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 $[n]$ and the q-factorial $[n]!$ are defined by

$$[n]:={[n]}_q=\{\begin{array}{ll}\frac{1-q^n}{1-q},& q\ne 1,\\ n,& q=1\end{array}\text{for }n\in \mathbb{N}$$

and

$$[n]!:=\{\begin{array}{ll}{[1]}_q{[2]}_q\cdots {[n]}_q,& n=1,2,\dots ,\\ 1,& n=0\end{array}\text{for }n\in \mathbb{N}\text{ and }[0]!=1,$$

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

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

The q-derivative of $f(z)$ is denoted by $D_{q}f(z)$ and defined as

$$D_{q}f(z):=\frac{f(qz)-f(z)}{(q-1)z},z\ne 0,D_{q}f(0)=f^{\prime }(0),$$

also

$$D_q^{0}f:=f,D_q^{n}f:=D_q\left(D_q^{n-1}f\right),n=1,2,\dots$$

q-Pochhammer formula is given by

$$\begin{array}{c}{(x,q)}_0=1,\hfill \\ {(x,q)}_n=\prod _{k=0}^{n-1}(1-q^{k}x)\hfill \end{array}$$

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

$$D_q(f(z)g(z))=f(z)D_q(g(z))+g(qz)D_q(f(z))$$

and

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

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

$$[x_0,x_1,\dots ,x_m;f\cdot g]=\sum _{i=0}^m[x_0,x_1,\dots ,x_i;f][x_i,x_{i+1},\dots ,x_m;g],$$

(1.1)

where $[x_0,x_1,\dots ,x_m;f]$ 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

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

where $x\ge 0$, $q>0$ and f is a real-valued continuous function on $[0,\mathrm{\infty })$. 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:

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

(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, $Z_n^q(f)(x)=W_{n,q}(f)(x)$ for $x\ge 0$ and $q>0$, so they proved that

$$[0,\frac{1}{[n]},\frac{[2]}{q[n]},\dots ,\frac{[j]}{q^{j-1}[n]};f]=\frac{q^{j(j-1)}{\mathrm{\nabla }}_q^{j}f(0)}{[j]!}{[n]}^j=\frac{f^{(j)}(\zeta )}{j!},\zeta \in (0,\frac{[j]}{q^{j-1}[n]}),$$

(1.3)

where ${\mathrm{\nabla }}_q^r$ stands for q-divided differences given by ${\mathrm{\nabla }}_q^{0}f(x_j)$,

$${\mathrm{\nabla }}_q^{r+1}f(x_j)=q^r{\mathrm{\nabla }}_q^{r}f(x_{j+1})-{\mathrm{\nabla }}_q^{r}f(x_j)$$

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

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 $Z_n^q(f)(x)$ for $0<q<1$. 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 $0<q<1$. 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

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

(1.4)

for $q>1$, $f:{\overline{D}}_R\cup [R,\mathrm{\infty })\to \mathbb{C}$, replacing x by z in the operator $W_{n,q}(f)(x)$ 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

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

(1.5)

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

Note that taking $\alpha =\beta =0$, $W_{n,q}^{\alpha ,\beta }(f)(z)$ reduces to the complex q-Baskakov operator $W_{n,q}(f)(z)$ 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 $e_k(z)=z^k$, $T_{n,k}^{\alpha ,\beta }(z):=W_{n,q}^{\alpha ,\beta }(e_k)(z)$, 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:

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

(2.1)

Hence

$$T_{n,1}^{\alpha ,\beta }(z)=\frac{[n]z+[\alpha ]}{[n]+[\beta ]},T_{n,2}^{\alpha ,\beta }(z)=\frac{z(1+\frac{z}{q})}{[n]+[\beta ]}\frac{[n]}{[n]+[\beta ]}+{\left(\frac{[n]z+[\alpha ]}{[n]+[\beta ]}\right)}^2$$

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

Proof Now we can write

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

(2.2)

