Smooth Logistic Real and Complex, Ordinary and Fractional Neural Network Approximations over Infinite Domains

Abstract

In this work, we study the univariate quantitative smooth approximations, including both real and complex and ordinary and fractional approximations, under different functions. The approximators presented here are neural network operators activated by Richard’s curve, a parametrized form of logistic sigmoid function. All domains used are obtained from the whole real line. The neural network operators used here are of the quasi-interpolation type: basic ones, Kantorovich-type ones, and those of the quadrature type. We provide pointwise and uniform approximations with rates. We finish with their applications.

1. Introduction

The author of [1,2], see Section 2, Section 3, Section 4 and Section 5, was the first to establish neural network approximation to continuous functions with rates via very specific neural network operators of Cardaliaguet–Euvrard and “Squashing” types by using the modulus of continuity of the engaged function or its high-order derivative and producing very tight Jackson-type inequalities. He treats both univariate and multivariate cases. These operators’ “bell-shaped” and “squashing” functions are assumed to provide compact support.

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

A parametrized activation function kills far fewer neurons than the original one.

Therefore, here, the author obtains parametrized Richard’s-curve-activated neural network approximations of differentiated functions, which are transformed from $\mathbb{R}$ into $\mathbb{R}$, going beyond the bounded domain functions.

We present real and complex ordinary and fractional quasi-interpolation quantitative approximations. The fractional case is extensively studied because of the applications of fractional calculus in the interpretation of many natural phenomena and in engineering. Therefore, we derive Jackson inequalities that are close to the sharp type.

Real feed-forward neural networks (FNNs) with one hidden layer, the ones we use here, are mathematically expressed by:

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

where, for $0\le j\le n$, $b_j\in \mathbb{R}$ are the thresholds, ${\alpha }_j\in {\mathbb{R}}^s$ are the connection weights, $c_j\in \mathbb{R}$ are the coefficients, $〈{\alpha }_j·x〉$ is the inner product of ${\alpha }_j$ and x, and $\sigma$ is the activation function of the network. For more information about neural networks in general, read [6,7,8]. Due to their efficiency, neural network approximations are widely used for many subjects, like differential equations, numerical analysis, statistics, and AI.

2. Preliminaries

The following come from [9], pp. 2–6.

The Richards’s curve is as follows:

$$\phi \left(x\right)=\frac{1}{1+e^{-\mu x}};x\in \mathbb{R},\mathrm{with}\mathrm{parameter}\mu >0,$$

(1)

which is of great interest when $0<\mu \le 1.$

The function $\phi \left(x\right)$ increases in terms of $\mathbb{R}$ and is a sigmoid function; specifically, this is a generalized logistic function [10].

It should be noted that

$$\underset{x\to +\infty }{lim}\phi \left(x\right)=1\mathrm{and}\underset{x\to -\infty }{lim}\phi \left(x\right)=0.$$

(2)

We consider the following activation function:

$$G\left(x\right)=\frac{1}{2}\left(\phi \left(x+1\right)-\phi \left(x-1\right)\right),x\in \mathbb{R},$$

(3)

which is $G\left(x\right)>0$, all $x\in \mathbb{R}$.

Function $\phi$ has many applications in epidemiology and especially in COVID-19 modeling of infection trajectories [11].

We can see that

$$\phi \left(0\right)=\frac{1}{2}\mathrm{and}\phi \left(x\right)=1-\phi \left(-x\right).$$

(4)

We notice that

$$G\left(-x\right):=\frac{1}{2}\left(\phi \left(-x+1\right)-\phi \left(-x-1\right)\right)$$

$$=\frac{1}{2}\left[1-\phi \left(x-1\right)-1+\phi \left(x+1\right)\right]$$

$$=\frac{1}{2}\left[\phi \left(x+1\right)-\phi \left(x-1\right)\right]=G\left(x\right),\forall x\in \mathbb{R}.$$

(5)

Therefore, G is an even function.

We can observe that

$$G\left(0\right)=\frac{1}{2}\left(\phi \left(1\right)-\phi \left(-1\right)\right)$$

$$=\frac{1}{2}\left(1-\phi \left(-1\right)-\phi \left(-1\right)\right)=\frac{1}{2}\left(1-2\phi \left(-1\right)\right)$$

$$=\frac{1}{2}\left(1-\frac{2}{1+e^{\mu }}\right)=\frac{1}{2}\left(\frac{1+e^{\mu }-2}{1+e^{\mu }}\right)=\frac{1}{2}\left(\frac{e^{\mu }-1}{e^{\mu }+1}\right),$$

that is

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

(6)

Let $x\ge 0$. We can observe that

$$G^{\prime }\left(x\right)=\frac{1}{2}\left({\phi }^{\prime }\left(x+1\right)-{\phi }^{\prime }\left(x-1\right)\right)=$$

$$\frac{1}{2}\left({\left({\left(1+e^{-\mu \left(x+1\right)}\right)}^{-1}\right)}^{\prime }-{\left({\left(1+e^{-\mu \left(x-1\right)}\right)}^{-1}\right)}^{\prime }\right)=$$

$$\frac{1}{2}\left(\left(\left(-1\right){\left(1+e^{-\mu \left(x+1\right)}\right)}^{-2}\right)e^{-\mu \left(x+1\right)}\left(-\mu \right)-\left(-1\right){\left(1+e^{-\mu \left(x-1\right)}\right)}^{-2}{e}^{-\mu \left(x-1\right)}\left(-\mu \right)\right)=$$

$$\frac{\mu }{2}\left[\frac{1}{e^{\mu \left(x+1\right)}+e^{-\mu \left(x+1\right)}+2}-\frac{1}{e^{\mu \left(x-1\right)}+e^{-\mu \left(x-1\right)}+2}\right]=$$

$$\frac{\mu }{4}\left[\frac{1}{\frac{e^{\mu \left(x+1\right)}+e^{-\mu \left(x+1\right)}}{2}+1}-\frac{1}{\frac{e^{\mu \left(x-1\right)}+e^{-\mu \left(x-1\right)}}{2}+1}\right]=$$

$$\frac{\mu }{4}\left[\frac{cosh\mu \left(x-1\right)-cosh\mu \left(x+1\right)}{\left(cosh\mu \left(x+1\right)+1\right)\left(cosh\mu \left(x-1\right)+1\right)}\right]<0,\forall x\ge 1.$$

(7)

Therefore, for $x\ge 1,$ $G^{\prime }\left(x\right)<0$ and $G\left(x\right)$ decrease.

Let $0<x<1;$ then, $1-x>0$ and $0<1-x<1+x,$ and then $cosh\mu (x-1)=cosh\mu (1-x)<cosh\mu (x+1),$ so that $G^{\prime }\left(x\right)<0$ and $G\left(x\right)$ decreases over $0<x<1.$

Thus, $G\left(x\right)$ decreases at $\left(0,+\infty \right)$.

Clearly, $G\left(x\right)$ increases at $\left(-\infty ,0\right)$, and $G^{\prime }\left(0\right)=0.$

We can observe that

$$\begin{array}{c}{\displaystyle \underset{x\to +\infty }{lim}}G\left(x\right)=\frac{1}{2}\left(\phi \left(+\infty \right)-\phi \left(+\infty \right)\right)=0,\\ \\ {\displaystyle \underset{x\to -\infty }{lim}}G\left(x\right)=\frac{1}{2}\left(\phi \left(-\infty \right)-\phi \left(-\infty \right)\right)=0.\end{array}$$

(8)

That is, the x axis is the horizontal asymptote for G.

In conclusion, G is a bell symmetric function with the following maximum:

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

(9)

We need to use the following theorems.

Theorem 1.

It holds that

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

(10)

Remark 1.

Because G is even, it holds that

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

Hence

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

and

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

(11)

Theorem 2.

It holds that

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

(12)

Proof. 

We can observe that

$${\int }_{-\infty }^{\infty }G\left(x\right)dx=\sum _{j=-\infty }^{\infty }{\int }_j^{j+1}G\left(x\right)dx=\sum _{j=-\infty }^{\infty }{\int }_0^{1}G\left(x+j\right)dx=$$

$${\int }_0^1\left(\sum _{j=-\infty }^{\infty }G\left(x+j\right)dx\right)={\int }_0^{1}1dx=1.$$

So, $G\left(x\right)$ is a density function. □

Remark 2.

We can obtain that

$$G\left(x\right)=\frac{1}{2}\left[\phi \left(x+1\right)-\phi \left(x-1\right)\right],\forall x\in \mathbb{R}.$$

Let $x\ge 1$. That is $0\le x-1<x+1$. Applying the mean value theorem, we obtain

$$G\left(x\right)=\frac{1}{2}2{\phi }^{\prime }\left(\xi \right)={\phi }^{\prime }\left(\xi \right)=\frac{\mu e^{-\mu \xi }}{{\left(1+e^{-\mu \xi }\right)}^2},$$

where $0\le x-1<\xi <x+1.$

Notice that

$$G\left(x\right)<\mu e^{-\mu \xi }<\mu e^{-\mu \left(x-1\right)},\forall x\ge 1.$$

(13)

We need the following definitions.

Definition 1.

In this article, we study the smooth approximation properties of the following quasi-interpolation neural network operators acting on $f\in C_B\left(\mathbb{R}\right)$ (continuous and bounded functions):

(i) 

The basic ones:

$$B_n\left(f,x\right):=\sum _{k=-\infty }^{\infty }f\left(\frac{k}{n}\right)G\left(nx-k\right),\forall x\in \mathbb{R},n\in \mathbb{N},$$

(14)

(ii) 

The Kantorovich-type operators:

$$C_n\left(f,x\right):=\sum _{k=-\infty }^{\infty }\left(n{\int }_{\frac{k}{n}}^{\frac{k+1}{n}}f\left(t\right)dt\right)G\left(nx-k\right),\forall x\in \mathbb{R},n\in \mathbb{N},$$

(15)

(iii) 

let $\theta \in \mathbb{N}$, $w_r\ge 0$, ${\displaystyle \sum _{r=0}^{\theta }}{w}_r=1$, $k\in \mathbb{Z}$, and

$${\delta }_{nk}\left(f\right):=\sum _{r=0}^{\theta }{w}_{r}f\left(\frac{k}{n}+\frac{r}{n\theta }\right),$$

(16)

We also consider the quadrature-type operators:

$$D_n\left(f,x\right):=\sum _{k=-\infty }^{\infty }{\delta }_{nk}\left(f\right)G\left(nx-k\right),\forall x\in \mathbb{R},n\in \mathbb{N}.$$

(17)

We will be using the first modulus of continuity:

$${\omega }_1\left(f,\delta \right):=\underset{\left|x-y\right|\le \delta }{\underset{x,y\in \mathbb{R}:}{sup}}\left|f\left(x\right)-f\left(y\right)\right|,\delta >0,$$

(18)

where $f\in C\left(\mathbb{R}\right),$ which is bounded and/or uniformly continuous.

We are motivated by the following result.

Theorem 3

([9], p. 13). Let $f\in C_B\left(\mathbb{R}\right)$, $0<\beta <1$, $\mu >0$, $n\in \mathbb{N}:n^{1-\beta }>2$, $x\in \mathbb{R}$. Then,

(i) 

$$\left|B_n\left(f,x\right)-f\left(x\right)\right|\le {\omega }_1\left(f,\frac{1}{n^{\beta }}\right)+\frac{4{∥f∥}_{\infty }}{e^{\mu \left(n^{1-\beta }-2\right)}}=:\gamma ,$$

(19)

and

(ii) 

$${∥B_n\left(f\right)-f∥}_{\infty }\le \gamma .$$

(20)

For $f\in C_{uB}\left(\mathbb{R}\right)$ (uniformly and bounded continuous functions), we can obtain $\underset{n\to \infty }{lim}{B}_n\left(f\right)=f$, both pointwise and uniformly.

3. Main Results

Here, we study the approximation properties of neural network operators $B_n$, $C_n$, $D_n$ under differentiation.

Theorem 4.

Here, $0<\beta <1$, $n\in \mathbb{N}:n^{1-\beta }>2$, $\mu >0,$ $N\in \mathbb{N},$ $f\in C^N\left(\mathbb{R}\right)$, with $f^{\left(i\right)}\in C_B\left(\mathbb{R}\right)$, $i=0,1,\dots ,N;$ $x\in \mathbb{R}$. Then,

(i) 

$$\left|B_n\left(f,x\right)-f\left(x\right)-\sum _{j=1}^N\frac{f^{\left(j\right)}\left(x\right)}{j!}{B}_n\left({\left(·-x\right)}^j\right)\left(x\right)\right|\le$$

$$\frac{{\omega }_1\left(f^{\left(N\right)},\frac{1}{n^{\beta }}\right)}{n^{\beta N}N!}+\left(\frac{{∥f^{\left(N\right)}∥}_{\infty }{e}^{\mu }{2}^{N+3}}{n^N{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}=:M,$$

(21)

(ii) 

Assume all $f^{\left(j\right)}\left(x\right)=0$, $j=1,\dots ,N;$ we have that

$$\left|B_n\left(f,x\right)-f\left(x\right)\right|\le M,$$

(22)

at high speed $n^{-\beta \left(N+1\right)},$

(iii) 

$$\left|B_n\left(f,x\right)-f\left(x\right)\right|\le \sum _{j=1}^N\frac{\left|f^{\left(j\right)}\left(x\right)\right|}{j!}\left[\frac{1}{n^{\beta j}}+\frac{1}{n^j}\left\{\frac{e^{\mu }{2}^{j+2}j!}{{\mu }^j}\right\}{e}^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right]+M,$$

(23)

and

(iv) 

$${∥B_n\left(f\right)-f∥}_{\infty }\le \sum _{j=1}^N\frac{{∥f^{\left(j\right)}∥}_{\infty }}{j!}\left[\frac{1}{n^{\beta j}}+\frac{1}{n^j}\left\{\frac{e^{\mu }{2}^{j+2}j!}{{\mu }^j}\right\}{e}^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right]+M.$$

(24)

Proof. 

