Strong right fractional calculus for Banach space valued functions

Transcription

1 Proyecciones Journal of Mathematics Vol. 36, N o, pp , March 207. Universidad Católica del Norte Antofagasta - Chile Strong right fractional calculus for Banach space valued functions George A. Anastassiou University of Memphis, U. S. A. Received : October 206. Accepted : November 206 Abstract We present here a strong right fractional calculus theory for Banach space valued functions of Caputo type. Then we establish many right fractional Bochner integral inequalities of various types. 200 AMS Subject Classification : 26A33, 26D0, 26D5, 46B25. Key Words and Phrases : Right Fractional derivative, Right Fractional Taylor s formula, Banach space valued functions, integral inequalities, Hausdorff measure, Bochner integral.

2 50 George A. Anastassiou. Introduction Here we use etensively the Bochner integral for Banach space valued functions, which is a direct generalization of Lebesgue integral to this case. The reader may read about Bochner integral and its properties from [2], [5], [6], [8], [0], [] and [2]. Using Bochner integral properties and the great article [2], we develop a right Caputo type strong fractional theory for the first time in the literature, which is the direct analog of the real one, but now dealing with Banach space valued functions. In the literature there are very few articles about the left weak fractional theory of Banach space valued functions with one of the best []. None eists about the right one. However we found the left weak theory, using Pettis integral and functionals, complicated, less clear, dificult and unnecessary. With this article we try to simplify matters and put the related right theory on its natural grounds and resemble the theory on real numbers. We define the right Riemann-Liouville fractional Bochner integral operator, see Definition 2, and we prove the right commutative semigroup property, see Theorem 7. We use the general Fundamental theorem of calculus for Bochner integration, see Theorem 0 here, from [2]. Based on the last we produce a related reverse general Taylor s formula for Banach valued functions. We introduce then the right Caputo type fractional derivative in our setting, see Definition 3. Then we are able to produce the related right fractional Taylor s formula in Banach space setting, which involves the Hausdorff measure. With this developed machinery we derive right fractional: Ostrowski type inequalities, Poincaré and Sobolev types, Opial type, Hilbert-Pachpatte type, and Landau type inequalities. All these right fractional inequalities for Banach space valued functions are using always the Hausdorff measure. We cover these inequalities to all possible directions, acting at the introductory basic level, which leaves big room for epansions later. 2. Main Results We mention

3 Strong right fractional calculus for Banach space valued functions 5 Definition. Let U R be an interval, and X is a Banach space, we denote by L (U, X) the Bochner integrable functions from U into X. We need Definition 2. Let α > 0, [a, b] R, X is a Banach space, and f L ([a, b],x). The Bochner integral operator I α (2.) b f () := (z ) α f (z) dz, [a, b], where Γ is the gamma function, is called the Riemann- Liouville right fractional Bochner integral operator of order α. For α =0,wesetIb 0 := I (the identity operator). We need Theorem 3. Let f L ([a, b],x), α>0. Then everywhere on [a, b], andib α f L ([a, b],x). ³I α b f () eists almost Proof. (2.2) Define k : Ω := [a, b] 2 R by k (z, ) =(z ) α +,thatis ( (z ) α, if a z b, k (z,) = 0, if a z b. R b z Then k is measurable on Ω, and we have R b a k (z, ) d = R z a k (z, ) d+ k (z,) d = Z z (z ) α (z a)α (2.3) d =. z α Let χ [a,b] () be the characteristic function, R. By [0], p. 0, Theorem 5.4, we get that f (z) χ [a,b] () =f (z) on[a, b] 2, is strongly (Bochner) measurable function on [a, b] 2.Clearlyk(z, ) isfinite a.e on [a, b] 2,andit is a real valued measurable function. By [7], p. 88, we get now that k(z, ) f (z) χ [a,b] () =k (z, ) f (z) on [a, b] 2 is strongly (Bochner) measurable function there. Net we work on the repeated integral R ³ b R b a a k (z,) kf (z)k d a kf (z)k dz = R b a kf (z)k ³ R b a k (z, ) d dz = (z a)α (b a)α dz α α a kf (z)k dz

4 52 George A. Anastassiou (2.4) (b a)α = kfk α L ([a,b],x) <. Therefore the function H : Ω X such that H (z.) :=k (z, ) f (z) is Bochner integrable over Ω by Tonelli s theorem, see [0], p. 00, Theorem 5.2. Hence by Fubini s theorem, [0], p. 93, Theorem 2, we obtain that R b a k (z,) f (z) dz is a Bochner integrable function on [a, b], as a function of [a, b]. That is ³I f b α () = R b Γ(α) (z )α f (z) dz is Bochner integrable on [a, b], and eists a.e. on [a, b]. 2 We further present and need Lemma 4. Let α and f L ([a, b],x), X a Banach space. Then Ib α f C ([a, b],x). Proof. (2.5) (i) Case of α =. Wehavethat ³ Ib f () = f (z) dz. Let, y [a, b] : y and y. We observe that ³ ³ Ib f () Ib f (y) = f (z) dz f (z) dz y ([5])= (2.6) Z Z f (z) dz f (z) dz f (z) dz y = f (z) dz y Z y kf (z)k dz = kf (z)k dz because R b kf (z)k dz is continuous in [a, b]. (ii) Case of α>. Let, y [a, b] : y and y. We observe that I α b f () I α b f (y) = y kf (z)k dz 0, (ζ ) α f (ζ) dζ (ζ y) α f (ζ) dζ y ([5])=

