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 $D{R}_{\mathrm{Î»}}^{m,n}$. 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 $U=\left\{zâˆˆ\mathbb{C}:|z|<1\right\}$ normalized by $f\left(0\right)={f}^{â€²}\left(0\right)âˆ’1=0$. Thus each $fâˆˆ\mathcal{A}$ has a Taylor series representation

$f\left(z\right)=z+\underset{j=2}{\overset{\mathrm{âˆž}}{âˆ‘}}{a}_{j}{z}^{j}.$
(1.1)

We denote by $\mathcal{S}\left(\mathrm{Î¾}\right)$ the well-known subclass of consisting of allanalytic functions which are, respectively, starlike of order Î¾[1, 2]

$\mathcal{S}\left(\mathrm{Î¾}\right)=\left\{fâˆˆ\mathcal{A}:Re\left(\frac{z{f}^{â€²}\left(z\right)}{f\left(z\right)}\right)>\mathrm{Î¾},zâˆˆU\right\},\phantom{\rule{1em}{0ex}}0â‰¤\mathrm{Î¾}<1.$

Let â„› be a class of all functions Ï• which are analytic andunivalent in U and for which $\mathrm{Ï•}\left(U\right)$ is convex with $\mathrm{Ï•}\left(0\right)=1$ and $Re\mathrm{Ï•}\left(z\right)>0$, $zâˆˆU$.

For two functions f and g analytic in U, we say that thefunction f is subordinate to g in U and write$f\left(z\right)â‰ºg\left(z\right)$, $zâˆˆU$, 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 $f\left(z\right)=g\left(w\left(z\right)\right)$, $zâˆˆU$.

Making use of the principle of subordination between analytic functions, denote by$\mathcal{S}\left(\mathrm{Î¾},\mathrm{Ï•}\right)$[3] a subclass of the class for$0â‰¤\mathrm{Î¾}<1$ and $\mathrm{Ï•}âˆˆ\mathcal{R}$ which are defined by

$\mathcal{S}\left(\mathrm{Î¾},\mathrm{Ï•}\right)=\left\{fâˆˆ\mathcal{A}:\frac{1}{1âˆ’\mathrm{Î¾}}\left(\frac{z{f}^{â€²}\left(z\right)}{f\left(z\right)}âˆ’\mathrm{Î¶}\right)â‰º\mathrm{Ï•}\left(z\right),zâˆˆU\right\}.$

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

$\left(fâˆ—g\right)\left(z\right)=z+\underset{j=2}{\overset{\mathrm{âˆž}}{âˆ‘}}{a}_{j}{b}_{j}{z}^{j}.$

Definition 1.1 (Al-Oboudi [4])

For $fâˆˆ\mathcal{A}$, $\mathrm{Î»}â‰¥0$ and $mâˆˆ\mathbb{N}$, the operator ${D}_{\mathrm{Î»}}^{m}$ is defined by ${D}_{\mathrm{Î»}}^{m}:\mathcal{A}â†’\mathcal{A}$,

$\begin{array}{c}{D}_{\mathrm{Î»}}^{0}f\left(z\right)=f\left(z\right),\hfill \\ {D}_{\mathrm{Î»}}^{1}f\left(z\right)=\left(1âˆ’\mathrm{Î»}\right)f\left(z\right)+\mathrm{Î»}z{f}^{â€²}\left(z\right)={D}_{\mathrm{Î»}}f\left(z\right),\hfill \\ â€¦,\hfill \\ {D}_{\mathrm{Î»}}^{m}f\left(z\right)=\left(1âˆ’\mathrm{Î»}\right){D}_{\mathrm{Î»}}^{mâˆ’1}f\left(z\right)+\mathrm{Î»}z{\left({D}_{\mathrm{Î»}}^{m}f\left(z\right)\right)}^{â€²}={D}_{\mathrm{Î»}}\left({D}_{\mathrm{Î»}}^{mâˆ’1}f\left(z\right)\right),\phantom{\rule{1em}{0ex}}zâˆˆU.\hfill \end{array}$

Remark 1.1 If $fâˆˆ\mathcal{A}$ and $f\left(z\right)=z+{âˆ‘}_{j=2}^{\mathrm{âˆž}}{a}_{j}{z}^{j}$, then ${D}_{\mathrm{Î»}}^{m}f\left(z\right)=z+{âˆ‘}_{j=2}^{\mathrm{âˆž}}{\left[1+\left(jâˆ’1\right)\mathrm{Î»}\right]}^{m}{a}_{j}{z}^{j}$, $zâˆˆU$.

