Читать книгу Generalized Ordinary Differential Equations in Abstract Spaces and Applications - Группа авторов - Страница 23

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,

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 in is denoted by (see Definition 1.42)

Theorem 1.54: Suppose and is bounded. Then and

(1.3)

If, in addition, , then .

By Theorem 1.47, the Perron–Stieltjes integral exists and the next corollary follows.

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)

The next theorem is due to C. S. Hönig (see [129]), and it concerns multipliers for Perron–Stieltjes integrals.

Theorem 1.57: Suppose and . Then, and Eqs. (1.3) and (1.4) hold.

Since and , it is immediate that if and , then . As a matter of fact, the next result gives us information about the multipliers for the Henstock vector integral. See [72, Theorem 7].

Theorem 1.58: Assume that and . Then, and equalities (1.3) and (1.4) hold.

Proof. Since , is continuous by Theorem 1.49. Thus, given , there exists such that

whenever , where . Moreover, there is a gauge on , with for , such that for every ‐fine ,

Thus,

But by Corollary 1.56, item (ii), and

and a similar formula also holds for every subinterval contained in . Hence, for and , we have

since for every , we have

The proof is then complete.

A proof of the next result, borrowed from [72, Theorem 8], follows from the definitions of the integrals.

Theorem 1.59: Let and . If is bounded, then and

(1.5)

If, in addition, , then .

Corollary 1.60: Suppose with and . Then, and (1.5) holds.

Proof. By Theorem 1.49, . Then, the result follows from Theorem 1.47, since .

The next corollaries follow from Theorems 1.49 and 1.53.

Corollary 1.61: Suppose with and . Then, , and we have

(1.6)

and the following integration by parts formula holds

(1.7)

Corollary 1.62: Consider functions and . Then, and equalities (1.6) and (1.7) hold.

The next two theorems generalize Corollary 1.62. For their proofs, the reader may want to consult [72].

Theorem 1.63: Consider . If respectively, , then respectively, and both (1.6) and (1.7) hold.

The next result is a Substitution Formula for Perron–Stieltjes integrals. A similar result holds for Riemann–Stieltjes integrals. For a proof of it, see [72, Theorem 11].

Theorem 1.64: Consider functions , , and . Let

Then, if and only if , in which case, we have

(1.8)

(1.9)

Using Theorem 1.53, one can prove the next corollary. See [72, Corollary 8]. From now on, and are Banach spaces.

Corollary 1.65: Consider functions , , and , and define

Then, and equality (1.8) and inequality (1.9) hold.

Another substitution formula for Perron–Stieltjes integrals is presented next. Its proof uses a very nice trick provided by Professor C. S. Hönig while advising M. Federson's Master Thesis. Such result is borrowed from [72, Theorem 12].

Theorem 1.66: Consider functions , and , that is,

Then, , if and only if and

(1.10)

Proof. Since , given , there exists a gauge on such that, for every ‐fine ,

Taking approximated sums for and , we obtain

But, if and , then

Hence, taking , , and , we obtain

where we applied the Saks–Henstock lemma (Lemma 1.45) to obtain

for every .

A proof of the next proposition follows similarly as the proof of Theorem 1.66.

Proposition 1.67: Let be any interval of the real line and , with . Consider functions and of locally bounded variation. Assume that is locally Perron–Stieltjes integrable with respect to , that is, the Perron–Stieltjes integral exists, for every compact interval . Assume, further, that , defined by

is also of locally bounded variation. Then, the Perron–Stieltjes integrals and exist and

(1.11)

Yet another substitution formula for Perron–Stieltjes integrals, borrowed from [72, Theorem 11], is brought up here and, again, another interesting trick provided by Professor Hönig is used in its proof. Such substitution formula will be used in Chapter 3 in order to guarantee the existence of some Perron–Stieltjes integrals. As a matter of fact, the corollary following Theorem 1.68 will do the job.

Theorem 1.68: Consider functions , , , that is,

and assume that . Thus, if and only if , in which case, we have

(1.12)

Proof. By hypothesis, . Therefore, for every , there is a gauge of such that for every ‐fine , we have

Taking approximated Riemannian‐type sums for the integrals and , we obtain

On the other hand, when and , we have

Then, taking , , , and , we get

because the Saks‐Henstock lemma (Lemma 1.45) yields , for every .

Corollary 1.69: Consider functions , , , and . Then, we have , , and Eq. (1.12) holds.

Proof. Theorem 1.47, item (i), yields . Then, the statement follows from Theorem 1.68.

The next result gives us an integration by parts formula for Perron–Stieltjes integrals. A proof of it can be found in [212, Theorem 13].

Proposition 1.70: Suppose and or and . Then, the Perron–Stieltjes integrals and exist, and the following equality holds:

where , , , and

As an immediate consequence of the previous proposition, we have the following result.

Corollary 1.71: If and is a nondecreasing function, then the integral exists.

We end this subsection by presenting a result, borrowed from [172] and [179, Theorem 5.4.5], which gives us a change of variable formula for Perron–Stieltjes integrals.

Theorem 1.72: Suppose is increasing and maps onto and consider functions . Then, both integrals

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