Approximation by Symmetrized and Perturbed Hyperbolic Tangent Activated Convolution-Type Operators

Abstract

In this article, for the first time, the univariate symmetrized and perturbed hyperbolic tangent activated convolution-type operators of three kinds are introduced. Their approximation properties are presented, i.e., the quantitative convergence to the unit operator via the modulus of continuity. It follows the global smoothness preservation of these operators. The related iterated approximation as well as the simultaneous approximation and their combinations, are also extensively presented. Including differentiability and fractional differentiability into our research produced higher rates of approximation. Simultaneous global smoothness preservation is also examined.

1. Organization

In Section 2, we give the preliminaries of our theory. In Section 3 are the basics, the introduction of our activated symmetrized and perturbed hyperbolic tangent convolution-type operators with properties. In Section 4 are the main approximation results. We also include the global smoothness preservation by our operators there. We further study the differentiation of these operators, as well as introducing their iterates and giving their basic properties. Next, we present the convergence of our operators under differentiability and Caputo fractional differentiability, achieving higher rates of approximation. It follows the simultaneous differential approximation and simultaneous global smoothness preservation in detail, as well as the iterated approximation. We finish with the combination of the simultaneous and iterated approximations.

We are motivated and inspired by [1,2,3,4,5,6,7,8,9,10,11,12].

2. About $\mathit{q}$-Deformed and $\mathit{\lambda }$-Parameterized Hyperbolic Tangent Function ${\mathit{g}}_{\mathit{q},\mathit{\lambda }}$

Here, all of this initial background comes from Chapter 18 of [1].

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

So, let us consider the hyperbolic tangent activation function

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

(1)

We have that

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

We notice also that

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

(2)

That is,

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

(3)

and

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

hence,

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

(4)

It is

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

i.e.,

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

(5)

Furthermore,

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

i.e.,

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

(6)

We find that

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

(7)

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

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

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

(8)

We observe that

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

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

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

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

By $1>-1$ and $x+1>x-1$, we consider the function

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

(9)

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

We have that

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

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

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

(10)

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

Thus,

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

(11)

a deformed symmetry.

Next, we have that

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

(12)

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

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

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

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

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

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

(13)

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

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

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

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

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

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

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

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

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

(14)

Thus,

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

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

(15)

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

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

(16)

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

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

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

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

(17)

We mention the following.

Theorem 1 

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

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

(18)

Also, the following holds.

Theorem 2 

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

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

(19)

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

Similarly, we see that

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

(20)

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

Furthermore, we observe the symmetry

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

(21)

Furthermore,

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

(22)

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

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

Clearly, then,

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

(23)

3. Basic

We give

Definition 1. 

Let $f\in C_B\left(\mathbb{R}\right)$ (continuous and bounded functions on $\mathbb{R}$), $n\in \mathbb{N}$. We define the following basic activated hyperbolic tangent perturbed convolution-type operators,

$$A_n\left(f\right)\left(x\right):={\int }_{-\infty }^{\infty }f\left({\displaystyle \frac{u}{n}}\right)\phi \left(nx-u\right)du,\forall x\in \mathbb{R}.$$

(24)

In this work, we examine the quantitative convergence of $A_n$ to the unit operator.

We study similarly the activated Kantorovich-type operators,

$$A_n^{*}\left(f\right)\left(x\right):=n{\int }_{-\infty }^{\infty }\left({\int }_{{\displaystyle \frac{u}{n}}}^{{\displaystyle \frac{u+1}{n}}}f\left(t\right)dt\right)\phi \left(nx-u\right)du$$

$$=n{\int }_{-\infty }^{\infty }\left({\int }_0^{{\displaystyle \frac{1}{n}}}f\left(t+{\displaystyle \frac{u}{n}}\right)dt\right)\phi \left(nx-u\right)du,$$

(25)

where $f\in C_B\left(\mathbb{R}\right)$, $n\in \mathbb{N}$, ∀$x\in \mathbb{R}$, and the activated Quadrature operators

$$\overline{A_n}\left(f\right)\left(x\right):={\int }_{-\infty }^{\infty }\left(\sum _{i=1}^r{w}_{i}f\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}\right)\right)\phi \left(nx-u\right)du,$$

(26)

where $w_i\ge 0$, ${\displaystyle \sum _{i=1}^r}{w}_i=1$; $f\in C_B\left(\mathbb{R}\right)$, $n\in \mathbb{N}$, ∀$x\in \mathbb{R}$.

An essential property follows.

Theorem 3. 

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

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

(27)

Proof. 

We have that $M_{q,\lambda }\left(x\right):={\displaystyle \frac{1}{4}}\left[g_{q,\lambda }\left(x+1\right)-g_{q,\lambda }\left(x-1\right)\right]$, ∀$x\in \mathbb{R}$.

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

$$M_{q,\lambda }\left(x\right)={\displaystyle \frac{1}{4}}·2·g_{q,\lambda }^{\prime }\left({\xi }_1\right)={\displaystyle \frac{1}{2}}{g}_{q,\lambda }^{\prime }\left({\xi }_1\right)={\displaystyle \frac{1}{2}}{\displaystyle \frac{4q\lambda e^{2\lambda {\xi }_1}}{{\left(e^{2\lambda {\xi }_1}+q\right)}^2}}<$$

(28)

$${\displaystyle \frac{2q\lambda \left(e^{2\lambda {\xi }_1+q}\right)}{{\left(e^{2\lambda {\xi }_1+q}\right)}^2}}={\displaystyle \frac{2q\lambda }{\left(e^{2\lambda {\xi }_1+q}\right)}}<2q\lambda e^{-2\lambda {\xi }_1},$$

where $0\le x-1<{\xi }_1<x+1$.

That is,

$$M_{q,\lambda }\left(x\right)<2q\lambda e^{-2\lambda \left(x-1\right)},\forall x\ge 1.$$

(29)

Similarly, it holds that

$$M_{{\displaystyle \frac{1}{q}},\lambda }\left(x\right)<{\displaystyle \frac{2}{q}}\lambda e^{-2\lambda \left(x-1\right)},\forall x\ge 1.$$

(30)

Hence, we see that

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

(31)

Consider the set

$$\mathsf{\Delta }:=\left\{u\in \mathbb{R}:\left|nx-u\right|\ge n^{1-\alpha }\right\}.$$

(32)

We have that

$${\int }_{\mathsf{\Delta }}\phi \left(nx-u\right)du={\int }_{\mathsf{\Delta }}\phi \left(\left|nx-u\right|\right)du<\left(q+{\displaystyle \frac{1}{q}}\right)\lambda {\int }_{\mathsf{\Delta }}{e}^{-2\lambda \left(\left|nx-u\right|-1\right)}du=$$

(33)

$$2\left(q+{\displaystyle \frac{1}{q}}\right)\lambda {\int }_{n^{1-\alpha }}^{\infty }{e}^{-2\lambda \left(x-1\right)}dx=2\left(q+{\displaystyle \frac{1}{q}}\right)\lambda {\int }_{n^{1-\alpha }-1}^{\infty }{e}^{-2\lambda z}dz=$$

$$\left(q+{\displaystyle \frac{1}{q}}\right){\int }_{n^{1-\alpha }-1}^{\infty }{e}^{-2\lambda z}d\left(2\lambda z\right)=\left(q+{\displaystyle \frac{1}{q}}\right){\int }_{n^{1-\alpha }-1}^{\infty }{e}^{-y}dy=$$

(34)

$$\left(q+{\displaystyle \frac{1}{q}}\right)\left\{e^{-2\lambda z}{|}_{\infty }^{n^{1-\alpha }-1}\right\}=\left(q+{\displaystyle \frac{1}{q}}\right)e^{-2\lambda \left(n^{1-\alpha }-1\right)}={\displaystyle \frac{\left(q+\frac{1}{q}\right)}{e^{2\lambda \left(n^{1-\alpha }-1\right)}}}.$$

(35)

The claim is proven. □

The first modulus of continuity here is

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

(36)

4. Main Results

We present the following approximation results.

Theorem 4. 

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

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

(37)

and

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

(38)

So, for $f\in C_{uB}\left(\mathbb{R}\right):=C_u\left(\mathbb{R}\right)\cap C_B\left(\mathbb{R}\right)$ (where $C_u\left(\mathbb{R}\right)$ the uniformly continuous functions on $\mathbb{R}$), we see that ${\displaystyle \underset{n\to \infty }{lim}}{A}_n\left(f\right)=f$, pointwise and uniformly.

Proof. 

Call

$${\mathsf{\Delta }}_1:=\left\{u\in \mathbb{R}:\left|{\displaystyle \frac{u}{n}}-x\right|<{\displaystyle \frac{1}{n^{\alpha }}}\right\},$$

(39)

and

$${\mathsf{\Delta }}_2:=\left\{u\in \mathbb{R}:\left|{\displaystyle \frac{u}{n}}-x\right|\ge {\displaystyle \frac{1}{n^{\alpha }}}\right\}.$$

(40)

That is, ${\mathsf{\Delta }}_1\cup {\mathsf{\Delta }}_2=\mathbb{R}$.

We have that

$$\left|A_n\left(f\right)\left(x\right)-f\left(x\right)\right|\stackrel{\left(23\right)}{=}$$

$$\left|{\int }_{-\infty }^{\infty }f\left({\displaystyle \frac{u}{n}}\right)\phi \left(nx-u\right)du-f\left(x\right){\int }_{-\infty }^{\infty }\phi \left(nx-u\right)du\right|=$$

$$\left|{\int }_{-\infty }^{\infty }\left(f\left({\displaystyle \frac{u}{n}}\right)-f\left(x\right)\right)\phi \left(nx-u\right)du\right|\le$$

(41)

$${\int }_{-\infty }^{\infty }\left|f\left({\displaystyle \frac{u}{n}}\right)-f\left(x\right)\right|\phi \left(nx-u\right)du=$$

$${\int }_{{\mathsf{\Delta }}_1}\left|f\left({\displaystyle \frac{u}{n}}\right)-f\left(x\right)\right|\phi \left(nx-u\right)du+{\int }_{{\mathsf{\Delta }}_2}\left|f\left({\displaystyle \frac{u}{n}}\right)-f\left(x\right)\right|\phi \left(nx-u\right)du\le$$

$${\int }_{{\mathsf{\Delta }}_1}{\omega }_1\left(f,\left|{\displaystyle \frac{u}{n}}-x\right|\right)\phi \left(nx-u\right)du+2{∥f∥}_{\infty }{\int }_{{\mathsf{\Delta }}_2}\phi \left(nx-u\right)du\le$$

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

(42)

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

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

□

Theorem 5. 

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

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

(43)

and

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

(44)

For $f\in C_{uB}\left(\mathbb{R}\right)$, we have ${\displaystyle \underset{n\to \infty }{lim}}{A}_n^{*}\left(f\right)=f$, pointwise and uniformly.

Proof. 

${\mathsf{\Delta }}_1$, ${\mathsf{\Delta }}_2$ as in (39), (40), respectively.

We have that

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

$$\left|n{\int }_{-\infty }^{\infty }\left({\int }_{{\displaystyle \frac{u}{n}}}^{{\displaystyle \frac{u+1}{n}}}f\left(t\right)dt\right)\phi \left(nx-u\right)du-{\int }_{-\infty }^{\infty }f\left(x\right)\phi \left(nx-u\right)du\right|=$$

$$\left|n{\int }_{-\infty }^{\infty }\left({\int }_{{\displaystyle \frac{u}{n}}}^{{\displaystyle \frac{u+1}{n}}}f\left(t\right)dt\right)\phi \left(nx-u\right)du-n{\int }_{-\infty }^{\infty }\left({\int }_{{\displaystyle \frac{u}{n}}}^{{\displaystyle \frac{u+1}{n}}}f\left(x\right)dt\right)\phi \left(nx-u\right)du\right|=$$

(45)

$$\left|n{\int }_{-\infty }^{\infty }\left({\int }_{{\displaystyle \frac{u}{n}}}^{{\displaystyle \frac{u+1}{n}}}\left(f\left(t\right)-f\left(x\right)\right)dt\right)\phi \left(nx-u\right)du\right|\le$$

$$n{\int }_{-\infty }^{\infty }\left({\int }_{{\displaystyle \frac{u}{n}}}^{{\displaystyle \frac{u+1}{n}}}\left|f\left(t\right)-f\left(x\right)\right|dt\right)\phi \left(nx-u\right)du=$$

$$n{\int }_{-\infty }^{\infty }\left({\int }_0^{{\displaystyle \frac{1}{n}}}\left|f\left(t+{\displaystyle \frac{u}{n}}\right)-f\left(x\right)\right|dt\right)\phi \left(nx-u\right)du\le$$

$${\int }_{{\mathsf{\Delta }}_1}\left(n{\int }_0^{{\displaystyle \frac{1}{n}}}\left|f\left(t+{\displaystyle \frac{u}{n}}\right)-f\left(x\right)\right|dt\right)\phi \left(nx-u\right)du+$$

$${\int }_{{\mathsf{\Delta }}_2}\left(n{\int }_0^{{\displaystyle \frac{1}{n}}}\left|f\left(t+{\displaystyle \frac{u}{n}}\right)-f\left(x\right)\right|dt\right)\phi \left(nx-u\right)du\le$$

(46)

$${\int }_{{\mathsf{\Delta }}_1}\left(n{\int }_0^{{\displaystyle \frac{1}{n}}}{\omega }_1\left(f,\left|t\right|+{\displaystyle \frac{1}{n^{\alpha }}}\right)dt\right)\phi \left(nx-u\right)du+2{∥f∥}_{\infty }{\int }_{{\mathsf{\Delta }}_2}\phi \left(nx-u\right)du\le$$

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

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

(47)

□

Theorem 6. 

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

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

(48)

and

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

(49)

For $f\in C_{uB}\left(\mathbb{R}\right)$, we have ${\displaystyle \underset{n\to \infty }{lim}}\overline{A_n}\left(f\right)=f$, pointwise and uniformly.

Proof. 

${\mathsf{\Delta }}_1$, ${\mathsf{\Delta }}_2$ as in (39), (40), respectively.

We have

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

$$\left|{\int }_{-\infty }^{\infty }\sum _{i=1}^r{w}_{i}f\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}\right)\phi \left(nx-u\right)du-{\int }_{-\infty }^{\infty }\left(\sum _{i=1}^r{w}_{i}f\left(x\right)\right)\phi \left(nx-u\right)du\right|=$$

$$\left|{\int }_{-\infty }^{\infty }\sum _{i=1}^r{w}_i\left(f\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}\right)-f\left(x\right)\right)\phi \left(nx-u\right)du\right|\le$$

$${\int }_{-\infty }^{\infty }\sum _{i=1}^r{w}_i\left|f\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}\right)-f\left(x\right)\right|\phi \left(nx-u\right)du\le$$

(50)

$${\int }_{A_1}\sum _{i=1}^r{w}_i\left|f\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}\right)-f\left(x\right)\right|\phi \left(nx-u\right)du+$$

$${\int }_{A_2}\sum _{i=1}^r{w}_i\left|f\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}\right)-f\left(x\right)\right|\phi \left(nx-u\right)du\le$$

$${\int }_{{\mathsf{\Delta }}_1}\sum _{i=1}^r{w}_i{\omega }_1\left(f,{\displaystyle \frac{1}{n^{\alpha }}}+{\displaystyle \frac{i}{nr}}\right)\phi \left(nx-u\right)du+2{∥f∥}_{\infty }{\int }_{{\mathsf{\Delta }}_2}\phi \left(nx-u\right)du\le$$

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

(51)

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

□

We need

Proposition 1. 

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

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

(52)

Proof. 

We can write

$${\int }_{-\infty }^{\infty }{\left|z\right|}^k\phi \left(z\right)dz=2{\int }_0^{\infty }{z}^k\phi \left(z\right)dz=$$

$$2\left[{\int }_0^1{z}^k\phi \left(z\right)dz+{\int }_1^{\infty }{z}^k\phi \left(z\right)dz\right]\stackrel{(\mathrm{by}\left(17\right),(31\left)\right)}{\le }$$

$$2\left[\left({\int }_0^1{z}^{k}dz\right){\displaystyle \frac{tanh\left(\lambda \right)}{2}}+\left(q+{\displaystyle \frac{1}{q}}\right)\lambda {\int }_1^{\infty }{z}^k{e}^{-2\lambda \left(z-1\right)}dz\right]\le$$

$$2\left[\left({\displaystyle \frac{1}{k+1}}\right){\displaystyle \frac{tanh\left(\lambda \right)}{2}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)}{2}}{\displaystyle \frac{e^{2\lambda }}{{\left(2\lambda \right)}^k}}{\int }_0^{\infty }{\left(2\lambda z\right)}^k{e}^{-2\lambda z}d\left(2\lambda z\right)\right]=$$

(53)

$$\left[{\displaystyle \frac{tanh\left(\lambda \right)}{\left(k+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }}{{\left(2\lambda \right)}^k}}{\int }_0^{\infty }{y}^k{e}^{-y}dy\right]=$$

$$\left[{\displaystyle \frac{tanh\left(\lambda \right)}{\left(k+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }}{{\left(2\lambda \right)}^k}}\mathsf{\Gamma }\left(k+1\right)\right]<\infty .$$

□

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

Theorem 7. 

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

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

(54)

If $f\in C_u\left(\mathbb{R}\right)$, then $A_n\left(f\right)\in C_u\left(\mathbb{R}\right)$.

Proof. 

We have that

$$A_n\left(f\right)\left(x\right)={\int }_{-\infty }^{\infty }f\left(x-{\displaystyle \frac{z}{n}}\right)\phi \left(z\right)dz.$$

(55)

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

$$A_n\left(f\right)\left(x\right)-A_n\left(f\right)\left(y\right)={\int }_{-\infty }^{\infty }\left(f\left(x-{\displaystyle \frac{z}{n}}\right)-f\left(y-{\displaystyle \frac{z}{n}}\right)\right)\phi \left(z\right)dz.$$

Thus,

$$\left|A_n\left(f\right)\left(x\right)-A_n\left(f\right)\left(y\right)\right|\le {\int }_{-\infty }^{\infty }\left|f\left(x-{\displaystyle \frac{z}{n}}\right)-f\left(y-{\displaystyle \frac{z}{n}}\right)\right|\phi \left(z\right)dz\le$$

$${\omega }_1\left(f,\left|x-y\right|\right){\int }_{-\infty }^{\infty }\phi \left(z\right)dz\stackrel{\left(22\right)}{=}{\omega }_1\left(f,\left|x-y\right|\right).$$

Let $\left|x-y\right|\le \delta$, $\delta >0$; then, we obtain (54). □

Remark 1. 

