1. Introduction and preliminaries

Let $\mathcal {H}$ denote the class of analytic functions in the open unit disc $\mathbb {D}=\{z\in \mathbb {C}:\ |z|<1\}$. Let $\mathcal {A}$ be the subclass of $\mathcal {H}$ comprising of functions $f$ normalized by $f(0)=0$, $f'(0)=1$, and let $\mathcal {S}\subset \mathcal {A}$ denote the class of functions which are univalent in $\mathbb {D}$. Let $f(z)\in \mathcal {H}$ and let $f(z)$ be univalent in $\mathbb {D}$. Then $f(z)$ maps $\mathbb {D}$ onto a convex domain if and only if

(1.1)\begin{equation} \mathfrak{Re}\left \{1+\frac{zf''(z)}{f'(z)}\right \} >0,\quad (z \in {\mathbb{D}}). \end{equation}

Such function $f$ is said to be convex in $\mathbb {D}$ (or briefly convex). Let ${\mathcal {K}}$ denote the subclass of ${\mathcal {H}}$ consisting of functions satisfying (1.1) and normalized by $f(0)=0$, $f'(0)=1$. A function $f \in {\mathcal {S}}$ is said to be starlike of order $\alpha$ if

(1.2)\begin{equation} \mathfrak{Re}\left\{\frac{zf' (z)}{f(z)}\right\}>\alpha,\quad (z \in {\mathbb{D}}), \end{equation}

for some $0\le \alpha <1$. The class of functions starlike of order $\alpha$ is denoted by ${\mathcal {S}^*(\alpha )}$.

Jackson in [Reference Jackson13, Reference Jackson14] introduced and studied the $q$-difference operator, $0< q<1$, as

(1.3)\begin{equation} {\rm d}_qf(z)=\frac{f(qz)-f(z)}{qz-z},\quad z\neq 0\ \text{and}\quad {\rm d}_qf(0)=f'(0). \end{equation}

This operator is the $q$-analogue of the derivative, which is also called the $q$-derivative or the Jackson derivative. Obviously, ${\rm d}_qf(z) \rightarrow f'(z)$, when $q \rightarrow 1^-$. This $q$-derivative is useful in the theory of hypergeometric series and quantum physics. Jackson's derivative is a part of a field called $q$-calculus (quantum calculus), which has many applications in combinatorics, number theory, fluid mechanics, quantum mechanics and physics. The quantum calculus has many applications in the fields of special functions and many other areas (see [Reference Abu-Risha, Annaby, Ismail and Mansour1–Reference Araci, Duran, Acikgoz and Srivastava6]). Further there is possibility of extension of the $q$-calculus to post quantum calculus denoted by the $p,q$-calculus. In [Reference Chakrabarti and Jagannathan15] Chakrabarti and Jagannathan introduced a consideration of the $p,q$-integer in order to generalize or unify several forms of $q$-oscillator algebras well known in the physics literature related to the representation theory of single-parameter quantum algebras (see also [Reference Annaby and Mansour3–Reference Aouf and Seoudy5, Reference Jagannathan and Rao16]).

1.1 $\xi,\eta$-difference operator

Let us begin by defining a basic number $[n]_{\xi,\eta }$ called a $\xi,\eta$ number by

\[ [n]_{\xi,\eta}=\frac{\xi^n-\eta^n}{\xi-\eta},\quad (\xi,\eta\in\mathbb{C},\ \xi\neq\eta,\ n\in\mathbb{N}\setminus\{1\}). \]

We consider the function

(1.4)\begin{equation} h_{\xi,\eta}(z)=\sum_{n=1}^{\infty}[n]_{\xi,\eta}z^{n} =z+\sum_{n=2}^{\infty}\frac{\xi^{n}-\eta^{n}}{\xi-\eta}z^{n}, \end{equation}

for $\xi,\eta \in \mathbb {D}$. It is easy to see that if $\xi,\eta \in \mathbb {D}$, then (1.4) converges for $|z|<1$. The $\xi,\eta$-difference operator on $f\in \mathcal {A}$ is

