Theory and Modern Applications

# Some properties of certain subclasses of analytic functions involving adifferential operator

## Abstract

In the present paper, we introduce and study certain subclasses of analyticfunctions in the open unit disk U which is defined by the differentialoperator $D{R}_{\lambda }^{m,n}$. We study and investigate some inclusionproperties of these classes. Furthermore, a generalizedBernardi-Libera-Livington integral operator is shown to be preserved for theseclasses.

MSC: 30C45.

## 1 Introduction

Let be aclass of functions f in the open unit disk $U=\left\{z\in \mathbb{C}:|z|<1\right\}$ normalized by $f\left(0\right)={f}^{\prime }\left(0\right)-1=0$. Thus each $f\in \mathcal{A}$ has a Taylor series representation

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

We denote by $\mathcal{S}\left(\xi \right)$ the well-known subclass of consisting of allanalytic functions which are, respectively, starlike of order ξ[1, 2]

$\mathcal{S}\left(\xi \right)=\left\{f\in \mathcal{A}:Re\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)>\xi ,z\in U\right\},\phantom{\rule{1em}{0ex}}0\le \xi <1.$

Let be a class of all functions ϕ which are analytic andunivalent in U and for which $\varphi \left(U\right)$ is convex with $\varphi \left(0\right)=1$ and $Re\varphi \left(z\right)>0$, $z\in U$.

For two functions f and g analytic in U, we say that thefunction f is subordinate to g in U and write$f\left(z\right)\prec g\left(z\right)$, $z\in U$, if there exists a Schwarz function$w\left(z\right)$ which is analytic in U with$w\left(0\right)=0$ and $|w\left(z\right)|<1$ such that $f\left(z\right)=g\left(w\left(z\right)\right)$, $z\in U$.

Making use of the principle of subordination between analytic functions, denote by$\mathcal{S}\left(\xi ,\varphi \right)$ a subclass of the class for$0\le \xi <1$ and $\varphi \in \mathcal{R}$ which are defined by

$\mathcal{S}\left(\xi ,\varphi \right)=\left\{f\in \mathcal{A}:\frac{1}{1-\xi }\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-\zeta \right)\prec \varphi \left(z\right),z\in U\right\}.$

Let $f,g\in \mathcal{A}$, where f and g are defined by$f\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}{a}_{j}{z}^{j}$ and $g\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}{b}_{j}{z}^{j}$. Then the Hadamard product (or convolution)$f\ast g$ of the functions f and g is definedby

$\left(f\ast g\right)\left(z\right)=z+\sum _{j=2}^{\mathrm{\infty }}{a}_{j}{b}_{j}{z}^{j}.$

Definition 1.1 (Al-Oboudi )

For $f\in \mathcal{A}$, $\lambda \ge 0$ and $m\in \mathbb{N}$, the operator ${D}_{\lambda }^{m}$ is defined by ${D}_{\lambda }^{m}:\mathcal{A}\to \mathcal{A}$,

$\begin{array}{c}{D}_{\lambda }^{0}f\left(z\right)=f\left(z\right),\hfill \\ {D}_{\lambda }^{1}f\left(z\right)=\left(1-\lambda \right)f\left(z\right)+\lambda z{f}^{\prime }\left(z\right)={D}_{\lambda }f\left(z\right),\hfill \\ \dots ,\hfill \\ {D}_{\lambda }^{m}f\left(z\right)=\left(1-\lambda \right){D}_{\lambda }^{m-1}f\left(z\right)+\lambda z{\left({D}_{\lambda }^{m}f\left(z\right)\right)}^{\prime }={D}_{\lambda }\left({D}_{\lambda }^{m-1}f\left(z\right)\right),\phantom{\rule{1em}{0ex}}z\in U.\hfill \end{array}$

Remark 1.1 If $f\in \mathcal{A}$ and $f\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}{a}_{j}{z}^{j}$, then ${D}_{\lambda }^{m}f\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}{\left[1+\left(j-1\right)\lambda \right]}^{m}{a}_{j}{z}^{j}$, $z\in U$.

