1 Introduction

Quantum calculus is ordinary calculus without limit. It is also referred as h-calculus, where h stands for Plank’s constant. Recently, quantum calculus attracted attention of many researcher due to its vast applications in many branches of mathematics and physics. Jackson (1909, 1910) introduced and studied q-derivative and q-integral in a systematic way. Ismail et al. (1990) generalized the class of starlike functions using quantum calculus. Mohammed and Darus (2013) studied geometric properties of q-operators in some classes of analytic functions. Sahoo and Sharma (2015) introduced studied q- close-to-convex functions. A comprehensive study of geometric properties of q-hypergeometric series can be found in Agarwal and Sahoo (2014). Recent work on q-calculus can be found in Gairola et al. (2017, 2016); Mishra et al. (2013, 2012).

We recall some basic concepts from quantum calculus.

Let A be the class of analytic functions defined on the open unit disc \(E=\left\{ z\in \mathbb {C} :\left| z\right| <1\right\}\) and is of the form:

$$\begin{aligned} f\left( z\right) =z+\sum \limits _{n=2}^{\infty }a_{n}z^{n}. \end{aligned}$$
(1.1)

The q-derivative of a function \(f\in A\) is defined by [see Jackson (1909)]

$$\begin{aligned} D_{q}f\left( z\right) =\frac{f\left( qz\right) -f\left( z\right) }{\left( q-1\right) z},\quad\left( z\ne 0\right) \end{aligned}$$
(1.2)

and \(D_{q}f\left( 0\right) =f^{\prime }\left( 0\right) ,\) where \(q\in \left( 0,1\right) .\)

For a function \(g\left( z\right) =z^{n},\) the q-derivative is

$$\begin{aligned} D_{q}g\left( z\right)&= \frac{1-q^{n}}{1-q}z^{n-1} \nonumber \\&= \left[ n\right] _{q}z^{n-1}, \end{aligned}$$
(1.3)

where

$$\begin{aligned} \left[ n\right] _{q}=\frac{1-q^{n}}{1-q}. \end{aligned}$$

We note that as \(q\rightarrow 1^{-},\) \(D_{q}f\left( z\right) \rightarrow f^{\prime }\left( z\right) ,\) where \(f^{\prime }\left( z\right)\) is ordinary derivative and \(\left[ n\right] _{q}\rightarrow n\) as \(q\rightarrow 1^{-}.\)

For \(f\in A\) defined in (1.1) and using (1.3), we conclude that

$$\begin{aligned} D_{q}f\left( z\right) =1+\sum \limits _{n=2}^{\infty }\left[ n\right] _{q}a_{n}z^{n-1}. \end{aligned}$$

As an inverse of q-derivative, Jackson (1910), introduced the q-integral of a function f given by

$$\begin{aligned} \int \limits _{0}^{z}f\left( t\right) d_{q}t=z\left( 1-q\right) \sum \limits _{n=0}^{\infty }q^{n}f\left( q^{n}z\right) , \end{aligned}$$
(1.4)

provided the series converges.

Let A be the class of analytic functions. Let C and \(S^{*}\) be the subclasses of univalent functions in E which respectively consists of convex and starlike functions. The q-analogous of these classes was introduced in Seoudy and Aouf (2016) which is defined as follows:

$$\begin{aligned} C_{q}=\left\{ f\in A:{\mathfrak{R}}\left( \frac{D_{q}\left( zD_{q}f\left( z\right) \right) }{D_{q}f\left( z\right) }\right) >0,\, z\in E\right\} , \end{aligned}$$
$$\begin{aligned} S_{q}^{*}=\left\{ f\in A:{\mathfrak{R}}\left( \frac{zD_{q}f\left( z\right) }{ f\left( z\right) }\right) >0,\, z\in E\right\} . \end{aligned}$$

We note that when \(q\rightarrow 1^{-}\), the above classes reduce to the class of convex and starlike functions.

For a function \(f\left( z\right)\) defined in (1.1) and \(g\left( z\right)\) be given by

$$\begin{aligned} g\left( z\right) =z+\sum \limits _{n=2}^{\infty }b_{n}z^{n}, \end{aligned}$$

the convolution (Hadamard product) is defined by

$$\begin{aligned} (f*g)\left( z\right) =z+\sum \limits _{n=2}^{\infty }a_{n}b_{n}z^{n}=\left( g*f\right) \left( z\right) ,\text { }z\in E. \end{aligned}$$

Let f and g be analytic in E,  then f is subordinate to g,  written as \(f\prec g\) or \(f\left( z\right) \prec g\left( z\right) ,\) \(z\in E,\) if there exist a Schwarz function \(\omega\) analytic in E with \(\omega \left( 0\right) =0\) and \(\left| \omega \left( z\right) \right| <1\) for \(z\in E\), such that

$$\begin{aligned} f\left( z\right) =g\left( \omega \left( z\right) \right) . \end{aligned}$$

If g is univalent in E,  then \(f\prec g\) if and only \(f\left( 0\right) =g\left( 0\right)\) and \(f\left( E\right) \subset g\left( E\right) .\)

For \(k\in [0,\infty ),\) the conic domain \(\Omega _{k}\) is defined in Kanas and Wisniaskawa (1999) as follows:

$$\begin{aligned} \Omega _{k}=\left\{ u+iv:u>k\sqrt{\left( u-1\right) ^{2}+v^{2}}\right\} . \end{aligned}$$
(1.5)

For fixed k,  \(\Omega _{k}\) represents the conic region bounded successively by the imaginary axis \(\left( k=0\right) ,\) the right branch of a hyperbola \(\left( 0<k<1\right) ,\) a parabola \(\left( k=1\right)\) and an ellipse \(\left( k>1\right) .\) In addition, we note that, for no choice of k \(\left( k>1\right) ,\) \(\Omega _{k}\) reduces to a disc, see Kanas (2003).

