# Approximation properties of complex q-Balázs-Szabados operators in compact disks

## Abstract

This paper deals with approximating properties and convergence results of the complex q-Balázs-Szabados operators attached to analytic functions on compact disks. The order of convergence and the Voronovskaja-type theorem with quantitative estimate of these operators and the exact degree of their approximation are given. Our study extends the approximation properties of the complex q-Balázs-Szabados operators from real intervals to compact disks in the complex plane with quantitative estimate.

MSC:30E10, 41A25.

## 1 Introduction

In the recent years, applications of q-calculus in the area of approximation theory and number theory have been an active area of research. Details on q-calculus can be found in [13]. Several researchers have purposed the q-analogue of Stancu, Kantorovich and Durrmeyer type operators. Gal [4] studied some approximation properties of the complex q-Bernstein polynomials attached to analytic functions on compact disks.

Also very recently, some authors [57] have studied the approximation properties of some complex operators on complex disks. Balázs [8] defined the Bernstein-type rational functions and gave some convergence theorems for them. In [9], Balázs and Szabados obtained an estimate that had several advantages with respect to that given in [8]. These estimates were obtained by the usual modulus of continuity. The q-form of these operator was given by Doğru. He investigated statistical approximation properties of q-Balázs-Szabados operators [10].

The rational complex Balázs-Szabados operators were defined by Gal [4] as follows:

${R}_{n}\left(f;z\right)=\frac{1}{{\left(1+{a}_{n}z\right)}^{n}}\sum _{j=0}^{n}f\left(\frac{j}{{b}_{n}}\right)\left(\begin{array}{c}n\\ j\end{array}\right){\left({a}_{n}z\right)}^{j},$

where ${D}_{R}=\left\{z\in \mathbb{C}:|z| with $R>\frac{1}{2}$, $f:{D}_{R}\cup \left[R,\mathrm{\infty }\right)\to \mathbb{C}$ is a function, ${a}_{n}={n}^{\beta -1}$, ${b}_{n}={n}^{\beta }$, $0<\beta \le \frac{2}{3}$, $n\in \mathbb{N}$, $z\in \mathbb{C}$ and $z\ne -\frac{1}{{a}_{n}}$.

He obtained the uniform convergence of ${R}_{n}\left(f;z\right)$ to $f\left(z\right)$ on compact disks and proved the upper estimate in approximation of these operators. Also, he obtained the Voronovskaja-type result and the exact degree of its approximation.

The goal of this paper is to obtain convergence results for the complex q-Balázs-Szabados operators given by

${R}_{n}\left(f;q,z\right)=\frac{1}{{\prod }_{s=0}^{n-1}\left(1+{q}^{s}{a}_{n}z\right)}\sum _{j=0}^{n}{q}^{j\left(j-1\right)/2}f\left(\frac{{\left[j\right]}_{q}}{{b}_{n}}\right){\left[\begin{array}{c}n\\ j\end{array}\right]}_{q}{\left({a}_{n}z\right)}^{j},$

where $f:{D}_{R}\cup \left[R,\mathrm{\infty }\right)\to \mathbb{C}$ is uniformly continuous and bounded on $\left[0,\mathrm{\infty }\right)$, ${a}_{n}={\left[n\right]}_{q}^{\beta -1}$, ${b}_{n}={\left[n\right]}_{q}^{\beta }$, $q\in \left(0,1\right]$, $0<\beta \le \frac{2}{3}$, $n\in \mathbb{N}$, $z\in \mathbb{C}$ and $z\ne -\frac{1}{{q}^{s}{a}_{n}}$ for $s=0,1,2,\dots$ .

These operators are obtained simply replacing x by z in the real form of the q-Balázs-Szabados operators introduced in Doğru [10].

The complex q-Balázs-Szabados operators ${R}_{n}\left(f;q,z\right)$ are well defined, linear, and these operators are analytic for all $n\ge {n}_{0}$ and $|z|\le r<{\left[{n}_{0}\right]}_{q}^{1-\beta }$ since $|-\frac{1}{{a}_{n}}|\le |-\frac{1}{q{a}_{n}}|\le \cdots \le |-\frac{1}{{q}^{n-1}{a}_{n}}|$.

In this paper, we obtain the following results:

• the order of convergence for the operators ${R}_{n}\left(f;q,z\right)$,

• the Voronovskaja-type theorem with quantitative estimate,

• the exact degree of the approximation for the operators ${R}_{n}\left(f;q,z\right)$.

Throughout the paper, we denote with ${\parallel f\parallel }_{r}=max\left\{|f\left(z\right)|\in \mathbb{R}:z\in {\overline{D}}_{r}\right\}$ the norm of f in the space of continuous functions on ${\overline{D}}_{r}$ and with ${\parallel f\parallel }_{B\left[0,\mathrm{\infty }\right)}=sup\left\{|f\left(x\right)|\in \mathbb{R}:x\in \left[0,\mathrm{\infty }\right)\right\}$ the norm of f in the space of bounded functions on $\left[0,\mathrm{\infty }\right)$.

Also, the many results in this study are obtained under the condition that $f:{D}_{R}\cup \left[R,\mathrm{\infty }\right)\to \mathbb{C}$ is analytic in ${D}_{R}$ for $r, which assures the representation $f\left(z\right)={\sum }_{k=0}^{\mathrm{\infty }}{c}_{k}{z}^{k}$ for all $z\in {D}_{R}$.

## 2 Convergence results

The following lemmas will help in the proof of convergence results.

Lemma 1 Let ${n}_{0}\ge 2$, $0<\beta \le \frac{2}{3}$ and $\frac{1}{2}. Let us define ${\alpha }_{k,n,q}\left(z\right)={R}_{n}\left({e}_{k};q,z\right)$ for all $z\in {\overline{D}}_{r}$, where ${e}_{k}\left(z\right)={z}^{k}$. If $f:{D}_{R}\cup \left[R,\mathrm{\infty }\right)\to \mathbb{C}$ is uniformly continuous, bounded on $\left[0,\mathrm{\infty }\right)$ and analytic in ${D}_{R}$, then we have the form

${R}_{n}\left(f;q,z\right)=\sum _{k=0}^{\mathrm{\infty }}{c}_{k}{\alpha }_{k,n,q}\left(z\right)$

for all $z\in {\overline{D}}_{r}$.

