# Stancu type q-Bernstein operators with shifted knots

## Abstract

In the present paper, Stancu type generalizations of the q-analog of Lupaş Bernstein operators with shifted knots are introduced. Some approximation results and rate of convergence for these operators are investigated. A Voronovskaja type theorem and local approximation results for the mentioned operators are studied. The extra parameters γ, δ, q, a and b provide more flexibility for approximation.

## 1 Introduction

Approximation theory basically deals with approximation of functions by simpler functions or more easily calculated functions. Broadly it is divided into theoretical and constructive approximation. In 1912 Bernstein [5] was the first to construct a sequence of positive linear operators as follows:

${B}_{m}\left(f;u\right)=\sum _{k=0}^{m}\left(\begin{array}{c}m\\ k\end{array}\right){u}^{k}{\left(1-u\right)}^{m-k}f\left(\frac{k}{m}\right),$
(1.1)

where $$u\in[0,1]$$, f is bounded on $$[0,1]$$.

A constructive proof of the well-known Weierstrass approximation theorem using a probabilistic approach was provided. Here $$C[0,1]$$ denotes the set of all continuous functions on $$[0,1]$$ which is equipped with the sup-norm $$\Vert\cdot \Vert$$. He showed that if $$f\in C[0,1]$$, then $$B_{m}(f;u)$$ converges to $$f(u)$$ uniformly on $$[0,1]$$. One can find a detailed monograph about the Bernstein polynomials in [12, 13].

Before proceeding, let us recall some basic definitions and notations of quantum calculus [9]. For any fixed real number $$q>0$$ satisfying the conditions $$0< q\leq1$$, the q-integer $$[k]_{q}$$, for $$k\in\mathbb{N}$$ is defined as

$$[k]_{q}:= \textstyle\begin{cases} \frac{(1-q^{k})}{(1-q)}, & q\neq1, \\ k, & q=1 \end{cases}$$

and the q-factorial by

$$[k]_{q}!:= \textstyle\begin{cases} [k]_{q}[k-1]_{q}\cdots [1]_{q}, & k\geq1, \\ 1, & k=0. \end{cases}$$

The q-binomial expansion is

$$(u+y)_{q}^{m}:=(u+y) (u+qy) \bigl(u+q^{2}y \bigr)\cdots\bigl(u+q^{m-1}y\bigr),$$

and the q-binomial coefficients are as follows:

${\left[\begin{array}{c}m\\ k\end{array}\right]}_{q}:=\frac{{\left[m\right]}_{q}!}{{\left[k\right]}_{q}!{\left[m-k\right]}_{q}!}.$

From the above

$$(u)_{q}^{m} = u^{m}$$
(1.2)

and

$$(y)_{q}^{m}:=(y) (qy) \bigl(q^{2}y\bigr) \cdots\bigl(q^{m-1}y\bigr)=q^{\frac{m(m-1)}{2}}y ^{m}.$$

Gauss-formula is defined as

${\left(u+y\right)}_{q}^{m}=\sum _{j=0}^{m}{\left[\begin{array}{c}m\\ k\end{array}\right]}_{q}{q}^{j\left(j-1\right)/2}{y}^{j}{u}^{m-j}.$

After development of q-calculus, Lupaş [14] in 1987 introduced the q-Lupaş operator (rational) as follows:

${L}_{m,q}\left(f;u\right)=\sum _{k=0}^{m}\frac{f\left(\frac{{\left[k\right]}_{q}}{{\left[m\right]}_{q}}\right){\left(\begin{array}{c}m\\ k\end{array}\right)}_{q}{q}^{\frac{k\left(k-1\right)}{2}}{u}^{k}{\left(1-u\right)}^{m-k}}{{\prod }_{j=1}^{m}\left\{\left(1-u\right)+{q}^{j-1}u\right\}},$
(1.3)

and studied its approximation properties.

Similarly, Phillips [23] in 1996 constructed another q-analog of Bernstein operators (polynomials) as follows:

${B}_{m,q}\left(f;u\right)=\sum _{k=0}^{m}{\left[\begin{array}{c}m\\ k\end{array}\right]}_{q}{u}^{k}\prod _{s=0}^{m-k-1}\left(1-{q}^{s}x\right)f\left(\frac{{\left[k\right]}_{q}}{{\left[m\right]}_{q}}\right),\phantom{\rule{1em}{0ex}}x\in \left[0,1\right],$
(1.4)

where $$B_{m,q}$$: $$C[0,1]\rightarrow C[0,1]$$ defined for any $$m\in\mathbb{N}$$ and any function $$f\in C[0,1]$$.

Bases of these operators have been used in computer aided geometric design (CAGD) to study curves and surfaces. From then onward it became an active area of research in approximation theory as well as CAGD. In the recent past, q-analogs of various operators were investigated by several researchers (see [6, 15, 19, 22, 24]). Also see [1, 2, 4, 10, 11, 1618, 21, 25] for other modifications.

In 1968 Stancu [26] showed that the Bernstein–Stancu polynomials

$\left({P}_{m}^{\left(\gamma ,\delta \right)}f\right)\left(u\right)=\sum _{k=0}^{m}\left(\begin{array}{c}m\\ k\end{array}\right){u}^{k}{\left(1-u\right)}^{m-k}f\left(\frac{k+\gamma }{m+\delta }\right),$
(1.5)

converge to continuous function $$f(u)$$ uniformly in $$[0,1]$$ for each real γ, δ such that $$0\leq\gamma\leq\delta$$.

In 2010, a new construction of Bernstein–Stancu type polynomials with shifted knots was introduced by Gadjiev and Gorhanalizadeh [8]:

$\begin{array}{rcl}{S}_{m,\gamma ,\delta }\left(f;u\right)& =& {\left(\frac{m+{\delta }_{2}}{m}\right)}^{m}\sum _{k=0}^{m}\left(\begin{array}{c}m\\ k\end{array}\right){\left(u-\frac{{\gamma }_{2}}{m+{\delta }_{2}}\right)}^{k}{\left(\frac{m+{\gamma }_{2}}{m+{\delta }_{2}}-u\right)}^{m-k}\\ & & ×f\left(\frac{k+{\gamma }_{1}}{m+{\delta }_{1}}\right),\end{array}$
(1.6)

where $$\frac{\gamma_{2}}{n+\delta_{2}}\leq u\leq\frac{n+\gamma_{2}}{n+ \delta_{2}}$$ and $$\gamma_{k}$$, $$\delta_{k}$$ ($$k=1,2$$) are positive real numbers provided $$0\leq\gamma_{1}\leq\gamma_{2}\leq\delta_{1} \leq\delta_{2}$$. It is clear that, for $$\gamma_{2}=\delta_{2}=0$$, the polynomials (1.6) turn into the Bernstein–Stancu polynomials (1.5) and if $$\gamma_{1}=\gamma_{2}=\delta_{1}=\delta _{2}=0$$ then these polynomials turn into the classical Bernstein polynomials.

Khalid et al. studied Bezier curves and surfaces using basis of shifted Bernstein polynomial in [11]. Recently, Mursaleen et al. [20] introduced and studied Lupaş Bernstein shifted operators based on q-integers as follows:

$\begin{array}{rcl}{S}_{m,q}^{\left(\gamma ,\delta \right)}\left(f;u\right)& =& \frac{1}{{\left(\frac{{\left[m\right]}_{q}}{{\left[m\right]}_{q}+\delta }\right)}_{q}^{m}}\sum _{j=0}^{m}{\left[\begin{array}{c}m\\ j\end{array}\right]}_{q}{q}^{\frac{j\left(j-1\right)}{2}}{\left(\frac{{\left[m\right]}_{q}+\gamma }{{\left[m\right]}_{q}+\delta }-u\right)}^{m-j}\\ & & ×{\left(u-\frac{a}{{\left[m\right]}_{q}+b}\right)}^{j}f\left(\frac{{\left[j\right]}_{q}}{{\left[m\right]}_{q}}\right)\end{array}$
(1.7)

or

${S}_{m,q}^{\left(\gamma ,\delta \right)}\left(f;u\right)=\frac{1}{{\left(\frac{{\left[m\right]}_{q}}{{\left[m\right]}_{q}+\delta }\right)}_{q}^{m}}\sum _{j=0}^{m}{\left[\begin{array}{c}m\\ j\end{array}\right]}_{q}{\left(\frac{{\left[m\right]}_{q}+\gamma }{{\left[m\right]}_{q}+\delta }-u\right)}^{m-j}{\left(u-\frac{a}{{\left[m\right]}_{q}+\delta }\right)}_{q}^{j}f\left(\frac{{\left[j\right]}_{q}}{{\left[m\right]}_{q}}\right),$

where $$\frac{\gamma}{[m]_{q}+\delta}\leq u\leq\frac{[m]_{q}+\gamma }{[m]_{q}+\delta}$$ and γ, δ are positive real numbers provided $$0\leq\gamma\leq\delta$$. In case $$\gamma=\delta=0$$, the above operators (1.7) reduce to Lupaş q-Bernstein operators [14].

