A-Statistical Convergence Properties of Kantorovich Type λ-Bernstein Operators Via (p, q)-Calculus

Abstract

In the present paper, Kantorovich type $\lambda$-Bernstein operators via (p, q)-calculus are constructed, and the first and second moments and central moments of these operators are estimated in order to achieve our main results. An A-statistical convergence theorem and the rate of A-statistical convergence theorems are obtained according to some analysis methods and the definitions of A-statistical convergence, the rate of A-statistical convergence and modulus of smoothness.

1. Introduction

As we know, one of the simplest and most elegant ways to prove the famous Weierstrass Approximation Theorem was given by S. N. Bernstein [1] in 1912 by constructing a sequence of polynomials which were defined as follows,

$$\begin{array}{c}\hfill B_n(f;x)=\sum _{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)x^k{(1-x)}^{n-k}f\left(\frac{k}{n}\right)\end{array}$$

for $f\in C[0,1]$ and $x\in [0,1]$. These polynomials are called Bernstein operators or Bernstein polynomials. Due to the fine properties of approximation, Bernstein operators play a significant role in Approximation Theory and Computer Aided Geometric Design (CAGD).

In 2016, Mursaleen et al. [2] defined the following (p, q)-analogue of Bernstein operators:

$$\begin{array}{c}\hfill B_{n,p,q}(f;x)=\sum _{k=0}^n{b}_{n,k}(x;p,q)f\left(\frac{{\left[k\right]}_{p,q}}{p^{k-n}{\left[n\right]}_{p,q}}\right),x\in [0,1],\end{array}$$

(1)

where $b_{n,k}(x;p,q)$$(k=0,1,...,n)$ are (p, q)-Bernstein basis functions and defined as

$$\begin{array}{c}\hfill b_{n,k}(x;p,q)=\frac{1}{p^{\frac{n(n-1)}{2}}}{\left[\begin{array}{c}n\\ k\end{array}\right]}_{p,q}{p}^{\frac{k(k-1)}{2}}{x}^k\prod _{s=0}^{n-k-1}\left(p^s-q^{s}x\right),x\in [0,1].\end{array}$$

(2)

Then, there are many papers mention about the approximation properties of (p, q)-type positive linear operators, such as [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22].

Very recently, Cai et al. [23] proposed the following positive linear $\lambda$-Bernstein operators based on (p, q)-integers as

$$\begin{array}{c}\hfill B_{n,p,q}^{\lambda }(f;x)=\sum _{k=0}^n{b}_{n,k}^{\lambda }(x;p,q)f\left(\frac{{\left[k\right]}_{p,q}}{p^{k-n}{\left[n\right]}_{p,q}}\right),x\in [0,1],\end{array}$$

(3)

where

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}{b}_{n,0}^{\lambda }(x;p,q)& =b_{n,0}(x;p,q)-\frac{\lambda }{p^{1-n}{\left[n\right]}_{p,q}+1}{b}_{n+1,1}(x;p,q),\hfill \\ b_{n,k}^{\lambda }(x;p,q)& =b_{n,k}(x;p,q)+\lambda \left(\frac{p^{1-n}{\left[n\right]}_{p,q}-2p^{1-k}{\left[k\right]}_{p,q}+1}{p^{2-2n}{\left[n\right]}_{p,q}^2-1}{b}_{n+1,k}(x;p,q)\right.\hfill \\ & \left.-\frac{p^{1-n}{\left[n\right]}_{p,q}-2q{p}^{-k}{\left[k\right]}_{p,q}-1}{p^{2-2n}{\left[n\right]}_{p,q}^2-1}{b}_{n+1,k+1}(x;p,q)\right),(k=1,2,...,n-1)\hfill \\ b_{n,n}^{\lambda }(x;p,q)& =b_{n,n}(x;p,q)-\frac{\lambda }{p^{1-n}{\left[n\right]}_{p,q}+1}{b}_{n+1,n}(x;p,q),\hfill \end{array}\right.\end{array}$$

(4)

$b_{n,k}(x;p,q)$$(k=0,1,...,n)$ are defined in Equation (2), $\lambda \in [-1,1]$, $n\ge 2$, $x\in [0,1]$ and $0<q<p\le 1$.

Inspired by the above research, based on Equation (3), we introduce Kantorovich type $\lambda$-Bernstein operators via (p, q)-calculus as

$$\begin{array}{c}\hfill K_{n,p,q}^{\lambda }(f;x)={[n+1]}_{p,q}\sum _{k=0}^n{b}_{n,k}^{\lambda }(x;p,q)p^{-k}{\int }_{\frac{q{\left[k\right]}_{p,q}}{{[n+1]}_{p,q}}}^{\frac{{[k+1]}_{p,q}}{{[n+1]}_{p,q}}}f\left(p^{-k}u\right)d_{p,q}u,\end{array}$$

(5)

where $x\in [0,1]$, $0<q<p\le 1$ and $b_{n,k}^{\lambda }(x;p,q)$$(k=0,1,...,n)$ are defined in Equation (4).

Firstly, we give some definitions of (p, q)-integers, which can be referred to [24,25,26,27,28]. For any fixed real number $p>0$ and $q>0$, ${\left[n\right]}_{p,q}$ are defined by ${\left[n\right]}_{p,q}=\left\{\begin{array}{cc}\frac{p^n-q^n}{p-q},& p\ne q\ne 1;\hfill \\ \frac{1-q^n}{1-q},& p=1;\hfill \\ n,& p=q=1.\hfill \end{array}\right.$ ${\left[n\right]}_{p,q}!$ and ${\left[\begin{array}{c}n\\ k\end{array}\right]}_{p,q}$ are defined as follows:

$$\begin{array}{c}\hfill {\left[n\right]}_{p,q}!=\left\{\begin{array}{cc}{\left[n\right]}_{p,q}{[n-1]}_{p,q}...{\left[1\right]}_{p,q},& n=1,2,\cdots ;\hfill \\ 1,& n=0,\hfill \end{array}\right.{\left[\begin{array}{c}n\\ k\end{array}\right]}_{p,q}=\frac{{\left[n\right]}_{p,q}!}{{\left[k\right]}_{p,q}!{[n-k]}_{p,q}!}.\end{array}$$

The (p, q)-power basis ${(x\oplus t)}_{p,q}^n$ and ${(x\ominus t)}_{p,q}^n$ are defined by

$${(x\oplus t)}_{p,q}^n=(x+t)(px+qt)\left(p^{2}x+q^{2}t\right)\cdots \left(p^{n-1}x+q^{n-1}t\right)$$

and

$${(x\ominus t)}_{p,q}^n=(x-t)(px-qt)\left(p^{2}x-q^{2}t\right)\cdots \left(p^{n-1}x-q^{n-1}t\right).$$

We also give the fundamental theorem of (p, q)-calculus, say, if $F\left(x\right)$ is an anti-derivative of $f\left(x\right)$ and $F\left(x\right)$ is continuous at $x=0$, then ${\int }_a^{b}f\left(x\right)d_{p,q}x=F\left(b\right)-F\left(a\right)$ holds, where $0\le a<b\le \infty$ and $F\left(x\right)$ is given by the formula

$$\begin{array}{c}\hfill F\left(x\right)=(p-q)x\sum _{j=0}^{\infty }\frac{q^j}{p^{j+1}}f\left(\frac{q^j}{p^{j+1}}x\right)+F\left(0\right),\end{array}$$

the infinite series here converges.

The main goal of the present work is to study the rate of A-statistical convergence of Kantorovich type $\lambda$-Bernstein operators based on (p, q)-integers by means of modulus of continuity. The rest of this paper are mainly organized as follows: in Section 2, some moments and central moments of $K_{n,p,q}^{\lambda }(f;x)$ are estimated; in Section 3, we prove $K_{n,p,q}^{\lambda }(f;x)$ is A-statistically convergent to $f\left(x\right)$ and investigate the rate of A-statistical convergence by means of the first and second modulus continuity.

