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Article

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

by
George A. Anastassiou
Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152, USA
Mathematics 2024, 12(20), 3302; https://doi.org/10.3390/math12203302
Submission received: 8 October 2024 / Revised: 15 October 2024 / Accepted: 19 October 2024 / Published: 21 October 2024
(This article belongs to the Special Issue Fractional Calculus and Mathematical Applications, 2nd Edition)

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 q -Deformed and λ -Parameterized Hyperbolic Tangent Function g q , λ

Here, all of this initial background comes from Chapter 18 of [1].
We use g q , λ (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 , λ x : = e λ x q e λ x e λ x + q e λ x , λ , q > 0 , x R .
We have that
g q , λ 0 = 1 q 1 + q .
We notice also that
g q , λ x = e λ x q e λ x e λ x + q e λ x = 1 q e λ x e λ x 1 q e λ x + e λ x = e λ x 1 q e λ x e λ x + 1 q e λ x = g 1 q , λ x .
That is,
g q , λ x = g 1 q , λ x , x R ,
and
g 1 q , λ x = g q , λ x ,
hence,
g 1 q , λ x = g q , λ x .
It is
g q , λ x = e 2 λ x q e 2 λ x + q = 1 q e 2 l x 1 + q e 2 λ x x + 1 ,
i.e.,
g q , λ + = 1 ,
Furthermore,
g q , λ x = e 2 λ x q e 2 λ x + q x q q = 1 ,
i.e.,
g q , λ = 1 .
We find that
g q , λ x = 4 q λ e 2 λ x e 2 λ x + q 2 > 0 ,
therefore, g q , λ is strictly increasing.
Next, we obtain ( x R )
g q , λ x = 8 q λ 2 e 2 λ x q e 2 λ x e 2 λ x + q 3 C R .
We observe that
q e 2 λ x 0 q e 2 λ x ln q 2 λ x x ln q 2 λ .
So, in case of x < ln q 2 λ , we have that g q , λ is strictly concave up, with g q , λ ln q 2 λ = 0 .
And, in case of x > ln q 2 λ , we have that g q , λ is strictly concave down.
Clearly, g q , λ is a shifted sigmoid function with g q , λ 0 = 1 q 1 + q , and g q , λ x = g q 1 , λ x , (a semi-odd function).
By 1 > 1 and x + 1 > x 1 , we consider the function
M q . λ x : = 1 4 g q , λ x + 1 g q , λ x 1 > 0 ,
x R ; q , λ > 0 . Notice that M q , λ ± = 0 , so the x-axis is horizontal asymptote.
We have that
M q , λ x = 1 4 g q , λ x + 1 g q , λ x 1 =
1 4 g q , λ x 1 g q , λ x + 1 =
1 4 g 1 q , λ x 1 + g 1 q , λ x + 1 =
1 4 g 1 q , λ x + 1 g 1 q , λ x 1 = M 1 q , λ x , x R .
Thus,
M q , λ x = M 1 q , λ x , x R ; q , λ > 0 ,
a deformed symmetry.
Next, we have that
M q , λ x = 1 4 g q , λ x + 1 g q , λ x 1 , x R .
Let x < ln q 2 λ 1 , then x 1 < x + 1 < ln q 2 λ and g q , λ x + 1 > g q , λ x 1 (by g q , λ being strictly concave up for x < ln q 2 λ ), i.e., M q , λ x > 0 . Hence, M q , λ is strictly increasing over , ln q 2 λ 1 .
Now, let x 1 > ln q 2 λ ; then, x + 1 > x 1 > ln q 2 λ , and g q , λ x + 1 < g q , λ x 1 , i.e., M q , λ x < 0 .
Therefore, M q , λ is strictly decreasing over ln q 2 λ + 1 , + .
Let us next consider ln q 2 λ 1 x ln q 2 λ + 1 . We have that
M q , λ x = 1 4 g q , λ x + 1 g q , λ x 1 =
2 q λ 2 e 2 λ x + 1 q e 2 λ x + 1 e 2 λ x + 1 + q 3 e 2 λ x 1 q e 2 λ x 1 e 2 λ x 1 + q 3 .
By ln q 2 λ 1 x ln q 2 λ x + 1 ln q 2 λ x + 1 q e 2 λ x + 1 q e 2 λ x + 1 0 .
By x ln q 2 λ + 1 x 1 ln q 2 λ 2 λ x 1 ln q e 2 λ x 1 q q e 2 λ β x 1 0 .
Clearly, by (13) we see that M q , λ x 0 , for x ln q 2 λ 1 , ln q 2 λ + 1 .
More precisely, M q , λ is concave down over ln q 2 λ 1 , ln q 2 λ + 1 , and strictly concave down over ln q 2 λ 1 , ln q 2 λ + 1 .
Consequently, M q , λ has a bell-type shape over R .
Of course, it holds that M q , λ ln q 2 λ < 0 .
At x = ln q 2 λ , we have
M q , λ x = 1 4 g q , λ x + 1 g q , λ x 1 =
q λ e 2 λ x + 1 e 2 λ x + 1 + q 2 e 2 λ x 1 e 2 λ x 1 + q 2 .
Thus,
M q , λ ln q 2 λ = q λ e 2 λ ln q 2 λ + 1 e 2 λ ln q 2 λ + 1 + q 2 e 2 λ ln q 2 λ 1 e 2 λ ln q 2 λ 1 + q 2 =
λ e 2 λ e 2 λ + 1 2 e 2 λ e 2 λ + 1 2 e 2 λ + 1 2 e 2 λ + 1 2 = 0 .
That is, ln q 2 λ is the only critical number of M q , λ over R . Hence, at x = ln q 2 λ , M q , λ achieves its global maximum, which is
M q , λ ln q 2 λ = 1 4 g q , λ ln q 2 λ + 1 g q , λ ln q 2 λ 1 =
1 4 e λ e λ e λ + e λ e λ e λ e λ + e λ =
1 4 2 e λ e λ e λ + e λ = 1 2 e λ e λ e λ + e λ = tanh λ 2 .
Conclusion: The maximum value of M q , λ is
M q , λ ln q 2 λ = tanh λ 2 , λ > 0 .
We mention the following.
Theorem 1 
([1], Ch. 18, p. 458). We have that
i = M q , λ x i = 1 , x R , λ , q > 0 .
Also, the following holds.
Theorem 2 
([1], Ch. 18, p. 459). It holds that
M q , λ x d x = 1 , λ , q > 0 .
Thus, M q , λ is a density function on R ; λ , q > 0 .
Similarly, we see that
M 1 q , λ x d x = 1 , λ , q > 0 ,
thus, M 1 q , λ is a density function.
Furthermore, we observe the symmetry
M q , λ + M 1 q , λ x = M q , λ + M 1 q , λ x , x R .
Furthermore,
φ = M q , λ + M 1 q , λ 2
is a new density function over R , i.e.,
φ x d x = 1 .
Clearly, then,
φ n x u d u = 1 , n N , x R .

3. Basic

We give
Definition 1. 
Let f C B R (continuous and bounded functions on R ), n N . We define the following basic activated hyperbolic tangent perturbed convolution-type operators,
A n f x : = f u n φ n x u d u , x R .
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 * f x : = n u n u + 1 n f t d t φ n x u d u
= n 0 1 n f t + u n d t φ n x u d u ,
where f C B R , n N , ∀ x R , and the activated Quadrature operators
A n ¯ f x : = i = 1 r w i f u n + i n r φ n x u d u ,
where w i 0 , i = 1 r w i = 1 ; f C B R , n N , ∀ x R .
An essential property follows.
Theorem 3. 
Let 0 < α < 1 , n N : n 1 α > 2 . Then,
u R : n x u n 1 α φ n x u d u < q + 1 q e 2 λ n 1 α 1 , q , λ > 0 .
Proof. 
We have that M q , λ x : = 1 4 g q , λ x + 1 g q , λ x 1 , ∀ x R .
Let x 1 , i.e., 0 x 1 < x + 1 . Applying the mean value theorem, we obtain
M q , λ x = 1 4 · 2 · g q , λ ξ 1 = 1 2 g q , λ ξ 1 = 1 2 4 q λ e 2 λ ξ 1 e 2 λ ξ 1 + q 2 <
2 q λ e 2 λ ξ 1 + q e 2 λ ξ 1 + q 2 = 2 q λ e 2 λ ξ 1 + q < 2 q λ e 2 λ ξ 1 ,
where 0 x 1 < ξ 1 < x + 1 .
That is,
M q , λ x < 2 q λ e 2 λ x 1 , x 1 .
Similarly, it holds that
M 1 q , λ x < 2 q λ e 2 λ x 1 , x 1 .
Hence, we see that
φ x < q + 1 q λ e 2 λ x 1 , x 1 .
Consider the set
Δ : = u R : n x u n 1 α .
We have that
Δ φ n x u d u = Δ φ n x u d u < q + 1 q λ Δ e 2 λ n x u 1 d u =
2 q + 1 q λ n 1 α e 2 λ x 1 d x = 2 q + 1 q λ n 1 α 1 e 2 λ z d z =
q + 1 q n 1 α 1 e 2 λ z d 2 λ z = q + 1 q n 1 α 1 e y d y =
q + 1 q e 2 λ z | n 1 α 1 = q + 1 q e 2 λ n 1 α 1 = q + 1 q e 2 λ n 1 α 1 .
The claim is proven. □
The first modulus of continuity here is
ω 1 f , δ : = sup x , y R , x y δ f x f y , δ > 0 .

