RACSAM https://doi.org/10.1007/s13398-018-0531-y ORIGINAL PAPER

Hadamard-type fractional calculus in Banach spaces Hussein A. H. Salem1

Received: 3 October 2017 / Accepted: 27 March 2018 © Springer-Verlag Italia S.r.l., part of Springer Nature 2018

Abstract In this pages, we present the definitions and some properties of the Hadamard-type fractional Pettis integrals (and corresponding fractional derivatives) for the functions that take values in Banach space. Further, we show that a well known properties of the Hadamard-type fractional calculus over the space of real-valued functions also hold in Banach spaces. Some emphasizes examples are demonstrated. Meanwhile, we construct an example of a function that has no pseudo derivative everywhere, but has a Hadamard-type fractional derivative. As far as we know, the topic of this paper was never investigated before, and so is new. Keywords Fractional calculus · Pettis integrals 2000 Mathematics Subject Classification 26A33 · 34G20

1 Introduction and preliminaries The topic fractional calculus in Banach spaces when endowed with their weak topology has been studied for the first time by the Salem and El-Sayed [28]. It is worth mentioning that, following the appearance of [28], there has been a significant interest in the study of this topic (see e.g. [1–3] and [29–36]). The above investigations were not complete, however, all the above investigations have focused on using Riemann-Liouville fractional calculus of a vector-valued functions, however, our destination here is to consider the so called Hadamardtype fractional calculus of a vector-valued functions (see [19,37] for the definition ) where the kernel of the integral (in the definition of Hadamard-type integral) contains logarithmic function of arbitrary exponent. Although the Hadamard-type fractional calculus is an old topic, it is go back to the works of Hadamard in 1892, this type of fractional calculus is not yet well studied and much exists to be done. However, interest in this topic has been

B 1

Hussein A. H. Salem [email protected] Department of Mathematics and Computer Science, Faculty of Sciences, Alexandria University, Alexandria, Egypt

H. A. H. Salem

rising over the last few years. Nevertheless, the Hadamard-type fractional calculus for realvalued functions has been treated by many authors (see e.g. [4,7,8,20,21,43] and references cited therein). As a pursuit of this, in what follows, we define the Hadamard-type fractional calculus of vector-valued functions in terms of Pettis-integral and the corresponding fractional derivatives. Also, we show that the well-known properties of the Hadamard-type fractional calculus over the domains of the Lebesgue integrable functions also hold in the Pettis space. Some illustrative examples are given. For convenience, here we present some notations and the main properties for Pettis integral and corresponding derivatives. For further background, unexplained terminology and details pertaining to this paper can be found in the monograph [10] by Diestel and Uhl Jr., and the paper [27] of J. Pettis. Through this paper, the triple (I, , μ), will denote a measure space where μ denotes the Lebesgue measure and I = [a, b], 0 < a < b < ∞, endowed with the Lebesgue σ −algebra (I ) (that is, the sigma-algebra generated by the topology induced by the set  τ , where τ is the topology induced by the Euclidean metric on R and  is the collection of null sets in R). Also, E considered to be a Banach space with norm · and dual E ∗ . By E ω we denote the space E when endowed with the weak topology generated by the continuous linear functionals on E. We will let C[I, E] denotes the Banach space of strong continuous functions x : I → E endowed by the norm x0 = supt∈I x(t). Further, denote by C[I, E ω ] the Banach space of weakly continuous functions on I , 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] and [27] for the definition). Recall that a map T : X −→ Y , X and Y are Banach spaces, is said to be weakly continuous if and only if T is continuous with respect to the weak topologies on X and Y , (i.e. if x α → x weakly, where (xα ) is a net in X , then T (xα ) → T (x) weakly on Y ( see [6] and [13] ). Finally we recall that every reflexive space is weakly sequentially complete, but the converse is not (in general) true. For convenience, we remember that (a) The space 1 has no copy of c0 but it is not reflexive, (b) The space L 1 (I ) is weakly complete and separable but it is not reflexive (c) he space 2 (I ) of countably nonzero functions on I is non-separable weakly complete space, (d) The space c0 is an example of a separable, but not weakly complete space. Definition 1.1 For 1 ≤ p ≤ ∞, we define the class H p (E) to be the class of all weakly measurable functions x : I → E having ϕx ∈ L p (I ) for every ϕ ∈ E ∗ . If p = ∞, the added condition l.u.b.ϕ=1 (ess supt∈I |ϕx(t)|) < ∞ must be satisfied by each x ∈ H∞ (E). The class H0 (E) is defined by p

p

H0 (E) := {x ∈ P[I, E] : ϕx ∈ L p (I ).}

In the following proposition, we summarize some important facts which are the main tools in carrying out our investigations (see [10,14,15] and [27]) Proposition 1.1 1. If E is reflexive, the weakly measurable function x : I → E is Pettis integrable on I if and only if ϕ(x(·)) ∈ L 1 (I ) for very ϕ ∈ E ∗ . That is, if E is reflexive p then H p (E) ≡ H0 (E) holds for every p ∈ [1, ∞], 2. If E contain no isometric copy of C0 . The strongly measurable x : I → E is Pettis integrable on I if and only if, ϕ(x(·)) ∈ L 1 (I ) for every ϕ ∈ E ∗ ,

Hadamard-type fractional…

3. If x(·) is Pettis integrable and u(·) is measurable and essentially bounded real-valued function, then x(·)u(·) is Pettis integrable, p 4. In order that x(·) be in H0 (E), it is necessary and sufficient that x(·)u(·) be Pettis integrable for every u(·) ∈ L q (I ). Proposition 1.2 Let p > 1. If the strongly measurable x : I → E is in H p (E) then, x is in p H0 (E). If E weakly sequentially complete , this is also true for p = 1. It is well-known that the absolutely continuous real-valued function is a.e. differentiable. This one of the few properties of the real-valued function that is not carry over in arbitrary Banach spaces. We remark that the a.e. weakly differentiable functions are pseudo-differentiable, but the converse implication is not true (this is a trivial consequence of the definitions [10,27,39]). It is worth to remark here that, while the weak derivative of a weakly-differentiable function is uniquely determined, the pseudo-derivative of a pseudo-differentiable function is not unique and two pseudo-derivatives need not be a.e. equal ( see [27, Example 9.1] and [39, p.2 ]). However, the pseudo-derivatives of a pseudo-differentiable function are weakly equivalent. Moreover, if the image of a pseudo-differentiable function lie in Banach space having a countable determining set, then any two pseudo derivatives of this function must be a.e. equal (see [39]). We would like to mention that, the weak absolute continuity of a function x : I → E yields ϕx is a.e. differentiable on I every ϕ ∈ E ∗ , but this gives no guarantee for existence of a pseudo-derivative of x on I . However, even when E is separable, and x is a Lipschitz function, the pseudo-derivative of x need not to exist [40]. It was cited in [27,39,40], that the weak (pesudo) derivative of an a.e. weakly (pesudo) differentiable function is strongly (weakly) measurable. The following results play a major rule in our analysis Proposition 1.3 (see [25], Theorem 5.1) The function y : [0, b] → E is an indefinite Pettis integrable, if and only if, y is weakly absolutely continuous on [a, b] and have a pseudo derivative on [a, b]. In this case, y is an indefinite Pettis integral of any of its pseudo derivatives. A fundamental property of Pettis integral contained in Proposition 1.4 ([27] Corollary 2.51). If x ∈ P[I, E], then for any bounded subset  of elements of E ∗ , the integrals  |ϕ(x(s))| ds, ϕ ∈  J

are weakly equi-absolutely continuous. Meaning that,  lim sup |ϕx(s)| ds = 0, x ∈ P[I, E]. μ(J )→0 ϕ∈

