Approximation of Kantorovich-type Generalization of (p,q) - Bernstein type Rational Functions Via Statistical Convergence

In this paper, we use the modulus of continuity to study the rate of A-statistical convergence of the Kantorovich-type (p,q) - analogue of the Balázs–Szabados operators by using the statistical notion of convergence.

Mathematics subject classification: Primary 4H6D1; Secondary 4H6R1; 4H6R5.

Bernstein type rational functions, $R_n\left(f;x\right)=\frac{1}{{\left(1+a_{n}x\right)}^n}{\displaystyle \sum _{k=0}^{n}f\left(\frac{k}{b_n}\right)}\left(\begin{array}{l}n\\ k\end{array}\right){\left(a_{n}x\right)}^k\left(n=1,\text{ 2, }\mathrm{...}\right)$ Balázs defined and investigated them in 1975, (see [1]). In this definition, f is a real and single-valued function defined on the interval [0,∞), a_n and b_n are real numbers that have been appropriately chosen and are independent of x. Seven years later, in 1982, Balázs and Szabados cooperated to improve the estimate in [1] by selecting appropriate parameters a_n and b_n under some restrictions for f(x), (see[2]).

Recently, different q - generalizations of Balázs-Szabados operators have been studied by several researchers, see [3-7]. In [8], the Kantorovich-type q - analogue of the Balázs-Szabados operators is defined by Hamal and Sabancigil as follows:

$R_{n,q}^{\ast }\left(f,x\right)={\displaystyle \sum _{k=0}^n{r}_{n,k}}\left(q,x\right){\displaystyle \underset{0}{\overset{1}{\int }}f\left(\frac{{\left[k\right]}_q+q^{k}t}{b_n}\right)}{d}_{q}t,$ (1)

where

$\begin{array}{l}f:\left[0,\infty \right)\to ℝ,q\in \left(0,1\right),a_n={\left[n\right]}_q^{\beta -1},\\ b_n={\left[n\right]}_q^{\beta },0<\beta \le \frac{2}{3},n\in \mathbb{N},x\ge 0,\end{array}$

and

$r_{n,k}\left(q,x\right)=\frac{1}{{\left(1+a_{n}x\right)}^n}{\left[\begin{array}{l}n\\ k\end{array}\right]}_q{\left(a_{n}x\right)}^k{\displaystyle \prod _{s=0}^{n-k-1}\left(1+\left(1-q\right){\left[s\right]}_q{a}_{n}x\right)}.$

Additionally, the fast rise of (p,q) - calculus has encouraged many mathematicians in this subject to discover different generalizations. In the last decade, Mursaleen et al. defined and studied the analogue of many operators (see [9-15]). The (p,q) - generalization of Szász–Mirakjan operators was studied by Acar (see [16]), (p,q) - Kantorovich modification of Bernstein operators was studied by Acar and Aral (see [17]).

In [18-20], recently, Hamal and Sabancigil introduced a new Kantorovich-type (p,q) - analogue of the Balázs–Szabados operators by generalizing the new Kantorovich-type q - analogue of Balázs–Szabados operators, given by (1), as follows:

$R_{n,p,q}^{\ast }\left(f,x\right)={\displaystyle \sum _{k=0}^n{r}_{n,k}^{\ast }\left(p,q,x\right){\displaystyle \underset{0}{\overset{1}{\int }}f}\left(\frac{p^{n-k}\left({\left[k\right]}_{p,q}+q^{k}t\right)}{b_n}\right)}{d}_{p,q}t,$ (2)

where

$\begin{array}{l}{r}_{n,k}^{\ast }\left(p,q,x\right)=\frac{1}{p^{n\left(n-1\right)/2}}{\left[\begin{array}{l}n\\ k\end{array}\right]}_{p,q}{p}^{k\left(k-1\right)/2}\\ {\left(\frac{a_{n}x}{1+a_{n}x}\right)}^k{\displaystyle \prod _{j=0}^{n-k-1}\left(p^j-q^j\frac{a_{n}x}{1+a_{n}x}\right)}\end{array}$

and

$\begin{array}{l}0<q<p\le 1,a_n={\left[n\right]}_{p,q}^{\beta -1},b_n={\left[n\right]}_{p,q}^{\beta },\\ \text{0<}\beta \le \frac{2}{3},n\in \mathbb{N},x\ge 0,f:\left[0,\infty \right)\to ℝ.\end{array}$

