# Approximation by a complex q-Baskakov-Stancu operator in compact disks

## Abstract

In this paper, we consider a complex q-Baskakov-Stancu operator and study some approximation properties. We give a quantitative estimate of the convergence, Voronovskaja-type result and exact order of approximation in compact disks.

MSC:30E10, 41A25, 41A28.

## 1 Introduction

Recently complex approximation operators have been studied intensively. For this approach, we refer to the book of Gal [1], where he considers approximation properties of several complex operators such as Bernstein, q-Bernstein, Favard-Szasz-Mirakjan, Baskakov and some others. Also we refer to the useful book of Aral, Gupta and Agarwal [2] who consider many applications of q-calculus in approximation theory. Now, for the construction of the new operators, we give some notations on q-analysis [3, 4].

Let $q>0$. The q-integer $\left[n\right]$ and the q-factorial $\left[n\right]!$ are defined by

and

respectively. For integers $n\ge r\ge 0$, the q-binomial coefficient is defined as

${\left[\begin{array}{c}n\\ r\end{array}\right]}_{q}=\frac{{\left[n\right]}_{q}!}{{\left[r\right]}_{q}!{\left[n-r\right]}_{q}!}.$

The q-derivative of $f\left(z\right)$ is denoted by ${D}_{q}f\left(z\right)$ and defined as

${D}_{q}f\left(z\right):=\frac{f\left(qz\right)-f\left(z\right)}{\left(q-1\right)z},\phantom{\rule{1em}{0ex}}z\ne 0,\phantom{\rule{2em}{0ex}}{D}_{q}f\left(0\right)={f}^{\prime }\left(0\right),$

also

${D}_{q}^{0}f:=f,\phantom{\rule{2em}{0ex}}{D}_{q}^{n}f:={D}_{q}\left({D}_{q}^{n-1}f\right),\phantom{\rule{1em}{0ex}}n=1,2,\dots$

q-Pochhammer formula is given by

$\begin{array}{c}{\left(x,q\right)}_{0}=1,\hfill \\ {\left(x,q\right)}_{n}=\prod _{k=0}^{n-1}\left(1-{q}^{k}x\right)\hfill \end{array}$

with $x\in \mathbb{R}$, $n\in \mathbb{N}\cup \left\{\mathrm{\infty }\right\}$. The q-derivative of the product and the quotient of two functions f and g are

${D}_{q}\left(f\left(z\right)g\left(z\right)\right)=f\left(z\right){D}_{q}\left(g\left(z\right)\right)+g\left(qz\right){D}_{q}\left(f\left(z\right)\right)$

and

${D}_{q}\left(\frac{f\left(z\right)}{g\left(z\right)}\right)=\frac{g\left(z\right){D}_{q}\left(f\left(z\right)\right)-f\left(z\right){D}_{q}\left(g\left(z\right)\right)}{g\left(z\right)g\left(qz\right)},$

respectively (see in [3]). Moreover, we have

$\left[{x}_{0},{x}_{1},\dots ,{x}_{m};f\cdot g\right]=\sum _{i=0}^{m}\left[{x}_{0},{x}_{1},\dots ,{x}_{i};f\right]\left[{x}_{i},{x}_{i+1},\dots ,{x}_{m};g\right],$
(1.1)

where $\left[{x}_{0},{x}_{1},\dots ,{x}_{m};f\right]$ denotes the divided difference of the function f on the knots ${x}_{0},{x}_{1},\dots ,{x}_{m}$ (see [4] also [5]).

In [6], Aral and Gupta constructed the q-Baskakov operator as

${Z}_{n}^{q}\left(f\right)\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}\left[\begin{array}{c}n+k-1\\ k\end{array}\right]{q}^{\frac{k\left(k-1\right)}{2}}{z}^{k}{\left(-x,q\right)}_{n+k}^{-1}f\left(\frac{\left[k\right]}{{q}^{k-1}\left[n\right]}\right),\phantom{\rule{1em}{0ex}}n\in \mathbb{N},$

where $x\ge 0$, $q>0$ and f is a real-valued continuous function on $\left[0,\mathrm{\infty }\right)$. The authors studied the rate of convergence in a polynomial weighted norm and gave a theorem related to monotonic convergence of the sequence of operators with respect to n. Not only they proved a kind of monotonicity by means of q-derivative but also they expressed the operator in terms of divided differences as follows:

${W}_{n,q}\left(f\right)\left(x\right)=\sum _{j=0}^{\mathrm{\infty }}\frac{\left[n+j-1\right]!}{\left[n-1\right]!}{q}^{\frac{-j\left(j-1\right)}{2}}\left[0,\frac{1}{\left[n\right]},\frac{\left[2\right]}{q\left[n\right]},\dots ,\frac{\left[j\right]}{{q}^{j-1}\left[n\right]};f\right]\frac{{x}^{j}}{{\left[n\right]}^{j}}$
(1.2)

$n\in \mathbb{N}$, similar to the case of classical Baskakov operators in the sense of Lupaş in [7]. That is to say, ${Z}_{n}^{q}\left(f\right)\left(x\right)={W}_{n,q}\left(f\right)\left(x\right)$ for $x\ge 0$ and $q>0$, so they proved that

$\left[0,\frac{1}{\left[n\right]},\frac{\left[2\right]}{q\left[n\right]},\dots ,\frac{\left[j\right]}{{q}^{j-1}\left[n\right]};f\right]=\frac{{q}^{j\left(j-1\right)}{\mathrm{\nabla }}_{q}^{j}f\left(0\right)}{\left[j\right]!}{\left[n\right]}^{j}=\frac{{f}^{\left(j\right)}\left(\zeta \right)}{j!},\phantom{\rule{1em}{0ex}}\zeta \in \left(0,\frac{\left[j\right]}{{q}^{j-1}\left[n\right]}\right),$
(1.3)

where ${\mathrm{\nabla }}_{q}^{r}$ stands for q-divided differences given by ${\mathrm{\nabla }}_{q}^{0}f\left({x}_{j}\right)$,

${\mathrm{\nabla }}_{q}^{r+1}f\left({x}_{j}\right)={q}^{r}{\mathrm{\nabla }}_{q}^{r}f\left({x}_{j+1}\right)-{\mathrm{\nabla }}_{q}^{r}f\left({x}_{j}\right)$

for $r\in \mathbb{N}\cup \left\{0\right\}$.

A different type of the q-Baskakov operator was also given by Aral and Gupta in [8]. In [9] Finta and Gupta studied the q-Baskakov operator ${Z}_{n}^{q}\left(f\right)\left(x\right)$ for $0. Using the second-order Ditzian-Totik modulus of smoothness, they gave direct estimates. They also introduced the limit q-Baskakov operator.

In [10] Gupta and Radu introduced a q-analogue of Baskakov-Kantorovich operators and studied weighted statistical approximation properties of them for $0. They also obtained some direct estimations for error with the help of weighted modulus of smoothness. Moreover, Durrmeyer-type modifications of q-Baskakov operators were studied in [11] and [12]. In [13], Söylemez, Tunca and Aral defined a complex form of q-Baskakov operators by

${W}_{n,q}\left(f\right)\left(z\right)=\sum _{j=0}^{\mathrm{\infty }}\frac{\left[n+j-1\right]!}{\left[n-1\right]!}{q}^{\frac{-j\left(j-1\right)}{2}}\left[0,\frac{1}{\left[n\right]},\frac{\left[2\right]}{q\left[n\right]},\dots ,\frac{\left[j\right]}{{q}^{j-1}\left[n\right]};f\right]\frac{{z}^{j}}{{\left[n\right]}^{j}}$
(1.4)

for $q>1$, $f:{\overline{D}}_{R}\cup \left[R,\mathrm{\infty }\right)\to \mathbb{C}$, replacing x by z in the operator ${W}_{n,q}\left(f\right)\left(x\right)$ given by (1.2). They obtained a quantitative estimate for simultaneous approximation, Voronovskaja-type result and degree of simultaneous approximation in compact disks.

In recent years, a Stancu-type generalization of the operators has been studied. Büyükyazıcı and Atakut considered a Stancu-type generalization of the real Baskakov operators in [14]. Also in [15], q-Baskakov-Beta-Stancu operators were introduced. In [16] Gupta-Verma studied the Stancu-type generalization of complex Favard-Szasz-Mirakjan operators and established some approximation results in the complex domain. In [17] Gal, Gupta, Verma and Agrawal introduced complex Baskakov-Stancu operators and studied Voronovskaja-type results with quantitative estimates for these operators attached to analytic functions on compact disks.

