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Strong right fractional calculus for Banach spacevalued functions

George A. AnastassiouUniversity of Memphis, U. S. A.

Received : October 2016. Accepted : November 2016

Proyecciones Journal of MathematicsVol. 36, No 1, pp. 149-186, March 2017.Universidad Catolica del NorteAntofagasta - Chile

Abstract

We present here a strong right fractional calculus theory for Ba-nach space valued functions of Caputo type. Then we establish manyright fractional Bochner integral inequalities of various types.

2010 AMS Subject Classification : 26A33, 26D10, 26D15, 46B25.

Key Words and Phrases : Right Fractional derivative, Right Frac-tional Taylor’s formula, Banach space valued functions, integral in-equalities, Hausdorff measure, Bochner integral.

150 George A. Anastassiou

1. Introduction

Here we use extensively the Bochner integral for Banach space valued func-tions, which is a direct generalization of Lebesgue integral to this case. Thereader may read about Bochner integral and its properties from [2], [5], [6],[8], [10], [11] and [12].

Using Bochner integral properties and the great article [12], we develop aright 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 Banachspace valued functions.

In the literature there are very few articles about the left weak fractionaltheory of Banach space valued functions with one of the best [1]. Noneexists about the right one.

However we found the left weak theory, using Pettis integral and func-tionals, complicated, less clear, dificult and unnecessary.

With this article we try to simplify matters and put the related righttheory on its natural grounds and resemble the theory on real numbers.

We define the right Riemann-Liouville fractional Bochner integral op-erator, see Definition 2, and we prove the right commutative semigroupproperty, see Theorem 7.

We use the general Fundamental theorem of calculus for Bochner inte-gration, see Theorem 10 here, from [12].

Based on the last we produce a related reverse general Taylor’s formulafor Banach valued functions.

We introduce then the right Caputo type fractional derivative in oursetting, see Definition 13. Then we are able to produce the related rightfractional Taylor’s formula in Banach space setting, which involves theHausdorff measure. With this developed machinery we derive right frac-tional: Ostrowski type inequalities, Poincare and Sobolev types, Opial type,Hilbert-Pachpatte type, and Landau type inequalities.

All these right fractional inequalities for Banach space valued functionsare using always the Hausdorff measure. We cover these inequalities to allpossible directions, acting at the introductory basic level, which leaves bigroom for expansions later.

2. Main Results

We mention

Strong right fractional calculus for Banach space valued functions 151

Definition 1. Let U ⊆ R be an interval, and X is a Banach space, wedenote by L1 (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 ∈L1 ([a, b] ,X). The Bochner integral operator

¡Iαb−f

¢(x) :=

1

Γ (α)

Z b

x(z − x)α−1 f (z) dz,(2.1)

∀ x ∈ [a, b] , where Γ is the gamma function, is called the Riemann-Liouville right fractional Bochner integral operator of order α.

For α = 0, we set I0b− := I (the identity operator).

We need

Theorem 3. Let f ∈ L1 ([a, b] ,X), α > 0. Then³Iαb−f

´(x) exists almost

everywhere on [a, b], and Iαb−f ∈ L1 ([a, b] ,X) .

Proof. Define k : Ω := [a, b]2 → R by k (z, x) = (z − x)α−1+ , that is

k (z, x) =

((z − x)α−1 , if a ≤ x ≤ z ≤ b,0, if a ≤ z ≤ x ≤ b.

(2.2)

Then k is measurable on Ω, and we haveR ba k (z, x) dx =

R za k (z, x) dx+R b

z k (z, x) dx = Z z

z(z − x)α−1 dx =

(z − a)α

α.(2.3)

Let χ[a,b] (x) be the characteristic function, x ∈ R. By [10], p. 101, The-orem 5.4, we get that f (z)χ[a,b] (x) = f (z) on [a, b]2, is strongly (Bochner)

measurable function on [a, b]2. Clearly k (z, x) is finite a.e on [a, b]2, and itis a real valued measurable function. By [7], p. 88, we get now that

k(z, x) f (z)χ[a,b] (x) = k (z, x) f (z) on [a, b]2

is strongly (Bochner) measurable function there.Next we work on the repeated integralR ba

³R ba k (z, x) kf (z)k dx

´dz =

R ba kf (z)k

³R ba k (z, x) dx

´dz =Z b

akf (z)k (z − a)α

αdz ≤ (b− a)α

α

Z b

akf (z)k dz


=(b− a)α

αkfkL1([a,b],X) <∞.(2.4)

Therefore the function H : Ω→ X such that H (z.x) := k (z, x) f (z) isBochner integrable over Ω by Tonelli’s theorem, see [10], p. 100, Theorem5.2.

Hence by Fubini’s theorem, [10], p. 93, Theorem 2, we obtain thatR ba k (z, x) f (z) dz is a Bochner integrable function on [a, b], as a function

of x ∈ [a, b]. That is³Iαb−f

´(x) = 1

Γ(α)

R bx (z − x)α−1 f (z) dz is Bochner

integrable on [a, b], and exists a.e. on [a, b] . 2We further present and need

Lemma 4. Let α ≥ 1 and f ∈ L1 ([a, b] ,X), X a Banach space. ThenIαb−f ∈ C ([a, b] ,X) .

Proof. (i) Case of α = 1. We have that³I1b−f

´(x) =

Z b

xf (z) dz.(2.5)

Let x, y ∈ [a, b] : x ≥ y and x→ y. We observe that

°°°³I1b−f´ (x)− ³I1b−f´ (y)°°° =°°°°°Z b

xf (z) dz −

Z b

yf (z) dz

°°°°° ([5])=°°°°°Z b

xf (z) dz −

Z x

yf (z) dz −

Z b

xf (z) dz

°°°°° =°°°°Z x

yf (z) dz

°°°° ≤(2.6)

Z x

ykf (z)k dz =

Z b

xkf (z)k dz −

Z b

ykf (z)k dz → 0,

becauseR bx kf (z)k dz is continuous in x ∈ [a, b] .

(ii) Case of α > 1. Let x, y ∈ [a, b] : x ≥ y and x→ y.

We observe that

°°Iαb−f (x)− Iαb−f (y)°° =

1

Γ (α)

°°°°°Z b

x(ζ − x)α−1 f (ζ) dζ −

Z b

y(ζ − y)α−1 f (ζ) dζ

°°°°° ([5])=


1

Γ (α)

°°°°°Z b

x(ζ − x)α−1 f (ζ) dζ −

Z x

y(ζ − y)α−1 f (ζ) dζ −

Z b

x(ζ − y)α−1 f (ζ) dζ

°°°°°(2.7)

(see [2], p. 426, Theorem 11.43)

≤ 1Γ(α)

hR bx

¯(ζ − x)α−1 − (ζ − y)α−1

¯kf (ζ)k dζ +

R xy (ζ − y)α−1 kf (ζ)k dζ

i≤ 1

Γ(α)

hR bx

¯(ζ − x)α−1 − (ζ − y)α−1

¯kf (ζ)k dζ + (x− y)α−1 kfkL1([a,b],X)

i.

As x→ y we get (ζ − x)α−1 → (ζ − y)α−1, thus¯(ζ − x)α−1 − (ζ − y)α−1

¯→ 0,

and also ¯(ζ − x)α−1 − (ζ − y)α−1

¯≤ 2 (b− a)α−1 .

Thus

¯(ζ − x)α−1 − (ζ − y)α−1

¯kf (ζ)k ≤ 2 (b− a)α−1 kf (ζ)k ∈ L1 ([a, b] ,X) ,

(2.8)

and also¯(ζ − x)α−1 − (ζ − y)α−1

¯kf (ζ)k → 0 as x → y, for almost all

ζ ∈ [a, b] .

Therefore by Dominated Convergence Theorem we conclude, as x→ y,that Z b

x

¯(ζ − x)α−1 − (ζ − y)α−1

¯kf (ζ)k dζ → 0.

Consequently, °°Iαb−f (x)− Iαb−f (y)°°→ 0 as x→ y.

