# On the Theory of Fractional Calculus in the Pettis-Function Spaces

Authors: Hussein A. H. Salem
Journal: Journal of Function Spaces
 (2018)

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## Abstract

Throughout this paper, we outline some aspects of fractional calculus in Banach spaces. Some examples are demonstrated. In our investigations, the integrals and the derivatives are understood as Pettis integrals and the corresponding derivatives. Our results here extended all previous contributions in this context and therefore are new. To encompass the full scope of our paper, we show that a weakly continuous solution of a fractional order integral equation, which is modeled off some fractional order boundary value problem (where the derivatives are taken in the usual definition of the Caputo fractional weak derivative), may not solve the problem.


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## Body

## 1. Introduction and Preliminaries

The issue of fractional calculus for the functions that take values in Banach space where the integrals and the derivatives are understood as Pettis integrals and the corresponding derivatives has been studied for the first time by the authors of [1, 2]. Following the appearance of [1], there has been a significant interest in the study of this topic (see, e.g., [3–6]; see also [7–9]). This paper is devoted to presenting general results and examples for the existence of the fractional integral (and corresponding fractional differential) operators in arbitrary Banach space where it is endowed with its weak topology. In our investigations, we show that the well-known properties of the fractional calculus for functions taking values in finite dimensional spaces also hold in infinite dimensional spaces. Our results extend all previous contributions of the same type in the Bochner integrability setting and in the Pettis integrability one.For the readers convenience, here we present some notations and the main properties for the Pettis integrals. For further background, unexplained terminology and details pertaining to this paper can be found in Diestel et al. [10, 11] and Pettis [12].Throughout this paper, we consider the measure space(I,Ω,μ), where I=[0,1], 0≤a<b<∞ denote a fixed interval of the real line, Ω denotes the Lebesgue σ-algebra L(I), and μ stands for the Lebesgue measure. E denotes a real Banach space with a norm · and E∗ is its dual. By Eω we denotes the space E when endowed with the weak topology generated by the continuous linear functionals on E. We will let C[I,Eω] denote the Banach space of weakly continuous functions I→E, with the topology of weak uniform convergence. And P[I,E] denotes the space of E-valued Pettis integrable functions in the interval I (see [10, 12] for the definition). Recall that (see, e.g., [10, 13–19]) the weakly measurable function x:I→E is said to be Dunford (or Gelfand) integrable on I if and only if φx is Lebesgue integrable on I for each φ∈E∗.Definition 1.
Letp∈[1,∞]. Define Hp(E) to be the class of all weakly measurable functions x:I→E having φx∈Lp(I) for every φ∈E∗.
Ifp=∞, the added condition(1)l.u.b.φ=1esssupt∈I⁡φxt<∞ must be satisfied by each x∈H∞(E). We also define the class H0p(E) by(2)H0pE≔x∈PI,E:φx∈LpI. Further, we define the space HλE, λ≥0 by(3)HλE≔x∈CI,Eω:φxis  Ho¨lderain  of  orderλonI,for  everyφ∈E∗. We also define H0(E)≔C[I,Eω].In the following proposition, we summarize some important facts which are the main tool in carrying out our investigations (see [10, 12, 16, 17]).Proposition 2.
Letp,q∈[1,∞] be of conjugate exponents (that is, 1/p+1/q=1 with the convention that 1/∞=0). If x:I→E is weakly measurable, then (1)
ifx(·)∈P[I,E] and u(·)∈L∞[I,R], then the following hold x(·)u(·)∈P[I,E],(2)
x∈H0p(E), p>1 if and only if x(·)u(·)∈P[I,E] for every u(·)∈Lq(I),(3)
ifE is reflexive (containing no isometric copy of c0), the weakly (strongly) measurable function x:I→E is Pettis integrable on I if and only if x is Dunford integrable on I,(4)
for anyp>1, we have {x∈Hp(E):xstrongly  measurable}⊆H0p(E). If E is weakly sequentially complete, this is also true for p=1.We remark that there is a bounded weakly measurable function which is not Pettis integrable (see, e.g., [19]).A fundamental property of Pettis integral is contained in the following.Proposition 3 (see [12] Corollary 2.51).
Ifx∈P[I,E], then for any bounded subset Ω of elements of E∗, the integrals(4)∫Jφxsds,φ∈Ω are weakly equi-absolutely continuous.Theorem 4 (mean value theorem for Pettis integral).
If the functionx:I→E is Pettis integrable on I, then(5)∫Jxsds∈Jconv¯xJ, where J⊂I, |J| is the length of J and conv¯(x(J)) is the closed convex hull of x(J).


## 2. Fractional Integrals of Vector-Valued Functions

In this section, we define and study the Riemann-Liouville fractional integral operators and the corresponding fractional derivatives in Banach spaces.Devoted by the definition of the Riemann-Liouville fractional integral of real-valued function, we introduce the following.Definition 5.
Letx:I→E. The Riemann-Liouville fractional Pettis integral (shortly RFPI) of x of order α>0 is defined by(6)Iαxt≔1Γα∫0tt-sα-1xsds,t>0.In the preceding definition, “∫” stands for the Pettis integral.WhenE=R, it is well known (see, e.g., [20, 21]) that the operator Iα sends Lq[0,b], b∈(0,∞) continuously to Lp[0,b] if p∈[1,∞] satisfy q>1/(α+(1/p)).This seems to be a good place to put the following.Example 6.
LetE be an infinite dimensional Banach space that fails cotype (see [22] and the references therein). Define the strongly measurable function x:[0,1]→E by(7)xt≔∑n=1∞∑k=12ncnχAkntμAknekn,μ:Lebesgue  measure,with similar notations as in ([13], Corollary 4) where we choose cn=2Kψ(23-n), K>1, ψ(t)≔tϵ,ϵ=0.9 and (8)Akn=ak3n,bk3n:n=0,1,2,…,k=1,2,3,…,2n to be the fat Cantor sets (that is, μ(Akn)=1/3n holds for every k∈{1,2,3,…,2n}).