Using Taylor’s theorem, we have ($x\in \mathbb{R}$)

$$f\left(\frac{k}{n}\right)=\sum _{j=0}^N\frac{f^{\left(j\right)}\left(x\right)}{j!}{\left(\frac{k}{n}-x\right)}^j+{\int }_x^{\frac{k}{n}}\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right)\frac{{\left(\frac{k}{n}-t\right)}^{N-1}}{\left(N-1\right)!}dt.$$

(25)

It follows that

$$f\left(\frac{k}{n}\right)G\left(nx-k\right)=\sum _{j=0}^N\frac{f^{\left(j\right)}\left(x\right)}{j!}G\left(nx-k\right){\left(\frac{k}{n}-x\right)}^j+$$

$$G\left(nx-k\right){\int }_x^{\frac{k}{n}}\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right)\frac{{\left(\frac{k}{n}-t\right)}^{N-1}}{\left(N-1\right)!}dt.$$

(26)

Hence,

$$B_n\left(f,x\right)=\sum _{k=-\infty }^{\infty }f\left(\frac{k}{n}\right)G\left(nx-k\right)=\sum _{j=0}^N\frac{f^{\left(j\right)}\left(x\right)}{j!}{B}_n\left({\left(·-x\right)}^j\right)\left(x\right)+$$

(27)

$$\sum _{k=-\infty }^{\infty }G\left(nx-k\right){\int }_x^{\frac{k}{n}}\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right)\frac{{\left(\frac{k}{n}-t\right)}^{N-1}}{\left(N-1\right)!}dt.$$

Use the following equation:

$$R:=\sum _{k=-\infty }^{\infty }G\left(nx-k\right){\int }_x^{\frac{k}{n}}\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right)\frac{{\left(\frac{k}{n}-t\right)}^{N-1}}{\left(N-1\right)!}dt.$$

(28)

(I) Let $\left|\frac{k}{n}-x\right|<\frac{1}{n^{\beta }}$. Then,

(i) case $\frac{k}{n}\ge x$:

$$\left|R\right|\le \sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|\frac{k}{n}-x\right|<\frac{1}{n^{\beta }}\end{array}\right.}^{\infty }G\left(nx-k\right){\int }_x^{\frac{k}{n}}\left|f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right|\frac{{\left(\frac{k}{n}-t\right)}^{N-1}}{\left(N-1\right)!}dt$$

(29)

$$\le \sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|\frac{k}{n}-x\right|<\frac{1}{n^{\beta }}\end{array}\right.}^{\infty }G\left(nx-k\right){\omega }_1\left(f^{\left(N\right)},\frac{1}{n^{\beta }}\right)\frac{{\left(\frac{k}{n}-x\right)}^N}{N!}$$

$$\le {\omega }_1\left(f^{\left(N\right)},\frac{1}{n^{\beta }}\right)\frac{1}{N!n^{\beta N}}.$$

We found that

$$\left|R\right|{|}_{\left|\frac{k}{n}-x\right|<\frac{1}{n^{\beta }}}\le {\omega }_1\left(f^{\left(N\right)},\frac{1}{n^{\beta }}\right)\frac{1}{N!n^{\beta N}}.$$

(30)

(ii) Case $\frac{k}{n}<x$: then,

$$\left|R\right|\le \sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|\frac{k}{n}-x\right|<\frac{1}{n^{\beta }}\end{array}\right.}^{\infty }G\left(nx-k\right)\left|{\int }_{\frac{k}{n}}^x\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right)\frac{{\left(\frac{k}{n}-t\right)}^{N-1}}{\left(N-1\right)!}dt\right|$$

(31)

$$\le \sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|\frac{k}{n}-x\right|<\frac{1}{n^{\beta }}\end{array}\right.}^{\infty }G\left(nx-k\right){\int }_{\frac{k}{n}}^x\left|f^{\left(N\right)}\left(x\right)-f^{\left(N\right)}\left(t\right)\right|\frac{{\left(t-\frac{k}{n}\right)}^{N-1}}{\left(N-1\right)!}dt$$

$$\le \sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|\frac{k}{n}-x\right|<\frac{1}{n^{\beta }}\end{array}\right.}^{\infty }G\left(nx-k\right){\omega }_1\left(f^{\left(N\right)},\frac{1}{n^{\beta }}\right)\frac{{\left(x-\frac{k}{n}\right)}^N}{N!}$$

$$\le {\omega }_1\left(f^{\left(N\right)},\frac{1}{n^{\beta }}\right)\frac{1}{N!n^{\beta N}}.$$

Consequently, we prove that

$$\left|R\right|{|}_{\left|\frac{k}{n}-x\right|<\frac{1}{n^{\beta }}}\le \frac{{\omega }_1\left(f^{\left(N\right)},\frac{1}{n^{\beta }}\right)}{N!n^{\beta N}}.$$

(32)

Next, we can observe ($\frac{k}{n}\ge x$)

$$\left|R\right|\le \sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|\frac{k}{n}-x\right|\ge \frac{1}{n^{\beta }}\end{array}\right.}^{\infty }G\left(nx-k\right)\left|{\int }_x^{\frac{k}{n}}\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right)\frac{{\left(\frac{k}{n}-t\right)}^{N-1}}{\left(N-1\right)!}dt\right|$$

$$\le \sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|\frac{k}{n}-x\right|\ge \frac{1}{n^{\beta }}\end{array}\right.}^{\infty }G\left(nx-k\right){\int }_x^{\frac{k}{n}}\left|f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right|\frac{{\left(\frac{k}{n}-t\right)}^{N-1}}{\left(N-1\right)!}dt$$

$$\le 2{∥f^{\left(N\right)}∥}_{\infty }\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|\frac{k}{n}-x\right|\ge \frac{1}{n^{\beta }}\end{array}\right.}^{\infty }G\left(nx-k\right)\frac{{\left(\frac{k}{n}-x\right)}^N}{N!}$$

(33)

$$=\frac{2{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\beta }\end{array}\right.}^{\infty }G\left(nx-k\right){\left(k-nx\right)}^N.$$

In case $\frac{k}{n}<x$, we obtain

$$\left|R\right|\le \sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|\frac{k}{n}-x\right|\ge \frac{1}{n^{\beta }}\end{array}\right.}^{\infty }G\left(nx-k\right)\left|{\int }_{\frac{k}{n}}^x\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right)\frac{{\left(t-\frac{k}{n}\right)}^{N-1}}{\left(N-1\right)!}dt\right|$$

$$\le \sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\beta }\end{array}\right.}^{\infty }G\left(nx-k\right){\int }_{\frac{k}{n}}^x\left|f^{\left(N\right)}\left(x\right)-f^{\left(N\right)}\left(t\right)\right|\frac{{\left(t-\frac{k}{n}\right)}^{N-1}}{\left(N-1\right)!}dt$$

$$\le 2{∥f^{\left(N\right)}∥}_{\infty }\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\beta }\end{array}\right.}^{\infty }G\left(nx-k\right)\frac{{\left(x-\frac{k}{n}\right)}^N}{N!}$$

(34)

$$=\frac{2{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\beta }\end{array}\right.}^{\infty }G\left(nx-k\right){\left(nx-k\right)}^N.$$

Consequently, it holds

$$\left|R\right|{|}_{\left|\frac{k}{n}-x\right|\ge \frac{1}{n^{\beta }}}\le \frac{2{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\beta }\end{array}\right.}^{\infty }G\left(\left|nx-k\right|\right){\left|nx-k\right|}^N.$$

(35)

Next, we treat

$$\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\beta }\end{array}\right.}^{\infty }G\left(\left|nx-k\right|\right){\left|nx-k\right|}^N\stackrel{\left(13\right)}{\le }$$

$$\mu \sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\beta }\end{array}\right.}^{\infty }{e}^{-\mu \left(\left|nx-k\right|-1\right)}{\left|nx-k\right|}^N=$$

$$\mu e^{\mu }\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\beta }\end{array}\right.}^{\infty }{e}^{-\mu \left|nx-k\right|}{\left|nx-k\right|}^N=:\left(\ast \right).$$

(36)

Notice that

$$e^{\frac{\mu \left|nx-k\right|}{2}}=\sum _{\lambda =0}^{\infty }\frac{{\left(\frac{\mu \left|nx-k\right|}{2}\right)}^{\lambda }}{\lambda !}\ge {\left(\frac{\mu \left|nx-k\right|}{2}\right)}^N\frac{1}{N!}.$$

(37)

Therefore, we have

$${\left(\frac{\mu \left|nx-k\right|}{2}\right)}^N\le N!e^{\frac{\mu \left|nx-k\right|}{2}},\mathrm{or}$$

$${\left(\mu \left|nx-k\right|\right)}^N\le 2^{N}N!e^{\frac{\mu \left|nx-k\right|}{2}},\mathrm{or}$$

(38)

$${\left|nx-k\right|}^N\le \frac{2^N}{{\mu }^N}N!e^{\frac{\mu \left|nx-k\right|}{2}}.$$

Hence, it holds that

$$\left(\ast \right)\le \frac{\mu }{{\mu }^N}{e}^{\mu }{2}^{N}N!\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\beta }\end{array}\right.}^{\infty }{e}^{-\mu \left|nx-k\right|}{e}^{\frac{\mu \left|nx-k\right|}{2}}=$$

$$\frac{e^{\mu }}{{\mu }^{N-1}}{2}^{N}N!\left(\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\beta }\end{array}\right.}^{\infty }{e}^{-\frac{\mu }{2}\left|nx-k\right|}\right)\le$$

$$\frac{2e^{\mu }}{{\mu }^{N-1}}{2}^{N}N!\left({\int }_{n^{1-\beta }-1}^{\infty }{e}^{-\frac{\mu }{2}x}dx\right)=$$

$$\frac{2}{\mu }\frac{2e^{\mu }}{{\mu }^{N-1}}{2}^{N}N!\left({\int }_{n^{1-\beta }-1}^{\infty }{e}^{-\frac{\mu }{2}x}d\frac{\mu x}{2}\right)=$$

(39)

$$\left(\frac{4e^{\mu }}{{\mu }^N}{2}^{N}N!\right)\left({\int }_{n^{1-\beta }-1}^{\infty }{e}^{-y}dy\right)=\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)\left(-e^{-y}{|}_{n^{1-\beta }-1}^{\infty }\right)=$$

$$\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)\left(e^{-y}{|}_{\infty }^{n^{1-\beta }-1}\right)=\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)\left(e^{-\frac{\mu x}{2}}{|}_{\infty }^{n^{1-\beta }-1}\right)=$$

$$\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}.$$

Thus, we have

$$\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\beta }\end{array}\right.}^{\infty }G\left(\left|nx-k\right|\right){\left|nx-k\right|}^N\le \left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}.$$

(40)

and the following holds:

$$\left|R\right|{|}_{\left|\frac{k}{n}-x\right|\ge \frac{1}{n^{\beta }}}\le \left(\frac{2{∥f^{\left(N\right)}∥}_{\infty }}{N!n^N}\right)\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}=$$

(41)

$$\left(\frac{{∥f^{\left(N\right)}∥}_{\infty }}{n^N{\mu }^N}\right)\left(e^{\mu }{2}^{N+3}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}.$$

We prove that

$$\left|R\right|{|}_{\left|\frac{k}{n}-x\right|\ge \frac{1}{n^{\beta }}}\le \left(\frac{{∥f^{\left(N\right)}∥}_{\infty }}{n^N}\right)\left(\frac{e^{\mu }{2}^{N+3}}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}.$$

(42)

and we derive that

$$\left|R\right|\le \left|R\right|{{|}_{\left|\frac{k}{n}-x\right|<\frac{1}{n^{\beta }}}+\left|R\right||}_{\left|\frac{k}{n}-x\right|\ge \frac{1}{n^{\beta }}}\le$$

(43)

$$\frac{{\omega }_1\left(f^{\left(N\right)},\frac{1}{n^{\beta }}\right)}{n^{\beta N}N!}+\left(\frac{{∥f^{\left(N\right)}∥}_{\infty }}{n^N}\right)\left(\frac{e^{\mu }{2}^{N+3}}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}.$$

Finally, we estimate

$$\left|B_n\left({\left(·-x\right)}^j\right)\left(x\right)\right|\le \sum _{k=-\infty }^{\infty }G\left(nx-k\right){\left|\frac{k}{n}-x\right|}^j=$$

$$\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|x-\frac{k}{n}\right|<\frac{1}{n^{\beta }}\end{array}\right.}^{\infty }G\left(nx-k\right){\left|\frac{k}{n}-x\right|}^j+$$

$$\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|x-\frac{k}{n}\right|\ge \frac{1}{n^{\beta }}\end{array}\right.}^{\infty }G\left(nx-k\right){\left|\frac{k}{n}-x\right|}^j\le$$

$$\frac{1}{n^{\beta j}}+\frac{1}{n^j}\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\beta }\end{array}\right.}^{\infty }G\left(\left|nx-k\right|\right){\left|nx-k\right|}^j\le$$

(44)

$$\frac{1}{n^{\beta j}}+\frac{1}{n^j}\left\{\frac{e^{\mu }{2}^{j+2}j!}{{\mu }^j}\right\}{e}^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}.$$

The theorem is proved. □

Next comes

Theorem 5.