Clearly, (54) is attained by $f=$ identity map $=:id$.

We have

$$\left|A_n\left(id\right)\left(x\right)-A_n\left(id\right)\left(y\right)\right|=\left|id\left(x\right)-id\left(y\right)\right|=\left|x-y\right|,$$

and

$${\omega }_1\left(A_n\left(id\right),\delta \right)={\omega }_1\left(id,\delta \right)=\delta >0.$$

(56)

We also see that

$$A_n\left(id\right)\left(x\right)={\int }_{-\infty }^{\infty }\left(x-{\displaystyle \frac{z}{n}}\right)\phi \left(z\right)dz=$$

$$x{\int }_{-\infty }^{\infty }\phi \left(z\right)dz-{\displaystyle \frac{1}{n}}{\int }_{-\infty }^{\infty }z\phi \left(z\right)dz=x-{\displaystyle \frac{1}{n}}{\int }_{-\infty }^{\infty }z\phi \left(z\right)dz.$$

So, for fixed $x\in \mathbb{R}$, we have

$$\left|A_n\left(id\right)\left(x\right)\right|\le \left|x\right|+{\displaystyle \frac{1}{n}}{\int }_{-\infty }^{\infty }\left|z\right|\phi \left(z\right)dz$$

(57)

$$\stackrel{\left(52\right)}{\le }\left|x\right|+{\displaystyle \frac{1}{n}}\left[{\displaystyle \frac{tanh\left(\lambda \right)}{2}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }}{\left(2\lambda \right)}}\right]<\infty .$$

Of course, $id\in C_u\left(\mathbb{R}\right)$.

Theorem 8. 

Again, $f\in C_B\left(\mathbb{R}\right)\cup C_u\left(\mathbb{R}\right)$. Then,

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

(58)

If $f\in C_u\left(\mathbb{R}\right)$, then $A_n^{*}\left(f\right)\in C_u\left(\mathbb{R}\right)$.

Inequality (58) is attained by $f\left(x\right)=id\left(x\right)$.

Proof. 

We notice that ($x,y\in \mathbb{R}$),

$$A_n^{*}\left(f\right)\left(x\right)=n{\int }_{-\infty }^{\infty }\left({\int }_0^{{\displaystyle \frac{1}{n}}}f\left(t+\left(x-{\displaystyle \frac{z}{n}}\right)\right)dt\right)\phi \left(z\right)dz,$$

(59)

and

$$A_n^{*}\left(f\right)\left(y\right)=n{\int }_{-\infty }^{\infty }\left({\int }_0^{{\displaystyle \frac{1}{n}}}f\left(t+\left(y-{\displaystyle \frac{z}{n}}\right)\right)dt\right)\phi \left(z\right)dz.$$

Thus,

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

$$n\left|\left[{\int }_{-\infty }^{\infty }\left({\int }_0^{{\displaystyle \frac{1}{n}}}\left(f\left(t+\left(x-{\displaystyle \frac{z}{n}}\right)\right)-f\left(t+\left(y-{\displaystyle \frac{z}{n}}\right)\right)\right)dt\right)\phi \left(z\right)dz\right]\right|\le$$

$$n{\int }_{-\infty }^{\infty }\left({\int }_0^{{\displaystyle \frac{1}{n}}}\left|f\left(t+\left(x-{\displaystyle \frac{z}{n}}\right)\right)-f\left(t+\left(y-{\displaystyle \frac{z}{n}}\right)\right)\right|dt\right)\phi \left(z\right)dz\le$$

(60)

$${\omega }_1\left(f,\left|x-y\right|\right){\int }_{-\infty }^{\infty }\phi \left(z\right)dz={\omega }_1\left(f,\left|x-y\right|\right),$$

proving Inequality (58).

Notice that

$$\left|A_n^{*}\left(id\right)\left(x\right)-A_n^{*}\left(id\right)\left(y\right)\right|=\left|x-y\right|,$$

and

$$\left|A_n^{*}\left(id\right)\left(x\right)\right|\le n{\int }_{-\infty }^{\infty }\left({\int }_0^{{\displaystyle \frac{1}{n}}}\left(\left|t\right|+\left|x\right|+{\displaystyle \frac{\left|z\right|}{n}}\right)dt\right)\phi \left(z\right)dz\le$$

(61)

$${\int }_{-\infty }^{\infty }\left({\displaystyle \frac{1}{n}}+\left|x\right|+{\displaystyle \frac{\left|z\right|}{n}}\right)\phi \left(z\right)dz={\displaystyle \frac{1}{n}}+\left|x\right|+{\displaystyle \frac{1}{n}}{\int }_{-\infty }^{\infty }\left|z\right|\phi \left(z\right)dz<\infty ,$$

proving the attainability of (58). □

Theorem 9. 

We consider $f\in C_B\left(\mathbb{R}\right)\cup C_u\left(\mathbb{R}\right)$. Then,

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

(62)

If $f\in C_u\left(\mathbb{R}\right)$, then $\overline{A_n}\left(f\right)\in C_u\left(\mathbb{R}\right)$.

Inequality (62) is attained by $f\left(x\right)=id\left(x\right)$.

Proof. 

For $x,y\in \mathbb{R}$, we have

$$\overline{A_n}\left(f\right)\left(x\right)={\int }_{-\infty }^{\infty }\left(\sum _{i=1}^r{w}_{i}f\left(\left(x-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{i}{nr}}\right)\right)\phi \left(z\right)dz,$$

(63)

and

$$\overline{A_n}\left(f\right)\left(y\right)={\int }_{-\infty }^{\infty }\left(\sum _{i=1}^r{w}_{i}f\left(\left(y-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{i}{nr}}\right)\right)\phi \left(z\right)dz.$$

Hence,

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

$$\left|{\int }_{-\infty }^{\infty }\sum _{i=1}^r{w}_i\left[f\left(\left(x-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{i}{nr}}\right)-f\left(\left(y-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{i}{nr}}\right)\right]\phi \left(z\right)dz\right|\le$$

(64)

$${\int }_{-\infty }^{\infty }\sum _{i=1}^r{w}_i\left|f\left(\left(x-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{i}{nr}}\right)-f\left(\left(y-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{i}{nr}}\right)\right|\phi \left(z\right)dz\le$$

$${\omega }_1\left(f,\left|x-y\right|\right){\int }_{-\infty }^{\infty }\phi \left(z\right)dz={\omega }_1\left(f,\left|x-y\right|\right).$$

That is, (62) true, and is attained by $f\left(x\right)=id\left(x\right)$.

Indeed, we observe that

$$\left|\overline{A_n}\left(id\right)\left(x\right)-\overline{A_n}\left(id\right)\left(y\right)\right|=\left|x-y\right|,$$

and, furthermore,

$$\left|\overline{A_n}\left(id\right)\left(x\right)\right|\le {\int }_{-\infty }^{\infty }\left(\sum _{i=1}^r{w}_i\left(\left|x\right|+{\displaystyle \frac{\left|z\right|}{n}}+{\displaystyle \frac{1}{n}}\right)\right)\phi \left(z\right)dz\le$$

$${\int }_{-\infty }^{\infty }\left(\left|x\right|+{\displaystyle \frac{\left|z\right|}{n}}+{\displaystyle \frac{1}{n}}\right)\phi \left(z\right)dz={\displaystyle \frac{1}{n}}+\left|x\right|+{\displaystyle \frac{1}{n}}{\int }_{-\infty }^{\infty }\left|z\right|\phi \left(z\right)dz<\infty ,$$

(65)

proving the attainability of (62). □

We make the following remark.

Remark 2. 

Let $i\in \mathbb{N}$ be fixed. Assume that $f\in C^{\left(i\right)}\left(\mathbb{R}\right)$, with $f^{\left(j\right)}\in C_B\left(\mathbb{R}\right)$, for $j=0,1,\dots ,i$.

We have that

$$A_n\left(f\right)\left(x\right)={\int }_{-\infty }^{\infty }f\left(x-{\displaystyle \frac{z}{n}}\right)\phi \left(z\right)dz.$$

By repeatedly applying Leibnitz’s rule, we obtain

$${\displaystyle \frac{{\partial }^i{A}_n\left(f\right)\left(x\right)}{\partial x^i}}={\int }_{-\infty }^{\infty }{f}^{\left(i\right)}\left(x-{\displaystyle \frac{z}{n}}\right)\phi \left(z\right)dz=$$

$${\int }_{-\infty }^{\infty }{f}^{\left(i\right)}\left({\displaystyle \frac{u}{n}}\right)\phi \left(nx-u\right)du=A_n\left(f^{\left(i\right)}\right)\left(x\right),\forall x\in \mathbb{R}.$$

(66)

Clearly, it is valid that

$${\displaystyle \frac{{\partial }^i{A}_n^{*}\left(f\right)\left(x\right)}{\partial x^i}}=A_n^{*}\left(f^{\left(i\right)}\right)\left(x\right),$$

(67)

and

$${\displaystyle \frac{{\partial }^i\overline{A_n}\left(f\right)\left(x\right)}{\partial x^i}}=\overline{A_n}\left(f^{\left(i\right)}\right)\left(x\right),$$

(68)

∀$x\in \mathbb{R}$.

So, all of our results in this work can be written in the simultaneous approximation context; see Remark 7 and Theorems 16–19.

We make the following remark about iterated convolution.

Remark 3. 

We have that

$$A_n\left(f\right)\left(x\right)={\int }_{-\infty }^{\infty }f\left(x-{\displaystyle \frac{z}{n}}\right)\phi \left(z\right)dz,$$

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

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

$$A_n\left(f\right)\left(x_N\right)-A_n\left(f\right)\left(x\right)={\int }_{-\infty }^{\infty }\left(f\left(x_N-{\displaystyle \frac{z}{n}}\right)-f\left(x-{\displaystyle \frac{z}{n}}\right)\right)\phi \left(z\right)dz.$$

(69)

We have that

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

Furthermore, it holds that

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

(70)

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

by dominated convergence theorem, because we have that

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

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

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

Furthermore, it holds that

$$\left|A_n\left(f\right)\left(x\right)\right|\le {∥f∥}_{\infty }{\int }_{-\infty }^{\infty }\phi \left(z\right)dz={∥f∥}_{\infty },$$

i.e.,

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

(71)

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

Clearly it holds that

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

(72)

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

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

(73)

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

Remark 4. 

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

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

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

(74)

Then,

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

$$+\dots +{∥A_n^{2}f-A_{n}f∥}_{\infty }+{∥A_{n}f-f∥}_{\infty }=$$

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

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

Therefore,

$${∥A_n^{r}f-f∥}_{\infty }\le r{∥A_{n}f-f∥}_{\infty }.$$

(75)

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

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

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

(76)

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

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

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

(77)

Next, we have

Remark 5. 

$$A_n^{*}\left(f\right)\left(x\right)=n{\int }_{-\infty }^{\infty }\left({\int }_0^{{\displaystyle \frac{1}{n}}}f\left(t+\left(x-{\displaystyle \frac{z}{n}}\right)\right)dt\right)\phi \left(z\right)dz,$$

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

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

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

$$\left|n{\int }_{-\infty }^{\infty }\left({\int }_0^{{\displaystyle \frac{1}{n}}}\left[f\left(t+\left(x_N-{\displaystyle \frac{z}{n}}\right)\right)-f\left(t+\left(x-{\displaystyle \frac{z}{n}}\right)\right)\right]dt\right)\phi \left(z\right)dz\right|\le$$

(78)

$${\int }_{-\infty }^{\infty }\left(n{\int }_0^{{\displaystyle \frac{1}{n}}}\left|f\left(t+\left(x_N-{\displaystyle \frac{z}{n}}\right)\right)-f\left(t+\left(x-{\displaystyle \frac{z}{n}}\right)\right)\right|dt\right)\phi \left(z\right)dz\to 0,\mathit{as}N\to \infty .$$

This is true by the bounded convergence theorem, and we see that

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

and

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

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

and $\left[0,{\displaystyle \frac{1}{n}}\right]$ is finite interval. Thus,

$$n{\int }_0^{{\displaystyle \frac{1}{n}}}\left|f\left(t+\left(x_N-{\displaystyle \frac{z}{n}}\right)\right)-f\left(t+\left(x-{\displaystyle \frac{z}{n}}\right)\right)\right|dt\to 0,\mathit{as}N\to \infty .$$

(79)

Therefore, it holds that

$$n{\int }_0^{{\displaystyle \frac{1}{n}}}f\left(t+\left(x_N-{\displaystyle \frac{z}{n}}\right)\right)dt\to n{\int }_0^{{\displaystyle \frac{1}{n}}}f\left(t+\left(x-{\displaystyle \frac{z}{n}}\right)\right)dt,\mathit{as}N\to \infty .$$

(80)

Further,

$$\left(n{\int }_0^{{\displaystyle \frac{1}{n}}}f\left(t+\left(x_N-{\displaystyle \frac{z}{n}}\right)\right)dt\right)\phi \left(z\right)\to \left(n{\int }_0^{{\displaystyle \frac{1}{n}}}f\left(t+\left(x-{\displaystyle \frac{z}{n}}\right)\right)dt\right)\phi \left(z\right),$$

(81)

as $N\to \infty$, ∀$z\in \mathbb{R}$.

Furthermore, we have

$$\left|\left(n{\int }_0^{{\displaystyle \frac{1}{n}}}f\left(t+\left(x_N-{\displaystyle \frac{z}{n}}\right)\right)dt\right)\phi \left(z\right)\right|\le {∥f∥}_{\infty }\phi \left(z\right),$$

(82)

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

Therefore, by the dominated convergence theorem,

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

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

The iterated facts hold for $A_n^{*}$, as in the $A_n\left(f\right)$ case, all the same! See (71)–(73), and all of Remark 4.

Remark 6. 

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

$$\overline{A_n}\left(f\right)\left(x\right)={\int }_{-\infty }^{\infty }\left(\sum _{i=1}^r{w}_{i}f\left(\left(x-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{i}{nr}}\right)\right)\phi \left(z\right)dz.$$

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

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

$$\left|{\int }_{-\infty }^{\infty }\left(\sum _{i=1}^r{w}_i\left(f\left(\left(x_N-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{i}{nr}}\right)-f\left(\left(x-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{i}{nr}}\right)\right)\right)\phi \left(z\right)dz\right|\le$$

(83)

$${\int }_{-\infty }^{\infty }\left|\sum _{i=1}^r{w}_i\left(f\left(\left(x_N-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{i}{nr}}\right)-f\left(\left(x-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{i}{nr}}\right)\right)\right|\phi \left(z\right)dz\to 0,\mathit{as}N\to \infty .$$

The last comes by the dominated convergence theorem,

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

and

$$\sum _{i=1}^r{w}_{i}f\left(\left(x_N-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{i}{nr}}\right)\to \sum _{i=1}^r{w}_{i}f\left(\left(x-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{i}{nr}}\right),$$

and

$$\left(\sum _{i=1}^r{w}_{i}f\left(\left(x_N-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{i}{nr}}\right)\right)\phi \left(z\right)\to \left(\sum _{i=1}^r{w}_{i}f\left(\left(x-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{i}{nr}}\right)\right)\phi \left(z\right),$$

as $N\to \infty$, ∀$z\in \mathbb{R}$.

Furthermore, it holds that

$$\left|\sum _{i=1}^r{w}_{i}f\left(\left(x_N-{\displaystyle \frac{z}{n}}\right)+{\displaystyle \frac{i}{nr}}\right)\right|\phi \left(z\right)\le {∥f∥}_{\infty }\phi \left(z\right),$$

(84)

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

Therefore,

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

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

Iterated stuff for $\overline{A_n}$: it is all the same, as with $A_n\left(f\right)$! See (71)–(73) and all of Remark 4.

See the related Theorems 20 and 21 later.

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

First, we treat the basic ones.

Theorem 10. 

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

(i)

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

$${\displaystyle \frac{{\omega }_1\left(f^{\left(N\right)},\frac{1}{n^{\alpha }}\right)}{n^{\alpha N}N!}}+{\displaystyle \frac{4{∥f^{\left(N\right)}∥}_{\infty }}{n^N}}\left(q+{\displaystyle \frac{1}{q}}\right){\displaystyle \frac{e^{2\lambda }}{{\lambda }^N}}{e}^{-\lambda n^{1-\alpha }}\to 0,\mathit{as}n\to \infty .$$

(85)

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

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

$${\displaystyle \frac{{\omega }_1\left(f^{\left(N\right)},\frac{1}{n^{\alpha }}\right)}{n^{\alpha N}N!}}+{\displaystyle \frac{4{∥f^{\left(N\right)}∥}_{\infty }}{n^N}}\left(q+{\displaystyle \frac{1}{q}}\right){\displaystyle \frac{e^{2\lambda }}{{\lambda }^N}}{e}^{-\lambda n^{1-\alpha }},$$

(86)

(iii)

$$\left|A_n\left(f\right)\left(x\right)-f\left(x\right)\right|\le \sum _{j=1}^N{\displaystyle \frac{\left|f^{\left(j\right)}\left(x\right)\right|}{j!}}{\displaystyle \frac{1}{n^j}}\left({\displaystyle \frac{tanh\left(\lambda \right)}{\left(j+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }j!}{{\left(2\lambda \right)}^j}}\right)$$

$$+{\displaystyle \frac{{\omega }_1\left(f^{\left(N\right)},\frac{1}{n^{\alpha }}\right)}{n^{\alpha N}N!}}+{\displaystyle \frac{4{∥f^{\left(N\right)}∥}_{\infty }}{n^N}}\left(q+{\displaystyle \frac{1}{q}}\right){\displaystyle \frac{e^{2\lambda }}{{\lambda }^N}}{e}^{-\lambda n^{1-\alpha }},$$

(87)

(iv)

$${∥A_n\left(f\right)-f∥}_{\infty }\le \sum _{j=1}^N{\displaystyle \frac{{∥f^{\left(j\right)}∥}_{\infty }}{j!}}{\displaystyle \frac{1}{n^j}}\left({\displaystyle \frac{tanh\left(\lambda \right)}{\left(j+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }j!}{{\left(2\lambda \right)}^j}}\right)$$

$$+{\displaystyle \frac{{\omega }_1\left(f^{\left(N\right)},\frac{1}{n^{\alpha }}\right)}{n^{\alpha N}N!}}+{\displaystyle \frac{4{∥f^{\left(N\right)}∥}_{\infty }}{n^N}}\left(q+{\displaystyle \frac{1}{q}}\right){\displaystyle \frac{e^{2\lambda }}{{\lambda }^N}}{e}^{-\lambda n^{1-\alpha }}.$$

(88)

Proof. 

