# Convergence of λ-Bernstein operators based on (p, q)-integers

## Abstract

In the present paper, we construct a new class of positive linear λ-Bernstein operators based on (p, q)-integers. We obtain a Korovkin type approximation theorem, study the rate of convergence of these operators by using the conception of K-functional and moduli of continuity, and also give a convergence theorem for the Lipschitz continuous functions.

## Introduction

In 2018, Cai et al.  introduced the following new family of Bernstein operators with parameter $$\lambda \in [-1,1]$$:

\begin{aligned} B_{n,\lambda }(f; x)=\sum_{k=0}^{n} \widetilde{b}_{n,k}(\lambda; x)f \biggl(\frac{k}{n} \biggr), \end{aligned}
(1)

where $$\widetilde{b}_{n,k}(\lambda; x)$$$$(k=0, 1,\ldots, n)$$ are the λ-Bernstein basis functions defined as

\begin{aligned} \textstyle\begin{cases} \widetilde{b}_{n, 0}(\lambda; x)=b_{n, 0}(x)-\frac{\lambda }{n+1}b _{n+1,1}(x), \\ \widetilde{b}_{n, k}(\lambda; x)=b_{n, k}(x)+\lambda (\frac{n-2k+1}{n ^{2}-1}b_{n+1, k}(x)-\frac{n-2k-1}{n^{2}-1}b_{n+1,k+1}(x) ), \\ \quad (k=1,2,\ldots,n-1), \\ \widetilde{b}_{n, n}(\lambda; x)=b_{n, n}(x)-\frac{\lambda }{n+1}b _{n+1,n}(x), \end{cases}\displaystyle \end{aligned}

and $bn,k(x)=(nk)xk(1−x)n−k$$$(k=0, 1,\ldots, n)$$ are the classical Bernstein basis functions. They call these operators (1) λ-Bernstein operators, they investigated some approximation theorems and also gave some numerical examples. In the same year, Acu et al.  studied some approximation properties of Kantorovich type of operators (1). Later, Özger  gave some statistical approximation results of (1). In 2019, Cai et al.  proposed new λ-Bernstein operators based on q-integers and established a statistical approximation theorem. Some other papers also mention λ-Bernstein operators, see [5, 6].

Recently, Mursaleen et al.  defined the following (p, q)-analogue of Bernstein operators:

\begin{aligned} B_{n,p,q}(f; x)=\sum_{k=0}^{n}b_{n,k}(x; p, q)f \biggl(\frac{[k]_{p,q}}{p ^{k-n}[n]_{p,q}} \biggr),\quad x\in [0, 1], \end{aligned}
(2)

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

\begin{aligned} b_{n,k}(x; p, q)=\frac{1}{p^{\frac{n(n-1)}{2}}} \begin{bmatrix} n \\ k \end{bmatrix}_{p,q}p^{\frac{k(k-1)}{2}}x^{k} \prod_{s=0}^{n-k-1} \bigl(p ^{s}-q^{s}x \bigr),\quad x\in [0, 1]. \end{aligned}
(3)

There are many papers about the research and application of (p, q) operators, we mention some of them .

Motivated by the above work, in this paper, we introduce positive linear λ-Bernstein operators based on (p, q)-integers as follows:

\begin{aligned} B_{n,p,q}^{\lambda }(f; x)=\sum_{k=0}^{n}b_{n,k}^{\lambda }(x; p, q)f \biggl(\frac{[k]_{p,q}}{p^{k-n}[n]_{p,q}} \biggr),\quad x\in [0, 1], \end{aligned}
(4)

where $$b_{n,k}^{\lambda }(x; p, q)$$$$(k=0, 1, \ldots, n)$$ are called (p, q)-analogue of λ-Bernstein basis functions and defined as

\begin{aligned} \textstyle\begin{cases} b_{n,0}^{\lambda }(x; p, q)=b_{n,0}(x; p, q)- \frac{\lambda }{p^{1-n}[n]_{p,q}+1}b_{n+1,1}(x; p, q), \\ b_{n,k}^{\lambda }(x; p, q)=b_{n,k}(x; p, q)+\lambda (\frac{p ^{1-n}[n]_{p,q}-2p^{1-k}[k]_{p,q}+1}{p^{2-2n}[n]_{p,q}^{2}-1}b_{n+1,k}(x; p, q) \\ \phantom{b_{n,k}^{\lambda }(x; p, q)=}{} -\frac{p^{1-n}[n]_{p,q}-2qp^{-k}[k]_{p,q}-1}{p^{2-2n}[n]_{p,q} ^{2}-1}b_{n+1,k+1}(x; p, q) ),\quad (k=1,2,\ldots,n-1), \\ b_{n,n}^{\lambda }(x; p, q)=b_{n,n}(x; p, q)- \frac{\lambda }{p^{1-n}[n]_{p,q}+1}b_{n+1,n}(x; p, q), \end{cases}\displaystyle \end{aligned}

$$b_{n,k}(x; p, q)$$$$(k=0, 1,\ldots, n)$$ are defined in (3), $$\lambda \in [-1,1]$$, $$n\geq 2$$, $$x\in [0,1]$$, and $$0< q< p\leq 1$$. Obviously, when $$p\rightarrow 1^{-}$$, $$B_{n,p,q}^{\lambda }(f; x)$$ reduces to q-analogue of λ-Bernstein operators in ; when $$p, q\rightarrow 1^{-}$$, $$B_{n,p,q}^{\lambda }(f; x)$$ reduces to (1).

Here we mention certain notations on (p, q)-calculus, details can be found in .

For any fixed real numbers $$p>0$$ and $$q>0$$, the (p, q)-integers $$[n]_{p,q}$$ are defined as follows:

\begin{aligned}{} [n]_{p,q}=p^{n-1}+p^{n-2}q+p^{n-3}q^{2}+ \cdots +pq^{n-2}+q^{n-1}= \textstyle\begin{cases} \frac{p^{n}-q^{n}}{p-q}, & p\neq q\neq 1; \\ \frac{1-q^{n}}{1-q}, & p=1; \\ n, & p=q=1. \end{cases}\displaystyle \end{aligned}

(p, q)-factorial and (p, q)-binomial coefficients are defined as follows:

\begin{aligned}{} [n]_{p,q}!=\textstyle\begin{cases} [n]_{p,q}[n-1]_{p,q}\cdots _{p,q}, & n=1,2,\ldots; \\ 1, & n=0, \end{cases}\displaystyle \qquad \begin{bmatrix} n \\ k \end{bmatrix}_{p,q}= \frac{[n]_{p,q}!}{[k]_{p,q}![n-k]_{p,q}!}. \end{aligned}

The aims of the present paper are to construct a new class of λ-Bernstein operators based on (p, q)-integers and give the rate of convergence of these operators.

The rest of this paper is mainly organized as follows. In Sect. 2, we compute some moments and central moments of $$B_{n,p,q} ^{\lambda }(f; x)$$ which are needed to prove our main results. In Sect. 3, we obtain a Korovkin approximation theorem and rate of convergence of operators $$B_{n,p,q}^{\lambda }(f; x)$$.

## Some lemmas

In order to obtain the main results, we need the following lemmas.

### Lemma 2.1

(see )

For$$x\in [0, 1]$$, $$0< q< p\leq 1$$, we have

\begin{aligned} B_{n,p,q}(1; x)=1;\qquad B_{n,p,q}(t; x)=x;\qquad B_{n,p,q} \bigl(t^{2}; x \bigr)=\frac{p ^{n-1}}{[n]_{p,q}}x+\frac{q[n-1]_{p,q}}{[n]_{p,q}}x^{2}. \end{aligned}

### Lemma 2.2

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

\begin{aligned} B_{n,p,q}^{\lambda }(1; x)=1. \end{aligned}
(5)

### Proof

Actually, by (4) and the fact that $$q[n-1]_{p,q}=[n]_{p,q}-p ^{n-1}$$, we have

\begin{aligned} &B_{n,p,q}^{\lambda }(1; x) \\ &\quad=\sum_{k=0}^{n}b_{n,k}^{\lambda }(x; p, q) \\ &\quad=\sum_{k=0}^{n}b_{n,k}(x; p, q)-\frac{\lambda }{p^{1-n}[n]_{p,q}+1}b _{n+1,1}(x; p, q)\\ &\qquad{}+\lambda \frac{p^{1-n}[n]_{p,q}-2_{p,q}+1}{p^{2-2n}[n]_{p,q} ^{2}-1}b_{n+1,1}(x; p, q) \\ &\qquad{}-\lambda \frac{p^{1-n}[n]_{p,q}-2qp^{-1}_{p,q}-1}{p^{2-2n}[n]_{p,q} ^{2}-1}b_{n+1,2}(x; p, q)\\ &\qquad{}+\lambda \frac{p^{1-n}[n]_{p,q}-2p^{-1}_{p,q}+1}{p ^{2-2n}[n]_{p,q}^{2}-1}b_{n+1,2}(x; p, q) \\ &\qquad{}-\lambda \frac{p^{1-n}[n]_{p,q}-2qp^{-2}_{p,q}-1}{p^{2-2n}[n]_{p,q} ^{2}-1}b_{n+1,3}(x; p, q)\\ &\qquad{}+\lambda \frac{p^{1-n}[n]_{p,q}-2p^{-2}_{p,q}+1}{p ^{2-2n}[n]_{p,q}^{2}-1}b_{n+1,3}(x; p, q) \\ &\qquad{}-\cdots \\ &\qquad{}-\lambda \frac{p^{1-n}[n]_{p,q}-2qp^{1-n}[n-1]_{p,q}-1}{p^{2-2n}[n]_{p,q} ^{2}-1}b_{n+1,n}(x; p, q)-\frac{\lambda }{p^{1-n}[n]_{p,q}+1}b_{n+1,n}(x; p, q) \\ &\quad =\sum_{k=0}^{n}b_{n,k}(x; p, q). \end{aligned}