Motivated by the above work, in the next section we present a Stancu type modification of Lupaş q-Bernstein shifted operators and will study its approximation properties.

## 2 Construction of Lupaş q-Bernstein–Stancu shifted operators

Let $$\gamma,\delta\in\mathbb{N}_{0}$$ (the set of all non-negative integers) be such that $$0\leq\gamma\leq\delta$$ and $$0\leq a \leq b$$, then we have

$\begin{array}{rl}{S}_{m,q}^{\left(\gamma ,\delta ,a,b\right)}\left(f;u\right)=& \frac{1}{{\left(\frac{{\left[m\right]}_{q}}{{\left[m\right]}_{q}+\delta }\right)}_{q}^{m}}\sum _{k=0}^{m}{\left[\begin{array}{c}m\\ k\end{array}\right]}_{q}{q}^{\frac{k\left(k-1\right)}{2}}{\left(u-\frac{\gamma }{{\left[m\right]}_{q}+\delta }\right)}^{k}\\ & ×{\left(\frac{{\left[m\right]}_{q}+\gamma }{{\left[m\right]}_{q}+\delta }-u\right)}^{m-k}f\left(\frac{{\left[k\right]}_{q}+a}{{\left[m\right]}_{q}+b}\right)\end{array}$
(2.1)

or

$\begin{array}{rl}& {S}_{m,q}^{\left(\gamma ,\delta ,a,b\right)}\left(f;u\right)\\ & \phantom{\rule{1em}{0ex}}=\frac{1}{{\left(\frac{{\left[m\right]}_{q}}{{\left[m\right]}_{q}+\delta }\right)}_{q}^{m}}\sum _{k=0}^{m}{\left[\begin{array}{c}m\\ k\end{array}\right]}_{q}{\left(u-\frac{\gamma }{{\left[m\right]}_{q}+\delta }\right)}_{q}^{k}{\left(\frac{{\left[m\right]}_{q}+\gamma }{{\left[m\right]}_{q}+\delta }-u\right)}^{m-k}f\left(\frac{{\left[k\right]}_{q}+a}{{\left[m\right]}_{q}+b}\right),\end{array}$

where $$\frac{\gamma}{[m]_{q}+\delta}\leq u\leq\frac{[m]_{q}+\gamma }{[m]_{q}+\delta}$$ and γ, δ are positive real numbers provided $$0\leq\gamma\leq\delta$$. In the case $$a= b=0$$, the above operators (2.1) reduces to (1.7).

## 3 Definitions and auxiliary results

### Lemma 3.1

Let $$S^{(\gamma,\delta,a,b)}_{m,q}(f;u)$$be given by (2.1). Then the following properties hold:

1. (1)

$$S^{(\gamma,\delta,a,b)}_{m,q}(1;u)=1$$,

2. (2)

$$S^{(\gamma,\delta,a,b)}_{m,q}(t;u)= \frac{[m]_{q}+\delta}{{[m]_{q}}+b} ( u-\frac{\gamma}{[m]_{q}+ \delta} ) +\frac{a}{{[m]_{q}}+b}$$,

3. (3)

$$S^{(\gamma,\delta,a,b)}_{m,q}(t^{2};u)= (\frac{q^{2}[m-1]_{q}}{ {[m]_{q}}+b} ) ( \frac{[m]_{q}+\delta}{{[m]_{q}}+b} ) \frac{ (u-\frac{\gamma}{[m]_{q}+\delta} )^{2}}{ \lbrace\frac{[m]_{q}+ \gamma}{[m]_{q}+\delta}-u+q ( u-\frac{\gamma}{[m]_{q}+\delta } ) \rbrace} + ( \frac{[m]_{q}+\delta}{{([m]_{q}}+b)^{2}} ) (u-\frac{ \gamma}{[m]_{q}+\delta} )+ \frac{a}{{([m]_{q}}+b)^{2}}$$,

4. (4)

$$S^{(\gamma,\delta,a,b)}_{m,q}(t^{3};u)= ( \frac{[m]_{q}+ \delta}{{[m]_{q}}+b} ) \frac{q^{6}[m-1]_{q}[m-2]_{q} (u-\frac{ \gamma}{[m]_{q}+\delta} )^{3}}{{([m]_{q}}+b)^{2} \lbrace\frac{[m]_{q}+ \gamma}{[m]_{q}+\delta}-u+q ( u-\frac{\gamma}{[m]_{q}+\delta } ) \rbrace \lbrace\frac{[m]_{q}+\gamma}{[m]_{q}+ \delta}-u+q^{2} ( u-\frac{\gamma}{[m]_{q}+\delta} ) \rbrace}+(2+3a+q) (\frac{q^{2}[m-1]_{q}}{{([m]_{q}}+b)^{2}} ) ( \frac{[m]_{q}+ \delta}{[m]_{q}+b} ) \frac{ (u-\frac{\gamma}{[m]_{q}+ \delta} )^{2}}{ \lbrace\frac{[m]_{q}+\gamma}{[m]_{q}+ \delta}-u+q ( u-\frac{\gamma}{[m]_{q}+\delta} ) \rbrace}+ {(1+3a^{2})} (\frac{[m]_{q}+\delta}{{([m]_{q}}+b)^{3}} ) (u-\frac{\gamma}{[m]_{q}+\delta} )+ \frac{a^{3}}{{([m]_{q}}+b)^{3}}+ \frac{3a^{2}([m]_{q}+\delta)}{{([m]_{q}}+b)^{3}}$$,

5. (5)

$$S^{(\gamma,\delta,a,b)}_{m,q}(t^{4};u)=\frac {q^{12}[m-1]_{q}[m-2]_{q}[m-3]_{q}}{ {([m]_{q}}+b)^{3}} (\frac{[m]_{q}+\delta}{{[m]_{q}}+b} ) \times\frac{1}{ \lbrace\frac{[m]_{q}+\gamma}{[m]_{q}+\delta}-u+q ( u-\frac{\gamma}{[m]_{q}+\delta} ) \rbrace} \frac{ (u-\frac{\gamma}{[m]_{q}+\delta} )^{4}}{ \lbrace\frac{[m]_{q}+ \gamma}{[m]_{q}+\delta}-u+q^{2} ( u-\frac{\gamma}{[m]_{q}+ \delta} ) \rbrace \lbrace\frac{[m]_{q}+\gamma}{[m]_{q}+ \delta}-u+q^{3} ( u-\frac{\gamma}{[m]_{q}+\delta} ) \rbrace} +\frac{[m-1]_{q}[m-2]_{q}}{{([m]_{q}}+b)^{3}} (\frac{[m]_{q} + \delta}{[m]_{q}+b} ) \frac{(q^{8}+2q^{7}+3q^{6}+4aq^{2}) (u-\frac{ \gamma}{[m]_{q}+\delta} )^{3}}{ \lbrace\frac{[m]_{q}+ \gamma}{[m]_{q}+\delta}-u+q ( u-\frac{\gamma}{[m]_{q}+\delta } ) \rbrace \lbrace\frac{[m]_{q}+\gamma}{[m]_{q}+ \delta}-u+q^{2} ( u-\frac{\gamma}{[m]_{q}+\delta} ) \rbrace} +(q^{4}+3q^{3}+3q^{2}+6a^{2}q^{2})\frac{[m-1]_{q}}{{([m]_{q}}+b)^{3}} ( \frac{[m]_{q}+\delta}{[m]_{q}+b} ) \frac{ (u-\frac{ \gamma}{[m]_{q}+\delta} )^{2}}{ \lbrace\frac{[m]_{q}+ \gamma}{[m]_{q}+\delta}-u+q ( u-\frac{\gamma}{[m]_{q}+\delta } ) \rbrace}+(1+4a^{3}) ( \frac{[m]_{q}+\delta}{ {([m]_{q}}+b)^{4}} ) (u-\frac{\gamma}{[m]_{q}+\delta} )+ \frac{a^{4}}{{([m]_{q}}+b)^{4}}+ \frac{6a^{2}([m]_{q}+\delta)}{{([m]_{q}}+b)^{4}}$$.

