Return to Article Details Multivariate error function based neural network approximations

Multivariate error function based neural network approximations

George A. Anastassiou $^{*}$

May 1st, 2014.

$^{*}$ Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152, U.S.A., e-mail: ganastss@memphis.edu.

Here we present multivariate quantitative approximations of real and complex valued continuous multivariate functions on a box or $R^{N},$ $N \in N$ , by the multivariate quasi-interpolation, Baskakov type and quadrature type neural network operators. We treat also the case of approximation by iterated operators of the last three types. These approximations are derived by establishing multidimensional Jackson type inequalities involving the multivariate modulus of continuity of the engaged function or its high order partial derivatives. Our multivariate operators are defined by using a multidimensional density function induced by the Gaussian error special function. The approximations are pointwise and uniform. The related feed-forward neural network is with one hidden layer.

MSC. 41A17, 41A25, 41A30, 41A36.

Keywords. error function, multivariate neural network approximation, quasi-interpolation operator, Baskakov type operator, quadrature type operator, multivariate modulus of continuity, complex approximation, iterated approximation.

1 Introduction

The author in [ 2 ] and [ 3 ] , 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 Cardaliagnet-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 ”bell-shaped” and ”squashing” functions are assumed to be of compact support. Also in [ 3 ] he gives the $N$ th 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 there.

For this article the author is motivated by the article [ 12 ] of Z. Chen and F. Cao, also by [ 4 ] , [ 5 ] , [ 6 ] , [ 7 ] , [ 8 ] , [ 9 ] , [ 10 ] , [ 13 ] , [ 14 ] .

The author here performs multivariate error function based neural network approximations to continuous functions over boxes or over the whole $R^{N}$ , $N \in N$ , then he extends his results to complex valued multivariate functions. Also he does iterated approximation. All convergences here are with rates expressed via the multivariate modulus of continuity of the involved function or its high order partial derivative and given by very tight multidimensional Jackson type inequalities.

The author here comes up with the ”right” precisely defined multivariate quasi-interpolation neural network operators related to boxes or $R^{N}$ , as well as Baskakov type and quadrature type related operators on $R^{N}$ . Our boxes are not necessarily symmetric to the origin. In preparation to prove our results we establish important properties of the basic multivariate density function induced by error function and defining our operators.

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

N_{n} (x) = \sum_{j = 0}^{n} c_{j} σ (⟨ a_{j} \cdot x ⟩ + b_{j}), \  \  \ x \in R^{s}, \  \ s \in N,

where for $0 \leq j \leq n$ , $b_{j} \in R$ are the thresholds, $a_{j} \in R^{s}$ are the connection weights, $c_{j} \in R$ are the coefficients, $⟨ a_{j} \cdot x ⟩$ is the inner product of $a_{j}$ and $x$ , and $σ$ is the activation function of the network. In many fundamental network models, the activation function is the error function. About neural networks read [ 15 ] , [ 16 ] , [ 17 ] .

2 Basics

We consider here the (Gauss) error special function ( [ 1 ] , [ 11 ] )

\erf (x) = \frac{2}{\sqrt{π}} \int_{0}^{x} e^{- t^{2}} d t, x \in R,

which is a sigmoidal type function and is a strictly increasing function.

It has the basic properties

\erf (0) = 0, \ \erf (- x) = - \erf (x), \  \ \erf (+ \infty) = 1, \ \erf (- \infty) = - 1.

We consider the activation function ( [ 10 ] )

χ (x) = \frac{1}{4} (\erf (x + 1) - \erf (x - 1)) > 0, \forall x \in R,

which is an even function.

Next we follow [ 10 ] on $χ$ . We got there $χ (0) ≃ 0.4215$ , and that $χ$ is strictly decreasing on $[0, \infty)$ and strictly increasing on $(- \infty, 0]$ , and the $x$ -axis is the horizontal asymptote on $χ,$ i.e. $χ$ is a bell symmetric function.

Theorem 1

[ 10 ] We have that

\begin{aligned} \sum_{i = - \infty}^{\infty} χ (x - i) & = 1, \forall x \in R, \\ \sum_{i = - \infty}^{\infty} χ (n x - i) & = 1, \forall x \in R, n \in N, \end{aligned}

and

\int_{- \infty}^{\infty} χ (x) d x = 1,

that is $χ (x)$ is a density function on $R .$

We need the important

Theorem 2

[ 10 ] Let $0 < α < 1,$ and $n \in N$ with $n^{1 - α} \geq 3$ . It holds

\sum_{\overset{k = - \infty}{| n x - k | \geq n^{1 - α}}}^{\infty} χ (n x - k) < \frac{1}{2 \sqrt{π} (n^{1 - α} - 2) e^{{(n^{1 - α} - 2)}^{2}}} .

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

Theorem 3

[ 10 ] Let $x \in [a, b] \subset R$ and $n \in N$ so that $⌈ n a ⌉ \leq ⌊ n b ⌋$ . It holds

\frac{1}{\sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} χ (n x - k)} < \frac{1}{χ (1)} ≃ 4.019, \forall x \in [a, b] .

Also from [ 10 ] we get

lim_{n \to \infty} \sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} χ (n x - k) \neq 1,

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

For large enough $n$ we always obtain $⌈ n a ⌉ \leq ⌊ n b ⌋$ . Also $a \leq \frac{k}{n} \leq b$ , iff $⌈ n a ⌉ \leq k \leq ⌊ n b ⌋$ . In general it holds by (4) that

\sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} χ (n x - k) \leq 1.

We introduce

Z (x_{1}, . . ., x_{N}) := Z (x) := \prod_{i = 1}^{N} χ (x_{i}), x = (x_{1}, . . ., x_{N}) \in R^{N}, N \in N .

It has the properties:

$Z (x) > 0$ , $\forall$ $x \in R^{N},$
$\sum_{k = - \infty}^{\infty} Z (x - k) := \sum_{k_{1} = - \infty}^{\infty} \sum_{k_{2} = - \infty}^{\infty} . . . \sum_{k_{N} = - \infty}^{\infty} Z (x_{1} - k_{1}, . . ., x_{N} - k_{N}) = 1,$
11

where $k := (k_{1}, . . ., k_{n}) \in Z^{N}$ , $\forall$ $x \in R^{N},$

hence
$\begin{aligned} \sum_{k = - \infty}^{\infty} Z (n x - k) = \\ = \sum_{k_{1} = - \infty}^{\infty} \sum_{k_{2} = - \infty}^{\infty} . . . \sum_{k_{N} = - \infty}^{\infty} Z (n x_{1} - k_{1}, . . ., n x_{N} - k_{N}) = 1, \forall x \in R^{N}; n \in N, \end{aligned}$

and
$\int_{R^{N}} Z (x) d x = 1,$
13

that is $Z$ is a multivariate density function.

Here ${‖ x ‖}_{\infty} := max {| x_{1} |, . . ., | x_{N} |}$ , $x \in R^{N}$ , also set $\infty := (\infty, . . ., \infty)$ , $- \infty := (- \infty, . . ., - \infty)$ upon the multivariate context, and

\begin{array}{rcl} ⌈ n a ⌉ & : & = (⌈ n a_{1} ⌉, . . ., ⌈ n a_{N} ⌉), \\ ⌊ n b ⌋ & : & = (⌊ n b_{1} ⌋, . . ., ⌊ n b_{N} ⌋), \end{array}

where $a := (a_{1}, . . ., a_{N})$ , $b := (b_{1}, . . ., b_{N}) .$

We obviously see that

\begin{aligned} \sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} Z (n x - k) = \sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} (\prod_{i = 1}^{N} χ (n x_{i} - k_{i})) \\ = \sum_{k_{1} = ⌈ n a_{1} ⌉}^{⌊ n b_{1} ⌋} . . . \sum_{k_{N} = ⌈ n a_{N} ⌉}^{⌊ n b_{N} ⌋} (\prod_{i = 1}^{N} χ (n x_{i} - k_{i})) = \prod_{i = 1}^{N} (\sum_{k_{i} = ⌈ n a_{i} ⌉}^{⌊ n b_{i} ⌋} χ (n x_{i} - k_{i})) . \end{aligned}

For $0 < β < 1$ and $n \in N$ , a fixed $x \in R^{N}$ , we have that

\begin{aligned} \sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} χ (n x - k) = \\ = \sum_{\overset{k = ⌈ n a ⌉}{{‖ \frac{k}{n} - x ‖}_{\infty} \leq \frac{1}{n^{β}}}}^{⌊ n b ⌋} χ (n x - k) + \sum_{\overset{k = ⌈ n a ⌉}{{‖ \frac{k}{n} - x ‖}_{\infty} > \frac{1}{n^{β}}}}^{⌊ n b ⌋} χ (n x - k) . \end{aligned}

In the last two sums the counting is over disjoint vector sets of $k$ ’s, because the condition ${‖ \frac{k}{n} - x ‖}_{\infty} > \frac{1}{n^{β}}$ implies that there exists at least one $| \frac{k_{r}}{n} - x_{r} | > \frac{1}{n^{β}}$ , where $r \in {1, . . ., N} .$

We treat

Unknown environment 'bgroup'

when $n^{1 - β} \geq 3.$

We have proved that

$\sum_{\overset{k = ⌈ n a ⌉}{{‖ \frac{k}{n} - x ‖}_{\infty} > \frac{1}{n^{β}}}}^{⌊ n b ⌋} Z (n x - k) \leq \frac{1}{2 \sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}},$
20

$0 < β < 1$ , $n \in N; n^{1 - β} \geq 3$ , $x \in (\prod_{i = 1}^{N} [a_{i}, b_{i}]) .$

By Theorem 3 clearly we obtain

\begin{aligned} 0 & < \frac{1}{\sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} Z (n x - k)} = \frac{1}{\prod_{i = 1}^{N} (\sum_{k_{i} = ⌈ n a_{i} ⌉}^{⌊ n b_{i} ⌋} χ (n x_{i} - k_{i}))} \\ < \frac{1}{{(χ (1))}^{N}} ≃ {(4.019)}^{N} . \end{aligned}

