Better Rate of Convergence by Modified Integral Type Operators

Abstract

In this article, we introduce Chlodowsky Integral type operators with the help of generalized exponential function with two unbounded and non-negative real number sequences $a_n$ and $b_n$. We study their basic estimates and investigate local and global approximation results with the aid of second-order modulus of continuity, Peetre’s K-functional, Lipschitz-type class and rth-order Lipschitz-type maximal function. In the last, statistical approximation results are studied.

1 Introduction

The Sz$\acute{a}$sz-type operators via generalized exponential functions was presented by Sucu [25] as

$$\begin{aligned} S_n(g;u):=\frac{1}{e_\nu (nu)}\sum \limits _{k=0}^{\infty }\frac{(nu)^k}{\gamma _\nu (k)}g\bigg (\frac{k+2\nu \theta _k}{n}\bigg ), \end{aligned}$$

(1)

where the generating function [23] is given as

$$\begin{aligned} e_\mu (x)=\sum _{\nu =0}^{\infty }\frac{x^\nu }{\gamma _\mu (\nu )}, \end{aligned}$$

(2)

with coefficients $\gamma _\mu (kgiven)$ are introduced as

For $\mu >-1/2$ and $k\in \mathbb {N}_0=\{0\}\bigcup \mathbb {N}$, we have

$$\begin{aligned} \gamma _\mu (2k)=\frac{2^{2k}k!\Gamma (k+\mu +1/2)}{\Gamma (\mu +1/2)} \text {, } \gamma _\mu (2k+1)=\frac{2^{2k+1}k!\Gamma (k+\mu +3/2)}{\Gamma (\mu +1/2)}. \end{aligned}$$

Recursive relation is given as

$$\begin{aligned} \gamma _{\mu }(k+1)=(k+1+2\mu \theta _{k+1})\gamma _\mu (k), \text { } k\in \mathbb {N}_0, \end{aligned}$$

(3)

with $\theta _k$ is given to be 0 if $k\in 2\mathbb {N}$ and 1 if $k\in 2\mathbb {N}+1$. The operators presented in (1) are restricted to approximate the continuous functions only. Wafi and Rao [8] constructed a sequence of positive linear operators to discuss the approximation results for the Lebesgue measurable functions as

$$\begin{aligned} D_n(f;x)=\frac{1}{e_\mu (nx)}\sum \limits _{k=0}^{\infty }\frac{(nx)^k}{\gamma _\mu (k)}\frac{n^{k+2\mu \theta _k+\lambda +1}}{\Gamma (k+2\mu \theta _k+\lambda +1)}\int _{0}^{\infty }t^{k+2\mu \theta _k+\lambda }e^{-nt}f(t)dt.\nonumber \\ \end{aligned}$$

(4)

Several mathematicians researched in this direction to approximate the continuous functions only and Lebesgue measurable functions, i.e. Wafi and Rao [7] and Mursaleen et al. [9,10,11], Karaisa et al. [14] and Icoz et al. [12, 13] Motivated by the above, we present a Chlodowsky Integral type operators via Dunkl analogue as

$$\begin{aligned} A_n(f;x)=\frac{1}{e_\mu (a_nx)}\sum \limits _{k=0}^{\infty }\frac{(a_nx)^k}{\gamma _\mu (k)}\frac{b_n^{k+2\mu \theta _k+\lambda +1}}{\Gamma (k+2\mu \theta _k+\lambda +1)}\int _{0}^{\infty }t^{k+2\mu \theta _k+\lambda }e^{-b_nt}f(t)dt,\nonumber \\ \end{aligned}$$

(5)

where $a_n$ and $b_n$ are unbounded and increasing sequences of real numbers such that

$$\begin{aligned} \lim \limits _{n\rightarrow \infty } b_n=\infty \text { and } \lim \limits _{n\rightarrow \infty } \frac{b_n}{n}=0. \end{aligned}$$

(6)

In the subsequent sections, we prove some basic lemmas and proposition which shows the uniform convergence of the operators (5). Further, we study the pointwise approximation results and global approximation results. In the last part of this manuscript, statistical approximation results are investigated.

2 Approximation Properties of $A_n(f;x)$

Lemma 2.1

Let $\mu \ge -\frac{1}{2}$ and $x\ge 0$. Then with the aid of generalized exponential function given in (2), one has

