# q-analogue of a new sequence of linear positive operators

## Abstract

This paper deals with Durrmeyer type generalization of q-Baskakov type operators using the concept of q-integral, which introduces a new sequence of positive q-integral operators. We show that this sequence is an approximation process in the polynomial weighted space of continuous functions defined on the interval $\left[0,\mathrm{\infty }\right)$. An estimate for the rate of convergence and weighted approximation properties are also obtained.

MSC:41A25, 41A36.

## 1 Introduction

In the year 2003 Agrawal and Mohammad  introduced a new sequence of linear positive operators by modifying the well-known Baskakov operators having weight functions of Szasz basis function as

${\mathcal{D}}_{n}\left(f,x\right)=n\sum _{k=1}^{\mathrm{\infty }}{p}_{n,k}\left(x\right){\int }_{0}^{\mathrm{\infty }}{s}_{n,k-1}\left(t\right)f\left(t\right)\phantom{\rule{0.2em}{0ex}}dt+{p}_{n,0}\left(x\right)f\left(0\right),\phantom{\rule{1em}{0ex}}x\in \left[0,\mathrm{\infty }\right),$
(1.1)

where

${p}_{n,k}\left(x\right)=\left(\begin{array}{c}n+k-1\\ k\end{array}\right)\frac{{x}^{k}}{{\left(1+x\right)}^{n+k}},\phantom{\rule{2em}{0ex}}{s}_{n,k}\left(t\right)={e}^{-nt}\frac{{\left(nt\right)}^{k}}{k!}.$

It is observed in  that these operators reproduce constant as well as linear functions. Later, some direct approximation results for the iterative combinations of these operators were studied in .

A lot of works on q-calculus are available in literature of different branches of mathematics and physics. For systematic study, we refer to the work of Ernst , Kim [10, 11], and Kim and Rim . The application of q-calculus in approximation theory was initiated by Phillips , who was the first to introduce q-Bernstein polynomials and study their approximation properties. Very recently the q-analogues of the Baskakov operators and their Kantorovich and Durrmeyer variants have been studied in [2, 3] and  respectively. We recall some notations and concepts of q-calculus. All of the results can be found in  and . In what follows, q is a real number satisfying $0.

For $n\in \mathbb{N}$, The q-binomial coefficients are given by

$\left[\begin{array}{c}n\\ k\end{array}{\right]}_{q}=\frac{{\left[n\right]}_{q}!}{{\left[k\right]}_{q}!{\left[n-k\right]}_{q}!},\phantom{\rule{1em}{0ex}}0\le k\le n.$

The q-Beta integral is defined by 

${\mathrm{\Gamma }}_{q}\left(t\right)={\int }_{0}^{\frac{1}{1-q}}{x}^{t-1}{E}_{q}\left(-qx\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}x,\phantom{\rule{1em}{0ex}}t>0,$
(1.2)

which satisfies the following functional equation:

${\mathrm{\Gamma }}_{q}\left(t+1\right)={\left[t\right]}_{q}{\mathrm{\Gamma }}_{q}\left(t\right),\phantom{\rule{2em}{0ex}}{\mathrm{\Gamma }}_{q}\left(1\right)=1.$

For $f\in C\left[0,\mathrm{\infty }\right)$$q>0$ and each positive integer n, the q-Baskakov operators  are defined as

$\begin{array}{rl}{\mathcal{B}}_{n,q}\left(f,x\right)& =\sum _{k=0}^{\mathrm{\infty }}\left[\begin{array}{c}n+k-1\\ k\end{array}{\right]}_{q}{q}^{\frac{k\left(k-1\right)}{2}}\frac{{x}^{k}}{{\left(1+x\right)}_{q}^{n+k}}f\left(\frac{{\left[k\right]}_{q}}{{q}^{k-1}{\left[n\right]}_{q}}\right)\\ =\sum _{k=0}^{\mathrm{\infty }}{p}_{n,k}^{q}\left(x\right)f\left(\frac{{\left[k\right]}_{q}}{{q}^{k-1}{\left[n\right]}_{q}}\right),\end{array}$
(1.3)

where

${\left(1+x\right)}_{q}^{n}:=\left\{\begin{array}{cc}\left(1+x\right)\left(1+qx\right)\cdots \left(1+{q}^{n-1}x\right),\hfill & n=1,2,\dots ,\hfill \\ 1,\hfill & n=0.\hfill \end{array}$

