A New Modification of Baskakov–Schurer–Stancu Operators: Weighted and Pointwise Approximation Theories

Abstract

The behavior of a new modification of operators of the Baskakov–Schurer–Stancu variant is discussed in this study. First, we establish certain necessary moment and central moment estimates. We then demonstrate the weighted approximation result of the suggested operators using a Korovkin-type theorem in weighted spaces. We also give the rate at which these operators converge. Next, we establish theorems of pointwise convergence. Finally, we show several graphical representations to illustrate the accuracy and functionality of the operators.

1. Introduction

A key component of mathematical analysis, approximation theory provides methodical ways to represent functions using more manageable and straightforward forms. The family of Bernstein polynomials, which S. N. Bernstein established in 1912 as part of his constructive demonstration of the Stone–Weierstrass theorem [1], is a fundamental concept in this discipline. Bernstein polynomials provide a reliable and easily understandable foundation for uniform approximation because of their positivity, continuity, and capacity to replicate endpoint values on the interval $[0,1]$. They are a common tool in the analysis of approximation processes because of their straightforward algebraic form and computational stability. Beyond their inherent theoretical value, Bernstein-type constructions represent more general topics in approximation theory, which has many uses in computational mathematics, signal processing, numerical analysis, and data fitting. Modern algorithms that need both accuracy and processing economy are based on techniques based on polynomials, splines, wavelets, and kernel approximations.

In [2], Schurer introduced a novel adaptation of Bernstein polynomials for $\rho \in \mathbb{N}\cup \left\{0\right\}$, $\mathsf{{\rm Y}}\in C[0,1+\rho ]$ and $\zeta \in [0,1]$ as follows:

$$S_{\xi ,\rho }(\mathsf{{\rm Y}};\zeta )=\sum _{\sigma =0}^{\xi +\rho }\mathsf{{\rm Y}}\left({\displaystyle \frac{\sigma }{\xi }}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho }{\sigma }}\right){\zeta }^{\sigma }{(1-\zeta )}^{\xi +\rho -\sigma }.$$

Stancu [3] made a proposal in 1969 for $\xi \in \mathbb{N}$ the operators $S_{\xi }^{\vartheta ,\psi }:C[0,1]\to C[0,1]$, real parameters $\vartheta$, $\psi$ which are satisfied $0\le \vartheta \le \psi$ as below:

$$S_{\xi }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )=\sum _{\sigma =0}^{\xi }{p}_{\xi ,\sigma }\left(\zeta \right)\mathsf{{\rm Y}}\left({\displaystyle \frac{\sigma +\vartheta }{\xi +\psi }}\right),$$

where the Bernstein basis functions defined by $p_{\xi ,\sigma }\left(\zeta \right)=\left({\displaystyle \genfrac{}{}{0pt}{}{\xi }{\sigma }}\right){\zeta }^{\sigma }{(1-\zeta )}^{\xi -\sigma }$.

It should be noted that Stancu-type operators reduce to the well-known Bernstein operators explored by S. N. Bernstein [1] in 1912 when $\vartheta =0=\psi$.

Inspired by the operator sequences mentioned above, many researchers have studied the results of various approximations in detail by creating modifications or generalizations of many different operator sequences. To gain a deeper understanding of this topic, numerous academic studies provide significant analyses and advancements; see [4,5,6,7,8,9,10,11,12,13,14]. These works examine the shape parameters extensively and offer detailed perspectives, practical applications in geometric modeling, and several new conceptual developments. Very recently, a novel class of Bernstein operators was derived by Usta [15] for $\mathsf{{\rm Y}}$ on $C\left(\right(0,1\left)\right)$, $\xi \in \mathbb{N}$ and $\zeta \in (0,1)$ as below:

$$S_{\xi }^{*}\left(\mathsf{{\rm Y}};\zeta \right)={\displaystyle \frac{1}{\xi }}\sum _{\sigma =0}^{\xi }\left({\displaystyle \genfrac{}{}{0pt}{}{\xi }{\sigma }}\right){(\sigma -\xi \zeta )}^2{\zeta }^{\sigma -1}{(1-\zeta )}^{\xi -\sigma -1}\mathsf{{\rm Y}}\left({\displaystyle \frac{\sigma }{\xi }}\right).$$

For $\mathsf{{\rm Y}}\in C[0,\infty )$, $\zeta \in (0,\infty )$ and $\xi \in {\mathbb{N}}_{+}$, Cheng et al. [16] proposed and investigated the following new modification of the Baskakov-type [17] operators:

$$V_{\xi }(\mathsf{{\rm Y}};\zeta )={\displaystyle \frac{1}{\xi }}\sum _{\sigma =0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma -1}}{{(1+\zeta )}^{\xi +\sigma +1}}}{(\xi \zeta -\sigma )}^2\mathsf{{\rm Y}}\left({\displaystyle \frac{\sigma }{\xi }}\right).$$

Motivated by [15,16], we aim to study the following new modification of Baskakov–Schurer–Stancu operators.

Let $\rho \ge 0$ be a fixed positive integer and $\vartheta ,\psi \in \mathbb{R}$ with $0\le \vartheta \le \psi$. For every function $\mathsf{{\rm Y}}\in C(0,\infty )$, the corresponding Baskakov–Schurer–Stancu operators are constructed as follows:

$${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )={\displaystyle \frac{1}{\xi +\rho }}\sum _{\sigma =0}^{\infty }{\mathit{v}}_{\xi ,\rho ,\sigma }\left(\zeta \right){\left((\xi +\rho )\zeta -\sigma \right)}^2\mathsf{{\rm Y}}\left({\displaystyle \frac{\sigma +\vartheta }{\xi +\psi }}\right),$$

where

$$\begin{array}{ccc}\hfill {\mathit{v}}_{\xi ,\rho ,\sigma }\left(\zeta \right)& =& \left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma -1}}{{(1+\zeta )}^{\xi +\rho +\sigma +1}}}.\hfill \end{array}$$

(1)

Note that, when $\vartheta =\rho =\psi =0$, operators ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }$ reduce to the operators $V_{\xi }$, which are constructed by Cheng et al. [16].

The layout of the paper is structured as follows: In Section 2, we present several essential moment calculations. Section 3 is devoted to the study of weighted approximation properties. In Section 4, we establish the rate of convergence of the proposed operators ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }$ by means of the weighted modulus of continuity. Subsequently, Section 5 provides a collection of pointwise approximation results together with certain direct local theorems. In Section 6, we derive a Voronovskaja-type theorem. Lastly, Section 7 includes various graphical illustrations based on different parameter choices in order to show the accuracy and performance of the operators ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }$.

2. Preliminaries

We begin our analysis in this section by examining basic estimates pertaining to moments and central moments, which are essential for determining the primary findings.

Lemma 1.

Let ${\sum }_{\sigma =0}^{\infty }{\sigma }^m{\mathit{v}}_{\xi ,\rho ,\sigma }\left(\zeta \right)$ be a series of ${\mathit{v}}_{\xi ,\rho ,\sigma }$ base functions defined in (1). Here, ${\sigma }^m$ represents the power of the summation index and is used to compute the test functions in Lemma 2. The cases $m=0,1,2,3,4$ are considered separately.

$$\begin{array}{cccc}\hfill \sum _{\sigma =0}^{\infty }{\sigma }^0{\mathit{v}}_{\xi ,\rho ,\sigma }\left(\zeta \right)& ={\displaystyle \frac{1}{\zeta (1+\zeta )}},\hfill & & \\ \hfill \sum _{\sigma =0}^{\infty }{\sigma }^1{\mathit{v}}_{\xi ,\rho ,\sigma }\left(\zeta \right)& ={\displaystyle \frac{\xi +\rho }{1+\zeta }},\hfill & & \\ \hfill \sum _{\sigma =0}^{\infty }{\sigma }^2{\mathit{v}}_{\xi ,\rho ,\sigma }\left(\zeta \right)& ={\displaystyle \frac{(\xi +\rho +1)(\xi +\rho )\zeta }{1+\zeta }}+{\displaystyle \frac{\xi +\rho }{1+\zeta }},\hfill & & \\ \hfill \sum _{\sigma =0}^{\infty }{\sigma }^3{\mathit{v}}_{\xi ,\rho ,\sigma }\left(\zeta \right)& ={\displaystyle \frac{(\xi +\rho +2)(\xi +\rho +1)(\xi +\rho ){\zeta }^2}{1+\zeta }}+{\displaystyle \frac{3(\xi +\rho +1)(\xi +\rho )\zeta }{1+\zeta }}+{\displaystyle \frac{\xi +\rho }{1+\zeta }},\hfill & & \\ \hfill \sum _{\sigma =0}^{\infty }{\sigma }^4{\mathit{v}}_{\xi ,\rho ,\sigma }\left(\zeta \right)& ={\displaystyle \frac{(\xi +\rho +3)(\xi +\rho +2)(\xi +\rho +1)(\xi +\rho ){\zeta }^3}{1+\zeta }}+{\displaystyle \frac{6(\xi +\rho +2)(\xi +\rho +1)(\xi +\rho ){\zeta }^2}{1+\zeta }}\hfill & & \\ & +{\displaystyle \frac{7(\xi +\rho +1)(\xi +\rho )\zeta }{1+\zeta }}+{\displaystyle \frac{\xi +\rho }{1+\zeta }}.\hfill & & \end{array}$$

Proof. 