$$\begin{aligned} \frac{1}{e_\mu (a_nx)}\sum \limits _{k=0}^{\infty }\frac{(a_nx)^k}{\gamma _\mu (k)}= & {} 1,\\ \frac{1}{e_\mu (a_nx)}\sum \limits _{k=0}^{\infty }\frac{(a_nx)^k}{\gamma _\mu (k)}(k+2\mu \theta _k)= & {} a_nx,\\ \frac{1}{e_\mu (a_nx)}\sum \limits _{k=0}^{\infty }\frac{(a_nx)^k}{\gamma _\mu (k)}(k+2\mu \theta _k)^2= & {} a_n^2x^2+\Bigg (1+2\mu \frac{e_\mu (-a_nx)}{e_\mu (a_nx)}\Bigg )a_nx,\\ \frac{1}{e_\mu (a_nx)}\sum \limits _{k=0}^{\infty }\frac{(a_nx)^k}{\gamma _\mu (k)}(k+2\mu \theta _k)^3= & {} a_n^3x^3+\Bigg (3-2\mu \frac{e_\mu (-a_nx)}{e_\mu (a_nx)}\Bigg )a_n^2x^2\\+ & {} \left( 1+4\mu ^2+2\mu \frac{e_\mu (-a_nx)}{e_\mu (a_nx)}\right) a_nx,\\ \frac{1}{e_\mu (a_nx)}\sum \limits _{k=0}^{\infty }\frac{(a_nx)^k}{\gamma _\mu (k)}(k+2\mu \theta _k)^4= & {} a_n^4x^4+\Bigg (6+4\mu \frac{e_\mu (-a_nx)}{e_\mu (a_nx)}\Bigg )a_n^3x^3\\+ & {} \left( 7+4\mu ^2-8\mu \frac{e_\mu (-a_nx)}{e_\mu (a_nx)}\right) a_n^2x^2\\+ & {} \left( 1+12\mu ^2+6\mu \frac{e_\mu (-a_nx)}{e_\mu (a_nx)}+8\mu ^3\frac{e_\mu (-a_nx)}{e_\mu (a_nx)}\right) a_nx. \end{aligned}$$

Proof

With the help of Eq. (2) and $\theta _{k+1}=(-1)^{k}+\theta _{k}$, one can easily prove Lemma 2.1.   $\square$

In order to discuss the basic properties of the operators introduced by the Eq. (5), we consider $e_{\nu }(t)=t^{\nu },\nu \in \{0,1,2,3,4\}$ and $\psi _x^{\nu }(t)=(t-x)^{\nu },\nu \in \{1,2,3,4\}$, respectively.

Lemma 2.2

Let $A_n(f;x)$ be the operators defined in (5). Then one has

$$\begin{aligned} A_n(e_0;x)= & {} 1,\\ A_n(e_1;x)= & {} \frac{a_n}{b_n} x+\frac{\lambda +1}{b_n},\\ A_n(e_2;x)= & {} \frac{a_n^2}{b_n^2}x^2+\left( 4+2\lambda +2\mu \frac{e_\mu (-a_nx)}{e_\mu (a_nx)}\right) \frac{a_nx}{b_n^2}+\frac{(\lambda +1)(\lambda +2)}{b_n^2},\\ A_n(e_3;x)= & {} \frac{a_n^3}{b_n^3}x^3+\left( 9+3\lambda -2\mu \frac{e_\mu (-a_nx)}{e_\mu (a_nx)}\right) \frac{a_n^2}{b_n^3}x^2+\Bigg (18+k\lambda (\lambda +5)+4\mu ^2\\+ & {} 2\mu (8+3\lambda )\frac{e_\mu (-a_nx)}{e_\mu (a_nx)}\Bigg )\frac{a_n}{b_n^3}x+\frac{\lambda ^3+6\lambda ^2+11\lambda +6}{n^3},\\ A_n(e_4;x)= & {} \frac{a_n^4}{b_n^4}x^4+\left( 16+4\lambda +4\mu \frac{e_\mu (-a_nx)}{e_\mu (a_nx)}\right) \frac{a_n^3}{b_n^4}+o\left( \frac{1}{b_n^2}\right) . \end{aligned}$$

Proof

Using Lemma 2.1, we have for $\nu =0$

$$\begin{aligned} A_n(e_0;x)= & {} \frac{1}{e_\mu (a_nx)}\sum \limits _{k=0}^{\infty }\frac{(a_nx)^k}{\gamma _\mu (k)}\frac{b_n^{k+2\mu \theta _k+\lambda +1}}{\Gamma (k+2\mu \theta _k+\lambda +1)}\int _{0}^{\infty }t^{k+2\mu \theta _k+\lambda }e^{-b_nt}dt\\= & {} \frac{1}{e_\mu (a_nx)}\sum \limits _{k=0}^{\infty }\frac{(a_nx)^k}{\gamma _\mu (k)}\frac{\Gamma (k+2\mu \theta _k+\lambda +1)}{\Gamma (k+2\mu \theta _k+\lambda +1)}\\= & {} 1. \end{aligned}$$

For $\nu =1$

$$\begin{aligned} A_n(e_1;x)= & {} \frac{1}{e_\mu (a_nx)}\sum \limits _{k=0}^{\infty }\frac{(a_nx)^k}{\gamma _\mu (k)}\frac{b_n^{k+2\mu \theta _k+\lambda +1}}{\Gamma (k+2\mu \theta _k+\lambda +1)}\int _{0}^{\infty }t^{k+2\mu \theta _k+\lambda +1}e^{-b_nt}dt,\\= & {} \frac{1}{e_\mu (a_nx)}\sum \limits _{k=0}^{\infty }\frac{(a_nx)^k}{\gamma _\mu (k)}\frac{\Gamma (k+2\mu \theta _k+\lambda +2)}{b_n\Gamma (k+2\mu \theta _k+\lambda +1)}\\= & {} \frac{1}{b_ne_\mu (a_nx)}\sum \limits _{k=0}^{\infty }\frac{(a_nx)^k}{\gamma _\mu (k)}(k+2\mu \theta _k+\lambda +1)\frac{\Gamma (k+2\mu \theta _k+\lambda +1)}{\Gamma (k+2\mu \theta _k+\lambda +1)}\\= & {} \frac{a_n}{b_n} x+\frac{\lambda +1}{b_n}. \end{aligned}$$

