Statistical $\fancyscript{A}$-summation process and Korovkin type approximation theorem on modular spaces

Abstract

In this paper, we obtain an extension of the classical Korovkin theorem for a sequence of positive linear operators on a modular space using a statistical $\fancyscript{A}$-summation process. Also, we give an example which satisfies this theorem.

1 Introduction

The Korovkin theorem is the object of study of many mathematicians. In the classical Korovkin theorem [1, 18] the uniform convergence in $C\left( \left[ a,b\right] \right)$, the space of all continuous real-valued functions defined on the compact interval $\left[ a,b\right]$, is proved for a sequence of positive linear operators, assuming the convergence only on the test functions 1, $x,$ $x^{2}.$ Recently some versions of Korovkin theorems were proved in the setting of modular spaces, which include as particular cases $L_{p},$ Orlicz and Musielak-Orlicz spaces [24]. Also, in [2], some versions of abstract Korovkin-type theorems in modular function spaces, with respect to filter convergence for linear positive operators, by considering several kinds of test functions have studied. Note that for certain function spaces, as for example $L_{p}$ spaces, in general it is not possible to get the convergence in $L_{p}$ of a sequence of positive linear operators for all the $L_{p}$ functions, but it is necessary to consider spaces, depending on the form of the operators involved. In the present paper, we give a modular version of the Korovkin theorem for a sequence of positive linear operators using a statistical $\fancyscript{A}$-summation process, is an extension of Theorem 2 in [24].

We now recall some basic definitions and notations used in the paper.

Let

$$\begin{aligned} \fancyscript{B}:=\left( b_{lk}\right) \quad (l,k\in {\mathbb {N}} :=\{1,2,3,\ldots \}) \end{aligned}$$

be an infinite summability matrix. For a given sequence of real numbers $x:=\{x_{k}\}$, the $\fancyscript{B}$ -transform of $x$, denoted by

$$\begin{aligned} \fancyscript{B}x:=\{(\fancyscript{B}x)_{l}\}, \end{aligned}$$

is given by

$$\begin{aligned} (\fancyscript{B}x)_{l}=\sum _{k=1}^{\infty }b_{lk}x_{k}, \end{aligned}$$

provided the series converges for each $l\in {\mathbb {N}}$. We say that $\fancyscript{B}$ is regular (see [15]) if

$$\begin{aligned} \lim \fancyscript{B}x=L\text { whenever }\lim x=L. \end{aligned}$$

Assume that $\fancyscript{B}$ is a non-negative regular summability matrix. Then the sequence $x=\{x_{k}\}$ is called $\fancyscript{B}$-statistically convergent to $L$ provided that, for every $\varepsilon >0,$

$$\begin{aligned} \lim _{l}\sum _{k:~\left| x_{k}-L\right| \ge \varepsilon }b_{lk}=0. \end{aligned}$$

(1)

We denote this limit as follows (cf. [11]; see also [7, 8, 17])

$$\begin{aligned} st_{\fancyscript{B}}-\lim x=L. \end{aligned}$$

Actually, this convergence method is based on the concept of $\fancyscript{B}$ -density. Recall the $\fancyscript{B}$ -density of a subset $K\subset {\mathbb {N}}$, denoted by

$$\begin{aligned} \delta _{\fancyscript{B}}\{K\}, \end{aligned}$$

is given by

$$\begin{aligned} \delta _{\fancyscript{B}}\{K\}=\lim _{l}\sum _{k=1}^{\infty }b_{lk}\chi _{K}\left( k\right) , \end{aligned}$$

provided the limit exists, where $\chi _{K}$ is the characteristic function of $K$; or equivalently

$$\begin{aligned} \delta _{\fancyscript{B}}\{K\}=\lim _{l}\sum _{k\in K}b_{lk}. \end{aligned}$$

So, by (1), we easily see that

$$\begin{aligned} st_{\fancyscript{B}}-\lim x=L\text { iff }\delta _{\fancyscript{B} }\{k:\left| x_{k}-L\right| \ge \varepsilon \}=0\text { for every }\varepsilon >0. \end{aligned}$$

We should note that if we take $\fancyscript{B}=C_{1}:=\left( c_{lk}\right)$, the Cesáro matrix defined by

$$\begin{aligned} c_{lk}:=\left\{ \begin{array}{l@{\quad }l} \dfrac{1}{l}, &{} \,\text {if }1\le k\le l, \\ 0, &{} \text {otherwise,} \end{array} \right. \end{aligned}$$

then $\fancyscript{B}$-statistical convergence reduces to the concept of statistical convergence (cf. [10]; see also [12–14]). In this case, we write

$$\begin{aligned} st-\lim x=L\text { instead of }st_{C_{1}}-\lim x=L. \end{aligned}$$

Further, taking $\fancyscript{B}=\fancyscript{I}$, the identity matrix, $\fancyscript{B}$ -statistical convergence coincides with the ordinary convergence, i.e.,

$$\begin{aligned} st_{\fancyscript{I}}-\lim x=\lim x=L. \end{aligned}$$

Observe that every convergent sequence (in the usual sense) is $\fancyscript{B}$ -statistically convergent to the same value for any non-negative regular matrix $\fancyscript{B}$, but its converse is not always true. Actually, in [17], Kolk proved that $\fancyscript{B}$-statistical convergence is stronger than convergence when $\,\fancyscript{B}=\left( b_{lk}\right)$ is a non-negative regular summability matrix such that

$$\begin{aligned} \lim _{l}\max _{k}\{b_{lk}\}=0. \end{aligned}$$

The concepts of statistical limit superior and limit inferior have been introduced by Fridy and Orhan [14]. ${\fancyscript{B}}$ -statistical analogs of these concepts have been examined by Connor and Kline [7], and Demirci [8] as follows. The ${\fancyscript{B}}$ -statistical limit superior of a number sequence $x=\{x_{k}\}$, denoted by

$$\begin{aligned} st_{\fancyscript{B}}-\lim \sup x, \end{aligned}$$

is defined by

$$\begin{aligned} st_{\fancyscript{B}}-\lim \sup x=\left\{ \begin{array}{l@{\quad }l} \sup B_{x}, &{} \text {if }B_{x}\ne \phi , \\ -\infty , &{} \text {if }B_{x}=\phi , \end{array} \right. \end{aligned}$$

where $B_{x}:=\left\{ b\in {\mathbb {R}} :\delta _{\fancyscript{B}}\left\{ k:x_{k}>b\right\} \ne 0\right\}$ and $\phi$ denotes the empty set. We note that by $\delta _{\fancyscript{B}}\{K\}\ne 0$ we mean either $\delta _{\fancyscript{B}}\{K\}>0$ or $K$ fails to have ${\fancyscript{B}}$ -density. Similarly, the ${\fancyscript{B}}$ -statistical limit inferior of $\{x_{k}\},$ denoted by

