Higher Order Sobolev-Type Spaces on the Real Line

Abstract

This paper gives a characterization of Sobolev functions on the real line by means of pointwise inequalities involving finite differences. This is also shown to apply to more general Orlicz-Sobolev, Lorentz-Sobolev, and Lorentz-Karamata-Sobolev spaces.

1. Introduction

The general opinion in the literature on Sobolev spaces and their generalizations is that the one-dimensional case may not be interesting. In the first order case , the theory is essentially contained in the fundamental theorem of calculus and the higher order case reduces recursively to the first order case according to the following definition.

Definition 1 (Sobolev spaces). Assume that is an open interval with finite Lebesgue measure in and let be a vector space of functions. One says that if there exists that agrees with almost everywhere and has weak derivatives of order in and the derivatives of order , , are absolutely continuous.
The main goal in this paper is to point out that there are interesting phenomena related to the Sobolev spaces already in the one-dimensional case. More precisely, we give a direct and relatively elementary proof of the following result.

Theorem 2. Suppose that is an open interval with finite measure. A real valued (continuous) function is in the Sobolev space   , if and only if for each and the th difference of the function at the point with step satisfies the inequality for some function .

Note that this result is false for domains with infinite measure as nonzero constant functions satisfy (3) with but are not in    (we leave it to the reader to formulate an appropriate version of Theorem 2 for homogeneous Sobolev spaces, which avoids this deficiency). We will also consider this result in the more general framework of Banach function spaces. However, the material is presented in such a manner that it is accessible to readers who are only interested in the classical case. Inequality (3) in Theorem 2 is one of the examples of pointwise inequalities characterising Sobolev functions, proved and propagated by Bojarski and his students in a series of papers over the last almost 30 years; see, for example, [1–6]. The case lies at the initial concepts and cornerstones of analysis on general measure metric spaces [7].

Let us give an overview of the paper. The proof of Theorem 2 is presented in a series of results. We first give a detailed proof of the fact that Sobolev functions satisfy the pointwise inequality (3). The idea is to consider the Taylor expansion of the function. Using the linearity of the finite differences, we only need to compute and bound the finite differences of polynomials and of terms like . The so-called Hermite-Genocchi formula is an integral representation of the finite differences (actually divided differences), which makes the computation rather easy. It turns out that the functions are multiples of the Hardy-Littlewood maximal function of the corresponding derivatives. Reference [8] studies the possibility to find whose definition does not involve derivatives at all. The other direction is easily shown to hold for smooth functions and the general result follows from this. Thereafter, we recall the theorems from literature necessary to be able to deduce that our main results also apply to Orlicz, Lorentz, and Lorentz-Karamata spaces.

2. Sobolev Functions Satisfy Pointwise Inequality

We refer the reader to Chapter 7 in [9] for the definition and properties of absolutely continuous functions. We write to express that the function is -times differentiable and that the derivative of order is absolutely continuous. The aim of this section is to show that membership in a Sobolev space implies (3), where the finite difference is replaced by divided differences.

Definition 3 (divided difference). Let be points in a domain . One assumes that if has multiplicity , then is -times differentiable in . First one considers the case where are ordered. One defines for We extend the definition to unordered tuples in the only possible way that makes the divided difference independent of the ordering.

Remark 4. In the recursion formula, we omitted in the nominator of the minuend and in the nominator of the subtrahend and then divided by . We claim that we obtain the same result if we replace by and by as long as . This can be seen by induction. We look here only at one of the two more difficult cases when one of the omitted points is either the largest or the smallest (and there are at least three points). The omitted points are denoted by a hat. We also omit the function. Consider We collect the first and third term and expand the second one: Collecting the first two terms and the second two, cancelling, and applying the induction hypothesis give

The following formula is a useful tool to compute divided differences.