J

We close this section by introducing the following result (proved by Pettis [27]) Theorem 1.1 A function x : I → E is strongly measurable if and only if, x is weakly measurable and almost separably valued. In particular, if E is separable, the strong and weak measurability are equivalent.

H. A. H. Salem

2 Hadamard-type fractional calculus of vector-valued functions Throughout this section, we outline some aspects of Hadamard-type fractional calculus in Banach spaces. Devoted by the definition of the Hadamard fractional integral of real-valued function, we introduce the following Definition 2.1 Let x : I → E. The Hadamard-type fractional Pettis-integral (shortly HFPI) of x of order α > 0 is defined by   t 1 t α−1 x(s) Jaα x(t) := log ds, t > a > 0. (1)

(α) a s s  In the preceding definition the sign ” ” stands the Pettis integral. Obviously, the HFPI of a weakly measurable function x : I → E exists on I if, there is an element J ∈ E corresponding to each J := [a, t] ⊂ I such that   t t α−1 ϕ(x(s)) 1 ϕ( J ) = ds, holds for all ϕ ∈ E ∗ , t ∈ I, log

(α) a s s where the integral on the right is supposed to exist in the sense of Lebesgue. The element J is called the Hadamard-type fractional Pettis-integral of x over J . The element J is called the HFPI of x over I . We use the notation Jaα x(t) := J .  α−1 x(s) The function Jaα x makes sense on I if and only if, the integral of s  → log st s exists for every ∈ I . In the case when E = R, it is well-known (see e.g.[19,37]) that the linear fractional integral operator Jaα is defined on the space L p (I ), p ∈ [1, ∞]. That is, for any x ∈ L p (I ) and t ∈ I , the function s  → [log(t/s)]α−1 x(s)/s is Lebesgue integrable on [a, t]. Also, as a consequence of Hölder’s inequality, it can be easily shown that the operator Jaα , sends L p (I ) continuously into L p (I ) for each p ∈ [1, ∞] . Now, we are in the position to state and prove a lemma describing particular sufficient conditions for the existence of the Hadamard-type fractional integral of the functions from H p (E). Clearly, the values of α > 0 and p ∈ [1, ∞] allow us to characterize the space E and to characterize the function x ∈ H p (E) for which Jaα x exists. Lemma 2.1 The HFPI of a function x : I → E of order α > 0 makes sense if at least one of the following cases holds: (a) E is reflexive, α > 0 and x ∈ H p (E), with p ∈ [1, ∞], (b) E is separable space and either weakly complete or contains no copy of c0 , α > 0 and x ∈ H p (E) with p ∈ [1, ∞]. p (c) α ≥ 1 and x ∈ H0 (E), p (d) α > 0 and x ∈ H0 (E), p > max{1, 1/α}. In all cases, ϕ(Jaα x) = Jaα ϕx holds for every ϕ ∈ E ∗ . Proof Firstly, let x ∈ H p (E), α > 0 and p ∈ [1, ∞]. Since the real-valued function ϕx ∈ L p [a, t], for every ϕ ∈ E ∗ and every t ∈ I , it follows that the measurable function   [log(t/s)]α−1 α−1 x(s) = ϕx(s) s  → ϕ [log(t/s)] s s is Lebesgue integrable on [a, t] for every t ∈ I . So, if E is reflexive, the assertion (a) follows immediately from Proposition 1.1 part (1).

Hadamard-type fractional…

Secondly, let α > 0 and x ∈ H p (E) with p ∈ [1, ∞]. Because of the separability of E, it follow in view of Theorem 1.1 that the function s  → [log(t/s)]α−1 x(s) s is strongly measurable on [a, t] for every t ∈ I . Moreover, for every ϕ ∈ E ∗ , the real-vaued function   x(s) 1 = [log(t/s)]α−1 ϕx(s) s  → ϕ [log(t/s)]α−1 s s is Lebesgue integrable on [a, t], for every t ∈ I . So the assertion (b) for the case when E contains no copy of c0 follows immediately from Proposition 1.1 part (2) and follows from Proposition 1.2 when E is weakly complete. p Thirdly, when α ≥ 1 and x ∈ H0 (E), p ∈ [1, ∞], the assertion (c) follows immediately α−1

∈ L ∞ [a, t], for every t ∈ I . from Proposition 1.1 part (3) because s  → [log(t/s)] s p Finally, let α > 0 and x ∈ H0 (E), p > max{1, 1/α}. When α ≥ 1, the assertion (d) follows immediately from part (c). It remains to consider the case when α ∈ (0, 1). To do this, let p ≥ 1 be the conjugate exponents to p (that is 1/ p + 1/ p = 1). Since p > 1/α, we have p ∈ [1, 1/(1 − α)). We claim s  → [log(t/s)]α−1 /s ∈ L p [a, t] for any t ∈ I . Once our claim is established, the Proposition 1.1 part (4) guarantees the existence of Jaα x p for every x ∈ H0 (E). It remains to prove our claim. To see this, let t ∈ I and observe that ⎛  ⎝ a

t

   t α−1   log  s

 p ⎞ p1    t t p (α−1) 1    log  ds ⎠ = s s a  1

≤ a

p −1 p

  1 p 1  ds  p −1 s s

   t t p (α−1)   log s a 

  1 p 1  ds  p −1 s s

⎛ ⎛  p (α−1)+1 ⎞ p1 (α−1)+ 1 ⎞ p log at 1 ⎝ log at 1 ⎠ ⎝  ⎠ = = 1 1

p

p (α − 1) + 1 p (α − 1) + 1 p p a a ⎛ ⎞  α− 1p ⎟ log ab 1 ⎜ ⎜ ⎟. ≤ 1 ⎝    ⎠

p 1 p a p α − p

Consequently, since α >

1 p,

the function s  →

[log(t/s)]α−1 s

∈ L p [a, t] for any t ∈ I . By

Proposition 1.1 part (4), it follows that the function s  → [log(t/s)]α−1 x(s) s is Pettis integrable on [a, t], for every t ∈ I . This completes the proof of the assertion (d). However, in all cases, the function s  → [log(t/s)]α−1 x(s) s is Pettis integrable on [a, b]. Therefore it follows, by the definition of Pettis integral that, for every t ∈ I there exists an element in E denoted by Jaα x(t) such that   t   t [log(t/s)]α−1 [log(t/s)]α−1 ϕ x(s) ds = ϕx(s) ds = Jaα ϕx(t), ϕ(Jaα x(t)) = s (α) s (α) a a holds for every for every ϕ ∈ E ∗ . This completes the proof.

