Statistical summability $(C,1)$ and a Korovkin type approximation theorem

Abstract

The concept of statistical summability $(C,1)$ has recently been introduced by Móricz [Jour. Math. Anal. Appl. 275, 277-287 (2002)]. In this paper, we use this notion of summability to prove the Korovkin type approximation theorem by using the test functions 1, $e^{-x}$, $e^{-2x}$. We also give here the rate of statistical summability $(C,1)$ and apply the classical Baskakov operator to construct an example in support of our main result.

MSC:41A10, 41A25, 41A36, 40A30, 40G15.

1 Introduction and preliminaries

In 1951, Fast [6] presented the following definition of statistical convergence for sequences of real numbers. Let $K\subseteq \mathbb{N}$, the set of all natural numbers and $K_n=\{k\le n:k\in K\}$. Then the natural density of K is defined by $\delta (K)={lim}_n{n}^{-1}|K_n|$ if the limit exists, where the vertical bars indicate the number of elements in the enclosed set. The sequence $x=(x_k)$ is said to be statistically convergent to L if for every $ϵ;0$, the set $K_{ϵ}:=\{k\in \mathbb{N}:|x_k-L|\ge ϵ\}$ has natural density zero, i.e., for each $ϵ;0$,

$$\underset{n}{lim}\frac{1}{n}|\{j\le n:|x_j-L|\ge ϵ\}|=0.$$

In this case, we write $L=\mathit{st}\text{-}limx$. Note that every convergent sequence is statistically convergent but not conversely. Define the sequence $w=(w_n)$ by

$$w_n=\{\begin{array}{ll}1& \text{if}n=k^2,k\in \mathbb{N},\\ 0& \text{otherwise.}\end{array}$$

(1.1)

Then x is statistically convergent to 0 but not convergent.

Recently, Móricz [12] has defined the concept of statistical summability $(C,1)$ as follows:

For a sequence $x=(x_k)$, let us write $t_n=\frac{1}{n+1}{\sum }_{k=0}^n{x}_k$. We say that a sequence $x=(x_k)$ is statistically summable$(C,1)$ if $\mathit{st}\text{-}{lim}_{n\to \mathrm{\infty }}{t}_n=L$. In this case, we write $L=C_1(\mathit{st})\text{-}limx$.

In the following example, we exhibit that a sequence is statistically summable $(C,1)$ but not statistically convergent. Define the sequence $x=(x_k)$ by

$$x_k=\{\begin{array}{ll}1& \text{if}k=m^2-m,m^2-m+1,\dots ,m^2-1,\\ -m& \text{if}k=m^2,m=2,3,4,\dots ,\\ 0& \text{otherwise.}\end{array}$$

(1.2)

Then

$$\begin{array}{rcl}{t}_n& =& \frac{1}{n+1}\sum _{k=0}^n{x}_k\\ =& \{\begin{array}{ll}\frac{s+1}{n+1}& \text{if}n=m^2-m+s;s=0,1,2,\dots ,m-1;m=2,3,\dots ,\\ 0& \text{otherwise.}\end{array}\end{array}$$

(1.3)

It is easy to see that ${lim}_{n\to \mathrm{\infty }}{t}_n=0$ and hence $\mathit{st}\text{-}{lim}_{n\to \mathrm{\infty }}{t}_n=0$, i.e., a sequence $x=(x_k)$ is statistically summable $(C,1)$ to 0. On the other hand $\mathit{st}\text{-}lim{inf}_{k\to \mathrm{\infty }}{x}_k=0$ and $\mathit{st}\text{-}lim{sup}_{k\to \mathrm{\infty }}{x}_k=1$, since the sequence ${(m^2)}_{m=2}^{\mathrm{\infty }}$ is statistically convergent to 0. Hence $x=(x_k)$ is not statistically 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 norm

$${\parallel f\parallel }_{\mathrm{\infty }}:=\underset{x\in [a,b]}{sup}|f(x)|,f\in C[a,b].$$

The classical Korovkin approximation theorem is stated as follows [9]:

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 $f_2(x)=x^2$.

Recently, Mohiuddine [10] has obtained an application of almost convergence for single sequences in Korovkin-type approximation theorem and proved some related results. For the function of two variables, such type of approximation theorems are proved in [1] by using almost convergence of double sequences. Quite recently, in [13] and [14] the Korovkin type theorem is proved for statistical λ-convergence and statistical lacunary summability, respectively. For some recent work on this topic, we refer to [5, 7, 8, 11, 15, 16]. Boyanov and Veselinov [3] 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 statistical summability $(C,1)$ and the same test functions 1, $e^{-x}$, $e^{-2x}$. We also give an example to justify that our result is stronger than that of Boyanov and Veselinov [3].

