The convergence of $(p,q)$-Bernstein operators for the Cauchy kernel with a pole via divided difference

Abstract

In this paper, some qualitative approximation results for the $(p,q)$-Bernstein operators $B_{p,q}^{n}(f;x)$ are obtained for the Cauchy kernel $\frac{1}{x-\alpha }$ with a pole $\alpha \in {}[ 0,1]$ for $q>p>1$. The main focus lies in the study of behavior of operators $B_{p,q}^{n}(f;x)$ for the function $f_{m}(x)=\frac{1}{x-p^{m}q^{-m}}$, $x\neq p^{m}q^{-m}$ and $f_{m}(p^{m}q ^{-m})=a$, $a\in \mathbb{R}$ and the extra parameter p provides flexibility for the approximation.

1 Introduction and preliminaries

The uniform convergence of a sequence of operators to a continuous function was introduced by Bohman [9] and Korovkin [16]. Through q-calculus various modifications of Bernstein operators [7] have been studied so far [10, 18, 31]. The $(p,q)$-integers are the generalization of the q-integers, which has an important role in the representation theory of quantum calculus in the physics literature. Recently, the approximation by the $(p,q)$-analog of a positive linear operator has become an active area of research. For the theory and numerical implementations of the $(p,q)$-analog of Bernstein operators introduced by Mursaleen et al. [22] and other $(p,q)$-analogs, the reader may refer to [1,2,3,4,5, 11,12,13,14,15, 19,20,21] and [32]. For most recent work on the $(p,q)$-approximation we refer to [8, 24, 26].

The $(p,q)$-integer, $(p,q)$-binomial expansion and the $(p,q)$-binomial coefficients are defined by

$$\begin{array}{c}{[m]}_{p,q}:=\frac{p^m-q^m}{p-q},m=0,1,2,\dots ,p>q\ge 1,\hfill \\ {(a+b)}_{p,q}^m:=(a+b)(pa+qb)(p^{2}a+q^{2}b)\cdots (p^{m-1}a+q^{m-1}b)\hfill \\ \phantom{{(a+b)}_{p,q}^m}=\sum _{r=0}^k{p}^{\frac{(m-r)(m-r-1)}{2}}{q}^{\frac{r(r-1)}{2}}{\left[\begin{array}{c}m\\ r\end{array}\right]}_{p,q}{a}^r,\hfill \\ {\left[\begin{array}{c}m\\ r\end{array}\right]}_{p,q}:=\frac{{[m]}_{p,q}!}{{[r]}_{p,q}!{[m-r]}_{p,q}!}.\hfill \end{array}$$

It can easily be verified by induction that

$$(1+a)(p+qa)(p^2+q^{2}a)\cdots (p^{n-1}+q^{n-1}a)=\sum _{r=0}^k{p}^{\frac{(m-r)(m-r-1)}{2}}{q}^{\frac{r(r-1)}{2}}{\left[\begin{array}{c}m\\ r\end{array}\right]}_{p,q}{a}^r.$$

The $(p,q)$-analog of Euler’s identity is defined by

$$\prod _{s=0}^{m-1}(p^s-q^{s}a):=\sum _{k=0}^m{p}^{\frac{(m-k)(m-k-1)}{2}}{q}^{\frac{k(k-1)}{2}}{\left[\begin{array}{c}m\\ k\end{array}\right]}_{p,q}{a}^k.$$

Let $f:[0,1]\longrightarrow \mathbb{R}$ and $q>p>1$. The $(p,q)$-Bernstein operators [22] of f is defined as

$$ B_{p,q}^{n}(f;x):=\sum_{k=0}^{n}f \biggl( \frac{[k]_{p,q}}{p^{k-n}[n]_{p,q}} \biggr) p_{n,k}(p,q;x),\quad n \in \mathbb{N}, $$

(1.1)

where the polynomial $p_{n,k}(p,q;x)$ is given by

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

(1.2)

For $p=1$, $B_{p,q}^{n} (f;x)$ turns into the q-Bernstein operator. We have

$$ B_{p,q}^{n}(f;0)=f(0), \qquad B_{p,q}^{n}(f;1)=f(1), \quad n \in \mathbb{N}. $$

(1.3)

The following $(p,q)$-difference form of Bernstein operators [25] is given by

$$ B_{p,q}^{n}(f;x):=\sum_{r=0}^{n} \lambda _{p,q}^{n} f \biggl[ 0,\frac{p ^{n-1}[1]_{p,q}}{[n]_{p,q}},\ldots, \frac{p^{n-r}[r]_{p,q}}{[n]_{p,q}} \biggr] x^{r}, $$

