Approximation by Parametric Extension of Szász-Mirakjan-Kantorovich Operators Involving the Appell Polynomials

Abstract

The purpose of this article is to introduce a Kantorovich variant of Szász-Mirakjan operators by including the Dunkl analogue involving the Appell polynomials, namely, the Szász-Mirakjan-Jakimovski-Leviatan-type positive linear operators. We study the global approximation in terms of uniform modulus of smoothness and calculate the local direct theorems of the rate of convergence with the help of Lipschitz-type maximal functions in weighted space. Furthermore, the Voronovskaja-type approximation theorems of this new operator are also presented.

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1. Introduction

In the year 1950, a famous mathematician Szász [1] invented the positive linear operators for the continuous function $f$ on $0,\infty$ and that were extensively searched rather than Bernstein operators [2]. For $z\in 0,\infty$ and $f\in C0,\infty$, Szász introduced the operators as follows: $$\begin{array}{}\text{(1)}& S_{r}f;z=e^{-rz}\sum _{k=0}^{\infty }\frac{{rz}^k}{k!}f\frac{k}{r},\end{array}$$where $C0,\infty$ is the space of continuous functions on $0,\infty$. In recent years, Szász-Mirakjan operators were introduced by Sucu [3] by proposing an exponential function on Dunkl generalization by including a nonnegative number $\eta \ge 0$, such that $$\begin{array}{}\text{(2)}& {\mathcal{S}}_r^{\ast }f;z=\frac{1}{e_{\eta }rz}\sum _{\kappa =0}^{\infty }\frac{{rz}^{\kappa }}{{\gamma }_{\eta }\kappa }f\frac{\kappa +2\eta {\theta }_{\kappa }}{r},\end{array}$$where $e_{\eta }z={\sum }_{\kappa =0}^{\infty }{z}^{\kappa }/{\gamma }_{\eta }\kappa$ and a recursion formula for $s=0,1,2,\cdots$. $$\begin{array}{c}\text{(3)}& \frac{{\gamma }_{\eta }\kappa +1}{\kappa +1+2\eta {\theta }_{\kappa +1}}={\gamma }_{\eta }\kappa ,{\theta }_{\kappa }=\begin{array}{cc}0\hfill & \text{if}\kappa =2r,r\in \mathbb{N}\cup 0,\hfill \\ 1\hfill & \text{if}\kappa =2r+1,r\in \mathbb{N}\cup 0.\hfill \end{array}\end{array}$$

In 1969, Jakimovski and Leviatan introduced the sequence of Szász-Mirakjan-type positive linear operators by the use of Appell polynomials [4], $Lu{e}^{uz}={\sum }_{\kappa =0}^{\infty }{H}_{\kappa }z{u}^{\kappa }$ such that $$\begin{array}{}\text{(4)}& J_{r}h;z=\frac{e^{-rz}}{L1}\sum _{\kappa =0}^{\infty }{H}_{\kappa }rzf\frac{\kappa }{r},\end{array}$$where $L1\ne 0$, $Lu={\sum }_{\kappa =0}^{\infty }{b}_{\kappa }{u}^{\kappa }$, $H_{\kappa }z={\sum }_{j=0}^{\kappa }{b}_j{z}^{\kappa -j}/\kappa -j!\kappa \in \mathbb{N}$. Note that, if $L1=1$ in (4), the Szász-Mirakjan operator (1) is obtained. Most recently, in [5], Nasiruzzaman and Aljohani have introduced the Szász-Mirakjan-Jakimovski-Leviatan-type operators involving the Dunkl generalization for the function $f\in C0,\infty$ by $$\begin{array}{}\text{(5)}& {\mathcal{P}}_{r,\eta }^{\ast }f;z=\frac{1}{L1e_{\eta }rz}\sum _{\kappa =0}^{\infty }{H}_{\kappa }rzf\frac{\kappa +2\eta {\theta }_{\kappa }}{r}.\end{array}$$

Lemma 1. [5].

For the test function ${\gamma }_j=t^j$, if $j=0,1,2,3,4$, the operators ${\mathcal{P}}_{r,\eta }^{\ast }$ have ${\mathcal{P}}_{r,\eta }^{\ast }{\gamma }_0;z=1,{\mathcal{P}}_{r,\eta }^{\ast }{\gamma }_1;z=z+1/r{L}^{\prime }1/\text{L}1+2\eta$, and the following identities: $$\begin{array}{c}\text{(6)}& {\mathcal{P}}_{r,\eta }^{\ast }{\gamma }_2;z=z^2+\frac{1}{r}\frac{2L^{\prime }1}{L1}+4\eta +1z+\frac{1}{r^2}\frac{L^{\prime \prime }1}{L1}+1+4\eta \frac{L^{\prime }1}{L1}+4{\eta }^2,{\mathcal{P}}_{r,\eta }^{\ast }{\gamma }_3;z=z^3+\frac{3}{r}\frac{L^{\prime }1}{L1}+2\eta +1z^2+\frac{1}{r^2}\frac{3L^{\prime \prime }1}{L1}+61+2\eta \frac{L^{\prime }1}{L1}+2+6\eta +12{\eta }^{2}z+\frac{1}{r^3}\frac{3L^{\prime \prime \prime }1}{L1}+31+2\eta \frac{L^{\prime \prime }1}{L1}+21+3\eta +6{\eta }^2\frac{L^{\prime }1}{L1}+8{\eta }^3,\\ {\mathcal{P}}_{r,\eta }^{\ast }{\gamma }_4;z=z^4+\frac{1}{r}\frac{4L^{\prime }1}{L1}+8\eta +6z^3+\frac{1}{r^2}\frac{6L^{\prime \prime }1}{L1}+81+3\eta \frac{L^{\prime }1}{L1}+11+24\eta +24{\eta }^2{z}^2+\frac{1}{r^3}\frac{4L^{\prime \prime \prime }1}{L1}+81+3\eta \frac{L^{\prime \prime }1}{L1}+211+24\eta +24{\eta }^2\frac{L^{\prime }1}{L1}+6+16\eta +24{\eta }^{+}32{\eta }^{3}z+\frac{1}{r^4}\frac{L^{\prime v}1}{L1}+23+4\eta \frac{L^{\prime \prime \prime }1}{L1}+11+24\eta +24{\eta }^2\frac{L^{\prime \prime }1}{L1}+6+16\eta +24{\eta }^2+32{\eta }^3\frac{L^{\prime }1}{L1}+16{\eta }^4.\end{array}$$

There are several research articles mentioned regarding the Szász-Mirakjan-type operators, for instance, [6–13]. For some further related concepts and approximation, we refer to see [9, 10, 14–20].

2. Kantorovich Operators Involving Appell Polynomials and Their Moments

In this section, we construct the generalized operators of recent investigation [5] including the Kantorovich polynomial. For this purpose, we let $f\in C_{\Phi }0,\infty =f\in C0,\infty \text{:}ft=O{t}^{\Phi }$ as $t\to \infty$; then, for all $z\in 0,\infty$, $\Phi >r$, $r\in \mathbb{N}$, $L1\ne 0$, and $\eta \ge 0$, we define the operators as follows: $$\begin{array}{}\text{(7)}& {\mathcal{R}}_{r,\eta }^{\ast }f;z=\frac{r}{L1e_{\eta }rz}\sum _{\kappa =0}^{\infty }{H}_{\kappa }rz{\int }_{\kappa +2\eta {\theta }_{\kappa }/r}^{\kappa +1+2\eta {\theta }_{\kappa }/r}ft\text{d}t.\end{array}$$

Lemma 2.

