Approximation on a class of Szász–Mirakyan operators via second kind of beta operators

Abstract

In the present article, we construct a new sequence of positive linear operators via Dunkl analogue of modified Szász–Durrmeyer operators. We study the moments and basic results. Further, we investigate the pointwise approximation and uniform approximation results in various functional spaces for these sequences of positive linear operators. Finally, we prove the global approximation and A-statistical convergence results for these operators.

1 Introduction

In recent past, the Szász–Mirakjan operators and Szász–Durrmeyer type operators have been intensively investigated in various functional spaces to approximate the continuous functions and Lebesgue measurable functions respectively. Mazhar and Totik [17] gave an integral modification of Szász–Mirakyan [37] operators to study the Lebesgue measurable functions as follows:

$$\begin{aligned} S_{n}^{*}(g;u)=ne^{-nu}\sum _{k=0}^{\infty }\frac{(nu)^{k}}{k!} \int _{0}^{\infty }e^{-nt}\frac{(nt)^{k}}{k!}g(t) \,dt,\quad u \geq 0, n\in \mathbb{N}=\{1,2,3,\dots \}. \end{aligned}$$

(1.1)

Researchers have obtained several approximations of Szász–Mirakyan type operators via Dunkl generalization; for instance, see [6, 18, 26, 28, 29, 32, 39]. Related to these results, more approximation results have been studied in different functional spaces (see [1, 2, 4, 5, 14, 38] and [3, 16, 27, 31]). Sucu [36] introduced Szász–Mirakyan type operators for $k\in {\mathbb{N}}_{0}= \mathbb{N}\cup \{0\}$ as follows:

$$\begin{aligned} S_{n}(g;u):=\frac{1}{e_{i}(nu)}\sum _{k=0}^{\infty }\frac{(nu)^{k}}{ \gamma _{i}(k)}g \biggl(\frac{k+2i\theta _{k}}{n} \biggr), \end{aligned}$$

(1.2)

using generalized exponential function [33] given by

$$\begin{aligned} e_{i}(t)=\sum_{j=0}^{\infty } \frac{t^{j}}{\gamma _{i}(j)},\quad t\in [0, \infty ), \end{aligned}$$

(1.3)

where the coefficients $\gamma _{i}$ are defined as follows:

$$ \gamma _{i}(2j)=\frac{2^{2j}j!\varGamma (j+i+\frac{1}{2})}{\varGamma (i+ \frac{1}{2} )} \quad\text{and}\quad \gamma _{i}(2j+1)=2^{2j+1}j!\frac{ \varGamma (j+i+\frac{3}{2})}{\varGamma (i+\frac{1}{2} )}. $$

Here, Γ is the gamma function and $i>-\frac{1}{2}$. In [36] the pointwise and uniform approximation results for the operators defined by (1.2) are obtained. Several extensions of the Szász–Mirakyan operators defined by (1.2) and their approximation results are studied by the authors viz İçöz et al. [11, 12]. In [40], we have constructed Szász–Durrmeyer type operators to approximate the Lebesgue measurable functions as follows:

$$\begin{aligned} D_{n}(g;u)=\frac{1}{e_{i}(nu)}\sum _{j=0}^{\infty }\frac{(nu)^{j}}{ \gamma _{i}(j)}\frac{n^{j+2i\theta _{j}+\lambda +1}}{\varGamma (j+2i\theta _{j}+\lambda +1)} \int _{0}^{\infty }t^{j+2i\theta _{j}+\lambda }e^{-nt}g(t) \,dt, \end{aligned}$$

(1.4)

where $\varGamma (t)=\int _{0}^{\infty }u^{t}e^{-u}\,du$ is the gamma function, $\lambda \geq 0$, and

$$ \theta _{j}= \textstyle\begin{cases} 0,&j\in 2{\mathbb{N}}, \\ 1,& j\in 2{\mathbb{N}}+1. \end{cases} $$

In order to obtain a better rate of convergence in comparison to (1.4), in this paper we introduce a modification with two nonnegative shifted nodes as follows:

$$ M_{n}^{\alpha,\beta }(g;u)= \sum _{k=0}^{\infty }\nu _{n}(u;k) \int _{0}^{\infty }t^{k+2i\theta _{k}+\lambda }e^{-nt}g \biggl(\frac{nt+ \alpha }{n+\beta } \biggr)\,dt, $$

(1.5)

where $\nu _{n}(u;k)=\frac{1}{e_{i}(nu)} \frac{(nu)^{k}}{\gamma _{i}(k)}\frac{n^{k+2i\theta _{k}+\lambda +1}}{ \varGamma (k+2i\theta _{k}+\lambda +1)}$, $0\leq \alpha \leq \beta$, and $f\in C[0,\infty )$ whenever the above series converges. One can notice that (i) for $i=\lambda =\alpha =\beta =0$, the sequence of operators introduced in (1.5) reduces to the operators defined in (1.1) and (ii) for $\alpha =\beta =0$, the sequence of positive linear operators defined in (1.5) reduces to the operators defined in [40].

In the subsequent sections, we deduce some basic results, direct approximation results. Further, global approximation results are studied in [19–21, 24, 25, 35].

