On q-Uniformly Mocanu Functions

Abstract

Let f be analytic in open unit disc $E=\{z:|z|<1\}$ with $f\left(0\right)=0$ and $f^{{}^{\prime }}\left(0\right)=1$. The q-derivative of f is defined by: $D_{q}f\left(z\right)=\frac{f\left(z\right)-f\left(qz\right)}{(1-q)z},q\in (0,1),z\in \mathcal{B}-\left\{0\right\},$ where $\mathcal{B}$ is a q-geometric subset of $\mathbb{C}$. Using operator $D_q$, q-analogue class $k-U{M}_q(\alpha ,\beta )$, k-uniformly Mocanu functions are defined as: For $k=0$ and $q\to 1^{-}$, $k-$ reduces to $M\left(\alpha \right)$ of Mocanu functions. Subordination is used to investigate many important properties of these functions. Several interesting results are derived as special cases.

1. Introduction

Let A denote the class of functions f that are analytic in the open unit disc E and are also normalized by the conditions $f\left(0\right)=0$, $f^{{}^{\prime }}\left(0\right)=1$. Let $f,g\in A.$ f is said to be subordinate to g (written as $f\prec g$), if there exists a Schwartz function $w\left(z\right)$ such that $f\left(z\right)=g\left(w\right(z\left)\right).$

q-calculus is ordinary calculus without a limit, and it has been used recently by many researchers in the field of geometric function theory. q-derivatives and q-integrals play an important and significant role in the study of quantum groups and q-deformed super-algebras, the study of fractal and multi-fractal measures and in chaotic dynamical systems. The name q-calculus also appears in other contexts; see [1,2]. The most sophisticated tool that derives functions in non-integer order is the long-known fractional calculus; see [1,2,3,4].

We recall here some basic concepts from q-calculus for which we refer to [5,6,7,8,9,10,11,12,13,14,15,16] and the references therein.

A subset $\beta \subset \mathbb{C}$ is called q-geometric, if $zq\in \beta ,$ whenever $z\in \mathcal{B}$, and it contains all the geometric sequences ${\left\{z{q}^m\right\}}_0^{\infty }$.

The q-derivative $D_q$ of a function $f\in A$ is defined by:

$$\begin{array}{c}\hfill D_{q}f\left(z\right)=\frac{f\left(z\right)-f\left(qz\right)}{(1-q)z},(z\in \mathcal{B}-\left\{0\right\})\end{array}$$

(1)

and $D_{q}f\left(0\right)=f^{{}^{\prime }}\left(0\right).$

Under this definition, we have the following rules for q-derivative $D_q.$

(i).$D_q{z}^m=\frac{1-q^m}{1-q}{z}^{m-1}=[n,q]z^{m-1},$ where $[m,q]=\frac{1-q^m}{1-q}.$

Let $f\left(z\right)$ and $g\left(z\right)$ be defined on a q-geometric set $\mathcal{B}\subset \mathbb{C}$ such that q-derivatives of $f\left(z\right)$ and $g\left(z\right)$ exist for all $z\in \mathcal{B}$. Then, for $a,b$ complex numbers, we have:

(ii).$D_q(af\left(z\right)\pm bg\left(z\right))=a{D}_{q}f\left(z\right)\pm b{D}_{q}g\left(z\right).$

(iii).$D_q(f\left(z\right).g\left(z\right))=g\left(z\right)D_{q}f\left(z\right)+f\left(qz\right)D_{q}g\left(z\right).$

(iv).$D_q\left(\frac{f\left(z\right)}{g\left(z\right)}\right)=\frac{g\left(z\right)D_{q}f\left(z\right)-f\left(z\right)D_{q}g\left(z\right)}{g\left(z\right)g\left(qz\right)},g\left(z\right).g\left(qz\right)\ne 0.$

(v).$D_q(logf\left(z\right))=\frac{D_{q}f\left(z\right)}{f\left(z\right)}.$

Let $P\left(z\right)$ be the class of functions $p\left(z\right)=1+c_{1}z+...,$ analytic in E and satisfying:

$$\begin{array}{c}\hfill |p\left(z\right)-\frac{1}{1-q}|\le \frac{1}{1-q},(z\in E,q\in (0,1)).\end{array}$$

(2)

It is known [9] that $p\in P\left(q\right)$ implies that $p\left(z\right)\prec \frac{1+z}{1-qz}$, where ≺ denotes subordination, and from this, it easily follows that $Rep\left(z\right)>0,z\in E.$

Now, we have:

Definition 1.

