Generalized Bihari Type Integral Inequalities and the Corresponding Integral Equations

Abstract

We study some special nonlinear integral inequalities and the corresponding integral equations in measure spaces. They are significant generalizations of Bihari type integral inequalities and Volterra and Fredholm type integral equations. The kernels of the integral operators are determined by concave functions. Explicit upper bounds are given for the solutions of the integral inequalities. The integral equations are investigated with regard to the existence of a minimal and a maximal solution, extension of the solutions, and the generation of the solutions by successive approximations.

1. Introduction and the Main Results

In this paper we study integral inequalities of the form

(1.1)

and the corresponding integral equations

(1.2)

where

is a measure space;

is a function from into such that the following properties hold:

for every ,

if , then ,

is -measurable;

is a function from into with the following conditions:

is concave,

;

the functions and belong to

(1.3)

It may be noted that under the condition () the function is increasing (see Lemma 2.1 for the justification).

always represents a -algebra in . The -integrable functions over a measurable set are considered to be -almost measurable on . The product of the measure space with itself is understood as in [1], and it is denoted by .

By we designate the set of nonnegative integers.

Special cases of (1.1) seem first to have been investigated by Lasalle [2] and Bihari [3]. Bihari's classical result gives an explicit upper bound for the solutions of the integral inequality

(1.4)

where , and are nonnegative continuous functions on , and is a positive continuous and increasing function on . A group of inequalities is now associated with Bihari's name. Results for the various forms of such inequalities and references to different works in this topic can be found in [4–6]. Bihari type inequalities have been widely studied because they can be applied in the theory of difference, differential and integral equations. Riemann or classical Lebesgue integral is used in most of the theorems in this area. There are relatively few papers using other types of integral. For generalizations to abstract Lebesgue integral; see [7–9]. The linear version of (1.1) is given in [7]. The special case , , of (1.1) is considered in [8], while the special case , , of (1.1) is discussed in [9]. It turns out to be useful to study Bihari type inequalities with abstract Lebesgue integral. It is motivated proceeding in this direction as follows. We can get new facts about the nature of Bihari type inequalities even in the finite dimensional environment; the results can be applied in the study of certain new classes of differential and integral equations (see [7–11]).

The traditional treatment assumes not only that , but also that the sets , are intervals, while the present treatment (it should be emphasized that the methods employed to establish our results are not usual in this topic) makes it possible to consider more general sets (examples for functions satisfying (A) and (A) can be found in [11]). Such results are not quite so easy to find in literature, although they can be used as powerful tools in many fields of mathematics.

Besides , the following function spaces will play an important role.

Definition 1.1.

If is a nonempty subset of such that for every , then let

(1.5)

Next, the basic concepts of the solutions of the inequalities (1.1) and

(1.6)

and the equation (1.2) are defined.

Definition 1.2.

We say that a function is a solution of (1.1), (1.6), or (1.2) if

(i) is a nonempty subset of such that for every ,

(ii),

(iii) satisfies (1.1), (1.6), or (1.2) for each .

It is easily verified (see Lemma 2.5) that if is a solution of (1.1), (1.6), or (1.2), then is -integrable over for all .

After these preparations we set ourselves the task of obtaining an upper bound for the solutions of (1.1). The following definition will be useful.

Definition 1.3.

• (a) For every with , let

(1.7)

Let

(1.8)

By (), is a nonnegative real number for every .

Now we are in a position to formulate the first main result.

Theorem 1.4.

Assume the conditions ()–().

(a)Every solution of (1.1) satisfies

(1.9)

(b)The function defined on by

(1.10)

belongs to .

In the second main result we test the scope of the previous theorem by applying it to prove the existence of a maximal and a minimal solution of the integral equation (1.2). At the same time, we show that every solution has maximal domain of existence , and we apply the method of successive approximations to (1.2). Moreover, the behavior of the solutions is studied in a special case. The considered integral equations are in a very general form, there are classical Volterra and Fredholm type integral equations among them.

Theorem 1.5.

Suppose the conditions ()–().

There exists a solution of (1.2), which is minimal in the sense that , whenever is a solution of (1.6).

There exists a solution of (1.2), which is maximal in the sense that , whenever is a solution of (1.1).

If is a solution of (1.2), then has an extension to that is a solution of (1.2) on .

(c)Let be a solution of (1.1). Then the successive approximations determined by

(1.11)

are well defined, , ; the sequence is increasing and converges pointwise on to a solution of (1.2).

(d)Let be a solution of (1.6). Then the successive approximations (1.11) determined by are well defined, , , and the sequence is decreasing. Moreover, if either is continuous (at ) or , then they converge pointwise on to a solution of (1.2).

