Conformable fractional approximations by max-product operators using convexity

Arabian Journal of Mathematics, Feb 2018

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.

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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. Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http:// creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. 1. Anastassiou , G.: Moments in Probability and Approximation Theory . 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George A. Anastassiou. Conformable fractional approximations by max-product operators using convexity, Arabian Journal of Mathematics, 2018, 1-16, DOI: 10.1007/s40065-018-0199-3