Multiple-Composite Quantitative Approximation by Multivariate Kantorovich–Choquet Neural Networks

Abstract

In this work we study the univariate and multivariate quantitative approximation by multi-composite Kantorovich–Choquet-type quasi-interpolation neural network operators with respect to the supremum norm. This is achieved with rates via the first univariate and multivariate moduli of continuity. We approximate continuous and bounded non-negative functions on ${\mathbb{R}}^N,$ $N\in \mathbb{N}$. When they are also uniformly continuous we have pointwise and uniform convergences, plus $L_p$ estimates. Our multi-composite activation functions are formed by general sigmoid functions.

1. Introduction

Multi-composition of activation functions refers to the technique of chaining, mixing, or stacking multiple activation functions, either in sequence or in parallel (concatenation), to create more complex, flexible, and often trainable non-linearities in neural networks. Instead of using a single activation (like ReLU) throughout the network, this approach leverages multiple functions to improve performance, enhance model expressiveness, and optimize learning.

Types of Activation Composition

• Sequential Composition: Applying one activation function after another (e.g., $f\left(g(x\right))$). A recent development includes “PolyCom” (Polynomial Composition), which combines polynomial functions with others like ReLU to accelerate convergence and improve accuracy. Another example is compounding functions to create “normalized cusp neural network operators”, which can reduce infinite domains to compact supports, enhancing approximation capabilities.
• Parallel Composition (Concatenation): Computing the outputs of different activation functions ($f_1\left(x\right),$ $f_2\left(x\right),\dots$) for the same input and concatenating them into a single vector. This allows the network to utilize multiple non-linearities at once.
• Dynamic Activation Composition (Dyn): A technique that uses learnable, normalized convex combinations of basis activation functions. This allows the network to adaptively “mix” activations during training, enhancing model adaptability and performance.
• Multivariate/Multi-dimensional Activations: Moving beyond simple scalar functions to functions that take multiple inputs and produce multiple outputs (e.g., generalizing ReLU to a second-order cone projection). These approaches are shown to have higher expressive power than traditional single-input, single-output activations.

Advantages and Applications

• Increased Expressiveness: Complex compositions allow networks to learn more intricate patterns and handle non-linearly separable data more effectively.
• Improved Accuracy: Studies have shown that combining different activation functions can lead to better performance compared to using a single standard activation, particularly for less predictable data distributions.
• Faster Convergence: Certain compositions, such as multi-kernel activation functions (multi-KAF) or dynamic mixtures, can help models converge faster by better adapting to the data.
• Controlled Output Range: Composing functions can limit the output to specific, desirable ranges, which helps in focusing on important information and filtering out noise.
• Better Gradient Flow: Some compositions, such as concatenating Swish and Tanh, can provide paths with non-zero derivatives, helping to mitigate vanishing gradient problems.

Key research Findings

• Flexibility: Research indicates that compositing activation functions—such as using a “cusp” function (a composition of two functions)—results in more flexible and powerful neural networks.
• Learned Activations: Instead of manually selecting the composition, techniques like “Trainable Adaptive Activation Function Structure (TAAFS)” learn the optimal combination of activation functions during training.
• Efficiency Concerns: While composing functions can improve accuracy, it may add to the computational load. However, the performance gains often justify the added complexity.

This approach is increasingly being explored to improve the performance of deep learning models, including Transformers and Large Language Models (LLMs).

The author in [1,2], see Chapters 2–5, was the first to establish neural network approximations to continuous functions with rates by very specifically defined neural network operators of Cardaliaguet–Euvrard and “Squashing” types by employing the modulus of continuity of the engaged function or its high-order derivative and producing very tight Jackson-type inequalities. He treats there both the univariate and multivariate cases. The defining these operators as “bell-shaped” and “squashing” functions assumes they are of compact support. Also in [2], he gives the Nth order asymptotic expansion for the error of weak approximation of these two operators to a special natural class of smooth functions; see Chapters 4–5.

The author, inspired by [3], continued his studies on neural network approximation by introducing and using the proper quasi-interpolation operators of sigmoidal and hyperbolic tangent-type which resulted in [4,5,6,7], by treating both the univariate and multivariate cases. He did also the corresponding fractional case [5,7].

The author here performs multi-composite univariate and multivariate general sigmoid activation functions-based neural network approximations to continuous non-negative functions over the whole ${\mathbb{R}}^N$, $N\in \mathbb{N}$; then he extends his results to complex-valued functions. $L_p$ approximations are also included. All convergences here are with rates expressed via the modulus of continuity of the involved function and given by very tight Jackson-type inequalities.

The author comes up with the “right” precisely defined flexible quasi-interpolation, Kantorovich–Choquet-type integral coefficient neural networks operators associated with multi-composite general sigmoid activation functions. In preparation to prove our results, we establish important properties of the multi-composite general density functions defining our operators.

Feed-forward neural networks (FNNs) with one hidden layer, the only type of networks we deal with in this work, are mathematically expressed as

$$N_n\left(x\right)=\sum _{j=0}^n{c}_j\sigma \left(〈a_j·x〉+b_j\right),x\in {\mathbb{R}}^s,s\in \mathbb{N},$$

where for $0\le j\le n$, $b_j\in \mathbb{R}$ are the thresholds, $a_j\in {\mathbb{R}}^s$ are the connection weights, $c_j\in \mathbb{R}$ are the coefficients, $〈a_j·x〉$ is the inner product of $a_j$ and x, and $\sigma$ is the activation function of the network. For neural networks in general, read [8,9,10,11,12]. For recent related works, see [13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32].

2. Background

2.1. Description of Choquet Integral [33]

We make

Definition 1. 

Consider $\mathrm{\Omega }\ne ⌀$ and let $\mathcal{C}$ be a σ-algebra of subsets in Ω.

