Convergence of Certain Baskakov Operators of Integral Type ^†

Abstract

In the present paper, we propose a Baskakov operator of integral type using a function $\phi$ on $[0,\infty )$ with the properties: $\phi \left(0\right)=0,$${\phi }^{\prime }>0$ on $[0,\infty )$ and ${lim}_{x\to \infty }\phi \left(x\right)=\infty$. The proposed operators reproduce the function $\phi$ and constant functions. For the constructed operator, some approximation properties are studied. Voronovskaja asymptotic type formulas for the proposed operator and its derivative are also considered. In the last section, the interest is focused on weighted approximation properties, and a weighted convergence theorem of Korovkin’s type on unbounded intervals is obtained. The results can be extended on the interval $(-\infty ,0]$ (the symmetric of the interval $[0,\infty )$ from the origin).

1. Introduction

Since 1912, when Bernstein introduced his famous polynomials, in order to prove Weierstrass’s fundamental theorem in the most elegant form, the field of approximation theory proved its usefulness many times with various applications.

Different kinds of generalizations of Bernstein operators have become a powerful tool in solving differential equations and highlight their applicability in domains such as numerical analysis, computer aided geometric design or artificial neural networks. It is a well known fact that approximation theory shines light on several scientific domains: physics, medicine and engineering sciences. To the readers, we suggest the following references [1,2].

Since 1950, the concept of linear and positive operators has gained great importance: the uniform convergence of a sequence of operators to a continuous function on a finite and closed interval can be proved. This theorem is now known as the famous Popoviciu–Bohman–Korovkin theorem. Therefore, we want to underline why the investigation of approximation properties of linear and positive operators has such an important role.

As an immediate consequence, the study was expanded for linear and positive operators defined in unlimited intervals, compact or complex intervals, and even more for non-positive operators. Furthermore, the study of these operators was extended towards quantum calculus.

In the present paper, we introduced and discussed the approximation properties of a new class of Baskakov operators of integral type, defined on right-unbounded interval $[0,\infty )$.

Starting from the operators defined by Bascanbaz-Tunca, Bodur and Söylemez in 2018 [3], in the most recently published paper by Gupta, Muraru and Radu [4] a hybrid operator of summation integral type was proposed as follows:

$$G_n^{\beta ,\eta }\left(f\right)\left(x\right)=\frac{n(\eta -1)}{\eta }\sum _{i=0}^{\infty }{\omega }_{\beta }(i,nx){\int }_0^{\infty }{m}_{n,i}^{\eta }\left(t\right)f\left(t\right)dt,x\in [0,\infty ),$$

(1)

for $0\le \beta <1$ and $\eta \in \left\{n\right\}\cup \{\infty \},n\in N$, where

$${\omega }_{\beta }(i,nx)=\frac{nx{(nx+1+i\beta )}_{i-1}}{2^i·i!}·2^{-(nx+i\beta )},m_{n,i}^{\eta }\left(t\right)=\frac{{\left(\eta \right)}_i}{i!}·\frac{{\left(\frac{nt}{\eta }\right)}^i}{{\left(1+\frac{nt}{\eta }\right)}^{\eta +i}}.$$

As special cases of (1), some of the important classes of integral type operators are obtained. If $\eta =n$, we immediately get Jain-Lupaş-Baskakov operators [4], defined for $x\in [0,\infty )$ by

$$G_n^{\beta ,n}\left(f\right)\left(x\right)=(n-1)\sum _{i=0}^{\infty }\frac{nx{(nx+1+i\beta )}_{i-1}}{2^i·i!}·2^{-(nx+i\beta )}{\int }_0^{\infty }\left(\genfrac{}{}{0pt}{}{n+i-1}{i}\right)\frac{t^i}{{(1+t)}^{n+i}}f\left(t\right)dt.$$

For $\beta =0$ and $\eta =n$, we get Lupas-Baskakov operators [5], defined for $x\in [0,\infty ]$ as

$$G_n^{0,n}\left(f\right)\left(x\right)=(n-1)\sum _{i=0}^{\infty }\frac{{\left(nx\right)}_i}{2^i·i!}·2^{-nx}{\int }_0^{\infty }\left(\genfrac{}{}{0pt}{}{n+i-1}{i}\right)\frac{t^i}{{(1+t)}^{n+i}}f\left(t\right)dt.$$

Let $\phi :[0,\infty )\to [0,\infty )$ be a differentiable function with the properties:

(i)

$\phi \left(0\right)=0,$

(ii)

${\phi }^{\prime }>0$ on $[0,\infty ),$

(iii)

${lim}_{x\to \infty }\phi \left(x\right)=\infty .$

Let $\mathcal{F}=\{f:[0,\infty )\to \mathbb{R}:f$ integrable on $[0,\infty )$ and continuous at the point 0$\}.$ For any function $f\in \mathcal{F}$, we define the following operator

$$G_n^{\phi }\left(f\right)\left(x\right)=2^{-a_n\phi \left(x\right)}\left(f\left(0\right)+\sum _{k=1}^{\infty }\frac{{\left(a_n\phi \left(x\right)\right)}_k}{k!2^k·B(a_n+1,k)}{\int }_0^{\infty }\frac{t^{k-1}}{{(1+t)}^{a_n+1+k}}f\left({\phi }^{-1}\left(t\right)\right)dt\right),x\ge 0$$

(2)

with ${\left\{a_n\right\}}_{n\ge 1}$ an increasing sequence of positive numbers such that ${lim}_{n\to \infty }{a}_n=\infty$. The proposed operators reproduce any function $\phi$, which has the properties i–iii from the above.

