ON A BOUNDARY VALUE PROBLEM FOR SCALAR LINEAR FUNCTIONAL DIFFERENTIAL EQUATIONS

Theorems on the Fredholm alternative and well-posedness of the linear boundary value problem u′(t)= (u)(t) + q(t), h(u)= c, where : C([a,b];R)→ L([a,b];R) and h : C([a, b];R)→R are linear bounded operators, q ∈ L([a,b];R), and c ∈R, are established even in the case when is not a strongly bounded operator. The question on the dimension of the solution space of the homogeneous equation u′(t)= (u)(t) is discussed as well.


Introduction
The following notation is used throughout: N is the set of all natural numbers; R is the set of all real numbers, R By a solution of (1.4) we understand a function u ∈ C([a,b];R) satisfying the equality (1.4) almost everywhere on [a,b].By a solution of the problem (1.4), (1.5), we understand a solution u of (1.4) which also satisfies the condition (1.5).Together with (1.4), (1.5), we will consider the corresponding homogeneous problem (1.6) h(u) = 0. (1.7) From the general theory of boundary value problems for functional differential equations, it is known that if ∈ ᏸ ab , then the problem (1.4), (1.5) has a Fredholm property (see, e.g., [1,2,7,8,10]).More precisely, the following assertion is valid.
Theorem 1.3.Let ∈ ᏸ ab and let the problem (1.6), (1.7) have only the trivial solution.Then the Green operator of the problem (1.6), (1.7) is a linear bounded operator.
In [7,8] the question on the well-posedness of linear boundary value problem for systems of functional differential equations is studied.Theorem 1.3 can also be derived as a consequence of more general results on well-posedness obtained therein.
Note that both Theorems 1.1 and 1.3 claim that ∈ ᏸ ab .This condition covers a quite wide class of linear operators; for example, the equation with a deviating argument More generally, it is known (see [6, page 317]) that ∈ ᏸ ab if and only if the operator admits the representation by means of a Stieltjes integral.
On the other hand, Schaefer proved that there exists an operator ∈ ᏸ ab such that ∈ ᏸ ab (see [9,Theorem 4]).Therefore, a question naturally arises to study boundary value problem (1.4), (1.5) without the additional requirement (1.1).In particular, the question whether Theorems 1.1 and 1.3 are valid for general operator ∈ ᏸ ab is interesting.
The first important step in this direction was made by Bravyi (see [3]), where Theorem 1.1 was proved for ∈ ᏸ ab (i.e., without the additional assumption ∈ ᏸ ab ).Bravyi's proof essentially uses Nikol'ski's theorem (see, e.g., [5, Theorem XIII.5.2, page 504]) and it is concentrated on the question of Fredholm property.The question whether Theorem 1.3 is valid for the case when ∈ ᏸ ab remains open.
In the present paper, among others, we answer this question affirmatively.More precisely, in Section 2 we prove that the operator Based on this result and Riesz-Schauder theory, we give an alternative proof (different from that in [3]) of Theorem 1.1 for ∈ ᏸ ab (see Theorem 2.1).
On the other hand, the compactness of the operator T allows us to study a question on the well-posedness of boundary value problem (1.4), (1.5).Section 3 is devoted to this question.As a special case of theorem on well-posedness, we obtain the validity of Theorem 1.3 for ∈ ᏸ ab (see Corollary 3.3).
In Section 4, the question on dimension of solution space U of homogeneous equation (1.6) is discussed.Proposition 4.6 shows that if dimU ≥ 2, then there exists q ∈ L([a,b];R) such that the nonhomogeneous equation (1.4) has no solution.This "pathological" behaviour of functional differential equations affirms the importance of the question whether the solution space of the homogeneous equation (1.6) is one dimensional.In Theorems 4.8 and 4.10, the nonimprovable effective sufficient conditions are established guaranteeing that dimU = 1.
Analogously as in Section 1, we can introduce the notion of the Green operator of the problem (1.6), (1.7).Evidently, it follows from Theorem 2.1 that the Green operator is well defined.
Remark 2.3.From the proof of Theorem 2.1 and Riesz-Schauder theory, it follows that if the problem (1.6), (1.7) has a nontrivial solution, then for every c ∈ R there exists q ∈ L([a,b];R), respectively, for every q ∈ L([a,b];R) there exists c ∈ R, such that the problem (1.4), (1.5) has no solution.
To prove Theorem 2.1 we will need several auxiliary propositions.First we recall some definitions.
Definition 2.4.Let X be a linear topological space, X * its dual space.A sequence {x n } +∞ n=1 ⊆ X is called weakly convergent if there exists x ∈ X such that ϕ(x)=lim n→+∞ ϕ(x n ) for every ϕ ∈ X * .The point x is called a weak limit of this sequence.
A set M ⊆ X is called weakly relatively compact if every sequence of points from M contains a subsequence which is weakly convergent in X.
A sequence {x n } +∞ n=1 ⊆ X is called weakly fundamental if for every ϕ ∈ X * , a sequence {ϕ(x n )} +∞ n=1 is fundamental.A space X is called weakly complete if every weakly fundamental sequence from X possesses a weak limit in X.
Let X and Y be Banach spaces and let T : X → Y be a linear bounded operator.The operator T is said to be weakly completely continuous if it maps a unit ball of X into weakly relatively compact subset of Y .Definition 2.5.A set M ⊆ L([a,b];R) has a property of absolutely continuous integral if for every ε > 0, there exists δ > 0 such that for an arbitrary measurable set E ⊆ [a,b] satisfying the condition mes E ≤ δ, the following inequality is true: E p(s)ds ≤ ε for every p ∈ M. (2.1) Proofs of the following three assertions can be found in [4].The following proposition plays a crucial role in the proof of Theorem 2.1.
Proposition 2.9.Let ∈ ᏸ ab .Then the operator and thus, since ∈ ᏸ ab and M is bounded, the set T(M) is bounded.Further, Lemmas 2.6 and 2.7 imply that the operator is weakly completely continuous, that is, a set (M) = { (v) : v ∈ M} is weakly relatively compact.Therefore, according R. Hakl et al. 49 to Lemma 2.8, for every ε > 0, there exists δ > 0 such that (2.4) On the other hand, which, together with (2.4), results in Consequently, the set T(M) is equicontinuous. Let and define a linear operator T : X → X by setting (2.9) Obviously, the problem (1.4), (1.5) is equivalent to the operator equation in the space X in the following sense: and u is a solution of (1.4), (1.5), and vice versa, if