That is,

it holds

$0 < \frac{1}{\sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} Z (n x - k)} < \frac{1}{{(χ (1))}^{N}} ≃ {(4.019)}^{N}, \forall x \in (\prod_{i = 1}^{N} [a_{i}, b_{i}]), n \in N .$
22

It is also clear that

$\sum_{\overset{k = - \infty}{{‖ \frac{k}{n} - x ‖}_{\infty} > \frac{1}{n^{β}}}}^{\infty} Z (n x - k) \leq \frac{1}{2 \sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}},$
23

$0 < β < 1$ , $n \in N : n^{1 - β} \geq 3$ , $x \in (\prod_{i = 1}^{N} [a_{i}, b_{i}]) .$

Also we get that

lim_{n \to \infty} \sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} Z (n x - k) \neq 1,

for at least some $x \in (\prod_{i = 1}^{N} [a_{i}, b_{i}]) .$

Let $f \in C (\prod_{i = 1}^{N} [a_{i}, b_{i}])$ and $n \in N$ such that $⌈ n a_{i} ⌉ \leq ⌊ n b_{i} ⌋$ , $i = 1, . . ., N .$

We introduce and define the multivariate positive linear neural network operator ( $x := (x_{1}, . . ., x_{N}) \in (\prod_{i = 1}^{N} [a_{i}, b_{i}])$ )

\begin{aligned} A_{n} (f, x_{1}, . . ., x_{N}) := A_{n} (f, x) := \frac{\sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} f (\frac{k}{n}) Z (n x - k)}{\sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} Z (n x - k)} \\ := \frac{\sum_{k_{1} = ⌈ n a_{1} ⌉}^{⌊ n b_{1} ⌋} \sum_{k_{2} = ⌈ n a_{2} ⌉}^{⌊ n b_{2} ⌋} . . . \sum_{k_{N} = ⌈ n a_{N} ⌉}^{⌊ n b_{N} ⌋} f (\frac{k_{1}}{n}, . . ., \frac{k_{N}}{n}) (\prod_{i = 1}^{N} χ (n x_{i} - k_{i}))}{\prod_{i = 1}^{N} (\sum_{k_{i} = ⌈ n a_{i} ⌉}^{⌊ n b_{i} ⌋} χ (n x_{i} - k_{i}))} . \end{aligned}

For large enough $n$ we always obtain $⌈ n a_{i} ⌉ \leq ⌊ n b_{i} ⌋$ , $i = 1, . . ., N$ . Also $a_{i} \leq \frac{k_{i}}{n} \leq b_{i}$ , iff $⌈ n a_{i} ⌉ \leq k_{i} \leq ⌊ n b_{i} ⌋$ , $i = 1, . . ., N$ .

For convenience we call

\begin{aligned} A_{n}^{*} (f, x) := \sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} f (\frac{k}{n}) Z (n x - k) \\ := \sum_{k_{1} = ⌈ n a_{1} ⌉}^{⌊ n b_{1} ⌋} \sum_{k_{2} = ⌈ n a_{2} ⌉}^{⌊ n b_{2} ⌋} . . . \sum_{k_{N} = ⌈ n a_{N} ⌉}^{⌊ n b_{N} ⌋} f (\frac{k_{1}}{n}, . . ., \frac{k_{N}}{n}) (\prod_{i = 1}^{N} χ (n x_{i} - k_{i})), \end{aligned}

$\forall$ $x \in (\prod_{i = 1}^{N} [a_{i}, b_{i}]) .$

That is

A_{n} (f, x) := \frac{A_{n}^{*} (f, x)}{\sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} Z (n x - k)},

$\forall$ $x \in (\prod_{i = 1}^{N} [a_{i}, b_{i}])$ , $n \in N .$

Hence

A_{n} (f, x) - f (x) = \frac{A_{n}^{*} (f, x) - f (x) (\sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} Z (n x - k))}{\sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} Z (n x - k)} .

Consequently we derive

| A_{n} (f, x) - f (x) | \leq {(4.019)}^{N} | A_{n}^{*} (f, x) - f (x) \sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} Z (n x - k) |,

$\forall$ $x \in (\prod_{i = 1}^{N} [a_{i}, b_{i}]) .$

We will estimate the right hand side of (29).

For the last we need, for $f \in C (\footnotesize \prod_{i = 1}^{N} [a_{i}, b_{i}])$ the first multivariate modulus of continuity

ω_{1} (f, h) := sup_{\overset{x, y \in \prod_{i = 1}^{N} [a_{i}, b_{i}]}{{‖ x - y ‖}_{\infty} \leq h}} | f (x) - f (y) |, h > 0.

It holds that

lim_{h \to 0} ω_{1} (f, h) = 0.

Similarly it is defined for $f \in C_{B} (R^{N})$ (continuous and bounded functions on $R^{N}$ ) the $ω_{1} (f, h)$ , and it has the property (31), given that $f \in C_{U} (R^{N})$ (uniformly continuous functions on $R^{N}$ ).

When $f \in C_{B} (R^{N})$ we define,

\begin{aligned} B_{n} (f, x) := B_{n} (f, x_{1}, . . ., x_{N}) := \sum_{k = - \infty}^{\infty} f (\frac{k}{n}) Z (n x - k) := \\ := \sum_{k_{1} = - \infty}^{\infty} \sum_{k_{2} = - \infty}^{\infty} . . . \sum_{k_{N} = - \infty}^{\infty} f (\frac{k_{1}}{n}, \frac{k_{2}}{n}, . . ., \frac{k_{N}}{n}) (\prod_{i = 1}^{N} χ (n x_{i} - k_{i})), \end{aligned}

$n \in N$ , $\forall$ $x \in R^{N},$ $N \in N$ , the multivariate quasi-interpolation neural network operator.

Also for $f \in C_{B} (R^{N})$ we define the multivariate Kantorovich type neural network operator

\begin{aligned} C_{n} (f, x) := C_{n} (f, x_{1}, . . ., x_{N}) := \sum_{k = - \infty}^{\infty} (n^{N} \int_{\frac{k}{n}}^{\frac{k + 1}{n}} f (t) d t) Z (n x - k) := \\ = \sum_{k_{1} = - \infty}^{\infty} \sum_{k_{2} = - \infty}^{\infty} . . . \sum_{k_{N} = - \infty}^{\infty} (n^{N} \int_{\frac{k_{1}}{n}}^{\frac{k_{1} + 1}{n}} \int_{\frac{k_{2}}{n}}^{\frac{k_{2} + 1}{n}} . . . \int_{\frac{k_{N}}{n}}^{\frac{k_{N} + 1}{n}} f (t_{1}, . . ., t_{N}) d t_{1} . . . d t_{N}) \cdot \\ \cdot (\prod_{i = 1}^{N} χ (n x_{i} - k_{i})), \end{aligned}

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

Again for $f \in C_{B} (R^{N}),$ $N \in N,$ we define the multivariate neural network operator of quadrature type $D_{n} (f, x)$ , $n \in N,$ as follows. Let $θ = (θ_{1}, . . ., θ_{N}) \in N^{N},$ $r = (r_{1}, . . ., r_{N}) \in Z_{+}^{N}$ , $w_{r} = w_{r_{1}, r_{2}, . . . r_{N}} \geq 0$ , such that $\sum_{r = 0}^{θ} w_{r} = \sum_{r_{1} = 0}^{θ_{1}} \sum_{r_{2} = 0}^{θ_{2}} . . . \sum_{r_{N} = 0}^{θ_{N}} w_{r_{1}, r_{2}, . . . r_{N}} = 1;$ $k \in Z^{N}$ and

\begin{aligned} δ_{n k} (f) := δ_{n, k_{1}, k_{2}, . . ., k_{N}} (f) := \sum_{r = 0}^{θ} w_{r} f (\frac{k}{n} + \frac{r}{n θ}) := \\ := \sum_{r_{1} = 0}^{θ_{1}} \sum_{r_{2} = 0}^{θ_{2}} . . . \sum_{r_{N} = 0}^{θ_{N}} w_{r_{1}, r_{2}, . . . r_{N}} f (\frac{k_{1}}{n} + \frac{r_{1}}{n θ_{1}}, \frac{k_{2}}{n} + \frac{r_{2}}{n θ_{2}}, . . ., \frac{k_{N}}{n} + \frac{r_{N}}{n θ_{N}}), \end{aligned}

where $\frac{r}{θ} := (\frac{r_{1}}{θ_{1}}, \frac{r_{2}}{θ_{2}}, . . ., \frac{r_{N}}{θ_{N}}) .$

We put

\begin{aligned} D_{n} (f, x) := D_{n} (f, x_{1}, . . ., x_{N}) := \sum_{k = - \infty}^{\infty} δ_{n k} (f) Z (n x - k) := \\ := \sum_{k_{1} = - \infty}^{\infty} \sum_{k_{2} = - \infty}^{\infty} . . . \sum_{k_{N} = - \infty}^{\infty} δ_{n, k_{1}, k_{2}, . . ., k_{N}} (f) (\prod_{i = 1}^{N} χ (n x_{i} - k_{i})), \end{aligned}

$\forall$ $x \in R^{N} .$

Let fixed $j \in N$ , $0 < β < 1$ , and $A, B > 0$ . For large enough $n \in N : n^{1 - β} \geq 3,$ in the linear combination $(\frac{A}{n^{β j}} + \frac{B}{(n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}}),$ the dominant rate of convergence, as $n \to \infty$ , is $n^{- β j} .$ The closer $β$ is to $1$ we get faster and better rate of convergence to zero.

Let $f \in C^{m} (\prod_{i = 1}^{N} [a_{i}, b_{i}])$ , $m, N \in N$ . Here $f_{α}$ denotes a partial derivative of $f$ , $α := (α_{1}, . . ., α_{N})$ , $α_{i} \in Z_{+}$ , $i = 1, . . ., N$ , and $| α | := \sum_{i = 1}^{N} α_{i} = l,$ where $l = 0, 1, . . ., m$ . We write also $f_{α} := \frac{\partial^{α} f}{\partial x^{α}}$ and we say it is of order $l$ .