5 Strong right fractional calculus for Banach space valued functions 53 (2.7) (ζ ) α f (ζ) dζ Z y (ζ y) α f (ζ) dζ (ζ y) α f (ζ) dζ (see [2], p. 426, Theorem.43) h R b Γ(α) (ζ ) α (ζ y) α R i kf (ζ)k dζ + y (ζ y)α kf (ζ)k dζ h R b (ζ ) α (ζ y) α i kf (ζ)k dζ +( y) α kfk L ([a,b],x). Γ(α) As y we get (ζ ) α (ζ y) α,thus (ζ ) α (ζ y) α 0, and also (ζ ) α (ζ y) α 2(b a) α. Thus (ζ ) α (ζ y) α kf (ζ)k 2(b a) α kf (ζ)k L ([a, b],x), (2.8) and also (ζ ) α (ζ y) α kf (ζ)k 0as y, for almost all ζ [a, b]. Therefore by Dominated Convergence Theorem we conclude, as y, that (ζ ) α (ζ y) α kf (ζ)k dζ 0. Consequently, I α b f () I α b f (y) 0as y. Therefore I α b f C ([a, b],x). 2 We give

6 54 George A. Anastassiou Theorem 5. Here [a, b] R, X is a Banach space, F :[a, b] X. Let r>0, F L ([a, b],x), and the Bochner integral (2.9) G (s) := s (t s) r F (t) dt, all s [a, b]. ThenG AC ([a, b],x) (absolutely continuous functions) for r and G C ([a, b],x) for r (0, ). Proof. Denote by kf k := kf k L ([a,b],x) := t [a, b]es sup kf (t)k X < +. Hence F L ([a, b],x). By [7], p. 88, (t s) r F (t) is a strongly measurable function in t, t [s, b], s [a, b]. So that (t s) r F (t) L ([s, b],x), see [6]. () Case r. We use the definition of absolute continuity. So for every ε>0 we need δ>0: whenever (a i,b i ), i =,...,n, are disjoint subintervals of [a, b], then nx nx (2.0) (b i a i ) <δ kg (b i ) G (a i )k <ε. i= i= If kf k =0,thenG (s) =0,foralls [a, b], the trivial case and all fulfilled. So we assume kf k 6= 0. Hence we have (see [5]) G (b i ) G (a i )= b i (t b i ) r F (t) dt a i (t a i ) r F (t) dt = b i (t b i ) r F (t) dt i (see [2], p. 426, Theorem.43) a i (t a i ) r F (t) dt (t a i ) r F (t) dt = b i (2.) Call b i ³(t r Z b i ) r bi (t a i ) F (t) dt (t a i ) r F (t) dt. a i (2.2) I i := (t b i ) r (t a i ) r dt. b i

7 Strong right fractional calculus for Banach space valued functions 55 Thus kg (b i ) G (a i )k I i + (b i a i ) r (2.3) kf k r := T i. If r =,theni i =0,and (2.4) kg (b i ) G (a i )k kf k (b i a i ), for all i :=,...,n. h If r>, then because (t a i ) r (t b i ) r i 0 for all t [b i,b], we find I i = R ³ r b b i (t a i ) r (t b i ) dt = (b a i) r (b i a i ) r (b b i ) r r = r (b ξ)r (b i a i ) (b i a i ) r (2.5), for some ξ (a i,b i ). r Therefore, it holds (2.6) and (2.7) That is (2.8) so that I i r (b a)r (b i a i ) (b i a i ) r, r µi i + (b i a i ) r (b a) r (b i a i ). r T i kf k (b a) r (b i a i ), kg (b i ) G (a i )k kf k (b a) r (b i a i ), for all i =,...,n. So in the case of r =, and by choosing δ := ε kf k,weget à nx n! X (2.9) kg (b i ) G (a i )k kf k (b i a i ) kf k δ = ε, i= i= proving for r =thatg is absolutely continuous. In the case of r>, and ε by choosing δ :=,weget kf k (b a) r

8 56 George A. Anastassiou (2.20) Ã nx n! kg (b i ) G (a i )k kf k (b a) r X (b i a i ) i= i= kf k (b a) r δ = ε, proving for r > thatg is absolutely continuous again. (2)Caseof0<r<. Let a i,b i [a, b] :a i b i. Then (t a i ) r (t b i ) r, for all t [b i,b]. Then I i = R ³ r b b i (t b i ) r (t a i ) dt = (b b i ) r r µ (b ai ) r (b i a i ) r (b i a i ) r (2.2), r r by (b b i ) r (b a i ) r < 0. Therefore (2.22) and I i (b i a i ) r r T i 2(b i a i ) r (2.23) kf k r, proving that µ 2 kf k (2.24) kg (b i ) G (a i )k (b i a i ) r, r which proves that G is continuous. This completes the proof. 2 We make Remark 6. Let [a, b] R and (X, k k) a Banach space. Let also f : [a, b] X. If f is continuous, i.e. f C ([a, b],x), thenf is strongly measurable, by [8], Corollary 2.3, p. 5. Furthermore f ([a, b]) X is compact, thus it is closed and bounded, hence f is bounded, i.e. kf (t)k M, t [a, b], M>0. Let n, [a, b] : n, asn,thenf ( n ) f () in k k, that is kf ( n )k kf ()k kf ( n ) f ()k 0, provingkfk is continuous, hence bounded, so that kfk L ([a,b],x) := t [a, b]es sup kf (t)k < +, that is f L ([a, b],x), and hence f L ([a, b],x). Consequently, f is Bochner integrable ([2], p. 426), given that f is continuous.

