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Invariant mean and a Korovkin-type approximation theorem

Journal of Inequalities and Applications20132013:103

https://doi.org/10.1186/1029-242X-2013-103

• Accepted: 24 February 2013
• Published:

Abstract

In this paper we apply this form of convergence to prove some Korovkin-type approximation theorem by using the test functions 1, ${e}^{-x}$, ${e}^{-2x}$, which generalizes the results of Boyanov and Veselinov (Bull. Math. Soc. Sci. Math. Roum. 14(62):9-13, 1970).

MSC:41A65, 46A03, 47H10, 54H25.

Keywords

• invariant mean
• σ-convergence
• Korovkin-type approximation theorem

1 Introduction and preliminaries

Let c and ${\ell }_{\mathrm{\infty }}$ denote the spaces of all convergent and bounded sequences, respectively, and note that $c\subset {\ell }_{\mathrm{\infty }}$. In the theory of sequence spaces, an application of the well-known Hahn-Banach extension theorem gave rise to the concept of the Banach limit. That is, the lim functional defined on c can be extended to the whole of ${\ell }_{\mathrm{\infty }}$ and this extended functional is known as the Banach limit. In 1948, Lorentz  used this notion of a generalized limit to define a new type of convergence, known as almost convergence. Later on, Raimi  gave a slight generalization of almost convergence and named it σ-convergence. Before proceeding further, we recall some notations and basic definitions used in this paper.

Let σ be a mapping of the set of positive integers into itself. A continuous linear functional φ defined on the space ${\ell }_{\mathrm{\infty }}$ of all bounded sequences is called an invariant mean (or a σ-mean; cf. ) if it is non-negative, normal and $\phi \left(x\right)=\phi \left(\left({x}_{\sigma \left(n\right)}\right)\right)$.

A sequence $x=\left({x}_{k}\right)$ is said to be σ-convergent to the number L if and only if all of its σ-means coincide with L, i.e., $\phi \left(x\right)=L$ for all φ. A bounded sequence $x=\left({x}_{k}\right)$ is σ-convergent (cf. ) to the number L if and only if ${lim}_{p\to \mathrm{\infty }}{t}_{pm}=L$ uniformly in m, where
${t}_{pm}=\frac{{x}_{m}+{x}_{\sigma \left(m\right)}+{x}_{{\sigma }^{2}\left(m\right)}+\cdots +{x}_{{\sigma }^{p}\left(m\right)}}{p+1}.$

We denote the set of all σ-convergent sequences by ${V}_{\sigma }$ and in this case we write ${x}_{k}\to L\left({V}_{\sigma }\right)$ and L is called the σ-limit of x. Note that a σ-mean extends the limit functional on c in the sense that $\phi \left(x\right)=limx$ for all $x\in c$ if and only if σ has no finite orbits (cf. ) and $c\subset {V}_{\sigma }\subset {\ell }_{\mathrm{\infty }}$.

If σ is a translation then the σ-mean is called a Banach limit and σ-convergence is reduced to the concept of almost convergence introduced by Lorentz .

In , the idea of statistical σ-convergence is defined which is further applied to prove some approximation theorems in  and .

If $m=1$, then we get $\left(C,1\right)$ convergence, and in this case we write ${x}_{k}\to \ell \left(C,1\right)$, where $\ell =\left(C,1\right)\text{-}limx$.

Remark 1.1 Note that
1. (a)

a convergent sequence is also σ-convergent;

2. (b)

a σ-convergent sequence implies $\left(C,1\right)$ convergence.

Example 1.1 Let $\sigma \left(n\right)=n+1$. Define the sequence $z=\left({z}_{n}\right)$ by

Then x is σ-convergent to 1/2 but not convergent.

Let $C\left[a,b\right]$ be the space of all functions f continuous on $\left[a,b\right]$. We know that $C\left[a,b\right]$ is a Banach space with the norm ${\parallel f\parallel }_{\mathrm{\infty }}:={sup}_{a\le x\le b}|f\left(x\right)|$, $f\in C\left[a,b\right]$. Suppose that ${T}_{n}:C\left[a,b\right]\to C\left[a,b\right]$. We write ${T}_{n}\left(f,x\right)$ for ${T}_{n}\left(f\left(t\right),x\right)$ and we say that T is a positive operator if $T\left(f,x\right)\ge 0$ for all $f\left(x\right)\ge 0$.