Remark 1 The first three moments of the q-Baskakov operators are given by As the operators ${\mathcal{D}}_{n}\left(f,x\right)$ have mixed basis functions in summation and integration and have an interesting property of reproducing linear functions, we were motivated to study these operators further. Here we define the q-analogue of the operators as

${\mathcal{D}}_{n}^{q}\left(f,x\right)={\left[n\right]}_{q}\sum _{k=1}^{\mathrm{\infty }}{p}_{n,k}^{q}\left(x\right){\int }_{0}^{q/\left(1-{q}^{n}\right)}{q}^{-k}{s}_{n,k-1}^{q}\left(t\right)f\left(t{q}^{-k}\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}t+{p}_{n,0}^{q}\left(x\right)f\left(0\right),$
(1.4)

where $x\in \left[0,\mathrm{\infty }\right)$ and

${p}_{n,k}^{q}\left(x\right)=\left[\begin{array}{c}n+k-1\\ k\end{array}{\right]}_{q}{q}^{\frac{k\left(k-1\right)}{2}}\frac{{x}^{k}}{{\left(1+x\right)}_{q}^{n+k}},\phantom{\rule{2em}{0ex}}{s}_{n,k}^{q}\left(t\right)={E}_{q}\left(-{\left[n\right]}_{q}t\right)\frac{{\left({\left[n\right]}_{q}t\right)}^{k}}{{\left[k\right]}_{q}!}.$

In case $q=1$, the above operators reduce to the operators (1.1). In the present paper, we estimate a local approximation theorem and the rate of convergence of these new operators as well as their weighted approximation properties.

## 2 Moment estimation

Lemma 1 The following equalities hold:

1. (i)

${\mathcal{D}}_{n}^{q}\left(1,x\right)=1$,

2. (ii)

${\mathcal{D}}_{n}^{q}\left(t,x\right)=x$,

3. (iii)

${\mathcal{D}}_{n}^{q}\left({t}^{2},x\right)={x}^{2}+\frac{x}{{\left[n\right]}_{q}}\left(1+q+\frac{x}{q}\right)$.

Proof The operators ${\mathcal{D}}_{n}^{q}$ are well defined on the function $1,t,{t}^{2}$. Then for every $x\in \left[0,\mathrm{\infty }\right)$, we obtain

${\mathcal{D}}_{n}^{q}\left(1,x\right)={\left[n\right]}_{q}\sum _{k=1}^{\mathrm{\infty }}{p}_{n,k}^{q}\left(x\right){\int }_{0}^{q/\left(1-{q}^{n}\right)}{q}^{-k}\frac{{\left({\left[n\right]}_{q}t\right)}^{k-1}}{{\left[k-1\right]}_{q}!}{E}_{q}\left(-{\left[n\right]}_{q}t\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}t+{p}_{n,0}^{q}\left(x\right).$

Substituting ${\left[n\right]}_{q}t=qy$ and using (1.2), we have

$\begin{array}{rcl}{\mathcal{D}}_{n}^{q}\left(1,x\right)& =& {\left[n\right]}_{q}\sum _{k=1}^{\mathrm{\infty }}{p}_{n,k}^{q}\left(x\right){\int }_{0}^{1/\left(1-q\right)}{q}^{-k}\frac{{\left(qy\right)}^{k-1}}{{\left[k-1\right]}_{q}!}{E}_{q}\left(-qy\right)\frac{q\phantom{\rule{0.2em}{0ex}}{d}_{q}y}{{\left[n\right]}_{q}}+{p}_{n,0}^{q}\left(x\right)\\ =& \sum _{k=1}^{\mathrm{\infty }}{p}_{n,k}^{q}\left(x\right)+{p}_{n,0}^{q}\left(x\right)={\mathcal{B}}_{n,q}\left(1,x\right)=1,\end{array}$

where ${\mathcal{B}}_{n,q}\left(f,x\right)$ is the q-Baskakov operator defined by (1.3).