Therefore Iαb−f ∈ C ([a, b] ,X) . 2We give


Theorem 5. Here [a, b] ⊂ R, X is a Banach space, F : [a, b] → X. Letr > 0, F ∈ L∞ ([a, b] ,X), and the Bochner integral

G (s) :=

Z b

s(t− s)r−1 F (t) dt,(2.9)

all s ∈ [a, b]. Then G ∈ AC ([a, b] ,X) (absolutely continuous functions) forr ≥ 1 and G ∈ C ([a, b] ,X) for r ∈ (0, 1) .

Proof. Denote by kFk∞ := kFkL∞([a,b],X) := t ∈ [a, b]es sup kF (t)kX <+∞. Hence F ∈ L1 ([a, b] ,X) .

By [7], p. 88, (t− s)r−1 F (t) is a strongly measurable function in t,t ∈ [s, b], s ∈ [a, b]. So that (t− s)r−1 F (t) ∈ L1 ([s, b] ,X), see [6].

(1) Case r ≥ 1. We use the definition of absolute continuity. So for everyε > 0 we need δ > 0: whenever (ai, bi), i = 1, ..., n, are disjoint subintervalsof [a, b], then

nXi=1

(bi − ai) < δ ⇒nXi=1

kG (bi)−G (ai)k < ε.(2.10)

If kFk∞ = 0, then G (s) = 0, for all s ∈ [a, b], the trivial case and allfulfilled.

So we assume kFk∞ 6= 0. Hence we have (see [5])

G (bi)−G (ai) =

Z b

bi

(t− bi)r−1 F (t) dt−

Z b

ai

(t− ai)r−1 F (t) dt =

Z b

bi(t− bi)

r−1 F (t) dt−Z bi

ai(t− ai)

r−1 F (t) dt−Z b

bi(t− ai)

r−1 F (t) dt =

(see [2], p. 426, Theorem 11.43)

Z b

bi

³(t− bi)

r−1 − (t− ai)r−1

´F (t) dt−

Z bi

ai(t− ai)

r−1 F (t) dt.(2.11)

Call

Ii :=

Z b

bi

¯(t− bi)

r−1 − (t− ai)r−1

¯dt.(2.12)


Thus

kG (bi)−G (ai)k ≤∙Ii +

(bi − ai)r

r

¸kFk∞ := Ti.(2.13)

If r = 1, then Ii = 0, and

kG (bi)−G (ai)k ≤ kFk∞ (bi − ai) ,(2.14)

for all i := 1, ..., n.

If r > 1, then becauseh(t− ai)

r−1 − (t− bi)r−1

i≥ 0 for all t ∈ [bi, b],

we findIi =

R bbi

³(t− ai)

r−1 − (t− bi)r−1

´dt = (b−ai)r−(bi−ai)r−(b−bi)r

r

=r (b− ξ)r−1 (bi − ai)− (bi − ai)

r

r, for some ξ ∈ (ai, bi) .(2.15)

Therefore, it holds

Ii ≤r (b− a)r−1 (bi − ai)− (bi − ai)

r

r,(2.16)

and µIi +

(bi − ai)r

r

¶≤ (b− a)r−1 (bi − ai) .(2.17)

That is

Ti ≤ kFk∞ (b− a)r−1 (bi − ai) ,(2.18)

so that

kG (bi)−G (ai)k ≤ kFk∞ (b− a)r−1 (bi − ai) , for all i = 1, ..., n.

So in the case of r = 1, and by choosing δ := εkFk∞

, we get

nXi=1

kG (bi)−G (ai)k ≤ kFk∞

ÃnXi=1

(bi − ai)

!≤ kFk∞ δ = ε,(2.19)

proving for r = 1 that G is absolutely continuous. In the case of r > 1, andby choosing δ := ε

kFk∞(b−a)r−1 , we get


nXi=1

kG (bi)−G (ai)k ≤ kFk∞ (b− a)rÃ

nXi=1

(bi − ai)

!

≤ kFk∞ (b− a)r−1 δ = ε,(2.20)

proving for r > 1 that G is absolutely continuous again.(2) Case of 0 < r < 1. Let ai∗, bi∗ ∈ [a, b] : ai∗ ≤ bi∗. Then (t− ai∗)

r−1 ≤(t− bi∗)

r−1, for all t ∈ [bi, b]. ThenIi∗ =

R bbi∗

³(t− bi∗)

r−1 − (t− ai∗)r−1

´dt = (b−bi∗)r

r −µ(b− ai∗)

r − (bi∗ − ai∗)r

r

¶≤ (bi∗ − ai∗)

r

r,(2.21)

by (b− bi∗)r − (b− ai∗)

r < 0.Therefore

Ii∗ ≤(bi∗ − ai∗)

r

r(2.22)

and

Ti∗ ≤2 (bi∗ − ai∗)

r

rkFk∞ ,(2.23)

proving that

kG (bi∗)−G (ai∗)k ≤µ2 kFk∞

r

¶(bi∗ − ai∗)

r ,(2.24)

which proves that G is continuous.This completes the proof. 2We 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), then f is stronglymeasurable, 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 xn, x ∈ [a, b] : xn → x, as n→ ∞, then f (xn)→ f (x) in k·k, thatis |kf (xn)k− kf (x)k| ≤ kf (xn)− f (x)k → 0, proving kfk is continuous,hence bounded, so that kfkL∞([a,b],X) := t ∈ [a, b]es sup kf (t)k < +∞, thatis f ∈ L∞ ([a, b] ,X), and hence f ∈ L1 ([a, b] ,X). Consequently, f isBochner integrable ([2], p. 426), given that f is continuous.


For the last we used the fact:Z[a,b]

kf (t)k dt ≤ kfkL∞([a,b],X) (b− a) < +∞,(2.25)

proving that f ∈ L1 ([a, b] ,X) .Also, clearly, absolute continuity of f : [a, b] → X, implies uniform

continuity and continuity of f .

We also have

Theorem 7. Let α, β ≥ 0, f ∈ L1 ([a, b] ,X). Then

Iαb−Iβb−f = Iα+βb− f = Iβb−I

αb−f,(2.26)

valid almost everywhere on [a, b] .If additionally f ∈ C ([a, b] ,X) or α+β ≥ 1, then we have identity true

on all of [a, b] .

Proof. Since I0b− := I (the identity operator), if α = 0 or β = 0 or bothare zero, then the statement of the theorem is trivially true. So we assumeα, β > 0.

We observe that

Iαb−Iβb−f (x) =

1

Γ (α)Γ (β)

Z b

x(t− x)α−1

ÃZ b

t(τ − t)β−1 f (τ) dτ

!dt

(2.27)

=1

Γ (α)Γ (β)

Z b

x

Z b

xχ[t,b] (τ) (t− x)α−1 (τ − t)β−1 f (τ) dτdt.

The above integrals in (2.26) exist a.e. on [a, b]. So if Iαb−Iβb−f (x) , I

α+βb− f (x)

exist we can apply Fubini’s theorem, see Theorem 2, p. 93, [10], to inter-change the order of integration and obtain

Iαb−Iβb−f (x) =

1

Γ (α)Γ (β)

Z b

x

µZ τ

x(t− x)α−1 (τ − t)β−1 f (τ) dt

¶dτ =


1

Γ (α)Γ (β)

Z b

xf (τ)

µZ τ

x(τ − t)β−1 (t− x)α−1 dt

¶dτ =(2.28)

1

Γ (α)Γ (β)

Z b

xf (τ)

Γ (α)Γ (β)

Γ (α+ β)(τ − x)α+β−1 dτ =

1

Γ (α+ β)

Z b

xf (τ) (τ − x)(α+β)−1 dτ = Iα+βb− f (x) .

That is

Iαb−Iβb−f (x) = Iα+βb− f (x)(2.29)

true a.e. on [a, b] .

By Theorem 5 and Remark 6, if f ∈ C ([a, b] ,X), then Iβb−f ∈ C ([a, b] ,X),

therefore Iαb−Iβb−af ∈ C ([a, b] ,X) and Iα+βb− f ∈ C ([a, b] ,X).

Since in (29.) two continuous functions coincide a.e., the must be equaleverywhere.