Remark 1.2 For $\mathrm{Î»}=1$ in the above definition, we obtain theSÄƒlÄƒgean differential operator [5].

Definition 1.2 (Ruscheweyh [6])

For $fâˆˆ\mathcal{A}$ and $nâˆˆ\mathbb{N}$, the operator ${R}^{n}$ is defined by ${R}^{n}:\mathcal{A}â†’\mathcal{A}$,

$\begin{array}{c}{R}^{0}f\left(z\right)=f\left(z\right),\hfill \\ {R}^{1}f\left(z\right)=z{f}^{â€²}\left(z\right),\hfill \\ â€¦,\hfill \\ \left(n+1\right){R}^{n+1}f\left(z\right)=z{\left({R}^{n}f\left(z\right)\right)}^{â€²}+n{R}^{n}f\left(z\right),\phantom{\rule{1em}{0ex}}zâˆˆU.\hfill \end{array}$

Remark 1.3 If $fâˆˆ\mathcal{A}$, $f\left(z\right)=z+{âˆ‘}_{j=2}^{\mathrm{âˆž}}{a}_{j}{z}^{j}$, then ${R}^{n}f\left(z\right)=z+{âˆ‘}_{j=2}^{\mathrm{âˆž}}\frac{\left(n+jâˆ’1\right)!}{n!\left(jâˆ’1\right)!}{a}_{j}{z}^{j}$, $zâˆˆU$.

Definition 1.3 ([7])

Let $\mathrm{Î»}â‰¥0$ and $n,mâˆˆ\mathbb{N}$. Denote by $D{R}_{\mathrm{Î»}}^{m,n}:\mathcal{A}â†’\mathcal{A}$ the operator given by the Hadamard product of thegeneralized SÄƒlÄƒgean operator ${D}_{\mathrm{Î»}}^{m}$ and the Ruscheweyh operator ${R}^{n}$,

$D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)=\left({D}_{\mathrm{Î»}}^{m}âˆ—{R}^{n}\right)f\left(z\right),$

for any $zâˆˆU$ and each nonnegative integer m,n.

Remark 1.4 If $fâˆˆ\mathcal{A}$ and $f\left(z\right)=z+{âˆ‘}_{j=2}^{\mathrm{âˆž}}{a}_{j}{z}^{j}$, then $D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)=z+{âˆ‘}_{j=2}^{\mathrm{âˆž}}{\left[1+\left(jâˆ’1\right)\mathrm{Î»}\right]}^{m}\frac{\left(n+jâˆ’1\right)!}{n!\left(jâˆ’1\right)!}{a}_{j}^{2}{z}^{j}$, $zâˆˆU$.

Remark 1.5 The operator $D{R}_{\mathrm{Î»}}^{m,n}$ was studied also in [8â€“10].

For $\mathrm{Î»}=1$, $m=n$, we obtain the Hadamard product$S{R}^{n}$[11] of the SÄƒlÄƒgean operator ${S}^{n}$ and the Ruscheweyh derivative ${R}^{n}$, which was studied in [12, 13].

For $m=n$, we obtain the Hadamard product$D{R}_{\mathrm{Î»}}^{n}$[14] of the generalized SÄƒlÄƒgean operator ${D}_{\mathrm{Î»}}^{n}$ and the Ruscheweyh derivative ${R}^{n}$, which was studied in [15â€“20].

Using a simple computation, one obtains the next result.