 

Remark 2.1 Let F is a Banach space having E as a sub-space and x : I → E. If the HFPI of x : I → F exists on I and x(I ) ⊂ E, then for every t ∈ I ,  t   [log(t/s)]α−1 ϕ Jaα x(t) = ϕx(s) ds, s (α) a

H. A. H. Salem

holds for every ϕ ∈ F ∗ , t ∈ I . In particular, if ϕ ∈ F ∗ and ϕ vanishes on E, then ϕ(Jaα x(t)) = 0 which implies Jaα x(t) lies on E for every t ∈ I . Thus, we arrive to the p following elementary fact: if the HFPI of x ∈ H0 (E) not exist on I , then it is not exist either even we ”enlarge” the space E into F. (cf. [14, Remark.2]). Example 2.1 Let α > 0 and assume E := 2 (I ) be the non-separable Hilbert space of countably nonzero functions on I := [1, 2] that are square-summable, under the 2 −norm. Define the function x : I → 2 (I ) by x(t) := et , where et (s) := 1, if s = t and vanishes otherwise. This function (cf. [17]) is weakly measurable, Pettis integrable on I and ϕx = 0 a.e. (with respect to the Lebesgue measure) for any ϕ ∈ 2 (I ). Thus, by part (a) of Lemma 2.1 with p = 1, the HFPI of x exists and Jα1 x(t) = 0. Example 2.2 Let α ∈ (0, 1). Define the function x : [1, 2] → c0 by xn (t) := n δ χ(1,1+ 1 ] (t),

x(t) := {xn (t)} ,

δ ∈ (0, α).

n

(2)

Clearly the function x is a countable-valued function, hence x is strongly measurable on [1, 2]. We claim that x ∈ H p (c0 ) for  any p ∈ (1/α, 1/δ]. To see this, let {λn } ∈ 1 be the corresponds to ϕ ∈ c0∗ , then ϕx = n λn xn . Since 1p − δ ≥ 0 we have ∞  

2

|λn xn (t)| dt p

 1p

=

1

n=1

∞   1

n=1

2

|λn | n χ(1,1+ 1 ] (t) dt p pδ

 1p

n

≤

∞  |λn | 1

n=1

np

−δ

≤

∞ 

|λn |.

n=1

With a bit work using Levi’s Theorem (or Lebesuge dominated convergence theorem) and Minkowski’s inequality we obtain 

2

|ϕx(t)| p dt

 1p



2

≤

1



1

⎡

∞ 

k→∞

|λn |xn (t)

 1p dt

⎡ = ⎣ lim

k  n=1

p⎤ 1

p

λn xn  p



k→∞ 1

n=1



≤ ⎣ lim

p

⎦ ≤

∞  n=1

2



k 

p |λn |xn (t)

⎤1

p

dt ⎦

n=1

λn xn  p ≤

∞ 

|λn | < ∞.

n=1

 p Indeed, the function sequence gk := ( kn=1 n |x n (t)) is monotone increasing, non|λ ∞ negative, and converges improperly to g = ( n=1 |λn |xn (t)) p . Hence, Levi’s monotone convergence theorem (or Fatou’s lemma) implies that if limk→∞ gk is finite then g is integrable (inparticular:  a.e. finite). Conversely, if g is integrable then the above limit is finite (because gk ≤ g for every k), and Levi’s theorem (or Lebesuge dominated convergence theorem with g as the majorant) implies that gk → g. Whence, for every ϕ ∈ c0∗ , ϕx(·) ∈ L p [1, 2] holds for any p ∈ (1/α, 1/δ] (meaning that x is strongly measurable function lies on H P (c0 )). Since p > α1 > 1, it follows by Proposition 1.2, x ∈ H0P (c0 ). Consequently, owing to part (c) of Lemma 2.1, HFPI of x of any order α ∈ (δ, 1) exists on the interval [1, 2]. To compute this integral, fix t ∈ In 0 := [1 + n 01+1 , 1 + n10 ] for some n 0 ∈ N and let ϕ ∈ c0∗ . Since |x α − y α | ≤ (x − y)α , for any x, y ≥ 0 with y < x, we have

Hadamard-type fractional…

      n 0  t    t   t α−1 n δ λn  t α−1 n δ λn   log 1 (s) ds ≤  log  χ  ds  s s  (1,1+ n ] s s  1  n n=1 1      1+ n1  t α−1 n δ λn  + log    1 s s  n>n 0

ds =

n0  n δ |λn | n=1

α

(log t)α

   n δ |λn |  t α + (log t) − log  α 1+ n>n 0    n 0 α  1 ≤ n δ |λn | log 1 + n0 n=1 α   n δ |λn |  t  +  log t − log α  1+ 1 n>n 0 n0  n δ |λn | ≤ n α0 α n=1  n δ |λn |

+

n

   1 α log 1 + n

α √ 2|λ2 | |λ1 | ≤ + + ··· + α αn 0 αn α0  |λn | + αn α−δ n>n 0 √ 2|λ2 | |λ1 | ≤ + + ··· + α αn 0 αn α0  |λn | + < ∞. α n>n n>n 0

|λn 0 |

αn α−δ 0

|λn 0 |

αn α−δ 0

0

By Beppo Levi Theorem it follows     t   t t α−1 x(s) t α−1 ϕ (x(s)) ϕ log ds = ds log s s s s 1 1   t t α−1  λn n δ χ(1,1+ 1 ] (s) ds log = n s s 1 n   t  t α−1 λn n δ log χ(1,1+ 1 ] (s) ds = n s s 1 n =

 n δ λn α α n>n 0 n=1   α  t = ϕ(g(t)), (log t)α − log 1 + n1 n0  n δ λn

(log t)α +

α     1 

n

H. A. H. Salem

where g(t) :=

n 0 δ (log t)α (n 0 + 1)δ (log t)α 2δ (log t)α , ,··· , , α α α α  α  !  t ,··· . (log t)α − log 1 + n 01+1

Consequently, we conclude that ⎧& (log t)α ⎪ ⎨ (1+α) , α J1 x(t) = & ⎪ ⎩ (log t)α ,

(1+α)

2δ (log t)α

(1+α)

,··· ,

2δ (log t)α

(1+α)

,

n 0 δ (log t)α

(1+α)

3δ (log t)α

(1+α)

,

) ,··· ,

(n 0 +1)δ

(1+α)

'

 (log t)α − log

t (n 0 +1) n 0 +2

α (

) , · · · , t ∈ In 0 , n 0 ∈ N, otherwise.

