# Approximation by a generalization of the Jakimovski -Leviatan operators

Didem AYDIN ARI Kırıkkale University  and  Sevilay KIRCI SERENBAY Başkent University
2015
###### Abstract.

In this paper, we introduce a Kantorovich type generalization of  Jakimovski-Leviatan operators constructed by A. Jakimovski and D. Leviatan (1969) and the theorems on convergence and the degrree of convergence are established. Furthermore, we study the convergence of these operators in a weighted space of functions on a positive semi-axis.

###### Key words and phrases:
Jakimovski-Leviatan operator, Lipschitz class, weighted modulus of continuity, weighted spaces,rate of convergence
###### 2000 Mathematics Subject Classification:
This paper is in final form and no version of it will be submitted for publication elsewhere.

## 1. Introduction

In approximation theory, Szàsz type operators and Chlodowsky type generalizations of these operators have been studied intensively [see [1], [2], [15], [12], [10], [16], [17], [11], [8] and many others]. Also orthogonal polynomials are important area of mathematical analysis, mathematical and theoretical physics. In mathematical analysis and in the positive approximation processes, the notion of orthogonal polynomials seldomly appears. Cheney and Sharma [7] established an operator

 P_{n}(f;x)=\left(1-x\right)^{n+1}\exp\left(\frac{tx}{1-x}\right){\displaystyle% \sum\limits_{k=0}^{\infty}}f\left(\frac{k}{k+n}\right)L_{k}^{\left(n\right)}(t% )x^{k} (1.1)

where t\leq 0 and L_{k}^{\left(n\right)} denotes the Laguerre polynomials. For the special case t=0, the operators given by (1.2) reduce to the well-known Meyer-König and Zeller operators [14].

In view of the relation between orthogonal polynomials and positive linear operators have been investigated by many researchers (see [17],[8]).

One of them is Jakimovski and Leviatan ’s study. In 1969, authors introduced Favard-Szàsz type operators P_{n}, by using Appell polynomials are given with

g(u)=\sum\limits_{n=0}^{\infty}a_{n}u^{n},g(1)\neq 1 be an analytic function in the disk \left|u<r\right| (r>1) and p_{k}(x)=\sum\limits_{i=0}^{k}a_{i}\frac{x^{k-i}}{(k-i)!}, (k\in\mathbb{N}) be the Appell polynomials defined by the identity

 g(u)e^{ux}\equiv\sum\limits_{k=0}^{\infty}p_{k}(x)u^{k}. (1.2)

Let E\left[0,\infty\right) denote the space of exponential type functions on \left[0,\infty\right) which satisfy the property \ \left|f(x)\right|\leq\beta e^{\alpha x} for some finite constants \alpha,\beta>0.

In [1], the authors considered the operator P_{n}, with

 P_{n}(f;x)=\frac{e^{-nx}}{g(1)}\sum\limits_{k=0}^{\infty}p_{k}(nx)f(\frac{k}{n}) (1.3)

for f\in E\left[0,\infty\right) and studied approximation properties of these operators, as well as the analogue to Szàsz’s results.

If g(u)\equiv 1,from (1.2) we obtain p_{k}(x)=\dfrac{x^{k}}{k!}and we obtain classical Szàsz-Mirakjan operator which is given by

 S_{n}(f;x)=e^{-nx}{\displaystyle\sum\limits_{k=0}^{\infty}}\dfrac{(nx)^{k}}{k!% }f(\frac{k}{n})\text{.}

In 1969, Wood [6] showed that the operators P_{n} are positive if and only if \frac{a_{k}}{g(1)}\geq 0, (k=0,1,...).In 1996, Ciupa [2] was studied the rate of convergence of these operators. In 1999, Abel and Ivan [18] showed an asimptotic expansion of the operators given by (1.3) and their derivatives. In 2003, İspir [15] showed that the approximation of continuous functions having polynomial growth at infinity by the operator in (1.3). In 2007, Ciupa [3] defined Modified Jakimovski-Leviatan operators and studied rate of convergence,order of approximation and Voronovskaya type theorem. Recently, Büyükyazıcı and et. al, [12] studied approximation properties of Chlodowsky type Jakimovski -Leviatan operators. They proved Voronovskaya-type theorem and studied the convergence of these operators in a weighted space by using a new type of weighted modulus of continuity.

