Multivariate Perturbed Hyperbolic Tangent-Activated Singular Integral Approximation

Abstract

Here we study the quantitative multivariate approximation of perturbed hyperbolic tangent-activated singular integral operators to the unit operator. The engaged neural network activation function is both parametrized and deformed, and the related kernel is a density function on ${\mathbb{R}}^N$. We exhibit uniform and $L_p$, $p\ge 1$ approximations via Jackson-type inequalities involving the first $L_p$ modulus of smoothness, $1\le p\le \infty$. The differentiability of our multivariate functions is covered extensively in our approximations. We continue by detailing the global smoothness preservation results of our operators. We conclude the paper with the simultaneous approximation and the simultaneous global smoothness preservation by our multivariate perturbed activated singular integrals.

1. Introduction

The classic theory of univariate approximation by singular integral operators is well-documented in the monograph [1]. The corresponding multivariate theory is also extensively presented in the monograph [2]. Earlier important works that motivate us are [3,4,5]. So, inspired by all of the above we attempt to expose a new theory: the multivariate approximation by activated singular integral operators. In this case, the kernel comes from a multivariate neural network activation function, and the parametrized and deformed hyperbolic tangent function. In recent intense mathematical activity, neural networks play a leading role in solving numerically univariate and multivariate differential equations, so multivariate activated singular integrals will play a central role.

What is surprising here is the use of the reverse process from applied mathematics to theoretical ones, which is very rare.

So, here we present multivariate pointwise and uniform and $L_p$, $p\ge 1$ approximations. We continue by detailing the global smoothness preservation property by our operators. We conclude with the simultaneous treatment of all of these. This is a seminal article, as it is the first of its kind. Other inspiring sources have been the articles [6,7,8,9,10,11,12,13,14,15].

2. Background

(I) Upon uniform approximation, see [2], Ch. 1.

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

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

(4)

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

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

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

(5)

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.

The operators ${\theta }_{r,n}^{\left[m\right]}$ are not in general positive, and they preserve the constant functions in N variables.

We need

Definition 1. 

Let $f\in C_B\left({\mathbb{R}}^N\right)$, the space of all bounded and continuous functions on ${\mathbb{R}}^N$. Then, the rth multivariate modulus of smoothness of f is given by

$${\omega }_r\left(f;h\right):=\underset{\sqrt{u_1^2+\dots +u_N^2}\le h}{sup}{∥{\mathsf{\Delta }}_{u_1,u_2,\dots ,u_N}^r\left(f\right)∥}_{\infty }<\infty ,h>0,$$

(6)

where ${∥·∥}_{\infty }$ is the sup-norm and

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

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

(7)

Let $m\in \mathbb{N}$ and let $f\in C^m\left({\mathbb{R}}^N\right)$.

Assume that all partial derivatives of f of order m are bounded, i.e.,

$${∥{\displaystyle \frac{{\partial }^{m}f\left(·,·,\dots ,·\right)}{\partial x_1^{{\alpha }_1}\dots \partial x_N^{{\alpha }_N}}}∥}_{\infty }<\infty ,$$

(8)

for all ${\alpha }_i\in {\mathbb{Z}}^{+},$ i $=1,\dots ,N;$ ${\displaystyle \sum _{i=1}^N}{\alpha }_i=m.$

We need

Theorem 1. 

Let $m\in C^m\left({\mathbb{R}}^N\right)$, $N\ge 1$, $x\in {\mathbb{R}}^N$. Assume ${∥{\displaystyle \frac{{\partial }^{m}f\left(·,·,\dots ,·\right)}{\partial {\displaystyle \underset{1}{\overset{{\alpha }_1}{x}}}\dots \partial x_N^{{\alpha }_N}}}∥}_{\infty }<\infty ,$ for all ${\alpha }_i\in {\mathbb{Z}}^{+},$ $i=1,\dots ,N;$ $\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}{\alpha }_i=m,$ $r\in \mathbb{N}.$

Let ${\mu }_{{\xi }_n}$ be a Borel probability measure on ${\mathbb{R}}^N$, for ${\xi }_n>0$, ${\left({\xi }_n\right)}_{n\in \mathbb{N}}$ bounded sequence.

Assume that for all $\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right)$, ${\alpha }_i\in {\mathbb{Z}}^{+},$ $i=1,\dots ,N;$ $\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}{\alpha }_i=m$ we have

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

(9)

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

$$c_{\alpha ,n}:=c_{\alpha ,n,\tilde{j}}:={\int }_{{\mathbb{R}}^N}{\displaystyle \prod _{i=1}^N}{s}_i^{{\alpha }_i}d{\mu }_{{\xi }_n}\left(s_1,\dots ,s_N\right).$$

(10)

Then

(i) 

$$E_{r,n}^{\left[m\right]}\left(x\right):=\left|{\theta }_{r,n}^{\left[m\right]}\left(f;x\right)-f\left(x\right)-{\displaystyle \sum _{\tilde{j}=1}^m}{\delta }_{\tilde{j},r}^{\left[m\right]}\left(\sum _{\begin{array}{c}{\alpha }_1,\dots ,{\alpha }_N\ge 0\\ \left|\alpha \right|=\tilde{j}\end{array}}{\displaystyle \frac{c_{\alpha ,n,\tilde{j}}{f}_{\alpha }\left(x\right)}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\right|\le$$

$$\sum _{\left({\hspace{0.17em}}_{\begin{array}{c}{\alpha }_1,\dots ,{\alpha }_N\ge 0\\ \left|\alpha \right|=m\end{array}}\right)}{\displaystyle \frac{\left({\omega }_r\left(f_{\alpha },{\xi }_n\right)\right)}{\left({\displaystyle \prod _{i=1}^N}{\alpha }_i!\right)}}\left({\int }_{{\mathbb{R}}^N}\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{{\alpha }_i}\right){\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^{r}d{\mu }_{{\xi }_n}\left(s\right)\right),$$

(11)

∀ $x\in {\mathbb{R}}^N$,

(ii) 

$${∥E_{r,n}^{\left[m\right]}∥}_{\infty }\le R.H.S.\left(11\right).$$

(12)

Given that ${\xi }_n\to 0$, as $n\to \infty$ and $u_{{\xi }_n}$ is uniformly bounded, then we obtain ${∥E_{r,n}^{\left[m\right]}∥}_{\infty }\to 0$ with rates.

(iii) 

It holds also that

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

$${\displaystyle \sum _{\tilde{j}=1}^m}\left|{\delta }_{\tilde{j},r}^{\left[m\right]}\right|\left(\sum _{\begin{array}{c}{\alpha }_1,\dots ,{\alpha }_N\ge 0\\ \left|\alpha \right|=\tilde{j}\end{array}}{{\displaystyle \frac{\left|c_{\alpha ,n,\tilde{j}}\right|∥f_{\alpha }∥}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}}_{\infty }\right)+R.H.S.\left(11\right),$$

(13)

given that ${∥f_{\alpha }∥}_{\infty }<\infty ,$ for all $\alpha :\left|\alpha \right|=\tilde{j}$, $\tilde{j}=1,\dots ,m.$ Furthermore, as ${\xi }_n\to 0,$ when $n\to \infty$, assuming that $c_{\alpha ,n,\tilde{j}}\to 0$, while $u_{{\xi }_n}$ is uniformly bounded, we conclude

$${∥{\theta }_{r,n}^{\left[m\right]}\left(f\right)-f∥}_{\infty }\to 0,$$

(14)

with rates.

Theorem 2. 

Let $f\in C_B\left({\mathbb{R}}^N\right)$, $N\ge 1$. Then

$${∥{\theta }_{r,n}^{\left[0\right]}f-f∥}_{\infty }\le \left({\int }_{{\mathbb{R}}^N}{\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^{r}d{\mu }_{{\xi }_n}\left(s\right)\right){\omega }_r\left(f,{\xi }_n\right).$$

(15)

under the assumption

$${\mathrm{\Phi }}_{{\xi }_n}:={\int }_{{\mathbb{R}}^N}{\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^{r}d{\mu }_{{\xi }_n}\left(s\right)<\infty .$$

(16)

As $n\to \infty$ and ${\xi }_n\to 0$, given that ${\mathrm{\Phi }}_{{\xi }_n}$ are uniformly bounded and f is uniformly continuous, we obtain

$${∥{\theta }_{r,n}^{\left[0\right]}f-f∥}_{\infty }\to 0,$$

(17)

with rates.

(II) On $L_p$, $p\ge 1$, approximation, see [2], Ch. 2.

Here, 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 ${\displaystyle \frac{{\partial }^{\tilde{j}}f\left(·,\dots ,·\right)}{\partial x_1^{{\alpha }_1}\dots \partial x_N^{{\alpha }_N}}}$, ${\alpha }_i\in {\mathbb{Z}}^{+}$, i $=1,\dots ,N:\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}{\alpha }_i=\tilde{j}$, $\tilde{j}=1,\dots ,m.$

We need

Definition 2. 

We call

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

(18)

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

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

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

(19)

$h>0$.

The following comes from ([16], Ch. 25).

Theorem 3. 

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 $p,q>1:{\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$. 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

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

(20)

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

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

(21)

Then

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

(22)

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

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

As $n\to \infty$ and ${\xi }_n\to 0$, by (22), we obtain ${∥E_{r,n}^{\left[m\right]}∥}_p\to 0$ with rates.

We also obtain, by (22),

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

$${\displaystyle \sum _{\tilde{j}=1}^m}\left|{\delta }_{\tilde{j},r}^{\left[m\right]}\right|\left(\sum _{\left|\alpha \right|=\tilde{j}}{\displaystyle \frac{\left|c_{\alpha ,n,\tilde{j}}\right|}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}{∥f_{\alpha }∥}_p\right)+R.H.S.\left(22\right),$$

(23)

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 obtain ${∥{\theta }_{r,n}^{\left[m\right]}\left(f\right)-f∥}_p\to 0$, that is ${\theta }_{r,n}^{\left[m\right]}\to I$ the unit operator, in $L_p$ norm, with rates.

Inequality (22) provides a correction in the constants of the inequality in (Theorem 4 in [2], p. 25).

In particular, we have

Theorem 4. 

Let $f\in \left(C\left({\mathbb{R}}^N\right)\cap L_p\left({\mathbb{R}}^N\right)\right)$; $N\ge 1$; $p,q>1:{\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$. Assume ${\mu }_{{\xi }_n}$ probability Borel measures on ${\mathbb{R}}^N$, ${\left({\xi }_n\right)}_{n\in \mathbb{N}}>0$ and bounded. Also suppose

$${\int }_{{\mathbb{R}}^N}{\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^{rp}d{\mu }_{{\xi }_n}\left(s\right)<\infty .$$

(24)

Then

$${∥{\theta }_{r,n}^{\left[0\right]}\left(f\right)-f∥}_p\le$$

(25)

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

As ${\xi }_n\to 0$, when $n\to \infty$, we obtain ${∥{\theta }_{r,n}^{\left[0\right]}\left(f\right)-f∥}_p\to 0$, i.e., ${\theta }_{r,n}^{\left[0\right]}\to I$, the unit operator, in $L_p$ norm.

Next, we mention

Theorem 5. 

Let $f\in C^m\left({\mathbb{R}}^N\right)$, $m,N\in \mathbb{N}$, with $f_{\alpha }\in L_1\left({\mathbb{R}}^N\right),$ $\left|\alpha \right|=m$, $x\in {\mathbb{R}}^N.$ Here, ${\mu }_{{\xi }_n}$ is a Borel probability measure on ${\mathbb{R}}^N$ for ${\xi }_n>0$, ${\left({\xi }_n\right)}_{n\in \mathbb{N}}$ is a 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

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

(26)

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

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

(27)

Then

$${∥E_{r,n}^{\left[m\right]}∥}_1={∥{\theta }_{r,n}^{\left[m\right]}\left(f;x\right)-f\left(x\right)-{\displaystyle \sum _{\tilde{j}=1}^m}{\delta }_{\tilde{j},r}^{\left[m\right]}\left(\sum _{\left|\alpha \right|=\tilde{j}}{\displaystyle \frac{c_{\alpha ,n,\tilde{j}}{f}_{\alpha }\left(x\right)}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)∥}_{1,x}\le$$

(28)

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

As ${\xi }_n\to 0$, we obtain ${∥E_{r,n}^{\left[m\right]}∥}_1\to 0$ with rates.