Using the negative binomial series ${\sum }_{\sigma =0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{n+\sigma -1}{\sigma }}\right)z^{\sigma }={(1-z)}^{-n},\left|z\right|\le 1$ with $n=\xi +\rho$ and $z={\displaystyle \frac{\zeta }{1+\zeta }},$ and multiplying by the factor ${(1-\zeta )}^{-(\xi +\rho )}$, we obtain that the above series equals 1.

$$\begin{array}{ccc}\hfill \sum _{\sigma =0}^{\infty }{\sigma }^0{\mathit{v}}_{\xi ,\rho ,\sigma }\left(\zeta \right)& =& \sum _{\sigma =0}^{\infty }{\sigma }^0\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma -1}}{{(1+\zeta )}^{\xi +\rho +\sigma +1}}}\hfill \\ & =& {\displaystyle \frac{1}{\zeta (1+\zeta )}}\sum _{\sigma =0}^{\infty }{\sigma }^0\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma }}{{(1+\zeta )}^{\xi +\rho +\sigma }}}\hfill \\ & =& {\displaystyle \frac{1}{\zeta (1+\zeta )}}.\hfill \end{array}$$

$$\begin{array}{ccc}\hfill \sum _{\sigma =0}^{\infty }{\sigma }^1{\mathit{v}}_{\xi ,\rho ,\sigma }\left(\zeta \right)& =& \sum _{\sigma =1}^{\infty }{\sigma }^1\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma -1}}{{(1+\zeta )}^{\xi +\rho +\sigma +1}}}\hfill \\ & =& {\displaystyle \frac{1}{\zeta (1+\zeta )}}\sum _{\sigma =1}^{\infty }\sigma \left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma }}{{(1+\zeta )}^{\xi +\rho +\sigma }}}.\hfill \end{array}$$

When we substitute $\sigma +1$ for $\sigma$ in the equation above, we obtain

$$\begin{array}{ccc}\hfill \sum _{\sigma =0}^{\infty }{\sigma }^1{\mathit{v}}_{\xi ,\rho ,\sigma }\left(\zeta \right)& =& {\displaystyle \frac{1}{\zeta (1+\zeta )}}\sum _{\sigma =0}^{\infty }(\sigma +1)\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma }{\sigma +1}}\right){\displaystyle \frac{{\zeta }^{\sigma +1}}{{(1+\zeta )}^{\xi +\rho +\sigma +1}}}.\hfill \end{array}$$

Using $\xi +\rho$ to multiply and divide the equation above, we get

$$\begin{array}{ccc}\hfill \sum _{\sigma =0}^{\infty }{\sigma }^1{\mathit{v}}_{\xi ,\rho ,\sigma }\left(\zeta \right)& =& {\displaystyle \frac{\xi +\rho }{1+\zeta }}\sum _{\sigma =0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma }{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma }}{{(1+\zeta )}^{\xi +\rho +\sigma +1}}}\hfill \\ & =& {\displaystyle \frac{\xi +\rho }{1+\zeta }}.\hfill \end{array}$$

$$\begin{array}{ccc}\hfill \sum _{\sigma =0}^{\infty }{\sigma }^2{\mathit{v}}_{\xi ,\rho ,\sigma }\left(\zeta \right)& =& \sum _{\sigma =0}^{\infty }{\sigma }^2\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma -1}}{{(1+\zeta )}^{\xi +\rho +\sigma +1}}}\hfill \\ & =& {\displaystyle \frac{1}{\zeta (1+\zeta )}}\sum _{\sigma =0}^{\infty }{\sigma }^2\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma }}{{(1+\zeta )}^{\xi +\rho +\sigma }}}.\hfill \end{array}$$

By substituting $\sigma (\sigma -1)+\sigma$ for ${\sigma }^2$, one has

$$\begin{array}{ccc}\hfill \sum _{\sigma =0}^{\infty }{\sigma }^2{\mathit{v}}_{\xi ,\rho ,\sigma }\left(\zeta \right)& =& {\displaystyle \frac{1}{\zeta (1+\zeta )}}\sum _{\sigma =2}^{\infty }\sigma (\sigma -1)\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma }}{{(1+\zeta )}^{\xi +\rho +\sigma }}}\hfill \\ & & +{\displaystyle \frac{1}{\zeta (1+\zeta )}}\sum _{\sigma =1}^{\infty }\sigma \left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma }}{{(1+\zeta )}^{\xi +\rho +\sigma }}}.\hfill \end{array}$$

When we substitute $\sigma +2$ for $\sigma$ in the first summation, we obtain

$$\begin{array}{ccc}\hfill \sum _{\sigma =0}^{\infty }{\sigma }^2{\mathit{v}}_{\xi ,\rho ,\sigma }\left(\zeta \right)& =& {\displaystyle \frac{1}{\zeta (1+\zeta )}}\sum _{\sigma =0}^{\infty }(\sigma +2)(\sigma +1)\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma +1}{\sigma +2}}\right){\displaystyle \frac{{\zeta }^{\sigma +2}}{{(1+\zeta )}^{\xi +\rho +\sigma +2}}}+\sum _{\sigma =0}^{\infty }{\sigma }^1{\mathit{v}}_{\xi ,\rho ,\sigma }\left(\zeta \right).\hfill \end{array}$$

Using $(\xi +\rho +1)(\xi +\rho )$ to multiply and divide the first summation, we get

$$\begin{array}{ccc}\hfill \sum _{\sigma =0}^{\infty }{\sigma }^2{\mathit{v}}_{\xi ,\rho ,\sigma }\left(\zeta \right)& =& {\displaystyle \frac{(\xi +\rho +1)(\xi +\rho )\zeta }{1+\zeta }}\sum _{\sigma =0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma +1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma }}{{(1+\zeta )}^{\xi +\rho +\sigma +2}}}+{\displaystyle \frac{\xi +\rho }{1+\zeta }}\hfill \\ & =& {\displaystyle \frac{(\xi +\rho +1)(\xi +\rho )\zeta }{1+\zeta }}+{\displaystyle \frac{\xi +\rho }{1+\zeta }}.\hfill \end{array}$$

The proof consists of five assertions. Assertions ${\sum }_{\sigma =0}^{\infty }{\sigma }^0{\mathit{v}}_{\xi ,\rho ,\sigma }\left(\zeta \right)$, ${\sum }_{\sigma =0}^{\infty }{\sigma }^1{\mathit{v}}_{\xi ,\rho ,\sigma }\left(\zeta \right)$ and ${\sum }_{\sigma =0}^{\infty }{\sigma }^2{\mathit{v}}_{\xi ,\rho ,\sigma }\left(\zeta \right)$ are proved explicitly above. For the proof of assertion ${\sum }_{\sigma =0}^{\infty }{\sigma }^3{\mathit{v}}_{\xi ,\rho ,\sigma }\left(\zeta \right)$, we replace ${\sigma }^3$ with $\sigma (\sigma -1)(\sigma -2)+3\sigma (\sigma -1)+\sigma$. Similarly, for the proof of assertion ${\sum }_{\sigma =0}^{\infty }{\sigma }^4{\mathit{v}}_{\xi ,\rho ,\sigma }\left(\zeta \right)$, we replace ${\sigma }^4$ with $\sigma (\sigma -1)(\sigma -2)(\sigma -3)+6\sigma (\sigma -1)(\sigma -2)+7\sigma (\sigma -1)+\sigma$. Once these substitutions are made, the proof can be completed in an analogous way. □

Lemma 2.

Let $\rho \ge 0$ be a fixed positive integer, and $\vartheta ,\psi \in \mathbb{R}$ with $0\le \vartheta \le \psi$. For every $\zeta \in (0,\infty )$, the operators ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }$ satisfy

$$\begin{array}{cccc}\hfill {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(1;\zeta )& =1,\hfill & & \\ \hfill {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\varrho ;\zeta )& ={\displaystyle \frac{(\xi +\rho +2)\zeta }{\xi +\psi }}+{\displaystyle \frac{\vartheta +1}{\xi +\psi }},\hfill & & \\ \hfill {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }({\varrho }^2;\zeta )& ={\displaystyle \frac{\left({(\xi +\rho )}^2+7(\xi +\rho )+6\right){\zeta }^2}{{(\xi +\psi )}^2}}+{\displaystyle \frac{\left((2\vartheta +5)(\xi +\rho )+4\vartheta +6\right)\zeta }{{(\xi +\psi )}^2}}+{\displaystyle \frac{{(\vartheta +1)}^2}{{(\xi +\psi )}^2}},\hfill & & \\ \hfill {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }({\varrho }^3;\zeta )& ={\displaystyle \frac{\left({(\xi +\rho )}^3+15{(\xi +\rho )}^2+38(\xi +\rho )+24\right){\zeta }^3}{{(\xi +\psi )}^3}}\hfill & & \\ & +{\displaystyle \frac{\left((3\vartheta +12){(\xi +\rho )}^2+(21\vartheta +48)(\xi +\rho )+18\vartheta +36\right){\zeta }^2}{{(\xi +\psi )}^3}}\hfill & & \\ & +{\displaystyle \frac{\left((3{\vartheta }^2+15\vartheta +13)(\xi +\rho )+6{\vartheta }^2+18\vartheta +14\right)\zeta }{{(\xi +\psi )}^3}}\hfill & & \\ & +{\displaystyle \frac{{(\vartheta +1)}^3}{{(\xi +\psi )}^3}},\hfill \end{array}$$

$$\begin{array}{cccc}\hfill {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }({\varrho }^4;\zeta )& ={\displaystyle \frac{\left({(\xi +\rho )}^4+26{(\xi +\rho )}^3+131{(\xi +\rho )}^2+226(\xi +\rho )+120\right){\zeta }^4}{{(\xi +\psi )}^4}}\hfill & & \\ & +{\displaystyle \frac{\left((4\vartheta +22){(\xi +\rho )}^3+(60\vartheta +186){(\xi +\rho )}^2+(152\vartheta +404)(\xi +\rho )+96\vartheta +240\right){\zeta }^3}{{(\xi +\psi )}^4}}\hfill & & \\ & +{\displaystyle \frac{\left((6{\vartheta }^2+48\vartheta +61){(\xi +\rho )}^2+(42{\vartheta }^2+48\vartheta +211)(\xi +\rho )+144\vartheta +150\right){\zeta }^2}{{(\xi +\psi )}^4}}\hfill & & \\ & +{\displaystyle \frac{\left((4{\vartheta }^3+30{\vartheta }^2+52\vartheta +29)(\xi +\rho )+8{\vartheta }^3+26{\vartheta }^2+56\vartheta +30\right)\zeta }{{(\xi +\psi )}^4}}\hfill & & \\ & +{\displaystyle \frac{{(\vartheta +1)}^4}{{(\xi +\psi )}^4}}.\hfill & & \end{array}$$

Proof. 