For $j=0,1,2,3,4$, let the test function be ${\gamma }_j=t^j$. Then, operators ${\mathcal{R}}_{r,\eta }^{\ast }·;·$ have the following identities: $$\begin{array}{c}\text{(8)}& {\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_0;z=1,{\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_1;z=z+\frac{1}{2r}2\frac{L^{\prime }1}{\text{L}1}+4\eta +1,\\ {\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_2;z=z^2+\frac{1}{r}2\frac{L^{\prime }1}{L1}+4\eta +2z+\frac{1}{3r^2}3\frac{L^{\prime \prime }1}{L1}+61+2\eta \frac{L^{\prime }1}{L1}+12{\eta }^2+6\eta +1,\\ {\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_3;z=z^3+\frac{3}{2r}2\frac{L^{\prime }1}{L1}+4\eta +3z^2+\frac{3}{2r^2}2\frac{L^{\prime \prime }1}{L1}+23+4\eta \frac{L^{\prime }1}{L1}+8{\eta }^2+8\eta +3z+\frac{1}{4r^3}12\frac{L^{\prime \prime \prime }1}{L1}+63+4\eta \frac{L^{\prime \prime }1}{L1}+68{\eta }^2+8\eta +3\frac{L^{\prime }1}{L1}+32{\eta }^3+8{\eta }^2+8\eta +1,\\ {\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_4;z=z^4+\frac{1}{r}4\frac{L^{\prime }1}{L1}+8\eta +8z^3+\frac{1}{r^2}6\frac{L^{\prime \prime }1}{L1}+14+24\eta \frac{L^{\prime }1}{L1}+24{\eta }^2+36\eta +9z^2+\frac{1}{r^3}4\frac{L^{\prime \prime \prime }1}{L1}+14+24\eta \frac{L^{\prime \prime }1}{L1}+38+72\eta +48{\eta }^2\frac{L^{\prime }1}{L1}+32{\eta }^3+48{\eta }^2+36\eta +13z+\frac{1}{r^4}\frac{L^{\prime v}1}{L1}+43+2\eta \frac{L^{\prime \prime \prime }1}{L1}+19+36\eta +24{\eta }^2\frac{L^{\prime \prime }1}{L1}+13+36\eta +48{\eta }^2+32{\eta }^3\frac{L^{\prime }1}{L1}+16{\eta }^4+16{\eta }^3+8{\eta }^2+2\eta +1.\end{array}$$

Proof.

To prove this Lemma, we take into account [5] Lemma 1. Thus, for all $j=0,1,2,3,4$ and ${\gamma }_j=t^j$, we can conclude that $$\begin{array}{}\text{(9)}& {\int }_{\kappa +2\eta {\theta }_{\kappa }/r}^{\kappa +1+2\eta {\theta }_{\kappa }/r}{t}^j\text{d}t=\begin{array}{cc}\frac{1}{r}\hfill & \hspace{1em}\text{for}j=0,\hfill \\ \frac{1}{2r^2}+\frac{1}{r}\frac{\kappa +2\eta {\theta }_{\kappa }}{r}\hfill & \hspace{1em}\text{for}j=1,\hfill \\ \frac{1}{3r^3}+\frac{1}{r^2}\frac{\kappa +2\eta {\theta }_{\kappa }}{r}+\frac{1}{r}{\frac{\kappa +2\eta {\theta }_{\kappa }}{r}}^2\hfill & \hspace{1em}\text{for}j=2,\hfill \\ \frac{1}{4r^4}+\frac{1}{r^3}\frac{\kappa +2\eta {\theta }_{\kappa }}{r}+\frac{3}{2r^2}{\frac{\kappa +2\eta {\theta }_{\kappa }}{r}}^2+\frac{1}{r}{\frac{\kappa +2\eta {\theta }_{\kappa }}{r}}^3\hfill & \hspace{1em}\text{for}j=3,\hfill \\ \frac{1}{5r^5}+\frac{1}{r^4}\frac{\kappa +2\eta {\theta }_{\kappa }}{r}+\frac{2}{r^3}{\frac{\kappa +2\eta {\theta }_{\kappa }}{r}}^2+\frac{2}{r^2}{\frac{\kappa +2\eta {\theta }_{\kappa }}{r}}^3+\frac{1}{r}{\frac{\kappa +2\eta {\theta }_{\kappa }}{r}}^4\hfill & \hspace{1em}\text{for}j=4.\hfill \end{array}\end{array}$$

Thus, from (7) and (9), clearly we can write $$\begin{array}{c}\text{(10)}& {\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_0;z={\mathcal{P}}_{r,\eta }^{\ast }{\gamma }_0;z=1,{\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_1;z={\mathcal{P}}_{r,\eta }^{\ast }{\gamma }_1;z+\frac{1}{2r}{\mathcal{P}}_{r,\eta }^{\ast }{\gamma }_0;z,\\ {\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_2;z={\mathcal{P}}_{r,\eta }^{\ast }{\gamma }_2;z+\frac{1}{r}{\mathcal{P}}_{r,\eta }^{\ast }{\gamma }_1;z+\frac{1}{3r^2}{\mathcal{P}}_{r,\eta }^{\ast }{\gamma }_0;z,\\ {\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_3;z={\mathcal{P}}_{r,\eta }^{\ast }{\gamma }_3;z+\frac{3}{2r}{\mathcal{P}}_{r,\eta }^{\ast }{\gamma }_2;z+\frac{1}{r^2}{\mathcal{P}}_{r,\eta }^{\ast }{\gamma }_1;z+\frac{1}{4r^3}{\mathcal{P}}_{r,\eta }^{\ast }{\gamma }_0;z,\\ {\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_4;z={\mathcal{P}}_{r,\eta }^{\ast }{\gamma }_4;z+\frac{2}{r}{\mathcal{P}}_{r,\eta }^{\ast }{\gamma }_3;z+\frac{2}{r^2}{\mathcal{P}}_{r,\eta }^{\ast }{\gamma }_2;z+\frac{1}{r^3}{\mathcal{P}}_{r,\eta }^{\ast }{\gamma }_1;z+\frac{1}{r^4}{\mathcal{P}}_{r,\eta }^{\ast }{\gamma }_0;z.\end{array}$$

Therefore, by applying Lemma 1, we get the required results.

Lemma 3.

