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 [1‐5, 11‐15, 19‐21] 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
where the polynomial \(p_{n,k}(p,q;x)\) is given by
For \(p=1\), \(B_{p,q}^{n} (f;x)\) turns into the q-Bernstein operator. We have
The following \((p,q)\)-difference form of Bernstein operators [25] is given by
where \(f[x_{0},x_{1},\ldots,x_{n}]\) indicates the nth order divided difference of f with pairwise distinct node, that is,
and \(\lambda _{p,q}^{n}\) is given by
and \(\lambda _{p,q}^{0}=\lambda _{p,q}^{1}=1\), \(0\leq \lambda _{p,q} ^{n}\leq 1\), \(r=0,1,\ldots,n\).
It can easily be verified by induction that
The \((p,q)\)-analog of Euler’s identity is defined by
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)
(1.2)
$$ B_{p,q}^{n}(f;0)=f(0), \qquad B_{p,q}^{n}(f;1)=f(1), \quad n \in \mathbb{N}. $$
(1.3)
$$ B_{p,q}^{n}(f;x):=\sum_{r=0}^{n} \lambda _{p,q}^{n} f \biggl[ 0,\frac{p ^{n-1}[1]_{p,q}}{[n]_{p,q}},\ldots, \frac{p^{n-r}[r]_{p,q}}{[n]_{p,q}} \biggr] x^{r}, $$
(1.4)
$$\begin{aligned}& f[x_{0}]=f(x_{0}), \qquad f[x_{0},x_{1}]= \frac{f(x_{1})-f(x_{0})}{x_{1}-x_{0}}, \\& f[x_{0},x_{1},\ldots,x_{n}]=\frac{f[x_{1},\ldots,x_{n}]-f[x_{0}, \ldots,x_{n-1}]}{[x_{n}-x_{0}]} \end{aligned}$$
(1.5)
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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 [27‐30] 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)
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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
\(m\in \mathbb{N}\),
Besides,
$$ \lim_{n\rightarrow \infty }B_{p,q}^{n}\bigl(f_{m};p^{j}q^{-j} \bigr)=f _{m}\bigl(p^{j}q^{-j}\bigr),\quad j\in \mathbb{N}_{0}\setminus \{m,m+1\}. $$
$$\begin{aligned}& \lim_{n\rightarrow \infty }B_{p,q}^{n}\bigl(f_{m};p^{m}q^{-m} \bigr)=- \infty, \quad \textit{and} \\& \lim_{n\rightarrow \infty }B_{p,q}^{n}\bigl(f_{m};p ^{-(m+1)}q^{-(m+1)}\bigr) =f_{m}\bigl(p^{m+1}q^{-(m+1)}\bigr)- \frac{p^{-m}q^{m}[m+1]_{p,q}}{{p^{-1}(q-1)}[m]_{p,q}}. \end{aligned}$$
(b)
For
\(m=0\),
i.e., \(B_{p,q}^{n}(f_{0};\cdot)\)
approximates
\(f_{0}\)
on
\(\mathbb{J}_{p,q}\).
$$ \lim_{n\rightarrow \infty }B_{p,q}^{n}\bigl(f_{0};p^{j}q^{-j} \bigr)=f _{0}\bigl(p^{j}q^{-j}\bigr), \quad j\in \mathbb{N}_{0} $$
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
Besides
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 \(k\neq m\).
$$ B_{p,q}^{n}\bigl(f;p^{-j}q^{j}\bigr)= \sum_{k=0}^{\min \{k,j\}}f \biggl( \frac{[n-k]_{p,q}}{[n]_{p,q}} \biggr) p_{n,n-k} \bigl(p,q;p^{j}q^{-j}\bigr). $$
(2.1)
$$ \lim_{n\rightarrow \infty }p_{n,n-k} \bigl(p,q;p^{j}q^{-j} \bigr)= \delta _{k,j} \quad \mbox{and} \quad \lim_{n\rightarrow \infty } \frac{[n-k]_{p,q}}{[n]_{p,q}}=p^{k}q^{-k} . $$
(2.2)
Now by easy calculation, we have
and combining with (2.1) and (2.2) yields the result.