We shall choose \(k\in \left[ 0,1\right] ,\) for these values of k,  the following functions \(p_{k}\left( z\right)\) are univalent in E,  continuous as regards to k,  have real coefficients and map E onto \(\Omega _{k}\), such that \(p_{k}\left( 0\right) =1,\) \(p_{k}^{\prime }\left( 0\right) >0\):

$$p_{k} \left( z \right) = \left\{ {\begin{array}{*{20}l} {\frac{{1 + z}}{{1 - z}},} \hfill & {\left( {k = 0} \right)} \hfill \\ {1 + \frac{2}{{\pi ^{2} }}\left( {\log \frac{{1 + \sqrt z }}{{1 - \sqrt z }}} \right)^{2} ,} \hfill & {\left( {k = 1} \right)} \hfill \\ {1 + \frac{2}{{1 - k^{2} }}\sinh ^{2} \left\{ {\frac{2}{\pi }\arccos \left( k \right)\arctan \left( {\sqrt z } \right)} \right\},} \hfill & {\left( {0 < ;k < ;1} \right).} \hfill \\ \end{array} } \right.$$
(1.6)

Utilizing the q-derivative and the conic domain given in (1.5), we now define the following subclasses of analytic functions.

Definition 1.1

Let \(f\in A,\) \(k\in \left[ 0,1\right]\) and \(0<q<1\). Then, \(f\in k\text {-}qST\), if

$$\begin{aligned} \frac{zD_{q}f\left( z\right) }{f\left( z\right) }\prec p_{k}\left( z\right) , \text { } \end{aligned}$$

where \(p_{k}\left( z\right)\) is given by (1.6).

Using the Alexander-type relation, the class \(k\text {-}qCV\) is defined as follows:

$$\begin{aligned} f\in k-\text{qCV}\text { }\iff zD_{q}f\in k-\text {qST}. \end{aligned}$$

As a special case, when \(q\rightarrow 1^{-},\) the above classes reduces to well-known classes of \(k\text {-}ST\) (\(k\text {-}\)uniformly starlike functions)and \(k\text {-}UCV\) (\(k-\)uniformly convex functions) respectively, see Kanas and Wisniaskawa (1999).

The dual set of a set V is defined as follows:

Definition 1.2

Ruscheweyh (1975). Let \(V\subset A,\) dual set \(V^{*}\) of V is defined as

$$\begin{aligned} V^{*}=\left\{ g\in A:\frac{\left( f*g\right) \left( z\right) }{z} \ne 0,\text{for} \, \text{all }\,\it {f}\in \it {V}\right\} . \end{aligned}$$
(1.7)

Our main focus in this paper is to discuss the geometric properties of subclasses of analytic functions using dual sets.

2 Main Results

Theorem 2.1

Let \(k\in \left[ 0,1\right]\) and \(0<q<1.\) Then, \(V^{*}=k\text {-}qST,\) where

$$\begin{aligned} V=\left\{ g\in A:g\left( z\right) =\frac{z\left( 1-L+Lqz\right) }{\left( 1-L\right) \left( 1-z\right) \left( 1-qz\right) }\right\} , \end{aligned}$$

and

$$\begin{aligned} L=L\left( \alpha \right) =k\alpha \pm \sqrt{\alpha ^{2}+\left( \alpha k-1\right) ^{2}},\text { }\alpha ^{2}+\left( \alpha k-1\right) ^{2}\ge 0. \end{aligned}$$
(2.1)

Proof

Let \(f\in A\) and is of the form (1.1) and let \(f\in k\text {-}qST.\)

Then

$$\begin{aligned} \frac{zD_{q}f\left( z\right) }{f\left( z\right) }\in \Omega _{k}, \end{aligned}$$
(2.2)

where \(\Omega _{k}\) is given in (1.5) and \(k\in \left[ 0,1\right]\).

Equivalently, (2.2) can be written as

$$\begin{aligned} \frac{zD_{q}f\left( z\right) }{f\left( z\right) }\notin \partial \Omega _{k}. \end{aligned}$$

Using parametric form of \(\partial \Omega _{k},\) we obtain

$$\begin{aligned} \frac{zD_{q}f\left( z\right) }{f\left( z\right) }\ne L\left( \alpha \right) , \end{aligned}$$
(2.3)

where \(L\left( \alpha \right)\) is given by (2.1).