The classical Korovkin approximation theorem states the following : Let $\left({T}_{n}\right)$ be a sequence of positive linear operators from $C\left[a,b\right]$ into $C\left[a,b\right]$. Then ${lim}_{n}{\parallel {T}_{n}\left(f,x\right)-f\left(x\right)\parallel }_{\mathrm{\infty }}=0$, for all $f\in C\left[a,b\right]$ if and only if ${lim}_{n}{\parallel {T}_{n}\left({f}_{i},x\right)-{f}_{i}\left(x\right)\parallel }_{\mathrm{\infty }}=0$, for $i=0,1,2$, where ${f}_{0}\left(x\right)=1$, ${f}_{1}\left(x\right)=x$ and ${f}_{2}\left(x\right)={x}^{2}$.

Quite recently, such type of approximation theorem has been studied in [8, 9] and  by using λ-statistical convergence, while in  lacunary statistical convergence has been used. Boyanov and Veselinov  have proved the Korovkin theorem on $C\left[0,\mathrm{\infty }\right)$ by using the test functions 1, ${e}^{-x}$, ${e}^{-2x}$. In this paper, we generalize the result of Boyanov and Veselinov by using the notion of σ-convergence. Our results also generalize the results of Mohiuddine , in which the author has used almost convergence and the test functions 1, x, ${x}^{2}$.

2 Korovkin-type approximation theorem

We prove the following σ-version of the classical Korovkin approximation theorem.

Theorem 2.1 Let $\left({T}_{k}\right)$ be a sequence of positive linear operators from $C\left(I\right)$ into $C\left(I\right)$. Then, for all $f\in C\left(I\right)$,
$\sigma \text{-}\underset{k\to \mathrm{\infty }}{lim}{\parallel {T}_{k}\left(f;x\right)-f\left(x\right)\parallel }_{\mathrm{\infty }}=0$
(2.1)
if and only if
$\sigma \text{-}\underset{k\to \mathrm{\infty }}{lim}{\parallel {T}_{k}\left(1;x\right)-1\parallel }_{\mathrm{\infty }}=0,$
(2.2)
$\sigma \text{-}\underset{k\to \mathrm{\infty }}{lim}{\parallel {T}_{k}\left({e}^{-s};x\right)-{e}^{-x}\parallel }_{\mathrm{\infty }}=0,$
(2.3)
$\sigma \text{-}\underset{k\to \mathrm{\infty }}{lim}{\parallel {T}_{k}\left({e}^{-2s};x\right)-{e}^{-2x}\parallel }_{\mathrm{\infty }}=0.$
(2.4)
Proof Since each 1, ${e}^{-x}$, ${e}^{-2x}$ belongs to $C\left(I\right)$, conditions (2.2)-(2.4) follow immediately from (2.1). Let $f\in C\left(I\right)$. Then there exists a constant $M>0$ such that $|f\left(x\right)|\le M$ for $x\in I$. Therefore,
$|f\left(s\right)-f\left(x\right)|\le 2M,\phantom{\rule{1em}{0ex}}-\mathrm{\infty }
(2.5)
It is easy to prove that for a given $\epsilon >0$ there is a $\delta >0$ such that
$|f\left(s\right)-f\left(x\right)|<\epsilon ,$
(2.6)

whenever $|{e}^{-s}-{e}^{-x}|<\delta$ for all $x\in I$.