Now we define a new type of the complex q-Baskakov-Stancu operator

$\begin{array}{rcl}{W}_{n,q}^{\alpha ,\beta }\left(f\right)\left(z\right)& =& \sum _{j=0}^{\mathrm{\infty }}\frac{\left[n+j-1\right]!}{\left[n-1\right]!}{q}^{\frac{-j\left(j-1\right)}{2}}\\ ×\left[\frac{\left[\alpha \right]}{\left[n\right]+\left[\beta \right]},\frac{\left[\alpha \right]+\left[1\right]}{\left[n\right]+\left[\beta \right]},\dots ,\frac{{q}^{j-1}\left[\alpha \right]+\left[j\right]}{{q}^{j-1}\left(\left[n\right]+\left[\beta \right]\right)};f\right]\frac{{z}^{j}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{j}},\end{array}$
(1.5)

where $0\le \alpha \le \beta$; for $j=0$, we take $\left[n\right]\left[n+1\right]\cdots \left[n+j-1\right]=1$. We suppose that f is analytic on the disk $|z|, $R>1$ and has exponential growth in the compact disk with all derivatives bounded in $\left[0,\mathrm{\infty }\right)$ by the same constant.

Note that taking $\alpha =\beta =0$, ${W}_{n,q}^{\alpha ,\beta }\left(f\right)\left(z\right)$ reduces to the complex q-Baskakov operator ${W}_{n,q}\left(f\right)\left(z\right)$ given in (1.4).

In this work, for such f and $q>1$, we study some approximation properties of the complex q-Baskakov-Stancu operator which is defined by forward differences.

## 2 Auxiliary results

In this section, we give some results which we shall use in the proof of theorems.

Lemma 1 Let us define ${e}_{k}\left(z\right)={z}^{k}$, ${T}_{n,k}^{\alpha ,\beta }\left(z\right):={W}_{n,q}^{\alpha ,\beta }\left({e}_{k}\right)\left(z\right)$, and ${\mathbb{N}}^{0}$ denotes the set of all nonnegative integers. Then, for all $n,k\in {\mathbb{N}}^{0}$, $0\le \alpha \le \beta$ and $z\in \mathbb{C}$, we have the following recurrence formula:

${T}_{n,k+1}^{\alpha ,\beta }\left(z\right)=\frac{qz\left(1+\frac{z}{q}\right)}{\left[n\right]+\left[\beta \right]}{D}_{q}{T}_{n,k}^{\alpha ,\beta }\left(\frac{z}{q}\right)+\frac{\left[n\right]z+\left[\alpha \right]}{\left(\left[n\right]+\left[\beta \right]\right)}{T}_{n,k}^{\alpha ,\beta }\left(z\right).$
(2.1)

Hence

${T}_{n,1}^{\alpha ,\beta }\left(z\right)=\frac{\left[n\right]z+\left[\alpha \right]}{\left[n\right]+\left[\beta \right]},\phantom{\rule{2em}{0ex}}{T}_{n,2}^{\alpha ,\beta }\left(z\right)=\frac{z\left(1+\frac{z}{q}\right)}{\left[n\right]+\left[\beta \right]}\frac{\left[n\right]}{\left[n\right]+\left[\beta \right]}+{\left(\frac{\left[n\right]z+\left[\alpha \right]}{\left[n\right]+\left[\beta \right]}\right)}^{2}$

for all $z\in \mathbb{C}$.

Proof Now we can write

$\begin{array}{rcl}{T}_{n,k}^{\alpha ,\beta }\left(z\right)& =& \sum _{j=0}^{\mathrm{\infty }}\frac{\left[n\right]\left[n+1\right]\cdots \left[n+j-1\right]}{{\left(\left[n\right]+\left[\beta \right]\right)}^{j}}{q}^{\frac{-j\left(j-1\right)}{2}}\\ ×\left[\frac{\left[\alpha \right]}{\left(\left[n\right]+\left[\beta \right]\right)},\dots ,\frac{{q}^{j-1}\left[\alpha \right]+\left[j\right]}{{q}^{j-1}\left(\left[n\right]+\left[\beta \right]\right)};{e}_{k}\right]{z}^{j}.\end{array}$
(2.2)

Using relation (1.1) and taking $f={e}_{k}$, $g={e}_{1}$ and ${x}_{j}=\frac{{q}^{j-1}\left[\alpha \right]+\left[j\right]}{{q}^{j-1}\left(\left[n\right]+\left[\beta \right]\right)}$, we obtain

$\begin{array}{r}\left[\frac{\left[\alpha \right]}{\left[n\right]+\left[\beta \right]},\dots ,\frac{{q}^{j-1}\left[\alpha \right]+\left[j\right]}{{q}^{j-1}\left(\left[n\right]+\left[\beta \right]\right)};{e}_{k+1}\right]\\ \phantom{\rule{1em}{0ex}}=\frac{{q}^{j-1}\left[\alpha \right]+\left[j\right]}{{q}^{j-1}\left(\left[n\right]+\left[\beta \right]\right)}\left[\frac{\left[\alpha \right]}{\left[n\right]+\left[\beta \right]},\dots ,\frac{{q}^{j-1}\left[\alpha \right]+\left[j\right]}{{q}^{j-1}\left(\left[n\right]+\left[\beta \right]\right)};{e}_{k}\right]\\ \phantom{\rule{2em}{0ex}}+\left[\frac{\left[\alpha \right]}{\left[n\right]+\left[\beta \right]},\dots ,\frac{{q}^{j-2}\left[\alpha \right]+\left[j-1\right]}{{q}^{j-2}\left(\left[n\right]+\left[\beta \right]\right)};{e}_{k}\right],\end{array}$
(2.3)

using this in ${T}_{n,k+1}^{\alpha ,\beta }\left(z\right)$ we reach

${T}_{n,k+1}^{\alpha ,\beta }\left(z\right)=\frac{qz\left(1+\frac{z}{q}\right)}{\left[n\right]+\left[\beta \right]}{D}_{q}{T}_{n,k}^{\alpha ,\beta }\left(\frac{z}{q}\right)+\frac{\left[n\right]z+\left[\alpha \right]}{\left[n\right]+\left[\beta \right]}{T}_{n,k}^{\alpha ,\beta }\left(z\right).$

□

Lemma 2 Let α and β satisfy $0\le \alpha \le \beta$. Denoting ${e}_{j}\left(z\right)={z}^{j}$ and ${W}_{n,q}^{0,0}\left({e}_{j}\right)$ by ${W}_{n,q}\left({e}_{j}\right)$ given in (1.4), for all $n,k\in {\mathbb{N}}^{0}$, we have the following recursive relation for the images of monomials ${e}_{k}$ under ${W}_{n,q}^{\alpha ,\beta }$ in terms of ${W}_{n,q}\left({e}_{j}\right)$, $j=0,1,\dots ,k$:

${T}_{n,k}^{\alpha ,\beta }\left(z\right)=\sum _{j=0}^{k}\left(\genfrac{}{}{0}{}{k}{j}\right)\frac{{\left[n\right]}^{j}{\left[\alpha \right]}^{k-j}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k}}{W}_{n,q}\left({e}_{j},z\right).$
(2.4)

Proof We can use mathematical induction with respect to k. For $k=0$, equality (2.4) holds. Let it be true for $k=m$, namely

${T}_{n,m}^{\alpha ,\beta }\left(z\right)=\sum _{j=0}^{m}\left(\genfrac{}{}{0}{}{m}{j}\right)\frac{{\left[n\right]}^{j}{\left[\alpha \right]}^{m-j}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{m}}{W}_{n,q}\left({e}_{j},z\right).$

Using (2.1), we have

$\begin{array}{rcl}{T}_{n,m+1}^{\alpha ,\beta }\left(z\right)& =& \frac{qz\left(1+\frac{z}{q}\right)}{\left[n\right]+\left[\beta \right]}\sum _{j=0}^{m}\left(\genfrac{}{}{0}{}{m}{j}\right)\frac{{\left[n\right]}^{j}{\left[\alpha \right]}^{m-j}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{m}}{D}_{q}{W}_{n,q}\left({e}_{j},\frac{z}{q}\right)\\ +\frac{\left[n\right]z+\left[\alpha \right]}{\left[n\right]+\left[\beta \right]}\sum _{j=0}^{m}\left(\genfrac{}{}{0}{}{m}{j}\right)\frac{{\left[n\right]}^{j}{\left[\alpha \right]}^{m-j}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{m}}{W}_{n,q}\left({e}_{j},z\right)\\ =& \sum _{j=0}^{m}\left(\genfrac{}{}{0}{}{m}{j}\right)\frac{{\left[n\right]}^{j+1}{\left[\alpha \right]}^{m-j}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{m+1}}\\ ×\left[\frac{qz\left(1+\frac{z}{q}\right)}{\left[n\right]}{D}_{q}{W}_{n,q}\left({e}_{j},\frac{z}{q}\right)+\frac{\left[n\right]z+\left[\alpha \right]}{\left[n\right]}{W}_{n,q}\left({e}_{j},z\right)\right].\end{array}$