We have that

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

(89)

Then,

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

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

(90)

Hence,

$$A_n\left(f\right)\left(x\right)={\int }_{-\infty }^{\infty }f\left({\displaystyle \frac{u}{n}}\right)\phi \left(nx-u\right)du=$$

$$\sum _{j=0}^N{\displaystyle \frac{f^{\left(j\right)}\left(x\right)}{j!}}{\int }_{-\infty }^{\infty }\phi \left(nx-u\right){\left({\displaystyle \frac{u}{n}}-x\right)}^{j}du+$$

$${\int }_{-\infty }^{\infty }\phi \left(nx-u\right)\left({\int }_x^{{\displaystyle \frac{u}{n}}}\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right){\displaystyle \frac{{\left(\frac{u}{n}-t\right)}^{N-1}}{\left(N-1\right)!}}dt\right)du.$$

(91)

Thus, it holds that

$$A_n\left(f\right)\left(x\right)-f\left(x\right)=\sum _{j=1}^N{\displaystyle \frac{f^{\left(j\right)}\left(x\right)}{j!}}{\int }_{-\infty }^{\infty }\phi \left(nx-u\right){\left({\displaystyle \frac{u}{n}}-x\right)}^{j}du$$

$$+{\int }_{-\infty }^{\infty }\phi \left(nx-u\right)\left({\int }_x^{{\displaystyle \frac{u}{n}}}\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right){\displaystyle \frac{{\left(\frac{u}{n}-t\right)}^{N-1}}{\left(N-1\right)!}}dt\right)du.$$

(92)

Call

$$R_n\left(x\right):={\int }_{-\infty }^{\infty }\phi \left(nx-u\right)\left({\int }_x^{{\displaystyle \frac{u}{n}}}\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right){\displaystyle \frac{{\left(\frac{u}{n}-t\right)}^{N-1}}{\left(N-1\right)!}}dt\right)du.$$

(93)

Call

$$\gamma \left(u\right):={\int }_x^{{\displaystyle \frac{u}{n}}}\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right){\displaystyle \frac{{\left(\frac{u}{n}-t\right)}^{N-1}}{\left(N-1\right)!}}dt.$$

(94)

Let $\left|{\displaystyle \frac{u}{n}}-x\right|<{\displaystyle \frac{1}{n^{\alpha }}}$.

(i) The case where ${\displaystyle \frac{u}{n}}\ge x$. Then,

$$\left|\gamma \left(u\right)\right|\le {\int }_x^{{\displaystyle \frac{u}{n}}}\left|f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right|{\displaystyle \frac{{\left(\frac{u}{n}-t\right)}^{N-1}}{\left(N-1\right)!}}dt\le$$

$${\omega }_1\left(f^{\left(N\right)},{\displaystyle \frac{1}{n^{\alpha }}}\right)\left({\int }_x^{{\displaystyle \frac{u}{n}}}{\displaystyle \frac{{\left(\frac{u}{n}-t\right)}^{N-1}}{\left(N-1\right)!}}dt\right)=$$

(95)

$${\omega }_1\left(f^{\left(N\right)},{\displaystyle \frac{1}{n^{\alpha }}}\right){\displaystyle \frac{{\left(\frac{u}{n}-t\right)}^N}{N!}}\le {\omega }_1\left(f^{\left(N\right)},{\displaystyle \frac{1}{n^{\alpha }}}\right){\displaystyle \frac{1}{n^{\alpha N}N!}}.$$

(ii) The case where ${\displaystyle \frac{u}{n}}<x$. Then,

$$\left|\gamma \left(u\right)\right|=\left|{\int }_{{\displaystyle \frac{u}{n}}}^x\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right){\displaystyle \frac{{\left(\frac{u}{n}-t\right)}^{N-1}}{\left(N-1\right)!}}dt\right|\le$$

$${\int }_{{\displaystyle \frac{u}{n}}}^x\left|f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right|{\displaystyle \frac{{\left(t-\frac{u}{n}\right)}^{N-1}}{\left(N-1\right)!}}dt\le$$

$${\omega }_1\left(f^{\left(N\right)},{\displaystyle \frac{1}{n^{\alpha }}}\right)\left({\int }_{{\displaystyle \frac{u}{n}}}^x{\displaystyle \frac{{\left(t-\frac{u}{n}\right)}^{N-1}}{\left(N-1\right)!}}dt\right)=$$

(96)

$${\omega }_1\left(f^{\left(N\right)},{\displaystyle \frac{1}{n^{\alpha }}}\right){\displaystyle \frac{{\left(x-\frac{u}{n}\right)}^N}{N!}}\le {\omega }_1\left(f^{\left(N\right)},{\displaystyle \frac{1}{n^{\alpha }}}\right){\displaystyle \frac{1}{n^{\alpha N}N!}}.$$

Therefore, it holds that

$$\left|\gamma \left(u\right)\right|\le {\omega }_1\left(f^{\left(N\right)},{\displaystyle \frac{1}{n^{\alpha }}}\right){\displaystyle \frac{1}{n^{\alpha N}N!}},$$

(97)

when $\left|{\displaystyle \frac{u}{n}}-x\right|<{\displaystyle \frac{1}{n^{\alpha }}}$.

Consequently, we see that

$$\left|\underset{\left|{\displaystyle \frac{u}{n}}-x\right|<{\displaystyle \frac{1}{n^{\alpha }}}}{\int }\phi \left(nx-u\right)\left({\int }_x^{{\displaystyle \frac{u}{n}}}\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right){\displaystyle \frac{{\left(\frac{u}{n}-t\right)}^{N-1}}{\left(N-1\right)!}}dt\right)du\right|\le$$

$$\underset{\left|{\displaystyle \frac{u}{n}}-x\right|<{\displaystyle \frac{1}{n^{\alpha }}}}{\int }\phi \left(nx-u\right)\left|\gamma \left(u\right)\right|du\le {\omega }_1\left(f^{\left(N\right)},{\displaystyle \frac{1}{n^{\alpha }}}\right){\displaystyle \frac{1}{n^{\alpha N}N!}}.$$

(98)

Next, we treat

$$\left|\underset{\left|{\displaystyle \frac{u}{n}}-x\right|\ge {\displaystyle \frac{1}{n^{\alpha }}}}{\int }\phi \left(nx-u\right)\left({\int }_x^{{\displaystyle \frac{u}{n}}}\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right){\displaystyle \frac{{\left(\frac{u}{n}-t\right)}^{N-1}}{\left(N-1\right)!}}dt\right)du\right|\le$$

(99)

$$\underset{\left|{\displaystyle \frac{u}{n}}-x\right|\ge {\displaystyle \frac{1}{n^{\alpha }}}}{\int }\phi \left(nx-u\right)\left|{\int }_x^{{\displaystyle \frac{u}{n}}}\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right){\displaystyle \frac{{\left(\frac{u}{n}-t\right)}^{N-1}}{\left(N-1\right)!}}dt\right|du=$$

$$\underset{\left|{\displaystyle \frac{u}{n}}-x\right|\ge {\displaystyle \frac{1}{n^{\alpha }}}}{\int }\phi \left(nx-u\right)\left|\gamma \left(u\right)\right|du=:\left({\xi }_1\right).$$

Let ${\displaystyle \frac{u}{n}}\ge x$; then,

$$\left|\gamma \left(u\right)\right|\le 2{∥f^{\left(N\right)}∥}_{\infty }{\displaystyle \frac{{\left(\frac{u}{n}-x\right)}^N}{N!}}.$$

If ${\displaystyle \frac{u}{n}}<x$, then

$$\left|\gamma \left(u\right)\right|=\left|{\int }_{{\displaystyle \frac{u}{n}}}^x\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right){\displaystyle \frac{{\left(\frac{u}{n}-t\right)}^{N-1}}{\left(N-1\right)!}}dt\right|\le$$

$$2{∥f^{\left(N\right)}∥}_{\infty }{\int }_{{\displaystyle \frac{u}{n}}}^x{\displaystyle \frac{{\left(t-\frac{u}{n}\right)}^{N-1}}{\left(N-1\right)!}}dt=2{∥f^{\left(N\right)}∥}_{\infty }{\displaystyle \frac{{\left(x-\frac{u}{n}\right)}^N}{N!}}.$$

(100)

Consequently, we see that

$$\left|\gamma \left(u\right)\right|\le 2{∥f^{\left(N\right)}∥}_{\infty }{\displaystyle \frac{{\left|x-\frac{u}{n}\right|}^N}{N!}}.$$

(101)

Furthermore, we obtain

$$\left({\xi }_1\right)\le {\displaystyle \frac{2{∥f^{\left(N\right)}∥}_{\infty }}{N!}}\underset{\left|{\displaystyle \frac{u}{n}}-x\right|\ge {\displaystyle \frac{1}{n^{\alpha }}}}{\int }\phi \left(nx-u\right){\left|x-{\displaystyle \frac{u}{n}}\right|}^{N}du=$$

$${\displaystyle \frac{2{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\underset{\left|nx-u\right|\ge n^{1-\alpha }}{\int }\phi \left(\left|nx-u\right|\right){\left|nx-u\right|}^{N}du\stackrel{\left(31\right)}{\le }$$

(here, it is $\mathsf{\Delta }:=\left\{u\in \mathbb{R}:\left|nx-u\right|\ge n^{1-\alpha }\right\}$)

$${\displaystyle \frac{2{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right)\lambda {\int }_{\mathsf{\Delta }}{e}^{2\lambda \left(\left|nx-u\right|-1\right)}{\left|nx-u\right|}^{N}du=$$

$${\displaystyle \frac{4{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right)\lambda {\int }_{n^{1-\alpha }}^{\infty }{e}^{-2\lambda \left(x-1\right)}{x}^{N}dx=$$

$${\displaystyle \frac{4{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right)\lambda e^{2\lambda }{\int }_{n^{1-\alpha }}^{\infty }{e}^{-2\lambda x}{x}^{N}dx=$$

$${\displaystyle \frac{4{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right)\lambda {\displaystyle \frac{e^{2\lambda }}{{\left(2\lambda \right)}^{N+1}}}{\int }_{n^{1-\alpha }}^{\infty }{e}^{-2\lambda x}{\left(2\lambda x\right)}^{N}d\left(2\lambda x\right)=$$

(102)

$${\displaystyle \frac{2{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right){\displaystyle \frac{e^{2\lambda }}{{\left(2\lambda \right)}^N}}{\int }_{2\lambda n^{1-\alpha }}^{\infty }{e}^{-y}{y}^{N}dy\le$$

$${\displaystyle \frac{2{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right){\displaystyle \frac{e^{2\lambda }}{{\left(2\lambda \right)}^N}}{2}^{N}N!{\int }_{2\lambda n^{1-\alpha }}^{\infty }{e}^{-y}{e}^{{\displaystyle \frac{y}{2}}}dy=$$

$${\displaystyle \frac{2^{N+1}{∥f^{\left(N\right)}∥}_{\infty }}{n^N}}\left(q+{\displaystyle \frac{1}{q}}\right){\displaystyle \frac{e^{2\lambda }}{{\left(2\lambda \right)}^N}}{\int }_{2\lambda n^{1-\alpha }}^{\infty }{e}^{-{\displaystyle \frac{y}{2}}}dy=$$

$${\displaystyle \frac{2{∥f^{\left(N\right)}∥}_{\infty }}{n^N}}\left(q+{\displaystyle \frac{1}{q}}\right){\displaystyle \frac{e^{2\lambda }}{{\lambda }^N}}\left(-2\right)\left(e^{-{\displaystyle \frac{y}{2}}}{|}_{2\lambda n^{1-\alpha }}^{\infty }\right)=$$

$${\displaystyle \frac{4{∥f^{\left(N\right)}∥}_{\infty }}{n^N}}\left(q+{\displaystyle \frac{1}{q}}\right){\displaystyle \frac{e^{2\lambda }}{{\lambda }^N}}\left(e^{-{\displaystyle \frac{y}{2}}}{|}_{\infty }^{2\lambda n^{1-\alpha }}\right)=$$

(103)

$${\displaystyle \frac{4{∥f^{\left(N\right)}∥}_{\infty }}{n^N}}\left(q+{\displaystyle \frac{1}{q}}\right){\displaystyle \frac{e^{2\lambda }}{{\lambda }^N}}{e}^{-\lambda n^{1-\alpha }}.$$

We have proved that

$$\left|\underset{\left|{\displaystyle \frac{u}{n}}-x\right|\ge {\displaystyle \frac{1}{n^{\alpha }}}}{\int }\phi \left(nx-u\right)\left({\int }_x^{{\displaystyle \frac{u}{n}}}\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right){\displaystyle \frac{{\left(\frac{u}{n}-t\right)}^{N-1}}{\left(N-1\right)!}}dt\right)du\right|\le$$

(104)

$${\displaystyle \frac{4{∥f^{\left(N\right)}∥}_{\infty }}{n^N}}\left(q+{\displaystyle \frac{1}{q}}\right){\displaystyle \frac{e^{2\lambda }}{{\lambda }^N}}{e}^{-\lambda n^{1-\alpha }}\to 0,\mathrm{as}n\to \infty .$$

Finally, we have that

$$\left|R_n\left(x\right)\right|\le \left|\underset{\left|{\displaystyle \frac{u}{n}}-x\right|<{\displaystyle \frac{1}{n^{\alpha }}}}{\int }\phi \left(nx-u\right)\left({\int }_x^{{\displaystyle \frac{u}{n}}}\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right){\displaystyle \frac{{\left(\frac{u}{n}-t\right)}^{N-1}}{\left(N-1\right)!}}dt\right)du\right|$$

$$+\left|\underset{\left|{\displaystyle \frac{u}{n}}-x\right|\ge {\displaystyle \frac{1}{n^{\alpha }}}}{\int }\phi \left(nx-u\right)\left({\int }_x^{{\displaystyle \frac{u}{n}}}\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right){\displaystyle \frac{{\left(\frac{u}{n}-t\right)}^{N-1}}{\left(N-1\right)!}}dt\right)du\right|\le$$

$${\omega }_1\left(f^{\left(N\right)},{\displaystyle \frac{1}{n^{\alpha }}}\right){\displaystyle \frac{1}{n^{\alpha N}N!}}+{\displaystyle \frac{4{∥f^{\left(N\right)}∥}_{\infty }}{n^N}}\left(q+{\displaystyle \frac{1}{q}}\right){\displaystyle \frac{e^{2\lambda }}{{\lambda }^N}}{e}^{-\lambda n^{1-\alpha }}\to 0,\mathrm{as}n\to \infty .$$

(105)

Let $j=1,\dots ,N$ and $z:=nx-u$. Then,

$$\left|A_n\left({\left(·-x\right)}^j\right)\left(x\right)\right|=\left|{\int }_{-\infty }^{\infty }{\left({\displaystyle \frac{u}{n}}-x\right)}^j\phi \left(nx-u\right)du\right|\le$$

$${\int }_{-\infty }^{\infty }{\left|{\displaystyle \frac{u}{n}}-x\right|}^j\phi \left(nx-u\right)du=$$

$${\displaystyle \frac{1}{n^j}}{\int }_{-\infty }^{\infty }{\left|nx-u\right|}^j\phi \left(nx-u\right)du={\displaystyle \frac{1}{n^j}}{\int }_{-\infty }^{\infty }{\left|z\right|}^j\phi \left(z\right)dz\stackrel{\left(52\right)}{\le }$$

$${\displaystyle \frac{1}{n^j}}\left[{\displaystyle \frac{tanh\left(\lambda \right)}{\left(j+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }j!}{{\left(2\lambda \right)}^j}}\right]\to 0,$$

(106)

as $n\to \infty$.

The theorem now is proven. □

We continue with the activated Kantorovich operators.

Theorem 11. 

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

(i)

$$\left|A_n^{*}\left(f\right)\left(x\right)-f\left(x\right)-\sum _{j=1}^N{\displaystyle \frac{f^{\left(j\right)}\left(x\right)}{j!}}{A}_n^{*}\left({\left(·-x\right)}^j\right)\left(x\right)\right|\le$$

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

(107)

$${\displaystyle \frac{2^{N+1}{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{N!}{{\lambda }^N}}\right)e^{-\lambda n^{1-\alpha }}\to 0,\mathit{as}n\to \infty .$$

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

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

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

$${\displaystyle \frac{2^{N+1}{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{N!}{{\lambda }^N}}\right)e^{-\lambda n^{1-\alpha }},$$

(108)

(iii)

$$\left|A_n^{*}\left(f\right)\left(x\right)-f\left(x\right)\right|\le \sum _{j=1}^N{\displaystyle \frac{\left|f^{\left(j\right)}\left(x\right)\right|}{j!}}{\displaystyle \frac{2^{j-1}}{n^j}}\left[1+\left({\displaystyle \frac{tanh\left(\lambda \right)}{\left(j+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }j!}{{\left(2\lambda \right)}^j}}\right)\right]$$

(109)

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

(iv)

$${∥A_n^{*}\left(f\right)-f∥}_{\infty }\le \sum _{j=1}^N{\displaystyle \frac{{∥f^{\left(j\right)}∥}_{\infty }}{j!}}{\displaystyle \frac{2^{j-1}}{n^j}}\left[1+\left({\displaystyle \frac{tanh\left(\lambda \right)}{\left(j+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }j!}{{\left(2\lambda \right)}^j}}\right)\right]$$

$$+{\omega }_1\left(f^{\left(N\right)},{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\alpha }}}\right){\displaystyle \frac{{\left(\frac{1}{n}+\frac{1}{n^{\alpha }}\right)}^N}{N!}}+{\displaystyle \frac{2^{N+1}{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{N!}{{\lambda }^N}}\right)e^{-\lambda n^{1-\alpha }}\to 0,$$

(110)

as $n\to \infty$.

