Approximation properties of q-Baskakov-Durrmeyer-Stancu operators

Abstract

Purpose

The purpose of this paper is to introduce and study the Baskakov- Durrmeyer- Stancu operators based on q-integers.

Methods

First we use property of q-calculus to estimate moments of these operators. Also study some approximation properties, asymptotic formula including q-derivative and point-wise estimation for the operators${\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}$.

Results

We studied better error estimations for these operators using King type approach.

Conclusions

The results proposed here are new and have a better rate of convergence.

MSC 2000

Primary 41A 25,41A 35.

Introduction

For f ∈ C[ 0,∞), a new type of Baskakov-Durrmeyer operators studied by Finta[1] is defined as

$${\mathcal{ℒ}}_n(f,x)=\sum _{k=1}^{\infty }{p}_{n,k}(x){\int }_0^{\infty }{b}_{n,k}(t)f(t)\mathit{\text{dt}}+p_{n,0}(x)f(0),$$

(1)

where$p_{n,k}(x)=\left(\genfrac{}{}{0ex}{}{n+k-1}{k}\right)\frac{x^k}{{(1+x)}^{n+k}}$ and$b_{n,k}(x)=\frac{1}{B(k,n+1)}\frac{x^{(k-1)}}{{(1+x)}^{n+k+1}}.$

Very recently in[2], Gupta introduce q analogue of (1) which is defined as

$$\begin{array}{l}{\mathcal{ℒ}}_n^q(f,x)=\sum _{k=1}^{\infty }{p}_{n,k}^q(x){\int }_0^{\infty /A}{q}^{-k}{b}_{n,k}^q(t)f(t)d_{q}t+p_{n,0}^q(x)f(0),\end{array}$$

(2)

where$p_{n,k}^q(x)={\left(\genfrac{}{}{0ex}{}{n+k-1}{k}\right)}_q{q}^{\frac{k^2}{2}}\frac{{(\mathit{\text{qx}})}^k}{{(1+\mathit{\text{qx}})}_q^{n+k}}$ and$b_{n,k}^q(x)=\frac{1}{B_q(k,n+1)}{q}^{\frac{k^2}{2}}\frac{x^{k-1}}{{(1+x)}_q^{n+k+1}}.$

Recently, Govil and Gupta[3] studied some approximation properties for the operators defined in (1) and estimated local result in terms of modulus of continuity. Also, further properties like point-wise convergence, asymptotic formula and inverse result in simultaneous approximation have been established in[4].

Starting with two parameters α, β satisfying the condition 0 ≤ α ≤ β in 1983, the generalization of Stancu operators was given in[5] and studied the linear positive operators$S_n^{\alpha ,\beta }:C[0,1]\to C[0,1]$ defined for any f ∈ C [ 0,1] as follows:

$$S_n^{\alpha }(f,x)=\sum _{k=0}^n{p}_{n,k}(x)f\left(\frac{k+\alpha }{n+\beta }\right),0\le x\le 1.$$

(3)

where$p_{n,k}(x)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)x^k{(1-x)}^{n-k}$ is the Bernstein basis function.

In 2010, Ibrahim[6] introduced Stancu-Chlodowsky Polynomials and investigated convergence and approximation properties of these operators. Motivated by such type operators Verma et al.[7] introduce the Baskakov-Durrmeyer-Stancu operators${\mathcal{ℒ}}_n^{(\alpha ,\beta )}$, which is a Stancu-type generalization of the Baskakov-Durrmeyer operators (1) as follows:

$$\begin{array}{ll}{\mathcal{ℒ}}_n^{(\alpha ,\beta )}(f,x)=& \sum _{k=1}^{\infty }{p}_{n,k}(x){\int }_0^{\infty }{b}_{n,k}(t)f\left(\frac{\mathit{\text{nt}}+\alpha }{n+\beta }\right)\mathit{\text{dt}}\\ +p_{n,0}(x)f\left(\frac{\alpha }{n+\beta }\right),\end{array}$$

(4)

where the Baskakov and beta basis functions are given in (1).

During last decade, q-calculus was extensively used for constructing various generalizations of many classical approximation operators. In 1987, Lapuş[8] introduced q-Bernstein polynomial whose approximation properties were studied in[9, 10]. The recent work on such type of operators can be found in[11–13].

A Lupaş-Phillips-type q-analog of the operators${\mathcal{ℒ}}_n^{(\alpha ,\beta )}$ is defined in (4) as follows:

$$\begin{array}{ll}{\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(f,x)=& \sum _{k=1}^{\infty }{p}_{n,k}^q(x){\int }_0^{\infty /A}{q}^{-k}{b}_{n,k}^q(t)f\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }\right)d_{q}t\\ +p_{n,0}^q(x)f\left(\frac{\alpha }{{\left[n\right]}_q+\beta }\right),\end{array}$$

(5)

where$p_{n,k}^q(x)$ and$b_{n,k}^q(x)$ are given in (2).

Notice that

$${\int }_o^{\infty /A}{q}^{-k}{b}_{n,k}^q(t)d_{q}t=q^{\frac{-k}{2}}\text{and}\sum _{k=0}^{\infty }{p}_{n,k}^q(x)q^{\frac{-k}{2}}=1.$$

The aim of this paper is to study the approximation properties of a new generalization of the Baskakov-Durrmeyer operators with two parameter α and β based on q-integers. We estimated moments for these operators and also studied the asymptotic formula based on q-derivative for the operators defined in (5). Finally, we give point-wise estimation and better error estimations for operators${\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}$ using King’s approach. First, we recall some definitions and notations of q-calculus. Such notations can be found in[14].

We consider q as a real number satisfying 0 < q < 1.

We have

$${\left[n\right]}_q=\left\{\begin{array}{ll}\frac{1-q^n}{1-q},& q\ne 1,\\ n,& q=1\end{array}\right.$$

and

$${\left[n\right]}_q!=\left\{\begin{array}{ll}{\left[n\right]}_q{[n-1]}_q{[n-2]}_q\dots {\left[1\right]}_q,& n=1,2,\dots ,\\ 1,& n=0\text{.}\end{array}\right.$$

Furthermore,

$$\begin{array}{ll}{(1+x)}_q^n& ={(-x;q)}_n\\ =\left\{\begin{array}{ll}(1+x)(1+\mathit{\text{qx}})(1+q^{2}x)\dots (1+q^{n-1}x),& n=1,2,\dots ,\\ 1,& n=0.\end{array}\right.\end{array}$$

Also, for any real number α, we have

$${(1+x)}_q^{\alpha }=\frac{{(1+x)}_q^{\infty }}{{(1+q^{\alpha }x)}_q^{\infty }}.$$

In special case when α is a whole number, this definition coincides with the above definition.

The q-binomial coefficients are given by

$${\left(\genfrac{}{}{0ex}{}{n}{k}\right)}_q=\frac{{\left[n\right]}_q!}{{\left[k\right]}_q!{[n-k]}_q!},0\le k\le \mathrm{n.}$$

The q-derivative D _q f of a function f is given by

$$D_q(f(x))=\frac{f(x)-f(\mathit{\text{qx}})}{(1-q)x}.$$

The q-Jackson integral and q-improper integral defined as

$${\int }_0^{a}f(x)d_{q}x=(1-q)a\sum _{n=0}^{\infty }f(a{q}^n)q^n$$

and

$${\int }_0^{\infty /A}f(x)d_{q}x=(1-q)a\sum _{n=0}^{\infty }f\left(\frac{q^n}{A}\right)\frac{q^n}{A},$$

provided sum converges absolutely.

De Sole and Kac[15] defined q-analog of beta function of second kind$B(t,s)={\int }_0^{\infty }\frac{x^{t-1}}{{(1+x)}^{t+s}}\mathit{\text{dx}}$ as follows:

$$B(t,s)=K(A,t){\int }_0^{\infty /A}\frac{x^{t-1}}{{(1+x)}_q^{t+s}}{d}_{q}x,$$

where$K(x,t)=\frac{1}{1+x}{x}^t{\left(1+\frac{1}{x}\right)}_q^t{\left(1+x\right)}_q^{1-t}$. This function is q-constant in x i.e. K(q x,t) = K(x,t).

In particular, for any positive integer n,

$$\begin{array}{ll}K(x,n)& =q^{\frac{n(n-1)}{2}},K(x,0)=1,\text{and}{B}_q(t,s)\\ =\frac{{[t-1]}_q!{[s-1]}_q!}{{[t+s-1]}_q!}.\end{array}$$

Moment Estimation

Remark 1.

Applying the product rule for differentiation, we easily obtain the relation

$$\begin{array}{l}{q}^k{{\varphi }^{\ast }}^2(x)D_q\left(p_{n,k}^q(x)\right)=\left({\left[k\right]}_q-q^{k+1}{\left[n\right]}_{q}x\right)p_{n,k}^q(\mathit{\text{qx}}),\\ q^{k-1}{\varphi }^2(x)D_q\left(b_{n,k}^q(x)\right)=\left({[k-1]}_q-q^{k-1}{[n+2]}_{q}x\right)b_{n,k}^q(\mathit{\text{qx}}),\end{array}$$

where ϕ^∗2(x) = x(1 + q x) and ϕ²(x) = x(1 + x).

Lemmma 1.

[2] The following hold:

1. ${\mathcal{ℒ}}_n^q(1,x)=1,$

2. ${\mathcal{ℒ}}_n^q(t,x)=x,$

3. ${\mathcal{ℒ}}_n^q(t^2,x)=\frac{q{\left[n\right]}_q+1}{q^2{[n-1]}_q}{x}^2+\frac{{\left[2\right]}_q}{q^2{[n-1]}_q}x,\mathit{\text{for}}n>1.$

Lemma 2.

