CUBO A Mathematical Journal Vol.10, N o ¯ 01, (117–142). March 2008 Converse Fractional Opial Inequalities for Several Functions George A. Anastassiou Department of Mathematical Sciences University of Memphis, Memphis, TN 38152 U.S.A. E-mail address: ganastss@memphis.edu Tel: 901-678-3144 – Fax: 901-678-2480 ABSTRACT A variety of very general Lp(0 < p < 1) form converse Opial type inequalities ([8]) is presented involving generalized fractional derivatives ([3],[6]) of several functions in different orders and powers. From the established results derive other particular results of special interest. RESUMEN Una variedad general de desigualdades inversas de tipo Opial en Lp(0 < p < 1) son presentadas, las cuales envuelven deridadas fraccionarias generalidades ([3],[6]) de varias funciones con diferentes ordenes y potencias. Deducimos algunos casos particulares de especial interes. Key Words and Phrases: Opial type inequality, Fractional derivative. Math Subj. Class.: 26A33, 26D10, 26D15. 118 George A. Anastassiou CUBO 10, 1 (2008) 1 Introduction Opial inequalities appeared for the first time in [8] and then many authors dealt with them in different directions and for various cases. For a complete recent account on the activity of this field see [1], and still it remains a very active area of research. One of their main attractions to these inequalities is their applications, especially to proving uniqueness and upper bounds of solution of initial value problems in differential equations. The author was the first to present Opial inequalities involving fractional derivatives of functions in [2], [3] with applications to fractional differential equations. Fractional derivatives come up naturally in a number of fields, especially in Physics, see the recent books [7], [9]. Here the author continues his study of fractional Opial inequalities now involving several different functions and produces a wide variety of converse results. To give an idea to the reader of the kind of inequalities we are dealing with, briefly we mention a specific one (see Corollary 15). ∫ x x0   M ∑ j=1 ∣ ∣ ( D γ1 x0 fj ) (w) ∣ ∣ λα ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ λv   dw ≥ C (x)   ∫ x x0   M ∑ j=1 ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ p  dw   ( λα+λv p ) , (∗) all x0 ≤ x ≤ b. In (∗) , C (x) is a constant that depends on x0, x, and the involved parameters, γ1 ≥ 0, 1 ≤ v − γ1 < 1 p , 0 < p < 1; D v x0 fj is of fixed sign on [x0, b] , j = 1, . . . , M ∈ N. Also λα ≥ 0, λv > p. Here f (i) j (x0) = 0, i = 0, 1, . . . , n − 1, n := [v] (integral part); j = 1, . . . , M. And D γ1 x0 fj , D v x0 fj are the generalized (of Canavati) type [6], [2] fractional derivatives of fj of orders γ1, v respectively. 2 Preliminaries In the sequel we follow [6]. Let g ∈ C ([0, 1]) . Let v be a positive number, n := [v] and α := v − n (0 < α < 1) . Define (Jvg) (x) := 1 Γ (v) ∫ x 0 (x − t) v−1 g (t) dt, 0 ≤ x ≤ 1, (1) CUBO 10, 1 (2008) Converse Fractional Opial Inequalities ... 119 the Riemann-Liouville integral, where Γ is the gamma function. We define the subspace C v ([0, 1]) of C n ([0, 1]) as follows: C v ([0, 1]) := { g ∈ Cn ([0, 1]) : J1−αD n g ∈ C1 ([0, 1]) } , where D := d dx . So for g ∈ Cv ([0, 1]) , we define the v−fractional derivative of g as D v g := DJ1−αD n g. (2) When v ≥ 1 we have the Taylor’s formula g (t) = g (0) + g ′ (0) t + g” (0) t 2 2! + . . . + g (n−1) (0) t n−1 (n − 1)! + (JvD v g) (t) , for all t ∈ [0, 1] . (3) When 0 < v < 1 we find g (t) = (JvD v g) (t) , for all t ∈ [0, 1] . (4) Next we carry above notions over to arbitrary [a, b] ⊆ R (see[3]). Let x, x0 ∈ [a, b] such that x > x0, where x0 < b is fixed. Let f ∈ C ([a, b]) and define (J x0 v f ) (x) := 1 Γ (v) ∫ x x0 (x − t) v−1 f (t) dt, x0 ≤ x ≤ b, (5) the generalized Riemann-Liouville integral. We define the subspace Cvx0 ([a, b]) of C n ([a, b]) : C v x0 ([a, b]) := { f ∈ Cn ([a, b]) : J x01−αD n f ∈ C1 ([x0, b]) } , clearly C 0 x0 ([a, b]) = C ([a, b]) , also C n x0 ([a, b]) = C n ([a, b]) , n ∈ N. For f ∈ Cvx0 ([a, b]) , we define the generalized v−fractional derivative of f over [x0, b] as D v x0 f := DJ x0 1−αf (n) ( f (n) := D n f ) . (6) Notice that ( J x0 1−αf (n) ) (x) = 1 Γ (1 − α) ∫ x x0 (x − t) −α f (n) (t) dt exists for f ∈ Cvx0 ([a, b]) . We recall the following generalization of Taylor’s formula (see [6], [3]). 