2 Preliminary results

For $i>-\frac{1}{2}, u \geq 0$, and $n\in \mathbb{N}$, we denote

$$ E_{n}(u)=\frac{e_{i}(-nu)}{e_{i}(nu)}. $$

(2.1)

Lemma 2.1

([40])

For the operators given by (1.4), we have

$$\begin{aligned} &D_{n}(1;u)=1, \\ &D_{n}(t;u)=u+\frac{\lambda +1}{n}, \\ &D_{n}\bigl(t^{2};u\bigr)=u^{2}+ \bigl(4+2 \lambda +2iE_{n}(u) \bigr) \frac{u}{n}+\frac{(\lambda +1)(\lambda +2)}{n^{2}}, \\ &D_{n}\bigl(t^{3};u\bigr)=u^{3}+ \bigl(9+3 \lambda -2iE_{n}(u) \bigr)\frac{u ^{2}}{n}+ \bigl(18+3\lambda ( \lambda +5)+4i^{2} \\ &\phantom{D_{n}\bigl(t^{3};u\bigr)=}{}+2i(8+3\lambda )E_{n}(u) \bigr)\frac{u}{n^{2}}+ \frac{\lambda ^{3}+6 \lambda ^{2}+11\lambda +6}{n^{3}}, \\ &D_{n}\bigl(t^{4};u\bigr)=u^{4}+{\mathcal{O}} \biggl(\frac{1}{n} \biggr)\quad \textit{for each compact subset of } [0,\infty ). \end{aligned}$$

Lemma 2.2

For operators (1.5), we have

$$\begin{aligned} M_{n}^{\alpha,\beta }\bigl(t^{r};u\bigr)=\sum _{i=0}^{r}\binom{r}{i}\frac{n^{i} \alpha ^{r-i}}{(n+\beta )^{r}}D_{n} \bigl(t^{i};u\bigr),\quad r\in \mathbb{N}. \end{aligned}$$

Proof

From (1.5), we get

$$\begin{aligned} M_{n}^{\alpha,\beta }\bigl(t^{r};u\bigr) &= \sum _{k=0}^{\infty }\nu _{n}(u;t) \int _{0}^{\infty }t^{k+2i\theta _{k}+ \lambda }e^{-nt} \biggl( \frac{nt+\alpha }{n+\beta }-u \biggr)^{r} \\ &= \sum_{j=0}^{r}\binom{r}{j} \frac{n^{j}\alpha ^{r-j}}{(n+\beta )^{r}} \sum_{k=0}^{\infty }\nu _{n}(u;t) \int _{0}^{\infty }t^{k+2i\theta _{k}+ \lambda }e^{-nt}t^{j} \,dt \\ &=\sum_{j=0}^{r}\binom{r}{j} \frac{n^{j}\alpha ^{r-j}}{(n+\beta )^{r}}D_{n}\bigl(t^{j};u\bigr). \end{aligned}$$

 □

Lemma 2.3

For$r\in \mathbb{N}$, we have the following relation:

$$\begin{aligned} M_{n}^{\alpha,\beta }\bigl((t-u)^{r};u\bigr)=\sum _{i=0}^{r}\binom{r}{i}(-u)^{r-i}M _{n}^{\alpha,\beta }\bigl(t^{i};u\bigr), \end{aligned}$$

where$M_{n}^{\alpha,\beta }$is defined in (1.5).

Proof

From relation (1.5), we get

$$\begin{aligned} M_{n}^{\alpha,\beta }\bigl((t-u)^{r};u\bigr) &= \sum _{k=0}^{\infty }\nu _{n}(u;k) \int _{0}^{\infty }t^{k+2i\theta _{k}+ \lambda }e^{-nt} \biggl( \frac{nt+\alpha }{n+\beta }-u \biggr)^{r}\,dt \\ &= \sum_{j=0}^{r}\binom{r}{j}(-u)^{r-j}M_{n}^{\alpha,\beta } \bigl(t^{j};u\bigr). \end{aligned}$$

 □

Lemma 2.4

Let$e_{r}(t)=t^{r}$for$r\in \{0,1,2,3,4\}$be the test functions and$E_{n}(u)$be defined in (2.1). Then

$$\begin{aligned} &M_{n}^{\alpha,\beta }(e_{0};u)=1, \\ &M_{n}^{\alpha,\beta }(e_{1};u)=\frac{n}{n+\beta }u+ \frac{\lambda + \alpha +1}{n+\beta }, \\ &M_{n}^{\alpha,\beta }(e_{2};u)=\frac{n^{2}}{(n+\beta )^{2}}u^{2}+ \bigl(4+2\lambda +2\alpha +2iE_{n}(u) \bigr)\frac{n}{(n+\beta )^{2}}u \\ &\phantom{M_{n}^{\alpha,\beta }(e_{2};u)=}{}+ \frac{(\lambda +1)(\lambda +2\alpha +2)+\alpha ^{2}}{(n+\beta )^{2}}, \\ &M_{n}^{\alpha,\beta }(e_{3};u)=\frac{n^{3}}{(n+\beta )^{3}}u^{3}+ \bigl(9+3\alpha +3\lambda -2iE_{n}(u) \bigr) \frac{n^{2}}{(n+\beta )^{3}}u^{2} \\ &\phantom{M_{n}^{\alpha,\beta }(e_{3};u)=}{}+ \bigl(18 + 12\alpha + 3\lambda (\lambda + 5 + 2\alpha ) + 3\alpha ^{2} + 4i^{2} + 2i(8+ 3\lambda + 3\alpha )E_{n}(u) \bigr) \\ &\phantom{M_{n}^{\alpha,\beta }(e_{3};u)=}{}\times \frac{n}{(n + \beta )^{3}}u + \frac{\lambda ^{3} + \alpha ^{3} + \lambda ^{2}(6+3\alpha ) + \lambda (11 + 9\alpha + 3\alpha ^{2}) + 6 + 6\alpha + 3\alpha ^{2}}{(n+\beta )^{3}}, \\ &M_{n}^{\alpha,\beta }(e_{4};u) =\frac{n^{4}}{(n+\beta )^{4}}u^{4}+ {\mathcal{O}} \biggl(\frac{1}{n+\beta } \biggr) \quad\textit{for each compact subset of } [0,\infty ). \end{aligned}$$