Let us define$S_{n,m}(x)={\mathcal{ℒ}}_n^q(e_m,x),e_m=t^m,m=0,1,2,\dots$. Then, we have

$$\begin{array}{ll}{S}_{n,m+1}(\mathit{\text{qx}})& =\frac{{{\varphi }^{\ast }}^2(x)D_q(S_{n,m}(x))+\{{\left[m\right]}_q+q{\left[n\right]}_{q}x\}{S}_{n,m}(\mathit{\text{qx}})}{q^m{[n-m]}_q},\\ n& \ge m+2.\end{array}$$

From this recurrence relation, we obtain${\mathcal{ℒ}}_n^q({(t-x)}^m,x)=O\left(\frac{1}{{\left[n\right]}_q^{\left[\frac{m+1}{2}\right]}}\right)$, where [α] denotes the integer part of α.

Proof.

Using remark 1, we obtain

$$\begin{array}{ll}{{\varphi }^{\ast }}^2& (x)D_q(S_{n,m}(x))\\ =\sum _{k=1}^{\infty }\left({{\varphi }^{\ast }}^2(x)\right)D_q(p_{n,k}^q(x)){\int }_0^{\infty /A}{q}^{-k}{b}_{n,k}^q(t)t^m{d}_{q}t\\ =\sum _{k=1}^{\infty }\left\{q^{-k}{\left[k\right]}_q-q{\left[n\right]}_{q}x\right\}{p}_{n,k}^q(\mathit{\text{qx}}){\int }_0^{\infty /A}{q}^{-k}{b}_{n,k}^q(t)t^m{d}_{q}t\\ =\sum _{k=1}^{\infty }{q}^{-k}{\left[k\right]}_q{p}_{n,k}^q(\mathit{\text{qx}}){\int }_0^{\infty /A}{q}^{-k}{b}_{n,k}^q(t)t^m{d}_{q}t-q{\left[n\right]}_{q}x{S}_{n,m}(\mathit{\text{qx}})\\ =\sum _{k=1}^{\infty }\left\{q^{-k}{[k-1]}_q+q^{-1}\right\}{p}_{n,k}^q(\mathit{\text{qx}}){\int }_0^{\infty /A}{q}^{-k}{b}_{n,k}^q(t)t^m{d}_{q}t\\ -q{\left[n\right]}_{q}x{S}_{n,m}(\mathit{\text{qx}})\\ =\sum _{k=1}^{\infty }{q}^{-k}{[k-1]}_q{p}_{n,k}^q(\mathit{\text{qx}}){\int }_0^{\infty /A}{q}^{-k}{b}_{n,k}^q(t)t^m{d}_{q}t\\ +(q^{-1}-q{\left[n\right]}_{q}x)S_{n,m}(\mathit{\text{qx}})\\ =q^{-2}\sum _{k=1}^{\infty }{p}_{n,k}^q(\mathit{\text{qx}}){\int }_0^{\infty /A}\left\{q^{-k+2}{[k-1]}_q-{[n+2]}_{q}t+{[n+2]}_{q}t\right\}\\ \times q^{-k}{b}_{n,k}^q(t)t^m{d}_{q}t+(q^{-1}-q{\left[n\right]}_{q}x)S_{n,m}(\mathit{\text{qx}})\\ =q^{-2}\sum _{k=1}^{\infty }{p}_{n,k}^q(\mathit{\text{qx}}){\int }_0^{\infty /A}\left\{q^{-k+2}{[k-1]}_q-{[n+2]}_{q}t\right\}\\ \times q^{-k}{b}_{n,k}^q(t)t^m{d}_{q}t+q^{-2}{[n+2]}_q{S}_{n,m+1}(\mathit{\text{qx}})\\ +(q^{-1}-q{\left[n\right]}_{q}x)S_{n,m}(\mathit{\text{qx}}).\end{array}$$

To simplify the integral, we make use of the chain rule (which is applicable only for this particular transformation) for the transformation t = q z, which gives d _q t = q d _q z (see page 3-4,[14]). Thus, in view of above remark, we get

$$\begin{array}{ll}{{\varphi }^{\ast }}^2(x)& D_q(S_{n,m}(x))\\ =q^m\sum _{k=1}^{\infty }{p}_{n,k}^q(\mathit{\text{qx}}){\int }_0^{\infty /A}\left\{q^{-k+1}{[k-1]}_q-{[n+2]}_{q}z\right\}\\ \times q^{-k}{b}_{n,k}^q(\mathit{\text{qz}})z^m{d}_{q}z+q^{-2}{[n+2]}_q{S}_{n,m+1}(\mathit{\text{qx}})\\ +(q^{-1}-q{\left[n\right]}_{q}x)S_{n,m}(\mathit{\text{qx}})\\ =q^m\sum _{k=1}^{\infty }{p}_{n,k}^q(\mathit{\text{qx}}){\int }_0^{\infty /A}{q}^{-k}(z^{m+1}+z^{m+2})D_q\left(b_{n,k}^q(z)\right)d_{q}z\\ +q^{-2}{[n+2]}_q{S}_{n,m+1}(\mathit{\text{qx}})+(q^{-1}-q{\left[n\right]}_{q}x)S_{n,m}(\mathit{\text{qx}})\\ =I_1+I_2+q^{-2}{[n+2]}_q{S}_{n,m+1}(\mathit{\text{qx}})+(q^{-1}-q{\left[n\right]}_{q}x)\\ \times S_{n,m}(\mathit{\text{qx}}).\end{array}$$

In order to obtain I₁ and I₂ we make use of the q integration by parts

$$\begin{array}{l}{\int }_a^{b}u(t)D_q(v(t))d_{q}t={\left[u(t)v(t)\right]}_a^b-{\int }_a^{b}v(\mathit{\text{qt}})D_q(u(t))d_q\mathrm{t.}\end{array}$$

Therefore, we get I₁ = -q^-1[ m+1] _q S_n,m(q x) and I₂ = -q^-2[ m + 2] _q S_n,m+1(q x). Combining the expression, we have

$$\begin{array}{l}{S}_{n,m+1}(\mathit{\text{qx}})=\frac{{{\varphi }^{\ast }}^2(x)D_q(S_{n,m}(x))+\{{\left[m\right]}_q+q{\left[n\right]}_{q}x\}{S}_{n,m}(\mathit{\text{qx}})}{q^m{[n-m]}_q}.\end{array}$$

□

Lemma 3.

The following hold:

1. ${\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(1,x)=1$

2. ${\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(t,x)=\frac{{\left[n\right]}_{q}x+\alpha }{{\left[n\right]}_q+\beta }$

3. $\begin{array}{l}{\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(t^2,x)=\frac{{\left[n\right]}_q^2(q{\left[n\right]}_q+1)}{q^2{[n-1]}_q{({\left[n\right]}_q+\beta )}^2}{x}^2\\ +\left(\frac{{\left[n\right]}_q^2{\left[2\right]}_q}{q^2{[n-1]}_q{({\left[n\right]}_q+\beta )}^2}+\frac{2{\left[n\right]}_q\alpha }{{({\left[n\right]}_q+\beta )}^2}\right)x\\ +\frac{{\alpha }^2}{{({\left[n\right]}_q+\beta )}^2},\mathit{\text{for}}n>1\end{array}$

Proof.

Using Lemma 1, for every n > 1 and x ∈ [ 0,∞), we have

$$\begin{array}{ll}{\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(1,x)& ={\mathcal{ℒ}}_n^q(1,x)=1,\\ {\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(t,x)& =\frac{{\left[n\right]}_q}{{\left[n\right]}_q+\beta }{\mathcal{ℒ}}_n^q(t,x)+\frac{\alpha }{{\left[n\right]}_q+\beta }{\mathcal{ℒ}}_n^q(1,x)\\ =\frac{{\left[n\right]}_{q}x+\alpha }{{\left[n\right]}_q+\beta }.\end{array}$$

Finally,

$$\begin{array}{ll}{\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(t^2,x)& =\frac{{\left[n\right]}_q^2}{{({\left[n\right]}_q+\beta )}^2}{\mathcal{ℒ}}_n^q(t^2,x)+\frac{2{\left[n\right]}_q\alpha }{{\left[n\right]}_q+{\beta }^2}{\mathcal{ℒ}}_n^q(t,x)\\ +\frac{{\alpha }^2}{{\left[n\right]}_q+{\beta }^2}{\mathcal{ℒ}}_n^q(1,x)\\ =\frac{{\left[n\right]}_q^2(q{\left[n\right]}_q+1)}{q^2{[n-1]}_q{({\left[n\right]}_q+\beta )}^2}{x}^2\\ +\left(\frac{{\left[n\right]}_q^2{\left[2\right]}_q}{q^2{[n-1]}_q{({\left[n\right]}_q+\beta )}^2}+\frac{2{\left[n\right]}_q\alpha }{{({\left[n\right]}_q+\beta )}^2}\right)x\\ +\frac{{\alpha }^2}{{({\left[n\right]}_q+\beta )}^2}.\end{array}$$

□

Remark 2.

If we put q = 1 and α =β = 0, we get the moments of Baskakov-Durrmeyer operators (1) as${\mathcal{ℒ}}_n(t,x)=x$ and${\mathcal{ℒ}}_n(t^2,x)=\frac{(n+1)x^2+2x}{n-1}.$

Lemma 4.

If we define the central moments as${\mu }_{n,m}^q(x)={\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}({(t-x)}^m,x)$, m ∈ N. Then

$$\begin{array}{ll}{\mu }_{n,1}^q(x)& ={\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(t-x,x)=\frac{\alpha -\beta x}{{\left[n\right]}_q+\beta },\\ {\mu }_{n,2}^q(x)& ={\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}({(t-x)}^2,x)\\ =\left(\frac{{\left[n\right]}_q^2(q{\left[n\right]}_q+1)}{q^2{({\left[n\right]}_q+\beta )}^2{[n-1]}_q}+1-\frac{2{\left[n\right]}_q}{{\left[n\right]}_q+\beta }\right)x^2\\ +\left(\frac{{\left[2\right]}_q{\left[n\right]}_q^2}{q^2{({\left[n\right]}_q+\beta )}^2{[n-1]}_q}+\frac{2{\left[n\right]}_q\alpha }{{({\left[n\right]}_q+\beta )}^2}-\frac{2\alpha }{{\left[n\right]}_q+\beta }\right)x\\ +\frac{{\alpha }^2}{{({\left[n\right]}_q+\beta )}^2}.\end{array}$$

Proof.

