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 [1–3]. 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 [5–7] 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:

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

where $D_R=\{z\in \mathbb{C}:|z|<R\}$ with $R>\frac{1}{2}$, $f:D_R\cup [R,\mathrm{\infty })\to \mathbb{C}$ 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 $R_n(f;z)$ to $f(z)$ 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

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

where $f:D_R\cup [R,\mathrm{\infty })\to \mathbb{C}$ is uniformly continuous and bounded on $[0,\mathrm{\infty })$, $a_n={[n]}_q^{\beta -1}$, $b_n={[n]}_q^{\beta }$, $q\in (0,1]$, $0<\beta \le \frac{2}{3}$, $n\in \mathbb{N}$, $z\in \mathbb{C}$ 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 $R_n(f;q,z)$ are well defined, linear, and these operators are analytic for all $n\ge n_0$ and $|z|\le r<{[n_0]}_q^{1-\beta }$ 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 $R_n(f;q,z)$,

• the Voronovskaja-type theorem with quantitative estimate,

• the exact degree of the approximation for the operators $R_n(f;q,z)$.

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

Also, the many results in this study are obtained under the condition that $f:D_R\cup [R,\mathrm{\infty })\to \mathbb{C}$ is analytic in $D_R$ for $r<R$, which assures the representation $f(z)={\sum }_{k=0}^{\mathrm{\infty }}{c}_k{z}^k$ 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 $\frac{1}{2}<r<R\le \frac{{[n_0]}_q^{1-\beta }}{2}$. Let us define ${\alpha }_{k,n,q}(z)=R_n(e_k;q,z)$ for all $z\in {\overline{D}}_r$, where $e_k(z)=z^k$. If $f:D_R\cup [R,\mathrm{\infty })\to \mathbb{C}$ is uniformly continuous, bounded on $[0,\mathrm{\infty })$ and analytic in $D_R$, then we have the form

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

for all $z\in {\overline{D}}_r$.

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

$$f_m(z)=\sum _{k=0}^m{c}_k{e}_k(z)\text{if }|z|\le r\text{and}{f}_m(z)=f(z)\text{if }z\in (r,\mathrm{\infty }).$$

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

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

that is all $R_n(f_m;q,z)$ with $n\ge n_0$, $r<\frac{{[n_0]}_q^{1-\beta }}{2}$, $m\in \mathbb{N}$ are well defined for all $z\in {\overline{D}}_r$.

Defining

$$f_{m,k}(z)=c_k{e}_k(z)\text{if }|z|\le r\text{and}{f}_{m,k}(z)=\frac{f(z)}{m+1}\text{if }z\in (r,\mathrm{\infty }),$$

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

From the linearity of $R_n(f;q,z)$, we have

$$R_n(f_m;q,z)=\sum _{k=0}^m{c}_k{\alpha }_{k,n,q}(z)\text{for all }|z|\le r.$$

It suffices to prove that

$$\underset{m\to \mathrm{\infty }}{lim}{R}_n(f_m;q,z)=R_n(f;q,z)$$

for any fixed $n\in \mathbb{N}$, $n\ge n_0$ and $|z|\le r$.

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

$$|R_n(f_m;q,z)-R_n(f;q,z)|\le M_{r,n,q}{\parallel f_m-f\parallel }_r,$$

(1)

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

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

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

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

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

Proof We can write ${(\beta +z)}_q^n$ as follows:

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

(2)

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

$$D_q\left[{(\beta +z)}_q^n\right]={[n]}_q{(\beta +z)}_q^{n-1}.$$

(3)

Using (2) and (3), we get

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

(4)

From (4), we obtain the result. □

Lemma 3 We have the following recurrence formula for the complex q-Balázs-Szabados operators $R_n(f;q,z)$:

$${\alpha }_{k+1,n,q}(z)=\frac{(1+q^n{a}_{n}z)z}{(1+a_{n}z)b_n}{D}_q[{\alpha }_{k,n,q}(z)]+\frac{z}{1+a_{n}z}{\alpha }_{k,n,q}(z),$$

