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# Convergence of the q-Stancu-Szász-Beta type operators

Journal of Inequalities and Applications20142014:354

https://doi.org/10.1186/1029-242X-2014-354

• Received: 2 April 2014
• Accepted: 27 August 2014
• Published:

## Abstract

In this paper, we study on q-Stancu-Szász-Beta type operators. We give these operators convergence properties and obtain a weighted approximation theorem in the interval $\left[0,\mathrm{\infty }\right)$.

MSC:41A25, 41A36.

## Keywords

• q-Stancu type operators
• Szász-Beta type operators
• weighted approximation

## 1 Introduction

In , Mahmudov constructed q-Szász operators and obtained rate of global convergence in the frame of weighted spaces and a Voronovskaja type theorem for these operators. In , Gupta and Mahmudov studied on the q-analog of the Szász-Beta type operators. In , Yüksel and Dinlemez gave a Voronovskaja type theorem for q-analog of a certain family Szász-Beta type operators. In , Govil and Gupta introduced the q-analog of certain Beta-Szász-Stancu operators. They estimated the moments and established direct results in terms of modulus of continuity and an asymptotic formula for the q-operators. In , interesting generalization about q-calculus were given. Our aims are to give approximation properties and a weighted approximation theorem for q-Stancu-Szász-Beta type operators. We use without further explanation the basic notations and formulas, from the theory of q-calculus as set out in . Let $A>0$ and f be a real valued continuous function defined on the interval $\left[0,\mathrm{\infty }\right)$. For $0, q-Stancu-Szász-Beta type operators are defined as
${B}_{n,q}^{\left(\alpha ,\beta \right)}\left(f,x\right)=\sum _{k=0}^{\mathrm{\infty }}{s}_{n,k}^{q}\left(x\right){\int }_{0}^{\mathrm{\infty }/A}{b}_{n,k}^{q}\left(t\right)f\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_{q}+\beta }\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}t,$
(1.1)
where
${s}_{n,k}^{q}\left(x\right)={\left({\left[n\right]}_{q}x\right)}^{k}\frac{{e}^{-{\left[n\right]}_{q}x}}{{\left[k\right]}_{q}!}$
and
${b}_{n,k}^{q}\left(x\right)=\frac{{q}^{{k}^{2}}{x}^{k}}{{B}_{q}\left(k+1,n\right){\left(1+x\right)}_{q}^{n+k+1}}.$

If we write $q=1$ and $\alpha =\beta =0$ in (1.1), then the operators ${B}_{n,q}^{\left(\alpha ,\beta \right)}\left(f,x\right)$ are reduced to Szász-Beta type operators studied in .

## 2 Auxiliary results

For the sake of brevity, the notation ${F}_{s}^{q}\left(n\right)={\prod }_{i=1}^{s}{\left[n-i\right]}_{q}$ and ${G}_{\beta }^{q}\left(n\right)=\left({\left[n\right]}_{q}+\beta \right)$ will be used throughout the article. Now we are ready to give the following lemma for the Korovkin test functions.

