A Study on Certain Subclasses of Analytic Functions Involving the Jackson q-Difference Operator

Abstract

We introduce two new subclasses of analytic functions in the open symmetric unit disc using a linear operator associated with the q-binomial theorem. In addition, we discuss inclusion relations and properties preserving integral operators for functions in these classes. This paper generalizes some known results, as well as provides some new ones.

1. Introduction

As a result of the pioneering work of Euler and Heine, Frank Hilton Jackson systematically developed q-calculus at the beginning of the previous century. In his work, Jackson created the concepts of the q-derivative (Jackson [1]), and the q-integral (Jackson [2]). Simply, q-calculus is ordinary classical calculus without the notion of limits. The symmetric q-calculus finds applications in different fields, specially in quantum mechanics; see [3,4]. Furthermore, the area of q-calculus is experiencing rapid growth due to its applications in mathematics, mechanics, and physics. This history of q-calculus may be illustrated by its wide variety of applications in quantum mechanics, analytic number theory, theta functions, hypergeometric functions, finite difference theory, gamma function theory, Bernoulli and Euler polynomials, mock theta functions, combinatorics, multiple hypergeometric functions, Sobolev spaces, operator theory, and more recently, in analytic and harmonic univalent functions. q-Bernstein polynomials are an application of q-calculus in approximation theory initiated by Lupas [5]. By generalizing the set of starlike functions into a q-analogue, called the set of q-starlike functions, Ismail et al. [6] were the first to apply q-calculus to geometric function theory (GFT). Srivastava’s work; see [7] discussed the operators of q-calculus and fractional q-calculus and their applications in the GFT of complex analysis were also significant in this direction. Following the same idea, the q-difference operator has been extensively investigated in the field of GFT by many authors. For some recent works related to this operator on the classes of analytic functions, we refer to [8,9,10,11,12,13,14,15]. The theory of q-series is based on the observation that

$$\underset{q\to 1}{lim}\frac{1-q^m}{1-q}=m\mathrm{for}m\in \mathbb{C},$$

where $\mathbb{C}$ is the set of complex numbers. For $0<\left|q\right|<1$ the number

$${\left[m\right]}_q:=\frac{1-q^m}{1-q}$$

is called a q-number (or basic number). The q-shifted factorial, see [16], is defined for $a\in \mathbb{C}$ by

$${(a;q)}_n=\left\{\begin{array}{cc}1& \mathrm{if}n=0\\ (1-a)(1-aq)(1-a{q}^2)\dots (1-a{q}^{n-1}),& \mathrm{if}n\in \mathbb{N}=\{1,2,\dots \}.\end{array}\right.$$

It is easy to see that

$$\underset{q\to 1}{lim}\frac{{(q^{\alpha };q)}_n}{{(1-q)}^n}={\left(\alpha \right)}_n,$$

where ${\left(\alpha \right)}_n$ is the familiar Pochhammer symbol given by

$${\left(\alpha \right)}_n=\left\{\begin{array}{cc}1\hfill & \left(n=0;\alpha \in {\mathbb{C}}^{*}=\mathbb{C}\backslash \left\{0\right\}\right)\\ \hfill \alpha \left(\alpha +1\right)...\left(\alpha +n-1\right)& \left(n\in \mathbb{N};\alpha \in \mathbb{C}\right).\end{array}\right.$$

The following formula is one of the most important summation formulas for hypergeometric series:

$${}_1{F}_0(a;-;z)=\sum _{n=0}^{\infty }\frac{{\left(a\right)}_n}{n!}{z}^n={(1-z)}^{-a}\left(\left|z\right|<1\right).$$

A q-analogue of this formula is called the q-binomial theorem:

$${}_1{\mathsf{\Phi }}_0\left(a;-;q,z\right)=\sum _{n=0}^{\infty }\frac{{(a;q)}_n}{{(q;q)}_n}{z}^n=\frac{{(az;q)}_{\infty }}{{(z;q)}_{\infty }}\left(\left|z\right|<1\right),$$