Proof

The proof follows immediately from Lemma 2.2. □

Lemma 2.5

Let$\psi _{u}^{r}(t)=(t-u)^{r}$, $r\in {\mathbb{N}}$, be the central moments of$M_{n}^{\alpha,\beta }$. Then

$$\begin{aligned} &M_{n}^{\alpha,\beta }\bigl(\psi _{u}^{0};u\bigr) =1, \\ &M_{n}^{\alpha,\beta }\bigl(\psi _{u}^{1};u\bigr) =\frac{1}{(n+\beta )}(1+ \lambda +\alpha -\beta u), \\ &M_{n}^{\alpha,\beta }\bigl(\psi _{u}^{2};u\bigr) = \frac{\beta ^{2}}{(n+\beta )^{2}}u^{2}+2 \biggl(1+iE_{n}(u)-2\beta \frac{1+ \lambda +\alpha }{n} \biggr)\frac{n}{(n+\beta )^{2}}u \\ &\phantom{M_{n}^{\alpha,\beta }\bigl(\psi _{u}^{2};u\bigr) = }{}+ \frac{\alpha ^{2}+(\lambda +1)(\lambda +2\alpha +2)}{(n+\beta )^{2}}, \\ &M_{n}^{\alpha,\beta }\bigl(\psi _{u}^{4};u\bigr) ={\mathcal{O}} \bigl((n+\beta )^{-1} \bigr)\quad \textit{for each compact subset of } [0,\infty ). \end{aligned}$$

Proof

In the light of Lemmas 2.2 and 2.3, we can easily prove Lemma 2.5. □

Definition 2.6

Let $g\in C[0,\infty )$. Then the modulus of continuity for a uniformly continuous function g is defined as follows:

$$\begin{aligned} \omega (g;\delta )=\sup_{ \vert t_{1}-t_{2} \vert \leq \delta } \bigl\vert g(t_{1})-g(t_{2}) \bigr\vert ,\quad t_{1},t_{2} \in [0,\infty ). \end{aligned}$$

For a uniformly continuous function g in $C[0,\infty )$ and $\delta >0$, we get

$$\begin{aligned} \bigl\vert g(t_{1})-g(t_{2}) \bigr\vert \leq \biggl(1+\frac{(t_{1}-t_{2})^{2}}{\delta ^{2}} \biggr)\omega (g;\delta ). \end{aligned}$$

(2.2)

Theorem 2.7

Let$M_{n}^{\alpha,\beta }$be the operators defined by (1.5) and$g\in C[0,\infty )\cap \{g: u\geq 0, \frac{g(u)}{1+u^{2}}\textit{ is convergent as }u\rightarrow \infty \}$. Then the operators are defined by (1.5), $M_{n}^{\alpha, \beta }\rightrightarrows g$on each compact subset of$[0,\infty )$, where ⇉ stands for uniform convergence.

Proof

For the proof of uniformity for the operators $M_{n}^{\alpha,\beta }$, we use the well-known Korovkin theorem. So, it is sufficient to show that

$$\begin{aligned} M_{n}^{\alpha,\beta }(e_{\nu };u)\rightarrow e_{\nu }(u) \quad\text{for } \nu =0,1,2. \end{aligned}$$

Using Lemma 2.4, it is obvious that $M_{n}^{\alpha,\beta }(e _{\nu };u)\rightarrow e_{\nu }(u)$ as $n\rightarrow \infty$, $\nu =0,1,2$, which proves Theorem 2.7. □

Theorem 2.8

(See [34])

Let$L:C([a,b])\rightarrow B([a,b])$be a linear and positive operator, and let$\varphi _{x}$be the function defined by

$$\begin{aligned} \varphi _{x}(t)= \vert t-x \vert , \quad(x,t)\in [a,b]\times [a,b]. \end{aligned}$$

If$f\in C_{B}([a,b])$for any$x\in [a,b]$and any$\delta >0$, the operatorLverifies

$$\begin{aligned} \bigl\vert (Lf) (x)-f(x) \bigr\vert \leq{}& \bigl\vert f(x) \bigr\vert \bigl\vert (Le_{0}) (x)-1 \bigr\vert \\ &{}+\Bigl\{ (Le_{0}) (x)+\delta ^{-1}\sqrt {(Le_{0}) (x) \bigl(L\varphi _{x}^{2} \bigr) (x)} \Bigr\} \omega _{f}(\delta ), \end{aligned}$$

where$C[a,b]$is the space of all continuous functions defined on$[a,b]$, and$C_{B}[a,b]$is the space of all continuous and bounded functions defined on$[a,b]$.

Theorem 2.9

For the operators$M_{n}^{\alpha,\beta }$given by (1.5) and$g\in C_{B}[0,\infty )$, we have

$$\begin{aligned} \bigl\vert M_{n}^{\alpha,\beta }(g;u)-g(u) \bigr\vert \leq 2\omega (g;\delta ), \end{aligned}$$

where$\delta =\sqrt{M_{n}^{\alpha,\beta }(\psi _{u}^{2};u)}$and$C_{B}[0,\infty )$is the space of continuous and bounded functions defined on$[0,\infty )$.