At last, if f ∈ L1 ([a, b] ,X) and α+β ≥ 1, we get Iα+βb− f ∈ C ([a, b] ,X)

by Lemma l4. Hence, since Iα+βb− f (x) is defined and existing for any x ∈[a, b], by Fubini’s theorem as before, equals to Iαb−I

βb−f (x), for all x ∈ [a, b],

proving the claim. 2The algebraic version of previous theorem follows:

Theorem 8. The Bochner integral operatorsnIαb− : L1 ([a, b] ,X)→ L1 ([a, b] ,X) ; α > 0 make a commutative semi-group with respect to composition. The identity operator I0b− = I is theneutral element of this semigroup.

We need

Definition 9. (see [12]) A definition of the Hausdorff measure hα goes asfollows: if (T, d) is a metric space, A ⊆ T and δ > 0, let Λ (A, δ) be the setof all arbitrary collections (C)i of subsets of T , such that A ⊆ ∪iCi anddiam (Ci) ≤ δ (diam =diameter) for every i. Now, for every α > 0 define

hδα (A) := infnX

(diamCi)α | (Ci)i ∈ Λ (A, δ)

o.(2.30)

Then there exists δ → 0limhδα (A) = δ > 0suphδα (A), and hα (A) :=δ → 0limhδα (A) gives an outer measure on the power set P (T ), which is


countably additive on the σ-field of all Borel subsets of T . If T = Rn, thenthe Hausdorff measure hn, restricted to the σ-field of the Borel subsets ofRn, equals the Lebesgue measure on Rn up to a constant multiple. Inparticular, h1 (C) = µ (C) for every Borel set C ⊆ R, where µ is theLebesgue measure.

We will use the following spectacular result

Theorem 10. ([12]) (Fundamental Theorem of Calculus for Bochner in-tegration)

Suppose that for the given function f : [a, b] → X, there exists F :[a, b] → X, which is continuous, the derivative F 0 (t) exists and F 0 (t) =f (t) outside a µ-null Borel set B ⊆ [a, b] such that h1 (F (B)) = 0.

Then f is µ-measurable (i.e. strongly measurable), and if we assumethe Bochner integrability of f ,

F (b)− F (a) =

Z b

af (t) dt.(2.31)

Notice here that the derivatives of a function f : [a, b] → X, whereX is a Banach space, are defined exactly as the numerical ones, see fordefinitions and properties, [11], pp. 83-86, and p. 93, that is they arestrong derivatives.

Notation 11. Let f ∈ L1 ([a, b] ,X). We denote byZ a

bf (t) dt = −

Z b

af (t) dt.(2.32)

We will use Theorem 10 to give a general Taylor’s formula for Banachspace valued functions with a Bochner integral remainder.

Theorem 12. Let n ∈ N and f ∈ Cn−1 ([a, b] ,X), where [a, b] ⊂ R andX is a Banach space. Set

F (x) :=n−1Xi=0

(a− x)i

i!f (i) (x) , x ∈ [a, b] .(2.33)

Assume that f (n) exists outside a µ-null Borel set B ⊆ [a, b] such that

h1 (F (B)) = 0.(2.34)

We further assume the Bochner integrability of f (n). Then


f (a) =n−1Xi=0

(a− b)i

i!f (i) (b) +

1

(n− 1)!

Z a

b(a− t)n−1 f (n) (t) dt.(2.35)

Proof. We get that F ∈ C ([a, b] ,X). Notice that F (a) = f (a), and

F (b) =n−1Xi=0

(a− b)i

i!f (i) (b) .

Clearly F 0 exists outside of B. Infact it holds

F 0 (x) =(a− x)n−1

(n− 1)! f (n) (x) , ∀x ∈ [a, b]−B.(2.36)

Also F 0 is Bochner integrable.By Theorem 10 now we get that

F (b)− F (a) =

Z b

aF 0 (t) dt.(2.37)

That is, we havePn−1

i=0(a−b)i

i! f (i) (b)− f (a) =R ba(a−t)n−1(n−1)! f

(n) (t) dt =

−Z a

b

(a− t)n−1

(n− 1)! f (n) (t) dt,(2.38)

proving (2.35). 2

We give

Definition 13. 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) ∈ L1 ([a, b] ,X), wheref : [a, b] → X. We call the Caputo-Bochner right fractional derivative oforder α:

¡Dαb−f

¢(x) := (−1)m Im−αb− f (m) (x) ,(2.39)

i.e.

¡Dαb−f

¢(x) :=

(−1)m

Γ (m− α)

Z b

x(J − x)m−α−1 f (m) (J) dJ, ∀ x ∈ [a, b] .

(2.40)


We observe that Dmb−f (x) = (−1)

m f (m) (x) , for m ∈ N, andD0b−f (x) = f (x) .

By Theorem 3³Dαb−f

´(x) exists almost everywhere on [a, b] and³

Dαb−f

´∈ L1 ([a, b] ,X).

If°°°f (m)°°°

L∞([a,b],X)<∞, and α /∈ N, then by Theorem 5,

Dαb−f ∈ C ([a, b] ,X) , hence

°°°Dαb−f

°°° ∈ C ([a, b]) .

We make

Remark 14. (to Definition 13) We notice that (by Theorem 7)

¡Iαb−D

αb−f

¢(x) = (−1)m

³Iαb−I

m−αb− f (m)

´(x) =

(−1)m³Iα+m−αb− f (m)

´(x) = (−1)m

³Imb−f

(m)´(x) ,(2.41)

almost everywhere on [a, b].I.e.

¡Iαb−D

αb−f

¢(x) = (−1)m

³Imb−f

(m)´(x) ,(2.42)

for almost all x ∈ [a, b] .Notice here that

³Imb−f

(m)´(x) =

1

(m− 1)!

Z b

x(z − x)m−1 f (m) (z) dz ∈ L1 ([a, b] ,X)

(2.43)

and exists for almost all x ∈ [a, b], by Theorem 3.We have proved, by (2.42), that

1

Γ (α)

Z b

x(z − x)α−1

¡Dαb−f

¢(z) dz =

1

(m− 1)!

Z x

b(x− z)m−1 f (m) (z) dz,

(2.44)for almost all x ∈ [a, b] .


We present the following right fractional Taylor’s formula

Theorem 15. Let [a, b] ⊂ R, X be a Banach space, α > 0, m = dαe,f ∈ Cm−1 ([a, b] ,X). Set

Fx (t) :=m−1Xi=0

(x− t)i

i!f (i) (t) , ∀ t ∈ [x, b] ,(2.45)

where x ∈ [a, b] .Assume that f (m) exists outside a µ-null Borel set Bx ⊆ [x, b], such that

h1 (Fx (Bx)) = 0, where x ∈ [a, b] .(2.46)

We also assume that f (m) ∈ L1 ([a, b] ,X). Then

f (x) =m−1Xi=0

(x− b)i

i!f (i) (b) +

1

Γ (α)

Z b

x(z − x)α−1

¡Dαb−f

¢(z) dz,

(2.47)

for x ∈ [a, b] .

Proof. We use Theorem 12.

Clearly it holds

µf (·)−Pm−1

i=0(·−b)ii! f (i) (b)

¶∈ C ([a, b] ,X), that is (by

(35) 1(m−1)!

R xb (x− t)m−1 f (m) (t) dt ∈ C ([a, b] ,X) as a function of x ∈

[a, b]. Hence (44) holds as an equality over [a, b] (by Tonelli’s theorem),

therefore 1Γ(α)

R bx (z − x)α−1

³Dαb−f

´(z) dz ∈ C ([a, b] ,X), as a function of

x ∈ [a, b]. Now (2.47) is valid. 2More generally we get

Theorem 16. Let [a, b] ⊂ R, X be a Banach space, α > 0, m = dαe,f ∈ Cm−1 ([a, b] ,X). Set

Fx (t) :=m−1Xi=0

(x− t)i

i!f (i) (t) , ∀ t ∈ [x, b] ,(2.48)

where x ∈ [a, b] .Assume that f (m) exists outside a µ-null Borel set Bx ⊆ [x, b], such that


h1 (Fx (Bx)) = 0, ∀ x ∈ [a, b] .(2.49)

We also assume that f (m) ∈ L1 ([a, b] ,X). Then

f (x) =m−1Xi=0

(x− b)i

i!f (i) (b) +

1

Γ (α)

Z b

x(z − x)α−1

¡Dαb−f

¢(z) dz,

(2.50)

∀ x ∈ [a, b] .