In this paper, we consider the following Kantorovich type generalization of Jakimovski-Leviatan operators given by

 L_{n}^{\ast}(f;x)=\frac{e^{-nx}}{g(1)}\frac{n}{b_{n}}\sum\limits_{k=0}^{\infty% }p_{k}(\frac{n}{b_{n}}x)\int\limits_{\frac{k}{n}b_{n}}^{\frac{k+1}{n}b_{n}}f(t% )dx (1.4)

with b_{n} a positive increasing sequence with the properties

 \underset{n\rightarrow\infty}{\lim}b_{n}=\infty,\text{ }\underset{n\rightarrow% \infty}{\lim}\frac{b_{n}}{n}=0 (1.5)

and p_{k} are Appell polynomials defined by (1.2). Recently, some generalizations of (1.3) operators have been considered in [2],[3] and [9].

## 2. Some approximation properties of L_{n}^{\ast}(f;x)

In approximation theory, the positive approximation processes discovered by Korovkin play a central role and arise in a natural way in many problems connected with functional analysis, harmonic analysis, measure theory, partial differential equations and probability theory.

Let C_{E}\left[0,\infty\right) denote the set of all continuous functions f on \left[0,\infty\right) with the property that \left|f(x)\right|\leq\beta e^{\alpha x} for all x\geq 0 and some positive finite \alpha and \beta. For a fixed r\in\mathbb{N}, we denote by

 C_{E}^{r}\left[0,\infty\right)=\left\{f\in C_{E}\left[0,\infty\right):f^{% \prime},f^{\prime\prime},...f^{\left(r\right)}\in C_{E}\left[0,\infty\right)% \right\}.
###### Lemma 1.

The operators L_{n}^{\ast}(f;x) defined by (1.4) satisfy the following equalities.

 L_{n}^{\ast}(1;x)=1, (2.1)
 L_{n}^{\ast}(t;x)=x+\frac{g^{\prime}(1)}{g(1)}\frac{b_{n}}{n}+\frac{b_{n}}{n}, (2.2)
 L_{n}^{\ast}(t^{2};x)=x^{2}+\frac{b_{n}}{n}x\left(\frac{g(1)+2g^{\prime}(1)}{g% (1)}+1\right)+\frac{b_{n}^{2}}{n^{2}}\left(\frac{2g^{\prime}(1)}{g(1)}+\frac{g% ^{\prime\prime}(1)}{g(1)}+\frac{1}{3}\right). (2.3)
###### Lemma 2.

The central moments of the operators L_{n}^{\ast}(f;x) are given by

 L_{n}^{\ast}(t-x;x)=\frac{g^{\prime}(1)}{g(1)}\frac{b_{n}}{n}+\frac{b_{n}}{n}
 L_{n}^{\ast}((t-x)^{2};x)=\frac{b_{n}^{2}}{n^{2}}\left(\frac{2g^{\prime}(1)}{g% (1)}+\frac{g^{\prime\prime}(1)}{g(1)}+1\right) (2.4)
###### Theorem 1.

If f\in C_{E}\left[0,\infty\right), then \lim L_{n}^{\ast}(f)=f uniformly on \left[0,a\right].

###### Proof.

From (2.1)-(2.3), we have

 \lim L_{n}^{\ast}(e_{i};x)=e_{i}\left(x\right),\text{ \ }i\in\left\{0,1,2% \right\},

where e_{i}\left(x\right)=t^{i} . Applying the Korovkin theorem [5], we obtain the desired result. ∎

In this section, we deal with the rate of convergence of the L_{n}^{\ast}(f;x) to f by means of a classical approach, the second modulus of continuity, and Peetre’s K-functional.