2 Main result

Let $C(I)$ be the Banach space with the uniform norm ${\parallel \cdot \parallel }_{\mathrm{\infty }}$ of all real-valued two dimensional continuous functions on $I=[0,\mathrm{\infty })$; provided that ${lim}_{x\to \mathrm{\infty }}f(x)$ is finite. Suppose that $L_n:C(I)\to C(I)$. We write $L_n(f;x)$ for $L_n(f(s);x)$; and we say that L is a positive operator if $L(f;x)\ge 0$ for all $f(x)\ge 0$.

The following statistical version of Boyanov and Veselinov’s result can be found in [4].

Theorem A Let$(T_k)$be a sequence of positive linear operators from$C(I)$into$C(I)$. Then for all$f\in C(I)$

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

if and only if

$$\begin{array}{c}\mathit{st}\text{-}\underset{k\to \mathrm{\infty }}{lim}{\parallel T_k(1;x)-1\parallel }_{\mathrm{\infty }}=0,\hfill \\ \mathit{st}\text{-}\underset{k\to \mathrm{\infty }}{lim}{\parallel T_k(e^{-s};x)-e^{-x}\parallel }_{\mathrm{\infty }}=0,\hfill \\ \mathit{st}\text{-}\underset{k\to \mathrm{\infty }}{lim}{\parallel T_k(e^{-2s};x)-e^{-2x}\parallel }_{\mathrm{\infty }}=0.\hfill \end{array}$$

Now we prove the following result by using the notion of statistical summability $(C,1)$.

Theorem 2.1 Let$(T_k)$be a sequence of positive linear operators from$C(I)$into$C(I)$. Then for all$f\in C(I)$

$$C_1(\mathit{st})\text{-}\underset{k\to \mathrm{\infty }}{lim}{\parallel T_k(f;x)-f(x)\parallel }_{\mathrm{\infty }}=0$$

(2.1)

if and only if

$$C_1(\mathit{st})\text{-}\underset{k\to \mathrm{\infty }}{lim}{\parallel T_k(1;x)-1\parallel }_{\mathrm{\infty }}=0,$$

(2.2)

$$C_1(\mathit{st})\text{-}\underset{k\to \mathrm{\infty }}{lim}{\parallel T_k(e^{-s};x)-e^{-x}\parallel }_{\mathrm{\infty }}=0,$$

(2.3)

$$C_1(\mathit{st})\text{-}\underset{k\to \mathrm{\infty }}{lim}{\parallel T_k(e^{-2s};x)-e^{-2x}\parallel }_{\mathrm{\infty }}=0.$$

(2.4)

Proof Since each of 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), and 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, applying the operator $T_k(1;x)$ to this inequality, since $T_k(f;x)$ is monotone and linear, we obtain

$$T_k(1;x)(-\epsilon -\frac{2M}{{\delta }^2}({\psi }_1))T_k(1;x)(f(s)-f(x))T_k(1;x)(\epsilon +\frac{2M}{{\delta }^2}({\psi }_1)).$$

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

$$-\epsilon T_k(1;x)-\frac{2M}{{\delta }^2}{T}_{j,k}({\psi }_1;x)T_k(f;x)-f(x)T_k(1;x)\epsilon T_k(1;x)+\frac{2M}{{\delta }^2}{T}_k({\psi }_1;x).$$

(2.7)

Also,

$$\begin{array}{rcl}{T}_k(f;x)-f(x)& =& T_k(f;x)-f(x)T_k(1;x)+f(x)T_k(1;x)-f(x)\\ =& T_k(f;x)-f(x)T_k(1;x)+f(x)[T_k(1;x)-1].\end{array}$$

(2.8)

It follows from (2.7) and (2.8) that

$$T_k(f;x)-f(x)\epsilon T_k(1;x)+\frac{2M}{{\delta }^2}{T}_k({\psi }_1;x)+f(x)[T_k(1;x)-1].$$

(2.9)

Now

$$\begin{array}{rcl}{T}_k({\psi }_1;x)& =& T_k({(e^{-s}-e^{-x})}^2;x)\\ =& T_k(e^{-2s}-2e^{-s}{e}^{-x}+e^{-2x};x)\\ =& T_k(e^{-2s};x)-2e^{-x}{T}_k(e^{-s};x)+e^{-2x}{T}_k(1;x)\\ =& [T_k(e^{-2s};x)-e^{-2x}]-2e^{-x}[T_k(e^{-s};x)-e^{-x}]+e^{-2x}[T_k(1;x)-1].\end{array}$$