From (28), we obtain

$${∥{\theta }_{r,n}^{\left[m\right]}f-f∥}_1\le$$

$${\displaystyle \sum _{\tilde{j}=1}^m}\left|{\delta }_{\tilde{j},r}^{\left[m\right]}\right|\left(\sum _{\left|\alpha \right|=\tilde{j}}{\displaystyle \frac{\left|c_{\alpha ,n,\tilde{j}}\right|}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}{∥f_{\alpha }∥}_1\right)+R.H.S.\left(28\right),$$

given that ${∥f_{\alpha }∥}_1<\infty ,$ $\left|\alpha \right|=\tilde{j}$, $\tilde{j}=1,\dots ,m.$

As $n\to \infty$, assuming ${\xi }_n\to 0$ and $c_{\alpha ,n,\tilde{j}}\to 0$, we obtain ${∥{\theta }_{r,n}^{\left[m\right]}f-f∥}_1\to 0$, that is ${\theta }_{r,n}^{\left[m\right]}\to I$ in $L_1$ norm, with rates.

Theorem 6. 

Let $f\in \left(C\left({\mathbb{R}}^N\right)\cap L_1\left({\mathbb{R}}^N\right)\right)$, $N\ge 1.$ Assume ${\mu }_{{\xi }_n}$ probability Borel measures on ${\mathbb{R}}^N$, ${\left({\xi }_n\right)}_{n\in \mathbb{N}}>0$ and bounded. Also suppose

$${\int }_{{\mathbb{R}}^N}{\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^{r}d{\mu }_{{\xi }_n}\left(s\right)<\infty .$$

(29)

Then

$${∥{\theta }_{r,n}^{\left[0\right]}\left(f\right)-f∥}_1\le$$

(30)

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

As ${\xi }_n\to 0$, we obtain ${\theta }_{r,n}^{\left[0\right]}\to I$ in $L_1$ norm.

(III) Global smoothness and simultaneous approximation ([2], Ch. 3).

Denoted by

$${\omega }_m{\left(f;h\right)}_{\infty }={\omega }_m\left(f,h\right).$$

We mention the following general global smoothness preservation result

Theorem 7. 

We suppose ${\theta }_{r,n}^{\left[\tilde{m}\right]}\left(f;x\right)\in \mathbb{R}$, ∀$x\in \mathbb{R}$. Let $h>0$, $f\in C\left({\mathbb{R}}^N\right)$, $N\ge 1$.

(i) 

Assume ${\omega }_m\left(f,h\right)<\infty$. Then

$${\omega }_m\left({\theta }_{r,n}^{\left[\tilde{m}\right]}f,h\right)\le \left({\displaystyle \sum _{\tilde{j}=0}^r}\left|{\alpha }_{\tilde{j},r}^{\left[\tilde{m}\right]}\right|\right){\omega }_m\left(f,h\right).$$

(31)

(ii) 

Assume $f\in \left(C\left({\mathbb{R}}^N\right)\cap L_1\left({\mathbb{R}}^N\right)\right).$ Then

$${\omega }_m{\left({\theta }_{r,n}^{\left[\tilde{m}\right]}f,h\right)}_1\le \left({\displaystyle \sum _{\tilde{j}=0}^r}\left|{\alpha }_{\tilde{j},r}^{\left[\tilde{m}\right]}\right|\right){\omega }_m{\left(f,h\right)}_1.$$

(32)

(iii) 

Assume $f\in \left(C\left({\mathbb{R}}^N\right)\cap L_p\left({\mathbb{R}}^N\right)\right)$, $p>1.$ Then

$${\omega }_m{\left({\theta }_{r,n}^{\left[\tilde{m}\right]}f,h\right)}_p\le \left({\displaystyle \sum _{\tilde{j}=0}^r}\left|{\alpha }_{\tilde{j},r}^{\left[\tilde{m}\right]}\right|\right){\omega }_m{\left(f,h\right)}_p.$$

(33)

We need

Theorem 8. 

Let $f\in C^l\left({\mathbb{R}}^N\right)$, $l,N\in \mathbb{N}$. Here ${\mu }_{{\xi }_n}$ is a Borel probability measure on ${\mathbb{R}}^N,$ ${\xi }_n>0$, ${\left({\xi }_n\right)}_{n\in \mathbb{N}}$ a bounded sequence. Let $\beta :=\left({\beta }_1,\dots ,{\beta }_N\right)$, ${\beta }_i\in {\mathbb{Z}}^{+}$, $i=1,\dots ,N;$ $\left|\beta \right|:={\displaystyle \sum _{i=1}^N}{\beta }_i=l.$ Here, $f\left(x+sj\right)$, $x,s\in {\mathbb{R}}^N$, is ${\mu }_{{\xi }_n}$-integrable with reference to s, for $j=1,\dots ,r.$ There exist ${\mu }_{{\xi }_n}$-integrable functions $h_{i_1,j},$ $h_{{\beta }_1,i_2,j},$ $h_{{\beta }_1,{\beta }_2,i_3,j},\dots ,h_{{\beta }_1,{\beta }_2,\dots ,{\beta }_{N-1},i_N,j}\ge 0$ ($j=1,\dots ,r$) on ${\mathbb{R}}^N$, such that

$$\left|{\displaystyle \frac{{\partial }^{i_1}f\left(x+sj\right)}{\partial x_1^{i_1}}}\right|\le h_{i_1,j}\left(s\right),i_1=1,\dots ,{\beta }_1,$$

(34)

$$\begin{array}{ccc}\hfill \left|{\displaystyle \frac{{\partial }^{{\beta }_1+i_2}f\left(x+sj\right)}{\partial x_2^{i_2}\partial x_1^{{\beta }_1}}}\right|& \le & h_{{\beta }_1,i_2,j}\left(s\right),i_2=1,\dots ,{\beta }_2,\hfill \\ & & ⋮\hfill \end{array}$$

$$\left|{\displaystyle \frac{{\partial }^{{\beta }_1+{\beta }_2+\dots +{\beta }_{N-1}+i_N}f\left(x+sj\right)}{\partial x_N^{i_N}\partial x_{N-1}^{{\beta }_{N-1}}\dots \partial x_2^{{\beta }_2}\partial x_1^{{\beta }_1}}}\right|\le h_{{\beta }_1,{\beta }_2,\dots ,{\beta }_{N-1},i_N,j}\left(s\right),i_N=1,\dots ,{\beta }_N,$$

∀$x,s\in {\mathbb{R}}^N$.

Then, both of the following exist, and

$${\left({\theta }_{r,n}^{\left[\tilde{m}\right]}\left(f;x\right)\right)}_{\beta }={\theta }_{r,n}^{\left[\tilde{m}\right]}\left(f_{\beta };x\right).$$

(35)

We detail the simultaneous global smoothness results.

Theorem 9. 

Let $h>0$ and assumptions of Theorem 8 are valid. Here, $\gamma =0,\beta$, ($0=\left(0,\dots ,0\right)$).

(i) 

Assume ${\omega }_m\left(f_{\gamma },h\right)<\infty$. Then

$${\omega }_m\left({\left({\theta }_{r,n}^{\left[\tilde{m}\right]}\left(f\right)\right)}_{\gamma },h\right)\le \left({\displaystyle \sum _{\tilde{j}=0}^r}\left|{\alpha }_{\tilde{j},r}^{\left[\tilde{m}\right]}\right|\right){\omega }_m\left(f_{\gamma },h\right).$$

(36)

(ii) 

Additionally, assume $f_{\gamma }\in L_1\left({\mathbb{R}}^N\right).$ Then

$${\omega }_m{\left({\left({\theta }_{r,n}^{\left[\tilde{m}\right]}\left(f\right)\right)}_{\gamma },h\right)}_1\le \left({\displaystyle \sum _{\tilde{j}=0}^r}\left|{\alpha }_{\tilde{j},r}^{\left[\tilde{m}\right]}\right|\right){\omega }_m{\left(f_{\gamma },h\right)}_1.$$

(37)

(iii) 

Additionally, assume $f_{\gamma }\in L_p\left({\mathbb{R}}^N\right)$, $p>1.$ Then

$${\omega }_m{\left({\left({\theta }_{r,n}^{\left[\tilde{m}\right]}\left(f\right)\right)}_{\gamma },h\right)}_p\le \left({\displaystyle \sum _{\tilde{j}=0}^r}\left|{\alpha }_{\tilde{j},r}^{\left[\tilde{m}\right]}\right|\right){\omega }_m{\left(f_{\gamma },h\right)}_p.$$

(38)

Next comes simultaneous approximation.

Theorem 10. 

Let $f\in C^{m+l}\left({\mathbb{R}}^N\right)$, $m,l,N\in \mathbb{N}$. The assumptions of Theorem 8 are valid. Call $\gamma =0,\beta$. Assume ${∥f_{\gamma +\alpha }∥}_{\infty }\le \infty$ and

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

for all ${\alpha }_i\in {\mathbb{Z}}^{+}$, i $=1,\dots ,N,$ $\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}{\alpha }_i=m$, where ${\mu }_{{\xi }_n}$ is a Borel probability measure on ${\mathbb{R}}^N$, for ${\xi }_n>0$, ${\left({\xi }_n\right)}_{n\in \mathbb{N}}$ bounded sequence.

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

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

(39)

Then

$${∥{\left({\theta }_{r,n}^{\left[m\right]}\left(f;·\right)\right)}_{\gamma }-f_{\gamma }\left(·\right)-{\displaystyle \sum _{\tilde{j}=1}^m}{\delta }_{\tilde{j},r}^{\left[m\right]}\left(\sum _{\begin{array}{c}{\alpha }_1,\dots ,{\alpha }_N\ge 0:\\ \left|\alpha \right|=\tilde{j}\end{array}}{\displaystyle \frac{c_{\alpha ,n,\tilde{j}}{f}_{\gamma +\alpha }\left(·\right)}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)∥}_{\infty }\le$$

(40)

$$\sum _{\left({\hspace{0.17em}}_{\begin{array}{c}{\alpha }_1,\dots ,{\alpha }_N\ge 0\\ \left|\alpha \right|=m\end{array}}\right)}{\displaystyle \frac{\left({\omega }_r\left(f_{\gamma +\alpha },{\xi }_n\right)\right)}{\left({\displaystyle \prod _{i=1}^N}{\alpha }_i!\right)}}\left({\int }_{{\mathbb{R}}^N}\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{{\alpha }_i}\right){\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^{r}d{\mu }_{{\xi }_n}\left(s\right)\right).$$

Theorem 11. 

Let $f\in C_B^l\left({\mathbb{R}}^N\right)$, $l,N\in \mathbb{N}$ (functions l-times continuously differentiable and bounded). The assumptions of Theorem 8 are valid. Call $\gamma =0,\beta$. Assume

$${\int }_{{\mathbb{R}}^N}{\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^{r}d{\mu }_{{\xi }_n}\left(s\right)<\infty .$$

(41)

Then

$${∥{\left({\theta }_{r,n}^{\left[0\right]}f\right)}_{\gamma }-f_{\gamma }∥}_{\infty }\le \left({\int }_{{\mathbb{R}}^N}{\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^{r}d{\mu }_{{\xi }_n}\left(s\right)\right){\omega }_r\left(f_{\gamma },{\xi }_n\right).$$

(42)

We mention more simultaneous approximation results.

Theorem 12. 

Let $f\in C^{m+l}\left({\mathbb{R}}^N\right),$ $m,l,N\in \mathbb{N}.$ The assumptions of Theorem 8 are valid. Call $\gamma =0,\beta$. Let $f_{\left(\gamma +\alpha \right)}\in L_p\left({\mathbb{R}}^N\right)$, $\left|\alpha \right|=m$, $x\in {\mathbb{R}}^N$, and $p,q>1:{\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1.$ 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 $\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right)$, ${\alpha }_i\in {\mathbb{Z}}^{+}$, $i=1,\dots ,N,$ $\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}{\alpha }_i=m$ we have

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

(43)

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

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

(44)

Then

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

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

(45)

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

Theorem 13. 

