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A basic problem of \((p,q)\)-Bernstein-type operators
Journal of Inequalities and Applications volume 2017, Article number: 140 (2017)
Abstract
In this note, we give an elaboration of a basic problem on convergence theorem of \((p, q)\)-analogue of Bernstein-type operators. By some classical analysis techniques, we derive an exact class of \((p_{n},q_{n})\)-integer satisfying \(\lim _{n\to\infty }[n]_{p_{n},q_{n}}=\infty\) with \(\lim _{n\to\infty}p_{n}=1\) and \(\lim _{n\to\infty}q_{n}=1\) under \(0< q_{n}< p_{n}\leq1\). Our results provide an erratum to corresponding results on \((p,q)\)-analogue of Bernstein-type operators that appeared in recent literature.
1 Introduction
During the last decades, the applications of q-calculus emerged as a new area in the field of approximation theory. The rapid development of q-calculus has led to the discovery of various generalizations of Bernstein polynomials involving q-integers. A detailed review of the results on q-Bernstein polynomials along with an extensive bibliography is given in [1]. The q-Bernstein polynomials are shown to be closely related to the q-deformed binomial distribution [2]. It plays an important role in the q-boson theory giving a q-deformation of the quantum harmonic formalism [3]. The q-analogue of the boson operator calculus has proved to be a powerful tool in theoretical physics. It provides explicit expressions for the representations of the quantum group \(SU_{q}(2)\) [4]. Meanwhile, the \((p,q)\)-integers were introduced in order to generalize or unify several forms of q-oscillator algebras well known in the earlier physics literature related to the representation theory of single parameter quantum algebras [5].
Recently, \((p,q)\)-integers have been introduced into classical linear positive operators to construct new approximation processes. A sequence of \((p,q)\)-analogue of Bernstein operators was first introduced by Mursaleen [6, 7]. Besides, \((p,q)\)-analogues of Szász-Mirakyan [8], Baskakov Kantorovich [9], Bleimann-Butzer-Hahn [10] and Kantorovich-type Bernstein-Stancu-Schurer [11] operators were also considered, see [12–15]. For further developments, one can also refer to [8, 16–18]. These operators are double parameters corresponding to p and q versus single parameter q-Bernstein-type operators [1, 19, 20]. The aim of these generalizations is to provide appropriate and powerful tools to application areas such as numerical analysis, computer-aided geometric design and solutions of differential equations (see, e.g., [21]).
For example, consider the \((p,q)\)-analogue of the Bernstein operators proposed in [7]. Given \(f\in C[0,1]\) and \(0< q< p\leq1\), operators \(B_{n,p,q}\) are defined as follows:
where, for any nonnegative integer k and \(0< q< p\leq1\), the \((p,q)\)-integer \([k]_{p,q}\) is defined by
and the \((p,q)\)-factorial \([k]_{p,q}!\) is defined by
For integers k, n with \(0\leq k\leq n\), the \((p,q)\)-binomial coefficient is defined by
In general, we expect \(B_{n,p,q}(f;x)\) to converge to \(f(x)\) as \(n\to \infty\). But we see that, for fixed value of p and q with \(q\in (0,1)\) and \(p\in(q,1]\),
To obtain a sequence of generalized \((p,q)\)-analogue Bernstein polynomials which converge, we let \(q_{n}\in(0,1)\) and \(p_{n}\in(q_{n},1]\) depend on n. We then choose a sequence \((p_{n},q_{n})\) such that \([n]_{p_{n},q_{n}}\to \infty\) as \(n\to\infty\), to ensure that \(B_{n,p,q}(f;x)\) converge to \(f(x)\).
The convergence theorems for \((p,q)\)-analogue Bernstein-type operators were established in some recent papers (see [6], Theorem 3.1 (Remark 3.1), [7], Theorem 1, and further reading [10], Theorem 2.2, [14], Theorem 3.1, [12], Theorem 3, and [11], Remark 2.3, see also [9, 15]). For example, Mursaleen [7] gives the following.
Theorem 1.1
Let \(0< q_{n}< p_{n}\leq1\) such that \(\lim _{n\rightarrow\infty }p_{n}=1\) and \(\lim _{n\rightarrow\infty}q_{n}=1\). Then, for each \(f\in C[0,1]\), \(B_{n,p_{n},q_{n}}(f;x)\) converge uniformly to f on \([0,1]\).
