Keywords

2010 Mathematics Subject Classification

1 Introduction

In the year 1912, Bernstein [5] introduced the Bernstein operators and provided the constructive proof of Weierstrass theorem. Later, several researchers have generalized Bernstein operators using different parameters and studied various convergence properties. For more (see [6, 7, 16]).

Recently, Chen et al. [7] defined a family of Bernstein operators, for the functions \(f \in \left[ {0,1} \right] \), \(\alpha \) is fixed and \(n \in \mathbb {N}\) are as follows:

$$\begin{aligned} {B_n^{(\alpha )}}(f;x) = \sum \limits _{k= 0}^n {{f_k}p_{n,k}^{(\alpha )}(x)}, \end{aligned}$$
(1.1)

where \({f_k} = f\left( {\frac{k}{n}} \right) \). For \(n>2\) the \(\alpha \)-Bernstein polynomial \({p_{n,k}^{(\alpha )}(x)}\) of degree n is defined by

$${p_{1,0}^{(\alpha )}(x) = 1 - x,\;p_{1,1}^{(\alpha )}(x) = x,}$$

and

$$\begin{aligned} p_{n,k}^{\left( \alpha \right) }(x)&= \left[ {\left( {\begin{array}{*{20}{c}} {n - 2} \\ k \end{array}} \right) \left( {1 - \alpha } \right) x} \right. + \left( {\begin{array}{*{20}{l}} {n - 2} \\ {k- 2} \end{array}} \right) \left( {1 - \alpha } \right) \left( {1 - x} \right) \\&\left. { + \left( {\begin{array}{*{20}{l}} n \\ k \end{array}} \right) \alpha x\left( {1 - x} \right) } \right] {x^{k - 1}}{\left( {1 - x} \right) ^{n - k - 1}},\;\;\;\;\;\;x \in \left[ {0,1} \right] . \end{aligned}$$

For the first time in 1987, Bernstein operators based on q-integers were introduced by Lupas [12] and they are rational functions. Again in 1997, Phillips [14] introduced the q-Bernstein polynomials known as Phillips q-Bernstein operators. In past decade, linear positive operators based on q-integers is an active area of research. For more (see [4, 8, 11]).

Chai et al. [8] have considered the q-analouge of (1.1) is as follows:

$$\begin{aligned} B_{n,q}^{(\alpha )}(f;x)=\sum \limits _{k = 0}^n {{f_k}p_{n,q,k}^{(\alpha )}(x)}, \end{aligned}$$
(1.2)

where

$$\begin{aligned} p_{n,q,k}^{\left( \alpha \right) }(x) =&{\left( {\left[ {\begin{array}{*{20}{c}} {n - 2}\\ k \end{array}} \right] } \right. _q}\left( {1 - \alpha } \right) x + {\left[ {\begin{array}{*{20}{c}} {n - 2}\\ {k - 2} \end{array}} \right] _q}\left( {1 - \alpha } \right) {q^{n - k - 2}}\left( {1 - {q^{n - k - 1}}x} \right) \\&+ \left. {{{\left[ {\begin{array}{*{20}{c}} n\\ k \end{array}} \right] }_q}\alpha x\left( {1 - {q^{n - k - 1}}x} \right) } \right) {x^{k - 1}}(1 - x)_q^{n - k - 1} , \end{aligned}$$

\(q \in (0,1]\) and \({\mathrm{{f}}_k} = f\left( {\frac{{{{[k]}_q}}}{{{{[n]}_q}}}} \right) \). For detailed explanation (see [3]).

Dhamija et al. [10] proposed the Kantorovich form of modified Szász–Mirakyan operators. Several researchers have also studied Kantorovich form of different linear positive operators and established local and global approximation results. More details (see [1, 2, 13, 15]).

Mohiuddine et al. [13] proposed the Kantorovich form of the operators (1.1), which is given as

$$\begin{aligned} K_n^{(\alpha )}(f;x) = (n + 1)\sum \limits _{k = 0}^n {p_{n,k}^{\left( \alpha \right) }(x)} \int \limits _{k/(n + 1)}^{(k + 1)/(n + 1)} {f(t)dt}, \end{aligned}$$
(1.3)

where \(p_{n,k}^{\left( \alpha \right) }(x)\) is defined in (1.1).

