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Some approximation properties of \((p,q)\)-Bernstein operators


This paper is concerned with the \((p,q)\)-analog of Bernstein operators. It is proved that, when the function is convex, the \((p,q)\)-Bernstein operators are monotonic decreasing, as in the classical case. Also, some numerical examples based on Maple algorithms that verify these properties are considered. A global approximation theorem by means of the Ditzian-Totik modulus of smoothness and a Voronovskaja type theorem are proved.

Introduction and preliminaries

During the last decade, the applications of q-calculus in the field of approximation theory has led to the discovery of new generalizations of classical operators. Lupaş [1] was first to observe the possibility of using q-calculus in this context. For more comprehensive details the reader should consult monograph of Aral et al. [2] and the recent references [39].

Nowadays, the generalizations of several operators in post-quantum calculus, namely the \((p,q)\)-calculus have been studied intensively. The \((p,q)\)-calculus has been used in many areas of sciences, such as oscillator algebra, Lie group theory, field theory, differential equations, hypergeometric series, physical sciences (see [10, 11]). Recently, Mursaleen et al. [12] defined \((p,q)\)-analog of Bernstein operators. The approximation properties for these operators based on Korovkin’s theorem and some direct theorems were considered. Also, many well-known approximation operators have been introduced using these techniques, such as Bleimann-Butzer-Hahn operators [13] and Szász-Mirakyan operators [14].

In the present paper, we prove new approximation properties of \((p,q)\)-analog of Bernstein operators. First of all, we recall some notations and definitions from the \((p,q)\)-calculus. Let \(0< q< p\leq1\). For each non-negative integer \(n\geq k\geq0\), the \((p,q)\)-integer \([k]_{p,q}\), \((p,q)\)-factorial \([k]_{p,q}!\), and \((p,q)\)-binomial are defined by

$$\begin{aligned}& [k]_{p,q}:= \frac{p^{k}-q^{k}}{p-q}, \\& [k]_{p,q}!:= \textstyle\begin{cases} [k]_{p,q}[k-1]_{p,q}\cdots[1]_{p,q}, & k\geq1, \\ 1,&k=0, \end{cases}\displaystyle \end{aligned}$$


$$\left[ \textstyle\begin{array}{@{}c@{}} n\\ k \end{array}\displaystyle \right]_{p,q}:= \frac{[n]_{p,q}!}{[n-k]_{p,q}![k]_{p,q}!}. $$

As a special case when \(p=1\), the above notations reduce to q-analogs.

The \((p,q)\)-power basis is defined as

$$(x\ominus a)_{p,q}^{n}=(x-a) (px-qa) \bigl(p^{2}x-q^{2}a \bigr)\cdots\bigl(p^{n-1}x-q^{n-1}a\bigr). $$

The \((p,q)\)-derivative of the function f is defined as

$$D_{p,q}f(x)= \frac{f(px)-f(qx)}{(p-q)x}, \quad x\ne0. $$

Let f be an arbitrary function and \(a\in\mathbb{R}\). The \((p,q)\)-integral of f on \([0,a]\) is defined as

$$\begin{aligned}& \int_{0}^{a} f(t)d_{p,q} t=(q-p)a\sum _{k=0}^{\infty}f \biggl( \frac {p^{k}}{q^{k+1}}a \biggr)\frac{p^{k}}{q^{k+1}},\quad \text{if } \biggl\vert \frac{p}{q}\biggr\vert < 1, \\& \int_{0}^{a} f(t)d_{p,q} t=(p-q)a\sum _{k=0}^{\infty}f \biggl( \frac {q^{k}}{p^{k+1}}a \biggr)\frac{q^{k}}{p^{k+1}},\quad \text{if } \biggl\vert \frac{q}{p}\biggr\vert < 1. \end{aligned}$$

The \((p,q)\)-analog of Bernstein operators for \(x\in[0,1]\) and \(0< q< p\leq1\) are introduced as follows:

$$B_{n}^{p,q}(f;x)= \sum_{k=0}^{n} b_{n,k}^{p,q}(x)f \biggl( \frac{p^{n-k}[k]_{p,q}}{[n]_{p,q}} \biggr), $$

where the \((p,q)\)-Bernstein basis is defined as

$$b_{n,k}^{p,q}(x)= \left[ \textstyle\begin{array}{@{}c@{}}n\\ k \end{array}\displaystyle \right]_{p,q}p^{[k(k-1)-n(n-1)]/2}x^{k}(1\ominus x)_{p,q}^{n-k}. $$