Notice that

$$\begin{array}{ll}{\mu }_{n,1}^q(x)& ={\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}((t-x),x)\\ ={\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(t,x)-x{\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(1,x)=\frac{{\left[n\right]}_{q}x+\alpha }{{\left[n\right]}_q+\beta }-x\\ =\frac{\alpha -\beta x}{{\left[n\right]}_q+\beta }.\\ {\mu }_{n,2}^q(x)& ={\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}({(t-x)}^2,x)={\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(t^2,x)-2x{\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(t,x)\\ +x^2{\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(t,x)\\ =\frac{{\left[n\right]}_q^2(q{\left[n\right]}_q+1)}{q^2{[n-1]}_q{({\left[n\right]}_q+\beta )}^2}{x}^2\\ +\left(\frac{{\left[n\right]}_q^2(1+q)}{q^2{[n-1]}_q{({\left[n\right]}_q+\beta )}^2}+\frac{2{\left[n\right]}_q\alpha }{{({\left[n\right]}_q+\beta )}^2}\right)x\\ +\frac{{\alpha }^2}{{({\left[n\right]}_q+\beta )}^2}-2x\left(\frac{{\left[n\right]}_{q}x+\alpha }{{\left[n\right]}_q+\beta }\right)+x^2\\ =\left(\frac{{\left[n\right]}_q^2(q{\left[n\right]}_q+1)}{q^2{({\left[n\right]}_q+\beta )}^2{[n-1]}_q}+1-\frac{2{\left[n\right]}_q}{{\left[n\right]}_q+\beta }\right)x^2\\ +\left(\frac{{\left[2\right]}_q{\left[n\right]}_q^2}{q^2{({\left[n\right]}_q+\beta )}^2{[n-1]}_q}+\frac{2{\left[n\right]}_q\alpha }{{({\left[n\right]}_q+\beta )}^2}-\frac{2\alpha }{{\left[n\right]}_q+\beta }\right)x\\ +\frac{{\alpha }^2}{{({\left[n\right]}_q+\beta )}^2}.\end{array}$$

□

Remmark 3.

For all m ∈ N,0 ≤ α ≤ β; we have the following recursive relation for the images of the monomials t^m under${\mathcal{ℒ}}_{n,q}^{\alpha ,\beta }(t^m,x)$ in terms of${\mathcal{ℒ}}_n^q(t^j,x),j=0,1,2,\dots ,m$ as

$${\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(t^m,x)=\sum _{j=0}^m\left(\genfrac{}{}{0ex}{}{m}{j}\right)\frac{{\left[n\right]}_q^j{\alpha }^{m-j}}{{({\left[n\right]}_q+\beta )}^m}{\mathcal{ℒ}}_n^q(t^j,x).$$

Also, we have

$$\begin{array}{ll}{\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}({(t-x)}^m,x)& =\sum _{k=0}^m\left(\genfrac{}{}{0ex}{}{m}{k}\right){(-x)}^{m-k}{\mathcal{ℒ}}_{n,q}^{\alpha ,\beta }(t^k,x).\\ =\sum _{k=0}^m\left(\genfrac{}{}{0ex}{}{m}{k}\right){(-x)}^{m-k}\sum _{j=0}^k\left(\genfrac{}{}{0ex}{}{k}{j}\right)\frac{{\left[n\right]}_q^j{\alpha }^{k-j}}{{({\left[n\right]}_q+\beta )}^k}\\ \times {\mathcal{ℒ}}_n^q(t^j,x).\end{array}$$

By Lemma 2 and above equality, we have${\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}({(t-x)}^m,x)=O\left(\frac{1}{{\left[n\right]}_q^{\left[\frac{m+1}{2}\right]}}\right)$, where [α] denotes the integer part of α.

Direct results

Let the space C _B [ 0,∞) of all continuous and bounded functions be endowed with the norm ∥ f∥ = sup{|f(x)| : x ∈ [ 0,∞)}. Further let us consider the following K-functional:

$$K_2(f,\delta )=\underset{g\in W^2}{\text{inf}}\{\parallel f-g\parallel +\delta \parallel g^{\prime \prime }\parallel \},$$

(6)

where δ > 0 and W² = {g ∈ C _B [ 0,∞) : g^′,g^′′ ∈ C _B [ 0,∞)}.

By the method as given in p. 177, Theorem 2.4 of[16], there exists an absolute constant C > 0 such that

$$K_2(f,\delta )\le C{\omega }_2(f,\sqrt{\delta }),$$

(7)

where

$${\omega }_2(f,\sqrt{\delta })=\underset{0<h<\sqrt{\delta }}{\text{sup}}\underset{x\in [0,\infty )}{\text{sup}}|f(x+2h)-2f(x+h)+f(x)|$$

(8)

is the second order modulus of smoothness of f ∈ C _B [ 0,∞). Also, we set

$$\omega (f,\sqrt{\delta })=\underset{0<h<\sqrt{\delta }}{\text{sup}}\underset{x\in [0,\infty )}{\text{sup}}|f(x+h)-f(x)|.$$

(9)

In what follows, we shall use notation$\varphi (x)=\sqrt{x(1+x)}$, where x ∈ [ 0,∞).

Theorem 1.

Let f ∈ C _B [ 0,∞) and 0 < q < 1. Then for all x ∈ [ 0,∞), there exists an absolute constant M > 0 such that

$$\begin{array}{l}\left|{\mathcal{ℒ}}_{n,q}^{\alpha ,\beta }(f,x)-f(x)\right|\le M{\omega }_2\left(f,\sqrt{{\mu }_{n,2}^q+{({\mu }_{n,1}^q)}^2}\right)+\omega (f,{\mu }_{n,1}^q).\end{array}$$

(10)

Proof.

Let g ∈ W² and x,t ∈ [ 0,∞). By Taylor’s expansion, we have

$$g(t)=g(x)+g^{\prime }(x)(t-x)+{\int }_x^t(t-u)g^{\prime \prime }(u)\mathrm{du.}$$

(11)

Let us define auxiliary operators${\widetilde{\mathcal{ℒ}}}_{n,q}^{(\alpha ,\beta )}$ as follows

$${\widetilde{\mathcal{ℒ}}}_{n,q}^{(\alpha ,\beta )}(f,x)={\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(f,x)+f(x)-f\left(x+\frac{\alpha -\beta x}{{\left[n\right]}_q+\beta }\right).$$

(12)

Now, we have

$${\widetilde{\mathcal{ℒ}}}_{n,q}^{(\alpha ,\beta )}(t-x,x)=0.$$

Applying${\widetilde{\mathcal{ℒ}}}_{n,q}^{\alpha ,\beta }$ on both side of (11), we get

$$\begin{array}{ll}{\widetilde{\mathcal{ℒ}}}_{n,q}^{(\alpha ,\beta )}(g,x)-g(x)& =g^{\prime }(x){\widetilde{\mathcal{ℒ}}}_{n,q}^{(\alpha ,\beta )}((t-x),x)+{\widetilde{\mathcal{ℒ}}}_{n,q}^{(\alpha ,\beta )}\\ \times \left({\int }_x^t(t-u)g^{\prime \prime }(u)\mathit{\text{du}},x\right)\\ ={\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}\left({\int }_x^t(t-u)g^{\prime \prime }(u)\mathit{\text{du}},x\right)\\ +{\int }_x^{x+\frac{\alpha -\beta x}{{\left[n\right]}_q+\beta }}\left(x+\frac{\alpha -\beta x}{{\left[n\right]}_q+\beta }-u\right)g^{\prime \prime }(u)\mathit{\text{du}}.\end{array}$$

On the other hand, we obtain$\left|{\int }_x^t(x-u)g^{\prime \prime }(u)\mathit{\text{du}}\right|\le \parallel g^{\prime \prime }\parallel {(t-x)}^2$ and

$$\begin{array}{ll}{\int }_x^{x+\frac{\alpha -\beta x}{{\left[n\right]}_q+\beta }}\left(x+\frac{\alpha -\beta x}{{\left[n\right]}_q+\beta }-u\right)g^{\prime \prime }(u)\mathit{\text{du}}& \le {\left(x+\frac{\alpha -\beta x}{{\left[n\right]}_q+\beta }-x\right)}^2\parallel g^{\prime \prime }\parallel \\ =\parallel g^{\prime \prime }\parallel {\left({\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(t-x,x)\right)}^2\\ =\parallel g^{\prime \prime }\parallel {\left({\mu }_{n,1}^q(x)\right)}^2.\end{array}$$

Notice that,

$$\begin{array}{ll}\left|{\widetilde{\mathcal{ℒ}}}_{n,q}^{(\alpha ,\beta )}(g,x)-g(x)\right|& \le \left|{\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}\left({\int }_x^t(t-u)g^{\prime \prime }(u)\mathit{\text{du}},x\right)\right|\\ +\left|{\int }_x^{x+\frac{\alpha -\beta x}{{\left[n\right]}_q+\beta }}\left(x+\frac{\alpha -\beta x}{{\left[n\right]}_q+\beta }-u\right)g^{\prime \prime }(u)\mathit{\text{du}}\right|\\ \le \parallel g^{\prime \prime }\parallel {\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}\left({(t-x)}^2,x\right)+\parallel g^{\prime \prime }\parallel {\left({\mu }_{n,1}^q(x)\right)}^2\\ \le \parallel g^{\prime \prime }\parallel \left[{\mu }_{n,2}^q(x)+{\left({\mu }_{n,1}^q(x)\right)}^2\right].\end{array}$$