Proof

Using Lemma 2.4, Lemma 2.5, and Theorem 2.8, we get

$$\begin{aligned} \bigl\vert M_{n}^{\alpha,\beta }(g;u)-g(u) \bigr\vert \leq \Bigl\{ 1+\delta ^{-1}\sqrt{M_{n} ^{\alpha,\beta }\bigl( \psi _{u}^{2};u\bigr)}\Bigr\} \omega (g;\delta ). \end{aligned}$$

On taking $\delta =\sqrt{M_{n}^{\alpha,\beta }(\psi _{u}^{2};u)}$, we arrive at the required result. □

3 Direct results

Let us equip the norm on g such as $\|g\|=\sup_{0\leq u< \infty }|g(u)|$. For any $g\in C_{B}[0,\infty )$ and $\delta >0$, Peetre’s K-functional is defined as follows:

$$ K_{2}(g,\delta )=\inf \bigl\{ \Vert g-h \Vert + \delta \bigl\Vert h'' \bigr\Vert : h\in C_{B}^{2}[0, \infty )\bigr\} , $$

(3.1)

where $C_{B}^{2}[0,\infty )=\{h\in C_{B}[0,\infty ):h', h''\in C_{B}[0, \infty )\}$. From DeVore and Lorentz ([8], p.177, Theorem 2.4), there exists an absolute constant $C>0$ in such a way that

$$\begin{aligned} K_{2}(g;\delta )\leq C\omega _{2}(g;\sqrt{\delta }). \end{aligned}$$

Lemma 3.1

Let the auxiliary operators be

$$\begin{aligned} \widehat{M}_{n}^{\alpha,\beta }(g;u)=M_{n}^{\alpha,\beta }(g;u)+g(u)-g \biggl(\frac{n}{n+\beta }u+\frac{\lambda +\alpha +1}{n+\beta } \biggr). \end{aligned}$$

(3.2)

For$h, g\in C_{B}^{2}[0,\infty )$, $u\geq 0$, and$\mu,\lambda \geq 0$, we get

$$\begin{aligned} \bigl\vert \widehat{M}_{n}^{\alpha,\beta }(g;u)-g(u) \bigr\vert \leq \xi _{n}^{u} \bigl\Vert h'' \bigr\Vert , \end{aligned}$$

where

$$\begin{aligned} \xi _{n}^{u}=M_{n}^{\alpha,\beta } \bigl(\psi _{u}^{2};u\bigr)+ \bigl(M_{n}^{ \alpha,\beta } \bigl(\psi _{u}^{1};u\bigr) \bigr)^{2}. \end{aligned}$$

Proof

From (3.2), we have

$$\begin{aligned} \widehat{M}_{n}^{\alpha,\beta }(1;u)=1,\qquad \widehat{M}_{n} ^{\alpha,\beta }(\psi _{u};u)=0\quad \text{and}\quad \bigl\vert \widehat{M}_{n}^{ \alpha,\beta }(g;u) \bigr\vert \leq 3 \Vert g \Vert . \end{aligned}$$

(3.3)

From Taylor’s series for $h\in C_{B}^{2}[0,\infty )$, we have

$$\begin{aligned} h(t)=h(u)+(t-u)h'(u)+ \int _{u}^{t} (t-v)h''(v) \,dv. \end{aligned}$$

(3.4)

Using $\widehat{M}_{n}^{\alpha,\beta }(g;u)$ in (3.4), we get

$$\begin{aligned} \widehat{M}_{n}^{\alpha,\beta }(h;u)-h(u)=h'(u) \widehat{M}_{n}^{ \alpha,\beta }(t-u;u)+\widehat{M}_{n}^{\alpha,\beta } \biggl( \int _{u}^{t}(t-v)h''(v) \,dv;u \biggr). \end{aligned}$$

From (3.2) and (3.3), we have

$$\begin{aligned} &\widehat{M}_{n}^{\alpha,\beta }(h;u)-h(u)=\widehat{M}_{n}^{\alpha, \beta } \biggl( \int _{u}^{t}(t-v)h''(v) \,dv;u \biggr) \\ &\phantom{\widehat{M}_{n}^{\alpha,\beta }(h;u)-h(u)}=M_{n}^{\alpha,\beta } \biggl( \int _{u}^{t}(t-v)h''(v) \,dv;u \biggr) \\ &\phantom{\widehat{M}_{n}^{\alpha,\beta }(h;u)-h(u)=}{}- \int _{u}^{\frac{n}{n+\beta }u+\frac{\lambda +\alpha +1}{n+\beta }} \biggl(\frac{n}{n+\beta }u+ \frac{\lambda +\alpha +1}{n+\beta }-v \biggr)g''(v)\,dv, \\ &\bigl\vert \widehat{M}_{n}^{\alpha,\beta }(h;u)-h(u) \bigr\vert \\ &\quad \leq \biggl\vert M_{n}^{\alpha ,\beta } \biggl( \int _{u}^{t}(t-v)h''(v) \,dv;u \biggr) \biggr\vert \\ &\qquad{}+ \biggl\vert \int _{u}^{\frac{n}{n+\beta }u+\frac{\lambda +\alpha +1}{n+ \beta }} \biggl(\frac{n}{n+\beta }u+ \frac{\lambda +\alpha +1}{n+\beta }-v \biggr)h''(v)\,dv \biggr\vert . \end{aligned}$$

(3.5)

Since

$$\begin{aligned} \biggl\vert \int _{u}^{t} (t-v)h''(v) \,dv \biggr\vert \leq (t-v)^{2} \bigl\Vert h'' \bigr\Vert , \end{aligned}$$