Here $0<\beta <1$, $n\in \mathbb{N}:n^{1-\beta }>2$, $\mu >0,$ $N\in \mathbb{N},$ $f\in C^N\left(\mathbb{R}\right)$, with $f^{\left(i\right)}\in C_B\left(\mathbb{R}\right)$, $i=0,1,\dots ,N;$ $x\in \mathbb{R}$. Then,

(i)

$$\left|C_n\left(f,x\right)-f\left(x\right)-\sum _{j=1}^N\frac{f^{\left(j\right)}\left(x\right)}{j!}{C}_n\left({\left(·-x\right)}^j\right)\left(x\right)\right|\le$$

$${\omega }_1\left(f^{\left(N\right)},\frac{1}{n}+\frac{1}{n^{\beta }}\right)\frac{{\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)}^N}{N!}+$$

$$\frac{2^N{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}\left[2e^{\mu }{e}^{-\mu \left(n^{1-\beta }-1\right)}+\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right]=:\Psi ,$$

(45)

(ii) assume all $f^{\left(j\right)}\left(x\right)=0$, $j=1,\dots ,N,$ then

$$\left|C_n\left(f,x\right)-f\left(x\right)\right|\le \Psi ,$$

(46)

at high speed $n^{-\beta \left(N+1\right)},$

(iii)

$$\left|C_n\left(f,x\right)-f\left(x\right)\right|\le \sum _{j=1}^N\frac{\left|f^{\left(j\right)}\left(x\right)\right|}{j!}$$

$$\left\{{\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)}^j+\frac{2^{j-1}}{n^j}\left[2e^{\mu }{e}^{-\mu \left(n^{1-\beta }-1\right)}+\left(\frac{e^{\mu }{2}^{j+2}j!}{{\mu }^j}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right]\right\}+\Psi ,$$

(47)

and

(iv)

$${∥C_n\left(f\right)-f∥}_{\infty }\le \sum _{j=1}^N\frac{{∥f^{\left(j\right)}∥}_{\infty }}{j!}$$

$$\left\{{\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)}^j+\frac{2^{j-1}}{n^j}\left[2e^{\mu }{e}^{-\mu \left(n^{1-\beta }-1\right)}+\left(\frac{e^{\mu }{2}^{j+2}j!}{{\mu }^j}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right]\right\}+\Psi .$$

(48)

Proof. 

One can write

$$C_n\left(f,x\right)=\sum _{k=-\infty }^{\infty }\left(n{\int }_0^{\frac{1}{n}}f\left(t+\frac{k}{n}\right)dt\right)G\left(nx-k\right).$$

(49)

Let now $f\in C^N\left(\mathbb{R}\right)$ with $f^{\left(i\right)}\in C_B\left(\mathbb{R}\right)$, $i=0,1,\dots ,N\in \mathbb{N}$.

We have that

$$f\left(t+\frac{k}{n}\right)=\sum _{j=0}^N\frac{f^{\left(j\right)}\left(x\right)}{j!}{\left(t+\frac{k}{n}-x\right)}^j+$$

(50)

$${\int }_x^{t+\frac{k}{n}}\left(f^{\left(N\right)}\left(s\right)-f^{\left(N\right)}\left(x\right)\right)\frac{{\left(t+\frac{k}{n}-s\right)}^{N-1}}{\left(N-1\right)!}ds,$$

and

$$n{\int }_0^{\frac{1}{n}}f\left(t+\frac{k}{n}\right)dt=\sum _{j=0}^N\frac{f^{\left(j\right)}\left(x\right)}{j!}n{\int }_0^{\frac{1}{n}}{\left(t+\frac{k}{n}-x\right)}^{j}dt+$$

$$n{\int }_0^{\frac{1}{n}}\left({\int }_x^{t+\frac{k}{n}}\left(f^{\left(N\right)}\left(s\right)-f^{\left(N\right)}\left(x\right)\right)\frac{{\left(t+\frac{k}{n}-s\right)}^{N-1}}{\left(N-1\right)!}ds\right)dt.$$

(51)

Hence,

$$C_n\left(f,x\right)=\sum _{k=-\infty }^{\infty }\left(n{\int }_0^{\frac{1}{n}}f\left(t+\frac{k}{n}\right)dt\right)G\left(nx-k\right)=$$

$$\sum _{j=0}^N\frac{f^{\left(j\right)}\left(x\right)}{j!}{C}_n\left({\left(·-x\right)}^j\right)\left(x\right)+$$

(52)

$$\sum _{k=-\infty }^{\infty }G\left(nx-k\right)\left(n{\int }_0^{\frac{1}{n}}\left({\int }_x^{t+\frac{k}{n}}\left(f^{\left(N\right)}\left(s\right)-f^{\left(N\right)}\left(x\right)\right)\frac{{\left(t+\frac{k}{n}-s\right)}^{N-1}}{\left(N-1\right)!}ds\right)dt\right).$$

Therefore, we can write

$$C_n\left(f,x\right)-f\left(x\right)-\sum _{j=1}^N\frac{f^{\left(j\right)}\left(x\right)}{j!}{C}_n\left({\left(·-x\right)}^j\right)\left(x\right)=R,$$

(53)

where

$$R:=\sum _{k=-\infty }^{\infty }G\left(nx-k\right)\left(n{\int }_0^{\frac{1}{n}}\left({\int }_x^{t+\frac{k}{n}}\left(f^{\left(N\right)}\left(s\right)-f^{\left(N\right)}\left(x\right)\right)\frac{{\left(t+\frac{k}{n}-s\right)}^{N-1}}{\left(N-1\right)!}ds\right)dt\right).$$

(54)

Call

$$\lambda \left(k\right):=n{\int }_0^{\frac{1}{n}}\left({\int }_x^{t+\frac{k}{n}}\left(f^{\left(N\right)}\left(s\right)-f^{\left(N\right)}\left(x\right)\right)\frac{{\left(t+\frac{k}{n}-s\right)}^{N-1}}{\left(N-1\right)!}ds\right)dt,$$

(55)

where $k\in \mathbb{Z}$.

(I) Let $\left|\frac{k}{n}-x\right|<\frac{1}{n^{\beta }}$ ($0<\beta <1$).

(i) if $t+\frac{k}{n}\ge x$, then

$$\left|\lambda \left(k\right)\right|\le n{\int }_0^{\frac{1}{n}}\left({\int }_x^{t+\frac{k}{n}}\left|f^{\left(N\right)}\left(s\right)-f^{\left(N\right)}\left(x\right)\right|\frac{{\left(t+\frac{k}{n}-s\right)}^{N-1}}{\left(N-1\right)!}ds\right)dt$$

$$\le n{\int }_0^{\frac{1}{n}}\left({\omega }_1\left(f^{\left(N\right)},\left|t+\frac{k}{n}-x\right|\right)\frac{{\left(t+\frac{k}{n}-x\right)}^N}{N!}\right)dt$$

(56)

$$\le n{\int }_0^{\frac{1}{n}}{\omega }_1\left(f^{\left(N\right)},\left|t\right|+\left|\frac{k}{n}-x\right|\right)\frac{{\left(\left|t\right|+\left|\frac{k}{n}-x\right|\right)}^N}{N!}dt$$

$$\le {\omega }_1\left(f^{\left(N\right)},\frac{1}{n}+\frac{1}{n^{\beta }}\right)\frac{{\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)}^N}{N!}.$$

(ii) if $t+\frac{k}{n}<x$, then

$$\left|\lambda \left(k\right)\right|\le n{\int }_0^{\frac{1}{n}}\left|{\int }_{t+\frac{k}{n}}^x\left(f^{\left(N\right)}\left(s\right)-f^{\left(N\right)}\left(x\right)\right)\frac{{\left(t+\frac{k}{n}-s\right)}^{N-1}}{\left(N-1\right)!}ds\right|dt$$

$$\le n{\int }_0^{\frac{1}{n}}\left({\int }_{t+\frac{k}{n}}^x\left|f^{\left(N\right)}\left(x\right)-f^{\left(N\right)}\left(s\right)\right|\frac{{\left(s-\left(t+\frac{k}{n}\right)\right)}^{N-1}}{\left(N-1\right)!}ds\right)dt$$

$$\le n{\int }_0^{\frac{1}{n}}{\omega }_1\left(f^{\left(N\right)},\left(x-\frac{k}{n}-t\right)\right)\frac{{\left(x-t-\frac{k}{n}\right)}^N}{N!}dt$$

(57)

$$\le n{\int }_0^{\frac{1}{n}}{\omega }_1\left(f^{\left(N\right)},\left|\frac{k}{n}-x\right|+\left|t\right|\right)\frac{{\left(\left|x-\frac{k}{n}\right|+\left|t\right|\right)}^N}{N!}dt$$

$$\le {\omega }_1\left(f^{\left(N\right)},\frac{1}{n^{\beta }}+\frac{1}{n}\right)\frac{{\left(\frac{1}{n^{\beta }}+\frac{1}{n}\right)}^N}{N!}.$$

Therefore, when $\left|\frac{k}{n}-x\right|<\frac{1}{n^{\beta }}$, then

$$\left|\lambda \left(k\right)\right|\le {\omega }_1\left(f^{\left(N\right)},\frac{1}{n^{\beta }}+\frac{1}{n}\right)\frac{{\left(\frac{1}{n^{\beta }}+\frac{1}{n}\right)}^N}{N!}.$$

(58)

Clearly, now the following holds:

$$\left|R\right|{|}_{\left|\frac{k}{n}-x\right|<\frac{1}{n^{\beta }}}\le {\omega }_1\left(f^{\left(N\right)},\frac{1}{n}+\frac{1}{n^{\beta }}\right)\frac{{\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)}^N}{N!}.$$

(59)

(II) Let $\left|\frac{k}{n}-x\right|\ge \frac{1}{n^{\beta }}$.

(i) if $t+\frac{k}{n}\ge x$, then

$$\left|\lambda \left(k\right)\right|\le 2{∥f^{\left(N\right)}∥}_{\infty }n{\int }_0^{\frac{1}{n}}\frac{{\left(t+\frac{k}{n}-x\right)}^N}{N!}dt$$

(60)

$$\le 2{∥f^{\left(N\right)}∥}_{\infty }n{\int }_0^{\frac{1}{n}}\frac{{\left(\left|\frac{k}{n}-x\right|+\left|t\right|\right)}^N}{N!}dt$$

$$\le 2{∥f^{\left(N\right)}∥}_{\infty }\frac{{\left(\left|\frac{k}{n}-x\right|+\frac{1}{n}\right)}^N}{N!}.$$

(ii) if $t+\frac{k}{n}<x$, then

$$\left|\lambda \left(k\right)\right|\le n{\int }_0^{\frac{1}{n}}\left({\int }_{t+\frac{k}{n}}^x\left|f^{\left(N\right)}\left(x\right)-f^{\left(N\right)}\left(s\right)\right|\frac{{\left(s-\left(t+\frac{k}{n}\right)\right)}^{N-1}}{\left(N-1\right)!}ds\right)dt$$

$$\le 2{∥f^{\left(N\right)}∥}_{\infty }n{\int }_0^{\frac{1}{n}}\frac{{\left(x-t-\frac{k}{n}\right)}^N}{N!}dt$$

(61)

$$\le 2{∥f^{\left(N\right)}∥}_{\infty }n{\int }_0^{\frac{1}{n}}\frac{{\left(\left|\frac{k}{n}-x\right|+\left|t\right|\right)}^N}{N!}dt\le 2{∥f^{\left(N\right)}∥}_{\infty }\frac{{\left(\left|\frac{k}{n}-x\right|+\frac{1}{n}\right)}^N}{N!}.$$

Hence, when $\left|\frac{k}{n}-x\right|\ge \frac{1}{n^{\beta }}$, then

$$\left|\lambda \left(k\right)\right|\le \frac{2{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}{\left(\left|nx-k\right|+1\right)}^N$$

(62)

$$\le \frac{2^N{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}\left(1+{\left|nx-k\right|}^N\right),\forall k\in \mathbb{Z}.$$

Clearly, then

$$\left|R\right|{|}_{\left|\frac{k}{n}-x\right|\ge \frac{1}{n^{\beta }}}\le \left(\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\beta }\end{array}\right.}^{\infty }G\left(nx-k\right)\left|\lambda \left(k\right)\right|\right)\le$$

$$\frac{2^N{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}\left(\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\beta }\end{array}\right.}^{\infty }G\left(\left|nx-k\right|\right)\left(1+{\left|nx-k\right|}^N\right)\right)\le$$

$$\frac{2^N{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}\left[\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\beta }\end{array}\right.}^{\infty }G\left(\left|nx-k\right|\right)+\right.$$

$$\left.\left(\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\beta }\end{array}\right.}^{\infty }G\left(\left|nx-k\right|\right){\left|nx-k\right|}^N\right)\right]\le$$

(63)

(by [9], p. 6, Theorem 1.4)

$$\frac{2^N{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}\left[\frac{2}{e^{\mu \left(n^{1-\beta }-2\right)}}+\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right].$$