Lemma 1.1

For \(x\in[0,1]\), \(0< q< p\leq1\), we have

$$\begin{aligned}& B_{n}^{p,q} (e_{0};x ) =1,\qquad B_{n}^{p,q} (e_{1};x )=x, \\& B_{n}^{p,q} (e_{2};x ) = \frac {p^{n-1}}{[n]_{p,q}}x+ \frac{q[n-1]_{p,q}}{[n]_{p,q}}x^{2}, \end{aligned}$$

where \(e_{i}(x)=x^{i}\) and \(i\in\{0,1,2\}\).

Lemma 1.2

Let n be a given natural number, then

$$B_{n}^{p,q} \bigl((t-x)^{2};x \bigr)= \frac {p^{n-1}}{[n]_{p,q}}\phi^{2}(x)\leq\frac{1}{[n]_{p,q}}\phi^{2}(x), $$

where \(\phi(x)=\sqrt{x(1-x)}\) and \(x\in[0,1]\).

Monotonicity for convex functions

Oru and Phillips [15] proved that when the function f is convex on \([0,1]\), its q-Bernstein operators are monotonic decreasing. In this section we will study the monotonicity of \((p,q)\)-Bernstein operators.

Theorem 2.1

If f is convex function on \([0,1]\), then

$$B_{n}^{p,q}(f;x)\geq f(x),\quad 0\leq x\leq1, $$

for all \(n\geq1\) and \(0< q< p\leq1\).


We consider the knots \(x_{k}= \frac{p^{n-k}[k]_{p,q}}{[n]_{p,q}}\), \(\lambda_{k}=\bigl [ \scriptsize{\begin{array}{@{}c@{}} n\\ k \end{array}} \bigl ]_{p,q}p^{[k(k-1)-n(n-1)]/2}x^{k}(1\ominus x)_{p,q}^{n-k}\), \(0\leq k\leq n\).

Using Lemma 1.1, it follows that

$$\begin{aligned}& \lambda_{0}+\lambda_{1}+\cdots+\lambda_{n}=1, \\& x_{0}\lambda _{0}+x_{1}+ \lambda_{1}+\cdots+x_{n}\lambda_{n}=x. \end{aligned}$$

From the convexity of the function f, we get

$$B_{n}^{p,q}(f;x)= \sum_{k=0}^{n} \lambda_{k}f(x_{k})\geq f \Biggl(\sum _{k=0}^{n}\lambda_{k} x_{k} \Biggr)=f(x). $$


Example 2.2

Let \(f:\mathbb{R}\to\mathbb{R}\), \(f(x)=xe^{x+1}\). Figure 1 illustrates that \(B_{n}^{p,q}(f;x)\geq f(x)\) for the convex function f and \(x\in[0,1]\).

Figure 1
figure 1

Approximation process by \(\pmb{B_{n}^{p,q}(f;x)}\) for \(\pmb{f(x)=xe^{x+1}}\) .

Theorem 2.3

Let f be convex on \([0,1]\). Then \(B_{n-1}^{p,q}(f;x)\geq B_{n}^{p,q}(f;x)\) for \(0< q< p\leq1\), \(0\leq x\leq 1\), and \(n\geq 2\). If \(f\in C[0,1]\) the inequality holds strictly for \(0< x<1\) unless f is linear in each of the intervals between consecutive knots \(\frac{p^{n-1-k}[k]_{p,q}}{[n-1]_{p,q}}\), \(0\leq k\leq n-1\), in which case we have the equality.