Proof. 

We have that

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

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

(111)

and

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

(112)

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

Hence,

$$n{\int }_{-\infty }^{\infty }\left({\int }_0^{{\displaystyle \frac{1}{n}}}f\left(t+{\displaystyle \frac{u}{n}}\right)dt\right)\phi \left(nx-u\right)du=$$

(113)

$$\sum _{j=0}^N{\displaystyle \frac{f^{\left(j\right)}\left(x\right)}{j!}}n{\int }_{-\infty }^{\infty }\left({\int }_0^{{\displaystyle \frac{1}{n}}}{\left(t+{\displaystyle \frac{u}{n}}-x\right)}^{j}dt\right)\phi \left(nx-u\right)du+$$

$$n{\int }_{-\infty }^{\infty }\left({\int }_0^{{\displaystyle \frac{1}{n}}}\left({\int }_x^{t+{\displaystyle \frac{u}{n}}}\left(f^{\left(N\right)}\left(s\right)-f^{\left(N\right)}\left(x\right)\right){\displaystyle \frac{{\left(t+\frac{u}{n}-s\right)}^{N-1}}{\left(N-1\right)!}}ds\right)dt\right)\phi \left(nx-u\right)du.$$

Furthermore, it holds that

$$A_n^{*}\left(f\right)\left(x\right)-f\left(x\right)=\sum _{j=1}^N{\displaystyle \frac{f^{\left(j\right)}\left(x\right)}{j!}}{A}_n^{*}\left({\left(·-x\right)}^j\right)\left(x\right)+R_n\left(x\right),$$

(114)

where

$$R_n\left(x\right):=n$$

$${\int }_{-\infty }^{\infty }\left({\int }_0^{{\displaystyle \frac{1}{n}}}\left({\int }_x^{t+{\displaystyle \frac{u}{n}}}\left(f^{\left(N\right)}\left(s\right)-f^{\left(N\right)}\left(x\right)\right){\displaystyle \frac{{\left(t+\frac{u}{n}-s\right)}^{N-1}}{\left(N-1\right)!}}ds\right)dt\right)\phi \left(nx-u\right)du.$$

(115)

Call

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

(116)

Let $\left|{\displaystyle \frac{u}{n}}-x\right|<{\displaystyle \frac{1}{n^{\alpha }}}$, ($0<\alpha <1$).

(i) The case where $t+{\displaystyle \frac{u}{n}}\ge x$. Then,

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

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

(117)

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

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

Therefore, it holds that

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

(118)

(ii) The case where $t+{\displaystyle \frac{u}{n}}<x$. Then,

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

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

(119)

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

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

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

(by $x-\left(t+{\displaystyle \frac{u}{n}}\right)=\left|x-t-{\displaystyle \frac{u}{n}}\right|\le \left|{\displaystyle \frac{u}{n}}-x\right|+\left|t\right|$)

$${\omega }_1\left(f^{\left(N\right)},{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\alpha }}}\right)n{\int }_0^{{\displaystyle \frac{1}{n}}}{\displaystyle \frac{{\left(\frac{1}{n}+\frac{1}{n^{\alpha }}\right)}^N}{N!}}dt=$$

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

(120)

So, when $\left|{\displaystyle \frac{u}{n}}-x\right|<{\displaystyle \frac{1}{n^{\alpha }}}$, we have proved that

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

(121)

Consequently, we see that

$$\left|n\underset{\left|{\displaystyle \frac{u}{n}}-x\right|<{\displaystyle \frac{1}{n^{\alpha }}}}{{\int }_{-\infty }^{\infty }}\left({\int }_0^{{\displaystyle \frac{1}{n}}}\left({\int }_x^{t+{\displaystyle \frac{u}{n}}}\left(f^{\left(N\right)}\left(s\right)-f^{\left(N\right)}\left(x\right)\right){\displaystyle \frac{{\left(t+\frac{u}{n}-s\right)}^{N-1}}{\left(N-1\right)!}}ds\right)dt\right)\phi \left(nx-u\right)du\right|$$

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

(122)

Next, we treat

$$\left|\underset{\left|{\displaystyle \frac{u}{n}}-x\right|\ge {\displaystyle \frac{1}{n^{\alpha }}}}{\int }n\left({\int }_0^{{\displaystyle \frac{1}{n}}}\left({\int }_x^{t+{\displaystyle \frac{u}{n}}}\left(f^{\left(N\right)}\left(s\right)-f^{\left(N\right)}\left(x\right)\right){\displaystyle \frac{{\left(t+\frac{u}{n}-s\right)}^{N-1}}{\left(N-1\right)!}}ds\right)dt\right)\phi \left(nx-u\right)du\right|$$

(123)

$$\le \underset{\left|{\displaystyle \frac{u}{n}}-x\right|\ge {\displaystyle \frac{1}{n^{\alpha }}}}{\int }\left|n\left({\int }_0^{{\displaystyle \frac{1}{n}}}\left({\int }_x^{t+{\displaystyle \frac{u}{n}}}\left(f^{\left(N\right)}\left(s\right)-f^{\left(N\right)}\left(x\right)\right){\displaystyle \frac{{\left(t+\frac{u}{n}-s\right)}^{N-1}}{\left(N-1\right)!}}ds\right)dt\right)\right|$$

$$\phi \left(nx-u\right)du=:\left({\psi }_1\right).$$

We also have that

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

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

(124)

(let $t+{\displaystyle \frac{u}{n}}\ge x$)

$$2{∥f^{\left(N\right)}∥}_{\infty }n\left({\int }_0^{{\displaystyle \frac{1}{n}}}\left({\int }_x^{t+{\displaystyle \frac{u}{n}}}{\displaystyle \frac{{\left(t+\frac{u}{n}-s\right)}^{N-1}}{\left(N-1\right)!}}ds\right)dt\right)=$$

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

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

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

So, when $t+{\displaystyle \frac{u}{n}}\ge x$, we obtain

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

(125)

If $t+{\displaystyle \frac{u}{n}}<x$, then

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

$$2{∥f^{\left(N\right)}∥}_{\infty }n\left({\int }_0^{{\displaystyle \frac{1}{n}}}\left({\int }_{t+{\displaystyle \frac{u}{n}}}^x{\displaystyle \frac{{\left(s-\left(t+\frac{u}{n}\right)\right)}^{N-1}}{\left(N-1\right)!}}ds\right)dt\right)=$$

(126)

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

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

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

Consequently, we see that

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

(127)

Furthermore, we obtain

$$\left({\psi }_1\right)\le \left(\underset{\left|{\displaystyle \frac{u}{n}}-x\right|\ge {\displaystyle \frac{1}{n^{\alpha }}}}{\int }{\left({\displaystyle \frac{1}{n}}+\left|{\displaystyle \frac{u}{n}}-x\right|\right)}^N\phi \left(nx-u\right)du\right){\displaystyle \frac{2{∥f^{\left(N\right)}∥}_{\infty }}{N!}}=$$

$${\displaystyle \frac{2^N{∥f^{\left(N\right)}∥}_{\infty }}{N!}}\underset{\left|nx-u\right|\ge n^{1-\alpha }}{\int }\left({\displaystyle \frac{1}{n^N}}+{\displaystyle \frac{{\left|nx-u\right|}^N}{n^N}}\right)\phi \left(nx-u\right)du\le$$

(128)

$${\displaystyle \frac{2^N{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(\underset{\left|nx-u\right|\ge n^{1-\alpha }}{\int }\left(1+{\left|nx-u\right|}^N\right)\phi \left(\left|nx-u\right|\right)du\right)\stackrel{\left(31\right)}{\le }$$

(here, it is $\mathsf{\Delta }:=\left\{u\in \mathbb{R}:\left|nx-u\right|\ge n^{1-\alpha }\right\}$)

$${\displaystyle \frac{2^N{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(\left(q+{\displaystyle \frac{1}{q}}\right)\lambda {\int }_{\mathsf{\Delta }}\left(1+{\left|nx-u\right|}^N\right)e^{-2\lambda \left(\left|nx-u\right|-1\right)}du\right)=$$

$${\displaystyle \frac{2^{N+1}{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right)\lambda {\int }_{n^{1-\alpha }}^{\infty }{e}^{-2\lambda \left(x-1\right)}\left(1+x^N\right)dx=$$

$${\displaystyle \frac{2^{N+1}{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right)\lambda e^{2\lambda }{\int }_{n^{1-\alpha }}^{\infty }{e}^{-2\lambda x}\left(1+x^N\right)dx=$$

$${\displaystyle \frac{2^{N+1}{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right)\lambda e^{2\lambda }\left[{\int }_{n^{1-\alpha }}^{\infty }{e}^{-2\lambda x}dx+{\int }_{n^{1-\alpha }}^{\infty }{e}^{-2\lambda x}{x}^{N}dx\right]=$$

(129)

$${\displaystyle \frac{2^{N+1}{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right)\lambda e^{2\lambda }\left[{\displaystyle \frac{e^{-2\lambda n^{1-\alpha }}}{2\lambda }}+{\int }_{n^{1-\alpha }}^{\infty }{e}^{-2\lambda x}{x}^{N}dx\right]=$$

$${\displaystyle \frac{2^{N+1}{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right)\lambda e^{2\lambda }\left[{\displaystyle \frac{e^{-2\lambda n^{1-\alpha }}}{2\lambda }}+{\displaystyle \frac{1}{{\left(2\lambda \right)}^{N+1}}}{\int }_{2\lambda n^{1-\alpha }}^{\infty }{e}^{-y}{y}^{N}dy\right]\le$$

$${\displaystyle \frac{2^{N+1}{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right)\lambda e^{2\lambda }\left[{\displaystyle \frac{e^{-2\lambda n^{1-\alpha }}}{2\lambda }}+{\displaystyle \frac{2^{N}N!}{{\left(2\lambda \right)}^{N+1}}}{\int }_{2\lambda n^{1-\alpha }}^{\infty }{e}^{-{\displaystyle \frac{y}{2}}}dy\right]=$$

(130)

$${\displaystyle \frac{2^{N+1}{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right)\lambda e^{2\lambda }\left[{\displaystyle \frac{e^{-2\lambda n^{1-\alpha }}}{2\lambda }}+{\displaystyle \frac{2^{N+1}N!}{{\left(2\lambda \right)}^{N+1}}}{e}^{-\lambda n^{1-\alpha }}\right]=$$

$${\displaystyle \frac{2^{N+1}{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right)\lambda e^{2\lambda }\left[{\displaystyle \frac{e^{-2\lambda n^{1-\alpha }}}{2\lambda }}+{\displaystyle \frac{N!}{{\lambda }^{N+1}}}{e}^{-\lambda n^{1-\alpha }}\right]\le$$

(131)

$$\le {\displaystyle \frac{2^{N+1}{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }\left[{\displaystyle \frac{1}{2}}+{\displaystyle \frac{N!}{{\lambda }^N}}\right]e^{-\lambda n^{1-\alpha }}\to 0,$$

as $n\to \infty$.

We have proved that

$$\left|\underset{\left|{\displaystyle \frac{u}{n}}-x\right|\ge {\displaystyle \frac{1}{n^{\alpha }}}}{\int }n\left({\int }_0^{{\displaystyle \frac{1}{n}}}\left({\int }_x^{t+{\displaystyle \frac{u}{n}}}\left(f^{\left(N\right)}\left(s\right)-f^{\left(N\right)}\left(x\right)\right){\displaystyle \frac{{\left(t+\frac{u}{n}-s\right)}^{N-1}}{\left(N-1\right)!}}ds\right)dt\right)\phi \left(nx-u\right)du\right|$$

(132)

$$\le {\displaystyle \frac{2^{N+1}{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }\left[{\displaystyle \frac{1}{2}}+{\displaystyle \frac{N!}{{\lambda }^N}}\right]e^{-\lambda n^{1-\alpha }}\to 0,\mathrm{as}n\to \infty .$$

Finally, we have that

$$\left|R_n\left(x\right)\right|\le {\omega }_1\left(f^{\left(N\right)},{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\alpha }}}\right){\displaystyle \frac{{\left(\frac{1}{n}+\frac{1}{n^{\alpha }}\right)}^N}{N!}}+$$

$${\displaystyle \frac{2^{N+1}{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{N!}{{\lambda }^N}}\right)e^{-\lambda n^{1-\alpha }}\to 0,\mathrm{as}n\to \infty .$$

(133)

We also observe that ($j=1,\dots ,N$)

$$\left|A_n^{*}\left({\left(·-x\right)}^j\right)\left(x\right)\right|=\left|n{\int }_{-\infty }^{\infty }\left({\int }_0^{{\displaystyle \frac{1}{n}}}{\left(t+{\displaystyle \frac{u}{n}}-x\right)}^{j}dt\right)\phi \left(nx-u\right)du\right|\le$$

$$n{\int }_{-\infty }^{\infty }\left({\int }_0^{{\displaystyle \frac{1}{n}}}{\left|t+{\displaystyle \frac{u}{n}}-x\right|}^{j}dt\right)\phi \left(nx-u\right)du\le$$

$$n{\int }_{-\infty }^{\infty }\left({\int }_0^{{\displaystyle \frac{1}{n}}}{\left(\left|t\right|+\left|{\displaystyle \frac{u}{n}}-x\right|\right)}^{j}dt\right)\phi \left(nx-u\right)du\le$$

$${\int }_{-\infty }^{\infty }{\left({\displaystyle \frac{1}{n}}+\left|{\displaystyle \frac{u}{n}}-x\right|\right)}^j\phi \left(nx-u\right)du=$$

(134)

$${\displaystyle \frac{1}{n^j}}{\int }_{-\infty }^{\infty }{\left(1+\left|nx-u\right|\right)}^j\phi \left(nx-u\right)du\le$$

$${\displaystyle \frac{2^{j-1}}{n^j}}\left[1+{\int }_{-\infty }^{\infty }{\left|nx-u\right|}^j\phi \left(nx-u\right)du\right]=$$

$${\displaystyle \frac{2^{j-1}}{n^j}}\left[1+{\int }_{-\infty }^{\infty }{\left|z\right|}^j\phi \left(z\right)dz\right]\stackrel{\left(52\right)}{\le }$$

$${\displaystyle \frac{2^{j-1}}{n^j}}\left[1+\left({\displaystyle \frac{tanh\left(\lambda \right)}{\left(j+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }j!}{{\left(2\lambda \right)}^j}}\right)\right]\to 0,\mathrm{as}n\to \infty .$$

The theorem is proven. □

We continue with the activated Quadrature operators.

Theorem 12. 

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

(i)

$$\left|\overline{A_n}\left(f\right)\left(x\right)-f\left(x\right)-\sum _{j=1}^N{\displaystyle \frac{f^{\left(j\right)}\left(x\right)}{j!}}\overline{A_n}\left({\left(·-x\right)}^j\right)\left(x\right)\right|\le$$

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

(135)

$${\displaystyle \frac{2^{N+1}{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{N!}{{\lambda }^N}}\right)e^{-\lambda n^{1-\alpha }}\to 0,\mathit{as}n\to \infty .$$

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

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

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

$${\displaystyle \frac{2^{N+1}{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{N!}{{\lambda }^N}}\right)e^{-\lambda n^{1-\alpha }},$$

(136)

(iii)

$$\left|\overline{A_n}\left(f\right)\left(x\right)-f\left(x\right)\right|\le \sum _{j=1}^N{\displaystyle \frac{\left|f^{\left(j\right)}\left(x\right)\right|}{j!}}{\displaystyle \frac{2^{j-1}}{n^j}}\left[1+\left({\displaystyle \frac{tanh\left(\lambda \right)}{\left(j+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }j!}{{\left(2\lambda \right)}^j}}\right)\right]$$

(137)

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

(iv)

$${∥\overline{A_n}\left(f\right)-f∥}_{\infty }\le \sum _{j=1}^N{\displaystyle \frac{{∥f^{\left(j\right)}∥}_{\infty }}{j!}}{\displaystyle \frac{2^{j-1}}{n^j}}\left[1+\left({\displaystyle \frac{tanh\left(\lambda \right)}{\left(j+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }j!}{{\left(2\lambda \right)}^j}}\right)\right]$$

(138)

$$+{\omega }_1\left(f^{\left(N\right)},{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\alpha }}}\right){\displaystyle \frac{{\left(\frac{1}{n}+\frac{1}{n^{\alpha }}\right)}^N}{N!}}+{\displaystyle \frac{2^{N+1}{∥f^{\left(N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{N!}{{\lambda }^N}}\right)e^{-\lambda n^{1-\alpha }}\to 0,$$

as $n\to \infty$.

Proof. 

We have that

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

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

(139)

and

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

(140)

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

Furthermore, it holds that

$$\overline{A_n}\left(f\right)\left(x\right)={\int }_{-\infty }^{\infty }\left(\sum _{i=1}^r{w}_{i}f\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}\right)\right)\phi \left(nx-u\right)du=$$

$$\sum _{j=0}^N{\displaystyle \frac{f^{\left(j\right)}\left(x\right)}{j!}}\left({\int }_{-\infty }^{\infty }\sum _{i=1}^r{w}_i{\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}-x\right)}^j\right)\phi \left(nx-u\right)du$$

$$+{\int }_{-\infty }^{\infty }\left(\sum _{i=1}^r{w}_i{\int }_x^{{\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}}\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right){\displaystyle \frac{{\left(\frac{u}{n}+\frac{i}{nr}-t\right)}^{N-1}}{\left(N-1\right)!}}dt\right)\phi \left(nx-u\right)du,$$

(141)

and it is

$$\overline{A_n}\left(f\right)\left(x\right)-f\left(x\right)=\sum _{j=1}^N{\displaystyle \frac{f^{\left(j\right)}\left(x\right)}{j!}}\left(\overline{A_n}\left({\left(·-x\right)}^j\right)\left(x\right)\right)+R_n\left(x\right),$$

(142)

where

$$R_n\left(x\right):=$$

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

(143)

Let

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

(144)

Let $\left|{\displaystyle \frac{u}{n}}-x\right|<{\displaystyle \frac{1}{n^{\alpha }}}$.

