Trigonometric Induced Multivariate Smooth Gauss–Weierstrass Singular Integrals Approximation

Abstract

In this article, we employ the uniform and $L_p$, $1\le p<\infty$ approximation properties of general smooth multivariate singular integral operators over ${\mathbb{R}}^N$, $N\ge 1$. It is a trigonometric relief approach with detailed applications to the corresponding smooth multivariate Gauss–Weierstrass singular integral operators. The results are quantitative via Jackson-type inequalities involving the first uniform and $L_p$ moduli of continuity.

1. Introduction

The degree of approximation by univariate and multivariate singular integral operators has been researched extensively in [1,2,3,4,5]. All these sources motivate our current work. In particular we studied the approximation properties of the smooth singular integral operators in [1,3,4]. These are not in general positive operators. Here, we use the uniform and $L_p$, $p\ge 1$, results of our multivariate general theory [6,7], to establish approximation properties of the smooth Gauss–Weierstrass singular integral operators. The degrees of approximation are given quantitatively by employing the uniform and $L_p$ first moduli of continuity. The fundamental tool here comes from [8], where a multivariate trigonometric Taylor formula is presented. For recent related work, see [9,10,11,12,13]. Other important articles on the topic are [14,15,16,17,18].

In the history of this topic, we mention the monograph [4] published in 2012, which was the first comprehensive source to exclusively address the classic theory of the approximation of singular integrals to the identity-unit operator. The authors there quantitatively studied the basic approximation properties of the general Picard, Gauss–Weierstrass and Poisson–Cauchy singular integral operators over the real line, which are not positive linear operators. In particular, they studied the rate of convergence of these operators to the unit operator, as well as the related simultaneous approximation. This is given via inequalities and with the use of a higher-order modulus of smoothness of the high-order derivative of the involved function. Some of these inequalities have been proved to be sharp. Also, they studied the global smoothness preservation property of these operators. Furthermore, they gave asymptotic expansions of Voronovskaya type for the error of approximation. They continued with the study of related properties of the general fractional Gauss–Weierstrass and Poisson–Cauchy singular integral operators. These properties were studied with respect to the Lp norm, $1⩽p⩽\infty$. The case of Lipschitz-type functions approximation was studied separately and in detail. Furthermore, they presented the corresponding general approximation theory of general singular integral operators with many applications to the previously under-explored domain of trigonometric singular integrals.

2. Background on General Theory

Here, $r\in \mathbb{N}$, $m\in {\mathbb{Z}}_{+}$, we define

$${\alpha }_j:={\alpha }_{j,r}^{\left[m\right]}:=\left\{\begin{array}{c}{\left(-1\right)}^{r-j}\left(\begin{array}{c}r\\ j\end{array}\right)j^{-m},\mathrm{if}j=1,2,\dots ,r,\hfill \\ 1-{\displaystyle \sum _{j=1}^r}{\left(-1\right)}^{r-j}\left(\begin{array}{c}r\\ j\end{array}\right)j^{-m},\mathrm{if}j=0.\hfill \end{array}\right.$$

(1)

and

$${\delta }_k:={\delta }_{k,r}^{\left[m\right]}:={\displaystyle \sum _{j=0}^r}{\alpha }_{j,r}^{\left[m\right]}{j}^k,k=1,2,\dots ,m\in \mathbb{N}.$$

(2)

See that

$${\displaystyle \sum _{j=0}^r}{\alpha }_{j,r}^{\left[m\right]}=1,$$

(3)

and

$$-{\displaystyle \sum _{j=1}^r}{\left(-1\right)}^{r-j}\left(\begin{array}{c}r\\ j\end{array}\right)={\left(-1\right)}^r\left(\begin{array}{c}r\\ 0\end{array}\right).$$

Let ${\mu }_{{\xi }_n}$ be a probability Borel measure on ${\mathbb{R}}^N$, $N\ge 1$, ${\xi }_n>0$, $n\in \mathbb{N}.$

We now define the multiple smooth singular integral operators

$${\theta }_n\left(f;x_1,\dots ,x_N\right):={\theta }_{r,n}^{\left[m\right]}\left(f;x_1,\dots ,x_N\right):=$$

$${\displaystyle \sum _{j=0}^r}{\alpha }_{j,r}^{\left[m\right]}{\int }_{{\mathbb{R}}^N}f\left(x_1+s_{1}j,x_2+s_{2}j,\dots ,x_N+s_{N}j\right)d{\mu }_{{\xi }_n}\left(s\right),$$

(4)

where $s:=\left(s_1,\dots ,s_N\right)$, $x:=\left(x_1,\dots ,x_N\right)\in {\mathbb{R}}^N;$ $n,r\in \mathbb{Z}$, $m\in {\mathbb{Z}}_{+}$, $f:{\mathbb{R}}^N\to \mathbb{R}$ is a Borel measurable function, and also ${\left({\xi }_n\right)}_{n\in \mathbb{N}}$ is a bounded sequence of positive real numbers; we take $0<{\xi }_n\le 1$.

Remark 1.

The operators ${\theta }_{r,n}^{\left[m\right]}$ are not generally positive (see [2], p. 2).

We observe that

Lemma 1.

It holds

$${\theta }_{r,n}^{\left[m\right]}\left(c;x_1,\dots ,x_n\right)=c,$$

where c is a constant.

We need

Definition 1.

Let $f\in C\left({\mathbb{R}}^N\right)$, $N\ge 1$. We define the first uniform modulus of continuity of f as

$${\omega }_1\left(f,\delta \right):=\underset{{∥x-y∥}_{\infty }\le \delta }{\underset{x,y\in {\mathbb{R}}^N:}{sup}}\left|f\left(x\right)-f\left(y\right)\right|,\delta >0,$$

(5)

where ${∥·∥}_{\infty }$ is the max norm in ${\mathbb{R}}^N$. The functional ${\omega }_1\left(f,\delta \right)$ is bounded for f being bounded or uniformly continuous, and ${\omega }_1\left(f,\delta \right)\to 0$ as $\delta \to 0$, in the case of f being uniformly continuous.

We mention the main uniform general approximation result regarding the operator ${\theta }_n$.

Theorem 1

([6]). Here, $f\in C^2\left({\mathbb{R}}^N\right)$ and let all ${\alpha }_i\in {\mathbb{Z}}^{+}$, $i=1,\dots ,N$, $N\ge 1$, $\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}{\alpha }_i=2$; $x\in {\mathbb{R}}^N$, and all the partials $f_{\alpha }$ of order 2, along with $f\in C_B\left({\mathbb{R}}^N\right)$ (continuous and bounded functions); or all $f_{\alpha }$ of order 2, $f\in C_U\left({\mathbb{R}}^N\right)$ (uniformly continuous functions). Let ${\mu }_{{\xi }_n}$ be a Borel probability measure on ${\mathbb{R}}^N$, for $0<{\xi }_n\le 1$, $n\in \mathbb{N}$.

Suppose that for all $\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right)$, ${\alpha }_i\in {\mathbb{Z}}^{+}$, $i=1,\dots ,N$, $\left|\alpha \right|={\displaystyle \sum _{i=1}^N}{\alpha }_i=2$, $j=0,1,\dots ,r$, we have that both

$$I_{1j}\left(\alpha \right):={\int }_{{\mathbb{R}}^N}\left(1+\frac{j{∥s∥}_1}{3{\xi }_n}\right)\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{{\alpha }_i}\right)d{\mu }_{{\xi }_n}\left(s\right),$$

(6)

$$I_{2j}\left(\alpha \right):={\int }_{{\mathbb{R}}^N}\left(1+\frac{j{∥s∥}_1}{3{\xi }_n}\right)d{\mu }_{{\xi }_n}\left(s\right),$$

(7)

are uniformly bounded in ${\xi }_n\in (0,1].$

Denote $(n\in \mathbb{N})$

$${\Delta }_n\left(x\right):={\theta }_n\left(f,x\right)-f\left(x\right)-\left({\displaystyle \sum _{j=0}^r}{\alpha }_{j}j\right)sin\left(1\right)\left[{\displaystyle \sum _{i=1}^N}\frac{\partial f\left(x\right)}{\partial x_i}\left({\int }_{{\mathbb{R}}^N}{s}_{i}d{\mu }_{{\xi }_n}\left(s\right)\right)\right]$$

$$-2\left({\displaystyle \sum _{j=0}^r}{\alpha }_j{j}^2\right){sin}^2\left(\frac{1}{2}\right)\left\{{\displaystyle \sum _{i=1}^N}\left({\int }_{{\mathbb{R}}^N}{s}_i^{2}d{\mu }_{{\xi }_n}\left(s\right)\right)\frac{{\partial }^{2}f\left(x\right)}{\partial x_i^2}+\right.$$

(8)

$$\left.\sum _{\begin{array}{c}i\ne j^{*},\\ i,j^{*}\in \left\{1,\dots ,N\right\}\end{array}}\left({\int }_{{\mathbb{R}}^N}{s}_i{s}_{j^{*}}d{\mu }_{{\xi }_n}\left(s\right)\right)\frac{{\partial }^{2}f\left(x\right)}{\partial x_i\partial x_{j^{*}}}\right\}.$$

Then

(i)

$$\left|{\Delta }_n\left(x\right)\right|\le {∥{\Delta }_n\left(x\right)∥}_{\infty }\le {\displaystyle \sum _{j=0}^r}\left|{\alpha }_j\right|$$

$$\left[\left[j^2\sum _{\begin{array}{c}{\alpha }_i\in {\mathbb{Z}}^{+},\\ \alpha :\left|\alpha \right|=2\end{array}}\left(\frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}\right){\omega }_1\left(f_{\alpha },{\xi }_n\right){\int }_{{\mathbb{R}}^N}\left(1+\frac{j{∥s∥}_{\infty }}{3{\xi }_n}\right)\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{{\alpha }_i}\right)d{\mu }_{{\xi }_n}\left(s\right)\right]+\right.$$

$$\left.\frac{1}{2}{\omega }_1\left(f,{\xi }_n\right){\int }_{{\mathbb{R}}^N}\left(1+\frac{j{∥s∥}_{\infty }}{3{\xi }_n}\right)d{\mu }_{{\xi }_n}\left(s\right)\right]=:{\phi }_{{\xi }_n}.$$

(9)

In case of all $f_{\alpha }$ of order 2 and $f\in C_U\left({\mathbb{R}}^N\right)$ and ${\xi }_n\to 0$, as $n\to \infty$, then ${\Delta }_n\left(x\right)$, ${∥{\Delta }_n\left(x\right)∥}_{\infty }\to 0$ with rates.

