Abstract

We present the space of functions of bounded -variation in the sense of Riesz-Korenblum, denoted by κBVφ[a,b], which is a combination of the notions of bounded φ-variation in the sense of Riesz and bounded κ-variation in the sense of Korenblum. Moreover, we prove that the space generated by this class of functions is a Banach space with a given norm and we prove that the uniformly bounded composition operator satisfies Matkowski's weak condition.

1. Introduction

The concept of functions of bounded variation has been well known since Jordan [1] gave the complete characterization of functions of a bounded variation as the difference of two increasing functions in 1881. This class of functions immediately proved to be important in connection with the rectification of curves and with Dirichlet’s theorem on the convergence of Fourier series. Functions of a bounded variation exhibit many interesting properties that make them a suitable class of functions in a variety of contexts with wide applications in pure and applied mathematics (see [24]).

Riesz [5] in 1910 generalized the notion of Jordan and introduced the concept of bounded -variation () and showed that, for , this class coincides with the class of functions absolutely continuous with the derivative in the space . On the other hand, this notion of bounded -variation was generalized by Medvedev [6] in 1953 who introduced the concept of bounded -variation in the sense of Riesz and also showed a Riesz’s lemma for this class of functions.

Korenblum [7] in 1975 introduced the notion of bounded -variation. This concept differs from others due to the fact that it introduces a distortion function that measures intervals in the domain of the function and not in the range. In 1985, Cyphert and Kelingos [8] showed that a function is of bounded -variation if it can be written as the difference of two -decreasing functions. In 1986, S. K. Kim and J. Kim [9] and Park [10], in 2010, introduced the notion of functions of -bounded variation on compact interval which is a combination of concepts of bounded -variation and bounded -variation in the sense of Schramm [11], and in 2011 Aziz et al. [12] showed that the space of bounded -variation satisfies Matkowski’s weak condition.

Recently in [13] Castillo et al. introduce the notion of bounded -variation in the sense of Riesz-Korenblum, which is a combination of the notions of bounded -variation in the sense of Riesz and bounded -variation in the sense of Korenblum.

The purpose of this paper is twofold. First, to introduce the concept of bounded -variation in the sense of Riesz-Korenblum, which is a combination of the notions of bounded -variation in the sense of Riesz and bounded -variation in the sense of Korenblum. We prove some properties of this class of functions and its relation with the functions of bounded -variation and bounded -variation in the sense of Riesz. Second we prove that the space generated by this class of functions is a Banach space with a given norm and that the uniformly bounded composition operator satisfies Matkowski’s weak condition in this space. The Matkowski property has been studied by several authors (see [1416]), and for Matkowski’s weak property, see also [3, 1721]. In [2224] Matkowski, Merentes, and others authors have been studying a weaker condition on the composition operator such as uniformly bounded and uniformly continuous composition operators.

2. Preliminaries

In this section we present some definitions and preliminary results related to the notion of functions of bounded -variation in the sense of Riesz-Korenblum.

Definition 1. Let be a function. For each partition of the interval , we define where the supremum is taken over all partitions of the interval . If  , we say that has bounded variation. We denote by the collection of all functions of bounded variation on .

Now, we will give some well-known properties of the space of functions . (1)If the function is monotone, then . (2)If , then is bounded on . (3)A function has bounded variation in an interval if and only if it can be decomposed as a difference of increasing functions. (4)Every function of bounded variation has left- and right-hand limits at each point of its domain. (5)If , then that is, . (6) is a Banach space endowed with the norm

In 1937, Young (see [25]) introduced the definition of -function as follows.

Definition 2. A function is said to be a -function if it satisfies the following properties.(a) is continuous on . (b) if and only if . (c) is strictly increasing. (d).

Definition 3 (conditions and ). Let be a convex -function, then(a) satisfies the condition if , (b) satisfies the condition if there is , such that

The notion of bounded variation due to Jordan (Definition 1) was generalized by Medvedev (see [6]) as follows.

Definition 4. Let be a -function and be a function. For each partition of the interval , we define where the supremum is taken over all partitions of the interval . If  , we say that has bounded -variation in the sense of Riesz. We denote by the class of all functions of bounded -variation in the sense of Riesz on .

Now we give some properties of the class of functions . (1)If , then ; that is, . (2)If is convex then and if , then . (3)If satisfies the condition and , then is bounded on . (4) is a symmetric convex set. (5)If is convex, then is a vector space if and only if satisfies the condition . (6)If is convex, then the set is a Banach space endowed with the norm (7)Let be a convex -function such that it satisfies the condition . A function has bounded -variation in an interval if and only if is absolutely continuous in and . Also,

Other generalization of the notion of bounded variation was introduced by Korenblum. Korenblum employed a function called -function. This function can be viewed as a rescaling of lengths of subintervals of such that the length of is 1 if .