(i) The case where ${\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}\ge x$. Then,

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

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

(145)

(ii) The case where ${\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}<x$. Then,

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

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

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

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

So, when $\left|{\displaystyle \frac{u}{n}}-x\right|<{\displaystyle \frac{1}{n^{\alpha }}}$, we see that

$$\left|\delta \left(u\right)\right|\le {\omega }_1\left(f^{\left(N\right)},{\displaystyle \frac{1}{n^{\alpha }}}+{\displaystyle \frac{1}{n}}\right){\displaystyle \frac{{\left(\frac{1}{n^{\alpha }}+\frac{1}{n}\right)}^N}{N!}}.$$

(146)

Consequently, we have that

$$\left|\underset{\left|{\displaystyle \frac{u}{n}}-x\right|<{\displaystyle \frac{1}{n^{\alpha }}}}{\int }\left(\sum _{i=1}^r{w}_i{\int }_x^{{\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}}\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right){\displaystyle \frac{{\left(\frac{u}{n}+\frac{i}{nr}-t\right)}^{N-1}}{\left(N-1\right)!}}dt\right)\phi \left(nx-u\right)du\right|$$

$$\le \underset{\left|{\displaystyle \frac{u}{n}}-x\right|<{\displaystyle \frac{1}{n^{\alpha }}}}{\int }\phi \left(nx-u\right)\left|\delta \left(u\right)\right|du\le {\omega }_1\left(f^{\left(N\right)},{\displaystyle \frac{1}{n^{\alpha }}}+{\displaystyle \frac{1}{n}}\right){\displaystyle \frac{{\left(\frac{1}{n^{\alpha }}+\frac{1}{n}\right)}^N}{N!}}.$$

(147)

Next, we treat

$$\left|\underset{\left|{\displaystyle \frac{u}{n}}-x\right|\ge {\displaystyle \frac{1}{n^{\alpha }}}}{\int }\phi \left(nx-u\right)\left(\sum _{i=1}^r{w}_i{\int }_x^{{\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}}\left(f^{\left(N\right)}\left(t\right)-f^{\left(N\right)}\left(x\right)\right){\displaystyle \frac{{\left(\frac{u}{n}+\frac{i}{nr}-t\right)}^{N-1}}{\left(N-1\right)!}}dt\right)du\right|$$

(148)

$$\le \underset{\left|{\displaystyle \frac{u}{n}}-x\right|\ge {\displaystyle \frac{1}{n^{\alpha }}}}{\int }\phi \left(nx-u\right)\left|\delta \left(u\right)\right|du=:\left({\phi }_1\right).$$

Let ${\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}\ge x$, then

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

(149)

Let ${\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}<x$, then

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

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

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

(150)

Consequently, we see that

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

(151)

(see also (127)).

The quantity

$$\left|\underset{\left|{\displaystyle \frac{u}{n}}-x\right|\ge {\displaystyle \frac{1}{n^{\alpha }}}}{\int }\phi \left(nx-u\right)\delta \left(u\right)du\right|\le \left({\phi }_1\right)\le$$

(152)

is estimated as in Theorem 11.

We also see that ($j=1,\dots ,N$)

$$\left|\overline{A_n}\left({\left(·-x\right)}^j\right)\left(x\right)\right|=\left|{\int }_{-\infty }^{\infty }\left(\sum _{i=1}^r{w}_i{\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}-x\right)}^j\right)\phi \left(nx-u\right)du\right|\le$$

$${\int }_{-\infty }^{\infty }\left(\sum _{i=1}^r{w}_i{\left|{\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}-x\right|}^j\right)\phi \left(nx-u\right)du\le$$

$${\int }_{-\infty }^{\infty }\left(\sum _{i=1}^r{w}_i{\left(\left|{\displaystyle \frac{u}{n}}-x\right|+{\displaystyle \frac{i}{nr}}\right)}^j\right)\phi \left(nx-u\right)du\le$$

(153)

$${\int }_{-\infty }^{\infty }{\left({\displaystyle \frac{1}{n}}+\left|{\displaystyle \frac{u}{n}}-x\right|\right)}^j\phi \left(nx-u\right)du=$$

$${\displaystyle \frac{1}{n^j}}{\int }_{-\infty }^{\infty }{\left(1+\left|nx-u\right|\right)}^j\phi \left(nx-u\right)du$$

(see (134); the rest goes as in Theorem 11).

The theorem is proven. □

We need the following.

Definition 2. 

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

Definition 3. 

Let $\nu \ge 0$, $n=⌈\nu ⌉$ ($⌈·⌉$ is the ceiling of the number), $f\in A{C}^n\left(\mathbb{R}\right)$. We call left Caputo fractional derivative ([13,14,15], pp. 49–52), the function

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

(154)

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

Notice that $D_{*a}^{\nu }f\in L_1\left(\left[a,b\right]\right)$ and $D_{*a}^{\nu }f$ exists almost everywhere on $\left[a,b\right]$, ∀$\left[a,b\right]\subset \mathbb{R}$.

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

We need the following.

Lemma 1 

(see also [16]). 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, $D_{*a}^{\nu }f\left(a\right)=0$ for any $a\in \mathbb{R}$.

Definition 4 

(see also [14,17,18]). 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

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

(155)

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

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

Lemma 2 

(see also [16]). 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 $b\in \mathbb{R}$.

We assume that

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

(156)

We mention the following.

Proposition 2 

(see also [16]). Let $f\in C^n\left(\mathbb{R}\right)$, $n=⌈\nu ⌉$, $\nu >0$. Then, $D_{*a}^{\nu }f\left(x\right)$ is continuous in $x\in [a,\infty )$, $a\in \mathbb{R}$.

Also we have the following.

Proposition 3 

(see also [16]). 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 the following.

Proposition 4 

(see also [16]). 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, $D_{*x_0}^{\alpha }f\left(x\right)$ is continuous in $x_0$.

Proposition 5 

(see also [16]). 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 [16]). Let $f\in C^m\left(\mathbb{R}\right)$, $m=⌈\alpha ⌉$, $\alpha >0$; $x,x_0\in \mathbb{R}$. Then, $D_{*x_0}^{\alpha }f\left(x\right)$, $D_{x_0-}^{\alpha }f\left(x\right)$ are jointly continuous functions in $\left(x,x_0\right)$ from $R^2\to \mathbb{R}$.

Here comes our first main fractional result.

Theorem 13. 

Let $\alpha >0$, $N=⌈\alpha ⌉$, $\alpha \notin \mathbb{N}$, $f\in A{C}^N\left(\mathbb{R}\right)$, $f^{\left(N\right)}\in L_{\infty }\left(\mathbb{R}\right)$, $0<\beta <1$, $x\in \mathbb{R}$, $n\in \mathbb{N}:n^{1-\beta }\ge max\left(3,{\displaystyle \frac{1}{2\lambda }}\right)$. Assume also that both ${\displaystyle \underset{x\in \mathbb{R}}{sup}}{∥D_{*x}^{\alpha }f∥}_{\infty }$, ${\displaystyle \underset{x\in \mathbb{R}}{sup}}{∥D_{x-}^{\alpha }f∥}_{\infty }<\infty$.

Then,

(i)

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

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

(157)

$$\left\{\left[{∥D_{x-}^{\alpha }f∥}_{\infty }+{∥D_{*x}^{\alpha }f∥}_{\infty }\right]{\displaystyle \frac{2^{N+1}N!\left(q+\frac{1}{q}\right)e^{2\lambda }}{\mathsf{\Gamma }\left(\alpha +1\right){\left(2\lambda \right)}^{\alpha }{n}^{\alpha }{e}^{\lambda n^{1-\beta }}}}\right\}\to 0,\mathit{as}n\to \infty .$$

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

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

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

(158)

$$\left\{\left[{∥D_{x-}^{\alpha }f∥}_{\infty }+{∥D_{*x}^{\alpha }f∥}_{\infty }\right]{\displaystyle \frac{2^{N+1}N!\left(q+\frac{1}{q}\right)e^{2\lambda }}{\mathsf{\Gamma }\left(\alpha +1\right){\left(2\lambda \right)}^{\alpha }{n}^{\alpha }{e}^{\lambda n^{1-\beta }}}}\right\}=:{\mathsf{\Phi }}_1\left(x\right),$$

(iii)

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

$${\displaystyle \frac{1}{n^j}}\left[\left({\displaystyle \frac{tanh\left(\lambda \right)}{\left(j+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }j!}{{\left(2\lambda \right)}^j}}\right)\right]+{\mathsf{\Phi }}_1\left(x\right),$$

(159)

(iv)

$${∥A_n\left(f\right)-f∥}_{\infty }\le \sum _{j=1}^{N-1}{\displaystyle \frac{{∥f^{\left(j\right)}∥}_{\infty }}{j!}}{\displaystyle \frac{1}{n^j}}\left[\left({\displaystyle \frac{tanh\left(\lambda \right)}{\left(j+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }j!}{{\left(2\lambda \right)}^j}}\right)\right]+$$

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

$$\left\{\left[\underset{x\in \mathbb{R}}{sup}{∥D_{x-}^{\alpha }f∥}_{\infty }+\underset{x\in \mathbb{R}}{sup}{∥D_{*x}^{\alpha }f∥}_{\infty }\right]{\displaystyle \frac{2^{N+1}N!\left(q+\frac{1}{q}\right)e^{2\lambda }}{\mathsf{\Gamma }\left(\alpha +1\right){\left(2\lambda \right)}^{\alpha }{n}^{\alpha }{e}^{\lambda n^{1-\beta }}}}\right\}.$$

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

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

Proof. 

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

From [13], p. 54, we obtain the left Caputo fractional Taylor formula that

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

(160)

$${\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_x^{{\displaystyle \frac{u}{n}}}{\left({\displaystyle \frac{u}{n}}-s\right)}^{\alpha -1}\left(D_{*x}^{\alpha }f\left(s\right)-D_{*x}^{\alpha }f\left(x\right)\right)ds,$$

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

Also, from [17], using the right Caputo fractional Taylor formula, we obtain

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

(161)

$${\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_{{\displaystyle \frac{u}{n}}}^x{\left(s-{\displaystyle \frac{u}{n}}\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\left(s\right)-D_{x-}^{\alpha }f\left(x\right)\right)ds,$$

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

Hence,

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

(162)

$${\displaystyle \frac{\phi \left(nx-u\right)}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_x^{{\displaystyle \frac{u}{n}}}{\left({\displaystyle \frac{u}{n}}-s\right)}^{\alpha -1}\left(D_{*x}^{\alpha }f\left(s\right)-D_{*x}^{\alpha }f\left(x\right)\right)ds,$$

for all $x\le {\displaystyle \frac{u}{n}}<\infty$, and

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

(163)

$${\displaystyle \frac{\phi \left(nx-u\right)}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_{{\displaystyle \frac{u}{n}}}^x{\left(s-{\displaystyle \frac{u}{n}}\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\left(s\right)-D_{x-}^{\alpha }f\left(x\right)\right)ds,$$

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

Therefore, we have

$${\int }_{nx}^{\infty }f\left({\displaystyle \frac{u}{n}}\right)\phi \left(nx-u\right)du=\sum _{j=0}^{N-1}{\displaystyle \frac{f^{\left(j\right)}\left(x\right)}{j!}}{\int }_{nx}^{\infty }\phi \left(nx-u\right){\left({\displaystyle \frac{u}{n}}-x\right)}^{j}du+$$

(164)

$${\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_{nx}^{\infty }\phi \left(nx-u\right)\left({\int }_x^{{\displaystyle \frac{u}{n}}}{\left({\displaystyle \frac{u}{n}}-s\right)}^{\alpha -1}\left(D_{*x}^{\alpha }f\left(s\right)-D_{*x}^{\alpha }f\left(x\right)\right)ds\right)du,$$

and

$${\int }_{-\infty }^{nx}f\left({\displaystyle \frac{u}{n}}\right)\phi \left(nx-u\right)du=\sum _{j=0}^{N-1}{\displaystyle \frac{f^{\left(j\right)}\left(x\right)}{j!}}{\int }_{-\infty }^{nx}\phi \left(nx-u\right){\left({\displaystyle \frac{u}{n}}-x\right)}^{j}du+$$

(165)

$${\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_{-\infty }^{nx}\phi \left(nx-u\right)\left({\int }_{{\displaystyle \frac{u}{n}}}^x{\left(s-{\displaystyle \frac{u}{n}}\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\left(s\right)-D_{x-}^{\alpha }f\left(x\right)\right)ds\right)du.$$

Adding the last two equalities, (164) and (165), we have

$$A_n\left(f\right)\left(x\right)=\sum _{j=0}^{N-1}{\displaystyle \frac{f^{\left(j\right)}\left(x\right)}{j!}}{A}_n\left({\left(·-x\right)}^j\right)\left(x\right)+$$

$${\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}\left[{\int }_{nx}^{\infty }\phi \left(nx-u\right)\left({\int }_x^{{\displaystyle \frac{u}{n}}}{\left({\displaystyle \frac{u}{n}}-s\right)}^{\alpha -1}\left(D_{*x}^{\alpha }f\left(s\right)-D_{*x}^{\alpha }f\left(x\right)\right)ds\right)du\right.$$

(166)

$$\left.+{\int }_{-\infty }^{nx}\phi \left(nx-u\right)\left({\int }_{{\displaystyle \frac{u}{n}}}^x{\left(s-{\displaystyle \frac{u}{n}}\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\left(s\right)-D_{x-}^{\alpha }f\left(x\right)\right)ds\right)du\right].$$

Consequently, it holds that

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

(167)

where

$$R_n\left(x\right):={\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}$$

$$\left[{\int }_{-\infty }^{nx}\phi \left(nx-u\right)\left({\int }_{{\displaystyle \frac{u}{n}}}^x{\left(s-{\displaystyle \frac{u}{n}}\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\left(s\right)-D_{x-}^{\alpha }f\left(x\right)\right)ds\right)du\right.$$

$$\left.+{\int }_{nx}^{\infty }\phi \left(nx-u\right)\left({\int }_x^{{\displaystyle \frac{u}{n}}}{\left({\displaystyle \frac{u}{n}}-s\right)}^{\alpha -1}\left(D_{*x}^{\alpha }f\left(s\right)-D_{*x}^{\alpha }f\left(x\right)\right)ds\right)du\right],$$

(168)

∀$x\in \mathbb{R}$.

Denote by

$$R_{n1}\left(x\right)={\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_{-\infty }^{nx}\phi \left(nx-u\right)\left({\int }_{{\displaystyle \frac{u}{n}}}^x{\left(s-{\displaystyle \frac{u}{n}}\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\left(s\right)-D_{x-}^{\alpha }f\left(x\right)\right)ds\right)du,$$

(169)

∀$x\in \mathbb{R}:x\ge {\displaystyle \frac{u}{n}}>-\infty$, and

$$R_{n2}\left(x\right)={\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_{nx}^{\infty }\phi \left(nx-u\right)\left({\int }_x^{{\displaystyle \frac{u}{n}}}{\left({\displaystyle \frac{u}{n}}-s\right)}^{\alpha -1}\left(D_{*x}^{\alpha }f\left(s\right)-D_{*x}^{\alpha }f\left(x\right)\right)ds\right)du,$$

(170)

∀$x\in \mathbb{R}:\infty >{\displaystyle \frac{u}{n}}\ge x$.

That is,

$$R_n\left(x\right)=R_{n1}\left(x\right)+R_{n2}\left(x\right),\forall x\in \mathbb{R}.$$

(171)

Let first $\left|{\displaystyle \frac{u}{n}}-x\right|<{\displaystyle \frac{1}{n^{\beta }}}$.

Call

$${\gamma }_{n1}\left(x\right):={\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_{{\displaystyle \frac{u}{n}}}^x{\left(s-{\displaystyle \frac{u}{n}}\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\left(s\right)-D_{x-}^{\alpha }f\left(x\right)\right)ds,$$

(172)

and

$${\gamma }_{n2}\left(x\right):={\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_x^{{\displaystyle \frac{u}{n}}}{\left({\displaystyle \frac{u}{n}}-s\right)}^{\alpha -1}\left(D_{*x}^{\alpha }f\left(s\right)-D_{*x}^{\alpha }f\left(x\right)\right)ds.$$

(173)

Let us assume that $x\ge {\displaystyle \frac{u}{n}}$; then,

$$\left|{\gamma }_{n1}\left(x\right)\right|\le {\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_{{\displaystyle \frac{u}{n}}}^x{\left(s-{\displaystyle \frac{u}{n}}\right)}^{\alpha -1}\left|D_{x-}^{\alpha }f\left(s\right)-D_{x-}^{\alpha }f\left(x\right)\right|ds\le$$

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

(174)

Hence,

$$\left|{\gamma }_{n1}\left(x\right)\right|\le {\omega }_1{\left(D_{x-}^{\alpha }f,{\displaystyle \frac{1}{n^{\beta }}}\right)}_{(-\infty ,x]}{\displaystyle \frac{1}{n^{\alpha \beta }\mathsf{\Gamma }\left(\alpha +1\right)}},\mathrm{for}x\ge {\displaystyle \frac{u}{n}}\mathrm{and}\left|{\displaystyle \frac{u}{n}}-x\right|<{\displaystyle \frac{1}{n^{\beta }}}.$$

(175)

Let us assume that $x\le {\displaystyle \frac{u}{n}}$; then,

$$\left|{\gamma }_{n2}\left(x\right)\right|\le {\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_x^{{\displaystyle \frac{u}{n}}}{\left({\displaystyle \frac{u}{n}}-s\right)}^{\alpha -1}\left|D_{*x}^{\alpha }f\left(s\right)-D_{*x}^{\alpha }f\left(x\right)\right|ds$$

$$\le {\omega }_1{\left(D_{*x}^{\alpha }f,{\displaystyle \frac{1}{n^{\beta }}}\right)}_{[x,\infty )}{\displaystyle \frac{{\left(\frac{u}{n}-x\right)}^{\alpha }}{\mathsf{\Gamma }\left(\alpha +1\right)}}.$$

(176)

Hence,

$$\left|{\gamma }_{n2}\left(x\right)\right|\le {\omega }_1{\left(D_{*x}^{\alpha }f,{\displaystyle \frac{1}{n^{\beta }}}\right)}_{[x,\infty )}{\displaystyle \frac{1}{n^{\alpha \beta }\mathsf{\Gamma }\left(\alpha +1\right)}},\mathrm{for}x\le {\displaystyle \frac{u}{n}}\mathrm{and}\left|{\displaystyle \frac{u}{n}}-x\right|<{\displaystyle \frac{1}{n^{\beta }}}.$$

(177)

Consequently, we see that

$$\left|R_{n1}\left(x\right)\right|{|}_{\left|{\displaystyle \frac{u}{n}}-x\right|<{\displaystyle \frac{1}{n^{\beta }}}}\le {\displaystyle \frac{{\omega }_1{\left(D_{x-}^{\alpha }f,\frac{1}{n^{\beta }}\right)}_{(-\infty ,x]}}{n^{\alpha \beta }\mathsf{\Gamma }\left(\alpha +1\right)}},$$

(178)

and

$$\left|R_{n2}\left(x\right)\right|{|}_{\left|{\displaystyle \frac{u}{n}}-x\right|<{\displaystyle \frac{1}{n^{\beta }}}}\le {\displaystyle \frac{{\omega }_1{\left(D_{*x}^{\alpha }f,\frac{1}{n^{\beta }}\right)}_{[x,\infty )}}{n^{\alpha \beta }\mathsf{\Gamma }\left(\alpha +1\right)}}.$$

(179)

Furthermore, we have that

$$\left|{\gamma }_{n1}\left(x\right)\right|\le {∥D_{x-}^{\alpha }f∥}_{\infty }{\displaystyle \frac{{\left(x-\frac{u}{n}\right)}^{\alpha }}{\mathsf{\Gamma }\left(\alpha +1\right)}},$$

(180)

where $x\ge {\displaystyle \frac{u}{n}}>-\infty$, and

$$\left|{\gamma }_{n2}\left(x\right)\right|\le {∥D_{*x}^{\alpha }f∥}_{\infty }{\displaystyle \frac{{\left(\frac{u}{n}-x\right)}^{\alpha }}{\mathsf{\Gamma }\left(\alpha +1\right)}},$$

(181)

where $\infty >{\displaystyle \frac{u}{n}}\ge x$.