It can be easily seen that Jα1 x(t) ∈ c0 for any t ∈ [1, 2]. Evidently, an explicit calculation using L’Hospital’s rule reveals that     t (n + 1) α lim n δ (log t)α − log = 0, n→∞ n+2 n 0 ∈ N. Also we have    1 1 α n δ (log t)α ≤ n δ log 1 + ≤ α−δ → 0, as n → ∞. n n

holds for any t ∈ [1 +

1 1 n 0 +1 , 1 + n 0 ],

This yields Jα1 x(t) ∈ c0 for any t ∈ [1, 2] (this is precisely what we would expect from Definition 2.1). The following example asserts that, the assumption of (Lemma 2.1, part (b) ) that the separable space E is weakly complete or contains no copy of c0 is really essential in the case when p = 1 and can not be omitted or extended to be only separable. However, if x is not Pettis integrable on I , then Jaα x fails to exists everywhere on I . Example 2.3 Let α ∈ (0, 1). Define the function x : [1, 2] → c0 by x(t) := {xn (t)} ,

xn (t) := nχ(1,1+ 1 ] (t). n

(3)

The function x is strongly measurable but it is not Pettis integrable on [1, 2] (see [38]). We claim that Jα1 x does not exists on [1, 2]. To see this, arguing similarly as in the Example 2 with δ = 1, we can show ϕx(·) ∈ L 1 [1, 2] (but it is not in L q [1, 2] for any q > 1) holds for every ϕ ∈ c0∗ . That is, x ∈ H1 (E) (but not in Hq (E) for any q > 1). Now, making use of Proposition 1.4, implies that the HFPI of x of any order α ∈ (0, 1) not exist on the interval [1, 2]: To ensure this, we define for each set In = [1, 1 + 1/n] the functionals ϕn ∈ c0∗ (required by Proposition 1.4 ) to be the corresponding to the element {λn } := (0, 0, · · · , 0, 1, 0, 0, · · · ) ∈ 1 where the non-zero coordinate is in the n th place. Then ϕn (x(t)) = xn (t) and thus |ϕn (x(s))| = nχ[1,1+1/n] (s). Clearly, the family {ϕn } runs through the unit ball of the dual of c0 . Also for each t ∈ [1.5, 2] we have     α−1   ϕ [log(t/s)]  ds x(s)   s (α) In  1+ 1 n [log(t/s)]α−1 |ϕ(x(s))| ds = s (α) 1

Hadamard-type fractional…



1+ n1

[log(t/s)]α−1 n ds s (α) 1   tn α n = [log t]α − [log ]

(1 + α) n+1

=

(4)

An explicit calculation using L’Hospital’s rule reveals that the right hand side of (4) tends to αt −1 [log t]α−1 as n → ∞ for each t ∈ [1.5, 2]. Therefore by Proposition 1.4, we deduce s  → [log(t/s)]α−1 x(s) / P ([1, 2], c0 ). Hence Jα1 x makes no sense. s ∈ Example 2.4 Let α > 0. Define the weakly measurable function x : [1, 2] → L ∞ [1, 2] by x(t) = χ[1,t] . This function is Pettis integrable on [1, 2] and for each ϕ ∈ L ∗∞ , ϕ(x) is a function of bounded variation (see e.g. [16] ). Since a functions of bounded variation are bounded, it follows that ϕx ∈ L ∞ for any ϕ ∈ L ∗∞ . Therefore, x ∈ H0∞ (L ∞ ) and it follow in view of Lemma 2.1 with p = ∞ that the HFPI of x exists on [1, 2]. To compute the HFPI of x, we let ρ ∈ L 1 [1, 2]. Assume that ϕ be the element in L ∗∞ corresponding to ρ and we note that for any t > 1, we have   t  [log(t/s)]α−1 x(s) ds ϕ s (α) 1  t [log(t/s)]α−1 ϕ(x(s)) ds = s (α) 1  t  [log(t/s)]α−1 2 = ρ(θ )χ[1,s] (θ )dθ ds s (α) 1 1   t [log(t/s)]α−1 s ρ(θ )dθ ds = s (α) 1 1  t s [log(t/s)]α−1 = ρ(θ )dθ ds s (α) 1 1  t t [log(t/s)]α−1 ρ(θ ) dsdθ = s (α) 1 θ  t [log(t/θ ]α ρ(θ )dθ = 1 (1 + α)  2 [log(t/θ ]α = ρ(θ )χ[1,t] (θ )dθ. 1 (1 + α) Thus  ϕ 1

t

[log(t/s]α−1 x(s) ds s (α)



 =

t

1

 ϕ

   [log(t/s]α−1 [log(t/s]α x(s) ds = ϕ χ[1,t] (s) . s (α)

(α + 1)

Consequently, we conclude that Jα1 x(t)(·) =

(log t − log ·)α χ[1,t] (·) ∈ L ∞ [1, 2].

(1 + α)

Evidently for any t ∈ [1, 2] we have ess supζ ∈[1,2] Jα1 x(t)(ζ ) = ess supζ ∈[1,t] This yields Jα1 x(t) ∈ L ∞ for every t ∈ [1, 2].

(log t − log ζ )α (log 2)α ≤ .

(1 + α)

(1 + α)

H. A. H. Salem

The statements revealing haw much the fractional integral Jaα is better than the function x ∈ H p (E) are more important. Indeed, based on Lemma 2.1 and the fact that the Hadamardtype fractional integral operator maps L p (I ) continuously into L p (I ) for each p ∈ [1, ∞], it can be easily seen that the following holds Lemma 2.2 For any α > 0 the following holds (a) If E is reflexive, the operator Jaα takes H p (E) into H0 (E), for ever p ∈ [1, ∞] p (b) If p > max{1, 1/α}, the operator Jaα sends H0 (E) into C[I, E ω ] (if we define Jaα x(a) := α 0). In particular, Ja : C[I, E ω ] → C[I, E ω ]. p

α Proof For the proof of the claim (a), we note in the  αview  ofαLemma 2.1 that, the Ja ∗is p defined on H0 (E), p ≥ 1 when E is reflexive and ϕ Ja x = Ja ϕx holds for every ϕ ∈ E . Moreover, since the Hadamard-type fractional integral operator maps L p (I ) into itself for each p ∈ [1, ∞], it follows that ϕ (I α x) ∈ L p for every ϕ ∈ E ∗ , meaning that Jaα x ∈ H p (E). p Now, the reflexivity of E together with Proposition 1.1 part (1) result in Jaα x ∈ H0 (E). Next, repeating the argument of part (d) of Lemma 2.1 yields Jaα : H p (E) → C[I, E ω ] p for any p > max{1, 1/α}, as follows: Take x ∈ H0 (E), p > max{(1/α), 1}. If p ∈ [1, ∞]

such that 1/ p + 1/ p = 1, then p (α − 1) > −1. Let a ≤ τ ≤ t ≤ b. Without loss of generality, assume Jaα x(t) − Jaα x(τ )  = 0. Then of the Hahn* there exists ( as a*consequence   Banach theorem) ϕ ∈ E ∗ with ϕ = 1 and *Jaα x(t) − Jaα x(τ )* = ϕ Jaα x(t) − Jaα x(τ ) . Putting the Hölder inequality in mind, one can write the following chain of inequalities