For the central moments ${{\gamma }_1-z}^i,i=1,2,4$, we have the following identities: $$\begin{array}{c}\text{(11)}& {\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_1-z;z=\frac{1}{2r}2\frac{L^{\prime }1}{\text{L}1}+4\eta +1,{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2;z=\frac{z}{r}+\frac{1}{3r^2}3\frac{L^{\prime \prime }1}{L1}+61+2\eta \frac{{\text{L}}^{\prime }1}{L1}+12{\eta }^2+6\eta +1,\\ {\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^4;z=-\frac{7}{r^2}{z}^2+\frac{1}{r^3}-8\frac{L^{\prime \prime \prime }1}{L1}-4\frac{L^{\prime \prime }1}{L1}+24\eta +20\frac{L^{\prime }1}{L1}+40{\eta }^2+28\eta +12z+\frac{1}{r^4}\frac{L^{\prime v}1}{L1}+43+2\eta \frac{L^{\prime \prime \prime }1}{L1}+19+36\eta +24{\eta }^2\frac{L^{\prime \prime }1}{L1}+13+36\eta +48{\eta }^2+32{\eta }^3\frac{L^{\prime }1}{L1}+16{\eta }^4+16{\eta }^3+8{\eta }^2+2\eta +1.\end{array}$$

3. Approximations in Weighted Space

In the present section, we follow the well-known results by Gadziev [21] and recall the results in weighted spaces with some additional conditions precisely, under the analogous of P.P. Korovkin’s theorem holds. In order to define the uniformly approximations, we take $z\to \phi z$ be the kind of functions which is continuous and strictly increasing with the assumptions $\Phi z=1+{\phi }^{2}z$ and ${\lim }_{z\to \infty }\Phi z=\infty$. For this reason, we let $B_{\Phi }0,\infty$ be a set of all such functions which are defined on $0,\infty$ and verifying the results $$\begin{array}{}\text{(12)}& B_{\Phi }0,\infty =f:\mid fz\mid \le K_f\Phi z,\end{array}$$where $K_f$ is a constant and depending only on function $f$ and $B_{\Phi }0,\infty$ equipped the norm with $$\begin{array}{}\text{(13)}& \parallel f{\parallel }_{\Phi }=\underset{z\in 0,\infty }{\sup }\frac{\mid fz\mid }{\Phi z}.\end{array}$$

Furthermore, we denote the set all continuous functions on $0,\infty$ by $C0,\infty$ and its subsets be $C_{\Phi }0,\infty$ defined as $C_{\Phi }0,\infty =B_{\Phi }0,\infty \cap C0,\infty$.

It is well known for the sequence of linear positive operators ${K_r}_{r\ge 1}$ (see [21]) maps $C_{\Phi }0,\infty$ into $B_{\Phi }0,\infty$ if and only if $$\begin{array}{}\text{(14)}& \mid K_r\Phi ;z\mid \le M\Phi z,\end{array}$$where $M$ is a positive constant. For $m\in \mathbb{N}$, let us denote $$\begin{array}{}\text{(15)}& C_{\Phi }^{m}0,\infty =f\in C_{\Phi }0,\infty \text{:}\underset{z\to \infty }{\lim }\frac{fz}{\Phi z}=c,\text{exists}\text{and}\text{is}\text{finite}.\end{array}$$

Theorem 4.

Let $H_f=f:\text{such}\text{that}fz/\Phi z\text{is}\text{convergent}\text{when}z\to \infty$. Then, for every $f\in H_f\cap C_{\Phi }0,\infty$, operators (7) are uniformly convergent on each compact subset of $0,\infty$ such that $$\begin{array}{}\text{(16)}& {\mathcal{R}}_{r,\eta }^{\ast }f;z\Rightarrow f,\end{array}$$where $\Rightarrow$ denotes the uniform convergence.

Proof.

In view of Lemma 2, we use Korovkin’s theorem by [22]; then, it is enough to see that for each $j=0,1,2$$$\begin{array}{}\text{(17)}& {\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_j;z\to z^j\end{array}$$uniformly. Thus obviously, we get ${\lim }_{r\to \infty }{\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_0;z=1$, ${\lim }_{r\to \infty }{\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_1;z=z$, and ${\lim }_{r\to \infty }{\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_2;z=z^2$, which completes the proof of Theorem 4.

Theorem 5 [21, 23].

Let the positive linear operators ${J_r}_{r\ge 1}$ acting from $C_{\Phi }0,\infty$ to $B_{\Phi }0,\infty$ and for $j=0,1,2$ if it verifies that ${\lim }_{r\to \infty }{J_r{\gamma }_j-z^j}_{\Phi }=0$, then for every $f\in C_{\Phi }^{m}0,\infty$ it satisfies $$\begin{array}{}\text{(18)}& \underset{r\to \infty }{\lim }{J_{r}f-f}_{\Phi }=0.\end{array}$$

Theorem 6.

For every $\phi \in C_{\Phi }^{m}0,\infty$, operators ${\mathcal{R}}_{r,\eta }^{\ast }$ satisfy $$\begin{array}{}\text{(19)}& \underset{r\to \infty }{\lim }{{\mathcal{R}}_{r,\eta }^{\ast }\phi -\phi }_{\Phi }=0.\end{array}$$

Proof.

It is enough to prove Theorem 6; we use the well-known Korovkin theorem and show $$\begin{array}{}\text{(20)}& \underset{r\to \infty }{\lim }\parallel {\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_j-z^j{\parallel }_{\Phi }=0,\hspace{1em}j=0,1,2.\end{array}$$

Taking into account Lemma 2, then it is easy to see that $$\begin{array}{}\text{(21)}& \parallel {\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_0-1{\parallel }_{\Phi }=\underset{z\in 0,\infty }{\sup }\frac{\mid {\mathcal{R}}_{r,\eta }^{\ast }1;z-1\mid }{\Phi z}=0.\end{array}$$

For $j=1$, we can write here $$\begin{array}{}\text{(22)}& \parallel {\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_1-z{\parallel }_{\Phi }=\underset{z\in 0,\infty }{\sup }\frac{{\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_1;z-z}{\Phi z}=\underset{z\in 0,\infty }{\sup }\frac{1}{\Phi z}\frac{1}{2r}\frac{2L^{\prime }1}{\text{L}1}+4\eta +1.\end{array}$$

If $r\to \infty$, then easily we get $\parallel {\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_1-z{\parallel }_{\Phi }\to 0$. Similarly, for $j=2$, we conclude that $$\begin{array}{}\text{(23)}& \parallel {\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_2-z^2{\parallel }_{\Phi }=\underset{z\in 0,\infty }{\sup }\frac{\mid {\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_2;z-z^2\mid }{\Phi z}=\underset{z\in 0,\infty }{\sup }\frac{z}{1+z^2}\frac{1}{r}2\frac{L^{\prime }1}{L1}+4\eta +2+\underset{z\in 0,\infty }{\sup }\frac{1}{\Phi z}\frac{1}{3r^2}3\frac{L^{\prime \prime }1}{L1}+61+2\eta \frac{L^{\prime }1}{L1}+12{\eta }^2+6\eta +1.\end{array}$$

Thus, we easily get $\parallel {\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_2-z^2{\parallel }_{\Phi }\to 0$, as $r\to \infty$.

Theorem 7.

If $\phi \in C_{\Phi }^{m}0,\infty$. Then, operators ${\mathcal{R}}_{r,\eta }^{\ast }$ follow that $$\begin{array}{}\text{(24)}& \underset{r\to \infty }{\lim }\underset{z\in 0,\infty }{\sup }\frac{\mid {\mathcal{R}}_{r,\eta }^{\ast }\phi ;z-\phi z\mid }{{\Phi z}^{1+\xi }}=0,\end{array}$$where the number $\xi \in 0,\infty$.