Let $f\in C^l\left({\mathbb{R}}^N\right),$ $l,N\in \mathbb{N}.$ The assumptions of Theorem 8 are valid. Call $\gamma =0,\beta$. Let $f_{\gamma }\in L_p\left({\mathbb{R}}^N\right)$, $x\in {\mathbb{R}}^N$; $p,q>1:{\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1.$ Assume ${\mu }_{{\xi }_n}$ probability Borel measures on ${\mathbb{R}}^N,$ ${\left({\xi }_n\right)}_{n\in \mathbb{N}}>0$ and bounded. Also suppose

$${\int }_{{\mathbb{R}}^N}{\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^{rp}d{\mu }_{{\xi }_n}\left(s\right)<\infty .$$

Then

$${∥{\left({\theta }_{r,n}^{\left[0\right]}f\right)}_{\gamma }-f_{\gamma }∥}_p\le {\left({\int }_{{\mathbb{R}}^N}{\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^{rp}d{\mu }_{{\xi }_n}\left(s\right)\right)}^{{\displaystyle \frac{1}{p}}}{\omega }_r{\left(f_{\gamma },{\xi }_n\right)}_p.$$

(46)

Theorem 14. 

Let $f\in C^l\left({\mathbb{R}}^N\right),$ $l,N\in \mathbb{N}.$ The assumptions of Theorem 8 are valid. Call $\gamma =0,\beta$. Let $f_{\gamma }\in L_1\left({\mathbb{R}}^N\right)$, $x\in {\mathbb{R}}^N.$ Assume ${\mu }_{{\xi }_n}$ probability Borel measures on ${\mathbb{R}}^N,$ ${\left({\xi }_n\right)}_{n\in \mathbb{N}}>0$ and bounded. Also suppose

$${\int }_{{\mathbb{R}}^N}{\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^{r}d{\mu }_{{\xi }_n}\left(s\right)<\infty .$$

Then

$${∥{\left({\theta }_{r,n}^{\left[0\right]}\left(f\right)\right)}_{\gamma }-f_{\gamma }∥}_1\le \left({\int }_{{\mathbb{R}}^N}{\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^{r}d{\mu }_{{\xi }_n}\left(s\right)\right){\omega }_r{\left(f_{\gamma },{\xi }_n\right)}_1.$$

(47)

The last supporting result is as follows:

Theorem 15. 

Let $f\in C^{m+l}\left({\mathbb{R}}^N\right)$, $m,l,N\in \mathbb{N}$. The assumptions of Theorem 8 are valid. Call $\gamma =0,\beta$. Let $f_{\left(\gamma +\alpha \right)}\in L_1\left({\mathbb{R}}^N\right),$ $\left|\alpha \right|=m$, $x\in {\mathbb{R}}^N$. Here ${\mu }_{{\xi }_n}$ is a Borel probability measure on ${\mathbb{R}}^N$ for ${\xi }_n>0$, ${\left({\xi }_n\right)}_{n\in \mathbb{N}}$ is a bounded sequence. Assume for all $\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right)$, ${\alpha }_i\in {\mathbb{Z}}^{+}$, $i=1,\dots ,N,$ $\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}{\alpha }_i=m,$ we have

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

(48)

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

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

(49)

Then

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

(50)

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

3. About the $\mathit{q}$-Deformed and $\mathit{\lambda }$-Parametrized Hyperbolic Tangent Function ${\mathit{g}}_{\mathit{q},\mathit{\lambda }}$

We consider the activation function $g_{q,\lambda }$ and we mention some of its related properties; most of these come from ([17], ch. 17).

Let the activation function

$$g_{q,\lambda }\left(x\right)={\displaystyle \frac{e^{\lambda x}-q{e}^{-\lambda x}}{e^{\lambda x}+q{e}^{-\lambda x}}},\lambda ,q>0,x\in \mathbb{R}.$$

(51)

It is

$$g_{q,\lambda }\left(0\right)={\displaystyle \frac{1-q}{1+q}},$$

and

$$g_{q,\lambda }\left(-x\right)=-g_{{\displaystyle \frac{1}{q}},\lambda }\left(x\right),\forall x\in \mathbb{R},$$

(52)

with

$$g_{q,\lambda }\left(+\infty \right)=1,g_{q,\lambda }\left(-\infty \right)=-1.$$

We consider the function

$$M_{q.\lambda }\left(x\right):={\displaystyle \frac{1}{4}}\left(g_{q,\lambda }\left(x+1\right)-g_{q,\lambda }\left(x-1\right)\right)>0,$$

(53)

$\forall x\in \mathbb{R}$; $q,\lambda >0$. We have $M_{q,\lambda }\left(\pm \infty \right)=0$, so the x-axis is the horizontal asymptote.

It holds

$$M_{q,\lambda }\left(-x\right)=M_{{\displaystyle \frac{1}{q}},\lambda }\left(x\right),\forall x\in \mathbb{R};q,\lambda >0,$$

(54)

and

$$M_{{\displaystyle \frac{1}{q}},\lambda }\left(-x\right)=M_{q,\lambda }\left(x\right),\forall x\in \mathbb{R}.$$

That is

$$\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left(-x\right)=\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left(x\right),$$

(55)

an even function.

The $M_{q,\lambda }$ maximum is

$$M_{q,\lambda }\left({\displaystyle \frac{lnq}{2\lambda }}\right)={\displaystyle \frac{tanh\left(\lambda \right)}{2}},\lambda >0.$$

(56)

In [18], we proved that

$$M_{q,\lambda }\left(x\right)<2q\lambda e^{2\lambda }{e}^{-2\lambda x},\forall x\ge 1.$$

(57)

Next we follow [17], pp. 432–433.

Let $x\in {\mathbb{R}}^N$, $N\in \mathbb{N}$; $x=\left(x_1,\dots ,x_N\right).$

We define

$$Z_1\left(x\right):=Z_{q,\lambda }\left(x_1,\dots ,x_N\right):={\displaystyle \prod _{i=1}^N}{M}_{q,\lambda }\left(x_i\right)>0,q,\lambda >0.$$

(58)

From (57), we have

$$M_{q,\lambda }\left(x_i\right)<2q\lambda e^{2\lambda }{e}^{-2\lambda x_i},\forall x_i\ge 1,i=1,\dots ,N.$$

(59)

That is

$$Z_1\left(x\right)<{\left(2q\lambda \right)}^N{e}^{2\lambda N}{e}^{-2\lambda \left({\displaystyle \sum _{i=1}^N}{x}_i\right)},x_i\ge 1,i=1,\dots ,N.$$

(60)

So, the last is $x\in {\left({\mathbb{R}}_{+}-[0,1)\right)}^N$.

We have

$${\int }_{{\mathbb{R}}^N}{Z}_{q,\lambda }\left(x\right)dx=1,$$

(61)

that is $Z_q$ is a multivariate density function.

Let $\xi >0$, then

$${\displaystyle \frac{1}{{\xi }^N}}{\int }_{{\mathbb{R}}^N}{Z}_{q,\lambda }\left({\displaystyle \frac{x}{\xi }}\right)dx=1,$$

(62)

so that ${\displaystyle \frac{1}{{\xi }^N}}{Z}_{q,\lambda }\left({\displaystyle \frac{x}{\xi }}\right)$ is a multivariate density function on ${\mathbb{R}}^N$, and let

$$d{\mu }_{{\xi }_n}:={\displaystyle \frac{1}{{\xi }^N}}{Z}_{q,\lambda }\left({\displaystyle \frac{x}{\xi }}\right)dx,\xi >0,dx=d{x}_{1}d{x}_2\dots d{x}_N.$$

(63)

Clearly, ${\mu }_{\xi }$ is a Borel probability measure in ${\mathbb{R}}^N$.

We have that

$$\begin{array}{c}{M}_{q,\lambda }\left(-x_i\right)=M_{{\displaystyle \frac{1}{q}},\lambda }\left(x_i\right),\\ \mathrm{and}\\ M_{{\displaystyle \frac{1}{q}},\lambda }\left(-x_i\right)=M_{q,\lambda }\left(x_i\right),\end{array}$$

(64)

∀$x_i\in \mathbb{R},$ $i=1,\dots ,N.$

Adding the above, we obtain

$$M_{q,\lambda }\left(-x_i\right)+M_{{\displaystyle \frac{1}{q}},\lambda }\left(-x_i\right)=M_{{\displaystyle \frac{1}{q}},\lambda }\left(x_i\right)+M_{q,\lambda }\left(x_i\right),$$

(65)

that is

$$\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left(-x_i\right)=\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left(x_i\right),$$

∀$x_i\in \mathbb{R},$ $i=1,\dots ,N.$

So, $\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left(x_i\right)$ is a symmetric function over $\mathbb{R}$, $i=1,\dots ,N$, and

$$0<M_{q,\lambda }\left(x_i\right)<\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left(x_i\right),$$

(66)

∀$x_i\in \mathbb{R},$ $i=1,\dots ,N.$

Therefore, we obtain

$$0<Z_{q,\lambda }\left(x\right)={\displaystyle \prod _{i=1}^N}{M}_{q,\lambda }\left(x_i\right)<{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left(x_i\right),$$

(67)

∀$x\in {\mathbb{R}}^N.$

Furthermore, it holds that

$$Z_{q,\lambda }\left(x\right)<{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left(x_i\right)<$$

$${\displaystyle \prod _{i=1}^N}\left(2q\lambda e^{2\lambda }{e}^{-2\lambda x_i}+{\displaystyle \frac{2}{q}}\lambda e^{2\lambda }{e}^{-2\lambda x_i}\right)={\displaystyle \prod _{i=1}^N}\left(\left(q+{\displaystyle \frac{1}{q}}\right)2\lambda e^{2\lambda }{e}^{-2\lambda x_i}\right)=$$

$${\left(q+{\displaystyle \frac{1}{q}}\right)}^N{\left(2\lambda \right)}^N{e}^{2\lambda N}{e}^{-2\lambda \left({\displaystyle \sum _{i=1}^N}{x}_i\right)},$$

(68)

for any $x_i\ge 1$, $i=1,\dots ,N;$ so that $x\in {\left({\mathbb{R}}_{+}-[0,1)\right)}^N$.

Furthemore, ($\xi >0$) we have

$${\displaystyle \frac{1}{{\xi }^N}}{Z}_{q,\lambda }\left({\displaystyle \frac{x}{\xi }}\right)<{\displaystyle \frac{1}{{\xi }^N}}{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left({\displaystyle \frac{x_i}{\xi }}\right)=$$

(69)

$${\displaystyle \prod _{i=1}^N}\left[{\displaystyle \frac{1}{\xi }}{M}_{q,\lambda }\left({\displaystyle \frac{x_i}{\xi }}\right)+{\displaystyle \frac{1}{\xi }}{M}_{{\displaystyle \frac{1}{q}},\lambda }\left({\displaystyle \frac{x_i}{\xi }}\right)\right],$$

∀$x\in {\mathbb{R}}^N$, $x=\left(x_1,\dots ,x_N\right).$

Above ${\displaystyle \frac{1}{\xi }}{M}_{q,\lambda }\left({\displaystyle \frac{x_i}{\xi }}\right)$ and ${\displaystyle \frac{1}{\xi }}{M}_{{\displaystyle \frac{1}{q}},\lambda }\left({\displaystyle \frac{x_i}{\xi }}\right)$ are univariate density functions.

Furthermore, ${\displaystyle \frac{1}{\xi }}\left({\displaystyle \frac{M_{q,\lambda }\left(\frac{x_i}{\xi }\right)+M_{\frac{1}{q},\lambda }\left(\frac{x_i}{\xi }\right)}{2}}\right)$ is also a density function on $\mathbb{R}$, and ${\displaystyle \frac{1}{{\xi }^N}}{\displaystyle \prod _{i=1}^N}\left({\displaystyle \frac{\left(M_{q,\lambda }+M_{\frac{1}{q},\lambda }\right)\left(\frac{x_i}{\xi }\right)}{2}}\right)$ is a density function on ${\mathbb{R}}^N.$

So, we can rewrite

$${\displaystyle \frac{1}{{\xi }^N}}{Z}_{q,\lambda }\left({\displaystyle \frac{x}{\xi }}\right)<2^N\left[{\displaystyle \frac{1}{{\xi }^N}}{\displaystyle \prod _{i=1}^N}\left({\displaystyle \frac{\left(M_{q,\lambda }+M_{\frac{1}{q},\lambda }\right)\left(\frac{x_i}{\xi }\right)}{2}}\right)\right],$$

(70)

∀$x\in {\mathbb{R}}^N$, $x=\left(x_1,\dots ,x_N\right).$

In Section 4, all proofs are based on our “Symmetrization Technique” in estimating integrals, as you will see.

4. Auxiliary Essential Results

We need the following:

Theorem 16. 

