Keywords

AMS MSC 2010

1 Introduction

Very recently, Varma and Taşdelen [12] introduced the Szász-type operators involving Charlier polynomials (1.1). Also, they estimated some results for the Kantorovich-type generalization of these operators and established the convergence properties for their operators with the help of Korovkin’s theorem and the order of approximation by using the classical modulus of continuity. The operators discussed in [12] are defined as

$$\begin{aligned} L_{n}(f;x,a)=e^{-1}\left( 1-\frac{1}{a}\right) ^{(a-1)nx}\sum _{k=0}^{\infty }\frac{C_k^{(a)}(-(a-1)nx)}{k!}f\left( \frac{k}{n}\right) \end{aligned}$$
(1.1)

where \(a>0, \,x\in [0,\infty )\) and \(C_k^{(a)}\) be the Charlier polynomials, which have the generating functions of the type

$$\begin{aligned} e^{t}\left( 1-\frac{t}{a}\right) ^{x}=\sum _{k=0}^{\infty }\frac{C_k^{(a)}(x)}{k!}t^k,\,\,\,\,\, |t|<a, \end{aligned}$$

and the explicit representation

$$\begin{aligned} C_k^{(a)}(u)=\sum _{r=0}^{k}\left( {\begin{array}{c}n\\ r\end{array}}\right) (-u)_r\left( \frac{1}{a}\right) _r, \end{aligned}$$

where \((\alpha )_k\) is the Pochhammer’s symbol given by

$$\begin{aligned} (\alpha )_0=1,\,\,\,\,\, (\alpha )_r=\alpha (\alpha +1)\ldots (\alpha +r-1)\,\,\,\,\,\,\,\,\,\,r=1,2,\ldots \end{aligned}$$

Note that Charlier polynomials are positive if \(a>0,\,\,u\le 0\).

In order to approximate Lebesgue integrable functions, several new modifications of the discrete operators were discovered by the researchers in the last five decades. We mention the recent book [8] for some of the work on the integral operators in this direction and the references therein. Some other integral operators we mention in the papers [5, 6, 9], etc.

Also, recently with an idea of generalization of the Phillips operators [11] (see also [2, 3, 7]), Pǎltǎnea in [10] proposed the modified form of the Phillips operators based on certain parameter \(\rho >0\), which provide the link with the well-known Szász–Mirakyan operators as \(\rho \rightarrow \infty \) for some class of functions. Motivated by such modifications we propose here for \(a>0, \rho \ge 0\) the integral-type generalization of the operator (1.1) as follows:

$$\begin{aligned} T_{n,\rho ,c}(f;x,a)= & {} e^{-1}\left( 1-\frac{1}{a}\right) ^{(a-1)nx}\bigg [C_0^{(a)}f(0)\\&+\sum _{k=1}^{\infty }\frac{C_k^{(a)}(-(a-1)nx)}{k!}\int _0^\infty \varTheta _{n,k}^\rho (t,c)f(t)dt\bigg ]\nonumber \end{aligned}$$
(1.2)

where \(C_k^{(a)}(u)\) is the Charlier polynomial and

$$\begin{aligned} \varTheta _{n,k}^\rho (t,c)=\left\{ \begin{array}{c} \frac{n\rho }{\varGamma {(k\rho )}}e^{-n\rho t}(n\rho t)^{k\rho -1}, c=0 \\ \frac{\varGamma \left( \frac{n\rho }{c}+k\rho \right) }{\varGamma (k\rho )\varGamma \left( \frac{n\rho }{c}\right) }\frac{c^{k\rho }t^{k\rho -1}}{(1+ct)^{\frac{n\rho }{c}+k\rho }},c=1,2,3,\ldots , \end{array} \right. \end{aligned}$$