We have found that

$$\left|R\right|{|}_{\left|\frac{k}{n}-x\right|\ge \frac{1}{n^{\beta }}}\le \frac{2^N{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}\left[2e^{\mu }{e}^{-\mu \left(n^{1-\beta }-1\right)}+\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right].$$

(64)

Therefore, the following holds:

$$\left|R\right|\le \left|R\right|{|}_{\left|\frac{k}{n}-x\right|<\frac{1}{n^{\beta }}}+\left|R\right|{|}_{\left|\frac{k}{n}-x\right|\ge \frac{1}{n^{\beta }}}\le$$

$${\omega }_1\left(f^{\left(N\right)},\frac{1}{n}+\frac{1}{n^{\beta }}\right)\frac{{\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)}^N}{N!}+$$

(65)

$$\frac{2^N{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}\left[2e^{\mu }{e}^{-\mu \left(n^{1-\beta }-1\right)}+\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right].$$

Finally, we estimate

$$\left|C_n\left({\left(·-x\right)}^j\right)\left(x\right)\right|\le \sum _{k=-\infty }^{\infty }G\left(nx-k\right)\left(n{\int }_0^{\frac{1}{n}}{\left|t+\frac{k}{n}-x\right|}^{j}dt\right)\le$$

$$\sum _{k=-\infty }^{\infty }G\left(nx-k\right)\left(n{\int }_0^{\frac{1}{n}}{\left(\left|\frac{k}{n}-x\right|+\left|t\right|\right)}^{j}dt\right)\le$$

$$\sum _{k=-\infty }^{\infty }G\left(nx-k\right){\left(\left|\frac{k}{n}-x\right|+\frac{1}{n}\right)}^j=$$

$$\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|\frac{k}{n}-x\right|<\frac{1}{n^{\beta }}\end{array}\right.}^{\infty }G\left(nx-k\right){\left(\left|\frac{k}{n}-x\right|+\frac{1}{n}\right)}^j+$$

$$\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|x-\frac{k}{n}\right|\ge \frac{1}{n^{\beta }}\end{array}\right.}^{\infty }G\left(nx-k\right){\left(\left|\frac{k}{n}-x\right|+\frac{1}{n}\right)}^j\le$$

(66)

$${\left(\frac{1}{n^{\beta }}+\frac{1}{n}\right)}^j+\frac{1}{n^j}\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\beta }\end{array}\right.}^{\infty }G\left(\left|nx-k\right|\right){\left(\left|nx-k\right|+1\right)}^j\le$$

$${\left(\frac{1}{n^{\beta }}+\frac{1}{n}\right)}^j+\frac{2^{j-1}}{n^j}\left(\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\beta }\end{array}\right.}^{\infty }G\left(\left|nx-k\right|\right)\left(1+{\left|nx-k\right|}^j\right)\right)$$

(… as earlier)

$$\le {\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)}^j+\frac{2^{j-1}}{n^j}\left[2e^{\mu }{e}^{-\mu \left(n^{1-\beta }-1\right)}+\left(\frac{e^{\mu }{2}^{j+2}j!}{{\mu }^j}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right].$$

(67)

Therefore,

$$\left|C_n\left({\left(·-x\right)}^j\right)\left(x\right)\right|\le {\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)}^j+$$

$$\frac{2^{j-1}}{n^j}\left[2e^{\mu }{e}^{-\mu \left(n^{1-\beta }-1\right)}+\left(\frac{e^{\mu }{2}^{j+2}j!}{{\mu }^j}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right].$$

(68)

The theorem is proved. □

Therefore, the following theorem holds.

Theorem 6.

Here, $0<\beta <1$, $n\in \mathbb{N}:n^{1-\beta }>2$, $\mu >0,$ $N\in \mathbb{N},$ $f\in C^N\left(\mathbb{R}\right)$, with $f^{\left(i\right)}\in C_B\left(\mathbb{R}\right)$, $i=0,1,\dots ,N;$ $x\in \mathbb{R}$. Then,

(i)

$$\left|D_n\left(f,x\right)-f\left(x\right)-\sum _{j=1}^N\frac{f^{\left(j\right)}\left(x\right)}{j!}{D}_n\left({\left(·-x\right)}^j\right)\left(x\right)\right|\le$$

$${\omega }_1\left(f^{\left(N\right)},\frac{1}{n}+\frac{1}{n^{\beta }}\right)\frac{{\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)}^N}{N!}+$$

(69)

$$\frac{2^N{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}\left[2e^{\mu }{e}^{-\mu \left(n^{1-\beta }-1\right)}+\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right]=:\Psi ,$$

(ii) Assume all $f^{\left(j\right)}\left(x\right)=0$, $j=1,\dots ,N;$ then,

$$\left|D_n\left(f,x\right)-f\left(x\right)\right|\le \Psi ,$$

(70)

at high speed $n^{-\beta \left(N+1\right)},$

(iii)

$$\left|D_n\left(f,x\right)-f\left(x\right)\right|\le \sum _{j=1}^N\frac{\left|f^{\left(j\right)}\left(x\right)\right|}{j!}$$

$$\left\{{\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)}^j+\frac{2^{j-1}}{n^j}\left[2e^{\mu }{e}^{-\mu \left(n^{1-\beta }-1\right)}+\left(\frac{e^{\mu }{2}^{j+2}j!}{{\mu }^j}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right]\right\}+\Psi ,$$

(71)

(iv)

$${∥D_n\left(f\right)-f∥}_{\infty }\le \sum _{j=1}^N\frac{{∥f^{\left(j\right)}∥}_{\infty }}{j!}$$

$$\left\{{\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)}^j+\frac{2^{j-1}}{n^j}\left[2e^{\mu }{e}^{-\mu \left(n^{1-\beta }-1\right)}+\left(\frac{e^{\mu }{2}^{j+2}j!}{{\mu }^j}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right]\right\}+\Psi .$$

(72)

Proof. 

We have that

$$f\left(\frac{k}{n}+\frac{i}{nr}\right)=\sum _{j=0}^N\frac{f^{\left(j\right)}\left(x\right)}{j!}{\left(\frac{k}{n}+\frac{i}{nr}-x\right)}^j+$$

(73)

$${\int }_x^{\frac{k}{n}+\frac{i}{nr}}\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right)\frac{{\left(\frac{k}{n}+\frac{i}{nr}-t\right)}^{N-1}}{\left(N-1\right)!}dt,$$

and

$$\sum _{i=1}^r{w}_{i}f\left(\frac{k}{n}+\frac{i}{nr}\right)=\sum _{j=0}^N\frac{f^{\left(j\right)}\left(x\right)}{j!}\sum _{i=1}^r{w}_i{\left(\frac{k}{n}+\frac{i}{nr}-x\right)}^j+$$

(74)

$$\sum _{i=1}^r{w}_i{\int }_x^{\frac{k}{n}+\frac{i}{nr}}\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right)\frac{{\left(\frac{k}{n}+\frac{i}{nr}-t\right)}^{N-1}}{\left(N-1\right)!}dt.$$

Furthermore, the following holds:

$$D_n\left(f,x\right)-f\left(x\right)-\sum _{j=1}^N\frac{f^{\left(j\right)}\left(x\right)}{j!}{D}_n\left({\left(·-x\right)}^j\right)\left(x\right)=R\left(x\right),$$

(75)

where

$$R\left(x\right)=\sum _{k=-\infty }^{\infty }G\left(nx-k\right)\left(\sum _{i=1}^r{w}_i{\int }_x^{\frac{k}{n}+\frac{i}{nr}}\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right)\frac{{\left(\frac{k}{n}+\frac{i}{nr}-t\right)}^{N-1}}{\left(N-1\right)!}dt\right).$$

(76)

Use the following equation:

$$\gamma \left(k\right):=\sum _{i=1}^r{w}_i{\int }_x^{\frac{k}{n}+\frac{i}{nr}}\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right)\frac{{\left(\frac{k}{n}+\frac{i}{nr}-t\right)}^{N-1}}{\left(N-1\right)!}dt.$$

(77)

(I) Let $\left|\frac{k}{n}-x\right|<\frac{1}{n^{\beta }}$.

(i) if $\frac{k}{n}+\frac{i}{nr}\ge x$; then,

$$\left|\gamma \left(k\right)\right|\le \sum _{i=1}^r{w}_i{\omega }_1\left(f^{\left(N\right)},\left|x-\frac{k}{n}\right|+\frac{i}{nr}\right)\frac{{\left(\left|\frac{k}{n}-x\right|+\frac{i}{nr}\right)}^N}{N!}$$

$$\le {\omega }_1\left(f^{\left(N\right)},\frac{1}{n^{\beta }}+\frac{1}{n}\right)\frac{{\left(\frac{1}{n^{\beta }}+\frac{1}{n}\right)}^N}{N!}.$$

(78)

(ii) If $\frac{k}{n}+\frac{i}{nr}<x$, then

$$\left|\gamma \left(k\right)\right|\le \sum _{i=1}^r{w}_i\left|{\int }_x^{\frac{k}{n}+\frac{i}{nr}}\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right)\frac{{\left(\frac{k}{n}+\frac{i}{nr}-t\right)}^{N-1}}{\left(N-1\right)!}dt\right|$$

$$\le \sum _{i=1}^r{w}_i{\int }_{\frac{k}{n}+\frac{i}{nr}}^x\left|\left(f^{\left(N\right)}\left(x\right)-f^{\left(N\right)}\left(t\right)\right)\right|\frac{{\left(t-\left(\frac{k}{n}+\frac{i}{nr}\right)\right)}^{N-1}}{\left(N-1\right)!}dt$$

$$\le \sum _{i=1}^r{w}_i{\omega }_1\left(f^{\left(N\right)},\left|x-\frac{k}{n}\right|+\frac{i}{nr}\right)\frac{{\left(x-\left(\frac{k}{n}+\frac{i}{nr}\right)\right)}^N}{N!}$$

$$\le {\omega }_1\left(f^{\left(N\right)},\frac{1}{n^{\beta }}+\frac{1}{n}\right)\frac{{\left(\frac{1}{n^{\beta }}+\frac{1}{n}\right)}^N}{N!}.$$

(79)

Therefore, when $\left|\frac{k}{n}-x\right|<\frac{1}{n^{\beta }}$, then

$$\left|\gamma \left(k\right)\right|\le {\omega }_1\left(f^{\left(N\right)},\frac{1}{n^{\beta }}+\frac{1}{n}\right)\frac{{\left(\frac{1}{n^{\beta }}+\frac{1}{n}\right)}^N}{N!}.$$

(80)

Clearly, now the following holds:

$$\left|R\right|{|}_{\left|\frac{k}{n}-x\right|<\frac{1}{n^{\beta }}}\le {\omega }_1\left(f^{\left(N\right)},\frac{1}{n}+\frac{1}{n^{\beta }}\right)\frac{{\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)}^N}{N!}.$$

(81)

(II) Let $\left|\frac{k}{n}-x\right|\ge \frac{1}{n^{\beta }}$.

(i) if $\frac{k}{n}+\frac{i}{nr}\ge x$, then

$$\left|\gamma \left(k\right)\right|\le \sum _{i=1}^r{w}_{i}2{∥f^{\left(N\right)}∥}_{\infty }\frac{{\left(\frac{k}{n}+\frac{i}{nr}-x\right)}^N}{N!}$$

$$\le 2{∥f^{\left(N\right)}∥}_{\infty }\frac{{\left(\left|\frac{k}{n}-x\right|+\frac{1}{n}\right)}^N}{N!}.$$

(82)

(ii) if $\frac{k}{n}+\frac{i}{nr}<x$, then

$$\left|\gamma \left(k\right)\right|\le \sum _{i=1}^r{w}_i{\int }_{\frac{k}{n}+\frac{i}{nr}}^x\left|f^{\left(N\right)}\left(x\right)-f^{\left(N\right)}\left(t\right)\right|\frac{{\left(t-\left(\frac{k}{n}+\frac{i}{nr}\right)\right)}^{N-1}}{\left(N-1\right)!}dt$$

$$\le 2{∥f^{\left(N\right)}∥}_{\infty }\sum _{i=1}^r{w}_i\frac{{\left(\left|x-\frac{k}{n}\right|+\frac{i}{nr}\right)}^N}{N!}$$

(83)

$$\le 2{∥f^{\left(N\right)}∥}_{\infty }\frac{{\left(\left|x-\frac{k}{n}\right|+\frac{1}{n}\right)}^N}{N!}.$$

So, in general, we obtain

$$\left|\gamma \left(k\right)\right|\le 2{∥f^{\left(N\right)}∥}_{\infty }\frac{{\left(\left|x-\frac{k}{n}\right|+\frac{1}{n}\right)}^N}{N!}$$

$$=\frac{2{∥f^{\left(N\right)}∥}_{\infty }}{n^N}\frac{{\left(\left|nx-k\right|+1\right)}^N}{N!}\le$$

$$\frac{2^N{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}\left(1+{\left|nx-k\right|}^N\right),\forall k\in \mathbb{Z}.$$

(84)

Clearly, then

$$\left|R\left(x\right)\right|{|}_{\left|\frac{k}{n}-x\right|\ge \frac{1}{n^{\beta }}}\le \left(\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\beta }\end{array}\right.}^{\infty }G\left(nx-k\right)\left|\gamma \left(k\right)\right|\right)\le$$

(85)

$$\frac{2^N{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}\left(\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\beta }\end{array}\right.}^{\infty }G\left(\left|nx-k\right|\right)\left(1+{\left|nx-k\right|}^N\right)\right)$$

$$\le \left(\mathrm{as}\mathrm{earlier}\right)$$

$$\le \frac{2^N{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}\left[2e^{\mu }{e}^{-\mu \left(n^{1-\beta }-1\right)}+\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right].$$

(86)

Therefore, the following holds:

$$\left|R\left(x\right)\right|\le \left|R\left(x\right)\right|{|}_{\left|\frac{k}{n}-x\right|<\frac{1}{n^{\beta }}}+\left|R\left(x\right)\right|{|}_{\left|\frac{k}{n}-x\right|\ge \frac{1}{n^{\beta }}}\le$$

$${\omega }_1\left(f^{\left(N\right)},\frac{1}{n}+\frac{1}{n^{\beta }}\right)\frac{{\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)}^N}{N!}+$$

(87)

$$\frac{2^N{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}\left[2e^{\mu }{e}^{-\mu \left(n^{1-\beta }-1\right)}+\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right].$$

Next, we estimate

$$\left|D_n\left({\left(·-x\right)}^j\right)\left(x\right)\right|\le \sum _{k=-\infty }^{\infty }G\left(nx-k\right)\left(\sum _{i=1}^r{w}_i{\left(\left|\frac{k}{n}-x\right|+\frac{i}{nr}\right)}^j\right)\le$$

$$\sum _{k=-\infty }^{\infty }G\left(nx-k\right){\left(\left|\frac{k}{n}-x\right|+\frac{1}{n}\right)}^j\le \left(\mathrm{as}\mathrm{earlier}\right)$$

(88)

$$\le {\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)}^j+\frac{2^{j-1}}{n^j}\left[2e^{\mu }{e}^{-\mu \left(n^{1-\beta }-1\right)}+\left(\frac{e^{\mu }{2}^{j+2}j!}{{\mu }^j}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right].$$

The theorem is proved. □

We need the following definition.

