Some Approximation Results on Two Parametric q-Stancu–Beta Operators

Abstract

In the present paper, we introduce a two parametric q-analogue of Stancu-Beta operators and establish some direct results in the polynomial weighted space of continuous functions defined on the interval $[0,\infty )$. We use Lipschitz-type maximal function to find pointwise estimate. Furthermore, we obtain a Voronovskaja-type theorem for these operators.

1 Introduction and Notations

In [18], Stancu introduced the sequence of Beta operators of second kind in order to approximate the Lebesgue integrable functions on the interval $(0,\infty )$. After the papers of Lupaş [11] and Phillips [17], the q-calculus has been extensively used in Approximation Theory which has become recently one of the most interesting areas of research. Ali and Gupta [1] constructed the q-analogue of the Stancu–Beta operators, and they established direct results in terms of modulus of continuity and also presented an asymptotic formula for Voronovskaja-type theorem.

Recently, q-generalizations of various operators and their approximation properties are studied in [12,13,14,15,16]. First, we recall certain notations of q-calculus.

For each nonnegative integer k, the q-integer $[k]_{q}$ and q-factorial $[k]_{q}!$ are defined by

$$\begin{aligned}{}[k]_{q}:=\left\{ \begin{array}{ll} \frac{1-q^{k}}{1-q}, &{} \quad q\ne 1 \\ ~n, &{} \quad ~q=1 \end{array} \right. \hbox {for}\,k\in {\mathbb {N}} \text { and } [0]_q=0, \end{aligned}$$

and

$$\begin{aligned}{}[k]_{q}!:=\left\{ \begin{array}{ll} [k]_{q}[k-1]_{q}\dots [1]_{q}, &{} \quad k\ge 1 \\ ~1, &{} \quad k=0 \end{array} \right. \end{aligned}$$

respectively. For the integers n,  k, $0\le k\le n$, the q-binomial coefficients are defined by

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

(1)

Note that $(a+b)_{q}^{n}=\prod _{j=0}^{n-1}(a+q^{j}b)$.

The q-improper integral is defined as (see [9])

$$\begin{aligned} \int _{0}^{\infty /A}f(x)\text{ d }_qx=(1-q)\sum \limits _{n=-\infty }^{\infty }f \left( \frac{q^n}{A}\right) \frac{q^n}{A},~~ A>0. \end{aligned}$$

(2)

For $t>0$, the q-gamma function is defined by

$$\begin{aligned} \Gamma _{q}(t)=\int _{0}^{\frac{1}{1-q}}x^{t-1}E_{q}(-qx)\text{ d }_{q}x, \end{aligned}$$

(3)

where $E_{q}(x)=\sum \limits _{n=0}^{\infty }q^{n(n-1)/2}\frac{x^{n}}{[n]_{q}!}$. Also $\Gamma _{q}(t+1)=[t]_{q}\Gamma _{q}(t)$, $\Gamma _{q}(1)=1$.

The q-Beta integral representations are given by

$$\begin{aligned} B_{q}(t,s)=K(A,t)\int _{0}^{\infty /A}\frac{x^{t-1}}{(1+x)_{q}^{t+s}}\text{ d } _{q}x, \end{aligned}$$

(4)

where

$$\begin{aligned} K(A,t+1)=q^{t}K(A,t)~ \text{ and } K(A,t)=\frac{1}{A+1}A^{t}\left( 1+\frac{1 }{A}\right) _{q}^{t}(1+A)_{q}^{1-t}, \end{aligned}$$

for $A>0$ (see [3]). In particular for any positive integer n,

$$\begin{aligned} K(A,n)= & {} q^{\frac{n(n-1)}{2}}, \quad K(A,0)=1 \quad \text{ and } \quad \nonumber \\ B_q(t;s)= & {} \frac{\Gamma _q(t)\Gamma _q(s)}{\Gamma _q(t+s)}. \end{aligned}$$

(5)

For more details on the q-calculus, one can refer to [8] and for the applications of q-calculus in approximation theory, we recommend the readers [2].

Let $B_m[0,\infty )$ be the set of all functions f satisfying the condition that $|f(x)|\le M_f(1+x^m),~x\in [0,\infty ),~m>0$ with some constant $M_f$ depending on f. Introduce

$$\begin{aligned}&\displaystyle C_m[0,\infty )=B_m[0,\infty )\cap C[0,\infty ),\\&\displaystyle C_m^*[0,\infty ){:}=\left\{ f\in C_m[0,\infty ): \exists \lim \limits _{x\rightarrow \infty } \frac{|f(x)|}{1+x^m}<\infty \right\} . \end{aligned}$$

These spaces are endowed with the norm

$$\begin{aligned} \Vert f\Vert _m:=\sup \limits _{x\in [0,\infty )}\frac{|f(x)|}{1+x^m}. \end{aligned}$$

The aim of this paper is to introduce a two parametric q-analogue of Stancu–Beta operators and establish some direct results in the polynomial weighted space of continuous functions defined on the interval $[0,\infty )$ . We use Lipschitz-type maximal function to find pointwise estimate. Furthermore, we obtain a Voronovskaja-type theorem for these operators.

2 Operators and Estimation of Moments

In order to introduce two parametric q-Stancu–Beta operators, we present a construction due to Aral and Gupta [1].

Definition 2.1

Let $q\in (0,1)$ and $n\in {\mathbb {N}}$. For $f{:}[0,\infty )\rightarrow {\mathbb {R}}$, the q-analogue of Stancu–Beta operators are defined as

$$\begin{aligned} L_{n}^{q}(f;x)=\frac{K(A,[n]_{q}x)}{B_{q}([n]_{q}x,[n]_{q}+1)} \int _{0}^{\infty /A}\frac{u^{[n]_{q}x-1}}{(1+u)_{q}^{[n]_{q}x+[n]_{q}+1}} f\left( q^{[n]_{q}x}u\right) \text{ d }_{q}u. \end{aligned}$$

(6)

Lemma 2.1

([1]) We have

$$\begin{aligned} L_{n}^{q}(e_{0};x)=1, \quad L_{n}^{q}(e_{1};x)=x, \quad L_{n}^{q}(e_{2};x)=\frac{ ([n]_{q}x+1)x}{q([n]_{q}-1)}. \end{aligned}$$

Now we define the q-analogue of Stancu–Beta operators with two parameters $\alpha$ and $\beta$ as follows:

Definition 2.2

Let $q\in (0,1)$ and $n\in {\mathbb {N}}$. For $f:[0,\infty )\rightarrow {\mathbb {R}}$, the q-analogue of Stancu–Beta operators are defined as

$$\begin{aligned} L_{n,q}^{(\alpha ,\beta )}(f;x)&=\frac{K(A,[n]_{q}x)}{ B_{q}([n]_{q}x,[n]_{q}+1)}\int _{0}^{\infty /A}\frac{u^{[n]_{q}x-1}}{ (1+u)_{q}^{[n]_{q}x+[n]_{q}+1}}\nonumber \\&\qquad f\left( \frac{[n]_{q}q^{[n]_{q}x}u+\alpha }{ [n]_{q}+\beta }\right) \text{ d }_{q}u. \end{aligned}$$

(7)

where $0\le \alpha \le \beta$. If we take $\alpha =\beta =0$ in the above operator, it reduces to the operator (6). Moments $L_{n,q}^{(\alpha ,\beta )}(e_{m};x)$ are of particular importance in approximation theory by positive operators. From (7) we easily derive the following formula for moments $L_{n,q}^{(\alpha ,\beta )}(e_{m};x),~m=0,1,2$.

