# Statistical summability $\left(C,1\right)$ and a Korovkin type approximation theorem

## Abstract

The concept of statistical summability $\left(C,1\right)$ 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 $\left(C,1\right)$ 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 ${K}_{n}=\left\{k\le n:k\in K\right\}$. Then the natural density of K is defined by $\delta \left(K\right)={lim}_{n}{n}^{-1}|{K}_{n}|$ if the limit exists, where the vertical bars indicate the number of elements in the enclosed set. The sequence $x=\left({x}_{k}\right)$ is said to be statistically convergent to L if for every $ϵ;0$, the set ${K}_{ϵ}:=\left\{k\in \mathbb{N}:|{x}_{k}-L|\ge ϵ\right\}$ has natural density zero, i.e., for each $ϵ;0$,

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

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

${w}_{n}=\left\{\begin{array}{ll}1& \text{if}n={k}^{2},k\in \mathbb{N},\\ 0& \text{otherwise.}\end{array}$
(1.1)

Then x is statistically convergent to 0 but not convergent.

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

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

In the following example, we exhibit that a sequence is statistically summable $\left(C,1\right)$ but not statistically convergent. Define the sequence $x=\left({x}_{k}\right)$ by

${x}_{k}=\left\{\begin{array}{ll}1& \text{if}k={m}^{2}-m,{m}^{2}-m+1,\dots ,{m}^{2}-1,\\ -m& \text{if}k={m}^{2},m=2,3,4,\dots ,\\ 0& \text{otherwise.}\end{array}$
(1.2)

Then

$\begin{array}{rcl}{t}_{n}& =& \frac{1}{n+1}\sum _{k=0}^{n}{x}_{k}\\ =& \left\{\begin{array}{ll}\frac{s+1}{n+1}& \text{if}n={m}^{2}-m+s;s=0,1,2,\dots ,m-1;m=2,3,\dots ,\\ 0& \text{otherwise.}\end{array}\end{array}$
(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 $x=\left({x}_{k}\right)$ is statistically summable $\left(C,1\right)$ 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 ${\left({m}^{2}\right)}_{m=2}^{\mathrm{\infty }}$ is statistically convergent to 0. Hence $x=\left({x}_{k}\right)$ is not statistically 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 norm

${\parallel f\parallel }_{\mathrm{\infty }}:=\underset{x\in \left[a,b\right]}{sup}|f\left(x\right)|,\phantom{\rule{1em}{0ex}}f\in C\left[a,b\right].$

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

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 ${f}_{2}\left(x\right)={x}^{2}$.

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 $C\left[0,\mathrm{\infty }\right)$ 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 statistical summability $\left(C,1\right)$ 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 $C\left(I\right)$ be the Banach space with the uniform norm ${\parallel \cdot \parallel }_{\mathrm{\infty }}$ of all real-valued two dimensional continuous functions on $I=\left[0,\mathrm{\infty }\right)$; provided that ${lim}_{x\to \mathrm{\infty }}f\left(x\right)$ is finite. Suppose that ${L}_{n}:C\left(I\right)\to C\left(I\right)$. We write ${L}_{n}\left(f;x\right)$ for ${L}_{n}\left(f\left(s\right);x\right)$; and we say that L is a positive operator if $L\left(f;x\right)\ge 0$ for all $f\left(x\right)\ge 0$.

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

Theorem A 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$f\in C\left(I\right)$

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

if and only if

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

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

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$f\in C\left(I\right)$

${C}_{1}\left(\mathit{st}\right)\text{-}\underset{k\to \mathrm{\infty }}{lim}{\parallel {T}_{k}\left(f;x\right)-f\left(x\right)\parallel }_{\mathrm{\infty }}=0$
(2.1)

if and only if

${C}_{1}\left(\mathit{st}\right)\text{-}\underset{k\to \mathrm{\infty }}{lim}{\parallel {T}_{k}\left(1;x\right)-1\parallel }_{\mathrm{\infty }}=0,$
(2.2)
${C}_{1}\left(\mathit{st}\right)\text{-}\underset{k\to \mathrm{\infty }}{lim}{\parallel {T}_{k}\left({e}^{-s};x\right)-{e}^{-x}\parallel }_{\mathrm{\infty }}=0,$
(2.3)
${C}_{1}\left(\mathit{st}\right)\text{-}\underset{k\to \mathrm{\infty }}{lim}{\parallel {T}_{k}\left({e}^{-2s};x\right)-{e}^{-2x}\parallel }_{\mathrm{\infty }}=0.$
(2.4)