Remark 1

For \(f\in \overline{\varPi }\), where \(\overline{\varPi }\) be the closure of the space of polynomials, we have

$$\begin{aligned} \lim _{\rho \rightarrow \infty }T_{n,\rho ,c}(f;x,a)=L_n(f;x,a),\,\,\, \text{ for } \text{ all } x\in [0,\infty ). \end{aligned}$$

Since,

$$\begin{aligned} \int _0^\infty \varTheta _{n,k}^\rho (t,c)t^rdt= & {} \frac{\varGamma (k\rho +r)}{\varGamma (k\rho )}\frac{1}{\prod _{i=1}^{r}(n\rho -ic)},\,\,\,\, n\rho >rc \end{aligned}$$

and

$$\begin{aligned} \lim _{\rho \rightarrow \infty }\frac{\varGamma (k\rho +r)}{\varGamma (k\rho )}\frac{1}{\prod _{i=1}^{r}(n\rho -ic)}=\left( \frac{k}{n}\right) ^r,\,\,\,\,n\rho >rc. \end{aligned}$$

From this the result follows immediately.

Remark 2

We obtain Szász–Mirakyan operators by applying, respectively, the following operations to the both sides of (1.2)

  1. (i)

    \(\rho \rightarrow \infty \),

  2. (ii)

    \(a\rightarrow \infty \) and write \(x-\frac{1}{n}\) instead of x.

In the present article, we first obtain the moments of the operators \(T_{n,\rho ,c}(f;x,a).\) Then we establish some direct results in ordinary approximation, which include the asymptotic formula, direct estimate in terms of modulus of continuity and the weighted approximation.

2 Auxiliary Results

In this section we provide the following set of lemmas.

Lemma 1

([12]) For \(L_{n}(t^m;x,a), \, m=0, 1, 2\), we have

$$\begin{aligned} L_{n}(1;x,a)= & {} 1,\\ L_{n}(t;x,a)= & {} x+\frac{1}{n}\\ L_{n}(t^2;x,a)= & {} x^2+\frac{x}{n}\left( 3+\frac{1}{a-1}\right) +\frac{2}{n^2}. \end{aligned}$$

Lemma 2

For \(T_{n,\rho ,c}(t^m;x,a), \, m=0, 1, 2\), we have

$$\begin{aligned} T_{n,\rho ,c}(1;x,a)= & {} 1,\\ T_{n,\rho ,c}(t;x,a)= & {} \frac{\rho (nx+1)}{n\rho -c}\\ T_{n,\rho ,c}(t^2;x,a)= & {} \frac{n^2}{(n\rho -c)(n\rho -2c)}\bigg [\rho ^2x^2+\frac{\rho x}{n}\left( 3\rho +1+\frac{\rho }{a-1}\right) +\frac{\rho (2\rho +1)}{n^2}\bigg ] \end{aligned}$$

Proof

It is easy to see

$$\begin{aligned} \int _0^\infty \varTheta _{n,k}^\rho (t,c)t^rdt= & {} \frac{\varGamma (k\rho +r)}{\varGamma (k\rho )}\frac{1}{\prod _{i=1}^{r}(n\rho -ic)}. \end{aligned}$$

In view of Lemma 1, the zeroth order moment is

$$\begin{aligned} T_{n,\rho ,c}(1;x,a)= & {} e^{-1}\left( 1-\frac{1}{a}\right) ^{(a-1)nx}\left( C_0^{(a)}f(0)+ \sum _{k=1}^{\infty }\frac{C_k^{(a)}(-(a-1)nx)}{k!}\int _0^\infty \varTheta _{n,k}^\rho (t)dt\right) \\= & {} L_n(1;x,a)=1. \end{aligned}$$