We denote

ω_{1, m}^{max} (f_{α}, h) := max_{α : | α | = m} ω_{1} (f_{α}, h) .

Call also

{‖ f_{α} ‖}_{\infty, m}^{max} := max_{| α | = m} {{‖ f_{α} ‖}_{\infty}},

${‖ \cdot ‖}_{\infty}$ is the supremum norm.

In this article we study the basic approximation properties of $A_{n}, B_{n}, C_{n}, D_{n}$ neural network operators and as well of their iterates. That is, the quantitative pointwise and uniform convergence of these operators to the unit operator $I$ . We study also the complex functions related approximation.

3 Multidimensional Real Neural Network Approximations

Here we present a series of neural network approximations to a function given with rates.

We give

Theorem 4

Let $f \in C (\prod_{i = 1}^{N} [a_{i}, b_{i}]),$ $0 < β < 1$ , $x \in (\prod_{i = 1}^{N} [a_{i}, b_{i}]),$ $N, n \in N$ with $n^{1 - β} \geq 3$ . Then

| A_{n} (f, x) - f (x) | \leq {(4.019)}^{N} [ω_{1} (f, \frac{1}{n^{β}}) + \frac{{‖ f ‖}_{\infty}}{\sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}}] =: λ_{1},

and

{‖ A_{n} (f) - f ‖}_{\infty} \leq λ_{1} .

We notice that $lim_{n \to \infty} A_{n} (f) = f$ , pointwise and uniformly.

Proof â–¼

We observe that

\begin{aligned} Δ (x) & := A_{n}^{*} (f, x) - f (x) \sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} Z (n x - k) = \end{aligned}

\begin{aligned} = \sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} f (\frac{k}{n}) Z (n x - k) - \sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} f (x) Z (n x - k) \\ = \sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} (f (\frac{k}{n}) - f (x)) Z (n x - k) . \end{aligned}

Thus

\begin{aligned} | Δ (x) | & \leq \sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} | f (\frac{k}{n}) - f (x) | Z (n x - k) = \\ = \sum_{\overset{k = ⌈ n a ⌉}{{‖ \frac{k}{n} - x ‖}_{\infty} \leq \frac{1}{n^{β}}}}^{⌊ n b ⌋} | f (\frac{k}{n}) - f (x) | Z (n x - k) \\ + \sum_{\overset{k = ⌈ n a ⌉}{{‖ \frac{k}{n} - x ‖}_{\infty} > \frac{1}{n^{β}}}}^{⌊ n b ⌋} | f (\frac{k}{n}) - f (x) | Z (n x - k) \\ \overset{(by (???))}{\leq} ω_{1} (f, \frac{1}{n^{β}}) + 2 {‖ f ‖}_{\infty} \sum_{\overset{k = ⌈ n a ⌉}{{‖ \frac{k}{n} - x ‖}_{\infty} > \frac{1}{n^{β}}}}^{⌊ n b ⌋} Z (n x - k) \\ \overset{(by (???))}{\leq} ω_{1} (f, \frac{1}{n^{β}}) + \frac{{‖ f ‖}_{\infty}}{\sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}} . \end{aligned}

So that

| Δ | \leq ω_{1} (f, \frac{1}{n^{β}}) + \frac{{‖ f ‖}_{\infty}}{\sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}} .

Now using (29) we finish proof.

Proof â–¼

We continue with

Theorem 5

Let $f \in C_{B} (R^{N}),$ $0 < β < 1$ , $x \in R^{N},$ $N, n \in N$ with $n^{1 - β} \geq 3$ . Then

| B_{n} (f, x) - f (x) | \leq ω_{1} (f, \frac{1}{n^{β}}) + \frac{{‖ f ‖}_{\infty}}{\sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}} =: λ_{2},

{‖ B_{n} (f) - f ‖}_{\infty} \leq λ_{2} .

Given that $f \in (C_{U} (R^{N}) \cap C_{B} (R^{N}))$ , we obtain $lim_{n \to \infty} B_{n} (f) = f$ , uniformly.

Proof â–¼

We have that

\begin{aligned} B_{n} (f, x) - f (x) & \overset{(???)}{=} \sum_{k = - \infty}^{\infty} f (\frac{k}{n}) Z (n x - k) - f (x) \sum_{k = - \infty}^{\infty} Z (n x - k) \\ = \sum_{k = - \infty}^{\infty} (f (\frac{k}{n}) - f (x)) Z (n x - k) . \end{aligned}

Hence

\begin{aligned} | B_{n} (f, x) - f (x) | & \leq \sum_{k = - \infty}^{\infty} | f (\frac{k}{n}) - f (x) | Z (n x - k) \\ = \sum_{\overset{k = - \infty}{{‖ \frac{k}{n} - x ‖}_{\infty} \leq \frac{1}{n^{β}}}}^{\infty} | f (\frac{k}{n}) - f (x) | Z (n x - k) \\ + \sum_{\overset{k = - \infty}{{‖ \frac{k}{n} - x ‖}_{\infty} > \frac{1}{n^{β}}}}^{\infty} | f (\frac{k}{n}) - f (x) | Z (n x - k) \\ \overset{(???)}{\leq} ω_{1} (f, \frac{1}{n^{β}}) + 2 {‖ f ‖}_{\infty} \sum_{\overset{k = - \infty}{{‖ \frac{k}{n} - x ‖}_{\infty} > \frac{1}{n^{β}}}}^{\infty} Z (n x - k) \\ \overset{(???)}{\leq} ω_{1} (f, \frac{1}{n^{β}}) + \frac{{‖ f ‖}_{\infty}}{\sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}}, \end{aligned}

proving the claim.

Proof â–¼

We give

Theorem 6

Let $f \in C_{B} (R^{N}),$ $0 < β < 1$ , $x \in R^{N},$ $N, n \in N$ with $n^{1 - β} \geq 3$ . Then

| C_{n} (f, x) - f (x) | \leq ω_{1} (f, \frac{1}{n} + \frac{1}{n^{β}}) + \frac{{‖ f ‖}_{\infty}}{\sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}} =: λ_{3},

{‖ C_{n} (f) - f ‖}_{\infty} \leq λ_{3} .

Given that $f \in (C_{U} (R^{N}) \cap C_{B} (R^{N})),$ we obtain $lim_{n \to \infty} C_{n} (f) = f$ , uniformly.

Proof â–¼

We notice that

\int_{\frac{k}{n}}^{\frac{k + 1}{n}} f (t) d t = \int_{\frac{k_{1}}{n}}^{\frac{k_{1} + 1}{n}} \int_{\frac{k_{2}}{n}}^{\frac{k_{2} + 1}{n}} . . . \int_{\frac{k_{N}}{n}}^{\frac{k_{N} + 1}{n}} f (t_{1}, t_{2}, . . ., t_{N}) d t_{1} d t_{2} . . . d t_{N} =

= \int_{0}^{\frac{1}{n}} \int_{0}^{\frac{1}{n}} . . . \int_{0}^{\frac{1}{n}} f (t_{1} + \frac{k_{1}}{n}, t_{2} + \frac{k_{2}}{n}, . . ., t_{N} + \frac{k_{N}}{n}) d t_{1} . . . d t_{N} = \int_{0}^{\frac{1}{n}} f (t + \frac{k}{n}) d t .

Thus it holds

C_{n} (f, x) = \sum_{k = - \infty}^{\infty} (n^{N} \int_{0}^{\frac{1}{n}} f (t + \frac{k}{n}) d t) Z (n x - k) .

We observe that

\begin{aligned} | C_{n} (f, x) - f (x) | = \\ = | \sum_{k = - \infty}^{\infty} (n^{N} \int_{0}^{\frac{1}{n}} f (t + \frac{k}{n}) d t) Z (n x - k) - \sum_{k = - \infty}^{\infty} f (x) Z (n x - k) | \\ = | \sum_{k = - \infty}^{\infty} ((n^{N} \int_{0}^{\frac{1}{n}} f (t + \frac{k}{n}) d t) - f (x)) Z (n x - k) | \\ = | \sum_{k = - \infty}^{\infty} (n^{N} \int_{0}^{\frac{1}{n}} (f (t + \frac{k}{n}) - f (x)) d t) Z (n x - k) | \\ \leq \sum_{k = - \infty}^{\infty} (n^{N} \int_{0}^{\frac{1}{n}} | f (t + \frac{k}{n}) - f (x) | d t) Z (n x - k) \\ = \sum_{\overset{k = - \infty}{{‖ \frac{k}{n} - x ‖}_{\infty} \leq \frac{1}{n^{β}}}}^{\infty} (n^{N} \int_{0}^{\frac{1}{n}} | f (t + \frac{k}{n}) - f (x) | d t) Z (n x - k) \\ + \sum_{\overset{k = - \infty}{{‖ \frac{k}{n} - x ‖}_{\infty} > \frac{1}{n^{β}}}}^{\infty} (n^{N} \int_{0}^{\frac{1}{n}} | f (t + \frac{k}{n}) - f (x) | d t) Z (n x - k) \\ \leq \sum_{\overset{k = - \infty}{{‖ \frac{k}{n} - x ‖}_{\infty} \leq \frac{1}{n^{β}}}}^{\infty} (n^{N} \int_{0}^{\frac{1}{n}} ω_{1} (f, {‖ t ‖}_{\infty} + {‖ \frac{k}{n} - x ‖}_{\infty}) d t) Z (n x - k) \\ + 2 {‖ f ‖}_{\infty} (\sum_{\overset{k = - \infty}{{‖ \frac{k}{n} - x ‖}_{\infty} > \frac{1}{n^{β}}}}^{\infty} Z (| n x - k |)) \\ \leq ω_{1} (f, \frac{1}{n} + \frac{1}{n^{β}}) + \frac{{‖ f ‖}_{\infty}}{\sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}}, \end{aligned}

proving the claim.