10).
According to Proposition 2.9, we have that the operator T is compact.From Riesz-Schauder theory, it follows that (2.10) is uniquely solvable if and only if the corresponding homogeneous equation has only the trivial solution (see, e.g., [11,Theorem 2,page 221]).On the other hand, (2.11) is equivalent to the problem (1.6), (1.7) in the above-mentioned sense.
Remark 2.12.The proof of Theorem 2.11 is omitted since it is completely the same as the proof of [8, Theorem 1.3.1](see also [7,Theorem 1.2]).The only difference is that instead of Theorem 1.1, Theorem 2.1 has to be used.
Theorem 2.11 implies the following corollary.
Corollary 2.13.Let ∈ ᏸ ab be a t 0 -Volterra operator.Then the problem To prove this corollary we need the following lemma.
In an analogous way, it can be shown that where (2.34) Hence, since ε < 1, it follows that (2.17) holds.
For t 0 -Volterra operators, Theorem 2.11 can be inverted.More precisely, the following assertion is valid.
Let u 0 be a solution of the problem the existence of which is guaranteed by Corollary 2.13.Obviously, since otherwise the function u 0 would be a nontrivial solution of the problem (1.6), (1.7).Let To continue this process, we obtain Therefore, there exist k 0 ∈ N and δ > 0 such that Hence, by virtue of (2.45), it follows that there exists ρ ∈ ]0,+∞[ such that (2.48) According to Lemma 2.14, there exist k > k 0 and m ∈ N such that Furthermore, in view of (2.14), we have which, together with (2.48) and (2.49), implies that (2.37) holds.

