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

Journal of Inequalities and Applications20192019:132

https://doi.org/10.1186/s13660-019-2090-y

  • Received: 4 February 2019
  • Accepted: 3 May 2019
  • Published:

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.

Keywords

  • \((p,q)\)-integer
  • \((p,q)\)-Bernstein operator
  • Convergence
  • Approximation of unbounded function
  • Cauchy kernel

MSC

  • 41A10
  • 41A25
  • 41A36

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 [15, 1115, 1921] 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
[ m ] p , q : = p m q m p q , m = 0 , 1 , 2 , , p > q 1 , ( a + b ) p , q m : = ( a + b ) ( p a + q b ) ( p 2 a + q 2 b ) ( p m 1 a + q m 1 b ) ( a + b ) p , q m = r = 0 k p ( m r ) ( m r 1 ) 2 q r ( r 1 ) 2 [ m r ] p , q a r , [ m r ] p , q : = [ m ] p , q ! [ r ] p , q ! [ m r ] p , q ! .
It can easily be verified by induction that
( 1 + a ) ( p + q a ) ( p 2 + q 2 a ) ( p n 1 + q n 1 a ) = r = 0 k p ( m r ) ( m r 1 ) 2 q r ( r 1 ) 2 [ m r ] p , q a r .
The \((p,q)\)-analog of Euler’s identity is defined by
s = 0 m 1 ( p s q s a ) : = k = 0 m p ( m k ) ( m k 1 ) 2 q k ( k 1 ) 2 [ m k ] 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 ) = 1 p n ( n 1 ) 2 [ n k ] p , q p k ( k 1 ) 2 x k s = 0 n k 1 ( p s q s x ) , x [ 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
λ p , q n = [ n r ] p , q [ r ] p , q ! [ n ] p , q r p ( n r ) ( n r 1 ) 2 q r ( r 1 ) 2 = ( 1 p n 1 [ 1 ] p , q [ n ] p , q ) ( 1 p n 2 [ 2 ] p , q [ n ] p , q ) ( 1 p n r + 1 [ r 1 ] p , q [ n ] p , q ) ,
(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 [2730] 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
  1. (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,
    $$\begin{aligned}& \lim_{n\rightarrow \infty }B_{p,q}^{n}\bigl(f_{m};p^{m}q^{-m} \bigr)=- \infty, \quad \textit{and} \\& \lim_{n\rightarrow \infty }B_{p,q}^{n}\bigl(f_{m};p ^{-(m+1)}q^{-(m+1)}\bigr) =f_{m}\bigl(p^{m+1}q^{-(m+1)}\bigr)- \frac{p^{-m}q^{m}[m+1]_{p,q}}{{p^{-1}(q-1)}[m]_{p,q}}. \end{aligned}$$
     
  2. (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
lim n C n m + j , n p , q C n m , n p , q = ( 1 ) j [ m j ] p , q p ( n j ) ( n j 1 ) 2 q j ( j 1 ) 2 .

Proof

Using (2.3) and (1.5), we get
C n m + j , n p , q = C n m , n p , q ( 1 p m [ n m ] p , q [ n ] p , q ) ( 1 p m j + 1 [ n m + j 1 ] p , q [ n ] p , q ) ( 1 p 1 q m [ n m + 1 ] [ n ] p , q ) ( 1 p j q m [ n m + j ] p , q [ n ] p , q ) , lim n C n m + j , n p , q C n m , n p , q = ( 1 p m q m ) ( 1 p m j + 1 q m j + 1 ) ( 1 q p ) ( 1 q j p j ) , lim n C n m + j , n p , q C n m , n p , q = ( 1 ) j [ m j ] p , q p ( n j ) ( n j 1 ) 2 q j ( j 1 ) 2 .
 □

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 = j = 0 m ( 1 ) j [ m j ] p , q p ( n j ) ( n j 1 ) 2 q 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. (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. (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].

Declarations

Funding

The third author would like to thank “Prince Sultan University for funding this work through research group Nonlinear Analysis Methods in Applied Mathematics (NAMAM)” group number RG-DES-2017-01-17.

Authors’ contributions

All the authors contributed equally and significantly in writing this paper. All of them read and approved the final manuscript.

Competing interests

The authors declare that they have no competing interests.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Department of Mathematics, Aligarh Muslim University, Aligarh, India
(2)
Department of Applied Mathematics, Aligarh Muslim University, Aligarh, India
(3)
Department of Mathematics and General Sciences, Prince Sultan University, Riyadh, Saudi Arabia

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