Proof â–¼

We also present

Theorem 7

Let $f \in C_{B} (R^{N}),$ $0 < β < 1$ , $x \in R^{N},$ $N, n \in N$ with $n^{1 - β} \geq 3$ . Then

| D_{n} (f, x) - f (x) | \leq ω_{1} (f, \frac{1}{n} + \frac{1}{n^{β}}) + \frac{{‖ f ‖}_{\infty}}{\sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}} = λ_{3},

{‖ D_{n} (f) - f ‖}_{\infty} \leq λ_{3} .

Given that $f \in (C_{U} (R^{N}) \cap C_{B} (R^{N})),$ we obtain $lim_{n \to \infty} D_{n} (f) = f$ , uniformly.

Proof â–¼

We have that

\begin{aligned} | D_{n} (f, x) - f (x) | \\ = | \sum_{k = - \infty}^{\infty} δ_{n k} (f) Z (n x - k) - \sum_{k = - \infty}^{\infty} f (x) Z (n x - k) | \\ = | \sum_{k = - \infty}^{\infty} (δ_{n k} (f) - f (x)) Z (n x - k) | \\ = | \sum_{k = - \infty}^{\infty} (\sum_{r = 0}^{θ} w_{r} (f (\frac{k}{n} + \frac{r}{n θ}) - f (x))) Z (n x - k) | \\ \leq \sum_{k = - \infty}^{\infty} (\sum_{r = 0}^{θ} w_{r} | f (\frac{k}{n} + \frac{r}{n θ}) - f (x) |) Z (n x - k) \\ = \sum_{\overset{k = - \infty}{{‖ \frac{k}{n} - x ‖}_{\infty} \leq \frac{1}{n^{β}}}}^{\infty} (\sum_{r = 0}^{θ} w_{r} | f (\frac{k}{n} + \frac{r}{n θ}) - f (x) |) Z (n x - k) \\ + \sum_{\overset{k = - \infty}{{‖ \frac{k}{n} - x ‖}_{\infty} > \frac{1}{n^{β}}}}^{\infty} (\sum_{r = 0}^{θ} w_{r} | f (\frac{k}{n} + \frac{r}{n θ}) - f (x) |) Z (n x - k) \\ \leq \sum_{\overset{k = - \infty}{{‖ \frac{k}{n} - x ‖}_{\infty} \leq \frac{1}{n^{β}}}}^{\infty} (\sum_{r = 0}^{θ} w_{r} ω_{1} (f, {‖ \frac{k}{n} - x ‖}_{\infty} + {‖ \frac{r}{n θ} ‖}_{\infty})) Z (n x - k) + \end{aligned}

\begin{aligned} + 2 {‖ f ‖}_{\infty} (\sum_{\overset{k = - \infty}{{‖ \frac{k}{n} - x ‖}_{\infty} > \frac{1}{n^{β}}}}^{\infty} Z (n x - k)) \\ \leq ω_{1} (f, \frac{1}{n} + \frac{1}{n^{β}}) + \frac{{‖ f ‖}_{\infty}}{\sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}}, \end{aligned}

proving the claim.

Proof â–¼

In the next we discuss high order of approximation by using the smoothness of $f$ .

We give

Theorem 8

Let $f \in C^{m} (\prod_{i = 1}^{N} [a_{i}, b_{i}])$ , $0 < β < 1$ , $n, m, N \in N$ , $n^{1 - β} \geq 3,$ $x \in (\prod_{i = 1}^{N} [a_{i}, b_{i}])$ . Then

\begin{aligned} | A_{n} (f, x) - f (x) - \sum_{j = 1}^{m} (\sum_{| α | = j} (\frac{f_{α} (x)}{\prod_{i = 1}^{N} α_{i}!}) A_{n} (\prod_{i = 1}^{N} {(\cdot - x_{i})}^{α_{i}}, x)) | \leq \\ \leq {(4.019)}^{N} \cdot {\frac{N^{m}}{m! n^{m β}} ω_{1, m}^{max} (f_{α}, \frac{1}{n^{β}}) + (\frac{{‖ b - a ‖}_{\infty}^{m} {‖ f_{α} ‖}_{\infty, m}^{max} N^{m}}{m!}) \frac{1}{\sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}}}, \end{aligned}

ii)

\begin{aligned} | A_{n} (f, x) - f (x) | \leq {(4.019)}^{N} \cdot \\ \cdot {\sum_{j = 1}^{m} (\sum_{| α | = j} (\frac{| f_{α} (x) |}{\prod_{i = 1}^{N} α_{i}!}) [\frac{1}{n^{β j}} + (\prod_{i = 1}^{N} {(b_{i} - a_{i})}^{α_{i}}) \cdot \frac{1}{2 \sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}}]) \\ + \frac{N^{m}}{m! n^{m β}} ω_{1, m}^{max} (f_{α}, \frac{1}{n^{β}}) + (\frac{{‖ b - a ‖}_{\infty}^{m} {‖ f_{α} ‖}_{\infty, m}^{max} N^{m}}{m!}) \frac{1}{\sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}}}, \end{aligned}

iii)

\begin{aligned} {‖ A_{n} (f) - f ‖}_{\infty} {(4.019)}^{N} \cdot \\ {\sum_{j = 1}^{m} (\sum_{| α | = j} (\frac{| f_{α} (x) |}{\prod_{i = 1}^{N} α_{i}!}) [\frac{1}{n^{β j}} + (\prod_{i = 1}^{N} {(b_{i} - a_{i})}^{α_{i}}) \cdot \frac{1}{2 \sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}}]) + \end{aligned}

\begin{aligned} + \frac{N^{m}}{m! n^{m β}} ω_{1, m}^{max} (f_{α}, \frac{1}{n^{β}}) + (\frac{{‖ b - a ‖}_{\infty}^{m} {‖ f_{α} ‖}_{\infty, m}^{max} N^{m}}{m!}) \frac{1}{\sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}}} =: K_{n}, \end{aligned}

iv) Assume $f_{α} (x_{0}) = 0$ , for all $α : | α | = 1, . . ., m;$ $x_{0} \in (\prod_{i = 1}^{N} [a_{i}, b_{i}])$ . Then

\begin{aligned} | A_{n} (f, x_{0}) - f (x_{0}) | \leq \\ {(4.019)}^{N} {\frac{N^{m}}{m! n^{m β}} ω_{1}^{max} (f_{α}, \frac{1}{n^{β}}) + (\frac{{‖ b - a ‖}_{\infty}^{m} {‖ f_{α} ‖}_{\infty, m}^{max} N^{m}}{m!}) \frac{1}{\sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}}}, \end{aligned}

notice in the last the extremely high rate of convergence at $n^{- β (m + 1)} .$

Proof â–¼

Consider

g_{z} (t) := f (x_{0} + t (z - x_{0}))

t \geq 0;

x_{0}, z \in \prod_{i = 1}^{N} [a_{i}, b_{i}] .

Then

g_{z}^{(j)} (t) = [{(\sum_{i = 1}^{N} (z_{i} - x_{0 i}) \frac{\partial}{\partial x_{i}})}^{j} f] (x_{01} + t (z_{1} - x_{01}), . . ., x_{0 N} + t (z_{N} - x_{0 N})),

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

We have the multivariate Taylor’s formula

\begin{aligned} f (z_{1}, . . ., z_{N}) = g_{z} (1) = \\ = \sum_{j = 0}^{m} \frac{g_{z}^{(j)} (0)}{j!} + \frac{1}{(m - 1)!} \int_{0}^{1} {(1 - θ)}^{m - 1} (g_{z}^{(m)} (θ) - g_{z}^{(m)} (0)) d θ . \end{aligned}

Notice $g_{z} (0) = f (x_{0})$ . Also for $j = 0, 1, . . ., m$ , we have

g_{z}^{(j)} (0) = \sum_{\begin{matrix} α := (α_{1}, . . ., α_{N}), α_{i} \in Z^{+}, \\ i = 1, . . ., N, | α | := \sum_{i = 1}^{N} α_{i} = j \end{matrix}} (\frac{j!}{\prod_{i = 1}^{N} α_{i}!}) (\prod_{i = 1}^{N} {(z_{i} - x_{0 i})}^{α_{i}}) f_{α} (x_{0}) .

Furthermore

g_{z}^{(m)} (θ) = \sum_{\begin{matrix} α := (α_{1}, . . ., α_{N}), α_{i} \in Z^{+}, \\ i = 1, . . ., N, | α | := \sum_{i = 1}^{N} α_{i} = m \end{matrix}} (\frac{m!}{\prod_{i = 1}^{N} α_{i}!}) (\prod_{i = 1}^{N} {(z_{i} - x_{0 i})}^{α_{i}}) f_{α} (x_{0} + θ (z - x_{0})),

$0 \leq θ \leq 1.$

So we treat $f \in C^{m} (\prod_{i = 1}^{N} [a_{i}, b_{i}])$ .

Thus, we have for $\frac{k}{n}, x \in (\prod_{i = 1}^{N} [a_{i}, b_{i}])$ that

Unknown environment 'bgroup'

where

R := m \int_{0}^{1} {(1 - θ)}^{m - 1} \sum_{\begin{matrix} α := (α_{1}, . . ., α_{N}), \\ α_{i} \in Z^{+}, i = 1, . . ., N, \\ | α | := \sum_{i = 1}^{N} α_{i} = m \end{matrix}} (\frac{1}{\prod_{i = 1}^{N} α_{i}!}) (\prod_{i = 1}^{N} {(\frac{k_{i}}{n} - x_{i})}^{α_{i}}) \cdot

\cdot [f_{α} (x + θ (\frac{k}{n} - x)) - f_{α} (x)] d θ .