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

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

$$\begin{array}{r}{D}_q[{\alpha }_{k,n,q}(z)]\\ =D_q\left[\frac{1}{{\prod }_{s=0}^{n-1}(1+q^s{a}_{n}z)}\right]\sum _{j=0}^n{q}^{j(j-1)/2}{\left(\frac{{[j]}_q}{b_n}\right)}^k{\left[\begin{array}{c}n\\ j\end{array}\right]}_q{(a_{n}z)}^j\\ +\frac{1}{{\prod }_{s=0}^{n-1}(1+q^{s+1}{a}_{n}z)}\sum _{j=0}^n{q}^{j(j-1)/2}{\left(\frac{{[j]}_q}{b_n}\right)}^k{\left[\begin{array}{c}n\\ j\end{array}\right]}_q{(a_n)}^j{D}_q\left[z^j\right].\end{array}$$

(5)

Considering Lemma 2 and using $D_q[z^j]={[j]}_q{z}^{j-1}$ in (5), we get

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

(6)

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

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

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

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

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

$$\left|D_q[f(z)]\right|\le \frac{R+r}{R-r}\cdot \frac{k}{r}{\parallel f\parallel }_r.$$

(7)

Lemma 4 Let $n_0\ge 2$, $0<\beta \le \frac{2}{3}$ and $\frac{1}{2}<r<R\le \frac{{[n_0]}_q^{1-\beta }}{2}$. For all $n\ge n_0$, $|z|\le r$ and $k=0,1,2,\dots$ , we have

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

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

$$|{\alpha }_{k+1,n,q}(z)|\le \frac{(1+q^n{a}_n|z|)|z|}{|1-a_n|z||b_n}\left|D_q[{\alpha }_{k,n,q}(z)]\right|+\frac{|z|}{|1-a_n|z||}|{\alpha }_{k,n,q}(z)|.$$

(8)

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

$$\left|D_q[{\alpha }_{k,n,q}(z)]\right|\le \frac{{[n_0]}_q^{1-\beta }+r}{{[n_0]}_q^{1-\beta }-r}\cdot \frac{k}{r}{\parallel {\alpha }_{k,n,q}\parallel }_r.$$

Under the condition $r<\frac{{[n_0]}_q^{1-\beta }}{2}$, it holds $\frac{{[n_0]}_q^{1-\beta }+r}{{[n_0]}_q^{1-\beta }-r}<3$, which implies

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

(9)

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

$${\parallel {\alpha }_{k+1,n,q}\parallel }_r\le \frac{(1+q^n{a}_{n}r)3k}{(1-a_{n}r)b_n}{\parallel {\alpha }_{k,n,q}\parallel }_r+\frac{r}{1-a_{n}r}{\parallel {\alpha }_{k,n,q}\parallel }_r.$$

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

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

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

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

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 $q=\{q_n\}$ be a sequence satisfying the following conditions:

$$\underset{n\to \mathrm{\infty }}{lim}{q}_n=1\text{and}\underset{n\to \mathrm{\infty }}{lim}{q}_n^n=c(0\le c<1).$$

(10)

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

Theorem 1 Let $\{q_n\}$ be a sequence satisfying the conditions (10) with $q_n\in (0,1]$ for all $n\in \mathbb{N}$, and let $n_0\ge 2$, $0<\beta \le \frac{2}{3}$ and $\frac{1}{2}<r<R\le \frac{{[n_0]}_{q_n}^{1-\beta }}{2}$. If $f:D_R\cup [R,\mathrm{\infty })\to \mathbb{C}$ is uniformly continuous, bounded on $[0,\mathrm{\infty })$ and analytic in $D_R$ and there exist $M>0$, $0<A<\frac{1}{20r}$ with $|c_k|\le M\frac{A^k}{k!}$ (which implies $|f(z)|\le M{e}^{A|z|}$ for all $z\in D_R$), then the sequence ${\{R_n(f;q_n,z)\}}_{n\ge n_0}$ 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

$$|R_n(f;q_n,z)|\le \sum _{k=0}^{\mathrm{\infty }}|c_k||{\alpha }_{k,n,q_n}(z)|\le \sum _{k=0}^{\mathrm{\infty }}M\frac{A^k}{k!}k!{(20r)}^k=M\sum _{k=0}^{\mathrm{\infty }}{(20Ar)}^k,$$

where the series ${\sum }_{k=0}^{\mathrm{\infty }}{(20Ar)}^k$ is convergent for $0<A<\frac{1}{20r}$.

