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 $[0,\mathrm{\infty })$. 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 [1] introduced a new sequence of linear positive operators by modifying the well-known Baskakov operators having weight functions of Szasz basis function as

$${\mathcal{D}}_n(f,x)=n\sum _{k=1}^{\mathrm{\infty }}{p}_{n,k}(x){\int }_0^{\mathrm{\infty }}{s}_{n,k-1}(t)f(t)dt+p_{n,0}(x)f(0),x\in [0,\mathrm{\infty }),$$

(1.1)

where

$$p_{n,k}(x)=\left(\begin{array}{c}n+k-1\\ k\end{array}\right)\frac{x^k}{{(1+x)}^{n+k}},s_{n,k}(t)=e^{-nt}\frac{{(nt)}^k}{k!}.$$

It is observed in [1] 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 [14].

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 [5], Kim [10, 11], and Kim and Rim [9]. The application of q-calculus in approximation theory was initiated by Phillips [13], 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 [7] respectively. We recall some notations and concepts of q-calculus. All of the results can be found in [5] and [8]. In what follows, q is a real number satisfying $0<q<1$.

For $n\in \mathbb{N}$,

The q-binomial coefficients are given by

$$[\begin{array}{c}n\\ k\end{array}{]}_q=\frac{{[n]}_q!}{{[k]}_q!{[n-k]}_q!},0\le k\le n.$$

The q-Beta integral is defined by [12]

$${\mathrm{\Gamma }}_q(t)={\int }_0^{\frac{1}{1-q}}{x}^{t-1}{E}_q(-qx)d_{q}x,t>0,$$

(1.2)

which satisfies the following functional equation:

$${\mathrm{\Gamma }}_q(t+1)={[t]}_q{\mathrm{\Gamma }}_q(t),{\mathrm{\Gamma }}_q(1)=1.$$

For $f\in C[0,\mathrm{\infty })$$q>0$ and each positive integer n, the q-Baskakov operators [2] are defined as

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

(1.3)

where

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

Remark 1 The first three moments of the q-Baskakov operators are given by

As the operators ${\mathcal{D}}_n(f,x)$ 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

$${\mathcal{D}}_n^q(f,x)={[n]}_q\sum _{k=1}^{\mathrm{\infty }}{p}_{n,k}^q(x){\int }_0^{q/(1-q^n)}{q}^{-k}{s}_{n,k-1}^q(t)f\left(t{q}^{-k}\right)d_{q}t+p_{n,0}^q(x)f(0),$$

(1.4)

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

$$p_{n,k}^q(x)=[\begin{array}{c}n+k-1\\ k\end{array}{]}_q{q}^{\frac{k(k-1)}{2}}\frac{x^k}{{(1+x)}_q^{n+k}},s_{n,k}^q(t)=E_q(-{[n]}_{q}t)\frac{{({[n]}_{q}t)}^k}{{[k]}_q!}.$$

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:

• (i) ${\mathcal{D}}_n^q(1,x)=1$,

• (ii) ${\mathcal{D}}_n^q(t,x)=x$,

• (iii) ${\mathcal{D}}_n^q(t^2,x)=x^2+\frac{x}{{[n]}_q}(1+q+\frac{x}{q})$.

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

$${\mathcal{D}}_n^q(1,x)={[n]}_q\sum _{k=1}^{\mathrm{\infty }}{p}_{n,k}^q(x){\int }_0^{q/(1-q^n)}{q}^{-k}\frac{{({[n]}_{q}t)}^{k-1}}{{[k-1]}_q!}{E}_q(-{[n]}_{q}t)d_{q}t+p_{n,0}^q(x).$$

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

$$\begin{array}{rcl}{\mathcal{D}}_n^q(1,x)& =& {[n]}_q\sum _{k=1}^{\mathrm{\infty }}{p}_{n,k}^q(x){\int }_0^{1/(1-q)}{q}^{-k}\frac{{(qy)}^{k-1}}{{[k-1]}_q!}{E}_q(-qy)\frac{q{d}_{q}y}{{[n]}_q}+p_{n,0}^q(x)\\ =& \sum _{k=1}^{\mathrm{\infty }}{p}_{n,k}^q(x)+p_{n,0}^q(x)={\mathcal{B}}_{n,q}(1,x)=1,\end{array}$$