Lemma 2.2

The operators defined at (2) verify the following identities

$$\begin{aligned} L_{n,q}^{(\alpha ,\beta )}(e_{0};x)= & {} 1, \quad L_{n,q}^{(\alpha ,\beta )}(e_{1};x)=\frac{[n]_{q}}{[n]_{q}+\beta }x+\frac{\alpha }{[n]_{q}+\beta },\\ L_{n,q}^{(\alpha ,\beta )}(e_{2};x)= & {} \frac{[n]_{q}^{2}([n]_{q}x+1)x}{ q([n]_{q}+\beta )^{2}([n]_{q}-1)}+\frac{2\alpha [n]_{q}x}{ ([n]_{q}+\beta )^{2}}+\frac{\alpha ^{2}}{([n]_{q}+\beta )^{2}}. \end{aligned}$$

Proof

By the definition of q-Stancu–Beta operators we have

$$\begin{aligned} L_{n,q}^{(\alpha ,\beta )}(e_{0};x)= & {} \frac{K(A,[n]_{q}x)}{ B_{q}([n]_{q}x,[n]_{q}+1)}\int _{0}^{\infty /A}\frac{u^{[n]_{q}x-1}}{ (1+u)_{q}^{[n]_{q}x+[n]_{q}+1}}\text{ d }_{q}u=L_{n}^{q}(e_{0};x) \\= & {} 1.\\ L_{n,q}^{(\alpha ,\beta )}(e_{1};x)= & {} \frac{K(A,[n]_{q}x)}{ B_{q}([n]_{q}x,[n]_{q}+1)}\int _{0}^{\infty /A}\frac{u^{[n]_{q}x-1}}{ (1+u)_{q}^{[n]_{q}x+[n]_{q}+1}}\frac{[n]_{q}q^{[n]_{q}x}u+\alpha }{ [n]_{q}+\beta }\text{ d }_{q}u \\= & {} \frac{[n]_{q}}{[n]_{q}+\beta }\frac{K(A,[n]_{q}x)q^{[n]_{q}x}}{ B_{q}([n]_{q}x,[n]_{q}+1)}\int _{0}^{\infty /A}\frac{u^{[n]_{q}x}}{ (1+u)_{q}^{[n]_{q}x+[n]_{q}+1}}\text{ d }_{q}u \\&+\frac{\alpha }{[n]_{q}+\beta }\frac{K(A,[n]_{q}x)}{ B_{q}([n]_{q}x,[n]_{q}+1)}\int _{0}^{\infty /A}\frac{u^{[n]_{q}x-1}}{ (1+u)_{q}^{[n]_{q}x+[n]_{q}+1}}\text{ d }_{q}u \\= & {} \frac{[n]_{q}}{[n]_{q}+\beta }L_{n}^{q}(e_{1};x)+\frac{\alpha }{ [n]_{q}+\beta }L_{n}^{q}(e_{0};x) \\= & {} \frac{[n]_{q}}{[n]_{q}+\beta }x+\frac{\alpha }{[n]_{q}+\beta }. \end{aligned}$$

Finally, by using the q-Beta integral, we have

$$\begin{aligned} L_{n,q}^{(\alpha ,\beta )}(e_{2};x)= & {} \frac{K(A,[n]_{q}x)}{ B_{q}([n]_{q}x,[n]_{q}+1)}\int _{0}^{\infty /A}\frac{u^{[n]_{q}x-1}}{ (1+u)_{q}^{[n]_{q}x+[n]_{q}+1}}\\&\left( \frac{[n]_{q}q^{[n]_{q}x}u+\alpha }{ [n]_{q}+\beta }\right) ^{2}\text{ d }_{q}u \\= & {} \frac{[n]_{q}^{2}}{([n]_{q}+\beta )^{2}}\frac{K(A,[n]_{q}x)q^{2[n]_{q}x}}{ B_{q}([n]_{q}x,[n]_{q}+1)}\int _{0}^{\infty /A}\frac{u^{[n]_{q}x+1}}{ (1+u)_{q}^{[n]_{q}x+[n]_{q}+1}}\text{ d }_{q}u \\&+\,\frac{2\alpha [n]_{q}}{([n]_{q}+\beta )^{2}}\frac{ K(A,[n]_{q}x)q^{[n]_{q}x}}{B_{q}([n]_{q}x,[n]_{q}+1)}\int _{0}^{\infty /A} \frac{u^{[n]_{q}x}}{(1+u)_{q}^{[n]_{q}x+[n]_{q}+1}}\text{ d }_{q}u \\&+\,\frac{\alpha ^{2}}{([n]_{q}+\beta )^{2}}\frac{K(A,[n]_{q}x)}{ B_{q}([n]_{q}x,[n]_{q}+1)}\int _{0}^{\infty /A}\frac{u^{[n]_{q}x-1}}{ (1+u)_{q}^{[n]_{q}x+[n]_{q}+1}}\text{ d }_{q}u \\= & {} \frac{[n]_{q}^{2}}{([n]_{q}+\beta )^{2}}L_{n}^{q}(e_{2};x) +\frac{2\alpha [n]_{q}}{([n]_{q}+\beta )^{2}}L_{n}^{q}(e_{1};x)\\&+\,\frac{\alpha ^{2}}{ ([n]_{q}+\beta )^{2}}L_{n}^{q}(e_{0};x) \\= & {} \frac{[n]_{q}^{2}([n]_{q}x+1)x}{q([n]_{q}+\beta )^{2}([n]_{q}-1)}+\frac{ 2\alpha [n]_{q}x}{([n]_{q}+\beta )^{2}}+\frac{\alpha ^{2}}{ ([n]_{q}+\beta )^{2}}. \end{aligned}$$

$\square$

Remark 2.1

Let $q\in (0,1)$. Then for $x\in [0,\infty )$, we can have the following formula for the mth order moment:

$$\begin{aligned} L_{n,q}^{(\alpha ,\beta )}(e_{m};x)=\sum \limits _{j=0}^{m}{\left( {\begin{array}{c}m\\ j\end{array}}\right) }\frac{ [n]_{q}^{m-j}\alpha ^{j}}{([n]_{q}+\beta )^{m}}\frac{\Gamma _{q}\bigl ( [n]_{q}x+m-j\bigl )\Gamma _{q}\bigl ([n]_{q}-m+j+1\bigl )}{ \Gamma _{q}\bigl ([n]_{q}x\bigl )\Gamma _{q}\bigl ([n]_{q}+1 \bigl )q^{(m-j)(m-j-1)/2}}. \end{aligned}$$

3 Direct Theorem

By $C_{B}[0,\infty )$ we mean the class of all real valued continuous bounded functions f on $[0,\infty )$ endowed with the norm $\Vert f\Vert =\sup \{|f(x)|:x\in [0,\infty )\}$. By

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

we denote the usual modulus of continuity of $f\in C_{B}[0,\infty )$.