Taking into account the recurrence relation for the complex q-Baskakov operator in Lemma 2 in [13], we get

${W}_{n,q}\left({e}_{j+1},z\right)=\frac{qz\left(1+\frac{z}{q}\right)}{\left[n\right]}{D}_{q}{W}_{n,q}\left({e}_{j},\frac{z}{q}\right)+z{W}_{n,q}\left({e}_{j},z\right),$

which implies

$\begin{array}{rcl}{T}_{n,m+1}^{\alpha ,\beta }\left(z\right)& =& \sum _{j=0}^{m}\left(\genfrac{}{}{0}{}{m}{j}\right)\frac{{\left[n\right]}^{j+1}{\left[\alpha \right]}^{m-j}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{m+1}}\left[{W}_{n,q}\left({e}_{j+1},z\right)+\frac{\left[\alpha \right]}{\left[n\right]}{W}_{n,q}\left({e}_{j},z\right)\right]\\ =& \sum _{j=1}^{m}\left(\genfrac{}{}{0}{}{m}{j-1}\right)\frac{{\left[n\right]}^{j}{\left[\alpha \right]}^{m-j+1}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{m+1}}{W}_{n,q}\left({e}_{j},z\right)\\ +\sum _{j=0}^{m}\left(\genfrac{}{}{0}{}{m}{j}\right)\frac{{\left[n\right]}^{j}{\left[\alpha \right]}^{m-j+1}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{m+1}}{W}_{n,q}\left({e}_{j},z\right)\\ =& \sum _{j=0}^{m+1}\left(\genfrac{}{}{0}{}{m+1}{j}\right)\frac{{\left[n\right]}^{j}{\left[\alpha \right]}^{m-j+1}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{m+1}}{W}_{n,q}\left({e}_{j},z\right),\end{array}$

which proves the lemma. □

## 3 Approximation by a complex q-Baskakov-Stancu operator

In this section, we give quantitative estimates concerning approximation with the following theorem.

Theorem 1 For $1, let

$f:{\overline{D}}_{R}\cup \left[R,\mathrm{\infty }\right)\to \mathbb{C}$

be a function with all its derivatives bounded in $\left[0,\mathrm{\infty }\right)$ by the same positive constant, analytic in ${D}_{R}$, namely $f\left(z\right)={\sum }_{k=0}^{\mathrm{\infty }}{c}_{k}{z}^{k}$ for all $z\in {D}_{R}$ and suppose that there exist $M>0$ and $A\in \left(\frac{1}{R},1\right)$, with the property $|{c}_{k}|\le \frac{M{A}^{k}}{k!}$ for all $k=0,1,\dots$ (which implies $|f\left(z\right)|\le M{e}^{A|z|}$ for all $z\in {D}_{R}$).

Let $0\le \alpha \le \beta$, $q>1$ and $1\le r<\frac{1}{A}$ be arbitrary but fixed. Then, for all $|z|\le r$ and $n\in \mathbb{N}$, we have

$\begin{array}{r}|{W}_{n,q}^{\alpha ,\beta }\left(f\right)\left(z\right)-f\left(z\right)|\\ \phantom{\rule{1em}{0ex}}\le \frac{{M}_{1,r}\left(f\right)}{\left[n\right]+\left[\beta \right]}+\frac{\left[\beta \right]}{\left[n\right]+\left[\beta \right]}{M}_{2,r}\left(f\right)+\frac{\left[\alpha \right]}{\left[n\right]+\left[\beta \right]}{M}_{3,r}\left(f\right)\\ \phantom{\rule{1em}{0ex}}={M}_{r,\alpha ,\beta }\left(f\right)\end{array}$

with

$\begin{array}{c}{M}_{1,r}\left(f\right)=6\sum _{k=2}^{\mathrm{\infty }}|{c}_{k}|\left(k+1\right)!\left(k-1\right){r}^{k}<\mathrm{\infty },\hfill \\ {M}_{2,r}\left(f\right)=\sum _{k=1}^{\mathrm{\infty }}|{c}_{k}|k{r}^{k}<\mathrm{\infty },\phantom{\rule{2em}{0ex}}{M}_{3,r}\left(f\right)=\sum _{k=1}^{\mathrm{\infty }}|{c}_{k}|k{r}^{k-1}<\mathrm{\infty }.\hfill \end{array}$

Proof Using (2.1), one can obtain

$\begin{array}{rcl}{T}_{n,k}^{\alpha ,\beta }\left(z\right)-{z}^{k}& =& \frac{qz\left(1+\frac{z}{q}\right)}{\left[n\right]+\left[\beta \right]}{D}_{q}\left({T}_{n,k-1}^{\alpha ,\beta }\left(\frac{z}{q}\right)\right)+\frac{\left[n\right]z+\left[\alpha \right]}{\left[n\right]+\left[\beta \right]}\left({T}_{n,k-1}^{\alpha ,\beta }\left(z\right)-{z}^{k-1}\right)\\ +\frac{\left[n\right]z+\left[\alpha \right]}{\left[n\right]+\left[\beta \right]}{z}^{k-1}-{z}^{k}\\ =& \frac{z\left(1+\frac{z}{q}\right)}{\left[n\right]+\left[\beta \right]}q{D}_{q}\left({T}_{n,k-1}^{\alpha ,\beta }\left(\frac{z}{q}\right)\right)+\frac{\left[n\right]z+\left[\alpha \right]}{\left[n\right]+\left[\beta \right]}\left({T}_{n,k-1}^{\alpha ,\beta }\left(z\right)-{z}^{k-1}\right)\\ +\left(\frac{\left[n\right]}{\left[n\right]+\left[\beta \right]}-1\right){z}^{k}+\frac{\left[\alpha \right]}{\left[n\right]+\left[\beta \right]}{z}^{k-1}.\end{array}$

Moreover, we have

$q{D}_{q}\left({T}_{n,k-1}^{\alpha ,\beta }\left(\frac{z}{q}\right)\right)=|{D}_{q}\left({T}_{n,k-1}^{\alpha ,\beta }\left(w\right)\right){|}_{w=\frac{z}{q}}.$
(3.1)

Now from (3.1) and the Bernstein inequality (see [1]), we have

$q{D}_{q}\left({T}_{n,k-1}^{\alpha ,\beta }\left(\frac{z}{q}\right)\right)=|{D}_{q}\left({T}_{n,k-1}^{\alpha ,\beta }\left(z\right)\right)|\le |{T}_{n,k-1}^{\prime \alpha ,\beta }\left(z\right)|\le \frac{k-1}{r}{\parallel {T}_{n,k-1}^{\alpha ,\beta }\parallel }_{r},$

where ${\parallel \cdot \parallel }_{r}$ is the standard maximum norm over ${D}_{r}=\left\{z\in \mathbb{C}:|z|\le r\right\}$. Passing to modulus for all $|z|\le r$ and $n\in \mathbb{N}$, we have that

$\begin{array}{rl}|{T}_{n,k}^{\alpha ,\beta }\left(z\right)-{z}^{k}|\le & \frac{r\left(1+r\right)}{\left[n\right]+\left[\beta \right]}\left(\frac{k-1}{r}\right){\parallel {T}_{n,k-1}^{\alpha ,\beta }\parallel }_{r}+\frac{\left[n\right]r+\left[\alpha \right]}{\left[n\right]+\left[\beta \right]}|{T}_{n,k-1}^{\alpha ,\beta }\left(z\right)-{z}^{k-1}|\\ +\left(\frac{\left[n\right]}{\left[n\right]+\left[\beta \right]}-1\right){r}^{k}+\frac{\left[\alpha \right]}{\left[n\right]+\left[\beta \right]}{r}^{k-1}.\end{array}$
(3.2)

In order to get an estimate for ${\parallel {T}_{n,k-1}^{\alpha ,\beta }\parallel }_{r}$ in (3.2), we use the following fact:

${T}_{n,k}^{\alpha ,\beta }\left(z\right)=\sum _{j=0}^{k}\frac{\left[n\right]\left[n+1\right]\cdots \left[n+j-1\right]}{{\left(\left[n\right]+\left[\beta \right]\right)}^{j}}{q}^{\frac{-j\left(j-1\right)}{2}}\left[\frac{\left[\alpha \right]}{\left[n\right]+\left[\beta \right]},\dots ,\frac{{q}^{j-1}\left[\alpha \right]+\left[j\right]}{{q}^{j-1}\left(\left[n\right]+\left[\beta \right]\right)};{e}_{k}\right]{z}^{j}$

for $k\in \mathbb{N}$. Taking into account Lemma 1 in [13] for $q>1$, $|z|\le r$, $r\ge 1$ and (1.3), we have