(see Gasper and Rahman [17] (p. 8)). Jackson’s q-derivative of a function f defined on a subset of $\mathbb{C}$ is given by (see [1,2])

$$D_{q}f\left(z\right):=\frac{f\left(z\right)-f\left(qz\right)}{(1-q)z}\left(z\ne 0\right).$$

Then, we have

$$D_q{z}^k={\left[k\right]}_q{z}^{k-1},$$

and

$$\underset{q\to 1}{lim}{D}_{q}f\left(z\right)=f^{{}^{\prime }}\left(z\right)\left(z\ne 0\right),$$

where $f^{{}^{\prime }}$ is the ordinary derivative. Under the definition of q-derivative, $D_q,$ we have the following rules.

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

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

and

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

Jackson [2] introduced the q-integral

$$\underset{0}{\overset{z}{\int }}f\left(t\right)d_{q}t=z(1-q)\sum _{k=0}^{\infty }{q}^{k}f\left(z{q}^k\right),$$

as long as the series converges. In the case of a function $f\left(z\right)=z^k$, we observe that

$$\underset{0}{\overset{z}{\int }}f\left(t\right)d_{q}t=\underset{0}{\overset{z}{\int }}{t}^k{d}_{q}t=\frac{1}{{[k+1]}_q}{z}^{k+1}(k\ne -1),$$

and

$$\underset{q\to 1}{lim}\underset{0}{\overset{z}{\int }}f\left(t\right)d_{q}t=\underset{q\to 1}{lim}\frac{1}{{[k+1]}_q}{z}^{k+1}=\frac{1}{k+1}{z}^{k+1}=\underset{0}{\overset{z}{\int }}f\left(t\right)dt,$$

where $\underset{0}{\overset{z}{\int }}f\left(t\right)dt$ is the ordinary integral. Let A denote the class of all functions of the form

$$f\left(z\right)=\sum _{n=0}^{\infty }{a}_{n+1}{z}^{n+1}(a_1=1)$$

(1)

which are analytic in the open symmetric unit disc $\mathbb{U}=\{z:z\in \mathbb{C}$ and $\left|z\right|<1\}$. Seoudy and Aouf [18] introduced the class $S_q^{*}\left(\alpha \right)$ of q-starlike functions of order $\alpha ,$ $0\le \alpha <1,$ consisting of all functions $f\in A$ satisfying the inequality

$$\Re \left(\frac{z{D}_{q}f\left(z\right)}{f\left(z\right)}\right)>\alpha \left(z\in \mathbb{U}\right).$$

For the functions $f_j\left(z\right)(j=1,2)$ defined by

$$f_j\left(z\right)=\sum _{n=0}^{\infty }{a}_{n+1,j}{z}^{n+1},$$

let $f_1\ast f_2$ denote the Hadamard product (or convolution) of $f_2$ and $f_2$ defined by

$$(f_1\ast f_2)\left(z\right)=\sum _{n=0}^{\infty }{a}_{n+1,1}{a}_{n+1,2}{z}^{n+1}.$$

In [19], Ruscheweyh introduced an operator $D^n:A\to A$ defined by convolution:

$$D^{n}f\left(z\right)=\frac{z}{{(1-z)}^{n+1}}\ast f\left(z\right)(n>-1,z\in \mathbb{U}),$$

which implies that

$$D^{n}f\left(z\right)=\frac{z{\left(z^{n-1}f\left(z\right)\right)}^{\left(n\right)}}{n!}(n\in N_0:=N\cup \left\{0\right\}).$$

Ruscheweyh considered the class:

$$R_n=\{f\in A:\Re \left(\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}\right)>\frac{1}{2}\},$$

and showed that $R_{n+1}\subset R_n$ for each $n\in N_0:=\mathbb{N}\cup \left\{0\right\}=\{0,1,2,\dots \}.$ In [20], Singh and Singh defined the subclass $K_n(n\in {\mathbb{N}}_0)$ of A whose members are characterized by the following condition

$$\Re \left(\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}\right)>\frac{n}{n+1}.$$