Remark 1.2 For $\lambda =1$ in the above definition, we obtain theSălăgean differential operator .

Definition 1.2 (Ruscheweyh )

For $f\in \mathcal{A}$ and $n\in \mathbb{N}$, the operator ${R}^{n}$ is defined by ${R}^{n}:\mathcal{A}\to \mathcal{A}$,

$\begin{array}{c}{R}^{0}f\left(z\right)=f\left(z\right),\hfill \\ {R}^{1}f\left(z\right)=z{f}^{\prime }\left(z\right),\hfill \\ \dots ,\hfill \\ \left(n+1\right){R}^{n+1}f\left(z\right)=z{\left({R}^{n}f\left(z\right)\right)}^{\prime }+n{R}^{n}f\left(z\right),\phantom{\rule{1em}{0ex}}z\in U.\hfill \end{array}$

Remark 1.3 If $f\in \mathcal{A}$, $f\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}{a}_{j}{z}^{j}$, then ${R}^{n}f\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}{a}_{j}{z}^{j}$, $z\in U$.

Definition 1.3 ()

Let $\lambda \ge 0$ and $n,m\in \mathbb{N}$. Denote by $D{R}_{\lambda }^{m,n}:\mathcal{A}\to \mathcal{A}$ the operator given by the Hadamard product of thegeneralized Sălăgean operator ${D}_{\lambda }^{m}$ and the Ruscheweyh operator ${R}^{n}$,

$D{R}_{\lambda }^{m,n}f\left(z\right)=\left({D}_{\lambda }^{m}\ast {R}^{n}\right)f\left(z\right),$

for any $z\in U$ and each nonnegative integer m,n.

Remark 1.4 If $f\in \mathcal{A}$ and $f\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}{a}_{j}{z}^{j}$, then $D{R}_{\lambda }^{m,n}f\left(z\right)=z+{\sum }_{j=2}^{\mathrm{\infty }}{\left[1+\left(j-1\right)\lambda \right]}^{m}\frac{\left(n+j-1\right)!}{n!\left(j-1\right)!}{a}_{j}^{2}{z}^{j}$, $z\in U$.

Remark 1.5 The operator $D{R}_{\lambda }^{m,n}$ was studied also in .

For $\lambda =1$, $m=n$, we obtain the Hadamard product$S{R}^{n}$ of the Sălăgean operator ${S}^{n}$ and the Ruscheweyh derivative ${R}^{n}$, which was studied in [12, 13].

For $m=n$, we obtain the Hadamard product$D{R}_{\lambda }^{n}$ of the generalized Sălăgean operator ${D}_{\lambda }^{n}$ and the Ruscheweyh derivative ${R}^{n}$, which was studied in .

Using a simple computation, one obtains the next result.

Proposition 1.1 ()

For$m,n\in \mathbb{N}$and$\lambda \ge 0$, we have

$D{R}_{\lambda }^{m+1,n}f\left(z\right)=\left(1-\lambda \right)D{R}_{\lambda }^{m,n}f\left(z\right)+\lambda z{\left(D{R}_{\lambda }^{m,n}f\left(z\right)\right)}^{\prime }$
(1.2)

and

$z{\left(D{R}_{\lambda }^{m,n}f\left(z\right)\right)}^{\prime }=\left(n+1\right)D{R}_{\lambda }^{m,n+1}f\left(z\right)-nD{R}_{\lambda }^{m,n}f\left(z\right).$
(1.3)

By using the operator $D{R}_{\lambda }^{m,n}f\left(z\right)$, we define the following subclasses of analyticfunctions for $0\le \zeta <1$ and $\varphi \in \mathcal{R}$:

$\begin{array}{c}{\mathcal{S}}_{\lambda }^{m,n}\left(\xi \right)=\left\{f\in \mathcal{A}:D{R}_{\lambda }^{m,n}f\in \mathcal{S}\left(\xi \right)\right\},\hfill \\ {\mathcal{S}}_{\lambda }^{m,n}\left(\xi ,\varphi \right)=\left\{f\in \mathcal{A}:D{R}_{\lambda }^{m,n}f\in \mathcal{S}\left(\xi ,\varphi \right)\right\}.\hfill \end{array}$