Proposition 1.1 ([7])

For$m,nâˆˆ\mathbb{N}$and$\mathrm{Î»}â‰¥0$, we have

$D{R}_{\mathrm{Î»}}^{m+1,n}f\left(z\right)=\left(1âˆ’\mathrm{Î»}\right)D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)+\mathrm{Î»}z{\left(D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)\right)}^{â€²}$
(1.2)

and

$z{\left(D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)\right)}^{â€²}=\left(n+1\right)D{R}_{\mathrm{Î»}}^{m,n+1}f\left(z\right)âˆ’nD{R}_{\mathrm{Î»}}^{m,n}f\left(z\right).$
(1.3)

By using the operator $D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)$, we define the following subclasses of analyticfunctions for $0â‰¤\mathrm{Î¶}<1$ and $\mathrm{Ï•}âˆˆ\mathcal{R}$:

$\begin{array}{c}{\mathcal{S}}_{\mathrm{Î»}}^{m,n}\left(\mathrm{Î¾}\right)=\left\{fâˆˆ\mathcal{A}:D{R}_{\mathrm{Î»}}^{m,n}fâˆˆ\mathcal{S}\left(\mathrm{Î¾}\right)\right\},\hfill \\ {\mathcal{S}}_{\mathrm{Î»}}^{m,n}\left(\mathrm{Î¾},\mathrm{Ï•}\right)=\left\{fâˆˆ\mathcal{A}:D{R}_{\mathrm{Î»}}^{m,n}fâˆˆ\mathcal{S}\left(\mathrm{Î¾},\mathrm{Ï•}\right)\right\}.\hfill \end{array}$

In particular, we set

${\mathcal{S}}_{\mathrm{Î»}}^{m,n}\left(\mathrm{Î¾},\frac{1+Az}{1+Bz}\right)={\mathcal{S}}_{\mathrm{Î»}}^{m,n}\left(\mathrm{Î¾},A,B\right),\phantom{\rule{1em}{0ex}}âˆ’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 $D{R}_{\mathrm{Î»}}^{m,n}$

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

Lemma 2.1 ([22, 23])

Let$\mathrm{Ï†}\left(\mathrm{Î¼},v\right)$be a complex function such that$\mathrm{Ï†}:Dâ†’\mathbb{C}$, $DâŠ†\mathbb{C}Ã—\mathbb{C}$, and let$\mathrm{Î¼}={\mathrm{Î¼}}_{1}+i{\mathrm{Î¼}}_{2}$, $v={v}_{1}+i{v}_{2}$. Suppose that$\mathrm{Ï†}\left(\mathrm{Î¼},v\right)$satisfies the following conditions:

1. 1.

$\mathrm{Ï†}\left(\mathrm{Î¼},v\right)$ is continuous in D,

2. 2.

$\left(1,0\right)âˆˆD$ and $Re\mathrm{Ï†}\left(1,0\right)>0$,

3. 3.

$Re\mathrm{Ï†}\left(i{\mathrm{Î¼}}_{2},{v}_{1}\right)â‰¤0$ for all $\left(i{\mathrm{Î¼}}_{2},{v}_{1}\right)âˆˆD$ such that ${v}_{1}â‰¤âˆ’\frac{1}{2}\left(1+{\mathrm{Î¼}}_{2}^{2}\right)$.

Let$h\left(z\right)=1+{c}_{1}z+{c}_{2}{z}^{2}+â‹¯$be analytic in U, such that$\left(h\left(z\right),z{h}^{â€²}\left(z\right)\right)âˆˆD$for all$zâˆˆU$. If$Re\left\{\mathrm{Ï†}h\left(z\right),z{h}^{â€²}\left(z\right)\right\}>0$, $zâˆˆU$, then$Re\left\{h\left(z\right)\right\}>0$.

Lemma 2.2 ([24])

Let Ï• be convex univalent in U with$\mathrm{Ï•}\left(0\right)=1$and$Re\left\{k\mathrm{Ï•}\left(z\right)+\mathrm{Î½}\right\}>0$, $k,\mathrm{Î½}âˆˆ\mathbb{C}$. If p is analytic in U with$p\left(0\right)=1$, then

$p\left(z\right)+\frac{z{p}^{â€²}\left(z\right)}{kp\left(z\right)+\mathrm{Î½}}â‰º\mathrm{Ï•}\left(z\right),\phantom{\rule{1em}{0ex}}zâˆˆU,$

implies$p\left(z\right)â‰º\mathrm{Ï•}\left(z\right)$, $zâˆˆU$.