  α    ϕ J x(t) − Jα x(τ )  (α) = Jα ϕx(t) − Jα ϕx(τ ) (α) a a a a  t    [log(t/s)]α−1 =  ϕx(s) ds s a    τ  [log(τ/s)]α−1 ϕx(s) ds  − s a   τ   [log(t/s)]α−1 − [log(τ/s)]α−1   |ϕx(s)| ds  ≤   s a   t α−1 [log(t/s)] + |ϕx(s)| ds s τ  τ   1 [log(t/s)]α−1 − [log(τ/s)]α−1  p ds ≤ 1 s a ap 1/ p   t

ds ϕx L p [0,b] + (log(t/s))(α−1) p s τ  1/ p  τ  1  p (α−1) p (α−1)  1 ≤ 1 − [log(τ/s)] [log(t/s)]  ds s a ap 1/ p   t

1 ϕx L p [0,b] + (log(t/s))(α−1) p ds s τ 1 '

≤ 1 [log(t/τ )] p (α−1)+1 − [log(t/a)] p (α−1)+1 ap

Hadamard-type fractional…

1/ p

 +[log(τ/a)] p (α−1)+1 

1/ p  ϕx L [0,b] 

p  . + (log(t/τ )) p (α−1)+1 p p (α − 1) + 1 Thus   α  ϕ J x(t) − Jα x(τ )  a a  1/ p

  = C(a, α, p) [log(t/τ )] p (α−1)+1 − [log(t/a)] p (α−1)+1 + [log(τ/a)] p (α−1)+1  ( α− 1 + (log(t/τ )) p ϕx L p [0,b] , (5) with some finite constant C(a, α, p), depending only on a, α and p. In particular, when α ∈ (0, 1), the above estimates yield   α 1  ϕ J x(t) − Jα x(τ )  ≤ C(a, α, p)| log t − log τ |α− p ϕx L [0,b] . a a p

(6)

This estimations shows that Jaα x is norm continuous. Thus it is clear, in view of Jaα x(a) = 0, p that Jaα maps H0 (E) with p > max{(1/α), 1} into C[I, E w ]. By noting that the weak continuity implies a strong measurability ([18] page 73), it follows that every x ∈ C[I, E ω ] is strongly measurable on I and x ∈ H∞ (E) which implies (in view p of Proposition 1.2) that C[I, E ω ] ⊆ H0 (E) for every p ∈ [1, ∞]. That is Jaα : C[I, E ω ] → C[I, E ω ]. Hence the required result.   Example 2.5 Obviously, the assertion of Lemma 2.2 holds for the function x : I → 2 (I ) defined as in Example 2.1, but to see that it is also holds for the function defined as in Example 2.4, we provide a proof: Define the function function x : [1, 2] → L ∞ [1, 2] by x(t) = χ[1,t] , we have (in view of Example 2.4 ) Jα1 x(t)(ζ ) =

(log t/ζ )α χ[1,t] (ζ ), t ∈ [1, 2].

(1 + α)

To show that Jα1 x is norm continuous on [1, 2], let 1 ≤ τ ≤ t ≤ 2. Then * α * *J x(t) − Jα x(τ )* 1

1

∞

  1 ess supζ ∈[1,2] (log t/ζ )α χ[1,t] (ζ ) − (log τ/ζ )α χ[1,τ ] (ζ )

(1 + α)   + 1 ess supζ ∈[1,2] (log t/ζ )α χ[1,t] (ζ ) − (log t/ζ )α χ[1,τ ] (ζ ) ≤

(1 + α)  , + ess supζ ∈[1,2] (log t/ζ )α χ[1,τ ] (ζ ) − (log τ/ζ )α χ[1,τ ] (ζ )   + 1 ≤ ess supζ ∈[1,2] (log t/ζ )α χ[τ,t] (ζ ) + ess supζ ∈[1,2]

(1 + α) , |(log t/ζ ) − (log τ/ζ )|α χ[1,τ ] (ζ ) 2 ≤ (log t/τ )α .

(1 + α)

=

Thus the HFPI of x is norm continuous on [1, 2]. Analogously, an explicit calculation reveals that Jα1 x lies in C([1, 2], c0 ), where x : [1, 2] → c0 defined by the formula (2). As a consequence of Lemma 2.2 we introduce the following observation

H. A. H. Salem

Corollary 2.1 Let E be a reflexive Banach space. The map Jaα : C[I, E ω ] → C[I, E ω ] is weakly compact operator (That is, it maps bounded subsets on C[I, E ω ] weakly continuously into relatively weakly compact ones). Proof Let Q be bounded subset of C[I, E ω ]. Firstly, we examine the weak continuity of the mapping Jaα : Q → C[I, E ω ]. The usual idea to show that Jaα : C[I, E ω ] → C[I, E ω ] is weakly continuous operator, is to apply the Lebesgue dominated convergence theorem for Pettis integral, however this theorem is certainly valid for sequences not for nets. If {x n } a net in Q converges weakly to x in Q. With a bit work using the Krasnosielski-type lemma ([42], Lemma 2), it follows  t   [log(t/s)]α−1 |λ| |ϕ Jaα xn (t) − Jaα x(t) | ≤ |ϕ(xn (s)) − x(s))| ds ≤ .

(α) a s So, Jaα is weakly continuous on I . The proof of the compactness of Jaα (Q) is a standard application of the Arzela-Ascoli Theorem using Eberlein Šmulian Theorem ([9] and [23]). We omit the details because the analysis is similar to that in [26] with (small) necessary changes: Arguing similarly as in proof of Theorem 2.2 in [26], we can show, by the aid of (5) that Jaα (Q) is relatively weakly compact. Indeed, the equicontinuity of the image of Q under Jaα is an immediate consequence of (5). The uniform boundedness also, follows because * * sup *Jaα x(t)* ≤ C(a, α, p)(log b/a)α sup x(t) , t∈I

t∈I

holds for every x ∈ C[I, E ω ].

  p

Proposition 2.1 If the two functions x, y ∈ H0 (E) are weakly equivalent on I (that is, for every ϕ ∈ E ∗ , ϕx(t) = ϕy(t) for a.e. t ∈ I ) such that Jaα x and Jaα y exist for some p ≥ 1 and α > 0, then Jaα x = Jaα y on I . Proof Fix t ∈ I . Since x and y are weakly equivalent on I , then for every ϕ ∈ E ∗ there exists a null set N depends on ϕ ∈ E ∗ such that [log(t/a)]1−α ϕx(s)/s = [log(t/a)]1−α ϕy(s)/s for every s ∈ [a, t]/N. Consequently     ϕ Jaα x(t) = Jaα ϕx(t) = Jaα ϕy(t) = ϕ Jaα y(t) holds for every ϕ ∈ E ∗ and every t ∈ I . Thus Jaα x = Jaα y on I .