Proof For any $m\in \mathbb{N}$, we define

From the hypothesis on f, it is clear that each ${f}_{m}$ is bounded on $\left[0,\mathrm{\infty }\right)$, that is, there exist $M\left({f}_{m}\right)>0$ with $|{f}_{m}\left(z\right)|\le M\left({f}_{m}\right)$, which implies that

$|{R}_{n}\left({f}_{m};q,z\right)|\le \frac{1}{|{\prod }_{s=0}^{n-1}\left(1+{q}^{s}{a}_{n}z\right)|}\sum _{j=0}^{n}{q}^{j\left(j-1\right)/2}M\left({f}_{m}\right){\left[\begin{array}{c}n\\ j\end{array}\right]}_{q}{\left({a}_{n}|z|\right)}^{j}<\mathrm{\infty },$

that is all ${R}_{n}\left({f}_{m};q,z\right)$ with $n\ge {n}_{0}$, $r<\frac{{\left[{n}_{0}\right]}_{q}^{1-\beta }}{2}$, $m\in \mathbb{N}$ are well defined for all $z\in {\overline{D}}_{r}$.

Defining

it is clear that each ${f}_{m,k}$ is bounded on $\left[0,\mathrm{\infty }\right)$ and that ${f}_{m}\left(z\right)={\sum }_{k=0}^{m}{f}_{m,k}\left(z\right)$.

From the linearity of ${R}_{n}\left(f;q,z\right)$, we have

It suffices to prove that

$\underset{m\to \mathrm{\infty }}{lim}{R}_{n}\left({f}_{m};q,z\right)={R}_{n}\left(f;q,z\right)$

for any fixed $n\in \mathbb{N}$, $n\ge {n}_{0}$ and $|z|\le r$.

We have the following inequality for all $|z|\le r$:

$|{R}_{n}\left({f}_{m};q,z\right)-{R}_{n}\left(f;q,z\right)|\le {M}_{r,n,q}{\parallel {f}_{m}-f\parallel }_{r},$
(1)

where ${M}_{r,n,q}={\prod }_{s=0}^{n-1}\frac{\left(1+{q}^{s}{a}_{n}r\right)}{\left(1-{q}^{s}{a}_{n}r\right)}$.

Using (1), ${lim}_{m\to \mathrm{\infty }}{\parallel {f}_{m}-f\parallel }_{r}=0$ and ${\parallel {f}_{m}-f\parallel }_{B\left[0,\mathrm{\infty }\right)}\le {\parallel {f}_{m}-f\parallel }_{r}$, the proof of the lemma is finished. □

Lemma 2 If we denote ${\left(\beta +z\right)}_{q}^{n}={\prod }_{s=0}^{n-1}\left(\beta +{q}^{s}z\right)$, then the following formula holds:

${D}_{q}\left[\frac{1}{{\left(\beta +z\right)}_{q}^{n}}\right]=-\frac{{\left[n\right]}_{q}}{{\left(\beta +z\right)}_{q}^{n+1}},$

where β is a fixed real number and $z\in \mathbb{C}$.

Proof We can write ${\left(\beta +z\right)}_{q}^{n}$ as follows:

${\left(\beta +z\right)}_{q}^{n}={q}^{n\left(n-1\right)/2}{\left(z+{q}^{-n+1}\beta \right)}_{q}^{n}.$
(2)

In [3] (see p.10, Proposition 3.3), we already have the following formula:

${D}_{q}\left[{\left(\beta +z\right)}_{q}^{n}\right]={\left[n\right]}_{q}{\left(\beta +z\right)}_{q}^{n-1}.$
(3)

Using (2) and (3), we get

$\begin{array}{rcl}{D}_{q}\left[{\left(\beta +z\right)}_{q}^{n}\right]& =& {q}^{n\left(n-1\right)/2}{\left[n\right]}_{q}{\left(z+{q}^{-n+1}\beta \right)}_{q}^{n-1}\\ =& {\left[n\right]}_{q}{q}^{n-1}{q}^{\left(n-1\right)\left(n-2\right)/2}{\left(z+{q}^{-n+2}\left({q}^{-1}\beta \right)\right)}_{q}^{n-1}\\ =& {\left[n\right]}_{q}{q}^{n-1}{\left({q}^{-1}\beta +z\right)}_{q}^{n-1}\\ =& {\left[n\right]}_{q}{\left(\beta +qz\right)}_{q}^{n-1}.\end{array}$
(4)

From (4), we obtain the result. □

Lemma 3 We have the following recurrence formula for the complex q-Balázs-Szabados operators ${R}_{n}\left(f;q,z\right)$:

${\alpha }_{k+1,n,q}\left(z\right)=\frac{\left(1+{q}^{n}{a}_{n}z\right)z}{\left(1+{a}_{n}z\right){b}_{n}}{D}_{q}\left[{\alpha }_{k,n,q}\left(z\right)\right]+\frac{z}{1+{a}_{n}z}{\alpha }_{k,n,q}\left(z\right),$

where ${\alpha }_{k,n,q}\left(z\right)={R}_{n}\left({e}_{k};q,z\right)$ for all $n\in \mathbb{N}$, $z\in \mathbb{C}$ and $k=0,1,2,\dots$ .

Proof Firstly, we calculate ${D}_{q}\left[{\alpha }_{k,n,q}\left(z\right)\right]$ as follows:

$\begin{array}{r}{D}_{q}\left[{\alpha }_{k,n,q}\left(z\right)\right]\\ \phantom{\rule{1em}{0ex}}={D}_{q}\left[\frac{1}{{\prod }_{s=0}^{n-1}\left(1+{q}^{s}{a}_{n}z\right)}\right]\sum _{j=0}^{n}{q}^{j\left(j-1\right)/2}{\left(\frac{{\left[j\right]}_{q}}{{b}_{n}}\right)}^{k}{\left[\begin{array}{c}n\\ j\end{array}\right]}_{q}{\left({a}_{n}z\right)}^{j}\\ \phantom{\rule{2em}{0ex}}+\frac{1}{{\prod }_{s=0}^{n-1}\left(1+{q}^{s+1}{a}_{n}z\right)}\sum _{j=0}^{n}{q}^{j\left(j-1\right)/2}{\left(\frac{{\left[j\right]}_{q}}{{b}_{n}}\right)}^{k}{\left[\begin{array}{c}n\\ j\end{array}\right]}_{q}{\left({a}_{n}\right)}^{j}{D}_{q}\left[{z}^{j}\right].\end{array}$
(5)

Considering Lemma 2 and using ${D}_{q}\left[{z}^{j}\right]={\left[j\right]}_{q}{z}^{j-1}$ in (5), we get

$\begin{array}{rcl}{D}_{q}\left[{\alpha }_{k,n,q}\left(z\right)\right]& =& -\frac{{b}_{n}}{1+{q}^{n}{a}_{n}z}\frac{1}{{\prod }_{s=0}^{n-1}\left(1+{q}^{s}{a}_{n}z\right)}{\alpha }_{k,n,q}\left(z\right)\\ +\frac{{b}_{n}\left(1+{a}_{n}z\right)}{z\left(1+{q}^{n}{a}_{n}z\right)}{\alpha }_{k+1,n,q}\left(z\right).\end{array}$
(6)

From (6), the proof of the lemma is finished. □

Corollary 1 ([11], p.143, Corollary 1.10.4)