In particular, we set

${\mathcal{S}}_{\lambda }^{m,n}\left(\xi ,\frac{1+Az}{1+Bz}\right)={\mathcal{S}}_{\lambda }^{m,n}\left(\xi ,A,B\right),\phantom{\rule{1em}{0ex}}-1

Next, we will investigate various inclusion relationships for the subclasses ofanalytic functions introduced above. Furthermore, we study the results of Faisalet al., Darus and Faisal .

## 2 Inclusion relationship associated with the operator $D{R}_{\lambda }^{m,n}$

First, we start with the following lemmas which we need for our main results.

Lemma 2.1 ([22, 23])

Let$\phi \left(\mu ,v\right)$be a complex function such that$\phi :D\to \mathbb{C}$, $D\subseteq \mathbb{C}×\mathbb{C}$, and let$\mu ={\mu }_{1}+i{\mu }_{2}$, $v={v}_{1}+i{v}_{2}$. Suppose that$\phi \left(\mu ,v\right)$satisfies the following conditions:

1. 1.

$\phi \left(\mu ,v\right)$ is continuous in D,

2. 2.

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

3. 3.

$Re\phi \left(i{\mu }_{2},{v}_{1}\right)\le 0$ for all $\left(i{\mu }_{2},{v}_{1}\right)\in D$ such that ${v}_{1}\le -\frac{1}{2}\left(1+{\mu }_{2}^{2}\right)$.

Let$h\left(z\right)=1+{c}_{1}z+{c}_{2}{z}^{2}+\cdots$be analytic in U, such that$\left(h\left(z\right),z{h}^{\prime }\left(z\right)\right)\in D$for all$z\in U$. If$Re\left\{\phi h\left(z\right),z{h}^{\prime }\left(z\right)\right\}>0$, $z\in U$, then$Re\left\{h\left(z\right)\right\}>0$.

Lemma 2.2 ()

Let ϕ be convex univalent in U with$\varphi \left(0\right)=1$and$Re\left\{k\varphi \left(z\right)+\nu \right\}>0$, $k,\nu \in \mathbb{C}$. If p is analytic in U with$p\left(0\right)=1$, then

$p\left(z\right)+\frac{z{p}^{\prime }\left(z\right)}{kp\left(z\right)+\nu }\prec \varphi \left(z\right),\phantom{\rule{1em}{0ex}}z\in U,$

implies$p\left(z\right)\prec \varphi \left(z\right)$, $z\in U$.

Theorem 2.1 Let$f\in \mathcal{A}$, $0\le \xi <1$, $m,n\in \mathbb{N}$, $\lambda >0$, then

${\mathcal{S}}_{\lambda }^{m,n+1}\left(\xi \right)\subseteq {\mathcal{S}}_{\lambda }^{m,n}\left(\xi \right)\subseteq {\mathcal{S}}_{\lambda }^{m,n-1}\left(\xi \right).$

Proof Let $f\in {\mathcal{S}}_{\lambda }^{m,n+1}\left(\xi \right)$ and suppose that

$\frac{z{\left(D{R}_{\lambda }^{m,n}f\left(z\right)\right)}^{\prime }}{D{R}_{\lambda }^{m,n}f\left(z\right)}=\xi +\left(1-\xi \right)h\left(z\right).$
(2.1)

Since from (1.3)