Using (2.9), we obtain

$$\begin{array}{rc}{T}_k(f;x)-f(x)& \epsilon T_k(1;x)+\frac{2M}{{\delta }^2}\{[T_k(e^{-2s};x)-e^{-2x}]-2e^{-x}[T_k(e^{-s};x)-e^{-x}]\\ +e^{-2x}[T_k(1;x)-1]\}+f(x)[T_k(1;x)-1]\\ =& \epsilon [T_k(1;x)-1]+\epsilon +\frac{2M}{{\delta }^2}\{[T_k(e^{-2s};x)-e^{-2x}]-2e^{-x}[T_k(e^{-s};x)-e^{-x}]\\ +e^{-2x}[T_k(1;x)-1]\}+f(x)[T_k(1;x)-1].\end{array}$$

Therefore,

$$\begin{array}{rcl}|T_k(f;x)-f(x)|& \le & \epsilon +(\epsilon +M)|T_k(1;x)-1|+\frac{2M}{{\delta }^2}|T_k(e^{-2s};x)-e^{-2x}|\\ +\frac{4M}{{\delta }^2}\left|e^{-x}\right||T_k(e^{-s};x)-e^{-x}|+\frac{2M}{{\delta }^2}\left|e^{-2x}\right||T_k(1;x)-1|\\ \le & \epsilon +(\epsilon +M+\frac{2M}{{\delta }^2})|T_k(1;x)-1|+\frac{2M}{{\delta }^2}|T_k(e^{-2s};x)-e^{-2x}|\\ +\frac{4M}{{\delta }^2}|T_k(e^{-s};x)-e^{-x}|,\end{array}$$

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

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

(2.10)

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

Now replacing $T_k(\cdot ,x)$ by $\frac{1}{m+1}{\sum }_{k=0}^m{T}_k(\cdot ,x)$ and then by $B_m(\cdot ,x)$ in (2.10) on both sides. For a given $r;0$ choose ${\epsilon }^{\mathrm{\prime }};0$ such that ${\epsilon }^{\mathrm{\prime }}r$ . Define the following sets

$$\begin{array}{c}D=\{m\le n:{\parallel B_m(f,x)-f(x)\parallel }_{\mathrm{\infty }}\ge r\},\hfill \\ D_1=\{m\le n:{\parallel B_m(1,x)-1\parallel }_{\mathrm{\infty }}\ge \frac{r-{\epsilon }^{\mathrm{\prime }}}{3K}\},\hfill \\ D_2=\{m\le n:{\parallel B_m(t,x)-e^{-x}\parallel }_{\mathrm{\infty }}\ge \frac{r-{\epsilon }^{\mathrm{\prime }}}{3K}\},\hfill \\ D_3=\{m\le n:{\parallel B_m(t^2,x)-e^{-2x}\parallel }_{\mathrm{\infty }}\ge \frac{r-{\epsilon }^{\mathrm{\prime }}}{3K}\}.\hfill \end{array}$$

Then $D\subset D_1\cup D_2\cup D_3$, and so $\delta (D)\le \delta (D_1)+\delta (D_2)+\delta (D_3)$. Therefore, using conditions (2.2)-(2.4), we get

$$C_1(\mathit{st})\text{-}\underset{n}{lim}{\parallel T_n(f,x)-f(x)\parallel }_{\mathrm{\infty }}=0.$$

This completes the proof of the theorem. □

3 Rate of statistical summability $(C,1)$

In this section, we study the rate of weighted statistical convergence of a sequence of positive linear operators defined from $C(I)$ into $C(I)$.

Definition 3.1 Let $(a_n)$ be a positive nonincreasing sequence. We say that the sequence $x=(x_k)$ is statistically summable $(C,1)$ to the number L with the rate $o(a_n)$ if for every $\epsilon ;0$,

$$\underset{n}{lim}\frac{1}{a_n}\left|\{m\le n:|t_m-L|\ge \epsilon \}\right|=0.$$

In this case, we write $x_k-L=C_1(\mathit{st})-o(a_n)$.

Now, we recall the notion of modulus of continuity. The modulus of continuity of $f\in C(I)$, denoted by $\omega (f,\delta )$ is defined by

$$\omega (f,\delta )=\underset{|x-y|\delta }{sup}|f(x)-f(y)|.$$

It is well known that

$$|f(x)-f(y)|\le \omega (f,\delta )(\frac{|e^{-y}-e^{-x}|}{\delta }+1).$$

(3.1)

Then we have the following result.