Proof. By Theorem 15. 2

Remark 17. (to Theorem 16) By (2.50), we have

¡Iαb−D

αb−f

¢(x) =

1

Γ (α)

Z b

x(z − x)α−1

¡Dαb−f

¢(z) dz ∈ C ([a, b] ,X)

(2.51)

as a function of x ∈ [a, b] .

We have also

Corollary 18. (to Theorem 16) All as in Theorem 16. Assume that f (i) (b) =0, i = 0, 1, ...,m− 1. Then

f (x) =1

Γ (α)

Z b

x(z − x)α−1

¡Dαb−f

¢(z) dz,(2.52)

∀ x ∈ [a, b] .

Next we present Ostrowski type inequalities at right fractional level forBanach valued functions. See also [3].

Theorem 19. Let α > 0, m = dαe. Here all as in Theorem t16. Assumef (k) (b) = 0, k = 1, ...,m− 1, and Dα

b−f ∈ L∞ ([a, b] ,X). Then

°°°°° 1

b− a

Z b

af (x) dx− f (b)

°°°°° ≤°°°Dα

b−f°°°L∞([a,b],X)

Γ (α+ 2)(b− a)α .(2.53)


Proof. Let x ∈ [a, b]. We have by (2.50) that

f (x)− f (b) =1

Γ (α)

Z b

x(J − x)α−1Dα

b−f (J) dJ .

Thus

kf (x)− f (b)k = 1

Γ (α)

Z b

x(J − x)α−1

°°Dαb−f (J)

°° dJ≤ 1

Γ (α)

ÃZ b

x(J − x)α−1 dJ

!°°Dαb−f

°°L∞([a,b],X)

=

1

Γ (α)

µ(J − x)α

α|bx¶°°Dα

b−f°°L∞([a,b],X)

=1

Γ (α+ 1)(b− x)α

°°Dαb−f

°°L∞([a,b],X)

.

(2.54)

Therefore

kf (x)− f (b)k ≤ (b− x)α

Γ (α+ 1)

°°Dαb−f

°°L∞([a,b],X)

, ∀ x ∈ [a, b] .(2.55)

Hence it holds°°°°° 1

b− a

Z b

af (x) dx− f (b)

°°°°° =°°°°° 1

b− a

Z b

a(f (x)− f (b)) dx

°°°°° ≤1

b− a

Z b

akf (x)− f (b)k dx ≤ 1

(b− a)

Z b

a

(b− x)α

Γ (α+ 1)

°°Dαb−f

°°L∞([a,b],X)

dx =

°°°Dαb−f

°°°L∞([a,b],X)

(b− a)Γ (α+ 1)

Z b

a(b− x)α dx =

°°°Dαb−f

°°°L∞([a,b],X)

(b− a)Γ (α+ 1)

Ã−Ã(b− x)α+1

α+ 1|ba

!!(2.56)

=

°°°Dαb−f

°°°L∞([a,b],X)

(b− a)Γ (α+ 1)(−1)

Ã0− (b− a)α+1

α+ 1

!=

°°°Dαb−f

°°°L∞([a,b],X)

(b− a)Γ (α+ 2)(b− a)α+1 =

°°°Dαb−f

°°°L∞([a,b],X)

(b− a)α

Γ (α+ 2),


proving the claim. 2We also give

Theorem 20. Let α ≥ 1, m = dαe. Here all as in Theorem 16. Assumethat f (k) (b) = 0, k = 1, ...,m− 1, and Dα

b−f ∈ L1 ([a, b] ,X). Then

°°°°° 1

b− a

Z b

af (x) dx− f (b)

°°°°° ≤°°°Dα

b−f°°°L1([a,b],X)

Γ (α+ 1)(b− a)α−1 .(2.57)

Proof. We have again

kf (x)− f (b)k ≤ 1

Γ (α)

Z b

x(J − x)α−1

°°Dαb−f (J)

°° dJ≤ 1

Γ (α)(b− x)α−1

Z b

x

°°Dαb−f (J)

°° dJ≤ 1

Γ (α)(b− x)α−1

°°Dαb−f

°°L1([a,b],X)

.(2.58)

Hence

kf (x)− f (b)k ≤

°°°Dαb−f

°°°L1([a,b],X)

Γ (α)(b− x)α−1 , ∀ x ∈ [a, b] .(2.59)

Therefore°°°°° 1

b− a

Z b

af (x) dx− f (b)

°°°°° ≤ 1

b− a

Z b

akf (x)− f (b)k dx

≤ 1

(b− a)

Z b

a

°°°Dαb−f

°°°L1([a,b],X)

Γ (α)(b− x)α−1 dx =(2.60)

°°°Dαb−f

°°°L1([a,b],X)

(b− a)Γ (α)

Z b

a(b− x)α−1 dx =

°°°Dαb−f

°°°L1([a,b],X)

(b− a)Γ (α)

(b− a)α

α

=

°°°Dαb−f

°°°L1([a,b],X)

Γ (α+ 1)(b− a)α−1 ,

proving the claim. 2We continue with


Theorem 21. Let p, q > 1 : 1p +

1q = 1, α > 1 − 1

p , m = dαe. Hereall as in Theorem 16. Assume that f (k) (b) = 0, k = 1, ...,m − 1, andDαb−f ∈ Lq ([a, b] ,X). Then

°°°°° 1

b− a

Z b

af (x) dx− f (b)

°°°°° ≤°°°Dα

b−f°°°Lq([a,b],X)

Γ (α) (p (α− 1) + 1)1p

³α+ 1

p

´ (b− a)α−1+1p .

(2.61)

Proof. We have again

kf (x)− f (b)k ≤ 1

Γ (α)

Z b

x(J − x)α−1

°°Dαb−f (J)

°° dJ≤ 1

Γ (α)

ÃZ b

x(J − x)p(α−1) dJ

! 1pÃZ b

x

°°Dαb−f (J)

°°q dJ! 1q

≤ 1

Γ (α)

(b− x)(α−1)+1p

(p (α− 1) + 1)1p

ÃZ b

x

°°Dαb−f (J)

°°q dJ! 1q

(2.62)

≤ 1

Γ (α)

(b− x)(α−1)+ 1

p

(p (α− 1) + 1)1p

°°Dαb−f

°°Lq([a,b],X)

.

Therefore

kf (x)− f (b)k ≤

°°°Dαb−f

°°°Lq([a,b],X)

Γ (α) (p (α− 1) + 1)1p

(b− x)α−1+1p , ∀ x ∈ [a, b] .

(2.63)

Hence°°°°° 1

b− a

Z b

af (x) dx− f (b)

°°°°° ≤ 1

b− a

Z b

akf (x)− f (b)k dx


≤

°°°Dαb−f

°°°Lq([a,b],X)

(b− a)Γ (α) (p (α− 1) + 1)1p

Z b

a(b− x)α−1+

1p dx =(2.64)

°°°Dαb−f

°°°Lq([a,b],X)

Γ (α) (p (α− 1) + 1)1p

(b− a)α−1+1p³

α+ 1p

´ .

2

Corollary 22. Let α > 12 , m = dαe. All as in Theorem 16. Assume

f (k) (b) = 0, k = 1, ...,m− 1, Dαb−f ∈ L2 ([a, b] ,X). Then

°°°°° 1

b− a

Z b

af (x) dx− f (b)

°°°°° ≤°°°Dα

b−f°°°L2([a,b],X)

Γ (α)³√2α− 1

´ ³α+ 1

2

´ (b− a)α−12 .

(2.65)

We give

Proposition 23. Inequality (2.53) is sharp; namely it is attained by

f (x) = (b− x)α−→i , α > 0, α /∈ N, x ∈ [a, b] ,(2.66)

−→i ∈ X, such that

°°°°−→i °°°° = 1.Proof. (see also [4], pp. 26-27) We see that

f0 (x) = −α (b− x)α−1−→i , f 00 (x) = (−1)2 α (α− 1) (b− x)α−2

−→i ,...,

f (m−1) (x) = (−1)m−1 α (α− 1) (α− 2) ... (α−m+ 2) (b− x)α−m+1−→i ,

(2.67)

andf(m) (x) = (−1)m α (α− 1) (α− 2) ... (α−m+ 2) (α−m+ 1) (b− x)α−m

−→i .