It is known that when \(0<q<1,\)

$$\begin{aligned} zD_{q}f\left( z\right) =f\left( z\right) *\frac{z}{\left( 1-z\right) \left( 1-qz\right) }, \end{aligned}$$
(2.4)

and

$$\begin{aligned} f\left( z\right) *\frac{z}{1-z}=f\left( z\right) . \end{aligned}$$
(2.5)

Using (2.4) and (2.5) in (2.3), we obtain

$$\begin{aligned} \frac{1}{z}\left[ f*\left\{ \frac{z\left( 1-L+Lqz\right) }{\left( 1-L\right) \left( 1-z\right) \left( 1-qz\right) }\right\} \right] \ne 0. \end{aligned}$$

Using dual set defined by (1.7), we obtain the required result. \(\quad \square\)

Theorem 2.2

Let \(k\in \left[ 0,1\right]\) and \(0<q<1.\) Then, \(W^{*}=k\text {-}qCV\), where

$$\begin{aligned} W=\left\{ g\in A:g\left( z\right) =\frac{z}{\left( 1-L\right) \left( 1-z\right) \left( 1-qz\right) }\left( \frac{1+qz}{1-q^{2}z}-L\right) \right\} , \end{aligned}$$

where L is given by (2.1).

Proof

Let \(f\in A\) and is of the form (1.1) and let \(f\in k-q \text {CV}.\)

Then, by definition

$$\begin{aligned} \frac{D_{q}\left( zD_{q}f\left( z\right) \right) }{D_{q}f\left( z\right) } \in \Omega _{k}, \end{aligned}$$
(2.6)

where \(\Omega _{k}\) is given in (1.5) and \(k\in \left[ 0,1\right] .\) we can write (2.6) as

$$\begin{aligned} \frac{D_{q}\left( zD_{q}f\left( z\right) \right) }{D_{q}f\left( z\right) } \notin \partial \Omega _{k}. \end{aligned}$$

Using parametric form of \(\partial \Omega _{k},\) we get

$$\begin{aligned} \frac{D_{q}\left( zD_{q}f\left( z\right) \right) }{D_{q}f\left( z\right) } \ne L\left( \alpha \right) , \end{aligned}$$
(2.7)

where \(L\left( \alpha \right)\) is given by (2.1). We know that for \(0<q<1,\)

$$\begin{aligned} zD_{q}f\left( z\right) =f\left( z\right) *\frac{z}{\left( 1-z\right) \left( 1-qz\right) }, \end{aligned}$$
(2.8)

and

$$\begin{aligned} zD_{q}\left( f\left( z\right) *\frac{z}{\left( 1-z\right) \left( 1-qz\right) }\right) =f\left( z\right) *\frac{z\left( 1+qz\right) }{ \left( 1-z\right) \left( 1-qz\right) \left( 1-q^{2}z\right) }. \end{aligned}$$
(2.9)

Using (2.8) and (2.9) in (2.7), we obtain

$$\begin{aligned} \frac{1}{z}\left[ f*\left\{ \frac{z}{\left( 1-L\right) \left( 1-z\right) \left( 1-qz\right) }\left( \frac{1+qz}{1-q^{2}z}-L\right) \right\} \right] \ne 0. \end{aligned}$$

Using the Definition 1.3, we obtain the required result. \(\quad \square\)

When \(q\rightarrow 1^{-},\) The above theorems reduce to following:

Corollary 2.3

[ Kanas and Wisniaskawa (1999)]. Let \(k\in \left[ 0,1\right]\) and L is given by (2.1). Then, \(V^{*}=k\text {-}ST\) and \(W^{*}=k\text {-}UCV,\) where

$$\begin{aligned} V=\left\{ g\in A:g\left( z\right) =\frac{z\left( 1-L+Lz\right) }{\left( 1-L\right) \left( 1-z\right) ^{2}}\right\} , \end{aligned}$$

and

$$\begin{aligned} W=\left\{ g\in A:g\left( z\right) =\frac{z}{\left( 1-L\right) \left( 1-z\right) ^{2}}\left( \frac{1+z}{1-z}-L\right) \right\} . \end{aligned}$$

We now discuss coefficient bounds of functions in set V and W.

Theorem 2.4

Let \(k\in \left[ 0,1\right]\) and \(0<q<1.\) Then, for all \(h\left( z\right) =z+\sum \nolimits _{n=2}^{\infty }c_{n}z^{n}\in V,\)

$$\begin{aligned} \left| c_{n}\right| \le \left[ n\right] _{q}+k\left( \left[ n\right] _{q}-1\right) \end{aligned}$$

and

$$\begin{aligned} \left| c_{n}\right| \ge \sqrt{1-k^{2}\left( \frac{\left[ n\right] _{q}-1}{\left[ n\right] _{q}+1}\right) }. \end{aligned}$$

Proof

Let \(h\in V,\) using series representation of \(h\left( z\right)\) and after some simplifications, we obtain

$$\begin{aligned} c_{n}=\frac{\left[ n\right] _{q}-L\left( \alpha \right) }{1-L\left( \alpha \right) }, \end{aligned}$$
(2.10)

where \(L\left( \alpha \right)\) is given in (2.1). Using (2.1) in (2.10), we obtain

$$\begin{aligned} \left| c_{n}\right| ^{2}=1+\frac{\left( \left[ n\right] _{q}\right) ^{2}-1}{\alpha ^{2}}+\frac{2k\left( 1-\left[ n\right] _{q}\right) }{\alpha } =g\left( \alpha \right) . \end{aligned}$$