Definition 2.

A function $f:\mathbb{R}\to \mathbb{R}$ is absolutely continuous over $\mathbb{R}$, if ${f|}_{\left[a,b\right]}$ is absolutely continuous, for every $\left[a,b\right]\subset \mathbb{R}$. We can write $f\in A{C}^n\left(\mathbb{R}\right)$, if $f^{\left(n-1\right)}\in AC\left(R\right)$ (absolutely continuous functions over $\mathbb{R}$) and $n\in \mathbb{N}$.

Definition 3.

Let $\nu \ge 0$, $n=⌈\nu ⌉$ ($⌈·⌉$ is the ceiling of the number); $f\in A{C}^n\left(\mathbb{R}\right)$. We use the left Caputo fractional derivative ([12,13,14], pp. 49–52) as the following function:

$$D_{\ast a}^{\nu }f\left(x\right)=\frac{1}{\Gamma \left(n-\nu \right)}{\int }_a^x{\left(x-t\right)}^{n-\nu -1}{f}^{\left(n\right)}\left(t\right)dt,$$

(89)

$\mathrm{\forall }x\in [a,\infty )$, $a\in \mathbb{R}$; where Γ is the gamma function.

Note that $D_{\ast a}^{\nu }f\in L_1\left(\left[a,b\right]\right)$ and $D_{\ast a}^{\nu }f$ exists, i.e., on $\left[a,b\right]$, ∀$\left[a,b\right]\subset \mathbb{R}$.

We set $D_{\ast a}^{0}f\left(x\right)=f\left(x\right)$, ∀$x\in [a,\infty ).$

Lemma 1

(See also [15]). Let $\nu >0$, $\nu \notin \mathbb{N}$, $n=⌈\nu ⌉$, $f\in C^{n-1}\left(\mathbb{R}\right)$ and $f^{\left(n\right)}\in L_{\infty }\left(\mathbb{R}\right)$. Then, $D_{\ast a}^{\nu }f\left(a\right)=0$ for any $a\in \mathbb{R}$.

Definition 4

(See also [13,16,17]). Let $f\in A{C}^m\left(\mathbb{R}\right)$, $m=⌈\alpha ⌉$, $\alpha >0$. The right Caputo fractional derivative of order $\alpha >0$ is given by

$$D_{b-}^{\alpha }f\left(x\right)=\frac{{\left(-1\right)}^m}{\Gamma \left(m-\alpha \right)}{\int }_x^b{\left(z-x\right)}^{m-\alpha -1}{f}^{\left(m\right)}\left(z\right)dz,$$

(90)

$\mathrm{\forall }x\in (-\infty ,b]$, $b\in \mathbb{R}$. We set $D_{b-}^{0}f\left(x\right)=f\left(x\right)$.

Note that $D_{b-}^{\alpha }f\in L_1\left(\left[a,b\right]\right)$ and $D_{b-}^{\alpha }f$ exists a.e. on $\left[a,b\right]$, ∀$\left[a,b\right]\subset \mathbb{R}$.

Lemma 2

(see also [15]). Let $f\in C^{m-1}\left(\mathbb{R}\right)$, $f^{\left(m\right)}\in L_{\infty }\left(\mathbb{R}\right)$, $m=⌈\alpha ⌉,$ $\alpha >0$. Then, $D_{b-}^{\alpha }f\left(b\right)=0,$ for any $b\in \mathbb{R}$.

Convention 1.

We assume that

$$\begin{array}{c}{D}_{\ast x_0}^{\alpha }f\left(x\right)=0,\mathit{for}x<x_0,\hfill \\ \mathit{and}\hfill \\ D_{x_0-}^{\alpha }f\left(x\right)=0,\mathit{for}x>x_0.\hfill \end{array}$$

(91)

Proposition 2

(See also [15]). Let $f\in C^n\left(\mathbb{R}\right)$, $n=⌈\nu ⌉$, $\nu >0$. Then $D_{\ast a}^{\nu }f\left(x\right)$ is continuous in $x\in [a,\infty )$, $a\in \mathbb{R}$.

Also we have

Proposition 3

(see also [15]). Let $f\in C^m\left(\mathbb{R}\right)$, $m=⌈\alpha ⌉$, $\alpha >0$. Then, $D_{b-}^{\alpha }f\left(x\right)$ is continuous in $x\in (-\infty ,b]$, $b\in \mathbb{R}$.

We further mention

Proposition 4

(see also [15]). Let $f\in C^{m-1}\left(\mathbb{R}\right)$, $f^{\left(m\right)}\in L_{\infty }\left(\mathbb{R}\right)$, $m=⌈\alpha ⌉$, $\alpha >0$ and let $x,x_0\in \mathbb{R}:x\ge x_0$. Then, $D_{\ast x_0}^{\alpha }f\left(x\right)$ is continuous in $x_0.$

Proposition 5

(See also [15]). Let $f\in C^{m-1}\left(\mathbb{R}\right)$, $f^{\left(m\right)}\in L_{\infty }\left(\mathbb{R}\right)$, $m=⌈\alpha ⌉$, $\alpha >0$ and let $x,x_0\in \mathbb{R}:x\le x_0$. Then, $D_{x_0-}^{\alpha }f\left(x\right)$ is continuous in $x_0.$

Proposition 6

(See also [15]). Let $f\in C^m\left(\mathbb{R}\right)$, $m=⌈\alpha ⌉$, $\alpha >0$; $x,x_0\in \mathbb{R}$. Then, $D_{\ast x_0}^{\alpha }f\left(x\right),$ $D_{x_0-}^{\alpha }f\left(x\right)$ are jointly continuous functions in $\left(x,x_0\right)$ from $R^2\to \mathbb{R}.$

Fractional results follow.

Theorem 7.

Let $\alpha >0$, $N=⌈\alpha ⌉$, $\alpha \notin \mathbb{N}$, $\mu >0$, $f\in A{C}^N\left(\mathbb{R}\right)$, $f^{\left(N\right)}\in L_{\infty }\left(\mathbb{R}\right)$, $0<\beta <1$, $x\in \mathbb{R}$, $n\in \mathbb{N}:n^{1-\beta }\ge 3$. Assume also that both ${\displaystyle \underset{x\in \mathbb{R}}{sup}}{∥D_{\ast x}^{\alpha }f∥}_{\infty },$ ${\displaystyle \underset{x\in \mathbb{R}}{sup}}{∥D_{x-}^{\alpha }f∥}_{\infty }<\infty$. Then,

(I)

$$\left|B_n\left(f,x\right)-f\left(x\right)-\sum _{j=1}^{N-1}\frac{f^{\left(j\right)}\left(x\right)}{j!}{B}_n\left({\left(·-x\right)}^j\right)\left(x\right)\right|\le$$

$$\frac{1}{n^{\alpha \beta }\Gamma \left(\alpha +1\right)}\left[{\omega }_1{\left(D_{x-}^{\alpha }f,\frac{1}{n^{\beta }}\right)}_{(-\infty ,x]}+{\omega }_1{\left(D_{\ast x}^{\alpha }f,\frac{1}{n^{\beta }}\right)}_{[x,\infty )}\right]+$$

$$\frac{1}{n^{\alpha }\Gamma \left(\alpha +1\right)}\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\left[{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}+{∥D_{\ast x}^{\alpha }f∥}_{\infty ,[x,\infty )}\right]=:\theta ,$$

(92)

(II) given $f^{\left(j\right)}\left(x\right)=0$, $j=1,\dots ,N-1$, we have

$$\left|B_n\left(f,x\right)-f\left(x\right)\right|\le \theta ,$$

(93)

(III)

$$\left|B_n\left(f,x\right)-f\left(x\right)\right|\le \sum _{j=1}^{N-1}\frac{\left|f^{\left(j\right)}\left(x\right)\right|}{j!}$$

$$\left\{\frac{1}{n^{\beta j}}+\frac{1}{n^j}\left[\frac{e^{\mu }{2}^{j+2}j!}{{\mu }^j}\right]e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right\}+\theta ,$$

(94)

and

(IV) Adding ${∥f^{\left(j\right)}∥}_{\infty }<\infty$, $j=1,\dots ,N-1,$ we obtain

$${∥B_n\left(f\right)-f∥}_{\infty }\le \sum _{j=1}^{N-1}\frac{{∥f^{\left(j\right)}∥}_{\infty }}{j!}\left\{\frac{1}{n^{\beta j}}+\frac{1}{n^j}\left[\frac{e^{\mu }{2}^{j+2}j!}{{\mu }^j}\right]e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right\}+$$

$$\frac{1}{n^{\alpha \beta }\Gamma \left(\alpha +1\right)}\left[\underset{x\in \mathbb{R}}{sup}{\omega }_1{\left(D_{x-}^{\alpha }f,\frac{1}{n^{\beta }}\right)}_{(-\infty ,x]}+\underset{x\in \mathbb{R}}{sup}{\omega }_1{\left(D_{\ast x}^{\alpha }f,\frac{1}{n^{\beta }}\right)}_{[x,\infty )}\right]+$$

$$\frac{1}{n^{\alpha }\Gamma \left(\alpha +1\right)}\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}$$

$$\left[\underset{x\in \mathbb{R}}{sup}{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}+\underset{x\in \mathbb{R}}{sup}{∥D_{\ast x}^{\alpha }f∥}_{\infty ,[x,\infty )}\right].$$

(95)

As shown above, when $N=1$, the sum ${\displaystyle \sum _{j=1}^{N-1}}·=0.$

As we can see here, we obtain fractional pointwise and uniform convergence with rates of $B_n\to I$, with the unit operator as $n\to \infty .$

Proof. 

Let $x\in \mathbb{R}$. We have that $D_{x-}^{\alpha }f\left(x\right)=D_{\ast x}^{\alpha }f\left(x\right)=0$.