(ii) If $\frac{\partial f\left(x\right)}{\partial x_i}=0$, $i=1,\dots ,N$, and $f_{\alpha }\left(x\right)=0$, ${\alpha }_i\in {\mathbb{Z}}^{+}$, $i=1,\dots ,N$, with $\left|\alpha \right|=2$, then

$$\left|{\theta }_n\left(f,x\right)-f\left(x\right)\right|\le {\phi }_{{\xi }_n}.$$

(10)

And ${\theta }_n\left(f,x\right)\to f\left(x\right)$ in the uniformly continuous case.

(iii) Additionally, assume all partials of order $\le 2$ are bounded. Hence,

$${∥{\theta }_n\left(f\right)-f∥}_{\infty }\le \left({\displaystyle \sum _{j=0}^r}\left|{\alpha }_j\right|j\right)\left(0.8414\right)\left[{\displaystyle \sum _{i=1}^N}{∥\frac{\partial f}{\partial x_i}∥}_{\infty }\left({\int }_{{\mathbb{R}}^N}\left|s_i\right|d{\mu }_{{\xi }_n}\left(s\right)\right)\right]+$$

$$\left({\displaystyle \sum _{j=0}^r}\left|{\alpha }_j\right|j^2\right)\left(0.4596\right)\left\{{\displaystyle \sum _{i=1}^N}\left({\int }_{{\mathbb{R}}^N}{s}_i^{2}d{\mu }_{{\xi }_n}\left(s\right)\right){∥\frac{{\partial }^{2}f}{\partial x_i^2}∥}_{\infty }+\right.$$

$$\left.\sum _{\begin{array}{c}i\ne j^{*},\\ i,j^{*}\in \left\{1,\dots ,N\right\}\end{array}}\left({\int }_{{\mathbb{R}}^N}\left|s_i\right|\left|s_{j^{*}}\right|d{\mu }_{{\xi }_n}\left(s\right)\right){∥\frac{{\partial }^{2}f}{\partial x_i\partial x_{j^{*}}}∥}_{\infty }\right\}+{\phi }_{{\xi }_n}.$$

(11)

If all ${\int }_{{\mathbb{R}}^N}{s}_i^{2}d{\mu }_{{\xi }_n}\left(s\right)$ and ${\int }_{{\mathbb{R}}^N}\left|s_i\right|\left|s_{j^{*}}\right|d{\mu }_{{\xi }_n}\left(s\right)$ converge to zero, as $n\to \infty$, with ${\xi }_n\to 0$, and all $f_{\alpha }$ of order 2, $f\in C_U\left({\mathbb{R}}^N\right)$, then

$${∥{\theta }_n\left(f\right)-f∥}_{\infty }\to 0withrates,as{\xi }_n\to 0,n\to +\infty .$$

Next, we deal with $f\in C^m\left({\mathbb{R}}^N\right)$, $m\in {\mathbb{Z}}^{+}$, with $f_{\alpha }\in L_p\left({\mathbb{R}}^N\right)$, $\left|\alpha \right|=m\in {\mathbb{Z}}^{+}$, $p\ge 1$; where $f_{\alpha }$ denotes the mixed partial $\frac{{\partial }^{\tilde{j}}f\left(·,\dots ,·\right)}{\partial x_1^{{\alpha }_1}\dots \partial x_N^{{\alpha }_N}}$, ${\alpha }_j\in {\mathbb{Z}}^{+}$, $j=1,\dots ,N:\left|\alpha \right|:={\displaystyle \sum _{j=1}^N}{\alpha }_j=\tilde{j}$, $\tilde{j}=1,\dots ,m.$

We need

Definition 2

(see also [2], p. 20). We call

$${\Delta }_u^{r}f\left(x\right):={\Delta }_{u_1,u_2,\dots ,u_N}^{r}f\left(x_1,\dots ,x_N\right):=$$

(12)

$${\displaystyle \sum _{j=0}^r}{\left(-1\right)}^{r-j}\left(\begin{array}{c}r\\ j\end{array}\right)f\left(x_1+j{u}_1,x_2+j{u}_2,\dots ,x_N+j{u}_N\right).$$

Let $p\ge 1$, the $L_p$ modulus of smoothness of order r is given by

$${\omega }_r{\left(f;h\right)}_p:=\underset{{∥u∥}_2\le h}{sup}{∥{\Delta }_u^r\left(f\right)∥}_p,$$

(13)

$h>0.$

We mention

Theorem 2

([7]). Let $f\in C^m\left({\mathbb{R}}^N\right)$, $m\in \mathbb{N}$, $N\ge 1$, with $f_{\alpha }\in L_p\left({\mathbb{R}}^N\right),$ $\left|\alpha \right|=m$, $x\in {\mathbb{R}}^N.$ Let $p,q>1:\frac{1}{p}+\frac{1}{q}=1$. Here, ${\mu }_{{\xi }_n}$ is a Borel probability measure on ${\mathbb{R}}^N$ for ${\xi }_n>0$, ${\left({\xi }_n\right)}_{n\in \mathbb{N}}$ bounded sequence. Assume for all $\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right),$ ${\alpha }_i\in {\mathbb{Z}}^{+}$, $i=1,\dots ,N,$ $\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}{\alpha }_i=m$ that we have

$${\int }_{{\mathbb{R}}^N}{\left(\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{{\alpha }_i}\right){\left(1+\frac{{∥s∥}_2}{{\xi }_n}\right)}^r\right)}^{p}d{\mu }_{{\xi }_n}\left(s\right)<\infty .$$

(14)

For $\tilde{j}=1,\dots ,m,$ and $\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right),$ ${\alpha }_i\in {\mathbb{Z}}^{+}$, $i=1,\dots ,N,$ $\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}{\alpha }_i=\tilde{j}$, call

$$c_{\alpha ,n,\tilde{j}}:={\int }_{{\mathbb{R}}^N}{\displaystyle \prod _{i=1}^N}{s}_i^{{\alpha }_i}d{\mu }_{{\xi }_n}\left(s\right).$$

(15)

Then

$${∥E_{r,n}^{\left[m\right]}∥}_p:={∥{\theta }_{r,n}^{\left[m\right]}\left(f;x\right)-f\left(x\right)-{\displaystyle \sum _{\tilde{j}=1}^m}{\delta }_{\tilde{j},r}^{\left[m\right]}\left(\sum _{\left|\alpha \right|=\tilde{j}}\frac{c_{\alpha ,n,\tilde{j}}{f}_{\alpha }\left(x\right)}{\left({\displaystyle \prod _{i=1}^N}{\alpha }_i!\right)}\right)∥}_{p,x}\le$$

(16)

$$\left[{\left(\begin{array}{c}\frac{N^m}{(m-1)!}\end{array}\right)}^{\frac{p}{q}}\left(\frac{m}{{\left(q\left(m-1\right)+1\right)}^{\frac{p}{q}}}\right)\left(\sum _{\left|\alpha \right|=m}\frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}·\right.\right.$$

$${\left.\left.\left[{\int }_{{\mathbb{R}}^N}{\left[\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{{\alpha }_i}\right){\left(1+\frac{{∥s∥}_2}{{\xi }_n}\right)}^r\right]}^{p}d{\mu }_{{\xi }_n}\left(s\right)\right]{\omega }_r{\left(f_{\alpha },{\xi }_n\right)}_p^p\right)\right]}^{\frac{1}{p}}.$$

As $n\to \infty$ and ${\xi }_n\to 0$, by (16), we obtain that ${∥E_{r,n}^{\left[m\right]}∥}_p\to 0$ with rates.

One also gets by (16) that

$${∥{\theta }_{r,n}^{\left[m\right]}\left(f;x\right)-f\left(x\right)∥}_{p,x}\le$$

$${\displaystyle \sum _{\tilde{j}=1}^m}\left|{\delta }_{\tilde{j},r}^{\left[m\right]}\right|\left(\sum _{\left|\alpha \right|=\tilde{j}}\frac{\left|c_{\alpha ,n,\tilde{j}}\right|}{{\displaystyle {\displaystyle \prod _{i=1}^N}}{\alpha }_i!}{∥f_{\alpha }∥}_p\right)+R.H.S.\left(16\right),$$

(17)

given that ${∥f_{\alpha }∥}_p<\infty ,$ $\left|\alpha \right|=\tilde{j}$, $\tilde{j}=1,\dots ,m.$

Assuming that $c_{\alpha ,n,\tilde{j}}\to 0$, ${\xi }_n\to 0,$ as $n\to \infty$, we get ${∥{\theta }_{r,n}^{\left[m\right]}\left(f\right)-f∥}_p\to 0$, that is ${\theta }_{r,n}^{\left[m\right]}\to I$ the unit operator, in $L_p$ norm, with rates.

We make

Remark 2.

Notice that ($p>1$)

$${\int }_{{\mathbb{R}}^N}\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{{\alpha }_i}\right){\left(1+\frac{{∥s∥}_2}{{\xi }_n}\right)}^{r}d{\mu }_{{\xi }_n}\left(s\right)\le$$

$${\left[{\int }_{{\mathbb{R}}^N}{\left(\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{{\alpha }_i}\right){\left(1+\frac{{∥s∥}_2}{{\xi }_n}\right)}^r\right)}^{p}d{\mu }_{{\xi }_n}\left(s\right)\right]}^{\frac{1}{p}}<\infty ,$$

(18)

as assumed in Theorem 2.

