q-Generalized Linear Operator on Bounded Functions of Complex Order

Abstract

This article presents a q-generalized linear operator in Geometric Function Theory (GFT) and investigates its application to classes of analytic bounded functions of complex order $S_q(c;M)$ and $C_q(c;M)$ where $0<q<1,$$0\ne c\in \mathbb{C}$, and $M>\frac{1}{2}.$ Integral inclusion of the classes related to the q-Bernardi operator is also proven.

1. Introduction

Quantum calculus or q-calculus is attributed to the great mathematicians L.Euler and C. Jacobi, but it became popular when Albert Einstein used it in quantum mechanics in his paper [1] published in 1905. F.H. Jackson [2,3] introduced and studied the q-derivative and q-integral in a proper way. Later, quantum groups gave the geometrical aspects to q-calculus. It is pertinent to mention that q-calculus can be considered an extension of classical calculus discovered by I. Newton and G.W. Leibniz. In fact, the operators defined as:

$$d_{h}f\left(z\right)=\frac{f(z+h)-f\left(z\right)}{h}$$

and:

$$d_{q}f\left(z\right)=\frac{f\left(z\right)-f\left(qz\right)}{(1-q)z},0<q<1,$$

where $z\in \mathbb{C}$ and $h>0$ are the h-derivative and q-derivative, respectively, where h is Planck’s constant, are related as: $q=e^{ih}=e^{2\pi i\overline{h}}$ where $\overline{h}=h/2\pi$. Srivastava [4] applied the concepts of q-calculus by using the basic (or q-) hypergeometric functions in Geometric Function Theory (GFT). Ismail [5] and Agarwal [6] introduced the class of q-starlike functions by using the q-derivative. The q-close-to-convex functions were defined in [7], and Sahoo and Sharma [8] obtained several interesting results for q-close-to-convex functions. Several convolution and fractional calculus q-operators were defined by the researchers, which were reposited by Srivastava in [9]. Darus [10] defined a new differential operator called the q-generalized operator by using q-hypergeometric functions. Let A be the class of functions of the form:

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

(1)

analytic in the open unit disc $E=\{z:|z|<1\}$.

Let $f\left(z\right)$ be given by (1) and $g\left(z\right)$ defined as:

$$g\left(z\right)=z+\sum _{k=2}^{\infty }{b}_k{z}^k.$$

The Hadamard product (or convolution) of f and g is defined by:

$$(f\ast g)\left(z\right)=z+\sum _{k=2}^{\infty }{a}_k{b}_k{z}^k.$$

Let $f,h$ be analytic functions. Then, f is subordinate to h, written as $f\prec h$ or $f\left(z\right)\prec h\left(z\right)$, $z\in E$, if there exists a Schwartz function $\omega \left(z\right)$ analytic in E with $\omega \left(0\right)=0$ and $\left|\omega \left(z\right)\right|<1$ for $z\in E$, such that $f\left(z\right)=h\left(\omega \right(z\left)\right)$. If h is univalent in E, then $f\prec h$, if and only if $f\left(0\right)=h\left(0\right)$ and $f\left(E\right)\subset h\left(E\right)$.

A sequence ${\left\{b_k\right\}}_{k=1}^{\infty }$ of complex numbers is a subordinating factor if, whenever $f\left(z\right)={\sum }_{k=1}^{\infty }{a}_k{z}^k,a_1=1$ is regular, univalent, and convex in $E,$ we have ${\sum }_{n=1}^{\infty }{b}_n{a}_n{z}^n\prec f\left(z\right),$$z\in E$ [11].

We recall some basic concepts from q-calculus that are used in our discussion and refer to [2,3,12] for more details.

A subset $B\subset \mathbb{C}$ is called q-geometric if $zq\in B$ whenever $z\in B$, and it contains all the geometric sequences ${\left\{z{q}^k\right\}}_0^{\infty }.$ In GFT, the q-derivative of $f\left(z\right)$ is defined as:

$$d_{q}f\left(z\right)=\frac{f\left(z\right)-f\left(qz\right)}{(1-q)z},q\in (0,1),(z\in B\setminus \left\{0\right\}),$$

and $d_{q}f\left(0\right)=f^{\prime }\left(0\right)$. For a function $g\left(z\right)=z^k$, the q-derivative is:

$$d_{q}g\left(z\right)=\left[k\right]z^{k-1},$$

where $\left[k\right]=\frac{1-q^k}{1-q}=1+q+q^2+....+q^{k-1}.$

We note that as $q\to 1^{-}$, $d_{q}f\left(z\right)\to f^{\prime }\left(z\right)$, which is the ordinary derivative. From (1), we deduce that:

$$d_{q}f\left(z\right)=1+\sum _{k=2}^{\infty }\left[k\right]a_k{z}^k.$$

Let $f\left(z\right)$ and $g\left(z\right)$ be defined on a q-geometric set B. Then, for complex numbers $a,b$, we have:

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

$$d_q\left(f\left(z\right)g\left(z\right)\right)=f\left(qz\right)d_{q}g\left(z\right)+g\left(z\right)d_{q}f\left(z\right).$$

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

$$d_q\left(logf\left(z\right)\right)=\frac{ln{q}^{-1}}{1-q}\frac{d_{q}f\left(z\right)}{f\left(z\right)}.$$

Jackson [2] introduced the q-integral of a function f, given by:

$${\int }_0^{z}f\left(t\right)d_{q}t=z(1-q)\sum _{k=0}^{\infty }{q}^{k}f\left(q^{k}z\right),$$

provided that the series converges.