The Peetre’s K-functional is defined by

$$\begin{aligned} K_{2}(f,\delta )=\inf \{\Vert f-g\Vert +\delta \Vert g^{\prime \prime }\Vert :g\in W^{2}\}, \end{aligned}$$

where $\delta >0$ and $W^{2}=\{g\in C_{B}[0,\infty ):g^{\prime },g^{\prime \prime }\in C_{B}[0,\infty )\}$. By [4], p. 177, Theorem 2.4], 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

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

is the second-order modulus of smoothness of $f\in C_{B}[0,\infty )$.

First we prove the following lemma which will be needed in proving the main result of this section.

Lemma 3.1

Let $n>1$ be a given number. For every $q\in (0,1)$ we have

$$\begin{aligned} L_{n,q}^{(\alpha ,\beta )}\bigl ((e_{1}-e_{0}x)^{2};x\bigl )\le & {} \left( \frac{[n]_{q}}{q([n]_{q}-1)}-\frac{[n]_{q}-\beta }{[n]_{q}+\beta }\right) x^{2}\\&\quad +\,s \frac{x}{q([n]_{q}-1)}+\frac{\alpha ^{2}}{([n]_{q}+\beta )^{2}} \\\le & {} \frac{2(\beta +1)^{2}x^{2}+x+\alpha ^{2}}{q([n]_{q}-1)}. \end{aligned}$$

where $x\in [0,\infty )$.

Proof

By using the linearity of the operator, we have

$$\begin{aligned}&L_{n,q}^{(\alpha ,\beta )}\bigl ((e_{1}-e_{0}x)^{2};x\bigl ) \\&\quad =\frac{[n]_{q}^{2}([n]_{q}x+1)x}{q([n]_{q}+\beta )^{2}([n]_{q}-1)}+\frac{ 2\alpha [n]_{q}x}{([n]_{q}+\beta )^{2}}+\frac{\alpha ^{2}}{ ([n]_{q}+\beta )^{2}}\\&\qquad -\,2x\left\{ \frac{[n]_{q}}{[n]_{q}+\beta }x+\frac{\alpha }{[n]_{q}+\beta }\right\} +x^{2} \\&\quad =\left\{ \frac{[n]_{q}^{3}}{q([n]_{q}+\beta )^{2}([n]_{q}-1)}-\frac{ 2[n]_{q}}{[n]_{q}+\beta }+1\right\} x^{2} \\&\qquad +\left\{ \frac{[n]_{q}^{2}}{q([n]_{q}+\beta )^{2}([n]_{q}-1)}+\frac{ 2\alpha [n]_{q}}{([n]_{q}+\beta )^{2}}-\frac{2\alpha }{[n]_{q}+\beta } \right\} x+\frac{\alpha ^{2}}{([n]_{q}+\beta )^{2}} \\&\quad \le \left( \frac{[n]_{q}}{q([n]_{q}-1)}-\frac{[n]_{q}-\beta }{ [n]_{q}+\beta }\right) x^{2}+\frac{x}{q([n]_{q}-1)}+\frac{\alpha ^{2}}{ ([n]_{q}+\beta )^{2}}\\&\quad =\frac{\bigl \{(1-q)[n]_{q}^{3}+([n]_{q}+q[n]_{q}-q)\beta ^{2}+(2\beta +q)[n]_{q}^{2}\bigl \}x^{2}+([n]_{q}+\beta )^{2}x+q([n]_{q}-1)\alpha ^{2}}{ q([n]_{q}-1)([n]_{q}+\beta )^{2}} \\&\quad =\frac{\bigl \{(1-q^{n})[n]_{q}^{2}+([n]_{q}+q[n]_{q}-q)\beta ^{2}+(2\beta +q)[n]_{q}^{2}\bigl \}x^{2}+([n]_{q}+\beta )^{2}x+q([n]_{q}-1)\alpha ^{2}}{q([n]_{q}-1)([n]_{q}+\beta )^{2}} \\&\quad \le \frac{\bigl \{[n]_{q}^{2}+(1+q)[n]_{q}^{2}\beta ^{2}+(2\beta +q)[n]_{q}^{2}\bigl \}x^{2}+([n]_{q}+\beta )^{2}x+[n]_{q}\alpha ^{2}}{ q([n]_{q}-1)([n]_{q}+\beta 2)^{2}} \\&\quad \le \frac{2(\beta ^{2}+\beta +1)x^{2}+x+\alpha ^{2}}{q([n]_{q}-1)}\le \frac{2(\beta +1)^{2}x^{2}+x+\alpha ^{2}}{q([n]_{q}-1)}. \end{aligned}$$

In the following theorem, we use the notation $\varphi (x)= \sqrt{x(1+x)}$.$\square$

Theorem 3.1

Let $f\in C_{B}[0,\infty )$, with $q\in (0,1)$ .Then for every $x\in [0,\infty )$ and $n\ge 2$, we have

$$\begin{aligned} \left| L_{n,q}^{(\alpha ,\beta )}(f;x)-f(x)\right| \le C\omega _{2}\left( f,\frac{\delta _{n}(x)}{\sqrt{q([n]_{q}-1)}}\right) +\omega \left( f,\frac{\alpha -\beta x}{[n]_{q}+\beta }\right) , \end{aligned}$$

where $\delta _{n}(x)=\sqrt{\varphi ^{2}(x)+\frac{2q\alpha ^{2}}{ [n]_{q}+\beta }}$ and C is an absolute constant.

Proof

We consider the auxiliary operators $\widehat{ L_{n,q}^{(\alpha ,\beta )}}$ which for $x\in [0,\infty )$ can be defined as

$$\begin{aligned} \widehat{L_{n,q}^{(\alpha ,\beta )}}(f;x)=L_{n,q}^{(\alpha ,\beta )}(f;x)-f\left( \frac{[n]_{q}x+\alpha }{[n]_{q}+\beta }\right) +f(x). \end{aligned}$$

(9)

By Lemma 2.2, it may be seen that the operators $\widehat{L_{n,q}^{(\alpha ,\beta )}}$ are linear and reproduce the linear functions:

$$\begin{aligned} \widehat{L_{n,q}^{(\alpha ,\beta )}}(e_{1}-e_{0}x;x)=0. \end{aligned}$$

(10)

Let $g\in W^{2}$. Then by Taylor’s theorem, we have

$$\begin{aligned} g(t)=g(x)+g^{\prime }(x)(t-x)+\int _{x}^{t}(t-u)g^{\prime \prime }(u)\hbox {d}u, \quad t\in [0,\infty ). \end{aligned}$$