Proof.

By the virtue of $\mid \phi z\mid \le {\phi }_{\Phi }1+z^2$ and for any positive real $z_0$, we easily obtain $$\begin{array}{}\text{(25)}& \underset{r\to \infty }{\lim }\underset{z\in 0,\infty }{\sup }\frac{\mid {\mathcal{R}}_{r,\eta }^{\ast }\phi ;z-\phi z\mid }{{\Phi z}^{1+\xi }}\le \underset{z\le z_0}{\sup }\frac{\mid {\mathcal{R}}_{r,\eta }^{\ast }\phi ;z-\phi z\mid }{{\Phi z}^{1+\xi }}+\underset{z\ge z_0}{\sup }\frac{\mid {\mathcal{R}}_{r,\eta }^{\ast }\phi ;z-\phi z\mid }{{\Phi z}^{1+\xi }}\le {{\mathcal{R}}_{r,\eta }^{\ast }\phi ;z-\phi z}_{C0,z_0}+{\phi }_{\Phi }\underset{z\ge z_0}{\sup }\frac{\mid {\mathcal{R}}_{r,\eta }^{\ast }1+t^2;z-\phi z\mid }{{\Phi z}^{1+\xi }}+\underset{z\ge z_0}{\sup }\frac{\mid \phi z\mid }{{\Phi z}^{1+\xi }}={\mathcal{J}}_1+{\mathcal{J}}_2+{\mathcal{J}}_3,\text{suppose}.\end{array}$$

Thus, $$\begin{array}{}\text{(26)}& {\mathcal{J}}_3=\underset{z\ge z_0}{\sup }\frac{\mid \phi z\mid }{{\Phi z}^{1+\xi }}\le \underset{z\ge z_0}{\sup }\frac{{\phi }_{\Phi }1+z^2}{{\Phi z}^{1+\xi }}\le \frac{{\phi }_{\Phi }}{{1+z_0^2}^{\xi }}.\end{array}$$

From Lemma 2, it follows that $$\begin{array}{}\text{(27)}& \underset{r\to \infty }{\lim }\underset{z\ge z_0}{\sup }\frac{{\mathcal{R}}_{r,\eta }^{\ast }1+t^2;z}{\Phi z}=1.\end{array}$$

Now, for each $\epsilon >0$, there exists $r_1\in \mathbb{N}$ for all $r\ge r_1$ such that $$\begin{array}{}\text{(28)}& \underset{z\ge z_0}{\sup }\frac{{\mathcal{R}}_{r,\eta }^{\ast }1+t^2;z}{\Phi z}\le \frac{{1+z_0^2}^{\xi }}{{\phi }_{\Phi }}\frac{\epsilon }{3}+1.\end{array}$$

Therefore, for all $r\ge r_1$$$\begin{array}{}\text{(29)}& {\mathcal{J}}_2={\phi }_{\Phi }\underset{z\ge z_0}{\sup }\frac{{\mathcal{R}}_{r,\eta }^{\ast }1+t^2;z}{{\Phi z}^{1+\xi }}\le \frac{{\phi }_{\Phi }}{{1+z_0^2}^{\xi }}+\frac{\epsilon }{3}.\end{array}$$

In view of (26) and (29), we get $$\begin{array}{}\text{(30)}& {\mathcal{J}}_2+{\mathcal{J}}_3\le 2\frac{{\phi }_{\Phi }}{{1+z_0^2}^{\xi }}+\frac{\epsilon }{3}.\end{array}$$

If we choose any $z_0$ so large, such that ${\phi }_{\Phi }/{1+z_0^2}^{\xi }\le \epsilon /6$, then we get $$\begin{array}{}\text{(31)}& {\mathcal{J}}_2+{\mathcal{J}}_3\le \frac{2\epsilon }{3},\hspace{1em}\text{for}\text{all}r\ge r_1.\end{array}$$

On the other hand, there exists $r_2\ge r$ such that $$\begin{array}{}\text{(32)}& {\mathcal{J}}_1={{\mathcal{R}}_{r,\eta }^{\ast }\phi ;z-\phi z}_{C0,z_0}\le \frac{\epsilon }{3}.\end{array}$$

Finally, take $r_3=\max r_1,r_2$ and on combining (31) and (32) with the above expression, we get $$\begin{array}{}\text{(33)}& \underset{z\in 0,\infty }{\sup }\frac{\mid {\mathcal{R}}_{r,\eta }^{\ast }\phi ;z-\phi z\mid }{{\Phi z}^{1+\xi }}<\epsilon .\end{array}$$

This completes the proof of Theorem 7.

Definition 8.

For every $\overline{\delta }>0$ and all $f\in C0,\infty$, the modulus of continuity of the uniformly continuous function $f$ on $0,\infty$ defined as $$\begin{array}{c}\text{(34)}& \overline{\omega }f;\overline{\delta }=\underset{\mid t_1-t_2\mid \le \overline{\delta }}{\sup }f{t}_1-f{t}_2,t_1,t_2\in 0,\infty ,\mid f{t}_1-f{t}_2\mid \le 1+\frac{\mid t_1-t_2\mid }{{\overline{\delta }}^2}\overline{\omega }f;\overline{\delta }.\end{array}$$

Theorem 9 [24].

Let the sequence of positive linear operators $K_{r\ge 1}:x,y\to Cu,v$ and $u,v\subseteq x,y$, then

(1) for any $f\in Cx,y$ and $z\in u,v$, it follows that

$$\begin{array}{}\text{(35)}& K_{r}f;z-fz\le fz{K}_{r}1;z-1+K_{r}1;z+\frac{1}{\overline{\delta }}\sqrt{K_r{t-z}^2;z}\sqrt{K_{r}1;z}\overline{\omega }f;\overline{\delta },\end{array}$$

(2) if any ${\phi }^{\prime }\in Cx,y$, then for all $z\in u,v$ one has

$$\begin{array}{}\text{(36)}& K_r\varphi ;z-\varphi z\le \varphi z{K}_{r}1;z-1+{\varphi }^{\prime }z{K}_{r}t-z;z+K_r{t-z}^2;z\sqrt{K_{r}1;z}+\frac{1}{\overline{\delta }}\sqrt{K_r{t-z}^2;z}\overline{\omega }{\varphi }^{\prime };\overline{\delta }.\end{array}$$

Theorem 10.

Let $f\in C_{\Phi }0,\infty$, then for all $z\in 0,\infty$ it follows the inequality $$\begin{array}{}\text{(37)}& \mid {\mathcal{R}}_{r,\eta }^{\ast }f;z-fz\mid \le 2\overline{\omega }f;\sqrt{{\overline{\delta }}_{r,\eta }^{\ast }z},\end{array}$$where $\overline{\delta }=\sqrt{{\overline{\delta }}_{r,\eta }^{\ast }z}=\sqrt{{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2;z}$.

Proof.