Let $f\left(z\right)=\frac{{p}_{k}\left(z\right)}{{\prod }_{j=1}^{k}\left(z-{a}_{j}\right)}$, where ${p}_{k}\left(z\right)$ is a polynomial of degreek, and we suppose that $|{a}_{j}|\ge R>1$ for all $j=1,2,\dots ,k$. If $1\le r, then for all $|z|\le r$ we have

$|{f}^{\mathrm{\prime }}\left(z\right)|\le \frac{R+r}{R-r}\cdot \frac{k}{r}{\parallel f\parallel }_{r}.$

Under hypothesis of the corollary above, by the mean value theorem [12] in complex analysis, we have

$|{D}_{q}\left[f\left(z\right)\right]|\le \frac{R+r}{R-r}\cdot \frac{k}{r}{\parallel f\parallel }_{r}.$
(7)

Lemma 4 Let ${n}_{0}\ge 2$, $0<\beta \le \frac{2}{3}$ and $\frac{1}{2}. For all $n\ge {n}_{0}$, $|z|\le r$ and $k=0,1,2,\dots$ , we have

$|{\alpha }_{k,n,q}\left(z\right)|\le k!{\left(20r\right)}^{k}.$

Proof Taking the absolute value of the recurrence formula in Lemma 3 and using the triangle inequality, we get

$|{\alpha }_{k+1,n,q}\left(z\right)|\le \frac{\left(1+{q}^{n}{a}_{n}|z|\right)|z|}{|1-{a}_{n}|z||{b}_{n}}|{D}_{q}\left[{\alpha }_{k,n,q}\left(z\right)\right]|+\frac{|z|}{|1-{a}_{n}|z||}|{\alpha }_{k,n,q}\left(z\right)|.$
(8)

In order to get an upper estimate for $|{D}_{q}\left[{\alpha }_{k,n,q}\left(z\right)\right]|$, by using (7), we obtain

$|{D}_{q}\left[{\alpha }_{k,n,q}\left(z\right)\right]|\le \frac{{\left[{n}_{0}\right]}_{q}^{1-\beta }+r}{{\left[{n}_{0}\right]}_{q}^{1-\beta }-r}\cdot \frac{k}{r}{\parallel {\alpha }_{k,n,q}\parallel }_{r}.$

Under the condition $r<\frac{{\left[{n}_{0}\right]}_{q}^{1-\beta }}{2}$, it holds $\frac{{\left[{n}_{0}\right]}_{q}^{1-\beta }+r}{{\left[{n}_{0}\right]}_{q}^{1-\beta }-r}<3$, which implies

$|{D}_{q}\left[{\alpha }_{k,n,q}\left(z\right)\right]|\le \frac{3k}{r}{\parallel {\alpha }_{k,n,q}\parallel }_{r}.$
(9)

Applying (9) to (8) and passing to norm, we get

${\parallel {\alpha }_{k+1,n,q}\parallel }_{r}\le \frac{\left(1+{q}^{n}{a}_{n}r\right)3k}{\left(1-{a}_{n}r\right){b}_{n}}{\parallel {\alpha }_{k,n,q}\parallel }_{r}+\frac{r}{1-{a}_{n}r}{\parallel {\alpha }_{k,n,q}\parallel }_{r}.$

From the hypothesis of the lemma, we have $\frac{1}{1-{a}_{n}r}<2$, $1+{q}^{n}{a}_{n}r<\frac{3}{2}$, and $\frac{1}{{b}_{n}}<1$, which implies

${\parallel {\alpha }_{k+1,n,q}\parallel }_{r}\le 20r\left(k+1\right){\parallel {\alpha }_{k,n,q}\parallel }_{r}.$

Taking step by step $k=0,1,2,\dots$ , we obtain

${\parallel {\alpha }_{k+1,n,q}\parallel }_{r}\le {\left(20r\right)}^{k+1}\left(k+1\right)!.$

Using $|{\alpha }_{k+1,n,q}|\le {\parallel {\alpha }_{k+1,n,q}\parallel }_{r}$ and replacing $k+1$ with k, the proof of the lemma is finished. □

Let $q=\left\{{q}_{n}\right\}$ be a sequence satisfying the following conditions:

$\underset{n\to \mathrm{\infty }}{lim}{q}_{n}=1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty }}{lim}{q}_{n}^{n}=c\phantom{\rule{1em}{0ex}}\left(0\le c<1\right).$
(10)

Now we are in a position to prove the following convergence result.

Theorem 1 Let $\left\{{q}_{n}\right\}$ be a sequence satisfying the conditions (10) with ${q}_{n}\in \left(0,1\right]$ for all $n\in \mathbb{N}$, and let ${n}_{0}\ge 2$, $0<\beta \le \frac{2}{3}$ and $\frac{1}{2}. If $f:{D}_{R}\cup \left[R,\mathrm{\infty }\right)\to \mathbb{C}$ is uniformly continuous, bounded on $\left[0,\mathrm{\infty }\right)$ and analytic in ${D}_{R}$ and there exist $M>0$, $0 with $|{c}_{k}|\le M\frac{{A}^{k}}{k!}$ (which implies $|f\left(z\right)|\le M{e}^{A|z|}$ for all $z\in {D}_{R}$), then the sequence ${\left\{{R}_{n}\left(f;{q}_{n},z\right)\right\}}_{n\ge {n}_{0}}$ is uniformly convergent to f in ${\overline{D}}_{r}$.

Proof From Lemma 2 and Lemma 6, for all $n\ge {n}_{0}$ and $|z|\le r$, we have

$|{R}_{n}\left(f;{q}_{n},z\right)|\le \sum _{k=0}^{\mathrm{\infty }}|{c}_{k}||{\alpha }_{k,n,{q}_{n}}\left(z\right)|\le \sum _{k=0}^{\mathrm{\infty }}M\frac{{A}^{k}}{k!}k!{\left(20r\right)}^{k}=M\sum _{k=0}^{\mathrm{\infty }}{\left(20Ar\right)}^{k},$

where the series ${\sum }_{k=0}^{\mathrm{\infty }}{\left(20Ar\right)}^{k}$ is convergent for $0.

Since ${lim}_{n\to \mathrm{\infty }}{R}_{n}\left(f;{q}_{n},x\right)=f\left(x\right)$ for all $x\in \left[0,r\right]$ (see [10]), by Vitali’s theorem (see [13], p.112, Theorem 3.2.10), it follows that $\left\{{R}_{n}\left(f;{q}_{n},z\right)\right\}$ uniformly converges to $f\left(z\right)$ in ${\overline{D}}_{r}$. □

We can give the following upper estimate in the approximation of ${R}_{n}\left(f;{q}_{n},z\right)$.