4. Main Results

We present the following approximation results.
Theorem 4. 
Let 0 < α < 1 , n N : n 1 α > 2 , f C B R , x R . Then,
A n f x f x ω 1 f , 1 n α + 2 q + 1 q f e 2 λ n 1 α 1 = : λ s ,
and
A n f f λ s .
So, for f C u B R : = C u R C B R (where C u R the uniformly continuous functions on R ), we see that lim n A n f = f , pointwise and uniformly.
Proof. 
Call
Δ 1 : = u R : u n x < 1 n α ,
and
Δ 2 : = u R : u n x 1 n α .
That is, Δ 1 Δ 2 = R .
We have that
A n f x f x = ( 23 )
f u n φ n x u d u f x φ n x u d u =
f u n f x φ n x u d u
f u n f x φ n x u d u =
Δ 1 f u n f x φ n x u d u + Δ 2 f u n f x φ n x u d u
Δ 1 ω 1 f , u n x φ n x u d u + 2 f Δ 2 φ n x u d u
ω 1 f , 1 n α Δ 1 φ n x u d u + 2 f q + 1 q e 2 λ n 1 α 1
ω 1 f , 1 n α φ n x u d u + 2 q + 1 q f e 2 λ n 1 α 1 =
ω 1 f , 1 n α + 2 q + 1 q f e 2 λ n 1 α 1 .
Theorem 5. 
Let 0 < α < 1 , n N : n 1 α > 2 , f C B R , x R . Then,
A n * f x f x ω 1 f , 1 n + 1 n α + 2 q + 1 q f e 2 λ n 1 α 1 = : ρ s ,
and
A n * f f ρ s .
For f C u B R , we have lim n A n * f = f , pointwise and uniformly.
Proof. 
Δ 1 , Δ 2 as in (39), (40), respectively.
We have that
A n * f x f x =
n u n u + 1 n f t d t φ n x u d u f x φ n x u d u =
n u n u + 1 n f t d t φ n x u d u n u n u + 1 n f x d t φ n x u d u =
n u n u + 1 n f t f x d t φ n x u d u
n u n u + 1 n f t f x d t φ n x u d u =
n 0 1 n f t + u n f x d t φ n x u d u
Δ 1 n 0 1 n f t + u n f x d t φ n x u d u +
Δ 2 n 0 1 n f t + u n f x d t φ n x u d u
Δ 1 n 0 1 n ω 1 f , t + 1 n α d t φ n x u d u + 2 f Δ 2 φ n x u d u
ω 1 f , 1 n + 1 n α + 2 f q + 1 q e 2 λ n 1 α 1 =
ω 1 f , 1 n + 1 n α + 2 q + 1 q f e 2 λ n 1 α 1 .
Theorem 6. 
Let 0 < α < 1 , n N : n 1 α > 2 , f C B R , x R . Then,
A n ¯ f x f x ω 1 f , 1 n + 1 n α + 2 q + 1 q f e 2 λ n 1 α 1 = ρ s ,
and
A n ¯ f f ρ s .
For f C u B R , we have lim n A n ¯ f = f , pointwise and uniformly.
Proof. 
Δ 1 , Δ 2 as in (39), (40), respectively.
We have
A n ¯ f x f x =
i = 1 r w i f u n + i n r φ n x u d u i = 1 r w i f x φ n x u d u =
i = 1 r w i f u n + i n r f x φ n x u d u
i = 1 r w i f u n + i n r f x φ n x u d u
A 1 i = 1 r w i f u n + i n r f x φ n x u d u +
A 2 i = 1 r w i f u n + i n r f x φ n x u d u
Δ 1 i = 1 r w i ω 1 f , 1 n α + i n r φ n x u d u + 2 f Δ 2 φ n x u d u
ω 1 f , 1 n + 1 n α + 2 f q + 1 q e 2 λ n 1 α 1 =
ω 1 f , 1 n + 1 n α + 2 q + 1 q f e 2 λ n 1 α 1 .
We need
Proposition 1. 
It holds that ( k N ), and
z k φ z d z tanh λ k + 1 + q + 1 q e 2 λ k ! 2 λ k < .
Proof. 
We can write
z k φ z d z = 2 0 z k φ z d z =
2 0 1 z k φ z d z + 1 z k φ z d z ( by ( 17 ) , ( 31 ) )
2 0 1 z k d z tanh λ 2 + q + 1 q λ 1 z k e 2 λ z 1 d z
2 1 k + 1 tanh λ 2 + q + 1 q 2 e 2 λ 2 λ k 0 2 λ z k e 2 λ z d 2 λ z =
tanh λ k + 1 + q + 1 q e 2 λ 2 λ k 0 y k e y d y =
tanh λ k + 1 + q + 1 q e 2 λ 2 λ k Γ k + 1 < .
Next we describe the global smoothness preservation property of our activated operators.
Theorem 7. 
Here, f C B R C u R . Then,
ω 1 A n f , δ ω 1 f , δ , δ > 0 .
If f C u R , then A n f C u R .
Proof. 
We have that
A n f x = f x z n φ z d z .
Let x , y R ; then,
A n f x A n f y = f x z n f y z n φ z d z .
Thus,
A n f x A n f y f x z n f y z n φ z d z
ω 1 f , x y φ z d z = ( 22 ) ω 1 f , x y .
Let x y δ , δ > 0 ; then, we obtain (54). □
Remark 1. 
Clearly, (54) is attained by f = identity map = : i d .
We have
A n i d x A n i d y = i d x i d y = x y ,
and
ω 1 A n i d , δ = ω 1 i d , δ = δ > 0 .
We also see that
A n i d x = x z n φ z d z =
x φ z d z 1 n z φ z d z = x 1 n z φ z d z .
So, for fixed x R , we have
A n i d x x + 1 n z φ z d z
( 52 ) x + 1 n tanh λ 2 + q + 1 q e 2 λ 2 λ < .
Of course, i d C u R .
Theorem 8. 
Again, f C B R C u R . Then,
ω 1 A n * f , δ ω 1 f , δ , δ > 0 .
If f C u R , then A n * f C u R .
Inequality (58) is attained by f x = i d x .
Proof. 
We notice that ( x , y R ),
A n * f x = n 0 1 n f t + x z n d t φ z d z ,
and
A n * f y = n 0 1 n f t + y z n d t φ z d z .
Thus,
A n * f x A n * f y =
n 0 1 n f t + x z n f t + y z n d t φ z d z
n 0 1 n f t + x z n f t + y z n d t φ z d z
ω 1 f , x y φ z d z = ω 1 f , x y ,
proving Inequality (58).
Notice that
A n * i d x A n * i d y = x y ,
and
A n * i d x n 0 1 n t + x + z n d t φ z d z
1 n + x + z n φ z d z = 1 n + x + 1 n z φ z d z < ,
proving the attainability of (58). □
Theorem 9. 
We consider f C B R C u R . Then,
ω 1 A n ¯ f , δ ω 1 f , δ , δ > 0 .
If f C u R , then A n ¯ f C u R .
Inequality (62) is attained by f x = i d x .
Proof. 
For x , y R , we have
A n ¯ f x = i = 1 r w i f x z n + i n r φ z d z ,
and
A n ¯ f y = i = 1 r w i f y z n + i n r φ z d z .
Hence,
A n ¯ f x A n ¯ f y =
i = 1 r w i f x z n + i n r f y z n + i n r φ z d z
i = 1 r w i f x z n + i n r f y z n + i n r φ z d z
ω 1 f , x y φ z d z = ω 1 f , x y .
That is, (62) true, and is attained by f x = i d x .
Indeed, we observe that
A n ¯ i d x A n ¯ i d y = x y ,
and, furthermore,
A n ¯ i d x i = 1 r w i x + z n + 1 n φ z d z
x + z n + 1 n φ z d z = 1 n + x + 1 n z φ z d z < ,
proving the attainability of (62). □
We make the following remark.
Remark 2. 
Let i N be fixed. Assume that f C i R , with f j C B R , for j = 0 , 1 , , i .
We have that
A n f x = f x z n φ z d z .
By repeatedly applying Leibnitz’s rule, we obtain
i A n f x x i = f i x z n φ z d z =
f i u n φ n x u d u = A n f i x , x R .
Clearly, it is valid that
i A n * f x x i = A n * f i x ,
and
i A n ¯ f x x i = A n ¯ f i x ,
x 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 f x = f x z n φ z d z ,
where f C B R .
Let x N x , as N , and
A n f x N A n f x = f x N z n f x z n φ z d z .
We have that
f x N z n φ z f x z n φ z , z R , as N .
Furthermore, it holds that
A n f x N A n f x
f x N z n f x z n φ z d z 0 , as N ,
by dominated convergence theorem, because we have that
f x N z n φ z f φ z ,
and f φ z is integrable over R , z R .
Hence, A n f C B R .
Furthermore, it holds that
A n f x f φ z d z = f ,
i.e.,
A n f f .
So, A n is a bounded positive linear operator.
Clearly it holds that
A n 2 f = A n A n f A n f f .
And, for k N , we obtain
A n k f A n k 1 f A n k 2 f f ,
so the contraction property valid, and A n k is a bounded linear operator.
Remark 4. 
Let r N . We observe that
A n r f f = A n r f A n r 1 f + A n r 1 f A n r 2 f + A n r 2 f A n r 3 f
+ + A n 2 f A n f + A n f f .
Then,
