Invariant mean and a Korovkin-type approximation theorem

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.

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 [1] used this notion of a generalized limit to define a new type of convergence, known as almost convergence. Later on, Raimi [2] 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. [2]) if it is non-negative, normal and $\phi (x)=\phi ((x_{\sigma (n)}))$.

A sequence $x=(x_k)$ is said to be σ-convergent to the number L if and only if all of its σ-means coincide with L, i.e., $\phi (x)=L$ for all φ. A bounded sequence $x=(x_k)$ is σ-convergent (cf. [3]) to the number L if and only if ${lim}_{p\to \mathrm{\infty }}{t}_{pm}=L$ uniformly in m, where

$$t_{pm}=\frac{x_m+x_{\sigma (m)}+x_{{\sigma }^2(m)}+\cdots +x_{{\sigma }^p(m)}}{p+1}.$$

We denote the set of all σ-convergent sequences by $V_{\sigma }$ and in this case we write $x_k\to L(V_{\sigma })$ and L is called the σ-limit of x. Note that a σ-mean extends the limit functional on c in the sense that $\phi (x)=limx$ for all $x\in c$ if and only if σ has no finite orbits (cf. [4]) 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 [1].

In [5], the idea of statistical σ-convergence is defined which is further applied to prove some approximation theorems in [6] and [7].

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

Remark 1.1 Note that

• (a) a convergent sequence is also σ-convergent;

• (b) a σ-convergent sequence implies $(C,1)$ convergence.

Example 1.1 Let $\sigma (n)=n+1$. Define the sequence $z=(z_n)$ by

$$z_n=\{\begin{array}{ll}1& \text{if }n\text{ is odd},\\ 0& \text{if }n\text{ is even}.\end{array}$$

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

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

The classical Korovkin approximation theorem states the following [7]: Let $(T_n)$ be a sequence of positive linear operators from $C[a,b]$ into $C[a,b]$. Then ${lim}_n{\parallel T_n(f,x)-f(x)\parallel }_{\mathrm{\infty }}=0$, for all $f\in C[a,b]$ if and only if ${lim}_n{\parallel T_n(f_i,x)-f_i(x)\parallel }_{\mathrm{\infty }}=0$, for $i=0,1,2$, where $f_0(x)=1$, $f_1(x)=x$ and $f_2(x)=x^2$.

Quite recently, such type of approximation theorem has been studied in [8, 9] and [10] by using λ-statistical convergence, while in [11] lacunary statistical convergence has been used. Boyanov and Veselinov [12] have proved the Korovkin theorem on $C[0,\mathrm{\infty })$ 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 [13], 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 $(T_k)$ be a sequence of positive linear operators from $C(I)$ into $C(I)$. Then, for all $f\in C(I)$,

$$\sigma \text{-}\underset{k\to \mathrm{\infty }}{lim}{\parallel T_k(f;x)-f(x)\parallel }_{\mathrm{\infty }}=0$$

(2.1)

if and only if

$$\sigma \text{-}\underset{k\to \mathrm{\infty }}{lim}{\parallel T_k(1;x)-1\parallel }_{\mathrm{\infty }}=0,$$

(2.2)

$$\sigma \text{-}\underset{k\to \mathrm{\infty }}{lim}{\parallel T_k(e^{-s};x)-e^{-x}\parallel }_{\mathrm{\infty }}=0,$$

(2.3)

$$\sigma \text{-}\underset{k\to \mathrm{\infty }}{lim}{\parallel T_k(e^{-2s};x)-e^{-2x}\parallel }_{\mathrm{\infty }}=0.$$

(2.4)

Proof Since each 1, $e^{-x}$, $e^{-2x}$ belongs to $C(I)$, conditions (2.2)-(2.4) follow immediately from (2.1). Let $f\in C(I)$. Then there exists a constant $M>0$ such that $|f(x)|\le M$ for $x\in I$. Therefore,

$$|f(s)-f(x)|\le 2M,-\mathrm{\infty }<s,x<\mathrm{\infty }.$$

(2.5)

It is easy to prove that for a given $\epsilon >0$ there is a $\delta >0$ such that

$$|f(s)-f(x)|<\epsilon ,$$

(2.6)