9 Strong right fractional calculus for Banach space valued functions 57 For the last we used the fact: Z (2.25) kf (t)k dt kfk L ([a,b],x) (b a) < +, [a,b] proving that f L ([a, b],x). Also,clearly,absolutecontinuityoff :[a, b] X, implies uniform continuity and continuity of f. We also have Theorem 7. Let α, β 0, f L ([a, b],x). Then (2.26) Ib I α β b f = Iα+β b f = Iβ b Iα b f, valid almost everywhere on [a, b]. If additionally f C ([a, b],x) or α+β, thenwehaveidentitytrue on all of [a, b]. Proof. Since Ib 0 := I (the identity operator), if α =0orβ =0orboth are zero, then the statement of the theorem is trivially true. So we assume α, β > 0. We observe that (2.27) I α b I β b f () = Γ (β) Z Ã b (t ) α t! (τ t) β f (τ) dτ dt = Γ (β) χ [t,b] (τ)(t ) α (τ t) β f (τ) dτdt. The above integrals in (2.26) eist a.e. on [a, b]. So if Ib α Iβ b f (),Iα+β b f () eist we can apply Fubini s theorem, see Theorem 2, p. 93, [0], to interchange the order of integration and obtain I α b I β b f () = Γ (β) µz τ (t ) α (τ t) β f (τ) dt dτ =

10 58 George A. Anastassiou (2.28) Γ (β) µz τ f (τ) (τ t) β (t ) α dt dτ = Γ (β) f (τ) Γ (β) Γ (α + β) (τ )α+β dτ = That is Γ (α + β) f (τ)(τ ) (α+β) dτ = I α+β b f (). (2.29) I α b I β b f () =Iα+β f () true a.e. on [a, b]. By Theorem 5 and Remark 6, if f C ([a, b],x), then I β b f C ([a, b],x), therefore Ib α Iβ b af C ([a, b],x)andiα+β b f C ([a, b],x). Since in (29.) two continuous functions coincide a.e., the must be equal everywhere. b At last, if f L ([a, b],x)andα + β, we get I α+β b f C ([a, b],x) by Lemma l4. Hence, since I α+β b f () isdefined and eisting for any [a, b], by Fubini s theorem as before, equals to Ib α Iβ b f (), for all [a, b], proving the claim. 2 The algebraic version of previous theorem follows: ntheorem 8. The Bochner integral operators Ib α : L ([a, b],x) L ([a, b],x); α>0} make a commutative semigroup with respect to composition. The identity operator Ib 0 = I is the neutral element of this semigroup. We need Definition 9. (see [2]) A definition of the Hausdorff measure h α goes as follows: if (T,d) is a metric space, A T and δ>0, letλ (A, δ) be the set of all arbitrary collections (C) i of subsets of T, such that A i C i and diam (C i ) δ (diam =diameter) for every i. Now,foreveryα>0 define o (2.30) h δ α (A) :=infnx (diamci ) α (C i ) i Λ (A, δ). Then there eists δ 0limh δ α (A) = δ>0suph δ α (A), and h α (A) := δ 0limh δ α (A) gives an outer measure on the power set P (T ), whichis

11 Strong right fractional calculus for Banach space valued functions 59 countably additive on the σ-field of all Borel subsets of T.IfT = R n,then the Hausdorff measure h n, restricted to the σ-field of the Borel subsets of R n, equals the Lebesgue measure on R n up to a constant multiple. In particular, h (C) = µ (C) for every Borel set C R, where µ is the Lebesgue measure. We will use the following spectacular result Theorem 0. ([2]) (Fundamental Theorem of Calculus for Bochner integration) Suppose that for the given function f :[a, b] X, there eists F : [a, b] X, which is continuous, the derivative F 0 (t) eists and F 0 (t) = f (t) outside a µ-null Borel set B [a, b] such that h (F (B)) = 0. Then f is µ-measurable (i.e. strongly measurable), and if we assume the Bochner integrability of f, (2.3) F (b) F (a) = a f (t) dt. Notice here that the derivatives of a function f :[a, b] X, where X is a Banach space, are defined eactly as the numerical ones, see for definitions and properties, [], pp , and p. 93, that is they are strong derivatives. Notation. Let f L ([a, b],x). Wedenoteby (2.32) Z a b f (t) dt = a f (t) dt. We will use Theorem 0 to give a general Taylor s formula for Banach space valued functions with a Bochner integral remainder. Theorem 2. Let n N and f C n ([a, b],x), where[a, b] R and X is a Banach space. Set (2.33) (2.34) F () := n X i=0 (a ) i f (i) (), [a, b]. i! Assume that f (n) eists outside a µ-null Borel set B [a, b] such that h (F (B)) = 0. We further assume the Bochner integrability of f (n).then

12 60 George A. Anastassiou (2.35) f (a) = n X i=0 (a b) i f (i) (b)+ i! (n )! Z a b (a t) n f (n) (t) dt. Proof. We get that F C ([a, b],x). Notice that F (a) =f (a), and F (b) = n X i=0 (a b) i f (i) (b). i! Clearly F 0 eists outside of B. Infactitholds F 0 (a )n (2.36) () = f (n) (), [a, b] B. (n )! Also F 0 is Bochner integrable. By Theorem 0 now we get that (2.37) (2.38) That is, we have P n i=0 proving (2.35). 2 We give F (b) F (a) = Z a b a F 0 (t) dt. (a b) i i! f (i) (b) f (a) = R b (a t) n a (n )! f (n) (t) dt = (a t) n f (n) (t) dt, (n )! Definition 3. Let [a, b] R, X be a Banach space, α>0, m := dαe, (d e the ceiling of the number). We assume that f (m) L ([a, b],x), where f :[a, b] X. We call the Caputo-Bochner right fractional derivative of order α: (2.39) i.e. D α b f () :=( ) m I m α b f (m) (), (2.40) D α b f () := ( )m Γ (m α) (J ) m α f (m) (J) dj, [a, b].

13 Strong right fractional calculus for Banach space valued functions 6 We observe that Db m f () =( )m f (m) (), for m N, and Db 0 f () =f (). By Theorem 3 ³D f b α () eistsalmosteverywhereon[a, b] and ³D f b α L ([a, b],x). If f (m) L <, andα/ N, then by Theorem 5, ([a,b],x) Db α D f C ([a, b],x), hence α b f C ([a, b]). We make Remark 4. (to Definition 3) We notice that (by Theorem 7) I α b D α b f () =( ) m ³ I α b I m α b f (m) () = ³ (m) ³ (m) (2.4) ( ) m Ib α+m α f () =( ) m Ib f m (), almost everywhere on [a, b]. I.e. I α b Db f α ³ (m) (2.42) () =( ) m Ib f m (), for almost all [a, b]. Notice here that (2.43) ³I m b f (m) () = (m )! (z ) m f (m) (z) dz L ([a, b],x) and eists for almost all [a, b], by Theorem 3. We have proved, by (2.42), that (z ) α D b f α (z) dz = (2.44) for almost all [a, b]. (m )! Z b ( z) m f (m) (z) dz,