We see that

\begin{aligned} | R | \leq & m \int_{0}^{1} {(1 - θ)}^{m - 1} \sum_{| α | = m} (\frac{1}{\prod_{i = 1}^{N} α_{i}!}) (\prod_{i = 1}^{N} {| \frac{k_{i}}{n} - x_{i} |}^{α_{i}}) \cdot \\ \cdot | f_{α} (x + θ (\frac{k}{n} - x)) - f_{α} (x) | d θ \leq m \int_{0}^{1} {(1 - θ)}^{m - 1} \cdot \\ \cdot (\sum_{| α | = m} (\frac{1}{\prod_{i = 1}^{N} α_{i}!}) (\prod_{i = 1}^{N} {| \frac{k_{i}}{n} - x_{i} |}^{α_{i}}) ω_{1} (f_{α}, θ {‖ \frac{k}{n} - x ‖}_{\infty})) d θ \leq (*) . \end{aligned}

Notice here that

{‖ \frac{k}{n} - x ‖}_{\infty} \leq \frac{1}{n^{β}} \Leftrightarrow | \frac{k_{i}}{n} - x_{i} | \leq \frac{1}{n^{β}}, \ i = 1, . . ., N .

We further see that

\begin{aligned} (*) & \leq m \cdot ω_{1, m}^{max} (f_{α}, \frac{1}{n^{β}}) \int_{0}^{1} {(1 - θ)}^{m - 1} (\sum_{| α | = m} (\frac{1}{\prod_{i = 1}^{N} α_{i}!}) (\prod_{i = 1}^{N} {(\frac{1}{n^{β}})}^{α_{i}})) d θ \\ = (\frac{ω_{1, m}^{max} (f_{α}, \frac{1}{n^{β}})}{(m!) n^{m β}}) (\sum_{| α | = m} \frac{m!}{\prod_{i = 1}^{N} α_{i}!}) = (\frac{ω_{1, m}^{max} (f_{α}, \frac{1}{n^{β}})}{(m!) n^{m β}}) N^{m} . \end{aligned}

Conclusion: When ${‖ \frac{k}{n} - x ‖}_{\infty} \leq \frac{1}{n^{β}}$ , we proved that

| R | \leq (\frac{N^{m}}{m! n^{m β}}) ω_{1, m}^{max} (f_{α}, \frac{1}{n^{β}}) .

In general we notice that

\begin{aligned} | R | & \leq m \int_{0}^{1} {(1 - θ)}^{m - 1} (\sum_{| α | = m} (\frac{1}{\prod_{i = 1}^{N} α_{i}!}) (\prod_{i = 1}^{N} {(b_{i} - a_{i})}^{α_{i}}) 2 {‖ f_{α} ‖}_{\infty}) d θ \\ = 2 \sum_{| α | = m} \frac{1}{\prod_{i = 1}^{N} α_{i}!} (\prod_{i = 1}^{N} {(b_{i} - a_{i})}^{α_{i}}) {‖ f_{α} ‖}_{\infty} \\ \leq (\frac{2 {‖ b - a ‖}_{\infty}^{m} {‖ f_{α} ‖}_{\infty, m}^{max}}{m!}) (\sum_{| α | = m} \frac{m!}{\prod_{i = 1}^{N} α_{i}!}) = \frac{2 {‖ b - a ‖}_{\infty}^{m} {‖ f_{α} ‖}_{\infty, m}^{max} N^{m}}{m!} . \end{aligned}

We proved in general that

| R | \leq \frac{2 {‖ b - a ‖}_{\infty}^{m} {‖ f_{α} ‖}_{\infty, m}^{max} N^{m}}{m!} := ρ .

Next we see that

\begin{aligned} U_{n} & := \sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} Z (n x - k) R \\ = \sum_{\overset{k = ⌈ n a ⌉}{{‖ \frac{k}{n} - x ‖}_{\infty} \leq \frac{1}{n^{β}}}}^{⌊ n b ⌋} Z (n x - k) R + \sum_{\overset{k = ⌈ n a ⌉}{{‖ \frac{k}{n} - x ‖}_{\infty} > \frac{1}{n^{β}}}}^{⌊ n b ⌋} Z (n x - k) R . \end{aligned}

Consequently

\begin{aligned} | U_{n} | & \leq (\sum_{\overset{k = ⌈ n a ⌉}{{‖ \frac{k}{n} - x ‖}_{\infty} \leq \frac{1}{n^{β}}}}^{⌊ n b ⌋} Z (n x - k)) \frac{N^{m}}{m! n^{m β}} ω_{1, m}^{max} (f_{α}, \frac{1}{n^{β}}) \\ + ρ \frac{1}{2 \sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}} \\ \leq \frac{N^{m}}{m! n^{m β}} ω_{1, m}^{max} (f_{α}, \frac{1}{n^{β}}) + ρ \frac{1}{2 \sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}} . \end{aligned}

We have established that

\begin{aligned} | U_{n} | & \leq \frac{N^{m}}{m! n^{m β}} ω_{1, m}^{max} (f_{α}, \frac{1}{n^{β}}) + (\frac{{‖ b - a ‖}_{\infty}^{m} {‖ f_{α} ‖}_{\infty, m}^{max} N^{m}}{m!}) \frac{1}{\sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}} . \end{aligned}

We observe that

\begin{aligned} \sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} f (\frac{k}{n}) Z (n x - k) - f (x) \sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} Z (n x - k) = \\ = \sum_{j = 1}^{m} (\sum_{| α | = j} (\frac{f_{α} (x)}{\prod_{i = 1}^{N} α_{i}!}) (\sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} Z (n x - k) (\prod_{i = 1}^{N} {(\frac{k_{i}}{n} - x_{i})}^{α_{i}}))) \\ + \sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} Z (n x - k) R . \end{aligned}

The last says

\begin{aligned} A_{n}^{*} (f, x) - f (x) (\sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} Z (n x - k)) - \\ - \sum_{j = 1}^{m} (\sum_{| α | = j} (\frac{f_{α} (x)}{\prod_{i = 1}^{N} α_{i}!}) A_{n}^{*} (\prod_{i = 1}^{N} {(\cdot - x_{i})}^{α_{i}}, x)) = U_{n} . \end{aligned}

Clearly $A_{n}^{*}$ is a positive linear operator.

Thus (here $α_{i} \in Z^{+} : | α | = \sum_{i = 1}^{N} α_{i} = j$ )

Unknown environment 'bgroup'

\begin{aligned} \leq \frac{1}{n^{β j}} + (\prod_{i = 1}^{N} {(b_{i} - a_{i})}^{α_{i}}) \frac{1}{2 \sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}} . \end{aligned}

So we have proved that

| A_{n}^{*} (\prod_{i = 1}^{N} {(\cdot - x_{i})}^{α_{i}}, x) | \leq \frac{1}{n^{β j}} + (\prod_{i = 1}^{N} {(b_{i} - a_{i})}^{α_{i}}) \frac{1}{2 \sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}},

for all $j = 1, . . ., m .$

At last we observe

\begin{aligned} | A_{n} (f, x) - f (x) - \sum_{j = 1}^{m} (\sum_{| α | = j} (\frac{f_{α} (x)}{\prod_{i = 1}^{N} α_{i}!}) A_{n} (\prod_{i = 1}^{N} {(\cdot - x_{i})}^{α_{i}}, x)) | \leq \\ \leq {(4.019)}^{N} \cdot | A_{n}^{*} (f, x) - f (x) \sum_{k = ⌈ n a ⌉}^{⌊ n b ⌋} Z (n x - k) - \\ - \sum_{j = 1}^{m} (\sum_{| α | = j} (\frac{f_{α} (x)}{\prod_{i = 1}^{N} α_{i}!}) A_{n}^{*} (\prod_{i = 1}^{N} {(\cdot - x_{i})}^{α_{i}}, x)) | . \end{aligned}

Putting all of the above together we prove theorem.

Proof â–¼

We make

Definition 9

Let $f \in C_{B} (R^{N})$ , $N \in N$ . We define the general neural network operator

\begin{aligned} F_{n} (f, x) & := \sum_{k = - \infty}^{\infty} l_{n k} (f) Z (n x - k) = \\ = {\begin{cases} B_{n} (f, x), \  if & l_{n k} (f) = f (\frac{k}{n}), \\ C_{n} (f, x), \  if & l_{n k} (f) = n^{N} \int_{\frac{k}{n}}^{\frac{k + 1}{n}} f (t) d t, \\ D_{n} (f, x), \  if & l_{n k} (f) = δ_{n k} (f) . \end{cases} \end{aligned}

Clearly $l_{n k} (f)$ is a positive linear functional such that $| l_{n k} (f) | \leq {‖ f ‖}_{\infty} .$

Hence $F_{n} (f)$ is a positive linear operator with ${‖ F_{n} (f) ‖}_{\infty} \leq {‖ f ‖}_{\infty}$ , a continuous bounded linear operator.

We need

Theorem 10

Let $f \in C_{B} (R^{N})$ , $N \geq 1$ . Then $F_{n} (f) \in C_{B} (R^{N}) .$

Proof â–¼

Clearly

F_{n} (f)

is a bounded function.

Next we prove the continuity of $F_{n} (f)$ . Notice for $N = 1$ , $Z = χ$ by (10).

We will use the Weierstrass $M$ test: If a sequence of positive constants $M_{1}, M_{2}, M_{3}, . . .,$ can be found such that in some interval

(a) $| u_{n} (x) | \leq M_{n}$ , $n = 1, 2, 3, . . .$

(b) $\sum M_{n}$ converges,

then $\sum u_{n} (x)$ is uniformly and absolutely convergent in the interval.

Also we will use:

If ${u_{n} (x)}$ , $n = 1, 2, 3, . . .$ are continuous in $[a, b]$ and if $\sum u_{n} (x)$ converges uniformly to the sum $S (x)$ in $[a, b]$ , then $S (x)$ is continuous in $[a, b]$ . I.e. a uniformly convergent series of continuous functions is a continuous function. First we prove claim for $N = 1$ .

We will prove that $\sum_{k = - \infty}^{\infty} l_{n k} (f) χ (n x - k)$ is continuous in $x \in R$ .

There always exists $λ \in N$ such that $n x \in [- λ, λ] .$

Since $n x \leq λ$ , then $- n x \geq - λ$ and $k - n x \geq k - λ \geq 0$ , when $k \geq λ$ . Therefore

\sum_{k = λ}^{\infty} χ (n x - k) = \sum_{k = λ}^{\infty} χ (k - n x) \leq \sum_{k = λ}^{\infty} χ (k - λ) = \sum_{k^{'} = 0}^{\infty} χ (k^{'}) \leq 1.

