Neural Networks as Positive Linear Operators

Abstract

Basic neural network operators are interpreted as positive linear operators and the related general theory applies to them. These operators are induced by a symmetrized density function deriving from the parametrized and deformed hyperbolic tangent activation function. I explore the space of continuous functions on a compact interval of the real line to the reals. I study quantitatively the rate of convergence of these neural network operators to the unit operator. The studied inequalities involve the modulus of continuity of the function under approximation or its derivative. I produce uniform and $L_p$, $p\ge 1$, approximation results via these inequalities. The convexity of functions is also taken into consideration.

1. Introduction

I have previously studied extensively the quantitative approximation of positive linear operators to the unit since 1985—see, for example, [1,2,3,4], which is used in this work. I began with the quantitative weak convergence of finite positive measures to the unit Dirac measure, having as a method the geometric moment theory (see [2]), and I produced the best upper bounds, leading to attained (i.e., sharp Jackson-type inequalities)—e.g., see [1,2]. These studies can be taken in many possible directions, univariate and multivariate, though in this work I focus only on the univariate approach.

Also, in 1997, I started studying the quantitative convergence of neural network operators to the unit; since them, I have written numerous articles and books—e.g., see [3,4,5], which I draw from here.

The wide range of neural network operators I have treated are by nature positive linear operators.

Here, for the first time in the literature, neural network operators are treated as positive linear operators.

So, all methods of positive linear operators apply here to the defined summation neural network operators, producing new and interesting results: point-wise, uniform, and $L_p$, $p\ge 1$, results. Via the Riesz representation theorem, neural networks are connected to measure theory. The convexity of functions produces optimal and sharper results.

The use of the perturbed hyperbolic tangent function as an activation function has been well established and supported in [5] (Chapter 18). The hyperbolic tangent activation function is one of the most commonly used activation functions in this area.

The symmetrization method presented here aims to have the operators converge at a high speed to the unit operator using half the relevant data feed.

This article establishes a bridge for the first time connecting neural network approximation (part of AI) to positive linear operators (an important part of functional analysis).

So, this is a theoretical work intended mainly for relevant experts.

This study was greatly inspired by [6,7]. For classic studies on neural networks, I also recommend [8,9,10,11,12,13].

For newer work on neural networks, I also refer the reader to [14,15,16,17,18,19,20,21,22,23]. For recent studies on positive linear operators, I commend [24,25,26,27,28,29] to you.

2. Basics

Here, we will follow [5] (pp. 455–460).

The activation function to be used is

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

(1)

Above, $\lambda$ is the parameter and q is the deformation coefficient, typically $0<\lambda ,q\le 1.$

For further reading, refer to Chapter 18 of [5], “q-Deformed and $\lambda$-Parametrized Hyperbolic Tangent based Banach space Valued Ordinary and Fractional Neural Network Approximation”.

This chapter motivates the current work.

The proposed “symmetrization method” aims to use half the data feed to the relevant neural networks.

We will employ the following density function:

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

(2)

$\forall x\in \mathbb{R}$; $q,\lambda >0$.

We have

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

(3)

and

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

(4)

Adding (3) and (4), we obtain

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

(5)

the key to this work.

The result is that

$$\mathsf{\Phi }\left(x\right):={\displaystyle \frac{M_{q,\lambda }\left(x\right)+M_{\frac{1}{q},\lambda }\left(x\right)}{2}}$$

(6)

is an even function, symmetric with respect to the y-axis.

By (18.18) of [5], we have

$$\begin{array}{c}{M}_{q,\lambda }\left({\displaystyle \frac{\ln q}{2\lambda }}\right)={\displaystyle \frac{\tanh \left(\lambda \right)}{2}},\hfill \\ \mathrm{and}\hfill \\ M_{{\displaystyle \frac{1}{q}},\lambda }\left(-{\displaystyle \frac{\ln q}{2\lambda }}\right)={\displaystyle \frac{\tanh \left(\lambda \right)}{2}},\lambda >0.\hfill \end{array}$$

(7)

sharing the same maximum at symmetric points.

By Theorem 18.1, p. 458 of [5], we have that

$$\begin{array}{c}{\displaystyle \sum _{i=-\infty }^{\infty }}{M}_{q,\lambda }\left(x-i\right)=1,\forall x\in \mathbb{R},\lambda ,q>0,\hfill \\ \mathrm{and}\hfill \\ {\displaystyle \sum _{i=-\infty }^{\infty }}{M}_{{\displaystyle \frac{1}{q}},\lambda }\left(x-i\right)=1,\forall x\in \mathbb{R},\lambda ,q>0.\hfill \end{array}$$

(8)

Consequently, we derive that

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

(9)

By Theorem 18.2, p. 459 of [5], we have that

$$\begin{array}{c}{\int }_{-\infty }^{\infty }{M}_{q,\lambda }\left(x\right)dx=1,\hfill \\ \mathrm{and}\hfill \\ {\int }_{-\infty }^{\infty }{M}_{{\displaystyle \frac{1}{q}},\lambda }\left(x\right)dx=1,\hfill \end{array}$$

(10)

so that

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

(11)

Thereforem $\mathsf{\Phi }$ is a density function.

By Theorem 18.3, p. 459 of [5], we have the following:

Let $0<\alpha <1$, and $n\in \mathbb{N}$ with $n^{1-\alpha }>2$; $q,\lambda >0$. Then,

$$\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\alpha }\end{array}\right.}^{\infty }{M}_{q,\lambda }\left(nx-k\right)<2\max \left\{q,{\displaystyle \frac{1}{q}}\right\}{e}^{4\lambda }{e}^{-2\lambda n^{\left(1-\alpha \right)}}=T{e}^{-2\lambda n^{\left(1-\alpha \right)}},$$

(12)

where $T:=2\max \left\{q,{\displaystyle \frac{1}{q}}\right\}{e}^{4\lambda }.$

Similarly, we obtain that

$$\sum _{k=-\infty }^{\infty }{M}_{{\displaystyle \frac{1}{q}},\lambda }\left(nx-k\right)<T{e}^{-2\lambda n^{\left(1-\alpha \right)}}.$$

(13)

Consequently, we obtain that

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

(14)

where $T:=2\max \left\{q,{\displaystyle \frac{1}{q}}\right\}{e}^{4\lambda }.$

Here, $⌈·⌉$ denotes the ceiling of the number, and $⌊·⌋$ its integral part.