(i) (see, e.g., [29], p. 63) The set function $\mu :\mathcal{C}\to \left[0,+\infty \right]$ is called a monotone set function (or capacity) if $\mu \left(⌀\right)=0$ and $\mu \left(A\right)\le \mu \left(B\right)$ for all $A,B\in \mathcal{C}$, with $A\subset B$. Also, μ is called submodular if

$$\mu \left(A\cup B\right)+\mu \left(A\cap B\right)\le \mu \left(A\right)+\mu \left(B\right),forallA,B\in \mathcal{C}.$$

μ is called bounded if $\mu \left(\mathrm{\Omega }\right)<+\infty$ and normalized if $\mu \left(\mathrm{\Omega }\right)=1.$

(ii) (see, e.g., [29], p. 233, or [2]) If μ is a monotone set function on $\mathcal{C}$ and if $f:\mathrm{\Omega }\to \mathbb{R}$ is $\mathcal{C}$-measurable (that is, for any Borel subset $B\subset \mathbb{R}$ it follows $f^{-1}\left(B\right)\in \mathcal{C}$), the for any $A\in \mathcal{C}$, the Choquet integral is defined by

$$\left(C\right){\int }_{A}fd\mu ={\int }_0^{+\infty }\mu \left(F_{\beta }\left(f\right)\cap A\right)d\beta +{\int }_{-\infty }^0\left[\mu \left(F_{\beta }\left(f\right)\cap A\right)-\mu \left(A\right)\right]d\beta ,$$

where we used the notation $F_{\beta }\left(f\right)=\left\{\omega \in \mathrm{\Omega }:f\left(\omega \right)\ge \beta \right\}$. Notice that if $f\ge 0$ on A, then in the above formula we get ${\int }_{-\infty }^0=0.$

The integrals on the right-hand side are the usual Riemann integral.

The function f will be called Choquet integrable on A if $\left(C\right){\int }_{A}fd\mu \in \mathbb{R}$.

Next, we list some well-known properties of the Choquet integral.

Remark 1. 

If $\mu :\mathcal{C}\to \left[0,+\infty \right]$ is a monotone set function, then the following properties hold:

(i) For all $a\ge 0$, we have $\left(C\right){\int }_{A}afd\mu =a·\left(C\right){\int }_{A}fd\mu$ (if $f\ge 0$ then see, e.g., [29], Theorem 11.2, (5), p. 228 and if f is arbitrary sign, then see, e.g., [34], p. 64, Proposition 5.1, (ii)).

(ii) For all $c\in \mathbb{R}$ and f of arbitrary sign, we have (see, e.g., [29], pp. 232–233, or [34], p. 65) $\left(C\right){\int }_A\left(f+c\right)d\mu =\left(C\right){\int }_{A}fd\mu +c·\mu \left(A\right).$

If μ is submodular too, then for all $f,g$ of arbitrary sign and lower bounded, we have (see, e.g., [34], p. 75, Theorem 6.3)

$$\left(C\right){\int }_A\left(f+g\right)d\mu \le \left(C\right){\int }_{A}fd\mu +\left(C\right){\int }_{A}gd\mu .$$

(iii) If $f\le g$ on A, then $\left(C\right){\int }_{A}fd\mu \le \left(C\right){\int }_{A}gd\mu$ (see, e.g., [29], p. 228, Theorem 11.2, (3) if $f,g\ge 0$ and p. 232 if $f,g$ are of arbitrary sign).

(iv) Let $f\ge 0$. If $A\subset B$ then $\left(C\right){\int }_{A}fd\mu \le \left(C\right){\int }_{B}fd\mu$. In addition, if μ is finitely subadditive, then

$$\left(C\right){\int }_{A\cup B}fd\mu \le \left(C\right){\int }_{A}fd\mu +\left(C\right){\int }_{B}fd\mu .$$

(v) It is immediate that $\left(C\right){\int }_{A}1·d\mu \left(t\right)=\mu \left(A\right).$

(vi) If μ is a countably additive bounded measure, then the Choquet integral $\left(C\right){\int }_{A}fd\mu$ reduces to the usual Lebesgue-type integral (see, e.g., [34], p. 62, or [29], p. 226).

(vii) If $f\ge 0$, then $\left(C\right){\int }_{A}fd\mu \ge 0$.

(viii) If $\mathrm{\Omega }={\mathbb{R}}^N$, $N\in \mathbb{N}$, we call μ strictly positive if $\mu \left(A\right)>0$, for any open subset $A\subseteq {\mathbb{R}}^N.$ See also [35].

2.2. On Multi-Composite Activation Functions

We mention:

Definition 2. 

Let $i=1,2,\dots ,$ and $h_i:\mathbb{R}\to \left[-1,1\right]$ be general sigmoid activation functions, such that they are strictly increasing, $h_i\left(0\right)=0$, $h_i\left(-x\right)=-h_i\left(x\right)$, $x\in \mathbb{R}$, $h_i\left(+\infty \right)=1$, $h_i\left(-\infty \right)=-1$. Also $h_i$ is strictly convex over $(-\infty ,0]$ and strictly concave over $[0,+\infty )$, with $h_i^{\left(2\right)}\in C\left(\mathbb{R}\right)$. Examples $tanhx,$ $erf\left(x\right)$, normalized $arctanx$, normalized Gudermanian$\left(x\right)$, etc.

Notice here $0<h_i\left(1\right)\le 1$, $i=1,2,\dots$. Any composition $h_1\circ h_2\circ h_3\circ \dots \circ h_{\lambda }$ is meant to be $h_1{|}_{\left[-1,1\right]}\circ h_2{|}_{\left[-1,1\right]}\circ h_3{{|}_{\left[-1,1\right]}\circ \dots \circ h_{\lambda -1}|}_{\left[-1,1\right]}\circ h_{\lambda }$, $\lambda \in \mathbb{N}$, and for convenience, we denote it by $G_{\lambda }:=h_1\circ h_2\circ h_3\circ \dots \circ h_{\lambda }$. We have for any $\lambda \in \mathbb{N}:0<h_{\lambda }\left(1\right)\le 1$, hence $0<h_{\lambda -1}\left(h_{\lambda }\left(1\right)\right)\le h_{\lambda -1}\left(1\right)\le 1,$ and $0<h_{\lambda -2}\left(h_{\lambda -1}\left(h_{\lambda }\left(1\right)\right)\right)\le h_{\lambda -2}\left(h_{\lambda -1}\left(1\right)\right)\le h_{\lambda -2}\left(1\right)\le 1.$

Inductively we derive that $0<G_{\lambda }\left(1\right)\le 1$, ∀ $\lambda \in \mathbb{N}$.