Lemma 5 (Hermite-Genocchi formula). Let be points in a domain . One supposes that is such that is absolutely continuous if not all coincide and that additionally exists if all points are the same. Then

Proof (see page 122 in [10] as well). If , we have that Assume now that the claim is true for some . We assume here that and leave the case when and coincide to the reader (computation (14) might be helpful). Then, evaluating the right most integral,

We use the Hermite-Genocchi formula to compute finite differences of polynomials.

Lemma 6. Let be a point in and . We set Then

Proof. As is smooth, we can use the Hermite-Genocchi formula, Lemma 5. If , then ; hence the Hermite-Genocchi formula gives the claim. If , then ; we compute the integral in the Hermite-Genocchi formula via induction over .
We let first for fixed We claim that For , this can easily be seen. Assume that the claim holds for some . Then The Hermite-Genocchi formula tells us , giving the proof.

The following notation is on page 15 in [11].

Definition 7. Set and for
The definition below is a variant of Definition 4.12 in [11]. However, as our starting point is different than Schumaker’s, there is a slight regularity issue when does not differ from at least two . However, we are only interested in cases where lives under an integral, and we agree here to set in points where we do not have enough regularity.

Definition 8 (B-spline). Let be a sequence of real numbers. Given integers and , one defines (note that is fixed so that the finite difference is with respect to ) We call the th order B-spline associated with the knots .
This is Theorem 2.2 in [11]. Schumaker does not give the proof but writes that it is similar to the one of Theorem 2.1 in his book.

Theorem 9 (Dual Taylor Expansion). Let . Then for all

Proof. The proof uses induction and integration by parts. Note that, by replacing by in the integral, we may delete the + in . The case reduces to the fundamental theorem of calculus for absolutely continuous functions; see, for example, Theorem 7.1.15 in [9]. Let us assume that the statement holds for some and fix . Then

We now cite a part of Theorem 4.23 in [11]. Note that is a positive integer and means that we take the right hand derivative of order . We only need and prove the equality for . We also assume that not all points coincide.

Theorem 10 (Peano representation). Fix , where is the maximum multiplicity of . Then for all .

Proof of the Peano representation for . We take the dual Taylor expansion from Theorem 9. Then, we apply on both sides and remember that we computed the divided differences of polynomials in Lemma 6.

Our goal is to use the Peano representation to obtain an upper bound for the divided differences. We start with establishing an upper bound of the factor in the integrand.

Lemma 11. Suppose that and . Then there is a constant such that

Proof. We want to estimate Assume . We claim that This claim is easily seen to be true for . That it is true also for follows by induction. We also note that is absolutely continuous on bounded intervals and that is a weak derivative that equals the derivative whenever . We first assume that not all coincide. An application of the Hermite-Genocchi formula, Lemma 5, gives We use now the fact that the integrand is bounded from above by and the computation (14) to obtain Suppose now that . Then is bounded from above by if and vanishes if . In the end, we find a constant such that

In the next result, we use the notation for the famous Hardy-Littlewood maximal function of introduced in the seminal paper [12] of Hardy-Littlewood in Acta Math. (1930).

Theorem 12. Suppose , and is in for some . Then there is some constant such that

Proof. We use the Peano representation, Theorem 10, with . Then, using the bound of found in Lemma 11, For , we have and further The first factors in both integrals are bounded from above by . We see that the integrals are bounded from above by the corresponding maximal operators. This gives the claim.

The finite differences defined in (2) are, up to a factor, divided differences for equispaced nodes.

Lemma 13. Let be two points in a domain . Let and , where . Then each satisfies

Proof. We use induction. For , we have As , , , and , we obtain the claim. Suppose that the claim is true for some . Then

We can now formulate Theorem 12 in terms of finite differences.

Corollary 14 (Corollary of Theorem 12). Let be a domain. Suppose for some ; then there is a constant such that In particular, we have the following special case. Suppose and are normed spaces of functions such that the maximal operator is bounded. If for some , then for some function depending on .

Proof. We first use Lemma 13 and then Theorem 12 with step . There is some constant such that

Proof of Theorem 2 (Sobolev functions satisfy pointwise inequality). We combine Corollary 14 with the boundedness of the maximal operator, which is by now classical and can be found, for example, in Theorem 3.3.10 in [13].