Next, we see that

$${\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}\left|{\int }_{\left|{\displaystyle \frac{u}{n}}-x\right|\ge {\displaystyle \frac{1}{n^{\beta }}}}\phi \left(nx-u\right)\left({\int }_{{\displaystyle \frac{u}{n}}}^x{\left(s-{\displaystyle \frac{u}{n}}\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\right)\left(s\right)ds\right)du\right|\le$$

$${\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_{\left|{\displaystyle \frac{u}{n}}-x\right|\ge {\displaystyle \frac{1}{n^{\beta }}}}\phi \left(nx-u\right)\left|{\int }_{{\displaystyle \frac{u}{n}}}^x{\left(s-{\displaystyle \frac{u}{n}}\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\right)\left(s\right)ds\right|du\le$$

(182)

$${\displaystyle \frac{{∥D_{x-}^{\alpha }f∥}_{\infty }}{\mathsf{\Gamma }\left(\alpha +1\right)}}{\int }_{\left|{\displaystyle \frac{u}{n}}-x\right|\ge {\displaystyle \frac{1}{n^{\beta }}}}\phi \left(nx-u\right){\left(x-{\displaystyle \frac{u}{n}}\right)}^{\alpha }du=$$

$${\displaystyle \frac{{∥D_{x-}^{\alpha }f∥}_{\infty }}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}{\int }_{\left|u-nx\right|\ge n^{1-\beta }}\phi \left(\left|nx-u\right|\right){\left|nx-u\right|}^{\alpha }du\stackrel{\left(31\right)}{\le }$$

$${\displaystyle \frac{{∥D_{x-}^{\alpha }f∥}_{\infty }}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right)\lambda {\int }_{\left|u-nx\right|\ge n^{1-\beta }}{e}^{-2\lambda \left(\left|nx-u\right|-1\right)}{\left|nx-u\right|}^{\alpha }du=$$

$${\displaystyle \frac{2{∥D_{x-}^{\alpha }f∥}_{\infty }}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right)\lambda {\int }_{n^{1-\beta }}^{\infty }{e}^{-2\lambda \left(x-1\right)}{x}^{\alpha }dx=$$

(183)

$${\displaystyle \frac{2{∥D_{x-}^{\alpha }f∥}_{\infty }}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right)\lambda {\displaystyle \frac{e^{2\lambda }}{{\left(2\lambda \right)}^{\alpha +1}}}{\int }_{n^{1-\beta }}^{\infty }{e}^{-2\lambda x}{\left(2\lambda x\right)}^{\alpha }d\left(x2\lambda \right)=$$

$${\displaystyle \frac{{∥D_{x-}^{\alpha }f∥}_{\infty }}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right){\displaystyle \frac{e^{2\lambda }}{{\left(2\lambda \right)}^{\alpha }}}{\int }_{2\lambda n^{1-\beta }}^{\infty }{e}^{-y}{y}^{\alpha }dy\le$$

$${\displaystyle \frac{{∥D_{x-}^{\alpha }f∥}_{\infty }}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right){\displaystyle \frac{e^{2\lambda }}{{\left(2\lambda \right)}^{\alpha }}}{\int }_{2\lambda n^{1-\beta }}^{\infty }{e}^{-y}{y}^{N}dy\le$$

$${\displaystyle \frac{{∥D_{x-}^{\alpha }f∥}_{\infty }}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right){\displaystyle \frac{e^{2\lambda }}{{\left(2\lambda \right)}^{\alpha }}}{2}^{N}N!{\int }_{2\lambda n^{1-\beta }}^{\infty }{e}^{-y}{e}^{{\displaystyle \frac{y}{2}}}dy=$$

$${\displaystyle \frac{{∥D_{x-}^{\alpha }f∥}_{\infty }}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right){\displaystyle \frac{e^{2\lambda }}{{\left(2\lambda \right)}^{\alpha }}}{2}^{N}N!{\int }_{2\lambda n^{1-\beta }}^{\infty }{e}^{-{\displaystyle \frac{y}{2}}}dy=$$

(184)

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

We have proved that

$${\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}\left|{\int }_{\left|{\displaystyle \frac{u}{n}}-x\right|\ge {\displaystyle \frac{1}{n^{\beta }}}}\phi \left(nx-u\right)\left({\int }_{{\displaystyle \frac{u}{n}}}^x{\left(s-{\displaystyle \frac{u}{n}}\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\right)\left(s\right)ds\right)du\right|\le$$

(185)

$${\displaystyle \frac{2^{N+1}N!{∥D_{x-}^{\alpha }f∥}_{\infty }}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right){\displaystyle \frac{e^{2\lambda }}{{\left(2\lambda \right)}^{\alpha }}}{e}^{-\lambda n^{1-\beta }}\to 0,\mathrm{as}n\to \infty .$$

Next, we see that

$${\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}\left|{\int }_{\left|{\displaystyle \frac{u}{n}}-x\right|\ge {\displaystyle \frac{1}{n^{\beta }}}}\phi \left(nx-u\right)\left({\int }_x^{{\displaystyle \frac{u}{n}}}{\left({\displaystyle \frac{u}{n}}-s\right)}^{\alpha -1}\left(D_{*x}^{\alpha }f\right)\left(s\right)ds\right)du\right|\le$$

$${\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_{\left|{\displaystyle \frac{u}{n}}-x\right|\ge {\displaystyle \frac{1}{n^{\beta }}}}\phi \left(nx-u\right)\left|{\int }_x^{{\displaystyle \frac{u}{n}}}{\left({\displaystyle \frac{u}{n}}-s\right)}^{\alpha -1}\left(D_{*x}^{\alpha }f\right)\left(s\right)ds\right|du\le$$

$${\displaystyle \frac{{∥D_{*x}^{\alpha }f∥}_{\infty }}{\mathsf{\Gamma }\left(\alpha +1\right)}}{\int }_{\left|{\displaystyle \frac{u}{n}}-x\right|\ge {\displaystyle \frac{1}{n^{\beta }}}}\phi \left(nx-u\right){\left({\displaystyle \frac{u}{n}}-x\right)}^{\alpha }du=$$

$${\displaystyle \frac{{∥D_{*x}^{\alpha }f∥}_{\infty }}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}{\int }_{\left|u-nx\right|\ge n^{1-\beta }}\phi \left(\left|nx-u\right|\right){\left|nx-u\right|}^{\alpha }du\le$$

(186)

(as before)

$${\displaystyle \frac{2^{N+1}N!{∥D_{*x}^{\alpha }f∥}_{\infty }}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right){\displaystyle \frac{e^{2\lambda }}{{\left(2\lambda \right)}^{\alpha }}}{e}^{-\lambda n^{1-\beta }}\to 0,\mathrm{as}n\to \infty .$$

Therefore, it holds that

$${\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}\left|{\int }_{\left|{\displaystyle \frac{u}{n}}-x\right|\ge {\displaystyle \frac{1}{n^{\beta }}}}\phi \left(nx-u\right)\left({\int }_x^{{\displaystyle \frac{u}{n}}}{\left({\displaystyle \frac{u}{n}}-s\right)}^{\alpha -1}\left(D_{*x}^{\alpha }f\right)\left(s\right)ds\right)du\right|\le$$

(187)

$${\displaystyle \frac{2^{N+1}N!{∥D_{*x}^{\alpha }f∥}_{\infty }}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right){\displaystyle \frac{e^{2\lambda }}{{\left(2\lambda \right)}^{\alpha }}}{e}^{-\lambda n^{1-\beta }}\to 0,\mathrm{as}n\to \infty .$$

Consequently, it holds that

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

(188)

$$\left\{\left[{∥D_{x-}^{\alpha }f∥}_{\infty }+{∥D_{*x}^{\alpha }f∥}_{\infty }\right]{\displaystyle \frac{2^{N+1}N!\left(q+\frac{1}{q}\right)e^{2\lambda }}{\mathsf{\Gamma }\left(\alpha +1\right){\left(2\lambda \right)}^{\alpha }{n}^{\alpha }{e}^{\lambda n^{1-\beta }}}}\right\}\to 0,\mathrm{as}n\to \infty .$$

(189)

We need $2\lambda n^{1-\beta }\ge 1$, iff $n^{1-\beta }\ge {\displaystyle \frac{1}{2\lambda }}$, iff $n\ge {\displaystyle \frac{1}{{\left(2\lambda \right)}^{\frac{1}{1-\beta }}}}$.

(i) In the case of $0<2\lambda \le 1$ (i.e., $0<\lambda \le {\displaystyle \frac{1}{2}}$), then ${\displaystyle \frac{1}{2\lambda }}\ge 1$, and ${\displaystyle \frac{1}{{\left(2\lambda \right)}^{\frac{1}{1-\beta }}}}\ge 1$. So, for large enough $n\in \mathbb{N}$, we can have $2\lambda n^{1-\beta }\ge 1$.

(ii) If $2\lambda >1$ (i.e., $\lambda >{\displaystyle \frac{1}{2}}$), then ${\displaystyle \frac{1}{2\lambda }}<1$ and ${\displaystyle \frac{1}{{\left(2\lambda \right)}^{\frac{1}{1-\beta }}}}<1$. So, for any $n\in \mathbb{N}$, we have that $2\lambda n^{1-\beta }\ge 1$.

See also the assumption.

We have that

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

(190)

and

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

(191)

∀$x,s\in \mathbb{R}$.

Therefore, it holds that

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

(192)

$s\ge x$, and

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

$s\le x$.

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

Consequently, it holds that

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

(193)

The theorem is now proven. □

We continue with the following results.

Theorem 14. 

Let $\alpha >0$, $N=⌈\alpha ⌉$, $\alpha \notin \mathbb{N}$, $f\in A{C}^N\left(\mathbb{R}\right)$, $f^{\left(N\right)}\in L_{\infty }\left(\mathbb{R}\right)$, $0<\beta <1$, $x\in \mathbb{R}$, $n\in \mathbb{N}:n^{1-\beta }\ge max\left(3,{\displaystyle \frac{1}{2\lambda }}\right)$. Assume also that both ${\displaystyle \underset{x\in \mathbb{R}}{sup}}{∥D_{*x}^{\alpha }f∥}_{\infty }$, ${\displaystyle \underset{x\in \mathbb{R}}{sup}}{∥D_{x-}^{\alpha }f∥}_{\infty }<\infty$.

Then,

(i)

$$\left|A_n^{*}\left(f\right)\left(x\right)-f\left(x\right)-\sum _{j=1}^{N-1}{\displaystyle \frac{f^{\left(j\right)}\left(x\right)}{j!}}{A}_n^{*}\left({\left(·-x\right)}^j\right)\left(x\right)\right|\le$$

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

(194)

$$\left[{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}+{∥D_{*x}^{\alpha }f∥}_{\infty ,[x,\infty )}\right]{\displaystyle \frac{2^N}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }\left[{\displaystyle \frac{1}{2}}+{\displaystyle \frac{N!}{{\lambda }^N}}\right]e^{-\lambda n^{1-\beta }}$$

$$=:{\mathsf{\Phi }}_1^{*}\left(x\right)\to 0,\mathit{as}n\to \infty ,$$

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

$$\left|A_n^{*}\left(f\right)\left(x\right)-f\left(x\right)\right|\le {\mathsf{\Phi }}_1^{*}\left(x\right),$$

(195)

(iii)

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

$${\displaystyle \frac{2^{j-1}}{n^j}}\left[1+\left({\displaystyle \frac{tanh\left(\lambda \right)}{\left(j+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }j!}{{\left(2\lambda \right)}^j}}\right)\right]+{\mathsf{\Phi }}_1^{*}\left(x\right),$$

(196)

(iv)

$${∥A_n^{*}\left(f\right)-f∥}_{\infty }\le \sum _{j=1}^{N-1}{\displaystyle \frac{{∥f^{\left(j\right)}∥}_{\infty }}{j!}}{\displaystyle \frac{2^{j-1}}{n^j}}\left[1+\left({\displaystyle \frac{tanh\left(\lambda \right)}{\left(j+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }j!}{{\left(2\lambda \right)}^j}}\right)\right]+$$

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

(197)

$$\left[\underset{x\in \mathbb{R}}{sup}{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}+\underset{x\in \mathbb{R}}{sup}{∥D_{*x}^{\alpha }f∥}_{\infty ,[x,\infty )}\right]{\displaystyle \frac{2^N}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }\left[{\displaystyle \frac{1}{2}}+{\displaystyle \frac{N!}{{\lambda }^N}}\right]e^{-\lambda n^{1-\beta }}.$$

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

So, we obtain the pointwise and uniform convergence with rates of $A_n^{*}\to I$, as $n\to \infty$.

Proof. 

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

We can write

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

(198)

$${\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_x^{t+{\displaystyle \frac{u}{n}}}{\left(t+{\displaystyle \frac{u}{n}}-s\right)}^{\alpha -1}\left(D_{*x}^{\alpha }f\left(s\right)-D_{*x}^{\alpha }f\left(x\right)\right)ds,$$

for all $x\le t+{\displaystyle \frac{u}{n}}<\infty$.

Also it holds that

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

(199)

$${\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_{t+{\displaystyle \frac{u}{n}}}^x{\left(s-\left(t+{\displaystyle \frac{u}{n}}\right)\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\left(s\right)-D_{x-}^{\alpha }f\left(x\right)\right)ds,$$

for all $-\infty <t+{\displaystyle \frac{u}{n}}\le x$.

Hence, we see

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

(200)

$${\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_0^{{\displaystyle \frac{1}{n}}}\left({\int }_x^{t+{\displaystyle \frac{u}{n}}}{\left(t+{\displaystyle \frac{u}{n}}-s\right)}^{\alpha -1}\left(D_{*x}^{\alpha }f\left(s\right)-D_{*x}^{\alpha }f\left(x\right)\right)ds\right)dt,$$

for all $x\le t+{\displaystyle \frac{u}{n}}<\infty$, iff $nx\le nt+u<\infty$.

Also, it holds that

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

(201)

$${\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_0^{{\displaystyle \frac{1}{n}}}\left({\int }_{t+{\displaystyle \frac{u}{n}}}^x{\left(s-\left(t+{\displaystyle \frac{u}{n}}\right)\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\left(s\right)-D_{x-}^{\alpha }f\left(x\right)\right)ds\right)dt,$$

for all $-\infty <t+{\displaystyle \frac{u}{n}}\le x$, iff $-\infty <nt+u\le nx$.

Therefore, we obtain

$$n{\int }_{nx}^{\infty }\left({\int }_0^{{\displaystyle \frac{1}{n}}}f\left(t+{\displaystyle \frac{u}{n}}\right)dt\right)\phi \left(nx-u\right)du=$$

$$\sum _{j=0}^{N-1}{\displaystyle \frac{f^{\left(j\right)}\left(x\right)}{j!}}n{\int }_{nx}^{\infty }\left({\int }_0^{{\displaystyle \frac{1}{n}}}{\left(t+{\displaystyle \frac{u}{n}}-x\right)}^{j}dt\right)\phi \left(nx-u\right)du+$$

(202)

$${\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}n{\int }_{nx}^{\infty }\phi \left(nx-u\right)$$

$$\left({\int }_0^{{\displaystyle \frac{1}{n}}}\left({\int }_x^{t+{\displaystyle \frac{u}{n}}}{\left(t+{\displaystyle \frac{u}{n}}-s\right)}^{\alpha -1}\left(D_{*x}^{\alpha }f\left(s\right)-D_{*x}^{\alpha }f\left(x\right)\right)ds\right)dt\right)du,$$

and

$$n{\int }_{-\infty }^{nx}\left({\int }_0^{{\displaystyle \frac{1}{n}}}f\left(t+{\displaystyle \frac{u}{n}}\right)dt\right)\phi \left(nx-u\right)du=$$

$$\sum _{j=0}^{N-1}{\displaystyle \frac{f^{\left(j\right)}\left(x\right)}{j!}}n{\int }_{-\infty }^{nx}\left({\int }_0^{{\displaystyle \frac{1}{n}}}{\left(t+{\displaystyle \frac{u}{n}}-x\right)}^{j}dt\right)\phi \left(nx-u\right)du+$$

(203)

$${\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}n{\int }_{-\infty }^{nx}\phi \left(nx-u\right)$$

$$\left({\int }_0^{{\displaystyle \frac{1}{n}}}\left({\int }_{t+{\displaystyle \frac{u}{n}}}^x{\left(s-\left(t+{\displaystyle \frac{u}{n}}\right)\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\left(s\right)-D_{x-}^{\alpha }f\left(x\right)\right)ds\right)dt\right)du.$$

Adding the last two equations, (193) and (194), we derive

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

(204)

where

$$R_n^{*}\left(x\right)=R_{n1}^{*}\left(x\right)+R_{n2}^{*}\left(x\right),$$

(205)

with

$$R_{n2}^{*}\left(x\right):={\displaystyle \frac{n}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_{nx}^{\infty }\phi \left(nx-u\right)$$

$$\left({\int }_0^{{\displaystyle \frac{1}{n}}}\left({\int }_x^{t+{\displaystyle \frac{u}{n}}}{\left(t+{\displaystyle \frac{u}{n}}-s\right)}^{\alpha -1}\left(D_{*x}^{\alpha }f\left(s\right)-D_{*x}^{\alpha }f\left(x\right)\right)ds\right)dt\right)du,$$

(206)

$nx\le nt+u<\infty$, and

$$R_{n1}^{*}\left(x\right):={\displaystyle \frac{n}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_{-\infty }^{nx}\phi \left(nx-u\right)$$

$$\left({\int }_0^{{\displaystyle \frac{1}{n}}}\left({\int }_{t+{\displaystyle \frac{u}{n}}}^x{\left(s-\left(t+{\displaystyle \frac{u}{n}}\right)\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\left(s\right)-D_{x-}^{\alpha }f\left(x\right)\right)ds\right)dt\right)du,$$

(207)

$-\infty <nt+u\le nx$.

