A generalization of Szász-type operators which preserves constant and quadratic test functions

Abstract

In the present article, we introduced a new form of Szász-type operators which preserves test functions $e_0$ and $e_2$$(e_i(t)=t^i,i=0,2)$. By these sequence of positive linear operators, we gave rate of convergence and better error estimation by means of modulus of continuity. Moreover, we have discussed order of approximation with the help of local results. In the last, weighted Korovkin theorem is established.

Keywords:

• Szász operators
• positive linear operators
• modulus of continuity
• Peetre’s K-functional

AMS Subject Classifications:

• 41A10
• 41A25
• 41A36

Public Interest Statement

The approximation of functions by positive linear operators plays a significant role in the areas of numerical analysis, computer-aided geometric design (CAGD), solutions of differential equations, etc. In this paper, we modified Szász type operators based on Charlier polynomials using King’s method to obtain better approximation results. Moreover, we discussed basic convergence theorem in terms of classical modulus of continuity, investigated pointwise approximation theorems and weighted approximation theorem.

1. Introduction

For $f\in C(0,+\infty )$ and $x\in (0,+\infty )$, Szász (1950) defined a sequence of positive linear operators as follows(1.1) $$\begin{array}{c}\hfill S_n(f;x)=\sum _{k=0}^{\infty }{p}_{n,k}(x)f\left(\frac{k}{n}\right),\end{array}$$(1.1)

where $p_{n,k}(x)=e^{-nx}\frac{{(nx)}^k}{k!},n\in \mathbb{N}$. These operators play important role in approximation theory. In this paper, Szász showed the manner in which the operators $S_n(f;x)$ tend to f(x). Various well-known positive linear operators $L_n$ preserve the constant as well as the linear functions i.e. $L_n(e_0;x)=e_0(x)$ and $L_n(e_1;x)=e_1(x)$ for the test functions $e_i(x)=x^i(i=0,1)$. But, these operators do not preserve $e_2(x)$ and it is rather difficult to approach $e_2(x)$ for large value of n (see also Rao & Wafi, 2015; Wafi & Rao, 2016). In King (2003) introduced a method by which every linear positive operators preserve $e_2(x)$ i.e. $L_n(e_2;x)=e_2(x)$ and provided the better error estimation. Several authors used this powerful tool to different type of positive linear operators and discussed the better error estimation for instance (Ali Özarslan & Duman, 2009; Deo & Bhardwaj, 2015; Duman & Ozarslan, 2007). Recently, Varma and Tasdelen (2012) gave a generalization of well-known Szász-Mirakjan operators using Charlier polynomials (Ismail, 2005) having the generating function of the form:(1.2) $$\begin{array}{c}\hfill e^t{\left(1-\frac{t}{a}\right)}^u=\sum _{k=0}^{\infty }{C}_k^{(a)}(u)\frac{t^k}{k!},|t|<a,\end{array}$$(1.2)

and the explicit representation$$\begin{array}{c}\hfill C_k^{(a)}(u)=\sum _{r=0}^k\left(\begin{array}{c}k\hfill \\ r\hfill \end{array}\right){(-u)}_r{\left(\frac{1}{a}\right)}^r,\end{array}$$

where ${(\mathit{\alpha })}_k$ is the Pochhammer’s symbol given by$$\begin{array}{c}\hfill {(\mathit{\alpha })}_0=1,{(\mathit{\alpha })}_k=\mathit{\alpha }(\mathit{\alpha }+1)\dots (\mathit{\alpha }+k-1),k\in \mathbb{N}.\end{array}$$

For $a>0$ and $u\le 0$, Charlier polynomials are positive. Using these polynomials, they (Varma & Taşdelen, 2012) defined the Szász-type operators as follows:(1.3) $$\begin{array}{c}\hfill L_n(f;x,a)=e^{-1}{\left(1-\frac{1}{a}\right)}^{(a-1)nx}\sum _{k=0}^{\infty }{C}_k^{(a)}\left(-(a-1)nx\right)f\left(\frac{k}{n}\right),n\in \mathbb{N},\end{array}$$(1.3)