2. Some Preliminary Results

In the sequel, consider sequences of functions $e_i\left(x\right)=x^i$$(i=0,1,2)$, and ${\varphi }_j(t,x)={(t-x)}^j$$(j=1,2)$, $x,t\in [0,1]$. Before we give our main theorems, we need the following lemmas.

Lemma 1.

The following statements are true:

$$\begin{array}{c}{\int }_{\frac{q{\left[k\right]}_{p,q}}{{[n+1]}_{p,q}}}^{\frac{{[k+1]}_{p,q}}{{[n+1]}_{p,q}}}{d}_{p,q}u=\frac{p^k}{{[n+1]}_{p,q}},\hfill \end{array}$$

(6)

$$\begin{array}{c}{\int }_{\frac{q{\left[k\right]}_{p,q}}{{[n+1]}_{p,q}}}^{\frac{{[k+1]}_{p,q}}{{[n+1]}_{p,q}}}{p}^{-k}u{d}_{p,q}u=\frac{2q{\left[k\right]}_{p,q}+p^k}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}^2},\hfill \end{array}$$

(7)

$$\begin{array}{c}{\int }_{\frac{q{\left[k\right]}_{p,q}}{{[n+1]}_{p,q}}}^{\frac{{[k+1]}_{p,q}}{{[n+1]}_{p,q}}}{\left(p^{-k}u\right)}^2{d}_{p,q}u=\frac{3p^{-k}{q}^2{\left[k\right]}_{p,q}^2+3q{\left[k\right]}_{p,q}+p^k}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}}.\hfill \end{array}$$

(8)

Proof. 

By the fundamental theorem of (p, q)-calculus given in Section 1, we have

$$\begin{array}{ccc}\hfill {\int }_{\frac{q{\left[k\right]}_{p,q}}{{[n+1]}_{p,q}}}^{\frac{{[k+1]}_{p,q}}{{[n+1]}_{p,q}}}{d}_{p,q}u& =& (p-q)\frac{{[k+1]}_{p,q}}{{[n+1]}_{p,q}}\sum _{j=0}^{\infty }\frac{q^j}{p^{j+1}}-(p-q)\frac{q{\left[k\right]}_{p,q}}{{[n+1]}_{p,q}}\sum _{j=0}^{\infty }\frac{q^j}{p^{j+1}}\hfill \\ & =& \frac{p-q}{p}\frac{1}{1-\frac{q}{p}}\left(\frac{{[k+1]}_{p,q}}{{[n+1]}_{p,q}}-\frac{q{\left[k\right]}_{p,q}}{{[n+1]}_{p,q}}\right)=\frac{p^k}{{[n+1]}_{p,q}}.\hfill \end{array}$$

Similarly,

$$\begin{array}{cc}& {\int }_{\frac{q{\left[k\right]}_{p,q}}{{[n+1]}_{p,q}}}^{\frac{{[k+1]}_{p,q}}{{[n+1]}_{p,q}}}{p}^{-k}u{d}_{p,q}u\hfill \\ =& p^{-k}\left(\frac{(p-q){[k+1]}_{p,q}}{{[n+1]}_{p,q}}\sum _{j=0}^{\infty }\frac{q^{2j}{[k+1]}_{p,q}}{p^{2j+2}{[n+1]}_{p,q}}-\frac{(p-q)q{\left[k\right]}_{p,q}}{{[n+1]}_{p,q}}\sum _{j=0}^{\infty }\frac{q^{2j+1}{\left[k\right]}_{p,q}}{p^{2j+2}{[n+1]}_{p,q}}\right)\hfill \\ =& \frac{p-q}{p^2}\frac{p^{-k}}{1-\frac{q^2}{p^2}}\frac{{[k+1]}_{p,q}^2-q^2{\left[k\right]}_{p,q}^2}{{[n+1]}_{p,q}^2}=\frac{{[k+1]}_{p,q}+q{\left[k\right]}_{p,q}}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}^2}=\frac{2q{\left[k\right]}_{p,q}+p^k}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}^2}.\hfill \end{array}$$

Finally,

$$\begin{array}{cc}& {\int }_{\frac{q{\left[k\right]}_{p,q}}{{[n+1]}_{p,q}}}^{\frac{{[k+1]}_{p,q}}{{[n+1]}_{p,q}}}{\left(p^{-k}u\right)}^2{d}_{p,q}u\hfill \\ =& p^{-2k}\left(\frac{(p-q){[k+1]}_{p,q}}{{[n+1]}_{p,q}}\sum _{j=0}^{\infty }\frac{q^{3j}{[k+1]}_{p,q}^2}{p^{3j+3}{[n+1]}_{p,q}^2}-\frac{(p-q)q{\left[k\right]}_{p,q}}{{[n+1]}_{p,q}}\sum _{j=0}^{\infty }\frac{q^{3j+2}{\left[k\right]}_{p,q}^2}{p^{3j+3}{[n+1]}_{p,q}^2}\right)\hfill \\ =& \frac{p-q}{p^3}\frac{p^{-2k}}{1-\frac{q^3}{p^3}}\frac{{[k+1]}_{p,q}^3-q^3{\left[k\right]}_{p,q}^3}{{[n+1]}_{p,q}^3}=\frac{3p^{-k}{q}^2{\left[k\right]}_{p,q}^2+3q{\left[k\right]}_{p,q}+p^k}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}}.\hfill \end{array}$$

Lemma 1 is proved. ☐

Lemma 2.

Let $\lambda \in [-1,1]$, $x\in [0,1]$ and $0<q<p\le 1$, for the operators $K_{n,p,q}^{\lambda }(f;x)$, we have

$$\begin{array}{ccc}\hfill K_{n,p,q}^{\lambda }(e_0;x)& =& 1,\hfill \end{array}$$

(9)

$$\begin{array}{ccc}\hfill K_{n,p,q}^{\lambda }(e_1;x)& =& \frac{2x}{{\left[2\right]}_{p,q}{p}^n}+\frac{1-2x}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}}+2\lambda \left\{\frac{\frac{2q}{p}\left(1-\frac{q}{p}\right)x^2\left(1-x^{n-1}\right)}{{\left[2\right]}_{p,q}\left(p{\left[n\right]}_{p,q}+p^n\right)}-\frac{\left(1-\frac{q}{p}\right)x\left(1-x^n\right)}{{\left[2\right]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}\right.\hfill \\ & & -\frac{2p^{n-1}q\left[x(1-x)+\frac{q}{p}{x}^2\left(1-x^{n-1}\right)\right]}{{\left[2\right]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}+\frac{1-x^{n+1}}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}\left(p^{1-n}{\left[n\right]}_{p,q}-1\right)}\hfill \\ & & \left.-\frac{{(1\ominus x)}_{p,q}^{n+1}}{{\left[2\right]}_{p,q}{p}^{\frac{n(n-1)}{2}}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}\right\}\doteq {\theta }_{p,q}^{\lambda }(n,x),\hfill \end{array}$$

