eu EXISTENCE OF WEAK SOLUTIONS FOR A FUNCTIONAL INTEGRAL INCLUSION

In this paper we study the existence of weak solution x ∈ C[I, E] for the nonlinear functional integral inclusion x(t) ∈ F (


Introduction
Let I = [0, T ], and let L 1 (I) be the class of all Lebesgue integrable functions defined on the interval I. Let E be a reflexive Banach space with norm .and dual E * .
Denote C[I, E] the Banach space of strongly continuous functions x : I → E with sup-norm x C = sup x(t) E .
Consider the functional integral inclusion where F : I × E → P (E) is a nonlinear set-valued mapping, and P (E) denote the family of nonempty subsets of the Banach space E.
Here we study the existence of weak solution x ∈ C[I, E] of the functional integral inclusion (1) in the reflexive Banach space E, under the assumption that the set-valued function F satisfy Lipschitz condition.
Indeed a set-valued functional equations have been extensively investigated by a number of authors and there are many interesting results concerning this problem (see [3], [8]- [10] and [12]- [13]).
The existence of weak solutions of the integral equations were studied by a number of authors such as (see for instance [1]- [2], [5] and [15]- [16]).

Preliminaries
Here, we present some auxiliary results that will be needed in this work.
Let E be a Banach space and let x : I → E, then: (1) x(.) is said to be weakly continuous (measurable) at t 0 ∈ I if for every φ ∈ E * , φ(x(.)) is continuous (measurable) at t 0 .
(2) A function h : E → E is said to be weakly sequentially continuous if h maps weakly convergent sequence in E to weakly convergent sequence in E .
If x is weakly continuous on I , then x is strongly measurable and hence weakly measurable (see [4] and [7]).Note that in reflexive Banach spaces weakly measurable functions are pettis integrable (see [7] and [11] for the definition) if and only if φ(x(.)) is Lebesgue integrable on I for every φ ∈ E * .Now we state a fixed point theorem and some propositions which will be used in the sequal (see [14]).
Theorem 1 (O'Regan Fixed Point Theorem).Let E be a Banach space and let Q be a nonempty, bounded, closed and convex subset of the space (C[0, T ], E) and let A : Q → Q be a weakly sequentially continuous and assume that AQ(t) is relatively weakly compact in E for each t ∈ [0, T ].Then A has a fixed point in the set Q. Proposition 2. A subset of a reflexive Banach space is weakly compact if and only if it is closed in the weak topology and bounded in the norm topology.Proposition 3. Let E be a normed space with y = 0. Then there exists a φ ∈ E * with φ = 1 and y = φ(y).Definition 4 (see [6]).A set-valued map F from I × E to the family of all nonempty closed subsets of E is called Lipschitzian if there exists L > 0 such that for all t 1 , t 2 ∈ I and all x 1 , x 2 ∈ E, we have where H(A, B) is the Hausdorff metric between the two subsets A, B ∈ I × E.
Denote S F = Lip(I, E) be the set of all Lipschitz selections of F .

Existence of Weak Solution
In this section, we present our main result by proving the existence of weak solution x ∈ C[I, E] of the functional integral inclusion (1) in the reflexive Banach space E, under the assumption that the set-valued function F satisfy Lipschitz condition.

Coupled System Approach
Consider now the functional integral inclusion (1) under the following assumptions: (H1) The set F (t, x) is compact and convex for all (t, x) ∈ I × E.
(H2) The set-valued map F is Lipschitzian with a Lipschitz constant L > 0.
(H3) The set of all Lipschitz selections S F is nonempty.
(H5) g(., x) is weakly measurable on I for every x ∈ E.
(H6) There exists a function a ∈ L 1 [0, T ] and a constant b > 0 such that Remark 5. From assumptions (H2) and (H3), there exists Then the solution of the functional integral equation ( 3), if it exists, is a solution of the functional integral inclusion (1).

Now let
then from (3) we have Then the functional integral equation ( 3) is equivalent to the coupled system (4) and (5).Consider now the coupled system (4) and ( 5).Now, we study the existence of a weak solution of the functional integral equation (3), which is a solution of the functional integral inclusion (1), by getting the weak solution of the coupled system (4) and (5).Definition 6.By a weak solution of the coupled system (4) and ( 5) we mean the ordered pair of functions (x, y), x, y ∈ C Now for the existence of a weak continuous solution of the coupled system (4) and ( 5) we have the following theorem.
Theorem 7. Let the assumptions (H1)-(H7) be satisfied.Then the coupled system (4) and ( 5) has at least one weak continuous solution. Proof.Let Let A be any operator defined by where Let the set Q r defined by Let U = (x, y) ∈ Q r be an arbitrary ordered pair, then we have from proposition 3 where x(m(s)) ds Then AU X ≤ r.
Hence, AU ∈ Q r , which proves that AQ r ⊂ Q r , i.e.A : Q r → Q r , and the class of functions {AQ r } is uniformly bounded.Now Q r is nonempty, closed, convex and uniformly bounded.As a consequence of proposition 2, then {AQ r } is relatively weakly compact.Now, we shall prove that A : X → X.Let t 1 , t 2 ∈ I, t 1 < t 2 (without loss of generality assume that AU (t 2 ) − AU (t 1 ) = 0), then firstly we have and Which proves that A : X → X.
Finally, we want to prove that A is weakly sequentially continuous.
Let {U n } be a sequence in Q r converges weakly to U ∀t ∈ I, then we have the two sequences {y n }, {x n }, such that {y n } converges strongly to y and {x n } converges weakly to x, i.e. y n (t) → y, x n (t) ⇀ x, ∀t ∈ I.

), and then
, and then Hence, A is weakly sequentially continuous (i.e.AU n (t) → AU (t), ∀t ∈ I weakly).
Since all conditions of O'Regan theorem are satisfied, then the operator A has at least one fixed point U ∈ Q r , and then the coupled system (4) and ( 5) has at least one weak solution (x, y) ∈ X, then there exists at least one weak solution x ∈ C[I, E] of the functional integral equation (3).
Consequently, there exists at least one weak continuous solution x ∈ C[I, E] of the functional integral inclusion (1).Then Ax(t) ≤ r.

Functional Integral Inclusion Approach
Finally, we want to prove that A is weakly sequentially continuous.Let {x n } be a sequence in Q r converges weakly to x ∀t ∈ I, i.e. x n (t) ⇀ x, ∀t ∈ I.
Thus φ(g(t, x n (m(t)))) converges strongly to φ(g(t, x(m(t)))).Also, Since all conditions of O'Regan theorem are satisfied, then the operator A has at least one fixed point x ∈ Q r , and then there exists at least one weak solution x ∈ C[I, E] of the functional integral equation (3).
Consequently, there exists at least one weak solution x ∈ C[I, E] of the functional integral inclusion (1).
and the class of functions {AQ r } is uniformly bounded.Now Q r is nonempty, closed, convex and uniformly bounded.As a consequence of proposition 1, then {AQ r } is relatively weakly compact.Now, we shall prove that A :C[I, E] → C[I, E].Let t 1 , t 2 ∈ I, t 1 < t 2 (without loss of generality assume that Ax(t 2 ) − Ax(t 1 ) = 0), then we have Then by applying Lebesgue dominated convergence theorem for Pettis integral, we getφ(Ax n (t)) = φ(f (t,Hence, A is weakly sequentially continuous (i.e.Ax n (t) → Ax(t), ∀t ∈ I weakly).