#### Conformable fractional approximations by max-product operators using convexity

Conformable fractional approximations by max-product operators using convexity
George A. Anastassiou 0
Mathematics Subject Classification 0
0 G. A. Anastassiou (
Here, we consider the approximation of functions by a large variety of max-product operators under conformable fractional differentiability and using convexity. These are positive sublinear operators. Our study relies on our general results about positive sublinear operators. We derive Jackson-type inequalities under conformable fractional initial conditions and convexity. So our approach is quantitative by obtaining inequalities where their right hand sides involve the modulus of continuity of a high-order conformable fractional derivative of the function under approximation. Due to the convexity assumptions, our inequalities are compact and elegant with small constants. In this article, we study under convexity quantitatively the conformable fractional approximation properties of max-product operators to the unit. These are special cases of positive sublinear operators. We first present results regarding the convergence to the unit of general positive sublinear operators under convexity. The focus of our study is approximation under the presence of conformable fractional smoothness. Under our convexity conditions, the derived conformable fractional convergence inequalities are elegant and compact with very small constants.
1 Background
Our work is inspired by [5].
We make
Remark 1.1 Let x , y ∈ [a, b] ⊆ [0, ∞), and g (x ) = x α, 0 < α ≤ 1. Then g (x ) = αx α−1 = x1α−α , for
x ∈ (0, ∞) . Since a ≤ x ≤ b, x1 ≥ b1 > 0 and x1α−α ≥ b1α−α > 0.
Assume y > x . By the mean value theorem, we get
A similar to (1) equality when x > y is true.
Then we obtain
Thus, it holds that Hence, we get
yα − x α
α
= ξ 1−α (y − x ) ,
where ξ ∈ (x , y).
α
b1−α |y − x | ≤ y
α − x α
α
= ξ 1−α |y − x |.
α
b1−α |y − x | ≤ |y
α − x α|.
|y − x | ≤
b1−α
α |yα − x α|,
∀ x , y ∈ [a, b] ⊂ [0, ∞), α ∈ (0, 1].
We also make
Remark 1.2 For 0 < α ≤ 1, x , y, t, s ≥ 0, we have
Assume that t > s. Then
and hence t α − sα ≤ (t − s)α .
Similarly, s > t ⇒ sα − t α ≤ (s − t )α.
Therefore it holds that
2α−1 x α + yα
≤ (x + y)α ≤ x α + yα.
t = t − s + s ⇒ t α = (t − s + s)α ≤ (t − s)α + sα,
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
If f is differentiable, then where f is the usual derivative. We define
t α − sα ≤ |t − s|α ,
∀ t, s ∈ [0, ∞).
Dα f (t ) := εl→im0
Dα f (0) = t →li m0+Dα f (t ).
ε
f t + εt 1−α
− f (t )
Definition 1.3 [
6,7
] Let f : [0, ∞) → R. The conformable α-fractional derivative for α ∈ (0, 1] is given by
If f : [0, ∞) → R is α-differentiable at t0 > 0, α ∈ (0, 1], then f is continuous at t0; see [7].
We will use
The case n = 0 follows.
Theorem 1.4 (See [
4
]) (Taylor formula) Let α ∈ (0, 1] and n ∈ N. Suppose f is (n + 1) times conformable
α-fractional differentiable on [0, ∞), and s, t ∈ [0, ∞), and Dn+1 f is assumed to be continuous on [0, ∞).
α
Then we have
Note 1.6 Theorem 1.4 and Corollary 1.5 are also true for f : [a, b] → R, [a, b] ⊆ [0, ∞), s, t ∈ [a, b].
Definition 1.7 Let f ∈ C([a, b]). We define the first modulus of continuity of f as:
(11)
(12)
(13)
(14)
(15)
(16)
(17)
(by ddττα = ατ α−1 ⇒ dτ α = ατ α−1dτ ⇒ α1 dτ α = τ α−1dτ )
=
Dn+1 f (s)
α
αn+1n!
s
t
tα − τ α n dτ α
(by t ≤ τ ≤ s ⇒ tα ≤ τ α(=: z) ≤ sα)
Therefore it holds that
tα
Dn+1 f (s)
α
αn+1n!