### Proof

(1) By using definition of q-binomial coefficients and Gauss-formula, we have

$\begin{array}{rl}& {S}_{m,q}^{\left(\gamma ,\delta ,a,b\right)}\left(1;u\right)\\ & \phantom{\rule{1em}{0ex}}=\frac{1}{{\left(\frac{{\left[m\right]}_{q}}{{\left[m\right]}_{q}+\delta }\right)}_{q}^{m}}\sum _{k=0}^{m}{\left[\begin{array}{c}m\\ k\end{array}\right]}_{q}{q}^{\frac{k\left(k-1\right)}{2}}{\left(u-\frac{\gamma }{{\left[m\right]}_{q}+\delta }\right)}^{k}{\left(\frac{{\left[m\right]}_{q}+\gamma }{{\left[m\right]}_{q}+\delta }-u\right)}^{m-k}=1.\end{array}$
(3.1)

(2) For $$f(t)=t$$, we have

(3) For $$f(t)=t^{2}$$, we have

$\begin{array}{rl}& {S}_{m,q}^{\left(\gamma ,\delta ,a,b\right)}\left({t}^{2};u\right)\\ & \phantom{\rule{1em}{0ex}}=\frac{1}{{\left(\frac{{\left[m\right]}_{q}}{{\left[m\right]}_{q}+\delta }\right)}_{q}^{m}}\sum _{k=0}^{m}{\left[\begin{array}{c}m\\ k\end{array}\right]}_{q}{q}^{\frac{k\left(k-1\right)}{2}}{\left(u-\frac{\gamma }{{\left[m\right]}_{q}+\delta }\right)}^{k}{\left(\frac{{\left[m\right]}_{q}+\gamma }{{\left[m\right]}_{q}+\delta }-u\right)}^{m-k}{\left(\frac{{\left[k\right]}_{q}+a}{{\left[m\right]}_{q}+b}\right)}^{2}\\ & \phantom{\rule{1em}{0ex}}=\frac{1}{{\left({\left[m\right]}_{q}+b\right)}^{2}{\left(\frac{{\left[m\right]}_{q}}{{\left[m\right]}_{q}+\delta }\right)}_{q}^{m}}\sum _{k=0}^{m}{\left[\begin{array}{c}m\\ k\end{array}\right]}_{q}{q}^{\frac{k\left(k-1\right)}{2}}{\left(u-\frac{\gamma }{{\left[m\right]}_{q}+\delta }\right)}^{k}\\ & \phantom{\rule{2em}{0ex}}×{\left(\frac{{\left[m\right]}_{q}+\gamma }{{\left[m\right]}_{q}+\delta }-u\right)}^{m-k}{\left({\left[k\right]}_{q}+a\right)}^{2}\\ & \phantom{\rule{1em}{0ex}}=\frac{1}{{\left({\left[m\right]}_{q}+b\right)}^{2}{\left(\frac{{\left[m\right]}_{q}}{{\left[m\right]}_{q}+\delta }\right)}_{q}^{m}}\sum _{k=0}^{m}{\left[\begin{array}{c}m\\ k\end{array}\right]}_{q}{q}^{\frac{k\left(k-1\right)}{2}}{\left(u-\frac{\gamma }{{\left[m\right]}_{q}+\delta }\right)}^{k}{\left(\frac{{\left[m\right]}_{q}+\gamma }{{\left[m\right]}_{q}+\delta }-u\right)}^{m-k}{{\left[k\right]}_{q}}^{2}\\ & \phantom{\rule{2em}{0ex}}+\frac{{a}^{2}}{{\left({\left[m\right]}_{q}+b\right)}^{2}{\left(\frac{{\left[m\right]}_{q}}{{\left[m\right]}_{q}+\delta }\right)}_{q}^{m}}\sum _{k=0}^{m}{\left[\begin{array}{c}m\\ k\end{array}\right]}_{q}{q}^{\frac{k\left(k-1\right)}{2}}{\left(u-\frac{\gamma }{{\left[m\right]}_{q}+\delta }\right)}^{k}{\left(\frac{{\left[m\right]}_{q}+\gamma }{{\left[m\right]}_{q}+\delta }-u\right)}^{m-k}\\ & \phantom{\rule{2em}{0ex}}+\frac{2a}{{\left({\left[m\right]}_{q}+b\right)}^{2}{\left(\frac{{\left[m\right]}_{q}}{{\left[m\right]}_{q}+\delta }\right)}_{q}^{m}}\sum _{k=0}^{m}{\left[\begin{array}{c}m\\ k\end{array}\right]}_{q}{q}^{\frac{k\left(k-1\right)}{2}}{\left(u-\frac{\gamma }{{\left[m\right]}_{q}+\delta }\right)}^{k}\\ & \phantom{\rule{2em}{0ex}}×{\left(\frac{{\left[m\right]}_{q}+\gamma }{{\left[m\right]}_{q}+\delta }-u\right)}^{m-k}{\left[k\right]}_{q}\\ & \phantom{\rule{1em}{0ex}}=A+B+C\phantom{\rule{1em}{0ex}}\left(\text{say}\right),\end{array}$

after calculating the values of A, B and C we get

\begin{aligned} & S^{(\gamma,\delta,a,b)}_{m,q}\bigl(t^{2};u\bigr) \\ &\quad = \biggl(\frac{[m]_{q}+\delta}{([m]+b)_{q})^{2}} \biggr) \biggl( u-\frac{ \gamma}{[m]_{q}+\delta} \biggr) \\ &\qquad {} + \biggl(\frac{q^{2}[m-1]_{q}}{[m]_{q}+b} \biggr) \biggl( \frac{[m]_{q}+ \delta}{[m]_{q}+b} \biggr) \frac{ (u-\frac{\gamma}{[m]_{q}+ \delta} )^{2}}{ \lbrace\frac{[m]_{q}+\gamma}{[m]_{q}+ \delta}-u+q ( u-\frac{\gamma}{[m]_{q}+\delta} ) \rbrace}+ \frac {a}{([m]+b)_{q})^{2}}, \end{aligned}

which is the required result.

(4) For $$f(t)=t^{3}$$, we have

$\begin{array}{rl}& {S}_{m,q}^{\left(\gamma ,\delta ,a,b\right)}\left({t}^{3};u\right)\\ & \phantom{\rule{1em}{0ex}}=\frac{1}{{\left(\frac{{\left[m\right]}_{q}}{{\left[m\right]}_{q}+\delta }\right)}_{q}^{m}}\sum _{k=0}^{m}{\left[\begin{array}{c}m\\ k\end{array}\right]}_{q}{q}^{\frac{k\left(k-1\right)}{2}}{\left(u-\frac{\gamma }{{\left[m\right]}_{q}+\delta }\right)}^{k}{\left(\frac{{\left[m\right]}_{q}+\gamma }{{\left[m\right]}_{q}+\delta }-u\right)}^{m-k}{\left(\frac{{\left[k\right]}_{q}+a}{{\left[m\right]}_{q}+b}\right)}^{3}\\ & \phantom{\rule{1em}{0ex}}=\frac{1}{{\left(\left[m\right]+b\right)}_{q}{\right)}^{3}{\left(\frac{{\left[m\right]}_{q}}{{\left[m\right]}_{q}+\delta }\right)}_{q}^{m}}\sum _{k=0}^{m}{\left[\begin{array}{c}m\\ k\end{array}\right]}_{q}{q}^{\frac{k\left(k-1\right)}{2}}{\left(u-\frac{\gamma }{{\left[m\right]}_{q}+\delta }\right)}^{k}\\ & \phantom{\rule{2em}{0ex}}×{\left(\frac{{\left[m\right]}_{q}+\gamma }{{\left[m\right]}_{q}+\delta }-u\right)}^{m-k}{\left({\left[k\right]}_{q}+a\right)}^{3}\\ & \phantom{\rule{1em}{0ex}}=\frac{1}{{\left(\left[m\right]+b\right)}_{q}{\right)}^{3}{\left(\frac{{\left[m\right]}_{q}}{{\left[m\right]}_{q}+\delta }\right)}_{q}^{m}}\sum _{k=0}^{m}{\left[\begin{array}{c}m\\ k\end{array}\right]}_{q}{q}^{\frac{k\left(k-1\right)}{2}}{\left(u-\frac{\gamma }{{\left[m\right]}_{q}+\delta }\right)}^{k}{\left(\frac{{\left[m\right]}_{q}+\gamma }{{\left[m\right]}_{q}+\delta }-u\right)}^{m-k}{\left[k\right]}_{q}^{3}\\ & \phantom{\rule{2em}{0ex}}+\frac{{b}^{3}}{{\left(\left[m\right]+b\right)}_{q}{\right)}^{3}{\left(\frac{{\left[m\right]}_{q}}{{\left[m\right]}_{q}+\delta }\right)}_{q}^{m}}\sum _{k=0}^{m}{\left[\begin{array}{c}m\\ k\end{array}\right]}_{q}{q}^{\frac{k\left(k-1\right)}{2}}{\left(u-\frac{\gamma }{{\left[m\right]}_{q}+\delta }\right)}^{k}{\left(\frac{{\left[m\right]}_{q}+\gamma }{{\left[m\right]}_{q}+\delta }-u\right)}^{m-k}\\ & \phantom{\rule{2em}{0ex}}+\frac{3a}{{\left(\left[m\right]+b\right)}_{q}{\right)}^{3}{\left(\frac{{\left[m\right]}_{q}}{{\left[m\right]}_{q}+\delta }\right)}_{q}^{m}}\sum _{k=0}^{m}{\left[\begin{array}{c}m\\ k\end{array}\right]}_{q}{q}^{\frac{k\left(k-1\right)}{2}}{\left(u-\frac{\gamma }{{\left[m\right]}_{q}+\delta }\right)}^{k}\\ & \phantom{\rule{2em}{0ex}}×{\left(\frac{{\left[m\right]}_{q}+\gamma }{{\left[m\right]}_{q}+\delta }-u\right)}^{m-k}{\left[k\right]}_{q}^{2}\\ & \phantom{\rule{2em}{0ex}}+\frac{3{a}^{2}}{{\left(\left[m\right]+b\right)}_{q}{\right)}^{3}{\left(\frac{{\left[m\right]}_{q}}{{\left[m\right]}_{q}+\delta }\right)}_{q}^{m}}\sum _{k=0}^{m}{\left[\begin{array}{c}m\\ k\end{array}\right]}_{q}{q}^{\frac{k\left(k-1\right)}{2}}{\left(u-\frac{\gamma }{{\left[m\right]}_{q}+\delta }\right)}^{k}\\ & \phantom{\rule{2em}{0ex}}×{\left(\frac{{\left[m\right]}_{q}+\gamma }{{\left[m\right]}_{q}+\delta }-u\right)}^{m-k}{\left[k\right]}_{q}\\ & \phantom{\rule{1em}{0ex}}=D+E+F+G\phantom{\rule{1em}{0ex}}\left(\mathrm{say}\right).\end{array}$