It follows immediately that $K_n\subset R_n$ for each $n\in \mathbb{N}.$ Furthermore, $K_{n+1}\subset K_n$ for every $n\in {\mathbb{N}}_0.$ For other classes defined by convolution, see [21,22]. In 2014, Aldweby and Darus in [23] introduced the q-analogue of the Ruschewewh differential operator by:

$$R_q^{n}f\left(z\right)=z+\sum _{k=2}^{\infty }\frac{{[n-1+k]}_q!}{{\left[n\right]}_q!{[k-1]}_q!}{a}_k{z}^n(n\ge 0),$$

where ${\left[n\right]}_q!$ is given by

$${\left[n\right]}_q!=\left\{\begin{array}{c}{\left[n\right]}_q{[n-1]}_q{[n-2]}_q\dots {\left[1\right]}_{q}n=1,2,\dots ,\\ 1n=0.\end{array}\right.$$

We observe that when $q\to 1^{-}$, we have $R_q^{n}f\left(z\right)=D^{n}f\left(z\right).$ For more details on the q-analogue Ruschewewh differential operators, see [24,25,26,27]. Now, we define the function $\phi (a,q,z)$ by

$$\phi (a,q;z)=z_1{\mathsf{\Phi }}_0\left(a;-;q,z\right)=\sum _{n=0}^{\infty }\frac{{(a;q)}_n}{{(q;q)}_n}{z}^{n+1}(z\in \mathbb{U}).$$

Corresponding to the function $\phi (a,q,z),$ we define a linear operator $L(a,q)$ on A by the convolution

$$L(a,q)f\left(z\right)=\phi (a,q;z)\ast f\left(z\right)=\sum _{n=0}^{\infty }\frac{{(a;q)}_n}{{(q;q)}_n}{a}_{n+1}{z}^{n+1}(a_1=1).$$

Remark 1.

For $f\left(z\right)\in A$

$$L(q^{\alpha +1},q)f\left(z\right)=\sum _{n=0}^{\infty }\frac{{(q^{\alpha +1};q)}_n}{{(q;q)}_n}{a}_{n+1}{z}^{n+1}=R_q^{\alpha }f\left(z\right).$$

For the operator $L(a,q)$, it is easy to verify the following identity

$$\frac{1-a}{1-q}L(aq,q)f\left(z\right)=\frac{q-a}{q(1-q)}L(a,q)f\left(z\right)+\frac{a}{q}z{D}_{q}L(a,q)f\left(z\right).$$

(2)

When $a=q^{\alpha +1}$ in (2), we get the identity given by Aldweby and Darus in [23] for the operator $R_q^n.$

The following definition is a generalization of the definition of the class $K_n$ given by Singh and Singh [20].

Definition 1.

Let $R(a,q),$ $(0\le a<q)$ denote the subclass of A whose members satisfy the following condition

$$\Re \left\{\frac{L(aq,q)f\left(z\right)}{L(a,q)f\left(z\right)}\right\}>\frac{q-a}{q(1-a)}(z\in \mathbb{U}).$$

Letting $a=q^{\alpha +1}$ in the above definition, we get the following remark.

Remark 2.

$\underset{q\to 1^{-}}{lim}$$R(q^{\alpha +1},q)=K_n$.

Definition 2.

Let $S_{\alpha }(a,q),$ $(0\le a<q,$ $0\le \alpha <1)$ be the subclass of A consisting of functions of the form (1) and satisfying the following condition

$$\Re \left\{\frac{L(a,q)f\left(z\right)}{z}\right\}>\alpha (z\in \mathbb{U}).$$

If we set $a=q^{n+1}$ and let $q\to 1^{-}$ in Definition 2, we get the class obtained by Singh and Singh [20].