We mention the following:

Theorem 18.4, p. 459, [5]: Let $x\in \left[a,b\right]\subset \mathbb{R}$ and $n\in \mathbb{N}$ so that $⌈na⌉\le ⌊nb⌋$. For $q,\lambda >0$, we consider ${\lambda }_q>z_0>0$, such that $M_{q,\lambda }\left(z_0\right)=M_{q,\lambda }\left(0\right)$, and ${\lambda }_q>1$. Then,

$${\displaystyle \frac{1}{{\displaystyle \sum _{k=⌈na⌉}^{⌊nb⌋}}{M}_{q,\lambda }\left(nx-k\right)}}<\max \left\{{\displaystyle \frac{1}{M_{q,\lambda }\left({\lambda }_q\right)}},{\displaystyle \frac{1}{M_{\frac{1}{q},\lambda }\left({\lambda }_{\frac{1}{q}}\right)}}\right\}=:\Delta \left(q\right).$$

(15)

Similarly, we consider ${\lambda }_{{\displaystyle \frac{1}{q}}}>z_1>0$, such that $M_{{\displaystyle \frac{1}{q}},\lambda }\left(z_1\right)=M_{{\displaystyle \frac{1}{q}},\lambda }\left(0\right)$, and ${\lambda }_{{\displaystyle \frac{1}{q}}}>1$. Thus,

$${\displaystyle \frac{1}{{\displaystyle \sum _{k=⌈na⌉}^{⌊nb⌋}}{M}_{\frac{1}{q},\lambda }\left(nx-k\right)}}<\max \left\{{\displaystyle \frac{1}{M_{\frac{1}{q},\lambda }\left({\lambda }_{\frac{1}{q}}\right)}},{\displaystyle \frac{1}{M_{q,\lambda }\left({\lambda }_q\right)}}\right\}=\Delta \left(q\right).$$

(16)

Hence,

$$\sum _{k=⌈na⌉}^{⌊nb⌋}{M}_{q,\lambda }\left(nx-k\right)>{\displaystyle \frac{1}{\Delta \left(q\right)}},$$

(17)

and

$$\sum _{k=⌈na⌉}^{⌊nb⌋}{M}_{{\displaystyle \frac{1}{q}},\lambda }\left(nx-k\right)>{\displaystyle \frac{1}{\Delta \left(q\right)}}.$$

(18)

Consequently, the following holds:

$$\sum _{k=⌈na⌉}^{⌊nb⌋}{\displaystyle \frac{\left(M_{q,\lambda }\left(nx-k\right)+M_{\frac{1}{q},\lambda }\left(nx-k\right)\right)}{2}}>{\displaystyle \frac{2}{2\Delta \left(q\right)}}={\displaystyle \frac{1}{\Delta \left(q\right)}},$$

(19)

so that

$${\displaystyle \frac{1}{{\displaystyle \sum _{k=⌈na⌉}^{⌊nb⌋}}\frac{\left(M_{q,\lambda }\left(nx-k\right)+M_{\frac{1}{q},\lambda }\left(nx-k\right)\right)}{2}}}<\Delta \left(q\right),$$

(20)

that is,

$${\displaystyle \frac{1}{{\displaystyle \sum _{k=⌈na⌉}^{⌊nb⌋}}\mathsf{\Phi }\left(nx-k\right)}}<\Delta \left(q\right).$$

(21)

We have proved the theorem.

Theorem 1.

Let $x\in \left[a,b\right]\subset \mathbb{R}$ and $n\in \mathbb{N}$ so that $⌈na⌉\le ⌊nb⌋$. For $q,\lambda >0$, we consider ${\lambda }_q>z_0>0$, such that $M_{q,\lambda }\left(z_0\right)=M_{q,\lambda }\left(0\right)$, and ${\lambda }_q>1$. Also consider ${\lambda }_{{\displaystyle \frac{1}{q}}}>z_1>0$, such that $M_{{\displaystyle \frac{1}{q}},\lambda }\left(z_1\right)=M_{{\displaystyle \frac{1}{q}},\lambda }\left(0\right)$, and ${\lambda }_{{\displaystyle \frac{1}{q}}}>1$. Then,

$${\displaystyle \frac{1}{{\displaystyle \sum _{k=⌈na⌉}^{⌊nb⌋}}\mathsf{\Phi }\left(nx-k\right)}}<\Delta \left(q\right).$$

(22)

We make the following remark:

Remark 1.

(I) By Remark 18.5, p. 460 of [5], we have

$$\underset{n\to \infty }{\lim }\sum _{k=⌈na⌉}^{⌊nb⌋}{M}_{q,\lambda }\left(n{x}_1-k\right)\ne 1,\mathit{for}\mathit{some}{x}_1\in \left[a,b\right],$$

(23)

and

$$\underset{n\to \infty }{\lim }\sum _{k=⌈na⌉}^{⌊nb⌋}{M}_{{\displaystyle \frac{1}{q}},\lambda }\left(n{x}_2-k\right)\ne 1,\mathit{for}\mathit{some}{x}_2\in \left[a,b\right].$$

(24)

Therefore, it holds that

$$\underset{n\to \infty }{\lim }\sum _{k=⌈na⌉}^{⌊nb⌋}{\displaystyle \frac{\left(M_{q,\lambda }\left(n{x}_1-k\right)+M_{\frac{1}{q},\lambda }\left(n{x}_2-k\right)\right)}{2}}\ne 1.$$

(25)

Hence,

$$\underset{n\to \infty }{\lim }\sum _{k=⌈na⌉}^{⌊nb⌋}{\displaystyle \frac{\left(M_{q,\lambda }\left(n{x}_1-k\right)+M_{\frac{1}{q},\lambda }\left(n{x}_1-k\right)\right)}{2}}\ne 1,$$

(26)

even if

$$\underset{n\to \infty }{\lim }\sum _{k=⌈na⌉}^{⌊nb⌋}{M}_{{\displaystyle \frac{1}{q}},\lambda }\left(n{x}_1-k\right)=1,$$

(27)

because then

$$\underset{n\to \infty }{\lim }\sum _{k=⌈na⌉}^{⌊nb⌋}{\displaystyle \frac{M_{q,\lambda }\left(n{x}_1-k\right)}{2}}+{\displaystyle \frac{1}{2}}\ne 1,$$

(28)

which is equivalent to

$$\underset{n\to \infty }{\lim }\sum _{k=⌈na⌉}^{⌊nb⌋}{\displaystyle \frac{M_{q,\lambda }\left(n{x}_1-k\right)}{2}}\ne {\displaystyle \frac{1}{2}},$$

(29)

which is true given that

$$\underset{n\to \infty }{\lim }\sum _{k=⌈na⌉}^{⌊nb⌋}{M}_{q,\lambda }\left(n{x}_1-k\right)\ne 1.$$

(30)

(II) Let $\left[a,b\right]\subset \mathbb{R}$. For large n, we always have $⌈na⌉\le ⌊nb⌋$. Also, $a\le {\displaystyle \frac{k}{n}}\le b$ iff $⌈na⌉\le k\le ⌊nb⌋$. So, in general, the following holds:

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

(31)

We need the following definition:

Definition 1.

Let $f\in C\left(\left[a,b\right]\right),$ $x\in \left[a,b\right]$ and $n\in \mathbb{N}:⌈na⌉\le ⌊nb⌋$. We introduce and define the following real-valued symmetrized and perturbed positive linear neural network operators:

$$H_n\left(f,x\right):={\displaystyle \frac{{\displaystyle \sum _{k=⌈na⌉}^{⌊nb⌋}}f\left(\frac{k}{n}\right)\mathsf{\Phi }\left(nx-k\right)}{{\displaystyle \sum _{k=⌈na⌉}^{⌊nb⌋}}\mathsf{\Phi }\left(nx-k\right)}}=$$

$${\displaystyle \frac{{\displaystyle \sum _{k=⌈na⌉}^{⌊nb⌋}}f\left(\frac{k}{n}\right)\left(M_{q,\lambda }\left(n{x}_1-k\right)+M_{\frac{1}{q},\lambda }\left(n{x}_1-k\right)\right)}{{\displaystyle \sum _{k=⌈na⌉}^{⌊nb⌋}}\left(M_{q,\lambda }\left(n{x}_1-k\right)+M_{\frac{1}{q},\lambda }\left(n{x}_1-k\right)\right)}}.$$

(32)

In fact, we have $H_n\left(f\right)\in C\left(\left[a,b\right]\right);$ and $H_n\left(1\right)=1.$ The modulus of continuity is defined by

$${\omega }_1\left(f,\delta \right):=\underset{\begin{array}{c}x,y\in \left[a,b\right]\\ \left|x-y\right|\le \delta \end{array}}{\sup }\left|f\left(x\right)-f\left(y\right)\right|,\delta >0.$$

(33)

In this work, $0<\alpha <1$, $n\in \mathbb{N}:n^{1-\alpha }>2;$ where ${∥·∥}_{\infty }$ is the supremum norm.