The results obtained for the operator $G_n^{\phi }$ from (2) are also valid on the interval $(-\infty ,0]$ (the symmetric of the interval $[0,\infty )$ from origin). In this case, the space of functions is ${\mathcal{F}}_{-}=\{f:(-\infty ,0]\to \mathbb{R}:f$ integrable on $(-\infty ,0]$ and continuous at the point $0\}$ and the function $\phi :(-\infty ,0]\to [0,\infty )$ is a differentiable function satisfying:

(i)

$\phi \left(0\right)=0,$

(ii)

${\phi }^{\prime }<0$ on $(-\infty ,0]$

(iii)

${lim}_{x\to -\infty }\phi \left(x\right)=\infty .$

Our purpose is to study the local approximation properties of the proposed operators of integral type. We start with the values of moments and central moments of these operators. Then, as main results, we present two Voronovskaja type asymptotic theorems for the sequences of the type operators and also for the sequences of their derivatives. Finally, we study the weighted approximation properties, implying the weighted modulus of continuity introduced by Ispir and Atakut in [6].

2. Convergence of the ${\mathit{G}}_{\mathit{n}}$ Operators

In the following, we provide basic approximation properties of our hybrid operators in order to help us in demonstrate some Voronovskaja asymptotic type results.

Theorem 1.

For every $x>0$ and every $n\in \mathbb{N}$, we have:

(i)

$G_n^{\phi }\left({\phi }^0\right)\left(x\right)=1$,

(ii)

$G_n^{\phi }\left(\phi \right)\left(x\right)=\phi \left(x\right)$,

(iii)

$G_n^{\phi }\left({\phi }^2\right)\left(x\right)=\frac{a_n}{a_n-1}{\phi }^2\left(x\right)+\frac{3}{a_n-1}\phi \left(x\right)$, $a_n>1.$

Proof. 

(i)

We have

$$G_n^{\phi }\left({\phi }^0\right)\left(x\right)=2^{-a_n\phi \left(x\right)}\left(1+\sum _{k=1}^{\infty }\frac{{\left(a_n\phi \left(x\right)\right)}_k}{k!2^k·B(a_n+1,k)}{\int }_0^{\infty }\frac{t^{k-1}}{{(1+t)}^{a_n+1+k}}dt\right).$$

As

$${\int }_0^{\infty }\frac{t^{k-1}}{{(1+t)}^{a_n+1+k}}dt=B(a_n+1,k)$$

we get

$$G_n^{\phi }\left({\phi }^0\right)\left(x\right)=2^{-a_n\phi \left(x\right)}\left(1+\sum _{k=1}^{\infty }\frac{{\left(a_n\phi \left(x\right)\right)}_k}{k!2^k}\right)=1.$$

(ii)

We have

$$G_n^{\phi }\left(\phi \right)\left(x\right)=2^{-a_n\phi \left(x\right)}\sum _{k=1}^{\infty }\frac{{\left(a_n\phi \left(x\right)\right)}_k}{k!2^k·B(a_n+1,k)}{\int }_0^{\infty }\frac{t^k}{{(1+t)}^{a_n+1+k}}dt.$$

Since

$${\int }_0^{\infty }\frac{t^k}{{(1+t)}^{a_n+1+k}}dt=B(a_n,k+1)$$

we get

$$G_n^{\phi }\left(\phi \right)\left(x\right)=2^{-a_n\phi \left(x\right)}\sum _{k=1}^{\infty }\frac{{\left(a_n\phi \left(x\right)\right)}_k}{k!2^k}·\frac{k}{a_n}.$$

Taking into account that ${\left(a_n\phi \left(x\right)\right)}_k=a_n\phi \left(x\right){(a_n\phi \left(x\right)+1)}_{k-1},$$k\ge 1$ and changing the summation index, we obtain

$$G_n^{\phi }\left(\phi \right)\left(x\right)=2^{-a_n\phi \left(x\right)-1}\phi \left(x\right)\sum _{k=0}^{\infty }\frac{{(a_n\phi \left(x\right)+1)}_k}{k!2^k}=\phi \left(x\right).$$

(iii)

We have

$$G_n^{\phi }\left({\phi }^2\right)\left(x\right)=2^{-a_n\phi \left(x\right)}\sum _{k=1}^{\infty }\frac{{\left(a_n\phi \left(x\right)\right)}_k}{k!2^k·B(a_n+1,k)}{\int }_0^{\infty }\frac{t^{k+1}}{{(1+t)}^{a_n+1+k}}dt.$$

From

$${\int }_0^{\infty }\frac{t^{k+1}}{{(1+t)}^{a_n+1+k}}dt=B(a_n-1,k+2)$$

we obtain

$$G_n^{\phi }\left({\phi }^2\right)\left(x\right)=2^{-a_n\phi \left(x\right)-1}\frac{1}{a_n-1}\phi \left(x\right)\sum _{k=0}^{\infty }\frac{{(a_n\phi \left(x\right)+1)}_k}{k!2^k}·(k+2).$$

It follows that

$$G_n^{\phi }\left({\phi }^2\right)\left(x\right)=2^{-a_n\phi \left(x\right)-2}\frac{1}{a_n-1}\phi \left(x\right)\sum _{k=1}^{\infty }\frac{(a_n\phi \left(x\right)+1){(a_n\phi \left(x\right)+2)}_{k-1}}{(k-1)!2^{k-1}}+\frac{2\phi \left(x\right)}{a_n-1}$$

and therefore

$$G_n^{\phi }\left({\phi }^2\right)\left(x\right)=\frac{a_n}{a_n-1}{\phi }^2\left(x\right)+\frac{3\phi \left(x\right)}{a_n-1}.$$

□

Lemma 1.

Let $m\in \mathbb{N},$$m\ge 1.$ For every $x>0$ and every $n\in \mathbb{N}$ such that $a_n>m-1$, we have:

$$G_n^{\phi }\left({\phi }^m\right)\left(x\right)=2^{-a_n\phi \left(x\right)}\sum _{k=1}^{\infty }\frac{{\left(a_n\phi \left(x\right)\right)}_k}{k!2^k·B(a_n+1,k)}B(a_n-m+1,k+m).$$

(3)

Proof. 

Considering the classical change of variable ${\displaystyle \frac{t}{1+t}=u}$ in the integral ${\displaystyle {\int }_0^{\infty }\frac{t^{k+m-1}}{{(1+t)}^{a_n+1+k}}dt}$, we arrive at the desired form of Beta function (3). □

Lemma 2.