For $\nu =2$

$$\begin{aligned} A_n(e_2;x)= & {} \frac{1}{e_\mu (a_nx)}\sum \limits _{k=0}^{\infty }\frac{(a_nx)^k}{\gamma _\mu (k)}\frac{b_n^{k+2\mu \theta _k+\lambda +1}}{\Gamma (k+2\mu \theta _k+\lambda +1)}\int _{0}^{\infty }t^{k+2\mu \theta _k+\lambda +2}e^{-b_nt}dt\\= & {} \frac{1}{e_\mu (a_nx)}\sum \limits _{k=0}^{\infty }\frac{(a_nx)^k}{\gamma _\mu (k)}\frac{\Gamma (k+2\mu \theta _k+\lambda +3)}{b_n^2\Gamma (k+2\mu \theta _k+\lambda +1)}\\= & {} \frac{1}{e_\mu (a_nx)}\sum \limits _{k=0}^{\infty }\frac{(a_nx)^k}{\gamma _\mu (k)}\frac{(k+2\mu \theta _k+\lambda +2)(k+2\mu \theta _k+\lambda +1)}{b_n^2}\\= & {} \frac{1}{b_n^2e_\mu (a_nx)}\sum \limits _{k=0}^{\infty }\frac{(a_nx)^k}{\gamma _\mu (k)}((k+2\mu \theta _k)^2+(2\lambda +3)(k+2\mu \theta _k)\\+ & {} (\lambda +1)(\lambda +2))\\= & {} \frac{a_n^2}{b_n^2}x^2+\left( 4+2\lambda +2\mu \frac{e_\mu (-a_nx)}{e_\mu (a_nx)}\right) \frac{a_n}{b_n^2}x+\frac{(\lambda +1)(\lambda +2)}{b_n^2}. \end{aligned}$$

Similarly, the rest part of the Lemma 2.2 can be easily proved.   $\square$

Lemma 2.3

Let the $A_n(f;x)$ be the operators given in (5). Then we have

$$\begin{aligned} A_n(\psi _x^1;x)= & {} \frac{\lambda +1}{n},\\ A_n(\psi _x^2;x)= & {} \left( 2+2i\frac{e_i(-nx)}{e_i(nx)}\right) \frac{x}{n}+\frac{(\lambda +1)(\lambda +2)}{n^2},\\ A_n(\psi _x^4;x)= & {} o\left( \frac{1}{n}\right) . \end{aligned}$$

Proof

In view of Lemma 2.2 and linearity property, one has

$$\begin{aligned} A_n(\psi _x^1;x)= & {} A_n(t;x)-xA_n(1;x),\\ A_n(\psi _x^2;x)= & {} A_n(t^2;x)-2xA_n(t;x)+x^2A_n(1;x),\\ A_n(\psi _x^4;x)= & {} A_n(t^4;x)-4xA_n(t^3;x)+6x^2A_n(t^2;x)-4x^3A_n(t;x)+x^4A_n(1;x). \end{aligned}$$

In the light of Lemma 2.2, we prove the Lemma 2.3.   $\square$

Proposition 2.4

For the operators $A_n$ given in (2) and for every $f\in C[0,\infty )$, $A_n$ converges to f uniformly on $[0,a], a>0$.

Proof

From Korovkin Theorem 4.1.4 in [1], it is sufficient to show that

$$\begin{aligned} A_n(e_{\nu };x)\rightarrow e_{\nu }(x), \text { for } \nu =0,1,2. \end{aligned}$$

Lemma 2.2 implies that $A_n(e_0;x)\rightarrow e_0(x)$ as $n\rightarrow \infty$. For $\nu =1$

$$\begin{aligned} \lim \limits _{n\rightarrow \infty } A_n(e_1;x)=\lim \limits _{n\rightarrow \infty }\left( \left( \frac{a_n}{b_n}-2\right) x+\frac{1}{b_n}\right) =e_1(x). \end{aligned}$$

In the similar manner, one can show that for $\nu =2$, $A_n(e_2;x)\rightarrow e_2$ which completes the proof of Proposition 2.4.   $\square$

3 Pointwise Approximation Results of $A_n$

Here, we recall some notations from DeVore and Lorentz [2] as $C_B[0,\infty )$ be the space of bounded and real valued continuous functions endowed with the norm $\Vert f\Vert =\sup \limits _{0\le x<\infty }|f(x)|$. Let the function $f\in C_B[0,\infty )$ and $\delta >0$. Then, the Peetre’s K-functional is given by

$$\begin{aligned} K_2(f,\delta )=inf\{\Vert f-g\Vert +\delta \Vert g''\Vert : g\in C_B^2[0,\infty )\}, \end{aligned}$$

where $C_B^2[0,\infty )=\{g\in C_B[0,\infty ):g', g''\in C_B[0,\infty )\}$. From DeVore and Lorentz [2], p.177, Theorem 2.4], there exits an absolute constant $C>0$ such that