$\left(n+1\right)\frac{D{R}_{\lambda }^{m,n+1}f\left(z\right)}{D{R}_{\lambda }^{m,n}f\left(z\right)}=n+\xi +\left(1-\xi \right)h\left(z\right),$

we obtain

$\begin{array}{c}\left(1-\xi \right){h}^{\prime }\left(z\right)=\left(n+1\right)\left[\frac{{\left(D{R}_{\lambda }^{m,n+1}f\left(z\right)\right)}^{\prime }}{D{R}_{\lambda }^{m,n}f\left(z\right)}-\frac{D{R}_{\lambda }^{m,n+1}f\left(z\right)}{D{R}_{\lambda }^{m,n}f\left(z\right)}\cdot \frac{{\left(D{R}_{\lambda }^{m,n}f\left(z\right)\right)}^{\prime }}{D{R}_{\lambda }^{m,n}f\left(z\right)}\right],\hfill \\ \left(1-\xi \right)z{h}^{\prime }\left(z\right)=\left(n+1\right)\frac{D{R}_{\lambda }^{m,n+1}f\left(z\right)}{D{R}_{\lambda }^{m,n}f\left(z\right)}\left[\frac{z{\left(D{R}_{\lambda }^{m,n+1}f\left(z\right)\right)}^{\prime }}{D{R}_{\lambda }^{m,n+1}f\left(z\right)}-\xi -\left(1-\xi \right)h\left(z\right)\right],\hfill \\ \frac{\left(1-\xi \right){h}^{\prime }\left(z\right)z}{n+\xi +\left(1-\xi \right)h\left(z\right)}=\frac{z{\left(D{R}_{\lambda }^{m,n+1}f\left(z\right)\right)}^{\prime }}{D{R}_{\lambda }^{m,n+1}f\left(z\right)}-\xi -\left(1-\xi \right)h\left(z\right),\hfill \\ \frac{z{\left(D{R}_{\lambda }^{m,n+1}f\left(z\right)\right)}^{\prime }}{D{R}_{\lambda }^{m,n+1}f\left(z\right)}-\xi =\left(1-\xi \right)h\left(z\right)+\frac{\left(1-\xi \right){h}^{\prime }\left(z\right)z}{n+\xi +\left(1-\xi \right)h\left(z\right)}.\hfill \end{array}$

Taking $h\left(z\right)=\mu ={\mu }_{1}+i{\mu }_{2}$ and $z{h}^{\prime }\left(z\right)=v={v}_{1}+i{v}_{2}$, we define $\phi \left(\mu ,v\right)$ by

$\phi \left(\mu ,v\right)=\left(1-\xi \right)\mu +\frac{\left(1-\xi \right)v}{n+\xi +\left(1-\xi \right)\mu }$

and

$\begin{array}{c}Re\left\{\phi \left(i{\mu }_{2},{v}_{1}\right)\right\}=\frac{\left(1-\xi \right)\left(n+\xi \right){v}_{1}}{{\left(n+\xi \right)}^{2}+{\left(1-\xi \right)}^{2}{\mu }_{2}^{2}},\hfill \\ Re\left\{\phi \left(i{\mu }_{2},{v}_{1}\right)\right\}\le -\frac{\left(1-\xi \right)\left(n+\xi \right)\left(1+{\mu }_{2}^{2}\right)}{2\left[{\left(n+\xi \right)}^{2}+{\left(1-\xi \right)}^{2}{\mu }_{2}^{2}\right]}<0.\hfill \end{array}$

Clearly, $\phi \left(\mu ,v\right)$ satisfies the conditions of Lemma 2.1. Hence$Re\left\{h\left(z\right)\right\}>0$, $z\in U$, implies $f\in {\mathcal{S}}_{\lambda }^{m,n}\left(\xi \right)$. □

Remark 2.1 Using relation (1.2) and the same techniques as to prove theearlier results, we can obtain a new similar result.

