# Generalization of Szász–Mirakjan–Kantorovich operators using multiple Appell polynomials

## Abstract

The purpose of the present paper is to introduce and study a sequence of positive linear operators defined on suitable spaces of measurable functions on $$[0,\infty )$$ and continuous function spaces with polynomial weights. These operators are Kantorovich type generalization of Jakimovski–Leviatan operators based on multiple Appell polynomials. Using these operators, we approximate suitable measurable functions by knowing their mean values on a sequence of subintervals of $$[0,\infty )$$ that do not constitute a subdivision of it. We also discuss the rate of convergence of these operators using moduli of smoothness.

## Introduction

A multiple polynomial system  $$\{s_{n_{1},n_{2}}(x)\}$$ is called multiple Appell if it has a generating function of the form

\begin{aligned} H(t_{1},t_{2})e^{x(t_{1}+t_{2})}= \sum _{n_{1}=0}^{ \infty } \sum_{n_{2}=0}^{\infty } \frac{s_{n_{1},n_{2}}(x)}{n_{1}!n_{2}!}t_{1}^{n_{1}}t_{2}^{n_{2}}, \end{aligned}
(1.1)

where $$H(t_{1},t_{2})$$ has a series expansion

\begin{aligned} H(t_{1},t_{2})= \sum _{n_{1}=0}^{\infty } \sum_{n_{2}=0}^{\infty } \frac{a_{n_{1},n_{2}}}{n_{1}!n_{2}!}t_{1}^{n_{1}}t_{2}^{n_{2}}, \end{aligned}
(1.2)

with $$H(0,0)=a_{0,0} \neq 0$$ and $$\frac{a_{n_{1},n_{2}}}{H(1,1)}\geq 0$$ for all $$n_{1},n_{2}\in \mathbb{N}$$. Also, (1.1) and (1.2) converge for $$\vert t_{1} \vert \leq R_{1}$$, $$\vert t_{2} \vert \leq R_{2}$$ ($$R_{1},R_{2}>1$$). $$s_{n_{1},n_{2}}$$ is a multiple polynomial system, and for every $$n_{1}+n_{2}\geq 1$$, this satisfies the following relationship:

\begin{aligned} s'_{n_{1},n_{2}}(x)=n_{1}s_{n_{1}-1,n_{2}}(x)+n_{2}s_{n_{1},n_{2}-1}(x). \end{aligned}

Also, for the multiple Appell polynomial systems $$s_{n_{1},n_{2}}(x)$$, there exists a sequence $$\{a_{n_{1},n_{2}}\}_{n_{1},n_{2}=0}^{\infty }$$ with $$a_{0,0}\neq 0$$ such that

\begin{aligned} s_{n_{1},n_{2}}(x)= \sum_{k_{1}=0}^{n_{1}} \sum_{k_{2}=0}^{n_{2}}\binom{n_{1}}{k_{1}} \binom{n_{2} }{k_{2}}a_{n_{1}-k_{1},n_{2}-k_{2}}x^{k_{1}+k_{2}}. \end{aligned}
(1.3)

Therefore, $$s_{n_{1},n_{2}}(x)$$ is a polynomial in x of degree $$n_{1}+n_{2}$$.

Using these Appell polynomials, Varma  defined a generalization of Szász operators  as follows:

\begin{aligned} S_{n}(f;x)=\frac{e^{-nx}}{H(1,1)} \sum _{n_{1}=0}^{ \infty } \sum_{n_{2}=0}^{\infty } \frac{s_{n_{1},n_{2}} (\frac{nx}{2})}{n_{1}!n_{2}!}f \biggl(\frac{n_{1}+n_{2}}{n} \biggr). \end{aligned}
(1.4)

He also defined the Kantorovich type modification of these operators as follows:

\begin{aligned} K_{n}^{*}(f;x)=\frac{ne^{-nx}}{H(1,1)} \sum_{n_{1}=0}^{ \infty } \sum _{n_{2}=0}^{\infty } \frac{s_{n_{1},n_{2}}(\frac{nx}{2})}{n_{1}!n_{2}!} \int _{ \frac{n_{1}+n_{2}}{n}}^{\frac{n_{1}+n_{2}+1}{n}}f(u)\,du, \end{aligned}
(1.5)

and obtained the rate of convergence of these operators in terms of the classical modulus of continuity. Alternatively, the operator given by (1.5) may be expressed as

\begin{aligned} K_{n}^{*}(f;x)= \int _{0}^{\infty }K_{n}(x,u)f(u)\,du, \end{aligned}

where $$K_{n}(x,u)=\frac{ne^{-nx}}{H(1,1)} \sum_{n_{1}=0}^{ \infty } \sum_{n_{2}=0}^{\infty } \frac{s_{n_{1},n_{2}}(\frac{nx}{2})}{n_{1}!n_{2}!} \chi _{[ \frac{n_{1}+n_{2}}{n},\frac{n_{1}+n_{2}+1}{n}]}(u)$$, $$\chi _{[\frac{n_{1}+n_{2}}{n},\frac{n_{1}+n_{2}+1}{n}]}(u)$$ being the characteristic function of $$[\frac{n_{1}+n_{2}}{n},\frac{n_{1}+n_{2}+1}{n} ]$$ on $$[0,\infty )$$.

The purpose of the present paper is to make a generalization of these operators that extends to the unbounded setting an idea given in  where the authors studied a modification of Kantorovich operators. We refer the reader to some of the related papers [1, 2, 79, 1114, 16].

