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

Journal of Inequalities and Applications20122012:172

https://doi.org/10.1186/1029-242X-2012-172

• Accepted: 19 July 2012
• Published:

## 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.

## Keywords

• statistical convergence
• statistical summability $\left(C,1\right)$
• positive linear operator
• Korovkin type approximation theorem

## 1 Introduction and preliminaries

In 1951, Fast  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  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 :

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  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  by using almost convergence of double sequences. Quite recently, in  and  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  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 .

## 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 .

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  and the conditions of Theorem A.

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 ${L}_{n}:C\left(I\right)\to C\left(I\right)$ 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  as well as Theorem A.

## Declarations

### 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.

## Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah, 21589, Saudi Arabia
(2)
Department of Mathematics, Aligarh Muslim University, Aligarh, 202002, India

## References 