Generalized Ordinary Differential Equations in Abstract Spaces and Applications

Lemma 1.45 (Saks–Henstock Lemma): The following assertions hold.

1 Let and . Given , let be a gauge on such that for every ‐fine ,Then, for every ‐fine , we have

2 Let and Given , let be a gauge on such that for every ‐fine ,Then, for every ‐fine , we have

Now, we define some important sets of functions.

Definition 1.46:Let be the set of all functions which are weakly continuous , that is, for every , the function is continuous, and we denote by the set of all weakly regulated functions , that is, for every , the function is r egulated.

Given and , let us define

By the Banach–Steinhaus theorem, the limits and exist and belong to . Then, by the Uniform Boundedness Principle, is a Banach space when equipped with the usual supremum norm. It is also clear that

The next result concerns the existence of Perron–Stieltjes integrals. A proof of its item (i) can be found in [210, Theorem 15]. A proof of item (ii) follows similarly as the proof of item (i). See also [212, Proposition 7].

Theorem 1.47: The following assertions hold.

1 If and , then .

2 If and , then we have .

The following consequence of Theorem 1.47will be used later in many chapters. The inequalities follow after some calculations. See, for instance, [210, Proposition 10].

Corollary 1.48: The following assertions hold.

1 If and , then the Perron–Stieltjes integral exists, and we haveSimilarly, if and is nondecreasing, then

2 If and , then the Perron–Stieltjes integral exists, and we have

The next result, borrowed from [74, Theorem 1.2], gives us conditions for indefinite Perron–Stieltjes integrals to be regulated functions.

Theorem 1.49: The following assertions hold:

1 if and , then ;

2 if and , then .

Proof. We prove (i). Item (ii) follows similarly. For item (i), it is enough to show that

because, in this case, the equality

follows in an analogous way. By hypothesis, . Hence, given , there is a gauge on such that for every ‐fine division ,

Now, let . Since , there exists , for every . In particular, there exists such that

If and , then by the Saks–Henstock lemma ( Lemma 1.45)