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

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

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      For a proof of the next lemma, it is enough to adapt the proof found in [107, Theorem 16] for the case of Banach space-valued functions.

      Lemma 1.95: script upper L 1 left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis subset-of italic upper K upper M upper S left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis.

      Now, we are able to prove the next inclusion.

      Theorem 1.96: script upper L 1 left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis subset-of italic upper H upper M upper S left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis.

      Proof. By Lemma 1.95, script upper L 1 left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis subset-of italic upper K upper M upper S left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis. Then, following the steps of the proof of Lemma 1.95 and using Lemma 1.94, we obtain the desired result.

      Lemma 1.97: If , then .

      Proof. It is enough to show that every xi element-of left-bracket a comma b right-bracket has a neighborhood where f overTilde is of bounded variation. By hypothesis, given epsilon greater-than 0, there exists a gauge delta on left-bracket a comma b right-bracket such that for every delta-fine semitagged division d equals left-parenthesis xi Subscript i Baseline comma left-bracket t Subscript i minus 1 Baseline comma t Subscript i Baseline right-bracket right-parenthesis of left-bracket a comma b right-bracket,

sigma-summation Underscript j equals 1 Overscript m Endscripts vertical-bar vertical-bar vertical-bar vertical-bar minus minus of ff tilde left-parenthesis right-parenthesis sj of ff tilde left-parenthesis right-parenthesis s minus minus j 1 less-than epsilon

      and the proof is complete.

      Lemma 1.98: Suppose . The following properties are equivalent:

      1  is absolutely

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