For \(0< q< p\leq1\) we begin by writing

$$\begin{aligned}& \prod_{s=0}^{n-1}\bigl(p^{s}-q^{s}x \bigr)^{-1} \bigl[B_{n-1}^{p,q}(f;x)-B_{n}^{p,q}(f;x) \bigr] \\& \quad = \prod_{s=0}^{n-1} \bigl(p^{s}-q^{s}x\bigr)^{-1}\Biggl[\sum _{k=0}^{n-1} \left[ \textstyle\begin{array}{@{}c@{}} n-1 \\ k \end{array}\displaystyle \right]_{p,q}p^{[k(k-1)-(n-2)(n-1)]/2}x^{k}(1 \ominus x)_{p,q}^{n-k-1}f \biggl( \frac{p^{n-1-k}[k]}{[n-1]} \biggr) \\& \qquad {}- \sum_{k=0}^{n-1} \left[ \textstyle\begin{array}{@{}c@{}} n \\ k \end{array}\displaystyle \right]_{p,q}p^{[k(k-1)-n(n-1)]/2}x^{k}(1\ominus x)_{p,q}^{n-k}f \biggl( \frac{p^{n-k}[k]}{[n]} \biggr)\Biggr] \\& \quad =\sum_{k=0}^{n-1} \left[ \textstyle\begin{array}{@{}c@{}} n-1 \\ k \end{array}\displaystyle \right]_{p,q}p^{[k(k-1)-(n-2)(n-1)]/2}x^{k}\prod _{s=n-k-1}^{n-1}\bigl(p^{s}-q^{s}x \bigr)^{-1}f \biggl( \frac {p^{n-1-k}[k]}{[n-1]} \biggr) \\& \qquad {}-\sum_{k=0}^{n} \left[ \textstyle\begin{array}{@{}c@{}} n \\ k \end{array}\displaystyle \right]_{p,q}p^{[k(k-1)-n(n-1)]/2}x^{k}\prod _{s=n-k}^{n-1}\bigl(p^{s}-q^{s}x \bigr)^{-1}f \biggl( \frac {p^{n-k}[k]}{[n]} \biggr). \end{aligned}$$


$$ \Psi_{k}(x)=p^{\frac{k(k-1)}{2}}x^{k}\prod _{s=n-k}^{n-1}\bigl(p^{s}-q^{s}x \bigr)^{-1}, $$

and using the following relation:

$$p^{n-1}p^{\frac{k(k-1)}{2}}x^{k}\prod _{s=n-k-1}^{n-1}\bigl(p^{s}-q^{s}x \bigr)^{-1}=p^{k}\Psi_{k}(x)+q^{n-k-1} \Psi_{k+1}(x), $$

we find

$$\begin{aligned}& \prod_{s=0}^{n-1}\bigl(p^{s}-q^{s}x \bigr)^{-1} \bigl[B_{n-1}^{p,q}(f;x)-B_{n}^{p,q}(f;x) \bigr] \\& \quad = \sum_{k=0}^{n-1} \left[ \textstyle\begin{array}{@{}c@{}}n-1\\ k \end{array}\displaystyle \right]_{p,q}p^{-\frac{(n-1)(n-2)}{2}}p^{-(n-1)} \bigl\{ p^{k}\Psi _{k}(x)+q^{n-k-1} \Psi_{k+1}(x) \bigr\} f \biggl(\frac {p^{n-1-k}[k]_{p,q}}{[n-1]_{p,q}} \biggr) \\& \qquad {} - \sum_{k=0}^{n} \left[ \textstyle\begin{array}{@{}c@{}}n\\ k \end{array}\displaystyle \right]_{p,q} p^{-\frac{n(n-1)}{2}} \Psi_{k}(x)f \biggl(\frac {p^{n-k}[k]_{p,q}}{[n]_{p,q}} \biggr) \\& \quad =p^{-\frac{n(n-1)}{2}}\Biggl\{ \sum_{k=0}^{n-1} \left[ \textstyle\begin{array}{@{}c@{}}n-1\\ k \end{array}\displaystyle \right]_{p,q} p^{k} \Psi_{k}(x)f \biggl(\frac{p^{n-1-k}[k]_{p,q}}{[n- 1]_{p,q}} \biggr) \\& \qquad {} +\sum_{k=1}^{n} \left[ \textstyle\begin{array}{@{}c@{}}n-1\\ k-1 \end{array}\displaystyle \right]_{p,q}q^{n-k} \Psi_{k}(x)f \biggl( \frac {p^{n-k}[k-1]_{p,q}}{[n-1]_{p,q}} \biggr) -\sum _{k=0}^{n}\left [ \textstyle\begin{array}{@{}c@{}}n\\ k \end{array}\displaystyle \right ]\Psi_{k}(x)f \biggl( \frac {p^{n-k}[k]_{p,q}}{[n]_{p,q}} \biggr)\Biggr\} \\& \quad =p^{-\frac{n(n-1)}{2}} \sum_{k=1}^{n-1} \biggl\{ \left[ \textstyle\begin{array}{@{}c@{}}n-1\\ k \end{array}\displaystyle \right]_{p,q}p^{k}f \biggl(\frac{p^{n-1-k}[k]_{p,q}}{[n-1]_{p,q}} \biggr) + \left[ \textstyle\begin{array}{@{}c@{}}n-1\\ k-1 \end{array}\displaystyle \right]_{p,q}q^{n-k}f \biggl(\frac {p^{n-k}[k-1]_{p,q}}{[n-1]_{p,q}} \biggr) \\& \qquad {} - \left[ \textstyle\begin{array}{@{}c@{}}n\\ k \end{array}\displaystyle \right]_{p,q}f \biggl( \frac{p^{n-k}[k]_{p,q}}{[n]_{p,q}} \biggr) \biggr\} \Psi_{k}(x) \\& \quad =p^{-\frac{n(n-1)}{2}}\sum_{k=1}^{n-1} \left[ \textstyle\begin{array}{@{}c@{}}n\\ k \end{array}\displaystyle \right]_{p,q}\biggl\{ \frac{[n-k]_{p,q}}{[n]_{p,q}}p^{k} f \biggl(\frac{p^{n-1-k}[k]_{p,q}}{[n-1]_{p,q}} \biggr) \\& \qquad {} +\frac{[k]_{p,q}}{[n]_{p,q}}q^{n-k}f \biggl(\frac {p^{n-k}[k-1]_{p,q}}{[n-1]_{p,q}} \biggr)-f \biggl(\frac {p^{n-k}[k]_{p,q}}{[n]_{p,q}} \biggr)\biggr\} \Psi_{k}(x) \\& \quad =p^{-\frac{n(n-1)}{2}}\sum_{k=1}^{n-1} \left[ \textstyle\begin{array}{@{}c@{}}n\\ k \end{array}\displaystyle \right]_{p,q}a_{k} \Psi_{k}(x), \end{aligned}$$