(3.6)

we get

$$\begin{aligned} \biggl\vert \int _{u}^{M_{n}^{\alpha,\beta }(e_{1};u)} \bigl(M_{n}^{\alpha ,\beta }(e_{1};u)-v \bigr)h''(v)\,dv \biggr\vert \leq \bigl(M_{n}^{\alpha, \beta }(t-v;u) \bigr)^{2} \bigl\Vert h'' \bigr\Vert . \end{aligned}$$

(3.7)

Combining (3.5), (3.6), and (3.7), we have

$$\begin{aligned} \widehat{M}_{n}^{\alpha,\beta }(h;u)-h(u)|\leq \xi _{n}^{u} \bigl\Vert h'' \bigr\Vert . \end{aligned}$$

 □

Theorem 3.2

Let$f\in C_{B}^{2}[0,\infty )$. Then there exists a constant$C>0$such that

$$\begin{aligned} \bigl| M_{n}^{\alpha,\beta }(g;u)-g(u)\bigr| \leq C\omega _{2} \biggl(g; \frac{1}{2}\sqrt{\xi _{n}^{u}} \biggr)+\omega (g; M_{n}^{\alpha, \beta }(\psi _{u};u), \end{aligned}$$

where$\xi _{n}^{u}$is defined in Lemma 3.1.

Proof

For $h\in C_{B}^{2}[0,\infty )$ and $g\in C_{B}[0,\infty )$ and by the definition of $\widehat{M}_{n}^{\alpha,\beta }$, we have

$$\begin{aligned} \bigl\vert M_{n}^{\alpha,\beta }(g;u)-g(u) \bigr\vert \leq {}&\bigl\vert \widehat{M}_{n}^{\alpha, \beta }(g-h;u) \bigr\vert + \bigl\vert (g-h) (u) \bigr\vert +\bigl|\widehat{M}_{n}^{\alpha,\beta }(h;u)-h(u)) \bigr| \\ &{}+\biggl\vert g \biggl(\frac{n}{n+\beta }u+\frac{\lambda +\alpha +1}{n+\beta } \biggr)-g(u) \biggr\vert . \end{aligned}$$

From Lemma 3.1 and relations in (3.3), one gets

$$\begin{aligned} \bigl\vert M_{n}^{\alpha,\beta }(g;u)-g(u) \bigr\vert \leq{}& 4 \Vert g-h \Vert + \bigl\vert \widehat{M}_{n} ^{\alpha,\beta }(h;u)-h(u) \bigr\vert \\ &{}+ \biggl\vert g \biggl(\frac{n}{n+\beta }u+\frac{\lambda +\alpha +1}{n+ \beta } \biggr)-g(u) \biggr\vert \\ \leq{}& 4 \Vert g-h \Vert +\xi _{n}^{u} \bigl\Vert h'' \bigr\Vert +\omega \bigl(g;M_{n}^{\alpha, \beta }( \psi _{u};u) \bigr). \end{aligned}$$

Using the definition of Peetre’s K-functional, we have

$$\begin{aligned} \bigl\vert M_{n}^{\alpha,\beta }(g;u)-g(u) \bigr\vert \leq C\omega _{2} \biggl(g; \frac{1}{2}\sqrt{\xi _{n}^{u}} \biggr)+\omega (g;M_{n}^{\alpha, \beta }({ \psi _{u}};u). \end{aligned}$$

This completes the proof of Theorem 3.2. □

We consider the Lipschitz type space [30]:

$$\begin{aligned} \operatorname{Lip}_{M}^{k_{1},k_{2}}(\rho ):= \biggl\{ g \in C _{B}[0,\infty ): \bigl\vert g(t) - g(u) \bigr\vert \leq M \frac{ \vert t - u \vert ^{\rho }}{(t + k_{1}u + k_{2}u^{2})^{\frac{\rho }{2}}}:u,t\in (0,\infty ) \biggr\} , \end{aligned}$$

where $M\geq 0$ is a constant, $k_{1},k_{2}>0$, $\rho >0$, and $\rho \in (0, 1]$.

Theorem 3.3

For$g\in \operatorname{Lip}_{M}^{k_{1},k_{2}}(\rho )$, we have

$$\begin{aligned} \bigl\vert M_{n}^{\alpha,\beta }(g;u)-g(u) \bigr\vert \leq M \biggl(\frac{\eta _{n}^{*}(u)}{k _{1}u+k_{2}u^{2}} \biggr)^{\frac{\rho }{2}}, \end{aligned}$$

(3.8)

where$u>0$and$\eta _{n}^{*}(u)=M_{n}^{\alpha,\beta }(\psi _{u}^{2};u)$.

Proof

For $\rho =1$, we have

$$\begin{aligned} \bigl\vert M_{n}^{\alpha,\beta }(g;u)-g(u) \bigr\vert &\leq M_{n}^{\alpha,\beta }{\bigl( \bigl\vert g(t)-g(u) \bigr\vert \bigr)}(u) \\ &\leq M M_{n}^{\alpha,\beta } \biggl(\frac{ \vert t-u \vert }{(t+k_{1}u+k_{2}u ^{2})^{\frac{1}{2}}};u \biggr). \end{aligned}$$

Since $\frac{1}{t+k_{1}u+k_{2}u^{2}}<\frac{1}{k_{1}u+k_{2}u^{2}}$ for all $t,u\in (0,\infty )$, we get

$$\begin{aligned} \bigl\vert M_{n}^{\alpha,\beta }(g;u)-g(u) \bigr\vert &\leq \frac{M}{(k_{1}u+k_{2}u^{2})^{ \frac{1}{2}}} \bigl(M_{n}^{\alpha,\beta }\bigl((t-u)^{2};x \bigr)\bigr)^{\frac{1}{2}} \\ &\leq M \biggl(\frac{\eta _{n}^{*}(u)}{k_{1}u+k_{2}u^{2}} \biggr)^{ \frac{1}{2}}. \end{aligned}$$