=
sα
Dn+1 f (s) tα − sα n+1
α
= (n + 1)! α
.
tα − z n dz =
Dn+1 f (s) (tα − sα)n+1
α
αn+1n! n + 1
1
n! s
n! s
∀ t ∈ [a, b] .
Using (4), we obtain
∀ t ∈ [a, b] .
Next, we estimate (20).
1. We observe that (t ≥ x0)
By the assumption Dαk f (x0) = 0, k = 1, . . . , n + 1, we can write
1 t tα − τ α n
f (t) − f (x0) = n! x0 α
Dαn+1 f (τ ) − Dαn+1 f (x0) τ α−1 dτ.
By assumption here Dn+1 f is convex over [a, b] ⊆ [0, ∞); x0 ∈ (a, b).
α
Let h : 0 < h ≤ min (x0 − a, b − x0), by Lemma 8.1.1, p. 243 of [
1
] we get that
Dαn+1 f (t) − Dαn+1 f (x0) ≤
ω1 Dαn+1 f, h
h
|t − x0| ,
Dαn+1 f (t) − Dαn+1 f (x0) ≤
ω1 Dαn+1 f, h b1−α
h
α
tα − x0α ,
(20) 1 t tα − τ α n
| f (t) − f (x0)| ≤ n! x0 α
(22)
Dαn+1 f (τ ) − Dαn+1 f (x0) τ α−1 dτ ≤
ω1 Dαn+1 f, h b1−α t
n!hαn+2
ω1 Dαn+1 f, h b1−α
n!hαn+2
n!hαn+2
n!hαn+2
ω1 Dαn+1 f, h b1−α
ω1 Dαn+1 f, h b1−α
ω1 Dαn+1 f, h b1−α
(n + 2)!hαn+2
x0
tα
xα
0
tα − τ α n τ α − x0α dτ α =
tα − z (n+1)−1 z − x0α 2−1 dz =
n!
(n + 2)!
tα − x0α n+2 .
(n + 1) (2) tα − x0α n+2
(n + 3)
=
tα − x0α n+2 =
We have proved that (case of t ≥ x0)
ω1 Dαn+1 f, h b1−α
(n + 2)!hαn+2
tα − x0α n+2 .
(18)
(19)
(20)
(21)
(22)
(23)
(24)
(25)
2. We observe that (t ≤ x0)
Dn+1 f (τ ) − Dαn+1 f (x0) τ α−1 dτ =
α
(20) 1
| f (t) − f (x0)| = n
x0α − τ α τ α − tα n dτ α =
n!
(n + 2)!
x0α − tα n+2 .
(2) (n + 1)
(n + 3)
x0α − tα n+2 =
x0α − tα n+2 =
We have proved (t ≤ x0) that
In conclusion, we have established that
tα − x0α n+2 , ∀t ∈ [a, b] .
By (6), we have
Thus by (29) and (30), the claim is proved.
tα − x0α ≤ |t − x0|α .
Theorem 2.2 Let α ∈ (0, 1] and n ∈ N. Suppose f is n times conformable α-fractional differentiable on
[a, b] ⊆ [0, ∞), and x ∈ (a, b), and Dαn f is continuous on [a, b]. Let 0 < h ≤ min (x − a, b − x) and
assume Dαn f is convex over [a, b]. Furthermore, assume that Dαk f (x) = 0, k = 1, . . . , n. Then, over [a, b],
we have
| f (·) − f (x)| ≤
(iii)
L N ( f ) ≤ L N (g) , ∀N ∈ N,
L N ( f + g) ≤ L N ( f ) + L N (g) , ∀ f, g ∈ C+ ([a, b]) .
We call {L N }N∈N positive sublinear operators.