$$\begin{aligned} st_{\fancyscript{B}}-\lim \inf x, \end{aligned}$$

is defined by

$$\begin{aligned} st_{\fancyscript{B}}-\lim \inf x=\left\{ \begin{array}{l@{\quad }l} \inf C_{x}, &{} \text {if }C_{x}\ne \phi , \\ +\infty , &{} \text {if }C_{x}=\phi , \end{array} \right. \end{aligned}$$

where $C_{x}:=\left\{ c\in {\mathbb {R}} :\delta _{\fancyscript{B}}\left\{ k:x_{k}<c\right\} \ne 0\right\}$. Of course, if we take ${\fancyscript{B}}=C_{1},$ then the above definitions reduce to the concepts of $st-\lim \sup x$ and $st-\lim \inf x$ given in [14], respectively. As in the ordinary limit superior or inferior, it was proved that

$$\begin{aligned} st_{\fancyscript{B}}-\lim \inf x\le st_{\fancyscript{B}}-\lim \sup x \end{aligned}$$

and also that, for any sequence $x=\{x_{k}\}$ satisfying $\delta _{\fancyscript{B}}\{k:\left| x_{k}\right| >M\}=0$ for some $M>0,$

$$\begin{aligned} st_{\fancyscript{B}}-\lim x=L\text { iff }st_{\fancyscript{B}}-\lim \inf x=st_{\fancyscript{B}}-\lim \sup x=L. \end{aligned}$$

We now focus on modular spaces.

Let $I=\left[ a,b\right]$ be a bounded interval of the real line ${\mathbb {R}}$ provided with the Lebesgue measure. Then, by $X\left( I\right)$ we denote the space of all real-valued measurable functions on $I$ provided with equality a.e. As usual, let $C\left( I\right)$ denote the space of all continuous real-valued functions, and $C^{\infty }\left( I\right)$ denote the space of all infinitely differentiable functions on $I$. In this case, we say that a functional $\rho :X\left( I\right) \rightarrow \left[ 0,+\infty \right]$ is a modular on $X\left( I\right)$ provided that the following conditions hold:

• (i) $\rho \left( f\right) =0$ if and only if $f=0$ a.e. in $I$,

• (ii) $\rho \left( -f\right) =\rho \left( f\right)$ for every $f\in X\left( I\right)$,

• (iii) $\rho \left( \alpha f+\beta g\right) \le \rho \left( f\right) +\rho \left( g\right)$ for every $f,g\in X(I)$ and for any $\alpha ,\beta \ge 0$ with $\alpha +\beta =1$.

A modular $\rho$ is said to be $N$ -quasi convex if there exists a constant $N\ge 1$ such that

$$\begin{aligned} \rho \left( \alpha f+\beta g\right) \le N\alpha \rho \left( Nf\right) +N\beta \rho \left( Ng\right) \end{aligned}$$

holds for every $f,g\in X\left( I\right)$, $\alpha ,\beta \ge 0$ with $\alpha +\beta =1$. In particular, if $N=1,$ then $\rho$ is called convex.

A modular $\rho$ is said to be $N$-quasi semiconvex if there exists a constant $N\ge 1$ such that

$$\begin{aligned} \rho (af)\le Na\rho (Nf) \end{aligned}$$

holds for every $f\in X\left( I\right)$ and $a\in (0,1].$

It is clear that every $N$-quasi semiconvex modular is $N$-quasi convex. We should recall that the above two concepts were introduced and discussed in details by Bardaro et. al. [4].

We now consider some appropriate vector subspaces of $X(I)$ by means of a modular $\rho$ as follows:

$$\begin{aligned} L^{\rho }\left( I\right) :=\left\{ f\in X\left( I\right) :\underset{\lambda \rightarrow 0^{+}}{\lim }\rho \left( \lambda f\right) =0\right\} \end{aligned}$$

and

$$\begin{aligned} E^{\rho }\left( I\right) :=\left\{ f\in L^{\rho }\left( I\right) :\rho \left( \lambda f\right) <+\infty \quad \text { for all }\lambda >0\right\} . \end{aligned}$$

Here, $L^{\rho }\left( I\right)$ is called the modular space generated by $\rho ;$ and $E^{\rho }\left( I\right)$ is called the space of the finite elements of $L^{\rho }\left( I\right) .$ Observe that if $\rho$ is $N$-quasi semiconvex, then the space

$$\begin{aligned} \left\{ f\in X\left( I\right) :\rho \left( \lambda f\right) <+\infty \quad \text { for some }\lambda >0\right\} \end{aligned}$$

coincides with $L^{\rho }\left( I\right)$. The notions about modulars are introduced in [23] and widely discussed in [4] (see also [19, 22]).

With the help of the notions of modular convergence and strong convergence, some approximation theorems have recently been introduced by Bardaro and Mantellini [5].

Now we recall the convergence methods in modular spaces.

$\bullet$ Let $\{f_{n}\}$ be a function sequence whose terms belong to $L^{\rho }\left( I\right) .$ Then, $\{f_{n}\}$ is modularly convergent to a function $f\in L^{\rho }\left( I\right)$ iff

$$\begin{aligned} \lim _{n}\rho \left( \lambda _{0}\left( f_{n}-f\right) \right) =0 \quad \text { for some }\lambda _{0}>0. \end{aligned}$$

(2)

$\bullet$ Also, $\{f_{n}\}$ is $F$-norm convergent (or, strongly convergent) to $f$ iff

$$\begin{aligned} \lim _{n}\rho \left( \lambda \left( f_{n}-f\right) \right) =0\quad \text { for every }\lambda >0. \end{aligned}$$

(3)

It is known from [22] that (2) and (3) are equivalent if and only if the modular $\rho$ satisfies the $\varDelta _{2}$-condition, i.e. there exists a constant $M>0$ such that $\rho \left( 2f\right) \le M\rho \left( f\right)$ for every $f\in X\left( I\right)$.

In this paper, we will need the following assumptions on a modular $\rho$:

• if $\rho (f)\le \rho (g)$ for $\left| f\right| \le \left| g\right| ,$ then $\rho$ is monotone,

• $\rho$ is finite if $\chi _{A}$ $\in$ $L^{\rho }\left( I\right)$ whenever $A$ is measurable subset of $I$ such that $\mu \left( A\right) <\infty ,$

• if $\rho$ is finite and, for every $\varepsilon >0$, $\lambda >0,$ there exists a $\delta >0$ such that $\rho \left( \lambda \chi _{B}\right) <\varepsilon$ for any measurable subset $B\subset I$ with $\mu \left( B\right) <\delta$, then $\rho$ is absolutely finite,

• if $\chi _{I}\in E^{\rho }\left( I\right) ,$ then $\rho$ is strongly finite

• $\rho$ is absolutely continuous provided that there exists an $\alpha >0$ such that, for every $f\in X\left( I\right)$ with $\rho \left( f\right) <+\infty$, the following condition holds: for every $\varepsilon >0$ there is $\delta >0$ such that $\rho \left( \alpha f\chi _{B}\right) <\varepsilon$ whenever $B$ is any measurable subset of $I$ with $\mu \left( B\right) <\delta$.