Since \(k\in \left[ 0,1\right] ,\) we note that \(\alpha \ge \frac{1}{k+1}.\) In addition, \(g\left( \alpha \right)\) attains its minimum at \(\alpha _{0}=\frac{ \left[ n\right] _{q}+1}{k}.\) Note that \(\alpha _{0}\ge \frac{1}{k+1}\) and \(g\left( \alpha \right) \le g\left( {\frac{1}{{k + 1}}} \right) = \left( {\left[ n \right]_{q} + k\left( {\left[ n \right]_{q} - 1} \right)} \right)^{2}\) for all \(\alpha \ge 0.\) Thus, we have

$$\begin{aligned} \left| c_{n}\right| \le \left[ n\right] _{q}+k\left( \left[ n\right] _{q}-1\right) . \end{aligned}$$
(2.11)

Since α0 is the minimum, therefore, \(g\left( \alpha \right) \ge g\left( \alpha _{0}\right) =1-k^{2}\left( \frac{\left[ n\right] _{q}-1}{ \left[ n\right] _{q}+1}\right) .\)

Which gives us

$$\begin{aligned} \left| c_{n}\right| \ge \sqrt{1-k^{2}\left( \frac{\left[ n\right] _{q}-1}{\left[ n\right] _{q}+1}\right) }. \end{aligned}$$
(2.12)

Note that (2.11) and (2.12) are valid for \(k\in \left[ 0,1\right]\) and \(0<q<1.\) \(\quad \square\)

Using Alexander-type relation between k-qST and k-qCV,  we have the following.

Corollary 2.5

Let \(k\in \left[ 0,1\right]\) and \(0<q<1.\) Then, for all \(h\left( z\right) =z+\sum \nolimits _{n=2}^{\infty }c_{n}z^{n}\in W\)

$$\begin{aligned} \left| c_{n}\right| \le \left[ n\right] _{q}\left( \left[ n\right] _{q}+k\left( \left[ n\right] _{q}-1\right) \right) \end{aligned}$$

and

$$\begin{aligned} \left| c_{n}\right| \ge \left[ n\right] _{q}\sqrt{1-k^{2}\left( \frac{\left[ n\right] _{q}-1}{\left[ n\right] _{q}+1}\right) }. \end{aligned}$$

Corollary 2.6

Let \(k\in \left[ 0,1\right] ,\) \(0<q<1\) and let \(f\left( z\right) =z+\lambda z^{n},\) \(n\ge 2.\) Then, \(f\in k-q\text{ST},\) if and only if

$$\begin{aligned} \left| \lambda \right| \le \frac{1}{\left[ n\right] _{q}+k\left( \left[ n\right] _{q}-1\right) }, \end{aligned}$$
(2.13)

and \(f\in k-qCV,\) if and only if

$$\begin{aligned} \left| \lambda \right| \le \frac{1}{\left[ n\right] _{q}\{\left[ n \right] _{q}+k\left( \left[ n\right] _{q}-1\right) \}}. \end{aligned}$$
(2.14)

Proof

Let \(f\left( z\right) =z+\lambda z^{n},\) with \(\lambda\) satisfies inequality (2.13).

Let \(g\in V\) and

$$\begin{aligned} \left| \frac{\left( f*g\right) \left( z\right) }{z}\right| \ge 1-\left| A\right| \left| c_{n}\right| \left| z\right| ^{n-1}>1-\left| z\right| >0. \end{aligned}$$

Applying Theorem 2.1, we obtain \(f\in k\text {-}qST.\) Conversely, let \(f\in k\text {-}qST\) and let \(g\left( z\right) =z+\sum \nolimits _{n=2}^{\infty }\left( \left[ n \right] _{q}+k\left( \left[ n\right] _{q}-1\right) \right) z^{n}.\)

Then

$$\begin{aligned} \frac{\left( f*g\right) \left( z\right) }{z}=1+\lambda \left( \left[ n \right] _{q}+k\left( \left[ n\right] _{q}-1\right) \right) z^{n-1}\ne 0. \end{aligned}$$

Let \(\left| \lambda \right| >\frac{1}{\left[ n\right] _{q}+k\left( \left[ n\right] _{q}-1\right) },\) then there exists \(u\in E,\) such that

$$\begin{aligned} \frac{\left( f*g\right) \left( u\right) }{u}=0,\text { }u\in E. \end{aligned}$$

which is a contradiction. Hence, \(\left| \lambda \right| \le \frac{1 }{\left[ n\right] _{q}+k\left( \left[ n\right] _{q}-1\right) }.\)

Using the Alexander-type relation between k-qST and k-qCV,  one can obtain condition given in (2.14). \(\quad \square\)

Using Theorem 2.1, we now obtain the following result which is a special case of theorem given in Dziok (2011).

