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# On the convergence of a kind of q-gamma operators

Journal of Inequalities and Applications20132013:105

https://doi.org/10.1186/1029-242X-2013-105

• Received: 15 November 2012
• Accepted: 25 February 2013
• Published:

## Abstract

In this paper, we introduce a kind of q-gamma operators based on the concept of a q-integral. We estimate moments of these operators and establish direct and local approximation theorems of the operators. The estimates on the rate of convergence and weighted approximation of the operators are obtained, a Voronovskaya asymptotic formula is also presented.

MSC:41A10, 41A25, 41A36.

## Keywords

• q-integral
• q-gamma operators
• weighted approximation
• rate of convergence

## 1 Introduction

In recent years, the applications of q-calculus in the approximation theory is one of the main areas of research. After q-Bernstein polynomials were introduced by Phillips  in 1997, many researchers have performed studies in this field; we mention some of them .

In 2007, Karsli  introduced and estimated the rate of convergence for functions with derivatives of bounded variation on $\left[0,\mathrm{\infty }\right)$ of new gamma type operators as follows:
${L}_{n}\left(f;x\right)=\frac{\left(2n+3\right)!{x}^{n+3}}{n!\left(n+2\right)!}{\int }_{0}^{\mathrm{\infty }}\frac{{t}^{n}}{{\left(x+t\right)}^{2n+4}}f\left(t\right)\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}x>0.$
(1)

In 2009, Karsli, Gupta and Izgi  gave an estimate of the rate of pointwise convergence of these operators (1) on a Lebesgue point of bounded variation function f defined on the interval $\left(0,\mathrm{\infty }\right)$. In 2010, Karsli and Özarslan  obtained some direct local and global approximation results and gave a Voronoskaya-type theorem for the operators (1). As the application of q-calculus in approximation theory is an active field, it seems there are no papers mentioning the q analogue of these operators defined in (1). Inspired by Aral and Gupta , they defined a generalization of q-Baskakov type operators using q-Beta integral and obtained some important approximation properties, which motivates us to introduce the q analogue of this kind of gamma operators.

Before introducing the operators, we mention certain definitions based on q-integers; details can be found in [8, 9]. For any fixed real number $0 and each nonnegative integer k, we denote q-integers by ${\left[k\right]}_{q}$, where
${\left[k\right]}_{q}=\left\{\begin{array}{ll}\frac{1-{q}^{k}}{1-q},& q\ne 1,\\ k,& q=1.\end{array}$
Also, q-factorial and q-binomial coefficients are defined as follows:
$\begin{array}{c}{\left[k\right]}_{q}!=\left\{\begin{array}{ll}{\left[k\right]}_{q}{\left[k-1\right]}_{q}\cdots {\left[1\right]}_{q},& k=1,2,\dots ,\\ 1,& k=0,\end{array}\hfill \\ {\left[\begin{array}{c}n\\ k\end{array}\right]}_{q}=\frac{{\left[n\right]}_{q}!}{{\left[k\right]}_{q}!{\left[n-k\right]}_{q}!}\phantom{\rule{1em}{0ex}}\left(n\ge k\ge 0\right).\hfill \end{array}$
The q-improper integrals are defined as (see )
${\int }_{0}^{\mathrm{\infty }/A}f\left(x\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}x=\left(1-q\right)\sum _{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(\frac{{q}^{n}}{A}\right)\frac{{q}^{n}}{A},\phantom{\rule{1em}{0ex}}A>0,$

provided the sums converge absolutely.

The q-beta integral is defined by
${B}_{q}\left(t;s\right)=K\left(A;t\right){\int }_{0}^{\mathrm{\infty }/A}\frac{{x}^{t-1}}{{\left(1+x\right)}_{q}^{t+s}}\phantom{\rule{0.2em}{0ex}}{d}_{q}x,$
(2)

where $K\left(x;t\right)=\frac{1}{x+1}{x}^{t}{\left(1+\frac{1}{x}\right)}_{q}^{t}{\left(1+x\right)}_{q}^{1-t}$ and ${\left(a+b\right)}_{q}^{\tau }={\prod }_{j=0}^{\tau -1}\left(a+{q}^{j}b\right)$, $\tau >0$.