Let $r,N\in \mathbb{N}$, $\overline{{\alpha }_j}\in {\mathbb{Z}}^{+}$, $j=1,\dots ,N:\left|\alpha \right|:={\displaystyle \sum _{j=1}^N}\overline{{\alpha }_j}=m\in \mathbb{N}$; ${\xi }_n\in (0,1]$, $n\in \mathbb{N}$; $s=\left(s_1,\dots ,s_N\right)\in {\mathbb{R}}^N$; $\lambda >0.$

Then

$$U_{{\xi }_n}\left(\alpha \right):={\displaystyle \frac{1}{{\left(2{\xi }_n\right)}^N}}{\int }_{{\mathbb{R}}^N}\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{\overline{{\alpha }_i}}\right){\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^r{Z}_{q,\lambda }\left({\displaystyle \frac{s}{{\xi }_n}}\right)d{s}_1\dots d{s}_N$$

(71)

$$\le {\xi }_n^m\left[{\left(1+N\right)}^r{\left(tanh\lambda \right)}^N+{\displaystyle \frac{{\left(q+\frac{1}{q}\right)}^N{e}^{2\lambda N}}{2^{m+r\left(N-1\right)}{\lambda }^{m+Nr}}}{\displaystyle \prod _{i=1}^N}\mathrm{\Gamma }\left(\overline{{\alpha }_i}+r+1,2\lambda \right)\right]<+\infty ,$$

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

Proof. 

We observe that

$$U_{{\xi }_n}\left(\alpha \right)\le {\displaystyle \frac{1}{{\left(2{\xi }_n\right)}^N}}{\int }_{{\mathbb{R}}^N}\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{\overline{{\alpha }_i}}\right){\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^r$$

$$\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)\right]d{s}_1\dots d{s}_N=$$

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

$$\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)\right]d{s}_1\dots d{s}_N\le$$

(72)

$${\displaystyle \frac{1}{{\xi }_n^N}}{\int }_{{\mathbb{R}}_{+}^N}\left({\displaystyle \prod _{i=1}^N}{s}_i^{\overline{{\alpha }_i}}\right){\left(1+{\displaystyle \frac{{\displaystyle \sum _{i=1}^N}{s}_i}{{\xi }_n}}\right)}^r$$

$$\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)\right]d{s}_1\dots d{s}_N=$$

$${\xi }_n^m{\int }_{{\mathbb{R}}_{+}^N}\left({\displaystyle \prod _{i=1}^N}{\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)}^{\overline{{\alpha }_i}}\right){\left(1+{\displaystyle \sum _{i=1}^N}\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)\right)}^r$$

$$\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)\right]d\left({\displaystyle \frac{s_1}{{\xi }_n}}\right)d\left({\displaystyle \frac{s_2}{{\xi }_n}}\right)\dots d\left({\displaystyle \frac{s_N}{{\xi }_n}}\right)=$$

$${\xi }_n^m{\int }_{{\mathbb{R}}_{+}^N}\left({\displaystyle \prod _{i=1}^N}{z}_i^{\overline{{\alpha }_i}}\right){\left(1+{\displaystyle \sum _{i=1}^N}{z}_i\right)}^r$$

$$\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left(z_i\right)\right]d{z}_1\dots d{z}_N=$$

(73)

$${\xi }_n^m\left[{\int }_{{[0,1)}^N}\left({\displaystyle \prod _{i=1}^N}{z}_i^{\overline{{\alpha }_i}}\right){\left(1+{\displaystyle \sum _{i=1}^N}{z}_i\right)}^r\right.$$

$$\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left(z_i\right)\right]d{z}_1\dots d{z}_N+$$

$${\int }_{{\left({\mathbb{R}}_{+}-[0,1)\right)}^N}\left({\displaystyle \prod _{i=1}^N}{z}_i^{\overline{{\alpha }_i}}\right){\left(1+{\displaystyle \sum _{i=1}^N}{z}_i\right)}^r$$

$$\left.\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left(z_i\right)\right]d{z}_1\dots d{z}_N\right]\le$$

$${\xi }_n^m\left[\left({\int }_{{[0,1)}^N}\left({\displaystyle \prod _{i=1}^N}{z}_i^{\overline{{\alpha }_i}}\right){\left(1+{\displaystyle \sum _{i=1}^N}{z}_i\right)}^{r}d{z}_1\dots d{z}_N\right){\left(tanh\left(\lambda \right)\right)}^N+\right.$$

$$\left(2^r{\left(q+{\displaystyle \frac{1}{q}}\right)}^N{\left(2\lambda \right)}^N{e}^{2\lambda N}\right)$$

$$\left.\underset{\left(N\mathrm{-times}\right)}{\underbrace{{\int }_1^{\infty }\dots {\int }_1^{\infty }}}\left({\displaystyle \prod _{i=1}^N}{z}_i^{\overline{{\alpha }_i}}\right){\left({\displaystyle \sum _{i=1}^N}{z}_i\right)}^r{e}^{-2\lambda \left({\displaystyle \sum _{i=1}^N}{z}_i\right)}d{z}_1\dots d{z}_N\right]\le$$

(74)

$${\xi }_n^m\left[{\left(1+N\right)}^r{\left(tanh\left(\lambda \right)\right)}^N+2^{r+N}{\left(\lambda \left(q+{\displaystyle \frac{1}{q}}\right)\right)}^N{e}^{2\lambda N}\right.$$

$$\left.{\int }_1^{\infty }\dots {\int }_1^{\infty }\left({\displaystyle \prod _{i=1}^N}{z}_i^{\overline{{\alpha }_i}}\right)\left({\displaystyle \prod _{i=1}^N}{z}_i^r\right)\left({\displaystyle \prod _{i=1}^N}{e}^{-2\lambda z_i}\right){\displaystyle \prod _{i=1}^N}d{z}_i\right]=$$

$${\xi }_n^m\left[{\left(1+N\right)}^r{\left(tanh\left(\lambda \right)\right)}^N+2^{r+N}{\left(\lambda \left(q+{\displaystyle \frac{1}{q}}\right)\right)}^N{e}^{2\lambda N}\right.$$

$$\left.{\displaystyle \prod _{i=1}^N}{\int }_1^{\infty }{z}_i^{\overline{{\alpha }_i}+r}{e}^{-2\lambda z_i}d{z}_i\right]=$$

$${\xi }_n^m\left[{\left(1+N\right)}^r{\left(tanh\lambda \right)}^N+2^{r+N}{\left(\lambda \left(q+{\displaystyle \frac{1}{q}}\right)\right)}^N{e}^{2\lambda N}\right.$$

$$\left.{\displaystyle \prod _{i=1}^N}{\displaystyle \frac{1}{{\left(2\lambda \right)}^{\overline{{\alpha }_i}+r+1}}}{\int }_{2\lambda }^{\infty }{y}_i^{\left(\overline{{\alpha }_i}+r+1\right)-1}{e}^{-y_i}d{y}_i\right]$$

(by [19], p. 348)

$$={\xi }_n^m\left[{\left(1+N\right)}^r{\left(tanh\lambda \right)}^N+2^{r+N}{\left(\lambda \left(q+{\displaystyle \frac{1}{q}}\right)\right)}^N{e}^{2\lambda N}\right.$$

$$\left.{\displaystyle \frac{1}{{\left(2\lambda \right)}^{m+N\left(r+1\right)}}}{\displaystyle \prod _{i=1}^N}\mathrm{\Gamma }\left(\overline{{\alpha }_i}+r+1,2\lambda \right)\right]=$$

(75)

$${\xi }_n^m\left[{\left(1+N\right)}^r{\left(tanh\lambda \right)}^N+{\displaystyle \frac{{\left(q+\frac{1}{q}\right)}^N{e}^{2\lambda N}}{2^{m+r\left(N-1\right)}{\lambda }^{m+Nr}}}{\displaystyle \prod _{i=1}^N}\mathrm{\Gamma }\left(\overline{{\alpha }_i}+r+1,2\lambda \right)\right]<+\infty .$$

□

We continue with

Theorem 17. 

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

$${\mathrm{\Phi }}_{{\xi }_n}^{\ast }:={\displaystyle \frac{1}{{\left(2{\xi }_n\right)}^N}}{\int }_{{\mathbb{R}}^N}{\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^r{Z}_{q,\lambda }\left({\displaystyle \frac{s}{{\xi }_n}}\right)d{s}_1\dots d{s}_N\le$$

(76)

$$\left[{\left(1+N\right)}^r{\left(tanh\lambda \right)}^N+{\displaystyle \frac{{\left(q+\frac{1}{q}\right)}^N{e}^{2\lambda N}}{2^{r\left(N-1\right)}{\lambda }^{Nr}}}{\mathrm{\Gamma }}^N\left(r+1,2\lambda \right)\right]<+\infty .$$

Proof. 

We observe that

$${\mathrm{\Phi }}_{{\xi }_n}^{\ast }={\displaystyle \frac{1}{{\left(2{\xi }_n\right)}^N}}{\int }_{{\mathbb{R}}^N}{\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^r{Z}_{q,\lambda }\left({\displaystyle \frac{s}{{\xi }_n}}\right)d{s}_1\dots d{s}_N\le$$

(77)

$${\displaystyle \frac{1}{{\left(2{\xi }_n\right)}^N}}{\int }_{{\mathbb{R}}^N}{\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^r\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)\right]d{s}_1\dots d{s}_N=$$

$${\displaystyle \frac{1}{{\xi }_n^N}}{\int }_{{\mathbb{R}}_{+}^N}{\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^r\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)\right]d{s}_1\dots d{s}_N\le$$

$${\displaystyle \frac{1}{{\xi }_n^N}}{\int }_{{\mathbb{R}}_{+}^N}{\left(1+{\displaystyle \frac{{\displaystyle \sum _{i=1}^N}{s}_i}{{\xi }_n}}\right)}^r\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)\right]d{s}_1\dots d{s}_N=$$

$${\int }_{{\mathbb{R}}_{+}^N}{\left(1+{\displaystyle \sum _{i=1}^N}\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)\right)}^r\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)\right]d{\displaystyle \frac{s_1}{{\xi }_n}}\dots d{\displaystyle \frac{s_N}{{\xi }_n}}=$$

$${\int }_{{\mathbb{R}}_{+}^N}{\left(1+{\displaystyle \sum _{i=1}^N}{z}_i\right)}^r\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left(z_i\right)\right]d{z}_1\dots d{z}_N=$$

(78)

$$\left[{\int }_{{[0,1)}^N}{\left(1+{\displaystyle \sum _{i=1}^N}{z}_i\right)}^r\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left(z_i\right)\right]d{z}_1\dots d{z}_N+\right.$$

$$\left.{\int }_{{\left({\mathbb{R}}_{+}-[0,1)\right)}^N}{\left(1+{\displaystyle \sum _{i=1}^N}{z}_i\right)}^r\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left(z_i\right)\right]d{z}_1\dots d{z}_N\right]\le$$

$$\left[\left({\int }_{{[0,1)}^N}{\left(1+{\displaystyle \sum _{i=1}^N}{z}_i\right)}^{r}d{z}_1\dots d{z}_n\right){\left(tanh\left(\lambda \right)\right)}^N+\right.$$

$$\left.\left(2^r{\left(q+{\displaystyle \frac{1}{q}}\right)}^N{\left(2\lambda \right)}^N{e}^{2\lambda N}\right)\underset{\left(N\mathrm{-times}\right)}{\underbrace{{\int }_1^{\infty }\dots {\int }_1^{\infty }}}{\left({\displaystyle \sum _{i=1}^N}{z}_i\right)}^r{e}^{-2\lambda \left({\displaystyle \sum _{i=1}^N}{z}_i\right)}d{z}_1\dots d{z}_N\right]\le$$

(79)

$$\left[{\left(1+N\right)}^r{\left(tanh\lambda \right)}^N+2^{r+N}{\left(\lambda \left(q+{\displaystyle \frac{1}{q}}\right)\right)}^N{e}^{2\lambda N}\right.$$

$$\left.{\int }_1^{\infty }\dots {\int }_1^{\infty }\left({\displaystyle \prod _{i=1}^N}{z}_i^r\right)\left({\displaystyle \prod _{i=1}^N}{e}^{-2\lambda z_i}\right){\displaystyle \prod _{i=1}^N}d{z}_i\right]=$$

$$\left[{\left(1+N\right)}^r{\left(tanh\lambda \right)}^N+2^{r+N}{\left(\lambda \left(q+{\displaystyle \frac{1}{q}}\right)\right)}^N{e}^{2\lambda N}{\displaystyle \prod _{i=1}^N}{\int }_1^{\infty }{z}_i^r{e}^{-2\lambda z_i}d{z}_i\right]=$$

$$\left[{\left(1+N\right)}^r{\left(tanh\lambda \right)}^N+2^{r+N}{\left(\lambda \left(q+{\displaystyle \frac{1}{q}}\right)\right)}^N{e}^{2\lambda N}\right.$$

$$\left.{\displaystyle \frac{1}{{\left(2\lambda \right)}^{N\left(r+1\right)}}}{\displaystyle \prod _{i=1}^N}{\int }_{2\lambda }^{\infty }{y}_i^{\left(r+1\right)-1}{e}^{-y_i}d{y}_i\right]=$$

$$\left[{\left(1+N\right)}^r{\left(tanh\lambda \right)}^N+2^{r+N}{\left(\lambda \left(q+{\displaystyle \frac{1}{q}}\right)\right)}^N{e}^{2\lambda N}\right.$$

$$\left.{\displaystyle \frac{1}{{\left(2\lambda \right)}^{N\left(r+1\right)}}}{\mathrm{\Gamma }}^N\left(r+1,2\lambda \right)\right]=$$

(80)

$$\left[{\left(1+N\right)}^r{\left(tanh\lambda \right)}^N+{\displaystyle \frac{{\left(q+\frac{1}{q}\right)}^N{e}^{2\lambda N}}{2^{r\left(N-1\right)}{\lambda }^{Nr}}}{\left(\mathrm{\Gamma }\left(r+1,2\lambda \right)\right)}^N\right]<+\infty .$$

□

It follows

Theorem 18. 