Next, we have

${\mathcal{D}}_{n}^{q}\left(t,x\right)={\left[n\right]}_{q}\sum _{k=1}^{\mathrm{\infty }}{p}_{n,k}^{q}\left(x\right){\int }_{0}^{q/\left(1-{q}^{n}\right)}{q}^{-k}\frac{{\left({\left[n\right]}_{q}t\right)}^{k-1}}{{\left[k-1\right]}_{q}!}{E}_{q}\left(-{\left[n\right]}_{q}t\right)t{q}^{-k}\phantom{\rule{0.2em}{0ex}}{d}_{q}t.$

Again substituting ${\left[n\right]}_{q}t=qy$ and using (1.2), we have

$\begin{array}{rcl}{\mathcal{D}}_{n}^{q}\left(t,x\right)& =& {\left[n\right]}_{q}\sum _{k=1}^{\mathrm{\infty }}{p}_{n,k}^{q}\left(x\right){\int }_{0}^{1/\left(1-q\right)}{q}^{-k}\frac{{\left(qy\right)}^{k}}{{\left[k-1\right]}_{q}!{\left[n\right]}_{q}}{E}_{q}\left(-qy\right)\frac{q\phantom{\rule{0.2em}{0ex}}{d}_{q}y}{{\left[n\right]}_{q}{q}^{k}}\\ =& \sum _{k=0}^{\mathrm{\infty }}{p}_{n,k}^{q}\left(x\right)q\frac{{\left[k\right]}_{q}}{{\left[n\right]}_{q}{q}^{k}}={\mathcal{B}}_{n,q}\left(t,x\right)=x.\end{array}$

Finally,

${\mathcal{D}}_{n}^{q}\left({t}^{2},x\right)={\left[n\right]}_{q}\sum _{k=0}^{\mathrm{\infty }}{p}_{n,k}^{q}\left(x\right){\int }_{0}^{q/\left(1-{q}^{n}\right)}{q}^{-k}\frac{{\left({\left[n\right]}_{q}t\right)}^{k-1}}{{\left[k-1\right]}_{q}!}{E}_{q}\left(-{\left[n\right]}_{q}t\right){t}^{2}{q}^{-2k}\phantom{\rule{0.2em}{0ex}}{d}_{q}t.$

Again substituting ${\left[n\right]}_{q}t=qy$, using (1.2) and ${\left[k+1\right]}_{q}={\left[k\right]}_{q}+{q}^{k}$, we have

$\begin{array}{rcl}{\mathcal{D}}_{n}^{q}\left({t}^{2},x\right)& =& {\left[n\right]}_{q}\sum _{k=1}^{\mathrm{\infty }}{p}_{n,k}^{q}\left(x\right){\int }_{0}^{1/\left(1-q\right)}{q}^{-k}\frac{{\left(qy\right)}^{k+1}}{{\left[k-1\right]}_{q}!{\left[n\right]}_{q}^{2}}{E}_{q}\left(-qy\right){q}^{-2k}\frac{q\phantom{\rule{0.2em}{0ex}}{d}_{q}y}{{\left[n\right]}_{q}}\\ =& \sum _{k=1}^{\mathrm{\infty }}{p}_{n,k}^{q}\left(x\right)\frac{{\left[k+1\right]}_{q}{\left[k\right]}_{q}}{{\left[n\right]}_{q}^{2}{q}^{2k-2}}\\ =& \sum _{k=1}^{\mathrm{\infty }}{p}_{n,k}^{q}\left(x\right)\frac{\left({\left[k\right]}_{q}+{q}^{k}\right){\left[k\right]}_{q}}{{\left[n\right]}_{q}^{2}{q}^{2k-2}}\\ =& {\mathcal{B}}_{n,q}\left({t}^{2},x\right)+\frac{q}{{\left[n\right]}_{q}}{\mathcal{B}}_{n,q}\left(t,x\right)={x}^{2}+\frac{x}{{\left[n\right]}_{q}}\left(1+q+\frac{x}{q}\right).\end{array}$

□

Remark 2 If we put $q=1$, we get the moments of a new sequence ${\mathcal{D}}_{n}\left(f,x\right)$ considered in  as operators as Lemma 2 Let$q\in \left(0,1\right)$, then for$x\in \left[0,\mathrm{\infty }\right)$we have

${\mathcal{D}}_{n}^{q}\left({\left(t-x\right)}^{2},x\right)=\frac{x\left(x+q{\left[2\right]}_{q}\right)}{q{\left[n\right]}_{q}}.$

## 3 Direct theorems

By ${C}_{B}\left[0,\mathrm{\infty }\right)$ we denote the space of real valued continuous bounded functions f on the interval $\left[0,\mathrm{\infty }\right)$; the norm-$\parallel \cdot \parallel$ on the space ${C}_{B}\left[0,\mathrm{\infty }\right)$ is given by