$$ \lim_{n\rightarrow \infty }f \biggl( \frac{[n-k]_{p,q}}{[n]_{p,q}} \biggr) p_{n,n-k}\bigl(p,q;p^{j}q^{-j}\bigr)= \textstyle\begin{cases} -\infty ,& j=m, \\ -\frac{p^{-m}q^{m}[m+1]_{p,q}}{{p^{-1}(q-1)} [m]_{p,q}},& j=m+1, \\ 0,& \geq m+2, \end{cases} $$
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
where
\(\lambda _{k,n}^{p,q}\)
are given by (1.5).
$$ C_{k,n}^{p,q}=-\frac{\lambda _{k,n}^{p,q}p^{-m(k+1)}q^{m(k+1)}}{ ( 1-\frac{p^{n-m-1}q^{m}[1]_{p,q}}{[n]_{p,q}} ) ( 1-\frac{p ^{n-m-2}q^{m}[2]_{p,q}}{[n]_{p,q}} ) \cdots ( 1-\frac{p ^{n-m-k}q^{m}[k]_{p,q}}{[n]_{p,q}} ) }, $$
(2.3)
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
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
By the residue theorem
Since \(f_{m} (z)=f_{m} (x)\) for \(z=x\in {}[ 0,1]\), the statement follows from the divided difference representation (1.4). □
$$ f[x_{0},x_{1},\ldots,x_{k}]=\frac{1}{2\pi i} \oint _{\mathcal{L}}\frac{f( \zeta)\,d\zeta }{(\zeta -x_{0})(\zeta -x_{1})\cdots (\zeta -x_{k})}, $$
$$ f \biggl[ 0,\frac{p^{n-1}[1]_{p,q}}{[n]_{p,q}},\ldots, \frac{p^{n-r}[r]_{p,q}}{[n]_{p,q}} \biggr] = \frac{1}{2\pi i} \oint _{\mathcal{L}}\frac{f_{m} (\zeta)\,d\zeta }{\zeta (\zeta -\frac{p ^{n-1}[1]_{p,q}}{[n]_{p,q}})\cdots (\zeta - \frac{p^{n-r}[r]_{p,q}}{[n]_{p,q}})}. $$
(2.4)
$$\begin{aligned} f \biggl[ 0,\frac{p^{n-1}[1]_{p,q}}{[n]_{p,q}},\ldots, \frac{p^{n-r}[r]_{p,q}}{[n]_{p,q}} \biggr] =&\sum _{j=0}^{k}\operatorname{Res}_{z=p ^{n-j}\frac{[j]_{p,q}}{[n]_{p,q}}} \frac{f_{m} (z)}{\prod_{j=0}^{k} ( z-p^{n-j}\frac{[j]_{p,q}}{[n]_{p,q}} ) } \\ =&-\operatorname{Res}_{z=p^{m}q^{-m}}\frac{f_{m} (z)}{\prod_{j=0}^{k} ( z-p^{n-j}\frac{[j]_{p,q}}{[n]_{p,q}} ) } \\ =&-\frac{p^{-m(k+1)} q^{m(k+1)}}{\prod_{j=1}^{k} ( 1-p^{n-m-j}\frac{[j]_{p,q}}{[n]_{p,q}} ) }. \end{aligned}$$
Now, we find the asymptotic estimates for the coefficient \(C_{k,n} ^{p,q}\) in the next lemma.
Lemma 2.3
We have
for
\(j>m\), \(q>p>1\).