120 George A. Anastassiou CUBO 10, 1 (2008) Theorem 1. Let f ∈ Cvx0 ([a, b]) , x0 ∈ [a, b] fixed. (i) If v ≥ 1 then f (x) = f (x0) + f ′ (x0) (x − x0) + f ” (x0) (x − x0) 2 2 + . . . + f (n−1) (x0) (x − x0) n−1 (n − 1)! + ( J x0 v D v x0 f ) (x) , for all x ∈ [a, b] : x ≥ x0. (7) (ii) If 0 < v < 1 then f (x) = ( J x0 v D v x0 f ) (x) , for all x ∈ [a, b] : x ≥ x0. (8) We make Remark 2. 1) ( D n x0 f ) = f (n) , n ∈ N. 2) Let f ∈ Cvx0 ([a, b]) , v ≥ 1 and f (i) (x0) = 0, i = 0, 1, . . . , n − 1, n := [v] . Then by (7) f (x) = ( J x0 v D v x0 f ) (x) . I.e. f (x) = 1 Γ (v) ∫ x x0 (x − t) v−1 ( D v x0 f ) (t) dt, (9) for all x ∈ [a, b] with x ≥ x0. Notice that (9) is true, also when 0 < v < 1. We need from [3] Lemma 3. Let f ∈ C ([a, b]) , µ, v > 0. Then J x0 µ (J x0 v f ) = J x0 µ+v (f ) . (10) We also make Remark 4. Let v ≥ γ + 1, γ ≥ 0, so that γ < v. Call n := [v] , α := v − n; m := [γ] , ρ := γ−m. Note that v−m ≥ 1 and n−m ≥ 1. Let f ∈ Cvx0 ([a, b]) be such that f (i) (x0) = 0, i = 0, 1, . . . , n − 1. Hence by (7) f (x) = ( J x0 v D v x0 f ) (x) , for all x ∈ [a, b] : x ≥ x0. Therefore by Leibnitz’s formula and Γ (p + 1) = pΓ (p) , p > 0, we get that f (m) (x) = ( J x0 v−mD v x0 f ) (x) , for all x ≥ x0. (11) CUBO 10, 1 (2008) Converse Fractional Opial Inequalities ... 121 It follows that f ∈ Cγx0 ([a, b]) and thus ( D γ x0 f ) (x) := ( DJ x0 1−ρf (m) ) (x) exists for all x ≥ x0. (12) Really by the use of (11) we have on [x0, b] J x0 1−ρ ( f (m) ) = J x0 1−ρ ( J x0 v−mD v x0 f ) = ( J x0 1−ρ ◦ J x0 v−m )( D v x0 f ) = J x0 v−m+1−ρ ( D v x0 f ) = J x0 v−γ+1 ( D v x0 f ) , by (10). That is, ( J x0 1−ρf (m) ) (x) = 1 Γ (v − γ + 1) ∫ x x0 (x − t) v−γ ( D v x0 f ) (t) dt. Therefore ( D γ x0 f ) (x) = D (( J x0 1−ρf (m) ) (x) ) = 1 Γ (v − γ) · ∫ x x0 (x − t) (v−γ)−1 ( D v x0 f ) (t) dt; (13) hence ( D γ x0 f ) (x) = ( J x0 v−γ ( D v x0 f )) (x) and is continuous in x on [x0, b] . In particular when v ≥ 2 we have ( D v−1 x0 f ) (x) = ∫ x x0 ( D v x0 f ) (t) dt, x ≥ x0. (14) That is ( D v−1 x0 f )′ = D v x0 f, ( D v−1 x0 f ) (x0) = 0. 3 Main Results 3.1 Results involving two functions We present our first main result Theorem 5. Let γj ≥ 0, 1 ≤ v − γj < 1/p, 0 < p < 1, j = 1, 2, and f1, f2 ∈ C v x0 ([a, b]) with f (i) 1 (x0) = f (i) 2 (x0) = 0, i = 0, 1, . . . , n − 1, n := [v] . Here x, x0 ∈ [a, b] : x ≥ x0. We assume here that D v x0 fj is of fixed sign on [x0, b] , j = 1, 2. Consider also p (t) > 0 and q (t) > 0 continuous functions on [x0, b] . Let λv > 0 and λα, λβ ≥ 0 such that λv > p. 122 George A. Anastassiou CUBO 10, 1 (2008) Set Pk (w) := ∫ w x0 (w − t) (v−γk−1)p p−1 (p (t)) − 1 p−1 dt, k = 1, 2, x0 ≤ x ≤ b; (15) A (w) := q (w) · (P1 (w)) λα( p−1 p ) · (P2 (w)) λβ( p−1 p ) (p (w)) − λv p (Γ (v − γ1)) λα (Γ (v − γ2)) λβ , (16) A0 (x) := ( ∫ x x0 (A (w)) p p−λv dw ) p−λv p , (17) and δ1 := 2 1−( λα+λv p ) (18) If λβ = 0, we obtain that, ∫ x x0 q (w) [ ∣ ∣ ( D γ1 x0 f1 ) (w) ∣ ∣ λα · ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ λv + ∣ ∣ ( D γ1 x0 f2 ) (w) ∣ ∣ λα · ∣ ∣ ( D v x0 f2 ) (w) ∣ ∣ λv ] dw ≥ ( A0 (x) |λβ =0 ) · ( λv λα + λv ) λv p · δ1 · [ ∫ x x0 p (w) [ ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ p + ∣ ∣ ( D v x0 f2 ) (w) ∣ ∣ p] dw ]( λα+λv p ) . (19) Proof. From (13) and assumption we have ∣ ∣ ( D γk x0 fj ) (w) ∣ ∣ = 1 Γ (v − γk) ∫ w x0 (w − t) v−γk−1 ∣ ∣ ( D v x0 fj ) (t) ∣ ∣dt, for k = 1, 2, j = 1, 2 and for all x0 ≤ w ≤ b. Next applying Hölder’s inequality with indices p, p p−1 we get ∣ ∣ ( D γk x0 fj ) (w) ∣ ∣ = 1 Γ (v − γk) ∫ w x0 (w − t) v−γk−1 (p (t)) − 1 p (p (t)) 1 p ∣ ∣ ( D v x0 fj ) (t) ∣ ∣ dt ≥ 1 Γ (v − γk) ( ∫ w x0 ( (w − t) v−γk−1 (p (t)) − 1 p ) p p−1 dt ) p−1 p ( ∫ w x0 p (t) ∣ ∣ ( D v x0 fj ) (t) ∣ ∣ p dt ) 1 p CUBO 10, 1 (2008) Converse Fractional Opial Inequalities ... 