(1.4)

where $f[x_{0},x_{1},\ldots,x_{n}]$ indicates the nth order divided difference of f with pairwise distinct node, that is,

$$\begin{aligned}& f[x_{0}]=f(x_{0}), \qquad f[x_{0},x_{1}]= \frac{f(x_{1})-f(x_{0})}{x_{1}-x_{0}}, \\& f[x_{0},x_{1},\ldots,x_{n}]=\frac{f[x_{1},\ldots,x_{n}]-f[x_{0}, \ldots,x_{n-1}]}{[x_{n}-x_{0}]} \end{aligned}$$

and $\lambda _{p,q}^{n}$ is given by

$$\begin{array}{rcl}{\lambda }_{p,q}^n& =& {\left[\begin{array}{c}n\\ r\end{array}\right]}_{p,q}\frac{{[r]}_{p,q}!}{{[n]}_{p,q}^r}{p}^{\frac{(n-r)(n-r-1)}{2}}{q}^{\frac{r(r-1)}{2}}\\ & =& (1-\frac{p^{n-1}{[1]}_{p,q}}{{[n]}_{p,q}})(1-\frac{p^{n-2}{[2]}_{p,q}}{{[n]}_{p,q}})\cdots (1-\frac{p^{n-r+1}{[r-1]}_{p,q}}{{[n]}_{p,q}}),\end{array}$$

(1.5)

and $\lambda _{p,q}^{0}=\lambda _{p,q}^{1}=1$, $0\leq \lambda _{p,q} ^{n}\leq 1$, $r=0,1,\ldots,n$.

In this paper, some qualitative approximation results for the $(p,q)$-Bernstein operators $B_{p,q}^{n}(f;x)$ have been obtained for the Cauchy kernel $\frac{1}{x-\alpha }$ with a pole $\alpha \in {}[ 0,1]$ for $q>p>1$. The main focus lies in the study of behavior of operators $B_{p,q}^{n}(f;x)$ for the function $f_{m}(x)=\frac{1}{x-p ^{m}q^{-m}}$, $x\neq p^{m}q^{-m}$ and $f_{m}(p^{m}q^{-m})=a$, $a\in \mathbb{R}$ and the extra parameter p provides flexibility for the approximation.

The time scale $\mathbb{J}_{p,q}$ for $q>p>1$ is denoted and defined as

$$ \mathbb{J}_{p,q}=\{0\}\cup \bigl\{ p^{k}q^{-k}\bigr\} _{k=0}^{\infty }. $$

(1.6)

Here, we consider the $(p,q)$-Bernstein operators with the Cauchy kernel $\frac{1}{x-\alpha }$, $\alpha \in [ 0,1]$. The previously obtained results [27,28,29,30] lead to the following conclusions.

• If $\alpha =0$, that is, $f(x)=\frac{1}{x}$, $x\neq 0$ and $f(0)=a$, then, for $q\geq 2$,

  $$ \lim_{n\rightarrow \infty }B_{p,q}^{n}(f;x)= \textstyle\begin{cases} f(x),& x\in \mathbb{J}_{p,q}, \\ \infty ,& x\notin \mathbb{J}_{p,q}. \end{cases} $$

  (1.7)

• If $\alpha \in \mathbb{J}_{p,q}\setminus [0,1]$ that is $f(x)=\frac{1}{(x- \alpha)}$ if $x\neq \alpha$ and $f(\alpha) = a$, then

  $$ \lim_{n\rightarrow \infty } B^{n}_{p,q}(f;x) = f(x),\quad x \in \mathbb{J}_{p,q}. $$

Furthermore, as $n\rightarrow \infty$, $B_{p,q}^{n}(f;x)\rightarrow f(x)$ uniformly on any compact subset of $(-\alpha,\alpha)$ and $B_{p,q}^{n}(f;x)\rightarrow \infty$ for $\vert x \vert > \alpha$, $x\notin \mathbb{J}_{p,q}$. Therefore, it is left to examine the case $\alpha \in \mathbb{J}_{p,q} \setminus \{0\}$ which is exactly the subject of the present paper. Let the function $f_{m}:\mathbb{R}\rightarrow \mathbb{R}$ be defined by

$$ f_{m}(x)= \textstyle\begin{cases} \frac{1}{(x-p^{m}q^{-m})^{j}},& x\in \mathbb{R}\setminus \{p^{m}q^{-m}\}, \\ a,& x=p^{m}q^{-m}, \end{cases}\displaystyle m\in \mathbb{N}_{0}, a \in \mathbb{R}. $$

(1.8)

2 Some auxiliary results

In this section, we prove some important lemmas.

Lemma 2.1

For the function $f_{m}$ defined by (1.8), we have

• (a) for $m\in \mathbb{N}$,

  $$ \lim_{n\rightarrow \infty }B_{p,q}^{n}\bigl(f_{m};p^{j}q^{-j} \bigr)=f _{m}\bigl(p^{j}q^{-j}\bigr),\quad j\in \mathbb{N}_{0}\setminus \{m,m+1\}. $$

  Besides,

  $$\begin{aligned}& \lim_{n\rightarrow \infty }B_{p,q}^{n}\bigl(f_{m};p^{m}q^{-m} \bigr)=- \infty, \quad \textit{and} \\& \lim_{n\rightarrow \infty }B_{p,q}^{n}\bigl(f_{m};p ^{-(m+1)}q^{-(m+1)}\bigr) =f_{m}\bigl(p^{m+1}q^{-(m+1)}\bigr)- \frac{p^{-m}q^{m}[m+1]_{p,q}}{{p^{-1}(q-1)}[m]_{p,q}}. \end{aligned}$$

• (b) For $m=0$,

  $$ \lim_{n\rightarrow \infty }B_{p,q}^{n}\bigl(f_{0};p^{j}q^{-j} \bigr)=f _{0}\bigl(p^{j}q^{-j}\bigr), \quad j\in \mathbb{N}_{0} $$

  i.e., $B_{p,q}^{n}(f_{0};\cdot)$ approximates $f_{0}$ on $\mathbb{J}_{p,q}$.