3. Main Results

The following approximation results are valid.

Theorem 2.

Let $f\in C\left(\left[a,b\right]\right)$. Then

$${∥H_n\left(f\right)-f∥}_{\infty }\le 2{\omega }_1\left(f,\sqrt{\Delta \left(q\right)}\sqrt{{\displaystyle \frac{1}{n^{2\alpha }}}+{\left(b-a\right)}^{2}T{e}^{-2\lambda n^{\left(1-\alpha \right)}}}\right)\to 0,\mathrm{as}n\to +\infty .$$

(34)

so that $\underset{n\to \infty }{\lim }{H}_n\left(f\right)=f,$ uniformly.

Proof. 

We estimate

$$\left(H_n\left({\left(t-x\right)}^2\right)\right)\left(x\right)={\displaystyle \frac{{\displaystyle \sum _{k=⌈na⌉}^{⌊nb⌋}}{\left(\frac{k}{n}-x\right)}^2\mathsf{\Phi }\left(nx-k\right)}{{\displaystyle \sum _{k=⌈na⌉}^{⌊nb⌋}}\mathsf{\Phi }\left(nx-k\right)}}\stackrel{\left(22\right)}{\le }$$

$$\Delta \left(q\right)\left(\sum _{k=⌈na⌉}^{⌊nb⌋}{\left({\displaystyle \frac{k}{n}}-x\right)}^2\mathsf{\Phi }\left(nx-k\right)\right)=$$

$$\Delta \left(q\right)\left[\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|{\displaystyle \frac{k}{n}}-x\right|\le {\displaystyle \frac{1}{n^{\alpha }}}\end{array}\right.}^{⌊nb⌋}{\left({\displaystyle \frac{k}{n}}-x\right)}^2\mathsf{\Phi }\left(nx-k\right)+\right.$$

(35)

$$\left.\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|{\displaystyle \frac{k}{n}}-x\right|>{\displaystyle \frac{1}{n^{\alpha }}}\end{array}\right.}^{⌊nb⌋}{\left({\displaystyle \frac{k}{n}}-x\right)}^2\mathsf{\Phi }\left(nx-k\right)\right]\stackrel{(\mathrm{by}\left(9),\right(14\left)\right)}{\le }$$

$$\Delta \left(q\right)\left[{\displaystyle \frac{1}{n^{2\alpha }}}+{\left(b-a\right)}^{2}T{e}^{-2\lambda n^{\left(1-\alpha \right)}}\right].$$

That is,

$$\left(H_n\left({\left(t-x\right)}^2\right)\right)\left(x\right)\le \Delta \left(q\right)\left\{{\displaystyle \frac{1}{n^{2\alpha }}}+{\left(b-a\right)}^{2}T{e}^{-2\lambda n^{\left(1-\alpha \right)}}\right\}.$$

(36)

Consequently, it holds that

$${\delta }_n:={∥\left(H_n\left({\left(t-x\right)}^2\right)\right)\left(x\right)∥}_{\infty }^{{\displaystyle \frac{1}{2}}}\le$$

$$\sqrt{\Delta \left(q\right)}\sqrt{{\displaystyle \frac{1}{n^{2\alpha }}}+{\left(b-a\right)}^{2}T{e}^{-2\lambda n^{\left(1-\alpha \right)}}}.$$

(37)

By Theorem 7.1.7, p. 203, [2], the Shisha–Mond inequality ([6]) for positive linear operators, and $H_n\left(1\right)=1$, we have that

$${∥H_n\left(f\right)-f∥}_{\infty }\le 2{\omega }_1\left(f,{\delta }_n\right).$$

(38)

Thus, it holds that

$${∥H_n\left(f\right)-f∥}_{\infty }\le 2{\omega }_1\left(f,\sqrt{\Delta \left(q\right)}\sqrt{{\displaystyle \frac{1}{n^{2\alpha }}}+{\left(b-a\right)}^{2}T{e}^{-2\lambda n^{\left(1-\alpha \right)}}}\right),$$

proving the claim. □

The following holds:

Theorem 3.

Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous and $2\pi$-periodic function with modulus of continuity ${\omega }_1$. Here, ${∥·∥}_{\infty }$ denotes the sup-norm over $\left[a,b\right]\subset \mathbb{R}$, and the operators $H_n$ are acting on such f over $\left[a,b\right]$; $n\in \mathbb{N}:n^{1-\alpha }>2$, $0<\alpha <1$. Then,

$${∥H_n\left(f\right)-f∥}_{\infty }\le 2{\omega }_1\left(f,\pi \sqrt{\Delta \left(q\right)}\sqrt{{\omega }_1\left({\sin }^2,{\displaystyle \frac{1}{2n^{\alpha }}}\right)+T{e}^{-2\lambda n^{\left(1-\alpha \right)}}}\right).$$

(39)

Proof. 

We want to estimate ($x\in \left[a,b\right]$):

$${\epsilon }_n:=\pi {∥\left(H_n\left({\sin }^2\left({\displaystyle \frac{t-x}{2}}\right)\right)\right)\left(x\right)∥}_{\infty }^{{\displaystyle \frac{1}{2}}}=$$

$$\pi {∥\left({\displaystyle \frac{{\displaystyle \sum _{k=⌈na⌉}^{⌊nb⌋}}{\sin }^2\left(\frac{\frac{k}{n}-x}{2}\right)\mathsf{\Phi }\left(nx-k\right)}{{\displaystyle \sum _{k=⌈na⌉}^{⌊nb⌋}}\mathsf{\Phi }\left(nx-k\right)}}\right)∥}_{\infty }^{{\displaystyle \frac{1}{2}}}\le$$

$$\pi {∥\Delta \left(q\right)\sum _{k=⌈na⌉}^{⌊nb⌋}{\sin }^2\left({\displaystyle \frac{\frac{k}{n}-x}{2}}\right)\mathsf{\Phi }\left(nx-k\right)∥}_{\infty }^{{\displaystyle \frac{1}{2}}}=$$

$$\pi \sqrt{\Delta \left(q\right)}∥\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|{\displaystyle \frac{k}{n}}-x\right|\le {\displaystyle \frac{1}{n^{\alpha }}}\end{array}\right.}^{⌊nb⌋}\left({\sin }^2\left({\displaystyle \frac{\frac{k}{n}-x}{2}}\right)-{\sin }^2\left(0\right)\right)\mathsf{\Phi }\left(nx-k\right)+$$