Clearly, it gives us $G_{\lambda }\left(0\right)=0$ and $G_{\lambda }$ is strictly increasing over $\mathbb{R}$. Furthermore, it holds

$$\begin{array}{c}{G}_{\lambda }\left(-x\right)=h_1\left(h_2\left(h_3\left(\dots \left(h_{\lambda -1}\left(h_{\lambda }\left(-x\right)\right)\right)\right)\right)\right)=h_1\left(h_2\left(h_3\left(\dots \left(h_{\lambda -1}\left(-h_{\lambda }\left(x\right)\right)\right)\right)\right)\right)\\ =\dots =-h_1\left(h_2\left(h_3\left(\dots \left(h_{\lambda -1}\left(h_{\lambda }\left(x\right)\right)\right)\right)\right)\right)=-G_{\lambda }\left(x\right),x\in \mathbb{R}.\end{array}$$

Clearly, it holds $G_{\lambda }^{\left(2\right)}\in C\left(\mathbb{R}\right).$

We notice that

$$\begin{array}{c}{G}_{\lambda }\left(+\infty \right)=h_1\left(h_2\left(h_3\left(\dots \left(h_{\lambda -1}\left(h_{\lambda }\left(+\infty \right)\right)\right)\right)\right)\right)=\\ h_1\left(h_2\left(h_3\left(\dots \left(h_{\lambda -1}\left(1\right)\right)\right)\right)\right)=G_{\lambda -1}\left(1\right),\end{array}$$

(1)

and

$$\begin{array}{c}{G}_{\lambda }\left(-\infty \right)=h_1\left(h_2\left(h_3\left(\dots \left(h_{\lambda -1}\left(h_{\lambda }\left(-\infty \right)\right)\right)\right)\right)\right)=h_1\left(h_2\left(h_3\left(\dots \left(h_{\lambda -1}\left(-1\right)\right)\right)\right)\right)\\ =-h_1\left(h_2\left(h_3\left(\dots \left(h_{\lambda -1}\left(1\right)\right)\right)\right)\right)=-G_{\lambda -1}\left(1\right).\end{array}$$

(2)

Consequently, it holds

$$-G_{\lambda -1}\left(1\right)\le G_{\lambda }\left(x\right)\le G_{\lambda -1}\left(1\right),\forall x\in \mathbb{R}.$$

(3)

Thus, $y=\pm G_{\lambda -1}\left(1\right)$ are asymptotes of $G_{\lambda }\left(x\right),$ any $\lambda \in \mathbb{N}.$

Next, we act over $(-\infty ,0]:$ let $\lambda ,\mu \ge 0:\lambda +\mu =1$. Then, by convexity of $h_2$, we have

$$h_2\left(\lambda x+\mu y\right)\le \lambda h_2\left(x\right)+\mu h_2\left(y\right),\forall x,y\in (-\infty ,0];$$

and

$$\begin{array}{c}{h}_1\left(h_2\left(\lambda x+\mu y\right)\right)\le h_1\left(\lambda h_2\left(x\right)+\mu h_2\left(y\right)\right)\le \\ \lambda \left(h_1\circ h_2\right)\left(x\right)+\mu \left(h_1\circ h_2\right)\left(y\right),\forall x,y\in (-\infty ,0].\end{array}$$

(4)

so that $h_1\circ h_2$ is convex over $(-\infty ,0].$

Now we work on $[0,+\infty ):$ and let $\lambda ,\mu \ge 0:\lambda +\mu =1$. Then, by concavity of $h_2$, we have

$$h_2\left(\lambda x+\mu y\right)\ge \lambda h_2\left(x\right)+\mu h_2\left(y\right),\forall x,y\in [0,+\infty );$$

and

$$\begin{array}{c}{h}_1\left(h_2\left(\lambda x+\mu y\right)\right)\ge h_1\left(\lambda h_2\left(x\right)+\mu h_2\left(y\right)\right)\ge \\ \lambda \left(h_1\circ h_2\right)\left(x\right)+\mu \left(h_1\circ h_2\right)\left(y\right),\forall x,y\in [0,+\infty ).\end{array}$$

(5)

Thus, $h_1\circ h_2$ is concave over $[0,+\infty ).$

Therefore, $G_2=h_1\circ h_2$ is a general sigmoid activation function with asymptotes $y=\pm h_1\left(1\right)$, and fulfilling the rest of conditions of Definition 2.

Arguing as above $h_2\circ h_3:\mathbb{R}\to \left[-1,1\right]$, fulfills Definition 2, and $h_1\circ h_2\circ h_3$ does the same with asymptotes $h=\pm h_1\left(h_2\left(1\right)\right)$.

Inductively, we prove that $G_{\lambda }$ fulfills Definition 2 with asymptotes $y=\pm G_{\lambda -1}\left(1\right).$

We have proved the following:

Theorem 1. 

Let $\lambda \in \mathbb{N}$. Then $G_{\lambda }:=h_1\circ h_2\circ h_3\circ \dots \circ h_{\lambda }$ fulfills all the properties of Definition 2 with asymptotes $y=\pm G_{\lambda -1}\left(1\right)$. That is $G_{\lambda }$ is a multi-composite general sigmoid activation function from $\mathbb{R}\to \left[-1,1\right].$

Corollary 1. 

${\displaystyle \frac{G_{\lambda }}{G_{\lambda -1}\left(1\right)}}$ fulfills Definition 2 with asymptotes $y=\pm 1$.

We call

$${\tilde{G}}_{\lambda }:={\displaystyle \frac{G_{\lambda }}{G_{\lambda -1}\left(1\right)}}.$$

(6)

Remark 2. 

Next, we consider the function

$$T_{\lambda }\left(x\right):={\displaystyle \frac{1}{4}}\left({\tilde{G}}_{\lambda }\left(x+1\right)-{\tilde{G}}_{\lambda }\left(x-1\right)\right)>0,\forall x\in \mathbb{R},\lambda \in \mathbb{N}.$$

(7)

We observe that

$$\begin{array}{c}{T}_{\lambda }\left(-x\right)={\displaystyle \frac{1}{4}}\left({\tilde{G}}_{\lambda }\left(-x+1\right)-{\tilde{G}}_{\lambda }\left(-x-1\right)\right)=\\ {\displaystyle \frac{1}{4}}\left({\tilde{G}}_{\lambda }\left(-\left(x-1\right)\right)-{\tilde{G}}_{\lambda }\left(-\left(x+1\right)\right)\right)={\displaystyle \frac{1}{4}}\left(-{\tilde{G}}_{\lambda }\left(x-1\right)+{\tilde{G}}_{\lambda }\left(x+1\right)\right)=\\ {\displaystyle \frac{1}{4}}\left({\tilde{G}}_{\lambda }\left(x+1\right)-{\tilde{G}}_{\lambda }\left(x-1\right)\right)=T_{\lambda }\left(x\right).\end{array}$$

(8)