This describes the behavior of $B_{p,q}^{n}(f_{m};\cdot)$ on the time scale $\mathbb{J}_{p,q}$.

Proof

(a) From (1.2), we can easily see that $p_{n,n-k}(p,q;p^{j}q^{-j})=0$ for $k>j$, whence

$$ B_{p,q}^{n}\bigl(f;p^{-j}q^{j}\bigr)= \sum_{k=0}^{\min \{k,j\}}f \biggl( \frac{[n-k]_{p,q}}{[n]_{p,q}} \biggr) p_{n,n-k} \bigl(p,q;p^{j}q^{-j}\bigr). $$

(2.1)

Besides

$$ \lim_{n\rightarrow \infty }p_{n,n-k} \bigl(p,q;p^{j}q^{-j} \bigr)= \delta _{k,j} \quad \mbox{and} \quad \lim_{n\rightarrow \infty } \frac{[n-k]_{p,q}}{[n]_{p,q}}=p^{k}q^{-k} . $$

(2.2)

Thus, $\lim_{n\rightarrow \infty }f_{m} ( \frac{[n-k]_{p,q}}{[n]_{p,q}} ) p_{n,n-k} (p,q;p^{j}q^{-j})=f _{m} (p^{k}q^{-k}) \delta _{j,k}$ for all $k\neq m$.

Now by easy calculation, we have

$$ \lim_{n\rightarrow \infty }f \biggl( \frac{[n-k]_{p,q}}{[n]_{p,q}} \biggr) p_{n,n-k}\bigl(p,q;p^{j}q^{-j}\bigr)= \textstyle\begin{cases} -\infty ,& j=m, \\ -\frac{p^{-m}q^{m}[m+1]_{p,q}}{{p^{-1}(q-1)} [m]_{p,q}},& j=m+1, \\ 0,& \geq m+2, \end{cases} $$

and combining with (2.1) and (2.2) yields the result.

(b) It can be obtained easily from (1.3) and (2.2) as $f_{0}$ is continuous at all points $p^{j}q^{-j}$, $j\in \mathbb{N}$. □

The next lemma is related to the coefficient of $B^{n}_{p,q}(f_{0};\cdot)$.

Lemma 2.2

Let $f_{m}$ be a function as in (1.8). If $B_{p,q}^{n}(f_{m};x)=\sum_{k=0}^{n}C_{k,n}^{p,q}x^{k}$ and $\frac{[k]_{p,q}}{[n]_{p,q}}\neq p^{m}q^{-m}$ for $k=0,1,2,\ldots,n$, then

$$ C_{k,n}^{p,q}=-\frac{\lambda _{k,n}^{p,q}p^{-m(k+1)}q^{m(k+1)}}{ ( 1-\frac{p^{n-m-1}q^{m}[1]_{p,q}}{[n]_{p,q}} ) ( 1-\frac{p ^{n-m-2}q^{m}[2]_{p,q}}{[n]_{p,q}} ) \cdots ( 1-\frac{p ^{n-m-k}q^{m}[k]_{p,q}}{[n]_{p,q}} ) }, $$

(2.3)

where $\lambda _{k,n}^{p,q}$ are given by (1.5).

Proof

Consider $f_{m} (z)=\frac{1}{z-p^{m}q^{-m}}$, which is analytic function in $\mathbb{C}\setminus \{p^{m}q^{-m}\}$. It is well known that [17] the kth order divided difference of f can be expressed as

$$ f[x_{0},x_{1},\ldots,x_{k}]=\frac{1}{2\pi i} \oint _{\mathcal{L}}\frac{f( \zeta)\,d\zeta }{(\zeta -x_{0})(\zeta -x_{1})\cdots (\zeta -x_{k})}, $$

where $\mathcal{L}$ is contour encircling $x_{0},\ldots,x_{k}$ and f is assumed to be analytic on and within $\mathcal{L}$. Hence, when the nodes $0,\frac{[1]_{p,q}}{[n]_{p,q}},\frac{[2]_{p,q}}{[n]_{p,q}}, \ldots,\frac{[k]_{p,q}}{[n]_{p,q}}$ are inside $\mathcal{L}$ and the pole $\alpha =p^{m}q^{-m}$ is outside, one has

$$ f \biggl[ 0,\frac{p^{n-1}[1]_{p,q}}{[n]_{p,q}},\ldots, \frac{p^{n-r}[r]_{p,q}}{[n]_{p,q}} \biggr] = \frac{1}{2\pi i} \oint _{\mathcal{L}}\frac{f_{m} (\zeta)\,d\zeta }{\zeta (\zeta -\frac{p ^{n-1}[1]_{p,q}}{[n]_{p,q}})\cdots (\zeta - \frac{p^{n-r}[r]_{p,q}}{[n]_{p,q}})}. $$