All linear positive operators mentioned in the articles cited above require that \(\lim _{n\to\infty} [n]_{p_{n},q_{n}}=\infty\); otherwise, these operators do not define approximation processes. However, the claim that both \(\lim _{n\to\infty}p_{n}=1\) and \(\lim _{n\to\infty}q_{n}=1\) with \(0< q_{n}< p_{n}\leq1\) imply that \(\lim _{n\to\infty} [n]_{p_{n},q_{n}}=\infty\), in general, is not true. A counterexample is presented below.
Example 1.2
Let \(p_{n}=1-1/\sqrt{n}\), then \([n]_{p_{n},q_{n}}\to0\) for any sequence \(\{ q_{n}\}\) satisfying \(0<{q_{n}<p_{n}}\). Indeed,
Later, the author [8] presented a more accurate assertion: Let \(q_{n}\), \(p_{n}\) such that \(0< q_{n}< p_{n}\leq1\) and \(q_{n}\to1\), \(p_{n}\to1\), \(q^{n}_{n}\to a\), \(p^{n}_{n}\to b\) (\(a< b\)) as \(n\to\infty\), then \(\lim _{n\to \infty}[n]_{p_{n},q_{n}}\to\infty\). It is natural to ask: What is the class of sequences \((p_{n},q_{n})\) satisfying \(\lim _{n\rightarrow \infty}[n]_{p_{n},q_{n}}=\infty\) when \(\lim _{n\rightarrow\infty }p_{n}=1\) and \(\lim _{n\rightarrow\infty}q_{n}=1\) under \(0< q_{n}< p_{n}\leq1\)? Undoubtedly, this is an important problem. In this note, we will solve this problem in Section 2.
2 Main results
For \(0< q_{n}< p_{n}\leq1\), set \(q_{n}:=1-\alpha_{n}\), \(p_{n}:=1-\beta_{n}\) such that \(0\leq\beta_{n}<\alpha_{n}<1\), \(\alpha_{n}\rightarrow0\), \(\beta _{n}\rightarrow0\) as \(n\rightarrow\infty\). In the sequel, we use notation \(a_{n} \sim b_{n}\text{ $(a_{n},b_{n}>0)$}\Leftrightarrow\lim _{n\to \infty}\frac{b_{n}}{a_{n}}=1\).
First, let us present the following auxiliary proposition.
Lemma 2.1
Let \(n\in\mathbb{N}\), then as \(n\to\infty\) we have
On the other hand,
Proof
We note that
Similarly, we have
Therefore, from (2.3) and (2.4), for sufficiently large n, there exist two positive real numbers \(C_{1}\) and \(C_{2}\) satisfying
This yields the proof. □
The main result of this work is expressed by the next assertion.
Theorem 2.1
The following statements are true:
- (\(\mathcal{A}\)):
-
If \(\lim _{n\to\infty}e^{n(\beta _{n}-\alpha_{n})}=1\) and \(e^{n\beta_{n}}/n\to0\), then \([n]_{p_{n},q_{n}}\to \infty\).
- (\(\mathcal{B}\)):
-
If \(\varlimsup _{n\to\infty }e^{n(\beta_{n}-\alpha_{n})}<1\) and \(e^{n\beta_{n}}(\alpha_{n}-\beta_{n})\to 0\), then \([n]_{p_{n},q_{n}}\to\infty\).
- (\(\mathcal{C}\)):
-
If \(\varliminf _{n\to\infty }e^{n(\beta_{n}-\alpha_{n})}<1\), \(\varlimsup _{n\to\infty }e^{n(\beta_{n}-\alpha_{n})}=1\) and \(\max\{e^{n\beta_{n}}/n,e^{n\beta _{n}}(\alpha_{n}-\beta_{n})\}\to0\), then \([n]_{p_{n},q_{n}}\to\infty\).
Conversely,
- (\(\mathcal{B'}\)):
-
If \(\varlimsup _{n\to\infty }e^{n(\beta_{n}-\alpha_{n})}<1\) and \([n]_{p_{n},q_{n}}\to\infty\), then \(e^{n\beta_{n}}(\alpha_{n}-\beta_{n})\to0\).
- (\(\mathcal{C'}\)):
-
If \(\varliminf _{n\to\infty }e^{n(\beta_{n}-\alpha_{n})}<1,\varlimsup _{n\to\infty }e^{n(\beta_{n}-\alpha_{n})}=1\) and \([n]_{p_{n},q_{n}}\to\infty\), then \(\max\{e^{n\beta_{n}}/n, e^{n\beta_{n}}(\alpha_{n}-\beta_{n})\}\to0\).