For every $x>0$, we have the moments of superior grades:

(i)

$G_n^{\phi }\left({\phi }^3\right)\left(x\right)={\displaystyle \frac{\phi \left(x\right)(a_n^2{\phi }^2\left(x\right)+9a_n\phi \left(x\right)+14)}{(a_n-1)(a_n-2)}},a_n>2,$

(ii)

$G_n^{\phi }\left({\phi }^4\right)\left(x\right)={\displaystyle \frac{\phi \left(x\right)(a_n^3{\phi }^3\left(x\right)+18a_n^2{\phi }^2\left(x\right)+83a_n\phi \left(x\right)+90)}{(a_n-1)(a_n-2)(a_n-3)}},a_n>3,$

(iii)

$G_n^{\phi }\left({\phi }^5\right)\left(x\right)={\displaystyle \frac{\phi \left(x\right)(a_n^4{\phi }^4\left(x\right)+30a_n^3{\phi }^3\left(x\right)+275a_n^2{\phi }^2\left(x\right)+870a_n\phi \left(x\right)+744)}{(a_n-1)(a_n-2)(a_n-3)(a_n-4)}},a_n>4,$

(iv)

$G_n^{\phi }\left({\phi }^6\right)\left(x\right)={\displaystyle \frac{\phi \left(x\right)(a_n^5{\phi }^5\left(x\right)+45a_n^4{\phi }^4\left(x\right)+685a_n^3{\phi }^3\left(x\right)+4275a_n^2{\phi }^2\left(x\right)+10474a_n\phi \left(x\right)+7560)}{(a_n-1)(a_n-2)(a_n-3)(a_n-4)(a_n-5)}},$$a_n>5.$

Proof. 

The proof of the above lemma follows easily, using Equation (2) from Lemma 1 and applying properties of the Beta functions. □

Remark 1.

From the well known Bohman–Korovkin theorem and using Theorem 1, we have the following: for every $b>0$ and $f\in C[0,b]$ we have

$$\underset{n\to \infty }{lim}{G}_n^{\phi }\left(f\right)=funiformlyon[0,b].$$

From the definition of function $\phi$ as being differentiable with the properties $\phi \left(0\right)=0,{\phi }^{\prime }>0$ on $[0,\infty )$ and ${lim}_{x\to \infty }\phi \left(x\right)=\infty$, it is clear that both linear and exponential functions fulfill these conditions.

Consequentially, from Theorem 1, Lemma 1 and Lemma 2, we obtain:

Lemma 3.

Let us denote $g_{n,m}^{\phi }\left(x\right)=G_n^{\phi }\left({(\phi (·)-\phi \left(x\right))}^m\right)\left(x\right),m=0,1,2,3,4$, then

(i)

$g_{n,0}^{\phi }\left(x\right)=1,$

(ii)

$g_{n,1}^{\phi }\left(x\right)=0,$

(iii)

$g_{n,2}^{\phi }\left(x\right)={\displaystyle \frac{\phi \left(x\right)\left(\phi \right(x)+3)}{a_n-1}},a_n>1,$

(iv)

$g_{n,3}^{\phi }\left(x\right)={\displaystyle \frac{2\phi \left(x\right)\left(\phi \right(x)+1)\left(2\phi \right(x)+7)}{(a_n-1)(a_n-2)}},a_n>2,$

(v)

$g_{n,4}^{\phi }\left(x\right)={\displaystyle \frac{3a_n{\phi }^2\left(x\right){(\phi \left(x\right)+3)}^2+6\phi \left(x\right)(3{\phi }^3\left(x\right)+18{\phi }^2\left(x\right)+28\phi \left(x\right)+15)}{(a_n-1)(a_n-2)(a_n-3)}},a_n>3,$

(vi)

$g_{n,6}^{\phi }\left(x\right)={\displaystyle \frac{15a_n^2{\phi }^3\left(x\right){(\phi \left(x\right)+3)}^3}{(a_n-1)(a_n-2)(a_n-3)(a_n-4)(a_n-5)}}+O\left(\frac{1}{a_n^4}\right),a_n>5.$

Using Lemma 3 and Mathematica software, the following results hold.

Lemma 4.

We have

(i)

${lim}_{n\to \infty }{a}_n{g}_{n,2}^{\phi }\left(x\right)=\phi \left(x\right)(\phi \left(x\right)+3),$

(ii)

${lim}_{n\to \infty }{a}_n^2{g}_{n,3}^{\phi }\left(x\right)=2\phi \left(x\right)(\phi \left(x\right)+1)(2\phi \left(x\right)+7),$

(iii)

${lim}_{n\to \infty }{a}_n^2{g}_{n,4}^{\phi }\left(x\right)=3{\phi }^2\left(x\right){(\phi \left(x\right)+3)}^2,$

(iv)

${lim}_{n\to \infty }{a}_n^4{g}_{n,8}^{\phi }\left(x\right)=105{\phi }^4\left(x\right){(\phi \left(x\right)+3)}^4,$

(v)

${lim}_{n\to \infty }{a}_n^8{g}_{n,16}^{\phi }\left(x\right)=2027025{\phi }^8\left(x\right){(\phi \left(x\right)+3)}^8.$

As a rapid consequence, considering $\phi =e_1$, where $e_1\left(x\right)=x$ for the sake of the readers, we have

Lemma 5.

Let $n_0\in \mathbb{N}$ such that $a_{n_0}>3$. Then, for every $x\ge 0$ and $n\ge n_0$ we have

$$\left|G_n^{e_1}\left({(·-x)}^4\right)\left(x\right)\right|=O\left(\frac{1}{a_n^2}\right)\left(x^4+x^3+x^2+x\right).$$

Our first important result is a Voronovskaja type theorem for our operators $G_n^{\phi }$ defined in (2).

Theorem 2.