Let f\in\widetilde{C}[0,\infty). If \delta>0, the modulus of continuity of f is defined by

 \omega(f,\delta)=\sup_{\underset{\left|x-y\right|\leq\delta}{x,y\in[0,\infty)}% }\left|f(x)-f(y)\right|,

where \widetilde{C}[0,\infty) denotes the space of uniformly continuous functions on [0,\infty) . It is also well known that, for any \delta>0 and each x\in[0,\infty) ,

 \left|f(x)-f(y)\right|\leq\omega(f,\delta)\left(\frac{\left|x-y\right|}{\delta% }+1\right).

The next result gives the rate of convergence of the sequence L_{n}^{\ast}(f;x) to f  by means of the modulus of continuity.

###### Theorem 2.

If f\in C_{E}\left[0,\infty\right), then for any x\in[0,a] we have

 \left|L_{n}^{\ast}(f;x)-f\left(x\right)\right|\leq\left\{1+\frac{1}{\delta}% \left(\sqrt{\theta_{n}}\right)\right\}\omega(f,\delta).

where

 \theta_{n}=\frac{b_{n}^{2}}{n^{2}}\left(\frac{2g^{\prime}(1)}{g(1)}+\frac{g^{% \prime\prime}(1)}{g(1)}+1\right).
###### Proof.
 \displaystyle\left|L_{n}^{\ast}(f;x)-f\left(x\right)\right| \displaystyle\leq\frac{e^{-nx}}{g(1)}\frac{n}{b_{n}}\sum\limits_{k=0}^{\infty}% p_{k}(\frac{n}{b_{n}}x)\int\limits_{\frac{k}{n}b_{n}}^{\frac{k+1}{n}b_{n}}% \left|f(s)-f\left(x\right)\right|ds \displaystyle\leq\frac{e^{-nx}}{g(1)}\frac{n}{b_{n}}\sum\limits_{k=0}^{\infty}% p_{k}(\frac{n}{b_{n}}x)\int\limits_{\frac{k}{n}b_{n}}^{\frac{k+1}{n}b_{n}}% \omega(f,\delta)\left(\frac{\left|s-x\right|}{\delta}+1\right)ds \displaystyle\leq\left\{1+\frac{e^{-nx}}{g(1)}\frac{n}{b_{n}}\sum\limits_{k=0}% ^{\infty}p_{k}(\frac{n}{b_{n}}x)\int\limits_{\frac{k}{n}b_{n}}^{\frac{k+1}{n}b% _{n}}\left|s-x\right|ds\right\}\omega(f,\delta)

By using the Cauchy-Schwarz inequality for integration, we get

 \int\limits_{\frac{k}{n}b_{n}}^{\frac{k+1}{n}b_{n}}\left|s-x\right|ds\leq\frac% {1}{\sqrt{n}}\left(\int\limits_{\frac{k}{n}b_{n}}^{\frac{k+1}{n}b_{n}}\left|s-% x\right|^{2}ds\right)^{1/2}

which holds that

 \sum\limits_{k=0}^{\infty}p_{k}(\frac{n}{b_{n}}x)\int\limits_{\frac{k}{n}b_{n}% }^{\frac{k+1}{n}b_{n}}\left|s-x\right|ds\leq\frac{1}{\sqrt{n}}\sum\limits_{k=0% }^{\infty}p_{k}(\frac{n}{b_{n}}x)\left(\int\limits_{\frac{k}{n}b_{n}}^{\frac{k% +1}{n}b_{n}}\left|s-x\right|^{2}ds\right)^{1/2}.

If we apply the Cauchy-Schwarz inequality, we get

 \displaystyle\left|L_{n}^{\ast}(f;x)-f\left(x\right)\right| \displaystyle\leq\left\{1+\frac{1}{\delta}\left(\frac{e^{-nx}}{g(1)}\frac{n}{b% _{n}}\sum\limits_{k=0}^{\infty}p_{k}(\frac{n}{b_{n}}x)\int\limits_{\frac{k}{n}% b_{n}}^{\frac{k+1}{n}b_{n}}\left|s-x\right|^{2}ds\right)\right\}\omega(f,\delta) \displaystyle=\left\{1+\frac{1}{\delta}\left(\sqrt{L_{n}^{\ast}((s-x)^{2};x)}% \right)\right\}\omega(f,\delta) \displaystyle\leq\left\{1+\frac{1}{\delta}\left(\sqrt{\theta_{n}}\right)\right% \}\omega(f,\delta).