(2.4)

By the residue theorem

$$\begin{aligned} f \biggl[ 0,\frac{p^{n-1}[1]_{p,q}}{[n]_{p,q}},\ldots, \frac{p^{n-r}[r]_{p,q}}{[n]_{p,q}} \biggr] =&\sum _{j=0}^{k}\operatorname{Res}_{z=p ^{n-j}\frac{[j]_{p,q}}{[n]_{p,q}}} \frac{f_{m} (z)}{\prod_{j=0}^{k} ( z-p^{n-j}\frac{[j]_{p,q}}{[n]_{p,q}} ) } \\ =&-\operatorname{Res}_{z=p^{m}q^{-m}}\frac{f_{m} (z)}{\prod_{j=0}^{k} ( z-p^{n-j}\frac{[j]_{p,q}}{[n]_{p,q}} ) } \\ =&-\frac{p^{-m(k+1)} q^{m(k+1)}}{\prod_{j=1}^{k} ( 1-p^{n-m-j}\frac{[j]_{p,q}}{[n]_{p,q}} ) }. \end{aligned}$$

Since $f_{m} (z)=f_{m} (x)$ for $z=x\in {}[ 0,1]$, the statement follows from the divided difference representation (1.4). □

Now, we find the asymptotic estimates for the coefficient $C_{k,n} ^{p,q}$ in the next lemma.

Lemma 2.3

We have

$$ \lim_{n\rightarrow \infty }\prod_{k=1}^{n-j} \biggl( 1-p^{n-m-j} q^{m}\frac{[k]_{p,q}}{[n]_{p,q}} \biggr) = \biggl( \frac{p^{2j-m}}{q ^{j-m}};\frac{p}{q} \biggr) _{\infty } $$

for $j>m$, $q>p>1$.

Proof

It is clear that

$$ \log \prod_{k=1}^{n-j} \biggl( 1-p^{n-m-j} q^{m} \frac{[k]_{p,q}}{[n]_{p,q}} \biggr) =\sum _{k=j}^{n-1}\log \biggl( 1-p ^{n-m-j} q^{m}\frac{[n-k]_{p,q}}{[n]_{p,q}} \biggr) =\sum_{k=j}^{ \infty }a_{k,n}^{p,q}, $$

where

$$ a_{k,n}^{p,q}= \textstyle\begin{cases} \log ( 1-p^{n-m-j} q^{m}\frac{[n-k]_{p,q}}{[n]_{p,q}} ),& k< n, \\ 0,& k\geq n. \end{cases} $$

Since

$$\begin{aligned} \bigl\vert a_{k,n}^{p,q} \bigr\vert \leq& \biggl\vert \log \biggl( 1-p ^{n-m-j} q^{m}\frac{[n-k]_{p,q}}{[n]_{p,q}} \biggr) \biggr\vert \\ \leq& \frac{q}{q-p}\frac{p^{n-m-k} q^{m} [n-k]_{p,q}}{[n]_{p,q}} \leq \frac{q}{q-p}p^{n-k-m} q^{m}, \end{aligned}$$

which gives $\sum_{k=j}^{\infty } \vert a_{k,n}^{p,q} \vert < \infty$, and by the Lebesgue dominated convergence theorem, we have

$$\begin{aligned} \lim_{n\rightarrow \infty }\sum_{k=j}^{n-1} \log \biggl( 1-p ^{n-m-j}q^{m}\frac{[n-k]_{p,q}}{[n]_{p,q}} \biggr) =&\sum _{k=j}^{ \infty } \biggl( \lim_{n\rightarrow \infty } \log \biggl( 1-p ^{n-m-j}q^{m}\frac{[n-k]_{p,q}}{[n]_{p,q}} \biggr) \biggr) \\ =&\sum_{k=j}^{\infty }\lim_{n\rightarrow \infty } \biggl( 1-p ^{n-k-m} \frac{q^{m}}{q^{k}} \biggr), \end{aligned}$$

as a result

$$ \lim_{n\rightarrow \infty }\log \prod_{k=1}^{n-j} \biggl( 1-p ^{n-m-j} q^{m} \frac{[n-k]_{p,q}}{[n]_{p,q}} \biggr) =\log \prod_{k=j}^{\infty } \biggl( 1-p^{n-k-m} \frac{q^{m}}{p^{k}} \biggr), $$

which completes the proof. □

The following lemma gives an upper bound for $n-m-1$.

Lemma 2.4

If $m\in \mathbb{N}$, $k=0,1,2,\ldots,n-m-1$, then

$$ \bigl\vert C_{k,n}^{p,q} \bigr\vert \leq \mathcal{C}_{m,p,q} p^{-mn} q^{mn}, $$

where $\mathcal{C}$ in RHS is a positive constant, whose value need not to be addressed.

