1 Introduction
In recent years, many researches focused their attention on the study of a generalized version in q-calculus of the well-known linear and positive operators [1‐6]. Lupaş [7] initiated in 1987 the convergence of Bernstein operators based on q-integers and in 1996 another generalization of these operators was introduced by Philips [8]. Also, in [9], Agratini introduced a new class of q-Bernstein-type operators, which fix certain polynomials. More results on q-Bernstein polynomials were obtained by Ostrovska [10]. Muraru [11] proposed and studied some approximation properties of the q-Bernstein-Schurer operators. In [12], Ren and Zeng introduced a modified version of the q-Bernstein-Schurer operators and investigated the statistical approximation properties. The Kantorovich-type generalization of these operators was given in [13] by Özarslan and Vedi. In [14], Agrawal et al. introduced a Stancu-type generalization of the Bernstein-Schurer operators based on q-integer. They obtained the rate of convergence of these operators in terms of the modulus of continuity and by a Voronovskaja-type theorem. Many generalizations and applications of the Stancu operators were considered in the last years [15‐17]. The goal of the present paper is to study some approximation properties of the q-analog of the Stancu-Kantorovich operators.
Before proceeding, we mention some basic definitions and notations from q-calculus. Let \(q>0\). For each nonnegative integer k, the q-integer \([k]_{q}\) and q-factorial \([k]_{q}!\) are defined by respectively.
$$\begin{aligned}& {[}k]_{q}:= \textstyle\begin{cases} {\frac{1-q^{k}}{1-q}}, & q\ne1, \\ k, & q=1, \end{cases}\displaystyle \\& {[}k]_{q}!:= \textstyle\begin{cases} [k]_{q}[k-1]_{q}\cdots[1]_{q}, & k\geq1, \\ 1, & k=0, \end{cases}\displaystyle \end{aligned}$$
Anzeige
For the integers n, k satisfying \(n\geq k\geq0\), the q-binomial coefficients are defined by We denote \((a+b)_{q}^{k}=\prod_{j=0}^{k-1}(a+bq^{j})\).
$$ \left .\begin{bmatrix} n \\ k\end{bmatrix} \right ._{q}:= {\frac{[n]_{q}!}{[k]_{q}![n-k]_{q}!}}. $$
The q-Jackson integral on the interval \([0,b]\) is defined as provided that the sums converge absolutely. Suppose that \(0< a< b\). The q-Jackson integral on the interval \([a,b]\) is defined as The Riemann-type q-integral (see [18]) is defined by The classical Stancu-Kantorovich operators \(S_{n}^{(\alpha,\beta)}\), \(n=1,2,\ldots\) , are defined by where \(p_{n,k}(x)= {\binom{n}{k}}x^{k}(1-x)^{n-k}\) and \(0\leq\alpha\leq\beta\).
$$ \int_{0}^{b}f(t)\,d_{q}t=(1-q)b\sum _{j=0}^{\infty}f\bigl(q^{j}b \bigr)q^{j},\quad 0< q< 1, $$
$$ \int_{a}^{b}f(t)\,d_{q}t=\int _{0}^{b} f(t)\,d_{q}t-\int _{0}^{a} f(t)\,d_{q}t,\quad 0< q< 1. $$
$$ \int_{a}^{b} f(t)\,d_{q}^{R}t=(1-q) (b-a)\sum_{j=0}^{\infty}f \bigl(a+(b-a)q^{j} \bigr)q^{j}. $$
$$ \begin{aligned}[b] S_{n}^{(\alpha,\beta)}(f,x)&:=(n+1) \sum _{k=0}^{n} p_{n,k}(x)\int _{\frac{k+\alpha}{n+1+\beta}}^{\frac{k+1+\alpha}{n+1+\beta}}f(t)\,dt \\ &= \sum_{k=0}^{n}p_{n,k}(x)\int _{0}^{1}f \biggl(\frac{t+k+\alpha}{n+1+\beta} \biggr)\,dt, \quad f:[0,1]\to\mathbb{R}, \end{aligned} $$
(1.1)
In [19], Ren and Zeng introduced two kinds of Kantorovich-type q-Bernstein-Stancu operators. The first version is defined using the q-Jackson integral as follows: where \(f\in C[0,1]\) and \(p_{n,k}(q;x)=\bigl [{\scriptsize\begin{matrix}{} n \cr k\end{matrix}} \bigr ]_{q}x^{k}(1-x)_{q}^{n-k}\).
$$ S_{n,q}^{(\alpha,\beta)}(f,x)= \bigl([n+1]_{q}+\beta \bigr) \sum_{k=0}^{n} q^{-k}p_{n,k}(q;x)\int_{\frac{[k]_{q}+\alpha}{[n+1]_{q}+\beta }}^{\frac{[k+1]_{q}+\alpha}{[n+1]_{q}+\beta}}f(t)\,d_{q}t, $$
(1.2)
To guarantee the positivity of the q-Bernstein-Stancu-Kantorovich operators, in [19] \(S_{n,q}^{(\alpha,\beta)}(f;x)\) is redefined by putting the Riemann-type q-integral into the operators instead of the q-Jackson integral as These operators verify the following.
$$ \tilde{S}_{n,q}^{(\alpha,\beta)}(f,x)= \bigl([n+1]_{q}+\beta \bigr) \sum_{k=0}^{n} q^{-k}p_{n,k}(q;x)\int_{\frac{[k]_{q}+\alpha}{[n+1]_{q}+\beta }}^{\frac{[k+1]_{q}+\alpha}{[n+1]_{q}+\beta}}f(t)\,d_{q}^{R}t. $$
(1.3)
Anzeige
Lemma 1.1
([19])
For
\(\tilde{S}_{n,q}^{(\alpha,\beta)}\), \(0< q<1\), and
\(0\leq\alpha\leq\beta\), we have
(i)
$$\tilde{S}_{n,q}^{(\alpha,\beta)}(1;x)=1, $$
(ii)
$$\tilde{S}_{n,q}^{(\alpha,\beta)}(t;x)= \frac{2q[n]_{q}}{[2]_{q} ([n+1]_{q}+\beta )}x+ \frac{1+[2]_{q}\alpha}{[2]_{q} ([n+1]_{q}+\beta )}, $$
(iii)
$$\begin{aligned} \tilde{S}_{n,q}^{\alpha,\beta)}\bigl(t^{2};x\bigr)={}& \frac{q[n]_{q}[n-1]_{q}}{ ([n+1]_{q}+\beta )^{2}} \biggl(1+ \frac{2(q-1)}{[2]_{q}}+\frac{(q-1)^{2}}{[3]_{q}} \biggr)x^{2} + \frac{[n]_{q}}{([n+1]_{q}+\beta)^{2}}\\ &{}\cdot \biggl(1+2\alpha+\frac{2(q-1)(1+\alpha)}{[2]_{q}} + \frac{2}{[2]_{q}}+\frac{2(q-1)}{[3]_{q}}+\frac {(q-1)^{2}}{[3]_{q}} \biggr)x \\ &{}+ \frac{1}{([n+1]_{q}+\beta)^{2}} \biggl(\frac{1}{[3]_{q}}+\frac{2\alpha }{[2]_{q}}+ \alpha^{2} \biggr). \end{aligned}$$
In [20], Mahmudov and Sabancigil introduced a q-type generalization of the Bernstein-Kantorovich operators as follows: where \(f\in C[0,1]\) and \(0< q\leq1\). In [21], inspired by (1.4) we introduced a q-type generalization of the Stancu-Kantorovich operators as follows: where \(0\leq\alpha\leq\beta\) and \(f\in C[0,1]\).