Due to the wide range of applications of q-calculus and to the significance of q-operators, many researchers have exhaustively studied several subclasses of analytic functions by using the q-derivative operator $D_q$, for a brief survey on these classes, readers may refer to [28,29,30,31,32,33,34,35,36,37,38,39].

The purpose of this paper is to establish inclusion relations for the new classes $R(a,q)$ and $S_{\alpha }(a,q)$. Moreover, we consider the class-preserving integral operators associated with functions belonging to these classes. Our study generalizes some of the earlier works obtained by other authors.

2. Main Results

In order to prove our main results, we shall require the following lemma to be used in the sequel.

Lemma 1

(q-Jack lemma [40]). Let $\omega \left(z\right)$ be analytic in $\mathbb{U}$ with $\omega \left(0\right)=0.$ Then, if $\left|\omega \left(z\right)\right|$ attains its maximum value on the circle $\left|z\right|=r(r<1)$ at a point $z_0,$ we can write

$$z_0{D}_q\omega \left(z_0\right)=k\omega \left(z_0\right),$$

where k is real and $k\ge 1.$

Theorem 1.

Let $0<q<1$ and $0\le a<q$, then

$$R(aq,q)\subset R(a,q).$$

Proof. 

Suppose $f\left(z\right)\in R(aq,q)$, then

$$\Re \left\{\frac{L(a{q}^2,q)f\left(z\right)}{L(aq,q)f\left(z\right)}\right\}>\frac{1-a}{1-aq}.$$

(3)

We have to show that (3) implies the following inequality

$$\Re \left\{\frac{L(aq,q)f\left(z\right)}{L(a,q)f\left(z\right)}\right\}>\frac{q-a}{q(1-a)}.$$

Define $\omega \left(z\right)$ in $\mathbb{U}$ by

$$\frac{L(aq,q)f\left(z\right)}{L(a,q)f\left(z\right)}=\frac{q-a}{q(1-a)}+\frac{a(1-q)}{q(1-a)}\frac{1-\omega \left(z\right)}{1+\omega \left(z\right)}.$$

(4)

Clearly, $\omega \left(0\right)=0.$ Equation (4) may be written as

$$\frac{L(aq,q)f\left(z\right)}{L(a,q)f\left(z\right)}=\frac{q(1-a)+\left[\right(q-a)-a(1-q\left)\right]\omega \left(z\right)}{q(1-a)[1+\omega (z\left)\right]}.$$

(5)

With the q-derivative rules and some simple calculations, (5) gives

$$\frac{z{D}_{q}L(aq,q)f\left(z\right)}{L(aq,q)f\left(z\right)}-\frac{z{D}_{q}L(a,q)f\left(z\right)}{L(a,q)f\left(z\right)}=-\frac{2a(1-q)z{D}_q\omega \left(z\right)}{\left[q\right(1-a)+[(q-a)-a(1-q)\left]\omega \right(z\left)\right][1+\omega (z\left)\right]}.$$

Using the identity (2) and Equation (4), we can conclude that

$$\begin{array}{ccc}\hfill \frac{L(a{q}^2,q)f\left(z\right)}{L(aq,q)f\left(z\right)}-\frac{1-a}{1-aq}& =& \frac{a(1-q)}{(1-aq)}\frac{1-\omega \left(z\right)}{1+\omega \left(z\right)}\hfill \\ & & -\frac{2a^2{(1-q)}^2}{(1-aq)}\frac{z{D}_q\omega \left(z\right)}{\left[q\right(1-a)+[(q-a)-a(1-q)\left]\omega \right(z\left)\right][1+\omega (z\left)\right]}.\hfill \end{array}$$

(6)