$$ \lim_{n\rightarrow \infty }\prod_{k=1}^{n-j} \biggl( 1-p^{n-m-j} q^{m}\frac{[k]_{p,q}}{[n]_{p,q}} \biggr) = \biggl( \frac{p^{2j-m}}{q ^{j-m}};\frac{p}{q} \biggr) _{\infty } $$
Proof
It is clear that
where
$$ \log \prod_{k=1}^{n-j} \biggl( 1-p^{n-m-j} q^{m} \frac{[k]_{p,q}}{[n]_{p,q}} \biggr) =\sum _{k=j}^{n-1}\log \biggl( 1-p ^{n-m-j} q^{m}\frac{[n-k]_{p,q}}{[n]_{p,q}} \biggr) =\sum_{k=j}^{ \infty }a_{k,n}^{p,q}, $$
$$ a_{k,n}^{p,q}= \textstyle\begin{cases} \log ( 1-p^{n-m-j} q^{m}\frac{[n-k]_{p,q}}{[n]_{p,q}} ),& k< n, \\ 0,& k\geq n. \end{cases} $$
Since
which gives \(\sum_{k=j}^{\infty } \vert a_{k,n}^{p,q} \vert < \infty \), and by the Lebesgue dominated convergence theorem, we have
as a result
which completes the proof. □
$$\begin{aligned} \bigl\vert a_{k,n}^{p,q} \bigr\vert \leq& \biggl\vert \log \biggl( 1-p ^{n-m-j} q^{m}\frac{[n-k]_{p,q}}{[n]_{p,q}} \biggr) \biggr\vert \\ \leq& \frac{q}{q-p}\frac{p^{n-m-k} q^{m} [n-k]_{p,q}}{[n]_{p,q}} \leq \frac{q}{q-p}p^{n-k-m} q^{m}, \end{aligned}$$
$$\begin{aligned} \lim_{n\rightarrow \infty }\sum_{k=j}^{n-1} \log \biggl( 1-p ^{n-m-j}q^{m}\frac{[n-k]_{p,q}}{[n]_{p,q}} \biggr) =&\sum _{k=j}^{ \infty } \biggl( \lim_{n\rightarrow \infty } \log \biggl( 1-p ^{n-m-j}q^{m}\frac{[n-k]_{p,q}}{[n]_{p,q}} \biggr) \biggr) \\ =&\sum_{k=j}^{\infty }\lim_{n\rightarrow \infty } \biggl( 1-p ^{n-k-m} \frac{q^{m}}{q^{k}} \biggr), \end{aligned}$$
$$ \lim_{n\rightarrow \infty }\log \prod_{k=1}^{n-j} \biggl( 1-p ^{n-m-j} q^{m} \frac{[n-k]_{p,q}}{[n]_{p,q}} \biggr) =\log \prod_{k=j}^{\infty } \biggl( 1-p^{n-k-m} \frac{q^{m}}{p^{k}} \biggr), $$
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
where
\(\mathcal{C}\)
in RHS is a positive constant, whose value need not to be addressed.
$$ \bigl\vert C_{k,n}^{p,q} \bigr\vert \leq \mathcal{C}_{m,p,q} p^{-mn} q^{mn}, $$
Proof
For \(n>m+1\) and from (2.3), we have
Further, we discuss the nature of \(C_{n-m+1,n},\ldots,C_{n,n}\) as follows. □
$$\begin{aligned}& \bigl\vert C_{k,n}^{p,q} \bigr\vert \leq \frac{p^{-m(k+1)}q^{m(k+1)}}{ ( 1-\frac{p^{n-m-1}q^{m}[1]_{p,q}}{[n]_{p,q}} ) ( 1-\frac{p ^{n-m-2}q^{m}[2]_{p,q}}{[n]_{p,q}} ) \cdots ( 1-\frac{p ^{n-m-k}q^{m}[k]_{p,q}}{[n]_{p,q}} ) } \\& \hphantom{ \big\vert C_{k,n}^{p,q}\big\vert } \leq \frac{p^{-m(n-m)}q^{m(n-m)}}{ ( 1-\frac{p^{2(n-m-1)}q ^{m}}{q^{n-1}} ) ( 1- \frac{p^{2(n-m-2)}q^{m}[2]_{p,q}}{[n]_{p,q}} ) \cdots ( 1-\frac{p ^{n-m-k}q^{m}[k]_{p,q}}{[n]_{p,q}} ) }, \\& \bigl\vert C_{k,n}^{p,q} \bigr\vert \leq \frac{p^{-mn}q^{mn}}{p^{2m}q ^{2m}(\frac{p}{q};\frac{p}{q})_{\infty }}. \end{aligned}$$
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
From Lemma 2.3, we have
The nature of the remaining coefficients \(C_{n-m+1,n},\ldots,C_{n,n}\) is given as follows. □
$$ \bigl\vert C_{n-m}^{p,q} \bigr\vert =\lambda _{n-m,n}^{p,q}\frac{p ^{-m(n-m+1)} q^{m(n-m+1)}}{ ( 1- \frac{p^{n-m-1}q^{m}[1]_{p,q}}{[n]_{p,q}} ) ( 1-\frac{p ^{n-m-2}q^{m}[2]_{p,q}}{[n]_{p,q}} ) \cdots ( 1-\frac{p ^{n-m-(n-m)}q^{m}[n-m]_{p,q}}{[n]_{p,q}} ) }. $$