A n r f f A n r f A n r 1 f + A n r 1 f A n r 2 f + A n r 2 f A n r 3 f
+ + A n 2 f A n f + A n f f =
A n r 1 A n f f + A n r 2 A n f f + + A n A n f f +
A n f f r A n f f .
Therefore,
A n r f f r A n f f .
Now, let m 1 , m 2 , , m r N : m 1 m 2 m r , and A m i , as above.
A m r A m r 1 A m 2 A m 1 f f = =
A m r A m r 1 A m 2 A m 1 f f + A m r A m r 1 A m 3 A m 2 f f +
A m r A m r 1 A m 4 A m 3 f f + + A m r A m r 1 f f + A m r f f .
Consequently, it holds, as in Chapter 2 of [1], that
A m r A m r 1 A m 2 A m 1 f f i = 1 r A m i f f .
Next, we have
Remark 5. 
A n * f x = n 0 1 n f t + x z n d t φ z d z ,
f C B R .
Let x N x , as N , and
A n * f x N A n * f x =
n 0 1 n f t + x N z n f t + x z n d t φ z d z
n 0 1 n f t + x N z n f t + x z n d t φ z d z 0 , as N .
This is true by the bounded convergence theorem, and we see that
x N x t + x N z n t + x z n
and
f t + x N z n f t + x z n , and
f t + x N z n f ,
and 0 , 1 n is finite interval. Thus,
n 0 1 n f t + x N z n f t + x z n d t 0 , as N .
Therefore, it holds that
n 0 1 n f t + x N z n d t n 0 1 n f t + x z n d t , as N .
Further,
n 0 1 n f t + x N z n d t φ z n 0 1 n f t + x z n d t φ z ,
as N , z R .
Furthermore, we have
n 0 1 n f t + x N z n d t φ z f φ z ,
with f φ z being integrable over R .
Therefore, by the dominated convergence theorem,
A n * f x N A n * f x , as N .
Hence, A n * f x is bounded and continuous in x R .
The iterated facts hold for A n * , as in the A n f case, all the same! See (71)–(73), and all of Remark 4.
Remark 6. 
Next, we observe the following. Let f C B R , and
A n ¯ f x = i = 1 r w i f x z n + i n r φ z d z .
Let x N x , as N . Then,
A n ¯ f x N A n ¯ f x =
i = 1 r w i f x N z n + i n r f x z n + i n r φ z d z
i = 1 r w i f x N z n + i n r f x z n + i n r φ z d z 0 , as N .
The last comes by the dominated convergence theorem,
x N z n + i n r x z n + i n r
and
i = 1 r w i f x N z n + i n r i = 1 r w i f x z n + i n r ,
and
i = 1 r w i f x N z n + i n r φ z i = 1 r w i f x z n + i n r φ z ,
as N , z R .
Furthermore, it holds that
i = 1 r w i f x N z n + i n r φ z f φ z ,
in which the function is integrable over R .
Therefore,
A n ¯ f x N A n ¯ f x , as N .
Hence, A n ¯ f x is bounded and continuous in x R .
Iterated stuff for A n ¯ : it is all the same, as with A n f ! 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 < α < 1 , n N : n 1 α > 2 ; x R , f C N R , N N , with f N C B R . Then,
(i)
A n f x f x j = 1 N f j x j ! A n · x j x
ω 1 f N , 1 n α n α N N ! + 4 f N n N q + 1 q e 2 λ λ N e λ n 1 α 0 , as n .
(ii) If f j x = 0 , j = 1 , , N , we have
A n f x f x
ω 1 f N , 1 n α n α N N ! + 4 f N n N q + 1 q e 2 λ λ N e λ n 1 α ,
(iii)
A n f x f x j = 1 N f j x j ! 1 n j tanh λ j + 1 + q + 1 q e 2 λ j ! 2 λ j
+ ω 1 f N , 1 n α n α N N ! + 4 f N n N q + 1 q e 2 λ λ N e λ n 1 α ,
(iv)
A n f f j = 1 N f j j ! 1 n j tanh λ j + 1 + q + 1 q e 2 λ j ! 2 λ j
+ ω 1 f N , 1 n α n α N N ! + 4 f N n N q + 1 q e 2 λ λ N e λ n 1 α .
Proof. 
We have that
f u n = j = 0 N f j x j ! u n x j + x u n f N t f N x u n t N 1 N 1 ! d t .
Then,
f u n φ n x u = j = 0 N f j x j ! φ n x u u n x j +
φ n x u x u n f N t f N x u n t N 1 N 1 ! d t .
Hence,
A n f x = f u n φ n x u d u =
j = 0 N f j x j ! φ n x u u n x j d u +
φ n x u x u n f N t f N x u n t N 1 N 1 ! d t d u .
Thus, it holds that
A n f x f x = j = 1 N f j x j ! φ n x u u n x j d u
+ φ n x u x u n f N t f N x u n t N 1 N 1 ! d t d u .
Call
R n x : = φ n x u x u n f N t f N x u n t N 1 N 1 ! d t d u .
Call
γ u : = x u n f N t f N x u n t N 1 N 1 ! d t .
Let u n x < 1 n α .
(i) The case where u n x . Then,
γ u x u n f N t f N x u n t N 1 N 1 ! d t
ω 1 f N , 1 n α x u n u n t N 1 N 1 ! d t =
ω 1 f N , 1 n α u n t N N ! ω 1 f N , 1 n α 1 n α N N ! .
(ii) The case where u n < x . Then,
γ u = u n x f N t f N x u n t N 1 N 1 ! d t
u n x f N t f N x t u n N 1 N 1 ! d t
ω 1 f N , 1 n α u n x t u n N 1 N 1 ! d t =
ω 1 f N , 1 n α x u n N N ! ω 1 f N , 1 n α 1 n α N N ! .
Therefore, it holds that
γ u ω 1 f N , 1 n α 1 n α N N ! ,
when u n x < 1 n α .
Consequently, we see that
u n x < 1 n α φ n x u x u n f N t f N x u n t N 1 N 1 ! d t d u
u n x < 1 n α φ n x u γ u d u ω 1 f N , 1 n α 1 n α N N ! .
Next, we treat
u n x 1 n α φ n x u x u n f N t f N x u n t N 1 N 1 ! d t d u
u n x 1 n α φ n x u x u n f N t f N x u n t N 1 N 1 ! d t d u =
u n x 1 n α φ n x u γ u d u = : ξ 1 .
Let u n x ; then,
γ u 2 f N u n x N N ! .
If u n < x , then
γ u = u n x f N t f N x u n t N 1 N 1 ! d t
2 f N u n x t u n N 1 N 1 ! d t = 2 f N x u n N N ! .
Consequently, we see that
γ u 2 f N x u n N N ! .
Furthermore, we obtain
ξ 1 2 f N N ! u n x 1 n α φ n x u x u n N d u =
2 f N n N N ! n x u n 1 α φ n x u n x u N d u ( 31 )
(here, it is Δ : = u R : n x u n 1 α )
2 f N n N N ! q + 1 q λ Δ e 2 λ n x u 1 n x u N d u =
4 f N n N N ! q + 1 q λ n 1 α e 2 λ x 1 x N d x =
4 f N n N N ! q + 1 q λ e 2 λ n 1 α e 2 λ x x N d x =
4 f N n N N ! q + 1 q λ e 2 λ 2 λ N + 1 n 1 α e 2 λ x 2 λ x N d 2 λ x =
2 f N n N N ! q + 1 q e 2 λ 2 λ N 2 λ n 1 α e y y N d y
2 f N n N N ! q + 1 q e 2 λ 2 λ N 2 N N ! 2 λ n 1 α e y e y 2 d y =
2 N + 1 f N n N q + 1 q e 2 λ 2 λ N 2 λ n 1 α e y 2 d y =
2 f N n N q + 1 q e 2 λ λ N 2 e y 2 | 2 λ n 1 α =
4 f N n N q + 1 q e 2 λ λ N e y 2 | 2 λ n 1 α =
4 f N n N q + 1 q e 2 λ λ N e λ n 1 α .
We have proved that
u n x 1 n α φ n x u x u n f N t f N x u n t N 1 N 1 ! d t d u
4 f N n N q + 1 q e 2 λ λ N e λ n 1 α 0 , as n .
Finally, we have that
R n x u n x < 1 n α φ n x u x u n f N t f N x u n t N 1 N 1 ! d t d u
+ u n x 1 n α φ n x u x u n f N t f N x u n t N 1 N 1 ! d t d u
ω 1 f N , 1 n α 1 n α N N ! + 4 f N n N q + 1 q e 2 λ λ N e λ n 1 α 0 , as n .
Let j = 1 , , N and z : = n x u . Then,
A n · x j x = u n x j φ n x u d u
u n x j φ n x u d u =
1 n j n x u j φ n x u d u = 1 n j z j φ z d z ( 52 )
1 n j tanh λ j + 1 + q + 1 q e 2 λ j ! 2 λ j 0 ,
as n .
The theorem now is proven. □
We continue with the activated Kantorovich operators.
Theorem 11. 
Let 0 < α < 1 , n N : n 1 α > 2 ; x R , f C N R , N N , with f N C B R . Then,
(i)
A n * f x f x j = 1 N f j x j ! A n * · x j x
ω 1 f N , 1 n + 1 n α 1 n + 1 n α N N ! +
2 N + 1 f N n N N ! q + 1 q e 2 λ 1 2 + N ! λ N e λ n 1 α 0 , as n .
(ii) If f j x = 0 , j = 1 , , N , we have
A n * f x f x