By using (10), we get

$$\begin{aligned} \widehat{L_{n,q}^{(\alpha ,\beta )}}(g;x)=g(x)+\widehat{L_{n,q}^{(\alpha ,\beta )}}\left( \int _{x}^{t}(t-u)g^{\prime \prime }(u)\hbox {d}u;x\right) . \end{aligned}$$

Now using (9), we have

$$\begin{aligned}&\left| \widehat{L_{n,q}^{(\alpha ,\beta )}}(g;x)-g(x)\right| \nonumber \\&\quad \le \left| L_{n,q}^{(\alpha ,\beta )}\left( \int _{x}^{t}(t-u)g^{\prime \prime }(u)\hbox {d}u;x\right) \right| +\left| \int _{x}^{\frac{ [n]_{q}x+\alpha }{[n]_{q}+\beta }}\left( \frac{[n]_{q}x+\alpha }{ [n]_{q}+\beta }-u\right) g^{\prime \prime }(u)\hbox {d}u\right| \nonumber \\&\quad \le L_{n,q}^{(\alpha ,\beta )}\left( \left| \int _{x}^{t}|t-u||g^{\prime \prime }(u)|\hbox {d}u\right| ;x\right) +\int _{x}^{ \frac{[n]_{q}x+\alpha }{[n]_{q}+\beta }}\left| \frac{[n]_{q}x+\alpha }{ [n]_{q}+\beta }-u\right| |g^{\prime \prime }(u)|\hbox {d}u\nonumber \\&\quad \le \left\{ L_{n,q}^{(\alpha ,\beta )}\bigl ((e_{1}-e_{0}x)^{2};x\bigl ) +\left( \frac{\alpha -\beta x}{[n]_{q}+\beta }\right) ^{2}\right\} \Vert g^{\prime \prime }\Vert . \end{aligned}$$

(11)

By Lemma 3.1, we get

$$\begin{aligned}&L_{n,q}^{(\alpha ,\beta )}\bigl ((e_{1}-e_{0}x)^{2};x\bigl )+\left( \frac{\alpha -\beta x}{[n]_{q}+\beta }\right) ^{2} \\&\quad \le \left( \frac{[n]_{q}}{q([n]_{q}-1)}-\frac{[n]_{q}-\beta }{ [n]_{q}+\beta }\right) x^{2}+\frac{x}{q([n]_{q}-1)}+\frac{\alpha ^{2}}{ ([n]_{q}+\beta )^{2}}+\left( \frac{\alpha -\beta x}{[n]_{q}+\beta }\right) ^{2} \\&\quad =\left( \frac{[n]_{q}}{q([n]_{q}-1)}+\frac{[n]_{q}^{2}+2\beta ^{2}}{ ([n]_{q}+\beta )^{2}}\right) x^{2}+\left( \frac{1}{q([n]_{q}-1)}-\frac{ 2\alpha \beta }{([n]_{q}+\beta )^{2}}\right) x\\&\quad \quad +\,\frac{2\alpha ^{2}}{ ([n]_{q}+\beta )^{2}} \\&\quad \le \frac{(1-q)[n]_{q}^{3}x^{2}}{q([n]_{q}+\beta )^{2}([n]_{q}-1)}+\frac{ \bigl ([n]_{q}^{2}+4q(1-[n]_{q})\alpha \beta \bigl )x}{ q([n]_{q}+\beta )^{2}([n]_{q}-1)}+\frac{2\alpha ^{2}}{([n]_{q}+\beta )^{2}}\\&\quad \le \frac{(1-q)[n]_{q}^{3}x^{2}+[n]_{q}^{2}x+2q([n]_{q}-1)\alpha ^{2}}{ q([n]_{q}-1)([n]_{q}+\beta )^{2}} \\&\quad =\frac{(1-q)[n]_{q}^{3}x^{2}+[n]_{q}^{2}x+2q([n]_{q}-1)\alpha ^{2}}{ q([n]_{q}-1)([n]_{q}+\beta )^{2}} \\&\quad =\frac{(1-q^{n})[n]_{q}^{2}x^{2}+[n]_{q}^{2}x+2q([n]_{q}-1)\alpha ^{2}}{ q([n]_{q}-1)([n]_{q}+\beta )^{2}} \\&\quad \le \frac{[n]_{q}^{2}x^{2}+[n]_{q}^{2}x+2q[n]_{q}\alpha ^{2}}{ q([n]_{q}-1)([n]_{q}+\beta )^{2}}\le \frac{[n]_{q}x(1+x)+2q\alpha ^{2}}{ q([n]_{q}-1)([n]_{q}+\beta )} \\&\quad \le \frac{1}{q([n]_{q}-1)}\left( \varphi ^{2}(x)+\frac{2q\alpha ^{2}}{ [n]_{q}+\beta }\right) =\frac{\delta _{n}^{2}(x)}{q([n]_{q}-1)}. \end{aligned}$$

Then using (11), we get

$$\begin{aligned} \left| \widehat{L_{n,q}^{(\alpha ,\beta )}}(g;x)-g(x)\right| \le \frac{\delta _{n}^{2}(x)}{q([n]_{q}-1)}\Vert g^{\prime \prime }\Vert . \end{aligned}$$

(12)

On the other hand by (9), we have

$$\begin{aligned} \left| \widehat{L_{n,q}^{(\alpha ,\beta )}}(f;x)\right| \le \left| L_{n,q}^{(\alpha ,\beta )}(f;x)\right| +2\Vert f\Vert \le 3\Vert f\Vert . \end{aligned}$$

(13)

Hence by (9), (12) and (13), we get

$$\begin{aligned}&\left| L_{n,q}^{(\alpha ,\beta )}(f;x)-f(x)\right| \\&\quad \le \left| \widehat{L_{n,q}^{(\alpha ,\beta )}}(f-g;x)-(f-g)(x)\right| +\left| \widehat{L_{n,q}^{(\alpha ,\beta )}}(g;x)-g(x)\right| \\&\qquad +\,\left| f\left( \frac{[n]_{q}x+\alpha }{[n]_{q}+\beta }\right) -f(x)\right| \\&\quad \le 4\Vert f-g\Vert +\frac{\delta _{n}^{2}(x)}{q([n]_{q}-1)}\Vert g^{\prime \prime }\Vert +\omega \left( f,\frac{\alpha -\beta x}{ [n]_{q}+\beta }\right) . \end{aligned}$$

Now by taking infimum on the right-hand side over all $g\in W^{2}$, we get

$$\begin{aligned} \left| L_{n,q}^{(\alpha ,\beta )}(f;x)-f(x)\right| \le CK_{2}\left( f,\frac{\delta _{n}^{2}(x)}{q([n]_{q}-1)}\right) +\omega \left( f,\frac{ \alpha -\beta x}{[n]_{q}+\beta }\right) . \end{aligned}$$