$$\begin{aligned}& \begin{aligned}& \bigl\vert C_{n-m}^{p,q} \bigr\vert \sim \frac{ ( \frac{p^{m+1}}{q ^{m+1}};\frac{p}{q} ) _{\infty }q^{m(n-m+1)} p^{-m(n-m+1)} p ^{m}(p^{n}-q^{n})}{ ( \frac{p}{q};\frac{p}{q} ) _{\infty }p ^{n} (p^{m}-q^{m})} \\ &\hphantom{ \vert C_{n-m}^{p,q} \vert }\sim \frac{ ( \frac{p^{m+1}}{q^{m+1}};\frac{p}{q} ) _{ \infty }q^{mn} p^{-mn}p^{m} (p^{n}-q^{n})}{ ( \frac{p}{q}; \frac{p}{q} ) _{\infty }q^{m(m-1)}p^{m(m-1)}(p^{m}-q^{m})}, \\ &C_{n-m,n}^{p,q}=\mathcal{C}_{m}^{p,q} p^{-n(m+1)}q^{n(m+1)}. \end{aligned} \end{aligned}$$
(2.5)
Lemma 2.6
For
\(j=1,2,\ldots,m\), we have
Corollary 2.7
The following estimate holds:
and
\(\mathcal{C}_{p,q,m}\)
is independent of both
k
and
n.
$$ \bigl\vert C_{k,n}^{p,q} \bigr\vert \leq \mathcal{C}_{p,q,m} p^{-(m+1)n}q ^{(m+1)n},\quad k=0,1,2, \ldots,n, $$
(2.6)
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],
□
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
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
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:
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. □
$$ B_{p,q}^{n}(f;x)=\sum_{k=0}^{n}f \biggl( \frac{[k]_{p,q}}{p^{k-n}[n]_{p,q}} \biggr) p_{n,k} (p, q;x),\quad n\in \mathbb{N},z \in \mathbb{C}, $$
(3.1)
$$ B_{p,q}^{n}(f_{m};z)=\sum _{k=0}^{n}C_{k,n}^{p,q}z^{k}, $$
$$ \bigl\vert B_{p,q}^{n}(f_{m};z) \bigr\vert \leq \sum_{k=0}^{n} \bigl\vert C _{k,n}^{p,q}\rho ^{k} \bigr\vert \leq \mathcal{C}_{p,q,m}\sum_{k=0} ^{n} \bigl(p^{-(m+1)}q^{(m+1)}\rho\bigr)^{k}\leq \mathcal{C}_{p,q,m}\frac{1}{(1-p ^{-(m+1)}q^{(m+1)}\rho)}. $$
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
Hence
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 \(x\notin \{ p^{(m+1)}q^{-(m+1)},\ldots,1 \} \). □
$$\begin{aligned} \Biggl\vert \sum_{k=0}^{\infty }C_{k,n}^{p,q}x^{k} \Biggr\vert \leq& \mathcal{C}_{m,p,q} \sum _{k=0}^{n-m-1}p^{-mk}q^{mk}x^{k}= \mathcal{C} _{m,p,q}\frac{(p^{-m}q^{m}x)^{n-m}-1}{p^{-m}q^{m}x-1} \\ =&o \bigl( \bigl(p^{-(m+1)}q^{(m+1)}x\bigr)^{n} \bigr), \quad n\rightarrow \infty. \end{aligned}$$
$$\begin{aligned} B_{p,q}^{n}(f_{m};x) =&\sum _{k=n-m}^{n}C_{k,n}^{p,q}x^{k}+o \bigl( \bigl(p ^{-(m+1)}q^{(m+1)}x\bigr)^{n} \bigr) \\ =&C_{n-m}^{p,q}x^{n-m}g_{n}(x)+o \bigl( \bigl(p^{-(m+1)}q^{(m+1)}x\bigr)^{n} \bigr),\quad n \rightarrow \infty. \end{aligned}$$
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
since by (1.3), \(B_{p,q}^{n}(f_{0};1)=f_{0}(1)=a\) and the statement is proved. □
$$ C_{k,n}^{p,q}=\frac{-\lambda _{k,n}^{p,q}}{\prod_{j=0}^{k} ( 1-p ^{n-j}\frac{[j]_{p,q}}{[n]} ) }=\frac{-1}{ ( 1-p^{n-k} \frac{[k]_{p,q}}{[n]_{p,q}} ) }, $$
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
\(x\in (-1,1]\),
and the convergence is uniform on any compact subset of
\((-1,1)\).
