Approximation Properties of Chlodowsky Variant of (p, q) Szász–Mirakyan–Stancu Operators

Abstract

In the present paper, we introduce the Chlodowsky variant of (p, q) Szász–Mirakyan–Stancu operators on the unbounded domain which is a generalization of (p, q) Szász–Mirakyan operators. We have also derived its Korovkin-type approximation properties and rate of convergence.

1 Introduction and Preliminaries

The applications of q-calculus emerged as a new area in the field of approximation theory from last two decades. The development of q-calculus has led to the discovery of various modifications of Bernstein polynomials involving q-integers. 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.

In 1987, Lupaş and in 1997, Phillips introduced a sequence of Bernstein polynomials based on q-integers and investigated its approximation properties.

Mursaleen et al. [12, 13, 18] introduced on Chlodowsky variant of Szász operators by Brenke-type polynomials, rate of convergence of Chlodowsky-type Durrmeyer Jakimovski–Leviatan operators and Dunkl generalization of q-parametric Szász–Mirakjan operators and shape preserving properties.

Several authors produced generalizations of well-known positive linear operators based on q-integers and studied them extensively. For instance, the approximation properties of A generalization of Szász–Mirakyan operators based on q-integers [5], convergence of the q-analogue of Szász-Beta operators [9], Dunkl generalization of q-parametric Szász–Mirakjan operators [18] and weighted statistical approximation by Kantorovich-type q-Szász–Mirakyan operators [6].

Recently, Mursaleen et al. introduced (p, q)-calculus in approximation theory and constructed the (p, q)-analogue of Bernstein operators [14], (p, q)-analogue of Bernstein–Stancu operators [15]. Further, Acar [1] has studied recently, (p, q)-generalization of Szász–Mirakyan operators.

In the present paper, we introduce the Chlodowsky variant of (p, q) Szász–Mirakyan–Stancu operators on the unbounded domain. Most recently, the (p, q)-analogue of some more operators has been studied in [2, 4, 10, 11, 14, 17, 19, 21].

The (p, q)-integer or in general the (p, q)-calculus was introduced to generalize or unify several forms of q-oscillator algebras well known in the Physics literature related to the representation theory of single-parameter quantum algebras. The (p, q)-integer is defined by

$$\begin{aligned} \left[ n\right] _{p,q}=p^{n-1}+qp^{n-2}+\cdots +pq^{n-2}+q^{n-1}=\left\{ \begin{array}{ll} \dfrac{p^{n}-q^{n}}{p-q} &{} \qquad (p\ne q\ne 1) \\ &{} \\ \dfrac{1-q^{n}}{1-q} &{} \qquad (p=1) \\ &{} \\ n &{} \qquad (p=q=1). \\ &{} \end{array} \right. \end{aligned}$$

(1)

The (p, q)-binomial expansion is

$$\begin{aligned} (ax+by)_{p,q}^{n}:=\sum \limits _{k=0}^{n}p^{\frac{(n-k)(n-k-1)}{2}}q^{\frac{ k(k-1)}{2}}\left[ \begin{array}{c} n \\ k \end{array} \right] _{p,q}a^{n-k}b^{k}x^{n-k}y^{k}, \end{aligned}$$

$$\begin{aligned} (x+y)_{p,q}^{n}:=(x+y)(px+qy)(p^{2}x+q^{2}y)\cdots (p^{n-1}x+q^{n-1}y), \end{aligned}$$

$$\begin{aligned} (1-x)_{p,q}^{n}:=(1-x)(p-qx)(p^{2}-q^{2}x)\cdots (p^{n-1}-q^{n-1}x). \end{aligned}$$

The (p, q)-binomial coefficients are defined by

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

The definite integrals of a function f is defined by

$$\begin{aligned} \int _{0}^{a}f(t)d_{p,q}t=(q-p)a\sum _{k=0}^{\infty }f\left( \frac{p^{k}}{ q^{k+1}}a\right) \frac{p^{k}}{q^{k+1}},\qquad if\left| \frac{p}{q} \right| <1, \end{aligned}$$

$$\begin{aligned} \int _{0}^{a}f(t)d_{p,q}t=(p-q)a\sum _{k=0}^{\infty }f\left( \frac{q^{k}}{ p^{k+1}}a\right) \frac{q^{k}}{p^{k+1}},\qquad if\left| \frac{q}{p} \right| <1. \end{aligned}$$

There are two (p, q)-analogues of the classical exponential function defined as follows

$$\begin{aligned} e_{p,q}(x)=\sum _{n=0}^{\infty }\frac{p^{\frac{n(n-1)}{2}}x^{n}}{[n]_{p,q}!}, \end{aligned}$$

and

$$\begin{aligned} E_{p,q}(x)=\sum _{n=0}^{\infty }\frac{q^{\frac{n(n-1)}{2}}x^{n}}{[n]_{p,q}!}, \end{aligned}$$

which satisfy the equality $e_{p,q}(x)E_{p,q}(-x)=1$. For $p=1$, $e_{p,q}(x)$ and $E_{p,q}(x)$ reduce to q-exponential functions.

For $0<q<1$, Aral [5] introduced the generalized q-Szász–Mirakyan operators as follows

$$\begin{aligned} S_{n,q}(f;x)=\sum _{k=0}^{\infty } s_{n,k}^{q}(x) f \bigg ( \dfrac{ [k]_{q}b_{n}}{[n]_{q}} \bigg ) \end{aligned}$$

(2)

where

$$\begin{aligned} s_{n,k}^{q}(x) = \dfrac{1}{E_{q}([n]_{q}\frac{x}{b_{n}})} \dfrac{ ([n]_{q}x)^{k}}{[k]_{q}! (b_{n})^{k}}. \end{aligned}$$

where $0 \le x <\alpha _{q}(n),$ $\alpha _{q}(n) :=\dfrac{b_{n}}{(1-q)[n]_{q} }$, $f\in C( \mathbb {R}_{0} )$ and $(b_{n})$ is a sequence of positive numbers such that $\lim _{n\rightarrow \infty } b_{n} =\infty .$

Mursaleen et al. [10] introduced the (p, q)-analogue of the Szász–Mirakyan operators as follows

$$\begin{aligned} {S}_{n,p,q}(f;x)= \sum _{k=0}^{\infty } \frac{p^{ \frac{k(k-1)}{2}}}{q^{ \frac{ k(k-1)}{2}}}\frac{([n]_{p,q}x)^{k} }{ [k]_{p,q}! }e_{p,q}(-[n]_{p,q}q ^{-k}x) f\left( \frac{[k]_{p,q}}{p^{k-1}[n]_{p,q}}\right) . \end{aligned}$$

(3)

Lemma 1.1

Let $0<q<p\le 1$ and $n\in \mathbb {N}$. We have

• (i) ${S}_{n,p,q}(1;x)=1$

• (ii) ${S}_{n,p,q}(t;x)=x$

• (iii) ${S}_{n,p,q}(t^{2};x)= \frac{x^{2}}{p}+\frac{ x}{[n]_{p,q} }$

• (iv) ${S}_{n,p,q}(t^{3};x)= \frac{x^{3}}{p^{3^{}}}+ \frac{ 2p+q}{ p^{2}[n]_{p,q} }x^{2} + \frac{ x}{[n]_{p,q}^{2} }$

• (v) ${S}_{n,p,q}(t^{4};x)= \frac{x^{4}}{p^{6^{}}}+ \frac{3p^{2}+ 2pq+q^{2}}{p^{5}[n]_{p,q} }x^{3} + \frac{3p^{2}+ 3pq+q^{2}}{ p^{3}[n]_{p,q}^{2} }x^{2}+ \frac{ x}{[n]_{p,q}^{3} }$ .