In particular for any positive integer m, n,
$K\left(x,n\right)={q}^{\frac{n\left(n-1\right)}{2}},\phantom{\rule{2em}{0ex}}K\left(x,0\right)=1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{B}_{q}\left(m;n\right)=\frac{{\mathrm{\Gamma }}_{q}\left(m\right){\mathrm{\Gamma }}_{q}\left(n\right)}{{\mathrm{\Gamma }}_{q}\left(m+n\right)},$
(3)
where ${\mathrm{\Gamma }}_{q}\left(t\right)$ is the q-gamma function satisfying the following functional equations:
${\mathrm{\Gamma }}_{q}\left(t+1\right)={\left[t\right]}_{q}{\mathrm{\Gamma }}_{q}\left(t\right),\phantom{\rule{2em}{0ex}}{\mathrm{\Gamma }}_{q}\left(1\right)=1$

(see ).

For $f\in C\left[0,\mathrm{\infty }\right)$, $q\in \left(0,1\right)$ and $n\in \mathbb{N}$, we introduce a kind of q-gamma operators ${G}_{n,q}\left(f;x\right)$ as follows:
${G}_{n,q}\left(f;x\right)=\frac{{\left[2n+3\right]}_{q}!{\left({q}^{n+\frac{3}{2}}x\right)}^{n+3}{q}^{\frac{n\left(n+1\right)}{2}}}{{\left[n\right]}_{q}!{\left[n+2\right]}_{q}!}{\int }_{0}^{\mathrm{\infty }/A}\frac{{t}^{n}}{{\left({q}^{n+\frac{3}{2}}x+t\right)}_{q}^{2n+4}}f\left(t\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}t.$
(4)

Note that for $q\to {1}^{-}$, ${G}_{n,{1}^{-}}\left(f;x\right)$ become the gamma operators defined in (1).

## 2 Some preliminary results

In order to obtain the approximation properties of the operators ${G}_{n,q}$, we need the following lemmas.

Lemma 1 For any $k\in \mathbb{N}$, $k\le n+2$ and $q\in \left(0,1\right)$, we have
${G}_{n,q}\left({t}^{k};x\right)=\frac{{\left[n+k\right]}_{q}!{\left[n-k+2\right]}_{q}!}{{\left[n\right]}_{q}!{\left[n+2\right]}_{q}!}{q}^{\frac{2k-{k}^{2}}{2}}{x}^{k}.$
(5)
Proof Using the properties of q-beta integral, we have
$\begin{array}{rcl}{G}_{n,q}\left({t}^{k};x\right)& =& \frac{{\left[2n+3\right]}_{q}!{\left({q}^{n+\frac{3}{2}}x\right)}^{n+3}{q}^{\frac{n\left(n+1\right)}{2}}}{{\left[n\right]}_{q}!{\left[n+2\right]}_{q}!}{\int }_{0}^{\mathrm{\infty }/A}\frac{{t}^{n+k}}{{\left({q}^{n+\frac{3}{2}}x+t\right)}_{q}^{2n+4}}\phantom{\rule{0.2em}{0ex}}{d}_{q}t\\ =& \frac{{x}^{k}{\left[2n+3\right]}_{q}!{q}^{\frac{n\left(n+1\right)}{2}}{q}^{k\left(n+\frac{3}{2}\right)}}{{\left[n\right]}_{q}!{\left[n+2\right]}_{q}!}{\int }_{0}^{\mathrm{\infty }/A}\frac{{\left(\frac{t}{{q}^{n+\frac{3}{2}}x}\right)}^{n+k}}{{\left(1+\frac{t}{{q}^{n+\frac{3}{2}}x}\right)}_{q}^{2n+4}}\phantom{\rule{0.2em}{0ex}}{d}_{q}\left(\frac{t}{{q}^{n+\frac{3}{2}}x}\right)\\ =& \frac{{x}^{k}{\left[2n+3\right]}_{q}!{q}^{\frac{n\left(n+1\right)}{2}}{q}^{k\left(n+\frac{3}{2}\right)}}{{\left[n\right]}_{q}!{\left[n+2\right]}_{q}!}\frac{{B}_{q}\left(n+k+1;n-k+3\right)}{K\left(A;n+k+1\right)}\\ =& \frac{{\left[n+k\right]}_{q}!{\left[n-k+2\right]}_{q}!}{{\left[n\right]}_{q}!{\left[n+2\right]}_{q}!}{q}^{\frac{2k-{k}^{2}}{2}}{x}^{k}.\end{array}$