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

(7)

Consider the auxiliary operator $\widehat{A}_n^*$ as

$$\begin{aligned} \widehat{A}_n(f;x)=A_n(f;x)+f(x)-f\left( \frac{a_n}{b_n}x+\frac{1}{b_n}\right) . \end{aligned}$$

(8)

Lemma 3.1

For $g\in C_B^2[0,\infty )$ and $x\ge 0$, one has

$$\begin{aligned} | \widehat{A}_n(g;x)-g(x)|\le & {} \xi _n(x)\Vert g''\Vert , \end{aligned}$$

where

$$\begin{aligned} \nonumber \xi _n(x)=A_n(\psi _x^2;x)+\left( A_n(\psi _x^1;x)\right) ^2. \end{aligned}$$

Proof

With the aid of auxiliary operators given in (8), we get

$$\begin{aligned} \widehat{A}_n(1;x)=1, \, \widehat{A}_n(\psi _x;x)=0 \text { and } |\widehat{A}_n(f;x)|\le & {} 3\Vert f\Vert . \end{aligned}$$

(9)

From Taylor’s series expansion, for every $g\in C_B^2[0,\infty )$, we obtain

$$\begin{aligned} g(t)=g(x)+(t-x)g'(x)+\int \limits _x^t (t-v)g''(v)dv. \end{aligned}$$

(10)

Applying auxiliary operators $\widehat{A}_n$ in Eq. 10, we have

$$\begin{aligned} \widehat{A}_n(g;x)-g(x)= & {} g'(x)\widehat{A}_n(t-x;x)+\widehat{A}_n^*\Big ( \int \limits _x^t (t-v)g''(v)dv;x\Big ). \end{aligned}$$

From (8) and (9), we have

$$\begin{aligned} \widehat{A}_n(g;x)-g(x)= & {} \widehat{A}_n\Big ( \int \limits _x^t (t-v)g''(v)dv;x\Big )\\= & {} A_n\Big ( \int \limits _x^t (t-v)g''(v)dv;x\Big )-\int \limits _x^{\frac{a_n}{b_n}x+\frac{1}{b_n}}\Bigg (\frac{a_n}{b_n}x+\frac{1}{b_n}-v\Bigg )g''(v)dv. \end{aligned}$$

$$\begin{aligned} |\widehat{A}_n(g;x)-g(x)|\le & {} \Bigg |A_n\Big ( \int \limits _x^t (t-v)g''(v)dv;x\Big )\Bigg |+\Bigg |\int \limits _x^{\frac{a_n}{b_n}x+\frac{1}{b_n}} \Bigg (\frac{a_n}{b_n}x+\frac{1}{b_n}-v\Bigg )g''(v)dv\Bigg |. \end{aligned}$$

(11)

Since

$$\begin{aligned} \Bigg | \int \limits _x^t (t-v)g''(v)dv\Bigg |\le (t-x)^2\parallel g''\parallel . \end{aligned}$$

(12)

Then

$$\begin{aligned} \Bigg |\int \limits _x^{\frac{a_n}{b_n}x+\frac{1}{b_n}} \Bigg (\frac{a_n}{b_n}x+\frac{1}{b_n}-v\Bigg )g''(v)dv\Bigg |\le \bigg (\frac{a_n}{b_n}x+\frac{1}{b_n}-x\bigg )^2\parallel g''\parallel . \end{aligned}$$

(13)

Using (12) and (13) in (11), we deduce

$$\begin{aligned} | \widehat{A}_n^*(g;x)-g(x)|\le & {} \Bigg \{A_n((t-x)^2;x)+\Bigg (\frac{a_n}{b_n}x+\frac{1}{b_n}-x\Bigg )^2\Bigg \}\Vert g''\Vert \\= & {} \xi _n(x)\Vert g''\Vert . \end{aligned}$$

Hence, the proof of Lemma 3.1 is completed.   $\square$

Theorem 3.2

For $f\in C_B^2[0,\infty )$, we have

$$\begin{aligned} \mid A_n(f;x)-f(x)\mid \le C\omega _2\big (f;\sqrt{\xi _n(x)}\big )+\omega (f; A_n(\psi _x;x)), \end{aligned}$$

where $\xi _n(x)$ is calculated in Lemma 3.1 and $C>0$ is a constant.