Proof
Case (\(\mathcal{A}\)):
If \(\lim _{n\to\infty}e^{n(\beta_{n}-\alpha_{n})}=1\), since \(\lim _{n\to\infty}e^{n\alpha_{n}}/n=\lim _{n\to \infty}e^{n\beta_{n}}/n/e^{n(\beta_{n}-\alpha_{n})}\to0\) and combined with (2.2) imply \([n]_{p_{n},q_{n}}\to\infty\) (see Remark 2.1).
Case (\(\mathcal{B}\)) and Case (\(\mathcal{B'}\)):
If \(\varlimsup _{n\to\infty}e^{n(\beta_{n}-\alpha_{n})}<1\). Note that, for sufficiently large n,
Since
it is not difficult to obtain from (2.7) and \(\varlimsup _{n\to\infty}e^{n(\beta_{n}-\alpha_{n})}<1\) that, for sufficiently large n, there exists \(c\in(0,1)\) such that
Set \(d_{n}:=1- (1-\frac{\alpha_{n}-\beta_{n}}{1-\beta_{n}} )^{n}\), then for sufficiently large n, \(d_{n}>1-c\). Thus, from (2.6), (2.7) and (2.8), we have
which entails that
This yields the proof of Case (\(\mathcal{B}\)) and Case (\(\mathcal{B'}\)).
Case (\(\mathcal{C}\)) and Case (\(\mathcal{C'}\)):
If \(\varliminf _{n\to\infty}e^{n(\beta_{n}-\alpha _{n})}<1\), \(\varlimsup _{n\to\infty}e^{n(\beta_{n}-\alpha_{n})}=1\). Since the sequence \(0< x_{n}:=e^{n(\beta_{n}-\alpha_{n})}<1\) is bounded, set \(E:=\{x|x\text{ is a limit point of }\{x_{n}\},n\geq1\}\), then \(\sup E=1\) from \(\varlimsup _{n\to\infty}e^{n(\beta_{n}-\alpha _{n})}=1\). Now, we are going to extract a subsequence \(\{x_{n_{k}}\}\) of \(\{ x_{n}\}\) such that
-
(a)
\(\lim _{k\to\infty}x_{n_{k}}=1\), and
-
(b)
\(E^{1}:=\{x|x\text{ is a limit point of }\{x_{n}\}\setminus\{x_{n_{k}}\}\}\), with \(\sup E^{1}<1\).
We verify that it is possible to extract such a subsequence \(\{x_{n_{k}}\} \). Since 1 is a limit point of \(A:=\{x_{n},n\geq1\}\), take a subsequence \(\{x_{n^{(1)}_{k}}\}\) of A such that \(\lim _{k\to \infty}x_{n^{(1)}_{k}}=1\), set \(A^{(1)}:=\{x_{n^{(1)}_{k}},k\geq1\}\) and let \(A_{1}:=A\setminus A^{(1)}\). If 1 is also a limit point of \(A_{1}\), take a subsequence \(\{ x_{n^{(2)}_{k}}\}\) of \(A_{1}\) such that \(\lim _{k\to\infty}x_{n^{(2)}_{k}}=1\), set \(A^{(2)}:=\{x_{n^{(2)}_{k}},k\geq1\}\) and let \(A_{2}:=A_{1}\setminus A^{(2)}\). Continuing this process, we obtain a series of sequences, i.e., \(A^{(1)},A^{(2)},\ldots\) , set \(A_{f}:=A\setminus\bigcup _{s\in\mathbb{I}\subseteq\mathbb{N}}A^{(s)}\). Since A is a countable set, this process will stop until 1 is not a limit point of \(A_{f}\) after finite or countable steps; otherwise, we will see that 1 is the only limit point of A, which contradicts to the assumptions of Case (\(\mathcal{C}\)). Then we can take the subsequence \(\{x_{n_{k}}\}=\bigcup _{s\in\mathbb {N}}A^{(s)}\) which satisfies (a) and (b).