2 Construction of the Operators

We construct the Chlodowsky variant of (p, q) Szász–Mirakyan–Stancu operators as

$$\begin{aligned} {S}_{n,p,q}^{(\alpha ,\beta )}(f;x)= \sum _{k=0}^{\infty } \frac{p^{ \frac{k(k-1) }{2}}}{q^{ \frac{k(k-1)}{2}}}\frac{([n]_{p,q}x)^{k} }{ [k]_{p,q}!(b_{n})^{k} }e_{p,q}(-[n]_{p,q}q ^{-k}\frac{x}{b_{n}}) f\left( \frac{p^{1-k} [k]_{p,q}+\alpha }{[n]_{p,q}+\beta }b_{n}\right) \end{aligned}$$

(4)

where $n \in \mathbb {N}, \alpha , \beta \in \mathbb {N}_{0}$ with $0 \le \alpha \le \beta , 0 \le x \le b_{n}, 0< q < p \le 1$ and $b_{n}$ is an increasing sequence of positive terms with the properties $b_{n}\longrightarrow \infty$ and $\dfrac{b_{n}}{[n]_{p,q}} \longrightarrow 0$ as $n \longrightarrow \infty$. We observe that ${S}_{n,p,q}^{(\alpha , \beta )}$ is positive and linear. Furthermore, in the case of $q =p= 1$ and $\alpha =\beta =0$, the operators (4) are similar to the classical Szász–Mirakyan operators.

Lemma 2.1

Let $0<q<p\le 1$ and $n\in \mathbb {N}$, $\alpha , \beta \in \mathbb {N}_{0}$ with $0 \le \alpha \le \beta$, $0 \le x \le b_{n}$, and integer $m\ge 0$, we have

$$\begin{aligned} {S}_{n,p,q}^{(\alpha ,\beta )}(t^{m};x)= \dfrac{b_{n}^{m}}{ ( [n]_{p,q}+ \beta )^{m}} \sum _{j=0}^{m}\left( \begin{array}{c} m \\ j \end{array} \right) \alpha ^{m-j} {S}_{n,p,q}\left( t^{j};q^{-1}\frac{x}{b_{n}}\right) . \end{aligned}$$

(5)

Proof

Using the identity

$$\begin{aligned}{}[k+1]_{p,q}=p^{k}+q[k]_{p,q}, \end{aligned}$$

we can write

$$\begin{aligned} {S}_{n,p,q}^{(\alpha ,\beta )}(t^{m};x)= & {} \sum _{k=0}^{\infty } \frac{p^{ \frac{k(k-1)}{2}}}{q^{ \frac{k(k-1)}{2}}}\frac{([n]_{p,q}x)^{k} }{ [k]_{p,q}! (b_{n})^{k} } e_{p,q}\left( -[n]_{p,q}q ^{-k}\frac{x}{b_{n}}\right) \left( \frac{p^{1-k}[k]_{p,q}+ \alpha }{[n]_{p,q}+ \beta }b_{n}\right) ^{m} \\= & {} \dfrac{b_{n}^{m}}{ ( [n]_{p,q}+ \beta )^{m} }\sum _{k=0}^{\infty } \frac{ p^{ \frac{k(k-1)}{2}}}{q^{ \frac{k(k-1)}{2}}}\frac{([n]_{p,q}x)^{k} }{ [k]_{p,q}! (b_{n})^{k} } e_{p,q}\left( -[n]_{p,q}q ^{-k}\frac{x}{b_{n}}\right) \left( p^{1-k}[k]_{p,q}+ \alpha \right) ^{m} \\= & {} \dfrac{b_{n}^{m}}{ ( [n]_{p,q}+ \beta )^{m} }\sum _{k=0}^{\infty } \frac{ p^{ \frac{k(k-1)}{2}}}{q^{ \frac{k(k-1)}{2}}}\frac{([n]_{p,q}x)^{k} }{ [k]_{p,q}! (b_{n})^{k} } e_{p,q}\left( -[n]_{p,q}q ^{-k}\frac{x}{b_{n}}\right) \\&\times \sum _{j=0}^{m}\left( \begin{array}{c} m \\ j \end{array} \right) \alpha ^{m-j} p^{j(1-k)}[k]_{p,q}^{j} \\= & {} \dfrac{b_{n}^{m}}{ ( [n]_{p,q}+ \beta )^{m}} \sum _{j=0}^{m}\left( \begin{array}{c} m \\ j \end{array} \right) \alpha ^{m-j} \sum _{k=0}^{\infty } \frac{p^{ \frac{k(k-1)}{2}}}{q^{ \frac{k(k-1)}{2}}}\frac{([n]_{p,q}x)^{k} }{ [k]_{p,q}! (b_{n})^{k} } \\&\times e_{p,q}(-[n]_{p,q}q ^{-k}\frac{x}{b_{n}}) p^{j(1-k)}[k]_{p,q}^{j} \\= & {} \dfrac{b_{n}^{m}}{ ( [n]_{p,q}+ \beta )^{m}} \sum _{j=0}^{m}\left( \begin{array}{c} m \\ j \end{array} \right) \alpha ^{m-j} {S}_{n,p,q}(t^{j};q^{-1}\frac{x}{b_{n}}) \end{aligned}$$

   $\square$

which is desired.

Lemma 2.2

Let ${S}_{n,p,q}^{(\alpha ,\beta )}(f;x)$ be given by (4). Then the following properties hold:

$$\begin{aligned} (i)\,{S}_{n,p,q}^{(\alpha ,\beta )}(1;x)= & {} 1 \\ (ii)\,{S}_{n,p,q}^{(\alpha ,\beta )}(t;x)= & {} \dfrac{ [n]_{p,q} }{[n]_{p,q}+ \beta }x + \dfrac{\alpha b_{n}}{[n]_{p,q} + \beta } \\ (iii)\,{S}_{n,p,q}^{(\alpha ,\beta )}(t^{2};x)= & {} \frac{ [n]_{p,q}^{2} }{p ( [n]_{p,q}+ \beta )^{2} }x^{2} +\frac{(1+2 \alpha ) b_{n} [n]_{p,q} }{ ( [n]_{p,q}+ \beta )^{2}}x+\dfrac{\alpha ^{2} b_{n}^{2}}{ ( [n]_{p,q}+ \beta )^{2}}\\ (iv)\,{S}_{n,p,q}^{(\alpha ,\beta )}(t^{3};x)= & {} \frac{ [n]_{p,q}^{3} }{p^{3} ([n]_{p,q} + \beta )^{3} }x^{3} +\frac{(3p\alpha +2p+q) b_{n}[n]_{p,q}^{2} }{p^{2} ([n]_{p,q}+ \beta )^{3} }x^{2} \\&+\frac{(1+3 \alpha ++3 \alpha ^{2} ) b_{n}^{2} [n]_{p,q} }{ ( [n]_{p,q}+ \beta )^{3}}x+\dfrac{\alpha ^{3} b_{n}^{3}}{ ( [n]_{p,q}+ \beta )^{3}}\\ \end{aligned}$$