ω 1 f N , 1 n + 1 n α 1 n + 1 n α N N ! +
2 N + 1 f N n N N ! q + 1 q e 2 λ 1 2 + N ! λ N e λ n 1 α ,
(iii)
A n * f x f x j = 1 N f j x j ! 2 j 1 n j 1 + tanh λ j + 1 + q + 1 q e 2 λ j ! 2 λ j
+ ω 1 f N , 1 n + 1 n α 1 n + 1 n α N N ! + 2 N + 1 f N n N N ! q + 1 q e 2 λ 1 2 + N ! λ N e λ n 1 α ,
(iv)
A n * f f j = 1 N f j j ! 2 j 1 n j 1 + tanh λ j + 1 + q + 1 q e 2 λ j ! 2 λ j
+ ω 1 f N , 1 n + 1 n α 1 n + 1 n α N N ! + 2 N + 1 f N n N N ! q + 1 q e 2 λ 1 2 + N ! λ N e λ n 1 α 0 ,
as n .
Proof. 
We have that
f t + u n = j = 0 N f j x j ! t + u n x j +
x t + u n f N s f N x t + u n s N 1 N 1 ! d s ,
and
0 1 n f t + u n d t = j = 0 N f j x j ! 0 1 n t + u n x j d t +
0 1 n x t + u n f N s f N x t + u n s N 1 N 1 ! d s d t .
Hence,
n 0 1 n f t + u n d t φ n x u d u =
j = 0 N f j x j ! n 0 1 n t + u n x j d t φ n x u d u +
n 0 1 n x t + u n f N s f N x t + u n s N 1 N 1 ! d s d t φ n x u d u .
Furthermore, it holds that
A n * f x f x = j = 1 N f j x j ! A n * · x j x + R n x ,
where
R n x : = n
0 1 n x t + u n f N s f N x t + u n s N 1 N 1 ! d s d t φ n x u d u .
Call
λ u : = n 0 1 n x t + u n f N s f N x t + u n s N 1 N 1 ! d s d t .
Let u n x < 1 n α , ( 0 < α < 1 ).
(i) The case where t + u n x . Then,
λ u n 0 1 n x t + u n f N s f N x t + u n s N 1 N 1 ! d s d t
n 0 1 n ω 1 f N , t + u n x x t + u n t + u n s N 1 N 1 ! d s d t
ω 1 f N , 1 n + 1 n α n 0 1 n t + u n x N N ! d t
ω 1 f N , 1 n + 1 n α N ! 1 n + 1 n α N .
Therefore, it holds that
λ u ω 1 f N , 1 n + 1 n α 1 n + 1 n α N N ! .
(ii) The case where t + u n < x . Then,
λ u = n 0 1 n x t + u n f N s f N x t + u n s N 1 N 1 ! d s d t =
n 0 1 n t + u n x f N x f N s s t + u n N 1 N 1 ! d s d t
n 0 1 n t + u n x f N x f N s s t + u n N 1 N 1 ! d s d t
n 0 1 n ω 1 f N , t + u n x t + u n x s t + u n N 1 N 1 ! d s d t
ω 1 f N , 1 n + 1 n α n 0 1 n x t + u n N N ! d t
(by x t + u n = x t u n u n x + t )
ω 1 f N , 1 n + 1 n α n 0 1 n 1 n + 1 n α N N ! d t =
ω 1 f N , 1 n + 1 n α 1 n + 1 n α N N ! .
So, when u n x < 1 n α , we have proved that
λ u ω 1 f N , 1 n + 1 n α 1 n + 1 n α N N ! .
Consequently, we see that
n u n x < 1 n α 0 1 n x t + u n f N s f N x t + u n s N 1 N 1 ! d s d t φ n x u d u
ω 1 f N , 1 n + 1 n α 1 n + 1 n α N N ! .
Next, we treat
u n x 1 n α n 0 1 n x t + u n f N s f N x t + u n s N 1 N 1 ! d s d t φ n x u d u
u n x 1 n α n 0 1 n x t + u n f N s f N x t + u n s N 1 N 1 ! d s d t
φ n x u d u = : ψ 1 .
We also have that
λ u = n 0 1 n x t + u n f N s f N x t + u n s N 1 N 1 ! d s d t
n 0 1 n x t + u n f N s f N x t + u n s N 1 N 1 ! d s d t
(let t + u n x )
2 f N n 0 1 n x t + u n t + u n s N 1 N 1 ! d s d t =
2 f N n 0 1 n t + u n x N N ! d t
2 f N n 0 1 n t + u n x N N ! d t
2 f N N ! 1 n + u n x N .
So, when t + u n x , we obtain
λ u 2 f N 1 n + u n x N N ! .
If t + u n < x , then
λ u = n 0 1 n t + u n x f N s f N x s t + u n N 1 N 1 ! d s d t
2 f N n 0 1 n t + u n x s t + u n N 1 N 1 ! d s d t =
2 f N n 0 1 n x t + u n N N ! d t
2 f N n 0 1 n t + u n x N N ! d t
2 f N N ! 1 n + u n x N .
Consequently, we see that
λ u 2 f N N ! 1 n + u n x N .
Furthermore, we obtain
ψ 1 u n x 1 n α 1 n + u n x N φ n x u d u 2 f N N ! =
2 N f N N ! n x u n 1 α 1 n N + n x u N n N φ n x u d u
2 N f N n N N ! n x u n 1 α 1 + n x u N φ n x u d u ( 31 )
(here, it is Δ : = u R : n x u n 1 α )
2 N f N n N N ! q + 1 q λ Δ 1 + n x u N e 2 λ n x u 1 d u =
2 N + 1 f N n N N ! q + 1 q λ n 1 α e 2 λ x 1 1 + x N d x =
2 N + 1 f N n N N ! q + 1 q λ e 2 λ n 1 α e 2 λ x 1 + x N d x =
2 N + 1 f N n N N ! q + 1 q λ e 2 λ n 1 α e 2 λ x d x + n 1 α e 2 λ x x N d x =
2 N + 1 f N n N N ! q + 1 q λ e 2 λ e 2 λ n 1 α 2 λ + n 1 α e 2 λ x x N d x =
2 N + 1 f N n N N ! q + 1 q λ e 2 λ e 2 λ n 1 α 2 λ + 1 2 λ N + 1 2 λ n 1 α e y y N d y
2 N + 1 f N n N N ! q + 1 q λ e 2 λ e 2 λ n 1 α 2 λ + 2 N N ! 2 λ N + 1 2 λ n 1 α e y 2 d y =
2 N + 1 f N n N N ! q + 1 q λ e 2 λ e 2 λ n 1 α 2 λ + 2 N + 1 N ! 2 λ N + 1 e λ n 1 α =
2 N + 1 f N n N N ! q + 1 q λ e 2 λ e 2 λ n 1 α 2 λ + N ! λ N + 1 e λ n 1 α
2 N + 1 f N n N N ! q + 1 q e 2 λ 1 2 + N ! λ N e λ n 1 α 0 ,
as n .
We have proved that
u n x 1 n α n 0 1 n x t + u n f N s f N x t + u n s N 1 N 1 ! d s d t φ n x u d u
2 N + 1 f N n N N ! q + 1 q e 2 λ 1 2 + N ! λ N e λ n 1 α 0 , as n .
Finally, we have that
R n x ω 1 f N , 1 n + 1 n α 1 n + 1 n α N N ! +
2 N + 1 f N n N N ! q + 1 q e 2 λ 1 2 + N ! λ N e λ n 1 α 0 , as n .
We also observe that ( j = 1 , , N )
A n * · x j x = n 0 1 n t + u n x j d t φ n x u d u
n 0 1 n t + u n x j d t φ n x u d u
n 0 1 n t + u n x j d t φ n x u d u
1 n + u n x j φ n x u d u =
1 n j 1 + n x u j φ n x u d u
2 j 1 n j 1 + n x u j φ n x u d u =
2 j 1 n j 1 + z j φ z d z ( 52 )
2 j 1 n j 1 + tanh λ j + 1 + q + 1 q e 2 λ j ! 2 λ j 0 , as n .
The theorem is proven. □
We continue with the activated Quadrature operators.
Theorem 12. 
Let 0 < α < 1 , n N : n 1 α > 2 ; x R , f C N R , N N , with f N C B R . Then,
(i)
A n ¯ f x f x j = 1 N f j x j ! A n ¯ · x j x
ω 1 f N , 1 n + 1 n α 1 n + 1 n α N N ! +
2 N + 1 f N n N N ! q + 1 q e 2 λ 1 2 + N ! λ N e λ n 1 α 0 , as n .
(ii) If f j x = 0 , j = 1 , , N , we have
A n ¯ f x f x
ω 1 f N , 1 n + 1 n α 1 n + 1 n α N N ! +
2 N + 1 f N n N N ! q + 1 q e 2 λ 1 2 + N ! λ N e λ n 1 α ,
(iii)
A n ¯ f x f x j = 1 N f j x j ! 2 j 1 n j 1 + tanh λ j + 1 + q + 1 q e 2 λ j ! 2 λ j
+ ω 1 f N , 1 n + 1 n α 1 n + 1 n α N N ! + 2 N + 1 f N n N N ! q + 1 q e 2 λ 1 2 + N ! λ N e λ n 1 α ,
(iv)
A n ¯ f f j = 1 N f j j ! 2 j 1 n j 1 + tanh λ j + 1 + q + 1 q e 2 λ j ! 2 λ j
+ ω 1 f N , 1 n + 1 n α 1 n + 1 n α N N ! + 2 N + 1 f N n N N ! q + 1 q e 2 λ 1 2 + N ! λ N e λ n 1 α 0 ,
as n .
Proof. 
We have that
f u n + i n r = j = 0 N f j x j ! u n + i n r x j +
x u n + i n r f N t f N x u n + i n r t N 1 N 1 ! d t ,
and
i = 1 r w i f u n + i n r = j = 0 N f j x j ! i = 1 r w i u n + i n r x j +
i = 1 r w i x u n + i n r f N t f N x u n + i n r t N 1 N 1 ! d t .
Furthermore, it holds that
A n ¯ f x = i = 1 r w i f u n + i n r φ n x u d u =
j = 0 N f j x j ! i = 1 r w i u n + i n r x j φ n x u d u
+ i = 1 r w i x u n + i n r f N t f N x u n + i n r t N 1 N 1 ! d t φ n x u d u ,
and it is
A n ¯ f x f x = j = 1 N f j x j ! A n ¯ · x j x + R n x ,
where
R n x : =
i = 1 r w i x u n + i n r f N t f N x u n + i n r t N 1 N 1 ! d t φ n x u d u .
Let
