Generalized Ordinary Differential Equations in Abstract Spaces and Applications. Группа авторов
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It is clear that
Theorem 1.4: Let and consider a function . The assertions below are equivalent:
1 is the uniform limit of step functions , with ;
2 ;
3 given , there exists a division such that
Proof. We will prove (i)
(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
See also [127, Theorem I.3.6].
If
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 1.4 in the case where
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 1.4 implies condition (iv). Now, assume that condition (iv) holds. Given