Let $\left|{\displaystyle \frac{u}{n}}-x\right|<{\displaystyle \frac{1}{n^{\beta }}}$.

Call

$${\delta }_{n1}\left(x\right):={\displaystyle \frac{n}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_0^{{\displaystyle \frac{1}{n}}}\left({\int }_{t+{\displaystyle \frac{u}{n}}}^x{\left(s-\left(t+{\displaystyle \frac{u}{n}}\right)\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\left(s\right)-D_{x-}^{\alpha }f\left(x\right)\right)ds\right)dt,$$

(208)

and

$${\delta }_{n2}\left(x\right):={\displaystyle \frac{n}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_0^{{\displaystyle \frac{1}{n}}}\left({\int }_x^{t+{\displaystyle \frac{u}{n}}}{\left(t+{\displaystyle \frac{u}{n}}-s\right)}^{\alpha -1}\left(D_{*x}^{\alpha }f\left(s\right)-D_{*x}^{\alpha }f\left(x\right)\right)ds\right)dt.$$

(209)

We have that

$$\left|{\delta }_{n1}\left(x\right)\right|\le {\displaystyle \frac{n}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_0^{{\displaystyle \frac{1}{n}}}{\omega }_1\left(D_{x-}^{\alpha }f,x-t-{\displaystyle \frac{u}{n}}\right){\displaystyle \frac{{\left(x-t-\frac{u}{n}\right)}^{\alpha }}{\alpha }}dt=$$

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

(210)

and

$$\left|{\delta }_{n2}\left(x\right)\right|\le {\displaystyle \frac{n}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_0^{{\displaystyle \frac{1}{n}}}{\omega }_1\left(D_{*x}^{\alpha }f,t+{\displaystyle \frac{u}{n}}-x\right){\displaystyle \frac{{\left(t+\frac{u}{n}-x\right)}^{\alpha }}{\alpha }}dt=$$

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

(211)

Consequently, we see that

$$|R_{n1}^{*}\left(x\right){||}_{\left|{\displaystyle \frac{u}{n}}-x\right|<{\displaystyle \frac{1}{n^{\beta }}}}\le {\displaystyle \frac{{\omega }_1{\left(D_{x-}^{\alpha }f,\frac{1}{n}+\frac{1}{n^{\beta }}\right)}_{(-\infty ,x]}{\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)}^{\alpha }}{\mathsf{\Gamma }\left(\alpha +1\right)}},$$

(212)

and

$$|R_{n2}^{*}\left(x\right)|{|}_{\left|{\displaystyle \frac{u}{n}}-x\right|<{\displaystyle \frac{1}{n^{\beta }}}}\le {\displaystyle \frac{{\omega }_1{\left(D_{*x}^{\alpha }f,\frac{1}{n}+\frac{1}{n^{\beta }}\right)}_{[x,+\infty )}{\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)}^{\alpha }}{\mathsf{\Gamma }\left(\alpha +1\right)}}.$$

(213)

We continue as follows:

$$\left|{\delta }_{n1}\left(x\right)\right|\le {\displaystyle \frac{{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}}{\mathsf{\Gamma }\left(\alpha \right)}}n{\int }_0^{{\displaystyle \frac{1}{n}}}{\displaystyle \frac{{\left(x-t-\frac{u}{n}\right)}^{\alpha }}{\alpha }}dt\le$$

(214)

$${\displaystyle \frac{{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}}{\mathsf{\Gamma }\left(\alpha +1\right)}}{\left(\left|x-{\displaystyle \frac{u}{n}}\right|+{\displaystyle \frac{1}{n}}\right)}^{\alpha },$$

and

$$\left|{\delta }_{n2}\left(x\right)\right|\le {\displaystyle \frac{{∥D_{*x}^{\alpha }f∥}_{\infty ,[x,\infty )}}{\mathsf{\Gamma }\left(\alpha +1\right)}}{\left(\left|x-{\displaystyle \frac{u}{n}}\right|+{\displaystyle \frac{1}{n}}\right)}^{\alpha }.$$

(215)

We have that

$$\left|R_{n1}^{*}\left(x\right)\right|{|}_{\left|x-{\displaystyle \frac{u}{n}}\right|\ge {\displaystyle \frac{1}{n^{\beta }}}}\le$$

$${\displaystyle \frac{{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}}{\mathsf{\Gamma }\left(\alpha +1\right)}}\underset{(-\infty ,nx]\cap \left|x-{\displaystyle \frac{u}{n}}\right|\ge {\displaystyle \frac{1}{n^{\beta }}}}{\int }\phi \left(nx-u\right){\left(\left|x-{\displaystyle \frac{u}{n}}\right|+{\displaystyle \frac{1}{n}}\right)}^{\alpha }du=$$

$${\displaystyle \frac{{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\underset{(-\infty ,nx]\cap \left|nx-u\right|\ge n^{1-\beta }}{\int }\phi \left(nx-u\right){\left(\left|nx-u\right|+1\right)}^{\alpha }du\le$$

(216)

$${\displaystyle \frac{{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\underset{\left|nx-u\right|\ge n^{1-\beta }}{\int }\phi \left(\left|nx-u\right|\right){\left(\left|nx-u\right|+1\right)}^{N}du\le$$

$${\displaystyle \frac{{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}{2}^{N-1}\underset{\left|nx-u\right|\ge n^{1-\beta }}{\int }\phi \left(\left|nx-u\right|\right){\left(1+\left|nx-u\right|\right)}^{N}du\le$$

(217)

$${\displaystyle \frac{2^{N-1}{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}$$

$$\left[\underset{\left|nx-u\right|\ge n^{1-\beta }}{\int }\phi \left(\left|nx-u\right|\right)du+\underset{\left|nx-u\right|\ge n^{1-\beta }}{\int }\phi \left(\left|nx-u\right|\right){\left|nx-u\right|}^{N}du\right]\stackrel{\left(31\right)}{\le }$$

$${\displaystyle \frac{2^{N-1}{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right)\lambda$$

$$\left[\underset{\left|nx-u\right|\ge n^{1-\beta }}{\int }{e}^{-2\lambda \left(\left|nx-u\right|-1\right)}du+\underset{\left|nx-u\right|\ge n^{1-\beta }}{\int }{e}^{-2\lambda \left(\left|nx-u\right|-1\right)}{\left|nx-u\right|}^{N}du\right]=$$

(218)

$${\displaystyle \frac{2^N{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right)\lambda \left[{\int }_{n^{1-\beta }}^{\infty }{e}^{-2\lambda \left(x-1\right)}dx+{\int }_{n^{1-\beta }}^{\infty }{e}^{-2\lambda \left(x-1\right)}{x}^{N}dx\right]=$$

$${\displaystyle \frac{2^N{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right)\lambda e^{2\lambda }\left[{\int }_{n^{1-\beta }}^{\infty }{e}^{-2\lambda x}dx+{\int }_{n^{1-\beta }}^{\infty }{e}^{-2\lambda x}{x}^{N}dx\right]=$$

$${\displaystyle \frac{2^N{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right)\lambda e^{2\lambda }\left[{\displaystyle \frac{1}{2\lambda }}{\int }_{2\lambda n^{1-\beta }}^{\infty }{e}^{-y}dy+{\displaystyle \frac{1}{{\left(2\lambda \right)}^{N+1}}}{\int }_{2\lambda n^{1-\beta }}^{\infty }{e}^{-y}{y}^{N}dy\right]\le$$

(219)

$${\displaystyle \frac{2^N{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right)\lambda e^{2\lambda }\left[{\displaystyle \frac{1}{2\lambda }}{e}^{-2\lambda n^{1-\beta }}+{\displaystyle \frac{2^{N}N!}{{\left(2\lambda \right)}^{N+1}}}{\int }_{2\lambda n^{1-\beta }}^{\infty }{e}^{-{\displaystyle \frac{y}{2}}}dy\right]=$$

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

(220)

$${\displaystyle \frac{2^N{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right)\lambda e^{2\lambda }\left[{\displaystyle \frac{1}{2\lambda }}+{\displaystyle \frac{2^{N+1}N!}{{\left(2\lambda \right)}^{N+1}}}\right]e^{-\lambda n^{1-\beta }}=$$

$${\displaystyle \frac{2^N{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right)\lambda e^{2\lambda }\left[{\displaystyle \frac{1}{2\lambda }}+{\displaystyle \frac{N!}{{\lambda }^{N+1}}}\right]e^{-\lambda n^{1-\beta }}=$$

(221)

$${\displaystyle \frac{2^N{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }\left[{\displaystyle \frac{1}{2}}+{\displaystyle \frac{N!}{{\lambda }^N}}\right]e^{-\lambda n^{1-\beta }}\to 0,\mathrm{as}n\to \infty .$$

So, it holds that

$$\left|R_{n1}^{*}\left(x\right)\right|{|}_{\left|x-{\displaystyle \frac{u}{n}}\right|\ge {\displaystyle \frac{1}{n^{\beta }}}}\le$$

(222)

$${\displaystyle \frac{2^N{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }\left[{\displaystyle \frac{1}{2}}+{\displaystyle \frac{N!}{{\lambda }^N}}\right]e^{-\lambda n^{1-\beta }}\to 0,\mathrm{as}n\to \infty .$$

Similarly, we obtain

$$\left|R_{n2}^{*}\left(x\right)\right|{|}_{\left|x-{\displaystyle \frac{u}{n}}\right|\ge {\displaystyle \frac{1}{n^{\beta }}}}\le$$

$${\displaystyle \frac{2^N{∥D_{*x}^{\alpha }f∥}_{\infty ,[x,\infty )}}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }\left[{\displaystyle \frac{1}{2}}+{\displaystyle \frac{N!}{{\lambda }^N}}\right]e^{-\lambda n^{1-\beta }}\to 0,\mathrm{as}n\to \infty .$$

(223)

At the end, we see that

$$\left|R_n^{*}\left(x\right)\right|\le {\displaystyle \frac{{\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)}^{\alpha }}{\mathsf{\Gamma }\left(\alpha +1\right)}}$$

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

(224)

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

$${\displaystyle \frac{2^N}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }\left[{\displaystyle \frac{1}{2}}+{\displaystyle \frac{N!}{{\lambda }^N}}\right]e^{-\lambda n^{1-\beta }}\to 0,\mathrm{as}n\to \infty .$$

The theorem is proven. □

We also present the following.

Theorem 15. 

Let $\alpha >0$, $N=⌈\alpha ⌉$, $\alpha \notin \mathbb{N}$, $f\in A{C}^N\left(\mathbb{R}\right)$, $f^{\left(N\right)}\in L_{\infty }\left(\mathbb{R}\right)$, $0<\beta <1$, $x\in \mathbb{R}$, $n\in \mathbb{N}:n^{1-\beta }\ge max\left(3,{\displaystyle \frac{1}{2\lambda }}\right)$. Assume also that both ${\displaystyle \underset{x\in \mathbb{R}}{sup}}{∥D_{*x}^{\alpha }f∥}_{\infty }$, ${\displaystyle \underset{x\in \mathbb{R}}{sup}}{∥D_{x-}^{\alpha }f∥}_{\infty }<\infty$.

Then,

(i)

$$\left|\overline{A_n}\left(f\right)\left(x\right)-f\left(x\right)-\sum _{j=1}^{N-1}{\displaystyle \frac{f^{\left(j\right)}\left(x\right)}{j!}}\overline{A_n}\left({\left(·-x\right)}^j\right)\left(x\right)\right|\le$$

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

(225)

$$\left[{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}+{∥D_{*x}^{\alpha }f∥}_{\infty ,[x,\infty )}\right]{\displaystyle \frac{2^N}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }\left[{\displaystyle \frac{1}{2}}+{\displaystyle \frac{N!}{{\lambda }^N}}\right]e^{-\lambda n^{1-\beta }}$$

$$=:\overline{{\mathsf{\Phi }}_1}\left(x\right)\to 0,\mathit{as}n\to \infty ,$$

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

$$\left|\overline{A_n}\left(f\right)\left(x\right)-f\left(x\right)\right|\le \overline{{\mathsf{\Phi }}_1}\left(x\right),$$

(226)

(iii)

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

$${\displaystyle \frac{2^{j-1}}{n^j}}\left[1+\left({\displaystyle \frac{tanh\left(\lambda \right)}{\left(j+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }j!}{{\left(2\lambda \right)}^j}}\right)\right]+{\overline{\mathsf{\Phi }}}_1\left(x\right),$$

(227)

(iv)

$${∥\overline{A_n}\left(f\right)-f∥}_{\infty }\le \sum _{j=1}^{N-1}{\displaystyle \frac{{∥f^{\left(j\right)}∥}_{\infty }}{j!}}{\displaystyle \frac{2^{j-1}}{n^j}}\left[1+\left({\displaystyle \frac{tanh\left(\lambda \right)}{\left(j+1\right)}}+{\displaystyle \frac{\left(q+\frac{1}{q}\right)e^{2\lambda }j!}{{\left(2\lambda \right)}^j}}\right)\right]+$$

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

(228)

$$\left[\underset{x\in \mathbb{R}}{sup}{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}+\underset{x\in \mathbb{R}}{sup}{∥D_{*x}^{\alpha }f∥}_{\infty ,[x,\infty )}\right]{\displaystyle \frac{2^N}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }\left[{\displaystyle \frac{1}{2}}+{\displaystyle \frac{N!}{{\lambda }^N}}\right]e^{-\lambda n^{1-\beta }}.$$

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

So, we obtain the pointwise and uniform convergence with rates of $\overline{A_n}\to I$, as $n\to \infty$.

Proof. 

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

We can write

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

(229)

$${\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_x^{{\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}}{\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}-s\right)}^{\alpha -1}\left(D_{*x}^{\alpha }f\left(s\right)-D_{*x}^{\alpha }f\left(x\right)\right)ds,$$

for all $x\le {\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}<\infty$, $i=1,\dots ,r$.

Also, it holds that

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

(230)

$${\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_{{\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}}^x{\left(s-\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}\right)\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\left(s\right)-D_{x-}^{\alpha }f\left(x\right)\right)ds,$$

for all $-\infty <{\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}\le x$, $i=1,\dots ,r$.

Hence,

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

(231)

$${\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}\sum _{i=1}^r{w}_i{\int }_x^{{\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}}{\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}-s\right)}^{\alpha -1}\left(D_{*x}^{\alpha }f\left(s\right)-D_{*x}^{\alpha }f\left(x\right)\right)ds,$$

and

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

(232)

$${\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}\sum _{i=1}^r{w}_i{\int }_{{\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}}^x{\left(s-\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}\right)\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\left(s\right)-D_{x-}^{\alpha }f\left(x\right)\right)ds.$$

Furthermore, it holds that

$${\int }_{nx}^{\infty }\left(\sum _{i=1}^r{w}_{i}f\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}\right)\right)\phi \left(nx-u\right)du=$$

(233)

$$\sum _{j=0}^{N-1}{\displaystyle \frac{f^{\left(j\right)}\left(x\right)}{j!}}{\int }_{nx}^{\infty }\left(\sum _{i=1}^r{w}_i{\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}-x\right)}^j\right)\phi \left(nx-u\right)du+$$

$${\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_{nx}^{\infty }\phi \left(nx-u\right)$$

$$\left(\sum _{i=1}^r{w}_i{\int }_x^{{\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}}{\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}-s\right)}^{\alpha -1}\left(D_{*x}^{\alpha }f\left(s\right)-D_{*x}^{\alpha }f\left(x\right)\right)ds\right)du,$$

and

$${\int }_{-\infty }^{nx}\left(\sum _{i=1}^r{w}_{i}f\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}\right)\right)\phi \left(nx-u\right)du=$$

(234)

$$\sum _{j=0}^{N-1}{\displaystyle \frac{f^{\left(j\right)}\left(x\right)}{j!}}{\int }_{-\infty }^{nx}\left(\sum _{i=1}^r{w}_i{\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}-x\right)}^j\right)\phi \left(nx-u\right)du+$$

$${\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_{-\infty }^{nx}\phi \left(nx-u\right)$$

$$\left(\sum _{i=1}^r{w}_i{\int }_{{\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}}^x{\left(s-\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}\right)\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\left(s\right)-D_{x-}^{\alpha }f\left(x\right)\right)ds\right)du.$$

Adding the last two equations, (233) and (234), we obtain

$$\overline{A_n}\left(f\right)\left(x\right)-f\left(x\right)-\sum _{j=1}^{N-1}{\displaystyle \frac{f^{\left(j\right)}\left(x\right)}{j!}}\overline{A_n}\left({\left(·-x\right)}^j\right)\left(x\right)=\overline{R_n}\left(x\right),$$

(235)

where

$$\overline{R_n}\left(x\right)=\overline{R_{n1}}\left(x\right)+\overline{R_{n2}}\left(x\right),$$

(236)

with

$$\overline{R_{n1}}\left(x\right)={\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_{-\infty }^{nx}\phi \left(nx-u\right)$$

$$\left(\sum _{i=1}^r{w}_i{\int }_{{\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}}^x{\left(s-\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}\right)\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\left(s\right)-D_{x-}^{\alpha }f\left(x\right)\right)ds\right)du,$$

(237)

and

$$\overline{R_{n2}}\left(x\right)={\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}{\int }_{nx}^{\infty }\phi \left(nx-u\right)$$

$$\left(\sum _{i=1}^r{w}_i{\int }_x^{{\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}}{\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}-s\right)}^{\alpha -1}\left(D_{*x}^{\alpha }f\left(s\right)-D_{*x}^{\alpha }f\left(x\right)\right)ds\right)du.$$

(238)

Let $\left|{\displaystyle \frac{u}{n}}-x\right|<{\displaystyle \frac{1}{n^{\beta }}}$.