(10)

$$\begin{array}{ccc}\hfill K_{n,p,q}^{\lambda }(e_2;x)& =& \frac{3x^2}{{\left[3\right]}_{p,q}{p}^{2n}}+\frac{3x-6x^2+\frac{3q}{p}x(1-x)}{{\left[3\right]}_{p,q}{p}^n{[n+1]}_{p,q}}+\frac{1-3\left(1+\frac{q}{p}\right)x(1-x)}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}^2}\hfill \\ & & +3\lambda \left\{\frac{\left(1-\frac{q^2}{p^2}\right)x^2\left[\frac{2q}{p}x\left(1-x^{n-2}\right)-1+x^{n-1}\right]}{{\left[3\right]}_{p,q}{p}^n\left(p{\left[n\right]}_{p,q}+p^n\right)}+\frac{\left(\frac{q}{p}-1\right)x\left(1-x^n\right)}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}\right.\hfill \\ & & +\frac{\left(1+\frac{q^2}{p^2}\right)x+\left(1-\frac{q^2}{p^2}\right)x^2\left(1-\frac{2q}{p}x\right)+2\left(\frac{q}{p}-1-\frac{q^3}{p^3}\right)x^{n+1}}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}+p^n\right)}\hfill \\ & & -\frac{\frac{2q^2}{p^2}\left(1+\frac{q}{p}\right)x^2\left(1-x^{n-1}\right)}{{\left[3\right]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}+\frac{2p^n\left(\frac{2q}{p}-1\right)x{(1\ominus x)}_{p,q}^n}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}{p}^{\frac{n(n-1)}{2}}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}\hfill \\ & & \left.+\frac{2p^{n}x\left[\frac{q^2}{p^2}\left(1+\frac{q}{p}\right)x\left(1-x^{n-1}\right)+\left(1-\frac{3q}{p}\right)\left(1-x^n\right)\right]}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}\right\}.\hfill \end{array}$$

(11)

Proof. 

We get Equation (9) easily by Equations (5) and (6) and [23]. From Equations (5), (7), (8) and (2), we have

$$\begin{array}{ccc}\hfill K_{n,p,q}^{\lambda }(e_1;x)& =& {[n+1]}_{p,q}\sum _{k=0}^n{b}_{n,k}^{\lambda }(x;p,q)p^{-k}{\int }_{\frac{q{\left[k\right]}_{p,q}}{{[n+1]}_{p,q}}}^{\frac{{[k+1]}_{p,q}}{{[n+1]}_{p,q}}}{p}^{-k}u{d}_{p,q}u\hfill \\ & =& \frac{2q{\left[n\right]}_{p,q}}{{\left[2\right]}_{p,q}{p}^n{[n+1]}_{p,q}}{B}_{n,p,q}^{\lambda }(e_1;x)+\frac{1}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}}{B}_{n,p,q}^{\lambda }(e_0;x),\hfill \end{array}$$

(12)

$$\begin{array}{ccc}\hfill K_{n,p,q}^{\lambda }(e_2;x)& =& {[n+1]}_{p,q}\sum _{k=0}^n{b}_{n,k}^{\lambda }(x;p,q)p^{-k}{\int }_{\frac{q{\left[k\right]}_{p,q}}{{[n+1]}_{p,q}}}^{\frac{{[k+1]}_{p,q}}{{[n+1]}_{p,q}}}{\left(p^{-k}u\right)}^2{d}_{p,q}u\hfill \\ & =& \frac{3q^2{\left[n\right]}_{p,q}^2}{{\left[3\right]}_{p,q}{p}^{2n}{[n+1]}_{p,q}^2}{B}_{n,p,q}^{\lambda }(e_2;x)+\frac{3q{\left[n\right]}_{p,q}}{{\left[3\right]}_{p,q}{p}^n{[n+1]}_{p,q}^2}{B}_{n,p,q}^{\lambda }(e_1;x)\hfill \\ & & +\frac{1}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}^2}{B}_{n,p,q}^{\lambda }(e_0;x).\hfill \end{array}$$

(13)

Therefore, Equations (11) and (12) can be obtained by Equations (12), (14), [23] and some tedious computations, here we omit those processes. ☐

Lemma 3.

Let $\lambda \in [-1,1]$, $x\in [0,1]$ and $0<q<p\le 1$, we have the following inequalities for the operators $K_{n,p,q}^{\lambda }(f;x)$.

$$\begin{array}{cc}& K_{n,p,q}^{\lambda }({\varphi }_1(t,x);x)\hfill \\ \le & \frac{2-{\left[2\right]}_{p,q}{p}^n}{{\left[2\right]}_{p,q}{p}^n}+\frac{1}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}}+\frac{4}{{\left[2\right]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}+\frac{5}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}\hfill \end{array}$$

(14)

$$\begin{array}{ll}& +\frac{2}{{\left[2\right]}_{p,q}{p}^{\frac{n(n-1)}{2}}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}\hfill \\ \doteq & \Theta (p,q;n),\hfill \\ & K_{n,p,q}^{\lambda }({\varphi }_2(t,x);x)\hfill \\ \le & \frac{3{\left[2\right]}_{p,q}-4{\left[3\right]}_{p,q}{p}^n+{\left[3\right]}_{p,q}{\left[2\right]}_{p,q}{p}^{2n}}{{\left[3\right]}_{p,q}{\left[2\right]}_{p,q}{p}^{2n}}+\frac{10}{{\left[3\right]}_{p,q}{p}^n{[n+1]}_{p,q}}+\frac{4}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}}+\frac{16}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}^2}\hfill \\ & +\frac{8}{{\left[2\right]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}+\frac{4}{{\left[2\right]}_{p,q}{p}^{\frac{n(n-1)}{2}}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}+\frac{16}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}\hfill \\ & +\frac{6}{{\left[3\right]}_{p,q}{p}^{\frac{n(n-1)}{2}}{[n+1]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}+\frac{18}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}\hfill \\ \doteq & \Phi (p,q;n).\hfill \end{array}$$

(15)

Proof. 

By Equations (9) and (11), we can get

$$\begin{array}{cc}& K_{n,p,q}^{\lambda }({\varphi }_1(t,x);x)\hfill \\ =& K_{n,p,q}^{\lambda }(e_1;x)-x\hfill \\ =& \left(\frac{2}{{\left[2\right]}_{p,q}{p}^n}-1\right)x+\frac{1-2x}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}}+\lambda \left\{\frac{\frac{4q}{p}\left(1-\frac{q}{p}\right)x^2\left(1-x^{n-1}\right)}{{\left[2\right]}_{p,q}\left(p{\left[n\right]}_{p,q}+p^n\right)}-\frac{2\left(1-\frac{q}{p}\right)x\left(1-x^n\right)}{{\left[2\right]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}\right.\hfill \\ & -\frac{4p^{n-1}q\left[x(1-x)+\frac{q}{p}{x}^2\left(1-x^{n-1}\right)\right]}{{\left[2\right]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}+\frac{2\left(1-x^{n+1}\right)}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}\left(p^{1-n}{\left[n\right]}_{p,q}-1\right)}\hfill \\ & \left.-\frac{2{(1\ominus x)}_{p,q}^{n+1}}{{\left[2\right]}_{p,q}{p}^{\frac{n(n-1)}{2}}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}\right\}\hfill \\ \le & \left\{\begin{array}{c}\left(\frac{2}{{\left[2\right]}_{p,q}{p}^n}-1\right)x+\frac{1-2x}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}}+\frac{\frac{4q}{p}\left(1-\frac{q}{p}\right)x^2\left(1-x^{n-1}\right)}{{\left[2\right]}_{p,q}\left(p{\left[n\right]}_{p,q}+p^n\right)}\\ +\frac{2\left(1-x^{n+1}\right)}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}\left(p^{1-n}{\left[n\right]}_{p,q}-1\right)},& for\lambda \in [0,1];\hfill \\ \left(\frac{2}{{\left[2\right]}_{p,q}{p}^n}-1\right)x+\frac{1-2x}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}}+\frac{2\left(1-\frac{q}{p}\right)x\left(1-x^n\right)}{{\left[2\right]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}\\ \frac{4p^{n-1}q\left[x(1-x)+\frac{q}{p}{x}^2\left(1-x^{n-1}\right)\right]}{{\left[2\right]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}+\frac{2{(1\ominus x)}_{p,q}^{n+1}}{{\left[2\right]}_{p,q}{p}^{\frac{n(n-1)}{2}}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)},& for\lambda \in [-1,0]\hfill \end{array}\right.\hfill \\ \le & \frac{2-{\left[2\right]}_{p,q}{p}^n}{{\left[2\right]}_{p,q}{p}^n}+\frac{1}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}}+\frac{4}{{\left[2\right]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}+\frac{5}{{\left[2\right]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}\hfill \\ & +\frac{2}{{\left[2\right]}_{p,q}{p}^{\frac{n(n-1)}{2}}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}.\hfill \end{array}$$

