I. INTRODUCTION AND BACKGROUND

The famous Szasz-Mirakjan perator is a positive linear operator on $C[0,\mathrm{\infty })$ (continuous function on $[0,\mathrm{\infty })$ ), defined as follows

It is easy to obtain by calculation

There are a lot of in-depth researches on its approximation, and the research results are very rich.[1-21]

The literature[17] introduce The Szasz-Mirakjan-Durrmeyer operator by Durrmeyer transformation of Szasz-Mirakjan operator

Of which

It is easily obtained by (1) and (2)

Obviously this is a positive linear operator on $L[0,\mathrm{\infty })$ (integrable function on $[0,\mathrm{\infty })$ ), and there has been a lot of research on it.[21-26].

Operator localization is one of the methods to transform operators. In order to be convenient and feasible, the localization of some operators is necessary. The idea of operator localization originates from the truncation of infinite sum. In 1980, literature [27] firstly introduced Szasz-Mirakjan localization operator as follows

In this paper, the same method is used to introduce the localized Szasz-Mirakjan-Durrmeyer operator as follows

Of which

Here we mainly study the approximation problem of localized Szasz-Mirakjan-Durrmeyer operator. The following part is used to discuss the convergence and approximation of ${\tilde{S}}_{n,N}(f)$ operator, resulting in theorems 1 and 2. The last part is used to study the approximation velocity of ${\tilde{S}}_{n,N}(f)$ operator, and theorem 3 is obtained.

II. THE CONVERGENCE OF LOCALIZED SZASZ-MIRAKJAN-DURRMeyer OPERATOR

Let's start with the following lemma.

Lemma 1 [28] Let and $y\in (-\mathrm{\infty },+\mathrm{\infty })$, Then

Where $A$ is a constant. If , change $\frac{y}{\sqrt{x}}$ to $\mathrm{\infty }$, then the conclusion is valid.

Using this lemma, we come to the following conclusion

Theorem 1: Let $f\in C[0,\mathrm{\infty })$ and $N=N(n,x)$, make

1. $(N-nx)/n$ uniformly bounded on $[x_1,x_2]$, where .
2. $\underset{n\to \mathrm{\infty }}{lim}\frac{N-nx}{\sqrt{n}}=C(x)$ preserving uniform convergence in $[x_1,x_2]$, Then, it is established that there is

uniformly in $[x_1,x_2]$.

Proof: Let

For , define the optical sliding mode

According to the property of the optical sliding mode, if , then

If $x\in [x_1,x_2]$, then

According to (6) and (4)

From lemma 1, we have

From (7) and (8), it can be concluded that (5) converges uniformly in $[x_1,x_2]$. In a similar way to Theorem 1, we get the following conclusion.

Theorem 2: Let $f\in C[0,\mathrm{\infty })$ and $N=N(n,x)$, make

1. $(N-nx)/n$ is uniformly bounded on $[x_1,x_2]$, where, .
2. maintain consistent convergence within $[x_1,x_2]$, then $\underset{n\to \mathrm{\infty }}{lim}{\tilde{S}}_{n,N}(f;x)=\frac{f(x)}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{\frac{C(x)}{\sqrt{x}}}{e}^{-\frac{1}{2}{u}^2}du$ is uniformly true on $[x_1,x_2]$. When $x=0$, think of $\frac{C(x)}{\sqrt{x}}$ as $\mathrm{\infty }$.

Corollary 1: Let $f\in C[0,\mathrm{\infty })$ and $N=N(n,x)$, make

1. $(N-nx)/n$ is uniformly bounded on $[x_1,x_2]$, where, .

• (ii) $\underset{n\to \mathrm{\infty }}{lim}\frac{N-nx}{\sqrt{n}}=\mathrm{\infty }$ preserving uniform convergence in $[x_1,x_2]$, then $\underset{n\to \mathrm{\infty }}{lim}{\tilde{S}}_{n,N}(f;x)=f(x)$ is uniformly true on $[x_1,x_2]$.

Corollary 2: Let $f\in C[0,\mathrm{\infty })$ and $x_0\in (0,\mathrm{\infty })$, If $(N-n{x}_0)/n$ is bounded, and when $n\to \mathrm{\infty }$, $(N-n{x}_0)/n$ does not converge to $\mathrm{\infty }$, Then $\underset{n\to \mathrm{\infty }}{lim}{\tilde{S}}_{n,N}(f;x_0)=f(x_0)$ must result in $f(x_0)=0$

Corollary 3: Let $x_0\in [0,\mathrm{\infty })$, If $(N-n{x}_0)/n$ is bounded, then

is true for any $f\in C[0,\mathrm{\infty })$,

If and only if

III. APPROXIMATION SPEED OF SZÁSZ-MIRAKJA-DURRMAYER DEFORMATION LOCALIZATION OPERATOR

A similar result is obtained for the localization Szasz-Mirakjan-Durrmeyer operator ${\tilde{S}}_{n,N}(f,x)$ as follows.

Theorem 3: Let $f\in C_{\alpha }$ for fixed , there is

Make $f$ conform to

in which $|t-x|\le \delta$, and $t\ge 0$. And $N=N(n,x)$,

is true, then

If $f^{\prime }(x)$ exists, and $N=N(n,x)$, (13) is true, then

Proof: (11) first. We note that

From (9) and (4) we have

• , 由 $f\in C_{\alpha }$, 可得