Читать книгу Generalized Ordinary Differential Equations in Abstract Spaces and Applications - Группа авторов - Страница 24

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

Next, we give an example, borrowed from [73], of a Banach space-valued function 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 and consider the function given by , where denotes the characteristic function of a measurable set .

Since (see Definition 1.24) and the function , , is an element of , the abstract Riemann–Stieltjes integral, , exists (see [127, Theorem 4.6], p. 24). Moreover, the Riemann–Stieltjes integral, , also exists and the integration by parts formula

holds (see Theorem 1.53). Hence, .

Consider the indefinite integral , , of . Then,

and, hence,

Thus, is absolutely continuous.

On the other hand, is nowhere differentiable (see [73], Example 3.1). Then, the Lebesgue theorem implies , where by , we denote the space of functions from to which are Lebesgue integrable with finite integral. See the appendix of this chapter. As a matter of fact, the Fundamental Theorem of Calculus for the Henstock integral (see Theorem 1.73) yields that . Optionally, one can verify that simply by noticing that

for every .

Claim. , that is, is Riemann–McShane integrable (see the appendix of this chapter).

Is is sufficient to prove that, given , we can find such that for every -fine (the reader may want to check the notation in the appendix of this chapter),

Consider and take a -fine .

 If and , then . Therefore, and, hence,

 If and , then . Therefore,and we obtain

Finally, we get

and the Claim is proved.

A less restrict version of the Fundamental Theorem of Calculus is stated next. A proof of it follows as in [108, Theorem 9.6].

Theorem 1.75 (Fundamental Theorem of Calculus): Suppose is a continuous function such that there exists the derivative , for nearly everywhere on i.e. except for a countable subset of . Then, and

Now, we present a class of functions , laying between absolute continuous and continuous functions, for which we can obtain a version of the Fundamental Theorem of Calculus for Henstock vector integrals. Let denote the Lebesgue measure.

Definition 1.76: A function satisfies the strong Lusin condition, and we write , if given and with , then there is a gauge on such that for every -fine with for all , we have

If we denote by the space of all absolutely continuous functions from to , then it is not difficult to prove that

In , we consider the usual supremum norm, , induced by .

The next two versions of the Fundamental Theorem of Calculus for Henstock vector integrals, as described in Definition 1.41, are borrowed from [70, Theorems 1 and 2]. We use the term almost everywhere in the sense of the Lebesgue measure .

Theorem 1.77: If and are both differentiable and is such that for almost every , then and , that is,

Theorem 1.78: If is differentiable and is bounded, then , and there exists the derivative for almost every , that is,

Corollary 1.79: Suppose is differentiable and nonconstant on any nondegenerate subinterval of and is bounded and such that . Then, almost everywhere in .

From Corollary 1.50, we know that if and , then . For the Henstock vector integral, we have the following analogue whose proof can be found in [70, Theorem 7].

Theorem 1.80: If and , then we have .

The next result is borrowed from [70, Theorem 5]. We reproduce its proof here.

Theorem 1.81: Suppose and is such that almost everywhere on . Then, and , that is,

Proof. Consider the sets

By hypothesis, , where denotes the Lebesgue measure. Hence, for every . In addition, . Then, for every and every , there exists a gauge on such that, for every -fine tagged partial division , with , for , we have

Consider a gauge of such that , whenever , and can assume any value in , otherwise. Then, for every -fine , we have

and we complete the proof.

Given a function , since , Theorem 1.81 holds for instead of . Then, next proposition follows easily (see, also, [70, Corollary after Theorem 5]).

Proposition 1.82: Suppose and . Assume, in addition, that is such that almost everywhere in . Then, and , for every . If, moreover, , then .

In view of Proposition 1.82, we can define equivalence classes of nonabsolute vector integrable functions.

Definition 1.83: Let us assume that . Two functions are equivalent if and only if their indefinite integrals coincide, that is, . By and we mean, respectively, the spaces of equivalence classes of functions of and of , and we endow these spaces with an Alexiewicz-type norm

From Example 1.74, we know that although may belong to the same equivalence class, that is, for all , one cannot conclude that almost everywhere in . It is known, however, that the space of all equivalence classes of real-valued Perron integrable functions , equipped with the usual Alexiewicz norm (see [5]) given by

is noncomplete (see [24], for instance). The same applies to Banach space-valued functions. Let us denote by the space of all equivalence classes of functions , equipped with the Alexiewicz norm The space is noncomplete (see [129]). However, is ultrabornological (see [105]) and, therefore, barrelled. In particular, good functional analytic properties hold, such as the Banach–Steinhaus theorem and the Uniform Boundedness Principle (see, for instance, [142]). The same applies to the space of equivalence classes of functions , endowed with the Alexiewicz norm .

The next example, borrowed from [73], exhibits a Cauchy sequence of Henstock integrable functions which is not convergent in the usual Alexiewicz norm, .

Example 1.84: Consider functions , defined by , where , whenever , , and otherwise.

Hence,

Then,

and, hence,

By induction, one can show that

for every . Then,

which goes to zero for sufficiently large , with . Thus, is a -Cauchy sequence. On the other hand,

for every . Hence, there is no function , with , such that .

The next result follows from Theorem 1.80. A proof of it can be found in [75, Theorem 5].

Theorem 1.85: Suppose is nonconstant on any nondegenerate subinterval of . Then, the mapping

is an isometry, that is onto a dense subspace of .

The next result, known as straddle Lemma, will be useful to prove that the space of regulated functions from to is dense in in the Alexiewicz norm . For a proof of the straddle Lemma, the reader may want to consult [130, 3.4] or [119].

Lemma 1.86 (Straddle Lemma): Suppose are functions such that is differentiable, with , for all . Then, given , there exists such that

whenever .

The next result is adapted from [75, Theorem 8].

Proposition 1.87: Suppose is differentiable and nonconstant on any nondegenerate subinterval of . Then, the Banach space is dense in under the Alexiewicz norm .

Proof. Assume that and let be given. We need to find a function such that , or equivalently,

(1.13)

By Corollary 1.50, . Let us denote by the subspace of of functions which are differentiable with continuous derivative. Hence, there is a function such that

(1.14)

Let be defined by , for all such that , and by . In particular, whenever . Therefore, for almost every , since is differentiable and nonconstant on any nondegenerate subinterval of . Hence, . Then, the Riemann integral exists and

(1.15)

where we applied the Fundamental Theorem of Calculus for the Riemann integral in order to obtain the last equality. Thus, replacing (1.15) in (1.14), we obtain

(1.16)

Now, in view of (1.13), it remains to prove that the Perron–Stieltjes integral exists and

(1.17)

Let be the gauge on from the definition of . Take , and for every , let be such that if , then, by the Straddle Lemma (Lemma 1.86), we have

(1.18)

Fix . We now define a gauge on by , for every . Hence, for every -fine , we have

by (1.18) and by the Riemann integrability of . Finally, (1.13) follows from (1.16) and (1.17) and the proof is complete.