After calculating the values of D, E, F and G, we get

\begin{aligned} S^{(\gamma,\delta,a,b)}_{m,q}\bigl(t^{3};u\bigr)={}& \biggl( \frac{[m]_{q}+ \delta}{[m]_{q}+b} \biggr) \\ &{} \times\frac{q^{6}[m-1]_{q}[m-2]_{q} (u-\frac{\gamma}{[m]_{q}+ \delta} )^{3}}{([m]_{q}+b)^{2} \lbrace\frac{[m]_{q}+\gamma }{[m]_{q}+\delta}-u+q ( u-\frac{\gamma}{[m]_{q}+\delta} ) \rbrace \lbrace\frac{[m]_{q}+\gamma}{[m]_{q}+\delta}-u+q^{2} ( u-\frac{ \gamma}{[m]_{q}+\delta} ) \rbrace} \\ & {}+(2+3a+q) \biggl(\frac{q^{2}[m-1]_{q}}{([m]_{q}+b)^{2}} \biggr) \biggl( \frac{[m]_{q}+\delta}{[m]_{q}+b} \biggr) \frac{ (u-\frac{ \gamma}{[m]_{q}+\delta} )^{2}}{ \lbrace\frac{[m]_{q}+ \gamma}{[m]_{q}+\delta}-u+q ( u-\frac{\gamma}{[m]_{q}+\delta } ) \rbrace} \\ &{} +{\bigl(1+3a^{2}\bigr)} \biggl(\frac{[m]_{q}+\delta }{{([m]_{q}}+b)^{3}} \biggr) \biggl(u-\frac{\gamma}{[m]_{q}+\delta} \biggr) \\ &{}+ \frac {a^{3}}{{([m]_{q}}+b)^{3}}+ \frac{3a^{2}([m]_{q}+\delta)}{{([m]_{q}}+b)^{3}}. \end{aligned}

(5) For $$f(t)=t^{4}$$, we have

$\begin{array}{rl}& {S}_{m,q}^{\left(\gamma ,\delta ,a,b\right)}\left({t}^{4};u\right)\\ & \phantom{\rule{1em}{0ex}}=\frac{1}{{\left(\frac{{\left[m\right]}_{q}}{{\left[m\right]}_{q}+\delta }\right)}_{q}^{m}}\sum _{k=0}^{m}{\left[\begin{array}{c}m\\ k\end{array}\right]}_{q}{q}^{\frac{k\left(k-1\right)}{2}}{\left(u-\frac{\gamma }{{\left[m\right]}_{q}+\delta }\right)}^{k}{\left(\frac{{\left[m\right]}_{q}+\gamma }{{\left[m\right]}_{q}+\delta }-u\right)}^{m-k}{\left(\frac{{\left[k\right]}_{q}+a}{{\left[m\right]}_{q}+b}\right)}^{4}\\ & \phantom{\rule{1em}{0ex}}=\frac{1}{{\left({\left[m\right]}_{q}+b\right)}^{4}{\left(\frac{{\left[m\right]}_{q}}{{\left[m\right]}_{q}+\delta }\right)}_{q}^{m}}\sum _{k=0}^{m}{\left[\begin{array}{c}m\\ k\end{array}\right]}_{q}{q}^{\frac{k\left(k-1\right)}{2}}{\left(u-\frac{\gamma }{{\left[m\right]}_{q}+\delta }\right)}^{k}\\ & \phantom{\rule{2em}{0ex}}×{\left(\frac{{\left[m\right]}_{q}+\gamma }{{\left[m\right]}_{q}+\delta }-u\right)}^{m-k}{\left({\left[k\right]}_{q}+a\right)}^{4}\\ & \phantom{\rule{1em}{0ex}}=\frac{1}{{\left({\left[m\right]}_{q}+b\right)}^{4}{\left(\frac{{\left[m\right]}_{q}}{{\left[m\right]}_{q}+\delta }\right)}_{q}^{m}}\sum _{k=0}^{m}{\left[\begin{array}{c}m\\ k\end{array}\right]}_{q}{q}^{\frac{k\left(k-1\right)}{2}}{\left(u-\frac{\gamma }{{\left[m\right]}_{q}+\delta }\right)}^{k}{\left(\frac{{\left[m\right]}_{q}+\gamma }{{\left[m\right]}_{q}+\delta }-u\right)}^{m-k}{\left[k\right]}_{q}^{4}\\ & \phantom{\rule{2em}{0ex}}+\frac{{a}^{4}}{{\left({\left[m\right]}_{q}+b\right)}^{4}{\left(\frac{{\left[m\right]}_{q}}{{\left[m\right]}_{q}+\delta }\right)}_{q}^{m}}\sum _{k=0}^{m}{\left[\begin{array}{c}m\\ k\end{array}\right]}_{q}{q}^{\frac{k\left(k-1\right)}{2}}{\left(u-\frac{\gamma }{{\left[m\right]}_{q}+\delta }\right)}^{k}{\left(\frac{{\left[m\right]}_{q}+\gamma }{{\left[m\right]}_{q}+\delta }-u\right)}^{m-k}\\ & \phantom{\rule{2em}{0ex}}+\frac{6{a}^{2}}{{\left({\left[m\right]}_{q}+b\right)}^{4}{\left(\frac{{\left[m\right]}_{q}}{{\left[m\right]}_{q}+\delta }\right)}_{q}^{m}}\sum _{k=0}^{m}{\left[\begin{array}{c}m\\ k\end{array}\right]}_{q}{q}^{\frac{k\left(k-1\right)}{2}}{\left(u-\frac{\gamma }{{\left[m\right]}_{q}+\delta }\right)}^{k}{\left(\frac{{\left[m\right]}_{q}+\gamma }{{\left[m\right]}_{q}+\delta }-u\right)}^{m-k}{\left[k\right]}_{q}^{2}\\ & \phantom{\rule{2em}{0ex}}+\frac{4a}{{\left({\left[m\right]}_{q}+b\right)}^{4}{\left(\frac{{\left[m\right]}_{q}}{{\left[m\right]}_{q}+\delta }\right)}_{q}^{m}}\sum _{k=0}^{m}{\left[\begin{array}{c}m\\ k\end{array}\right]}_{q}{q}^{\frac{k\left(k-1\right)}{2}}{\left(u-\frac{\gamma }{{\left[m\right]}_{q}+\delta }\right)}^{k}{\left(\frac{{\left[m\right]}_{q}+\gamma }{{\left[m\right]}_{q}+\delta }-u\right)}^{m-k}{\left[k\right]}_{q}^{3}\\ & \phantom{\rule{2em}{0ex}}+\frac{4{a}^{3}}{{\left({\left[m\right]}_{q}+b\right)}^{4}{\left(\frac{{\left[m\right]}_{q}}{{\left[m\right]}_{q}+\delta }\right)}_{q}^{m}}\sum _{k=0}^{m}{\left[\begin{array}{c}m\\ k\end{array}\right]}_{q}{q}^{\frac{k\left(k-1\right)}{2}}{\left(u-\frac{\gamma }{{\left[m\right]}_{q}+\delta }\right)}^{k}{\left(\frac{{\left[m\right]}_{q}+\gamma }{{\left[m\right]}_{q}+\delta }-u\right)}^{m-k}{\left[k\right]}_{q}\\ & \phantom{\rule{1em}{0ex}}=H+I+J+K+L\phantom{\rule{1em}{0ex}}\left(\mathrm{say}\right).\end{array}$

