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

Abstract

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.

1 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 [3–9].

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

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

and

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

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 $x\in[0,1]$.

2 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 $0< q< p\leq1$.

Proof

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

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.

Proof

For $0< q< p\leq1$ we begin by writing

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

Denote

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

(2.1)

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}$$

where

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

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

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

3 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

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

(3.1)

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

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

(3.2)

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.

Proof

Using the representation

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

we get

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

(3.3)

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

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

(3.4)

Further,

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

(3.5)

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. □

4 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:

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

• (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.

Proof

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

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

Therefore,

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

In view of Lemma 1.1 and Lemma 1.2, we get

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

(4.1)

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:

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

(4.2)

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

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

(4.3)

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

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

(4.4)

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

5 Better approximation

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

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

(5.1)

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:

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

where

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

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

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