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
152 George A. Anastassiou
=(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])=
Strong right fractional calculus for Banach space valued functions 153
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
154 George A. Anastassiou
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)
Strong right fractional calculus for Banach space valued functions 155
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
156 George A. Anastassiou
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.
Strong right fractional calculus for Banach space valued functions 157
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 =
158 George A. Anastassiou
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
Strong right fractional calculus for Banach space valued functions 159
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
160 George A. Anastassiou
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)
Strong right fractional calculus for Banach space valued functions 161
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] .
162 George A. Anastassiou
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
Strong right fractional calculus for Banach space valued functions 163
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)
164 George A. Anastassiou
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),
Strong right fractional calculus for Banach space valued functions 165
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
166 George A. Anastassiou
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
Strong right fractional calculus for Banach space valued functions 167
≤
°°°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 .
168 George A. Anastassiou
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:
Strong right fractional calculus for Banach space valued functions 169
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)
170 George A. Anastassiou
∀ 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
Strong right fractional calculus for Banach space valued functions 171
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:
Theorem 26. Let p, q > 1 : 1p +
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] .
172 George A. Anastassiou
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
Strong right fractional calculus for Banach space valued functions 173
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:
174 George A. Anastassiou
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)
Strong right fractional calculus for Banach space valued functions 175
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)
176 George A. Anastassiou
∀ 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] .
Strong right fractional calculus for Banach space valued functions 177
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)
178 George A. Anastassiou
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 ,
Strong right fractional calculus for Banach space valued functions 179
(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
180 George A. Anastassiou
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)
Strong right fractional calculus for Banach space valued functions 181
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.
Theorem 36. Let p, q > 1 : 1p +
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)
182 George A. Anastassiou
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 ,
Strong right fractional calculus for Banach space valued functions 183
(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)
·
184 George A. Anastassiou
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
¶·
Strong right fractional calculus for Banach space valued functions 185
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.
186 George A. Anastassiou
[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]