 

The HFPI enjoys the following commutative property which is an immediate consequence of the properties of Hadamard-type fractional integrals in the space of real-valued functions Lemma 2.3 Let α, β > 0 and p > max{1, 1/α, 1/β}. Then Jaα Jaβ x = Jaβ Jaα x = Jaα+β x p

holds for every x ∈ H0 (E). If E is reflexive, this is also true for every p ≥ 1. Proof Observe that our assumption p > max{1, 1/α, 1/β} yields p > max{1, 1/(α + β)}. β α+β Thus, in view of Lemma 2.1 and Lemma 2.2, it follows that Jaα x, Ja x and Ja x exist on I p for every x ∈ H0 (E) as a weakly continuous functions from I to E. Precisely, Jaα x, Jβ x and α+β Ja x are in H0∞ (E). Consequently, for any ϕ ∈ E ∗ we have (by the aid of ( [19] Property 2.26))       ϕ Jaα Jaβ x(t) = Jaα ϕ Jaβ x(t) = Jaα Jaβ ϕ(x(t)) = Jaα+β ϕ(x(t)) = ϕ Jaα+β x(t) ,

Hadamard-type fractional…

that is

  ϕ Jaα Jaβ x(t) − Jaα+β x(t) = 0, for every ϕ ∈ E ∗ . β

α+β

β

Hence Jaα Ja x(t) = Ja x(t), t ∈ I . Similarly, we are able to show that Ja Jaα x(t) = α+β Ja x(t). When E is reflexive, the result follows as a direct consequence of Lemma 2.2 (part (a)).   In what follows, let α ∈ (0, 1) and x : I → E be weakly measurable on I . Define the function x1−α : I → E by  t [log(t/s)]−α 1 x1−α (t) := Ja1−α x(t) = x(s) ds, 0 < α < 1, (7)

(1 − α) a s Obviously, the weak absolute continuity of x1−α , is necessarily (but not sufficient ) condition for the existence of a pseudo derivative of x1−α . However, in view of Lemma 2.2, p if the pseudo derivative of x1−α would exists on I , then necessarily x lies in H0 (E) with p > 1/(1 − α). Lemma 2.4 Let p > 1/(1 − α). If the function x : I → E is weakly absolutely continuous p on I and have a pseudo derivative in H0 (E), then the function x1−α is weakly absolutely continuous and have a pseudo derivative on I . In this case. the pseudo-derivatives of x1−α does not depend on the choice of a pseudo derivative of the function x. If E is reflexive, this is also true for every p ≥ 1. Proof At the beginning we notice, in view of Proposition 1.3, that our assumptions imposed on x yield  t x(t) = x(a) + y(s) ds, for t ∈ I, a

where y is a Pettis integrable pseudo derivatives of x ( occasionally, we note that the pseudoderivative of x is not unique and two pseudo-derivatives need not be a.e. equal). If y ∈ p p H0 (E), then (·)y(·) ∈ H0 (E) as well: Evidently, ϕ(sy(s)) = sϕ(y(s)) ∈ L p (I ) holds for ∗ all ϕ ∈ E and (·)y(·) ∈ P[I, E] (this follows by Proposition 1.1 (part 3)). Thus  t  t sy(s) y(s) ds = ds = Ja1 t y(t), s a a p

where the right hand side exists by part (c) of Lemma 2.1, since (·)y(·) ∈ H0 (E). Consequently x(t) = x(a) + Ja1 t y(t). (8) Now insert (8) into (7) we can show, in view of p > 1/(1 − α) with the aid of Lemma 2.3, that , [log(t/a)]1−α x(a) + x1−α (t) = Ja1−α x(a) + Ja1 t y(t) = + Ja1−α Ja1 t y(t)

(2 − α) [log(t/a)]1−α x(a) = + Ja1 Ja1−α t y(t).

(2 − α) That is x1−α (t) =

[log(t/a)]1−α x(a) +

(2 − α)

 a

t

Ja1−α sy(s) ds. s

(9)

(10)

H. A. H. Salem

Obviously, since



[log(t/a)]

1−α

x(a) = a

t

(1 − α)

[log(t/s)]−α x(a) ds, s

the first term of (10) is weakly absolutely continuous having pseudo derivatives on I . Moreover, the second term in (10) is also weakly absolutely continuous having pseudo derivatives on I : This an immediate consequence of the fact that (cf. [27], see also ( [18] page 88), the indefinite integral of Pettis integrable function is weakly absolutely continuous and possesses a pseudo derivative on I . This is also true, in view of Lemma 2.3, when E is reflexive for all p ≥ 1. In order to prove the uniqueness of the pseudo-derivative of x1−α we start with the case when E is reflexive and p ≥ 1. We note, in view of Equation (9) by the aid of Lemma 2.3 that [log(t/a)]1−α x(a) x1−α (t) = (11) + Ja2−α t y(t).

(2 − α) Thus, for any ϕ ∈ E ∗ we have ϕ (x1−α (t)) =

[log(t/a)]1−α ϕ (x(a)) + Ja2−α tϕ(y(t)).

(2 − α)