For \(\alpha =1\) and \(q=1\) the operators (1.4) reduces to Bernstein–Kantorovich operators.

Motivated from the above stated work, we consider the q-analogue of the operators (1.3) as follows:

$$\begin{aligned} K_{n,q}^{(\alpha )}(f;x) = {\left[ {n + 1} \right] _q}\sum \limits _{k = 0}^n {p_{n,q,k}^{\left( \alpha \right) }(x)} \int \limits _{\frac{{q{{\left[ k \right] }_q}}}{{{{\left[ {n + 1} \right] }_q}}}}^{\frac{{{{\left[ {k + 1} \right] }_q}}}{{{{\left[ {n + 1} \right] }_q}}}} {f(t){d_q}t}, \end{aligned}$$
(1.4)

and \(p_{n,q,k}^{\left( \alpha \right) }(x)\) is given in (1.2).

In this paper, we estimated the moments of the proposed operators and discuss the rate of convergence using modulus of continuity.

2 Basic Results

In this section, we prove some auxiliary result to prove our main results.

Lemma 2.1

From [8], we have \(B_{n,q}^{(\alpha )}(1;x) = 1\), \(B_{n,q}^{(\alpha )}(t;x) = x\) and

$$\begin{aligned} B_{n,q}^{(\alpha )}({t^2};x) = {x^2} + \frac{{x(1 - x)}}{{{{\left[ n \right] }_q}}} + \frac{{(1 - \alpha ){q^{n - 1}}{{\left[ 2 \right] }_q}x(1 - x)}}{{\left[ n \right] _q^2}}. \end{aligned}$$

Lemma 2.2

  1. (i)

    \(K_{n,q}^{(\alpha )}(1;x) = 1\);

  2. (ii)

    \(K_{n,q}^{(\alpha )}(t;x) = \frac{{2q{{[n]}_q}}}{{{{\left[ 2 \right] }_q}{{[n + 1]}_q}}}x +\frac{1}{{{{\left[ 2 \right] }_q}{{\left[ {n + 1} \right] }_q}}}\);

  3. (iii)

    \(K_{n,q}^{(\alpha )}({t^2};x) = \frac{{3{q^2}[n]_q^2}}{{{{\left[ 3 \right] }_q}[n + 1]_q^2}}{x^2} + \frac{{3{q^2}}}{{{{\left[ 3 \right] }_q}[n + 1]_q^2}}\left( {{{[n]}_q} + (1 - \alpha ){q^{n - 1}}{[2]}_q} \right) x(1 - x)\) \(+\,\frac{{3q{{[n]}_q}x}}{{{{\left[ 3 \right] }_q}\left[ {n + 1} \right] _q^2}} + \frac{1}{{{{\left[ 3 \right] }_q}\left[ {n + 1} \right] _q^2}}.\)

Proof

From [15], \(\int \limits _{\frac{{q{{\left[ k \right] }_q}}}{{{{\left[ {n + 1} \right] }_q}}}}^{\frac{{{{\left[ {k + 1} \right] }_q}}}{{{{\left[ {n + 1} \right] }_q}}}} {1{d_q}t = \frac{1}{{{{\left[ {n + 1} \right] }_q}}}}\), \(\int \limits _{\frac{{q{{\left[ k \right] }_q}}}{{{{\left[ {n + 1} \right] }_q}}}}^{\frac{{{{\left[ {k + 1} \right] }_q}}}{{{{\left[ {n + 1} \right] }_q}}}} {t{d_q}t = \frac{{2q{{\left[ k \right] }_q}}}{{{{\left[ 2 \right] }_q}\left[ {n + 1} \right] _q^2}}} + \frac{1}{{{{\left[ 2 \right] }_q}\left[ {n + 1} \right] _q^2}}\) and

$$\begin{aligned} \int \limits _{\frac{{q{{\left[ k \right] }_q}}}{{{{\left[ {n + 1} \right] }_q}}}}^{\frac{{{{\left[ {k + 1} \right] }_q}}}{{{{\left[ {n + 1} \right] }_q}}}} {{t^2}{d_q}t = \frac{{3{q^2}\left[ k \right] _q^2}}{{{{\left[ 3 \right] }_q}\left[ {n + 1} \right] _q^3}}} + \frac{{3q{{\left[ k \right] }_q}}}{{{{\left[ 3 \right] }_q}\left[ {n + 1} \right] _q^3}} + \frac{1}{{{{\left[ 3 \right] }_q}\left[ {n + 1} \right] _q^3}}. \end{aligned}$$

It is easy to say that \(K_{n,q}^{(\alpha )}(1;x) = 1\).