$$a_{k}= \frac{[n-k]_{p,q}}{[n]_{p,q}}p^{k} f \biggl(\frac{p^{n-1-k}[k]_{p,q}}{[n-1]_{p,q}} \biggr)+\frac {[k]_{p,q}}{[n]_{p,q}}q^{n-k}f \biggl(\frac {p^{n-k}[k-1]_{p,q}}{[n-1]_{p,q}} \biggr)-f \biggl(\frac {p^{n-k}[k]_{p,q}}{[n]_{p,q}} \biggr). $$

From (2.1) it is clear that each \(\Psi_{k}(x)\) is non-negative on \([0,1]\) for \(0< q< p\leq1\) and, thus, it suffices to show that each \(a_{k}\) is non-negative.

Since f is convex on \([0,1]\), then for any \(t_{0},t_{1}\in[0,1]\) and \(\lambda\in[0,1]\), it follows that

$$f\bigl(\lambda t_{0}+(1-\lambda)t_{1}\bigr)\leq\lambda f(t_{0})+(1-\lambda)f(t_{1}). $$

If we choose \(t_{0}=\frac{p^{n-k}[k-1]_{p,q}}{[n-1]_{p,q}}\), \(t_{1}=\frac{p^{n-1-k}[k]_{p,q}}{[n-1]_{p,q}}\), and \(\lambda=\frac{[k]_{p,q}}{[n]_{p,q}}q^{n-k}\), then \(t_{0},t_{1}\in[0,1]\) and \(\lambda\in(0,1)\) for \(1\leq k\leq n-1\), and we deduce that

$$a_{k}=\lambda f(t_{0})+(1-\lambda)f(t_{1})-f \bigl(\lambda t_{0}+(1-\lambda )t_{1}\bigr)\geq0. $$

Thus \(B_{n-1}^{p,q}(f;x)\geq B_{n}^{p,q}(f;x)\).

We have equality for \(x=0\) and \(x=1\), since the Bernstein polynomials interpolate f on these end-points. The inequality will be strict for \(0< x<1\) unless when f is linear in each of the intervals between consecutive knots \(\frac{p^{n-1-k}[k]_{p,q}}{[n-1]_{p,q}}\), \(0\leq k\leq n-1\), then we have \(B_{n-1}^{p,q}(f;x)=B_{n}^{p,q}(f;x)\) for \(0\leq x\leq1\). □

Example 2.4

Let \(f(x)=\sin(2\pi x)\), \(x\in[0,1]\). Figure 2 illustrates the monotonicity of \((p,q)\)-Bernstein operators for \(p=0.95\) and \(q=0.9\). We note that if f is increasing (decreasing) on \([0,1]\), then the operators is also increasing (decreasing) on \([0,1]\).