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

Next, we have

$${\mathcal{D}}_n^q(t,x)={[n]}_q\sum _{k=1}^{\mathrm{\infty }}{p}_{n,k}^q(x){\int }_0^{q/(1-q^n)}{q}^{-k}\frac{{({[n]}_{q}t)}^{k-1}}{{[k-1]}_q!}{E}_q(-{[n]}_{q}t)t{q}^{-k}{d}_{q}t.$$

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

$$\begin{array}{rcl}{\mathcal{D}}_n^q(t,x)& =& {[n]}_q\sum _{k=1}^{\mathrm{\infty }}{p}_{n,k}^q(x){\int }_0^{1/(1-q)}{q}^{-k}\frac{{(qy)}^k}{{[k-1]}_q!{[n]}_q}{E}_q(-qy)\frac{q{d}_{q}y}{{[n]}_q{q}^k}\\ =& \sum _{k=0}^{\mathrm{\infty }}{p}_{n,k}^q(x)q\frac{{[k]}_q}{{[n]}_q{q}^k}={\mathcal{B}}_{n,q}(t,x)=x.\end{array}$$

Finally,

$${\mathcal{D}}_n^q(t^2,x)={[n]}_q\sum _{k=0}^{\mathrm{\infty }}{p}_{n,k}^q(x){\int }_0^{q/(1-q^n)}{q}^{-k}\frac{{({[n]}_{q}t)}^{k-1}}{{[k-1]}_q!}{E}_q(-{[n]}_{q}t)t^2{q}^{-2k}{d}_{q}t.$$

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

$$\begin{array}{rcl}{\mathcal{D}}_n^q(t^2,x)& =& {[n]}_q\sum _{k=1}^{\mathrm{\infty }}{p}_{n,k}^q(x){\int }_0^{1/(1-q)}{q}^{-k}\frac{{(qy)}^{k+1}}{{[k-1]}_q!{[n]}_q^2}{E}_q(-qy)q^{-2k}\frac{q{d}_{q}y}{{[n]}_q}\\ =& \sum _{k=1}^{\mathrm{\infty }}{p}_{n,k}^q(x)\frac{{[k+1]}_q{[k]}_q}{{[n]}_q^2{q}^{2k-2}}\\ =& \sum _{k=1}^{\mathrm{\infty }}{p}_{n,k}^q(x)\frac{({[k]}_q+q^k){[k]}_q}{{[n]}_q^2{q}^{2k-2}}\\ =& {\mathcal{B}}_{n,q}(t^2,x)+\frac{q}{{[n]}_q}{\mathcal{B}}_{n,q}(t,x)=x^2+\frac{x}{{[n]}_q}(1+q+\frac{x}{q}).\end{array}$$

 □

Remark 2 If we put $q=1$, we get the moments of a new sequence ${\mathcal{D}}_n(f,x)$ considered in [1] as operators as

Lemma 2 Let$q\in (0,1)$, then for$x\in [0,\mathrm{\infty })$we have

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

3 Direct theorems

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

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

The Peetre’s K-functional is defined by

$$K_2(f,\delta )=inf\{\parallel f-g\parallel +\delta \parallel g^{″}\parallel :g\in W_{\mathrm{\infty }}^2\},$$

where $W_{\mathrm{\infty }}^2=\{g\in C_B[0,\mathrm{\infty }):g^{\prime },g^{″}\in C_B[0,\mathrm{\infty })\}$. By [4], pp.177], there exists a positive constant $C>0$ such that $K_2(f,\delta )\le C{\omega }_2(f,{\delta }^{1/2})$$\delta >0$ and the second order modulus of smoothness is given by

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

Also, for $f\in C_B[0,\mathrm{\infty })$ a usual modulus of continuity is given by

$$\omega (f,\delta )=\underset{0<h\le \delta }{sup}\underset{0\le x<\mathrm{\infty }}{sup}|f(x+h)-f(x)|.$$