14 62 George A. Anastassiou We present the following right fractional Taylor s formula Theorem 5. Let [a, b] R, X be a Banach space, α>0, m = dαe, f C m ([a, b],x). Set (2.45) F (t) := m X i=0 ( t) i f (i) (t), t [, b], i! where [a, b]. Assume that f (m) eists outside a µ-null Borel set B [, b], such that (2.46) h (F (B )) = 0, where [a, b]. We also assume that f (m) L ([a, b],x). Then (2.47) f () = m X i=0 ( b) i i! f (i) (b)+ (z ) α D α b f (z) dz, for [a, b]. Proof. We use Theorem µ 2. Clearly it holds f ( ) P m i=0 ( b) i i! f (i) (b) C ([a, b],x), that is (by R (35) (m )! b ( t)m f (m) (t) dt C ([a, b],x) as a function of [a,b]. Hence (44) holds as an equality over [a, b] (by Tonelli s theorem), therefore Γ(α) R b (z )α ³ D α b f (z) dz C ([a, b],x), as a function of [a, b]. Now (2.47) is valid. 2 More generally we get Theorem 6. Let [a, b] R, X be a Banach space, α>0, m = dαe, f C m ([a, b],x). Set (2.48) F (t) := m X i=0 ( t) i f (i) (t), t [, b], i! where [a, b]. Assume that f (m) eists outside a µ-null Borel set B [, b], such that

15 Strong right fractional calculus for Banach space valued functions 63 (2.49) h (F (B )) = 0, [a, b]. We also assume that f (m) L ([a, b],x). Then (2.50) f () = m X i=0 ( b) i i! f (i) (b)+ (z ) α D α b f (z) dz, [a, b]. Proof. By Theorem 5. 2 Remark 7. (to Theorem 6) By (2.50), we have I α b Db f α () = (2.5) (z ) α D α b f (z) dz C ([a, b],x) as a function of [a, b]. We have also Corollary 8. (to Theorem 6) All as in Theorem 6. Assume that f (i) (b) = 0, i =0,,...,m. Then (2.52) [a, b]. f () = (z ) α D α b f (z) dz, Net we present Ostrowski type inequalities at right fractional level for Banach valued functions. See also [3]. Theorem 9. Let α>0, m = dαe. Here all as in Theorem t6. Assume f (k) (b) =0, k =,...,m, anddb α f L ([a, b],x). Then Z b D α f () d f (b) b a b f L ([a,b],x) (2.53) (b a) α. Γ (α +2) a

16 64 George A. Anastassiou Proof. Thus Let [a, b]. We have by (2.50) that f () f (b) = (J ) α D α b f (J) dj. kf () f (b)k = (J ) α D b f α (J) dj à Z! b (J ) α dj D b f α = L ([a,b],x) µ (J ) α b D α α b f L ([a,b],x) = Γ (α +) (b )α Db f α L ([a,b],x). (2.54) Therefore (2.55) (b )α kf () f (b)k Γ (α +) Db f α L ([a,b],x), [a, b]. Hence it holds Z b Z f () d f (b) b a = b (f () f (b)) d b a b a a a kf () f (b)k d Db α f L ([a,b],x) (b a) Γ (α +) a (2.56) = (b ) α d = a Z b (b ) α (b a) a Γ (α +) Db f α d = L ([a,b],x) Db α f L à ([a,b],x) (b a) Γ (α +) ( ) 0 D α b f L ([a,b],x) (b a) Γ (α +2) (b a)α+ = Db α f L à Ã!! ([a,b],x) (b ) α+ b a (b a) Γ (α +) α +! (b a)α+ = α + Db α f L ([a,b],x) Γ (α +2) (b a)α,

17 Strong right fractional calculus for Banach space valued functions 65 proving the claim. 2 We also give Theorem 20. Let α, m = dαe. Here all as in Theorem 6. Assume that f (k) (b) =0, k =,...,m, andd α b f L ([a, b],x). Then (2.57) Proof. Z b f () d f (b) b a a We have again kf () f (b)k Db α f L ([a,b],x) (b a) α. Γ (α +) (J ) α Db f α (J) dj (2.58) Hence (b )α Db f α (J) dj (b )α D α b f L ([a,b],x). (2.59) kf () f (b)k D α b f L ([a,b],x) (b ) α, [a, b]. Therefore Z b f () d f (b) b a a kf () f (b)k d b a a Z b Db α f L ([a,b],x) (2.60) (b ) α d = (b a) a D α b f L ([a,b],x) (b a) proving the claim. 2 We continue with = D (b ) α b α f L ([a,b],x) (b a) α d = a (b a) α Db α f L ([a,b],x) (b a) α, Γ (α +)

18 66 George A. Anastassiou Theorem 2. Let p, q > : p + q =, α> p, m = dαe. Here all as in Theorem 6. Assume that f (k) (b) =0, k =,...,m, and Db α f L q ([a, b],x). Then Z b f () d f (b) b a a (2.6) Db α f Lq ([a,b],x) ³ (b a) α + p. α + p (p (α ) + ) p Proof. We have again kf () f (b)k à Z! à b (J ) p(α ) p dj (J ) α D α b f (J) dj D b f α (J) q dj! q (2.62) (b ) (α )+ p (p (α ) + ) p à Db f α (J) q dj! q (b ) (α )+ p (p (α ) + ) p D α b f Lq([a,b],X). Therefore (2.63) kf () f (b)k D α b f Lq ([a,b],x) (p (α ) + ) p (b ) α + p, [a, b]. Hence Z b f () d f (b) b a a kf () f (b)k d b a a

