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

Generalized Ordinary Differential Equations in Abstract Spaces and Applications

and consider functions . Then, both integrals

integral Subscript alpha Superscript beta Baseline g left-parenthesis tau right-parenthesis d h left-parenthesis tau right-parenthesis comma integral Subscript a Superscript b Baseline g left-parenthesis phi left-parenthesis s right-parenthesis right-parenthesis d left-bracket h left-parenthesis phi left-parenthesis s right-parenthesis right-parenthesis right-bracket

exists, whenever one of the integrals exists, in which case, we have

integral Subscript alpha Superscript beta Baseline g left-parenthesis tau right-parenthesis d h left-parenthesis tau right-parenthesis equals integral Subscript a Superscript b Baseline g left-parenthesis phi left-parenthesis s right-parenthesis right-parenthesis d left-bracket h left-parenthesis phi left-parenthesis s right-parenthesis right-parenthesis right-bracket period

1.3.4 The Fundamental Theorem of Calculus

The first result we present in this section is the Fundamental Theorem of Calculus for the variational Henstock integral. The proof follows standard steps (see [172], p. 43, for instance) adapted to Banach space-valued functions.

Theorem 1.73 (Fundamental Theorem of Calculus): Suppose is a function such that there exists the derivative , for every . Then, and

integral Subscript a Superscript t Baseline f left-parenthesis s right-parenthesis d s equals upper F left-parenthesis t right-parenthesis minus upper F left-parenthesis a right-parenthesis comma t element-of left-bracket a comma b right-bracket period

Next, we give an example, borrowed from [73], of a Banach space-valued function f colon left-bracket 0 comma 1 right-bracket right-arrow upper X which is Riemann–McShane integrable (see the appendix to this chapter for this definition). However, is not variationally Henstock integrable, nor it is integrable in the sense of Bochner–Lebesgue.

Example 1.74: Let upper X equals upper G Superscript minus Baseline left-parenthesis left-bracket 0 comma 1 right-bracket comma double-struck upper R right-parenthesis equals StartSet f element-of upper G left-parenthesis left-bracket 0 comma 1 right-bracket comma double-struck upper R right-parenthesis colon f is left hyphen continuous EndSet and consider the function f colon left-bracket 0 comma 1 right-bracket right-arrow upper X given by f left-parenthesis t right-parenthesis equals chi Subscript left-bracket t comma 1 right-bracket , where chi Subscript upper A denotes the characteristic function of a measurable set upper A subset-of left-bracket 0 comma 1 right-bracket .

Since f element-of upper S upper V left-parenthesis left-bracket 0 comma 1 right-bracket comma upper L left-parenthesis double-struck upper R comma upper X right-parenthesis right-parenthesis (see Definition 1.24) and the function phi left-parenthesis t right-parenthesis equals t , t element-of left-bracket 0 comma 1 right-bracket , is an element of upper C left-parenthesis left-bracket 0 comma 1 right-bracket comma double-struck upper R right-parenthesis , the abstract Riemann–Stieltjes integral, integral Subscript 0 Superscript 1 Baseline d f left-parenthesis t right-parenthesis phi left-parenthesis t right-parenthesis , exists (see [127, Theorem 4.6], p. 24). Moreover, the Riemann–Stieltjes integral, , also exists and the integration by parts formula

integral Subscript 0 Superscript 1 Baseline f left-parenthesis t right-parenthesis d t equals integral Subscript 0 Superscript 1 Baseline f left-parenthesis t right-parenthesis d phi left-parenthesis t right-parenthesis equals left-parenthesis f left-parenthesis t right-parenthesis dot t right-parenthesis vertical-bar Subscript 0 Baseline Superscript 1 Baseline minus integral Subscript 0 Superscript 1 Baseline d f left-parenthesis t right-parenthesis phi left-parenthesis t right-parenthesis

holds (see Theorem 1.53). Hence, f element-of upper R left-parenthesis left-bracket 0 comma 1 right-bracket comma upper X right-parenthesis subset-of upper K left-parenthesis left-bracket 0 comma 1 right-bracket comma upper X right-parenthesis .

Consider the indefinite integral ModifyingAbove f With tilde left-parenthesis t right-parenthesis equals integral Subscript 0 Superscript t Baseline f left-parenthesis r right-parenthesis d r , t element-of left-bracket 0 comma 1 right-bracket , of . Then,

left-parenthesis integral Subscript 0 Superscript t Baseline f left-parenthesis r right-parenthesis d r right-parenthesis left-parenthesis s right-parenthesis equals left-parenthesis integral Subscript 0 Superscript t Baseline chi Subscript left-bracket r comma 1 right-bracket Baseline d r right-parenthesis left-parenthesis s right-parenthesis equals integral Subscript 0 Superscript t Baseline chi Subscript left-bracket r comma 1 right-bracket Baseline left-parenthesis s right-parenthesis d r equals integral Subscript 0 Superscript t logical-and s Baseline d r equals t logical-and s

and, hence,

ModifyingAbove f With tilde left-parenthesis t right-parenthesis left-parenthesis s right-parenthesis equals t logical-and s equals inf left-brace right-brace comma t comma s period

Thus, f overTilde is absolutely continuous.

On the other hand, f overTilde is nowhere differentiable (see [73], Example 3.1). Then, the Lebesgue theorem implies f not-an-element-of script upper L 1 left-parenthesis left-bracket 0 comma 1 right-bracket comma upper X right-parenthesis , where by , we denote the space of functions from

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