3. Pointwise Inequality Implies Membership in Sobolev Space

In this section, our aim is to show that the pointwise inequality (3) implies the existence of the highest order derivative in the correct space. We will take care of the intermediate derivatives in the next section. In this section, the properties of the Lebesgue spaces play an important role. Our point of view is however to extract the properties of the Lebesgue spaces and base our proofs solely on these properties. We hope that the disadvantage of being slightly more abstract is later on compensated when we generalize the results. We start with some auxiliary results.

Lemma 15. Let be a domain and ; that is, has compact support in and is -times continuously differentiable. Then uniformly in as .

Proof. By Taylor’s formula, for , there is such that We apply on both sides (with variable ) and use the computation of the divided differences of polynomials carried out in Lemma 6 to obtain As is uniformly continuous on compact intervals, the claim follows.

Lemma 16. Assume is open and is compact. Then there is a neighborhood of that is a subset of .

Proof. For each , we find a ball . Suppose the statement of the lemma is false. Then for every there is with . Thus, for each , there is with . Since is compact, there is a subsequence of that converges to some . Now, lies in one of the balls . For large enough, we have that and are in . This is a contradiction.

Lemma 17. Suppose is open and compact. Then there exists such that for .

Proof. We first show that any member of the left hand side is contained in the right hand side; this is true for any . If is a member of the left hand side, then it is in , and since is a subset of , it is also a member of the right hand side. To prove the other inclusion, note that, by Lemma 16, there is such that the -neighborhood of lies in . Now suppose that is in the right hand side and . Then for , we have implying that . That is clear. This gives the proof.

Next, we state basically Definition 3.1.1 in [14]. Readers following [13] might want to have a look at Definitions 1.1.1 and 1.1.3 in their book. To gain familiarity with the concept of Banach function norm, the reader may want to verify that Lebesgue spaces are examples of Banach function spaces. As we are here only interested in Lebesgue measure, we adapt the notion to this case. The set denotes the cone of measurable functions whose values lie in .

Definition 18 (Banach function norm; Banach function space). Assume that . A Banach function norm on is a map from to such that, for all , , and () in , all scalars , and all measurable subsets of , the following are true, where is the Lebesgue measure:(P1) if, and only if,   -almost everywhere, , ;(P2)if   -almost everywhere, then ;(P3)if   -almost everywhere, we have ;(P4)if , then ;(P5)if , then there is a constant such that .Given such a function norm , the set of all extended measurable functions (identifying functions equal -almost everywhere) such that is called a Banach function space (BFS for short), and we define

We leave it to the reader to verify that Lebesgue spaces satisfy the following properties ( can be chosen as the dual space).

Properties 19. Given a Banach function space , we require the existence of a Banach function space such that (B) contains each essentially bounded function whose support has finite measure,(HI)Hölder’s inequality: ,(C) is complete,(D),(AC)absolutely continuous norm: if is a sequence of characteristic functions converging pointwise almost everywhere to , then converges to zero for all ,(A)each function can be approximated by simple functions (in -norm),(NC)if and are step functions with , then

Lemma 20. Suppose is open and is a Banach function space fulfilling the items listed in Properties 19. Assume that is in . We let be defined as If for some , then has a continuous extension to with norm bounded from above by .

Proof. Our goal is to show because extends then by the theorem of Hahn-Banach to a bounded functional on with norm bounded from above by . Without loss of generality, we may assume that is nonnegative. If were be smooth, we could just use integration by parts and use as bound the limiting case of the pointwise inequality (48). As is not necessarily smooth, we have to operate on the level of finite differences.
Let us choose and denote its support by . Lemma 17 ensures the existence of some positive such that for as long as . We assume in the following that has been chosen accordingly.
The following integrals exist by Hölder’s inequality because and thus are bounded with compact support and is integrable. Consider We compute the following integral: We have chosen such that . We continue With the aid of (52), we can now rewrite (50) as Inequality (48) leads to Using Hölder’s inequality (HI) for the first summand, we are left to show that the second summand tends to as approaches . We note first that since the second factor has support in a set of finite measures, we can replace the integration domain by a set of finite measures. Then we apply a -Hölder inequality and the uniform convergence of the divided difference to established in Lemmas 13 and 15.