As cited in ([13], Corollary 4), x is Pettis integrable functions on [0,1] whose indefinite integral is nowhere weakly differentiable on [0,1]. Here we will show that x has RFPI of all order α≥3/4 and(9)Iαxt=∑n=1∞∑k=12ncneknμAknΓα∫0,t∩Aknt-sα-1ds∈E.Arguing similarly as in ([13], page 368), we have in view of ∑k=12nεknuknl∞2n=1 that ∑k=12nφekn≤2φ holds for every φ∈E∗.
Also, for anyt∈[0,1] and fixed n∈N, we have for some n0<n that(10)∑k=12n∫0tt-sα-1χAknsμAknds2=∑k=12n32n∫0,t∩Aknt-sα-1ds2=32n∑k=12n0∫Aknt-sα-1ds2+32n∫a2n0+1tt-sα-1ds2≤32nα∑k=12n0t-ak3nα-t-bk3nα2+32nα2t-a2n0+13n2α≤32nα∑k=12n0t-ak3n-t-bk3n2α+32nα2b2n0+12n-a2n0+12n2α≤32n2nα32nα=2nα232n1-α. Therefore, for any measurable J⊂[0,1], we arrive at (11)∫0,t∩J∑n=1∞∑k=12ncnΓαt-sα-1χAknsμAknφeknds≤∫0t∑n=1∞∑k=12ncnΓαt-sα-1χAknsμAknφeknds≤∑n=1∞cnΓα∫0t∑k=12nt-sα-1χAknsμAknφeknds≤∑n=1∞cnΓα∑k=12nφekn∫0tt-sα-1χAknsμAknds≤∑n=1∞cnΓα∑k=12nφekn21/2∑k=12n∫0tt-sα-1χAknsμAknds21/2≤2φ∑n=1∞2n3n1-αcnαΓα=4K23ϵφ∑n=1∞2n1/2-ϵ3n1-αΓ1+α. Obviously, the latter series converges whenever α≥3/4, ϵ=0.9 which allows us to interchange the integral and summation below to see that (12)1Γα∫0,t∩Jφt-sα-1xs=1Γα∫0,t∩Jt-sα-1φxs=∫0,t∩J∑n=1∞∑k=12ncnΓαt-sα-1χAknsμAknφeknds=∑n=1∞∑k=12n∫0,t∩JcnΓαt-sα-1χAknsμAknφeknds=φ∑n=1∞∑k=12ncneknμAknΓα∫0,t∩Aknt-sα-1ds. It remains to prove that(13)∑n=1∞∑k=12ncneknμAknΓα∫0,t∩Aknt-sα-1ds∈E. Evidently, we have (14)∑n=lm∑k=12nekncn∫0,t∩Aknt-sα-1μAknΓαdsE=∑n=lm∑k=12ncn∫0,t∩Aknt-sα-1μAknΓαdsTnuknE≤2∑n=lmcn∑k=12n∫0,t∩Aknt-sα-1μAknΓαdsuknl∞2n=2Γα∑n=lm3ncnmax1≤k≤2n⁡∫0,t∩Aknt-sα-1χAknsds≤2Γα∑n=lm3ncnmax1≤k≤2n⁡∫ak/3nbk/3nt-sα-1ds≤4K23ϵΓ1+α∑n=lm3n1-α2ϵn, which approaches zero as l,m→∞ as needed for (9).Remark 7.
Observe Example6. We remark the following:(1)
There is a reflexive Banach space for which the indefinite Pettis integral of the functionx defined by (7) is nowhere weakly differentiable on [0,1] (see [13], the remark below Corollary 4). Meanwhile, x has a RFPI of all order α≥3/4.(2)
The functionIαx(·) is weakly continuous on [0,1]: this follows easily from the definition of the Pettis integral. In fact, we have in view of (9) that(15)φIαxt=Iαφxt=∑n=1∞∑k=12ncnφeknμAknΓα∫0,t∩Aknt-sα-1dsholds for everyφ∈E∗. Since (16)∑k=12ncnφeknμAknΓα∫0,t∩Aknt-sα-1ds≤cn∑k=12nφekn21/2∑k=12n∫0,t∩Aknt-sα-1μAknΓαds21/2≤4K23ϵφ2n1/2-ϵ3n1-αΓ1+α,then the infinite series of continuous functions in the left hand side of (15) converges uniformly in [0,1]. Hence, the function φIαx(·) is continuous on [0,1] (this yields the weak continuity of Iαx(·) on [0,1]).In the following lemma, we gather together some simple particular sufficient conditions that ensure the existence of the Riemann-Liouville fractional integral of the functions fromHp(E).Lemma 8.
Letx:I→E be weakly measurable function. The RFPI of the function x of order α>0 makes sense a.e. on I if at least one of the following cases holds: (a)
x∈H0p(E), p∈[1,∞] and α≥1.(b)
x∈H0p(E), p>max{1,1/α}.(c)
x is strongly measurable which lies in Hp(E), where p∈[1,∞] provided that E is weakly complete or contains no copy of c0.
IfE is reflexive, this is also true for any x∈Hp(E), p∈[1,∞] and α>0.
In all cases,φ(Iαx)=Iαφx holds for every φ∈E∗.In the assertions ((a) and (b)), we find sufficient conditions needed for the existence of Iαx in the situation in which no restriction is placed on E. In the third assertion, the properties of E allow us to characterize a function x∈Hp(E) for which Iαx exists.Proof.
Firstly, assertion(a) is direct consequence of Proposition 2 (part (1)) since s→t-sα-1∈L∞[0,t] holds for almost every t∈I whenever α≥1.
Secondly, to prove assertion(b), let p>max{1,1/α} and q be the conjugate exponents to p. Since q(α-1)+1>0, we have that s→t-sα-1∈Lq[0,t] holds for every t∈I. Thus, the assertion (b) is now an easy consequence of Proposition 2 (part (2)).