Theorem 2 Let $\left\{{q}_{n}\right\}$ be a sequence satisfying the conditions (10) with ${q}_{n}\in \left(0,1\right]$ for all $n\in \mathbb{N}$, and let ${n}_{0}\ge 2$, $0<\beta \le \frac{2}{3}$ and $\frac{1}{2}. If $f:{D}_{R}\cup \left[R,\mathrm{\infty }\right)\to \mathbb{C}$ is uniformly continuous, bounded on $\left[0,\mathrm{\infty }\right)$ and analytic in ${D}_{R}$ and there exist $M>0$, $0 with $|{c}_{k}|\le M\frac{{A}^{k}}{k!}$ (which implies $|f\left(z\right)|\le M{e}^{A|z|}$ for all $z\in {D}_{R}$), then the following upper estimate holds:

$|{R}_{n}\left(f;{q}_{n},z\right)-f\left(z\right)|\le {C}_{r}^{1}\left(f\right)\left({a}_{n}+\frac{1}{{b}_{n}}\right),$

where ${C}_{r}^{1}\left(f\right)=max\left\{9MA{\sum }_{k=1}^{\mathrm{\infty }}\left(k-1\right){\left(20Ar\right)}^{k-1},2{r}^{2}MA{e}^{2Ar}\right\}$ and ${\sum }_{k=1}^{\mathrm{\infty }}\left(k-1\right){\left(20Ar\right)}^{k-1}<\mathrm{\infty }$.

Proof Using the recurrence formula in Lemma 4, we have

$\begin{array}{rcl}|{\alpha }_{k+1,n,{q}_{n}}\left(z\right)-{z}^{k+1}|& \le & \frac{\left(1+{q}_{n}^{n}{a}_{n}|z|\right)|z|}{|1-{a}_{n}|z||{b}_{n}}|{D}_{{q}_{n}}\left[{\alpha }_{k,n,{q}_{n}}\left(z\right)-{z}^{k}\right]|\\ +\frac{|z|}{|1-{a}_{n}|z||}|{\alpha }_{k,n,{q}_{n}}\left(z\right)-{z}^{k}|+\frac{1}{{b}_{n}}\frac{\left(1+{q}_{n}^{n}{a}_{n}|z|\right)}{|1-{a}_{n}|z||}{\left[k\right]}_{{q}_{n}}|z{|}^{k}\\ +\frac{{a}_{n}}{|1-{a}_{n}|z||}|z{|}^{k+2}.\end{array}$

For $|z|\le r$, we get

$\begin{array}{rcl}|{\alpha }_{k+1,n,{q}_{n}}\left(z\right)-{z}^{k+1}|& \le & \frac{\left(1+{q}_{n}^{n}{a}_{n}r\right)r}{\left(1-{a}_{n}r\right){b}_{n}}|{D}_{{q}_{n}}\left[{\alpha }_{k,n,{q}_{n}}\left(z\right)\right]|+\frac{r}{1-{a}_{n}r}|{\alpha }_{k,n,{q}_{n}}\left(z\right)-{z}^{k}|\\ +\frac{2}{{b}_{n}}\frac{\left(1+{q}_{n}^{n}{a}_{n}r\right)}{\left(1-{a}_{n}r\right)}{\left[k\right]}_{{q}_{n}}{r}^{k}+\frac{{a}_{n}}{1-{a}_{n}r}{r}^{k+2}.\end{array}$

Using (9), $\frac{1}{1-{a}_{n}r}<2$, and $1+{q}_{n}^{n}{a}_{n}r<\frac{3}{2}$, we obtain

$|{\alpha }_{k+1,n,{q}_{n}}\left(z\right)-{z}^{k+1}|\le \frac{9k\cdot k!}{{b}_{n}}{\left(20r\right)}^{k}+2r|{\alpha }_{k,n,{q}_{n}}\left(z\right)-{z}^{k}|+\frac{6}{{b}_{n}}{\left[k\right]}_{{q}_{n}}{r}^{k}+2{a}_{n}{r}^{k+2}.$

Since $6{\left[k\right]}_{{q}_{n}}{r}^{k}\le 9k\cdot k!{\left(20r\right)}^{k}$ for all $k=0,1,2,\dots$ , we can write

$|{\alpha }_{k+1,n,{q}_{n}}\left(z\right)-{z}^{k+1}|\le \frac{18k\cdot k!}{{b}_{n}}{\left(20r\right)}^{k}+2r|{\alpha }_{k,n,{q}_{n}}\left(z\right)-{z}^{k}|+2{a}_{n}{r}^{k+2}.$

Taking $k=0,1,2,\dots$ step by step, finally we arrive at

$|{\alpha }_{k,n,{q}_{n}}\left(z\right)-{z}^{k}|\le \frac{9}{{b}_{n}}\left(k-1\right)k!{\left(20r\right)}^{k-1}+2{a}_{n}{r}^{2}k{\left(2r\right)}^{k-1},$
(11)

which implies

$\begin{array}{rcl}|{R}_{n}\left(f;{q}_{n},z\right)-f\left(z\right)|& \le & \sum _{k=1}^{\mathrm{\infty }}|{c}_{k}||{\alpha }_{k,n,{q}_{n}}\left(z\right)-{z}^{k}|\\ \le & \sum _{k=1}^{\mathrm{\infty }}M\frac{{A}^{k}}{k!}\left\{\frac{9}{{b}_{n}}\left(k-1\right)k!{\left(20r\right)}^{k-1}+2{a}_{n}{r}^{2}k{\left(2r\right)}^{k-1}\right\}\\ =& \frac{9MA}{{b}_{n}}\sum _{k=1}^{\mathrm{\infty }}\left(k-1\right){\left(20Ar\right)}^{k-1}+2{a}_{n}{r}^{2}MA\sum _{k=1}^{\mathrm{\infty }}\frac{{\left(20Ar\right)}^{k-1}}{\left(k-1\right)!}\\ =& \frac{9MA}{{b}_{n}}\sum _{k=1}^{\mathrm{\infty }}\left(k-1\right){\left(20Ar\right)}^{k-1}+2{a}_{n}{r}^{2}MA{e}^{2Ar}.\end{array}$

Choosing ${C}_{r}^{1}\left(f\right)=max\left\{9MA{\sum }_{k=1}^{\mathrm{\infty }}\left(k-1\right){\left(20Ar\right)}^{k-1},2{r}^{2}MA{e}^{2Ar}\right\}$, we obtain the desired result.

Here the series ${\sum }_{k=0}^{\mathrm{\infty }}{\left(20Ar\right)}^{k}$ is convergent for $0 and the series is absolutely convergent in ${\overline{D}}_{r}$, it easily follows that ${\sum }_{k=1}^{\mathrm{\infty }}\left(k-1\right){\left(20Ar\right)}^{k-1}<\mathrm{\infty }$. □

The following lemmas will help in the proof of the next theorem.