Lemma 1 is proved. □

Lemma 2 The following equalities hold:
${G}_{n,q}\left(1;x\right)=1,\phantom{\rule{2em}{0ex}}{G}_{n,q}\left(t;x\right)=\frac{\sqrt{q}{\left[n+1\right]}_{q}}{{\left[n+2\right]}_{q}}x,\phantom{\rule{2em}{0ex}}{G}_{n,q}\left({t}^{2};x\right)={x}^{2},$
(6)
${G}_{n,q}\left({t}^{3};x\right)=\frac{{\left[n+3\right]}_{q}}{{q}^{3/2}{\left[n\right]}_{q}}{x}^{3},\phantom{\rule{2em}{0ex}}{G}_{n,q}\left({t}^{4};x\right)=\frac{{\left[n+3\right]}_{q}{\left[n+4\right]}_{q}}{{q}^{4}{\left[n\right]}_{q}{\left[n-1\right]}_{q}}{x}^{4},$
(7)
${G}_{n,q}\left({\left(t-x\right)}^{2};x\right)=2{x}^{2}\left(1-\frac{\sqrt{q}{\left[n+1\right]}_{q}}{{\left[n+2\right]}_{q}}\right),$
(8)
${G}_{n,q}\left({\left(t-x\right)}^{4};x\right)=\left(1+\frac{{\left[n+3\right]}_{q}{\left[n+4\right]}_{q}}{{q}^{4}{\left[n\right]}_{q}{\left[n-1\right]}_{q}}-\frac{4{\left[n+3\right]}_{q}}{{q}^{3/2}{\left[n\right]}_{q}}-\frac{4\sqrt{q}{\left[n+1\right]}_{q}}{{\left[n+2\right]}_{q}}\right){x}^{4}.$
(9)

Proof From Lemma 1, taking $k=0,1,2,3,4$, we get (6) and (7). Since ${G}_{n,q}\left({\left(t-x\right)}^{2};x\right)={G}_{n,q}\left({t}^{2};x\right)-2x{G}_{n,q}\left(t;x\right)+{x}^{2}$ and ${G}_{n,q}\left({\left(t-x\right)}^{4};x\right)={G}_{n,q}\left({t}^{4};x\right)-4x{G}_{n,q}\left({t}^{3};x\right)+6{x}^{2}{G}_{n,q}\left({t}^{2};x\right)-4{x}^{3}{G}_{n,q}\left(t;x\right)+{x}^{4}$, using (6), (7), we obtain (8) and (9) easily. □

Remark 1 Note that for $q\to {1}^{-}$, from Lemma 2, we have
$\begin{array}{c}{G}_{n,{1}^{-}}\left(1;x\right)=1,\phantom{\rule{2em}{0ex}}{G}_{n,{1}^{-}}\left(t;x\right)=\frac{n+1}{n+2}x,\phantom{\rule{2em}{0ex}}{G}_{n,{1}^{-}}\left({t}^{2};x\right)={x}^{2},\hfill \\ {G}_{n,{1}^{-}}\left({\left(t-x\right)}^{2};x\right)=\frac{2}{n+2}{x}^{2},\hfill \end{array}$

which is the moments and central moments of the operators defined in (1).

## 3 Local approximation

In this section we establish direct and local approximation theorems in connection with the operators ${G}_{n,q}\left(f,x\right)$.

We denote the space of all real-valued continuous bounded functions f defined on the interval $\left[0,\mathrm{\infty }\right)$ by ${C}_{B}\left[0,\mathrm{\infty }\right)$. The norm $\parallel \cdot \parallel$ on the space ${C}_{B}\left[0,\mathrm{\infty }\right)$ is given by $\parallel f\parallel =sup\left\{|f\left(x\right)|:x\in \left[0,\mathrm{\infty }\right)\right\}$.