These newly defined operators have some advantages when they are compared with the other (p,q) - analogues given in the other studies. The first advantage is that they are positive for all continuous and real-valued functions on the half-open interval [0,∞). The second advantage is that they can be used to approximate also the integrable functions. If p = 1, these polynomials reduce to the new Kantorovich-type analogue of the Balázs–Szabados operators, which are defined by Hamal and Sabancigil in [8]. Moreover, we considered the following two special cases:

• If 0<p<q≤1 or 1≤p<q<∞ or when the positivity property of the operators fails.

• If 1≤p<q<∞ then approximation by the new operators $R_{n,p,q}^{\ast }\left(f,x\right)$ becomes difficult because if p is large enough then the sequence ${\left\{R_{n,p,q}^{\ast }\right\}}_{n\in \mathbb{N}}$ may diverge.

Before stating the main result for these operators, we give some notations and definitions of (p,q) - calculus. For any p>0, q>0 non-negative integer n, the (p,q) - integer of the number n is defined as follows:

${\left[n\right]}_{p,q}=p^{n-1}+p^{n-2}q+p^{n-3}{q}^2+\mathrm{...}+p{q}^{n-2}+q^{n-1}=\{\begin{array}{l}\frac{p^n-q^n}{p-q}ifp\ne q\ne 1\\ n{p}^{n-1}ifp=q\ne 1\\ {\left[n\right]}_{q}ifp=1\\ nifp=q=1\end{array},$

the (p,q) - factorial is defined by

${\left[n\right]}_{p,q}!={\displaystyle \prod _{k=1}^n{\left[k\right]}_{p,q}}\text{, }n\ge 1\text{ and }{\left[0\right]}_{p,q}!=1,$

and (p,q) - binomial coefficient is defined by

${\left[\begin{array}{l}n\\ k\end{array}\right]}_{p,q}=\frac{{\left[n\right]}_{p,q}!}{{\left[k\right]}_{p,q}!{\left[n-k\right]}_{p,q}!}\text{, }0\le k\le n.$

The formula of (p,q) - binomial expansion is defined by

${\left(ax+by\right)}_{p,q}^n={\displaystyle \sum _{k=0}^n{p}^{\frac{(n-k)(n-k-1)}{2}}{q}^{\frac{k(k-1)}{2}}{a}^{n-k}{b}^k{x}^{n-k}{y}^k}=\left(ax+by\right)\left(pax+qby\right)\left(p^{2}ax+q^{2}by\right)\mathrm{...}\left(p^{n-1}ax+q^{n-1}by\right).$

Let $f:C[0,a]\to ℝ,$

Let $f:C[0,a]\to ℝ,$ the (p,q) - integral of is defined by:

${\displaystyle \underset{0}{\overset{a}{\int }}f(t)d_{p,q}t=(p-q)a{\displaystyle \sum _{k=0}^{\infty }f\left(\frac{q^k}{p^{k+1}}a\right)\frac{q^k}{p^{k+1}}\text{ if }\left|\frac{p}{q}\right|>1}}$

Fast [21] and Fridy [22] provided the following notions.

Suppose that $E\subseteq \mathbb{N}=\left\{1,\text{ 2, }\mathrm{...}\right\}$ and $E_n=\left\{k\le n:k\in E\right\}.$ . Then $\delta \left(E\right)=\underset{n\to \infty }{\lim }\frac{1}{n}\left|E_n\right|$ is called the natural density of E provided that the limit exists.