After calculating the values of H, I, J, K and L, we obtain

\begin{aligned} &S^{(\gamma,\delta,a,b)}_{m,q}\bigl(t^{4};u\bigr) \\ &\quad = \frac{q^{12}[m-1]_{q}[m-2]_{q}[m-3]_{q}}{{([m]_{q}}+b)^{3}} \biggl(\frac{[m]_{q} + \delta}{[m]_{q}+b} \biggr) \frac{1}{ \lbrace\frac{[m]_{q}+\gamma }{[m]_{q}+\delta}-u+q ( u-\frac{\gamma}{[m]_{q}+\delta} ) \rbrace} \\ &\qquad {} \times\frac{ (u-\frac{\gamma}{[m]_{q}+\delta} )^{4}}{ \lbrace\frac{[m]_{q}+\gamma}{[m]_{q}+\delta}-u+q^{2} ( u-\frac{ \gamma}{[m]_{q}+\delta} ) \rbrace \lbrace\frac{[m]_{q}+ \gamma}{[m]_{q}+\delta}-u+q^{3} ( u-\frac{\gamma}{[m]_{q}+ \delta} ) \rbrace} \\ &\qquad {} +\frac{[m-1]_{q}[m-2]_{q}}{{([m]_{q}}+b)^{3}} \biggl(\frac{[m]_{q} +\delta}{[m]_{q}+b} \biggr) \\ &\qquad {} \times\frac{(q^{8}+2q^{7}+3q^{5}+4aq^{2}) (u-\frac{ \gamma}{[m]_{q}+\delta} )^{3}}{ \lbrace\frac{[m]_{q}+ \gamma}{[m]_{q}+\delta}-u+q ( u-\frac{\gamma}{[m]_{q}+\delta } ) \rbrace \lbrace\frac{[m]_{q}+\gamma}{[m]_{q}+ \delta}-u+q^{2} ( u-\frac{\gamma}{[m]_{q}+\delta} ) \rbrace} \\ &\qquad {} +\bigl(a^{4}+3q^{3}+3q^{2}+6a^{2}q^{2} \bigr) \frac{[m-1]_{q}}{{([m]_{q}}+b)^{3}} \biggl( \frac {[m]_{q}+\delta}{[m]_{q}+b} \biggr) \frac{ (u-\frac{\gamma }{[m]_{q}+\delta} )^{2}}{ \lbrace\frac{[m]_{q}+\gamma}{[m]_{q}+ \delta}-u+q ( u-\frac{\gamma}{[m]_{q}+\delta} ) \rbrace} \\ & \qquad {}+\bigl(1+4a^{3}\bigr) \biggl( \frac{[m]_{q}+\delta }{{([m]_{q}}+b)^{4}} \biggr) \biggl(u-\frac{\gamma}{[m]_{q}+\delta} \biggr)+ \frac {a^{4}}{{([m]_{q}}+b)^{4}}+ \frac{6a^{2}([m]_{q}+\delta)}{{([m]_{q}}+b)^{4}}. \end{aligned}

□

### Lemma 3.2

By using the linearity of operators $$S^{(\gamma,\delta,a,b)}_{m,q}(f;u)$$and by Lemma 3.1, for all $$u\in [\frac{\gamma}{[m]_{q}+\delta},\frac{[m]_{q}+\gamma }{[m]_{q}+\delta} ]$$, we can acquire the central moments as

1. (1)

$$S^{(\gamma,\delta,a,b)}_{m,q}((t-u);u)=\frac{[m]_{q}+\delta}{ {[m]_{q}}+b} ( u-\frac{\gamma}{[m]_{q}+\delta} ) +\frac{a}{ {[m]_{q}}+b}-u$$,

2. (2)

$$S^{(\gamma,\delta,a,b)}_{m,q}((t-u)^{2};u)= ( \frac{q^{2}[m-1]_{q}}{[m]_{q}+b} ) ( \frac{[m]_{q}+\delta}{[m]_{q}+b} ) \frac{ (u-\frac{\gamma }{[m]_{q}+\delta} )^{2}}{ \lbrace\frac{[m]_{q}+\gamma}{[m]_{q}+ \delta}-u+q ( u-\frac{\gamma}{[m]_{q}+\delta} ) \rbrace} + ( \frac{1}{[m]_{q}+b}-2u ) ( \frac{[m]_{q}+\delta }{[m]_{q}+b} ) (u-\frac{\gamma}{[m]_{q}+\delta} )+ ( \frac{a}{[m]_{q}+b} ) ( \frac{1}{[m]_{q}+b}-2u ) +u^{2}$$,

3. (3)

$$S^{(\gamma,\delta,a,b)}_{m,q}((t-u)^{3};u) = ( \frac{[m]_{q}+ \delta}{{[m]_{q}}+b} ) \frac{q^{6}[m-1]_{q}[m-2]_{q} (u-\frac{ \gamma}{[m]_{q}+\delta} )^{3}}{{([m]_{q}}+b)^{2} \lbrace\frac{[m]_{q}+ \gamma}{[m]_{q}+\delta}-u+q ( u-\frac{\gamma}{[m]_{q}+\delta } ) \rbrace \lbrace\frac{[m]_{q}+\gamma}{[m]_{q}+ \delta}-u+q^{2} ( u-\frac{\gamma}{[m]_{q}+\delta} ) \rbrace} + (\frac{2+3a+q}{({[m]_{q}}+b)}-3u ) (\frac{q^{2}[m-1]_{q}}{ {([m]_{q}}+b)} ) ( \frac{[m]_{q}+\delta}{[m]_{q}+b} ) \frac{ (u-\frac{\gamma}{[m]_{q}+\delta} )^{2}}{ \lbrace\frac{[m]_{q}+ \gamma}{[m]_{q}+\delta}-u+q ( u-\frac{\gamma}{[m]_{q}+\delta } ) \rbrace} + (\frac{1+3a^{2}}{{([m]_{q}}+b)^{2}}-\frac{3u}{{[m]_{q}}+b}+3u ^{2} ) (\frac{[m]_{q}+\delta}{{[m]_{q}}+b} ) (u-\frac{ \gamma}{[m]_{q}+\delta} )+\frac{a^{3}}{{([m]_{q}}+b)^{3}} +3a^{2} (\frac{[m]_{q}+\delta}{{([m]_{q}}+b)^{3}} ) -\frac{3u ^{2}a}{{([m]_{q}}+b)^{2}}+3u^{2} (\frac{[m]_{q}+\delta}{{[m]_{q}}+b} ) (u-\frac{\gamma}{[m]_{q}+\delta} )+ \frac{3a^{2}}{{([m]_{q}}+b)}-u^{3}$$,