Lemma 1 Let ${e}_{m}\left(t\right)={t}^{m}$, $m=0,1,2$, we get
$\begin{array}{rc}\text{(i)}& {B}_{n,q}^{\left(\alpha ,\beta \right)}\left({e}_{0},x\right)=1,\\ \text{(ii)}& {B}_{n,q}^{\left(\alpha ,\beta \right)}\left({e}_{1},x\right)=\frac{{\left[n\right]}_{q}^{2}x}{{q}^{2}{G}_{\beta }^{q}\left(n\right){F}_{1}^{q}\left(n\right)}+\frac{{\left[n\right]}_{q}}{q{G}_{\beta }^{q}\left(n\right){F}_{1}^{q}\left(n\right)}+\frac{\alpha }{{G}_{\beta }^{q}\left(n\right)},\\ \text{(iii)}& {B}_{n,q}^{\left(\alpha ,\beta \right)}\left({e}_{2},x\right)=\frac{{\left[n\right]}_{q}^{4}{x}^{2}}{{q}^{6}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{2}^{q}\left(n\right)}+\left\{\frac{{\left[n\right]}_{q}^{3}}{{q}^{5}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{2}^{q}\left(n\right)}\\ \phantom{{B}_{n,q}^{\left(\alpha ,\beta \right)}\left({e}_{2},x\right)=}+\frac{\left(1+{\left[2\right]}_{q}\right){\left[n\right]}_{q}^{3}}{{q}^{4}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{2}^{q}\left(n\right)}+\frac{2\alpha {\left[n\right]}_{q}^{2}}{{q}^{2}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{1}^{q}\left(n\right)}\right\}x\\ \phantom{{B}_{n,q}^{\left(\alpha ,\beta \right)}\left({e}_{2},x\right)=}+\frac{{\left[2\right]}_{q}{\left[n\right]}_{q}^{2}}{{q}^{3}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{2}^{q}\left(n\right)}+\frac{2\alpha {\left[n\right]}_{q}}{q{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{1}^{q}\left(n\right)}+\frac{{\alpha }^{2}}{{G}_{\beta }^{q}{\left(n\right)}^{2}}.\end{array}$
Proof Using the q-Gamma and q-Beta functions in [15, 24], we obtain the following equality:
$\begin{array}{c}{q}^{{k}^{2}}{\int }_{0}^{\mathrm{\infty }/A}\frac{1}{B\left(k+1,n\right)}\frac{{t}^{k+m}}{{\left(1+t\right)}_{q}^{n+k+1}}\phantom{\rule{0.2em}{0ex}}{d}_{q}t\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{{\left[m+k\right]}_{q}!{\left[n-m-1\right]}_{q}!{q}^{\left\{2{k}^{2}-\left(k+m\right)\left(k+m+1\right)\right\}/2}}{{\left[k\right]}_{q}!{\left[n-1\right]}_{q}!}.\hfill \end{array}$
(2.1)
Then, using (2.1), for $m=0$, we get
$\begin{array}{rcl}{B}_{n,q}^{\left(\alpha ,\beta \right)}\left({e}_{0},x\right)& =& {e}^{-{\left[n\right]}_{q}x}\sum _{k=0}^{\mathrm{\infty }}\frac{{\left({\left[n\right]}_{q}x\right)}^{k}}{{\left[k\right]}_{q}!}{q}^{k\left(k-1\right)/2}\\ =& {e}^{-{\left[n\right]}_{q}x}{E}_{q}^{{\left[n\right]}_{q}x}=1,\end{array}$
and the proof of (i) is finished. With a direct computation, we obtain (ii) as follows:
$\begin{array}{rcl}{B}_{n,q}^{\left(\alpha ,\beta \right)}\left({e}_{1},x\right)& =& \frac{{\left[n\right]}_{q}}{{G}_{\beta }^{q}\left(n\right){F}_{1}^{q}\left(n\right)}\sum _{k=1}^{\mathrm{\infty }}\frac{{\left({\left[n\right]}_{q}x\right)}^{k}}{{\left[k-1\right]}_{q}!}{q}^{k\left(k-3\right)-2/2}{e}^{-{\left[n\right]}_{q}x}\\ +\frac{{\left[n\right]}_{q}}{{G}_{\beta }^{q}\left(n\right){F}_{1}^{q}\left(n\right)}\sum _{k=0}^{\mathrm{\infty }}\frac{{\left({\left[n\right]}_{q}x\right)}^{k}}{{\left[k\right]}_{q}!}{q}^{k\left(k-1\right)-2/2}{e}^{-{\left[n\right]}_{q}x}\\ +\frac{\alpha }{{G}_{\beta }^{q}\left(n\right)}\sum _{k=0}^{\mathrm{\infty }}\frac{{\left({\left[n\right]}_{q}x\right)}^{k}}{{\left[k\right]}_{q}!}{q}^{k\left(k-1\right)/2}{e}^{-{\left[n\right]}_{q}x}\\ =& \frac{{\left[n\right]}_{q}^{2}x}{{q}^{2}{G}_{\beta }^{q}\left(n\right){F}_{1}^{q}\left(n\right)}{E}_{q}^{{\left[n\right]}_{q}x}{e}^{-{\left[n\right]}_{q}x}+\frac{{\left[n\right]}_{q}}{q{G}_{\beta }^{q}\left(n\right){F}_{1}^{q}\left(n\right)}{E}_{q}^{{\left[n\right]}_{q}x}{e}^{-{\left[n\right]}_{q}x}\\ +\frac{\alpha }{{G}_{\beta }^{q}\left(n\right)}{E}_{q}^{{\left[n\right]}_{q}x}{e}^{-{\left[n\right]}_{q}x}\\ =& \frac{{\left[n\right]}_{q}^{2}x}{{q}^{2}{G}_{\beta }^{q}\left(n\right){F}_{1}^{q}\left(n\right)}+\frac{{\left[n\right]}_{q}}{q{G}_{\beta }^{q}\left(n\right){F}_{1}^{q}\left(n\right)}+\frac{\alpha }{{G}_{\beta }^{q}\left(n\right)}.\end{array}$
Using the equality
${\left[n\right]}_{q}={\left[s\right]}_{q}+{q}^{s}{\left[n-s\right]}_{q},\phantom{\rule{1em}{0ex}}0\le s\le n,$
(2.2)
we get
$\begin{array}{rcl}{B}_{n,q}^{\left(\alpha ,\beta \right)}\left({e}_{2},x\right)& =& \frac{{\left[n\right]}_{q}^{4}{x}^{2}}{{q}^{6}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{2}^{q}\left(n\right)}\\ +\left\{\frac{{\left[n\right]}_{q}^{3}}{{q}^{5}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{2}^{q}\left(n\right)}+\frac{\left(1+{\left[2\right]}_{q}\right){\left[n\right]}_{q}^{3}}{{q}^{4}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{2}^{q}\left(n\right)}+\frac{2\alpha {\left[n\right]}_{q}^{2}}{{q}^{2}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{1}^{q}\left(n\right)}\right\}x\\ +\frac{{\left[2\right]}_{q}{\left[n\right]}_{q}^{2}}{{q}^{3}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{2}^{q}\left(n\right)}+\frac{2\alpha {\left[n\right]}_{q}}{q{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{1}^{q}\left(n\right)}+\frac{{\alpha }^{2}}{{G}_{\beta }^{q}{\left(n\right)}^{2}},\end{array}$