Now if we choose \delta=\sqrt{\theta_{n}}, it completes the proof. ∎

Now we remember the second modulus of continuity of f\in C_{B}[0,\infty ) which is defined by

 \omega_{2}\left(f;\delta\right)=\sup_{0

where C_{B}[0,\infty ) is the class of real valued functions defined on [0,\infty ) which are bounded and uniformly continuous with the norm \left\|f\right\|_{C_{B}}=\sup_{x\in[0,\infty)}\left|f\left(x\right)\right|.

Peetre’s K-functional of the function f\in C_{B}[0,\infty ) is defined by

 K(f;\delta)=\inf\left\{\left\|f-g\right\|_{C_{B}}+\delta\left\|g\right\|_{C_{B% }^{2}}\right\}, (2.5)

where

 C_{B}^{2}[0,\infty)=\left\{g\in C_{B}[0,\infty):g^{{}^{\prime}},g^{{}^{\prime% \prime}}\in C_{B}[0,\infty)\right\},

and the norm \left\|g\right\|_{C_{B}^{2}}=\left\|g\right\|_{C_{B}^{2}}+\left\|g^{{}^{\prime% }}\right\|_{C_{B}}+\left\|g^{{}^{\prime\prime}}\right\|_{C_{B}}.  The following inequality

 K(f;\delta)\leq M\left\{\omega_{2}\left(f;\sqrt{\delta}\right)+\min\left(1,% \delta\right)\left\|f\right\|_{C_{B}}\right\}, (2.6)

holds for all \delta>0 and the constant M is independent of f and \delta.

###### Theorem 3.

If f\in C_{B}^{2}[0,\infty), then we have

 \left|L_{n}^{\ast}(f;x)-f\left(x\right)\right|\leq\xi\left\|f\right\|_{C_{B}^{% 2}},

where

 \xi:=\xi_{n}\left(x\right)=\left\{\frac{g^{\prime}(1)}{g(1)}\frac{b_{n}}{n}+% \frac{b_{n}}{n}+\frac{b_{n}^{2}}{n^{2}}\left(\frac{2g^{\prime}(1)}{g(1)}+\frac% {g^{\prime\prime}(1)}{g(1)}+1\right)\right\}.
###### Proof.

From the Taylor expansion of f, the linearity of the operators L_{n}^{\ast} and (2.1), we have

 L_{n}^{\ast}(f;x)-f\left(x\right)=f^{\prime}\left(x\right)L_{n}^{\ast}(s-x;x)+% \frac{1}{2}f^{\prime\prime}\left(\eta\right)L_{n}^{\ast}(\left(s-x\right)^{2};% x),\eta\in\left(x,s\right). (2.7)

Since

 L_{n}^{\ast}(s-x;x)=\frac{g^{\prime}(1)}{g(1)}\frac{b_{n}}{n}+\frac{b_{n}}{n}\geq 0

for s\geq x, by considering Lemma and (2.7), we can write

 \displaystyle\left|L_{n}^{\ast}(f;x)-f\left(x\right)\right| \displaystyle\leq\left\{\frac{g^{\prime}(1)}{g(1)}\frac{b_{n}}{n}+\frac{b_{n}}% {n}\right\}\left\|f^{\prime}\right\|_{C_{B}}+\left\{\frac{b_{n}^{2}}{n^{2}}% \left(\frac{2g^{\prime}(1)}{g(1)}+\frac{g^{\prime\prime}(1)}{g(1)}+1\right)% \right\}\left\|f^{{}^{\prime\prime}}\right\|_{C_{B}} \displaystyle\leq\left\{\frac{g^{\prime}(1)}{g(1)}\frac{b_{n}}{n}+\frac{b_{n}}% {n}+\frac{b_{n}^{2}}{n^{2}}\left(\frac{2g^{\prime}(1)}{g(1)}+\frac{g^{\prime% \prime}(1)}{g(1)}+1\right)\right\}\left\|f\right\|_{C_{B}^{2}}.