where $a>1$ and $x\ge 0$. Now, we introduce a new sequence of Szász-type operators which preserves constant and quadratic test functions and provides better estimates. Let $T_{n,a}:C[0,+\infty )\to C[0,+\infty )$. Then we have(1.4) $$\begin{array}{c}\hfill T_{n,a}(f;r_{n,a}(x))=e^{-1}{\left(1-\frac{1}{a}\right)}^{\left(a-1\right)n{r}_{n,a}(x)}\sum _{k=0}^{\infty }{C}_k^{(a)}(-(a-1)n{r}_{n,a}(x))f\left(\frac{k}{n}\right),\end{array}$$(1.4)

for any function $f\in C_{\mathit{\beta }}[0,\infty )=\{f\in C[0,\infty ):|f(x)|\le M{(1+x)}^{\mathit{\beta }}$ for some $M>0$ and $\mathit{\beta }>0\}$ and(1.5) $$\begin{array}{c}\hfill r_{n,a}(x)=\frac{-\left(3+\frac{1}{a-1}\right)+\sqrt{{\left(3+\frac{1}{a-1}\right)}^2+4\left(n^2{x}^2-2\right)}}{2n},\end{array}$$(1.5)

is a sequence of real-valued continuous functions which is defined on $[0,+\infty )$. We observe that $r_{n,a}(x)\ge 0$ for $x\ge \frac{\sqrt{2}}{n}$. One can notice that

(i)	if $r_{n,a}(x)\to x$ as $n\to +\infty$, the sequence of operators defined in (1.4) reduces to operators (1.3) and
(ii)	for $r_{n,a}(x)=x$, $a\to +\infty$ and $x-\frac{1}{n}$ in place of x, these operators tend to classical Szász operators defined by (1.1).

In this paper, we have discussed rate of convergence, local approximation results and Korovkin-type approximation theorem in polynomial weighted space and obtained better estimates for the operators (1.4).

2. Basic estimates

Lemma 2.1

For the operators $T_{n,a}$ defined by (1.4), we have$$\begin{array}{cc}\hfill T_{n,a}(1;r_{n,a}(x))& =1,\hfill \\ \hfill T_{n,a}(t;r_{n,a}(x))& =\frac{-\left(1+\frac{1}{a-1}\right)+\sqrt{{\left(3+\frac{1}{a-1}\right)}^2+4\left(n^2{x}^2-2\right)}}{2n},\hfill \\ \hfill T_{n,a}(t^2;r_{n,a}(x))& =x^2,\hfill \end{array}$$

for $x\ge \frac{\sqrt{2}}{n}$.

Proof

Using $t=1,u=-(a-1)n{r}_{n,a}(x)$ in (1.2) and by simple differentiation, we get$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }\frac{C_k^{(u)}\left(-\left(a-1\right)n{r}_{n,a}(x)\right)}{k!}& =e{\left(1-\frac{1}{a}\right)}^{-(a-1)n{r}_{n,a}(x)},\hfill \\ \hfill \sum _{k=0}^{\infty }k\frac{C_k^{(u)}\left(-(a-1)n{r}_{n,a}(x)\right)}{k!}& =e{\left(1-\frac{1}{a}\right)}^{-(a-1)n{r}_{n,a}(x)}\left(1+n{r}_{n,a}(x)\right),\hfill \\ \hfill \sum _{k=0}^{\infty }{k}^2\frac{C_k^{(u)}\left(-(a-1)n{r}_{n,a}(x)\right)}{k!}& =e{\left(1-\frac{1}{a}\right)}^{-(a-1)n{r}_{n,a}(x)}\left(2+\left(3+\frac{1}{a-1}\right)n{r}_{n,a}(x)+n^2{r}_{n,a}^2(x)\right).\hfill \end{array}$$

Using these equalities and operators (1.4), we can easily prove Lemma 2.1. $\square$

Lemma 2.2

Let ${\mathit{\psi }}_x^i(t)={(t-x)}^i,i=0,1,2$. Then, for the operators (1.4), we have$$\begin{array}{cc}\hfill T_{n,a}({\mathit{\psi }}_x^0;r_{n,a}(x))& =1,\hfill \\ \hfill T_{n,a}({\mathit{\psi }}_x^1;r_{n,a}(x))& =\frac{-\left(1+\frac{1}{a-1}\right)+\sqrt{{\left(3+\frac{1}{a-1}\right)}^2+4\left(n^2{x}^2-2\right)}}{2n}-x,\hfill \\ \hfill T_{n,a}({\mathit{\psi }}_x^2;r_{n,a}(x))& =2x^2+\frac{\left(1+\frac{1}{a-1}\right)}{n}x-\frac{x\sqrt{{\left(3+\frac{1}{a-1}\right)}^2+4\left(n^2{x}^2-2\right)}}{n}.\hfill \end{array}$$

Proof

Using Lemma 2.1, we can easily prove Lemma 2.2.

