Multivariate Approximation Using Symmetrized and Perturbed Hyperbolic Tangent-Activated Multidimensional Convolution-Type Operators

Abstract

In this article, we introduce, for the first time, multivariate symmetrized and perturbed hyperbolic tangent-activated convolution-type operators in three forms. We present their approximation properties, that is, their quantitative convergence to the unit operator via the multivariate modulus of continuity. We continue with the multivariate global smoothness preservation of these operators. We present, in detail, the related multivariate iterative approximation, as well as, multivariate simultaneous approximation, and their combinations. Using differentiability in our research, we produce higher rates of approximation, and multivariate simultaneous global smoothness preservation is also achieved.

1. Organization

In Section 2 we give the preliminaries of our theory. In Section 3, we present the basics, as well as an introduction to our multivariate symmetrized and perturbed hyperbolic tangent-activated convolution-type operators and their properties. In Section 4, we present the main multivariate approximation results. We also include the multivariate global smoothness preservation achieved by our operators. We further study the differentiation of these operators, and we introduce their iterations and give their basic properties. Next, we present the convergence of our operators under differentiability, achieving higher rates of approximation. Following this, we present the multivariate simultaneous differential approximation and, in detail, the multivariate simultaneous global smoothness preservation, as well as the multivariate iterative approximation. We finish with the combination of multivariate simultaneous and iterative approximations.

We are motivated and inspired by [1,2,3].

Other related recent interesting developments in this field are given in the works [4,5,6,7,8,9,10,11,12,13,14,15,16,17].

2. Symmetrization Related to $\mathit{q}$-Deformed and $\mathit{\lambda }$-Parametrized Hyperbolic Tangent Function ${\mathit{g}}_{\mathit{q},\mathit{\lambda }}$

In this section, all of the initial background information comes from Chapter 18 in [1].

We use $g_{q,\lambda }$ (see (1)), and exhibit that it is a sigmoid function and we will present several of its properties related to approximation by neural network operators.

So, let us consider the activation function

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

(1)

We determine that

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

We also notice that

$$g_{q,\lambda }\left(-x\right)={\displaystyle \frac{e^{-\lambda x}-q{e}^{\lambda x}}{e^{-\lambda x}+q{e}^{\lambda x}}}=-{\displaystyle \frac{\left(e^{\lambda x}-\frac{1}{q}{e}^{-\lambda x}\right)}{e^{\lambda x}+\frac{1}{q}{e}^{-\lambda x}}}=-g_{{\displaystyle \frac{1}{q}},\lambda }\left(x\right).$$

(2)

That is,

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

(3)

and

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

Hence,

$$g_{{\displaystyle \frac{1}{q}},\lambda }^{\prime }\left(x\right)=g_{q,\lambda }^{\prime }\left(-x\right).$$

(4)

We obtain

$$g_{q,\lambda }\left(x\right)={\displaystyle \frac{e^{2\lambda x}-q}{e^{2\lambda x}+q}}={\displaystyle \frac{1-\frac{q}{e^{2^{lx}}}}{1+\frac{q}{e^{2\lambda x}}}}\underset{\left(x\to +\infty \right)}{\to }1,$$

i.e.,

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

(5)

Furthermore,

$$g_{q,\lambda }\left(x\right)={\displaystyle \frac{e^{2\lambda x}-q}{e^{2\lambda x}+q}}\underset{\left(x\to -\infty \right)}{\to }{\displaystyle \frac{-q}{q}}=-1,$$

i.e.,

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

(6)

We find that

$$g_{q,\lambda }^{\prime }\left(x\right)={\displaystyle \frac{4q\lambda e^{2\lambda x}}{{\left(e^{2\lambda x}+q\right)}^2}}>0,$$

(7)

Therefore, $g_{q,\lambda }$ is strictly increasing.

Next, we obtain ($x\in \mathbb{R}$):

$$g_{q,\lambda }^{″}\left(x\right)=8q{\lambda }^2{e}^{2\lambda x}\left({\displaystyle \frac{q-e^{2\lambda x}}{{\left(e^{2\lambda x}+q\right)}^3}}\right)\in C\left(\mathbb{R}\right).$$

(8)

We observe that

$$q-e^{2\lambda x}\gtrless 0\iff q\gtrless e^{2\lambda x}\iff lnq\gtrless 2\lambda x\iff x\lessgtr {\displaystyle \frac{lnq}{2\lambda }}.$$

So, in the case of $x<{\displaystyle \frac{lnq}{2\lambda }}$, we determine that $g_{q,\lambda }$ is strictly concave up, with $g_{q,\lambda }^{″}\left({\displaystyle \frac{lnq}{2\lambda }}\right)=0.$

And in the case of $x>{\displaystyle \frac{lnq}{2\lambda }}$, we determine that $g_{q,\lambda }$ is strictly concave down.

Clearly, $g_{q,\lambda }$ is a shifted sigmoid function with $g_{q,\lambda }\left(0\right)={\displaystyle \frac{1-q}{1+q}}$, and $g_{q,\lambda }\left(-x\right)=-g_{q^{-1},\lambda }\left(x\right)$ (a semi-odd function).

Based on $1>-1$, $x+1>x-1$, we consider the function

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

(9)

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

We determine that

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

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

(10)

$${\displaystyle \frac{1}{4}}\left(g_{{\displaystyle \frac{1}{q}},\lambda }\left(x+1\right)-g_{{\displaystyle \frac{1}{q}},\lambda }\left(x-1\right)\right)=M_{{\displaystyle \frac{1}{q}},\lambda }\left(x\right),\forall x\in \mathbb{R}.$$

Thus,

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

(11)

a deformed symmetry.

Next, we determine that

$$M_{q,\lambda }^{\prime }\left(x\right)={\displaystyle \frac{1}{4}}\left(g_{q,\lambda }^{\prime }\left(x+1\right)-g_{q,\lambda }^{\prime }\left(x-1\right)\right),\forall x\in \mathbb{R}.$$

(12)

Let $x<{\displaystyle \frac{lnq}{2\lambda }}-1$; then, $x-1<x+1<{\displaystyle \frac{lnq}{2\lambda }}$ and $g_{q,\lambda }^{\prime }\left(x+1\right)>g_{q,\lambda }^{\prime }\left(x-1\right)$ (based on $g_{q,\lambda }$ being strictly concave up for $x<{\displaystyle \frac{lnq}{2\lambda }}$), that is, $M_{q,\lambda }^{\prime }\left(x\right)>0$. Hence $M_{q,\lambda }$ is strictly increasing over $\left(-\infty ,{\displaystyle \frac{lnq}{2\lambda }}-1\right).$

Now, let $x-1>{\displaystyle \frac{lnq}{2\lambda }}$; then, $x+1>x-1>{\displaystyle \frac{lnq}{2\lambda }}$, and $g_{q,\lambda }^{\prime }\left(x+1\right)<g_{q,\lambda }^{\prime }\left(x-1\right)$, that is, $M_{q,\lambda }^{\prime }\left(x\right)<0.$

Therefore $M_{q,\lambda }$ is strictly decreasing over $\left({\displaystyle \frac{lnq}{2\lambda }}+1,+\infty \right).$

Let us next consider ${\displaystyle \frac{lnq}{2\lambda }}-1\le x\le {\displaystyle \frac{lnq}{2\lambda }}+1.$ We determine that

$$M_{q,\lambda }^{″}\left(x\right)={\displaystyle \frac{1}{4}}\left(g_{q,\lambda }^{″}\left(x+1\right)-g_{q,\lambda }^{″}\left(x-1\right)\right)=$$

$$2q{\lambda }^2\left[e^{2\lambda \left(x+1\right)}\left({\displaystyle \frac{q-e^{2\lambda \left(x+1\right)}}{{\left(e^{2\lambda \left(x+1\right)}+q\right)}^3}}\right)-e^{2\lambda \left(x-1\right)}\left({\displaystyle \frac{q-e^{2\lambda \left(x-1\right)}}{{\left(e^{2\lambda \left(x-1\right)}+q\right)}^3}}\right)\right].$$

(13)

${\displaystyle \frac{lnq}{2\lambda }}-1\le x\iff {\displaystyle \frac{lnq}{2\lambda }}\le x+1\iff lnq\le 2\lambda \left(x+1\right)\iff q\le e^{2\lambda \left(x+1\right)}\iff q-e^{2\lambda \left(x+1\right)}\le 0.$

$x\le {\displaystyle \frac{lnq}{2\lambda }}+1\iff x-1\le {\displaystyle \frac{lnq}{2\lambda }}\iff 2\lambda \left(x-1\right)\le lnq\iff e^{2\lambda \left(x-1\right)}\le q\iff q-e^{2\lambda \beta \left(x-1\right)}\ge 0.$

Clearly, according to (13), we determine that $M_{q,\lambda }^{″}\left(x\right)\le 0$ for $x\in \left[{\displaystyle \frac{lnq}{2\lambda }}-1,{\displaystyle \frac{lnq}{2\lambda }}+1\right].$

More precisely, $M_{q,\lambda }$ is concave down over $\left[{\displaystyle \frac{lnq}{2\lambda }}-1,{\displaystyle \frac{lnq}{2\lambda }}+1\right]$, and strictly concave down over $\left({\displaystyle \frac{lnq}{2\lambda }}-1,{\displaystyle \frac{lnq}{2\lambda }}+1\right).$

Consequently, $M_{q,\lambda }$ has a bell-type shape over $\mathbb{R}.$

Of course, it holds that $M_{q,\lambda }^{″}\left({\displaystyle \frac{lnq}{2\lambda }}\right)<0.$

At $x={\displaystyle \frac{lnq}{2\lambda }}$, we have

$$M_{q,\lambda }^{\prime }\left(x\right)={\displaystyle \frac{1}{4}}\left(g_{q,\lambda }^{\prime }\left(x+1\right)-g_{q,\lambda }^{\prime }\left(x-1\right)\right)=$$

$$q\lambda \left({\displaystyle \frac{e^{2\lambda \left(x+1\right)}}{{\left(e^{2\lambda \left(x+1\right)}+q\right)}^2}}-{\displaystyle \frac{e^{2\lambda \left(x-1\right)}}{{\left(e^{2\lambda \left(x-1\right)}+q\right)}^2}}\right).$$

(14)

Thus,

$$M_{q,\lambda }^{\prime }\left({\displaystyle \frac{lnq}{2\lambda }}\right)=q\lambda \left({\displaystyle \frac{e^{2\lambda \left(\frac{lnq}{2\lambda }+1\right)}}{{\left(e^{2\lambda \left(\frac{lnq}{2\lambda }+1\right)}+q\right)}^2}}-{\displaystyle \frac{e^{2\lambda \left(\frac{lnq}{2\lambda }-1\right)}}{{\left(e^{2\lambda \left(\frac{lnq}{2\lambda }-1\right)}+q\right)}^2}}\right)=$$

$$\lambda \left({\displaystyle \frac{e^{2\lambda }{\left(e^{-2\lambda }+1\right)}^2-e^{-2\lambda }{\left(e^{2\lambda }+1\right)}^2}{{\left(e^{2\lambda }+1\right)}^2{\left(e^{-2\lambda }+1\right)}^2}}\right)=0.$$

(15)

That is, ${\displaystyle \frac{lnq}{2\lambda }}$ is the only critical number of $M_{q,\lambda }$ over $\mathbb{R}$. Hence, at $x={\displaystyle \frac{lnq}{2\lambda }},$ $M_{q,\lambda }$ achieves its global maximum, which is

$$M_{q,\lambda }\left({\displaystyle \frac{lnq}{2\lambda }}\right)={\displaystyle \frac{1}{4}}\left[g_{q,\lambda }\left({\displaystyle \frac{lnq}{2\lambda }}+1\right)-g_{q,\lambda }\left({\displaystyle \frac{lnq}{2\lambda }}-1\right)\right]=$$

(16)

$${\displaystyle \frac{1}{4}}\left[{\displaystyle \frac{2\left(e^{\lambda }-e^{-\lambda }\right)}{e^{\lambda }+e^{-\lambda }}}\right]={\displaystyle \frac{tanh\left(\lambda \right)}{2}}.$$

Conclusion: The maximum value of $M_{q,\lambda }$ is

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

(17)

We mention the following:

Theorem 1 

([1], Ch. 18, p. 458). We determine that

$$\sum _{i=-\infty }^{\infty }{M}_{q,\lambda }\left(x-i\right)=1,\forall x\in \mathbb{R},\forall \lambda ,q>0.$$

(18)

Also, the following holds:

Theorem 2 

([1], Ch. 18, p. 459). It holds that

$${\int }_{-\infty }^{\infty }{M}_{q,\lambda }\left(x\right)dx=1,\lambda ,q>0.$$

(19)

So that $M_{q,\lambda }$ is a density function on $\mathbb{R};$ $\lambda ,q>0.$

Similarly, we determine that

$${\int }_{-\infty }^{\infty }{M}_{{\displaystyle \frac{1}{q}},\lambda }\left(x\right)dx=1,\lambda ,q>0,$$

(20)

so that $M_{{\displaystyle \frac{1}{q}},\lambda }$ is a density function.

Furthermore, we observe the important symmetry

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

(21)

Furthermore

$$\phi ={\displaystyle \frac{M_{q,\lambda }+M_{\frac{1}{q},\lambda }}{2}}$$

(22)

is a new density function over $\mathbb{R}$, i.e.,

$${\int }_{-\infty }^{\infty }\phi \left(x\right)dx=1.$$

(23)

Clearly, then,

$${\int }_{-\infty }^{\infty }\phi \left(nx-u\right)du=1,\forall n\in \mathbb{N},x\in \mathbb{R}.$$

(24)

An essential property follows:

Theorem 3 

([2]). Let $0<\alpha <1$, $n\in \mathbb{N}:n^{1-\alpha }>2$. Then,

$$\underset{\left\{u\in \mathbb{R}:\left|nx-u\right|\ge n^{1-\alpha }\right\}}{\int }\phi \left(nx-u\right)du<{\displaystyle \frac{\left(q+\frac{1}{q}\right)}{e^{2\lambda \left(n^{1-\alpha }-1\right)}}},q,\lambda >0.$$

(25)

We need the following:

Proposition 1 

([2]). It holds for ($k\in \mathbb{N}$) that

$${\int }_{-\infty }^{\infty }{\left|z\right|}^k\phi \left(z\right)dz\le \left[{\displaystyle \frac{tanh\left(\lambda \right)}{\left(k+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }k!}{{\left(2\lambda \right)}^k}}\right]<\infty .$$

(26)

We mention the following:

Definition 1. 

The modulus of continuity here is defined by

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

(27)

where $f:{\mathbb{R}}^N\to \mathbb{R}$ is bounded and continuous, denoted by $f\in C_B\left({\mathbb{R}}^N\right)$, $N\in \mathbb{N}$. Similarly, ${\omega }_1$ is defined for $f\in C_U\left({\mathbb{R}}^N\right)$ (uniformly continuous functions). We determine that $f\in C_U\left({\mathbb{R}}^N\right)$, iff ${\omega }_1\left(f,\delta \right)\to 0$ as $\delta \downarrow 0.$

Denote ${∥x∥}_{\infty }:=max\left\{\left|x_1\right|,\dots ,\left|x_N\right|\right\}$, $x\in {\mathbb{R}}^N.$

3. Basics

We establish the following:

Remark 1. 

We introduce

$$Z\left(x_1,\dots ,x_N\right):=Z\left(x\right):=\prod _{i=1}^N\phi \left(x_i\right),x=\left(x_1,\dots ,x_N\right)\in {\mathbb{R}}^N,N\in \mathbb{N}.$$

(28)

It has the following properties:

(i) 

$Z\left(x\right)>0$, ∀$x\in {\mathbb{R}}^N;$$Z\left(-x\right)=Z\left(x\right),$

(ii) 

$${\int }_{{\mathbb{R}}^N}Z\left(x-u\right)du={\int }_{-\infty }^{\infty }\dots {\int }_{-\infty }^{\infty }Z\left(x_1-u_1,\dots ,x_N-u_N\right)d{u}_1\dots d{u}_N=$$

$$\prod _{i=1}^N{\int }_{-\infty }^{\infty }\phi \left(x_i-u_i\right)d{u}_i\stackrel{\left(24\right)}{=}1,\forall x\in {\mathbb{R}}^N,$$

(29)

Hence,

(iii) 

$${\int }_{{\mathbb{R}}^N}Z\left(nx-u\right)du=1,\forall x\in {\mathbb{R}}^N,n\in \mathbb{N},\mathit{and}$$

(30)

(iv) 

according to (23),

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

(31)

That is, Z is a multivariate density function.

