Fractional differential inclusions with anti-periodic boundary conditions in Banach spaces

The main purpose of this paper is to provide the theory of differential inclusions by new existence results of solutions for boundary value problems of differential inclusions with fractional order and with anti-periodic boundary conditions in Banach spaces. We prove existence theorems of solutions under both convexity and nonconvexity conditions on the multivalued side. Meanwhile, the compactness of the set solutions is also established.


Introduction
During the past two decades, fractional differential equations and fractional differential inclusions have gained considerable importance due to their applications in various fields, such as physics, mechanics and engineering.For some of these applications, one can see [16,22,27,33] and the references therein.El Sayed et al. [15] initiated the study of fractional multivalued differential inclusions.For some recent development on initial value problems for differential equations and inclusions of fractional order we refer the reader to the references [1,32,[34][35][36][37][38].
Some applied problems in physics require fractional differential equations and inclusions with boundary conditions.Recently, many authors have studied differential inclusions with various boundary conditions.Some of these works have been done in finite dimensional spaces and of positive integer order, for example, Ibrahim et al. [25] and Gomma [17] considered a functional multivalued three-point boundary value problem of second-order.Gomma [18] studied four-point boundary value problems for non-convex differential inclusions.
Several results have been obtained for fractional differential equations and inclusions with various boundary value conditions in finite dimensional spaces.We refer, for example, to Agarwal et al. [1] who established conditions for the existence of solutions for various classes A. G. Ibrahim of initial and boundary value problems for fractional differential equations and inclusions involving the Caputo derivative, Ouahab [31] studied a fractional differential inclusion with Dirichlet boundary conditions under both convexity and nonconvexity conditions on the multivalued right-hand side and Ntouyas et al. [30] discussed the existence of solutions for a boundary value problem of fractional differential inclusions with three-point integral boundary conditions involving convex and non-convex multivalued maps.
For some recent works on boundary value problems for fractional differential inclusions in infinite dimensional spaces, we refer to Benchohra et al. [8] who established the existence of solutions of nonlinear fractional differential inclusions with two point boundary conditions.
Anti-periodic boundary conditions appear in a variety of situation, see [2,3,12,13] and the references therein.
In [28], Lan et al. pointed out that we can neither prove that if h ∈ C([0, 1], R), then x given in (1.4) is a solution of (1.3) nor show that if h ∈ C(J, R) satisfies (1.3), then x and h satisfy (1.4) although these results have been widely used in some papers such as [2, Lemma 2.1], [3,Lemma 1.2] and [12,Lemma 2.7].Due to the requirement h ∈ AC(J, E), the continuity assumption on f is not sufficient.To overcome this difficulty we shall impose a Lipschitz type condition on f .Motivated by the work of Lan et al. [28] we introduce a correct formula of solutions of (1.1).
Fractional differential inclusions with anti-periodic boundary conditions 3 Also, due to the requirement h ∈ AC(J, E), a problem arises when we study the problem (1.1) in the case when f is a multifunction.This problem arises because we do not know the conditions that guarantee the closedness of the set of absolutely continuous selections for a multifunction.To avoid this problem we consider the Caputo derivative in the generalized sense when we study the problem (1.2).We shall show that if x ∈ AC 3 (J, E) is a solutions of (1.2), then the following fractional boundary value problem holds.We note that Ahmed [2] considered (1.1) in infinite dimensional Banach spaces while Cerna [12] considered (1.2) in finite dimensional spaces.It is important to note that, based on the remark of Lan et al. [28], and mentioned in the preceding paragraph, the proofs in the paper of both Ahmed [2] and Cerna [12] require additional assumptions.
The present paper is organized as follows: in Section 2 we collect some background material and basic results from multivalued analysis and fractional calculus to be used later.In Section 3 we prove an existence result for (1.1) and in Section 4 we establish various existence results for (1.2) and we prove the compactness of the set of solutions.At the end of the paper we give examples in order to illustrate the feasibility of our assumptions.
The proofs rely on the methods and results for boundary value fractional differential inclusions, the properties of noncompact measure and fixed point techniques.