Theorem 2.2 Let $f\in \mathcal{A}$ and $\varphi \in \mathcal{R}$ with

$Re\left\{\varphi \left(z\right)\right\}<\frac{\xi -1+\frac{1}{\lambda }}{1-\xi }.$

Then

${\mathcal{S}}_{\lambda }^{m+1,n}\left(\xi ,\varphi \right)\subset {\mathcal{S}}_{\lambda }^{m,n}\left(\xi ,\varphi \right)\subset {\mathcal{S}}_{\lambda }^{m-1,n}\left(\xi ,\varphi \right).$

Proof Let $f\left(z\right)\in {\mathcal{S}}_{\lambda }^{m+1,n}\left(\xi ,\varphi \right)$ and set

$p\left(z\right)=\frac{1}{1-\xi }\left(\frac{z{\left(D{R}_{\lambda }^{m,n}f\left(z\right)\right)}^{\prime }}{D{R}_{\lambda }^{m,n}f\left(z\right)}-\xi \right),$
(2.2)

where p is analytic in U with $p\left(0\right)=1$.

By using (1.2) we have

$\frac{z{\left(D{R}_{\lambda }^{m,n}f\left(z\right)\right)}^{\prime }}{D{R}_{\lambda }^{m,n}f\left(z\right)}=\frac{1}{\lambda }\frac{D{R}_{\lambda }^{m+1,n}f\left(z\right)}{D{R}_{\lambda }^{m,n}f\left(z\right)}-\frac{1-\lambda }{\lambda }.$

Now, by using (2.2) we get

$\begin{array}{c}{p}^{\prime }\left(z\right)=\frac{1}{1-\xi }\left(\frac{1}{\lambda }\frac{D{R}_{\lambda }^{m+1,n}f\left(z\right)}{D{R}_{\lambda }^{m,n}f\left(z\right)}-\frac{1-\lambda }{\lambda }-\xi \right),\hfill \\ \frac{1}{\lambda }\frac{D{R}_{\lambda }^{m+1,n}f\left(z\right)}{D{R}_{\lambda }^{m,n}f\left(z\right)}=\xi +\frac{1-\lambda }{\lambda }+\left(1-\xi \right)p\left(z\right).\hfill \end{array}$
(2.3)

By using (2.2) and (2.3), we obtain

$\begin{array}{c}z{p}^{\prime }\left(z\right)=\frac{1}{1-\xi }\cdot \frac{1}{\lambda }\left[\frac{z{\left(D{R}_{\lambda }^{m+1,n}f\left(z\right)\right)}^{\prime }}{D{R}_{\lambda }^{m,n}f\left(z\right)}-\frac{D{R}_{\lambda }^{m+1,n}f\left(z\right)}{D{R}_{\lambda }^{m,n}f\left(z\right)}\cdot \frac{z{\left(D{R}_{\lambda }^{m,n}f\left(z\right)\right)}^{\prime }}{D{R}_{\lambda }^{m,n}f\left(z\right)}\right],\hfill \\ \left(1-\xi \right)z{p}^{\prime }\left(z\right)=\frac{1}{\lambda }\cdot \frac{D{R}_{\lambda }^{m+1,n}f\left(z\right)}{D{R}_{\lambda }^{m,n}f\left(z\right)}\left[\frac{z{\left(D{R}_{\lambda }^{m+1,n}f\left(z\right)\right)}^{\prime }}{D{R}_{\lambda }^{m+1,n}f\left(z\right)}-\frac{z{\left(D{R}_{\lambda }^{m,n}f\left(z\right)\right)}^{\prime }}{D{R}_{\lambda }^{m,n}f\left(z\right)}\right],\hfill \\ \left(1-\xi \right)z{p}^{\prime }\left(z\right)=\left[\zeta -1+\frac{1}{\lambda }+\left(1-\xi \right)p\left(z\right)\right]\left[\frac{z{\left(D{R}_{\lambda }^{m+1,n}f\left(z\right)\right)}^{\prime }}{D{R}_{\lambda }^{m+1,n}f\left(z\right)}-\left(1-\xi \right)p\left(z\right)-\xi \right],\hfill \\ \frac{\left(1-\xi \right)z{p}^{\prime }\left(z\right)}{\left(1-\xi \right)p\left(z\right)+\zeta -1+\frac{1}{\lambda }}=\frac{z{\left(D{R}_{\lambda }^{m+1,n}f\left(z\right)\right)}^{\prime }}{D{R}_{\lambda }^{m+1,n}f\left(z\right)}-\xi -\left(1-\xi \right)p\left(z\right).\hfill \end{array}$