Similarly, by Lemma 2, we have

$$\begin{array}{cc}& K_{n,p,q}^{\lambda }({\varphi }_2(t,x);x)\hfill \\ =& K_{n,p,q}^{\lambda }(e_2;x)-2x{K}_{n,p,q}^{\lambda }(e_1;x)+x^2{K}_{n,p,q}^{\lambda }(e_0;x)\hfill \\ =& \left(\frac{3}{{\left[3\right]}_{p,q}{p}^{2n}}-\frac{4}{{\left[2\right]}_{p,q}{p}^n}+1\right)x^2+\frac{3x-6x^2+\frac{3q}{p}x(1-x)}{{\left[3\right]}_{p,q}{p}^n{[n+1]}_{p,q}}-\frac{2x(1-2x)}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}}+\frac{1-3\left(1+\frac{q}{p}\right)x(1-x)}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}^2}\hfill \\ & +\lambda \left\{\frac{3\left(1-\frac{q^2}{p^2}\right)x^2\left[\frac{2q}{p}x\left(1-x^{n-2}\right)-1+x^{n-1}\right]}{{\left[3\right]}_{p,q}{p}^n\left(p{\left[n\right]}_{p,q}+p^n\right)}-\frac{\frac{8q}{p}\left(1-\frac{q}{p}\right)x^3\left(1-x^{n-1}\right)}{{\left[2\right]}_{p,q}\left(p{\left[n\right]}_{p,q}+p^n\right)}\right.\hfill \\ & +\frac{4\left(1-\frac{q}{p}\right)x^2\left(1-x^n\right)}{{\left[2\right]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}+\frac{3\left(\frac{q}{p}-1\right)x\left(1-x^n\right)}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}+\frac{8p^{n-1}q\left[x^2(1-x)+\frac{q}{p}{x}^3\left(1-x^{n-1}\right)\right]}{{\left[2\right]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}\hfill \\ & -\frac{4x\left(1-x^{n+1}\right)}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}\left(p^{1-n}{\left[n\right]}_{p,q}-1\right)}+\frac{4x{(1\ominus x)}_{p,q}^{n+1}}{{\left[2\right]}_{p,q}{p}^{\frac{n(n-1)}{2}}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}\hfill \\ & +\frac{3\left(1+\frac{q^2}{p^2}\right)x+3\left(1-\frac{q^2}{p^2}\right)x^2\left(1-\frac{2q}{p}x\right)+6\left(\frac{q}{p}-1-\frac{q^3}{p^3}\right)x^{n+1}}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}+p^n\right)}-\frac{\frac{6q^2}{p^2}\left(1+\frac{q}{p}\right)x^2\left(1-x^{n-1}\right)}{{\left[3\right]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}\hfill \\ & \left.+\frac{6p^n\left(\frac{2q}{p}-1\right)x{(1\ominus x)}_{p,q}^n}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}{p}^{\frac{n(n-1)}{2}}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}+\frac{6p^{n}x\left[\frac{q^2}{p^2}\left(1+\frac{q}{p}\right)x\left(1-x^{n-1}\right)+\left(1-\frac{3q}{p}\right)\left(1-x^n\right)\right]}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}\right\}.\hfill \end{array}$$

Next, we will discuss in two cases:

Case 1: for $\lambda \in [0,1]$, we have

$$\begin{array}{cc}& K_{n,p,q}^{\lambda }({\varphi }_2(t,x);x)\hfill \\ \le & \left(\frac{3}{{\left[3\right]}_{p,q}{p}^{2n}}-\frac{4}{{\left[2\right]}_{p,q}{p}^n}+1\right)x^2+\frac{3x-6x^2+\frac{3q}{p}x(1-x)}{{\left[3\right]}_{p,q}{p}^n{[n+1]}_{p,q}}-\frac{2x(1-2x)}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}}+\frac{1-3\left(1+\frac{q}{p}\right)x(1-x)}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}^2}\hfill \\ & +\lambda \left\{\frac{6\left(1-\frac{q^2}{p^2}\right)\frac{q}{p}{x}^3\left(1-x^{n-2}\right)}{{\left[3\right]}_{p,q}{p}^n\left(p{\left[n\right]}_{p,q}+p^n\right)}+\frac{4\left(1-\frac{q}{p}\right)x^2\left(1-x^n\right)}{{\left[2\right]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}+\frac{8p^{n-1}q\left[x^2(1-x)+\frac{q}{p}{x}^3\left(1-x^{n-1}\right)\right]}{{\left[2\right]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}\right.\hfill \\ & +\frac{4x{(1\ominus x)}_{p,q}^{n+1}}{{\left[2\right]}_{p,q}{p}^{\frac{n(n-1)}{2}}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}+\frac{3\left(1+\frac{q^2}{p^2}\right)x+3\left(1-\frac{q^2}{p^2}\right)x^2\left(1-\frac{q}{p}x\right)}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}+p^n\right)}\hfill \\ & \left.+\frac{6p^n\frac{q}{p}x{(1\ominus x)}_{p,q}^n}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}{p}^{\frac{n(n-1)}{2}}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}+\frac{6p^{n}x\left[\frac{q^2}{p^2}\left(1+\frac{q}{p}\right)x\left(1-x^{n-1}\right)+\left(1-\frac{q}{p}\right)\left(1-x^n\right)\right]}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}\right\}\hfill \\ \le & \frac{3}{{\left[3\right]}_{p,q}{p}^{2n}}-\frac{4}{{\left[2\right]}_{p,q}{p}^n}+1+\frac{4}{{\left[3\right]}_{p,q}{p}^n{[n+1]}_{p,q}}+\frac{4}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}}+\frac{1}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}^2}\hfill \\ & +\frac{6}{{\left[3\right]}_{p,q}{p}^n\left(p{\left[n\right]}_{p,q}+p^n\right)}+\frac{4}{{\left[2\right]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}+\frac{10}{{\left[2\right]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}\hfill \\ & +\frac{4}{{\left[2\right]}_{p,q}{p}^{\frac{n(n-1)}{2}}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}+\frac{9}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}+p^n\right)}\hfill \\ & +\frac{6}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}{p}^{\frac{n(n-1)}{2}}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}+\frac{18}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}.\hfill \end{array}$$

(16)