We need a Hölder’s type inequality; see next:
Theorem 2.4 (See [
2
]) Let L : C+ ([a, b]) → C+([a, b]), be a positive sublinear operator and f, g ∈
1 1
C+([a, b]), furthermore let p, q > 1 : p + q = 1. Assume that L ( f (·))p (s∗), L (g (·))q (s∗) > 0 for
some s∗ ∈ [a, b]. Then
1 1
L ( f (·) g (·)) (s∗) ≤ L ( f (·))p (s∗) p L (g (·))q (s∗) q .
Remark 2.5 By [5, p. 17], we get: let f, g ∈ C+ ([a, b]). Then
Furthermore, we also have that
|L N ( f ) (x) − L N (g) (x)| ≤ L N (| f − g|) (x) , ∀ x ∈ [a, b] ⊆ [0, ∞).
|L N ( f ) (x) − f (x)| ≤ L N (| f (·) − f (x)|) (x) + | f (x)| |L N (e0) (x) − 1| ,
(33)
(34)
(35)
(36)
(37)
(38)
(40)
(41)
L N |· − x|(n+1)α (x) ≤
α
L N |· − x|(n+1)(α+1) (x) α+1 ,
∀ x ∈ [a, b] ⊆ [0, ∞); e0 (t) = 1.
From now on, we assume that L N (1) = 1. Hence, it holds that
|L N ( f ) (x) − f (x)| ≤ L N (| f (·) − f (x)|) (x) , ∀ x ∈ [a, b] ⊆ [0, ∞).
eTnhtieaobrleemon2[.6a,Lbe]t⊆α [∈0,(∞0, )1,]aannddxn∈∈(Na,. bS)u,papnodseDαnf f∈isCcontinuous on [a, b]. Let 0 < h ≤ min (x − a, b − x)
+ ([a, b]) is n times conformable α-fractional
differand assume Dαn f is convex over [a, b]. Furthermore, assume that Dαk f (x) = 0, k = 1, . . . , n. Let {L N }N∈N
from C+ ([a, b]) into itself, positive sublinear operators such that: L N (1) = 1, ∀ N ∈ N. Then,
|L N ( f ) (x) − f (x)| ≤
Theorem 2.8 Let {L N }N ∈N from C+ ([a, b]) into itself, positive sublinear operators, such that: L N (1) = 1, ∀
N ∈ N. Additionally, assume that L N |· − x |(n+1)(α+1) (x ) > 0, ∀ N ∈ N; x ∈ (a, b) . Here, α ∈ (0, 1] and
n ∈ N. Suppose f ∈ C+ ([a, b]) is n times conformable α-fractional differentiable on [a, b] ⊆ [0, ∞), and
α
Dαn f is continuous on [a, b]. Assume here that 0 < L N |· − x |(n+1)(α+1) (x ) α+1 ≤ min (x − a, b − x ) ,
∀ N ∈ N : N ≥ N ∗ ∈ N, and assume Dαn f is convex over [a, b]. Furthermore assume that Dαk f (x ) = 0,
k = 1, . . . , n. Then,
|L N ( f ) (x ) − f (x )| ≤
α
b1−αω1 Dαn f, L N |· − x |(n+1)(α+1) (x ) α+1
Theorem 2.9 Let {L N }N ∈N from C+ ([a, b]) into itself, positive sublinear operators: L N (1) = 1, ∀ N ∈ N.
Also L N |· − x |(n+1)α (x ) > 0, ∀ N ∈ N. Here α ∈ (0, 1], n ∈ N and x ∈ (a, b); [a, b] ⊆ [0, ∞). Suppose
f ∈ C+ ([a, b]) is n times conformable α-fractional differentiable on [a, b], and Dαn f is continuous on [a, b].