For any non-negative integer n, the q-number shift factorial is defined as:

$$\left[n\right]!=\left\{\begin{array}{cc}\left[1\right]\left[2\right]...\left[n\right]& \mathrm{if}n\ne 0,\\ 1& \mathrm{if}n=0\end{array}\right.$$

Let $\lambda \in \mathbb{R}$ and $n\in \mathbb{N}$; the q-generalized Pochhammer symbol is defined as:

$${\left[\lambda \right]}_n=\left[\lambda \right]\left[\lambda +1\right]\left[\lambda +2\right]....\left[\lambda +n-1\right].$$

The q-Gamma function is defined for $\lambda >0$ as:

$${\Gamma }_q(\lambda +1)=\left[\lambda \right]{\Gamma }_q\left(\lambda \right)\mathrm{and}{\Gamma }_q\left(1\right)=1.$$

For complex parameters $a_i$$(1\le i\le l),b_j\ne 0,-1,-2,...(1\le j\le m)$ with $l\le m+1,$ the basic q-hypergeometric function is defined as,

$${}_l{F}_m(a_1,...a_l;b_1,...,b_m,z)={\displaystyle \sum _{k=0}^{\infty }}\frac{{\left(a_1\right)}_k...{\left(a_l\right)}_k}{{\left(q\right)}_k{\left(b_1\right)}_k...{\left(b_m\right)}_k}{\left[{\left(-1\right)}^n{q}^{\left(\genfrac{}{}{0pt}{}{n}{2}\right)}\right]}^{1+m-l}{z}^k.$$

(2)

with $\left(\genfrac{}{}{0pt}{}{n}{2}\right)=\frac{n(n-1)}{2}$ and $l,m\in {\mathbb{N}}_0=\mathbb{N}\cup \left\{0\right\}$. Here, the q-shifted factorial is defined for $a\in \mathbb{C}$ as:

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

Let $l=m+1,$$a_1=q^{\lambda +1}(\lambda >-1)$, $a_i=q$$(\forall$$2\le i\le l)$, and $b_j=q$$\left(\forall 1\le j\le m\right)$, and by using the property ${\left(q^a\right)}_k={\Gamma }_q(a+k){\left(1-q\right)}^k/{\Gamma }_q\left(a\right),$ from (2), we get the function,

$$F_{q,\lambda +1}\left(z\right)=z+\sum _{k=2}^{\infty }\frac{{\Gamma }_q(\lambda +k)}{\left[k-1\right]!{\Gamma }_q(\lambda +1)}{z}^k=z+\sum _{k=2}^{\infty }\frac{{\left[\lambda +1\right]}_{k-1}}{\left[k-1\right]!}{z}^k,z\in E.$$

In [13], the q-Srivastava–Attiya convolution operator is defined as:

$$G_{q,a}^s\left(z\right)=z+\sum _{k=2}^{\infty }{\left(\frac{\left[1+a\right]}{\left[k+a\right]}\right)}^s{z}^k,z\in E,$$

$(a\in \mathbb{C}\setminus {\mathbb{Z}}_0^{-};s\in \mathbb{C}$ when $\left|z\right|<1;$$Re\left(s\right)>1$ when $\left|z\right|=1).$

Using convolution, the operator $D_{q,a,\lambda }^s$ for $\lambda >-1$ is defined as:

$$\begin{array}{cc}\hfill D_{q,a,\lambda }^{s}f\left(z\right)& =J_{q,a,\lambda }^s\left(z\right)\ast f\left(z\right)\hfill \\ \hfill & =z+\sum _{k=2}^{\infty }{\left(\frac{\left[k+a\right]}{\left[1+a\right]}\right)}^s\frac{{\left[\lambda +1\right]}_{k-1}}{\left[k-1\right]!}{a}_k{z}^k,z\in E,\hfill \end{array}$$

where:

$$J_{q,a,\lambda }^s\left(z\right)={\left(G_{q,a}^s\left(z\right)\right)}^{-1}\ast F_{q,\lambda +1}\left(z\right)=z+\sum _{k=2}^{\infty }{\left(\frac{\left[k+a\right]}{\left[1+a\right]}\right)}^s\frac{{\left[\lambda +1\right]}_{k-1}}{\left[k-1\right]!}{z}^k.$$

It is a convergent series with a radius of convergence of one. We observe that $D_{q,a,0}^{0}f\left(z\right)=f\left(z\right)$ and $D_{q,0,0}^{1}f\left(z\right)=z{d}_{q}f\left(z\right)$. The operator $D_{q,a,\lambda }^s$ reduces to known linear operators for different values of parameters $a,s$, and $\lambda$ as:

(i)

If $q\to 1^{-},$ it reduces to the operator $D_{a,\lambda }^s$ discussed by Noor et al. in [14].

(ii)

For $s=0$, it is a q-Ruscheweyh differential operator [15].

(iii)

If $s=-1,$$\lambda =0$, and $q\to 1^{-}$, it is an Owa–Srivastava integral operator [16].

(iv)

If $s\in {\mathbb{N}}_0,$$a=1,$$\lambda =0$, and $q\to 1^{-},$ it reduces to the generalized Srivastava–Attiya integral operator [17].

(v)

If $s\in {\mathbb{N}}_0,$$a=0,$$\lambda =0,$ it is a q-Salagean differential operator [18].

(vi)

For $s,\lambda \in {\mathbb{N}}_0$, and $a=0,$ it is the operator defined in [19].

The following identities hold for the operator $D_{q,a,\lambda }^{s}f\left(z\right),$

$$\begin{array}{cc}\hfill z{d}_q\left(D_{q,a,\lambda }^{s}f\left(z\right)\right)& =\left(\frac{\left[1+a\right]}{q^a}\right)D_{q,a,\lambda }^{s+1}f\left(z\right)-\frac{\left[a\right]}{q^a}{D}_{q,a,\lambda }^{s}f\left(z\right)\hfill \end{array}$$

(3)

$$\begin{array}{cc}\hfill z{d}_q(D_{q,a,\lambda }^{s}f\left(z\right))& =\left(\frac{\left[1+\lambda \right]}{q^{\lambda }}\right)D_{q,a,\lambda +1}^{s}f\left(z\right)-\frac{\left[\lambda \right]}{q^{\lambda }}{D}_{q,a,\lambda }^{s}f\left(z\right).\hfill \end{array}$$

(4)

Let $P\left(q\right)$ be the class of functions of the form $p\left(z\right)=1+c_{1}z+c_2{z}^2+....,$ analytic in E, and satisfying:

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

It is known from [20] that $p\in P\left(q\right)$ implies $p\left(z\right)\prec \frac{1+z}{1-qz}.$ It follows immediately that Re$p\left(z\right)>0,$$z\in E.$