Observe that

$$\begin{array}{ll}\left|{\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(f,x)-f(x)\right|& \le \left|{\widetilde{\mathcal{ℒ}}}_{n,q}^{(\alpha ,\beta )}(f-g,x)-(f-g)(x)\right|\\ +\left|{\widetilde{\mathcal{ℒ}}}_{n,q}^{(\alpha ,\beta )}(g,x)-g(x)\right|\\ +\left|f(x)-f\left(x+\frac{\alpha -\beta x}{{\left[n\right]}_q+\beta }\right)\right|\\ \le 4\parallel f-g\parallel +\parallel g^{\prime \prime }\parallel \left[{\mu }_{n,2}^q(x)+{\left({\mu }_{n,1}^q(x)\right)}^2\right]\\ +\omega (f,{\mu }_{n,1}^q(x)).\end{array}$$

Now, taking infimum on the right-hand side over all$g\in C_B^2[0,\infty )$ and from (7), we get

$$\begin{array}{ll}\left|{\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(f,x)-f(x)\right|& \le 4K_2\left(f,{\mu }_{n,2}^q(x)+{\left({\mu }_{n,1}^q(x)\right)}^2\right)\\ +\omega (f,{\mu }_{n,1}^q(x))\\ \le M{\omega }_2\left(f,\sqrt{{\mu }_{n,2}^q(x)+{\left({\mu }_{n,1}^q(x)\right)}^2}\right)\\ +\omega (f,{\mu }_{n,1}^q(x)).\end{array}$$

This complete the proof of Theorem 1. □

Central moments and asymptotic formula

In this section, we observe that it is not possible to estimate recurrence formula${\mathcal{ℒ}}_{n,q_n}^{(\alpha ,\beta )}$ in q calculus; however, there may be some techniques, but at the moment it can be considered as an open problem. Here we establish the recurrence relation for the central moments and obtain asymptotic formula.

Let$B_{x^2}[0,\infty )=\{f:\text{for every}x\in [0,\infty ),|f(x)|\le M_f(1+x^2)\},M_f$ being a constant depending on f. By$C_{x^2}[0,\infty )$, we denote the subspace of all continuous functions belonging to$B_{x^2}[0,\infty )$. Also,$C_{x^2}^{\ast }[0,\infty )$ is subspace of all functions$f\in C_{x^2}[0,\infty )$, for which$\underset{x\to \infty }{\text{lim}}\frac{f(x)}{1+x^2}$ is finite. The norm on$C_{x^2}^{\ast }[0,\infty )$ is

$$\parallel f{\parallel }_{x^2}=\underset{x\in [0,\infty )}{\text{sup}}\frac{|f(x)|}{1+x^2}.$$

Lemma 5.

If we define the central moments as

$$\begin{array}{ll}{T}_{n,m}(x)=& {\mathcal{ℒ}}_{n,q_n}^{(\alpha ,\beta )}({(t-x)}_q^m,x)=\sum _{k=1}^{\infty }{p}_{n,k}^q(x){\int }_0^{\infty }{q}^{-k}{b}_{n,k}^q(t)\\ \times {\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^m{d}_{q}t+p_{n,0}^q(x){\left(\frac{\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^m,\end{array}$$

then${\mathcal{ℒ}}_{n,q_n}^{(\alpha ,\beta )}(1,x)=1$,${\mathcal{ℒ}}_{n,q_n}^{(\alpha ,\beta )}(t-x,x)=\frac{\alpha -\beta x}{{\left[n\right]}_q+\beta }$and for n > m + 2, we have the following recurrence relation:

$$\begin{array}{ll}\frac{{\left[n\right]}_q+\beta }{{\left[n\right]}_q}& \left(q^{-3}\frac{({\left[n\right]}_q+\beta ){[m+2]}_q}{{\left[n\right]}_q}-q^{-1}{[n+2]}_q\right)T_{n,m+1}(\mathit{\text{qx}})\\ =-x(1+\mathit{\text{qx}})\left[D_q(T_{n,m}(x))+{\left[m\right]}_q{T}_{n,m-1}(\mathit{\text{qx}})\right]\\ +\left[-q^{-3}\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\left\{\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}{\left[3\right]}_q{q}^{m}x+q-\frac{2\alpha }{{\left[n\right]}_q}-x\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right\}\right.\\ \times {[m+1]}_q+q^{-1}-q{\left[n\right]}_{q}x+q^{-1}{[n+2]}_q\\ \left.\times \left\{{\left[2\right]}_q{q}^{m}x-x-\frac{\alpha }{{\left[n\right]}_q+\beta }\right\}\right]T_{n,m}(\mathit{\text{qx}})+\left[-\left\{\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right.\right.\\ \times \left({\left[2\right]}_q{q}^{m}x\left(\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}{\left[3\right]}_q{q}^{m}x+q-\frac{2\alpha }{{\left[n\right]}_q}-x\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right)\right.\\ -\left.{\left[3\right]}_q{q}^{2m+1}{x}^2\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right)-\left(x\left(q-\frac{2\alpha }{{\left[n\right]}_q}\right)\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right.\\ +\left.\left.\frac{{\alpha }^2+\mathrm{q\alpha }{\left[n\right]}_q}{{\left[n\right]}_q^2}\right)\right\}{q}^{-3}{\left[m\right]}_q+\left(1-q{\left[n\right]}_{q}x+q^{-1}{[n+2]}_q\right.\\ \times \left.\left.\left(q^{m}x-\frac{\alpha }{{\left[n\right]}_q+\beta }\right)\right)(q^m-1)x\right]T_{n,m-1}(\mathit{\text{qx}})+\left[\left\{q^{m}x\right.\right.\\ \times \left\{\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\left({\left[2\right]}_q{q}^{m}x\left(\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}{\left[3\right]}_q{q}^{m}x+q-\frac{2\alpha }{{\left[n\right]}_q}\right.\right.\right.\\ -\left.\left.x\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right)-{\left[3\right]}_q{q}^{2m+1}{x}^2\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right)\\ -\left.\left(x\left(q-\frac{2\alpha }{{\left[n\right]}_q}\right)\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}+\frac{{\alpha }^2+\mathrm{q\alpha }{\left[n\right]}_q}{{\left[n\right]}_q^2}\right)\right\}\\ -\left\{q^{2m+1}{x}^2\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\left(q^{m+2}x\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}+q-\frac{2\alpha }{{\left[n\right]}_q}-x\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right)\right.\\ -\left.\left.\left.x\frac{{\alpha }^2+\mathrm{q\alpha }{\left[n\right]}_q}{{\left[n\right]}_q^2}\right\}\right\}{q}^{-3}{[m-1]}_q\right]T_{q,m-2}(\mathit{\text{qx}})\\ +{(1+q^{2}x)}_q^{-n}\left\{\frac{{\left[n\right]}_q+\beta }{{\left[n\right]}_q}\left(q^{-3}\frac{({\left[n\right]}_q+\beta ){[m+2]}_q}{{\left[n\right]}_q}-q^{-1}[n+2]\right)\right.\\ \times {\left(\frac{\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^{m+1}+\left[q^{-3}\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\left\{\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}{\left[3\right]}_q{q}^{m}x\right.\right.\\ +\left.q-\frac{2\alpha }{{\left[n\right]}_q}-x\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right\}{[m+1]}_q-q^{-1}-q^{-1}{[n+2]}_q\\ \times \left.\left\{{\left[2\right]}_q{q}^{m}x-x-\frac{\alpha }{{\left[n\right]}_q+\beta }\right\}\right]{\left(\frac{\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^m\\ +\left[\left\{\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\left({\left[2\right]}_q{q}^{m}x\left(\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}{\left[3\right]}_q{q}^{m}x+q-\frac{2\alpha }{{\left[n\right]}_q}\right.\right.\right.\right.\\ -\left.\left.x\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right)-{\left[3\right]}_q{q}^{2m+1}{x}^2\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right)-\left(x\left(q-\frac{2\alpha }{{\left[n\right]}_q}\right)\right.\\ \times \left.\left.\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}+\frac{{\alpha }^2+\mathrm{q\alpha }{\left[n\right]}_q}{{\left[n\right]}_q^2}\right)\right\}{q}^{-3}{\left[m\right]}_q-\left(1+q^{-1}{[n+2]}_q\right.\\ \times \left.\left.\left(q^{m}x\frac{\alpha }{{\left[n\right]}_q+\beta }\right)\right)(q^m-1)x\right]{\left(\frac{\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^{m-1}\\ +\left[\left\{q^{m}x\left\{\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\left({\left[2\right]}_q{q}^{m}x\right.\right.\right.\right.\\ \times \left(\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}{\left[3\right]}_q{q}^{m}x+q-\frac{2\alpha }{{\left[n\right]}_q}-x\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right)\\ -\left.{\left[3\right]}_q{q}^{2m+1}{x}^2\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right)-\left(x\left(q-\frac{2\alpha }{{\left[n\right]}_q}\right)\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right.\\ +\left.\left.\frac{{\alpha }^2+\mathrm{q\alpha }{\left[n\right]}_q}{{\left[n\right]}_q^2}\right)\right\}-\left\{q^{2m+1}{x}^2\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\left(q^{m+2}x\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right.\right.\\ +\left.\left.\left.q-\frac{2\alpha }{{\left[n\right]}_q}-x\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right)-x\frac{{\alpha }^2+\mathrm{q\alpha }{\left[n\right]}_q}{{\left[n\right]}_q^2}\right\}\right\}\\ \times \left.\left.q^{-3}{[m-1]}_q\right]{\left(\frac{\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^{m-2}\right\}.\end{array}$$

Proof.