Here f (m) is continuous on [a, b), and f (m) ∈ L1 ([a, b] ,X) . All assump-tions of Theorem 16 are easily fulfilled.

Thus

Dαb−f (x) =

−→i (−1)2m

Γ (m− α)α (α− 1) ... (α−m+ 1)

Z b

x(J − x)m−α−1 (b− J)α−m dJ

=

−→i α (α− 1) ... (α−m+ 1)

Γ (m− α)

Z b

x(b− J)(α−m+1)−1 (J − x)(m−α)−1 dJ

(2.68)

=

−→i α (α− 1) ... (α−m+ 1)

Γ (m− α)

Γ (α−m+ 1)Γ (m− α)

Γ (1)

=−→i α (α− 1) ... (α−m+ 1)Γ (α−m+ 1) = Γ (α+ 1)

−→i .

That is

Dαb−f (x) = Γ (α+ 1)

−→i , ∀ x ∈ [a, b] .(2.69)

Also we see that f (k) (b) = 0, k = 0, 1, ...,m−1, andDαb−f ∈ L∞ ([a, b] ,X).

So f fulfills all assumptions of Theorem 19. Next we see

R.H.S. (2.53) =Γ (α+ 1)

Γ (α+ 2)(b− a)α =

(b− a)α

(α+ 1).(2.70)

L.H.S. (2.53) =1

b− a

Z b

a(b− x)α dx =

1

b− a

(b− a)α+1

(α+ 1)=(b− a)α

(α+ 1),

(2.71)

proving attainability and sharpness of (2.53). 2We continue with a Poincare like right fractional inequality:


Theorem 24. Let p, q > 1 : 1p +1q = 1, and α > 1

q , m = dαe. Here all asin Theorem 16.

Assume that f (k) (b) = 0, k = 0, 1, ...,m− 1, and Dαb−f ∈ Lq ([a, b] ,X),

where X is a Banach space. Then

kfkLq([a,b],X) ≤(b− a)α

°°°Dαb−f

°°°Lq([a,b],X)

Γ (α) (p (α− 1) + 1)1p (qα)

1q

.(2.72)

Proof. We have that (by (2.52))

f (x) =1

Γ (α)

Z b

x(z − x)α−1

¡Dαb−f

¢(z) dz, ∀ x ∈ [a, b] .(2.73)

Hence

kf (x)k = 1

Γ (α)

°°°°°Z b

x(z − x)α−1

¡Dαb−f

¢(z) dz

°°°°° ≤1

Γ (α)

Z b

x(z − x)α−1

°°Dαb−f (z)

°° dz ≤1

Γ (α)

ÃZ b

x(z − x)p(α−1) dz

! 1pÃZ b

x

°°¡Dαb−f

¢(z)°°q dz! 1

q

≤(2.74)

1

Γ (α)

(b− x)p(α−1)+1

p

(p (α− 1) + 1)1p

°°Dαb−f

°°Lq([a,b],X)

.

We have proved that

kf (x)k ≤ 1

Γ (α)

(b− x)p(α−1)+1

p

(p (α− 1) + 1)1p

°°Dαb−f

°°Lq([a,b],X)

, ∀ x ∈ [a, b] .

(2.75)

Then

kf (x)kq ≤ (b− x)(p(α−1)+1)qp

(Γ (α))q (p (α− 1) + 1)qp

°°Dαb−f

°°qLq([a,b],X)

,(2.76)


∀ x ∈ [a, b] .Hence it holds

Z b

akf (x)kq dx ≤ (b− a)qα

(Γ (α))q (p (α− 1) + 1)qp qα

°°Dαb−f

°°qLq([a,b],X)

.(2.77)

The last inequality implies

ÃZ b

akf (x)kq dx

! 1q

≤(b− a)α

°°°Dαb−f

°°°Lq([a,b],X)

Γ (α) (p (α− 1) + 1)1p (qα)

1q

,(2.78)

proving the claim. 2Next comes a right Sobolev like fractional inequality:

Theorem 25. All as in the last Theorem 24. Let r > 0. Then

kfkLr([a,b],X) ≤(b− a)α−

1q+ 1r

°°°Dαb−f

°°°Lq([a,b],X)

Γ (α) (p (α− 1) + 1)1p

³r³α− 1

q

´+ 1

´ 1r

.(2.79)

Proof. As in the last theorem’s proof we get that

kf (x)k ≤ 1

Γ (α)

(b− x)α−1q

(p (α− 1) + 1)1p

°°Dαb−f

°°Lq([a,b],X)

, ∀ x ∈ [a, b] .

(2.80)

Since r > 0, we get

kf (x)kr ≤ 1

(Γ (α))r(b− x)r

¡α− 1

q

¢(p (α− 1) + 1)

rp

°°Dαb−f

°°rLq([a,b],X)

, ∀ x ∈ [a, b] .

(2.81)

Hence it holds


Z b

akf (x)kr dx ≤ (b− a)r

¡α− 1

q

¢+1

(Γ (α))r (p (α− 1) + 1)rp

³r³α− 1

q

´+ 1

´ °°Dαb−f

°°rLq([a,b],X)

.

(2.82)That is

ÃZ b

akf (x)kr dx

! 1r

≤ (b− a)α−1q+ 1r

Γ (α) (p (α− 1) + 1)1p

³r³α− 1

q

´+ 1

´ 1r

°°Dαb−f

°°Lq([a,b],X)

,

(2.83)proving the claim. 2

We give the following Opial type right fractional inequality:


1q = 1, and α > 1

q , m := dαe. Let

[a, b] ⊂ R, X a Banach space, and f ∈ Cm−1 ([a, b] ,X). Set

Fx (t) :=m−1Xi=0

(x− t)i

i!f (i) (t) , ∀ t ∈ [x, b] , where x ∈ [a, b] .(2.84)

Assume that f (m) exists outside a µ-null Borel set Bx ⊆ [x, b], such that

h1 (Fx (Bx)) = 0, ∀ x ∈ [a, b] .(2.85)

We assume that f (m) ∈ L∞ ([a, b] ,X). Assume also that f (k) (b) = 0,k = 0, 1, ...,m− 1. ThenZ b

xkf (w)k

°°¡Dαb−f

¢(w)

°° dw ≤(b− x)α−1+

2p

21qΓ (α) ((p (α− 1) + 1) (p (α− 1) + 2))

1p

ÃZ b

x

°°¡Dαb−f

¢(z)°°q dz! 2

q

,

(2.86)

∀ x ∈ [a, b] .


Proof. By (2.52) we get

f (x) =1

Γ (α)

Z b

x(z − x)α−1

¡Dαb−f

¢(z) dz, ∀ x ∈ [a, b] .(2.87)

Let x ≤ w ≤ b, then we have

f (w) =1

Γ (α)

Z b

w(z −w)α−1

¡Dαb−f

¢(z) dz.(2.88)

Furthermore it holds

kf (w)k ≤ 1

Γ (α)

Z b

w(z −w)α−1

°°¡Dαb−f

¢(z)°° dz ≤

1

Γ (α)

ÃZ b

w(z − w)p(α−1) dz

! 1pÃZ b

w

°°¡Dαb−f

¢(z)°°q dz! 1

q

=(2.89)

1

Γ (α)

(b− w)(p(α−1)+1)

p

(p (α− 1) + 1)1p

ÃZ b

w

°°¡Dαb−f

¢(z)°°q dz! 1

q

=

1

Γ (α)

(b− w)α−1q

(p (α− 1) + 1)1p

(z (w))1q ,

where

z (w) :=

Z b

w

°°¡Dαb−f

¢(z)°°q dz,(2.90)

all x ≤ w ≤ b, z (b) = 0.Thus

− z (w) :=

Z w

b

°°¡Dαb−f

¢(z)°°q dz,(2.91)

and

(−z (w))0 =°°¡Dα

b−f¢(w)

°°q ≥ 0,(2.92)

and

°°¡Dαb−f

¢(w)

°° = ³(−z (w))0

´ 1q =

³− (z (w))0

´ 1q .(2.93)

Therefore we obtain


kf (w)k°°¡Dα

b−f¢(w)

°° ≤ (b− w)α−1q

Γ (α) (p (α− 1) + 1)1p

³z (w)

³− (z (w))0

´´ 1q ,

(2.94)

all x ≤ w ≤ b.