All as in Theorem 16, $p>1$. Then

$$V_{{\xi }_n}\left(\alpha \right):={\displaystyle \frac{1}{{\left(2{\xi }_n\right)}^N}}{\int }_{{\mathbb{R}}^N}{\left(\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{\overline{{\alpha }_i}}\right){\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^r\right)}^p{Z}_{q,\lambda }\left({\displaystyle \frac{s}{{\xi }_n}}\right)d{s}_1\dots d{s}_N\le$$

(81)

$${\xi }_n^{mp}\left[{\left(1+N\right)}^{rp}{\left(tanh\lambda \right)}^N+{\displaystyle \frac{{\left(q+\frac{1}{q}\right)}^N{e}^{2\lambda N}}{2^{\left(m+r\left(N-1\right)\right)p}{\lambda }^{\left(m+Nr\right)p}}}{\displaystyle \prod _{i=1}^N}\mathrm{\Gamma }\left(\left(\overline{{\alpha }_i}+r\right)p+1,2\lambda \right)\right]<+\infty .$$

Proof. 

We observe that

$$V_{{\xi }_n}\left(\alpha \right)={\displaystyle \frac{1}{{\left(2{\xi }_n\right)}^N}}{\int }_{{\mathbb{R}}^N}{\left(\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{\overline{{\alpha }_i}}\right){\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^r\right)}^p{Z}_{q,\lambda }\left({\displaystyle \frac{s}{{\xi }_n}}\right)d{s}_1\dots d{s}_N$$

$$\le {\displaystyle \frac{1}{{\left(2{\xi }_n\right)}^N}}{\int }_{{\mathbb{R}}^N}{\left(\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{\overline{{\alpha }_i}}\right){\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^r\right)}^p$$

$$\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)\right]d{s}_1\dots d{s}_N=$$

(82)

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

$$\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)\right]d{s}_1\dots d{s}_N\le$$

$${\displaystyle \frac{1}{{\xi }_n^N}}{\int }_{{\mathbb{R}}_{+}^N}\left({\displaystyle \prod _{i=1}^N}{s}_i^{\overline{{\alpha }_i}p}\right){\left(1+{\displaystyle \frac{{\displaystyle \sum _{i=1}^N}{s}_i}{{\xi }_n}}\right)}^{rp}$$

$$\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)\right]d{s}_1\dots d{s}_N=$$

$${\xi }_n^{mp}{\int }_{{\mathbb{R}}_{+}^N}{\displaystyle \prod _{i=1}^N}{\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)}^{\overline{{\alpha }_i}p}{\left(1+{\displaystyle \sum _{i=1}^N}{\displaystyle \frac{s_i}{{\xi }_n}}\right)}^{rp}$$

$$\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)\right]d{\displaystyle \frac{s_1}{{\xi }_n}}\dots d{\displaystyle \frac{s_N}{{\xi }_n}}=$$

$${\xi }_n^{mp}{\int }_{{\mathbb{R}}_{+}^N}\left({\displaystyle \prod _{i=1}^N}{z}_i^{\overline{{\alpha }_i}p}\right){\left(1+{\displaystyle \sum _{i=1}^N}{z}_i\right)}^{rp}$$

$$\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left(z_i\right)\right]d{z}_1\dots d{z}_N=$$

$${\xi }_n^{mp}\left[{\int }_{{[0,1)}^N}\left({\displaystyle \prod _{i=1}^N}{z}_I^{\overline{{\alpha }_i}p}\right){\left(1+{\displaystyle \sum _{i=1}^N}{z}_i\right)}^{rp}\right.$$

(83)

$$\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left(z_i\right)\right]d{z}_1\dots d{z}_N+$$

$${\int }_{{\left({\mathbb{R}}_{+}-[0,1)\right)}^N}\left({\displaystyle \prod _{i=1}^N}{z}_i^{\overline{{\alpha }_i}p}\right){\left(1+{\displaystyle \sum _{i=1}^N}{z}_i\right)}^{rp}$$

$$\left.\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left(z_i\right)\right]d{z}_1\dots d{z}_N\right]\le$$

$${\xi }_n^{mp}\left[{\left(1+N\right)}^{rp}{\left(tanh\lambda \right)}^N+2^{rp}{\left(q+{\displaystyle \frac{1}{q}}\right)}^N{\left(2\lambda \right)}^N{e}^{2\lambda N}\right.$$

$$\left.\underset{\left(N\mathrm{-times}\right)}{\underbrace{{\int }_1^{\infty }\dots {\int }_1^{\infty }}}\left({\displaystyle \prod _{i=1}^N}{z}_i^{\overline{{\alpha }_i}p}\right){\left({\displaystyle \sum _{i=1}^N}{z}_i\right)}^{rp}{e}^{-2\lambda \left({\displaystyle \sum _{i=1}^N}{z}_i\right)}d{z}_1\dots d{z}_N\right]\le$$

$${\xi }_n^{mp}\left[{\left(1+N\right)}^{rp}{\left(tanh\lambda \right)}^N+2^{rp}{\left(q+{\displaystyle \frac{1}{q}}\right)}^N{\left(2\lambda \right)}^N{e}^{2\lambda N}\right.$$

$$\left.{\int }_1^{\infty }\dots {\int }_1^{\infty }\left({\displaystyle \prod _{i=1}^N}{z}_i^{\overline{{\alpha }_i}p}\right)\left({\displaystyle \prod _{i=1}^N}{z}_i^{rp}\right)\left({\displaystyle \prod _{i=1}^N}{e}^{-2\lambda z_i}\right){\displaystyle \prod _{i=1}^N}d{z}_i\right]=$$

$${\xi }_n^{mp}\left[{\left(1+N\right)}^{rp}{\left(tanh\lambda \right)}^N+2^{rp}{\left(q+{\displaystyle \frac{1}{q}}\right)}^N{\left(2\lambda \right)}^N{e}^{2\lambda N}\right.$$

$$\left.{\displaystyle \prod _{i=1}^N}{\int }_1^{\infty }{z}_i^{\left(\overline{{\alpha }_i}+r\right)p}{e}^{-2\lambda z_i}d{z}_i\right]=$$

(84)

$${\xi }_n^{mp}\left[{\left(1+N\right)}^{rp}{\left(tanh\lambda \right)}^N+2^{rp}{\left(q+{\displaystyle \frac{1}{q}}\right)}^N{\left(2\lambda \right)}^N{e}^{2\lambda N}\right.$$

$$\left.{\displaystyle \prod _{i=1}^N}{\displaystyle \frac{1}{{\left(2\lambda \right)}^{\left(\overline{{\alpha }_i}+r\right)p+1}}}{\int }_{2\lambda }^{\infty }{y}_i^{\left(\left(\overline{{\alpha }_i}+r\right)p+1\right)-1}{e}^{-y_i}d{y}_i\right]=$$

$${\xi }_n^{mp}\left[{\left(1+N\right)}^{rp}{\left(tanh\lambda \right)}^N+2^{rp}{\left(q+{\displaystyle \frac{1}{q}}\right)}^N{\left(2\lambda \right)}^N{e}^{2\lambda N}\right.$$

(85)

$$\left.{\displaystyle \frac{1}{{\left(2\lambda \right)}^{mp+N\left(rp+1\right)}}}{\displaystyle \prod _{i=1}^N}\mathrm{\Gamma }\left(\left(\overline{{\alpha }_i}+r\right)p+1,2\lambda \right)\right]=$$

$${\xi }_n^{mp}\left[{\left(1+N\right)}^{rp}{\left(tanh\lambda \right)}^N+{\displaystyle \frac{{\left(q+\frac{1}{q}\right)}^N{e}^{2\lambda N}}{2^{\left(m+r\left(N-1\right)\right)p}{\lambda }^{\left(m+Nr\right)p}}}{\displaystyle \prod _{i=1}^N}\mathrm{\Gamma }\left(\left(\overline{{\alpha }_i}+r\right)p+1,2\lambda \right)\right]<+\infty .$$

□

We continue with

Theorem 19. 

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

$$E_{{\xi }_n}:={\displaystyle \frac{1}{{\left(2{\xi }_n\right)}^N}}{\int }_{{\mathbb{R}}^N}{\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^{rp}{Z}_{q,\lambda }\left({\displaystyle \frac{s}{{\xi }_n}}\right)d{s}_1\dots d{s}_N\le$$

(86)

$${\left(1+N\right)}^{rp}{\left(tanh\lambda \right)}^N+{\displaystyle \frac{{\left(q+\frac{1}{q}\right)}^N{e}^{2\lambda N}}{2^{rp\left(N-1\right)}{\lambda }^{Nrp}}}{\mathrm{\Gamma }}^N\left(rp+1,2\lambda \right)<+\infty .$$

Proof. 

We observe that

$$E_{{\xi }_n}={\displaystyle \frac{1}{{\left(2{\xi }_n\right)}^N}}{\int }_{{\mathbb{R}}^N}{\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^{rp}{Z}_{q,\lambda }\left({\displaystyle \frac{s}{{\xi }_n}}\right)d{s}_1\dots d{s}_N\le$$

$${\displaystyle \frac{1}{{\left(2{\xi }_n\right)}^N}}{\int }_{{\mathbb{R}}^N}{\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^{rp}\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)\right]d{s}_1\dots d{s}_N=$$

(87)

$${\displaystyle \frac{1}{{\xi }_n^N}}{\int }_{{\mathbb{R}}_{+}^N}{\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^{rp}\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)\right]d{s}_1\dots d{s}_N\le$$

$${\displaystyle \frac{1}{{\xi }_n^N}}{\int }_{{\mathbb{R}}_{+}^N}{\left(1+{\displaystyle \frac{{\displaystyle \sum _{i=1}^N}{s}_i}{{\xi }_n}}\right)}^{rp}\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)\right]d{s}_1\dots d{s}_N=$$

$${\int }_{{\mathbb{R}}_{+}^N}{\left(1+{\displaystyle \sum _{i=1}^N}\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)\right)}^{rp}\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)\right]d{\displaystyle \frac{s_1}{{\xi }_n}}\dots d{\displaystyle \frac{s_N}{{\xi }_n}}=$$

$${\int }_{{\mathbb{R}}_{+}^N}{\left(1+{\displaystyle \sum _{i=1}^N}{z}_i\right)}^{rp}\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left(z_i\right)\right]d{z}_1\dots d{z}_N=$$