and so we have the proof of (iii). □

To obtain our main results we need to compute the second moment.

Lemma 2 Let $q\in \left(0,1\right)$ and $n>2$. Then we have the following inequality:
${B}_{n,q}^{\left(\alpha ,\beta \right)}\left({\left(t-x\right)}^{2},x\right)\le \left(\frac{2\left(1-{q}^{4}\right)}{{q}^{6}}+\frac{164{\left(\alpha +\beta +1\right)}^{2}{\left[n\right]}_{q}}{{q}^{6}{F}_{2}^{q}\left(n\right)}\right)x\left(x+1\right)+\frac{6{\left(\alpha +1\right)}^{2}}{{q}^{3}{G}_{\beta }^{q}\left(n\right)}.$
Proof From the linearity of the ${B}_{n,q}^{\left(\alpha ,\beta \right)}$ operators and Lemma 1, we write the second moment as
$\begin{array}{c}{B}_{n,q}^{\left(\alpha ,\beta \right)}\left({\left(t-x\right)}^{2},x\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left\{\frac{{\left[n\right]}_{q}^{4}}{{q}^{6}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{2}^{q}\left(n\right)}-\frac{2{\left[n\right]}_{q}^{2}}{{q}^{2}{G}_{\beta }^{q}\left(n\right){F}_{1}^{q}\left(n\right)}+1\right\}{x}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\left\{\frac{\left\{1+\left(1+{\left[2\right]}_{q}\right)q\right\}{\left[n\right]}_{q}^{3}}{{q}^{5}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{2}^{q}\left(n\right)}+\frac{2\alpha {\left[n\right]}_{q}^{2}}{{q}^{2}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{1}^{q}\left(n\right)}-\frac{2{\left[n\right]}_{q}}{q{G}_{\beta }^{q}\left(n\right){F}_{1}^{q}\left(n\right)}-\frac{2\alpha }{{G}_{\beta }^{q}\left(n\right)}\right\}x\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\frac{{\left[2\right]}_{q}{\left[n\right]}_{q}^{2}}{{q}^{3}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{2}^{q}\left(n\right)}+\frac{2\alpha {\left[n\right]}_{q}}{q{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{1}^{q}\left(n\right)}+\frac{{\alpha }^{2}}{{G}_{\beta }^{q}{\left(n\right)}^{2}}\hfill \\ \phantom{\rule{1em}{0ex}}\le \left\{\frac{{\left[n\right]}_{q}^{4}}{{q}^{6}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{2}^{q}\left(n\right)}-\frac{2{\left[n\right]}_{q}^{2}}{{q}^{2}{G}_{\beta }^{q}\left(n\right){F}_{1}^{q}\left(n\right)}+1+\frac{\left\{1+\left(1+{\left[2\right]}_{q}\right)q\right\}{\left[n\right]}_{q}^{3}}{{q}^{5}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{2}^{q}\left(n\right)}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\frac{2\alpha {\left[n\right]}_{q}^{2}}{{q}^{2}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{1}^{q}\left(n\right)}\right\}x\left(x+1\right)+\frac{{\left[2\right]}_{q}{\left[n\right]}_{q}^{2}}{{q}^{3}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{2}^{q}\left(n\right)}+\frac{2\alpha {\left[n\right]}_{q}}{q{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{1}^{q}\left(n\right)}+\frac{{\alpha }^{2}}{{G}_{\beta }^{q}{\left(n\right)}^{2}}\hfill \\ \phantom{\rule{1em}{0ex}}\le \left\{\frac{{\left[n\right]}_{q}^{4}\left(1+{q}^{6}\right)-2{q}^{4}{\left[n-2\right]}_{q}^{4}+2\beta {q}^{6}{\left[n\right]}_{q}{\left[n-1\right]}_{q}{\left[n-2\right]}_{q}}{{q}^{6}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{2}^{q}\left(n\right)}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\frac{\left(q+{q}^{2}+{\left[2\right]}_{q}{q}^{2}\right){\left[n\right]}_{q}^{3}}{{q}^{6}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{2}^{q}\left(n\right)}+\frac{{q}^{6}{\beta }^{2}{\left[n-1\right]}_{q}{\left[n-2\right]}_{q}}{{q}^{6}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{2}^{q}\left(n\right)}+\frac{2\alpha {q}^{4}{\left[n\right]}_{q}^{2}{\left[n-2\right]}_{q}}{{q}^{6}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{2}^{q}\left(n\right)}\right\}x\left(x+1\right)\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\frac{\left\{{\left[2\right]}_{q}+2\alpha {q}^{2}+{\alpha }^{2}{q}^{3}\right\}{\left[n\right]}_{q}}{{q}^{3}{G}_{\beta }^{q}\left(n\right){F}_{2}^{q}\left(n\right)}.\hfill \end{array}$
From (2.2), we have
$\begin{array}{c}{B}_{n,q}^{\left(\alpha ,\beta \right)}\left({\left(t-x\right)}^{2},x\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \left\{\frac{{\left[n-2\right]}_{q}^{4}\left({q}^{14}+{q}^{8}-2{q}^{4}\right)}{{q}^{6}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{2}^{q}\left(n\right)}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\frac{\left(1+{q}^{6}\right)\left\{4{\left[2\right]}_{q}{q}^{6}{\left[n-2\right]}_{q}^{3}+6{\left[2\right]}_{q}^{2}{q}^{4}{\left[n-2\right]}_{q}^{2}+4{\left[2\right]}_{q}^{3}{q}^{2}{\left[n-2\right]}_{q}+{\left[2\right]}_{q}^{4}\right\}}{{q}^{6}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{2}^{q}\left(n\right)}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\frac{\left(q+{q}^{2}+{\left[2\right]}_{q}{q}^{2}+2\beta {q}^{6}+2\alpha {q}^{4}\right){\left[n\right]}_{q}^{3}+{\beta }^{2}{q}^{6}{\left[n\right]}_{q}^{2}}{{q}^{6}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{2}^{q}\left(n\right)}\right\}x\left(x+1\right)\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\frac{\left({\left[2\right]}_{q}+{q}^{2}\right)\left({\left[2\right]}_{q}+2\alpha {q}^{2}+{\alpha }^{2}{q}^{3}\right)}{{q}^{3}{G}_{\beta }^{q}\left(n\right){F}_{1}^{q}\left(n\right)}\hfill \\ \phantom{\rule{1em}{0ex}}\le \left(\frac{2\left(1-{q}^{4}\right)}{{q}^{6}}+\frac{164{\left(\alpha +\beta +1\right)}^{2}{\left[n\right]}_{q}}{{q}^{6}{F}_{2}^{q}\left(n\right)}\right)x\left(x+1\right)+\frac{6{\left(\alpha +1\right)}^{2}}{{q}^{3}{G}_{\beta }^{q}\left(n\right)}.\hfill \end{array}$

