Theory and Modern Applications

# Approximation by modified Kantorovich–Szász type operators involving Charlier polynomials

## Abstract

In this paper, we give some direct approximation results by modified Kantorovich–Szász type operators involving Charlier polynomials. Further, approximation results are also developed in polynomial weighted spaces. Moreover, for the functions of bounded variation, approximation results are proved. Finally, some graphical examples are provided to show comparisons of convergence between old and modified operators towards a function under different parameters and conditions.

## Introduction

In 1950, Szász  introduced positive linear operators in the sense of exponential growth on nonnegative semiaxes and exhaustively investigated them. These operators later became known as Szász operators. The Szász type operators involving Charlier polynomials were defined in  as

$$L_{n}(f;y,b)=e^{-1} \biggl( 1- \frac{1}{b} \biggr)^{(b-1)ny}\sum_{l=0}^{ \infty }\frac{C_{l}^{(b)}(-(b-1)ny)}{l!}f \biggl(\frac{l}{n} \biggr),$$
(1.1)

where $$b > 1$$ and $$y \geq 0$$, having the generating functions  of the form

$$e^{t} \biggl( 1-\frac{t}{b} \biggr)^{u} =\sum_{l=0}^{\infty }C_{l}^{(b)}(u) \frac{t^{l}}{l!},\quad \vert t \vert < b,$$
(1.2)

where ${C}_{l}^{\left(b\right)}\left(u\right)={\sum }_{r=0}^{l}\left(\begin{array}{c}l\\ r\end{array}\right){\left(-u\right)}_{r}{\left(\frac{1}{b}\right)}^{r}$ and $$(k)_{0} = 1$$, $$(k)_{m} = k(k+1) \cdots (k+m-1)$$, for $$m \geq 1$$.

Motivated by the work done in , we define the Kantorovich generalization  of (1.2) as follows:

$$\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)=\nu _{n}e^{-1} \biggl( 1- \frac{1}{b} \biggr) ^{(b-1)\mu _{n}y}\sum _{l=0}^{\infty } \frac{C_{l}^{(b)}(-(b-1)\mu _{n}y)}{l!} \int _{l/\nu _{n}}^{l+1/\nu _{n}}f(t)\,dt$$
(1.3)

where $$\mu _{n}$$ and $$\nu _{n}$$ are sequences of positive numbers which are increasing and unbounded such that

$$\lim_{n\rightarrow \infty }\frac{1}{\nu _{n}}=0,\qquad \frac{\mu _{n}}{\nu _{n}}=1+O \biggl( \frac{1}{\nu _{n}} \biggr) .$$
(1.4)

If we take $$\mu _{n}=\nu _{n}=n$$, we will have the operators defined in .

For some recent and interesting results on the various generalizations and corresponding approximation results, we refer to [1, 3, 68, 1417, 22].

## Auxiliary results

We first present some auxiliary results.

### Lemma 2.1

Let$$\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}$$be defined by (1.3). Then, we have

1. 1.

$$\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(1;y)= 1$$,

2. 2.

$$\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(t;y) = \frac{\mu _{n}}{\nu _{n}}y+\frac{3}{2\nu _{n}}$$

3. 3.

$$\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(t^{2};y) = \frac{\mu _{n}^{2}}{\nu _{n}^{2}}y^{2}+\frac{\mu _{n}}{\nu _{n}^{2}} (4+ \frac{1}{b-1} )y+\frac{10}{3\nu _{n}^{2}}$$.

4. 4.

$$\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(t^{3};y) = \frac{\mu _{n}^{3}}{\nu _{n}^{3}}y^{3}+ \frac{\mu _{n}^{2}}{\nu _{n}^{3}} (\frac{15}{2}+\frac{3}{b-1} )y^{2}+\frac{\mu _{n}}{\nu _{n}^{3}} ( \frac{31}{2}+\frac{15}{2(b-1)} +\frac{2}{(b-1)^{2}} )y+\frac{37}{4\nu _{n}^{3}}$$.

5. 5.

$$\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(t^{4};y) = \frac{\mu _{n}^{4}}{\nu _{n}^{4}}y^{4}+\frac{\mu _{n}^{3}}{\nu _{n}^{4}} (12+\frac{6}{b-1} )y^{3}+ \frac{\mu _{n}^{2}}{\nu _{n}^{4}} (46+ \frac{36}{b-1}+\frac{11}{(b-1)^{2}} )y^{2} + \frac{\mu _{n}}{\nu _{n}^{4}} (64+\frac{46}{b-1} + \frac{24}{(b-1)^{2}} +\frac{6}{(b-1)^{3}} )y+ \frac{151}{5\nu _{n}^{4}}$$.

### Proof

With the help of the Charlier polynomials’ generating function given by (1.2), after some simple calculations, we obtain

\begin{aligned}& \sum_{l=0}^{\infty }\frac{C_{l}^{(b)}(-(b-1)\mu _{n}y)}{l!} = e \biggl( 1-\frac{1}{b} \biggr)^{-(b-1)\mu _{n}y}, \\& \sum_{l=0}^{\infty }\frac{l C_{l}^{(b)}(-(b-1)\mu _{n}y)}{l!} = e \biggl( 1-\frac{1}{b} \biggr)^{-(b-1)\mu _{n}y} (1+ \mu _{n}y ), \\& \sum_{l=0}^{\infty }\frac{l^{2} C_{l}^{(b)}(-(b-1)\mu _{n}y)}{l!} = e \biggl( 1-\frac{1}{b} \biggr)^{-(b-1)\mu _{n}y} \biggl( \mu _{n}^{2}y^{2} +\mu _{n}y\biggl(3+ \frac{1}{b-1}\biggr)+2 \biggr), \\& \begin{aligned} \sum_{l=0}^{\infty } \frac{l^{3} C_{l}^{(b)}(-(b-1)\mu _{n}y)}{l!} &= e \biggl( 1-\frac{1}{b} \biggr)^{-(b-1)\mu _{n}y} \biggl( \mu _{n}^{3}y^{3} +\mu _{n}^{2}\biggl(6+ \frac{3}{b-1} \biggr)y^{2} \\ &\quad {}+ \mu _{n}\biggl(10+\frac{6}{b-1}+\frac{2}{(b-1)^{2}} \biggr)y +5 \biggr), \end{aligned} \\& \begin{aligned} \sum_{l=0}^{\infty } \frac{l^{4} C_{l}^{(b)}(-(b-1)\mu _{n}y)}{l!} &= e \biggl( 1-\frac{1}{b} \biggr)^{-(b-1)\mu _{n}y} \biggl( \mu _{n}^{4}y^{4} +\mu _{n}^{3}\biggl(10+ \frac{6}{b-1} \biggr)y^{3} \\ &\quad {}+ \mu _{n}^{2} \biggl(32+\frac{30}{b-1}+ \frac{11}{(b-1)^{2}} \biggr)y^{2} \\ &\quad {}+ \mu _{n}\biggl(37+\frac{32}{b-1}+\frac{20}{(b-1)^{2}}+ \frac{6}{(b-1)^{3}}\biggr)y +15 \biggr). \end{aligned} \end{aligned}

From the above equalities, the claims of the lemma can be obtained. □

### Lemma 2.2

For the operator$$\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}$$given by (1.3), we have the following equalities:

\begin{aligned}& (i)\quad \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(t-y;y) = \biggl( \frac{\mu _{n}}{\nu _{n}}-1 \biggr)y+\frac{1}{\nu _{n}}, \\& (\mathit{ii}) \quad \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})} \bigl((t-y)^{2};y\bigr) = \biggl( \frac{\mu _{n}}{\nu _{n}}-1 \biggr)^{2}y^{2}+ \biggl(\frac{\mu _{n}}{\nu _{n}^{2}} \biggl(3+ \frac{1}{b-1} \biggr)-\frac{2}{\nu _{n}} \biggr)y + \frac{2}{\nu _{n}^{2}}, \\& \begin{aligned} (\mathit{iii}) \quad& \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})} \bigl((t-y)^{4};y\bigr) \\ &\quad = \biggl( \frac{\mu _{n}}{\nu _{n}}-1 \biggr)^{4}y^{4} \\ &\qquad {}+2 \biggl(\frac{\mu _{n}^{3}}{\nu _{n}^{4}} \biggl(5+\frac{3}{b-1} \biggr)- \frac{6\mu _{n}^{2}}{\nu _{n}^{3}} \biggl(2+\frac{1}{b-1} \biggr)+ \frac{3\mu _{n}}{\nu _{n}^{2}} \biggl(3+\frac{1}{b-1} \biggr)-\frac{2}{\nu _{n}} \biggr)y^{3} \\ &\qquad {}+ \biggl(\frac{\mu _{n}^{2}}{\nu _{n}^{4}} \biggl(32+\frac{30}{b-1}+ \frac{11}{(b-1)^{2}} \biggr)-\frac{4\mu _{n}}{\nu _{n}^{3}} \biggl(10+ \frac{6}{b-1}+ \frac{2}{(b-1)^{2}} \biggr)+\frac{12}{\nu _{n}^{2}} \biggr)y^{2} \\ &\qquad {}+ \biggl(\frac{\mu _{n}}{\nu _{n}^{4}} \biggl(37+\frac{32}{b-1}+ \frac{20}{(b-1)^{2}}+\frac{6}{(b-1)^{3}} \biggr)-\frac{20}{\nu _{n}^{3}} \biggr)y + \frac{15}{\nu _{n}^{4}}. \end{aligned} \end{aligned}

## Local approximation results

In what follows, let $$\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(t - y; y)=\chi_{\mu _{n},\nu _{n}}(y)$$ and $$\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}((t - y)^{2}; y)=\xi_{\mu _{n},\nu _{n}}(y)$$. We will now give two theorems on the uniform convergence and the order of approximation.