Using relation (1.1) and taking $f=e_k$, $g=e_1$ and $x_j=\frac{q^{j-1}[\alpha ]+[j]}{q^{j-1}([n]+[\beta ])}$, we obtain

$$\begin{array}{r}[\frac{[\alpha ]}{[n]+[\beta ]},\dots ,\frac{q^{j-1}[\alpha ]+[j]}{q^{j-1}([n]+[\beta ])};e_{k+1}]\\ =\frac{q^{j-1}[\alpha ]+[j]}{q^{j-1}([n]+[\beta ])}[\frac{[\alpha ]}{[n]+[\beta ]},\dots ,\frac{q^{j-1}[\alpha ]+[j]}{q^{j-1}([n]+[\beta ])};e_k]\\ +[\frac{[\alpha ]}{[n]+[\beta ]},\dots ,\frac{q^{j-2}[\alpha ]+[j-1]}{q^{j-2}([n]+[\beta ])};e_k],\end{array}$$

(2.3)

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

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

 □

Lemma 2 Let α and β satisfy $0\le \alpha \le \beta$. Denoting $e_j(z)=z^j$ and $W_{n,q}^{0,0}(e_j)$ by $W_{n,q}(e_j)$ 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 $W_{n,q}(e_j)$, $j=0,1,\dots ,k$:

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

(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

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

Using (2.1), we have

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

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

$$W_{n,q}(e_{j+1},z)=\frac{qz(1+\frac{z}{q})}{[n]}{D}_q{W}_{n,q}(e_j,\frac{z}{q})+z{W}_{n,q}(e_j,z),$$

which implies

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

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 $1<R<\mathrm{\infty }$, let

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

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

$$\begin{array}{r}|W_{n,q}^{\alpha ,\beta }(f)(z)-f(z)|\\ \le \frac{M_{1,r}(f)}{[n]+[\beta ]}+\frac{[\beta ]}{[n]+[\beta ]}{M}_{2,r}(f)+\frac{[\alpha ]}{[n]+[\beta ]}{M}_{3,r}(f)\\ =M_{r,\alpha ,\beta }(f)\end{array}$$

with

$$\begin{array}{c}{M}_{1,r}(f)=6\sum _{k=2}^{\mathrm{\infty }}|c_k|(k+1)!(k-1)r^k<\mathrm{\infty },\hfill \\ M_{2,r}(f)=\sum _{k=1}^{\mathrm{\infty }}|c_k|k{r}^k<\mathrm{\infty },M_{3,r}(f)=\sum _{k=1}^{\mathrm{\infty }}|c_k|k{r}^{k-1}<\mathrm{\infty }.\hfill \end{array}$$

Proof Using (2.1), one can obtain

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

Moreover, we have

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

(3.1)

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

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

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

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

(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:

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

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

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

(3.3)

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

$$\begin{array}{r}|T_{n,k}^{\alpha ,\beta }(z)-z^k|\\ \le \frac{r(1+r)}{[n]+[\beta ]}{r}^{k-2}(k+1)!+\frac{[n]r+[\alpha ]}{[n]+[\beta ]}|T_{n,k-1}^{\alpha ,\beta }(z)-z^{k-1}|\\ +(\frac{[n]}{[n]+[\beta ]}-1)r^k+\frac{[\alpha ]}{[n]+[\beta ]}{r}^{k-1}\\ \le \frac{[n]r+[\alpha ]}{[n]+[\beta ]}|T_{n,k-1}^{\alpha ,\beta }(z)-z^{k-1}|+\frac{r(1+r)}{[n]+[\beta ]}{r}^{k-2}(k+1)!\\ +\frac{[\beta ]}{[n]+[\beta ]}{r}^k+\frac{[\alpha ]}{[n]+[\beta ]}{r}^{k-1}\\ \le r|T_{n,k-1}^{\alpha ,\beta }(z)-z^{k-1}|+\frac{2r^k}{[n]+[\beta ]}(k+1)!\\ +\frac{[\beta ]}{[n]+[\beta ]}{r}^k+\frac{[\alpha ]}{[n]+[\beta ]}{r}^{k-1}.\end{array}$$

(3.4)

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

$$\begin{array}{r}|T_{n,k}^{\alpha ,\beta }(z)-z^k|\\ \le \frac{2r^k}{[n]+[\beta ]}\sum _{j=2}^k(j+1)!+\frac{[\beta ]}{[n]+[\beta ]}k{r}^k+\frac{[\alpha ]}{[n]+[\beta ]}k{r}^{k-1}\\ \le \frac{6r^k}{[n]+[\beta ]}(k+1)!(k-1)+\frac{[\beta ]}{[n]+[\beta ]}k{r}^k+\frac{[\alpha ]}{[n]+[\beta ]}k{r}^{k-1}.\end{array}$$

