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$\mathcal{S}\left(\xi ,\varphi \right)$[3] 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 [4])

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 [5].

Definition 1.2 (Ruscheweyh [6])

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 ([7])

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 [810].

For $\lambda =1$, $m=n$, we obtain the Hadamard product$S{R}^{n}$[11] 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$D{R}_{\lambda }^{n}$[14] of the generalized Sălăgean operator ${D}_{\lambda }^{n}$ and the Ruscheweyh derivative ${R}^{n}$, which was studied in [1520].

Using a simple computation, one obtains the next result.

Proposition 1.1 ([7])

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.[21], Darus and Faisal [3].

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 ([24])

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 [25] 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.

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Andrei, L. Some properties of certain subclasses of analytic functions involving adifferential operator. Adv Differ Equ 2014, 142 (2014). https://doi.org/10.1186/1687-1847-2014-142