Using (2.5), (2.6), putting ${\psi }_{1}={\psi }_{1}\left(s,x\right)={\left({e}^{-s}-{e}^{-x}\right)}^{2}$, we get
$|f\left(s\right)-f\left(x\right)|<\epsilon +\frac{2M}{{\delta }^{2}}\left({\psi }_{1}\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }|s-x|<\delta .$
This is,
$-\epsilon -\frac{2M}{{\delta }^{2}}\left({\psi }_{1}\right)
Now, we operate ${T}_{{\sigma }^{k}\left(n\right)}\left(1,x\right)$ for all n to this inequality since ${T}_{{\sigma }^{k}\left(n\right)}\left(f,x\right)$ is monotone and linear. We obtain
$\begin{array}{rcl}{T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)\left(-\epsilon -\frac{2M}{{\delta }^{2}}\left({\psi }_{1}\right)\right)& <& {T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)\left(f\left(s\right)-f\left(x\right)\right)\\ <& {T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)\left(\epsilon +\frac{2M}{{\delta }^{2}}\left({\psi }_{1}\right)\right).\end{array}$
Note that x is fixed and so $f\left(x\right)$ is a constant number. Therefore
$\begin{array}{rl}-\epsilon {T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)-\frac{2M}{{\delta }^{2}}{T}_{{\sigma }^{k}\left(n\right)}\left({\psi }_{1};x\right)& <{T}_{{\sigma }^{k}\left(n\right)}\left(f;x\right)-f\left(x\right){T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)\\ <\epsilon {T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)+\frac{2M}{{\delta }^{2}}{T}_{{\sigma }^{k}\left(n\right)}\left({\psi }_{1};x\right).\end{array}$
(2.7)
But
$\begin{array}{r}{T}_{{\sigma }^{k}\left(n\right)}\left(f;x\right)-f\left(x\right)\\ \phantom{\rule{1em}{0ex}}={T}_{{\sigma }^{k}\left(n\right)}\left(f;x\right)-f\left(x\right){T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)+f\left(x\right){T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)-f\left(x\right)\\ \phantom{\rule{1em}{0ex}}=\left[{T}_{{\sigma }^{k}\left(n\right)}\left(f;x\right)-f\left(x\right){T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)\right]+f\left(x\right)\left[{T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)-1\right].\end{array}$
(2.8)
Using (2.7) and (2.8), we have
$\begin{array}{rl}{T}_{{\sigma }^{k}\left(n\right)}\left(f;x\right)-f\left(x\right)<& \epsilon {T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)+\frac{2M}{{\delta }^{2}}{T}_{{\sigma }^{k}\left(n\right)}\left({\psi }_{1};x\right)\\ +f\left(x\right)\left({T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)-1\right).\end{array}$
(2.9)
Now
$\begin{array}{rl}{T}_{{\sigma }^{k}\left(n\right)}\left({\psi }_{1};x\right)=& {T}_{{\sigma }^{k}\left(n\right)}\left({\left({e}^{-s}-{e}^{-x}\right)}^{2};x\right)={T}_{{\sigma }^{k}\left(n\right)}\left({e}^{-2s}-2{e}^{-s}{e}^{-x}+{e}^{-2x};x\right)\\ =& {T}_{{\sigma }^{k}\left(n\right)}\left({e}^{-2s};x\right)-2{e}^{-x}{T}_{{\sigma }^{k}\left(n\right)}\left({e}^{-s};x\right)+\left({e}^{-2x}\right){T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)\\ =& \left[{T}_{{\sigma }^{k}\left(n\right)}\left({e}^{-2s};x\right)-{e}^{-2x}\right]-2{e}^{-x}\left[{T}_{{\sigma }^{k}\left(n\right)}\left({e}^{-s};x\right)-{e}^{-x}\right]\\ +{e}^{-2x}\left[{T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)-1\right].\end{array}$
Using (2.9), we obtain