Thirdly, to prove(c) we let α>0, p∈[1,∞] and x∈Hp(E) be strongly measurable. Since the strong measurability is preserved under multiplication operation on functions, the product t-·α-1x(·):[0,t]→E is strongly measurable on [0,t] for almost every t∈I. In view of Young’s inequality, for every φ∈E∗, the real-valued function,(17)s⟼φt-sα-1xs=t-sα-1φxs, is Lebesgue integrable on [0,t], for almost every t∈I. So the assertion (c) follows immediately from Proposition 2 (parts (3, 4)).
Similarly, whenE is reflexive, the result follows from part (3) of Proposition 2.
However, in all cases, the functions↦(t-s)α-1x(s) is Pettis integrable on [0,t] for almost every t∈I. That is, for almost every t∈I, there exists an element in E denoted by Iαx(t) such that(18)φIαxt=∫0tφt-sα-1Γαxsds=∫0tt-sα-1Γαφxsds=Iαφxt holds for every φ∈E∗. This completes the proof.Remark 9.
Ifx∈H0p(E) such that Iαx(t) does not exist for some t∈I, then it does not exist even when we “enlarge” the space E into F. To see this, let x:I→F such that x(I)⊂E. If Iαx(t) exists for some t∈I, then t-·α-1x(·)∈P([0,t],F). Since x assumes only values in E, it follows by the mean value theorem for Pettis integral (Theorem 4) that the RFPI of x should lie in E.Before we come to a deep study of the mathematical properties of the RFPI operator, let us take a look at the following miscellaneous examples.Example 10.
Letα>0. Define the function x:[0,1]→L∞[0,1] by x(t)=χ[0,t].
This function is weakly measurable, Pettis integrable on[0,1], and φx is a function of bounded variation (see, e.g., [18]). That is, x∈H0∞(L∞). Hence, in view of Lemma 8 with p=∞, the RFPI of x exists on [0,1]. Further, calculations (cf. [4]) show that(19)Iαxt·=t-·αΓ1+αχ0,t·∈L∞0,1.Example 11.
Letα>0. Define the function x from the interval [0,1] into the Hilbert space l2 as(20)x:t⟼11+t,12+t,13+t,…,t∈0,1. We note that(21)xtl22=∑n1n+t2≤∑n1n2<∞,t∈0,1. Thus, the function x is well defined. We claim that x is Dunford integrable on [0,1]. Once our claim is established, Lemma 8 guarantees the existence of Iαx on [0,1]. It remains to prove this claim and to calculate Iαx.
To see this, letφ∈l2∗=l2. According to the Riesz representation theorem on Hilbert spaces there exists a uniquely determined λ≔{λn}∈l2 such that φx(t)=∑nλn/(n+t). A standard arguments using Beppo Levi’s Theorem yields(22)∫01φxtdt≤∑n∫01λnn+tdt=∑nλnln⁡1+1n≤∑nλnn≤∑nλn<∞. So x is Dunford integrable on [0,1] and hence Pettis integrable on [0,1] since l2 is reflexive. Consequently, in view of Lemma 8, Iαx exists on [0,1]. To calculate the RFPI of x fix t∈[0,1] and let φ∈l2∗. We have(23)∫0tφt-sα-1Γαxsds=∫0tt-sα-1Γαφxsds=Iα∑n=1∞λnn+t. Since the series ∑nλn/(n+t) is uniformly convergent on [0,1], it follows in view of the generalized linearity of the fractional integrals [23, Lemma 5], that (cf. [21, Table 9.1])(24)∫0tφt-sα-1Γαxsds=∑n=1∞λntαnΓ1+αF121,1,α+1,-tn=φgt, where F12 is the Gauss hypergeometric function evaluated at (1,1,α+1,-t/n) and(25)gt≔tαnΓ1+αF121,1,α+1,-tn. Since(26)gtl22=∑ntαΓ1+αF121,1,α+1,-tn2≤F1221,1,α+1,-1Γ21+α∑n1n2<∞,t∈0,1, we see that g∈l2. Thus,(27)Iαxt=tαnΓ1+αF121,1,α+1,-tn,t∈0,1.Example 12.
Letα∈1/2,1. Define the countable-valued function x:[0,1]→c0 by(28)xt≔xnt,xn≔nχ1/n+1,1/n. This function is strongly measurable, Pettis integrable function on [0,1] (see, e.g., [10, 16]). We claim that x∈H0P(c0) with p∈(1/α,2]. Once our claim is established, Lemma 8 guarantees the existence of Iαx, α∈1/2,1 on [0,1]. It remains to prove this claim by showing firstly that HP(c0) and to calculate Iαx. To do this, let φ∈c0∗. Then there corresponds to φ a unique {λn}∈l1 such that φx=∑nλnxn. By noting that(29)λnxnp=∫01λnxntpdt1/p=λn∫1/n+11/nnpdt1/p=λnn1/qn+11/p≤λnn1/qn1/p≤λnn2/p-1≤λn holds for any p∈(1/α,2], a standard argument using Levi’s Theorem (or Lebesgue dominated convergence theorem) and Minkowski’s inequality yields (30)∫01φxtpdt1/p≤∫01∑n=1∞λnxntpdt1/p=limk→∞⁡∫01∑n=1kλnxntpdt1/p≤limk→∞⁡∑n=1kλnxnpp1/p≤∑n=1∞λnxnp≤∑n=1∞λn<∞. Thus, φx(·)∈Lp[0,1], p∈(1/α,2] holds for every φ∈c0∗ (that is, x∈HP(c0)). Since p>1/α>1, it follows by Proposition 2, in view of the strong measurability of x, that x∈H0P(c0). Owing to Lemma 8, we infer that the RFPI of x of any order α∈1/2,1 exists on the interval [0,1]. To compute this integral, fix t∈In0≔1/n0+1,1/n0 for some n0∈N, the set of natural numbers, and let φ∈c0∗. Then (31)∑n∫0tt-sα-1λnxnsds≤λn0∫1/n0+1tt-sα-1n0ds+∑n>n0λn∫1/n+11/nt-sα-1nds=n0λn0αt-1n0+1α+∑n>n0nλnαt-1n+1α-t-1nα≤n0λn0αt-1n0+1α+∑n>n0nλnαt-1n+1-t-1nα≤n0λn0αt-1n0+1α+∑n>n0nλnα1nαn+1α≤n0λn0αt-1n0+1α+∑n>n0λnα1n2α-1≤n0λn0αt-1n0+1α+∑n>n0λnα<∞.becauseα∈12,1. Therefore, by Beppo Levi Theorem it follows that(32)∫0tφt-sα-1Γαxsds=∫0tt-sα-1Γαφxsds=∫0tt-sα-1Γα∑nλnnχ1/n+1,1/nsds=∑n∫0tt-sα-1Γαλnnχ1/n+1,1/nsds=n0λn0Γ1+αt-1n0+1α+∑n>n0nλnΓ1+αt-1n+1α-t-1nα.