Theorem 2.1 Let$fâˆˆ\mathcal{A}$, $0â‰¤\mathrm{Î¾}<1$, $m,nâˆˆ\mathbb{N}$, $\mathrm{Î»}>0$, then

${\mathcal{S}}_{\mathrm{Î»}}^{m,n+1}\left(\mathrm{Î¾}\right)âŠ†{\mathcal{S}}_{\mathrm{Î»}}^{m,n}\left(\mathrm{Î¾}\right)âŠ†{\mathcal{S}}_{\mathrm{Î»}}^{m,nâˆ’1}\left(\mathrm{Î¾}\right).$

Proof Let $fâˆˆ{\mathcal{S}}_{\mathrm{Î»}}^{m,n+1}\left(\mathrm{Î¾}\right)$ and suppose that

$\frac{z{\left(D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)\right)}^{â€²}}{D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)}=\mathrm{Î¾}+\left(1âˆ’\mathrm{Î¾}\right)h\left(z\right).$
(2.1)

Since from (1.3)

$\left(n+1\right)\frac{D{R}_{\mathrm{Î»}}^{m,n+1}f\left(z\right)}{D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)}=n+\mathrm{Î¾}+\left(1âˆ’\mathrm{Î¾}\right)h\left(z\right),$

we obtain

$\begin{array}{c}\left(1âˆ’\mathrm{Î¾}\right){h}^{â€²}\left(z\right)=\left(n+1\right)\left[\frac{{\left(D{R}_{\mathrm{Î»}}^{m,n+1}f\left(z\right)\right)}^{â€²}}{D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)}âˆ’\frac{D{R}_{\mathrm{Î»}}^{m,n+1}f\left(z\right)}{D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)}â‹\dots \frac{{\left(D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)\right)}^{â€²}}{D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)}\right],\hfill \\ \left(1âˆ’\mathrm{Î¾}\right)z{h}^{â€²}\left(z\right)=\left(n+1\right)\frac{D{R}_{\mathrm{Î»}}^{m,n+1}f\left(z\right)}{D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)}\left[\frac{z{\left(D{R}_{\mathrm{Î»}}^{m,n+1}f\left(z\right)\right)}^{â€²}}{D{R}_{\mathrm{Î»}}^{m,n+1}f\left(z\right)}âˆ’\mathrm{Î¾}âˆ’\left(1âˆ’\mathrm{Î¾}\right)h\left(z\right)\right],\hfill \\ \frac{\left(1âˆ’\mathrm{Î¾}\right){h}^{â€²}\left(z\right)z}{n+\mathrm{Î¾}+\left(1âˆ’\mathrm{Î¾}\right)h\left(z\right)}=\frac{z{\left(D{R}_{\mathrm{Î»}}^{m,n+1}f\left(z\right)\right)}^{â€²}}{D{R}_{\mathrm{Î»}}^{m,n+1}f\left(z\right)}âˆ’\mathrm{Î¾}âˆ’\left(1âˆ’\mathrm{Î¾}\right)h\left(z\right),\hfill \\ \frac{z{\left(D{R}_{\mathrm{Î»}}^{m,n+1}f\left(z\right)\right)}^{â€²}}{D{R}_{\mathrm{Î»}}^{m,n+1}f\left(z\right)}âˆ’\mathrm{Î¾}=\left(1âˆ’\mathrm{Î¾}\right)h\left(z\right)+\frac{\left(1âˆ’\mathrm{Î¾}\right){h}^{â€²}\left(z\right)z}{n+\mathrm{Î¾}+\left(1âˆ’\mathrm{Î¾}\right)h\left(z\right)}.\hfill \end{array}$

Taking $h\left(z\right)=\mathrm{Î¼}={\mathrm{Î¼}}_{1}+i{\mathrm{Î¼}}_{2}$ and $z{h}^{â€²}\left(z\right)=v={v}_{1}+i{v}_{2}$, we define $\mathrm{Ï†}\left(\mathrm{Î¼},v\right)$ by