(88)

$${\int }_{{[0,1)}^N}{\left(1+{\displaystyle \sum _{i=1}^N}{z}_i\right)}^{rp}\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left(z_i\right)\right]{\displaystyle \prod _{i=1}^N}d{z}_i+$$

$${\int }_{{\left({\mathbb{R}}_{+}-[0,1)\right)}^N}{\left(1+{\displaystyle \sum _{i=1}^N}{z}_i\right)}^{rp}\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left(z_i\right)\right]{\displaystyle \prod _{i=1}^N}d{z}_i\le$$

$${\left(1+N\right)}^{rp}{\left(tanh\lambda \right)}^N+2^{rp}{\left(q+{\displaystyle \frac{1}{q}}\right)}^N{\left(2\lambda \right)}^N{e}^{2\lambda N}$$

$${\int }_1^{\infty }\dots {\int }_1^{\infty }{\left({\displaystyle \sum _{i=1}^N}{z}_i\right)}^{rp}{e}^{-2\lambda \left({\displaystyle \sum _{i=1}^N}{z}_i\right)}{\displaystyle \prod _{i=1}^N}d{z}_i\le$$

$${\left(1+N\right)}^{rp}{\left(tanh\lambda \right)}^N+2^{rp}{\left(q+{\displaystyle \frac{1}{q}}\right)}^N{\left(2\lambda \right)}^N{e}^{2\lambda N}{\displaystyle \prod _{i=1}^N}{\int }_1^{\infty }{z}_i^{rp}{e}^{-2\lambda z_i}d{z}_i=$$

$${\left(1+N\right)}^{rp}{\left(tanh\lambda \right)}^N+2^{rp}{\left(q+{\displaystyle \frac{1}{q}}\right)}^N{\left(2\lambda \right)}^N{e}^{2\lambda N}$$

$${\displaystyle \prod _{i=1}^N}{\displaystyle \frac{1}{{\left(2\lambda \right)}^{rp+1}}}{\int }_{2\lambda }^{\infty }{y}_i^{\left(rp+1\right)-1}{e}^{-y_i}d{y}_i=$$

$${\left(1+N\right)}^{rp}{\left(tanh\lambda \right)}^N+2^{rp}{\left(q+{\displaystyle \frac{1}{q}}\right)}^N{\left(2\lambda \right)}^N{e}^{2\lambda N}$$

$$\left.{\displaystyle \frac{1}{{\left(2\lambda \right)}^{N\left(rp+1\right)}}}{\mathrm{\Gamma }}^N\left(rp+1,2\lambda \right)\right]=$$

(89)

$${\left(1+N\right)}^{rp}{\left(tanh\lambda \right)}^N+{\displaystyle \frac{{\left(q+\frac{1}{q}\right)}^N{e}^{2\lambda N}}{2^{rp\left(N-1\right)}{\lambda }^{Nrp}}}{\mathrm{\Gamma }}^N\left(rp+1,2\lambda \right)<+\infty .$$

□

The last supporting result for $Z_{q,\lambda }$ uses the following:

Theorem 20. 

Let $N\in \mathbb{N}$, ${\xi }_n\in (0,1]$, $n\in \mathbb{N},$ $\overline{{\alpha }_j}\in {\mathbb{Z}}^{+}$, $j=1,\dots ,N:\left|\alpha \right|:={\displaystyle \sum _{j=1}^N}\overline{{\alpha }_j}=m\in \mathbb{N}$, $\lambda >0.$

Then

$${\mathsf{\Delta }}_{{\xi }_n}\left(\alpha \right):={\xi }_n^{-m}\left({\displaystyle \frac{1}{{\left(2{\xi }_n\right)}^N}}{\int }_{{\mathbb{R}}^N}\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{\overline{{\alpha }_i}}\right)Z_{q,\lambda }\left({\displaystyle \frac{s}{{\xi }_n}}\right)d{s}_1\dots d{s}_N\right)\le$$

(90)

$$\left[{\left(tanh\lambda \right)}^N+{\displaystyle \frac{{\left(q+\frac{1}{q}\right)}^N{e}^{2\lambda N}}{{\left(2\lambda \right)}^m}}{\displaystyle \prod _{i=1}^N}\mathrm{\Gamma }\left(\overline{{\alpha }_i}+1,2\lambda \right)\right]<+\infty .$$

Proof. 

We have

$${\mathsf{\Delta }}_{{\xi }_n}\left(\alpha \right):={\xi }_n^{-m}\left({\displaystyle \frac{1}{{\left(2{\xi }_n\right)}^N}}{\int }_{{\mathbb{R}}^N}\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{\overline{{\alpha }_i}}\right)Z_{q,\lambda }\left({\displaystyle \frac{s}{{\xi }_n}}\right)d{s}_1\dots d{s}_N\right)\le$$

$${\xi }_n^{-m}\left({\displaystyle \frac{1}{{\left(2{\xi }_n\right)}^N}}{\int }_{{\mathbb{R}}^N}\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{\overline{{\alpha }_i}}\right)\right.$$

$$\left.\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)\right]d{s}_1\dots d{s}_N\right)=$$

$${\xi }_n^{-m}\left({\displaystyle \frac{1}{{\xi }_n^N}}{\int }_{{\mathbb{R}}_{+}^N}\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{\overline{{\alpha }_i}}\right)\right.$$

$$\left.\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)\right]d{s}_1\dots d{s}_N\right)=$$

(91)

$${\int }_{{\mathbb{R}}_{+}^N}\left({\displaystyle \prod _{i=1}^N}{\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)}^{\overline{{\alpha }_i}}\right)\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left({\displaystyle \frac{s_i}{{\xi }_n}}\right)\right]d{\displaystyle \frac{s_1}{{\xi }_n}}\dots d{\displaystyle \frac{s_N}{{\xi }_n}}=$$

$${\int }_{{\mathbb{R}}_{+}^N}\left({\displaystyle \prod _{i=1}^N}{z}_i^{\overline{{\alpha }_i}}\right)\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left(z_i\right)\right]{\displaystyle \prod _{i=1}^N}d{z}_i=$$

$${\int }_{{[0,1)}^N}\left({\displaystyle \prod _{i=1}^N}{z}_i^{\overline{{\alpha }_i}}\right)\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left(z_i\right)\right]{\displaystyle \prod _{i=1}^N}d{z}_i+$$

$${\int }_{{\left({\mathbb{R}}_{+}-[0,1)\right)}^N}\left({\displaystyle \prod _{i=1}^N}{z}_i^{\overline{{\alpha }_i}}\right)\left[{\displaystyle \prod _{i=1}^N}\left(M_{q,\lambda }+M_{{\displaystyle \frac{1}{q}},\lambda }\right)\left(z_i\right)\right]{\displaystyle \prod _{i=1}^N}d{z}_i\le$$

$${\left(tanh\lambda \right)}^N+{\left(q+{\displaystyle \frac{1}{q}}\right)}^N{\left(2\lambda \right)}^N{e}^{2\lambda N}$$

$${\int }_1^{\infty }\dots {\int }_1^{\infty }\left({\displaystyle \prod _{i=1}^N}{z}_i^{\overline{{\alpha }_i}}\right)\left({\displaystyle \prod _{i=1}^N}{e}^{-2\lambda z_i}\right){\displaystyle \prod _{i=1}^N}d{z}_i=$$

(92)

$${\left(tanh\lambda \right)}^N+{\left(q+{\displaystyle \frac{1}{q}}\right)}^N{\left(2\lambda \right)}^N{e}^{2\lambda N}{\displaystyle \prod _{i=1}^N}{\int }_1^{\infty }{z}_i^{\overline{{\alpha }_i}}{e}^{-2\lambda z_i}d{z}_i=$$

$${\left(tanh\lambda \right)}^N+{\left(q+{\displaystyle \frac{1}{q}}\right)}^N{\left(2\lambda \right)}^N{e}^{2\lambda N}{\displaystyle \prod _{i=1}^N}{\displaystyle \frac{1}{{\left(2\lambda \right)}^{\overline{{\alpha }_i}+1}}}{\int }_{2\lambda }^{\infty }{y}_i^{\left(\overline{{\alpha }_i}+1\right)-1}{e}^{-y_i}d{y}_i=$$

$${\left(tanh\lambda \right)}^N+{\left(q+{\displaystyle \frac{1}{q}}\right)}^N{\left(2\lambda \right)}^N{e}^{2\lambda N}{\displaystyle \frac{1}{{\left(2\lambda \right)}^{m+N}}}{\displaystyle \prod _{i=1}^N}\mathrm{\Gamma }\left(\overline{{\alpha }_i}+1,2\lambda \right)=$$

$${\left(tanh\lambda \right)}^N+{\displaystyle \frac{{\left(q+\frac{1}{q}\right)}^N{e}^{2\lambda N}}{{\left(2\lambda \right)}^m}}{\displaystyle \prod _{i=1}^N}\mathrm{\Gamma }\left(\overline{{\alpha }_i}+1,2\lambda \right)<+\infty .$$

□

5. Main Results

We construct

Definition 3. 

Let $f:{\mathbb{R}}^N\to \mathbb{R}$ be a Borel measurable function, we follow (1)–(5). We define

$${\theta }_{1n}\left(f;x_1,\dots ,x_N\right):={\theta }_{1,r,n}^{\left[m\right]}\left(f;x_1,\dots ,x_N\right):=$$

$${\displaystyle \sum _{j=0}^r}{\alpha }_{j,r}^{\left[m\right]}{\displaystyle \frac{1}{{\xi }_n^N}}{\int }_{{\mathbb{R}}^N}f\left(x_1+s_{1}j,x_2+s_{2}j,\dots ,x_N+s_{N}j\right)Z_{q,\lambda }\left({\displaystyle \frac{x}{{\xi }_n}}\right)dx.$$

(93)

The above multiple smooth singular integral operators are a special case of (5).

In this section, we study the approximation properties of ${\theta }_{1n}$, $n\in \mathbb{N}$.

So, we will apply the results of Section 2 using Section 3 and Section 4.

We present

Theorem 21. 

Let $m\in \mathbb{N}$, $f\in C^m\left({\mathbb{R}}^N\right)$, $N\ge 1$, $x\in {\mathbb{R}}^N$. Assume ${∥{\displaystyle \frac{{\partial }^{m}f\left(·,\dots ,·\right)}{\partial x_1^{\overline{{\alpha }_1}}\dots \partial x_N^{\overline{{\alpha }_N}}}}∥}_{\infty }<\infty ,$ for all $\overline{{\alpha }_j}\in {\mathbb{Z}}^{+},$ $j=1,\dots ,N;$ ${\displaystyle \sum _{j=1}^N}\overline{{\alpha }_j}=m.$

Let ${\xi }_n\in (0,1]$, $n\in \mathbb{N}$, and the Borel probability measure on ${\mathbb{R}}^N$ denoted by ${\overline{\mu }}_{{\xi }_n}$, such that

$$d{\overline{\mu }}_{{\xi }_n}\left(x\right)={\displaystyle \frac{1}{{\xi }_n^N}}{Z}_{q,\lambda }\left({\displaystyle \frac{x}{{\xi }_n}}\right)dx,\forall x\in {\mathbb{R}}^N.$$

For all $\alpha :=\left(\overline{{\alpha }_1},\dots ,\overline{{\alpha }_N}\right)$, $\overline{{\alpha }_i}\in {\mathbb{Z}}^{+},$ $i=1,\dots ,N;$ $\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}\overline{{\alpha }_i}=m,$, we denote

$$U_{{\xi }_n}\left(\alpha \right):={\displaystyle \frac{1}{{\left(2{\xi }_n\right)}^N}}{\int }_{{\mathbb{R}}^N}\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{\overline{{\alpha }_i}}\right){\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^r{Z}_{q,\lambda }\left({\displaystyle \frac{s}{{\xi }_n}}\right)d{s}_1\dots d{s}_N.$$

(94)

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

$${\overline{c}}_{\alpha ,n}:={\overline{c}}_{\alpha ,n,\tilde{j}}:={\displaystyle \frac{1}{{\xi }_n^N}}{\int }_{{\mathbb{R}}^N}{\displaystyle \prod _{i=1}^N}{s}_i^{\overline{{\alpha }_i}}{Z}_{q,\lambda }\left({\displaystyle \frac{s}{{\xi }_n}}\right)ds.$$

(95)

Then

(i) 

$$E_{1,r,n}^{\left[m\right]}\left(x\right):=\left|{\theta }_{1,r,n}^{\left[m\right]}\left(f;x\right)-f\left(x\right)-{\displaystyle \sum _{\tilde{j}=1}^m}{\delta }_{\tilde{j},r}^{\left[m\right]}\left(\sum _{\begin{array}{c}{\overline{\alpha }}_1,\dots ,{\overline{\alpha }}_N\ge 0:\\ \left|\alpha \right|=\tilde{j}\end{array}}{\displaystyle \frac{{\overline{c}}_{\alpha ,n,\tilde{j}}{f}_{\alpha }\left(x\right)}{{\displaystyle \prod _{i=1}^N}{\overline{\alpha }}_i!}}\right)\right|\le$$

$$2^N\sum _{\left(\begin{array}{c}{\overline{\alpha }}_1,\dots ,{\overline{\alpha }}_N\ge 0\\ \left|\alpha \right|=m\end{array}\right)}{\displaystyle \frac{\left({\omega }_r\left(f_{\alpha },{\xi }_n\right)\right)}{\left({\displaystyle \prod _{i=1}^N}{\overline{\alpha }}_i!\right)}}{U}_{{\xi }_n}\left(\alpha \right),$$

(96)

∀$x\in {\mathbb{R}}^N$,

(ii) 

$${∥E_{1,r,n}^{\left[m\right]}∥}_{\infty }\le R.H.S.\left(96\right).$$

(97)

Given that ${\xi }_n\to 0$, as $n\to \infty$, we have ${∥E_{1,r,n}^{\left[m\right]}∥}_{\infty }\to 0$ with rates.