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

Using (2.5), (2.6), putting ${\psi }_1={\psi }_1(s,x)={(e^{-s}-e^{-x})}^2$, we get

$$|f(s)-f(x)|<\epsilon +\frac{2M}{{\delta }^2}({\psi }_1),\mathrm{\forall }|s-x|<\delta .$$

This is,

$$-\epsilon -\frac{2M}{{\delta }^2}({\psi }_1)<f(s)-f(x)<\epsilon +\frac{2M}{{\delta }^2}({\psi }_1).$$

Now, we operate $T_{{\sigma }^k(n)}(1,x)$ for all n to this inequality since $T_{{\sigma }^k(n)}(f,x)$ is monotone and linear. We obtain

$$\begin{array}{rcl}{T}_{{\sigma }^k(n)}(1;x)(-\epsilon -\frac{2M}{{\delta }^2}({\psi }_1))& <& T_{{\sigma }^k(n)}(1;x)(f(s)-f(x))\\ <& T_{{\sigma }^k(n)}(1;x)(\epsilon +\frac{2M}{{\delta }^2}({\psi }_1)).\end{array}$$

Note that x is fixed and so $f(x)$ is a constant number. Therefore

$$\begin{array}{rl}-\epsilon T_{{\sigma }^k(n)}(1;x)-\frac{2M}{{\delta }^2}{T}_{{\sigma }^k(n)}({\psi }_1;x)& <T_{{\sigma }^k(n)}(f;x)-f(x)T_{{\sigma }^k(n)}(1;x)\\ <\epsilon T_{{\sigma }^k(n)}(1;x)+\frac{2M}{{\delta }^2}{T}_{{\sigma }^k(n)}({\psi }_1;x).\end{array}$$

(2.7)

But

$$\begin{array}{r}{T}_{{\sigma }^k(n)}(f;x)-f(x)\\ =T_{{\sigma }^k(n)}(f;x)-f(x)T_{{\sigma }^k(n)}(1;x)+f(x)T_{{\sigma }^k(n)}(1;x)-f(x)\\ =[T_{{\sigma }^k(n)}(f;x)-f(x)T_{{\sigma }^k(n)}(1;x)]+f(x)[T_{{\sigma }^k(n)}(1;x)-1].\end{array}$$

(2.8)

Using (2.7) and (2.8), we have

$$\begin{array}{rl}{T}_{{\sigma }^k(n)}(f;x)-f(x)<& \epsilon T_{{\sigma }^k(n)}(1;x)+\frac{2M}{{\delta }^2}{T}_{{\sigma }^k(n)}({\psi }_1;x)\\ +f(x)(T_{{\sigma }^k(n)}(1;x)-1).\end{array}$$

(2.9)

Now

$$\begin{array}{rl}{T}_{{\sigma }^k(n)}({\psi }_1;x)=& T_{{\sigma }^k(n)}({(e^{-s}-e^{-x})}^2;x)=T_{{\sigma }^k(n)}(e^{-2s}-2e^{-s}{e}^{-x}+e^{-2x};x)\\ =& T_{{\sigma }^k(n)}(e^{-2s};x)-2e^{-x}{T}_{{\sigma }^k(n)}(e^{-s};x)+\left(e^{-2x}\right)T_{{\sigma }^k(n)}(1;x)\\ =& [T_{{\sigma }^k(n)}(e^{-2s};x)-e^{-2x}]-2e^{-x}[T_{{\sigma }^k(n)}(e^{-s};x)-e^{-x}]\\ +e^{-2x}[T_{{\sigma }^k(n)}(1;x)-1].\end{array}$$