Proof

For $n>m+1$ and from (2.3), we have

$$\begin{aligned}& \bigl\vert C_{k,n}^{p,q} \bigr\vert \leq \frac{p^{-m(k+1)}q^{m(k+1)}}{ ( 1-\frac{p^{n-m-1}q^{m}[1]_{p,q}}{[n]_{p,q}} ) ( 1-\frac{p ^{n-m-2}q^{m}[2]_{p,q}}{[n]_{p,q}} ) \cdots ( 1-\frac{p ^{n-m-k}q^{m}[k]_{p,q}}{[n]_{p,q}} ) } \\& \hphantom{ \big\vert C_{k,n}^{p,q}\big\vert } \leq \frac{p^{-m(n-m)}q^{m(n-m)}}{ ( 1-\frac{p^{2(n-m-1)}q ^{m}}{q^{n-1}} ) ( 1- \frac{p^{2(n-m-2)}q^{m}[2]_{p,q}}{[n]_{p,q}} ) \cdots ( 1-\frac{p ^{n-m-k}q^{m}[k]_{p,q}}{[n]_{p,q}} ) }, \\& \bigl\vert C_{k,n}^{p,q} \bigr\vert \leq \frac{p^{-mn}q^{mn}}{p^{2m}q ^{2m}(\frac{p}{q};\frac{p}{q})_{\infty }}. \end{aligned}$$

Further, we discuss the nature of $C_{n-m+1,n},\ldots,C_{n,n}$ as follows. □

Lemma 2.5

For $m\in \mathbb{N}$, $q>p>1$,

$$ \bigl\vert C_{n-m,n}^{p,q} \bigr\vert \sim \mathcal{C}_{p,q,m} p ^{-(m+1)n}q^{(m+1)n}, \quad n\rightarrow \infty. $$

Proof

Using (2.3), we obtain the following

$$ \bigl\vert C_{n-m}^{p,q} \bigr\vert =\lambda _{n-m,n}^{p,q}\frac{p ^{-m(n-m+1)} q^{m(n-m+1)}}{ ( 1- \frac{p^{n-m-1}q^{m}[1]_{p,q}}{[n]_{p,q}} ) ( 1-\frac{p ^{n-m-2}q^{m}[2]_{p,q}}{[n]_{p,q}} ) \cdots ( 1-\frac{p ^{n-m-(n-m)}q^{m}[n-m]_{p,q}}{[n]_{p,q}} ) }. $$

From Lemma 2.3, we have

$$\begin{aligned}& \begin{aligned}& \bigl\vert C_{n-m}^{p,q} \bigr\vert \sim \frac{ ( \frac{p^{m+1}}{q ^{m+1}};\frac{p}{q} ) _{\infty }q^{m(n-m+1)} p^{-m(n-m+1)} p ^{m}(p^{n}-q^{n})}{ ( \frac{p}{q};\frac{p}{q} ) _{\infty }p ^{n} (p^{m}-q^{m})} \\ &\hphantom{ \vert C_{n-m}^{p,q} \vert }\sim \frac{ ( \frac{p^{m+1}}{q^{m+1}};\frac{p}{q} ) _{ \infty }q^{mn} p^{-mn}p^{m} (p^{n}-q^{n})}{ ( \frac{p}{q}; \frac{p}{q} ) _{\infty }q^{m(m-1)}p^{m(m-1)}(p^{m}-q^{m})}, \\ &C_{n-m,n}^{p,q}=\mathcal{C}_{m}^{p,q} p^{-n(m+1)}q^{n(m+1)}. \end{aligned} \end{aligned}$$

(2.5)

The nature of the remaining coefficients $C_{n-m+1,n},\ldots,C_{n,n}$ is given as follows. □

Lemma 2.6

For $j=1,2,\ldots,m$, we have

$$\underset{n\to \mathrm{\infty }}{lim}\frac{C_{n-m+j,n}^{p,q}}{C_{n-m,n}^{p,q}}={(-1)}^j{\left[\begin{array}{c}m\\ j\end{array}\right]}_{p,q}{p}^{\frac{(n-j)(n-j-1)}{2}}{q}^{\frac{j(j-1)}{2}}.$$