(e)If in addition and are bounded on for all , then every solution of (1.2) is bounded on for all .

We conclude this section with some remarks.

Remark 1.6.

The next example shows that the concavity of alone does not imply neither the existence of an upper bound for the solutions of the integral inequality (1.1) nor the existence of a solution of the integral equation (1.2). Consider the integral inequality

(1.12)

and the corresponding integral equation

(1.13)

where is the unit mass at defined on the -algebra of Borel subsets of , and

(1.14)

Then the conditions ()–() are satisfied without (). It is obvious that the functions

(1.15)

are solutions of (1.12), showing that there are no either global or local upper bounds for the solutions of (1.12). It is easy to check that (1.13) has no solution.

Remark 1.7.

It is illustrated by an example that under the conditions ()–() the maximal domain of existence of a solution of (1.1) may be a proper subset of . Let , let be the Lebesgue measurable subsets of , and let be the Lebesgue measure on . The function is defined on by

(1.16)

The functions and are defined on by . Suppose , . Then ()–() are satisfied. Let be a non (Lebesgue) measurable function. Then is a solution of (1.1) which has no extension to .

Remark.

The following example makes it clear that some extra conditions for are necessary in Theorem 1.5(d). Let

(1.17)

let be the power set of , and let the measure be defined on by

(1.18)

where the measure is the unit mass at defined on . We consider the integral equation

(1.19)

where , , and

(1.20)

The conditions ()–() can be immediately verified. Define the function by . Noting that leads to the inequality

(1.21)

A few easy calculations imply that, for every

(1.22)

and therefore

(1.23)

Since (1.19) has the unique solution , , then the successive approximations do not converge to the solution of (1.19).

2. Preliminaries

This section is devoted to some preparatory results. In the following three lemmas we establish some useful properties of concave functions.

Lemma 2.1.

If the function is concave, then is increasing.

Proof.

Suppose that there exist for which . By the concavity of , the points of the graph of are below or on the ray from through for all , and therefore

(2.1)

Since

(2.2)

it follows from (2.1) that if is large enough. This contradicts the range of .

Lemma 2.2.

Suppose the function is concave, , and . Associate to the nonnegative real number

(2.3)

Then

(a) is strictly decreasing on ;

(b) for all ;

(c)If , and there is a such that , then for all .

Proof.

The hypotheses on (since is concave, is continuous on ) guarantee that exactly one of the following three cases holds:

(i) and for all ;

(ii)there exists a such that for all and for all ;

(iii)there is a unique such that , for all and for all .

It follows that if (i) is satisfied, and otherwise. At the same time (b) and (c) are proved.

It remains to show (a). If , then (i)–(iii) show that . Assume . By the concavity of ,

(2.4)

and hence

(2.5)

The proof is complete.

Lemma 2.3.

Suppose the function is concave, and . Associate to and to each of the nonnegative real numbers , the function

(2.6)

Then

(a) is concave, and ;

(b)If , , then , where

(2.7)

(c)If and , then uniformly on as in ;

(d)the function defined on is upper semicontinuous.

Proof.

• (a) It is obvious.

By (a), Lemma 2.2(b) and (a) give the result.

The triangle inequality insures that

(2.8)

Then from Lemma 2.1

(2.9)

and this gives the result.

To prove this, choose , and . The definition of and Lemma 2.2(a) imply that

(2.10)

By (b), uniformly on as in , and hence there exists a neighborhood of in such that

(2.11)

It now follows from Lemma 2.2 (a) that

(2.12)

and the proof is complete.

The next result was proved in [8, Lemma (b)].

Lemma 2.4.

Suppose that () and () hold. Let such that for every . Suppose is -integrable over , is -almost measurable on , and there exists a measurable subset of such that is -finite and for all . Then the function

(2.13)

is -almost measurable on .

Lemma 2.5.

Assume the conditions ()–(). If is a nonempty subset of such that for every and , then .

Proof.

Let be fixed. Since is increasing it is Borel measurable. Consequently, since is -almost measurable on , is -almost measurable on . By (A), we can find such that for all . Hence, note that is increasing:

(2.14)

It now follows from the definition of and from () that is -integrable over . The proof is complete.

A consequence of the previous results that will be important later on is follows.

Lemma 2.6.

Suppose that ()–() hold. If is a nonempty subset of such that for every and , then the function

(2.15)

belongs to .

Proof.

Let .