Let 0 < L N |· − x |(n+1)α (x ) ≤ min (x − a, b − x ) , ∀ N ≥ N ∗; N , N ∗ ∈ N, and assume Dαn f is convex
over [a, b]. Furthermore, assume that Dαk f (x ) = 0, k = 1, . . . , n. Then,
|L N ( f ) (x ) − f (x )| ≤
b1−αω1 Dαn f, L N |· − x |(n+1)α (x )
,
x k (1 − x )N −k and f : [
0, 1
] → R+ = [0, ∞) is
∀ N ≥ N ∗, where N , N ∗ ∈ N.
If L N |· − x |(n+1)α (x ) → 0, then L N ( f ) (x ) → f (x ), as N → + ∞.
Proof By (39) and choosing h := L N |· − x |(n+1)α (x ) .
3 Applications
where ∨ stands for maximum, and pN,k (x ) =
continuous.
These are nonlinear and piecewise rational operators.
We have B(M) (1) = 1, and
N
N
k
6
N + 1
B(M) (|· − x |) (x ) ≤ √
N
, ∀ x ∈ [
0, 1
] , ∀ N ∈ N;
see [5, p. 31].
B(M) are positive sublinear operators and thus they possess the monotonicity property, also since |· − x | ≤
N
1, then |· − x |β ≤ 1, ∀ x ∈ [
0, 1
], ∀ β > 0.
(43)
(44)
(45)
Therefore it holds that
Furthermore, clearly it holds that
The operator BN(M) maps C+ ([
0, 1
]) into itself.
,
∀ N ≥ N ∗, N ∈ N.
It holds that limN→+ ∞ BN(M) ( f ) (x) = f (x) .
Proof By (47), we get that BN(M) |· − x|(n+1)(α+1) (x) > 0, ∀ N ∈ N.
By (40), (46), we obtain that
B(M) ( f ) (x) − f (x) ≤
N
Remark 3.3 The truncated Favard–Szász–Mirakjan operators are given by
T (M) ( f ) (x) =
N
N
k=0 sN,k (x) f Nk , x ∈ [
0, 1
] , N ∈ N, f ∈ C+ ([
0, 1
]) ,
N
k=0 sN,k (x)
sN,k (x) = (Nkx)k ; see also [5, p. 11].
By [5, p. 1!78–179], we get that
Clearly, it holds that
T (M) (|· − x|) (x) ≤ √
N
, ∀ x ∈ [
0, 1
] , ∀ N ∈ N.
TN(M) |· − x|1+β (x) ≤ √
, ∀ x ∈ [
0, 1
] , ∀ N ∈ N, ∀ β > 0.
(46)
(47)
(48)
(49)
(50)
(51)
(52)
TN(M) |· − x |λ (x ) =
,
Theorem 3.4 Let α ∈ (0, 1] and n ∈ N. Suppose f ∈ C+ ([
0, 1
]) is n times conformable α-fractional
differentiable on [
0, 1
], x ∈ (0, 1), and Dαn f is continuous on [
0, 1
]. Let N ∗ ∈ N such that (N ∗) 12(αα+1) ≤
min (x , 1 − x ) and assume Dαn f is convex over [
0, 1
]. Furthermore, assume that Dαk f (x ) = 0, k = 1, . . . , n.
Then,
T (M) ( f ) (x ) − f (x ) ≤
N
,
∀ N ≥ N ∗, N ∈ N.
T (M) ( f ) (x ) = f (x ) .
It holds that limN →+∞ N
Proof By (53), we have TN(M) |· − x |(n+1)(α+1) (x ) > 0, ∀ N ∈ N.
By (40), (52), we get that
T (M) ( f ) (x ) − f (x ) ≤
N
=
where
U (M) ( f ) (x ) =
N
N
k=0 bN,k (x ) f
N
k=0 bN,k (x )
From [5, pp. 217–218], we get (x ∈ [
0, 1
])
Let λ ≥ 1, clearly then it holds that
Remark 3.5 Next, we study the truncated max-product Baskakov operators (see [5, p. 11])
k
N , x ∈ [
0, 1
] , f ∈ C+ ([
0, 1
]) , N ∈ N,
U (M) (|· − x |) (x ) ≤
N
,
N ≥ 2, N ∈ N.