Further, let us consider Peetre’s K-functional
${K}_{2}\left(f,\delta \right)=\underset{g\in {W}^{2}}{inf}\left\{\parallel f-g\parallel +\delta \parallel {g}^{″}\parallel \right\},$

where $\delta >0$ and ${W}^{2}=\left\{g\in {C}_{B}\left[0,\mathrm{\infty }\right):{g}^{\prime },{g}^{″}\in {C}_{B}\left[0,\mathrm{\infty }\right)\right\}$.

For $f\in {C}_{B}\left[0,\mathrm{\infty }\right)$, the modulus of continuity of second order is defined by
${\omega }_{2}\left(f,\delta \right)=\underset{0
By [, p.177] there exists an absolute constant $C>0$ such that
${K}_{2}\left(f,\delta \right)\le C{\omega }_{2}\left(f,\sqrt{\delta }\right),\phantom{\rule{1em}{0ex}}\delta >0.$
(10)

Our first result is a direct local approximation theorem for the operators ${G}_{n,q}\left(f,x\right)$.

Theorem 1 For $q\in \left(0,1\right)$, $x\in \left[0,\mathrm{\infty }\right)$ and $f\in {C}_{B}\left[0,\mathrm{\infty }\right)$, we have
$|{G}_{n,q}\left(f;x\right)-f\left(x\right)|\le C{\omega }_{2}\left(f;\sqrt{{\alpha }_{n,q}\left(x\right)}\right)+\omega \left(f;{\beta }_{n,q}\left(x\right)\right),$
(11)
where C is a positive constant,
${\alpha }_{n,q}\left(x\right)=\left(\frac{3}{4}-\frac{3\sqrt{q}{\left[n+1\right]}_{q}}{4{\left[n+2\right]}_{q}}\right){x}^{2},\phantom{\rule{2em}{0ex}}{\beta }_{n,q}\left(x\right)=\left(1-\frac{\sqrt{q}{\left[n+1\right]}_{q}}{{\left[n+2\right]}_{q}}\right)x.$
(12)

Proof

Let us define the auxiliary operators
${\stackrel{˜}{G}}_{n,q}\left(f;x\right)={G}_{n,q}\left(f;x\right)-f\left(\frac{\sqrt{q}{\left[n+1\right]}_{q}}{{\left[n+2\right]}_{q}}x\right)+f\left(x\right),$
(13)
$x\in \left[0,\mathrm{\infty }\right)$
. The operators ${\stackrel{˜}{G}}_{n,q}\left(f;x\right)$ are linear and preserve the linear functions
${\stackrel{˜}{G}}_{n,q}\left(t-x;x\right)=0$
(14)

(see (6)).