Then we can obtain (5) since $$B_{n,p,q}(1;x)=\sum_{k=0}^{n}b _{n,k}(x; p, q)=1$$. □

### Lemma 2.3

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

\begin{aligned} B_{n,p,q}^{\lambda }(t;x) ={}&x+\frac{2\lambda [n+1]_{p,q}x^{2} (1-x ^{n-1} )}{p^{n} (p^{2-2n}[n]_{q}^{2}-1 )} \biggl(1- \frac{q}{p} \biggr)+\frac{\lambda p^{n} (1-x^{n+1} )}{q[n]_{p,q} (p^{1-n}[n]_{p,q}-1 )} \\ &{}+\frac{\lambda [n+1]_{p,q}x (1-x^{n} )}{[n]_{p,q} (p ^{1-n}[n]_{p,q}-1 )} \biggl(\frac{1}{p}-\frac{1}{q}- \frac{2}{p (p^{1-n}[n]_{p,q}+1 )} \biggr) \\ &{}-\frac{\lambda \prod_{s=0}^{n} (p^{s}-q^{s}x )}{p^{ \frac{n(n-1)}{2}}q[n]_{p,q} (p^{1-n}[n]_{p,q}-1 )}. \end{aligned}
(6)

### Proof

From (4), we have

\begin{aligned} &B_{n,p,q}^{\lambda }(t;x) \\ &\quad=\sum_{k=0}^{n}b_{n,k}^{\lambda }(x; p, q) \frac{[k]_{p,q}}{p^{k-n}[n]_{p,q}} \\ &\quad=\sum_{k=1}^{n-1}\frac{[k]_{p,q}}{p^{k-n}[n]_{p,q}} \biggl[b_{n,k}(x; p, q)+\lambda \biggl(\frac{p^{1-n}[n]_{p,q}-2p^{1-k}[k]_{p,q}+1}{p ^{2-2n}[n]_{p,q}^{2}-1}b_{n+1,k}(x; p, q) \\ &\qquad{}-\frac{p^{1-n}[n]_{p,q}-2qp^{-k}[k]_{p,q}-1}{p^{2-2n}[n]_{p,q} ^{2}-1}b_{n+1,k+1}(x; p, q) \biggr) \biggr]+b_{n,n}(x; p, q) \\ &\qquad{}-\frac{\lambda }{p^{1-n}[n]_{p,q}+1}b_{n+1,n}(x; p, q) \\ &\quad=\sum_{k=0}^{n}\frac{[k]_{p,q}}{p^{k-n}[n]_{p,q}}b_{n,k}(x; p, q) \\ &\qquad{}+\lambda \sum_{k=0}^{n} \frac{[k]_{p,q}}{p^{k-n}[n]_{p,q}}\frac{p ^{1-n}[n]_{p,q}-2p^{1-k}[k]_{p,q}+1}{p^{2-2n}[n]_{p,q}^{2}-1}b_{n+1,k}(x; p, q) \\ &\qquad{}-\lambda \sum_{k=1}^{n-1} \frac{[k]_{p,q}}{p^{k-n}[n]_{p,q}}\frac{p ^{1-n}[n]_{p,q}-2qp^{-k}[k]_{p,q}-1}{p^{2-2n}[n]_{p,q}^{2}-1}b_{n+1,k+1}(x; p, q) \\ &\quad=\triangle _{1,1}+\triangle _{1,2}+\triangle _{1,3}, \end{aligned}
(7)

where

\begin{aligned} &\triangle _{1,1} =\sum_{k=0}^{n} \frac{[k]_{p,q}}{p^{k-n}[n]_{p,q}}b _{n,k}(x; p, q); \\ &\triangle _{1,2} =\lambda \sum_{k=0}^{n} \frac{[k]_{p,q}}{p^{k-n}[n]_{p,q}}\frac{p^{1-n}[n]_{p,q}-2p^{1-k}[k]_{p,q}+1}{p ^{2-2n}[n]_{p,q}^{2}-1}b_{n+1,k}(x; p, q); \\ &\triangle _{1,3} =-\lambda \sum_{k=1}^{n-1} \frac{[k]_{p,q}}{p^{k-n}[n]_{p,q}}\frac{p^{1-n}[n]_{p,q}-2qp^{-k}[k]_{p,q}-1}{p ^{2-2n}[n]_{p,q}^{2}-1}b_{n+1,k+1}(x; p, q). \end{aligned}

By Lemma 2.1, we get

\begin{aligned} \triangle _{1,1}=B_{n, p, q}(t; x)=x. \end{aligned}
(8)

Next, since $$[k]_{p,q}^{2}=q[k]_{p,q}[k-1]_{p,q}+p^{k-1}[k]_{p,q}$$, we have

\begin{aligned} &\triangle _{1,2} \\ &\quad=\lambda \sum_{k=0}^{n} \frac{[k]_{p,q}}{p^{k-n}[n]_{p,q}} \biggl(\frac{1}{p ^{1-n}[n]_{p,q}-1}-\frac{2p^{1-k}[k]_{p,q}}{p^{2-2n}[n]_{p,q}^{2}-1} \biggr)b _{n+1,k}(x; p, q) \\ &\quad= \frac{\lambda [n+1]_{p,q}x}{p[n]_{p,q} (p^{1-n}[n]_{p,q}-1 )} \sum_{k=0}^{n-1}b_{n,k}(x; p, q)-\frac{2\lambda [n+1]_{p,q}x}{p[n]_{p,q} (p^{2-2n}[n]_{p,q}^{2}-1 )}\sum_{k=0}^{n-1}b_{n,k}(x; p, q) \\ &\qquad{}-\frac{2q\lambda [n+1]_{p,q}x^{2}}{p^{n+1} (p^{2-2n}[n]_{p,q} ^{2}-1 )}\sum_{k=0}^{n-2}b_{n-1,k}(x; p, q) \\ &\quad= \frac{\lambda [n+1]_{p,q}x}{p[n]_{p,q} (p^{1-n}[n]_{p,q}-1 )} \bigl(1-x^{n} \bigr)-\frac{2\lambda [n+1]_{p,q}x}{p[n]_{p,q} (p ^{2-2n}[n]_{p,q}^{2}-1 )} \bigl(1-x^{n} \bigr) \\ &\qquad{}-\frac{2q\lambda [n+1]_{p,q}x^{2}}{p^{n+1} (p^{2-2n}[n]_{p,q} ^{2}-1 )} \bigl(1-x^{n-1} \bigr). \end{aligned}
(9)

Finally, using the following two identities:

\begin{aligned}{} [k]_{p,q}=\frac{[k+1]_{p,q}}{q}-\frac{p^{k}}{q},\qquad [k]_{p,q}^{2}=\frac{[k+1]_{p,q}[k]_{p,q}}{q}-\frac{p^{k}[k]_{p,q}}{q}, \end{aligned}