Let y1 and y2 be two pseudo derivatives of the function x. Since y1 and y2 are weakly equivalent on I , it follows Ja2−α tϕ(y1 (t)) = Ja2−α tϕ(y2 (t)), holds for every ϕ ∈ E ∗ and every t ∈ I . Hence Ja2−α t y1 (t) = Ja2−α t y2 (t). Thus the pseudoderivatives of x1−α does not depend on the choice of a pseudo derivative of the function x. This also true for arbitrary Banach space E provided p > 1/(1 − α). To see this, let y1 and y2 be two pseudo derivatives of the function x. Then y1 and y2 are weakly equivalent on I . Arguing similarly as in the preceding proof, it can be seen that Ja1−α t y1 (t) = Ja1−α t y2 (t), t ∈ I . Putting Lemma 2.2 in mind, we conclude that the functions Ja1−α (·)y1 ((·)) 1 1−α 1 1−α and Ja1−α (·)y2 ((·)) (hence (·) Ja (·)y1 (·) and (·) Ja (·)y2 (·)) are weakly continuous (hence strongly measurable ) on I . Recall ( cf. [27], Theorem 5.2) that the Pettis integral of two strongly measurable functions coincide over every Lebesgue measurable set on I if, and only if, the two strongly measurable function are coincide a.e. on I . Consequently, by looking at Equation (10), we conclude that the pseudo-derivatives of x1−α independent on the choice of y1 and y2 . This completes the proof.   Remark 2.2 When p > 1/(1 − α) with α ∈ (0, 1), the pseudo differentiable function x 1−α defined in Lemma 2.4 is ” in fact” a.e. weakly differentiable on I (that is, the null set where the differentiability of ϕx1−α fails invariant with ϕ ∈ E ∗ ): To see this, we notice, in view of (Lemma 2.2, part (b)), that the function s → Ja1−α sy(s) (hence the function s → 1s Ja1−α sy(s) ) is weakly continuous on I . Recalling [24] that, the indefinite Pettis integral of a weakly continuous function is weakly absolutely continuous and possesses a weak derivative on I . This yields the weak differentiability of the second term in (9). Also, the null set (where the derivative of t → [log(t/a)]1−α ϕx(a) fails) invariant with ϕ ∈ E ∗ , meaning that x1−α is weakly absolutely continuous and a.e. weakly differentiable on I . It is worth mentioning that ( cf. e.g. [22]), a function defined on a compact real interval and having values in a weakly sequentially complete space is absolutely continuous if, and only if, it is weakly absolutely continuous.

Hadamard-type fractional…

Also, if the range space E of a function x has the the Radon-Nikodym property (cf. e.g. [11] and [12]), then every absolutely continuous function has a Pettis integrable pseudo derivative. Since the weakly sequentially complete spaces and the spaces with Radon-Nikodym property include reflexive spaces, the following result follows immediately as a direct consequence of Lemma 2.4 Lemma 2.5 Let α ∈ (0, 1) and E be reflexive. If the function x : I → E is weakly absolutely continuous on I , then the function x1−α defined by (7) is weakly absolutely continuous and have a pseudo derivative on I . And the pseudo-derivatives of x 1−α does not depend on the choice of a pseudo derivative of the function x. In what follows, we show that the assertion of Lemma 2.4 still valid for all p ≥ 1 if E is weakly sequentially complete. Lemma 2.6 Let α ∈ (0, 1) and E be weakly sequentially complete. If the function x : I → E is weakly absolutely continuous on I and have a weakly continuous pseudo derivative on I , then, the function x1−α defined by (7) is weakly absolutely continuous and have a.e. weak derivative on I . Proof Let E be weakly sequentially complete. Arguing similarly as in the proof of Lemma 2.4, we arrive, at x1−α (t) =

[log(t/a)]1−α x(a) + Ja1−α Ja1 t y(t).

(2 − α)

Thus, for any ϕ ∈ E ∗ we have ϕ (x1−α (t)) =

[log(t/a)]1−α ϕ (x(a)) + Ja1−α Ja1 tϕ(y(t)).

(2 − α)

Since, tϕ(y(·)) ∈ C[I, R] (so is Ja1−α tϕ(y(·)) ) for every ϕ ∈ E ∗ , it follows by the properties of the Hadamard fractional integrals of the real-functions that ϕ (x1−α (t)) =

[log(t/a)]1−α ϕ (x(a)) + Ja1 Ja1−α tϕ(y(t)).

(2 − α)

In other words, ϕ (x1−α (·)) is a.e. differentiable on I (the null set invariant for every ϕ ∈ E ∗ ). Since E is weakly sequentially complete and ϕx1−α a.e. differentiable on I for every ϕ ∈ E ∗ , the result is straightforward consequence of ([38], Theorem 7.3.3).   After the notation of the Hadamard-type fractional integral, that of fractional derivative becomes a natural requirement: We shall therefore now discuss a concept of a Hadamard-type fractional derivatives of vectorvalued functions Definition 2.2 Let x : I → E. For the positive integer m such that α ∈ (m − 1, m), m ∈ N0 := {0, 1, 2, · · · we define the Hadamard fractional-pesudo (-weak) derivative ”shortly HFPD (HFWD)” of x of order α by Daα x(t) := δ m Jam−α x(t),

δ := t D.

(12)

α and Dα Here D denote the pseudo-(weak-)differential operator. We use the notation Da, p a,w to characterize the Hadamard fractional-pesudo derivatives and Hadamard fractional-weak derivatives respectively.

H. A. H. Salem α x ( Dα x) makes sense, if x : I → E is a function such that Jm−α x is Of course, Da, p a,w a α x exists a.e. on I , then pseudo (weakly) differentiable on I . It is easy to see that, if Da,w α x exists on I and Dα x = Dα x a.e. on I . Da, p a, p a,w Clearly, in infinite dimension Banach spaces, the weak absolute continuity of x 1−α , is necessarily (but not sufficient) condition for the existence of HFPD (in particular HFWD) of x. However, in view of Lemma 2.2, if the HFPD (in particular HFWD) of x would exists p on I , then necessarily x lies in H0 (E) with p > 1/(1 − α). We start with following example. In this example, we construct a strong continuous function that has HFPD of all orders α ∈ (0, 1), but has no pseudo derivatives of order one.

Example 2.6 Define x : [1, 2] → L 1 [1, 2] by x(t) := χ[1,t] (·). This function satisfies  * * x(t2 ) − x(t1 ) = *χ[t1 ,t2 ] * =

2 1

|χ[t1 ,t2 ] (s)| ds = |t1 − t2 |,

t1 , t2 ∈ [1, 2], t1 ≤ t2 ,

and so x is strongly (trivially weakly) absolutely continuous on [1, 2]. We will show that function x is not weakly differentiable everywhere on [1, 2]. We start by showing that the strong derivative of x does not exists on [1, 2]. To show this, let h > 0 and t ∈ [1, 2] and note that * *   t+h   * x(t + h) − x(t) * 1 2 1 * *= |χ (s)| ds = ds = 1. [t,t+h] * * h h h 1

t

Therefore, if the strong derivative y would exist at some t ∈ [1, 2], then necessarily y = 1 and * * * x(t + h) − x(t) * * * → 0 as h → 0 − y(t) * * h where the term on the left-hand side is given by    2  2    x(t + h)(s) − x(t)(s)  χ[t,t+h] (s)    − y(t)(s) ds = − y(t)(s) ds   h h 1 1  t  2 |y(t)(s)| ds + |y(t)(s)| ds ≥ 1

 =

2

t+h t+h

 |y(t)(s)| ds −

1

|y(t)(s)| ds

t

→ y(t) as h → 0. Combining the preceding relations, we find y(t) = 0, which is a contradiction. By a result due to Alexiewicz ([5], Theorem 4), x is not everywhere weakly differentiable on [1, 2]. The function x does not even everywhere pseudo differentiable on [1, 2]: To see this, let ψ ∈ L ∞ [1, 2] corresponding to ϕ ∈ L ∗1 [1, 2] and observe that  t  2 ψ(ζ )χ[1,t] (ζ )dζ = ψ(ζ )dζ. ϕx(t) = 1