Let $g\in {C}_{B}^{2}$. By Taylor’s expansion
$g\left(t\right)=g\left(x\right)+{g}^{\prime }\left(x\right)\left(t-x\right)+{\int }_{x}^{t}\left(t-u\right){g}^{″}\left(u\right)\phantom{\rule{0.2em}{0ex}}du,\phantom{\rule{1em}{0ex}}t\in \left[0,\mathrm{\infty }\right),$
and (14), we get
${\stackrel{˜}{G}}_{n,q}\left(g;x\right)=g\left(x\right)+{\stackrel{˜}{G}}_{n,q}\left({\int }_{x}^{t}\left(t-u\right){g}^{″}\left(u\right)\phantom{\rule{0.2em}{0ex}}du;x\right).$
Hence, by (13) and (8), we have
$\begin{array}{r}|{\stackrel{˜}{G}}_{n,q}\left(g;x\right)-g\left(x\right)|\\ \phantom{\rule{1em}{0ex}}\le |{G}_{n,q}\left({\int }_{x}^{t}\left(t-u\right){g}^{″}\left(u\right)\phantom{\rule{0.2em}{0ex}}du;x\right)|+|{\int }_{\frac{\sqrt{q}{\left[n+1\right]}_{q}}{{\left[n+2\right]}_{q}}x}^{x}\left(u-\frac{\sqrt{q}{\left[n+1\right]}_{q}}{{\left[n+2\right]}_{q}}x\right){g}^{″}\left(u\right)\phantom{\rule{0.2em}{0ex}}du|\\ \phantom{\rule{1em}{0ex}}\le {G}_{n,q}\left(|{\int }_{x}^{t}\left(t-u\right)|{g}^{″}\left(u\right)|\phantom{\rule{0.2em}{0ex}}du|;x\right)+{\int }_{\frac{\sqrt{q}{\left[n+1\right]}_{q}}{{\left[n+2\right]}_{q}}x}^{x}|u-\frac{\sqrt{q}{\left[n+1\right]}_{q}}{{\left[n+2\right]}_{q}}x||{g}^{″}\left(u\right)|\phantom{\rule{0.2em}{0ex}}du\\ \phantom{\rule{1em}{0ex}}\le \left[{G}_{n,q}\left({\left(t-x\right)}^{2};x\right)+{\left(1-\frac{\sqrt{q}{\left[n+1\right]}_{q}}{{\left[n+2\right]}_{q}}\right)}^{2}{x}^{2}\right]\parallel {g}^{″}\parallel \\ \phantom{\rule{1em}{0ex}}=\left(3-\frac{\sqrt{q}{\left[n+1\right]}_{q}}{{\left[n+2\right]}_{q}}\right)\left(1-\frac{\sqrt{q}{\left[n+1\right]}_{q}}{{\left[n+2\right]}_{q}}\right){x}^{2}\parallel {g}^{″}\parallel \\ \phantom{\rule{1em}{0ex}}\le 3\left(1-\frac{\sqrt{q}{\left[n+1\right]}_{q}}{{\left[n+2\right]}_{q}}\right){x}^{2}\parallel {g}^{″}\parallel .\end{array}$
On the other hand, by (13), (4) and (6), we have
$|{\stackrel{˜}{G}}_{n,q}\left(f;x\right)|\le |{G}_{n,q}\left(f;x\right)|+2\parallel f\parallel \le \parallel f\parallel {G}_{n,q}\left(1;x\right)+2\parallel f\parallel \le 3\parallel f\parallel .$
(15)
Now (13) and (15) imply
$\begin{array}{r}|{G}_{n,q}\left(f;x\right)-f\left(x\right)|\\ \phantom{\rule{1em}{0ex}}\le |{\stackrel{˜}{G}}_{n,q}\left(f-g;x\right)-\left(f-g\right)\left(x\right)|+|{\stackrel{˜}{G}}_{n,q}\left(g;x\right)-g\left(x\right)|+|f\left(\frac{\sqrt{q}{\left[n+1\right]}_{q}}{{\left[n+2\right]}_{q}}x\right)-f\left(x\right)|\\ \phantom{\rule{1em}{0ex}}\le 4\parallel f-g\parallel +3\left(1-\frac{\sqrt{q}{\left[n+1\right]}_{q}}{{\left[n+2\right]}_{q}}\right){x}^{2}\parallel {g}^{″}\parallel +\omega \left[f;\left(1-\frac{\sqrt{q}{\left[n+1\right]}_{q}}{{\left[n+2\right]}_{q}}\right)x\right].\end{array}$
Hence taking infimum on the right-hand side over all $g\in {W}^{2}$, we get
$|{G}_{n,q}\left(f;x\right)-f\left(x\right)|\le 4{K}_{2}\left[f;\left(\frac{3}{4}-\frac{3\sqrt{q}{\left[n+1\right]}_{q}}{4{\left[n+2\right]}_{q}}\right){x}^{2}\right]+\omega \left[f;\left(1-\frac{\sqrt{q}{\left[n+1\right]}_{q}}{{\left[n+2\right]}_{q}}\right)x\right].$
By (10), for every $q\in \left(0,1\right)$, we have
$|{G}_{n,q}\left(f;x\right)-f\left(x\right)|\le C{\omega }_{2}\left(f;\sqrt{{\alpha }_{n,q}\left(x\right)}\right)+\omega \left(f;{\beta }_{n,q}\left(x\right)\right),$

where ${\alpha }_{n,q}\left(x\right)$ and ${\beta }_{n,q}\left(x\right)$ are defined in (12). This completes the proof of Theorem 1. □

Remark 2 Let $q=\left\{{q}_{n}\right\}$ be a sequence satisfying $0<{q}_{n}<1$ and ${lim}_{n}{q}_{n}=1$, we have ${lim}_{n}{\alpha }_{n,{q}_{n}}=0$ and ${lim}_{n}{\beta }_{n,{q}_{n}}\left(x\right)=0$, these give us the pointwise rate of convergence of the operators ${G}_{n,{q}_{n}}\left(f;x\right)$ to $f\left(x\right)$.