For \(f(t)=t\) and using Lemma 2.1, we have

$$\begin{aligned} K_{n,q}^{(\alpha )}(t;x)&= {\left[ {n + 1} \right] _q}\sum \limits _{k = 0}^n {p_{n,q,k}^{\left( \alpha \right) }(x)} \int \limits _{\frac{{q{{\left[ k \right] }_q}}}{{{{\left[ {n + 1} \right] }_q}}}}^{\frac{{{{\left[ {k + 1} \right] }_q}}}{{{{\left[ {n + 1} \right] }_q}}}} {t{d_q}t}\\&= {\left[ {n + 1} \right] _q}\sum \limits _{k = 0}^n {p_{n,q,k}^{\left( \alpha \right) }(x)} \left( {\frac{{2q{{[k]}_q}}}{{{[2]}_q}[n + 1]_q^2}} + \frac{1}{{{{[2]}_q}[n + 1]_q^2}} \right) \\&= \frac{{{{[n]}_q}}}{{{{[n + 1]}_q}}}\left( {\frac{{2q}}{{{{[2]}_q}}}\sum \limits _{k = 0}^n {p_{n,q,k}^{\left( \alpha \right) }(x)\frac{{{{[k]}_q}}}{{{{[n]}_q}}} + \frac{1}{{{{[2]}_q}{{[n]}_q}}}\sum \limits _{k = 0}^n {p_{n,q,k}^{\left( \alpha \right) }(x)} } } \right) \\&= \frac{{2q{{[n]}_q}x + 1}}{{{{[2]}_q}{{[n + 1]}_q}}}. \end{aligned}$$

Similarly, for \(f(t)=t^2\), we can estimate. So here we skip.    \(\square \)

Lemma 2.3

The central moments for the operators (1.4) are as follows:

  1. (i)

    \(K_{n,q}^{(\alpha )}(t - x;x) = \frac{{2q{{[n]}_q}}}{{{{[2]}_q}{{[n + 1]}_q}}}x + \frac{1}{{{{[2]}_q}{{[n + 1]}_q}}}\);

  2. (ii)

    \(K_{n,q}^{(\alpha )}({(t - x)^2};x) = \left( {\frac{{3{q^2}{{[n]}_q^2}}}{{{{[3]}_q}[n + 1]_q^2}} - \frac{{4q{{[n]}_q}}}{{{{[2]}_q}{{[n + 1]}_q}}} + 1} \right) {x^2}\) \(+\,\frac{{3{q^2}}}{{{{[3]}_q}{{[n + 1]}_q}}}\left( {{{[n]}_q} + {{[2]}_q}(1 - \alpha ){q^{n - 1}}} \right) x(1 - x) + \left( {\frac{{3q{{[n]}_q}}}{{{{[3]}_q}[n + 1]_q^2}} - \frac{2}{{{{[3]}_q}{{[n + 1]}_q}}}} \right) x\) \( +\, \frac{1}{{{{[3]}_q}[n + 1]_q^2}}\).

Proof

Using linearity property of the operators (1.4) and Lemma 2.2, we get the required results.    \(\square \)

Lemma 2.4

Let \(0<q<1\) and \(c \in [0,q d]\), \(d>0\). Then the inequality

$$\begin{aligned} \int \limits _c^d {\left| {t - x} \right| {d_q}t} \le {\left( {\int \limits _c^d {{{(t - x)}^2}{d_q}t} } \right) ^{\frac{1}{2}}}{\left( {\int \limits _c^d {{d_q}t} } \right) ^{\frac{1}{2}}}. \end{aligned}$$

Proof

For the proof of the Lemma (see [15]).    \(\square \)