Integrating (2.94) we get

Z b

xkf (w)k

°°¡Dαb−f

¢(w)

°° dw ≤1

Γ (α) (p (α− 1) + 1)1p

Z b

x(b− w)α−

1q¡z (w)

¡−z0 (w)

¢¢ 1q dw ≤

1

Γ (α) (p (α− 1) + 1)1p

ÃZ b

x(b− w)

¡α−1

q

¢p dw

! 1pÃZ b

x

¡z (w)

¡−z0 (w)

¢¢dw

! 1q

=

(2.95)

1

Γ (α) (p (α− 1) + 1)1p

(b− x)α−1+2p

(p (α− 1) + 2)1p

(z (x))2q

21q

=

(b− x)α−1+2p

21qΓ (α) [(p (α− 1) + 1) (p (α− 1) + 2)]

1p

ÃZ b

x

°°¡Dαb−f

¢(z)°°q dz! 2

q

,

(2.96)

∀ x ∈ [a, b] , proving the claim. 2Next we present a Hilbert-Pachpatte right fractional inequality:


Theorem 27. Let p, q > 1 : 1p +1q = 1, and α1 >

1q , α2 >

1p , mi := dαie,

i = 1, 2. Here [ai, bi] ⊂ R, i = 1, 2; X is a Banach space. Let fi ∈Cmi−1 ([ai, bi] ,X), i = 1, 2. Set

Fxi (ti) :=mi−1Xji=0

(xi − ti)ji

ji!f(ji)i (ti) ,(2.97)

∀ ti ∈ [xi, bi], where xi ∈ [ai, bi]; i = 1, 2. Assume that f (mi)i exists outside

a µ-null Borel set Bxi ⊆ [xi, bi], such that

h1 (Fxi (Bxi)) = 0, ∀ xi ∈ [ai, bi] ; i = 1, 2.(2.98)

We also assume that f(mi)i ∈ L1 ([ai, bi] ,X), and

f(ki)i (bi) = 0, ki = 0, 1, ..., ,mi − 1; i = 1, 2,(2.99)

and

³Da1b1−f1

´∈ Lq ([a1, b1] ,X) ,

³Dα2b2−f2

´∈ Lp ([a2, b2] ,X) .(2.100)

Then Z b1

a1

Z b2

a2

kf1 (x1)k kf2 (x2)k dx1dx2µ(b1−x1)p(α1−1)+1p(p(α1−1)+1) + (b2−x2)q(α2−1)+1

q(q(α2−1)+1)

¶ ≤

(b1 − a1) (b2 − a2)

Γ (α1)Γ (α2)

°°°Dα1b1−f1

°°°Lq([a1,b1],X)

°°°Dα2b2−f2

°°°Lp([a2,b2],X)

.

(2.101)

Proof. We have that (by (2.52))

fi (xi) =1

Γ (αi)

Z bi

xi

(zi − xi)αi−1

³Dαibi−fi

´(zi) dzi, ∀ xi ∈ [ai, bi] , i = 1, 2.

(2.102)


Then

kfi (xi)k ≤1

Γ (αi)

Z bi

xi

(zi − xi)αi−1

°°°³Dαibi−fi

´(zi)

°°° dzi,(2.103)

∀ xi ∈ [ai, bi] , i = 1, 2.We get as before,

kf1 (x1)k ≤1

Γ (α1)

(b1 − x1)p(α1−1)+1

p

(p (α1 − 1) + 1)1p

°°°Dα1b1−f1

°°°Lq([a1,b1],X)

,(2.104)

and

kf2 (x2)k ≤1

Γ (α2)

(b2 − x2)q(α2−1)+1

q

(q (α2 − 1) + 1)1q

°°°Dα2b2−f2

°°°Lp([a2,b2],X)

.(2.105)

Hence we have

kf1 (x1)k kf2 (x2)k ≤1

Γ (α1)Γ (α2) (p (α1 − 1) + 1)1p (q (α2 − 1) + 1)

1q

·

(b1 − x1)p(α1−1)+1

p (b2 − x2)q(α2−1)+1

q

°°°Dα1b1−f1

°°°Lq([a1,b1],X)

°°°Dα2b2−f2

°°°Lp([a2,b2],X)

(2.106)

(using Young’s inequality for a, b ≥ 0, a1p b

1q ≤ a

p +bq )

≤ 1

Γ (α1)Γ (α2)

Ã(b1 − x1)

p(α1−1)+1

p (p (α1 − 1) + 1)+(b2 − x2)

q(α2−1)+1

q (q (α2 − 1) + 1)

!·(2.107)

°°°Dα1b1−f1

°°°Lq([a1,b1],X)

°°°Dα2b2−f2

°°°Lp([a2,b2],X)

,

∀ xi ∈ [ai, bi]; i = 1, 2.So far we have

kf1 (x1)k kf2 (x2)kµ(b1−x1)p(α1−1)+1p(p(α1−1)+1) + (b2−x2)q(α2−1)+1

q(q(α2−1)+1)

¶ ≤°°°Dα1

b1−f1°°°Lq([a1,b1],X)

°°°Dα2b2−f2

°°°Lp([a2,b2],X)

Γ (α1)Γ (α2),(2.108)


∀ xi ∈ [ai, bi]; i = 1, 2.The denominator in (2.108) can be zero only when x1 = b1 and x2 = b2.

Integrating (2.108) over [a1, b1] × [a2, b2] we derive inequality (2.101).2

When 0 < α ≤ 1, Definition 13 becomes

Definition 28. Let [a, b] ⊂ R, X be a Banach space, 0 < α ≤ 1. Weassume that f 0 ∈ L1 ([a, b] ,X), where f : [a, b]→ X.

We define the Caputo-Bochner right fractional derivative of order α:

¡Dαb−f

¢(x) := −I1−αb− f 0 (x) ,(2.109)

i.e.

¡Dαb−f

¢(x) =

−1Γ (1− α)

Z b

x(J − x)−α f 0 (J) dJ, ∀ x ∈ [a, b] .(2.110)

Clearly D1b−f (x) = −f 0 (x).

Remark 29. Let [A,B] ⊂ R, X be a Banach space, 0 < α ≤ 1, f :[A,B]→ X. We assume that f (2) ∈ L1 ([A,B] ,X). Then dα+ 1e = 2, and

DαB−f

0 (x) =−1

Γ (1− α)

Z B

x(J − x)−α f 00 (J) dJ =(2.111)

- (−1)2

Γ(2−(α+1))R Bx(J−x)2−(α+1)−1f 00(J)dJ=−Dα+1

B− f(x),

i.e.

DαB−f

0 (x) = −Dα+1B− f (x) ,(2.112)

and

°°DαB−f

0 (x)°° = °°°Dα+1

B− f (x)°°° , ∀ x ∈ [A,B] .(2.113)

We apply Theorem 19 when 0 < α ≤ 1.

Theorem 30. Let 0 < α ≤ 1, [A,B] ⊂ R, X a Banach space, f ∈C ([A,B] ,X). Assume that f 0 exists outside a µ-null Borel set Bx ⊆ [x,B],such that

h1 (f (Bx)) = 0, ∀ x ∈ [A,B] .


We assume that f 0 ∈ L1 ([A,B] ,X), and DαB−f ∈ L∞ ([A,B] ,X).

Then

°°°°° 1

B −A

Z B

Af (x) dx− f (B)

°°°°° ≤°°°Dα

B−f°°°L∞([A,B],X)

Γ (α+ 2)(B −A)α .

(2.114)

We present the following right Caputo-Bochner fractional Landau in-equality for k·k∞ .