Case 2: for $\lambda \in [-1,0]$, we have

$$\begin{array}{cc}& K_{n,p,q}^{\lambda }({\varphi }_2(t,x);x)\hfill \\ \le & \left(\frac{3}{{\left[3\right]}_{p,q}{p}^{2n}}-\frac{4}{{\left[2\right]}_{p,q}{p}^n}+1\right)x^2+\frac{3x-6x^2+\frac{3q}{p}x(1-x)}{{\left[3\right]}_{p,q}{p}^n{[n+1]}_{p,q}}-\frac{2x(1-2x)}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}}+\frac{1-3\left(1+\frac{q}{p}\right)x(1-x)}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}^2}\hfill \\ & -\lambda \left\{\frac{3\left(1-\frac{q^2}{p^2}\right)x^2\left(1-x^{n-1}\right)}{{\left[3\right]}_{p,q}{p}^n\left(p{\left[n\right]}_{p,q}+p^n\right)}+\frac{\frac{8q}{p}\left(1-\frac{q}{p}\right)x^3\left(1-x^{n-1}\right)}{{\left[2\right]}_{p,q}\left(p{\left[n\right]}_{p,q}+p^n\right)}+\frac{3\left(1-\frac{q}{p}\right)x\left(1-x^n\right)}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}\right.\hfill \\ & +\frac{4x\left(1-x^{n+1}\right)}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}\left(p^{1-n}{\left[n\right]}_{p,q}-1\right)}+\frac{3\left(1-\frac{q^2}{p^2}\right)\frac{q}{p}{x}^3+6\left(1+\frac{q^3}{p^3}-\frac{q}{p}\right)x^{n+1}}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}+p^n\right)}\hfill \\ & +\frac{\frac{6q^2}{p^2}\left(1+\frac{q}{p}\right)x^2\left(1-x^{n-1}\right)}{{\left[3\right]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}+\frac{6p^n\left(1-\frac{q}{p}\right)x{(1\ominus x)}_{p,q}^n}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}{p}^{\frac{n(n-1)}{2}}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}\hfill \\ & \left.+\frac{12p^{n}x\frac{q}{p}\left(1-x^n\right)}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}\right\}\hfill \\ \le & \frac{3}{{\left[3\right]}_{p,q}{p}^{2n}}-\frac{4}{{\left[2\right]}_{p,q}{p}^n}+1+\frac{4}{{\left[3\right]}_{p,q}{p}^n{[n+1]}_{p,q}}+\frac{4}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}}+\frac{1}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}^2}\hfill \\ & +\frac{3}{{\left[3\right]}_{p,q}{p}^n\left(p{\left[n\right]}_{p,q}+p^n\right)}+\frac{8}{{\left[2\right]}_{p,q}\left(p{\left[n\right]}_{p,q}+p^n\right)}+\frac{3}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}\hfill \\ & +\frac{4}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}\left(p^{1-n}{\left[n\right]}_{p,q}-1\right)}+\frac{15}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}+p^n\right)}+\frac{12}{{\left[3\right]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}\hfill \\ & +\frac{6}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}{p}^{\frac{n(n-1)}{2}}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}+\frac{12}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}.\hfill \end{array}$$

(17)

Combining Equations (17) and (18), we obtain

$$\begin{array}{cc}& K_{n,p,q}^{\lambda }({\varphi }_2(t,x);x)\hfill \\ \le & \frac{3}{{\left[3\right]}_{p,q}{p}^{2n}}-\frac{4}{{\left[2\right]}_{p,q}{p}^n}+1+\frac{4}{{\left[3\right]}_{p,q}{p}^n{[n+1]}_{p,q}}+\frac{4}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}}+\frac{1}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}^2}\hfill \\ & \frac{6}{{\left[3\right]}_{p,q}{p}^n\left(p{\left[n\right]}_{p,q}+p^n\right)}+\frac{8}{{\left[2\right]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}+\frac{12}{{\left[2\right]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}\hfill \\ & +\frac{4}{{\left[2\right]}_{p,q}{p}^{\frac{n(n-1)}{2}}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}+\frac{4}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}\hfill \\ & +\frac{15}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}+p^n\right)}+\frac{6}{{\left[3\right]}_{p,q}{p}^{\frac{n(n-1)}{2}}{[n+1]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}\hfill \\ & +\frac{18}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}\hfill \\ \le & \frac{3}{{\left[3\right]}_{p,q}{p}^{2n}}-\frac{4}{{\left[2\right]}_{p,q}{p}^n}+1+\frac{10}{{\left[3\right]}_{p,q}{p}^n{[n+1]}_{p,q}}+\frac{4}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}}+\frac{16}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}^2}\hfill \\ & +\frac{8}{{\left[2\right]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}+\frac{4}{{\left[2\right]}_{p,q}{p}^{\frac{n(n-1)}{2}}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}+\frac{16}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}\hfill \\ & +\frac{6}{{\left[3\right]}_{p,q}{p}^{\frac{n(n-1)}{2}}{[n+1]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}+\frac{18}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}\hfill \end{array}$$

with the fact that $\frac{1}{p{\left[n\right]}_{p,q}+p^n}\le \frac{1}{q{\left[n\right]}_{p,q}+p^n}=\frac{1}{{[n+1]}_{p,q}}$. Thus, Lemma 3 is proved. ☐

3. $\mathbf{A}$-Statistical Convergence Properties

Let $C[0,1]$ be the space of all real-valued continuous bounded functions f on $[0,1]$, endowed with the norm ${\left|\right|f\left|\right|}_{C[0,1]}={sup}_{x\in [0,1]}\left|f\left(x\right)\right|$. In this section, we will give some A-statistical convergence properties for positive linear operators $K_{n,p,q}^{\lambda }(f;x)$ by the following definition of A-statistical convergence and the first and second modulus of continuity.

Definition 1.

(See [29]) For a given non-negative infinite summability matrix $A=\left(a_{nk}\right)$, $n,k\in \mathbb{N}$, A-transform of x denoted by $Ax:=\left\{{\left(Ax\right)}_n\right\}$ is defined as ${\left(Ax\right)}_n={\sum }_{k=0}^{\infty }{a}_{nk}{x}_k$ provided the series converges for each n. We say that A is regular if ${lim}_n{\left(Ax\right)}_n=L$ whenever $limx=1$. Assume that A is non-negative regular summability matrix, a sequence $x=\left\{x_k\right\}$ is called A-statistically convergent to L provided that for every $ϵ>0$, ${lim}_n{\sum }_{k:|x_k-L|\ge ϵ}{a}_{nk}=0$. We denote this limit by $s{t}_A-limx=L$.

As we know, A-statistical convergence becomes ordinary statistical convergence when $A=\left(C_1\right)$, the Cesaro matrix of order one, and it becomes classical convergence when $A=I$, the identity matrix. There is also a conclusion, every convergent sequence is statistically convergent to the same limit but not conversely.

We need the following Korovkin theorem via the conception of A-statistical convergence to prove our main results.

Theorem 2.

(See [29]) Let $A=\left(a_{nk}\right)$ be a non-negative regular summability matrix and $L_n(f;x)$ be positive linear operators over $C[a,b]$. Then the following two statements are equivalent:

$$\begin{array}{c}\left(i\right)s{t}_A-\underset{n}{lim}{\left|\left|L_n\left(f\right)-f\right|\right|}_{C[a,b]}=0forf\in C[a,b];\hfill \\ \left(ii\right)s{t}_A-\underset{n}{lim}{\left|\left|L_n\left(e_i\right)-e_i\right|\right|}_{C[a,b]}=0fori=0,1,2.\hfill \end{array}$$

Consider sequences $p:=\left\{p_n\right\}$, $q:=\left\{q_n\right\}$ for $0<q_n<p_n\le 1$ satisfying the following conditions

$$\begin{array}{c}\hfill s{t}_A-\underset{n\to \infty }{lim}{p}_n=s{t}_A-\underset{n\to \infty }{lim}{q}_n=s{t}_A-\underset{n\to \infty }{lim}{p}_n^n=1.\end{array}$$

(18)

We now give main results related to statistical convergence of operators in Equation (5).

Theorem 3.

Let $A=\left(a_{jn}\right)$ be a weighted non-negative regular summability matrix for $n,k\in \mathbb{N}$, $\lambda \in [-1,1]$, $x\in [0,1]$ and $0<q<p\le 1$. For any $f\in C[0,1]$, we have

$$\begin{array}{c}\hfill s{t}_A-\underset{n\to \infty }{lim}{\left|\left|K_{n,p,q}^{\lambda }\left(f\right)-f\right|\right|}_{C[0,1]}=0.\end{array}$$

(19)

Proof. 