(40)

$${\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|{\displaystyle \frac{k}{n}}-x\right|>{\displaystyle \frac{1}{n^{\alpha }}}\end{array}\right.}^{⌊nb⌋}{\sin }^2\left({\displaystyle \frac{\frac{k}{n}-x}{2}}\right)\mathsf{\Phi }\left(nx-k\right)∥}_{\infty }^{{\displaystyle \frac{1}{2}}}\le$$

$$\pi \sqrt{\Delta \left(q\right)}∥\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|{\displaystyle \frac{k}{n}}-x\right|\le {\displaystyle \frac{1}{n^{\alpha }}}\end{array}\right.}^{⌊nb⌋}{\omega }_1\left({\sin }^2,{\displaystyle \frac{\left|\frac{k}{n}-x\right|}{2}}\right)\mathsf{\Phi }\left(nx-k\right)+$$

$${1·\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|{\displaystyle \frac{k}{n}}-x\right|>{\displaystyle \frac{1}{n^{\alpha }}}\end{array}\right.}^{⌊nb⌋}\mathsf{\Phi }\left(nx-k\right)∥}_{\infty }^{{\displaystyle \frac{1}{2}}}\le$$

$$\pi \sqrt{\Delta \left(q\right)}\sqrt{{\omega }_1\left({\sin }^2,{\displaystyle \frac{1}{2n^{\alpha }}}\right)+T{e}^{-2\lambda n^{\left(1-\alpha \right)}}}.$$

We have proved that

$${\epsilon }_n\le \pi \sqrt{\Delta \left(q\right)}\sqrt{{\omega }_1\left({\sin }^2,{\displaystyle \frac{1}{2n^{\alpha }}}\right)+T{e}^{-2\lambda n^{\left(1-\alpha \right)}}}.$$

(41)

Therefore, by [7], the following inequalities hold:

$${∥H_n\left(f\right)-f∥}_{\infty }\le 2{\omega }_1\left(f,{\epsilon }_n\right)\le$$

$$2{\omega }_1\left(f,\pi \sqrt{\Delta \left(q\right)}\sqrt{{\omega }_1\left({\sin }^2,{\displaystyle \frac{1}{2n^{\alpha }}}\right)+T{e}^{-2\lambda n^{\left(1-\alpha \right)}}}\right),$$

(42)

hence proving the claim. □

We need the following definition.

Definition 2.

Let $f\in C\left(\left[a,b\right]\right)$. The Peetre [30] $K_1$ functional is defined as follows:

$$K_1\left(f,t\right):=\inf \left({∥f-g∥}_{\infty }+t{∥g^{\prime }∥}_{\infty }\right),g\in C^1\left(\left[a,b\right]\right),t\ge 0.$$

(43)

We now present the following theorem.

Theorem 4.

Let $f\in C\left(\left[a,b\right]\right)$. Then,

$${∥H_n\left(f\right)-f∥}_{\infty }\le 2K_1\left(f,{\displaystyle \frac{\Delta \left(q\right)}{2}}\left[{\displaystyle \frac{1}{n^{\alpha }}}+\left(b-a\right)T{e}^{-2\lambda n^{\left(1-\alpha \right)}}\right]\right)\to 0,$$

(44)

as $n\to +\infty .$

The result is that $\underset{n\to +\infty }{\lim }{H}_n\left(f\right)=f$, uniformly.

For the proof of (44), we use the following:

Corollary 1

([1], Point-wise Approximation). Let $f\in C\left(\left[a,b\right]\right)$ and let L be a positive linear operator acting on $C\left(\left[a,b\right]\right)$ satisfying $L\left(1,x\right)=1$, all $x\in \left[a,b\right]$.

Then, we have the attainable (i.e., sharp) inequality

$$\left|L\left(f,x\right)-f\left(x\right)\right|\le 2K_1\left(f,{\displaystyle \frac{1}{2}}L\left(\left|y-x\right|,x\right)\right),\forall x\in \left[a,b\right].$$

(45)

Proof of Theorem 4.

We apply Inequality (45). Clearly, we have that

$$\left|H_n\left(f,x\right)-f\left(x\right)\right|\le 2K_1\left(f,{\displaystyle \frac{1}{2}}{H}_n\left(\left|t-x\right|\right)\left(x\right)\right),$$

(46)

which is attainable (i.w., sharp).

We estimate

$$H_n\left(\left|t-x\right|\right)\left(x\right)={\displaystyle \frac{{\displaystyle \sum _{k=⌈na⌉}^{⌊nb⌋}}\left|\frac{k}{n}-x\right|\mathsf{\Phi }\left(nx-k\right)}{{\displaystyle \sum _{k=⌈na⌉}^{⌊nb⌋}}\mathsf{\Phi }\left(nx-k\right)}}\le$$

$$\Delta \left(q\right)\sum _{k=⌈na⌉}^{⌊nb⌋}\left|{\displaystyle \frac{k}{n}}-x\right|\mathsf{\Phi }\left(nx-k\right)=$$

$$\Delta \left(q\right)\left[\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|{\displaystyle \frac{k}{n}}-x\right|\le {\displaystyle \frac{1}{n^{\alpha }}}\end{array}\right.}^{⌊nb⌋}\left|{\displaystyle \frac{k}{n}}-x\right|\mathsf{\Phi }\left(nx-k\right)+\right.$$

(47)

$$\left.\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|{\displaystyle \frac{k}{n}}-x\right|>{\displaystyle \frac{1}{n^{\alpha }}}\end{array}\right.}^{⌊nb⌋}\left|{\displaystyle \frac{k}{n}}-x\right|\mathsf{\Phi }\left(nx-k\right)\right]\le$$

$$\Delta \left(q\right)\left[{\displaystyle \frac{1}{n^{\alpha }}}+\left(b-a\right)T{e}^{-2\lambda n^{\left(1-\alpha \right)}}\right].$$

That is,

$$H_n\left(\left|t-x\right|\right)\left(x\right)\le \Delta \left(q\right)\left[{\displaystyle \frac{1}{n^{\alpha }}}+\left(b-a\right)T{e}^{-2\lambda n^{\left(1-\alpha \right)}}\right].$$

(48)

We have proved

$$\left|H_n\left(f,x\right)-f\left(x\right)\right|\le 2K_1\left(f,{\displaystyle \frac{\Delta \left(q\right)}{2}}\left[{\displaystyle \frac{1}{n^{\alpha }}}+\left(b-a\right)T{e}^{-2\lambda n^{\left(1-\alpha \right)}}\right]\right),$$

(49)

∀$x\in \left[a,b\right].$ Now, the validity of (44) is clear. □

A trigonometric result follows.

Theorem 5.