(1.5)\begin{equation} {\rm d}_{\xi,\eta}f(z)=\frac{f(\xi z)-f(\eta z)}{\xi z-\eta z},\quad z\neq 0 \quad \quad \text{and}\quad {\rm d}_{\xi,\eta}f(0)=f'(0). \end{equation}

It is easy to find that for $f\in \mathcal {A}$

(1.6)\begin{equation} {\rm d}_{\xi,\eta}f(z)=\frac{1}{z}\sum_{n=1}^{\infty}[n]_{\xi,\eta}a_nz^{n}= \frac{1}{z}\left\{h_{\xi,\eta}(z)*f(z)\right\},\quad z\in\mathbb{D}, \end{equation}

where $*$ denotes the Hadamard product of power series (1.4) and

(1.7)\begin{equation} f(z)=z+a_2z^2+a_3z^3+\cdots. \end{equation}

Lemma 1.1 [Reference Ruscheweyh and Sheil-Small19] If $f\in \mathcal {S}^*(1/2)$ and $g\in \mathcal {S}^*(1/2)$ $[$ or if $f\in \mathcal {K}$ and $g\in \mathcal {S}^*$ $]$, then

(1.8)\begin{equation} \frac{f(z)*g(z)F(z)}{f(z)*g(z)}\in\overline{{\rm co}} \{F(\mathbb{D})\}, \quad z\in \mathbb{D}, \end{equation}

where $F\in \mathcal {H}$ and $\overline {{\rm co}} \{F(\mathbb {D})\}$ denotes the closed convex hull of $F(\mathbb {D})$.

1.2 Other known difference operators

If $\xi$ and $\eta$ are real then ${\rm d}_{\xi,\eta }$ becomes a $p,q$-difference operator. Corcino [Reference Corcino7] developed the theory of a $p,q$-extension of the binomial coefficients and also established some properties parallel to those of the ordinary and $q$-binomial coefficients, which comprised horizontal generating function, the triangular, vertical and the horizontal recurrence relations and the inverse and the orthogonality relations. Sadjang [Reference Sadjang20] investigated some properties of the $p,q$-derivatives and the $p,q$-integrations. Sadjang [Reference Sadjang20] also provided two suitable polynomial bases for the $p,q$-derivative and gave various properties of these bases.

The $p,q$-derivative operator

(1.9)\begin{equation} D_{p,q} f(x) = \frac{f(px)-f(qx)}{(p-q)x},\quad (p \neq q, x \neq 0) \end{equation}

was perhaps first used in [Reference Duren8]. The operator is also mentioned in the paper [Reference Gupta, Rassis, Agarwal and Acu9].

In [Reference Hahn10] Hahn introduced the Hahn difference operator:

(1.10)\begin{equation} D_{q,\omega} f(x) = \frac{f(qx+\omega)-f(x)}{(q-1)x+\omega},\quad (0< q<1, \omega>0) \end{equation}

and defined the $q$-analogues of the trigonometric functions. When $\omega =0$, (1.10) becomes Jackson's derivative operator. On the other hand, when $q\rightarrow 1$, (1.10) becomes the familiar forward difference operator of difference calculus. The operator (1.10) is a useful tool in constructing families of orthogonal polynomials and investigating certain approximation problems, see, for instance the paper [Reference Alvarez-Nodarse4].

If $E$ is a given interval and $\beta :E\rightarrow E$, $\beta \neq x$, is strictly monotonically increasing function, i.e., $x>t \Rightarrow \beta (x)>\beta (t)$, then $\beta$ derivative of a function $f(x)$ is defined by

(1.11)\begin{equation} D_\beta f(x) = \frac{f(\beta (x))-f(x)}{ \beta(x)-x}. \end{equation}

If $\beta (x)=qx$, then (1.11) becomes the $q$-derivative operator. Further, for $\beta (x) = qx+\omega$, the operator (1.11) reduces to the Hahn difference operator (1.10). The operator (1.11) is found in the papers [Reference Hamza, Sarhan, Shehata and Aldwoah11].

