CUBO A Mathematical Journal Vol.21, No¯ 02, (01–13). August 2019 http: // dx. doi. org/ 10. 4067/ S0719-06462019000200001 Caputo fractional Iyengar type Inequalities George A. Anastassiou Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152, U.S.A. ganastss@memphis.edu ABSTRACT Here we present Caputo fractional Iyengar type inequalities with respect to Lp norms, with 1 ≤ p ≤ ∞. The method is based on the right and left Caputo fractional Taylor’s formulae. RESUMEN Aqúı presentamos desigualdades de tipo Caputo fraccional Iyengar con respecto a las normas Lp, con 1 ≤ p ≤ ∞. El método se basa en las fórmulas de Taylor fraccionales de Caputo derecha e izquierda. Keywords and Phrases: Iyengar inequality, right and left Caputo fractional, Taylor formulae, Caputo fractional derivative. 2010 AMS Mathematics Subject Classification: 26A33, 26D10, 26D15. http://dx.doi.org/10.4067/S0719-06462019000200001 2 George A. Anastassiou CUBO 21, 2 (2019) 1 Introduction We are motivated by the following famous Iyengar inequality (1938), [4]. Theorem 1. Let f be a differentiable function on [a, b] and |f′ (x)| ≤ M. Then ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − 1 2 (b − a) (f (a) + f (b)) ∣ ∣ ∣ ∣ ∣ ≤ M (b − a) 4 2 − (f (b) − f (a)) 2 4M . (1) We need Definition 2. ([1], p. 394) Let ν > 0, n = ⌈ν⌉ (⌈·⌉ the ceiling of the number), f ∈ ACn ([a, b]) (i.e. f(n−1) is absolutely continuous on [a, b]). The left Caputo fractional derivative of order ν is defined as Dν ∗af (x) = 1 Γ (n − ν) ∫x a (x − t) n−ν−1 f(n) (t) dt, (2) ∀ x ∈ [a, b], and it exists almost everywhere over [a, b] . We need Definition 3. ([2], p. 336-337) Let ν > 0, n = ⌈ν⌉, f ∈ ACn ([a, b]). The right Caputo fractional derivative of order ν is defined as Dνb−f (x) = (−1) n Γ (n − ν) ∫b x (z − x) n−ν−1 f(n) (z) dz, (3) ∀ x ∈ [a, b], and exists almost everywhere over [a, b] . 2 Main Results We present the following Caputo fractional Iyengar type inequalities: Theorem 4. Let ν > 0, n = ⌈ν⌉ (⌈·⌉ is the ceiling of the number), and f ∈ ACn ([a, b]) (i.e. f(n−1) is absolutely continuous on [a, b]). We assume that Dν ∗af, D ν b−f ∈ L∞ ([a, b]). Then i) ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − n−1∑ k=0 1 (k + 1) ! [ f(k) (a) (t − a) k+1 + (−1)kf(k) (b) (b − t) k+1 ] ∣ ∣ ∣ ∣ ∣ ≤ max { ‖Dν ∗af‖L∞([a,b]) , ∥ ∥Dνb−f ∥ ∥ L∞([a,b]) } Γ (ν + 2) [ (t − a) ν+1 + (b − t) ν+1 ] , (4) ∀ t ∈ [a, b] , CUBO 21, 2 (2019) Caputo fractional Iyengar type Inequalities 3 ii) at t = a+b 2 , the right hand side of (4) is minimized, and we get: ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − n−1∑ k=0 1 (k + 1) ! (b − a) k+1 2k+1 [ f(k) (a) + (−1)kf(k) (b) ] ∣ ∣ ∣ ∣ ∣ ≤ max { ‖Dν ∗af‖L∞([a,b]) , ∥ ∥Dνb−f ∥ ∥ L∞([a,b]) } Γ (ν + 2) (b − a) ν+1 2ν , (5) iii) if f(k) (a) = f(k) (b) = 0, for all k = 0, 1, ..., n − 1, we obtain ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx ∣ ∣ ∣ ∣ ∣ ≤ max { ‖Dν ∗af‖L∞([a,b]) , ∥ ∥Dνb−f ∥ ∥ L∞([a,b]) } Γ (ν + 2) (b − a) ν+1 2ν , (6) which is a sharp inequality, iv) more generally, for j = 0, 1, 2, ..., N ∈ N, it holds ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − n−1∑ k=0 1 (k + 1) ! ( b − a N )k+1 [ jk+1f(k) (a) + (−1)k (N − j) k+1 f(k) (b) ] ∣ ∣ ∣ ∣ ∣ ≤ max { ‖Dν ∗af‖L∞([a,b]) , ∥ ∥Dνb−f ∥ ∥ L∞([a,b]) } Γ (ν + 2) ( b − a N )ν+1 [ jν+1 + (N − j) ν+1 ] , (7) v) if f(k) (a) = f(k) (b) = 0, k = 1, ..., n − 1, from (7) we get: ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − ( b − a N ) [jf (a) + (N − j) f (b)] ∣ ∣ ∣ ∣ ∣ ≤ max { ‖Dν ∗af‖L∞([a,b]) , ∥ ∥Dνb−f ∥ ∥ L∞([a,b]) } Γ (ν + 2) ( b − a N )ν+1 [ jν+1 + (N − j) ν+1 ] , (8) j = 0, 1, 2, ..., N, vi) when N = 2 and j = 1, (8) turns to ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − ( b − a 2 ) (f (a) + f (b)) ∣ ∣ ∣ ∣ ∣ ≤ max { ‖Dν ∗af‖L∞([a,b]) , ∥ ∥Dνb−f ∥ ∥ L∞([a,b]) } Γ (ν + 2) (b − a) ν+1 2ν , (9) vii) when 0 < ν ≤ 1, inequality (9) is again valid without any boundary conditions. 4 George A. Anastassiou CUBO 21, 2 (2019) Proof. Let ν > 0, n = ⌈ν⌉, and f ∈ ACn ([a, b]). Then by ([3], p. 54) left Caputo fractional Taylor’s formula we have f (x) − n−1∑ k=0 f(k) (a) k! (x − a) k = 1 Γ (ν) ∫x a (x − t) ν−1 Dν ∗af (t) dt, (10) ∀ x ∈ [a, b] . Also by ([2], p. 341) right Caputo fractional Taylor’s formula we get: f (x) − n−1∑ k=0 f(k) (b) k! (x − b) k = 1 Γ (ν) ∫b x (z − x) ν−1 Dνb−f (z) dz, (11) ∀ x ∈ [a, b] . By (10) we derive ∣ ∣ ∣ ∣ ∣ f (x) − n−1∑ k=0 f(k) (a) k! (x − a) k ∣ ∣ ∣ ∣ ∣ ≤ ‖Dν ∗af‖L∞([a,b]) Γ (ν + 1) (x − a) ν , (12) and by (11) we obtain ∣ ∣ ∣ ∣ ∣ f (x) − n−1∑ k=0 f(k) (b) k! (x − b) k ∣ ∣ ∣ ∣ ∣ ≤ ∥ ∥Dνb−f ∥ ∥ L∞([a,b]) Γ (ν + 1) (b − x) ν , (13) ∀ x ∈ [a, b] . Call γ1 := ‖Dν ∗af‖L∞([a,b]) Γ (ν + 1) , (14) and γ2 := ∥ ∥Dνb−f ∥ ∥ L∞([a,b]) Γ (ν + 1) . (15) Set γ := max (γ1, γ2) . (16) That is ∣ ∣ ∣ ∣ ∣ f (x) − n−1∑ k=0 f(k) (a) k! (x − a) k ∣ ∣ ∣ ∣ ∣ ≤ γ (x − a) ν , (17) and ∣ ∣ ∣ ∣ ∣ f (x) − n−1∑ k=0 f(k) (b) k! (x − b) k ∣ ∣ ∣ ∣ ∣ ≤ γ (b − x) ν , (18) ∀ x ∈ [a, b] . CUBO 21, 2 (2019) Caputo fractional Iyengar type Inequalities 5 Hence it holds n−1∑ k=0 f(k) (a) k! (x − a) k − γ (x − a) ν ≤ f (x) ≤ n−1∑ k=0 f(k) (a) k! (x − a) k + γ (x − a) ν (19) and n−1∑ k=0 f(k) (b) k! (x − b) k − γ (b − x) ν ≤ f (x) ≤ n−1∑ k=0 f(k) (b) k! (x − b) k + γ (b − x) ν , (20) ∀ x ∈ [a, b] . Let any t ∈ [a, b], then by integration over [a, t] and [t, b], respectively, we obtain n−1∑ k=0 f(k) (a) (k + 1) ! (t − a) k+1 − γ (ν + 1) (t − a) ν+1 ≤ ∫t a f (x) dx ≤ n−1∑ k=0 f(k) (a) (k + 1) ! (t − a) k+1 + γ (ν + 1) (t − a) ν+1 , (21) and − n−1∑ k=0 f(k) (b) (k + 1) ! (t − b) k+1 − γ (ν + 1) (b − t) ν+1 ≤ ∫b t f (x) dx ≤ − n−1∑ k=0 f(k) (b) (k + 1) ! (t − b) k+1 + γ (ν + 1) (b − t) ν+1 . (22) Adding (21) and (22), we obtain { n−1∑ k=0 1 (k + 1) ! [ f(k) (a) (t − a) k+1 − f(k) (b) (t − b) k+1 ] } − γ (ν + 1) [ (t − a) ν+1 + (b − t) ν+1 ] ≤ ∫b a f (x) dx ≤ { n−1∑ k=0 1 (k + 1) ! [ f(k) (a) (t − a) k+1 − f(k) (b) (t − b) k+1 ] } + γ (ν + 1) [ (t − a) ν+1 + (b − t) ν+1 ] , (23) ∀ t ∈ [a, b] . Consequently we derive: ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − n−1∑ k=0 1 (k + 1) ! [ f(k) (a) (t − a) k+1 + (−1)kf(k) (b) (b − t) k+1 ] ∣ ∣ ∣ ∣ ∣ ≤ γ (ν + 1) [ (t − a) ν+1 + (b − t) ν+1 ] , (24) 6 George A. Anastassiou CUBO 21, 2 (2019) ∀ t ∈ [a, b] . Let us consider g (t) := (t − a) ν+1 + (b − t) ν+1 , ∀ t ∈ [a, b] . Hence g′ (t) = (ν + 1) [ (t − a) ν − (b − t) ν ] = 0, giving (t − a) ν = (b − t) ν and t − a = b − t, that is t = a+b 2 the only critical number here. We have g (a) = g (b) = (b − a) ν+1 , and g ( a+b 2 ) = (b−a) ν+1 2ν , which the minimum of g over [a, b]. Consequently the right hand side of (24) is minimized when t = a+b 2 , with value γ (ν+1) (b−a) ν+1 2ν . Assuming f(k) (a) = f(k) (b) = 0, for k = 0, 1, ..., n − 1, then we obtain that ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx ∣ ∣ ∣ ∣ ∣ ≤ γ (ν + 1) (b − a) ν+1 2ν , (25) which is a sharp inequality. When t = a+b 2 , then (24) becomes ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − n−1∑ k=0 1 (k + 1) ! (b − a) k+1 2k+1 [ f(k) (a) + (−1)kf(k) (b) ] ∣ ∣ ∣ ∣ ∣ ≤ γ (ν + 1) (b − a) ν+1 2ν . (26) Next let N ∈ N, j = 0, 1, 2, ..., N and tj = a + j ( b−a N ) , that is t0 = a, t1 = a + b−a N , ..., tN = b. Hence it holds tj − a = j ( b − a N ) , (b − tj) = (N − j) ( b − a N ) , j = 0, 1, 2, ..., N. (27) We notice that (tj − a) ν+1 + (b − tj) ν+1 = ( b − a N )ν+1 [ jν+1 + (N − j) ν+1 ] , (28) j = 0, 1, 2, ..., N, and (for k = 0, 1, ..., n − 1) [ f(k) (a) (tj − a) k+1 + (−1)kf(k) (b) (b − tj) k+1 ] = [ f(k) (a) jk+1 ( b − a N )k+1 + (−1)kf(k) (b) (N − j) k+1 ( b − a N )k+1 ] = CUBO 21, 2 (2019) Caputo fractional Iyengar type Inequalities 7 ( b − a N )k+1 [ f(k) (a) jk+1 + (−1)kf(k) (b) (N − j) k+1 ] , (29) j = 0, 1, 2, ..., N. By (24) we get ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − n−1∑ k=0 1 (k + 1) ! ( b − a N )k+1 [ f(k) (a) jk+1 + (−1)kf(k) (b) (N − j) k+1 ] ∣ ∣ ∣ ∣ ∣ ≤ γ (ν + 1) ( b − a N )ν+1 [ jν+1 + (N − j) ν+1 ] , (30) j = 0, 1, 2, ..., N. If f(k) (a) = f(k) (b) = 0, k = 1, ..., n − 1, then (30) becomes ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − ( b − a N ) [jf (a) + (N − j) f (b)] ∣ ∣ ∣ ∣ ∣ ≤ γ (ν + 1) ( b − a N )ν+1 [ jν+1 + (N − j) ν+1 ] , (31) j = 0, 1, 2, ..., N. When N = 2 and j = 1, then (31) becomes ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − ( b − a 2 ) (f (a) + f (b)) ∣ ∣ ∣ ∣ ∣ ≤ γ (ν + 1) 2 ( b − a 2 )ν+1 = γ (ν + 1) (b − a) ν+1 2ν . (32) Let 0 < ν ≤ 1, then n = ⌈ν⌉ = 1. In that case, without any boundary conditions, we derive from (32) again that ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − ( b − a 2 ) (f (a) + f (b)) ∣ ∣ ∣ ∣ ∣ ≤ γ (ν + 1) (b − a) ν+1 2ν . (33) The theorem is proved in all cases. We give Theorem 5. Let ν ≥ 1, n = ⌈ν⌉, and f ∈ ACn ([a, b]). We assume that Dν ∗af, D ν b−f ∈ L1 ([a, b]). Then i) ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − n−1∑ k=0 1 (k + 1) ! [ f(k) (a) (t − a) k+1 + (−1)kf(k) (b) (b − t) k+1 ] ∣ ∣ ∣ ∣ ∣ ≤ 8 George A. Anastassiou CUBO 21, 2 (2019) max { ‖Dν ∗af‖L1([a,b]) , ∥ ∥Dνb−f ∥ ∥ L1([a,b]) } Γ (ν + 1) [ (t − a) ν + (b − t) ν ] , (34) ∀ t ∈ [a, b] , ii) when ν = 1, from (34), we have ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − [f (a) (t − a) + f (b) (b − t)] ∣ ∣ ∣ ∣ ∣ ≤ ‖f′‖ L1([a,b]) (b − a) , ∀ t ∈ [a, b] , (35) iii) from (35), we obtain (ν = 1 case) ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − ( b − a 2 ) (f (a) + f (b)) ∣ ∣ ∣ ∣ ∣ ≤ ‖f′‖ L1([a,b]) (b − a) , (36) iv) at t = a+b 2 , ν > 1, the right hand side of (34) is minimized, and we get: ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − n−1∑ k=0 1 (k + 1) ! (b − a) k+1 2k+1 [ f(k) (a) + (−1)kf(k) (b) ] ∣ ∣ ∣ ∣ ∣ ≤ max { ‖Dν ∗af‖L1([a,b]) , ∥ ∥Dνb−f ∥ ∥ L1([a,b]) } Γ (ν + 1) (b − a) ν 2ν−1 , (37) v) if f(k) (a) = f(k) (b) = 0, for all k = 0, 1, ..., n − 1; ν > 1, from (37), we obtain ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx ∣ ∣ ∣ ∣ ∣ ≤ max { ‖Dν ∗af‖L1([a,b]) , ∥ ∥Dνb−f ∥ ∥ L1([a,b]) } Γ (ν + 1) (b − a) ν 2ν−1 , (38) which is a sharp inequality, vi) more generally, for j = 0, 1, 2, ..., N ∈ N, it holds ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − n−1∑ k=0 1 (k + 1) ! ( b − a N )k+1 [ jk+1f(k) (a) + (−1)k (N − j) k+1 f(k) (b) ] ∣ ∣ ∣ ∣ ∣ ≤ max { ‖Dν ∗af‖L1([a,b]) , ∥ ∥Dνb−f ∥ ∥ L1([a,b]) } Γ (ν + 1) ( b − a N )ν [ jν + (N − j) ν ] , (39) vii) if f(k) (a) = f(k) (b) = 0, k = 1, ..., n − 1, from (39) we get: ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − ( b − a N ) [jf (a) + (N − j) f (b)] ∣ ∣ ∣ ∣ ∣ ≤ CUBO 21, 2 (2019) Caputo fractional Iyengar type Inequalities 9 max { ‖Dν ∗af‖L1([a,b]) , ∥ ∥Dνb−f ∥ ∥ L1([a,b]) } Γ (ν + 1) ( b − a N )ν [ jν + (N − j) ν ] , (40) j = 0, 1, 2, ..., N, viii) when N = 2 and j = 1, (40) turns to ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − (b − a) 2 (f (a) + f (b)) ∣ ∣ ∣ ∣ ∣ ≤ max { ‖Dν ∗af‖L1([a,b]) , ∥ ∥Dνb−f ∥ ∥ L1([a,b]) } Γ (ν + 1) (b − a) ν 2ν−1 . (41) Proof. Here ν ≥ 1 and Dν ∗af, D ν b−f ∈ L1 ([a, b]). By (10) we get ∣ ∣ ∣ ∣ ∣ f (x) − n−1∑ k=0 f(k) (a) k! (x − a) k ∣ ∣ ∣ ∣ ∣ ≤ 1 Γ (ν) (x − a) ν−1 ∫x a |Dν ∗af (t)| dt ≤ (x − a) ν−1 Γ (ν) ‖Dν ∗af‖L1([a,b]) , (42) ∀ x ∈ [a, b] . That is ∣ ∣ ∣ ∣ ∣ f (x) − n−1∑ k=0 f(k) (a) k! (x − a) k ∣ ∣ ∣ ∣ ∣ ≤ ‖Dν ∗af‖L1([a,b]) Γ (ν) (x − a) ν−1 , (43) ∀ x ∈ [a, b] . By (11) we get ∣ ∣ ∣ ∣ ∣ f (x) − n−1∑ k=0 f(k) (b) k! (x − b) k ∣ ∣ ∣ ∣ ∣ ≤ 1 Γ (ν) (b − x) ν−1 ∫b x ∣ ∣Dνb−f (z) ∣ ∣dz ≤ (b − x) ν−1 Γ (ν) ∥ ∥Dνb−f ∥ ∥ L1([a,b]) , (44) ∀ x ∈ [a, b] . That is ∣ ∣ ∣ ∣ ∣ f (x) − n−1∑ k=0 f(k) (b) k! (x − b) k ∣ ∣ ∣ ∣ ∣ ≤ ∥ ∥Dνb−f ∥ ∥ L1([a,b]) Γ (ν) (b − x) ν−1 , (45) ∀ x ∈ [a, b] . Call δ1 := ‖Dν ∗af‖L1([a,b]) Γ (ν) , (46) 10 George A. Anastassiou CUBO 21, 2 (2019) and δ2 := ∥ ∥Dνb−f ∥ ∥ L1([a,b]) Γ (ν) . (47) Set δ := max (δ1, δ2) . (48) That is ∣ ∣ ∣ ∣ ∣ f (x) − n−1∑ k=0 f(k) (a) k! (x − a) k ∣ ∣ ∣ ∣ ∣ ≤ δ (x − a) ν−1 , (49) and ∣ ∣ ∣ ∣ ∣ f (x) − n−1∑ k=0 f(k) (b) k! (x − b) k ∣ ∣ ∣ ∣ ∣ ≤ δ (b − x) ν−1 , (50) ∀ x ∈ [a, b] . As in