In view of the property of the K-functional for $q\in (0,1)$, we get

$$\begin{aligned} \left| L_{n,q}^{(\alpha ,\beta )}(f;x)-f(x)\right| \le C\omega _{2}\left( f,\frac{\delta _{n}(x)}{\sqrt{q([n]_{q}-1)}}\right) +\omega \left( f,\frac{\alpha -\beta x}{[n]_{q}+\beta }\right) . \end{aligned}$$

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

4 Rate of Approximation

By $H_{x^{2}}[0,\infty ),$ we denote the set of all functions f defined on $[0,\infty )$ satisfying the condition $|f(x)|\le M_{f}(1+x^{2})$, where $M_{f}$ is a constant depending only on f. Let $C_{x^{2}}[0,\infty )$ be the subspace of all continuous functions belonging to $H_{x^{2}}[0,\infty ).$ Let

$$\begin{aligned} C_{x^{2}}^{*}[0,\infty ):=\left\{ f\in C_{x^{2}}[0,\infty ):\lim \limits _{|x|\rightarrow \infty }\frac{f(x)}{1+x^{2}}<\infty ,\text { } \Vert f\Vert _{x^{2}}=\sup \limits _{x\in [0,\infty )}\frac{|f(x)|}{ 1+x^{2}}\right\} . \end{aligned}$$

We denote the modulus of continuity of f on the closed interval $[0,a],~a>0$ by

$$\begin{aligned} \omega _{a}(f,\delta )=\sup \limits _{|t-x|\le \delta }~\sup \limits _{x,t\in [0,a]}|f(t)-f(x)|. \end{aligned}$$

Note that if $f\in C_{x^{2}}[0,\infty )$, then $\omega _{a}(f,\delta )\rightarrow 0$.

Now we prove a result on rate of convergence for the operator $L_{n,q}^{(\alpha ,\beta )}(f;x)$.

Theorem 4.1

Let $f\in C_{x^{2}}[0,\infty ).$ Then for every $n\ge 2$

$$\begin{aligned} \left\| L_{n,q}^{(\alpha ,\beta )}(f)-f\right\| _{C[0,a]}\le & {} \frac{ 4M_{f}(1+a^{2})\big (2(\beta +1)^{2}a^{2}+\alpha ^{2}+a\big )}{q([n]_{q}-1)}\\+ & {} 2\omega _{a+1}\left( f,\sqrt{\frac{2(\beta +1)^{2}a^{2}+\alpha ^{2}+a}{ q([n]_{q}-1)}}~\right) \end{aligned}$$

where$~q\in (0,1)$ and $\omega _{a+1}(f,\delta )$ is the modulus of continuity on the finite interval $[0,a+1]\subset [0,\infty )$, $a>0$.

Proof

Since $t-x>1$ for $x\in [0,a]$ and $t>a+1$, we have

$$\begin{aligned} |f(t)-f(x)|\le & {} M_{f}(2+x^{2}+t^{2})\le M_{f}\bigl (2+3x^{2}+2(t-x)^{2} \bigl )\nonumber \\\le & {} M_{f}(4+3x^{2})(t-x)^{2}\le 4M_{f}(1+a^{2})(t-x)^{2}. \end{aligned}$$

(14)

Now, let $x\in [0,a]$ and $t\le a+1$. Then for $\delta >0$, we have

$$\begin{aligned} |f(t)-f(x)|\le \omega _{a+1}(f,|t-x|)\le \left( 1+\frac{|t-x|}{\delta } \right) \omega _{a+1}(f,\delta ). \end{aligned}$$

(15)

First, we will prove that the second part of the inequality (15) is valid. We have two relations (see [19]) as:

• (i) $\forall \delta >0 ~\text {and}~ n\in {\mathbb {N}}, \omega (f,n\delta )\le n\omega (f,\delta )$,

• (ii) $\forall \delta >0$ and $r>0, \omega (f,r\delta )\le (1+[r])\omega (f,\delta )$, where [a] is the integral part of a.

We have to prove now that

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

As we have $|t-x|\le \delta$ implies $\frac{|t-x|}{\delta }\le 1$, that means 0 is the integral part of the fraction $\frac{|t-x|}{\delta }$, then

$$\begin{aligned} \omega (f,|t-x|) = \omega \left( f,\frac{|t-x|}{\delta }\delta \right) = (1+0) \omega (f,\delta )\le \left( 1+\frac{|t-x|}{\delta }\right) \omega (f,\delta ). \end{aligned}$$

Using (14) and (15), we may write

$$\begin{aligned} |f(t)-f(x)|\le 4M_{f}(1+a^{2})(t-x)^{2}+\left( 1+\frac{|t-x|}{\delta } \right) \omega _{a+1}(f,\delta ), \end{aligned}$$

(16)

for $x\in [0,a]$ and $t\ge 0$. As we know that if L is a positive linear operator, then for every $f\in X$, we have $|Lf|\le L(|f|)$. Also we know the Hölder’s inequality for positive linear operators as follows:

Let $L:X\rightarrow Y$ be a positive linear operator and let $p,q>1$ be real numbers such that $1/p+1/q=1$. Then

$$\begin{aligned} L(|f\cdot g|)\le (L(|f|^p))^{\frac{1}{p}}\cdot (L(|g|^q))^{\frac{1}{q}},~~\text {for every }f,g\in X. \end{aligned}$$

An important particular case is the Cauchy–Schwarz inequality for positive linear operators, which is obtained from Hölder’s inequality for $p=q=2$:

$$\begin{aligned} |L(f\cdot g;x)|\le \sqrt{L(f^2;x)}\cdot \sqrt{L(g^2;x)}. \end{aligned}$$

Hence

$$\begin{aligned}&\left| L_{n,q}^{(\alpha ,\beta )}(f;x)-f(x)\right| \le L_{n,q}^{(\alpha ,\beta )}\bigl (|f(t)-f(x)|;x\bigl ) \\&\quad \le 4M_{f}(1+a^{2})L_{n,q}^{(\alpha ,\beta )}\bigl ((t-x)^{2};x\bigl ) +\omega _{a+1}(f,\delta )L_{n,q}^{(\alpha ,\beta )}\left( 1+\frac{|t-x|}{ \delta }\right) . \end{aligned}$$

Now using Cauchy–Schwarz inequality on the second term of the above inequality in right-hand side, we get

$$\begin{aligned} L_{n,q}^{(\alpha ,\beta )}\left( 1+\frac{|t-x|}{\delta }\right)&= L_{n,q}^{(\alpha ,\beta )}(1;x)+\frac{1}{\delta }L_{n,q}^{(\alpha ,\beta )}\left( |t-x|;x\right) \\&= 1+\frac{1}{\delta }L_{n,q}^{(\alpha ,\beta )}\left( 1.|t-x|;x\right) \\&\le 1+\frac{1}{\delta }\sqrt{L_{n,q}^{(\alpha ,\beta )}(1^2;x)}.\sqrt{ L_{n,q}^{(\alpha ,\beta )}\big ((t-x)^2;x\big )} \\&=1+\frac{1}{\delta }\sqrt{L_{n,q}^{(\alpha ,\beta )}\big ((t-x)^2;x\big )}. \end{aligned}$$