123 = 1 Γ (v − γk) (Pk (w)) p−1 p ( ∫ w x0 p (t) ∣ ∣ ( D v x0 fj ) (t) ∣ ∣ p dt ) 1 p . I.e., it holds ∣ ∣ ( D γk x0 fj ) (w) ∣ ∣ ≥ 1 Γ (v − γk) (Pk (w)) p−1 p ( ∫ w x0 p (t) ∣ ∣ ( D v x0 fj ) (t) ∣ ∣ p dt ) 1 p . (20) Put zj (w) := ∫ w x0 p (t) ∣ ∣ ( D v x0 fj ) (t) ∣ ∣ p dt, thus, z ′ j (w) = p (w) ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ p , zj (x0) = 0; j = 1, 2. Hence, we have ∣ ∣ ( D γk x0 fj ) (w) ∣ ∣ ≥ 1 Γ (v − γk) (Pk (w)) p−1 p (zj (w)) 1 p , and ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ λv = p (w) − λv p ( z ′ j (w) ) λv p , j = 1, 2. Therefore we obtain q (w) ∣ ∣ ( D γ1 x0 f1 ) (w) ∣ ∣ λα ∣ ∣ ( D γ2 x0 f2 ) (w) ∣ ∣ λβ ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ λv ≥ q (w) 1 (Γ (v − γ1)) λα (P1 (w)) λα( p−1 p ) (z1 (w)) λα p 1 (Γ (v − γ2)) λβ (P2 (w)) λβ ( p−1 p ) (z2 (w)) λβ p (p (w)) − λv p ( z ′ 1 (w) ) λv p = A (w) (z1 (w)) λα p (z2 (w)) λβ p ( z ′ 1 (w) ) λv p . Consequently, by another Hölder’s inequality application, we find (by p λv < 1) ∫ x x0 q (w) ∣ ∣ ( D γ1 x0 f1 ) (w) ∣ ∣ λα ∣ ∣ ( D γ2 x0 f2 ) (w) ∣ ∣ λβ ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ λv dw ≥ A0 (x) [ ∫ x x0 (z1 (w)) λα λv (z2 (w)) λβ λv z ′ 1 (w) dw ] λv p . (21) Similary one finds ∫ x x0 q (w) ∣ ∣ ( D γ2 x0 f1 ) (w) ∣ ∣ λβ ∣ ∣ ( D γ1 x0 f2 ) (w) ∣ ∣ λα ∣ ∣ ( D v x0 f2 ) (w) ∣ ∣ λv dw 124 George A. Anastassiou CUBO 10, 1 (2008) ≥ A0 (x) [ ∫ x x0 (z1 (w)) λβ λv (z2 (w)) λα λv z ′ 2 (w) dw ] λv p . (22) Taking λβ = 0 and adding (21) and (22) we obtain ∫ x x0 q (w) [ ∣ ∣ ( D γ1 x0 f1 ) (w) ∣ ∣ λα · ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ λv + ∣ ∣ ( D γ1 x0 f2 ) (w) ∣ ∣ λα · ∣ ∣ ( D v x0 f2 ) (w) ∣ ∣ λv ] dw ≥ ( A0 (x) |λβ =0 ) { [ ∫ x x0 (z1 (w)) λα λv z ′ 1 (w) dw ] λv p + [ ∫ x x0 (z2 (w)) λα λv z ′ 2 (w) dw ] λv p } = ( A0 (x) |λβ =0 ) { (z1 (x)) (λα+λv ) p + (z2 (x)) (λα+λv ) p } ( λv λα + λv ) λv p = ( A0 (x) |λβ =0 ) ( λv λα + λv ) λv p      ( ∫ x x0 p (t) ∣ ∣ ( D v x0 f1 ) (t) ∣ ∣ p dt ) (λα+λv ) p + ( ∫ x x0 p (t) ∣ ∣ ( D v x0 f2 ) (t) ∣ ∣ p dt ) (λα+λv ) p      =: (∗) . In this article we are using frequently the basic inequalities 2 r−1 (a r + b r ) ≤ (a + b) r ≤ ar + br, a, b ≥ 0, 0 ≤ r ≤ 1, (23) a r + b r ≤ (a + b) r ≤ 2r−1 (ar + br) , a, b ≥ 0, r ≥ 1. (24) Finally using (23) , (24) and (18) we get (∗) ≥ ( A0 (x) |λβ =0 ) · ( λv λα + λv ) λv p · δ1 CUBO 10, 1 (2008) Converse Fractional Opial Inequalities ... 125 { ∫ x x0 p (t) [ ∣ ∣ ( D v x0 f1 ) (t) ∣ ∣ p + ∣ ∣ ( D v x0 f2 ) (t) ∣ ∣ p] dt } (λα+λv ) p . Inequality (19) has been established. Here we see that ( p p−1 ) (v − γi − 1) + 1 > 0, − 1 (p−1) > 0 and p (t) ∈ C ([x0, b]) , thus (see (15)) Pi (w) ∈ R for every w ∈ [x0, b] , also Pi (w) is continuous and bounded on [x0, b] for i = 1, 2. By λv > p > 0, we have 0 < p λv < 1, p p−λv < 0. We observe that 1 A (w) = 1 q (w) (Γ (v − γ1)) λα (Γ (v − γ2)) λβ (p (w)) λv /p (P1 (w)) λα( 1−p p ) (P2 (w)) λβ ( 1−p p ) ∈ C ([x0, b]) , and 1 A(w) > 0 on (x0, b] , 1 A(x0) = 0. Therefore 0 < A0 (x) < ∞, and all we have done in this proof are valid. � It follows the counterpart of the last theorem. Theorem 6. All here as in Theorem 5. Further assume λβ ≥ λv. Denote δ2 := 2 1−(λβ /λv ), δ3 := (δ2 − 1) 2 −(λβ /λv ). (25) If λα = 0, then it holds ∫ x x0 q (w) [ ∣ ∣ ( D γ2 x0 f2 ) (w) ∣ ∣ λβ · ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ λv + ∣ ∣ ( D γ2 x0 f1 ) (w) ∣ ∣ λβ · ∣ ∣ ( D v x0 f2 ) (w) ∣ ∣ λv ] dw ≥ (A0 (x) |λα=0) 2 p−λv p ( λv λβ + λv ) λv p δ λv p 3 · ( ∫ x x0 p (w) [ ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ p + ∣ ∣ ( D v x0 f2 ) (w) ∣ ∣ p] dw ) ( λv +λβ p ) , all x0 ≤ x ≤ b. (26) Proof. When λα = 0 from (21) and (22) we obtain ∫ x x0 q (w) ∣ ∣ ( D γ2 x0 f2 ) (w) ∣ ∣ λβ ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ λv dw 126 George A. Anastassiou CUBO 10, 1 (2008) ≥ (A0 (x) |λα=0) [ ∫ x x0 (z2 (w)) λβ λv z ′ 1 (w) dw ] λv p , (27) and ∫ x x0 q (w) ∣ ∣ ( D γ2 x0 f1 ) (w) ∣ ∣ λβ ∣ ∣ ( D v x0 f2 ) (w) ∣ ∣ λv dw ≥ (A0 (x) |λα=0) [ ∫ x x0 (z1 (w)) λβ λv z ′ 2 (w) dw ] λv p , (28) all x0 ≤ x ≤ b. Adding (27) and (28) we get ∫ x x0 q (w) [ ∣ ∣ ( D γ2 x0 f2 ) (w) ∣ ∣ λβ · ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ λv + ∣ ∣ ( D γ2 x0 f1 ) (w) ∣ ∣ λβ · ∣ ∣ ( D v x0 f2 ) (w) ∣ ∣ λv ] dw ≥ (A0 (x) |λα=0) { [ ∫ x x0 (z2 (w)) λβ λv z ′ 1 (w) dw ] λv p + [ ∫ x x0 (z1 (w)) λβ λv z ′ 2 (w) dw ] λv p } ≥ (A0 (x) |λα=0) · 2 p−λv p · (M (x)) λv p =: (∗) , (29) by λv p > 1 and (24) , where M (x) := ∫ x x0 (z2 (w)) λβ λv z ′ 1 (w) + (z1 (w)) λβ λv z ′ 2 (w) dw. (30) Next we work on M (x) . We have that M (x) = ∫ x x0 ( (z1 (w)) λβ λv + (z2 (w)) λβ λv ) ( z ′ 1 (w) + z ′ 2 (w) ) dw − ∫ x x0 [ (z1 (w)) λβ λv z ′ 1 (w) + (z2 (w)) λβ λv z ′ 2 (w) ] dw (by (24)) ≥ δ2 ∫ x x0 (z1 (w) + z2 (w)) λβ λv (z1 (w) + z2 (w)) ′ dw − ( λv λβ + λv )[ (z1 (x)) ( λv +λβ λv ) + (z2 (x)) ( λv +λβ λv ) ] = δ2 (z1 (x) + z2 (x)) ( λv +λβ λv ) ( λv λv + λβ ) − ( λv λβ + λv ) CUBO 10, 1 (2008) Converse Fractional Opial Inequalities ... 127 [ (z1 (x)) ( λv +λβ λv ) + (z2 (x)) ( λv +λβ λv ) ] = ( λv λβ + λv )[ δ2 (z1 (x) + z2 (x)) ( λv +λβ λv ) − ( (z1 (x)) ( λv +λβ λv ) + (z2 (x)) ( λv +λβ λv ) )] (24) ≥ ( λv λβ + λv )[ δ2 ( (z1 (x)) ( λv +λβ λv ) + (z2 (x)) ( λv +λβ λv ) ) − ( (z1 (x)) ( λv +λβ λv ) + (z2 (x)) ( λv +λβ λv ) )] = ( λv λβ + λv ) (δ2 − 1) [ (z1 (x)) ( λv +λβ λv ) + (z2 (x)) ( λv +λβ λv ) ] (24) ≥ ( λv λβ + λv ) δ3 (z1 (x) + z2 (x)) ( λv +λβ λv ) . I.e. we present that M (x) ≥ ( λv λβ + λv ) δ3 (z1 (x) + z2 (x)) ( λv +λβ λv ) . (31) Consequently, by (29) and (31) we get (∗) ≥ (A0 (x) |λα=0) 2 p−λv p ( λv λβ + λv ) λv p δ λv p 3 (z1 (x) + z2 (x)) ( λv +λβ p ) = (A0 (x) |λα=0) 2 p−λv p ( λv λβ + λv ) λv p δ λv p 3 ( ∫ x x0 p (t) [ ∣ ∣ ( D v x0 f1 ) (t) ∣ ∣ p + ∣ ∣ ( D v x0 f2 ) (t) ∣ ∣ p] dt ) ( λv +λβ p ) . We have established (26) . � A special important case follows. Theorem 7. Let v ≥ 2 and γ1 ≥ 0 such that 2 ≤ v − γ1 < 1 p , 0 < p < 1. Let f1, f2 ∈ C v x0 ([a, b]) with f (i) 1 (x0) = f (i) 2 (x0) = 0, i = 0, 1, . . . , n − 1, n := [v] . Here x, x0 ∈ [a, b] : x ≥ x0. We assume here that D v x0 fj is of fixed sign on [x0, b] , j = 1, 2. Consider also p (t) > 0 and q (t) > 0 continuous functions on [x0, b] . Let λα ≥ λα+1 > 1. 128 George A. Anastassiou CUBO 10, 1 (2008) Denote θ3 := ( 2 1−(λα /λα+1) − 1 ) 2 −λα/λα+1 , (32) L (x) := ( 2 ∫ x x0 (q (w)) ( 1 1−λα+1 ) dw )(1−λα+1) ( θ3λα+1 λα + λα+1 )λα+1 , (33) and P1 (x) := ∫ x x0 (x − t) (v−γ1−1)p p−1 (p (t)) − 1 p−1 dt, (34) T (x) := L (x) · ( P1 (x) ( p−1 p ) Γ (v − γ1) )(λα+λα+1) , ω1 := 2 ( p−1 p )(λα+λα+1), (35) and Φ (x) := T (x) ω1. (36) Then ∫ x x0 q (w) [ ∣ ∣ ( D γ1 x0 f1 ) (w) ∣ ∣ λα ∣ ∣ ( D γ1+1 x0 f2 ) (w) ∣ ∣ λα+1 + ∣ ∣ ( D γ1 x0 f2 ) (w) ∣ ∣ λα ∣ ∣ ( D γ1+1 x0 f1 ) (w) ∣ ∣ λα+1 ] dw (37) ≥ Φ (x) [ ∫ x x0 p (w) ( ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ p + ∣ ∣ ( D v x0 f2 ) (w) ∣ ∣ p) dw ] (λα+λα+1) p , all x0 ≤ x ≤ b. Proof. For convenience we set γ2 := γ1 + 1. From (13) and assumption we obtain ∣ ∣ ( D γk x0 fj ) (w) ∣ ∣ = 1 Γ (v − γk) ∫ w x0 (w − t) v−γk−1 ∣ ∣ ( D v x0 fi ) (t) ∣ ∣dt =: gj,γk (w) , (38) where j = 1, 2, k = 1, 2, all x0 ≤ x ≤ b. We observe that (( D γ1 x0 fj ) (x) )′ = ( D γ1+1 x0 fj ) (x) = ( D γ2 x0 fj ) (x) , (39) CUBO 10, 1 (2008) Converse Fractional Opial Inequalities ... 