[4,5] Let $f\in A$. Then, it is said to belong to the class $S_q^{\ast }\left(\alpha \right)$ of q-starlike functions of order α, $0\le \alpha \le 1,$ if and only if,

$$\begin{array}{c}\hfill \frac{1}{1-\alpha }\left(\frac{z{D}_{q}f\left(z\right)}{f\left(z\right)}-\alpha \right)\prec \frac{1+z}{1-qz}.\end{array}$$

(3)

We can write (3) as:

$$\begin{array}{c}\hfill |\frac{z{D}_{q}f\left(z\right)}{f\left(z\right)}-\frac{1-\alpha q}{1-q}|\le \frac{1-\alpha }{1-q}.\end{array}$$

(4)

By taking $a=\frac{1-\alpha }{1-q}$, $b=\frac{1-\alpha }{1-q}$ in 4, it has been shown in [17] that $f\in S_q^{\ast }\left(\alpha \right),$ if and only if,

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \frac{z{D}_{q}f\left(z\right)}{f\left(z\right)}\prec \frac{1+Az}{1+Bz},-1<B<0\le A\le 1,\end{array}\end{array}$$

(5)

where $A=1-(1+q)\alpha$ and $B=-q$.

As a special case, we note that:

$$\underset{q\to 1^{-}}{lim}{S}_q^{\ast }\left(\alpha \right)=S^{\ast }\left(\alpha \right)\mathrm{with}A=1-2\alpha ,$$

which is the class of starlike functions denoted as $S^{\ast }\left(\alpha \right)$.

Furthermore, for $\alpha =0$, we obtain the class $S_q^{\ast }$ of q-starlike functions introduced and studied in [10].

Definition 2.

Let $f\in A$ and $k\ge 0,0\le \alpha ,\beta \le 1$, $q\in (0,1)$. Then,

$$f\in k-U{M}_q(\alpha ,\beta ),$$

if and only if, for $z\in E,$

$$\begin{array}{c}\hfill Re\left[(1-\alpha )\frac{z{D}_{q}f\left(z\right)}{f\left(z\right)}+\alpha \frac{D_q\left(z{D}_{q}f\left(z\right)\right)}{D_{q}f\left(z\right)}\right]>k|(1-\beta )\frac{z{D}_{q}f\left(z\right)}{f\left(z\right)}+\beta \frac{D_q\left(z{D}_{q}f\left(z\right)\right)}{D_{q}f\left(z\right)}-1|.\end{array}$$

Selecting special values of parameters $\alpha ,\beta$ and k and letting $q\to 1^{-}$, we obtain a number of known classes of analytic functions; see [5,9,18,19,20,21]. We list some of these as follows:

(i)

Choosing $k=0$, we get $limq\to 1^{-}{M}_q\left(\alpha \right)=M\left(\alpha \right)$, the class of $\alpha$-convex functions; see [22].

(ii)

For $\beta =0$, $k=1$, and $q\to 1^{-}$, we have the class $MN$; see [23].

(iii)

Choosing $\beta =0,q\to 1^{-}$, we get the class $k-MN$ introduced in [18,19].

(iv)

$k-U{M}_q(0,0)=k-sT,k-U{M}_q(1,1)=k-UC{V}_q$.

Throughout this paper, we shall assume that $q\in (0,1),0\le \alpha <1$ and $z\in E,$ unless otherwise mentioned.

2. Preliminary Results

Lemma 1.

[4]. Let $\varphi \left(z\right)$ be analytic with $\varphi \left(0\right)=0$. If $\left|\varphi \right(z_0\left)\right|$ attains its maximum value on the circle $\left|z\right|=r$ at a point $z_0\in E$, then we have:

$$z_0{D}_q\varphi (z_0=m\varphi \left(z_0\right),m\ge 1,realnumber.$$

Lemma 2.

[24]. Let $\alpha \ge 0$ and $0\le r<1$. Let $p\left(z\right)$ be analytic in E with $p\left(0\right)=1$.