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

We can give the following upper estimate in the approximation of $R_n(f;q_n,z)$.

Theorem 2 Let $\{q_n\}$ be a sequence satisfying the conditions (10) with $q_n\in (0,1]$ for all $n\in \mathbb{N}$, and let $n_0\ge 2$, $0<\beta \le \frac{2}{3}$ and $\frac{1}{2}<r<R\le \frac{{[n_0]}_{q_n}^{1-\beta }}{2}$. If $f:D_R\cup [R,\mathrm{\infty })\to \mathbb{C}$ is uniformly continuous, bounded on $[0,\mathrm{\infty })$ and analytic in $D_R$ and there exist $M>0$, $0<A<\frac{1}{20r}$ with $|c_k|\le M\frac{A^k}{k!}$ (which implies $|f(z)|\le M{e}^{A|z|}$ for all $z\in D_R$), then the following upper estimate holds:

$$|R_n(f;q_n,z)-f(z)|\le C_r^1(f)(a_n+\frac{1}{b_n}),$$

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

Proof Using the recurrence formula in Lemma 4, we have

$$\begin{array}{rcl}|{\alpha }_{k+1,n,q_n}(z)-z^{k+1}|& \le & \frac{(1+q_n^n{a}_n|z|)|z|}{|1-a_n|z||b_n}\left|D_{q_n}[{\alpha }_{k,n,q_n}(z)-z^k]\right|\\ +\frac{|z|}{|1-a_n|z||}|{\alpha }_{k,n,q_n}(z)-z^k|+\frac{1}{b_n}\frac{(1+q_n^n{a}_n|z|)}{|1-a_n|z||}{[k]}_{q_n}|z{|}^k\\ +\frac{a_n}{|1-a_n|z||}|z{|}^{k+2}.\end{array}$$

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

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

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

$$|{\alpha }_{k+1,n,q_n}(z)-z^{k+1}|\le \frac{9k\cdot k!}{b_n}{(20r)}^k+2r|{\alpha }_{k,n,q_n}(z)-z^k|+\frac{6}{b_n}{[k]}_{q_n}{r}^k+2a_n{r}^{k+2}.$$

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

$$|{\alpha }_{k+1,n,q_n}(z)-z^{k+1}|\le \frac{18k\cdot k!}{b_n}{(20r)}^k+2r|{\alpha }_{k,n,q_n}(z)-z^k|+2a_n{r}^{k+2}.$$

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

$$|{\alpha }_{k,n,q_n}(z)-z^k|\le \frac{9}{b_n}(k-1)k!{(20r)}^{k-1}+2a_n{r}^{2}k{(2r)}^{k-1},$$

(11)

which implies

$$\begin{array}{rcl}|R_n(f;q_n,z)-f(z)|& \le & \sum _{k=1}^{\mathrm{\infty }}|c_k||{\alpha }_{k,n,q_n}(z)-z^k|\\ \le & \sum _{k=1}^{\mathrm{\infty }}M\frac{A^k}{k!}\{\frac{9}{b_n}(k-1)k!{(20r)}^{k-1}+2a_n{r}^{2}k{(2r)}^{k-1}\}\\ =& \frac{9MA}{b_n}\sum _{k=1}^{\mathrm{\infty }}(k-1){(20Ar)}^{k-1}+2a_n{r}^{2}MA\sum _{k=1}^{\mathrm{\infty }}\frac{{(20Ar)}^{k-1}}{(k-1)!}\\ =& \frac{9MA}{b_n}\sum _{k=1}^{\mathrm{\infty }}(k-1){(20Ar)}^{k-1}+2a_n{r}^{2}MA{e}^{2Ar}.\end{array}$$

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

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

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

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

$$R_n(e_0;q,z)=1,$$

(12)

$$R_n(e_1;q,z)=\frac{z}{1+a_{n}z},$$

(13)

$$R_n(e_2;q,z)=\frac{(1-\frac{a_n}{b_n})q{z}^2}{(1+a_{n}z)(1+a_{n}qz)}+\frac{z}{b_n(1+a_{n}z)},$$