$$\begin{aligned} (v)\,{S}_{n,p,q}^{(\alpha ,\beta )}(t^{4};x)= & {} \frac{ [n]_{p,q}^{4} }{p^{6} ([n]_{p,q} + \beta )^{4} }x^{4}+ \frac{(3p^{2}+2pq+q^{2}+4p\alpha ) b_{n}[n]_{p,q}^{3} }{p^{5} ([n]_{p,q} + \beta )^{4} }x^{3} \\&+ \frac{( 3p^{2}+3pq+q^{2}+4pq\alpha +8p^{2}\alpha +6p^{2}\alpha ^{2} ) b_{n}^{2}[n]_{p,q}^{2} }{p^{3} ([n]_{p,q} + \beta )^{4} }x^{2} \\&+\frac{(1+4\alpha +6\alpha ^{2} +4\alpha ^{3} ) b_{n}^{3}[n]_{p,q} }{ ([n]_{p,q}+ \beta )^{4} }x +\dfrac{\alpha ^{4} b_{n}^{4}}{ ( [n]_{p,q}+ \beta )^{4}}.\\ \end{aligned}$$

Proof

• (i)
  $$\begin{aligned} {S}_{n,p,q}^{(\alpha ,\beta )}(1;x)= & {} {S}_{n,p,q}(1;q^{-1}\frac{x}{b_{n}}) \\= & {} 1. \end{aligned}$$
• (ii)
  $$\begin{aligned} {S}_{n,p,q}^{(\alpha ,\beta )}(t;x)= & {} \dfrac{b_{n}}{ ( [n]_{p,q}+ \beta )} \sum _{j=0}^{1}\left( \begin{array}{c} 1 \\ j \end{array} \right) \alpha ^{1-j} {S}_{n,p,q}\left( t^{j};q^{-1}\frac{x}{b_{n}}\right) \\= & {} \dfrac{b_{n}}{ ( [n]_{p,q}+ \beta )} \bigg \{ \alpha + {S} _{n,p,q}\left( t;q^{-1}\frac{x}{b_{n}}\right) \bigg \} \\= & {} \dfrac{b_{n}}{ ( [n]_{p,q}+ \beta )} \bigg \{ \alpha + \dfrac{ [n]_{p,q} }{b_{n}} x \bigg \} \\= & {} \dfrac{ [n]_{p,q} }{[n]_{p,q}+ \beta }x + \dfrac{\alpha b_{n}}{[n]_{p,q} + \beta }. \end{aligned}$$
• (iii)
  $$\begin{aligned} {S}_{n,p,q}^{(\alpha ,\beta )}(t^{2};x)= & {} \dfrac{b_{n}^{2}}{ ( [n]_{p,q}+ \beta )^{2}} \sum _{j=0}^{2}\left( \begin{array}{c} 2 \\ j \end{array} \right) \alpha ^{2-j} {S}_{n,p,q}(t^{j};q^{-1}\frac{x}{b_{n}}) \\= & {} \dfrac{b_{n}^{2}}{ ( [n]_{p,q}+ \beta )^{2}} \bigg \{ \alpha ^{2}+2\alpha {S}_{n,p,q}(t;q^{-1}\frac{x}{b_{n}}) + {S}_{n,p,q}(t^{2};q^{-1} \frac{x}{b_{n}}) \bigg \} \\= & {} \dfrac{b_{n}^{2}}{ ( [n]_{p,q}+ \beta )^{2}} \bigg \{ \alpha ^{2}+2\alpha \dfrac{ [n]_{p,q} }{b_{n}} x +\dfrac{ [n]_{p,q} }{b_{n}} x + \dfrac{ [n]_{p,q} ^{2}}{pb_{n}^{2}} x^{2} \bigg \} \\= & {} \frac{ [n]_{p,q}^{2} }{p ( [n]_{p,q}+ \beta )^{2} }x^{2} +\frac{(1+2 \alpha ) b_{n} [n]_{p,q} }{ ( [n]_{p,q}+ \beta )^{2}}x+\dfrac{\alpha ^{2} b_{n}^{2}}{ ( [n]_{p,q}+ \beta )^{2} }. \end{aligned}$$
• (iv)
  $$\begin{aligned}&{S}_{n,p,q}^{(\alpha ,\beta )}(t^{3};x) = \dfrac{b_{n}^{3}}{ ( [n]_{p,q}+ \beta )^{3}} \sum _{j=0}^{3}\left( \begin{array}{c} 3 \\ j \end{array} \right) \alpha ^{3-j} {S}_{n,p,q}(t^{j};q^{-1}\frac{x}{b_{n}}) \\&\qquad \qquad \,= \dfrac{b_{n}^{3}}{ ( [n]_{p,q}+ \beta )^{3}} \bigg \{ \alpha ^{3}+3\alpha ^{2} {S}_{n,p,q}(t;q^{-1}\frac{x}{b_{n}}) +3\alpha {S}_{n,p,q}(t^{2};q^{-1} \frac{x}{b_{n}})\\&\qquad \qquad \,\,\,\,\,+ {S}_{n,p,q}(t^{3};q^{-1}\frac{x}{b_{n}})\bigg \}\\&\qquad \,\,\,= \dfrac{b_{n}^{3}}{ ( [n]_{p,q}+ \beta )^{3}} \bigg \{ \alpha ^{3}+3\alpha ^{2} \dfrac{ [n]_{p,q} }{b_{n}} x +3\alpha \bigg ( \dfrac{ [n]_{p,q} }{b_{n}} x + \dfrac{ [n]_{p,q} ^{2}}{pb_{n}^{2}} x^{2} \bigg ) \\&\qquad \quad \,\,+ \dfrac{ [n]_{p,q} }{b_{n}} x+ \dfrac{2 [n]_{p,q}^{2} }{pb_{n}^{2}} x^{2}+\dfrac{ q[n]_{p,q}^{2} }{p^{2}b_{n}^{2}} x^{2} +\dfrac{ [n]_{p,q}^{3} }{p^{3}b_{n}^{3}} x^{3} \bigg \} \\&\qquad \,\,\,= \frac{ [n]_{p,q}^{3} }{p^{3} ([n]_{p,q} + \beta )^{3} }x^{3} +\frac{(3p\alpha +2p+q) b_{n}[n]_{p,q}^{2} }{p^{2} ([n]_{p,q}+ \beta )^{3} }x^{2} \\&\qquad \quad \,\,+\frac{(1+3 \alpha ++3 \alpha ^{2} ) b_{n}^{2} [n]_{p,q} }{ ( [n]_{p,q}+ \beta )^{3}}x+\dfrac{\alpha ^{3} b_{n}^{3}}{ ( [n]_{p,q}+ \beta )^{3}}. \end{aligned}$$
• (v)
  $$\begin{aligned} {S}_{n,p,q}^{(\alpha ,\beta )}(t^{4};x)= & {} \dfrac{b_{n}^{4}}{ ( [n]_{p,q}+ \beta )^{4}} \sum _{j=0}^{4}\left( \begin{array}{c} 4 \\ j \end{array} \right) \alpha ^{4-j} {S}_{n,p,q}(t^{j};q^{-1}\frac{x}{b_{n}}) \\= & {} \dfrac{b_{n}^{4}}{ ( [n]_{p,q}+ \beta )^{4}} \bigg \{ \alpha ^{4}+4\alpha ^{3} {S}_{n,p,q}(t;q^{-1}\frac{x}{b_{n}}) +6\alpha ^{2} {S}_{n,p,q}(t^{2};q^{-1} \frac{x}{b_{n}})\\&+ 4\alpha {S}_{n,p,q}(t^{3};q^{-1}\frac{x}{b_{n}}) + {S}_{n,p,q}(t^{4};q^{-1}\frac{x}{b_{n}}) \bigg \} \\= & {} \dfrac{b_{n}^{4}}{ ( [n]_{p,q}+ \beta )^{4}} \bigg \{ \alpha ^{4}+4\alpha ^{3} \dfrac{ [n]_{p,q} }{b_{n}} x +6\alpha ^{2} \bigg ( \dfrac{ [n]_{p,q} }{b_{n}} x + \dfrac{ [n]_{p,q} ^{2}}{pb_{n}^{2}} x^{2} \bigg ) \\&+4\alpha \bigg ( \dfrac{ [n]_{p,q} }{b_{n}} x+ \dfrac{2 [n]_{p,q}^{2} }{pb_{n}^{2}} x^{2}+\dfrac{ q[n]_{p,q}^{2} }{p^{2}b_{n}^{2}} x^{2} +\dfrac{ [n]_{p,q}^{3} }{p^{3}b_{n}^{3}} x^{3} \bigg ) + \dfrac{ [n]_{p,q}^{4} }{p^{6}b_{n}^{4}} x^{4}\\&+\dfrac{q^{2} [n]_{p,q}^{3} }{p^{5}b_{n}^{3}} x^{3} +\dfrac{2q [n]_{p,q}^{3} }{p^{4}b_{n}^{3}} x^{3}+ \dfrac{q^{2} [n]_{p,q}^{2} }{p^{3}b_{n}^{2}} x^{2}+ \dfrac{3 [n]_{p,q}^{3} }{p^{3}b_{n}^{3}} x^{3}+\dfrac{3q [n]_{p,q}^{2} }{p^{2}b_{n}^{2}} x^{2}\\&+\dfrac{ 3[n]_{p,q}^{2} }{pb_{n}^{2}} x^{2}+ \dfrac{ [n]_{p,q} }{b_{n}} x\bigg \} \\= & {} \frac{ [n]_{p,q}^{4} }{p^{6} ([n]_{p,q} + \beta )^{4} }x^{4}+ \frac{(3p^{2}+2pq+q^{2}+4p\alpha ) b_{n}[n]_{p,q}^{3} }{p^{5} ([n]_{p,q} + \beta )^{4} }x^{3} \\&+ \frac{( 3p^{2}+3pq+q^{2}+4pq\alpha +8p^{2}\alpha +6p^{2}\alpha ^{2} ) b_{n}^{2}[n]_{p,q}^{2} }{p^{3} ([n]_{p,q} + \beta )^{4} }x^{2} \\&+\frac{(1+4\alpha +6\alpha ^{2} +4\alpha ^{3} ) b_{n}^{3}[n]_{p,q} }{ ([n]_{p,q}+ \beta )^{4} }x +\dfrac{\alpha ^{4} b_{n}^{4}}{ ( [n]_{p,q}+ \beta )^{4}}. \end{aligned}$$