which completes the proof. ∎

###### Theorem 4.

Let f\in C_{B}[0,\infty), then

 \left|L_{n}^{\ast}(f;x)-f\left(x\right)\right|\leq 2M\left\{\omega_{2}\left(f;% \sqrt{\delta}\right)+\min\left(1,\delta\right)\left\|f\right\|_{C_{B}}\right\},

where

 \delta:=\delta_{n}\left(x\right)=\frac{1}{2}\varsigma_{n}\left(x\right)

and M>0 is a constant which is independent of the functions f and \delta. Also, \varsigma_{n}\left(x\right) is the same as in the Theorem 3.

###### Proof.

Suppose that g\in C_{B}^{2}[0,\infty). From Theorem 3, we can write

 \displaystyle\left|L_{n}^{\ast}(f;x)-f\left(x\right)\right| \displaystyle\leq\left|L_{n}^{\ast}(f-g;x)\right|+\left|L_{n}^{\ast}(g;x)-g% \left(x\right)\right|+\left|g(x)-f\left(x\right)\right| (2.8) \displaystyle\leq 2\left\|f-g\right\|_{C_{B}}+\varsigma\left\|g\right\|_{C_{B}% ^{2}} \displaystyle=2\left[\left\|f-g\right\|_{C_{B}}+\delta\left\|g\right\|_{C_{B}^% {2}}\right]

The left-hand side of inequality (2.8) does not depend on the function g\in C_{B}^{2}[0,\infty), so

 \left|L_{n}^{\ast}(f;x)-f\left(x\right)\right|\leq 2K(f,\delta)

holds where K \left(f;\delta\right) is Peetre’s K-functional defined by (2.5) . By the relation between Peetre’s K functional and the second modulus of smoothness given by (2.6) , inequality (2.8) becomes

 \left|L_{n}^{\ast}(f;x)-f\left(x\right)\right|\leq 2M\left\{\omega_{2}\left(f;% \sqrt{\delta}\right)+\min\left(1,\delta\right)\left\|f\right\|_{C_{B}}\right\}

hence we have the result. ∎

Now let us consider the Lipschitz type space with two parameters (see [13]).

 Lip_{M}^{\left(\alpha_{1},\alpha_{2}\right)}\left(\alpha\right):=\left\{f\in C% _{B}\left[0,\infty\right):\left|f(t)-f\left(x\right)\right|\leq M\frac{\left|t% -x\right|^{\alpha}}{\left(t+\alpha_{1}x^{2}+\alpha_{2}x\right)^{\frac{\alpha}{% 2}}};x,t\in\left[0,\infty\right)\right\}

for \alpha_{1},\alpha_{2}>0, M is a positive constant and \alpha\in\left[0,1\right).

###### Theorem 5.

Let f\in Lip_{M}^{\left(\alpha_{1},\alpha_{2}\right)}\left(\alpha\right). For all x>0, we have

 \left|L_{n}^{\ast}(f;x)-f\left(x\right)\right|\leq M\left(\frac{\frac{b_{n}^{2% }}{n^{2}}\left(\frac{2g^{\prime}(1)}{g(1)}+\frac{g^{\prime\prime}(1)}{g(1)}+1% \right)}{\alpha_{1}x^{2}+\alpha_{2}x}\right)^{\frac{\alpha}{2}}
###### Proof.

Let \alpha=1.