2. Main results

Theorem 2.1 If $f$ is in the class $\mathcal {K}$ of convex univalent functions, then for $\xi,\eta$ such that $\xi,\eta \in \mathbb {D}$, the function given by

(2.1)\begin{equation} z{\rm d}_{\xi,\eta}f(z)=z+\sum_{n=2}^\infty[n]_{\xi,\eta}a_nz^n,\quad z\in\mathbb{D}, \end{equation}

is in the class $\mathcal {S}^*(\alpha )$ of starlike univalent functions of order $\alpha =\frac {1-|\xi \eta |}{(1+|\xi |)(1+|\eta |)}$. The order $\alpha$ is the best possible.

Proof. It is easy to check that

(2.2)\begin{equation} h_{\xi,\eta}(z)=\sum_{n=1}^{\infty}[n]_{\xi,\eta}z^{n}=\frac{z}{(1-\xi z)(1-\eta z)},\quad z\in\mathbb{D}. \end{equation}

Hence

\[ \frac{zh'_{\xi,\eta}(z)}{h_{\xi,\eta}(z)} = 1+\frac{\xi z}{1-\xi z}+\frac{\eta z}{1-\eta z}. \]

Because

\[ \mathfrak{Re}\frac{\xi z}{1-\xi z }>\frac{-|\xi|}{1+|\xi|},\quad \mathfrak{Re}\frac{\eta z}{1-\eta z}>\frac{-|\eta|}{1+|\eta|}, \quad z\in\mathbb{D}, \]

we have

\begin{align*} \mathfrak{Re}\frac{zh'_{\xi,\eta}(z)}{h_{\xi,\eta}(z)}& >1-\frac{|\xi|}{1+|\xi|}-\frac{|\eta|}{1+|\eta|}\\ & =\frac{1-|\xi||\eta|}{(1+|\xi|)(1+|\eta|)},\quad z\in\mathbb{D}. \end{align*}

Therefore, if $\xi,\eta \in \mathbb {D}$, then $h_{\xi,\eta }$ is starlike univalent in $\mathbb {D}$ of order $\frac {1-|\xi ||\eta |}{(1+|\xi |)(1+|\eta |)}$. On the other hand

(2.3)\begin{equation} z{\rm d}_{\xi,\eta}f(z) = \left\{z+\sum_{n=2}^\infty a_nz^n\right\}*\sum_{n=1}^\infty[n]_{\xi,\eta}z^n=f(z)*\frac{z}{(1-\xi z)(1-\eta z)}. \end{equation}

Hence $z{\rm d}_{\xi,\eta }f$ is a convolution of $f$ with a starlike function of order $\alpha =\frac {1-|\xi ||\eta |}{(1+|\xi |)(1+|\eta |)}$. Because of the famous result [Reference Ruscheweyh and Sheil-Small19] that $\mathcal {K} *\mathcal {S}^*(\delta )=\mathcal {S}^*(\delta )$, $\delta \in [0,1)$, we finally obtain that the function in (2.1) is in the class $\mathcal {S}^*(\frac {1-|\xi ||\eta |}{(1+|\xi |)(1+|\eta |)})$. The order $\alpha$ is the best possible because if we take convex function $f(z)=z/(1-z)$, then

\[ z{\rm d}_{\xi,\eta}f(z)=h_{\xi,\eta}(z) \]

and the order of starlikeness of $z{\rm d}_{\xi,\eta }f(z)$ is equal to the order of starlikeness of $h_{\xi,\eta }(z)$ which is $\alpha$ as we just have proved.

It is known that

\[ \forall f\in\mathcal{S}^*\exists\ g\in\mathcal{K}:\quad f(z)=zg'(z). \]

A question worth considering here is the following:

Is it true that

\[ \forall f\in\mathcal{S}^*\ \exists\ g\in\mathcal{K}\ \exists\xi,\eta\in\mathbb{D}:\quad f(z)=zd_{\xi,\eta}g(z)\quad? \]

In terms of the convolution, this problem becomes: Are there $\xi,\eta \in \mathbb {D}$ such that for given $f(z)=z+a_2z^2+\cdots \in \mathcal {S}^*$, the function