3 Main Results

Let C[0, 1] be the space of all continuous functions on [0, 1] with sup-norm \(\left\| f \right\| : = {\sup _{x \in [0,1]}}\left| {f(x)} \right| \). Let \(f \in C\left[ {0,1} \right] \) and \(\delta > 0\). Then the modulus of continuity \(\omega \left( {f,\delta } \right) \) is given as:

$$\omega \left( {f,\delta } \right) = \mathop {\sup }\limits _{\begin{array}{*{20}{c}} {\left| {v - w} \right| \le \delta }\\ {v,w \in \left[ {0,1} \right] } \end{array}} \left| {f(v) - f(w)} \right| .$$

It is well-known \(\mathop {\lim }\limits _{\delta \rightarrow 0} \omega (f;\delta ) = 0\). For \(f \in C[0,1]\) and \(x,t \in [0,1]\), we have

$$\begin{aligned} \left| {f(t) - f(x)} \right| \le \omega (f;\delta )\left( {1 + \frac{{\left| {t - x} \right| }}{\delta }} \right) \end{aligned}$$
(3.1)

For \(f \in C[0,1]\) the Peetre K-functional is given by

$${K_2}(f;\delta ) = {\inf _{g \in {W^2}}}\left\{ {\left| {f - g} \right| + \delta \left\| {{g^{\prime \prime }}} \right\| } \right\} ,$$

where \(\delta >0\) and \({W^2} = \left\{ {g \in C [0,1]:{g^{\prime }},{g^{\prime \prime }} \in C [0,1]} \right\} \). In [9], there exists an absolute constant \(\lambda >0\), such that

$$\begin{aligned} {K_2}(f;\delta ) \le \lambda {\omega _2}(f;\sqrt{\delta }). \end{aligned}$$
(3.2)

and the second-order modulus of continuity \({\omega _2}(.;\delta )\) for \(f\in C [0,1]\) as follows:

$$\begin{aligned} {\omega _2}(f;\delta ) = {\sup _{0 < h \le \delta }}\mathop {\sup }\nolimits _{_{x,x + h,x + 2h \in [0,1]}} \left| {f(x + 2h) - 2f(x + h) + f(x)} \right| . \end{aligned}$$

Theorem 3.1

For \(0 <q\le 1\), \(q=\left\{ {{q_n}} \right\} \) be a sequence converging to 1 as \(n\rightarrow \infty \). Then, for all \(f \in C [0,1]\) and \(\alpha \in [0,1]\), it implies \(K_{n,q}^{(\alpha )}(f;x)\) converges to f(x) uniformly on [0, 1] for sufficiently large n.

Proof

From Lemma 2.2, \(\mathop {\lim }\nolimits _{n \rightarrow \infty } {q_n} = 1\), we have \(\mathop {\lim }\nolimits _{n\rightarrow \infty } K_{n,q}^{(\alpha )}(1;x) = 1\), \(\mathop {\lim }\nolimits _{n\rightarrow \infty } K_{n,q}^{(\alpha )}(t;x) = x\) and \(\mathop {\lim }\nolimits _{n\rightarrow \infty } K_{n,q}^{(\alpha )}(t^2;x) = x^2\). Then by Bohaman–Korovkin theorem \(\mathop {\lim }\nolimits _{n\rightarrow \infty } K_{n,q}^{(\alpha )}(f(t);x) = f(x)\) converges uniformly on [0, 1].    \(\square \)

Theorem 3.2

For \(f \in C [0,1]\), \(q \in (0,1)\) and \(\alpha \in [0,1]\), we have

$$\begin{aligned} \left| {K_{n,q}^{(\alpha )}(f;x) - f(x)} \right| \le \lambda {\omega _2}\left( {f;\sqrt{\mu _{n,2}^q(x) + \mu {{_{n,1}^q}^2}(x)} } \right) + \omega \left( {f;\omega _{n,1}^q(x)} \right) , \end{aligned}$$

where \(\mu _{n,2}^q(x)\) and \(\mu _{n,1}^q(x)\) are second- and first-central moments of the operators (1.4).