Figure 2
figure 2

Monotonicity of \(\pmb{(p,q)}\) -Bernstein operators.

A global approximation theorem

In the following we establish a global approximation theorem by means of Ditzian-Totik modulus of smoothness. In order to prove our next result, we recall the definitions of the Ditzian-Totik first order modulus of smoothness and the K-functional [16]. Let \(\phi(x) =\sqrt{x(1-x)}\) and \(f\in C[0,1]\). The first order modulus of smoothness is given by

$$ \omega_{\phi}(f;t)=\sup_{0< h\leq t} \biggl\{ \biggl\vert f \biggl(x+\frac{h\phi(x)}{2} \biggr)-f \biggl(x-\frac{h\phi (x)}{2} \biggr)\biggr\vert ,x\pm\frac{h\phi(x)}{2}\in[ 0,1] \biggr\} . $$

The corresponding K-functional to (3.1) is defined by

$$ {K}_{\phi}(f;t)=\inf_{g\in W_{\phi}[0,1]}\bigl\{ \Vert f-g\Vert +t \bigl\Vert \phi g^{\prime}\bigr\Vert \bigr\} \quad (t>0), $$

where \(W_{\phi}[0,1]=\{g:g\in AC_{\mathrm{loc}}[0,1],\|\phi g^{\prime }\|<\infty\}\) and \(g\in AC_{\mathrm{loc}}[0,1]\) means that g is absolutely continuous on every interval \([a,b]\subset[0,1]\). It is well known ([16], p.11) that there exists a constant \(C>0\) such that

$$ {K}_{\phi}(f;t)\leq C\omega_{\phi}(f;t). $$

Theorem 3.1

Let \(f\in C[0,1] \) and \(\phi(x) =\sqrt{x(1-x)}\), then for every \(x\in[0,1]\), we have

$$ \bigl\vert B_{n}^{p,q}(f;x)-f(x)\bigr\vert \leq C \omega_{\phi}\biggl(f;\frac{1}{\sqrt{[n]_{p,q}}} \biggr), $$

where C is a constant independent of n and x.


Using the representation

$$g(t)=g(x)+ \int_{x}^{t}g^{\prime}(u)\,du, $$

we get

$$ \bigl\vert B_{n}^{p,q}(g;x)-g(x)\bigr\vert = \biggl\vert B_{n}^{p,q} \biggl( \int_{x}^{t}g^{\prime}(u)\,du;x \biggr)\biggr\vert . $$

For any \(x\in(0,1)\) and \(t\in[0,1]\) we find that

$$ \biggl\vert \int_{x}^{t}g^{\prime}(u)\,du\biggr\vert \leq\bigl\Vert \phi g'\bigr\Vert \biggl\vert \int_{x}^{t}\frac{1}{\phi(u)}\,du\biggr\vert . $$


$$\begin{aligned} \biggl\vert \int_{x}^{t}\frac{1}{\phi(u)}\,du\biggr\vert &= \biggl\vert \int_{x}^{t}\frac {1}{\sqrt{u(1-u)}}\,du\biggr\vert \\ &\leq \biggl\vert \int_{x}^{t}\biggl(\frac{1}{\sqrt{u}}+ \frac{1}{\sqrt {1-u}}\biggr)\,du\biggr\vert \\ &\leq2\bigl(|\sqrt{t}-\sqrt{x}|+|\sqrt{1-t}-\sqrt {1-x}|\bigr) \\ &= 2\vert t-x\vert \biggl(\frac{1}{\sqrt{t}+\sqrt{x}}+\frac{1}{\sqrt {1-t}+\sqrt{1-x}}\biggr) \\ &< 2\vert t-x\vert \biggl(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{1-x}}\biggr)\leq \frac{2\sqrt{2} \vert t-x\vert }{\phi(x)}. \end{aligned}$$

From (3.3)-(3.5) and using the Cauchy-Schwarz inequality, we obtain

$$\begin{aligned} \bigl\vert B_{n}^{p,q}(g;x)-g(x)\bigr\vert &< 2\sqrt{2} \bigl\Vert \phi g'\bigr\Vert \phi ^{-1}(x)B_{n}^{p,q} \bigl(\vert t-x\vert ;x\bigr) \\ &\leq2\sqrt{2}\bigl\Vert \phi g'\bigr\Vert \phi^{-1}(x) \bigl( B_{n}^{p,q}\bigl((t-x)^{2};x \bigr) \bigr)^{1/2}. \end{aligned}$$