That is $T_{\lambda }$ is an even function,

$$T_{\lambda }\left(-x\right)=T_{\lambda }\left(x\right),\forall x\in \mathbb{R},\lambda \in \mathbb{N}.$$

(9)

We see that

$$T_{\lambda }\left(0\right)={\displaystyle \frac{{\tilde{G}}_{\lambda }\left(1\right)}{2}}.$$

(10)

Let $x>1$, we have that

$$T_{\lambda }^{\prime }\left(x\right)={\displaystyle \frac{1}{4}}\left({\tilde{G}}_{\lambda }^{\prime }\left(x+1\right)-{\tilde{G}}_{\lambda }^{\prime }\left(x-1\right)\right)<0,$$

by ${\tilde{G}}_{\lambda }^{\prime }$ being strictly decreasing over $[0,+\infty ).$

Let now $0<x<1$, then $1-x>0$ and $0<1-x<1+x$. It holds ${\tilde{G}}_{\lambda }^{\prime }\left(x-1\right)={\tilde{G}}_{\lambda }^{\prime }\left(1-x\right)>{\tilde{G}}_{\lambda }^{\prime }\left(x+1\right)$, so that again $T_{\lambda }^{\prime }\left(x\right)<0.$ Consequently, $T_{\lambda }$ is strictly decreasing on $\left(0,+\infty \right).$

Clearly, $T_{\lambda }$ is strictly increasing on $(-\infty ,0)$, and $T_{\lambda }^{\prime }\left(0\right)=0.$

We observe that

$$\underset{x\to +\infty }{lim}{T}_{\lambda }\left(x\right)={\displaystyle \frac{1}{4}}\left({\tilde{G}}_{\lambda }\left(+\infty \right)-{\tilde{G}}_{\lambda }\left(+\infty \right)\right)=0,$$

(11)

and

$$\underset{x\to -\infty }{lim}{T}_{\lambda }\left(x\right)={\displaystyle \frac{1}{4}}\left({\tilde{G}}_{\lambda }\left(-\infty \right)-{\tilde{G}}_{\lambda }\left(-\infty \right)\right)=0.$$

(12)

That is the x-axis is the horizontal asymptote on $T_{\lambda }$.

As a result, $T_{\lambda }$ is a bell-shaped symmetric function with maximum

$$T_{\lambda }\left(0\right)={\displaystyle \frac{{\tilde{G}}_{\lambda }\left(1\right)}{2}}.$$

(13)

We need

Theorem 2. 

It holds

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

(14)

Proof. 

As similar to [6], p. 286 is omitted. □

Theorem 3. 

We have that

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

(15)

Proof. 

As similar to [6], p. 287, it is omitted. □

So that $T_{\lambda }\left(x\right)$ can serve as a density function in general.

We need

Theorem 4 

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

$$\sum _{\left\{\begin{array}{c}k=-\infty \hfill \\ :\left|nx-k\right|\ge n^{1-\alpha }\hfill \end{array}\right.}^{\infty }{T}_{\lambda }\left(nx-k\right)<\left(1-{\tilde{G}}_{\lambda }\left(n^{1-\alpha }-2\right)\right),$$

(16)

and

$$\underset{n\to +\infty }{lim}\left(1-{\tilde{G}}_{\lambda }\left(n^{1-\alpha }-2\right)\right)=0.$$

(17)

Denote by $⌊·⌋$ the integral part of the number and by $⌈·⌉$ the ceiling of the number.

We also need

Theorem 5. 

Let $x\in \left[a,b\right]\subset \mathbb{R}$ and $n\in \mathbb{N}$ so that $⌈na⌉\le ⌊nb⌋$. It holds

$${\displaystyle \frac{1}{{\sum }_{k=⌈na⌉}^{⌊nb⌋}{T}_{\lambda }\left(nx-k\right)}}<{\displaystyle \frac{1}{T_{\lambda }\left(1\right)}},\forall x\in \left[a,b\right].$$

(18)

Proof. 

As similar to [6], p. 289 is omitted. □

Remark 3. 

We have that

$$\underset{n\to \infty }{lim}\sum _{k=⌈na⌉}^{⌊nb⌋}{T}_{\lambda }\left(nx-k\right)\ne 1,$$

(19)

for at least some $x\in \left[a,b\right].$

See [6], p. 290, same reasoning.

Note 1. 

For large enough n, we always obtain $⌈na⌉\le ⌊nb⌋$. Also, $a\le {\displaystyle \frac{k}{n}}\le b$, if $⌈na⌉\le k\le ⌊nb⌋$. In general, it holds (by (14))

$$\sum _{k=⌈na⌉}^{⌊nb⌋}{T}_{\lambda }\left(nx-k\right)\le 1.$$

(20)

We make

Remark 4. 

We define

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

(21)

It has the properties:

(i)

$$\widehat{Z}\left(x\right)>0,\forall x\in {\mathbb{R}}^N,$$

(22)

(ii)

$$\begin{array}{c}{\displaystyle \sum _{k=-\infty }^{\infty }}\widehat{Z}\left(x-k\right):={\displaystyle \sum _{k_1=-\infty }^{\infty }}{\displaystyle \sum _{k_2=-\infty }^{\infty }}\dots {\displaystyle \sum _{k_N=-\infty }^{\infty }}\widehat{Z}\left(x_1-k_1,\dots ,x_N-k_N\right)=\\ {\displaystyle \sum _{k_1=-\infty }^{\infty }}{\displaystyle \sum _{k_2=-\infty }^{\infty }}\dots {\displaystyle \sum _{k_N=-\infty }^{\infty }}{\displaystyle \prod _{i=1}^N}{T}_{\lambda }\left(x_i-k_i\right)={\displaystyle \prod _{i=1}^N}\left({\displaystyle \sum _{k_i=-\infty }^{\infty }}{T}_{\lambda }\left(x_i-k_i\right)\right)\stackrel{\left(\mathit{14}\right)}{=}1.\end{array}$$

Hence

$$\sum _{k=-\infty }^{\infty }\widehat{Z}\left(x-k\right)=1.$$

(23)

That is

(iii)

$$\sum _{k=-\infty }^{\infty }\widehat{Z}\left(nx-k\right)=1,\forall x\in {\mathbb{R}}^N;n\in \mathbb{N}.$$

(24)

and

(iv)

$${\int }_{{\mathbb{R}}^N}\widehat{Z}\left(x\right)dx={\int }_{{\mathbb{R}}^N}\left({\displaystyle \prod _{i=1}^N}{T}_{\lambda }\left(x_i\right)\right)d{x}_1\dots d{x}_N={\displaystyle \prod _{i=1}^N}\left({\int }_{-\infty }^{\infty }{T}_{\lambda }\left(x_i\right)d{x}_i\right)\stackrel{\left(\mathit{15}\right)}{=}1,$$

(25)

Thus

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

(26)

That is, $\widehat{Z}$ is a multivariate density function.

