Approximation properties of the modification of q-Stancu-Beta operators which preserve $x^2$

Abstract

In this paper, we introduce a new kind of modification of q-Stancu-Beta operators which preserve $x^2$ based on the concept of q-integer. We investigate the moments and central moments of the operators by computation, obtain a local approximation theorem, and get the pointwise convergence rate theorem and also a weighted approximation theorem.

MSC:41A10, 41A25, 41A36.

1 Introduction

In 2012, Aral and Gupta [1] introduced the q analog of Stancu-Beta operators as

$$L_{n,q}^{\ast }(f;x)=\frac{K(A;{[n]}_{q}x)}{B_q({[n]}_{q}x;{[n]}_q+1)}{\int }_0^{\mathrm{\infty }/A}\frac{u^{{[n]}_{q}x-1}}{{(1+u)}_q^{{[n]}_{q}x+{[n]}_q+1}}f\left(q^{{[n]}_{q}x}u\right)d_{q}u,$$

(1)

for every $n\in \mathbb{N}$, $q\in (0,1)$, $x\in [0,\mathrm{\infty })$. They estimated moments, established direct result in terms of modulus of continuity and present an asymptotic formula.

Since the types of operators which preserve $x^2$ are important in approximation theory, in this paper, we will introduce a modification of q-Stancu-Beta operators which will be defined in (5). The advantage of these new operators is that they reproduce not only constant functions but also $x^2$.

Firstly, we recall some concepts of q-calculus. All of the results can be found in [2]. For any fixed real number $0<q\le 1$ and each nonnegative integer k, we denote q-integers by ${[k]}_q$, where

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

Also the q-factorial and q-binomial coefficients are defined as follows:

$${[k]}_q!=\{\begin{array}{ll}{[k]}_q{[k-1]}_q\cdots {[1]}_q,& k=1,2,\dots ;\\ 1,& k=0\end{array}$$

and

$${\left[\begin{array}{c}n\\ k\end{array}\right]}_q=\frac{{[n]}_q!}{{[k]}_q!{[n-k]}_q!}(n\ge k\ge 0).$$

The q-improper integrals are defined as

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

(2)

provided the sums converge absolutely.

The q-Beta integral is defined as

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

(3)

where $K(x;t)=\frac{1}{x+1}{x}^t{(1+\frac{1}{x})}_q^t{(1+x)}_q^{1-t}$, and ${(1+x)}_q^{\tau }=\frac{(1+x)(1+qx)(1+q^{2}x)\cdots }{(1+q^{\tau }x)(1+q^{\tau +1}x)(1+q^{\tau +2}x)\cdots }$, $\tau >0$ ($\tau =t+s$).

In particular for any positive integer n,

$$K(x;n)=q^{\frac{n(n-1)}{2}},K(x;0)=1\text{and}{B}_q(t;s)=\frac{{\Gamma }_q(t){\Gamma }_q(s)}{{\Gamma }_q(t+s)}.$$

(4)

For $f\in C[0,\mathrm{\infty })$, $q\in (0,1)$, and $n\in \mathbb{N}$, we introduce the new modification of q-Stancu-Beta operators $L_{n,q}(f,x)$ as

$$L_{n,q}(f;x)=\frac{K(A;{[n]}_q{v}_n(x))}{B_q({[n]}_q{v}_n(x);{[n]}_q+1)}{\int }_0^{\mathrm{\infty }/A}\frac{u^{{[n]}_q{v}_n(x)-1}}{{(1+u)}_q^{{[n]}_q{v}_n(x)+{[n]}_q+1}}f\left(q^{{[n]}_q{v}_n(x)}u\right)d_{q}u,$$

(5)

where

$$v_n(x)=\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}}-\frac{1}{2{[n]}_q}.$$

(6)

2 Some preliminary results

In this section we give the following lemmas, which we need to prove our theorems.