Definition 1: A sequence x = (x_n) is statistically convergent to the number L if for every ε>0, we have $\delta \left\{k\in \mathbb{N}:\left|x_k-L\right|\ge \epsilon \right\}=0$ is denoted by $s{t}_A-\underset{n\to \infty }{\lim }{x}_n=L$ .

Because all finite subsets of the natural numbers have density zero, any convergent sequence is statistically convergent, but not contrariwise.

For example, consider the sequence A = {a_n, n = 1,2,3…} whose terms are

$a_n=\{\begin{array}{l}\sqrt{n}\text{ when }n=m^2,\forall m=1\text{, 2, 3,}\mathrm{...}\\ 1\text{ otherwise }\end{array}$

We can see that the sequence is divergent in the ordinary sense, but it is statistically convergent to 1.

Let C_B [a,b] denote the space of all functions f which are continuous in every point of the interval [a,b] and bounded on the entire positive real line, $\left|f\left(x\right)\right|\le M_f,\forall x\in \left(0,\infty \right)$ .

Lemma 1 ([10]): For all Let $n\in \mathbb{N},x\in \left[0,\infty \right)$ and 0

$R_{n,p,q}^{\ast }\left(1,x\right)=1.$

$R_{n,p,q}^{\ast }\left(t,x\right)=\frac{p^n}{{\left[2\right]}_{p,q}{b}_n}+\frac{2q}{{\left[2\right]}_{p,q}}\left(\frac{x}{1+a_{n}x}\right)$

$\begin{array}{l}{R}_{n,p,q}^{\ast }\left(t^2,x\right)=\frac{p^{2n}}{{\left[3\right]}_{p,q}{b}_n^2}+\frac{\left(4q^3+5q^{2}p+3q{p}^2\right)p^{n-1}}{{\left[2\right]}_{p,q}{\left[3\right]}_{p,q}{b}_n}\left(\frac{x}{1+a_{n}x}\right)\\ +\frac{q{\left[n-1\right]}_{p,q}}{{\left[n\right]}_{p,q}}\frac{4q^3+q^{2}p+q{p}^2}{{\left[2\right]}_{p,q}{\left[3\right]}_{p,q}}{\left(\frac{x}{1+a_{n}x}\right)}^2.\end{array}$

Lemma 2 ([10]): For all $n\in \mathbb{N},x\in \left[0,\infty \right)$

${\left(R_{n,p,q}^{\ast }\left((t-x),x\right)\right)}^2\le \frac{1}{b_n}\left\{\frac{1}{b_n}+\frac{{\left(p^n-q^n\right)}^2}{b_n}{\left(\frac{1}{p+q}+\frac{1}{p-q}\left(a_{n}x\right)\right)}^2\right\},x\in \left[0,\infty \right),$ (3)

$R_{n,p,q}^{\ast }\left({\left(t-x\right)}^2,x\right)\le \frac{A_1}{b_n}{\varphi }_n\left(p,q\right){\left(1+x\right)}^2,x\in \left[0,\infty \right),$ (4)

$R_{n,p,q}^{\ast }\left({\left(t-x\right)}^4,x\right)\le \frac{A_2}{b_n^2}{\left(1+x\right)}^2\text{, }x\in \left[0,\infty \right),$ (5)

Where A₁ > 0, A₂ > 0 and ${\varphi }_n\left(p,q\right)=\max \left\{p^{n-1},b_n-a_n{p}^{n-1},\frac{1}{{\left[3\right]}_{p,q}{b}_n}\right\}.$

In the following theorem, the Bohman –Korovkin type statistical approximation theorem was proved by Gadjiev and Orhan [23].