Thus

$$\begin{aligned}&\left| L_{n,q}^{(\alpha ,\beta )}(f;x)-f(x)\right| \\&\quad \le 4M_{f}(1+a^{2})L_{n,q}^{(\alpha ,\beta )}\bigl ((t-x)^{2};x\bigl ) +\omega _{a+1}(f,\delta )\left( 1+\frac{1}{\delta }\sqrt{L_{n,q}^{(\alpha ,\beta )}\bigl ((t-x)^{2};x\bigl )}\right) . \end{aligned}$$

So by Lemma 3.1, for every $q\in (0,1)$ and $x\in [0,a]$, we get

$$\begin{aligned}&\left| L_{n,q}^{(\alpha ,\beta )}(f;x)-f(x)\right| \\&\quad \le 4M_{f}(1+a^{2})\left( \frac{2(\beta +1)^{2}x^{2}+x+\alpha ^{2}}{ q([n]_{q}-1)}\right) \\&\qquad +\,\omega _{a+1}(f,\delta )\left( 1+\frac{1}{\delta } \sqrt{\frac{2(\beta +1)^{2}x^{2}+x+\alpha ^{2}}{q([n]_{q}-1)}}~\right) \\&\quad \le \frac{4M_{f}(1+a^{2})\big (2(\beta +1)^{2}a^{2}+\alpha ^{2}+a\big )}{ q([n]_{q}-1)}\\&\quad \quad +\,\omega _{a+1}\left( 1+\frac{1}{\delta }\sqrt{\frac{2(\beta +1)^{2}a^{2}+\alpha ^{2}+a}{q([n]_{q}-1)}}~\right) . \end{aligned}$$

Choosing $\delta =\sqrt{\frac{2(\beta +1)^{2}a^{2}+\alpha ^{2}+a}{ q([n]_{q}-1)}}$, we arrived at the desired result.$\square$

In the next result, we discuss the weighted approximation for the operators $L_{n,q}^{(\alpha ,\beta )}(f;x)$, where the approximation formula holds true on $[0,\infty )$.

Theorem 4.2

Let $q=q_{n}$ such that $0<q_{n}<1$ and $q_{n}\rightarrow 1$$(n\rightarrow \infty )$. Then for each $f\in C_{x^{2}}^{*}[0,\infty )$, we have

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\Vert L_{n,q_{n}}^{(\alpha ,\beta )}(f)-f\Vert _{x^{2}}=0. \end{aligned}$$

Proof

As in [5] and [7] it is sufficient to verify the following three conditions

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\left\| L_{n,q_{n}}^{(\alpha ,\beta )}(e_{m};x)-e_{m}\right\| _{x^{2}}=0, \quad m=0,1,2. \end{aligned}$$

(17)

Since $L_{n,q_{n}}^{(\alpha ,\beta )}(e_{0};x)=1$, (17) holds true for $m=0$. Using Lemma 2.2 for $n>1$, we have

$$\begin{aligned} \left\| L_{n,q_{n}}^{(\alpha ,\beta )}(e_{1};x)-e_{1}\right\| _{x^{2}}&= \sup \limits _{x\in [0,\infty )}\frac{\left| L_{n,q_{n}}^{(\alpha ,\beta )}(t;x)-x\right| }{1+x^{2}}\\&=\sup \limits _{x\in [0,\infty )}\frac{1}{1+x^{2}} \left| \frac{[n]_{q}x+\alpha }{[n]_{q}+\beta }-x\right| \\&\le \frac{\beta }{[n]_{q}+\beta }\sup \limits _{x\in [0,\infty )} \frac{x}{1+x^{2}}+\frac{\alpha }{[n]_{q}+\beta }\le \frac{\beta }{ [n]_{q}+\beta }\\&\quad +\frac{\alpha }{[n]_{q}+\beta }. \end{aligned}$$

In the case $m=1$, (17) is also true when $n\rightarrow \infty$. Similarly, for $n\ge 2,$ we can write

$$\begin{aligned} \left\| L_{n,q_{n}}^{(\alpha ,\beta )}(e_{2};x)-e_{2}\right\| _{x^{2}}= & {} \sup \limits _{x\in [0,\infty )}\frac{\left| L_{n,q_{n}}^{(\alpha ,\beta )}(t^{2};x)-x^{2}\right| }{1+x^{2}} \\\le & {} \left( \frac{[n]_{q}^{3}}{q([n]-1)([n]_{q}+\beta )^{2}}-1\right) \sup \limits _{x\in [0,\infty )}\frac{x^{2}}{1+x^{2}} \\&+\left( \frac{[n]_{q}^{2}+2q[n]_{q}([n]_{q}-1)\alpha }{q([n]-1)([n]_{q}+ \beta )^{2}}\right) \sup \limits _{x\in [0,\infty )}\frac{x}{1+x^{2}}\\&+\,\frac{\alpha ^{2}}{([n]_{q}+\beta )^{2}} \\\le & {} \frac{(1-q^{n})[n]_{q}^{2}-q(2\beta -1)[n]_{q}^{2}-q\beta (\beta -1)[n]_{q}+q\beta ^{2}}{q([n]-1)([n]_{q}+\beta )^{2}} \\&+\quad \left( \frac{[n]_{q}^{2}+2q[n]_{q}([n]_{q}-1)\alpha }{q([n]-1)([n]_{q}+ \beta )^{2}}\right) +\frac{\alpha ^{2}}{([n]_{q}+\beta )^{2}}, \end{aligned}$$

which implies that

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\left\| L_{n,q_{n}}^{(\alpha ,\beta )}(e_{2};x)-e_{2}\right\| _{x^{2}}=0. \end{aligned}$$

This completes the proof.$\square$

In the following result, we approximate the functions $f\in C_{x^{2}}[0,\infty )$. Such type of results are given in [6] for locally integrable functions.

Theorem 4.3

Let $q=q_{n}$ such that $0<q_{n}<1$ and $q_{n}\rightarrow 1$$(n\rightarrow \infty )$. Then for each $f\in C_{x^{2}}[0,\infty )$ and $\alpha >0$, we have

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\sup \limits _{x\in [0,\infty )}\frac{ \left| L_{n,q_{n}}^{(\alpha ,\beta )}(f;x)-f(x)\right| }{ (1+x^{2})^{1+\alpha ^{2}}}=0. \end{aligned}$$