Definition 5. A function is said to be a -function if it satisfies the following properties:(a) is continuous with and , (b) is concave (down), increasing, and (c).

The set of all -functions will be denoted by . Note that, every -function is subadditive; that is,

Then, for all partition of , we have

Korenblum (see [7]) introduces the definition of bounded -variation as follows.

Definition 6. A real function on is said to be of bounded -variation, if where the supremum is taken over all partitions of the interval . We denote by the collection of all functions of bounded -variation on .

Next, some properties of the space are exposed (see [8]). (1)If the function is monotone, then . (2)If , then is bounded on . (3)If , then ; that is, . (4)A function has bounded -variation in an interval if and only if it can be decomposed as a difference of -decreasing functions. (5)Every function of bounded -variation has left- and right-hand limits at each point of its domain. (6) is a Banach space endowed with the norm

3. Main Results

In this section we present the principal results of this paper. Next, we introduce the definition of function of bounded -variation in the sense of Riesz-Korenblum for the function .

Definition 7. Let be a -function, , and be a function. For each partition of the interval , we define where the supremum is taken over all partitions of the interval . If , we say that has bounded -variation in the sense of Riesz-Korenblum. We will denote by the class of all functions of bounded -variation in the sense of Riesz-Korenblum on .

Remark 8. Note that the class is not empty since for an affine function is defined by , where are fixed real numbers. For a given partition of we have
Taking the supremum over all partitions of the interval , the greater value of the right side of the above expression is obtain for the partition and in this case we get
Therefore,

In the following proposition, we prove two important properties of the space .

Proposition 9. Let be a convex -function, then (a). (b)If , then .

Proof. (a) Let , be a partition of the interval and
Since is a convex -function and , we have so .
Hence, for we get multiplied by and applying the sum on both sides of the above inequality, we have then
Then
Considering the supremum over all partitions of the interval in the above expression, we get therefore .
Now, we will show part (b). If then there exist and , such that
Let us consider the partition of the interval , and then thus,
Then by considering the supremum over all partitions of the interval of the left side, we get that is,
Therefore, from part (a) and (28) we have .

Proposition 10. Let be the Banach space of all Lipschitz functions . Then .

Proof. Let be a partition of the interval and , then there exists such that for any we have . Hence, considering the supremum over all partitions of the interval , we get
Therefore, .

The class of functions of a bounded -variation has many interesting properties as the following proposition showes.

Proposition 11. Let be a -function, and be a function, then(a)The function is an even function, that is, . (b) if and only if is constant. (c)If then is bounded on . (d) is convex if and only if the function defined by is convex.

Proof. (a) From Definition 4, we have
(b) If is constant, then . Now if , we get that for some . From Definitions 2 and 5, we have hence Since if and only if , we get therefore, . So, is constant.
(c) Suppose that is unbounded on . Then, there exists a sequence , , for all , such that . Let be a subsequence of such that converges to point . Then, is a subsequence of . So, .
Case  1. Suppose that . Since for all and since is continuous, we have On the other hand, tends to infinity as . Then, since as , we get then , which is a contradiction.
Case  2. Suppose that . Then, since for all , and since is continuous, we have Since as and , we get so which is a contradiction. In both cases a contradiction is reached. Hence is bounded.
(d) Let and such that . Suppose that is convex, then where the supremum is taken over all partitions of the interval .
Since is convex, we have So for all such that , and therefore is convex.
In order to prove the order direction, let us consider the following functions. defined by and defined by . Therefore, we have Since is convex, we get which implies that Therefore, is convex, and the proof of the theorem is completed.

Remark 12. The part (c) of Proposition 11 is a consequence of the part (a) of Proposition 9 if the -function is convex.

Proposition 13. Let be a -function, , and be functions, then

Proof. Let such that and . Since is nondecreasing and nonnegative and is one of the segment joining point with , then we have From the inequality above, we deduce that thus Therefore, for all such that .

Remark 14. From Propositions 11 and 13, we have to be convex and symmetric.

The following lemma allows us to give a characterization of the space .

Lemma 15. Let be a vector space and a nonempty convex and symmetric set. Then one has the following.(a). (b)The vector space associated by is iqual to:

Proof. (a) For all , we have that .
(b) By definition, . In order to show the other inclusion, we have to prove that is a vector space. Indeed, if , we get
Also, if and , , then
Therefore, is a vector space.
Let us now prove that . Indeed, if , then . Conversely, if , for some , then .

As a consequence of Lemma 15 and since is a convex and symmetric set, we have the following corollary.

Corollary 16. Let be a -function and . Then, the vector space generated by the class is equal to the following

Theorem 17. Let be a -function and , then

Proof. First we prove that . Let , then there exists such that and is a partition of the interval , then from inequality (8) we have
Thus,
Then, considering the supremum of the left side, we get therefore, and . On the other hand, by part (a) of Proposition 9 we get that .