By (18), we get that

$${\int }_{{\mathbb{R}}^N}{\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{{\alpha }_i}d{\mu }_{{\xi }_n}\left(s\right)<\infty .$$

(19)

Hence, $c_{\alpha ,n,\tilde{j}}\in \mathbb{R}.$

We mention also the following trigonometric induced alternative $L_p$ approximation result for ${\theta }_n$ operators.

Theorem 3

([7]). Let $p,q>1:\frac{1}{p}+\frac{1}{q}=1$, $0<{\xi }_n\le 1$, $n\in \mathbb{N}$. Here, we deal with $f\in C^2\left({\mathbb{R}}^N\right)$, $N\ge 1$, with $f,$$f_{\alpha }\in L_p\left({\mathbb{R}}^N\right)$, $\left|\alpha \right|=2$, where ${\alpha }_i\in {\mathbb{Z}}^{+}$, $i=1,\dots ,N$, and $\left|\alpha \right|={\displaystyle \sum _{i=1}^N}{\alpha }_i$; $x\in {\mathbb{R}}^N$. Let ${\mu }_{\xi n}$ be a Borel probability measure on ${\mathbb{R}}^N$. Suppose that for all $\alpha :\left|\alpha \right|=2$, $j=0,1,\dots ,r$, we have that both

$$I_{1j}\left(\alpha \right):={\int }_{{\mathbb{R}}^N}\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{p{\alpha }_i}\right)\left(1+\frac{j^p{∥s∥}_2^p}{{\xi }_n^p}\right)d{\mu }_{{\xi }_n}\left(s\right),$$

(20)

$$I_2:={\int }_{{\mathbb{R}}^N}{\left(\frac{{∥s∥}_2}{{\xi }_n}\right)}^{p}d{\mu }_{{\xi }_n}\left(s\right),$$

(21)

are uniformly bounded in ${\xi }_n\in (0,1].$

Denote $(n\in \mathbb{N})$

$${\Delta }_n\left(x\right):={\theta }_n\left(f,x\right)-f\left(x\right)-\left({\displaystyle \sum _{j=0}^r}{\alpha }_{j}j\right)sin\left(1\right)\left[{\displaystyle \sum _{i=1}^N}\frac{\partial f\left(x\right)}{\partial x_i}\left({\int }_{{\mathbb{R}}^N}{s}_{i}d{\mu }_{{\xi }_n}\left(s\right)\right)\right]$$

$$-2\left({\displaystyle \sum _{j=0}^r}{\alpha }_j{j}^2\right){sin}^2\left(\frac{1}{2}\right)\left\{{\displaystyle \sum _{i=1}^N}\left({\int }_{{\mathbb{R}}^N}{s}_i^{2}d{\mu }_{{\xi }_n}\left(s\right)\right)\frac{{\partial }^{2}f\left(x\right)}{\partial x_i^2}+\right.$$

(22)

$$\left.\sum _{\begin{array}{c}i\ne j^{*},\\ i,j^{*}\in \left\{1,\dots ,N\right\}\end{array}}\left({\int }_{{\mathbb{R}}^N}{s}_i{s}_{j^{*}}d{\mu }_{{\xi }_n}\left(s\right)\right)\frac{{\partial }^{2}f\left(x\right)}{\partial x_i\partial x_{j^{*}}}\right\}.$$

Then, it holds

$${∥{\Delta }_n\left(x\right)∥}_p\le {\left({\displaystyle \sum _{j=0}^r}{\left|{\alpha }_j\right|}^q\right)}^{\frac{1}{q}}\left\{\frac{2}{{\left(q+1\right)}^{\frac{1}{q}}}{\left(\frac{N\left(N+1\right)+2}{2}\right)}^{\frac{1}{q}}\right\}$$

$$\left\{{\displaystyle \sum _{j=0}^r}\left[j^{2p}\sum _{\left|\alpha \right|=2}{\left(\frac{1}{{\displaystyle {\displaystyle \prod _{i=1}^N}}{\alpha }_i!}\right)}^p{\omega }_1{\left(f_{\alpha },{\xi }_n\right)}_p^p\left({\int }_{{\mathbb{R}}^N}\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{p{\alpha }_i}\right)\left(1+\frac{j^p{∥s∥}_2^p}{{\xi }_n^p}\right)d{\mu }_{{\xi }_n}\left(s\right)\right)\right.\right.$$

(23)

$${\left.\left.+{\omega }_1{\left(f,{\xi }_n\right)}_p^p\left({\int }_{{\mathbb{R}}^N}\left(1+\frac{j^p{∥s∥}_2^p}{{\xi }_n^p}\right)d{\mu }_{{\xi }_n}\left(s\right)\right)\right]\right\}}^{\frac{1}{p}}.$$

As $n\to \infty$ and ${\xi }_n\to 0$, by (23), we obtain that ${∥{\Delta }_n∥}_p\to 0$ with rates. One also gets by (23) that

$${∥{\theta }_n\left(f,x\right)-f\left(x\right)∥}_{p,x}\le$$

$$\left({\displaystyle \sum _{j=0}^r}\left|{\alpha }_j\right|j\right)sin\left(1\right)\left[{\displaystyle \sum _{i=1}^N}{∥\frac{\partial f}{\partial x_i}∥}_p\left({\int }_{{\mathbb{R}}^N}\left|s_i\right|d{\mu }_{{\xi }_n}\left(s\right)\right)\right]$$

$$+2\left({\displaystyle \sum _{j=0}^r}\left|{\alpha }_j\right|j^2\right){sin}^2\left(\frac{1}{2}\right)\left\{{\displaystyle \sum _{i=1}^N}\left({\int }_{{\mathbb{R}}^N}{s}_i^{2}d{\mu }_{{\xi }_n}\left(s\right)\right){∥\frac{{\partial }^{2}f}{\partial x_i^2}∥}_p+\right.$$

$$\left.\sum _{\begin{array}{c}i\ne j^{*},\\ i,j^{*}\in \left\{1,\dots ,N\right\}\end{array}}\left({\int }_{{\mathbb{R}}^N}\left|s_i{s}_{j^{*}}\right|d{\mu }_{{\xi }_n}\left(s\right)\right){∥\frac{{\partial }^{2}f}{\partial x_i\partial x_{j^{*}}}∥}_p\right\}+R.H.S.(23),$$

(24)

given that ${∥f_{\alpha }∥}_p<\infty$, $\left|\alpha \right|=\tilde{j}$, $\tilde{j}=1,2.$ Assuming that ${\int }_{{\mathbb{R}}^N}{s}_i^{2}d{\mu }_{{\xi }_m}\left(s\right),$${\int }_{{\mathbb{R}}^N}\left|s_i{s}_{j^{*}}\right|d{\mu }_{{\xi }_n}\left(s\right)$, $i,j^{*}\in \left\{1,\dots ,N\right\}$, $i\ne j^{*}$, converge to zero as ${\xi }_n\to 0$, we get ${∥{\theta }_n\left(f,x\right)-f\left(x\right)∥}_p\to 0$, that is ${\theta }_n\to I$ the unit operator, in $L_p$ norm, with rates.

Furthermore, we mention the following trigonometric-based alternative $L_1$ approximation result for ${\theta }_n$ operators.

Theorem 4

([7]). Let $0<{\xi }_n\le 1$, $n\in \mathbb{N}$, $x\in {\mathbb{R}}^N$. Here, we deal with $f\in C^2\left({\mathbb{R}}^N\right)$, $N\ge 1$, with $f,$ $f_{\alpha }\in L_1\left({\mathbb{R}}^N\right)$, $\left|\alpha \right|=2$, where ${\alpha }_i\in {\mathbb{Z}}^{+}$, $i=1,\dots ,N$, and $\left|\alpha \right|={\displaystyle \sum _{i=1}^N}{\alpha }_i$. Let ${\mu }_{\xi n}$ be a Borel probability measure on ${\mathbb{R}}^N$. Suppose that for all $\alpha :\left|\alpha \right|=2$, $j=0,1,\dots ,r$, we have that both

$$I_{1j}^{*}\left(\alpha \right):={\int }_{{\mathbb{R}}^N}\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{{\alpha }_i}\right)\left(1+\frac{j{∥s∥}_2}{{\xi }_n}\right)d{\mu }_{{\xi }_n}\left(s\right),$$

(25)

and

$$I_2^{*}:={\int }_{{\mathbb{R}}^N}\frac{{∥s∥}_2}{{\xi }_n}d{\mu }_{{\xi }_n}\left(s\right),$$

(26)

are uniformly bounded in ${\xi }_n\in (0,1].$

Here, ${\Delta }_n\left(x\right)$ is as in (22).