 \displaystyle\left|L_{n}^{\alpha,\ast}(f;x)-f\left(x\right)\right| \displaystyle\leq\frac{e^{-nx}}{g(1)}\frac{n}{b_{n}}\sum\limits_{k=0}^{\infty}% p_{k}(\frac{n}{b_{n}}x)\int\limits_{\frac{k}{n}b_{n}}^{\frac{k+1}{n}b_{n}}% \left|f(t)-f\left(x\right)\right|dt \displaystyle\leq M\frac{e^{-nx}}{g(1)}\frac{n}{b_{n}}\sum\limits_{k=0}^{% \infty}p_{k}(\frac{n}{b_{n}}x)\int\limits_{\frac{k}{n}b_{n}}^{\frac{k+1}{n}b_{% n}}\left|t-x\right|dt \displaystyle\leq\frac{M}{\sqrt{\alpha_{1}x^{2}+\alpha_{2}x}}L_{n}^{\alpha,% \ast}(\left|t-x\right|;x) \displaystyle\leq M\left(\sqrt{\frac{L_{n}^{\alpha,\ast}((t-x)^{2};x)}{\alpha_% {1}x^{2}+\alpha_{2}x}}\right) \displaystyle\leq M\frac{b_{n}}{n}\sqrt{\frac{\frac{2g^{\prime}(1)}{g(1)}+% \frac{g^{\prime\prime}(1)}{g(1)}+1}{\alpha_{1}x^{2}+\alpha_{2}x}}.

Let \alpha\in\left(0,1\right).By applying Hölder inequality with p=\frac{1}{\alpha} and q=\frac{1}{1-\alpha}

 \displaystyle\left|L_{n}^{\alpha,\ast}(f;x)-f\left(x\right)\right| \displaystyle\leq\frac{e^{-nx}}{g(1)}\frac{n}{b_{n}}\sum\limits_{k=0}^{\infty}% p_{k}(\frac{n}{b_{n}}x)\int\limits_{\frac{k}{n}b_{n}}^{\frac{k+1}{n}b_{n}}% \left|f(t)-f\left(x\right)\right|dt \displaystyle\leq\left\{\frac{e^{-nx}}{g(1)}\sum\limits_{k=0}^{\infty}p_{k}(% \frac{n}{b_{n}}x)\left(\frac{n}{b_{n}}\int\limits_{\frac{k}{n}b_{n}}^{\frac{k+% 1}{n}b_{n}}\left|f(t)-f\left(x\right)\right|dt\right)^{\frac{1}{\alpha}}\right% \}^{\alpha} \displaystyle\leq\left\{\frac{n}{b_{n}}\frac{e^{-nx}}{g(1)}\sum\limits_{k=0}^{% \infty}p_{k}(\frac{n}{b_{n}}x)\int\limits_{\frac{k}{n}b_{n}}^{\frac{k+1}{n}b_{% n}}\left|f(t)-f\left(x\right)\right|^{\frac{1}{\alpha}}dt\right\}^{\alpha} \displaystyle\leq M\left\{\frac{n}{b_{n}}\frac{e^{-nx}}{g(1)}\sum\limits_{k=0}% ^{\infty}p_{k}(\frac{n}{b_{n}}x)\int\limits_{\frac{k}{n}b_{n}}^{\frac{k+1}{n}b% _{n}}\frac{\left|t-x\right|}{\sqrt{t+\alpha_{1}x^{2}+\alpha_{2}x}}dt\right\}^{\alpha} \displaystyle\leq\frac{M}{\left(\alpha_{1}x^{2}+\alpha_{2}x\right)^{\frac{% \alpha}{2}}}\left\{\frac{n}{b_{n}}\frac{e^{-nx}}{g(1)}\sum\limits_{k=0}^{% \infty}p_{k}(\frac{n}{b_{n}}x)\int\limits_{\frac{k}{n}b_{n}}^{\frac{k+1}{n}b_{% n}}\left|t-x\right|dt\right\}^{\alpha} \displaystyle\leq\frac{M}{\left(\alpha_{1}x^{2}+\alpha_{2}x\right)^{\frac{% \alpha}{2}}}\left(L_{n}^{\alpha,\ast}(\left|t-x\right|;x)\right)^{\alpha} \displaystyle\leq M\left(\frac{L_{n}^{\alpha,\ast}((t-x)^{2};x)}{\alpha_{1}x^{% 2}+\alpha_{2}x}\right)^{\frac{\alpha}{2}}=M\left(\frac{\frac{b_{n}^{2}}{n^{2}}% \left(\frac{2g^{\prime}(1)}{g(1)}+\frac{g^{\prime\prime}(1)}{g(1)}+1\right)}{% \alpha_{1}x^{2}+\alpha_{2}x}\right)^{\frac{\alpha}{2}}