4. (4)

$$S^{(\gamma,\delta,a,b)}_{m,q}((t-u)^{4};u)=\frac {q^{12}[m-1]_{q}[m-2]_{q}[m-3]_{q}}{ {([m]_{q}}+b)^{3}} (\frac{[m]_{q} +\delta}{{[m]_{q}}+b} ) \frac{1}{ \lbrace\frac{[m]_{q}+\gamma}{[m]_{q}+\delta}-u+q ( u-\frac{\gamma}{[m]_{q}+\delta} ) \rbrace} \frac{ (u-\frac{\gamma}{[m]_{q}+\delta} )^{4}}{ \lbrace\frac{[m]_{q}+ \gamma}{[m]_{q}+\delta}-u+q^{2} ( u-\frac{\gamma}{[m]_{q}+ \delta} ) \rbrace \lbrace\frac{[m]_{q}+\gamma}{[m]_{q}+ \delta}-u+q^{3} ( u-\frac{\gamma}{[m]_{q}+\delta} ) \rbrace} +\frac{[m-1]_{q}[m-2]_{q}}{{([m]_{q}}+b)^{2}} (\frac{[m]_{q} + \delta}{{[m]_{q}}+b} ) \frac{ ( \frac{q^{8}+3q^{5}+2q^{6}+4aq ^{2}}{[m]_{q}}-4uq^{6} ) (u-\frac{\gamma}{[m]_{q}+\delta } )^{3}}{ \lbrace\frac{[m]_{q}+\gamma}{[m]_{q}+\delta}-u+q ( u-\frac{\gamma}{[m]_{q}+\delta} ) \rbrace \lbrace\frac{[m]_{q}+ \gamma}{[m]_{q}+\delta}-u+q^{2} ( u-\frac{\gamma}{[m]_{q}+ \delta} ) \rbrace} + ( \frac{q^{4}+3q^{3}+3q^{2}+6a^{2}q^{2}}{{([m]_{q}}+b)^{2}}-\frac{8uq ^{2}+4uq^{3}+12auq^{2}}{[m]_{q}+b}-2u^{2}q^{2} ) \frac{[m-1]_{q}}{[m]_{q}+b} ( \frac{[m]_{q}+\delta}{[m]_{q}+b} )\frac{ (u-\frac{\gamma}{[m]_{q}+\delta} )^{2}}{ \lbrace\frac{[m]_{q}+ \gamma}{[m]_{q}+\delta}-u+q ( u-\frac{\gamma}{[m]_{q}+\delta } ) \rbrace} + ( \frac{1+4a^{3}}{{([m]_{q}}+b)^{3}}- \frac{4u+12a^{2}u}{{([m]_{q}}+b)^{2}}+\frac{2u^{2}}{{[m]_{q}}+b}-4u ^{3} ) ( \frac{[m]_{q}+\delta}{{[m]_{q}}+b} ) (u-\frac{\gamma}{[m]_{q}+\delta} ) + ( \frac{3a^{2}}{ {[m]_{q}}+b}-12a^{2}u ) ( \frac{[m]_{q}+\delta}{{([m]_{q}}+b)^{3}} )+ \frac{a^{4}}{{([m]_{q}}+b)^{4}}- \frac{4a^{3}u}{{([m]_{q}}+b)^{3}}- \frac{4au ^{2}}{{([m]_{q}}+b)^{2}} \frac{4au^{3}}{{[m]_{q}}+b}+u^{4}$$.

We can easily see that $$S^{(\gamma,\delta,a,b)}_{m,q}(f;u)$$are positive linear operators.

## 4 Main results

Firstly, we prove some theorems on the convergence of $$S^{(\gamma, \delta,a,b)}_{m,q}(f;u)$$ to $$f(u)$$.

### Theorem 4.1

Let $$f \in C[0,1]$$and the sequence $$q_{m}$$satisfying $$0< q_{m}<1$$such that $$q_{m}\to1$$as $$m\to\infty$$. Then

$$\lim_{m\to\infty}\max_{\frac{\gamma}{[m]_{q_{m}}+\delta}\leq u\leq\frac{[m]_{q _{m}}+\gamma}{[m]_{q_{m}}+\delta}} \bigl\vert S^{(\gamma,\delta,a,b)} _{m,q_{m}}(f;u)-f(u) \bigr\vert =0.$$

### Proof

From Lemma 3.1, it follows that

$$\lim_{m\to\infty}\max_{\frac{\gamma }{[m]_{q_{m}}+\delta}\leq u\leq\frac{[m]_{q _{m}}+\gamma}{[m]_{q_{m}}+\delta}} \bigl\vert S^{(\gamma,\delta,a,b)} _{m,q_{m}}\bigl(t^{i};u \bigr)-u^{i} \bigr\vert =0, \quad i=0,1,2.$$
(4.1)

Consider the sequence of operators

$$S^{*}_{m}(f;u)= \textstyle\begin{cases} S^{(\gamma,\delta,a,b)}_{m,q_{m}}, & \text{if }{\frac{\gamma}{[m]_{q _{m}}+\delta}\leq u\leq\frac{[m]_{q_{m}}+\gamma}{[m]_{q_{m}}+ \delta}} ; \\ f(u), & \text{if }u\in [0,\frac{\gamma}{[m]_{q_{m}}+\delta} ]\cup [\frac{[m]_{q_{m}}+\gamma}{[m]_{q_{m}}+\delta},1 ]. \end{cases}$$

Then obviously

$$\bigl\Vert S^{*}_{m}f-f \bigr\Vert =\max_{\frac{\gamma}{[m]_{q_{m}}+\delta}\leq u\leq\frac{[m]_{q _{m}}+\gamma}{[m]_{q_{m}}+\delta}} \bigl\vert S^{(\gamma,\delta,a,b)} _{m,q_{m}}(f;u)-f(u) \bigr\vert$$
(4.2)

and using (4.1) we obtain

$$\lim_{m\to\infty} \bigl\Vert S^{*}_{m}t^{i}-t^{i} \bigr\Vert _{C[0,1]}=0, \quad i=0,1,2.$$

Now, by applying the Korovkin theorem [12] (see also [3]) to the sequence of positive linear operators $$S^{*}_{m}$$, we obtain

$$\lim_{m\to\infty} \bigl\Vert S^{*}_{m}(f;u)-f(u) \bigr\Vert _{C[0,1]}=0$$

for every continuous function f. Therefore (4.2) gives

$$\lim_{m\to\infty}\max_{\frac{\gamma}{[m]_{q_{m}}+\delta}\leq u\leq\frac{[m]_{q _{m}}+\gamma}{[m]_{q_{m}}+\delta}} \bigl\vert S^{(\gamma,\delta,a,b)} _{m,q_{n}}(f;u)-f(u) \bigr\vert =0,$$

and the proof is completed. □

### Theorem 4.2

Iffbe a continuous function on $$[0,1]$$and taking $$0< q<1$$, then

$$\bigl\Vert S^{(\gamma,\delta,a,b)}_{m,q}f-f \bigr\Vert \leq 2\omega _{f}( \sigma_{m}),$$

where

$$\sigma_{m}=\sqrt{S^{(\gamma,\delta,a,b)}_{m,q} \bigl((t-u)^{2};u \bigr) )}.$$

### Proof

For any $$u,y\in[a,b]$$, it is well known that

$$\bigl\vert f(y)-f(u) \bigr\vert \leq\omega_{f}(\sigma) \biggl(1+\frac{(y-u)^{2}}{\sigma ^{2}} \biggr).$$

Therefore, we get

\begin{aligned} \bigl\vert S^{(\gamma,\delta,a,b)}_{m,q}(f;u)-f(u) \bigr\vert \leq& S^{(\gamma,\delta ,a,b)}_{m,q}\bigl( \bigl\vert f(t)-f(u) \bigr\vert ;u \bigr) \\ \leq& \omega_{f}(\sigma) \biggl(1+\frac{1}{\sigma^{2}}S^{(\gamma, \delta,a,b)}_{m,q} \bigl((t-u)^{2};u \bigr) \biggr). \end{aligned}

Choosing $$\sigma=\sigma_{m}=\sqrt{S^{(\gamma,\delta,a,b)}_{m,q} ((t-u)^{2};u ) )}$$, we have

$$\bigl\Vert S^{(\gamma,\delta,a,b)}_{m,q}f-f \bigr\Vert \leq 2\omega _{f}( \sigma_{m}).$$

Thus, we obtain the desired result. □

### Theorem 4.3

(Voronovskaja type theorem)

Let $$f^{\prime\prime}\in C[0,1]$$and $${(q_{m})}_{m\in\mathbb{N}}\subseteq(0,1)$$be a sequence such that $$q_{m}\to1$$as $$m\to\infty$$and $$q^{m}_{m}\to0$$as $$m\to\infty$$. Then

$$\lim_{m\to\infty}[m]_{q_{m}} \bigl(S^{(\gamma,\delta,a,b)} _{m,q_{m}}(f;u)-f(u) \bigr)=(\delta u-\gamma)f^{\prime}(u)+ \frac{1}{2} \bigl(u-\delta u^{2} \bigr)f^{\prime\prime}(u),$$

uniformly on $$[\frac{\gamma}{[m]_{q_{m}}+\delta},\frac{[m]_{q _{m}}+\gamma}{[m]_{q_{m}}+\delta} ]$$.