We claim that $\left|\omega \left(z\right)\right|<1$ for $z\in \mathbb{U}$, as otherwise by the above lemma, there exists a point $z_0\in \mathbb{U}$ such that

$$z_0{D}_q\omega \left(z_0\right)=k\omega \left(z_0\right)$$

where $\left|\omega \left(z_0\right)\right|=1$ and $k\ge 1.$ Then,

$$\begin{array}{ccc}& & \Re \left\{\frac{L(a{q}^2,q)f\left(z_0\right)}{L(aq,q)f\left(z_0\right)}\right\}-\frac{1-a}{1-aq}\hfill \\ & =& -\frac{2a^2{(1-q)}^2}{(1-aq)}\frac{k\omega \left(z_0\right)}{[q(1-a)+[(q-a)-a(1-q)]\omega \left(z_0\right)][1+\omega \left(z_0\right)]}.\hfill \end{array}$$

Letting $\omega \left(z_0\right)=e^{i\theta },$ then we have

$$\begin{array}{ccc}& & \Re \left\{\frac{L(a{q}^2,q)f\left(z_0\right)}{L(aq,q)f\left(z_0\right)}\right\}-\frac{1-a}{1-aq}\hfill \\ & =& -\frac{4a^{2}q{(1-q)}^2}{(1-aq)}\frac{k(q-a)(1+cos\theta )}{{\left|[q(1-a)+[(q-a)-a(1-q)]\omega \left(z_0\right)][1+\omega \left(z_0\right)]\right|}^2}\le 0\hfill \end{array}$$

which contradicts (3). Hence, $\left|\omega \left(z\right)\right|<1$ and from (4), it follows that $f\left(z\right)\in R(aq,q).$ □

If we set $a=q^{\alpha +1}$ in Definition 1 of the function class $R(a,q)$, we obtain a new class ${\tilde{R}}_{\alpha }$ given by Definition 3 below.

Definition 3.

Let ${\tilde{R}}_{\alpha },\alpha \ge 0$ denote the class of functions $f\in A$ which satisfy the condition

$$\Re \left\{\frac{R_q^{\alpha +1}f\left(z\right)}{R_q^{\alpha }f\left(z\right)}\right\}>\frac{{\left[\alpha \right]}_q}{{[\alpha +1]}_q}(z\in \mathbb{U}).$$

If we set $a=q^{\alpha +1}$ in Theorem 1, we get the following corollary.

Corollary 1.

Let $\alpha \ge 0$, then

$${\tilde{R}}_{\alpha +1}\subset {\tilde{R}}_{\alpha }.$$

(7)

It follows from (7) that all functions in ${\tilde{R}}_{\alpha }$ are q-starlike.

Remark 3.

Let $q\to 1^{-}$ in Corollary 1, we get the result obtained by Singh and Singh [20].

Theorem 2.

Let $0\le a<q$ and $0\le \alpha <1,$ then

$$S_{\alpha }(aq,q)\subset S_{\alpha }(a,q).$$

Proof. 

Let $f\left(z\right)\in S_{\alpha }(aq,q)$ and $\omega \left(z\right)$ be a regular function in $\mathbb{U}$ defined by

$$\frac{L(a,q)f\left(z\right)}{z}=\frac{1+(2\alpha -1)\omega \left(z\right)}{1+\omega \left(z\right)}.$$

(8)

Clearly, $\omega \left(0\right)=0$ and $\omega \left(z\right)\ne -1$ for $z\in \mathbb{U}.$ It is sufficient to show that $\left|\omega \left(z\right)\right|<1$ for $z\in \mathbb{U}.$ By applying the q-derivative rules, (8) gives

$$D_{q}L(a,q)f\left(z\right)=\frac{1+(2\alpha -1)\omega \left(z\right)}{1+\omega \left(z\right)}-2q(1-\alpha )\frac{z{D}_q\omega \left(z\right)}{[1+\omega (z\left)\right][1+\omega (qz\left)\right]}.$$

(9)