Lemma 5 For all $n\in \mathbb{N}$, we have

${R}_{n}\left({e}_{0};q,z\right)=1,$
(12)
${R}_{n}\left({e}_{1};q,z\right)=\frac{z}{1+{a}_{n}z},$
(13)
${R}_{n}\left({e}_{2};q,z\right)=\frac{\left(1-\frac{{a}_{n}}{{b}_{n}}\right)q{z}^{2}}{\left(1+{a}_{n}z\right)\left(1+{a}_{n}qz\right)}+\frac{z}{{b}_{n}\left(1+{a}_{n}z\right)},$
(14)

where ${e}_{k}\left(z\right)={z}^{k}$ for $k=0,1,2$.

Proof (12) and (13) are obtained simply replacing x by z in Lemma 3.1 and Lemma 3.2 in [10]. Also, using ${\left[n\right]}_{q}=1+q{\left[n-1\right]}_{q}$ and $\frac{{a}_{n}}{{b}_{n}}=\frac{1}{{\left[n\right]}_{q}}$ and replacing x by z in Lemma 3.3 in [10], (14) is obtained. □

Lemma 6 For all $n\in \mathbb{N}$, the following equalities for the operators ${R}_{n}\left(f;q,z\right)$ hold:

${\psi }_{n,q}^{1}\left(z\right)=\frac{-{a}_{n}{z}^{2}}{1+{a}_{n}z},$
(15)
$\begin{array}{l}{\psi }_{n,q}^{2}\left(z\right)=\frac{z}{{b}_{n}\left(1+{a}_{n}z\right)\left(1+{a}_{n}qz\right)}-\frac{\left(1-q\right){z}^{2}}{\left(1+{a}_{n}z\right)\left(1+{a}_{n}qz\right)}\\ \phantom{{\psi }_{n,q}^{2}\left(z\right)=}-\frac{{a}_{n}\left(1-q\right){z}^{3}}{\left(1+{a}_{n}z\right)\left(1+{a}_{n}qz\right)}+\frac{{a}_{n}^{2}q{z}^{4}}{\left(1+{a}_{n}z\right)\left(1+{a}_{n}qz\right)},\end{array}$
(16)

where ${\psi }_{n,q}^{i}\left(z\right)={R}_{n}\left({\left(t-{e}_{1}\right)}^{i};q,z\right)$ for $i=1,2$.

Proof From Lemma 5, the proof can be easily got, so we omit it. □

Now, we present a quantitative Voronovskaja-type formula.

Let us define

${A}_{k,n,{q}_{n}}\left(z\right)={R}_{n}\left(f;{q}_{n},z\right)-f\left(z\right)-{\psi }_{n,q}^{1}\left(z\right){f}^{\mathrm{\prime }}\left(z\right)-\frac{1}{2}{\psi }_{n,q}^{2}\left(z\right){f}^{\mathrm{\prime }\mathrm{\prime }}\left(z\right).$
(17)

Theorem 3 Let $\left\{{q}_{n}\right\}$ be a sequence satisfying the conditions (10) with ${q}_{n}\in \left(0,1\right]$ for all $n\in \mathbb{N}$, ${n}_{0}\ge 2$, $0<\beta \le \frac{2}{3}$ and $\frac{1}{2}. If $f:{D}_{R}\cup \left[R,\mathrm{\infty }\right)\to \mathbb{C}$ is uniformly continuous, bounded on $\left[0,\mathrm{\infty }\right)$ and analytic in ${D}_{R}$ and there exist $M>0$, $0 with $|{c}_{k}|\le M\frac{{A}^{k}}{k!}$ (which implies $|f\left(z\right)|\le M{e}^{A|z|}$ for all $z\in {D}_{R}$), then for all $n\ge {n}_{0}$ and $|z|\le r$, we have

$|{A}_{k,n,{q}_{n}}\left(z\right)|\le {C}_{r}^{2}\left(f\right){\left({a}_{n}+\frac{1}{{b}_{n}}\right)}^{2},$

where ${C}_{r}^{2}\left(f\right)={C}_{\ast }M{r}^{3}{\sum }_{k=3}^{\mathrm{\infty }}\left(k-2\right)\left(k-1\right)k\left(k+1\right){\left(20rA\right)}^{k-3}<\mathrm{\infty }$ and ${C}_{\ast }$ is a fixed real number.

Proof From Lemma 1 and the analyticity of f, we can write

$|{A}_{k,n,{q}_{n}}\left(z\right)|\le \sum _{k=2}^{\mathrm{\infty }}|{c}_{k}||{E}_{k,n,{q}_{n}}\left(z\right)|,$
(18)

where

$\begin{array}{rcl}{E}_{k,n,{q}_{n}}\left(z\right)& =& {\alpha }_{k,n,{q}_{n}}\left(z\right)-{z}^{k}+\frac{{a}_{n}k{z}^{k+1}}{1+{a}_{n}z}-\frac{\left(k-1\right)k{z}^{k-1}}{2{b}_{n}\left(1+{a}_{n}z\right)\left(1+{a}_{n}{q}_{n}z\right)}\\ +\frac{\left(1-{q}_{n}\right)\left(k-1\right)k{z}^{k}}{2\left(1+{a}_{n}z\right)\left(1+{a}_{n}{q}_{n}z\right)}+\frac{{a}_{n}\left(1-{q}_{n}\right)\left(k-1\right)k{z}^{k+1}}{2\left(1+{a}_{n}z\right)\left(1+{a}_{n}{q}_{n}z\right)}\\ -\frac{{a}_{n}^{2}{q}_{n}\left(k-1\right)k{z}^{k+2}}{2\left(1+{a}_{n}z\right)\left(1+{a}_{n}{q}_{n}z\right)}.\end{array}$
(19)

Using Lemma 5, we easily obtain that ${E}_{0,n,q}\left(z\right)={E}_{1,n,q}\left(z\right)={E}_{2,n,q}\left(z\right)=0$.

Combining (19) with the recurrence formula in Lemma 3, a simple calculation leads us to the following recurrence formula:

${E}_{k+1,n,{q}_{n}}\left(z\right)=\frac{\left(1+{q}_{n}^{n}{a}_{n}z\right)z}{{b}_{n}\left(1+{a}_{n}z\right)}{D}_{{q}_{n}}\left[{E}_{k,n,{q}_{n}}\left(z\right)\right]+\frac{z}{1+{a}_{n}z}{E}_{k,n,{q}_{n}}\left(z\right)+{F}_{k,n,{q}_{n}}\left(z\right),$
(20)

where

$\begin{array}{rcl}{F}_{k,n,{q}_{n}}\left(z\right)& =& -\frac{\left(k-{\left[k\right]}_{{q}_{n}}\right){z}^{k}}{{b}_{n}{\left(1+{a}_{n}z\right)}^{2}\left(1+{a}_{n}{q}_{n}z\right)}+\frac{{a}_{n}^{2}k{z}^{k+3}}{{\left(1+{a}_{n}z\right)}^{2}}-\frac{\left(1-{q}_{n}\right)k{z}^{k+1}}{{\left(1+{a}_{n}z\right)}^{2}\left(1+{a}_{n}{q}_{n}z\right)}\\ +\frac{{a}_{n}\left(1-{q}_{n}\right)k{z}^{k+2}}{{\left(1+{a}_{n}z\right)}^{2}\left(1+{a}_{n}{q}_{n}z\right)}-\frac{{a}_{n}^{2}{q}_{n}k{z}^{k+3}}{{\left(1+{a}_{n}z\right)}^{2}\left(1+{a}_{n}{q}_{n}z\right)}\\ -\frac{{a}_{n}k\left(k+1\right){z}^{k+1}}{2{b}_{n}{\left(1+{a}_{n}z\right)}^{2}\left(1+{a}_{n}{q}_{n}z\right)}+\frac{{a}_{n}\left(1-{q}_{n}\right)k\left(k+1\right){z}^{k+2}}{2{\left(1+{a}_{n}z\right)}^{2}\left(1+{a}_{n}{q}_{n}z\right)}\\ +\frac{{a}_{n}^{2}\left(1-{q}_{n}\right)k\left(k+1\right){z}^{k+3}}{2{\left(1+{a}_{n}z\right)}^{2}\left(1+{a}_{n}{q}_{n}z\right)}-\frac{{a}_{n}^{3}{q}_{n}k\left(k+1\right){z}^{k+4}}{2{\left(1+{a}_{n}z\right)}^{2}\left(1+{a}_{n}{q}_{n}z\right)}\\ -\frac{{a}_{n}\left(1+{q}_{n}^{n}{a}_{n}z\right)\left(\left(k-1\right){\left[k+1\right]}_{{q}_{n}}-{q}_{n}{\left[k-1\right]}_{{q}_{n}}\right){z}^{k+1}}{{b}_{n}{\left(1+{a}_{n}z\right)}^{2}\left(1+{a}_{n}{q}_{n}z\right)}\\ -\frac{{a}_{n}^{2}\left(1+{q}_{n}^{n}{a}_{n}z\right)\left(k-1\right){q}_{n}{\left[k\right]}_{{q}_{n}}{z}^{k+2}}{{b}_{n}{\left(1+{a}_{n}z\right)}^{2}\left(1+{a}_{n}{q}_{n}z\right)}+\frac{{a}_{n}{q}_{n}^{n}{\left[k\right]}_{{q}_{n}}{z}^{k+1}}{{b}_{n}{\left(1+{a}_{n}z\right)}^{2}\left(1+{a}_{n}{q}_{n}z\right)}\\ -\frac{\left(1+{q}_{n}^{n}{a}_{n}z\right){\left[k-1\right]}_{{q}_{n}}\left(k-1\right)k{z}^{k-1}}{2{b}_{n}^{2}\left(1+{a}_{n}z\right)\left(1+{a}_{n}{q}_{n}z\right)\left(1+{a}_{n}{q}_{n}^{2}z\right)}\\ +\frac{\left(1-{q}_{n}\right)\left(1+{q}_{n}^{n}{a}_{n}z\right){\left[k\right]}_{{q}_{n}}\left(k-1\right)k{z}^{k}}{2{b}_{n}\left(1+{a}_{n}z\right)\left(1+{a}_{n}{q}_{n}z\right)\left(1+{a}_{n}{q}_{n}^{2}z\right)}\\ +\frac{{a}_{n}\left(1-{q}_{n}\right)\left(1+{q}_{n}^{n}{a}_{n}z\right){\left[k+1\right]}_{{q}_{n}}\left(k-1\right)k{z}^{k+1}}{2{b}_{n}\left(1+{a}_{n}z\right)\left(1+{a}_{n}{q}_{n}z\right)\left(1+{a}_{n}{q}_{n}^{2}z\right)}\\ -\frac{{a}_{n}^{2}{q}_{n}\left(1+{q}_{n}^{n}{a}_{n}z\right){\left[k+2\right]}_{{q}_{n}}\left(k-1\right)k{z}^{k+2}}{2{b}_{n}\left(1+{a}_{n}z\right)\left(1+{a}_{n}{q}_{n}z\right)\left(1+{a}_{n}{q}_{n}^{2}z\right)}\\ +\frac{{a}_{n}\left(1+{q}_{n}^{n}{a}_{n}z\right)\left(1+{q}_{n}\right)\left(k-1\right)k{z}^{k}}{2{b}_{n}^{2}{\left(1+{a}_{n}z\right)}^{2}\left(1+{a}_{n}{q}_{n}z\right)\left(1+{a}_{n}{q}_{n}^{2}z\right)}\\ -\frac{{a}_{n}\left(1-{q}_{n}\right)\left(1+{q}_{n}\right)\left(1+{q}_{n}^{n}{a}_{n}z\right)\left(k-1\right)k{z}^{k+1}}{2{b}_{n}{\left(1+{a}_{n}z\right)}^{2}\left(1+{a}_{n}{q}_{n}z\right)\left(1+{a}_{n}{q}_{n}^{2}z\right)}\\ -\frac{{a}_{n}^{2}\left(1-{q}_{n}\right)\left(1+{q}_{n}\right)\left(1+{q}_{n}^{n}{a}_{n}z\right)\left(k-1\right)k{z}^{k+2}}{2{b}_{n}{\left(1+{a}_{n}z\right)}^{2}\left(1+{a}_{n}{q}_{n}z\right)\left(1+{a}_{n}{q}_{n}^{2}z\right)}\\ -\frac{{a}_{n}^{3}{q}_{n}\left(1+{q}_{n}\right)\left(1+{q}_{n}^{n}{a}_{n}z\right)\left(k-1\right)k{z}^{k+3}}{2{b}_{n}{\left(1+{a}_{n}z\right)}^{2}\left(1+{a}_{n}{q}_{n}z\right)\left(1+{a}_{n}{q}_{n}^{2}z\right)}.\end{array}$

In the following results, ${C}_{i}$ will denote fixed real numbers for $i=1,2,3$.