Thus Theorem 3.3 holds for $\rho =1$. Next, for $\rho \in (0,1)$ and from Hölder’s inequality with $p=\frac{2}{ \rho }$ and $q=\frac{2}{2-\rho }$, we get

$$\begin{aligned} \bigl\vert M_{n}^{\alpha,\beta }(g;u)-g(u) \bigr\vert &\leq \bigl(M_{n}^{\alpha,\beta }\bigl(\bigl( \bigl\vert g(t)-g(u) \bigr\vert \bigr)^{\frac{2}{ \rho }};u\bigr) \bigr)^{\frac{\rho }{2}} \\ &\leq M \biggl(M_{n}^{\alpha,\beta } \biggl(\frac{ \vert t-u \vert ^{2}}{(t+k_{1}u+k _{2}u^{2})};u \biggr) \biggr)^{\frac{\rho }{2}}. \end{aligned}$$

Since $\frac{1}{t+k_{1}u+k_{2}u^{2}}<\frac{1}{k_{1}u+k_{2}u^{2}}$ for all $t,u\in (0,\infty )$, we obtain

$$\begin{aligned} \bigl\vert M_{n}^{\alpha,\beta }(g;u)-g(u) \bigr\vert &\leq M \biggl(\frac{M_{n}^{\alpha, \beta } ( \vert t-u \vert ^{2};u )}{k_{1}u+k_{2}u^{2}} \biggr)^{\frac{ \rho }{2}} \\ &\leq M \biggl(\frac{\eta _{n}^{*}(u)}{k_{1}u+k_{2}u^{2}} \biggr)^{\frac{ \rho }{2}}. \end{aligned}$$

Hence, we prove Theorem 3.3. □

4 Global approximation

From [10], we recall some notation to prove the global approximation results.

For the weight function $1+u^{2}$ and $0\leq u<\infty$, we have

$$ B_{1+u^{2}}[0,\infty )=\bigl\{ f(u): \bigl\vert f(u) \bigr\vert \leq M_{f} \bigl(1+u^{2}\bigr)\bigr\} , $$

(4.1)

where $M_{f}$ is a constant depending on f and $C_{1+u^{2}}[0, \infty )\subset B_{1+u^{2}}[0,\infty )$ is a space of continuous functions equipped with the norm $\|f\|_{1+u^{2}}=\sup_{u\in [0,\infty )}\frac{|f|}{1+u^{2}}$.

Denote

$$ C_{1+u^{2}}^{k}[0,\infty )=\biggl\{ f\in C_{1+u^{2}}: \lim_{u\rightarrow \infty }\frac{f(u)}{1+u^{2}}=k, \text{ where } k \text{ is a constant}\biggr\} . $$

(4.2)

Theorem 4.1

Let the operators be defined by (1.5) acting from$C_{{1+u^{2}}} ^{k}[0,\infty )$to$B_{1+u^{2}}[0,\infty )$. Then, for every$f\in C_{{1+u^{2}}}^{k}[0,\infty )$, we have

$$\begin{aligned} \lim_{n\rightarrow \infty } \bigl\Vert M_{n}^{\alpha,\beta }(g;u)-g \bigr\Vert _{1+u^{2}}=0, \end{aligned}$$

where$B_{1+u^{2}}$and$C_{1+u^{2}}^{k}$are defined in (4.1) and (4.2).

Proof

To prove this result, it is sufficient to show that

$$\begin{aligned} \lim_{n\rightarrow \infty } \bigl\Vert M_{n}^{\alpha,\beta }(e_{i};u)-u ^{i} \bigr\Vert _{1+u^{2}}=0,\quad i=0,1,2. \end{aligned}$$

Using Lemma 2.4, we obtain

$$\begin{aligned} \bigl\Vert M_{n}^{\alpha,\beta }(e_{0};u)-u^{0} \bigr\Vert _{1+u^{2}}=\sup_{u\in [0,\infty )}\frac{ \vert M_{n}^{\alpha,\beta }(1;u)-1 \vert }{1+u ^{2}}=0 \quad\text{for }i=0. \end{aligned}$$

For $i=1$,

$$\begin{aligned} \bigl\Vert M_{n}^{\alpha,\beta }(e_{1};u) - u \bigr\Vert _{1+u^{2}} & = \sup_{u\in [0,\infty )}\frac{ (\frac{n}{n+\beta }-1 )u+\frac{ \lambda +\alpha +1}{n+\beta }}{1+u^{2}} \\ & = \biggl(\frac{n}{n+\beta } - 1 \biggr)\sup_{u\in [0,\infty )} \frac{u}{1 + u^{2}} + \frac{\lambda + \alpha + 1}{n+\beta }\sup_{u\in [0,\infty )} \frac{1}{1+u ^{2}}. \end{aligned}$$