Consequently, we conclude that(33)Iαxt=0,…,0︸n0-1,n0Γ1+αt-1n0+1α,n0+1Γ1+αt-1n0+2α-t-1n0+1α,…, where the nonzero coordinate started from the n0th place. It can be easily seen that Iαx(t)∈c0 for any t∈[0,1]. Evidently, we have that(34)nt-1n+1α-t-1nα≤nt-1n+1-t-1nα≤n1nn+1α≤1n2α-1⟶0asn⟶∞,n0Γ1+αt-1n0+1α≤n0Γ1+α1n0-1n0+1α≤1n02α-1Γ1+α⟶0asn0⟶∞ hold for any t∈[0,1]. This yields I1αx(t)∈c0 for any t∈[0,1] (this is precisely what we would expect from Definition 5).Look at part(c) of Lemma 8, with α∈(0,1) and E=c0. Below we give an example showing that the strongly measurability hypothesis imposed on a function x∈H1(E)/H01(E) is not sufficient for the existence of Iαx even when x is Denjoy-Pettis integrable (cf. [16]).Example 13.
Letα∈(0,1). Define the strongly measurable function x:[0,1]→c0 by(35)xt≔xnt,xnt≔nn+1χInt-χJnt,where(36)In≔1n+1,n+1/2nn+1,Jn≔n+1/2nn+1,1n.
It is immediate that (cf. [16]) x is a well-defined, Denjoy-Pettis (but it is not Pettis) integrable on [0,1].
In what follows, we will show that the RFPI ofx does not exist on a subinterval of positive measure on [0,1].
To see this, we make use of Proposition3 as follows: define for each interval In=1/n+1,1/n the functionals φn∈c0∗ (required by Proposition 3) to be the corresponding to the element {λn}≔(0,0,…,0,1,0,0,…)∈l1 where the nonzero coordinate is in the nth place. Then φn(x(t))=xn(t) and thus |φn(x(s))|=n(n+1)[χIn(t)-χJn(t)]. Clearly, the family {φn} runs through the unit ball of the dual of c0. Evidently, the isometric isomorphism between c0∗ and l1 yields φnc0∗={λn}l1=1.
Now taket∈1/2,1 and define the continuous function f:1/2,1→R+ by(37)ft≔t-1n+1α-t-1nα,n≥2,t∈12,1. It can be easily seen that, in view of 0<α<1, f′ is negative on (0,1). By standard results from (classical) calculus, it follows that f is strictly decreasing on [0,1], in particular f(1)<f(t) for all t∈0,1. Thus, for any n≥2 and any φ∈{φn} we have(38)∫Inφt-sα-1Γαxsds=∫Int-sα-1Γαφxsds=∫1/n+11/nt-sα-1Γαnn+1ds=nn+1Γ1+αt-1n+1α-t-1nα≥n2Γ1+α1-1n+1α-1-1nα. An explicit calculation using L’Hospital’s rule two times reveals (39)limn→∞1-1/n+1α-1-1/nαn-2=-α2limn→∞n3nn+1α-11n+12-1-1nα-11n2=-α2limn→∞nnn+1α-1n2n+12-1-1nα-1=-α2limn→∞nnn+1α+1-1nα-1=-α2limn→∞n2α+1nn+1α1n+12-α-11-1nα-21n2=-α2limn→∞α+1nn+1α+2-α-11-1nα-2=α,
from which it follows that(40)∫Inφt-sα-1Γαxsds⟶0asn⟶∞.
Therefore, in view of Proposition3, s↦(t-s)α-1x(s)∉P[0,1],c0. Hence, Iαx does not exist, which is what we wished to show.The following theorem provides a useful characterization of the spaceHp(E), for which the statements reveal how much the fractional integral Iα isbetter than the function x∈Hp(E). Indeed, based on Lemma 8 using an inequality of Young, we can easily prove the following.Lemma 14.
For anyα>0, the following holds.(a)
IfE is reflexive, then for every p∈[1,∞], the operator Iα maps Hp(E) into H0p(E). In particular, if 0<α<1, then Iα:H1(E)→H01/(1-α)-ϵ(E) however small ϵ>0 is.(b)
IfE contains no copy of c0, then for every p∈[1,∞], the operator Iα maps {x∈H0p(E):xstrongly  measurable} into H0p(E). In particular, if α≥1, the operator Iα maps H0p(E) into itself.(c)
Ifp>max{1,1/α}, the operator Iα sends H0p(E) to C[I,Eω] (if we define Iαx(0)≔0). In particular, if α∈(0,1), Iα:H0p(E)→Hα-1/p(E).Proof.