And the proof of Lemma 2 is now finished. □

## 3 Direct estimates

Now in our considerations, ${C}_{B}\left[0,\mathrm{\infty }\right)$ denotes the set of all bounded-continuous functions from $\left[0,\mathrm{\infty }\right)$ to . ${C}_{B}\left[0,\mathrm{\infty }\right)$ is a normed space with the norm ${\parallel f\parallel }_{B}=sup\left\{|f\left(x\right)|:x\in \left[0,\mathrm{\infty }\right)\right\}$. We denote the first modulus of continuity on the finite interval $\left[0,b\right]$, $b>0$,
${\omega }_{\left[0,b\right]}\left(f,\delta \right)=\underset{0
(3.1)
The Peetre K-functional is defined by
${K}_{2}\left(f,\delta \right)=inf\left\{{\parallel f-g\parallel }_{B}+\delta {\parallel {g}^{″}\parallel }_{B}:g\in {W}_{\mathrm{\infty }}^{2}\right\},\phantom{\rule{1em}{0ex}}\delta >0,$
where ${W}_{\mathrm{\infty }}^{2}=\left\{g\in {C}_{B}\left[0,\mathrm{\infty }\right):{g}^{\prime },{g}^{″}\in {C}_{B}\left[0,\mathrm{\infty }\right)\right\}$. By Theorem 2.4 in , p.177, there exists a positive constant C such that
${K}_{2}\left(f,\delta \right)\le C{\omega }_{2}\left(f,\sqrt{\delta }\right),$
(3.2)
where
${\omega }_{2}\left(f,\sqrt{\delta }\right)=\underset{0

Gadzhiev proved the weighted Korovkin-type theorems in . We give the Gadzhiev results in weighted spaces. Let $\rho \left(x\right)=1+{x}^{2}$ and the weighted spaces ${C}_{\rho }\left[0,\mathrm{\infty }\right)$ denote the space of all continuous functions f, satisfying $|f\left(x\right)|\le {M}_{f}\rho \left(x\right)$, where ${M}_{f}$ is a constant depending only on f. ${C}_{\rho }\left[0,\mathrm{\infty }\right)$ is a normed space with the norm ${\parallel f\parallel }_{\rho }=sup\left\{\frac{|f\left(x\right)|}{\rho \left(x\right)}:x\in {\mathbb{R}}^{+}\cup \left\{0\right\}\right\}$ and ${C}_{\rho }^{\ast }\left[0,\mathrm{\infty }\right)$ denotes the subspace of all functions $f\in {C}_{\rho }\left[0,\mathrm{\infty }\right)$ for which ${lim}_{|x|\to \mathrm{\infty }}\frac{|f\left(x\right)|}{\rho \left(x\right)}$ exists finitely.

Thus we are ready to give direct results. The following lemma is routine and its proof is omitted.

Lemma 3 Let
${\overline{B}}_{n,q}^{\left(\alpha ,\beta \right)}\left(f,x\right)={B}_{n,q}^{\left(\alpha ,\beta \right)}\left(f,x\right)-f\left({D}_{n,q}^{\left(\alpha ,\beta \right)}\left(x\right)\right)+f\left(x\right).$
(3.3)
Then the following assertions hold for the operators (3.3):
$\begin{array}{rc}\text{(i)}& {\overline{B}}_{n,q}^{\left(\alpha ,\beta \right)}\left(1,x\right)=1,\\ \text{(ii)}& {\overline{B}}_{n,q}^{\left(\alpha ,\beta \right)}\left(t,x\right)=x,\\ \text{(iii)}& {\overline{B}}_{n,q}^{\left(\alpha ,\beta \right)}\left(t-x,x\right)=0,\end{array}$

where ${D}_{n,q}^{\left(\alpha ,\beta \right)}\left(x\right)=\frac{{\left[n\right]}_{q}^{2}x}{{q}^{2}{G}_{\beta }^{q}\left(n\right){F}_{1}^{q}\left(n\right)}+\frac{{\left[n\right]}_{q}}{q{G}_{\beta }^{q}\left(n\right){F}_{1}^{q}\left(n\right)}+\frac{\alpha }{{G}_{\beta }^{q}\left(n\right)}$.