     $\square$

Lemma 2.3

Let $p, q \in (0,1)$. Then for, $x \in [0,\infty )$, we have:

$$\begin{aligned} (i)\,{S}_{n,p,q}^{(\alpha ,\beta )}(t-x;x)= & {} \bigg ( \dfrac{ [n]_{p,q} }{ [n]_{p,q}+ \beta }-1\bigg ) x + \dfrac{\alpha b_{n}}{[n]_{p,q} + \beta } \\ (ii)\,{S}_{n,p,q}^{(\alpha ,\beta )}((t-x)^{2};x)= & {} \frac{((1-p)[n]_{p,q}^{2}{+}p\beta ^{2})x^{2} \,{+}\,( [n]_{p,q}\,{+}\,2\alpha \beta )pb_{n}x\,{+}\,p\alpha ^{2}b_{n}^{2}}{p ( [n]_{p,q}\,{+}\, \beta )^{2} }. \end{aligned}$$

3 Korovkin-Type Approximation Theorem

Suppose $C_{\rho }$ is the space of all continuous functions f such that $|f(x)|\le M\rho (x)$, $-\infty< x <\infty$. Then $C_{\rho }$ is a Banach space with the norm $\Vert f\Vert _{\rho }=\sup \limits _{-\infty< x <\infty } \frac{|f(x)|}{\rho (x)}$. The subsequent results are used for proving Korovkin approximation theorem on unbounded sets.

Theorem 3.1

(See [8]) There exists a sequence of positive linear operators $T_{n}$, acting from $C_{\rho }$ to $B_{\rho }$ , satisfying the conditions

• (i)
  $$ \lim \limits _{n\rightarrow \infty } \Vert T_{n} (1;x)-1 \Vert _{\rho } = 0 $$
• (ii)
  $$ \lim \limits _{n\rightarrow \infty } \Vert T_{n}(\varphi ;x)-\varphi \Vert _{\rho } = 0 $$
• (iii)
  $$ \lim \limits _{n\rightarrow \infty } \Vert T_{n} (\varphi ^{2};x)-\varphi ^{2} \Vert _{\rho } = 0, $$

where $\varphi (x)$ is a continuous and increasing function on $(-\infty ,\infty )$, such that $\lim \limits _{x\rightarrow \pm \infty } \varphi (x) = \pm \infty$, $\rho (x)=1+\varphi ^{2}$, and there exists a function $f^{*}\in C_{\rho }$, for which $\overline{ \lim \limits _{n \rightarrow \infty } } \Vert T_{n} f^{*}-f^{*} \Vert _{\rho } >0.$

Theorem 3.2

(See [8]) Conditions (i), (ii), (iii) of above theorem implies that

$$\begin{aligned} \lim \limits _{n\rightarrow \infty } \Vert T_{n} f-f \Vert _{\rho }= & {} 0, \end{aligned}$$

for any function f belonging to the subset

$$\begin{aligned} C_{\rho }^{0} =\bigg \{f\in C_{\rho }[0,\infty ):\lim \limits _{x\rightarrow \infty }\frac{\mid f(x)\mid }{\rho (x)}<\infty \bigg \}. \end{aligned}$$

Consider the weight function $\rho (x) = 1 + x^{2}$ and operators:

$$\begin{aligned} T_{n,p,q}^{(\alpha ,\beta )} (f;x)=\left\{ \begin{array}{ll} {S}_{n,p,q}^{(\alpha ,\beta )}(f;x), &{} \qquad x \in [0, b_{n}] \\ &{} \\ f(x), &{} \qquad x \in [0, \infty ) /[0, b_{n}]. \end{array} \right. \end{aligned}$$

(6)

Thus for $f \in C_{1+x^{2} }$ , we have

$$\begin{aligned} \Vert T_{n,p,q}^{(\alpha ,\beta )} (f;x) \Vert _{1+x^{2}}\le & {} \sup \limits _{x\in [0,b_{n} ]} \dfrac{|T_{n,p,q}^{(\alpha ,\beta )} (f;x) |}{1+x^{2}}+ \sup \limits _{b_{n}<x<\infty } \dfrac{|f(x) |}{1+x^{2}} \\\le & {} \Vert f \Vert _{1+x^{2} } \bigg ( \sup \limits _{x\in [0,\infty )} \dfrac{|T_{n,p,q}^{(\alpha ,\beta )} (1+t^{2};x) |}{1+x^{2}}+1 \bigg ). \end{aligned}$$

Now we will obtain,

$$\begin{aligned} \Vert T_{n,p,q}^{(\alpha ,\beta )} (f;x) \Vert _{1+x^{2}}\le & {} M \Vert f \Vert _{1+x^{2} } \end{aligned}$$

if $p := (p_{n})$ and $q := (q_{n})$ satisfy $0<q_{n}<p_{n} \le 1$ and for n sufficiently large $p_{n} \rightarrow 1$, $q_{n} \rightarrow 1$ and $p_{n}^{n} \rightarrow N$, $q_{n}^{n} \rightarrow N$, $N < \infty$ and $\lim \limits _{n\rightarrow \infty } \dfrac{b_{n}}{ [n]_{p,q} }=0$.