(iii) 

It also holds that

$${∥{\theta }_{1,r,n}^{\left[m\right]}\left(f\right)-f∥}_{\infty }\le$$

$${\displaystyle \sum _{\tilde{j}=1}^m}\left|{\delta }_{\tilde{j},r}^{\left[m\right]}\right|\left(\sum _{\left(\begin{array}{c}{\overline{\alpha }}_1,\dots ,{\overline{\alpha }}_N\ge 0\\ \left|\alpha \right|=\tilde{j}\end{array}\right)}{{\displaystyle \frac{\left|{\overline{c}}_{\alpha ,n,\tilde{j}}\right|{∥f_{\alpha }∥}_{\infty }}{{\displaystyle \prod _{i=1}^N}{\overline{\alpha }}_i!}}}_{\infty }\right)+R.H.S.\left(96\right),$$

(98)

given that ${∥f_{\alpha }∥}_{\infty }<\infty ,$ for all $\alpha :\left|\alpha \right|=\tilde{j}$, $\tilde{j}=1,\dots ,m.$. Also, it holds that

$${∥{\theta }_{r,n}^{\left[m\right]}\left(f\right)-f∥}_{\infty }\to 0$$

with rates, as ${\xi }_n\to 0,$ $n\to \infty$.

Proof. 

By Theorems 1, 16, 20. □

We continue with

Theorem 22. 

Let $f\in C_B\left({\mathbb{R}}^N\right)$, $N\ge 1$. Then

$${∥{\theta }_{1,r,n}^{\left[0\right]}f-f∥}_{\infty }\le 2^N{\mathrm{\Phi }}_{{\xi }_n}^{\ast }{\omega }_r\left(f,{\xi }_n\right),$$

(99)

where

$${\mathrm{\Phi }}_{{\xi }_n}^{\ast }:={\displaystyle \frac{1}{{\left(2{\xi }_n\right)}^N}}{\int }_{{\mathbb{R}}^N}{\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^r{Z}_{q,\lambda }\left({\displaystyle \frac{s}{{\xi }_n}}\right)ds.$$

(100)

Clearly, we have

$${∥{\theta }_{1,r,n}^{\left[0\right]}f-f∥}_{\infty }\to 0,$$

(101)

with rates, given that f is uniformly continuous.

Proof. 

By Theorem 2 and Theorem 17. □

Next comes $L_p$, $p>1$ approximation.

Theorem 23. 

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 $p,q>1:{\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$. Let ${\xi }_n\in (0,1]$, $n\in \mathbb{N}$, and the Borel probability measure on ${\mathbb{R}}^N$, denoted by ${\overline{\mu }}_{{\xi }_n}$, such that

$$d{\overline{\mu }}_{{\xi }_n}\left(x\right)={\displaystyle \frac{1}{{\xi }_n^N}}{Z}_{q,\lambda }\left({\displaystyle \frac{x}{{\xi }_n}}\right)dx,\forall x\in {\mathbb{R}}^N.$$

For all $\alpha :=\left({\overline{\alpha }}_1,\dots ,{\overline{\alpha }}_N\right),$ ${\overline{\alpha }}_i\in {\mathbb{Z}}^{+}$, $i=1,\dots ,N,$ $\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}{\overline{\alpha }}_i=m$, we denote this by

$$V_{{\xi }_n}\left(\alpha \right):={\displaystyle \frac{1}{{\left(2{\xi }_n\right)}^N}}{\int }_{{\mathbb{R}}^N}{\left(\left({\displaystyle \prod _{i=1}^N}{\left|s_i\right|}^{{\alpha }_i}\right){\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^r\right)}^p{Z}_{q,\lambda }\left({\displaystyle \frac{s}{{\xi }_n}}\right)ds.$$

(102)

Here, ${\overline{c}}_{\alpha ,n}:={\overline{c}}_{\alpha ,n,\tilde{j}}$ is as in (95).

Then

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

$${\left({\displaystyle \frac{N^m}{(m-1)!}}\right)}^{{\displaystyle \frac{1}{q}}}\left({\displaystyle \frac{m^{\frac{1}{p}}}{{\left(q\left(m-1\right)+1\right)}^{\frac{1}{q}}}}\right){\left\{\sum _{\left|\alpha \right|=m}{\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\overline{\alpha }}_i!}}{2}^N{V}_{{\xi }_n}\left(\alpha \right){\omega }_r{\left(f_{\alpha },{\xi }_n\right)}_p^p\right\}}^{{\displaystyle \frac{1}{p}}}.$$

(103)

As $n\to \infty$ and ${\xi }_n\to 0$, by (103), we obtain ${∥E_{1,r,n}^{\left[m\right]}∥}_p\to 0$ with rates.

By (103), furthermore, we obtain

$${∥{\theta }_{1,r,n}^{\left[m\right]}\left(f;x\right)-f\left(x\right)∥}_{p,x}\le$$

$${\displaystyle \sum _{\tilde{j}=1}^m}\left|{\delta }_{\tilde{j},r}^{\left[m\right]}\right|\left(\sum _{\left|\alpha \right|=\tilde{j}}{\displaystyle \frac{\left|{\overline{c}}_{\alpha ,n,\tilde{j}}\right|}{{\displaystyle \prod _{i=1}^N}{\overline{\alpha }}_i!}}{∥f_{\alpha }∥}_p\right)+R.H.S.\left(103\right),$$

(104)

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

Indeed, here, ${∥{\theta }_{1,r,n}^{\left[m\right]}\left(f\right)-f∥}_p\to 0$, that is ${\theta }_{1,r,n}^{\left[m\right]}\to I$ the unit operator, in $L_p$ norm, with rates.

Proof. 

By Theorems 3, 18, 20. □

We especially have

Theorem 24. 

Let $f\in \left(C\left({\mathbb{R}}^N\right)\cap L_p\left({\mathbb{R}}^N\right)\right)$; $N\ge 1$; $p,q>1:{\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$; ${\xi }_n\in (0,1]$, $n\in \mathbb{N}$. Set

$$E_{{\xi }_n}:={\displaystyle \frac{1}{{\left(2{\xi }_n\right)}^n}}{\int }_{{\mathbb{R}}^N}{\left(1+{\displaystyle \frac{{∥s∥}_2}{{\xi }_n}}\right)}^{rp}{Z}_{q,\lambda }\left({\displaystyle \frac{s}{{\xi }_n}}\right)ds.$$

(105)

Then

$${∥{\theta }_{1,r,n}^{\left[0\right]}\left(f\right)-f∥}_p\le 2^{{\displaystyle \frac{N}{p}}}{E}_{{\xi }_n}^{{\displaystyle \frac{1}{p}}}{\omega }_r{\left(f,{\xi }_n\right)}_p.$$

(106)

As ${\xi }_n\to 0$, when $n\to \infty$, we obtain ${∥{\theta }_{1,r,n}^{\left[0\right]}\left(f\right)-f∥}_p\to 0$, that is ${\theta }_{1,r,n}^{\left[0\right]}\to I$, the unit operator, in $L_p$ norm.

Proof. 

By Theorems 4, 19. □

The $L_1$ approximation follows.

Theorem 25. 

Let $f\in C^m\left({\mathbb{R}}^N\right)$, $m,N\in \mathbb{N}$, with $f_{\alpha }\in L_1\left({\mathbb{R}}^N\right),$ $\left|\alpha \right|=m$, $x\in {\mathbb{R}}^N.$ Here ${\xi }_n\in (0,1]$, $n\in \mathbb{N}$, and the Borel probability measure on ${\mathbb{R}}^N$, denoted by ${\overline{\mu }}_{{\xi }_n}$:

$$d{\overline{\mu }}_{{\xi }_n}\left(x\right)={\displaystyle \frac{1}{{\xi }_n^N}}{Z}_{q,\lambda }\left({\displaystyle \frac{x}{{\xi }_n}}\right)dx,\forall x\in {\mathbb{R}}^N.$$

The $U_{{\xi }_n}\left(\alpha \right)$ is as in (94), and ${\overline{c}}_{\alpha ,n,\tilde{j}}$, as in (95).

Then

$${∥E_{1,r,n}^{\left[m\right]}∥}_1={∥{\theta }_{1,r,n}^{\left[m\right]}\left(f;x\right)-f\left(x\right)-{\displaystyle \sum _{\tilde{j}=1}^m}{\delta }_{\tilde{j},r}^{\left[m\right]}\left(\sum _{\left|\alpha \right|=\tilde{j}}{\displaystyle \frac{{\overline{c}}_{\alpha ,n,\tilde{j}}{f}_{\alpha }\left(x\right)}{{\displaystyle \prod _{i=1}^N}{\overline{\alpha }}_i!}}\right)∥}_{1,x}\le$$

(107)

$$2^N\sum _{\left|\alpha \right|=m}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\overline{\alpha }}_i!}}\right){\omega }_r{\left(f_{\alpha },{\xi }_n\right)}_1{U}_{{\xi }_n}\left(\alpha \right).$$

As ${\xi }_n\to 0$, we find ${∥E_{1,r,n}^{\left[m\right]}∥}_1\to 0$ with rates.

From (107), we derive

$${∥{\theta }_{1,r,n}^{\left[m\right]}f-f∥}_1\le$$

(108)

$${\displaystyle \sum _{\tilde{j}=1}^m}\left|{\delta }_{\tilde{j},r}^{\left[m\right]}\right|\left(\sum _{\left|\alpha \right|=\tilde{j}}{\displaystyle \frac{\left|{\overline{c}}_{\alpha ,n,\tilde{j}}\right|}{{\displaystyle \prod _{i=1}^N}{\overline{\alpha }}_i!}}{∥f_{\alpha }∥}_1\right)+R.H.S.\left(107\right),$$

given that ${∥f_{\alpha }∥}_1<\infty ,$ $\left|\alpha \right|=\tilde{j}$, $\tilde{j}=1,\dots ,m.$

As $n\to \infty$, and ${\xi }_n\to 0$, we derive that ${∥{\theta }_{1,r,n}^{\left[m\right]}f-f∥}_1\to 0$, that is ${\theta }_{1,r,n}^{\left[m\right]}\to I$ in $L_1$ norm, with rates.

Proof. 

By Theorems 5, 16, 20. □

Theorem 26. 

Let $f\in \left(C\left({\mathbb{R}}^N\right)\cap L_1\left({\mathbb{R}}^N\right)\right)$, $N\ge 1;$ ${\xi }_n\in (0,1]$, $n\in \mathbb{N}.$ Here, ${\mathrm{\Phi }}_{{\xi }_n}^{\ast }$ is as in (76).

Then

$${∥{\theta }_{1,r,n}^{\left[0\right]}\left(f\right)-f∥}_1\le 2^N{\mathrm{\Phi }}_{{\xi }_n}^{\ast }{\omega }_r{\left(f,{\xi }_n\right)}_1.$$

(109)

As ${\xi }_n\to 0$, we obtain ${\theta }_{1,r,n}^{\left[0\right]}\to I$ in $L_1$ norm.

Proof. 

By Theorems 6, 17. □

We continue with global smoothness preservation.

Theorem 27. 

We suppose ${\theta }_{1,r,n}^{\left[\tilde{m}\right]}\left(f;x\right)\in \mathbb{R}$, ∀$x\in \mathbb{R}$. Let $h>0$, $f\in C\left({\mathbb{R}}^N\right)$, $N\ge 1$.