The classes of bounded q-starlike functions $S_q(c,M)$ and bounded q-convex functions $C_q(c,M)$ of complex order c were defined in [21], respectively, as:

$$\begin{array}{cc}\hfill S_q(c,M)& =\left\{f\in A:\left|\frac{c-1+\frac{z{d}_{q}f\left(z\right)}{f\left(z\right)}}{c}-M\right|<M\right\},\hfill \\ \hfill & \left(c\in {\mathbb{C}}^{\ast };M>\frac{1}{2},z\in E\right),\hfill \end{array}$$

or equivalently,

$$\begin{array}{cc}\hfill S_q(c,M)& =\left\{f\in A:\frac{z{d}_{q}f\left(z\right)}{f\left(z\right)}\prec \frac{1+\left\{c\right(1+m)-m\}z}{1-mz}\right\},\hfill \\ \hfill & \left(c\in {\mathbb{C}}^{\ast };m=1-\frac{1}{M};M>\frac{1}{2}\right).\hfill \end{array}$$

The class of bounded q-convex functions $C_q(c,M)$ of complex order c is defined as:

$$\begin{array}{cc}\hfill C_q(c,M)& =\left\{f\in A:\left|\frac{c-1+\frac{d_q\left(z{d}_{q}f\left(z\right)\right)}{d_{q}f\left(z\right)}}{c}-M\right|<M\right\},\hfill \\ \hfill & \left(c\in {\mathbb{C}}^{\ast };M>\frac{1}{2},z\in E\right),\hfill \end{array}$$

or equivalently,

$$\begin{array}{cc}\hfill C_q(c,M)& =\left\{f\in A:\frac{d_q\left(z{d}_{q}f\left(z\right)\right)}{d_{q}f\left(z\right)}\prec \frac{1+\left\{c\right(1+m)-m\}z}{1-mz}\right\}\hfill \\ \hfill & \left(c{\in }^{\ast };m=1-\frac{1}{M};M>\frac{1}{2}\right).\hfill \end{array}$$

Using the operator $D_{q,a,\lambda }^{s}f\left(z\right),$ we now define the following new classes $S_{q,a,s,\lambda }(c,M)$ and $C_{q,a,s,\lambda }(c,M)$ as:

$$\begin{array}{cc}\hfill S_{q,a,s,\lambda }(c,M)& =\left\{f\in A:\frac{z(d_q{D}_{q,a,\lambda }^s\left(f\left(z\right)\right))}{D_{q,a,\lambda }^s\left(f\left(z\right)\right)}\prec \frac{1+\left\{c\right(1+m)-m\}z}{1-mz},z\in E\right\},\hfill \\ \hfill & \left(0<q<1,c{\in }^{\ast };m=1-\frac{1}{M};M>\frac{1}{2}\right).\hfill \end{array}$$

Special cases:

(i)

If $c=1,$ $m=1$, and $q\to 1^{-},$ then $S_{q,a,s,\lambda }(c,M)$ reduces to class $S^s(a,\lambda )$ discussed in [22].

(ii)

If $c=1$, $s=0,$ $\lambda =0,$ $m=-q,$ then $S_{q,a,s,\lambda }(c,M)$ reduces to class $S_q^{\ast }$ introduced by Noor et al. [23].

(iii)

If $s=0,$ $c=\frac{m}{1+m}$ $(-1<m<0),$ $m=-q$, then $S_{q,a,s,\lambda }(c,M)$ reduces to class $S{T}_q$ studied by Noor [24].

(iv)

If $s=0,\lambda =0$, $c=a{e}^{-i\beta }cos\beta$ $(a\in {\mathbb{C}}^{\ast },$ $\left|\beta \right|<\frac{\pi }{2})$, and $q\to 1^{-}$, then $S_{q,a,s,\lambda }(c,M)$ becomes special cases of Janowski $\beta$-spiral like functions of complex order $S^{\beta }(A,B,a)$ discussed in [25].

(v)

If $s\in {\mathbb{N}}_0,$ $\lambda =0,a=0$, and $q\to 1^{-},$ then $S_{q,a,s,\lambda }(c,M)$ reduces to class $H_n(c,M)$ discussed by Aouf et al. in [26].

(vi)

If $0<c\le 1$, $-1<m<0$, and $q\to 1^{-}$, then $S_{q,a,s,\lambda }(c,M)$ becomes a special case of the class $S_{a,\lambda }^s(\eta ,$ $A,$ $B)$ with $\eta =0$ discussed in [19].

A function $f\in A$ is in the class $S_{q,a,s,\lambda }(c,M)$ if and only if:

$$\left|\frac{\frac{z{d}_q(D_{q,a,\lambda }^{s}f\left(z\right))}{D_{q,a,\lambda }^{s}f\left(z\right)}-1}{A-B\left\{\frac{z{d}_q\left(D_{q,a,\lambda }^{s}f\left(z\right)\right)}{D_{q,a,\lambda }^{s}f\left(z\right)}\right\}}\right|<1,$$

(5)

where $A=c(1+m)-m$ and $B=-m.$

The class $C_{q,a,s,\lambda }(c,M)$ is defined as:

$$\begin{array}{cc}\hfill C_{q,a,s,\lambda }(c,M)& =\left\{f\in A:\frac{d_q(z{d}_q(D_{q,a,\lambda }^{s}f\left(z\right))}{d_q(D_{q,a,\lambda }^{s}f\left(z\right))}\prec \frac{1+\left\{c\right(1+m)-m\}z}{1-mz},z\in E\right\},\hfill \\ \hfill & \left(0<q<1,c\in {\mathbb{C}}^{\ast };m=1-\frac{1}{M};M>\frac{1}{2}\right).\hfill \end{array}$$

It is easy to see that $f\in C_{q,a,s,\lambda }(c,M)\iff z{d}_{q}f\in S_{q,a,s,\lambda }(c,M).$ In order to develop results for the classes $S_{q,a,s,\lambda }(c,M)$ and $C_{q,a,s,\lambda }(c,M)$, we need the following:

Lemma 1

([27]). Let β and γ be complex numbers with $\beta \ne 0,$ and let $h\left(z\right)$ be regular in E with $h\left(0\right)=1$ and $Re\left[\beta h\right(z)+\gamma ]>0.$ If $p\left(z\right)=1+p_{1}z+p_2{z}^2+...$ is analytic in $E,$ then $p\left(z\right)+\frac{z{d}_{q}p\left(z\right)}{\beta p\left(z\right)+\gamma }\prec h\left(z\right)\Rightarrow p\left(z\right)\prec h\left(z\right).$

Lemma 2

([11]). The sequence ${\left\{b_n\right\}}_{n=1}^{\infty }$ is a subordinating factor sequence if and only if:

$$Re\left\{1+2\sum _{k=1}^{\infty }{b}_k{z}^k\right\}>0,z\in E.$$

2. Properties of Classes ${\mathit{S}}_{\mathit{q},\mathit{a},\mathit{s},\mathit{\lambda }}(\mathit{c},\mathit{M})$ and ${\mathit{C}}_{\mathit{q},\mathit{a},\mathit{s},\mathit{\lambda }}(\mathit{c},\mathit{M})$

We start the section with the necessary and sufficient condition for a function to be in the class $S_{q,a,s,\lambda }(c,M).$

Theorem 1.

Let $f\in A.$ Then, $f\in S_{q,a,s,\lambda }(c,M)$ if and only if:

$${\sum }_{k=2}^{\infty }\left\{\left[k\right]-1+\left|c(1+m)+m(\left[k\right]-1)\right|\right\}\frac{{\left[\lambda +1\right]}_{k-1}}{\left[k-1\right]!}\left|{\left(\frac{\left[k+a\right]}{\left[1+a\right]}\right)}^s\right|\left|a_k\right|<\left|c(1+m)\right|,$$

(6)

where $m=1-\frac{1}{M},$ $(M>\frac{1}{2})$.

Proof. 

Let us assume first that Inequality (6) holds. To show $f\in S_{q,a,s,\lambda }(c,M),$ we need to prove Inequality (5).

$$\begin{array}{cc}\hfill \left|\frac{\frac{z(d_q\left(D_{q,a,\lambda }^{s}f\left(z\right)\right)}{D_{q,a,\lambda }^{s}f\left(z\right)}-1}{A-B\left\{\frac{z{d}_q\left(D_{q,a,\lambda }^{s}f\left(z\right)\right)}{D_{q,a,\lambda }^{s}f\left(z\right)}\right\}}\right|& =\left|\frac{{\sum }_{k=2}^{\infty }{\left(\frac{\left[k+a\right]}{\left[1+a\right]}\right)}^s.\frac{{\left[\lambda +1\right]}_{k-1}}{\left[k-1\right]!}(\left[k\right]-1)a_k{z}^k}{(A-B)z+{\sum }_{k=2}^{\infty }\left(A-B\left[k\right]\right){\left(\frac{\left[1+a\right]}{\left[k+a\right]}\right)}^s.\frac{{\left[\lambda +1\right]}_{k-1}}{\left[k-1\right]!}{a}_k{z}^k}\right|\hfill \\ \hfill & \le \frac{{\sum }_{k=2}^{\infty }\left|{\left(\frac{\left[k+a\right]}{\left[1+a\right]}\right)}^s\right|.\frac{{\left[\lambda +1\right]}_{k-1}}{\left[k-1\right]!}(\left[k\right]-1)\left|a_k\right|}{\left|A-B\right|-\left|{\sum }_{k=2}^{\infty }\left(A-B\left[k\right]\right){\left(\frac{\left[k+a\right]}{\left[1+a\right]}\right)}^s.\frac{{\left[\lambda +1\right]}_{k-1}}{\left[k-1\right]!}{a}_k\right|}\hfill \\ \hfill & \le \frac{{\sum }_{k=2}^{\infty }\left|{\left(\frac{\left[k+a\right]}{\left[1+a\right]}\right)}^s\right|.\frac{{\left[\lambda +1\right]}_{k-1}}{\left[k-1\right]!}(\left[k\right]-1)\left|a_k\right|}{\left|c(1+m)\right|-{\sum }_{k=2}^{\infty }\left|c(1+m)+m(\left[k\right]-1)\right|\frac{{\left[\lambda +1\right]}_{k-1}}{\left[k-1\right]!}\left|{\left(\frac{\left[k+a\right]}{\left[1+a\right]}\right)}^s\right|\left|a_k\right|}\hfill \\ \hfill & <1.\hfill \end{array}$$

Hence, $f\in S_{q,a,s,\lambda }(c,M)$ by using Inequality (6). Conversely, let $f\in S_{q,a,s,\lambda }(c,M)$ be of the form (1), then:

$$\left|\frac{\frac{z(d_q(D_{q,a,\lambda }^{s}f\left(z\right)))}{D_{q,a,\lambda }^s\left(f\left(z\right)\right)}-1}{A-B\left\{\frac{z{d}_q(D_{q,a,\lambda }^{s}f\left(z\right))}{D_{q,a,\lambda }^{s}f\left(z\right)}\right\}}\right|=\left|\frac{{\sum }_{k=2}^{\infty }{\left(\frac{\left[k+a\right]}{\left[1+a\right]}\right)}^s.\frac{{\left[\lambda +1\right]}_{k-1}}{\left[k-1\right]!}(\left[k\right]-1)a_k{z}^k}{(A-B)z+{\sum }_{k=2}^{\infty }\left(A-B\left[k\right]\right){\left(\frac{\left[k+a\right]}{\left[1+a\right]}\right)}^s.\frac{{\left[\lambda +1\right]}_{k-1}}{\left[k-1\right]!}{a}_k{z}^k}\right|.$$

Since $\left|Rez\right|\le \left|z\right|,$, we have:

$$Re\left\{\left|\frac{{\sum }_{k=2}^{\infty }{\left(\frac{\left[k+a\right]}{\left[1+a\right]}\right)}^s.\frac{{\left[\lambda +1\right]}_{k-1}}{\left[k-1\right]!}(\left[k\right]-1)a_k{z}^k}{(A-B)z+{\sum }_{k=2}^{\infty }\left(A-B\left[k\right]\right){\left(\frac{\left[k+a\right]}{\left[1+a\right]}\right)}^s.\frac{{\left[\lambda +1\right]}_{k-1}}{\left[k-1\right]!}{a}_k{z}^k}\right|\right\}<1.$$

Now, we choose values of z on the real axis such that $z{d}_q\left(D_{q,a,\lambda }^{s}f\left(z\right)\right)/D_{q,a,\lambda }^{s}f\left(z\right)$ is real. Letting $z\to 1^{-}$ through real values, after some calculations, we obtain Inequality (6). □

Remark 1.