Set \(\{x_{n'_{k}}\}:=\{x_{n}\}\setminus\{x_{n_{k}}\}\), it is obvious that \(\{ n_{k},k\geq0\}\cup\{n'_{k},k\geq0\}=\mathbb{N}\). Then from (\(\mathcal{A}\)) and (a), we have seen that \([n]_{p_{n_{k}},q_{n_{k}}}\to\infty\Leftrightarrow e^{n_{k}\beta _{n_{k}}}/n_{k}\to0\) as \(k\to\infty\). And since \(\varlimsup _{n\to\infty}x_{n'_{k}}<1\) from (b), we also have seen from (\(\mathcal{B}\)) that \([n]_{p_{n'_{k}},q_{n'_{k}}}\to\infty\Leftrightarrow e^{n'_{k}\beta _{n'_{k}}}(\alpha_{n'_{k}}-\beta_{n'_{k}})\to0\) as \(k\to\infty\). In summary,
-
(i)
\(\lim _{n\to\infty}[n]_{p_{n},q_{n}}=\infty\Leftrightarrow\) for any subsequence \(\{N_{k}\}\) of a natural number set such that \(\lim _{k\to\infty}N_{k}=\infty\), we have \(\lim _{k\to \infty}[n]_{p_{N_{k}},q_{N_{k}}}=\infty\).
-
(ii)
Since \(\{n_{k}\}_{k\geq0}\) and \(\{n'_{k}\}_{k\geq0}\) are two subsequences of a natural number set such that \(\lim _{k\to\infty}n_{k}=\infty\) and \(\lim _{k\to \infty}n'_{k}=\infty\), thus as \(k\to\infty\)
$$\lim _{n\to\infty}[n]_{p_{n},q_{n}}=\infty\quad\Rightarrow\quad \textstyle\begin{cases} [n]_{p_{n_{k}},q_{n_{k}}}\to\infty\quad\Leftrightarrow\quad e^{n_{k}\beta_{n_{k}}}/n_{k}\to0, \\ [n]_{p_{n'_{k}},q_{n'_{k}}}\to\infty\quad\Leftrightarrow\quad e^{n'_{k}\beta _{n'_{k}}}(\alpha_{n'_{k}}-\beta_{n'_{k}})\to0. \end{cases} $$
We assert that the inverse proposition of (ii) also holds. Indeed, for each sufficiently large \(N>0\), there exists a positive integer \(K_{1}\) such that for every natural number \(k>K_{1}\), we have \([n]_{p_{n_{k}},q_{n_{k}}}>N\); meanwhile, for the previous \(N>0\), there exists a positive integer \(K_{2}\) such that for every natural number \(k>K_{2}\), we have \([n]_{p_{n'_{k}},q_{n'_{k}}}>N\); take \(n_{0}=\max\{ n_{K_{1}},n'_{K_{2}}\}\), for every natural number \(n>n_{0}\), we have \([n]_{p_{n},q_{n}}>N\), i.e., \(\lim _{n\to\infty }[n]_{p_{n},q_{n}}=\infty\). Now we have proved that as \(k\to\infty\)
Next, we infer that as \(k\to\infty\)
‘⇐’ of (2.12) is straightforward. Now we show ‘⇒’. On the one hand, by Remark (2.2), we know from (d) of Δ that \(e^{n'_{k}\beta_{n'_{k}}}/n'_{k} \to0\), and combined with (c) \(e^{n_{k}\beta_{n_{k}}}/n_{k}\to0\), we can show \(e^{n\beta_{n}}/n\to0\) by using a similar method as in the previous paragraph. On the other hand, \(e^{n_{k}\beta_{n_{k}}}(\alpha_{n_{k}}-\beta_{n_{k}})\to0\) is straightforward (since \(\lim _{k\to\infty}x_{n_{k}}=\lim _{k\to\infty}e^{n_{k}(\beta_{n_{k}}-\alpha_{n_{k}})}=1\), and note (c) in Δ), and combined with \(e^{n'_{k}\beta_{n'_{k}}}(\alpha _{n'_{k}}-\beta_{n'_{k}})\to0\), we can also deduce that \(e^{n\beta _{n}}(\alpha_{n}-\beta_{n})\to0\). This yields the proof of ‘⇒’ in (2.12).
Therefore, (\(\mathcal{C}\)) and (\(\mathcal{C'}\)) follow from (2.11) and (2.12). □
Remark 2.1
In general, from (2.1) we have only \([n]_{p_{n},q_{n}}\to\infty \Rightarrow e^{n\beta_{n}}/n\to0\). However, if \(\lim _{n\to \infty}e^{n(\beta_{n}-\alpha_{n})}=1\) holds, then we have the equivalent relation \(e^{n\beta_{n}}/n\to0\Leftrightarrow[n]_{p_{n},q_{n}}\to\infty\) from (\(\mathcal{A}\)). Thus, we have:
- (\(\mathcal{A}_{0}\)):
-
If \(\lim _{n\to\infty}e^{n(\beta _{n}-\alpha_{n})}=1\), then \(e^{n\beta_{n}}/n\to0\Leftrightarrow [n]_{p_{n},q_{n}}\to\infty\).