Observe now that (see [5]) if a modular $\rho$ is monotone and finite, then we have $C(I)\subset$ $L^{\rho }\left( I\right) .$ In a similar manner, if $\rho$ is monotone and strongly finite, then $C(I)\subset$ $E^{\rho }\left( I\right)$. Also, if $\rho$ is monotone, absolutely finite and absolutely continuous, then $\overline{C^{\infty }\left( I\right) }=$ $L^{\rho }\left( I\right) .$ Some important relations between the above properties may be found in [3, 4, 21, 23].

2 Korovkin type theorems

Let $\fancyscript{A}:=\left\{ A^{n}\right\} _{n\ge 1}$, $A^{n}=\left( a_{kj}^{\left( n\right) }\right) _{k,j\in }$ be a sequence of infinite non-negative real matrices. For a sequence of real numbers, $x=\left\{ x_{j}\right\} _{j\in }$, the double sequence

$$\begin{aligned} \fancyscript{A}x:=\left\{ \left( Ax\right) _{k}^{n}:k,n\in \right\} \end{aligned}$$

defined by $\left( Ax\right) _{k}^{n}:= \sum \nolimits _{j=1}^{\infty } a_{kj}^{\left( n\right) }x_{j}$ is called the $\fancyscript{A}$-transform of $x$ whenever the series converges for all $k$ and $n$. A sequence $x$ is said to be $\fancyscript{A}$-summable to $L$ if

$$\begin{aligned} \underset{k}{\lim }\underset{j=1}{\overset{\infty }{\sum }}a_{kj}^{\left( n\right) }x_{j}=L \end{aligned}$$

uniformly in $n$ ([6, 26]).

If $A^{n}=\fancyscript{B}$ for some matrix $\fancyscript{B}$, then $\fancyscript{A}$-summability is the ordinary matrix summability by $\fancyscript{B}$. If, $a_{kj}^{\left( n\right) }=\frac{1}{k+1}$, for $n\le j\le k+n$, $\left( n=1,2,\ldots \right)$, and $a_{kj}^{\left( n\right) }=0$ otherwise, then $\fancyscript{A}$-summability reduces to almost convergence [20].

Let $\rho$ be a monotone and finite modular on $X\left( I\right)$, and let $\fancyscript{B}=\left( b_{lk}\right)$ be a non-negative regular summability matrix. Assume that $D$ is a set satisfying $C^{\infty }\left( I\right) \subset D\subset L^{\rho }\left( I\right) .$ We can construct such a subset $D$ when $\rho$ is monotone and finite (see [5]). Here, $D$ is domain of the operator ${\mathbb {T}}$. We will assume that ${\mathbb {T}}:=\{T_{j}\}$ is a sequence of positive linear operators from $D$ into $X\left( I\right)$ and for all $k,n\in ,$ $f\in D$ the series

$$\begin{aligned} A_{k,n}^{\mathbb {T}}\left( f\right) :=\overset{\infty }{\underset{j=1}{\sum } }a_{kj}^{\left( n\right) }T_{j}f, \end{aligned}$$

is absolutely convergent almost everywhere with respect to Lebesgue measure. Also, assume that there exists a subset $X_{\mathbb {T}}\subset D$ with $C^{\infty }\left( I\right) \subset X_{\mathbb {T}}$ and a constant $P>0$ such that,

$$\begin{aligned} st_{\fancyscript{B}}-\limsup \limits _{k}\rho \left( \lambda \left( A_{k,n}^{\mathbb {T}}\left( f\right) \right) \right) \le P\rho \left( \lambda f\right) ,\text { uniformly in }n \end{aligned}$$

(4)

holds for every $f\in X_{\mathbb {T}},$ $\lambda >0.$

A sequence ${\mathbb {T}}:=\{T_{j}\}$ of positive linear operators of $D$ into $X\left( I\right)$ is called an $\fancyscript{A}$-summation process on $D$ if $\left\{ T_{j}\left( f\right) \right\}$ is $\fancyscript{A}$-summable to $f$ (with respect to modular $\rho$) for every $f\in D$ , i.e.,

$$\begin{aligned} \underset{k}{\lim }\rho \left[ \lambda \left( \overset{\infty }{\underset{j=1 }{\sum }}a_{kj}^{\left( n\right) }T_{j}f-f\right) \right] =0, \text { uniformly in }n. \end{aligned}$$

(5)

A different definition is given by, (see [16])

$$\begin{aligned} \underset{k}{\lim }\overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }\rho \left[ \lambda \left( T_{j}f-f\right) \right] =0,\text { uniformly in }n \end{aligned}$$

(6)

for all $f\in D$ where it is assumed that ${\sup \nolimits _{n,k}}{\sum \nolimits _{j=1}^{\infty }}a_{kj}^{\left( n\right) }<\infty$ holds.

In this paper, we establish a theorem of the Korovkin type with respect to the convergence behavior (5) for a sequence of positive linear operators of $D$ into $X\left( I\right) .$ So the results of type (5 ) are extensions of type (6). Also, the following theorem is an extension of Theorem 2 in [24]. Some results concerning summation processes in the space $L_{p}\left[ a,b\right]$ of Lebesgue integrable functions on a compact interval may be found [24, 25].

Throughout the paper we use the test functions $e_{i}$ defined by

$$\begin{aligned} e_{i}(x)=x^{i} \quad (i=0,1,2,\ldots ). \end{aligned}$$

Also, we denote the value of $T_{j}f$ at a point $x\in I$ by $T_{j}\left( f\left( y\right) ;x\right)$ or, briefly, $T_{j}\left( f;x\right)$.