Theorem 3.3

Let $p := (p_{n})$ and $q := (q_{n})$ satisfy $0<q_{n}<p_{n} \le 1$ and for n sufficiently large $p_{n} \rightarrow 1$, $q_{n} \rightarrow 1$ and $p_{n}^{n} \rightarrow N$, $q_{n}^{n} \rightarrow N$, $N < \infty$ and $\lim \limits _{n\rightarrow \infty } \dfrac{b_{n}}{ [n]_{p,q} }=0$. Then, for any $f \in C^{0}_{1+x^{2}}$, we have

$$\begin{aligned} \Vert T_{n,p_{n},q_{n}}^{(\alpha ,\beta )} (f;\cdot )-f(\cdot ) \Vert _{1+x^{2}}= & {} 0. \end{aligned}$$

Using the results of Theorem 3.1, Lemma 2.2, we will obtain the following assessments, respectively:

$$\begin{aligned} \sup \limits _{x\in [0,\infty )} \dfrac{|T_{n,p_{n},q_{n}}^{(\alpha ,\beta )} (1;x) -1|}{1+x^{2}}= & {} \sup \limits _{0\le x \le b_{n}} \dfrac{| {S} _{n,p_{n},q_{n}}^{(\alpha ,\beta )} (1;x)-1 |}{1+x^{2}}=0. \end{aligned}$$

$$\begin{aligned} \sup \limits _{x\in [0,\infty )} \dfrac{|T_{n,p_{n},q_{n}}^{(\alpha ,\beta )} (t;x)-t |}{1+x^{2}}= & {} \sup \limits _{0\le x \le b_{n}} \dfrac{|{S} _{n,p_{n},q_{n}}^{(\alpha ,\beta )} (t;x) -x|}{1+x^{2}} \\\le & {} \sup \limits _{0\le x \le b_{n}} \dfrac{\bigg ( \dfrac{ [n]_{p,q} }{ [n]_{p,q}+ \beta }-1\bigg ) x + \dfrac{\alpha b_{n}}{[n]_{p,q} + \beta } }{ 1+x^{2}} \\\le & {} \dfrac{\alpha b_{n}}{[n]_{p,q} + \beta } + \bigg | \dfrac{ [n]_{p,q} }{[n]_{p,q}+ \beta }-1 \bigg | \longrightarrow 0, \end{aligned}$$

$$\begin{aligned} \sup \limits _{x\in [0,\infty )} \dfrac{|T_{n,p_{n},q_{n}}^{(\alpha ,\beta )} (t^{2};x)-t^{2} |}{1+x^{2}}= & {} \sup \limits _{0\le x \le b_{n}} \dfrac{|{S} _{n,p_{n},q_{n}}^{(\alpha ,\beta )} (t^{2};x) -x^{2}|}{1+x^{2}} \\\le & {} \sup \limits _{0\le x \le b_{n}} \dfrac{1}{1+x^{2}}\bigg ( \dfrac{ ((1-p)[n]_{p,q}^{2}+p\beta ^{2})x^{2}}{p ( [n]_{p,q}+ \beta )^{2} } \\&+\dfrac{( [n]_{p,q}+2\alpha \beta )b_{n}x}{ ( [n]_{p,q}+ \beta )^{2} } + \dfrac{\alpha ^{2}b_{n}^{2}}{ ( [n]_{p,q}+ \beta )^{2} } \bigg ) \\\le & {} \dfrac{\alpha ^{2}b_{n}^{2}}{ ( [n]_{p,q}+ \beta )^{2} } + \bigg | \dfrac{(1-p)[n]_{p,q}^{2}+p\beta ^{2}}{p ( [n]_{p,q} + \beta )^{2} } \bigg | \\&+ \bigg | \dfrac{( [n]_{p,q}+2\alpha \beta )b_{n}}{ ( [n]_{p,q}+ \beta )^{2} }\bigg | \longrightarrow 0, \end{aligned}$$

whenever $n \longrightarrow \infty$, because we have $\lim \limits _{n \rightarrow \infty } q_{n}=\lim \limits _{n\rightarrow \infty } p_{n}= 1$, $\lim \limits _{n\rightarrow \infty } \dfrac{b_{n}}{ [n]_{p,q} }=0$, as $n \longrightarrow \infty$.

Theorem 3.4

Assuming C as a positive and real number independent of n and f as a continuous function which vanishes on $[C,\infty )$. Let $p := (p_{n})$ and $q := (q_{n})$ satisfy $0<q_{n}<p_{n} \le 1$ and for n sufficiently large $p_{n} \rightarrow 1$, $q_{n} \rightarrow 1$ and $p_{n}^{n} \rightarrow N$, $q_{n}^{n} \rightarrow N$, $N < \infty$ and $\lim \limits _{n\rightarrow \infty } \dfrac{b_{n}}{ [n]_{p,q} }=0$. Then

$$\begin{aligned} \lim \limits _{n\rightarrow \infty } \sup \limits _{0\le x \le b_{n}} | {S} _{n,p_{n},q_{n}}^{(\alpha ,\beta )} (f;x)-f(x) |= & {} 0. \end{aligned}$$

Proof

From the hypothesis on f, it is bounded, i.e. $| f(x) | \le M (M>0)$. For any $\varepsilon > 0$, we have

$$\begin{aligned} \bigg | f\left( \frac{p^{1-k} [k]_{p,q}+\alpha }{[n]_{p,q}+\beta } b_{n}\right) -f(x) \bigg |< & {} \varepsilon +\dfrac{2M}{ \delta ^{2} } \bigg ( \frac{p^{1-k} [k]_{p,q}+\alpha }{[n]_{p,q}+\beta }b_{n}-x \bigg )^{2} \end{aligned}$$

where $x \in [0, b_{n}]$ and $\delta = \delta (\varepsilon )$ are independent of n. Now since we know,

$$\begin{aligned} {S}_{n,p_{n},q_{n}}^{(\alpha ,\beta )} ((t-x)^{2};x)= & {} {S} _{n,p_{n},q_{n}}^{(\alpha ,\beta )} (t^{2};x)-2x {S}_{n,p_{n},q_{n}}^{(\alpha , \beta )} (t;x)+x^{2} {S}_{n,p_{n},q_{n}}^{(\alpha ,\beta )} (1;x). \end{aligned}$$