Preliminaries and notation
Let C(J, E) be the space of E-valued continuous functions on J with the uniform norm x = sup{ x(t) , x ∈ J}, L 1 (J, E) be the quotient space of all E-valued Bochner integrable functions on J with the norm Let G : J → 2 E be a multifunction.By S 1 G we will denote the set of integrable selections of G; i.e. S 1 If the multifunction G is completely continuous with nonempty compact values, then G is u.s.c.if and only if G is closed.Lemma 2.2 ([26, Theorem 1.3.5]).Let X 0 , X be (not necessarily separable) Banach spaces, and let F : J × X 0 → P k (X) be such that (i) for every x ∈ X 0 the multifunction F(•, x) has a strongly measurable selection; (ii) for a.e.t ∈ J the multifunction F(t, •) is upper semicontinuous.
Then for every strongly measurable function z : J → X 0 there exists a strongly measurable function f : J → X such that f (t) ∈ F(t, z(t)), a.e.Remark 2.3 ([26,Theorem 1.3.1]).For single-valued or compact-valued multifunctions acting on a separable Banach space the notions measurability and strongly measurable coincide.So, if X 0 , X are separable Banach spaces we can replace strongly measurable with measurable in the above lemma.
is absolutely continuous with compact values, t 0 ∈ [a, b] and x 0 ∈ G(t 0 ).Then G admits an absolutely continuous selection g satisfying g(t 0 ) = x 0 .
The Hausdorff measure of noncompactness is defined on P b (E) as Definition 2.14 ([6]).The Riemann-Liouville fractional integral of order q > 0 with lower limit zero for a function f ∈ L P (J, E), P ∈ [1, ∞) is defined as follows: where the integration is in the sense of Bochner, Γ is the Euler gamma function defined by Γ(q) = ∞ 0 t q−1 e −t dt, g q (t) = t q−1 Γ(q) , for t > 0, g q (t) = 0, for t ≤ 0 and * denotes the convolution of functions.For q = 0, we set I 0 f (t) = f (t).It is known that I q I β f (t) = I q+β f (t), β, q ≥ 0.
Note that by applying the Young inequality, it follows that Then I q maps L P (J, E) to L P (J, E).
Definition 2.15 ([6]).Let q > 0 and m the smallest integer greater than or equal to q.The Riemann-Liouville fractional derivative of order q for a function A. G. Ibrahim Lemma 2.16 ([6, Lemma 1.8]).Let q > 0, m be the smallest integer greater than or equal to q and 1 < P < ∞.Let L q be an operator with domain L P (J, E), defined by L q ( f ) = I q f = g q * f and L q be an operator with domain R m,P 0 (J, E) = f ∈ L P (J, E) : g m−q * f ∈ W m,P 0 (J, E) and defined by L q ( f ) = D q f .Then L q = L −1 q .
To know more about fractional calculus see [7,27,33].The proof of the following lemma is the same way as in the scalar case (see [27,33]).Lemma 2.17.Let AC(J, E) be the space of absolutely continuous functions defined on J to E and q ∈ (0, 1).E) and E is separable, then I q (D q f (t)) = f (t), a.e.t ∈ J.
(ii) I q maps AC(J, E) to AC(J, E).
Proof.(i) Let f ∈ AC(J, E).From Lemma 2.16, it suffices to show that g 1−q * f ∈ W m,1 0 (J, E).Since E is separable, f has a Bochner integrable derivative f (1) almost everywhere and The first term is an absolutely continuous function because t 1−q = (1 − q) t 0 x −q dx.The second term is also a primitive of a Bochner integrable function and hence it is absolutely continuous.Moreover, (g (ii) Let f ∈ AC(J, E).By arguing as in (i), we get We denote by C m (J, E) the Banach space of m times continuously differentiable functions with the norm

Definition 2.18 ([6]
).Let q > 0 and m be the smallest integer greater than or equal to q.The Caputo derivative of order q for given function f ∈ AC m (J, E) is defined by Fractional differential inclusions with anti-periodic boundary conditions 7 We need the following lemma.Lemma 2.19.Let E be a separable Banach space and f ∈ AC 3 (J, E).Then where b 1 , b 2 and b 3 are elements in E.
Proof.In view of Definition 2.18 we have The following lemma is essential and its proof is similar to the proofs of Theorems 2.4 and 2.7 in [28].
Lemma 2.20.Let E be a separable Banach space.
We need also to the following auxiliary results: Lemma 2.21 (Kakutani-Glicksberg-Fan theorem [19]).Let W be a nonempty compact and convex subset of a locally convex topological vector space.If R : W → P P cl,cv (W) is an u.s.c.multifunction, then it has a fixed point.
Fractional differential inclusions with anti-periodic boundary conditions 9 Lemma 2.22 ([26,Prop. 3.5.1]).Let W be a closed subset of a Banach space X and R : W → P k (X) be a closed multifunction which is γ-condensing on every bounded subset of W, where γ is a monotone measure of noncompactness defined on X.If the set of fixed points for R is a bounded subset of X then it is compact.
The following fixed point theorem for contraction multivalued is proved by Govitz and Nadler [14].
Lemma 2.23.Let (X, d) be a complete metric space.If R : X → P cl (X) is contraction, then R has a fixed point.