$\parallel f\parallel =\underset{0\le x<\mathrm{\infty }}{sup}|f\left(x\right)|.$

The Peetre’s K-functional is defined by

${K}_{2}\left(f,\delta \right)=inf\left\{\parallel f-g\parallel +\delta \parallel {g}^{″}\parallel :g\in {W}_{\mathrm{\infty }}^{2}\right\},$

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 , pp.177], there exists a positive constant $C>0$ such that ${K}_{2}\left(f,\delta \right)\le C{\omega }_{2}\left(f,{\delta }^{1/2}\right)$$\delta >0$ and the second order modulus of smoothness is given by

${\omega }_{2}\left(f,\sqrt{\delta }\right)=\underset{0

Also, for $f\in {C}_{B}\left[0,\mathrm{\infty }\right)$ a usual modulus of continuity is given by

$\omega \left(f,\delta \right)=\underset{0

Theorem 1 Let$f\in {C}_{B}\left[0,\mathrm{\infty }\right)$and$0. Then for all$x\in \left[0,\mathrm{\infty }\right)$and$n\in N$, there exists an absolute constant$C>0$such that

$|{\mathcal{D}}_{n}^{q}\left(f,x\right)-f\left(x\right)|\le C{\omega }_{2}\left(f,\sqrt{\frac{x\left(x+q{\left[2\right]}_{q}\right)}{q{\left[n\right]}_{q}}}\right).$

Proof Let $g\in {W}_{\mathrm{\infty }}^{2}$ and $x,t\in \left[0,\mathrm{\infty }\right)$. By Taylor’s expansion, we have

$g\left(t\right)=g\left(x\right)+{g}^{\prime }\left(x\right)\left(t-x\right)+{\int }_{x}^{t}\left(t-u\right){g}^{″}\left(u\right)\phantom{\rule{0.2em}{0ex}}du.$

Applying Lemma 2, we obtain

${\mathcal{D}}_{n}^{q}\left(g,x\right)-g\left(x\right)={\mathcal{D}}_{n}^{q}\left({\int }_{x}^{t}\left(t-u\right){g}^{″}\left(u\right)\phantom{\rule{0.2em}{0ex}}du,x\right).$

Obviously, we have $|{\int }_{x}^{t}\left(t-u\right){g}^{″}\left(u\right)\phantom{\rule{0.2em}{0ex}}du|\le {\left(t-x\right)}^{2}\parallel {g}^{″}\parallel$. Therefore,

$|{\mathcal{D}}_{n}^{q}\left(g,x\right)-g\left(x\right)|\le {\mathcal{D}}_{n}^{q}\left({\left(t-x\right)}^{2},x\right)\parallel {g}^{″}\parallel =\frac{x\left(x+q{\left[2\right]}_{q}\right)}{q{\left[n\right]}_{q}}\parallel {g}^{″}\parallel .$

Using Lemma 1, we have

$|{\mathcal{D}}_{n}^{q}\left(f,x\right)|\le {\left[n\right]}_{q}\sum _{k=1}^{\mathrm{\infty }}{p}_{n,k}^{q}\left(x\right){\int }_{0}^{q/\left(1-{q}^{n}\right)}{q}^{-k}{s}_{n,k-1}^{q}\left(t\right)|f\left(t{q}^{-k}\right)|\phantom{\rule{0.2em}{0ex}}{d}_{q}t+{p}_{n,0}^{q}\left(x\right)|f\left(0\right)|\le \parallel f\parallel .$

Thus

$\begin{array}{rcl}|{\mathcal{D}}_{n}^{q}\left(f,x\right)-f\left(x\right)|& \le & |{\mathcal{D}}_{n}^{q}\left(f-g,x\right)-\left(f-g\right)\left(x\right)|+|{\mathcal{D}}_{n}^{q}\left(g,x\right)-g\left(x\right)|\\ \le & 2\parallel f-g\parallel +\frac{x\left(x+q{\left[2\right]}_{q}\right)}{q{\left[n\right]}_{q}}\parallel {g}^{″}\parallel .\end{array}$

Finally, taking the infimum over all $g\in {W}_{\mathrm{\infty }}^{2}$ and using the inequality ${K}_{2}\left(f,\delta \right)\le C{\omega }_{2}\left(f,{\delta }^{1/2}\right)$, $\delta >0$, we get the required result. This completes the proof of Theorem 1. □