If we consider Lemma 2 and Theorem 9, then we can obtain $$\begin{array}{}\text{(38)}& {\mathcal{R}}_{r,\eta }^{\ast }f;z-fz\le fz{\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_0;z-1+{\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_0;z+\frac{1}{\overline{\delta }}\sqrt{{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_0-z}^2;z}\sqrt{{\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_0;z}\overline{\omega }f;\overline{\delta },\end{array}$$where if we take $\overline{\delta }=\sqrt{{\overline{\delta }}_{r,\eta }^{\ast }z}=\sqrt{{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2;z}$ then we are easily denumerable to get results.

Theorem 11.

For any $z\in 0,\infty$, if $\varphi \in C_{\Phi }^{\prime }0,\infty$, then we have the inequality $$\begin{array}{}\text{(39)}& {\mathcal{R}}_{r,\eta }^{\ast }\varphi ;z-\varphi z\le \frac{1}{2r}2\frac{L^{\prime }1}{\text{L}1}+4\eta +1{\varphi }^{\prime }z+2{\overline{\delta }}_{r,\eta }^{\ast }z\overline{\omega }{\varphi }^{\prime };\sqrt{{\overline{\delta }}_{r,\eta }^{\ast }z},\end{array}$$where $\overline{\delta }=\sqrt{{\overline{\delta }}_{r,\eta }^{\ast }z}=\sqrt{{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2;z}$.

Proof.

If we consider Lemmas 2 and 3 and (2) of Theorem 9, then it is obvious to get that $$\begin{array}{}\text{(40)}& {\mathcal{R}}_{r,\eta }^{\ast }\varphi ;z-\varphi z\le {\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_0;z-1\varphi z+{\varphi }^{\prime }z{\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_1-z;z+{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2;z\sqrt{{\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_0;z}+\frac{1}{\overline{\delta }}\sqrt{{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2;z}\overline{\omega }{\varphi }^{\prime };\overline{\delta }.\end{array}$$

Put $\overline{\delta }=\sqrt{{\overline{\delta }}_{r,\eta }^{\ast }z}=\sqrt{{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2;z}$, then we easily get our desired results of Theorem 11.

From [25] for an arbitrary $f\in C_{\Phi }^{m}0,\infty ,m\in \mathbb{N}\cup 0$, the weighted modulus of continuity introduced such that $$\begin{array}{}\text{(41)}& \overline{\Omega }f;\overline{\delta }=\underset{z\in 0,\infty ,\mid h\mid \le \overline{\delta }}{\sup }\frac{\mid fz+h-fz\mid }{1+h^{2}1+z^2}.\end{array}$$

Two main properties of this modulus of continuity are ${\lim }_{\overline{\delta }\to 0}\overline{\Omega }f;\overline{\delta }=0$ and $$\begin{array}{}\text{(42)}& \mid ft-fz\mid \le 21+\frac{\mid t-z\mid }{\overline{\delta }}1+{\overline{\delta }}^{2}1+z^{2}1+{t-z}^2\overline{\Omega }f;\overline{\delta },\end{array}$$where $t,z\in 0,\infty$ and $\overline{\Omega }$ weighted modulus of continuity of the function for $f\in C_{\Phi }^{m}0,\infty$.

Theorem 12.

Let $f\in C_{\Phi }^{m}0,\infty$, then for all $z\in 0,\infty$ we have the inequality $$\begin{array}{}\text{(43)}& \underset{z\in 0,{\alpha }_r\eta }{\sup }\frac{{\mathcal{R}}_{r,\eta }^{\ast }f;z-fz}{1+z^2}\le M1+{\alpha }_r\eta \Omega f;\sqrt{{\alpha }_r\eta },\end{array}$$where $M=22+M_1+\sqrt{M_2}>0$, for $M_1,M_2>0$ and $$\begin{array}{}\text{(44)}& {\alpha }_r\eta =\max \frac{1}{r},\frac{1}{3r^2}3\frac{L^{\prime \prime }1}{L1}+61+2\eta \frac{{\text{L}}^{\prime }1}{L1}+12{\eta }^2+6\eta +1.\end{array}$$

Proof.

We use expressions (41) and (42) and applying the Cauchy-Schwarz inequality to operators ${\mathcal{R}}_{r,\eta }^{\ast }$, we get $$\begin{array}{}\text{(45)}& {\mathcal{R}}_{r,\eta }^{\ast }f;z-fz\le 21+{\overline{\delta }}^{2}1+z^2\overline{\Omega }f;\overline{\delta }1+{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2;z+{\mathcal{R}}_{r,\eta }^{\ast }1+{{\gamma }_1-z}^2\frac{\mid {\gamma }_1-z\mid }{\overline{\delta }};z.\end{array}$$

We know the expression $$\begin{array}{}\text{(46)}& {\mathcal{R}}_{r,\eta }^{\ast }1+{{\gamma }_1-z}^2\frac{\mid {\gamma }_1-z\mid }{\overline{\delta }};z=\frac{1}{\overline{\delta }}{\mathcal{R}}_{r,\eta }^{\ast }\mid {\gamma }_1-z\mid ;z+{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2\frac{\mid {\gamma }_1-z\mid }{\overline{\delta }};z\le \frac{1}{\overline{\delta }}{{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2;z}^{1/2}+{{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^4;z}^{1/2}{{\mathcal{R}}_{r,\eta }^{\ast }\frac{{{\gamma }_1-z}^2}{{\overline{\delta }}^2};z}^{1/2}=\frac{1}{\overline{\delta }}{{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2;z}^{1/2}1+\sqrt{{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^4;z}.\end{array}$$

In view of Lemma 3, we can obtain $$\begin{array}{c}\text{(47)}& {\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2;z\le {\alpha }_r\eta z+1\le M_{1}z+1\text{as}r\to \infty ,{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^4;z\le {\beta }_r\eta z^2+z+1\le M_2{z}^2+z+1\text{as}r\to \infty ,\end{array}$$where $M_1$ and $M_2$ are positive constant and $$\begin{array}{c}\text{(48)}& {\alpha }_r\eta =\max \frac{1}{r},\frac{1}{3r^2}3\frac{L^{\prime \prime }1}{L1}+61+2\eta \frac{L^{\prime }1}{L1}+12{\eta }^2+6\eta +1,{\beta }_r\eta =\max -\frac{7}{r^2},\frac{1}{r^3}-8\frac{L^{\prime \prime \prime }1}{L1}-4\frac{L^{\prime \prime }1}{L1}+24\eta +20\frac{L^{\prime }1}{L1}+40{\eta }^2+28\eta +12,\frac{1}{r^4}\frac{L^{\prime v}1}{L1}+43+2\eta \frac{L^{\prime \prime \prime }1}{L1}+19+36\eta +24{\eta }^2\frac{L^{\prime \prime }1}{L1}+13+36\eta +48{\eta }^2+32{\eta }^3\frac{L^{\prime }1}{L1}+16{\eta }^4+16{\eta }^3+8{\eta }^2+2\eta +1.\end{array}$$

Thus, from inequality (45), we get $$\begin{array}{}\text{(49)}& {\mathcal{R}}_{r,\eta }^{\ast }f;z-fz\le 21+{\overline{\delta }}^{2}1+z^2\overline{\Omega }f;\overline{\delta }1+{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2;z+\frac{1}{\overline{\delta }}{{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2;z}^{1/2}1+\sqrt{{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^4;z}\le 21+{\overline{\delta }}^{2}1+z^2\overline{\Omega }f;\overline{\delta }1+M_{1}z+1+\frac{1}{\overline{\delta }}\sqrt{{\alpha }_r\eta z+1}1+\sqrt{M_2{z}^2+z+1}.\end{array}$$

If we choose $\overline{\delta }=\sqrt{{\alpha }_r\eta }$ and taking supremum $z\in 0,{\alpha }_r\eta$, then we easily get the result.