So for $k \geq λ$ we get

| l_{n k} (f) | χ (n x - k) \leq {‖ f ‖}_{\infty} χ (k - λ),

and

{‖ f ‖}_{\infty} \sum_{k = λ}^{\infty} χ (k - λ) \leq {‖ f ‖}_{\infty} .

Hence by Weierstrass $M$ test we obtain that $\sum_{k = λ}^{\infty} l_{n k} (f) χ (n x - k)$ is uniformly and absolutely convergent on $[- \frac{λ}{n}, \frac{λ}{n}] .$

Since $l_{n k} (f) χ (n x - k)$ is continuous in $x$ , then $\sum_{k = λ}^{\infty} l_{n k} (f) χ (n x - k)$ is continuous on $[- \frac{λ}{n}, \frac{λ}{n}] .$

Because $n x \geq - λ$ , then $- n x \leq λ$ , and $k - n x \leq k + λ \leq 0$ , when $k \leq - λ$ . Therefore

\sum_{k = - \infty}^{- λ} χ (n x - k) = \sum_{k = - \infty}^{- λ} χ (k - n x) \leq \sum_{k = - \infty}^{- λ} χ (k + λ) = \sum_{k^{'} = - \infty}^{0} χ (k^{'}) \leq 1.

So for $k \leq - λ$ we get

| l_{n k} (f) | χ (n x - k) \leq {‖ f ‖}_{\infty} χ (k + λ),

and

{‖ f ‖}_{\infty} \sum_{k = - \infty}^{- λ} χ (k + λ) \leq {‖ f ‖}_{\infty} .

Hence by Weierstrass $M$ test we obtain that $\sum_{k = - \infty}^{- λ} l_{n k} (f) χ (n x - k)$ is uniformly and absolutely convergent on $[- \frac{λ}{n}, \frac{λ}{n}] .$

Since $l_{n k} (f) χ (n x - k)$ is continuous in $x$ , then $\sum_{k = - \infty}^{- λ} l_{n k} (f) χ (n x - k)$ is continuous on $[- \frac{λ}{n}, \frac{λ}{n}] .$

So we proved that $\sum_{k = λ}^{\infty} l_{n k} (f) χ (n x - k)$ and $\sum_{k = - \infty}^{- λ} l_{n k} (f) χ (n x - k)$ are continuous on $R$ . Since $\sum_{k = - λ + 1}^{λ - 1} l_{n k} (f) χ (n x - k)$ is a finite sum of continuous functions on $R$ , it is also a continuous function on $R$ .

Writing

\begin{aligned} \sum_{k = - \infty}^{\infty} l_{n k} (f) χ (n x - k) & = \sum_{k = - \infty}^{- λ} l_{n k} (f) χ (n x - k) + \\ + \sum_{k = - λ + 1}^{λ - 1} l_{n k} (f) χ (n x - k) + \sum_{k = λ}^{\infty} l_{n k} (f) χ (n x - k) \end{aligned}

we have it as a continuous function on $R$ . Therefore $F_{n} (f)$ , when $N = 1$ , is a continuous function on $R$ .

When $N = 2$ we have

\begin{aligned} F_{n} (f, x_{1}, x_{2}) & = \sum_{k_{1} = - \infty}^{\infty} \sum_{k_{2} = - \infty}^{\infty} l_{n k} (f) χ (n x_{1} - k_{1}) χ (n x_{2} - k_{2}) = \\ = \sum_{k_{1} = - \infty}^{\infty} χ (n x_{1} - k_{1}) (\sum_{k_{2} = - \infty}^{\infty} l_{n k} (f) χ (n x_{2} - k_{2})) \end{aligned}

(there always exist $λ_{1}, λ_{2} \in N$ such that $n x_{1} \in [- λ_{1}, λ_{1}]$ and $n x_{2} \in [- λ_{2}, λ_{2}]$ )

\begin{aligned} = \sum_{k_{1} = - \infty}^{\infty} χ (n x_{1} - k_{1}) [\sum_{k_{2} = - \infty}^{- λ_{2}} l_{n k} (f) χ (n x_{2} - k_{2}) + \\ + \sum_{k_{2} = - λ_{2} + 1}^{λ_{2} - 1} l_{n k} (f) χ (n x_{2} - k_{2}) + \sum_{k_{2} = λ_{2}}^{\infty} l_{n k} (f) χ (n x_{2} - k_{2})] \\ = \sum_{k_{1} = - \infty}^{\infty} \sum_{k_{2} = - \infty}^{- λ_{2}} l_{n k} (f) χ (n x_{1} - k_{1}) χ (n x_{2} - k_{2}) + \\ + \sum_{k_{1} = - \infty}^{\infty} \sum_{k_{2} = - λ_{2} + 1}^{λ_{2} - 1} l_{n k} (f) χ (n x_{1} - k_{1}) χ (n x_{2} - k_{2}) + \\ + \sum_{k_{1} = - \infty}^{\infty} \sum_{k_{2} = λ_{2}}^{\infty} l_{n k} (f) χ (n x_{1} - k_{1}) χ (n x_{2} - k_{2}) =: (*) . \end{aligned}

(For convenience call

F (k_{1}, k_{2}, x_{1}, x_{2}) := l_{n k} (f) χ (n x_{1} - k_{1}) χ (n x_{2} - k_{2}) .)

Thus

\begin{aligned} (*) & = \sum_{k_{1} = - \infty}^{- λ_{1}} \sum_{k_{2} = - \infty}^{- λ_{2}} F (k_{1}, k_{2}, x_{1}, x_{2}) + \sum_{k_{1} = - λ_{1} + 1}^{λ_{1} - 1} \sum_{k_{2} = - \infty}^{- λ_{2}} F (k_{1}, k_{2}, x_{1}, x_{2}) + \\ + \sum_{k_{1} = λ_{1}}^{\infty} \sum_{k_{2} = - \infty}^{- λ_{2}} F (k_{1}, k_{2}, x_{1}, x_{2}) + \sum_{k_{1} = - \infty}^{- λ_{1}} \sum_{k_{2} = - λ_{2} + 1}^{λ_{2} - 1} F (k_{1}, k_{2}, x_{1}, x_{2}) + \\ + \sum_{k_{1} = - λ_{1} + 1}^{λ_{1} - 1} \sum_{k_{2} = - λ_{2} + 1}^{λ_{2} - 1} F (k_{1}, k_{2}, x_{1}, x_{2}) + \sum_{k_{1} = λ_{1}}^{\infty} \sum_{k_{2} = - λ_{2} + 1}^{λ_{2} - 1} F (k_{1}, k_{2}, x_{1}, x_{2}) + \\ + \sum_{k_{1} = - \infty}^{- λ_{1}} \sum_{k_{2} = λ_{2}}^{\infty} F (k_{1}, k_{2}, x_{1}, x_{2}) + \sum_{k_{1} = - λ_{1} + 1}^{λ_{1} - 1} \sum_{k_{2} = λ_{2}}^{\infty} F (k_{1}, k_{2}, x_{1}, x_{2}) + \\ + \sum_{k_{1} = λ_{1}}^{\infty} \sum_{k_{2} = λ_{2}}^{\infty} F (k_{1}, k_{2}, x_{1}, x_{2}) . \end{aligned}

Notice that the finite sum of continuous functions $F (k_{1}, k_{2}, x_{1}, x_{2})$ ,

\sum_{k_{1} = - λ_{1} + 1}^{λ_{1} - 1} \sum_{k_{2} = - λ_{2} + 1}^{λ_{2} - 1} F (k_{1}, k_{2}, x_{1}, x_{2})

is a continuous function.

The rest of the summands of $F_{n} (f, x_{1}, x_{2})$ are treated all the same way and similarly to the case of $N = 1$ . The method is demonstrated as follows.

We will prove that

\sum_{k_{1} = λ_{1}}^{\infty} \sum_{k_{2} = - \infty}^{- λ_{2}} l_{n k} (f) χ (n x_{1} - k_{1}) χ (n x_{2} - k_{2})

is continuous in $(x_{1}, x_{2}) \in R^{2}$ .

The continuous function

| l_{n k} (f) | χ (n x_{1} - k_{1}) χ (n x_{2} - k_{2}) \leq {‖ f ‖}_{\infty} χ (k_{1} - λ_{1}) χ (k_{2} + λ_{2}),

and

\begin{aligned} {‖ f ‖}_{\infty} \sum_{k_{1} = λ_{1}}^{\infty} \sum_{k_{2} = - \infty}^{- λ_{2}} χ (k_{1} - λ_{1}) χ (k_{2} + λ_{2}) = \\ = {‖ f ‖}_{\infty} (\sum_{k_{1} = λ_{1}}^{\infty} χ (k_{1} - λ_{1})) (\sum_{k_{2} = - \infty}^{- λ_{2}} χ (k_{2} + λ_{2})) \\ \leq {‖ f ‖}_{\infty} (\sum_{k_{1}^{'} = 0}^{\infty} χ (k_{1}^{'})) (\sum_{k_{2}^{'} = - \infty}^{0} χ (k_{2}^{'})) \leq {‖ f ‖}_{\infty} . \end{aligned}

So by the Weierstrass $M$ test we get that

\sum_{k_{1} = λ_{1}}^{\infty} \sum_{k_{2} = - \infty}^{- λ_{2}} l_{n k} (f) χ (n x_{1} - k_{1}) χ (n x_{2} - k_{2})

is uniformly and absolutely convergent. Therefore it is continuous on $R^{2} .$

Next we prove continuity on $R^{2}$ of

\sum_{k_{1} = - λ_{1} + 1}^{λ_{1} - 1} \sum_{k_{2} = - \infty}^{- λ_{2}} l_{n k} (f) χ (n x_{1} - k_{1}) χ (n x_{2} - k_{2}) .