U N(M) |· − x |λ
(x ) ≤
, ∀ N ≥ 2, N ∈ N.
2√3 √
√
N + 1
2 + 2
2√3 √
√
N + 1
2 + 2
Also, it holds that U (M) (1) = 1, and U (M) are positive sublinear operators from C+ ([
0, 1
]) into itself.
N N
Furthermore, it holds that
U N(M) |· − x |λ (x ) > 0,
∀ x ∈ (0, 1], ∀ λ ≥ 1, ∀ N ∈ N.
Theorem 3.6 Let α ∈ (0, 1] and n ∈ N. Suppose f ∈ C+ ([
0, 1
]) is n times conformable α-fractional
differentiable on [
0, 1
], x ∈ (0, 1), and Dαn f is continuous on [
0, 1
]. Let N ∗ ∈ N−{1} such that (N ∗+11) 2(αα+1) ≤
min (x , 1 − x ) and assume Dαn f is convex over [
0, 1
]. Furthermore, assume that Dαk f (x ) = 0, k = 1, . . . , n.
Then,
U (M) ( f ) (x ) − f (x ) ≤ ⎜
N ⎝
⎛
,
∀ N ∈ N : N ≥ N ∗ ≥ 2.
It holds that N →lim+∞U N(M) ( f ) (x ) = f (x ) .
Proof By (60), we have that U N(M) |· − x |(n+1)(α+1) (x ) > 0, ∀ N ∈ N.
By (40) and (59), we get
U (M) ( f ) (x ) − f (x ) ≤
N
As before, we get that (for λ ≥ 1)
Z (NM) (|· − x |) (x ) ≤
8 1 + √5 √x (1 − x )
3 √N
, ∀ x ∈ [
0, 1
] , ∀ N ≥ 4, N ∈ N.
Z (NM) |· − x |λ (x ) ≤
8 1 + √5 √x (1 − x )
3 √N
,
∀ x ∈ [
0, 1
], N ≥ 4, N ∈ N.
Also, it holds that Z (NM) (1) = 1, and Z (NM) are positive sublinear operators from C+ ([
0, 1
]) into itself.
Also, it holds that
Z (NM) |· − x |λ (x ) > 0,
∀ x ∈ (0, 1), ∀ λ ≥ 1, ∀ N ∈ N.
(61)
(62)
(63)
(64)
(65)
(66)
Theorem 3.8 Let α ∈ (0, 1] and n ∈ N. Suppose f ∈ C+ ([
0, 1
]) is n times conformable α-fractional
differentiable on [
0, 1
], x ∈ (0, 1), and Dαn f is continuous on [
0, 1
]. Let N ∗ ∈ N, N ∗ ≥ 4, such that
(N∗) 12(αα+1) ≤ min (x, 1 − x) and assume Dαn f is convex over [
0, 1
]. Furthermore, assume that Dαk f (x) = 0,
Z (NM) ( f ) (x) − f (x) ≤ ⎛⎝ 8 1 + √5 √x (1 − x)⎠
3
⎜
⎜
⎝
Z (NM) ( f ) (x) − f (x) ≤
proving the claim.
Remark 3.9 Here, we deal with the max-product truncated sampling operators (see [5, p. 13]) defined by
(67)
(68)
(69)
(70)
,
W N(M) (1) = 1.
∀ x ∈ [0, π ], f : [0, π ] → R+ a continuous function.
Following [5, p. 343], and making the convention sin0(0) = 1 and denoting sN,k (x) = sinN(Nx −x−kπkπ) , we
= 1, and sN,k jNπ = 0, if k = j , and furthermore W N(M) ( f ) jNπ = f jNπ , for all
get that sN,k kπ
N
j ∈ {0, . . . , N } .
Clearly, W N(M) ( f ) is a well-defined function for all x ∈ [0, π ], and it is continuous on [0, π ], also
By [5, p. 344], W N(M) are positive sublinear operators.
