CUBO A Mathematical Journal
Vol.15, No¯ 01, (01–12). March 2013

Weak Solutions of Fractional Order Pettis Integral
Inclusions with Multiple Time Delay in Banach Spaces

Mouffak Benchohra

Université de Sidi Bel-Abbès

Laboratoire de Mathématiques,

BP 89, 22000 Sidi Bel-Abbès, Algérie

benchohra@univ-sba.dz

Fatima-Zohra Mostefai

Université de Saida

Département de Mathématiques,

BP 138 Cité Ennasr, 20000, Saida, Algérie

f.z.mostefai@gmail.com

ABSTRACT

We study the existence of weak solutions for nonlinear integral inclusion with multiple

time delay. The main result of the paper is based on the fixed point theorem of Mönch

type and the technique of measure of weak noncompactness.

RESUMEN

Estudiamos la existencia de soluciones débiles de la inclusión integral no lineal con

retardos temporales múltiples. El resultado principal del art́ıculo se basa en el Teorema

de Punto Fijo de tipo Mönch y la técnica de medida de la no-compacidad débil

Keywords and Phrases: Hyperbolic differential inclusion, measure of weak noncompactness,

left sided mixed Pettis integral, weak solution, Banach space.

2010 AMS Mathematics Subject Classification: 26A33, 35H10, 35D30



2 Mouffak Benchohra and Fatima-Zohra Mostefai CUBO
15, 1 (2013)

1 Introduction

Fractional differential equations have been of great interest recently. It is due to the development

of the theory of fractional calculus itself and by application of such constructions in various fields of

science and engineering such as control theory, physics, mechanics, electrochemistry, porous media,

etc. There are many works discussing the solvability of nonlinear fractional differential equations

and inclusions, see the monographs of Abbas et al. [2], Kilbas et al. [14], Lakshmikantham et al.

[15], Podlubny [18], Tarasov [20], the papers of Agarwal et al. [3, 4, 5], Benchohra et al. [7, 8],

Kilbas and Marzan [13], Salem [19], Vityuk and Golushkov [21], and the references therein.

In [12], R. W. Ibrahim and H. A. Jalab studied the existence of solutions of the following

fractional integral inclusion

u(t) −

m∑

i=1

bi(t)u(t − τi) ∈ I
αF(t, u(t)); if t ∈ [0, T], (1)

where τi < t ∈ [0, T], bi : [0, T] → R, i = 1 . . . , n are continuous functions, and F : [0, T]×R → P(R)
is a given multivalued map.

In [1], Abbas and Benchohra considered the following fractional integral equation with delay

u(x, y) =

m∑

i=1

gi(x, y)u(x − ξi, y − µi) + I
r
θf(x, y, u(x, y)); if (x, y) ∈ J := [0, a] × [0, b], (2)

u(x, y) = Φ(x, y); if (x, y) ∈ J̃ := [−ξ, a] × [−µ, b]\(0, a] × (0, b], (3)

where a, b > 0, θ = (0, 0), ξi, µi ≥ 0; i = 1 . . . , m, ξ = maxi=1...,m{ξi}, µ = maxi=1...,m{µi}, I
r
θ

is the left-sided mixed Riemann-Liouville integral of order r = (r1, r2) ∈ (0, ∞) × (0, ∞), f :
J × Rn → Rn, gi : J → R; i = 1 . . . m are given continuous functions, and Φ : J̃ → Rn is a given
continuous function such that

Φ(x, 0) =

m∑

i=1

gi(x, 0)Φ(x − ξi, −µi); x ∈ [0, a],

and

Φ(0, y) =

m∑

i=1

gi(0, y)Φ(−ξi, y − µi); y ∈ [0, b].