In this paper we will study the following sequence $$(P_{n})$$ of positive linear operators:

\begin{aligned} P_{n}(h;x)=\frac{ne^{-nx}}{(b_{n}-c_{n})H(1,1)} \sum _{n_{1}=0}^{ \infty } \sum_{n_{2}=0}^{\infty } \frac{s_{n_{1},n_{2}} (\frac{nx}{2})}{n_{1}!n_{2}!} \int _{\frac{n_{1}+n_{2}+c_{n}}{n}}^{\frac{n_{1}+n_{2}+b_{n}}{n}}h(u)\,du\quad (n \geq 1, x\geq 0). \end{aligned}
(1.6)

For every $$h\in \Im [0,\infty )$$ (the space of all Borel measurable locally integrable functions $$g:[0,\infty )\rightarrow \mathbb{R}$$ such that the antiderivative $$G(x)=\int _{0}^{x} g(t) \,dt$$ ($$x\geq 0$$) belongs to $$\ell ([0,\infty ))$$, the space of all functions $$b:[0,\infty )\rightarrow \mathbb{R}$$ such that $$\vert b(x) \vert \leq Me^{rx}$$ ($$x\geq 0$$) for some $$M\geq 0$$ and $$r\in \mathbb{R}$$).

Here $$(c_{n})_{n\geq 1}$$ and $$(b_{n})_{n\geq 1}$$ are two sequences of real numbers satisfying $$0\leq c_{n}< b_{n}\leq 1$$ for every $$n\geq 1$$. If $$c_{n}=0$$ and $$b_{n}=1$$ for all $$n\geq 1$$, then the $$P_{n}$$’s (operators in (1.6) turn into (1.5). By using $$P_{n}$$’s we can reconstruct some suitable continuous or integrable functions by knowing their mean values on subinterval of $$[0,\infty )$$ which do not necessarily constitute a subdivision of $$[0,\infty )$$.

We will study the approximation properties of $$(P_{n})$$ for every $$n\geq 1$$ on several continuous and weighted continuous function spaces as well as on Lebesgue spaces. We also discuss the rate of convergence of these operators by using appropriate moduli of smoothness.

## Generalizing Jakimovski–Leviatan operators

Throughout this paper the following notations are used:

$$\wp [0,\infty )$$::

The space of all continuous real-valued functions on $$[0,\infty )$$.

$$\wp _{b}[0,\infty )$$::

The subspace of all functions in $$\wp [0,\infty )$$ which are bounded. This space endowed with the sup-norm and the natural pointwise ordering is a Banach lattice.

$$\wp _{\ast }[0,\infty )$$::

The space of all continuous functions converging at infinity. This space is a Banach sublattice of $$\wp _{b}([0,\infty ))$$.

$$\wp _{0}[0,\infty )$$::

subspace of $$\wp _{\ast }[0,\infty )$$, consisting of all those functions that vanish at infinity.

Moreover, for every $$m\geq 1$$, we set $$r_{m}(y)=(1+y^{m})^{-1}$$ ($$y\geq 0$$) and

\begin{aligned} G_{m}:= \Bigl\{ h\in \wp [0,\infty )\big| \sup_{y\geq 0}r_{m}(y) \bigl\vert h(y) \bigr\vert \in \mathbb{R} \Bigr\} ; \end{aligned}

$$G_{m}$$ is a Banach lattice endowed with the pointwise ordering and the weighted norm

\begin{aligned} \Vert h \Vert _{m}:= \sup_{y\geq 0}r_{m}(y) \bigl\vert h(y) \bigr\vert \quad (h \in G_{m}). \end{aligned}

The following space is the Banach sublattices of $$G_{m}$$:

\begin{aligned} G_{m}^{\ast }:= \Bigl\{ h\in G_{m}\big| \lim _{y\rightarrow \infty } r_{m}(y)h(y)\in \mathbb{R} \Bigr\} . \end{aligned}

For a given $$g\in \Im ([0,\infty ))$$ and $$G(x)=\int _{0}^{x} g(t)\,dt$$, the antiderivative of $$g(x)$$, operators given in (1.6) may be expressed as

\begin{aligned} P_{n}(g;x) =&\frac{ne^{-nx}}{(b_{n}-c_{n})H(1,1)} \\ &{}\times\sum _{n_{1}=0}^{ \infty } \sum_{n_{2}=0}^{\infty } \frac{s_{n_{1},n_{2}} (\frac{nx}{2})}{n_{1}!n_{2}!} \biggl[G \biggl(\frac{n_{1}+n_{2}+b_{n}}{n} \biggr)-G \biggl( \frac{n_{1}+n_{2}+c_{n}}{n} \biggr) \biggr] \end{aligned}
(2.1)
\begin{aligned} =&\frac{n}{(b_{n}-c_{n})}S_{n}\bigl(\tau _{n}(G)\bigr) (x), \end{aligned}
(2.2)

where $$S_{n}$$ is given by (1.4) and the mapping $$\tau _{n}$$ is defined as

\begin{aligned} \tau _{n}(G) (x):=G \biggl(\frac{n_{1}+n_{2}+b_{n}}{n} \biggr)-G \biggl( \frac{n_{1}+n_{2}+c_{n}}{n} \biggr)\quad (x \geq 0), \end{aligned}

$$P_{n}(g)$$ can also be written as

\begin{aligned} P_{n}(g) (x)= \int _{0}^{+\infty }g \,d\mu _{n,x}\quad (n \geq 1,x\geq 0), \end{aligned}
(2.3)

where

\begin{aligned} \mu _{n,x}=\frac{ne^{-nx}}{(b_{n}-c_{n})H(1,1)} \sum_{n_{1}=0}^{ \infty } \sum_{n_{2}=0}^{\infty } \frac{s_{n_{1},n_{2}} (\frac{nx}{2})}{n_{1}!n_{2}!}\mu _{n,k}, \end{aligned}

and each $$\mu _{n,k}$$ denotes the Borel measure on $$[0,\infty )$$ having density, the characteristic function of $$[\frac{n_{1}+n_{2}+c_{n}}{n}, \frac{n_{1}+n_{2}+b_{n}}{n} ]$$ w.r.t. the Borel–Lebesgue measure on $$[0,\infty )$$. Throughout the paper the symbol $$c_{m}$$ denotes the function $$x^{m}$$ by setting $$c_{m}(x)=x^{m}$$ for every $$m\geq 0$$ and for every $$x\geq 0$$. Particularly $$c_{0}=\mathbf{1}$$, where 1 denotes the constant function on $$[0,\infty )$$ of constant value 1. Finally, we shall set $$\rho _{x}(y)=y-x$$ ($$y\geq 0$$) and x is a fixed nonnegative real number.

