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Generalized Ordinary Differential Equations in Abstract Spaces and Applications


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left-parenthesis t Superscript plus Baseline right-parenthesis"/> and alpha left-parenthesis t Superscript minus Baseline right-parenthesis also exist. We prove (i). The proof of (ii) follows analogously.

      Suppose s greater-than t. Then, property (V2) implies v a r Subscript a Superscript s Baseline left-parenthesis alpha right-parenthesis equals v a r Subscript a Superscript t Baseline left-parenthesis alpha right-parenthesis plus v a r Subscript t Superscript s Baseline left-parenthesis alpha right-parenthesis period Thus,

parallel-to alpha left-parenthesis s right-parenthesis minus alpha left-parenthesis t right-parenthesis parallel-to less-than-or-slanted-equals v a r Subscript t Superscript s Baseline left-parenthesis alpha right-parenthesis equals v a r Subscript a Superscript s Baseline left-parenthesis alpha right-parenthesis minus v a r Subscript a Superscript t Baseline left-parenthesis alpha right-parenthesis

      and, hence, parallel-to alpha left-parenthesis t Superscript plus Baseline right-parenthesis minus alpha left-parenthesis t right-parenthesis parallel-to less-than-or-slanted-equals v left-parenthesis t Superscript plus Baseline right-parenthesis minus v left-parenthesis t right-parenthesis period

      Conversely, for any given d element-of upper D Subscript left-bracket a comma t right-bracket, let v Subscript d Baseline left-parenthesis t right-parenthesis equals v a r Subscript d Baseline left-parenthesis alpha right-parenthesis. Then for every epsilon greater-than 0, there exists delta greater-than 0 such that v left-parenthesis t plus sigma right-parenthesis minus v left-parenthesis t Superscript plus Baseline right-parenthesis less-than-or-slanted-equals epsilon and parallel-to alpha left-parenthesis t plus sigma right-parenthesis minus alpha left-parenthesis t Superscript plus Baseline right-parenthesis parallel-to less-than-or-slanted-equals epsilon, and there exists a division d colon a equals t 0 less-than t 1 less-than midline-horizontal-ellipsis less-than t Subscript n Baseline equals t less-than t Subscript n plus 1 Baseline equals t plus sigma such that

v left-parenthesis t plus sigma right-parenthesis minus v Subscript d Baseline left-parenthesis t plus sigma right-parenthesis less-than-or-slanted-equals epsilon comma for all 0 less-than sigma less-than-or-slanted-equals delta period

      Then,

StartLayout 1st Row 1st Column v left-parenthesis t plus sigma right-parenthesis minus v left-parenthesis t right-parenthesis 2nd Column less-than-or-slanted-equals v Subscript d Baseline left-parenthesis t plus sigma right-parenthesis plus epsilon minus v Subscript d Baseline left-parenthesis t right-parenthesis equals parallel-to alpha left-parenthesis t plus sigma right-parenthesis minus alpha left-parenthesis t right-parenthesis parallel-to plus epsilon 2nd Row 1st Column Blank 2nd Column less-than-or-slanted-equals parallel-to alpha left-parenthesis t Superscript plus Baseline right-parenthesis minus alpha left-parenthesis t right-parenthesis parallel-to plus 2 epsilon EndLayout

      and, hence, v left-parenthesis t Superscript plus Baseline right-parenthesis minus v left-parenthesis t right-parenthesis less-than-or-slanted-equals parallel-to alpha left-parenthesis t Superscript plus Baseline right-parenthesis minus alpha left-parenthesis t right-parenthesis parallel-to which completes the proof.

      Using the fact that parallel-to alpha left-parenthesis t right-parenthesis parallel-to less-than-or-slanted-equals parallel-to alpha left-parenthesis a right-parenthesis parallel-to plus v a r Subscript a Superscript t Baseline left-parenthesis alpha right-parenthesis, the following corollary follows immediately from Lemma 1.30.

      Corollary 1.31: Let . Then the sets

StartSet t element-of left-bracket a comma b right-parenthesis colon parallel-to alpha left-parenthesis t Superscript plus Baseline right-parenthesis minus alpha left-parenthesis t right-parenthesis parallel-to greater-than-or-slanted-equals epsilon EndSet and StartSet t element-of left-parenthesis a comma b right-bracket colon parallel-to alpha left-parenthesis t right-parenthesis minus alpha left-parenthesis t Superscript minus Baseline right-parenthesis parallel-to greater-than-or-slanted-equals epsilon EndSet

       are finite for every .

      Thus, we have the next result which can be found in [126, Proposition I.2.10].

      Proposition 1.32: Let . Then the set of points of discontinuity of is countable.

      Let us define

upper B upper V Subscript a Superscript plus Baseline left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis equals left-brace alpha element-of upper B upper V Subscript a Baseline left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis colon alpha left-parenthesis t Superscript plus Baseline right-parenthesis equals alpha left-parenthesis t right-parenthesis comma t element-of left-parenthesis a comma b right-parenthesis right-brace period

      A proof that upper B upper V Subscript a Superscript plus Baseline left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis equipped with the variation norm, parallel-to dot parallel-to Subscript upper B upper V Baseline, is complete can be found in [126, Theorem I.2.11]. We reproduce it in the next theorem.

      Theorem 1.33: , equipped with the variation norm, is a Banach space.