Under the hypothesis of Theorem 3, we have

(21)
${a}_{n}r<\frac{1}{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}1+{q}_{n}^{n}{a}_{n}r<\frac{3}{2},$
(22)
$1-{q}_{n}\le \frac{{a}_{n}}{{b}_{n}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}k-{\left[k\right]}_{{q}_{n}}\le \frac{{a}_{n}}{{b}_{n}}\frac{\left(k-1\right)k}{2},$
(23)
${\left[k\right]}_{{q}_{n}}\le k,\phantom{\rule{2em}{0ex}}{a}_{n}<1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{1}{{b}_{n}}<1.$
(24)

Using (21)-(24), for $|z|\le r$, we get

$\begin{array}{rcl}|{F}_{k,n,{q}_{n}}\left(z\right)|& \le & {C}_{1}\left({a}_{n}^{2}+\frac{{a}_{n}}{{b}_{n}}+\frac{1}{{b}_{n}^{2}}\right)k\left(k+1\right)\left(k+2\right)\\ ×max\left\{{r}^{k-1},{r}^{k},{r}^{k+1},{r}^{k+2},{r}^{k+3},{r}^{k+4}\right\}\\ \le & {C}_{1}{\left({a}_{n}+\frac{1}{{b}_{n}}\right)}^{2}k\left(k+1\right)\left(k+2\right){\left(2r\right)}^{k+4}.\end{array}$
(25)

On the other hand, for $|z|\le r$, we have

$\begin{array}{rcl}|\frac{\left(1+{q}_{n}^{n}{a}_{n}z\right)z}{{b}_{n}\left(1+{a}_{n}z\right)}{D}_{{q}_{n}}\left[{E}_{k,n,{q}_{n}}\left(z\right)\right]|& \le & \frac{\left(1+{q}_{n}^{n}{a}_{n}r\right)r}{{b}_{n}\left(1-{a}_{n}r\right)}\frac{3k}{r}{\parallel {E}_{k,n,{q}_{n}}\parallel }_{r}\\ \le & \frac{3k\left(1+{q}_{n}^{n}{a}_{n}r\right)}{{b}_{n}\left(1-{a}_{n}r\right)}\left\{{\parallel {\alpha }_{k,n,{q}_{n}}-{e}_{k}\parallel }_{r}\\ +\frac{{a}_{n}k{r}^{k+1}}{1-{a}_{n}r}+\frac{\left(k-1\right)k{r}^{k-1}}{2{b}_{n}\left(1-{a}_{n}r\right)\left(1-{a}_{n}{q}_{n}r\right)}\\ +\frac{\left(1-{q}_{n}\right)\left(k-1\right)k{r}^{k}}{2\left(1-{a}_{n}r\right)\left(1-{a}_{n}{q}_{n}r\right)}\\ +\frac{{a}_{n}\left(1-{q}_{n}\right)\left(k-1\right)k{r}^{k+1}}{2\left(1-{a}_{n}r\right)\left(1-{a}_{n}{q}_{n}r\right)}+\frac{{a}_{n}^{2}{q}_{n}\left(k-1\right)k{r}^{k+2}}{2\left(1-{a}_{n}r\right)\left(1-{a}_{n}{q}_{n}r\right)}\right\}.\end{array}$

Taking into account (11) in the proof of Theorem 2, we obtain

$\begin{array}{rcl}|\frac{\left(1+{q}_{n}^{n}{a}_{n}z\right)z}{{b}_{n}\left(1+{a}_{n}z\right)}{D}_{{q}_{n}}\left[{E}_{k,n,{q}_{n}}\left(z\right)\right]|& \le & {C}_{2}\frac{1}{{b}_{n}}\left({a}_{n}+\frac{1}{{b}_{n}}\right)\left(k-1\right)k\left(k+1\right)\\ ×\left(k!\right){\left(20r\right)}^{k+2}\\ \le & {C}_{2}{\left({a}_{n}+\frac{1}{{b}_{n}}\right)}^{2}\left(k-1\right)k\left(k+1\right)\left(k!\right){\left(20r\right)}^{k+2}.\end{array}$
(26)

Considering (25) and (26) in (20), we get

$|{E}_{k+1,n,{q}_{n}}\left(z\right)|\le 2r|{E}_{k,n,{q}_{n}}\left(z\right)|+{C}_{3}{\left({a}_{n}+\frac{1}{{b}_{n}}\right)}^{2}k\left(k+1\right)\left(k+2\right)\left(k+1\right)!{\left(20r\right)}^{k+4}.$

Since ${E}_{0,n,q}\left(z\right)={E}_{1,n,q}\left(z\right)={E}_{2,n,q}\left(z\right)=0$, taking $k=2,3,4,\dots$ in the last inequality step by step, finally we arrive at

$|{E}_{k,n,{q}_{n}}\left(z\right)|\le {C}_{3}{\left({a}_{n}+\frac{1}{{b}_{n}}\right)}^{2}\left(k-2\right)\left(k-1\right)k\left(k+1\right)\left(k!\right){\left(20r\right)}^{k+3}.$
(27)

Finally, considering (27) in (18) and using $20rA<1$, the proof of the theorem is complete. □

Remark 1 For $0, since $\frac{1}{{\left[n\right]}_{q}}\to 1-q$ as $n\to \mathrm{\infty }$, therefore ${a}_{n}={\left(\frac{1}{{\left[n\right]}_{q}}\right)}^{1-\beta }\to {\left(1-q\right)}^{1-\beta }$ and $\frac{1}{{b}_{n}}={\left(\frac{1}{{\left[n\right]}_{q}}\right)}^{\beta }\to {\left(1-q\right)}^{\beta }$ as $n\to \mathrm{\infty }$. If a sequence $\left\{{q}_{n}\right\}$ satisfies the conditions (10), then $\frac{1}{{\left[n\right]}_{q}}\to 0$ as $n\to \mathrm{\infty }$; therefore ${a}_{n}={\left(\frac{1}{{\left[n\right]}_{q}}\right)}^{1-\beta }\to 0$ and $\frac{1}{{b}_{n}}={\left(\frac{1}{{\left[n\right]}_{q}}\right)}^{\beta }\to 0$ as $n\to \mathrm{\infty }$.

Under the conditions (10), Theorem 2 and Theorem 3 show that ${\left\{{R}_{n}\left(f;{q}_{n},z\right)\right\}}_{n\ge {n}_{0}}$ uniformly converges to $f\left(z\right)$ in ${\overline{D}}_{r}$.

From Theorem 2 and Theorem 3, we get the following consequence.

Theorem 4 Let $\left\{{q}_{n}\right\}$ be a sequence satisfying the conditions (10) with ${q}_{n}\in \left(0,1\right]$ for all $n\in \mathbb{N}$, ${n}_{0}\ge 2$, $0<\beta \le \frac{2}{3}$, $\beta \ne \frac{1}{2}$ and $\frac{1}{2}. Suppose that $f:{D}_{R}\cup \left[R,\mathrm{\infty }\right)\to \mathbb{C}$ is uniformly continuous, bounded on $\left[0,\mathrm{\infty }\right)$ and analytic in ${D}_{R}$ and there exist $M>0$, $0 with $|{c}_{k}|\le M\frac{{A}^{k}}{k!}$ (which implies $|f\left(z\right)|\le M{e}^{A|z|}$ for all $z\in {D}_{R}$). If f is not a polynomial of degree ≤1, then for all $n\ge {n}_{0}$ we have

${\parallel {R}_{n}\left(f;{q}_{n},\cdot \right)-f\parallel }_{r}\sim \left({a}_{n}+\frac{1}{{b}_{n}}\right).$