(14)

where $e_k(z)=z^k$ 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 ${[n]}_q=1+q{[n-1]}_q$ and $\frac{a_n}{b_n}=\frac{1}{{[n]}_q}$ 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 $R_n(f;q,z)$ hold:

$${\psi }_{n,q}^1(z)=\frac{-a_n{z}^2}{1+a_{n}z},$$

(15)

$$\begin{array}{l}{\psi }_{n,q}^2(z)=\frac{z}{b_n(1+a_{n}z)(1+a_{n}qz)}-\frac{(1-q)z^2}{(1+a_{n}z)(1+a_{n}qz)}\\ \phantom{{\psi }_{n,q}^2(z)=}-\frac{a_n(1-q)z^3}{(1+a_{n}z)(1+a_{n}qz)}+\frac{a_n^{2}q{z}^4}{(1+a_{n}z)(1+a_{n}qz)},\end{array}$$

(16)

where ${\psi }_{n,q}^i(z)=R_n({(t-e_1)}^i;q,z)$ 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

$$A_{k,n,q_n}(z)=R_n(f;q_n,z)-f(z)-{\psi }_{n,q}^1(z)f^{\mathrm{\prime }}(z)-\frac{1}{2}{\psi }_{n,q}^2(z)f^{\mathrm{\prime }\mathrm{\prime }}(z).$$

(17)

Theorem 3 Let $\{q_n\}$ be a sequence satisfying the conditions (10) with $q_n\in (0,1]$ for all $n\in \mathbb{N}$, $n_0\ge 2$, $0<\beta \le \frac{2}{3}$ and $\frac{1}{2}<r<R\le \frac{{[n_0]}_{q_n}^{1-\beta }}{2}$. If $f:D_R\cup [R,\mathrm{\infty })\to \mathbb{C}$ is uniformly continuous, bounded on $[0,\mathrm{\infty })$ and analytic in $D_R$ and there exist $M>0$, $0<A<\frac{1}{20r}$ with $|c_k|\le M\frac{A^k}{k!}$ (which implies $|f(z)|\le M{e}^{A|z|}$ for all $z\in D_R$), then for all $n\ge n_0$ and $|z|\le r$, we have

$$|A_{k,n,q_n}(z)|\le C_r^2(f){(a_n+\frac{1}{b_n})}^2,$$

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

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

$$|A_{k,n,q_n}(z)|\le \sum _{k=2}^{\mathrm{\infty }}|c_k||E_{k,n,q_n}(z)|,$$

(18)

where

$$\begin{array}{rcl}{E}_{k,n,q_n}(z)& =& {\alpha }_{k,n,q_n}(z)-z^k+\frac{a_{n}k{z}^{k+1}}{1+a_{n}z}-\frac{(k-1)k{z}^{k-1}}{2b_n(1+a_{n}z)(1+a_n{q}_{n}z)}\\ +\frac{(1-q_n)(k-1)k{z}^k}{2(1+a_{n}z)(1+a_n{q}_{n}z)}+\frac{a_n(1-q_n)(k-1)k{z}^{k+1}}{2(1+a_{n}z)(1+a_n{q}_{n}z)}\\ -\frac{a_n^2{q}_n(k-1)k{z}^{k+2}}{2(1+a_{n}z)(1+a_n{q}_{n}z)}.\end{array}$$

(19)

Using Lemma 5, we easily obtain that $E_{0,n,q}(z)=E_{1,n,q}(z)=E_{2,n,q}(z)=0$.