$\mathrm{Ï†}\left(\mathrm{Î¼},v\right)=\left(1âˆ’\mathrm{Î¾}\right)\mathrm{Î¼}+\frac{\left(1âˆ’\mathrm{Î¾}\right)v}{n+\mathrm{Î¾}+\left(1âˆ’\mathrm{Î¾}\right)\mathrm{Î¼}}$

and

$\begin{array}{c}Re\left\{\mathrm{Ï†}\left(i{\mathrm{Î¼}}_{2},{v}_{1}\right)\right\}=\frac{\left(1âˆ’\mathrm{Î¾}\right)\left(n+\mathrm{Î¾}\right){v}_{1}}{{\left(n+\mathrm{Î¾}\right)}^{2}+{\left(1âˆ’\mathrm{Î¾}\right)}^{2}{\mathrm{Î¼}}_{2}^{2}},\hfill \\ Re\left\{\mathrm{Ï†}\left(i{\mathrm{Î¼}}_{2},{v}_{1}\right)\right\}â‰¤âˆ’\frac{\left(1âˆ’\mathrm{Î¾}\right)\left(n+\mathrm{Î¾}\right)\left(1+{\mathrm{Î¼}}_{2}^{2}\right)}{2\left[{\left(n+\mathrm{Î¾}\right)}^{2}+{\left(1âˆ’\mathrm{Î¾}\right)}^{2}{\mathrm{Î¼}}_{2}^{2}\right]}<0.\hfill \end{array}$

Clearly, $\mathrm{Ï†}\left(\mathrm{Î¼},v\right)$ satisfies the conditions of Lemma 2.1. Hence$Re\left\{h\left(z\right)\right\}>0$, $zâˆˆU$, implies $fâˆˆ{\mathcal{S}}_{\mathrm{Î»}}^{m,n}\left(\mathrm{Î¾}\right)$.â€ƒâ–¡

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 $fâˆˆ\mathcal{A}$ and $\mathrm{Ï•}âˆˆ\mathcal{R}$ with

$Re\left\{\mathrm{Ï•}\left(z\right)\right\}<\frac{\mathrm{Î¾}âˆ’1+\frac{1}{\mathrm{Î»}}}{1âˆ’\mathrm{Î¾}}.$

Then

${\mathcal{S}}_{\mathrm{Î»}}^{m+1,n}\left(\mathrm{Î¾},\mathrm{Ï•}\right)âŠ‚{\mathcal{S}}_{\mathrm{Î»}}^{m,n}\left(\mathrm{Î¾},\mathrm{Ï•}\right)âŠ‚{\mathcal{S}}_{\mathrm{Î»}}^{mâˆ’1,n}\left(\mathrm{Î¾},\mathrm{Ï•}\right).$

Proof Let $f\left(z\right)âˆˆ{\mathcal{S}}_{\mathrm{Î»}}^{m+1,n}\left(\mathrm{Î¾},\mathrm{Ï•}\right)$ and set

$p\left(z\right)=\frac{1}{1âˆ’\mathrm{Î¾}}\left(\frac{z{\left(D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)\right)}^{â€²}}{D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)}âˆ’\mathrm{Î¾}\right),$
(2.2)

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

By using (1.2) we have

$\frac{z{\left(D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)\right)}^{â€²}}{D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)}=\frac{1}{\mathrm{Î»}}\frac{D{R}_{\mathrm{Î»}}^{m+1,n}f\left(z\right)}{D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)}âˆ’\frac{1âˆ’\mathrm{Î»}}{\mathrm{Î»}}.$

Now, by using (2.2) we get

$\begin{array}{c}{p}^{â€²}\left(z\right)=\frac{1}{1âˆ’\mathrm{Î¾}}\left(\frac{1}{\mathrm{Î»}}\frac{D{R}_{\mathrm{Î»}}^{m+1,n}f\left(z\right)}{D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)}âˆ’\frac{1âˆ’\mathrm{Î»}}{\mathrm{Î»}}âˆ’\mathrm{Î¾}\right),\hfill \\ \frac{1}{\mathrm{Î»}}\frac{D{R}_{\mathrm{Î»}}^{m+1,n}f\left(z\right)}{D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)}=\mathrm{Î¾}+\frac{1âˆ’\mathrm{Î»}}{\mathrm{Î»}}+\left(1âˆ’\mathrm{Î¾}\right)p\left(z\right).\hfill \end{array}$
(2.3)