Using Lemma 1.2, we get

$$ \bigl\vert B_{n}^{p,q}(g;x)-g(x)\bigr\vert \leq \frac{2\sqrt{2}}{\sqrt{[n]_{p,q}}}\bigl\Vert \phi g^{\prime}\bigr\Vert . $$

Now, using the above inequality we can write

$$\begin{aligned} \bigl\vert B_{n}^{p,q}(f;x)-f(x)\bigr\vert &\leq\bigl\vert B_{n}^{p,q}(f-g;x)\bigr\vert +\bigl\vert f(x)-g(x) \bigr\vert +\bigl\vert B_{n}^{p,q}(g;x)-g(x)\bigr\vert \\ &\leq 2\sqrt{2} \biggl(\Vert f-g\Vert +\frac{1}{\sqrt{[n]_{p,q}}}\bigl\Vert \phi g'\bigr\Vert \biggr). \end{aligned}$$

Taking the infimum on the right-hand side of the above inequality over all \(g\in W_{\phi}[0,1]\), we get

$$\bigl\vert B_{n}^{p,q}(f;x)-f(x)\bigr\vert \leq C {K}_{\phi}\biggl(f;\frac{1}{\sqrt {[n]_{p,q}}} \biggr). $$

Using equation (3.2) this theorem is proven. □

Voronovskaja type theorem

Using the first order Ditzian-Totik modulus of smoothness, we prove a quantitative Voronovskaja type theorem for the \((p,q)\)-Bernstein operators.

Theorem 4.1

For any \(f\in C^{2}[0,1]\) the following inequalities hold:

  1. (i)

    \(|[n]_{p,q} [B_{n}^{p,q}(f;x)-f(x) ]-\frac {p^{n-1}\phi^{2}(x)}{2}f^{\prime\prime}(x) |\leq C\omega_{\phi} (f^{\prime\prime},\phi(x)n^{-1/2} )\),

  2. (ii)

    \(|[n]_{p,q} [B_{n}^{p,q}(f;x)-f(x) ]-\frac {p^{n-1}\phi^{2}(x)}{2}f^{\prime\prime}(x) |\leq C\phi(x)\omega_{\phi} (f^{\prime\prime},n^{-1/2} )\),

where C is a positive constant.


Let \(f\in C^{2}[0,1]\) be given and \(t,x\in[0,1]\). Using Taylor’s expansion, we have

$$f(t)-f(x)=(t-x)f^{\prime}(x)+ \int_{x}^{t}(t-u)f^{\prime \prime}(u)\,du. $$


$$\begin{aligned} \begin{aligned} f(t)-f(x)-(t-x)f^{\prime}(x)- \frac {1}{2}(t-x)^{2}f^{\prime\prime}(x) &= \int_{x}^{t}(t-u)f^{\prime\prime}(u)\,du- \int _{x}^{t}(t-u)f^{\prime\prime}(x)\,du \\ &= \int_{x}^{t}(t-u)\bigl[f^{\prime\prime}(u)-f^{\prime\prime}(x) \bigr]\,du. \end{aligned} \end{aligned}$$

In view of Lemma 1.1 and Lemma 1.2, we get

$$ \biggl\vert B_{n}^{p,q}(f;x)-f(x)- \frac{p^{n-1}}{2[n]_{p,q}}\phi^{2}(x)f^{\prime\prime}(x)\biggr\vert \leq B_{n}^{p,q} \biggl(\biggl\vert \int_{x}^{t}|t-u|\bigl\vert f^{\prime\prime}(u)-f^{\prime \prime}(x) \bigr\vert \,du\biggr\vert ;x \biggr). $$