Proof

Using (2.3) and (1.5), we get

$$\begin{array}{c}{C}_{n-m+j,n}^{p,q}=C_{n-m,n}^{p,q}\frac{(1-\frac{p^m{[n-m]}_{p,q}}{{[n]}_{p,q}})\cdots (1-p^{m-j+1}\frac{{[n-m+j-1]}_{p,q}}{{[n]}_{p,q}})}{(1-p^{-1}{q}^m\frac{[n-m+1]}{{[n]}_{p,q}})\cdots (1-p^{-j}{q}^m\frac{{[n-m+j]}_{p,q}}{{[n]}_{p,q}})},\hfill \\ \underset{n\to \mathrm{\infty }}{lim}\frac{C_{n-m+j,n}^{p,q}}{C_{n-m,n}^{p,q}}=\frac{(1-\frac{p^m}{q^m})\cdots (1-\frac{p^{m-j+1}}{q^{m-j+1}})}{(1-\frac{q}{p})\cdots (1-\frac{q^j}{p^j})},\hfill \\ \underset{n\to \mathrm{\infty }}{lim}\frac{C_{n-m+j,n}^{p,q}}{C_{n-m,n}^{p,q}}={(-1)}^j{\left[\begin{array}{c}m\\ j\end{array}\right]}_{p,q}{p}^{\frac{(n-j)(n-j-1)}{2}}{q}^{\frac{j(j-1)}{2}}.\hfill \end{array}$$

 □

Corollary 2.7

The following estimate holds:

$$ \bigl\vert C_{k,n}^{p,q} \bigr\vert \leq \mathcal{C}_{p,q,m} p^{-(m+1)n}q ^{(m+1)n},\quad k=0,1,2, \ldots,n, $$

(2.6)

and $\mathcal{C}_{p,q,m}$ is independent of both k and n.

Corollary 2.8

We have the following:

$$ \lim_{n\rightarrow \infty }\frac{C_{n-m,n}+\cdots +C_{n-m+j,n}x ^{j}+\cdots +C_{n,n}x^{n}}{C_{n-m,n}}=(x;p,q)_{m}. $$

(2.7)

Proof

The statement follows from Rothe’s identity [6],

$${(x;p,q)}_m=\sum _{j=0}^m{(-1)}^j{\left[\begin{array}{c}m\\ j\end{array}\right]}_{p,q}{p}^{\frac{(n-j)(n-j-1)}{2}}{q}^{\frac{j(j-1)}{2}}.$$

 □

3 Main results

• First we consider the case when pole $\alpha \in \mathbb{J}_{p,q} \setminus \{0,1\}$.

Now, we obtain the results that concern with the uniform approximation of $f_{m}(x)$, $m\in \mathbb{N}$ by its $(p,q)$-Bernstein operators. It may be noted that, while the case when $\alpha \in {}[ 0,1]\setminus \mathbb{J}_{p,q}$ can easily be examined by using the result and method of [27], the condition $\alpha \in \mathbb{J}_{p,q}$ requires a different approach.

Theorem 3.1

If $m\in \mathbb{N}$, then $B_{p,q}^{n}(f_{m};x) \rightarrow f_{m}(x)$ as $n\rightarrow \infty$ uniformly on any compact subset of $(-p^{(m+1)}{q^{(m+1)}},p^{(m+1)}{q^{(m+1)}})$.

Proof

We consider the complex $(p,q)$-Bernstein operators given by

$$ B_{p,q}^{n}(f;x)=\sum_{k=0}^{n}f \biggl( \frac{[k]_{p,q}}{p^{k-n}[n]_{p,q}} \biggr) p_{n,k} (p, q;x),\quad n\in \mathbb{N},z \in \mathbb{C}, $$

(3.1)

and the function $f_{m}(z)=\frac{1}{(z-p^{m}q^{-m})}$, $z\in \mathbb{C}$. Let n be large enough to satisfy the condition $\frac{[k]_{p,q}}{[n]_{p,q}}\neq p^{m}q^{-m}$. Then

$$ B_{p,q}^{n}(f_{m};z)=\sum _{k=0}^{n}C_{k,n}^{p,q}z^{k}, $$

where $C_{k,n}^{p,q}$ is given by (2.3). Let $\rho \in (0, p ^{(m+1)}q^{-(m+1)})$. Therefore for $\vert z \vert \leq \rho$ the following estimate is valid by Corollary 2.7:

$$ \bigl\vert B_{p,q}^{n}(f_{m};z) \bigr\vert \leq \sum_{k=0}^{n} \bigl\vert C _{k,n}^{p,q}\rho ^{k} \bigr\vert \leq \mathcal{C}_{p,q,m}\sum_{k=0} ^{n} \bigl(p^{-(m+1)}q^{(m+1)}\rho\bigr)^{k}\leq \mathcal{C}_{p,q,m}\frac{1}{(1-p ^{-(m+1)}q^{(m+1)}\rho)}. $$

Hence it follows that the operators $\{B_{p,q}^{n}(f_{m},z)\}$ are uniformly bounded in the disk $\{z: \vert z \vert \leq \rho \}$ and convergent on the sequence $\{p^{j}q^{-j}\}_{j=m+2}^{ \infty }$ having an accumulation point at 0 to the function $f_{m}(z)$ analytic in this disc. Using Vitali’s convergence theorem, we have $B_{p,q}^{n}(f_{m};z)\rightarrow f_{m}(z)$ ($n\rightarrow \infty$) uniformly on any compact set in $\{z: \vert z \vert \leq \rho \}$ as $\rho \in (0,p^{(m+1)}q^{-(m+1)})$ was arbitrary. This completes the proof. □

Next we demonstrate that, outside of the interval, operators diverge everywhere except a finite number of points.