Here, we denote $x=\left(x_1,\dots ,x_N\right),$ ${∥x∥}_{\infty }:=max\left\{\left|x_1\right|,\dots ,\left|x_N\right|\right\}$, $x\in {\mathbb{R}}^N$, also set $\infty :=\left(\infty ,\dots ,\infty \right)$, $-\infty :=\left(-\infty ,\dots ,-\infty \right)$ upon the multivariate context, $0<\beta <1,$

(v) We have

$$\begin{array}{c}{\displaystyle \sum _{\left\{\begin{array}{c}k=-\infty \\ {∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }>{\displaystyle \frac{1}{n^{\beta }}}\end{array}\right.}^{\infty }}\widehat{Z}\left(nx-k\right)=\underset{{∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }>{\displaystyle \frac{1}{n^{\beta }}}}{{\displaystyle \sum _{k_1=-\infty }^{\infty }}\dots {\displaystyle \sum _{k_N=-\infty }^{\infty }}}\left({\displaystyle \prod _{i=1}^N}{T}_{\lambda }\left(n{x}_i-k_i\right)\right)=\\ {\displaystyle \prod _{i=1}^N}\left({\displaystyle \sum _{\left\{\begin{array}{c}{k}_i=-\infty \\ {∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }>{\displaystyle \frac{1}{n^{\beta }}}\end{array}\right.}^{\infty }}{T}_{\lambda }\left(n{x}_i-k_i\right)\right)\le (forsomer\in \left\{1,\dots ,N\right\})\\ \left({\displaystyle \prod _{\begin{array}{c}i=1\\ i\ne r\end{array}}^N}\left({\displaystyle \sum _{k_i=-\infty }^{\infty }}{T}_{\lambda }\left(n{x}_i-k_i\right)\right)\right)\left({\displaystyle \sum _{\left\{\begin{array}{c}{k}_r=-\infty \\ \left|{\displaystyle \frac{k_r}{n}}-x_r\right|>{\displaystyle \frac{1}{n^{\beta }}}\end{array}\right.}^{\infty }}{T}_{\lambda }\left(n{x}_r-k_r\right)\right)=\\ {\displaystyle \sum _{\left\{\begin{array}{c}{k}_r=-\infty \\ \left|{\displaystyle \frac{k_r}{n}}-x_r\right|>{\displaystyle \frac{1}{n^{\beta }}}\end{array}\right.}^{\infty }}{T}_{\lambda }\left(n{x}_r-k_r\right)={\displaystyle \sum _{\left\{\begin{array}{c}{k}_r=-\infty \\ \left|n{x}_r-k_r\right|>n^{1-\beta }\end{array}\right.}^{\infty }}{T}_{\lambda }\left(n{x}_r-k_r\right)\stackrel{\left(16\right)}{<}\\ 1-{\tilde{G}}_{\lambda }\left(n^{1-\beta }-2\right).\end{array}$$

(27)

That is

$$\sum _{\left\{\begin{array}{c}k=-\infty \\ {∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }>{\displaystyle \frac{1}{n^{\beta }}}\end{array}\right.}^{\infty }\widehat{Z}\left(nx-k\right)<1-{\tilde{G}}_{\lambda }\left(n^{1-\beta }-2\right).$$

(28)

$0<\beta <1$, $n\in \mathbb{N}:n^{1-\beta }>2$, $\forall x\in {\prod }_{i=1}^N\left[a_i,b_i\right].$

We denote by

$${\widehat{\delta }}_N\left(\beta ,n\right):=1-{\tilde{G}}_{\lambda }\left(n^{1-\beta }-2\right).$$

(29)

$0<\beta <1.$

For $f\in C_B^{+}\left({\mathbb{R}}^N\right)$ (continuous and bounded functions from ${\mathbb{R}}^N$ into ${\mathbb{R}}_{+}$), we define the first modulus of continuity

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

(30)

Given that $f\in C_U^{+}\left({\mathbb{R}}^N\right)$ (uniformly continuous from ${\mathbb{R}}^N$ into ${\mathbb{R}}_{+}$, the same definition for ${\omega }_1$), we have that

$$\underset{h\to 0}{lim}{\omega }_1\left(f,h\right)=0.$$

(31)

When $N=1$, ${\omega }_1$ is defined as in (30) with ${∥·∥}_{\infty }$ collapsing to $\left|·\right|$ and has the property (31).

3. Main Results

We need

Definition 3. 

Let $\mathcal{L}$ be the Lebesgue σ-algebra on ${\mathbb{R}}^N$, $N\in \mathbb{N}$, and the set function $\mu :\mathcal{L}\to [0,+\infty )$, which is assumed to be monotone, submodular and strictly positive. For $f\in C_B^{+}\left({\mathbb{R}}^N\right)$, we define the general multi-composite multivariate Kantorovich–Choquet-type neural network operators for any $x\in {\mathbb{R}}^N$:

$$\begin{array}{c}{\widehat{K}}_n^{\mu }\left(f,x\right)={\widehat{K}}_n^{\mu }\left(f,x_1,\dots ,x_N\right):=\\ {\displaystyle \sum _{k=-\infty }^{\infty }}\left({\displaystyle \frac{\left(C\right){\int }_{{\left[0,\frac{1}{n}\right]}^N}f\left(t+\frac{k}{n}\right)d\mu \left(t\right)}{\mu \left({\left[0,\frac{1}{n}\right]}^N\right)}}\right)\widehat{Z}\left(nx-k\right)=\\ {\displaystyle \sum _{k_1=-\infty }^{\infty }}{\displaystyle \sum _{k_2=-\infty }^{\infty }}\dots {\displaystyle \sum _{k_N=-\infty }^{\infty }}\left({\displaystyle \frac{\left(C\right){\int }_0^{\frac{1}{n}}\dots {\int }_0^{\frac{1}{n}}f\left(t_1+\frac{k_1}{n},t_2+\frac{k_2}{n},\dots ,t_N+\frac{k_N}{n}\right)d\mu \left(t_1,\dots ,t_N\right)}{\mu \left({\left[0,\frac{1}{n}\right]}^N\right)}}\right)\\ \left({\displaystyle \prod _{i=1}^N}{T}_{\lambda }\left(n{x}_i-k_i\right)\right),\end{array}$$

(32)

where $x=\left(x_1,\dots ,x_N\right)\in {\mathbb{R}}^N$, $k=\left(k_1,\dots ,k_N\right)$, $t=\left(t_1,\dots ,t_N\right)$, $n\in \mathbb{N}.$

Clearly, here $\mu \left({\left[0,{\displaystyle \frac{1}{n}}\right]}^N\right)>0$, $\forall n\in \mathbb{N}$.