First-order moment is

$$\begin{aligned} T_{n,\rho ,c}(t;x,a)= & {} e^{-1}\left( 1-\frac{1}{a}\right) ^{(a-1)nx}\sum _{k=1}^{\infty }\frac{C_k^{(a)}(-(a-1)nx)}{k!}\int _0^\infty \varTheta _{n,k}^\rho (t)tdt\\= & {} e^{-1}\left( 1-\frac{1}{a}\right) ^{(a-1)nx}\sum _{k=1}^{\infty }\frac{C_k^{(a)}(-(a-1)nx)}{k!} \frac{\varGamma (k\rho +1)}{\varGamma (k\rho )}\frac{1}{(n\rho -c)}\\= & {} \frac{n\rho }{(n\rho -c)}e^{-1}\left( 1-\frac{1}{a}\right) ^{(a-1)nx}\sum _{k=0}^{\infty }\frac{C_k^{(a)}(-(a-1)nx)}{k!} \frac{k}{n}\\= & {} \frac{n\rho }{(n\rho -c)}L_n(t;x,a)\\= & {} \frac{\rho (nx+1)}{n\rho -c}. \end{aligned}$$

Second-order moment is

$$\begin{aligned} T_{n,\rho ,c}(t^2;x,a)= & {} e^{-1}\left( 1-\frac{1}{a}\right) ^{(a-1)nx}\sum _{k=1}^{\infty }\frac{C_k^{(a)}(-(a-1)nx)}{k!}\int _0^\infty \varTheta _{n,k}^\rho (t)t^2dt\\= & {} e^{-1}\left( 1-\frac{1}{a}\right) ^{(a-1)nx}\sum _{k=1}^{\infty }\frac{C_k^{(a)}(-(a-1)nx)}{k!} \frac{\varGamma (k\rho +2)}{\varGamma (k\rho )}\frac{1}{(n\rho -c)(n\rho -2c)}\\= & {} \frac{n^2\rho ^2}{(n\rho -c)(n\rho -2c)}e^{-1}\left( 1-\frac{1}{a}\right) ^{(a-1)nx}\sum _{k=0}^{\infty }\frac{C_k^{(a)}(-(a-1)nx)}{k!}\frac{k^2}{n^2}\\&+\frac{n\rho }{(n\rho -c)(n\rho -2c)}e^{-1}\left( 1-\frac{1}{a}\right) ^{(a-1)nx}\sum _{k=0}^{\infty }\frac{C_k^{(a)}(-(a-1)nx)}{k!}\frac{k}{n}\\= & {} \frac{n^2\rho ^2}{(n\rho -c)(n\rho -2c)}L_n(t^2;x,a)+\frac{n\rho }{(n\rho -c)(n\rho -2c)}L_n(t;x,a)\\= & {} \frac{n^2}{(n\rho -c)(n\rho -2c)}\bigg [\rho ^2x^2+\frac{\rho x}{n}\left( 3\rho +1+\frac{\rho }{a-1}\right) +\frac{\rho (2\rho +1)}{n^2}\bigg ]. \end{aligned}$$

Remark 3

By simple computation, we have

$$\begin{aligned} T_{n,\rho ,c}(t-x;x,a)= & {} \frac{cx+\rho }{n\rho -c}\\ T_{n,\rho ,c}((t-x)^2;x,a)= & {} \frac{1}{(n\rho -c)(n\rho -2c)}\bigg [(n\rho +2c)cx^2+n\rho x\left( \rho +1+\frac{\rho }{a-1}\right) \\&\, +\,4\rho cx+{\rho (2\rho +1)}\bigg ] \end{aligned}$$

3 Direct Result and Asymptotic Formula

In this section we discuss the direct result and Voronovskaja-type asymptotic formula.