129 all x0 ≤ x ≤ b. And also (gj,γ1 (w)) ′ = gj,γ2 (w) ; gj,γk (x0) = 0. (40) Notice that if v − γ2 = 1, then gj,γ2 (w) = ∫ w x0 ∣ ∣ ( D v x0 fj ) (t) ∣ ∣ dt. Next we apply Hölder’s inequality with indices 1 λα+1 < 1, 1 (1−λα+1) < 0, we obtain ∫ x x0 q (w) ∣ ∣ ( D γ1 x0 f1 ) (w) ∣ ∣ λα ∣ ∣ ( D γ1+1 x0 f2 ) (w) ∣ ∣ λα+1 dw = ∫ x x0 q (w) (g1,γ1 (w)) λα ( (g2,γ1 (w)) ′)λα+1 dw (41) ≥ ( ∫ x x0 (q (w)) ( 1 1−λα+1 ) dw )(1−λα+1) ( ∫ x x0 (g1,γ1 (w)) λα λα+1 (g2,γ1 (w)) ′ dw ) λα+1 . Similarly we get ∫ x x0 q (w) ∣ ∣ ( D γ1 x0 f2 ) (w) ∣ ∣ λα ∣ ∣ ( D γ1+1 x0 f1 ) (w) ∣ ∣ λα+1 dw ≥ ( ∫ x x0 (q (w)) ( 1 1−λα+1 ) dw )(1−λα+1) ( ∫ x x0 (g2,γ1 (w)) λα λα+1 (g1,γ1 (w)) ′ dw ) λα+1 . (42) Adding (41) and (42) we observe ∫ x x0 q (w) [ ∣ ∣ ( D γ1 x0 f1 ) (w) ∣ ∣ λα ∣ ∣ ( D γ1+1 x0 f2 ) (w) ∣ ∣ λα+1 + ∣ ∣ ( D γ1 x0 f2 ) (w) ∣ ∣ λα ∣ ∣ ( D γ1+1 x0 f1 ) (w) ∣ ∣ λα+1 ] dw ≥ ( ∫ x x0 (q (w)) ( 1 1−λα+1 ) dw )(1−λα+1) 130 George A. Anastassiou CUBO 10, 1 (2008) [ ( ∫ x x0 (g1,γ1 (w)) λα λα+1 (g2,γ1 (w)) ′ dw ) λα+1 + ( ∫ x x0 (g2,γ1 (w)) λα λα+1 (g1,γ1 (w)) ′ dw ) λα+1 ] (24) ≥ ( 2 ∫ x x0 (q (w)) ( 1 1−λα+1 ) dw )(1−λα+1) · [ ∫ x x0 [ (g1,γ1 (w)) λα λα+1 (g2,γ1 (w)) ′ + (g2,γ1 (w)) λα λα+1 (g1,γ1 (w)) ′ ] dw ]λα+1 (notice (30) and the proof of (31) , accordingly here we have) ≥ ( 2 ∫ x x0 (q (w)) ( 1 1−λα+1 ) dw )(1−λα+1) ( λα+1θ3 λα + λα+1 )λα+1 (g1,γ1 (x) + g2,γ1 (x)) (λα+λα+1) = L (x) (g1,γ1 (x) + g2,γ1 (x)) (λα+λα+1) = L (x) (Γ (v − γ1)) (λα+λα+1) { ∫ x x0 (x − t) v−γ1−1 (p (t)) − 1 p (p (t)) 1 p [ ∣ ∣ ( D v x0 f1 ) (t) ∣ ∣ + ∣ ∣ ( D v x0 f2 ) (t) ∣ ∣ ] dt }(λα+λα+1) (applying Hölder’s inequality with indices p p − 1 and p we find) ≥ L (x) (Γ (v − γ1)) (λα+λα+1) · ( ∫ x x0 (x − t) (v−γ1−1)p p−1 (p (t)) − 1 p−1 dt )( p−1 p )(λα+λα+1) ( ∫ x x0 p (t) [ ∣ ∣ ( D v x0 f1 ) (t) ∣ ∣ + ∣ ∣ ( D v x0 f2 ) (t) ∣ ∣ ]p dt ) ( λα+λα+1 p ) CUBO 10, 1 (2008) Converse Fractional Opial Inequalities ... 131 = T (x) · [ ∫ x x0 p (t) ( ∣ ∣ ( D v x0 f1 ) (t) ∣ ∣ + ∣ ∣ ( D v x0 f2 ) (t) ∣ ∣ )p dt ] ( λα+λα+1 p ) ≥ Φ (x) · [ ∫ x x0 p (t) ( ∣ ∣ ( D v x0 f1 ) (t) ∣ ∣ p + ∣ ∣ ( D v x0 f2 ) (t) ∣ ∣ p) dt ] ( λα+λα+1 p ) . We have proved (37) . � Next we treat the case of exponents λβ = λα + λv. Theorem 8. All here as in Theorem 5. Consider the special case of λβ = λα + λv. Assume here for j = 1, 2 that zj (x) := ∫ x x0 p (t) ∣ ∣ ( D v x0 fj ) (t) ∣ ∣ p dt ∈ [H, Ψ] , 0 < H < Ψ, h := Ψ H > 1, Mh (1) := (h − 1) h 1 h−1 e ln h . (43) Denote T̃ (x) := A0 (x) ( λv λα + λv ) λv p 2 p−2λα−3λv p (Mh (1)) −2(λα+λv)/p (44) Then ∫ x x0 q (w) [ ∣ ∣ ( D γ1 x0 f1 ) (w) ∣ ∣ λα ∣ ∣ ( D γ2 x0 f2 ) (w) ∣ ∣ λα+λv ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ λv + ∣ ∣ ( D γ2 x0 f1 ) (w) ∣ ∣ λα+λv ∣ ∣ ( D γ1 x0 f2 ) (w) ∣ ∣ λα ∣ ∣ ( D v x0 f2 ) (w) ∣ ∣ λv ] dw ≥ T̃ (x) ( ∫ x x0 p (w) ( ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ p + ∣ ∣ ( D v x0 f2 ) (w) ∣ ∣ p) dw )2( λα+λv p ) , (45) all x0 ≤ x ≤ b. Proof. We apply (21) and (22) for λβ = λα + λv and add to get ∫ x x0 q (w) [ ∣ ∣ ( D γ1 x0 f1 ) (w) ∣ ∣ λα ∣ ∣ ( D γ2 x0 f2 ) (w) ∣ ∣ λα+λv ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ λv 132 George A. Anastassiou CUBO 10, 1 (2008) + ∣ ∣ ( D γ2 x0 f1 ) (w) ∣ ∣ λα+λv ∣ ∣ ( D γ1 x0 f2 ) (w) ∣ ∣ λα ∣ ∣ ( D v x0 f2 ) (w) ∣ ∣ λv ] dw ≥ A0 (x) { [ ∫ x x0 (z1 (w)) λα λv (z2 (w)) λα λv +1 z ′ 1 (w) dw ] λv p + [ ∫ x x0 (z1 (w)) λα λv +1 (z2 (w)) λα λv z ′ 2 (w) dw ] λv p } (24) ≥ A0 (x) 2 1− λv p { ∫ x x0 [ (z1 (w)) λα λv (z2 (w)) λα λv +1 z ′ 1 (w) + (z1 (w)) λα λv +1 (z2 (w)) λα λv z ′ 2 (w) ] dw } λv p = A0 (x) 2 1− λv p { ∫ x x0 (z1 (w) z2 (w)) λα λv [z2 (w) z ′ 1 (w) + z1 (w) z ′ 2 (w)] dw} λv p = A0 (x) 2 1− λv p { ∫ x x0 (z1 (w) z2 (w)) λα λv (z1 (w) z2 (w)) ′ dw } λv p = A0 (x) 2 1− λv p ( (z1 (x) z2 (x)) λα λv +1 λα λv + 1 ) λv p = A0 (x) 2 p−λv p ( λv λα + λv ) λv p (z1 (x) z2 (x)) (λα+λv ) p (see [10]) ≥ A0 (x) 2 p−λv p ( λv λα + λv ) λv p ( z1 (x) + z2 (x) 2Mh (1) ) 2(λα+λv ) p = T̃ (x) (z1 (x) + z2 (x)) 2(λα+λv ) p = T̃ (x) ( ∫ x x0 p (w) ( ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ p + ∣ ∣ ( D v x0 f2 ) (w) ∣ ∣ p) dw )2( λα+λv p ) . CUBO 10, 1 (2008) Converse Fractional Opial Inequalities ... 