Applying q derivatives of product rule, we have

$$\begin{array}{ll}{D}_q(T_{n,m}(x))=& -{\left[m\right]}_q\sum _{k=1}^{\infty }{p}_{n,k}^q(\mathit{\text{qx}}){\int }_0^{\infty }{q}^{-k}{b}_{n,k}^q(t)\\ \times {\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^{m-1}{d}_{q}t-{\left[m\right]}_q{p}_{n,0}^q(\mathit{\text{qx}})\\ \times {\left(\frac{\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^{m-1}\\ +\sum _{k=1}^{\infty }{D}_q(p_{n,k}^q(x)){\int }_0^{\infty }{q}^{-k}{b}_{n,k}^q(t){\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^m\\ \times d_t+D_q(p_{n,0}^q(x)){\left(\frac{\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^m.\end{array}$$

Using the identity

$$q^{k}x(1+\mathit{\text{qx}})D_q\left(p_{n,k}^q(x)\right)=\left({\left[k\right]}_q-q^{k+1}{\left[n\right]}_{q}x\right)p_{n,k}^q(\mathit{\text{qx}}),$$

$$\begin{array}{l}x(1+\mathit{\text{qx}})\left[D_q(T_{n,m}(x))+{\left[m\right]}_q{T}_{n,m-1}(\mathit{\text{qx}})\right]\\ =\sum _{k=1}^{\infty }x(1+\mathit{\text{qx}})D_q(p_{n,k}^q(x)){\int }_0^{\infty }{q}^{-k}{b}_{n,k}^q(t)\\ \times {\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^m\mathit{\text{dt}}+x(1+\mathit{\text{qx}})D_q(p_{n,0}^q(x))\\ \times {\left(\frac{\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^m\\ =\sum _{k=1}^{\infty }\left(q^{-k}{\left[k\right]}_q-q{\left[n\right]}_{q}x\right)p_{n,k}^q(\mathit{\text{qx}}){\int }_0^{\infty }{q}^{-k}{b}_{n,k}^q(t)\\ \times {\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^m\mathit{\text{dt}}+\left(-q{\left[n\right]}_{q}x\right)p_{n,0}^q(\mathit{\text{qx}})\\ \times {\left(\frac{\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^m,\\ =\left[\sum _{k=1}^{\infty }{q}^{-k}{\left[k\right]}_q{p}_{n,k}^q(\mathit{\text{qx}}){\int }_0^{\infty }{q}^{-k}{b}_{n,k}^q(t){\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^m{d}_{q}t\right]\\ -q{\left[n\right]}_{q}x\left[\sum _{k=1}^{\infty }{p}_{n,k}^q(\mathit{\text{qx}}){\int }_0^{\infty }{q}^{-k}{b}_{n,k}^q(t){\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^m\right.\\ \times \left.d_{q}t+p_{n,0}^q(\mathit{\text{qx}}){\left(\frac{\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^m\right],\\ :=I_1-q{\left[n\right]}_{q}x{I}_2.\end{array}$$

(13)

We can write I₁ as

$$\begin{array}{ll}{I}_1=& \sum _{k=1}^{\infty }{q}^{-k}\left({[k-1]}_q+q^{k-1}\right)p_{n,k}^q(\mathit{\text{qx}}){\int }_0^{\infty }{q}^{-k}{b}_{n,k}^q(t)\\ \times {\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^m{d}_{q}t\\ =& \sum _{k=1}^{\infty }{q}^{-k}{p}_{n,k}^q(\mathit{\text{qx}}){\int }_0^{\infty }{q}^{-k}\left({[k-1]}_q-q^{k-1}{[n+2]}_{q}t\right.\\ +q^{k-1}{[n+2]}_{q}t+\left.q^{k-1}\right)b_{n,k}^q(t){\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^m{d}_{q}t\\ =& \sum _{k=1}^{\infty }{q}^{-k}{p}_{n,k}^q(\mathit{\text{qx}}){\int }_0^{\infty }{q}^{-k}\left({[k-1]}_q-q^{k-1}{[n+2]}_{q}t\right)\\ \times b_{n,k}^q(t){\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^m{d}_{q}t\\ +\sum _{k=1}^{\infty }{q}^{-1}{p}_{n,k}^q(\mathit{\text{qx}}){\int }_0^{\infty }{q}^{-k}\left({[n+2]}_{q}t+1\right)b_{n,k}^q(t)\\ \times {\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^m{d}_q\mathrm{t.}\end{array}$$

Now using the identity

$$\begin{array}{l}{q}^{k-1}x(x+1)D_q\left(b_{n,k}^q(x)\right)\\ =\left({[k-1]}_q-q^{k-1}{[n+2]}_{q}x\right)b_{n,k}^q(\mathit{\text{qx}}),\end{array}$$

we have

$$\begin{array}{ll}{I}_1& =\sum _{k=1}^{\infty }{q}^{-k}{p}_{n,k}^q(\mathit{\text{qx}}){\int }_0^{\infty }{q}^{-k}{q}^{k-1}\frac{t}{q}\left(1+\frac{t}{q}\right)\\ \times D_q\left(b_{n,k}^q\left(\frac{t}{q}\right)\right){\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^m{d}_{q}t\\ +\sum _{k=1}^{\infty }{q}^{-1}{p}_{n,k}^q(\mathit{\text{qx}}){\int }_0^{\infty }{q}^{-k}\left({[n+2]}_{q}t+1\right)b_{n,k}^q(t)\\ \times {\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^m{d}_{q}t\\ =\sum _{k=1}^{\infty }{q}^{-3}{p}_{n,k}^q(\mathit{\text{qx}}){\int }_0^{\infty }{q}^{-k}\left(\mathit{\text{tq}}+t^2\right)D_q\left(b_{n,k}^q\left(\frac{t}{q}\right)\right)\\ \times {\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^m{d}_{q}t+{[n+2]}_q\sum _{k=1}^{\infty }{q}^{-1}{p}_{n,k}^q(\mathit{\text{qx}})\\ \times {\int }_0^{\infty }{q}^{-k}t{b}_{n,k}^q(t){\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^m{d}_{q}t\\ +\sum _{k=1}^{\infty }{q}^{-1}{p}_{n,k}^q(\mathit{\text{qx}}){\int }_0^{\infty }{q}^{-k}{b}_{n,k}^q(t){\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^m{d}_{q}t\\ =I_3+I_4+I_5.\end{array}$$

(14)

Notice that

$$\begin{array}{l}\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-q^{m}x\right)\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-q^{m+1}x\right)\\ ={\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }\right)}^2-{\left[2\right]}_q{q}^{m}x\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }\right)+q^{2m+1}{x}^2\end{array}$$

and

$$\begin{array}{l}\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-q^{m}x\right)\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-q^{m+1}x\right)\\ \times \left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-q^{m+2}x\right)\\ =& {\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }\right)}^3-{\left[3\right]}_q{q}^{m}x{\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }\right)}^2\\ +{\left[3\right]}_q{q}^{2m+1}{x}^2\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }\right)-q^{3m+3}{x}^3.\end{array}$$

We obtain the following identity after some computation

$$\begin{array}{l}(\mathit{\text{tq}}+t^2){\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^m\\ =(\mathit{\text{tq}}+t^2)\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-x\right){\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^{m-1}\\ =\left(\frac{{({\left[n\right]}_q+\beta )}^2}{{\left[n\right]}_q^2}{\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }\right)}^2+\left(q-\frac{2\alpha }{{\left[n\right]}_q}\right)\right.\\ \times \left.\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }\right)-\frac{{\alpha }^2+\mathrm{q\alpha }{\left[n\right]}_q}{{\left[n\right]}_q^2}\right)\\ \times \left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-x\right){\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^{m-1}\\ =\left\{\frac{{({\left[n\right]}_q+\beta )}^2}{{\left[n\right]}_q^2}{\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }\right)}^3+\left(q-\frac{2\alpha }{{\left[n\right]}_q}\right.\right.\\ -\left.x\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right)\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}{\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }\right)}^2\\ -\left(x\left(q-\frac{2\alpha }{{\left[n\right]}_q}\right)\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}+\frac{{\alpha }^2+\mathrm{q\alpha }{\left[n\right]}_q}{{\left[n\right]}_q^2}\right)\\ \times \left.\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }\right)+x\frac{{\alpha }^2+\mathrm{q\alpha }{\left[n\right]}_q}{{\left[n\right]}_q^2}\right\}{\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^{m-1}\\ =\frac{{({\left[n\right]}_q+\beta )}^2}{{\left[n\right]}_q^2}{\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^{m+2}\\ +\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\left\{\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}{\left[3\right]}_q{q}^{m}x+q-\frac{2\alpha }{{\left[n\right]}_q}\right.\\ -\left.x\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right\}{\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^{m+1}\\ +\left\{\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\left({\left[2\right]}_q{q}^{m}x\left(\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}{\left[3\right]}_q{q}^{m}x+q\right.\right.\right.\\ -\left.\left.\frac{2\alpha }{{\left[n\right]}_q}-x\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right)-{\left[3\right]}_q{q}^{2m+1}{x}^2\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right)\\ \left.-\left(x\left(q-\frac{2\alpha }{{\left[n\right]}_q}\right)\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}+\frac{{\alpha }^2+\mathrm{q\alpha }{\left[n\right]}_q}{{\left[n\right]}_q^2}\right)\right\}\\ \times {\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^m\\ +\left[q^{m}x\left\{\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\left({\left[2\right]}_q{q}^{m}x\left(\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}{\left[3\right]}_q{q}^{m}x\right.\right.\right.\right.\\ +\left.\left.q-\frac{2\alpha }{{\left[n\right]}_q}-x\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right)-{\left[3\right]}_q{q}^{2m+1}{x}^2\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right)\\ -\left.\left(x\left(q-\frac{2\alpha }{{\left[n\right]}_q}\right)\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}+\frac{{\alpha }^2+\mathrm{q\alpha }{\left[n\right]}_q}{{\left[n\right]}_q^2}\right)\right\}\\ -\left\{q^{2m+1}{x}^2\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\left(q^{m+2}x\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}+q-\frac{2\alpha }{{\left[n\right]}_q}\right.\right.\\ -\left.\left.\left.x\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right)-x\frac{{\alpha }^2+\mathrm{q\alpha }{\left[n\right]}_q}{{\left[n\right]}_q^2}\right\}\right]\\ \times {\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^{m-1}.\end{array}$$