(3.5)

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

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

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

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

which implies

$$\begin{array}{r}|W_{n,q}^{\alpha ,\beta }(f)(z)-f(z)|\\ \le \sum _{k=1}^{\mathrm{\infty }}|c_k||T_{n,k}^{\alpha ,\beta }(z)-z^k|\\ \le \frac{6}{[n]+[\beta ]}\sum _{k=1}^{\mathrm{\infty }}|c_k|(k+1)!(k-1)r^k+\frac{[\beta ]}{[n]+[\beta ]}\sum _{k=1}^{\mathrm{\infty }}|c_k|k{r}^k\\ +\frac{[\alpha ]}{[n]+[\beta ]}\sum _{k=1}^{\mathrm{\infty }}|c_k|k{r}^{k-1}\\ =\frac{M_{1,r}(f)}{[n]+[\beta ]}+\frac{[\beta ]}{[n]+[\beta ]}{M}_{2,r}(f)+\frac{[\alpha ]}{[n]+[\beta ]}{M}_{3,r}(f).\end{array}$$

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

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

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

$$\begin{array}{r}|W_{n,q}^{\alpha ,\beta }(f)(z)-f(z)-\frac{[\alpha ]-[\beta ]z}{[n]+[\beta ]}{f}^{\prime }(z)-\frac{z}{2[n]}(1+\frac{z}{q})f^{″}(z)|\\ \le \frac{K_{1,r}(f)}{{[n]}^2}+\frac{{\sum }_{j=2}^6{K}_{j,r}(f)}{{([n]+[\beta ])}^2}\end{array}$$

holds with

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

Proof For all $z\in D_R$, let us consider

$$\begin{array}{r}{W}_{n,q}^{\alpha ,\beta }(f)(z)-f(z)-\frac{[\alpha ]-[\beta ]z}{[n]+[\beta ]}{f}^{\prime }(z)-\frac{z}{2[n]}(1+\frac{z}{q})f^{″}(z)\\ =W_{n,q}(f)(z)-f(z)-\frac{z}{2[n]}(1+\frac{z}{q})f^{″}(z)\\ +W_{n,q}^{\alpha ,\beta }(f)(z)-W_{n,q}(f)(z)-\frac{[\alpha ]-[\beta ]z}{[n]+[\beta ]}{f}^{\prime }(z).\end{array}$$

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

$$\begin{array}{r}{W}_{n,q}^{\alpha ,\beta }(f)(z)-f(z)-\frac{[\alpha ]-[\beta ]z}{[n]+[\beta ]}{f}^{\prime }(z)-\frac{z}{2[n]}(1+\frac{z}{q})f^{″}(z)\\ =\sum _{k=2}^{\mathrm{\infty }}{c}_k(W_{n,q}(e_k;z)-z^k-\frac{z}{2[n]}(1+\frac{z}{q})k(k-1)z^{k-2})\\ +\sum _{k=2}^{\mathrm{\infty }}{c}_k(T_{n,k}^{\alpha ,\beta }(z)-W_{n,q}(e_k;z)-\frac{[\alpha ]-[\beta ]z}{[n]+[\beta ]}k{z}^{k-1}).\end{array}$$

From Theorem 2 in [13], we have

$$\begin{array}{r}|W_{n,q}(f)(z)-f(z)-\frac{z}{2[n]}(1+\frac{z}{q})f^{″}(z)|\\ \le \frac{16}{{[n]}^2}\sum _{k=3}^{\mathrm{\infty }}|c_k|(k-1){(k-2)}^{2}k!r^k.\end{array}$$

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

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

Also it is clear that

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

(3.6)

which implies

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

(3.7)

Now from (3.3) we obtain

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

(3.8)

Also, we need to prove the following inequality:

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

(3.9)

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

$$|W_{n,q}(e_k;z)-z^k|\le \frac{6}{[n]}{r}^k(k+1)!(k-1).$$

(3.10)