Then, it holds

$${∥{\Delta }_n\left(x\right)∥}_1\le \left\{{\displaystyle \sum _{j=0}^r}\left|{\alpha }_j\right|\left[j^2\sum _{\left|\alpha \right|=2}\left(\frac{2}{{\displaystyle {\displaystyle \prod _{i=1}^N}}{\alpha }_i!}\right){\omega }_1{\left(f_{\alpha },{\xi }_n\right)}_1\left({\int }_{{\mathbb{R}}^N}\left({\displaystyle \prod _{i=1}^N}{\left(\left|s_i\right|\right)}^{{\alpha }_i}\right)\right.\right.\right.$$

$$\left.\left.\left.\left(1+\frac{j{∥s∥}_2}{{\xi }_n}\right)d{\mu }_{{\xi }_n}\left(s\right)\right)+{\omega }_1{\left(f,{\xi }_n\right)}_1\left({\int }_{{\mathbb{R}}^N}\left(1+\frac{j{∥s∥}_2}{{\xi }_n}\right)d{\mu }_{{\xi }_n}\left(s\right)\right)\right]\right\}.$$

(27)

As $n\to \infty$ and ${\xi }_n\to 0$, by (27), we obtain that ${∥{\Delta }_n∥}_1\to 0$ with rates. One also obtains by (27) that

$${∥{\theta }_n\left(f\right)-f∥}_1\le$$

$$\left({\displaystyle \sum _{j=0}^r}\left|{\alpha }_j\right|j\right)sin\left(1\right)\left[{\displaystyle \sum _{i=1}^N}{∥\frac{\partial f}{\partial x_i}∥}_1\left({\int }_{{\mathbb{R}}^N}\left|s_i\right|d{\mu }_{{\xi }_n}\left(s\right)\right)\right]$$

$$+2\left({\displaystyle \sum _{j=0}^r}\left|{\alpha }_j\right|j^2\right){sin}^2\left(\frac{1}{2}\right)\left\{{\displaystyle \sum _{i=1}^N}\left({\int }_{{\mathbb{R}}^N}{s}_i^{2}d{\mu }_{{\xi }_n}\left(s\right)\right){∥\frac{{\partial }^{2}f}{\partial x_i^2}∥}_1+\right.$$

$$\left.\sum _{\begin{array}{c}i\ne j^{*},\\ i,j^{*}\in \left\{1,\dots ,N\right\}\end{array}}\left({\int }_{{\mathbb{R}}^N}\left|s_i{s}_{j^{*}}\right|d{\mu }_{{\xi }_n}\left(s\right)\right){∥\frac{{\partial }^{2}f}{\partial x_i\partial x_{j^{*}}}∥}_1\right\}+R.H.S.(27),$$

(28)

given that ${∥f_{\alpha }∥}_1<\infty$, $\left|\alpha \right|=\tilde{j}$, $\tilde{j}=1,2.$ Assuming that ${\int }_{{\mathbb{R}}^N}{s}_i^{2}d{\mu }_{{\xi }_n}\left(s\right),$${\int }_{{\mathbb{R}}^N}\left|s_i{s}_{j^{*}}\right|d{\mu }_{{\xi }_n}\left(s\right)$, $i,j^{*}\in \left\{1,\dots ,N\right\}$, $i\ne j^{*}$, converge to zero as ${\xi }_n\to 0$, we derive ${∥{\theta }_n\left(f\right)-f∥}_1\to 0$; that is ${\theta }_n\to I$ in $L_1$ norm, with rates.

3. Auxiliary Essential Results

We need

Theorem 5.

Let $N\in \mathbb{N};$ ${\alpha }_i\in {\mathbb{Z}}^{+},$ $i=1,\dots ,N:\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}{\alpha }_i=2$, ${\xi }_n\in (0,1]$, $n\in \mathbb{N};$ $j=0,1,\dots ,r\in \mathbb{N}.$ Then

$$I_{1j}^{*}\left(\alpha \right):=\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}{\int }_{{\mathbb{R}}^N}\left(1+\frac{j{∥s∥}_1}{3{\xi }_n}\right)\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{{\alpha }_i}\right)e^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\le$$

(29)

$$\sqrt{{\xi }_n}{\left(\frac{2}{\sqrt{\pi }}\right)}^N\left[\left(1+jN\right)+\left(1+j\right){\left(\frac{⌊6e⌋}{e}\right)}^N\right]\le$$

$${\left(\frac{2}{\sqrt{\pi }}\right)}^N\left[\left(1+jN\right)+\left(1+j\right){\left(\frac{⌊6e⌋}{e}\right)}^N\right]<+\infty ,$$

are uniformly bounded, and $I_{1j}^{*}\left(\alpha \right)\to 0$, as ${\xi }_n\to 0$.

Above, $⌊·⌋$ denotes the integral part.

Proof. 

We have that

$$I_{1j}^{*}\left(\alpha \right)=\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}{\int }_{{\mathbb{R}}^N}\left(1+\frac{j{∥s∥}_1}{3{\xi }_n}\right)\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{{\alpha }_i}\right)e^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N=$$

$$\frac{2^N}{{\left(\sqrt{\pi }\right)}^N{\left(\sqrt{{\xi }_n}\right)}^N}{\int }_{{\mathbb{R}}_{+}^N}\left(1+\frac{j{∥s∥}_1}{3{\xi }_n}\right)\left({\displaystyle \prod _{i=1}^N}{s}_i^{{\alpha }_i}\right)e^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N=$$

(30)

$$\frac{2^N}{{\left(\sqrt{\pi }\right)}^N{\left(\sqrt{{\xi }_n}\right)}^N}{\int }_{{\mathbb{R}}_{+}^N}\left(1+\frac{j\left({\displaystyle \sum _{i=1}^N}{s}_i\right)}{3{\xi }_n}\right)\left({\displaystyle \prod _{i=1}^N}{s}_i^{{\alpha }_i}\right)e^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\le$$

$$\frac{2^N{\left(\sqrt{{\xi }_n}\right)}^2}{{\left(\sqrt{\pi }\right)}^N}{\int }_{{\mathbb{R}}_{+}^N}\left(\frac{1}{\sqrt{{\xi }_n}}+j\frac{1}{\sqrt{{\xi }_n}}\left({\displaystyle \sum _{i=1}^N}\frac{s_i}{\sqrt{{\xi }_n}}\right)\right)$$

$$\left({\displaystyle \prod _{i=1}^N}{\left(\frac{s_i}{\sqrt{{\xi }_n}}\right)}^{{\alpha }_i}\right)e^{-{\displaystyle \sum _{i=1}^N}{\left(\frac{s_i}{\sqrt{{\xi }_n}}\right)}^2}\frac{d{s}_1}{\sqrt{{\xi }_n}}\dots \frac{d{s}_N}{\sqrt{{\xi }_n}}=$$

$$\frac{2^N\sqrt{{\xi }_n}}{{\left(\sqrt{\pi }\right)}^N}{\int }_{{\mathbb{R}}_{+}^N}\left(1+j\left({\displaystyle \sum _{i=1}^N}{z}_i\right)\right)\left({\displaystyle \prod _{i=1}^N}{z}_i^{{\alpha }_i}\right)e^{-\left({\displaystyle \sum _{i=1}^N}{z}_i^2\right)}d{z}_1\dots d{z}_N=$$

(31)

$$\frac{2^N\sqrt{{\xi }_n}}{{\left(\sqrt{\pi }\right)}^N}\left[{\int }_{{\left[0,1\right]}^N}\left(1+j\left({\displaystyle \sum _{i=1}^N}{z}_i\right)\right)\left({\displaystyle \prod _{i=1}^N}{z}_i^{{\alpha }_i}\right)e^{-\left({\displaystyle \sum _{i=1}^N}{z}_i^2\right)}d{z}_1\dots d{z}_N+\right.$$

$$\left.{\int }_{{\left({\mathbb{R}}_{+}-\left[0,1\right]\right)}^N}\left(1+j\left({\displaystyle \sum _{i=1}^N}{z}_i\right)\right)\left({\displaystyle \prod _{i=1}^N}{z}_i^{{\alpha }_i}\right)e^{-\left({\displaystyle \sum _{i=1}^N}{z}_i^2\right)}d{z}_1\dots d{z}_N\right]\le$$

$$\frac{2^N\sqrt{{\xi }_n}}{{\left(\sqrt{\pi }\right)}^N}\left[\left(1+jN\right)+\left(1+j\right){\int }_{{\left({\mathbb{R}}_{+}-\left[0,1\right]\right)}^N}\left({\displaystyle \sum _{i=1}^N}{z}_i\right)\left({\displaystyle \prod _{i=1}^N}{z}_i^{{\alpha }_i}\right)e^{-{\displaystyle \sum _{i=1}^N}{z}_i}d{z}_1\dots d{z}_N\right]\le$$

$$\frac{2^N\sqrt{{\xi }_n}}{{\left(\sqrt{\pi }\right)}^N}\left[\left(1+jN\right)+\left(1+j\right){\int }_{{\left({\mathbb{R}}_{+}-\left[0,1\right]\right)}^N}\left({\displaystyle \prod _{i=1}^N}{z}_i\right)\left({\displaystyle \prod _{i=1}^N}{z}_i^{{\alpha }_i}\right){\displaystyle \prod _{i=1}^N}{e}^{-z_i}{\displaystyle \prod _{i=1}^N}d{z}_i\right]=$$

$$\frac{2^N\sqrt{{\xi }_n}}{{\left(\sqrt{\pi }\right)}^N}\left[\left(1+jN\right)+\left(1+j\right){\displaystyle \prod _{i=1}^N}{\int }_1^{\infty }{z}_i^{\left({\alpha }_i+2\right)-1}{e}^{-z_i}d{z}_i\right]=$$

(32)

(by [19], p. 348)

$$\frac{2^N\sqrt{{\xi }_n}}{{\left(\sqrt{\pi }\right)}^N}\left[\left(1+jN\right)+\left(1+j\right){\displaystyle \prod _{i=1}^N}\mathrm{\Gamma }\left(\left({\alpha }_i+2\right),1\right)\right]=$$

(where $\mathrm{\Gamma }\left(·,·\right)$ is the upper incomplete gamma function)

$$\frac{2^N\sqrt{{\xi }_n}}{{\left(\sqrt{\pi }\right)}^N}\left[\left(1+jN\right)+\left(1+j\right){\displaystyle \prod _{i=1}^N}\frac{⌊e\left({\alpha }_i+1\right)!⌋}{e}\right]\le$$

$$\frac{2^N\sqrt{{\xi }_n}}{{\left(\sqrt{\pi }\right)}^N}\left[\left(1+jN\right)+\left(1+j\right){\left(\frac{⌊6e⌋}{e}\right)}^N\right]<+\infty .$$

That is,

$$I_{1j}^{*}\left(\alpha \right)\le \sqrt{{\xi }_n}{\left(\frac{2}{\sqrt{\pi }}\right)}^N\left[\left(1+jN\right)+\left(1+j\right){\left(\frac{⌊6e⌋}{e}\right)}^N\right]\le$$

(33)

$${\left(\frac{2}{\sqrt{\pi }}\right)}^N\left[\left(1+jN\right)+\left(1+j\right){\left(\frac{⌊6e⌋}{e}\right)}^N\right]<+\infty ,$$

are uniformly bounded; furthermore, $I_{1j}^{*}\left(\alpha \right)\to 0$, as ${\xi }_n\to 0.$ □

We make

Remark 3.