By Lemma 2.4, the function

(2.16)

is -almost measurable on . Hence it follows from the -integrability of and over (the latter can be seen from Lemma 2.5), combined with the inequality

(2.17)

that the function (2.16) is -integrable over . We conclude that the function (2.15) is -integrable over .

The proof is complete.

We need the concept of AL-space, which is of fundamental significance in the proof of Theorem 1.5.

Definition 2.7.

Suppose (), and let be a nonempty set from .

(a)Let

(2.18)

For a given , the symbol is defined by .

(b)Let , and let . For every , let be the equivalence class containing , and we set .

(c)We introduce the canonical ordering on : for , , and means that -almost everywhere on .

Remark 2.8.

• (a)
  

  is a complete pseudometric space.

is an -normed Banach lattice, briefly, -space (see [12]).

If is a function and is a subset of the domain of , then the restriction of to is denoted by .

Lemma 2.9.

Suppose (), and let and be nonempty sets from such that . If and are nonempty majorized subsets of such that , then

(2.19)

Proof.

Since is an -space, it is order complete (see [12]), and hence and exist. Let and . Then -almost everywhere on , thus the function

(2.20)

is an upper bound of . It follows that , that is -almost everywhere on . An argument entirely similar to the preceding part gives that , where

(2.21)

and therefore -almost everywhere on .

The proof is now complete.

The next result can be found in [11, Lemma ].

Lemma 2.10.

Assume that the hypotheses (), (), (), and () are satisfied. Let . Suppose we are given solutions , of (1.2) such that for each , with . Then there exists exactly one solution of (1.2) for which , .

3. Proofs of the Main Results

Consider now the proof of Theorem 1.4.

Proof.

• (a) If such that , then (1.9) follows directly from (1.1).

Now, fix a point with . To estimate the second term on the right of (1.1), we can apply Jensen's inequality (see [13]), by ():

(3.1)

and therefore

(3.2)

This inequality, together with Definition 1.3(a), implies that the expression

(3.3)

satisfies the inequality

(3.4)

By Lemma 2.3 (b) and Definition 1.3 (b), , and hence (1.9) can be deduced from

(3.5)

The properties defining are trivial if with . So assume such that .

First, we show that the function is -almost measurable on . It is an easy consequence of () and Lemma 2.4 that the functions

(3.6)

(3.7)

are -almost measurable on . This means that there exists a measurable subset of such that and (3.6), (3.7) are measurable on . Further, since is -almost measurable on , it can be supposed that is measurable on . Thus we need to show that the function

(3.8)

is measurable on . To prove this let

(3.9)

The measurability of (3.6) on implies that . Since

(3.10)

it is enough to show that (3.8) is measurable on . It follows from the definitions of and that the function

(3.11)

is measurable. Hence, by Lemma 2.3 (d), (3.8) is measurable on .

It is now clear that is -almost measurable on .

Next, we prove that is -integrable over .

To prove this, it is enough to show that the function

(3.12)

is bounded on . Since

(3.13)

we need to verify that (3.12) is bounded on . By (), we can find a such that

(3.14)

It therefore follows from

(3.15)

that

(3.16)

for all , and hence

(3.17)

Thus

(3.18)

and therefore another application of (3.14) gives

(3.19)

and from this the claim follows. Consequently, is -integrable over , as required.

The result is completely proved.

Now we are in a position to prove Theorem 1.5.

Proof.

We begin with the proof of (c) and (d).

• (c) To prove that we use induction on . Clearly belongs to . Let such that the assertion holds. Then Lemma 2.6 yields that . We show now that the sequence is increasing. By our hypotheses on , it follows that , and we again complete the proof by induction. Suppose such that . Then, by Lemma 2.1 and the induction hypothesis

(3.20)

Since

(3.21)

Theorem 1.4(a) implies that

(3.22)

Because of (3.22) and the fact that is increasing, there exists a function such that converges pointwise on to . Then is a consequence of together with (3.22), Lemma 2.1(a) and, Theorem 1.4(b). If with , then according to the continuity of on , converges to . Let with . As we have seen is increasing, and therefore for all . Consequently

(3.23)

Now (1.11) and the monotone convergence theorem give that is a solution of (1.2).

• (d) We can see exactly as in the proof of (c) that , and the sequence is decreasing. It follows from an easy induction argument that is nonnegative for all . Linking up with the foregoing, there exists a function such that converges pointwise on to . can be shown as in the proof of (c).

The continuity of on implies that converges to for every .

Assume now that . If such that for every large enough , then

(3.24)

If such that for every , then

(3.25)

which leads to . In both cases converges to for every .

According to (1.11) and the monotone convergence theorem is a solution of (1.2).