Let $f\in \mathcal{F}$ be a bounded function and $f^{″},{\phi }^{″}$ exist for $x\in [0,\infty )$. Then,

$${\displaystyle \underset{n\to \infty }{lim}{a}_n(G_n^{\phi }\left(f\right)\left(x\right)-f\left(x\right))=\frac{\phi \left(x\right)\left(\phi \right(x)+3)}{2}\left(-\frac{{\phi }^{″}\left(x\right)}{{\phi }^{{\prime }^3}\left(x\right)}{f}^{\prime }\left(x\right)+\frac{1}{{\phi }^{{\prime }^2}\left(x\right)}{f}^{″}\left(x\right)\right).}$$

Proof. 

By the Taylor’s expansion of f, we have

$$f\left(t\right)=f\left(x\right)+f^{\prime }\left(x\right)(t-x)+\frac{1}{2}{f}^{″}\left(x\right){(t-x)}^2+r(t,x){(t-x)}^2,$$

where ${\displaystyle \underset{t\to x}{lim}r(t,x)=0.}$ Operating $G_n^{e_1}$ to the above identity, we obtain

$$G_n^{e_1}\left(f\right)\left(x\right)-f\left(x\right)=g_{n,1}^{e_1}\left(x\right)f^{\prime }\left(x\right)+g_{n,2}^{e_1}\left(x\right)\frac{f^{″}\left(x\right)}{2}+G_n^{e_1}\left(r\left(·,x\right){\left(·-x\right)}^2\right)\left(x\right).$$

Using the Cauchy–Schwarz inequality, we have

$$G_n^{e_1}\left(r\left(·,x\right){\left(·-x\right)}^2\right)\left(x\right)\le \sqrt{G_n^{e_1}\left(r^2\left(·,x\right)\left)\right(x\right)}\sqrt{g_{n,4}^{e_1}\left(x\right)}.$$

In view of Remark 1, we have

$$\underset{n\to \infty }{lim}{G}_n^{e_1}\left(r^2\left(·,x\right)\right)\left(x\right)=r^2\left(x,x\right)=0.$$

From Lemma 4, we get

$$\underset{n\to \infty }{lim}{a}_n{G}_n^{e_1}\left(r\left(·,x\right){\left(·-x\right)}^2\right)\left(x\right)=0.$$

Thus,

$$\underset{n\to \infty }{lim}{a}_n\left(G_n^{e_1}\left(f\right)\left(x\right)-f\left(x\right)\right)=\underset{n\to \infty }{lim}{a}_n\left[g_{n,1}^{e_1}\left(x\right)f^{\prime }\left(x\right)+\frac{1}{2}{f}^{″}\left(x\right)g_{n,2}^{e_1}\left(x\right)+G_n^{e_1}\left(r\left(·,x\right){\left(·-x\right)}^2\right)\left(x\right)\right].$$

Applying Lemma 4, we obtain

$${\displaystyle \underset{n\to \infty }{lim}{a}_n(G_n^{e_1}\left(f\right)\left(x\right)-f\left(x\right))=\frac{x(x+3)}{2}{f}^{″}\left(x\right).}$$

Taking $f:=f\circ {\phi }^{-1}$ and $x:=\phi \left(x\right)$ the result follows immediately. □

For our next result, we will need the values of the derivatives of operators $G_n^{\phi }$ applied on the test functions.

Lemma 6.

Let $e_s\left(t\right)=t^s,s=0,1,2,3,4$, then we have

(i)

${\left(G_n^{e_1}\left(e_0\right)\right)}^{\prime }\left(x\right)=0,$

(ii)

${\left(G_n^{e_1}\left(e_1\right)\right)}^{\prime }\left(x\right)=1,$

(iii)

${\left(G_n^{e_1}\left(e_2\right)\right)}^{\prime }\left(x\right)={\displaystyle \frac{2a_{n}x+3}{a_n-1}},a_n>1,$

(iv)

${\left(G_n^{e_1}\left(e_3\right)\right)}^{\prime }\left(x\right)={\displaystyle \frac{3a_n^2{x}^2+18a_{n}x+14}{(a_n-1)(a_n-2)}},a_n>2,$

(v)

${\left(G_n^{e_1}\left(e_4\right)\right)}^{\prime }\left(x\right)={\displaystyle \frac{4a_n^3{x}^3+54a_n^2{x}^2+166a_{n}x+90}{(a_n-1)(a_n-2)(a_n-3)}},a_n>3,.$

Next, following the proof’s idea of Theorem 3.7. from paper [7], we can provide a Voronovskaja type asymptotic formula for the derivative of our operators.

Theorem 3.

Let $f\in \mathcal{F}$ be a bounded function and $f^{\left(4\right)},{\phi }^{‴}$ exists for $x\in [0,\infty )$. Then,

$$\begin{array}{ccc}\hfill \underset{n\to \infty }{lim}{a}_n\left[{\left(G_n^{\phi }\left(f\right)\right)}^{\prime }\left(x\right)-f^{\prime }\left(x\right)\right]& =& \frac{2\phi \left(x\right)+3}{2}\left(-\frac{{\phi }^{″}\left(x\right)}{{\phi }^{{\prime }^2}\left(x\right)}{f}^{\prime }\left(x\right)+\frac{1}{{\phi }^{\prime }\left(x\right)}{f}^{″}\left(x\right)\right)\hfill \\ & +& \frac{\phi \left(x\right)\left(\phi \right(x)+3)}{2}\left(-\frac{{\phi }^{\prime }\left(x\right)·{\phi }^{‴}\left(x\right)+3{\phi }^{″}{}^2\left(x\right)}{{\phi }^{\prime }{}^4\left(x\right)}{f}^{\prime }\left(x\right)-\frac{3{\phi }^{″}\left(x\right)}{{\phi }^{\prime }{}^3\left(x\right)}{f}^{″}\left(x\right)+\frac{1}{{\phi }^{\prime }{}^2\left(x\right)}{f}^{‴}\left(x\right)\right).\hfill \end{array}$$

Proof. 