(i) If $q\to 1^{-},$ $s\in {\mathbb{N}}_0,$ $a=0$, and $\lambda =0,$ the above result reduces to the sufficient condition for $f\left(z\right)$ to be in class $H_n(c,M)$ ($c\in {\mathbb{C}}^{\ast }$, $M>\frac{1}{2})$ discussed in [26]. (ii) If $c=1-\alpha$ $(\alpha \in [0,1\left)\right),m=0,\lambda =0$, and $q\to 1^{-},$ the above result reduces to the sufficient condition for $f\left(z\right)$ to be in class $S_{s,a}^{\ast }\left(\alpha \right)$ discussed in [28].

Theorem 2.

Let $f_i\in S_{q,a,s,\lambda }(c,M)$ having the form:

$$f_i\left(z\right)=z+\sum _{k=2}^{\infty }{a}_{k,i}{z}^k,\mathrm{for}i=1,2,3,..,l.$$

Then, $F\in S_{q,a,s,\lambda }(c,M)$, where $F\left(z\right)={\sum }_{i=1}^l{c}_i{f}_i\left(z\right)$ with ${\sum }_{i=1}^{ł}{c}_i=1.$

Proof. 

From Theorem 1, we can write:

$${\sum }_{k=2}^{\infty }\left\{\frac{\left\{\left[k\right]-1+\left|b(1+m)+m(\left[k\right]-1)\right|\right\}\frac{{\left[\lambda +1\right]}_{k-1}}{\left[k-1\right]!}\left|{\left(\frac{\left[k+a\right]}{\left[1+a\right]}\right)}^s\right|}{\left|b(1+m)\right|}\right\}{a}_{k,i}<1.$$

(7)

Therefore:

$$\begin{array}{cc}\hfill F\left(z\right)& =\sum _{i=1}^l{c}_i\left(z+\sum _{k=2}^{\infty }{a}_{k,i}{z}^k\right)\hfill \\ \hfill & =z+\sum _{k=2}^{\infty }\left(\sum _{i=1}^l{c}_i{a}_{k,i}\right)z^k;\hfill \end{array}$$

where however due to (7), we have:

$$\begin{array}{cc}\hfill & {\sum }_{k=2}^{\infty }\frac{\left\{\left[k\right]-1+\left|b(1+m)+m(\left[k\right]-1)\right|\right\}\frac{{\left[\lambda +1\right]}_{k-1}}{\left[k-1\right]!}\left|{\left(\frac{\left[k+a\right]}{\left[1+a\right]}\right)}^s\right|}{\left|b(1+m)\right|}\left(\sum _{i=2}^l{c}_i{a}_{k,i}\right)\hfill \\ \hfill & =\sum _{i=2}^l\left[\frac{\left\{\left[k\right]-1+\left|b(1+m)+m(\left[k\right]-1)\right|\right\}\frac{{\left[\lambda +1\right]}_{k-1}}{\left[k-1\right]!}\left|{\left(\frac{\left[k+a\right]}{\left[1+a\right]}\right)}^s\right|}{\left|b(1+m)\right|}\right]c_i\le 1;\hfill \end{array}$$

Therefore, $F\in S_{q,a,s,\lambda }(c,M).$  □

Theorem 3.

Let $f_i$ with $i=1,2,...,\nu$ belong to the class $S_{q,a,s,\lambda }(c,M).$ The arithmetic mean h of $f_i$ is given by:

$$h\left(z\right)=\frac{1}{v}\sum _{i=1}^v{f}_i\left(z\right)$$

(8)

belonging to class $S_{q,a,s,\lambda }(c,M).$

Proof. 

From (8), we can write:

$$h\left(z\right)=\frac{1}{v}\sum _{i=1}^v\left(z+\sum _{k=2}^{\infty }{a}_{k,i}{z}^k\right)=z+\sum _{k=2}^{\infty }\left(\frac{1}{v}\sum _{i=1}^v{a}_{k,i}\right)z^k.$$

(9)

Since $f_i\in S_{q,a,s,\lambda }(c,M)$ for every $i=1,2,...,v,$ using (6) and (9), we have:

$$\begin{array}{cc}\hfill & \sum _{k=2}^{\infty }\left\{\left[k\right]-1+\left|b(1+m)+m(\left[k\right]-1)\right|\right\}\frac{{\left[\lambda +1\right]}_{k-1}}{\left[k-1\right]!}\left|{\left(\frac{\left[k+a\right]}{\left[1+a\right]}\right)}^s\right|\left(\frac{1}{v}\sum _{i=1}^v{a}_{k,i}\right)\hfill \\ \hfill & =\frac{1}{v}\sum _{i=1}^v\left(\sum _{k=2}^{\infty }\left\{\left[k\right]-1+\left|b(1+m)+m(\left[k\right]-1)\right|\right\}\frac{{\left[\lambda +1\right]}_{k-1}}{\left[k-1\right]!}\left|{\left(\frac{\left[k+a\right]}{\left[1+a\right]}\right)}^s\right|a_{k,i}\right)\hfill \\ \hfill & \le \frac{1}{v}\sum _{i=1}^v\left(\left|b(1+m)\right|\right)=\left|b(1+m)\right|,\hfill \end{array}$$

and this completes the proof.  □

Now, we give the subordination relation for the functions in class $S_{q,a,s,\lambda }(c,M)$ by using the subordination theorem.

Theorem 4.