We can conclude by Theorem 3.3,

$$\begin{aligned} \sup \limits _{0\le x \le b_{n}} | {S}_{n,p_{n},q_{n}}^{(\alpha ,\beta )} (f;x)-f(x) |\le & {} \varepsilon +\dfrac{2M}{ \delta ^{2} } \bigg ( \dfrac{ \alpha ^{2}b_{n}^{2}}{ ( [n]_{p,q}+ \beta )^{2} } + \dfrac{ ((1-p)[n]_{p,q}^{2}+p\beta ^{2})x^{2}}{p ( [n]_{p,q} + \beta )^{2} } \\&+ \dfrac{( [n]_{p,q}+2\alpha \beta )b_{n}x}{ ( [n]_{p,q}+ \beta )^{2} } \bigg ). \end{aligned}$$

Since $\dfrac{b_{n}}{ [n]_{p,q} }=0$, as $n\rightarrow \infty$, we have the desired result.    $\square$

4 Rate of Convergence

Now we give the rate of convergence of the operators ${S}_{n,p,q}^{(\alpha , \beta )}(f;x)$ in terms of the elements of the usual Lipschitz class $Lip_{M}(\gamma )$.

Let $f\in C_{B}[0,\infty )$, $M>0$ and $0<\gamma \le 1$. We recall that f belongs to the class $Lip_{M}(\gamma )$ if the inequality

$$\begin{aligned} \mid f(t)-f(x)\mid \le M\mid t-x\mid ^{\gamma }~~~t,x\in [ 0,\infty ) \end{aligned}$$

is satisfied.

Theorem 4.1

Let $0<q<p\le 1$. Then for each $f\in Lip_{M}(\gamma ),$ we have

$$\begin{aligned} \mid {S}_{n,p,q}^{(\alpha ,\beta )}(f;x)-f(x)\mid \le M (\delta _{n}(x))^{ \frac{\gamma }{2} } \end{aligned}$$

where

$$\begin{aligned} \delta _{n}(x)={S}_{n,p,q}^{(\alpha ,\beta )}((t-x)^{2};x). \end{aligned}$$

Proof

For $f\in Lip_{M}(\gamma )$, we obtain

$$\begin{aligned}&\mid {S}_{n,p,q}^{(\alpha ,\beta )}(f;x)-f(x)\mid \\&\quad = \bigg | \sum _{k=0}^{\infty } \frac{p^{ \frac{k(k-1)}{2}}}{q^{ \frac{ k(k-1)}{2}}}\frac{([n]_{p,q}x)^{k} }{ [k]_{p,q}!(b_{n})^{k} } e_{p,q}(-[n]_{p,q}q ^{-k}\frac{x}{b_{n}}) \bigg ( f\left( \frac{p^{1-k} [k]_{p,q}+\alpha }{[n]_{p,q}+\beta }b_{n}\right) -f(x)\bigg ) \bigg | \\&\quad \le \sum _{k=0}^{\infty } \frac{p^{ \frac{k(k-1)}{2}}}{q^{ \frac{k(k-1)}{2} }}\frac{([n]_{p,q}x)^{k} }{ [k]_{p,q}!(b_{n})^{k} }e_{p,q}\left( -[n]_{p,q}q ^{-k} \frac{x}{b_{n}}\right) \bigg | f\left( \frac{p^{1-k} [k]_{p,q}+\alpha }{ [n]_{p,q}+\beta }b_{n}\right) -f(x)\bigg | \\&\quad \le M\sum _{k=0}^{\infty } \frac{p^{ \frac{k(k-1)}{2}}}{q^{ \frac{k(k-1)}{2}}} \frac{([n]_{p,q}x)^{k} }{ [k]_{p,q}!(b_{n})^{k} }e_{p,q}\left( -[n]_{p,q}q ^{-k} \frac{x}{b_{n}}\right) \bigg | \frac{p^{1-k} [k]_{p,q}+\alpha }{[n]_{p,q}+\beta } b_{n} -x\bigg | ^{\gamma }. \end{aligned}$$

Applying Hölder’s inequality with the values $p=\frac{2}{\gamma }$ and $q=\frac{2}{2-\gamma }$, we get following inequality,

$$\begin{aligned}&\mid {S}_{n,p,q}^{(\alpha ,\beta )}(f;x)-f(x)\mid \\&\quad \le M\left( \sum _{k=0}^{\infty } \frac{p^{ \frac{k(k-1)}{2}}}{q^{ \frac{ k(k-1)}{2}}}\frac{([n]_{p,q}x)^{k} }{ [k]_{p,q}!(b_{n})^{k} } e_{p,q}\left( -[n]_{p,q}q ^{-k}\frac{x}{b_{n}}\right) \bigg ( \frac{p^{1-k} [k]_{p,q}+\alpha }{[n]_{p,q}+\beta }b_{n} -x\bigg ) ^{2}\right) ^{\frac{ \gamma }{2}} \\&\quad \quad \left( \sum _{k=0}^{\infty } \frac{p^{ \frac{k(k-1)}{2}}}{q^{ \frac{k(k-1)}{2 }}}\frac{([n]_{p,q}x)^{k} }{ [k]_{p,q}!(b_{n})^{k} }e_{p,q}\left( -[n]_{p,q}q ^{-k} \frac{x}{b_{n}}\right) \right) ^{\frac{2-\gamma }{2}}. \end{aligned}$$

From Lemma 2.2, we get

$$\begin{aligned}= & {} M\left( {S}_{n,p,q}^{(\alpha ,\beta )}\left( (t-x)^{2};x\right) \right) ^{ \frac{\gamma }{2} } \left( {S}_{n,p,q}^{(\alpha ,\beta )}\left( 1;x\right) \right) ^{\frac{2-\gamma }{2}} \\= & {} M\left( {S}_{n,p,q}^{(\alpha ,\beta )}\left( (t-x)^{2};x\right) \right) ^{ \frac{\gamma }{2} }. \end{aligned}$$

Choosing $\delta :\delta _{n}(x)={S}_{n,p,q}^{(\alpha ,\beta )}\left( (t-x)^{2};x\right)$,

we obtain

$$\begin{aligned} \mid {S}_{n,p,q}^{(\alpha ,\beta )}(f;x)-f(x)\mid \le M (\delta _{n}(x))^{ \frac{\gamma }{2}}. \end{aligned}$$

Hence, the desired result is obtained.    $\square$

We will estimate the rate of convergence in terms of modulus of continuity. Let $f \in C_{B}[0,\infty )$, and the modulus of continuity of f denoted by $\omega (f,\delta )$ gives the maximum oscillation of f in any interval of length not exceeding $\delta >0$ and it is given by the relation

$$\begin{aligned} \omega (f,\delta )=\max _{\mid y-x \mid \le \delta } \mid f(y)-f(x) \mid ,~~~x,y \in [0,\infty ). \end{aligned}$$

It is known that $\lim _{\delta \rightarrow 0+}\omega (f,\delta )=0$ for $f \in C_{B}[0, \infty )$ and for any $\delta >0$ one has

$$\begin{aligned} \mid f(y)- f(x) \mid \le \left( \frac{\mid y-x \mid }{\delta } +1\right) \omega (f,\delta ). \end{aligned}$$

(7)

Theorem 4.2

If $f\in C_{B}[0,\infty )$, then

$$\begin{aligned} \mid {S}_{n,p,q}^{(\alpha ,\beta )}(f;x)-f(x)\mid \le 2{\omega }(f;(\sqrt{ \delta _{n}(x)}), \end{aligned}$$

where $\omega (f; \cdot )$ is modulus of continuity of f and $\delta _{n}(x)$ be the same as in Theorem 4.1.