Lemma 1 (see [[1], Lemma 1])

The following equalities hold:

$$L_{n,q}^{\ast }(1;x)=1,L_{n,q}^{\ast }(t;x)=x,L_{n,q}^{\ast }(t^2;x)=\frac{({[n]}_{q}x+1)x}{q({[n]}_q-1)}.$$

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

$$L_{n,q}(1;x)=1,L_{n,q}(t;x)=\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}}-\frac{1}{2{[n]}_q},L_{n,q}(t^2;x)=x^2.$$

(7)

Proof From Lemma 1, we get $L_{n,q}(1;x)=1$ and $L_{n,q}(t;x)=\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}}-\frac{1}{2{[n]}_q}$ easily. Finally, we have

$$\begin{array}{rcl}{L}_{n,q}(t^2;x)& =& \frac{({[n]}_q{v}_n(x)+1)v_n(x)}{q({[n]}_q-1)}\\ =& \frac{{[n]}_q}{q{[n]}_q-q}(\frac{1}{4{[n]}_q^2}+\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}-\frac{1}{{[n]}_q}\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}})\\ +\frac{1}{q{[n]}_q-q}(\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}}-\frac{1}{2{[n]}_q})=x^2.\end{array}$$

Lemma 2 is proved. □

Remark 1 Let $n\in \mathbb{N}$ and $x\in [0,\mathrm{\infty })$, then for every $q\in (0,1)$, by Lemma 2, we have

$$L_{n,q}(1+t^2;x)=1+x^2.$$

(8)

Lemma 3 For every $q\in (0,1)$ and $x\in [0,\mathrm{\infty })$, we have

$$L_{n,q}({(t-x)}^2;x)=2x^2-2x\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}}+\frac{x}{{[n]}_q}.$$

(9)

Proof Since $L_{n,q}({(t-x)}^2;x)=L_{n,q}(t^2;x)-2x{L}_{n,q}(t;x)+x^2$ and from Lemma 2, we get Lemma 3 easily. □

Remark 2 Let the sequence $q=\{q_n\}$ satisfy that $q_n\in (0,1)$ and $q_n\to 1$ as $n\to \mathrm{\infty }$, then for any fixed $x\in [0,\mathrm{\infty })$, by Lemma 3, we have

$$\underset{n\to \mathrm{\infty }}{lim}{L}_{n,q_n}({(t-x)}^2;x)=0.$$

(10)

3 Local approximation

In this section we establish direct local approximation theorem in connection with the operators $L_{n,q}(f;x)$.

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

Further let us consider Peetre’s K-functional:

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

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

For $f\in C_B[0,\mathrm{\infty })$, the modulus of continuity of second order is defined by

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

by [[3], p.177] there exists an absolute constant $C>0$ such that

$$K_2(f;\delta )\le C{\omega }_2(f;\sqrt{\delta }),\delta >0.$$

(11)

For $f\in C_B[0,\mathrm{\infty })$, the modulus of continuity is defined by

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

Our first result is a direct local approximation theorem for the operators $L_{n,q}(f;x)$.

Theorem 1 For $q\in (0,1)$, $x\in [0,\mathrm{\infty })$, $n\in \mathbb{N}$, and $f\in C_B[0,\mathrm{\infty })$, we have

$$\begin{array}{c}|L_{n,q}(f,x)-f(x)|\hfill \\ \le C{\omega }_2(f;\sqrt{2x^2-2x\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}}+\frac{x}{{[n]}_q}+{(x+\frac{1}{2{[n]}_q}-\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}})}^2})\hfill \\ +\omega (f;x+\frac{1}{2{[n]}_q}-\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}}).\hfill \end{array}$$

(12)

Proof For $x\in (0,\mathrm{\infty }]$, we define the auxiliary operators $\overline{L_{n,q}}(f;x)$

$$\overline{L_{n,q}}(f;x)=L_{n,q}(f;x)-f(\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}}-\frac{1}{2{[n]}_q})+f(x).$$

(13)

Obviously, we have

$$\overline{L_{n,q}}(t-x;x)=0.$$

(14)

Let $g\in W^2$, by Taylor’s expansion, we have

$$g(t)=g(x)+g^{\prime }(x)(t-x)+{\int }_x^t(t-u)g^{″}(u)du,x,t\in [0,\mathrm{\infty }).$$