Let $m=1-\frac{1}{M}$$(M>\frac{1}{2}).$ Furthermore, $c\ne 0$ with $Re\left(c\right)>\frac{-m}{2(1+m)}$ when $m>0$ and $Re\left(c\right)<\frac{-m}{2(1+m)}$ when $m<0$ and $\lambda \ge 0.$ If $f\in S_{q,a,s,\lambda }(c,M),$ then:

$$\begin{array}{cc}\hfill \frac{\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)}{2[\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)+\left|c(1+m)\right|]}(f\ast g)\left(z\right)& \prec g\left(z\right)\hfill \end{array}$$

(10)

where $g\left(z\right)$ is a convex function in E, $C_{\lambda ,k}=\frac{{\left[\lambda +1\right]}_{k-1}}{\left[k-1\right]!}$, $B_{s,a}\left(k\right)=\left|{\left(\frac{\left[k+a\right]}{\left[1+a\right]}\right)}^s\right|$, and:

$$Ref\left(z\right)>-1-\frac{(1+m)\left|c\right|}{\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)}.$$

(11)

The constant $\frac{\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)}{2[\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)+\left|c(1+m)\right|]}$ is the best estimate.

Proof. 

Let $f\left(z\right)\in S_{q,a,s,\lambda }(c,M)$ and $g\left(z\right)=z+{\sum }_{k=2}^{\infty }{c}_k{z}^k.$ Then:

$$\begin{array}{cc}\hfill & \frac{\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)}{2[\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)+\left|c(1+m)\right|]}(f\ast g)\left(z\right)\hfill \\ \hfill & =\frac{\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)}{2[\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)+\left|c(1+m)\right|]}\left(z+\sum _{k=2}^{\infty }{a}_k{c}_k{z}^k\right).\hfill \end{array}$$

(12)

Thus, (10) holds true if:

$${\left\{\frac{\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)}{2[\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)+\left|c(1+m)\right|]}{a}_k\right\}}_{k=1}^{\infty }$$

(13)

is a subordinating factor sequence with $a_1=1.$ From Lemma 2, it suffices to show:

$$Re\left\{1+{\sum }_{k=1}^{\infty }\frac{\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)}{[\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)+\left|c(1+m)\right|]}{a}_k{z}^k\right\}>0.$$

(14)

Now, as $\left\{\left[k\right]-1+\left|c(1+m)+m(\left[k\right]-1)\right|\right\}{C}_{\lambda ,k}{B}_{s,a}\left(k\right)$ is an increasing function of k$(k\ge 2),$ we have:

$$\begin{array}{cc}\hfill & Re\left\{1+{\sum }_{k=1}^{\infty }\frac{\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)}{[\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)+\left|c(1+m)\right|]}{a}_k{z}^k\right\}\hfill \\ \hfill & =Re\left\{1+\frac{\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)}{[\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)+\left|c(1+m)\right|]}z+\right.\hfill \\ \hfill & +\left.\frac{{\sum }_{k=2}^{\infty }\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)a_k{z}^k}{[\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)+\left|c(1+m)\right|]}\right\}\hfill \\ \hfill & \ge 1-\frac{\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)}{[\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)+\left|c(1+m)\right|]}r-\hfill \\ \hfill & \frac{{\sum }_{k=2}^{\infty }\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)\left|a_k\right|r^k}{[\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)+\left|c(1+m)\right|]}\hfill \\ \hfill & >1-\frac{\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)}{[\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)+\left|c(1+m)\right|]}r-\hfill \\ \hfill & \frac{(1+m)\left|c\right|}{[\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)+\left|c(1+m)\right|]}r\hfill \\ \hfill & >0.\left(\left|z\right|=r<1\right)\hfill \end{array}$$

Hence, (14) holds true in E, and the subordination result (10) is affirmed by Theorem 4. The inequality (11) follows by taking $g\left(z\right)=\frac{z}{1-z}={\sum }_{k=1}^{\infty }{z}^k$ in (10).

Let us consider the function:

$$\varphi \left(z\right)=z-\frac{\left|c(1+m)\right|}{[\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)+\left|c(1+m)\right|]}{z}^2(z\in E)$$

which is a member of $S_{q,a,s,\lambda }(c,M).$ Then. by using (10), we have:

$$\frac{\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)}{2[\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)+\left|c(1+m)\right|]}\varphi \left(z\right)\prec \frac{z}{1-z}.$$

It is easily verified that:

$$minRe\left\{\frac{\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)}{2[\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)+\left|c(1+m)\right|]}\varphi \left(z\right)\right\}=-\frac{1}{2}\left(z\in E\right),$$

then the constant $\frac{\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)}{2[\{q+\left|c(1+m)+mq\right|\}{C}_{\lambda ,2}{B}_{s,a}\left(2\right)+\left|c(1+m)\right|]}$ cannot be replaced by a larger one. □

Remark 2.

If $s\in {\mathbb{N}}_0,$$a=0,$$\lambda =0$, and $q\to 1^{-},$ Theorem 4 reduces to the subordination result proven in [29].

Now, we discuss the inclusion results pertaining to classes $S_{q,a,s,\lambda }(c,M)$ and $C_{q,a,s,\lambda }(c,M)$ in reference to parameters s and $\lambda .$

Theorem 5.

For any complex number s, $S_{q,a,s+1,\lambda }(c,M)\subset S_{q,a,s,\lambda }(c,M)$ if $Re(\frac{1+\left\{c\right(1+m)-m\}z}{1-mz})>\frac{1}{q^{a_1}(1-q)}\left\{1-cos(a_{2}lnq)\right\}$ where $a=a_1+i{a}_2$.

Proof. 