(2.4)\begin{equation} z+\sum_{n=2}^\infty\frac{a_n}{[n]_{\xi,\eta}}z^n\in\mathcal{K}\quad? \end{equation}

The answer of the question (2.4) is ‘no’. Namely, for the starlike function

\[ \frac{z}{(1-z)^2}=z+\sum_{n=2}^\infty nz^n,\quad z\in\mathbb{D}, \]

$a_n=n$ and the function in (2.4) becomes

\[ z+\sum_{n=2}^\infty\frac{n}{[n]_{\xi,\eta}}z^n, \]

which is not in the class $\mathcal {K}$ because it has the coefficients $n/[n]_{\xi,\eta }$ and

\[ \left|\frac{n}{[n]_{\xi,\eta}}\right| =\frac{n}{|\xi^{n-1}+\xi^{n-2}\eta+\cdots+\xi\eta^{n-2}+\eta^{n-1}|}>1. \]

Theorem 2.2 If $f$ is in the class $\mathcal {K}$ of convex univalent functions, then

(2.5)\begin{equation} \mathfrak{Re} \left\{\frac{1}{1-\eta}\frac{{\rm d}_{\xi,\eta} f(z)}{{\rm d}_\xi f(\eta z)}\right\} =\mathfrak{Re} \left\{\frac{1}{1-\eta}\frac{{\rm d}_{\xi,\eta} f(z)}{{\rm d}_{\xi\eta,\eta} f(z)}\right\} >\mathfrak{Re} \left\{\frac{1+\eta}{2(1-\eta)}\right\},\quad z\in\mathbb{D}, \end{equation}

for all $\xi$, $\eta \in \mathbb {D}$.

Proof. It is known from [Reference Ruscheweyh and Sheil-Small19], [p.10], that if $f\in \mathcal {K}$, then for all $\zeta,v$ and $w\in \mathbb {D}$, we have

(2.6)\begin{equation} \mathfrak{Re}\left\{\frac{\zeta}{\zeta-v}\frac{v-w}{\zeta-w}\frac{f(\zeta)-f(w)}{f(v)-f(w)}-\frac{v}{\zeta-v}\right\}>\frac{1}{2}. \end{equation}

If we put $\zeta =\xi z$, $w=\eta z$ and $v=\xi \eta z$ in (2.6), then we obtain

\[ \mathfrak{Re} \left\{\frac{1}{1-\eta}\frac{f(\xi z)-f(\eta z)}{z(\xi-\eta)}\left[\frac{f(\xi\eta z)-f(\eta z)}{\eta z(\xi-1)}\right]^{{-}1}\right\}-\mathfrak{Re}\left\{\frac{\eta}{1-\eta}\right\} >\frac{1}{2},\quad z\in\mathbb{D}. \]

Trivial calculations give

\[ \mathfrak{Re} \left\{\frac{1}{1-\eta}\frac{f(\xi z)-f(\eta z)}{z(\xi-\eta)}\left[\frac{f(\xi\eta z)-f(\eta z)}{\eta z(\xi-1)}\right]^{{-}1}\right\} >\mathfrak{Re} \left\{\frac{1+\eta}{2(1-\eta)}\right\},\quad z\in\mathbb{D}. \]

This gives (2.5).

Theorem 2.3 If $f$ is in the class $\mathcal {K}$ of convex univalent functions, then, we have

(2.7)\begin{equation} \mathfrak{Re} \left\{\frac{\xi^2}{\eta-\xi^2}\frac{{\rm d}_{\xi^2,\xi} f(z)}{{\rm d}_{\xi,\eta} f(z)}\right\} =\mathfrak{Re}\left\{\frac{\xi^2}{\eta-\xi^2}\frac{{\rm d}_{\xi} f(\xi z)}{{\rm d}_{\xi,\eta} f(z)}\right\} <\mathfrak{Re} \left\{\frac{\eta+\xi^2}{2(\eta-\xi^2)}\right\},\quad z\in\mathbb{D}, \end{equation}

for all $\xi,\eta \in \mathbb {D}$.