Theorem 1

Let $\fancyscript{A}=\left\{ A^{n}\right\} _{n\ge 1}$ be a sequence of infinite non-negative real matrices and let $\rho$ be a monotone, strongly finite, absolutely continuous and $N$-quasi semiconvex modular on $X\left( I\right)$, also $\fancyscript{B}=\left( b_{lk}\right)$ be a non-negative regular summability matrix. Let ${\mathbb {T}}:=\{T_{j}\}$ be a sequence of positive linear operators from $D$ into $X\left( I\right)$ satisfying (4) for each $f\in D.$ Suppose that

$$\begin{aligned} st_{\fancyscript{B}}-\lim _{k}\rho \left( \lambda \left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}e_{i}-e_{i}\right) \right) =0,\text { uniformly in }n\, \end{aligned}$$

(7)

for every $\lambda >0$ and $i=0,1,2.$ Now let $f$ be any function belonging to $L^{\rho }\left( I\right)$ such that $f-g\in X_{\mathbb {T}}$ for every $g\in C^{\infty }\left( I\right) .$ Then, we have

$$\begin{aligned} st_{\fancyscript{B}}-\lim _{k}\rho \left( \lambda _{0}\left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}f-f\right) \right) =0, \text { uniformly in }n \end{aligned}$$

(8)

for some $\lambda _{0}>0.$

Proof

We first claim that

$$\begin{aligned} st_{\fancyscript{B}}-\lim _{k}\rho \left( \eta \left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}g-g\right) \right) =0 \,\textit{ uniformly in} \, n \end{aligned}$$

(9)

for every $g\in C(I)\cap D$ and every $\eta >0.$ To see this assume that $g$ belongs to $C\left( I\right) \cap D$. By the continuity of $g$ on $I$, given $\varepsilon >0,$ there exists a number $\delta >0$ such that for all $x,y\in I$ satisfying $\left| y-x\right| <\delta$ we have

$$\begin{aligned} \left| g\left( y\right) -g\left( x\right) \right| <\varepsilon \end{aligned}$$

(10)

Also we get for all $x,y\in I$ satisfying $\left| y-x\right| >\delta$ that

$$\begin{aligned} \left| g\left( y\right) -g\left( x\right) \right| \le \frac{2M}{\delta ^{2}}\left( y-x\right) ^{2} \end{aligned}$$

(11)

where $M:=\sup \nolimits _{x\in I}\left| g(x)\right| .$Combining ( 10) and (11) we have for $x,y\in I$ that

$$\begin{aligned} \left| g\left( y\right) -g\left( x\right) \right| <\varepsilon +\frac{2M}{\delta ^{2}}\left( y-x\right) ^{2}. \end{aligned}$$

Since $T_{j}$ is a positive linear operator, we get

$$\begin{aligned}&\left| \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}\left( g;x\right) -g\left( x\right) \right| \\&\quad =\left| \overset{\infty }{\sum \limits _{j=1}}a_{kj}^{\left( n\right) }T_{j}\left( g\left( y\right) ;x\right) -\overset{\infty }{\sum \limits _{j=1} }a_{kj}^{\left( n\right) }T_{j}\left( g\left( x\right) ;x\right) +\overset{\infty }{\sum \limits _{j=1}}a_{kj}^{\left( n\right) }T_{j}\left( g\left( x\right) ;x\right) -g\left( x\right) \right| \\&\quad =\left| \overset{\infty }{\sum \limits _{j=1}}a_{kj}^{\left( n\right) }T_{j}\left( g\left( y\right) -g\left( x\right) ;x\right) +g\left( x\right) \left( \overset{\infty }{\sum \limits _{j=1}}a_{kj}^{\left( n\right) }T_{j}\left( 1;x\right) -1\right) \right| \\&\quad \le \overset{\infty }{\sum \limits _{j=1}}a_{kj}^{\left( n\right) }T_{j}\left( \left| g\left( y\right) -g\left( x\right) \right| ;x\right) +\left| g\left( x\right) \right| \left| \overset{\infty }{\sum \limits _{j=1}}a_{kj}^{\left( n\right) }T_{j}\left( 1;x\right) -1\right| \\&\quad \le \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}\left( \varepsilon +\frac{2M}{\delta ^{2}}\left( y-x\right) ^{2};x\right) +\left| g\left( x\right) \right| \left| \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}\left( 1;x\right) -1\right| \\&\quad \le \varepsilon +\varepsilon \left| \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}\left( 1;x\right) -1\right| +\frac{2M}{\delta ^{2}}\overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}\left( \left( y-x\right) ^{2};x\right) \\&\qquad +M\left| \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}\left( 1;x\right) -1\right| \\&\quad =\varepsilon +\left( \varepsilon +M\right) \left| \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}\left( 1;x\right) -1\right| +\frac{2M}{\delta ^{2}}\left[ \left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}\left( y^{2};x\right) -x^{2}\right) \right. \\&\qquad \left. -2x\left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}\left( y;x\right) -x\right) +x^{2}\left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}\left( 1;x\right) -1\right) \right] \\&\quad \le \varepsilon +\left( \varepsilon +M+\frac{2Mc^{2}}{\delta ^{2}}\right) \left| \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}\left( 1;x\right) -1\right| +\frac{4Mc}{\delta ^{2}}\left| \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}\left( y;x\right) -x\right| \\&\qquad +\frac{2M}{\delta ^{2}}\left| \overset{\infty }{\underset{j=1}{\sum }} a_{kj}^{\left( n\right) }T_{j}\left( y^{2};x\right) -x^{2}\right| \end{aligned}$$

where $c:=\max \left\{ \left| a\right| ,\left| b\right| \right\}$. So, the last inequality gives, for any $\eta >0$ that

$$\begin{aligned} \eta \left| \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}\left( g;x\right) -g\left( x\right) \right| \!&\le \! \left\{ \eta \varepsilon \!+\!\eta K\left| \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}\left( 1;x\right) -1\right| \right. \\&\left. +\left| \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}\left( y;x\right) \!-\! x\right| \!+\! \left| \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}\left( y^{2};x\right) \!-\! x^{2}\right| \right\} \end{aligned}$$

where $K:=\max \left\{ \varepsilon +M+\frac{2Mc^{2}}{\delta ^{2}},\frac{4Mc}{\delta ^{2}},\frac{2M}{\delta ^{2}}\right\} .$ Applying the modular $\rho$ in both-sides of the above inequality, since $\rho$ is monotone, we have

$$\begin{aligned} \rho \left( \eta \left( \overset{\infty }{\underset{j=1}{\sum }} a_{kj}^{\left( n\right) }T_{j}g-g\right) \right)&\le \rho \left( \eta \varepsilon +\eta K\left| \overset{\infty }{\underset{j=1}{\sum }} a_{kj}^{\left( n\right) }T_{j}e_{0}-e_{0}\right| \right. \\&\left. +\,\eta K\left| \overset{\infty }{\underset{j=1}{\sum }} a_{kj}^{\left( n\right) }T_{j}e_{1}-e_{1}\right| \right. \\&\left. +\,\eta K\left| \overset{\infty }{\underset{j=1}{\sum }} a_{kj}^{\left( n\right) }T_{j}e_{2}-e_{2}\right| \right) . \end{aligned}$$

So, we may write that

$$\begin{aligned}&\rho \left( \eta \left( \overset{\infty }{\underset{j=1}{\sum }} a_{kj}^{\left( n\right) }T_{j}g-g\right) \right) \\&\quad \le \rho (4\eta \varepsilon )+\rho \left( 4\eta K\left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}e_{0}-e_{0}\right) \right) \\&\qquad +\rho \left( 4\eta K\left( \overset{\infty }{\underset{j=1}{\sum }} a_{kj}^{\left( n\right) }T_{j}e_{1}-e_{1}\right) \right) +\rho \left( 4\eta K\left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}e_{2}-e_{2}\right) \right) . \end{aligned}$$