4. Direct Approximation Results of ${\mathcal{R}}_{r,\eta }^{\ast }$

The present section gives some direct approximation results in space of $K$-functional and in Lipschitz spaces. We take $C_{b}0,\infty$ be the set of all continuous and bounded functions defined on $0,\infty$.

Definition 13.

For every $\overline{\delta }>0$ and $f\in C0,\infty$ the $K$-functional is defined such that $$\begin{array}{}\text{(50)}& {\mathcal{K}}_{\psi }f;\overline{\delta }=\inf {f-\psi }_{C_{b}0,\infty }+\overline{\delta }{{\psi }^{\prime \prime }}_{C_{b}0,\infty }\text{:}\psi ,{\psi }^{\prime }\in C_b^{2}0,\infty ,\text{(51)}& C_b^{k}0,\infty =f:f\in C_{b}0,\infty ,k\in \mathbb{N};\text{such}\text{that}\underset{z\to \infty }{\lim }\frac{fz}{1+z^2}=m_f<\infty .\end{array}$$

For an absolute constant $M>0$, one has $$\begin{array}{}\text{(52)}& {\mathcal{K}}_{\psi }f;\overline{\delta }\le M{\overline{\omega }}_{2}f;\sqrt{\overline{\delta }}+\min 1,\overline{\delta }{f}_{C_{b}0,\infty }.\end{array}$$

Let ${\overline{\omega }}_{2}f;\overline{\delta }$ denote the modulus of continuity of order two such that $$\begin{array}{}\text{(53)}& {\overline{\omega }}_{2}f;\overline{\delta }=\underset{0<h<\overline{\delta }}{\sup }\underset{z\in 0,\infty }{\sup }fz+2h-2fz+h+fz,\end{array}$$while the classical modulus of continuity is given by $$\begin{array}{}\text{(54)}& \overline{\omega }f;\overline{\delta }=\underset{0<h<\overline{\delta }}{\sup }\underset{z\in 0,\infty }{\sup }fz+h-fz.\end{array}$$

Theorem 14.

For an arbitrary $\phi \in C_b^{2}0,\infty$, let an auxiliary operator ${\mathcal{S}}_{r,\eta }^{\ast }$ be such that $$\begin{array}{}\text{(55)}& {\mathcal{K}}_{r,\eta }^{\ast }\phi ;z={\mathcal{R}}_{r,\eta }^{\ast }\phi ;z+\phi z-\phi z+\frac{1}{2r}2\frac{L^{\prime }1}{\text{L}1}+4\eta +1.\end{array}$$

Then, for any $\varphi \in C_b^{2}0,\infty$ operators (55), verify the inequality $$\begin{array}{}\text{(56)}& {\mathcal{K}}_{r,\eta }^{\ast }\varphi ;z-\varphi y\le {\overline{\delta }}_{r,\eta }^{\ast }z++\frac{1}{4r^2}{2\frac{L^{\prime }1}{\text{L}1}+4\eta +1}^2\parallel {\psi }^{\prime \prime }\parallel ,\end{array}$$where ${\overline{\delta }}_{r,\eta }^{\ast }z$ is defined by Theorem 10.

Proof.

For any $\varphi \in C_b^{2}0,\infty$, it is easy to verify that ${\mathcal{K}}_{r,\eta }^{\ast }{\gamma }_0;z=1$ and $$\begin{array}{}\text{(57)}& {\mathcal{K}}_{r,\eta }^{\ast }{\gamma }_1;z={\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_1;z+z-z+\frac{1}{2r}2\frac{L^{\prime }1}{\text{L}1}+4\eta +1=z.\end{array}$$

We have $$\begin{array}{c}\text{(58)}& {\mathcal{R}}_{r,\eta }^{\ast }\phi ;z\le \phi ,\mid {\mathcal{K}}_{r,\eta }^{\ast }\phi ;z\mid \le \mid {\mathcal{R}}_{r,\eta }^{\ast }\phi ;z\mid +\mid \phi z\mid +\mid \phi z+\frac{1}{2r}2\frac{L^{\prime }1}{\text{L}1}+4\eta +1\mid \le 3\parallel \phi \parallel .\end{array}$$

For any $\varphi \in C_b^{2}0,\infty$, the Taylor series expression gives us $$\begin{array}{}\text{(59)}& \varphi t=\varphi z+t-z{\varphi }^{\prime }z+{\int }_z^{t}t-\chi {\varphi }^{\prime \prime }\chi \text{d}\chi .\end{array}$$

Therefore, after applying the operators ${\mathcal{K}}_{r,\eta }^{\ast }$, on both sides we get $$\begin{array}{}\text{(60)}& {\mathcal{K}}_{r,\eta }^{\ast }\varphi ;z-\varphi z={\varphi }^{\prime }z{\mathcal{K}}_{r,\eta }^{\ast }{\gamma }_1-z;z+{\mathcal{K}}_{r,\eta }^{\ast }{\int }_z^{{\gamma }_1}{\gamma }_1-\chi {\varphi }^{\prime \prime }\chi \text{d}\chi ;z={\mathcal{K}}_{r,\eta }^{\ast }{\int }_z^t{\gamma }_1-\chi {\varphi }^{\prime \prime }\chi \text{d}\chi ;z={\mathcal{R}}_{r,\eta }^{\ast }{\int }_z^{{\gamma }_1}{\gamma }_1-\chi {\varphi }^{\prime \prime }\chi \text{d}\chi ;z+{\int }_z^{z}z-\chi {\varphi }^{\prime \prime }\chi \text{d}\chi ;z-{\int }_z^{z+1/2r2L^{\prime }1/\text{L}1+4\eta +1}z+\frac{1}{2r}2\frac{L^{\prime }1}{\text{L}1}+4\eta +1-\chi {\varphi }^{\prime \prime }\chi \text{d}\chi ;{\mathcal{K}}_{r,\eta }^{\ast }\varphi ;z-\varphi z\le {\mathcal{R}}_{r,\eta }^{\ast }{\int }_z^{{\gamma }_1}{\gamma }_1-\chi {\varphi }^{\prime \prime }\chi \text{d}\chi ;z+{\int }_z^{z+1/2r2L^{\prime }1/\text{L}1+4\eta +1}z+\frac{1}{2r}2\frac{L^{\prime }1}{\text{L}1}+4\eta +1-\chi {\varphi }^{\prime \prime }\chi \text{d}\chi .\end{array}$$

We know the inequality $$\begin{array}{c}\text{(61)}& {\int }_z^{t}t-\chi {\varphi }^{\prime \prime }\chi \text{d}\chi \le {t-z}^2{\varphi }^{\prime \prime },{\int }_z^{z+1/2r2L^{\prime }1/\text{L}1+4\eta +1}z+\frac{1}{2r}2\frac{L^{\prime }1}{\text{L}1}+4\eta +1-\chi {\varphi }^{\prime \prime }\chi \text{d}\chi \le {\frac{1}{2r}2\frac{L^{\prime }1}{\text{L}1}+4\eta +1}^2{\varphi }^{\prime \prime }.\end{array}$$

Thus, we get $$\begin{array}{}\text{(62)}& {\mathcal{K}}_{r,\eta }^{\ast }\varphi ;z-\varphi y\le {\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2;z+\frac{1}{4r^2}{2\frac{L^{\prime }1}{\text{L}1}+4\eta +1}^2\parallel {\varphi }^{\prime \prime }\parallel .\end{array}$$

This gives the complete proof.