## Preliminaries

### Lemma 1

For operators (1.6), the estimates of moments are as follows :

\begin{aligned}& (\mathrm{i}) \quad P_{n}(1;y)=1; \\& (\mathrm{ii}) \quad P_{n}(t;y)=y+ \frac{1}{2} \frac{b_{n}+c_{n}}{n}+ \frac{H_{t_{1}}(1,1)+H_{t_{2}}(1,1)}{nH(1,1)}; \\& \begin{aligned} (\mathrm{iii}) \quad P_{n}\bigl(t^{2};y \bigr)={}&y^{2}+ \frac{y}{n} \biggl(1+ \frac{b_{n}+c_{n}}{n}+2 \frac{(H_{t_{1}}(1,1)+H_{t_{2}}(1,1))}{H(1,1)} \biggr) +\frac{b_{n}^{2}+c_{n}^{2}+b_{n}c_{n}}{3n^{2}} \\ &{}\times \frac{1}{n^{2}H(1,1)}\bigl\{ H_{t_{1}}(1,1) (1+b_{n}+c_{n})+H_{t_{2}}(1,1) (1+b_{n}+c_{n}) \\ &{}+H_{t_{1}t_{1}}(1,1)+H_{t_{2}t_{2}}(1,1)+2H_{t_{1}t_{2}}(1,1) \bigr\} ; \end{aligned} \\& (\mathrm{iv}) \quad P_{n}\bigl(\rho _{y}(t);y\bigr)= \frac{1}{2}\frac{b_{n}+c_{n}}{n}+ \frac{H_{t_{1}}(1,1)+H_{t_{2}}(1,1)}{nH(1,1)}; \\& \begin{aligned} (\mathrm{v}) \quad P_{n}\bigl(\rho _{y}^{2}(t);y\bigr)= {}&\frac{y}{n}+ \frac{1}{n^{2}H(1,1)}\bigl\{ H_{t_{1}}(1,1)+H_{t_{2}}(1,1)+H_{t_{1}t_{1}}(1,1) \\ &{}+H_{t_{2}t_{2}}(1,1)+2H_{t_{1}t_{2}}(1,1) \bigr\} \\ &{}+ \frac{b_{n}^{2}+c_{n}^{2}+b_{n}c_{n}}{n^{2}}+\frac{1}{n}(b_{n}+c_{n}) \biggl(\frac{H_{t_{1}}(1,1)+H_{t_{2}}(1,1)}{nH(1,1)} \biggr). \end{aligned} \end{aligned}

### Proposition 1

For every$$\rho >0$$, let$$f_{\rho }(x)=e^{-\rho x}$$where$$x\geq 0$$. Then

\begin{aligned} S_{n}(f_{\rho }) (x)= \frac{H(e^{\frac{-\rho }{n}},e^{\frac{-\rho }{n}})}{H(1,1)}\exp \bigl(nx \bigl(e^{\frac{-\rho }{n}}-1 \bigr) \bigr). \end{aligned}
(3.1)

### Proof

\begin{aligned} S_{n}(f_{\rho }) (x)= \frac{e^{-nx}}{H(1,1)} \sum_{n_{1}=0}^{ \infty } \sum _{n_{2}=0}^{\infty } \frac{s_{n_{1},n_{2}} (\frac{nx}{2})}{n_{1}!n_{2}!}e^{-\rho (\frac{n_{1}+n_{2}}{n} )}. \end{aligned}
(3.2)

Putting $$x= \frac{nx}{2}$$ and $$t_{1}=t_{2}=e^{(\frac{-\rho }{n})}$$ in (1.1), we get

\begin{aligned} \sum_{n_{1}=0}^{\infty } \sum_{n_{2}=0}^{ \infty }\frac{s_{n_{1},n_{2}} (\frac{nx}{2})}{n_{1}!n_{2}!}e^{-\rho (\frac{n_{1}+n_{2}}{n} )}=H \bigl(e^{\frac{-\rho }{n}},e^{\frac{-\rho }{n}}\bigr)e^{\frac{nx}{2} ( e^{\frac{-\rho }{n}}+e^{\frac{-\rho }{n}} )}. \end{aligned}
(3.3)

Using (3.3) in (3.2), we have the result. □

### Proposition 2

\begin{aligned} P_{n}(f_{\rho })=\frac{n}{\rho (b_{n}-c_{n})} \bigl(e^{ \frac{-\rho c_{n}}{n}}-e^{\frac{-\rho b_{n}}{n}} \bigr)S_{n}(f_{\rho }) \end{aligned}
(3.4)

for every$$n\geq 1$$.

Moreover, for every$$n\geq 1$$and$$\rho >0$$,

\begin{aligned} P_{n}(f_{\rho })\leq S_{n}(f_{\rho }). \end{aligned}
(3.5)

### Proof

(3.4) holds after a straightforward computation, and the proof of (3.5) is as follows:

\begin{aligned} \frac{n}{\rho (b_{n}-c_{n})} \bigl(e^{\frac{-\rho c_{n}}{n}}-e^{ \frac{-\rho b_{n}}{n}} \bigr)\leq \frac{n}{\rho (b_{n}-c_{n})} \bigl(1-e^{-( \frac{\rho b_{n}}{n}-\frac{\rho c_{n}}{n})} \bigr)\leq 1 \end{aligned}

since $$1-e^{-x}\leq x$$ ($$x\geq 0$$). □

### Theorem 1

The operator$$P_{n}$$for every$$n\geq 1$$defined by (1.6) has the following properties:

1. (i)

$$P_{n}$$is a positive and continuous linear operator from$$\wp _{b}([0,\infty ))$$to$$\wp _{b}([0,\infty ))$$and$$\Vert P_{n} \Vert _{\wp _{b}([0,\infty ))}=1$$.