$\begin{array}{rcl}{\parallel {T}_{n,k}^{\alpha ,\beta }\left(z\right)\parallel }_{r}& \le & {r}^{j}\sum _{j=0}^{k}\frac{\left[n\right]\left[n+1\right]\cdots \left[n+j-1\right]}{{\left[n\right]}^{j}}{q}^{\frac{-j\left(j-1\right)}{2}}\\ ×\left[\frac{\left[\alpha \right]}{\left[n\right]+\left[\beta \right]},\dots ,\frac{{q}^{j-1}\left[\alpha \right]+\left[j\right]}{{q}^{j-1}\left(\left[n\right]+\left[\beta \right]\right)};{e}_{k}\right]\\ \le & \sum _{j=0}^{k}j!\frac{kk-1\cdots k-j+1}{j!}{r}^{k-j}\cdot {r}^{j}\\ =& {r}^{k}\sum _{j=0}^{k}kk-1\cdots k-j+1\le {r}^{k}\left(k+1\right)!.\end{array}$
(3.3)

Now, considering (3.3) in (3.2), for all $|z|\le r$, $r\ge 1$, with $q>1$ and $0\le \alpha \le \beta$,

$\begin{array}{r}|{T}_{n,k}^{\alpha ,\beta }\left(z\right)-{z}^{k}|\\ \phantom{\rule{1em}{0ex}}\le \frac{r\left(1+r\right)}{\left[n\right]+\left[\beta \right]}{r}^{k-2}\left(k+1\right)!+\frac{\left[n\right]r+\left[\alpha \right]}{\left[n\right]+\left[\beta \right]}|{T}_{n,k-1}^{\alpha ,\beta }\left(z\right)-{z}^{k-1}|\\ \phantom{\rule{2em}{0ex}}+\left(\frac{\left[n\right]}{\left[n\right]+\left[\beta \right]}-1\right){r}^{k}+\frac{\left[\alpha \right]}{\left[n\right]+\left[\beta \right]}{r}^{k-1}\\ \phantom{\rule{1em}{0ex}}\le \frac{\left[n\right]r+\left[\alpha \right]}{\left[n\right]+\left[\beta \right]}|{T}_{n,k-1}^{\alpha ,\beta }\left(z\right)-{z}^{k-1}|+\frac{r\left(1+r\right)}{\left[n\right]+\left[\beta \right]}{r}^{k-2}\left(k+1\right)!\\ \phantom{\rule{2em}{0ex}}+\frac{\left[\beta \right]}{\left[n\right]+\left[\beta \right]}{r}^{k}+\frac{\left[\alpha \right]}{\left[n\right]+\left[\beta \right]}{r}^{k-1}\\ \phantom{\rule{1em}{0ex}}\le r|{T}_{n,k-1}^{\alpha ,\beta }\left(z\right)-{z}^{k-1}|+\frac{2{r}^{k}}{\left[n\right]+\left[\beta \right]}\left(k+1\right)!\\ \phantom{\rule{2em}{0ex}}+\frac{\left[\beta \right]}{\left[n\right]+\left[\beta \right]}{r}^{k}+\frac{\left[\alpha \right]}{\left[n\right]+\left[\beta \right]}{r}^{k-1}.\end{array}$
(3.4)

Using the above inequalities beginning from $k=2,3,\dots$ and using the mathematical induction with respect to k, we arrive at

$\begin{array}{r}|{T}_{n,k}^{\alpha ,\beta }\left(z\right)-{z}^{k}|\\ \phantom{\rule{1em}{0ex}}\le \frac{2{r}^{k}}{\left[n\right]+\left[\beta \right]}\sum _{j=2}^{k}\left(j+1\right)!+\frac{\left[\beta \right]}{\left[n\right]+\left[\beta \right]}k{r}^{k}+\frac{\left[\alpha \right]}{\left[n\right]+\left[\beta \right]}k{r}^{k-1}\\ \phantom{\rule{1em}{0ex}}\le \frac{6{r}^{k}}{\left[n\right]+\left[\beta \right]}\left(k+1\right)!\left(k-1\right)+\frac{\left[\beta \right]}{\left[n\right]+\left[\beta \right]}k{r}^{k}+\frac{\left[\alpha \right]}{\left[n\right]+\left[\beta \right]}k{r}^{k-1}.\end{array}$
(3.5)

Also we obtain the following: for $k=1$ it is not difficult to see that

$|{T}_{n,1}^{\alpha ,\beta }\left(z\right)-z|=|\frac{\left[\alpha \right]-\left[\beta \right]z}{\left[n\right]+\left[\beta \right]}|\le \frac{\left[\alpha \right]+\left[\beta \right]r}{\left[n\right]+\left[\beta \right]}.$

Now, taking into account the proof of Theorem 1 in [13], we can write, for $q>1$, $|z|\le r$, $r\ge 1$, that

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

which implies

$\begin{array}{r}|{W}_{n,q}^{\alpha ,\beta }\left(f\right)\left(z\right)-f\left(z\right)|\\ \phantom{\rule{1em}{0ex}}\le \sum _{k=1}^{\mathrm{\infty }}|{c}_{k}||{T}_{n,k}^{\alpha ,\beta }\left(z\right)-{z}^{k}|\\ \phantom{\rule{1em}{0ex}}\le \frac{6}{\left[n\right]+\left[\beta \right]}\sum _{k=1}^{\mathrm{\infty }}|{c}_{k}|\left(k+1\right)!\left(k-1\right){r}^{k}\phantom{\rule{0.25em}{0ex}}+\frac{\left[\beta \right]}{\left[n\right]+\left[\beta \right]}\sum _{k=1}^{\mathrm{\infty }}|{c}_{k}|k{r}^{k}\\ \phantom{\rule{2em}{0ex}}+\frac{\left[\alpha \right]}{\left[n\right]+\left[\beta \right]}\sum _{k=1}^{\mathrm{\infty }}|{c}_{k}|k{r}^{k-1}\\ \phantom{\rule{1em}{0ex}}=\frac{{M}_{1,r}\left(f\right)}{\left[n\right]+\left[\beta \right]}+\frac{\left[\beta \right]}{\left[n\right]+\left[\beta \right]}{M}_{2,r}\left(f\right)+\frac{\left[\alpha \right]}{\left[n\right]+\left[\beta \right]}{M}_{3,r}\left(f\right).\end{array}$

Here from the analyticity of f we have ${M}_{2,r}\left(f\right)<\mathrm{\infty }$ and ${M}_{3,r}\left(f\right)<\mathrm{\infty }$. Also from the hypotheses of the theorem, one can get

${M}_{1,r}\left(f\right)=6\sum _{k=1}^{\mathrm{\infty }}|{c}_{k}|\left(k+1\right)!\left(k-1\right){r}^{k}\le 6M\sum _{k=1}^{\mathrm{\infty }}\left(k+1\right)\left(k-1\right){\left(rA\right)}^{k}$

for all $|z|\le r$, $1\le r\le \frac{1}{A}$, $n\in \mathbb{N}$. □

Theorem 2 Let $0\le \alpha \le \beta$, $1\le r\le \frac{1}{A}$ and $q>1$. Under the hypotheses of Theorem  1, for all $|z|\le r$ and $n\in \mathbb{N}$, the following Voronovskaja-type result