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

$$|{\mathcal{D}}_n^q(f,x)-f(x)|\le C{\omega }_2(f,\sqrt{\frac{x(x+q{[2]}_q)}{q{[n]}_q}}).$$

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

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

Applying Lemma 2, we obtain

$${\mathcal{D}}_n^q(g,x)-g(x)={\mathcal{D}}_n^q({\int }_x^t(t-u)g^{″}(u)du,x).$$

Obviously, we have $|{\int }_x^t(t-u)g^{″}(u)du|\le {(t-x)}^2\parallel g^{″}\parallel$. Therefore,

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

Using Lemma 1, we have

$$|{\mathcal{D}}_n^q(f,x)|\le {[n]}_q\sum _{k=1}^{\mathrm{\infty }}{p}_{n,k}^q(x){\int }_0^{q/(1-q^n)}{q}^{-k}{s}_{n,k-1}^q(t)\left|f\left(t{q}^{-k}\right)\right|d_{q}t+p_{n,0}^q(x)|f(0)|\le \parallel f\parallel .$$

Thus

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

Finally, taking the infimum over all $g\in W_{\mathrm{\infty }}^2$ and using the inequality $K_2(f,\delta )\le C{\omega }_2(f,{\delta }^{1/2})$, $\delta >0$, we get the required result. This completes the proof of Theorem 1. □

We consider the following class of functions:

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

$${\omega }_a(f,\delta )=\underset{|t-x|\le \delta }{sup}\underset{x,t\in [0,a]}{sup}|f(t)-f(x)|.$$

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

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

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

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

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

(3.1)

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

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

(3.2)

with $\delta >0$.

From (3.1) and (3.2) we can write

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

(3.3)

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

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

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

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

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

4 Higher order moments and an asymptotic formula

Lemma 3 ([6])

Let$0<q<1$, we have

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

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

Lemma 4 Let$0<q<1$, we have

The proof of Lemma 4 can be obtained by using Lemma 3.

We consider the following classes of functions:

Theorem 3 Let$q_n\in (0,1)$, then the sequence$\{{\mathcal{D}}_n^{q_n}(f)\}$converges to f uniformly on$[0,A]$for each$f\in C_2^{\ast }[0,\mathrm{\infty })$if and only if${lim}_{n\to \mathrm{\infty }}{q}_n=1$.

Theorem 4 Assume that$q_n\in (0,1)$, $q_n\to 1$and$q_n^n\to a$as$n\to \mathrm{\infty }$. For any$f\in C_2^{\ast }[0,\mathrm{\infty })$such that$f^{\mathrm{\prime }},f^{\mathrm{\prime }\mathrm{\prime }}\in C_2^{\ast }[0,\mathrm{\infty })$the following equality holds

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

uniformly on any$[0,A]$, $A>0$.

Proof Let $f,f^{\mathrm{\prime }},f^{\mathrm{\prime }\mathrm{\prime }}\in C_2^{\ast }[0,\mathrm{\infty })$ and $x\in [0,\mathrm{\infty })$ be fixed. By using Taylor’s formula, we may write

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

(4.1)

where $r(t;x)$ is the Peano form of the remainder, $r(\cdot ;x)\in C_2^{\ast }[0,\mathrm{\infty })$ and ${lim}_{t\to x}r(t;x)=0$. Applying ${\mathcal{D}}_n^{q_n}$ to (4.1), we obtain

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

By the Cauchy-Schwarz inequality, we have

$${\mathcal{D}}_n^{q_n}(r(t;x){(t-x)}^2;x)\le \sqrt{{\mathcal{D}}_n^{q_n}(r^2(t;x);x)}\sqrt{{\mathcal{D}}_n^{q_n}({(t-x)}^4;x)}.$$

(4.2)

Observe that $r^2(x;x)=0$ and $r^2(\cdot ;x)\in C_2^{\ast }[0,\mathrm{\infty })$. Then it follows from Theorem 3 and Lemma 4, that

$$\underset{n\to \mathrm{\infty }}{lim}{\mathcal{D}}_n^{q_n}(r^2(t;x);x)=r^2(x;x)=0$$

(4.3)

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

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

Then, we get the following

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