Corollary 2.7

Let \(k\in \left[ 0,1\right] ,\) \(0<q<1\) and let \(f\in A\) and is of the form (1.1). If

$$\begin{aligned} \sum \limits _{n=2}^{\infty }\left( \left[ n\right] _{q}+k\left( \left[ n \right] _{q}-1\right) \right) \left| a_{n}\right| \le 1, \end{aligned}$$
(2.15)

then \(f\in k-qST.\) In addition, if

$$\begin{aligned} \sum \limits _{n=2}^{\infty }\left[ n\right] _{q}\left( \left[ n\right] _{q}+k\left( \left[ n\right] _{q}-1\right) \right) \left| a_{n}\right| \le 1, \end{aligned}$$
(2.16)

then \(f\in k-qCV.\)

Proof

Let \(f\left( z\right) =z+\sum \nolimits _{n=2}^{\infty }a_{n}z^{n}\) and \(g\left( z\right) =z+\sum \limits _{n=2}^{\infty }c_{n}z^{n}\in V.\) The convolution

$$\begin{aligned} \frac{\left( f*g\right) \left( z\right) }{z}=1+\sum \limits _{n=2}^{ \infty }c_{n}a_{n}z^{n-1},\quad z\in E. \end{aligned}$$
(2.17)

It is known from Theorem 2.1 that \(f\in k-qST\) if and only if \(\frac{\left( f*g\right) \left( z\right) }{z}\ne 0,\) for all \(g\in V\). Using Theorem 2.4 in (2.17), we have

$$\begin{aligned} \left| \frac{\left( f*g\right) \left( z\right) }{z}\right| \ge 1-\sum \limits _{n=2}^{\infty }\left( \left[ n\right] _{q}+k\left( \left[ n \right] _{q}-1\right) \right) \left| a_{n}\right| \left| z\right| ^{n-1}>0,\quad z\in E. \end{aligned}$$
(2.18)

Which can be written as

$$\begin{aligned} \sum \limits _{n=2}^{\infty }\left( \left[ n\right] _{q}+k\left( \left[ n \right] _{q}-1\right) \right) \left| a_{n}\right| \le 1. \end{aligned}$$

For (2.16), we apply the same method with \(g\in W.\) \(\quad \square\)

As a special case when \(q\rightarrow 1^{-},\) we have the following special case.

Corollary 2.8

Kanas and Wisniaskawa (1999). Let \(k\in \left[ 0,1\right]\) and let \(f\in A.\)

$$\begin{aligned} If \sum \limits _{n=2}^{\infty }\left( n+k\left( n-1\right) \right) \left| a_{n}\right| \le 1, then f\in k \text{-}ST, \end{aligned}$$

and

$$\begin{aligned} If \sum \limits _{n=2}^{\infty }n\left( n+k\left( n-1\right) \right) \left| a_{n}\right| \le 1,\text { then}\,\it {f}\in \it {k}\text{-UCV}. \end{aligned}$$

Let \(k-S_{q}\) be the class of satisfying the condition:

$$\begin{aligned} \sum \limits _{n=2}^{\infty }\left( \left[ n\right] _{q}+k\left( \left[ n \right] _{q}-1\right) \right) \left| a_{n}\right| <1. \end{aligned}$$
(2.19)

From Corollary 2.7, we note \(k-S_{q}\subset k-qST.\) We prove the following theorem.

Theorem 2.9

Let \(k\in \left[ 0,1\right] ,\) \(0<q<1\). If \(f\in k-S_{q}^{*}\) and \(g\in C,\,\)then \(\left( f*g\right) \in k-qST.\)

Proof

Let \(f\left( z\right) =z+\sum \nolimits _{n=2}^{\infty }a_{n}z^{n}\in k-S_{q}\) and \(g\left( z\right) =z+\sum \nolimits _{n=2}^{\infty }b_{n}z^{n}\in C\) and \(h\left( z\right) =z+\sum \nolimits _{n=2}^{\infty }c_{n}z^{n}\in V.\)

Since \(f\in k-S_{q}^{*}\), therefore, by definition

$$\begin{aligned} 1-\sum \limits _{n=2}^{\infty }\left( \left[ n\right] _{q}+k\left( \left[ n \right] _{q}-1\right) \right) \left| a_{n}\right| >0. \end{aligned}$$
(2.20)

To prove that \((f*g)\in k-qST,\) it is enough to show that

$$\begin{aligned} \frac{\left( f*g*h\right) \left( z\right) }{z}\ne 0. \end{aligned}$$

Consider

$$\begin{aligned} \left| \frac{\left( f*g*h\right) \left( z\right) }{z}\right| \ge 1-\sum \limits _{n=2}^{\infty }\left| a_{n}\right| \left| b_{n}\right| \left| c_{n}\right| \left| z\right| ^{n-1}. \end{aligned}$$

As \(z\in E\) and \(g\in C,\) therefore, \(\left| b_{n}\right| \le 1.\) Using coefficient bounds of \(g\left( z\right)\) from Theorem 2.4 and (2.20), we obtain

$$\begin{aligned} \left| \frac{\left( f*g*h\right) \left( z\right) }{z}\right|>1-\sum \limits _{n=2}^{\infty }\left( \left[ n\right] _{q}+k\left( \left[ n \right] _{q}-1\right) \right) \left| a_{n}\right| >0,\, z\in E. \end{aligned}$$

Thus, \(f*g\in k-qST.\)

Similarly, we can construct k-\(C_{q}\) using (2.16) and prove the following.