$$ B_{n,q}^{*}(f,x):= \sum_{k=0}^{n}p_{n,k}(q;x) \int_{0}^{1}f \biggl(\frac{[k]_{q}+q^{k}t}{[n+1]_{q}} \biggr)\,d_{q}t, $$
(1.4)
$$ S_{n,q}^{*(\alpha,\beta)}(f,x)= \sum_{k=0}^{n}p_{n,k}(q;x)\int_{0}^{1}f \biggl(\frac{[k]_{q}+q^{k}t+\alpha}{[n+1]_{q}+\beta} \biggr)\,d_{q}t, $$
(1.5)
Lemma 1.2
([21])
For all
\(n\in\mathbb{N}\), \(x\in [0,1]\), and
\(0< q\leq 1\), we have
$$\begin{aligned} & S_{n,q}^{*(\alpha,\beta)}(1,x)=1,\qquad S_{n,q}^{*(\alpha,\beta)}(t,x)={ \frac{2q}{[2]_{q}}\frac{[n]_{q}}{[n+1]_{q}+\beta}x+\frac{\alpha}{[n+1]_{q}+\beta}+\frac{1}{[2]_{q}([n+1]_{q}+\beta)}}, \\ &S_{n,q}^{*(\alpha,\beta)}\bigl(t^{2},x\bigr) = \frac{1}{([n + 1]_{q} + \beta)^{2}} \biggl\{ \frac{q^{2}(q + 2)}{[3]_{q}}[n]_{q}[n - 1]_{q}x^{2} + \frac{q[n]_{q}}{[2]_{q}} \biggl(4\alpha + \frac{4 + 7q + q^{2}}{[3]_{q}} \biggr)x \\ &\hphantom{S_{n,q}^{*(\alpha,\beta)}\bigl(t^{2},x\bigr) =}{} + \frac{2\alpha}{[2]_{q}} + \frac {1}{[3]_{q}} + \alpha^{2} \biggr\} . \end{aligned}$$
Lemma 1.3
([21])
For all
\(n\in\mathbb{N}\), \(x\in[0,1]\), and
\(0< q\leq 1\), we have
where
C
is a positive absolute constant.
$$\begin{aligned} &\begin{aligned}[b] &S_{n,q}^{*(\alpha,\beta)} \bigl((t-x)^{2},x \bigr)\\ &\quad\leq \frac{2[n+1]_{q}^{2}}{([n+1]_{q}+\beta)^{2}} \biggl\{ \frac{4}{[n]_{q}} \biggl(x(1-x)+ \frac{1}{[n]_{q}} \biggr)+ \biggl(\frac{\alpha}{[n+1]_{q}}-\frac {\beta }{[n+1]_{q}}x \biggr)^{2} \biggr\} , \end{aligned}\\ &S_{n,q}^{*(\alpha,\beta)} \bigl((t-x)^{4},x \bigr)\\ &\quad\leq \frac{8[n+1]_{q}^{2}}{([n+1]_{q}+\beta)^{2}} \biggl\{ \frac{C}{[n]_{q}^{2}} \biggl(x(1-x)+\frac{1}{[n]_{q}^{2}} \biggr)+ \biggl(\frac{\alpha}{[n+1]_{q}}-\frac{\beta}{[n+1]_{q}}x \biggr)^{4} \biggr\} , \end{aligned}$$
Also, in [21] a Voronovskaja-type theorem for the \(S_{n,q}^{*(\alpha,\beta)}\) was established.
Theorem 1.4
([21])
Let
\(f^{\prime\prime}\in C[0,1]\), \(q_{n}\in (0,1)\), \(q_{n}\to1\), and
\(q_{n}^{n}\to a\), \(a\in[0,1)\)
as
\(n\to\infty\). Then we have
$$ \lim_{n\to\infty}[n]_{q}\bigl(S_{n,q_{n}}^{*(\alpha,\beta)}(f,x)-f(x) \bigr)= \biggl(-\frac{1+a+2\beta}{2}x+\alpha+\frac{1}{2} \biggr) f^{\prime}(x)+\frac{1}{2} \biggl(-\frac{2a+1}{3}x^{2}+x \biggr)f^{\prime \prime}(x). $$
The paper is organized as follows. In Section 2 we prove a Voronovskaja-type asymptotic formula for \(\tilde{S}_{n,q}^{\alpha,\beta}\). In Section 3 we establish some approximation properties of the q-Stancu-Kantorovich operators \(\tilde{S}_{n,q}^{\alpha,\beta}\) and \({S}_{n,q}^{*\alpha,\beta}\). In the final section we give statistical approximation results for the q-Stancu-Kantorovich operators.
2 A Voronovskaya theorem for q-Stancu-Kantorovich operators
In this section we shall establish a Voronovskaja-type theorem for the q-Stancu-Kantorovich operators \(\tilde{S}_{n,q}^{(\alpha,\beta)}\). First, we need the auxiliary result contained in the following lemma.
Lemma 2.1
Assume that
\(0< q_{n}<1\), \(q_{n}\to1\), and
\(q_{n}^{n}\to a\), \(a\in[0,1)\)
as
\(n\to\infty\). Then we have
$$\begin{aligned} & \lim_{n\to\infty}[n]_{q_{n}}\tilde{S}_{n,q_{n}}^{(\alpha,\beta)}(t-x,x)=- \frac{1+a+2\beta}{2}x+\alpha+\frac{1}{2}, \\ & \lim_{n\to\infty}[n]_{q_{n}}\tilde{S}_{n,q_{n}}^{(\alpha,\beta)} \bigl((t-x)^{2},x\bigr)=x(1-x). \end{aligned}$$
Proof
To prove the lemma we use the formulas for \(\tilde{S}_{n,q_{n}}^{(\alpha ,\beta)}(t^{i},x)\), \(i=0,1,2\), given in Lemma 1.1. We have □
$$\begin{aligned} & \lim_{n\to\infty}[n]_{q_{n}}\tilde{S}_{n,q_{n}}^{(\alpha,\beta)}(t-x;x) \\ &\quad= \lim_{n\to\infty}[n]_{q_{n}} \biggl\{ \biggl( \frac{2q_{n}}{[2]_{q_{n}}} \frac{[n]_{q_{n}}}{[n + 1]_{q_{n}} + \beta} - 1 \biggr)x + \frac{\alpha}{[n + 1]_{q_{n}} + \beta} + \frac {1}{[2]_{q_{n}}([n + 1]_{q_{n}} + \beta} \biggr\} \\ &\quad= \lim_{n\to\infty} \biggl\{ \frac{[n]_{q_{n}}}{[2]_{q_{n}}([n+1]_{q_{n}}+\beta)} \bigl(-1-q_{n}^{n+1}-[2]_{q_{n}}\beta\bigr)x \\ &\qquad{}+ \frac {\alpha[n]_{q_{n}}}{[n+1]_{q_{n}}+\beta}+\frac{[n]_{q_{n}}}{[2]_{q_{n}}([n+1]_{q_{n}}+\beta )} \biggr\} \\ &\quad= -\frac{1+a+2\beta}{2}x+\alpha+\frac{1}{2}, \\ & \lim_{n\to\infty}[n]_{q_{n}}\tilde{S}_{n,q_{n}}^{(\alpha,\beta)} \bigl((t-x)^{2};x\bigr)\\ &\quad= \lim_{n\to\infty}[n]_{q_{n}} \bigl\{ \tilde{S}_{n,q_{n}}^{(\alpha,\beta)}\bigl(t^{2},x \bigr)-x^{2} -2x\tilde{S}_{n,q_{n}}^{(\alpha,\beta)}(t-x,x) \bigr\} \\ &\quad= \lim_{n\to\infty}[n]_{q_{n}} \biggl(\frac{q_{n}[n]_{q_{n}}[n-1]_{q_{n}}}{([n+1]_{q_{n}}+\beta)^{2}} \cdot\frac{4q_{n}^{3}+q_{n}^{2}+q_{n}}{[2]_{q_{n}}[3]_{q_{n}}} -1 \biggr)x^{2} \\ &\qquad{} + \lim_{n\to\infty} \frac{[n]_{q_{n}}^{2}}{([n+1]_{q_{n}}+\beta)^{2}} \biggl(1+2\alpha+ \frac{2(q_{n}-1)(1+\alpha )}{[2]_{q_{n}}}\\ &\qquad{}+\frac{2}{[2]_{q_{n}}}+\frac{2(q_{n}-1)}{[3]_{q_{n}}}+ \frac {(q_{n}-1)^{2}}{[3]_{q_{n}}} \biggr)x \\ &\qquad{} + \lim_{n\to\infty}\frac{[n]_{q_{n}}}{([n+1]_{q_{n}}+\beta)^{2}} \biggl(\frac{1}{[3]_{q_{n}}}+\frac{2\alpha}{[2]_{q_{n}}}+\alpha^{2} \biggr)- \lim_{n\to\infty}2x[n]_{q_{n}}\tilde{S}_{n,q_{n}}^{(\alpha,\beta)}(t-x,x) \\ &\quad= \lim_{n\to\infty}[n]_{q_{n}} \biggl( \frac{(4q_{n}^{3}+q_{n}^{2}+q_{n})\cdot ([n]_{q_{n}}^{2}-[n]_{q_{n}})}{[2]_{q_{n}}[3]_{q_{n}}([n+1]_{q_{n}}+\beta)^{2}}-1 \biggr)x^{2} \\ &\qquad{}+ (2\alpha+2)x - 2x \biggl(-\frac{1 + a + 2\beta}{2}x + \alpha + \frac{1}{2} \biggr) \\ &\quad= \lim_{n\to\infty}\frac{[n]_{q_{n}}}{[2]_{q_{n}}[3]_{q_{n}}([n+1]_{q_{n}}+\beta)^{2}} \bigl\{ \bigl(4q_{n}^{3}+q_{n}^{2}+q_{n} \bigr)[n]_{q_{n}}^{2}-\bigl(4q_{n}^{3}+q_{n}^{2}+q_{n} \bigr)[n]_{q_{n}} \\ &\qquad{} -[2]_{q_{n}}[3]_{q_{n}}\bigl([n+1]_{q_{n}}+ \beta\bigr)^{2} \bigr\} x^{2}+x+(1+a+2\beta)x^{2} \\ &\quad= \lim_{n\to\infty}\frac{[n]_{q_{n}}}{[2]_{q_{n}}[3]_{q_{n}}([n+1]_{q_{n}}+\beta)^{2}} \bigl\{ \bigl(4q_{n}^{3}+q_{n}^{2}+q_{n} \bigr)[n]_{q_{n}}^{2}-\bigl(4q_{n}^{3}+q_{n}^{2}+q_{n} \bigr)[n]_{q_{n}} \\ &\qquad{} -[2]_{q_{n}}[3]_{q_{n}}\bigl(1+q_{n}[n]_{q_{n}}+ \beta\bigr)^{2} \bigr\} x^{2}+x+(1+a+2\beta)x^{2} \\ &\quad= \lim_{n\to\infty}\frac{[n]_{q_{n}}}{[2]_{q_{n}}[3]_{q_{n}}([n+1]_{q_{n}}+\beta)^{2}} \bigl\{ \bigl(4q_{n}^{3}+q_{n}^{2}+q_{n}-[2]_{q_{n}}[3]_{q_{n}}q_{n}^{2} \bigr)[n]_{q_{n}}^{2} \\ &\qquad{} - \bigl(4q_{n}^{3}+q_{n}^{2}+q_{n}+2q_{n}(1+ \beta)[2]_{q_{n}}[3]_{q_{n}} \bigr)[n]_{q_{n}}-[2]_{q_{n}}[3]_{q_{n}}(1+ \beta)^{2} \bigr\} x^{2}\\ &\qquad{}+x+(1+a+2\beta)x^{2}\\ &\quad=x(1-x). \end{aligned}$$
The main result of this section is the following Voronovskaja-type theorem.
Theorem 2.2
Let
\(f^{\prime\prime}\in C[0,1]\), \(q_{n}\in(0,1)\), \(q_{n}\to1 \), and
\(q_{n}^{n}\to a\), \(a\in[0,1)\)
as
\(n\to\infty\). Then we have
$$ \lim_{n\to\infty}[n]_{q_{n}}\bigl(\tilde{S}_{n,q_{n}}^{(\alpha,\beta)}(f,x)-f(x) \bigr)= \biggl(-\frac{1+a+2\beta}{2}x+\alpha+\frac{1}{2} \biggr) f^{\prime}(x)+\frac{1}{2}x(1-x)f^{\prime\prime}(x). $$
Proof
From the Taylor theorem, we have where ξ lies between t and x.
$$ f(t)=f(x)+(t-x)f^{\prime}(x)+\frac{1}{2}(t-x)^{2}f^{\prime\prime}(x)+ \frac {1}{2}(t-x)^{2}\bigl(f^{\prime\prime}(\xi)-f^{\prime\prime}(x) \bigr), $$
(2.1)
Applying \(\tilde{S}_{n,q_{n}}^{(\alpha,\beta)}\) on both sides of (2.1), we obtain For all \(x,t\in[0,1]\), \(| f^{\prime\prime}(\xi)-f^{\prime\prime}(x)|\leq \omega_{f^{\prime\prime}}(\delta) (1+ \frac{(t-x)^{2}}{\delta^{2}} ) \) for any \(\delta>0\). Therefore, it follows that Let \(B_{n,q}^{(\alpha,\beta)}(f,x)= \sum_{k=0}^{n}p_{n,k}(q;x)f (\frac{[k]_{q}+\alpha}{[n]_{q}+\beta} )\) be q-Bernstein-Stancu operators. It follows that From [14], Corrollary 1, we have \(B_{n,q}^{(\alpha,\beta)}((t-x)^{r},x)=O (\frac{1}{[n]_{q}^{[\frac{r+1}{2}]}} )\), where \(x\in[0,1]\) and \([\alpha]\) denotes the integer part of α. Also, we have Therefore
$$ \begin{aligned}[b] [n]_{q_{n}} \bigl(\tilde{S}_{n,q_{n}}^{(\alpha,\beta)}(f,x)-f(x) \bigr)={}& [n]_{q_{n}}f^{\prime}(x)\tilde{S}_{n,q_{n}}^{(\alpha,\beta)}(t-x,x)+ \frac {1}{2}[n]_{q_{n}}f^{\prime\prime}(x) \tilde{S}_{n,q_{n}}^{(\alpha,\beta)} \bigl((t-x)^{2},x \bigr) \\ &{}+[n]_{q_{n}}\tilde{S}_{n,q_{n}}^{(\alpha,\beta)} \biggl( \frac{(t-x)^{2}}{2}\bigl(f^{\prime\prime}(\xi)-f^{\prime\prime}(x)\bigr),x \biggr). \end{aligned} $$
(2.2)
$$ \bigl\vert [n]_{q_{n}}\tilde{S}_{n,q_{n}}^{(\alpha,\beta)} \bigl((t-x)^{2}\bigl(f^{\prime \prime}(\xi)-f^{\prime\prime}(x)\bigr),x \bigr)\bigr\vert \leq\omega_{f^{\prime\prime}}(\delta)[n]_{q_{n}}\tilde {S}_{n,q_{n}}^{(\alpha,\beta)} \biggl((t-x)^{2}+ \frac{(t-x)^{4}}{\delta^{2}},x \biggr). $$
(2.3)
$$\begin{aligned} &\tilde{S}_{n,q_{n}}^{(\alpha,\beta)} \bigl((t-x)^{4},x \bigr) \\ &\quad= \bigl([n+1]_{q_{n}}+\beta \bigr) \sum _{k=0}^{n}q_{n}^{-k}p_{n,k}(q_{n};x) \int_{\frac{[k]_{q_{n}}+\alpha}{[n+1]_{q_{n}}+\beta}}^{\frac {[k+1]_{q_{n}}+\alpha}{[n+1]_{q_{n}}+\beta}}(t-x)^{4}d_{q_{n}}^{R}t \\ &\quad=\bigl([n+1]_{q_{n}}+\beta\bigr) \sum_{k=0}^{n}q_{n}^{-k}p_{n,k}(q_{n};x) (1-q_{n})\frac{[k+1]_{q_{n}}-[k]_{q_{n}}}{[n+1]_{q_{n}}+\beta} \\ & \qquad{}\times\sum_{j=0}^{\infty} \biggl( \frac{[k]_{q_{n}}+\alpha }{[n+1]_{q_{n}}+\beta}+\frac{[k+1]_{q_{n}}-[k]_{q_{n}}}{[n+1]_{q_{n}}+\beta} q_{n}^{j}-x \biggr)^{4}q_{n}^{j} \\ &\quad=(1-q_{n}) \sum_{k=0}^{n}p_{n,k}(q_{n};x) \sum_{j=0}^{\infty} \biggl( \frac{[k]_{q_{n}}+\alpha}{[n+1]_{q_{n}}+\beta }+\frac{q_{n}^{k}}{[n+1]_{q_{n}}+\beta}q_{n}^{j}-x \biggr)^{4}q_{n}^{j} \\ &\quad\leq8(1-q_{n}) \sum_{k=0}^{n}p_{n,k}(q_{n};x) \sum_{j=0}^{\infty} \biggl( \frac{[k]_{q_{n}}+\alpha}{[n+1]_{q_{n}}+\beta}-x \biggr)^{4}q_{n}^{j} \\ &\qquad{} +8 (1-q_{n}) \sum_{k=0}^{n}p_{n,k}(q_{n};x) \sum_{j=0}^{\infty} \biggl( \frac{q_{n}^{k}}{[n+1]_{q_{n}}+\beta} \biggr)^{4}q_{n}^{5j} \\ &\quad= 8 \sum_{k=0}^{n}p_{n,k}(q_{n};x) \biggl(\frac{[k]_{q_{n}}+\alpha}{[n+1]_{q_{n}}+\beta}-x \biggr)^{4} \\ &\qquad{}+ \frac{8}{1+q_{n}+q_{n}^{2}+q_{n}^{3}+q_{n}^{4}}\sum_{k=0}^{n}p_{n,k}(q_{n};x) \biggl(\frac{q_{n}^{k}}{[n+1]_{q_{n}}+\beta} \biggr)^{4} \\ &\quad\leq8 \sum_{k=0}^{n}p_{n,k}(q_{n};x) \biggl(\frac{[k]_{q_{n}}+\alpha}{[n+1]_{q_{n}}+\beta}-\frac{[k]_{q_{n}}+\alpha}{[n]_{q_{n}}+\beta}+\frac{[k]_{q_{n}}+\alpha}{[n]_{q_{n}}+\beta}-x \biggr)^{4} \\ &\qquad{} + \frac{8}{1+q_{n}+q_{n}^{2}+q_{n}^{3}+q_{n}^{4}} \sum_{k=0}^{n}p_{n,k}(q_{n};x) \biggl(\frac{q_{n}^{k}}{[n]_{q_{n}}+\beta} \biggr)^{4} \\ &\quad\leq64 \sum_{k=0}^{n}p_{n,k}(q_{n};x) \biggl(\frac{[k]_{q_{n}}+\alpha}{[n]_{q_{n}}+\beta} \biggr)^{4} \biggl( \frac{q_{n}^{n}}{[n+1]_{q_{n}}+\beta} \biggr)^{4}+64B_{n,q_{n}}^{(\alpha,\beta)} \bigl((t-x)^{4},x \bigr) \\ &\qquad{} +8 \sum_{k=0}^{n}p_{n,k}(q_{n};x) \biggl(\frac{q_{n}^{k}}{[n]_{q_{n}}+\beta} \biggr)^{4} \\ &\quad\leq64 \biggl( \frac{q_{n}^{n}}{[n+1]_{q_{n}}+\beta} \biggr)^{4}+64B_{n,q_{n}}^{(\alpha,\beta)} \bigl((t-x)^{4},x \bigr)+8 \sum _{k=0}^{n}p_{n,k}(q_{n};x) \biggl( \frac{q_{n}^{k}}{[n]_{q_{n}}+\beta} \biggr)^{4}. \end{aligned}$$
$$\begin{aligned} & \sum_{k=0}^{n}p_{n,k}(q_{n};x) \biggl(\frac{q_{n}^{k}}{[n]_{q_{n}}+\beta} \biggr)^{4} \\ &\quad= \frac{1}{([n]_{q_{n}}+\beta)^{2}} \sum_{k=0}^{n}p_{n,k}(q_{n};x) \biggl(\frac{1-(1-q_{n})[k]_{q_{n}}-(1-q_{n})\alpha+(1-q_{n})\alpha}{[n]_{q_{n}}+\beta} \biggr)^{2} \\ &\quad\leq\frac{2}{([n]_{q_{n}}+\beta)^{2}} \Biggl\{ \frac{(1+(1-q_{n})\alpha)^{2}}{([n]_{q_{n}}+\beta)^{2}}+(1-q_{n})^{2} \sum_{k=0}^{n}p_{n,k}(q_{n};x) \biggl(\frac{[k]+\alpha}{[n]_{q_{n}}+\beta} \biggr)^{2} \Biggr\} \\ &\quad\leq\frac{2}{([n]_{q_{n}}+\beta)^{2}} \biggl\{ \frac{(1+(1-q_{n})\alpha)^{2}}{([n]_{q_{n}}+\beta)^{2}}+(1-q_{n})^{2} B_{n,q_{n}}^{(\alpha,\beta)}\bigl(t^{2},x\bigr) \biggr\} \\ &\quad= \frac{2}{([n]_{q_{n}} + \beta)^{4}} \bigl\{ \bigl(1 + (1 - q_{n})\alpha \bigr)^{2} + (1 - q_{n})^{2} \bigl[[n]_{q_{n}}^{2}x^{2} + [n]_{q_{n}}x(1 - x) + 2\alpha[n]_{q_{n}}x + \alpha ^{2} \bigr] \bigr\} \\ &\quad= O \biggl( \frac{1}{[n]_{q_{n}}^{2}} \biggr). \end{aligned}$$
$$ \tilde{S}_{n,q_{n}}^{(\alpha,\beta)}\bigl((t-x)^{4},x\bigr)=O \biggl( \frac{1}{[n]_{q_{n}}^{2}} \biggr). $$
(2.4)
In view of the Lemma 2.1 and the relation (2.4), we have Choosing \(\delta=[n]_{q_{n}}^{-1/2}\), we get Hence In view of Lemma 2.1, we obtain □
$$\begin{aligned} \bigl\vert [n]_{q_{n}}\tilde{S}_{n,q_{n}}^{(\alpha,\beta)} \bigl((t-x)^{2}\bigl(f^{\prime \prime}(\xi)-f^{\prime\prime}(x)\bigr),x \bigr)\bigr\vert &\leq\omega_{f^{\prime\prime}}(\delta) \biggl(O(1)+ \frac{1}{\delta ^{2}}O \biggl(\frac{1}{[n]_{q_{n}}} \biggr) \biggr). \end{aligned}$$
$$ \bigl\vert [n]_{q_{n}}\tilde{S}_{n,q_{n}}^{(\alpha,\beta)} \bigl((t-x)^{2}\bigl(f^{\prime \prime}(\xi)-f^{\prime\prime}(x)\bigr),x \bigr) \bigr\vert =\omega_{f^{\prime \prime}} \bigl([n]_{q_{n}}^{-1/2} \bigr) O(1). $$
$$ \lim_{n\to\infty}\bigl\vert [n]_{q_{n}}\tilde{S}_{n,q_{n}}^{(\alpha,\beta)} \bigl((t-x)^{2}\bigl(f^{\prime\prime}(\xi)-f^{\prime\prime}(x)\bigr),x \bigr) \bigr\vert =0. $$
$$ \lim_{n\to\infty}[n]_{q_{n}}\bigl(\tilde{S}_{n,q_{n}}^{(\alpha,\beta)}(f,x)-f(x) \bigr)= \biggl(-\frac{1+a+2\beta}{2}x+\alpha+\frac{1}{2} \biggr) f^{\prime}(x)+\frac{1}{2} \bigl(x(1-x) \bigr)f^{\prime\prime}(x). $$
3 Approximation properties of q-Stancu-Kantorovich operators
Recall that the first and second modulus of continuity of \(f\in C[0,1]\) are defined, respectively, by and Let us consider the following K-functional: It is well known (see [22]) that there exists an absolute constant \(C>0\) such that Denote \(a_{n}= \frac{2q}{[2]_{q}}\frac{[n]_{q}}{[n+1]_{q}+\beta}\), \(b_{n}= \frac{\alpha}{[n+1]_{q}+\beta}+\frac{1}{[2]_{q}([n+1]_{q}+\beta)}\), \(\delta_{n}(x)=x(1-x)+ \frac{1}{[n]_{q}}\), \(\nu(\alpha,\beta,x)=(2\alpha^{2}+2\beta^{2}+2\alpha+4) \frac{[n+1]_{q}^{2}}{([n+1]_{q}+\beta)^{2}} \frac{1}{[n]_{q}}\delta_{n}(x)\), \(\tilde{\nu}(\alpha,\beta,x)=(2\alpha^{2}+\beta^{2}+4\alpha+4) \frac{[n]_{q}}{([n]_{q}+\beta)^{2}}\delta_{n}(x)\).
$$ \omega(f,\delta):= \sup_{0< h< \delta; x,x+h\in[0,1]} \bigl|f(x+h)-f(x)\bigr| $$
$$ \omega_{2}(f,\delta):= \sup_{0< h< \delta; x,x+2h\in[0,1]} \bigl|f(x+2h)-2f(x+h)+f(x)\bigr|, \quad\mbox{where } \delta>0. $$
$$ K_{2}(f,\delta):=\inf \bigl\{ \|f-g\|+\delta\bigl\| g^{\prime\prime}\bigr\| , g \in C^{2}[0,1] \bigr\} , \quad\mbox{where } \delta\geq0. $$
$$ K_{2}(f,\delta)\leq C\omega_{2}(f,\sqrt{\delta}). $$
(3.1)
Theorem 3.1
There exists an absolute constant
\(C>0\)
such that
where
\(f\in C[0,1]\)
and
\(0< q<1\).