At the beginning, letα>0 and x∈Hp(E) with p∈[1,∞] and define the real-valued function g:I→R by g(t)≔tα-1/Γ(α). If E is reflexive, it follows, in view of Lemma 8, that Jαx exists a.e. in I and φ(Iαx)=Iαφx for every φ∈E∗. Young’s inequality guarantees that the convolution product g∗φx(·) lies in Lp[I]. Consequently φ(Iαx)=Iαφx=g∗φx∈Lp[I] for every φ∈E∗. The reflexivity of E together with Proposition 2 yields Iαx∈H0p(E).
In particular, if0<α<1, it can be easily seen that ∫Igtqdt<∞ for every q∈[1,1/(1-α)). That is, g∈L1/(1-α)-ϵ[I],ϵ>0. By Young’s inequality, it follows that φ(Iαx)=g∗φx∈L1/(1-α)-ϵ[I] for every φ∈E∗ however small ϵ>0 is. Now, the assertion (a) follows because of the reflexivity of E.
Next, in order to prove the assertion(b) let x∈H0p(E) and note, in view of Lemma 8, that Iαx exists on I. Define y:I→E by y(t)≔Iαx(t),x∈H0p(E). Since φy(·)∈Lp[I] for every φ∈E∗, it follows that y∈Hp(E). Moreover, for any a,b∈I(a<b), we have (41)∫abφysdt=∫abIφxtdt=1Γα∫ab∫0tt-sα-1φxsdsdt=1Γα∫ab∫sbt-sα-1φxsdtds+∫0a∫abt-sα-1φxsdtds=1Γα∫abb-sαφxsds+∫0ab-sα-a-sαφxsds=1Γα∫0ab-sαφxsds+∫abb-sαφxsds-∫0aa-sαφxsds=∫0bb-sαΓ1+αφxsds-∫0aa-sαΓ1+αφxsds=φxa,b, where(42)xa,b≔P∫0bb-sαΓ1+αxsds-P∫0aa-sαΓ1+αxsds. Since x∈P[I,E], owing to Proposition 2 (part (1)), it follows that b-·αx(s) and a-·αx(s) are Pettis integrable on I and so x[a,b]∈E.
A combination of these results yieldsφy∈Lp(I) for every φ∈E∗ and there exists an element x[a,b]∈E such that ϕx[a,b]=∫abφy(t)dt for every a,b∈I. Since E contains no copy of c0, it follows that (cf. [17, Theorem 23.]) y∈P[I,E]. Consequently y(·)∈H0p(E). This is the claim (b).
To prove the assertion(c), let p>max{1,1/α} and x∈H0p(E). By Lemma 8, we deduce that Iαx exists a.e. on I. Now, let q∈[1,∞] be such that 1/p+1/q=1. Since q(α-1)>-1, g≔·α-1/Γ(α)∈Lq[I]. Therefore, as a direct consequence of Young’ inequality it follows that(43)φIαx·=Iαφx·=g∗φx·∈CI,R holds for every φ∈E∗. Now, we claim that Iαx(t)→0 in E as t→0. Once our claim is established, the definition Iαx(0)≔0 guarantees that Iαx(·)∈C[I,Eω], which is what we wished to show. It remains to prove our claim: without loss of generality, assume that Iαx(t)≠0. Then there exists (as a consequence of the Hahn-Banach theorem) φ∈E∗ with φ=1 such that Iαx(t)=φIαx(t)=Iαφx(t). By Hölder’s inequality we obtain (44)φIαxt≤1Γα∫0tt-sqα-1ds1/q∫0tφxspds1/p≤1Γαtα-1q+1α-1q+11/q∫Iφxspds1/p=tα-1+1/qΓαqα-1+11/qφxLp=tα-1/pφxLpΓαqα-1+11/q. This is equivalent with the following estimate:(45)Iαxt≤tα-1/pφxLpΓαqα-1+11/q.
Owing toα>1/p we get Iαx(t)→0 in E as t→0 and consequently, in view of Iαx(0)≔0, Iαx(·)∈C[I,Eω]. This proves the first part of the assertion (c).
Finally, letα∈(0,1) and t,τ∈I. Without loss of generality, assume τ<t. Then for any φ∈E∗ we have by Hölder inequality with p>1/α, in view q(α-1)>-1, that(46)φIαxt-IαxτΓα=∫0tt-sα-1φxsds-∫0ττ-sα-1φxsds≤∫0τt-sα-1-τ-sα-1φxsds+∫τtt-sα-1φxsds=∫0τt-sα-1-τ-sα-1qds1/q+∫τtt-sα-1qds1/qφxLpI≤∫0τt-sqα-1-τ-sqα-1ds1/q+t-τα-1q+1α-1q+11/qφxLpI=∫0τt-sqα-1-τ-sqα-1ds1/q+t-τα-1/pqα-1+11/qφxLpI. By noting that (α-1)q∈(-1,0) when α∈(0,1), it can be easily seen that(47)∫0τt-sqα-1-τ-sqα-1ds≤1qα-1+1t-τqα-1+1.That is,(48)∫0τt-sqα-1-τ-sqα-1ds1/q≤t-τα-1/pqα-1+11/q.A combination of these results yields(49)φIαxt-φIαxτ=φIαxt-Iαxτ≤φxLpIΓα2t-τα-1/pqα-1+11/q. Thus, Iαx∈Hα-1/p(E). This completes the proof.Corollary 15.
For anyα>0, Iα:C[I,Eω]→C[I,Eω].Proof.
Letx∈C[I,Eω], then φx∈L∞ holds for every φ∈E∗. By noting that the weak continuity implies a strong measurability ([24] page 73), it follows that x is strongly measurable on I. Hence, in view of Proposition 2, x∈H0∞(E). Consequently C[I,Eω]⊆H0p(E) for every p∈[1,∞]. That is, Iα:C[I,Eω]→C[I,Eω]. Hence, the desired result is obtained.Example 16.