### Theorem 3.1

Let$$f\in C[0,\infty )\cap G$$. Then$$\lim_{n\to \infty }\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)=f(y)$$, the sequence of operators Eq. (1.3) converges uniformly in each compact subset of$$[0,\infty )$$, where

$$G:= \biggl\{ f:\mathbb{R}_{0}^{+}\to \mathbb{R}, \bigl\vert F(y) \bigr\vert = \biggl\vert \int _{0}^{y}f(s)\,ds \biggr\vert \le Ke^{By}, K\in \mathbb{R^{+}}\textit{ and } B\in \mathbb{R} \biggr\} .$$

### Proof

From Lemma 2.1(1)–(3), we get

$$\lim_{n\to \infty }\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})} \bigl(s^{k};y\bigr)=y^{k},\quad k=0,1,2.$$

The proof of the theorem is established by taking advantage of the above uniform convergence in each compact subset of $$[0,\infty )$$ and the famous Korovkin’s theorem. □

Suppose $$f\in \tilde{C}[0,\infty )$$, i.e., f belongs to the space of uniformly continuous functions on $$[0,\infty )$$. If $$\delta >0$$, then the modulus of continuity $$\omega (f,\delta )$$ is defined by

$$\omega (f,\delta ):=\sup_{ \substack{ \sigma ,\varsigma \in [0,\infty ) \\ \vert \sigma -\varsigma \vert \leq \delta }} \bigl\vert f( \sigma )-f( \varsigma ) \bigr\vert .$$

### Theorem 3.2

Let$$f\in \tilde{C}[0,\infty )\cap E$$. For the operators$$\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)$$given by (1.3) the following estimate holds:

\begin{aligned} & \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert \\ &\quad \leq \biggl\{ 1+\sqrt{(\mu _{n}-\nu _{n})^{2}y^{2}+ \biggl( \biggl(4+ \frac{1}{b-1} \biggr)\mu _{n}-3\nu _{n} \biggr)y+\frac{10}{3}} \biggr\} \omega \biggl(f, \frac{1}{\nu _{n}} \biggr). \end{aligned}
(3.1)

### Proof

From (1.3) and the property of modulus of continuity, the left-hand side of (3.1) leads to

\begin{aligned} & \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert \\ &\quad \leq \nu _{n}e^{-1} \biggl( 1-\frac{1}{b} \biggr)^{(b-1)\mu _{n}y} \sum_{l=0}^{\infty } \frac{C_{l}^{(b)}(-(b-1)\mu _{n}y)}{l!} \int _{l/\nu _{n}}^{l+1/\nu _{n}} \bigl\vert f(t)-f(y) \bigr\vert \,dt \\ &\quad \leq \Biggl\{ 1+\frac{1}{\delta }\nu _{n}e^{-1} \biggl( 1-\frac{1}{b} \biggr)^{(b-1)\mu _{n}y}\sum _{l=0}^{\infty } \frac{C_{l}^{(b)}(-(b-1)\mu _{n}y)}{l!}\int _{l/\nu _{n}}^{l+1/\nu _{n}} \vert t-y \vert \,dt \Biggr\} \omega (f,\delta ). \end{aligned}

Using Cauchy–Schwarz inequality for the integral, we get

\begin{aligned} & \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert \\ & \quad \leq \Biggl\{ 1+\frac{1}{\delta }e^{-1} \biggl( 1-\frac{1}{b} \biggr)^{(b-1)\mu _{n}y} \\ &\qquad {}\times\sum_{l=0}^{\infty } \frac{C_{l}^{(b)}(-(b-1)\mu _{n}y)}{l!} \biggl(\nu _{n} \int _{l/\nu _{n}}^{l+1/\nu _{n}}(t-y)^{2}\,dt \biggr)^{1/2} \Biggr\} \omega (f,\delta ). \end{aligned}
(3.2)

In the above sum, we apply Cauchy–Schwarz inequality, and then in view of Lemma 2.1, (3.2) becomes

\begin{aligned} & \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert \\ &\quad \leq \Biggl\{ 1+\frac{1}{\delta } \Biggl(e^{-1} \biggl( 1- \frac{1}{b} \biggr)^{(b-1)\mu _{n}y}\sum_{l=0}^{\infty } \frac{C_{l}^{(b)}(-(b-1)\mu _{n}y)}{l!} \Biggr)^{1/2} \\ &\qquad {}\times \Biggl(\nu _{n}e^{-1} \biggl(1- \frac{1}{b} \biggr)^{(b-1)\mu _{n}y}\sum_{l=0}^{\infty } \frac{C_{l}^{(b)}(-(b-1)\mu _{n}y)}{l!} \int _{l/\nu _{n}}^{l+1/\nu _{n}}(t-y)^{2}\,dt \Biggr)^{1/2} \Biggr\} \omega (f,\delta ) \\ &\quad \leq \biggl\{ 1+\frac{1}{\delta } \bigl(\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})} \bigl((t-y)^{2};y,b\bigr) \bigr)^{1/2} \biggr\} \omega (f, \delta ) \\ &\quad \leq \biggl\{ 1+\frac{1}{\delta }\frac{1}{\nu _{n}}\sqrt{(\mu _{n}- \nu _{n})^{2}y^{2}+ \biggl( \biggl(4+ \frac{1}{b-1} \biggr)\mu _{n}-3\nu _{n} \biggr)y+ \frac{10}{3}} \biggr\} \omega \biggl(f,\frac{1}{\delta } \biggr), \end{aligned}

where, taking $$\delta =\frac{1}{\nu _{n}}$$, we get (3.1). □

Let $$a_{1},a_{2} > 0$$ be fixed. We now consider the following space of Lipschitz type (see ):

$$\mathrm{Lip}_{M}^{(a_{1},a_{2})}(r):= \biggl\{ f\in C[0,\infty ): \bigl\vert f( \kappa )-f(\lambda ) \bigr\vert \leq M\frac{ \vert \kappa -\lambda \vert ^{r}}{(\kappa +a_{1}\lambda ^{2}+a_{2}\lambda )^{r/2}}; \lambda ,\kappa >0 \biggr\} ,$$
(3.3)

where M is a positive constant and $$0< r\leq 1$$.

### Theorem 3.3

Let$$f\in \mathrm{Lip}_{M}^{(a_{1},a_{2})}(r)$$and$$r\in (0,1]$$, then$$\forall y>0$$, we have

$$\bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert \leq M \biggl( \frac{\xi _{\mu _{n},\nu _{n}}(y)}{(a_{1}y^{2}+a_{2}y)} \biggr)^{r/2}.$$

### Proof

Since

$$\bigl\vert f(t)-f(y) \bigr\vert \leq M\frac{ \vert t-y \vert ^{r}}{(t+a_{1}y^{2}+a_{2}y)^{r/2}},$$

one has

$$\bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert \leq M \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})} \biggl( \frac{ \vert t-y \vert ^{r}}{(t+a_{1}y^{2}+a_{2}y)^{r/2}};y \biggr).$$

Applying Hölder’s inequality with $$p=\frac{2}{r}$$ and $$\frac{2}{2-r}$$, we find that

$$\bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert \leq M \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})} \biggl( \frac{(t-y)^{2}}{t+a_{1}y^{2}+a_{2}y};y \biggr)^{r/2}.$$

Since $$f\in \mathrm{Lip}_{M}^{(a_{1},a_{2})}(r)$$ and $$\frac{1}{t+a_{1}y^{2}+a_{2}y}<\frac{1}{a_{1}y^{2}+a_{2}y}$$, $$\forall y\in (0,\infty )$$, we have

\begin{aligned} \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert & \leq M \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})} \biggl( \frac{(t-y)^{2}}{a_{1}y^{2}+a_{2}y};y \biggr)^{r/2} \\ &\leq \frac{M}{(a_{1}y^{2}+a_{2}y)^{r/2}}\mathcal{Q}_{n,b}^{(\mu _{n}, \nu _{n})} \bigl((t-y)^{2};y \bigr)^{r/2} \\ &= M \biggl(\frac{\xi _{\mu _{n},\nu _{n}}(y)}{(a_{1}y^{2}+a_{2}y)} \biggr)^{r/2}. \end{aligned}

Our proof is now completed. □

We denote the space of all functions h on $$[0,1)$$ which are real-valued, uniformly continuous, as well as bounded by $$\tilde{C}_{B}[0,\infty )$$ and endow it with the norm $$\Vert h \Vert _{\infty }=\sup_{y\in [0,1)}|h(y)|$$. Further, we obtain a local direct estimate for the operators (1.3), using the Lipschitz maximal function of order r introduced by Lenze  as:

$$\tilde{\omega }_{r}(h,y)=\sup_{t\ne y, t\in [0,\infty )} \frac{ \vert h(t)-h(y) \vert }{ \vert t-y \vert ^{r}},$$
(3.4)

where $$y \in [0,1)$$ and $$r\in (0,1]$$.