(v) 

Let $0<\beta <1$, $n\in \mathbb{N}:n^{1-\beta }>2$. Then, using Theorem 3 and (24), we derive that

$$\underset{\left\{u\in {\mathbb{R}}^N:{∥nx-u∥}_{\infty }\ge n^{1-\beta }\right\}}{\int }Z\left(nx-u\right)du<{\displaystyle \frac{\left(q+\frac{1}{q}\right)}{e^{e^{2\lambda }\left(n^{1-\beta }-1\right)}}},q,\lambda >0.$$

(32)

The latter is true, because the condition any $u\in {\mathbb{R}}^N:{∥nx-u∥}_{\infty }\ge n^{1-\beta }$, implies that there exists at least one $u_r:\left|n{x}_r-u_r\right|\ge n^{1-\beta }$, where $r\in \left\{1,\dots ,N\right\}.$

Indeed, the following holds for some $r^{*}\in \left\{1,\dots ,N\right\}.$

$$\left\{u\in {\mathbb{R}}^N:{∥nx-u∥}_{\infty }\ge n^{1-\beta }\right\}\subset$$

(33)

$${\cup }_{r=1}^N\left\{{\mathbb{R}}^{N-1}\cup \left\{u_r\in \mathbb{R}:\left|n{x}_r-u_r\right|\ge n^{1-\beta }\right\}\right\}\subset$$

$${\mathbb{R}}^{N-1}\cup \left\{u_{r^{*}}\in \mathbb{R}:\left|n{x}_{r^{*}}-u_{r^{*}}\right|\ge n^{1-\beta }\right\},$$

We also mention a useful related result.

Theorem 4. 

It holds for ($k\in \mathbb{N}$) that

$${\int }_{{\mathbb{R}}^N}{∥x∥}_{\infty }^{k}Z\left(x\right)dx\le N^k\left[{\displaystyle \frac{tanh\left(\lambda \right)}{\left(k+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }k!}{{\left(2\lambda \right)}^k}}\right]<\infty .$$

(34)

When $k=0$, (34) is again valid.

Proof. 

We determine that

$${\int }_{{\mathbb{R}}^N}{∥x∥}_{\infty }^{k}Z\left(x\right)dx\le {\int }_{{\mathbb{R}}^N}{\left(\sum _{j=1}^N\left|x_j\right|\right)}^{k}Z\left(x\right)dx\le$$

(according to a convexity argument)

$${\int }_{{\mathbb{R}}^N}{N}^{k-1}\left(\sum _{j=1}^N{\left|x_j\right|}^k\right)Z\left(x\right)dx=N^{k-1}\left(\sum _{j=1}^N{\int }_{{\mathbb{R}}^N}{\left|x_j\right|}^{k}Z\left(x\right)dx\right)=$$

(35)

$$N^{k-1}\left(\sum _{j=1}^N{\int }_{{\mathbb{R}}^N}{\left|x_j\right|}^k\prod _{i=1}^N\phi \left(x_i\right)dx\right)=$$

$$N^{k-1}\left(\sum _{j=1}^N\left[\left({\int }_{-\infty }^{\infty }{\left|x_j\right|}^k\phi \left(x_j\right)d{x}_j\right)\prod _{\begin{array}{c}i=1\\ i\ne j\end{array}}^N{\int }_{-\infty }^{\infty }\phi \left(x_i\right)dx\right]\right)\stackrel{\left(23\right)}{=}$$

$$N^{k-1}\left(\sum _{j=1}^N{\int }_{-\infty }^{\infty }{\left|x_j\right|}^k\phi \left(x_j\right)d{x}_j\right)\stackrel{\left(26\right)}{\le }$$

$$N^{k-1}\left(\sum _{j=1}^N\left[{\displaystyle \frac{tanh\left(\lambda \right)}{\left(k+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }k!}{{\left(2\lambda \right)}^k}}\right]\right)=$$

$$N^k\left[{\displaystyle \frac{tanh\left(\lambda \right)}{\left(k+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }k!}{{\left(2\lambda \right)}^k}}\right],$$

(36)

proving the claim. □

We will use the following differentiation result.

Theorem 5 

(H. Bauer [18], pp. 103–104). Let $\left(\Omega ,A,\mu \right)$ be a measure space. Let U be an open subset of ${\mathbb{R}}^d$, $d\ge 1$, and let $f:U\times \Omega \to \mathbb{R}$ be a function with the following properties:

(a) 

$\omega \to f\left(x,\omega \right)$ is μ-integrable for all $x\in U$.

(b) 

$x\to f\left(x,\omega \right)$ is, at each point in U, partially differentiable with respect to $x_i$.

(c) 

There exists a μ-integrable function $h\ge 0$ on Ω such that

$$\left|{\displaystyle \frac{\partial f}{\partial x_i}}\left(x,\omega \right)\right|\le h\left(\omega \right),\mathit{for}\mathit{all}\left(x,\omega \right)\in U\times \Omega .$$

Then, the function $\overline{\phi }$, defined on U as

$$\overline{\phi }\left(x\right)={\int }_{\Omega }f\left(x,\omega \right)\mu \left(d\omega \right)$$

is partially differentiable with respect to $x_i$ on all of U. The mapping $\omega \to {\displaystyle \frac{\partial f}{\partial x_i}}\left(x,\omega \right)$ is μ-integrable, and we have

$${\displaystyle \frac{\partial \overline{\phi }}{\partial x_i}}\left(x\right)={\int }_{\Omega }{\displaystyle \frac{\partial f}{\partial x_i}}\left(x,\omega \right)\mu \left(d\omega \right),\mathit{all}x\in U.$$

We give the following:

Definition 2. 

Let $f\in C_B\left({\mathbb{R}}^N\right)$, $N\in \mathbb{N}$. We define the following activated symmetrized and perturbed hyperbolic tangent multivariate convolution-type operators:

The basic one:

$$S_n\left(f\right)\left(x\right):={\int }_{{\mathbb{R}}^N}f\left({\displaystyle \frac{u}{n}}\right)Z\left(nx-u\right)du,\forall x\in {\mathbb{R}}^N,$$

(37)

The activated Kantorovich type:

$$S_n^{*}\left(f\right)\left(x\right):=n^N{\int }_{{\mathbb{R}}^N}\left({\int }_{{\displaystyle \frac{u}{n}}}^{{\displaystyle \frac{u+1}{n}}}f\left(t\right)dt\right)Z\left(nx-u\right)du,\forall x\in {\mathbb{R}}^N.$$

(38)

Now, let $\theta =\left({\theta }_1,\dots ,{\theta }_N\right)\in {\mathbb{N}}^N$, $r=\left(r_1,\dots ,r_N\right)\in {\mathbb{Z}}_{+}^N$, $w_r=w_{r_1{r}_2\dots r_N}\ge 0$, such that

$$\sum _{r=0}^{\theta }{w}_r=\sum _{r_1=0}^{{\theta }_1}\sum _{r_2=0}^{{\theta }_2}\dots \sum _{r_N=0}^{{\theta }_N}{w}_{r_1{r}_2\dots r_N}=1;u\in {\mathbb{R}}^N,$$

and

$${\delta }_n\left(f\right)\left(u\right):={\delta }_n\left(f\right)\left(u_1,\dots ,u_N\right):=\sum _{r=0}^{\theta }{w}_{r}f\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{r}{n\theta }}\right)=$$

$$\sum _{r_1=0}^{{\theta }_1}\sum _{r_2=0}^{{\theta }_2}\dots \sum _{r_N=0}^{{\theta }_N}{w}_{r_1{r}_2\dots r_N}f\left({\displaystyle \frac{u_1}{n}}+{\displaystyle \frac{r_1}{n{\theta }_1}},{\displaystyle \frac{u_2}{n}}+{\displaystyle \frac{r_2}{n{\theta }_2}},\dots ,{\displaystyle \frac{u_N}{n}}+{\displaystyle \frac{r_N}{n{\theta }_N}}\right),$$

(39)

where ${\displaystyle \frac{r}{\theta }}:=\left({\displaystyle \frac{r_1}{{\theta }_1}},{\displaystyle \frac{r_2}{{\theta }_2}},\dots ,{\displaystyle \frac{r_N}{{\theta }_N}}\right).$

We define the activated quadrature operators

$$\overline{S_n}\left(f\right)\left(x\right):=\overline{S_n}\left(f,x_1,\dots ,x_N\right):={\int }_{{\mathbb{R}}^N}{\delta }_n\left(f\right)\left(u\right)Z\left(nx-u\right)du=$$

(40)

$${\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }\dots {\int }_{-\infty }^{\infty }{\delta }_n\left(f\right)\left(u_1,\dots ,u_N\right)\left(\prod _{i=1}^N\phi \left(n{x}_i-u_i\right)\right)d{u}_1\dots d{u}_N,\forall x\in {\mathbb{R}}^N.$$

One can rewrite

$$S_n\left(f\right)\left(x\right)={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }\dots {\int }_{-\infty }^{\infty }f\left({\displaystyle \frac{u_1}{n}},{\displaystyle \frac{u_2}{n}},\dots ,{\displaystyle \frac{u_N}{n}}\right)\left(\prod _{i=1}^N\phi \left(n{x}_i-u_i\right)\right)d{u}_1\dots d{u}_N,$$

(41)

and

$$S_n^{*}\left(f\right)\left(x\right)=n^N{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }\dots {\int }_{-\infty }^{\infty }\left({\int }_{{\displaystyle \frac{u_1}{n}}}^{{\displaystyle \frac{u_1+1}{n}}}{\int }_{{\displaystyle \frac{u_2}{n}}}^{{\displaystyle \frac{u_2+1}{n}}}\dots {\int }_{{\displaystyle \frac{u_N}{n}}}^{{\displaystyle \frac{u_N+1}{n}}}f\left(t_1,\dots ,t_N\right)d{t}_1\dots d{t}_N\right)$$

$$\left(\prod _{i=1}^N\phi \left(n{x}_i-u_i\right)\right)d{u}_1\dots d{u}_N,$$

(42)

$n\in \mathbb{N}$, ∀$x\in {\mathbb{R}}^N.$

For some $f\in C_U\left({\mathbb{R}}^N\right)$, the above operators can exist.

In this work, we study the approximation properties of the operators $S_n$, $S_n^{*}$ and $\overline{S_n}$, especially their convergence to the unit operator $I.$

4. Main Results

We present the following approximation results:

Theorem 6. 

Let $0<\beta <1$, $n\in \mathbb{N}:n^{1-\beta }>2$, $f\in C_B\left({\mathbb{R}}^N\right)$, $n\in \mathbb{N},$ $x\in {\mathbb{R}}^N$. Then,

$$\left|S_n\left(f\right)\left(x\right)-f\left(x\right)\right|\le {\omega }_1\left(f,{\displaystyle \frac{1}{n^{\beta }}}\right)+{\displaystyle \frac{2\left(q+\frac{1}{q}\right){∥f∥}_{\infty }}{e^{2\lambda \left(n^{1-\beta }-1\right)}}}=:{\lambda }^s,$$

(43)

and

$${∥S_n\left(f\right)-f∥}_{\infty }\le {\lambda }^s,$$

(44)

where ${∥·∥}_{\infty }$ is the supremum norm.

So, for $f\in C_{UB}\left({\mathbb{R}}^N\right):=C_U\left({\mathbb{R}}^N\right)\cap C_B\left({\mathbb{R}}^N\right)$, we determine that ${\displaystyle \underset{n\to \infty }{lim}}{S}_n\left(f\right)=f$, pointwise and uniformly.

Proof. 

Call

$$A_1:=\left\{u\in {\mathbb{R}}^N:{∥{\displaystyle \frac{u}{n}}-x∥}_{\infty }<{\displaystyle \frac{1}{n^{\beta }}}\right\},$$

(45)

and

$$A_2:=\left\{u\in {\mathbb{R}}^N:{∥{\displaystyle \frac{u}{n}}-x∥}_{\infty }\ge {\displaystyle \frac{1}{n^{\beta }}}\right\}.$$

(46)

That is, $A_1\cup A_2={\mathbb{R}}^N.$

We determine that

$$\left|S_n\left(f\right)\left(x\right)-f\left(x\right)\right|\stackrel{\left(31\right)}{=}$$

$$\left|{\int }_{{\mathbb{R}}^N}f\left({\displaystyle \frac{u}{n}}\right)Z\left(nx-u\right)du-f\left(x\right){\int }_{{\mathbb{R}}^N}Z\left(nx-u\right)du\right|=$$

$$\left|{\int }_{{\mathbb{R}}^N}\left(f\left({\displaystyle \frac{u}{n}}\right)-f\left(x\right)\right)Z\left(nx-u\right)du\right|\le$$

(47)

$${\int }_{{\mathbb{R}}^N}\left|f\left({\displaystyle \frac{u}{n}}\right)-f\left(x\right)\right|Z\left(nx-u\right)du=$$

$${\int }_{A_1}\left|f\left({\displaystyle \frac{u}{n}}\right)-f\left(x\right)\right|Z\left(nx-u\right)du+{\int }_{A_2}\left|f\left({\displaystyle \frac{u}{n}}\right)-f\left(x\right)\right|Z\left(nx-u\right)du\le$$

$${\int }_{A_1}{\omega }_1\left(f,{∥{\displaystyle \frac{u}{n}}-x∥}_{\infty }\right)Z\left(nx-u\right)du+2{∥f∥}_{\infty }{\int }_{A_2}Z\left(nx-u\right)du\stackrel{\left(32\right)}{\le }$$

(48)

$${\omega }_1\left(f,{\displaystyle \frac{1}{n^{\beta }}}\right){\int }_{A_1}Z\left(nx-u\right)du+2{∥f∥}_{\infty }{\displaystyle \frac{\left(q+\frac{1}{q}\right)}{e^{2\lambda \left(n^{1-\beta }-1\right)}}}\le$$

$${\omega }_1\left(f,{\displaystyle \frac{1}{n^{\beta }}}\right){\int }_{{\mathbb{R}}^N}Z\left(nx-u\right)du+{\displaystyle \frac{2\left(q+\frac{1}{q}\right){∥f∥}_{\infty }}{e^{2\lambda \left(n^{1-\beta }-1\right)}}}=$$

$${\omega }_1\left(f,{\displaystyle \frac{1}{n^{\beta }}}\right)+{\displaystyle \frac{2\left(q+\frac{1}{q}\right){∥f∥}_{\infty }}{e^{2\lambda \left(n^{1-\beta }-1\right)}}}.$$

□

We continue with the following.

Theorem 7. 

This theorem is all as in Theorem 6. Then,

$$\left|S_n^{*}\left(f\right)\left(x\right)-f\left(x\right)\right|\le {\omega }_1\left(f,{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+{\displaystyle \frac{2\left(q+\frac{1}{q}\right){∥f∥}_{\infty }}{e^{2\lambda \left(n^{1-\beta }-1\right)}}}=:{\rho }^s,$$

(49)

and

$${∥S_n^{*}\left(f\right)-f∥}_{\infty }\le {\rho }^s.$$

(50)

For $f\in C_{UB}\left({\mathbb{R}}^N\right),$ we determine that ${\displaystyle \underset{n\to \infty }{lim}}{S}_n^{*}\left(f\right)=f$, pointwise and uniformly.

Proof. 