According to Theorem 2, it is sufficient fo satisfy

$$\begin{array}{c}\hfill s{t}_A-\underset{n}{lim}{\left|\left|K_{n,p,q}^{\lambda }\left(e_i\right)-e_i\right|\right|}_{C[0,1]}=0,i=0,1,2.\end{array}$$

(20)

From Lemma 2, it is clear that Equation (20) is true for $i=0$. For $i=1$, by Equation (14), we have

$$\begin{array}{cc}& {\left|\left|K_{n,p,q}^{\lambda }\left(e_1\right)-e_1\right|\right|}_{C[0,1]}=\underset{x\in [0,1]}{sup}\left|K_{n,p,q}^{\lambda }({\varphi }_1(t,x);x)\right|\le \Theta (p,q;n)\hfill \\ =& \frac{2-{\left[2\right]}_{p,q}{p}^n}{{\left[2\right]}_{p,q}{p}^n}+\frac{1}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}}+\frac{4}{{\left[2\right]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}+\frac{5}{{\left[2\right]}_{p,q}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}\hfill \\ & +\frac{2}{{\left[2\right]}_{p,q}{p}^{\frac{n(n-1)}{2}}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}.\hfill \end{array}$$

Given $ϵ>0$, we define the following sets:

$$\begin{array}{c}D=\left\{n:{\left|\left|K_{n,p,q}^{\lambda }\left(e_1\right)-e_1\right|\right|}_{C[0,1]}\ge ϵ\right\},D_1=\left\{n:\frac{2-{\left[2\right]}_{p,q}{p}^n}{{\left[2\right]}_{p,q}{p}^n}\ge \frac{ϵ}{5}\right\},\hfill \\ D_2=\left\{n:\frac{1}{{[n+1]}_{p,q}}\ge \frac{{\left[2\right]}_{p,q}ϵ}{5}\right\},D_3=\left\{n:\frac{1}{p{\left[n\right]}_{p,q}-p^n}\ge \frac{{\left[2\right]}_{p,q}ϵ}{20}\right\},\hfill \\ D_4=\left\{n:\frac{1}{{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}\ge \frac{{\left[2\right]}_{p,q}ϵ}{25}\right\},\hfill \\ D_5=\left\{n:\frac{1}{p^{\frac{n(n-1)}{2}}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}\ge \frac{{\left[2\right]}_{p,q}ϵ}{10}\right\}.\hfill \end{array}$$

Then it is clear that $D\subseteq D_1\cup D_2\cup D_3\cup D_4\cup D_5$. Hence, for every $l\in \mathbb{N}$, we have

$$\begin{array}{c}\hfill \sum _{n\in D}{a}_{jn}\le \sum _{n\in D_1}{a}_{jn}+\sum _{n\in D_2}{a}_{jn}+\sum _{n\in D_3}{a}_{jn}+\sum _{n\in D_4}{a}_{jn}+\sum _{n\in D_5}{a}_{jn}.\end{array}$$

Letting $j\to \infty$, we get $s{t}_A-{lim}_n{\left|\left|K_{n,p,q}^{\lambda }(e_1;x)-e_1\right|\right|}_{C[0,1]}=0$ with the help of Equation (18) and

$$\begin{array}{c}\hfill s{t}_A-\underset{n}{lim}{\left[2\right]}_{p,q}{p}^n=2,s{t}_A-\underset{n}{lim}\frac{1}{{[n+1]}_{p,q}}=0,s{t}_A-\underset{n}{lim}\frac{1}{p{\left[n\right]}_{p,q}-p^n}=0.\end{array}$$

For $i=2$, by Lemma 2, we have

$$\begin{array}{cc}& {\left|\left|K_{n,p,q}^{\lambda }\left(e_2\right)-e_2\right|\right|}_{C[0,1]}\hfill \\ =& \underset{x\in [0,1]}{sup}\left|\frac{3-{\left[3\right]}_{p,q}{p}^{2n}}{{\left[3\right]}_{p,q}{p}^{2n}}{x}^2+\frac{3x-6x^2+\frac{3q}{p}x(1-x)}{{\left[3\right]}_{p,q}{p}^n{[n+1]}_{p,q}}+\frac{1-3\left(1+\frac{q}{p}\right)x(1-x)}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}^2}\right.\hfill \\ & +3\lambda \left\{\frac{\left(1-\frac{q^2}{p^2}\right)x^2\left[\frac{2q}{p}x\left(1-x^{n-2}\right)-1+x^{n-1}\right]}{{\left[3\right]}_{p,q}{p}^n\left(p{\left[n\right]}_{p,q}+p^n\right)}+\frac{\left(\frac{q}{p}-1\right)x\left(1-x^n\right)}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}\right.\hfill \\ & +\frac{\left(1+\frac{q^2}{p^2}\right)x+\left(1-\frac{q^2}{p^2}\right)x^2\left(1-\frac{2q}{p}x\right)+2\left(\frac{q}{p}-1-\frac{q^3}{p^3}\right)x^{n+1}}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}+p^n\right)}\hfill \\ & -\frac{\frac{2q^2}{p^2}\left(1+\frac{q}{p}\right)x^2\left(1-x^{n-1}\right)}{{\left[3\right]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}+\frac{2p^n\left(\frac{2q}{p}-1\right)x{(1\ominus x)}_{p,q}^n}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}{p}^{\frac{n(n-1)}{2}}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}\hfill \\ & \left.\left.+\frac{2p^{n}x\left[\frac{q^2}{p^2}\left(1+\frac{q}{p}\right)x\left(1-x^{n-1}\right)+\left(1-\frac{3q}{p}\right)\left(1-x^n\right)\right]}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}\left(p^2{\left[n\right]}_{p,q}^2-p^{2n}\right)}\right\}\right|\hfill \\ \le & \frac{3-{\left[3\right]}_{p,q}{p}^{2n}}{{\left[3\right]}_{p,q}{p}^{2n}}+\frac{15}{2{\left[3\right]}_{p,q}{p}^n{[n+1]}_{p,q}}+\frac{6}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}^2}+\frac{7}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}\hfill \\ & +\frac{6}{{\left[3\right]}_{p,q}{[n+1]}_{p,q}^2\left(p{\left[n\right]}_{p,q}-p^n\right)}+\frac{2}{{\left[3\right]}_{p,q}{p}^{\frac{n(n-1)}{2}}{[n+1]}_{p,q}^2\left(p{\left[n\right]}_{p,q}-p^n\right)}.\hfill \end{array}$$

For a given $ϵ>0$, let us define the following sets

$$\begin{array}{c}U=\left\{n:{\left|\left|K_{n,p,q}^{\lambda }\left(e_2\right)-e_2\right|\right|}_{C[0,1]}\ge ϵ\right\},U_1=\left\{n:\frac{3-{\left[3\right]}_{p,q}{p}^{2n}}{{\left[3\right]}_{p,q}{p}^{2n}}\ge \frac{ϵ}{6}\right\},\hfill \\ U_2=\left\{n:\frac{1}{p^n{[n+1]}_{p,q}}\ge \frac{{\left[3\right]}_{p,q}ϵ}{45}\right\},U_3=\left\{n:\frac{1}{{[n+1]}_{p,q}^2}\ge \frac{{\left[3\right]}_{p,q}ϵ}{36}\right\},\hfill \\ U_4=\left\{n:\frac{1}{{[n+1]}_{p,q}\left(p{\left[n\right]}_{p,q}-p^n\right)}\ge \frac{{\left[3\right]}_{p,q}ϵ}{42}\right\},\hfill \\ U_5=\left\{n:\frac{1}{{[n+1]}_{p,q}^2\left(p{\left[n\right]}_{p,q}-p^n\right)}\ge \frac{{\left[3\right]}_{p,q}ϵ}{36}\right\},\hfill \\ U_6=\left\{n:\frac{1}{p^{\frac{n(n-1)}{2}}{[n+1]}_{p,q}^2\left(p{\left[n\right]}_{p,q}-p^n\right)}\ge \frac{{\left[3\right]}_{p,q}ϵ}{12}\right\}.\hfill \end{array}$$