Theorem 3.2 Let$(T_k)$be a sequence of positive linear operators from$C(I)$into$C(I)$. Suppose that

• (i) ${\parallel T_k(1;x)-1\parallel }_{\mathrm{\infty }}=C_1(\mathit{st})-o(a_n)$,

• (ii) $\omega (f,{\lambda }_k)=C_1(\mathit{st})-o(b_n)$, where ${\lambda }_k=\sqrt{T_k({\phi }_x;x)}$ and ${\phi }_x(y)={(e^{-y}-e^{-x})}^2$.

Then for all$f\in C(I)$, we have

$${\parallel T_k(f;x)-f(x)\parallel }_{\mathrm{\infty }}=C_1(\mathit{st})-o(c_n),$$

where$c_n=max\{a_n,b_n\}$.

Proof Let $f\in C(I)$ and $x\in I$. From (2.8) and (3.1), we can write

$$\begin{array}{rcl}|T_k(f;x)-f(x)|& \le & T_k(|f(y)-f(x)|;x)+|f(x)||T_k(1;x)-1|\\ \le & T_k(\frac{|e^{-y}-e^{-x}|}{\delta }+1;x)\omega (f,\delta )+|f(x)||T_k(1;x)-1|\\ \le & T_k(1+\frac{1}{{\delta }^2}{(e^{-y}-e^{-x})}^2;x)\omega (f,\delta )+|f(x)||T_k(1;x)-1|\\ \le & (T_k(1;x)+\frac{1}{{\delta }^2}{T}_k({\phi }_x;x))\omega (f,\delta )+|f(x)||T_k(1;x)-1|\\ \le & \omega (f,\delta )|T_k(1;x)-1|+|f||T_k(1;x)-1|+\omega (f,\delta )\\ +\frac{1}{{\delta }^2}\omega (f,\delta )T_k({\phi }_x;x).\end{array}$$

Put $\delta ={\lambda }_k=\sqrt{T_k({\phi }_x;x)}$. Hence, we get

$$\begin{array}{rcl}{\parallel T_k(f;x)-f(x)\parallel }_{\mathrm{\infty }}& \le & \parallel f\parallel {\parallel T_k(1;x)-1\parallel }_{\mathrm{\infty }}+2\omega (f,{\lambda }_k)+\omega (f,{\lambda }_k){\parallel T_k(1;x)-1\parallel }_{\mathrm{\infty }}\\ \le & K\{{\parallel T_k(1;x)-1\parallel }_{\mathrm{\infty }}+\omega (f,{\lambda }_k)+\omega (f,{\lambda }_k){\parallel T_k(1;x)-1\parallel }_{\mathrm{\infty }}\},\end{array}$$

where $K=max\{\parallel f\parallel ,2\}$. Now replacing $T_k(\cdot ;x)$ by $\frac{1}{n+1}{\sum }_{k=0}^n{T}_k(\cdot ;x)=L_n(\cdot ;x)$

$${\parallel L_n(f;x)-f(x)\parallel }_{\mathrm{\infty }}\le K\{{\parallel L_n(1;x)-1\parallel }_{\mathrm{\infty }}+\omega (f,{\lambda }_k)+\omega (f,{\lambda }_k){\parallel T_k(1;x)-1\parallel }_{\mathrm{\infty }}\}.$$

Using the Definition 3.1, and conditions (i) and (ii), we get the desired result. □

This completes the proof of the theorem.

4 Example and the concluding remark

In the following, we construct an example of a sequence of positive linear operators satisfying the conditions of Theorem 2.1 but does not satisfy the conditions of the Korovkin approximation theorem due to of Boyanov and Veselinov [3] and the conditions of Theorem A.

Consider the sequence of classical Baskakov operators [2]

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

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

Let $L_n:C(I)\to C(I)$ be defined by

$$L_n(f;x)=(1+x_n)V_n(f;x),$$

where the sequence $x=(x_n)$ is defined by (1.2). Note that this sequence is statistically summable $(C,1)$ to 0 but neither convergent nor statistically convergent. Now

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

we have that the sequence $(L_n)$ satisfies the conditions (2.2), (2.3), and (2.4). Hence, by Theorem 2.1, we have

$$C_1(\mathit{st})\text{-}\underset{n\to \mathrm{\infty }}{lim}{\parallel L_n(f)-f\parallel }_{\mathrm{\infty }}=0.$$

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

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

We see that $(L_n)$ does not satisfy the conditions of the theorem of Boyanov and Veselinov as well as of Theorem A, since $(x_n)$ is neither convergent nor statistically convergent.

Hence, our Theorem 2.1 is stronger than that of Boyanov and Veselinov [3] as well as Theorem A.