Motivated by the above papers, in this paper, we consider the following fractional integral

inclusion with multiple time delay:

u(x, y) −

m∑

i=1

gi(x, y)u(x − ξi, y − µi) ∈ I
α
θ F(x, y, u(x, y)); (x, y) ∈ Ja × Jb. (4)

u(x, y) = Ψ(x, y); (x, y) ∈ J̃ = [−ξ, a] × [−µ, b]\(0, a] × (0, b], (5)



CUBO
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Weak Solutions of Fractional Order Pettis Integral Inclusions ... 3

where Ja = [0, a], Jb = [0, b] for a, b > 0, θ = (0, 0), ξ = maxi=1...m{ξi}, µ = maxi=1...m{µi}, I
α
θ

is the left sided mixed Pettis integral of order α, α = (α1, α2) ∈ (0, ∞) × (0, ∞), F : Ja × Jb × E →
P(E) is a multivalued map (P(E) is the family of all nonempty subsets of E), gi : Ja × Jb → R; i =
1, . . . m are given continuous functions, and Ψ : J̃ → E is a given continuous function such that

Ψ(0, y) =

m∑

i=1

gi(0, y)Ψ(−ξi, y − µi); y ∈ [0, b],

and

Ψ(x, 0) =

m∑

i=1

gi(x, 0)Ψ(x − ξi, −µi); x ∈ [0, a].

E is a Banach space with norm ‖.‖. Our result is based on fixed point theorem of Mönch type

and the technique of measure of weak noncompactness. Let us mention that other tools like the

nonlinear alternative of Leray-Schauder type, the Banach fixed point theorem and Schauder’s fixed

point theorem, such have been used to analyze the above problem in the scalar case [1, 2]. The

present results complement and extend those considered in the scalar case.

2 Preliminaries

In this section, we introduce the notation, definitions, and preliminary facts that will be used in

the remainder of this survey paper. Let R denote the real line and let Ja = [0, a] and Jb = [0, b]

be two closed and bounded intervals in R for some real numbers a > 0 and b > 0. Throughout

the paper, E is a Banach space with norm ‖.‖ and dual E∗. Also (E, w) = (E, σ(E, E∗)) denotes

the space E with its weak topology. We take C(Ja × Jb, E) to be the Banach space of continuous

functions u : Ja × Jb → E, with the usual supremum norm

‖u‖
∞

= sup{‖u(x, y)‖, (x, y) ∈ Ja × Jb}.

Definition 2.1. [17] The function x : Ja × Jb → E is said to be Pettis integrable on Ja × Jb if and
only if there is an element xI×J ∈ E corresponding to each I×J ⊂ Ja ×Jb (I and J are measurable),

such that ϕ(xI×J) =
∫
I

∫
J
ϕ(x(s, t))dsdt for all ϕ ∈ E∗ where the integral on the right is assumed

to exist in the sense of Lebesgue (by definition, xI×J =
∫
I

∫
J
x(s, t)dsdt).

We let L1(Ja × Jb, E) denote the Banach space of measurable functions u : Ja × Jb → E that
are Pettis integrable, equipped with the norm

‖u‖L1 =

∫a

0

∫b

0

‖u(x, y)‖dxdy.

Let P(E) is the family of all nonempty subsets of E.

A multivalued map G : E → P(E) has convex (closed) valued if G(x) is convex (closed) for all
x ∈ E. We say that G is bounded on bounded sets if G(B) is bounded in E for each bounded set



4 Mouffak Benchohra and Fatima-Zohra Mostefai CUBO
15, 1 (2013)

B of E (i.e. supx∈B{sup{‖y‖ : y ∈ G(x)}} < ∞). The map G is upper semicontinuous (u.s.c) on E
if for each x0 ∈ E, the set G(x0) is a nonempty closed subset of E and for each open set N of E

containing G(x0) there exists an open neighborhood M of x0 such that G(M) ⊆ N. The mapping

G has a fixed point if there exists x ∈ E such that x ∈ G(x).

In what follows Pcl(E) = {Y ∈ P(E) : Y is closed}, Pb(E) = {Y ∈ P(E) : Y is bounded},

Pcp(E) = {Y ∈ P(E) : Y is compact}, and Pcp,cv(E) = {Y ∈ P(E) : Y is compact and convex}.

A multivalued map G : Ja × Jb → Pcl(E) is said to be measurable if for each ω ∈ E the
function

(x, y) → d(ω, G(x, y)) = inf{|ω − υ| : υ ∈ G(x, y)}

is measurable. For more details on multivalued maps see the books of Aubin and Cellina [6],

Deimling [10].