Here, $f\in C^2\left(\left[a,b\right]\right)$, $x_0\in \left[a,b\right]$, $r>0$, ${\overline{D}}_3\left(x_0\right):={\left(H_n\left({\left|·-x_0\right|}^3\right)\left(x_0\right)\right)}^{{\displaystyle \frac{1}{3}}}.$ Then,

(i) 

$$\left|H_n\left(f\right)\left(x_0\right)-f\left(x_0\right)\right|\le \left|f^{\prime }\left(x_0\right)\right|\left|H_n\left(\sin \left(·-x_0\right)\right)\left(x_0\right)\right|+$$

$$2\left|f^{″}\left(x_0\right)\right|\left(H_n\left({\sin }^2\left({\displaystyle \frac{·-x_0}{2}}\right)\right)\left(x_0\right)\right)+$$

$${\displaystyle \frac{{\omega }_1\left(f^{″}+f,r{\overline{D}}_3\left(x_0\right)\right)}{2}}{\overline{D}}_3^2\left(x_0\right)\left[1+{\displaystyle \frac{1}{3r}}\right],$$

(50)

(ii) 

$${∥H_n\left(f\right)-f∥}_{\infty }\le {∥f^{\prime }∥}_{\infty }{∥H_n\left(\sin \left(·-x_0\right)\right)\left(x_0\right)∥}_{\infty }+$$

$$2{∥f^{″}∥}_{\infty }{∥H_n\left({\sin }^2\left({\displaystyle \frac{·-x_0}{2}}\right)\right)\left(x_0\right)∥}_{\infty }+$$

$${\displaystyle \frac{{\omega }_1\left(f^{″}+f,r{∥{\overline{D}}_3∥}_{\infty }\right)}{2}}{∥{\overline{D}}_3∥}_{\infty }^2\left(1+{\displaystyle \frac{1}{3r}}\right),$$

(51)

(iii) 

If $f^{\prime }\left(x_0\right)=f^{″}\left(x_0\right)=0$, we obtain

$$\left|H_n\left(f\right)\left(x_0\right)-f\left(x_0\right)\right|\le {\displaystyle \frac{{\omega }_1\left(f^{″}+f,r{\overline{D}}_3\left(x_0\right)\right)}{2}}{\overline{D}}_3^2\left(x_0\right)\left(1+{\displaystyle \frac{1}{3r}}\right),$$

(52)

and

(iv) 

$$\left|H_n\left(f\right)\left(x_0\right)-f\left(x_0\right)-f^{\prime }\left(x_0\right)H_n\left(\sin \left(·-x_0\right)\right)\left(x_0\right)-2f^{″}\left(x_0\right)H_n\left({\sin }^2\left({\displaystyle \frac{·-x_0}{2}}\right)\right)\left(x_0\right)\right|$$

$$\le {\displaystyle \frac{{\omega }_1\left(f^{″}+f,r{\overline{D}}_3\left(x_0\right)\right)}{2}}{\overline{D}}_3^2\left(x_0\right)\left(1+{\displaystyle \frac{1}{3r}}\right).$$

(53)

Proof. 

Direct application of Theorem 12.3, p. 384, [4]. □

Next, we give the hyperbolic version of the last result.

Theorem 6.

All as in Theorem 5. Then,

(i) 

$$\left|H_n\left(f\right)\left(x_0\right)-f\left(x_0\right)\right|\le \left|f^{\prime }\left(x_0\right)\right|\left|H_n\left(\sinh \left(·-x_0\right)\right)\left(x_0\right)\right|+$$

(54)

$$2\left|f^{″}\left(x_0\right)\right|\left(H_n\left({\sinh }^2\left({\displaystyle \frac{·-x_0}{2}}\right)\right)\left(x_0\right)\right)+$$

$${\displaystyle \frac{\cosh \left(b-a\right){\omega }_1\left(f^{″}-f,r{\overline{D}}_3\left(x_0\right)\right)}{2}}{\overline{D}}_3^2\left(x_0\right)\left[1+{\displaystyle \frac{1}{3r}}\right],$$

(ii) 

$${∥H_n\left(f\right)-f∥}_{\infty }\le {∥f^{\prime }∥}_{\infty }{∥H_n\left(\sinh \left(·-x_0\right)\right)\left(x_0\right)∥}_{\infty }+$$

$$2{∥f^{″}∥}_{\infty }{∥H_n\left({\sinh }^2\left({\displaystyle \frac{·-x_0}{2}}\right)\right)\left(x_0\right)∥}_{\infty }+$$

$${\displaystyle \frac{\cosh \left(b-a\right){\omega }_1\left(f^{″}-f,r{∥{\overline{D}}_3∥}_{\infty }\right)}{2}}{∥{\overline{D}}_3∥}_{\infty }^2\left(1+{\displaystyle \frac{1}{3r}}\right),$$

(55)

(iii) 

If $f^{\prime }\left(x_0\right)=f^{″}\left(x_0\right)=0$, $x_0\in \left[a,b\right],$ we obtain

$$\left|H_n\left(f\right)\left(x_0\right)-f\left(x_0\right)\right|\le {\displaystyle \frac{\cosh \left(b-a\right){\omega }_1\left(f^{″}-f,r{\overline{D}}_3\left(x_0\right)\right)}{2}}{\overline{D}}_3^2\left(x_0\right)\left(1+{\displaystyle \frac{1}{3r}}\right),$$

(56)

and

(iv) 

$$\left|H_n\left(f\right)\left(x_0\right)-f\left(x_0\right)-f^{\prime }\left(x_0\right)H_n\left(\sinh \left(·-x_0\right)\right)\left(x_0\right)-2f^{″}\left(x_0\right)H_n\left({\sinh }^2\left({\displaystyle \frac{·-x_0}{2}}\right)\right)\left(x_0\right)\right|$$

$$\le {\displaystyle \frac{\cosh \left(b-a\right){\omega }_1\left(f^{″}-f,r{\overline{D}}_3\left(x_0\right)\right)}{2}}{\overline{D}}_3^2\left(x_0\right)\left(1+{\displaystyle \frac{1}{3r}}\right).$$

(57)

Proof. 

Direct application of Theorem 12.5, p. 390 of [4]. □

Remark 2.

We similarly have that

$${\left(H_n\left({\left|·-x_0\right|}^3\right)\left(x_0\right)\right)}^{{\displaystyle \frac{1}{3}}}\le \sqrt[3]{\Delta \left(q\right)}\sqrt[3]{\left\{{\displaystyle \frac{1}{n^{3\alpha }}}+{\left(b-a\right)}^{3}T{e}^{-2\lambda n^{\left(1-\alpha \right)}}\right\}}\to 0,$$

(58)

as $n\to +\infty .$

Hence, by each of the above Theorems 5 and 6, we obtain that $H_n\left(f\right)\to f$ is uniformly convergent as $n\to +\infty$, ∀$f\in C^2\left(\left[a,b\right]\right).$

These results are valid by Theorem 12.4 (p. 390) and Theorem 12.6 (p. 395) from [4], respectively.

We make the following remark:

Remark 3.

Let $x_0\in \left(a,b\right)$, $f\in C\left(\left[a,b\right]\right)$; here, $H_n$ is a positive linear operator from $C\left(\left[a,b\right]\right)$ into itself.

Also, $H_n\left(1\right)=1$. By the Riesz representation theorem, there exists a probability measure ${\mu }_{x_0}$ on $\left[a,b\right]$ such that

$$H_n\left(f\right)\left(x_0\right)={\int }_{\left[a,b\right]}f\left(t\right)d{\mu }_{x_0}\left(t\right),\forall f\in C\left(\left[a,b\right]\right).$$

(59)

Assume here that $\left|f\left(t\right)-f\left(x_0\right)\right|$ is convex in t. We consider

$$d_2\left(x_0\right):={\left({\int }_{\left[a,b\right]}{\left(t-x_0\right)}^{2}d{\mu }_{x_0}\left(t\right)\right)}^{{\displaystyle \frac{1}{2}}}$$

$$={\left(\left(H_n\left({\left(t-x_0\right)}^2\right)\right)\left(x_0\right)\right)}^{{\displaystyle \frac{1}{2}}}(\mathit{as}\mathit{earlier})$$

(60)

$$\le \sqrt{\Delta \left(q\right)}\sqrt{{\displaystyle \frac{1}{n^{2\alpha }}}+{\left(b-a\right)}^{2}T{e}^{-2\lambda n^{\left(1-\alpha \right)}}}=\left(\ast \right).$$

Clearly, here, we have $d_2\left(x_0\right)>0$ (see (32)).

