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Generalized Ordinary Differential Equations in Abstract Spaces and Applications. Группа авторов
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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