Using the above identity and q integral by parts

$${\int }_a^{b}u(t)D_q[v(t)]d_{q}t={[u(t)v(t)]}_a^b-{\int }_a^{b}v(\mathit{\text{qt}})D_q[u(t)]d_{q}t,$$

we obtain

$$\begin{array}{l}{q}^3{I}_3=-{[m+2]}_q\frac{{({\left[n\right]}_q+\beta )}^2}{{\left[n\right]}_q^2}\left(T_{n,m+1}(\mathit{\text{qx}})-{(1+q^{2}x)}_q^{-n}\right.\\ \times \left.{\left(\frac{\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^{m+1}\right)-\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\\ \times \left\{\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}{\left[3\right]}_q{q}^{m}x+q-\frac{2\alpha }{{\left[n\right]}_q}-x\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right\}\\ \times {[m+1]}_q\left(T_{n,m}(\mathit{\text{qx}})-{(1+q^{2}x)}_q^{-n}{\left(\frac{\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^m\right)\\ -\left\{\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\left({\left[2\right]}_q{q}^{m}x\left(\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}{\left[3\right]}_q{q}^{m}x+q-\frac{2\alpha }{{\left[n\right]}_q}\right.\right.\right.\\ -\left.\left.x\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right)-{\left[3\right]}_q{q}^{2m+1}{x}^2\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right)\\ \left.-\left(x\left(q-\frac{2\alpha }{{\left[n\right]}_q}\right)\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}+\frac{{\alpha }^2+\mathrm{q\alpha }{\left[n\right]}_q}{{\left[n\right]}_q^2}\right)\right\}{\left[m\right]}_q\\ \times \left(T_{n,m-1}(\mathit{\text{qx}})-{(1+q^{2}x)}_q^{-n}{\left(\frac{\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^{m-1}\right)\\ -\left[q^{m}x\left\{\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\left({\left[2\right]}_q{q}^{m}x\left(\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}{\left[3\right]}_q{q}^{m}x+q\right.\right.\right.\right.\\ -\left.\left.\frac{2\alpha }{{\left[n\right]}_q}-x\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right)-{\left[3\right]}_q{q}^{2m+1}{x}^2\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right)\\ \left.-\left(x\left(q-\frac{2\alpha }{{\left[n\right]}_q}\right)\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}+\frac{{\alpha }^2+\mathrm{q\alpha }{\left[n\right]}_q}{{\left[n\right]}_q^2}\right)\right\}\\ -\left\{q^{2m+1}{x}^2\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\left(q^{m+2}x\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}+q-\frac{2\alpha }{{\left[n\right]}_q}\right.\right.\\ -\left.\left.\left.x\frac{({\left[n\right]}_q+\beta )}{{\left[n\right]}_q}\right)-x\frac{{\alpha }^2+\mathrm{q\alpha }{\left[n\right]}_q}{{\left[n\right]}_q^2}\right\}\right]{[m-1]}_q\\ \times \left(T_{n,m-2}(\mathit{\text{qx}})-{(1+q^{2}x)}_q^{-n}{\left(\frac{\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^{m-2}\right).\end{array}$$

(15)

Finally, using

$$\begin{array}{ll}\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }\right.& {\left.-x\right)}_q^m\\ =\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-x\right){\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^{m-1}\\ =\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-q^{m}x+q^{m}x-x\right)\\ \times {\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^{m-1}\\ ={\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^m+(q^m-1)x\\ \times {\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^{m-1},\end{array}$$

we get

$$\begin{array}{ll}{I}_5& =q^{-1}\sum _{k=1}^{\infty }{p}_{n,k}^q(\mathit{\text{qx}}){\int }_0^{\infty }{q}^{-k}{b}_{n,k}^q(t){\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^m{d}_{q}t\\ =q^{-1}\left\{T_{n,m}(\mathit{\text{qx}})-{(1+q^{2}x)}_q^{-n}{\left(\frac{\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^m\right.\\ +(q^m-1)x\left(T_{n,m-1}(\mathit{\text{qx}})-{(1+q^{2}x)}_q^{-n}\right.\\ \times \left.\left.{\left(\frac{\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^{m-1}\right)\right\}.\end{array}$$

(16)

Also,

$$-q{\left[n\right]}_{q}x{I}_2=-q{\left[n\right]}_{q}x\left\{T_{n,m}(\mathit{\text{qx}})+(q^m-1)x\left(T_{n,m-1}(\mathit{\text{qx}})\right)\right\}.$$

To estimate I₄ we use

$$\begin{array}{ll}t\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }\right.& {\left.-x\right)}_q^m\\ =\frac{{\left[n\right]}_q+\beta }{{\left[n\right]}_q}\left\{\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-q^{m}x+q^{m}x-\frac{\alpha }{{\left[n\right]}_q+\beta }\right\}\\ \times {\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^m\\ =\frac{{\left[n\right]}_q+\beta }{{\left[n\right]}_q}\left\{{\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^{m+1}\right.\\ +\left.\left(q^{m}x-\frac{\alpha }{{\left[n\right]}_q+\beta }\right){\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-x\right)}_q^m\right\}\\ =\frac{{\left[n\right]}_q+\beta }{{\left[n\right]}_q}\left\{{\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^{m+1}\right.\\ +(q^{m+1}-1)x{\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^m\\ +\left(q^{m}x-\frac{\alpha }{{\left[n\right]}_q+\beta }\right)\left[{\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^m\right.\\ +\left.\left.(q^m-1)x{\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^{m-1}\right]\right\}\\ =\frac{{\left[n\right]}_q+\beta }{{\left[n\right]}_q}\left\{{\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^{m+1}\right.\\ +\left({\left[2\right]}_q{q}^{m}x-x-\frac{\alpha }{{\left[n\right]}_q+\beta }\right){\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^m\\ +\left(q^{m}x-\frac{\alpha }{{\left[n\right]}_q+\beta }\right)(q^m-1)x\\ \times \left.{\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^{m-1}\right\}.\end{array}$$

Therefore,

$$\begin{array}{ll}{I}_4=& q^{-1}{[n+2]}_q\left\{\frac{{\left[n\right]}_q+\beta }{{\left[n\right]}_q}\left(T_{n,m+1}(\mathit{\text{qx}})-{(1+q^{2}x)}_q^{-n}\right.\right.\\ \times \left.{\left(\frac{\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^{m+1}\right)+\left({\left[2\right]}_q{q}^{m}x-x-\frac{\alpha }{{\left[n\right]}_q+\beta }\right)\\ \times \left(T_{n,m}(\mathit{\text{qx}})-{(1+q^{2}x)}_q^{-n}{\left(\frac{\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^m\right)\\ +\left(q^{m}x-\frac{\alpha }{{\left[n\right]}_q+\beta }\right)(q^m-1)x\left(T_{n,m-1}(\mathit{\text{qx}})\right.\\ -\left.\left.{(1+q^{2}x)}_q^{-n}{\left(\frac{\alpha }{{\left[n\right]}_q+\beta }-\mathit{\text{qx}}\right)}_q^{m-1}\right)\right\}.\end{array}$$

(17)

Combining (13) -(17), we get required result. □

Theorem 2.

Let f ∈ C [ 0,∞) be a bounded function and (q _n ) denote a sequence such that 0 < q _n  < 1 and q _n  → 1 as n → ∞. Then we have for a point x ∈ (0,∞)

$$\begin{array}{l}\underset{n\to \infty }{\text{lim}}{\left[n\right]}_q\left({\mathcal{ℒ}}_{n,q_n}^{(\alpha ,\beta )}(f,x)-f(x)\right)\\ =(\alpha -\beta x)D_{q_n}f(x)+x(1+x)D_{q_n}^{2}f(x).\end{array}$$

Proof.