We consider the following class of functions:

Let ${H}_{{x}^{2}}\left[0,\mathrm{\infty }\right)$ be the set of all functions f defined on $\left[0,\mathrm{\infty }\right)$ satisfying the condition $|f\left(x\right)|\le {M}_{f}\left(1+{x}^{2}\right)$, where ${M}_{f}$ is a constant depending only on f. By ${C}_{{x}^{2}}\left[0,\mathrm{\infty }\right)$, we denote the subspace of all continuous functions belonging to ${H}_{{x}^{2}}\left[0,\mathrm{\infty }\right)$. Also, let ${C}_{{x}^{2}}^{\ast }\left[0,\mathrm{\infty }\right)$ be the subspace of all functions $f\in {C}_{{x}^{2}}\left[0,\mathrm{\infty }\right)$, for which ${lim}_{|x|\to \mathrm{\infty }}\frac{f\left(x\right)}{1+{x}^{2}}$ is finite. The norm on ${C}_{{x}^{2}}^{\ast }\left[0,\mathrm{\infty }\right)$ is ${\parallel f\parallel }_{{x}^{2}}={sup}_{x\in \left[0,\mathrm{\infty }\right)}\frac{|f\left(x\right)|}{1+{x}^{2}}$. We denote the modulus of continuity of f on closed interval $\left[0,a\right]$, $a>0$ as by

${\omega }_{a}\left(f,\delta \right)=\underset{|t-x|\le \delta }{sup}\underset{x,t\in \left[0,a\right]}{sup}\phantom{\rule{0.25em}{0ex}}|f\left(t\right)-f\left(x\right)|.$

We observe that for function $f\in {C}_{{x}^{2}}\left[0,\mathrm{\infty }\right)$, the modulus of continuity ${\omega }_{a}\left(f,\delta \right)$ tends to zero.

Theorem 2 Let$f\in {C}_{{x}^{2}}\left[0,\mathrm{\infty }\right)$, $q\in \left(0,1\right)$and${\omega }_{a+1}\left(f,\delta \right)$be its modulus of continuity on the finite interval$\left[0,a+1\right]\subset \left[0,\mathrm{\infty }\right)$, where$a>0$. Then for every$n>2$,

${\parallel {\mathcal{D}}_{n}^{q}\left(f\right)-f\parallel }_{C\left[0,a\right]}\le \frac{6{M}_{f}a\left(1+{a}^{2}\right)\left(2+a\right)}{q{\left[n\right]}_{q}}+2\omega \left(f,\sqrt{\frac{a\left(a+q{\left[2\right]}_{q}\right)}{q{\left[n\right]}_{q}}}\right).$

Proof For $x\in \left[0,a\right]$ and $t>a+1$, since $t-x>1$, we have

$\begin{array}{rcl}|f\left(t\right)-f\left(x\right)|& \le & {M}_{f}\left(2+{x}^{2}+{t}^{2}\right)\\ \le & {M}_{f}\left(2+3{x}^{2}+2{\left(t-x\right)}^{2}\right)\\ \le & 6{M}_{f}\left(1+{a}^{2}\right){\left(t-x\right)}^{2}.\end{array}$
(3.1)

For $x\in \left[0,a\right]$ and $t\le a+1$, we have

$|f\left(t\right)-f\left(x\right)|\le {\omega }_{a+1}\left(f,|t-x|\right)\le \left(1+\frac{|t-x|}{\delta }\right){\omega }_{a+1}\left(f,\delta \right)$
(3.2)

with $\delta >0$.

From (3.1) and (3.2) we can write

$|f\left(t\right)-f\left(x\right)|\le 6{M}_{f}\left(1+{a}^{2}\right){\left(t-x\right)}^{2}+\left(1+\frac{|t-x|}{\delta }\right){\omega }_{a+1}\left(f,\delta \right)$
(3.3)

for $x\in \left[0,a\right]$ and $t\ge 0$. Thus

$\begin{array}{rcl}|{\mathcal{D}}_{n}^{q}\left(f,x\right)-f\left(x\right)|& \le & {\mathcal{D}}_{n}^{q}\left(|f\left(t\right)-f\left(x\right)|,x\right)\\ \le & 6{M}_{f}\left(1+{a}^{2}\right){\mathcal{D}}_{n}^{q}\left({\left(t-x\right)}^{2},x\right)\\ +{\omega }_{a+1}\left(f,\delta \right){\left(1+\frac{1}{\delta }{\mathcal{D}}_{n}^{q}\left({\left(t-x\right)}^{2},x\right)\right)}^{\frac{1}{2}}.\end{array}$

