Generalized Ordinary Differential Equations in Abstract Spaces and Applications

(ii) Note that for all . We need to show that , see Remark 1.2. Let . We will only prove that exists, because the existence of follows analogously. Consider a sequence in such that , that is, , for every , and converges to as . Consider the sequence of step functions from to such that uniformly as . Then, given , there exists such that , for all . In addition, since is a step function, there exists such that , for all . Therefore, for , we haveThen, once is a Banach space, exists.

(iii) Let be given. Since , it follows that (see Remark 1.2). Thus, for every , there exists such thatSimilarly, there are such thatNotice that the set of intervals is an open cover of the interval and, hence, there is a division of , with , such that is a finite subcover of for and, moreover,

(i). Given , let , , be a division of such thatand , . Definewhere denotes the characteristic function of a measurable set . Note that for all and all . Moreover, is a sequence of step functions which converge uniformly to , as .

It is a consequence of Theorem 1.4(with ) that the closure of is . Therefore, is a Banach space when equipped with the usual supremum norm

See also [127, Theorem I.3.6].

If denotes the Banach space of bounded functions from to , equipped with the supremum norm, then the inclusion

follows from Theorem 1.4, items (i) and (ii), taking the limit of step functions which are constant on each subinterval of continuity.

Recently, D. Franková established a fourth assertion equivalent to those assertions of Theorem 1.4in the case where . See [97, Theorem 2.3]. One can note, however, that such result also holds for any open set . This is the content of the next lemma.

Lemma 1.5: Let and be a function. Then the assertions of Theorem 1.4 are also equivalent to the following assertion:

1 (iv) for every , there is a division such that

Proof. Note that condition (iii) from Theorem 1.4implies condition (iv). Now, assume that condition (iv) holds. Given , there is a division such that , for all and According to [97, Theorem 2.3], take and consider a step function given by

Hence, .

The next result, borrowed from [7, Lemma 2.3], specifies the supremum of a function .

Proposition 1.6: Let . Then where either , for some , or , for some , or for some .

Proof. Let . Since , . By the definition of the supremum, for all , one can choose such that which implies