After having verified that has a continuous extension to , we want to show that this extension has a representation as integral.

Lemma 21. Assume is open. Suppose the properties listed in Properties 19 are met. Let be defined as in (47) and bounded on . Then there exists with such that for all , especially for all .

Proof. The proof is heavily influenced by the proof of Theorem 1.4.1 in [13]. We first want to find a signed measure such that for each . Then, we want to apply the Lebesgue-Radon-Nikodym theorem to replace by for some integrable function for which we then show that it is contained in .
We start by assuming that the measure of is finite. For each -measurable set , we set We start by proving the -additivity of (for the -measurable sets). In the first step, we just show the additivity. By induction, it is enough to consider only two sets.
Let and be two disjoint -measurable sets. Then
We move to the verification of the -additivity. Let be countably many pairwise disjoint -measurable sets. We first want to show that converges in .
We let and . Then, since is increasing, As has finite measure, given , there exists such that By the disjointness of the sets , we see that It follows that We have for By the considerations before and the fact that is absolutely continuous as stipulated in (AC), we see that is a Cauchy sequence. Now, as is complete by (C), the Cauchy sequence converges in . We have, by the continuity of , Thus is -additive.
In order to apply the theorem of Lebesgue-Radon-Nikodym, see, for example, Section 6.9 in [15], and concluding the existence of such that we need to show that is absolutely continuous with respect to . To do so, assume that . Then . By the absolute continuity of the norm of required in (AC), we see that , and therefore . Replacing by if necessary, we obtain Now, we assume that does not necessarily have finite measure. We write as union of open sets with finite measure such that . By the considerations above, we find for each a such that for each measurable set . It is easily seen that and agree almost everywhere on . This gives rise to a function satisfying By linearity, we have for each simple function : To show that , we want to use property (D).
Let be a nonnegative simple function with support in some . Then is also a simple function with support in . It is a finite linear combination of characteristic functions of sets in . Hence Using the boundedness of , we see that
If is an arbitrary function in , then we may construct a sequence of simple functions, each having support in , such that   -almost everywhere. By the monotone convergence theorem, we have by (71) that By (P3) in Definition 18, we see that and Property (D) gives that . In the last step, we want to upgrade the integral representation (69) to hold for arbitrary functions in . As before, we approximate by simple functions with support contained in and in -almost everywhere point. Define . By (NC), converges to in and as both and “its integral representation” are continuous and agree on the set of step functions, they agree on .

Having shown the existence of the weak derivatives of highest order, we are left to show the existence of the intermediate derivatives, which we do in Proposition 22 in the next section. To keep the flow going, we give here the proof that the pointwise inequality implies membership in the corresponding Sobolev space under the assumption that we already know that Proposition 22 holds.

Proof of Theorem 2 (pointwise inequality implies membership in Sobolev space). We combine Lemmas 20 and 21 and apply Proposition 22.

4. Equivalent Definitions of Sobolev Spaces

The aim of this section is to prove the following equivalence.

Proposition 22. Let be a domain with finite Lebesgue measure in . Suppose is a Banach function space satisfying (B) and (D). Assume that . Then the following sets coincide: (a) there exists , where agrees almost everywhere with and has weak derivatives up to order in , where the derivatives of order , , are absolutely continuous,(b), where are the weak derivatives of ,(c), where is the weak derivative of of order .

We see that . We now show that .

Let us give an outline of the proof. Given (we will choose ), by integrating, we construct in Lemma 23 an absolutely continuous function together with its weak derivatives , which are absolutely continuous, and . However, and might not agree almost everywhere. We verify however in the proof of Theorem 26 (with the help of Theorem 24) that we find a polynomial such that agrees with almost everywhere and shares with the other required properties. We conclude the proof of Theorem 26 by settling the membership questions with the aid of Proposition 25.