Proof We can write

${R}_{n}\left(f;{q}_{n},z\right)-f\left(z\right)=\left({a}_{n}+\frac{1}{{b}_{n}}\right)\left\{G\left(z\right)+{H}_{n}\left(z\right)\right\},$
(28)

where

$\begin{array}{rcl}G\left(z\right)& =& -\frac{{a}_{n}}{{a}_{n}+1/{b}_{n}}\frac{{z}^{2}{f}^{\mathrm{\prime }}\left(z\right)}{1+{a}_{n}z}\\ +\frac{1}{{a}_{n}{b}_{n}+1}\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}\left(z\right)}{2\left(1+{a}_{n}z\right)\left(1+{a}_{n}{q}_{n}z\right)}\\ -\frac{1-{q}_{n}}{{a}_{n}+1/{b}_{n}}\frac{{z}^{2}{f}^{\mathrm{\prime }\mathrm{\prime }}\left(z\right)}{2\left(1+{a}_{n}z\right)\left(1+{a}_{n}{q}_{n}z\right)}\\ -\frac{{a}_{n}\left(1-{q}_{n}\right)}{{a}_{n}+1/{b}_{n}}\frac{{z}^{3}{f}^{\mathrm{\prime }\mathrm{\prime }}\left(z\right)}{2\left(1+{a}_{n}z\right)\left(1+{a}_{n}{q}_{n}z\right)}\\ +\frac{{a}_{n}^{2}}{{a}_{n}+1/{b}_{n}}\frac{{q}_{n}{z}^{4}{f}^{\mathrm{\prime }\mathrm{\prime }}\left(z\right)}{2\left(1+{a}_{n}z\right)\left(1+{a}_{n}{q}_{n}z\right)}\end{array}$
(29)

and

${H}_{n}\left(z\right)=\left({a}_{n}+\frac{1}{{b}_{n}}\right)\left[\frac{1}{{\left({a}_{n}+\frac{1}{{b}_{n}}\right)}^{2}}{A}_{k,n,{q}_{n}}\left(z\right)\right],$
(30)

and also ${\left({H}_{n}\left(z\right)\right)}_{n\in \mathbb{N}}$ is a sequence of analytic functions uniformly convergent to zero for all $|z|\le r$.

Since ${a}_{n}+\frac{1}{{b}_{n}}\to 0$ as $n\to \mathrm{\infty }$, and taking into account Theorem 3, it remains only to show that for sufficiently large n and for all $|z|\le r$, we have $|G\left(z\right)|>\rho >0$, where ρ is independent of n.

If $2\beta -1<0$, then the term $\frac{1}{{a}_{n}{b}_{n}+1}\to 1$ as $n\to \mathrm{\infty }$, while the other terms converge to zero, so there exists a natural number ${n}_{1}\in \mathbb{N}$ with ${n}_{1}\ge {n}_{0}$ so that for all $n\ge {n}_{1}$ and $|z|\le r$, we have

$|G\left(z\right)|\ge \frac{1}{2}|\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}\left(z\right)}{2\left(1+{a}_{n}z\right)\left(1+{a}_{n}{q}_{n}z\right)}|\ge \frac{1}{4}\frac{|z{f}^{\mathrm{\prime }\mathrm{\prime }}\left(z\right)|}{{\left(1+r\right)}^{2}}.$
(31)

If $2\beta -1>0$, then the term $\frac{{a}_{n}}{{a}_{n}+1/{b}_{n}}\to 1$ as $n\to \mathrm{\infty }$, while the other terms converge to zero. So, there exists a natural number ${n}_{2}\in \mathbb{N}$ with ${n}_{2}\ge {n}_{0}$ so that for all $n\ge {n}_{2}$ and $|z|\le r$, we have

$|G\left(z\right)|\ge \frac{1}{2}|\frac{{z}^{2}{f}^{\mathrm{\prime }}\left(z\right)}{1+{a}_{n}z}|\ge \frac{1}{2}\frac{|{z}^{2}{f}^{\mathrm{\prime }}\left(z\right)|}{1+r}.$
(32)

In the case of $2\beta -1=0$, that is, $\beta =\frac{1}{2}$, we obtain$\frac{{a}_{n}^{2}}{{a}_{n}+1/{b}_{n}}={\left[n\right]}_{{q}_{n}}^{1/2}\to \mathrm{\infty }$, as $n\to \mathrm{\infty }$, so that the case $\beta =\frac{1}{2}$ remains unsettled.

Choosing ${n}_{3}=max\left\{{n}_{1},{n}_{2}\right\}$, considering (31) and (32), for all $n\ge {n}_{3}$, we get

${\parallel {R}_{n}\left(f;{q}_{n},\cdot \right)-f\parallel }_{r}\ge \left({a}_{n}+\frac{1}{{b}_{n}}\right)|{\parallel G\parallel }_{r}-{\parallel {H}_{n}\parallel }_{r}|\ge \left({a}_{n}+\frac{1}{{b}_{n}}\right)\frac{1}{2}{\parallel G\parallel }_{r}.$

For all $n\in \left\{{n}_{0},\dots ,{n}_{3}-1\right\}$, we get

${\parallel {R}_{n}\left(f;{q}_{n},\cdot \right)-f\parallel }_{r}\ge \left({a}_{n}+\frac{1}{{b}_{n}}\right){M}_{r,n,{q}_{n}}\left(z\right)$

with ${M}_{r,n,{q}_{n}}\left(z\right)=\frac{1}{{a}_{n}+1/{b}_{n}}{\parallel {R}_{n}\left(f;{q}_{n},\cdot \right)-f\parallel }_{r}>0$, which finally implies

${\parallel {R}_{n}\left(f;{q}_{n},\cdot \right)-f\parallel }_{r}\ge \left({a}_{n}+\frac{1}{{b}_{n}}\right){C}_{r}\left(f\right)$
(33)

for all $n\ge {n}_{0}$, with ${C}_{r}\left(f\right)=min\left\{{M}_{r,{n}_{0},{q}_{n}}\left(z\right),\dots ,{M}_{r,{n}_{3}-1,{q}_{n}}\left(z\right),\frac{1}{2}{\parallel G\parallel }_{r}\right\}$.

From (33) and Theorem 3, the proof is complete. □

Remark 2 Recently, it is much more interesting to study these operators in the case $q>1$. Authors continue to study that case.

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## Acknowledgements

The authors are grateful to the editor and the reviewers for making valuable suggestions, leading to a better presentation of the work.

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Correspondence to Esma Yıldız Özkan.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

The main idea of this paper is proposed by NI. All authors contributed equally in writing this article and read and approved the final manuscript.

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İspir, N., Yıldız Özkan, E. Approximation properties of complex q-Balázs-Szabados operators in compact disks. J Inequal Appl 2013, 361 (2013). https://doi.org/10.1186/1029-242X-2013-361