By Theorem 5, $j=0,1,\dots ,r\in \mathbb{N}$; $i,j^{*}\in \left\{1,\dots ,N\right\},$ $i\ne j^{*}$, we have that

$$\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}{\int }_{{\mathbb{R}}^N}\left|s_i\right|\left|s_{j^{*}}\right|e^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N,$$

$$\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}{\int }_{{\mathbb{R}}^N}{s}_i^2{e}^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\le I_{1j}^{*}<\infty ,$$

(34)

and

$$\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}{\int }_{{\mathbb{R}}^N}\left|s_i\right|e^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\le$$

(35)

$$\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}{\int }_{{\mathbb{R}}^N}{s}_i^2{e}^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N<\infty ,$$

and all these integrals are uniformly bounded in ${\xi }_n\in (0,1]$. And all integrals converge to zero, as ${\xi }_n\to 0.$

We continue with

Theorem 6.

Let $N\ge 1$, ${\xi }_n\in (0,1]$, $n\in \mathbb{N}$. Then

$$I_2^{*}:=\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}{\int }_{{\mathbb{R}}^N}\frac{{∥s∥}_1}{{\xi }_n}{e}^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\le$$

$$\frac{1}{\sqrt{{\xi }_n}}{\left(\frac{2}{\sqrt{\pi }}\right)}^N\left[N+{\left(\frac{⌊e⌋}{e}\right)}^N\right].$$

(36)

Proof. 

We observe that

$$I_2^{*}=\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}{\int }_{{\mathbb{R}}^N}\frac{{∥s∥}_1}{{\xi }_n}{e}^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N=$$

$$\frac{2^N}{{\left(\sqrt{\pi }\right)}^N{\left(\sqrt{{\xi }_n}\right)}^N}{\int }_{{\mathbb{R}}_{+}^N}\frac{{\displaystyle \sum _{i=1}^N}{s}_i}{{\xi }_n}{e}^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N=$$

$$\frac{2^N}{{\left(\sqrt{\pi }\right)}^N}\frac{1}{\sqrt{{\xi }_n}}{\int }_{{\mathbb{R}}_{+}^N}\left({\displaystyle \sum _{i=1}^N}\left(\frac{s_i}{\sqrt{{\xi }_n}}\right)\right)e^{-{\displaystyle \sum _{i=1}^N}{\left(\frac{s_i}{\sqrt{{\xi }_n}}\right)}^2}d\frac{s_1}{\sqrt{{\xi }_n}}\dots d\frac{s_N}{\sqrt{{\xi }_n}}=$$

(37)

$$\frac{2^N}{{\left(\sqrt{\pi }\right)}^N}\frac{1}{\sqrt{{\xi }_n}}{\int }_{{\mathbb{R}}_{+}^N}\left({\displaystyle \sum _{i=1}^N}{z}_i\right)e^{-{\displaystyle \sum _{i=1}^N}{z}_i^2}d{z}_1\dots d{z}_N=$$

$${\left(\frac{2}{\sqrt{\pi }}\right)}^N\frac{1}{\sqrt{{\xi }_n}}\left[{\int }_{{\left[0,1\right]}^N}\left({\displaystyle \sum _{i=1}^N}{z}_i\right)e^{-{\displaystyle \sum _{i=1}^N}{z}_i^2}d{z}_1\dots d{z}_N+\right.$$

$$\left.{\int }_{{\left({\mathbb{R}}_{+}-\left[0,1\right]\right)}^N}\left({\displaystyle \sum _{i=1}^N}{z}_i\right)e^{-{\displaystyle \sum _{i=1}^N}{z}_i^2}d{z}_1\dots d{z}_N\right]\le$$

$${\left(\frac{2}{\sqrt{\pi }}\right)}^N\frac{1}{\sqrt{{\xi }_n}}\left[N+{\int }_{{\left({\mathbb{R}}_{+}-\left[0,1\right]\right)}^N}\left({\displaystyle \prod _{i=1}^N}{z}_i\right)e^{-{\displaystyle \sum _{i=1}^N}{z}_i}d{z}_1\dots d{z}_N\right]=$$

$${\left(\frac{2}{\sqrt{\pi }}\right)}^N\frac{1}{\sqrt{{\xi }_n}}\left[N+{\displaystyle \prod _{i=1}^N}{\int }_1^{\infty }{z}_i{e}^{-z_i}d{z}_i\right]=$$

$${\left(\frac{2}{\sqrt{\pi }}\right)}^N\frac{1}{\sqrt{{\xi }_n}}\left[N+{\left({\int }_1^{\infty }{z}^{2-1}{e}^{-z}dz\right)}^N\right]=$$

(38)

(by [19], p. 348)

$${\left(\frac{2}{\sqrt{\pi }}\right)}^N\frac{1}{\sqrt{{\xi }_n}}\left[N+{\mathrm{\Gamma }}^N\left(2,1\right)\right]=$$

$${\left(\frac{2}{\sqrt{\pi }}\right)}^N\frac{1}{\sqrt{{\xi }_n}}\left[N+{\left(\frac{⌊e⌋}{e}\right)}^N\right],$$

proving the claim. □

We need the following.

Theorem 7

([3], p. 403). Let $r,N,m\in \mathbb{N}$, with $m>r$; ${\alpha }_i\in {\mathbb{Z}}^{+}$, $i=1,\dots ,N:\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}{\alpha }_i=m$, ${\xi }_n\in (0,1]$, $n\in \mathbb{N}$; $p>1$. Then

$${\tilde{A}}_{{\xi }_n}\left(\alpha \right):=\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}{\int }_{{\mathbb{R}}^N}{\left(\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{{\alpha }_i}\right){\left(1+\frac{{∥s∥}_2}{{\xi }_n}\right)}^r\right)}^p{e}^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\le$$

(39)

$${\sqrt{{\xi }_n}}^{p\left(m-r\right)}{\left(\frac{2}{\sqrt{\pi }}\right)}^N\left[{\left(1+N\right)}^{rp}+2^{rp}{\mathrm{\Gamma }}^N\left(\left(m+r\right)p+1,1\right)\right]\le$$

$${\left(\frac{2}{\sqrt{\pi }}\right)}^N\left[{\left(1+N\right)}^{rp}+2^{rp}{\mathrm{\Gamma }}^N\left(\left(m+r\right)p+1,1\right)\right]<+\infty ,$$

are uniformly bounded, where $m>r.$

Also, ${\tilde{A}}_{{\xi }_n}\left(\alpha \right)\to 0$, as ${\xi }_n\to 0$, $n\to +\infty .$

Above, $\mathrm{\Gamma }\left(·,·\right)$ is the upper incomplete gamma function.

Clearly, for ($m>r$),

$$B_{{\xi }_n}\left(\alpha \right):=\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}{\int }_{{\mathbb{R}}^N}\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{{\alpha }_i}\right)e^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N<\infty ,$$

(40)

uniformly bounded in ${\xi }_n\in (0,1]$. And, clearly, $B_{{\xi }_n}\left(\alpha \right)\to 0$, as ${\xi }_n\to 0$, $n\to \infty$, by (39).

We need the following

Theorem 8.

Let ${\alpha }_i\in {\mathbb{Z}}^{+}$, $i=1,\dots ,N\in \mathbb{N}:\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}{\alpha }_i=2$, ${\xi }_n\in (0,1]$, $n\in \mathbb{N}$, $p\ge 1$, and $j=0,1,\dots ,r\in \mathbb{N}.$ Then

$${\tilde{A}}_{j{\xi }_n}\left(\alpha \right):=\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}{\int }_{{\mathbb{R}}^N}\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{p{\alpha }_i}\right)\left(1+\frac{j^p{∥s∥}_2^p}{{\xi }_n^p}\right)e^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\le$$

(41)

$${\left(\sqrt{{\xi }_n}\right)}^p{\left(\frac{2}{\sqrt{\pi }}\right)}^N\left[\left(1+j^p{N}^p\right)+\left(1+j^p\right){\mathrm{\Gamma }}^N\left(3p+1,1\right)\right]\le$$

$${\left(\frac{2}{\sqrt{\pi }}\right)}^N\left[\left(1+j^p{N}^p\right)+\left(1+j^p\right){\mathrm{\Gamma }}^N\left(3p+1,1\right)\right]<+\infty ,$$

are uniformly bounded; furthermore, ${\tilde{A}}_{j{\xi }_n}\left(\alpha \right)\to 0$, as ${\xi }_n\to 0$, $n\to \infty$.

Let ${\tilde{A}}_{j{\xi }_n}\left(\alpha \right)$ be denoted as ${\tilde{\tilde{A}}}_{j{\xi }_n}\left(\alpha \right)$ when $p=1$.

Proof. 