2. (ii)

$$P_{n}(\wp _{0}([0,\infty )))\subset \wp _{0}([0,\infty ))$$.

### Proof

(i) For any $$f\in \wp _{b}([0,\infty ))$$, there exists an $$M_{f}$$ depending on f such that $$\vert f \vert \leq M_{f}$$. Therefore, for every $$n\geq 1$$,

\begin{aligned} \bigl\vert P_{n}(f) \bigr\vert \leq & \frac{ne^{-nx}}{(b_{n}-c_{n})H(1,1)} \sum _{n_{1}=0}^{\infty } \sum_{n_{2}=0}^{\infty } \frac{s_{n_{1},n_{2}} (\frac{nx}{2})}{n_{1}!n_{2}!} \int _{\frac{n_{1}+n_{2}+c_{n}}{n}}^{\frac{n_{1}+n_{2}+b_{n}}{n}} \bigl\vert f(u) \bigr\vert \,du \\ \leq & M_{f}, \end{aligned}

and $$P_{n}(1)=1$$. Hence $$\Vert P_{n} \Vert _{\wp _{b}([0,\infty ))}=1$$.

(ii) For fixed $$h\in \wp _{0}([0,\infty ))$$ and $$\epsilon \geq 0$$, there exists $$u\geq 0$$ such that $$\vert h(x) \vert \leq \frac{\epsilon }{4}$$ for any $$x\geq u-1$$. Now, since $$T_{m}(x)e^{-nx}\rightarrow 0$$ as $$x\rightarrow \infty$$, where $$T_{m}(x)$$ is any polynomial of degree $$m\in \mathbb{N}$$ in x for every m, therefore there exists $$v>u$$ such that, for every $$x\geq v$$,

\begin{aligned} T_{m}(x)e^{-nx}\leq \frac{\epsilon }{4 \Vert h \Vert _{\infty }(n[u]+1)^{2}} \end{aligned}

for any $$m=0,1,\ldots,2n[u]$$, where $$[u]$$ denotes the integer part of u.

Using (1.3), we have $$\frac{s_{n_{1},n_{2}}(\frac{nx}{2})}{H(1,1)}$$ is a polynomial of degree $$n_{1}+n_{2}$$. Therefore, for every $$x\geq v$$, we have

\begin{aligned} \bigl\vert P_{n}(h;x) \bigr\vert \leq & \frac{n}{b_{n}-c_{n}} \sum _{n_{1}=0}^{n[u]} \sum _{n_{2}=0}^{n[u]} e^{-nx} \frac{s_{n_{1},n_{2}}(\frac{nx}{2})}{n_{1}!n_{2}!H(1,1)} \int _{ \frac{n_{1}+n_{2}+c_{n}}{n}}^{\frac{n_{1}+n_{2}+b_{n}}{n}} \bigl\vert h(t) \bigr\vert \,dt \\ &{}+ \frac{n}{b_{n}-c_{n}} \sum_{n_{1}=n[u]+1}^{\infty } \sum_{n_{2}=0}^{n[u]} e^{-nx} \frac{s_{n_{1},n_{2}}(\frac{nx}{2})}{n_{1}!n_{2}!H(1,1)} \int _{ \frac{n_{1}+n_{2}+c_{n}}{n}}^{\frac{n_{1}+n_{2}+b_{n}}{n}} \bigl\vert h(t) \bigr\vert \,dt \\ &{}+ \frac{n}{b_{n}-c_{n}} \sum_{n_{1}=0}^{n[u]} \sum_{n_{2}=n[u]+1}^{\infty } e^{-nx} \frac{s_{n_{1},n_{2}}(\frac{nx}{2})}{n_{1}!n_{2}!H(1,1)} \int _{ \frac{n_{1}+n_{2}+c_{n}}{n}}^{\frac{n_{1}+n_{2}+b_{n}}{n}} \bigl\vert h(t) \bigr\vert \,dt \\ &{}+ \frac{n}{b_{n}-c_{n}} \sum_{n_{1}=n[u]+1}^{\infty } \sum_{n_{2}=n[u]+1}^{\infty } e^{-nx} \frac{s_{n_{1},n_{2}}(\frac{nx}{2})}{n_{1}!n_{2}!H(1,1)} \int _{ \frac{n_{1}+n_{2}+c_{n}}{n}}^{\frac{n_{1}+n_{2}+b_{n}}{n}} \bigl\vert h(t) \bigr\vert \,dt \\ \leq & \frac{\epsilon }{4}+ \frac{\epsilon }{4} \sum _{n_{1}=0}^{ \infty } \sum_{n_{2}=0}^{\infty } e^{-nx} \frac{s_{n_{1},n_{2}}(\frac{nx}{2})}{n_{1}!n_{2}!H(1,1)} \\ &{}+\frac{\epsilon }{4} \sum_{n_{1}=0}^{\infty } \sum _{n_{2}=0}^{\infty } e^{-nx} \frac{s_{n_{1},n_{2}}(\frac{nx}{2})}{n_{1}!n_{2}!H(1,1)} \\ &{}+\frac{\epsilon }{4} \sum_{n_{1}=0}^{\infty } \sum _{n_{2}=0}^{\infty } e^{-nx} \frac{s_{n_{1},n_{2}}(\frac{nx}{2})}{n_{1}!n_{2}!H(1,1)} \\ =&\epsilon . \end{aligned}

□

### Theorem 2

If$$f\in \wp _{*}([0,\infty ))$$, then$$\lim_{n\rightarrow \infty } P_{n}(f)=f$$uniformly on$$[0,\infty )$$.