## 3. Approximation properties in weighted spaces

Now we present the weighted spaces of functions that appear in the paper. With this purpose we firstly introduce the function

We give approximation properties of the operators L_{n}^{\ast} of the weighted spaces of continuous functions with exponential growth on R_{0}^{+}=\left[0,\infty\right) with the help of the weighted Korovkin type theorem proved by Gadjiev in [4], [5]. Therefore we consider the following weighted spaces of functions which are defined on the R_{0}^{+}.

Let \rho(x) be the weight function and M_{f} be a positive constant, we define

 B_{\rho}\left(R_{0}^{+}\right)=\left\{f\in E\left(R_{0}^{+}\right):\left|f(x)% \right|\leq M_{f}\rho(x)\right\},
 C_{\rho}\left(R_{0}^{+}\right)=\left\{f\in B\left(R_{0}^{+}\right):f\text{ is % continuous}\right\},
 C_{\rho}^{k}\left(R_{0}^{+}\right)=\left\{f\in C\left(R_{0}{}^{+}\right):% \underset{x\rightarrow\infty}{\lim}\frac{f(x)}{\rho(x)}=K_{f}<\infty\right\}.

It is obvious that C_{\rho}^{k}\left(R_{0}^{+}\right)\subset C_{\rho}\left(R_{0}^{+}\right)% \subset B_{\rho}\left(R_{0}^{+}\right). B_{\rho}\left(R_{0}^{+}\right) is a linear normed space with the norm

 \left\|f\right\|_{\rho}=\underset{x\in R_{0}^{+}}{\sup}\frac{\left|f(x)\right|% }{\rho(x)}.

The following results on the sequence of positive linear operators in these spaces are given in [4], [5].

###### Lemma 3.

The sequence of positive linear operators \left(L_{n}\right)_{n\geq 1} which act from C_{\rho}\left(R_{0}^{+}\right) to B_{\rho}\left(R_{0}^{+}\right) if and only if there exists a positive constant k such that L_{n}(\rho;x)\leq k\rho(x), i.e. \left\|L_{n}(\rho;x)\right\|_{\rho}\leq k.

###### Theorem 6.

Let \left(L_{n}\right)_{n\geq 1} be the space of positive linear operators which act from C_{\rho}\left(R_{0}^{+}\right) to B_{\rho}\left(R_{0}^{+}\right) satisfying the conditions

 \underset{n\rightarrow\infty}{\lim}\left\|L_{n}(t^{v};x)-x^{v}\right\|_{\rho}=% 0,\text{ }v=0,1,2,

then for any function f\in C_{\rho}^{k}\left(R_{0}^{+}\right)

 \underset{n\rightarrow\infty}{\lim}\left\|L_{n}f-f\right\|_{\rho}=0.

###### Lemma 4.

Let \rho(x)=1+x^{2} be a weight function. If f\in C_{\rho}\left(R_{0}^{+}\right), then

 \left\|L_{n}^{\ast}(\rho;x)\right\|_{\rho}\leq 1+M.
###### Proof.

