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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rel="nofollow" href="#fb3_img_img_424cabda-8b2d-5f19-b675-c9e0ee50e747.png" alt="upper H Subscript f Baseline left-parenthesis left-bracket a comma b right-bracket comma double-struck upper R right-parenthesis subset-of upper K Subscript f Baseline left-parenthesis left-bracket a comma b right-bracket comma double-struck upper R right-parenthesis period"/> In order to prove that upper K Subscript f Baseline left-parenthesis left-bracket a comma b right-bracket comma double-struck upper R right-parenthesis subset-of upper H Subscript f Baseline left-parenthesis left-bracket a comma b right-bracket comma double-struck upper R right-parenthesis, it is enough to write the usual Riemannian‐type sum as two sums (with the positive and negative parts of the sum):

StartLayout 1st Row 1st Column Blank 2nd Column sigma-summation Underscript i equals 1 Overscript StartAbsoluteValue d EndAbsoluteValue Endscripts StartAbsoluteValue alpha left-parenthesis tau Subscript i Baseline right-parenthesis left-bracket f left-parenthesis t Subscript i Baseline right-parenthesis minus f left-parenthesis t Subscript i minus 1 Baseline right-parenthesis right-bracket minus left-bracket ModifyingAbove alpha With tilde left-parenthesis t Subscript i Baseline right-parenthesis minus ModifyingAbove alpha With tilde left-parenthesis t Subscript i minus 1 Baseline right-parenthesis right-bracket EndAbsoluteValue 2nd Row 1st Column Blank 2nd Column equals sigma-summation Underscript i equals 1 Overscript StartAbsoluteValue d EndAbsoluteValue Endscripts left-brace alpha left-parenthesis tau Subscript i Baseline right-parenthesis left-bracket f left-parenthesis t Subscript i Baseline right-parenthesis minus f left-parenthesis t Subscript i minus 1 Baseline right-parenthesis right-bracket minus left-bracket ModifyingAbove alpha With tilde left-parenthesis t Subscript i Baseline right-parenthesis minus ModifyingAbove alpha With tilde left-parenthesis t Subscript i minus 1 Baseline right-parenthesis right-bracket right-brace Subscript plus Baseline 3rd Row 1st Column Blank 2nd Column plus sigma-summation Underscript i equals 1 Overscript StartAbsoluteValue d EndAbsoluteValue Endscripts left-brace alpha left-parenthesis tau Subscript i Baseline right-parenthesis left-bracket f left-parenthesis t Subscript i Baseline right-parenthesis minus f left-parenthesis t Subscript i minus 1 Baseline right-parenthesis right-bracket minus left-bracket ModifyingAbove alpha With tilde left-parenthesis t Subscript i Baseline right-parenthesis minus ModifyingAbove alpha With tilde left-parenthesis t Subscript i minus 1 Baseline right-parenthesis right-bracket right-brace Subscript minus 4th Row 1st Column Blank 2nd Column less-than-or-slanted-equals epsilon plus epsilon equals 2 epsilon comma EndLayout

      for every delta‐fine d equals left-parenthesis tau Subscript i Baseline comma left-bracket t Subscript i minus 1 Baseline comma t Subscript i Baseline right-bracket right-parenthesis element-of upper T upper D Subscript left-bracket a comma b right-bracket corresponding to a given epsilon greater-than 0.

integral Subscript a Superscript t Baseline alpha left-parenthesis s right-parenthesis d f left-parenthesis s right-parenthesis comma and integral Subscript a Superscript t Baseline d alpha left-parenthesis s right-parenthesis f left-parenthesis s right-parenthesis comma t element-of left-bracket a comma b right-bracket comma

      as Perron–Stieltjes integrals, where alpha colon left-bracket a comma b right-bracket right-arrow upper L left-parenthesis upper X comma upper Y right-parenthesis and f colon left-bracket a comma b right-bracket right-arrow upper X.

      As it should be expected, the above integrals are linear and additive over nonoverlapping intervals. These facts will be put aside for a while, because in Chapter 2 they will be proved for the more general form of the Kurzweil integral. In the meantime, we present a simple example of a function which is Riemann improper integrable (and, hence, also Perron integrable, due to Theorem 2.9), but it is not Lebesgue integrable (because it is not absolutely integrable).

      Example 1.43: Let f colon left-bracket 0 comma infinity right-parenthesis right-arrow double-struck upper R be given by f left-parenthesis t right-parenthesis equals StartFraction sine t Over t EndFraction, for t element-of left-parenthesis 0 comma infinity right-parenthesis, and f left-parenthesis 0 right-parenthesis equals upper L, for some upper L element-of double-struck upper R period Then, it is not difficult to prove that limit Underscript t right-arrow infinity Endscripts integral Subscript 0 Superscript t Baseline f left-parenthesis s right-parenthesis d s exists, but integral Subscript 0 Superscript infinity Baseline StartAbsoluteValue StartFraction sine s Over s EndFraction EndAbsoluteValue d s equals infinity, once

integral Subscript 0 Superscript infinity Baseline StartAbsoluteValue StartFraction sine s Over s EndFraction EndAbsoluteValue d s greater-than-or-slanted-equals sigma-summation Underscript k equals 1 Overscript n Endscripts integral Subscript left-parenthesis k minus 1 right-parenthesis pi Superscript k pi Baseline StartAbsoluteValue StartFraction sine s Over s EndFraction EndAbsoluteValue d s greater-than-or-slanted-equals sigma-summation Underscript k equals 1 Overscript n Endscripts StartFraction 1 Over k pi EndFraction integral Subscript left-parenthesis k minus 1 right-parenthesis pi Superscript k pi Baseline StartAbsoluteValue sine s EndAbsoluteValue d s equals StartFraction 2 Over pi EndFraction sigma-summation Underscript k equals 1 Overscript n Endscripts StartFraction 1 Over k EndFraction period

      Another example is also needed at this point. Borrowed from [73, example 2.1], the example below exhibits a function f element-of upper R left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis minus upper H left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis (that is, f belongs to upper R left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis, but not to upper H left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis), satisfying f overTilde equals 0. However, f left-parenthesis t right-parenthesis not-equals 0 for almost every t element-of left-bracket a comma b right-bracket. Thus, such a function is also an element

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