We notice that

$${\int }_{{\displaystyle \frac{u}{n}}}^{{\displaystyle \frac{u+1}{n}}}f\left(t\right)dt={\int }_{{\displaystyle \frac{u_1}{n}}}^{{\displaystyle \frac{u_1+1}{n}}}{\int }_{{\displaystyle \frac{u_2}{n}}}^{{\displaystyle \frac{u_2+1}{n}}}\dots {\int }_{{\displaystyle \frac{u_N}{n}}}^{{\displaystyle \frac{u_N+1}{n}}}f\left(t_1,\dots ,t_N\right)d{t}_1\dots d{t}_N=$$

$${\int }_0^{{\displaystyle \frac{1}{n}}}{\int }_0^{{\displaystyle \frac{1}{n}}}\dots {\int }_0^{{\displaystyle \frac{1}{n}}}f\left(t_1+{\displaystyle \frac{u_1}{n}},t_2+{\displaystyle \frac{u_2}{n}},\dots ,t_N+{\displaystyle \frac{u_N}{n}}\right)d{t}_1\dots d{t}_N={\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}f\left(t+{\displaystyle \frac{u}{n}}\right)dt.$$

(51)

Thus, it holds that

$$S_n^{*}\left(f\right)\left(x\right)=n^N{\int }_{{\mathbb{R}}^N}\left({\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}f\left(t+{\displaystyle \frac{u}{n}}\right)dt\right)Z\left(nx-u\right)du.$$

(52)

We observe that

$$\left|S_n^{*}\left(f\right)\left(x\right)-f\left(x\right)\right|=$$

$$\left|{\int }_{{\mathbb{R}}^N}\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}f\left(t+{\displaystyle \frac{u}{n}}\right)dt\right)Z\left(nx-u\right)du-f\left(x\right){\int }_{{\mathbb{R}}^N}Z\left(nx-u\right)du\right|=$$

$$\left|{\int }_{{\mathbb{R}}^N}\left(\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}f\left(t+{\displaystyle \frac{u}{n}}\right)dt\right)-f\left(x\right)\right)Z\left(nx-u\right)du\right|=$$

$$\left|{\int }_{{\mathbb{R}}^N}\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left(f\left(t+{\displaystyle \frac{u}{n}}\right)-f\left(x\right)\right)dt\right)Z\left(nx-u\right)du\right|\le$$

$${\int }_{{\mathbb{R}}^N}\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left|f\left(t+{\displaystyle \frac{u}{n}}\right)-f\left(x\right)\right|dt\right)Z\left(nx-u\right)du=$$

(53)

$${\int }_{A_1}\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left|f\left(t+{\displaystyle \frac{u}{n}}\right)-f\left(x\right)\right|dt\right)Z\left(nx-u\right)du+$$

$${\int }_{A_2}\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left|f\left(t+{\displaystyle \frac{u}{n}}\right)-f\left(x\right)\right|dt\right)Z\left(nx-u\right)du\le$$

$${\int }_{A_1}\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}{\omega }_1\left(f,{∥t∥}_{\infty }+{∥{\displaystyle \frac{u}{n}}-x∥}_{\infty }\right)dt\right)Z\left(nx-u\right)du+$$

$$2{∥f∥}_{\infty }\left({\int }_{A_2}Z\left(nx-u\right)du\right)\le$$

$${\omega }_1\left(f,{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right){\int }_{A_1}Z\left(nx-u\right)du+2{∥f∥}_{\infty }{\displaystyle \frac{\left(q+\frac{1}{q}\right)}{e^{2\lambda \left(n^{1-\beta }-1\right)}}}\le$$

(54)

$${\omega }_1\left(f,{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+{\displaystyle \frac{2\left(q+\frac{1}{q}\right){∥f∥}_{\infty }}{e^{2\lambda \left(n^{1-\beta }-1\right)}}}.$$

□

Thus, the following are true:

Theorem 8. 

This theorem is all as in Theorem 6. Then,

$$\left|\overline{S_n}\left(f\right)\left(x\right)-f\left(x\right)\right|\le {\omega }_1\left(f,{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+{\displaystyle \frac{2\left(q+\frac{1}{q}\right){∥f∥}_{\infty }}{e^{2\lambda \left(n^{1-\beta }-1\right)}}}={\rho }^s,$$

(55)

and

$${∥\overline{S_n}\left(f\right)-f∥}_{\infty }\le {\rho }^s.$$

(56)

For $f\in C_{UB}\left({\mathbb{R}}^N\right),$ we determine that ${\displaystyle \underset{n\to \infty }{lim}}\overline{S_n}\left(f\right)=f$, pointwise and uniformly.

Proof. 

We determine that

$$\left|\overline{S_n}\left(f\right)\left(x\right)-f\left(x\right)\right|=$$

$$\left|{\int }_{{\mathbb{R}}^N}{\delta }_n\left(f\right)\left(u\right)Z\left(nx-u\right)du-f\left(x\right){\int }_{{\mathbb{R}}^N}Z\left(nx-u\right)du\right|=$$

$$\left|{\int }_{{\mathbb{R}}^N}\left({\delta }_n\left(f\right)\left(u\right)-f\left(x\right)\right)Z\left(nx-u\right)du\right|=$$

$$\left|{\int }_{{\mathbb{R}}^N}\left(\sum _{r=0}^{\theta }{w}_r\left(f\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{r}{n\theta }}\right)-f\left(x\right)\right)\right)Z\left(nx-u\right)du\right|\le$$

(57)

$${\int }_{{\mathbb{R}}^N}\left(\sum _{r=0}^{\theta }{w}_r\left|f\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{r}{n\theta }}\right)-f\left(x\right)\right|\right)Z\left(nx-u\right)du=$$

$${\int }_{A_1}\left(\sum _{r=0}^{\theta }{w}_r\left|f\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{r}{n\theta }}\right)-f\left(x\right)\right|\right)Z\left(nx-u\right)du+$$

$${\int }_{A_2}\left(\sum _{r=0}^{\theta }{w}_r\left|f\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{r}{n\theta }}\right)-f\left(x\right)\right|\right)Z\left(nx-u\right)du\le$$

$${\int }_{A_1}\left(\sum _{r=0}^{\theta }{w}_r{\omega }_1\left(f,{∥{\displaystyle \frac{u}{n}}-x∥}_{\infty }+{\displaystyle \frac{1}{n}}{∥{\displaystyle \frac{r}{\theta }}∥}_{\infty }\right)\right)Z\left(nx-u\right)du+$$

$$2{∥f∥}_{\infty }{\int }_{A_2}Z\left(nx-u\right)du\le$$

$${\omega }_1\left(f,{\displaystyle \frac{1}{n^{\beta }}}+{\displaystyle \frac{1}{n}}\right){\int }_{A_1}Z\left(nx-u\right)du+2{∥f∥}_{\infty }{\displaystyle \frac{\left(q+\frac{1}{q}\right)}{e^{2\lambda \left(n^{1-\beta }-1\right)}}}\le$$

(58)

$${\omega }_1\left(f,{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+{\displaystyle \frac{2\left(q+\frac{1}{q}\right){∥f∥}_{\infty }}{e^{2\lambda \left(n^{1-\beta }-1\right)}}}.$$

□

Next, we describe the global smoothness preservation property of our activated multivariate operators.

Theorem 9. 

Here, $f\in C_B\left({\mathbb{R}}^N\right)\cup C_U\left({\mathbb{R}}^N\right)$. Then,

$${\omega }_1\left(S_n\left(f\right),\delta \right)\le {\omega }_1\left(f,\delta \right),\delta >0.$$

(59)

If $f\in C_U\left({\mathbb{R}}^N\right)$, then $S_n\left(f\right)\in C_U\left({\mathbb{R}}^N\right)$.

Proof. 

We determine that

$$S_n\left(f\right)\left(x\right)={\int }_{{\mathbb{R}}^N}f\left(x-{\displaystyle \frac{z}{n}}\right)Z\left(z\right)dz.$$

(60)

Let $x,y\in {\mathbb{R}}^N$; then,

$$S_n\left(f\right)\left(x\right)-S_n\left(f\right)\left(y\right)={\int }_{{\mathbb{R}}^N}\left(f\left(x-{\displaystyle \frac{z}{n}}\right)-f\left(y-{\displaystyle \frac{z}{n}}\right)\right)Z\left(z\right)dz.$$

(61)

Thus,

$$\left|S_n\left(f\right)\left(x\right)-S_n\left(f\right)\left(y\right)\right|\le {\int }_{{\mathbb{R}}^N}\left|f\left(x-{\displaystyle \frac{z}{n}}\right)-f\left(y-{\displaystyle \frac{z}{n}}\right)\right|Z\left(z\right)dz\le$$

$${\omega }_1\left(f,{∥x-y∥}_{\infty }\right){\int }_{{\mathbb{R}}^N}Z\left(z\right)dz\stackrel{\left(31\right)}{=}{\omega }_1\left(f,{∥x-y∥}_{\infty }\right).$$

(62)

Let ${∥x-y∥}_{\infty }\le \delta$, $\delta >0$; then, we obtain (59). □

Remark 2. 

Let f be the projection function onto the $x_i$ coordinate, call it $p{r}_i\left(x\right):=x_i$, $i\in \left\{1,\dots ,N\right\}$, where $x=\left(x_1,\dots ,x_i,\dots ,x_N\right)\in {\mathbb{R}}^N.$ Then, it holds that

$$S_n\left(p{r}_i\right)\left(x\right)={\int }_{{\mathbb{R}}^N}\left(x_i-{\displaystyle \frac{z_i}{n}}\right)Z\left(z\right)dz=x_i-{\displaystyle \frac{1}{n}}{\int }_{{\mathbb{R}}^N}{z}_{i}Z\left(z\right)dz.$$

(63)

Hence,

$$\left|S_n\left(p{r}_i\right)\left(x\right)-S_n\left(p{r}_i\right)\left(y\right)\right|=\left|x_i-y_i\right|=\left|p{r}_i\left(x\right)-p{r}_i\left(y\right)\right|,$$

proving that

$${\omega }_1\left(S_n\left(p{r}_i\right),\delta \right)={\omega }_1\left(p{r}_i,\delta \right),$$

(64)

for any $\delta >0.$

So, (59) is an attained sharp inequality.

Furthermore,

$$S_n\left(p{r}_i\right)\left(x\right)=x_i-{\displaystyle \frac{1}{n}}\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^N{\int }_{-\infty }^{\infty }\phi \left(z_j\right)d{z}_j\right)\left({\int }_{-\infty }^{\infty }{z}_i\phi \left(z_i\right)d{z}_i\right)$$

(65)

$$=x_i-{\displaystyle \frac{1}{n}}{\int }_{-\infty }^{\infty }{z}_i\phi \left(z_i\right)d{z}_i,$$

and

$$\left|S_n\left(p{r}_i\right)\left(x\right)\right|\le \left|x_i\right|+{\displaystyle \frac{1}{n}}{\int }_{-\infty }^{\infty }\left|z_i\right|\phi \left(z_i\right)d{z}_i$$

$$\stackrel{\left(26\right)}{\le }\left|x_i\right|+{\displaystyle \frac{1}{n}}\left[{\displaystyle \frac{tanh\left(\lambda \right)}{2}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }}{2\lambda }}\right]<\infty .$$

(66)

Thus, $S_n\left(p{r}_i\right)\left(x\right)$ is well defined.

Theorem 10. 

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

$${\omega }_1\left(S_n^{*}\left(f\right),\delta \right)\le {\omega }_1\left(f,\delta \right),\delta >0.$$

(67)

If $f\in C_U\left({\mathbb{R}}^N\right)$, then $S_n^{*}\left(f\right)\in C_U\left({\mathbb{R}}^N\right)$.

Inequality (67) is an attained sharp inequality according to $f\left(x\right)=p{r}_i\left(x\right)$, $i\in \left\{1,\dots ,N\right\}.$

Proof. 

According to (52), one can write

$$B_n^{*}\left(f\right)\left(x\right)={\int }_{{\mathbb{R}}^N}\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}f\left(t+{\displaystyle \frac{u}{n}}\right)dt\right)Z\left(nx-u\right)du=$$

(68)

$${\int }_{{\mathbb{R}}^N}\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}f\left(t+\left(x-{\displaystyle \frac{z}{n}}\right)\right)dt\right)Z\left(z\right)dz,$$

and

$$S_n^{*}\left(f\right)\left(y\right)={\int }_{{\mathbb{R}}^N}\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}f\left(t+\left(y-{\displaystyle \frac{z}{n}}\right)\right)dt\right)Z\left(z\right)dz,$$

(69)

where $x,y\in {\mathbb{R}}^N.$

Thus,

$$\left|S_n^{*}\left(f\right)\left(x\right)-S_n^{*}\left(f\right)\left(y\right)\right|=$$

$$\left|{\int }_{{\mathbb{R}}^N}\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left(f\left(t+\left(x-{\displaystyle \frac{z}{n}}\right)\right)-f\left(t+\left(y-{\displaystyle \frac{z}{n}}\right)\right)\right)dt\right)Z\left(z\right)dz\right|\le$$

$${\int }_{{\mathbb{R}}^N}\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left|f\left(t+\left(x-{\displaystyle \frac{z}{n}}\right)\right)-f\left(t+\left(y-{\displaystyle \frac{z}{n}}\right)\right)\right|dt\right)Z\left(z\right)dz\le$$

$${\omega }_1\left(f,{∥x-y∥}_{\infty }\right){\int }_{{\mathbb{R}}^N}Z\left(z\right)dz={\omega }_1\left(f,{∥x-y∥}_{\infty }\right),$$

(70)

proving inequality (67).

We do have

$$\left|S_n^{*}\left(p{r}_i\right)\left(x\right)\right|\le {\int }_{{\mathbb{R}}^N}\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left|\left(t_i+\left(x_i-{\displaystyle \frac{z_i}{n}}\right)\right)\right|dt\right)Z\left(z\right)dz\le$$

$${\int }_{{\mathbb{R}}^N}\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left(\left|t_i\right|+\left|x_i\right|+{\displaystyle \frac{\left|z_i\right|}{n}}\right)dt\right)Z\left(z\right)dz\le$$

(71)

$${\int }_{{\mathbb{R}}^N}\left({\displaystyle \frac{1}{n}}+\left|x_i\right|+{\displaystyle \frac{\left|z_i\right|}{n}}\right)Z\left(z\right)dz={\int }_{-\infty }^{\infty }\left({\displaystyle \frac{1}{n}}+\left|x_i\right|+{\displaystyle \frac{\left|z_i\right|}{n}}\right)\phi \left(z_i\right)d{z}_i=$$

$${\displaystyle \frac{1}{n}}+\left|x_i\right|+{\displaystyle \frac{1}{n}}{\int }_{-\infty }^{\infty }\left|z_i\right|\phi \left(z_i\right)d{z}_i\stackrel{\left(26\right)}{\le }$$

(72)

$${\displaystyle \frac{1}{n}}+\left|x_i\right|+{\displaystyle \frac{1}{n}}\left[{\displaystyle \frac{tanh\left(\lambda \right)}{2}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }}{2\lambda }}\right]<\infty .$$

Thus, $S_n^{*}\left(p{r}_i\right)\left(x\right)$ is well defined. □

Theorem 11. 

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

$${\omega }_1\left(\overline{S_n}\left(f\right),\delta \right)\le {\omega }_1\left(f,\delta \right),\delta >0.$$

(73)

If $f\in C_U\left({\mathbb{R}}^N\right)$, then $\overline{S_n}\left(f\right)\in C_U\left({\mathbb{R}}^N\right)$.

Inequality (73) is an attained sharp inequality according to $f\left(x\right)=p{r}_i\left(x\right)$, $i\in \left\{1,\dots ,N\right\}.$

Proof. 