Proof. From (2.6), we have that for $f\in \mathcal {K}$, for all $t,v$ and $w\in \mathbb {D}$, we have

(2.8)\begin{equation} \mathfrak{Re} \left\{\frac{t}{t-v}\frac{v-w}{t-w}\frac{f(t)-f(w)}{f(v)-f(w)}-\frac{v}{t-v}\right\}>\frac{1}{2}. \end{equation}

If we put $v=\eta z$, $w=\xi z$ and $t=\xi ^2z$ in (2.8), then we obtain

\[ \mathfrak{Re}\left\{\frac{\xi^2z}{\xi^2z-\eta z}\frac{\eta z-\xi z}{\xi^2z-\xi z}\frac{f(\xi^2z)-f(\xi z)}{f(\eta z)-f(\xi z)}-\frac{\eta z}{\xi^2z-\eta z}\right\} >\frac{1}{2}. \]

After some calculations, we obtain

\[ \mathfrak{Re}\left\{\frac{\xi^2}{\xi^2-\eta}\frac{f(\xi^2z)-f(\xi z)}{\xi^2z-\xi z}\frac{\eta z-\xi z}{f(\eta z)-f(\xi z)}-\frac{\eta}{\xi^2-\eta}\right\} >\frac{1}{2} \]

or

\[ \mathfrak{Re}\left\{\frac{\xi^2}{\xi^2-\eta}\frac{{\rm d}_{\xi^2,\xi} f(z)}{{\rm d}_{\xi,\eta} f(z)}-\frac{\eta}{\xi^2-\eta}\right\} >\frac{1}{2}. \]

This proves (2.7).

Corollary 2.4 If $f$ is in the class $\mathcal {K}$ of convex univalent functions, then we have

\[ \mathfrak{Re}\left\{\frac{\xi^2}{\eta-\xi^2}\frac{\frac{1}{1-\xi z}*{\rm d}_{\xi} f(z)}{{\rm d}_{\xi,\eta} f(z)}\right\} <\mathfrak{Re} \left\{\frac{\eta+\xi^2}{2(\eta-\xi^2)}\right\},\quad z\in\mathbb{D}, \]

for all $\xi,\eta \in \mathbb {D}$, $\eta \neq \xi ^2$.

Corollary 2.5 If $f$ is in the class $\mathcal {K}$ of convex univalent functions, then we have

(2.9)\begin{equation} \mathfrak{Re} \left\{\frac{\eta z{\rm d}_{\xi,\eta} f(z)}{ f(\eta z)}\right\} >\frac{1}{2},\quad z\in\mathbb{D} \end{equation}

and

(2.10)\begin{equation} \mathfrak{Re} \left\{\frac{\xi}{\xi-\eta}\left({\rm d}_{\xi,\eta} f(z)-\frac{\eta}{\xi}\right)\right\} >\frac{1}{2},\quad z\in\mathbb{D}, \end{equation}

for all $\xi,\eta \in \mathbb {D}$.

Proof. If we put $v=0$, $w=\eta z$ and $t=\xi z$ in (2.8), then we obtain

\[ \mathfrak{Re}\left\{\frac{\eta z}{ f(\eta z)}\frac{f(\xi z)-f(\eta z)}{\xi z-\eta z}\right\} >\frac{1}{2},\quad z\in\mathbb{D}. \]

After some calculations, we obtain (2.9). If we put $v\rightarrow w=\eta z$ and $t=\xi z$ in (2.8), then we obtain (2.10), in the same way.

The class of starlike functions of order $\alpha$ is defined by condition (1.2). We want to consider here this condition with the operator ${\rm d}_{\xi,\eta }$ instead of derivative $f'$. For the purpose of this paper, we represent by $\mathcal {S}^*_{\xi,\eta }(\alpha )$ a class which is defined by

Definition 2.6 Let $f\in \mathcal {A}$. For given $\xi,\eta \in \mathbb {D}$, we say that $f$ is in the class $\mathcal {S}^*_{\xi,\eta }(\alpha )$ of $\xi,\eta$-starlike functions of order $\alpha$, $0\leq \alpha <1$, if

(2.11)\begin{equation} \mathfrak{Re}\left\{\frac{z{\rm d}_{\xi,\eta}f(z)}{f(z)}\right\}>\alpha,\quad z\in\mathbb{D}, \end{equation}

where the operator ${\rm d}_{\xi,\eta }$ is defined in (1.5).