and some computations, we have

\begin{aligned} &\triangle _{1,3} \\ &\quad=-\lambda \sum_{k=1}^{n-1} \frac{[k]_{p,q}}{p^{k-n}[n]_{p,q}} \biggl(\frac{1}{p ^{1-n}[n]_{p,q}+1}-\frac{2qp^{-k}[k]_{p,q}}{p^{2-2n}[n]_{p,q}^{2}-1} \biggr)b _{n+1,k+1}(x; p, q) \\ &\quad=\frac{-\lambda }{q (p^{1-n}[n]_{p,q}+1 )}\sum_{k=1}^{n-1} \frac{[k+1]_{p,q}-p ^{k}}{p^{k-n}[n]_{p,q}}b_{n+1,k+1}(x; p, q) \\ &\qquad{}+\frac{2\lambda }{[n]_{p,q} (p^{2-2n}[n]_{p,q}^{2}-1 )} \sum_{k=1}^{n-1} \frac{[k+1]_{p,q}[k]_{p,q}-p^{k}[k]_{p,q}}{p^{2k-n}}b _{n+1,k+1}(x; p, q) \\ &\quad= \frac{-\lambda [n+1]_{p,q}x}{q[n]_{p,q} (p^{1-n}[n]_{p,q}+1 )} \sum_{k=1}^{n-1}b_{n,k}(x; p, q)+\frac{p^{n}\lambda }{q[n]_{p,q} (p ^{1-n}[n]_{p,q}+1 )}\sum_{k=1}^{n-1}b_{n+1,k+1}(x; p, q) \\ &\qquad{}+ \frac{2\lambda [n+1]_{p,q}x^{2}}{p^{n} (p^{2-2n}[n]_{p,q}^{2}-1 )} \sum_{k=0}^{n-2}b_{n-1,k}(x; p, q)+\frac{-2\lambda [n+1]_{p,q}x}{q[n]_{p,q} (p^{2-2n}[n]_{p,q}^{2}-1 )}\sum_{k=1}^{n-1}b_{n,k}(x; p, q) \\ &\qquad{}+ \frac{2p^{n}\lambda }{q[n]_{p,q} (p^{2-2n}[n]_{p,q}^{2}-1 )} \sum_{k=1}^{n-1}b_{n+1,k+1}(x; p, q) \\ &\quad= \frac{-\lambda [n+1]_{p,q}x}{q[n]_{p,q} (p^{1-n}[n]_{p,q}-1 )} \biggl(1-\frac{\prod_{s=0}^{n-1} (p^{s}-q^{s}x )}{p^{ \frac{n(n-1)}{2}}}-x^{n} \biggr)+\frac{2\lambda [n+1]_{p,q}x^{2}}{p ^{n} (p^{2-2n}[n]_{p,q}^{2}-1 )} \bigl(1-x^{n-1} \bigr) \\ &\qquad{}+\frac{p^{n}\lambda }{q[n]_{p,q} (p^{1-n}[n]_{p,q}-1 )} \biggl(1-\frac{\prod_{s=0}^{n} (p^{s}-q^{s}x )}{p^{ \frac{n(n+1)}{2}}}-\frac{[n+1]_{p,q}x\prod_{s=0}^{n-1} (p^{s}-q ^{s}x )}{p^{\frac{n(n+1)}{2}}}-x^{n+1} \biggr) \\ &\quad= \frac{-\lambda [n+1]_{p,q}x}{q[n]_{p,q} (p^{1-n}[n]_{p,q}-1 )} \bigl(1-x^{n} \bigr)+\frac{2\lambda [n+1]_{p,q}x^{2}}{p^{n} (p ^{2-2n}[n]_{p,q}^{2}-1 )} \bigl(1-x^{n-1} \bigr) \\ &\qquad{}+\frac{p^{n}\lambda }{q[n]_{p,q} (p^{1-n}[n]_{p,q}-1 )} \biggl(1-\frac{\prod_{s=0}^{n} (p^{s}-q^{s}x )}{p^{ \frac{n(n+1)}{2}}}-x^{n+1} \biggr). \end{aligned}
(10)

Combining (7)–(10), we can obtain (6), Lemma 2.2 is proved. □

### Lemma 2.4

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

\begin{aligned} &B_{n,p,q}^{\lambda } \bigl(t^{2}; x \bigr) \\ &\quad=x^{2}+\frac{p^{n-1}x(1-x)}{[n]_{p,q}}+\frac{\lambda (q^{2}-p ^{2} )[n+1]_{p,q}x^{2} (1-x^{n-1} )}{p^{2}q[n]_{p,q} (p^{1-n}[n]_{p,q}+1 )} \\ &\qquad{}-\frac{p^{2n}\lambda (1-x^{n+1} )}{q^{2}[n]_{p,q}^{2} (p^{1-n}[n]_{p,q}-1 )}+\frac{2\lambda (p^{2}q-p^{3}-pq ^{2}-q^{3} )[n+1]_{p,q}x^{2} (1-x^{n-1} )}{p^{3}q[n]_{p,q} (p^{2-2n}[n]_{p,q}^{2}-1 )} \\ &\qquad{}+\frac{p^{n-2} (p^{2}+q^{2} )\lambda [n+1]_{p,q}x (1-x ^{n} )}{q^{2}[n]_{p,q}^{2} (p^{1-n}[n]_{p,q}+1 )}+\frac{p \lambda \prod_{s=0}^{n} (p^{s}-q^{s}x )}{p^{ \frac{(n-1)(n-2)}{2}}q^{2}[n]_{p,q}^{2} (p^{1-n}[n]_{p,q}-1 )} \\ &\qquad{}+\frac{2q (p^{2}-q^{2} )\lambda [n+1]_{p,q}[n-1]_{p,q}x ^{3} (1-x^{n-2} )}{p^{n+2}[n]_{p,q} (p^{2-2n}[n]_{p,q} ^{2}-1 )} \\ &\qquad{}+\frac{2p^{n-1}(2q-p)\lambda [n+1]_{p,q}x}{q^{2}[n]_{p,q}^{2} (p ^{2-2n}[n]_{p,q}^{2}-1 )} \biggl(\frac{\prod_{s=0}^{n-1} (p ^{s}-q^{s}x )}{p^{\frac{n(n-1)}{2}}}- \bigl(1-x^{n} \bigr) \biggr). \end{aligned}
(11)

### Proof

From (4), we have

\begin{aligned} &B_{n,p,q}^{\lambda } \bigl(t^{2}; x \bigr) \\ &\quad=\sum_{k=0}^{n}b_{n,k}^{\lambda }(x; p, q)\frac{[k]_{p,q}^{2}}{p ^{2k-2n}[n]_{p,q}^{2}} \\ &\quad=\sum_{k=1}^{n-1}\frac{[k]_{p,q}^{2}}{p^{2k-2n}[n]_{p,q}^{2}} \biggl[b _{n,k}(x; p, q)+\lambda \biggl(\frac{p^{1-n}[n]_{p,q}-2p^{1-k}[k]_{p,q}+1}{p ^{2-2n}[n]_{p,q}^{2}-1}b_{n+1,k}(x; p, q) \\ &\qquad{}-\frac{p^{1-n}[n]_{p,q}-2qp^{-k}[k]_{p,q}-1}{p^{2-2n}[n]_{p,q} ^{2}-1}b_{n+1,k+1}(x; p, q) \biggr) \biggr]+b_{n,n}(x; p, q) \\ &\qquad{}-\frac{\lambda }{p^{1-n}[n]_{p,q}+1}b_{n+1,n}(x; p, q) \\ &\quad=\sum_{k=0}^{n}\frac{[k]_{p,q}^{2}}{p^{2k-2n}[n]_{p,q}^{2}}b_{n,k}(x; p, q) \\ &\qquad{}+\lambda \sum_{k=0}^{n} \frac{[k]_{p,q}^{2}}{p^{2k-2n}[n]_{p,q}^{2}}\frac{p ^{1-n}[n]_{p,q}-2p^{1-k}[k]_{p,q}+1}{p^{2-2n}[n]_{p,q}^{2}-1}b_{n+1,k}(x; p, q) \\ &\qquad{}-\lambda \sum_{k=1}^{n-1} \frac{[k]_{p,q}^{2}}{p^{2k-2n}[n]_{p,q} ^{2}}\frac{p^{1-n}[n]_{p,q}-2qp^{-k}[k]_{p,q}-1}{p^{2-2n}[n]_{p,q} ^{2}-1}b_{n+1,k+1}(x; p, q) \\ &\quad=\triangle _{2,1}+\triangle _{2,2}+\triangle _{2,3}, \end{aligned}
(12)

where

\begin{aligned} &\triangle _{2,1} =\sum_{k=0}^{n} \frac{[k]_{p,q}^{2}}{p^{2k-2n}[n]_{p,q} ^{2}}b_{n,k}(x; p, q); \\ &\triangle _{2,2} =\lambda \sum_{k=0}^{n} \frac{[k]_{p,q}^{2}}{p^{2k-2n}[n]_{p,q} ^{2}}\frac{p^{1-n}[n]_{p,q}-2p^{1-k}[k]_{p,q}+1}{p^{2-2n}[n]_{p,q} ^{2}-1}b_{n+1,k}(x; p, q); \\ &\triangle _{2,3} =-\lambda \sum_{k=1}^{n-1} \frac{[k]_{p,q}^{2}}{p^{2k-2n}[n]_{p,q} ^{2}}\frac{p^{1-n}[n]_{p,q}-2qp^{-k}[k]_{p,q}-1}{p^{2-2n}[n]_{p,q} ^{2}-1}b_{n+1,k+1}(x; p, q). \end{aligned}

By Lemma 2.1, we have

\begin{aligned} \triangle _{2,1}=B_{n,p,q} \bigl(t^{2}; x \bigr)=x^{2}+ \frac{p^{n-1}x(1-x)}{[n]_{p,q}}. \end{aligned}
(13)