Hence,

$\frac{1}{1-\xi }\left[\frac{z{\left(D{R}_{\lambda }^{m+1,n}f\left(z\right)\right)}^{\prime }}{D{R}_{\lambda }^{m+1,n}f\left(z\right)}-\xi \right]=p\left(z\right)+\frac{z{p}^{\prime }\left(z\right)}{\left(1-\zeta \right)p\left(z\right)+\zeta -1+\frac{1}{\lambda }}.$
(2.4)

Since $Re\left\{\varphi \left(z\right)\right\}<\frac{\xi -1+\frac{1}{\lambda }}{1-\xi }$ implies $Re\left\{\left(1-\xi \right)p\left(z\right)+\xi -1+\frac{1}{\lambda }\right\}>0$, applying Lemma 2.2 to (2.4) we have that$f\left(z\right)\in {\mathcal{S}}_{\lambda }^{m,n}\left(\xi ,\varphi \right)$, as required. □

Remark 2.2 By using relation (1.3) and the same techniques as to prove theearlier results, we can obtain a new similar result.

Corollary 2.3 Let$\frac{1+A}{1+B}<\frac{\xi -1+\frac{1}{\lambda }}{1-\xi }$for$-1, then

${\mathcal{S}}_{\lambda }^{m+1,n}\left(\xi ,A,B\right)\subset {\mathcal{S}}_{\lambda }^{m,n}\left(\xi ,A,B\right)\subset {\mathcal{S}}_{\lambda }^{m-1,n}\left(\xi ,A,B\right).$

Proof Taking $\varphi \left(z\right)=\frac{1+Az}{1+Bz}$, $-1 in Theorem 2.2, we get thecorollary. □

## 3 Integral-preserving properties

In this section, we present several integral-preserving properties for the subclassesof analytic functions defined above. We recall the generalizedBernardi-Libera-Livington integral operator  defined by

${F}_{c}\left[f\left(z\right)\right]=\frac{c+1}{{z}^{c}}{\int }_{0}^{z}{t}^{c-1}f\left(t\right)\phantom{\rule{0.2em}{0ex}}dt=z+\sum _{j=2}^{\mathrm{\infty }}\frac{c+1}{j+c}{a}_{j}{z}^{c},\phantom{\rule{1em}{0ex}}f\in \mathcal{A},c>-1,$
(3.1)

which satisfies the following equality:

$cD{R}_{\lambda }^{m,n}{F}_{c}\left[f\left(z\right)\right]+z{\left[D{R}_{\lambda }^{m,n}{F}_{c}\left(f\left(z\right)\right)\right]}^{\prime }=\left(c+1\right)D{R}_{\lambda }^{m,n}f\left(z\right).$
(3.2)

Theorem 3.1 Let$c>-1$, $0\le \xi <1$. If$f\in {\mathcal{S}}_{\lambda }^{m,n}\left(\xi \right)$, then${F}_{c}f\in {\mathcal{S}}_{\lambda }^{m,n}\left(\xi \right)$.

Proof Let $f\in {\mathcal{S}}_{\lambda }^{m,n}\left(\xi \right)$. By using (3.2), we get

$\frac{z{\left[D{R}_{\lambda }^{m,n}{F}_{c}\left[f\left(z\right)\right]\right]}^{\prime }}{D{R}_{\lambda }^{m,n}{F}_{c}\left[f\left(z\right)\right]}=\left(c+1\right)\frac{D{R}_{\lambda }^{m,n}f\left(z\right)}{D{R}_{\lambda }^{m,n}{F}_{c}\left[f\left(z\right)\right]}-c.$

Let

$\frac{z{\left[D{R}_{\lambda }^{m,n}{F}_{c}\left[f\left(z\right)\right]\right]}^{\prime }}{D{R}_{\lambda }^{m,n}{F}_{c}\left[f\left(z\right)\right]}=\xi +\left(1-\xi \right)h\left(z\right),\phantom{\rule{1em}{0ex}}h\left(z\right)=1+{c}_{1}z+{c}_{2}{z}^{2}+\cdots .$