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

$$E_{k+1,n,q_n}(z)=\frac{(1+q_n^n{a}_{n}z)z}{b_n(1+a_{n}z)}{D}_{q_n}[E_{k,n,q_n}(z)]+\frac{z}{1+a_{n}z}{E}_{k,n,q_n}(z)+F_{k,n,q_n}(z),$$

(20)

where

$$\begin{array}{rcl}{F}_{k,n,q_n}(z)& =& -\frac{(k-{[k]}_{q_n})z^k}{b_n{(1+a_{n}z)}^2(1+a_n{q}_{n}z)}+\frac{a_n^{2}k{z}^{k+3}}{{(1+a_{n}z)}^2}-\frac{(1-q_n)k{z}^{k+1}}{{(1+a_{n}z)}^2(1+a_n{q}_{n}z)}\\ +\frac{a_n(1-q_n)k{z}^{k+2}}{{(1+a_{n}z)}^2(1+a_n{q}_{n}z)}-\frac{a_n^2{q}_{n}k{z}^{k+3}}{{(1+a_{n}z)}^2(1+a_n{q}_{n}z)}\\ -\frac{a_{n}k(k+1)z^{k+1}}{2b_n{(1+a_{n}z)}^2(1+a_n{q}_{n}z)}+\frac{a_n(1-q_n)k(k+1)z^{k+2}}{2{(1+a_{n}z)}^2(1+a_n{q}_{n}z)}\\ +\frac{a_n^2(1-q_n)k(k+1)z^{k+3}}{2{(1+a_{n}z)}^2(1+a_n{q}_{n}z)}-\frac{a_n^3{q}_{n}k(k+1)z^{k+4}}{2{(1+a_{n}z)}^2(1+a_n{q}_{n}z)}\\ -\frac{a_n(1+q_n^n{a}_{n}z)((k-1){[k+1]}_{q_n}-q_n{[k-1]}_{q_n})z^{k+1}}{b_n{(1+a_{n}z)}^2(1+a_n{q}_{n}z)}\\ -\frac{a_n^2(1+q_n^n{a}_{n}z)(k-1)q_n{[k]}_{q_n}{z}^{k+2}}{b_n{(1+a_{n}z)}^2(1+a_n{q}_{n}z)}+\frac{a_n{q}_n^n{[k]}_{q_n}{z}^{k+1}}{b_n{(1+a_{n}z)}^2(1+a_n{q}_{n}z)}\\ -\frac{(1+q_n^n{a}_{n}z){[k-1]}_{q_n}(k-1)k{z}^{k-1}}{2b_n^2(1+a_{n}z)(1+a_n{q}_{n}z)(1+a_n{q}_n^{2}z)}\\ +\frac{(1-q_n)(1+q_n^n{a}_{n}z){[k]}_{q_n}(k-1)k{z}^k}{2b_n(1+a_{n}z)(1+a_n{q}_{n}z)(1+a_n{q}_n^{2}z)}\\ +\frac{a_n(1-q_n)(1+q_n^n{a}_{n}z){[k+1]}_{q_n}(k-1)k{z}^{k+1}}{2b_n(1+a_{n}z)(1+a_n{q}_{n}z)(1+a_n{q}_n^{2}z)}\\ -\frac{a_n^2{q}_n(1+q_n^n{a}_{n}z){[k+2]}_{q_n}(k-1)k{z}^{k+2}}{2b_n(1+a_{n}z)(1+a_n{q}_{n}z)(1+a_n{q}_n^{2}z)}\\ +\frac{a_n(1+q_n^n{a}_{n}z)(1+q_n)(k-1)k{z}^k}{2b_n^2{(1+a_{n}z)}^2(1+a_n{q}_{n}z)(1+a_n{q}_n^{2}z)}\\ -\frac{a_n(1-q_n)(1+q_n)(1+q_n^n{a}_{n}z)(k-1)k{z}^{k+1}}{2b_n{(1+a_{n}z)}^2(1+a_n{q}_{n}z)(1+a_n{q}_n^{2}z)}\\ -\frac{a_n^2(1-q_n)(1+q_n)(1+q_n^n{a}_{n}z)(k-1)k{z}^{k+2}}{2b_n{(1+a_{n}z)}^2(1+a_n{q}_{n}z)(1+a_n{q}_n^{2}z)}\\ -\frac{a_n^3{q}_n(1+q_n)(1+q_n^n{a}_{n}z)(k-1)k{z}^{k+3}}{2b_n{(1+a_{n}z)}^2(1+a_n{q}_{n}z)(1+a_n{q}_n^{2}z)}.\end{array}$$

In the following results, $C_i$ will denote fixed real numbers for $i=1,2,3$.