Taylor’s theorem provides us

$$f\left(t\right)=\sum _{k=0}^4\frac{{(t-x)}^k}{k!}{f}^{\left(k\right)}\left(x\right)+\psi (t,x){(t-x)}^4,t\in [0,\infty ),$$

(4)

where ${\displaystyle \underset{t\to x}{lim}\psi (t,x)=0.}$

For operators $G_n^{e_1}$, from Equation (4), we obtain

$$\begin{array}{ccc}\hfill {\left(G_n^{e_1}\left(f\right)\right)}^{\prime }\left(x\right)& =& {\left(\frac{\partial }{\partial s}{G}_n^{e_1}\left(f\right)\left(s\right)\right)}_{s=x}=f^{\prime }\left(x\right){\left(\frac{\partial }{\partial s}\left(G_n^{e_1}\left(e_1\right)\left(s\right)-x\right)\right)}_{s=x}\hfill \\ & & +\frac{f^{″}\left(x\right)}{2}{\left(\frac{\partial }{\partial s}\left(G_n^{e_1}\left(e_2\right)\left(s\right)-2x{G}_n^{e_1}\left(e_1\right)\left(s\right)+x^2\right)\right)}_{s=x}\hfill \\ & & +\frac{f^{‴}\left(x\right)}{3!}{\left(\frac{\partial }{\partial s}\left(G_n^{e_1}\left(e_3\right)\left(s\right)-3x{G}_n^{e_1}\left(e_2\right)\left(s\right)+3x^2{G}_n^{e_1}\left(e_1\right)\left(s\right)-x^3\right)\right)}_{s=x}\hfill \\ & & +\frac{f^{\prime }{}^{\left(4\right)}\left(x\right)}{4!}{\left(\frac{\partial }{\partial s}\left(G_n^{e_1}\left(e_4\right)\left(s\right)-4x{G}_n^{e_1}\left(e_3\right)\left(s\right)+6x^2{G}_n^{e_1}\left(e_2\right)\left(s\right)-4x^3{G}_n^{e_1}\left(e_1\right)\left(s\right)+x^4\right)\right)}_{s=x}\hfill \\ & & +{\left(\frac{\partial }{\partial s}{G}_n^{e_1}\left(\psi (·,x){(·-x)}^4\right)(s\right)}_{s=x}.\hfill \end{array}$$

Consequently,

$$\begin{array}{ccc}\hfill {\left(G_n^{e_1}\left(f\right)\right)}^{\prime }\left(x\right)& =& f^{\prime }\left(x\right){\left(\frac{\partial }{\partial s}{G}_n^{e_1}\left(e_1\right)\left(s\right)\right)}_{s=x}+\frac{f^{″}\left(x\right)}{2}{\left(\frac{\partial }{\partial s}{G}_n^{e_1}\left(e_2\right)\left(s\right)-2x\frac{\partial }{\partial s}{G}_n^{e_1}\left(e_1\right)\left(s\right)\right)}_{s=x}\hfill \\ & & +\frac{f^{‴}\left(x\right)}{3!}{\left(\frac{\partial }{\partial s}{G}_n^{e_1}\left(e_3\right)\left(s\right)-3x\frac{\partial }{\partial s}{G}_n^{e_1}\left(e_2\right)\left(s\right)+3x^2\frac{\partial }{\partial s}{G}_n^{e_1}\left(e_1\right)\left(s\right)\right)}_{s=x}\hfill \\ & & +\frac{f^{\prime }{}^{\left(4\right)}\left(x\right)}{4!}{\left(\frac{\partial }{\partial s}{G}_n^{e_1}\left(e_4\right)\left(s\right)-4x\frac{\partial }{\partial s}{G}_n^{e_1}\left(e_3\right)\left(s\right)+6x^2\frac{\partial }{\partial s}{G}_n^{e_1}\left(e_2\right)\left(s\right)-4x^3\frac{\partial }{\partial s}{G}_n^{e_1}\left(e_1\right)\left(s\right)\right)}_{s=x}\hfill \\ & & +{\left(\frac{\partial }{\partial s}{G}_n^{e_1}\left(\psi (·,x){(·-x)}^4\right)\left(s\right)\right)}_{s=x}.\hfill \end{array}$$

Using Lemma 6, we get

$$\begin{array}{ccc}\hfill {\left(G_n^{e_1}\left(f\right)\right)}^{\prime }\left(x\right)& =& f^{\prime }\left(x\right)+\frac{2x+3}{a_n-1}·\frac{f^{″}\left(x\right)}{2}++\frac{3a_{n}x(x+3)+2(7+9x+3x^2)}{(a_n-1)(a_n-2)}·\frac{f^{″}\left(x\right)}{3!}\hfill \\ & & +\frac{2a_n(55x+63x^2+14x^3)+6(15+28x+18x^2+4x^3)}{(a_n-1)(a_n-2)(a_n-2)}·\frac{f^{\left(4\right)}\left(x\right)}{4!}\hfill \\ & & +{\left(\frac{\partial }{\partial s}{G}_n^{e_1}\left(\psi (·,x){(·-x)}^4\right)\left(s\right)\right)}_{s=x}.\hfill \end{array}$$

Taking limit as $n\to \infty$ on both sides of the above equation, we have

$$\begin{array}{ccc}\hfill \underset{n\to \infty }{lim}{a}_n\left[{\left(G_n^{e_1}\left(f\right)\right)}^{\prime }\left(x\right)-f^{\prime }\left(x\right)\right]& =& \frac{2x+3}{2}{f}^{″}\left(x\right)+\frac{x(x+3)}{2}{f}^{‴}\left(x\right)\hfill \\ & & {\displaystyle +\underset{n\to \infty }{lim}{a}_n{\left(\frac{\partial }{\partial s}{G}_n^{e_1}\left(\psi (·,x){(·-x)}^4\right)\left(s\right)\right)}_{s=x}.}\hfill \end{array}$$

Next, we show that

$${\displaystyle \underset{n\to \infty }{lim}{a}_n{\left(\frac{\partial }{\partial s}{G}_n^{e_1}\left(\psi (·,x){(·-x)}^4\right)\left(s\right)\right)}_{s=x}=0.}$$