Let the space \(C_B[0,\infty )\) of all continuous and bounded functions be endowed with the norm \(\Vert f\Vert =\sup \{|f(x)|:x\in [0,\infty )\}.\) Further let us consider the following K-functional:

$$\begin{aligned} K_2(f,\delta )=\inf _{g\in W^2}\{\Vert f-g\Vert +\delta \Vert g^{\prime \prime }\Vert \}, \end{aligned}$$

where \(\delta >0\) and \(W^2=\{g\in C_B[0,\infty ):g^\prime ,g^{\prime \prime }\in C_B[0,\infty )\}.\) By ([1] p. 177, Th. 2.4), there exists an absolute constant \(C>0\) such that

$$\begin{aligned} K_2(f,\delta )\le C\, \omega _2(f,\sqrt{\delta }), \end{aligned}$$
(3.1)

where

$$\begin{aligned} \omega _2(f,\sqrt{\delta })=\sup _{0<h\le \sqrt{\delta }}\sup _{x\in [0,\infty )} |f(x+2h)-2f(x+h)+f(x)| \end{aligned}$$

is the second-order modulus of smoothness of \(f\in C_B[0,\infty ).\)

Theorem 1

For \(f\in C_B[0,\infty )\) and \(a>1\), we have

$$\begin{aligned} |T_{n,\rho ,c}(f;x,a)-f(x)|\le & {} C\omega _2(f,\sqrt{\delta })+\omega \left( f,\left| \frac{cx+\rho }{n\rho -c}\right| \right) \end{aligned}$$

where C is a positive constant and \(\delta =\left| T_{n,\rho ,c}((t-x)^2;x,a)\right| +\frac{1}{2}\left( \frac{cx+\rho }{n\rho -c}\right) ^2.\) Also, the both \(\omega (f, \delta )\) and \(\omega _2(f,\sqrt{\delta })\) tends to zero as \(\delta \rightarrow 0.\)

Proof

We introduce auxiliary operators \(\overline{T}_{n,\rho ,c}\) as follows:

$$\begin{aligned} \overline{T}_{n,\rho ,c}(f;x,a)= & {} T_{n,\rho ,c}(f;x,a)-f\left( x+\frac{cx+\rho }{n\rho -c}\right) +f(x). \end{aligned}$$

These operators are linear and preserve the linear functions in view of Lemma 2. Let \(g\in W^2.\) From the Taylor’s expansion of g we have

$$\begin{aligned} g(t) = g(x)+(t-x)g'(x)+\int _x^t(t-u)g''(u)du. \end{aligned}$$

Applying the operator \(\overline{T}_{n,\rho ,c}\) on above

$$\begin{aligned} \overline{T}_{n,\rho ,c}(g;x,a) = g(x)+g'(x)\overline{T}_{n,\rho ,c}((t-x);x,a)+\overline{T}_{n,\rho ,c}\left( \int _x^t(t-u)g''(u)du;x,a\right) \end{aligned}$$
$$\begin{aligned} |\overline{T}_{n,\rho ,c}(g;x,a)-g(x)|&=\left| \overline{T}_{n,\rho ,c}\left( \int _x^t(t-u)g''(u)du;x,a\right) \right| \nonumber \\&\le \left| T_{n,\rho ,c}\left( \int _x^t(t-u)g''(u)du;x,a\right) \right| \nonumber \\&\quad +\left| \int _x^{x+\frac{cx+\rho }{n\rho -c}}\left( x+\frac{cx+\rho }{n\rho -c}-u)g''(u)\right) du\right| \nonumber \\&\le \left[ \left| T_{n,\rho ,c}\left( \big |\int _x^t|t-u|\,du\big |;x,a\right) \right| \right. \nonumber \\&\quad +\left. \left| \int _x^{x+\frac{cx+\rho }{n\rho -c}}\left| x+\frac{cx+\rho }{n\rho -c}-u\right| du\right| \right] \Vert g^{\prime \prime }\Vert \nonumber \\&\le \left[ \left| T_{n,\rho ,c}((t-x)^2;x,a)\right| +\frac{1}{2}\left( \frac{cx+\rho }{n\rho -c}\right) ^2\right] \Vert g^{\prime \prime }\Vert \end{aligned}$$
(3.2)
$$\begin{aligned}&=\delta \Vert g^{\prime \prime }\Vert , \end{aligned}$$
(3.3)