Lemma 4 Let $q\in \left(0,1\right)$ and $n>2$. Then for every $x\in \left[0,\mathrm{\infty }\right)$ and ${f}^{″}\in {C}_{B}\left[0,\mathrm{\infty }\right)$, we have the inequality
$|{\overline{B}}_{n,q}^{\left(\alpha ,\beta \right)}\left(f,x\right)-f\left(x\right)|\le {\delta }_{n,q}^{\left(\alpha ,\beta \right)}\left(x\right){\parallel {f}^{″}\parallel }_{B},$

where ${\delta }_{n,q}^{\left(\alpha ,\beta \right)}\left(x\right)=\left(\frac{2\left(1-{q}^{4}\right)}{{q}^{6}}+\frac{263{\left(\alpha +\beta +1\right)}^{2}}{{q}^{6}{F}_{1}^{q}\left(n\right)}\right)x\left(x+1\right)+\frac{5{\left(\alpha +1\right)}^{2}}{{q}^{3}{G}_{\beta }^{q}\left(n\right)}$.

Proof Using Taylor’s expansion
$f\left(t\right)=f\left(x\right)+\left(t-x\right){f}^{\prime }\left(x\right)+{\int }_{x}^{t}\left(t-u\right){f}^{″}\left(u\right)\phantom{\rule{0.2em}{0ex}}du$
and Lemma 3, we obtain
${\overline{B}}_{n,q}^{\left(\alpha ,\beta \right)}\left(f,x\right)-f\left(x\right)={\overline{B}}_{n,q}^{\left(\alpha ,\beta \right)}\left({\int }_{x}^{t}\left(t-u\right){f}^{″}\left(u\right)\phantom{\rule{0.2em}{0ex}}du,x\right).$
Then, using Lemma 1 and the inequality
$|{\int }_{x}^{t}\left(t-u\right){f}^{″}\left(u\right)\phantom{\rule{0.2em}{0ex}}du|\le {\parallel {f}^{″}\parallel }_{B}\frac{{\left(t-x\right)}^{2}}{2},$
we get
$\begin{array}{c}|{\overline{B}}_{n,q}^{\left(\alpha ,\beta \right)}\left(f,x\right)-f\left(x\right)|\hfill \\ \phantom{\rule{1em}{0ex}}\le |{B}_{n,q}^{\left(\alpha ,\beta \right)}\left({\int }_{x}^{t}\left(t-u\right){f}^{″}\left(u\right)\phantom{\rule{0.2em}{0ex}}du,x\right)-{\int }_{x}^{{D}_{n,q}^{\left(\alpha ,\beta \right)}\left(x\right)}\left\{{D}_{n,q}^{\left(\alpha ,\beta \right)}\left(x\right)-u\right\}{f}^{″}\left(u\right)\phantom{\rule{0.2em}{0ex}}du|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{{\parallel {f}^{″}\parallel }_{B}}{2}\left\{\left(\frac{2\left(1-{q}^{4}\right)}{{q}^{6}}+\frac{164{\left(\alpha +\beta +1\right)}^{2}{\left[n\right]}_{q}}{{q}^{6}{F}_{2}^{q}\left(n\right)}+{\left(\frac{{\left[n\right]}_{q}^{2}}{{q}^{2}{G}_{\beta }^{q}\left(n\right){F}_{1}^{q}\left(n\right)}-1\right)}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\frac{2{\left[n\right]}_{q}^{3}}{{q}^{3}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{1}^{q}{\left(n\right)}^{2}}+\frac{2{\left[n\right]}_{q}^{2}\alpha }{{q}^{2}{G}_{\beta }^{q}{\left(n\right)}^{2}{F}_{1}^{q}\left(n\right)}\right)x\left(x+1\right)+{\left(\frac{{\left[n\right]}_{q}+\alpha q{\left[n-1\right]}_{q}}{q{G}_{\beta }^{q}\left(n\right){F}_{1}^{q}\left(n\right)}\right)}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\frac{6{\left(\alpha +1\right)}^{2}}{{q}^{3}{G}_{\beta }^{q}\left(n\right){F}_{1}^{q}\left(n\right)}\right\}\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{{\parallel {f}^{″}\parallel }_{B}}{2}\left\{\left(\frac{4\left(1-{q}^{4}\right)}{{q}^{6}}+\frac{526{\left(\alpha +\beta +1\right)}^{2}}{{q}^{6}{F}_{1}^{q}\left(n\right)}\right)x\left(x+1\right)+\frac{10{\left(\alpha +1\right)}^{2}}{{q}^{3}{G}_{\beta }^{q}\left(n\right)}\right\}.\hfill \end{array}$