Proof Since each of 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,

$|f\left(s\right)-f\left(x\right)|\le 2M,\phantom{\rule{1em}{0ex}}-\mathrm{\infty };s,x;\mathrm{\infty }.$
(2.5)

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

$|f\left(s\right)-f\left(x\right)|\epsilon ,$
(2.6)

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

Using (2.5), (2.6), and 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,

$-\epsilon -\frac{2M}{{\delta }^{2}}\left({\psi }_{1}\right)f\left(s\right)-f\left(x\right)\epsilon +\frac{2M}{{\delta }^{2}}\left({\psi }_{1}\right).$

Now, applying the operator ${T}_{k}\left(1;x\right)$ to this inequality, since ${T}_{k}\left(f;x\right)$ is monotone and linear, we obtain

${T}_{k}\left(1;x\right)\left(-\epsilon -\frac{2M}{{\delta }^{2}}\left({\psi }_{1}\right)\right){T}_{k}\left(1;x\right)\left(f\left(s\right)-f\left(x\right)\right){T}_{k}\left(1;x\right)\left(\epsilon +\frac{2M}{{\delta }^{2}}\left({\psi }_{1}\right)\right).$

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

$-\epsilon {T}_{k}\left(1;x\right)-\frac{2M}{{\delta }^{2}}{T}_{j,k}\left({\psi }_{1};x\right){T}_{k}\left(f;x\right)-f\left(x\right){T}_{k}\left(1;x\right)\epsilon {T}_{k}\left(1;x\right)+\frac{2M}{{\delta }^{2}}{T}_{k}\left({\psi }_{1};x\right).$
(2.7)

Also,

$\begin{array}{rcl}{T}_{k}\left(f;x\right)-f\left(x\right)& =& {T}_{k}\left(f;x\right)-f\left(x\right){T}_{k}\left(1;x\right)+f\left(x\right){T}_{k}\left(1;x\right)-f\left(x\right)\\ =& {T}_{k}\left(f;x\right)-f\left(x\right){T}_{k}\left(1;x\right)+f\left(x\right)\left[{T}_{k}\left(1;x\right)-1\right].\end{array}$
(2.8)

It follows from (2.7) and (2.8) that

${T}_{k}\left(f;x\right)-f\left(x\right)\epsilon {T}_{k}\left(1;x\right)+\frac{2M}{{\delta }^{2}}{T}_{k}\left({\psi }_{1};x\right)+f\left(x\right)\left[{T}_{k}\left(1;x\right)-1\right].$
(2.9)

Now

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

Using (2.9), we obtain

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

Therefore,

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

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

$\begin{array}{rcl}{\parallel {T}_{k}\left(f;x\right)-f\left(x\right)\parallel }_{\mathrm{\infty }}& \le & \epsilon +K\left({\parallel {T}_{k}\left(1;x\right)-1\parallel }_{\mathrm{\infty }}+{\parallel {T}_{k}\left({e}^{-s};x\right)-{e}^{-x}\parallel }_{\mathrm{\infty }}\\ +{\parallel {T}_{k}\left({e}^{-2s};x\right)-{e}^{-2x}\parallel }_{\mathrm{\infty }}\right),\end{array}$
(2.10)

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

Now replacing ${T}_{k}\left(\cdot ,x\right)$ by $\frac{1}{m+1}{\sum }_{k=0}^{m}{T}_{k}\left(\cdot ,x\right)$ and then by ${B}_{m}\left(\cdot ,x\right)$ 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

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

Then $D\subset {D}_{1}\cup {D}_{2}\cup {D}_{3}$, and so $\delta \left(D\right)\le \delta \left({D}_{1}\right)+\delta \left({D}_{2}\right)+\delta \left({D}_{3}\right)$. Therefore, using conditions (2.2)-(2.4), we get

${C}_{1}\left(\mathit{st}\right)\text{-}\underset{n}{lim}{\parallel {T}_{n}\left(f,x\right)-f\left(x\right)\parallel }_{\mathrm{\infty }}=0.$

This completes the proof of the theorem. □

## 3 Rate of statistical summability $\left(C,1\right)$

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

Definition 3.1 Let $\left({a}_{n}\right)$ be a positive nonincreasing sequence. We say that the sequence $x=\left({x}_{k}\right)$ is statistically summable $\left(C,1\right)$ to the number L with the rate $o\left({a}_{n}\right)$ 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 ${x}_{k}-L={C}_{1}\left(\mathit{st}\right)-o\left({a}_{n}\right)$.