19 Strong right fractional calculus for Banach space valued functions 67 (2.64) 2 Db α f Lq([a,b],X) (b ) α + (b a) (p(α ) + ) p d = p a Db α f Lq ([a,b],x) (b a) α + p ³. (p(α ) + ) p α + p Corollary 22. Let α > 2, m = dαe. All as in Theorem 6. Assume f (k) (b) =0, k =,...,m, Db α f L 2 ([a, b],x). Then Z b f () d f (b) b a a Db α f L2 ³ ([a,b],x) (b a) 2α ³ α α (2.65) We give Proposition 23. Inequality (2.53) is sharp; namely it is attained by (2.66) f () =(b ) α i, α>0, α/ N, [a, b], i X, such that i =. Proof. (see also [4], pp ) We see that f 0 () = α (b ) α i, f 00 () =( ) 2 α (α ) (b ) α 2 i,..., f (m ) () =( ) m α (α ) (α 2)... (α m +2)(b ) α m+ i, (2.67) and f (m) () =( ) m α (α ) (α 2)... (α m +2)(α m +)(b ) α m i.

20 68 George A. Anastassiou Here f (m) is continuous on [a, b), and f (m) L ([a, b],x). All assumptions of Theorem 6 are easily fulfilled. Thus D α b f () = i ( ) 2m Γ (m α) α (α )... (α m +) (J ) m α (b J) α m dj = iα(α )... (α m +) Γ (m α) (b J) (α m+) (J ) (m α) dj (2.68) = iα(α )... (α m +) Γ (m α) Γ (α m +)Γ (m α) Γ () (2.69) That is = iα(α )... (α m +)Γ (α m +)=Γ (α +) i. D α b f () =Γ (α +) i, [a, b]. Also we see that f (k) (b) =0,k =0,,...,m, and D α b f L ([a, b],x). So f fulfills all assumptions of Theorem 9. Net we see (2.70) R.H.S. (2.53) = Γ (α +) Γ (α +2) (b (b a)α a)α = (α +). L.H.S. (2.53) = b a a (b ) α d = (b a) α+ b a (α +) = (b a)α (α +), (2.7) proving attainability and sharpness of (2.53). 2 We continue with a Poincaré like right fractional inequality:

21 Strong right fractional calculus for Banach space valued functions 69 Theorem 24. Let p, q > : p + q =,andα> q, m = dαe. Here all as in Theorem 6. Assume that f (k) (b) =0, k =0,,...,m, anddb α f L q ([a, b],x), where X is a Banach space. Then (b a) α D α kfk b f Lq ([a,b],x) (2.72) Lq([a,b],X). (p(α ) + ) p (qα) q Proof. We have that (by (2.52)) (2.73) f () = Hence kf ()k = (z ) α D α b f (z) dz, [a, b]. (z ) α D b f α (z) dz (z ) α D α b f (z) dz (2.74) Ã Z! Ã b (z ) p(α ) p dz Db f α (z) q dz! q (b ) p(α )+ p (p (α ) + ) p We have proved that D α b f Lq([a,b],X). (2.75) kf ()k (b ) p(α )+ p (p (α ) + ) p D b f α Lq ([a,b],x), [a, b]. Then (2.76) kf ()k q (b ) (p(α )+) q p () q (p (α ) + ) q p D α b f q L q ([a,b],x),

22 70 George A. Anastassiou [a, b]. Hence it holds (2.77) a kf ()k q d (b a) qα () q (p (α ) + ) q p qα D α b f q L q ([a,b],x). (2.78) The last inequality implies à kf ()k q d a! q (b a) α D α b f Lq ([a,b],x), (p(α ) + ) p (qα) q proving the claim. 2 Net comes a right Sobolev like fractional inequality: Theorem 25. All as in the last Theorem 24. Let r>0. Then (2.79) (b a) α q + r kfk Lr([a,b],X) (p(α ) + ) p D α b f Lq ([a,b],x) ³ ³ r α r q +. Proof. As in the last theorem s proof we get that (2.80) kf ()k (b ) α q (p (α ) + ) p D b f α Lq, [a, b]. ([a,b],x) Since r>0, we get kf ()k r (b ) r α q () r (p (α ) + ) r p D b f α r L q, [a, b]. ([a,b],x) (2.8) Hence it holds

23 Strong right fractional calculus for Banach space valued functions 7 a kf ()k r d (2.82) That is (b a) r α q + ³ ³ D b f α r () r (p (α ) + ) r p r α L q +. q([a,b],x) à kf ()k r d a! r (b a) α q + r ³ (p(α ) + ) p r ³ α q r + (2.83) proving the claim. 2 We give the following Opial type right fractional inequality: D α b f Lq([a,b],X), Theorem 26. Let p, q > : p + q =, and α > q, m := dαe. Let [a, b] R, X a Banach space, and f C m ([a, b],x). Set (2.84) F (t) := m X i=0 ( t) i f (i) (t), t [, b], where [a, b]. i! (2.85) Assume that f (m) eists outside a µ-null Borel set B [, b], such that h (F (B )) = 0, [a, b]. We assume that f (m) L ([a, b],x). Assume also that f (k) (b) =0, k =0,,...,m. Then kf (w)k D α b f (w) dw (b ) α + 2 p 2 q ((p (α ) + ) (p (α ) + 2)) p à Db f α (z) q dz! 2 q, (2.86) [a, b].