Let $x,y\in {\mathbb{R}}^N$; then,

$$\overline{S_n}\left(f\right)\left(x\right)={\int }_{{\mathbb{R}}^N}\left(\sum _{r=0}^{\theta }{w}_{r}f\left(\left(x-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{r}{n\theta }}\right)\right)Z\left(z\right)dz,$$

(74)

and

$$\overline{S_n}\left(f\right)\left(y\right)={\int }_{{\mathbb{R}}^N}\left(\sum _{r=0}^{\theta }{w}_{r}f\left(\left(y-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{r}{n\theta }}\right)\right)Z\left(z\right)dz.$$

(75)

Hence,

$$\left|\overline{S_n}\left(f\right)\left(x\right)-\overline{S_n}\left(f\right)\left(y\right)\right|=$$

$$\left|{\int }_{{\mathbb{R}}^N}\left(\sum _{r=0}^{\theta }{w}_r\left(f\left(\left(x-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{r}{n\theta }}\right)-f\left(\left(y-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{r}{n\theta }}\right)\right)\right)Z\left(z\right)dz\right|\le$$

$${\int }_{{\mathbb{R}}^N}\left(\sum _{r=0}^{\theta }{w}_r\left|f\left(\left(x-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{r}{n\theta }}\right)-f\left(\left(y-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{r}{n\theta }}\right)\right|\right)Z\left(z\right)dz\le$$

(76)

$${\omega }_1\left(f,{∥x-y∥}_{\infty }\right){\int }_{{\mathbb{R}}^N}Z\left(z\right)dz={\omega }_1\left(f,{∥x-y∥}_{\infty }\right).$$

That is, (73) is true, and it is an attained sharp inequality according to $f\left(x\right)=p{r}_i\left(x\right)$, $i\in \left\{1,\dots ,N\right\}.$

Indeed, we determine that

$$\left|\overline{S_n}\left(p{r}_i\right)\left(x\right)\right|=\left|{\int }_{{\mathbb{R}}^N}\left(\sum _{r=0}^{\theta }{w}_r\left[\left(x_i-{\displaystyle \frac{z_i}{n}}\right)+{\displaystyle \frac{r_i}{n{\theta }_i}}\right]\right)Z\left(z\right)dz\right|\le$$

$${\int }_{{\mathbb{R}}^N}\left(\sum _{r=0}^{\theta }{w}_r\left|\left(x_i-{\displaystyle \frac{z_i}{n}}\right)+{\displaystyle \frac{r_i}{n{\theta }_i}}\right|\right)Z\left(z\right)dz\le$$

$${\int }_{{\mathbb{R}}^N}\left(\sum _{r=0}^{\theta }{w}_r\left[\left|x_i\right|+{\displaystyle \frac{\left|z_i\right|}{n}}+{\displaystyle \frac{1}{n}}\right]\right)Z\left(z\right)dz=$$

(77)

$${\int }_{{\mathbb{R}}^N}\left[\left|x_i\right|+{\displaystyle \frac{\left|z_i\right|}{n}}+{\displaystyle \frac{1}{n}}\right]Z\left(z\right)dz=$$

(as in (72))

$${\int }_{-\infty }^{\infty }\left[\left|x_i\right|+{\displaystyle \frac{\left|z_i\right|}{n}}+{\displaystyle \frac{1}{n}}\right]\phi \left(z_i\right)d{z}_i<\infty .$$

Thus, $\overline{S_n}\left(p{r}_i\right)\left(x\right)$ is well defined. □

We establish the following:

Remark 3. 

Let $i\in \mathbb{N}$ be fixed. Assume that $f\in C^{\left(i\right)}\left({\mathbb{R}}^N\right)$, $N\in \mathbb{N}$. Here, $f_{\alpha }$ denotes a partial derivative of f, $\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right)$, ${\alpha }_j\in {\mathbb{Z}}_{+}$, $j=1,\dots ,N$, and $\left|\alpha \right|:={\displaystyle \sum _{j=1}^N}{\alpha }_j=l$, where $l=0,1,\dots ,i.$

We also write $f_{\alpha }:={\displaystyle \frac{{\partial }^{\alpha }f}{\partial x^{\alpha }}}$, and we say it is of the order l.

We assume that any partial $f_{\alpha }\in C_B\left({\mathbb{R}}^N\right)$ for all $\alpha :\left|\alpha \right|=l$, $l=0,1,\dots ,i.$

Through the repeated application of Theorem 5, we obtain

$${\left(S_n\left(f\right)\right)}_{\alpha }\left(x\right)={\int }_{{\mathbb{R}}^N}{f}_{\alpha }\left(x-{\displaystyle \frac{z}{n}}\right)Z\left(z\right)dz=$$

(78)

$${\int }_{{\mathbb{R}}^N}{f}_{\alpha }\left({\displaystyle \frac{u}{n}}\right)Z\left(nx-u\right)du=\left(S_n\left(f_{\alpha }\right)\right)\left(x\right),\forall x\in {\mathbb{R}}^N.$$

Similarly, we determine that

$${\left(S_n^{*}\left(f\right)\right)}_{\alpha }\left(x\right)=\left(S_n^{*}\left(f_{\alpha }\right)\right)\left(x\right),$$

(79)

and

$${\left(\overline{S_n}\left(f\right)\right)}_{\alpha }\left(x\right)=\left(\overline{S_n}\left(f_{\alpha }\right)\right)\left(x\right),\forall x\in {\mathbb{R}}^N;$$

(80)

for all $\alpha :\left|\alpha \right|=l$, $l=0,1,\dots ,i.$

So, all of our results in this work can be written in the simultaneous approximation context (see Theorems 15–17).

We establish the following:

Remark 4. 

Activated Iterative Multivariate Convolution

We determine that

$$S_n\left(f\right)\left(x\right)={\int }_{{\mathbb{R}}^N}f\left(x-{\displaystyle \frac{z}{n}}\right)Z\left(z\right)dz,$$

∀$x\in {\mathbb{R}}^N$, where $f\in C_B\left({\mathbb{R}}^N\right).$

Let $x_N\to x$, as $N\to \infty$, and

$$S_n\left(f\right)\left(x_N\right)-S_n\left(f\right)\left(x\right)={\int }_{{\mathbb{R}}^N}\left(f\left(x_N-{\displaystyle \frac{z}{n}}\right)-f\left(x-{\displaystyle \frac{z}{n}}\right)\right)Z\left(z\right)dz.$$

(81)

We determine that

$$f\left(x_N-{\displaystyle \frac{z}{n}}\right)Z\left(z\right)\to f\left(x-{\displaystyle \frac{z}{n}}\right)Z\left(z\right),\forall z\in {\mathbb{R}}^N,\mathit{as}N\to \infty .$$

Furthermore, it holds that

$$\left|S_n\left(f\right)\left(x_N\right)-S_n\left(f\right)\left(x\right)\right|\le$$

(82)

$${\int }_{{\mathbb{R}}^N}\left|f\left(x_N-{\displaystyle \frac{z}{n}}\right)-f\left(x-{\displaystyle \frac{z}{n}}\right)\right|Z\left(z\right)dz\to 0,\mathit{as}N\to \infty ,$$

according to the dominated convergence theorem, because we determine that

$$\left|f\left(x_N-{\displaystyle \frac{z}{n}}\right)\right|Z\left(z\right)\le {∥f∥}_{\infty }Z\left(z\right),$$

and ${∥f∥}_{\infty }Z\left(z\right)$ is integrable over ${\mathbb{R}}^N$, ∀$z\in {\mathbb{R}}^N$.

Hence, $S_n\left(f\right)\in C_B\left({\mathbb{R}}^N\right)$.

Furthermore it holds that

$$\left|S_n\left(f\right)\left(x\right)\right|\le {∥f∥}_{\infty }{\int }_{{\mathbb{R}}^N}Z\left(z\right)dz={∥f∥}_{\infty },$$

i.e.,

$${∥S_n\left(f\right)∥}_{\infty }\le {∥f∥}_{\infty }.$$

(83)

So, $S_n$ is a bounded positive linear operator.

Clearly, it holds that

$${∥S_n^2\left(f\right)∥}_{\infty }={∥S_n\left(S_n\left(f\right)\right)∥}_{\infty }\le {∥S_n\left(f\right)∥}_{\infty }\le {∥f∥}_{\infty }.$$

(84)

And for $k\in \mathbb{N}$, we obtain

$${∥S_n^k\left(f\right)∥}_{\infty }\le {∥S_n^{k-1}\left(f\right)∥}_{\infty }\le {∥S_n^{k-2}\left(f\right)∥}_{\infty }\le \dots \le {∥f∥}_{\infty },$$

(85)

so the contraction property is valid and $S_n^k$ is a bounded linear operator.

Remark 5. 

Let $r\in \mathbb{N}$. We observe that

$$S_n^{r}f-f=\left(S_n^{r}f-S_n^{r-1}f\right)+\left(S_n^{r-1}f-S_n^{r-2}f\right)+\left(S_n^{r-2}f-S_n^{r-3}f\right)$$

$$+\dots +\left(S_n^{2}f-S_{n}f\right)+\left(S_{n}f-f\right).$$

(86)

Then,

$${∥S_n^{r}f-f∥}_{\infty }\le {∥S_n^{r}f-S_n^{r-1}f∥}_{\infty }+{∥S_n^{r-1}f-S_n^{r-2}f∥}_{\infty }+{∥S_n^{r-2}f-S_n^{r-3}f∥}_{\infty }$$

$$+\dots +{∥S_n^{2}f-S_{n}f∥}_{\infty }+{∥S_{n}f-f∥}_{\infty }=$$

$${∥S_n^{r-1}\left(S_{n}f-f\right)∥}_{\infty }+{∥S_n^{r-2}\left(S_{n}f-f\right)∥}_{\infty }+\dots +{∥S_n\left(S_{n}f-f\right)∥}_{\infty }+$$

$${∥S_{n}f-f∥}_{\infty }\le r{∥S_{n}f-f∥}_{\infty }.$$

Therefore,

$${∥S_n^{r}f-f∥}_{\infty }\le r{∥S_{n}f-f∥}_{\infty }.$$

(87)

Now, let $m_1,m_2,\dots ,m_r\in \mathbb{N}:m_1\le m_2\le \dots \le m_r$, and $S_{m_i}$ as above.

$$S_{m_r}\left(S_{m_{r-1}}\left(\dots S_{m_2}\left(S_{m_1}f\right)\right)\right)-f=\dots =$$

$$S_{m_r}\left(S_{m_{r-1}}\left(\dots S_{m_2}\right)\right)\left(S_{m_1}f-f\right)+S_{m_r}\left(S_{m_{r-1}}\left(\dots S_{m_3}\right)\right)\left(S_{m_2}f-f\right)+$$

(88)

$$S_{m_r}\left(S_{m_{r-1}}\left(\dots S_{m_4}\right)\right)\left(S_{m_3}f-f\right)+\dots +S_{m_r}\left(S_{m_{r-1}}f-f\right)+S_{m_r}f-f.$$

Consequently, it holds, as in [1], Chapter 2, that

$${∥S_{m_r}\left(S_{m_{r-1}}\left(\dots S_{m_2}\left(S_{m_1}f\right)\right)\right)-f∥}_{\infty }\le \sum _{i=1}^r{∥S_{m_i}f-f∥}_{\infty }.$$

(89)

Next, we have

Remark 6. 

We obtain

$$S_n^{*}\left(f\right)\left(x\right)=n^N{\int }_{{\mathbb{R}}^N}\left({\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}f\left(t+\left(x-{\displaystyle \frac{z}{n}}\right)\right)dt\right)Z\left(z\right)dz,$$

$f\in C_B\left({\mathbb{R}}^N\right).$

Let $x_N\to x$, as $N\to \infty$, and

$$\left|S_n^{*}\left(f\right)\left(x_N\right)-S_n^{*}\left(f\right)\left(x\right)\right|=$$

$$\left|n^N{\int }_{{\mathbb{R}}^N}\left({\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left[f\left(t+\left(x_N-{\displaystyle \frac{z}{n}}\right)\right)-f\left(t+\left(x-{\displaystyle \frac{z}{n}}\right)\right)\right]dt\right)Z\left(z\right)dz\right|\le$$

(90)

$${\int }_{{\mathbb{R}}^N}\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left|f\left(t+\left(x_N-{\displaystyle \frac{z}{n}}\right)\right)-f\left(t+\left(x-{\displaystyle \frac{z}{n}}\right)\right)\right|dt\right)Z\left(z\right)dz\to 0,$$

as $N\to \infty .$

This is true according to the bounded convergence theorem, and we determine that

$$x_N\to x⇝t+\left(x_N-{\displaystyle \frac{z}{n}}\right)\to t+\left(x-{\displaystyle \frac{z}{n}}\right)$$

and

$$f\left(t+\left(x_N-{\displaystyle \frac{z}{n}}\right)\right)\to f\left(t+\left(x-{\displaystyle \frac{z}{n}}\right)\right),\mathrm{and}$$

$$\left|f\left(t+\left(x_N-{\displaystyle \frac{z}{n}}\right)\right)\right|\le {∥f∥}_{\infty },$$

where ${\left[0,{\displaystyle \frac{1}{n}}\right]}^N$ is a cube. Thus,

$$n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left|f\left(t+\left(x_N-{\displaystyle \frac{z}{n}}\right)\right)-f\left(t+\left(x-{\displaystyle \frac{z}{n}}\right)\right)\right|dt\to 0,\mathit{as}N\to \infty .$$

(91)

Therefore, it holds that

$$n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}f\left(t+\left(x_N-{\displaystyle \frac{z}{n}}\right)\right)dt\to n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}f\left(t+\left(x-{\displaystyle \frac{z}{n}}\right)\right)dt,\mathit{as}N\to \infty .$$

(92)

and we obtain

$$\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}f\left(t+\left(x_N-{\displaystyle \frac{z}{n}}\right)\right)dt\right)Z\left(z\right)\to \left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}f\left(t+\left(x-{\displaystyle \frac{z}{n}}\right)\right)dt\right)Z\left(z\right),$$

(93)

as $N\to \infty ,$∀$z\in {\mathbb{R}}^N$.

Furthermore, we have

$$\left|\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}f\left(t+\left(x_N-{\displaystyle \frac{z}{n}}\right)\right)dt\right)Z\left(z\right)\right|\le {∥f∥}_{\infty }Z\left(z\right),$$

(94)

with ${∥f∥}_{\infty }Z\left(z\right)$ being integrable over ${\mathbb{R}}^N$.

Therefore according to the dominated convergence theorem,

$$S_n^{*}\left(f\right)\left(x_N\right)\to S_n^{*}\left(f\right)\left(x\right),\mathit{as}N\to \infty .$$

Hence, $S_n^{*}\left(f\right)\left(x\right)$ is a bounded and continuous in $x\in {\mathbb{R}}^N$.

The iterated facts hold for $S_n^{*}$ as in the $S_n\left(f\right)$ case, all the same! See (83)–(85) and all of Remark 5.

Remark 7. 

Next, we observe the following: Let $f\in C_B\left({\mathbb{R}}^N\right)$, and

$$\overline{S_n}\left(f\right)\left(x\right)={\int }_{{\mathbb{R}}^N}\left(\sum _{r=0}^{\theta }{w}_{r}f\left(\left(x-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{r}{n\theta }}\right)\right)Z\left(z\right)dz.$$

Let $x_N\to x$, as $N\to \infty$. Then,

$$\left|\overline{S_n}\left(f\right)\left(x_N\right)-\overline{S_n}\left(f\right)\left(x\right)\right|=$$

$$\left|{\int }_{{\mathbb{R}}^N}\left(\sum _{r=0}^{\theta }{w}_r\left(f\left(\left(x_N-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{r}{n\theta }}\right)-f\left(\left(x-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{r}{n\theta }}\right)\right)\right)Z\left(z\right)dz\right|\le$$

(95)

$${\int }_{{\mathbb{R}}^N}\left|\sum _{r=0}^{\theta }{w}_r\left(f\left(\left(x_N-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{r}{n\theta }}\right)-f\left(\left(x-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{r}{n\theta }}\right)\right)\right|Z\left(z\right)dz\to 0,$$

as $N\to \infty .$

The latter is obtained according go the dominated convergence theorem:

$$\left(x_N-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{r}{n\theta }}\to \left(x-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{r}{n\theta }}$$

and

$$\sum _{r=0}^{\theta }{w}_{r}f\left(\left(x_N-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{r}{n\theta }}\right)\to \sum _{r=0}^{\theta }{w}_{r}f\left(\left(x-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{r}{n\theta }}\right),$$

and

$$\left(\sum _{r=0}^{\theta }{w}_{r}f\left(\left(x_N-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{r}{n\theta }}\right)\right)Z\left(z\right)\to \left(\sum _{r=0}^{\theta }{w}_{r}f\left(\left(x-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{r}{n\theta }}\right)\right)Z\left(z\right),$$

as $N\to \infty ,$∀$z\in {\mathbb{R}}^N$.

Furthermore, it holds that

$$\left|\sum _{r=0}^{\theta }{w}_{r}f\left(\left(x_N-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{r}{n\theta }}\right)\right|Z\left(z\right)\le {∥f∥}_{\infty }Z\left(z\right),$$

(96)

in which the last function is integrable over ${\mathbb{R}}^N$.