Since $\rho$ is $N$-quasi semiconvex and strongly finite, we have, assuming

$0<\varepsilon \le 1$

$$\begin{aligned}&\rho \left( \eta \left( \overset{\infty }{\underset{j=1}{\sum }} a_{kj}^{\left( n\right) }T_{j}g-g\right) \right) \\&\quad \le N\varepsilon \rho \left( 4\eta N\right) +\rho \left( 4\eta K\left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}e_{0}-e_{0}\right) \right) \\&\qquad +\rho \left( 4\eta K\left( \overset{\infty }{\underset{j=1}{\sum }} a_{kj}^{\left( n\right) }T_{j}e_{1}-e_{1}\right) \right) +\rho \left( 4\eta K\left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}e_{2}-e_{2}\right) \right) \text {.} \end{aligned}$$

For a given $r>0,$ choose an $\varepsilon \in (0,1]$ such that $N\varepsilon \rho \left( 4\eta N\right) <r.$ Now define the following sets:

$$\begin{aligned} S_{\eta }&:= \left\{ k:\rho \left( \eta \left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{(n) }T_{j}g-g\right) \right) \ge r\right\} , \\ S_{\eta ,i}&:= \left\{ k:\rho \left( 4\eta K\left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{(n) }T_{j}e_{i}-e_{i}\right) \right) \ge \frac{r-N\varepsilon \rho \left( 4\eta N\right) }{3}\right\} , \end{aligned}$$

where $i=0,1,2.$ Then, it is easy to see that $S_{\eta }\subseteq \bigcup \nolimits _{i=0}^{2}S_{\eta ,i}$. So we can write, for all $l\in {\mathbb {N}},$ that

$$\begin{aligned} \sum _{k\in S_{\eta }}b_{lk}\le \sum _{k\in S_{\eta ,0}}b_{lk}+\sum _{k\in S_{\eta ,1}}b_{lk}+\sum _{k\in S_{\eta ,2}}b_{lk}\text {.} \end{aligned}$$

(12)

Taking limit as $l\rightarrow \infty$ in (12) and using the hypothesis (7), we get

$$\begin{aligned} \lim _{l}\sum _{k\in S_{\eta }}b_{lk}=0, \end{aligned}$$

which proves our claim (9). Observe that (9) also holds for every

$g\in C^{\infty }(I)$ because of $C^{\infty }(I)\subset C(I)\cap D.$ Now let $f\in L^{\rho }\left( I\right)$ satisfying $f-g\in X_{\mathbb {T}}$ for every $g\in C^{\infty }\left( I\right)$. Since $\mu \left( I\right) <\infty$ and $\rho$ is strongly finite and absolutely continuous, we can see that $\rho$ is also absolutely finite on $X(I)$ (see [3]). Using these properties of the modular $\rho ,$ it is known from [4, 21] that the space $C^{\infty }(I)$ is modularly dense in $L^{\rho }\left( I\right) ,$ i.e., there exists a sequence $\{g_{k}\}\subset C^{\infty }\left( I\right)$ such that

$$\begin{aligned} \lim _{k}\rho \left[ 3\lambda _{0}^{*}\left( g_{k}-f\right) \right] =0\quad \text { for some }\lambda _{0}^{*}>0. \end{aligned}$$

This means that, for every $\varepsilon >0,$ there is a positive number $k_{0}=k_{0}(\varepsilon )$ so that

$$\begin{aligned} \rho \left[ 3\lambda _{0}^{*}\left( g_{k}-f\right) \right] <\varepsilon \quad \text { for every }k\ge k_{0}. \end{aligned}$$

(13)

On the other hand, by the linearity and positivity of the operators $T_{j}$, we may write that

$$\begin{aligned} \lambda _{0}^{*}\left| \overset{\infty }{\underset{j=1}{\sum }} a_{kj}^{\left( n\right) }T_{j}(f;x)-f(x)\right|&\le \lambda _{0}^{*}\left| \overset{\infty }{\underset{j=1}{\sum }} a_{kj}^{\left( n\right) }T_{j}(f-g_{k_{0}};x)\right| \\&+\lambda _{0}^{*}\left| \overset{\infty }{\underset{j=1}{\sum }} a_{kj}^{\left( n\right) }T_{j}(g_{k_{0}};x)-g_{k_{0}}(x)\right| \\&+\lambda _{0}^{*}\left| g_{k_{0}}(x)-f(x)\right| \end{aligned}$$

holds for every $x\in I$ and $n\in {\mathbb {N}}$. Applying the modular $\rho$ in the last inequality and using the monotonicity of $\rho ,$ we have

$$\begin{aligned} \rho \left( \lambda _{0}^{*}\left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}f-f\right) \right)&\le \rho \left( 3\lambda _{0}^{*}\left( \overset{\infty }{\underset{j=1}{\sum }} a_{kj}^{\left( n\right) }T_{j}\left( f-g_{k_{0}}\right) \right) \right) \nonumber \\&+\rho \left( 3\lambda _{0}^{*}\left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}g_{k_{0}}-g_{k_{0}}\right) \right) \nonumber \\&+\rho \left( 3\lambda _{0}^{*}\left( g_{k_{0}}-f\right) \right) \text {.} \end{aligned}$$

(14)

Then, it follows from (13) and (14) that

$$\begin{aligned} \rho \left( \lambda _{0}^{*}\left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}f-f\right) \right)&\le \varepsilon +\rho \left( 3\lambda _{0}^{*}\overset{\infty }{\underset{j=1}{\sum }} a_{kj}^{\left( n\right) }T_{j}\left( f-g_{k_{0}}\right) \right) \nonumber \\&+\rho \left( 3\lambda _{0}^{*}\left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}g_{k_{0}}-g_{k_{0}}\right) \right) . \end{aligned}$$

(15)

So, taking $\fancyscript{B}$-statistical limit superior as $k\rightarrow \infty$ in the both-sides of (15) and also using the facts that $g_{k_{0}}\in C^{\infty }(I)$ and $f-g_{k_{0}}\in X_{\mathbb {T}},$ we obtained from (4) that

$$\begin{aligned}&st_{\fancyscript{B}}-\limsup _{k}\rho \left( \lambda _{0}^{*}\left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}f-f\right) \right) \\&\quad \le \varepsilon +P\rho \left( 3\lambda _{0}^{*}(f-g_{k_{0}})\right) \\&\qquad +st_{\fancyscript{B}}-\limsup _{k}\rho \left( 3\lambda _{0}^{*}\left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}g_{k_{0}}-g_{k_{0}}\right) \right) , \end{aligned}$$

which gives

$$\begin{aligned}&st_{\fancyscript{B}}-\limsup _{k}\rho \left( \lambda _{0}^{*}\left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}f-f\right) \right) \nonumber \\&\quad \le \varepsilon (P+1)+st_{\fancyscript{B}}-\limsup _{k}\rho \left( 3\lambda _{0}^{*}\left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}g_{k_{0}}-g_{k_{0}}\right) \right) . \end{aligned}$$