Corollary 2.10

Let \(k\in \left[ 0,1\right]\) and \(0<q<1.\) Then

$$\begin{aligned} \left( k-C_{q}\right) *C\subset k-qCV. \end{aligned}$$

We now prove the generalization of Theorem 2.9.

Theorem 2.11

Let \(V\subset A\) and \(V^{*}\) is a dual set of V. Let \(V_{1}\subset V^{*}\) consist of functions satisfying the condition:

$$\begin{aligned} \sum \limits _{n=2}^{\infty }\left| a_{n}\right| \left| c_{n}\right| <1. \end{aligned}$$

Then

$$\begin{aligned} C*V_{1}\subset V^{*}. \end{aligned}$$

Proof

Let \(f\left( z\right) =z+\sum \nolimits _{n=2}^{\infty }a_{n}z^{n}\in V^{*},\) \(g\left( z\right) =z+\sum \nolimits _{n=2}^{\infty }b_{n}z^{n}\in C\) and \(h\left( z\right) =z+\sum \nolimits _{n=2}^{\infty }c_{n}z^{n}\in V.\)

Then, by definition of \(V^{*}\)

$$\begin{aligned} \frac{\left( f*h\right) \left( z\right) }{z}\ne 0,\quad z\in E. \end{aligned}$$

Let \(f\in V_{1}\). Then

$$\begin{aligned} 1-\sum \limits _{n=2}^{\infty }\left| a_{n}\right| \left| c_{n}\right| >0. \end{aligned}$$
(2.21)

Now, consider

$$\begin{aligned} \left| \frac{\left( f*g*h\right) \left( z\right) }{z}\right| \ge 1-\sum \limits _{n=2}^{\infty }\left| a_{n}\right| \left| b_{n}\right| \left| c_{n}\right| \left| z\right| ^{n-1}, \quad z\in E. \end{aligned}$$
(2.22)

Since \(g\in C\), it is known that \(\left| b_{n}\right| \le 1.\) Using coefficient bounds of \(g\left( z\right)\) and (2.21) in (2.22), we obtain

$$\begin{aligned} \left| \frac{\left( f*g*h\right) \left( z\right) }{z}\right| >0,\quad z\in E. \end{aligned}$$

Thus, by definition of dual set \(f*g\in V^{*}.\) This completes the proof. \(\quad \square\)

3 Applications of Theorem 2.9

Consider the following operators:

  1. (1)

    \(\ f_{1}\left( z\right) =\int \limits _{0}^{z}\frac{f\left( \xi \right) }{\xi }d\xi =\left( f_{1}*\phi _{1}\right) \left( z\right) .\)

  2. (2)

    \(\ f_{2}\left( z\right) =\frac{2}{z}\int \limits _{0}^{z}f\left( \xi \right) d\xi =\left( f_{2}*\phi _{2}\right) \left( z\right) .\)

  3. (3)

    \(f_{2}\left( z\right) =\int \limits _{0}^{z}\frac{f\left( \xi \right) -f\left( x\xi \right) }{\xi -x\xi }d\xi =\left( f_{3}*\phi _{3}\right) \left( z\right), \left| x\right| \le 1,x\ne 1.\)

  4. (4)

    \(f_{4}\left( z\right) =\frac{1+c}{z^{c}}\int \limits _{0}^{z}\xi ^{c-1}f\left( \xi \right) d\xi =\left( f_{4}*\phi _{4}\right) \left( z\right), {\mathfrak{R}}\left( c\right) >0.\)

Where

$$\begin{aligned} \phi _{1}= & {} -\log \left( 1-z\right) , \\ \phi _{2}= & {} -2\left[ z+\log \left( 1-z\right) \right] , \\ \phi _{3}= & {} \frac{1}{1-x}\log \left[ \frac{1-xz}{1-z}\right] ,\quad \left| x\right| =1, x\ne 1, \\ \phi _{4}= & {} \sum \limits _{n=1}^{\infty }\frac{1+c}{n+c}z^{n},\text { } {\mathfrak{R}}\left( c\right) >0, \end{aligned}$$

\(\phi _{i},\) \(1\le i\le 3,\) can easily be verified to be convex in E and for \(\phi _{4}\in C,\) we refer to Ruscheweyh (1975). If \(f\in k-S_{q}\) or \(k-C_{q}\), then \(f_{i}\in k-qST\) or \(k-qCV.\) For more applications, see Ruscheweyh and Sheil-Small (1973).

4 Neighborhood of Analytic Functions

For \(f\in A\) and is of form (1.1) and \(\delta \ge 0,\) the \(T_{\delta }\) neighborhood of function f is defined as follows:

$$\begin{aligned} T_{\delta }\left( f\right) =\left\{ g\left( z\right) =z+\sum \limits _{n=2}^{\infty }b_{n}z^{n}\in A:\sum \limits _{n=2}^{\infty }t_{n}\left| b_{n}-a_{n}\right| \le \delta \right\} . \end{aligned}$$
(2.23)

Ruscheweyh Ruscheweyh (1981) proved many inclusion properties including when \(t_{n}=n,\) especially \(T_{\frac{1}{4}}\left( f\right) \subset S^{*}\) for all \(f\in C.\)

We prove the following.