$$ \lim_{n\rightarrow \infty } B^{n}_{p,q}(f_{0};x)= f_{0}(x) $$
(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
Apart from that,
whence
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.
$$ \Biggl\vert \sum_{k=0}^{n-1}C_{k,n}^{p,q}z^{k} \Biggr\vert \leq \frac{ \mathcal{C}_{p,q}}{1-\rho }. $$
$$ \bigl\vert C_{n,n}^{p,q}z^{k} \bigr\vert \leq \vert a \vert + \sum_{k=0}^{n-1} \frac{1}{1-p^{n-k}\frac{[k]_{p,q}}{[n]_{p,q}}}\leq \vert a \vert +n \bigl\vert C_{k,n}^{p,q} \bigr\vert \leq \vert a \vert +\frac{nq}{q-p}, $$
$$ \bigl\vert C_{n,n}^{p,q}z^{n} \bigr\vert \leq \vert a \vert \rho ^{n}+\rho ^{n}\frac{nq}{q-p}\leq \mathcal{C}_{p,q,\rho }. $$
(2) Given that x satisfies \(\vert x \vert >1\), by Able’s inequality, we have
Meanwhile,
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 \(n\rightarrow \infty \).
$$ \Biggl\vert \sum_{k=0}^{n-1}C_{k,n}^{p,q}x^{k} \Biggr\vert \leq \frac{ \vert x \vert ^{n}-1}{ \vert x \vert -1}\bigl( \bigl\vert C _{0,n}^{p,q} \bigr\vert +2 \bigl\vert C_{n-1,n}^{p,q} \bigr\vert \bigr) \leq \frac{ \vert x \vert ^{n}}{ \vert x \vert -1} \biggl( 1+\frac{2p}{p-q} \biggr) = \mathcal{C}_{p,q,x} \vert x \vert ^{n}. $$
$$ \bigl\vert C_{n,n}^{p,q}x^{n} \bigr\vert \geq \Biggl( \sum_{k=0} ^{n-1}\frac{1}{1-p^{n-k}\frac{[k]_{p,q}}{[n]_{p,q}}} \Biggr) \vert x \vert ^{n}- \vert a \vert \cdot \vert x \vert ^{n}\geq \bigl(n- \vert a \vert \bigr) \vert x \vert ^{n}. $$
At \(x=-1\), we have
and again applying Able’s inequality,
On the other hand
which implies that
□
$$ B_{p,q}^{n}(f_{0};-1)=\sum _{k=0}^{n-1}C_{k,n}^{p,q}(-1)^{k}+ \Biggl( a+ \sum_{k=0}^{n-1}\frac{1}{1-p^{n-k}\frac{[k]_{p,q}}{[n]_{p,q}}} \Biggr) (-1)^{n}, $$
$$ \Biggl\vert \sum_{k=0}^{n-1}C_{k,n,}^{p,q}(-1)^{k} \Biggr\vert \leq \vert C_{0,n} \vert +2 \vert C_{n-1,n} \vert \leq 1+\frac{2p}{p-q}. $$
$$ \Biggl\vert \Biggl( a+\sum_{k=0}^{n-1} \frac{1}{1-p^{n-k} \frac{[k]_{p,q}}{[n]_{p,q}}} \Biggr) (-1)^{n} \Biggr\vert \geq n- \vert a \vert , $$
$$ \bigl\vert B_{p,q}^{n}(f_{0};-1) \bigr\vert \geq n- \vert a \vert - \biggl( 1+\frac{2p}{p-q} \biggr) \rightarrow \infty,\quad n\rightarrow \infty. $$
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].
Competing interests
The authors declare that they have no competing interests.
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