$\square$

Lemma 2.3

For the operators $T_{n,a}$$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}n{T}_{n,a}\left({\mathit{\psi }}_x^1;r_{n,a}(x)\right)=-\frac{\left(1+\frac{1}{a-1}\right)}{2},\\ \hfill \underset{n\to \infty }{lim}{T}_{n,a}\left({\mathit{\psi }}_x^2;r_{n,a}(x)\right)=\left(1+\frac{1}{a-1}\right)x.\end{array}$$

3. Rate of convergence

Theorem 3.1

Let $f\in C_{\mathit{\beta }}[0,+\infty )$ and $x\ge \frac{\sqrt{2}}{n}$. Then for the operators $T_{n,a}$ defined by (1.4), we have$$\begin{array}{c}\hfill |T_{n,a}(f;r_{n,a}(x)-f(x))|\le 2\mathit{\omega }\left(f;{\mathit{\delta }}_n\right),\end{array}$$

where ${\mathit{\delta }}_n={\left(T_{n,a}\left({\mathit{\psi }}_x^2;r_{n,a}(x)\right)\right)}^{\frac{1}{2}}$.

Proof

We have the difference$$\begin{array}{cc}\hfill \left|T_{n,a}\left(f;r_{n,a}(x)\right)-f(x)\right|& \le e^{-1}{\left(1-\frac{1}{a}\right)}^{(a-1)n{r}_{n,a}(x)}\sum _{k=0}^{\infty }\frac{C_k^{(a)}\left(-(a-1)n{r}_{n,a}(x)\right)}{k!}\left|f\left(\frac{k}{n}\right)-f(x)\right|\hfill \\ \hfill & \le \left\{1+\frac{1}{{\mathit{\delta }}_n}{e}^{-1}{\left(1-\frac{1}{a}\right)}^{(a-1)n{r}_{n,a}(x)}\sum _{k=0}^{\infty }\frac{C_k^{(a)}\left(-(a-1)n{r}_{n,a}(x)\right)}{k!}\left|\frac{k}{n}-x\right|\right\}\mathit{\omega }\left(f;{\mathit{\delta }}_n\right)\hfill \\ \hfill & \le \left\{1+\frac{1}{{\mathit{\delta }}_n}\sqrt{e^{-1}{\left(1-\frac{1}{a}\right)}^{(a-1)n{r}_{n,a}(x)}\sum _{k=0}^{\infty }\frac{C_k^{(a)}\left(-(a-1)n{r}_{n,a}(x)\right)}{k!}{\left(\frac{k}{n}-x\right)}^2}\right\}\mathit{\omega }\left(f;{\mathit{\delta }}_n\right)\hfill \\ \hfill & =\left\{1+\frac{1}{{\mathit{\delta }}_n}\sqrt{T_{n,a}\left({\mathit{\psi }}_x^2;x\right)}\right\}\mathit{\omega }\left(f;{\mathit{\delta }}_n\right),\hfill \end{array}$$

which proves the Theorem 3.1.

$\square$

Remark 3.2

For the Szász-type operators $L_n$ given by (1.3), and for every $f\in C[0,\infty )\cap E$, we have(3.1) $$\begin{array}{c}\hfill |L_n(f;x)-f(x))|\le \left\{1+\sqrt{x\left(1+\frac{1}{a-1}\right)+\frac{2}{n}}\right\}\mathit{\omega }\left(f;\frac{1}{\sqrt{n}}\right),\end{array}$$(3.1)

where $E=\{f:[0,\infty )\to R,|f(x)|\le M{e}^{Ax},$$A\in R$ and $M\in (0,\infty )\}$ Here, we show that our operators $T_{n,a}$ has the better approximation than the operators $L_n$.