$\begin{array}{rl}{T}_{{\sigma }^{k}\left(n\right)}\left(f;x\right)-f\left(x\right)<& \epsilon {T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)+\frac{2M}{{\delta }^{2}}\left\{\left[{T}_{{\sigma }^{k}\left(n\right)}\left(\left({e}^{-2s}\right);x\right)-{e}^{-2x}\right]\\ -2{e}^{-x}\left[{T}_{{\sigma }^{k}\left(n\right)}\left({e}^{-s};x\right)-{e}^{-x}\right]+{e}^{-2x}\left[{T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)-1\right]\right\}\\ +f\left(x\right)\left({T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)-1\right)\\ =& \epsilon \left[{T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)-1\right]+\epsilon +\frac{2M}{{\delta }^{2}}\left\{\left[{T}_{{\sigma }^{k}\left(n\right)}\left(\left({e}^{-2s}\right);x\right)-{e}^{-2x}\right]\\ -2{e}^{-x}\left[{T}_{{\sigma }^{k}\left(n\right)}\left({e}^{-s};x\right)-{e}^{-x}\right]+{e}^{-2x}\left[{T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)-1\right]\right\}\\ +f\left(x\right)\left({T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)-1\right).\end{array}$
Since ε is arbitrary, we can write
$\begin{array}{rl}{T}_{{\sigma }^{k}\left(n\right)}\left(f;x\right)-f\left(x\right)\le & \epsilon \left[{T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)-1\right]+\frac{2M}{{\delta }^{2}}\left\{\left[{T}_{{\sigma }^{k}\left(n\right)}\left(\left({e}^{-2s}\right);x\right)-{e}^{-2x}\right]\\ -2{e}^{-x}\left[{T}_{{\sigma }^{k}\left(n\right)}\left({e}^{-s};x\right)-{e}^{-x}\right]+{e}^{-2x}\left[{T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)-1\right]\right\}\\ +f\left(x\right)\left[{T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)-1\right].\end{array}$
Therefore
$\begin{array}{r}|{T}_{{\sigma }^{k}\left(n\right)}\left(f;x\right)-f\left(x\right)|\\ \phantom{\rule{1em}{0ex}}\le \epsilon +\left(\epsilon +M\right)|{T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)-1|+\frac{2M}{{\delta }^{2}}|{e}^{-2x}||{T}_{{\sigma }^{k}\left(n\right)}\left(1;x,y\right)-1|\\ \phantom{\rule{2em}{0ex}}+\frac{2M}{{\delta }^{2}}|{T}_{{\sigma }^{k}\left(n\right)}\left({e}^{-2s};x\right)||-{e}^{-2x}|+\frac{4M}{{\delta }^{2}}|{e}^{-x}||{T}_{{\sigma }^{k}\left(n\right)}\left({e}^{-s};x\right)-{e}^{-x}|\\ \phantom{\rule{1em}{0ex}}\le \epsilon +\left(\epsilon +M+\frac{4M}{{\delta }^{2}}\right)|{T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)-1|+\frac{2M}{{\delta }^{2}}|{e}^{-2x}||{T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)-1|\\ \phantom{\rule{2em}{0ex}}+\frac{2M}{{\delta }^{2}}|{T}_{{\sigma }^{k}\left(n\right)}\left({e}^{-2s};x\right)-{e}^{-2x}|+\frac{4M}{{\delta }^{2}}|{T}_{{\sigma }^{k}\left(n\right)}\left({e}^{-s};x\right)-{e}^{-x}|\end{array}$
since $|{e}^{-x}|\le 1$ for all $x\in I$. Now, taking ${sup}_{x\in I}$
$\begin{array}{rl}{\parallel {T}_{{\sigma }^{k}\left(n\right)}\left(f;x\right)-f\left(x\right)\parallel }_{\mathrm{\infty }}\le & \epsilon +K\left({\parallel {T}_{{\sigma }^{k}\left(n\right)}\left(1;x\right)-1\parallel }_{\mathrm{\infty }}+{\parallel {T}_{{\sigma }^{k}\left(n\right)}\left({e}^{-s};x\right)-{e}^{-x}\parallel }_{\mathrm{\infty }}\\ +{\parallel {T}_{{\sigma }^{k}\left(n\right)}\left({e}^{-2s};x\right)-{e}^{-2x}\parallel }_{\mathrm{\infty }}\right),\end{array}$
where $K=max\left\{\epsilon +M+\frac{4M}{{\delta }^{2}},\frac{2M}{{\delta }^{2}}\right\}$. Now writing
${D}_{n,p}\left(f,x\right)=\frac{1}{p}\sum _{k=0}^{p-1}{T}_{{\sigma }^{k}\left(n\right)}\left(f,x\right),$
we get
$\begin{array}{rl}{\parallel {D}_{n,p}\left(f,x\right)-f\left(x\right)\parallel }_{\mathrm{\infty }}\le & \left(ϵ+\frac{2M{b}^{2}}{{\delta }^{2}}+M\right){\parallel {D}_{n,p}\left(1,x\right)-1\parallel }_{\mathrm{\infty }}\\ +\frac{4Mb}{{\delta }^{2}}{\parallel {D}_{n,p}\left(t,x\right)-{e}^{-x}\parallel }_{\mathrm{\infty }}+\frac{2M}{{\delta }^{2}}{\parallel {D}_{n,p}\left({t}^{2},x\right)-{e}^{-2x}\parallel }_{\mathrm{\infty }}.\end{array}$
Letting $p\to \mathrm{\infty }$ and using (2.2), (2.3), (2.4), we get