Existence of solutions for the problem (1.1)
In the rest of the paper, E will denote a separable Banach space.In this section, we give an existence result of solutions of (1.1).Theorem 3.1.Let E be a separable Banach space, f : J × E → E be a function.We assume the following hypothesis: (H 1 ) For each ρ > 0, there exists L ρ > 0 such that for all s, t ∈ J and x, y Then, the problem (1.1) has a solution provided that there is a positive real number r such that Proof.According to condition (H 1 ), there exists L r > 0 such that for all s, t ∈ J and x, y ∈ E with x ≤ r, y ≤ r we have Let us introduce a function T : C(J, E) → C(J, E) defined by Firstly we prove, by using Schauder's fixed point theorem, that T has a fixed point.The proof will be given in several steps.
Step 1.Let B 0 = {x ∈ C(J, E) : x ≤ r}.Obviously, B 0 is a bounded, closed and convex subset of C(J, E).We claim that T(B 0 ) ⊆ B 0 .Let x ∈ B 0 .We note that, by ( Let y = T(x).Then (3.3) and (3.4) imply that for t ∈ J y(t) ≤ (L r max{b, r} + f (0, 0) ) Step 2. Let Z = T(B 0 ).We claim that Z is equicontinuous.Let y ∈ Z. Then there is x ∈ B r with y = T(x).Therefore, for t, t + λ ∈ J we have This inequality implies y(t By induction, the sequence (B n ), n ≥ 1 is a decreasing sequence of nonempty, closed convex and bounded subsets of C(J, E).Our goal is to show that the subset B = ∩ ∞ n=1 B n is nonempty and compact in C(J, E).By Lemma 2.8, it is enough to show that lim where χ C(J,E) is the Hausdorff measure of noncompactness on C(J, E).
Step 3. Our aim in this step is to show that the relation (3.5) is satisfied.Let n ≥ 1 be a fixed natural number and ε > 0. In view of Lemma 2.9, there exists a sequence From Step 2, B n−1 is equicontinuous.This together with Lemma 2.10 and by using the nonsingularity of χ, the above inequality becomes Let t ∈ J be fixed.Note that from (3.2) for every natural number m, n we have From (3.7) and the properties of χ, for t ∈ J we get Therefore, Now in order to estimate the quantity χ b 0 (b − s) α−3 f (s, x k (s)) ds : k ≥ 1 , we note that (3.4) implies for any k ≥ 1 and for t ∈ J f (s, x k (s)) ≤ L r max{b, r} + f (0, 0) . (3.9) Then, by (3.7), (3.9) and Lemma 2.12, there exist a compact K ⊆ E, a measurable set J ⊂ J with measure less than , and a sequence of functions {z k } ⊂ L 1 (J, E) such that for all s ∈ J, {z k (s for every k ≥ 1 and every s ∈ J − J .Therefore, for any k ≥ 1 (3.10) Also, for any k ≥ 1 This together with (3.10) we have From this inequality and by taking into account that ε is arbitrary, we get where By means of a finite number of steps, we obtain from (3.12) for every n ∈ N, Fractional differential inclusions with anti-periodic boundary conditions 13 Observe that (3.1) implies Then ζ < 1.By passing to the limit as n → +∞ in (3.13) we obtain (3.5) and so our aim in this step is verified.Therefore, the set B = ∩ ∞ n=1 B n is a nonempty and compact subset of C(J, E).Moreover, every B n being bounded, closed and convex, B is also bounded closed and convex.
Step 4. Let us verify that T(B) ⊆ B.
Step 5.The function T| B : B → 2 B is continuous.Consider a sequence {x n } n≥1 in B with x n → x in B and let y n = T(x n ).We have to show that lim n→∞ y n = T(x).For any n ≥ 1 and Note that for every t ∈ J f (t, x n (t))(t) ≤ L r max{b, r} + f (0, 0) .