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

$\begin{array}{c}z{p}^{â€²}\left(z\right)=\frac{1}{1âˆ’\mathrm{Î¾}}â‹\dots \frac{1}{\mathrm{Î»}}\left[\frac{z{\left(D{R}_{\mathrm{Î»}}^{m+1,n}f\left(z\right)\right)}^{â€²}}{D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)}âˆ’\frac{D{R}_{\mathrm{Î»}}^{m+1,n}f\left(z\right)}{D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)}â‹\dots \frac{z{\left(D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)\right)}^{â€²}}{D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)}\right],\hfill \\ \left(1âˆ’\mathrm{Î¾}\right)z{p}^{â€²}\left(z\right)=\frac{1}{\mathrm{Î»}}â‹\dots \frac{D{R}_{\mathrm{Î»}}^{m+1,n}f\left(z\right)}{D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)}\left[\frac{z{\left(D{R}_{\mathrm{Î»}}^{m+1,n}f\left(z\right)\right)}^{â€²}}{D{R}_{\mathrm{Î»}}^{m+1,n}f\left(z\right)}âˆ’\frac{z{\left(D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)\right)}^{â€²}}{D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)}\right],\hfill \\ \left(1âˆ’\mathrm{Î¾}\right)z{p}^{â€²}\left(z\right)=\left[\mathrm{Î¶}âˆ’1+\frac{1}{\mathrm{Î»}}+\left(1âˆ’\mathrm{Î¾}\right)p\left(z\right)\right]\left[\frac{z{\left(D{R}_{\mathrm{Î»}}^{m+1,n}f\left(z\right)\right)}^{â€²}}{D{R}_{\mathrm{Î»}}^{m+1,n}f\left(z\right)}âˆ’\left(1âˆ’\mathrm{Î¾}\right)p\left(z\right)âˆ’\mathrm{Î¾}\right],\hfill \\ \frac{\left(1âˆ’\mathrm{Î¾}\right)z{p}^{â€²}\left(z\right)}{\left(1âˆ’\mathrm{Î¾}\right)p\left(z\right)+\mathrm{Î¶}âˆ’1+\frac{1}{\mathrm{Î»}}}=\frac{z{\left(D{R}_{\mathrm{Î»}}^{m+1,n}f\left(z\right)\right)}^{â€²}}{D{R}_{\mathrm{Î»}}^{m+1,n}f\left(z\right)}âˆ’\mathrm{Î¾}âˆ’\left(1âˆ’\mathrm{Î¾}\right)p\left(z\right).\hfill \end{array}$

Hence,

$\frac{1}{1âˆ’\mathrm{Î¾}}\left[\frac{z{\left(D{R}_{\mathrm{Î»}}^{m+1,n}f\left(z\right)\right)}^{â€²}}{D{R}_{\mathrm{Î»}}^{m+1,n}f\left(z\right)}âˆ’\mathrm{Î¾}\right]=p\left(z\right)+\frac{z{p}^{â€²}\left(z\right)}{\left(1âˆ’\mathrm{Î¶}\right)p\left(z\right)+\mathrm{Î¶}âˆ’1+\frac{1}{\mathrm{Î»}}}.$
(2.4)

Since $Re\left\{\mathrm{Ï•}\left(z\right)\right\}<\frac{\mathrm{Î¾}âˆ’1+\frac{1}{\mathrm{Î»}}}{1âˆ’\mathrm{Î¾}}$ implies $Re\left\{\left(1âˆ’\mathrm{Î¾}\right)p\left(z\right)+\mathrm{Î¾}âˆ’1+\frac{1}{\mathrm{Î»}}\right\}>0$, applying Lemma 2.2 to (2.4) we have that$f\left(z\right)âˆˆ{\mathcal{S}}_{\mathrm{Î»}}^{m,n}\left(\mathrm{Î¾},\mathrm{Ï•}\right)$, as required.â€ƒâ–¡