Theorem 1 ([13]): Let ${\left({\ell }_n\right)}_{n\in \mathbb{N}}$ be a sequence of positive linear operators acting from C_B [a,b] to B [a,b] that is, ${\ell }_n:C_B\left[a,b\right]\to B\left[a,b\right]$ satisfies the conditions that

$s{t}_A-\underset{}{\lim \Vert {\ell }_n\left(e_i\right)-e_i\Vert }=0\text{ with }{e}_i\left(t\right)=t^i\text{ and }\forall i=0,\text{ 1, 2}\text{.}$ (6)

Then, we have

$s{t}_A-\underset{n}{\lim }\Vert {\ell }_{n}f-f\Vert =0\text{ ,}\forall f\in C_B\left(\left[a,b\right]\right).$

Now, we give the main result of this research is to use the modulus of continuity to study the rate of A-statistical convergence of Kantorovich-type (p,q) - analogue of the Balázs–Szabados operators $R_{n,p,q}^{\ast }\left(f,x\right)$ .

Theorem 2: Let $q=\left(q_n\right),p=\left(p_n\right),\text{ 0}<q_n<p_n\le 1$ such that $s{t}_A-\underset{n}{\lim }{q}_n=1,s{t}_A-\underset{n}{\lim }{p}_n=1$ and $s{t}_A-\underset{n}{\lim }{p}_n^n=1$ . Then for each compact interval $\left[0,b\right]\subset \left[0,\infty \right)$ , we have $s{t}_A-\underset{n}{\lim }\Vert R_{n,p,q}^{\ast }\left(f,x\right)-f\left(x\right)\Vert =0\text{ , }\forall f\in C\left(\left[0,b\right]\right)$ .

Proof: According to Theorem 1, it is sufficient to show that it satisfies (6). By using Lemma 1, it is clear that

$s{t}_A-\underset{n}{\lim }\Vert R_{n,p_{n,}{q}_n}^{\ast }\left(e_0;x\right)-e_0\Vert =0\text{ , since }{R}_{n,p_{n,}{q}_n}^{\ast }\left(e_0;x\right)=1.$ (7)

Again by Lemma 1, we have

$\begin{array}{l}\left|R_{n,p_{n,}{q}_n}^{\ast }\left(e_1;x\right)-e_1\right|=\left|\frac{p_n^n}{{\left[2\right]}_{p,q}{b}_n}+\frac{2q_n}{{\left[2\right]}_{p_{n,}{q}_n}}\left(\frac{x}{1+a_{n,p_{n,}{q}_n}x}\right)-x\right|\\ =\frac{p_n^n}{{\left[2\right]}_{p_{n,}{q}_n}{b}_n}+\frac{\left(p_n-q_n\right)}{{\left[2\right]}_{p_{n,}{q}_n}}\frac{x}{1+a_{n,p_{n,}{q}_n}x}+\frac{a_{n,p_{n,}{q}_n}{x}^2}{1+a_{n,p_{n,}{q}_n}x}.\end{array}$

By taking the maximum of both sides of the last equality on [0,b] with $0<b<\frac{1}{a_{n,p_n,q_n}}$ , we obtain

$\Vert R_{n,p_{n,}{q}_n}^{\ast }\left(e_1;x\right)-e_1\Vert \le \frac{p_n^n}{{\left[2\right]}_{p_{n,}{q}_n}{b}_n}+\frac{\left(p_n-q_n\right)}{{\left[2\right]}_{p_{n,}{q}_n}}\frac{b}{1+a_{n,p_{n,}{q}_n}b}+\frac{a_{n,p_{n,}{q}_n}{b}^2}{1+a_{n,p_{n,}{q}_n}b}.$

By using the limits $s{t}_A-\underset{n}{\lim }{q}_n=1,s{t}_A-\underset{n}{\lim }{p}_n=1$ , we have

herefore,

$\Vert R_{n,p_{n,}{q}_n}^{\ast }\left(e_1;x\right)-e_1\Vert <\epsilon .$

For ε>0, we define the sets

$A:=\left\{n\in N:\Vert R_{n,p_{n,}{q}_n}^{\ast }\left(e_1;.\right)-e_1\Vert \ge \epsilon \right\},$ (8)