δ u : = i = 1 r w i x u n + i n r f N t f N x u n + i n r t N 1 N 1 ! d t .
Let u n x < 1 n α .
(i) The case where u n + i n r x . Then,
δ u i = 1 r w i ω 1 f N , u n + i n r x u n + i n r x N N !
ω 1 f N , 1 n α + 1 n 1 n α + 1 n N N ! .
(ii) The case where u n + i n r < x . Then,
δ u = i = 1 r w i u n + i n r x f N t f N x t u n + i n r N 1 N 1 ! d t
i = 1 r w i u n + i n r x f N t f N x t u n + i n r N 1 N 1 ! d t
i = 1 r w i ω 1 f N , x u n + i n r x u n + i n r N N !
ω 1 f N , 1 n α + 1 n 1 n α + 1 n N N ! .
So, when u n x < 1 n α , we see that
δ u ω 1 f N , 1 n α + 1 n 1 n α + 1 n N N ! .
Consequently, we have that
u n x < 1 n α i = 1 r w i x u n + i n r f N t f N x u n + i n r t N 1 N 1 ! d t φ n x u d u
u n x < 1 n α φ n x u δ u d u ω 1 f N , 1 n α + 1 n 1 n α + 1 n N N ! .
Next, we treat
u n x 1 n α φ n x u i = 1 r w i x u n + i n r f N t f N x u n + i n r t N 1 N 1 ! d t d u
u n x 1 n α φ n x u δ u d u = : φ 1 .
Let u n + i n r x , then
δ u 2 f N N ! i = 1 r w i u n + i n r x N .
Let u n + i n r < x , then
δ u = i = 1 r w i u n + i n r x f N t f N x t u n + i n r N 1 N 1 ! d t
i = 1 r w i u n + i n r x f N t f N x t u n + i n r N 1 N 1 ! d t
2 f N N ! i = 1 r w i x u n + i n r N .
Consequently, we see that
δ u 2 f N N ! x u n + 1 n N
(see also (127)).
The quantity
u n x 1 n α φ n x u δ u d u φ 1
is estimated as in Theorem 11.
We also see that ( j = 1 , , N )
A n ¯ · x j x = i = 1 r w i u n + i n r x j φ n x u d u
i = 1 r w i u n + i n r x j φ n x u d u
i = 1 r w i u n x + i n r j φ n x u d u
1 n + u n x j φ n x u d u =
1 n j 1 + n x u j φ n x u d u
(see (134); the rest goes as in Theorem 11).
The theorem is proven. □
We need the following.
Definition 2. 
A function f : R R is absolutely continuous over R , iff f | a , b is absolutely continuous, for every a , b R . We write f A C n R , iff f n 1 A C R (absolutely continuous functions over R ), n N .
Definition 3. 
Let ν 0 , n = ν ( · is the ceiling of the number), f A C n R . We call left Caputo fractional derivative ([13,14,15], pp. 49–52), the function
D * a ν f x = 1 Γ n ν a x x t n ν 1 f n t d t ,
x [ a , ) , a R , where Γ is the gamma function.
Notice that D * a ν f L 1 a , b and D * a ν f exists almost everywhere on a , b , a , b R .
We set D * a 0 f x = f x , x [ a , ) .
We need the following.
Lemma 1 
(see also [16]). Let ν > 0 , ν N , n = ν , f C n 1 R and f n L R . Then, D * a ν f a = 0 for any a R .
Definition 4 
(see also [14,17,18]). Let f A C m R , m = α , α > 0 . The right Caputo fractional derivative of order α > 0 is given by
D b α f x = 1 m Γ m α x b z x m α 1 f m z d z ,
x ( , b ] , b R . We set D b 0 f x = f x .
Notice that D b α f L 1 a , b and D b α f exists almost everywhere on a , b , a , b R .
Lemma 2 
(see also [16]). Let f C m 1 R , f m L R , m = α , α > 0 . Then, D b α f b = 0 , for any b R .
We assume that
D * x 0 α f x = 0 , for x < x 0 , and D x 0 α f x = 0 , for x > x 0 .
We mention the following.
Proposition 2 
(see also [16]). Let f C n R , n = ν , ν > 0 . Then, D * a ν f x is continuous in x [ a , ) , a R .
Also we have the following.
Proposition 3 
(see also [16]). Let f C m R , m = α , α > 0 . Then, D b α f x is continuous in x ( , b ] , b R .
We further mention the following.
Proposition 4 
(see also [16]). Let f C m 1 R , f m L R , m = α , α > 0 and let x , x 0 R : x x 0 . Then, D * x 0 α f x is continuous in x 0 .
Proposition 5 
(see also [16]). Let f C m 1 R , f m L R , m = α , α > 0 , and let x , x 0 R : x x 0 . Then, D x 0 α f x is continuous in x 0 .
Proposition 6 
(see also [16]). Let f C m R , m = α , α > 0 ; x , x 0 R . Then, D * x 0 α f x , D x 0 α f x are jointly continuous functions in x , x 0 from R 2 R .
Here comes our first main fractional result.
Theorem 13. 
Let α > 0 , N = α , α N , f A C N R , f N L R , 0 < β < 1 , x R , n N : n 1 β max 3 , 1 2 λ . Assume also that both sup x R D * x α f , sup x R D x α f < .
Then,
(i)
A n f x f x j = 1 N 1 f j x j ! A n · x j x
1 n α β Γ α + 1 ω 1 D x α f , 1 n β ( , x ] + ω 1 D * x α f , 1 n β [ x , ) +
D x α f + D * x α f 2 N + 1 N ! q + 1 q e 2 λ Γ α + 1 2 λ α n α e λ n 1 β 0 , as n .
(ii) Given that f j x = 0 , j = 1 , , N 1 , we have
A n f x f x
1 n α β Γ α + 1 ω 1 D x α f , 1 n β ( , x ] + ω 1 D * x α f , 1 n β [ x , ) +
D x α f + D * x α f 2 N + 1 N ! q + 1 q e 2 λ Γ α + 1 2 λ α n α e λ n 1 β = : Φ 1 x ,
(iii)
A n f x f x j = 1 N 1 f j x j !
1 n j tanh λ j + 1 + q + 1 q e 2 λ j ! 2 λ j + Φ 1 x ,
(iv)
A n f f j = 1 N 1 f j j ! 1 n j tanh λ j + 1 + q + 1 q e 2 λ j ! 2 λ j +
1 n α β Γ α + 1 sup x R ω 1 D x α f , 1 n β ( , x ] + sup x R ω 1 D * x α f , 1 n β [ x , ) +
sup x R D x α f + sup x R D * x α f 2 N + 1 N ! q + 1 q e 2 λ Γ α + 1 2 λ α n α e λ n 1 β .
Above, when N = 1 , the sum j = 1 N 1 · = 0 .
As we see here, we obtain fractional pointwise and uniform convergence with rates of A n I the unit operator, as n .
Proof. 
Let x R . We have that D x α f x = D * x α f x = 0 .
From [13], p. 54, we obtain the left Caputo fractional Taylor formula that
f u n = j = 0 N 1 f j x j ! u n x j +
1 Γ α x u n u n s α 1 D * x α f s D * x α f x d s ,
for all x u n < .
Also, from [17], using the right Caputo fractional Taylor formula, we obtain
f u n = j = 0 N 1 f j x j ! u n x j +
1 Γ α u n x s u n α 1 D x α f s D x α f x d s ,
for all < u n x .
Hence,
f u n φ n x u = j = 0 N 1 f j x j ! φ n x u u n x j +
φ n x u Γ α x u n u n s α 1 D * x α f s D * x α f x d s ,
for all x u n < , and
f u n φ n x u = j = 0 N 1 f j x j ! φ n x u u n x j +
φ n x u Γ α u n x s u n α 1 D x α f s D x α f x d s ,
for all < u n x .
Therefore, we have
n x f u n φ n x u d u = j = 0 N 1 f j x j ! n x φ n x u u n x j d u +
1 Γ α n x φ n x u x u n u n s α 1 D * x α f s D * x α f x d s d u ,
and
n x f u n φ n x u d u = j = 0 N 1 f j x j ! n x φ n x u u n x j d u +
1 Γ α n x φ n x u u n x s u n α 1 D x α f s D x α f x d s d u .
Adding the last two equalities, (164) and (165), we have
A n f x = j = 0 N 1 f j x j ! A n · x j x +
1 Γ α n x φ n x u x u n u n s α 1 D * x α f s D * x α f x d s d u
+ n x φ n x u u n x s u n α 1 D x α f s D x α f x d s d u .
Consequently, it holds that
A n f x f x j = 1 N 1 f j x j ! A n · x j x = R n x ,
where
R n x : = 1 Γ α
n x φ n x u u n x s u n α 1 D x α f s D x α f x d s d u