Call IN+ (x) = k ∈ {0, 1, . . . , N } ; sN,k (x) > 0 , and set xN,k := kNπ , k ∈ {0, 1, . . . , N }.
We see that By [5, p. 346], we have
.
π
W N(M) (|· − x|) (x) ≤ 2N , ∀ N ∈ N, ∀ x ∈ [0, π ] .
such that x = xN,k , for any k ∈ {0, 1, . . . , N }.
[0, π ]. Furthermore, assume that Dαk f (x) = 0, k = 1, . . . , n. Then,
Theorem 3.10 Let α ∈ (0, 1] and n ∈ N. Suppose f ∈ kCNπ+, (k[0∈,π{]0),i1s,n. .t.im,Nes},co∀nfNorm∈abNle, aαn-fdraDctαniofn aisl
differentiable on [0, π ], and x ∈ (0, π ), such that x =
continuous on [0, π ]. Let N ∗ ∈ N such that (N∗)1α+α1 ≤ min (x, π − x) and assume Dαn f is convex over
,
proving the claim.
We make
W N(M) ( f ) (x) − f (x) ≤
ω1 Dαn f, h π1−α
(n + 1)!αn+1h
πnα+1ω1 Dαn f, h
,
=
(71)
(72)
(73)
(74)
(75)
(76)
Remark 3.11 Here, we continue with the max-product truncated sampling operators (see [5, p. 13]) defined
by
,
is a well-defined function for all x ∈ [0, π ], and it is continuous on [0, π ], also K N(M) (1) = 1. By [5, p. 350],
K N(M) are positive sublinear operators.
Denote xN,k := kNπ , k ∈ {0, 1, . . . , N }.
By [5, p. 352], we have
Notice also xN,k − x ≤ π , ∀ x ∈ [0, π ] .
Therefore, (λ ≥ 1) it holds that
π
K N(M) (|· − x|) (x) ≤ 2N , ∀ N ∈ N, ∀ x ∈ [0, π ] .
K N(M) |· − x|λ (x) ≤
such that x = xN,k , for any k ∈ {0, 1, . . . , N } .
[0, π ]. Furthermore, assume that Dαk f (x) = 0, k = 1, . . . , n. Then,
dTihffeeorernemtiab3l.1e2onLe[t0,απ ∈], a(0n,d1]x a∈nd(0n, π∈),Ns.uScuhppthoaste xf =∈ kCNπ+, (k[0∈,π{]0),i1s,n. .t.im,Nes},co∀nfNorm∈abNle, aαn-fdraDctαniofn aisl
continuous on [0, π ]. Let N ∗ ∈ N such that (N∗)1α+α1 ≤ min (x, π − x) and assume Dαn f is convex over
,
(II) Here, we apply Theorem 2.6 to the well-known max-product operators in the case of (n + 1) α ≥ 1, that
1
is, when n+1 ≤ α ≤ 1, where n ∈ N.
Theorem 3.13 Let α ∈ (0, 1] and n ∈ N. Suppose f ∈ C+ ([
0, 1
]) is n times conformable α-fractional
differentiable on [
0, 1
], x ∈ (0, 1), and Dαn f is continuous on [
0, 1
]. Let N ∗ ∈ N such that √N1∗+1 ≤
(setting h := √N1+1 )
proving the claim.
ω1 Dαn f, h
BN(M) ( f ) (x) − f (x) ≤ (n + 1)!αn+1h
BN(M) |· − x|(n+1)α (x) ≤
ω1 Dαn f, h 6
(n + 1)!αn+1h √N + 1 =
Theorem 3.14 Let α ∈ (0, 1] and n ∈ N. Suppose f ∈ C+ ([
0, 1
]) is n times conformable α-fractional
differ1
entiable on [
0, 1
], x ∈ (0, 1), and Dαn f is continuous on [
0, 1
]. Let N ∗ ∈ N such that √N∗ ≤ min (x, 1 − x)
and assume Dαn f is convex over [
0, 1
]. Furthermore, assume that Dαk f (x) = 0, k = 1, . . . , n. Then,
T (M) ( f ) (x) − f (x) ≤
N
3
(n + 1)!αn+1
Proof By (39), (52), we get that
(setting h := √1N )
proving the claim.