If:

$$\begin{array}{c}\hfill \left\{p\left(z\right)+\alpha \frac{z{p}^{{}^{\prime }}\left(z\right)}{p\left(z\right)}\right\}\prec \frac{1+(1-2r)z}{1-z},\end{array}$$

then:

$$p\left(z\right)\prec \frac{1+(1-2\S )z}{1-z},$$

where:

$$\delta =\frac{1}{4}\left[(2r-\alpha )+\sqrt{{(2r-\alpha )}^2+8\alpha }\right].$$

3. Main Results

Theorem 1.

Let $p\left(z\right)$ be analytic in E with $p\left(0\right)=1.$ Let, for $k>\frac{1+q}{q}$,

$$\begin{array}{c}\hfill \begin{array}{c}\hfill Re\left[1+\frac{z{D}_{q}p\left(z\right)}{p\left(z\right)}\right]>k\left|\frac{z{D}_{q}p\left(z\right)}{p\left(z\right)}\right|,z\in E.\end{array}\end{array}$$

Then, $p\left(z\right)$ is subordinate to $\frac{1}{1-qz}$, that is, $p\left(z\right)\prec \frac{1}{1-qz}$ in $E.$

Proof. 

Let $p\left(z\right)=\frac{1}{1-q\varphi \left(z\right)}$. It can easily be seen that $\varphi \left(z\right)$ is analytic in E and $\varphi \left(0\right)=0$. We shall show that $\left|\varphi \right(z\left)\right|<1$ for all $z\in E.$ We suppose on the contrary that there exists a $z_0\in E$ such that $\left|\varphi \right(z_0\left)\right|=1.$

Then:

$$\begin{array}{c}\hfill Re\left[1+\frac{z_{\circ }{D}_{q}p\left(z_{\circ }\right)}{p\left(z_{\circ }\right)}\right]-k\left|\frac{z_{\circ }{D}_{q}p\left(z_{\circ }\right)}{p\left(z_{\circ }\right)}\right|=Re\left[1+\frac{q{z}_{\circ }{D}_q\varphi \left(z_{\circ }\right)}{1-q\varphi \left(z_{\circ }\right)}\right]-k\left|\frac{q{z}_{\circ }{D}_q\varphi \left(z_{\circ }\right)}{1-q\varphi \left(z_{\circ }\right)}\right|.\end{array}$$

(6)

Now, by Lemma 1, $z_{\circ }{D}_q\varphi \left(z_{\circ }\right)=m\varphi \left(z_{\circ }\right)=m{e}^{i\theta },m\ge 1,$ and we use it in (6) for:

$$\begin{array}{c}\hfill Re\left[1+\frac{mq{e}^{i\theta }}{1-q{e}^{i\theta }}\right]>k\left|\frac{mq{e}^{i\theta }}{1-q{e}^{i\theta }}\right|>k\left|\frac{q{e}^{i\theta }}{1-q{e}^{i\theta }}\right|.\end{array}$$

(7)

From (6), (7), and choosing $\theta =\pi$, we have:

$$\begin{array}{ccc}\hfill Re\left[1+\frac{z_{\circ }{D}_{q}p\left(z_{\circ }\right)}{p\left(z_{\circ }\right)}\right]-k\left|\frac{z_{\circ }{D}_{q}p\left(z_{\circ }\right)}{p\left(z_{\circ }\right)}\right|& =& 1-\frac{mq}{1+q}-\frac{kq}{1+q}<0fork>\frac{1+q}{q}.\hfill \end{array}$$

This is a contradiction, and hence, $\left|\varphi \right(z\left)\right|<1$ for all $z\in E$. This proves that:

$$p\left(z\right)\prec \frac{1}{1-qz}inE.$$

 □

We apply Theorem 1 to have the following results.

Corollary 1.

Let $p\left(z\right)=f^{{}^{\prime }}\left(z\right)$, $q\to 1^{-}$, and $k>2$. Then, from Theorem 1, it follows that:

$$Re\left[1+\frac{z{f}^{{}^{″}}\left(z\right)}{zf{}^{\prime }\left(z\right)}\right]>k\left|\frac{z{f}^{{}^{″}}\left(z\right)}{zf{}^{\prime }\left(z\right)}\right|,$$

which implies $f\in k-UCV$, and so, $Re{f}^{{}^{\prime }}\left(z\right)>\frac{1}{2}$ in E.

Corollary 2.

For $k>\frac{q+1}{q}$, let $f\in k-S{T}_q.$ Then, $\frac{f\left(z\right)}{z}\prec \frac{1}{1-qz}$ in E.

The proof is immediate when we take $p\left(z\right)=\frac{f\left(z\right)}{z}$ in Theorem 1.