This shows that $\|M_{n}^{\alpha,\beta }(e_{1};u)-u\|_{1+u^{2}} \rightarrow 0$, $n\rightarrow \infty$. For $i=2$,

$$\begin{aligned} \bigl\Vert M_{n}^{\alpha,\beta }(e_{2};u)-u^{2} \bigr\Vert _{1+u^{2}}={}&\sup_{u\in [0,\infty )}\frac{ \vert M_{n}^{\alpha,\beta }(e_{2};u)-u ^{2} \vert }{1+u^{2}} \\ \leq{}& \biggl(\frac{n^{2}}{(n+\beta )^{2}}-1 \biggr)\sup_{u\in [0,\infty )} \frac{u^{2}}{1+u^{2}} \\ &{}+ \bigl(4+2\lambda +2\alpha +2iE_{n}(u) \bigr) \frac{n}{(n+\beta )^{2}} \sup_{u\in [0,\infty )}\frac{u}{1+u ^{2}} \\ &{}+ \frac{(\lambda +1)(\lambda +2\alpha +2)+\alpha ^{2}}{(n+\beta )^{2}} \sup_{u\in [0,\infty )}\frac{1}{1+u^{2}}, \end{aligned}$$

which shows that $\|M_{n}^{\alpha,\beta }(e_{2};u)-u^{2}\|_{1+u^{2}} \rightarrow 0$, $n\rightarrow \infty$. □

Let $f\in C_{\rho }^{k}[0,\infty )$, Yüksel and Ispir [41] introduced the weighted modulus of continuity as follows:

$$\begin{aligned} \varOmega (g;\delta )=\sup_{u\in [0,\infty ),0< h\leq \delta } \frac{ \vert g(u+h)-g(u) \vert }{1+(u+h)^{2}}. \end{aligned}$$

Lemma 4.2

([41])

Let$f\in C_{1+x^{2}}^{k}[0,\infty )$. Then

• (i) $\varOmega (f;\delta )$is a monotone increasing function ofδ;

• (ii) $\lim_{\delta \rightarrow 0^{+}}\varOmega (f;\delta )=0$;

• (iii) for each$\lambda \in [0,\infty )$, $\varOmega (f;\lambda \delta ) \leq (1+\lambda )(1+\delta ^{2})\varOmega (f;\delta )$.

For$t,x\in [0,\infty )$, one gets

$$\begin{aligned} \bigl\vert f(t)-f(x) \bigr\vert \leq 2 \biggl(1+ \frac{ \vert t-x \vert }{\delta } \biggr) \bigl(1+\delta ^{2} \bigr) \bigl(1+x^{2}\bigr) \bigl(1+(t-x)^{2} \bigr)\varOmega (f;\delta ). \end{aligned}$$

(4.3)

Theorem 4.3

Let$g\in C_{1+u^{2}}^{k}[0,\infty )$. Then we have

$$\begin{aligned} \sup_{u\in [0,\infty )}\frac{ \vert M_{n}^{\alpha,\beta }(g;u)-g(x) \vert }{(1+u ^{2})^{3}}\leq C(n) \biggl(1+ \frac{1}{n+\beta } \biggr)\varOmega \biggl(g;\frac{1}{\sqrt{n+ \beta }} \biggr), \end{aligned}$$

where$C(n)=1+\frac{C_{1}}{n+\beta }\cdot \frac{3\sqrt{(}3)}{16}+\frac{\sqrt{C _{1}}}{2}+\frac{1}{4}\sqrt{\frac{C_{1}C_{2}}{n+\beta }}$and$C_{1}, C_{2}\in (0,\infty )$.

Proof

Using (4.3) and $x,t\in (0,\infty )$, we have

$$\begin{aligned} \bigl\vert M_{n}^{\alpha,\beta }(g;u)-g(u) \bigr\vert \leq{}& 2 \biggl(1+\frac{M_{n}^{\alpha ,\beta }( \vert t-u \vert ;u)}{\delta } \biggr) \bigl(1+\delta ^{2} \bigr) \bigl(1+u ^{2}\bigr) \\ &{}\times \bigl(1+M_{n}^{\alpha,\beta }\bigl((t-u)^{2};u \bigr) \bigr)\varOmega (g; \delta ). \end{aligned}$$

(4.4)

Applying the Cauchy–Schwarz inequality for (4.4), we get

$$\begin{aligned} & \bigl\vert M_{n}^{\alpha,\beta }(g;u)-g(u) \bigr\vert \\ &\quad \leq 2 \bigl(1+\delta ^{2} \bigr) \bigl(1+u ^{2}\bigr)\varOmega (g;\delta ) \biggl(1+M_{n}^{\alpha,\beta } \bigl((t-u)^{2};u\bigr) \\ &\qquad{}+\frac{\sqrt{M_{n}^{\alpha,\beta }((t-u)^{2};u)}}{\delta }+\frac{\sqrt{M _{n}^{\alpha,\beta }((t-u)^{2};u)M_{n}^{\alpha,\beta }((t-u)^{4};u)}}{ \delta } \biggr). \end{aligned}$$

(4.5)

By elementary calculations, we get

$$\begin{aligned} & \lim_{n\to \infty }(n+\beta )M_{n}^{\alpha,\beta } \bigl( \psi _{u} ^{2};u \bigr)=2(1+i)u; \\ & \lim_{n\to \infty }(n+\beta )M_{n}^{\alpha,\beta } \bigl( \psi _{u} ^{4};u \bigr)=3(8i+1)u^{3}. \end{aligned}$$

It follows that, for each fixed point $u>0$, there exists $N_{u} \in \mathbb{N}$ such that, for all $n>N_{u}$,

$$\begin{aligned} & M_{n}^{\alpha,\beta }\bigl(\psi ^{2};u\bigr) \leq C_{1}\frac{u}{n+\beta }, \end{aligned}$$

(4.6)

$$\begin{aligned} &M_{n}^{\alpha,\beta }\bigl(\psi ^{4};u\bigr)\leq C_{2}\frac{u^{3}}{n+\beta }, \end{aligned}$$