We estimate ($p\ge 1$)

$${\tilde{A}}_{j{\xi }_n}\left(\alpha \right)=\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}{\int }_{{\mathbb{R}}^N}\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{p{\alpha }_i}\right)\left(1+\frac{j^p{∥s∥}_2^p}{{\xi }_n^p}\right)e^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N=$$

$$\frac{2^N}{{\left(\sqrt{\pi }\right)}^N{\left(\sqrt{{\xi }_n}\right)}^N}{\int }_{{\mathbb{R}}_{+}^N}\left({\displaystyle \prod _{i=1}^N}{s}_i^{p{\alpha }_i}\right)\left(1+\frac{j^p{∥s∥}_2^p}{{\xi }_n^p}\right)e^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\le$$

(42)

$$\frac{2^N{\left(\sqrt{{\xi }_n}\right)}^{2p}}{{\left(\sqrt{\pi }\right)}^N}{\int }_{{\mathbb{R}}_{+}^N}\left({\displaystyle \prod _{i=1}^N}{\left(\frac{s_i}{\sqrt{{\xi }_n}}\right)}^{p{\alpha }_i}\right)$$

$$\left(\frac{1}{{\left(\sqrt{{\xi }_n}\right)}^p}+{\left(\frac{j}{\sqrt{{\xi }_n}}\right)}^p{\left({\displaystyle \sum _{i=1}^N}\frac{s_i}{\sqrt{{\xi }_n}}\right)}^p\right)e^{-{\displaystyle \sum _{i=1}^N}{\left(\frac{s_i}{\sqrt{{\xi }_n}}\right)}^2}d\frac{s_1}{\sqrt{{\xi }_n}}\dots d\frac{s_N}{\sqrt{{\xi }_n}}=$$

$${\left(\frac{2}{\sqrt{\pi }}\right)}^N{\left(\sqrt{{\xi }_n}\right)}^p{\int }_{{\mathbb{R}}_{+}^N}\left({\displaystyle \prod _{i=1}^N}{z}_i^{p{\alpha }_i}\right)\left(1+j^p{\left({\displaystyle \sum _{i=1}^N}{z}_i\right)}^p\right)e^{-{\displaystyle \sum _{i=1}^N}{z}_i^2}d{z}_1\dots d{z}_N=$$

$${\left(\sqrt{{\xi }_n}\right)}^p{\left(\frac{2}{\sqrt{\pi }}\right)}^N\left[{\int }_{{\left[0,1\right]}^N}\left({\displaystyle \prod _{i=1}^N}{z}_i^{p{\alpha }_i}\right)\left(1+j^p{\left({\displaystyle \sum _{i=1}^N}{z}_i\right)}^p\right)e^{-{\displaystyle \sum _{i=1}^N}{z}_i^2}d{z}_1\dots d{z}_N+\right.$$

(43)

$$\left.{\int }_{{\left({\mathbb{R}}_{+}-\left[0,1\right]\right)}^N}\left({\displaystyle \prod _{i=1}^N}{z}_i^{p{\alpha }_i}\right)\left(1+j^p{\left({\displaystyle \sum _{i=1}^N}{z}_i\right)}^p\right)e^{-{\displaystyle \sum _{i=1}^N}{z}_i^2}d{z}_1\dots d{z}_N\right]\le$$

$${\left(\sqrt{{\xi }_n}\right)}^p{\left(\frac{2}{\sqrt{\pi }}\right)}^N$$

$$\left[\left(1+j^p{N}^p\right)+\left(1+j^p\right){\int }_{{\left({\mathbb{R}}_{+}-\left[0,1\right]\right)}^N}\left({\displaystyle \prod _{i=1}^N}{z}_i^{p{\alpha }_i}\right){\left({\displaystyle \sum _{i=1}^N}{z}_i\right)}^p{e}^{-{\displaystyle \sum _{i=1}^N}{z}_i}d{z}_1\dots d{z}_N\right]\le$$

$${\left(\sqrt{{\xi }_n}\right)}^p{\left(\frac{2}{\sqrt{\pi }}\right)}^N$$

$$\left[\left(1+j^p{N}^p\right)+\left(1+j^p\right){\int }_{{\left({\mathbb{R}}_{+}-\left[0,1\right]\right)}^N}\left({\displaystyle \prod _{i=1}^N}{z}_i^{p{\alpha }_i}\right)\left({\displaystyle \prod _{i=1}^N}{z}_i^p\right)\left({\displaystyle \prod _{i=1}^N}{e}^{-z_i}\right){\displaystyle \prod _{i=1}^N}d{z}_i\right]=$$

$${\left(\sqrt{{\xi }_n}\right)}^p{\left(\frac{2}{\sqrt{\pi }}\right)}^N$$

$$\left[\left(1+j^p{N}^p\right)+\left(1+j^p\right){\int }_{{\left({\mathbb{R}}_{+}-\left[0,1\right]\right)}^N}\left({\displaystyle \prod _{i=1}^N}{z}_i^{p\left({\alpha }_i+1\right)}\right)\left({\displaystyle \prod _{i=1}^N}{e}^{-z_i}\right){\displaystyle \prod _{i=1}^N}d{z}_i\right]=$$

$${\left(\sqrt{{\xi }_n}\right)}^p{\left(\frac{2}{\sqrt{\pi }}\right)}^N\left[\left(1+j^p{N}^p\right)+\left(1+j^p\right){\displaystyle \prod _{i=1}^N}\left({\int }_1^{\infty }{z}_i^{p\left({\alpha }_i+1\right)}{e}^{-z_i}d{z}_i\right)\right]=$$

(44)

(by [19], p. 348)

$${\left(\sqrt{{\xi }_n}\right)}^p{\left(\frac{2}{\sqrt{\pi }}\right)}^N\left[\left(1+j^p{N}^p\right)+\left(1+j^p\right){\displaystyle \prod _{i=1}^N}\mathrm{\Gamma }\left(\left(p\left({\alpha }_i+1\right)+1\right),1\right)\right]\le$$

(by [19], p. 909)

$${\left(\sqrt{{\xi }_n}\right)}^p{\left(\frac{2}{\sqrt{\pi }}\right)}^N\left[\left(1+j^p{N}^p\right)+\left(1+j^p\right){\mathrm{\Gamma }}^N\left(3p+1,1\right)\right]\le$$

(45)

$${\left(\frac{2}{\sqrt{\pi }}\right)}^N\left[\left(1+j^p{N}^p\right)+\left(1+j^p\right){\mathrm{\Gamma }}^N\left(3p+1,1\right)\right]<+\infty ,$$

are uniformly bounded. □

We need the following.

Theorem 9.

Let $p\ge 1,$ ${\xi }_n\in (0,1]$, $n\in \mathbb{N}$, $N\ge 1$. Then

$$K_{p{\xi }_n}:=\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}{\int }_{{\mathbb{R}}^N}{\left(\frac{{∥s∥}_2}{{\xi }_n}\right)}^p{e}^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\le$$

(46)

$$\frac{1}{{\left(\sqrt{{\xi }_n}\right)}^p}{\left(\frac{2}{\sqrt{\pi }}\right)}^N\left[N^p+{\mathrm{\Gamma }}^N\left(p+1,1\right)\right].$$

We set $K_{p{\xi }_n}$ as $K_{{\xi }_n}^{*}$ when $p=1.$

Proof. 

We have ($p\ge 1$)

$$K_{p{\xi }_n}=\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}{\int }_{{\mathbb{R}}^N}{\left(\frac{{∥s∥}_2}{{\xi }_n}\right)}^p{e}^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N=$$

$$\frac{2^N}{{\left(\sqrt{\pi }\right)}^N{\left(\sqrt{{\xi }_n}\right)}^N}{\int }_{{\mathbb{R}}_{+}^N}{\left(\frac{{∥s∥}_2}{{\xi }_n}\right)}^p{e}^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\le$$

$$\frac{2^N}{{\left(\sqrt{\pi }\right)}^N{\left(\sqrt{{\xi }_n}\right)}^N}{\int }_{{\mathbb{R}}_{+}^N}{\left(\frac{{\displaystyle \sum _{i=1}^N}{s}_i}{{\xi }_n}\right)}^p{e}^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N=$$

(47)

$$\frac{2^N}{{\left(\sqrt{\pi }\right)}^N}\frac{1}{{\left(\sqrt{{\xi }_n}\right)}^p}{\int }_{{\mathbb{R}}_{+}^N}{\left({\displaystyle \sum _{i=1}^N}\left(\frac{s_i}{\sqrt{{\xi }_n}}\right)\right)}^p{e}^{-{\displaystyle \sum _{i=1}^N}{\left(\frac{s_i}{\sqrt{{\xi }_n}}\right)}^2}d\frac{s_1}{\sqrt{{\xi }_n}}\dots d\frac{s_N}{\sqrt{{\xi }_n}}=$$

$$\frac{2^N}{{\left(\sqrt{\pi }\right)}^N}{\left(\sqrt{{\xi }_n}\right)}^{-p}{\int }_{{\mathbb{R}}_{+}^N}{\left({\displaystyle \sum _{i=1}^N}{z}_i\right)}^p{e}^{-{\displaystyle \sum _{i=1}^N}{z}_i^2}d{z}_1\dots d{z}_N=$$

$$\frac{2^N}{{\left(\sqrt{\pi }\right)}^N}{\left(\sqrt{{\xi }_n}\right)}^{-p}\left[{\int }_{{\left[0,1\right]}^N}{\left({\displaystyle \sum _{i=1}^N}{z}_i\right)}^p{e}^{-{\displaystyle \sum _{i=1}^N}{z}_i^2}d{z}_1\dots d{z}_N+\right.$$

$$\left.{\int }_{{\left({\mathbb{R}}_{+}-\left[0,1\right]\right)}^N}{\left({\displaystyle \sum _{i=1}^N}{z}_i\right)}^p{e}^{-{\displaystyle \sum _{i=1}^N}{z}_i^2}d{z}_1\dots d{z}_N\right]\le$$

$$\frac{2^N}{{\left(\sqrt{\pi }\right)}^N}{\left(\sqrt{{\xi }_n}\right)}^{-p}\left[N^p+{\int }_{{\left({\mathbb{R}}_{+}-\left[0,1\right]\right)}^N}\left({\displaystyle \prod _{i=1}^N}{z}_i^p\right)e^{-{\displaystyle \sum _{i=1}^N}{z}_i}d{z}_1\dots d{z}_N\right]=$$

$$\frac{2^N}{{\left(\sqrt{\pi }\right)}^N}{\left(\sqrt{{\xi }_n}\right)}^{-p}\left[N^p+{\displaystyle \prod _{i=1}^N}{\int }_1^{\infty }{z}_i^p{e}^{-z_i}d{z}_i\right]=$$

(48)

$$\frac{2^N}{{\left(\sqrt{\pi }\right)}^N}{\left(\sqrt{{\xi }_n}\right)}^{-p}\left[N^p+{\left({\int }_1^{\infty }{z}^p{e}^{-z}dz\right)}^N\right]=$$

(by [19], p. 348)

$$\frac{2^N}{{\left(\sqrt{\pi }\right)}^N}{\left(\sqrt{{\xi }_n}\right)}^{-p}\left[N^p+{\mathrm{\Gamma }}^N\left(p+1,1\right)\right].$$

□

4. Main Results

The general smooth multivariate Gauss–Weierstrass singular integral operators are defined as:

$$W_n\left(f;x_1,\dots ,x_N\right):=W_{r,n}^{\left[m\right]}\left(f;x_1,\dots ,x_N\right):=$$

$$\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}{\displaystyle \sum _{j=0}^r}{\alpha }_{j,r}^{\left[m\right]}{\int }_{{\mathbb{R}}^N}f\left(x_1+s_{1}j,x_2+s_{2}j,\dots ,x_N+s_{N}j\right)e^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N.$$

(49)

Observe that

$$\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}{\int }_{{\mathbb{R}}^N}{e}^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N=1,$$

(50)

see [2], p. 15.