(16)

By (9), since

$$\begin{aligned} st_{\fancyscript{B}}-\underset{k}{\lim }\rho \left( 3\lambda _{0}^{*}\left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}g_{k_{0}}-g_{k_{0}}\right) \right) =0,\text { uniformly in }n, \end{aligned}$$

we get

$$\begin{aligned} st_{\fancyscript{B}}-\limsup _{k}\rho \left( 3\lambda _{0}^{*}\left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}g_{k_{0}}-g_{k_{0}}\right) \right) =0,\text { uniformly in }n. \end{aligned}$$

(17)

Combining (16) with (17), we conclude that

$$\begin{aligned} st_{\fancyscript{B}}-\limsup _{k}\rho \left( \lambda _{0}^{*}\left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}f-f\right) \right) \le \varepsilon (P+1). \end{aligned}$$

Since $\varepsilon >0$ was arbitrary, we find

$$\begin{aligned} st_{\fancyscript{B}}-\limsup _{k}\rho \left( \lambda _{0}^{*}\left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}f-f\right) \right) =0\text { uniformly in }n. \end{aligned}$$

Furthermore, since $\rho \left( \lambda _{0}^{*}\left( {\sum _{j=1}^{\infty }}a_{kj}^{\left( n\right) }T_{j}f-f\right) \right)$ is non-negative for all $k,n\in {\mathbb {N}}$, we can easily show that

$$\begin{aligned} st_{\fancyscript{B}}-\lim _{k}\rho \left( \lambda _{0}^{*}\left( \overset{\infty }{\underset{j=1}{\sum }}a_{kj}^{\left( n\right) }T_{j}f-f\right) \right) =0,\text { uniformly in }n \end{aligned}$$

which completes the proof.

If the modular $\rho$ satisfies the $\varDelta _{2}$-condition, then one can get the following result from Theorem 1 at once.

Theorem 2

Let $\fancyscript{A}=\left\{ A^{n}\right\} _{n\ge 1}$ be a sequence of infinite non-negative real matrices, $\fancyscript{B}=\left( b_{lk}\right)$ be a non-negative regular summability matrix and ${\mathbb {T}}:=\{T_{j}\},$ $\rho$ be the same as in Theorem 1$.$ If $\rho$ satisfies the $\varDelta _{2}$-condition, then the following statements are equivalent:

• (a) $st_{\fancyscript{B}}-\lim \limits _{k}\rho \left( \lambda \left( {\sum _{j=1}^{\infty }}a_{kj}^{\left( n\right) }T_{j}e_{i}-e_{i}\right) \right) =0$ uniformly in $n$ for every $\lambda >0$ and $i=0,1,2,$

• (b) $st_{\fancyscript{B}}-\lim \limits _{k}\rho \left( \lambda \left( {\sum _{j=1}^{\infty }}a_{kj}^{\left( n\right) }T_{j}f-f\right) \right) =0$ uniformly in $n$ for every $\lambda >0$ provided that $f$ is any function belonging to $L^{\rho }(I)$ such that $f-g\in X_{\mathbb {T}}$ for every $g\in C^{\infty }\left( I\right)$.

If one replaces the matrices $\fancyscript{B}$ and $A^{n}\left( n\ge 1\right)$ by the identity matrix, then the condition (4) reduces to

$$\begin{aligned} st-\limsup \limits _{k}\rho \left( \lambda \left( T_{j}h\right) \right) \le P\rho \left( \lambda h\right) \end{aligned}$$

(18)

for every $h\in X_{\mathbb {T}},$ $\lambda >0$ and for an absolute positive constant $P.$ In this case, the next results which were obtained by Bardaro and Mantellini [5] immediately follows from our Theorems 1 and 2.

Corollary 1

[5] Let $\rho$ be a monotone, strongly finite, absolutely continuous and $N$-quasi semiconvex modular on $X\left( I\right)$. Let ${\mathbb {T}}:=\{T_{j}\}$ be a sequence of positive linear operators from $D$ into $X\left( I\right)$ satisfying (18). If $\{T_{j}e_{i}\}$ is strongly convergent to $e_{i}$ for each $i=0,1,2,$ then $\{T_{j}f\}$ is modularly convergent to $f$ provided that $f$ is any function belonging to $L^{\rho }\left( I\right)$ such that $f-g\in X_{\mathbb {T}}$ for every $g\in C^{\infty }\left( I\right)$.

Corollary 2

[5] ${\mathbb {T}}:=\{T_{j}\}$ and $\rho$ be the same as in Corollary 1$.$ If $\rho$ satisfies the $\varDelta _{2}$-condition, then the following statements are equivalent:

• (a) $\{T_{j}e_{i}\}$ is strongly convergent to $e_{i}$ for each $i=0,1,2,$

• (b) $\{T_{j}f\}$ is strongly convergent to $f$ provided that $f$ is any function belonging to $L^{\rho }(I)$ such that $f-g\in X_{\mathbb {T}}$ for every $g\in C^{\infty }\left( I\right)$.

3 Application

In this section we give an example of positive linear operators which satisfy the conditions of Theorem 1.

Example 1

Take $I=[0,1]$ and let $\varphi :\left[ 0,\infty \right) \rightarrow \left[ 0,\infty \right)$ be a continuous function for which the following conditions hold:

• $\varphi$ is convex,

• $\varphi \left( 0\right) =0$, $\varphi \left( u\right) >0$ for $u>0$ and $\lim \nolimits _{u\rightarrow \infty } \varphi \left( u\right) =\infty$.