$$ \bigl\vert S_{n,q}^{*(\alpha,\beta)} (f,x )-f(x)\bigr\vert \leq C \omega_{2} \bigl(f,\sqrt{\nu(\alpha,\beta,x)} \bigr) +\omega \bigl(f,\bigl|(a_{n}-1)x+b_{n}\bigr|\bigr), $$
Proof
Let
$$ {T}_{n,q}^{*(\alpha,\beta)}(f,x)=S_{n,q}^{*(\alpha,\beta)}(f,x)+f(x)-f(a_{n}x+b_{n}), \quad\mbox{where } f\in C[0,1]. $$
(3.2)
Using the Taylor formula it follows that Therefore Using the above relation we obtain and using (3.1) the theorem is proved. □
$$ g(t)=g(x)+g^{\prime}(x) (t-x)+ \int_{x}^{t}(t-s)g^{\prime \prime}(s) \,ds, \quad g\in C^{2}[0,1], $$
(3.3)
$$\begin{aligned} &{T}_{n,q}^{*(\alpha,\beta)}(g,x)=g(x)+S_{n,q}^{*(\alpha,\beta)} \biggl(\int_{x}^{t}(t-s)g^{\prime\prime}(s) \,ds,x \biggr)- \int_{x}^{a_{n}x+b_{n}} (a_{n}x+b_{n}-s)g^{\prime\prime}(s)\,ds,\\ &\quad g\in C^{2}[0,1]. \end{aligned}$$
$$\begin{aligned} &\bigl\vert {T}_{n,q}^{*(\alpha,\beta)}(g;x)-g(x)\bigr\vert \\ &\quad\leq S_{n,q}^{*(\alpha,\beta)} \biggl(\biggl\vert \int _{x}^{t}(t-s)g^{\prime \prime}(s)\,ds\biggr\vert ,x \biggr)+ \biggl| \int_{x}^{a_{n}x+b_{n}}|a_{n}x+b_{n}-s|\bigl|g^{\prime\prime}(s)\bigr| \,ds \biggr| \\ &\quad \leq\bigl\| g^{\prime\prime}\bigr\| S_{n,q}^{*(\alpha,\beta)} \bigl((t-x)^{2},x \bigr)+\bigl\| g^{\prime\prime}\bigr\| (a_{n}x+b_{n}-x)^{2} \\ &\quad\leq\bigl\| g^{\prime\prime}\bigr\| \frac{2[n+1]_{q}^{2}}{([n+1]_{q}+\beta)^{2}} \biggl\{ \frac{4}{[n]_{q}} \biggl(x(1-x)+\frac{1}{[n]_{q}} \biggr) + \biggl( \frac{\alpha}{[n+1]_{q}}-\frac{\beta}{[n+1]_{q}}x \biggr)^{2} \biggr\} \\ &\qquad{} + \bigl\| g^{\prime\prime}\bigr\| \biggl( \frac{2q}{[2]_{q}}\frac{[n]_{q}}{[n+1]_{q}+\beta}x+ \frac{\alpha}{[n+1]_{q}+\beta} +\frac {1}{[2]_{q}([n+1]_{q}+\beta)}-x \biggr)^{2} \\ &\quad\leq\frac{[n+1]_{q}^{2}}{([n+1]_{q}+\beta)^{2}}\bigl\| g^{\prime\prime}\bigr\| \biggl\{ \frac{8}{[n]_{q}} \biggl(x(1-x)+\frac{1}{[n]_{q}} \biggr)+\frac{4\alpha^{2}}{[n]_{q}^{2}}+ \frac{4\beta ^{2}}{[n]_{q}^{2}}x^{2} \\ &\qquad{} + \frac{1}{[n]_{q}^{2}} \biggl[2 \biggl( \frac{2q}{[2]_{q}}[n]_{q}-[n+1]_{q} \biggr)^{2}x^{2}+2 \biggl(\alpha+\frac{1}{[2]_{q}}-\beta x \biggr)^{2} \biggr] \biggr\} \\ &\quad\leq\frac{[n+1]_{q}^{2}}{([n+1]_{q}+\beta)^{2}}\frac{1}{[n]_{q}}\bigl\| g^{\prime\prime}\bigr\| \biggl\{ 8 \biggl(x(1-x)+\frac{1}{[n]_{q}} \biggr) \\ &\qquad{} + \frac{4\alpha^{2}}{[n]_{q}}+\frac{4\beta^{2}}{[n]_{q}}x^{2}+ \frac {2}{[n]_{q}} \biggl(\frac{1+q^{n+1}}{1+q} \biggr)^{2}x^{2}+ \frac{4}{[n]_{q}} \biggl(\alpha +\frac{1}{[2]_{q}} \biggr)^{2} + \frac{4}{[n]_{q}}\beta^{2}x^{2} \biggr\} \\ &\quad\leq\frac{[n+1]_{q}^{2}}{([n+1]_{q}+\beta)^{2}}\frac{1}{[n]_{q}}\bigl\| g^{\prime\prime}\bigr\| \biggl\{ 8 \biggl(x(1-x)+\frac{1}{[n]_{q}} \biggr)+\frac{4\alpha^{2}}{[n]_{q}} + \frac{4\beta^{2}}{[n]_{q}}x^{2} \\ &\qquad{} +\frac{2}{[n]_{q}}x^{2}+\frac{4}{[n]_{q}}( \alpha+1)^{2}+\frac {4}{[n]_{q}}\beta^{2}x^{2} \biggr\} \\ &\quad\leq4\nu(\alpha,\beta,x)\bigl\| g^{\prime\prime}\bigr\| . \end{aligned}$$
$$\begin{aligned} &\bigl\vert S_{n,q}^{*(\alpha,\beta)}(f;x) - f(x)\bigr\vert \\ &\quad\leq\bigl\vert {T}_{n,q}^{*(\alpha,\beta)}(f - g;x)\bigr\vert + \bigl\vert {T}_{n,q}^{*(\alpha,\beta)}(g;x) - g(x)\bigr\vert + \bigl|f(x) - g(x)\bigr| + \bigl|f(a_{n}x + b_{n}) - f(x)\bigr| \\ &\quad\leq4\| f-g\|+4\nu(\alpha,\beta,x)\bigl\| g^{\prime\prime}\bigr\| +\omega \bigl(f,\bigl|(a_{n}-1)x+b_{n}\bigr| \bigr) \\ &\quad\leq4K_{2} \bigl(f, {\nu}(\alpha,\beta,x) \bigr)+\omega \bigl(f,\bigl|(a_{n}-1)x+b_{n}\bigr| \bigr) \end{aligned}$$
Theorem 3.2
There exists an absolute constant
\(C>0\)
such that
where
\(f\in C[0,1]\)
and
\(0< q<1\).