the proof of Theorem 4, we get: ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − n−1∑ k=0 1 (k + 1) ! [ f(k) (a) (t − a) k+1 + (−1)kf(k) (b) (b − t) k+1 ] ∣ ∣ ∣ ∣ ∣ ≤ δ ν [ (t − a) ν + (b − t) ν ] , (51) ∀ t ∈ [a, b] . The rest of the proof is similar to the proof of Theorem 4. We continue with Theorem 6. Let p, q > 1 : 1 p + 1 q = 1, ν > 1 q , n = ⌈ν⌉ ; f ∈ ACn ([a, b]), with Dν ∗af, D ν b−f ∈ Lq ([a, b]). Then i) ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − n−1∑ k=0 1 (k + 1) ! [ f(k) (a) (t − a) k+1 + (−1)kf(k) (b) (b − t) k+1 ] ∣ ∣ ∣ ∣ ∣ ≤ max { ‖Dν ∗af‖Lq([a,b]) , ∥ ∥Dνb−f ∥ ∥ Lq([a,b]) } Γ (ν) ( ν + 1 p ) (p (ν − 1) + 1) 1 p [ (t − a) ν+ 1 p + (b − t) ν+ 1 p ] , (52) ∀ t ∈ [a, b] , ii) at t = a+b 2 , the right hand side of (52) is minimized, and we get: ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − n−1∑ k=0 1 (k + 1) ! (b − a) k+1 2k+1 [ f(k) (a) + (−1)kf(k) (b) ] ∣ ∣ ∣ ∣ ∣ ≤ CUBO 21, 2 (2019) Caputo fractional Iyengar type Inequalities 11 max { ‖Dν ∗af‖Lq([a,b]) , ∥ ∥Dνb−f ∥ ∥ Lq([a,b]) } Γ (ν) ( ν + 1 p ) (p (ν − 1) + 1) 1 p (b − a) ν+ 1 p 2 ν− 1 q , (53) iii) if f(k) (a) = f(k) (b) = 0, for all k = 0, 1, ..., n − 1, we obtain ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx ∣ ∣ ∣ ∣ ∣ ≤ max { ‖Dν ∗af‖Lq([a,b]) , ∥ ∥Dνb−f ∥ ∥ Lq([a,b]) } Γ (ν) ( ν + 1 p ) (p (ν − 1) + 1) 1 p (b − a) ν+ 1 p 2 ν− 1 q , (54) which is a sharp inequality, iv) more generally, for j = 0, 1, 2, ..., N ∈ N, it holds ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − n−1∑ k=0 1 (k + 1) ! ( b − a N )k+1 [ jk+1f(k) (a) + (−1)k (N − j) k+1 f(k) (b) ] ∣ ∣ ∣ ∣ ∣ ≤ max { ‖Dν ∗af‖Lq([a,b]) , ∥ ∥Dνb−f ∥ ∥ Lq([a,b]) } Γ (ν) ( ν + 1 p ) (p (ν − 1) + 1) 1 p ( b − a N )ν+ 1 p [ j ν+ 1 p + (N − j) ν+ 1 p ] , (55) v) if f(k) (a) = f(k) (b) = 0, k = 1, ..., n − 1, from (55) we get: ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − ( b − a N ) [jf (a) + (N − j) f (b)] ∣ ∣ ∣ ∣ ∣ ≤ max { ‖Dν ∗af‖Lq([a,b]) , ∥ ∥Dνb−f ∥ ∥ Lq([a,b]) } Γ (ν) ( ν + 1 p ) (p (ν − 1) + 1) 1 p ( b − a N )ν+ 1 p [ j ν+ 1 p + (N − j) ν+ 1 p ] , (56) for j = 0, 1, 2, ..., N, vi) when N = 2 and j = 1, (56) turns to ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − ( b − a 2 ) (f (a) + f (b)) ∣ ∣ ∣ ∣ ∣ ≤ max { ‖Dν ∗af‖Lq([a,b]) , ∥ ∥Dνb−f ∥ ∥ Lq([a,b]) } Γ (ν) ( ν + 1 p ) (p (ν − 1) + 1) 1 p (b − a) ν+ 1 p 2 ν− 1 q , (57) vii) when 1/q < ν ≤ 1, inequality (57) is again valid but without any boundary conditions. Proof. Here ν > 0, n = ⌈ν⌉, f ∈ ACn ([a, b]) ; p, q > 1 : 1 p + 1 q = 1, with Dν ∗af, D ν b−f ∈ Lq ([a, b]). By (10) we have ∣ ∣ ∣ ∣ ∣ f (x) − n−1∑ k=0 f(k) (a) k! (x − a) k ∣ ∣ ∣ ∣ ∣ ≤ 1 Γ (ν) ∫x a (x − t) ν−1 |Dν ∗af (t)| dt ≤ 12 George A. Anastassiou CUBO 21, 2 (2019) 1 Γ (ν) (∫x a (x − t) p(ν−1) dt ) 1 p (∫x a |Dν ∗af (t)| q dt ) 1 q ≤ 1 Γ (ν) (x − a) p(ν−1)+1 p (p (ν − 1) + 1) 1 p ‖Dν ∗af‖Lq([a,b]) . (58) Here we assume that ν > 1 q ⇔ p (ν − 1) + 1 > 0. So, we get ∣ ∣ ∣ ∣ ∣ f (x) − n−1∑ k=0 f(k) (a) k! (x − a) k ∣ ∣ ∣ ∣ ∣ ≤ ‖Dν ∗af‖Lq([a,b]) Γ (ν) (p (ν − 1) + 1) 1 p (x − a) ν− 1 q , (59) ∀ x ∈ [a, b] . By (11) we have ∣ ∣ ∣ ∣ ∣ f (x) − n−1∑ k=0 f(k) (b) k! (x − b) k ∣ ∣ ∣ ∣ ∣ ≤ 1 Γ (ν) ∫b x (z − x) ν−1 ∣ ∣Dνb−f (z) ∣ ∣dz ≤ 1 Γ (ν) (∫b x (z − x) p(ν−1) dz ) 1 p (∫b x ∣ ∣Dνb−f (z) ∣ ∣ q dz ) 1 q ≤ 1 Γ (ν) (b − x) p(ν−1)+1 p (p (ν − 1) + 1) 1 p ∥ ∥Dνb−f ∥ ∥ Lq([a,b]) . (60) So, we get ∣ ∣ ∣ ∣ ∣ f (x) − n−1∑ k=0 f(k) (b) k! (x − b) k ∣ ∣ ∣ ∣ ∣ ≤ ∥ ∥Dνb−f ∥ ∥ Lq([a,b]) Γ (ν) (p (ν − 1) + 1) 1 p (b − x) ν− 1 q , (61) ∀ x ∈ [a, b] . Call ρ1 := ‖Dν ∗af‖Lq([a,b]) Γ (ν) (p (ν − 1) + 1) 1 p , (62) and ρ2 := ∥ ∥Dνb−f ∥ ∥ Lq([a,b]) Γ (ν) (p (ν − 1) + 1) 1 p . (63) Set ρ := max (ρ1, ρ2) , m := ν − 1 q > 0. (64) That is ∣ ∣ ∣ ∣ ∣ f (x) − n−1∑ k=0 f(k) (a) k! (x − a) k ∣ ∣ ∣ ∣ ∣ ≤ ρ (x − a) m , (65) and ∣ ∣ ∣ ∣ ∣ f (x) − n−1∑ k=0 f(k) (b) k! (x − b) k ∣ ∣ ∣ ∣ ∣ ≤ ρ (b − x) m , (66) CUBO 21, 2 (2019) Caputo fractional Iyengar type Inequalities 13 ∀ x ∈ [a, b] . As in the proof of Theorem 4, we obtain: ∣ ∣ ∣ ∣ ∣ ∫b a f (x) dx − n−1∑ k=0 1 (k + 1) ! [ f(k) (a) (t − a) k+1 + (−1)kf(k) (b) (b − t) k+1 ] ∣ ∣ ∣ ∣ ∣ ≤ ρ (m + 1) [ (t − a) m+1 + (b − t) m+1 ] = max { ‖Dν ∗af‖Lq([a,b]) , ∥ ∥Dνb−f ∥ ∥ Lq([a,b]) } Γ (ν) (p (ν − 1) + 1) 1 p ( ν + 1 p ) [ (t − a) ν+ 1 p + (b − t) ν+ 1 p ] , (67) ∀ t ∈ [a, b] . The rest of the proof is similar to the proof of Theorem 4. References [1] George A. Anastassiou, Fractional Differentiation Inequalities, Springer, Heidelberg, New York, 2009. [2] George A. Anastassiou, Intelligent Mathematical Computational Analysis, Springer, Heidelberg, New York, 2011. [3] K. Diethelm, The Analysis of Fractional Differential Equations, Springer, Heidelberg, New York, 2010. [4] K.S.K. Iyengar, Note on an inequality, Math. Student, 6 (1938), 75-76. Introduction Main Results