For large enough $n\in \mathbb{N}$, we obtain

$$\left(\ast \right)\le \min \left(x_0-a,b-x_0\right)\le \max \left(x_0-a,b-x_0\right).$$

That is,

$$d_2\left(x_0\right)\le \min \left(x_0-a,b-x_0\right).$$

(61)

By Corollary 8.1.1, p. 245 of [2], we obtain the following:

Theorem 7.

The following holds:

$$\left|H_n\left(f\right)\left(x_0\right)-f\left(x_0\right)\right|\le {\omega }_1\left(f,\sqrt{\Delta \left(q\right)}\sqrt{{\displaystyle \frac{1}{n^{2\alpha }}}+{\left(b-a\right)}^{2}T{e}^{-2\lambda n^{\left(1-\alpha \right)}}}\right),$$

(62)

for large enough $n\in \mathbb{N}$.

Remark 4.

We have that

$$d_1\left(x_0\right):={\int }_{\left[a,b\right]}\left|t-x_0\right|d{\mu }_{x_0}\left(t\right)=H_n\left(\left|t-x_0\right|\right)\left(x_0\right)\stackrel{\left(48\right)}{\le }$$

(63)

$$\Delta \left(q\right)\left[{\displaystyle \frac{1}{n^{\alpha }}}+\left(b-a\right)T{e}^{-2\lambda n^{\left(1-\alpha \right)}}\right]\le \min \left(x_0-a,b-x_0\right),$$

for large enough $n\in \mathbb{N}$.

By (8.1.8) (p. 245, [2]), we obtain the following:

Theorem 8.

Let $r\ge 1$; then,

$$\left|H_n\left(f\right)\left(x_0\right)-f\left(x_0\right)\right|\le r{\omega }_1\left(f,{\displaystyle \frac{1}{r}}\Delta \left(q\right)\left\{{\displaystyle \frac{1}{n^{\alpha }}}+\left(b-a\right)T{e}^{-2\lambda n^{\left(1-\alpha \right)}}\right\}\right),$$

(64)

for large enough $n\in \mathbb{N}.$

By Theorem 8.1.2, p. 248 of [2], we can derive the following result.

Theorem 9.

Here, we consider $\left[a,b\right]\subset \mathbb{R}$, and $x_0\in \left(a,b\right)$. Let L be a positive linear operator from $C\left(\left[a,b\right]\right)$ into itself, such that $L\left(1\right)=1$.

Let

$$d_{m+1}\left(x_0\right):={\left(L\left({\left|t-x_0\right|}^{m+1}\right)\left(x_0\right)\right)}^{{\displaystyle \frac{1}{m+1}}},m\in \mathbb{N}.$$

Consider $f\in C^m\left(\left[a,b\right]\right)$ such that $\left|f^{\left(m\right)}\left(t\right)-f^{\left(m\right)}\left(x_0\right)\right|$ is convex in t.

Assume that

$$0<d_{m+1}^{m+1}\left(x_0\right)\le \left(\min \left(x_0-a,b-x_0\right)\right)\left(m+1\right)!.$$

(65)

Then,

$$\left|L\left(f\right)\left(x_0\right)-f\left(x_0\right)\right|-\sum _{k=1}^m{\displaystyle \frac{\left|f^{\left(k\right)}\left(x_0\right)\right|}{k!}}\left|L\left({\left(t-x_0\right)}^k\right)\left(x_0\right)\right|\le$$

(66)

$${\omega }_1\left(f^{\left(m\right)},{\displaystyle \frac{L\left({\left|t-x_0\right|}^{m+1}\right)\left(x_0\right)}{\left(m+1\right)!}}\right).$$

We make the following remark.

Remark 5.

We have that

$$0<\left(H_n\left({\left|t-x_0\right|}^{\left(m+1\right)}\right)\right)\left(x_0\right)={\displaystyle \frac{{\displaystyle \sum _{k=⌈na⌉}^{⌊nb⌋}}{\left|\frac{k}{n}-x_0\right|}^{m+1}\mathsf{\Phi }\left(nx-k\right)}{{\displaystyle \sum _{k=⌈na⌉}^{⌊nb⌋}}\mathsf{\Phi }\left(nx-k\right)}}$$

$$\le \Delta \left(q\right){\displaystyle \sum _{k=⌈na⌉}^{⌊nb⌋}}{\left|{\displaystyle \frac{k}{n}}-x_0\right|}^{m+1}\mathsf{\Phi }\left(nx-k\right)$$

(67)

$$\dots (\mathit{as}\mathit{earlier})$$

$$\Delta \left(q\right)\left\{{\displaystyle \frac{1}{n^{\left(m+1\right)\alpha }}}+{\left(b-a\right)}^{m+1}T{e}^{-2\lambda n^{\left(1-\alpha \right)}}\right\}\le \min \left(x_0-a,b-x_0\right)\left(m+1\right)!,$$

(68)

for large enough $n\in \mathbb{N}$.

By Theorem 9 and Remark 5, we have proved the following:

Theorem 10.

Here, $\left[a,b\right]\subset \mathbb{R},$ and $x_0\in \left(a,b\right)$. Consider $f\in C^m\left(\left[a,b\right]\right):\left|f^{\left(m\right)}\left(t\right)-f^{\left(m\right)}\left(x_0\right)\right|$ which is convex in t, $m\in \mathbb{N}$.

Then, for sufficiently large enough $n\in \mathbb{N}$, we derive that

$$\left|H_n\left(f\right)\left(x_0\right)-f\left(x_0\right)\right|-\sum _{k=1}^m{\displaystyle \frac{\left|f^{\left(k\right)}\left(x_0\right)\right|}{k!}}\left|H_n\left({\left(t-x_0\right)}^k\right)\left(x_0\right)\right|\le$$

(69)

$${\omega }_1\left(f^{\left(m\right)},{\displaystyle \frac{\Delta \left(q\right)}{\left(m+1\right)!}}\left\{{\displaystyle \frac{1}{n^{\left(m+1\right)\alpha }}}+{\left(b-a\right)}^{m+1}T{e}^{-2\lambda n^{\left(1-\alpha \right)}}\right\}\right).$$

Next, we use Theorem 18.1, p. 419 of [3].

Theorem 11.