(5)

We have

$$\begin{array}{ccc}& & a_n{\left(\frac{\partial }{\partial s}{G}_n^{e_1}\left(\psi (·,x){(·-x)}^4\right)\left(s\right)\right)}_{s=x}\hfill \\ & & =2^{-a_{n}x}(-a_n^2)ln2\left(\psi (0,x)x^4+\sum _{k=1}^{\infty }\frac{{\left(a_{n}x\right)}_k}{k!2^{k}B(a_n+1,k)}{\int }_0^{\infty }\frac{t^{k-1}}{{(1+t)}^{a_n+1+k}}\psi (t,x){(t-x)}^{4}dt\right)\hfill \\ & & +2^{-a_{n}x}{a}_n\sum _{k=1}^{\infty }\frac{{\left({\left(a_{n}x\right)}_k\right)}^{\prime }}{k!2^{k}B(a_n+1,k)}{\int }_0^{\infty }\frac{t^{k-1}}{{(1+t)}^{a_n+1+k}}\psi (t,x){(t-x)}^{4}dt\hfill \\ & =& I_1+I_2+I_3,\hfill \end{array}$$

where

$$I_1=2^{-a_{n}x}(-a_n^2)ln2\psi (0,x)x^4,$$

$$I_2=2^{-a_{n}x}(-a_n^2)ln2\sum _{k=1}^{\infty }\frac{{\left(a_{n}x\right)}_k}{k!2^{k}B(a_n+1,k)}{\int }_0^{\infty }\frac{t^{k-1}}{{(1+t)}^{a_n+1+k}}\psi (t,x){(t-x)}^{4}dt,$$

$$I_3=2^{-a_{n}x}{a}_n\sum _{k=1}^{\infty }\frac{{\left({\left(a_{n}x\right)}_k\right)}^{\prime }}{k!2^{k}B(a_n+1,k)}{\int }_0^{\infty }\frac{t^{k-1}}{{(1+t)}^{a_n+1+k}}\psi (t,x){(t-x)}^{4}dt.$$

We get

$$\begin{array}{ccc}\hfill \left|I_2\right|& \le & 2^{-a_{n}x}{a}_n^{2}ln2\sum _{k=1}^{\infty }\frac{{\left(a_{n}x\right)}_k}{k!2^{k}B(a_n+1,k)}{\left({\int }_0^{\infty }\frac{t^{k-1}}{{(1+t)}^{a_n+1+k}}{\psi }^2(t,x)dt\right)}^{\frac{1}{2}}{\left({\int }_0^{\infty }\frac{t^{k-1}}{{(1+t)}^{a_n+1+k}}{(t-x)}^{8}dt\right)}^{\frac{1}{2}}\hfill \\ & \le & ln2{\left(2^{-a_{n}x}\sum _{k=1}^{\infty }\frac{{\left(a_{n}x\right)}_k}{k!2^{k}B(a_n+1,k)}{\int }_0^{\infty }\frac{t^{k-1}}{{(1+t)}^{a_n+1+k}}{\psi }^2(t,x)dt\right)}^{\frac{1}{2}}\hfill \\ & & \times {\left(a_n^4·2^{-a_{n}x}\sum _{k=1}^{\infty }\frac{{\left(a_{n}x\right)}_k}{k!2^{k}B(a_n+1,k)}{\int }_0^{\infty }\frac{t^{k-1}}{{(1+t)}^{a_n+1+k}}{(t-x)}^{8}dt\right)}^{\frac{1}{2}}\hfill \\ & \le & ln2·\sqrt{G_n^{e_1}\left({\psi }^2(·,x)\right)\left(x\right)}·\sqrt{a_n^4{g}_{n,8}^{e_1}\left(x\right)}.\hfill \end{array}$$

As

$${\left({\left(a_{n}x\right)}_k\right)}^{\prime }={\left(\prod _{l=1}^k(a_{n}x+l-1)\right)}^{\prime }=\sum _{j=1}^k\frac{a_n{\left(a_{n}x\right)}_k}{a_{n}x+j-1}\le \frac{k·{\left(a_{n}x\right)}_k}{x}$$

it follows that

$$\begin{array}{ccc}\hfill \left|I_3\right|& \le & 2^{-a_{n}x}\frac{a_n}{x}\sum _{k=1}^{\infty }\frac{k·{\left(a_{n}x\right)}_k}{k!2^{k}B(a_n+1,k)}{\int }_0^{\infty }\frac{t^{k-1}}{{(1+t)}^{a_n+1+k}}\left|\psi (t,x)\right|{(t-x)}^{4}dt\hfill \\ & \le & 2^{-a_{n}x}\frac{a_n}{x}\sum _{k=1}^{\infty }\frac{k·{\left(a_{n}x\right)}_k}{k!2^{k}B(a_n+1,k)}{\left({\int }_0^{\infty }\frac{t^{k-1}}{{(1+t)}^{a_n+1+k}}dt\right)}^{\frac{1}{2}}{\left({\int }_0^{\infty }\frac{t^{k-1}}{{(1+t)}^{a_n+1+k}}{\psi }^2(t,x){(t-x)}^{8}dt\right)}^{\frac{1}{2}}\hfill \\ & \le & \frac{a_n}{x}{\left(2^{-a_{n}x}\sum _{k=1}^{\infty }\frac{k^2·{\left(a_{n}x\right)}_k}{k!2^k}\right)}^{\frac{1}{2}}{\left(2^{-a_{n}x}\sum _{k=1}^{\infty }\frac{{\left(a_{n}x\right)}_k}{k!2^{k}B(a_n+1,k)}{\int }_0^{\infty }\frac{t^{k-1}}{{(1+t)}^{a_n+1+k}}{\psi }^2(t,x){(t-x)}^{8}dt\right)}^{\frac{1}{2}}\hfill \\ & \le & \frac{a_n}{x}\sqrt{a_{n}x(2+a_{n}x)}{\left(2^{-a_{n}x}\sum _{k=1}^{\infty }\frac{{\left(a_{n}x\right)}_k}{k!2^{k}B(a_n+1,k)}{\int }_0^{\infty }\frac{t^{k-1}}{{(1+t)}^{a_n+1+k}}{\psi }^4(t,x)dt\right)}^{\frac{1}{4}}\hfill \\ & & \times {\left(2^{-a_{n}x}\sum _{k=1}^{\infty }\frac{{\left(a_{n}x\right)}_k}{k!2^{k}B(a_n+1,k)}{\int }_0^{\infty }\frac{t^{k-1}}{{(1+t)}^{a_n+1+k}}{(t-x)}^{16}dt\right)}^{\frac{1}{4}}\hfill \\ & \le & \sqrt{\frac{2}{a_{n}x}+1}·{\left(G_n^{e_1}\left({\psi }^4(·,x)\right)\left(x\right)\right)}^{\frac{1}{4}}·{\left(a_n^8{g}_{n,16}^{e_1}\left(x\right)\right)}^{\frac{1}{4}}.\hfill \end{array}$$