Since $2x=\frac{\sqrt{4x^2{n}^2}}{n}$ and ${\left(3+\frac{1}{a-1}\right)}^2-8>0$ for all values of $a>1$, then $2x^2<x\frac{\sqrt{{\left(3+\frac{1}{a-1}\right)}^2+4\left(x^2{n}^2-2\right)}}{n}$. This implies that $2x^2-x\frac{\sqrt{{\left(3+\frac{1}{a-1}\right)}^2+4\left(x^2{n}^2-2\right)}}{n}+\frac{\left(1+\frac{1}{a-1}\right)}{n}x<\frac{\left(1+\frac{1}{a-1}\right)}{n}x+\frac{2}{n^2}$. Hence $\sqrt{T_{n,a}\left({\mathit{\psi }}_x^2;r_{n,a}(x)\right)}<\sqrt{L_n\left({\mathit{\psi }}_x^2;x\right)}$.

4. Local approximation results

In this section, we deal order of approximation locally in $C_B[0,\infty )$ (space of real-valued continuous and bounded functions f on $[0,\infty )$) with the norm $\Vert f\Vert =\underset{0\le x<\infty }{sup}|f(x)|$. Then, for any $f\in C_B[0,\infty )$ and $\mathit{\delta }>0$, Peeter’s K-functional is defined as$$\begin{array}{c}\hfill K_2\left(f,\mathit{\delta }\right)=inf\left\{\Vert f-g\Vert +\mathit{\delta }\Vert g^{″}\Vert :g\in C_B^2[0,\infty )\right\},\end{array}$$

where $C_B^2[0,\infty )=\left\{g\in C_B[0,\infty ):g^{\prime },g^{″}\in C_B[0,\infty )\right\}$. By DeVore and Lorentz (1993, p. 177, Theorem 2.4), there exits an absolute constant $C>0$ such that$$\begin{array}{c}\hfill K_2\left(f;\mathit{\delta }\right)\le C{\mathit{\omega }}_2\left(f;\sqrt{\mathit{\delta }}\right),\end{array}$$

where ${\mathit{\omega }}_2(f;\mathit{\delta })$ is the second-order modulus of continuity is defined as$$\begin{array}{c}\hfill {\mathit{\omega }}_2\left(f;\sqrt{\mathit{\delta }}\right)=\underset{0<h\le \sqrt{\mathit{\delta }}}{sup}\underset{x\in [0,\infty )}{sup}\left|f(x+2h)-2f(x+h)+f(x)\right|.\end{array}$$

Theorem 4.1

Let $f\in C_B^2[0,\infty )$. Then for all $x\ge \frac{\sqrt{2}}{n}$ there exists a constant $C>0$ such that$$\begin{array}{c}\hfill \left|T_{n,a}\left(f;r_{n,a}(x)\right)-f(x)\right|\le C{\mathit{\omega }}_2\left(f;\sqrt{{\mathit{\gamma }}_{n,a}(x)}\right)+\mathit{\omega }\left(f;T_{n,a}({\mathit{\psi }}_x;r_{n,a}(x)\right),\end{array}$$

where ${\mathit{\gamma }}_{n,a}(x)=T_{n,a}\left({\mathit{\psi }}_x^2;r_{n,a}(x)\right)+{\left(T_{n,a}\left({\mathit{\psi }}_x;r_{n,a}(x)\right)\right)}^2.$

Proof

First, we consider the auxiliary operators as follows(4.1) $$\begin{array}{c}\hfill {\widehat{T}}_{n,a}\left(f;r_{n,a}(x)\right)=T_{n,a}\left(f;r_{n,a}(x)\right)+f(x)-f\left({\mathit{\eta }}_{n,a}(x)\right),\end{array}$$(4.1)

where ${\mathit{\eta }}_{n,a}(x)=T_{n,a}\left({\mathit{\psi }}_x;r_{n,a}(x)\right)+x$. By the Equation (4.1), we get(4.2) $$\begin{array}{cc}\hfill {\widehat{T}}_{n,a}\left(1;r_{n,a}(x)\right)& =1,\hfill \\ \hfill {\widehat{T}}_{n,a}\left({\mathit{\psi }}_x(t);r_{n,a}(x)\right)& =0,\hfill \\ \hfill \left|{\widehat{T}}_{n,a}\left(f;r_{n,a}(x)\right)\right|& \le 3\Vert f\Vert .\hfill \end{array}$$(4.2)