Definition 2.2. A function h : E → E is said to be weakly sequentially continuous if h takes each
weakly convergent sequence in E to weakly convergent sequence in E (ie for any (xn)n in E with

xn → x in (E, w), h(xn) → h((x)) in (E, w)).

Definition 2.3. A function F : Q → Pcl,cv(Q) has weakly sequentially closed graph if for any
sequence (xn, yn) ∈ Q × Q, where yn ∈ F(xn) for n ∈ {1, 2, ...},and where both xn → x in (E, ω)
and yn → y in (E, ω) then y ∈ F(x).

Proposition 2.4. [17, 11] If x(.) is Pettis integrable and h(.) is a measurable and essentially

bounded real-valued function, then x(.)h(.) is Pettis integrable.

Definition 2.5. [9] Let E be a Banach space, ΩE the bounded subsets of E and B1 the unit ball of

E. The De Blasi measure of weak noncompactness is the map β : ΩE → [0, ∞) defined by
β(X) = inf{ǫ > 0 : there exists a weakly compact subset Ω of E : X ⊂ ǫB1 + Ω}

Properties: De Blasi measure of noncompactness satisfies some properties

(a) A ⊂ B ⇒ β(A) ≤ β(B),

(b) β(A) = 0 ⇔ A is relatively compact,

(c) β(A ∪ B) = max{β(A), β(B)},

(d) β(A
ω
) = β(A), (A

ω
denotes the weak clo-

sure of A),

(e) β(A + B) ≤ β(A) + β(B),

(f) β(λA) = |λ|β(A),

(g) β(conv(A)) = β(A),

(h) β(∪|λ|≤hλA = hβ(A).

The following result follows directly from the Hahn-Banach theorem.

Proposition 2.6. Let E be a normed space with x0 6= 0 then there exists ϕ ∈ E
∗ with ‖ϕ‖ = 1

and ϕ(x0) = ‖x0‖.



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Weak Solutions of Fractional Order Pettis Integral Inclusions ... 5

For a given set V of functions v : Ja × Jb → E let us denote by

V(x, y) = {v(x, y), v ∈ V}, (x, y) ∈ Ja × Jb

and

V(Ja × Jb) = {v(x, y) : v ∈ V, (x, y) ∈ Ja × Jb}.

For completeness, we recall the definition of the fractional Pettis-integral of order α > 0. Let

α1, α2 > 0 and α = (α1, α2). For h ∈ L
1(Ja × Jb, E), the expression

(Iα0 h)(x, y) =
1

Γ(α1)Γ(α2)

∫x

0

∫y

0

(x − s)α1−1(y − t)α2−1h(s, t)dsdt

where the sign "
∫
" denotes the Pettis integral and Γ(.) is the Euler gamma function, is called the

left sided mixed Pettis integral of order α.

For our purpose we will need the following fixed point theorem.

Theorem 2.7. [16] Let E be a Banach space with Q a nonempty, bounded, closed, convex and

equicontinuous subset of metrizable locally convex vector space C(J, E) such that 0 ∈ Q. Suppose

that T : Q → Pcl,cv(Q) has weakly-sequentially closed graph. If the implication

V = conv({0} ∪ T(V)) ⇒ V is relatively weakly compact, (6)

holds for every subset V ⊂ Q, then the operator T has a fixed point.

3 Main Results

we first define what we mean by solution of the problem (4)-(5).

Definition 3.1. A function u ∈ C(J, E) is said to be solution of problem (4)-(5) if there exists a

function v ∈ L1(Ja × Jb, E) with v(x, y) ∈ F(x, y, u(x, y)) and such that

u(x, y) =

m∑

i=1

gi(x, y)u(x − ξi, y − µi) +
1

Γ(α1)Γ(α2)

∫x

0

∫y

0

(x − s)α1−1(y − t)α2−1v(s, t)dsdt

and the function u satisfies condition (5) on J̃.

