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Generalized Ordinary Differential Equations in Abstract Spaces and Applications


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right-parenthesis t times times left-bracket right-bracket minus minus times times alpha left-parenthesis right-parenthesis plus plus xi rho of alpha alpha left-parenthesis right-parenthesis xi of ff left-parenthesis right-parenthesis xi 4th Row 1st Column Blank 2nd Column plus vertical-bar vertical-bar vertical-bar vertical-bar minus minus times times left-bracket right-bracket minus minus times times alpha left-parenthesis right-parenthesis plus plus xi rho of alpha alpha left-parenthesis right-parenthesis xi of ff left-parenthesis right-parenthesis xi times times left-bracket right-bracket minus minus of alpha alpha left-parenthesis right-parenthesis plus plus Ì‚ xi of alpha alpha left-parenthesis right-parenthesis xi of ff left-parenthesis right-parenthesis xi less-than epsilon EndLayout"/>

      and the proof is complete.

      With Theorem 1.49 at hand, the next corollary follows immediately.

      Corollary 1.50: The following statements hold:

      1 If and , then .

      2 If and , then .

      Next, we state a uniform convergence theorem for Perron–Stieltjes vector integrals. A proof of such result can be found in [210, Theorem 11].

      

      Theorem 1.51: Let and , , be such that the Perron–Stieltjes integral exists for every and uniformly in . Then, exists and

integral Subscript a Superscript b Baseline d alpha left-parenthesis t right-parenthesis f left-parenthesis t right-parenthesis equals limit Underscript n right-arrow infinity Endscripts integral Subscript a Superscript b Baseline d alpha left-parenthesis t right-parenthesis f Subscript n Baseline left-parenthesis t right-parenthesis period

      We finish this subsection by presenting a Grönwall‐type inequality for Perron–Stieltjes integrals. For a proof of it, we refer to [209, Corollary 1.43].

      

      Theorem 1.52 (Grönwall Inequality): Let be a nondecreasing left‐continuous function, and . Assume that is bounded and satisfies

f left-parenthesis t right-parenthesis less-than-or-slanted-equals k plus script l integral Subscript a Superscript t Baseline f left-parenthesis s right-parenthesis d g left-parenthesis s right-parenthesis comma t element-of left-bracket a comma b right-bracket period

       Then,

f left-parenthesis t right-parenthesis less-than-or-slanted-equals k e Superscript script l left-bracket g left-parenthesis t right-parenthesis minus g left-parenthesis a right-parenthesis right-bracket Baseline comma t element-of left-bracket a comma b right-bracket period

      Other properties of Perron–Stieltjes integrals can be found in Chapter 2, where they appear within the consequences of the main results presented there.

      1.3.3 Integration by Parts and Substitution Formulas

      The first result of this section is an Integration by Parts Formula for Riemann–Stieltjes integrals. It is a particular consequence of Proposition 1.70 presented in the end of this section. A proof of it can be found in [126, Theorem II.1.1].

      Theorem 1.53 (Integration by Parts): Let be a BT. Suppose

      1 either and ;

      2 or and .

       Then, and , that is, the Riemann–Stieltjes integrals and exist, and moreover,

integral Subscript a Superscript b Baseline d alpha left-parenthesis t right-parenthesis f left-parenthesis t right-parenthesis equals alpha left-parenthesis b right-parenthesis f left-parenthesis b right-parenthesis minus alpha left-parenthesis a right-parenthesis f left-parenthesis a right-parenthesis minus integral Subscript a Superscript b Baseline alpha left-parenthesis t right-parenthesis d f left-parenthesis t right-parenthesis period

      Next, we state a result which is not difficult to prove using the definitions involved in the statement. See [72, Theorem 5]. Recall that the indefinite integral of a function f in upper K left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis is denoted by (see Definition 1.42)

ModifyingAbove f With tilde left-parenthesis t right-parenthesis equals integral Subscript a Superscript t Baseline f left-parenthesis s right-parenthesis d s comma t element-of left-bracket a comma b right-bracket period

      Theorem 1.54: Suppose and is bounded. Then and

      (1.3)integral Subscript a Superscript b Baseline alpha left-parenthesis t right-parenthesis f left-parenthesis t right-parenthesis d t equals integral Subscript a Superscript b Baseline alpha left-parenthesis t right-parenthesis d ModifyingAbove f With tilde left-parenthesis t right-parenthesis period

       If, in addition, , then .

      Corollary 1.55: Let and be such that . Then, and (1.3) holds.

      A second corollary of Theorem 1.54 follows by the fact that Riemann–Stieltjes integrals are special cases of Perron–Stieltjes integrals. Then, it suffices to apply Theorems 1.49 and 1.53.

      Corollary 1.56: Suppose the following conditions hold:

      1 either and , with ;

      2 or and .

       Then, , equality (1.3) holds, and we have

      (1.4)integral Subscript a Superscript b Baseline alpha left-parenthesis t right-parenthesis d ModifyingAbove f With tilde left-parenthesis t right-parenthesis equals alpha left-parenthesis b right-parenthesis ModifyingAbove f With tilde left-parenthesis b right-parenthesis minus alpha left-parenthesis a right-parenthesis ModifyingAbove </p>
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