Therefore,

$$\overline{S_n}\left(f\right)\left(x_N\right)\to \overline{S_n}\left(f\right)\left(x\right),\mathit{as}N\to \infty .$$

Hence, $\overline{S_n}\left(f\right)\left(x\right)$ is bounded and continuous in $x\in {\mathbb{R}}^N$.

The iterated facts hold for $\overline{S_n}$ in the same manner as with $S_n\left(f\right)$! See (83)–(85) and all of Remark 5.

See the related Theorems 18 and 19, which we describe later.

Next, we greatly improve the speed of convergence of our activated multivariate operators by using the differentiation of functions.

Notation 1. 

Let $f\in C^m\left({\mathbb{R}}^N\right)$, $m,N\in \mathbb{N}$. Here, $f_{\alpha }$ denotes a partial derivative of f, $\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right)$, ${\alpha }_i\in {\mathbb{Z}}_{+}$, $i=1,\dots ,N,$ and $\left|\alpha \right|={\displaystyle \sum _{i=1}^N}{\alpha }_i=l$, where $l=0,1,\dots ,m$. We also write $f_{\alpha }:={\displaystyle \frac{{\partial }^{\alpha }f}{\partial x^{\alpha }}}$, and we say it is of the order l.

We denote

$${\omega }_{1,m}^{max}\left(f_{\alpha },h\right):=\underset{\alpha :\left|\alpha \right|=m}{max}{\omega }_1\left(f_{\alpha },h\right),h>0.$$

(97)

also written as

$${∥f_{\alpha }∥}_{\infty ,m}^{max}:=\underset{\left|\alpha \right|=m}{max}\left\{{∥f_{\alpha }∥}_{\infty }\right\},$$

(98)

where ${∥·∥}_{\infty }$ is the supremum norm.

Theorem 12. 

Let $0<\beta <1$, $n\in \mathbb{N}:n^{1-\beta }>2$; $x\in {\mathbb{R}}^N,$$f\in C^m\left({\mathbb{R}}^N\right)$, $m,N\in \mathbb{N}$, with $f_{\alpha }\in C_B\left({\mathbb{R}}^N\right)$ for all $\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right)$, ${\alpha }_i\in {\mathbb{Z}}_{+}$, $i=1,\dots ,N,$ and $\left|\alpha \right|={\displaystyle \sum _{i=1}^N}{\alpha }_i=l$, where $l=0,1,\dots ,m.$ Then,

(i) 

$$\left|S_n\left(f\right)\left(x\right)-f\left(x\right)-\right.$$

$$\left.\sum _{j=1}^m\sum _{\left\{\begin{array}{c}\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right),{\alpha }_i\in {\mathbb{Z}}_{+}\\ i=1,\dots ,N,\left|\alpha \right|:=\sum _{i=1}^N{\alpha }_i=j\end{array}\right\}}\left({\displaystyle \frac{1}{\prod _{i=1}^N{\alpha }_i!}}\right)f_{\alpha }\left(x\right)S_n\left(\left(\prod _{i=1}^N{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)\right|$$

$$\le \left({\displaystyle \frac{N^m}{m!n^{m\beta }}}\right){\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n^{\beta }}}\right)+$$

$$\left({\displaystyle \frac{2{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^{2m}}{n^{m}m!}}\right)\left[{\displaystyle \frac{tanh\left(\lambda \right)}{\left(m+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }m!}{{\left(2\lambda \right)}^m}}\right]=:{\Phi }_1^s,$$

(99)

(ii) 

Assume that $f_{\alpha }\left(x\right)=0$ for all $\alpha :\left|\alpha \right|=j$, $j=1,\dots ,m$. We have

$$\left|S_n\left(f\right)\left(x\right)-f\left(x\right)\right|\le {\Phi }_1^s,$$

(100)

with a high speed of $n^{-\beta \left(m+1\right)}.$

(iii) 

$$\left|S_n\left(f\right)\left(x\right)-f\left(x\right)\right|\le$$

$$\sum _{j=1}^N\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}\right.}\left({\displaystyle \frac{1}{\prod _{i=1}^N{\alpha }_i!}}\right)\left|f_{\alpha }\left(x\right)\right|{\displaystyle \frac{1}{n^j}}\left\{\prod _{i=1}^N\left[{\displaystyle \frac{tanh\left(\lambda \right)}{\left({\alpha }_i+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }{\alpha }_i!}{{\left(2\lambda \right)}^{{\alpha }_i}}}\right]\right\}$$

$$+{\Phi }_1^s,$$

(101)

and

(iv) 

$${∥S_n\left(f\right)-f∥}_{\infty }\le$$

$$\sum _{j=1}^N\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}\right.}\left({\displaystyle \frac{1}{\prod _{i=1}^N{\alpha }_i!}}\right){∥f_{\alpha }∥}_{\infty }{\displaystyle \frac{1}{n^j}}\left\{\prod _{i=1}^N\left[{\displaystyle \frac{tanh\left(\lambda \right)}{\left({\alpha }_i+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }{\alpha }_i!}{{\left(2\lambda \right)}^{{\alpha }_i}}}\right]\right\}+{\Phi }_1^s.$$

(102)

We determine that $S_n\left(f\right)\to f$, as $n\to \infty$, pointwise and uniformly.

Proof. 

Consider $g_z\left(t\right):=f\left(x_0+t\left(z-x_0\right)\right)$, $t\ge 0$; $x_0,z\in {\mathbb{R}}^N$. Then,

$$g_z^{\left(j\right)}\left(t\right)=\left[{\left(\sum _{i=1}^N\left(z_i-x_{0i}\right){\displaystyle \frac{\partial }{\partial x_i}}\right)}^{j}f\right]\left(x_{01}+t\left(z_1-x_{01}\right),\dots ,x_{0N}+t\left(z_N-x_{0N}\right)\right),$$

(103)

for all $j=0,1,\dots ,m.$

We have the multivariate Taylor’s formula

$$f\left(z_1,\dots ,z_N\right)=g_z\left(1\right)=\sum _{j=0}^m{\displaystyle \frac{g_z^{\left(j\right)}\left(0\right)}{j!}}+$$

$${\displaystyle \frac{1}{\left(m-1\right)!}}{\int }_0^1{\left(1-\theta \right)}^{m-1}\left(g_z^{\left(m\right)}\left(\theta \right)-g_z^{\left(m\right)}\left(0\right)\right)d\theta .$$

(104)

Notice that $g_z\left(0\right)=f\left(z_0\right)$. Also, for $j=0,1,\dots ,m,$ we have

$$g_z^{\left(j\right)}\left(0\right)=\sum _{\left\{\begin{array}{c}\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right),{\alpha }_i\in {\mathbb{Z}}_{+}\\ i=1,\dots ,N,\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}{\alpha }_i=j\end{array}\right\}}\left({\displaystyle \frac{j!}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

(105)

$$\left(\prod _{i=1}^N{\left(z_i-x_{0i}\right)}^{{\alpha }_i}\right)f_{\alpha }\left(x_0\right).$$

Furthermore we obtain

$$g_z^{\left(m\right)}\left(\theta \right)=\sum _{\left\{\begin{array}{c}\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right),{\alpha }_i\in {\mathbb{Z}}_{+}\\ i=1,\dots ,N,\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}{\alpha }_i=m\end{array}\right\}}\left({\displaystyle \frac{m!}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

$$\left(\prod _{i=1}^N{\left(z_i-x_{0i}\right)}^{{\alpha }_i}\right)f_{\alpha }\left(x_0+\theta \left(z-x_0\right)\right),$$

(106)

$0\le \theta \le 1.$

So, $f\in C^m\left({\mathbb{R}}^N\right).$

Thus, we determine, for $u,x\in {\mathbb{R}}^N$, that

$$f\left({\displaystyle \frac{u_1}{n}},\dots ,{\displaystyle \frac{u_N}{n}}\right)-f\left(x\right)=$$

$$\sum _{j=1}^m\sum _{\left\{\begin{array}{c}\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right),{\alpha }_i\in {\mathbb{Z}}_{+}\\ i=1,\dots ,N,\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}{\alpha }_i=j\end{array}\right\}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

$$\left(\prod _{i=1}^N{\left({\displaystyle \frac{u_i}{n}}-x_i\right)}^{{\alpha }_i}\right)f_{\alpha }\left(x\right)+R,$$

(107)

where

$$R:=m{\int }_0^1{\left(1-\theta \right)}^{m-1}\sum _{\left\{\begin{array}{c}\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right),{\alpha }_i\in {\mathbb{Z}}_{+}\\ i=1,\dots ,N,\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}{\alpha }_i=m\end{array}\right\}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

$$\left(\prod _{i=1}^N{\left({\displaystyle \frac{u_i}{n}}-x_i\right)}^{{\alpha }_i}\right)\left[f_{\alpha }\left(x+\theta \left({\displaystyle \frac{u}{n}}-x\right)\right)-f_{\alpha }\left(x\right)\right]d\theta .$$

(108)

We see that

$$\left|R\right|\le m{\int }_0^1{\left(1-\theta \right)}^{m-1}\sum _{\left|\alpha \right|=m}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

$$\left(\prod _{i=1}^N{\left|{\displaystyle \frac{u_i}{n}}-x_i\right|}^{{\alpha }_i}\right)\left|f_{\alpha }\left(x+\theta \left({\displaystyle \frac{u}{n}}-x\right)\right)-f_{\alpha }\left(x\right)\right|d\theta \le$$

$$m{\int }_0^1{\left(1-\theta \right)}^{m-1}\left(\sum _{\left|\alpha \right|=m}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left(\prod _{i=1}^N{\left|{\displaystyle \frac{u_i}{n}}-x_i\right|}^{{\alpha }_i}\right)\right.$$

(109)

$$\left.{\omega }_1\left(f_{\alpha },\theta {∥{\displaystyle \frac{u}{n}}-x∥}_{\infty }\right)\right)d\theta \le \left(*\right).$$

Notice here that

$${∥{\displaystyle \frac{u}{n}}-x∥}_{\infty }\le {\displaystyle \frac{1}{n^{\beta }}},\mathrm{iff}\left|{\displaystyle \frac{u_i}{n}}-x_i\right|\le {\displaystyle \frac{1}{n^{\beta }}},i=1,\dots ,N.$$

(110)

We further see that

$$\left(*\right)\le m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n^{\beta }}}\right){\int }_0^1{\left(1-\theta \right)}^{m-1}\left(\sum _{\left|\alpha \right|=m}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left(\prod _{i=1}^N{\left({\displaystyle \frac{1}{n^{\beta }}}\right)}^{{\alpha }_i}\right)\right)d\theta$$

$$=\left({\displaystyle \frac{{\omega }_{1,m}^{max}\left(f_{\alpha },\frac{1}{n^{\beta }}\right)}{\left(m!\right)n^{m\beta }}}\right)\left(\sum _{\left|\alpha \right|=m}{\displaystyle \frac{m!}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)=$$

$$\left({\displaystyle \frac{{\omega }_{1,m}^{max}\left(f_{\alpha },\frac{1}{n^{\beta }}\right)}{\left(m!\right)n^{m\beta }}}\right)N^m.$$

(111)

Conclusion: When ${∥{\displaystyle \frac{u}{n}}-x∥}_{\infty }\le {\displaystyle \frac{1}{n^{\beta }}}$, we prove that

$$\left|R\right|\le \left({\displaystyle \frac{N^m}{m!n^{m\beta }}}\right){\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n^{\beta }}}\right).$$

(112)

According to (108) we determine that

$$\left|R\right|\le m\left({\int }_0^1{\left(1-\theta \right)}^{m-1}\sum _{\left\{\begin{array}{c}\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right),{\alpha }_i\in {\mathbb{Z}}_{+}\\ i=1,\dots ,N,\left|\alpha \right|=m\end{array}\right\}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\right.$$

$$\left.\left({\displaystyle \prod _{i=1}^N}\left({\left|{\displaystyle \frac{u_i}{n}}-x_i\right|}^{{\alpha }_i}\right)2{∥f_{\alpha }∥}_{\infty }\right)d\theta \right)=$$

$$2\sum _{\left|\alpha \right|=m}{\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\left({\displaystyle \prod _{i=1}^N}{\left|{\displaystyle \frac{u_i}{n}}-x_i\right|}^{{\alpha }_i}\right){∥f_{\alpha }∥}_{\infty }\le$$

$$\left({\displaystyle \frac{2{∥\frac{u}{n}-x∥}_{\infty }^m{∥f_{\alpha }∥}_{\infty ,m}^{max}}{m!}}\right)\left(\sum _{\left|\alpha \right|=m}{\displaystyle \frac{m!}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)=$$

$${\displaystyle \frac{2{∥\frac{u}{n}-x∥}_{\infty }^m{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{m!}}.$$

(113)

We prove, in general, that

$$\left|R\right|\le {\displaystyle \frac{2{∥\frac{u}{n}-x∥}_{\infty }^m{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{m!}}.$$

(114)

Next, let

$$\overline{U_n}:={\int }_{{\mathbb{R}}^N}RZ\left(nx-u\right)du,$$

(115)

Then,

$$\left|\overline{U_n}\right|\le {\int }_{{\mathbb{R}}^N}\left|R\right|Z\left(nx-u\right)du=$$

$${\int }_{\left\{u\in {\mathbb{R}}^N:{∥nx-u∥}_{\infty }<n^{1-\beta }\right\}}\left|R\right|Z\left(nx-u\right)du+$$

$${\int }_{\left\{u\in {\mathbb{R}}^N:{∥nx-u∥}_{\infty }\ge n^{1-\beta }\right\}}\left|R\right|Z\left(nx-u\right)du\stackrel{\left\{\right(112), (114\left)\right\}}{\le }$$

$$\left({\displaystyle \frac{N^m}{m!n^{m\beta }}}\right){\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n^{\beta }}}\right)+$$

(116)

$$\left({\displaystyle \frac{2{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{m!}}\right)\left({\int }_{\left\{u\in {\mathbb{R}}^N:{∥nx-u∥}_{\infty }\ge n^{1-\beta }\right\}}{∥{\displaystyle \frac{u}{n}}-x∥}_{\infty }^{m}Z\left(nx-u\right)du\right)\le$$

$$\left({\displaystyle \frac{N^m}{m!n^{m\beta }}}\right){\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n^{\beta }}}\right)+$$

$$\left({\displaystyle \frac{2{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{n^{m}m!}}\right){\int }_{{\mathbb{R}}^N}{∥nx-u∥}_{\infty }^{m}Z\left(nx-u\right)du=$$

$$\left({\displaystyle \frac{N^m}{m!n^{m\beta }}}\right){\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n^{\beta }}}\right)+$$

(117)

$$\left({\displaystyle \frac{2{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{n^{m}m!}}\right){\int }_{{\mathbb{R}}^N}{∥x∥}_{\infty }^{m}Z\left(x\right)dx\stackrel{\left(34\right)}{\le }$$

$$\left({\displaystyle \frac{N^m}{m!n^{m\beta }}}\right){\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n^{\beta }}}\right)+$$

$$\left({\displaystyle \frac{2{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{n^{m}m!}}\right)N^m\left[{\displaystyle \frac{tanh\left(\lambda \right)}{\left(m+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }m!}{{\left(2\lambda \right)}^m}}\right].$$

(118)

We prove that

$$\left|\overline{U_n}\right|\le \left({\displaystyle \frac{N^m}{m!n^{m\beta }}}\right){\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n^{\beta }}}\right)+$$

$$\left({\displaystyle \frac{2{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^{2m}}{n^{m}m!}}\right)\left[{\displaystyle \frac{tanh\left(\lambda \right)}{\left(m+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }m!}{{\left(2\lambda \right)}^m}}\right].$$

(119)