Condition (2.11) may be written as

(2.12)\begin{equation} \mathfrak{Re}\left\{\frac{f(\xi z)-f(\eta z)}{\xi z-\eta z}\frac{z}{f(z)}\right\}>\alpha, \quad z\in\mathbb{D}. \end{equation}

Remark 2.7 For $\xi \rightarrow \eta$ condition (2.11) becomes

(2.13)\begin{equation} \mathfrak{Re}\left\{\frac{zf'(z)}{f(z)}\right\}>\alpha,\quad z\in\mathbb{D}, \end{equation}

and the class $\mathcal {S}^*_{\xi,\eta }(\alpha )$ tends to the well-known class $\mathcal {S}^*(\alpha )$ of starlike functions of order $\alpha$.

Remark 2.8 It is known that condition (2.13) implies the univalence of $f$, whenever $f\in \mathcal {A}$. Notice that the condition (2.11) does not imply that $f$ is univalent in $\mathbb {D}$. For example, it is known that $f(z)=z+(3/4)z^2$ is not univalent in $\mathbb {D}$, while $f\in \mathcal {S}^*_{1/2,1/4}(0)$ because for this function $f$ we have

\[ \mathfrak{Re}\left\{ \frac{z{\rm d}_{\zeta}f(z)}{f(z)}\right\} =\mathfrak{Re}\left\{ \frac{1+(3/4)^2z}{1+(3/4)z}\right\}>\frac{25}{28}, \quad z\in\mathbb{D}. \]

Theorem 2.9 The function $g(z)=z+cz^2$ is in the class $\mathcal {S}^*_{\xi,\eta }(\alpha )$, if and only if,

(2.14)\begin{equation} \mathfrak{Re} \left\{ \frac{1-|c|^2(\xi+\eta)-|c||\xi+\eta-1|}{1-|c|^2} \right\} >\alpha. \end{equation}

Proof. We have

\[ \mathfrak{Re}\left\{\frac{z{\rm d}_{\xi,\eta}g(z)}{g(z)}\right\}=\mathfrak{Re}\left\{\frac{1+c[2]_{\xi,\eta}z}{1+cz}\right\}= \mathfrak {Re}\left\{\frac{1+c(\xi+\eta)z}{1+cz}\right\}. \]

The function

\[ z \mapsto \frac{1+c(\xi+\eta)z}{1+cz} \]

maps $\mathbb {D}$ onto a disc centred at $S$ with radius $R$, where

\[ S=\frac{1-|c|^2(\xi+\eta)}{1-|c|^2},\quad R=\frac{|c||\xi+\eta-1|}{1-|c|^2}. \]

Therefore, $g\in \mathcal {S}^*_{\xi,\eta }(\alpha )$, if and only if, $\mathfrak {Re}(S-R)>\alpha$, which gives (2.14).

Proof. It suffices to consider here a function $h(z)$ defined by

\[ h(z)=z+\frac{1}{\xi+\eta}z^2,\quad z\in\mathbb{D}. \]

In view of the definition 2.6, we have

\[ \mathfrak{Re}\left\{\frac{z{\rm d}_{\xi,\eta}h(z)}{h(z)}\right\}= \mathfrak{Re}\left\{\frac{1+z}{1+\frac{1}{\xi+\eta}z}\right\}>0,\quad z\in\mathbb{D}, \]

which implies that $h\in \mathcal {S}^*_{\xi,\eta }(0)$. But we observe that for the function $h(z)$ we have $|1/(\xi +\eta )|>1/2$, therefore the function $h(z)$ is not univalent which validates the assertion of the corollary 2.10.