For any u ∈ C(Ja × Jb, E), we define the set

SF,u = {v ∈ L
1(Ja × Jb, E), v(x, y) ∈ F(x, y, u(x, y)), (x, y) ∈ Ja × Jb}

This is known as the set of selection function. Set

G = max
i=1...m

{ sup
(x,y)∈Ja×Jb

|gi(x, y)|}.

We are now in the position to state and prove our existence result for the problem (4)-(5). We

first list the following hypotheses.



6 Mouffak Benchohra and Fatima-Zohra Mostefai CUBO
15, 1 (2013)

(H1) F : Ja × Jb × E → Pcp,cl,cv(E), has weakly sequentially closed graph.

(H2) For each u ∈ C(Ja × Jb, E), there exists a measurable function v : Ja × Jb → E with v(x, y) ∈
F(x, y, u(x, y)) a.e. on Ja × Jb and v is Pettis integrable on Ja × Jb.

(H3) There exists p ∈ L∞(Ja × Jb, R+) such that

‖F(x, y, u)‖P = sup{‖v‖ : v ∈ F(x, y, u)} ≤ p(x, y),

for (x, y) ∈ Ja × Jb and each u ∈ E.

(H4) There exists a number R > 0 such that

p∗aα1bα2

Γ(α1 + 1)Γ(α2 + 1)(1 − mG)
< R, (7)

where p∗ = ‖p‖
∞
.

(H5) Let r0 > 0 be arbitrary (but fixed). For any ǫ > 0 and for any subset X ⊂ Br0, there exists

a closed subset Iǫ ⊂ Ja × Jb such that µ(Ja × Jb\Iǫ) < ǫ and

β(F(T × X)) ≤ sup
(x,y)∈T

p(x, y)β(X),

for each closed subset T of Iǫ, where µ denotes the Lebesgue measure in R
2.

The main result in this paper reads as follows.

Theorem 3.2. Assume that assumptions (H1)-(H5) hold. If

mG +
p∗aα1bα2

Γ(α1 + 1)Γ(α2 + 1)
< 1, (8)

then problem(4)-(5) has at least one solution on J.

Proof. To transform problem (4)-(5) into a fixed point problem, we define a multivalued map

Ω : C(J, E) → Pcl(C(J, E)) as

Ω(u) = {h ∈ C(J, E) such that

h(x, y) =






Ψ(x, y) if (x, y) ∈ J̃,

∑m
i=1

gi(x, y)u(x − ξi, y − µi) if υ ∈ SF,u,

+ 1
Γ(α1)Γ(α2)

∫x
0

∫y
0
(x − s)α1−1(y − t)α2−1υ(s, t)dsdt; (x, y) ∈ Ja × Jb.

where Ψ(·, ·) is the function defined by (5). Now, we prove that Ω satisfies all the assumptions of

the Theorem 2.7 and thus Ω has a fixed point which is a solution of problem (4)-(5).



CUBO
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Weak Solutions of Fractional Order Pettis Integral Inclusions ... 7

First notice that, for all u ∈ C(J, E), there exists a Pettis integral υ : Ja × Jb → E such that
υ(x, y) ∈ F(x, y, u(x, y)) for a.e. (x, y) ∈ Ja × Jb (Assumption (H2)) then ϕ(υ(x, y)) ∈ L

1(Ja × Jb)

for any ϕ ∈ E∗. From the definition of the integral of fractional order we have

Iαϕ(υ(x, y, )) =

∫x

0

∫y

0

(x − s)α1−1(y − t)α2−1

Γ(α1)Γ(α2)
ϕ(υ(s, t))dsdt

=

∫x

0

∫y

0

ϕ

(

(x − s)α1−1(y − t)α2−1

Γ(α1)Γ(α2)
υ(s, t)

)

dsdt

exists for almost every (x, y) ∈ Ja ×Jb and is an element from L
1(Ja ×Jb), that is, for almost every

(x, y) ∈ Ja × Jb, s ∈ (0, x), t ∈ (0, y) the measurable function

ϕ

(

(x − s)α1−1(y − t)α2−1

Γ(α1)Γ(α2)
υ(s, t)

)

=
(x − s)α1−1(y − t)α2−1

Γ(α1)Γ(α2)
ϕ(υ(s, t))

is Lebesgue integrable, hence the function (s, t) → (x−s)
α
1
−1

(y−t)
α
2
−1

Γ(α1)Γ(α2)
υ(s, t) is Pettis integrable

on Ja × Jb, and thus the operator Ω is well defined.