From Lemma 1, we arrive at

$$\begin{array}{cccc}\hfill {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(1;\zeta )& ={\displaystyle \frac{1}{\xi +\rho }}\sum _{\sigma =0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma -1}}{{(1+\zeta )}^{\xi +\rho +\sigma +1}}}{\left((\xi +\rho )\zeta -\sigma \right)}^2\hfill & & \\ & ={\displaystyle \frac{1}{\xi +\rho }}\sum _{\sigma =0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma -1}}{{(1+\zeta )}^{\xi +\rho +\sigma +1}}}\left({(\xi +\rho )}^2{\zeta }^2-2(\xi +\rho )\zeta \sigma +{\sigma }^2\right)\hfill & & \\ & ={\displaystyle \frac{{(\xi +\rho )}^2{\zeta }^2}{\xi +\rho }}\sum _{\sigma =0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma -1}}{{(1+\zeta )}^{\xi +\rho +\sigma +1}}}\hfill & & \\ & -{\displaystyle \frac{2(\xi +\rho )\zeta }{\xi +\rho }}\sum _{\sigma =0}^{\infty }\sigma \left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma -1}}{{(1+\zeta )}^{\xi +\rho +\sigma +1}}}\hfill & & \\ & +{\displaystyle \frac{1}{\xi +\rho }}\sum _{\sigma =0}^{\infty }{\sigma }^2\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma -1}}{{(1+\zeta )}^{\xi +\rho +\sigma +1}}}\hfill & & \\ & ={\displaystyle \frac{{(\xi +\rho )}^2{\zeta }^2}{\xi +\rho }}\left({\displaystyle \frac{1}{\zeta (1+\zeta )}}\right)-{\displaystyle \frac{2(\xi +\rho )\zeta }{\xi +\rho }}\left({\displaystyle \frac{\xi +\rho }{1+\zeta }}\right)+{\displaystyle \frac{1}{\xi +\rho }}\left({\displaystyle \frac{(\xi +\rho +1)(\xi +\rho )\zeta }{1+\zeta }}+{\displaystyle \frac{\xi +\rho }{1+\zeta }}\right)\hfill & & \\ & ={\displaystyle \frac{(\xi +\rho )\zeta }{(1+\zeta )}}-{\displaystyle \frac{2(\xi +\rho )\zeta }{(1+\zeta )}}+{\displaystyle \frac{(\xi +\rho +1)\zeta }{1+\zeta }}+{\displaystyle \frac{1}{1+\zeta }}\hfill & & \\ & ={\displaystyle \frac{(\xi +\rho -2\xi -2\rho +\xi +\rho +1)\zeta }{(1+\zeta )}}+{\displaystyle \frac{1}{1+\zeta }}\hfill & & \\ & ={\displaystyle \frac{1+\zeta }{1+\zeta }}\hfill & & \\ & =1.\hfill \end{array}$$

$$\begin{array}{cccc}\hfill {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\varrho ;\zeta )& ={\displaystyle \frac{1}{\xi +\rho }}\sum _{\sigma =0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma -1}}{{(1+\zeta )}^{\xi +\rho +\sigma +1}}}{\left((\xi +\rho )\zeta -\sigma \right)}^2\left({\displaystyle \frac{\sigma +\vartheta }{\xi +\psi }}\right)\hfill & & \\ & ={\displaystyle \frac{1}{(\xi +\rho )(\xi +\psi )}}\sum _{\sigma =0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma -1}}{{(1+\zeta )}^{\xi +\rho +\sigma +1}}}{\left((\xi +\rho )\zeta -\sigma \right)}^2\sigma +{\displaystyle \frac{\vartheta }{\xi +\psi }}\hfill & & \\ & ={\displaystyle \frac{1}{(\xi +\rho )(\xi +\psi )}}\sum _{\sigma =0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma -1}}{{(1+\zeta )}^{\xi +\rho +\sigma +1}}}\left({(\xi +\rho )}^2{\zeta }^2\sigma -2(\xi +\rho )\zeta {\sigma }^2+{\sigma }^3\right)\hfill & & \\ & +{\displaystyle \frac{\vartheta }{\xi +\psi }}\hfill & & \\ & ={\displaystyle \frac{{(\xi +\rho )}^2{\zeta }^2}{(\xi +\rho )(\xi +\psi )}}\sum _{\sigma =0}^{\infty }\sigma \left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma -1}}{{(1+\zeta )}^{\xi +\rho +\sigma +1}}}\hfill & & \\ & -{\displaystyle \frac{2(\xi +\rho )\zeta }{(\xi +\rho )(\xi +\psi )}}\sum _{\sigma =0}^{\infty }{\sigma }^2\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma -1}}{{(1+\zeta )}^{\xi +\rho +\sigma +1}}}\hfill & & \\ & +{\displaystyle \frac{1}{(\xi +\rho )(\xi +\psi )}}\sum _{\sigma =0}^{\infty }{\sigma }^3\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma -1}}{{(1+\zeta )}^{\xi +\rho +\sigma +1}}}+{\displaystyle \frac{\vartheta }{\xi +\psi }}\hfill & & \\ & ={\displaystyle \frac{{(\xi +\rho )}^2{\zeta }^2}{(\xi +\rho )(\xi +\psi )}}\left({\displaystyle \frac{\xi +\rho }{1+\zeta }}\right)\hfill & & \end{array}$$

$$\begin{array}{cccc}& -{\displaystyle \frac{2(\xi +\rho )\zeta }{(\xi +\rho )(\xi +\psi )}}\left({\displaystyle \frac{(\xi +\rho +1)(\xi +\rho )\zeta }{1+\zeta }}+{\displaystyle \frac{\xi +\rho }{1+\zeta }}\right)\hfill & & \\ & +{\displaystyle \frac{1}{(\xi +\rho )(\xi +\psi )}}\left({\displaystyle \frac{(\xi +\rho +2)(\xi +\rho +1)(\xi +\rho ){\zeta }^2}{1+\zeta }}+{\displaystyle \frac{3(\xi +\rho +1)(\xi +\rho )\zeta }{1+\zeta }}+{\displaystyle \frac{\xi +\rho }{1+\zeta }}\right)\hfill & & \\ & +{\displaystyle \frac{\vartheta }{\xi +\psi }}\hfill \\ & ={\displaystyle \frac{{(\xi +\rho )}^2{\zeta }^2}{(\xi +\psi )(1+\zeta )}}-{\displaystyle \frac{2(\xi +\rho +1)(\xi +\rho ){\zeta }^2}{(\xi +\psi )(1+\zeta )}}-{\displaystyle \frac{2(\xi +\rho )\zeta }{(\xi +\psi )(1+\zeta )}}+{\displaystyle \frac{(\xi +\rho +2)(\xi +\rho +1){\zeta }^2}{(\xi +\psi )(1+\zeta )}}\hfill & & \\ & +{\displaystyle \frac{3(\xi +\rho +1)\zeta }{(\xi +\psi )(1+\zeta )}}+{\displaystyle \frac{1}{(\xi +\psi )(1+\zeta )}}+{\displaystyle \frac{\vartheta }{\xi +\psi }}\hfill & & \\ & ={\displaystyle \frac{\xi {\zeta }^2+\xi \zeta +\rho {\zeta }^2+\rho \zeta +2{\zeta }^2+2\zeta +\zeta +1}{(\xi +\psi )(1+\zeta )}}+{\displaystyle \frac{\vartheta }{\xi +\psi }}\hfill & & \\ & ={\displaystyle \frac{\xi \zeta (1+\zeta )+\rho \zeta (1+\zeta )+2\zeta (1+\zeta )+1+\zeta }{(\xi +\psi )(1+\zeta )}}+{\displaystyle \frac{\vartheta }{\xi +\psi }}\hfill & & \\ & ={\displaystyle \frac{(\xi \zeta +\rho \zeta +2\zeta +1)(1+\zeta )}{(\xi +\psi )(1+\zeta )}}+{\displaystyle \frac{\vartheta }{\xi +\psi }}\hfill & & \\ & ={\displaystyle \frac{(\xi +\rho +2)\zeta }{\xi +\psi }}+{\displaystyle \frac{\vartheta +1}{\xi +\psi }}.\hfill \end{array}$$