For any $g\in C_B^2[0,\infty )$ and by the Taylor’s theorem, we have(4.3) $$\begin{array}{c}\hfill g(t)=g(x)+(t-x)g^{\prime }(x)+\underset{x}{\overset{t}{\int }}(t-v)g^{″}(v)dv.\end{array}$$(4.3)

Applying auxiliary operators defined by (4.1) in Equation (4.3), we get$$\begin{array}{cc}\hfill {\widehat{T}}_{n,a}\left(g;r_{n,a}(x)\right)-g(x)& =g^{\prime }(x){\widehat{T}}_{n,a}\left(t-x;r_{n,a}(x)\right)+{\widehat{T}}_{n,a}\left(\underset{x}{\overset{t}{\int }}(t-v)g^{″}(v)dv;r_{n,a}(x)\right)\hfill \\ \hfill & ={\widehat{T}}_{n,a}\left(\underset{x}{\overset{t}{\int }}(t-v)g^{″}(v)dv;r_{n,a}(x)\right)\hfill \\ \hfill & =T_{n,a}\left(\underset{x}{\overset{t}{\int }}(t-v)g^{″}(v)dv;r_{n,a}(x)\right)-\underset{x}{\overset{{\mathit{\eta }}_{n,a}(x)}{\int }}\left({\mathit{\eta }}_{n,a}(x)-v\right)g^{″}(v)dv.\hfill \end{array}$$

Therefore(4.4) $$\begin{array}{cc}\hfill \left|{\widehat{T}}_{n,a}\left(g;r_{n,a}(x)\right)-g(x)\right|& \le \left|T_{n,a}\left(\underset{x}{\overset{t}{\int }}(t-v)g^{″}(v)dv;r_{n,a}(x)\right)\right|\hfill \\ \hfill & +\left|\underset{x}{\overset{{\mathit{\eta }}_{n,a}(x)}{\int }}\left({\mathit{\eta }}_{n,a}(x)-v\right)g^{″}(v)dv\right|.\hfill \end{array}$$(4.4)

Since, we have(4.5) $$\begin{array}{c}\hfill \left|\underset{x}{\overset{t}{\int }}(t-v)g^{″}(v)dv\right|\le {(t-x)}^2\Vert g^{″}\Vert ,\end{array}$$(4.5)

this implies that(4.6) $$\begin{array}{c}\hfill \left|\underset{x}{\overset{{\mathit{\eta }}_{n,a}(x)}{\int }}({\mathit{\eta }}_{n,a}(x)-v)g^{″}(v)dv\right|\le {\left({\mathit{\eta }}_{n,a}(x)-x\right)}^2\Vert g^{″}\Vert .\end{array}$$(4.6)

Using (4.4), (4.5) and (4.6), we have(4.7) $$\begin{array}{cc}\hfill \left|{\widehat{T}}_{n,a}\left(g;r_{n,a}(x)\right)-g(x)\right|& \le \left\{T_{n,a}\left({(t-x)}^2;r_{n,a}(x)\right)+{\left({\mathit{\eta }}_{n,a}(x)-x\right)}^2\right\}\Vert g^{″}\Vert \hfill \\ \hfill & ={\mathit{\gamma }}_{n,a}(x)\Vert g^{″}\Vert .\hfill \end{array}$$(4.7)

Now, we have$$\begin{array}{cc}\hfill \left|T_{n,a}\left(f;r_{n,a}(x)\right)-f(x)\right|& \le \left|{\widehat{T}}_{n,a}\left(f-g;r_{n,a}(x)\right)\right|+\left|(f-g)(x)\right|\hfill \\ \hfill & +\left|{\widehat{T}}_{n,a}\left(g;r_{n,a}(x)\right)-g(x)\right|+\left|f({\mathit{\eta }}_{n,a}(x))-f(x)\right|,\hfill \end{array}$$

using (4.7), we get$$\begin{array}{cc}\hfill \left|T_{n,a}\left(f;r_{n,a}(x)\right)-f(x)\right|& \le 4\Vert f-g\Vert +\left|{\widehat{T}}_{n,a}\left(g;r_{n,a}(x)\right)-g(x)\right|+\left|f\left({\mathit{\eta }}_{n,a}(x)\right)-f(x)\right|\hfill \\ \hfill & \le 4\Vert f-g\Vert +{\mathit{\gamma }}_{n,a}(x)\Vert g^{″}\Vert +\mathit{\omega }\left(f;T_{n,a}\left({\mathit{\psi }}_x;r_{n,a}(x)\right)\right).\hfill \end{array}$$