Theorem 15.

If $\varphi \in C_b^{2}0,\infty$, then for any $f\in C_{b}0,\infty$ operators ${\mathcal{R}}_{r,\eta }^{\ast }$ by (7) satisfying $$\begin{array}{}\text{(63)}& {\mathcal{R}}_{r,\eta }^{\ast }f;z-fz\le \mathcal{M}{\overline{\omega }}_{2}f;\frac{1}{2}\sqrt{{\overline{\delta }}_{r,\eta }^{\ast }z+\frac{1}{4r^2}{2\frac{L^{\prime }1}{\text{L}1}+4\eta +1}^2}+\min 1;\frac{1}{4}{\overline{\delta }}_{r,\eta }^{\ast }z+\frac{1}{4r^2}{2\frac{L^{\prime }1}{\text{L}1}+4\eta +1}^2{f}_{C_{b}0,\infty }+\overline{\omega }f;\frac{1}{2r}2\frac{L^{\prime }1}{\text{L}1}+4\eta +1,\end{array}$$where ${\overline{\delta }}_{r,\eta }^{\ast }z$ is defined by Theorem 10.

Proof.

We prove Theorem 15 in view of Theorem 14. Therefore, for all $f\in C_{b}0,\infty$ and $\varphi \in C_b^{2}0,\infty$, we get $$\begin{array}{}\text{(64)}& {\mathcal{R}}_{r,\eta }^{\ast }f;z-fz={\mathcal{K}}_{r,\eta }^{\ast }f;z-fz+fz+\frac{1}{2r}2\frac{L^{\prime }1}{\text{L}1}+4\eta +1-fz\le \mid {\mathcal{K}}_{r,\eta }^{\ast }f-\varphi ;z\mid +\mid {\mathcal{K}}_{r,\eta }^{\ast }\varphi ;z-\varphi z\mid +\mid \varphi \text{z}-fz\mid +fz+\frac{1}{2r}2\frac{L^{\prime }1}{\text{L}1}+4\eta +1-fz\le 4\parallel f-\varphi \parallel +\omega f;\frac{1}{2r}2\frac{L^{\prime }1}{\text{L}1}+4\eta +1+{\overline{\delta }}_{r,\eta }^{\ast }z+\frac{1}{4r^2}{2\frac{L^{\prime }1}{\text{L}1}+4\eta +1}^2\parallel {\psi }^{\prime \prime }\parallel .\end{array}$$

If we take infimum for all $\varphi \in C_b^{2}0,\infty$, then in view of (50) it is easy to conclude that $$\begin{array}{}\text{(65)}& {\mathcal{R}}_{r,\eta }^{\ast }f;z-fz\le 4K_{2}f;\frac{1}{4}{\overline{\delta }}_{r,\eta }^{\ast }z+\frac{1}{4r^2}{2\frac{L^{\prime }1}{\text{L}1}+4\eta +1}^2+\overline{\omega }f;\frac{1}{2r}2\frac{L^{\prime }1}{\text{L}1}+4\eta +1\le \mathcal{M}{\overline{\omega }}_{2}f;\frac{1}{2}\sqrt{{\overline{\delta }}_{r,\eta }^{\ast }z+\frac{1}{4r^2}{2\frac{L^{\prime }1}{\text{L}1}+4\eta +1}^2}+\min 1;\frac{1}{4}{\overline{\delta }}_{r,\eta }^{\ast }z+\frac{1}{4r^2}{2\frac{L^{\prime }1}{\text{L}1}+4\eta +1}^2{f}_{C_{b}0,\infty }+\overline{\omega }f;\frac{1}{2r}2\frac{L^{\prime }1}{\text{L}1}+4\eta +1.\end{array}$$

The proof is completed here.

Now, we give the local direct estimate for the operators ${\mathcal{R}}_{r,\eta }^{\ast }$ defined by (7) via the well-known Lipschitz-type maximal function involving the parameters $\mu ,\nu >0$ and number $\lambda \in 0,1$. Thus, from [26], we recall that $$\begin{array}{}\text{(66)}& \text{Li}{\text{p}}_{\mathcal{L}}^{\lambda }=f\in C_{b}0,\infty \text{:}\mid ft-fz\mid \le \mathcal{L}\frac{{t-z}^{\lambda }}{{\mu z^2+\nu z+t}^{\lambda /2}};z,t\in 0,\infty ,\end{array}$$where $\mathcal{L}$ is a positive constant.

Theorem 16.

For any $f\in \text{Li}{\text{p}}_{\mathcal{L}}^{\lambda }$ satisfied by (70), operators ${\mathcal{R}}_{r,\eta }^{\ast }$ hold the inequality $$\begin{array}{}\text{(67)}& \mid {\mathcal{R}}_{r,\eta }^{\ast }f;z-fz\mid \le \mathcal{L}{\frac{{\overline{\delta }}_{r,\eta }^{\ast }z}{\mu {\text{z}}^2+\nu z}}^{\lambda /2},\end{array}$$where ${\overline{\delta }}_{r,\eta }^{\ast }z$ is obtained by Theorem 10.

Proof.

Let $f\in \text{Li}{\text{p}}_{\mathcal{L}}^{\lambda }$ for $0<\lambda \le 1$; then, first we verify the results are true when $\lambda =1$. For any $\mu ,\nu \ge 0$, it is easy to use the result ${\mu z^2+\nu z+t}^{-1/2}\le {\mu z^2+\nu z}^{-1/2}$ and then we apply the Cauchy-Schwarz inequality. Thus, we can write $$\begin{array}{}\text{(68)}& {\mathcal{R}}_{r,\eta }^{\ast }f;z-fz\le {\mathcal{R}}_{r,\eta }^{\ast }ft-fz;z+fz1;z-1\le {\mathcal{R}}_{r,\eta }^{\ast }\frac{\mid t-z\mid }{{\mu z^2+\nu z+t}^{1/2}};z\le \mathcal{L}{\mu z^2+\nu z}^{-1/2}{\mathcal{R}}_{r,\eta }^{\ast }t-z;z\le \mathcal{L}{\mu z^2+\nu z}^{-1/2}{\mathcal{R}}_{r,\eta }^{\ast }{{{\gamma }_1-z}^2;z}^{1/2}.\end{array}$$