$$\begin{array}{cccc}\hfill {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }({\varrho }^2;\zeta )& ={\displaystyle \frac{1}{\xi +\rho }}\sum _{\sigma =0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma -1}}{{(1+\zeta )}^{\xi +\rho +\sigma +1}}}{\left((\xi +\rho )\zeta -\sigma \right)}^2{\left({\displaystyle \frac{\sigma +\vartheta }{\xi +\psi }}\right)}^2\hfill & & \\ & ={\displaystyle \frac{1}{(\xi +\rho ){\left(\xi +\psi \right)}^2}}\sum _{\sigma =0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma -1}}{{(1+\zeta )}^{\xi +\rho +\sigma +1}}}{\left((\xi +\rho )\zeta -\sigma \right)}^2{(\sigma +\vartheta )}^2\hfill & & \\ & ={\displaystyle \frac{1}{(\xi +\rho ){\left(\xi +\psi \right)}^2}}\sum _{\sigma =0}^{\infty }{\sigma }^4\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma -1}}{{(1+\zeta )}^{\xi +\rho +\sigma +1}}}\hfill & & \\ & +{\displaystyle \frac{2(\vartheta -(\xi +\rho \left)\zeta \right)}{(\xi +\rho ){\left(\xi +\psi \right)}^2}}\sum _{\sigma =0}^{\infty }{\sigma }^3\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma -1}}{{(1+\zeta )}^{\xi +\rho +\sigma +1}}}\hfill & & \\ & +{\displaystyle \frac{\left({(\xi +\rho )}^2{\zeta }^2+{\vartheta }^2-4\vartheta (\xi +\rho )\zeta \right)}{(\xi +\rho ){\left(\xi +\psi \right)}^2}}\sum _{\sigma =0}^{\infty }{\sigma }^2\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma -1}}{{(1+\zeta )}^{\xi +\rho +\sigma +1}}}\hfill & & \\ & +{\displaystyle \frac{\left(2\vartheta {(\xi +\rho )}^2{\zeta }^2-2{\vartheta }^2(\xi +\rho )\zeta \right)}{(\xi +\rho ){\left(\xi +\psi \right)}^2}}\sum _{\sigma =0}^{\infty }\sigma \left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma -1}}{{(1+\zeta )}^{\xi +\rho +\sigma +1}}}\hfill & & \\ & +{\displaystyle \frac{{\vartheta }^2{(\xi +\rho )}^2{\zeta }^2}{(\xi +\rho ){\left(\xi +\psi \right)}^2}}\sum _{\sigma =0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{\xi +\rho +\sigma -1}{\sigma }}\right){\displaystyle \frac{{\zeta }^{\sigma -1}}{{(1+\zeta )}^{\xi +\rho +\sigma +1}}}\hfill & & \\ & ={\displaystyle \frac{1}{(\xi +\rho ){\left(\xi +\psi \right)}^2}}({\displaystyle \frac{(\xi +\rho +3)(\xi +\rho +2)(\xi +\rho +1)(\xi +\rho ){\zeta }^3}{1+\zeta }}\hfill & & \\ & +{\displaystyle \frac{6(\xi +\rho +2)(\xi +\rho +1)(\xi +\rho ){\zeta }^2}{1+\zeta }}+{\displaystyle \frac{7(\xi +\rho +1)(\xi +\rho )\zeta }{1+\zeta }}+{\displaystyle \frac{\xi +\rho }{1+\zeta }})\hfill & & \\ & +{\displaystyle \frac{2(\vartheta -(\xi +\rho \left)\zeta \right)}{(\xi +\rho ){\left(\xi +\psi \right)}^2}}\left({\displaystyle \frac{(\xi +\rho +2)(\xi +\rho +1)(\xi +\rho ){\zeta }^2}{1+\zeta }}+{\displaystyle \frac{3(\xi +\rho +1)(\xi +\rho )\zeta }{1+\zeta }}+{\displaystyle \frac{\xi +\rho }{1+\zeta }}\right)\hfill & & \\ & +{\displaystyle \frac{\left({(\xi +\rho )}^2{\zeta }^2+{\vartheta }^2-4\vartheta (\xi +\rho )\zeta \right)}{(\xi +\rho ){\left(\xi +\psi \right)}^2}}\left({\displaystyle \frac{(\xi +\rho +1)(\xi +\rho )\zeta }{1+\zeta }}+{\displaystyle \frac{\xi +\rho }{1+\zeta }}\right)\hfill & & \\ & +{\displaystyle \frac{\left(2\vartheta {(\xi +\rho )}^2{\zeta }^2-2{\vartheta }^2(\xi +\rho )\zeta \right)}{(\xi +\rho ){\left(\xi +\psi \right)}^2}}\left({\displaystyle \frac{\xi +\rho }{1+\zeta }}\right)\hfill & & \\ & +{\displaystyle \frac{{\vartheta }^2{(\xi +\rho )}^2{\zeta }^2}{(\xi +\rho ){\left(\xi +\psi \right)}^2}}\left({\displaystyle \frac{1}{\zeta (1+\zeta )}}\right)\hfill & & \\ & ={\displaystyle \frac{\left({(\xi +\rho )}^2+7(\xi +\rho )+6\right){\zeta }^2}{{(\xi +\psi )}^2}}+{\displaystyle \frac{\left((2\vartheta +5)(\xi +\rho )+4\vartheta +6\right)\zeta }{{(\xi +\psi )}^2}}+{\displaystyle \frac{{(\vartheta +1)}^2}{{(\xi +\psi )}^2}}.\hfill \end{array}$$

Hence, we have established the proof equations ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(1;\zeta )$, ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\varrho ;\zeta )$ and ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }({\varrho }^2;\zeta )$. In the same manner, the expressions for ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }({\varrho }^3;\zeta )$ and ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }({\varrho }^4;\zeta )$ can be derived directly. For this reason, we omit the detailed steps here. □

Lemma 3.

Let $\rho \ge 0$ be a fixed positive integer, and $\vartheta ,\psi \in \mathbb{R}$ with $0\le \vartheta \le \psi$. For every $\zeta \in (0,\infty )$, operators ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }$ satisfy

$$\begin{array}{cccc}\hfill {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\varrho -\zeta ;\zeta )& ={\displaystyle \frac{(\rho -\psi +2)\zeta }{\xi +\psi }}+{\displaystyle \frac{\vartheta +1}{\xi +\psi }},\hfill & & \\ \hfill {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }({(\varrho -\zeta )}^2;\zeta )& ={\displaystyle \frac{\left((\xi +\rho +6)(\xi +\rho +1)-2(\xi +\rho +2)(\xi +\psi )+{(\xi +\psi )}^2\right){\zeta }^2}{{(\xi +\psi )}^2}}\hfill & & \\ & +{\displaystyle \frac{\left((2\vartheta +5)(\xi +\rho )+4\vartheta +6-2(\vartheta +1)(\xi +\psi )\right)\zeta }{{(\xi +\psi )}^2}}+{\displaystyle \frac{{(\vartheta +1)}^2}{{(\xi +\psi )}^2}},\hfill & & \\ \hfill {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }({(\varrho -\zeta )}^4;\zeta )& =({\displaystyle \frac{(\xi +\rho +20)(\xi +\rho +3)(\xi +\rho +2)(\xi +\rho +1)}{{(\xi +\psi )}^4}}\hfill & & \\ & -{\displaystyle \frac{4(\xi +\rho +2)(\xi +\rho +1)(\xi +\rho +12)(\xi +\psi )}{{(\xi +\psi )}^4}}\hfill & & \\ & +{\displaystyle \frac{6(\xi +\rho +6)(\xi +\rho +1){(\xi +\psi )}^2}{{(\xi +\psi )}^4}}\hfill & & \\ & -{\displaystyle \frac{4(\xi +\rho +2){(\xi +\psi )}^3+{(\xi +\psi )}^4}{{(\xi +\psi )}^4}}){\zeta }^4\hfill & & \\ & +({\displaystyle \frac{\left((4\vartheta +22)(\xi +\rho )+48\vartheta +120\right)(\xi +\rho +1)(\xi +\rho +2)}{{(\xi +\psi )}^4}}\hfill & & \\ & -{\displaystyle \frac{12\left((\vartheta +4)(\xi +\rho )+6\vartheta +12\right)(\xi +\rho +1)(\xi +\psi )}{{(\xi +\psi )}^4}}\hfill & & \\ & +{\displaystyle \frac{6\left((2\vartheta +5)(\xi +\rho )+4\vartheta +6\right){(\xi +\psi )}^2-4(\vartheta +1){(\xi +\psi )}^3}{{(\xi +\psi )}^4}}){\zeta }^3\hfill & & \\ & +({\displaystyle \frac{\left((6{\vartheta }^2+48\vartheta +61)(\xi +\rho )+36{\vartheta }^2+144\vartheta +150\right)(\xi +\rho +1)}{{(\xi +\psi )}^4}}\hfill & & \\ & -{\displaystyle \frac{4\left((3{\vartheta }^2+15\vartheta +13)(\xi +\rho )+6{\vartheta }^2+18\vartheta +14\right)(\xi +\psi )}{{(\xi +\psi )}^4}}\hfill & & \\ & +{\displaystyle \frac{6{(\vartheta +1)}^2{(\xi +\psi )}^2}{{(\xi +\psi )}^4}}){\zeta }^2\hfill & & \\ & +({\displaystyle \frac{(4\vartheta +30{\vartheta }^2+52\vartheta +29)(\xi +\rho )+8{\vartheta }^3+3{\vartheta }^2+56\vartheta +30}{{(\xi +\psi )}^4}}\hfill & & \\ & -{\displaystyle \frac{4{(\vartheta +1)}^3(\xi +\psi )}{{(\xi +\psi )}^4}})\zeta \hfill & & \\ & +{\displaystyle \frac{{(\vartheta +1)}^4}{{(\xi +\psi )}^4}}.\hfill & & \end{array}$$

The proof follows from Lemma 2 and the linearity and positivity of ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }$.

3. Weighted Approximation of Operators ${\mathit{V}}_{\mathit{\xi },\mathit{\rho }}^{\mathit{\vartheta },\mathit{\psi }}$

All functions on $(0,\infty )$ that satisfy the inequality $\left|\mathsf{{\rm Y}}\left(\zeta \right)\right|\le S_{\mathsf{{\rm Y}}}\nu \left(\zeta \right)$ form the set $B_{\nu }(0,\infty )$. Here, $S_{\mathsf{{\rm Y}}}$ is a positive constant depending only on the function $\mathsf{{\rm Y}}$ and $\nu \left(\zeta \right)=1+{\zeta }^2$. Additionally, let us denote these spaces as follows:

$$C_{\nu }(0,\infty )=B_{\nu }(0,\infty )\cap C(0,\infty )$$

and

$$C_{\nu }^{*}(0,\infty )=\{\mathsf{{\rm Y}}\in C_{\nu }(0,\infty ):\underset{\xi \to \infty }{\lim }{\displaystyle \frac{\mathsf{{\rm Y}}\left(\zeta \right)}{\nu \left(\zeta \right)}}<\infty \}$$

and equipped with the norm

$$\parallel \mathsf{{\rm Y}}{\parallel }_{\nu }=\sup \left\{{\displaystyle \frac{\left|\mathsf{{\rm Y}}\right(\zeta \left)\right|}{\nu \left(\zeta \right)}}:\zeta \in (0,\infty )\right\}.$$

Theorem 1.