And the proof of the Lemma 4 is now completed. □

Theorem 1 Let $\left({q}_{n}\right)\subset \left(0,1\right)$ a sequence such that ${q}_{n}\to 1$ as $n\to \mathrm{\infty }$. Then for every $n>2$, $x\in \left[0,\mathrm{\infty }\right)$ and $f\in {C}_{B}\left[0,\mathrm{\infty }\right)$, we have the inequality
$|{B}_{n,{q}_{n}}^{\left(\alpha ,\beta \right)}\left(f,x\right)-f\left(x\right)|\le 2M{\omega }_{2}\left(f,\sqrt{{\delta }_{n,{q}_{n}}^{\left(\alpha ,\beta \right)}\left(x\right)}\right)+w\left(f,{\eta }_{n,{q}_{n}}^{\left(\alpha ,\beta \right)}\left(x\right)\right),$

where ${\eta }_{n,{q}_{n}}^{\left(\alpha ,\beta \right)}\left(x\right)=\left(\frac{{\left[n\right]}_{{q}_{n}}^{2}}{{q}_{n}^{2}{G}_{\beta }^{{q}_{n}}\left(n\right){F}_{1}^{{q}_{n}}\left(n\right)}-1\right)x+\frac{{\left[n\right]}_{{q}_{n}}}{{q}_{n}{G}_{\beta }^{{q}_{n}}\left(n\right){F}_{1}^{{q}_{n}}\left(n\right)}+\frac{\alpha }{{G}_{\beta }^{{q}_{n}}\left(n\right)}$.

Proof Using (3.3) for any $g\in {W}_{\mathrm{\infty }}^{2}$, we obtain the following inequality:
$\begin{array}{rcl}|{B}_{n,{q}_{n}}^{\left(\alpha ,\beta \right)}\left(f,x\right)-f\left(x\right)|& \le & |{\overline{B}}_{n,{q}_{n}}^{\left(\alpha ,\beta \right)}\left(f-g,x\right)-\left(f-g\right)\left(x\right)+{\overline{B}}_{n,{q}_{n}}^{\left(\alpha ,\beta \right)}\left(g,x\right)-g\left(x\right)|\\ +|f\left(\frac{{\left[n\right]}_{{q}_{n}}^{2}}{{q}_{n}^{2}{G}_{\beta }^{{q}_{n}}\left(n\right){F}_{1}^{{q}_{n}}\left(n\right)}x+\frac{{\left[n\right]}_{{q}_{n}}}{{q}_{n}{G}_{\beta }^{{q}_{n}}\left(n\right){F}_{1}^{{q}_{n}}\left(n\right)}+\frac{\alpha }{{G}_{\beta }^{{q}_{n}}\left(n\right)}\right)-f\left(x\right)|.\end{array}$
From Lemma 4, we get
$\begin{array}{rcl}|{B}_{n,{q}_{n}}^{\left(\alpha ,\beta \right)}\left(f,x\right)-f\left(x\right)|& \le & 2{\parallel f-g\parallel }_{B}+{\delta }_{n,{q}_{n}}^{\left(\alpha ,\beta \right)}\left(x\right)\parallel {g}^{″}\parallel \\ +|f\left(\frac{{\left[n\right]}_{{q}_{n}}^{2}}{{q}_{n}^{2}{G}_{\beta }^{{q}_{n}}\left(n\right){F}_{1}^{{q}_{n}}\left(n\right)}x+\frac{{\left[n\right]}_{{q}_{n}}}{{q}_{n}{G}_{\beta }^{{q}_{n}}\left(n\right){F}_{1}^{{q}_{n}}\left(n\right)}+\frac{\alpha }{{G}_{\beta }^{{q}_{n}}\left(n\right)}\right)-f\left(x\right)|.\end{array}$
By using equality (3.1) we have
$|{B}_{n,{q}_{n}}^{\left(\alpha ,\beta \right)}\left(f,x\right)-f\left(x\right)|\le 2{\parallel f-g\parallel }_{B}+{\delta }_{n,{q}_{n}}^{\left(\alpha ,\beta \right)}\left(x\right){\parallel {g}^{″}\parallel }_{B}+w\left(f,{\eta }_{n,{q}_{n}}^{\left(\alpha ,\beta \right)}\left(x\right)\right).$

Taking the infimum over $g\in {W}_{\mathrm{\infty }}^{2}$ on the right-hand side of the above inequality and using the inequality (3.2), we get the desired result. □