Letα∈1/2,1. Define the function x:[0,1]→c0 by x(t)=nχ1/n+1,1/n. We have (in view of Example 12) that(50)Iαxt=0,0,…,0,n0t-1/n0+1αΓ1+α,n0+1Γ1+αt-1n0+2α-t-1n0+1α,….To show that Iαx is norm continuous on [0,1], let τ,t∈[0,1]. With no loss of generality, we may assume that t,τ∈In0, for some n0∈N. Since the nonzero terms of the sequence in (50) are nonincreasing, then, in view of(51)t-τ≤1n0-1n0+1=1n0n0+1≤1n02, we have(52)Iαxt-I1αxτc0=n0Γ1+αt-1n+1α-τ-1n+1α≤n0Γ1+αt-1n+1-τ-1n+1α≤n0Γ1+αt-τα≤1Γ1+αt-ταt-τ=t-τα-1/2Γ1+α. Thus, the RFPI of x is norm continuous on [0,1]. Precisely, since 0<α-1/2<α-1/p<1, for any p≤2, then Iαx∈Hα-1/2(E)⊂Hα-1/p(E). This is precisely what one would expect from Lemma 14 (part (c)).
Analogously, an explicit calculation reveals thatIαx lies in C([0,1], L∞[0,1]), where x:[0,1]→L∞[0,1] defined by Example 10.We now consider additional mapping properties of the operatorIα. Precisely, we will show that the RFPI enjoys the following commutative property which is folklore in case E=R. However, the proof is completely similar to that of [8], Lemma 3.5.Lemma 17.
Letα,β>0 and p>max1,1/α,1/β. Then(53)IαIβx=IβIαx=Iα+βx holds for every x∈H0p(E). If E is reflexive, this is also true for every p≥1.Proof.
Sincep>max1,1/α,1/β, it follows that p>max1,1/(α+β). So, by Lemmas 8 and 14, the mappings Iαx, Iβx, and Iα+βx belong to H0∞(E) for every x∈H0p(E). Now repeating the same process as in ([8], Lemma 3.5) implies the claim.
WhenE is reflexive, the result follows as a direct application of Lemma 14.


## 3. Fractional Derivatives of Vector-Valued Functions

After the notation of the fractional integrals of vector-valued functions, the fractional derivatives become a natural requirement. Before we come to the definitions and a detailed study of the mathematical properties of fractional differential operators, we recall the following.Definition 18.
Consider the vector-valued functionx:I→E:(1)
Letφx be differentiable on I for every φ∈E∗. The function x is said to be weakly differentiable on I if there exists y:I→E such that for every φ∈E∗ we have(54)dφxtdt=φyt,for  everyt∈I.The functiony is called the weak derivative of the function x.(2)
Letφx be differentiable a.e. on I for every φ∈E∗ (the null set may vary with φ∈E∗). The function x is said to be pseudo differentiable on I if there exists a function y:I→E such that for every φ∈E∗ there exists a null set N(φ)⊂I such that(55)φxt′=φyt,for  everyt∈INφ.In this case, the functiony is called the pseudo derivative of x.Ifx is pseudo differentiable on I and the null set invariant for every φ∈E∗, then x is a.e. weakly differentiable on I.Clearly, ifx is a.e. weakly differentiable on I, then φ(x) is a.e. differentiable on I. The converse holds in a weakly sequentially complete space (see [25], Theorem 7.3.3).For more details of the derivatives of vector-valued functions we refer to [10, 12, 26].The following results play a major role in our analysisProposition 19 (see [27], Theorem 5.1).
The functiony:I→E is an indefinite Pettis integrable, if and only if y is weakly absolutely continuous on I and have a pseudo derivative on I. In this case, y is an indefinite Pettis integral of any of its pseudo derivatives.Now we are in the position to define the fractional-type derivatives of vector-valued functions.Definition 20.
Letx:I→E. For the positive integer m such that α∈(m-1,m), m∈N0≔0,1,2,… we define the Caputo fractional-pseudo (weak) derivative “shortly CFPD (CFWD)” of x of order α by(56)dαdtαxt≔Im-αDmxt, where the sign “D” denotes the pseudo (or weak) differential operator. We use the notation dpα/dtα and dωα/dtα to characterize the Caputo fractional-pseudo derivatives and Caputo fractional weak derivatives, respectively.It is well known that, although the weak derivative of a weakly differentiable function is uniquely determined, the pseudo derivative of the pseudo differentiable function is not unique. Also, although any two pseudo derivativesy, z of a function x:I→E need not be a.e. equal (see [12, Example 9.1] and [26, p. 2]), the functions y, z are weakly equivalent on I (that is, φy=φz holds a.e. for every φ∈E∗). The next lemma provides a useful characterization property of the CFPD. Really, it can be easily seen that the CFPD of a Caputo fractional-pseudo differentiable function does not depend on the choice of a pseudo derivative of the function.Lemma 21.
Letx:I→E be pseudo differentiable function where pseudo derivatives lie in H0p(E). If dpα/dtαx exists on I, then the CFPD of x depends on the choice of the pseudo derivatives of x.Proof.
Lety,z∈H0p(E) be two pseudo derivatives of the pseudo differentiable function x. Since x and y are weakly equivalent on I, then for every φ∈E∗ there exists a null set N which depends on φ∈E∗ such that t-s-αφy(s)=t-s-αφz(s) for every s∈[0,t]/N. Consequently, for almost every t∈I,(57)φI1-αyt=I1-αφyt=I1-αφzt=φI1-αzt holds for every φ∈E∗. Thus, (58)dpαdtαx=I1-αDpx=I1-αy=I1-αzonI,which is what we wished to show.We consider the following examples.Example 22.