24 72 George A. Anastassiou Proof. By (2.52) we get (2.87) f () = (z ) α D α b f (z) dz, [a, b]. Let w b, thenwehave (2.88) f (w) = Furthermore it holds kf (w)k w w (z w) α D α b f (z) dz. (z w) α D α b f (z) dz (2.89) à Z! à b (z w) p(α ) p dz w w Db f α (z) q dz! q = (b w) (p(α )+) p (p (α ) + ) p à Db f α (z) q dz w! q = where (b w) α q (p (α ) + ) p (z (w)) q, (2.90) all w b, z (b) =0. Thus z (w) := w D α b f (z) q dz, (2.9) and (2.92) and (2.93) z (w) := Z w b D α b f (z) q dz, ( z (w)) 0 = D α b f (w) q 0, Db f α (w) ³ 0 q ³ 0 q = ( z (w)) = (z (w)). Therefore we obtain

25 Strong right fractional calculus for Banach space valued functions 73 kf (w)k D α b f (w) (2.94) (b w) α q (p (α ) + ) p ³ ³ 0 q z (w) (z (w)), all w b. Integrating (2.94) we get kf (w)k D α b f (w) dw (p (α ) + ) p (b w) α q z (w) z 0 (w) q dw (p(α ) + ) p (2.95) Ã! à (b w) α p q p dw z (w) z 0 (w) dw! q = (p(α ) + ) p (b ) α + 2 p (p (α ) + 2) p (z ()) 2 q 2 q = (b ) α + 2 p 2 q [(p (α ) + ) (p (α ) + 2)] p à Db f α (z) q dz! 2 q, (2.96) [a, b], proving the claim. 2 Net we present a Hilbert-Pachpatte right fractional inequality:

26 74 George A. Anastassiou Theorem 27. Let p, q > : p + q =,andα > q, α 2 > p, m i := dα i e, i =, 2. Here [a i,b i ] R, i =, 2; X is a Banach space. Let f i C m i ([a i,b i ],X), i =, 2. Set (2.97) F i (t i ):= m i X j i =0 ( i t i ) j i j i! f (j i) i (t i ), t i [ i,b i ],where i [a i,b i ]; i =, 2. Assume that f (m i) i a µ-null Borel set B i [ i,b i ], such that eists outside (2.98) h (F i (B i )) = 0, i [a i,b i ]; i =, 2. We also assume that f (m i) i L ([a i,b i ],X), and (2.99) and f (k i) i (b i )=0, k i =0,,...,,m i ; i =, 2, (2.00) ³ D a b f L q ([a,b ],X), ³ D α 2 b 2 f 2 L p ([a 2,b 2 ],X). Then 2 a a 2 (b ) p(α )+ µ kf ( )kkf 2 ( 2 )k d d 2 p(p(α )+) q(α 2 )+ + (b 2 2 ) q(q(α 2 )+) (b a )(b 2 a 2 ) Γ (α ) Γ (α 2 ) D α b f Lq([a,b ],X) D α 2 b 2 f 2. Lp([a 2,b 2 ],X) (2.0) Proof. We have that (by (2.52)) f i ( i )= Γ (α i ) (2.02) i i ³ (z i i ) α i D α i b i f i (z i ) dz i, i [a i,b i ], i =, 2.

27 Strong right fractional calculus for Banach space valued functions 75 Then kf i ( i )k i (z i i ) α i ³ D α i b Γ (α i ) i f (2.03) i (z i ) dz i, i i [a i,b i ],i=, 2. We get as before, (2.04) and kf ( )k (b ) p(α )+ p Γ (α ) (p (α ) + ) p D α b f, Lq ([a,b ],X) (2.05) kf 2 ( 2 )k (b 2 2 ) q(α 2 )+ q Γ (α 2 ) (q (α 2 ) + ) q Hence we have D α 2 b 2 f 2. Lp ([a 2,b 2 ],X) kf ( )kkf 2 ( 2 )k Γ (α ) Γ (α 2 )(p(α ) + ) p (q (α 2 ) + ) q (b ) p(α )+ p (b 2 2 ) q(α 2 )+ q D α b f Lq ([a,b ],X) (2.06) (using Young s inequality for a, b 0, a p b q a p + b q ) (2.07) D α 2 b 2 f 2 Lp ([a 2,b 2 ],X) Ã (b ) p(α )+ Γ (α ) Γ (α 2 ) p (p (α ) + ) + (b 2 2 ) q(α2 )+! q (q (α 2 ) + ) D α b f D α 2 Lq ([a,b ],X) b 2 f 2, Lp ([a 2,b 2 ],X) i [a i,b i ]; i =, 2. So far we have (2.08) kf ( )kkf 2 ( 2 )k µ (b ) p(α )+ p(p(α )+) + (b q(α 2 2 ) 2 )+ q(q(α 2 )+) b f D α 2 Lq([a,b ],X) b 2 f 2 Lp([a 2,b 2 ],X) D α Γ (α ) Γ (α 2 ),

28 76 George A. Anastassiou i [a i,b i ]; i =, 2. The denominator in (2.08) can be zero only when = b and 2 = b 2. Integrating (2.08) over [a,b ] [a 2,b 2 ] we derive inequality (2.0). 2 When 0 <α, Definition 3 becomes Definition 28. Let [a, b] R, X be a Banach space, 0 <α. We assume that f 0 L ([a, b],x), wheref :[a, b] X. We define the Caputo-Bochner right fractional derivative of order α: (2.09) i.e. D α b f () := I α b f 0 (), D α (2.0) b f () = Γ ( α) Clearly D b f () = f 0 (). (J ) α f 0 (J) dj, [a, b]. Remark 29. Let [A, B] R, X be a Banach space, 0 < α, f : [A, B] X. We assume that f (2) L ([A, B],X). Thendα +e =2,and (2.) D α B f 0 () = Z B (J ) α f 00 (J) dj = Γ ( α) - i.e. (2.2) and (2.3) R ( ) 2 B Γ(2 (α+)) (J )2 (α+) f 00 (J)dJ= D α+ B f(), D α B f 0 () = D α+ B f (), DB f α 0 () = DB α+ f (), [A, B]. We apply Theorem 9 when 0 <α. Theorem 30. Let 0 < α, [A, B] R, X a Banach space, f C ([A, B],X). Assume that f 0 eists outside a µ-null Borel set B [, B], such that h (f (B )) = 0, [A, B].