Let $f\in S_{q,a,s+1,\lambda }(c,M),$ then:

$$\frac{z{d}_q(D_{q,a,\lambda }^{s+1}f\left(z\right))}{D_{q,a,\lambda }^{s+1}f\left(z\right)}\prec \frac{1+\left\{c\right(1+m)-m\}z}{1-mz},$$

(15)

Let:

$$h\left(z\right)=\frac{1+\left\{c\right(1+m)-m\}z}{1-mz}$$

and:

$$r\left(z\right)=\frac{z{d}_q(D_{q,a,\lambda }^{s}f\left(z\right))}{D_{q,a,\lambda }^{s}f\left(z\right)}.$$

We will show:

$$r\left(z\right)\prec h\left(z\right),$$

which would prove $S_{q,a,s,\lambda }(c,M)\subset S_{q,a,s+1,\lambda }(c,M).$ From the identity relation (3), after a few calculations, we have:

$$\frac{z{d}_q(D_{q,a,\lambda }^{s}f\left(z\right))}{D_{q,a,\lambda }^{s}f\left(z\right)}=\frac{\left[1+a\right]}{q^a}.\frac{D_{q,a,\lambda }^{s+1}f\left(z\right)}{D_{q,a,\lambda }^{s}f\left(z\right)}-\frac{\left[a\right]}{q^a}.$$

After some calculations, we have:

$$\begin{array}{cc}\hfill \frac{D_{q,a,\lambda }^{s+1}f\left(z\right)}{D_{q,a,\lambda }^{s}f\left(z\right)}& =\frac{1}{\left[1+a\right]}\left\{\frac{q^{a}z{d}_q(D_{q,a,\lambda }^{s}f\left(z\right))}{D_{q,a,\lambda }^{s}f\left(z\right)}+\left[a\right]\right\}\hfill \\ \hfill & =\frac{1}{\left[1+a\right]}\left\{q^{a}r\left(z\right)+\left[a\right]\right\}.\hfill \end{array}$$

Applying logarithmic q-differentiation, we have:

$$\frac{z{d}_q(D_{q,a,\lambda }^{s+1}f\left(z\right))}{D_{q,a,\lambda }^{s+1}f\left(z\right)}=r\left(z\right)+\frac{z{d}_{q}r\left(z\right)}{r\left(z\right)+q^{-a}\left[a\right]}.$$

(16)

From (15) and (16), we have:

$$r\left(z\right)+\frac{z\left[d_{q}r\left(z\right)\right]}{r\left(z\right)+q^{-a}\left[a\right]}\prec \frac{1+\left\{c\right(1+m)-m\}z}{1-mz}.$$

If $Re\left(h\left(z\right)\right)>\frac{1}{q^{a_1}(1-q)}\left\{1-cos(a_{2}lnq)\right\}$, then from Lemma 1, it implies:

$$r\left(z\right)\prec h\left(z\right),$$

which implies $f\left(z\right)\in S_{q,a,s,\lambda }(c,M).$ Therefore, $S_{q,a,s,\lambda }(c,M)\subset S_{q,a,s+1,\lambda }(c,M)$.  □

Theorem 6.

For any complex number s, $C_{q,a,s+1,\lambda }(c,M)\subset C_{q,a,s,\lambda }(c,M)$ if $Re(\frac{1+\left\{c\right(1+m)-m\}z}{1-mz})>\frac{1}{q^{a_1}(1-q)}\left\{1-cos(a_{2}lnq)\right\}$ where $a=a_1+i{a}_2.$

Proof. 

It is obvious from the fact $f\in C_{q,a,s,\lambda }(c,M)\iff z{d}_{q}f\in S_{q,a,s,\lambda }(c,M).$  □

Theorem 7.

For any complex number s, $S_{q,a,s,\lambda +1}(c,M)\subset S_{q,a,s,\lambda }(c,M)$ if $Re(\frac{1+\left\{c\right(1+m)-m\}z}{1-mz})>\frac{1-q^{-\lambda }}{1-q},$ $\lambda >-1.$

Proof. 

Let $f\in S_{q,a,s,\lambda +1}(c,M)$, then:

$$\frac{z{d}_q(D_{q,a,\lambda +1}^{s}f\left(z\right))}{D_{q,a,\lambda +1}^{s}f\left(z\right)}\prec \frac{1+\left\{c\right(1+m)-m\}z}{1-mz}.$$

(17)

Consider:

$$h\left(z\right)=\frac{1+\left\{c\right(1+m)-m\}z}{1-mz}$$

and:

$$q\left(z\right)=\frac{z{d}_q(D_{q,a,\lambda }^{s}f\left(z\right))}{D_{q,a,\lambda }^{s}f\left(z\right)}.$$

We will show:

$$q\left(z\right)\prec h\left(z\right),$$

which would conveniently prove $S_{q,a,s,\lambda +1}(c,M)\subset S_{q,a,s,\lambda }(c,M)$. From the identity relation (4), after a few calculations, we have:

$$\frac{z{d}_q(D_{q,a,\lambda }^{s}f\left(z\right))}{D_{q,a,\lambda }^{s}f\left(z\right)}=\frac{\left[1+\lambda \right]}{q^{\lambda }}\frac{D_{q,a,\lambda +1}^{s}f\left(z\right)}{D_{q,a,\lambda }^{s}f\left(z\right)}-\frac{\left[\lambda \right]}{q^{\lambda }}.$$

After some calculations, we have:

$$\begin{array}{cc}\hfill \frac{D_{q,a,\lambda +1}^{s}f\left(z\right)}{D_{q,a,\lambda }^{s}f\left(z\right)}& =\frac{1}{\left[1+\lambda \right]}\left\{\frac{q^a.z{d}_q(D_{q,a,\lambda }^{s}f\left(z\right))}{D_{q,a,\lambda }^{s}f\left(z\right)}+\left[\lambda \right]\right\}\hfill \\ \hfill & =\frac{1}{\left[1+\lambda \right]}\left\{q^{\lambda }q\left(z\right)+\left[\lambda \right]\right\}.\hfill \end{array}$$

Applying logarithmic q-differentiation, we have:

$$\frac{z{d}_q(D_{q,a,\lambda +1}^{s}f\left(z\right))}{D_{q,a,\lambda +1}^{s}f\left(z\right)}=q\left(z\right)+\frac{z{d}_{q}q\left(z\right)}{q\left(z\right)+q^{-\lambda }\left[\lambda \right]}$$

(18)

From (17) and (18), we have:

$$q\left(z\right)+\frac{z\left[d_{q}q\left(z\right)\right]}{q\left(z\right)+q^{-\lambda }\left[\lambda \right]}\prec \frac{1+\left\{c\right(1+m)-m\}z}{1-mz}.$$

If $Re\left(h\left(z\right)\right)>\frac{1-q^{-\lambda }}{1-q}$ for any value of $\lambda >-1,$ so by Lemma 1, we have $q\left(z\right)\prec h\left(z\right)$, which implies $f\left(z\right)\in S_{q,a,s,\lambda }(c,M).$ Therefore, $S_{q,a,s,\lambda +1}(c,M)\subset S_{q,a,s,\lambda }(c,M).$  □

Remark 3.