$A_1=\left\{n\in \mathbb{N}:\frac{p_n^n}{{\left[2\right]}_{p_{n,}{q}_n}{b}_n}\ge \epsilon \right\},A_2=\left\{n\in \mathbb{N}:\frac{\left(p_n-q_n\right)}{{\left[2\right]}_{p_{n,}{q}_n}}\frac{b}{1+a_{n,p_{n,}{q}_n}b}\ge \epsilon \right\},$ and

$A_3=\left\{n\in \mathbb{N}:\frac{a_{n,p_{n,}{q}_n}{b}^2}{1+a_{n,p_{n,}{q}_n}b}\ge \epsilon \right\},$ thus from (8), we can see that $A\subseteq A_1\cup A_2\cup A_3$ ,

$\begin{array}{l}\delta \left\{n\in N:\Vert R_{n,p_{n,}{q}_n}^{\ast }\left(e_1;.\right)-e_1\Vert \ge \epsilon \right\}\le \delta \left\{n\in \mathbb{N}:\frac{p_n^{n-1}}{b_{n,p_{n,}{q}_n}}\frac{b}{1+a_{n,p_{n,}{q}_n}b}\ge \frac{\epsilon }{3}\right\}\\ +\delta \left\{n\in \mathbb{N}:\left(1-\frac{1}{{\left(1+a_{n,p_{n,}{q}_n}b\right)}^2}\right)b^2\ge \frac{\epsilon }{3}\right\}\end{array}$

$\text{+}\delta \left\{n\in \mathbb{N}:\frac{p_n^{n-1}{}^{`}}{{\left[n\right]}_{p_{n,}{q}_n}}\frac{b^2}{{\left(1+a_{n,p_{n,}{q}_n}b\right)}^2}\ge \frac{\epsilon }{3}\right\}.$ (9)

By taking the limit of both sides of the above inequality (9), It is obvious that

$s{t}_A-\underset{n}{\lim }\frac{p_n^{n-1}}{b_{n,p_{n,}{q}_n}}\frac{b}{1+a_{n,p_{n,}{q}_n}b}=0,s{t}_A-\underset{n}{\lim }\frac{1}{{\left(1+a_{n,p_{n,}{q}_n}b\right)}^2}=1,$

$s{t}_A-\underset{n}{\lim }\frac{p_n^{n-1}{}^{`}}{{\left[n\right]}_{p_{n,}{q}_n}}\frac{b^2}{{\left(1+a_{n,p_{n,}{q}_n}b\right)}^2}=0.$

Which implies

$s{t}_A-\underset{n}{\lim }\Vert R_{n,p_{n,}{q}_n}^{\ast }\left(e_1;x\right)-e_1\Vert =0$ . (10)

Also, by using Lemma 1, we may write

$\begin{array}{l}\left|R_{n,p_{n,}{q}_n}^{\ast }\left(e_2;x\right)-e_2\right|\le \left|\frac{p_n^{2n}}{{\left[3\right]}_{p_n,q_n}{b}_n^2}+\frac{\left(4q_n^3+5q_n^2{p}_n+3q_n{p}_n^2\right)p_n^{n-1}}{{\left[2\right]}_{p_n,q_n}{\left[3\right]}_{p_n,q_n}{b}_n}\left(\frac{x}{1+a_{n,p_n,q_n}x}\right)\right|\\ \left|+\frac{q_n{\left[n-1\right]}_{p_n,q_n}}{{\left[n\right]}_{p_n,q_n}}\frac{4q_n^3+q_n^2{p}_n+q_n{p}_n^2}{{\left[2\right]}_{p_n,q_n}{\left[3\right]}_{p_n,q_n}}{\left(\frac{x}{1+a_{n,p_n,q_n}x}\right)}^2-x^2\right|\end{array}$ .