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

$$F_N:={∥\left(H_n\left({\left|t-·\right|}^N\right)\right)\left(·\right)∥}_{\infty }^{{\displaystyle \frac{1}{N}}}<+\infty ,$$

(70)

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

Let $f\in C^N\left(\left[a,b\right]\right)$. Then,

$${∥H_{n}f-f∥}_{\infty }\le \sum _{\overline{k}=1}^N{\displaystyle \frac{{∥f^{\left(\overline{k}\right)}∥}_{\infty }}{\overline{k}!}}{∥\left(H_n{\left(t-·\right)}^{\overline{k}}\right)\left(·\right)∥}_{\infty }+$$

(71)

$${\omega }_1\left(f^{\left(N\right)},F_N\right)F_N^{N-1}\left({\displaystyle \frac{\left(b-a\right)}{\left(N+1\right)!}}+{\displaystyle \frac{F_N}{2N!}}+{\displaystyle \frac{F_N^2}{8\left(b-a\right)\left(N-1\right)!}}\right),\forall n\in \mathbb{N}.$$

Furthermore, the following holds:

$$F_N\le \sqrt[N]{\Delta \left(q\right)}\sqrt[N]{{\displaystyle \frac{1}{n^{N\alpha }}}+{\left(b-a\right)}^{N}T{e}^{-2\lambda n^{\left(1-\alpha \right)}}},$$

(72)

and for $\overline{k}=1,\dots ,N$, we have, similarly, that

$${∥\left(H_n{\left(t-·\right)}^{\overline{k}}\right)\left(·\right)∥}_{\infty }\le \Delta \left(q\right)\left\{{\displaystyle \frac{1}{n^{\overline{k}\alpha }}}+{\left(b-a\right)}^{\overline{k}}T{e}^{-2\lambda n^{\left(1-\alpha \right)}}\right\}.$$

(73)

By Corollary 18.1, p. 421 of [3], we obtain the following:

Corollary 2.

Here, $F_1:={∥\left(H_n\left(t-·\right)\right)\left(·\right)∥}_{\infty }<\infty$. Let $f\in C^1\left(\left[a,b\right]\right)$. Then,

$${∥H_{n}f-f∥}_{\infty }\le {∥f^{\prime }∥}_{\infty }{∥\left(H_n\left(t-·\right)\right)\left(·\right)∥}_{\infty }+$$

(74)

$${\displaystyle \frac{1}{2}}{\omega }_1\left(f^{\prime },F_1\right)\left(\left(b-a\right)+F_1+{\displaystyle \frac{F_1^2}{4\left(b-a\right)}}\right).$$

Here, we have

$$F_1\le \Delta \left(q\right)\left({\displaystyle \frac{1}{n^{\alpha }}}+\left(b-a\right)T{e}^{-2\lambda n^{\left(1-\alpha \right)}}\right).$$

(75)

By Corollary 18.2, p. 421 of [3], we have the following:

Corollary 3.

Here, $F_2:={∥\left(H_n{\left(t-·\right)}^2\right)\left(·\right)∥}_{\infty }^{{\displaystyle \frac{1}{2}}}<+\infty$. Let $f\in C^2\left(\left[a,b\right]\right)$. Then,

$${∥H_{n}f-f∥}_{\infty }\le {∥f^{\prime }∥}_{\infty }{∥\left(H_n\left(t-·\right)\right)\left(·\right)∥}_{\infty }+{\displaystyle \frac{{∥f^{″}∥}_{\infty }}{2}}{∥\left(H_n{\left(t-·\right)}^2\right)\left(·\right)∥}_{\infty }$$

(76)

$${\displaystyle \frac{1}{2}}{\omega }_1\left(f^{″},F_2\right)F_2\left({\displaystyle \frac{\left(b-a\right)}{3}}+{\displaystyle \frac{F_2}{2}}+{\displaystyle \frac{F_2^2}{4\left(b-a\right)}}\right).$$

Now, we have

$$F_2\le \sqrt{\Delta \left(q\right)}\sqrt{{\displaystyle \frac{1}{n^{2\alpha }}}+{\left(b-a\right)}^{2}T{e}^{-2\lambda n^{\left(1-\alpha \right)}}},$$

(77)

and

$${∥\left(H_n{\left(t-·\right)}^2\right)\left(·\right)∥}_{\infty }\le \Delta \left(q\right)\left({\displaystyle \frac{1}{n^{2\alpha }}}+{\left(b-a\right)}^{2}T{e}^{-2\lambda n^{\left(1-\alpha \right)}}\right).$$

(78)

By Theorem 18.2, p. 422 of [3]:

Theorem 12.

Let $\phi :={∥\left(H_n{\left(t-x_0\right)}^2\right)\left(x_0\right)∥}_{\infty }^{{\displaystyle \frac{1}{2}}}<+\infty$, and $r>0$. Let also $f\in C^1\left(\left[a,b\right]\right)$. Then,

$${∥H_{n}f-f∥}_{\infty }-{∥f^{\prime }∥}_{\infty }{∥\left(H_n\left(t-x_0\right)\right)\left(x_0\right)∥}_{\infty }\le$$

(79)

$$\left\{\begin{array}{c}{\displaystyle \frac{1}{8r}}{\left(2+r\right)}^2{\omega }_1\left(f^{\prime },r\phi \right)\phi ,\mathit{if}r\le 2;\hfill \\ \\ {\omega }_1\left(f^{\prime },r\phi \right)\phi ,\mathit{if}r>2.\hfill \end{array}\right.$$

The next results are with respect to the ${∥·∥}_1$, ${∥·∥}_2$ norms, which are with respect to the Lebesgue measure.

By Theorem 18.3, p. 424 of [3], we obtain the following:

Theorem 13.

Let

$${\gamma }_N:={∥\left(H_n{\left(t-x\right)}^N\right)\left(x\right)∥}_2^{{\displaystyle \frac{1}{N}}}<+\infty ,$$

(80)

$n,N\in \mathbb{N}$. Let also $f\in C^N\left(\left[a,b\right]\right)$. Then,

$${∥H_{n}f-f∥}_1\le \sum _{\overline{k}=1}^N{\displaystyle \frac{1}{\overline{k}!}}{∥f^{\left(\overline{k}\right)}∥}_2{∥\left(H_n{\left(t-·\right)}^{\overline{k}}\right)\left(·\right)∥}_2+{\displaystyle \frac{1}{2}}{\omega }_1\left(f^{\left(N\right)},{\gamma }_N\right){\gamma }_N^{N-1}$$

(81)

$$\left\{\left({\displaystyle \frac{b-a}{\left(N+1\right)!}}\right)·\sqrt{{\displaystyle \frac{7}{3}}\left(b-a\right)}+{\displaystyle \frac{\sqrt{b-a}}{N!}}{\gamma }_N+{\displaystyle \frac{{\gamma }_N^2}{4\left(N-1\right)!}}\sqrt{{\displaystyle \frac{2}{b-a}}}\right\}<+\infty .$$

By Corollary 18.3, p. 426 of [3], we have the following:

Corollary 4.

Set

$${\gamma }_1:={∥H_n\left(\left|t-x\right|\right)\left(x\right)∥}_2<+\infty .$$

(82)

Let $f\in C^1\left(\left[a,b\right]\right)$. Then,

$${∥H_n\left(f\right)-f∥}_1\le {∥f^{\prime }∥}_2{∥H_n\left(t-·\right)\left(·\right)∥}_2+{\displaystyle \frac{{\omega }_1\left(f^{\prime },{\gamma }_1\right)}{2}}$$

$$\left\{\left({\displaystyle \frac{b-a}{2}}\right)\sqrt{{\displaystyle \frac{7}{3}}\left(b-a\right)}+\left(\sqrt{b-a}\right){\gamma }_1+{\displaystyle \frac{{\gamma }_1^2}{4}}\sqrt{{\displaystyle \frac{2}{b-a}}}\right\}.$$

(83)

By Corollary 18.4, p. 427 of [3], we obtain

Corollary 5.