That is, $W_{r,n}^{\left[m\right]}$ are the ${\theta }_{r,n}^{\left[m\right]}$ operators applied for the Borel probability measures on ${\mathbb{R}}^N$, $N\ge 1,$

$$d{\mu }_{{\xi }_n}\left(s\right)=\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}{e}^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N,s\in {\mathbb{R}}^N,$$

(51)

where $0<{\xi }_n\le 1$, $n\in \mathbb{N}$.

We will apply to $W_{r,n}^{\left[m\right]}$ the Theorems 1–4, with the help of all of Section 3. This section presents an approximation of properties of $W_n.$

Here, first apply Theorem 1 to the $W_n$ operators.

Theorem 10.

We consider $f\in C^2\left({\mathbb{R}}^N\right)$ and let all ${\alpha }_i\in {\mathbb{Z}}^{+}$, $i=1,\dots ,N$, $N\ge 1$, $\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}{\alpha }_i=2$; $x\in {\mathbb{R}}^N$, and all the partials $f_{\alpha }$ of order 2, along with $f\in C_B\left({\mathbb{R}}^N\right)$; or all $f_{\alpha }$ of order 2, $f\in C_U\left({\mathbb{R}}^N\right)$; $0<{\xi }_n\le 1$, $n\in \mathbb{N}$.

Denote $(n\in \mathbb{N})$

$${\Delta }_n^{*}\left(x\right):=W_n\left(f,x\right)-f\left(x\right)-\left({\displaystyle \sum _{j=0}^r}{\alpha }_{j}j\right)sin\left(1\right)$$

$$\left[{\displaystyle \sum _{i=1}^N}\frac{\partial f\left(x\right)}{\partial x_i}\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}\left({\int }_{{\mathbb{R}}^N}{s}_i{e}^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\right)\right]$$

$$-2\left({\displaystyle \sum _{j=0}^r}{\alpha }_j{j}^2\right){sin}^2\left(\frac{1}{2}\right)\left\{{\displaystyle \sum _{i=1}^N}\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}\left({\int }_{{\mathbb{R}}^N}{s}_i^2{e}^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\right)\frac{{\partial }^{2}f\left(x\right)}{\partial x_i^2}\right.$$

$$\left.+\sum _{\begin{array}{c}i\ne j^{*},\\ i,j^{*}\in \left\{1,\dots ,N\right\}\end{array}}\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}\left({\int }_{{\mathbb{R}}^N}{s}_i{s}_{j^{*}}{e}^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\right)\frac{{\partial }^{2}f\left(x\right)}{\partial x_i\partial x_{j^{*}}}\right\}.$$

(52)

Then

(i)

$$\left|{\Delta }_n^{*}\left(x\right)\right|\le {∥{\Delta }_n^{*}∥}_{\infty }\le {\displaystyle \sum _{j=0}^r}\left|{\alpha }_j\right|$$

$$\left[\left[j^2\sum _{\begin{array}{c}{\alpha }_i\in {\mathbb{Z}}^{+},\\ \alpha :\left|\alpha \right|=2\end{array}}\left(\frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}\right){\omega }_1\left(f_{\alpha },{\xi }_n\right)I_{1j}^{*}\left(\alpha \right)\right]+\right.$$

$$\left.\frac{1}{2}{\omega }_1\left(f,{\xi }_n\right)\left[1+\frac{j}{3}{I}_2^{*}\right]\right]=:{\phi }_{{\xi }_n}^{*},$$

(53)

where $I_{1j}^{*}\left(\alpha \right)$ is as in (29), and $I_2^{*}$ is as in (36).

In the case of all $f_{\alpha }$ of order 2 and $f\in C_U\left({\mathbb{R}}^N\right)$ and ${\xi }_n\to 0$, as $n\to \infty$, then ${\Delta }_n^{*}\left(x\right)$, ${∥{\Delta }_n^{*}∥}_{\infty }\to 0$ with rates.

(ii) If $\frac{\partial f\left(x\right)}{\partial x_i}=0$, $i=1,\dots ,N$, and $f_{\alpha }\left(x\right)=0$, ${\alpha }_i\in {\mathbb{Z}}^{+}$, $i=1,\dots ,N$, with $\left|\alpha \right|=2$, then

$$\left|W_n\left(f,x\right)-f\left(x\right)\right|\le {\phi }_{{\xi }_n}^{*}.$$

(54)

And $W_n\left(f,x\right)\to f\left(x\right)$ in the uniformly continuous case.

(iii) Additionally, assume all partials of order $\le 2$ are bounded. Hence,

$${∥W_n\left(f\right)-f∥}_{\infty }\le \left({\displaystyle \sum _{j=0}^r}\left|{\alpha }_j\right|j\right)\left(0.8414\right)$$

$$\left[{\displaystyle \sum _{i=1}^N}{∥\frac{\partial f}{\partial x_i}∥}_{\infty }\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}\left({\int }_{{\mathbb{R}}^N}\left|s_i\right|e^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\right)\right]+$$

$$\left({\displaystyle \sum _{j=0}^r}\left|{\alpha }_j\right|j^2\right)\left(0.4596\right)\left\{{\displaystyle \sum _{i=1}^N}\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}\left({\int }_{{\mathbb{R}}^N}{s}_i^2{e}^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\right){∥\frac{{\partial }^{2}f}{\partial x_i^2}∥}_{\infty }+\right.$$

$$\left.\sum _{\begin{array}{c}i\ne j^{*},\\ i,j^{*}\in \left\{1,\dots ,N\right\}\end{array}}\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}\left({\int }_{{\mathbb{R}}^N}\left|s_i\right|\left|s_{j^{*}}\right|e^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\right){∥\frac{{\partial }^{2}f}{\partial x_i\partial x_{j^{*}}}∥}_{\infty }\right\}+{\phi }_{{\xi }_n}^{*}.$$

(55)

If all $f_{\alpha }$ of order 2, $f\in C_U\left({\mathbb{R}}^N\right)$, then

$${∥W_n\left(f\right)-f∥}_{\infty }\to 0withrates,\mathrm{as}{\xi }_n\to 0,n\to +\infty .$$

Proof. 

By Theorems 1, 5, 6 and Remark 3. □

Next, we apply Theorem 2 to $W_n$ operators.

Theorem 11.

Let $f\in C^m\left({\mathbb{R}}^N\right)$, $m\in \mathbb{N}$, $m>r$, $N\ge 1$, with $f_{\alpha }\in L_p\left({\mathbb{R}}^N\right),$ $\left|\alpha \right|=m$, $x\in {\mathbb{R}}^N.$ Let $p,q>1:\frac{1}{p}+\frac{1}{q}=1;$ $0<{\xi }_n\le 1$, $n\in \mathbb{N}.$ For $\tilde{j}=1,\dots ,m,$ and $\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right),$ ${\alpha }_i\in {\mathbb{Z}}^{+}$, $i=1,\dots ,N,$ $\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}{\alpha }_i=\tilde{j}$, call

$$c_{\alpha ,n,\tilde{j}}^{*}:=\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}{\int }_{{\mathbb{R}}^N}\left({\displaystyle \prod _{i=1}^N}{s}_i^{{\alpha }_i}\right)e^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N.$$

(56)

Then

$${∥E_{r,n}^{*\left[m\right]}∥}_p:={∥W_{r,n}\left(f;x\right)-f\left(x\right)-{\displaystyle \sum _{\tilde{j}=1}^m}{\delta }_{\tilde{j},r}^{\left[m\right]}\left(\sum _{\left|\alpha \right|=\tilde{j}}\frac{c_{\alpha ,n,\tilde{j}}{f}_{\alpha }\left(x\right)}{\left({\displaystyle \prod _{i=1}^N}{\alpha }_i!\right)}\right)∥}_{p,x}\le$$

$${\left[{\left(\begin{array}{c}\frac{N^m}{(m-1)!}\end{array}\right)}^{\frac{p}{q}}\left(\frac{m}{{\left(q\left(m-1\right)+1\right)}^{\frac{p}{q}}}\right)\left(\sum _{\left|\alpha \right|=m}\frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}{\tilde{A}}_{{\xi }_n}\left(\alpha \right){\omega }_r{\left(f_{\alpha },{\xi }_n\right)}_p^p\right)\right]}^{\frac{1}{p}},$$

(57)

where ${\tilde{A}}_{{\xi }_n}\left(\alpha \right)$ as in (39).

As $n\to \infty$ and ${\xi }_n\to 0$, by (57), we obtain that ${∥E_{r,n}^{*\left[m\right]}∥}_p\to 0$ with rates.