$\le \frac{p_n^{2n}}{{\left[3\right]}_{p_n,q_n}{b}_{n,p_n,q_n}^2}+\frac{\left(4q_n^3+5q_n^2{p}_n+3q_n{p}_n^2\right)p_n^{n-1}}{{\left[2\right]}_{p_n,q_n}{\left[3\right]}_{p_n,q_n}{b}_{n,p_n,q_n}}\left(\frac{x}{1+a_{n}x}\right)$

$+\left\{1-\frac{4q_n^3+q_n^2{p}_n+q_n{p}_n^2}{{\left[2\right]}_{p_n,q_n}{\left[3\right]}_{p_n,q_n}}\frac{1}{{\left(1+a_{n,p_n,q_n}\right)}^2}\right\}{x}^2+\frac{p_n^{n-1}}{{\left[n\right]}_{p_n,q_n}}\frac{4q_n^3+q_n^2{p}_n+q_n{p}_n^2}{{\left[2\right]}_{p_n,q_n}{\left[3\right]}_{p_n,q_n}}{\left(\frac{x}{1+a_{n}x}\right)}^2$

By taking the maximum of both sides of the last equality on [0,b] with $0<b<\frac{1}{a_{n,p_n,q_n}}$ , we get

$\Vert R_{n,p_{n,}{q}_n}^{\ast }\left(e_2;x\right)-e_2\Vert \le \frac{p_n^{2n}}{{\left[3\right]}_{p_n,q_n}{b}_{n,p_n,q_n}^2}+\frac{\left(4q_n^3+5q_n^2{p}_n+3q_n{p}_n^2\right)p_n^{n-1}}{{\left[2\right]}_{p_n,q_n}{\left[3\right]}_{p_n,q_n}{b}_{n,p_n,q_n}}\left(\frac{b}{1+a_{n,p_n,q_n}b}\right)$

$+\left\{1-\frac{4q_n^3+q_n^2{p}_n+q_n{p}_n^2}{{\left[2\right]}_{p_n,q_n}{\left[3\right]}_{p_n,q_n}}\frac{1}{{\left(1+a_{n,p_n,q_n}b\right)}^2}\right\}{b}^2+\frac{p_n^{n-1}}{{\left[n\right]}_{p_n,q_n}}\frac{4q_n^3+q_n^2{p}_n+q_n{p}_n^2}{{\left[2\right]}_{p_n,q_n}{\left[3\right]}_{p_n,q_n}}{\left(\frac{b}{1+a_{n,p_n,q_n}b}\right)}^2$

By using the limits $s{t}_A-\underset{n}{\lim }{q}_n=1,s{t}_A-\underset{n}{\lim }{p}_n=1$ , we have

$s{t}_A-\underset{n}{\lim }\frac{4q_n^3+q_n^2{p}_n+q_n{p}_n^2}{{\left[2\right]}_{p_n,q_n}{\left[3\right]}_{p_n,q_n}}=1,s{t}_A-\underset{n}{\lim }\frac{p_n^{n-1}}{{\left[n\right]}_{p_n,q_n}}=0,s{t}_A-\underset{n}{\lim }\frac{p_n^{2n}}{{\left[3\right]}_{p_n,q_n}{b}_{n,p_n,q_n}^2}=0.$

Therefore,

$\Vert R_{n,p_{n,}{q}_n}^{\ast }\left(e_2;x\right)-e_2\Vert <\epsilon .$

Now, for given ε > 0, we introduce the following sets;

$D:=\left\{n\in N:\Vert R_{n,p_{n,}{q}_n}^{\ast }\left(e_2;.\right)-e_2\Vert \ge \epsilon \right\},$