### Theorem 3.4

Let$$f\in \tilde{C}_{B}[0,\infty )$$and$$0< r\le 1$$, then$$\forall y\in [0,\infty )$$

$$\bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert \leq \tilde{\omega }_{r}(f,y) \bigl(\xi _{\mu _{n},\nu _{n}}(y) \bigr)^{r}.$$

### Proof

By equation (3.4),

$$\bigl\vert f(t)-f(y) \bigr\vert \leq \tilde{\omega _{r}}(f,y) \vert t-y \vert ^{r}.$$

Applying $$\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}$$ on both sides of the above inequality, then using Lemma 2.1, as well as Hölder’s inequality with $$p=2/r$$, $$q=2/(2-r)$$, we obtain

\begin{aligned} \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert & \leq \tilde{\omega }_{r}(f,y)\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})} \bigl( \vert t-y \vert ^{r};y\bigr) \\ & \leq \tilde{\omega }_{r}(f,y) \bigl(\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})} \bigl((t-y)^{2};y\bigr) \bigr)^{r/2} \\ & =\tilde{\omega }_{r}(f,y) \bigl(\xi _{\mu _{n},\nu _{n}}(y) \bigr)^{r/2}. \end{aligned}

Thus, we have our desired result. □

The Peetre’s K-functional is given by

$$K(g,\delta )=\inf_{h\in \tilde{C}_{B}^{2}[0,\infty )} \bigl\{ \Vert g-h \Vert _{\infty }+ \delta \Vert h \Vert _{\tilde{C}_{B}^{2}} \bigr\} ,$$

where $$\tilde{C}_{B}^{2}[0,\infty )= \{ h\in \tilde{C}_{B}[0,\infty ):h^{ \prime },h^{\prime \prime }\in \tilde{C}_{B}[0,\infty ) \}$$ with the norm $$\Vert h\Vert _{\tilde{C}_{B}^{2}}=\Vert h\Vert _{\infty }+\Vert h^{ \prime }\Vert _{\infty }+\Vert h^{\prime \prime }\Vert _{\infty }$$. Also, the inequality

$$K(g,\delta )\leq M \bigl( \omega _{2}(g,\sqrt{\delta })+\min \{ 1, \delta \} \Vert g \Vert _{\infty } \bigr)$$

holds for all $$\delta >0$$, where $$\omega _{2}$$ is the second-order modulus of smoothness of $$g\in \tilde{C}_{B}[0,\infty )$$, which is defined by

$$\omega _{2}(g,\delta )=\sup_{0< \vert h \vert \leq \delta }\sup _{y,y+2h \in {}[ 0,\infty )} \bigl\vert g(y+2h)-2g(y+h)+g(y) \bigr\vert .$$

### Theorem 3.5

If$$f\in \tilde{C}_{B}[0,\infty )$$, then

$$\bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert \leq 4K(f,\zeta_{\mu _{n},\nu _{n}}(y))+ \omega (f,\chi_{\mu _{n},\nu _{n}}(y)),$$

where$$\zeta_{\mu _{n},\nu _{n}}(y)=(\xi_{\mu _{n},\nu _{n}}(y)+\chi^{2}_{\mu _{n},\nu _{n}}(y))/4$$. Furthermore,

$$\bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert \leq M \bigl( \omega _{2} \bigl( f, \sqrt{\zeta_{\mu _{n},\nu _{n}}(y)}\bigr) +\min \{ 1,\zeta_{\mu _{n},\nu _{n}}(y) \} \Vert f \Vert _{\infty } \bigr) +\omega \bigl( f, \bigl\vert \chi_{\mu _{n},\nu _{n}}(y)\bigr\vert \bigr) .$$

### Proof

For $$f\in \tilde{C}_{B}[0,\infty )$$, we define the auxiliary operator as follows:

$$\tilde{\mathcal{Q}}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)= \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f \biggl( \frac{\mu _{n}}{\nu _{n}}y+ \frac{3}{2\nu _{n}} \biggr)+f(y).$$
(3.5)

After taking the absolute value of both sides,

\begin{aligned} \bigl\vert \tilde{\mathcal{Q}}_{n,b}^{(\mu _{n},\nu _{n})}(f;y) \bigr\vert & \leq \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y) \bigr\vert + \biggl\vert f \biggl( \frac{\mu _{n}}{\nu _{n}}y+\frac{3}{2\nu _{n}} \biggr) \biggr\vert + \bigl\vert f(y) \bigr\vert \\ &\leq \Vert f \Vert _{\infty } \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(1;y,b) \bigr\vert + \Vert f \Vert _{\infty }+ \Vert f \Vert _{\infty } \\ &\leq 3 \Vert f \Vert _{\infty }. \end{aligned}
(3.6)

By Lemma 2.1, we have $$\tilde{\mathcal{Q}}_{n,b}^{(\mu _{n},\nu _{n})}(t;y)=y$$, and therefore $$\tilde{\mathcal{Q}}_{n,b}^{(\mu _{n},\nu _{n})}(t-y;y)=0$$.

Let $$g\in \tilde{C}_{B}^{2}[0,\infty )$$, using Taylor’s theorem, we can write

$$g(t)=g(y)+g^{\prime }(y) (t-y)+ \int _{y}^{t}(t-u)g^{\prime \prime }(u)\,du.$$

Applying operator $$\tilde{\mathcal{Q}}_{n,b}^{(\mu _{n},\nu _{n})}$$ to the above equation, we get

\begin{aligned} &\tilde{\mathcal{Q}}_{n,b}^{(\mu _{n},\nu _{n})}(g;y)-g(y) \\ &\quad =\tilde{\mathcal{Q}}_{n,b}^{(\mu _{n},\nu _{n})} \biggl( \int _{y}^{t}(t-u)g^{ \prime \prime }(u)\,du;y \biggr) \\ &\quad =\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})} \biggl( \int _{y}^{t}(t-u)g^{ \prime \prime }(u)\,du;y \biggr)- \int _{y}^{\frac{\mu _{n}}{\nu _{n}}y+ \frac{3}{2\nu _{n}}} \biggl(\frac{\mu _{n}}{\nu _{n}}y+\frac{3}{2\nu _{n}}-u \biggr)g^{\prime \prime }(u)\,du. \end{aligned}

Now taking the absolute value of both sides, we obtain

\begin{aligned} & \bigl\vert \tilde{\mathcal{Q}}_{n,b}^{(\mu _{n},\nu _{n})}(g;y)-g(y) \bigr\vert \\ &\quad \leq \biggl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})} \biggl( \int _{y}^{t}(t-u)g^{ \prime \prime }(u)\,du;y \biggr) \biggr\vert + \biggl\vert \int _{y}^{ \frac{\mu _{n}}{\nu _{n}}y+\frac{3}{2\nu _{n}}} \biggl(\frac{\mu _{n}}{\nu _{n}}y+ \frac{3}{2\nu _{n}}-u \biggr)g^{\prime \prime }(u)\,du \biggr\vert \\ &\quad \leq \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})} \biggl( \biggl\vert \int _{y}^{t} \bigl\vert (t-u) \bigr\vert \bigl\vert g^{\prime \prime }(u) \bigr\vert \,du \biggr\vert ;y \biggr) + \biggl\vert \biggl( \int _{y}^{ \frac{\mu _{n}}{\nu _{n}}y+\frac{3}{2\nu _{n}}} \biggl\vert \frac{\mu _{n}}{\nu _{n}}y+ \frac{3}{2\nu _{n}}-u \biggr\vert \bigl\vert g^{\prime \prime }(u) \bigr\vert \,du \biggr) \biggr\vert \\ &\quad \leq \bigl\Vert g^{\prime \prime } \bigr\Vert _{\infty } \biggl\{ \mathcal{Q}_{n,b}^{( \mu _{n},\nu _{n})} \biggl( \biggl\vert \int _{y}^{t} \bigl\vert (t-u) \bigr\vert \,du \biggr\vert ;y \biggr) + \biggl\vert \biggl( \int _{y}^{\frac{\mu _{n}}{\nu _{n}}y+\frac{3}{2\nu _{n}}} \biggl\vert \frac{\mu _{n}}{\nu _{n}}y+ \frac{3}{2\nu _{n}}-u \biggr\vert \,du \biggr) \biggr\vert \biggr\} . \end{aligned}

Therefore, by using the norm on g, we have

\begin{aligned} \bigl\vert \tilde{\mathcal{Q}}_{n,b}^{(\mu _{n},\nu _{n})}(g;y)-g(y) \bigr\vert &\leq \Vert g \Vert _{\tilde{C}_{B}^{2}} \biggl\{ \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})} \bigl((t-y)^{2};y \bigr) + \biggl(\frac{\mu _{n}}{\nu _{n}}y+ \frac{3}{2\nu _{n}}-y \biggr)^{2} \biggr\} \\ &\leq \Vert g \Vert _{\tilde{C}_{B}^{2}} \bigl\{ \mathcal{Q}_{n,b}^{(\mu _{n}, \nu _{n})} \bigl((t-y)^{2};y \bigr)+ \bigl(\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})} (t-y;y ) \bigr)^{2} \bigr\} \\ &\leq \Vert g \Vert _{\tilde{C}_{B}^{2}} \bigl\{ \xi _{\mu _{n},\nu _{n}}(y)+ \chi ^{2}_{\mu _{n},\nu _{n}}(y) \bigr\} . \end{aligned}
(3.7)

Now, using the definition of auxiliary operators (3.5), we get

\begin{aligned} & \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert \\ &\quad \leq \biggl\vert \tilde{\mathcal{Q}}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y)+f \biggl(\frac{\mu _{n}}{\nu _{n}}y+\frac{3}{2\nu _{n}} \biggr)-f(y) \biggr\vert \\ &\quad \leq \bigl\vert \tilde{\mathcal{Q}}_{n,b}^{(\mu _{n},\nu _{n})}(f-g;y) \bigr\vert + \bigl\vert \tilde{\mathcal{Q}}_{n,b}^{(\mu _{n},\nu _{n})}(g;y)-g(y) \bigr\vert \\ &\qquad {}+ \bigl\vert g(y)-f(y) \bigr\vert + \biggl\vert f \biggl( \frac{\mu _{n}}{\nu _{n}}y+ \frac{3}{2\nu _{n}} \biggr)-f(y) \biggr\vert . \end{aligned}