For every $\mathsf{{\rm Y}}\in C_{\nu }^{*}(0,\infty )$, we have

$$\underset{\xi \to \infty }{\lim }\parallel {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left(\mathsf{{\rm Y}}\right)-\mathsf{{\rm Y}}{\parallel }_{\nu }=0.$$

Proof. 

It is sufficient to verify the following three requirements from [18,19,20,21].

$$\begin{array}{c}\hfill \underset{\xi \to \infty }{\lim }{\parallel {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }({\varrho }^{\eta };\zeta )-{\zeta }^{\eta }\parallel }_{\nu }=0,\eta =0,1,2.\end{array}$$

(2)

Since ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(1;\zeta )=1$, condition (2) holds for $\eta =0$. The following inequality can be derived using Lemma 2.

$$\begin{array}{ccc}\hfill {\parallel {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\varrho ;\zeta )-\zeta \parallel }_{\nu }& \le & \underset{\zeta \ge 0}{\sup }{\displaystyle \frac{\left|\frac{(\xi +\rho +2)\zeta }{\xi +\psi }+\frac{\vartheta +1}{\xi +\psi }-\zeta \right|}{1+{\zeta }^2}}\hfill \\ & =& \underset{\zeta \ge 0}{\sup }{\displaystyle \frac{\left|\frac{(\rho -\psi +2)\zeta }{\xi +\psi }+\frac{\vartheta +1}{\xi +\psi }\right|}{1+{\zeta }^2}}\hfill \\ & \le & {\displaystyle \frac{|\rho -\psi +2|+\vartheta +1}{\xi +\psi }}.\hfill \end{array}$$

The condition in (2) is therefore satisfied to be true for $\eta =1$ for ${\lim }_{\xi \to \infty }{\parallel {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\varrho ;\zeta )-\zeta \parallel }_{\nu }=0$. Likewise, this can be analyzed using Lemma 2

$$\begin{array}{ccc}\hfill {\parallel {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }({\varrho }^2;\zeta )-{\zeta }^2\parallel }_{\nu }& \le & \underset{\zeta \ge 0}{\sup }{\displaystyle \frac{\left|\frac{\left({(\xi +\rho )}^2+7(\xi +\rho )+6\right){\zeta }^2}{{(\xi +\psi )}^2}+\frac{\left((2\vartheta +5)(\xi +\rho )+4\vartheta +6\right)\zeta }{{(\xi +\psi )}^2}+\frac{{(\vartheta +1)}^2}{{(\xi +\psi )}^2}-{\zeta }^2\right|}{1+{\zeta }^2}}\hfill \\ & =& \underset{\zeta \ge 0}{\sup }{\displaystyle \frac{\left|\frac{\left(\left(2(\rho -\psi \right)+7)\xi +{\rho }^2-{\psi }^2+7\rho +6\right){\zeta }^2}{{(\xi +\psi )}^2}+\frac{\left((2\vartheta +5)(\xi +\rho )+4\vartheta +6\right)\zeta }{{(\xi +\psi )}^2}+\frac{{(\vartheta +1)}^2}{{(\xi +\psi )}^2}\right|}{1+{\zeta }^2}}\hfill \\ & \le & {\displaystyle \frac{\left|\left(2\right|\rho -\psi |+7)\xi +{\rho }^2+7\rho +6+(2\vartheta +5)(\xi +\rho )+4\vartheta +6+{(\vartheta +1)}^2\right|}{{(\xi +\psi )}^2}}.\hfill \end{array}$$

Then, if ${\lim }_{\xi \to \infty }{\parallel {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }({\varrho }^2;\zeta )-{\zeta }^2\parallel }_{\nu }=0$, it follows that condition (2) holds for $\eta =2$. Therefore, the assertion is proved. □

4. Rate of Convergence of Operators ${\mathit{V}}_{\mathit{\xi },\mathit{\rho }}^{\mathit{\vartheta },\mathit{\psi }}$

Let

$$\mathsf{\Omega }(\mathsf{{\rm Y}};\delta )=\underset{\zeta \in (0,\infty ),\left|l\right|\le \delta }{\sup }{\displaystyle \frac{\left|\mathsf{{\rm Y}}\right(\zeta +l)-\mathsf{{\rm Y}}(\zeta \left)\right|}{(1+{\zeta }^2)(1+l^2)}},$$

(3)

for every $\mathsf{{\rm Y}}\in C_{\nu }^{*}(0,\infty )$. The weighted modulus of continuity of the function $\mathsf{{\rm Y}}$ is defined by $\mathsf{\Omega }(\mathsf{{\rm Y}};\delta )$ [22]. Moreover, the properties of $\mathsf{\Omega }(\mathsf{{\rm Y}};\delta )$ defined in (3) are comparable to those of the usual modulus of continuity.

Theorem 2.

For $\mathsf{{\rm Y}}\in C_{\nu }^{*}(0,\infty )$, $S>0$ and $\overline{\nu }\left(\zeta \right)=1+{\zeta }^5$, we derive

$$\parallel {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )-\mathsf{{\rm Y}}{\parallel }_{\overline{\nu }}\le S\mathsf{\Omega }\left(\mathsf{{\rm Y}};{\displaystyle \frac{1}{\sqrt{\xi +\psi }}}\right).$$

Proof. 

By the definition in (3), we get

$$\begin{array}{c}|\mathsf{{\rm Y}}\left(\varrho \right)-\mathsf{{\rm Y}}\left(\zeta \right)|\le \left(1+{(\varrho -\zeta )}^2\right)\left(1+{\zeta }^2\right)\left(1+{\displaystyle \frac{|\varrho -\zeta |}{\delta }}\right)\mathsf{\Omega }(\mathsf{{\rm Y}};\delta ).\end{array}$$

Therefore, we have

$$\begin{array}{cc}\hfill |V_{\xi ,\rho }^{\vartheta ,\psi }(\mathrm{Y}\left(\varrho \right)-\mathrm{Y}\left(\zeta \right);\zeta )|& ⩽V_{\xi ,\rho }^{\vartheta ,\psi }\left(\right|\mathrm{Y}\left(\varrho \right)-\mathrm{Y}\left(\zeta \right)|;\zeta )\hfill \\ \hfill & ⩽\mathsf{\Omega }(\mathrm{Y};\delta )\left(1+{\zeta }^2\right)V_{\xi ,\rho }^{\vartheta ,\psi }\left(\left(1+{(\varrho -\zeta )}^2\right)\left(1+{\displaystyle \frac{|\varrho -\zeta |}{\delta }}\right);\zeta \right)\hfill \\ \hfill & =\mathsf{\Omega }(\mathrm{Y};\delta )\left(1+{\zeta }^2\right)\left[V_{\xi ,\rho }^{\vartheta ,\psi }(1+{(\varrho -\zeta )}^2;\zeta )\right.\hfill \\ \hfill & \left.+V_{\xi ,\rho }^{\vartheta ,\psi }\left(\left(1+{(\varrho -\zeta )}^2\right){\displaystyle \frac{|\varrho -\zeta |}{\delta }};\zeta \right)\right].\hfill \end{array}$$

(4)

By applying the Cauchy–Schwarz inequality to the second term of (4), we obtain

$$\begin{array}{c}\hfill {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left(\left(1+{(\varrho -\zeta )}^2\right){\displaystyle \frac{|\varrho -\zeta |}{\delta }};\zeta \right)\le \sqrt{{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left({\left(1+{(\varrho -\zeta )}^2\right)}^2;\zeta \right)}\sqrt{{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left({\displaystyle \frac{{|\varrho -\zeta |}^2}{{\delta }^2}};\zeta \right)}.\end{array}$$

(5)

Due to (4) and (5), we get

$$\begin{array}{ccc}\hfill {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left(\right|\mathsf{{\rm Y}}\left(\varrho \right)-\mathsf{{\rm Y}}\left(\zeta \right)|;\zeta )& \le & \mathsf{\Omega }(\mathsf{{\rm Y}};\delta )(1+{\zeta }^2)[{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }{\left(1+(\varrho -\zeta \right)}^2;\zeta )\hfill \\ & & +\sqrt{{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left({\left(1+{(\varrho -\zeta )}^2\right)}^2;\zeta \right)}\sqrt{{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left({\displaystyle \frac{{|\varrho -\zeta |}^2}{{\delta }^2}};\zeta \right)}].\hfill \end{array}$$

There are positive constants $S_1$ and $S_2$ such that

$$\begin{array}{c}\hfill {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left((1+{(\varrho -\zeta )}^2;\zeta \right)\le S_1(1+{\zeta }^2),\end{array}$$

(6)

$$\begin{array}{c}\hfill \sqrt{{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left({\left(1+{(\varrho -\zeta )}^2\right)}^2;\zeta \right)}\le S_2(1+{\zeta }^2).\end{array}$$

(7)

Here, $S_1$ and $S_2$ provide bounds, independent of $\xi$ for large $\xi$ for ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left((1+{(\varrho -\zeta )}^2;\zeta \right)$ and $\sqrt{{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left({\left(1+{(\varrho -\zeta )}^2\right)}^2;\zeta \right)}$.

Note that, from Lemma 3, we get

$$\begin{array}{ccc}\hfill \sqrt{{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left({\displaystyle \frac{{|\varrho -\zeta |}^2}{{\delta }^2}};\zeta \right)}& \le & {\displaystyle \frac{1}{\delta }}({\displaystyle \frac{\left((\xi +\rho +6)(\xi +\rho +1)-2(\xi +\rho +2)(\xi +\psi )+{(\xi +\psi )}^2\right){\zeta }^2}{{(\xi +\psi )}^2}}\hfill \\ & & +{\displaystyle \frac{\left((2\vartheta +5)(\xi +\rho )+4\vartheta +6-2(\vartheta +1)(\xi +\psi )\right)\zeta }{{(\xi +\psi )}^2}}\hfill \\ & & +{\displaystyle \frac{{(\vartheta +1)}^2}{{(\xi +\psi )}^2}}{)}^{{\displaystyle \frac{1}{2}}}\hfill \\ & \le & {\displaystyle \frac{1}{\delta }}O\left({\displaystyle \frac{1}{\sqrt{\xi +\psi }}}\right)\sqrt{{\zeta }^2+\zeta +1}\hfill \\ & \le & S_3{\displaystyle \frac{1+\zeta }{\delta \sqrt{\xi +\psi }}},\hfill \end{array}$$

(8)

where the notation O is Big O.