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$\frac{1+A}{1+B}<\frac{\mathrm{Î¾}âˆ’1+\frac{1}{\mathrm{Î»}}}{1âˆ’\mathrm{Î¾}}$for$âˆ’1, then

${\mathcal{S}}_{\mathrm{Î»}}^{m+1,n}\left(\mathrm{Î¾},A,B\right)âŠ‚{\mathcal{S}}_{\mathrm{Î»}}^{m,n}\left(\mathrm{Î¾},A,B\right)âŠ‚{\mathcal{S}}_{\mathrm{Î»}}^{mâˆ’1,n}\left(\mathrm{Î¾},A,B\right).$

Proof Taking $\mathrm{Ï•}\left(z\right)=\frac{1+Az}{1+Bz}$, $âˆ’1 in Theorem 2.2, we get thecorollary.â€ƒâ–¡

## 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

${F}_{c}\left[f\left(z\right)\right]=\frac{c+1}{{z}^{c}}{âˆ«}_{0}^{z}{t}^{câˆ’1}f\left(t\right)\phantom{\rule{0.2em}{0ex}}dt=z+\underset{j=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{c+1}{j+c}{a}_{j}{z}^{c},\phantom{\rule{1em}{0ex}}fâˆˆ\mathcal{A},c>âˆ’1,$
(3.1)

which satisfies the following equality:

$cD{R}_{\mathrm{Î»}}^{m,n}{F}_{c}\left[f\left(z\right)\right]+z{\left[D{R}_{\mathrm{Î»}}^{m,n}{F}_{c}\left(f\left(z\right)\right)\right]}^{â€²}=\left(c+1\right)D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right).$
(3.2)

Theorem 3.1 Let$c>âˆ’1$, $0â‰¤\mathrm{Î¾}<1$. If$fâˆˆ{\mathcal{S}}_{\mathrm{Î»}}^{m,n}\left(\mathrm{Î¾}\right)$, then${F}_{c}fâˆˆ{\mathcal{S}}_{\mathrm{Î»}}^{m,n}\left(\mathrm{Î¾}\right)$.

Proof Let $fâˆˆ{\mathcal{S}}_{\mathrm{Î»}}^{m,n}\left(\mathrm{Î¾}\right)$. By using (3.2), we get

$\frac{z{\left[D{R}_{\mathrm{Î»}}^{m,n}{F}_{c}\left[f\left(z\right)\right]\right]}^{â€²}}{D{R}_{\mathrm{Î»}}^{m,n}{F}_{c}\left[f\left(z\right)\right]}=\left(c+1\right)\frac{D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)}{D{R}_{\mathrm{Î»}}^{m,n}{F}_{c}\left[f\left(z\right)\right]}âˆ’c.$

Let

$\frac{z{\left[D{R}_{\mathrm{Î»}}^{m,n}{F}_{c}\left[f\left(z\right)\right]\right]}^{â€²}}{D{R}_{\mathrm{Î»}}^{m,n}{F}_{c}\left[f\left(z\right)\right]}=\mathrm{Î¾}+\left(1âˆ’\mathrm{Î¾}\right)h\left(z\right),\phantom{\rule{1em}{0ex}}h\left(z\right)=1+{c}_{1}z+{c}_{2}{z}^{2}+â‹¯.$

We obtain

$\frac{z{\left[D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)\right]}^{â€²}}{D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)}âˆ’\mathrm{Î¾}=\left(1âˆ’\mathrm{Î¾}\right)h\left(z\right)+\frac{\left(1âˆ’\mathrm{Î¾}\right)z{h}^{â€²}\left(z\right)}{\mathrm{Î¾}+\left(1âˆ’\mathrm{Î¾}\right)h\left(z\right)+c}.$