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

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

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 $1\le r<R$, $0\le \alpha \le \beta$. If f is not a polynomial of degree ≤0, then the lower estimate

$${\parallel W_{n,q}^{\alpha ,\beta }(f)-f\parallel }_r\ge \frac{C_r^{\alpha ,\beta }(f)}{[n]}$$

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

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

$$\begin{array}{r}{W}_{n,q}^{\alpha ,\beta }(f)(z)-f(z)\\ =\frac{1}{[n]}\{\frac{[n]}{[n]+[\beta ]}([\alpha ]-[\beta ]z)f^{\prime }(z)+\frac{z}{2}(1+\frac{z}{q})f^{″}(z)\\ +\frac{1}{[n]}{[n]}^2(W_{n,q}^{\alpha ,\beta }(f)(z)-f(z)-\frac{[\alpha ]-[\beta ]z}{[n]+[\beta ]}{f}^{\prime }(z)-\frac{z}{2[n]}(1+\frac{z}{q})f^{″}(z))\}\\ =\frac{1}{[n]}\{([\alpha ]-[\beta ]z)f^{\prime }(z)+\frac{z}{2}(1+\frac{z}{q})f^{″}(z)\\ +\frac{1}{[n]}{[n]}^2(W_{n,q}^{\alpha ,\beta }(f)(z)-f(z)-\frac{[\alpha ]-[\beta ]z}{[n]+[\beta ]}{f}^{\prime }(z))\\ +\frac{1}{[n]}{[n]}^2(-\frac{z}{2[n]}(1+\frac{z}{q})f^{″}(z)-\frac{[\beta ]([\alpha ]-[\beta ]z)}{[n]([n]+[\beta ])}{f}^{\prime }(z))\}.\end{array}$$

We set $E_{k,n}(z)$ by

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

(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

$${\parallel W_{n,q}^{\alpha ,\beta }(f)-f\parallel }_r\ge \frac{1}{[n]}{\parallel ([\alpha ]-[\beta ]e_1)f^{\prime }+\frac{e_1}{2}(1+\frac{e_1}{q})f^{″}\parallel }_r-\frac{1}{[n]}{[n]}^2{\parallel E_{k,n}\parallel }_r.$$

Since f is not a polynomial of degree ≤0 in $D_R$, we have

$${\parallel ([\alpha ]-[\beta ]e_1)f^{\prime }+\frac{e_1}{2}(1+\frac{e_1}{q})f^{″}\parallel }_r>0.$$

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

$$\begin{array}{rl}{[n]}^2{\parallel E_{k,n}\parallel }_r\le & {[n]}^2{\parallel W_{n,q}^{\alpha ,\beta }(f)-f-\left(\frac{[\alpha ]-[\beta ]e_1}{[n]+[\beta ]}\right)f^{\prime }-\frac{e_1}{2[n]}(1+\frac{e_1}{q})f^{″}\parallel }_r\\ +{\parallel [\beta ]([\alpha ]-[\beta ]e_1)f^{\prime }\parallel }_r\\ \le & \sum _{j=1}^6{M}_{j,r}(f)+[\beta ]([\alpha ]+[\beta ]r){\parallel f^{\prime }\parallel }_r.\end{array}$$

Since $\frac{1}{[n]}\to 0$ 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 $n\ge n_0$,

$$\begin{array}{r}\frac{1}{[n]}{\parallel ([\alpha ]-[\beta ]e_1)f^{\prime }+\frac{e_1}{2}(1+\frac{e_1}{q})f^{″}\parallel }_r-\frac{1}{[n]}{[n]}^2{\parallel E_{k,n}\parallel }_r\\ \ge \frac{1}{2}{\parallel ([\alpha ]-[\beta ]e_1)f^{\prime }+\frac{e_1}{2}(1+\frac{e_1}{q})f^{″}\parallel }_r,\end{array}$$

which implies

$${\parallel W_{n,q}^{\alpha ,\beta }(f)-f\parallel }_r\ge \frac{1}{2[n]}{\parallel ([\alpha ]-[\beta ]e_1)f^{\prime }+\frac{e_1}{2}(1+\frac{e_1}{q})f^{″}\parallel }_r$$