From [12], p. 54, we can derive the left Caputo fractional Taylor’s formula:

$$f\left(\frac{k}{n}\right)=\sum _{j=0}^{N-1}\frac{f^{\left(j\right)}\left(x\right)}{j!}{\left(\frac{k}{n}-x\right)}^j+$$

(96)

$$\frac{1}{\Gamma \left(\alpha \right)}{\int }_x^{\frac{k}{n}}{\left(\frac{k}{n}-J\right)}^{\alpha -1}\left(D_{\ast x}^{\alpha }f\left(J\right)-D_{\ast x}^{\alpha }f\left(x\right)\right)dJ,$$

for all $x\le \frac{k}{n}<\infty .$

Also, from [16], using the right Caputo fractional Taylor’s formula, we can obtain the following:

$$f\left(\frac{k}{n}\right)=\sum _{j=0}^{N-1}\frac{f^{\left(j\right)}\left(x\right)}{j!}{\left(\frac{k}{n}-x\right)}^j+$$

(97)

$$\frac{1}{\Gamma \left(\alpha \right)}{\int }_{\frac{k}{n}}^x{\left(J-\frac{k}{n}\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\left(J\right)-D_{x-}^{\alpha }f\left(x\right)\right)dJ,$$

for all $-\infty <\frac{k}{n}\le x.$

Hence, the following holds:

$$f\left(\frac{k}{n}\right)G\left(nx-k\right)=\sum _{j=0}^{N-1}\frac{f^{\left(j\right)}\left(x\right)}{j!}G\left(nx-k\right){\left(\frac{k}{n}-x\right)}^j+$$

(98)

$$\frac{G\left(nx-k\right)}{\Gamma \left(\alpha \right)}{\int }_x^{\frac{k}{n}}{\left(\frac{k}{n}-J\right)}^{\alpha -1}\left(D_{\ast x}^{\alpha }f\left(J\right)-D_{\ast x}^{\alpha }f\left(x\right)\right)dJ,$$

for all $x\le \frac{k}{n}<\infty ,$ iff $⌈nx⌉\le k<\infty ,$

and

$$f\left(\frac{k}{n}\right)G\left(nx-k\right)=\sum _{j=0}^{N-1}\frac{f^{\left(j\right)}\left(x\right)}{j!}G\left(nx-k\right){\left(\frac{k}{n}-x\right)}^j+$$

(99)

$$\frac{G\left(nx-k\right)}{\Gamma \left(\alpha \right)}{\int }_{\frac{k}{n}}^x{\left(J-\frac{k}{n}\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\left(J\right)-D_{x-}^{\alpha }f\left(x\right)\right)dJ,$$

for all $-\infty <\frac{k}{n}\le x$, iff $-\infty <k\le ⌊nx⌋.$

We have that $⌈nx⌉\le ⌊nx⌋+1.$

Therefore, the following holds:

$$\sum _{k=⌊nx⌋+1}^{\infty }f\left(\frac{k}{n}\right)G\left(nx-k\right)=\sum _{j=0}^{N-1}\frac{f^{\left(j\right)}\left(x\right)}{j!}\sum _{k=⌊nx⌋+1}^{\infty }G\left(nx-k\right){\left(\frac{k}{n}-x\right)}^j+$$

(100)

$$\frac{{\displaystyle \sum _{k=⌊nx⌋+1}^{\infty }}G\left(nx-k\right)}{\Gamma \left(\alpha \right)}{\int }_x^{\frac{k}{n}}{\left(\frac{k}{n}-J\right)}^{\alpha -1}\left(D_{\ast x}^{\alpha }f\left(J\right)-D_{\ast x}^{\alpha }f\left(x\right)\right)dJ,$$

and

$$\sum _{k=-\infty }^{⌊nx⌋}f\left(\frac{k}{n}\right)G\left(nx-k\right)=\sum _{j=0}^{N-1}\frac{f^{\left(j\right)}\left(x\right)}{j!}\sum _{k=-\infty }^{⌊nx⌋}G\left(nx-k\right){\left(\frac{k}{n}-x\right)}^j+$$

(101)

$$\frac{{\displaystyle \sum _{k=-\infty }^{⌊nx⌋}}G\left(nx-k\right)}{\Gamma \left(\alpha \right)}{\int }_{\frac{k}{n}}^x{\left(J-\frac{k}{n}\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\left(J\right)-D_{x-}^{\alpha }f\left(x\right)\right)dJ.$$

Adding the last two equalities (100), (101), we can obtain:

$$B_n\left(f,x\right)-f\left(x\right)-\sum _{j=1}^{N-1}\frac{f^{\left(j\right)}\left(x\right)}{j!}{B}_n\left({\left(·-x\right)}^j\right)\left(x\right)=R_n\left(x\right),$$

(102)

where

$$R_n\left(x\right):=R_{1n}\left(x\right)+R_{2n}\left(x\right),$$

(103)

with

$$R_{1n}\left(x\right):=\frac{{\displaystyle \sum _{k=⌊nx⌋+1}^{\infty }}G\left(nx-k\right)}{\Gamma \left(\alpha \right)}{\int }_x^{\frac{k}{n}}{\left(\frac{k}{n}-J\right)}^{\alpha -1}\left(D_{\ast x}^{\alpha }f\left(J\right)-D_{\ast x}^{\alpha }f\left(x\right)\right)dJ,$$

(104)

and

$$R_{2n}\left(x\right):=\frac{{\displaystyle \sum _{k=-\infty }^{⌊nx⌋}}G\left(nx-k\right)}{\Gamma \left(\alpha \right)}{\int }_{\frac{k}{n}}^x{\left(J-\frac{k}{n}\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\left(J\right)-D_{x-}^{\alpha }f\left(x\right)\right)dJ.$$

(105)

Furthermore, let

$${\delta }_{1n}\left(x\right):=\frac{1}{\Gamma \left(\alpha \right)}{\int }_x^{\frac{k}{n}}{\left(\frac{k}{n}-J\right)}^{\alpha -1}\left(D_{\ast x}^{\alpha }f\left(J\right)-D_{\ast x}^{\alpha }f\left(x\right)\right)dJ,$$

(106)

for $k=⌊nx⌋+1,\dots ,\infty ,$

and

$${\delta }_{2n}\left(x\right):=\frac{1}{\Gamma \left(\alpha \right)}{\int }_{\frac{k}{n}}^x{\left(J-\frac{k}{n}\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\left(J\right)-D_{x-}^{\alpha }f\left(x\right)\right)dJ,$$

(107)

for $k=-\infty ,\dots ,⌊nx⌋.$

Let $\left|\frac{k}{n}-x\right|<\frac{1}{n^{\beta }}$; we derive the following:

$$\left|{\delta }_{1n}\left(x\right)\right|\le {\omega }_1{\left(D_{\ast x}^{\alpha }f,\frac{1}{n^{\beta }}\right)}_{[x,\infty )}\frac{1}{n^{\alpha \beta }\Gamma \left(\alpha +1\right)},$$

(108)

$k=⌊nx⌋+1,\dots ,\infty ,$

and

$$\left|{\delta }_{2n}\left(x\right)\right|\le {\omega }_1{\left(D_{x-}^{\alpha }f,\frac{1}{n^{\beta }}\right)}_{(-\infty ,x]}\frac{1}{n^{\alpha \beta }\Gamma \left(\alpha +1\right)},$$

(109)

$k=-\infty ,\dots ,⌊nx⌋.$

Also, we obtain the following:

$$\left|{\delta }_{1n}\left(x\right)\right|\le {∥D_{\ast x}^{\alpha }f∥}_{\infty ,[x,\infty )}\frac{{\left(\frac{k}{n}-x\right)}^{\alpha }}{\Gamma \left(\alpha +1\right)},k=⌊nx⌋+1,\dots ,\infty ,$$

(110)

and

$$\left|{\delta }_{2n}\left(x\right)\right|\le {∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}\frac{{\left(x-\frac{k}{n}\right)}^{\alpha }}{\Gamma \left(\alpha +1\right)},k=-\infty ,\dots ,⌊nx⌋.$$

(111)

Therefore, the following holds:

$$\left|R_{1n}\left(x\right)\right|{|}_{\left|x-\frac{k}{n}\right|<\frac{1}{n^{\beta }}}\le \frac{{\omega }_1{\left(D_{\ast x}^{\alpha }f,\frac{1}{n^{\beta }}\right)}_{[x,\infty )}}{n^{\alpha \beta }\Gamma \left(\alpha +1\right)},$$

(112)

$k=⌊nx⌋+1,\dots ,\infty ,$

and

$$\left|R_{2n}\left(x\right)\right|{|}_{\left|x-\frac{k}{n}\right|<\frac{1}{n^{\beta }}}\le \frac{{\omega }_1{\left(D_{x-}^{\alpha }f,\frac{1}{n^{\beta }}\right)}_{(-\infty ,x]}}{n^{\alpha \beta }\Gamma \left(\alpha +1\right)},$$

(113)

$k=-\infty ,\dots ,⌊nx⌋.$

Next, we estimate

$$\left|R_{1n}\left(x\right)\right|{|}_{\left|x-\frac{k}{n}\right|\ge \frac{1}{n^{\beta }}}\le$$

$$\frac{{∥D_{\ast x}^{\alpha }f∥}_{\infty ,[x,\infty )}}{\Gamma \left(\alpha +1\right)}\sum _{\left\{\begin{array}{c}k=⌊nx⌋+1\\ :\left|x-\frac{k}{n}\right|\ge \frac{1}{n^{\beta }}\end{array}\right.}^{\infty }G\left(nx-k\right){\left(\frac{k}{n}-x\right)}^{\alpha }\le$$

(by $N=⌈\alpha ⌉$)

$$\frac{{∥D_{\ast x}^{\alpha }f∥}_{\infty ,[x,\infty )}}{n^{\alpha }\Gamma \left(\alpha +1\right)}\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\beta }\end{array}\right.}^{\infty }G\left(\left|nx-k\right|\right){\left|nx-k\right|}^N\le$$

(114)

$$\frac{{∥D_{\ast x}^{\alpha }f∥}_{\infty ,[x,\infty )}}{n^{\alpha }\Gamma \left(\alpha +1\right)}\mu e^{\mu }\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\beta }\end{array}\right.}^{\infty }{e}^{-\mu \left|nx-k\right|}{\left|nx-k\right|}^N\le \left(\mathrm{as}\mathrm{earlier}\right)$$

$$\le \left(\frac{{∥D_{\ast x}^{\alpha }f∥}_{\infty ,[x,\infty )}}{n^{\alpha }\Gamma \left(\alpha +1\right)}\right)\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}.$$

(115)

Hence, the following holds:

$$\left|R_{1n}\left(x\right)\right|{|}_{\left|x-\frac{k}{n}\right|\ge \frac{1}{n^{\beta }}}\le$$

(116)

$$\left(\frac{{∥D_{\ast x}^{\alpha }f∥}_{\infty ,[x,\infty )}}{n^{\alpha }\Gamma \left(\alpha +1\right)}\right)\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}.$$

Furthermore, we can obtain:

$$\left|R_{2n}\left(x\right)\right|{|}_{\left|x-\frac{k}{n}\right|\ge \frac{1}{n^{\beta }}}\le$$

$$\frac{{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x])}}{\Gamma \left(\alpha +1\right)}\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|x-\frac{k}{n}\right|\ge \frac{1}{n^{\beta }}\end{array}\right.}^{⌊nx⌋}G\left(nx-k\right){\left(x-\frac{k}{n}\right)}^{\alpha }\le$$

$$\frac{{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x])}}{n^{\alpha }\Gamma \left(\alpha +1\right)}\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\beta }\end{array}\right.}^{\infty }G\left(\left|nx-k\right|\right){\left|nx-k\right|}^{\alpha }\le$$

$$\left(\frac{{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x])}}{n^{\alpha }\Gamma \left(\alpha +1\right)}\right)\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\beta }\end{array}\right.}^{\infty }G\left(\left|nx-k\right|\right){\left|nx-k\right|}^N\le \left(\mathrm{as}\mathrm{earlier}\right)$$

(117)

$$\le \left(\frac{{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x])}}{n^{\alpha }\Gamma \left(\alpha +1\right)}\right)\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}.$$

That is

$$\left|R_{2n}\left(x\right)\right|{|}_{\left|x-\frac{k}{n}\right|\ge \frac{1}{n^{\beta }}}\le$$

$$\left(\frac{{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x])}}{n^{\alpha }\Gamma \left(\alpha +1\right)}\right)\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}.$$

(118)

We have proved that

$$\left|R_n\left(x\right)\right|\le \frac{1}{n^{\alpha \beta }\Gamma \left(\alpha +1\right)}\left[{\omega }_1{\left(D_{x-}^{\alpha }f,\frac{1}{n^{\beta }}\right)}_{(-\infty ,x]}+{\omega }_1{\left(D_{\ast x}^{\alpha }f,\frac{1}{n^{\beta }}\right)}_{[x,\infty )}\right]+$$

(119)

$$\frac{1}{n^{\alpha }\Gamma \left(\alpha +1\right)}\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\left[{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}+{∥D_{\ast x}^{\alpha }f∥}_{\infty ,[x,\infty )}\right].$$

As was shown earlier, we can obtain that

$$\left|B_n\left({\left(·-x\right)}^j\right)\left(x\right)\right|\le \frac{1}{n^{\beta j}}+\frac{1}{n^j}\left\{\frac{e^{\mu }{2}^{j+2}j!}{{\mu }^j}\right\}{e}^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}.$$

(120)

We have that

$$\left(D_{\ast x}^{\alpha }f\right)\left(s\right)=\frac{1}{\Gamma \left(N-\alpha \right)}{\int }_x^s{\left(s-t\right)}^{N-\alpha -1}{f}^{\left(N\right)}\left(t\right)dt,(s\ge x)$$

(121)

and

$$\left(D_{x-}^{\alpha }f\right)\left(s\right)=\frac{{\left(-1\right)}^N}{\Gamma \left(N-\alpha \right)}{\int }_s^x{\left(t-s\right)}^{N-\alpha -1}{f}^{\left(N\right)}\left(t\right)dt,(s\le x)$$

(122)

$\mathrm{\forall }x,s\in \mathbb{R}$.

Therefore, the following holds:

$$\left|\left(D_{\ast x}^{\alpha }f\right)\left(s\right)\right|\le \frac{1}{\Gamma \left(N-\alpha \right)}{∥f^{\left(N\right)}∥}_{\infty }\frac{{\left(s-x\right)}^{N-\alpha }}{N-\alpha }=\frac{{∥f^{\left(N\right)}∥}_{\infty }{\left(s-x\right)}^{N-\alpha }}{\Gamma \left(N-\alpha +1\right)},$$

(123)

$s\ge x,$

and

$$\left|\left(D_{x-}^{\alpha }f\right)\left(s\right)\right|\le \frac{{∥f^{\left(N\right)}∥}_{\infty }}{\Gamma \left(N-\alpha \right)}\frac{{\left(x-s\right)}^{N-\alpha }}{N-\alpha }=\frac{{∥f^{\left(N\right)}∥}_{\infty }}{\Gamma \left(N-\alpha +1\right)}{\left(x-s\right)}^{N-\alpha },$$

(124)

$s\le x.$

Thus, it is reasonable to assume that ${\displaystyle \underset{x\in \mathbb{R}}{sup}}{∥D_{\ast x}^{\alpha }f∥}_{\infty },$ ${\displaystyle \underset{x\in \mathbb{R}}{sup}}{∥D_{x-}^{\alpha }f∥}_{\infty }<\infty$.