Proof

For any fixed $x_{0}>0$, we have

$$\begin{aligned}&\sup \limits _{x\in [0,\infty )}\frac{\left| L_{n,q_{n}}^{(\alpha ,\beta )}(f;x)-f(x)\right| }{(1+x^{2})^{1+\alpha ^{2}}} \\&\quad =\sup \limits _{x\le x_{0}}\frac{\left| L_{n,q_{n}}^{(\alpha ,\beta )}(f;x)-f(x)\right| }{(1+x^{2})^{1+\alpha ^{2}}}+\sup \limits _{x\ge x_{0}}\frac{\left| L_{n,q_{n}}^{(\alpha ,\beta )}(f;x)-f(x)\right| }{(1+x^{2})^{1+\alpha ^{2}}} \\&\quad \le \Vert L_{n,q_{n}}^{(\alpha ,\beta )}(f)-f\Vert _{C[0,a]}+\Vert f\Vert _{x^{2}}\sup \limits _{x\ge x_{0}}\frac{\left| L_{n,q_{n}}^{(\alpha ,\beta )}\big (1+t^{2};x\big )\right| }{(1+x^{2})^{1+\alpha ^{2}}}+\sup \limits _{x\ge x_{0}}\frac{|f(x)|}{(1+x^{2})^{1+\alpha ^{2}}}. \end{aligned}$$

By Theorem 4.1, the first term of the above inequality tends to zero. It follows from Lemma 2.2, that $\sup \limits _{x\ge x_{0}}\frac{ |L_{n,q_{n}}^{(\alpha ,\beta )}(1+t^{2};x)|}{(1+x^{2})^{1+\alpha ^{2}}} \rightarrow 0$ as $n\rightarrow \infty ,$ for any fixed $x_{0}>0$. We can choose $x_{0}>0$ to be sufficiently large so that the last part of the above inequality can be made arbitrarily small and this proves the theorem.$\square$

5 Pointwise Estimates

In this section, we study some pointwise estimates of the rate of convergence of the q-Stancu–Beta operators.

We know that a function $f\in C[0,\infty )$ is Lip$(\alpha )$ on $E,~\alpha \in (0,1],~E\subset [0,\infty )$ if it satisfies the condition

$$\begin{aligned} |f(t)-f(x)|\le M_{f}|t-x|^{\alpha },\,\,\,t\in [0,\infty ) \text { and }x\in E, \end{aligned}$$

(18)

where $M_{f}$ is a constant depending only on $\alpha$ and f.

First, we give the relationship between the local smoothness of f and local approximation.

Theorem 5.1

Let $f\in \text{ Lip }(\alpha ),~E\subset [0,\infty ) \text{ and } \alpha \in (0,1]$. Then we have

$$\begin{aligned}&|L_{n,q}^{(\alpha ,\beta )}(f;x)-f(x)| \\&\quad \le M_{f}\left\{ \left[ \left( \frac{[n]_{q}}{q([n]_{q}-1)}-\frac{ [n]_{q}-\beta }{[n]_{q}+\beta }\right) x^{2}+\frac{x}{q([n]_{q}-1)}+\frac{ \alpha ^{2}}{([n]_{q}+\beta ^{2})}\right] \right. ^{\frac{\alpha }{2}}\\&\quad \quad +\,2\big (d(x,E)\big )^{\alpha }\left. \right\} , \end{aligned}$$

$x\in [0,\infty ),$ where d(x, E) is defined as

$$\begin{aligned} d(x,E)=\inf \{|x-y|:y\in E\}. \end{aligned}$$

Proof

For $x_{0}\in \overline{E}$, the closure of the set E in $[0,\infty )$, we have

$$\begin{aligned} |f(t)-f(x)|\le |f(t)-f(x_{0})|+|f(x_{0})-f(x)|, \quad x\in [0,\infty ). \end{aligned}$$

By (18), we get

$$\begin{aligned} |L_{n,q}^{(\alpha ,\beta )}(f;x)-f(x)|= & {} |L_{n,q}^{(\alpha ,\beta )}(f;x)-L_{n,q}^{(\alpha ,\beta )}\big (f(x);x\big )|\le L_{n,q}^{(\alpha ,\beta )}\big (|f(t)-f(x)|;x\big ) \\\le & {} L_{n,q}^{(\alpha ,\beta )}\big (|f(t)-f(x_{0})|;x\big ) +L_{n,q}^{(\alpha ,\beta )}\big (|f(x)-f(x_{0})|;x\big ) \\\le & {} L_{n,q}^{(\alpha ,\beta )}\big (|f(t)-f(x_{0})|;x\big )+|f(x)-f(x_{0})| \\\le & {} M_{f}\big \{L_{n,q}^{(\alpha ,\beta )}\big (|t-x_{0}|^{\alpha };x\big ) +|x-x_{0}|^{\alpha }\big \} \\\le & {} M_{f}\big \{L_{n,q}^{(\alpha ,\beta )}\big (|t-x|^{\alpha };x\big ) +|x-x_{0}|^{\alpha }+|x-x_{0}|^{\alpha }\big \} \\= & {} M_{f}\big \{L_{n,q}^{(\alpha ,\beta )}\big (|t-x|^{\alpha };x\big ) +2|x-x_{0}|^{\alpha }\big \}. \end{aligned}$$

Using the Hölder’s inequality with $p=\frac{2}{\alpha },~q=\frac{2}{ 2-\alpha }$, and Lemma 5, we find that

$$\begin{aligned}&|L_{n,q}^{(\alpha ,\beta )}(f;x)-f(x)| \\&\quad \le M_{f}\Big \{\Big (L_{n,q}^{(\alpha ,\beta )}\big (|t-x|^{\alpha p};x\big ) \Big )^{\frac{1}{p}}\Big (L_{n,q}^{(\alpha ,\beta )}\big (1^{q};x\big )\Big )^{ \frac{1}{q}}+2\big (d(x,E)\big )^{\alpha }\Big \} \\&\quad =M_{f}\Big \{\Big (L_{n,q}^{(\alpha ,\beta )}\big (|t-x|^{2};x\big )\Big )^{ \frac{\alpha }{2}}+2\big (d(x,E)\big )^{\alpha }\Big \} \\&\quad =M_{f}\left\{ \left[ \left( \frac{[n]_{q}}{q([n]_{q}-1)}-\frac{ [n]_{q}-\beta }{[n]_{q}+\beta }\right) \right. x^{2}+\frac{x}{q([n]_{q}-1)}+\frac{ \alpha ^{2}}{([n]_{q}+\beta ^{2})}\right] ^{\frac{\alpha }{2}}\\&\qquad +\,2\big (d(x,E)\big )^{\alpha }\left. \right\} \\&\quad =M_{f}\left\{ \delta _{n}(x)^{\frac{\alpha }{2}}+2\big (d(x,E)\big )^{\alpha }\right\} . \end{aligned}$$

$\square$

In the next result, we give local direct estimate for the q-Stancu–Beta operators using the Lipschitz-type maximal function of order $\alpha$ introduced by B. Lenze [10] as

$$\begin{aligned} \tilde{\omega }_{\alpha }(f,x)=\sup \limits _{t\ne x,~t\in [0,\infty )} \frac{|f(t)-f(x)|}{|t-x|^{\alpha }}, \quad x\in [0,\infty ) \text{ and } \alpha \in (0,1]. \end{aligned}$$

(19)