Using (14), we get

$$\overline{L_{n,q}}(g;x)=g(x)+\overline{L_{n,q}}({\int }_x^t(t-u)g^{″}(u)du;x),$$

hence, by Lemma 3, we have

$$\begin{array}{c}|\overline{L_{n,q}}(g;x)-g(x)|\hfill \\ =\left|L_{n,q}({\int }_x^t(t-u)g^{″}(u)du;x)\right|\hfill \\ +|{\int }_{\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}}-\frac{1}{2{[n]}_q}}^x[u-(\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}}-\frac{1}{2{[n]}_q})]g^{″}(u)du|\hfill \\ \le L_{n,q}\left(\right|{\int }_x^t(t-u)|g^{″}(u)|du|;x)\hfill \\ +{\int }_{\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}}-\frac{1}{2{[n]}_q}}^x|u-(\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}}-\frac{1}{2{[n]}_q})||g^{″}(u)|du\hfill \\ \le [2x^2-2x\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}}+\frac{x}{{[n]}_q}\hfill \\ +{(x+\frac{1}{2{[n]}_q}-\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}})}^2]\parallel g^{″}\parallel .\hfill \end{array}$$

On the other hand, using (13) and Lemma 2, we have

$$\begin{array}{rcl}|\overline{L_{n,q}}(f;x)|& \le & |L_{n,q}(f;x)|+2\parallel f\parallel \\ \le & \parallel f\parallel L_{n,q}(1;x)+2\parallel f\parallel \\ \le & 3\parallel f\parallel .\end{array}$$

(15)

Thus,

$$\begin{array}{c}|L_{n,q}(f;x)-f(x)|\hfill \\ \le |\overline{L_{n,q}}(f-g;x)-(f-g)(x)|+|\overline{L_{n,q}}(g;x)-g(x)|\hfill \\ +|f(\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}}-\frac{1}{2{[n]}_q})-f(x)|\hfill \\ \le 4\parallel f-g\parallel +|f(\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}}-\frac{1}{2{[n]}_q})-f(x)|\hfill \\ +[2x^2-2x\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}}+\frac{x}{{[n]}_q}\hfill \\ +{(x+\frac{1}{2{[n]}_q}-\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}})}^2]\parallel g^{″}\parallel .\hfill \end{array}$$

Hence taking the infimum on the right-hand side over all $g\in W^2$, we get

$$\begin{array}{c}|L_{n,q}(f;x)-f(x)|\hfill \\ \le 4K_2(f;2x^2-2x\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}}+\frac{x}{{[n]}_q}\hfill \\ +{(x+\frac{1}{2{[n]}_q}-\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}})}^2)\hfill \\ +\omega (f;x+\frac{1}{2{[n]}_q}-\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}}).\hfill \end{array}$$

By (11), for every $q\in (0,1)$, we have

$$\begin{array}{c}|L_{n,q}(f,x)-f(x)|\hfill \\ \le C{\omega }_2(f;\sqrt{2x^2-2x\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}}+\frac{x}{{[n]}_q}+{(x+\frac{1}{2{[n]}_q}-\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}})}^2})\hfill \\ +\omega (f;x+\frac{1}{2{[n]}_q}-\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{x}^2+\frac{1}{4{[n]}_q^2}}).\hfill \end{array}$$

This completes the proof of Theorem 1. □

4 Rate of convergence

Let $B_{x^2}[0,\mathrm{\infty })$ 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. We denote the subspace of all continuous functions belonging to $B_{x^2}[0,\mathrm{\infty })$ by $C_{x^2}[0,\mathrm{\infty })$. 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 usual modulus of continuity of f on the closed interval $[0,a]$ ($a>0$) by

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

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

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

$$\begin{array}{rcl}{\parallel L_{n,q}(f)-f\parallel }_{C[0,a]}& \le & 4M_f(1+a^2)(2a^2-2a\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{a}^2+\frac{1}{4{[n]}_q^2}}+\frac{a}{{[n]}_q})\\ +2{\omega }_{a+1}(f;\sqrt{2a^2-2a\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{a}^2+\frac{1}{4{[n]}_q^2}}+\frac{a}{{[n]}_q}}).\end{array}$$

(16)