Let R > 0 and consider the set

Q = {u ∈ C(J, E) : ‖u‖
∞

≤ R

and ‖u(x2, y2) − u(x1, y1)‖ ≤ R

m∑

i=1

|gi(x2, y2) − gi(x1, y1)|

+
p∗

Γ(α1 + 1)Γ(α2 + 1)
[x

α1
2

y
α2
2

− x
α1
1

y
α2
1

]; for (x1, y1), (x2, y2) ∈ Ja × Jb}

Clearly, the subset Q is closed, bounded, convex and equicontinuous subset of a metrisable locally

convex vector space C(J, E). The remainder of the proof will be given in four steps.

Step 1: Ω(u) is convex for each u ∈ Q.

For that, let 0 < λ < 1, h1, h2 ∈ Ω(u), obviously if (x, y) ∈ J̃ then λh1(x, y)+(1−λ)h2(x, y) ∈

Ω(u). Now if (x, y) ∈ Ja × Jb, then there exists υ1, υ2 ∈ SF,u such that

hi(x, y) =

m∑

i=1

gi(x, y)u(x − ξi, y − µi) +
1

Γ(α1)Γ(α2)

∫x

0

∫y

0

(x − s)α1−1(y − t)α2−1υi(s, t)dsdt;

i = 1, 2, Then for each (x, y) ∈ Ja × Jb we have

(λh1 + (1 − λ)h2)(x, y) =

m∑

i=1

gi(x, y)u(x − ξi, y − µi)

+
1

Γ(α1)Γ(α2)

∫x

0

∫y

0

(x − s)α1−1(y − t)α2−1(λυ1(s, t) + (1 − λ)υ2(s, t))dsdt.



8 Mouffak Benchohra and Fatima-Zohra Mostefai CUBO
15, 1 (2013)

Since SF,u is convex ( because F has convex values), it follows that λh1 + (1 − λ)h2 ∈ Ω(u).

Step 2: Ω maps Q into Q.

To see this, take h ∈ ΩQ. Then there exists u ∈ Q with h ∈ Ωu. And there exists υ :

Ja × Jb → E Pettis integrable with υ(x, y) ∈ F(x, y, u(x, y))

h(x, y) =






Ψ(x, y) if (x, y) ∈ J̃,

∑m
i=1

gi(x, y)u(x − ξi, y − µi) if υ ∈ SF,u,

+ 1
Γ(α1)Γ(α2)

∫x
0

∫y
0
(x − s)α1−1(y − t)α2−1υ(s, t)dsdt; (x, y) ∈ Ja × Jb.

We can consider that h(x, y) 6= 0 and by Proposition 2.6 there exists ϕ ∈ E∗ with ‖ϕ‖ = 1 and

ϕ(h(x, y)) = ‖h(x, y)‖ for (x, y) ∈ Ja × Jb, we have

‖h(x, y)‖ = ϕ(h(x, y))

= ϕ
(

m∑

i=1

gi(x, y)u(x − ξi, y − µi)

+
1

Γ(α1)Γ(α2)

∫x

0

∫y

0

(x − s)α1−1(y − t)α2−1υ(s, t)dsdt
)

= ϕ

(

m∑

i=1

gi(x, y)u(x − ξi, y − µi)

)

+ ϕ

(

1

Γ(α1)Γ(α2)

∫x

0

∫y

0

(x − s)α1−1(y − t)α2−1υ(s, t)dsdt

)

≤ mGR +
p∗

Γ(α1)Γ(α2)

∫x

0

∫y

0

(x − s)α1−1(y − t)α2−1dsdt

≤ mGR +
p∗aα1bα2

Γ(α1 + 1)Γ(α2 + 1)
≤ R.

on the other hand, for (x, y) ∈ J̃, we have

‖h(x, y)‖ = ϕ(h(x, y)) ≤ R.