By the definition of Peetre’s K-functional, we find$$\begin{array}{c}\hfill \left|T_{n,a}\left(f;r_{n,a}(x)\right)-f(x)\right|\le C{\mathit{\omega }}_2\left(f;\sqrt{{\mathit{\gamma }}_{n,a}(x)}\right)+\mathit{\omega }\left(f;T_{n,a}\left({\mathit{\psi }}_x;r_{n,a}(x)\right)\right).\end{array}$$

This proves Theorem 4.1. $\square$

Here, we introduce a local result in Lipschitz class$$\begin{array}{c}\hfill Li{p}_M^{\ast }(\mathit{\alpha })=\left\{f\in C[0,\infty ):\left|f(t)-f(x)\right|\le M\frac{{\left|t-x\right|}^{\mathit{\alpha }}}{{\left(t+x\right)}^{\frac{\mathit{\alpha }}{2}}}:x,t\in (0,\infty )\right\},\end{array}$$

where M is a constant and $0<\mathit{\alpha }\le 1$.

Theorem 4.2

For $x\in \left[\frac{\sqrt{2}}{n},+\infty \right)$ and $f\in Li{p}_M^{\ast }(\mathit{\alpha })$, we have$$\begin{array}{c}\hfill \left|T_{n,a}\left(f;r_{n,a}(x)\right)-f(x)\right|\le M{\left[\frac{{\mathrm{\Theta }}_{n,a}(x)}{x}\right]}^{\frac{\mathit{\alpha }}{2}},\end{array}$$

where ${\mathrm{\Theta }}_{n,a}(x)=T_{n,a}\left({(t-x)}^2;r_{n,a}(x)\right)$.

Proof

Let $\mathit{\alpha }=1$ and $x\in (0,\infty )$. Then, for $f\in Li{p}_M^{\ast }(1)$, we have$$\begin{array}{cc}\hfill \left|T_{n,a}\left(f;r_{n,a}(x)\right)-f(x)\right|& \le e^{-1}{\left(1-\frac{1}{a}\right)}^{(a-1)n{r}_{n,a}(x)}\sum _{k=0}^{\infty }\frac{C_k^{(a)}\left(-(a-1)n{r}_{n,a}(x)\right)}{k!}\left|f\left(\frac{k}{n}\right)-f(x)\right|\text{d}t\hfill \\ \hfill & \le M{e}^{-1}{\left(1-\frac{1}{a}\right)}^{(a-1)n{r}_{n,a}(x)}\sum _{k=0}^{\infty }\frac{C_k^{(a)}\left(-(a-1)n{r}_{n,a}(x)\right)}{k!}\frac{\left|\frac{k}{n}-x\right|}{\sqrt{\frac{k}{n}+x}}.\hfill \\ \hfill & \le \frac{M}{\sqrt{x}}{e}^{-1}{\left(1-\frac{1}{a}\right)}^{(a-1)n{r}_{n,a}(x)}\sum _{k=0}^{\infty }\frac{C_k^{(a)}\left(-(a-1)n{r}_{n,a}(x)\right)}{k!}\left|\frac{k}{n}-x\right|\hfill \\ \hfill & \le \frac{M}{\sqrt{x}}{T}_{n,a}\left(\left|t-x\right|;r_{n,a}(x)\right)\hfill \\ \hfill & \le M\frac{\sqrt{T_{n,a}\left({(t-x)}^2;r_{n,a}(x)\right)}}{\sqrt{x}}\hfill \\ \hfill & =M{\left(\frac{{\mathrm{\Theta }}_{n,a}(x)}{x}\right)}^{\frac{1}{2}}.\hfill \end{array}$$