29 Strong right fractional calculus for Banach space valued functions 77 We assume that f 0 Then (2.4) Z B f () d f (B) B A A L ([A, B],X), and D α B f L ([A, B],X). DB α f L ([A,B],X) (B A) α. Γ (α +2) We present the following right Caputo-Bochner fractional Landau inequality for k k. Theorem 3. Let f C ((,B 0 ],X), whereb 0 R is fied, 0 <α, X is a Banach space. For any A, B (,B 0 ]:A B, we assume that f fulfills: assume that f 00 eists outside a µ-null Borel set B [, B], such that (2.5) h f 0 (B ) =0, [A, B]. We assume that f 00 L ([A, B],X), anddb α+ f L ([A, B],X). We further assume that (2.6) D α+ B f L ((,B],X) D α+ B 0 f L ((,B 0 ],X) <, B B 0. (the last left inequality is obvious when α =), and (2.7) Then kfk,(,b0 ] := t (,B 0]sup kf (t)k <. f 0,(,B0 ] := t (,B 0]sup f 0 (t) µ α 2 ( α+) α (α +) (Γ (α +2)) (α+) (2.8) ³ α µ D (α+) α+ kfk,(,b0 ] B 0 f (α+). L ((,B 0 ],X)

30 78 George A. Anastassiou Proof. We have that (by Theorem 30) (2.9) Z B f 0 () d f 0 (B) B A A DB α+ f L ([A,B],X) (B A) α, Γ (α +2) A, B (,B 0 ]:A B. Subsequently by Theorem 0 we derive f (B) f (A) (2.20) B A A, B (,B 0 ], A B. Hence it holds f 0 (B) f 0 (B) kf (B) f (A)k B A (2.2) DB α+ f L ([A,B],X) (B A) α, Γ (α +2) DB α+ f L ([A,B],X) (B A) α, Γ (α +2) and f 0 (B) kf (B) f (A)k DB α+ f L ([A,B],X) + (B A) α, B A Γ (α +2) (2.22) A, B (,B 0 ]:A B. Therefore we obtain f 0 (B) 2 kfk,(,b 0 ] + B A DB α+ 0 f L ((,B 0 ],X) (B A) α, Γ (α +2)

31 Strong right fractional calculus for Banach space valued functions 79 (2.23) A, B (,B 0 ]:A B. The right hand side of (2.23) depends only on B A. Consequently, it holds f 0,(,B0 ] 2 kfk D,(,B 0 ] B α+ 0 f L ((,B + 0 ],X) (B A) α. B A Γ (α +2) (2.24) We may call t = B A>0. Thus by (2.24), f 0,(,B0 ] 2 kfk,(,b 0 ] + t D α+ B 0 f L ((,B 0 ],X) Γ (α +2) t α, t>0. (2.25) Set µ := 2 kfk,(,b0 ], DB α+ 0 f L ((,B θ := 0 ],X) (2.26), Γ (α +2) both are greater than 0. We consider the function (2.27) y (t) = µ t + θtα, 0 <α, t>0. As in [4], pp. 8-82, y has a global minimum at (2.28) t 0 = µ µ αθ (α+), which is

32 80 George A. Anastassiou (2.29) y (t 0 )=(θµ α ) (α+) (α +)α ( α Consequently it is y (t 0 )= D α+ B 0 f L ((,B 0 ],X) Γ (α +2) (2.30) We have proved that kf 0 k,(,b0 ] (α +) ³ 2 α α+). (α+) ³ 2 kfk,(,b0] ( α α+) (Γ (α +2)) (α+) ( α α+) (α +)α ( α α+). (2.3) µ D ³ α+ B 0 f L (α+) kfk α+) ((,B 0 ],X),(,B0 ] ( α, establishing the claim. 2 The case B 0 =0comesnet Corollary 32. (to Theorem 3) All as in Theorem 3 for B 0 =0. Then kf 0 k,r (α +)³ ( 2 α+) α α (Γ (α +2)) (α+) (2.32) ³kfk α+) µ,r ( α D α+ 0 f L (α+). (R,X) When α =weget Corollary 33. (to Theorem 3) Let f C ((,B 0 ],X), whereb 0 R is fied, X is a Banach space. For any A, B (,B 0 ]:A B, weassume that f fulfills: assume that f 00 eists outside a µ-null Borel set B [, B], such that (2.33) h f 0 (B ) =0, [A, B]. We assume that f 00 L ((,B 0 ],X), andkfk,(,b0 ] <. Then (2.34) f 0,(,B0 ] 2 kfk 2,(,B0 ] f 00 2 L ((,B 0 ],X). Net case of B 0 =0. Corollary 34. All as in Corollary 33. It holds (2.35) f 0,R 2 kfk 2,R f 00 2 L (R,X).

33 Strong right fractional calculus for Banach space valued functions 8 See also [9]. We apply Theorem 2 when 0 <α. Theorem 35. Let p, q > : p + q =, q <α. Let[A, B] R, X be a Banach space, f C ([A, B],X). Assume that f 0 eists outside a µ-null Borel set B [, B], such that (2.36) h (f (B )) = 0, [A, B]. We also assume that f 0 L ([A, B],X). Assume D α B f L q ([A, B],X). Then Z B f () d f (B) B A A (2.37) DB α f Lq([A,B],X) ³ (B A) α q. α + p (p (α ) + ) p We present the following right Caputo-Bochner fractional L q Landau inequality. Theorem 36. Let p, q > : p + q =, and q < α. Let f C ((,B 0 ],X), where B 0 R is fied, X is a Banach space. For any A, B (,B 0 ]:A B, we suppose that f fulfills: assume that f 00 eists outside a µ-null Borel set B [, B], such that (2.38) h f 0 (B ) =0, [A, B]. We also assume that f 00 L ([A, B],X) and DB α+ f L q ([A, B],X). We further assume that (2.39) D α+ B f Lq((,B],X) D α+ B 0 f Lq((,B 0 ],X) <, B B 0, (the last left inequality is obvious when α =), and (2.40) kfk,(,b0 ] <.