$$ \bigl\vert \tilde{ S}_{n,q}^{(\alpha,\beta)} (f,x )-f(x)\bigr\vert \leq C\omega_{2} \bigl(f,\sqrt{\tilde{\nu}(\alpha,\beta,x)} \bigr) + \omega\bigl(f,\bigl|(a_{n}-1)x+b_{n}\bigr|\bigr), $$
Proof
We have Let \(B_{n,k}^{(\alpha,\beta)}\) be q-Bernstein-Stancu operators. From [14], Lemma 1 and Lemma 4 the following properties of the q-Bernstein-Stancu operators hold: Therefore
$$\begin{aligned} &\tilde{S}_{n,q}^{(\alpha,\beta)} \bigl((t-x)^{2},x \bigr) \\ &\quad= \bigl([n+1]_{q}+\beta\bigr) \sum _{k=0}^{n} q^{-k}p_{n,k}(q;x) \int _{\frac{[k]_{q}+\alpha}{[n+1]_{q}+\beta}}^{\frac{[k+1]_{q}+\alpha}{[n+1]_{q}+\beta}}(t-x)^{2}d_{q}^{R}t \\ &\quad=\bigl([n+1]_{q}+\beta\bigr) \sum_{k=0}^{n}q^{-k}p_{n,k}(q;x) (1-q)\frac{[k+1]_{q}-[k]_{q}}{[n+1]_{q}+\beta} \\ &\qquad{} \times\sum_{j=0}^{\infty} \biggl( \frac{[k]_{q}+\alpha}{[n+1]_{q}+\beta}+ \frac{[k+1]_{q}-[k]_{q}}{[n+1]_{q}+\beta} q^{j}-x \biggr)^{2} q^{j} \\ &\quad=(1-q) \sum_{k=0}^{n}p_{n,k}(q;x) \sum_{j=0}^{\infty} \biggl( \frac{[k]_{q}+\alpha}{[n+1]_{q}+\beta}+ \frac{q^{k}}{[n+1]_{q}+\beta} q^{j}-x \biggr)^{2}q^{j} \\ &\quad\leq2(1-q) \sum_{k=0}^{n}p_{n,k}(q;x) \sum_{j=0}^{\infty} \biggl(\frac{[k]_{q}+\alpha}{[n+1]_{q}+\beta}-x \biggr)^{2}q^{j} \\ &\qquad{} +2(1-q) \sum_{k=0}^{n}p_{n,k}(q;x) \sum_{j=0}^{\infty} \biggl(\frac{q^{k}}{[n+1]_{q}+\beta} \biggr)^{2}q^{3j} \\ &\quad=2 \sum_{k=0}^{n}p_{n,k}(q;x) \biggl( \frac{[k]_{q}+\alpha }{[n+1]_{q}+\beta}-x \biggr)^{2} +\frac{2}{1+q+q^{2}} \sum_{k=0}^{n}p_{n,k}(q;x) \biggl(\frac{q^{k}}{[n+1]_{q}+\beta} \biggr)^{2} \\ &\quad\leq2 \sum_{k=0}^{n}p_{n,k}(q;x) \biggl( \frac{[k]_{q}+\alpha}{[n]_{q}+\beta}-x +\frac{[k]_{q}+\alpha}{[n+1]_{q}+\beta}-\frac{[k]_{q}+\alpha}{[n]_{q}+\beta} \biggr)^{2} \\ &\qquad{} +2 \sum_{k=0}^{n}p_{n,k}(q;x) \frac{1}{([n+1]_{q}+\beta)^{2}}. \end{aligned}$$
$$\begin{aligned} &B_{n,q}^{(\alpha,\beta)}\bigl(t^{2},x\bigr)= \frac{1}{([n]_{q}+\beta)^{2}} \bigl([n]_{q}^{2}x^{2}+[n]_{q}x(1-x)+2 \alpha[n]_{q}x+\alpha^{2} \bigr), \\ &B_{n,q}^{(\alpha,\beta)}\bigl((t-x)^{2},x\bigr)\leq \frac{[n]_{q}}{([n]_{q}+\beta)^{2}}x(1-x). \end{aligned}$$
$$\begin{aligned} &\tilde{S}_{n,q}^{(\alpha,\beta)}\bigl((t - x)^{2};x\bigr) \\ &\quad\leq 4B_{n,q}^{(\alpha,\beta)}\bigl((t - x)^{2};x\bigr) + 4 \sum_{k=0}^{n}p_{n,k}(q;x) \biggl( \frac{[k]_{q} + \alpha}{[n]_{q} + \beta} \biggr)^{2} \biggl( \frac{q^{n}}{[n + 1]_{q} + \beta} \biggr)^{2} \\ &\qquad{}+ \frac{2}{([n + 1]_{q} + \beta)^{2}} \\ &\quad\leq4 \frac{[n]_{q}}{([n]_{q}+\beta)^{2}}x(1-x)+ \frac{4}{([n+1]_{q}+\beta)^{2}}B_{n,q}^{(\alpha,\beta)} \bigl(t^{2};x\bigr)+ \frac{2}{([n+1]_{q}+\beta)^{2}} \\ &\quad= \frac{[n]_{q}}{([n]_{q}+\beta)^{2}} \biggl\{ 4x(1-x)+\frac{4}{[n]_{q}}B_{n,q}^{(\alpha,\beta)} \bigl(t^{2};x\bigr)+ \frac{2}{[n]_{q}} \biggr\} \\ & \quad= \frac{[n]_{q}}{([n]_{q} + \beta)^{2}} \biggl\{ 4x(1 - x) + \frac{4}{[n]_{q}} \cdot \frac{1}{([n]_{q} + \beta)^{2}} \bigl([n]_{q}^{2}x^{2} + [n]_{q}x(1 - x) \\ &\qquad{}+ \alpha^{2} + 2\alpha[n]_{q}x \bigr) + \frac{2}{[n]_{q}} \biggr\} \\ &\quad\leq\frac{[n]_{q}}{([n]_{q}+\beta)^{2}} \biggl\{ 4x(1-x)+ \frac{4}{[n]_{q}}x^{2}+ \frac{4}{[n]_{q}}x(1-x)+ \frac{4\alpha^{2}}{[n]_{q}}+\frac{8\alpha}{[n]_{q}}x+ \frac{2}{[n]_{q}} \biggr\} \\ &\quad= \frac{[n]_{q}}{([n]_{q}+\beta)^{2}} \biggl\{ 4\delta_{n}(x)+\frac{1}{[n]_{q}} \bigl(4x^{2}+4\alpha^{2}+8\alpha x+2 \bigr) \biggr\} \\ &\quad\leq\frac{[n]_{q}}{([n]_{q}+\beta)^{2}}\delta_{n}(x) \bigl(4 \alpha^{2}+8\alpha+10\bigr). \end{aligned}$$
Let
$$ \tilde{T}_{n,q}^{(\alpha,\beta)}(f,x)=\tilde{S}_{n,q}^{(\alpha,\beta)}(f,x)+f(x)-f(a_{n}x+b_{n}), \quad\mbox{where } f\in C[0,1]. $$
(3.4)
Using the Taylor formula (3.3) it follows that Therefore Using the above relation we obtain and using (3.1) the theorem is proved. □
$$\begin{aligned} &\tilde{T}_{n,q}^{(\alpha,\beta)}(g,x)=g(x)+\tilde{S}_{n,q}^{(\alpha ,\beta)} \biggl(\int_{x}^{t}(t-s)g^{\prime\prime}(s)\,ds,x \biggr)- \int_{x}^{a_{n}x+b_{n}} (a_{n}x+b_{n}-s)g^{\prime\prime}(s)\,ds,\\ &\quad g\in C^{2}[0,1]. \end{aligned}$$
$$\begin{aligned} &\bigl\vert \tilde{T}_{n,q}^{(\alpha,\beta)}(g;x)-g(x)\bigr\vert \\ &\quad\leq\tilde{S}_{n,q}^{(\alpha,\beta)} \biggl(\biggl\vert \int_{x}^{t}(t-s)g^{\prime \prime}(s)\,ds \biggr\vert ,x \biggr)+ \biggl| \int_{x}^{a_{n}x+b_{n}}|a_{n}x+b_{n}-s|\bigl|g^{\prime\prime}(s)\bigr| \,ds \biggr| \\ &\quad \leq\bigl\| g^{\prime\prime}\bigr\| \tilde{S}_{n,q}^{(\alpha,\beta)} \bigl((t-x)^{2},x \bigr)+\bigl\| g^{\prime\prime} \bigr\| (a_{n}x+b_{n}-x)^{2} \\ &\quad\leq\frac{[n]_{q}}{([n]_{q}+\beta)^{2}}\delta_{n}(x) \bigl(4 \alpha^{2}+8\alpha+10\bigr)\bigl\| g^{\prime\prime}\bigr\| \\ &\qquad{} +\bigl\| g^{\prime\prime}\bigr\| \biggl( \frac{2q}{[2]_{q}} \frac{[n]_{q}}{[n+1]_{q}+\beta}x+\frac{\alpha}{[n+1]_{q}+\beta}+\frac {1}{[2]_{q}([n+1]_{q}+\beta)}-x \biggr)^{2} \\ &\quad\leq\frac{[n]_{q}}{([n]_{q}+\beta)^{2}}\delta_{n}(x) \bigl(4 \alpha^{2}+8\alpha+10\bigr)\bigl\| g^{\prime\prime}\bigr\| \\ &\qquad{} + \bigl\| g^{\prime\prime}\bigr\| \frac{2}{([n]_{q}+\beta)^{2}} \biggl[ \biggl( \frac{1+q^{n+1}}{1+q} \biggr)^{2}x^{2} +2 \biggl( \alpha+\frac{1}{[2]_{q}} \biggr)^{2}+2\beta^{2}x^{2} \biggr] \\ &\quad\leq\frac{[n]_{q}}{([n]_{q} + \beta)^{2}}\delta_{n}(x) \bigl(4\alpha^{2} + 8\alpha + 10\bigr)\bigl\| g^{\prime\prime}\bigr\| + \bigl\| g^{\prime\prime}\bigr\| \frac{2}{([n]_{q} + \beta)^{2}} \bigl[x^{2} + 2(\alpha + 1)^{2} + 2 \beta^{2}x^{2} \bigr] \\ &\quad\leq\frac{[n]_{q}}{([n]_{q}+\beta)^{2}}\delta_{n}(x) \bigl(8 \alpha^{2}+4\beta^{2}+16\alpha+16\bigr) \bigl\| g^{\prime\prime}\bigr\| =4\tilde{\nu}(\alpha,\beta,x) \bigl\| g^{\prime\prime}\bigr\| . \end{aligned}$$
$$\begin{aligned} &\bigl\vert \tilde{S}_{n,q}^{(\alpha,\beta)}(f;x) - f(x)\bigr\vert \\ &\quad\leq\bigl\vert \tilde{T}_{n,q}^{(\alpha,\beta)}(f - g;x)\bigr\vert + \bigl\vert \tilde{T}_{n,q}^{(\alpha,\beta)}(g;x) - g(x)\bigr\vert + \bigl|f(x) - g(x)\bigr| + \bigl|f(a_{n}x + b_{n}) - f(x)\bigr| \\ &\quad\leq4\| f-g\|+4\tilde{\nu}(\alpha,\beta,x)\bigl\| g^{\prime\prime}\bigr\| + \omega \bigl(f,\bigl|(a_{n}-1)x+b_{n}\bigr| \bigr) \\ &\quad\leq4K_{2} \bigl(f,\tilde{\nu}(\alpha,\beta,x) \bigr)+ \omega \bigl(f,\bigl|(a_{n}-1)x+b_{n}\bigr| \bigr). \end{aligned}$$
In order to start the next result we need the second order Ditzian-Totik modulus [22] defined by in which \(\Phi:[0,1]\to\mathbb{R}\) is an admissible step-weight function.