Theorem 18. Let be a convex -function, then is a symmetric convex absorbent subset of .

Proof. First we show the convexity. Let such that and . Then, by Proposition 11 we get thus .
Now let and . If by Proposition 11, we have
For the case , by the symmetric and convexity of the functional given in Proposition 11 we get that
Hence, we have shown that is balance. Now we will show that is absorbent. Let then there exist such that . If then . On the other hand, if we have in this case, . Hence, is absorbent.

Remark 19. As a consequence of Theorem 18, the Minkowski functional associated to the set defines a seminorm on and is defined by

Theorem 20. Let be a convex -function. Then, , where the functional , defined by is a normed space.

Proof. Let , . Then, we have the following.(a) since and . (b)Therefore, . (c)Thus, . (d)Let us now prove that if and only if . Indeed, suppose that , that is, then and . This implies that for each positive integer , there exists such that and . As the function is convex, we have . Taking the limit as , we get . Moreover, by part (b) of Proposition 11 we have that is constant; that is, , therefore .Now, suppose that . Then and . Hence . Therefore is a normed space.

Lemma 21. Let . If , then if and only if .

Proof. Case  1. Let and . Then, by the infimum property, there is such that . Hence, by the convexity of ,
Case  2. Let and . Then, by the infimum property, there exists a sequence such that
Since pointwise converges to on as , by the lower semicontinuity of , we obtain that
As, by the definition of the infimum, the converse is obvious, this completes the proof.

In the next theorem, we prove that is a Banach space.

Theorem 22. Let be a convex -function such that satisfies the condition , then the space is a Banach space.

Proof. Let be a Cauchy sequence in , then given that , there is , such that for we have that is,
Then, by Lemma 21 and the last inequality, we have
Then, for a partition we get that Hence, we have therefore
Since satisfies the condition and is continuous, we obtain that is bounded. Let be a upper bound of (80), then
As a consequence, the sequence is a uniformly Cauchy sequence, on the interval , and by the completeness of exist for all .
We defined on the function . We claim that . In fact, let be a partition of the interval . Then, for and one has
 Since for , then
Thus,
Therefore, for all . As for all , and is a vector space, then .
Finally, let us prove that converge in norm to . Let be arbitrary, then
Therefore, the sequence converges to in the norm and thus is a Banach space.

3.1. Uniformly Bounded Composition Operator

In this section, we present the other main result of this paper; namely, we show that any uniformly bounded composition operator that maps the space the into itself necessarily satisfies the so-called Matkowski’s weak condition.

Definition 23. For a given function , the composition operator generated by is defined by Here denotes the family of all functions .

Remark 24. Since , then every function of bounded -variation in the sense of Riesz-Korenblum has left- and right-hand limits at each point of its domain (see [8]).

Now, we will give the definition of left regularization of a function.

Definition 25. Let , one defined its left regularization of mapping by the following:

We will denote by the subset in which consists of those functions that are left continuous on .

Lemma 26. If , then .

Thus, if a function has bounded -variation in the sense of Riesz-Korenblum, then its left regularization is a left continuous function.

Definition 27 (see [24, Definition  1]). Let and be two metric (or normed) spaces. One says that a mapping is uniformly bounded if, for any , there exists a nonnegative real number such that for any nonempty set we have

Theorem 28. Let be a -function, , and be a function continuous with respect to the second variable. Suppose that the composition operator generated by maps into itself and satisfies the following inequality: for some function . Then, there exist functions such that where is the left regularization of for all .

Proof. By hypothesis, for fixed the constant function belongs to . Since maps into itself, we have that the function . By Lemma 26, the left regularization for every .
From inequality (90) and the definition of the norm we obtain for that
From the inequality (92) and Lemma 21, if , then
On the other hand, if , then from the definitions of the operator , the functional , and inequality (93), we have whence
Let and consider the functions defined by then . From (96), we have therefore,
Moreover,
From (95) we get then
Since satisfies the condition , we have then from the definition of and letting tend to in (101), we obtain therefore which proves that for every fixed the function satisfies the Jensen functional equation in (see [26, page 315]). The continuity of with respect to the second variable implies that for every there exist such that
Since for all in particular for , therefore .
On the other hand, considering we obtain that as is a vector space which implies that .

Theorem 29. Let be a function and . Suppose that a function is continuous with respect to the second variable. If the composition operator generated by maps the space into itself and is uniformly bounded, then there exist the functions , such that

Proof. Take any and such that
Since , by the uniform boundedness of , we have that is, therefore, by the Theorem 28 we have

Remark 30. Observe that similar results hold for the right regularization of defined by

Acknowledgments

Thanks are due to the referee for the useful suggestion and comments to improve this paper. This research has been partly supported by the Central Bank of Venezuela. The authors want to thanks the library staff of B.C.V for compiling the references.