Consequently, the following holds:

$$\underset{x\in \mathbb{R}}{sup}{\omega }_1{\left(D_{x-}^{\alpha }f,\frac{1}{n^{\beta }}\right)}_{(-\infty ,x]},\underset{x\in \mathbb{R}}{sup}{\omega }_1{\left(D_{\ast x}^{\alpha }f,\frac{1}{n^{\beta }}\right)}_{[x,\infty )}<\infty .$$

(125)

The theorem is now proved. □

4. Applications for $\mathit{N}=\mathbf{1}$

We obtain the following results:

Corollary 1.

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

(I)

$$\left|B_n\left(f,x\right)-f\left(x\right)-f^{\prime }\left(x\right)B_n\left(\left(.-x\right)\right)\left(x\right)\right|\le$$

(126)

$$\frac{{\omega }_1\left(f^{\prime },\frac{1}{n^{\beta }}\right)}{n^{\beta }}+\left(\frac{16{∥f^{\prime }∥}_{\infty }{e}^{\mu }}{n\mu }\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}=:M_1,$$

(II) assume $f^{\prime }\left(x\right)=0$; we have that

$$\left|B_n\left(f,x\right)-f\left(x\right)\right|\le M_1,$$

(127)

at high speed $n^{-2\beta },$

(III)

$$\left|B_n\left(f,x\right)-f\left(x\right)\right|\le \left|f^{\prime }\left(x\right)\right|\left[\frac{1}{n^{\beta }}+\frac{1}{n}\left\{\frac{8e^{\mu }}{\mu }\right\}{e}^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right]+M_1,$$

(128)

(IV)

$${∥B_n\left(f\right)-f∥}_{\infty }\le {∥f^{\prime }∥}_{\infty }\left[\frac{1}{n^{\beta }}+\frac{1}{n}\left(\frac{8e^{\mu }}{\mu }\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right]+M_1.$$

(129)

Proof. 

Use Theorem 4 for $N=1.$ □

Corollary 2.

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

(I)

$$\left\{\begin{array}{c}\left|C_n\left(f,x\right)-f\left(x\right)-f^{\prime }\left(x\right)C_n\left(\left(·-x\right)\right)\left(x\right)\right|,\\ \left|D_n\left(f,x\right)-f\left(x\right)-f^{\prime }\left(x\right)D_n\left(\left(·-x\right)\right)\left(x\right)\right|\end{array}\right\}\le$$

$${\omega }_1\left(f^{\prime },\frac{1}{n}+\frac{1}{n^{\beta }}\right)\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)+$$

(130)

$$\frac{2{∥f^{\prime }∥}_{\infty }}{n}\left[2e^{\mu }{e}^{-\mu \left(n^{1-\beta }-1\right)}+\left(\frac{8e^{\mu }}{\mu }\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right]=:{\psi }_1,$$

(II) assume $f^{\prime }\left(x\right)=0$; we have that

$$\left\{\begin{array}{c}\left|C_n\left(f,x\right)-f\left(x\right)\right|,\\ \left|D_n\left(f,x\right)-f\left(x\right)\right|\end{array}\right\}\le {\psi }_1,$$

(131)

at high speed $n^{-2\beta },$

(III)

$$\left\{\begin{array}{c}\left|C_n\left(f,x\right)-f\left(x\right)\right|,\\ \left|D_n\left(f,x\right)-f\left(x\right)\right|\end{array}\right\}\le$$

$$\left|f^{\prime }\left(x\right)\right|\left\{\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)+\frac{1}{n}\left[2e^{\mu }{e}^{-\mu \left(n^{1-\beta }-1\right)}+\left(\frac{8e^{\mu }}{\mu }\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right]\right\}+{\psi }_1,$$

(132)

and

(IV)

$$\left\{\begin{array}{c}{∥C_n\left(f\right)-f∥}_{\infty },\\ {∥D_n\left(f\right)-f∥}_{\infty }\end{array}\right\}\le$$

$${∥f^{\prime }∥}_{\infty }\left\{\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)+\frac{1}{n}\left[2e^{\mu }{e}^{-\mu \left(n^{1-\beta }-1\right)}+\left(\frac{8e^{\mu }}{\mu }\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right]\right\}+{\psi }_1.$$

(133)

Proof. 

Use Theorems 5 and 6 for $N=1.$ □

Corollary 3.

Let $0<\alpha <1$, $\mu >0$, $f\in A{C}^1\left(\mathbb{R}\right)$, $f^{\prime }\in L_{\infty }\left(\mathbb{R}\right)$, $0<\beta <1$, $x\in \mathbb{R}$, $n\in \mathbb{N}:n^{1-\beta }\ge 3$. Assume also that ${\displaystyle \underset{x\in \mathbb{R}}{sup}}{∥D_{\ast x}^{\alpha }f∥}_{\infty },$ ${\displaystyle \underset{x\in \mathbb{R}}{sup}}{∥D_{x-}^{\alpha }f∥}_{\infty }<\infty$. Then,

(I)

$$\left|B_n\left(f,x\right)-f\left(x\right)\right|\le$$

$$\frac{1}{n^{\alpha \beta }\Gamma \left(\alpha +1\right)}\left[{\omega }_1{\left(D_{x-}^{\alpha }f,\frac{1}{n^{\beta }}\right)}_{(-\infty ,x]}+{\omega }_1{\left(D_{\ast x}^{\alpha }f,\frac{1}{n^{\beta }}\right)}_{[x,\infty )}\right]+$$

$$\frac{1}{n^{\alpha }\Gamma \left(\alpha +1\right)}\left(\frac{8e^{\mu }}{\mu }\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\left[{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}+{∥D_{\ast x}^{\alpha }f∥}_{\infty ,[x,\infty )}\right],$$

(134)

and

(II)

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

$$\frac{1}{n^{\alpha \beta }\Gamma \left(\alpha +1\right)}\left[\underset{x\in \mathbb{R}}{sup}{\omega }_1{\left(D_{x-}^{\alpha }f,\frac{1}{n^{\beta }}\right)}_{(-\infty ,x]}+\underset{x\in \mathbb{R}}{sup}{\omega }_1{\left(D_{\ast x}^{\alpha }f,\frac{1}{n^{\beta }}\right)}_{[x,\infty )}\right]+$$

$$\frac{1}{n^{\alpha }\Gamma \left(\alpha +1\right)}\left(\frac{8e^{\mu }}{\mu }\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\left[\underset{x\in \mathbb{R}}{sup}{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}+\underset{x\in \mathbb{R}}{sup}{∥D_{\ast x}^{\alpha }f∥}_{\infty ,[x,\infty )}\right].$$

(135)

Proof. 

Use Theorem 7 for $N=1.$ □

Next is the case of $\alpha =\frac{1}{2}.$

Corollary 4.

Let $\mu >0$, $f\in A{C}^1\left(\mathbb{R}\right)$, $f^{\prime }\in L_{\infty }\left(\mathbb{R}\right)$, $0<\beta <1$, $x\in \mathbb{R}$, $n\in \mathbb{N}:n^{1-\beta }\ge 3$. Assume that ${\displaystyle \underset{x\in \mathbb{R}}{sup}}{∥D_{\ast x}^{\frac{1}{2}}f∥}_{\infty },$ ${\displaystyle \underset{x\in \mathbb{R}}{sup}}{∥D_{x-}^{\frac{1}{2}}f∥}_{\infty }<\infty$. Then,

(I)

$$\left|B_n\left(f,x\right)-f\left(x\right)\right|\le$$

$$\frac{2}{n^{\frac{\beta }{2}}\sqrt{\pi }}\left[{\omega }_1{\left(D_{x-}^{\frac{1}{2}}f,\frac{1}{n^{\beta }}\right)}_{(-\infty ,x]}+{\omega }_1{\left(D_{\ast x}^{\frac{1}{2}}f,\frac{1}{n^{\beta }}\right)}_{[x,\infty )}\right]+$$

$$\frac{1}{\sqrt{\pi n}}\left(\frac{16e^{\mu }}{\mu }\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\left[{∥D_{x-}^{\frac{1}{2}}f∥}_{\infty ,(-\infty ,x]}+{∥D_{\ast x}^{\frac{1}{2}}f∥}_{\infty ,[x,\infty )}\right],$$

(136)

and

(II)

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

$$\frac{2}{n^{\frac{\beta }{2}}\sqrt{\pi }}\left[\underset{x\in \mathbb{R}}{sup}{\omega }_1{\left(D_{x-}^{\frac{1}{2}}f,\frac{1}{n^{\beta }}\right)}_{(-\infty ,x]}+\underset{x\in \mathbb{R}}{sup}{\omega }_1{\left(D_{\ast x}^{\frac{1}{2}}f,\frac{1}{n^{\beta }}\right)}_{[x,\infty )}\right]+$$

$$\frac{1}{\sqrt{\pi n}}\left(\frac{16e^{\mu }}{\mu }\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\left[\underset{x\in \mathbb{R}}{sup}{∥D_{x-}^{\frac{1}{2}}f∥}_{\infty ,(-\infty ,x]}+\underset{x\in \mathbb{R}}{sup}{∥D_{\ast x}^{\frac{1}{2}}f∥}_{\infty ,[x,\infty )}\right].$$

(137)

Proof. 

Use Corollary 3. □

5. Complex Neural Network Approximation

Remark 3.

Let $f:\mathbb{R}\to \mathbb{C}$, with real and imaginary parts $f_1,f_2:f=f_1+i{f}_2$, $i=\sqrt{-1}$. Clearly, f is continuous if $f_1$ and $f_2$ are continuous.

Also,

$$f^{\left(j\right)}\left(x\right)=f_1^{\left(j\right)}\left(x\right)+i{f}_2^{\left(j\right)}\left(x\right),$$

(138)

holds for all $j=1,\dots ,N$, given that $f_1,f_2\in C^N\left(\mathbb{R}\right)$, $N\in \mathbb{N}$.

Here, the following are defined:

$$\begin{array}{c}{B}_n\left(f,x\right):=B_n\left(f_1,x\right)+i{B}_n\left(f_2,x\right);\\ C_n\left(f,x\right):=C_n\left(f_1,x\right)+i{C}_n\left(f_2,x\right);\\ D_n\left(f,x\right):=D_n\left(f_1,x\right)+i{D}_n\left(f_2,x\right).\end{array}$$

(139)

We observe here that

$$\left|B_n\left(f,x\right)-f\left(x\right)\right|\le \left|B_n\left(f_1,x\right)-f_1\left(x\right)\right|+\left|B_n\left(f_2,x\right)-f_2\left(x\right)\right|,$$

(140)

and

$${∥B_n\left(f\right)-f∥}_{\infty }\le {∥B_n\left(f_1\right)-f_1∥}_{\infty }+{∥B_n\left(f_2\right)-f_2∥}_{\infty };$$

(141)

$$\left|C_n\left(f,x\right)-f\left(x\right)\right|\le \left|C_n\left(f_1,x\right)-f_1\left(x\right)\right|+\left|C_n\left(f_2,x\right)-f_2\left(x\right)\right|,$$

(142)

$${∥C_n\left(f\right)-f∥}_{\infty }\le {∥C_n\left(f_1\right)-f_1∥}_{\infty }+{∥C_n\left(f_2\right)-f_2∥}_{\infty };$$

(143)

and

$$\left|D_n\left(f,x\right)-f\left(x\right)\right|\le \left|D_n\left(f_1,x\right)-f_1\left(x\right)\right|+\left|D_n\left(f_2,x\right)-f_2\left(x\right)\right|,$$

(144)

$${∥D_n\left(f\right)-f∥}_{\infty }\le {∥D_n\left(f_1\right)-f_1∥}_{\infty }+{∥D_n\left(f_2\right)-f_2∥}_{\infty }.$$

(145)

Using $C_B\left(\mathbb{R},\mathbb{C}\right),$ we can denote the space of continuous and bounded functions $f:\mathbb{R}\to \mathbb{C}$. Clearly, f is bounded if both $f_1,f_2$ are bounded from $\mathbb{R}$ into $\mathbb{R}$, where $f=f_1+i{f}_2.$

Theorem 8.