Hence, by using Schwarz inequality and Lemma 2, for every $q\in \left(0,1\right)$ and $x\in \left[0,a\right]$

$\begin{array}{rcl}|{\mathcal{D}}_{n}^{q}\left(f,x\right)-f\left(x\right)|& \le & \frac{6{M}_{f}\left(1+{a}^{2}\right)x\left(q{\left[2\right]}_{q}+x\right)}{q{\left[n\right]}_{q}}\\ +{\omega }_{a+1}\left(f,\delta \right)\left(1+\frac{1}{\delta }\sqrt{\frac{x\left(q{\left[2\right]}_{q}+x\right)}{q{\left[n\right]}_{q}}}\right)\\ \le & \frac{6{M}_{f}a\left(1+{a}^{2}\right)\left(2+a\right)}{q{\left[n\right]}_{q}}+{\omega }_{a+1}\left(f,\delta \right)\left(1+\frac{1}{\delta }\sqrt{\frac{a\left(a+q{\left[2\right]}_{q}\right)}{q{\left[n\right]}_{q}}}\right).\end{array}$

By taking $\delta =\sqrt{\frac{a\left(q{\left[2\right]}_{q}+a\right)}{q{\left[n\right]}_{q}}}$ we get the assertion of our theorem. □

## 4 Higher order moments and an asymptotic formula

Lemma 3 ()

Let$0, we have

$\begin{array}{rcl}{\mathcal{B}}_{n,q}\left({t}^{3},x\right)& =& \frac{1}{{\left[n\right]}_{q}}x+\frac{1+2q}{{q}^{2}}\frac{{\left[n+1\right]}_{q}}{{\left[n\right]}_{q}^{2}}{x}^{2}+\frac{1}{{q}^{3}}\frac{{\left[n+1\right]}_{q}{\left[n+2\right]}_{q}}{{\left[n\right]}_{q}^{2}}{x}^{3},\\ {\mathcal{B}}_{n,q}\left({t}^{4},x\right)& =& \frac{1}{{\left[n\right]}_{q}^{3}}x+\frac{1}{{q}^{3}}\left(1+3q+3{q}^{2}\right)\frac{{\left[n+1\right]}_{q}}{{\left[n\right]}_{q}^{3}}{x}^{2}\\ +\frac{1}{{q}^{5}{\left[2\right]}_{q}}\left(1+3q+5{q}^{2}+3{q}^{3}\right)\frac{{\left[n+1\right]}_{q}{\left[n+2\right]}_{q}}{{\left[n\right]}_{q}^{3}}{x}^{3}\\ +\frac{1}{{q}^{6}{\left[2\right]}_{q}{\left[3\right]}_{q}{\left[4\right]}_{q}}\left(1+3q+5{q}^{2}+6{q}^{3}+5{q}^{4}+3{q}^{5}+{q}^{6}\right)\\ ×\frac{{\left[n+1\right]}_{q}{\left[n+2\right]}_{q}{\left[n+3\right]}_{q}}{{\left[n\right]}_{q}^{3}}{x}^{4}.\end{array}$

Now, we present higher order moments for the operators (1.4).

Lemma 4 Let$0, we have The proof of Lemma 4 can be obtained by using Lemma 3.

We consider the following classes of functions: Theorem 3 Let${q}_{n}\in \left(0,1\right)$, then the sequence$\left\{{\mathcal{D}}_{n}^{{q}_{n}}\left(f\right)\right\}$converges to f uniformly on$\left[0,A\right]$for each$f\in {C}_{2}^{\ast }\left[0,\mathrm{\infty }\right)$if and only if${lim}_{n\to \mathrm{\infty }}{q}_{n}=1$.