By using the identity (2) and Equation (9), we obtain

$$\frac{L(aq,q)f\left(z\right)}{z}-\alpha =(1-\alpha )\frac{1-\omega \left(z\right)}{1+\omega \left(z\right)}-\frac{2a(1-q)(1-\alpha )}{1-a}\frac{z{D}_q\omega \left(z\right)}{[1+\omega (z\left)\right][1+\omega (qz\left)\right]}.$$

(10)

We now claim that $\left|\omega \left(z\right)\right|<1$ for all $z\in \mathbb{U}$. Otherwise, by Lemma 1, there exists a point $z_0=qz$ such that

$$z_0{D}_q\omega \left(z_0\right)=k\omega \left(z_0\right),$$

(11)

where $k\ge 1.$ (10) in conjunction with (11) yields

$$\frac{L(aq,q)f\left(z_0\right)}{z_0}-\alpha =(1-\alpha )\frac{1-\omega \left(z_0\right)}{1+\omega \left(z_0\right)}-\frac{2a(1-q)(1-\alpha )}{1-a}\frac{k\omega \left(z_0\right)}{{[1+\omega \left(z_0\right)]}^2}.$$

since $\Re \left(\frac{1-\omega \left(z_0\right)}{1+\omega \left(z_0\right)}\right)=0,$ $k\ge 1$ and $\frac{\omega \left(z_0\right)}{{[1+\omega \left(z_0\right)]}^2}$ is real and positive, we see that $\Re \left(\frac{L(aq,q)f\left(z_0\right)}{z}\right)<\alpha .$ This contradicts the hypothesis that $f\left(z\right)\in S_{\alpha }(aq,q).$ Hence, $\left|\omega \left(z\right)\right|<1$ for all $z\in \mathbb{U}$ and by (8) $f\left(z\right)\in S_{\alpha }(a,q).$ □

If we set $a=q^{n+1}$ in Definition 3 of the function class $S_{\alpha }(a,q)$, then we obtain a new class $S_q^n\left(\alpha \right)$ given by Definition 4 below.

Definition 4.

Let $S_q^n\left(\alpha \right),$ $0\le \alpha <1,$ denote the class of functions $f\in A$ which satisfy the condition

$$\Re \left\{\frac{R_q^{n}f\left(z\right)}{z}\right\}>\alpha (z\in \mathbb{U}).$$

If we set $a=q^{n+1}$ in Theorem 1, we get the following corollary.

Corollary 2.

Let $0\le \alpha <1$, then

$$S_q^{n+1}\left(\alpha \right)\subset S_q^n\left(\alpha \right).$$

Remark 4.

Let $q\to 1^{-}$ in Corollary 2, we get the result obtained by Goel and Sohi [41].

3. $\mathbf{q}$-Bernardi Integral Operator

We recall here the following q-analogue definition given by Noor [27] of the Bernardi integral operator [42]. Let $f\in A,$ the q-Bernardi integral operator $I_{c,q}f:A\to A$ is given by

$$F\left(z\right)=I_{c,q}f\left(z\right)=\frac{{[c+1]}_q}{z^c}\underset{0}{\overset{z}{\int }}{t}^{c-1}f\left(t\right)d_{q}t,c\in {\mathbb{N}}_0.$$

(12)

Theorem 3.

If the function $f\in A$ given by (1) belongs to the class $R(a,q),$ then so does the function $F\left(z\right)$ defined by (12).

Proof. 