Well-posedness
Together with the problem (1.4), (1.5), for every k ∈ N, consider the perturbed boundary value problem where The question on well-posedness of general linear boundary value problem for functional differential equation under the assumptions ∈ ᏸ ab and k ∈ ᏸ ab is studied in [7,8] (see also references in [8, page 70]).In this section we will show that the theorems on well-posedness established in [7,8] are valid also for the case when ∈ ᏸ ab and k ∈ ᏸ ab .Then the Green operator of the problem (1.6), (1.7) is continuous.
To prove Theorem 3.2, we need two lemmas, the first of them immediately follows from Arzelà-Ascoli lemma and Proposition 2.9.
where Assume on the contrary that the lemma is not valid.Then there exist an increasing sequence of natural numbers {k m } +∞ m=1 and a sequence of functions  Consequently, y 0 is a nontrivial solution of (1.6).
On the other hand, from (3.12) and (3.16), we get Proof of Theorem 3.2.Let r and k 0 be numbers, the existence of which is guaranteed by Lemma 3.5.Then, obviously, for every k > k 0 , the problem has only the trivial solution.According to Theorem 2.1, for every k > k 0 , the problem (3.1) is uniquely solvable.We will show that if u and u k are solutions of the problems (1.4), (1.5), and (3.1), respectively, then (3.8) holds.Let (3.34) Then, for every k > k 0 , where Now, by virtue of (3.4), (3.5), (3.6), and (3.7), we have According to Theorem 2.1, we have U = {0}, that is, dimU ≥ 1.Moreover, the following assertion is valid.
Evidently, the operator T is linear.According to Proposition 2.9, the operator T is compact as well.Obviously, ( On the other hand, since T is a linear compact operator, from Riesz-Schauder theory, it follows that the solution space of (2.11) is finite-dimensional.Therefore, dim U < +∞.
has only the trivial solution.
Proof.Let dim U = 1 and let problem (4.2) have a nontrivial solution u ξ for every ξ ∈ [a,b].Choose t 0 ∈]a, b] such that u a (t 0 ) = 0.Then, obviously, functions u a and u t0 are linearly independent solutions of (1.6), which contradicts the assumption dim U = 1.Now assume that there exists ξ ∈ [a,b] such that the problem (4.2) has only the trivial solution and dim U ≥ 2. Let u 1 ,u 2 ∈ U be linearly independent.Obviously, Then u is a solution of the problem (4.2), and so However, the last equality, together with (4.3), contradicts the linear independence of u 1 and u 2 .
Remark 4.5.From Proposition 4.4 and Theorem 2.1, it follows that if dim U = 1, then for every q ∈ L([a,b]R), the nonhomogeneous equation (1.4) has at least one solution (in fact, it possesses infinitely many solutions).If dim U ≥ 2, the situation is substantially different.More precisely, the following assertion holds.On the other hand, the integration of (1.6) from a to t M and from t m to b, on account of (4.15), (4.16), and the assumption 0 , 1 ∈ ᏼ ab , results in   Proof of Theorem 4.8.Assume that dim U ≥ 2.Then, according to Remark 4.7, the boundary value problems    Remark 4.12.It is known (see [6, Theorem VII.1.27,page 234]) that ∈ ᏸ ab if and only if admits the representation = 0 − 1 , where 0 , 1 ∈ ᏼ ab .Therefore Theorems 4.8 and 4.10 actually claim that ∈ ᏸ ab .Let u be an arbitrary solution of (5.3).Clearly,

Examples
The integration of these equalities from t i−1 to t, with respect to u i (t i−1 ) = 0, yields u(t) = u t i−1 + u ξ i u i (t) for t ∈ t i−1 ,t i , i = 1,n. (5.5) Hence, for t = ξ i , since u i (ξ i ) = 1, we obtain u(t i−1 ) = 0 (i = 1,n), and, consequently, u(t) = u ξ i u i (t) for t ∈ t i−1 ,t i , i = 1,n. (5.6) Now it is obvious that where α i = u(ξ i ).Thus the solution space of (5.3) has dimension n.

4 .
to Lemma 3.5,(3.35),and (3.37),v k C ≤ r c k + δ k for k > k 0 .(3.39) R. Hakl et al. 59 Hence, in view of (3.37) and (3.38)On dimension of the solution set of homogeneous equation Notation 4.1.Let U be the solution set of the homogeneous equation (1.6).Obviously, U is a linear vector space.
) is equivalent to the operator equation(2.11)  in the following sense: if u ∈ C([a,b];R) is a solution of (1.6), then x = u is a solution of (2.11), and vice versa, if x ∈ C([a,b];R) is a solution of (2.11), then x ∈ C([a,b];R) and u = x is a solution of (1.6).In other words, the set U is also a solution set of the operator equation(2.11).

Remark 4 . 3 .Proposition 4 . 4 .
Example 5.1 below shows that dim U can be any natural number, even in the case when ∈ ᏸ ab .The equality dimU = 1 holds if and only if there exists ξ ∈ [a,b] such that the problem

Proposition 4 . 6 .
Let dimU ≥ 2. Then there exists q ∈ L([a,b];R) such that the nonhomogeneous equation (1.4) has no solution.Proof.According to Proposition 4.4, and the condition dim U ≥ 2, for every ξ ∈ [a,b] the problem (4.2) has a nontrivial solution u ξ .Therefore, by virtue of Remark 2.3, for every ξ ∈ [a,b], there exists q ξ ∈ L([a,b];R) such that the problem