$\begin{array}{r}|{W}_{n,q}^{\alpha ,\beta }\left(f\right)\left(z\right)-f\left(z\right)-\frac{\left[\alpha \right]-\left[\beta \right]z}{\left[n\right]+\left[\beta \right]}{f}^{\prime }\left(z\right)-\frac{z}{2\left[n\right]}\left(1+\frac{z}{q}\right){f}^{″}\left(z\right)|\\ \phantom{\rule{1em}{0ex}}\le \frac{{K}_{1,r}\left(f\right)}{{\left[n\right]}^{2}}+\frac{{\sum }_{j=2}^{6}{K}_{j,r}\left(f\right)}{{\left(\left[n\right]+\left[\beta \right]\right)}^{2}}\end{array}$

holds with

$\begin{array}{c}{K}_{1,r}\left(f\right)=16\sum _{k=3}^{\mathrm{\infty }}|{c}_{k}|\left(k-1\right){\left(k-2\right)}^{2}k!{r}^{k}<\mathrm{\infty },\hfill \\ {K}_{2,r}\left(f\right)={\left[\alpha \right]}^{2}\sum _{k=2}^{\mathrm{\infty }}|{c}_{k}|\frac{\left(k-1\right)k!}{2}{r}^{k-2}<\mathrm{\infty },\hfill \\ {K}_{3,r}\left(f\right)=6\left[\alpha \right]\sum _{k=2}^{\mathrm{\infty }}|{c}_{k}|{k}^{2}k!{r}^{k-1}<\mathrm{\infty },\hfill \\ {K}_{4,r}\left(f\right)=\left(\frac{{\left[\beta \right]}^{2}}{2}+6\left[\beta \right]\right)\sum _{k=0}^{\mathrm{\infty }}|{c}_{k}|{k}^{2}\left(k+1\right)!{r}^{k}<\mathrm{\infty },\hfill \\ {K}_{5,r}\left(f\right)=\left[\alpha \right]\left[\beta \right]\sum _{k=0}^{\mathrm{\infty }}|{c}_{k}|k\left(k-1\right){r}^{k-1}<\mathrm{\infty },\hfill \\ {K}_{6,r}\left(f\right)={\left[\beta \right]}^{2}\sum _{k=0}^{\mathrm{\infty }}|{c}_{k}|k\left(k-1\right){r}^{k}<\mathrm{\infty }.\hfill \end{array}$

Proof For all $z\in {D}_{R}$, let us consider

$\begin{array}{r}{W}_{n,q}^{\alpha ,\beta }\left(f\right)\left(z\right)-f\left(z\right)-\frac{\left[\alpha \right]-\left[\beta \right]z}{\left[n\right]+\left[\beta \right]}{f}^{\prime }\left(z\right)-\frac{z}{2\left[n\right]}\left(1+\frac{z}{q}\right){f}^{″}\left(z\right)\\ \phantom{\rule{1em}{0ex}}={W}_{n,q}\left(f\right)\left(z\right)-f\left(z\right)-\frac{z}{2\left[n\right]}\left(1+\frac{z}{q}\right){f}^{″}\left(z\right)\\ \phantom{\rule{2em}{0ex}}+{W}_{n,q}^{\alpha ,\beta }\left(f\right)\left(z\right)-{W}_{n,q}\left(f\right)\left(z\right)-\frac{\left[\alpha \right]-\left[\beta \right]z}{\left[n\right]+\left[\beta \right]}{f}^{\prime }\left(z\right).\end{array}$

Using the fact that $f\left(z\right)={\sum }_{k=0}^{\mathrm{\infty }}{c}_{k}{z}^{k}$, we get

$\begin{array}{r}{W}_{n,q}^{\alpha ,\beta }\left(f\right)\left(z\right)-f\left(z\right)-\frac{\left[\alpha \right]-\left[\beta \right]z}{\left[n\right]+\left[\beta \right]}{f}^{\prime }\left(z\right)-\frac{z}{2\left[n\right]}\left(1+\frac{z}{q}\right){f}^{″}\left(z\right)\\ \phantom{\rule{1em}{0ex}}=\sum _{k=2}^{\mathrm{\infty }}{c}_{k}\left({W}_{n,q}\left({e}_{k};z\right)-{z}^{k}-\frac{z}{2\left[n\right]}\left(1+\frac{z}{q}\right)k\left(k-1\right){z}^{k-2}\right)\\ \phantom{\rule{2em}{0ex}}+\sum _{k=2}^{\mathrm{\infty }}{c}_{k}\left({T}_{n,k}^{\alpha ,\beta }\left(z\right)-{W}_{n,q}\left({e}_{k};z\right)-\frac{\left[\alpha \right]-\left[\beta \right]z}{\left[n\right]+\left[\beta \right]}k{z}^{k-1}\right).\end{array}$

From Theorem 2 in [13], we have

$\begin{array}{r}|{W}_{n,q}\left(f\right)\left(z\right)-f\left(z\right)-\frac{z}{2\left[n\right]}\left(1+\frac{z}{q}\right){f}^{″}\left(z\right)|\\ \phantom{\rule{1em}{0ex}}\le \frac{16}{{\left[n\right]}^{2}}\sum _{k=3}^{\mathrm{\infty }}|{c}_{k}|\left(k-1\right){\left(k-2\right)}^{2}k!{r}^{k}.\end{array}$

Furthermore, in order to estimate the second sum, using Lemma 2, we obtain

$\begin{array}{r}{T}_{n,k}^{\alpha ,\beta }\left(z\right)-{W}_{n,q}\left({e}_{k};z\right)-\frac{\left[\alpha \right]-\left[\beta \right]z}{\left[n\right]+\left[\beta \right]}k{z}^{k-1}\\ \phantom{\rule{1em}{0ex}}=\sum _{j=0}^{k}\left(\genfrac{}{}{0}{}{k}{j}\right)\frac{{\left[n\right]}^{j}{\left[\alpha \right]}^{k-j}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k}}{W}_{n,q}\left({e}_{j};z\right)-{W}_{n,q}\left({e}_{k};z\right)-\frac{\left[\alpha \right]-\left[\beta \right]z}{\left[n\right]+\left[\beta \right]}k{z}^{k-1}\\ \phantom{\rule{1em}{0ex}}=\sum _{j=0}^{k-1}\left(\genfrac{}{}{0}{}{k}{j}\right)\frac{{\left[n\right]}^{j}{\left[\alpha \right]}^{k-j}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k}}{W}_{n,q}\left({e}_{j};z\right)\\ \phantom{\rule{2em}{0ex}}+\left(\frac{{\left[n\right]}^{k}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k}}-1\right){W}_{n,q}\left({e}_{k};z\right)-\frac{\left[\alpha \right]-\left[\beta \right]z}{\left[n\right]+\left[\beta \right]}k{z}^{k-1}.\end{array}$

Also it is clear that

$1-\frac{{\left[n\right]}^{k}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k}}=\sum _{j=1}^{k-1}\left(\genfrac{}{}{0}{}{k}{j}\right)\frac{{\left[n\right]}^{j}{\left[\beta \right]}^{k-j}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k}}\le \sum _{j=1}^{k-1}\left(1-\frac{\left[n\right]}{\left[n\right]+\left[\beta \right]}\right)=\frac{k\left[\beta \right]}{\left[n\right]+\left[\beta \right]},$
(3.6)

which implies

$\begin{array}{r}{T}_{n,k}^{\alpha ,\beta }\left(z\right)-{W}_{n,q}\left({e}_{k};z\right)-\frac{\left[\alpha \right]-\left[\beta \right]z}{\left[n\right]+\left[\beta \right]}k{z}^{k-1}\\ \phantom{\rule{1em}{0ex}}=\sum _{j=0}^{k-2}\left(\genfrac{}{}{0}{}{k}{j}\right)\frac{{\left[n\right]}^{j}{\left[\alpha \right]}^{k-j}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k}}{W}_{n,q}\left({e}_{j};z\right)+\frac{k{\left[n\right]}^{k-1}\left[\alpha \right]}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k}}{W}_{n,q}\left({e}_{k-1};z\right)\\ \phantom{\rule{2em}{0ex}}-\sum _{j=0}^{k-2}\left(\genfrac{}{}{0}{}{k}{j}\right)\frac{{\left[n\right]}^{j}{\left[\beta \right]}^{k-j}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k}}{W}_{n,q}\left({e}_{k};z\right)-\frac{\left[\alpha \right]-\left[\beta \right]z}{\left[n\right]+\left[\beta \right]}k{z}^{k-1}\\ \phantom{\rule{1em}{0ex}}=\sum _{j=0}^{k-2}\left(\genfrac{}{}{0}{}{k}{j}\right)\frac{{\left[n\right]}^{j}{\left[\alpha \right]}^{k-j}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k}}{W}_{n,q}\left({e}_{j};z\right)+\frac{k{\left[n\right]}^{k-1}\left[\alpha \right]}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k}}\left({W}_{n,q}\left({e}_{k-1};z\right)-{z}^{k-1}\right)\\ \phantom{\rule{2em}{0ex}}-\sum _{j=0}^{k-2}\left(\genfrac{}{}{0}{}{k}{j}\right)\frac{{\left[n\right]}^{j}{\left[\beta \right]}^{k-j}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k}}{W}_{n,q}\left({e}_{k};z\right)-\frac{k{\left[n\right]}^{k-1}\left[\beta \right]}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k}}\left({W}_{n,q}\left({e}_{k};z\right)-{z}^{k}\right)\\ \phantom{\rule{2em}{0ex}}+\left(\frac{{\left[n\right]}^{k-1}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k-1}}-1\right)\frac{k\left[\alpha \right]}{\left[n\right]+\left[\beta \right]}{z}^{k-1}\\ \phantom{\rule{2em}{0ex}}+\left(1-\frac{{\left[n\right]}^{k-1}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k-1}}\right)\frac{k\left[\beta \right]}{\left[n\right]+\left[\beta \right]}{z}^{k}.\end{array}$
(3.7)