It is obvious that $U\subseteq U_1\cup U_2\cup U_3\cup U_4\cup U_5\cup U_6$, then we obtain

$$\begin{array}{c}\hfill \sum _{n\in U}{a}_{jn}\le \sum _{n\in U_1}{a}_{jn}+\sum _{n\in U_2}{a}_{jn}+\sum _{n\in U_3}{a}_{jn}+\sum _{n\in U_4}{a}_{jn}+\sum _{n\in U_5}{a}_{jn}+\sum _{n\in U_6}{a}_{jn}.\end{array}$$

(21)

Letting $j\to \infty$ in Equation (21), using Equation (18) and

$$\begin{array}{c}\hfill s{t}_A-\underset{n}{lim}{\left[3\right]}_{p,q}{p}^{2n}=3,s{t}_A-\underset{n}{lim}\frac{1}{{[n+1]}_{p,q}}=0,s{t}_A-\underset{n}{lim}\frac{1}{p{\left[n\right]}_{p,q}-p^n}=0.\end{array}$$

We obtain $s{t}_A-{lim}_n{\left|\left|K_{n,p,q}^{\lambda }\left(e_2\right)-e_2\right|\right|}_{C[0,1]}=0$. Therefore, Equation (20) is proved, which yields the result of Theorem 3. ☐

We now need the following definitions to estimate the rate of A-statistical convergence of $K_{n,p,q}^{\lambda }(f;x)$.

Definition 4.

(See [29]) Let $A=\left(a_{nk}\right)$ be a non-negative regular summability matrix and let $\left(u_n\right)$ be a positive non-increasing sequence. The sequence $x=\left\{x_k\right\}$ is A-statistically convergent to the number L with the rate of $o\left(u_n\right)$ if for every $ϵ>0$,

$$\begin{array}{c}\hfill \underset{n}{lim}\frac{1}{u_n}\sum _{k:\left|x_k-L\right|\ge ϵ}{a}_{nk}=0.\end{array}$$

In this case we write $x_k-L=s{t}_A-o\left(u_k\right)$ as $k\to \infty$.

Definition 5.

(See [30]) Let $f\in C[0,1]$, Peetre’s K-functional is defined by

$$\begin{array}{c}\hfill K_2(f;\delta )=\underset{g\in C_{[0,1]}^2}{inf}\left\{\right||f-{g\left|\right|}_{C[0,1]}+\delta \left|\right|g^{{}^{\prime \prime }}{\left|\right|}_{C[0,1]}\},\end{array}$$

where $\delta >0$ and $C_{[0,1]}^2=\left\{g\in C[0,1]:g^{{}^{\prime }},g^{{}^{\prime \prime }}\in C[0,1]\right\}$. The second order modulus of smoothness of $f\in C[0,1]$ is defined by

$$\begin{array}{c}\hfill {\omega }_2(f;\delta )=\underset{0<h\le \delta }{sup}\underset{x\in [0,1]}{sup}|f(x+2h)-2f(x+h)+f\left(x\right)|.\end{array}$$

There exist an absolute constant $C>0$ such that $K_2(f;\delta )\le C{\omega }_2\left(f;\sqrt{\delta }\right)$. We also denote the usual of modulus of continuity by

$$\begin{array}{c}\hfill \omega (f;\delta )=\underset{0<h\le \delta }{sup}\underset{x,x+h\in [0,1]}{sup}|f(x+h)-f\left(x\right)|.\end{array}$$

Theorem 6.

Let $A=\left(a_{jn}\right)$ be a non-negative regular summability matrix. Assume that $\omega \left(f;\sqrt{\Phi (p,q;n)}\right)=s{t}_A-o\left(u_n\right)$, where $\Phi (p,q;n)$ is defined in Equation (16). Then for $f\in C[0,1]$, we have

$$\begin{array}{c}\hfill {\left|\left|K_{n,p,q}^{\lambda }\left(f\right)-f\right|\right|}_{C[0,1]}=s{t}_A-o\left(u_n\right).\end{array}$$

Proof. 

Since

$$\begin{array}{c}\hfill |f\left(t\right)-f\left(x\right)|\le \omega (f;|t-x\left|\right)\le \left(1+\frac{|t-x|}{\delta }\right)\omega (f;\delta ).\end{array}$$

Applying $K_{n,p,q}^{\lambda }(f;x)$ to both ends and using Cauchy–Schwarz inequality, we have

$$\begin{array}{ccc}\hfill \left|K_{n,p,q}^{\lambda }(f;x)-f\left(x\right)\right|& \le & K_{n,p,q}^{\lambda }\left(\right|f\left(t\right)-f\left(x\right)|;x)\le \left(1+\frac{K_{n,p,q}^{\lambda }\left(\left|{\varphi }_1(t,x)\right|;x\right)}{\delta }\right)\omega (f;\delta )\hfill \\ & \le & \left(1+\frac{\sqrt{K_{n,p,q}^{\lambda }\left({\varphi }_2(t,x)\right)}}{\delta }\right)\omega (f;\delta )\hfill \end{array}$$

Letting $\delta =\sqrt{K_{n,p,q}^{\lambda }\left({\varphi }_2(t,x)\right)}$, we get

$$\begin{array}{c}\hfill \left|K_{n,p,q}^{\lambda }(f;x)-f\left(x\right)\right|\le 2\omega \left(f;\sqrt{K_{n,p,q}^{\lambda }\left({\varphi }_2(t,x)\right)}\right)\le 2\omega \left(f;\sqrt{\Phi (p,q;n)}\right),\end{array}$$

where $\Phi (p,q;n)$ is defined in Equation (16). Taking supremum over $[0,1]$ on both sides, we obtain

$$\begin{array}{c}\hfill {\left|\left|K_{n,p,q}^{\lambda }(f;x)-f\left(x\right)\right|\right|}_{C[0,1]}\le 2\omega \left(f;\sqrt{\Phi (p,q;n)}\right).\end{array}$$

For a given $ϵ>0$, consider following sets

$$\begin{array}{c}\hfill V=\left\{n:{\left|\left|K_{n,p,q}^{\lambda }(f;x)-f\left(x\right)\right|\right|}_{C[0,1]}\ge ϵ\right\},V_1=\left\{n:\omega \left(f;\sqrt{\Phi (p,q;n)}\right)\ge \frac{ϵ}{2}\right\}.\end{array}$$

Obviously, we have $V\subseteq V_1$ and we also can obtain

$$\begin{array}{c}\hfill \frac{1}{u_j}\sum _{n\in V}{a}_{jn}\le \frac{1}{u_j}\sum _{n\in V_1}{a}_{jn}.\end{array}$$

Thus, let $j\to \infty$, by hypothesis we are led to the fact that ${\left|\left|K_{n,p,q}^{\lambda }\left(f\right)-f\right|\right|}_{C[0,1]}=s{t}_A-o\left(u_n\right)$. Theorem 6 is proved. ☐

Theorem 7.

Let $A=\left(a_{jn}\right)$ be a non-negative regular summability matrix. Assume that $\omega \left(f;\Theta (p,q;n)\right)=s{t}_A-o\left(a_n\right)$, ${\omega }_2\left(f;\sqrt{\Phi (p,q;n)+\Theta {(p,q;n)}^2}/2\right)=s{t}_A-o\left(b_n\right)$, where $\Theta (p,q;n)$ and $\Phi (p,q;n)$ are defined in Equations (15) and (16). Then for $f\in C[0,1]$, we have

$$\begin{array}{c}\hfill {\left|\left|K_{n,p,q}^{\lambda }\left(f\right)-f\right|\right|}_{C[0,1]}=s{t}_A-o(max\{a_n,b_n\}).\end{array}$$

Proof. 