□

In the following example we construct a sequence of positive linear operators satisfying the conditions of Theorem 2.1 but not satisfying the conditions of the Korovkin theorem of Boyanov and Veselinov .

Example 2.1 Consider the sequence of classical Baskakov operators 
${V}_{n}\left(f;x\right):=\sum _{k=0}^{\mathrm{\infty }}f\left(\frac{k}{n}\right)\left(\genfrac{}{}{0}{}{n-1+k}{k}\right){x}^{k}{\left(1+x\right)}^{-n-k},$

where $0\le x,y<\mathrm{\infty }$.

Let the sequence $\left({L}_{n}\right)$ be defined by ${L}_{n}:C\left(I\right)\to C\left(I\right)$ with ${L}_{n}\left(f;x\right)=\left(1+{z}_{n}\right){V}_{n}\left(f;x\right)$, where ${z}_{n}$ is defined as above. Since
$\begin{array}{c}{L}_{n}\left(1;x\right)=1,\hfill \\ {L}_{n}\left({e}^{-s};x\right)={\left(1+x-x{e}^{-\frac{1}{n}}\right)}^{-n},\hfill \\ {L}_{n}\left({e}^{-2s};x\right)={\left(1+{x}^{2}-{x}^{2}{e}^{-\frac{1}{n}}\right)}^{-n},\hfill \end{array}$
and the sequence $\left({P}_{n}\right)$ satisfies the conditions (2.1), (2.2) and (2.3). Hence we have
$\sigma \text{-}lim{\parallel {L}_{n}\left(f,x\right)-f\left(x\right)\parallel }_{\mathrm{\infty }}=0.$
On the other hand, we get ${L}_{n}\left(f,0\right)=\left(1+{z}_{n}\right)f\left(0\right)$ since ${L}_{n}\left(f,0\right)=f\left(0\right)$, and hence
${\parallel {L}_{n}\left(f,x\right)-f\left(x\right)\parallel }_{\mathrm{\infty }}\ge |{L}_{n}\left(f,0\right)-f\left(0\right)|={z}_{n}|f\left(0\right)|.$

We see that $\left({L}_{n}\right)$ does not satisfy the classical Korovkin theorem since $lim{sup}_{n\to \mathrm{\infty }}{z}_{n}$ does not exist. Hence our Theorem 2.1 is stronger than that of Boyanov and Veselinov .

3 A consequence

Now we present a slight general result.

Theorem 3.1 Let $\left({T}_{n}\right)$ be a sequence of positive linear operators on $C\left(I\right)$ such that
$\underset{n}{lim}\underset{m}{sup}\frac{1}{n}\sum _{k=0}^{n-1}\parallel {T}_{n}-{T}_{{\sigma }^{k}\left(m\right)}\parallel =0.$
If
$\sigma \text{-}\underset{n}{lim}{\parallel {T}_{n}\left({e}^{-\nu s},x\right)-{e}^{-\nu x}\parallel }_{\mathrm{\infty }}=0\phantom{\rule{1em}{0ex}}\left(\nu =0,1,2\right),$
(3.1)
then, for any function $f\in C\left(I\right)$ bounded on the real line, we have
$\underset{n}{lim}{\parallel {T}_{n}\left(f,x\right)-f\left(x\right)\parallel }_{\mathrm{\infty }}=0.$
(3.2)
Proof From Theorem 2.1, we have that if (3.1) holds, then
$\sigma \text{-}\underset{n}{lim}{\parallel {T}_{n}\left(f,x\right)-f\left(x\right)\parallel }_{\mathrm{\infty }}=0,$
which is equivalent to
$\underset{n}{lim}{\parallel \underset{m}{sup}{D}_{m,n}\left(f,x\right)-f\left(x\right)\parallel }_{\mathrm{\infty }}=0.$
Now
$\begin{array}{rl}{T}_{n}-{D}_{m,n}& ={T}_{n}-\frac{1}{n}\sum _{k=0}^{n-1}{T}_{{\sigma }^{k}\left(m\right)}\\ =\frac{1}{n}\sum _{k=0}^{n-1}\left({T}_{n}-{T}_{{\sigma }^{k}\left(m\right)}\right).\end{array}$
Therefore
${T}_{n}-\underset{m}{sup}{D}_{m,n}=\underset{m}{sup}\frac{1}{n}\sum _{k=0}^{n-1}\left({T}_{n}-{T}_{{\sigma }^{k}\left(m\right)}\right).$
Hence, using the hypothesis, we get
$\underset{n}{lim}{\parallel {T}_{n}\left(f,x\right)-f\left(x\right)\parallel }_{\mathrm{\infty }}=\underset{n}{lim}{\parallel \underset{m}{sup}{D}_{m,n}\left(f,x\right)-f\left(x\right)\parallel }_{\mathrm{\infty }}=0,$

that is, (3.2) holds. □

Authors’ Affiliations

(1)
Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah, 21589, Saudi Arabia

References 