Next, we estimate

$${\int }_{{\mathbb{R}}^N}\left({\displaystyle \prod _{i=1}^N}{\left|{\displaystyle \frac{u_i}{n}}-x_i\right|}^{{\alpha }_i}\right)Z\left(nx-u\right)du=$$

$${\displaystyle \prod _{i=1}^N}{\int }_{-\infty }^{\infty }{\left|{\displaystyle \frac{u_i}{n}}-x_i\right|}^{{\alpha }_i}\phi \left(n{x}_i-u_i\right)d{u}_i=$$

$${\displaystyle \frac{1}{n^j}}{\displaystyle \prod _{i=1}^N}{\int }_{-\infty }^{\infty }{\left|u_i-n{x}_i\right|}^{{\alpha }_i}\phi \left(u_i-n{x}_i\right)d{u}_i=$$

(120)

$${\displaystyle \frac{1}{n^j}}{\displaystyle \prod _{i=1}^N}\left({\int }_{-\infty }^{\infty }{\left|x\right|}^{{\alpha }_i}\phi \left(x\right)dx\right)\stackrel{\left(26\right)}{\le }$$

$${\displaystyle \frac{1}{n^j}}\left({\displaystyle \prod _{i=1}^N}\left[{\displaystyle \frac{tanh\left(\lambda \right)}{\left({\alpha }_i+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }{\alpha }_i!}{{\left(2\lambda \right)}^{{\alpha }_i}}}\right]\right).$$

so that the following holds:

$${\int }_{{\mathbb{R}}^N}\left({\displaystyle \prod _{i=1}^N}{\left|{\displaystyle \frac{u_i}{n}}-x_i\right|}^{{\alpha }_i}\right)Z\left(nx-u\right)du\le$$

$${\displaystyle \frac{1}{n^j}}\left\{{\displaystyle \prod _{i=1}^N}\left[{\displaystyle \frac{tanh\left(\lambda \right)}{\left({\alpha }_i+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }{\alpha }_i!}{{\left(2\lambda \right)}^{{\alpha }_i}}}\right]\right\}.$$

(121)

According to (107), we can write

$${\int }_{{\mathbb{R}}^N}f\left({\displaystyle \frac{u}{n}}\right)Z\left(nx-u\right)du-f\left(x\right)-$$

$${\displaystyle \sum _{j=1}^m}\sum _{\left\{\begin{array}{c}\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right),{\alpha }_i\in {\mathbb{Z}}_{+}\\ i=1,\dots ,N,\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}{\alpha }_i=j\end{array}\right\}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

$$\left({\int }_{{\mathbb{R}}^N}\left({\displaystyle \prod _{i=1}^N}{\left({\displaystyle \frac{u_i}{n}}-x_i\right)}^{{\alpha }_i}\right)Z\left(nx-u\right)du\right)f_{\alpha }\left(x\right)={\int }_{{\mathbb{R}}^N}RZ\left(nx-u\right)du.$$

(122)

The theorem is proven. □

We continue with the activated multivariate Kantorovich operators under differentiation.

Theorem 13. 

Let $0<\beta <1$, $n\in \mathbb{N}:n^{1-\beta }>2$; $x\in {\mathbb{R}}^N,$$f\in C^m\left({\mathbb{R}}^N\right)$, $m,N\in \mathbb{N}$, with $f_{\alpha }\in C_B\left({\mathbb{R}}^N\right):\left|\alpha \right|=l$, $l=0,1,\dots ,m.$ Then,

(i) 

$$\left|S_n^{*}\left(f\right)\left(x\right)-f\left(x\right)-{\displaystyle \sum _{j=1}^N}\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}\right.}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)f_{\alpha }\left(x\right)S_n^{*}\left(\left({\displaystyle \prod _{i=1}^N}{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)\right|$$

$$\le {\displaystyle \frac{N^m}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+$$

$$\left({\displaystyle \frac{{\left(2N\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}}{n^{m}m!}}\right)\left\{1+N^m\left[{\displaystyle \frac{tanh\left(\lambda \right)}{\left(m+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }m!}{{\left(2\lambda \right)}^m}}\right]\right\}=:\overline{{\Psi }_1},$$

(123)

(ii) 

Assume that $f_{\alpha }\left(x\right)=0$ for all $\alpha :\left|\alpha \right|=j$, $j=1,\dots ,m$; we have

$$\left|S_n^{*}\left(f\right)\left(x\right)-f\left(x\right)\right|\le \overline{{\Psi }_1},$$

(124)

with a high speed of ${\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^{m+1}.$

(iii) 

$$\left|S_n^{*}\left(f\right)\left(x\right)-f\left(x\right)\right|\le$$

$${\displaystyle \sum _{j=1}^N}\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}\right.}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left|f_{\alpha }\left(x\right)\right|{\left({\displaystyle \frac{2}{n}}\right)}^j$$

$$\left({\displaystyle \prod _{i=1}^N}\left[1+{\displaystyle \frac{tanh\left(\lambda \right)}{\left({\alpha }_i+2\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }\left({\alpha }_i+1\right)!}{{\left(2\lambda \right)}^{{\alpha }_i+1}}}\right]\right)+\overline{{\Psi }_1},$$

(125)

and

(iv) 

$${∥S_n^{*}\left(f\right)-f∥}_{\infty }\le$$

$${\displaystyle \sum _{j=1}^N}\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}\right.}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right){∥f_{\alpha }∥}_{\infty }{\left({\displaystyle \frac{2}{n}}\right)}^j$$

$$\left({\displaystyle \prod _{i=1}^N}\left[1+{\displaystyle \frac{tanh\left(\lambda \right)}{\left({\alpha }_i+2\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }\left({\alpha }_i+1\right)!}{{\left(2\lambda \right)}^{{\alpha }_i+1}}}\right]\right)+\overline{{\Psi }_1}.$$

(126)

We determine that $S_n^{*}\left(f\right)\to f$, as $n\to \infty$, pointwise and uniformly.

Proof. 

It holds that

$$f\left(t+{\displaystyle \frac{u}{n}}\right)-f\left(x\right)-{\displaystyle \sum _{j=1}^m}\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}\right.}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left({\displaystyle \prod _{i=1}^N}{\left(t_i+{\displaystyle \frac{u_i}{n}}-x_i\right)}^{{\alpha }_i}\right)f_{\alpha }\left(x\right)=R,$$

(127)

where

$$R:=m{\int }_0^1{\left(1-\theta \right)}^{m-1}\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=m\end{array}\right.}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

(128)

$$\left({\displaystyle \prod _{i=1}^N}{\left(t_i+{\displaystyle \frac{u_i}{n}}-x_i\right)}^{{\alpha }_i}\right)\left[f_{\alpha }\left(x+\theta \left(t+{\displaystyle \frac{u}{n}}-x\right)\right)-f_{\alpha }\left(x\right)\right]d\theta .$$

We see that

$$\left|R\right|\le m{\int }_0^1{\left(1-\theta \right)}^{m-1}\sum _{\left|\alpha \right|=m}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

$$\left({\displaystyle \prod _{i=1}^N}{\left|t_i+{\displaystyle \frac{u_i}{n}}-x_i\right|}^{{\alpha }_i}\right)\left|f_{\alpha }\left(x+\theta \left(t+{\displaystyle \frac{u}{n}}-x\right)\right)-f_{\alpha }\left(x\right)\right|d\theta \le$$

(129)

$$m{\int }_0^1{\left(1-\theta \right)}^{m-1}\left(\sum _{\left|\alpha \right|=m}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left({\displaystyle \prod _{i=1}^N}{\left|t_i+{\displaystyle \frac{u_i}{n}}-x_i\right|}^{{\alpha }_i}\right)\right.$$

$$\left.{\omega }_1\left(f_{\alpha },\theta {∥t+{\displaystyle \frac{u}{n}}-x∥}_{\infty }\right)\right)d\theta \le \left(\ast \right).$$

Notice that

$${∥{\displaystyle \frac{u}{n}}-x∥}_{\infty }\le {\displaystyle \frac{1}{n^{\beta }}},\mathrm{iff}\left|{\displaystyle \frac{u_i}{n}}-x_i\right|\le {\displaystyle \frac{1}{n^{\beta }}},i=1,\dots ,N.$$

Here we consider $0\le t_i\le {\displaystyle \frac{1}{n}}$, $i=1,\dots ,N.$

We further see that

$$\left(\ast \right)\le m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right){\int }_0^1{\left(1-\theta \right)}^{m-1}$$

$$\left(\sum _{\left|\alpha \right|=m}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left({\displaystyle \prod _{i=1}^N}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^{{\alpha }_i}\right)\right)d\theta =$$

$${\displaystyle \frac{{\omega }_{1,m}^{max}\left(f_{\alpha },\frac{1}{n}+\frac{1}{n^{\beta }}\right)}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m\left(\sum _{\left|\alpha \right|=m}{\displaystyle \frac{m!}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)=$$

$${\displaystyle \frac{{\omega }_{1,m}^{max}\left(f_{\alpha },\frac{1}{n}+\frac{1}{n^{\beta }}\right)}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{N}^m.$$

(130)

Conclusion: When ${∥{\displaystyle \frac{u}{n}}-x∥}_{\infty }\le {\displaystyle \frac{1}{n^{\beta }}}$, we prove that

$$\left|R\right|\le {\displaystyle \frac{N^m}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right).$$

(131)

According to (128), we determine that

$$\left|R\right|\le m{\int }_0^1{\left(1-\theta \right)}^{m-1}\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=m\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

$$\left({\displaystyle \prod _{i=1}^N}{\left(\left|t_i\right|+\left|{\displaystyle \frac{u_i}{n}}-x_i\right|\right)}^{{\alpha }_i}2{∥f_{\alpha }∥}_{\infty }\right)d\theta \le$$

$$m{\int }_0^1{\left(1-\theta \right)}^{m-1}\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=m\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

$$\left({\displaystyle \prod _{i=1}^N}{\left({\displaystyle \frac{1}{n}}+\left|{\displaystyle \frac{u_i}{n}}-x_i\right|\right)}^{{\alpha }_i}2{∥f_{\alpha }∥}_{\infty }\right)d\theta =$$

(132)

$$\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=m\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left({\displaystyle \prod _{i=1}^N}{\left({\displaystyle \frac{1}{n}}+\left|{\displaystyle \frac{u_i}{n}}-x_i\right|\right)}^{{\alpha }_i}\right)2{∥f_{\alpha }∥}_{\infty }\le$$

$$\left({\displaystyle \frac{2{\left({∥\frac{u}{n}-x∥}_{\infty }+\frac{1}{n}\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}}{m!}}\right)\left(\sum _{\left|\alpha \right|=m}{\displaystyle \frac{m!}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)=$$

$$\left({\displaystyle \frac{2{\left({∥\frac{u}{n}-x∥}_{\infty }+\frac{1}{n}\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}}{m!}}\right)N^m.$$

(133)

We prove, in general, that

$$\left|R\right|\le {\displaystyle \frac{2{\left({∥\frac{u}{n}-x∥}_{\infty }+\frac{1}{n}\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{m!}}.$$

(134)

Next, we see that

$$n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}f\left(t+{\displaystyle \frac{u}{n}}\right)dt-f\left(x\right)-{\displaystyle \sum _{j=1}^m}\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)f_{\alpha }\left(x\right)$$

(135)

$$n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left({\displaystyle \prod _{i=1}^N}{\left(t_i+{\displaystyle \frac{u_i}{n}}-x_i\right)}^{{\alpha }_i}\right)dt=n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}Rdt.$$

So, when ${∥{\displaystyle \frac{u}{n}}-x∥}_{\infty }\le {\displaystyle \frac{1}{n^{\beta }}}$, we obtain

$$n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left|R\right|dt\le {\displaystyle \frac{N^m}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right).$$

(136)

and, in general, it holds that

$$n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left|R\right|dt\le$$

(137)

$${\displaystyle \frac{2{\left({∥\frac{u}{n}-x∥}_{\infty }+\frac{1}{n}\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{m!}}.$$

Furthermore, it holds that

$$S_n^{*}\left(f\right)\left(x\right)-f\left(x\right)-{\displaystyle \sum _{j=1}^N}\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)f_{\alpha }\left(x\right)S_n^{*}\left(\left({\displaystyle \prod _{i=1}^N}{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)=$$

$${\int }_{{\mathbb{R}}^N}\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}Rdt\right)Z\left(nx-u\right)du.$$

(138)

Let

$$U_n^{*}:={\int }_{{\mathbb{R}}^N}\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}Rdt\right)Z\left(nx-u\right)du.$$

(139)

Here, $A_1$ is as in (45), and $A_2$ is as in (46).

We do have, under ${∥{\displaystyle \frac{u}{n}}-x∥}_{\infty }<{\displaystyle \frac{1}{n^{\beta }}}$,

$$|U_n^{*}|{|}_{A_1}\le {\displaystyle \frac{N^m}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right).$$

(140)

Furthermore, we determine that

$$|U_n^{*}{||}_{A_2}\le \left({\displaystyle \frac{2{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{m!}}\right){\int }_{A_2}{\left({∥{\displaystyle \frac{u}{n}}-x∥}_{\infty }+{\displaystyle \frac{1}{n}}\right)}^{m}Z\left(nx-u\right)du.$$

(141)

or, better still,

$$\left|U_n^{*}\right|{|}_{A_2}\le \left({\displaystyle \frac{2{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{n^{m}m!}}\right){\int }_{A_2}{\left({∥u-nx∥}_{\infty }+1\right)}^{m}Z\left(u-nx\right)du=$$

(142)

$$\left({\displaystyle \frac{2{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{n^{m}m!}}\right)\left({\int }_{A_2}{\left({∥z∥}_{\infty }+1\right)}^{m}Z\left(z\right)dz\right)\le$$

$$\left({\displaystyle \frac{2^m{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{n^{m}m!}}\right)\left({\int }_{A_2}\left(1+{∥z∥}_{\infty }^m\right)Z\left(z\right)dz\right)\le$$

(143)

$$\left({\displaystyle \frac{{\left(2N\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}}{n^{m}m!}}\right)\left[1+{\int }_{A_2}{∥z∥}_{\infty }^{m}Z\left(z\right)dz\right]\stackrel{\left(34\right)}{\le }$$

$$\left({\displaystyle \frac{{\left(2N\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}}{n^{m}m!}}\right)\left\{1+N^m\left[{\displaystyle \frac{tanh\left(\lambda \right)}{\left(m+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }m!}{{\left(2\lambda \right)}^m}}\right]\right\}.$$

We prove that

$$\left|U_n^{*}\right|{|}_{A_2}\le \left({\displaystyle \frac{{\left(2N\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}}{n^{m}m!}}\right)$$

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

(144)

Consequently, we derive that

$$\left|U_n^{*}\right|\le \left|U_n^{*}\right|{{|}_{A_1}+\left|U_n^{*}\right||}_{A_2}\le$$

$${\displaystyle \frac{N^m}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+$$

$$\left({\displaystyle \frac{{\left(2N\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}}{n^{m}m!}}\right)\left\{1+N^m\left[{\displaystyle \frac{tanh\left(\lambda \right)}{\left(m+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }m!}{{\left(2\lambda \right)}^m}}\right]\right\}.$$

(145)

Finally, we estimate

$$\left|S_n^{*}\left(\left({\displaystyle \prod _{i=1}^N}{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)\right|=$$

$$\left|{\int }_{{\mathbb{R}}^N}\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left({\displaystyle \prod _{i=1}^N}{\left(t_i+{\displaystyle \frac{u_i}{n}}-x_i\right)}^{{\alpha }_i}\right)dt\right)Z\left(nx-u\right)du\right|\le$$

(146)

$${\int }_{{\mathbb{R}}^N}\left({\displaystyle \prod _{i=1}^N}{\left({\displaystyle \frac{1}{n}}+\left|{\displaystyle \frac{u_i}{n}}-x_i\right|\right)}^{{\alpha }_i}\right)Z\left(nx-u\right)du=$$

$${\displaystyle \prod _{i=1}^N}{\int }_{-\infty }^{\infty }{\left({\displaystyle \frac{1}{n}}+\left|{\displaystyle \frac{u_i}{n}}-x_i\right|\right)}^{{\alpha }_i}\phi \left(n{x}_i-u_i\right)d{u}_i=$$

$${\displaystyle \frac{1}{n^j}}\left({\displaystyle \prod _{i=1}^N}{\int }_{-\infty }^{\infty }{\left(1+\left|u_i-n{x}_i\right|\right)}^{{\alpha }_i}\phi \left(u_i-n{x}_i\right)d{u}_i\right)=$$

$${\displaystyle \frac{1}{n^j}}\left({\displaystyle \prod _{i=1}^N}{\int }_{-\infty }^{\infty }{\left(1+\left|x\right|\right)}^{{\alpha }_i}\phi \left(x\right)dx\right)\le$$

$${\displaystyle \frac{1}{n^j}}\left({\displaystyle \prod _{i=1}^N}{\int }_{-\infty }^{\infty }{\left(1+\left|x\right|\right)}^{{\alpha }_i+1}\phi \left(x\right)dx\right)\le$$

(147)

$${\displaystyle \frac{1}{n^j}}\left({\displaystyle \prod _{i=1}^N}{2}^{{\alpha }_i}{\int }_{-\infty }^{\infty }\left(1+{\left|x\right|}^{{\alpha }_i+1}\right)\phi \left(x\right)dx\right)=$$

$${\left({\displaystyle \frac{2}{n}}\right)}^j\left({\displaystyle \prod _{i=1}^N}\left[1+{\int }_{-\infty }^{\infty }{\left|x\right|}^{{\alpha }_i+1}\phi \left(x\right)dx\right]\right)\stackrel{\left(26\right)}{\le }$$

$${\left({\displaystyle \frac{2}{n}}\right)}^j\left({\displaystyle \prod _{i=1}^N}\left[1+{\displaystyle \frac{tanh\left(\lambda \right)}{\left({\alpha }_i+2\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }\left({\alpha }_i+1\right)!}{{\left(2\lambda \right)}^{{\alpha }_i+1}}}\right]\right).$$

(148)

We derive that

$$\left|S_n^{*}\left(\left({\displaystyle \prod _{i=1}^N}{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)\right|\le$$

$${\left({\displaystyle \frac{2}{n}}\right)}^j\left({\displaystyle \prod _{i=1}^N}\left[1+{\displaystyle \frac{tanh\left(\lambda \right)}{\left({\alpha }_i+2\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }\left({\alpha }_i+1\right)!}{{\left(2\lambda \right)}^{{\alpha }_i+1}}}\right]\right).$$

(149)

The theorem is proven. □

We continue with the activated multivariate quadrature operators under differentiation.