Indeed, by using

\begin{aligned} &[k]_{p,q}^{2}=q[k]_{p,q}[k-1]_{p,q}+p^{k-1}[k]_{p,q}; \\ &[k]_{p,q}^{3}=q^{3}[k]_{p,q}[k-1]_{p,q}[k-2]_{p,q}+qp^{k-2}(q+2p)[k]_{p,q}[k-1]_{p,q}+p ^{2k-2}[k]_{p,q}, \end{aligned}

$$\sum_{k=0}^{n}b_{n,k}(x; p, q)=1$$, and some computations, we obtain

\begin{aligned} &\triangle _{2,2} \\ &\quad=\lambda \sum_{k=0}^{n} \frac{[k]_{p,q}^{2}}{p^{2k-2n}[n]_{p,q}^{2}} \biggl(\frac{1}{p^{1-n}[n]_{p,q}-1}-\frac{2p^{1-k}[k]_{p,q}}{p^{2-2n}[n]_{p,q} ^{2}-1} \biggr)b_{n+1,k}(x; p, q) \\ &\quad=\frac{\lambda }{p^{1-n}[n]_{p,q}-1}\sum_{k=0}^{n} \frac{q[k]_{p,q}[k-1]_{p,q}+p ^{k-1}[k]_{p,q}}{p^{2k-2n}[n]_{p,q}^{2}}b_{n+1,k}(x; p, q) \\ &\qquad{}+\frac{-2p\lambda }{[n]_{p,q}^{2}}\sum_{k=0}^{n} \frac{q^{3}[k]_{p,q}[k-1]_{p,q}[k-2]_{p,q}+qp ^{k-2}(q+2p)[k]_{p,q}[k-1]_{p,q}+p^{2k-2}[k]_{p,q}}{ (p^{2-2n}[n]_{p,q} ^{2}-1 )p^{3k-2n}} \\ &\quad= \frac{q\lambda [n+1]_{p,q}x^{2}}{p^{2}[n]_{p,q} (p^{1-n}[n]_{p,q}-1 )} \sum_{k=0}^{n-2}b_{n-1,k}(x; p, q)+\frac{p^{n-2}\lambda [n+1]_{p,q}x}{[n]_{p,q} ^{2} (p^{1-n}[n]_{p,q}-1 )}\sum_{k=0}^{n-1}b_{n,k}(x; p, q) \\ &\qquad{}-\frac{2q^{3}\lambda [n+1]_{p,q}[n-1]_{p,q}x^{3}}{p^{n+2}[n]_{p,q} (p^{2-2n}[n]_{p,q}^{2}-1 )}\sum_{k=0}^{n-3}b_{n-2,k}(x; p, q) \\ &\qquad{}-\frac{2q(q+2p)\lambda [n+1]_{p,q}x^{2}}{p^{3}[n]_{p,q} (p^{2-2n}[n]_{p,q} ^{2}-1 )}\sum_{k=0}^{n-2}b_{n-1,k}(x; p, q) \\ &\qquad{}-\frac{2p^{n-2} \lambda [n+1]_{p,q}x}{[n]_{p,q}^{2} (p^{2-2n}[n]_{p,q}^{2}-1 )} \sum_{k=0}^{n-1}b_{n,k}(x; p, q) \\ &\quad=\frac{q\lambda [n+1]_{p,q}x^{2} (1-x^{n-1} )}{p^{2}[n]_{p,q} (p^{1-n}[n]_{p,q}+1 )}-\frac{2_{p,q}q\lambda [n+1]_{p,q}x ^{2} (1-x^{n-1} )}{p^{3}[n]_{p,q} (p^{2-2n}[n]_{p,q} ^{2}-1 )} \\ &\qquad{}+\frac{p^{n-2}\lambda [n+1]_{p,q}x (1-x^{n} )}{[n]_{p,q} ^{2} (p^{1-n}[n]_{p,q}+1 )}-\frac{2q^{3}\lambda [n+1]_{p,q}[n-1]_{p,q}x ^{3} (1-x^{n-2} )}{p^{n+2}[n]_{p,q} (p^{2-2n}[n]_{p,q} ^{2}-1 )}. \end{aligned}
(14)

Finally, using the following identities:

\begin{aligned} &[k]_{p,q}^{2}=\frac{1}{q}[k+1]_{p,q}[k]_{p,q}- \frac{p^{k}}{q^{2}}[k+1]_{p,q}+\frac{p ^{2k}}{q^{2}}; \\ &[k]_{p,q}^{3}=[k+1]_{p,q}[k]_{p,q}[k-1]_{p,q}+ \frac{p^{k-1}}{q} \biggl(1-\frac{p}{q} \biggr)[k+1]_{p,q}[k]_{p,q} \\ &\phantom{[k]_{p,q}^{3}=}{} -\frac{p^{2k-1}}{q^{2}} \biggl(2-\frac{p}{q} \biggr)[k+1]_{p,q}- \frac{p ^{3k}}{q^{3}}, \end{aligned}

$$\sum_{k=0}^{n}b_{n,k}(x; p, q)=1$$, and some computations, we have

\begin{aligned} \triangle _{2,3} ={}&{-}\lambda \sum_{k=1}^{n-1} \frac{[k]_{p,q}^{2}}{p^{2k-2n}[n]_{p,q} ^{2}} \biggl(\frac{1}{p^{1-n}[n]_{p,q}+1}-\frac{2qp^{-k}[k]_{p,q}}{p ^{2-2n}[n]_{p,q}^{2}-1} \biggr)b_{n+1,k+1}(x; p, q) \\ ={}&{- }\frac{\lambda [n+1]_{p,q}x^{2}}{q[n]_{p,q} (p^{1-n}[n]_{p,q}+1 )} \sum_{k=0}^{n-2}b_{n-1,k}(x; p, q) \\ &{}+\frac{p^{n}\lambda [n+1]_{p,q}x}{q ^{2}[n]_{p,q}^{2} (p^{1-n}[n]_{p,q}+1 )}\sum_{k=1}^{n-1}b _{n,k}(x; p, q) \\ &{}- \frac{p^{2n}\lambda }{q^{2}[n]_{p,q}^{2} (p^{1-n}[n]_{p,q}+1 )} \sum_{k=1}^{n-1}b_{n+1,k+1}(x; p, q) \\ &{}+\frac{2\lambda q[n+1]_{p,q}[n-1]_{p,q}x ^{3}}{p^{n}[n]_{p,q} (p^{2-2n}[n]_{p,q}^{2}-1 )}\sum_{k=0} ^{n-3}b_{n-2,k}(x; p, q) \\ &{}+\frac{2(q-p)\lambda [n+1]_{p,q}x^{2}}{pq[n]_{p,q} (p^{2-2n}[n]_{p,q} ^{2}-1 )}\sum_{k=0}^{n-2}b_{n-1,k}(x; p, q) \\ &{}-\frac{2p^{n-1}(2q-p) \lambda [n+1]_{p,q}x}{q^{2}[n]_{p,q}^{2} (p^{2-2n}[n]_{p,q}^{2}-1 )} \sum_{k=1}^{n-1}b_{n,k}(x; p, q) \\ &{}-\frac{2\lambda p^{2n}}{q^{2}[n]_{p,q}^{2} (p^{2-2n}[n]_{p,q} ^{2}-1 )}\sum_{k=1}^{n-1}b_{n+1,k+1}(x; p, q) \\ ={}&{-}\frac{\lambda [n+1]_{p,q}x^{2} (1-x^{n-1} )}{q[n]_{p,q} (p^{1-n}[n]_{p,q}+1 )}+\frac{p^{n}\lambda [n+1]_{p,q}x (1-x^{n} )}{q^{2}[n]_{p,q}^{2} (p^{1-n}[n]_{p,q}+1 )} \\ &{}- \frac{p^{2n}\lambda }{q^{2}[n]_{p,q}^{2} (p^{1-n}[n]_{p,q}-1 )} \biggl(1-\frac{\prod_{s=0}^{n} (p^{s}-q^{s}x )}{p^{ \frac{n(n+1)}{2}}}-x^{n+1} \biggr) \\ &{}+\frac{2\lambda q[n+1]_{p,q}[n-1]_{p,q}x^{3} (1-x^{n-2} )}{p ^{n}[n]_{p,q} (p^{2-2n}[n]_{p,q}^{2}-1 )}+\frac{2(q-p) \lambda [n+1]_{p,q}x^{2} (1-x^{n-1} )}{pq[n]_{p,q} (p ^{2-2n}[n]_{p,q}^{2}-1 )} \\ &{}-\frac{2p^{n-1}(2q-p)\lambda [n+1]_{p,q}x}{q^{2}[n]_{p,q}^{2} (p ^{2-2n}[n]_{p,q}^{2}-1 )} \biggl(1-\frac{\prod_{s=0}^{n-1} (p ^{s}-q^{s}x )}{p^{\frac{n(n-1)}{2}}}-x^{n} \biggr). \end{aligned}
(15)