Now from (3.3) we obtain

$\begin{array}{r}|\sum _{j=0}^{k-2}\left(\genfrac{}{}{0}{}{k}{j}\right)\frac{{\left[n\right]}^{j}{\left[\alpha \right]}^{k-j}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k}}{W}_{n,q}\left({e}_{j};z\right)|\\ \phantom{\rule{1em}{0ex}}\le \sum _{j=0}^{k-2}\left(\genfrac{}{}{0}{}{k}{j}\right)\frac{{\left[n\right]}^{j}{\left[\alpha \right]}^{k-j}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k}}|{W}_{n,q}\left({e}_{j};z\right)|\\ \phantom{\rule{1em}{0ex}}=\sum _{j=0}^{k-2}\frac{k\left(k-1\right)}{\left(k-j\right)\left(k-j-1\right)}\left(\genfrac{}{}{0}{}{k-2}{j}\right)\frac{{\left[n\right]}^{j}{\left[\alpha \right]}^{k-j}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k}}|{W}_{n,q}\left({e}_{j};z\right)|\\ \phantom{\rule{1em}{0ex}}\le \frac{k\left(k-1\right)}{2}\frac{{\left[\alpha \right]}^{2}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{2}}{r}^{k-2}\left(k-1\right)!\sum _{j=0}^{k-2}\left(\genfrac{}{}{0}{}{k-2}{j}\right)\frac{{\left[n\right]}^{j}{\left[\alpha \right]}^{k-2-j}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k-2}}\\ \phantom{\rule{1em}{0ex}}\le \frac{k\left(k-1\right)}{2}\frac{{\left[\alpha \right]}^{2}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{2}}{r}^{k-2}\left(k-1\right)!.\end{array}$
(3.8)

Also, we need to prove the following inequality:

$\begin{array}{rl}\sum _{j=0}^{k-2}\left(\genfrac{}{}{0}{}{k-2}{j}\right)\frac{{\left[n\right]}^{j}{\left[\alpha \right]}^{k-2-j}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k-2}}& =\sum _{j=0}^{k-2}\left(\genfrac{}{}{0}{}{k-2}{j}\right)\frac{{\left[n\right]}^{j}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{j}}\frac{{\left[\alpha \right]}^{k-2-j}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k-2-j}}\\ ={\left(\frac{\left[n\right]+\left[\alpha \right]}{\left[n\right]+\left[\beta \right]}\right)}^{k-2}\le 1.\end{array}$
(3.9)

Moreover, taking $\alpha =\beta =0$ in Theorem 1, we have

$|{W}_{n,q}\left({e}_{k};z\right)-{z}^{k}|\le \frac{6}{\left[n\right]}{r}^{k}\left(k+1\right)!\left(k-1\right).$
(3.10)

Writing (3.8), (3.6), (3.9) and (3.10) in (3.7), we have

$\begin{array}{r}|{T}_{n,k}^{\alpha ,\beta }\left(z\right)-{W}_{n,q}\left({e}_{k};z\right)-\frac{\left[\alpha \right]-\left[\beta \right]z}{\left[n\right]+\left[\beta \right]}k{z}^{k-1}|\\ \phantom{\rule{1em}{0ex}}\le |\sum _{j=0}^{k-2}\left(\genfrac{}{}{0}{}{k}{j}\right)\frac{{\left[n\right]}^{j}{\left[\alpha \right]}^{k-j}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k}}{W}_{n,q}\left({e}_{j};z\right)|+\frac{k{\left[n\right]}^{k-1}\left[\alpha \right]}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k}}|{W}_{n,q}\left({e}_{k-1};z\right)-{z}^{k-1}|\\ \phantom{\rule{2em}{0ex}}+|\sum _{j=0}^{k-2}\left(\genfrac{}{}{0}{}{k}{j}\right)\frac{{\left[n\right]}^{j}{\left[\beta \right]}^{k-j}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k}}{W}_{n,q}\left({e}_{k};z\right)|+\frac{k{\left[n\right]}^{k-1}\left[\beta \right]}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k}}|{W}_{n,q}\left({e}_{k};z\right)-{z}^{k}|\\ \phantom{\rule{2em}{0ex}}+|\frac{{\left[n\right]}^{k-1}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k-1}}-1|\frac{k\left[\alpha \right]}{\left[n\right]+\left[\beta \right]}|z{|}^{k-1}+|1-\frac{{\left[n\right]}^{k-1}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k-1}}|\frac{k\left[\beta \right]}{\left[n\right]+\left[\beta \right]}|z{|}^{k}\\ \phantom{\rule{1em}{0ex}}\le \frac{\left(k-1\right)k!}{2}\frac{{\left[\alpha \right]}^{2}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{2}}{r}^{k-2}+\frac{k{\left[n\right]}^{k-1}\left[\alpha \right]}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k}}\frac{6}{\left[n\right]}{r}^{k-1}k!\left(k-2\right)\\ \phantom{\rule{2em}{0ex}}+{r}^{k}\left(k+1\right)!\sum _{j=0}^{k-2}\left(\genfrac{}{}{0}{}{k}{j}\right)\frac{{\left[n\right]}^{j}{\left[\beta \right]}^{k-j}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k}}\\ \phantom{\rule{2em}{0ex}}+\frac{k{\left[n\right]}^{k-1}\left[\beta \right]}{{\left(\left[n\right]+\left[\beta \right]\right)}^{k}}\frac{6}{\left[n\right]}{r}^{k}\left(k+1\right)!\left(k-1\right)+\frac{k\left(k-1\right)\left[\alpha \right]\left[\beta \right]}{{\left(\left[n\right]+\left[\beta \right]\right)}^{2}}{r}^{k-1}+\frac{k\left(k-1\right){\left[\beta \right]}^{2}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{2}}{r}^{k}\\ \phantom{\rule{1em}{0ex}}\le \frac{\left(k-1\right)k!}{2}\frac{{\left[\alpha \right]}^{2}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{2}}{r}^{k-2}+6\frac{{k}^{2}\left[\alpha \right]}{{\left(\left[n\right]+\left[\beta \right]\right)}^{2}}{r}^{k-1}k!+\frac{{k}^{2}{\left[\beta \right]}^{2}\left(k+1\right)!}{2{\left(\left[n\right]+\left[\beta \right]\right)}^{2}}{r}^{k}\\ \phantom{\rule{2em}{0ex}}+6\frac{{k}^{2}\left(k+1\right)!\left[\beta \right]}{{\left(\left[n\right]+\left[\beta \right]\right)}^{2}}{r}^{k}+\frac{k\left(k-1\right)\left[\alpha \right]\left[\beta \right]}{{\left(\left[n\right]+\left[\beta \right]\right)}^{2}}{r}^{k-1}+\frac{k\left(k-1\right){\left[\beta \right]}^{2}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{2}}{r}^{k}\\ \phantom{\rule{1em}{0ex}}\le \frac{\left(k-1\right)k!}{2}\frac{{\left[\alpha \right]}^{2}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{2}}{r}^{k-2}+6\frac{{k}^{2}\left[\alpha \right]}{{\left(\left[n\right]+\left[\beta \right]\right)}^{2}}{r}^{k-1}k!\\ \phantom{\rule{2em}{0ex}}+\left(\frac{{\left[\beta \right]}^{2}}{2}+6\left[\beta \right]\right)\frac{{k}^{2}\left(k+1\right)!}{{\left(\left[n\right]+\left[\beta \right]\right)}^{2}}{r}^{k}\\ \phantom{\rule{2em}{0ex}}+\frac{k\left(k-1\right)\left[\alpha \right]\left[\beta \right]}{{\left(\left[n\right]+\left[\beta \right]\right)}^{2}}{r}^{k-1}+\frac{k\left(k-1\right){\left[\beta \right]}^{2}}{{\left(\left[n\right]+\left[\beta \right]\right)}^{2}}{r}^{k}.\end{array}$

Thus the proof is completed. □

Now, let us give a lower estimate for the exact degree in approximation by ${W}_{n,q}^{\alpha ,\beta }$.

Theorem 3 Suppose that $q>1$ and suppose that the hypotheses on f and on the constants R, M, A in the statement of Theorem  1 hold, and let $1\le r, $0\le \alpha \le \beta$. If f is not a polynomial of degree ≤0, then the lower estimate

${\parallel {W}_{n,q}^{\alpha ,\beta }\left(f\right)-f\parallel }_{r}\ge \frac{{C}_{r}^{\alpha ,\beta }\left(f\right)}{\left[n\right]}$

holds for all n, where the constant ${C}_{r}^{\alpha ,\beta }\left(f\right)$ depends on f, α, β, q and r.