Combining (3.6) and (3.7) with the above equation, we get

\begin{aligned} & \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert \\ &\quad \leq 3 \Vert f-g \Vert _{\infty }+ \Vert g \Vert _{\tilde{C}_{B}^{2}} \bigl\{ \xi _{\mu _{n}, \nu _{n}}(y)+\chi ^{2}_{\mu _{n},\nu _{n}}(y) \bigr\} + \Vert f-g \Vert _{\infty }+\omega \biggl(f, \biggl\vert \frac{\mu _{n}}{\nu _{n}}y+ \frac{3}{2\nu _{n}}-y \biggr\vert \biggr) \\ &\quad \leq 4 \Vert f-g \Vert _{\infty }+ \Vert g \Vert _{\tilde{C}_{B}^{2}}4 \zeta _{\mu _{n}, \nu _{n}}(y)+ \omega \bigl(f, \bigl\vert \chi _{\mu _{n},\nu _{n}}(y) \bigr\vert \bigr), \end{aligned}

and after taking the infimum on the right-hand side over all $$g\in {\tilde{C}_{B}^{2}}$$, we have

\begin{aligned} & \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert \\ &\quad \leq 4K \bigl(f,\zeta _{\mu _{n},\nu _{n}}(y) \bigr)+\omega \bigl(f, \bigl\vert \chi _{\mu _{n},\nu _{n}}(y) \bigr\vert \bigr) \\ &\quad \leq M \bigl(\omega _{2} \bigl(f,\sqrt{\zeta _{\mu _{n},\nu _{n}}(y)} \bigr)+\min \bigl\{ 1,\zeta _{\mu _{n},\nu _{n}}(y) \bigr\} \Vert f \Vert _{ \infty } \bigr) +\omega \bigl(f, \bigl\vert \chi _{\mu _{n},\nu _{n}}(y) \bigr\vert \bigr). \end{aligned}

This completes the proof of the theorem. □

### Theorem 3.6

Let$$f\in \tilde{C}^{1}_{B}[0,\infty )$$, then$$\forall y\geq 0$$and$$\delta >0$$,

$$\bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert \leq \bigl\{ \bigl\vert f^{\prime }(y) \bigr\vert +2\omega \bigl(f^{\prime },\delta _{n}(y)\bigr) \bigr\} .$$

### Proof

Since $$f\in \tilde{C}^{1}_{B}[0,\infty )$$, we can write

$$f(t)-f(y)=f^{\prime }(y) (t-y)+ \int _{y}^{t} \bigl(f^{\prime }(u)-f^{ \prime }(y) \bigr)\,du.$$
(3.8)

Now, using the well-known property of the modulus of continuity for $$\delta >0$$ and $$f\in \tilde{C}^{1}_{B}[0,\infty )$$,

$$\bigl\vert f^{\prime }(u)-f^{\prime }(y) \bigr\vert \leq \biggl( \frac{ \vert u-y \vert }{\delta }+1 \biggr)\omega \bigl(f^{\prime },\delta \bigr),$$

hence

$$\biggl\vert \int _{y}^{t} \bigl(f^{\prime }(u)-f^{\prime }(y) \bigr)\,du \biggr\vert \leq \biggl(\frac{(t-y)^{2}}{\delta }+ \vert t-y \vert \biggr)\omega \bigl(f^{\prime },\delta \bigr)$$

Therefore, from (3.8) and the above equation, we have

\begin{aligned} & \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert \\ &\quad \leq \bigl\vert f^{\prime }(y) \bigr\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})} \bigl( \vert t-y \vert ;y\bigr)+ \biggl(\frac{1}{\delta } \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})} \bigl((t-y)^{2};y\bigr)+ \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}\bigl( \vert t-y \vert ;y \bigr) \biggr)\omega \bigl(f^{\prime }, \delta \bigr). \end{aligned}

After applying the Cauchy–Schwarz inequality, we get

\begin{aligned} & \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert \\ &\quad \leq \bigl( \bigl\vert f^{\prime }(y) \bigr\vert +\omega \bigl(f^{\prime },\delta \bigr) \bigr) \sqrt{\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})} \bigl((t-y)^{2};y\bigr)}\sqrt{\mathcal{Q}_{n,b}^{( \mu _{n},\nu _{n})}(1;y)} \\ &\qquad {}+ \biggl(\frac{1}{\delta }\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})} \bigl((t-y)^{2};y\bigr) \biggr)\omega \bigl(f^{\prime },\delta \bigr) \\ &\quad = \bigl( \bigl\vert f^{\prime }(y) \bigr\vert +\omega \bigl(f^{\prime },\delta \bigr) \bigr)\delta _{n}(y)+ \biggl(\frac{\delta _{n}^{2}(y)}{\delta } \biggr) \omega \bigl(f^{\prime },\delta \bigr). \end{aligned}

Choosing $$\delta =\delta _{n}(y)$$, we get our desired result. □

For $$f\in \tilde{C}_{B}[0,\infty )$$, the Ditzian–Totik modulus of smoothness  of the first order is given by

$$\omega _{\varphi }(f,\delta )=\sup_{0< h\leq \delta } \biggl\{ \biggl\vert f \biggl(y+\frac{h\varphi (y)}{2} \biggr)-f \biggl(y- \frac{h\varphi (y)}{2} \biggr) \biggr\vert ; y\pm \frac{h\varphi (y)}{2}\in [0,\infty ) \biggr\} ,$$

and an appropriate Peetre’s K-functional is defined by

$$K_{\varphi }(f,\delta )=\inf_{g\in W_{\varphi }[0,\infty )} \bigl\{ \Vert f-g \Vert _{\infty }+\delta \Vert \varphi g^{\prime } \Vert _{\infty } \bigr\} ,\quad \delta >0,$$

where $$W_{\varphi }[0,\infty ):=\{g: g\in \mathrm{AC}_{\mathrm{loc}}[0,\infty ), \Vert \varphi g^{\prime } \Vert _{\infty }<\infty \}$$ where $$g\in \mathrm{AC}_{\mathrm{loc}}[0,\infty )$$ means g is absolutely continuous on every compact subset $$[a,b]$$ of $$[0,\infty )$$. It is known from  that there exists a constant M such that

$$M^{-1}\omega _{\varphi }(f,\delta )\leq K_{\varphi }(f,\delta )\leq M \omega _{\varphi }(f,\delta ).$$
(3.9)

Now, we find the order of approximation of the sequence of operators (1.3) by means of Ditzian–Totik modulus of smoothness.

### Theorem 3.7

For any$$f\in \tilde{C}_{B}[0,\infty )$$and$$y\in [0,\infty )$$,

$$\bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert \leq M \omega _{\varphi } \biggl(f,\frac{\delta _{n}(y)}{\sqrt{y}} \biggr).$$

### Proof

Let $$\varphi (y)=\sqrt{y}$$, then by Taylor’s theorem, for any $$g\in W_{\varphi }[0,\infty )$$, we get

$$g(t)=g(y)+ \int _{y}^{t}g^{\prime }(u)\,du =g(y)+ \int _{y}^{t} \frac{g^{\prime }(u)\varphi (u)}{\varphi (u)}\,du,$$

therefore,

\begin{aligned} \bigl\vert g(t)-g(y) \bigr\vert &= \bigl\Vert \varphi g^{\prime } \bigr\Vert _{\infty } \biggl\vert \int _{y}^{t} \frac{1}{\varphi (u)\,du} \biggr\vert \\ &=2 \bigl\Vert \varphi g^{\prime } \bigr\Vert _{\infty } \vert \sqrt{t}-\sqrt{y} \vert \\ &=2 \bigl\Vert \varphi g^{\prime } \bigr\Vert _{\infty } \frac{ \vert t-y \vert }{\sqrt{t}+\sqrt{y}}, \end{aligned}

which gives

$$\bigl\vert g(t)-g(y) \bigr\vert \leq 2 \bigl\Vert \varphi g^{\prime } \bigr\Vert _{\infty }\frac{ \vert t-y \vert }{\sqrt{y}} =2 \bigl\Vert \varphi g^{\prime } \bigr\Vert _{\infty }\frac{ \vert t-y \vert }{\varphi (y)}.$$

Using Lemma 2.1 and the above equation, for any $$g\in W_{\varphi }[0,\infty )$$, we get

\begin{aligned} \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert & \leq \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f-g;y) \bigr\vert + \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(g;y)-g(y) \bigr\vert + \bigl\vert g(y)-f(y) \bigr\vert \\ &\leq 2 \Vert f-g \Vert _{\infty }+ \frac{2 \Vert \varphi g^{\prime } \Vert _{\infty }}{\varphi (y)}\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}\bigl( \vert t-y \vert ;y\bigr). \end{aligned}

Applying the Cauchy–Schwarz inequality yields

\begin{aligned} \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert & \leq 2 \Vert f-g \Vert _{\infty }+ \frac{2 \Vert \varphi g^{\prime } \Vert _{\infty }}{\varphi (y)}\sqrt {\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})} \bigl((t-y)^{2};y\bigr)} \\ &=2 \Vert f-g \Vert _{\infty }+ \frac{2 \Vert \varphi g^{\prime } \Vert _{\infty }}{\varphi (y)}\delta _{n}(y). \end{aligned}

Taking infimum on the right-hand side over all $$g\in W_{\varphi }[0,\infty )$$, we get

$$\bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert \leq 2K_{\varphi } \biggl(f,\frac{\delta _{n}(y)}{\sqrt{y}} \biggr),$$

which leads to the required result with the help of the relation between Peetre’s K-functional and Ditzian–Totik modulus of smoothness as given by the relation (3.9). □

## Approximation results in weighted spaces

Let $$\nu >0$$. We denote $$C_{\nu }[0,\infty ):=\{f\in C[0,\infty ):|f(t)|\leq M_{f}(1+t^{\nu }), \forall t\geq 0\}$$ equipped with the norm

$$\Vert f \Vert _{\nu }=\sup_{t\in [0,\infty )} \frac{ \vert f(t) \vert }{1+t^{\nu }}.$$
(4.1)

Further, let $$C_{2}^{*}[0,\infty )$$ be the subspace of $$C_{2}[0,\infty )$$ consisting of functions f such that $$\lim_{t\to \infty }\frac{f(t)}{1+t^{2}}$$ exists.