Proof

Using triangular inequality, we get

$$\begin{aligned} \mid {S}_{n,p,q}^{(\alpha ,\beta )}(f;x)-f(x)\mid= & {} \bigg | \sum _{k=0}^{\infty } \frac{p^{ \frac{k(k-1)}{2}}}{q^{ \frac{k(k-1)}{2}}}\frac{ ([n]_{p,q}x)^{k} }{ [k]_{p,q}!(b_{n})^{k} }e_{p,q}\left( -[n]_{p,q}q ^{-k}\frac{x}{ b_{n}}\right) \\&\times \bigg ( f\left( \frac{p^{1-k} [k]_{p,q}+\alpha }{[n]_{p,q}+\beta } b_{n}\right) -f(x)\bigg ) \bigg | \\\le & {} \sum _{k=0}^{\infty } \frac{p^{ \frac{k(k-1)}{2}}}{q^{ \frac{k(k-1)}{2 }}}\frac{([n]_{p,q}x)^{k} }{ [k]_{p,q}!(b_{n})^{k} }e_{p,q}\left( -[n]_{p,q}q ^{-k} \frac{x}{b_{n}}\right) \\&\times \bigg | f\left( \frac{p^{1-k} [k]_{p,q}+\alpha }{[n]_{p,q}+\beta } b_{n}\right) -f(x) \bigg | . \end{aligned}$$

Now using inequality (7), Hölder’s inequality and Lemma 2.2, we get

$$\begin{aligned} \mid {S}_{n,p,q}^{(\alpha ,\beta )}(f;x)-f(x)\mid= & {} \sum _{k=0}^{\infty } \frac{p^{ \frac{k(k-1)}{2}}}{q^{ \frac{k(k-1)}{2}}}\frac{([n]_{p,q}x)^{k} }{ [k]_{p,q}!(b_{n})^{k} }e_{p,q}\left( -[n]_{p,q}q ^{-k}\frac{x}{b_{n}}\right) \\&\times \left( \frac{\mid \frac{p^{1-k} [k]_{p,q}+\alpha }{[n]_{p,q}+\beta } b_{n}-x \mid }{\delta } +1\right) \omega (f,\delta ) \\\le & {} \omega (f,\delta ) \sum _{k=0}^{\infty } \frac{p^{ \frac{k(k-1)}{2}}}{ q^{ \frac{k(k-1)}{2}}}\frac{([n]_{p,q}x)^{k} }{ [k]_{p,q}!(b_{n})^{k} } e_{p,q}\left( -[n]_{p,q}q ^{-k}\frac{x}{b_{n}}\right) \\&+\dfrac{\omega (f,\delta )}{\delta }\sum _{k=0}^{\infty } \frac{p^{ \frac{ k(k-1)}{2}}}{q^{ \frac{k(k-1)}{2}}}\frac{([n]_{p,q}x)^{k} }{ [k]_{p,q}!(b_{n})^{k} }e_{p,q}\left( -[n]_{p,q}q ^{-k}\frac{x}{b_{n}}\right) \\&\times \bigg | \frac{p^{1-k} [k]_{p,q}+\alpha }{[n]_{p,q}+\beta }b_{n}-x \bigg | \\= & {} \omega (f,\delta ) +\dfrac{\omega (f,\delta )}{\delta } \bigg ( \sum _{k=0}^{\infty } \frac{p^{ \frac{k(k-1)}{2}}}{q^{ \frac{k(k-1)}{2}}}\frac{ ([n]_{p,q}x)^{k} }{ [k]_{p,q}!(b_{n})^{k} } \\&\times e_{p,q}(-[n]_{p,q}q ^{-k}\frac{x}{b_{n}}) \bigg ( \frac{p^{1-k} [k]_{p,q}+\alpha }{[n]_{p,q}+\beta }b_{n}-x \bigg ) ^{2} \bigg ) ^{ \frac{1}{2}} \\= & {} \omega (f,\delta ) +\dfrac{\omega (f,\delta )}{\delta } \bigg ( {S} _{n,p,q}^{(\alpha ,\beta )}((t-x)^{2};x) \bigg ) ^{\frac{1}{2}}. \end{aligned}$$

Now choosing $\delta =\delta _{n}(x)$ as in Theorem 4.1, we have

$$\begin{aligned} \mid {S}_{n,p,q}^{(\alpha ,\beta )}(f;x)-f(x)\mid \le 2{\omega }(f;(\sqrt{ \delta _{n}(x)}). \qquad \qquad \qquad \qquad \qquad \qquad \qquad \,\quad \square \end{aligned}$$

Now let us denote by $C_{B}^{2}[0,\infty ),$ the space of all functions $f \in C_{B}[0,\infty ),$ such that $f^{^{\prime }},f^{^{\prime \prime }}\in C_{B}[0,\infty )$. Let $\parallel f \parallel$ denote the usual supremum norm of f. Classical Peetre’s K-functional and the second modulus of smoothness of the function $f \in C_{B}[0,\infty )$ are defined, respectively, as

$$\begin{aligned} K_{2}(f,\delta )=\inf _{g\in C_{B}^{2}[0,\infty )}\{\parallel f-g\parallel +\delta \parallel g^{^{\prime \prime }}\parallel \}, \end{aligned}$$

where $\delta >0$ and $g\in C_{B}^{2}[0,\infty )$. By Theorem 2.4 of [7], there exists an absolute constant $C>0$ such that

$$\begin{aligned} K_{2}(f,\delta )\le C\omega _{2}(f,\sqrt{\delta }) \end{aligned}$$

(8)

where

$$\begin{aligned} \omega _{2} (f,\sqrt{\delta })=\sup _{0<h \le \sqrt{\delta }} \sup _{x, x+h\in I}\mid f(x+2h)-2f(x+h)+f(x)\mid \end{aligned}$$

is the second-order modulus of smoothness of $f\in C_{B}^{2} [0,\infty )$. The usual modulus of continuity of $f\in C_{B}^{2} [0,\infty )$ is defined by

$$\begin{aligned} \omega (f,\delta )=\sup _{0<h\le \delta } \sup _{x\in [0,\infty )}\mid f(x+h)-f(x)\mid . \end{aligned}$$

Theorem 4.3

Let $x \in [0, b_{n}],$ $f \in C_{B}[0,\infty )$ and  $0< q < p \le 1$, $0\le \alpha \le \beta$. Then for all $n\in \mathbb {N},$ there exists a positive constant $C > 0$ such that

$$\begin{aligned} \mid {S}_{n,p,q}^{(\alpha ,\beta )}(f;x)-f(x) \mid\le & {} C \omega _{2}(f,\delta _{n}(x) )+\omega (f,\alpha _{n}(x) ), \end{aligned}$$

where

$$\begin{aligned} \delta _{n}(x)=\sqrt{{S}_{n,p,q}^{(\alpha ,\beta )}((t-x)^{2}; x)+(\alpha _{n}(x) })^{2} , \quad \alpha _{n}(x)=\bigg ( \dfrac{ [n]_{p,q} }{[n]_{p,q}+ \beta } -1\bigg ) x + \dfrac{\alpha b_{n}}{[n]_{p,q} + \beta } . \end{aligned}$$

