Generalized Ordinary Differential Equations in Abstract Spaces and Applications

because the Saks‐Henstock lemma ( Lemma 1.45) yields , for every .

Corollary 1.69: Consider functions , , , and . Then, we have , , and Eq. (1.12) holds.

Proof. Theorem 1.47, item (i), yields . Then, the statement follows from Theorem 1.68.

The next result gives us an integration by parts formula for Perron–Stieltjes integrals. A proof of it can be found in [212, Theorem 13].

Proposition 1.70: Suppose and or and . Then, the Perron–Stieltjes integrals and exist, and the following equality holds:

where , , , and

As an immediate consequence of the previous proposition, we have the following result.

Corollary 1.71: If and is a nondecreasing function, then the integral exists.

We end this subsection by presenting a result, borrowed from [172] and [179, Theorem 5.4.5], which gives us a change of variable formula for Perron–Stieltjes integrals.

Theorem 1.72: Suppose is increasing and maps onto and consider functions . Then, both integrals

exists, whenever one of the integrals exists, in which case, we have

The first result we present in this section is the Fundamental Theorem of Calculus for the variational Henstock integral. The proof follows standard steps (see [172], p. 43, for instance) adapted to Banach space-valued functions.

Theorem 1.73 (Fundamental Theorem of Calculus): Suppose is a function such that there exists the derivative , for every . Then, and

Next, we give an example, borrowed from [73], of a Banach space-valued function which is Riemann–McShane integrable (see the appendix to this chapter for this definition). However, is not variationally Henstock integrable, nor it is integrable in the sense of Bochner–Lebesgue.

Example 1.74:Let and consider the function given by , where denotes the characteristic function of a measurable set .

Since (see Definition 1.24) and the function , , is an element of , the abstract Riemann–Stieltjes integral, , exists (see [127, Theorem 4.6], p. 24). Moreover, the Riemann–Stieltjes integral, , also exists and the integration by parts formula

holds (see Theorem 1.53). Hence, .

Consider the indefinite integral , , of . Then,