() For convergence of the successive approximations

(3.26)

to a solution of (1.2) it suffices, in view of (c), to show that is a solution of (1.1), which is evident. It remains to prove that if is a solution of (1.6), then for every . To this end, it is enough to show that

(3.27)

This is true for , since the functions and are nonnegative. Let for which the result holds. Then, because of the nonnegativity of and the fact that is increasing,

(3.28)

and the proof of the induction step is complete.

() Let

(3.29)

Choose . The set of the upper bounds from for the solutions of (1.1) on is denoted by . By Theorem 1.4, is not empty. Let

(3.30)

Then is a minorized subset of (the elements of are nonnegative). Since this space is order complete (see [12]),

(3.31)

exists. Let , and let be a solution of (1.1) such that . , gives that

(3.32)

It follows that

(3.33)

because is increasing. Since the proof of Lemma 2.5 shows that , and hence by (), the function defined by

(3.34)

belongs to , and therefore . It now comes from (3.33) and the definition of that

(3.35)

Since is increasing,

(3.36)

and thus

(3.37)

We can see that

(3.38)

If we set

(3.39)

then (3.38) shows that is a solution of (1.2). (3.33) gives that

(3.40)

for every solutions of (1.1) for which .

Repeat the preceding construction for every . We thus obtain a set of functions , each a solution of (1.2) on its domain. Moreover, if is a solution of (1.1), then for every we have

(3.41)

Introduce the next functions: for every with let the function be defined on by

(3.42)

Obviously, these functions are also solutions of (1.2) on their domains.

Now let , such that and . Using (3.41), it is easy to verify that

(3.43)

and therefore Lemma 2.10 is applicable to the solutions . This gives a unique solution of (1.2) for which

(3.44)

It remains to prove that is maximal. Let be a solution of (1.1), and let . If , then

(3.45)

while if , then by (3.41)

(3.46)

so that Lemma 2.1 implies that

(3.47)

We first show that every solution of (1.2) with has an extension that is a solution of (1.2) such that is a proper subset of . This follows from (c) as soon as it is realized that every solution of (1.2) with has an extension that is a solution of (1.1) such that is a proper subset of . Really, in this case the successive approximations determined by converge to a solution of (1.2), which is obviously an extension of .

That realization can be reached in finitely many steps.

Let be a solution of (1.2) with , and let .

• (i) Suppose .

If , then

(3.48)

is an appropriate solution.

If , then

(3.49)

is a solution of (1.1) that agrees with on .

• (ii) Suppose and . Let

(3.50)

If , then the function

(3.51)

is a solution of (1.1) (, , thus ) that agrees with on .

If , then we introduce the functions , ,

(3.52)

which all lie in , and they are all solutions of (1.1). By Theorem 1.4 (b), is a majorized subset of . Since this space is order complete (see [12]),

(3.53)

exists. Choose . Then -almost everywhere on for every , and therefore Lemma 2.1 yields that

(3.54)

Since the proof of Lemma 2.5 shows that , and hence by (A), the function

(3.55)

belongs to too. It now follows from and (3.54) that

(3.56)

We set

(3.57)

By (3.56)

(3.58)

Fix . From Lemma 2.9 (with and there being and resp.) we get that -almost everywhere on , and hence

(3.59)

Consequently, , . If with , then . We can see that

(3.60)

By what we have already proved that

(3.61)

is a solution of (1.1) that agrees with on .

• (iii) Suppose and . By an argument entirely similar to that for the case (ii), we can get a solution of (1.1) that agrees with on .

After these preparations, the proof can be concluded quickly. Let be a solution of (1.2), and let be the set of all solutions of (1.2) which agree with on . Since , is not empty. Partially order by declaring to mean that the restriction of to the domain of agrees with . By Hausdorff's maximality theorem, there exists a maximal totally ordered subcollection of . Let be the union of the domains of all members of , and define by , where occurs in . It is easy to check that is well defined, and it is a solution of (1.2). If were a proper subset of , then the first part of the proof would give a further extension of , and this would contradict the maximality of .

• (e) Let be a solution of (1.2), and let .

If , then , so that is bounded on .

Suppose . Since and are bounded on and is finite, Theorem 1.4 (a) implies that it is enough to prove the boundedness of the function

(3.62)

on . Moreover, we have only to observe that the function

(3.63)

is bounded on . To prove this, let

(3.64)

If and , , then and for every , and therefore by Lemma 2.3(d), the function (3.63) is bounded on . If , then the definition of the function gives that . The claim about is therewith confirmed.

The proof of the theorem is now complete.