Finally let $S_6=S_1{S}_5+S_2{S}_3{S}_4$ where $S_5={\sup }_{\zeta \ge 0}{\displaystyle \frac{1+{\zeta }^2}{1+{\zeta }^3}}$, $S_4={\sup }_{\zeta \ge 0}{\displaystyle \frac{(1+{\zeta }^2)(1+\zeta )}{1+{\zeta }^3}}$ and $S=S_6{S}_7$ where $S_7={\sup }_{\zeta \ge 0}{\displaystyle \frac{(1+{\zeta }^2)(1+{\zeta }^3)}{1+{\zeta }^5}}$. Moreover, let $\delta ={\displaystyle \frac{1}{\sqrt{\xi +\psi }}}$. Combining the results (4) to (8), we obtain

$$\begin{array}{c}\hfill |{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )-\mathsf{{\rm Y}}\left(\zeta \right)|\le S(1+{\zeta }^5)\mathsf{\Omega }\left(\mathsf{{\rm Y}};{\displaystyle \frac{1}{\sqrt{\xi +\psi }}}\right)\end{array}$$

as anticipated. □

5. Pointwise Approximation Properties by ${\mathit{V}}_{\mathit{\xi },\mathit{\rho }}^{\mathit{\vartheta },\mathit{\psi }}$

The usual modulus of continuity of $\mathsf{{\rm Y}}\in {\widehat{C}}_B(0,\infty )$ can be defined as follows:

$$\omega (\mathsf{{\rm Y}};\delta )=\underset{|\zeta -y|\le \delta }{\sup }|\mathsf{{\rm Y}}\left(\zeta \right)-\mathsf{{\rm Y}}\left(y\right)|,$$

(9)

where ${\widehat{C}}_B(0,\infty )$ is the space of bounded uniformly continuous functions on $(0,\infty )$.

Theorem 3.

Let $\mathsf{{\rm Y}}\in {\widehat{C}}_B(0,\infty )$; hence, one has

$$|{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )-\mathsf{{\rm Y}}\left(\zeta \right)|\le 2\omega \left(\mathsf{{\rm Y}};\sqrt{{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left({(\varrho -\zeta )}^2;\zeta \right)}\right).$$

Proof. 

Lemma 3 and the Shisha Mond Theorem [23] give us

$$\begin{array}{ccc}\hfill |{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )-\mathsf{{\rm Y}}\left(\zeta \right)|& \le & \omega (\mathsf{{\rm Y}};\delta )\left(1+{\displaystyle \frac{1}{\delta }}\sqrt{{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left({(\varrho -\zeta )}^2;\zeta \right)}\right)\hfill \\ & \le & 2\omega \left(\mathsf{{\rm Y}};\sqrt{{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left({(\varrho -\zeta )}^2;\zeta \right)}\right),\hfill \end{array}$$

where $\delta =\sqrt{{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left({(\varrho -\zeta )}^2;\zeta \right)}$. □

Let us construct the space ${\widehat{C}}_B^2(0,\infty )=\{\mathsf{{\rm Y}}\in {\widehat{C}}_B(0,\infty ):{\mathsf{{\rm Y}}}^{\prime },{\mathsf{{\rm Y}}}^{″}\in {\widehat{C}}_B(0,\infty )\}$ with the norm

$$\begin{array}{c}\hfill {\parallel \mathsf{{\rm Y}}\parallel }_{{\widehat{C}}_B^2(0,\infty )}={\parallel \mathsf{{\rm Y}}\parallel }_{{\widehat{C}}_B(0,\infty )}+{\parallel {\mathsf{{\rm Y}}}^{\prime }\parallel }_{{\widehat{C}}_B(0,\infty )}+{\parallel {\mathsf{{\rm Y}}}^{″}\parallel }_{{\widehat{C}}_B(0,\infty )},\end{array}$$

and

$$\begin{array}{c}\hfill {\parallel \mathsf{{\rm Y}}\parallel }_{{\widehat{C}}_B(0,\infty )}=\sup \left\{\right|\mathsf{{\rm Y}}\left(\zeta \right)|:\zeta \in (0,\infty )\}.\end{array}$$

The $\mathcal{K}$-functional of Peetre’s [24] for $\delta >0$, $b\in {\widehat{C}}_B^2(0,\infty )$, and $\mathsf{{\rm Y}}\in {\widehat{C}}_B(0,\infty )$ is given by

$$\begin{array}{c}\hfill \mathcal{K}(\mathsf{{\rm Y}};\delta )=\underset{b\in {\widehat{C}}_B^2(0,\infty )}{\inf }{\parallel \mathsf{{\rm Y}}-b\parallel }_{{\widehat{C}}_B(0,\infty )}+\delta {\parallel b^{″}\parallel }_{{\widehat{C}}_B^2(0,\infty )}.\end{array}$$

For a constant $S\ge 0$, we have

$$\mathcal{K}(\mathsf{{\rm Y}};\delta )\le S{\omega }_2(\mathsf{{\rm Y}};\sqrt{\delta }),$$

where ${\omega }_2(\mathsf{{\rm Y}};\sqrt{\delta })$ denotes the second-order modulus of continuity of $\mathsf{{\rm Y}}\in {\widehat{C}}_B(0,\infty )$ [25] and it is defined as

$$\begin{array}{c}\hfill {\omega }_2(\mathsf{{\rm Y}};\sqrt{\delta })\underset{0<\left|l\right|\le \sqrt{\delta }}{\sup }\underset{\zeta ,\zeta +l\in (0,\infty )}{\sup }|\mathsf{{\rm Y}}(\zeta +2l)-2\mathsf{{\rm Y}}(\zeta +l)+\mathsf{{\rm Y}}\left(\zeta \right)|.\end{array}$$

Theorem 4.

Let $\mathsf{{\rm Y}}\in {\widehat{C}}_B(0,\infty )$. For $\zeta \in (0,\infty )$, we have

$$\begin{array}{ccc}\hfill |{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )-\mathsf{{\rm Y}}\left(\zeta \right)|& \le & 4S{\omega }_2\left(\mathsf{{\rm Y}};\sqrt{{\delta }_{\xi }\left(\zeta \right)}\right)+\omega \left(\mathsf{{\rm Y}};{\displaystyle \frac{(\rho -\psi +2)\zeta }{\xi +\psi }}+{\displaystyle \frac{\vartheta +1}{\xi +\psi }}\right),\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill {\delta }_{\xi }\left(\zeta \right)& =& {\displaystyle \frac{(2{\psi }^2-4\psi \rho +2{\rho }^2-8\psi +3\xi +11\rho +10){\zeta }^2}{{(\xi +\psi )}^2}}\hfill \\ & & +{\displaystyle \frac{(4\vartheta \rho -4\vartheta \psi +8\vartheta -4\psi +3\xi +7\rho +10)\zeta }{{(\xi +\psi )}^2}}+{\displaystyle \frac{2{(\vartheta +1)}^2}{{(\xi +\psi )}^2}}.\hfill \end{array}$$

Proof. 

Using Taylor’s series, we can express for $\mathsf{{\rm Y}}\in {\widehat{C}}_B(0,\infty )$ and $\varrho \in (0,\infty )$,

$$\begin{array}{c}\hfill b\left(\varrho \right)-b\left(\zeta \right)=(\varrho -\zeta )b^{\prime }\left(\zeta \right)+{\int }_{\zeta }^{\varrho }(\varrho -c)b^{″}\left(c\right)dc.\end{array}$$

(10)

Upon applying ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }$ to (10), we obtain ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\varrho -\zeta ;\zeta )\ne 0$. Consequently, we must define an operator in this way:

$$\begin{array}{c}\hfill {\check{\mathit{V}}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )={\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )-\mathsf{{\rm Y}}\left({\displaystyle \frac{(\xi +\rho +2)\zeta }{\xi +\psi }}+{\displaystyle \frac{\vartheta +1}{\xi +\psi }}\right)+\mathsf{{\rm Y}}\left(\zeta \right).\end{array}$$

(11)

Applying ${\check{\mathit{V}}}_{\xi ,\rho }^{\vartheta ,\psi }$ to both sides of (10) now yields

$$\begin{array}{c}\hfill {\check{\mathit{V}}}_{\xi ,\rho }^{\vartheta ,\psi }(b;\zeta )-b\left(\zeta \right)={\check{\mathit{V}}}_{\xi ,\rho }^{\vartheta ,\psi }(\varrho -\zeta ;\zeta )b^{\prime }\left(\zeta \right)+{\check{\mathit{V}}}_{\xi ,\rho }^{\vartheta ,\psi }\left({\int }_{\zeta }^{\varrho }(\varrho -c)b^{″}\left(c\right)dc;\zeta \right).\end{array}$$

From the definition of (11), we have

$$\begin{array}{ccc}\hfill {\check{\mathit{V}}}_{\xi ,\rho }^{\vartheta ,\psi }(b;\zeta )-b\left(\zeta \right)& =& {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left({\int }_{\zeta }^{\varrho }(\varrho -c)b^{″}\left(c\right)dc;\zeta \right)\hfill \\ & & -{\int }_{\zeta }^{{\displaystyle \frac{(\xi +\rho +2)\zeta }{\xi +\psi }}+{\displaystyle \frac{\vartheta +1}{\xi +\psi }}}\left({\displaystyle \frac{(\xi +\rho +2)\zeta }{\xi +\psi }}+{\displaystyle \frac{\vartheta +1}{\xi +\psi }}-c\right)b^{″}\left(c\right)dc\hfill \\ & & +{\int }_{\zeta }^{\zeta }(\zeta -c)b^{″}\left(c\right)dc.\hfill \end{array}$$