where \(\delta =\left| T_{n,\rho ,c}((t-x)^2;x,a)\right| +\frac{1}{2}\left( \frac{cx+\rho }{n\rho -c}\right) ^2.\)

$$\begin{aligned} |T_{n,\rho ,c}(f;x,a)-f(x)|\le & {} |\overline{T}_{n,\rho ,c}(f-g;x,a)-(f-g)(x)|+|\overline{T}_{n,\rho ,c}(g;x,a)-g(x)|\\&+\left| f\left( x+\frac{cx+\rho }{n\rho -c}\right) -f(x)\right| \\\le & {} 2\Vert f-g\Vert +\delta \Vert g^{\prime \prime }\Vert +\omega \left( f,\left| \frac{cx+\rho }{n\rho -c}\right| \right) . \end{aligned}$$

Taking infimum over all \(g\in W^2\), we get

$$\begin{aligned} |T_{n,\rho ,c}(f;x,a)-f(x)|\le & {} 2K_2(f,\delta )+\omega \left( f,\left| \frac{cx+\rho }{n\rho -c}\right| \right) . \end{aligned}$$

In view of (3.1), we obtain

$$\begin{aligned} |T_{n,\rho ,c}(f;x,a)-f(x)|\le & {} C\omega _2(f,\sqrt{\delta })+\omega \left( f,\left| \frac{cx+\rho }{n\rho -c}\right| \right) , \end{aligned}$$

which proves the theorem.

Our next result in this section is the Voronovskaja-type asymptotic formula:

Theorem 2

For any function \(f\in C_{B}[0,\infty )\) and \(a>1\) such that \(f^\prime ,\,f^{\prime \prime }\in C_{B}[0,\infty )\), we have

$$\begin{aligned} \lim _{n\rightarrow \infty }n[T_{n,\rho ,c}(f;x,a)-f(x)]=\frac{cx+\rho }{\rho }f^\prime (x)+\frac{x}{2\rho }\left( cx+\rho +1+\frac{\rho }{a-1}\right) f^{\prime \prime }(x) \end{aligned}$$

for every \(x\ge 0.\)

Proof

Let \(f,f^\prime ,\,f^{\prime \prime }\in C_B[0,\infty )\) and \(x\in [0,\infty )\) be fixed. By Taylor expansion we can write

$$\begin{aligned} f(t)=f(x)+(t-x)f^\prime (x)+\frac{(t-x)^2}{2!}f^{\prime \prime }(x)+r(t,x)(t-x)^2, \end{aligned}$$

where r(t, x) is the Peano form of the remainder, \(r(t,x)\in C_B[0,\infty )\) and \(\displaystyle \lim _{t\rightarrow x}r(t,x)=0.\) Applying \(T_{n,\rho ,c},\) we get

$$\begin{aligned} n[T_{n,\rho ,c}(f;x,a)-f(x)]= & {} f^\prime (x)nT_{n,\rho ,c}(t-x;x,a)+\frac{f^{\prime \prime }(x)}{2!}nT_{n,\rho ,c}((t-x)^2;x,a)\\&+\,nT_{n,\rho ,c}(r(t,x)(t-x)^2;x,a)\\ \end{aligned}$$
$$\begin{aligned} \lim _{n\rightarrow \infty }n[T_{n,\rho ,c}(f;x,a)-f(x)]&=f^\prime (x)\lim _{n\rightarrow \infty }nT_{n,\rho ,c}(t-x;x,a)\nonumber \\&\quad +\frac{f^{\prime \prime }(x)}{2!}\lim _{n\rightarrow \infty }(x) T_{n,\rho ,c}((t-x)^2;x,a)\nonumber \\&\quad +\lim _{n\rightarrow \infty }nT_{n,\rho ,c}(r(t,x)(t-x)^2;x,a)\nonumber \\&=\frac{cx+\rho }{\rho }f^\prime (x) +\frac{x}{2\rho }\left( cx+\rho +1+\frac{\rho }{a-1}\right) f^{\prime \prime }(x) \nonumber \\&\quad +\lim _{n\rightarrow \infty }nT_{n,\rho ,c}\left( r(t,x)(t-x)^2;x,a\right) \nonumber \\&=\frac{cx+\rho }{\rho }f^\prime (x)+\frac{x}{2\rho }\left( cx+\rho +1+\frac{\rho }{a-1}\right) f^{\prime \prime }(x)+E. \end{aligned}$$