Theorem 31. Let f ∈ C1 ((−∞, B0],X), where B0 ∈ R is fixed, 0 < α ≤1, X is a Banach space. For any A,B ∈ (−∞, B0] : A ≤ B, we assume thatf fulfills: assume that f 00 exists outside a µ-null Borel set Bx ⊆ [x,B], suchthat

h1¡f 0 (Bx)

¢= 0, ∀ x ∈ [A,B] .(2.115)

We assume that f 00 ∈ L1 ([A,B] ,X), and Dα+1B− f ∈ L∞ ([A,B] ,X). We

further assume that

°°°Dα+1B− f

°°°L∞((−∞,B],X)

≤°°°Dα+1

B0−f°°°L∞((−∞,B0],X)

<∞, ∀ B ≤ B0.

(2.116)

(the last left inequality is obvious when α = 1), and

kfk∞,(−∞,B0]:= t ∈ (−∞, B0]sup kf (t)k <∞.(2.117)

Then

°°f 0°°∞,(−∞,B0]:= t ∈ (−∞, B0]sup

°°f 0 (t)°° ≤ (α+ 1)µ 2α

¶( αα+1)

(Γ (α+ 2))− 1(α+1) ·

³kfk∞,(−∞,B0]

´ α(α+1)

µ°°°Dα+1B0−f

°°°L∞((−∞,B0],X)

¶ 1(α+1)

.(2.118)


Proof. We have that (by Theorem 30)

°°°°° 1

B −A

Z B

Af 0 (x) dx− f 0 (B)

°°°°° ≤°°°Dα+1

B− f°°°L∞([A,B],X)

Γ (α+ 2)(B −A)α ,

(2.119)

∀ A,B ∈ (−∞, B0] : A ≤ B.Subsequently by Theorem 10 we derive

°°°°f (B)− f (A)

B −A− f 0 (B)

°°°° ≤°°°Dα+1

B− f°°°L∞([A,B],X)

Γ (α+ 2)(B −A)α ,(2.120)

∀ A,B ∈ (−∞, B0], A ≤ B.

Hence it holds

°°f 0 (B)°°− 1

B −Akf (B)− f (A)k ≤

°°°Dα+1B− f

°°°L∞([A,B],X)

Γ (α+ 2)(B −A)α ,

(2.121)

and

°°f 0 (B)°° ≤ kf (B)− f (A)kB −A

+

°°°Dα+1B− f

°°°L∞([A,B],X)

Γ (α+ 2)(B −A)α ,

(2.122)

∀ A,B ∈ (−∞, B0] : A ≤ B.

Therefore we obtain

°°f 0 (B)°° ≤ 2 kfk∞,(−∞,B0]

B −A+

°°°Dα+1B0−f

°°°L∞((−∞,B0],X)

Γ (α+ 2)(B −A)α ,


(2.123)

∀ A,B ∈ (−∞, B0] : A ≤ B.

The right hand side of (2.123) depends only on B − A. Consequently,it holds

°°f 0°°∞,(−∞,B0]≤2 kfk∞,(−∞,B0]

B −A+

°°°Dα+1B0−f

°°°L∞((−∞,B0],X)

Γ (α+ 2)(B −A)α .

(2.124)

We may call t = B −A > 0. Thus by (2.124),

°°f 0°°∞,(−∞,B0]≤2 kfk∞,(−∞,B0]

t+

°°°Dα+1B0−f

°°°L∞((−∞,B0],X)

Γ (α+ 2)tα, ∀ t > 0.

(2.125)

Set

µ := 2 kfk∞,(−∞,B0],

θ :=

°°°Dα+1B0−f

°°°L∞((−∞,B0],X)

Γ (α+ 2),(2.126)

both are greater than 0.

We consider the function

y (t) =µ

t+ θtα, 0 < α ≤ 1, t > 0.(2.127)

As in [4], pp. 81-82, y has a global minimum at

t0 =

µµ

αθ

¶ 1(α+1)

,(2.128)

which is


y (t0) = (θµα)

1(α+1) (α+ 1)α−(

αα+1).(2.129)

Consequently it is

y (t0) =

⎛⎜⎝°°°Dα+1

B0−f°°°L∞((−∞,B0],X)

Γ (α+ 2)

⎞⎟⎠1

(α+1) ³2 kfk∞,(−∞,B0]

´( αα+1) (α+ 1)α−(

αα+1).

(2.130)We have proved that

kf 0k∞,(−∞,B0]≤ (α+ 1)

³2α

´( αα+1) (Γ (α+ 2))

− 1(α+1) ·

µ°°°Dα+1B0−f

°°°L∞((−∞,B0],X)

¶ 1(α+1) ³

kfk∞,(−∞,B0]

´( αα+1) ,(2.131)

establishing the claim. 2The case B0 = 0 comes next

Corollary 32. (to Theorem 31) All as in Theorem 31 for B0 = 0.

Then kf 0k∞,R− ≤ (α+ 1)³2α

´( αα+1) (Γ (α+ 2))

− 1(α+1) ·

³kfk∞,R−

´( αα+1)

µ°°°Dα+10− f

°°°L∞(R−,X)

¶ 1(α+1)

.(2.132)

When α = 1 we get

Corollary 33. (to Theorem 31) Let f ∈ C1 ((−∞, B0],X), where B0 ∈ Ris fixed, X is a Banach space. For any A,B ∈ (−∞, B0] : A ≤ B, we assumethat f fulfills: assume that f 00 exists outside a µ-null Borel set Bx ⊆ [x,B],such that

h1¡f 0 (Bx)

¢= 0, ∀ x ∈ [A,B] .(2.133)

We assume that f 00 ∈ L∞ ((−∞, B0],X), and kfk∞,(−∞,B0]<∞. Then

°°f 0°°∞,(−∞,B0]≤ 2 kfk

12

∞,(−∞,B0]

°°f 00°° 12L∞((−∞,B0],X).(2.134)

Next case of B0 = 0.

Corollary 34. All as in Corollary 33. It holds°°f 0°°∞,R−≤ 2 kfk

12

∞,R−

°°f 00°° 12L∞(R−,X) .(2.135)


See also [9].We apply Theorem 21 when 0 < α ≤ 1.

Theorem 35. Let p, q > 1 : 1p +1q = 1,

1q < α ≤ 1. Let [A,B] ⊂ R, X be

a Banach space, f ∈ C ([A,B] ,X). Assume that f 0 exists outside a µ-nullBorel set Bx ⊆ [x,B], such that

h1 (f (Bx)) = 0, ∀ x ∈ [A,B] .(2.136)

We also assume that f 0 ∈ L1 ([A,B] ,X). AssumeDαB−f ∈ Lq ([A,B] ,X).

Then

°°°°° 1

B −A

Z B

Af (x) dx− f (B)

°°°°° ≤°°°Dα

B−f°°°Lq([A,B],X)

Γ (α) (p (α− 1) + 1)1p

³α+ 1

p

´ (B −A)α−1q .

(2.137)

We present the following right Caputo-Bochner fractional Lq Landauinequality.


1q = 1, and 1

q < α ≤ 1. Let f ∈C1 ((−∞, B0],X) , where B0 ∈ R is fixed, X is a Banach space. For anyA,B ∈ (−∞, B0] : A ≤ B, we suppose that f fulfills: assume that f 00 existsoutside a µ-null Borel set Bx ⊆ [x,B], such that

h1¡f 0 (Bx)

¢= 0, ∀ x ∈ [A,B] .(2.138)

We also assume that f 00 ∈ L1 ([A,B] ,X) and Dα+1B− f ∈ Lq ([A,B] ,X).