## 4 Rate of convergence

Let ${B}_{{x}^{2}}\left[0,\mathrm{\infty }\right)$ 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. We denote the subspace of all continuous functions belonging to ${B}_{{x}^{2}}\left[0,\mathrm{\infty }\right)$ by ${C}_{{x}^{2}}\left[0,\mathrm{\infty }\right)$. 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 usual modulus of continuity of f on the closed interval $\left[0,a\right]$ ($a>0$) by
${\omega }_{a}\left(f,\delta \right)=\underset{|t-x|\le \delta }{sup}\underset{x,t\in \left[0,a\right]}{sup}|f\left(t\right)-f\left(x\right)|.$

Obviously, 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 the modulus of continuity on the finite interval $\left[0,a+1\right]\subset \left[0,\mathrm{\infty }\right)$, where $a>0$. Then we have
$\begin{array}{r}{\parallel {G}_{n,q}\left(f\right)-f\parallel }_{C\left[0,a\right]}\\ \phantom{\rule{1em}{0ex}}\le 12{M}_{f}{a}^{2}\left(1+{a}^{2}\right)\left(1-\frac{\sqrt{q}{\left[n+1\right]}_{q}}{{\left[n+2\right]}_{q}}\right)+2{\omega }_{a+1}\left(f,\sqrt{2}a\sqrt{1-\frac{\sqrt{q}{\left[n+1\right]}_{q}}{{\left[n+2\right]}_{q}}}\right).\end{array}$
(16)
Proof For $x\in \left[0,a\right]$ and $t>a+1$, we have
$|f\left(t\right)-f\left(x\right)|\le {M}_{f}\left(2+{x}^{2}+{t}^{2}\right)\le {M}_{f}\left[2+3{x}^{2}+2{\left(t-x\right)}^{2}\right],$
hence we obtain
$|f\left(t\right)-f\left(x\right)|\le 6{M}_{f}\left(1+{a}^{2}\right){\left(t-x\right)}^{2}.$
(17)
For $x\in \left[0,a\right]$ and $t\le a+1$, we have
$|f\left(t\right)-f\left(x\right)|\le {\omega }_{a+1}\left(f,|t-x|\right)\le \left(1+\frac{|t-x|}{\delta }\right){\omega }_{a+1}\left(f,\delta \right),\phantom{\rule{1em}{0ex}}\delta >0.$
(18)
From (17) and (18), we get
$|f\left(t\right)-f\left(x\right)|\le 6{M}_{f}\left(1+{a}^{2}\right){\left(t-x\right)}^{2}+\left(1+\frac{|t-x|}{\delta }\right){\omega }_{a+1}\left(f,\delta \right).$
(19)
For $x\in \left[0,a\right]$ and $t\ge 0$, by Schwarz’s inequality and Lemma 2, we have
$\begin{array}{r}|{G}_{n,q}\left(f,x\right)-f\left(x\right)|\\ \phantom{\rule{1em}{0ex}}\le {G}_{n,q}\left(|f\left(t\right)-f\left(x\right)|,x\right)\\ \phantom{\rule{1em}{0ex}}\le 6{M}_{f}\left(1+{a}^{2}\right){G}_{n,q}\left({\left(t-x\right)}^{2},x\right)+{\omega }_{a+1}\left(f,\delta \right)\left(1+\frac{1}{\delta }\sqrt{{G}_{n,q}\left({\left(t-x\right)}^{2},x\right)}\right)\\ \phantom{\rule{1em}{0ex}}\le 12{M}_{f}{a}^{2}\left(1+{a}^{2}\right)\left(1-\frac{\sqrt{q}{\left[n+1\right]}_{q}}{{\left[n+2\right]}_{q}}\right)+{\omega }_{a+1}\left(f,\delta \right)\left(1+\frac{\sqrt{2}a}{\delta }\sqrt{1-\frac{\sqrt{q}{\left[n+1\right]}_{q}}{{\left[n+2\right]}_{q}}}\right).\end{array}$

By taking $\delta =\sqrt{2}a\sqrt{1-\frac{\sqrt{q}{\left[n+1\right]}_{q}}{{\left[n+2\right]}_{q}}}$, we get the assertion of Theorem 2. □

## 5 Weighted approximation and Voronovskaya-type asymptotic formula

Now we will discuss the weighted approximation theorem.