The quantity \(\vert \int_{x}^{t}\vert f^{\prime\prime}(u)-f^{\prime \prime}(x)\vert |t-u|\,du\vert \) was estimated in [17], p.337, as follows:

$$ \biggl\vert \int_{x}^{t}\bigl\vert f^{\prime\prime}(u)-f^{\prime\prime }(x) \bigr\vert |t-u|\,du\biggr\vert \leq 2\bigl\Vert f^{\prime\prime}-g\bigr\Vert (t-x)^{2}+2\bigl\Vert \phi g^{\prime}\bigr\Vert \phi^{-1}(x)|t-x|^{3}, $$

where \(g\in W_{\phi}[0,1]\). On the other hand, for any \(m=1,2,\ldots\) and \(0< q< p\leq1\), there exists a constant \(C_{m}>0\) such that

$$ \bigl\vert B_{n}^{p,q} \bigl((t-x)_{p,q}^{m};x \bigr)\bigr\vert \leq C_{m} \frac{\phi ^{2}(x)}{[n]_{p,q}^{\lfloor\frac{m+1}{2}\rfloor}}, $$

where \(x\in[0,1]\) and \(\lfloor a\rfloor\) is the integer part of \(a\geq0\).

Throughout this proof, C denotes a constant not necessarily the same at each occurrence.

Now, combining (4.1)-(4.3) and applying Lemma 1.2, the Cauchy-Schwarz inequality, we get

$$\begin{aligned}& \biggl\vert B_{n}^{p,q}(f;x)-f(x)- \frac {p^{n-1}\phi^{2}(x)}{2[n]_{p,q}}f^{\prime\prime}(x) \biggr\vert \\& \quad \leq2\bigl\Vert f^{\prime\prime}-g\bigr\Vert B_{n}^{p,q} \bigl((t-x)^{2};x \bigr)+2\bigl\Vert \phi g^{\prime}\bigr\Vert \phi^{-1}(x)B_{n}^{p,q} \bigl(|t-x|^{3};x \bigr) \\& \quad \leq2\bigl\Vert f^{\prime\prime}-g\bigr\Vert \frac{\phi ^{2}(x)}{[n]_{p,q}}+2\bigl\Vert \phi g^{\prime}\bigr\Vert \phi^{-1}(x) \bigl\{ B_{n}^{p,q}(t-x)^{2};x \bigr\} ^{1/2} \bigl\{ B_{n}^{p,q} \bigl((t-x)^{4};x \bigr) \bigr\} ^{1/2} \\& \quad \leq2\bigl\Vert f^{\prime\prime}-g\bigr\Vert \frac{\phi ^{2}(x)}{[n]_{p,q}}+2 \frac{C}{[n]_{p,q}}\bigl\Vert \phi g^{\prime}\bigr\Vert \frac{\phi(x)}{\sqrt{[n]_{p,q}}} \\& \quad \leq \frac{C}{[n]_{p,q}} \bigl\{ \phi^{2}(x)\bigl\Vert f^{\prime\prime}-g\bigr\Vert +[n]_{p,q}^{-1/2}\phi(x)\bigl\Vert \phi g^{\prime}\bigr\Vert \bigr\} . \end{aligned}$$

Since \(\phi^{2}(x)\leq\phi(x)\leq1\), \(x\in[0,1]\), we obtain

$$ \biggl\vert [n]_{p,q} \bigl[B_{n}^{p,q}(f;x)-f(x) \bigr]- \frac{p^{n-1}\phi^{2}(x)}{2}f^{\prime\prime}(x)\biggr\vert \leq C \bigl\{ \bigl\Vert f^{\prime\prime}-g\bigr\Vert +[n]_{p,q}^{-1/2}\phi(x) \bigl\Vert \phi g^{\prime}\bigr\Vert \bigr\} . $$

Also, the following inequality can be obtained:

$$ \biggl\vert [n]_{p,q} \bigl[B_{n}^{p,q}(f;x)-f(x) \bigr]-\frac {p^{n-1}\phi^{2}(x)}{2}f^{\prime\prime}(x)\biggr\vert \leq C\phi(x) \bigl\{ \bigl\Vert f^{\prime\prime}-g\bigr\Vert +[n]_{p,q}^{-1/2} \bigl\Vert \phi g^{\prime}\bigr\Vert \bigr\} . $$

Taking the infimum on the right-hand side of the above relations over \(g\in W_{\phi}[0,1]\), we get

$$ \biggl\vert [n]_{p,q} \bigl[B_{n}^{p,q}(f;x)- f(x) \bigr]- \frac{p^{n-1}\phi^{2}(x)}{2}f^{\prime \prime}(x)\biggr\vert \leq \left \{ \textstyle\begin{array}{l} C K_{\phi} (f^{\prime\prime};\phi (x)[n]_{p,q}^{-1/2} ), \\ C \phi(x)K_{\phi} (f^{\prime\prime};[n]_{p,q}^{-1/2} ). \end{array}\displaystyle \right . $$