Under the hypothesis of Theorem 3, we have

$$\left|\frac{1}{1+q_n^s{a}_{n}z}\right|\le \frac{1}{1-q_n^s{a}_{n}r}<2\text{for }s=0,1,2,$$

(21)

$$a_{n}r<\frac{1}{2}\text{and}1+q_n^n{a}_{n}r<\frac{3}{2},$$

(22)

$$1-q_n\le \frac{a_n}{b_n}\text{and}k-{[k]}_{q_n}\le \frac{a_n}{b_n}\frac{(k-1)k}{2},$$

(23)

$${[k]}_{q_n}\le k,a_n<1\text{and}\frac{1}{b_n}<1.$$

(24)

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

$$\begin{array}{rcl}|F_{k,n,q_n}(z)|& \le & C_1(a_n^2+\frac{a_n}{b_n}+\frac{1}{b_n^2})k(k+1)(k+2)\\ \times max\{r^{k-1},r^k,r^{k+1},r^{k+2},r^{k+3},r^{k+4}\}\\ \le & C_1{(a_n+\frac{1}{b_n})}^{2}k(k+1)(k+2){(2r)}^{k+4}.\end{array}$$

(25)

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

$$\begin{array}{rcl}\left|\frac{(1+q_n^n{a}_{n}z)z}{b_n(1+a_{n}z)}{D}_{q_n}[E_{k,n,q_n}(z)]\right|& \le & \frac{(1+q_n^n{a}_{n}r)r}{b_n(1-a_{n}r)}\frac{3k}{r}{\parallel E_{k,n,q_n}\parallel }_r\\ \le & \frac{3k(1+q_n^n{a}_{n}r)}{b_n(1-a_{n}r)}\{{\parallel {\alpha }_{k,n,q_n}-e_k\parallel }_r\\ +\frac{a_{n}k{r}^{k+1}}{1-a_{n}r}+\frac{(k-1)k{r}^{k-1}}{2b_n(1-a_{n}r)(1-a_n{q}_{n}r)}\\ +\frac{(1-q_n)(k-1)k{r}^k}{2(1-a_{n}r)(1-a_n{q}_{n}r)}\\ +\frac{a_n(1-q_n)(k-1)k{r}^{k+1}}{2(1-a_{n}r)(1-a_n{q}_{n}r)}+\frac{a_n^2{q}_n(k-1)k{r}^{k+2}}{2(1-a_{n}r)(1-a_n{q}_{n}r)}\}.\end{array}$$

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

$$\begin{array}{rcl}\left|\frac{(1+q_n^n{a}_{n}z)z}{b_n(1+a_{n}z)}{D}_{q_n}[E_{k,n,q_n}(z)]\right|& \le & C_2\frac{1}{b_n}(a_n+\frac{1}{b_n})(k-1)k(k+1)\\ \times (k!){(20r)}^{k+2}\\ \le & C_2{(a_n+\frac{1}{b_n})}^2(k-1)k(k+1)(k!){(20r)}^{k+2}.\end{array}$$

(26)

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

$$|E_{k+1,n,q_n}(z)|\le 2r|E_{k,n,q_n}(z)|+C_3{(a_n+\frac{1}{b_n})}^{2}k(k+1)(k+2)(k+1)!{(20r)}^{k+4}.$$

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

$$|E_{k,n,q_n}(z)|\le C_3{(a_n+\frac{1}{b_n})}^2(k-2)(k-1)k(k+1)(k!){(20r)}^{k+3}.$$

(27)

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

Remark 1 For $0<q\le 1$, since $\frac{1}{{[n]}_q}\to 1-q$ as $n\to \mathrm{\infty }$, therefore $a_n={(\frac{1}{{[n]}_q})}^{1-\beta }\to {(1-q)}^{1-\beta }$ and $\frac{1}{b_n}={(\frac{1}{{[n]}_q})}^{\beta }\to {(1-q)}^{\beta }$ as $n\to \mathrm{\infty }$. If a sequence $\{q_n\}$ satisfies the conditions (10), then $\frac{1}{{[n]}_q}\to 0$ as $n\to \mathrm{\infty }$; therefore $a_n={(\frac{1}{{[n]}_q})}^{1-\beta }\to 0$ and $\frac{1}{b_n}={(\frac{1}{{[n]}_q})}^{\beta }\to 0$ as $n\to \mathrm{\infty }$.