Theorem 14. 

Let $0<\beta <1$, $n\in \mathbb{N}:n^{1-\beta }>2$; $x\in {\mathbb{R}}^N,$$f\in C^m\left({\mathbb{R}}^N\right)$, $m,N\in \mathbb{N}$, with $f_{\alpha }\in C_B\left({\mathbb{R}}^N\right):\left|\alpha \right|=l,$$l=0,1,\dots ,m.$ Then,

(i) 

$$\left|\overline{S_n}\left(f\right)\left(x\right)-f\left(x\right)-{\displaystyle \sum _{j=1}^N}\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}\right.}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)f_{\alpha }\left(x\right)\overline{S_n}\left(\left({\displaystyle \prod _{i=1}^N}{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)\right|$$

$$\le {\displaystyle \frac{N^m}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+$$

$$\left({\displaystyle \frac{{\left(2N\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}}{n^{m}m!}}\right)\left\{1+N^m\left[{\displaystyle \frac{tanh\left(\lambda \right)}{\left(m+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }m!}{{\left(2\lambda \right)}^m}}\right]\right\}=\overline{{\Psi }_1},$$

(ii) 

Assume that $f_{\alpha }\left(x\right)=0$ for all $\alpha :\left|\alpha \right|=j$, $j=1,\dots ,m$; we have

$$\left|\overline{S_n}\left(f\right)\left(x\right)-f\left(x\right)\right|\le \overline{{\Psi }_1},$$

(150)

with a high speed of ${\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^{m+1},$

(iii) 

$$\left|\overline{S_n}\left(f\right)\left(x\right)-f\left(x\right)\right|\le$$

$${\displaystyle \sum _{j=1}^N}\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}\right.}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left|f_{\alpha }\left(x\right)\right|{\left({\displaystyle \frac{2}{n}}\right)}^j$$

$$\left({\displaystyle \prod _{i=1}^N}\left[1+{\displaystyle \frac{tanh\left(\lambda \right)}{\left({\alpha }_i+2\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }\left({\alpha }_i+1\right)!}{{\left(2\lambda \right)}^{{\alpha }_i+1}}}\right]\right)+\overline{{\Psi }_1},$$

(151)

and

(iv) 

$${∥\overline{S_n}\left(f\right)-f∥}_{\infty }\le$$

$${\displaystyle \sum _{j=1}^N}\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}\right.}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right){∥f_{\alpha }∥}_{\infty }{\left({\displaystyle \frac{2}{n}}\right)}^j$$

$$\left({\displaystyle \prod _{i=1}^N}\left[1+{\displaystyle \frac{tanh\left(\lambda \right)}{\left({\alpha }_i+2\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }\left({\alpha }_i+1\right)!}{{\left(2\lambda \right)}^{{\alpha }_i+1}}}\right]\right)+\overline{{\Psi }_1}.$$

(152)

We determine that $\overline{S_n}\left(f\right)\to f$, as $n\to \infty$, pointwise and uniformly.

Proof. 

We determine that

$$f\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{r}{n\theta }}\right)-f\left(x\right)-{\displaystyle \sum _{j=1}^m}\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left({\displaystyle \prod _{i=1}^N}{\left({\displaystyle \frac{u_i}{n}}+{\displaystyle \frac{r_i}{n{\theta }_i}}-x_i\right)}^{{\alpha }_i}\right)f_{\alpha }\left(x\right)=R,$$

(153)

where

$$R:=m{\int }_0^1{\left(1-\theta \right)}^{m-1}\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=m\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

(154)

$$\left({\displaystyle \prod _{i=1}^N}{\left({\displaystyle \frac{u_i}{n}}+{\displaystyle \frac{r_i}{n{\theta }_i}}-x_i\right)}^{{\alpha }_i}\right)\left[f_{\alpha }\left(x+\theta \left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{r}{n\theta }}-x\right)\right)-f_{\alpha }\left(x\right)\right]d\theta .$$

We see that

$$\left|R\right|\le m{\int }_0^1{\left(1-\theta \right)}^{m-1}\sum _{\left|\alpha \right|=m}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

$$\left({\displaystyle \prod _{i=1}^N}{\left|{\displaystyle \frac{u_i}{n}}+{\displaystyle \frac{r_i}{n{\theta }_i}}-x_i\right|}^{{\alpha }_i}\right)\left|f_{\alpha }\left(x+\theta \left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{r}{n\theta }}-x\right)\right)-f_{\alpha }\left(x\right)\right|d\theta \le$$

(155)

$$m{\int }_0^1{\left(1-\theta \right)}^{m-1}\left(\sum _{\left|\alpha \right|=m}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left({\displaystyle \prod _{i=1}^N}{\left|{\displaystyle \frac{u_i}{n}}+{\displaystyle \frac{r_i}{n{\theta }_i}}-x_i\right|}^{{\alpha }_i}\right)\right.$$

$$\left.{\omega }_1\left(f_{\alpha },\theta {∥{\displaystyle \frac{u}{n}}+{\displaystyle \frac{r}{n\theta }}-x∥}_{\infty }\right)\right)d\theta \le \left(\ast \right).$$

Notice that

$${∥{\displaystyle \frac{u}{n}}-x∥}_{\infty }\le {\displaystyle \frac{1}{n^{\beta }}},\mathrm{iff}\left|{\displaystyle \frac{u_i}{n}}-x_i\right|\le {\displaystyle \frac{1}{n^{\beta }}},i=1,\dots ,N.$$

We further see that

$$\left(\ast \right)\le m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right){\int }_0^1{\left(1-\theta \right)}^{m-1}$$

$$\left(\sum _{\left|\alpha \right|=m}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left({\displaystyle \prod _{i=1}^N}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^{{\alpha }_i}\right)\right)d\theta =$$

$${\displaystyle \frac{{\omega }_{1,m}^{max}\left(f_{\alpha },\frac{1}{n}+\frac{1}{n^{\beta }}\right)}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m\left(\sum _{\left|\alpha \right|=m}{\displaystyle \frac{m!}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)=$$

$${\displaystyle \frac{{\omega }_{1,m}^{max}\left(f_{\alpha },\frac{1}{n}+\frac{1}{n^{\beta }}\right)}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{N}^m.$$

(156)

Conclusion: When ${∥{\displaystyle \frac{u}{n}}-x∥}_{\infty }\le {\displaystyle \frac{1}{n^{\beta }}}$, we prove that

$$\left|R\right|\le {\displaystyle \frac{N^m}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right).$$

(157)

According to (154), we determine that

$$\left|R\right|\le m{\int }_0^1{\left(1-\theta \right)}^{m-1}\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=m\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

$$\left({\displaystyle \prod _{i=1}^N}{\left(\left|{\displaystyle \frac{u_i}{n}}-x_i\right|+{\displaystyle \frac{1}{n}}\right)}^{{\alpha }_i}2{∥f_{\alpha }∥}_{\infty }\right)d\theta =$$

(158)

$$\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=m\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left({\displaystyle \prod _{i=1}^N}{\left(\left|{\displaystyle \frac{u_i}{n}}-x_i\right|+{\displaystyle \frac{1}{n}}\right)}^{{\alpha }_i}\right)2{∥f_{\alpha }∥}_{\infty }\le$$

$$\left({\displaystyle \frac{2{\left({∥\frac{u}{n}-x∥}_{\infty }+\frac{1}{n}\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}}{m!}}\right)\left(\sum _{\left|\alpha \right|=m}{\displaystyle \frac{m!}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)=$$

$$\left({\displaystyle \frac{2{\left({∥\frac{u}{n}-x∥}_{\infty }+\frac{1}{n}\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}}{m!}}\right)N^m.$$

We establish, in general, that

$$\left|R\right|\le {\displaystyle \frac{2{\left({∥\frac{u}{n}-x∥}_{\infty }+\frac{1}{n}\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{m!}}.$$

(159)

Next, we observe that

$${\displaystyle \sum _{r=0}^{\theta }}{w}_{r}f\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{r}{n\theta }}\right)-f\left(x\right)-{\displaystyle \sum _{j=1}^m}\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)f_{\alpha }\left(x\right)$$

$$\left({\displaystyle \sum _{r=0}^{\theta }}{w}_r\left({\displaystyle \prod _{i=1}^N}{\left({\displaystyle \frac{u_i}{n}}+{\displaystyle \frac{r_i}{n{\theta }_i}}-x_i\right)}^{{\alpha }_i}\right)\right)={\displaystyle \sum _{r=0}^{\theta }}{w}_{r}R.$$

(160)

So, when ${∥{\displaystyle \frac{u}{n}}-x∥}_{\infty }\le {\displaystyle \frac{1}{n^{\beta }}}$, we obtain

$$\left|{\displaystyle \sum _{r=0}^{\theta }}{w}_{r}R\right|\le {\displaystyle \sum _{r=0}^{\theta }}{w}_r\left|R\right|\le {\displaystyle \frac{N^m}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right).$$

(161)

and it holds, in general, that

$$\left|{\displaystyle \sum _{r=0}^{\theta }}{w}_{r}R\right|\le {\displaystyle \frac{2{\left({∥\frac{u}{n}-x∥}_{\infty }+\frac{1}{n}\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{m!}}.$$

(162)

Furthermore, it holds that

$$\overline{S_n}\left(f\right)\left(x\right)-f\left(x\right)-{\displaystyle \sum _{j=1}^N}\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)f_{\alpha }\left(x\right)\overline{S_n}\left(\left({\displaystyle \prod _{i=1}^N}{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)=$$

$${\int }_{{\mathbb{R}}^N}\left({\displaystyle \sum _{r=0}^{\theta }}{w}_{r}R\right)Z\left(nx-u\right)du.$$

(163)

Let

$$\overline{E_n}:={\int }_{{\mathbb{R}}^N}\left({\displaystyle \sum _{r=0}^{\theta }}{w}_{r}R\right)Z\left(nx-u\right)du.$$

(164)

Here, $A_1$ is as in (45), and $A_2$ is as in (46).

We derive, under ${∥{\displaystyle \frac{u}{n}}-x∥}_{\infty }<{\displaystyle \frac{1}{n^{\beta }}}$,

$$|\overline{E_n}|{|}_{A_1}\le {\displaystyle \frac{N^m}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right).$$

(165)

Furthermore, we determine that

$$|\overline{E_n}|{|}_{A_2}\le \left({\displaystyle \frac{2{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{m!}}\right){\int }_{A_2}{\left({∥{\displaystyle \frac{u}{n}}-x∥}_{\infty }+{\displaystyle \frac{1}{n}}\right)}^{m}Z\left(nx-u\right)du.$$

(166)

As in the proof of Theorem 13, we obtain

$$|\overline{E_n}|{|}_{A_2}\le \left({\displaystyle \frac{{\left(2N\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}}{n^{m}m!}}\right)$$

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

(167)

Consequently, we derive that

$$\left|\overline{E_n}\right|\le \left|\overline{E_n}\right|{{|}_{A_1}+\left|\overline{E_n}\right||}_{A_2}\le$$

$${\displaystyle \frac{N^m}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+$$

$$\left({\displaystyle \frac{{\left(2N\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}}{n^{m}m!}}\right)\left\{1+N^m\left[{\displaystyle \frac{tanh\left(\lambda \right)}{\left(m+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }m!}{{\left(2\lambda \right)}^m}}\right]\right\}.$$

(168)

At the end, we estimate

$$\left|\overline{S_n}\left(\left({\displaystyle \prod _{i=1}^N}{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)\right|=$$

$$\left|{\int }_{{\mathbb{R}}^N}\left({\displaystyle \sum _{r=0}^{\theta }}{w}_r\left({\displaystyle \prod _{i=1}^N}{\left({\displaystyle \frac{u_i}{n}}+{\displaystyle \frac{r_i}{n{\theta }_i}}-x_i\right)}^{{\alpha }_i}\right)\right)Z\left(nx-u\right)du\right|\le$$

$${\int }_{{\mathbb{R}}^N}\left({\displaystyle \sum _{r=0}^{\theta }}{w}_r\left({\displaystyle \prod _{i=1}^N}{\left(\left|{\displaystyle \frac{u_i}{n}}-x_i\right|+{\displaystyle \frac{1}{n}}\right)}^{{\alpha }_i}\right)\right)Z\left(nx-u\right)du=$$

(169)

$${\int }_{{\mathbb{R}}^N}\left({\displaystyle \prod _{i=1}^N}{\left(\left|{\displaystyle \frac{u_i}{n}}-x_i\right|+{\displaystyle \frac{1}{n}}\right)}^{{\alpha }_i}\right)Z\left(nx-u\right)du$$

(as in the proof of Theorem 13)

$$\le {\left({\displaystyle \frac{2}{n}}\right)}^j\left({\displaystyle \prod _{i=1}^N}\left[1+{\displaystyle \frac{tanh\left(\lambda \right)}{\left({\alpha }_i+2\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }\left({\alpha }_i+1\right)!}{{\left(2\lambda \right)}^{{\alpha }_i+1}}}\right]\right).$$

(170)

The theorem is proven. □

Next comes the simultaneous multivariate activated approximation.

Theorem 15. 

Let $i\in \mathbb{N}$ be fixed, with $f\in C^{\left(i\right)}\left({\mathbb{R}}^N\right)$, $N\in \mathbb{N}$. We assume that $f_{\alpha }\in C_B\left({\mathbb{R}}^N\right)$ for $\alpha :\left|\alpha \right|=l$, $l=0,1,\dots ,i$. Here, $0<\beta <1$, $n\in \mathbb{N}:n^{1-\beta }>2$, $x\in {\mathbb{R}}^N$.