Using (2.1) and (2.2), we have

 L_{n}^{\ast}(\rho;x)=1+x^{2}+\frac{b_{n}}{n}x\left(\frac{g(1)+2g^{\prime}(1)}{% g(1)}+1\right)+\frac{2b_{n}^{2}}{n^{2}}\frac{g^{\prime}(1)}{g(1)}+\frac{b_{n}^% {2}}{n^{2}}\frac{g^{\prime\prime}(1)}{g(1)}+\frac{b_{n}^{2}}{3n^{2}}
 \displaystyle\left\|L_{n}^{\ast}(\rho;x)\right\|_{\rho} \displaystyle=\underset{x\in R_{0}^{+}}{\sup}\frac{1}{1+x^{2}}\left[1+x^{2}+% \frac{b_{n}}{n}x\left(\frac{g(1)+2g^{\prime}(1)}{g(1)}+1\right)+\frac{2b_{n}^{% 2}}{n^{2}}\frac{g^{\prime}(1)}{g(1)}+\frac{b_{n}^{2}}{n^{2}}\frac{g^{\prime% \prime}(1)}{g(1)}+\frac{b_{n}^{2}}{3n^{2}}\right] \displaystyle\leq 1+\frac{b_{n}}{n}\left(\frac{g(1)+2g^{\prime}(1)}{g(1)}+1% \right)+\frac{2b_{n}^{2}}{n^{2}}\frac{g^{\prime}(1)}{g(1)}+\frac{b_{n}^{2}}{n^% {2}}\frac{g^{\prime\prime}(1)}{g(1)}+\frac{b_{n}^{2}}{3n^{2}}

since \underset{n\rightarrow\infty}{\lim}\frac{b_{n}}{n}=0, there exists a positive constant M such that

 \left\|L_{n}^{\ast}(\rho;x)\right\|_{\rho}\leq 1+M.

By using Lemma 3, we can easily see that the operators L_{n}^{\ast} defined by (1.4) act from C_{\rho}\left(R_{0}^{+}\right) to B_{\rho}\left(R_{0}^{+}\right).

###### Theorem 7.

Let L_{n}^{\ast} be the sequence of linear positive operators defined by (1.4) and \rho(x)=1+x^{2}, then for each f\in C_{\rho}^{k}\left(R_{0}^{+}\right)

 \underset{n\rightarrow\infty}{\lim}\left\|L_{n}^{\ast}(\rho;x)-f(x)\right\|_{% \rho}=0.
###### Proof.

It is enough to show that the conditions of the weighted Korovkin type theorem given by Theorem 6. From (2.1), we can write

 \underset{n\rightarrow\infty}{\lim}\left\|L_{n}^{\ast}(1;x)-1\right\|_{\rho}=0. (3.1)

Using (2.2), we have

 \left\|L_{n}^{\ast}(e_{1};x)-e_{1}(x)\right\|_{\rho}=\frac{b_{n}}{n}\frac{g^{% \prime}(1)}{g(1)}+\frac{b_{n}}{n}

this implies that

 \underset{n\rightarrow\infty}{\lim}\left\|L_{n}^{\ast}(e_{1};x)-e_{1}(x)\right% \|_{\rho}=0. (3.2)

From (2.3),

 \displaystyle\left\|L_{n}^{\ast}(e_{2};x)-e_{2}(x)\right\|_{\rho} \displaystyle=\underset{x\in R_{0}^{+}}{\sup}\frac{1}{1+x^{2}}\left\{\frac{b_{% n}}{n}x\left(\frac{g(1)+2g^{\prime}(1)}{g(1)}+1\right)+\frac{b_{n}^{2}}{n^{2}}% \left(\frac{2g^{\prime}(1)}{g(1)}+\frac{g^{\prime\prime}(1)}{g(1)}+\frac{1}{3}% \right)\right\} \displaystyle\leq\frac{b_{n}}{n}\left(\frac{g(1)+2g^{\prime}(1)}{g(1)}+1\right% )+\frac{b_{n}^{2}}{n^{2}}\left(\frac{2g^{\prime}(1)}{g(1)}+\frac{g^{\prime% \prime}(1)}{g(1)}+\frac{1}{3}\right).

Using the conditions (1.5), it follows that

 \underset{n\rightarrow\infty}{\lim}\left\|L_{n}^{\ast}(e_{2};x)-e_{2}(x)\right% \|_{\rho}=0. (3.3)

From (3.1), (3.2) and (3.3) for v=0,1,2, we have

 \underset{n\rightarrow\infty}{\lim}\left\|L_{n}^{\ast}(e_{v};x)-e_{v}(x)\right% \|_{\rho}=0.

If we apply Theorem 6, we obtain desired result. ∎

## References

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