Under the conditions (10), Theorem 2 and Theorem 3 show that ${\{R_n(f;q_n,z)\}}_{n\ge n_0}$ uniformly converges to $f(z)$ in ${\overline{D}}_r$.

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

Theorem 4 Let $\{q_n\}$ be a sequence satisfying the conditions (10) with $q_n\in (0,1]$ for all $n\in \mathbb{N}$, $n_0\ge 2$, $0<\beta \le \frac{2}{3}$, $\beta \ne \frac{1}{2}$ and $\frac{1}{2}<r<R\le \frac{{[n_0]}_{q_n}^{1-\beta }}{2}$. Suppose that $f:D_R\cup [R,\mathrm{\infty })\to \mathbb{C}$ is uniformly continuous, bounded on $[0,\mathrm{\infty })$ and analytic in $D_R$ and there exist $M>0$, $0<A<\frac{1}{20r}$ with $|c_k|\le M\frac{A^k}{k!}$ (which implies $|f(z)|\le M{e}^{A|z|}$ for all $z\in D_R$). If f is not a polynomial of degree ≤1, then for all $n\ge n_0$ we have

$${\parallel R_n(f;q_n,\cdot )-f\parallel }_r\sim (a_n+\frac{1}{b_n}).$$

Proof We can write

$$R_n(f;q_n,z)-f(z)=(a_n+\frac{1}{b_n})\{G(z)+H_n(z)\},$$

(28)

where

$$\begin{array}{rcl}G(z)& =& -\frac{a_n}{a_n+1/b_n}\frac{z^2{f}^{\mathrm{\prime }}(z)}{1+a_{n}z}\\ +\frac{1}{a_n{b}_n+1}\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{2(1+a_{n}z)(1+a_n{q}_{n}z)}\\ -\frac{1-q_n}{a_n+1/b_n}\frac{z^2{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{2(1+a_{n}z)(1+a_n{q}_{n}z)}\\ -\frac{a_n(1-q_n)}{a_n+1/b_n}\frac{z^3{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{2(1+a_{n}z)(1+a_n{q}_{n}z)}\\ +\frac{a_n^2}{a_n+1/b_n}\frac{q_n{z}^4{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{2(1+a_{n}z)(1+a_n{q}_{n}z)}\end{array}$$

(29)

and

$$H_n(z)=(a_n+\frac{1}{b_n})[\frac{1}{{(a_n+\frac{1}{b_n})}^2}{A}_{k,n,q_n}(z)],$$

(30)

and also ${(H_n(z))}_{n\in \mathbb{N}}$ 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 $|G(z)|>\rho >0$, 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

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

(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

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

(32)

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

Choosing $n_3=max\{n_1,n_2\}$, considering (31) and (32), for all $n\ge n_3$, we get

$${\parallel R_n(f;q_n,\cdot )-f\parallel }_r\ge (a_n+\frac{1}{b_n})|{\parallel G\parallel }_r-{\parallel H_n\parallel }_r|\ge (a_n+\frac{1}{b_n})\frac{1}{2}{\parallel G\parallel }_r.$$

For all $n\in \{n_0,\dots ,n_3-1\}$, we get

$${\parallel R_n(f;q_n,\cdot )-f\parallel }_r\ge (a_n+\frac{1}{b_n})M_{r,n,q_n}(z)$$

with $M_{r,n,q_n}(z)=\frac{1}{a_n+1/b_n}{\parallel R_n(f;q_n,\cdot )-f\parallel }_r>0$, which finally implies

$${\parallel R_n(f;q_n,\cdot )-f\parallel }_r\ge (a_n+\frac{1}{b_n})C_r(f)$$

(33)

for all $n\ge n_0$, with $C_r(f)=min\{M_{r,n_0,q_n}(z),\dots ,M_{r,n_3-1,q_n}(z),\frac{1}{2}{\parallel G\parallel }_r\}$.

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.