+ n x φ n x u x u n u n s α 1 D * x α f s D * x α f x d s d u ,
x R .
Denote by
R n 1 x = 1 Γ α n x φ n x u u n x s u n α 1 D x α f s D x α f x d s d u ,
x R : x u n > , and
R n 2 x = 1 Γ α n x φ n x u x u n u n s α 1 D * x α f s D * x α f x d s d u ,
x R : > u n x .
That is,
R n x = R n 1 x + R n 2 x , x R .
Let first u n x < 1 n β .
Call
γ n 1 x : = 1 Γ α u n x s u n α 1 D x α f s D x α f x d s ,
and
γ n 2 x : = 1 Γ α x u n u n s α 1 D * x α f s D * x α f x d s .
Let us assume that x u n ; then,
γ n 1 x 1 Γ α u n x s u n α 1 D x α f s D x α f x d s
1 Γ α ω 1 D x α f , 1 n β ( , x ] x u n α α .
Hence,
γ n 1 x ω 1 D x α f , 1 n β ( , x ] 1 n α β Γ α + 1 , for x u n and u n x < 1 n β .
Let us assume that x u n ; then,
γ n 2 x 1 Γ α x u n u n s α 1 D * x α f s D * x α f x d s
ω 1 D * x α f , 1 n β [ x , ) u n x α Γ α + 1 .
Hence,
γ n 2 x ω 1 D * x α f , 1 n β [ x , ) 1 n α β Γ α + 1 , for x u n and u n x < 1 n β .
Consequently, we see that
R n 1 x | u n x < 1 n β ω 1 D x α f , 1 n β ( , x ] n α β Γ α + 1 ,
and
R n 2 x | u n x < 1 n β ω 1 D * x α f , 1 n β [ x , ) n α β Γ α + 1 .
Furthermore, we have that
γ n 1 x D x α f x u n α Γ α + 1 ,
where x u n > , and
γ n 2 x D * x α f u n x α Γ α + 1 ,
where > u n x .
Next, we see that
1 Γ α u n x 1 n β φ n x u u n x s u n α 1 D x α f s d s d u
1 Γ α u n x 1 n β φ n x u u n x s u n α 1 D x α f s d s d u
D x α f Γ α + 1 u n x 1 n β φ n x u x u n α d u =
D x α f n α Γ α + 1 u n x n 1 β φ n x u n x u α d u ( 31 )
D x α f n α Γ α + 1 q + 1 q λ u n x n 1 β e 2 λ n x u 1 n x u α d u =
2 D x α f n α Γ α + 1 q + 1 q λ n 1 β e 2 λ x 1 x α d x =
2 D x α f n α Γ α + 1 q + 1 q λ e 2 λ 2 λ α + 1 n 1 β e 2 λ x 2 λ x α d x 2 λ =
D x α f n α Γ α + 1 q + 1 q e 2 λ 2 λ α 2 λ n 1 β e y y α d y
D x α f n α Γ α + 1 q + 1 q e 2 λ 2 λ α 2 λ n 1 β e y y N d y
D x α f n α Γ α + 1 q + 1 q e 2 λ 2 λ α 2 N N ! 2 λ n 1 β e y e y 2 d y =
D x α f n α Γ α + 1 q + 1 q e 2 λ 2 λ α 2 N N ! 2 λ n 1 β e y 2 d y =
2 N + 1 N ! D x α f n α Γ α + 1 q + 1 q e 2 λ 2 λ α e λ n 1 β .
We have proved that
1 Γ α u n x 1 n β φ n x u u n x s u n α 1 D x α f s d s d u
2 N + 1 N ! D x α f n α Γ α + 1 q + 1 q e 2 λ 2 λ α e λ n 1 β 0 , as n .
Next, we see that
1 Γ α u n x 1 n β φ n x u x u n u n s α 1 D * x α f s d s d u
1 Γ α u n x 1 n β φ n x u x u n u n s α 1 D * x α f s d s d u
D * x α f Γ α + 1 u n x 1 n β φ n x u u n x α d u =
D * x α f n α Γ α + 1 u n x n 1 β φ n x u n x u α d u
(as before)
2 N + 1 N ! D * x α f n α Γ α + 1 q + 1 q e 2 λ 2 λ α e λ n 1 β 0 , as n .
Therefore, it holds that
1 Γ α u n x 1 n β φ n x u x u n u n s α 1 D * x α f s d s d u
2 N + 1 N ! D * x α f n α Γ α + 1 q + 1 q e 2 λ 2 λ α e λ n 1 β 0 , as n .
Consequently, it holds that
R n x 1 n α β Γ α + 1 ω 1 D x α f , 1 n β ( , x ] + ω 1 D * x α f , 1 n β [ x , ) +
D x α f + D * x α f 2 N + 1 N ! q + 1 q e 2 λ Γ α + 1 2 λ α n α e λ n 1 β 0 , as n .
We need 2 λ n 1 β 1 , iff n 1 β 1 2 λ , iff n 1 2 λ 1 1 β .
(i) In the case of 0 < 2 λ 1 (i.e., 0 < λ 1 2 ), then 1 2 λ 1 , and 1 2 λ 1 1 β 1 . So, for large enough n N , we can have 2 λ n 1 β 1 .
(ii) If 2 λ > 1 (i.e., λ > 1 2 ), then 1 2 λ < 1 and 1 2 λ 1 1 β < 1 . So, for any n N , we have that 2 λ n 1 β 1 .
See also the assumption.
We have that
D * x α f s = 1 Γ N α x s s t N α 1 f N t d t , ( s x )
and
D x α f s = 1 N Γ N α s x t s N α 1 f N t d t , ( s x )
x , s R .
Therefore, it holds that
D * x α f s 1 Γ N α f N s x N α N α = f N s x N α Γ N α + 1 ,
s x , and
D x α f s f N Γ N α x s N α N α = f N Γ N α + 1 x s N α ,
s x .
Thus, it is reasonable to assume that sup x R D * x α f , sup x R D x α f < .
Consequently, it holds that
sup x R ω 1 D x α f , 1 n β ( , x ] , sup x R ω 1 D * x α f , 1 n β [ x , ) < .
The theorem is now proven. □
We continue with the following results.
Theorem 14. 
Let α > 0 , N = α , α N , f A C N R , f N L R , 0 < β < 1 , x R , n N : n 1 β max 3 , 1 2 λ . Assume also that both sup x R D * x α f , sup x R D x α f < .
Then,
(i)
A n * f x f x j = 1 N 1 f j x j ! A n * · x j x
1 n + 1 n β α Γ α + 1 ω 1 D x α f , 1 n + 1 n β ( , x ] + ω 1 D * x α f , 1 n + 1 n β [ x , ) +
D x α f , ( , x ] + D * x α f , [ x , ) 2 N n α Γ α + 1 q + 1 q e 2 λ 1 2 + N ! λ N e λ n 1 β
= : Φ 1 * x 0 , as n ,
(ii) Given that f j x = 0 , j = 1 , , N 1 , we have
A n * f x f x Φ 1 * x ,
(iii)
A n * f x f x j = 1 N 1 f j x j !
2 j 1 n j 1 + tanh λ j + 1 + q + 1 q e 2 λ j ! 2 λ j + Φ 1 * x ,
(iv)
A n * f f j = 1 N 1 f j j ! 2 j 1 n j 1 + tanh λ j + 1 + q + 1 q e 2 λ j ! 2 λ j +
1 n + 1 n β α Γ α + 1 sup x R ω 1 D x α f , 1 n + 1 n β ( , x ] + sup x R ω 1 D * x α f , 1 n + 1 n β [ x , ) +
sup x R D x α f , ( , x ] + sup x R D * x α f , [ x , ) 2 N n α Γ α + 1 q + 1 q e 2 λ 1 2 + N ! λ N e λ n 1 β .
Above, when N = 1 , the sum j = 1 N 1 · = 0 .
So, we obtain the pointwise and uniform convergence with rates of A n * I , as n .
Proof. 
Let x R . We have that D x α f x = D * x α f x = 0 .
We can write
f t + u n = j = 0 N 1 f j x j ! t + u n x j +
1 Γ α x t + u n t + u n s α 1 D * x α f s D * x α f x d s ,
for all x t + u n < .
Also it holds that
f t + u n = j = 0 N 1 f j x j ! t + u n x j +
1 Γ α t + u n x s t + u n α 1 D x α f s D x α f x d s ,
for all < t + u n x .
Hence, we see
0 1 n f t + u n d t = j = 0 N 1 f j x j ! 0 1 n t + u n x j d t +
1 Γ α 0 1 n x t + u n t + u n s α 1 D * x α f s D * x α f x d s d t ,
for all x t + u n < , iff n x n t + u < .
Also, it holds that
0 1 n f t + u n d t = j = 0 N 1 f j x j ! 0 1 n t + u n x j d t +
1 Γ α 0 1 n t + u n x s t + u n α 1 D x α f s D x α f x d s d t ,
for all < t + u n x , iff < n t + u n x .
Therefore, we obtain
n n x 0 1 n f t + u n d t φ n x u d u =
j = 0 N 1 f j x j ! n n x 0 1 n t + u n x j d t φ n x u d u +
1 Γ α n n x φ n x u
0 1 n x t + u n t + u n s α 1 D * x α f s D * x α f x d s d t d u ,
and
n n x 0 1 n f t + u n d t φ n x u d u =
j = 0 N 1 f j x j ! n n x 0 1 n t + u n x j d t φ n x u d u +
1 Γ α n n x φ n x u
0 1 n t + u n x s t + u n α 1 D x α f s D x α f x d s d t d u .
Adding the last two equations, (193) and (194), we derive
A n * f x f x j = 1 N 1 f j x j ! A n * · x j x = R n * x ,
where
R n * x = R n 1 * x + R n 2 * x ,
with
R n 2 * x : = n Γ α n x φ n x u
0 1 n x t + u n t + u n s α 1 D * x α f s D * x α f x d s d t d u ,
n x n t + u < , and
R n 1 * x : = n Γ α n x φ n x u