From the above, we notice that

$${∥{\widehat{K}}_n^{\mu }\left(f\right)∥}_{\infty }\le {∥f∥}_{\infty },$$

(33)

so that ${\widehat{K}}_n^{\mu }\left(f,x\right)$ is well-defined.

We make

Remark 5. 

Let $f\in C_B^{+}\left({\mathbb{R}}^N\right),$ $t\in {\left[0,{\displaystyle \frac{1}{n}}\right]}^N$ and $x\in {\mathbb{R}}^N$, then

$$f\left(t+{\displaystyle \frac{k}{n}}\right)=f\left(t+{\displaystyle \frac{k}{n}}\right)-f\left(x\right)+f\left(x\right)\le \left|f\left(t+{\displaystyle \frac{k}{n}}\right)-f\left(x\right)\right|+f\left(x\right),$$

Hence

$$\begin{array}{c}\left(C\right){\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}f\left(t+{\displaystyle \frac{k}{n}}\right)d\mu \left(t\right)\le \\ \left(C\right){\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left|f\left(t+{\displaystyle \frac{k}{n}}\right)-f\left(x\right)\right|d\mu \left(t\right)+\left(C\right){\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}f\left(x\right)d\mu \left(t\right)=\\ \left(C\right){\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left|f\left(t+{\displaystyle \frac{k}{n}}\right)-f\left(x\right)\right|d\mu \left(t\right)+f\left(x\right)\mu \left({\left[0,{\displaystyle \frac{1}{n}}\right]}^N\right).\end{array}$$

(34)

That is

$$\begin{array}{c}\left(C\right){\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}f\left(t+{\displaystyle \frac{k}{n}}\right)d\mu \left(t\right)-f\left(x\right)\mu \left({\left[0,{\displaystyle \frac{1}{n}}\right]}^N\right)\le \\ \left(C\right){\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left|f\left(t+{\displaystyle \frac{k}{n}}\right)-f\left(x\right)\right|d\mu \left(t\right).\end{array}$$

(35)

Similarly, we have that

$$f\left(x\right)=f\left(x\right)-f\left(t+{\displaystyle \frac{k}{n}}\right)+f\left(t+{\displaystyle \frac{k}{n}}\right)\le \left|f\left(t+{\displaystyle \frac{k}{n}}\right)-f\left(x\right)\right|+f\left(t+{\displaystyle \frac{k}{n}}\right).$$

Hence

$$\begin{array}{c}\left(C\right){\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}f\left(x\right)\mu \left(dt\right)\le \\ \left(C\right){\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left|f\left(t+{\displaystyle \frac{k}{n}}\right)-f\left(x\right)\right|d\mu \left(t\right)+\left(C\right){\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}f\left(t+{\displaystyle \frac{k}{n}}\right)\mu \left(dt\right),\end{array}$$

and

$$\begin{array}{c}f\left(x\right)\mu \left({\left[0,{\displaystyle \frac{1}{n}}\right]}^N\right)-\left(C\right){\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}f\left(t+{\displaystyle \frac{k}{n}}\right)\mu \left(dt\right)\le \\ \left(C\right){\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left|f\left(t+{\displaystyle \frac{k}{n}}\right)-f\left(x\right)\right|d\mu \left(t\right).\end{array}$$

(36)

By (35) and (36), we derive that

$$\begin{array}{c}\left|\left(C\right){\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}f\left(t+{\displaystyle \frac{k}{n}}\right)\mu \left(dt\right)-f\left(x\right)\mu \left({\left[0,{\displaystyle \frac{1}{n}}\right]}^N\right)\right|\le \\ \left(C\right){\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left|f\left(t+{\displaystyle \frac{k}{n}}\right)-f\left(x\right)\right|d\mu \left(t\right).\end{array}$$

(37)

In particular, it holds

$$\left|\left({\displaystyle \frac{\left(C\right){\int }_{{\left[0,\frac{1}{n}\right]}^N}f\left(t+\frac{k}{n}\right)\mu \left(dt\right)}{\mu \left({\left[0,\frac{1}{n}\right]}^N\right)}}\right)-f\left(x\right)\right|\le {\displaystyle \frac{\left(C\right){\int }_{{\left[0,\frac{1}{n}\right]}^N}\left|f\left(t+\frac{k}{n}\right)-f\left(x\right)\right|d\mu \left(t\right)}{\mu \left({\left[0,\frac{1}{n}\right]}^N\right)}}.$$

(38)

We present the following approximation result:

Theorem 6. 

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

(i)

$$\underset{\mu }{sup}\left|{\widehat{K}}_n^{\mu }\left(f,x\right)-f\left(x\right)\right|\le {\omega }_1\left(f,{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+2{∥f∥}_{\infty }{\widehat{\delta }}_N\left(\beta ,n\right)=:{\sigma }_n,$$

(39)

and

(ii)

$$\underset{\mu }{sup}{∥{\widehat{K}}_n^{\mu }\left(f\right)-f∥}_{\infty }\le {\sigma }_n.$$

(40)

Given that $f\in \left(C_U^{+}\left({\mathbb{R}}^N\right)\cap C_B^{+}\left({\mathbb{R}}^N\right)\right),$ we obtain ${\displaystyle \underset{n\to \infty }{lim}}$${\widehat{K}}_n^{\mu }\left(f\right)=f$, uniformly.

Proof. 