Set

$${\gamma }_2:={∥H_n\left({\left(t-x\right)}^2\right)\left(x\right)∥}_2^{{\displaystyle \frac{1}{2}}}<+\infty .$$

(84)

Let $f\in C^2\left(\left[a,b\right]\right)$. Then,

$${∥H_{n}f-f∥}_1\le {∥f^{\prime }∥}_2{∥H_n\left(t-·\right)\left(·\right)∥}_2+{\displaystyle \frac{{∥f^{″}∥}_2}{2}}{\gamma }_2^2+{\displaystyle \frac{1}{4}}{\omega }_1\left(f^{″},{\gamma }_2\right){\gamma }_2$$

(85)

$$\left\{\left({\displaystyle \frac{b-a}{3}}\right)·\sqrt{{\displaystyle \frac{7}{3}}\left(b-a\right)}+\left(\sqrt{b-a}\right){\gamma }_2+{\displaystyle \frac{{\gamma }_2^2}{2}}\sqrt{{\displaystyle \frac{2}{b-a}}}\right\}.$$

In the next Theorems 14 and 15, we use the following:

Let $\left(\left[a,b\right],\mathcal{B},\mu \right)$, $a\ne b$, be a measure space, where $\mathcal{B}$ is the Borel $\sigma$-algebra on $\left[a,b\right]$ and $\mu$ is a positive finite measure on $\left[a,b\right]$. (Note that $C\left(\left[a,b\right]\right)\subset L_p\left(\left[a,b\right],\mathcal{B},\mu \right)$, for any $p>1.$) Here, ${∥·∥}_p$ stands for the related $L_p$ norm with respect to $\mu$. Let $p,q>1$ such that ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$.

Next, we apply Theorem 18.4 from p. 428 of [3].

Theorem 14.

Here, $n,N\in \mathbb{N}$. Set

$${\delta }_N:={∥\left(H_n\left({\left|t-·\right|}^N\right)\right)\left(·\right)∥}_p^{{\displaystyle \frac{1}{N}}}<+\infty .$$

(86)

Let $f\in C^N\left(\left[a,b\right]\right)$. Then,

$${∥H_{n}f-f∥}_p\le \sum _{\overline{k}=1}^N{\displaystyle \frac{1}{\overline{k}!}}{∥f^{\left(\overline{k}\right)}∥}_{pq}{∥\left(H_n{\left(t-·\right)}^{\overline{k}}\right)\left(·\right)∥}_{p^2}+$$

(87)

$${\omega }_1\left(f^{\left(N\right)},{\delta }_N\right){\delta }_N^{N-1}\left[{\displaystyle \frac{\left(b-a\right)}{\left(N+1\right)!}}+{\displaystyle \frac{{\delta }_N}{2N!}}+{\displaystyle \frac{{\delta }_N^2}{4\left(N-1\right)!\left(b-a\right)}}\right].$$

Remark 6.

We have that

$$\left(H_n\left({\left|t-x_0\right|}^N\right)\right)\left(x_0\right)\le \Delta \left(q\right)\left\{{\displaystyle \frac{1}{n^{N\alpha }}}+{\left(b-a\right)}^{N}T{e}^{-2\lambda n^{\left(1-\alpha \right)}}\right\}=:\tau ,$$

(88)

and

$${∥\left(H_n\left({\left|t-·\right|}^N\right)\right)\left(·\right)∥}_p=$$

$${\left({\int }_{\left[a,b\right]}{\left(\left(H_n\left({\left|t-x_0\right|}^N\right)\right)\left(x_0\right)\right)}^p\mu \left(d{x}_0\right)\right)}^{{\displaystyle \frac{1}{p}}}\le {\left({\int }_{\left[a,b\right]}{\tau }^{p}d\mu \right)}^{{\displaystyle \frac{1}{p}}}=\mu {\left(\left[a,b\right]\right)}^{{\displaystyle \frac{1}{p}}}·\tau .$$

(89)

Therefore, we have

$${\delta }_N\le \mu {\left(\left[a,b\right]\right)}^{{\displaystyle \frac{1}{Np}}}{\tau }^{{\displaystyle \frac{1}{N}}}.$$

(90)

By Theorem 18.5 from p. 431 of [3], we have the following:

Theorem 15.

Here, $n,N\in \mathbb{N}$. Let

$${\delta }_N:={∥H_n\left({\left|t-x\right|}^N,x\right)∥}_p^{{\displaystyle \frac{1}{N}}}<+\infty .$$

(91)

Let $f\in C^N\left(\left[a,b\right]\right)$. Then,

$${∥H_{n}f-f∥}_p\le \sum _{\overline{k}=1}^N{\displaystyle \frac{{∥f^{\left(\overline{k}\right)}∥}_{\infty }}{\overline{k}!}}{∥\left(H_n{\left(t-x\right)}^{\overline{k}},x\right)∥}_p+$$

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$${\omega }_1\left(f^{\left(N\right)},{\delta }_N\right){\delta }_N^{N-1}\left[{\displaystyle \frac{\left(b-a\right)}{\left(N+1\right)!}}+{\displaystyle \frac{{\delta }_N}{2N!}}+{\displaystyle \frac{{\delta }_N^2}{4\left(N-1\right)!\left(b-a\right)}}\right].$$

We now need the following definition.

Definition 3.

Let $f,g\in C\left(\left[a,b\right]\right)$ such that $\left|f\left(x\right)\right|\le \left|g\left(x\right)\right|$, ∀$x\in \left[a,b\right]$. A norm $∥·∥$ on $C\left(\left[a,b\right]\right)$ is called monotone iff $∥f∥\le ∥g∥$. We denote a monotone norm by ${∥·∥}_m$, e.g., $L_p$ norms in general ($1\le p\le +\infty$), Orlicz norms, etc.

Finally, we apply Corollary 18.5 from p. 432 of [3].

Corollary 6.

Let ${∥·∥}_m$ be a monotone norm on $C\left(\left[a,b\right]\right)$. Define

$${\epsilon }_N:={∥H_n\left({\left|t-x\right|}^N,x\right)∥}_m^{{\displaystyle \frac{1}{N}}}<+\infty ,n,N\in \mathbb{N}.$$

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Let $f\in C^N\left(\left[a,b\right]\right)$. Then,

$${∥H_{n}f-f∥}_m\le \sum _{\overline{k}=1}^N{\displaystyle \frac{{∥f^{\left(\overline{k}\right)}∥}_{\infty }}{\overline{k}!}}{∥\left(H_n{\left(t-x\right)}^{\overline{k}},x\right)∥}_m+$$

$${\omega }_1\left(f^{\left(N\right)},{\epsilon }_N\right){\epsilon }_N^{N-1}\left[{\displaystyle \frac{\left(b-a\right)}{\left(N+1\right)!}}+{\displaystyle \frac{{\epsilon }_N}{2N!}}+{\displaystyle \frac{{\epsilon }_N^2}{4\left(N-1\right)!\left(b-a\right)}}\right].$$

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4. Conclusions

The symmetrization technique presented here aims at having operators converge at a high speed to the unit operator using half the relevant data feed. My forthcoming work documents numerical work and programming related to this result. This article builds a bridge for the first time connecting neural networks approximation (part of ai) to positive linear operators (an important part of functional analysis). My work has helped pioneer the study of positive linear operators using geometric theory since 1985 and also helped found quantitative approximation theory using neural networks since 1997. The list of complete references supporting the above is provided below.