We obtain

$\frac{z{\left[D{R}_{\lambda }^{m,n}f\left(z\right)\right]}^{\prime }}{D{R}_{\lambda }^{m,n}f\left(z\right)}-\xi =\left(1-\xi \right)h\left(z\right)+\frac{\left(1-\xi \right)z{h}^{\prime }\left(z\right)}{\xi +\left(1-\xi \right)h\left(z\right)+c}.$

This implies

$\phi \left(\mu ,v\right)=\left(1-\xi \right)\mu +\frac{\left(1-\xi \right)v}{c+\xi +\left(1-\xi \right)\mu }$

(same as Theorem 2.1) and

$\begin{array}{c}Re\left\{\phi \left(i{\mu }_{2},{v}_{1}\right)\right\}=\frac{\left(1-\xi \right)\left(c+\xi \right){v}_{1}}{{\left(c+\xi \right)}^{2}+{\left(1-\xi \right)}^{2}{\mu }_{2}^{2}},\hfill \\ Re\left\{\phi \left(i{\mu }_{2},{v}_{1}\right)\right\}\le -\frac{\left(1-\xi \right)\left(c+\xi \right){\left(1+{\mu }_{2}\right)}^{2}}{2\left[{\left(c+\xi \right)}^{2}+{\left(1-\xi \right)}^{2}{\mu }_{2}^{2}\right]}<0.\hfill \end{array}$

After using Lemma 2.1 and Theorem 2.1, we have

${F}_{c}f\in {\mathcal{S}}_{\lambda }^{m,n}\left(\xi \right).$

□

Theorem 3.2 Let $c>-1$ and $\varphi \in \mathcal{R}$ with

$Re\left\{\varphi \left(z\right)\right\}<\frac{c+\xi }{1-\xi }.$

If$f\in {\mathcal{S}}_{\lambda }^{m,n}\left(\xi ,\varphi \right)$, then${F}_{c}f\in {\mathcal{S}}_{\lambda }^{m,n}\left(\xi ,\varphi \right)$.

Proof Let $f\left(z\right)\in {\mathcal{S}}_{\lambda }^{m,n}\left(\xi ,\varphi \right)$ and set

$p\left(z\right)=\frac{1}{1-\xi }\left(\frac{z{\left[D{R}_{\lambda }^{m,n}{F}_{c}\left[f\left(z\right)\right]\right]}^{\prime }}{D{R}_{\lambda }^{m,n}{F}_{c}\left[f\left(z\right)\right]}-\xi \right),$
(3.3)

where p is analytic in U with $p\left(0\right)=1$.

Using (3.2) and (3.3), we have

$\left(c+1\right)\frac{z\left[D{R}_{\lambda }^{m,n}f\left(z\right)\right]}{D{R}_{\lambda }^{m,n}{F}_{c}\left[f\left(z\right)\right]}=c+\xi +\left(1-\xi \right)p\left(z\right).$
(3.4)

Then, using (3.2), (3.3) and (3.4), we obtain

$\frac{1}{1-\xi }\left(\frac{z{\left[D{R}_{\lambda }^{m,n}f\left(z\right)\right]}^{\prime }}{D{R}_{\lambda }^{m,n}f\left(z\right)}-\xi \right)=p\left(z\right)+\frac{z{p}^{\prime }\left(z\right)}{\left(1-\xi \right)p\left(z\right)+c+\xi }.$
(3.5)

Applying Lemma 2.2 to (3.5), we conclude that

${F}_{c}f\in {\mathcal{S}}_{\lambda }^{m,n}\left(\xi ,\varphi \right).$

□

## Author’s contributions

The author drafted the manuscript, read and approved the final manuscript.

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The author thanks the referee for his/her valuable suggestions to improve thepresent article.

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Correspondence to Loriana Andrei. 