We observe that

$$\left|{\widehat{K}}_n^{\mu }\left(f,x\right)-f\left(x\right)\right|=$$

$$\left|\sum _{k=-\infty }^{\infty }\left({\displaystyle \frac{\left(C\right){\int }_{{\left[0,\frac{1}{n}\right]}^N}f\left(t+\frac{k}{n}\right)d\mu \left(t\right)}{\mu \left({\left[0,\frac{1}{n}\right]}^N\right)}}\right)\widehat{Z}\left(nx-k\right)-\sum _{k=-\infty }^{\infty }f\left(x\right)\widehat{Z}\left(nx-k\right)\right|$$

$$=\left|\sum _{k=-\infty }^{\infty }\left(\left({\displaystyle \frac{\left(C\right){\int }_{{\left[0,\frac{1}{n}\right]}^N}f\left(t+\frac{k}{n}\right)d\mu \left(t\right)}{\mu \left({\left[0,\frac{1}{n}\right]}^N\right)}}\right)-f\left(x\right)\right)\widehat{Z}\left(nx-k\right)\right|\le$$

(41)

$$\sum _{k=-\infty }^{\infty }\left|\left({\displaystyle \frac{\left(C\right){\int }_{{\left[0,\frac{1}{n}\right]}^N}f\left(t+\frac{k}{n}\right)d\mu \left(t\right)}{\mu \left({\left[0,\frac{1}{n}\right]}^N\right)}}\right)-f\left(x\right)\right|\widehat{Z}\left(nx-k\right)\stackrel{\left(38\right)}{\le }$$

$$\sum _{k=-\infty }^{\infty }\left({\displaystyle \frac{\left(C\right){\int }_{{\left[0,\frac{1}{n}\right]}^N}\left|f\left(t+\frac{k}{n}\right)-f\left(x\right)\right|d\mu \left(t\right)}{\mu \left({\left[0,\frac{1}{n}\right]}^N\right)}}\right)\widehat{Z}\left(nx-k\right)=$$

$$\sum _{\left\{\begin{array}{c}k=-\infty \\ {∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }\le {\displaystyle \frac{1}{n^{\beta }}}\end{array}\right.}^{\infty }\left({\displaystyle \frac{\left(C\right){\int }_{{\left[0,\frac{1}{n}\right]}^N}\left|f\left(t+\frac{k}{n}\right)-f\left(x\right)\right|d\mu \left(t\right)}{\mu \left({\left[0,\frac{1}{n}\right]}^N\right)}}\right)\widehat{Z}\left(nx-k\right)+$$

(42)

$$\sum _{\left\{\begin{array}{c}k=-\infty \\ {∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }>{\displaystyle \frac{1}{n^{\beta }}}\end{array}\right.}^{\infty }\left({\displaystyle \frac{\left(C\right){\int }_{{\left[0,\frac{1}{n}\right]}^N}\left|f\left(t+\frac{k}{n}\right)-f\left(x\right)\right|d\mu \left(t\right)}{\mu \left({\left[0,\frac{1}{n}\right]}^N\right)}}\right)\widehat{Z}\left(nx-k\right)\le$$

$$\sum _{\left\{\begin{array}{c}k=-\infty \\ {∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }\le {\displaystyle \frac{1}{n^{\beta }}}\end{array}\right.}^{\infty }\left({\displaystyle \frac{\left(C\right){\int }_{{\left[0,\frac{1}{n}\right]}^N}{\omega }_1\left(f,{∥t∥}_{\infty }+{∥\frac{k}{n}-x∥}_{\infty }\right)d\mu \left(t\right)}{\mu \left({\left[0,\frac{1}{n}\right]}^N\right)}}\right)\widehat{Z}\left(nx-k\right)+$$

$$2{∥f∥}_{\infty }\left(\sum _{\left\{\begin{array}{c}k=-\infty \\ {∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }>{\displaystyle \frac{1}{n^{\beta }}}\end{array}\right.}^{\infty }\widehat{Z}\left(\left|nx-k\right|\right)\right)$$

$$\stackrel{\left(28\right)}{\le }{\omega }_1\left(f,{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+2{∥f∥}_{\infty }{\widehat{\delta }}_N\left(\beta ,n\right),$$

(43)

proving the claim. □

Additionally, we give

Definition 4. 

Denote $C_B^{+}\left({\mathbb{R}}^N,\mathbb{C}\right)=\{f:{\mathbb{R}}^N\to \mathbb{C}|f=f_1+i{f}_2$, where $f_1,f_2\in C_B^{+}\left({\mathbb{R}}^N\right)\}$. We set for $f\in C_B^{+}\left({\mathbb{R}}^N,\mathbb{C}\right)$ that

$${\widehat{K}}_n^{\mu }\left(f,x\right):={\widehat{K}}_n^{\mu }\left(f_1,x\right)+i{\widehat{K}}_n^{\mu }\left(f_2,x\right),$$

(44)

$\forall n\in \mathbb{N}$, $x\in {\mathbb{R}}^N,$$i=\sqrt{-1}.$

We give

Theorem 7. 

Let $f\in C_B^{+}\left({\mathbb{R}}^N,\mathbb{C}\right)$, $f=f_1+i{f}_2$, $N\in \mathbb{N},$ $0<\beta <1$, $x\in {\mathbb{R}}^N$, $n\in \mathbb{N}$ with $n^{1-\beta }>2$. Then

(i)

$$\begin{array}{c}{\displaystyle \underset{\mu }{sup}}\left|{\widehat{K}}_n^{\mu }\left(f,x\right)-f\left(x\right)\right|\le \left({\omega }_1\left(f_1,{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+{\omega }_1\left(f_2,{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)\right)\\ +2\left({∥f_1∥}_{\infty }+{∥f_2∥}_{\infty }\right){\widehat{\delta }}_N\left(\beta ,n\right)=:{\phi }_n,\end{array}$$

(45)

and

(ii)

$$\underset{\mu }{sup}{∥{\widehat{K}}_n^{\mu }\left(f\right)-f∥}_{\infty }\le {\phi }_n.$$

Proof. 

We have that

$$\begin{array}{c}\left|{\widehat{K}}_n^{\mu }\left(f,x\right)-f\left(x\right)\right|=\left|{\widehat{K}}_n^{\mu }\left(f_1,x\right)+i{\widehat{K}}_n^{\mu }\left(f_2,x\right)-f_1\left(x\right)-i{f}_2\left(x\right)\right|=\\ \left|\left({\widehat{K}}_n^{\mu }\left(f_1,x\right)-f_1\left(x\right)\right)+i\left({\widehat{K}}_n^{\mu }\left(f_2,x\right)-f_2\left(x\right)\right)\right|\le \\ \left|{\widehat{K}}_n^{\mu }\left(f_1,x\right)-f_1\left(x\right)\right|+\left|{\widehat{K}}_n^{\mu }\left(f_2,x\right)-f_2\left(x\right)\right|\stackrel{\left(39\right)}{\le }\\ \left({\omega }_1\left(f_1,{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+2{∥f_1∥}_{\infty }{\widehat{\delta }}_N\left(\beta ,n\right)\right)+\\ \left({\omega }_1\left(f_2,{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+2{∥f_2∥}_{\infty }{\widehat{\delta }}_N\left(\beta ,n\right)\right),\end{array}$$

(46)

proving the claim. □

We need

Definition 5. 