Theorem 2.12

Let \(f\in T_{1}\left( e\right)\) and is of the form (1.1) with \(t_{n}= \left[ n\right] _{q}\) and \(e\left( z\right) =z.\) Then

$$\begin{aligned} \left| \frac{zD_{q}f\left( z\right) }{f\left( z\right) }-1\right| <1, \end{aligned}$$
(2.24)

where \(z\in E\) and \(0<q<1.\)

Proof

Let \(f\in A\) and \(f\left( z\right) =z+\sum \nolimits _{n=2}^{\infty }a_{n}z^{n}.\) Consider

$$\begin{aligned} \left| zD_{q}f\left( z\right) -f\left( z\right) \right|= & {} \left| \sum \limits _{n=2}^{\infty }\left( \left[ n\right] _{q}-1\right) a_{n}z^{n-1}\right| \\\le & {} \sum \limits _{n=2}^{\infty }\left[ n\right] _{q}\left| a_{n}\right| -\sum \limits _{n=2}^{\infty }\left[ n\right] _{q}\left| a_{n}\right| \left| z\right| ^{n-1} \\\le & {} \left| z\right| -\sum \limits _{n=2}^{\infty }\left[ n\right] _{q}\left| a_{n}\right| \left| z\right| ^{n-1}. \\\le & {} \left| f\left( z\right) \right| ,\,\ z\in E. \end{aligned}$$

This gives us the required result.

The procedure already described in Theorem 2.12 leads to the following new result.\(\quad \square\)

Theorem 2.13

If \(f\in T_{1}\left( e\right)\) with \(t_{n}=\left( \left[ n\right] _{q}\right) ^{2}\) and \(e\left( z\right) =z,\) then

$$\begin{aligned} \left| \frac{D_{q}\left( zD_{q}f\left( z\right) \right) }{D_{q}f\left( z\right) }-1\right| <1. \end{aligned}$$
(2.25)

The inequalities (2.24) and (2.25) describe subclasses of q- starlike and q-convex functions with \(\phi \left( z\right) =1+z,\) for more details, see Seoudy and Aouf (2016).

We now discuss the geometric properties of \(T_{\delta }\left( f\right)\) defined in (2.23). Consider a class of functions

$$\begin{aligned} F_{\alpha }\left( z\right) =\frac{f\left( z\right) +\alpha z}{1+\alpha }, \end{aligned}$$
(2.26)

where \(\alpha\) is a non-zero complex number. We note that if \(f\in A,\) then \(F_{\alpha }\) \(\in A.\) We discuss the relationship between \(f\left( z\right)\) and \(F_{\alpha }\left( z\right)\) in the following Lemma.

Lemma 2.14

Let \(k\in \left[ 0,1\right] ,\)  \(f\in A\) and \(\delta >0.\) If \(F_{\alpha }\in k-qST\) for all \(\alpha \in \mathbb {C} \backslash \{0\},\) then \(f\in k-qST.\) Furthermore, for all \(g\in V\)

$$\begin{aligned} \left| \frac{\left( f*g\right) \left( z\right) }{z}\right| >\delta , \end{aligned}$$

where \(\left| \alpha \right| <\delta\) and \(z\in E.\)

Proof

Since \(F_{\alpha }\in k-qST,\) Therefore, by Theorem 2.1, we have

$$\begin{aligned} \frac{\left( F_{\alpha }*g\right) \left( z\right) }{z}\ne 0\text {, for all } g\in \it {V} \,\text{and}\, \it {z} \in \it {E}. \end{aligned}$$

Using (2.26) and after some simplifications, we obtain

$$\begin{aligned} \frac{\left( f*g\right) \left( z\right) }{z}\ne -\alpha ,\text { for all }\alpha \in \mathbb {C} \backslash \{0\}. \end{aligned}$$

For \(\left| \alpha \right| <\delta\)

$$\begin{aligned} \left| \frac{\left( f*g\right) \left( z\right) }{z}\right| >\delta . \end{aligned}$$
(2.27)

Applying Theorem 2.1, we obtain that \(f\in k\text{-}q\text{ST}.\)\(\quad \square\)

Applying the similar method, we have the following Lemma.

Lemma 2.15

Let \(k\in \left[ 0,1\right] ,\)  \(f\in A\) and \(\delta >0\) and let \(\alpha\) be a non-zero complex number. if \(F_{\alpha }\in k-qCV,\) then \(f\in k-qCV,\) furthermore for all \(g\in W\):

$$\begin{aligned} \left| \frac{\left( f*g\right) \left( z\right) }{z}\right| >\delta , \end{aligned}$$

where \(\left| \alpha \right| <\delta\) and \(z\in E.\)

We now prove the following.