One also obtains by (57) that

$${∥W_n\left(f;x\right)-f\left(x\right)∥}_{p,x}\le$$

$${\displaystyle \sum _{\tilde{j}=1}^m}\left|{\delta }_{\tilde{j},r}^{\left[m\right]}\right|\left(\sum _{\left|\alpha \right|=\tilde{j}}\frac{\left|c_{\alpha ,n,\tilde{j}}^{*}\right|}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}{∥f_{\alpha }∥}_p\right)+R.H.S.(57),$$

(58)

given that ${∥f_{\alpha }∥}_p<\infty ,$$\left|\alpha \right|=\tilde{j}$, $\tilde{j}=1,\dots ,m.$

Finally, we get ${∥W_n\left(f\right)-f∥}_p\to 0$ as ${\xi }_n\to 0,$ and $n\to \infty$. That is, $W_n\to I$ the unit operator, in $L_p$ norm, with rates.

Proof. 

By Theorems 2, 7 and (40). □

Next, we apply Theorem 3 to $W_n$ operators.

Theorem 12.

Let $p,q>1:\frac{1}{p}+\frac{1}{q}=1$, $0<{\xi }_n\le 1$, $n\in \mathbb{N}$. Here, we deal with $f\in C^2\left({\mathbb{R}}^N\right)$, $N\ge 1$, with $f,$ $f_{\alpha }\in L_p\left({\mathbb{R}}^N\right)$, $\left|\alpha \right|=2$, where ${\alpha }_i\in {\mathbb{Z}}^{+}$, $i=1,\dots ,N$, and $\left|\alpha \right|={\displaystyle \sum _{i=1}^N}{\alpha }_i$; $x\in {\mathbb{R}}^N$; $j=0,1,\dots ,r.$

Denote ($n\in \mathbb{N}$)

$${\overline{\Delta }}_n\left(x\right):=W_n\left(f,x\right)-f\left(x\right)-\left({\displaystyle \sum _{j=0}^r}{\alpha }_{j}j\right)sin\left(1\right)$$

$$\left[{\displaystyle \sum _{i=1}^N}\frac{\partial f\left(x\right)}{\partial x_i}\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}\left({\int }_{{\mathbb{R}}^N}{s}_i{e}^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\right)\right]$$

$$-2\left({\displaystyle \sum _{j=0}^r}{\alpha }_j{j}^2\right){sin}^2\left(\frac{1}{2}\right)\left\{{\displaystyle \sum _{i=1}^N}\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}\left({\int }_{{\mathbb{R}}^N}{s}_i^2{e}^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\right)\frac{{\partial }^{2}f\left(x\right)}{\partial x_i^2}+\right.$$

(59)

$$\left.\sum _{\begin{array}{c}i\ne j^{*},\\ i,j^{*}\in \left\{1,\dots ,N\right\}\end{array}}\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}\left({\int }_{{\mathbb{R}}^N}{s}_i{s}_{j^{*}}{e}^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\right)\frac{{\partial }^{2}f\left(x\right)}{\partial x_i\partial x_{j^{*}}}\right\}.$$

Then

$${∥{\overline{\Delta }}_n∥}_p\le {\left({\displaystyle \sum _{j=0}^r}{\left|{\alpha }_j\right|}^q\right)}^{\frac{1}{q}}\left\{\frac{2}{{\left(q+1\right)}^{\frac{1}{q}}}{\left(\frac{N\left(N+1\right)+2}{2}\right)}^{\frac{1}{q}}\right\}$$

$${\left\{{\displaystyle \sum _{j=0}^r}\left[j^{2p}\sum _{\left|\alpha \right|=2}{\left(\frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}\right)}^p{\omega }_1{\left(f_{\alpha },{\xi }_n\right)}_p^p{\tilde{A}}_{j{\xi }_n}\left(\alpha \right)+{\omega }_1{\left(f,{\xi }_n\right)}_p^p\left[1+j^p{K}_{p{\xi }_n}\right]\right]\right\}}^{\frac{1}{p}},$$

(60)

where ${\tilde{A}}_{j{\xi }_n}\left(\alpha \right)$ as in (41), and $K_{p{\xi }_n}$ as in (46).

One also obtains from (60) that

$${∥W_n\left(f\right)-f∥}_p\le$$

$$\left({\displaystyle \sum _{j=0}^r}\left|{\alpha }_j\right|j\right)sin\left(1\right)\left[{\displaystyle \sum _{i=1}^N}{∥\frac{\partial f}{\partial x_i}∥}_p\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}\left({\int }_{{\mathbb{R}}^N}\left|s_i\right|e^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\right)\right]$$

$$+2\left({\displaystyle \sum _{j=0}^r}\left|{\alpha }_j\right|j^2\right){sin}^2\left(\frac{1}{2}\right)\left\{{\displaystyle \sum _{i=1}^N}\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}\left({\int }_{{\mathbb{R}}^N}{s}_i^2{e}^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\right){∥\frac{{\partial }^{2}f}{\partial x_i^2}∥}_p+\right.$$

$$\left.\sum _{\begin{array}{c}i\ne j^{*},\\ i,j^{*}\in \left\{1,\dots ,N\right\}\end{array}}\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}\left({\int }_{{\mathbb{R}}^N}\left|s_i{s}_{j^{*}}\right|e^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\right){∥\frac{{\partial }^{2}f}{\partial x_i\partial x_{j^{*}}}∥}_p\right\}+R.H.S.(60),$$

(61)

given that ${∥f_{\alpha }∥}_p<\infty$, $\left|\alpha \right|=\tilde{j}$, $\tilde{j}=1,2.$

Additionally, assuming that ${\omega }_1{\left(f,{\xi }_n\right)}_p\le \lambda {\xi }_n$, $\lambda >0,$ then as $n\to \infty$ and ${\xi }_n\to 0$, by (60), we obtain that ${∥{\overline{\Delta }}_n∥}_p\to 0$ with rates, and furthermore by (61), we derive that ${∥W_n\left(f\right)-f∥}_p\to 0$; that is, $W_n\to I$ the unit operator, in $L_p$ norm, with rates.

Proof. 

By Theorems 3, 8, 9 and Remark 3. □

We finish with an application of Theorem 4 to $W_n$ operators.

Theorem 13.

Let $0<{\xi }_n\le 1$, $n\in \mathbb{N}$, $x\in {\mathbb{R}}^N$. Here, we deal with $f\in C^2\left({\mathbb{R}}^N\right)$, $N\ge 1$, with $f,$ $f_{\alpha }\in L_1\left({\mathbb{R}}^N\right)$, $\left|\alpha \right|=2$, where ${\alpha }_i\in {\mathbb{Z}}^{+}$, $i=1,\dots ,N$, and $\left|\alpha \right|={\displaystyle \sum _{i=1}^N}{\alpha }_i$; $j=0,1,\dots ,r.$ Here ${\overline{\Delta }}_n$ as in (59). Then

$${∥{\overline{\Delta }}_n∥}_1\le \left\{{\displaystyle \sum _{j=0}^r}\left|{\alpha }_j\right|\left[j^2\sum _{\left|\alpha \right|=2}\left(\frac{2}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}\right){\omega }_1{\left(f_{\alpha },{\xi }_n\right)}_1{\tilde{\tilde{A}}}_{j{\xi }_n}\left(\alpha \right)\right.\right.$$

$$\left.\left.+{\omega }_1{\left(f,{\xi }_n\right)}_1\left[1+j{K}_{{\xi }_n}^{*}\right]\right]\right\},$$

(62)

where ${\tilde{\tilde{A}}}_{j{\xi }_n}\left(\alpha \right)$ as in Theorem 8, and $K_{{\xi }_n}^{*}$ as in Theorem 9.

We also derive that

$${∥W_n\left(f\right)-f∥}_1\le$$

$$\left({\displaystyle \sum _{j=0}^r}\left|{\alpha }_j\right|j\right)sin\left(1\right)\left[{\displaystyle \sum _{i=1}^N}{∥\frac{\partial f}{\partial x_i}∥}_1\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}\left({\int }_{{\mathbb{R}}^N}\left|s_i\right|e^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\right)\right]$$

$$+2\left({\displaystyle \sum _{j=0}^r}\left|{\alpha }_j\right|j^2\right){sin}^2\left(\frac{1}{2}\right)\left\{{\displaystyle \sum _{i=1}^N}\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}\left({\int }_{{\mathbb{R}}^N}{s}_i^2{e}^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\right){∥\frac{{\partial }^{2}f}{\partial x_i^2}∥}_1+\right.$$

(63)

$$\left.\sum _{\begin{array}{c}i\ne j^{*},\\ i,j^{*}\in \left\{1,\dots ,N\right\}\end{array}}\frac{1}{{\left(\sqrt{\pi {\xi }_n}\right)}^N}\left({\int }_{{\mathbb{R}}^N}\left|s_i{s}_{j^{*}}\right|e^{-\frac{{\displaystyle \sum _{i=1}^N}{s}_i^2}{{\xi }_n}}d{s}_1\dots d{s}_N\right){∥\frac{{\partial }^{2}f}{\partial x_i\partial x_{j^{*}}}∥}_1\right\}+R.H.S.(62),$$

given that ${∥f_{\alpha }∥}_1<\infty$, $\left|\alpha \right|=\tilde{j}$, $\tilde{j}=1,2.$ Additionally, assuming that ${\omega }_1{\left(f,{\xi }_n\right)}_1\le {\lambda }^{*}{\xi }_n$, ${\lambda }^{*}>0,$ then as $n\to \infty$ and ${\xi }_n\to 0$, by (62), we obtain that ${∥{\overline{\Delta }}_n∥}_1\to 0$ with rates. Furthermore, by (63), we derive that ${∥W_n\left(f\right)-f∥}_1\to 0$; that is, $W_n\to I$ the unit operator, in $L_1$ norm, with rates.

Proof. 

By Theorems 4, 8, 9 and Remark 3. □

5. Conclusions

A new type of approximation was introduced for singular integrals. This approximation is an interesting trigonometric-based one.