Let $f:\mathbb{R}\to \mathbb{C}$, such that $f=f_1+i{f}_2$. Assume $f_1,f_2\in C^N\left(\mathbb{R}\right)$, $N\in \mathbb{R}$, with $f_1^{\left(i\right)},f_2^{\left(i\right)}\in C_B\left(\mathbb{R}\right)$, $i=0,1,\dots ,N;$ $x\in \mathbb{R}$. Here, $0<\beta <1$, $n\in \mathbb{N}:n^{1-\beta }>2$, $\mu >0$. Then,

(I)

$$\left|B_n\left(f,x\right)-f\left(x\right)\right|\le \sum _{j=1}^N\frac{\left(\left|f_1^{\left(j\right)}\left(x\right)\right|+\left|f_2^{\left(j\right)}\left(x\right)\right|\right)}{j!}$$

$$\left[\frac{1}{n^{\beta j}}+\frac{1}{n^j}\left(\frac{e^{\mu }{2}^{j+2}j!}{{\mu }^j}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right]+$$

(146)

$$\left[\frac{\left({\omega }_1\left(f_1^{\left(N\right)},\frac{1}{n^{\beta }}\right)+{\omega }_1\left(f_2^{\left(N\right)},\frac{1}{n^{\beta }}\right)\right)}{n^{\beta N}N!}+\right.$$

$$\left.\frac{\left({∥f_1^{\left(N\right)}∥}_{\infty }+{∥f_2^{\left(N\right)}∥}_{\infty }\right)e^{\mu }{2}^{N+3}{e}^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}}{n^N{\mu }^N}\right],$$

(II) Given that $f_1^{\left(j\right)}\left(x\right)=f_2^{\left(j\right)}\left(x\right)=0$, $j=1,\dots ,N,$ we have

$$\left|B_n\left(f,x\right)-f\left(x\right)\right|\le \left[\frac{\left({\omega }_1\left(f_1^{\left(N\right)},\frac{1}{n^{\beta }}\right)+{\omega }_1\left(f_2^{\left(N\right)},\frac{1}{n^{\beta }}\right)\right)}{n^{\beta N}N!}+\right.$$

(147)

$$\left.\frac{\left({∥f_1^{\left(N\right)}∥}_{\infty }+{∥f_2^{\left(N\right)}∥}_{\infty }\right)e^{\mu }{2}^{N+3}{e}^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}}{n^N{\mu }^N}\right],$$

(III)

$${∥B_n\left(f\right)-f∥}_{\infty }\le \sum _{j=1}^N\frac{\left({∥f_1^{\left(j\right)}∥}_{\infty }+{∥f_2^{\left(j\right)}∥}_{\infty }\right)}{j!}$$

$$\left[\frac{1}{n^{\beta j}}+\frac{1}{n^j}\left\{\frac{e^{\mu }{2}^{j+2}j!}{{\mu }^j}\right\}{e}^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right]+$$

(148)

$$\left[\frac{\left({\omega }_1\left(f_1^{\left(N\right)},\frac{1}{n^{\beta }}\right)+{\omega }_1\left(f_2^{\left(N\right)},\frac{1}{n^{\beta }}\right)\right)}{n^{\beta N}N!}+\right.$$

$$\left.\frac{\left({∥f_1^{\left(N\right)}∥}_{\infty }+{∥f_2^{\left(N\right)}∥}_{\infty }\right)e^{\mu }{2}^{N+3}{e}^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}}{n^N{\mu }^N}\right].$$

Proof. 

By Theorem 4. □

Theorem 9.

All are presented as in Theorem 8. Then,

(I)

$$\left\{\begin{array}{c}\left|C_n\left(f,x\right)-f\left(x\right)\right|,\\ \left|D_n\left(f,x\right)-f\left(x\right)\right|\end{array}\right\}\le \sum _{j=1}^N\frac{\left(\left|f_1^{\left(j\right)}\left(x\right)\right|+\left|f_2^{\left(j\right)}\left(x\right)\right|\right)}{j!}$$

$$\left\{{\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)}^j+\frac{2^{j-1}}{n^j}\left[2e^{\mu }{e}^{-\mu \left(n^{1-\beta }-1\right)}+\left(\frac{e^{\mu }{2}^{j+2}j!}{{\mu }^j}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right]\right\}+$$

$$\left\{\left({\omega }_1\left(f_1^{\left(N\right)},\frac{1}{n}+\frac{1}{n^{\beta }}\right)+{\omega }_1\left(f_2^{\left(N\right)},\frac{1}{n}+\frac{1}{n^{\beta }}\right)\right)\frac{{\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)}^N}{N!}+\right.$$

$$\left.\frac{2^N\left({∥f_1^{\left(N\right)}∥}_{\infty }+{∥f_2^{\left(N\right)}∥}_{\infty }\right)}{n^N{\mu }^N}\left[2e^{\mu }{e}^{-\mu \left(n^{1-\beta }-1\right)}+\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right]\right\},$$

(149)

(II) Given that $f_1^{\left(j\right)}\left(x\right)=f_2^{\left(j\right)}\left(x\right)=0$, $j=1,\dots ,N,$ we have

$$\left\{\begin{array}{c}\left|C_n\left(f,x\right)-f\left(x\right)\right|,\\ \left|D_n\left(f,x\right)-f\left(x\right)\right|\end{array}\right\}\le$$

$$\left({\omega }_1\left(f_1^{\left(N\right)},\frac{1}{n}+\frac{1}{n^{\beta }}\right)+{\omega }_1\left(f_2^{\left(N\right)},\frac{1}{n}+\frac{1}{n^{\beta }}\right)\right)\frac{{\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)}^N}{N!}+$$

(150)

$$\frac{2^N\left({∥f_1^{\left(N\right)}∥}_{\infty }+{∥f_2^{\left(N\right)}∥}_{\infty }\right)}{n^{N}N!}\left[2e^{\mu }{e}^{-\mu \left(n^{1-\beta }-1\right)}+\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right],$$

(III)

$$\left\{\begin{array}{c}{∥C_n\left(f\right)-f∥}_{\infty },\\ {∥D_n\left(f\right)-f∥}_{\infty }\end{array}\right\}\le \sum _{j=1}^N\frac{\left({∥f_1^{\left(j\right)}∥}_{\infty }+{∥f_2^{\left(j\right)}∥}_{\infty }\right)}{j!}$$

$$\left\{{\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)}^j+\frac{2^{j-1}}{n^j}\left[2e^{\mu }{e}^{-\mu \left(n^{1-\beta }-1\right)}+\left(\frac{e^{\mu }{2}^{j+2}j!}{{\mu }^j}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right]\right\}+$$

$$\left\{\left({\omega }_1\left(f_1^{\left(N\right)},\frac{1}{n}+\frac{1}{n^{\beta }}\right)+{\omega }_1\left(f_2^{\left(N\right)},\frac{1}{n}+\frac{1}{n^{\beta }}\right)\right)\frac{{\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)}^N}{N!}+\right.$$

$$\left.\frac{2^N\left({∥f_1^{\left(N\right)}∥}_{\infty }+{∥f_2^{\left(N\right)}∥}_{\infty }\right)}{n^{N}N!}\left[2e^{\mu }{e}^{-\mu \left(n^{1-\beta }-1\right)}+\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right]\right\}.$$

(151)

Proof. 

Use Theorems 5 and 6. □

We finish with the following fractional result.

Theorem 10.

Let $f:\mathbb{R}\to \mathbb{C}$, such that $f=f_1+i{f}_2$. Assume $f_1,f_2\in A{C}^N\left(\mathbb{R}\right)$, $f_1^{\left(N\right)},f_2^{\left(N\right)}\in L_{\infty }\left(\mathbb{R}\right)$, $N\in \mathbb{N}$. Here $\alpha >0$, $N=⌈\alpha ⌉$, $\alpha \notin \mathbb{N}$, $\mu >0$, $0<\beta <1$, $x\in \mathbb{R}$, $n\in \mathbb{N}:n^{1-\beta }\ge 3$. Suppose also that ${\displaystyle \underset{x\in \mathbb{R}}{sup}}{∥D_{\ast x}^{\alpha }{f}_1∥}_{\infty },$ ${\displaystyle \underset{x\in \mathbb{R}}{sup}}{∥D_{\ast x}^{\alpha }{f}_2∥}_{\infty },$ ${\displaystyle \underset{x\in \mathbb{R}}{sup}}{∥D_{x-}^{\alpha }{f}_1∥}_{\infty },$ ${\displaystyle \underset{x\in \mathbb{R}}{sup}}{∥D_{x-}^{\alpha }{f}_2∥}_{\infty }<\infty$. Then,

(I)

$$\left|B_n\left(f,x\right)-f\left(x\right)\right|\le \sum _{j=1}^{N-1}\frac{\left(\left|f_1^{\left(j\right)}\left(x\right)\right|+\left|f_2^{\left(j\right)}\left(x\right)\right|\right)}{j!}$$

$$\left\{\frac{1}{n^{\beta j}}+\frac{1}{n^j}\left[\frac{e^{\mu }{2}^{j+2}j!}{{\mu }^j}\right]e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right\}+$$

(152)

$$\frac{1}{n^{\alpha \beta }\Gamma \left(\alpha +1\right)}\left[{\omega }_1{\left(D_{x-}^{\alpha }{f}_1,\frac{1}{n^{\beta }}\right)}_{(-\infty ,x]}+{\omega }_1{\left(D_{x-}^{\alpha }{f}_2,\frac{1}{n^{\beta }}\right)}_{(-\infty ,x]}+\right.$$

$$\left.{\omega }_1{\left(D_{\ast x}^{\alpha }{f}_1,\frac{1}{n^{\beta }}\right)}_{[x,\infty )}+{\omega }_1{\left(D_{\ast x}^{\alpha }{f}_2,\frac{1}{n^{\beta }}\right)}_{[x,\infty )}\right]+$$

$$\frac{1}{n^{\alpha }\Gamma \left(\alpha +1\right)}\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}$$

$$\left[{∥D_{x-}^{\alpha }{f}_1∥}_{\infty ,(-\infty ,x]}+{∥D_{x-}^{\alpha }{f}_2∥}_{\infty ,(-\infty ,x]}+{∥D_{\ast x}^{\alpha }{f}_1∥}_{\infty ,[x,\infty )}+{∥D_{\ast x}^{\alpha }{f}_2∥}_{\infty ,[x,\infty )}\right],$$

(II) Given $f_1^{\left(j\right)}\left(x\right)=f_2^{\left(j\right)}\left(x\right)=0$, $j=1,\dots ,N-1$, we have

$$\left|B_n\left(f,x\right)-f\left(x\right)\right|\le$$

(153)

$$\frac{1}{n^{\alpha \beta }\Gamma \left(\alpha +1\right)}\left[{\omega }_1{\left(D_{x-}^{\alpha }{f}_1,\frac{1}{n^{\beta }}\right)}_{(-\infty ,x]}+{\omega }_1{\left(D_{x-}^{\alpha }{f}_2,\frac{1}{n^{\beta }}\right)}_{(-\infty ,x]}+\right.$$

$$\left.{\omega }_1{\left(D_{\ast x}^{\alpha }{f}_1,\frac{1}{n^{\beta }}\right)}_{[x,\infty )}+{\omega }_1{\left(D_{\ast x}^{\alpha }{f}_2,\frac{1}{n^{\beta }}\right)}_{[x,\infty )}\right]+$$

$$\frac{1}{n^{\alpha }\Gamma \left(\alpha +1\right)}\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}$$

$$\left[{∥D_{x-}^{\alpha }{f}_1∥}_{\infty ,(-\infty ,x]}+{∥D_{x-}^{\alpha }{f}_2∥}_{\infty ,(-\infty ,x]}+{∥D_{\ast x}^{\alpha }{f}_1∥}_{\infty ,[x,\infty )}+{∥D_{\ast x}^{\alpha }{f}_2∥}_{\infty ,[x,\infty )}\right],$$

and

(III)

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

$$\sum _{j=1}^{N-1}\frac{\left({∥f_1^{\left(j\right)}∥}_{\infty }+{∥f_2^{\left(j\right)}∥}_{\infty }\right)}{j!}\left\{\frac{1}{n^{\beta j}}+\frac{1}{n^j}\left[\frac{e^{\mu }{2}^{j+2}j!}{{\mu }^j}\right]e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}\right\}+$$

$$\frac{1}{n^{\alpha \beta }\Gamma \left(\alpha +1\right)}\left[\underset{x\in \mathbb{R}}{sup}{\omega }_1{\left(D_{x-}^{\alpha }{f}_1,\frac{1}{n^{\beta }}\right)}_{(-\infty ,x]}+\underset{x\in \mathbb{R}}{sup}{\omega }_1{\left(D_{x-}^{\alpha }{f}_2,\frac{1}{n^{\beta }}\right)}_{(-\infty ,x]}+\right.$$

$$\left.\underset{x\in \mathbb{R}}{sup}{\omega }_1{\left(D_{\ast x}^{\alpha }{f}_1,\frac{1}{n^{\beta }}\right)}_{[x,\infty )}+\underset{x\in \mathbb{R}}{sup}{\omega }_1{\left(D_{\ast x}^{\alpha }{f}_2,\frac{1}{n^{\beta }}\right)}_{[x,\infty )}\right]+$$

$$\frac{1}{n^{\alpha }\Gamma \left(\alpha +1\right)}\left(\frac{e^{\mu }{2}^{N+2}N!}{{\mu }^N}\right)e^{-\frac{\mu }{2}\left(n^{1-\beta }-1\right)}$$

(154)

$$\left[\underset{x\in \mathbb{R}}{sup}{∥D_{x-}^{\alpha }{f}_1∥}_{\infty ,(-\infty ,x]}+\underset{x\in \mathbb{R}}{sup}{∥D_{x-}^{\alpha }{f}_2∥}_{\infty ,(-\infty ,x]}+\right.$$

$$\left.\underset{x\in \mathbb{R}}{sup}{∥D_{\ast x}^{\alpha }{f}_1∥}_{\infty ,[x,\infty )}+\underset{x\in \mathbb{R}}{sup}{∥D_{\ast x}^{\alpha }{f}_2∥}_{\infty ,[x,\infty )}\right].$$

Proof. 

By Theorem 7. □

Conclusions: The author used parametrized Richard’s curve activated neural network approximations to differentiate functions from $\mathbb{R}$ into $\mathbb{R}$, going beyond the bounded domain functions. He presented real and complex, ordinary and fractional quasi-interpolation quantitative approximations. The results are totally new.