As a special case, when $q\to 1^{-},k>2,f\in k-ST$ implies $Re\frac{f\left(z\right)}{z}>\frac{1}{2}$ in $E.$

Using a similar technique, we can prove the following results.

Theorem 2.

Let $k\ge 0,\alpha ,\beta \in [0,1],qk<1$, and let $p\left(z\right)$ be analytic in E with $p\left(0\right)=1.$

If:

$$\begin{array}{c}\hfill \begin{array}{c}\hfill Re\left[p\left(z\right)+\frac{\alpha z{D}_{q}p\left(z\right)}{p\left(z\right)}-\frac{1-kq}{1+q}\right]>k|p\left(z\right)+\frac{\beta z{D}_{q}p\left(z\right)}{p\left(z\right)}-1|,\end{array}\end{array}$$

(8)

then $p\left(z\right)\prec \frac{1}{1-qz},z\in E.$

We can easily deduce some special cases of Theorem 2 as given below.

Corollary 3.

Let $\beta =0,p\left(z\right)=\frac{z{D}_{q}f\left(z\right)}{f\left(z\right)}$ in (8). Then:

$$\begin{array}{c}\hfill Re\left[(1-\alpha )\frac{z{D}_{q}f\left(z\right)}{f\left(z\right)}+\alpha \frac{D_q\left(z{D}_{q}f\left(z\right)\right)}{D_{q}f\left(z\right)}-\frac{1-kq}{1+q}\right]-k|\frac{z{D}_{q}f\left(z\right)}{f\left(z\right)}-1|>0\end{array}$$

implies:

$$f\in S_q^{\ast }\left(\frac{1}{1+q}\right),z\in E.$$

As a special case of this corollary, we observe that $UST\subset S^{\ast }\left(\frac{1}{2}\right),$ when we choose $k=1,\alpha =0$, and let $q\to 1^{-}.$

Corollary 4.

Let $q\to 1^{-}$ and $p\left(z\right)=f^{{}^{\prime }}\left(z\right).$ Then:

$$\begin{array}{ccc}\hfill Re\left[f^{{}^{\prime }}\left(z\right)+\alpha \frac{{\left(z{f}^{{}^{\prime }}\left(z\right)\right)}^{{}^{\prime }}}{f^{{}^{\prime }}\left(z\right)}-(\alpha +\frac{1-k}{2})\right]& >& k|f^{{}^{\prime }}\left(z\right)+\beta \frac{{\left(z{f}^{{}^{\prime }}\left(z\right)\right)}^{{}^{\prime }}}{f^{{}^{\prime }}\left(z\right)}-(1+\beta )|\hfill \\ & =& k|(1+\beta )-f^{{}^{\prime }}\left(z\right)-\beta \frac{{\left(z{f}^{{}^{\prime }}\left(z\right)\right)}^{{}^{\prime }}}{f^{{}^{\prime }}\left(z\right)}|\hfill \\ & \ge & Re\left[(1+\beta )-f^{{}^{\prime }}\left(z\right)-\beta \frac{{\left(z{f}^{{}^{\prime }}\left(z\right)\right)}^{{}^{\prime }}}{f^{{}^{\prime }}\left(z\right)}\right].\hfill \end{array}$$

This gives us:

$$\begin{array}{c}\hfill Re\left[f^{{}^{\prime }}\left(z\right)+\left(\frac{\alpha +\beta }{1+k}\right)\frac{{\left(z{f}^{{}^{\prime }}\left(z\right)\right)}^{{}^{\prime }}}{f^{{}^{\prime }}\left(z\right)}\right]\ge \frac{k(1+\beta )+(\alpha +\gamma )}{1+k}=\eta ,(\gamma =\frac{1-k}{2}).\end{array}$$

Now, using Lemma 2 together with Theorem 2 when $q\to 1^{-},$ we obtain the result that:

$$\begin{array}{c}\hfill Re{f}^{{}^{\prime }}\left(z\right)>\delta =\frac{1}{4}[(2\eta -\rho )+\sqrt{{(2\eta -\rho )}^2+8\rho }],\rho =\frac{\alpha +\beta }{1+k}.\end{array}$$

Corollary 5.

