Abstract

We present a counterexample to the main result of the abovementioned paper showing that this result is false and cannot be improved in a simple way.

1. Introduction

In [1] author considers the nonlinear Volterra integral equation (VIE)and the nonlinear functional Volterra integral equation (FVIE)

Theorem 2.1 of [1] states the following.

Theorem 1. Let and let be given. Suppose that (C1)–(C4) are fulfilled.(C1) is continuous.(C2)For each , the function is Lebesgue measurable. For all and for almost all , where is a Lebesgue integrable function.(C3)For each (C4)Let , where . For every and all the functions are equicontinuous and tend to zero as .
Under the above assumptions VIE expressed by (1) has extremal solutions in the interval .

In the following we present a counterexample showing that this result is false.

2. Comment on the Assumption (C4)

Define(C4) states that is equicontinuous in and tends to zero as . In fact the last seems to be superfluous since it follows from equicontinuity (or from (C2)).

Assume that does not depend on ; that is, we set (with a small violation of notation). Now (C2) gives in for some . We also have

Proposition 2. If is Lipschitz continuous in , that is, if there exists such thatfor all then (C4) is satisfied.

Proof. For all and , , we haveThis gives equicontinuity of .

3. The Counterexample

Our example is a modification of this given in [2].

Consider VIEwhere for any (see [2]).

Consider VIE

Proposition 3. Set , , and , . A function is a solution of (10) if and only if is a solution of (11).

Proof. Suppose that is a solution of (10). Set , , in (10). We haveBy making a substitution in the integral we getHence, satisfies (11). Similarly setting , , in (11) we obtain that satisfies (10) if satisfies (11).

Corollary 4. VIE (10) has a maximal (minimal) solution if and only if (11) has a maximal (minimal) solution.

Consider VIEwhere for .

Proposition 5. VIE’s (11) and (14) have the same (nonempty) sets of solutions.

Proof. The statement follows from the fact that every solution of (11) and (14) takes its values in the interval where . Indeed, if satisfies (11) and then we havewhich implies and . Since a similar estimation holds for (14). Of course a zero function is a solution of both equations.

Set . Of course, is Lipschitz continuous in . It is not difficult to verify (see Proposition 2) the following.

Remark 6. VIE (14) satisfies all the assumptions of Theorem 1.

Proposition 7. VIE (14) has no extremal solution in .

Proof. In view of Corollary 4 and Proposition 5 we only need to show that (10) has no extremal solutions. This was in fact done in [2] where the proof is rather long and complicated. For the reader’s convenience, we present an original and short explanation.
Suppose that is not a trivial solution of the problemSuch solution exists since this problem, in view of the classical theory, has many solutions. Suppose that is a maximal solution of (10). Since , , are all solutions of (10) we have ; hence and it is not identically zero. This gives for some . This leads to a contradiction. We finish the proof by observing that the negative of a minimal solution of (10) must be its maximal solution.

Remark 8. Of course, we can improve Theorem 1 by assuming that is nondecreasing in . In this case however, (C3) is not necessary and (C4) can be reduced to a simpler one and the result is well-known.

Remark 9. Theorem 3.1 [1] (FVIE (2)) and Theorem 4.1 [1] (system of Volterra integral equation) are false since they generalize Theorem 1 (Theorem 2.1 [1]).

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.