Combining (12)–(15) and some computations, we can get (11), Lemma 2.4 is proved. □

### Corollary 2.5

Let$$\lambda \in [-1,1]$$, $$x\in [0,1]$$, $$n>1$$, and$$0< q< p\leq 1$$, we have

\begin{aligned} &B_{n,p,q}^{\lambda }(t-x; x) \\ &\quad=\frac{2\lambda [n+1]_{p,q}x^{2} (1-x^{n-1} )}{p^{n} (p^{2-2n}[n]_{q}^{2}-1 )} \biggl(1-\frac{q}{p} \biggr)+ \frac{ \lambda p^{n} (1-x^{n+1} )}{q[n]_{p,q} (p^{1-n}[n]_{p,q}-1 )} \\ &\qquad{}+\frac{\lambda [n+1]_{p,q}x (1-x^{n} )}{[n]_{p,q} (p ^{1-n}[n]_{p,q}-1 )} \biggl(\frac{1}{p}-\frac{1}{q}- \frac{2}{p (p^{1-n}[n]_{p,q}+1 )} \biggr) \\ &\qquad{}-\frac{\lambda \prod_{s=0}^{n} (p^{s}-q^{s}x )}{p^{ \frac{n(n-1)}{2}}q[n]_{p,q} (p^{1-n}[n]_{p,q}-1 )}:= \gamma _{n,p,q}^{\lambda }(x) \end{aligned}
(16)
\begin{aligned} &\quad\leq \frac{2}{p^{1-n}[n]_{p,q}-1}+\frac{4}{p^{2-2n}[n]_{p,q}^{2}-1}+\frac{2}{q[n]_{p,q} (p^{1-n}[n]_{p,q}-1 )} \\ &\qquad{}+\frac{2}{p[n]_{p,q} (p^{2-2n}[n]_{p,q}^{2}-1 )} \\ &\quad:=\alpha (n; p, q); \end{aligned}
(17)
\begin{aligned} &B_{n,p,q}^{\lambda } \bigl((t-x)^{2}; x \bigr) \\ &\quad\leq \frac{1}{4[n]_{p,q}}+ \frac{5}{p^{2} (p^{1-n}[n]_{p,q}-1 )}+\frac{6}{q[n]_{p,q} (p^{1-n}[n]_{p,q}-1 )}+ \frac{6}{p^{2-2n}[n]_{p,q}^{2}-1} \\ &\qquad{}+\frac{3}{q^{2}[n]_{p,q}^{2} (p^{1-n}[n]_{p,q}-1 )}+\frac{8}{q[n]_{p,q} (p^{2-2n}[n]_{p,q}^{2}-1 )}+\frac{2}{q^{2}[n]_{p,q}^{2} (p^{2-2n}[n]_{p,q}^{2}-1 )} \\ &\quad :=\beta (n; p, q). \end{aligned}
(18)

### Proof

By Lemmas 2.22.3, we have

\begin{aligned} &B_{n,p,q}^{\lambda }(t-x; x) \\ &\quad=\frac{2\lambda [n+1]_{p,q}x^{2} (1-x^{n-1} )}{p^{n} (p^{2-2n}[n]_{p,q}^{2}-1 )} \biggl(1-\frac{q}{p} \biggr)+ \frac{ \lambda p^{n} (1-x^{n+1} )}{q[n]_{p,q} (p^{1-n}[n]_{p,q}-1 )} \\ &\qquad{}+\frac{\lambda [n+1]_{p,q}x (1-x^{n} )}{[n]_{p,q} (p ^{1-n}[n]_{p,q}-1 )} \biggl(\frac{1}{p}-\frac{1}{q}- \frac{2}{p (p^{1-n}[n]_{p,q}+1 )} \biggr)-\frac{\lambda \prod_{s=0} ^{n} (p^{s}-q^{s}x )}{p^{\frac{n(n-1)}{2}}q[n]_{p,q} (p ^{1-n}-1 )} \\ &\quad\leq \textstyle\begin{cases} \frac{2[n+1]_{p,q}x^{2} (1-x^{n-1} )}{p^{n} (p^{2-2n}[n]_{p,q} ^{2}-1 )} (1-\frac{q}{p} )+\frac{p^{n} (1-x^{n+1} )}{q[n]_{p,q} (p^{1-n}[n]_{p,q}-1 )}, & \lambda \in [0,1], \\ \frac{[n+1]_{p,q}x (1-x^{n} )}{[n]_{p,q} (p^{1-n}[n]_{p,q}-1 )} (\frac{1}{q}+\frac{2}{p (p^{1-n}[n]_{p,q}+1 )} )+\frac{ \prod_{s=0}^{n} (p^{s}-q^{s}x )}{p^{\frac{n(n-1)}{2}}q[n]_{p,q} (p^{1-n}[n]_{p,q}-1 )}, & \lambda \in [-1,0]. \end{cases}\displaystyle \end{aligned}

Since

\begin{aligned} \prod_{s=0}^{n} \bigl(p^{s}-q^{s}x \bigr) =p^{\frac{n(n+1)}{2}}(1-x) \biggl(1-\frac{q}{p}x \biggr)\cdots \biggl(1-\frac{q^{n}}{p^{n}}x \biggr) \leq p^{\frac{n(n+1)}{2}}, \end{aligned}

we have

\begin{aligned} \frac{\prod_{s=0}^{n} (p^{s}-q^{s}x )}{p^{\frac{n(n-1)}{2}}q[n]_{p,q} (p^{1-n}[n]_{p,q}-1 )}\leq \frac{p^{\frac{n(n+1)}{2}}}{p ^{\frac{n(n-1)}{2}}q[n]_{p,q} (p^{1-n}[n]_{p,q}-1 )}=\frac{p ^{n}}{q[n]_{p,q} (p^{1-n}[n]_{p,q}-1 )}. \end{aligned}

We also have

\begin{aligned} \frac{2[n+1]_{p,q}}{p^{n} (p^{2-2n}[n]_{p,q}^{2}-1 )} &=\frac{2q[n]_{p,q}+2p ^{n}}{p^{n} (p^{2-2n}[n]_{p,q}^{2}-1 )} \\ &\leq \frac{2 (p[n]_{p,q}-p^{n} )+2p^{n}}{ (p^{1-n}[n]_{p,q}+1 ) (p[n]_{p,q}-p^{n} )}+\frac{2}{p^{2-2n}[n]_{p,q}^{2}-1} \\ &=\frac{2}{p^{1-n}[n]_{p,q}+1}+\frac{4}{p^{2-2n}[n]_{p,q}^{2}-1} \end{aligned}

and

\begin{aligned} \frac{[n+1]_{p,q}}{q[n]_{p,q} (p^{1-n}[n]_{p,q}-1 )}=\frac{1}{p ^{1-n}[n]_{p,q}-1}+ \frac{p^{n}}{q[n]_{p,q} (p^{1-n}[n]_{p,q}-1 )}. \end{aligned}

Then we obtain

\begin{aligned} &B_{n,p,q}^{\lambda }(t-x; x) \\ &\quad\leq \textstyle\begin{cases} \frac{2}{p^{1-n}[n]_{p,q}-1}+\frac{4}{p^{2-2n}[n]_{p,q}^{2}-1}+\frac{1}{q[n]_{p,q} (p^{1-n}-1 )}, & \lambda \in [0,1], \\ \frac{1}{p^{1-n}-1}+\frac{2}{p^{2-2n}[n]_{p,q}^{2}-1}+\frac{2}{q[n]_{p,q} (p^{1-n}[n]_{p,q}-1 )}+\frac{2}{p[n]_{p,q} (p^{2-2n}[n]_{p,q} ^{2}-1 )}, & \lambda \in [-1,0] \end{cases}\displaystyle \\ &\quad\leq \frac{2}{p^{1-n}[n]_{p,q}-1}+\frac{4}{p^{2-2n}[n]_{p,q}^{2}-1}+\frac{2}{q[n]_{p,q} (p^{1-n}[n]_{p,q}-1 )} \\ &\qquad{}+\frac{2}{p[n]_{p,q} (p^{2-2n}[n]_{p,q}^{2}-1 )}. \end{aligned}

Therefore, (16) and (17) are obtained. Similarly, by using Lemmas 2.22.4 and some computations, we can get (18). □

### Lemma 2.6

Let$$\lambda \in [-1, 1]$$, $$x\in [0, 1]$$, $$n>1$$, and$$0< q< p\leq 1$$, then$$B_{n,p,q}^{\lambda }(f; x)$$are positive linear operators.