$$ \omega_{2,\Phi}(f,\delta):= \sup_{0< h\leq\delta}\sup _{x\pm h\Phi(x)\in[0,1]}\bigl|f\bigl(x-\Phi(x)h\bigr)-2f(x)+f(x+\Phi(x)h\bigr|,\quad f\in C[0,1], $$
The weighted K-functional of second order for \(f\in C[0,1]\) is defined by where and It is well known that the K-functional \(K_{2,\Phi}(f,\delta)\) and the Ditzian-Totik modulus \(\omega_{2,\Phi}(f,\sqrt{\delta})\) are equivalent (see [22]).
$$ K_{2,\Phi}(f,\delta):= \inf\bigl\{ \| f-g\|+\delta\bigl\| \Phi^{2} g^{\prime\prime}\bigr\| , g\in W^{2}(\Phi) \bigr\} ,\quad \delta\geq0, $$
$$ W^{2}(\Phi):=\bigl\{ g\in C[0,1]\mid g^{\prime}\in AC[0,1], \Phi^{2}g^{\prime\prime }\in C[0,1]\bigr\} $$
$$ AC[0,1]:=\bigl\{ h\mid h \mbox{ is absolutely continuous in } [a,b], \mbox{ for every } 0< a< b< 1\bigr\} . $$
Denote
$$ \overrightarrow{\omega}_{\psi}(f,\delta)= \sup_{|h|\leq \delta} \sup_{x,x+h\psi(x)\in[0,1]}\bigl|f\bigl(x+\psi(x)h\bigr)-f(x)\bigr|. $$
Theorem 3.3
Let Φ be an admissible step-weight function of the Ditzian-Totik modulus of smoothness such that
\(\Phi^{2}\)
is concave and
\(\Phi\ne0\). Then there exists an absolute constant
\(C>0\)
such that
where
\(f\in C[0,1]\), \(0< q<1\), and
\(\psi(x)=(2+[2]_{q}\beta)x+[2]_{q}\alpha +1\), \(x\in[0,1]\).
$$ \bigl\vert S_{n,q}^{*(\alpha,\beta)}(f;x)-f(x)\bigr\vert \leq C \omega_{2,\Phi} \biggl(f,\frac{\sqrt{\nu(\alpha,\beta,x)}}{\Phi(x)} \biggr) +\overrightarrow{\omega}_{\psi} \biggl(f, \frac{1}{[2]_{q}([n+1]_{q}+\beta)} \biggr), $$
Proof
Applying the operators \({T}_{n,q}^{*(\alpha,\beta)}\) defined in (3.2) to Taylor’s formula in a similar way to the proof of Theorem 3.1 we obtain Let \(s=\tau x+(1-\tau)t\), \(\tau\in[0,1]\). Since \(\Phi^{2}\) is concave on \([0,1]\) it follows that and Therefore Using the above relation we obtain Also, we have Therefore Using the equivalence of the K-functional and the Ditzian-Totik modulus we get the desired estimate. □
$$\begin{aligned} &\bigl\vert {T}_{n,q}^{*(\alpha,\beta)}(g,x)-g(x) \bigr\vert \\ &\quad\leq{S}_{n,q}^{*(\alpha,\beta)} \biggl(\biggl\vert \int _{x}^{t}(t-s)g^{\prime\prime}(s)\,ds \biggr\vert ,x \biggr)+ \biggl| \int_{x}^{a_{n}x+b_{n}}|a_{n}x+b_{n}-s| \bigl|g^{\prime\prime}(s)\bigr| \,ds \biggr| \\ &\quad\leq\bigl\| \Phi^{2}g^{\prime\prime}\bigr\| {S}_{n,q}^{*(\alpha,\beta)} \biggl(\biggl\vert \int_{x}^{t} \frac{|t-s|}{\Phi^{2}(s)}\,ds\biggr\vert ,x \biggr) +\bigl\| \Phi^{2}g^{\prime\prime} \bigr\| \biggl| \int_{x}^{a_{n}x+b_{n}}\frac{|a_{n}x+b_{n}-s|}{\Phi^{2}(s)} \,ds \biggr|. \end{aligned}$$
$$ \Phi^{2}(s)\geq\tau\Phi^{2}(x)+(1-\tau)\Phi^{2}(t) $$
$$ \frac{|t-s|}{\Phi^{2}(s)}=\frac{\tau|x-t|}{\Phi^{2}(s)}\leq\frac{\tau|x-t|}{\tau\Phi^{2}(x)+(1-\tau)\Phi^{2}(t)}\leq \frac{|x-t|}{\Phi^{2}(x)}. $$
$$\begin{aligned} \bigl\vert {T}_{n,q}^{*(\alpha,\beta)}(g,x)-g(x)\bigr\vert &\leq \frac{\|\Phi^{2} g^{\prime\prime}\|}{\Phi^{2}(x)} \bigl[S_{n,q}^{*(\alpha,\beta)} \bigl((t-x)^{2};x \bigr)+(a_{n}x+b_{n}-x)^{2} \bigr]\\ &\leq4 \frac{\|\Phi^{2} g^{\prime\prime}\|}{\Phi^{2}(x)}\nu_{n}(\alpha,\beta,x). \end{aligned}$$
$$\begin{aligned} &\bigl\vert S_{n,q}^{*(\alpha,\beta)}(f;x)-f(x) \bigr\vert \\ &\quad\leq\bigl\vert {T}_{n,q}^{*(\alpha,\beta)}(f - g;x)\bigr\vert + \bigl\vert {T}_{n,q}^{*(\alpha,\beta)}(g;x) - g(x)\bigr\vert + \bigl|f(x) - g(x)\bigr| + \bigl|f(a_{n}x + b_{n}) - f(x)\bigr| \\ &\quad\leq4\| f-g\|+4\frac{\|\Phi^{2}g^{\prime\prime}\|}{\Phi^{2}(x)}\nu_{n}(\alpha, \beta,x)+\bigl|f(a_{n}x + b_{n}) - f(x)\bigr| \\ &\quad=4K_{2,\Phi} \biggl(f,\frac{\nu(\alpha,\beta,x)}{\Phi^{2}(x)} \biggr)+\bigl|f(a_{n}x+b_{n})-f(x)\bigr|. \end{aligned}$$
$$\begin{aligned} \bigl|f(a_{n}x+b_{n})-f(x)\bigr|&=\biggl\vert f \biggl(x+\psi(x) \frac{(a_{n}-1)x+b_{n}}{\psi(x)} \biggr)-f(x) \biggr\vert \\ &\leq \sup \biggl|f \biggl(x+\psi(x)\frac{S_{n}^{*(\alpha,\beta)} ((t-x),x )}{\psi(x)} \biggr)-f(x) \biggr| \\ &\leq\overrightarrow{\omega}_{\psi} \biggl(f, \frac{1}{\psi(x)}\biggl\vert \frac{-1-q^{n+1}-[2]_{q}\beta}{[2]_{q} ([n+1]_{q}+\beta )}x +\frac{[2]_{q}\alpha+1}{[2]_{q}([n+1]_{q}+\beta)}\biggr\vert \biggr) \\ &\leq\overrightarrow{\omega}_{\psi} \biggl(f,\frac{1}{[2]_{q} ([n+1]_{q}+\beta )} \biggr). \end{aligned}$$
$$ \bigl\vert S_{n,q}^{*(\alpha,\beta)}(f,x)-f(x) \bigr\vert \leq4K_{2,\Phi} \biggl(f, \frac{\nu(\alpha,\beta,x)}{\Phi^{2}(x)} \biggr) + \overrightarrow{ \omega}_{\psi} \biggl(f,\frac{1}{[2]_{q} ([n+1]_{q}+\beta )} \biggr). $$
In a similar way can be obtained the following result for the q-Stancu-Kantorovich operators \(\tilde{S}_{n,q}^{(\alpha,\beta)}\).