The linearity and positivity of ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }$ are used to establish the inequality,

$$\begin{array}{ccc}\hfill |{\check{\mathit{V}}}_{\xi ,\rho }^{\vartheta ,\psi }(b;\zeta )-b\left(\zeta \right)|& \le & \left|{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left({\int }_{\zeta }^{\varrho }(\varrho -c)b^{″}\left(c\right)dc;\zeta \right)\right|\hfill \\ & & +\left|{\int }_{\zeta }^{{\displaystyle \frac{(\xi +\rho +2)\zeta }{\xi +\psi }}+{\displaystyle \frac{\vartheta +1}{\xi +\psi }}}\left({\displaystyle \frac{(\xi +\rho +2)\zeta }{\xi +\psi }}+{\displaystyle \frac{\vartheta +1}{\xi +\psi }}-c\right)b^{″}\left(c\right)dc\right|\hfill \\ & \le & {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left({\int }_{\zeta }^{\varrho }|\varrho -c||b^{″}\left(c\right)|dc;\zeta \right)\hfill \\ & & +{\int }_{\zeta }^{{\displaystyle \frac{(\xi +\rho +2)\zeta }{\xi +\psi }}+{\displaystyle \frac{\vartheta +1}{\xi +\psi }}}\left|{\displaystyle \frac{(\xi +\rho +2)\zeta }{\xi +\psi }}+{\displaystyle \frac{\vartheta +1}{\xi +\psi }}-c\right|\left|b^{″}\left(c\right)\right|dc.\hfill \end{array}$$

Then, we get

$$\begin{array}{ccc}\hfill |{\check{\mathit{V}}}_{\xi ,\rho }^{\vartheta ,\psi }(b;\zeta )-b\left(\zeta \right)|& \le & \parallel b^{″}{\parallel }_{{\widehat{C}}_B(0,\infty )}{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }({(\varrho -\zeta )}^2;\zeta )+{\parallel b^{″}\parallel }_{{\widehat{C}}_B(0,\infty )}{\left({\displaystyle \frac{(\rho -\psi +2)\zeta }{\xi +\psi }}+{\displaystyle \frac{\vartheta +1}{\xi +\psi }}\right)}^2.\hfill \end{array}$$

From Lemma 3, we have

$$\begin{array}{ccc}\hfill |{\check{\mathit{V}}}_{\xi ,\rho }^{\vartheta ,\psi }(b;\zeta )-b\left(\zeta \right)|& \le & \parallel b^{″}{\parallel }_{{\widehat{C}}_B(0,\infty )}({\displaystyle \frac{\left((\xi +\rho +6)(\xi +\rho +1)-2(\xi +\rho +2)(\xi +\psi )+{(\xi +\psi )}^2\right){\zeta }^2}{{(\xi +\psi )}^2}}\hfill \\ & & +{\displaystyle \frac{\left((2\vartheta +5)(\xi +\rho )+4\vartheta +6-2(\vartheta +1)(\xi +\psi )\right)\zeta }{{(\xi +\psi )}^2}}+{\displaystyle \frac{{(\vartheta +1)}^2}{{(\xi +\psi )}^2}})\hfill \\ & & +\parallel b^{″}{\parallel }_{{\widehat{C}}_B(0,\infty )}{\left({\displaystyle \frac{(\rho -\psi +2)\zeta }{\xi +\psi }}+{\displaystyle \frac{\vartheta +1}{\xi +\psi }}\right)}^2\hfill \\ & \le & \parallel b^{″}{\parallel }_{{\widehat{C}}_B(0,\infty )}{\delta }_{\xi }\left(\zeta \right),\hfill \end{array}$$

(12)

where

$$\begin{array}{ccc}\hfill {\delta }_{\xi }\left(\zeta \right)& =& {\displaystyle \frac{(2{\psi }^2-4\psi \rho +2{\rho }^2-8\psi +3\xi +11\rho +10){\zeta }^2}{{(\xi +\psi )}^2}}\hfill \\ & & +{\displaystyle \frac{(4\vartheta \rho -4\vartheta \psi +8\vartheta -4\psi +3\xi +7\rho +10)\zeta }{{(\xi +\psi )}^2}}+{\displaystyle \frac{2{(\vartheta +1)}^2}{{(\xi +\psi )}^2}}.\hfill \end{array}$$

From Equation (11) for $\mathsf{{\rm Y}}\in {\widehat{C}}_B(0,\infty )$, we yield

$$\begin{array}{ccc}\hfill |{\check{\mathit{V}}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )|& \le & |{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )|+\left|\mathsf{{\rm Y}}\left({\displaystyle \frac{(\xi +\rho +2)\zeta }{\xi +\psi }}+{\displaystyle \frac{\vartheta +1}{\xi +\psi }}\right)\right|+\left|\mathsf{{\rm Y}}\left(\zeta \right)\right|\hfill \\ & \le & {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left(\right|\mathsf{{\rm Y}}|;\zeta )|+{\parallel \mathsf{{\rm Y}}\parallel }_{{\widehat{C}}_B(0,\infty )}+{\parallel \mathsf{{\rm Y}}\parallel }_{{\widehat{C}}_B(0,\infty )}\hfill \\ & \le & {3\parallel \mathsf{{\rm Y}}\parallel }_{{\widehat{C}}_B(0,\infty )}.\hfill \end{array}$$

(13)

Considering Equation (11), we get

$$\begin{array}{ccc}\hfill |{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )-\mathsf{{\rm Y}}\left(\zeta \right)|& \le & |{\check{\mathit{V}}}_{\xi ,\rho }^{\vartheta ,\psi }((\mathsf{{\rm Y}}-b);\zeta )-(\mathsf{{\rm Y}}-b)\left(\zeta \right)|+|{\check{\mathit{V}}}_{\xi ,\rho }^{\vartheta ,\psi }(b;\zeta )-b\left(\zeta \right)|\hfill \\ & & +\left|\mathsf{{\rm Y}}\left({\displaystyle \frac{(\xi +\rho +2)\zeta }{\xi +\psi }}+{\displaystyle \frac{\vartheta +1}{\xi +\psi }}\right)-\mathsf{{\rm Y}}\left(\zeta \right)\right|.\hfill \end{array}$$

(14)

By (9) and (11)–(14), we have

$$\begin{array}{ccc}\hfill |{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )-\mathsf{{\rm Y}}\left(\zeta \right)|& \le & {4\parallel \mathsf{{\rm Y}}-b\parallel }_{{\widehat{C}}_B(0,\infty )}+{\parallel b^{″}\parallel }_{{\widehat{C}}_B(0,\infty )}{\delta }_{\xi }\left(\zeta \right)+\omega \left(\mathsf{{\rm Y}};{\displaystyle \frac{(\rho -\psi +2)\zeta }{\xi +\psi }}+{\displaystyle \frac{\vartheta +1}{\xi +\psi }}\right).\hfill \end{array}$$

Taking the infimum of the right-hand side of the aforementioned inequalities over $b\in {\widehat{C}}_B^2(0,\infty )$ yields

$$\begin{array}{ccc}\hfill |{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )-\mathsf{{\rm Y}}\left(\zeta \right)|& \le & 4\mathcal{K}(\mathsf{{\rm Y}};{\delta }_{\xi }\left(\zeta \right))+\omega \left(\mathsf{{\rm Y}};{\displaystyle \frac{(\rho -\psi +2)\zeta }{\xi +\psi }}+{\displaystyle \frac{\vartheta +1}{\xi +\psi }}\right).\hfill \end{array}$$

Accordingly, we conclude that

$$\begin{array}{ccc}\hfill |{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )-\mathsf{{\rm Y}}\left(\zeta \right)|& \le & 4S{\omega }_2(\mathsf{{\rm Y}};\sqrt{{\delta }_{\xi }\left(\zeta \right)})+\omega \left(\mathsf{{\rm Y}};{\displaystyle \frac{(\rho -\psi +2)\zeta }{\xi +\psi }}+{\displaystyle \frac{\vartheta +1}{\xi +\psi }}\right).\hfill \end{array}$$

□

Theorem 5.

Let $\mathsf{{\rm Y}}\in Li{p}_S\left(\gamma \right)$ with $S>0$ and $0<\gamma \le 1$; then,

$$\begin{array}{ccc}\hfill |{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )-\mathsf{{\rm Y}}\left(\zeta \right)|& \le & S{\left({\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left({(\varrho -\zeta )}^2;\zeta \right)\right)}^{{\displaystyle \frac{\gamma }{2}}},\hfill \end{array}$$

where $Li{p}_S\left(\gamma \right):=\{\mathsf{{\rm Y}}\in {\widehat{C}}_B(0,\infty ):|\mathsf{{\rm Y}}\left(\varrho \right)-\mathsf{{\rm Y}}\left(\zeta \right)|\le S{|\varrho -\zeta |}^{\gamma },\varrho ,\zeta \in (0,\infty )\}.$

Proof. 

Since $\mathsf{{\rm Y}}\in Li{p}_S\left(\gamma \right)$, we have

$$\begin{array}{ccc}\hfill \left|\mathsf{{\rm Y}}\right(\varrho )-\mathsf{{\rm Y}}(\zeta \left)\right|& \le & S{|\varrho -\zeta |}^{\gamma }.\hfill \end{array}$$

(15)

Using the linearity and positivity of ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }$ and (15), we derive the inequality

$$\begin{array}{c}\hfill {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left(\right|\mathsf{{\rm Y}}\left(\varrho \right)-\mathsf{{\rm Y}}\left(\zeta \right)|;\zeta )\le {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(S{|\varrho -\zeta |}^{\gamma };\zeta ).\end{array}$$

Lemma 3 and the H$\ddot{o}$lder inequality give us

$$\begin{array}{ccc}\hfill |{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )-\mathsf{{\rm Y}}\left(\zeta \right)|& \le & S{\left({\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left({(\varrho -\zeta )}^2;\zeta \right)\right)}^{{\displaystyle \frac{\gamma }{2}}}.\hfill \end{array}$$

□

6. Voronovskaja Type Theorem

Theorem 6.