Taking into account that

$$\underset{n\to \infty }{lim}{G}_n^{e_1}\left({\psi }^j(·,x)\right)\left(x\right)={\psi }^j(x,x)=0,j=2,4$$

and Lemma 4, we get

$$\underset{n\to \infty }{lim}{I}_j=0,j=2,3.$$

Since

$$\underset{n\to \infty }{lim}{I}_1=0$$

it follows the limit in (5) holds, and therefore

$$\underset{n\to \infty }{lim}{a}_n\left[{\left(G_n^{e_1}\left(f\right)\right)}^{\prime }\left(x\right)-f^{\prime }\left(x\right)\right]=\frac{2x+3}{2}{f}^{″}\left(x\right)+\frac{x(x+3)}{2}{f}^{‴}\left(x\right).$$

Taking $f:=f\circ {\phi }^{-1}$and $x:=\phi \left(x\right)$, the result follows immediately and the proof is completed. □

3. Weighted Approximation

Now we shall discuss the weighted approximation theorem, where the approximation formula holds true on the interval $[0,\infty )$.

First, we set the $B_{\omega }$ the space of all functions f, defined on $[0,\infty )$

$$\left|f\left(x\right)\right|\le M_f\omega \left(x\right),$$

where $\omega \left(x\right)=1+{\phi }^2\left(x\right).$

The space is endowed with the norm

$${∥f∥}_{\omega }=\underset{x\in R^{+}}{sup}\frac{\left|f\left(x\right)\right|}{\omega \left(x\right)},f\in B_{\omega }.$$

The space of all continuous functions belonging to $B_{\omega }$ is denoted by $C_{\omega }$. We will also consider the subspace $C_{\omega }^{*}$ of all continuous functions belonging to $B_{\omega },$ for which

$$\genfrac{}{}{0pt}{}{lim}{x\to \infty }\frac{f\left(x\right)}{\omega \left(x\right)}=k_f,$$

where $k_f\in R$ and depending on f.

For $f\in C_{\omega }^{*},$ the weighted modulus of continuity introduced by Ispir and Atakut [6] is denoted as follows:

$$\begin{array}{c}\hfill \Omega (f,\delta )=\underset{\left|h\left(x\right)\right|\le \delta ,x\in [0,\infty )}{sup}\frac{\left|f\left(x+h\right)-f\left(x\right)\right|}{\omega \left(x\right)\left(1+h^2\right)},f\in B_{\omega },\end{array}$$

(6)

which tends to zero as $\delta \to 0$ on an infinite interval. For each $\lambda >0$,

$$\begin{array}{c}\hfill \Omega (f,\lambda \delta )\le 2(1+{\lambda }^2)((1+{\delta }^2)\Omega (f,\delta ).\end{array}$$

(7)

The weighted modulus of continuity defined by (6) has the following properties, in addition of (7):

(i)

$\Omega (f,\delta )$ is a monotonically increasing function with respect to $\delta ,$$\delta \ge 0;$

(ii)

for $f\in C_{\omega }^{*}$ and $x,t\in [0,\infty ),$

$$\left|f\left(x\right)-f\left(t\right)\right|\le 2(1+\frac{\left|t-x\right|}{\delta })(1+{\delta }^2)\Omega (f,\delta )(1+x^2)(1+{(t-x)}^2).$$

The space $C_{\omega }$ and $B_{\omega },$ were defined by Gadzhiev in [8,9] and showed that the classical Korovkin theorem does not hold on these spaces of functions defined on unbounded sets.

In order to prove our main result we will recall the result proved by Gadzhiev in [9] on unbounded intervals, which is a weighted convergence theorem of Korovkin’s type.

Theorem 4

([9]). For any function φ, there is a sequence of positive linear operators ${\left\{L\right\}}_n,$ mapping from $C_{\omega }$ to $B_{\omega }$ satisfying the conditions

$${∥L_n\left({\phi }^i\right)-{\phi }^i∥}_{\omega }\to 0,i=0,1,2.$$

Then, for any function $f\in C_{\omega }^{*}$

$${∥L_n\left(f\right)-f∥}_{\omega }\to 0,n\to \infty .$$

In the light of Theorem 4, we obtain the next result applied on our operators.

Theorem 5.

Let $\left\{G_n^{\phi }\right\}$ be the sequence of linear positive operators defined by (2), then for each $f\in C_{\omega }^{*},$

$$\underset{n\to \infty }{lim}{∥G_n^{\phi }\left(f\right)-f∥}_{\omega }=0.$$

Proof. 

In the first place, we will underline that positive operator $G_n^{\phi }$ acts from the space of all continuous functions on $[0,\infty ),$ belonging to $B_{\omega }$ to $B_{\omega },$ if and only if

$${∥G_n^{\phi }\left(\omega \right)∥}_{\omega }\le k_{\omega },$$

(8)

where $k_{\omega }$ is a positive constant (see [8]).