Theorem 2.11 If $f$ is in the class $\mathcal {S}^*(1/2)$ of starlike functions of order $1/2$, then

\[ \frac{z{\rm d}_{\xi,\eta}f(z)}{f(z)} \in\overline{{\rm co}} \{F(\mathbb{D})\}, \]

for all $\xi,\eta \in \mathbb {D}$ where

\[ F(z)=\frac{1-z}{(1-\xi z)(1-\eta z)},\quad z\in\mathbb{D}. \]

Proof. Note that

\[ g(z)=\frac{z}{1-z}\in\mathcal{S}^*(1/2), \quad z\in\mathbb{D}. \]

Therefore, lemma 1.1 gives

(2.15)\begin{equation} \frac{z{\rm d}_{\xi,\eta}f(z)}{f(z)}=\frac{f(z)*\frac{z}{1-z}\frac{\frac{z}{(1-\xi z)(1-\eta z)} }{\frac{z}{1-z}}}{f(z)*\frac{z}{1-z}}\in\overline{{\rm co}} \{F(\mathbb{D})\}, \end{equation}

where

\[ F(z)=\frac{\frac{z}{(1-\xi z)(1-\eta z)}}{\frac{z}{1-z}},\quad z\in\mathbb{D}, \]

for all $\xi,\eta \in \mathbb {D}$.

Corollary 2.12 If $f$ is in the class $\mathcal {S}^*(1/2)$ of starlike functions of order $1/2$, then $f\in \mathcal {S}^*_{\xi,\eta }(0)$ for all real $\xi,\eta \in (-1,1)$, such that

(2.16)\begin{equation} 1+\xi+\eta -3\xi\eta\geq 0\quad {and}\quad \xi\eta\geq0 \end{equation}

or

(2.17)\begin{equation} 1+\xi+\eta +\xi\eta\geq 0\quad {and}\quad \xi\eta<0. \end{equation}

Proof. From theorem 2.11, we have for $\xi \neq \eta$

\[ F(z)=\frac{1-z}{(1-\xi z)(1-\eta z)}=\frac{\frac{\xi-1}{\xi-\eta}}{1-\xi z }+\frac{\frac{1-\eta}{\xi-\eta}}{1-\eta z} \]

\[ \mathfrak {Re} \left\{\frac{z{\rm d}_{\xi,\eta}f(z)}{f(z)}\right\}> \min_{z\in\mathbb{D}}\mathfrak {Re} \left\{ \frac{\frac{\xi-1}{\xi-\eta}}{1-\xi z }+\frac{\frac{1-\eta}{\xi-\eta}}{1-\eta z}\right\}. \]

Assume that $z=\cos \phi +i\sin \phi$, then after some calculations, we have

\begin{align*} \mathfrak{Re} \left\{\frac{\frac{\xi-1}{\xi-\eta}}{1-\xi z } \right\} & = \frac{\xi-1}{\xi-\eta} \frac{1-\xi\cos\phi}{(1-\xi\cos\phi)^2+(\xi\sin\phi)^2}\\ & = \frac{\xi-1}{\xi-\eta} \frac{1-\xi\cos\phi}{1+\xi^2-2\xi\cos\phi} \end{align*}

also

\begin{align*} \mathfrak{Re} \left\{\frac{\frac{1-\eta}{\xi-\eta}}{1-\eta z}\right\} & = \frac{1-\eta}{\xi-\eta} \frac{1-\eta\cos\phi}{(1-\eta\cos\phi)^2+(\eta\sin\phi)^2}\\ & = \frac{1-\eta}{\xi-\eta}\frac{1-\eta\cos\phi}{1+\eta^2-2\eta\cos\phi}. \end{align*}

Therefore, we have

\begin{align*} & \mathfrak{Re} \left\{\frac{1-z}{(1-\xi z)(1-\eta z)}\right\} \\ & = \frac{1}{\xi-\eta}~\frac{\begin{array}{l}(\xi-1)(1-\xi\cos\phi)(1+\eta^2-2\eta\cos\phi)\\\quad-(\eta-1)(1-\eta\cos\phi)(1+\xi^2-2\xi\cos\phi)\end{array}}{(1+\xi^2-2\xi\cos\phi)(1+\eta^2-2\eta\cos\phi)} \\ & = \frac{(1-\cos\phi)\left(1+\xi+\eta -\xi\eta-2\xi\eta\cos\phi\right)}{(1+\xi^2-2\xi\cos\phi)(1+\eta^2-2\eta\cos\phi)}. \end{align*}

