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

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alt="delta"/>-fine tagged division left-parenthesis xi Subscript i Baseline comma left-bracket t Subscript i minus 1 Baseline comma t Subscript i Baseline right-bracket right-parenthesis).

      The next result is a consequence of the fact that italic upper H upper M upper S left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis subset-of upper H left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis and Lemmas 1.97 and 1.98. A proof of it can be found in [132, Theorem 10.3].

      Corollary 1.99: All functions of are absolutely.integrable

      The reader can find a proof of the next lemma in [35, Theorem 9].

      Lemma 1.100: All functions of are.measurable

      Finally, we can prove the following inclusion.

      Theorem 1.101: italic upper H upper M upper S left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis subset-of script upper L 1 left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis.

      Proof. The result follows from the facts that all functions of upper H left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis and, hence, of italic upper H upper M upper S left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis are measurable (Lemma 1.100), and all functions of italic upper H upper M upper S left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis are absolutely integrable by Corollary 1.99.

      Proposition 1.102 (Hönig): If is an infinite dimensional Banach space, then there exists .

      Proof. Let dimension upper X denote the dimension of upper X. If dimension upper X equals infinity, then the Theorem of Dvoretsky–Rogers (see [60] and also [57]) implies there exists a sequence left-brace x Subscript n Baseline right-brace Subscript n element-of double-struck upper N in upper X which is summable but not absolutely summable. Thus, if we define a function f colon left-bracket 1 comma infinity right-bracket right-arrow upper X by f left-parenthesis t right-parenthesis equals x Subscript n, for n less-than-or-slanted-equals t less-than n plus 1, then left-parenthesis italic upper K upper M upper S right-parenthesis integral Subscript a Superscript b Baseline f left-parenthesis t right-parenthesis d t equals sigma-summation Underscript n element-of double-struck upper N Endscripts x Subscript n Baseline element-of upper X comma whenever the integral exists. However, f not-an-element-of script upper L 1 left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis, since

integral Subscript a Superscript b Baseline vertical-bar vertical-bar vertical-bar vertical-bar of ff left-parenthesis right-parenthesis t d t equals vertical-bar vertical-bar vertical-bar vertical-bar x 1 plus vertical-bar vertical-bar vertical-bar vertical-bar x 2 plus vertical-bar vertical-bar vertical-bar vertical-bar x 3 plus midline-horizontal-ellipsis equals infinity period

      and this completes the proof.

      In the next example, borrowed from [73, Example 3.4], we exhibit a Banach space-valued function which is integrable in the variational Henstock sense and also in the sense of Kurzweil–McShane. Nevertheless, it is not absolutely integrable.

      Example 1.103: Let f colon left-bracket 0 comma 1 right-bracket right-arrow l 2 left-parenthesis double-struck upper N right-parenthesis be given by

f left-parenthesis t right-parenthesis equals StartFraction 2 Superscript i Baseline Over i EndFraction e Subscript i Baseline comma for StartFraction 1 Over 2 Superscript i Baseline EndFraction less-than-or-slanted-equals t less-than StartFraction 1 Over 2 Superscript i minus 1 Baseline EndFraction and i element-of double-struck upper N period

      Then,

integral Subscript StartFraction 1 Over 2 Superscript i Baseline EndFraction Superscript StartFraction 1 Over 2 Superscript i minus 1 Baseline EndFraction Baseline StartFraction 2 Superscript i Baseline Over i EndFraction e Subscript i Baseline d t equals StartFraction 1 Over i EndFraction e Subscript i

      which is summable in l 2 left-parenthesis double-struck upper N right-parenthesis. Since the Henstock integral contains its improper integrals (and the

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