0 1 n t + u n x s t + u n α 1 D x α f s D x α f x d s d t d u ,
< n t + u n x .
Let u n x < 1 n β .
Call
δ n 1 x : = n Γ α 0 1 n t + u n x s t + u n α 1 D x α f s D x α f x d s d t ,
and
δ n 2 x : = n Γ α 0 1 n x t + u n t + u n s α 1 D * x α f s D * x α f x d s d t .
We have that
δ n 1 x n Γ α 0 1 n ω 1 D x α f , x t u n x t u n α α d t =
1 Γ α + 1 ω 1 D x α f , 1 n + 1 n β ( , x ] 1 n + 1 n β α ,
and
δ n 2 x n Γ α 0 1 n ω 1 D * x α f , t + u n x t + u n x α α d t =
1 Γ α + 1 ω 1 D * x α f , 1 n + 1 n β [ x , + ) 1 n + 1 n β α .
Consequently, we see that
| R n 1 * x | | u n x < 1 n β ω 1 D x α f , 1 n + 1 n β ( , x ] 1 n + 1 n β α Γ α + 1 ,
and
| R n 2 * x | | u n x < 1 n β ω 1 D * x α f , 1 n + 1 n β [ x , + ) 1 n + 1 n β α Γ α + 1 .
We continue as follows:
δ n 1 x D x α f , ( , x ] Γ α n 0 1 n x t u n α α d t
D x α f , ( , x ] Γ α + 1 x u n + 1 n α ,
and
δ n 2 x D * x α f , [ x , ) Γ α + 1 x u n + 1 n α .
We have that
R n 1 * x | x u n 1 n β
D x α f , ( , x ] Γ α + 1 ( , n x ] x u n 1 n β φ n x u x u n + 1 n α d u =
D x α f , ( , x ] n α Γ α + 1 ( , n x ] n x u n 1 β φ n x u n x u + 1 α d u
D x α f , ( , x ] n α Γ α + 1 n x u n 1 β φ n x u n x u + 1 N d u
D x α f , ( , x ] n α Γ α + 1 2 N 1 n x u n 1 β φ n x u 1 + n x u N d u
2 N 1 D x α f , ( , x ] n α Γ α + 1
n x u n 1 β φ n x u d u + n x u n 1 β φ n x u n x u N d u ( 31 )
2 N 1 D x α f , ( , x ] n α Γ α + 1 q + 1 q λ
n x u n 1 β e 2 λ n x u 1 d u + n x u n 1 β e 2 λ n x u 1 n x u N d u =
2 N D x α f , ( , x ] n α Γ α + 1 q + 1 q λ n 1 β e 2 λ x 1 d x + n 1 β e 2 λ x 1 x N d x =
2 N D x α f , ( , x ] n α Γ α + 1 q + 1 q λ e 2 λ n 1 β e 2 λ x d x + n 1 β e 2 λ x x N d x =
2 N D x α f , ( , x ] n α Γ α + 1 q + 1 q λ e 2 λ 1 2 λ 2 λ n 1 β e y d y + 1 2 λ N + 1 2 λ n 1 β e y y N d y
2 N D x α f , ( , x ] n α Γ α + 1 q + 1 q λ e 2 λ 1 2 λ e 2 λ n 1 β + 2 N N ! 2 λ N + 1 2 λ n 1 β e y 2 d y =
2 N D x α f , ( , x ] n α Γ α + 1 q + 1 q λ e 2 λ 1 2 λ e 2 λ n 1 β + 2 N + 1 N ! 2 λ N + 1 e λ n 1 β
2 N D x α f , ( , x ] n α Γ α + 1 q + 1 q λ e 2 λ 1 2 λ + 2 N + 1 N ! 2 λ N + 1 e λ n 1 β =
2 N D x α f , ( , x ] n α Γ α + 1 q + 1 q λ e 2 λ 1 2 λ + N ! λ N + 1 e λ n 1 β =
2 N D x α f , ( , x ] n α Γ α + 1 q + 1 q e 2 λ 1 2 + N ! λ N e λ n 1 β 0 , as n .
So, it holds that
R n 1 * x | x u n 1 n β
2 N D x α f , ( , x ] n α Γ α + 1 q + 1 q e 2 λ 1 2 + N ! λ N e λ n 1 β 0 , as n .
Similarly, we obtain
R n 2 * x | x u n 1 n β
2 N D * x α f , [ x , ) n α Γ α + 1 q + 1 q e 2 λ 1 2 + N ! λ N e λ n 1 β 0 , as n .
At the end, we see that
R n * x 1 n + 1 n β α Γ α + 1
ω 1 D x α f , 1 n + 1 n β ( , x ] + ω 1 D * x α f , 1 n + 1 n β [ x , ) +
D x α f , ( , x ] + D * x α f , [ x , )
2 N n α Γ α + 1 q + 1 q e 2 λ 1 2 + N ! λ N e λ n 1 β 0 , as n .
The theorem is proven. □
We also present the following.
Theorem 15. 
Let α > 0 , N = α , α N , f A C N R , f N L R , 0 < β < 1 , x R , n N : n 1 β max 3 , 1 2 λ . Assume also that both sup x R D * x α f , sup x R D x α f < .
Then,
(i)
A n ¯ f x f x j = 1 N 1 f j x j ! A n ¯ · x j x
1 n + 1 n β α Γ α + 1 ω 1 D x α f , 1 n + 1 n β ( , x ] + ω 1 D * x α f , 1 n + 1 n β [ x , ) +
D x α f , ( , x ] + D * x α f , [ x , ) 2 N n α Γ α + 1 q + 1 q e 2 λ 1 2 + N ! λ N e λ n 1 β
= : Φ 1 ¯ x 0 , as n ,
(ii) Given that f j x = 0 , j = 1 , , N 1 , we have
A n ¯ f x f x Φ 1 ¯ x ,
(iii)
A n ¯ f x f x j = 1 N 1 f j x j !
2 j 1 n j 1 + tanh λ j + 1 + q + 1 q e 2 λ j ! 2 λ j + Φ ¯ 1 x ,
(iv)
A n ¯ f f j = 1 N 1 f j j ! 2 j 1 n j 1 + tanh λ j + 1 + q + 1 q e 2 λ j ! 2 λ j +
1 n + 1 n β α Γ α + 1 sup x R ω 1 D x α f , 1 n + 1 n β ( , x ] + sup x R ω 1 D * x α f , 1 n + 1 n β [ x , ) +
sup x R D x α f , ( , x ] + sup x R D * x α f , [ x , ) 2 N n α Γ α + 1 q + 1 q e 2 λ 1 2 + N ! λ N e λ n 1 β .
Above, when N = 1 , the sum j = 1 N 1 · = 0 .
So, we obtain the pointwise and uniform convergence with rates of A n ¯ I , as n .
Proof. 
Let x R . We have that D x α f x = D * x α f x = 0 .
We can write
f u n + i n r = j = 0 N 1 f j x j ! u n + i n r x j +
1 Γ α x u n + i n r u n + i n r s α 1 D * x α f s D * x α f x d s ,
for all x u n + i n r < , i = 1 , , r .
Also, it holds that
f u n + i n r = j = 0 N 1 f j x j ! u n + i n r x j +
1 Γ α u n + i n r x s u n + i n r α 1 D x α f s D x α f x d s ,
for all < u n + i n r x , i = 1 , , r .
Hence,
i = 1 r w i f u n + i n r = j = 0 N 1 f j x j ! i = 1 r w i u n + i n r x j +
1 Γ α i = 1 r w i x u n + i n r u n + i n r s α 1 D * x α f s D * x α f x d s ,
and
i = 1 r w i f u n + i n r = j = 0 N 1 f j x j ! i = 1 r w i u n + i n r x j +
1 Γ α i = 1 r w i u n + i n r x s u n + i n r α 1 D x α f s D x α f x d s .
Furthermore, it holds that
n x i = 1 r w i f u n + i n r φ n x u d u =
j = 0 N 1 f j x j ! n x i = 1 r w i u n + i n r x j φ n x u d u +
1 Γ α n x φ n x u
i = 1 r w i x u n + i n r u n + i n r s α 1 D * x α f s D * x α f x d s d u ,
and
n x i = 1 r w i f u n + i n r φ n x u d u =
j = 0 N 1 f j x j ! n x i = 1 r w i u n + i n r x j φ n x u d u +
1 Γ α n x φ n x u
i = 1 r w i u n + i n r x s u n + i n r α 1 D x α f s D x α f x d s d u .
Adding the last two equations, (233) and (234), we obtain
A n ¯ f x f x j = 1 N 1 f j x j ! A n ¯ · x j x = R n ¯ x ,
where
R n ¯ x = R n 1 ¯ x + R n 2 ¯ x ,
with
R n 1 ¯ x = 1 Γ α n x φ n x u
i = 1 r w i u n + i n r x s u n + i n r α 1 D x α f s D x α f x d s d u ,
and
R n 2 ¯ x = 1 Γ α n x φ n x u
i = 1 r w i x u n + i n r u n + i n r s α 1 D * x α f s D * x α f x d s d u .
Let u n x < 1 n β .
Call
ε n 1 x : = 1 Γ α i = 1 r w i u n + i n r x s u n + i n r α 1 D x α f s D x α f x d s ,
and
ε n 2 x : = 1 Γ α i = 1 r w i x u n + i n r u n + i n r s α 1 D * x α f s D * x α f x d s .
Then,
ε n 1 x 1 Γ α i = 1 r w i ω 1 D x α f , x u n + i n r ( , x ] x u n i n r α α
1 Γ α + 1 ω 1 D x α f , 1 n β + 1 n ( , x ] 1 n β + 1 n α .
Therefore, it holds that
ε n 1 x ω 1 D x α f , 1 n + 1 n β ( , x ] Γ α + 1 1 n + 1 n β α .
Furthermore, wee see that
ε n 2 x 1 Γ α i = 1 r w i ω 1 D * x α f , x u n + i n r [ x , ) u n + i n r x α α
1 Γ α + 1 ω 1 D * x α f , 1 n β + 1 n [ x , ) 1 n β + 1 n α .
That is,
ε n 2 x ω 1 D * x α f , 1 n + 1 n β [ x , ) Γ α + 1 1 n + 1 n β α .
Consequently, we see that
| R n 1 ¯ x | | u n x < 1 n β ω 1 D x α f , 1 n + 1 n β ( , x ] 1 n + 1 n β α Γ α + 1 ,
and
| R n 2 ¯ x | | u n x < 1 n β ω 1 D * x α f , 1 n + 1 n β [ x , + ) 1 n + 1 n β α Γ α + 1 .