Theorem 3 Let the sequence $q=\left\{{q}_{n}\right\}$ satisfy $0<{q}_{n}<1$ and ${q}_{n}\to 1$ as $n\to \mathrm{\infty }$, for $f\in {C}_{{x}^{2}}^{\ast }\left[0,\mathrm{\infty }\right)$, we have
$\underset{n\to \mathrm{\infty }}{lim}{\parallel {G}_{n,{q}_{n}}\left(f\right)-f\parallel }_{{x}^{2}}=0.$
(20)
Proof By using the Korovkin theorem in , we see that it is sufficient to verify the following three conditions.
$\underset{n\to \mathrm{\infty }}{lim}{\parallel {G}_{n,{q}_{n}}\left({t}^{v};x\right)-{x}^{v}\parallel }_{{x}^{2}},\phantom{\rule{1em}{0ex}}v=0,1,2.$
(21)

Since ${G}_{n,{q}_{n}}\left(1;x\right)=1$ and ${G}_{n,{q}_{n}}\left({t}^{2};x\right)={x}^{2}$, (20) holds true for $v=0$ and $v=2$.

Finally, for $v=1$, we have
$\begin{array}{rcl}{\parallel {G}_{n,{q}_{n}}\left(t;x\right)-x\parallel }_{{x}^{2}}& =& \underset{x\in \left[0,\mathrm{\infty }\right)}{sup}\frac{|{G}_{n,{q}_{n}}\left(t;x\right)-x|}{1+{x}^{2}}\\ =& \left(1-\frac{\sqrt{{q}_{n}}{\left[n+1\right]}_{{q}_{n}}}{{\left[n+2\right]}_{{q}_{n}}}\right)\underset{x\in \left[0,\mathrm{\infty }\right)}{sup}\frac{x}{1+{x}^{2}}\\ \le & 1-\frac{\sqrt{{q}_{n}}{\left[n+1\right]}_{{q}_{n}}}{{\left[n+2\right]}_{{q}_{n}}},\end{array}$

since ${lim}_{n\to \mathrm{\infty }}{q}_{n}=1$, we get ${lim}_{n\to \mathrm{\infty }}\frac{\sqrt{q}{\left[n+1\right]}_{q}}{{\left[n+2\right]}_{q}}=1$, so the condition of (21) holds for $v=1$ as $n\to \mathrm{\infty }$, then the proof of Theorem 3 is completed. □

Finally, we give a Voronovskaya-type asymptotic formula for ${G}_{n,q}\left(f;x\right)$ by means of the second and fourth central moments.

Theorem 4 Let $q:=\left\{{q}_{n}\right\}$ be a sequence satisfying $0<{q}_{n}<1$, ${lim}_{n}{q}_{n}=1$ and ${lim}_{n}{q}_{n}^{n}=1$. For $f\in {C}_{{x}^{2}}^{2}\left[0,\mathrm{\infty }\right)$, ($f\left(x\right)$ is a twice differentiable function in $\left[0,\mathrm{\infty }\right)$), the following equality holds
$\underset{n\to \mathrm{\infty }}{lim}{\left[n+2\right]}_{q}\left({G}_{n,q}\left(f;x\right)-f\left(x\right)\right)=-{f}^{\prime }\left(x\right)x+{f}^{″}\left(x\right){x}^{2}$
(22)

for every $x\in \left[0,A\right]$, $A>0$.

