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Generalized Ordinary Differential Equations in Abstract Spaces and Applications


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and every t element-of left-parenthesis a comma a plus delta right-bracket, we have

parallel-to f left-parenthesis t right-parenthesis minus f left-parenthesis a right-parenthesis parallel-to less-than-or-slanted-equals parallel-to f left-parenthesis t right-parenthesis minus f left-parenthesis a Superscript plus Baseline right-parenthesis parallel-to plus parallel-to f left-parenthesis a Superscript plus Baseline right-parenthesis minus f left-parenthesis a right-parenthesis parallel-to less-than-or-slanted-equals 1 plus gamma Subscript a Baseline equals upper K Subscript left-parenthesis a plus delta right-parenthesis Baseline period

      Then, left-parenthesis a comma a plus delta right-bracket subset-of upper B.

      Let tau 0 equals sup upper B. The equiregulatedness of script í’œ implies that there exists delta prime greater-than 0 such that parallel-to f left-parenthesis t right-parenthesis minus f left-parenthesis tau 0 Superscript minus Baseline right-parenthesis parallel-to less-than-or-slanted-equals 1 comma for every f element-of script í’œ and t element-of left-bracket tau 0 minus delta Superscript prime Baseline comma tau 0 right-parenthesis. Take tau element-of upper B intersection left-bracket tau 0 minus delta Superscript prime Baseline comma tau 0 right-parenthesis. Thus, for every f element-of script í’œ,

parallel-to f left-parenthesis tau 0 Superscript minus Baseline right-parenthesis minus f left-parenthesis a right-parenthesis parallel-to less-than-or-slanted-equals parallel-to f left-parenthesis tau 0 Superscript minus Baseline right-parenthesis minus f left-parenthesis tau right-parenthesis parallel-to plus parallel-to f left-parenthesis tau right-parenthesis minus f left-parenthesis a right-parenthesis parallel-to less-than-or-slanted-equals 1 plus upper K Subscript tau Baseline comma t element-of left-parenthesis tau comma tau 0 right-parenthesis comma

      which, together with the hypotheses, yield

parallel-to f left-parenthesis tau 0 right-parenthesis minus f left-parenthesis a right-parenthesis parallel-to less-than-or-slanted-equals parallel-to f left-parenthesis tau 0 right-parenthesis minus f left-parenthesis tau 0 Superscript minus Baseline right-parenthesis parallel-to plus parallel-to f left-parenthesis tau 0 Superscript minus Baseline right-parenthesis minus f left-parenthesis a right-parenthesis parallel-to less-than-or-slanted-equals gamma Subscript tau 0 Baseline plus 1 plus upper K Subscript tau Baseline period

      Hence, tau 0 element-of upper B.

      Let tau 0 less-than b. Since script í’œ is equiregulated, there exists delta double-prime greater-than 0 such that, for every f element-of script í’œ, parallel-to f left-parenthesis t right-parenthesis minus f left-parenthesis tau 0 Superscript plus Baseline right-parenthesis parallel-to less-than-or-slanted-equals 1, for all t element-of left-parenthesis tau 0 comma tau 0 plus delta Superscript double-prime Baseline right-bracket period Therefore, for every f element-of script í’œ, we have

StartLayout 1st Row 1st Column parallel-to f left-parenthesis t right-parenthesis minus f left-parenthesis a right-parenthesis parallel-to 2nd Column less-than-or-slanted-equals parallel-to f left-parenthesis t right-parenthesis minus f left-parenthesis tau 0 Superscript plus Baseline right-parenthesis parallel-to plus parallel-to f left-parenthesis tau 0 Superscript plus Baseline right-parenthesis minus f left-parenthesis tau 0 right-parenthesis parallel-to plus parallel-to f left-parenthesis tau 0 right-parenthesis minus f left-parenthesis a right-parenthesis parallel-to 2nd Row 1st Column Blank 2nd Column less-than-or-slanted-equals 1 plus gamma Subscript tau 0 Baseline plus upper K Subscript tau 0 Baseline equals upper K Subscript left-parenthesis tau 0 plus delta Sub Superscript double-prime Subscript right-parenthesis Baseline comma EndLayout

      for t element-of left-parenthesis tau 0 comma tau 0 plus delta double-prime right-bracket, where upper K Subscript tau 0 Baseline equals gamma Subscript tau 0 Baseline plus 1 plus upper K Subscript tau. Note that tau 0 plus delta double-prime element-of upper B which contradicts the fact that tau 0 equals sup upper B. Hence, tau 0 equals b and the statement follows.

      1.1.3 Uniform Convergence

      This subsection brings a few results borrowed from [177]. In particular, Lemma 1.13 describes an interesting and useful property of equiregulated converging sequences of Banach space‐valued functions and it is used later in the proof of a version of Arzelà–Ascoli theorem for Banach space‐valued regulated functions.

      Lemma 1.13: Let left-brace f Subscript k Baseline right-brace Subscript k element-of double-struck upper N be a sequence of functions from left-bracket a comma b right-bracket to upper X. If the sequence left-brace f Subscript k Baseline right-brace Subscript k element-of double-struck upper N converges pointwisely to f 0 and is equiregulated, then it converges uniformly to f 0.

      Proof. By hypothesis, the sequence of functions left-brace f Subscript k Baseline right-brace Subscript k element-of double-struck upper N is equiregulated. Then, Theorem 1.11 yields that, for every