Proof

For $f\in C_B[0,\infty )$ and $g\in C_B^2[0,\infty )$ and the operators $\widehat{A}_n$, we get

$$\begin{aligned} |A_n(f;x)-f(x)|\le & {} |\widehat{A}_n(f-g;x)|+|(f-g)(x)|+|\widehat{A}_n(g;x)-g(x)|\\+ & {} \Bigg |f\Big (\frac{a_n}{b_n}x+\frac{1}{b_n}\Big )-f(x)\Bigg |. \end{aligned}$$

From Lemma 3.1 and identities given by (9), we deduce

$$\begin{aligned} |A_n(f;x)-f(x)|\le & {} 4\Vert f-g\Vert +|\widehat{A}_n(g;x)-g(x)|+\Bigg |f\Big (\frac{a_n}{b_n}x+\frac{1}{b_n}\Big )-f(x)\Bigg |\\\le & {} 4\Vert f-g\Vert +\xi _n(x)\Vert g''\Vert +\omega \Big (f;A_n(\psi _x;x)\Big ). \end{aligned}$$

Using Peetre’s K-functional, one obtains

$$\begin{aligned} |A_n(f;x)-f(x)|\le C\omega _2\big (f;\sqrt{\xi _n(x)}\big )+\omega (f;A_n(\psi _x;x)), \end{aligned}$$

which completes the required result.   $\square$

Consider the Lipschitz-type space [22] as

$$\begin{aligned} Lip_M^{\beta _1,\beta _2}(\gamma ):=\Big \{f\in C_B[0,\infty ) :|f(t)-f(x)|\le M\frac{|t-x|^{\gamma }}{(t+\beta _1x+\beta _2x^2)^{\frac{\gamma }{2}}}:x,t\in (0,\infty )\Big \}, \end{aligned}$$

where $\beta _1,\beta _2>0$ are two fixed real numbers, M is a positive constant and $0<\gamma \le 1$.

Theorem 3.3

For the operators defined by (6) and for every $f\in Lip_M^{\beta _1,\beta _2}(\gamma )$, $0<x<\infty )$, we have

$$\begin{aligned} |A_n(f;x)-f(x)|\le M\Bigg (\frac{\eta _n(x)}{\beta _1x+\beta _2x^2}\Bigg )^{\frac{\gamma }{2}}, \end{aligned}$$

(14)

where $\gamma \in (0,1]$ and $\eta _n(x)=A_n(\psi _x^2;x)$.

Proof

For $\gamma =1$ and $x\in (0,\infty )$, we have

$$\begin{aligned} |A_n(f;x)-f(x)|\le & {} A_n(|f(t)-f(x)|;x)\\\le & {} M A_n\Bigg (\frac{|t-x|}{(t+\beta _1x+\beta _2x^2)^{\frac{1}{2}}};x\Bigg ). \end{aligned}$$

It is obvious that $\frac{1}{t+\beta _1x+\beta _2x^2}<\frac{1}{\beta _1x+\beta _2x^2}$ for all $0\le x<\infty$, we obtain

$$\begin{aligned} |A_n(f;x)-f(x)|\le & {} \frac{M}{(\beta _1x+\beta _2x^2)^{\frac{1}{2}}} (A_n((t-x)^2;x))^{\frac{1}{2}}\\\le & {} M\Bigg (\frac{\eta _n(x)}{\beta _1x+\beta _2x^2}\Bigg )^{\frac{1}{2}}. \end{aligned}$$

This shows that the Theorem 3.3 satisfies for $\gamma =1$. Next, for $\gamma \in (0,\infty )$ and with the aid of Hölder’s inequality with $p=\frac{2}{\gamma }$ and $q=\frac{2}{2-\gamma }$, we deduce

$$\begin{aligned} |A_n(f;x)-f(x)|\le & {} \big (A_n(|f(t)-f(x)|^{\frac{2}{\gamma }};x)\big )^{\frac{\gamma }{2}}\\\le & {} M\Bigg (A_n\Bigg (\frac{|t-x|^2}{(t+\beta _1x+\beta _2x^2)};x\Bigg )\Bigg )^{\frac{\gamma }{2}}. \end{aligned}$$

Since $\frac{1}{t+\beta _1x+\beta _2x^2}<\frac{1}{\beta _1x+\beta _2x^2}$ for all $x\in (0,\infty )$, we have

$$\begin{aligned} |A_n(f;x)-f(x)|\le & {} M \Bigg (\frac{A_n(|t-x|^2;x)}{\beta _1x+\beta _2x^2}\Bigg )^{\frac{\gamma }{2}}\\\le & {} M \Big (\frac{\eta _n(x)}{\beta _1x+\beta _2x^2}\Big )^{\frac{\gamma }{2}}. \end{aligned}$$

Hence, the proof of the Theorem 3.3 is completed.   $\square$

Here, we recall rth-order Lipschitz-type maximal function introduced by Lenze [16] as

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

(15)

Then, we get the next result

Theorem 3.4

For $f\in C_B[0,\infty )$ and $0<r\le 1, 0\le x< \infty$, we have

$$\begin{aligned} |A_n(f;x)-f(x)|\le & {} \widetilde{\omega }_r(f;x)\Big (\eta _n(x)\Big )^{\frac{r}{2}}. \end{aligned}$$

Proof

It is clear that

$$\begin{aligned} |A_n(f;x)-f(x)|\le & {} A_n(|f(t)-f(x)|;x). \end{aligned}$$

From Eq. (15), we have

$$\begin{aligned} |A_n(f;x)-f(x)|\le & {} \widetilde{\omega }_r(f;x)A_n(|t-x|^r;x). \end{aligned}$$