Proof For $x\in [0,a]$ and $t>a+1$, we have $t-x\ge t-a>1$. Hence ${(t-x)}^2>1$. Thus $2+3x^2+2{(t-x)}^2\le (2+3x^2){(t-x)}^2+2{(t-x)}^2=(4+3x^2){(t-x)}^2\le (4+3a^2){(t-x)}^2\le 4(1+a^2){(t-x)}^2$. Hence, we obtain

$$|f(t)-f(x)|\le 4M_f(1+a^2){(t-x)}^2.$$

(17)

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

$$|f(t)-f(x)|\le {\omega }_{a+1}(f;|t-x|)\le (1+\frac{|t-x|}{\delta }){\omega }_{a+1}(f;\delta ),\delta >0.$$

(18)

From (17) and (18), we get

$$|f(t)-f(x)|\le 4M_f(1+a^2){(t-x)}^2+(1+\frac{|t-x|}{\delta }){\omega }_{a+1}(f;\delta ).$$

(19)

For $x\in [0,a]$ and $t\ge 0$, by Schwarz’s inequality, Lemma 2, and Lemma 3, we have

$$\begin{array}{c}|L_{n,q}(f;x)-f(x)|\hfill \\ \le L_{n,q}\left(\right|f(t)-f(x)|;x)\hfill \\ \le 4M_f(1+a^2)L_{n,q}({(t-x)}^2;x)+{\omega }_{a+1}(f;\delta )(1+\frac{1}{\delta }\sqrt{L_{n,q}({(t-x)}^2;x)})\hfill \\ \le 4M_f(1+a^2)(2a^2-2a\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{a}^2+\frac{1}{4{[n]}_q}}+\frac{a}{{[n]}_q})\hfill \\ +{\omega }_{a+1}(f,\delta )(1+\frac{1}{\delta }\sqrt{2a^2-2a\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{a}^2+\frac{1}{4{[n]}_q}}+\frac{a}{{[n]}_q}}).\hfill \end{array}$$

By taking $\delta =\sqrt{2a^2-2a\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}{a}^2+\frac{1}{4{[n]}_q}}+\frac{a}{{[n]}_q}}$, we get the assertion of Theorem 2. □

5 Weighted approximation

Now we will discuss the weighted approximation theorems.

Theorem 3 Let the sequence $\{q_n\}$ satisfy $0<q_n<1$ and $q_n\to 1$ as $n\to \mathrm{\infty }$, for $f\in C_{x^2}^{\ast }[0,\mathrm{\infty })$, we have

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel L_{n,q_n}(f)-f\parallel }_{x^2}=0.$$

(20)

Proof By using the Korovkin theorem in [4], we see that it is sufficient to verify the following three conditions:

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel L_{n,q_n}(t^v;x)-x^v\parallel }_{x^2},v=0,1,2.$$

(21)

Since $L_{n,q_n}(1;x)=1$ and $L_{n,q_n}(t^2;x)=x^2$ (see Lemma 2), (21) holds true for $v=0$ and $v=2$.

Finally, for $v=1$, we have

$$\begin{array}{rcl}{\parallel L_{n,q_n}(t;x)-x\parallel }_{x^2}& =& \underset{x\in [0,\mathrm{\infty })}{sup}\frac{|L_{n,q_n}(t;x)-x|}{1+x^2}\\ \le & (1-\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}})\underset{x\in [0,\mathrm{\infty })}{sup}\frac{x}{1+x^2}+\frac{1}{2{[n]}_q}\underset{x\in [0,\mathrm{\infty })}{sup}\frac{1}{1+x^2}\\ \le & 1-\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}}+\frac{1}{2{[n]}_q},\end{array}$$

since ${lim}_{n\to \mathrm{\infty }}{q}_n=1$, we get ${lim}_{n\to \mathrm{\infty }}(1-\sqrt{\frac{q{[n]}_q-q}{{[n]}_q}})=0$ and ${lim}_{n\to \mathrm{\infty }}\frac{1}{2{[n]}_q}=0$, so the second condition of (21) holds for $v=1$ as $n\to \mathrm{\infty }$, then the proof of Theorem 3 is completed. □