In (8), if we take $\beta =0,\alpha =1,k=1$ and $p\left(z\right)=\frac{z{D}_{q}f\left(z\right)}{f\left(z\right)}$, then:

$$\begin{array}{c}\hfill Re\left[\frac{D_q\left(z{D}_{q}f\left(z\right)\right)}{D_q\left(f\left(z\right)\right)}-\frac{1-q}{1+q}\right]>|\frac{z{D}_{q}f\left(z\right)}{f\left(z\right)}-1|.\end{array}$$

implies

$$f\in S_q^{\ast }\left(\frac{1}{1+q}\right)inE.$$

Furthermore, with $\beta =1,\alpha =1,k=1$ and $p\left(z\right)=\frac{z{D}_{q}f\left(z\right)}{f\left(z\right)}$ in (8), it follows that:

$$\begin{array}{c}\hfill Re\left[\frac{D_q\left(z{D}_{q}f\left(z\right)\right)}{D_{q}f\left(z\right)}-\frac{1-q}{1+q}\right]>|\frac{D_q\left(z{D}_{q}f\left(z\right)\right)}{D_{q}f\left(z\right)}-1|\end{array}$$

implies $f\in S_q^{\ast }\left(\frac{1}{1+q}\right).$

Next, we prove the following:

Theorem 3.

Let $p\left(z\right)$ be analytic in E with $p\left(0\right)=1.$ Let:

$$\begin{array}{c}\hfill Re\left[p\left(z\right)+\frac{\left(z{D}_{q}p\left(z\right)\right)}{\lambda p\left(z\right)+c}-r\right]-k|p\left(z\right)+\frac{z{D}_{q}p\left(z\right)}{\lambda p\left(z\right)+c}-1|>0,\end{array}$$

(9)

where $r=\frac{1}{1+q},\lambda$, and c are positive real. Then, $p\left(z\right)\prec \frac{1}{1-qz}$ in $E.$

Proof. 

We shall follow the same procedure to prove this result as was used in Theorem 1. Let $p\left(z\right)=\frac{1}{1-q\varphi \left(z\right)}$. Clearly, $\varphi \left(0\right)=0$, and $\varphi \left(z\right)$ is analytic. We prove that $\varphi \left(z\right)$ is a Schwartz function, that is $\left|\varphi \right(z\left)\right|<1,\forall z\in E$. Suppose on the contrary that there exists $z_{\circ }\in E$ such that $|\varphi \left(z_{\circ }\right)|=1=|e^{i\theta }|,0\le \theta \le 2\pi$.

Now, with some computations, we have:

$$\begin{array}{c}\hfill p\left(z\right)+\frac{z{D}_{q}p\left(z\right)}{\lambda p\left(z\right)+c}=\frac{1}{1-q\varphi \left(z\right)}+\frac{\left(\frac{q}{\lambda }\right)z{D}_q\varphi \left(z\right)}{1-q\varphi \left(z\right)}-\frac{\left(\frac{q}{\lambda }\right)cz{D}_q\varphi \left(z\right)}{(\lambda +c)-qc\varphi \left(z\right)}.\end{array}$$

(10)

We apply Lemma 1 to have $z_{\circ }{D}_q\varphi \left(z_{\circ }\right)=m\varphi \left(z_{\circ }\right),m\ge 1$, and note that:

$$\begin{array}{ccc}\hfill Re\left\{\frac{\frac{q}{\lambda }{z}_{\circ }{D}_q\varphi \left(z_{\circ }\right)}{1-q\varphi \left(z_{\circ }\right)}\right\}=Re\left\{\frac{\frac{mq}{\lambda }\varphi \left(z_{\circ }\right)}{1-q\varphi \left(z_{\circ }\right)}\right\}& =& Re\left\{\frac{\frac{mq}{\lambda }{e}^{i\theta }}{1-q{e}^{i\theta }}\right\}\hfill \\ & =& \frac{\frac{mq}{\lambda }(cos\theta -q)}{|1-q{e}^{i\theta }{|}^2},\hfill \end{array}$$

(11)

$$\begin{array}{c}\hfill Re\left\{\frac{\frac{q}{\lambda }c{z}_{\circ }{D}_q\varphi \left(z_{\circ }\right)}{(\lambda +c)-qc\varphi \left(z_{\circ }\right)}\right\}=\frac{\frac{q}{\lambda }cm(\lambda +c)cos\theta -\frac{q^2{c}^{2}m}{\lambda }}{|(\lambda +c)-qc{e}^{i\theta }{|}^2},\end{array}$$