Notice here that

\begin{aligned} | l_{n k} (f) | χ (n x_{1} - k_{1}) χ (n x_{2} - k_{2}) & \leq {‖ f ‖}_{\infty} χ (n x_{1} - k_{1}) χ (k_{2} + λ_{2}) \\ \leq {‖ f ‖}_{\infty} χ (0) χ (k_{2} + λ_{2}) \\ = 0.4215 \cdot {‖ f ‖}_{\infty} χ (k_{2} + λ_{2}), \end{aligned}

and

\begin{aligned} 0.4215 \cdot {‖ f ‖}_{\infty} (\sum_{k_{1} = - λ_{1} + 1}^{λ_{1} - 1} 1) (\sum_{k_{2} = - \infty}^{- λ_{2}} χ (k_{2} + λ_{2})) = \\ = 0.4215 \cdot {‖ f ‖}_{\infty} (2 λ_{1} - 1) (\sum_{k_{2}^{'} = - \infty}^{0} χ (k_{2}^{'})) \leq 0.4215 \cdot (2 λ_{1} - 1) {‖ f ‖}_{\infty} . \end{aligned}

So the double series under consideration is uniformly convergent and continuous. Clearly $F_{n} (f, x_{1}, x_{2})$ is proved to be continuous on $R^{2} .$

Similarly reasoning one can prove easily now, but with more tedious work, that $F_{n} (f, x_{1}, . . ., x_{N})$ is continuous on $R^{N}$ , for any $N \geq 1$ . We choose to omit this similar extra work.

Proof â–¼

Remark 1

By (25) it is obvious that ${‖ A_{n} (f) ‖}_{\infty} \leq {‖ f ‖}_{\infty} < \infty$ , and $A_{n} (f) \in C (\prod_{i = 1}^{N} [a_{i}, b_{i}])$ , given that $f \in C (\prod_{i = 1}^{N} [a_{i}, b_{i}]) .$

Call $L_{n}$ any of the operators $A_{n}, B_{n}, C_{n}, D_{n} .$

Clearly then

{‖ L_{n}^{2} (f) ‖}_{\infty} = {‖ L_{n} (L_{n} (f)) ‖}_{\infty} \leq {‖ L_{n} (f) ‖}_{\infty} \leq {‖ f ‖}_{\infty},

etc.

Therefore we get

{‖ L_{n}^{k} (f) ‖}_{\infty} \leq {‖ f ‖}_{\infty}, \ \forall k \in N,

the contraction property.

Also we see that

{‖ L_{n}^{k} (f) ‖}_{\infty} \leq {‖ L_{n}^{k - 1} (f) ‖}_{\infty} \leq . . . \leq {‖ L_{n} (f) ‖}_{\infty} \leq {‖ f ‖}_{\infty} .

Also $L_{n} (1) = 1$ , $L_{n}^{k} (1) = 1$ , $\forall$ $k \in N .$

Here $L_{n}^{k}$ are positive linear operators.â–¡

Notation 11

Here $N \in N$ , $0 < β < 1.$ Denote by

\begin{aligned} c_{N} & := {\begin{cases} {(4.019)}^{N}, & if\ L_{n} = A_{n}, \\ 1, & if \ L_{n} = B_{n}, C_{n}, D_{n}, \end{cases} \\ φ (n) & := {\begin{cases} \frac{1}{n^{β}}, & if\ L_{n} = A_{n}, B_{n}, \\ \frac{1}{n} + \frac{1}{n^{β}}, & if\ L_{n} = C_{n}, D_{n}, \end{cases} \\ Ω & := {\begin{cases} C (\prod_{i = 1}^{N} [a_{i}, b_{i}]), & if\ L_{n} = A_{n}, \\ C_{B} (R^{N}), & if\ L_{n} = B_{n}, C_{n}, D_{n}, \end{cases} \end{aligned}

and

Y := {\begin{cases} \prod_{i = 1}^{N} [a_{i}, b_{i}], & if\ L_{n} = A_{n}, \\ R^{N}, & if\ L_{n} = B_{n}, C_{n}, D_{n} . \end{cases}

104

We give the condensed

Theorem 12

Let $f \in Ω$ , $0 < β < 1$ , $x \in Y;$ $n,$ $N \in N$ with $n^{1 - β} \geq 3$ . Then

(i)

| L_{n} (f, x) - f (x) | \leq c_{N} [ω_{1} (f, φ (n)) + \frac{{‖ f ‖}_{\infty}}{\sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}}] =: τ,

107

(ii)

{‖ L_{n} (f) - f ‖}_{\infty} \leq τ .

108

For $f$ uniformly continuous and in $Ω$ we obtain

lim_{n \to \infty} L_{n} (f) = f,

pointwise and uniformly.

Proof â–¼

By Theorems 4-7.

Proof â–¼

Next we do iterated neural network approximation (see also [ 9 ] ).

We make

Remark 2

Let $r \in N$ and $L_{n}$ as above. We observe that

\begin{aligned} L_{n}^{r} f - f & = (L_{n}^{r} f - L_{n}^{r - 1} f) + (L_{n}^{r - 1} f - L_{n}^{r - 2} f) + \\ + (L_{n}^{r - 2} f - L_{n}^{r - 3} f) + . . . + (L_{n}^{2} f - L_{n} f) + (L_{n} f - f) . \end{aligned}

Then

\begin{aligned} {‖ L_{n}^{r} f - f ‖}_{\infty} & \leq {‖ L_{n}^{r} f - L_{n}^{r - 1} f ‖}_{\infty} + {‖ L_{n}^{r - 1} f - L_{n}^{r - 2} f ‖}_{\infty} + \\ + {‖ L_{n}^{r - 2} f - L_{n}^{r - 3} f ‖}_{\infty} + . . . + {‖ L_{n}^{2} f - L_{n} f ‖}_{\infty} + {‖ L_{n} f - f ‖}_{\infty} = \\ = {‖ L_{n}^{r - 1} (L_{n} f - f) ‖}_{\infty} + {‖ L_{n}^{r - 2} (L_{n} f - f) ‖}_{\infty} + {‖ L_{n}^{r - 3} (L_{n} f - f) ‖}_{\infty} \\ + . . . + {‖ L_{n} (L_{n} f - f) ‖}_{\infty} + {‖ L_{n} f - f ‖}_{\infty} \leq r {‖ L_{n} f - f ‖}_{\infty} . \end{aligned}

That is

Unsupported use of \hfil

110

We give

Theorem 13

All here as in Theorem 12 and $r \in N$ , $τ$ as in (107). Then

{‖ L_{n}^{r} f - f ‖}_{\infty} \leq r τ .

111

So that the speed of convergence to the unit operator of $L_{n}^{r}$ is not worse than of $L_{n} .$

Proof â–¼

By (110) and (108).

Proof â–¼

We make

Remark 3

Let $m_{1}, . . ., m_{r} \in N : m_{1} \leq m_{2} \leq . . . \leq m_{r}$ , $0 < β < 1$ , $f \in Ω$ . Then $φ (m_{1}) \geq φ (m_{2}) \geq . . . \geq φ (m_{r})$ , $φ$ as in (100).

Therefore

ω_{1} (f, φ (m_{1})) \geq ω_{1} (f, φ (m_{2})) \geq . . . \geq ω_{1} (f, φ (m_{r})) .

112

Assume further that $m_{i}^{1 - β} \geq 3$ , $i = 1, . . ., r$ . Then

\begin{aligned} \frac{1}{(m_{1}^{1 - β} - 2) e^{{(m_{1}^{1 - β} - 2)}^{2}}} & \geq \frac{1}{(m_{2}^{1 - β} - 2) e^{{(m_{2}^{1 - β} - 2)}^{2}}} \\ \geq . . . \geq \frac{1}{(m_{r}^{1 - β} - 2) e^{{(m_{r}^{1 - β} - 2)}^{2}}} . \end{aligned}

Let $L_{m_{i}}$ as above, $i = 1, . . ., r,$ all of the same kind.

We write

\begin{aligned} L_{m_{r}} (L_{m_{r - 1}} (. . . L_{m_{2}} (L_{m_{1}} f))) - f = \\ = L_{m_{r}} (L_{m_{r - 1}} (. . . L_{m_{2}} (L_{m_{1}} f))) - L_{m_{r}} (L_{m_{r - 1}} (. . . L_{m_{2}} f)) + \\ + L_{m_{r}} (L_{m_{r - 1}} (. . . L_{m_{2}} f)) - L_{m_{r}} (L_{m_{r - 1}} (. . . L_{m_{3}} f)) + \\ + L_{m_{r}} (L_{m_{r - 1}} (. . . L_{m_{3}} f)) - L_{m_{r}} (L_{m_{r - 1}} (. . . L_{m_{4}} f)) + . . . + \\ + L_{m_{r}} (L_{m_{r - 1}} f) - L_{m_{r}} f + L_{m_{r}} f - f = \\ = L_{m_{r}} (L_{m_{r - 1}} (. . . L_{m_{2}})) (L_{m_{1}} f - f) + L_{m_{r}} (L_{m_{r - 1}} (. . . L_{m_{3}})) (L_{m_{2}} f - f) + \\ + L_{m_{r}} (L_{m_{r - 1}} (. . . L_{m_{4}})) (L_{m_{3}} f - f) + . . . + L_{m_{r}} (L_{m_{r - 1}} f - f) + L_{m_{r}} f - f . \end{aligned}

Hence by the triangle inequality property of ${‖ \cdot ‖}_{\infty}$ we get

\begin{aligned} {‖ L_{m_{r}} (L_{m_{r - 1}} (. . . L_{m_{2}} (L_{m_{1}} f))) - f ‖}_{\infty} \leq \\ \leq {‖ L_{m_{r}} (L_{m_{r - 1}} (. . . L_{m_{2}})) (L_{m_{1}} f - f) ‖}_{\infty} + {‖ L_{m_{r}} (L_{m_{r - 1}} (. . . L_{m_{3}})) (L_{m_{2}} f - f) ‖}_{\infty} + \\ + {‖ L_{m_{r}} (L_{m_{r - 1}} (. . . L_{m_{4}})) (L_{m_{3}} f - f) ‖}_{\infty} + . . . + \\ + {‖ L_{m_{r}} (L_{m_{r - 1}} f - f) ‖}_{\infty} + {‖ L_{m_{r}} f - f ‖}_{\infty} \end{aligned}