133 We have established (45) . � Next follow special cases of the above theorems. Corollary 9. (to Theorem 5; λβ = 0, p (t) = q (t) = 1). Then ∫ x x0 [ ∣ ∣ ( D γ1 x0 f1 ) (w) ∣ ∣ λα ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ λv + ∣ ∣ ( D γ1 x0 f2 ) (w) ∣ ∣ λα ∣ ∣ ( D v x0 f2 ) (w) ∣ ∣ λv ] dw ≥ C1 (x) · ( ∫ x x0 ( ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ p + ∣ ∣ ( D v x0 f2 ) (w) ∣ ∣ p) dw )( λα+λv p ) , (46) all x0 ≤ x ≤ b, where C1 (x) := ( A0 (x) |λβ =0 ) · ( λv λα + λv ) λv p · δ1, (47) δ1 := 2 1−( λα+λv p ) (48) We have that ( A0 (x) |λβ =0 ) = {( (p − 1)( λαp−λα p ) (Γ (v − γ1)) λα (vp − γ1p − 1) ( λαp−λα p ) ) (49) · ( (p − λv) ( p−λv p ) (λαvp − λαγ1p − λα + p − λv) ( p−λv p ) )} · (x − x0) ( λαvp−λαγ1p−λα+p−λv p ) . Proof. By Theorem 5. The constant ( A0 (x) |λβ =0 ) was calculated in [4]. � Corollary 10. (to Theorem 6; λα = 0, p (t) = q (t) = 1, λβ ≥ λv). Then ∫ x x0 [ ∣ ∣ ( D γ2 x0 f2 ) (w) ∣ ∣ λβ ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ λv + ∣ ∣ ( D γ2 x0 f1 ) (w) ∣ ∣ λβ ∣ ∣ ( D v x0 f2 ) (w) ∣ ∣ λv ] dw ≥ C2 (x) ( ∫ x x0 [ ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ p + ∣ ∣ ( D v x0 f2 ) (w) ∣ ∣ p] dw ) ( λv +λβ p ) , (50) all x0 ≤ x ≤ b, where 134 George A. Anastassiou CUBO 10, 1 (2008) C2 (x) := (A0 (x) |λα=0) 2 p−λv p ( λv λβ + λv ) λv p δ λv p 3 . (51) We have that (A0 (x) |λα=0) =      (p − 1) ( λβ p−λβ p ) (Γ (v − γ2)) λβ (vp − γ2p − 1) ( λβ p−λβ p )   (52) · ( (p − λv) ( p−λv p ) (λβ vp − λβ γ2p − λβ + p − λv) ( p−λv p ) )} · (x − x0) ( λβ vp−λβ γ2p−λβ +p−λv p ) . Proof. By Theorem 6. The constant (A0 (x) |λα=0) was calculated in [4]. � 3.2 Results involving several functions Here we use the following basic inequality. Let α1, ..., αn ≥ 0, n ∈ N, then a r 1 + . . . + a r n ≤ (a1 + . . . + an) r ≤ nr−1 ( n ∑ i=1 a r i ) , r ≥ 1, (53) We present Theorem 11. Let γ1, γ2 ≥ 0 such that 1 ≤ v − γi < 1 p , 0 < p < 1, i = 1, 2, and fj ∈ C v x0 ([a, b]) with f (i) j (x0) = 0, i = 0, 1, . . . , n − 1, n := [v] , j = 1, . . . , M ∈ N. Here x, x0 ∈ [a, b] : x ≥ x0. We assume that D v x0 fj is of fixed sign on [x0, b] , j = 1, . . . , M. Consider also p (t) > 0, and q (t) > 0 continuous functions on [x0, b] . Let λv > 0 and λα, λβ ≥ 0 such that λv > p. Set Pk (w) := ∫ w x0 (w − t) (v−γk −1)p p−1 (p (t)) − 1 p−1 dt, k = 1, 2; x0 ≤ w ≤ b; (54) A (w) := q (w) (P1 (w)) λα( p−1 p ) (P2 (w)) λβ( p−1 p ) (p (w)) − λv p (Γ (v − γ1)) λα (Γ (v − γ2)) λβ ; (55) CUBO 10, 1 (2008) Converse Fractional Opial Inequalities ... 135 A0 (x) := ( ∫ x x0 (A (w)) p p−λv dw ) p−λv p . (56) Call ϕ1 (x) := ( A0 (x) |λβ =0 ) · ( λv λα + λv ) λv p , (57) δ ∗ 1 := M 1−( λα+λv p ). (58) If λβ = 0, we obtain that ∫ x x0 q (w)   M ∑ j=1 ∣ ∣ ( D γ1 x0 fj ) (w) ∣ ∣ λα ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ λv   dw ≥ δ∗1 · ϕ1 (x) ·   ∫ x x0 p (w)   M ∑ j=1 ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ p  dw   ( λα+λv p ) , (59) all x0 ≤ x ≤ b. Proof. By Theorem 5 we get ∫ x x0 q (w) [ ∣ ∣ ( D γ1 x0 fj ) (w) ∣ ∣ λα ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ λv + ∣ ∣ ( D γ1 x0 fj+1 ) (w) ∣ ∣ λα ∣ ∣ ( D v x0 fj+1 ) (w) ∣ ∣ λv ] dw (60) ≥ δ1ϕ1 (x) [ ∫ x x0 p (w) [ ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ p + ∣ ∣ ( D v x0 fj+1 ) (w) ∣ ∣ p] dw ]( λα+λv p ) , j = 1, 2, . . . , M − 1. Hence by adding all the above we find ∫ x x0 q (w)   M−1 ∑ j=1 [ ∣ ∣ ( D γ1 x0 fj ) (w) ∣ ∣ λα ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ λv + ∣ ∣ ( D γ1 x0 fj+1 ) (w) ∣ ∣ λα ∣ ∣ ( D v x0 fj+1 ) (w) ∣ ∣ λv ]) dw (61) 136 George A. Anastassiou CUBO 10, 1 (2008) ≥ δ1ϕ1 (x) ·   M−1 ∑ j=1 [ ∫ x x0 p (w) [ ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ p + ∣ ∣ ( D v x0 fj+1 ) (w) ∣ ∣ p] dw ]( λα+λv p )   . Also it holds ∫ x x0 q (w) [ ∣ ∣ ( D γ1 x0 f1 ) (w) ∣ ∣ λα ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ λv + ∣ ∣ ( D γ1 x0 fM ) (w) ∣ ∣ λα ∣ ∣ ( D v x0 fM ) (w) ∣ ∣ λv ] dw (62) ≥ δ1ϕ1 (x) [ ∫ x x0 p (w) [ ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ p + ∣ ∣ ( D v x0 fM ) (w) ∣ ∣ p] dw ]( λα+λv p ) . Adding (61) and (62) , and using (53) we have 2 ∫ x x0 q (w)   M ∑ j=1 ∣ ∣ ( D γ1 x0 fj ) (w) ∣ ∣ λα ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ λv  dw ≥ δ1ϕ1 (x)       M−1 ∑ j=1 [ ∫ x x0 p (w) [ ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ p (63) + ∣ ∣ ( D v x0 fj+1 ) (w) ∣ ∣ p] dw ]( λα+λv p ) } + { ∫ x x0 p (w) [ ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ p + ∣ ∣ ( D v x0 fM ) (w) ∣ ∣ p] dw }( λα+λv p ) } ≥ M 1−( λα+λv p )δ1ϕ1 (x)    ∫ x x0 p (w)  2 M ∑ j=1 ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ p  dw    ( λα+λv p ) . (64) We have proved ∫ x x0 q (w)   M ∑ j=1 ∣ ∣ ( D γ1 x0 fj ) (w) ∣ ∣ λα ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ λv   dw ≥ (65) M 1−( λα+λv p )δ1 ( 2 ( λα+λv p )−1 ) ϕ1 (x) ·    ∫ x x0 p (w)   M ∑ j=1 ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ p  dw    ( λα+λv p ) . CUBO 10, 1 (2008) Converse Fractional Opial Inequalities ... 137 That is proving (59) . � Next we give Theorem 12. All here as in Theorem 11. Assume λβ ≥ λv. Denote ϕ2 (x) := (A0 (x) |λα=0) 2 (p−λv ) p ( λv λβ + λv ) λv p δ λv p 3 . (66) If λα = 0, then ∫ x x0 q (w)       M−1 ∑ j=1 [ ∣ ∣ ( D γ2 x0 fj+1 ) (w) ∣ ∣ λβ ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ λv + ∣ ∣ ( D γ2 x0 fj ) (w) ∣ ∣ λβ ∣ ∣ ( D v x0 fj+1 ) (w) ∣ ∣ λv ]} + [ ∣ ∣ ( D γ2 x0 fM ) (w) ∣ ∣ λβ ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ λv + ∣ ∣ ( D γ2 x0 f1 ) (w) ∣ ∣ λβ ∣ ∣ ( D v x0 fM ) (w) ∣ ∣ λv ]} dw ≥ M 1− ( λv +λβ p ) 2 ( λv +λβ p ) ϕ2 (x) · { ∫ x x0 p (w)   M ∑ j=1 ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ p  dw    ( λv +λβ p ) , x ≥ x0. (67) Proof. From Theorem 6 we have ∫ x x0 q (w) [ ∣ ∣ ( D γ2 x0 fj+1 ) (w) ∣ ∣ λβ ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ λv + ∣ ∣ ( D γ2 x0 fj ) (w) ∣ ∣ λβ ∣ ∣ ( D v x0 fj+1 ) (w) ∣ ∣ λv ] dw ≥ ϕ2 (x) ( ∫ x x0 p (w) [ ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ p + ∣ ∣ ( D v x0 fj+1 ) (w) ∣ ∣ p] dw ) ( λv +λβ p ) , (68) for j = 1, . . . , M − 1. Hence by adding all of the above we get 138 George A. Anastassiou CUBO 10, 1 (2008) ∫ x x0 q (w)   M−1 ∑ j=1 [ ∣ ∣ ( D γ2 x0 fj+1 ) (w) ∣ ∣ λβ ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ λv + ∣ ∣ ( D γ2 x0 fj ) (w) ∣ ∣ λβ ∣ ∣ ( D v x0 fj+1 ) (w) ∣ ∣ λv ]) dw ≥ ϕ2 (x)    M−1 ∑ j=1 ( ∫ x x0 p (w) [ ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ p + ∣ ∣ ( D v x0 fj+1 ) (w) ∣ ∣ p] dw ) ( λv +λβ p )} . (69) Similarly it holds ∫ x x0 q (w) [ ∣ ∣ ( D γ2 x0 fM ) (w) ∣ ∣ λβ ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ λv + ∣ ∣ ( D γ2 x0 f1 ) (w) ∣ ∣ λβ ∣ ∣ ( D v x0 fM ) (w) ∣ ∣ λv ] dw ≥ ϕ2 (x) ( ∫ x x0 p (w) [ ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ p + ∣ ∣ ( D v x0 fM ) (w) ∣ ∣ p] dw ) ( λv +λβ p ) . (70) Adding (69) , (70) and using (53) we derive (67) . � We continue with Theorem 13. Let v ≥ 2 and γ1 ≥ 0 such that 2 ≤ v − γ1 < 1/p, 0 < p < 1. Let fj ∈ C v x0 ([a, b]) with f (i) j (x0) = 0, i = 0, 1, . . . , n − 1, n := [v] , j = 1, . . . , M ∈ N. Here x, x0 ∈ [a, b] : x ≥ x0.Assume that D v x0 fj is of fixed sign on [x0, b] , j = 1, . . . , M. Consider also p (t) > 0, and q (t) > 0 continuous functions on [x0, b] . Let λα ≥ λα+1 > 1, Φ is as in Theorem 7. Then ∫ x x0 q (w)       M−1 ∑ j=1 [ ∣ ∣ ( D γ1 x0 fj ) (w) ∣ ∣ λα ∣ ∣D γ1+1 x0 fj+1 (w) ∣ ∣ λα+1 + ∣ ∣ ( D γ1 x0 fj+1 ) (w) ∣ ∣ λα ∣ ∣D γ1+1 x0 fj (w) ∣ ∣ λα+1 ]} + [ ∣ ∣ ( D γ1 x0 f1 ) (w) ∣ ∣ λα ∣ ∣D γ1+1 x0 fM (w) ∣ ∣ λα+1 + ∣ ∣ ( D γ1 x0 fM ) (w) ∣ ∣ λα ∣ ∣D γ1+1 x0 f1 (w) ∣ ∣ λα+1 ]} dw ≥ CUBO 10, 1 (2008) Converse Fractional Opial Inequalities ... 