Thus, the assertion holds for $\mathit{\alpha }=1$. Now, we will prove for $\mathit{\alpha }\in (0,1)$. From the Hölder’s Inequality with $p=\frac{1}{\mathit{\alpha }}$ and $q=\frac{1}{1-\mathit{\alpha }}$, we have$$\begin{array}{cc}\hfill \left|T_{n,a}\left(f;r_{n,a}(x)\right)-f(x)\right|& ={\left(e^{-1}{\left(1-\frac{1}{a}\right)}^{(a-1)n{r}_{n,a}(x)}\sum _{k=0}^{\infty }\frac{C_k^{(a)}\left(-(a-1)n{r}_{n,a}(x)\right)}{k!}{\left(\left|f\left(\frac{k}{n}\right)-f(x)\right|\right)}^{\frac{1}{\mathit{\alpha }}}\right)}^{\mathit{\alpha }}\hfill \\ \hfill & \times {\left(e^{-1}{\left(1-\frac{1}{a}\right)}^{(a-1)n{r}_{n,a}(x)}\sum _{k=0}^{\infty }\frac{C_k^{(a)}\left(-(a-1)n{r}_{n,a}(x)\right)}{k!}\right)}^{1-\mathit{\alpha }}\hfill \\ \hfill & \le {\left(e^{-1}{\left(1-\frac{1}{a}\right)}^{(a-1)n{r}_{n,a}(x)}\sum _{k=0}^{\infty }\frac{C_k^{(a)}\left(-(a-1)n{r}_{n,a}(x)\right)}{k!}{\left(\left|f\left(\frac{k}{n}\right)-f(x)\right|\right)}^{\frac{1}{\mathit{\alpha }}}\right)}^{\mathit{\alpha }}.\hfill \end{array}$$

Since $f\in Li{p}_M^{\ast }(\mathit{\alpha })$, we obtain$$\begin{array}{cc}\hfill \left|T_{n,a}\left(f;r_{n,a}(x)\right)-f(x)\right|& \le M{\left(e^{-1}{\left(1-\frac{1}{a}\right)}^{(a-1)n{r}_{n,a}(x)}\sum _{k=0}^{\infty }\frac{C_k^{(a)}\left(-(a-1)n{r}_{n,a}(x)\right)}{k!}\frac{\left|\frac{k}{n}-x\right|}{\sqrt{\frac{k}{n}+x}}dt\right)}^{\mathit{\alpha }}\hfill \\ \hfill & \le \frac{M}{x^{\frac{\mathit{\alpha }}{2}}}{\left(e^{-1}{\left(1-\frac{1}{a}\right)}^{(a-1)n{r}_{n,a}(x)}\sum _{k=0}^{\infty }\frac{C_k^{(a)}\left(-(a-1)n{r}_{n,a}(x)\right)}{k!}\left|\frac{k}{n}-x\right|\right)}^{\mathit{\alpha }}\hfill \\ \hfill & =\frac{M}{x^{\frac{\mathit{\alpha }}{2}}}{\left(T_{n,a}\left(\left|t-x\right|;r_{n,a}(x)\right)\right)}^{\mathit{\alpha }}\hfill \\ \hfill & \le M{\left(\frac{{\mathrm{\Theta }}_{n,a}(x)}{x}\right)}^{\frac{\mathit{\alpha }}{2}}.\hfill \end{array}$$

This completes the proof of Theorem 4.2.

5. Weighted Korovkin-type theorem

In this section, we introduce $T_{n,a}$ in polynomial weighted spaces of continuous and unbounded functions defined on $[0,\infty )$. In Gadzhiev (1976) gave the weighted Korovkin-type theorems. Here we recall some symbols and notions from Gadziev (1976). Let $\mathit{\rho }(x)=1+x^2$, $-\infty <x<\infty$ and $B_{\mathit{\rho }}[0,\infty )=\{f(x):|f(x)|\le M_f\mathit{\rho }(x)$, $\mathit{\rho }(x)$ is weight function, $M_f$ is a constant depending on f and $x\in [0,\infty )\}$, $C_{\mathit{\rho }}[0,\infty )$ is the space of continuous function in $B_{\mathit{\rho }}[0,\infty )$ with the norm ${\Vert f(x)\Vert }_{\mathit{\rho }}=\underset{x\in [0,\infty )}{sup}\frac{|f(x)|}{\mathit{\rho }(x)}$ and $C_{\mathit{\rho }}^k=\{f\in C_{\mathit{\rho }}:\underset{|x|\to \infty }{lim}\frac{f(x)}{\mathit{\rho }(x)}=k,$ where k is a constant depending on $f\}$.

