac{{I}_{\mu ,p}\left(a,b,c\right)\left(f\left(z\right)\right)}{{z}^{p}}={h}_{1}\left(z\right)+\gamma \frac{z{h}_{1}^{\prime }\left(z\right)}{p+\eta }.$
(2.7)
From (2.7), it follows that
${h}_{1}\left(z\right)+\gamma \frac{z{h}_{1}^{\prime }\left(z\right)}{p+\eta }\prec {q}_{k,\delta }\left(z\right),\phantom{\rule{1em}{0ex}}z\in E.$

Using Lemma 1.5, for ${\beta }_{1}=\frac{p+\eta }{\gamma }$, $n=1$ and $a=1$, we obtain ${h}_{1}\left(z\right)\prec h\left(z\right)\prec {q}_{k,\delta }\left(z\right)$. That is, $\frac{{I}_{\mu ,p}\left(a,b,c\right){F}_{\eta ,p}\left(f\left(z\right)\right)}{{z}^{p}}\prec {q}_{k,\delta }\left(z\right)$. □

Theorem 2.6 Let $f\in U{B}_{\mu ,p}^{\alpha }\left(a,b,c;0,0,\delta \right)$ for $a,b,c\in \mathbb{R}\mathrm{\setminus }{\mathbb{Z}}_{0}^{-}$, $\mu >-p$, $p\in \mathbb{N}$, $\alpha ,\gamma >0$, $0\le \delta <1$. Then $f\in U{B}_{\mu ,p}^{\alpha }\left(a,b,c;\gamma ,0,\delta \right)$, for $|z|<{r}_{0}$, where
${r}_{0}=\frac{\alpha \left(p}{+}\mu \right)+\gamma -\sqrt{{\gamma }^{2}+2\alpha \gamma \left(p+\mu \right)}\alpha \left(p+\mu \right).$
(2.8)
Proof Let $f\in U{B}_{\mu ,p}^{\alpha }\left(a,b,c;0,0,\delta \right)$. Then we have
${\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)}^{\alpha }=\left(1-\delta \right)h\left(z\right)+\delta ,$
(2.9)
where $g\in {A}_{p}$ satisfies the condition
$q\left(z\right)=\frac{{I}_{\mu +1,p}\left(a,b,c\right)g\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\in P,\phantom{\rule{1em}{0ex}}z\in E$
and $h\in P$. Differentiating (2.9) and then using (1.7), we obtain
$\gamma \frac{{I}_{\mu +1,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu +1,p}\left(a,b,c\right)g\left(z\right)}{\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)}^{\alpha -1}-\gamma {\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)}^{\alpha }=\frac{\gamma \left(p-\delta \right)z{h}^{\prime }\left(z\right)}{\alpha \left(p+\mu \right)q\left(z\right)}.$
This implies that
$\begin{array}{r}\frac{1}{1-\delta }\left\{\left(1-\gamma \right){\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)}^{\alpha }+\gamma \frac{{I}_{\mu +1,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu +1,p}\left(a,b,c\right)g\left(z\right)}{\left(\frac{{I}_{\mu ,p}\left(a,b,c\right)f\left(z\right)}{{I}_{\mu ,p}\left(a,b,c\right)g\left(z\right)}\right)}^{\alpha -1}-\delta \right\}\\ \phantom{\rule{1em}{0ex}}=h\left(z\right)+\frac{\gamma z{h}^{\prime }\left(z\right)}{\alpha \left(p+\mu \right)q\left(z\right)},\phantom{\rule{1em}{0ex}}z\in E.\end{array}$
(2.10)
Now, using the well-known distortion result for class P, we have
$|z{h}^{\prime }\left(z\right)|\le \frac{2rReh\left(z\right)}{1-{r}^{2}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}Reh\left(z\right)\ge \frac{1-r}{1+r},\phantom{\rule{1em}{0ex}}|z|
Thus, due to the applications of these inequalities, we have
$\begin{array}{rl}Re\left(h\left(z\right)+\frac{\gamma z{h}^{\prime }\left(z\right)}{\alpha \left(p+\mu \right)q\left(z\right)}\right)& \ge Reh\left(z\right)-\frac{\gamma |z{h}^{\prime }\left(z\right)|}{\alpha \left(p+\mu \right)|q\left(z\right)|}\\ \ge Reh\left(z\right)\left(1-\frac{2\gamma r}{\alpha \left(p+\mu \right){\left(1-r\right)}^{2}}\right)\\ =Reh\left(z\right)\left(\frac{\alpha \left(p+\mu \right){\left(1-r\right)}^{2}-2\gamma r}{\alpha \left(p+\mu \right){\left(1-r\right)}^{2}}\right).\end{array}$

For $|z|<{r}_{0}$, where ${r}_{0}$ is given in (2.8), the inequality above is positive. Sharpness of the result follows by taking $h\left(z\right)=\frac{1+z}{1-z}$. Hence from (2.10), we have the required result. □

## Authors’ Affiliations

(1)
Department of Mathematics, GC University, Faisalabad, Pakistan
(2)
Department of Mathematics, COMSATS Institute of Information Technology, Defence Road, Lahore, Pakistan

## References 