Hence, consider the functional $\rho ^{\varphi }$ on $X(I)$ defined by

$$\begin{aligned} \rho ^{\varphi }(f):=\int \limits _{0}^{1}\varphi \left( \left| f\left( x\right) \right| \right) dx\,\quad \textit{for} \, f\in X\left( I\right) . \end{aligned}$$

(19)

In this case, $\rho ^{\varphi }$ is a convex modular on $X\left( I\right) ,$ which satisfies all assumptions listed in Sect. 1 (see [5]). Consider the Orlicz space generated by $\varphi$ as follows:

$$\begin{aligned} L_{\varphi }^{\rho }(I):=\left\{ f\in X\left( I\right) :\rho ^{\varphi }\left( \lambda f\right) <+\infty \,\quad \textit{for some}\, \lambda >0\right\} . \end{aligned}$$

Then consider the following classical Bernstein-Kantorovich operator $\mathbf {U}:=\{U_{j}\}$ on the space $L_{\varphi }^{\rho }\left( I\right)$ (see [5]which is defined by:

$$\begin{aligned} U_{j}(f;x):=\overset{j}{\underset{k=0}{\sum }}\left( {\begin{array}{c}j\\ k\end{array}}\right) x^{k}\left( 1-x\right) ^{j-k}\left( j+1\right) \int \limits _{k/\left( j+1\right) }^{\left( k+1\right) /\left( j+1\right) }f\left( t\right) dt \text { for } x\in I. \end{aligned}$$

Observe that the operators $U_{j}$ map the Orlicz space $L_{\varphi }^{\rho }\left( I\right)$ into itself. Moreover, property (18) is satisfied with the choice of $X_{\mathbf {U} }:=L_{\varphi }^{\rho }(I).$ Then, by Corollary 1, we know that, for any function $f\in L_{\varphi }^{\rho }\left( I\right)$ such that $f-g\in X_{\mathbf {U}}$ for every $g\in C^{\infty }\left( I\right)$, $\left\{ U_{j}f\right\}$ is modularly convergent to $f.$

If $\varphi \left( x\right) =x^{p}$ for $1\le p<\infty ,$ $x\ge 0,$ then $L_{\varphi }^{\rho }(I)=L_{p}\left( I\right)$ Moreover we have

$$\begin{aligned} \rho ^{\varphi }(f)=\left\| f\right\| _{L_{p}\left( I\right) }^{p}. \end{aligned}$$

Now take $\fancyscript{B}=C_{1}=\left( c_{kj}\right) ,$ the Ces áro matrix of order one. In this case, we know that $C_{1}$ -statistical convergence coincides with statistical convergence, and its limit is denoted by $st-\lim$ Assume that $\fancyscript{A}:=\left\{ A^{n}\right\} _{n\ge 1}=\small \left\{ \left( a_{kj}^{\left( n\right) }\right) _{k,j\in {\mathbb {N}}}\right\} _{n\ge 1}$ is a sequence of infinite matrices defined by $a_{kj}^{\left( n\right) }=\frac{1}{k+1}$ if $n\le j\le n+k,$ $\left( n=1,2,\ldots \right)$ and $a_{kj}^{\left( n\right) }=0$ otherwise. Since, for positive constant $C,$ $\left\| U_{j}\left( f;x\right) \right\| _{L_{p}}\le C\left\| f\right\| _{L_{p}}$ [9], we can easily see that

$$\begin{aligned} st-\limsup \limits _{k}\left\| \overset{\infty }{\underset{j=1}{\sum }} a_{kj}^{\left( n\right) }U_{j}f\right\| _{L_{p}}^{p}\le C\left\| f\right\| _{L_{p}}^{p},\,\,\textit{uniformly in}\,\,n. \end{aligned}$$

We now claim that

$$\begin{aligned} st-\lim _{k}\left\| \overset{\infty }{\sum \limits _{j=1}}a_{kj}^{\left( n\right) }U_{j}e_{i}-e_{i}\right\| _{L_{p}}^{p}=0, \, \textit{uniformly in} \,n,\quad i=0,1,2. \end{aligned}$$

(20)

Observe that $U_{j}\left( e_{0};x\right) =e_{0},$ $U_{j}\left( e_{1};x\right) =\frac{jx}{j+1}+\frac{1}{2\left( j+1\right) }$ and $U_{j}\left( e_{2};x\right) =\frac{j\left( j-1\right) x^{2}}{\left( j+1\right) ^{2}}+\frac{2jx}{\left( j+1\right) ^{2}}+\frac{1}{3\left( j+1\right) ^{2}}.$ So, we can see,

$$\begin{aligned} \left\| \overset{\infty }{\sum \limits _{j=1}}a_{kj}^{\left( n\right) }U_{j}\left( e_{0};x\right) -e_{0}(x)\right\| _{L_{p}}&= \left\| \overset{n+k}{\sum \limits _{j=n}}\frac{1}{k+1}U_{j}\left( e_{0};x\right) -e_{0}(x)\right\| _{L_{p}} \\&= \left\| 1-1\right\| _{_{L_{p}}}=0, \end{aligned}$$

we get

$$\begin{aligned} st-\underset{k}{\lim }\left\| \overset{\infty }{\sum \limits _{j=1}} a_{kj}^{\left( n\right) }U_{j}e_{0}-e_{0}\right\| _{L_{p}}^{p}=0,\,\textit{uniformly in}\,n. \end{aligned}$$

which guarantees that ( 20 ) holds true for $i=0.$ Also, we have

$$\begin{aligned} \left\| \overset{\infty }{\sum \limits _{j=1}}a_{kj}^{\left( n\right) }U_{j}\left( e_{1};x\right) -e_{1}(x)\right\| _{L_{p}}&= \left\| \overset{n+k}{\sum \limits _{j=n}}\frac{1}{k+1}U_{j}\left( e_{1};x\right) -e_{1}(x)\right\| _{L_{p}} \\&= \left\| x\frac{1}{k+1}\overset{n+k}{\sum \limits _{j=n}}\frac{j}{j+1}+ \frac{1}{2\left( k+1\right) }\overset{n+k}{\sum \limits _{j=n}}\frac{1}{j+1} -x\right\| _{_{L_{p}}} \\&\le \left( \frac{1}{k+1}\overset{n+k}{\sum \limits _{j=n}}\frac{j}{j+1} -1\right) \left\| e_{1}(x)\right\| _{_{_{L_{p}}}} \\&+\left( \frac{1}{2\left( k+1\right) }\overset{n+k}{\sum \limits _{j=n}} \frac{1}{j+1}\right) \left\| e_{0}(x)\right\| _{_{_{L_{p}}}} \\&= \frac{1}{\left( p+1\right) ^{^{1/p}}}\left( -\frac{1}{k+1}\overset{n+k}{\sum \limits _{j=n}}\frac{1}{j+1}\right) \\&+\left( \frac{1}{2\left( k+1\right) }\overset{n+k}{\sum \limits _{j=n}} \frac{1}{j+1}\right) \\&= \left( \frac{1}{k+1}\overset{n+k}{\sum \limits _{j=n}}\frac{1}{j+1}\right) \left( \frac{1}{2}-\frac{1}{\left( p+1\right) ^{^{1/p}}}\right) \end{aligned}$$

