A Stancu type extension
of the Campiti-Metafune operator

Abstract.

We consider an extension of the Campiti-Metafune operator using a Stancu type technique. We study some properties of the new obtained operator.

1. Introduction

In 1982, D.D. Stancu [6], introduced a new Bernstein type operator given by

(1) $$L_{n,r}(f;x)=\sum _{k=0}^{n-r}{b}_{n-r,k}(x)\left[(1-x)f\left(\frac{k}{n}\right)+xf\left(\frac{k+r}{n}\right)\right],$$

where $b_{n,k}$ denote the basis Bernstein polynomials of degree $n$,

$$b_{n,k}=\left(\genfrac{}{}{0pt}{}{n}{k}\right)x^k{(1-x)}^{n-k}\text{, }k=0,1,\mathrm{\dots },n,$$

for $f\in C[0,1]$, $n,r\in \mathbb{N}$ such that $n>2r$.

In 1996, M. Campiti and G. Metafune [3], introduced and studied a new Bernstein type operator that now bears their names. For introducing the operator, we need two sequences $\lambda ={({\lambda }_n)}_{n\ge 1}$ and $\rho ={({\rho }_n)}_{n\ge 1}$, and the numbers ${\alpha }_{n,k}$ defined by

(2) $${\alpha }_{n,0}={\lambda }_n,{\alpha }_{n,n}={\rho }_n,{\alpha }_{n+1,k}={\alpha }_{n,k}+{\alpha }_{n,k-1},\text{for }k=\overline{1,n},n\in \mathbb{N}.$$

The Campiti-Metafune operator $A_n:C([0,1])\to C([0,1])$, is given by [3]

(3) $$(A_{n}f)(x)=\sum _{k=0}^n{\alpha }_{n,k}{x}^k{(1-x)}^{n-k}f(\frac{k}{n}).$$

Remark 1.

If $\lambda$ and $\rho$ are constant sequences of term $1$, then ${\alpha }_{n,k}=\left(\genfrac{}{}{0pt}{}{n}{k}\right),$ for $k=\overline{0,n}$ and $A_n$ becomes Bernstein operator $B_n$, given by

$$(B_{n}f)(x)=\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)x^k{(1-x)}^{n-k}f(\frac{k}{n}).$$

For the sequences $(\lambda ,\rho )$ we consider the following choices, as in [1]:

(i) Case $\lambda ={\delta }_m$, $m\in \mathbb{N}$, $\rho =0$, where ${\delta }_m={({\delta }_{n,m})}_{n\ge 1}$, We denote $A_n^{({\delta }_m,0)}$ by $L_{m,n}$ and it is called the elementary left operator of order $m$ associated to $A_n.$

The coefficients of the operators $L_{m,n}$ are denoted by $l_{m,n,k}$ and they are defined by

(ii) Case $\lambda =0$, $\rho ={\delta }_m$, $m\in \mathbb{N}$. We denote $A_n^{(0,{\delta }_m)}$ by $R_{m,n}$ and it is called elementary right operator of order $m$ associated to $A_n.$

The coefficients of the operators $R_{m,n}$ are denoted by $r_{m,n,k}$ and they are defined by

Consequently, for any $(m,n)\in \mathbb{N}\times \mathbb{N}$ we get

(4) $$(L_{m,n}f)(x)=\sum _{k=1}^{n-m}{a}_{m,n,k}(x)f(\frac{k}{n}),$$

with

(5) $$a_{m,n,k}(x)=\left(\genfrac{}{}{0pt}{}{n-m-1}{k-1}\right)x^k{(1-x)}^{n-k},$$

for and

$$(L_{n,n}f)(x)={(1-x)}^{n}f(0).$$

We have

(6) $$(R_{m,n}f)(x)=\sum _{k=m}^{n-1}{b}_{m,n,k}(x)f(\frac{k}{n}),$$

with

(7) $$b_{m,n,k}(x)=\left(\genfrac{}{}{0pt}{}{n-m-1}{k-m}\right)x^k{(1-x)}^{n-k},$$

for and

$$(R_{n,n}f)(x)=x^{n}f(1).$$

Using these elementary operators, the operator $A_n$ is decomposed as

(8) $$A_n=\sum _{m=1}^n{\lambda }_m{L}_{m,n}+\sum _{m=1}^n{\rho }_m{R}_{m,n}.$$

Remark 2.