34 82 George A. Anastassiou Then kf 0 k,(,b0 ] µ 2 α+ p α q µ α q α+ p (Γ(α)) (α+ p) (p(α )+) (pα+) µ α q ³ µ α+ D p kfk α+,(,b0 ] B 0 f Lq (α+ p).(2.4) ((,B 0 ],X) Proof. ByTheorem35wehavethat Z B D α+ f 0 () d f 0 (B) B A A B f Lq ([A,B],X) ³ (B A) α (p(α ) + ) q, p α + p (2.42) A, B (,B 0 ]:A B. From Theorem 0 we derive f (B) f (A) f 0 (B) B A (2.43) A, B (,B 0 ]:A B. Therefore we obtain f 0 (B) 2 kfk,(,b 0 ] + B A (2.44) A, B (,B 0 ]:A B. D α+ B f Lq([A,B],X) (p (α ) + ) p D α+ B 0 f Lq ((,B 0 ],X) (p (α ) + ) p ³ (B A) α q, α + p ³ (B A) α q, α + p The R.H.S. of (2.44) depends only on B A. Therefore f 0,(,B0 ] 2 kfk D,(,B 0 ] B α+ 0 f Lq((,B + 0 ],X) ³ B A (p(α ) + ) p α + p (2.45) We may call t = B A>0. Thus f 0,(,B0 ] 2 kfk,(,b 0 ] + t D α+ B 0 f Lq((,B 0 ],X) (p (α ) + ) p (B A) α q. ³ t α q, α + p

35 Strong right fractional calculus for Banach space valued functions 83 (2.46) t (0, ). Notice that 0 <α q <. Call (2.47) (2.48) eθ := both are positive, and eµ := 2 kfk,(,b0 ], D α+ B 0 f Lq ((,B 0 ],X) (p (α ) + ) p ³, α + p (2.49) eν := α q (0, ). We consider the function (2.50) ey (t) = eµ t + e θt eα, t (0, ). The only critical number here is µ eµ (2.5) et 0 = eα+ eα θ e, and ey has a global minimum at et 0,whichis (2.52) ³ ³ ey t e 0 eα ³ = θeµ e eα+ (eα +)eα eα eα+. Consequently, we derive ³ D ey t e 0 α+ B 0 f Lq((,B = 0 ],X) (2.53) Γ(α)(p(α )+) p α+ p µ ³ 2 kfk,(,b0 ] We have proved that kf 0 k,(,b0 ] µ 2 α+ p α q α q α+ p µ α q α+ p (α+ p) µ µ α + µ α α q α+ p. p q (Γ(α)) (α+ p) (p(α )+) (pα+)

36 84 George A. Anastassiou (2.54) µ α ³ q µ α+ D p kfk α+,(,b0 ] B 0 f Lq (α+ p), ((,B 0 ],X) establishing the claim. 2 Corollary 37. (to Theorem 36, µ B 0 =0) All as in Theorem 36. α µ q Then kf 0 2 α+ α+ p p k,r (2.55) kf 0 k,r α q (Γ(α)) (α+ p) (p(α )+) (pα+) µ α ³kfk,R q µ α+ D p α+ 0 f (α+ p) Lq(R.,X) Corollary 38. (to Theorem 36, B 0 =0, p = q =2) All as in Theorem 36, 2 <α. Then µ ³ α 2 2(α+ α+ 2) 2 (2.56) α 2 (Γ(α)) (α+ 2) (2α ) (2α+) α ³kfk,R ³ 2 µd α+ 2 α+ 0 f L2 (R,X) Case of α =follows: (α+ 2). Corollary 39. Let p, q > : p + q =, f C ((,B 0 ],X), where B 0 R is fied, X is a Banach space. For any A, B (,B 0 ]:A B, we suppose that f fulfills: assume that f 00 eists outside a µ-null Borel set B [, B], such that (2.57) h f 0 (B ) =0, [A, B]. We further assume that f 00 L q ((,B 0 ],X), and kfk,(,b0 ] <. Then µ ³+ f 0,(,B0 ] 2 p q q + p

37 Strong right fractional calculus for Banach space valued functions 85 (2.58) µ ³ q ³ + p kfk,(,b0 ] f 00 Lq((,B0],X) (+ p). Corollary 40. (to Corollary 39) Assume B 0 =0.Then We finish article with ³+ f 0,R 2 ( p q Corollary 4. (to Corollaries 39, 40) Assume B 0 =0and p = q =2. Then (2.60) f 0,R ³kfk,R ³ f L2 (R,X). Note 42. Many variations and generalizations of the above inequalities are possible, however due to lack of space we stop here. References [] R.P. Agarwal, V. Lupulescu, D. O Regan, G. Rahman, Multi-term fractional differential equations in a nonrefleive Banach space, Advances in Difference Equation, (203), 203:302. [2] C.D. Aliprantis and K.C. Border, Infinite Dimensional Analysis, Springer, New York, (2006). [3] G. Anastassiou, Fractional Differentiation Inequalities, Springer, New York, (2009). [4] G. Anastassiou, Advances on fractional inequalities, Springer, New York, (20). [5] Appendi F, The Bochner integral and vector-valued L p -spaces,

38 86 George A. Anastassiou [6] Bochner integral. Encyclopedia of Mathematics. URL: inde.php?title=bochner integral&oldid= [7] R.F. Curtain and A.J. Pritchard, Functional Analysis in Modern Applied Mathematics, Academic Press, London, New York, (977). [8]M.Kreuter,Sobolev Spaces of Vector-valued functions, Ulm Univ., Master Thesis in Math., Ulm, Germany, (205). [9] E. Landau, Einige Ungleichungen für zweimal differentzierban funktionen, Proc. London Math. Soc. 3, pp , (93). [0] J. Mikusinski, The Bochner integral, Academic Press, New York, (978). [] G.E. Shilov, Elementary Functional Analysis, Dover Publications, Inc., New York, (996). [2] C. Volintiru, A proof of the fundamental theorem of Calculus using Hausdorff measures, Real Analysis Echange, 26 (), 2000/200, pp George A. Anastassiou Department of Mathematical Sciences University of Memphis Memphis, TN 3852, U.S.A. ganastss@memphis.edu