139 M 1− ( λα+λα+1 p ) 2 ( λα+λα+1 p ) Φ (x)   ∫ x x0 p (w)   M ∑ j=1 ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ p  dw   ( λα+λα+1 p ) , (71) all x0 ≤ x ≤ b. Proof. From Theorem 7 we get ∫ x x0 q (w) M−1 ∑ j=1 [ ∣ ∣ ( D γ1 x0 fj ) (w) ∣ ∣ λα ∣ ∣D γ1+1 x0 fj+1 (w) ∣ ∣ λα+1 + ∣ ∣ ( D γ1 x0 fj+1 ) (w) ∣ ∣ λα ∣ ∣D γ1+1 x0 fj (w) ∣ ∣ λα+1 ] dw ≥ Φ (x) M−1 ∑ j=1 [ ∫ x x0 p (w) ( ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ p + ∣ ∣ ( D v x0 fj+1 ) (w) ∣ ∣ p) dw ] ( λα+λα+1 p ) (72) all x0 ≤ x ≤ b. Also it holds ∫ x x0 q (w) [ ∣ ∣ ( D γ1 x0 f1 ) (w) ∣ ∣ λα ∣ ∣D γ1+1 x0 fM (w) ∣ ∣ λα+1 + ∣ ∣ ( D γ1 x0 fM ) (w) ∣ ∣ λα ∣ ∣D γ1+1 x0 f1 (w) ∣ ∣ λα+1 ] dw ≥ Φ (x) [ ∫ x x0 p (w) ( ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ p + ∣ ∣ ( D v x0 fM ) (w) ∣ ∣ p) dw ] ( λα+λα+1 p ) , (73) all x0 ≤ x ≤ b. Adding (72) and (73) , along with (53) we derive (71) . � Next it comes Theorem 14. All here as in Theorem 11. Consider the special case of λβ = λα + λv. Here T̃ (x) as in (44) . Assume here for j = 1, . . . , M that zj (x) := ∫ x x0 p (t) ∣ ∣D v x0 fj (t) ∣ ∣ p dt ∈ [H, Ψ] , 0 < H < Ψ. Then 140 George A. Anastassiou CUBO 10, 1 (2008) ∫ x x0 q (w)       M−1 ∑ j=1 [ ∣ ∣ ( D γ1 x0 fj ) (w) ∣ ∣ λα ∣ ∣ ( D γ2 x0 fj+1 ) (w) ∣ ∣ λα+λv ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ λv + ∣ ∣ ( D γ2 x0 fj ) (w) ∣ ∣ λα+λv ∣ ∣ ( D γ1 x0 fj+1 ) (w) ∣ ∣ λα ∣ ∣ ( D v x0 fj+1 ) (w) ∣ ∣ λv ]} + [ ∣ ∣ ( D γ1 x0 f1 ) (w) ∣ ∣ λα ∣ ∣ ( D γ2 x0 fM ) (w) ∣ ∣ λα+λv ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ λv + ∣ ∣ ( D γ2 x0 f1 ) (w) ∣ ∣ λα+λv ∣ ∣ ( D γ1 x0 fM ) (w) ∣ ∣ λα ∣ ∣ ( D v x0 fM ) (w) ∣ ∣ λv ]} dw ≥ (74) M (1− 2(λα+λv ) p )2 2( λα+λv p )T̃ (x)   ∫ x x0 p (w)   M ∑ j=1 ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ p  dw   (2( λα+λv p )) , all x0 ≤ x ≤ b. Proof. Based on Theorem 8. The rest as in the proof of Theorem 13. � We continue with Corollary 15. (to Theorem 11, λβ = 0, p (t) = q (t) = 1). Then ∫ x x0   M ∑ j=1 ∣ ∣ ( D γ1 x0 fj ) (w) ∣ ∣ λα ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ λv   dw ≥ δ∗1 ϕ1 (x)   ∫ x x0   M ∑ j=1 ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ p  dw   ( λα+λv p ) (75) all x0 ≤ x ≤ b. In (75) , ( A0 (x) |λβ =0 ) of ϕ1 (x) is given by (49) . Proof. Based on Theorem 11. � Corollary 16. (to Theorem 12, λα = 0, p (t) = q (t) = 1). It holds CUBO 10, 1 (2008) Converse Fractional Opial Inequalities ... 141 ∫ x x0       M−1 ∑ j=1 [ ∣ ∣ ( D γ2 x0 fj+1 ) (w) ∣ ∣ λβ ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ λv + ∣ ∣ ( D γ2 x0 fj ) (w) ∣ ∣ λβ ∣ ∣ ( D v x0 fj+1 ) (w) ∣ ∣ λv ]} + [ ∣ ∣ ( D γ2 x0 fM ) (w) ∣ ∣ λβ ∣ ∣ ( D v x0 f1 ) (w) ∣ ∣ λv + ∣ ∣ ( D γ2 x0 f1 ) (w) ∣ ∣ λβ ∣ ∣ ( D v x0 fM ) (w) ∣ ∣ λv ]} dw ≥ ( M 1− ( λv +λβ p ) ) 2 ( λv +λβ p ) ϕ2 (x)    ∫ x x0   M ∑ j=1 ∣ ∣ ( D v x0 fj ) (w) ∣ ∣ p   dw    ( λv +λβ p ) , (76) all x0 ≤ x ≤ b. In (76) , (A0 (x) |λα=0) of ϕ2 (x) is given by (52) . Proof. Based on Theorem 12. � Received: October 2007. Revised: December 2007. References [1] R.P. Agarwal and P.Y.H. Pang, Opial Inequalities with Applications in Differential and Difference Equations, Kluwer Academic Publishers,Dordrecht, Boston, London, 1995. [2] G.A. 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Podlubny, Fractional Differential Equations, Academic Press, San Diego, 1999. [10] W. Specht, Zur Theorie der elementaren Mittel, Math. Z., 74 (1960), 91–98. [11] E.T. Whittaker and G.N. Watson, A Course in Modern Analysis, Cambridge University Press, 1927. gaConverseFractOpial_modified3.pdf