For the particular case of $\lambda =\rho =1,$ we get

$$B_n=\sum _{m=1}^n(L_{m,n}+R_{m,n}).$$

2. A Stancu type extension of the Campiti-Metafune operator

Now we introduce the Stancu type extension of the Campiti-Metafune operator, based on an idea from [2], also used, for example, in [4], [5]. Using the Stancu type operator (1) and the operator (8), we get

(9) $$(A_n^{S}f)(x):=\sum _{m=1}^n{\lambda }_m(L_{m,n,r}^{S}f)(x)+\sum _{m=1}^n{\rho }_m(R_{m,n,r}^{S}f)(x)$$

with

$$(L_{m,n,r}^{S}f)(x)=\sum _{k=1}^{n-m-r}{a}_{m,n-r,k}(x)[(1-x)f(\frac{k}{n})+xf(\frac{k+r}{n})]$$

and

$$(R_{m,n,r}^{S}f)(x)=\sum _{k=m}^{n-r-1}{b}_{m,n-r,k}(x)[(1-x)f(\frac{k}{n})+xf(\frac{k+r}{n})],$$

where $a_{m,n-r,k}(x)$ and $b_{m,n-r,k}(x)$ are given by (5) and (7), for $f\in C[0,1]$ and $n,r\in \mathbb{N}$ such that $n>2r$.

Remark 3.

For $r=0,$ it is obtained $A_n^S$ as the Campiti-Metafune operator $A_n$.

We are going to calculate the moments of the new operators and to study some approximation properties.

3. Properties of the Campiti-Metafune operator

We give first some results regarding the Campiti-Metafune operator that will be used in the sequel in order to prove some properties of the new constructed operator.

Lemma 4 ([1]).

The operators defined by (4) and (6) verify the following relations:

$1^{\circ }$

$2^{\circ }$

$3^{\circ }$

Now we study some properties for the new operator, $A_n^S$ introduced in (9).

Theorem 5.

For every $x\in [0,1]$, $n,r\in \mathbb{N}$ such that $n>2r$, we have the following results:

i)

$(A_n^S{e}_0)(x)$ $={\displaystyle \sum _{m=1}^n}{\lambda }_m(L_{m,n,r}^S{e}_0)(x)+{\displaystyle \sum _{m=1}^n}{\rho }_m(R_{m,n,r}^S{e}_0)(x)$

$={\displaystyle \sum _{m=1}^{n-r-1}}({\lambda }_{m}x{(1-x)}^m+{\rho }_m{x}^m(1-x))+{\lambda }_{n-r}{(1-x)}^{n-r}+{\rho }_{n-r}{x}^{n-r}$

ii)

$$(A_n^S{e}_1)(x):=\sum _{m=1}^n{\lambda }_m(L_{m,n,r}^S{e}_1)(x)+\sum _{m=1}^n{\rho }_m(R_{m,n,r}^S{e}_1)(x)$$

iii)

$$(A_n^S{e}_2)(x):=\sum _{m=1}^n{\lambda }_m(L_{m,n,r}^S{e}_2)(x)+\sum _{m=1}^n{\rho }_m(R_{m,n,r}^S{e}_2)(x)$$

with

and

and

$(L_{m,n,r}^S{e}_2)(x)(x)=$

$(R_{m,n,r}^S$ $e_2)(x)(x)=$

Proof.

(10) $$(A_n^S{e}_0)(x):=\sum _{m=1}^n{\lambda }_m(L_{m,n,r}^S{e}_0)(x)+\sum _{m=1}^n{\rho }_m(R_{m,n,r}^S{e}_0)(x)$$

with

$(L_{m,n,r}^S{e}_0)(x)$ $={\displaystyle \sum _{k=1}^{n-m-r}}{a}_{m-r,n,k}(x)=(L_{m,n-r}{e}_0)(x)$

and

$(R_{m,n,r}^S{e}_0)(x)$ $={\displaystyle \sum _{k=m}^{n-r-1}}{b}_{m-r,n,k}(x)=(R_{m,n-r}{e}_0)(x)$

So, we have

$$(A_n^S{e}_0)(x):=\sum _{m=1}^n{\lambda }_m(L_{m,n-r}{e}_0)(x)+\sum _{m=1}^n{\rho }_m(R_{m,n-r}{e}_0)(x)=(A_{n-r}{e}_0)(x)$$

By (10) we get

$$(A_n^S{e}_0)(x)=\sum _{m=1}^{n-r-1}({\lambda }_{m}x{(1-x)}^m+{\rho }_m{x}^m(1-x))+{\lambda }_{n-r}{(1-x)}^{n-r}+{\rho }_{n-r}{x}^{n-r}$$

ii) We have

$$(A_n^S{e}_1)(x):=\sum _{m=1}^n{\lambda }_m(L_{m,n,r}^S{e}_1)(x)+\sum _{m=1}^n{\rho }_m(R_{m,n,r}^S{e}_1)(x)$$

with

$(L_{m,n,r}^S{e}_1)(x)$ $={\displaystyle \sum _{k=1}^{n-m-r}}{a}_{m,n-r,k}(x)[(1-x)\frac{k}{n}+x\frac{k+r}{n}]$

$=(L_{m,n-r}{e}_1)(x)+\frac{r}{n}x(L_{m.n-r}{e}_0)(x)$

and

$(R_{m,n,r}^S{e}_1)(x)$ $={\displaystyle \sum _{k=m}^{n-r-1}}{b}_{m,n-r,k}(x)[(1-x)\frac{k}{n}+x\frac{k+r}{n}]$