From Hölder’s inequality with $p=\frac{2}{r}$ and $q=\frac{2}{2-r}$, we have

$$\begin{aligned} |A_n(f;x)-f(x)|\le & {} \widetilde{\omega }_r(f;x)\big (A_n(|t-x|^2;x)\big )^{\frac{r}{2}}, \end{aligned}$$

which proves the desired result.   $\square$

4 Global Approximation Results

Here, we recall some notations from [5] to prove next result. Let $B_{1+x^2}[0,\infty )=\{f(x):|f(x)|\le M_f (1+x^2),1+x^2$ is weight function, $M_f$ is a constant depending on f and $x\in [0,\infty ) \}$, $C_{1+x^2}[0,\infty )$ is the space of continuous function in $B_{1+x^2}[0,\infty )$ with the norm $\Vert f(x)\Vert _{1+x^2}=\sup \limits _{x\in [0,\infty )}\frac{|f(x)|}{1+x^2}$ and $C_{1+x^2}^{k}[0,\infty )=\{f\in C_{1+x^2}: \lim \limits _{|x|\rightarrow \infty }\frac{f(x)}{1+x^2}=k,$ where k is a constant depending on $f\}$.

Theorem 4.1

Let $A_n$ be the operators defined by (6) from $C_{{1+x^2}}^k[0,\infty )$ to $B_{1+x^2}[0,\infty )$ satisfying the conditions

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\Vert A_n(e_i;x)-x^i\Vert _{1+x^2}=0, \text { } i=0,1,2. \end{aligned}$$

Then for each $C_{{1+x^2}}^k[0,\infty )$

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\Vert A_n(f;x)-f\Vert _{1+x^2}=0. \end{aligned}$$

Proof

In order to prove this result, it is sufficient to show that

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\Vert A_n(e_i;x)-x^i\Vert _{1+x^2}=0, \text { } i=0,1,2. \end{aligned}$$

From Lemma 2.2, we have

$$\begin{aligned} \Vert A_n(e_0;x)-x^0\Vert _{1+x^2}=\sup \limits _{x\in [0,\infty )}\frac{|A_n(1;x)-1|}{1+x^2}=0 \text { for }i=0. \end{aligned}$$

For $i=1$

$$\begin{aligned} \Vert A_n(e_1;x)-x^1\Vert _{1+x^2}= & {} \sup \limits _{x\in [0,\infty )}\frac{|\frac{a_n}{b_n}x+\frac{1}{b_n}-x|}{1+x^2}\\= & {} \left( \frac{a_n}{b_n}-1\right) \sup \limits _{x\in [0,\infty )}\frac{x}{1+x^2}+\frac{1}{b_n}\sup \limits _{x\in [0,\infty )}\frac{1}{1+x^2}. \end{aligned}$$

This implies that $\Vert A_n(e_1;x)-x^1\Vert _{1+x^2}\rightarrow 0$ an $n\rightarrow \infty$.

For $i=2$

$$\begin{aligned} \Vert A_n(e_2;x)-x^2\Vert _{1+x^2}= & {} \sup \limits _{x\in [0,\infty )}\frac{\Bigg |\left( \frac{a_n^2}{b_n^2}-1\right) x^2+\frac{a_n}{b_n^2}\left( 2+2\mu \frac{e_\mu (-a_nx)}{e_\mu (a_nx)}\right) x+\frac{1}{3b_n^2}\Bigg |}{1+x^2}\\= & {} \left( \frac{a_n^2}{b_n^2}-1\right) \sup \limits _{x\in [0,\infty )}\frac{x^2}{1+x^2}+\frac{a_n}{b_n^2}\left( 2+2\mu \frac{e_\mu (-a_nx)}{e_\mu (a_nx)}\right) \sup \limits _{x\in [0,\infty )}\frac{x}{1+x^2}\\+ & {} \frac{1}{3b_n^2}\sup \limits _{x\in [0,\infty )}\frac{1}{1+x^2}. \end{aligned}$$

Which shows that $\Vert A_n(e_2;x)-x^2\Vert _{1+x^2}\rightarrow 0$ an $n\rightarrow \infty$.   $\square$

In the next result, we discuss a result to approximate each function belongs to $C_{1+x^2}^k[0,\infty )$.