### Proof

By the Taylor formula we may write

$$f(t)=f(u)+f^{\prime}(u) (t-u)+\frac{1}{2}f^{\prime \prime}(u) (t-u)^{2}+r(t,u) {(t-u)}^{2},$$
(4.3)

where $$r(t,u)$$ is the remainder term and $$\lim_{t\rightarrow u}r(t,u)=0$$. Applying $$S^{(\gamma,\delta ,a,b)}_{m,q_{m}}(f;u)$$ to (4.3), we obtain

\begin{aligned} {}[ m]_{q_{m}} \bigl( S^{(\gamma,\delta,a,b)}_{m,q_{m}}(f;u)-f(u) \bigr) =& [m]_{q_{m}} S^{(\gamma,\delta,a,b)}_{m,q_{m}} \bigl((t-u);u \bigr)f^{\prime}(u) \\ &{}+[m]_{q_{m}}S^{(\gamma,\delta,a,b)}_{m,q_{m}} \bigl({(t-u)}^{2};u \bigr) \frac{f^{\prime\prime}(u)}{2} \\ &{} +[m]_{q_{m}}S^{(\gamma,\delta,a,b)}_{m,q_{m}} \bigl( r(t,u){(t-u)} ^{2};u \bigr). \end{aligned}

By the Cauchy–Schwartz inequality, we have

$$S^{(\gamma,\delta,a,b)}_{m,q_{m}} \bigl( r(t,u){(}t-u{)}^{2};u \bigr) \leq\sqrt{ S^{(\gamma,\delta,a,b)}_{m,q_{m}} \bigl( r^{2}(t,u);u \bigr) }\times\sqrt{S^{(\gamma,\delta,a,b)}_{m,q _{m}} \bigl( {(}t-u{)}^{4};u \bigr)}.$$
(4.4)

Observe that $$r^{2}(u,u)=0$$ and $$r^{2}(\cdot ,u)\in C[0,1]$$, then it follows from Theorem 4.2 that

$$\lim_{m\rightarrow\infty}S^{(\gamma,\delta,a,b)}_{m,q_{m}} \bigl( r^{2}(t,u);u \bigr) =r^{2}(u,u)=0$$
(4.5)

uniformly with respect to $$u\in [\frac{\gamma}{[m]_{q_{m}}+ \delta},\frac{[m]_{q_{m}}+\gamma}{[m]_{q_{m}}+\delta} ]$$. From (4.4) and (4.5), we get

$$\lim_{m\rightarrow\infty}[m]_{q_{m}}S^{(\gamma,\delta,a,b)} _{m,q_{m}} \bigl( r(t,u){ (t-u)}^{2};q;u \bigr)=0.$$

Now we compute the following:

$$\lim_{m\to\infty}[m]_{q_{m}}S^{(\gamma,\delta,a,b)}_{m,q _{m}} \bigl((t-u);u \bigr)=\delta u-\gamma,$$

uniformly in $$[\frac{\gamma}{[m]_{q_{m}}+\delta},\frac{[m]_{q _{m}}+\gamma}{[m]_{q_{m}}+\delta} ]$$.

Finally using the above two equalities, we have

\begin{aligned} \lim_{m\rightarrow\infty}[m]_{q_{m}} \bigl( S^{(\gamma, \delta,a,b)}_{m,q_{m}}(f;u)-f(u) \bigr) = {}& \lim_{m\rightarrow\infty}[m]_{q_{m}} S^{(\gamma,\delta,a,b)} _{m,q_{m}} \bigl((t-u);u \bigr)f^{\prime}(u) \\ &{} +\lim_{m\rightarrow\infty}[m]_{q_{m}} S^{(\gamma,\delta ,a,b)}_{m,q_{m}} \bigl({(t-u)}^{2};u \bigr) \frac{f^{\prime\prime}(u)}{2} \\ &{} +\lim_{m\rightarrow\infty}[m]_{q_{m}}S^{(\gamma,\delta,a,b)} _{m,q_{m}} \bigl( r(t,u){ (t-u)}^{2};u \bigr) \\ = {}&(\delta u-\gamma)f^{\prime}(u)+\frac{1}{2} \bigl(u-\delta u^{2} \bigr)f^{\prime\prime}(u). \end{aligned}

This completes the proof of the theorem. □

## 5 Local approximation

If $$\sigma>0$$ and $$W^{2}=\{s\in C[0,1]; s^{\prime},s^{\prime\prime }\in C[0,1]\}$$, then the K-functional is defined as

$$K_{2}(f,\sigma)=\inf\bigl\{ \Vert f-s \Vert + \sigma\bigl\Vert s^{\prime\prime} \bigr\Vert \bigr\} .$$
(5.1)

By [7], p. 177, Theorem 2.4, there exists an absolute constant $$C>0$$ such that

$$K_{2}(f,\sigma)\leq C\omega_{2}(f, \sqrt{\sigma}),$$
(5.2)

where the second order modulus of smoothness for $$f\in C[0,1]$$ is defined as

$$\omega_{2}(f,\sqrt{\sigma})=\sup _{0< h\leq\sqrt{\sigma}u} \sup_{u+2h\in[0,1]} \bigl\vert f(u+2h)-2f(u+h)+f(u) \bigr\vert .$$
(5.3)

The usual modulus of continuity for $$f\in C[0,1]$$ is defined as

$$\omega(f,\sigma)=\sup_{0< h\leq\sqrt{\sigma}u}\sup_{u+h\in[0,1]} \bigl\vert f(u+h)-f(u) \bigr\vert .$$

Our next main result is the following local approximation theorem.

### Theorem 5.1

Letfbe a continuous function on $$[0,1]$$with $$0< q<1$$. Then, for every $$u\in [\frac{\gamma}{[m]_{q}+\delta},\frac{[m]_{q}+\gamma}{[m]_{q}+ \delta} ]$$, we have

$$\bigl\vert S^{(\gamma,\delta,a,b)}_{m,q}(f;u)-f(u) \bigr\vert \leq C \omega_{2}\bigl(f, \sigma_{m}(u)\bigr)+\omega\biggl(f, \biggl\vert \frac{\delta u-\gamma}{[m]_{q}+b} \biggr\vert \biggr),$$

where

\begin{aligned} \sigma^{2}_{m}(u) =& \biggl(\frac{ q^{2}[m-1]_{q}}{[m]_{q}+b} \biggr) \biggl( \frac{[m]_{q}+\delta}{[m]_{q}+b} \biggr) \frac{ (u-\frac{ \gamma}{[m]_{q}+\delta} )^{2}}{ \lbrace\frac{ [m]_{q}+ \gamma}{[m]_{q}+\delta}-u+q ( u-\frac{\gamma}{[m]_{q}+\delta } ) \rbrace} \\ &{}+ \biggl( \frac{1}{[m]_{q}+b}-2u \biggr) \biggl( \frac{[m]_{q}+ \delta}{[m]_{q}+b} \biggr) \biggl(u-\frac{\gamma}{[m]_{q}+\delta} \biggr) \\ &{} + \biggl( \frac{a}{[m]_{q}+b} \biggr) \biggl( \frac{1}{[m]_{q}+b}-2u \biggr) +u ^{2} \\ &{}+ \biggl(\frac{[m]_{q}+\delta}{[m]_{q}} \biggl( u-\frac{\gamma}{[m]_{q}+ \delta} \biggr)-u \biggr)^{2}. \end{aligned}

### Proof

We define

$$\bar{S}^{(\gamma,\delta,a,b)}_{m,q}(f;u)=S^{(\gamma ,\delta,a,b)} _{m,q}(f;u)+f(u)-f \biggl(\frac{[m]_{q}+\delta}{[m]_{q}+b} \biggl( u- \frac{ \gamma}{[m]_{q}+\delta} \biggr) \biggr).$$
(5.4)

From Lemma 3.1, we find

$$\bar{S}^{(\gamma,\delta,a,b)}_{m,q}(1;u)=S^{(\gamma ,\delta,a,b)} _{m,q}(1;u)=1$$
(5.5)

and

$$\bar{S}^{(\gamma,\delta,a,b)}_{m,q}(t;u)=S^{(\gamma ,\delta,a,b)} _{m,q}(t;u)+u- \biggl(\frac{[m]_{q}+\delta}{[m]_{q}+b} \biggr) \biggl( u- \frac{\gamma}{[m]_{q}+\delta} \biggr)=u.$$
(5.6)