If we consider $q\to 1^{-}$ with $Rea\ge 0,$$c=1,m=1$ in Theorem 5 and $\lambda \ge 0,c=1,m=1$ in Theorem 7, we obtain the special cases of the inclusion results, Theorems 2.4 and 2.5 in [19].

In [30], the q-Bernardi integral operator $L_{b}f\left(z\right)$ is defined as:

$$\begin{array}{cc}\hfill L_{b}f\left(z\right)& =\frac{\left[1+b\right]}{z^b}{\int }_0^z{t}^{b-1}f\left(t\right)d_{q}t\hfill \\ \hfill & =z+\sum _{k=2}^{\infty }\left(\frac{\left[1+b\right]}{\left[k+b\right]}\right)a_k{z}^k,b=1,2,3,....\hfill \end{array}$$

Now, we apply the generalized operator $D_{q,a,\lambda }^s$ on $L_{b}f\left(z\right)$ as:

$$D_{q,a,\lambda }^s\left(L_{b}f\left(z\right)\right)=z+\sum _{k=2}^{\infty }{\left(\frac{\left[k+a\right]}{\left[1+a\right]}\right)}^s.\frac{{\left[\lambda +1\right]}_{k-1}}{\left[k-1\right]!}\left(\frac{\left[1+b\right]}{\left[k+b\right]}\right)a_k{z}^k.$$

The identity relation of $D_{q,a,\lambda }^s\left(L_{b}f\left(z\right)\right)$ is given as:

$$z{d}_q\left[D_{q,a,\lambda }^s\left\{L_{b}f\left(z\right)\right\}\right]=\left(\frac{\left[1+b\right]}{q^b}\right)D_{q,a,\lambda }^{s}f\left(z\right)-\frac{\left[b\right]}{q^b}{D}_{q,a,\lambda }^s\left\{L_{b}f\left(z\right)\right\}.$$

(19)

The following theorems are the integral inclusions of the classes $S_{q,a,s,\lambda }(c,M)$ and $C_{q,a,s,\lambda }(c,M)$ with respect to the q-Bernardi integral operator.

Theorem 8.

If $f\left(z\right)\in S_{q,a,s,\lambda }(c,M)$ then $L_{b}f\left(z\right)\in S_{q,a,s,\lambda }(c,M)$ if $Re\left(\frac{1+\left\{c\right(1+m)-m\}z}{1-mz}\right)>\frac{1-q^{-b}}{1-q}$ for any complex number $s.$

Proof. 

Let $g\left(z\right)\in S_{q,a,s,\lambda }(c,M)$, then:

$$\frac{z{d}_q(D_{q,a,\lambda }^{s}g\left(z\right))}{D_{q,a,\lambda }^{s}g\left(z\right)}\prec \frac{1+\left\{c\right(1+m)-m\}z}{1-mz}.$$

(20)

Consider:

$$h\left(z\right)=\frac{1+\left\{c\right(1+m)-m\}z}{1-mz}$$

and:

$$u\left(z\right)=\frac{z{d}_q(D_{q,a,\lambda }^s{L}_{b}g\left(z\right))}{D_{q,a,\lambda }^s{L}_{b}g\left(z\right)}.$$

We will show:

$$u\left(z\right)\prec h\left(z\right),$$

which would prove $L_{b}g\left(z\right)\in S_{q,a,s,\lambda }(c,M).$ From the identity relation (19), after some calculations, we have:

$$\frac{z{d}_q(D_{q,a,\lambda }^s{L}_{b}g\left(z\right))}{D_{q,a,\lambda }^s{L}_{b}g\left(z\right)}=\left(\frac{\left[1+b\right]}{q^b}\right)\frac{D_{q,a,\lambda }^{s}g\left(z\right)}{(D_{q,a,\lambda }^s{L}_{b}g\left(z\right))}-\frac{\left[b\right]}{q^b}.$$

After some calculations, we have:

$$\frac{D_{q,a,\lambda }^{s}g\left(z\right)}{D_{q,a,\lambda }^s{L}_{b}g\left(z\right)}=\frac{1}{\left[1+b\right]}\left[\frac{q^b.z{d}_q(D_{q,a,\lambda }^s{L}_{b}g\left(z\right))}{D_{q,a,\lambda }^s{L}_{b}g\left(z\right)}+\left[b\right]\right]$$

Applying logarithmic q-differentiation, we have:

$$\frac{z{d}_q(D_{q,a,\lambda }^{s}g\left(z\right))}{D_{q,a,\lambda }^{s}g\left(z\right)}=u\left(z\right)+\frac{z\left[d_{q}u\left(z\right)\right]}{u\left(z\right)+q^{-b}\left[b\right]}$$

(21)

From (20) and (21), we have:

$$u\left(z\right)+\frac{z\left[d_{q}u\left(z\right)\right]}{u\left(z\right)+q^{-b}\left[b\right]}\prec \frac{1+\left\{c\right(1+m)-m\}z}{1-mz}$$

If $Re\left(h\left(z\right)\right)>\frac{1-q^{-b}}{1-q}$, so by Lemma 1, we have $u\left(z\right)\prec h\left(z\right)$, which implies $L_{b}g\left(z\right)\in S_{q,a,s,\lambda }(c,M).$  □

Theorem 9.

If $f\left(z\right)\in C_{q,a,s,\lambda }(c,M)$, then $L_{b}f\left(z\right)\in C_{q,a,s,\lambda }(c,M)$ for any complex number $s.$

Proof. 

It is an immediate consequence of the fact $C_{q,a,s,\lambda }(c,M)\iff z{d}_{q}f\in S_{q,a,s,\lambda }(c,M).$  □