Theorem 5.2

Let $\alpha \in (0,1]$ and $f\in C_{B}[0,\infty )$. Then for all $x\in [0,\infty )$, we have

$$\begin{aligned}&|L_{n,q}^{(\alpha ,\beta )}(f;x)-f(x)|\\&\quad \le \tilde{\omega }_{\alpha }(f,x)\left\{ \left( \frac{[n]_{q}}{q([n]_{q}-1)}-\frac{[n]_{q}-\beta }{[n]_{q}+\beta }\right) x^{2}+\frac{x}{q([n]_{q}-1)} +\,\frac{\alpha ^{2}}{([n]_{q}+\beta ^{2})}\right\} ^{\frac{\alpha }{2}}. \end{aligned}$$

Proof

From (19) we have

$$\begin{aligned} |f(t)-f(x)|\le \tilde{\omega }_{\alpha }(f,x)|t-x|^{\alpha } \end{aligned}$$

and

$$\begin{aligned} |L_{n,q}^{(\alpha ,\beta )}(f;x)-f(x)|\le L_{n,q}^{(\alpha ,\beta )}\big ( |f(t)-f(x)|;x\big )\le \tilde{\omega }_{\alpha }(f,x)L_{n,q}^{(\alpha ,\beta )}\big (|t-x|^{\alpha };x\big ). \end{aligned}$$

Applying Hölder’s inequality with $p=\frac{2}{\alpha },~q=\frac{2}{ 2-\alpha }$, and Lemma 3.1, we have

$$\begin{aligned} |L_{n,q}^{(\alpha ,\beta )}(f;x)-f(x)|\le \tilde{\omega }_{\alpha }(f,x)L_{n,q}^{(\alpha ,\beta )}\big (|t-x|^{2};x\big )^{\frac{\alpha }{2}}. \end{aligned}$$

By Lemma 5, we get our assertion.$\square$

6 Voronovskaja-Type Theorem

In this section, we prove Voronovskaja-type results for q-Stancu–Beta type operators.

Lemma 6.1

Assume that $q_{n}\in (0,1)$ and $q_{n}^{n}\rightarrow a,~(0\le a<1)$ as $n\rightarrow \infty$. For every $x\in [0,\infty )$ there hold

$$\begin{aligned}&\lim \limits _{n\rightarrow \infty }[n]_{q_{n}}L_{n,q_{n}}^{(\alpha ,\beta )}(e_{1}-e_{0}x;x)=\alpha -\beta x, \\&\lim \limits _{n\rightarrow \infty }[n]_{q_{n}}L_{n,q_{n}}^{(\alpha ,\beta )} \big ((e_{1}-e_{0}x)^{2};x\big )=(2-a)x^{2}+x. \end{aligned}$$

Theorem 6.1

Assume that $q_{n}\in (0,1),~q_{n}\rightarrow 1$ and $q_{n}^{n}\rightarrow a$$(0\le a<1)$ as $n\rightarrow \infty$. For any $f\in C_{2}^{*}[0,\infty )$ such that $f^{\prime },f^{\prime \prime }\in C_{2}^{*}[0,\infty )$ the following equality holds

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }[n]_{q_{n}}\left( L_{n,q_{n}}^{(\alpha ,\beta )}(f;x)-f(x)\right) =(\alpha -\beta x)f^{\prime }(x)+\frac{ (2-a)x^{2}+x}{2}f^{\prime \prime }(x) \end{aligned}$$

uniformly on any $[0,A],~A>0$.

Proof

Let $f,f^{\prime },f^{\prime \prime }\in C_{2}^{*}[0,\infty )$ and $x\in [0,\infty )$ be fixed. By the Taylor formula we may write

$$\begin{aligned} f(t)=f(x)+(t-x)f^{\prime }(x)+\frac{1}{2}(t-x)^{2}f^{\prime \prime 2}(x)+r(t;x)(t-x)^{2}, \end{aligned}$$

(20)

where r(t; x) is the Peano form of remainder, $r(.;x)\in C_{2}^{*}[0,\infty )$ and $\lim _{t\rightarrow x}r(t;x)=0$. Applying $L_{n,q_{n}}^{(\alpha ,\beta )}$ to (20), we obtain

$$\begin{aligned}{}[n]_{q_{n}}\left( L_{n,q_{n}}^{(\alpha ,\beta )}(f;x)-f(x)\right)= & {} [n]_{q_{n}}L_{n,q_{n}}^{(\alpha ,\beta )}(e_{1}-e_{0}x;x)f^{\prime }(x) \\&+\,\frac{1}{2}[n]_{q_{n}}L_{n,q_{n}}^{(\alpha ,\beta )}\left( (e_{1}-e_{0}x)^{2};x\right) f^{\prime \prime }(x)\\&+\,[n]_{q_{n}}L_{n,q_{n}}^{(\alpha ,\beta )}\left( r(.;x)(.-x)^{2};x\right) . \end{aligned}$$

By the Cauchy–Schwartz inequality, we have

$$\begin{aligned} L_{n,q_{n}}^{(\alpha ,\beta )}\left( r(.;x)(.-x)^{2};x\right) \le \sqrt{ L_{n,q_{n}}^{(\alpha ,\beta )}\big (r^{2}(.;x);x\big )}\sqrt{ L_{n,q_{n}}^{(\alpha ,\beta )}\big ((.-x)^{4};x\big )}. \end{aligned}$$

(21)

Observe that $r^{2}(x;x)=0$ and $r^{2}(.;x)\in C_{2}^{*}[0,\infty )$. Then it follows from Theorem 3.1 that

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }L_{n,q_{n}}^{(\alpha ,\beta )}\left( r^{2}(.;x);x\right) =r^{2}(x;x)=0 \end{aligned}$$

(22)

uniformly with respect to $x\in [0,A]$, in view of the fact that $\Big (L_{n,q_{n}}^{(\alpha ,\beta )}(.-x)^{4}\Big )(x)=O\left( \frac{1}{[n]^2_{q_n}}\right)$. Now from (21), (22) and Lemma above, we immediately get

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }[n]_{q_{n}}L_{n,q_{n}}^{(\alpha ,\beta )}\left( r(.;x)(.-x)^{2};x\right) =0. \end{aligned}$$

Then we get the following

$$\begin{aligned}&\lim \limits _{n\rightarrow \infty }[n]_{q_{n}}\big (L_{n,q_{n}}^{(\alpha ,\beta )}(f;x)-f(x)\big ) \\&\quad =\lim \limits _{n\rightarrow \infty }\Big (f^{\prime }(x)[n]_{q_{n}}L_{n,q_{n}}^{(\alpha ,\beta )}(e_{1}-e_{0}x;x)+\frac{1}{2} f^{\prime \prime }(x)[n]_{q_{n}}L_{n,q_{n}}^{(\alpha ,\beta )}\big ( (e_{1}-e_{0}x)^{2};x)\big ) \\&\quad \quad +\,[n]_{q_{n}}L_{n,q_{n}}^{(\alpha ,\beta )}\big (r(.;x)(.-x)^{2};x)\Big ) \\&\quad =(\alpha -\beta x)f^{\prime }(x)+\frac{(2-a)x^{2}+x}{2}f^{\prime \prime }(x). \end{aligned}$$

This completes the proof.$\square$