### Theorem 4.1

For each$$f\in C_{2}^{*}[0,\infty )$$and$$r>0$$, the following relation holds:

$$\lim_{n\to \infty }\sup_{x\in [0,\infty )} \frac{ \vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \vert }{(1+y^{2})^{1+r}}=0.$$

### Proof

Let $$y_{0}>0$$ be arbitrary but fixed, then by (4.1), we can write

\begin{aligned} & \sup_{y\in [0,\infty )} \frac{ \vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \vert }{(1+y^{2})^{1+r}} \\ &\quad \leq \sup_{y\leq y_{0}} \frac{ \vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \vert }{(1+y^{2})^{1+r}} +\sup _{y>y_{0}} \frac{ \vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \vert }{(1+y^{2})^{1+r}} \\ &\quad \leq \sup_{y\leq y_{0}}\bigl\{ \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert \bigr\} +\sup_{y> y_{0}} \frac{ \vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)+f(y) \vert }{(1+y^{2})^{1+r}}. \end{aligned}

Since $$|f(t)|\leq \Vert f \Vert _{2}(1+y^{2})$$, we get

\begin{aligned} &\sup_{y\in [0,\infty )} \frac{ \vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \vert }{(1+y^{2})^{1+r}} \end{aligned}
(4.2)
\begin{aligned} & \quad \leq \bigl\Vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\Vert _{C[0,y_{0}]}+ \Vert f \Vert _{2}\sup _{y> y_{0}} \frac{ \vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(1+t^{2};y) \vert }{(1+y^{2})^{1+r}} +\sup_{y>y_{0}} \frac{ \Vert f(y) \Vert }{(1+y^{2})^{1+r}} \\ &\quad =I_{1}+I_{2}+I_{3} \quad (\text{say}). \end{aligned}
(4.3)

By Korovkin’s theorem, we can see that the sequence of operators $$\{\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)\}$$ converges uniformly to the function f on every closed interval $$[0,a]$$ as $$n\to \infty$$, (cf. [12, p. 149]). Therefore, for a given $$\epsilon >0$$, $$\exists n_{1}\in \mathbb{N}$$ such that

$$I_{1}= \bigl\Vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\Vert _{C[0,y_{0}]}< \frac{\epsilon }{3},\quad \forall n\geq n_{1}.$$
(4.4)

By using Lemma 2.1, we can find $$n_{2}\in \mathbb{N}$$ such that

\begin{aligned} \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}\bigl(1+t^{2};y \bigr)- \bigl(1+y^{2}\bigr) \bigr\vert & < \frac{\epsilon }{3 \Vert f \Vert _{2}},\quad \forall n\geq n_{2}, \\ \text{or}\quad \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}\bigl(1+t^{2};y \bigr) &< \bigl(1+y^{2}\bigr)+ \frac{\epsilon }{3 \Vert f \Vert _{2}}, \quad \forall n \geq n_{2}. \end{aligned}

Hence

\begin{aligned} I_{2} & = \Vert f \Vert _{2}\sup _{y> y_{0}} \frac{ \vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(1+t^{2};y) \vert }{(1+y^{2})^{1+r}} \\ & < \Vert f \Vert _{2}\sup_{y> y_{0}} \frac{1}{(1+y^{2})^{1+r}} \biggl( \bigl(1+y^{2}\bigr)+ \frac{\epsilon }{3 \Vert f \Vert _{2}} \biggr) \\ & < \Vert f \Vert _{2}\sup_{y> y_{0}} \biggl( \frac{1}{(1+y^{2})^{r}} \biggr)+\frac{\epsilon }{3} \\ & < \frac{ \Vert f \Vert _{2}}{(1+y_{0}^{2})^{r}}+\frac{\epsilon }{3} , \quad \forall n\geq n_{2}. \end{aligned}
(4.5)

Now, using (4.1),

$$I_{3}=\sup_{y> y_{0}}\frac{ \vert f(y) \vert }{(1+y^{2})^{1+r}} \leq \frac{ \Vert f \Vert _{2}}{(1+y_{0}^{2})^{r}}.$$
(4.6)

Let us denote $$n_{0}=\max \{n_{1},n_{2}\}$$, then by (4.4), (4.5), and (4.6), we get

$$I_{1}+I_{2}+I_{3}< 2 \frac{ \Vert f \Vert _{2}}{(1+y_{0}^{2})^{r}}+ \frac{2\epsilon }{3},\quad \forall n\geq n_{0}.$$
(4.7)

Choose $$y_{0}$$ so large that

$$2\frac{ \Vert f \Vert _{2}}{(1+y_{0}^{2})^{r}}< \frac{\epsilon }{3}.$$
(4.8)

Then, combining (4.3), (4.7), and (4.8), we obtain

$$\sup_{y\in [0,\infty )} \frac{ \vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \vert }{(1+y^{2})^{1+r}}< \epsilon ,\quad \forall n \geq n_{0}.$$

Hence, the proof is completed. □

Now, we will obtain the rate of convergence of the operators $$\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)$$ defined by (1.3) for the functions having derivatives of bounded variation. Let $$\operatorname{DBV}[0,\infty )$$ be the space of functions in $$C_{2}[0,\infty )$$, which have the derivative of bounded variation on every finite subinterval of $$[0,\infty )$$. Here, we show at the point y, where $$f^{\prime }(y+)$$ and $$f^{\prime }(y-)$$ exist, the operators $$\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)$$ converge to the function $$f(y)$$. A function $$f\in \operatorname{DBV}[0,\infty )$$ can be represented as

$$f(y)= \int _{0}^{y}g(t)\,dt+f(0),$$

where g denotes a function of bounded variation on every finite subinterval $$[0,\infty )$$. Many researchers studied in this direction and their work pertaining to this area is described in the papers [2, 11, 19], etc.

In order to study the order of convergence of the operators $$\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)$$ for the functions having a derivative of bounded variation, we rewrite the operator (1.3) as follows:

$$\mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)= \int _{0}^{\infty }W(t,y)f(t)\,dt,$$
(4.9)

where $$W(t,y)$$ is a kernel given by

$$W(t,y)=\nu _{n}e^{-1} \biggl( 1-\frac{1}{b} \biggr)^{(b-1)\mu _{n}y} \sum_{l=0}^{\infty } \frac{C_{l}^{(b)}(-(b-1)\mu _{n}y)}{l!}\chi _{I}(t),$$

$$\chi _{I}(t)$$ being the characteristic function of $$I= [\frac{l}{\nu _{n}},\frac{l+1}{\nu _{n}} ]$$.

### Lemma 4.2

Let for all$$x>0$$and sufficiently largen,

1. (1)

$$\lambda _{\mu _{n},\nu _{n}}(t,y)=\int _{0}^{t}W(u,y)\,du\leq \frac{\xi _{\mu _{n},\nu _{n}}^{2}(x)}{(y-t)^{2}}$$, $$0\leq t< y$$,

2. (2)

$$1-\lambda _{\mu _{n},\nu _{n}}(t,y)=\int _{t}^{\infty }W(u,y)\,du\leq \frac{\xi _{\mu _{n},\nu _{n}}^{2}(y)}{(t-y)^{2}}$$, $$y\leq t< \infty$$.