Proof

For $x \in [0,\infty )$, we consider the auxiliary operators ${\bar{S}} _{n}^{*}$ defined by

$$\begin{aligned} {\bar{S}}_{n}^{*}(f; x)= & {} {S}_{n,p,q}^{(\alpha ,\beta )}(f; x)+f(x)-f\left( \frac{[n]_{p,q}}{[n]_{p,q}+\beta }x +\frac{\alpha b_{n} }{ [n]_{p,q}+\beta } \right) . \end{aligned}$$

From Lemma 2.2 (i) (ii) and Lemma 2.3 (i), we observe that the operators ${\bar{S}}_{n}^{*}(f; x)$ are linear and reproduce the linear functions. Hence,

$$\begin{aligned} {\bar{S}}_{n}^{*}(1; x)= & {} {S}_{n,p,q}^{(\alpha ,\beta )}(1; x)+1-1=1 \\ {\bar{S}}_{n}^{*}(t; x)= & {} {S}_{n,p,q}^{(\alpha ,\beta )}(t; x)+x- \left( \frac{[n]_{p,q}}{[n]_{p,q}+\beta }x +\frac{\alpha b_{n} }{[n]_{p,q}+\beta } \right) =x \\ {\bar{S}}_{n}^{*}((t-x); x)= & {} {\bar{S}}_{n}^{*}(t; x)-x{\bar{S}} _{n}^{*}(1; x)=0. \end{aligned}$$

Let $x \in [0,\infty )$ and $g \in C_{B}^{2}[0,\infty ).$ Using Taylor’s formula

$$\begin{aligned} g(t) = g(x)+ g^{\prime }(x)(t-x)+\int _{x}^t(t-u)g^{\prime \prime }(u) \mathrm {d}u. \end{aligned}$$

Applying ${\bar{S}}_{n}^{*}$ to both sides of the above equation, we have

$$\begin{aligned} {\bar{S}}_{n}^{*}(g;x)-g(x)= & {} g^{\prime }(x){\bar{S}}_{n}^{*}((t-x); x)+ {\bar{S}}_{n}^{*}\left( \int _{x}^t(t-u)g^{\prime \prime }(u) \mathrm {d }u;x\right) \\= & {} {S}_{n,p,q}^{(\alpha ,\beta )} \left( \int _{x}^t(t-u)g^{\prime \prime }(u) \mathrm {d}u;x\right) \\&- \int _{x}^{\frac{[n]_{p,q}}{[n]_{p,q}+\beta }x +\frac{\alpha b_{n} }{ [n]_{p,q}+\beta }} \left( \frac{[n]_{p,q}}{[n]_{p,q}+\beta }x +\frac{\alpha b_{n} }{[n]_{p,q}+\beta } -u\right) g^{\prime \prime }(u) \mathrm {d}u. \end{aligned}$$

On the other hand, since

$$\begin{aligned} \bigg | \int _{x}^t(t-u)g^{\prime \prime }(u) \mathrm {d}u \bigg |\le & {} \int _{x}^t\mid t-u\mid \mid g^{\prime \prime }(u)\mid \mathrm {d}u \le \parallel g^{\prime \prime } \parallel \bigg | \int _{x}^t\mid t-u\mid \mathrm {d}u\bigg | \le (t-x)^{2}\parallel g^{\prime \prime } \parallel \end{aligned}$$

and

$$\begin{aligned} \bigg | \int _{x}^{\frac{[n]_{p,q}}{[n]_{p,q}+\beta }x +\frac{\alpha b_{n} }{ [n]_{p,q}+\beta }} \left( \frac{[n]_{p,q}}{[n]_{p,q}+\beta }x +\frac{\alpha b_{n} }{[n]_{p,q}+\beta } -u\right) g^{\prime \prime }(u) \mathrm {d}u \bigg | \end{aligned}$$

$$\begin{aligned} \le \left( \frac{[n]_{p,q}}{[n]_{p,q}+\beta }x+\frac{\alpha b_{n}}{ [n]_{p,q}+\beta }-x\right) ^{2}\parallel g^{\prime \prime }\parallel . \end{aligned}$$

We conclude that

$$\begin{aligned} \bigg | {\bar{S}}_{n}^{*}(g;x)-g(x)\bigg |\le & {} \bigg | {S} _{n,p,q}^{(\alpha ,\beta )}\left( \int _{x}^{t}(t-u)g^{\prime \prime }(u) \mathrm {d}u;x\right) \bigg | \\&+\bigg | \int _{x}^{\frac{[n]_{p,q}}{[n]_{p,q}+\beta }x+\frac{\alpha b_{n}}{ [n]_{p,q}+\beta }}\left( \frac{[n]_{p,q}}{[n]_{p,q}+\beta }x+\frac{\alpha b_{n}}{[n]_{p,q}+\beta }-u\right) g^{\prime \prime }(u)\mathrm {d}u\bigg | \\\le & {} \parallel g^{\prime \prime }\parallel {S}_{n,p,q}^{(\alpha ,\beta )}((t-x)^{2};x)+\parallel g^{\prime \prime }\parallel \left( \frac{[n]_{p,q}}{[n]_{p,q}+\beta }x+\frac{\alpha b_{n}}{[n]_{p,q}+\beta }-x\right) ^{2} \\= & {} \parallel g^{\prime \prime }\parallel \delta _{n}^{2}(x). \end{aligned}$$

Now, taking into account Lemma 2.2 (i), we have

$$\begin{aligned} \mid {\bar{S}}_{n}^{*}(f;x)\mid \le \mid {S}_{n,p,q}^{(\alpha ,\beta )}(f;x)\mid +2\parallel f\parallel \le 3\parallel f\parallel . \end{aligned}$$

Therefore,

$$\begin{aligned} \mid {S}_{n,p,q}^{(\alpha ,\beta )}(f;x)-f(x)\mid\le & {} \mid {\bar{S}} _{n}^{*}(f-g;x)-(f-g)(x)\mid \\&+\bigg | f\left( \frac{[n]_{p,q}}{[n]_{p,q}+\beta }x+\frac{\alpha b_{n}}{ [n]_{p,q}+\beta }\right) -f(x)\bigg | +\mid {\bar{S}}_{n}^{*}(g;x)-g(x)\mid \\\le & {} 4\parallel f-g\parallel +\omega (f,\alpha _{n}(x))+\delta _{n}^{2}(x)\parallel g^{\prime \prime }\parallel . \end{aligned}$$

Hence, taking the infimum on the right-hand side over all $g\in C_{B}^{2}[0,\infty ),$ we have the following result

$$\begin{aligned} \mid {S}_{n,p,q}^{(\alpha ,\beta )}(f;x)-f(x)\mid \le 4K_{2}(f,\delta _{n}^{2}(x))+\omega (f,\alpha _{n}(x)). \end{aligned}$$

In view of the property of K-functional, we get

$$\begin{aligned} \mid {S}_{n,p,q}^{(\alpha ,\beta )}(f;x)-f(x)\mid \le C\omega _{2}(f,\delta _{n}(x))+\omega (f,\alpha _{n}(x)). \end{aligned}$$

This completes the proof of the theorem.    $\square$