### Proof

Apparently, $$B_{n,p,q}^{\lambda }(f; x)$$ are linear operators, we only need to prove the positive property. Actually, we have

\begin{aligned} b_{n,0}^{\lambda }(x; p, q) &=b_{n,0}(x; p, q)- \frac{\lambda }{p^{1-n}+1}b_{n+1,1}(x; p, q) \\ &=b_{n,0}(x; p, q) \biggl(1-\frac{\lambda [n+1]_{p,q}x}{ (p^{1-n}[n]_{p,q}+1 )p ^{n}} \biggr), \end{aligned}

since $$b_{n,0}(x; p, q)\geq 0$$ and $$1-\frac{\lambda [n+1]_{p,q}x}{p[n]_{p,q}+p ^{n}}\geq 0$$ with the fact that $$[n+1]_{p,q}=p[n]_{p,q}+q^{n}< p[n]_{p,q}+p ^{n}$$, then we obtain $$b_{n,0}^{\lambda }(x; p, q)\geq 0$$. Similarly, we can prove $$b_{n,n}^{\lambda }(x; p, q)\geq 0$$. Finally, we will prove $$b_{n,k}^{\lambda }(x; p, q)\geq 0$$$$(1\leq k\leq n-1)$$. Indeed,

\begin{aligned} &b_{n,k}^{\lambda }(x; p, q) \\ &\quad=b_{n,k}(x; p, q) \biggl[1+\lambda \biggl(\frac{p^{1-n}[n]_{p,q}-2p ^{1-k}[k]_{p,q}+1}{p^{2-2n}[n]_{p,q}^{2}-1} \frac{[n+1]_{p,q} (p ^{n-k}-q^{n-k}x )}{p^{n}[n+1-k]_{p,q}} \\ &\qquad{}-\frac{p^{1-n}[n]_{p,q}-2qp^{-k}[k]_{p,q}-1}{p^{2-2n}[n]_{p,q} ^{2}-1}\frac{p^{k}[n+1]_{p,q}x}{p^{n}[k+1]_{p,q}} \biggr) \biggr], \end{aligned}

we need to prove

\begin{aligned} &\biggl\vert \frac{p^{1-n}[n]_{p,q}-2p^{1-k}[k]_{p,q}+1}{p^{2-2n}[n]_{p,q} ^{2}-1} \frac{[n+1]_{p,q} (p^{n-k}-q^{n-k}x )}{p^{n}[n+1-k]_{p,q}} \\ &\quad{}-\frac{p^{1-n}[n]_{p,q}-2qp^{-k}[k]_{p,q}-1}{p^{2-2n}[n]_{p,q} ^{2}-1}\frac{p^{k}[n+1]_{p,q}x}{p^{n}[k+1]_{p,q}}\biggr\vert \leq 1. \end{aligned}
(19)

We can prove the above inequality in three cases:

\begin{aligned} &\text{Case 1}:\quad \text{for }p^{k-1}\leq [k]_{p,q}\leq \min \biggl\{ \frac{p^{k}}{2q} \bigl(p^{1-n}[n]_{p,q}-1 \bigr), \frac{1}{2} \bigl(p ^{k-n}[n]_{p,q}+p^{k-1} \bigr) \biggr\} ; \\ &\text{Case 2}:\quad \text{for } \max \biggl\{ \frac{p^{k}}{2q} \bigl(p^{1-n}[n]_{p,q}-1 \bigr), \frac{1}{2} \bigl(p^{k-n}[n]_{p,q}+p^{k-1} \bigr) \biggr\} \\ &\phantom{\text{Case 2}:\quad \text{for }}\quad \leq [k]_{p,q}\leq p^{1-k-n}[n-1]_{p,q}; \\ &\text{Case 3}: \quad\text{for }\min \biggl\{ \frac{p^{k}}{2q} \bigl(p^{1-n}[n]_{p,q}-1 \bigr), \frac{1}{2} \bigl(p^{k-n}[n]_{p,q}+p^{k-1} \bigr) \biggr\} \\ &\phantom{\text{Case 3}: \quad\text{for }}\quad \leq [k]_{p,q}\leq \max \biggl\{ \frac{p^{k}}{2q} \bigl(p^{1-n}[n]_{p,q}-1 \bigr), \frac{1}{2} \bigl(p^{k-n}[n]_{p,q}+p^{k-1} \bigr) \biggr\} . \end{aligned}

We mainly prove case 1. In fact, under the condition of case 1, we have

\begin{aligned} 0\leq p^{1-n}[n]_{p,q}-2p^{1-k}[k]_{p,q}+1 \leq p^{1-n}[n]_{p,q}-1, \end{aligned}

then we get

\begin{aligned} 0 &\leq \frac{p^{1-n}[n]_{p,q}-2p^{1-k}[k]_{p,q}+1}{p^{2-2n}[n]_{p,q} ^{2}-1}\frac{[n+1]_{p,q}}{p^{n}[n+1-k]_{p,q}} \\ &\leq \frac{[n+1]_{p,q}}{ (p^{1-n}[n]_{p,q}+1 )p^{n}} \frac{1}{[n+1-k]_{p,q}}\leq \frac{1}{p^{n-k}} \end{aligned}

with the help of $$[n+1-k]_{p,q}\geq p^{n-k}$$ and $$[n+1]_{p,q}\leq p[n]_{p,q}+p ^{n}$$. That is to say,

\begin{aligned} 0\leq \frac{p^{1-n}[n]_{p,q}-2p^{1-k}[k]_{p,q}+1}{p^{2-2n}[n]_{p,q} ^{2}-1}\frac{[n+1]_{p,q}p^{n-k}}{p^{n}[n+1-k]_{p,q}}\leq 1. \end{aligned}

Also we have

\begin{aligned} 0\leq \frac{p^{1-n}[n]_{p,q}-2p^{1-k}[k]_{p,q}+1}{p^{2-2n}[n]_{p,q} ^{2}-1}\frac{[n+1]_{p,q}q^{n-k}x}{p^{n}[n+1-k]_{p,q}}\leq 1. \end{aligned}

On the other hand, since

\begin{aligned} 0\leq p^{1-n}[n]_{p,q}-2qp^{-k}[k]_{p,q}-1 \leq p^{1-n}[n]_{p,q}-1, \end{aligned}

we obtain

\begin{aligned} 0\leq \frac{p^{1-n}[n]_{p,q}-2qp^{-k}-1}{p^{2-2n}[n]_{p,q}^{2}-1}\frac{p ^{k}[n+1]_{p,q}x}{p^{n}[k+1]_{p,q}}\leq \frac{[n+1]_{p,q}}{ (p ^{1-n}[n]_{p,q}+1 )p^{n}} \frac{p^{k}}{[k+1]_{p,q}}\leq 1 \end{aligned}

by the fact that $$x\in [0, 1]$$, $$[n+1]_{p,q}\leq p[n]_{p,q}+p^{n}$$, and $$[k+1]_{p,q}\geq p^{k}$$. Hence, (19) is proved in case 1. Case 2 and case 3 are similar. Lemma 2.6 is proved. □

### Lemma 2.7

(see )

Let the sequences$$q:=\{q_{n}\}=\{1-\alpha _{n} \}$$, $$p:=\{p_{n}\}=\{1-\beta _{n}\}$$such that$$0\leq \beta _{n}<\alpha _{n}<1$$, $$\alpha _{n}\rightarrow 0$$, $$\beta _{n}\rightarrow 0$$as$$n\rightarrow \infty$$. The following statements are true:

1. (A)

If$$\lim_{n\rightarrow \infty }e^{n(\beta _{n}-\alpha _{n})}=1$$and$$e^{n\beta _{n}}/n\rightarrow 0$$, then$$[n]_{p_{n},q_{n}}\rightarrow \infty$$.

2. (B)

If$$\overline{\lim }_{n\rightarrow \infty }e^{n(\beta _{n}- \alpha _{n})}<1$$and$$e^{n\beta _{n}}(\alpha _{n}-\beta _{n})\rightarrow 0$$, then$$[n]_{p_{n}, q_{n}}\rightarrow \infty$$.

3. (C)

If$$\underline{\lim }_{n\rightarrow \infty }e^{n(\beta _{n}- \alpha _{n})}<1, \overline{\lim }_{n\rightarrow \infty }e^{n(\beta _{n}-\alpha _{n})}=1$$, and$$\max \bigl\{e^{n\beta _{n}}/n, e^{n\beta _{n}}(\alpha _{n}- \beta _{n})\bigr\}\rightarrow 0$$, then$$[n]_{p_{n}, q_{n}}\rightarrow \infty$$.

## Main results

In the sequel, let the sequences $$q:=\{q_{n}\}$$ and $$p:=\{p_{n}\}$$ satisfy the conditions of Lemma 2.7 and $$p^{n}\rightarrow a \in (0, b]$$, b is a finite positive real number.

Now we give a Korovkin type approximation theorem for $$B_{n,p,q}^{ \lambda }(f; x)$$.