This implies

$\mathrm{Ï†}\left(\mathrm{Î¼},v\right)=\left(1âˆ’\mathrm{Î¾}\right)\mathrm{Î¼}+\frac{\left(1âˆ’\mathrm{Î¾}\right)v}{c+\mathrm{Î¾}+\left(1âˆ’\mathrm{Î¾}\right)\mathrm{Î¼}}$

(same as Theorem 2.1) and

$\begin{array}{c}Re\left\{\mathrm{Ï†}\left(i{\mathrm{Î¼}}_{2},{v}_{1}\right)\right\}=\frac{\left(1âˆ’\mathrm{Î¾}\right)\left(c+\mathrm{Î¾}\right){v}_{1}}{{\left(c+\mathrm{Î¾}\right)}^{2}+{\left(1âˆ’\mathrm{Î¾}\right)}^{2}{\mathrm{Î¼}}_{2}^{2}},\hfill \\ Re\left\{\mathrm{Ï†}\left(i{\mathrm{Î¼}}_{2},{v}_{1}\right)\right\}â‰¤âˆ’\frac{\left(1âˆ’\mathrm{Î¾}\right)\left(c+\mathrm{Î¾}\right){\left(1+{\mathrm{Î¼}}_{2}\right)}^{2}}{2\left[{\left(c+\mathrm{Î¾}\right)}^{2}+{\left(1âˆ’\mathrm{Î¾}\right)}^{2}{\mathrm{Î¼}}_{2}^{2}\right]}<0.\hfill \end{array}$

After using Lemma 2.1 and Theorem 2.1, we have

${F}_{c}fâˆˆ{\mathcal{S}}_{\mathrm{Î»}}^{m,n}\left(\mathrm{Î¾}\right).$

â€ƒâ–¡

Theorem 3.2 Let $c>âˆ’1$ and $\mathrm{Ï•}âˆˆ\mathcal{R}$ with

$Re\left\{\mathrm{Ï•}\left(z\right)\right\}<\frac{c+\mathrm{Î¾}}{1âˆ’\mathrm{Î¾}}.$

If$fâˆˆ{\mathcal{S}}_{\mathrm{Î»}}^{m,n}\left(\mathrm{Î¾},\mathrm{Ï•}\right)$, then${F}_{c}fâˆˆ{\mathcal{S}}_{\mathrm{Î»}}^{m,n}\left(\mathrm{Î¾},\mathrm{Ï•}\right)$.

Proof Let $f\left(z\right)âˆˆ{\mathcal{S}}_{\mathrm{Î»}}^{m,n}\left(\mathrm{Î¾},\mathrm{Ï•}\right)$ and set

$p\left(z\right)=\frac{1}{1âˆ’\mathrm{Î¾}}\left(\frac{z{\left[D{R}_{\mathrm{Î»}}^{m,n}{F}_{c}\left[f\left(z\right)\right]\right]}^{â€²}}{D{R}_{\mathrm{Î»}}^{m,n}{F}_{c}\left[f\left(z\right)\right]}âˆ’\mathrm{Î¾}\right),$
(3.3)

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

Using (3.2) and (3.3), we have

$\left(c+1\right)\frac{z\left[D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)\right]}{D{R}_{\mathrm{Î»}}^{m,n}{F}_{c}\left[f\left(z\right)\right]}=c+\mathrm{Î¾}+\left(1âˆ’\mathrm{Î¾}\right)p\left(z\right).$
(3.4)

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

$\frac{1}{1âˆ’\mathrm{Î¾}}\left(\frac{z{\left[D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)\right]}^{â€²}}{D{R}_{\mathrm{Î»}}^{m,n}f\left(z\right)}âˆ’\mathrm{Î¾}\right)=p\left(z\right)+\frac{z{p}^{â€²}\left(z\right)}{\left(1âˆ’\mathrm{Î¾}\right)p\left(z\right)+c+\mathrm{Î¾}}.$
(3.5)

Applying Lemma 2.2 to (3.5), we conclude that

${F}_{c}fâˆˆ{\mathcal{S}}_{\mathrm{Î»}}^{m,n}\left(\mathrm{Î¾},\mathrm{Ï•}\right).$

â€ƒâ–¡

## Authorâ€™s contributions

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

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## Acknowledgements

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

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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