From the definition of $F,$ we obtain

$$q^{c}z{D}_{q}L(a,q)F\left(z\right)={[c+1]}_{q}L(a,q)f\left(z\right)-{\left[c\right]}_{q}L(a,q)F\left(z\right).$$

(13)

Using the identities (13) and (2), the condition

$$\Re \left\{\frac{L(aq,q)f\left(z\right)}{L(a,q)f\left(z\right)}\right\}>\frac{q-a}{q(1-a)}$$

may be written in terms of F as follows

$$\Re \left\{\frac{q^c(1-aq)\frac{L(a{q}^2,q)F\left(z\right)}{L(aq,q)F\left(z\right)}+(a-q^c)}{q^{c+1}(1-a)+(a-q^{c+1})\frac{L(a,q)F\left(z\right)}{L(aq,q)F\left(z\right)}}\right\}>\frac{q-a}{q(1-a)}(z\in \mathbb{U}).$$

(14)

We have to prove that (14) implies the inequality

$$\Re \left\{\frac{L(aq,q)F\left(z\right)}{L(a,q)F\left(z\right)}\right\}>\frac{q-a}{q(1-a)}(z\in \mathbb{U}).$$

Define $\omega \left(z\right)$ in $\mathbb{U}$ by

$$\frac{L(aq,q)F\left(z\right)}{L(a,q)F\left(z\right)}=\frac{q-a}{q(1-a)}+\frac{a(1-q)}{q(1-a)}\frac{1-\omega \left(z\right)}{1+\omega \left(z\right)}.$$

(15)

Evidently, $\omega \left(0\right)=0.$ Equation (15) may be written as

$$\frac{L(aq,q)F\left(z\right)}{L(a,q)F\left(z\right)}=\frac{q(1-a)+\left[\right(q-a)-a(1-q\left)\right]\omega \left(z\right)}{q(1-a)[1+\omega (z\left)\right]}.$$

(16)

Now, using the q-derivative rules and some simple calculations, (16) yields

$$\frac{z{D}_{q}L(aq,q)F\left(z\right)}{L(aq,q)F\left(z\right)}-\frac{z{D}_{q}L(a,q)F\left(z\right)}{L(a,q)F\left(z\right)}=-\frac{2a(1-q)z{D}_q\omega \left(z\right)}{\left[q\right(1-a)+[(q-a)-a(1-q)\left]\omega \right(z\left)\right][1+\omega (z\left)\right]}.$$

Using identity (2) and Equation (15), we obtain

$$\begin{array}{ccc}& & \frac{q^c(1-aq)\frac{L(a{q}^2,q)F\left(z\right)}{L(aq,q)F\left(z\right)}+(a-q^c)}{q^{c+1}(1-a)+(a-q^{c+1})\frac{L(a,q)F\left(z\right)}{L(aq,q)F\left(z\right)}}\hfill \\ & =& \frac{q(1-a)+\left[\right(q-a)-a(1-q\left)\right]\omega \left(z\right)}{q(1-a)[1+\omega (z\left)\right]}\hfill \\ & & -\frac{2a{q}^c{(1-q)}^2}{q(1-a)}\frac{z{D}_q\omega \left(z\right)}{[(1-q^{c+1})+[(1-q^c)-q^c(1-q)]\omega \left(z\right)][1+\omega \left(z\right)]}.\hfill \end{array}$$

(17)

We claim that $\left|\omega \left(z\right)\right|<1.$ If otherwise, by Lemma 1, there exists a point $z_0\in \mathbb{U}$ where

$$z_0{D}_q\omega \left(z_0\right)=k\omega \left(z_0\right)=k{e}^{i\theta }(k\ge 1).$$

(18)

Combining (17) and (18), we obtain

$$\begin{array}{ccc}& & \frac{q^c(1-aq)\frac{L(a{q}^2,q)F\left(z\right)}{L(aq,q)F\left(z\right)}+(a-q^c)}{q^{c+1}(1-a)+((a-q^{c+1})\frac{L(a,q)F\left(z\right)}{L(aq,q)F\left(z\right)}}\hfill \\ & =& \frac{q-a}{q(1-a)}-\frac{2a{q}^c{(1-q)}^2}{q(1-a)}\frac{z{D}_q\omega \left(z_0\right)}{[(1-q^{c+1})+[(1-q^c)-q^c(1-q)]\omega \left(z_0\right)][1+\omega \left(z_0\right)]}\hfill \\ & =& \frac{q-a}{q(1-a)}-\frac{4a{q}^c{(1-q)}^2(1-q^c)}{q(1-a)}\frac{k(1+cos\theta )}{{\left|[(1-q^{c+1})+[(1-q^c)-q^c(1-q)]\omega \left(z_0\right)][1+\omega \left(z_0\right)]\right|}^2}\hfill \\ & \le & \frac{q-a}{q(1-a)}\hfill \end{array}$$

which contradicts (14). Hence, $\left|\omega \left(z\right)\right|<1$ and from (15), it follows that $F\left(z\right)\in R(a,q).$ □