Theorem 4.2

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

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

Proof

For $x_0>0$, a fixed number, we have

$$\begin{aligned} \nonumber \sup \limits _{x\in [0,\infty )}\frac{|A_n(f;x)|-f(x)}{(1+x^2)^{1+\gamma }}\le & {} \sup \limits _{x\le x_0}\frac{|A_n(f;x)|-f(x)}{(1+x^2)^{1+\gamma }}+\sup \limits _{x\ge x_0}\frac{|A_n(f;x)|-f(x)}{(1+x^2)^{1+\gamma }}\\ \nonumber\le & {} \Vert A_n(f;x)|-f(x)\Vert _{C[0,x_0]}\\ \nonumber+ & {} \Vert f\Vert _{1+x^2}\sup \limits _{x\ge x_0}\frac{|A_n(1+t^2;x)|}{(1+x^2)^{1+\gamma }}+\sup \limits _{x\ge x_0}\frac{|f(x)|}{(1+x^2)^{1+\gamma }}\\= & {} J_1+J_2+J_3, say. \end{aligned}$$

(16)

Since $|f(x)|\le \Vert f\Vert _{1+x^2}(1+x^2)$, we have

$$\begin{aligned} J_3= & {} \sup \limits _{x\ge x_0}\frac{|f(x)|}{(1+x^2)^{1+\gamma }}\\\le & {} \sup \limits _{x\ge x_0}\frac{\Vert f\Vert _{1+x^2}(1+x^2)}{(1+x^2)^{1+\gamma }}\le \frac{\Vert f\Vert _{1+x^2}}{(1+x^2)^{\gamma }}. \end{aligned}$$

Let $\epsilon >0$ be arbitrary real number. Then, from Proposition 2.4, there exists $n_1\in \mathbb {N}$ such that

$$\begin{aligned} J_2< & {} \frac{1}{(1+x^2)^{\gamma }}\Vert f\Vert _{1+x^2}\Big (1+x^2+\frac{\epsilon }{3\Vert f\Vert _{1+x^2}}\Big )\text { for all } n_1\ge n,\\< & {} \frac{\Vert f\Vert _{1+x^2}}{(1+x^2)^{\gamma }}+\frac{\epsilon }{3}\text { for all } n_1\ge n. \end{aligned}$$

This implies that

$$\begin{aligned} J_2+J_3<2\frac{\Vert f\Vert _{1+x^2}}{(1+x^2)^{\gamma }}+\frac{\epsilon }{3}. \end{aligned}$$

Next, let for a large value of $x_0$, we have $\frac{\Vert f\Vert _{1+x^2}}{(1+x^2)^{\gamma }}<\frac{\epsilon }{6}$.

$$\begin{aligned} J_2+J_3<\frac{2\epsilon }{3} \text { for all } n_1\ge n. \end{aligned}$$

(17)

Using Theorem 4.1, there exists $n_2>n$ such that

$$\begin{aligned} J_1=\Vert A_n(f)-f\Vert _{C[0,x_0]}<\frac{\epsilon }{3} \text { for all } n_2\ge n. \end{aligned}$$

(18)

For $n_3=max(n_1,n_2)$, using (16), (17) and (18), we obtain

$$\begin{aligned} \sup \limits _{x\in [0,\infty )}\frac{|A_n(f;x)|-f(x)}{(1+x^2)^{1+\gamma }}<\epsilon . \end{aligned}$$

Hence, we arrive at the desired results.   $\square$

5 A-Statistical Approximation Results

Here, we recall some notations [5, 6] as Let $A=(a_{nk})$ be a non-negative infinite sumability matrix. For a given sequence $x:=(x_k)$, the A-transform of x denoted by $Ax:((Ax)_n)$ is and defined as

$$\begin{aligned} (Ax)_n=\sum \limits _{k=1}^{\infty }a_{nk}x_k, \end{aligned}$$

provided the series converges for each n. A is said to be regular if $\lim \limits (Ax)_{n}=L$ whenever $\lim x=L$. Then $x=(x_n)$ is said to be a A-statistically convergent to L, i.e. $st_{A}-\lim$ $x=L$ if for every $\epsilon >0$, $\lim _{n}\sum _{k:|x_k-L|\ge \epsilon }a_{nk}=0$.

Theorem 5.1

For $A=(a_{nk})$, a non-negative regular sumability matrix and for all $f\in C_{1+x^{2+\lambda }}^k[0,\infty )$ with $\lambda >0$, we have

$$\begin{aligned} st_{A}-\lim _{n}\Vert A_n(f;x)-f\Vert _{1+x^{2+\lambda }}=0. \end{aligned}$$

Proof

In the light of [3], p. 191, Th. 3, it is enough to show that for $\lambda =0$

$$\begin{aligned} st_{A}-\lim _{n}\Vert A_n(e_i;x)-e_i\Vert _{1+x^{2}}=0, \text { for } i\in \{0,1,2\}. \end{aligned}$$

(19)

In view of Lemma 2.3, we get

$$\begin{aligned} \Vert A_n(e_1;x)-x\Vert _{1+x^{2}}= & {} \sup \limits _{x\in [0,\infty )}\frac{1}{1+x^2}\Big |\frac{a_n}{b_n}x+\frac{1}{b_n}-x\Big |\\\le & {} \left( \frac{a_n}{b_n}+\frac{1}{b_n}-1\right) . \end{aligned}$$

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

$$\begin{aligned} E_1:= & {} \Bigg \{n: \Vert A_n(e_1;x)-x\Vert \ge \epsilon \Bigg \}\\ E_2:= & {} \Bigg \{n:\left( \frac{a_n}{b_n}x+\frac{1}{b_n}-1\right) \ge \epsilon \Bigg \}. \end{aligned}$$