Let $$s\in W^{2}$$. By using Taylor’s formula we have

$$s(t)=s(u)+(t-u)s^{\prime}(u)(t-v)s^{\prime\prime}(v)\,dv,$$

we get

\begin{aligned} &\bar{S}^{(\gamma,\delta,a,b)}_{m,q}(s;u) \\ &\quad =s(u)+s^{\prime}(u)\bar{S}^{(\gamma,\delta,a,b)}_{m,q} \bigl((t-u);u\bigr)+ \bar{S}^{(\gamma,\delta,a,b)}_{m,q} \biggl( \int^{t}_{u}(t-v)s^{ \prime\prime}(v)\,dv;u \biggr) \\ &\quad = s(u)+S^{(\gamma,\delta,a,b)}_{m,q} \biggl( \int^{t}_{u}(t-v)s^{ \prime\prime}(v)\,dv;u \biggr) \\ &\qquad {} - \int^{\frac{[m]_{q}+\delta}{[m]_{q}+b} ( u-\frac{\gamma}{[m]_{q}+ \delta} )}_{u} \biggl[ \frac{[m]_{q}+\delta}{[m]_{q}+b} \biggl( u-\frac{\gamma}{[m]_{q}+\delta} \biggr)-v \biggr] s^{\prime\prime }(v)\,dv, \end{aligned}

which implies that

\begin{aligned}& \bigl\vert \bar{S}^{(\gamma,\delta,a,b)}_{m,q}(s;u)-s(u) \bigr\vert \\& \quad \leq S^{(\gamma,\delta,a,b)}_{m,q} \biggl( \biggl\vert \int^{t}_{u}(t-v)s ^{\prime\prime}(v)\,dv \biggr\vert ;u \biggr) \\& \qquad {} + \biggl\vert \int^{\frac{[m]_{q}+\delta}{[m]_{q}+b} ( u-\frac{\gamma}{[m]_{q}+ \delta} )}_{u} \biggl\vert \biggl[ \frac{[m]_{q}+\delta}{[m]_{q}+b} \biggl( u-\frac{\gamma}{[m]_{q}+ \delta} \biggr)-v \biggr] \biggr\vert \big\vert s^{\prime\prime}(v) \big\vert \,dv \biggr\vert \\& \quad \leq S^{(\gamma,\delta,a,b)}_{m,q} \bigl((t-u)^{2};u \bigr) \bigl\Vert s^{ \prime\prime} \bigr\Vert + \biggl[ \frac{[m]_{q}+\delta}{[m]_{q}+b} \biggl( u-\frac{ \gamma}{[m]_{q}+\delta} \biggr)-u \biggr] ^{2} \bigl\Vert s^{\prime\prime} \bigr\Vert . \end{aligned}

From Lemma 3.2(1), we have

\begin{aligned}& \bigl\vert \bar{S}^{(\gamma,\delta,a,b)}_{m,q}(s;u)-s(u) \bigr\vert \\& \quad \leq\biggl( \biggl(\frac{q^{2}[m-1]_{q}}{[m]_{q}+b} \biggr) \biggl( \frac{[m]_{q}+ \delta}{[m]_{q}+b} \biggr) \frac{ (u-\frac{\gamma}{[m]_{q}+ \delta} )^{2}}{ \lbrace\frac{[m]_{q}+\gamma}{[m]_{q}+ \delta}-u+q ( u-\frac{\gamma}{[m]_{q}+\delta} ) \rbrace} \\& \qquad {} + \biggl( \frac{1}{[m]_{q}+b}-2u \biggr) \biggl( \frac{[m]_{q}+ \delta}{[m]_{q}+b} \biggr) \biggl(u-\frac{\gamma}{[m]_{q}+\delta} \biggr) \\& \qquad {}+ \biggl( \frac{a}{[m]_{q}+b} \biggr) \biggl( \frac{1}{[m]_{q}+b}-2u \biggr)+u ^{2} \biggr) \bigl\Vert s^{\prime\prime} \bigr\Vert \\& \qquad {} + \biggl(\frac{[m]_{q}+\delta}{[m]_{q}+b} \biggl( u-\frac{ \gamma}{[m]_{q}+\delta} \biggr)-u \biggr)^{2} \bigl\Vert s^{\prime\prime} \bigr\Vert \\& \quad \leq\biggl\{ \biggl(\frac{q^{2}[m-1]_{q}}{[m]_{q}+b} \biggr) \biggl( \frac{[m]_{q}+\delta}{[m]_{q}+b} \biggr) \frac{ (u-\frac{\gamma }{[m]_{q}+\delta} )^{2}}{ \lbrace\frac{[m]_{q}+\gamma}{[m]_{q}+ \delta}-u+q ( u-\frac{\gamma}{[m]_{q}+\delta} ) \rbrace} \\& \qquad {} + \biggl( \frac{1}{[m]_{q}+b}-2u \biggr) \biggl( \frac{[m]_{q}+ \delta}{[m]_{q}+b} \biggr) \biggl(u-\frac{\gamma}{[m]_{q}+\delta} \biggr) \\& \qquad {}+ \biggl( \frac{a}{[m]_{q}+b} \biggr) \biggl( \frac{1}{[m]_{q}+b}-2u \biggr)+u ^{2} ) \\& \qquad {} + \biggl(\frac{[m]_{q}+\delta}{[m]_{q}+b} \biggl( u-\frac{ \gamma}{[m]_{q}+\delta} \biggr)-u \biggr)^{2} \biggr\} \bigl\Vert s^{\prime \prime} \bigr\Vert \\& \quad =\sigma^{2}_{m}(u) \bigl\Vert s^{\prime\prime} \bigr\Vert . \end{aligned}
(5.7)

On the other hand, we have

$$\bigl\vert \bar{S}^{(\gamma,\delta,a,b)}_{m,q}(f;u) \bigr\vert \leq3 \Vert f \Vert .$$
(5.8)

Now, for $$f\in C[0,1]$$ and $$s\in W^{2}$$, by using (5.7) and (5.8), we get

\begin{aligned}& \bigl\vert {S}^{(\gamma,\delta,a,b)}_{m,q}(f;u)-f(u) \bigr\vert \\& \quad \leq\biggl\vert \bar{S}^{(\gamma,\delta,a,b)}_{m,q}(f;u)-f(u) +f \biggl(\frac{[m]_{q}+\delta}{[m]_{q}+b} \biggl( u-\frac{\gamma}{[m]_{q}+ \delta} \biggr) \biggr)-f(u) \biggr\vert \\& \quad \leq\bigl\vert \bar{S}^{(\gamma,\delta,a,b)}_{m,q}(f-s;u) \bigr\vert + \bigl\vert \bar{S}^{(\gamma,\delta,a,b)}_{m,q}(s;u)-s(u) \bigr\vert \\& \qquad {} + \bigl\vert s(u)-f(u) \bigr\vert + \biggl\vert f \biggl( \frac{([m]_{q}+\delta)u-\gamma}{[m]_{q}+b} \biggr)-f(u) \biggr\vert \\& \quad \leq4 \Vert f-s \Vert +\sigma^{2}_{m}(u) \bigl\Vert s^{\prime\prime} \bigr\Vert + \biggl\vert f \biggl( \frac{([m]_{q}+\delta)u-\gamma}{[m]_{q}+b} \biggr)-f(u) \biggr\vert . \end{aligned}

Taking the infimum on the right hand side over all $$s\in W^{2}$$, we obtain

\begin{aligned} \bigl\vert {S}^{(\gamma,\delta,a,b)}_{m,q}(f;u)-f(u) \bigr\vert \leq & 4K _{2} \bigl(f,\sigma^{2}_{m}(u) \bigr) + \omega\biggl(f, \biggl\vert \frac{([m]_{q}+ \delta)u-\gamma}{[m]_{q}+b}-u \biggr\vert \biggr) \\ =& 4K_{2} \bigl(f,\sigma^{2}_{m}(u) \bigr) +\omega\biggl(f, \biggl\vert \frac{ \delta u-\gamma}{[m]_{q}+b} \biggr\vert \biggr). \end{aligned}

Now using (5.3), we have

$$\bigl\vert S^{(\gamma,\delta,a,b)}_{m,q}(f;u)-f(u) \bigr\vert \leq C \omega_{2}\bigl(f, \sigma_{m}(u)\bigr)+\omega\biggl(f, \biggl\vert \frac{\delta u-\gamma}{[m]_{q}+b} \biggr\vert \biggr).$$

This completes the proof. □

## 6 Conclusion

It can be concluded that the parameters γ, δ, q, a and b will provide more modeling flexibility for approximation of functions and bases of these operators can be used to draw curves and surfaces in CAGD. Also the approximation results derived for shifted intervals will be very helpful when it comes to implementation using computers for simulation purposes.

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### Acknowledgements

The authors are extremely thankful to the referees for their valuable comments and suggestion. The second author is also grateful to Council of Scientific and Industrial Research (CSIR), India, for providing the Senior Research Fellowship (09/1172(0001)/2017-EMR-I).

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Mursaleen, M., Qasim, M., Khan, A. et al. Stancu type q-Bernstein operators with shifted knots. J Inequal Appl 2020, 28 (2020). https://doi.org/10.1186/s13660-020-2303-4