Theorem 3.2

If $m \in \mathbb{N}$, then $\lim_{n\rightarrow \infty }B^{n}_{p,q} (f_{m};x)=\infty$ for $\vert x \vert > p^{(m+1)}q^{-(m+1)}$, $x \neq p^{(m+1)} q^{-(m+1)}$, $x \neq p^{(m-1)}q^{-(m-1)}$, $x \neq p^{(m-2)}q^{-(m-2)},\ldots,1$.

Proof

For exceptional points $p^{(m-1)}q^{-(m-1)},p^{(m-2)}q ^{-(m-2)},\ldots,1$, the situation has been analyzed in Lemma 2.1(a). We take x satisfying $\vert x \vert >p ^{(m-1)}q^{-(m-1)}$ different from these values. Let $n>m$ be sufficiently large such that (2.3) holds. By Lemma 2.4, we obtain

$$\begin{aligned} \Biggl\vert \sum_{k=0}^{\infty }C_{k,n}^{p,q}x^{k} \Biggr\vert \leq& \mathcal{C}_{m,p,q} \sum _{k=0}^{n-m-1}p^{-mk}q^{mk}x^{k}= \mathcal{C} _{m,p,q}\frac{(p^{-m}q^{m}x)^{n-m}-1}{p^{-m}q^{m}x-1} \\ =&o \bigl( \bigl(p^{-(m+1)}q^{(m+1)}x\bigr)^{n} \bigr), \quad n\rightarrow \infty. \end{aligned}$$

Hence

$$\begin{aligned} B_{p,q}^{n}(f_{m};x) =&\sum _{k=n-m}^{n}C_{k,n}^{p,q}x^{k}+o \bigl( \bigl(p ^{-(m+1)}q^{(m+1)}x\bigr)^{n} \bigr) \\ =&C_{n-m}^{p,q}x^{n-m}g_{n}(x)+o \bigl( \bigl(p^{-(m+1)}q^{(m+1)}x\bigr)^{n} \bigr),\quad n \rightarrow \infty. \end{aligned}$$

By Lemma 2.5, $\vert C_{n-m}^{p,q} \vert \sim \mathcal{C} _{p,q,m}(x) (p^{-(m+1)}q^{(m+1)}x)^{n}$ as $n\rightarrow \infty$ whenever $\vert x \vert >p^{(m+1)}q^{-(m+1)}$, since $\lim_{n\rightarrow \infty }g_{n} (x)=(x;p,q)_{m}\neq 0$, when $x\notin \{ p^{(m+1)}q^{-(m+1)},\ldots,1 \}$. □

Lemma 3.1

Let $f_{0}$ be given by putting $m=0$ in (1.8). If $B_{p,q}^{n}(f_{0};x)=\sum_{k=0}^{n}C_{k,n}^{p,q}x ^{k}$ then

$$ C_{k,n}^{p,q}=\frac{-1}{(1-p^{n-k}\frac{[k]_{p,q}}{[n]_{p,q}})},\quad k=0,1,2, \ldots,n-1, \qquad C_{n,n}^{p,q}=a+\sum_{k=0}^{n-1} \frac{1}{(1-p^{n-k} \frac{[k]_{p,q}}{[n]_{p,q}})}. $$

Proof

For $k=0,1,\ldots,n-1$, on a specific choice of the contour $\mathcal{L}$, such that the nodes $0, \frac{[1]_{p,q}}{[n]_{p,q}},\ldots,\frac{[k]_{p,q}}{[n]_{p,q}}$ are inside $\mathcal{L}$ while the pole $\alpha =1$ is outside, formula (2.4) implies

$$ C_{k,n}^{p,q}=\frac{-\lambda _{k,n}^{p,q}}{\prod_{j=0}^{k} ( 1-p ^{n-j}\frac{[j]_{p,q}}{[n]} ) }=\frac{-1}{ ( 1-p^{n-k} \frac{[k]_{p,q}}{[n]_{p,q}} ) }, $$

since by (1.3), $B_{p,q}^{n}(f_{0};1)=f_{0}(1)=a$ and the statement is proved. □

Corollary 3.2

For $k = 0 , 1, 2, \ldots, n-1$ with $q > p > 1$ we have the following result:

$$ \bigl\vert C^{p,q}_{k,n} \bigr\vert \leq \frac{q}{q-p}. $$

• Now, we consider the case when pole $\alpha =1$.

Here the point of singularity $x=1$ is one of the nodes $\frac{[k]_{p,q}}{[n]_{p,q}}$. Consider the function $f_{0}$

$$ f_{0}(x)= \textstyle\begin{cases} \frac{1}{x-1}, & x\in \mathbb{R}\setminus \{1\}, \\ a,& x=1. \end{cases} $$

(3.2)

Theorem 3.3

If $f_{0}$ is given by (3.2), then the following holds:

• (1) For all $x\in (-1,1]$,

  $$ \lim_{n\rightarrow \infty } B^{n}_{p,q}(f_{0};x)= f_{0}(x) $$

  and the convergence is uniform on any compact subset of $(-1,1)$.