### Proof

Using Lemma 2.1 and the definition of the kernel, we get

\begin{aligned} \lambda _{\mu _{n},\nu _{n}}(t,y)& = \int _{0}^{t}W(u,y)\,du \\ & \leq \int _{0}^{t} \biggl( \frac{y-u}{y-t} \biggr) ^{2}W(u,y)\,du \\ & \leq \frac{1}{(y-t)^{2}} \int _{0}^{t}(u-y)^{2}W(u,y)\,du. \end{aligned}

Hence, we have

\begin{aligned} \begin{aligned} \lambda _{\mu _{n},\nu _{n}}(t,y)& \leq \frac{1}{(y-t)^{2}} \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})} \bigl((u-y)^{2};y,b\bigr) \\ & \leq \frac{1}{(y-t)^{2}}\xi _{\mu _{n},\nu _{n}}^{2}(y). \end{aligned} \end{aligned}

In the same fashion, we can prove the other inequality, therefore, we omit the details. □

Let $$\bigvee_{a}^{b}f$$ be the total variation of f on $$[a,b]$$, i.e.,

$$\bigvee_{a}^{b}f=V \bigl(f;[a,b]\bigr)=\sup_{P\in \mathbb{P}} \Biggl(\sum _{i=1}^{n} \bigl\vert f(y_{i})-f(y_{i-1}) \bigr\vert \Biggr),$$
(4.10)

where $$\mathbb{P}$$ is the set of all partitions $$P=\{a=y_{0},y_{1},\dots ,y_{n}=b\}$$ of $$[a,b]$$, whick also has the property

$$\bigvee_{a}^{b}f=\bigvee _{a}^{c}f+\bigvee_{c}^{b}f.$$

Let

$$f_{y}^{\prime }(t)=\textstyle\begin{cases} f^{\prime }(t)-f^{\prime }(y-), & 0\leq t< y, \\ 0, & t=y, \\ f^{\prime }(t)-f^{\prime }(y+), & y< t< \infty .\end{cases}$$
(4.11)

### Theorem 4.3

Let$$f\in \operatorname{DBV}[0,\infty )$$, $$y>0$$, andnbe sufficiently large, then we get

\begin{aligned} & \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert \\ &\quad \leq \biggl\vert \frac{1}{2} \bigl( f^{\prime }(y+)+f^{\prime }(y-) \bigr) \biggr\vert \bigl\vert \psi _{\mu _{n},\nu _{n}}(y) \bigr\vert + \frac{\xi _{\mu _{n},\nu _{n}}^{2}(y)}{y}\sum_{k=1}^{[\sqrt{n}]} \Biggl( \bigvee_{y-y/k}^{y+y/k}f_{y}^{\prime } \Biggr) + \frac{y}{\sqrt{n}} \Biggl( \bigvee _{y-y/k}^{y+y/k}f_{y}^{\prime } \Biggr) \\ & \qquad {}+\frac{\xi _{\mu _{n},\nu _{n}}^{2}(y)}{y^{2}} \bigl\vert f(2y)-f(y)-yf^{ \prime }(y+) \bigr\vert + \biggl( \frac{M_{f}+ \vert f(y) \vert }{y^{2}}+4M_{f} \biggr) \xi _{\mu _{n},\nu _{n}}^{2}(y) \\ & \qquad {}+ \bigl\vert f^{\prime }(y+) \bigr\vert \bigl\vert \psi _{\mu _{n},\nu _{n}}(y) \bigr\vert + \biggl\vert \frac{1}{2} \bigl( f^{\prime }(y+)-f^{\prime }(y-) \bigr) \biggr\vert \xi _{ \mu _{n},\nu _{n}}(y). \end{aligned}

### Proof

By (4.11), we obtain

\begin{aligned} f^{\prime }(t)& =\frac{1}{2}\bigl(f^{\prime }(y+)+f^{\prime }(y-) \bigr)+f_{y}^{ \prime }(t)+\frac{1}{2} \bigl(f^{\prime }(y+)-f^{\prime }(y-) \bigr)\operatorname{sgn}(t-y) \\ & \quad {}+\delta _{y}(t) \biggl( f^{\prime }(t)- \frac{1}{2}\bigl(f^{ \prime }(y+)+f^{\prime }(y-)\bigr) \biggr) , \end{aligned}
(4.12)

where

$$\delta _{y}(t)=\textstyle\begin{cases} 1, & t=y, \\ 0, & t\neq y.\end{cases}$$

Now using Lemma 2.1, equations (4.9) and (4.12), we get

\begin{aligned} \begin{aligned} & \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \\ &\quad = \int _{0}^{\infty }\bigl(f(t)-f(y)\bigr)W(t,y)\,dt \\ &\quad = \int _{0}^{\infty } \biggl( \int _{y}^{t}f^{\prime }(u)\,du \biggr) W(t,y)\,dt \\ &\quad = \int _{0}^{\infty } \biggl[ \int _{y}^{t} \biggl\{ \frac{1}{2} \bigl(f^{ \prime }(y+)+f^{\prime }(y-)\bigr)+f_{y}^{\prime }(u)+ \frac{1}{2}\bigl(f^{ \prime }(y+)-f^{\prime }(y-)\bigr) \operatorname{sgn}(u-y) \\ & \qquad{} +\delta _{y}(u) \biggl(f^{\prime }(u)- \frac{1}{2}\bigl(f^{\prime }(y+)+f^{ \prime }(y-)\bigr) \biggr) \biggr\} \,du \biggr]W(t,y)\,dt. \end{aligned} \end{aligned}

Since $$\int _{y}^{t}\delta _{y}(u)\,du=0$$, we have

\begin{aligned} & \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \\ &\quad =\frac{1}{2}\bigl(f^{\prime }(y+)+f^{\prime }(y-)\bigr) \int _{0}^{\infty }(t-y)W(t,y)\,dt+ \int _{0}^{\infty } \biggl( \int _{y}^{t}f_{y}^{\prime }(u)\,du \biggr) W(t,y)\,dt \\ & \qquad{} +\frac{1}{2}\bigl(f^{\prime }(y+)-f^{\prime }(y-) \bigr) \int _{0}^{ \infty } \vert t-y \vert W(t,y)\,dt. \end{aligned}
(4.13)

Now, we break the second term on the right-hand side of the above equation as follows:

\begin{aligned} & \int _{0}^{\infty } \biggl( \int _{y}^{t}f_{y}^{\prime }(u)\,du \biggr) W(t,y)\,dt \\ &\quad =- \int _{0}^{y} \biggl( \int _{t}^{y}f_{y}^{\prime }(u)\,du \biggr) W(t,y)\,dt+ \int _{y}^{\infty } \biggl( \int _{y}^{t}f_{y}^{\prime }(u)\,du \biggr) W(t,y)\,dt \\ &\quad =-I_{1}+I_{2}, \end{aligned}

where

\begin{aligned} I_{1}& = \int _{0}^{y} \biggl( \int _{t}^{y}f_{y}^{\prime }(u)\,du \biggr) W(t,y)\,dt, \\ I_{2}& = \int _{y}^{\infty } \biggl( \int _{y}^{t}f_{y}^{\prime }(u)\,du \biggr) W(t,y)\,dt. \end{aligned}

Taking the absolute value on both sides of (4.13), we have

\begin{aligned} & \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert \\ &\quad \leq \biggl\vert \frac{1}{2}\bigl(f^{\prime }(y+)+f^{\prime }(y-) \bigr) \biggr\vert \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(t-y;y) \bigr\vert + \vert I_{1} \vert + \vert I_{2} \vert \\ &\qquad {} + \biggl\vert \frac{1}{2}\bigl(f^{\prime }(y+)-f^{\prime }(y-) \bigr) \biggr\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}\bigl( \vert t-y \vert ;y\bigr). \end{aligned}

After applying the Cauchy–Schwarz inequality, we obtain

\begin{aligned} \begin{aligned}[b] & \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert \\ &\quad \leq \biggl\vert \frac{1}{2}\bigl(f^{\prime }(y+)+f^{\prime }(y-) \bigr) \biggr\vert \bigl\vert \psi _{\mu _{n},\nu _{n}}(y) \bigr\vert + \vert I_{1} \vert + \vert I_{2} \vert \\ &\qquad {} + \biggl\vert \frac{1}{2}\bigl(f^{\prime }(y+)-f^{\prime }(y-) \bigr) \biggr\vert \sqrt{\mathcal{Q}_{n,b}^{( \mu _{n},\nu _{n})} \bigl((t-y)^{2};y\bigr)} \\ &\quad = \biggl\vert \frac{1}{2}\bigl(f^{\prime }(y+)+f^{\prime }(y-) \bigr) \biggr\vert \bigl\vert \psi _{\mu _{n},\nu _{n}}(y) \bigr\vert + \vert I_{1} \vert + \vert I_{2} \vert + \biggl\vert \frac{1}{2}\bigl(f^{\prime }(y+)-f^{\prime }(y-) \bigr) \biggr\vert \xi _{\mu _{n}, \nu _{n}}(y). \end{aligned} \end{aligned}
(4.14)

Now applying Lemma 4.2 and integration by parts, $$I_{1}$$ can be written as

\begin{aligned} I_{1}& = \int _{0}^{y} \biggl( \int _{t}^{y}f_{y}^{\prime }(u)\,du \biggr) W(t,y)\,dt \\ & = \int _{0}^{y} \biggl( \int _{t}^{y}f_{y}^{\prime }(u)\,du \biggr) \frac{\partial }{\partial t}\lambda _{\mu _{n},\nu _{n}}(t,y)\,dt \\ & = \int _{0}^{y}f_{y}^{\prime }(t) \lambda _{\mu _{n},\nu _{n}}(t,y)\,dt. \end{aligned}

On taking the absolute value of $$I_{1}$$, we have

\begin{aligned} \vert I_{1} \vert & = \int _{0}^{y} \bigl\vert f_{y}^{\prime }(t) \bigr\vert \lambda _{\mu _{n},\nu _{n}}(t,y)\,dt \\ & \leq \int _{0}^{y-y/\sqrt{n}} \bigl\vert f_{y}^{\prime }(t) \bigr\vert \lambda _{\mu _{n}, \nu _{n}}(t,y)\,dt+ \int _{y-y/\sqrt{n}}^{y} \bigl\vert f_{y}^{\prime }(t) \bigr\vert \lambda _{ \mu _{n},\nu _{n}}(t,y)\,dt \\ & =K_{1}+K_{2},\quad \text{say}. \end{aligned}

Since $$f_{y}^{\prime }(y)=0$$, by (4.11), we have

$$K_{1}= \int _{0}^{y-y/\sqrt{n}} \bigl\vert f_{y}^{\prime }(t)-f_{y}^{\prime }(y) \bigr\vert \lambda _{\mu _{n},\nu _{n}}(t,y)\,dt.$$

