A NOTE ON THE THEOREM ON DIFFERENTIAL INEQUALITIES

It is proved that if a linear operator ` : C([a, b]; R) → L([a, b]; R) is nonpositive and for the initial value problem u′′(t) = `(u)(t) + q(t), u(a) = c1, u ′(a) = c2 the theorem on differential inequalities is valid, then ` is an a−Volterra operator. 2000 Mathematics Subject Classification. 34K06, 34K12.

P ab is the set of operators ∈ L ab transforming the set C([a, b]; R + ) into the set L([a, b]; R + ).
We will say that ∈ L ab is an a−Volterra operator if for arbitrary b 0 ∈ ]a, b] and v ∈ C([a, b]; R) satisfying the condition In what follows, the equalities and inequalities with integrable functions are understood to hold almost everywhere.
Consider the problem on the existence and uniqueness of a solution of the equation satisfying the initial conditions where ∈ L ab , q ∈ L([a, b]; R) and c 0 , c 1 ∈ R . By a solution of the equation (1) we understand a function u ∈ C ([a, b]; R) satisfying this equation (almost everywhere) in [a, b]. Along with the problem (1), (2) consider the corresponding homogeneous problem The following result is well-known from the general theory of boundary value problems for functional differential equations (see, e.g., [1,2,4,5,8]). Theorem 1. The problem (1), (2) is uniquely solvable iff the corresponding homogeneous problem (1 0 ), (2 0 ) has only the trivial solution. Definition 1. We will say that an operator ∈ L ab belongs to the set H ab (a) if for every function u ∈ C ([a, b]; R) satisfying and (2 0 ), the inequality holds.
Remark 1. It follows from Definition 1 that if ∈ H ab (a), then the homogeneous problem (1 0 ), (2 0 ) has only the trivial solution. Therefore, according to Theorem 1 the problem (1), (2) is uniquely solvable. Moreover, the inclusion ∈ H ab (a) guarantees that if q ∈ L([a, b]; R + ), then the unique solution of the problem (1), (2 0 ) is nonnegative. Note also that ∈ H ab (a) iff a certain theorem on differential inequalities hold. More precisely, whenever u, v ∈ C ([a, b]; R) satisfy the inequalities In the paper [7], sufficient conditions are established guaranteeing the inclusion ∈ H ab (a). In particular, in [7, Theorem 1.3] the following proposition is proved. Proposition 1. Let − ∈ P ab be an a−Volterra operator and let there exist Then ∈ H ab (a).
Below we will prove (see Theorem 3) that in Proposition 1 the condition on to be a−Volterra operator is necessary. Analogous result for first order functional differential equations is proved in [3].
Before we formulate the main results, let us introduce the following definition.   Theorem 2. Let − ∈ P ab and ∈ H ab (a). Then Ω is an a−Volterra operator.
Proof. Let t 0 ∈ ]a, b[ and let the function q ∈ L([a, b]; R) be such that We will show that Ω(q)(t) = 0 for t ∈ [a, t 0 ].
According to Remark 1 (see also Remark 4) and the assumption − ∈ H ab (a), we have Since − ∈ P ab , it follows from (10) and (12) that Hence, on account of (8), (11) and (12), we obtain On the other hand, by virtue of (1), (10), (13), and the assumption − ∈ P ab , we get Hence in view of (8) and (14) we get The latter inequality, together with (13), (14) and (2 0 ), implies Consequently (since u(t) = Ω(q)(t) for t ∈ [a, b]), the equality (9) is fulfilled. Without loss of generality we can assume that First we will show that where Ω is the operator introduced in Definition 2.
Choose a sequence of functions v k ∈ C ([a, b]; R), k ∈ N, such that According to Remark 3 and (17), we get where Consequently, It follows from (20) Hence, taking into account the fact that Ω is an a−Volterra operator (see Theorem 2) and the condition (18), we obtain Thus, in view of (19), we get the equality (16). Let u be a solution of the problem (1), (2 0 ), where It is evident that and Moreover, on account of the assumption − ∈ P ab , the inequality holds. Consequently, due to (15) and (24) mes{t ∈ [a, b 0 ] : q(t) > 0} > 0.
According to Theorem 2, Ω is an a−Volterra operator. Hence by virtue of (16) and (24) we get from (25) that On the other hand, the inequality (26) and the assumption ∈ H ab (a) imply In view of (29) and the assumption − ∈ P ab , it follows from (1) that Hence, on account of (24), we obtain The latter inequality, together with (28) and (29), yields Therefore, it follows from (1) that q ≡ 0, which contradicts (27).