Moreover, if$$f\in \wp _{b}([0,\infty ))$$, then$$\lim_{n\rightarrow \infty } P_{n}(f)=f$$uniformly on every compact subset of$$[0,\infty )$$.

### Proof

To prove the first part, it suffices to show that $$P_{n}(f)\rightarrow f$$ for every $$f\in \wp _{0}([0,\infty ))$$ or, in fact, for each function $$f_{\rho }$$ defined in Proposition 1 since the subspace generated by them is dense in $$\wp _{0}([0,\infty ))$$ and the sequence $$(P_{n})_{n\geq 1}$$ is equibounded on $$\wp _{0}([0,\infty ))$$. Now, by using (3.1), for every $$x\geq 0$$ and $$n\geq 1$$, we get

\begin{aligned}& \bigl\vert P_{n}(f_{\rho }) (x)-f_{\rho }(x) \bigr\vert \\& \quad \leq \biggl\vert \frac{n}{\rho (b_{n}-c_{n})} \bigl(e^{-\rho \frac{c_{n}}{n}}-e^{-\rho \frac{b_{n}}{n}} \bigr)-1 \biggr\vert S_{n}(f_{ \rho }) (x)+ \bigl\vert S_{n}(f_{\rho }) (x)-f_{\rho }(x) \bigr\vert . \end{aligned}

Now since

\begin{aligned} S_{n}(f_{\rho }) (x) =&\frac{e^{-nx}}{H(1,1)} \sum _{n_{1}=0}^{ \infty } \sum_{n_{2}=0}^{\infty } \frac{s_{n_{1},n_{2}} (\frac{nx}{2})}{n_{1}!n_{2}!}e^{-\rho (\frac{n_{1}+n_{2}}{n} )} \\ \leq &\frac{e^{-nx}}{H(1,1)} \sum_{n_{1}=0}^{\infty } \sum_{n_{2}=0}^{\infty } \frac{s_{n_{1},n_{2}} (\frac{nx}{2})}{n_{1}!n_{2}!} \\ =&1. \end{aligned}

We have

\begin{aligned}& \bigl\vert P_{n}(f_{\rho }) (x)-f_{\rho }(x) \bigr\vert \\& \quad \leq \biggl\vert \frac{n}{\rho (b_{n}-c_{n})} \bigl(e^{-\rho \frac{c_{n}}{n}}-e^{-\rho \frac{b_{n}}{n}} \bigr)-1 \biggr\vert + \bigl\vert S_{n}(f_{ \rho }) (x)-f_{\rho }(x) \bigr\vert , \end{aligned}

and by using the inequality $$\frac{n}{\rho (b_{n}-c_{n})} (e^{-\rho \frac{c_{n}}{n}}-e^{- \rho \frac{b_{n}}{n}} )\leq 1$$ (proved in Proposition 2), we get

\begin{aligned}& \bigl\vert P_{n}(f_{\rho }) (x)-f_{\rho }(x) \bigr\vert \\& \quad \leq \biggl(1-\frac{n}{\rho (b_{n}-c_{n})} \bigl(e^{-\rho \frac{c_{n}}{n}}-e^{-\rho \frac{b_{n}}{n}} \bigr) \biggr) + \bigl\vert S_{n}(f_{ \rho }) (x)-f_{\rho }(x) \bigr\vert \\& \quad \leq \biggl(1-\frac{n}{\rho (b_{n}-c_{n})} \bigl(e^{-\rho \frac{c_{n}}{n}}-e^{-\rho \frac{b_{n}}{n}} \bigr) \biggr)+ \bigl\Vert S_{n}(f_{ \rho })-f_{\rho } \bigr\Vert _{\infty }. \end{aligned}

Now from [, p. 845], we have

$$\biggl(1-\frac{n}{\rho (b_{n}-c_{n})} \bigl(e^{-\rho \frac{c_{n}}{n}}-e^{- \rho \frac{b_{n}}{n}} \bigr) \biggr)\leq \frac{\rho }{n}.$$

Therefore

\begin{aligned} \bigl\vert P_{n}(f_{\rho }) (x)-f_{\rho }(x) \bigr\vert \leq \frac{\rho }{n}+ \bigl\Vert S_{n}(f_{ \rho })-f_{\rho } \bigr\Vert _{\infty }. \end{aligned}

Now since the sequence $$(S_{n}(f_{\rho }))_{n\geq 1}$$ converges uniformly to $$f_{\rho }$$ by Proposition 1, the result is achieved.

For the second part of the theorem, we see that, from Lemma 1, $$\lim_{n\rightarrow \infty }P_{n}(g)=g$$ uniformly on compact subsets of $$[0,\infty )$$ for every $$g\in \{\mathbf{1},c_{1},c_{2}\}\subset G_{2}^{\ast }$$, the result holds from [, Theorem 3.5]. □

## Estimating the rate of convergence

We now present some estimates of the rate of convergence of $$(P_{n}(h))_{n\geq 1}$$ to h by using the moduli of smoothness of first and second order $$\omega (h,\gamma )$$ and $$\omega _{2}(h,\gamma )$$. For the definitions of $$\omega (h,\gamma )$$ and $$\omega _{2}(h,\gamma )$$, we refer the reader to [, Sect. 5.1].