Now, using Lemma 4.2,

$$K_{1}\leq \xi _{\mu _{n},\nu _{n}}^{2}(y) \int _{0}^{y-y/\sqrt{n}} \bigl\vert f_{y}^{\prime }(t)-f_{y}^{\prime }(y) \bigr\vert \frac{dt}{(y-t)^{2}}.$$

By the definition of total variation (4.10) and taking $$t=y-y/u$$, we obtain

\begin{aligned} K_{1}& \leq \xi _{\mu _{n},\nu _{n}}^{2}(y) \int _{0}^{y-y/\sqrt{n}} \Biggl( \bigvee _{t}^{y}f_{y}^{\prime } \Biggr) \frac{dt}{(y-t)^{2}} \\ & =\xi _{\mu _{n},\nu _{n}}^{2}(y) \int _{1}^{\sqrt{n}} \Biggl( \bigvee _{y-y/u}^{y}f_{y}^{\prime } \Biggr) \frac{du}{y}. \end{aligned}

Now, after breaking the integral into a sum, we have

\begin{aligned} K_{1}& \leq \frac{\xi _{\mu _{n},\nu _{n}}^{2}(y)}{y}\sum_{k=1}^{[ \sqrt{n}]} \int _{k}^{k+1} \Biggl( \bigvee _{y-y/u}^{y}f_{y}^{\prime } \Biggr) \,du \\ & \leq \frac{\xi _{\mu _{n},\nu _{n}}^{2}(y)}{y}\sum_{k=1}^{[\sqrt{n}]} \Biggl( \bigvee_{y-y/k}^{y}f_{y}^{\prime } \Biggr) \biggl( \int _{k}^{k+1}\,du \biggr) \\ & =\frac{\xi _{\mu _{n},\nu _{n}}^{2}(y)}{y}\sum_{k=1}^{[\sqrt{n}]} \Biggl( \bigvee_{y-y/k}^{y}f_{y}^{\prime } \Biggr) . \end{aligned}

Since by Lemma 4.2, $$\lambda _{\mu _{n},\nu _{n}}(t,y)\leq 1$$ and using (4.11), we get

\begin{aligned} K_{2}& = \int _{y-y/n}^{y} \Biggl( \bigvee _{t}^{y}f_{y}^{\prime } \Biggr) \,dt \\ & \leq \Biggl( \bigvee_{y-y/n}^{y}f_{y}^{\prime } \Biggr) \int _{y-y/n}^{y}\,dt \\ & =\frac{y}{\sqrt{n}} \Biggl( \bigvee_{y-y/n}^{y}f_{y}^{\prime } \Biggr) . \end{aligned}

Thus, we get

$$\vert I_{1} \vert \leq \frac{\xi _{\mu _{n},\nu _{n}}^{2}(y)}{y}\sum _{k=1}^{[ \sqrt{n}]} \Biggl( \bigvee _{y-y/k}^{y}f_{y}^{\prime } \Biggr) + \frac{y}{\sqrt{n}} \Biggl( \bigvee_{y-y/n}^{y}f_{y}^{\prime } \Biggr) .$$
(4.15)

Using Lemma 4.2, we can write

\begin{aligned} \vert I_{2} \vert & = \biggl\vert \int _{y}^{\infty } \biggl( \int _{y}^{t}f_{y}^{ \prime }(u)\,du \biggr) W(t,y)\,dt \biggr\vert \\ & \leq \biggl\vert \int _{y}^{2y} \biggl( \int _{y}^{t}f_{y}^{\prime }(u)\,du \biggr) \frac{\partial }{\partial t}\bigl(1-\lambda _{\mu _{n},\nu _{n}}(t,y)\bigr)\,dt \biggr\vert + \biggl\vert \int _{2y}^{\infty } \biggl( \int _{y}^{t}f_{y}^{ \prime }(u)\,du \biggr) W(t,y)\,dt \biggr\vert . \end{aligned}

Now, applying integration by parts and (4.11), we get

\begin{aligned} \vert I_{2} \vert & = \biggl\vert \int _{y}^{2y}f_{y}^{\prime }(u)\,du \bigl(1-\lambda _{ \mu _{n},\nu _{n}}(2y,y)\bigr)- \int _{y}^{2y}f_{y}^{\prime }(t) \bigl(1-\lambda _{ \mu _{n},\nu _{n}}(t,y)\bigr)\,dt \biggr\vert \\ & \quad + \biggl\vert \int _{2y}^{\infty } \biggl( \int _{y}^{t}\bigl(f^{ \prime }(u)-f^{\prime }(y+) \bigr)\,du \biggr) W(t,y)\,dt \biggr\vert \\ & \leq \biggl\vert \int _{y}^{2y}f_{y}^{\prime }(u)\,du \biggr\vert . \frac{\xi _{\mu _{n},\nu _{n}}^{2}(y)}{y^{2}}+ \int _{y}^{2y} \bigl\vert f_{y}^{ \prime }(t) \bigr\vert (1- \lambda _{\mu _{n},\nu _{n}}) (t,y)\,dt \\ & \quad + \biggl\vert \int _{2y}^{\infty }\bigl(f(t)-f(y)\bigr)W(t,y)\,dt \biggr\vert + \bigl\vert f^{\prime }(y+) \bigr\vert \biggl\vert \int _{2y}^{\infty }(t-y)W(t,y)\,dt \biggr\vert \\ & =P_{1}+P_{2}+P_{3}+P_{4},\quad \text{say}. \end{aligned}

Now, by (4.11), we get

\begin{aligned} P_{1}& =\frac{\xi _{\mu _{n},\nu _{n}}^{2}(y)}{y^{2}} \biggl\vert \int _{y}^{2y}f_{y}^{\prime }(u)\,du \biggr\vert \\ & =\frac{\xi _{\mu _{n},\nu _{n}}^{2}(y)}{y^{2}} \biggl\vert \int _{y}^{2y}\bigl(f^{ \prime }(u)-f^{\prime }(y+) \bigr)\,du \biggr\vert \\ & \leq \frac{\xi _{\mu _{n},\nu _{n}}^{2}(y)}{y^{2}} \bigl\vert f(2y)-f(y)-yf^{ \prime }(y+) \bigr\vert \end{aligned}

and

\begin{aligned} P_{2}& = \int _{y}^{2y} \bigl\vert f_{y}^{\prime }(t) \bigr\vert . \bigl(1-\lambda _{\mu _{n},\nu _{n}}(t,y)\bigr)\,dt \\ & = \int _{y}^{y+y/\sqrt{n}} \bigl\vert f_{y}^{\prime }(t) \bigr\vert . \bigl(1-\lambda _{\mu _{n}, \nu _{n}}(t,y)\bigr)\,dt \\ & \quad + \int _{y+y/\sqrt{n}}^{2y} \bigl\vert f_{y}^{\prime }(t) \bigr\vert . \bigl(1-\lambda _{ \mu _{n},\nu _{n}}(t,y)\bigr)\,dt \\ & =J_{1}+J_{2},\quad \text{say}. \end{aligned}

Using Lemma 4.2, $$1-\lambda _{\mu _{n},\nu _{n}}(t,y)\leq 1$$ and (4.11), we get

\begin{aligned} J_{1}& = \int _{y}^{y+y/\sqrt{n}} \bigl\vert f_{y}^{\prime }(t) \bigr\vert . \bigl(1-\lambda _{ \mu _{n},\nu _{n}}(t,y)\bigr)\,dt \\ & \leq \int _{y}^{y+y/\sqrt{n}} \bigl\vert f_{y}^{\prime }(t)-f_{y}^{\prime }(y) \bigr\vert \,dt \\ & \leq \int _{y}^{y+y/\sqrt{n}} \Biggl( \bigvee _{y}^{t}f_{y}^{\prime } \Biggr) \,dt \\ & \leq \Biggl( \bigvee_{y}^{y+y/\sqrt{n}}f_{y}^{\prime } \Biggr) \int _{y}^{y+y/\sqrt{n}}\,dt \\ & \leq \frac{y}{\sqrt{n}} \Biggl( \bigvee_{y}^{y+y/\sqrt{n}}f_{y}^{ \prime } \Biggr) . \end{aligned}

Now, again with the help of Lemma 4.2 and (4.11), we obtain

\begin{aligned} J_{2}& = \int _{y+y/\sqrt{n}}^{2y} \bigl\vert f_{y}^{\prime }(t) \bigr\vert \bigl(1-\lambda _{ \mu _{n},\nu _{n}}(t,y)\bigr)\,dt \\ & \leq \xi _{\mu _{n},\nu _{n}}^{2}(y) \int _{y+y/\sqrt{n}}^{2y} \bigl\vert f_{y}^{ \prime }(t)-f_{y}^{\prime }(y) \bigr\vert \frac{dt}{(t-y)^{2}}. \end{aligned}