Proof For all $|z|\le r$ and $n\in \mathbb{N}$, we get

$\begin{array}{r}{W}_{n,q}^{\alpha ,\beta }\left(f\right)\left(z\right)-f\left(z\right)\\ \phantom{\rule{1em}{0ex}}=\frac{1}{\left[n\right]}\left\{\frac{\left[n\right]}{\left[n\right]+\left[\beta \right]}\left(\left[\alpha \right]-\left[\beta \right]z\right){f}^{\prime }\left(z\right)+\frac{z}{2}\left(1+\frac{z}{q}\right){f}^{″}\left(z\right)\\ \phantom{\rule{2em}{0ex}}+\frac{1}{\left[n\right]}{\left[n\right]}^{2}\left({W}_{n,q}^{\alpha ,\beta }\left(f\right)\left(z\right)-f\left(z\right)-\frac{\left[\alpha \right]-\left[\beta \right]z}{\left[n\right]+\left[\beta \right]}{f}^{\prime }\left(z\right)-\frac{z}{2\left[n\right]}\left(1+\frac{z}{q}\right){f}^{″}\left(z\right)\right)\right\}\\ \phantom{\rule{1em}{0ex}}=\frac{1}{\left[n\right]}\left\{\left(\left[\alpha \right]-\left[\beta \right]z\right){f}^{\prime }\left(z\right)+\frac{z}{2}\left(1+\frac{z}{q}\right){f}^{″}\left(z\right)\\ \phantom{\rule{2em}{0ex}}+\frac{1}{\left[n\right]}{\left[n\right]}^{2}\left({W}_{n,q}^{\alpha ,\beta }\left(f\right)\left(z\right)-f\left(z\right)-\frac{\left[\alpha \right]-\left[\beta \right]z}{\left[n\right]+\left[\beta \right]}{f}^{\prime }\left(z\right)\right)\\ \phantom{\rule{2em}{0ex}}+\frac{1}{\left[n\right]}{\left[n\right]}^{2}\left(-\frac{z}{2\left[n\right]}\left(1+\frac{z}{q}\right){f}^{″}\left(z\right)-\frac{\left[\beta \right]\left(\left[\alpha \right]-\left[\beta \right]z\right)}{\left[n\right]\left(\left[n\right]+\left[\beta \right]\right)}{f}^{\prime }\left(z\right)\right)\right\}.\end{array}$

We set ${E}_{k,n}\left(z\right)$ by

$\begin{array}{rcl}{E}_{k,n}\left(z\right)& :=& {W}_{n,q}^{\alpha ,\beta }\left(f\right)\left(z\right)-f\left(z\right)-\frac{\left[\alpha \right]-\left[\beta \right]z}{\left[n\right]+\left[\beta \right]}{f}^{\prime }\left(z\right)\\ -\frac{z}{2\left[n\right]}\left(1+\frac{z}{q}\right){f}^{″}\left(z\right)-\frac{\left[\beta \right]\left(\left[\alpha \right]-\left[\beta \right]z\right)}{\left[n\right]\left(\left[n\right]+\left[\beta \right]\right)}{f}^{\prime }\left(z\right).\end{array}$
(3.11)

Passing to the norm and using the inequality

${\parallel F+G\parallel }_{r}\ge |{\parallel F\parallel }_{r}-{\parallel G\parallel }_{r}|\ge {\parallel F\parallel }_{r}-{\parallel G\parallel }_{r},$

we get

${\parallel {W}_{n,q}^{\alpha ,\beta }\left(f\right)-f\parallel }_{r}\ge \frac{1}{\left[n\right]}{\parallel \left(\left[\alpha \right]-\left[\beta \right]{e}_{1}\right){f}^{\prime }+\frac{{e}_{1}}{2}\left(1+\frac{{e}_{1}}{q}\right){f}^{″}\parallel }_{r}-\frac{1}{\left[n\right]}{\left[n\right]}^{2}{\parallel {E}_{k,n}\parallel }_{r}.$

Since f is not a polynomial of degree ≤0 in ${D}_{R}$, we have

${\parallel \left(\left[\alpha \right]-\left[\beta \right]{e}_{1}\right){f}^{\prime }+\frac{{e}_{1}}{2}\left(1+\frac{{e}_{1}}{q}\right){f}^{″}\parallel }_{r}>0.$

It can also be seen in [[1], pp.75-76]. Now, from Theorem 2 it follows that

$\begin{array}{rl}{\left[n\right]}^{2}{\parallel {E}_{k,n}\parallel }_{r}\le & {\left[n\right]}^{2}{\parallel {W}_{n,q}^{\alpha ,\beta }\left(f\right)-f-\left(\frac{\left[\alpha \right]-\left[\beta \right]{e}_{1}}{\left[n\right]+\left[\beta \right]}\right){f}^{\prime }-\frac{{e}_{1}}{2\left[n\right]}\left(1+\frac{{e}_{1}}{q}\right){f}^{″}\parallel }_{r}\\ +{\parallel \left[\beta \right]\left(\left[\alpha \right]-\left[\beta \right]{e}_{1}\right){f}^{\prime }\parallel }_{r}\\ \le & \sum _{j=1}^{6}{M}_{j,r}\left(f\right)+\left[\beta \right]\left(\left[\alpha \right]+\left[\beta \right]r\right){\parallel {f}^{\prime }\parallel }_{r}.\end{array}$

Since $\frac{1}{\left[n\right]}\to 0$ as $n\to \mathrm{\infty }$, for $q>1$, there exists an ${n}_{0}$ depending on f, r, α, β and q such that for all $n\ge {n}_{0}$,

$\begin{array}{r}\frac{1}{\left[n\right]}{\parallel \left(\left[\alpha \right]-\left[\beta \right]{e}_{1}\right){f}^{\prime }+\frac{{e}_{1}}{2}\left(1+\frac{{e}_{1}}{q}\right){f}^{″}\parallel }_{r}-\frac{1}{\left[n\right]}{\left[n\right]}^{2}{\parallel {E}_{k,n}\parallel }_{r}\\ \phantom{\rule{1em}{0ex}}\ge \frac{1}{2}{\parallel \left(\left[\alpha \right]-\left[\beta \right]{e}_{1}\right){f}^{\prime }+\frac{{e}_{1}}{2}\left(1+\frac{{e}_{1}}{q}\right){f}^{″}\parallel }_{r},\end{array}$

which implies

${\parallel {W}_{n,q}^{\alpha ,\beta }\left(f\right)-f\parallel }_{r}\ge \frac{1}{2\left[n\right]}{\parallel \left(\left[\alpha \right]-\left[\beta \right]{e}_{1}\right){f}^{\prime }+\frac{{e}_{1}}{2}\left(1+\frac{{e}_{1}}{q}\right){f}^{″}\parallel }_{r}$

for all $n\ge {n}_{0}$. Now, for $n\in \left\{1,\dots ,{n}_{0}-1\right\}$, we have

${\parallel {W}_{n,q}^{\alpha ,\beta }\left(f\right)-f\parallel }_{r}\ge \frac{{A}_{r}\left(f\right)}{\left[n\right]}$

with

${A}_{r}\left(f\right)=\left[n\right]{\parallel {W}_{n,q}^{\alpha ,\beta }\left(f\right)-f\parallel }_{r}>0,$

which finally implies

${\parallel {W}_{n,q}^{\alpha ,\beta }\left(f\right)-f\parallel }_{r}\ge \frac{{C}_{r}^{\alpha ,\beta }\left(f\right)}{\left[n\right]}$

for all $n\ge {n}_{0}$ with

${C}_{r}^{\alpha ,\beta }\left(f\right)=min\left\{{A}_{r,1}\left(f\right),\dots ,{A}_{r,{n}_{0}-1}\left(f\right),\frac{1}{2}{\parallel \left(\left[\alpha \right]-\left[\beta \right]{e}_{1}\right){f}^{\prime }+\frac{{e}_{1}}{2}\left(1+\frac{{e}_{1}}{q}\right){f}^{″}\parallel }_{r}\right\}.$

This proves the theorem. □

Combining now Theorem 3 with Theorem 1, we immediately get the following equivalence result.

Remark 1 Suppose that $q>1$, $0\le \alpha \le \beta$ and that the hypotheses on f and on the constants R, M, A in the statement of Theorem 1 hold, and let $1\le r<\frac{1}{A}$ be fixed. If f is not a polynomial of degree ≤0, then we have the following equivalence:

${\parallel {W}_{n,q}^{\alpha ,\beta }\left(f\right)-f\parallel }_{r}\sim \frac{1}{\left[n\right]}$

for all n, where the constants in the equivalence depend on f, α, β, q and r.