Theorem 2.16

Let \(k\in \left[ 0,1\right]\) and \(\delta >0.\) If \(F_{\alpha }\in k-qST\) for all \(\alpha \in \mathbb {C} \backslash \{0\},\) then \(T_{\delta _{1}}\left( f\right) \subset k-qST\) with \(t_{n}=\left[ n\right] _{q}\) and

$$\begin{aligned} \delta _{1}=\frac{\delta }{k+1}, \end{aligned}$$

where \(\left| \alpha \right| <\delta .\)

Proof

Let \(g\left( z\right) =z+\sum \nolimits _{n=2}^{\infty }b_{n}z^{n}\in T_{\delta _{1}}\left( f\right) .\) Using Theorem 2.1, \(g\in k-qST,\) if and only if

$$\begin{aligned} \frac{\left( g*h\right) \left( z\right) }{z}\ne 0,\text { } \end{aligned}$$

for all \(h\in V.\)

Consider

$$\begin{aligned} \left| \frac{\left( g*h\right) \left( z\right) }{z}\right|= & {} \left| \frac{\left( f*h\right) \left( z\right) }{z}+\frac{\left( \left( g-f\right) *h\right) \left( z\right) }{z}\right| \nonumber \\\ge & {} \left| \frac{\left( f*h\right) \left( z\right) }{z} \right| -\left| \frac{\left( \left( g-f\right) *h\right) \left( z\right) }{z}\right| . \end{aligned}$$
(2.28)

Using Lemma 2.14 and series representations of \(f\left( z\right)\) and \(g\left( z\right)\) with \(h\left( z\right) =\sum \nolimits _{n=2}^{\infty }\left( \left[ n\right] _{q}+k\left( \left[ n\right] _{q}-1\right) \right) z^{n},\) we obtain

$$\begin{aligned} \left| \frac{\left( g*h\right) \left( z\right) }{z}\right| >\delta -\sum \limits _{n=2}^{\infty }\left( \left[ n\right] _{q}+k\left( \left[ n\right] _{q}-1\right) \right) \left| b_{n}-a_{n}\right| . \end{aligned}$$
(2.29)

Now

$$\begin{aligned} \sum \limits _{n=2}^{\infty }\left( \left[ n\right] _{q}+k\left( \left[ n \right] _{q}-1\right) \right) \left| b_{n}-a_{n}\right| \le \left( k+1\right) \sum \limits _{n=2}^{\infty }\left[ n\right] _{q}\left| b_{n}-a_{n}\right| \le \left( k+1\right) \delta _{1}. \end{aligned}$$
(2.30)

Using (2.29) and (2.30) in (2.28), we obtain

$$\begin{aligned} \left| \frac{\left( g*h\right) \left( z\right) }{z}\right|>\delta -\left( k+1\right) \delta _{1}>0. \end{aligned}$$

Where

$$\begin{aligned} \delta _{1}=\frac{\delta }{k+1}. \end{aligned}$$

Remark 2.17

In Theorem 2.16, we can replace \(t_{n}\) by n and obtain same result as \(\left[ n\right] _{q}<n\) when \(0<q<1.\)

Theorem 2.18

Let \(f\in k-qST.\) Then, \(T_{\delta _{1}}\left( f\right) \subset k-qST\) with

$$\begin{aligned} \delta _{1}<\frac{c}{k+1}, \end{aligned}$$
(2.31)

where c is a non-zero real number with \(c\le \left| \frac{\left( f*h\right) \left( z\right) }{z}\right| ,\) \(z\in E.\)

Proof

Let \(g\in T_{\delta _{1}}\left( f\right)\), with \(t_{n}=\left[ n\right] _{q}\) and let \(h\in V.\)

Consider

$$\begin{aligned} \left| \frac{\left( g*h\right) \left( z\right) }{z}\right| \ge \left| \frac{\left( f*h\right) \left( z\right) }{z}\right| -\left| \frac{\left( \left( g-f\right) *h\right) \left( z\right) }{z} \right| . \end{aligned}$$
(2.32)

Since \(f\in k-qST,\) therefore applying Theorem 2.1, we obtain

$$\begin{aligned} \left| \frac{\left( f*g\right) \left( z\right) }{z}\right| \ge c,\text { } \end{aligned}$$
(2.33)

where c is a non-zero real number and \(z\in E.\)

Now

$$\begin{aligned} \left| \frac{\left( \left( g-f\right) *h\right) \left( z\right) }{z} \right|= & {} \sum \limits _{n=2}^{\infty }\left| c_{n}\right| \left| b_{n}-a_{n}\right| \left| z\right| ^{n-1} \nonumber \\\le & {} \sum \limits _{n=2}^{\infty }\left( \left[ n\right] _{q}+k\left( \left[ n\right] _{q}-1\right) \right) \left| b_{n}-a_{n}\right| \nonumber \\\le & {} \left( k+1\right) \delta _{1}, \end{aligned}$$
(2.34)

where we have used Theorem 2.4. Using (2.33) and (2.34) in (2.32), we obtain

$$\begin{aligned} \left| \frac{\left( g*h\right) \left( z\right) }{z}\right|>c-\left( k+1\right) \delta _{1}>0. \end{aligned}$$

where \(\delta _{1}\) is given in (2.31). This completes the proof. \(\quad \square\)