Theorem 2 Let $\left({q}_{n}\right)\subset \left(0,1\right)$ a sequence such that ${q}_{n}\to 1$ as $n\to \mathrm{\infty }$. Then $f\in {C}_{\rho }^{\ast }\left[0,\mathrm{\infty }\right)$, and we have
$\underset{n\to \mathrm{\infty }}{lim}{\parallel {B}_{n,{q}_{n}}^{\left(\alpha ,\beta \right)}\left(f\right)-f\parallel }_{\rho }=0.$
Proof From Lemma 1, it is obvious that ${\parallel {B}_{n,{q}_{n}}^{\left(\alpha ,\beta \right)}\left({e}_{0}\right)-{e}_{0}\parallel }_{\rho }=0$. Since $|\frac{{\left[n\right]}_{{q}_{n}}^{2}}{{q}_{n}^{2}{G}_{\beta }^{{q}_{n}}\left(n\right){F}_{1}^{{q}_{n}}\left(n\right)}x+\frac{{\left[n\right]}_{{q}_{n}}}{{q}_{n}{G}_{\beta }^{{q}_{n}}\left(n\right){F}_{1}^{{q}_{n}}\left(n\right)}+\frac{\alpha }{{G}_{\beta }^{{q}_{n}}\left(n\right)}-x|\le \left(x+1\right)o\left(1\right)$ and $\frac{x+1}{1+{x}^{2}}$ is positive and bounded from above for each $x\ge 0$, we obtain
${\parallel {B}_{n,{q}_{n}}^{\left(\alpha ,\beta \right)}\left({e}_{1}\right)-{e}_{1}\parallel }_{\rho }\le \frac{x+1}{1+{x}^{2}}o\left(1\right).$

And then ${lim}_{n\to \mathrm{\infty }}{\parallel {B}_{n,{q}_{n}}^{\left(\alpha ,\beta \right)}\left({e}_{1}\right)-{e}_{1}\parallel }_{\rho }=0$.

Similarly for every $n>2$, we write
$\begin{array}{rcl}{\parallel {B}_{n,{q}_{n}}^{\left(\alpha ,\beta \right)}\left({e}_{2}\right)-{e}_{2}\parallel }_{\rho }& =& \underset{x\in \left[0,\mathrm{\infty }\right)}{sup}\left\{\frac{|\left(\frac{{\left[n\right]}_{{q}_{n}}^{4}}{{q}_{n}^{6}{G}_{\beta }^{{q}_{n}}{\left(n\right)}^{2}{F}_{2}^{{q}_{n}}\left(n\right)}-1\right){x}^{2}}{1+{x}^{2}}\\ +\frac{\left\{\frac{\left(1+\left(1+{\left[2\right]}_{{q}_{n}}\right){q}_{n}\right){\left[n\right]}_{{q}_{n}}^{3}+2\alpha {q}^{2}{\left[n\right]}_{{q}_{n}}^{2}{\left[n-1\right]}_{{q}_{n}}}{{q}_{n}^{5}{G}_{\beta }^{{q}_{n}}{\left(n\right)}^{2}{F}_{2}^{{q}_{n}}\left(n\right)}\right\}x+\frac{{\left[2\right]}_{{q}_{n}}{\left[n\right]}_{{q}_{n}}^{2}}{{q}_{n}^{3}{G}_{\beta }^{{q}_{n}}{\left(n\right)}^{2}{F}_{2}^{{q}_{n}}\left(n\right)}}{1+{x}^{2}}\\ +\frac{\frac{+2\alpha {q}_{n}^{2}{\left[n\right]}_{{q}_{n}}{\left[n-2\right]}_{{q}_{n}}}{{q}_{n}^{3}{G}_{\beta }^{{q}_{n}}{\left(n\right)}^{2}{F}_{2}^{{q}_{n}}\left(n\right)}+\frac{{\alpha }^{2}}{{G}_{\beta }^{{q}_{n}}{\left(n\right)}^{2}}|}{1+{x}^{2}}\right\}\\ \le & \underset{x\in \left[0,\mathrm{\infty }\right)}{sup}\frac{1+x+{x}^{2}}{1+{x}^{2}}o\left(1\right),\end{array}$

we get ${lim}_{n\to \mathrm{\infty }}{\parallel {B}_{n,{q}_{n}}^{\left(\alpha ,\beta \right)}\left({e}_{2}\right)-{e}_{2}\parallel }_{\rho }=0$. Thus, from AD Gadzhiev’s theorem in , we obtain the desired result of Theorem 2. □

## Declarations

### Acknowledgements

The author would like thank the referee for many helpful comments.

## Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, Gazi University, Teknikokullar, Ankara, 06500, Turkey

## References 