Theorem 4 Assume that${q}_{n}\in \left(0,1\right)$, ${q}_{n}\to 1$and${q}_{n}^{n}\to a$as$n\to \mathrm{\infty }$. For any$f\in {C}_{2}^{\ast }\left[0,\mathrm{\infty }\right)$such that${f}^{\mathrm{\prime }},{f}^{\mathrm{\prime }\mathrm{\prime }}\in {C}_{2}^{\ast }\left[0,\mathrm{\infty }\right)$the following equality holds

$\underset{n\to \mathrm{\infty }}{lim}{\left[n\right]}_{{q}_{n}}\left({\mathcal{D}}_{n}^{{q}_{n}}\left(f;x\right)-f\left(x\right)\right)=\left({x}^{2}+2x\right){f}^{\mathrm{\prime }\mathrm{\prime }}\left(x\right)$

uniformly on any$\left[0,A\right]$, $A>0$.

Proof Let $f,{f}^{\mathrm{\prime }},{f}^{\mathrm{\prime }\mathrm{\prime }}\in {C}_{2}^{\ast }\left[0,\mathrm{\infty }\right)$ and $x\in \left[0,\mathrm{\infty }\right)$ be fixed. By using Taylor’s formula, we may write

$f\left(t\right)=f\left(x\right)+{f}^{\mathrm{\prime }}\left(x\right)\left(t-x\right)+\frac{1}{2}{f}^{\mathrm{\prime }\mathrm{\prime }}\left(x\right){\left(t-x\right)}^{2}+r\left(t;x\right){\left(t-x\right)}^{2},$
(4.1)

where $r\left(t;x\right)$ is the Peano form of the remainder, $r\left(\cdot ;x\right)\in {C}_{2}^{\ast }\left[0,\mathrm{\infty }\right)$ and ${lim}_{t\to x}r\left(t;x\right)=0$. Applying ${\mathcal{D}}_{n}^{{q}_{n}}$ to (4.1), we obtain

${\left[n\right]}_{{q}_{n}}\left({\mathcal{D}}_{n}^{{q}_{n}}\left(f;x\right)-f\left(x\right)\right)=\frac{1}{2}{f}^{\mathrm{\prime }\mathrm{\prime }}\left(x\right){\left[n\right]}_{{q}_{n}}{\mathcal{D}}_{n}^{{q}_{n}}\left({\left(t-x\right)}^{2};x\right)+{\left[n\right]}_{{q}_{n}}{\mathcal{D}}_{n}^{{q}_{n}}\left(r\left(t;x\right){\left(t-x\right)}^{2};x\right).$

By the Cauchy-Schwarz inequality, we have

${\mathcal{D}}_{n}^{{q}_{n}}\left(r\left(t;x\right){\left(t-x\right)}^{2};x\right)\le \sqrt{{\mathcal{D}}_{n}^{{q}_{n}}\left({r}^{2}\left(t;x\right);x\right)}\sqrt{{\mathcal{D}}_{n}^{{q}_{n}}\left({\left(t-x\right)}^{4};x\right)}.$
(4.2)

Observe that ${r}^{2}\left(x;x\right)=0$ and ${r}^{2}\left(\cdot ;x\right)\in {C}_{2}^{\ast }\left[0,\mathrm{\infty }\right)$. Then it follows from Theorem 3 and Lemma 4, that

$\underset{n\to \mathrm{\infty }}{lim}{\mathcal{D}}_{n}^{{q}_{n}}\left({r}^{2}\left(t;x\right);x\right)={r}^{2}\left(x;x\right)=0$
(4.3)

uniformly with respect to $x\in \left[0,A\right]$. Now from (4.2), (4.3) and Remark 2, we get immediately

$\underset{n\to \mathrm{\infty }}{lim}{\left[n\right]}_{{q}_{n}}{\mathcal{D}}_{n}^{{q}_{n}}\left(r\left(t;x\right){\left(t-x\right)}^{2};x\right)=0.$

Then, we get the following □

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## Acknowledgements

The work was done while the first author visited Division of General Education-mathematics, Kwangwoon University, Seoul, South Korea for collaborative research during June 15-25, 2010.

## Author information

Authors

### Corresponding author

Correspondence to Taekyun Kim.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed in preparing the manuscript equally.

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Gupta, V., Kim, T. & Lee, SH. q-analogue of a new sequence of linear positive operators. J Inequal Appl 2012, 144 (2012). https://doi.org/10.1186/1029-242X-2012-144

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• DOI: https://doi.org/10.1186/1029-242X-2012-144

### Keywords

• Durrmeyer type operators
• weighted approximation
• rate of convergence
• q-integral 