Concerning the approximation by the derivatives of complex q-Baskakov-Stancu operators, we can state the following theorem.

Theorem 4 Suppose that $q>1$ and that the hypotheses on f and on the constants R, M, A in the statement of Theorem  1 hold, and let $0\le \alpha \le \beta$, $1\le r<{r}_{1}<\frac{1}{A}$ and $p\in \mathbb{N}$ be fixed. If f is not a polynomial of degree $\le p-1$, then we have the following equivalence:

${\parallel {\left[{W}_{n,q}^{\alpha ,\beta }\left(f\right)\right]}^{\left(p\right)}-{f}^{\left(p\right)}\parallel }_{r}\sim \frac{1}{\left[n\right]}$

for all n, where the constants in the equivalence depend on f (that is, on M, A), r, ${r}_{1}q$ and p.

Proof Denote by Γ the circle of radius ${r}_{1}$ with $1\le r<{r}_{1}<\frac{1}{A}$ centered at 0. Since $|z|\le r$ and $\gamma \in \mathrm{\Gamma }$, we have $|\gamma -z|\ge {r}_{1}-r$ and from Cauchy’s formulas and Theorem 1 we obtain, for all $|z|\le r$ and $n\in \mathbb{N}$, that

$\begin{array}{rcl}|{\left[{W}_{n,q}^{\alpha ,\beta }\left(f,z\right)\right]}^{\left(p\right)}-{f}^{\left(p\right)}\left(z\right)|& \le & \frac{p!}{2\pi }|{\int }_{\mathrm{\Gamma }}\frac{{W}_{n,q}^{\alpha ,\beta }f\left(\gamma \right)-f\left(\gamma \right)}{{\left(\gamma -z\right)}^{p+1}}\phantom{\rule{0.2em}{0ex}}d\gamma |\\ \le & \frac{{M}_{{r}_{1,}\alpha ,\beta }\left(f\right)}{\left[n\right]}\frac{p!}{2\pi }\frac{2\pi {r}_{1}}{{\left({r}_{1}-r\right)}^{p+1}}\\ =& \frac{{M}_{{r}_{1,}\alpha ,\beta }\left(f\right)}{\left[n\right]}\frac{p!{r}_{1}}{{\left({r}_{1}-r\right)}^{p+1}},\end{array}$

which proves one of the inequalities in the equivalence.

Now we need to prove the lower estimate. From Cauchy’s formula we get

${\left[{W}_{n,q}^{\alpha ,\beta }\left(f,z\right)\right]}^{\left(p\right)}-{f}^{\left(p\right)}\left(z\right)=\frac{p!}{2\pi i}{\int }_{\mathrm{\Gamma }}\frac{{W}_{n,q}^{\alpha ,\beta }f\left(\gamma \right)-f\left(\gamma \right)}{{\left(\gamma -z\right)}^{p+1}}\phantom{\rule{0.2em}{0ex}}d\gamma .$

Furthermore, using (3.11) one can have

$\begin{array}{r}{W}_{n,q}^{\alpha ,\beta }f\left(\gamma \right)-f\left(\gamma \right)\\ \phantom{\rule{1em}{0ex}}=\frac{1}{\left[n\right]}\left\{\left(\left[\alpha \right]-\left[\beta \right]\gamma \right){f}^{\prime }\left(\gamma \right)+\frac{\gamma }{2}\left(1+\frac{\gamma }{q}\right){f}^{″}\left(\gamma \right)+{\left[n\right]}^{2}{E}_{k,n}\left(\gamma \right)\right\}\end{array}$

for all $\gamma \in \mathrm{\Gamma }$ and $n\in \mathbb{N}$. Applications of Cauchy’s formula imply

$\begin{array}{r}{\left[{W}_{n,q}^{\alpha ,\beta }\left(f,z\right)\right]}^{\left(p\right)}-{f}^{\left(p\right)}\left(z\right)\\ \phantom{\rule{1em}{0ex}}=\left\{\frac{1}{\left[n\right]}\frac{p!}{2\pi i}{\int }_{\mathrm{\Gamma }}\frac{\left(\left[\alpha \right]-\left[\beta \right]\gamma \right){f}^{\prime }\left(\gamma \right)+\frac{\gamma }{2}\left(1+\frac{\gamma }{q}\right){f}^{″}\left(\gamma \right)}{{\left(\gamma -z\right)}^{p+1}}\phantom{\rule{0.2em}{0ex}}d\gamma \\ \phantom{\rule{2em}{0ex}}+\frac{1}{\left[n\right]}\frac{p!}{2\pi i}{\int }_{\mathrm{\Gamma }}\frac{{\left[n\right]}^{2}{E}_{k,n}\left(\gamma \right)}{{\left(\gamma -z\right)}^{p+1}}\phantom{\rule{0.2em}{0ex}}d\gamma \right\}\\ \phantom{\rule{1em}{0ex}}=\frac{1}{\left[n\right]}\left\{{\left[\left(\left[\alpha \right]-\left[\beta \right]\gamma \right){f}^{\prime }\left(\gamma \right)+\frac{z}{2}\left(1+\frac{z}{q}\right){f}^{″}\left(z\right)\right]}^{\left(p\right)}+\frac{p!}{2\pi i}{\int }_{\mathrm{\Gamma }}\frac{{\left[n\right]}^{2}{E}_{k,n}\left(\gamma \right)}{{\left(\gamma -z\right)}^{p+1}}\phantom{\rule{0.2em}{0ex}}d\gamma \right\}.\end{array}$

Now passing to the norm ${\parallel \cdot \parallel }_{r}$ we obtain

$\begin{array}{rcl}{\parallel {\left[{W}_{n,q}^{\alpha ,\beta }\left(f\right)\right]}^{\left(p\right)}-{f}^{\left(p\right)}\parallel }_{r}& \ge & \frac{1}{\left[n\right]}\left\{{\parallel {\left[\left(\left[\alpha \right]-\left[\beta \right]{e}_{1}\right){f}^{\prime }+\frac{{e}_{1}}{2}\left(1+\frac{{e}_{1}}{q}\right){f}^{″}\right]}^{\left(p\right)}\parallel }_{r}\\ -\frac{1}{\left[n\right]}{\parallel \frac{p!}{2\pi }{\int }_{\mathrm{\Gamma }}\frac{{\left[n\right]}^{2}{E}_{k,n}\left(\gamma \right)}{{\left(\gamma -z\right)}^{p+1}}\phantom{\rule{0.2em}{0ex}}d\gamma \parallel }_{r}\right\},\end{array}$

and from Theorem 2 we have

$\begin{array}{rcl}{\parallel \frac{p!}{2\pi }{\int }_{\mathrm{\Gamma }}\frac{{\left[n\right]}^{2}{E}_{k,n}\left(\gamma \right)}{{\left(\gamma -z\right)}^{p+1}}\phantom{\rule{0.2em}{0ex}}d\gamma \parallel }_{r}& \le & \frac{p!}{2\pi }\frac{2\pi {r}_{1}}{{\left({r}_{1}-r\right)}^{p+1}}{\left[n\right]}^{2}{\parallel {E}_{k,n}\parallel }_{{r}_{1}}\\ \le & {K}_{1,{r}_{1}}\left(f\right)+{\left[n\right]}^{2}\frac{{\sum }_{j=2}^{6}{K}_{j,{r}_{1}}\left(f\right)}{{\left(\left[n\right]+\left[\beta \right]\right)}^{2}}+\left[\beta \right]\left(\left[\alpha \right]+\left[\beta \right]{r}_{1}\right){\parallel {f}^{\prime }\parallel }_{{r}_{1}}.\end{array}$

Since f is not a polynomial of degree ≤0 in ${D}_{R}$, we have

${\parallel {\left[\left(\left[\alpha \right]-\left[\beta \right]{e}_{1}\right){f}^{\prime }+\frac{{e}_{1}}{2}\left(1+\frac{{e}_{1}}{q}\right){f}^{″}\right]}^{\left(p\right)}\parallel }_{r}>0$

(see [[1], pp.77-78]). The rest of the proof is obtained similarly to that of Theorem 3. □

Remark 2 Note that if we take $\alpha =\beta =0$, then Theorems 1, 2, 3 and 4 reduce to the results in [13].

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Correspondence to Didem Aydın Arı.

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Özden, D.S., Arı, D.A. Approximation by a complex q-Baskakov-Stancu operator in compact disks. J Inequal Appl 2014, 249 (2014). https://doi.org/10.1186/1029-242X-2014-249