### Theorem 3

Let$$h\in \wp _{b}([0,\infty ))$$, $$n\geq 1$$, and$$y\geq 0$$. Then

\begin{aligned}& \bigl\vert P_{n}(h) (y)-h(y) \bigr\vert \\& \quad \leq \biggl(\frac{b_{n}+c_{n}}{2\sqrt{n}}+ \frac{H_{t_{1}}(1,1)+H_{t_{2}}(1,1)}{\sqrt{n}H(1,1)} \biggr)\omega \biggl(h, \frac{1}{\sqrt{n}} \biggr) \\& \quad \quad {} + \biggl[1+\frac{1}{2} \biggl(y+\frac{1}{nH(1,1)}\bigl\{ H_{t_{1}}(1,1)+H_{t_{2}}(1,1)+H_{t_{1}t_{1}}(1,1) \\& \quad\quad{} +H_{t_{2}t_{2}}(1,1)+2H_{t_{1}t_{2}}(1,1) \bigr\} \\& \quad \quad {} +\frac{b_{n}^{2}+c_{n}^{2}+b_{n}c_{n}}{n}+(b_{n}+c_{n}) \biggl( \frac{H_{t_{1}}(1,1)+H_{t_{2}}(1,1)}{H(1,1)} \biggr) \biggr) \biggr]\omega _{2} \biggl(h, \frac{1}{\sqrt{n}} \biggr). \end{aligned}

### Proof

Since (2.3) holds, by [, Theorem 2.2.1] and Lemma 1, for every $$\gamma >0$$,

\begin{aligned}& \bigl\vert P_{n}(h) (y)-h(y) \bigr\vert \\& \quad \leq \bigl\vert P_{n}(1) (y)-1 \bigr\vert \bigl\vert h(y) \bigr\vert +\frac{1}{\gamma } \bigl\vert P_{n}(\rho _{y}) (y) \bigr\vert \omega (h,\gamma )+ \biggl[P_{n}(1) (y)+\frac{1}{2\gamma ^{2}}P_{n}\bigl( \rho _{y}^{2} \bigr)y) \biggr]\omega _{2}(h,\gamma ) \\& \quad =\frac{1}{\gamma } \biggl(\frac{b_{n}+c_{n}}{2n}+ \frac{H_{t_{1}}(1,1)+H_{t_{2}}(1,1)}{nH(1,1)} \biggr) \omega (h,\gamma ) \\& \quad \quad {}+ \biggl[1+\frac{1}{2\gamma ^{2}} \biggl(\frac{y}{n}+ \frac{1}{n^{2}H(1,1)}\bigl\{ H_{t_{1}}(1,1)+H_{t_{2}}(1,1)+H_{t_{1}t_{1}}(1,1) \\& \quad\quad{} +H_{t_{2}t_{2}}(1,1)+2H_{t_{1}t_{2}}(1,1) \bigr\} \\& \quad \quad {} +\frac{b_{n}^{2}+c_{n}^{2}+b_{n}c_{n}}{n^{2}}+\frac{1}{n}(b_{n}+c_{n}) \biggl(\frac{H_{t_{1}}(1,1)+H_{t_{2}}(1,1)}{nH(1,1)} \biggr) \biggr) \biggr] \omega _{2}(h,\gamma ). \end{aligned}

Putting $$\gamma =\frac{1}{\sqrt{n}}$$, we get the result. □

## Quantitative estimates

### Lemma 2

Let$$0\leq c_{n}\leq b_{n}\leq 1$$ ($$n\geq 1$$), $$h\in \wp _{b}([0,\infty ))$$, and$$G(x)=\int _{0}^{x} g(t)\,dt$$ (≥0). Then, for every$$x\geq 0$$and$$n\geq 1$$,

\begin{aligned} \biggl\vert \frac{n}{c_{n}-a_{n}}\tau _{n}(G) (x)-g(x) \biggr\vert \leq \omega \biggl(g,\frac{b_{n}-c_{n}}{n} \biggr). \end{aligned}
(5.1)

Moreover, for every$$\gamma >0$$,

\begin{aligned} \omega \bigl(\tau _{n}(G),\gamma \bigr)\leq \frac{b_{n}-c_{n}}{n}\omega \biggl(g, \gamma +\frac{b_{n}-c_{n}}{n} \biggr). \end{aligned}
(5.2)

### Proof

Let $$x\geq 0$$ be fixed and $$n\geq 1$$; then by applying Lagrange’s theorem to the function G in the interval $$[x+\frac{c_{n}}{n},x+\frac{b_{n}}{n} ]$$, we have

\begin{aligned} \frac{n}{b_{n}-c_{n}}\tau _{n}(G) (x)=g(\xi _{n,x}), \end{aligned}

where $$\xi _{n,x}$$ is a point in $$[x+\frac{c_{n}}{n},x+\frac{b_{n}}{n} ]$$.

Now,

\begin{aligned} \biggl\vert \frac{n}{b_{n}-c_{n}}\tau _{n}(G) (x)-g(x) \biggr\vert = \bigl\vert g(\xi _{n,x})-g(x) \bigr\vert \leq \omega \bigl(g, \vert \xi _{n,x}-x \vert \bigr)\leq \omega \biggl(g, \frac{b_{n}-c_{n}}{n} \biggr). \end{aligned}

Now, for $$x,y\geq 0$$ such that $$\vert x-y \vert \leq \gamma$$ where $$\gamma \geq 0$$ is a fixed number, again by Lagrange’s theorem,

\begin{aligned} \bigl\vert \tau _{n}(G) (x)-\tau _{n}(G) (y) \bigr\vert =\frac{b_{n}-c_{n}}{n} \bigl\vert g(\xi _{n,x})-g( \zeta _{n,y}) \bigr\vert \leq \frac{b_{n}-c_{n}}{n} \omega \bigl(g, \vert \xi _{n,x}-\zeta _{n,y} \vert \bigr), \end{aligned}

where $$\zeta _{n,y}$$ is some element in the interval $$[y+\frac{c_{n}}{n},y+\frac{b_{n}}{n} ]$$, and hence the result since

\begin{aligned} \vert \xi _{n,x}-\zeta _{n,y} \vert \leq \vert x-y \vert +\frac{b_{n}-c_{n}}{n}\leq \gamma + \frac{b_{n}-c_{n}}{n}. \end{aligned}