Theorem 5.1

Let $T_{n,a}$ be the sequence of linear positive operators defined by (1.4). Then for $f\in C_{\mathit{\rho }}^k$,$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}{\Vert T_{n,a}\left(f;r_{n,a}(x)\right)-f(x)\Vert }_{\mathit{\rho }}=0.\end{array}$$

Proof

To prove the theorem, it is sufficient to show that$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}{\Vert T_{n,a}(t^i;x)-x^i\Vert }_{\mathit{\rho }}=0,\text{for}i=0,1,2.\end{array}$$

It is obvious that $\underset{n\to \infty }{lim}{\Vert T_{n,a}(1;x)-1\Vert }_{\mathit{\rho }}=0$ and $\underset{n\to \infty }{lim}{\Vert T_{n,a}(x^2;x)-x^2\Vert }_{\mathit{\rho }}=0$. Now, from the Lemma 2.1, we have$$\begin{array}{cc}\hfill \underset{x\in [0,\infty )}{sup}\frac{\left|T_{n,a}(t;x)-x\right|}{1+x^2}& =\underset{x\in [0,\infty )}{sup}\frac{\left|\frac{-\left(1+\frac{1}{a-1}\right)+\sqrt{{\left(3+\frac{1}{a-1}\right)}^2+4\left(n^2{x}^2-2\right)}}{2n}-x\right|}{1+x^2}\hfill \\ \hfill & \le \frac{\left(1+\frac{1}{a-1}\right)}{2n}\underset{x\in [0,\infty )}{sup}\frac{1}{\left(1+x^2\right)}+\underset{x\in [0,\infty )}{sup}\frac{\sqrt{{\left(3+\frac{1}{a-1}\right)}^2+4\left(n^2{x}^2-2\right)}-2nx}{2n\left(1+x^2\right)}\hfill \\ \hfill & =\frac{\left(1+\frac{1}{a-1}\right)}{2n}\underset{x\in [0,\infty )}{sup}\frac{1}{\left(1+x^2\right)}+\underset{x\in [0,\infty )}{sup}\frac{\sqrt{\left(1+\frac{1}{{\left(a-1\right)}^2}+\frac{6}{a-1}\right)+n^2{x}^2}-2nx}{2n\left(1+x^2\right)}.\hfill \end{array}$$

Since $\sqrt{a+b}<\sqrt{a}+\sqrt{b}$, then$$\begin{array}{cc}\hfill \underset{x\in [0,\infty )}{sup}\frac{\left|T_{n,a}(t;x)-x\right|}{1+x^2}& \le \frac{\left(1+\frac{1}{a-1}\right)}{2n}\underset{x\in [0,\infty )}{sup}\frac{1}{\left(1+x^2\right)}+\underset{x\in [0,\infty )}{sup}\frac{\sqrt{\left(1+\frac{1}{{(a-1)}^2}+\frac{6}{a-1}\right)}+\sqrt{4n^2{x}^2}-2nx}{2n\left(1+x^2\right)}\hfill \\ \hfill & =\frac{\left(1+\frac{1}{a-1}\right)}{2n}\underset{x\in [0,\infty )}{sup}\frac{1}{\left(1+x^2\right)}+\underset{x\in [0,\infty )}{sup}\frac{\sqrt{1+\frac{1}{{(a-1)}^2}+\frac{6}{a-1}}}{2n\left(1+x^2\right)}\hfill \end{array}$$

which shows that as $n\to \infty$, $\Vert T_{n,a}{(t;x)-x\Vert }_{\mathit{\rho }}\to 0$.

Hence the theorem is proved. $\square$

Acknowledgements

The authors would like to express their deep gratitude to the anonymous learned referee(s) and the editor for their valuable suggestions and constructive comments which resulted in the subsequent improvement of this research article.

Additional information

Funding

The second author Nadeem Rao is thankful to University Grants Commission [grant number UGC-BSR fellowship] for providing the financial support under the UGC-BSR(Basic Scientific Research) scheme.

Notes on contributors

Abdul Wafi

Abdul Wafi received his PhD from Aligarh Muslim University. He worked as Associate Professor abroad. He is working as a full professor in the Department of Mathematics, Jamia Millia Islamia, New Delhi-110025. His field of research is approximation theory and operators theory. He guided many PhD students.

Nadeem Rao

Nadeem Rao received his master degree in Mathematics with Computer Science from Jamia Millia Islamia, New Delhi-110025. He is pursuing his PhD under the supervision of Prof. Abdul Wafi.