Definex:[0,1]→l2 by(59)xt≔ln⁡1+tn,n∈N.We claim that the CFWD of x exists on [0,1]. To see this, we let φ∈(l2)∗=l2. Then there corresponds to φ a sequence {λn}∈l2 such that(60)φxt=∑nλnln⁡1+tn.Since the series in (60) converges uniformly on [0,1], the formal differentiation of φx yields(61)φxt′=∑nλnn+t=φ1t+n.It is not hard to justify the differentiation by noting that the series in the right hand side of (61) is uniformly and absolutely convergent on [0,1]. Consequently, φx is differentiable on [0,1] for φ∈(l2)∗, meaning that x is weakly differentiable on [0,1] (because l2 is weakly complete (cf. [25], Theorem 7.3.3) and Dωx(t)=1/t+n. Owing to Lemma 25, the CFWD of any order α∈(0,1) of x exists on [0,1]. To calculate the CFWD of x on [0,1] we observe, in view of Example 11, that(62)dωαdtαxt=I1-αDωxt=t1-αnΓ2-αF121,1,2-α,-tn.Example 23.
LetE be an infinite dimension Banach space that fails cotype. Define y:[0,1]→E by y(t)≔∫0tx(s)ds, where x:[0,1]→E given by formula (7). As cited in ([13], Corollary 4), y is nowhere weakly differentiable on [0,1]. Consequently the CFWD of any order α∈3/4,1 of y loses its meaning on [0,1].Remark 24.
As shown in Example23, there is an infinite dimension Banach space E and weakly absolutely continuous function y:I→E which is nowhere weakly differentiable (hence the CFWD of y does not exist). Also, even when E is separable, and y is Lipschitz function, the pseudo derivatives (hence the CFPD) of y need not to exist [28].
However, Definition20 of the CFPD (CFWD) has the disadvantage that it completely loses its meaning if the function x fails to be pseudo (weakly) differentiable. Precisely, the CFPD (in particular the CFWD) of a function x loses its meaning if x is not weakly absolutely continuous.The next lemma gives sufficient conditions that ensure the existence of the Caputo fractional derivatives of a functionx∈Hp(E).Lemma 25.
Letα∈(0,1). For the function x:I→E, the following hold: (a)
Ifx has a pseudo derivative Dpx∈H0p(E), where p>1/(1-α), then the CFPD of x of order α exists on I. Moreover, dpα/dtαx∈C[I,Eω].(b)
IfE is weakly complete or contains no copy of c0 and if x has a weak derivative Dωx∈Hp(E), where p∈[1,∞], then the CFWD of x of order α exists on I. Moreover, dωα/dtαx∈C[I,Eω].
This holds for anyx∈Hp(E) with p∈[1,∞] if E is reflexive.
In all cases,φdpα/dtαx=dpα/dtαφx  φdωα/dtαx=dωα/dtαφx holds for every φ∈E∗.Proof.
Since the weak (pseudo) derivative of an a.e. weakly (pseudo) differentiable function is strongly (weakly) measurable [12, 26, 28], the proof is readily available, in view of the definition of Caputo fractional derivatives and Lemma 8. Moreover, since Dωx(Dpx)∈H0p(E) with p>1/(1-α), it follows, in view of Lemma 14, that I1-αDωx(I1-αDpx)∈C[I,Eω]. This completes the proof.Besides the Caputo fractional-pseudo (weak) derivatives, we define the Riemann-Liouville fractional-pseudo (weak) derivatives.Definition 26.
Letx:I→E. For the positive integer m such that α∈(m-1,m), m∈N0≔0,1,2,… we define the Riemann-Liouville fractional-pseudo (weak) derivative “shortly RFPD (RFWD)” of x of order α by(63)Dαxt≔DmIm-αxt,where D is defined as in Definition 20. We use the notation Dpα (Dωα) to characterize the Riemann-Liouville fractional-pseudo (weak) derivatives.Clearly, in infinite dimension Banach spaces, the weak absolute continuity ofIm-αx, is necessarily (but not sufficient) condition for the existence of RFPD (in particular RFWD) of x.Lemma 27.
Let0<α<1. For any x∈H0p(E) with p>max{1/α,1/(1-α)}, we have(64)DpαIαx=xa.e.If E is reflexive, this is also true for every p≥1.Proof.
Our assumptionx∈H0p(E) with p>max1/α,1/1-α yields, in view of Lemma 17, that(65)DpαIαx=DpI1-αIαx=DpI1x.The claim now follows immediately, since (cf. Proposition 19) the indefinite integral of Pettis integrable function is weakly absolutely continuous and it is pseudo differentiable with respect to the right endpoint of the integration interval and its pseudo derivative equals the integrand at that point.Remark 28.
When we replaceDpα by Dωα, then Lemma 27 is no longer necessarily true for arbitrary x∈H01(E) even when E is reflexive: evidently, in [13, remark below Corollary 4] the existence of a reflexive Banach space E and a strong measurable Pettis integrable function x:I→E was proved such that x has nowhere weakly differentiable integral. In this case, DωI1x lost its meaning. This gives a reason to believe that (65) (hence (64)) with Dωαx could not happen.However, we have the following result.Lemma 29.
Let0<α<1. For any x∈C[I,Eω], we have(66)DωαIαx=xonI.Proof.
Our assumptionx∈C[I,Eω]⊂H0∞(E) (see the proof of Corollary 15) yields, in view of Lemma 17, that(67)DωαIαx=DωI1-αIαx=DωI1x.The claim now follows immediately, since the indefinite integral of weakly contentious function is weakly absolutely continuous and it is weakly differentiable with respect to the right endpoint of the integration interval and its weak derivative equals the integrand at that point.The following lemma is folklore in caseE=R, but to see that it also holds in the vector-valued case, we provide a proof.Lemma 30.
Let0<α<1 and p>max{1/α,1/(1-α)}. If the function x:I→E is weakly absolutely continuous on I and passes a pseudo derivative in H0p(E), then(68)dpαxdtαt=Dpαxt-tαΓ1+αx0.In particular, if x passes a weak derivative in C[I,Eω], then(69)dωαxdtαt=Dωαxt-tαΓ1+αx0.
IfE is reflexive, this is also true for every p≥1.Proof.
