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 $U=\{z\in \mathbb{C}:|z|<1\}$ normalized by $f(0)=f^{\prime }(0)-1=0$. Thus each $f\in \mathcal{A}$ has a Taylor series representation

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

(1.1)

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

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

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

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

Making use of the principle of subordination between analytic functions, denote by$\mathcal{S}(\xi ,\varphi )$[3] a subclass of the class for$0\le \xi <1$ and $\varphi \in \mathcal{R}$ which are defined by

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

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

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

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 $D_{\lambda }^m:\mathcal{A}\to \mathcal{A}$,

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

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

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 $R^n:\mathcal{A}\to \mathcal{A}$,

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

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

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 $R^n$,

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

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

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

Remark 1.5 The operator $D{R}_{\lambda }^{m,n}$ was studied also in [8–10].

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 [15–20].

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

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

(1.2)

and

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

(1.3)

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

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

In particular, we set

$${\mathcal{S}}_{\lambda }^{m,n}(\xi ,\frac{1+Az}{1+Bz})={\mathcal{S}}_{\lambda }^{m,n}(\xi ,A,B),-1<B<A\le 1.$$

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$\phi (\mu ,v)$be a complex function such that$\phi :D\to \mathbb{C}$, $D\subseteq \mathbb{C}\times \mathbb{C}$, and let$\mu ={\mu }_1+i{\mu }_2$, $v=v_1+i{v}_2$. Suppose that$\phi (\mu ,v)$satisfies the following conditions:

• 1. $\phi (\mu ,v)$ is continuous in D,

• 2. $(1,0)\in D$ and $Re\phi (1,0)>0$,

• 3. $Re\phi (i{\mu }_2,v_1)\le 0$ for all $(i{\mu }_2,v_1)\in D$ such that $v_1\le -\frac{1}{2}(1+{\mu }_2^2)$.

Let$h(z)=1+c_{1}z+c_2{z}^2+\cdots$be analytic in U, such that$(h(z),z{h}^{\prime }(z))\in D$for all$z\in U$. If$Re\{\phi h(z),z{h}^{\prime }(z)\}>0$, $z\in U$, then$Re\{h(z)\}>0$.

Lemma 2.2 ([24])

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

$$p(z)+\frac{z{p}^{\prime }(z)}{kp(z)+\nu }\prec \varphi (z),z\in U,$$

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

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

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

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

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

(2.1)

Since from (1.3)

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

we obtain

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

Taking $h(z)=\mu ={\mu }_1+i{\mu }_2$ and $z{h}^{\prime }(z)=v=v_1+i{v}_2$, we define $\phi (\mu ,v)$ by

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

and

$$\begin{array}{c}Re\{\phi (i{\mu }_2,v_1)\}=\frac{(1-\xi )(n+\xi )v_1}{{(n+\xi )}^2+{(1-\xi )}^2{\mu }_2^2},\hfill \\ Re\{\phi (i{\mu }_2,v_1)\}\le -\frac{(1-\xi )(n+\xi )(1+{\mu }_2^2)}{2[{(n+\xi )}^2+{(1-\xi )}^2{\mu }_2^2]}<0.\hfill \end{array}$$

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

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

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

Then

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

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

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

(2.2)

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

By using (1.2) we have

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

Now, by using (2.2) we get

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

(2.3)

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

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

Hence,

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

(2.4)

Since $Re\{\varphi (z)\}<\frac{\xi -1+\frac{1}{\lambda }}{1-\xi }$ implies $Re\{(1-\xi )p(z)+\xi -1+\frac{1}{\lambda }\}>0$, applying Lemma 2.2 to (2.4) we have that$f(z)\in {\mathcal{S}}_{\lambda }^{m,n}(\xi ,\varphi )$, 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$-1<B<A\le 1$, then

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

Proof Taking $\varphi (z)=\frac{1+Az}{1+Bz}$, $-1<B<A\le 1$ 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

$$F_c[f(z)]=\frac{c+1}{z^c}{\int }_0^z{t}^{c-1}f(t)dt=z+\sum _{j=2}^{\mathrm{\infty }}\frac{c+1}{j+c}{a}_j{z}^c,f\in \mathcal{A},c>-1,$$

(3.1)

which satisfies the following equality:

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

(3.2)

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

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

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

Let

$$\frac{z{[D{R}_{\lambda }^{m,n}{F}_c[f(z)]]}^{\prime }}{D{R}_{\lambda }^{m,n}{F}_c[f(z)]}=\xi +(1-\xi )h(z),h(z)=1+c_{1}z+c_2{z}^2+\cdots .$$

We obtain

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

This implies

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

(same as Theorem 2.1) and

$$\begin{array}{c}Re\{\phi (i{\mu }_2,v_1)\}=\frac{(1-\xi )(c+\xi )v_1}{{(c+\xi )}^2+{(1-\xi )}^2{\mu }_2^2},\hfill \\ Re\{\phi (i{\mu }_2,v_1)\}\le -\frac{(1-\xi )(c+\xi ){(1+{\mu }_2)}^2}{2[{(c+\xi )}^2+{(1-\xi )}^2{\mu }_2^2]}<0.\hfill \end{array}$$

After using Lemma 2.1 and Theorem 2.1, we have

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

 □

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

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

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

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

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

(3.3)

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

Using (3.2) and (3.3), we have

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

(3.4)

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

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

(3.5)

Applying Lemma 2.2 to (3.5), we conclude that

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

 □

Author’s contributions

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