Thus, for our operators, using Theorem 1, the condition (8) becomes

$$\underset{n\to \infty }{lim}\underset{x\ge 0}{sup}\frac{\left|G_n^{\phi }\left(\omega \right)\left(x\right)\right|}{\omega \left(x\right)}=\underset{n\to \infty }{lim}\underset{x\ge 0}{sup}\frac{\left|\frac{a_n}{a_n-1}{\phi }^2\left(x\right)+\frac{3}{a_n-1}\phi \left(x\right)\right|}{\omega \left(x\right)}.$$

Taking into account the following inequalities

$$\frac{\phi \left(x\right)}{1+{\phi }^2\left(x\right)}\le \frac{1}{2}\mathrm{and}\frac{{\phi }^2\left(x\right)}{1+{\phi }^2\left(x\right)}\le 1$$

(9)

we find that

$$\underset{n\to \infty }{lim}{∥G_n^{\phi }\left(\omega \right)∥}_{\omega }\le 1.$$

To complete the proof of Theorem 5, it is sufficient to prove the following

$$\underset{n\to \infty }{lim}{∥G_n^{\phi }\left({\phi }^i\right)-{\phi }^i∥}_{\omega }=0,\mathrm{for}i=0,1,2.$$

It is obvious that ${∥G_n^{\phi }\left(e_0\right)-e_0∥}_{\omega }\to 0$ as $n\to \infty$ on $[0,\infty ).$

From Theorem 1 also, we easily find that

$$\underset{n\to \infty }{lim}{∥G_n^{\phi }\left(\phi \right)-\phi ∥}_{\omega }=0.$$

The third condition becomes

$$\underset{n\to \infty }{lim}{∥G_n^{\phi }\left({\phi }^2\right)-{\phi }^2∥}_{\omega }=\underset{n\to \infty }{lim}\underset{x\ge 0}{sup}\frac{\left|\frac{a_n}{a_n-1}{\phi }^2\left(x\right)+\frac{3}{a_n-1}\phi \left(x\right)-{\phi }^2\left(x\right)\right|}{\omega \left(x\right)}.$$

Furthermore, considering the inequalities (9), the following relation holds

$$\underset{n\to \infty }{lim}{∥G_n^{\phi }\left({\phi }^2\right)-{\phi }^2∥}_{\omega }=\underset{n\to \infty }{lim}\left(\frac{a_n}{a_n-1}-1\right)\underset{x\ge 0}{sup}\frac{{\phi }^2\left(x\right)}{1+{\phi }^2\left(x\right)}+\underset{n\to \infty }{lim}\frac{3}{a_n-1}\underset{x\ge 0}{sup}\frac{\phi \left(x\right)}{1+{\phi }^2\left(x\right)}.$$

Thus,

$$\underset{n\to \infty }{lim}{∥G_n^{\phi }\left({\phi }^2\right)-{\phi }^2∥}_{\omega }=0.$$

□

Another main result of the present section is regarding the rate of convergence for the operators (2), in the weighted spaces by means of the weighted modulus of continuity $\Omega (f,\delta )$, defined at (6).

Theorem 6.

Let $n_0\in \mathbb{N}$ such that $a_{n_0}>3$. Then, for every $f\in C_{\rho }^{*}$ and $n\ge n_0$, we have

$$\underset{x\ge 0}{sup}\frac{\left|G_n^{\phi }\left(f\right)\left(x\right)-f\left(x\right)\right|}{{(1+{\phi }^2\left(x\right))}^3}\le M\Omega (f\circ {\phi }^{-1},a_n^{-1/2})$$

where M is a constant independent of $a_n.$

Proof. 

Let $0<{\delta }_n<1/3.$ Using property (7), we have

$$\left|G_n^{e_1}\left(f\right)\left(x\right)-f\left(x\right)\right|\le 2(1+{\delta }_n^2)\Omega (f,{\delta }_n)(1+x^2)G_n^{e_1}\left(\left(1+\frac{\left|·-x\right|}{{\delta }_n}\right)\left(1+{(·-x)}^2\right)\right)\left(x\right).$$

Since

$$\left(1+\frac{\left|t-x\right|}{{\delta }_n}\right)\left(1+{(t-x)}^2\right)\le 2(1+{\delta }_n^2)\left(1+\frac{{(t-x)}^4}{{\delta }_n^4}\right),x,t\ge 0$$

we get

$$\left|G_n^{e_1}\left(f\right)\left(x\right)-f\left(x\right)\right|<\frac{400}{81}\Omega (f,{\delta }_n)(1+x^2)\left(1+\frac{1}{{\delta }_n^4}{G}_n^{e_1}\left({(·-x)}^4\right)\left(x\right)\right)$$

From Lemma 6 it follows, for $n\ge n_0,$

$$\left|G_n^{e_1}\left(f\right)\left(x\right)-f\left(x\right)\right|<\frac{400}{81}\Omega (f,{\delta }_n)(1+x^2)\left(1+\frac{1}{{\delta }_n^4}O\left(\frac{1}{a_n^2}\right)\left(x^4+x^3+x^2+x\right)\right).$$

Taking $f:=f\circ {\phi }^{-1}$and $x:=\phi \left(x\right)$, we obtain

$$\left|G_n^{\phi }\left(f\right)\left(x\right)-f\left(x\right)\right|<\frac{400}{81}\Omega (f\circ {\phi }^{-1},{\delta }_n)(1+{\phi }^2\left(x\right))\left(1+\frac{1}{{\delta }_n^4}O\left(\frac{1}{a_n^2}\right)\left({\phi }^4\left(x\right)+{\phi }^3\left(x\right)+{\phi }^2\left(x\right)+\phi \left(x\right)\right)\right)$$

Choosing ${\delta }_n=a_n^{-1/2}$ and sufficient large n it follows that

$$\underset{x\ge 0}{sup}\frac{\left|G_n^{\phi }\left(f\right)\left(x\right)-f\left(x\right)\right|}{{(1+{\phi }^2\left(x\right))}^3}\le M\Omega (f\circ {\phi }^{-1},{\delta }_n).$$

Thus, the proof is ended. □