Similarly, from (\(\mathcal{B}\)) and (\(\mathcal {B'}\)), (\(\mathcal{C}\)) and (\(\mathcal{C'}\)) we have
- (\(\mathcal{B}_{0}\)):
-
If \(\varlimsup _{n\to\infty }e^{n(\beta_{n}-\alpha_{n})}<1\), then \(e^{n\beta_{n}}(\alpha_{n}-\beta _{n})\to0\Leftrightarrow[n]_{p_{n},q_{n}}\to\infty\).
- (\(\mathcal{C}_{0}\)):
-
If \(\varliminf _{n\to\infty }e^{n(\beta_{n}-\alpha_{n})}<1\), \(\varlimsup _{n\to\infty }e^{n(\beta_{n}-\alpha_{n})}=1\), then \(\max\{e^{n\beta_{n}}/n,e^{n\beta _{n}}(\alpha_{n}-\beta_{n})\}\to0\Leftrightarrow[n]_{p_{n},q_{n}}\to\infty\).
Remark 2.2
In Case (\(\mathcal{B}\)), we can also deduce directly from \(\varlimsup _{n\to\infty}e^{n(\beta_{n}-\alpha_{n})}<1\) and \(e^{n\beta_{n}}(\alpha_{n}-\beta_{n})\to0\) that \(e^{n\beta_{n}}/n\to 0\) as \(n\to\infty\). Indeed, since \(\varlimsup _{n\to\infty }e^{n(\beta_{n}-\alpha_{n})}<1\), and combined with the classical inequality on upper (lower) limit
we have \(\varliminf _{n\to\infty}e^{n(\alpha_{n}-\beta _{n})}>1\). Thus, for sufficiently large n, there exists \(c_{0}>1\) such that \(e^{n(\alpha_{n}-\beta_{n})}>c_{0}\). This means that \(0<(\log c_{0}) e^{n\beta_{n}}/n<e^{n\beta_{n}}/n\cdot(n(\alpha_{n}-\beta_{n}))\to0\), and we have seen that \(e^{n\beta_{n}}/n\to0\).
Remark 2.3
Now we utilize Theorem 2.1 (Remark 2.1) to elaborate Example 1.2 again. In the example, \(\beta_{n}=1/\sqrt {n}\), while \(e^{n\beta_{n}}/n=e^{\sqrt{n}}/n \nrightarrow0\) as \(n\to \infty\), thus \([n]_{p_{n},q_{n}}\nrightarrow\infty\).
For \(p_{n}=1\), \(\beta_{n}=0\), \(q_{n}=1-\alpha_{n}\), then \(e^{n\beta_{n}}/n=1/n\to0\) and \(e^{n\beta_{n}}(\alpha_{n}-\beta_{n})=\alpha_{n}\to0\). Thus any case of (\(\mathcal{A}_{0}\))-(\(\mathcal{C}_{0}\)) is straightforward. In this case, \((p_{n}, q_{n})\)-integer reduces to \(q_{n}\)-integer, and it is known that \([n]_{q_{n}}\to\infty \Leftrightarrow q_{n}\to1\) as \(n\to\infty\). See [19], Theorem 2, and [22], formula (2.7).
3 Conclusion
In this note, we mainly obtain the sufficient and necessary conditions for \((p,q)\)-integer \([n]_{p,q}\) tending to infinity as \(n\to\infty\). The conclusion guarantees the \((p,q)\)-analogue of Bernstein-type operators to be approximation processes as \(n \to\infty\).
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Acknowledgements
The paper has been greatly improved by the efforts of the anonymous referees through their diligent reading of and perceptive comments on an initial draft.
This work is supported by the National Natural Science Foundation of China (Grant No. 61572020,11601266), the China Postdoctoral Science Foundation funded project (Grant No. 2015M582036) and the Natural Science Foundation of Fujian Province of China (Grant No. 2016J05017). We also thank Fujian Provincial Key Laboratory of Data Intensive Computing and Key Laboratory of Intelligent Computing and Information Processing of Fujian Province University.
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XWX carried out the proof of the main results. QBC read and approved the final manuscript.
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Cai, QB., Xu, XW. A basic problem of \((p,q)\)-Bernstein-type operators. J Inequal Appl 2017, 140 (2017). https://doi.org/10.1186/s13660-017-1413-0
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DOI: https://doi.org/10.1186/s13660-017-1413-0