Next, suppose that (x1, y1), (x2, y2) ∈ Ja × Jb with x1 < x2 and y1 < y2, and let h ∈ Ωu, so

h(x1, y1) − h(x2, y2) 6= 0. Then there exists ϕ ∈ E
∗ such that

‖h(x1, y1) − h(x2, y2)‖ = ϕ(h(x1, y1) − h(x2, y2)),

and ‖ϕ‖ = 1. Thus



CUBO
15, 1 (2013)

Weak Solutions of Fractional Order Pettis Integral Inclusions ... 9

‖h(x2, y2) − h(x1, y1)‖

= ϕ(

m∑
i=1

gi(x2, y2)u(x2 − ξi, y2 − µi)

+
1

Γ(α1)Γ(α2)

∫x2
0

∫y2
0

(x2 − s)
α1−1(y2 − t)

α2−1υ(s, t)dsdt

−

m∑
i=1

gi(x1, y1)u(x1 − ξi, y1 − µi)

+
1

Γ(α1)Γ(α2)

∫x1
0

∫y1
0

(x1 − s)
α1−1(y1 − t)

α2−1υ(s, t)dsdt)

= ϕ(

m∑
i=1

gi(x2, y2)u(x2 − ξi, y2 − µi) −

m∑
i=1

gi(x1, y1)u(x1 − ξi, y1 − µi))

+ϕ(
1

Γ(α1)Γ(α2)

∫x2
x1

∫y2
y1

(x2 − s)
α1−1(y2 − t)

α2−1υ(s, t)dsdt

+
1

Γ(α1)Γ(α2)

∫x1
0

∫y1
0

[(x2 − s)
α1−1(y2 − t)

α2−1 − (x1 − s)
α1−1(y1 − t)

α2−1]

×υ(s, t)dsdt +
1

Γ(α1)Γ(α2)

∫x1
0

∫y2
y1

(x2 − s)
α1−1(y2 − t)

α2−1υ(s, t)dsdt

+
1

Γ(α1)Γ(α2)

∫x2
x1

∫y1
0

(x2 − s)
α1−1(y2 − t)

α2−1υ(s, t)dsdt)

≤
m∑
i=1

‖gi(x2, y2)u(x2 − ξi, y2 − µi) − gi(x1, y1)u(x1 − ξi, y1 − µi)‖

+
p∗

Γ(α1)Γ(α2)

∫x1
0

∫y1
0

[(x2 − s)
α1−1(y2 − t)

α2−1 − (x1 − s)
α1−1(y1 − t)

α2−1]dsdt

+
p∗

Γ(α1)Γ(α2)

∫x2
x1

∫y2
y1

(x2 − s)
α1−1(y2 − t)

α2−1dsdt

+
p∗

Γ(α1)Γ(α2)

∫x1
0

∫y2
y1

(x2 − s)
α1−1(y2 − t)

α2−1dsdt

+
p∗

Γ(α1)Γ(α2)

∫x2
x1

∫y1
0

(x2 − s)
α1−1(y2 − t)

α2−1dsdt

≤ R
m∑
i=1

‖gi(x2, y2) − gi(x1, y1)‖ +
p∗

Γ(α1 + 1)Γ(α2 + 1)
[x

α1
2

y
α2
2

− x
α1
1

y
α2
1

].

This implies that h ∈ Q, hence ΩQ ⊂ Q

Step 3: Ω has weakly sequentially closed graph.

Let (un, wn) be a sequence in Q×Q with un(x, y) → u(x, y) in (E, w) for each (x, y) ∈ Ja×Jb,
wn(x, y) → w(x, y) in (E, w) for each (x, y) ∈ Ja ×Jb and wn ∈ Ω(un) for n ∈ {1, 2, . . .}. We show
that w ∈ Ω(u).

Since wn ∈ Ω(un), there exists υn ∈ SF,un such that

wn(x, y) =

m∑

i=1

gi(x, y)un(x − ξi, y − µi) +
1

Γ(α1)Γ(α2)

∫x

0

∫y

0

(x − s)α1−1(y − t)α2−1υn(s, t)dsdt.