Using q-Taylor’s expansion[15] of f, we can write

$$\begin{array}{l}f(t)-f(x)=(t-x)D_{q}f(x)+\frac{{(t-x)}_q^2}{{\left[2\right]}_q}{D}_q^{2}f(x)+{\Phi }_q(x,t){(t-x)}_q^2,\end{array}$$

for 0 < q < 1, where

$${\Phi }_q(x,t)=\left\{\begin{array}{ll}\frac{f(t)-f(x)-D_{q}f(x)(t-x)-\frac{1}{{\left[2\right]}_q}{D}_q^{2}f(x){(t-x)}_q^2}{{(t-x)}_q^2},& \text{if}x\ne t,\\ 0,& \text{if}x=\mathrm{t.}\end{array}\right.$$

(18)

We know that for n large enough

$$\underset{t\to x}{\text{lim}}{\Phi }_{q_n}(x,t)=0,$$

(19)

that is for any ϵ > 0, there exists a δ > 0 such that

$$|{\Phi }_{q_n}(x,t)|<ϵ$$

(20)

for |t - x| < δ and n sufficiently large. Using (18), we can write

$$\begin{array}{ll}{\mathcal{ℒ}}_{n,q_n}^{(\alpha ,\beta )}(f,x)-f(x)=& D_{q_n}f(x){\mathcal{ℒ}}_{n,q_n}^{(\alpha ,\beta )}({(t-x)}_{q_n},x)\\ +\frac{D_{q_n}^2}{{\left[2\right]}_q}{\mathcal{ℒ}}_{n,q_n}^{(\alpha ,\beta )}({(t-x)}_{q_n}^2,x)+E_n^{q_n}(x),\end{array}$$

where

$$E_n^{q_n}(x)={\mathcal{ℒ}}_{n,q_n}^{(\alpha ,\beta )}({\Phi }_{q_n}(t,x){(t-x)}_{q_n}^2,x).$$

We can easily see that$\underset{n\to \infty }{\text{lim}}{\left[n\right]}_{q_n}{\mathcal{ℒ}}_{n,q_n}^{(\alpha ,\beta )}({(t-x)}_{q_n},x)=\alpha -\beta x$ and

$$\underset{n\to \infty }{\text{lim}}{\left[n\right]}_{q_n}{\mathcal{ℒ}}_{n,q_n}^{(\alpha ,\beta )}({(t-x)}_{q_n}^2,x)=2x(1+x).$$

In order to complete the proof of the theorem, it is sufficient to show that$\underset{n\to \infty }{\text{lim}}{\left[n\right]}_{q_n}{E}_n^{q_n}(x)=0.$ We proceed as follows:

Let

$$R_{n,1}^{q_n}(x)={\left[n\right]}_{q_n}{\mathcal{ℒ}}_{n,q_n}^{(\alpha ,\beta )}({\Phi }_{q_n}(x,t){(t-x)}_{q_n}^2{\chi }_x(t),t)$$

and

$$R_{n,2}^{q_n}(x)={\left[n\right]}_{q_n}{\mathcal{ℒ}}_{n,q_n}^{(\alpha ,\beta )}({\Phi }_{q_n}(x,t){(t-x)}_{q_n}^2(1-{\chi }_x(t)),t),$$

so that

$${\left[n\right]}_{q_n}{E}_n^{q_n}(x)=R_{n,1}^{q_n}(x)+R_{n,2}^{q_n}(x),$$

where χ _x (t) is the characteristic function of the interval$\left\{\right.t:\left|\frac{{\left[n\right]}_{q_n}t+\alpha }{{\left[n\right]}_{q_n}+\beta }-x\right|<\delta \left\}\right.$.

It follows from (18)

$$R_{n,1}^{q_n}(x)<ϵ2x(x+1)\text{as}n\to \mathrm{\infty .}$$

If$\left|\frac{{\left[n\right]}_{q_n}t+\alpha }{{\left[n\right]}_{q_n}+\beta }-x\right|\ge \delta$, then$|{\Phi }_{q_n}(x,t)|\le \frac{M}{{\delta }^2}{\left(\frac{{\left[n\right]}_{q_n}t+\alpha }{{\left[n\right]}_{q_n}+\beta }-x\right)}^2,$ where M > 0 is a constant. Since

$$\begin{array}{ll}{\left(\frac{{\left[n\right]}_{q_n}t+\alpha }{{\left[n\right]}_{q_n}+\beta }-x\right)}^2& =\left(\frac{{\left[n\right]}_{q_n}t+\alpha }{{\left[n\right]}_{q_n}+\beta }-x\right)\left(\frac{{\left[n\right]}_{q_n}t+\alpha }{{\left[n\right]}_{q_n}+\beta }-x\right)\\ =\left(\frac{{\left[n\right]}_{q_n}t+\alpha }{{\left[n\right]}_{q_n}+\beta }-{q_n}^{2}x+{q_n}^{2}x-x\right)\\ \times \left(\frac{{\left[n\right]}_{q_n}t+\alpha }{{\left[n\right]}_{q_n}+\beta }-{q_n}^{3}x+{q_n}^{3}x-x\right)\\ =\left(\frac{{\left[n\right]}_{q_n}t+\alpha }{{\left[n\right]}_{q_n}+\beta }-{q_n}^{2}x\right)\left(\frac{{\left[n\right]}_{q_n}t+\alpha }{{\left[n\right]}_{q_n}+\beta }-{q_n}^{3}x\right)\\ +x({q_n}^3-1)\left(\frac{{\left[n\right]}_{q_n}t+\alpha }{{\left[n\right]}_{q_n}+\beta }-{q_n}^{2}x\right)\\ +x({q_n}^2-1)\left(\frac{{\left[n\right]}_{q_n}t+\alpha }{{\left[n\right]}_{q_n}+\beta }-{q_n}^{3}x\right)\\ +x^2({q_n}^2-1)({q_n}^3-1),\end{array}$$

we have

$$\begin{array}{ll}{R}_{n,2}^{q_n}(x)=& \frac{M}{{\delta }^2}{\left[n\right]}_{q_n}{\mathcal{ℒ}}_{n,q_n}^{(\alpha ,\beta )}({(t-x)}_{q_n}^4,t)+\frac{M}{{\delta }^2}x{\left[n\right]}_{q_n}\\ \times \left(|(q_n^3-1)+(q_n^2-1)|\right){\mathcal{ℒ}}_{n,q_n}^{(\alpha ,\beta )}({(t-x)}_{q_n}^3,t)\\ +\frac{M}{{\delta }^2}{x}^2{\left[n\right]}_{q_n}{\left(q_n^2-1\right)}^2{\mathcal{ℒ}}_{n,q_n}^{(\alpha ,\beta )}({(t-x)}_{q_n}^2,t)\end{array}$$

and

$$\begin{array}{ll}|R_{n,2}^{q_n}(x)|\le & \frac{M}{{\delta }^2}\left\{{\left[n\right]}_{q_n}{\mathcal{ℒ}}_{n,q_n}^{(\alpha ,\beta )}({(t-x)}_{q_n}^4,t)+x{\left[n\right]}_{q_n}\right.\\ \times \left(2-q_n^3-q_n^2\right){\mathcal{ℒ}}_{n,q_n}^{(\alpha ,\beta )}({(t-x)}_{q_n}^3,t)\\ \left.+x^2{\left[n\right]}_{q_n}{\left(q_n^2-1\right)}^2{\mathcal{ℒ}}_{n,q_n}^{(\alpha ,\beta )}({(t-x)}_{q_n}^2,t)\right\}.\end{array}$$

Using Lemma 5, we have

$$\begin{array}{ll}{\mathcal{ℒ}}_{n,q_n}^{(\alpha ,\beta )}({(t-x)}_{q_n}^4,t)\le & \frac{C_{1,x}}{{\left[n\right]}_{q_n}^2},{\mathcal{ℒ}}_{n,q_n}^{(\alpha ,\beta )}({(t-x)}_{q_n}^3,t)\le \frac{C_{2,x}}{{\left[n\right]}_{q_n}^2},\\ {\mathcal{ℒ}}_{n,q_n}^{(\alpha ,\beta )}({(t-x)}_{q_n}^2,t)\le & \frac{C_{3,x}}{{\left[n\right]}_{q_n}}.\end{array}$$

Thus, for n sufficiently large$R_{n,2}^{q_n}\to 0$. This completes the proof of theorem. □

Corollary 1.

[17]Let f ∈ C[ 0,∞) be a bounded function and (q _n ) denote a sequence such that 0 < q _n  < 1 and q _n  → 1 as n → ∞. Suppose that the first and second derivative f^′(x) and f^′′(x) exist at a point x ∈ [ 0,∞), we have

$$\begin{array}{l}\underset{n\to \infty }{\text{lim}}{\left[n\right]}_{q_n}({\mathcal{ℒ}}_{n,q_n}^{(\alpha ,\beta )}(f,x)-f(x))\\ =(\alpha -\beta x)f^{\prime }(x)+x(1+x)f^{\prime \prime }(x).\end{array}$$

Point-wise Estimation

Now, we establish some point-wise estimxates of the rate of convergence of the q-Baskakov-Durrmeyer-Stancu Operators. First we give the relationship between the local smoothness of f and local approximation.

We know that a function f ∈ C[ 0,∞) is Lip γ on D, γ ∈ (0,1], D ⊂ [ 0,∞) if it satisfies the condition

$$|f(t)-f(x)|\le M_f|t-x{|}^{\gamma },t\in [0,\infty )\text{and}x\in D,$$

(21)

where M _f  is a constant depending on γ and f.

Theorem 3.

Let f ∈ Lip γ, D ⊂ [ 0,∞) and γ∈[0,1]. We have

$$\begin{array}{ll}|{\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(f,x)-f(x)|\le & M_f\left({\left[{\mu }_{n,2}^q(x)\right]}^{\gamma /2}+2d^{\gamma }(x,D)\right),\\ x\in & [0,\infty )\end{array}$$

where d(x,D) represents the distance between x and D.

Proof.

For $x_0\in \overline{D}$, the closure of the set D in [ 0,∞), we have

$$|f(t)-f(x)|\le |f(t)-f(x_0)|+|f(x_0)-f(x)|,x\in [0,\infty ).$$

Using (21) we get

$$\begin{array}{ll}|{\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(f,x)-f(x)|& \le {\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(|f(t)-f(x_0)|,x)+|f(x_0)-f(x)|\\ \le M_f{\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(|t-x_0{|}^{\gamma },x)+M_f|x_0-x{|}^{\gamma }.\end{array}$$

(22)

Then by the Holder’s inequality with$p=\frac{2}{\gamma }$ and$\frac{1}{r}=1-\frac{1}{p}$, we have

$$\begin{array}{ll}{\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(|t-x_0{|}^{\gamma },x)\le & {\left({\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(|t-x_0{|}^2,x)\right)}^{\gamma /2}\\ \times {\left({\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(1,x)\right)}^{1-\gamma /2}.\end{array}$$

(23)

Also,${\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}$ is monotone,i.e.

$${\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(|t-x_0{|}^{\gamma },x)\le {\left({\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(|t-x{|}^2,x)\right)}^{\gamma /2}+|x_0-x{|}^{\gamma }.$$

(24)

Using (22)-(24), we have desired result. □

Now, we give local direct estimate for the q-Baskakov-Durrmeyer-Stancu operators using the Lipschitz type maximal function of order γ introduced by B. Lenze[18] as

$${\tilde{\omega }}_{\gamma }(f,x)=\underset{t\ne x,t\in [0,\infty )}{\text{sup}}\frac{|f(t)-f(x)|}{|t-x{|}^{\gamma }},x\in [0,\infty ),\gamma \in (0,1]$$

(25)

Theorem 4.