Furthermore, it holds that
ε n 1 x D x α f , ( , x ] Γ α + 1 x u n + 1 n α ,
and
ε n 2 x D * x α f , [ x , ) Γ α + 1 x u n + 1 n α .
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 N be fixed. Assume that f C i R , with f j C B R , for j = 0 , 1 , , i . We derive from Remark 2 that
A n f j x = A n f j x , A n * f j x = A n * f j x , A n ¯ f j x = A n ¯ f j x , x R ,
for all j = 1 , , i .
We give the following.
Theorem 16. 
Let 0 < α < 1 , n N : n 1 α > 2 , x R , f j C B R , for j = 0 , 1 , , i N . Then,
(I)
A n f j x f j x ω 1 f j , 1 n α + 2 q + 1 q f j e 2 λ n 1 α 1 = : λ j s ,
and
A n f j f j λ j s ,
for all j = 0 , 1 , , i N ;
(II)
A n * f j x f j x ω 1 f j , 1 n + 1 n α + 2 q + 1 q f j e 2 λ n 1 α 1 = : ρ j s ,
and
A n * f j f j ρ j s ,
for j = 0 , 1 , , i ;
(III)
A n ¯ f j x f j x ω 1 f j , 1 n + 1 n α + 2 q + 1 q f j e 2 λ n 1 α 1 = ρ j s ,
and
A n ¯ f j f j ρ j s ,
for j = 0 , 1 , , i .
Proof. 
Based on Theorems 4–6. □
We continue with simultaneous global smoothness preservation.
Theorem 17. 
Here, f j C B R C u R , for j = 0 , 1 , , i N . Then,
(I)
ω 1 A n f j , δ ω 1 f j , δ , δ > 0 ;
and if f j C u R , then A n f j C u R ;
(II)
ω 1 A n * f j , δ ω 1 f j , δ , δ > 0 ;
and if f j C u R , then A n * f j C u R ;
(III)
ω 1 A n ¯ f j , δ ω 1 f j , δ , δ > 0 ;
and if f j C u R , then A n ¯ f j C u R .
Proof. 
By Theorems 7–9. □
The related simultaneous differentiation results follow.
Theorem 18. 
Let 0 < α < 1 , n N : n 1 α > 2 ; x R , j = 0 , 1 , , i N ; and f j C N R , N N , with f N + j C B R . Then,
(I)
A n f j x f j x λ = 1 N f j + λ x λ ! A n · x λ x
ω 1 f j + N , 1 n α n α N N ! + 4 f j + N n N q + 1 q e 2 λ λ N e λ n 1 α ,
(II)
A n * f j x f j x λ = 1 N f j + λ x λ ! A n * · x λ x
ω 1 f j + N , 1 n + 1 n α 1 n + 1 n α N N ! +
2 N + 1 f j + N n N N ! q + 1 q e 2 λ 1 2 + N ! λ N e λ n 1 α ,
(III)
A n ¯ f j x f j x λ = 1 N f j + λ x λ ! A n ¯ · x λ x
ω 1 f j + N , 1 n + 1 n α 1 n + 1 n α N N ! +
2 N + 1 f j + N n N N ! q + 1 q e 2 λ 1 2 + N ! λ N e λ n 1 α .
Proof. 
By Theorems 10–12. □
We continue with simultaneous fractional results.
Theorem 19. 
Let α > 0 , N = α , α N , j = 0 , 1 , 2 , , i N ; f j A C N R , f j + N L R , 0 < β < 1 , x R , n N : n 1 β max 3 , 1 2 λ . Assume also that both sup x R D * x α f j , sup x R D x α f j < .
Then,
(I)
A n f j x f j x λ = 1 N 1 f j + λ x λ ! A n · x λ x
1 n α β Γ α + 1 ω 1 D x α f j , 1 n β ( , x ] + ω 1 D * x α f j , 1 n β [ x , ) +
D x α f j + D * x α f j 2 N + 1 N ! q + 1 q e 2 λ Γ α + 1 2 λ α n α e λ n 1 β ,
(II)
A n * f j x f j x λ = 1 N 1 f j + λ x λ ! A n * · x λ x
1 n + 1 n β α Γ α + 1 ω 1 D x α f j , 1 n + 1 n β ( , x ] + ω 1 D * x α f j , 1 n + 1 n β [ x , ) +
D x α f j , ( , x ] + D * x α f j , [ x , ) 2 N n α Γ α + 1 q + 1 q e 2 λ 1 2 + N ! λ N e λ n 1 β ,
(III)
A n ¯ f j x f j x λ = 1 N 1 f j + λ x λ ! A n ¯ · x λ x
1 n + 1 n β α Γ α + 1 ω 1 D x α f j , 1 n + 1 n β ( , x ] + ω 1 D * x α f j , 1 n + 1 n β [ x , ) +
D x α f j , ( , x ] + D * x α f j , [ x , ) 2 N n α Γ α + 1 q + 1 q e 2 λ 1 2 + N ! λ N e λ n 1 β .
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 < α < 1 , n N : n 1 α > 2 , r N , μ > 0 , f C B R . Then,
(I)
A n r f f r A n f f r ω 1 f , 1 n α + 2 q + 1 q f e 2 λ n 1 α 1 ,
(II)
A n * r f f r A n * f f r ω 1 f , 1 n + 1 n α + 2 q + 1 q f e 2 λ n 1 α 1 ,
(III)
A n ¯ r f f r A n ¯ f f r ω 1 f , 1 n + 1 n α + 2 q + 1 q f e 2 λ n 1 α 1 .
So, the speed of convergence of A n r , A n * r , A n ¯ r to unit I is not worse than the speed of convergence of A n , A n * , A n ¯ to I.
Proof. 
By Theorems 4–6 and (75). □
We continue with the following.
Theorem 21. 
Let 0 < α < 1 ; m 1 , m 2 , , m r N : m 1 m 2 m r , with m i 1 α > 2 , i = 1 , , r ; μ > 0 ; f C B R . Then,
(I)
A m r A m r 1 A m 2 A m 1 f f i = 1 r A m i f f
i = 1 r ω 1 f , 1 m i α + 2 q + 1 q f e 2 λ m i 1 α 1 r ω 1 f , 1 m 1 α + 2 q + 1 q f e 2 λ m 1 1 α 1 ,
(II)
A m r * A m r 1 * A m 2 * A m 1 * f f i = 1 r A m i * f f
i = 1 r ω 1 f , 1 m i + 1 m i α + 2 q + 1 q f e 2 λ m i 1 α 1
r ω 1 f , 1 m 1 + 1 m 1 α + 2 q + 1 q f e 2 λ m 1 1 α 1 ,
(III)
A m r ¯ A m r 1 ¯ A m 2 ¯ A m 1 ¯ f f i = 1 r A m i ¯ f f
i = 1 r ω 1 f , 1 m i + 1 m i α + 2 q + 1 q f e 2 λ m i 1 α 1
r ω 1 f , 1 m 1 + 1 m 1 α + 2 q + 1 q f e 2 λ m 1 1 α 1 .
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 * , 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 N be fixed. Assume that f C i R , with f j C B R , for j = 0 , 1 , , i ; r N . Then, by (75), we obtain
A n r f j f j r A n f j f j .
By (249) and inductively, we obtain
A n r f j f j r A n f j f j ,
for j = 0 , 1 , , i .
Similarly, we derive that
A n * r f j f j r A n * f j f j ,
and
A n ¯ r f j f j r A n ¯ f j f j ,
for j = 0 , 1 , , i .
Now, let m 1 , m 2 , , m r N : m 1 m 2 m r . Then, based on (77), we find that
A m r A m r 1 A m 2 A m 1 f j f j i * = 1 r A m i * f j f j ,
for j = 0 , 1 , , i .
Similarly, we see that
A m r * A m r 1 * A m 2 * A m 1 * f j f j i * = 1 r A m i * * f j f j ,
for j = 0 , 1 , , i , and
A m r ¯ A m r 1 ¯ A m 2 ¯ A m 1 ¯ f j f j i * = 1 r A m i * ¯ f j f j ,
for j = 0 , 1 , , 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.

Funding

This research received no external funding.

Data Availability Statement

No new data were created or analyzed in this study.

Conflicts of Interest

The authors declare no conflicts of interest.

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Anastassiou, G.A. Approximation by Symmetrized and Perturbed Hyperbolic Tangent Activated Convolution-Type Operators. Mathematics 2024, 12, 3302. https://doi.org/10.3390/math12203302

AMA Style

Anastassiou GA. Approximation by Symmetrized and Perturbed Hyperbolic Tangent Activated Convolution-Type Operators. Mathematics. 2024; 12(20):3302. https://doi.org/10.3390/math12203302

Chicago/Turabian Style

Anastassiou, George A. 2024. "Approximation by Symmetrized and Perturbed Hyperbolic Tangent Activated Convolution-Type Operators" Mathematics 12, no. 20: 3302. https://doi.org/10.3390/math12203302

APA Style

Anastassiou, G. A. (2024). Approximation by Symmetrized and Perturbed Hyperbolic Tangent Activated Convolution-Type Operators. Mathematics, 12(20), 3302. https://doi.org/10.3390/math12203302

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