□

### Theorem 4

Consider$$g\in \wp _{b}([0,\infty ))$$, $$n\geq 1$$, and$$x\geq 0$$. Then

\begin{aligned}& \bigl\vert P_{n}(g) (x)-g(x) \bigr\vert \\& \quad \leq \biggl(2+\sqrt{x+\frac{1}{nH(1,1)} \bigl(H_{t_{1}}(1,1)+H_{t_{2}}(1,1)+H_{t_{1}t_{1}}(1,1)+H_{t_{2}t_{2}}(1,1)+2H_{t_{1}t_{2}}(1,1)} \bigr) \biggr) \\& \quad\quad{}\times \omega \biggl(g,\frac{\sqrt{n}+b_{n}-c_{n}}{n} \biggr). \end{aligned}
(5.3)

Furthermore, ifgis differentiable on$$[0,\infty )$$and$$g'\in \wp _{b}([0,\infty ))$$, then

\begin{aligned}& \bigl\vert P_{n}(g) (x)-g(x) \bigr\vert \\& \quad \leq \biggl(\sqrt{\frac{x}{n}+ \frac{1}{n^{2}H(1,1)}\bigl(H_{t_{1}}(1,1)+H_{t_{2}}(1,1)+H_{t_{1}t_{1}}(1,1)+H_{t_{2}t_{2}}(1,1)+2H_{t_{1}t_{2}}(1,1)} \bigr) \biggr) \\& \quad \quad {}\times \biggl(1+\sqrt{x+\frac{1}{nH(1,1)}\bigl(H_{t_{1}}(1,1)+H_{t_{2}}(1,1)+H_{t_{1}t_{1}}(1,1)+H_{t_{2}t_{2}}(1,1)+2H_{t_{1}t_{2}}(1,1) \bigr)} \biggr) \\& \quad \quad {}\times \omega \biggl(g',\frac{\sqrt{n}+b_{n}-c_{n}}{n} \biggr)+ \bigl\Vert g' \bigr\Vert _{\infty } \frac{b_{n}-c_{n}}{n}. \end{aligned}
(5.4)

### Proof

From [, Theorem 5.2.4], it follows that, for every $$\gamma >0$$,

\begin{aligned}& \bigl\vert S_{n}(g) (x)-g(x) \bigr\vert \\& \quad \leq \biggl(1+\frac{1}{\gamma }\sqrt{\frac{x}{n}+ \frac{1}{n^{2}H(1,1)}\bigl(H_{t_{1}}(1,1)+H_{t_{2}}(1,1)+H_{t_{1}t_{1}}(1,1)+H_{t_{2}t_{2}}(1,1)+2H_{t_{1}t_{2}}(1,1) \bigr)} \biggr) \\& \quad\quad{}\times \omega (g,\gamma ). \end{aligned}
(5.5)

From this and from (5.1) and (5.2), we have

\begin{aligned}& \bigl\vert P_{n}(g) (x)-g(x) \bigr\vert \\& \quad \leq \frac{n}{b_{n}-c_{n}} \bigl\vert S_{n}\bigl(\tau _{n}(G)\bigr) (x)-\tau _{n}(G) (x) \bigr\vert + \biggl\vert \frac{n}{b_{n}-c_{n}}\tau _{n}(G) (x)-g(x) \biggr\vert \\& \quad \leq \biggl(1+\frac{1}{\gamma }\sqrt{\frac{x}{n}+ \frac{1}{n^{2}H(1,1)}\bigl(H_{t_{1}}(1,1)+H_{t_{2}}(1,1)+H_{t_{1}t_{1}}(1,1)+H_{t_{2}t_{2}}(1,1)+2H_{t_{1}t_{2}}(1,1) \bigr)} \biggr) \\& \quad \quad {}\times\frac{n}{b_{n}-c_{n}}\omega \bigl(\tau _{n}(G),\gamma \bigr)+ \omega \biggl(g,\frac{b_{n}-c_{n}}{n} \biggr) \\& \quad \leq \biggl(1+ \frac{1}{\gamma }\sqrt{\frac{x}{n}+ \frac{1}{n^{2}H(1,1)}\bigl(H_{t_{1}}(1,1)+H_{t_{2}}(1,1)+H_{t_{1}t_{1}}(1,1)+H_{t_{2}t_{2}}(1,1)+2H_{t_{1}t_{2}}(1,1) \bigr)} \biggr) \\& \quad \quad {}\times\omega \biggl(g,\gamma +\frac{b_{n}-c_{n}}{n} \biggr)+\omega \biggl(g, \frac{b_{n}-c_{n}}{n} \biggr) \\& \quad \leq \biggl(2+\frac{1}{\gamma }\sqrt{\frac{x}{n}+ \frac{1}{n^{2}H(1,1)}\bigl(H_{t_{1}}(1,1)+H_{t_{2}}(1,1)+H_{t_{1}t_{1}}(1,1)+H_{t_{2}t_{2}}(1,1)+2H_{t_{1}t_{2}}(1,1) \bigr)} \biggr) \\& \quad \quad{}\times \omega \biggl(g,\gamma +\frac{b_{n}-c_{n}}{n} \biggr). \end{aligned}

Taking $$\gamma = \frac{1}{\sqrt{n}}$$, we have (5.3).