This implies that $E_1\subseteq E_2$, which shows that $\sum _{k\in E_1}a_{nk}\le \sum _{k\in E_2}a_{nk}$. Therefore, we get

$$\begin{aligned} st_{A}-\lim _{n}\Vert A_n(e_1;x)-x\Vert _{1+x^{2}}=0. \end{aligned}$$

(20)

For $i=2$ and using Lemma 2.3, we have

$$\begin{aligned} \Vert A_n(e_2;x)-x^2\Vert _{1+x^{2}}= & {} \sup \limits _{x\in [0,\infty )}\frac{1}{1+x^2}\Bigg |\left( \frac{a_n^2}{b_n^2}-1\right) x^2+\frac{a_n}{b_n^2}\left( 2+2\mu \frac{e_\mu (-a_nx)}{e_\mu (a_nx)}\right) x+\frac{1}{3b_n^2}\Bigg |\\\le & {} \left( \frac{a_n^2}{b_n^2}-1\right) +\frac{a_n}{b_n^2}\left( 2+2\mu \frac{e_\mu (-a_nx)}{e_\mu (a_nx)}\right) +\frac{1}{3b_n^2}. \end{aligned}$$

For a given $\varepsilon >0$, we have the following sets

$$\begin{aligned} H_1:= & {} \Bigg \{n:\Bigg \Vert A_n(e_2;x)-x^2\Bigg \Vert \ge \epsilon \Bigg \}\\ H_2:= & {} \Bigg \{n:\left( \frac{a_n^2}{b_n^2}-1\right) \ge \frac{\epsilon }{3}\Bigg \}\\ H_3:= & {} \Bigg \{n:\frac{a_n}{b_n^2}\left( 2+2\mu \frac{e_\mu (-a_nx)}{e_\mu (a_nx)}\right) \ge \frac{\epsilon }{3}\Bigg \}\\ H_4:= & {} \Bigg \{n:\frac{1}{3b_n^2}\ge \frac{\epsilon }{3}\Bigg \}. \end{aligned}$$

This implies that $H_1\subseteq H_2\bigcup H_3\bigcup H_4$. By which, we obtained

$$\begin{aligned} \sum _{k\in H_1}a_{nk}\le \sum _{k\in H_2}a_{nk}+\sum _{k\in H_3}a_{nk}+\sum _{k\in H_4}a_{nk}. \end{aligned}$$

As $n\rightarrow \infty$, we get

$$\begin{aligned} st_{A}-\lim _{n}\Vert A_n(e_2;x)-x^2\Vert _{1+x^{2}}=0. \end{aligned}$$

(21)

Therefore, the proof of Theorem 5.1 is completed.   $\square$

Theorem 5.2

Let $f\in C_B^2[0,\infty )$. Then

$$\begin{aligned} st_A-\lim \limits _{n}\Vert A_n(f)-f\Vert _{C_B[0,\infty )}=0. \end{aligned}$$

Proof

With the aid of Taylor’s series expansion, we have

$$\begin{aligned} f(t)=f(x)+f'(x)(t-x)+\frac{1}{2}f''(\xi )(t-x)^2, \end{aligned}$$

where $t\le \xi \le x$. Applying $A_n$, we have

$$\begin{aligned} A_n(f;x)-f(x)=f'(x)A_n(\psi _x;x)+\frac{1}{2}f''(\xi )A_n(\psi _x^2;x). \end{aligned}$$

This implies that

$$\begin{aligned} \nonumber \Vert A_n(f)-f\Vert _{C_B[0,\infty )}\le & {} \Vert f'\Vert _{C_B[0,\infty )}\Vert A_n(e_1-,.)\Vert _{C_B[0,\infty )}\\ \nonumber+ & {} \Vert f''\Vert _{C_B[0,\infty )}\Vert A_n(e_1-,.)^2\Vert _{C_B[0,\infty )}\\= & {} I_1+I_2, say. \end{aligned}$$

(22)

From (19), one has

$$\begin{aligned} \lim \limits _{n}\sum \limits _{k\in \mathbb {N}:I_1\ge \frac{\epsilon }{2}}a_{nk}= & {} 0,\\ \lim \limits _{n}\sum \limits _{k\in \mathbb {N}:I_2\ge \frac{\epsilon }{2}}a_{nk}= & {} 0. \end{aligned}$$

From (22), we have

$$\begin{aligned} \lim \limits _{n}\sum \limits _{k\in \mathbb {N}:\Vert A_n(f)-f\Vert _{C_B[0,\infty )}\ge \epsilon }a_{nk}\le \lim \limits _{n}\sum \limits _{k\in \mathbb {N}:I_1\ge \frac{\epsilon }{2}}a_{nk}+\lim \limits _{n}\sum \limits _{k\in \mathbb {N}:I_2\ge \frac{\epsilon }{2}}a_{nk}. \end{aligned}$$

Thus $st_A-\lim \limits _{n}\Vert A_n(f)-f\Vert _{C_B[0,\infty )}\rightarrow 0.$ as $n\rightarrow \infty .$ Hence, we arrive at the required result.   $\square$