$\begin{array}{l}{D}_1=\left\{n\in \mathbb{N}:\frac{p_n^{2n}}{{\left[3\right]}_{p_n,q_n}{b}_{n,p_n,q_n}^2}\ge \frac{\epsilon }{4}\right\},\\ D_2=\left\{n\in \mathbb{N}:\frac{\left(4q_n^3+5q_n^2{p}_n+3q_n{p}_n^2\right)p_n^{n-1}}{{\left[2\right]}_{p_n,q_n}{\left[3\right]}_{p_n,q_n}{b}_{n,p_n,q_n}}\left(\frac{b}{1+a_{n,p_n,q_n}b}\right)\ge \frac{\epsilon }{4}\right\},\end{array}$

$D_3=\left\{n\in \mathbb{N}:\left\{1-\frac{4q_n^3+q_n^2{p}_n+q_n{p}_n^2}{{\left[2\right]}_{p_n,q_n}{\left[3\right]}_{p_n,q_n}}\frac{1}{{\left(1+a_{n,p_n,q_n}b\right)}^2}\right\}{b}^2\ge \frac{\epsilon }{4}\right\},$

$D_4=\left\{n\in \mathbb{N}:\frac{p_n^{n-1}}{{\left[n\right]}_{p_n,q_n}}\frac{4q_n^3+q_n^2{p}_n+q_n{p}_n^2}{{\left[2\right]}_{p_n,q_n}{\left[3\right]}_{p_n,q_n}}{\left(\frac{b}{1+a_{n,p_n,q_n}b}\right)}^2\ge \frac{\epsilon }{4}\right\}.$

Then, from (11) we may write $D\subseteq D_1\cup D_2\cup D_3\cup D_4,$

$\begin{array}{l}\delta \left\{n\in N:\Vert R_{n,p_{n,}{q}_n}^{\ast }\left(e_2;.\right)-e_2\Vert \ge \epsilon \right\}\le \delta \left\{n\in \mathbb{N}:\frac{p_n^{2n}}{{\left[3\right]}_{p_n,q_n}{b}_{n,p_n,q_n}^2}\ge \frac{\epsilon }{4}\right\}\\ +\delta \left\{n\in \mathbb{N}:\frac{\left(4q_n^3+5q_n^2{p}_n+3q_n{p}_n^2\right)p_n^{n-1}}{{\left[2\right]}_{p_n,q_n}{\left[3\right]}_{p_n,q_n}{b}_{n,p_n,q_n}}\left(\frac{b}{1+a_{n,p_n,q_n}b}\right)\ge \frac{\epsilon }{4}\right\}\\ \text{ + }\delta \left\{n\in \mathbb{N}:\left\{1-\frac{4q_n^3+q_n^2{p}_n+q_n{p}_n^2}{{\left[2\right]}_{p_n,q_n}{\left[3\right]}_{p_n,q_n}}\frac{1}{{\left(1+a_{n,p_n,q_n}b\right)}^2}\right\}{b}^2\ge \frac{\epsilon }{4}\right\}\end{array}$

$+\delta \left\{n\in \mathbb{N}:\frac{p_n^{n-1}}{{\left[n\right]}_{p_n,q_n}}\frac{4q_n^3+q_n^2{p}_n+q_n{p}_n^2}{{\left[2\right]}_{p_n,q_n}{\left[3\right]}_{p_n,q_n}}{\left(\frac{b}{1+a_{n,p_n,q_n}b}\right)}^2\ge \frac{\epsilon }{4}\right\},$

by taking the limit of both sides of the above inequality, It is obvious that

$\delta \left(D\right)\le \delta \left(D_1\right)+\delta \left(D_2\right)+\delta \left(D_3\right)+\delta \left(D_4\right)=0$ , which implies

$s{t}_A-\underset{n}{\lim }\Vert R_{n,p_{n,}{q}_n}^{\ast }\left(e_2;x\right)-e_2\Vert =0.$ . As a result, Equation (6) is proven, yielding the desired result.

In this paper, by using the notion of (p,q) - calculus and statistical convergence, we give the main result of this research to use the modulus of continuity to study the rate of A-statistical convergence of Kantorovich type (p,q) - analogue of the Balázs–Szabados operators.

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