From these conclusions, we get that the statement holds for $\lambda =1$. Now, we check if the statement is valid if $0<\lambda <1$. For this reason, we use monotonicity property to ${\mathcal{R}}_{r,\eta }^{\ast }$ and apply the well-known Hölder inequality $$\begin{array}{}\text{(69)}& {\mathcal{R}}_{r,\eta }^{\ast }f;z-fz\le {\mathcal{R}}_{r,\eta }^{\ast }\mid ft-fz\mid ;z\le {{\mathcal{R}}_{r,\eta }^{\ast }{ft-fz}^{2/\lambda };z}^{\lambda /2}{{\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_0;z}^{2-\lambda /2}\le \mathcal{L}{\frac{{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2;z}{t+\mu z^2+\nu z}}^{\lambda /2}\le \mathcal{L}{\mu z^2+\nu z}^{-\lambda /2}{{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2;z}^{\lambda /2}\le \mathcal{L}{\mu z^2+\nu z}^{-\lambda /2}{{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2;z}^{\lambda /2}=\mathcal{L}{\frac{{\overline{\delta }}_{r,\eta }^{\ast }z}{\mu z^2+\nu z}}^{\lambda /2},\end{array}$$which completes the proof.

Here, we obtain the other local approximation results of ${\mathcal{R}}_{r,\eta }^{\ast }$ in Lipschitz spaces. For all Lipschitz maximal function $f\in C_{b}0,\infty ,0<\lambda \le 1$ and $t,z\in 0,\infty$, from [27] we recall that $$\begin{array}{}\text{(70)}& {\overline{\omega }}_{\lambda }f;z=\underset{t\ne z,t\in 0,\infty }{\sup }\frac{\mid ft-fz\mid }{{t-z}^{\lambda }}\end{array}$$

Theorem 17.

Let $f\in C_{b}0,\infty$, then for all $z\in 0,\infty$, $$\begin{array}{}\text{(71)}& {\mathcal{R}}_{r,\eta }^{\ast }f;z-fz\le {{\overline{\delta }}_{r,\eta }^{\ast }z}^{\lambda /2}{\overline{\omega }}_{\lambda }f;z,\end{array}$$where ${\overline{\omega }}_{\lambda }f;z$ is obtained in and ${\overline{\delta }}_{r,\eta }^{\ast }z$ is defined by Theorem 10.

Proof.

From the well-known Hölder inequality, we get $$\begin{array}{}\text{(72)}& {\mathcal{R}}_{r,\eta }^{\ast }f;z-\text{f}z\le {\mathcal{R}}_{r,\eta }^{\ast }ft-fz;z\le {\overline{\omega }}_{\lambda }f;z\mid {\mathcal{R}}_{r,\eta }^{\ast }{t-z}^{\lambda };z\le {\overline{\omega }}_{\lambda }f;z{{\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_0;z}^{2-\lambda /2}{{\mathcal{R}}_{r,\eta }^{\ast }{t-z}^2;z}^{\lambda /2}={\overline{\omega }}_{\lambda }f;z{{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2;z}^{\lambda /2}.\end{array}$$

Thus, we get the proof.

5. Voronovskaja-Type Approximation Theorems

In this section, we establish a quantitative Voronovskaja-type theorem for the operators ${\mathcal{R}}_{r,\eta }^{\ast }f;z$.

Theorem 18.

Let $f\in C_{b}0,\infty$, then for each $z\in 0,\infty$$$\begin{array}{}\text{(73)}& \underset{r\to \infty }{\lim }r{\mathcal{R}}_{r,\eta }^{\ast }\psi ;z-\psi z=2\frac{L^{\prime }1}{\text{L}1}+4\eta +1\frac{{\psi }^{\prime }z}{2}+\frac{z{\psi }^{\prime \prime }z}{2},\end{array}$$where ${\psi }^{\prime }z,{\psi }^{\prime \prime }z\in C_{b}0,\infty$.

Proof.

From the expression of Taylor’s expansion of function $\psi z$ in $C_{b}0,\infty$, we write $$\begin{array}{}\text{(74)}& \psi t=\psi z+t-z{\psi }^{\prime }z+\frac{1}{2}{t-z}^2{\psi }^{\prime \prime }z+{t-z}^2{Q}_{z}t,\end{array}$$where $Q_{z}t$ is the remainder term and $Q_z\in 0,\infty$ with $Q_{z}t\to 0$ as $t\to z$. On applying the operators ${\mathcal{R}}_{r,\eta }^{\ast }·;z$ to (74), then use the Cauchy-Schwarz inequality. Thus, we get $$\begin{array}{}\text{(75)}& {\mathcal{R}}_{r,\eta }^{\ast }\psi ;z-\psi z={\psi }^{\prime }z{\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_1-z;z+\frac{{\psi }^{\prime \prime }z}{2}{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2;z+{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2{Q}_z{\gamma }_1;z\le {\psi }^{\prime }\text{z}{\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_1-z;z+\frac{{\psi }^{\prime \prime }z}{2}{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2;z+\sqrt{{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^4;z}\sqrt{{\mathcal{R}}_{r,\eta }^{\ast }{Q}_z^2{\gamma }_1;z}.\end{array}$$

Since we have ${\lim }_{r\to \infty }{\mathcal{R}}_{r,\eta }^{\ast }{Q}_z^2{\gamma }_1;z=0$, therefore $$\begin{array}{}\text{(76)}& \underset{r\to \infty }{\lim }r{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2{Q}_z{\gamma }_1;z=0.\end{array}$$

Thus, we have $$\begin{array}{}\text{(77)}& \underset{r\to \infty }{\lim }r{\mathcal{R}}_{r,\eta }^{\ast }\psi ;z-\psi z=\underset{r\to \infty }{\lim }r{\mathcal{R}}_{r,\eta }^{\ast }{\gamma }_1-z;z{\psi }^{\prime }z+\frac{{\psi }^{\prime \prime }z}{2}{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2;z+{\mathcal{R}}_{r,\eta }^{\ast }{{\gamma }_1-z}^2{Q}_z{\gamma }_1;z,\end{array}$$which completes the proof.

As a consequence of Theorem 18, we immediately get the corollary.

Corollary 19.

For any $\psi \in C0,\infty$, we have $$\begin{array}{}\text{(78)}& \underset{r\to \infty }{\lim }r{\mathcal{R}}_{r,\eta }^{\ast }\psi ;z-\psi z-\frac{1}{2r}2\frac{L^{\prime }1}{\text{L}1}+4\eta +1{\psi }^{\prime }z-\frac{{\overline{\delta }}_{r,\eta }^{\ast }z}{2}{\psi }^{\prime \prime }z=0.\end{array}$$

6. Conclusion

Motivated by article [5], we have introduced a Kantorovich generalization of the Szász-Mirakjan operators by Dunkl analogue involving the Appell polynomials. These types of generalizations enable to give the generalized results rather than the earlier study demonstrations by [3, 5, 7]. Lastly, we have also discussed the Voronovskaja-type approximation theorems of these new operators.

Authors’ Contributions

All authors read and agreed to the contents of this research article.

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