Let’s define the following auxiliary operators

$$\begin{array}{c}\hfill {\widehat{K}}_{n,p,q}^{\lambda }(f;x)=K_{n,p,q}^{\lambda }(f;x)-f\left({\theta }_{p,q}^{\lambda }(n,x)\right)+f\left(x\right),\end{array}$$

(22)

where $x\in [0,1]$, ${\theta }_{p,q}^{\lambda }(n,x)$ is defined in Equation (11). Thus, we get

$$\begin{array}{c}\hfill {\widehat{K}}_{n,p,q}^{\lambda }({\varphi }_1(t,x);x)=0\end{array}$$

(23)

by Lemma 2. Letting $g\in C_{[0,1]}^2$, $t\in [0,1]$, by Taylor’s expansion, we have

$$\begin{array}{c}\hfill g\left(t\right)=g\left(x\right)+g^{{}^{\prime }}\left(x\right)(t-x)+{\int }_x^t(t-u)g^{{}^{\prime \prime }}\left(u\right)du.\end{array}$$

(24)

Applying ${\widehat{K}}_{n,p,q}^{\lambda }$ on both sides for Equation (24) and using Equation (23), we obtain

$$\begin{array}{c}\hfill {\widehat{K}}_{n,p,q}^{\lambda }(g;x)=g\left(x\right)+{\widehat{K}}_{n,p,q}^{\lambda }\left({\int }_x^t(t-u)g^{{}^{\prime \prime }}\left(u\right)du;x\right).\end{array}$$

Therefore, by Equations (22), (15) and (16), we have

$$\begin{array}{cc}& \left|{\widehat{K}}_{n,p,q}^{\lambda }(g;x)-f\left(x\right)\right|\hfill \\ \le & \left|K_{n,p,q}^{\lambda }\left({\int }_x^t(t-u)g^{{}^{\prime \prime }}\left(u\right)du;x\right)\right|+\left|{\int }_x^{{\theta }_{p,q}^{\lambda }(n,x)}\left({\theta }_{p,q}^{\lambda }(n,x)-u\right)g^{{}^{\prime \prime }}\left(u\right)du\right|\hfill \\ \le & K_{n,p,q}^{\lambda }\left(\left|{\int }_x^t(t-u)\left|g^{{}^{\prime \prime }}\left(u\right)\right|du\right|;x\right)+{\int }_x^{{\theta }_{p,q}^{\lambda }(n,x)}\left|{\theta }_{p,q}^{\lambda }(n,x)-u\right|\left|g^{{}^{\prime \prime }}\left(u\right)\right|du\hfill \\ \le & \left(\Phi (p,q;n)+\Theta {(p,q;n)}^2\right){\left|\left|g^{{}^{\prime \prime }}\right|\right|}_{C[0,1]}.\hfill \end{array}$$

(25)

Besides, by Equations (22) and (9), we get

$$\begin{array}{c}\hfill \left|{\widehat{K}}_{n,p,q}^{\lambda }(f;x)\right|\le {\left|\right|f\left|\right|}_{C[0,1]}{K}_{n,p,q}^{\lambda }(e_0;x)+{2\left|\right|f\left|\right|}_{C[0,1]}={3\left|\right|f\left|\right|}_{C[0,1]}.\end{array}$$

(26)

By Equations (22), (26) and (25), we have

$$\begin{array}{cc}& \left|K_{n,p,q}^{\lambda }(f;x)-f\left(x\right)\right|\hfill \\ =& \left|{\widehat{K}}_{n,p,q}^{\lambda }(f-g;x)-(f-g)\left(x\right)+{\widehat{K}}_{n,p,q}^{\lambda }(g;x)-g\left(x\right)+f\left({\theta }_{p,q}^{\lambda }(n,x)\right)-f\left(x\right)\right|\hfill \\ \le & \left|{\widehat{K}}_{n,p,q}^{\lambda }(f-g;x)-(f-g)\left(x\right)\right|+\left|{\widehat{K}}_{n,p,q}^{\lambda }(g;x)-g\left(x\right)\right|+\left|f\left({\theta }_{p,q}^{\lambda }(n,x)\right)-f\left(x\right)\right|\hfill \\ \le & 4\left|\right|f-{g\left|\right|}_{C[0,1]}+\left(\Phi (p,q;n)+\Theta {(p,q;n)}^2\right){\left|\left|g^{{}^{\prime \prime }}\right|\right|}_{C[0,1]}+\omega \left(f;\Theta (p,q;n)\right).\hfill \end{array}$$

Taking infimum on the right hand side over all $f\in C_{[0,1]}^2$, we obtain

$$\begin{array}{c}\hfill \left|K_{n,p,q}^{\lambda }(f;x)-f\left(x\right)\right|\le 4K_2\left(f;\left(\Phi (p,q;n)+\Theta {(p,q;n)}^2\right)/4\right)+\omega \left(f;\Theta (p,q;n)\right).\end{array}$$

Hence, we get

$$\begin{array}{c}\hfill \left|K_{n,p,q}^{\lambda }(f;x)-f\left(x\right)\right|\le C{\omega }_2\left(f;\sqrt{\Phi (p,q;n)+\Theta {(p,q;n)}^2}/2\right)+\omega \left(f;\Theta (p,q;n)\right).\end{array}$$

Taking supremum over $[0,1]$ on both sides, we have

$$\begin{array}{c}\hfill {\left|\left|K_{n,p,q}^{\lambda }\left(f\right)-f\right|\right|}_{C[0,1]}\le C{\omega }_2\left(f;\sqrt{\Phi (p,q;n)+\Theta {(p,q;n)}^2}/2\right)+\omega \left(f;\Theta (p,q;n)\right).\end{array}$$

For a given $ϵ>0$, set

$$\begin{array}{c}M=\left\{n:{\left|\left|K_{n,p,q}^{\lambda }\left(f\right)-f\right|\right|}_{C[0,1]}\ge ϵ\right\},\hfill \\ M_1=\left\{n:{\omega }_2\left(f;\sqrt{\Phi (p,q;n)+\Theta {(p,q;n)}^2}/2\right)\ge \frac{ϵ}{2C}\right\},\hfill \\ M_2=\left\{n:\omega \left(f;\Theta (p,q;n)\right)\ge \frac{ϵ}{2}\right\}.\hfill \end{array}$$

Then we have $M\subseteq M_1\cup M_2$ and

$$\begin{array}{c}\hfill \frac{1}{max\{a_j,b_j\}}\sum _{j\in M}{a}_{jn}\le \frac{1}{b_j}\sum _{j\in M_1}{a}_{jn}+\frac{1}{a_j}\sum _{j\in M_2}{a}_{jn}.\end{array}$$

According to the assumptions of Theorem 7, we have

$$\begin{array}{c}\hfill \underset{j\to \infty }{lim}\frac{1}{b_j}\sum _{j\in M_1}{a}_{jn}+\underset{j\to \infty }{lim}\frac{1}{a_j}\sum _{j\in M_2}{a}_{jn}=0.\end{array}$$

Thus,

$$\begin{array}{c}\hfill \underset{j\to \infty }{lim}\frac{1}{max\{a_j,b_j\}}\sum _{j\in M}{a}_{jn}=0.\end{array}$$

Therefore, we get the desire result of Theorem 7. ☐

4. Conclusions

In this paper, we introduced a kind of Kantorovich type $\lambda$-Bernstein operators $K_{n,p,q}^{\lambda }(f;x)$ via (p, q)-calculus, we estimated the moments and central moments and used these results to obtain an A-statistical convergence theorem and the rate of A-statistical convergence of $K_{n,p,q}^{\lambda }(f;x)$ to $f\left(x\right)$. In the future research work, we will continue to investigate some approximation properties of Durrmeyer type $\lambda$-Bernstein operators via (p, q)-calculus.