Since $st-\lim \nolimits _{k} \small \left( {\sup \nolimits _{n}}\frac{1}{k+1} {\sum \nolimits _{j=n}^{n+k}}\frac{1}{j+1}\right) =0$, we have,

$$\begin{aligned}&st-\underset{k}{\lim }\left( \underset{n}{\sup }\left\| \overset{\infty }{\sum \limits _{j=1}}a_{kj}^{\left( n\right) }U_{j}\left( e_{1};x\right) -e_{1}\right\| _{L_{p}}\right) \\&\quad \le \left( \frac{1}{2}-\frac{1}{\left( p+1\right) ^{^{1/p}}}\right) st- \underset{k}{\lim }\left( \underset{n}{\sup }\frac{1}{k+1}\overset{n+k}{\sum \limits _{j=n}}\frac{1}{j+1}\right) \end{aligned}$$

which gives

$$\begin{aligned} st-\underset{k}{\lim }\left\| \overset{\infty }{\sum \limits _{j=1}} a_{kj}^{\left( n\right) }U_{j}e_{1}-e_{1}\right\| _{L_{p}}=0, \,\textit{uniformly in} \, n. \end{aligned}$$

So, we have

$$\begin{aligned} st-\underset{k}{\lim }\left\| \overset{\infty }{\sum \limits _{j=1}} a_{kj}^{\left( n\right) }U_{j}e_{1}-e_{1}\right\| _{L_{p}}^{p}=0,\,\textit{uniformly in}\,n. \end{aligned}$$

Finally, since

$$\begin{aligned}&\left\| \overset{\infty }{\sum \limits _{j=1}}a_{kj}^{\left( n\right) }U_{j}\left( e_{2};x\right) -e_{2}\left( x\right) \right\| _{L_{p}} \\&\quad =\left\| \overset{n+k}{\sum \limits _{j=n}}\frac{1}{k+1}U_{j}\left( e_{2};x\right) -e_{2}\left( x\right) \right\| _{L_{p}} \\&\quad =\left\| \overset{n+k}{\sum \limits _{j=n}}\frac{1}{k+1}\left( \frac{j\left( j-1\right) x^{2}}{\left( j+1\right) ^{2}}+\frac{2jx}{\left( j+1\right) ^{2}}+\frac{1}{3\left( j+1\right) ^{2}}\right) -x^{2}\right\| _{_{L_{p}}} \\&\quad \le \left( \frac{1}{k+1}\overset{n+k}{\sum \limits _{j=n}}\frac{j\left( j-1\right) }{\left( j+1\right) ^{2}}-1\right) \left\| e_{2}(x)\right\| _{_{_{L_{p}}}}+\frac{1}{k+1}\overset{n+k}{\sum \limits _{j=n}}\frac{2j}{\left( j+1\right) ^{2}}\left\| e_{1}(x)\right\| _{_{_{L_{p}}}} \\&\qquad +\frac{1}{k+1}\overset{n+k}{\sum \limits _{j=1}}\frac{1}{3\left( j+1\right) ^{2}}\left\| e_{0}(x)\right\| _{_{_{L_{p}}}}\\&\quad =\frac{1}{\left( 2p+1\right) ^{^{1/p}}}\left( \frac{1}{k+1}\overset{n+k}{\sum \limits _{j=n}}\frac{j\left( j-1\right) }{\left( j+1\right) ^{2}}-1\right) \\&\qquad +\frac{1}{\left( p+1\right) ^{^{1/p}}}\left( \frac{1}{k+1}\overset{n+k}{\sum \limits _{j=n}}\frac{2j}{\left( j+1\right) ^{2}}\right) +\frac{1}{k+1} \overset{n+k}{\sum \limits _{j=n}}\frac{1}{3\left( j+1\right) ^{2}} \\&\quad =\left( \frac{1}{k+1}\overset{n+k}{\sum \limits _{j=n}}\frac{j}{\left( j+1\right) ^{2}}\right) \left( \frac{2}{\left( p+1\right) ^{^{1/p}}}-\frac{3}{\left( 2p+1\right) ^{^{1/p}}}\right) \\&\qquad +\left( \frac{1}{k+1}\overset{n+k}{\sum \limits _{j=n}}\frac{1}{\left( j+1\right) ^{2}}\right) \left( \frac{1}{3}-\frac{1}{\left( 2p+1\right) ^{^{1/p}}}\right) , \end{aligned}$$

Since $st-{\lim \nolimits _{k}}\small \left( {\sup \nolimits _{n}}\frac{1}{k+1} {\sum \nolimits _{j=n}^{n+k}}\frac{j}{\left( j+1\right) ^{2}}\right) =0$ and $st-\lim \nolimits _{k} \small \left( {\sup \nolimits _{n}}\frac{1}{k+1}{\sum \nolimits _{j=n}^{n+k}}\frac{1}{\left( j+1\right) ^{2}}\right) =0$, we have,

$$\begin{aligned}&st-\underset{k}{\lim }\left( \underset{n}{\sup }\left\| \overset{\infty }{\sum \limits _{j=1}}a_{kj}^{\left( n\right) }U_{j}\left( e_{2};x\right) -e_{2}\left( x\right) \right\| _{L_{p}}\right) \\&\quad \le \left( \frac{2}{\left( p+1\right) ^{^{1/p}}}-\frac{3}{\left( 2p+1\right) ^{^{1/p}}}\right) st-\underset{k}{\lim }\left( \underset{n}{\sup }\frac{1}{k+1}\overset{n+k}{\sum \limits _{j=n}}\frac{j}{\left( j+1\right) ^{2}}\right) \\&\qquad +\left( \frac{1}{3}-\frac{1}{\left( 2p+1\right) ^{^{1/p}}}\right) st-\underset{k}{\lim }\left( \underset{n}{\sup }\frac{1}{k+1}\overset{n+k}{\sum \limits _{j=n}}\frac{1}{\left( j+1\right) ^{2}}\right) \end{aligned}$$

which gives

$$\begin{aligned} st-\underset{k}{\lim }\left( \left\| \overset{\infty }{\sum \limits _{j=1}} a_{kj}^{\left( n\right) }U_{j}e_{2}-e_{2}\right\| _{L_{p}}\right) =0, \textit{uniformly in}\,n. \end{aligned}$$

We get

$$\begin{aligned} st-\underset{k}{\lim }\left\| \overset{\infty }{\sum \limits _{j=1}} a_{kj}^{\left( n\right) }U_{j}e_{2}-e_{2}\text { }\right\| _{L_{p}}^{p}=0\, \textit{uniformly in}\,\, n. \end{aligned}$$

So, our claim ( 20 ) holds true for each $i=0,1,2.$ $\left\{ U_{j}\right\}$ satisfies all hypothesis of Theorem 1 and we immediately see that,

$$\begin{aligned} st-\underset{k}{\lim }\left\| \overset{\infty }{\sum \limits _{j=1}} a_{kj}^{\left( n\right) }U_{j}f-f\text { }\right\| _{L_{p}}^{p}=0, \,\textit{uniformly in}\,n \,\,on\, \left[ 0,1\right] \,\,\,\textit{for all}\, f\in L_{p}(I). \end{aligned}$$