Furthermore, lim
Therefore, by passing to the limit as n → ∞ in (3.14) we get lim n→∞ y n = T(x).
As a consequence of Steps 1-5 and Schauder's fixed point theorem, there is x ∈ B such that x = T(x).That is, for any t ∈ J where h(t) = f (t, x(t)).Obviously h is continuous, and hence x (1) (t) exists.Then there is a positive number η such that x (1) (t) ≤ η, t ∈ J. Thus, by (H 1 ) for t, s ∈ J This means that h ∈ AC(J, E) and hence by Lemma 2.20(2) the function x is a solution for (1.1).
In the following corollary we simplify the condition (3.1).Corollary 3.2.Assume that the assumption (H 1 ) is satisfied with L ρ ≤ σ, for any ρ > 0 then the problem (1.1) has a solution provided that Proof.From (3.15) we can take r such that max We need only to check that T(B 0 ) ⊆ B 0 .As in Step 1, let x ∈ B 0 and y = T(x).Then by (3.16) for any t ∈ J y(t) ≤ (σr Proof.In virtue of Lemma 2.7 there is h ∈ AC(J, E) such that h(t) ∈ G(t), a.e.We define x : J → E, by By Lemma 2.20 (2), x ∈ AC 3 (J, E) and satisfies (3.17).

Existence of solutions for the problem (1.2)
To give existence results of solutions for the problem (1.2) we present the definition of the Caputo derivative in the generalized sense [27, Sec.2.4].
Definition 4.1.Let q > 0 and m the smallest integer greater than or equal to q.The generalized or weak Caputo derivative of order q with lower limit zero for a function f ∈ J → E is defined by So, the generalized or weak Caputo derivative c D q g f (t) is defined for function f for which the Riemann-Liouville fractional derivative exists.In particular, when q ∈ (0, 1), we have As in the scalar case (see Theorems 2.1 and 2.2 [27]) Lemma 4.2.Let q > 0 and m be the smallest integer greater than or equal to q.If f ∈ AC m (J, E), then c D q g f (t) exists almost everywhere on J and if q is not natural number, then Next, we need the following auxiliary lemma.
Proof.(1) Since x is a solution of (4.2), then This equation implies From this equation together the condition x (2) (0) = −x (2) (b) we get By integrating both sides in this equation we get for a.e.
Again, by integrating both sides in this equation we have for a.e.t ∈ J Applying the condition x(0) = −x(b) we get for a.e.
The two functions on both sides of this equation are continuous, thus it holds for every t ∈ J.
This means that if z ∈ AC(J, E), then the function x given by (4.3) is a solution for the fractional boundary value problem As a result of Lemma 4.3 we can give the concept of solutions for (1.2) in the following definition.

Definition 4.5. A function
Now we are in position to give existence results of solutions of (1.2).We consider first the case when the values of the multifunction F are convex.
Then the problem (1.2) has a solution provided that there is r > 0 such that Proof.At first, in view of (H 2 ), Lemma 2.2 and Remark 2.3, for every x ∈ C(J, E), the multifunction t → F(t, x(t)) has a measurable selection and by (H 3 ) this selection belongs to S 1 F(•,x(•)) .So, we can introduce the multifunction R : C(J, E) → 2 C(J,E) which is defined as: let It is easy to see that any fixed point for R is a mild solution for (1.2).So our goal is to prove, by using Lemma 2.21, that R has a fixed point.The proof will be given in several steps.