Using (4.4) and (3.2) the theorem is proved. □

Better approximation

In 2003, King [18] proposed a technique to obtain a better approximation for the well-known Bernstein operators as follows:

$$ \bigl((B_{n}f)\circ r_{n} \bigr) (x)= \sum _{k=0}^{n} f \biggl( \frac{k}{n} \biggr){n\choose k}\bigl(r_{n}(x)\bigr)^{k} \bigl(1-r_{n}(x)\bigr)^{n-k}, $$

where \(r_{n}\) is a sequence of continuous functions defined on \([0,1]\) with \(0\leq r_{n}(x)\leq1\) for each \(x\in[0,1]\) and \(n\in\{ 1,2,\ldots\}\). The modified Bernstein operators (5.1) preserve \(e_{0}\) and \(e_{2}\) and present a degree of approximation at least as good. In [19], the authors consider the sequence of linear Bernstein-type operators defined for \(f\in C[0,1]\) by \({B}_{n}(f\circ\tau^{-1})\circ\tau\), τ being any function that is continuously differentiable ∞ times on \([0,1]\), such that \(\tau(0)=0\), \(\tau(1)=1\), and \(\tau^{\prime}(x)>0\) for \(x\in[0,1]\).

So, using the technique proposed in [19], we modify the \((p,q)\)-Bernstein operators as follows:

$$\overline{B}_{n}^{p,q}(f;x)= \sum _{k=0}^{n}\overline {b}_{n,k}^{p,q}(x) \bigl(f\circ\tau^{-1} \bigr) \biggl(\frac {p^{n-k}[k]_{p,q}}{[n]_{p,q}} \biggr), $$


$$\overline{b}_{n}^{p,q}(x)= \left[ \textstyle\begin{array}{@{}c@{}}n\\ k \end{array}\displaystyle \right]_{p,q}p^{[k(k-1)-n(n-1)]/2}\tau(x)^{k}\bigl(1\ominus \tau (x)\bigr)_{p,q}^{n-k}. $$

Then we have

$$\begin{aligned}& \overline{B}_{n}^{p,q}(e_{0};x)=1,\qquad \overline{B}_{n}^{p,q}\bigl(\tau (t);x\bigr)=\tau(x), \\& \overline{B}_{n}^{p,q}\bigl(\tau^{2}(t);x\bigr)= \frac {p^{n-1}}{[n]_{p,q}}\tau(x)+\frac{q[n-1]_{p,q}}{[n]_{p,q}}\tau ^{2}(x), \\& \overline{B}_{n}^{p,q} \bigl(\bigl(\tau(t)-\tau(x) \bigr)^{2};x \bigr)= \frac{p^{n-1}}{[n]_{p,q}}\phi_{\tau}^{2}(x), \end{aligned}$$

where \(\phi_{\tau}^{2}(x):=\tau(x)(1-\tau(x))\).

Example 5.1

We compare the convergence of \((p,q)\)-analog of Bernstein operators \(B_{n}^{p,q}f\) with the modified operators \(\overline{B}_{n}^{p,q}f\). We have considered the function \(f(x)=\sin(10x)\) and \(\tau(x)= \frac{x^{2}+x}{2}\). For \(x\in [\frac{1}{2},1 ]\), \(p=0.95\), \(q=0.9\), \(n=100\), the convergence of the operators \(B_{n}^{p,q}\) and \(\overline{B}_{n}^{p,q}\) to the function f is illustrated in Figure 3. Note that the approximation by \(\overline{B}_{n}^{p,q}f\) is better than using \((p,q)\)-Bernstein operators \(B_{n}^{p,q}f\).

Figure 3
figure 3

Approximation process by \(\pmb{B_{n}^{p,q}}\) and  \(\pmb{\overline{B}_{n}^{p,q}}\) .


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The authors would like to thank the editor and the referees for useful comments and suggestions. This work was supported by the Dong-A University research fund.

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Kang, S.M., Rafiq, A., Acu, AM. et al. Some approximation properties of \((p,q)\)-Bernstein operators. J Inequal Appl 2016, 169 (2016).

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  • 41A10
  • 41A25
  • 41A35


  • \((p,q)\)-Bernstein operators
  • \((p,q)\)-calculus
  • Voronovskaja type theorem
  • K-functional
  • Ditzian-Totik first order modulus of smoothness