• (2) For all $x\in \mathbb{R}\setminus (-1,1]$,

  $$ \lim_{n\rightarrow \infty } B^{n}_{p,q}(f_{0};x)= \infty. $$

Proof

(1) Since $B_{p,q}^{n}(f_{0};1)=f_{0}(1)$, we need to prove only the uniform convergence of the compact subset of $(-1,1)$. For any $\rho \in (0,1)$ and $\vert z \vert \leq \rho$. From Corollary 3.2, we have

$$ \Biggl\vert \sum_{k=0}^{n-1}C_{k,n}^{p,q}z^{k} \Biggr\vert \leq \frac{ \mathcal{C}_{p,q}}{1-\rho }. $$

Apart from that,

$$ \bigl\vert C_{n,n}^{p,q}z^{k} \bigr\vert \leq \vert a \vert + \sum_{k=0}^{n-1} \frac{1}{1-p^{n-k}\frac{[k]_{p,q}}{[n]_{p,q}}}\leq \vert a \vert +n \bigl\vert C_{k,n}^{p,q} \bigr\vert \leq \vert a \vert +\frac{nq}{q-p}, $$

whence

$$ \bigl\vert C_{n,n}^{p,q}z^{n} \bigr\vert \leq \vert a \vert \rho ^{n}+\rho ^{n}\frac{nq}{q-p}\leq \mathcal{C}_{p,q,\rho }. $$

Therefore, we conclude that the operators $B_{p,q}^{n}(f;z)$ are uniformly bounded in any disk $\{ z: \vert z \vert \leq \rho \}$ where $\rho \in (0,1)$. From Lemma 2.1(b) and Vitali’s convergence theorem we arrive at our result.

(2) Given that x satisfies $\vert x \vert >1$, by Able’s inequality, we have

$$ \Biggl\vert \sum_{k=0}^{n-1}C_{k,n}^{p,q}x^{k} \Biggr\vert \leq \frac{ \vert x \vert ^{n}-1}{ \vert x \vert -1}\bigl( \bigl\vert C _{0,n}^{p,q} \bigr\vert +2 \bigl\vert C_{n-1,n}^{p,q} \bigr\vert \bigr) \leq \frac{ \vert x \vert ^{n}}{ \vert x \vert -1} \biggl( 1+\frac{2p}{p-q} \biggr) = \mathcal{C}_{p,q,x} \vert x \vert ^{n}. $$

Meanwhile,

$$ \bigl\vert C_{n,n}^{p,q}x^{n} \bigr\vert \geq \Biggl( \sum_{k=0} ^{n-1}\frac{1}{1-p^{n-k}\frac{[k]_{p,q}}{[n]_{p,q}}} \Biggr) \vert x \vert ^{n}- \vert a \vert \cdot \vert x \vert ^{n}\geq \bigl(n- \vert a \vert \bigr) \vert x \vert ^{n}. $$

Thus, $\vert B_{p,q}^{n}(f_{0};x) \vert \geq n \vert x \vert ^{n}-(\mathcal{C}_{p,q,x}+ \vert a \vert ) \vert x \vert ^{n}\rightarrow \infty$ as $n\rightarrow \infty$.

At $x=-1$, we have

$$ B_{p,q}^{n}(f_{0};-1)=\sum _{k=0}^{n-1}C_{k,n}^{p,q}(-1)^{k}+ \Biggl( a+ \sum_{k=0}^{n-1}\frac{1}{1-p^{n-k}\frac{[k]_{p,q}}{[n]_{p,q}}} \Biggr) (-1)^{n}, $$

and again applying Able’s inequality,

$$ \Biggl\vert \sum_{k=0}^{n-1}C_{k,n,}^{p,q}(-1)^{k} \Biggr\vert \leq \vert C_{0,n} \vert +2 \vert C_{n-1,n} \vert \leq 1+\frac{2p}{p-q}. $$

On the other hand

$$ \Biggl\vert \Biggl( a+\sum_{k=0}^{n-1} \frac{1}{1-p^{n-k} \frac{[k]_{p,q}}{[n]_{p,q}}} \Biggr) (-1)^{n} \Biggr\vert \geq n- \vert a \vert , $$

which implies that

$$ \bigl\vert B_{p,q}^{n}(f_{0};-1) \bigr\vert \geq n- \vert a \vert - \biggl( 1+\frac{2p}{p-q} \biggr) \rightarrow \infty,\quad n\rightarrow \infty. $$

 □

Remark

For justification of the statement that the extra parameter p provides flexibility for approximation, one can see Remark 1 of [23].

Moreover, since for $q>p=1$ we recapture the q-Bernstein operators studied in [30], it is clear that the interval of uniform convergence for $B_{p,q}^{n}$ in Theorem 3.1, i.e. $(-p^{m+1}q ^{m+1},p^{m+1}q^{m+1})$, is larger than the interval of uniform convergence $(-q^{m+1},q^{m+1})$, obtained by Theorem 2.1 in [30].