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In this subsection, our goal is to investigate important properties of equiregulated sets. In addition to [97], the reference [40] also deals with a characterization of subsets of equiregulated functions.

Definition 1.10: A set is called equiregulated, if it has the following property: for every and every , there exists such that

1 if , and , then ;

2 if , and , then .

The next result, which can be found in [97, Proposition 3.2], gives a characterization of equiregulated functions taking values in a Banach space.

Theorem 1.11: A set is equiregulated if and only if for every , there is a division of such that

(1.1)

for every and , for .

Proof. Let be given and let be the set of all such that there is a division , that is, for which (1.1) holds with instead of . Since is equiregulated, there is such that for every and . Let and . Thus, for and , we have

Hence, .

Let be the supremum of the set . Since , is regulated. Thus, there exists such that for every and . Take and a division , say, , such that (1.1) holds with instead of . Denote . Then, for and , we have

which implies . Thus, we have two possibilities: either or . In the first case, the proof is finished. In the second case, one can use a similar argument as the one we used before in order to find such that , and this contradicts the fact that . Thus, , and we finish the proof of the sufficient condition.

Now, we prove the necessary condition. Given , there exists a division , say, such that the inequality (1.1) is fulfilled, for every and every , with . Then, for every , take and such that . Thus, (1.1) is satisfied, for all . In particular, if either and , or and , then the inequality (1.1) holds. Thus, is equiregulated.

The next result describes an interesting property of equiregulated sets of . Such result can be found in [97, Proposition 3.8].

Theorem 1.12: Assume that a set is equiregulated and, for any , there is a number such that

Then, there is a constant such that, for every ,

Proof. Take as the set of all numbers fulfilling the condition that there exists a positive number for which we have , for every and every Since is an equiregulated set, there exists such that for every and every . From this fact and the hypotheses, we can infer that for every and every , we have

Then, .

Let . The equiregulatedness of implies that there exists such that for every and . Take . Thus, for every ,

which, together with the hypotheses, yield

Hence, .

Let . Since is equiregulated, there exists such that, for every , , for all Therefore, for every , we have

for , where . Note that which contradicts the fact that . Hence, and the statement follows.