### Theorem 3.1

For$$f\in C_{[0, 1]}$$, $$\lambda \in [-1,1]$$, $$x\in [0,1]$$, and$$n>1$$, $$B_{n,p,q}^{\lambda }(f; x)$$converge uniformly tofon$$[0, 1]$$.

### Proof

The well-known Korovkin theorem (see , pp. 8–9) implies that positive linear operators $$B_{n,p,q}^{\lambda }(f; x)$$ converge to $$f(x)$$ uniformly on $$[0, 1]$$ as $$n\rightarrow \infty$$ for any $$f\in C_{[0,1]}$$ if and only if

\begin{aligned} B_{n,p,q}^{\lambda } \bigl(t^{i}; x \bigr) \rightarrow x^{i}, \quad i=0, 1, 2. \end{aligned}
(20)

Obviously, (20) can be easily obtained by Lemmas 2.22.4 as $$n\rightarrow \infty$$. Therefore, Theorem 3.1 is proved. □

Next, in order to get the rate of convergence of $$B_{n,p,q}^{\lambda }(f; x)$$, we give the following definitions about Peetre’s K-functional and modulus of smoothness. Let $$f\in C_{[0, 1]}$$, endowed with the norm $$\Vert f \Vert =\sup_{x\in [0,1]}|f(x)|$$.

Peetre’s K-functional is defined by

\begin{aligned} K_{2}(f; \delta )=\inf_{g\in C^{2}} \bigl\{ \Vert f-g \Vert +\delta \bigl\Vert g'' \bigr\Vert \bigr\} , \end{aligned}

where $$\delta >0$$ and $$C^{2}= \{ g\in C_{[0,1]}: g',g''\in C _{[0,1]} \}$$. It is known that

\begin{aligned} K_{2}(f; \delta )\leq C\omega _{2} (f; \sqrt{\delta } ), \end{aligned}

where C is a positive constant, and

\begin{aligned} \omega _{2}(f; \delta )=\sup_{0< h\leq \delta }\sup _{x,x+h,x+2h\in [0,1]} \bigl\vert f(x+2h)-2f(x+h)+f(x) \bigr\vert \end{aligned}

is the second order modulus of smoothness. We also denote the usual of modulus of continuity by

\begin{aligned} \omega (f; \delta )=\sup_{0< h\leq \delta }\sup_{x,x+h\in [0,1]} \bigl\vert f(x+h)-f(x) \bigr\vert . \end{aligned}

### Theorem 3.2

For$$f\in C_{[0, 1]}$$, $$\lambda \in [-1,1]$$, $$x\in [0,1]$$, and$$n>1$$, we have

\begin{aligned} \bigl\vert B_{n,p,q}^{\lambda }(f; x)-f(x) \bigr\vert \leq 2 \omega \bigl(f; \sqrt{ \beta (n; p, q)} \bigr), \end{aligned}

where$$\beta (n; p, q)$$is defined in (18).

### Proof

Since $$|f(t)-f(x)|\leq \omega (f; |t-x|)\leq (1+\frac{|t-x|}{ \delta } )\omega (f; \delta )$$, applying $$B_{n,p,q}^{\lambda }(f; x)$$ to both ends and using Lemma 2.2, we have

\begin{aligned} \bigl\vert B_{n,p,q}^{\lambda }(f; x)-f(x) \bigr\vert \leq B_{n,p,q}\bigl( \bigl\vert f(t)-f(x) \bigr\vert ; x\bigr)\leq \biggl(1+\frac{1}{ \delta }B_{n,p,q}^{\lambda }\bigl( \vert t-x \vert ; x\bigr) \biggr)\omega (f; \delta ). \end{aligned}

By the Cauchy–Schwarz inequality, we obtain

\begin{aligned} \bigl\vert B_{n,p,q}^{\lambda }(f; x)-f(x) \bigr\vert \leq \biggl(1+\frac{1}{ \delta }\sqrt{B_{n,p,q}^{\lambda } \bigl((t-x)^{2}; x \bigr)} \biggr) \omega (f; \delta ). \end{aligned}

Then we have

\begin{aligned} \bigl\vert B_{n,p,q}^{\lambda }(f; x)-f(x) \bigr\vert \leq 2 \omega \bigl(f; \sqrt{B _{n,p,q}^{\lambda } \bigl((t-x)^{2}; x \bigr)} \bigr)\leq 2\omega \bigl(f; \sqrt{\beta (n; p, q)} \bigr) \end{aligned}

by taking $$\delta = \sqrt{B_{n,p,q}^{\lambda } ((t-x)^{2}; x )}$$, $$\beta (n; p, q)$$ is defined in (18). Theorem 3.2 is proved. □

### Theorem 3.3

For$$f\in C_{[0, 1]}$$, $$\lambda \in [-1,1]$$, $$x\in [0,1]$$, and$$n>1$$, we have

\begin{aligned} \bigl\vert B_{n,p,q}^{\lambda }(f; x)-f(x) \bigr\vert \leq C \omega _{2} \bigl(f; \sqrt{\alpha (n; p, q)^{2}+\beta (n; p, q)}/2 \bigr)+\omega \bigl(f; \alpha (n; p, q) \bigr), \end{aligned}

whereCis a positive constant, $$\alpha (n; p, q)$$and$$\beta (n; p, q)$$are defined in (17) and (18).

### Proof

Let us define the auxiliary operators $$\overline{B}_{n,p,q}^{\lambda }(f; x)$$,

\begin{aligned} \overline{B}_{n,p,q}^{\lambda }(f; x)=B_{n,p,q}^{\lambda }(f; x)-f \bigl(x+\gamma _{n,p,q}^{\lambda }(x) \bigr)+f(x), \end{aligned}
(21)

where $$\gamma _{n,p,q}^{\lambda }(x)$$ is defined in (16). Obviously, operators $$\overline{B}_{n,p,q}^{\lambda }(f; x)$$ preserve not only constant functions but also linear functions, say,

\begin{aligned} \overline{B}_{n,p,q}^{\lambda }(t-x; x)=0. \end{aligned}
(22)

Letting $$g\in C^{2}$$, $$x, t\in [0, 1]$$, then by Taylor’s expansion

\begin{aligned} g(t)=g(x)+g^{\prime }(t-x)+ \int _{x}^{t}(t-u)g^{\prime \prime }(u)\,du \end{aligned}
(23)

and (22), applying $$\overline{B}_{n,p,q}^{\lambda }(g; x)$$ to (23), we have

\begin{aligned} \overline{B}_{n,p,q}^{\lambda }(g; x)-g(x)=\overline{B}_{n,p,q}^{ \lambda } \biggl( \int _{x}^{t}(t-u)g^{\prime \prime }(u)\,du; x \biggr). \end{aligned}

Therefore, using (21), we get

\begin{aligned} & \bigl\vert \overline{B}_{n,p,q}^{\lambda }(g; x) \bigr\vert \\ &\quad\leq \biggl\vert B_{n,p,q}^{\lambda } \biggl( \int _{x}^{t}(t-u)g^{\prime \prime }(u)\,du; x \biggr) \biggr\vert + \biggl\vert \int _{x}^{x+\gamma _{n,p,q}^{\lambda }(x)} \bigl(x+\gamma _{n,p,q}^{\lambda }(x)-u \bigr)g^{\prime \prime }(u)\,du \biggr\vert \\ &\quad\leq \biggl\vert B_{n,p,q}^{\lambda } \biggl( \int _{x}^{t}(t-u)g^{\prime \prime }(u)\,du; x \biggr) \biggr\vert + \int _{x}^{x+\gamma _{n,p,q}^{\lambda }(x)} \bigl\vert x+ \gamma _{n,p,q}^{\lambda }(x)-u \bigr\vert \bigl\vert g^{\prime \prime }(u) \bigr\vert \,du \\ &\quad\leq \bigl(B_{n,p,q}^{\lambda } \bigl((t-x)^{2}; x \bigr)+ \bigl(\gamma _{n,p,q}^{\lambda }(x) \bigr)^{2} \bigr) \bigl\vert g^{\prime \prime }\bigr\vert \\ &\quad\leq \bigl(\alpha (n; p, q)^{2}+\beta (n; p, q) \bigr)\bigl\vert g ^{\prime \prime }\bigr\vert , \end{aligned}
(24)

where $$\alpha (n; p, q)$$ and $$\beta (n; p, q)$$ are defined in (17) and (18). We also have the following fact by Lemma 2.2 and (21):

\begin{aligned} \bigl\vert \overline{B}_{n,p,q}^{\lambda }(f; x) \bigr\vert \leq \bigl\vert B_{n,p,q} ^{\lambda }(f; x) \bigr\vert +2 \Vert f \Vert \leq 3 \Vert f \Vert . \end{aligned}
(25)