By using (4.10) and $$t=y+y/u$$, we get

\begin{aligned} J_{2}& \leq \xi _{\mu _{n},\nu _{n}}^{2}(y) \int _{y+y/\sqrt{n}}^{2y} \Biggl( \bigvee _{y}^{t}f_{y}^{\prime } \Biggr) \frac{dt}{(t-y)^{2}} \\ & =\xi _{\mu _{n},\nu _{n}}^{2}(y) \int _{1}^{\sqrt{n}} \Biggl( \bigvee _{y}^{y+y/u}f_{y}^{\prime } \Biggr) \frac{du}{y} \\ & \leq \frac{\xi _{\mu _{n},\nu _{n}}^{2}(y)}{y}\sum_{k=1}^{[\sqrt{n}]} \int _{k}^{k+1} \Biggl( \bigvee _{y}^{y+y/u}f_{y}^{\prime } \Biggr) \,du \\ & \leq \frac{\xi _{\mu _{n},\nu _{n}}^{2}(y)}{y}\sum_{k=1}^{[\sqrt{n}]} \Biggl( \bigvee_{y}^{y+y/k}f_{y}^{\prime } \Biggr) \biggl( \int _{k}^{k+1}\,du \biggr) \\ & =\frac{\xi _{\mu _{n},\nu _{n}}^{2}(y)}{y}\sum_{k=1}^{[\sqrt{n}]} \Biggl( \bigvee_{y}^{y+y/k}f_{y}^{\prime } \Biggr) . \end{aligned}

Hence, we derive

$$P_{2}\leq \frac{y}{\sqrt{n}} \Biggl( \bigvee _{y}^{y+y/\sqrt{n}}f_{y}^{ \prime } \Biggr) + \frac{\xi _{\mu _{n},\nu _{n}}^{2}(y)}{y}\sum_{k=1}^{[ \sqrt{n}]} \Biggl( \bigvee_{y}^{y+y/k}f_{y}^{\prime } \Biggr) .$$

Now, we estimate $$P_{3}$$. As $$t\geq 2y$$, then using $$2(t-y)\geq t$$ and $$t-y\geq y$$, we get

\begin{aligned} P_{3}& = \biggl\vert \int _{2y}^{\infty }\bigl(f(t)-f(y)\bigr)W(t,y)\,dt \biggr\vert \\ & \leq \int _{2y}^{\infty } \bigl\vert f(t) \bigr\vert W(t,y)\,dt+ \int _{2y}^{\infty } \bigl\vert f(y) \bigr\vert W(t,y)\,dt \\ & \leq M_{f} \int _{2y}^{\infty }\bigl(1+t^{2} \bigr)W(t,y)\,dt+ \bigl\vert f(y) \bigr\vert \int _{2y}^{ \infty }W(t,y)\,dt \\ & =\bigl(M_{f}+ \bigl\vert f(y) \bigr\vert \bigr) \int _{2y}^{\infty }W(t,y)\,dt+M_{f} \int _{2y}^{ \infty }t^{2}W(t,y)\,dt \\ & \leq \bigl(M_{f}+ \bigl\vert f(y) \bigr\vert \bigr) \int _{2y}^{\infty }\frac{(t-y)^{2}}{y^{2}}W(t,y)\,dt+M_{f} \int _{2y}^{\infty }4(t-y)^{2}W(t,y)\,dt \\ & \leq \biggl( \frac{M_{f}+ \vert f(y) \vert }{y^{2}}+4M_{f} \biggr) \int _{2y}^{ \infty }(t-y)^{2}W(t,y)\,dt \\ & \leq \biggl( \frac{M_{f}+ \vert f(y) \vert }{y^{2}}+4M_{f} \biggr) \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}\bigl((t-y)^{2};y \bigr) \\ & = \biggl( \frac{M_{f}+ \vert f(y) \vert }{y^{2}}+4M_{f} \biggr) \xi _{\mu _{n}, \nu _{n}}^{2}(y). \end{aligned}

We can compute $$P_{4}$$ as follows:

\begin{aligned} P_{4}& = \bigl\vert f^{\prime }(y+) \bigr\vert \biggl\vert \int _{2y}^{\infty }(t-y)W(t,y)\,dt \biggr\vert \\ & \leq \bigl\vert f^{\prime }(y+) \bigr\vert \biggl\vert \int _{0}^{\infty }(t-y)W(t,y)\,dt \biggr\vert \\ & = \bigl\vert f^{\prime }(y+) \bigr\vert \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(t-y;y) \bigr\vert \\ & = \bigl\vert f^{\prime }(y+) \bigr\vert \bigl\vert \psi _{\mu _{n},\nu _{n}}(y) \bigr\vert . \end{aligned}

Hence, we get

\begin{aligned} \vert I_{2} \vert & \leq \frac{\xi _{\mu _{n},\nu _{n}}^{2}(y)}{y^{2}} \bigl\vert f(2y)-f(y)-yf^{\prime }(y+) \bigr\vert \\ & \quad +\frac{y}{\sqrt{n}} \Biggl( \bigvee_{y}^{y+y/\sqrt{n}}f_{y}^{ \prime } \Biggr) +\frac{\xi _{\mu _{n},\nu _{n}}^{2}(y)}{y}\sum_{k=1}^{[ \sqrt{n}]} \Biggl( \bigvee_{y}^{y+y/k}f_{y}^{\prime } \Biggr) \\ & \quad + \biggl( \frac{M_{f}+ \vert f(y) \vert }{y^{2}}+4M_{f} \biggr) \xi _{\mu _{n}, \nu _{n}}^{2}(y)+ \bigl\vert f^{\prime }(y+) \bigr\vert \bigl\vert \psi _{\mu _{n},\nu _{n}}(y) \bigr\vert . \end{aligned}
(4.16)

Now, from (4.14)–(4.16), we obtain

\begin{aligned} \bigl\vert \mathcal{Q}_{n,b}^{(\mu _{n},\nu _{n})}(f;y)-f(y) \bigr\vert & \leq \biggl\vert \frac{1}{2}\bigl(f^{\prime }(y+)+f^{\prime }(y-) \bigr) \biggr\vert \bigl\vert \psi _{\mu _{n}, \nu _{n}}(y) \bigr\vert + \vert I_{1} \vert + \vert I_{2} \vert \\ & \quad + \biggl\vert \frac{1}{2}\bigl(f^{\prime }(y+)-f^{\prime }(y-) \bigr) \biggr\vert \xi _{\mu _{n},\nu _{n}}(y) \\ & \leq \biggl\vert \frac{1}{2} \bigl( f^{\prime }(y+)+f^{\prime }(y-) \bigr) \biggr\vert \bigl\vert \psi _{\mu _{n},\nu _{n}}(y) \bigr\vert \\ & \quad +\frac{\xi _{\mu _{n},\nu _{n}}^{2}(y)}{y}\sum_{k=1}^{[\sqrt{n}]} \Biggl( \bigvee_{y-y/k}^{y}f_{y}^{\prime } \Biggr) + \frac{y}{\sqrt{n}} \Biggl( \bigvee _{y-y/\sqrt{n}}^{y}f_{y}^{\prime } \Biggr) \\ & \quad +\frac{\xi _{\mu _{n},\nu _{n}}^{2}(y)}{y^{2}} \bigl\vert f(2y)-f(y)-yf^{ \prime }(y+) \bigr\vert \\ & \quad +\frac{y}{\sqrt{n}} \Biggl( \bigvee_{y}^{y+y/\sqrt{n}}f_{y}^{\prime } \Biggr) +\frac{\xi _{\mu _{n},\nu _{n}}^{2}(y)}{y}\sum_{k=1}^{[\sqrt{n}]} \Biggl( \bigvee_{y}^{y+y/k}f_{y}^{ \prime } \Biggr) \\ & \quad + \biggl( \frac{M_{f}+ \vert f(y) \vert }{y^{2}}+4M_{f} \biggr) \xi _{\mu _{n}, \nu _{n}}^{2}(y)+ \bigl\vert f^{\prime }(y+) \bigr\vert \bigl\vert \psi _{\mu _{n},\nu _{n}}(y) \bigr\vert \\ & \quad + \biggl\vert \frac{1}{2} \bigl( f^{\prime }(y+)-f^{\prime }(y-) \bigr) \biggr\vert \xi _{\mu _{n},\nu _{n}}(y), \end{aligned}

which gives the desired result. □

## Graphical examples

### Example 5.1

Let us take $$f(x)=5x^{4}-11x^{3}+2x^{2}$$. The convergence of the sequence of operators defined by Eq. (1.3) when $$\mu _{n}=\nu _{n}=n$$ towards the function $$f(x)$$ (cyan) is shown for $$n=10,50,100$$, respectively, in Figs. 13 taking $$b=2$$ (blue), $$b=6$$ (black), and $$b=15$$ (red). Figures 46 illustrate the convergence of the sequence of operators defined by Eq. (1.3) taking $$\mu _{n}=n+\sqrt{n+1}$$, $$\nu _{n}=n+12$$ towards the function $$f(x)$$ (cyan) for $$n=10,50,100$$, keeping the value of b the same.

Also, a direct comparison between the convergence of the old operator applied to f (when $$\mu _{n}=\nu _{n}=n$$ discussed in ) (blue) and the new operator (red) defined in Eq. (1.3) towards $$f(x)$$ (cyan) is shown in Figs. 79, respectively, for $$n=10,50,100$$, and $$b=10$$. It is clear that the new operator exhibits faster convergence towards the limit than the old operator. Also, the new operator is giving flexibility in choosing parameters in the form of the sequences $$\mu _{n}$$ and $$\nu _{n}$$.

## Conclusions

We have modified the sequence of operators discussed in  and developed many approximation properties such as direct theorems, rate of convergence in weighted spaces, and approximation for functions of bounded variation. Moreover, we have also shown the convergence of old and modified new operators graphically.

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### Acknowledgements

The author (K.J. Ansari) extends his appreciation to the Deanship of Scientific Research at King Khalid University for funding this work through research groups program under grant number G.R.P-93-41.

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The authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.

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Correspondence to M. Mursaleen.

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