Let $\mathsf{{\rm Y}}\in {\widehat{C}}_B^2(0,\infty )$. Then,

$$\underset{\xi \to \infty }{\lim }\xi \left({\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )-\mathsf{{\rm Y}}\left(\zeta \right)\right)=\left((\rho -\psi +2)\zeta +\vartheta +1\right){\mathsf{{\rm Y}}}^{\prime }\left(\zeta \right)+{\displaystyle \frac{3}{2}}\left({\zeta }^2+\zeta \right){\mathsf{{\rm Y}}}^{″}\left(\zeta \right).$$

Proof. 

From Lemma 3, we have

$$\begin{array}{ccc}\hfill \underset{\xi \to \infty }{\lim }\xi {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\varrho -\zeta ;\zeta )& =& (\rho -\psi +2)\zeta +\vartheta +1,\hfill \end{array}$$

(16)

$$\begin{array}{ccc}\hfill \underset{\xi \to \infty }{\lim }\xi {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }({(\varrho -\zeta )}^2;\zeta )& =& 3{\zeta }^2+3\zeta ,\hfill \end{array}$$

(17)

$$\begin{array}{ccc}\hfill \underset{\xi \to \infty }{\lim }{\xi }^2{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }({(\varrho -\zeta )}^4;\zeta )& =& 15{\zeta }^4+30{\zeta }^3+15{\zeta }^2.\hfill \end{array}$$

(18)

With Taylor’s expansion in mind, we have

$$\begin{array}{c}\hfill \mathsf{{\rm Y}}\left(\varrho \right)=\mathsf{{\rm Y}}\left(\zeta \right)+{\mathsf{{\rm Y}}}^{\prime }\left(\zeta \right)(\varrho -\zeta )+{\displaystyle \frac{1}{2}}{\mathsf{{\rm Y}}}^{″}\left(\zeta \right){(\varrho -\zeta )}^2+\alpha (\varrho ;\zeta ){(\varrho -\zeta )}^2,\end{array}$$

(19)

where $\alpha (\varrho ;\zeta )$ denotes the remainder term in Peano’s form, satisfying ${\lim }_{\varrho \to \zeta }\alpha (\varrho ;\zeta )=0$.

Applying ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }$ to (19) and using its linearity, we obtain

$$\begin{array}{cccc}\hfill {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )-\mathsf{{\rm Y}}\left(\zeta \right)& ={\mathsf{{\rm Y}}}^{\prime }\left(\zeta \right){\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left((\varrho -\zeta );\zeta \right)+{\displaystyle \frac{1}{2}}{\mathsf{{\rm Y}}}^{″}\left(\zeta \right){\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left({(\varrho -\zeta )}^2;\zeta \right)\hfill & & \\ & +{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left(\alpha (\varrho ;\zeta ){(\varrho -\zeta )}^2;\zeta \right).\hfill \end{array}$$

(20)

By multiplying (20) by $\xi$, considering the limit $\xi \to \infty$ and applying (16) and (17), we get

$$\begin{array}{cccc}\hfill \underset{\xi \to \infty }{\lim }\xi \left({\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )-\mathsf{{\rm Y}}\left(\zeta \right)\right)& =\left((\rho -\psi +2)\zeta +\vartheta +1\right){\mathsf{{\rm Y}}}^{\prime }\left(\zeta \right)+{\displaystyle \frac{3}{2}}\left({\zeta }^2+\zeta \right){\mathsf{{\rm Y}}}^{″}\left(\zeta \right)\hfill & & \hfill \\ & +\underset{\xi \to \infty }{\lim }\xi {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left(\alpha (\varrho ;\zeta ){(\varrho -\zeta )}^2;\zeta \right).\hfill \end{array}$$

(21)

The Cauchy–Schwarz inequality applied to the second term on the right-hand side of (21) allows us to obtain

$$\begin{array}{c}\hfill \xi {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left(\alpha (\varrho ;\zeta ){(\varrho -\zeta )}^2;\zeta \right)\le \sqrt{{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left({\alpha }^2(\varrho ;\zeta );\zeta \right)}\sqrt{{\xi }^2{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left({(\varrho -\zeta )}^4;\zeta \right)}.\end{array}$$

Let $\mathsf{\Phi }(\varrho ;\zeta )={\alpha }^2(\varrho ;\zeta )$. Given that $\mathsf{\Phi }(.;\zeta )$ is continuous at $\varrho \in (0,\infty )$ and ${\lim }_{\varrho \to \zeta }\alpha (\varrho ;\zeta )=0$, we can write

$$\begin{array}{c}\hfill \underset{\varrho \to \zeta }{\lim }\mathsf{\Phi }(\varrho ;\zeta )=\underset{\varrho \to \zeta }{\lim }\mathsf{\Phi }(\zeta ;\zeta )=0.\end{array}$$

(22)

Using (18) and (22), we have

$$\begin{array}{c}\hfill \underset{\xi \to \infty }{\lim }\xi {\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }\left(\alpha (\varrho ;\zeta ){(\varrho -\zeta )}^2;\zeta \right)=0.\end{array}$$

(23)

Finally, we obtain from (21) and (23),

$$\begin{array}{c}\hfill \underset{\xi \to \infty }{\lim }\xi \left({\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )-\mathsf{{\rm Y}}\left(\zeta \right)\right)=\left((\rho -\psi +2)\zeta +\vartheta +1\right){\mathsf{{\rm Y}}}^{\prime }\left(\zeta \right)+{\displaystyle \frac{3}{2}}\left({\zeta }^2+\zeta \right){\mathsf{{\rm Y}}}^{″}\left(\zeta \right).\end{array}$$

This brings the proof to a close. □

7. Graphical Simulations

Graphical validation plays a crucial role in supporting theoretical approximation results. In this section, we perform a set of computational experiments to demonstrate the accuracy and effectiveness of the considered operators ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )$. Their performance is assessed using various test functions and compared against well-known approximation methods. The experimental findings confirm the theoretical analysis and highlight the practical benefits of the related approach.

Example 1.

Consider the function $\mathsf{{\rm Y}}\left(\zeta \right)=\left(\zeta +1\right)\left(\zeta -3\right)\left(\zeta -5\right)e^{-\zeta }.$ In Figure 1a, we present the convergence behavior of operators ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )$ to $\mathsf{{\rm Y}}\left(\zeta \right)$ for $0\le \zeta \le 3,\rho =2,\vartheta =0.1,\psi =0.2$ and $\xi =10,20,60,$ respectively. Also, in Figure 1b, we define with $E_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )=|{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )-\mathsf{{\rm Y}}\left(\zeta \right)|$ the absolute error of the approximation of related operators to the considered function. It is clear that larger values of ξ enhance the performance of the operators ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )$, resulting in a more accurate approximation of $\mathsf{{\rm Y}}\left(\zeta \right)$.

Figure 1. (a) Convergence and (b) error of approximation of operators ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )$ to $\mathsf{{\rm Y}}\left(\zeta \right)$ for $\rho =2,\vartheta =0.1,\psi =0.2$, $\xi =10,20,60$.

Example 2.

Consider the function $\mathsf{{\rm Y}}\left(\zeta \right)=\sin \left(0.4\pi \zeta \right)e^{-\zeta }.$ In Figure 2a, we show the convergence behavior of operators ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )$ to $\mathsf{{\rm Y}}\left(\zeta \right)$ for $0\le \zeta \le 4,\rho =3,\vartheta =0.1,\psi =0.2$ and $\xi =20,40,80,$ respectively. Also, in Figure 2b, we define with $E_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )=|{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )-\mathsf{{\rm Y}}\left(\zeta \right)|$ the absolute error of approximation of related operators to considered function. One can easily check from Figure 2 that, when ξ values increase, proposed operators ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )$ provide better approximation convergence.

Figure 2. (a) Convergence and (b) error of approximation of operators ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )$ to $\mathsf{{\rm Y}}\left(\zeta \right)$ for $\rho =3,\vartheta =0.1,\psi =0.2$, $\xi =20,40,80$.

Example 3.

Consider the function $\mathsf{{\rm Y}}\left(\zeta \right)={\zeta }^2{e}^{-3\zeta }.$ In Figure 3a, we compare the convergence of operators ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )$ with operators ${\mathit{V}}_{\xi }(\mathsf{{\rm Y}};\zeta )$ for the fixed parameters $0\le \zeta \le 5,\rho =1,$$\vartheta =0.1,\psi =0.2$ and $\xi =20$, respectively. Also, in Figure 3b, we denote with $E_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )=|{\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )-\mathsf{{\rm Y}}\left(\zeta \right)|$ the absolute error of approximation of operators ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )$ and ${\mathit{V}}_{\xi }(\mathsf{{\rm Y}};\zeta )$ to the considered function. In Figure 3, it is clearly seen that operators ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )$ provide better approximation than operators ${\mathit{V}}_{\xi }(\mathsf{{\rm Y}};\zeta )$.

Figure 3. (a) Comparison of convergence and (b) error of approximation of operators ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }(\mathsf{{\rm Y}};\zeta )$ with ${\mathit{V}}_{\xi }(\mathsf{{\rm Y}};\zeta )$ to $\mathsf{{\rm Y}}\left(\zeta \right)$ for $\rho =1,\vartheta =0.1,\psi =0.2$ and $\xi =20$.

8. Conclusions and Remarks

A novel variation of the Baskakov–Schurer–Stancu operators is presented in this paper. We derived and analyzed test functions and central moments. A Korovkin-type approximation theorem is developed based on these fundamental findings, and the weighted modulus of continuity in weighted spaces is used to analyze the rate of convergence. Additionally, functions on the Lipschitz-type class, Peetre’s K-functional and the classical modulus of continuity are used to analyze the pointwise approximation behavior. The theorem of Voronovskaja’s type is proved. Numerical examples are also provided to demonstrate the application and validity of the theoretical results. Consequently, we found that a more broad operator was developed in comparison to earlier investigations. Also, parameters $\vartheta ,\rho ,\psi$ used in the operators ${\mathit{V}}_{\xi ,\rho }^{\vartheta ,\psi }$ provided more flexibility in approximation.