Using (2.9), we obtain

$$\begin{array}{rl}{T}_{{\sigma }^k(n)}(f;x)-f(x)<& \epsilon T_{{\sigma }^k(n)}(1;x)+\frac{2M}{{\delta }^2}\{[T_{{\sigma }^k(n)}(\left(e^{-2s}\right);x)-e^{-2x}]\\ -2e^{-x}[T_{{\sigma }^k(n)}(e^{-s};x)-e^{-x}]+e^{-2x}[T_{{\sigma }^k(n)}(1;x)-1]\}\\ +f(x)(T_{{\sigma }^k(n)}(1;x)-1)\\ =& \epsilon [T_{{\sigma }^k(n)}(1;x)-1]+\epsilon +\frac{2M}{{\delta }^2}\{[T_{{\sigma }^k(n)}(\left(e^{-2s}\right);x)-e^{-2x}]\\ -2e^{-x}[T_{{\sigma }^k(n)}(e^{-s};x)-e^{-x}]+e^{-2x}[T_{{\sigma }^k(n)}(1;x)-1]\}\\ +f(x)(T_{{\sigma }^k(n)}(1;x)-1).\end{array}$$

Since ε is arbitrary, we can write

$$\begin{array}{rl}{T}_{{\sigma }^k(n)}(f;x)-f(x)\le & \epsilon [T_{{\sigma }^k(n)}(1;x)-1]+\frac{2M}{{\delta }^2}\{[T_{{\sigma }^k(n)}(\left(e^{-2s}\right);x)-e^{-2x}]\\ -2e^{-x}[T_{{\sigma }^k(n)}(e^{-s};x)-e^{-x}]+e^{-2x}[T_{{\sigma }^k(n)}(1;x)-1]\}\\ +f(x)[T_{{\sigma }^k(n)}(1;x)-1].\end{array}$$

Therefore

$$\begin{array}{r}|T_{{\sigma }^k(n)}(f;x)-f(x)|\\ \le \epsilon +(\epsilon +M)|T_{{\sigma }^k(n)}(1;x)-1|+\frac{2M}{{\delta }^2}\left|e^{-2x}\right||T_{{\sigma }^k(n)}(1;x,y)-1|\\ +\frac{2M}{{\delta }^2}\left|T_{{\sigma }^k(n)}(e^{-2s};x)\right||-e^{-2x}|+\frac{4M}{{\delta }^2}\left|e^{-x}\right||T_{{\sigma }^k(n)}(e^{-s};x)-e^{-x}|\\ \le \epsilon +(\epsilon +M+\frac{4M}{{\delta }^2})|T_{{\sigma }^k(n)}(1;x)-1|+\frac{2M}{{\delta }^2}\left|e^{-2x}\right||T_{{\sigma }^k(n)}(1;x)-1|\\ +\frac{2M}{{\delta }^2}|T_{{\sigma }^k(n)}(e^{-2s};x)-e^{-2x}|+\frac{4M}{{\delta }^2}|T_{{\sigma }^k(n)}(e^{-s};x)-e^{-x}|\end{array}$$

since $|e^{-x}|\le 1$ for all $x\in I$. Now, taking ${sup}_{x\in I}$

$$\begin{array}{rl}{\parallel T_{{\sigma }^k(n)}(f;x)-f(x)\parallel }_{\mathrm{\infty }}\le & \epsilon +K({\parallel T_{{\sigma }^k(n)}(1;x)-1\parallel }_{\mathrm{\infty }}+{\parallel T_{{\sigma }^k(n)}(e^{-s};x)-e^{-x}\parallel }_{\mathrm{\infty }}\\ +{\parallel T_{{\sigma }^k(n)}(e^{-2s};x)-e^{-2x}\parallel }_{\mathrm{\infty }}),\end{array}$$

where $K=max\{\epsilon +M+\frac{4M}{{\delta }^2},\frac{2M}{{\delta }^2}\}$. Now writing

$$D_{n,p}(f,x)=\frac{1}{p}\sum _{k=0}^{p-1}{T}_{{\sigma }^k(n)}(f,x),$$

we get

$$\begin{array}{rl}{\parallel D_{n,p}(f,x)-f(x)\parallel }_{\mathrm{\infty }}\le & (ϵ+\frac{2M{b}^2}{{\delta }^2}+M){\parallel D_{n,p}(1,x)-1\parallel }_{\mathrm{\infty }}\\ +\frac{4Mb}{{\delta }^2}{\parallel D_{n,p}(t,x)-e^{-x}\parallel }_{\mathrm{\infty }}+\frac{2M}{{\delta }^2}{\parallel D_{n,p}(t^2,x)-e^{-2x}\parallel }_{\mathrm{\infty }}.\end{array}$$

Letting $p\to \mathrm{\infty }$ and using (2.2), (2.3), (2.4), we get

$$\underset{p\to \mathrm{\infty }}{lim}{\parallel D_{n,p}(f,x)-f(x)\parallel }_{\mathrm{\infty }}=0,\text{uniformly in }n.$$

 □

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 [12].