10 Mouffak Benchohra and Fatima-Zohra Mostefai CUBO
15, 1 (2013)

We show that there exists υ ∈ SF,u such that

w(x, y) =

m∑

i=1

gi(x, y)u(x − ξi, y − µi) +
1

Γ(α1)Γ(α2)

∫x

0

∫y

0

(x − s)α1−1(y − t)α2−1υ(s, t)dsdt.

Since F(·, ·, ·) has compact values, there exists a subsequence υnm ∈ SF,un such that υnm is Pettis

integrable and

υnm(x, y) ∈ F(x, y, un(x, y)) a.e.(x, y) ∈ Ja × Jb

and

υnm(·, ·) → υ(·, ·) in (E, w) as m → ∞.

As F(x, y, ·) has weakly sequentially closed graph, υ(x, y) ∈ F(x, y, u(x, y)). Then Lebesgue

Dominated Convergence theorem for Pettis integral implies that

ϕ(wn(x, y)) → ϕ

(

m∑

i=1

gi(x, y)u(x − ξi, y − µi) +
1

Γ(α1)Γ(α2)

∫x

0

∫y

0

(x − s)α1−1(y − t)α2−1υ(s, t)dsdt

)

i.e. wn(x, y) → Ωu(x, y) in (E, w). Since this holds, for each (x, y) ∈ Ja × Jb, we have w ∈ Ωu.

Step 4: the implication (6) holds.

Let V be a subset of Q such that V = conv(Ω(V)∪{0}). Obviously V(x, y) ⊂ conv(Ω(V(x, y))∪

{0}), ∀(x, y) ∈ J. Further, as V is bounded and equicontinuous, the function (x, y) → υ(x, y) =
β(V(x, y)) is continuous on J.

If (x, y) ∈ J̃ then

ΩV(x, y) = {Ωu(x, y) : u ∈ V} = {Ψ(x, y) : (x, y) ∈ J̃}.

and since Ψ is continuous on [−ξ, 0] × [−µ, 0], the set {Ψ(x, y), (x, y) ∈ [−ξ, 0] × [−µ, 0]} ⊂ E is

compact. Now by (H3) and the properties of the measure β, for any (x, y) ∈ Ja × Jb, we have

υ(x, y) ≤ β((ΩV)(x, y) ∪ {0})

≤ β((ΩV)(x, y))

≤ β{Ωu(x, y) : u ∈ V}

≤ β
{∑m

i=1
gi(x, y)u(x − ξi, y − µi); u ∈ V

}

+β
{

1
Γ(α1)Γ(α2)

∫x
0

∫y
0
(x − s)α1−1(y − t)α2−1υ(s, t)dsdt; υ(x, y) ∈ F(x, y, u), u ∈ V

}

≤
∑m

i=1
β({gi(x, y)u(x − ξi, y − µi); u ∈ V})



CUBO
15, 1 (2013)

Weak Solutions of Fractional Order Pettis Integral Inclusions ... 11

+ 1
Γ(α1)Γ(α2)

β
{∫x

0

∫y
0
(x − s)α1−1(y − t)α2−1υ(s, t)dsdt; υ(x, y) ∈ F(x, y, u), u ∈ V

}

≤
∑m

i=1
|gi(x, y)|β(V(x, y))

+ 1
Γ(α1)Γ(α2)

∫x
0

∫y
0
(x − s)α1−1(y − t)α2−1p(s, t)β(V(s, t))dsdt

≤ mG‖υ‖
∞

+
p

∗

a
α
1 b

α
2

Γ(α1+1)Γ(α2+1)
‖υ‖

∞

In particular,

‖υ‖
∞

≤ ‖υ‖
∞

(

mG +
p∗aα1bα2

Γ(α1 + 1)Γ(α2 + 1)

)

.

By (8) it follows that ‖υ‖
∞

= 0, that is υ(x, y) = β(V(x, y)) = 0 for each (x, y) ∈ J and then V is

weakly relatively compact in C(J, E). Applying now Theorem 2.7 we conclude that T has a fixed

point which is a solution of problem (4)-(5). 2

Received: October 2012. Revised: February 2013.

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