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

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

It is well known that

$|f\left(x\right)-f\left(y\right)|\le \omega \left(f,\delta \right)\left(\frac{|{e}^{-y}-{e}^{-x}|}{\delta }+1\right).$
(3.1)

Then we have the following result.

Theorem 3.2 Let$\left({T}_{k}\right)$be a sequence of positive linear operators from$C\left(I\right)$into$C\left(I\right)$. Suppose that

1. (i)

${\parallel {T}_{k}\left(1;x\right)-1\parallel }_{\mathrm{\infty }}={C}_{1}\left(\mathit{st}\right)-o\left({a}_{n}\right)$,

2. (ii)

$\omega \left(f,{\lambda }_{k}\right)={C}_{1}\left(\mathit{st}\right)-o\left({b}_{n}\right)$, where ${\lambda }_{k}=\sqrt{{T}_{k}\left({\phi }_{x};x\right)}$ and ${\phi }_{x}\left(y\right)={\left({e}^{-y}-{e}^{-x}\right)}^{2}$.

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

${\parallel {T}_{k}\left(f;x\right)-f\left(x\right)\parallel }_{\mathrm{\infty }}={C}_{1}\left(\mathit{st}\right)-o\left({c}_{n}\right),$

where${c}_{n}=max\left\{{a}_{n},{b}_{n}\right\}$.

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

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

Put $\delta ={\lambda }_{k}=\sqrt{{T}_{k}\left({\phi }_{x};x\right)}$. Hence, we get

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

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

${\parallel {L}_{n}\left(f;x\right)-f\left(x\right)\parallel }_{\mathrm{\infty }}\le K\left\{{\parallel {L}_{n}\left(1;x\right)-1\parallel }_{\mathrm{\infty }}+\omega \left(f,{\lambda }_{k}\right)+\omega \left(f,{\lambda }_{k}\right){\parallel {T}_{k}\left(1;x\right)-1\parallel }_{\mathrm{\infty }}\right\}.$

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]

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

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

Let ${L}_{n}:C\left(I\right)\to C\left(I\right)$ be defined by

${L}_{n}\left(f;x\right)=\left(1+{x}_{n}\right){V}_{n}\left(f;x\right),$

where the sequence $x=\left({x}_{n}\right)$ is defined by (1.2). Note that this sequence is statistically summable $\left(C,1\right)$ to 0 but neither convergent nor statistically convergent. Now

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

we have that the sequence $\left({L}_{n}\right)$ satisfies the conditions (2.2), (2.3), and (2.4). Hence, by Theorem 2.1, we have

${C}_{1}\left(\mathit{st}\right)\text{-}\underset{n\to \mathrm{\infty }}{lim}{\parallel {L}_{n}\left(f\right)-f\parallel }_{\mathrm{\infty }}=0.$

On the other hand, we get ${L}_{n}\left(f;0\right)=\left(1+{x}_{n}\right)f\left(0\right)$, since ${V}_{n}\left(f;0\right)=f\left(0\right)$, and hence

${\parallel {L}_{n}\left(f;x\right)-f\left(x\right)\parallel }_{\mathrm{\infty }}\ge |{L}_{n}\left(f;0\right)-f\left(0\right)|={x}_{n}|f\left(0\right)|.$

We see that $\left({L}_{n}\right)$ does not satisfy the conditions of the theorem of Boyanov and Veselinov as well as of Theorem A, since $\left({x}_{n}\right)$ 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.

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## Acknowledgements

The authors would like to thank the Deanship of Scientific Research at King Abdulaziz University, Saudi Arabia, for its financial support under Grant No. 409/130/1432.

## Author information

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Correspondence to Abdullah Alotaibi.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All the authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.

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Mohiuddine, S.A., Alotaibi, A. & Mursaleen, M. Statistical summability $\left(C,1\right)$ and a Korovkin type approximation theorem. J Inequal Appl 2012, 172 (2012). https://doi.org/10.1186/1029-242X-2012-172