Proof

We define an auxiliary operators

$$\begin{aligned} \hat{K}_{n,q}^{(\alpha )}(f;x) = K_{n,q}^{(\alpha )}(f;x) - f\left( {\frac{{2q{{[n + 1]}_q}x + 1}}{{{{[2]}_q}{{[n + 1]}_q}}}} \right) + f(x). \end{aligned}$$
(3.3)

For the operators \(\hat{K}_{n,q}^{(\alpha )}(.;x)\), we get

$$\begin{aligned} \hat{K}_{n,q}^{(\alpha )}(t-x;x) = 0. \end{aligned}$$
(3.4)

Suppose, \(g \in W^2\), \(x,t \in [0,1]\). Then by Tylor’s expansion, we have

$$\begin{aligned} g(t) = g(x) + (t - x){g^{\prime }}(x) + \int \limits _x^t {(t - u){g^{\prime \prime }}} (u)du. \end{aligned}$$

Applying \(\hat{K}_{n,q}^{(\alpha )}(.;x)\) in above equation, we have

$$\begin{aligned} \hat{K}_{n,q}^{(\alpha )}(g;x) = g(x) + \hat{K}_{n,q}^{(\alpha )}\left( {\int \limits _x^t {(t - u){g^{''}}} (u)du;x} \right) . \end{aligned}$$

Therefore,

$$\begin{aligned} \nonumber \left| {\hat{K}_{n,q}^{(\alpha )}(g;x) - g(x)} \right|&\le \left| {K_{n,q}^{(\alpha )}\left( {\int \limits _x^t {(t - u){g^{\prime \prime }}} (u)du;x} \right) } \right| \\ \nonumber&\quad + \left| {\left( {\int \limits _x^{\frac{{2q{{[n + 1]}_q}x + 1}}{{{{[2]}_q}{{[n + 1]}_q}}}} {\left( {\frac{{2q{{[n + 1]}_q}x + 1}}{{{{[2]}_q}{{[n + 1]}_q}}} - x} \right) {g^{''}}} (u)du;x} \right) } \right| \\ \nonumber&\le K_{n,q}^{(\alpha )}\left( {\int \limits _x^t {\left| {t - x} \right| {g^{''}}} (u)du;x} \right) \\ \nonumber&\quad + \left| {\left( {\int \limits _x^{\frac{{2q{{[n + 1]}_q}x + 1}}{{{{[2]}_q}{{[n + 1]}_q}}}} {\left| {\frac{{2q{{[n + 1]}_q}x + 1}}{{{{[2]}_q}{{[n + 1]}_q}}} - u} \right| } \left| {{g^{\prime \prime }}(x)} \right| du;x} \right) } \right| \\&\le \left[ {K_{n,q}^{(\alpha )}({{(t - x)}^2};x) + {{\left( {\frac{{2q{{[n + 1]}_q}x + 1}}{{{{[2]}_q}{{[n + 1]}_q}}} - x} \right) }^2}} \right] \left\| {{g^{\prime \prime }}} \right\| . \end{aligned}$$
(3.5)

From (3.3), we have

$$\begin{aligned} \left| {K_{n,q}^{(\alpha )}(f;x) \le \left\| f \right\| } \right| K_{n,q}^{(\alpha )}(1;x) + 2\left\| f \right\| = 3\left\| f \right\| . \end{aligned}$$
(3.6)

From (3.3), (3.5) and (3.6), we have

$$\begin{aligned} \nonumber \left| {K_{n,q}^{(\alpha )}(f;x) - f(x)} \right|&\le \left| {K_{n,q}^{(\alpha )}(f - g;x)} \right| + \left| {f - g} \right| \\ \nonumber&\quad + \left| {f\left( {\frac{{2q{{[n + 1]}_q}x + 1}}{{{{[2]}_q}{{[n + 1]}_q}}}} \right) - f(x)} \right| \\ \nonumber&\le 4\left\| {f - g} \right\| + \left( {\mu _{n,2}^q(x) + \mu {{_{n,1}^q}^2}(x)} \right) \\&\quad + \left| {f\left( {\frac{{2q{{[n + 1]}_q}x + 1}}{{{{[2]}_q}{{[n + 1]}_q}}}} \right) - f(x)} \right| \end{aligned}$$

Now taking infimum on the right-hand side of the above inequality over \(g \in W^2\), we get

$$\begin{aligned} \le 4{K_2}\left( {f;\mu _{n,2}^q(x) + \mu {{_{n,1}^q}^2}(x)} \right) + \omega \left( {f;\mu _{n,1}^q(x)} \right) \end{aligned}$$