Example 2.1 Consider the sequence of classical Baskakov operators [14]

$$V_n(f;x):=\sum _{k=0}^{\mathrm{\infty }}f\left(\frac{k}{n}\right)\left(\genfrac{}{}{0ex}{}{n-1+k}{k}\right)x^k{(1+x)}^{-n-k},$$

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

Let the sequence $(L_n)$ be defined by $L_n:C(I)\to C(I)$ with $L_n(f;x)=(1+z_n)V_n(f;x)$, where $z_n$ is defined as above. Since

$$\begin{array}{c}{L}_n(1;x)=1,\hfill \\ L_n(e^{-s};x)={(1+x-x{e}^{-\frac{1}{n}})}^{-n},\hfill \\ L_n(e^{-2s};x)={(1+x^2-x^2{e}^{-\frac{1}{n}})}^{-n},\hfill \end{array}$$

and the sequence $(P_n)$ satisfies the conditions (2.1), (2.2) and (2.3). Hence we have

$$\sigma \text{-}lim{\parallel L_n(f,x)-f(x)\parallel }_{\mathrm{\infty }}=0.$$

On the other hand, we get $L_n(f,0)=(1+z_n)f(0)$ since $L_n(f,0)=f(0)$, and hence

$${\parallel L_n(f,x)-f(x)\parallel }_{\mathrm{\infty }}\ge |L_n(f,0)-f(0)|=z_n|f(0)|.$$

We see that $(L_n)$ 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 [12].

3 A consequence

Now we present a slight general result.

Theorem 3.1 Let $(T_n)$ be a sequence of positive linear operators on $C(I)$ such that

$$\underset{n}{lim}\underset{m}{sup}\frac{1}{n}\sum _{k=0}^{n-1}\parallel T_n-T_{{\sigma }^k(m)}\parallel =0.$$

If

$$\sigma \text{-}\underset{n}{lim}{\parallel T_n(e^{-\nu s},x)-e^{-\nu x}\parallel }_{\mathrm{\infty }}=0(\nu =0,1,2),$$

(3.1)

then, for any function $f\in C(I)$ bounded on the real line, we have

$$\underset{n}{lim}{\parallel T_n(f,x)-f(x)\parallel }_{\mathrm{\infty }}=0.$$

(3.2)

Proof From Theorem 2.1, we have that if (3.1) holds, then

$$\sigma \text{-}\underset{n}{lim}{\parallel T_n(f,x)-f(x)\parallel }_{\mathrm{\infty }}=0,$$

which is equivalent to

$$\underset{n}{lim}{\parallel \underset{m}{sup}{D}_{m,n}(f,x)-f(x)\parallel }_{\mathrm{\infty }}=0.$$

Now

$$\begin{array}{rl}{T}_n-D_{m,n}& =T_n-\frac{1}{n}\sum _{k=0}^{n-1}{T}_{{\sigma }^k(m)}\\ =\frac{1}{n}\sum _{k=0}^{n-1}(T_n-T_{{\sigma }^k(m)}).\end{array}$$

Therefore

$$T_n-\underset{m}{sup}{D}_{m,n}=\underset{m}{sup}\frac{1}{n}\sum _{k=0}^{n-1}(T_n-T_{{\sigma }^k(m)}).$$

Hence, using the hypothesis, we get

$$\underset{n}{lim}{\parallel T_n(f,x)-f(x)\parallel }_{\mathrm{\infty }}=\underset{n}{lim}{\parallel \underset{m}{sup}{D}_{m,n}(f,x)-f(x)\parallel }_{\mathrm{\infty }}=0,$$

that is, (3.2) holds. □