By Cauchy–Schwarz inequality, we have

$$\begin{aligned} |E|\le & {} \lim _{n\rightarrow \infty }nT_{n,\rho ,c}(r^2(t,x);x,a)^{1/2}T_{n,\rho ,c}\left( (t-x)^4;x,a\right) ^{1/2}. \end{aligned}$$
(3.4)

It is easy to show that \(T_{n,\rho ,c}\left( (t-x)^4;x,a\right) ^{1/2}\) is bounded for \(x\in [0,A].\) Also, observe that \(r^2(x,x)=0\) and \(r^2(.,x)\in C_{B}[0,\infty ).\) Then, it follows that

$$\begin{aligned} \lim _{n\rightarrow \infty }nT_{n,\rho ,c}(r^2(t,x);x,a)=r^2(x,x)=0 \end{aligned}$$
(3.5)

uniformly with respect to \(x\in [0,A].\) Now from (3.4), (3.5) we obtain

$$\begin{aligned} \lim _{n\rightarrow \infty }nT_{n,\rho ,c}(r(t,x)(t-x)^2;x,a)=0. \end{aligned}$$

Hence, \(E=0\). Thus, we have

$$\begin{aligned} \lim _{n\rightarrow \infty }n[T_{n,\rho ,c}(f;x,a)-f(x)]=\frac{cx+\rho }{\rho }f^\prime (x)+\frac{x}{2\rho }\left( cx+\rho +1+\frac{\rho }{a-1}\right) f^{\prime \prime }(x), \end{aligned}$$

which completes the proof.

4 Weighted Approximation

Let \(B_{x^2}[0,\infty )\,{=}\,\{f : \text{ for } \text{ every }\,\, x\in [0,\infty ), |f(x)|\le M_f(1+x^2), \,M_f\) being a constant depending on \(f\}.\) By \(C_{x^2}[0,\infty ),\) we denote the subspace of all continuous functions belonging to \(B_{x^2}[0,\infty ).\) Also,\( C_{x^2}^*[0,\infty )\) is subspace of all functions \(f\in C_{x^2}[0,\infty )\) for which \(\displaystyle \lim _{x\rightarrow {\infty }}\frac{f(x)}{1+x^2}\) is finite. The norm on \( C_{x^2}^*[0,\infty )\) is \(\displaystyle \Vert f\Vert _{x^2}=\sup _{x\in [0,\infty )}\frac{|f(x)|}{1+x^2}.\)

Theorem 3

For each \(f\in C_{x^2}^*[0,\infty ),\) we have

$$\begin{aligned} \lim _{n\rightarrow {\infty }}\Vert T_{n,\rho ,c}(f;.,a)-f\Vert _{x^2}=0 \end{aligned}$$

Proof

Using [4] we see that it is sufficient to verify the following conditions

$$\begin{aligned} \lim _{n\rightarrow {\infty }}\Vert T_{n,\rho ,c}(t^\nu ;x,a)-x^\nu \Vert _{x^2}=0, \nu =0,1,2. \end{aligned}$$
(4.1)

Since \(T_{n,\rho ,c}(1;x,a)=1\),therefore for \(\nu =0\,\) (4.1) holds.