$=(R_{m,n-r}{e}_1)(x)+\frac{r}{n}x(R_{m.n-r}{e}_0)(x)$

and by Lemma 4, we get

$(L_{m,n,r}^S{e}_1)(x)(x)$ $=(L_{m,n-r}{e}_1)+\frac{r}{n}x(L_{m.n-r}{e}_0)(x)$

and

$(R_{m,n,r}^S{e}_1)(x)$ $=(R_{m,n-r}{e}_1)(x)+\frac{r}{n}x(R_{m.n-r}{e}_0)(x)$

iii) We have

$$(A_n^S{e}_2)(x):=\sum _{m=1}^n{\lambda }_m(L_{m,n,r}^S{e}_2)(x)+\sum _{m=1}^n{\rho }_m(R_{m,n,r}^S{e}_2)(x)$$

with

$(L_{m,n,r}^S{e}_2)(x)(x)$ $={\displaystyle \sum _{k=1}^{n-m-r}}{a}_{m,n-r,k}(x)\left[(1-x)\frac{k^2}{n^2}+x\frac{{(k+r)}^2}{n^2}\right]$

$=(L_{m,n-r}{e}_2)(x)+\frac{2r}{n}x(L_{m.n-r}{e}_1)(x)+\frac{r^2{x}^2}{n^2}(L_{m.n-r}{e}_0)(x)$

and

$(R_{m,n,r}^S{e}_2)(x)(x)$ $={\displaystyle \sum _{k=1}^{n-m-r}}{a}_{m,n-r,k}(x)\left[(1-x)\frac{k^2}{n^2}+x\frac{{(k+r)}^2}{n^2}\right]$

$=(R_{m,n-r}{e}_2)(x)+\frac{2r}{n}x(R_{m.n-r}{e}_1)(x)+\frac{r^2{x}^2}{n^2}(R_{m.n-r}{e}_0)(x).$

By Lemma 4, we get

$(L_{m,n,r}^S{e}_2)(x)(x)=$

$=(L_{m,n-r}{e}_2)(x)+\frac{2r}{n}x(L_{m.n-r}{e}_1)(x)+\frac{r^2{x}^2}{n^2}(L_{m.n-r}{e}_0)(x)$

and

$(R_{m,n,r}^S{e}_2)(x)(x)=$

$=(R_{m,n-r}{e}_2)(x)+\frac{2r}{n}x(R_{m.n-r}{e}_1)(x)+\frac{r^2{x}^2}{n^2}(R_{m.n-r}{e}_0)(x)$

∎

Theorem 6.

For every $f\in C[0,1]$, we have

$$\Vert L_{m,n,r}^S\Vert \le \Vert f\Vert \text{ and }\Vert R_{m,n,r}^S\Vert \le \Vert f\Vert .$$

Proof.

Considering the expression of $L_{m,n,r}^{S}f$ and $R_{m,n,r}^{S}f,$ and Lemma 4, we get

$|(L_{m,n,r}^{S}f)(x)|$ $=\left|{\displaystyle \sum _{k=1}^{n-m-r}}{a}_{m,n-r,k}(x)\left[(1-x)f(\frac{k}{n})+xf(\frac{k+r}{n})\right]\right|$

$\le {\displaystyle \sum _{k=1}^{n-m-r}}{a}_{m,n-r,k}(x)\left|\left[(1-x)f(\frac{k}{n})+xf(\frac{k+r}{n})\right]\right|$

$\le {\displaystyle \sum _{k=1}^{n-m-r}}{a}_{m,n-r,k}(x)\left[(1-x)\left|f(\frac{k}{n})\right|+x\left|f(\frac{k+r}{n})\right|\right]$

$\le \Vert f\Vert {\displaystyle \sum _{k=1}^{n-m-r}}{a}_{m,n-r,k}(x)$

$\le \Vert f\Vert (L_{m,n,r}^S{e}_0)(x)$

$\le \Vert f\Vert$

$|(R_{m,n,r}^{S}f)(x)|$ $=\left|{\displaystyle \sum _{k=m}^{n-r-1}}{b}_{m,n-r,k}(x)\left[(1-x)f(\frac{k}{n})+xf(\frac{k+r}{n})\right]\right|$

$\le {\displaystyle \sum _{k=1}^{n-m-r}}{b}_{m,n-r,k}(x)\left|\left[(1-x)f(\frac{k}{n})+xf(\frac{k+r}{n})\right]\right|$

$\le {\displaystyle \sum _{k=1}^{n-m-r}}{b}_{m,n-r,k}(x)\left[(1-x)\left|f(\frac{k}{n})\right|+x\left|f(\frac{k+r}{n})\right|\right]$

$\le \Vert f\Vert (R_{m,n,r}^S{e}_0)(x)$

$\le \Vert f\Vert$

∎