Combining (21), (24), and (25), we have

\begin{aligned} & \bigl\vert B_{n,p,q}^{\lambda }(f; x)-f(x) \bigr\vert \\ &\quad\leq \bigl\vert \overline{B}_{n,p,q}^{\lambda }(f-g; x)-(f-g) (x) \bigr\vert + \bigl\vert \overline{B}_{n,p,q}^{\lambda }(g; x)-g(x) \bigr\vert + \bigl\vert f \bigl(x+\gamma _{n,p,q}^{\lambda }(x) \bigr)-f(x) \bigr\vert \\ &\quad \leq 4 \Vert f-g \Vert + \bigl(\alpha (n; p, q)^{2}+\beta (n; p, q) \bigr) \bigl\vert g^{\prime \prime }\bigr\vert +\omega \bigl(f; \alpha (n; p, q) \bigr). \end{aligned}

Taking the infimum on the right-hand side over all $$g\in C^{2}$$ and using the relationship between K-functional and the second order modulus of smoothness, we can obtain

\begin{aligned} \bigl\vert B_{n,p,q}^{\lambda }(f; x)-f(x) \bigr\vert \leq C \omega _{2} \bigl(f; \sqrt{\alpha (n; p, q)^{2}+\beta (n; p, q)}/2 \bigr)+\omega \bigl(f; \alpha (n; p, q) \bigr). \end{aligned}

This completes the proof of Theorem 3.3. □

### Theorem 3.4

For$$f^{\prime }\in C_{[0, 1]}$$, $$\lambda \in [-1,1]$$, $$x\in [0,1]$$, and$$n>1$$, we have

\begin{aligned} \bigl\vert B_{n,p,q}^{\lambda }(f; x)-f(x) \bigr\vert \leq \alpha (n; p, q) \bigl\vert f^{\prime }(x) \bigr\vert +2\sqrt{\beta (n; p, q)}\omega \bigl(f^{\prime }; \sqrt{\beta (n; p, q)} \bigr), \end{aligned}

where$$\alpha (n; p, q)$$, $$\beta (n; p, q)$$are defined in (17) and (18), respectively.

### Proof

Applying $$B_{n,p,q}^{\lambda }(f; x)$$ to both sides of $$f(t)=f(x)+f ^{\prime }(x)(t-x)+f(t)-f(x)-f^{\prime }(x)(t-x)$$, we have

\begin{aligned} & \bigl\vert B_{n,p,q}^{\lambda }(f; x)-f(x) \bigr\vert \\ &\quad\leq \bigl\vert f^{\prime }(x) \bigr\vert \bigl\vert B_{n,p,q}^{\lambda }(t-x; x) \bigr\vert +B _{n,p,q}^{\lambda } \bigl( \bigl\vert f(t)-f(x)-f^{\prime }(x) (t-x) \bigr\vert ; x \bigr) \\ &\quad\leq \bigl\vert f^{\prime }(x) \bigr\vert \bigl\vert B_{n,p,q}^{\lambda }(t-x; x) \bigr\vert +B _{n,p,q}^{\lambda } \biggl( \vert t-x \vert \biggl(1+\frac{ \vert t-x \vert }{\delta } \biggr) \omega \bigl(f^{\prime }; \delta \bigr); x \biggr) \\ &\quad\leq \bigl\vert f^{\prime }(x) \bigr\vert \bigl\vert B_{n,p,q}^{\lambda }(t-x; x) \bigr\vert +\sqrt{B _{n,p,q}^{\lambda } \bigl((t-x)^{2}; x \bigr)} \biggl(1+\frac{1}{ \delta }\sqrt{B_{n,p,q}^{\lambda } \bigl((t-x)^{2}; x \bigr)} \biggr) \\ &\qquad{}\times \omega \bigl(f^{\prime }; \delta \bigr) \end{aligned}

with the help of mean value theorem and the Cauchy–Schwarz inequality. Taking $$\delta = \sqrt{B_{n,p,q}^{\lambda } ((t-x)^{2}; x )}$$ and by Corollary 2.5, we can get the desired result. Theorem 3.4 is proved. □

For $$x, y\in [0, 1]$$, a function belongs to Lipschitz class $$\operatorname{Lip}_{M}(\xi )$$ if

\begin{aligned} \bigl\vert f(y)-f(x) \bigr\vert \leq M \vert y-x \vert ^{\xi }, \end{aligned}
(26)

where $$M>0$$ and $$\xi \in (0, 1]$$. We end up giving the rate of convergence of $$B_{n,p,q}^{\alpha }(f; x)$$ on $$f\in \operatorname{Lip}_{M}(\xi )$$.

### Theorem 3.5

Let$$f\in \operatorname{Lip}_{M}(\xi )$$, $$\xi \in (0,1]$$, then for$$\lambda \in [-1,1]$$, $$x\in [0,1]$$and$$n>1$$, we have

\begin{aligned} \bigl\vert B_{n,p,q}^{\lambda }(f; x)-f(x) \bigr\vert \leq M \bigl(\sqrt{ \beta (n; p, q)} \bigr)^{\xi }, \end{aligned}

where$$\beta (n; p, q)$$is defined in (18).

### Proof

Since $$f\in \operatorname{Lip}_{M}(\xi )$$, we have

\begin{aligned} & \bigl\vert B_{n,p,q}^{\lambda }(f; x)-f(x) \bigr\vert \\ &\quad\leq B_{n,p,q}^{\lambda } \bigl( \bigl\vert f(t)-f(x) \bigr\vert ; x \bigr) \\ &\quad\leq M\sum_{k=0}^{n}b_{n,k}^{\lambda }(x; p, q) \biggl\vert \frac{[k]_{p,q}}{p ^{k-n}[n]_{p,q}}-x \biggr\vert ^{\xi } \\ &\quad\leq M\sum_{k=0}^{n} \biggl(b_{n,k}^{\lambda }(x; p, q) \biggl(\frac{[k]_{p,q}}{p ^{k-n}[n]_{p,q}}-x \biggr)^{2} \biggr)^{\frac{\xi }{2}} \bigl(b_{n,k} ^{\lambda }(x; p, q) \bigr)^{\frac{2-\xi }{2}}. \end{aligned}

Applying Hölder’s inequality for sums, we have

\begin{aligned} & \bigl\vert B_{n,p,q}^{\lambda }(f; x)-f(x) \bigr\vert \\ &\quad\leq M \Biggl(\sum_{k=0}^{n}b_{n,k}^{\lambda }(x; p, q) \biggl(\frac{[k]_{p,q}}{p ^{k-n}[n]_{p,q}}-x \biggr)^{2} \Biggr)^{\frac{\xi }{2}} \Biggl(\sum_{k=0}^{n}b_{n,k}^{\lambda }(x; p, q) \Biggr)^{\frac{2-\xi }{2}} \\ &\quad=M \bigl(B_{n,p,q}^{\lambda } \bigl((t-x)^{2}; x \bigr) \bigr) ^{\frac{\xi }{2}} \\ &\quad \leq M \bigl(\beta (n; p, q) \bigr)^{\frac{\xi }{2}}, \end{aligned}

where $$\beta (n; p, q)$$ is defined in (18). Theorem 3.5 is proved. □

### Remark 3.6

Apparently, if the sequences $$q:=\{q_{n}\}$$ and $$p:=\{p_{n}\}$$ satisfy the conditions of Lemma 2.7 and $$p^{n}\rightarrow a\in (0,b]$$, b is finite, then $$\alpha (n; p, q), \beta (n; p, q)\rightarrow 0$$ as $$n\rightarrow \infty$$. Therefore, by Theorems 3.23.5, the convergence rate of $$B_{n,p,q}^{\lambda }(f; x)$$ to f is obtained.

## Conclusion

In this paper, λ-Bernstein operators based on (p, q)-integers are constructed, a Korovkin type approximation theorem is established, and also the rate of convergence of $$B_{n,p,q}^{\lambda }(f; x)$$ to $$f(x)$$ is obtained.

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### Acknowledgements

We thank Fujian Provincial Key Laboratory of Data-Intensive Computing, Fujian University Laboratory of Intelligent Computing and Information Processing and Fujian Provincial Big Data Research Institute of Intelligent Manufacturing of China.

### Availability of data and materials

All data generated or analysed during this study are included in this published article.

## Funding

This work is supported by the National Natural Science Foundation of China (Grant No. 11601266 and Grant No. 11626031), the Project for High-level Talent Innovation and Entrepreneurship of Quanzhou (Grant No. 2018C087R), the Program for New Century Excellent Talents in Fujian Province University and Sponsoring Agreement for Overseas Studies in Fujian Province, the Key Natural Science Research Project in Universities of Anhui Province (Grant No. KJ2019A0572), the Philosophy and Social Sciences General Planning Project of Anhui Province of China (Grant No. AHSKYG2017D153) and the Natural Science Foundation of Anhui Province of China (Grant No. 1908085QA29).

## Author information

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### Contributions

The authors carried out the whole manuscript. All authors read and approved the final manuscript.

### Corresponding author

Correspondence to Wen-Tao Cheng.

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### Competing interests

The authors declare that they have no competing interests. 