By Lemma 2 for \(n>\frac{c}{\rho },\) we have

$$\begin{aligned} \Vert T_{n,\rho ,c}(t;x,a)-x\Vert _{x^2}= & {} \sup _{x\in [0,\infty )}\frac{|T_{n,\rho ,c}(t;x,a)-x|}{1+x^2}\\\le & {} \bigg (\frac{n\rho }{n\rho -c}-1\bigg )\sup _{x\in [0,\infty )}\frac{x}{1+x^2}+\frac{\rho }{n\rho -c}\\\le & {} \bigg (\frac{c+2\rho }{2(n\rho -c)}\bigg ), \end{aligned}$$

the condition (4.1) holds for \(\nu =1\) as \({n\rightarrow {\infty }}.\)

Again by Lemma 2 for \(n>\frac{2c}{\rho },\) we have

$$\begin{aligned} \Vert T_{n,\rho ,c}(t^2;x,a)-x^2\Vert _{x^2}&=\sup _{x\in [0,\infty )}\frac{|T_{n,\rho ,c}(t^2;x,a)-x^2|}{1+x^2}\nonumber \\&\le \bigg |\frac{n^2\rho ^2}{(n\rho -c)(n\rho -2c)}-1\bigg |\sup _{x\in [0,\infty )}\frac{x^2}{1+x^2}\nonumber \\&\quad +\frac{n\rho }{(n\rho -c)(n\rho -2c)}\left( 3\rho +1+\frac{\rho }{a-1}\right) \sup _{x\in [0,\infty )}\frac{x}{1+x^2}\nonumber \\&\quad +\frac{\rho (2\rho +1)}{(n\rho -c)(n\rho -2c)}\bigg ]\\&\le \bigg |\frac{n^2\rho ^2}{(n\rho -c)(n\rho -2c)}-1\bigg |\\&\quad +\frac{n\rho }{(n\rho -c)(n\rho -2c)}\left( 3\rho +1+\frac{\rho }{a-1}\right) +\frac{\rho (2\rho +1)}{(n\rho -c)(n\rho -2c)}, \end{aligned}$$

the condition (4.1) holds for \(\nu =2\) as \({n\rightarrow {\infty }}.\)

Hence the theorem.

Corollary 1

For each \(f\in C_{x^2}[0,\infty ),\, a>1\) and \(\alpha >0,\) we have

$$\begin{aligned} \lim _{n\rightarrow {\infty }}\sup _{x\in [0,\infty )}\frac{|T_{n,\rho ,c}(f;x,a)-f(x)|}{(1+x^2)^\alpha }=0. \end{aligned}$$

Proof

For any fixed \(x_0>0\),

$$\begin{aligned} \sup _{x\in [0,\infty )}\frac{|T_{n,\rho ,c}(f;x,a)-f(x)|}{(1+x^2)^{1+\alpha }}&\le \sup _{x\le x_0}\frac{|T_{n,\rho ,c}(f;x,a)-f(x)|}{(1+x^2)^{1+\alpha }} +\sup _{x\ge x_0}\frac{|T_{n,\rho ,c}(f;x,a)-f(x)|}{(1+x^2)^{1+\alpha }}\\&\le \Vert T_{n,\rho ,c}(f;.,a)-f\Vert _{C[0,x_0]}+\Vert f\Vert _{x^2}\sup _{x\ge x_0}\frac{|T_{n,\rho ,c}(1+t^2;x,a)|}{(1+x^2)^{1+\alpha }}\\&\quad +\sup _{x\ge x_0}\frac{|f(x)|}{(1+x^2)^{1+\alpha }}. \end{aligned}$$

The first term of the above inequality tends to zero from Theorem 1. By Lemma 2 for any fixed \(x_0\) it is easily seen that \(\displaystyle \sup _{x\ge x_0}\frac{|T_{n,\rho ,c}(1+t^2; x,a)|}{(1+x^2)^{1+\alpha }}\le \frac{M}{(1+x_0^2)^{\alpha }}\) for some positive constant M independent of x. We can choose \(x_0\) so large that the right-hand side of the former inequality and last part of above inequality can be made small enough.

Thus the proof is completed.