(repeatedly applying (92))

\begin{aligned} \leq {‖ L_{m_{1}} f - f ‖}_{\infty} + {‖ L_{m_{2}} f - f ‖}_{\infty} + {‖ L_{m_{3}} f - f ‖}_{\infty} + . . . + \\ + {‖ L_{m_{r - 1}} f - f ‖}_{\infty} + {‖ L_{m_{r}} f - f ‖}_{\infty} = \sum_{i = 1}^{r} {‖ L_{m_{i}} f - f ‖}_{\infty} . \end{aligned}

That is, we proved

Unsupported use of \hfil

116

We give

Theorem 14

Let $f \in Ω$ ; $N,$ $m_{1}, m_{2}, . . ., m_{r} \in N : m_{1} \leq m_{2} \leq . . . \leq m_{r},$ $0 < β < 1;$ $m_{i}^{1 - β} \geq 3$ , $i = 1, . . ., r,$ $x \in Y,$ and let $(L_{m_{1}}, . . ., L_{m_{r}})$ as $(A_{m_{1}}, . . ., A_{m_{r}})$ or $(B_{m_{1}}, . . ., B_{m_{r}})$ or $(C_{m_{1}}, . . ., C_{m_{r}})$ or $(D_{m_{1}}, . . ., D_{m_{r}}) .$ Then

\begin{aligned} | L_{m_{r}} (L_{m_{r - 1}} (. . . L_{m_{2}} (L_{m_{1}} f))) (x) - f (x) | \leq \\ \leq {‖ L_{m_{r}} (L_{m_{r - 1}} (. . . L_{m_{2}} (L_{m_{1}} f))) - f ‖}_{\infty} \\ \leq \sum_{i = 1}^{r} {‖ L_{m_{i}} f - f ‖}_{\infty} \\ \leq c_{N} \sum_{i = 1}^{r} [ω_{1} (f, φ (m_{i})) + \frac{{‖ f ‖}_{\infty}}{\sqrt{π} (m_{i}^{1 - β} - 2) e^{{(m_{i}^{1 - β} - 2)}^{2}}}] \leq \end{aligned}

\begin{aligned} \leq r c_{N} [ω_{1} (f, φ (m_{1})) + \frac{{‖ f ‖}_{\infty}}{\sqrt{π} (m_{1}^{1 - β} - 2) e^{{(m_{1}^{1 - β} - 2)}^{2}}}] . \end{aligned}

Clearly, we notice that the speed of convergence to the unit operator of the multiply iterated operator is not worse than the speed of $L_{m_{1}} .$

Proof â–¼

Using (116), (112), (113) and (107), (108).

Proof â–¼

We continue with

Theorem 15

Let all as in Theorem 8, and $r \in N$ . Here $K_{n}$ is as in (58). Then

{‖ A_{n}^{r} f - f ‖}_{\infty} \leq r {‖ A_{n} f - f ‖}_{\infty} \leq r K_{n} .

118

Proof â–¼

By (110) and (58).

Proof â–¼

4 Complex Multivariate Neural Network Approximations

We make

Remark 4

Let $Y = \prod_{i = 1}^{n} [a_{i}, b_{i}]$ or $R^{N},$ and $f : Y \to C$ with real and imaginary parts $f_{1}, f_{2} : f = f_{1} + i f_{2}$ , $i = \sqrt{- 1}$ . Clearly $f$ is continuous iff $f_{1}$ and $f_{2}$ are continuous.

Given that $f_{1}, f_{2} \in C^{m} (Y)$ , $m \in N$ , it holds

f_{α} (x) = f_{1, α} (x) + i f_{2, α} (x),

119

where $α$ indicates a partial derivative of any order and arrangement.

We denote by $C_{B} (R^{N}, C)$ the space of continuous and bounded functions $f : R^{N} \to C$ . Clearly $f$ is bounded, iff both $f_{1}, f_{2}$ are bounded from $R^{N}$ into $R$ , where $f = f_{1} + i f_{2} .$

Here $L_{n}$ is any of $A_{n}, B_{n}, C_{n}, D_{n}$ , $n \in N .$

We define

L_{n} (f, x) := L_{n} (f_{1}, x) + i L_{n} (f_{2}, x), \ \forall \ x \in Y .

120

We observe that

| L_{n} (f, x) - f (x) | \leq | L_{n} (f_{1}, x) - f_{1} (x) | + | L_{n} (f_{2}, x) - f_{2} (x) |,

121

and

Unsupported use of \hfil

122

We present

Theorem 16

Let $f \in C (Y, C)$ which is bounded, $f = f_{1} + i f_{2}$ , $0 < β < 1,$ $n, N \in N : n^{1 - β} \geq 3,$ , $x \in Y$ . Then

\begin{aligned} | L_{n} (f, x) - f (x) | \leq \\ \leq c_{N} [ω_{1} (f_{1}, φ (n)) + ω_{1} (f_{2}, φ (n_{2})) + \frac{({‖ f_{1} ‖}_{\infty} + {‖ f_{2} ‖}_{\infty})}{\sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}}] =: ε, \end{aligned}

ii)

{‖ L_{n} (f) - f ‖}_{\infty} \leq ε .

124

Proof â–¼

Use of (107).

Proof â–¼

In the next we discuss high order of complex approximation by using the smoothness of $f$ .

We give

Theorem 17

Let $f : \prod_{i = 1}^{n} [a_{i}, b_{i}] \to C$ , such that $f = f_{1} + i f_{2.}$ Assume $f_{1}, f_{2} \in C^{m} (\prod_{i = 1}^{n} [a_{i}, b_{i}]),$ $0 < β < 1$ , $n, m, N \in N$ , $n^{1 - β} \geq 3,$
$x \in (\prod_{i = 1}^{n} [a_{i}, b_{i}])$ . Then

\begin{aligned} | A_{n} (f, x) - f (x) - \sum_{j = 1}^{m} (\sum_{| α | = j} (\frac{f_{α} (x)}{\prod_{i = 1}^{N} α_{i}!}) A_{n} (\prod_{i = 1}^{N} {(\cdot - x_{i})}^{α_{i}}, x)) | \leq \\ \leq {(4.019)}^{N} \cdot {\frac{N^{m}}{m! n^{m β}} (ω_{1, m}^{max} (f_{1, α}, \frac{1}{n^{β}}) + ω_{1, m}^{max} (f_{2, α}, \frac{1}{n^{β}})) + \\ + (\frac{{‖ b - a ‖}_{\infty}^{m} ({‖ f_{1, α} ‖}_{\infty, m}^{max} + {‖ f_{2, α} ‖}_{\infty, m}^{max}) N^{m}}{m!}) \frac{1}{\sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}}}, \end{aligned}

ii)

\begin{aligned} | A_{n} (f, x) - f (x) | \leq {(4.019)}^{N} \cdot \\ {\sum_{j = 1}^{m} (\sum_{| α | = j} (\frac{| f_{1, α} (x) | + | f_{2, α} (x) |}{\prod_{i = 1}^{N} α_{i}!}) [\frac{1}{n^{β j}} + (\prod_{i = 1}^{N} {(b_{i} - a_{i})}^{α_{i}}) \frac{1}{2 \sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}}]) \\ + \frac{N^{m}}{m! n^{m β}} (ω_{1, m}^{max} (f_{1, α}, \frac{1}{n^{β}}) + ω_{1, m}^{max} (f_{2, α}, \frac{1}{n^{β}})) \\ + (\frac{{‖ b - a ‖}_{\infty}^{m} ({‖ f_{1, α} ‖}_{\infty, m}^{max} + {‖ f_{2, α} ‖}_{\infty, m}^{max}) N^{m}}{m!}) \frac{1}{\sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}}}, \end{aligned}

iii)

\begin{aligned} {‖ A_{n} (f) - f ‖}_{\infty} \leq {(4.019)}^{N} \cdot \\ \cdot {\sum_{j = 1}^{m} (\sum_{| α | = j} (\frac{{‖ f_{1, α} (x) ‖}_{\infty} + {‖ f_{2, α} (x) ‖}_{\infty}}{\prod_{i = 1}^{N} α_{i}!}) [\frac{1}{n^{β j}} + (\prod_{i = 1}^{N} {(b_{i} - a_{i})}^{α_{i}}) \frac{1}{2 \sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}}]) \\ + \frac{N^{m}}{m! n^{m β}} (ω_{1, m}^{max} (f_{1, α}, \frac{1}{n^{β}}) + ω_{1, m}^{max} (f_{2, α}, \frac{1}{n^{β}})) \\ + (\frac{{‖ b - a ‖}_{\infty}^{m} ({‖ f_{1, α} ‖}_{\infty, m}^{max} + {‖ f_{2, α} ‖}_{\infty, m}^{max}) N^{m}}{m!}) \frac{1}{\sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}}}, \end{aligned}

iv) Assume $f_{α} (x_{0}) = 0$ , for all $α : | α | = 1, . . ., m;$ $x_{0} \in (\prod_{i = 1}^{N} [a_{i}, b_{i}])$ . Then

\begin{aligned} | A_{n} (f, x_{0}) - f (x_{0}) | \leq {(4.019)}^{N} \cdot \\ {\frac{N^{m}}{m! n^{m β}} (ω_{1, m}^{max} (f_{1, α}, \frac{1}{n^{β}}) + ω_{1, m}^{max} (f_{2, α}, \frac{1}{n^{β}})) + \\ (\frac{{‖ b - a ‖}_{\infty}^{m} ({‖ f_{1, α} ‖}_{\infty, m}^{max} + {‖ f_{2, α} ‖}_{\infty, m}^{max}) N^{m}}{m!}) \frac{1}{\sqrt{π} (n^{1 - β} - 2) e^{{(n^{1 - β} - 2)}^{2}}}}, \end{aligned}

notice in the last the extremely high rate of convergence at $n^{- β (m + 1)} .$

Proof â–¼

By Theorem 8 and Remark 4.

Proof â–¼

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