By arguing as in
Step 2 in the proof of Theorem 3.1, for any t, t + λ ∈ J we have Not that by the Hölder's inequality we have The previous two inequalities imply that y(t Clearly the sequence (D n ), n ≥ 1 is a decreasing sequence of nonempty, closed, convex and bounded subsets of C(J, E).Our goal is to use Lemma 2.8 to show that D is nonempty and compact in C(J, E).So we show that lim where χ C(J,E) is the Hausdorff measure of noncompactness on C(J, E).
Step 3. Our aim in this step is to show that the relation (4.10) is satisfied.Let n ≥ 1 be a fixed natural number and ε > 0. In view of Lemma 2.9, there exists a sequence (y By applying Lemma 2.10 and by using the nonsingularity of χ, the above inequality becomes By recalling the definition of R, repeating the same procedure as in Step 3 in the proof of Theorem 3.1, applying Lemma 2.11 and using (H 4 ), for every k ≥ 1 there is Now, we use the same procedure as in Step 3 in the proof of Theorem 3.1, to estimate the quantity χ b 0 (b − s) α−3 f k (s) ds : k ≥ 1 .We note that from (H 2 ), for any k ≥ 1 and for a.e.
Note that γ ∈ L 3 q (J, R + ).Then, by virtue of Lemma 2.12, there exists a compact K ⊆ E, a measurable set J ⊂ J, with measure less than , and a sequence of functions Then, using Hölder's inequality we obtain Fractional differential inclusions with anti-periodic boundary conditions 21 Also, by Hölder's inequality we get for any k ≥ 1 By (4.14) and (4.15), we have From this inequality with (4.13) and by taking into account that ε is arbitrary, we get Again from the fact that ε is arbitrary, this inequality with (4.12) and (4.13) gives us .
By means of a finite number of steps, we can write Since this inequality is true for every n ∈ N, by (4.8) and by passing to the limit as n → +∞, we obtain (4.10) and so our aim in this step is verified.Hence, D is a nonempty and compact subset of C(J, E).Moreover, it is convex.Note that R(D) ⊆ D.
Step 4. The graph of the multivalued function R| D : D → 2 D is closed.Consider a sequence {x n } n≥1 in D with x n → x in B and let y n ∈ R(x n ) with y n → y in C(J, E).We have to show that y ∈ R(x).By recalling the definition of R, for any n ≥ 1, there is Observe that for every n ≥ 1 and for a.e.t ∈ J This shows that the set { f n : n ≥ 1} is integrably bounded.In addition, the set { f n (t) : n ≥ 1} is relatively compact for a.e.t ∈ J because assumption (H 3 ) together with the convergence of {x n } n≥1 implies that Hence, the sequence{ f n } n≥1 is semicompact, therefore, by Lemma 2.5, it is weakly compact in L 1 (J, E).So, without loss of generality we can assume that f n converges weakly to a function f ∈ L 1 (J, E).From Mazur's theorem, there is a sequence (z n ), n ≥ 1 of convex combinations of f n such that for a.e.
and z n converges strongly to f ∈ L 1 (J, E).Then, for a.e.
But from the upper semicontinuity of F(t, •) with Lemma 2.12, we get Then f (t) ∈ F(t, x(t)) for a.e.t ∈ J. Now, for any n ≥ 1 we define Therefore, by passing to the limit as n → ∞ in (4.16),we obtain from the Lebesgue dominated convergence theorem that, for every t ∈ J This shows that the graph of R is closed.
Observe that, by repeating the same procedure in the previous step we can deduce that the values of R is closed.
As a result of the Steps 1-4, the multivalued R| D : D → 2 D is an u.s.c.multifunction with nonempty convex compact values.By applying Lemma 2.21 there is x ∈ D and x ∈ R(x).
Remark 4.7.The preceding theorem extends Theorem 3.2 in [12] to infinite dimensional spaces.Moreover, it gives a correct formula for the solutions.
In the following corollary we simplify the condition (4.9).Corollary 4.8.Assume that the assumptions (H 2 )-(H 4 ) are satisfied.If the function ϕ in (H 2 ) is constant, that is, there exists a positive constant µ such that ϕ(t) = µ for all t ∈ J, then the problem (1.2) has a solution provided that there is an r > 0 such that Proof.We need only to check that R(D 0 ) ⊆ D 0 .Let x ∈ D 0 and y ∈ R(x).As in Step 2 of Theorem 4.6, for any t ∈ J Then the problem (1.2) has a solution provided that Proof.By(H 5 ) and (H 6 ) we conclude, from Lemma 2.2, that for any x ∈ C(J, E) the set S 1 is nonempty.Then we can consider a multifunction map R : C(J, E) → 2 C(J,E) defined as in Theorem 4.6.We shall show that R satisfies the assumptions of Lemma 2.23.The proof will be given in two steps.
Step 1.The values of R are nonempty and closed.Since S 1 F(•,x(•)) is nonempty, the values of R are nonempty.In order to prove the values of R are closed, let x ∈ C(J, E) and (y n ), n ≥ 1 be a sequence in R(x) such that y n → y in C(J, E).Then, according to the definition of R, there is a sequence Let t ∈ J be a fixed.In view of (H 5 ), for every n ≥ 1, and for a.e. .
Then, for every n ≥ ).This shows that the set { f n : n ≥ 1} is integrably bounded.Arguing as in Step 4 in the proof of Theorem 4.6, we can show that the values of R are closed.
Step 2. R is a contraction.
Let z 1 , z 2 ∈ C(J, E) and y 1 ∈ R(z 1 ).Then there is f ∈ S 1 F(•,z 1 (•)) such that for any t ∈ J y Consider the multifunction Z : J → 2 E defined by Let us define Obviously y 2 ∈ R(z 2 ).Furthermore, we get from the definitions of y 1 and y 2 , (4.19), (4.20) and Hölder's inequality By interchanging the role of y 2 and y 1 we obtain Therefore, the multivalued function R is a contraction and thus, by Lemma 2.23, R has a fixed point which is a solution for (1.2).
In the following corollary we simplify the condition (4.19).
Remark 4.11.The previous corollary extends Theorem 3.1 in [2] to a multivalued version and Theorem 3.7 in [12] to infinite dimensional Banach spaces.In addition, it gives a correct formula for the solutions.Corollary 4.12.Assume that the assumptions (H 5 ) and (H 6 ) are satisfied.If there exists a positive constant ν such that ς(t) = ν, for all t ∈ J, then the problem (1.2) has a solution provided that Then R is a contraction.