And for (5.4), assume that g is differentiable on $$[0,\infty )$$ and $$g'\in \wp _{b}([0,\infty ))$$; then

\begin{aligned} \omega \biggl(g,\frac{b_{n}-c_{n}}{n} \biggr)\leq \bigl\Vert g' \bigr\Vert _{\infty } \frac{b_{n}-c_{n}}{n}. \end{aligned}
(5.6)

Moreover, from [, Theorem 5.2.4], we have, for $$\gamma >0$$,

\begin{aligned}& \bigl\vert S_{n}(g) (x)-g(x) \bigr\vert \\& \quad \leq \sqrt{\frac{x}{n}+\frac{1}{n^{2}H(1,1)}\bigl(H_{t_{1}}(1,1)+H_{t_{2}}(1,1)+H_{t_{1}t_{1}}(1,1)+H_{t_{2}t_{2}}(1,1)+2H_{t_{1}t_{2}}(1,1) \bigr)} \\& \quad \quad {}\times\biggl(1+\frac{1}{\gamma }\sqrt{\frac{x}{n}+ \frac{1}{n^{2}H(1,1)}\bigl(H_{t_{1}}(1,1)+H_{t_{2}}(1,1)+H_{t_{1}t_{1}}(1,1)+H_{t_{2}t_{2}}(1,1)+2H_{t_{1}t_{2}}(1,1) \bigr)} \biggr) \\& \quad \quad {}\times\omega \bigl(g',\gamma \bigr). \end{aligned}
(5.7)

Since g is differentiable, $$\tau _{n}(G)$$ is also differentiable with bounded and continuous derivative, and for every $$x\geq 0$$ and $$n\geq 1$$,

\begin{aligned} \tau _{n}(G)'(x)=g \biggl(x+\frac{b_{n}}{n} \biggr)-g \biggl(x+ \frac{c_{n}}{n} \biggr). \end{aligned}

Now, if $$x,y\geq 0$$ are two elements such that $$\vert x-y \vert \leq \gamma$$, then by Lagrange’s theorem, there exist $$\eta _{n,x}\in [x+\frac{c_{n}}{n},x+\frac{b_{n}}{n} ]$$ and $$\xi _{n,y}\in [y+\frac{c_{n}}{n},y+\frac{b_{n}}{n} ]$$ such that

\begin{aligned} \bigl\vert \tau _{n}(G)'(x)-\tau _{n}(G)'(y) \bigr\vert =& \biggl\vert g \biggl(x+\frac{b_{n}}{n} \biggr)-g \biggl(x+ \frac{c_{n}}{n} \biggr)-g \biggl(y+\frac{b_{n}}{n} \biggr)+g \biggl(y+ \frac{c_{n}}{n} \biggr) \biggr\vert \\ =&\frac{b_{n}-c_{n}}{n} \bigl\vert g'(\eta _{n,x})-g'( \xi _{n,y}) \bigr\vert \leq \frac{b_{n}-c_{n}}{n}\omega \biggl(g',\gamma +\frac{b_{n}-c_{n}}{n} \biggr). \end{aligned}

Hence,

\begin{aligned} \omega \bigl(\tau _{n}(G)', \gamma \bigr)\leq \frac{b_{n}-c_{n}}{n}\omega \biggl(g', \gamma + \frac{b_{n}-c_{n}}{n} \biggr), \end{aligned}
(5.8)

and by (5.6), (5.7), and (5.8),

\begin{aligned}& \bigl\vert P_{n}(g) (x)-g(x) \bigr\vert \\& \quad \leq \frac{n}{b_{n}-c_{n}} \bigl\vert S_{n}\bigl(\tau _{n}(G)\bigr) (x)-\tau _{n}(G) (x) \bigr\vert + \biggl\vert \frac{n}{b_{n}-c_{n}}\tau _{n}(G) (x)-g(x) \biggr\vert \\& \quad \leq \frac{n}{b_{n}-c_{n}} \\& \quad\quad{}\times \sqrt{\frac{x}{n}+\frac{1}{n^{2}H(1,1)} \bigl(H_{t_{1}}(1,1)+H_{t_{2}}(1,1)+H_{t_{1}t_{1}}(1,1)+H_{t_{2}t_{2}}(1,1)+2H_{t_{1}t_{2}}(1,1) \bigr)} \\& \quad \quad {}\times\biggl(1+\frac{1}{\gamma }\sqrt{\frac{x}{n}+ \frac{1}{n^{2}H(1,1)}\bigl(H_{t_{1}}(1,1)+H_{t_{2}}(1,1)+H_{t_{1}t_{1}}(1,1)+H_{t_{2}t_{2}}(1,1)+2H_{t_{1}t_{2}}(1,1) \bigr)} \biggr) \\& \quad \quad {}\times\omega \bigl(\tau _{n}(G)',\gamma \bigr)+ \omega \biggl(g,\frac{n}{b_{n}-c_{n}} \biggr) \\& \quad \leq \sqrt{\frac{x}{n}+\frac{1}{n^{2}H(1,1)}\bigl(H_{t_{1}}(1,1)+H_{t_{2}}(1,1)+H_{t_{1}t_{1}}(1,1)+H_{t_{2}t_{2}}(1,1)+2H_{t_{1}t_{2}}(1,1) \bigr)} \\& \quad \quad {}\times\biggl(1+\frac{1}{\gamma }\sqrt{\frac{x}{n}+ \frac{1}{n^{2}H(1,1)}\bigl(H_{t_{1}}(1,1)+H_{t_{2}}(1,1)+H_{t_{1}t_{1}}(1,1)+H_{t_{2}t_{2}}(1,1)+2H_{t_{1}t_{2}}(1,1) \bigr)} \biggr) \\& \quad \quad {}\times\omega \biggl(g',\gamma +\frac{b_{n}-c_{n}}{n} \biggr)+ \bigl\Vert g' \bigr\Vert _{\infty } \frac{b_{n}-c_{n}}{n}. \end{aligned}

In particular, for $$\gamma = \frac{1}{n^{\frac{1}{2}}}$$, we have (5.4). □

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### Acknowledgements

Pooja Gupta and Ramu Dubey gratefully acknowledge the Department of Mathematics, J.C. Bose University of Science and Technology, YMCA, Faridabad-121 006, Haryana, India. Also, all the authors would like to express sincere thanks to the referees for their valuable suggestions and remarks towards the improvement of the paper.

Nil.

Nil.

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Correspondence to Vishnu Narayan Mishra.

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