Let γ ∈ (0,1] and f ∈ C _B (0,∞). Then for all x ∈ [ 0,∞), we have

$$|{\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(f,x)-f(x)|\le {\tilde{\omega }}_{\gamma }(f,x){\left({\mu }_{n,2}^q\right)}^{\gamma /2}$$

(26)

Proof.

Form (25), we have

$$|f(t)-f(x)|\le {\tilde{\omega }}_{\gamma }(f,x)|t-x{|}^{\gamma }$$

and

$$\begin{array}{ll}\left|{\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(f,x)-f(x)\right|\le & {\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(|f(t)-f(x)|,x)\le {\tilde{\omega }}_{\gamma }(f,x)\\ \times {\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(|t-x{|}^{\gamma },x)\end{array}$$

Applying Holder’s inequality with$p:=\frac{2}{\gamma }$ and$\frac{1}{r}=1-\frac{1}{p}$, we have

$$\left|{\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(f,x)-f(x)\right|\le {\tilde{\omega }}_{\gamma }(f,x){({\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}(|t-x{|}^2,x))}^{\gamma /2},$$

which is required result. □

Better estimation

It is well known that the operators preserve constant as well as linear functions. To make the convergence faster, King[19] proposed an approach to modify the classical Bernstein polynomials so that this sequence preserves two test functions e₀ and e₂. After this, several researchers have studied that many approximating operators L possess these properties i.e., L(e _i ,x) = e _i (x), where e _i (x) = xⁱ(i = 0,1), like Bernstein, Baskakov, and Baskakov-Durrmeyer-Stancu-type operators.

As the operators${\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}$ introduced in (5) preserve only the constant functions, further modification of these operators is proposed to be made so that the modified operators preserve the constant as well as linear functions. For this purpose, the modification of${\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}$ will be as follows:

$$\begin{array}{ll}{{\mathcal{ℒ}}^{\ast }}_{n,q}^{(\alpha ,\beta )}(f,x)=& \sum _{k=1}^{\infty }{p}_{n,k}^q(r_n(x)){\int }_0^{\infty }{b}_{n,k}(t)f\left(\frac{{\left[n\right]}_{q}t+\alpha }{{\left[n\right]}_q+\beta }\right)\\ \times d_{q}t+p_{n,0}(r_n(x))f\left(\frac{\alpha }{{\left[n\right]}_q+\beta }\right),\end{array}$$

where$r_n(x)=\frac{({\left[n\right]}_q+\beta )x-\alpha }{{\left[n\right]}_q}$ and$x\in I_n=\left[\right.\frac{\alpha }{{\left[n\right]}_q+\beta },\infty \left)\right.$.

Lemma 6.

For each x ∈ I _n , we have

$$\begin{array}{ll}{{\mathcal{ℒ}}^{\ast }}_{n,q}^{(\alpha ,\beta )}(1,x)=& 1,{{\mathcal{ℒ}}^{\ast }}_{n,q}^{(\alpha ,\beta )}(t,x)=\frac{{\left[n\right]}_q{r}_n(x)+\alpha }{{\left[n\right]}_q+\beta },\\ {{\mathcal{ℒ}}^{\ast }}_{n,q}^{(\alpha ,\beta )}(t^2,x)=& \frac{{[n+1]}_q}{q^2{[n-1]}_q}{x}^2+\frac{{\left[2\right]}_q({\left[n\right]}_q-2\alpha )}{({\left[n\right]}_q+\beta ){[n-1]}_q}x\\ +\frac{{\left[2\right]}_q({\alpha }^2-{\left[n\right]}_q\alpha )}{{({\left[n\right]}_q+\beta )}^2{[n-1]}_q}.\end{array}$$

Lemma 7.

For x ∈ I _n , the following hold,

$$\begin{array}{ll}{\tilde{\mu }}_{n,1}^q(x)=& {{\mathcal{ℒ}}^{\ast }}_{n,q}^{(\alpha ,\beta )}(t-x,x)=0,\\ {\tilde{\mu }}_{n,2}^q(x)=& {{\mathcal{ℒ}}^{\ast }}_{n,q}^{(\alpha ,\beta )}({(t-x)}^2,x)=\frac{{\left[2\right]}_q}{q^2{[n-1]}_q}{x}^2\\ +\frac{{\left[2\right]}_q({\left[n\right]}_q-2\alpha )}{({\left[n\right]}_q+\beta ){[n-1]}_q}x+\frac{{\left[2\right]}_q({\alpha }^2-{\left[n\right]}_q\alpha )}{{({\left[n\right]}_q+\beta )}^2{[n-1]}_q}.\end{array}$$

Theorem 5.

Let f ∈ C _B (I _n ), x ∈ I _n  and 0 < q < 1. Then, for n > 1, there exist an absolute constant C > 0 such that

$$\left|{\mathcal{ℒ}}_{n,q}^{\ast (\alpha ,\beta )}(f,x)-f(x)\right|\le C{\omega }_2\left(f,\sqrt{{\tilde{\mu }}_{n,2}^q(x)}\right).$$

Proof.

Let g ∈ C _B (I _n ) and x,t ∈ I _n . By Taylor’s expansion, we have

$$g(t)=g(x)+(t-x)g^{\prime }(x)+{\int }_x^t(t-u)g^{\prime \prime }(u)\mathit{\text{du}}.$$

(27)

Applying${{\mathcal{ℒ}}^{\ast }}_{n,q}^{(\alpha ,\beta )}$, we get

$$\begin{array}{ll}{{\mathcal{ℒ}}^{\ast }}_{n,q}^{(\alpha ,\beta )}(g,x)-g(x)=& g^{\prime }(x){{\mathcal{ℒ}}^{\ast }}_{n,q}^{(\alpha ,\beta )}((t-x),x)+{{\mathcal{ℒ}}^{\ast }}_{n,q}^{(\alpha ,\beta )}\\ \times \left({\int }_x^t(t-u)g^{\prime \prime }(u)\mathit{\text{du}},x\right).\end{array}$$

Obviously, we have$\left|\right.{\int }_x^t(t-x)g^{\prime \prime }(u)\mathit{\text{du}}\left|\right.\le {(t-x)}^2\parallel g^{\prime \prime }\parallel$,

$$\begin{array}{ll}\left|{{\mathcal{ℒ}}^{\ast }}_{n,q}^{(\alpha ,\beta )}(g,x)-g(x)\right|& \le {{\mathcal{ℒ}}^{\ast }}_{n,q}^{(\alpha ,\beta )}({(t-x)}^2,x)\parallel g^{\prime \prime }\parallel \\ ={\tilde{\mu }}_{n,2}^q\parallel g^{\prime \prime }\parallel .\end{array}$$

Since$\left|{{\mathcal{ℒ}}^{\ast }}_{n,q}^{(\alpha ,\beta )}(f,x)\right|\le \parallel f\parallel$,

$$\begin{array}{ll}\left|{{\mathcal{ℒ}}^{\ast }}_{n,q}^{(\alpha ,\beta )}(f,x)-f(x)\right|& \le \left|{{\mathcal{ℒ}}^{\ast }}_{n,q}^{(\alpha ,\beta )}(f-g,x)-(f-g)(x)\right|\\ +\left|{{\mathcal{ℒ}}^{\ast }}_{n,q}^{(\alpha ,\beta )}(g,x)-g(x)\right|\\ \le 2\parallel f-g\parallel +{\tilde{\mu }}_{n,2}^q\parallel g^{\prime \prime }\parallel .\end{array}$$

Taking infimum overall g ∈ C²(I _n ), we obtain

$$\left|{{\mathcal{ℒ}}^{\ast }}_{n,q}^{(\alpha ,\beta )}(f,x)-f(x)\right|\le K_2(f,{\tilde{\mu }}_{n,2}^q).$$

In view of (7), we have

$$\left|{{\mathcal{ℒ}}^{\ast }}_{n,q}^{(\alpha ,\beta )}(f,x)-f(x)\right|\le C{\omega }_2\left(f,\sqrt{{\tilde{\mu }}_{n,2}^q}\right),$$

which proves the theorem. □

Theorem 6.

Assume that q _n  ∈ (0,1),q _n  → 1 as n → ∞. Then for any$f\in C_{x^2}^{\ast }(I_n)$ such as$f^{\prime },f^{\prime \prime }\in C_{x^2}^{\ast }(I_n)$, we have

$$\begin{array}{ll}\underset{n\to \infty }{\text{lim}}{\left[n\right]}_{q_n}\left[{{\mathcal{ℒ}}^{\ast }}_{n,q_n}^{(\alpha ,\beta )}(f,x)-f(x)\right]=& x(1+x)f^{\prime \prime }(x),\\ \text{for every}x\in I_n.\end{array}$$

Remark 4.

One can discuss rate of approximation in weighted spaces for the operators${\mathcal{ℒ}}_{n,q}^{(\alpha ,\beta )}$. we are omitting the details as it is similar to Theorem 3 and 4 in[20].

Authors’ information

Dr. VNM is an Assistant professor at the Sardar Vallabhbhai National Institute of Technology, Ichchhanath Mahadev Road, Surat (Gujarat), India and he is a very active researcher in various filed of Mathematics. PP is a research scholar at the Sardar Vallabhbhai National Institute of Technology, Ichchhanath Mahadev Road, Surat (Gujarat), India under the guidance of Dr. VNM.