From (3.2), we get

$$\begin{aligned} \left| {K_{n,q}^{(\alpha )}(f;x) - f(x)} \right| \le \lambda {\omega _2}\left( {f;\sqrt{\mu _{n,2}^q(x) + \mu {{_{n,1}^q}^2}(x)} } \right) + \omega \left( {f;\omega _{n,1}^q(x)} \right) . \end{aligned}$$

Hence, this is our required result.    \(\square \)

Theorem 3.3

Let \(q_{n} \in (0,1)\) be a sequence converging to 1 and \(\alpha \) is fixed. Then for \(f \in C [0,1]\), we have

$$\begin{aligned} \left| {K_{n,q}^{(\alpha )}(f;x) - f(x)} \right| \le 2\omega (f;{\delta _n}(x)), \end{aligned}$$

where \({\delta _n}(x) = {\left( {K_{n,q}^{(\alpha )}({{(t - x)}^2};x)} \right) ^{\frac{1}{2}}}\).

Proof

For nondecreasing function \(f \in C [0,1]\). Using linearity and monotonicity of \(K_{n,q}^{(\alpha )}\), we have

$$\begin{aligned} \left| {K_{n,q}^{(\alpha )}(f;x) - f(x)} \right|&\le K_{n,q}^{(\alpha )}\left( {\left| {f(t) - f(x)} \right| ;x} \right) \\&\le \omega (f;\delta )\left( {1 + \frac{1}{\delta }K_{n,q}^{(\alpha )}\left( {\left| {t - x} \right| ;x} \right) } \right) \end{aligned}$$

Applying Lemma 2.4 with \(c = \frac{{q{{[k]}_q}}}{{{{[n + 1]}_q}}}\) and \(d = \frac{{{{[k+1]}_q}}}{{{{[n + 1]}_q}}}\), we get

$$\begin{aligned} \left| {K_{n,q}^{(\alpha )}(f;x) - f(x)} \right|&\le \omega (f;x)\left\{ {1 + \frac{{{{[n + 1]}_q}}}{\delta }\sum \limits _{k = 0}^n {p_{n,q,k}^{(\alpha )}(x){{\left( {\int \limits _{\frac{{q{{[k]}_q}}}{{{{[n + 1]}_q}}}}^{\frac{{{{[k + 1]}_q}}}{{{{[n + 1]}_q}}}} {{{(t - x)}^2}{d_q}t} } \right) }^{\frac{1}{2}}}} } \right. \\&\quad \times \left. {{{\left( {\int \limits _{\frac{{q{{[k]}_q}}}{{{{[n + 1]}_q}}}}^{\frac{{{{[k + 1]}_q}}}{{{{[n + 1]}_q}}}} {{d_q}t} } \right) }^{\frac{1}{2}}}} \right\} \end{aligned}$$

Using H\(\ddot{o}\)lder’s inequality for sums, we have

$$\begin{aligned}&= \omega (f;x)\left\{ {1 + \frac{1}{\delta }{{\left( {{{[n + 1]}_q}\sum \limits _{k = 0}^n {p_{n,q,k}^{(\alpha )}(x)} \int \limits _{\frac{{q{{[k]}_q}}}{{{{[n + 1]}_q}}}}^{\frac{{{{[k + 1]}_q}}}{{{{[n + 1]}_q}}}} {{{(t - x)}^2}{d_q}t} } \right) }^{\frac{1}{2}}}} \right. \\&\quad \times \left. {{{\left( {{{[n + 1]}_q}\sum \limits _{k = 0}^n {p_{n,q,k}^{(\alpha )}(x)} \int \limits _{\frac{{q{{[k]}_q}}}{{{{[n + 1]}_q}}}}^{\frac{{{{[k + 1]}_q}}}{{{{[n + 1]}_q}}}} {{d_q}t} } \right) }^{\frac{1}{2}}}} \right\} \\&= \omega (f;x)\left\{ {1 + \frac{1}{\delta }{{\left( {K_{n,q}^{(\alpha )}({{(t - x)}^2};x)} \right) }^{\frac{1}{2}}}} \right\} . \end{aligned}$$

By choosing \(\delta = \delta _{n}(x)\), we get the required result.    \(\square \)