Then,

(i) 

$$\left|{\left(S_n\left(f\right)\right)}_{\alpha }\left(x\right)-f_{\alpha }\left(x\right)\right|\le {\omega }_1\left(f_{\alpha },{\displaystyle \frac{1}{n^{\beta }}}\right)+{\displaystyle \frac{2\left(q+\frac{1}{q}\right){∥f_{\alpha }∥}_{\infty }}{e^{2\lambda \left(n^{1-\beta }-1\right)}}}=:{\lambda }_{\alpha }^s,$$

(171)

and

$${∥{\left(S_n\left(f\right)\right)}_{\alpha }-f_{\alpha }∥}_{\infty }\le {\lambda }_{\alpha }^s,$$

(172)

(ii) 

$$\left|{\left(S_n^{*}\left(f\right)\right)}_{\alpha }\left(x\right)-f_{\alpha }\left(x\right)\right|\le {\omega }_1\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+{\displaystyle \frac{2\left(q+\frac{1}{q}\right){∥f_{\alpha }∥}_{\infty }}{e^{2\lambda \left(n^{1-\beta }-1\right)}}}=:{\rho }_{\alpha }^s,$$

(173)

and

$${∥{\left(S_n^{*}\left(f\right)\right)}_{\alpha }-f_{\alpha }∥}_{\infty }\le {\rho }_{\alpha }^s,$$

(174)

and

(iii) 

$$\left|{\left(\overline{S_n}\left(f\right)\right)}_{\alpha }\left(x\right)-f_{\alpha }\left(x\right)\right|\le {\omega }_1\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+{\displaystyle \frac{2\left(q+\frac{1}{q}\right){∥f_{\alpha }∥}_{\infty }}{e^{2\lambda \left(n^{1-\beta }-1\right)}}}={\rho }_{\alpha }^s,$$

(175)

and

$${∥{\left(\overline{S_n}\left(f\right)\right)}_{\alpha }-f_{\alpha }∥}_{\infty }\le {\rho }_{\alpha }^s.$$

(176)

Proof. 

We prove Theorems 6–8 and Remark 3. □

Next comes simultaneous global smoothness preservation.

Theorem 16. 

Let $i\in \mathbb{N}$ be fixed, with $f\in C^{\left(i\right)}\left({\mathbb{R}}^N\right)$, $N\in \mathbb{N}$. We assume that $f_{\alpha }\in C_B\left({\mathbb{R}}^N\right)\cup C_U\left({\mathbb{R}}^N\right)$ for $\alpha :\left|\alpha \right|=l$, $l=0,1,\dots ,i$.

Then,

$${\omega }_1\left({\left(S_n\left(f\right)\right)}_{\alpha },\delta \right)\le {\omega }_1\left(f_{\alpha },\delta \right),\delta >0$$

(177)

$${\omega }_1\left({\left(S_n^{*}\left(f\right)\right)}_{\alpha },\delta \right)\le {\omega }_1\left(f_{\alpha },\delta \right),$$

(178)

and

$${\omega }_1\left({\left(\overline{S_n}\left(f\right)\right)}_{\alpha },\delta \right)\le {\omega }_1\left(f_{\alpha },\delta \right).$$

(179)

If $f_{\alpha }\in C_U\left({\mathbb{R}}^N\right)$, then ${\left(S_n\left(f\right)\right)}_{\alpha }$, ${\left(S_n^{*}\left(f\right)\right)}_{\alpha }$ and ${\left(\overline{S_n}\left(f\right)\right)}_{\alpha }\in C_U\left({\mathbb{R}}^N\right).$

Proof. 

We prove Theorems 9–11 and Remark 3. □

Under simultaneous activated multivariate extended differentiation, we derive the following result.

Theorem 17. 

Let $0<\beta <1,$$n\in \mathbb{N}:n^{1-\beta }>2$; $x\in {\mathbb{R}}^N$, $N\in \mathbb{N}$. Let $f\in C^{\left(\overline{i}\right)}\left({\mathbb{R}}^N\right)$, $\overline{i}\in \mathbb{N};$$f_{\gamma }$ denote a partial derivative of f, $\gamma :=\left({\gamma }_1,\dots ,{\gamma }_N\right)$, ${\gamma }_j\in {\mathbb{Z}}_{+}$, $j=1,\dots ,N$, and $\left|\gamma \right|:={\displaystyle \sum _{j=1}^N}{\gamma }_j=r$, where $r=0,1,\dots ,\overline{i}$. We assume any $f_{\gamma }\in C_B\left({\mathbb{R}}^N\right)$ for all $\gamma :\left|\gamma \right|=r$, $r=0,1,\dots ,\overline{i}$.

We further assume that $f_{\gamma }\in C^m\left({\mathbb{R}}^N\right)$, $m\in \mathbb{N}$, with ${\left(f_{\gamma }\right)}_{\alpha }\in C_B\left({\mathbb{R}}^N\right):\left|\alpha \right|=l$, $l=0,1,\dots ,m$. Then,

(i) 

$$\left|{\left(S_n\left(f\right)\right)}_{\gamma }\left(x\right)-\left(f_{\gamma }\right)\left(x\right)-\right.$$

$$\left.{\displaystyle \sum _{j=1}^m}\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right){\left(f_{\gamma }\right)}_{\alpha }\left(x\right)S_n\left(\left({\displaystyle \prod _{i=1}^N}{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)\right|\le$$

$$\left({\displaystyle \frac{N^m}{m!n^{m\beta }}}\right){\omega }_{1,m}^{max}\left({\left(f_{\gamma }\right)}_{\alpha },{\displaystyle \frac{1}{n^{\beta }}}\right)+$$

$$\left({\displaystyle \frac{2{∥{\left(f_{\gamma }\right)}_{\alpha }∥}_{\infty ,m}^{max}{N}^{2m}}{n^{m}m!}}\right)\left[{\displaystyle \frac{tanh\left(\lambda \right)}{\left(m+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }m!}{{\left(2\lambda \right)}^m}}\right],$$

(180)

(ii) 

$$\left|{\left(S_n^{*}\left(f\right)\right)}_{\gamma }\left(x\right)-\left(f_{\gamma }\right)\left(x\right)-\right.$$

$$\left.{\displaystyle \sum _{j=1}^N}\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right){\left(f_{\gamma }\right)}_{\alpha }\left(x\right)S_n^{*}\left(\left({\displaystyle \prod _{i=1}^N}{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)\right|\le$$

$${\displaystyle \frac{N^m}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{\omega }_{1,m}^{max}\left({\left(f_{\gamma }\right)}_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+$$

$$\left({\displaystyle \frac{{\left(2N\right)}^m{∥{\left(f_{\gamma }\right)}_{\alpha }∥}_{\infty ,m}^{max}}{n^{m}m!}}\right)\left\{1+N^m\left[{\displaystyle \frac{tanh\left(\lambda \right)}{\left(m+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }m!}{{\left(2\lambda \right)}^m}}\right]\right\}=:{\left({\Psi }_{\gamma }^s\right)}_{\alpha },$$

(181)

and

(iii) 

$$\left|{\left(\overline{S_n}\left(f\right)\right)}_{\gamma }\left(x\right)-\left(f_{\gamma }\right)\left(x\right)-\right.$$

$$\left.{\displaystyle \sum _{j=1}^N}\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right){\left(f_{\gamma }\right)}_{\alpha }\left(x\right)\overline{S_n}\left(\left({\displaystyle \prod _{i=1}^N}{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)\right|\le {\left({\Psi }_{\gamma }^s\right)}_{\alpha }.$$

(182)

Proof. 

We prove Theorems 12–14 and Remark 3. □

In the final part of this work, we present our results related to activated iterative approximation. This is a continuation of Remarks 4–6.

Theorem 18. 

Let $0<\beta <1$, $n\in \mathbb{N}:n^{1-\beta }>2$, $r\in \mathbb{N},$$f\in C_B\left({\mathbb{R}}^N\right)$. Then,

(I) 

$${∥S_n^{r}f-f∥}_{\infty }\le r{∥S_{n}f-f∥}_{\infty }\le r\left[{\omega }_1\left(f,{\displaystyle \frac{1}{n^{\beta }}}\right)+{\displaystyle \frac{2\left(q+\frac{1}{q}\right){∥f∥}_{\infty }}{e^{2\lambda \left(n^{1-\beta }-1\right)}}}\right],$$

(183)

(II) 

$${∥S_n^{{*}^r}\left(f\right)-f∥}_{\infty }\le r{∥S_n^{*}f-f∥}_{\infty }\le r\left[{\omega }_1\left(f,{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+{\displaystyle \frac{2\left(q+\frac{1}{q}\right){∥f∥}_{\infty }}{e^{2\lambda \left(n^{1-\beta }-1\right)}}}\right],$$

(184)

and

(III) 

$${∥{\overline{S_n}}^r\left(f\right)-f∥}_{\infty }\le r{∥\overline{S_n}f-f∥}_{\infty }\le r\left[{\omega }_1\left(f,{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+{\displaystyle \frac{2\left(q+\frac{1}{q}\right){∥f∥}_{\infty }}{e^{2\lambda \left(n^{1-\beta }-1\right)}}}\right].$$

(185)

So, the speed of convergence of $S_n^r$, $S_n^{{*}^r}$, ${\overline{S_n}}^r$ to unit I is not worse than the speed of convergence of $S_n$, $S_n^{*}$, $\overline{S_n}$ to I.

Proof. 

We prove Theorems 6–8 and (87). □

We continue with the following:

Theorem 19. 

Let $0<\beta <1;$$m_1,m_2,\dots ,m_r\in \mathbb{N}:m_1\le m_2\le \dots \le m_r$, with $m_i^{1-\beta }>2$, $i=1,\dots ,r$; $f\in C_B\left({\mathbb{R}}^N\right)$. Then,

(I) 

$${∥S_{m_r}\left(S_{m_{r-1}}\left(\dots S_{m_2}\left(S_{m_1}f\right)\right)\right)-f∥}_{\infty }\le {\displaystyle \sum _{i=1}^r}{∥S_{m_i}f-f∥}_{\infty }\le$$

$${\displaystyle \sum _{i=1}^r}\left[{\omega }_1\left(f,{\displaystyle \frac{1}{m_i^{\beta }}}\right)+{\displaystyle \frac{2\left(q+\frac{1}{q}\right){∥f∥}_{\infty }}{e^{2\lambda \left(m_i^{1-\beta }-1\right)}}}\right]\le r\left[{\omega }_1\left(f,{\displaystyle \frac{1}{m_1^{\beta }}}\right)+{\displaystyle \frac{2\left(q+\frac{1}{q}\right){∥f∥}_{\infty }}{e^{2\lambda \left(m_1^{1-\beta }-1\right)}}}\right],$$

(186)

(II) 

$${∥S_{m_r}^{*}\left(S_{m_{r-1}}^{*}\left(\dots S_{m_2}^{*}\left(S_{m_1}^{*}f\right)\right)\right)-f∥}_{\infty }\le {\displaystyle \sum _{i=1}^r}{∥S_{m_i}^{*}f-f∥}_{\infty }\le$$

(187)

$${\displaystyle \sum _{i=1}^r}\left[{\omega }_1\left(f,{\displaystyle \frac{1}{m_i}}+{\displaystyle \frac{1}{m_i^{\beta }}}\right)+{\displaystyle \frac{2\left(q+\frac{1}{q}\right){∥f∥}_{\infty }}{e^{2\lambda \left(m_i^{1-\beta }-1\right)}}}\right]\le$$

$$r\left[{\omega }_1\left(f,{\displaystyle \frac{1}{m_1}}+{\displaystyle \frac{1}{m_1^{\beta }}}\right)+{\displaystyle \frac{2\left(q+\frac{1}{q}\right){∥f∥}_{\infty }}{e^{2\lambda \left(m_1^{1-\beta }-1\right)}}}\right],$$

and

(III) 

$${∥\overline{S_{m_r}}\left(\overline{S_{m_{r-1}}}\left(\dots \overline{S_{m_2}}\left(\overline{S_{m_1}}f\right)\right)\right)-f∥}_{\infty }\le {\displaystyle \sum _{i=1}^r}{∥\overline{S_{m_i}}f-f∥}_{\infty }\le$$

$${\displaystyle \sum _{i=1}^r}\left[{\omega }_1\left(f,{\displaystyle \frac{1}{m_i}}+{\displaystyle \frac{1}{m_i^{\beta }}}\right)+{\displaystyle \frac{2\left(q+\frac{1}{q}\right){∥f∥}_{\infty }}{e^{2\lambda \left(m_i^{1-\beta }-1\right)}}}\right]\le$$

(188)

$$r\left[{\omega }_1\left(f,{\displaystyle \frac{1}{m_1}}+{\displaystyle \frac{1}{m_1^{\beta }}}\right)+{\displaystyle \frac{2\left(q+\frac{1}{q}\right){∥f∥}_{\infty }}{e^{2\lambda \left(m_1^{1-\beta }-1\right)}}}\right].$$

Clearly, we notice that the speed of convergence to the unit operator of the above activated multiply iterated operator is not worse than the speed of operators $S_{m_1}$, $S_{m_1}^{*}$, and $\overline{S_{m_1}}$ to the unit, respectively.

Proof. 

We prove Theorems 6–8 and (89). □

We finish our work with multivariate simultaneous iterations.

Remark 8. 

Let $i\in \mathbb{N}$ be fixed. Assume that $f\in C^{\left(i\right)}\left({\mathbb{R}}^N\right)$, with $f_{\alpha }\in C_B\left({\mathbb{R}}^N\right)$, with $\alpha :\left|\alpha \right|=l$, $l=0,1,\dots ,i$; $r\in \mathbb{N}$. Then, according to (87), we obtain

$${∥S_n^r\left(f_{\alpha }\right)-f_{\alpha }∥}_{\infty }\le r{∥S_n\left(f_{\alpha }\right)-f_{\alpha }∥}_{\infty }.$$

(189)

According to (78) and inductively, we obtain

$${∥{\left(S_n^r\left(f\right)\right)}_{\alpha }-f_{\alpha }∥}_{\infty }\le r{∥{\left(S_n\left(f\right)\right)}_{\alpha }-f_{\alpha }∥}_{\infty },$$

(190)

Similarly, we derive that

$${∥{\left(S_n^{{*}^r}\left(f\right)\right)}_{\alpha }-f_{\alpha }∥}_{\infty }\le r{∥{\left(S_n^{*}\left(f\right)\right)}_{\alpha }-f_{\alpha }∥}_{\infty },$$

(191)

and

$${∥{\left({\overline{S_n}}^r\left(f\right)\right)}_{\alpha }-f_{\alpha }∥}_{\infty }\le r{∥{\left(\overline{S_n}\left(f\right)\right)}_{\alpha }-f_{\alpha }∥}_{\infty }.$$

(192)

Now, let $m_1,m_2,\dots ,m_r\in \mathbb{N}:m_1\le m_2\le \dots \le m_r$. Then, based on (89), we find that

$${∥{\left(S_{m_r}\left(S_{m_{r-1}}\left(\dots S_{m_2}\left(S_{m_1}f\right)\right)\right)\right)}_{\alpha }-f_{\alpha }∥}_{\infty }\le {\displaystyle \sum _{i^{*}=1}^r}{∥{\left(S_{m_{i^{*}}}\left(f\right)\right)}_{\alpha }-f_{\alpha }∥}_{\infty }.$$

(193)

Similarly, we determine that

$${∥{\left(S_{m_r}^{*}\left(S_{m_{r-1}}^{*}\left(\dots S_{m_2}^{*}\left(S_{m_1}^{*}f\right)\right)\right)\right)}_{\alpha }-f_{\alpha }∥}_{\infty }\le {\displaystyle \sum _{i^{*}=1}^r}{∥{\left(S_{m_{i^{*}}}^{*}\left(f\right)\right)}_{\alpha }-f_{\alpha }∥}_{\infty },$$

(194)

and

$${∥{\left(\overline{S_{m_r}}\left(\overline{S_{m_{r-1}}}\left(\dots \overline{S_{m_2}}\left(\overline{S_{m_1}}f\right)\right)\right)\right)}_{\alpha }-f_{\alpha }∥}_{\infty }\le {\displaystyle \sum _{i^{*}=1}^r}{∥{\left(\overline{S_{m_{i^{*}}}}\left(f\right)\right)}_{\alpha }-f_{\alpha }∥}_{\infty }.$$

(195)

All the above inequalities (189)–(195) prove that our implied multivariate iterative simultaneous approximations do not have a speed worse than our basic simultaneous approximations using activated convolution operators.

Conclusion: Here, we presented a new idea of going from neural networks’ main tools, the activation functions, to multivariate convolution integral approximation. This represents a rare case of applying applied mathematics to theoretical mathematics.