(12)

and:

$$\begin{array}{cc}\hfill & |\frac{1}{1-q{e}^{i\theta }}+\left\{\frac{q}{\lambda }\frac{m{e}^{i\theta }}{(1-q{e}^{i\theta })}\right\}-\{\frac{\frac{q}{\lambda }cm{e}^{i\theta }}{(\lambda +c)-qc{e}^{i\theta }}-1\}{|}_{\theta =\pi }\hfill \\ =& |\frac{1-q}{1+q}-\frac{\frac{mq}{\lambda }}{1+q}+\frac{\frac{qcm}{\lambda }}{(\lambda +c)+qc}|\hfill \end{array}$$

(13)

Using (10), (11), (12), and (13), we get a contradiction to the given hypothesis (9), when we assume $\left|\varphi \right(z_{\circ }\left)\right|=1$ for some $z_{\circ }\in E$. Hence $\left|\varphi \right(z\left)\right|<1$ for all $z\in E$ and:

$$p\left(z\right)\prec \frac{1}{1-qz},z\in E.$$

This completes the proof. □

In order to develop some applications of Theorem 3, we need the following.

Let the operator $D_q^n:A\to A$ be defined as:

$$\begin{array}{ccc}\hfill D_q^{n}f\left(z\right)& =& F_{n+1,q}\left(z\right)\ast f\left(z\right)\hfill \\ & =& z+\sum _{m=2}^{\infty }\frac{[m+n-1,q]!}{[n,q]![m-1,q]!}{a}_m{z}^n,\hfill \end{array}$$

(14)

where:

$$f\left(z\right)=z+\sum _{m=2}^{\infty }{a}_m{z}^m,$$

and:

$$F_{n+1,q}\left(z\right)=z+\sum \frac{[m+n-1,q]!}{[n,q]![m-1,q]!}{z}^m.$$

This series is absolutely convergent in E, and * denotes convolution. The operator $D_q^n$ is called the q-Ruscheweyh derivative of order n; see [25].

It can easily be seen that $D_q^{\circ }f\left(z\right)=f\left(z\right)$ and $D_q^{{}^{\prime }}f\left(z\right)=z{D}_{q}f\left(z\right).$

The relation (14) can be expressed as:

$$\begin{array}{c}\hfill D_q^{n}f\left(z\right)=\frac{z{D}_q^n\left(z^{n-1}f\left(z\right)\right)}{[n,q]!},n\in N.\end{array}$$

Furthermore,

$$li{m}_{q\to 1}{D}_q^{n}f\left(z\right)={\frac{z}{(1-z)}}^{n+1}\ast f\left(z\right)=D^{n}f\left(z\right),$$

which is called the Ruscheweyh derivative of order n; see [25].

Let $f\in A$. Then, f is said to belong to the class $S_q^{\ast }(n,\alpha ),$ if and only if, $D_q^{n}f\in S_q^{\ast }\left(\alpha \right),z\in E.$

The following identity can easily be obtained:

$$\begin{array}{c}\hfill z{D}_q\left(D_q^{n}f\left(z\right)\right)=\left(1+\frac{[n,q]}{q^n}\right)D_q^{n+1}f\left(z\right)-\frac{[n,q]}{q^n}{D}_q^{n}f\left(z\right)\end{array}$$

(15)

We now take $p\left(z\right)=\frac{z{D}_q\left(D_q^{n}f\left(z\right)\right)}{D_q^{n}f\left(z\right)}$ in relation (9) of Theorem 3 to have:

Theorem 4.

Let $D_q^{n}f=F_n$ denote the q-Ruscheweyh derivative of order n for $f\in A.$ Let:

$$\begin{array}{c}\hfill Re\left[\frac{z{D}_q{F}_{n+1}\left(z\right)}{F_{n+1}\left(z\right)}-\frac{1}{1+q}\right]>k|\frac{z{D}_q{F}_{n+1}\left(z\right)}{F_{n+1}\left(z\right)}-1|,k\ge 0.\end{array}$$

Then:

$$\begin{array}{c}\hfill \frac{z{D}_q{F}_n\left(z\right)}{F_n\left(z\right)}\prec \frac{1}{1-qz},z\in E.\end{array}$$

That is, $f\in S_q^{\ast }(n,\alpha ),\alpha =\frac{1}{1+q}.$

Proof. 

