Comment on ( Existence Theorem for Integral and Functional Integral Equations with Discontinuous Kernels )

and Applied Analysis 3 Remark 8. Of course, we can improveTheorem 1 by assuming that f is nondecreasing in x. 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. References [1] E. R. Hassan, “Existence theorem for integral and functional integral equations with discontinuous kernels,” Abstract and Applied Analysis, vol. 2012, Article ID 232314, 14 pages, 2012. [2] H. E. Gollwitzer and R. A. Hager, “The nonexistence of maximumsolutions ofVolterra integral equations,”Proceedings of the American Mathematical Society, vol. 26, pp. 301–304, 1970.


Introduction
In [1] Theorem 2.1 of [1] states the following.are equicontinuous and tend to zero as  ↓ 0.

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.(6) (C4) states that ℎ , () is equicontinuous in [0, 1] and tends to zero as  ↓ 0. In fact the last seems to be superfluous since it follows from equicontinuity (or from (C2)).

Comment on the Assumption (C4)
Assume that  does not depend on ; that is, we set (, , ) = (, ) (with a small violation of notation).Now This gives equicontinuity of ℎ , ().

The Counterexample
Our example is a modification of this given in [2].
Proof.The statement follows from the fact that every solution of ( 11) and ( 14) takes its values in the interval [−4, 4] where () = .Indeed, if  satisfies (11) and ‖‖ = max{|()|;  ∈ [0, 1]} then we have which implies ‖‖ ≤ 2‖‖  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 V is not a trivial solution of the problem Such solution exists since this problem, in view of the classical theory, has many solutions.Suppose that   () is a maximal solution of (10).Since 0, V(), −V() are all solutions of (10) we have   () ≥ max{0, V(), −V()}; hence   () ≥ 0 and it is not identically zero.This gives   () = − ∫  −1   () 1/2  < 0 for some  ∈ (0, 1].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.