Call

$${\epsilon }_{n1}\left(x\right):={\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}\sum _{i=1}^r{w}_i{\int }_{{\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}}^x{\left(s-\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}\right)\right)}^{\alpha -1}\left(D_{x-}^{\alpha }f\left(s\right)-D_{x-}^{\alpha }f\left(x\right)\right)ds,$$

(239)

and

$${\epsilon }_{n2}\left(x\right):={\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}\sum _{i=1}^r{w}_i{\int }_x^{{\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}}{\left({\displaystyle \frac{u}{n}}+{\displaystyle \frac{i}{nr}}-s\right)}^{\alpha -1}\left(D_{*x}^{\alpha }f\left(s\right)-D_{*x}^{\alpha }f\left(x\right)\right)ds.$$

(240)

Then,

$$\left|{\epsilon }_{n1}\left(x\right)\right|\le {\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}\sum _{i=1}^r{w}_i{\omega }_1{\left(D_{x-}^{\alpha }f,\left|x-{\displaystyle \frac{u}{n}}\right|+{\displaystyle \frac{i}{nr}}\right)}_{(-\infty ,x]}{\displaystyle \frac{{\left(x-\frac{u}{n}-\frac{i}{nr}\right)}^{\alpha }}{\alpha }}\le$$

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

(241)

Therefore, it holds that

$$\left|{\epsilon }_{n1}\left(x\right)\right|\le {\displaystyle \frac{{\omega }_1{\left(D_{x-}^{\alpha }f,\frac{1}{n}+\frac{1}{n^{\beta }}\right)}_{(-\infty ,x]}}{\mathsf{\Gamma }\left(\alpha +1\right)}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^{\alpha }.$$

(242)

Furthermore, wee see that

$$\left|{\epsilon }_{n2}\left(x\right)\right|\le {\displaystyle \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}}\sum _{i=1}^r{w}_i{\omega }_1{\left(D_{*x}^{\alpha }f,\left|x-{\displaystyle \frac{u}{n}}\right|+{\displaystyle \frac{i}{nr}}\right)}_{[x,\infty )}{\displaystyle \frac{{\left(\frac{u}{n}+\frac{i}{nr}-x\right)}^{\alpha }}{\alpha }}\le$$

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

(243)

That is,

$$\left|{\epsilon }_{n2}\left(x\right)\right|\le {\displaystyle \frac{{\omega }_1{\left(D_{*x}^{\alpha }f,\frac{1}{n}+\frac{1}{n^{\beta }}\right)}_{[x,\infty )}}{\mathsf{\Gamma }\left(\alpha +1\right)}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^{\alpha }.$$

(244)

Consequently, we see that

$$|\overline{R_{n1}}\left(x\right)|{|}_{\left|{\displaystyle \frac{u}{n}}-x\right|<{\displaystyle \frac{1}{n^{\beta }}}}\le {\displaystyle \frac{{\omega }_1{\left(D_{x-}^{\alpha }f,\frac{1}{n}+\frac{1}{n^{\beta }}\right)}_{(-\infty ,x]}{\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)}^{\alpha }}{\mathsf{\Gamma }\left(\alpha +1\right)}},$$

(245)

and

$$|\overline{R_{n2}}\left(x\right)|{|}_{\left|{\displaystyle \frac{u}{n}}-x\right|<{\displaystyle \frac{1}{n^{\beta }}}}\le {\displaystyle \frac{{\omega }_1{\left(D_{*x}^{\alpha }f,\frac{1}{n}+\frac{1}{n^{\beta }}\right)}_{[x,+\infty )}{\left(\frac{1}{n}+\frac{1}{n^{\beta }}\right)}^{\alpha }}{\mathsf{\Gamma }\left(\alpha +1\right)}}.$$

(246)

Furthermore, it holds that

$$\left|{\epsilon }_{n1}\left(x\right)\right|\le {\displaystyle \frac{{∥D_{x-}^{\alpha }f∥}_{\infty ,(-\infty ,x]}}{\mathsf{\Gamma }\left(\alpha +1\right)}}{\left(\left|x-{\displaystyle \frac{u}{n}}\right|+{\displaystyle \frac{1}{n}}\right)}^{\alpha },$$

(247)

and

$$\left|{\epsilon }_{n2}\left(x\right)\right|\le {\displaystyle \frac{{∥D_{*x}^{\alpha }f∥}_{\infty ,[x,\infty )}}{\mathsf{\Gamma }\left(\alpha +1\right)}}{\left(\left|x-{\displaystyle \frac{u}{n}}\right|+{\displaystyle \frac{1}{n}}\right)}^{\alpha }.$$

(248)

The rest of the proof is similar to Theorem 14. As such, it is omitted. □

Next, we discuss briefly simultaneous approximation.

Remark 7. 

Let $i\in \mathbb{N}$ be fixed. Assume that $f\in C^{\left(i\right)}\left(\mathbb{R}\right)$, with $f^{\left(j\right)}\in C_B\left(\mathbb{R}\right)$, for $j=0,1,\dots ,i$. We derive from Remark 2 that

$$\begin{array}{c}{\left(A_n\left(f\right)\right)}^{\left(j\right)}\left(x\right)=A_n\left(f^{\left(j\right)}\right)\left(x\right),\\ \\ {\left(A_n^{*}\left(f\right)\right)}^{\left(j\right)}\left(x\right)=A_n^{*}\left(f^{\left(j\right)}\right)\left(x\right),\\ \\ {\left(\overline{A_n}\left(f\right)\right)}^{\left(j\right)}\left(x\right)=\overline{A_n}\left(f^{\left(j\right)}\right)\left(x\right),\forall x\in \mathbb{R},\end{array}$$

(249)

for all $j=1,\dots ,i$.

We give the following.

Theorem 16. 

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

(I)

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

(250)

and

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

(251)

for all $j=0,1,\dots ,i\in \mathbb{N}$;

(II)

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

(252)

and

$${∥{\left(A_n^{*}\left(f\right)\right)}^{\left(j\right)}-f^{\left(j\right)}∥}_{\infty }\le {\rho }_j^s,$$

(253)

for $j=0,1,\dots ,i$;

(III)

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

(254)

and

$${∥{\left(\overline{A_n}\left(f\right)\right)}^{\left(j\right)}-f^{\left(j\right)}∥}_{\infty }\le {\rho }_j^s,$$

(255)

for $j=0,1,\dots ,i$.

Proof. 

Based on Theorems 4–6. □

We continue with simultaneous global smoothness preservation.

Theorem 17. 

Here, $f^{\left(j\right)}\in C_B\left(\mathbb{R}\right)\cup C_u\left(\mathbb{R}\right)$, for $j=0,1,\dots ,i\in \mathbb{N}$. Then,

(I)

$${\omega }_1\left({\left(A_n\left(f\right)\right)}^{\left(j\right)},\delta \right)\le {\omega }_1\left(f^{\left(j\right)},\delta \right),\delta >0;$$

(256)

and if $f^{\left(j\right)}\in C_u\left(\mathbb{R}\right)$, then ${\left(A_n\left(f\right)\right)}^{\left(j\right)}\in C_u\left(\mathbb{R}\right)$;

(II)

$${\omega }_1\left({\left(A_n^{*}\left(f\right)\right)}^{\left(j\right)},\delta \right)\le {\omega }_1\left(f^{\left(j\right)},\delta \right),\delta >0;$$

(257)

and if $f^{\left(j\right)}\in C_u\left(\mathbb{R}\right)$, then ${\left(A_n^{*}\left(f\right)\right)}^{\left(j\right)}\in C_u\left(\mathbb{R}\right)$;

(III)

$${\omega }_1\left({\left(\overline{A_n}\left(f\right)\right)}^{\left(j\right)},\delta \right)\le {\omega }_1\left(f^{\left(j\right)},\delta \right),\delta >0;$$

(258)

and if $f^{\left(j\right)}\in C_u\left(\mathbb{R}\right)$, then ${\left(\overline{A_n}\left(f\right)\right)}^{\left(j\right)}\in C_u\left(\mathbb{R}\right)$.

Proof. 

By Theorems 7–9. □

The related simultaneous differentiation results follow.

Theorem 18. 

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

(I)

$$\left|{\left(A_n\left(f\right)\right)}^{\left(j\right)}\left(x\right)-f^{\left(j\right)}\left(x\right)-\sum _{\lambda =1}^N{\displaystyle \frac{f^{\left(j+\lambda \right)}\left(x\right)}{\lambda !}}\left(A_n\left({\left(·-x\right)}^{\lambda }\right)\right)\left(x\right)\right|\le$$

$${\displaystyle \frac{{\omega }_1\left(f^{\left(j+N\right)},\frac{1}{n^{\alpha }}\right)}{n^{\alpha N}N!}}+{\displaystyle \frac{4{∥f^{\left(j+N\right)}∥}_{\infty }}{n^N}}\left(q+{\displaystyle \frac{1}{q}}\right){\displaystyle \frac{e^{2\lambda }}{{\lambda }^N}}{e}^{-\lambda n^{1-\alpha }},$$

(259)

(II)

$$\left|{\left(A_n^{*}\left(f\right)\right)}^{\left(j\right)}\left(x\right)-f^{\left(j\right)}\left(x\right)-\sum _{\lambda =1}^N{\displaystyle \frac{f^{\left(j+\lambda \right)}\left(x\right)}{\lambda !}}\left(A_n^{*}\left({\left(·-x\right)}^{\lambda }\right)\right)\left(x\right)\right|\le$$

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

$${\displaystyle \frac{2^{N+1}{∥f^{\left(j+N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{N!}{{\lambda }^N}}\right)e^{-\lambda n^{1-\alpha }},$$

(260)

(III)

$$\left|{\left(\overline{A_n}\left(f\right)\right)}^{\left(j\right)}\left(x\right)-f^{\left(j\right)}\left(x\right)-\sum _{\lambda =1}^N{\displaystyle \frac{f^{\left(j+\lambda \right)}\left(x\right)}{\lambda !}}\left(\overline{A_n}\left({\left(·-x\right)}^{\lambda }\right)\right)\left(x\right)\right|\le$$

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

$${\displaystyle \frac{2^{N+1}{∥f^{\left(j+N\right)}∥}_{\infty }}{n^{N}N!}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{N!}{{\lambda }^N}}\right)e^{-\lambda n^{1-\alpha }}.$$

(261)

Proof. 

By Theorems 10–12. □

We continue with simultaneous fractional results.

Theorem 19. 

Let $\alpha >0$, $N=⌈\alpha ⌉$, $\alpha \notin \mathbb{N}$, $j=0,1,2,\dots ,i\in \mathbb{N}$; $f^{\left(j\right)}\in A{C}^N\left(\mathbb{R}\right)$, $f^{\left(j+N\right)}\in L_{\infty }\left(\mathbb{R}\right)$, $0<\beta <1$, $x\in \mathbb{R}$, $n\in \mathbb{N}:n^{1-\beta }\ge max\left(3,{\displaystyle \frac{1}{2\lambda }}\right)$. Assume also that both ${\displaystyle \underset{x\in \mathbb{R}}{sup}}{∥D_{*x}^{\alpha }{f}^{\left(j\right)}∥}_{\infty }$, ${\displaystyle \underset{x\in \mathbb{R}}{sup}}{∥D_{x-}^{\alpha }{f}^{\left(j\right)}∥}_{\infty }<\infty$.

Then,

(I)

$$\left|{\left(A_n\left(f\right)\right)}^{\left(j\right)}\left(x\right)-f^{\left(j\right)}\left(x\right)-\sum _{\lambda =1}^{N-1}{\displaystyle \frac{f^{\left(j+\lambda \right)}\left(x\right)}{\lambda !}}\left(A_n\left({\left(·-x\right)}^{\lambda }\right)\right)\left(x\right)\right|\le$$

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

(262)

$$\left\{\left[{∥D_{x-}^{\alpha }{f}^{\left(j\right)}∥}_{\infty }+{∥D_{*x}^{\alpha }{f}^{\left(j\right)}∥}_{\infty }\right]{\displaystyle \frac{2^{N+1}N!\left(q+\frac{1}{q}\right)e^{2\lambda }}{\mathsf{\Gamma }\left(\alpha +1\right){\left(2\lambda \right)}^{\alpha }{n}^{\alpha }{e}^{\lambda n^{1-\beta }}}}\right\},$$

(II)

$$\left|{\left(A_n^{*}\left(f\right)\right)}^{\left(j\right)}\left(x\right)-f^{\left(j\right)}\left(x\right)-\sum _{\lambda =1}^{N-1}{\displaystyle \frac{f^{\left(j+\lambda \right)}\left(x\right)}{\lambda !}}\left(A_n^{*}\left({\left(·-x\right)}^{\lambda }\right)\right)\left(x\right)\right|\le$$

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

(263)

$$\left[{∥D_{x-}^{\alpha }{f}^{\left(j\right)}∥}_{\infty ,(-\infty ,x]}+{∥D_{*x}^{\alpha }{f}^{\left(j\right)}∥}_{\infty ,[x,\infty )}\right]{\displaystyle \frac{2^N}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }\left[{\displaystyle \frac{1}{2}}+{\displaystyle \frac{N!}{{\lambda }^N}}\right]e^{-\lambda n^{1-\beta }},$$

(III)

$$\left|{\left(\overline{A_n}\left(f\right)\right)}^{\left(j\right)}\left(x\right)-f^{\left(j\right)}\left(x\right)-\sum _{\lambda =1}^{N-1}{\displaystyle \frac{f^{\left(j+\lambda \right)}\left(x\right)}{\lambda !}}\left(\overline{A_n}\left({\left(·-x\right)}^{\lambda }\right)\right)\left(x\right)\right|\le$$

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

(264)

$$\left[{∥D_{x-}^{\alpha }{f}^{\left(j\right)}∥}_{\infty ,(-\infty ,x]}+{∥D_{*x}^{\alpha }{f}^{\left(j\right)}∥}_{\infty ,[x,\infty )}\right]{\displaystyle \frac{2^N}{n^{\alpha }\mathsf{\Gamma }\left(\alpha +1\right)}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }\left[{\displaystyle \frac{1}{2}}+{\displaystyle \frac{N!}{{\lambda }^N}}\right]e^{-\lambda n^{1-\beta }}.$$

Proof. 

By Theorems 13–15. □

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

Theorem 20. 

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

(I)

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

(265)

(II)

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

(266)

(III)

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

(267)

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

Proof. 

By Theorems 4–6 and (75). □

We continue with the following.

Theorem 21. 

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

(I)

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

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

(268)

(II)

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

(269)

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

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

(III)

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

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

(270)

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

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

Proof. 

By Theorems 4–6 and (77). □

We finish our work with simultaneous iterations.

Remark 8. 

Let $i\in \mathbb{N}$ be fixed. Assume that $f\in C^{\left(i\right)}\left(\mathbb{R}\right)$, with $f^{\left(j\right)}\in C_B\left(\mathbb{R}\right)$, for $j=0,1,\dots ,i$; $r\in \mathbb{N}$. Then, by (75), we obtain

$${∥A_n^r\left(f^{\left(j\right)}\right)-f^{\left(j\right)}∥}_{\infty }\le r{∥A_n\left(f^{\left(j\right)}\right)-f^{\left(j\right)}∥}_{\infty }.$$

(271)

By (249) and inductively, we obtain

$${∥{\left(A_n^r\left(f\right)\right)}^{\left(j\right)}-f^{\left(j\right)}∥}_{\infty }\le r{∥{\left(A_n\left(f\right)\right)}^{\left(j\right)}-f^{\left(j\right)}∥}_{\infty },$$

(272)

for $j=0,1,\dots ,i$.

Similarly, we derive that

$${∥{\left(A_n^{{*}^r}\left(f\right)\right)}^{\left(j\right)}-f^{\left(j\right)}∥}_{\infty }\le r{∥{\left(A_n^{*}\left(f\right)\right)}^{\left(j\right)}-f^{\left(j\right)}∥}_{\infty },$$

(273)

and

$${∥{\left({\overline{A_n}}^r\left(f\right)\right)}^{\left(j\right)}-f^{\left(j\right)}∥}_{\infty }\le r{∥{\left(\overline{A_n}\left(f\right)\right)}^{\left(j\right)}-f^{\left(j\right)}∥}_{\infty },$$

(274)

for $j=0,1,\dots ,i$.

Now, let $m_1,m_2,\dots ,m_r\in \mathbb{N}:m_1\le m_2\le \dots \le m_r$. Then, based on (77), we find that

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

(275)

for $j=0,1,\dots ,i$.

Similarly, we see that

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

(276)

for $j=0,1,\dots ,i$, and

$${∥{\left(\overline{A_{m_r}}\left(\overline{A_{m_{r-1}}}\left(\dots \overline{A_{m_2}}\left(\overline{A_{m_1}}f\right)\right)\right)\right)}^{\left(j\right)}-f^{\left(j\right)}∥}_{\infty }\le \sum _{i^{*}=1}^r{∥{\left(\overline{A_{m_{i^{*}}}}\left(f\right)\right)}^{\left(j\right)}-f^{\left(j\right)}∥}_{\infty },$$

(277)

for $j=0,1,\dots ,i$.

All of the above inequalities (271)–(277) prove that our implied iterated simultaneous approximations do not have a speed worse than our basic simultaneous approximations by the activated convolution operators.

To keep our work short, we stop here.

5. Conclusions

Here, we presented a new idea for going from neural networks’ main tools, the activation functions, to convolution integrals approximation. This is a rare case of employing applied mathematics to theoretical ones.