for all $n\ge n_0$. Now, for $n\in \{1,\dots ,n_0-1\}$, we have

$${\parallel W_{n,q}^{\alpha ,\beta }(f)-f\parallel }_r\ge \frac{A_r(f)}{[n]}$$

with

$$A_r(f)=[n]{\parallel W_{n,q}^{\alpha ,\beta }(f)-f\parallel }_r>0,$$

which finally implies

$${\parallel W_{n,q}^{\alpha ,\beta }(f)-f\parallel }_r\ge \frac{C_r^{\alpha ,\beta }(f)}{[n]}$$

for all $n\ge n_0$ with

$$C_r^{\alpha ,\beta }(f)=min\{A_{r,1}(f),\dots ,A_{r,n_0-1}(f),\frac{1}{2}{\parallel ([\alpha ]-[\beta ]e_1)f^{\prime }+\frac{e_1}{2}(1+\frac{e_1}{q})f^{″}\parallel }_r\}.$$

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:

$${\parallel W_{n,q}^{\alpha ,\beta }(f)-f\parallel }_r\sim \frac{1}{[n]}$$

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:

$${\parallel {[W_{n,q}^{\alpha ,\beta }(f)]}^{(p)}-f^{(p)}\parallel }_r\sim \frac{1}{[n]}$$

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

$$\begin{array}{rcl}|{[W_{n,q}^{\alpha ,\beta }(f,z)]}^{(p)}-f^{(p)}(z)|& \le & \frac{p!}{2\pi }\left|{\int }_{\mathrm{\Gamma }}\frac{W_{n,q}^{\alpha ,\beta }f(\gamma )-f(\gamma )}{{(\gamma -z)}^{p+1}}d\gamma \right|\\ \le & \frac{M_{r_{1,}\alpha ,\beta }(f)}{[n]}\frac{p!}{2\pi }\frac{2\pi r_1}{{(r_1-r)}^{p+1}}\\ =& \frac{M_{r_{1,}\alpha ,\beta }(f)}{[n]}\frac{p!r_1}{{(r_1-r)}^{p+1}},\end{array}$$

which proves one of the inequalities in the equivalence.

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

$${[W_{n,q}^{\alpha ,\beta }(f,z)]}^{(p)}-f^{(p)}(z)=\frac{p!}{2\pi i}{\int }_{\mathrm{\Gamma }}\frac{W_{n,q}^{\alpha ,\beta }f(\gamma )-f(\gamma )}{{(\gamma -z)}^{p+1}}d\gamma .$$

Furthermore, using (3.11) one can have

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

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

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

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

$$\begin{array}{rcl}{\parallel {[W_{n,q}^{\alpha ,\beta }(f)]}^{(p)}-f^{(p)}\parallel }_r& \ge & \frac{1}{[n]}\{{\parallel {[([\alpha ]-[\beta ]e_1)f^{\prime }+\frac{e_1}{2}(1+\frac{e_1}{q})f^{″}]}^{(p)}\parallel }_r\\ -\frac{1}{[n]}{\parallel \frac{p!}{2\pi }{\int }_{\mathrm{\Gamma }}\frac{{[n]}^2{E}_{k,n}(\gamma )}{{(\gamma -z)}^{p+1}}d\gamma \parallel }_r\},\end{array}$$

and from Theorem 2 we have

$$\begin{array}{rcl}{\parallel \frac{p!}{2\pi }{\int }_{\mathrm{\Gamma }}\frac{{[n]}^2{E}_{k,n}(\gamma )}{{(\gamma -z)}^{p+1}}d\gamma \parallel }_r& \le & \frac{p!}{2\pi }\frac{2\pi r_1}{{(r_1-r)}^{p+1}}{[n]}^2{\parallel E_{k,n}\parallel }_{r_1}\\ \le & K_{1,r_1}(f)+{[n]}^2\frac{{\sum }_{j=2}^6{K}_{j,r_1}(f)}{{([n]+[\beta ])}^2}+[\beta ]([\alpha ]+[\beta ]r_1){\parallel f^{\prime }\parallel }_{r_1}.\end{array}$$

Since f is not a polynomial of degree ≤0 in $D_R$, we have

$${\parallel {[([\alpha ]-[\beta ]e_1)f^{\prime }+\frac{e_1}{2}(1+\frac{e_1}{q})f^{″}]}^{(p)}\parallel }_r>0$$

(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].