We further assume that

°°°Dα+1B− f

°°°Lq((−∞,B],X)

≤°°°Dα+1

B0−f°°°Lq((−∞,B0],X)

<∞, ∀ B ≤ B0,

(2.139)

(the last left inequality is obvious when α = 1), and

kfk∞,(−∞,B0]<∞.(2.140)


Then kf 0k∞,(−∞,B0]≤µ2¡α+ 1

p

¢α− 1

q

¶µα− 1qα+1

p

¶1

(Γ(α))

1

(α+1p) (p(α−1)+1)

1(pα+1)

·

³kfk∞,(−∞,B0]

´µα− 1qα+1

p

¶ µ°°°Dα+1B0−f

°°°Lq((−∞,B0],X)

¶ 1

(α+1p) .(2.141)

Proof. By Theorem 35 we have that

°°°°° 1

B −A

Z B

Af 0 (x) dx− f 0 (B)

°°°°° ≤°°°Dα+1

B− f°°°Lq([A,B],X)

Γ (α) (p (α− 1) + 1)1p

³α+ 1

p

´ (B −A)α−1q ,

(2.142)∀ A,B ∈ (−∞, B0] : A ≤ B.

From Theorem 10 we derive°°°°f (B)− f (A)

B −A− f 0 (B)

°°°° ≤°°°Dα+1

B− f°°°Lq([A,B],X)

Γ (α) (p (α− 1) + 1)1p

³α+ 1

p

´ (B −A)α−1q ,

(2.143)∀ A,B ∈ (−∞, B0] : A ≤ B.

Therefore we obtain

°°f 0 (B)°° ≤ 2 kfk∞,(−∞,B0]

B −A+

°°°Dα+1B0−f

°°°Lq((−∞,B0],X)

Γ (α) (p (α− 1) + 1)1p

³α+ 1

p

´ (B −A)α−1q ,

(2.144)∀ A,B ∈ (−∞, B0] : A ≤ B.

The R.H.S. of (2.144) depends only on B −A. Therefore

°°f 0°°∞,(−∞,B0]≤2 kfk∞,(−∞,B0]

B −A+

°°°Dα+1B0−f

°°°Lq((−∞,B0],X)

Γ (α) (p (α− 1) + 1)1p

³α+ 1

p

´ (B −A)α−1q .

(2.145)

We may call t = B −A > 0. Thus

°°f 0°°∞,(−∞,B0]≤2 kfk∞,(−∞,B0]

t+

°°°Dα+1B0−f

°°°Lq((−∞,B0],X)

Γ (α) (p (α− 1) + 1)1p

³α+ 1

p

´tα−1q ,


(2.146)

∀ t ∈ (0,∞) .Notice that 0 < α− 1

q < 1. Call

eµ := 2 kfk∞,(−∞,B0],(2.147)

eθ :=°°°Dα+1

B0−f°°°Lq((−∞,B0],X)

Γ (α) (p (α− 1) + 1)1p

³α+ 1

p

´ ,(2.148)

both are positive, and

eν := α− 1q∈ (0, 1) .(2.149)

We consider the function

ey (t) = eµt+ eθteα, t ∈ (0,∞) .(2.150)

The only critical number here is

et0 = µ eµeαeθ¶ 1eα+1

,(2.151)

and ey has a global minimum at et0, which isey ³et0´ = ³eθeµeα´ 1eα+1 (eα+ 1) eα−

³ eαeα+1´.(2.152)

Consequently, we derive

ey ³et0´ =⎛⎜⎝°°°Dα+1

B0−f

°°°Lq((−∞,B0],X)

Γ(α)(p(α−1)+1)1p¡α+ 1

p

¢⎞⎟⎠

1

(α+1p)

·

³2 kfk∞,(−∞,B0]

´µα− 1qα+1

p

¶ µα+

1

p

¶µα− 1

q

¶−µα− 1qα+1

p

¶.(2.153)

We have proved that

kf 0k∞,(−∞,B0]≤µ2¡α+ 1

p

¢α− 1

q

¶µα− 1qα+1

p

¶1

(Γ(α))

1

(α+1p) (p(α−1)+1)

1(pα+1)

·


³kfk∞,(−∞,B0]

´µα− 1qα+1

p

¶ µ°°°Dα+1B0−f

°°°Lq((−∞,B0],X)

¶ 1

(α+1p) ,(2.154)

establishing the claim. 2

Corollary 37. (to Theorem 36, B0 = 0) All as in Theorem 36.

Then kf 0k∞,R− ≤µ2¡α+ 1

p

¢α− 1

q

¶µα− 1qα+1

p

¶1

(Γ(α))

1

(α+1p) (p(α−1)+1)

1(pα+1)

·

³kfk∞,R−

´µα− 1qα+1

p

¶ µ°°°Dα+10− f

°°°Lq(R−,X)

¶ 1

(α+1p) .(2.155)

Corollary 38. (to Theorem 36, B0 = 0, p = q = 2) All as in Theorem 36,12 < α ≤ 1. Then

kf 0k∞,R− ≤µ2(α+ 1

2)α− 1

2

¶³α− 12α+1

2

´1

(Γ(α))

1

(α+12) (2α−1)

1(2α+1)

·

³kfk∞,R−

´³α− 12α+1

2

´ µ°°°Dα+10− f

°°°L2(R−,X)

¶ 1

(α+12) .(2.156)

Case of α = 1 follows:

Corollary 39. Let p, q > 1 : 1p +1q = 1, f ∈ C1 ((−∞, B0],X) , where

B0 ∈ R is fixed, X is a Banach space. For any A,B ∈ (−∞, B0] : A ≤ B,we suppose that f fulfills: assume that f 00 exists outside a µ-null Borel setBx ⊆ [x,B], such that

h1¡f 0 (Bx)

¢= 0, ∀ x ∈ [A,B] .(2.157)

We further assume thatf00 ∈ Lq ((−∞, B0],X) ,

andkfk∞,(−∞,B0]

<∞.Then

°°f 0°°∞,(−∞,B0]≤

⎛⎝2³1 + 1

p

´1− 1

q

⎞⎠µ1− 1q1+ 1

p

¶·


³kfk∞,(−∞,B0]

´µ 1− 1q1+1

p

¶ ³°°f 00°°Lq((−∞,B0],X)

´ 1

(1+1p) .(2.158)

Corollary 40. (to Corollary 39) Assume B0 = 0. Then

°°f 0°°∞,R−≤

⎛⎝2³1 + 1

p

´1− 1

q

⎞⎠(

We finish article with

Corollary 41. (to Corollaries 39, 40) Assume B0 = 0 and p = q = 2.Then

°°f 0°°∞,R−≤ 3√6³kfk∞,R−

´ 13³°°f 00°°L2(R−,X)´ 23 .(2.160)

Note 42. Many variations and generalizations of the above inequalitiesare possible, however due to lack of space we stop here.

References

[1] R.P. Agarwal, V. Lupulescu, D. O’Regan, G. Rahman, Multi-termfractional differential equations in a nonreflexive Banach space, Ad-vances in Difference Equation, (2013), 2013:302.

[2] C.D. Aliprantis and K.C. Border, Infinite Dimensional Analysis,Springer, New York, (2006).

[3] G. Anastassiou, Fractional Differentiation Inequalities, Springer, NewYork, (2009).

[4] G. Anastassiou, Advances on fractional inequalities, Springer, NewYork, (2011).

[5] Appendix F, The Bochner integral and vector-valued Lp-spaces,https://isem.math.kit.edu/images/f/f7/AppendixF.pdf.


[6] Bochner integral. Encyclopedia of Mathematics. URL:http://www.encyclopediaofmath.org/index.php?title=Bochner integral&oldid=38659.

[7] R.F. Curtain and A.J. Pritchard, Functional Analysis in Modern Ap-plied Mathematics, Academic Press, London, New York, (1977).

[8] M. Kreuter, Sobolev Spaces of Vector-valued functions, Ulm Univ.,Master Thesis in Math., Ulm, Germany, (2015).

[9] E. Landau, Einige Ungleichungen fur zweimal differentzierban funktio-nen, Proc. London Math. Soc. 13, pp. 43-49, (1913).

[10] J. Mikusinski, The Bochner integral, Academic Press, New York,(1978).

[11] G.E. Shilov, Elementary Functional Analysis, Dover Publications, Inc.,New York, (1996).

[12] C. Volintiru, A proof of the fundamental theorem of Calculus usingHausdorff measures, Real Analysis Exchange, 26 (1), 2000/2001, pp.381-390.

George A. AnastassiouDepartment of Mathematical SciencesUniversity of MemphisMemphis, TN 38152,U.S.A.e-mail : [email protected]

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