(i) 

Assume ${\omega }_m\left(f,h\right)<\infty$. Then

$${\omega }_m\left({\theta }_{1,r,n}^{\left[\tilde{m}\right]}f,h\right)\le \left({\displaystyle \sum _{\tilde{j}=0}^r}\left|{\alpha }_{\tilde{j},r}^{\left[\tilde{m}\right]}\right|\right){\omega }_m\left(f,h\right).$$

(110)

(ii) 

Assume $f\in \left(C\left({\mathbb{R}}^N\right)\cap L_1\left({\mathbb{R}}^N\right)\right).$ Then

$${\omega }_m{\left({\theta }_{1,r,n}^{\left[\tilde{m}\right]}f,h\right)}_1\le \left({\displaystyle \sum _{\tilde{j}=0}^r}\left|{\alpha }_{\tilde{j},r}^{\left[\tilde{m}\right]}\right|\right){\omega }_m{\left(f,h\right)}_1.$$

(111)

(iii) 

Assume $f\in \left(C\left({\mathbb{R}}^N\right)\cap L_p\left({\mathbb{R}}^N\right)\right)$, $p>1.$ Then

$${\omega }_m{\left({\theta }_{1,r,n}^{\left[\tilde{m}\right]}f,h\right)}_p\le \left({\displaystyle \sum _{\tilde{j}=0}^r}\left|{\alpha }_{\tilde{j},r}^{\left[\tilde{m}\right]}\right|\right){\omega }_m{\left(f,h\right)}_p.$$

(112)

Proof. 

By Theorem 7. □

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

Let the Borel probability measure on ${\mathbb{R}}^N$, denoted by ${\overline{\mu }}_{{\xi }_n}$:

$$d{\overline{\mu }}_{{\xi }_n}\left(x\right)={\displaystyle \frac{1}{{\xi }_n^N}}{Z}_{q,\lambda }\left({\displaystyle \frac{x}{{\xi }_n}}\right)dx,\forall x\in {\mathbb{R}}^N.$$

We detail the simultaneous smoothness results.

Theorem 28. 

Let $h>0$, and the assumptions of Theorem 8 regarding ${\overline{\mu }}_{{\xi }_n}$ are valid. Here, $\gamma =0,\beta$, ($0=\left(0,\dots ,0\right)$).

(i) 

Assume ${\omega }_m\left(f_{\gamma },h\right)<\infty$. Then

$${\omega }_m\left({\left({\theta }_{1,r,n}^{\left[\tilde{m}\right]}\left(f\right)\right)}_{\gamma },h\right)\le \left({\displaystyle \sum _{\tilde{j}=0}^r}\left|{\alpha }_{\tilde{j},r}^{\left[\tilde{m}\right]}\right|\right){\omega }_m\left(f_{\gamma },h\right).$$

(113)

(ii) 

Additionally, assume $f_{\gamma }\in L_1\left({\mathbb{R}}^N\right).$ Then

$${\omega }_m{\left({\left({\theta }_{1,r,n}^{\left[\tilde{m}\right]}\left(f\right)\right)}_{\gamma },h\right)}_1\le \left({\displaystyle \sum _{\tilde{j}=0}^r}\left|{\alpha }_{\tilde{j},r}^{\left[\tilde{m}\right]}\right|\right){\omega }_m{\left(f_{\gamma },h\right)}_1.$$

(114)

(iii) 

Additionally, assume $f_{\gamma }\in L_p\left({\mathbb{R}}^N\right)$, $p>1.$ Then

$${\omega }_m{\left({\left({\theta }_{1,r,n}^{\left[\tilde{m}\right]}\left(f\right)\right)}_{\gamma },h\right)}_p\le \left({\displaystyle \sum _{\tilde{j}=0}^r}\left|{\alpha }_{\tilde{j},r}^{\left[\tilde{m}\right]}\right|\right){\omega }_m{\left(f_{\gamma },h\right)}_p.$$

(115)

Proof. 

By Theorem 9 □

Simultaneous approximation follows.

Theorem 29. 

Let $f\in C^{m+l}\left({\mathbb{R}}^N\right)$, $m,l,N\in \mathbb{N}$. The assumptions of Theorem 8 are valid with respect to ${\overline{\mu }}_{{\xi }_n}$, ${\xi }_n\in (0,1]$, $n\in \mathbb{N}$. Call $\gamma =0,\beta$, assume ${∥f_{\gamma +\alpha }∥}_{\infty }\le \infty ;$ $\left|\alpha \right|:={\displaystyle \sum _{j=1}^N}{\overline{\alpha }}_j=m.$

Here, ${\overline{c}}_{\alpha ,n,\tilde{j}}$, $\tilde{j}=1,\dots ,m$, is as in (95); and $U_{{\xi }_n}$ is as in (71).

Then

$${∥{\left({\theta }_{1,r,n}^{\left[m\right]}\left(f;·\right)\right)}_{\gamma }-f_{\gamma }\left(·\right)-{\displaystyle \sum _{\tilde{j}=1}^m}{\delta }_{\tilde{j},r}^{\left[m\right]}\left(\sum _{\begin{array}{c}{\overline{\alpha }}_1,\dots ,{\overline{\alpha }}_N\ge 0:\\ \left|\alpha \right|=\tilde{j}\end{array}}{\displaystyle \frac{{\overline{c}}_{\alpha ,n,\tilde{j}}{f}_{\gamma +\alpha }\left(·\right)}{{\displaystyle \prod _{i=1}^N}{\overline{\alpha }}_i!}}\right)∥}_{\infty }$$

$$\le 2^N\sum _{\left(\begin{array}{c}{\overline{\alpha }}_1,\dots ,{\overline{\alpha }}_N\ge 0\\ \left|\alpha \right|=m\end{array}\right)}{\displaystyle \frac{{\omega }_r\left(f_{\gamma +\alpha },{\xi }_n\right)}{\left({\displaystyle \prod _{i=1}^N}{\overline{\alpha }}_i!\right)}}{U}_{{\xi }_n}\left(\alpha \right).$$

(116)

Proof. 

By Theorems 8, 10, 16, 20. □

Theorem 30. 

Let $f\in C_B^l\left({\mathbb{R}}^N\right)$, $l,N\in \mathbb{N}$. The assumptions of Theorem 8 are valid for ${\overline{\mu }}_{{\xi }_n}$, ${\xi }_n\in (0,1]$, $n\in \mathbb{N}$. Call $\gamma =0,\beta$. Here, ${\mathrm{\Phi }}_{{\xi }_n}^{\ast }$ is as in (76).

Then

$${∥{\left({\theta }_{1,r,n}^{\left[0\right]}f\right)}_{\gamma }-f_{\gamma }∥}_{\infty }\le 2^N{\mathrm{\Phi }}_{{\xi }_n}^{\ast }{\omega }_r\left(f_{\gamma },{\xi }_n\right).$$

(117)

Proof. 

By Theorems 8, 11, 17. □

The $L_p$, $p>1$ simultaneous approximation follows:

Theorem 31. 

Let $f\in C^{m+l}\left({\mathbb{R}}^N\right),$ $m,l,N\in \mathbb{N}.$ The assumptions of Theorem 8 are valid regarding ${\overline{\mu }}_{{\xi }_n}$, ${\xi }_n\in (0,1]$, $n\in \mathbb{N}$. Call $\gamma =0,\beta$. Let $f_{\left(\gamma +\alpha \right)}\in L_p\left({\mathbb{R}}^N\right)$, $\left|\alpha \right|=m$, $x\in {\mathbb{R}}^N$, and $p,q>1:{\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1.$ Here, $V_{{\xi }_n}\left(\alpha \right)$ is as in (81), and ${\overline{c}}_{\alpha ,n,\tilde{j}},$ $\tilde{j}=1,\dots ,m$, is as in (95).

Then

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

(118)

$${\left({\displaystyle \frac{N^m}{\left(m-1\right)!}}\right)}^{{\displaystyle \frac{1}{q}}}\left({\displaystyle \frac{m^{\frac{1}{p}}}{{\left(q\left(m-1\right)+1\right)}^{\frac{1}{q}}}}\right)$$

$${\left(\sum _{\left|\alpha \right|=m}{\displaystyle \frac{1}{\left({\displaystyle \prod _{i=1}^N}{\overline{\alpha }}_i!\right)}}{V}_{{\xi }_n}\left(\alpha \right){\omega }_r{\left(f_{\gamma +\alpha },{\xi }_n\right)}_p^p\right)}^{{\displaystyle \frac{1}{p}}}.$$

Proof. 

By Theorems 8, 12, 18, 20. □

Theorem 32. 

Let $f\in C^l\left({\mathbb{R}}^N\right),$ $l,N\in \mathbb{N}.$ The assumptions of Theorem 8 are valid with respect to ${\overline{\mu }}_{{\xi }_n}$, ${\xi }_n\in (0,1]$, $n\in \mathbb{N}$. Call $\gamma =0,\beta$. Let $f_{\gamma }\in L_p\left({\mathbb{R}}^N\right)$, $x\in {\mathbb{R}}^N$; $p,q>1:{\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1.$ Here, $E_{{\xi }_n}$ is as in (86). Then

$${∥{\left({\theta }_{1,r,n}^{\left[0\right]}f\right)}_{\gamma }-f_{\gamma }∥}_p\le 2^{{\displaystyle \frac{N}{p}}}{\left(E_{{\xi }_n}\right)}^{{\displaystyle \frac{1}{p}}}{\omega }_r{\left(f_{\gamma },{\xi }_n\right)}_p.$$

(119)

Proof. 

By Theorems 8, 13, 19. □

We conclude with $L_1$ simultaneous approximation.

Theorem 33. 

Let $f\in C^l\left({\mathbb{R}}^N\right),$ $l,N\in \mathbb{N}.$ The assumptions of Theorem 8 are valid for ${\overline{\mu }}_{{\xi }_n}$, ${\xi }_n\in (0,1]$, $n\in \mathbb{N}$. Call $\gamma =0,\beta$. Let $f_{\gamma }\in L_1\left({\mathbb{R}}^N\right)$, $x\in {\mathbb{R}}^N.$ Here ${\mathrm{\Phi }}_{{\xi }_n}^{\ast }$ is as in (76). Then

$${∥{\left({\theta }_{1,r,n}^{\left[0\right]}\left(f\right)\right)}_{\gamma }-f_{\gamma }∥}_1\le 2^N{\mathrm{\Phi }}_{{\xi }_n}^{\ast }{\omega }_r{\left(f_{\gamma },{\xi }_n\right)}_1.$$

(120)

Proof. 

By Theorems 8, 14, 17. □

Theorem 34. 

Let $f\in C^{m+l}\left({\mathbb{R}}^N\right)$, $m,l,N\in \mathbb{N}$. The assumptions of Theorem 8 are valid for ${\overline{\mu }}_{{\xi }_n}$, ${\xi }_n\in (0,1]$, $n\in \mathbb{N}$. Call $\gamma =0,\beta$. Let $f_{\left(\gamma +\alpha \right)}\in L_1\left({\mathbb{R}}^N\right),$ $\left|\alpha \right|=m$, $x\in {\mathbb{R}}^N$. Here $U_{{\xi }_n}\left(\alpha \right)$, $\left|\alpha \right|=m$, is as in (71), and ${\overline{c}}_{\alpha ,n,\tilde{j}}$ is as in (95), $\tilde{j}=1,\dots ,m.$

Then

$${∥{\left({\theta }_{1,r,n}^{\left[m\right]}\left(f;x\right)\right)}_{\gamma }-f_{\gamma }\left(x\right)-{\displaystyle \sum _{\tilde{j}=1}^m}{\delta }_{\tilde{j},r}^{\left[m\right]}\left(\sum _{\left|\alpha \right|=\tilde{j}}{\displaystyle \frac{{\overline{c}}_{\alpha ,n,\tilde{j}}{f}_{\left(\gamma +\alpha \right)}\left(x\right)}{{\displaystyle \prod _{i=1}^N}{\overline{\alpha }}_i!}}\right)∥}_{1,x}\le$$

$$2^N\sum _{\left|\alpha \right|=m}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\overline{\alpha }}_i!}}\right){\omega }_r{\left(f_{\gamma +\alpha },{\xi }_n\right)}_1{U}_{{\xi }_n}\left(\alpha \right).$$

(121)

Proof. 

By Theorems 8, 15, 16, 20. □

6. Conclusions

Here we presented the novel idea of going from the neural networks main tools, the activation functions, to multivariate singular integral approximation. This is a rare case of employing applied mathematics to theoretical ones.