Let ${\mathcal{L}}^{*}$ be the Lebesgue σ-algebra on $\mathbb{R}$, and the set function ${\mu }^{*}:{\mathcal{L}}^{*}\to \left[0,+\infty \right]$, which is assumed to be monotone, submodular and strictly positive. For $f\in C_B^{+}\left(\mathbb{R}\right)$, we define the general multi-composite univariate Kantorovich–Choquet-type neural network operator for any $x\in \mathbb{R}$:

$${\widehat{M}}_n^{{\mu }^{*}}\left(f,x\right)=\sum _{k=-\infty }^{\infty }\left({\displaystyle \frac{\left(C\right){\int }_0^{\frac{1}{n}}f\left(t+\frac{k}{n}\right)d{\mu }^{*}\left(t\right)}{{\mu }^{*}\left(\left[0,\frac{1}{n}\right]\right)}}\right)T_{\lambda }\left(nx-k\right).$$

(47)

Clearly, here ${\mu }^{*}\left(\left[0,{\displaystyle \frac{1}{n}}\right]\right)>0$, $\forall n\in \mathbb{N}$.

From the above, we notice that

$${∥{\widehat{M}}_n^{{\mu }^{*}}\left(f\right)∥}_{\infty }\le {∥f∥}_{\infty },$$

(48)

so that ${\widehat{M}}_n^{{\mu }^{*}}\left(f,x\right)$ is well-defined.

Notice that ${\widehat{K}}_n^{\mu }$, when $N=1$, collapses to ${\widehat{M}}_n^{{\mu }^{*}}.$

This follows another related result.

Corollary 2. 

(to Theorem 6 when $N=1$)

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

(i)

$$\underset{{\mu }^{*}}{sup}\left|{\widehat{M}}_n^{{\mu }^{*}}\left(f,x\right)-f\left(x\right)\right|\le {\omega }_1\left(f,{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+2{∥f∥}_{\infty }\left(1-{\tilde{G}}_{\lambda }\left(n^{1-\beta }-2\right)\right)=:{\tau }_n,$$

(49)

and

(ii)

$$\underset{\mu }{sup}{∥{\widehat{M}}_n^{{\mu }^{*}}\left(f\right)-f∥}_{\infty }\le {\tau }_n.$$

(50)

Given that $f\in \left(C_U^{+}\left(\mathbb{R}\right)\cap C_B^{+}\left(\mathbb{R}\right)\right),$ we obtain ${\displaystyle \underset{n\to \infty }{lim}}$${\widehat{M}}_n^{{\mu }^{*}}\left(f\right)=f$, uniformly.

Proof. 

Similarly to Theorem 6, it is omitted. □

We need

Definition 6. 

Let $f\in C_B^{+}\left(\mathbb{R},\mathbb{C}\right)$, where $f=f_1+i{f}_2$ with $f_1,f_2\in C_B^{+}\left(\mathbb{R}\right)$. We set

$${\widehat{M}}_n^{{\mu }^{*}}\left(f,x\right):={\widehat{M}}_n^{{\mu }^{*}}\left(f_1,x\right)+i{\widehat{M}}_n^{{\mu }^{*}}\left(f_2,x\right),$$

(51)

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

We continue with

Corollary 3. 

(to Theorem 7 when $N=1$) Let $f\in C_B^{+}\left(\mathbb{R},\mathbb{C}\right)$, $f=f_1+i{f}_2$, $0<\beta <1$, $x\in \mathbb{R}$, $n\in \mathbb{N}$ with $n^{1-\beta }>2$. Then

(i)

$$\begin{array}{c}{\displaystyle \underset{{\mu }^{*}}{sup}}\left|{\widehat{M}}_n^{{\mu }^{*}}\left(f,x\right)-f\left(x\right)\right|\le \left({\omega }_1\left(f_1,{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+{\omega }_1\left(f_2,{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)\right)+\\ 2\left({∥f_1∥}_{\infty }+{∥f_2∥}_{\infty }\right)\left(1-{\tilde{G}}_{\lambda }\left(n^{1-\beta }-2\right)\right)=:{\psi }_n,\end{array}$$

(52)

and

(ii)

$$\underset{{\mu }^{*}}{sup}{∥{\widehat{M}}_n^{{\mu }^{*}}\left(f\right)-f∥}_{\infty }\le {\psi }_n.$$

Proof. 

Similarly to Theorem 7, it is omitted. □

We finish with $L_p$ estimates.

Theorem 8. 

Let all as in Theorem 7, $p>1$, $\widehat{\Lambda }$ a compact subset of ${\mathbb{R}}^N$. Then

$${∥{\widehat{K}}_n^{\mu }\left(f\right)-f∥}_{p,\widehat{\Lambda }}\le {\phi }_n{\left|\widehat{\Lambda }\right|}^{{\displaystyle \frac{1}{p}}},$$

(53)

where $\left|\widehat{\Lambda }\right|<\infty$, is the Lebesgue measure of $\widehat{\Lambda }$, where ${\phi }_n$ as in (45).

Proof. 

By (45), etc. □

Theorem 9. 

All as in Theorem 8, $N=1$ case. Then

$${∥{\widehat{M}}_n^{{\mu }^{*}}\left(f\right)-f∥}_{p,\widehat{\Lambda }}\le {\psi }_n{\left|\widehat{\Lambda }\right|}^{{\displaystyle \frac{1}{p}}},$$

(54)

where $\widehat{\Lambda }$ now is a closed interval of $\mathbb{R}$, where ${\psi }_n$ as in (52).

Proof. 

By (52), etc. □

4. Conclusions

We studied the multi-composite univariate and multivariate general approximation by the Kantorovich–Choquet-type quasi-interpolation integral neural network operators.

The functions under approximation are non-negative-valued, continuous, and bounded. We established a series of multi-composite Jackson-type inequalities giving the corresponding convergences via the modulus of continuity. We extended our results to complex-valued functions. We finished with related $L_p$ results.