We observe that, under the assumption imposed onDpx together with Proposition 3, the weakly absolutely continuity of x is equivalent to(70)xt=x0+∫0tDpxsds. Hence, owing to Lemma 27, it follows that (71)dpαxdtα=I1-αDpx=DpαIαI1-αDpx=DpαIαI1-αDpx=DpI1-αI1Dpx=DpI1-αxt-x0=Dpαxt-tαΓ1+αx0, which is what we wished to show.
The proof of (69) is very similar to that in (68); therefore, we omit the details.Remark 31.
As we remark above, the definition of the CFPD of a functionx loses completely its meaning if x is not weakly absolutely continuous. For this reason, we are able to use Lemma 30 to define the Caputo fractional derivative in general; that is, we put(72)dpαxdtαt≔Dpαxt-tαΓ1+αx0,0<α<1.Similarly, we define(73)dωαxdtαt≔Dωαxt-tαΓ1+αx0.However, Lemma 30 claims that, for the weakly absolutely continuous functions having Pettis integrable pseudo [weak] derivatives, definitions (72) and (73) of the Caputo fractional-pseudo [weak] derivatives coincide with Definition 20.


## 4. An Application

Letf:I×E→E be given function. Consider the boundary value problem of the fractional type(74)dωαdtαyt=λft,yt,α∈0,1,t∈0,1,λ∈R,y0+by1=h,with certain constants b,  h∈R,  b≠-1. To obtain the integral equation modeled off the problem (74), we let y be a weakly continuous solution to problem (74); thenformally we have(75)Iαdωαdtαyt=λIαft,yt, that is(76)IαI1-αDωyt=λIαft,yt⟹I1Dωyt=λIαft,yt. This reads (cf. [29])(77)yt=y0+λIαft,yt,α∈0,1,t∈I,λ∈R,with some (presently unknown) quantity y(0).Now, we solve (77) for y(0) by y(0)+by(1)=h, and it follows that(78)1+by0+λbΓα∫011-sα-1fs,ysds=h. Thus,(79)y0=h1+b-λb1+bIαf1,y1.Assume that the functionf is weakly-weakly continuous function such that(1)
for anyr>0, there is a constant M>0 such that f(t,y)≤M for all t∈I and y≤r,(2)
there exists a nondecreasing continuous functionψ:0,∞→[0,∞), ψ(0)=0 and ψ(t)<t for all t>0 such that βf(T×A)≤ψ(βA) for every bounded set A⊂E, where β stands for De Blasi’s weak measure of noncompactness (see [30]).Occasionally, iff:I×E→E is weakly-weakly continuous and E is reflexive, then the assumptions (1) and (2) are automatically satisfied (see, e.g., [31]).Theorem 32 (see [6], Theorem 3.3).
Letα∈(0,1) and |λ|<Γ(1+α). If f:I×E→E is weakly-weakly continuous and satisfies the assumptions (1) and (2), then the integral equation (77) has a weakly continuous solution y defined on [0,1].Proof.
We omit the proof since it is almost identical to that in the proof in ([6], Theorem 3.3) with (small) necessary changes.In the following example we assume thaty∈C[I,Eω] solves (77) and we will show that, not only do we have that y no longer necessarily solves (74) (when the Caputo fractional weak derivative is taken in the sense of Definition 20), but even worse, it could happen that the problem (74) is “meaningless” on I.Example 33.
Letα∈0,1/4. Let E be an indefinite dimensional reflexive Banach space that fails cotype. Define the weakly-weakly continuous function f:[0,1]×E→E by f≔I1-αx, where x:I→E is defined by formula (7). Obviously (cf. Example 6, in view of Remark 7), f satisfies the assumptions of Theorem 32.
Now, consider the integral equation (77) with f=I1-αx. Namely, we consider the integral equation(80)yt=y0+λIαI1-αxt,α∈0,14,t∈I,λ∈R.Obviously, the solution y is weakly continuous on [0,1] (this is an immediate consequence of Remark 7 and Lemma 17).
Sinceφx∈L1[0,1] holds for every φ∈E∗, it follows by the commutative property of the fractional integral operators calculus over the space of Lebesgue integrable functions (see [21], Section 2.3) that(81)φIαI1-αxt=IαφI1-αxt=IαI1-αφxt=I1φxt=φI1xt, that is(82)φIαI1-αxt-I1xt=0,for  everyφ∈E∗. Hence, IαI1-αx=I1x. Consequently, problem (80) becomes(83)yt=y0+λI1xt,α∈0,14,t∈I. As showed in ([13], Corollary 4), the indefinite integral t→∫0tx(s)ds is nowhere weakly differentiable on [0,1]. Thus, y is nowhere weakly differentiable on [0,1]; hence the CFWD of y (hence the boundary value problem) completely loses its meaning if the CFWD is taken in the sense of Definition 20. This is what we wished to show.Now, we are in the position to state and prove the following existence theorem.Theorem 34.
Iff:I×E→E is a function such that all conditions from Theorem 32 hold, then problem (74) (where the Caputo fractional weak derivative is taken in the sense (73)) has a weak solution on [0,1].Proof.
Lety∈C[I,Eω] be a solution to (77). Equip E andI ×E with weak topology and note that t↦(t,y(t)) is continuous as a mapping from I into I×E. Since f:I×E→E is weakly-weakly continuous on [0,1], then f(·,y(·)) is a composition of this mapping with f and thus f(·,y(·)) is weakly continuous on [0,1].
Now, operating by the operatorDωα on both sides of (77), it follows by Lemma 29 that(84)Dωαy=Dωαy0+λDωαIαf=t-αΓ1-αy0+λf.
Now, insert definition (73) of the Caputo fractional derivative, and we get(85)dωαdtαyt=λft,yt. With some further efforts, one can get the boundary condition y(0)+by(1)=h.
Therefore,y satisfies problem (74) which is what we wished to show.



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