Let p be analytic in E with $p\left(0\right)=0$, and let:

$$\begin{array}{c}\hfill p\left(z\right)=\frac{z{D}_q\left(D_q^{n}f\left(z\right)\right)}{D_q^{n}f\left(z\right)}=\frac{z{D}_q{F}_n\left(z\right)}{F_n\left(z\right)}.\end{array}$$

Using identity (15) and some computation, we have:

$$\begin{array}{c}\hfill Re\left[p\left(z\right)+\frac{z{D}_{q}p\left(z\right)}{p\left(z\right)+n}-\frac{1}{1+q}\right]-k|p\left(z\right)+\frac{z{D}_{q}p\left(z\right)}{p\left(z\right)}-1|>0.\end{array}$$

Now, the required result follows immediately from Theorem 3. □

Corollary 6.

In Theorem 4, we take $k=0$. Then, it gives us:

$$\begin{array}{c}\hfill S_q^{\ast }(n+1,\alpha )\subset S_q^{\ast }(n,\alpha )\subset ...\subset S_q^{\ast }\left(\alpha \right),\alpha =\frac{1}{1+q}.\end{array}$$

When $q\to 1^{-},\frac{1}{1+q}\to \frac{1}{2}$, and we have:

$$\begin{array}{c}\hfill S_q^{\ast }(n+1,\frac{1}{2})\subset S^{\ast }(n,\frac{1}{2})\subset ...\subset S^{\ast }\left(\frac{1}{2}\right).\end{array}$$

Corollary 7.

Let $f\in A$, and let:

$$\begin{array}{c}\hfill Re\left[\frac{z{D}_{q}f\left(z\right)}{f\left(z\right)}-\frac{1}{1+q}\right]>k|\frac{z{D}_{q}f\left(z\right)}{f\left(z\right)}-1|,k\ge 0.\end{array}$$

(16)

Define:

$$\begin{array}{c}\hfill L_B\left(f\right)=F_c\left(z\right)=\frac{{[c+1]}_q}{z^c}\underset{0}{\overset{z}{\int }}{t}^{c-1}f\left(t\right)d_{q}t,c\in N_{\circ }.\end{array}$$

(17)

Then:

$$\frac{z{D}_q{F}_c\left(z\right)}{F_c\left(z\right)}\prec \frac{1}{1-qz},z\in E.$$

Proof. 

The integral operator $L_B:A\to A$ defined in (16) is known as the q-Bernardi integral operator $L_B\left(f\right)=F_c.$ When $q\to 1^{-},$ (16) reduces to the well-known Bernardi operator; see [7].

Let,

$$\begin{array}{c}\hfill \frac{z{D}_q{F}_c\left(z\right)}{F_c\left(z\right)}=p\left(z\right).\end{array}$$

(18)

Then, from (16), (17), (18), and some computations, this leads us to:

$$\begin{array}{ccc}\hfill Re\left[\frac{z{D}_{q}f\left(z\right)}{f\left(z\right)}-\frac{1}{1+q}\right]-k|\frac{z{D}_{q}f\left(z\right)}{f\left(z\right)}-1|& =& Re\left[p\left(z\right)+\frac{z{D}_{q}p\left(z\right)}{p\left(z\right)+q^c[c,q]}\right]\hfill \\ & -& k|p\left(z\right)+\frac{z{D}_{q}p\left(z\right)}{p\left(z\right)+q^c[c,q]}-1|>0,z\in E.\hfill \end{array}$$

We now apply Theorem 3, and it follows that:

$$p\left(z\right)=\frac{z{D}_q{F}_c\left(z\right)}{F_c\left(z\right)}\prec \frac{q}{1-qz}\mathrm{in}E.$$

That is, $F_c\in S_q^{\ast }\left(\frac{1}{1+q}\right).$ □

As a special case, when $q\to 1^{-},$ then $f\in K-UST\left(\frac{1}{2}\right)$, and then, $F_c,$ defined by 17, belongs to $S^{\ast }\left(\frac{1}{2}\right)$ in $E.$

4. Conclusions

In this paper, we have used q-calculus, conic domains, and subordination to define and study some new subclasses involving Mocanu functions. Some interesting inclusion and subordination properties of these new classes have been derived. The q-analogue of the Ruscheweyh derivative has been used to obtain a new subordination result for q-Mocanu functions. Some special cases have been discussed as applications of our main results. The technique and ideas of this paper may stimulate further research in this dynamic field.