T (M) ( f ) (x) − f (x) ≤
N
(83)
(84)
(85)
(86)
Theorem 3.15 Let α ∈ (0, 1] and n ∈ N. Suppose f ∈ C+ ([
0, 1
]) is n times conformable α-fractional
differentiable on [
0, 1
], x ∈ (0, 1), and Dαn f is continuous on [
0, 1
]. Let N ∗ ∈ N − {1} such that √N1∗+1 ≤
(setting h := √N1+1 )
ω1 Dαn f, h ⎛ 2√3 √2 + 2 ⎞
UN(M) ( f ) (x) − f (x) ≤ (n + 1)!αn+1h ⎝
√N + 1 ⎠ =
2√3 √2 + 2
Theorem 3.17 Let α ∈ (0, 1] and n ∈ N. Suppose f ∈ kCNπ+, (k[0∈,π{]0),i1s,n. .t.im,Nes},co∀nfNorm∈abNle, aαn-fdraDctαniofn aisl
differentiable on [0, π ], and x ∈ (0, π ), such that x =
continuous on [0, π ]. Let N ∗ ∈ N such that N1∗ ≤ min (x, π − x) and assume Dαn f is convex over [0, π ].
Furthermore, assume that Dαk f (x) = 0, k = 1, . . . , n. Then,
W N(M) ( f ) (x) − f (x) ≤
∀ N ∈ N : N ≥ N ∗.
It holds that limN→+∞W N(M) ( f ) (x) = f (x) .
(91)
(setting h := √1N )
proving the claim.
proving the claim.
Theorem 3.16 Let α ∈ (0, 1] and n ∈ N. Suppose f ∈ C+ ([
0, 1
]) is n times conformable α-fractional
differentiable on [
0, 1
], x ∈ (0, 1), and Dαn f is continuous on [
0, 1
]. Let N ∗ ∈ N, N ∗ ≥ 4, such that
1
√N∗ ≤ min (x, 1 − x) and assume Dαn f is convex over [
0, 1
]. Furthermore assume that Dαk f (x) = 0,
Z (NM) ( f ) (x) − f (x) ≤ ⎛⎝ 8 1 +3 √5 √
⎞ ⎛ ω1 Dαn f, √1N ⎞
x (1 − x)⎠ ⎝ (n + 1)!αn+1 ⎠ ,
∀ N ≥ N ∗ ≥ 4, N ∈ N.
It holds that limN→+∞ Z (NM) ( f ) (x) = f (x) .
Proof By (39) and (65), we obtain
ω1 Dαn f, h ⎛ 8 1 + √5 √x (1 − x)⎠⎞ √1N =
Z(NM) ( f ) (x) − f (x) ≤ (n + 1)!αn+1h ⎝ 3
8 1 +3 √5 √x (1 − x) ⎛⎝ ω(1n +D1αn)f!,α√n1+N1 ⎞⎠ ,
Proof By (39) and (73), we obtain
(setting h := N1 )
proving the claim.
,
,
(92)
(93)
(94)
Furthermore, assume that Dαk f (x ) = 0, k = 1, . . . , n. Then,
Theorem 3.18 Let α ∈ (0, 1] and n ∈ N. Suppose f ∈ C ([0, π ]) is n times conformable α-fractional
differentiable on [0, π ], and x ∈ (0, π ), such that x = kNπ+, k ∈ {0, 1, . . . , N }, ∀ N ∈ N, and Dαn f is
continuous on [0, π ]. Let N ∗ ∈ N such that N1∗ ≤ min (x , π − x ) and assume Dαn f is convex over [0, π ].
(M)
K N
( f ) (x ) − f (x ) ≤
2 (n + 1)!αn+1h
(setting h := N1 )
proving the claim.
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