If we set $a=q^{\alpha +1}$ in Theorem 3, then we get the following corollary.

Corollary 3.

If $f\in {\tilde{R}}_{\alpha },$ then the function defined by (12) belongs to ${\tilde{R}}_{\alpha }.$

Remark 5.

Let $q\to 1^{-}$ in Corollary 3, then we get the result obtained by Singh and Singh [20].

Remark 6.

If we set $c=1$ in Remark 5, then we get the result obtained by Libera [43].

Theorem 4.

If $f\in S_{\alpha }(a,q),$ then the function $F\left(z\right)$ defined by (12) belongs to $S_{\alpha }(a,q).$

Proof. 

Suppose that $\omega \left(z\right)$ is a regular function in $\mathbb{U}$ with $\omega \left(0\right)=0$ and $\omega \left(z\right)\ne -1$ defined by the equation

$$\frac{L(a,q)F\left(z\right)}{z}=\frac{1+(2\alpha -1)\omega \left(z\right)}{1+\omega \left(z\right)}.$$

(19)

It is sufficient to show that $\left|\omega \left(z\right)\right|<1$ for $z\in \mathbb{U}.$ Using the q-derivative rules, we get from (19) that

$$D_{q}L(a,q)F\left(z\right)=\frac{1+(2\alpha -1)\omega \left(z\right)}{1+\omega \left(z\right)}-2q(1-\alpha )\frac{z{D}_q\omega \left(z\right)}{[1+\omega (z\left)\right][1+\omega (qz\left)\right]}.$$

Using the identity (13), we get

$$\frac{L(a,q)f\left(z\right)}{z}-\alpha =(1-\alpha )\frac{1-\omega \left(z\right)}{1+\omega \left(z\right)}-\frac{2q^{c+1}(1-\alpha )}{{[c+1]}_q}\frac{z{D}_q\omega \left(z\right)}{[1+\omega (z\left)\right][1+\omega (qz\left)\right]}.$$

By applying the same method and technique as in our proof of Theorem 2 in conjunction with Lemma 1, we see that $F\left(z\right)\in S_{\alpha }(a,q).$ This completes the proof of the theorem. □

We obtain the following corollary when we set $a=q^{n+1}$ in Theorem 4.

Corollary 4.

If $f\in S_q^n\left(\alpha \right),$ then the function defined by (12) belongs to $S_q^n\left(\alpha \right).$

Remark 7.

Let $q\to 1^{-}$ in Corollary 4, then we get the result obtained by Goel and Sohi [41].

4. Conclusions

Quantum calculus is ordinary classical calculus without the notion of limits. Researchers have recently focused their attention on the field of q-calculus. This extraordinary interest is due to its application in various branches of mathematics and physics. Jackson [1,2] was among the first few researchers who defined the q-analogue of the derivative and integral operators and provided some of their applications. In geometric function theory, a number of subclasses of normalized analytic functions in the open symmetric unit disc that are associated with the q-derivative have been studied already from different viewpoints. In this paper, we introduced two new classes of analytic functions in the open symmetric unit disc using a linear operator associated with the q-binomial theorem. In addition, we discussed inclusion relations and properties-preserving integral operators for functions in these classes. In addition to providing some new results, this paper generalizes some known ones. For future work, one can apply q-calculus to differential subordinations for some subclasses of analytic functions.