For all $\phi \in [0,2\pi )$, we have

\[ 1-\cos\phi \geq 0,\quad 1+\xi^2-2\xi\cos\phi >0,\quad 1+\eta^2-2\eta\cos\phi > 0. \]

Furthermore,

\[ \left[1+\xi+\eta -3\xi\eta\geq 0\quad {\rm and}\quad \xi\eta\geq0\right] \quad\Rightarrow\quad1+\xi+\eta -\xi\eta-2\xi\eta\cos\phi \geq 0 \]

for all $\phi \in [0,2\pi )$. Next,

\[ \left[1+\xi+\eta +\xi\eta\geq 0\quad {\rm and}\quad \xi\eta<0\right]\quad\Rightarrow\quad1+\xi+\eta -\xi\eta-2\xi\eta\cos\phi \geq 0 \]

for all $\phi \in [0,2\pi )$. Finally,

\[ \mathfrak{Re} \left\{\frac{1-z}{(1-\xi z)(1-\eta z)}\right\} =\frac{(1-\cos\phi)\left(1+\xi+\eta -\xi\eta-2\xi\eta\cos\phi\right)}{(1+\xi^2-2\xi\cos\phi)(1+\eta^2-2\eta\cos\phi)}\geq0 \]

for all $\phi \in [0,2\pi )$ and for all real $\xi,\eta \in (-1,1)$, $\xi \neq \eta$, satisfying (2.16) or (2.17). This means that in this case $f\in \mathcal {S}^*_{\xi,\eta }(0)$. For $\xi =\eta$, in the same way as above, we can obtain for $z=\cos \phi +i\sin \phi$

\begin{align*} \mathfrak{Re} \left\{F(z)\right\} & = \mathfrak{Re} \left\{\frac{1-z}{(1-\xi z)^2}\right\} \\ & = \frac{(1-\cos\phi)\left(1+2\xi -\xi^2-2\xi^2\cos\phi\right)}{(1+\xi^2-2\xi\cos\phi)^2} \\ & \geq 0 \end{align*}

for all $\phi \in [0,2\pi )$ and for all real $\xi \in (-1,1)$, satisfying (2.16) or (2.17) with $\xi =\eta$. This means that also in this case $f\in \mathcal {S}^*_{\xi,\eta }(0)$.

Corollary 2.12 provides some examples of functions in the class $\mathcal {S}^*_{\xi,\eta }(0)$ for all real $\xi,\eta \in (-1,1)$, satisfying (2.16) or (2.17). It is known that $\mathcal {K}\subset \mathcal {S}^*(1/2)$, therefore corollary 2.12 leads to the following result.

Recall here another definition of $q$-starlike functions of order $\alpha$. Namely, making use of $q$-derivative (1.3), Argawal and Sahoo in [Reference Agrawal and Sahoo2] introduced the class ${\mathcal {S}}^*_q(\alpha )$. A function $f\in \mathcal {A}$ belongs to the class ${\mathcal {S}}^*_q(\alpha )$, $0\leq \alpha <1$, if

(2.18)\begin{equation} \left|\frac{z{\rm d}_qf(z)}{f(z)}-\frac{1-\alpha q}{1-q}\right|\leq \frac{1-\alpha}{1-q},\quad z\in\mathbb{D}. \end{equation}

If $q\rightarrow 1^-$ the class ${\mathcal {S}}^*_q(\alpha )$ reduces to the class ${\mathcal {S}}^*(\alpha )$. If $\alpha =0$ the class ${\mathcal {S}}^*_q(\alpha )$ coincides with the class ${\mathcal {S}}^*_q(0)={\mathcal {S}}^*_q$, which was first introduced in [Reference Ismail, Merkes and Styer12] by Ismail et al. and was considered in [Reference Abu-Risha, Annaby, Ismail and Mansour1, Reference Annaby and Mansour3, Reference Aouf and Seoudy5, Reference Raghavendar and Swaminathan17, Reference Rønning18].