Nonconvex case
Now we present an existence result for the problem (1.1) when the values of the multivalued function are not necessarily convex.The proof is based on a selection theorem due to Bressan and Colombo [9] for lower semicontinuous maps with decomposable values.Our hypothesis on the orient field is the following: (H 7 ) F : J × E → P cl (E) is a multifunction such that (i) (t, x) → F(t, x) is graph measurable and x → F(t, x) is lower semicontinuous.
Theorem 4.13.If the hypotheses (H 4 ), (H 5 ) and (H 7 ) hold, then the problem (1.2) has a solution provided that there is r > 0 such that the condition (4.9) is satisfied.
Proof.Consider the multivalued Nemitsky operator N : C(J, E) → 2 L 1 (J,E) , defined by  We shall show that, for any λ ≥ 0, the set u λ = {x ∈ C(J, E) : d(u, N(x)) ≥ λ} is closed.For this purpose, let (x n ) be a sequence in u λ such that x n → x in C(J, E).Then, for all t ∈ J, x n (t) → x(t) in E. By virtue of (H 7 )(i) the function z → d(u(t), F(t, z)) is upper semicontinuous.So, via Fatou's lemma and (4.Therefore x ∈ u λ and hence N is lower semicontinuous.By applying Theorem 3 of [9], there is a continuous map Z : C(J, E) → L 1 (J, E) such that Z(x) ∈ N(x), for every x ∈ C(J, E).Then, Z(x)(s) ∈ F(s, x(s)), a.e.s ∈ J. Consider a map π : C(J, E) → C(J, E) defined by (πx)(t) =  Arguing as in the proof of Theorem 4.6, we can show that π satisfies all the conditions of Schauder's fixed point theorem.Thus, there is x ∈ C(J, E) such that x(t) = (πx)(t).This means that x is a solution for (1.2).

Examples
The following examples illustrate the feasibility of our assumptions.
where λ is a positive constant such that sup{ z : z ∈ K} ≤ λ.

Conclusion
In this paper, existence problems for fractional differential inclusions with anti-periodic boundary conditions have been considered in infinite dimensional Banach spaces.Some sufficient conditions have been obtained, as pointed in the first section, these conditions are strictly 30 A. G. Ibrahim weaker than the most of the existing ones.We have considered the convex as well as the nonconvex case.The obtained results extend those of [3,12] to infinite dimensional Banach spaces.Moreover, our technique allows to consider many boundary value problems in infinite dimensional Banach spaces.
nonempty and bounded}, P cl (E) = {B ⊆ E : B is nonempty and closed}, P k (E) = {B ⊆ E : B is nonempty and compact}, P ck (E) = {B ⊆ E : B is nonempty, convex and compact}, P cl,cv (E) = {B ⊆ E : B is nonempty, closed and convex}, Conv(B) (respectively, Conv(B)) be the convex hull (respectively, convex closed hull in E) of a subset B.

Lemma 2.13. Let
(K n ) be a sequence of subsets of E. Suppose there is a compact convex subset K ⊆ E such that for any neighborhood U of K there is a natural number N so that for any m ≥ N : K m ⊆ U. Then ∩ j≥1 Conv ∪{K j : n ≥ j} ⊆ K.