1

Then there exists a null set N depends on ψ (hence on ϕ) such that x is differentiable on [1, 2]/N (ψ) and d (13) ϕx(t) = ψ(t), for t ∈ Jϕ := [1, 2]/N (ψ), dt

Hadamard-type fractional…

Consequently, x is pseudo differentiable on [1, 2] if there exist a function y(·) : [1, 2] → L 1 [1, 2] so that (ϕx(t)) = ϕy(t), for every t ∈ Jϕ . Meaning, in view of (13), that a function y with the required property exists such that  2 ψ(t) = ψ(ζ )y(t)(ζ )dζ, 1

holds on Jϕ for all ψ ∈ L ∞ [1, 2]. In order to show that x is not pseudo differentiable on [1, 2], we assume for the sake of contradiction, that a function y with the required property exists, and put  2 K ψ(t) := ψ(s)y(t)(s) ds, t ∈ Jϕ . 1

Choose the sequence {ψn } in the unit ball of L ∞ [1, 2] corresponding to the family {ϕn } ⊂ L ∗1 [1, 2], where{ψn } converges to 0 in measure (but not almost everywhere) on [1, 2] (e.g. choose {ψn } to be the well-known typewriter sequence). It follows, by Lebesgue’s dominated convergence theorem, that the image sequence {K ψn } converges to 0 almost everywhere, which is a contradiction because, it is not possible that ψn = K ψn a.e. for every n ∈ N (note that the union of null sets corresponding to each {ψn } is again a null set). On the other hand, we will prove that the HFWD of x of any order α ∈ (0, 1) exists on [1, 2]. To see this, we notice that, the HFPI of x of order 1 − α exists on [1, 2] (this is, because of the continuity of x on [1, 2]) and y(t) := J1−α 1 x(t) =

(log t/·)1−α χ[1,t] (·).

(2 − α)

We omit the calculations since it is almost identical to our argumentations in Example 2.4 with (small) necessary changes. By a usual calculations or as a direct consequence of Lemma 2.2 it follows that y : [1, 2] → L 1 [1, 2] is weakly continuous on the interval [1, 2]. Further, for any t ∈ [1, 2] and any ϕ ∈ L ∗1 [1, 2], there exists ψ ∈ L ∞ [1, 2] such that  t  2 (log t/θ )1−α (log t/θ )1−α χ[1,t] (θ )dθ = dθ ψ(θ ) ψ(θ ) ϕy(t) =

(2 − α)

(2 − α) 1 1 Consequently, for every ϕ ∈ L ∗1 [1, 2] there exists a null set N (ϕ) such that the function ϕy(·) is differentiable on [1, 2]/N (ϕ) and for every t ∈ [1, 2]/N (ϕ) we have    t (log t/·)−α dϕy(t) (log t/θ )−α ψ(θ ) = dθ = ϕ χ[1,t] (·) = ϕ (g(t)) , dt t (1 − α) t (1 − α) 1 −α

t/·) where g(t)(·) := (log t (1−α) χ[1,t] (·), t ∈ [1, 2]. Since   t  2   (log t/s)−α 2 (log t/s)−α 2(log 2)1−α  ds ≤  (s) χ ds ≤ < ∞, [1,t]   t (1 − α) t (1 − α) 1 s

(2 − α) 1

then g(t) ∈ L 1 [1, 2] for a.e. t ∈ [1, 2]. That is, for almost every t ∈ [1, 2] there exists g : [1, 2] → L 1 [1, 2] such that (ϕy(t)) = ϕ (g(t)) a.e. , for every ϕ ∈ L ∗1 [1, 2]. Thus, y is a pseudo differentiable on [1, 2] with a pseudo derivative g(t)(·). By the definition the Hadamard fractional-pesudo derivative we obtain Dα1, p x(t)(s) =

(log t/s)−α χ[1,t] (s), 0 < α < 1.

(1 − α)

H. A. H. Salem

In what follows, we consider the most interesting case when m = 1. We investigate the existence and the main properties of the Hadamard fractional differential operators in the p space H0 (E). In the following lemma we gather together some particular sufficient conditions for the p existence of the Hadamard-type fractional derivatives of the functions x ∈ H0 (E). Lemma 2.7 The HFPD of a function x : I → E of order α ∈ (0, 1), makes sense if at least one of the following cases holds: p

(a) If x is weakly absolutely continuous function having a pseudo derivative in H0 (E), with p > 1/(1 − α), (b) If E be reflexive and x is weakly absolutely continuous on I , (c) If E be weakly sequentially complete and x is weakly absolutely continuous function having a weakly continuous pseudo derivative on I . α x does not depend on the choice of the pseudo derivatives of x. Also In all cases, Da, p α Da, p x(t)

1 [log(t/a)]−α x(a) + =

(1 − α)

(1 − α)

 t a

t log s

−α D p x(s) ds,

t > a > 0. (14)

Proof The proof of this lemma is an immediate consequence of Lemmas 2.4, 2.5 and 2.6 respectively. And (14) comes from (9). This because δ p Ja1 f (t) = f (t) holds for any f ∈ P[I, E]: To see we note, by Proposition 1.1(3), that the function f (·)/(·) is Pettis integrable on I for all f ∈ P[I, E]. Since the indefinite integral of the Pettis integrable function s → x(s)/s is a pseudo differentiable and its pseudo derivative equals the integrand at right t x(t) endpoint of the integration interval. That is D p a x(s) s ds = t . Consequently formula (14) holds.   The following Lemma is folklore in case E = R, but to see that it is also holds in the vector-valued case, we provide a proof p

Lemma 2.8 Let 0 < α ≤ β < 1. For every x ∈ H0 (E) with p > max{1, 1/α, 1/(1 − β)}, 0 x = x and we have Da, p β α β−α (15) Da, p Ja x = Da, p x a.e., and either both sides of (15) are defined or none. If E is reflexive, this is also true for every p ≥ 1. α Jα is defined on in H p (E) In particular, when α = β (15) means that the operator Da, p a 0 α α and that Da, p is the left-inverse of Ja . Proof Arguing similarly the first claim,  t as in the proof of Lemma 2.7, we are able to prove p i.e. δ p Ja1 x(t) = t D p a x(s) s ds = x(t). Also, our assumption x ∈ H0 (E) with p > 1/α implies, in view of Lemma 2.2, Jaα x exists and it is weakly continuous on I . By Lemma 2.3 it follows β α 1−β α 1−(α−β) β−α x = Da, Da, p Ja x = t D p Ja Ja x = t D p Ja p x, 0 < α ≤ β.

When α = β, the result follows by first claim.

 

Acknowledgements I would like to express my gratitude to Prof. Martin Väth, for his advise, guidance, patience and continuous support.

Hadamard-type fractional…

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