Theorem 3.4
Let Φ be an admissible step-weight function of the Ditzian-Totik modulus of smoothness such that
\(\Phi^{2}\)
is concave and
\(\Phi\ne0\). Then there exists an absolute constant
\(C>0\)
such that
where
\(f\in C[0,1]\), \(0< q<1\), and
\(\psi(x)=(2+[2]_{q}\beta)x+[2]_{q}\alpha +1\), \(x\in[0,1]\).
$$ \bigl\vert \tilde{S}_{n,q}^{(\alpha,\beta)}(f;x)-f(x)\bigr\vert \leq C\omega_{2,\Phi} \biggl(f,\frac{\sqrt{\tilde{\nu}(\alpha,\beta,x)}}{\Phi (x)} \biggr) + \overrightarrow{\omega}_{\psi} \biggl(f, \frac{1}{[2]_{q}([n+1]_{q}+\beta)} \biggr), $$
4 Statistical approximation of Korovkin type
The concept of statistical convergence was introduced by Fast [23] and Steinhaus [24] and recently has become an important area in approximation theory. The goal of this section is to obtain the statistical convergence properties of the Stancu-Kantorovich operators (1.3) and (1.5).
Let set \(K\subseteq N\) and \(K_{n}=\{k\leq n;k\in K\}\), the natural density of K is defined by \(\delta(K):= \lim_{n\to\infty}\frac{1}{n}|K_{n}|\) if the limit exists, where \(|K_{n}|\) denote the cardinality of the set \(K_{n}\).
A sequence \(x=\{{x_{n}}\}\) is called statistically convergent to a number L, if for every \(\epsilon>0\), \(\delta\{k\in N: |x_{k}-L|\geq\epsilon\}=0\). This convergence is denoted as \(\mbox{st-}\!\lim_{k}x_{k}=L\).
In [25] Gadjiev and Orhan proved the following Bohman-Korovkin-type approximation theorem for statistical convergence.
Theorem 4.1
([25])
If the sequence of positive linear operators
\(A_{n}:C[a,b]\to C[a,b] \)
satisfies the conditions
\(\textit{st-}\!\lim_{n}\|A_{n}(e_{i})-e_{i}\|_{C[a,b]}=0\)
with
\(e_{i}(t)=t^{i}\), \(i=0,1,2\), then for any function
\(f\in C[a,b]\)
we have
$$ \textit{st-}\!\lim_{n}\bigl\| A_{n}(f)-f \bigr\| _{C[a,b]}=0. $$
Theorem 4.2
Let
\((q_{n})\), \(0< q_{n}<1\)
be a sequence that satisfies
\(\textit{st-}\!\lim_{n}q_{n}=1\), \(\textit{st-}\!\lim_{n}q_{n}^{n}=a\in(0,1)\). Then for all
\(f\in C[0,1]\)
we have
$$ \textit{st-}\!\lim_{n}\bigl\| S_{n,q_{n}}^{*(\alpha,\beta)}(f, \cdot)-f\bigr\| _{C[0,1]}=0. $$
Proof
It is necessary to prove \(\mbox{st-}\!\lim_{n}\|S_{n,q_{n}}^{*(\alpha,\beta)}(e_{i},\cdot)-e_{i}\|_{C[0,1]}=0\), for \(i=0,1,2\), and the proof follows from Theorem 4.1. For the first equality it is clear from Lemma 1.2 that For the second equality we have We denote \(\nu_{n}=\vert \frac{2q_{n}}{[2]_{q}}\cdot\frac{[n]_{q}}{[n+1]_{q}+\beta}-1\vert \) and \(\mu_{n}= \frac{\alpha}{[n+1]_{q}+\beta}+ \frac{1}{[2]_{q} ([n+1]_{q}+\beta )}\).
$$ \mbox{st-}\!\lim_{n}\bigl\| S_{n,q_{n}}^{*(\alpha,\beta)}(e_{0}, \cdot)-e_{0}\bigr\| _{C[0,1]}=0. $$
(4.1)
$$ \bigl\Vert S_{n,q_{n}}^{*(\alpha,\beta)}(e_{1}, \cdot)-e_{1}\bigr\Vert _{C[0,1]}\leq \biggl\vert \frac{2q_{n}}{[2]_{q}}\cdot\frac{[n]_{q}}{[n+1]_{q}+\beta}-1\biggr\vert +\biggl\vert \frac{\alpha}{[n+1]_{q}+\beta}+ \frac{1}{[2]_{q} ([n+1]_{q}+\beta )}\biggr\vert . $$
From \(\mbox{st-}\!\lim_{n}q_{n}=1\) and \(\mbox{st-}\!\lim_{n}q_{n}^{n}=a\in(0,1)\) we have Now, for a given \(\epsilon>0\), we define the following sets: It is obvious that \(A\subseteq A_{1}\cup A_{2}\), which implies that \(\delta(A)\leq \delta(A_{1})+\delta(A_{2})\). From (4.2) we find that the right hand side of the above inequality is zero and we get finally In a similar way it can be proved that From (4.1), (4.3), and (4.4), the statement of our theorem follows from the Korovkin-type statistical approximation theorem. □
$$ \mbox{st-}\!\lim_{n}\nu_{n}=\mbox{st-}\!\lim _{n}\mu_{n}=0. $$
(4.2)
$$\begin{aligned}& A:= \bigl\{ n\in\mathbb{N}\mid\bigl\| S_{n,q_{n}}^{*(\alpha,\beta)}(e_{1}, \cdot)-e_{1} \bigr\| _{C[0,1]}\geq\epsilon \bigr\} , \\& A_{1}:= \biggl\{ n\in\mathbb{N} \Bigm|\nu_{n}\geq \frac{\epsilon}{2} \biggr\} \quad\mbox{and}\quad A_{2}:= \biggl\{ n\in\mathbb{N}\Bigm|\mu_{n}\geq \frac{\epsilon}{2} \biggr\} . \end{aligned}$$
$$ \mbox{st-}\!\lim_{n}\bigl\| S_{n,q_{n}}^{*(\alpha,\beta)}(e_{1}, \cdot)-e_{1}\bigr\| _{C[0,1]}=0. $$
(4.3)
$$ \mbox{st-}\!\lim_{n}\bigl\| S_{n,q_{n}}^{*(\alpha,\beta)}(e_{2}, \cdot)-e_{2}\bigr\| _{C[0,1]}=0. $$
(4.4)
A statistical approximation property of the q-Kantorovich-Stancu operators \(\tilde{S}^{(\alpha,\beta)}\) is obtained in the following theorem.
Theorem 4.3
Let
\((q_{n})\), \(0< q_{n}<1\)
be a sequence that satisfies
\(\textit{st-}\!\lim_{n}q_{n}=1\), \(\textit{st-}\!\lim_{n}q_{n}^{n}=a\in(0,1)\). Then for all
\(f\in C[0,1]\)
we have
$$ \textit{st-}\!\lim_{n}\bigl\| \tilde{S}_{n,q_{n}}^{(\alpha,\beta)}(f, \cdot)-f\bigr\| _{C[0,1]}=0. $$
Acknowledgements
The authors would like to thank the editor and referees for useful comments and suggestions. This study was supported by research funds from Dong-A University.
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The authors declare that they have no competing interests.
Authors’ contributions
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