Proof Let $x\in \left[0,\mathrm{\infty }\right)$ be fixed. By the Taylor formula, we may write
$f\left(t\right)=f\left(x\right)+{f}^{\prime }\left(x\right)\left(t-x\right)+\frac{1}{2}{f}^{″}\left(x\right){\left(t-x\right)}^{2}+r\left(t;x\right){\left(t-x\right)}^{2},$
(23)
where $r\left(t;x\right)$ is the Peano form of the remainder, $r\left(t;x\right)\in {C}_{{x}^{2}}\left[0,\mathrm{\infty }\right)$ and using L’Hopital’s rule, we have
$\begin{array}{rcl}\underset{t\to x}{lim}r\left(t;x\right)& =& \underset{t\to x}{lim}\frac{f\left(t\right)-f\left(x\right)-{f}^{\prime }\left(x\right)\left(t-x\right)-\frac{1}{2}{f}^{″}\left(x\right){\left(t-x\right)}^{2}}{{\left(t-x\right)}^{2}}\\ =& \underset{t\to x}{lim}\frac{{f}^{\prime }\left(t\right)-{f}^{\prime }\left(x\right)-{f}^{″}\left(x\right)\left(t-x\right)}{2\left(t-x\right)}=\underset{t\to x}{lim}\frac{{f}^{″}\left(t\right)-{f}^{″}\left(x\right)}{2}=0.\end{array}$
Applying ${G}_{n,q}\left(f;x\right)$ to (23), we obtain
$\begin{array}{rl}{\left[n+2\right]}_{q}\left({G}_{n,q}\left(f;x\right)-f\left(x\right)\right)=& {f}^{\prime }\left(x\right){\left[n+2\right]}_{q}{G}_{n,q}\left(t-x;x\right)+\frac{{f}^{″}\left(x\right)}{2}{\left[n+2\right]}_{q}{G}_{n,q}\left({\left(t-x\right)}^{2};x\right)\\ +{\left[n+2\right]}_{q}{G}_{n,q}\left(r\left(t;x\right){\left(t-x\right)}^{2};x\right).\end{array}$
By the Cauchy-Schwarz inequality, we have
${G}_{n,q}\left(r\left(t;x\right){\left(t-x\right)}^{2};x\right)\le \sqrt{{G}_{n,q}\left({r}^{2}\left(t;x\right);x\right)}\sqrt{{G}_{n,q}\left({\left(t-x\right)}^{4};x\right)}.$
(24)
Since ${r}^{2}\left(x;x\right)=0$, then it follows from Theorem 3 that
$\underset{n\to \mathrm{\infty }}{lim}{G}_{n,q}\left({r}^{2}\left(t;x\right);x\right)={r}^{2}\left(x;x\right)=0.$
(25)
Now, from (24), (25) and Lemma 2, we get immediately
$\underset{n\to \mathrm{\infty }}{lim}{\left[n+2\right]}_{q}{G}_{n,q}\left(r\left(t;x\right){\left(t-x\right)}^{2};x\right)=0,\phantom{\rule{2em}{0ex}}\underset{n\to \mathrm{\infty }}{lim}{\left[n+2\right]}_{q}{G}_{n,q}\left(t-x;x\right)=-x,$
and since ${q}^{n+1}={\left[n+2\right]}_{q}-{\left[n+1\right]}_{q}\le {\left[n+2\right]}_{q}-\sqrt{q}{\left[n+1\right]}_{q}\le {\left[n+2\right]}_{q}-q{\left[n+1\right]}_{q}=1$, we have
$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim}{\left[n+2\right]}_{q}{G}_{n,q}\left({\left(t-x\right)}^{2};x\right)& =& \underset{n\to \mathrm{\infty }}{lim}{\left[n+2\right]}_{q}\left(1-\frac{\sqrt{q}{\left[n+1\right]}_{q}}{{\left[n+2\right]}_{q}}\right)2{x}^{2}\\ =& \underset{n\to \mathrm{\infty }}{lim}\left({\left[n+2\right]}_{q}-\sqrt{q}{\left[n+1\right]}_{q}\right)2{x}^{2}=2{x}^{2}.\end{array}$

Theorem 4 is proved. □

## Declarations

### Acknowledgements

This work is supported by the National Natural Science Foundation of China (Grant No. 61170324), the Natural Science Foundation of Fujian Province of China (Grant No. 2010J01012) and the Project of the Educational Office of Fujian Province of China (Grant No. JK2011041).

## Authors’ Affiliations

(1)
School of Mathematics and Computer Science, Quanzhou Normal University, Quanzhou, 362000, P.R. China
(2)
Department of Mathematics, Xiamen University, Xiamen, 361005, P.R. China

## References

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