(4.7)

where $C_{1},C_{2}\in (0,\infty )$. From (4.5)–(4.7) and choosing $\delta =\frac{1}{\sqrt{n+ \beta }}$, we get the required result. □

5 A-statistical convergence

From [7, 15] we first introduce convergence approximation theorems in operators theory. Here, we recall same notation from [7, 15]. Let $A=(a_{nk})$ be a nonnegative infinite summability matrix. For a given sequence $x:=(x_{k})$, the A-transform of x denoted by $Ax:((Ax)_{n})$ is defined as follows:

$$\begin{aligned} (Ax)_{n}=\sum_{k=1}^{\infty }a_{nk}x_{k}, \end{aligned}$$

provided the series converges for each n. A is said to be regular if $\lim (Ax)_{n}=L$ whenever $\lim x=L$. Then $x=(x_{n})$ is said to be an A-statistical convergence to L i.e. $st_{A}-\lim$$x=L$ if, for every $\epsilon >0$, $\lim_{n}\sum_{k:|x_{k}-L|\geq \epsilon }a_{nk}=0$. In addition, A-statistical convergence results are studied in [13, 22, 23].

Theorem 5.1

Let$A=(a_{nk})$be a nonnegative regular summability matrix and$x\geq 0$. Then we have

$$\begin{aligned} st_{A}-\lim_{n} \bigl\Vert M_{n}^{\alpha,\beta }(g;u)-g \bigr\Vert _{1+u^{2}}=0\quad \textit{for all }g\in C_{1+u^{2}}^{k}[0, \infty ). \end{aligned}$$

Proof

From ([9], p. 191, Th. 3), it is sufficient to show that, for $\lambda _{1}=0$,

$$\begin{aligned} st_{A}-\lim_{n} \bigl\Vert M_{n}^{\alpha,\beta }(e_{i};u)-e_{i} \bigr\Vert _{1+u^{2}}=0 \quad\text{for } i\in \{0,1,2\}. \end{aligned}$$

(5.1)

Using Lemma 2.4, we have

$$\begin{aligned} \bigl\Vert M_{n}^{\alpha,\beta }(e_{1};u)-u \bigr\Vert _{1+u^{2}} &=\sup_{u\in [0,\infty )}\frac{1}{1+u^{2}} \biggl\vert \biggl(\frac{n}{n+ \beta }-1 \biggr)u+\frac{\lambda +\alpha +1}{n+\beta } \biggr\vert \\ &\leq \biggl\vert \frac{n}{n+\beta } - 1 \biggr\vert \sup _{u\in [0,\infty )}\frac{u}{1+u^{2}} + \biggl\vert \frac{\lambda + \alpha + 1}{n+\beta } \biggr\vert \sup_{u\in [0,\infty )}\frac{1}{1+u ^{2}} \\ &\leq \frac{1}{2} \biggl\vert \frac{n}{n+\beta } - 1 \biggr\vert + \biggl\vert \frac{ \lambda + \alpha + 1}{n+\beta } \biggr\vert . \end{aligned}$$

We have

$$ st_{A}- \lim_{n}\frac{1}{2} \biggl\vert \frac{n}{n+\beta } - 1 \biggr\vert =st_{A}- \lim _{n} \biggl\vert \frac{\lambda + \alpha + 1}{n+ \beta } \biggr\vert =0. $$

(5.2)

Now, for given $\epsilon >0$, we define the following sets:

$$\begin{aligned} &J_{1}:= \bigl\{ n: \bigl\Vert M_{n}^{\alpha,\beta }(e_{1};u)-u \bigr\Vert \geq \epsilon \bigr\} , \\ &J_{2}:= \biggl\{ n:\frac{1}{2} \biggl\vert \frac{n}{n+\beta }-1 \biggr\vert \geq \frac{\epsilon }{2} \biggr\} , \\ &J_{3}:= \biggl\{ n: \biggl\vert \frac{\lambda +\alpha +1}{n+\beta } \biggr\vert \geq \frac{\epsilon }{2} \biggr\} . \end{aligned}$$

This implies that $J_{1}\subseteq J_{2}\cup J_{3}$, which shows that $\sum_{k_{1}\in J_{1}}a_{nk_{1}}\leq \sum_{k_{1}\in J_{2}}a_{nk}+ \sum_{k_{1}\in J_{3}}a_{nk}$. Hence, from (5.2) we get

$$\begin{aligned} st_{A}-\lim_{n} \bigl\Vert M_{n}^{\alpha,\beta }(e_{1};u)-u \bigr\Vert _{1+u^{2}}=0. \end{aligned}$$

(5.3)

In a similar way the following can be proved:

$$\begin{aligned} st_{A}-\lim_{n} \bigl\Vert M_{n}^{\alpha,\beta }(e_{2};u)-u^{2} \bigr\Vert _{1+u^{2}}=0. \end{aligned}$$

(5.4)

This completes the proof of Theorem 5.1. □

6 Conclusion

The motive of the present paper is to give a generalized error estimation of convergence by modification of Szász–Mirakyan integral operators via Dunkl analogue. We have defined the Szász–Mirakyan integral operators based on Dunkl analogue with the aid of two nonnegative parameters $0\leq \alpha \leq \beta$. This type of modification enables us to give a generalied error estimation for a certain function in comparison to the Szász–Mirakyan integral operators based on Dunkl analogue. We investigated some approximation results by means of the well-known Korovkin type theorem. We have also calculated the rate of convergence of operators by means of Peetre’s K-functional and second order modulus of continuity. Lastly, we studied the global approximation results.