Lemma 23. Assume is a domain. We further assume that is in . One fixes and defines where the integrals are Lebesgue integrals, and we use the notation Then (a) is locally absolutely continuous for all , especially locally integrable,(b)if , and , one has

Proof. Let us prove (a) by induction. Assume . Note that . Thus, we have for As , we see that is absolutely continuous. The case is treated similarly.
Assume now that the claim is true for some . Now is the integral of a continuous function and thus locally absolutely continuous.
Let us have a look at (b). Note that the statement is true for ; thus we may assume that . We make an induction over . The case is clear. Let us choose an interval containing a neighborhood of the support of . As and are both absolutely continuous in , we have by the integration by part formula for absolutely continuous functions (see, for example, Theorem 7.1.47 in [9]) and the fact that Noting that almost everywhere, we see that Using the induction hypothesis, the claim follows.

We cite Theorem 2.5.3 in [10] without giving its proof.

Theorem 24. Let , , or and . If the function satisfies for all , then there is a polynomial of degree such that almost everywhere on .

To show that is in , we need to verify that (if there is danger of confusion with respect to which variable we take the norm, we use notations as ) under the assumption that is in . Actually, bounding from above by its absolute value, this means that we have first to take the -norm and then the norm in . We however would like to switch the norms.

We adapt Proposition 2.1 in [16] to our setting.

Proposition 25. Let be a Banach function norm satisfying property (D). If is a measurable function, and one has and functions on such that , then

The cases are easy to prove. If for some , then the result can be proven by using Fubini’s theorem together with property (D). For the general results we refer to Schep’s proof.

The next result builds upon Theorem 2.5.4 in [10].

Theorem 26. Let be a domain, and let . Suppose is a Banach function space satisfying (B) and (D). If is locally in and has a generalized th derivative locally in , then there is agreeing almost everywhere with , having absolutely continuous intermediate weak derivatives, and the weak derivative of order agrees almost everywhere with . Furthermore, the derivatives of are locally in . If has finite measure, one obtains the result with “locally” replaced by “globally.”

We thank Pilar Silvestre for pointing out how to prove the membership of the intermediate derivatives in .

Proof. Choose and let be defined as in Lemma 23. Assume . Then, using the fact that is a weak derivative of of order and (b) in Lemma 23 Using Theorem 24, we see that almost everywhere, where is a polynomial of degree . Thus almost everywhere and for and Thus is the th generalized derivative of . As it is locally absolutely continuous (if ) it lies locally in . In the case the same conclusion holds by the assumption on .
Moreover, there exists , for example, , that agrees with almost everywhere, such that its generalized derivatives up to order are locally absolutely continuous and the th generalized derivative equals almost everywhere.
As is locally in and the other derivatives are locally absolutely continuous, they are locally in and thus locally in by (B).
Suppose now that has additionally finite measure. Assume that and are in globally. We want to show that the intermediate derivatives are in as well.
Note in our setting that, by (P5) in Definition 18, membership in implies membership in . We proceed by induction. Suppose that . We will apply Proposition 25. We set, denoting by the closed interval with endpoints and , Then

5. Banach Function Space Setting

Here, our main focus is a more advanced readership that likely knows already Banach function spaces. Therefore, we adopt a much briefer style, pointing merely to the literature where the corresponding definitions and results can be found. We start with giving generalized versions of the necessity and sufficiency of Theorem 2.

We list the results and prove them simultaneously by splitting the proofs into parts according to the function spaces.

Theorem 27. Assume is a domain, is a normed space, and . If the maximal function is bounded, then there is a constant such that for each continuous there is with and In particular such exists if(i) and is the Lebesgue space ,(ii) , , and is the Lorentz space ,(iii), , is slowly varying, and is the Lorentz-Karamata space ,(iv) is an -function with , where is the density function of and is the Orlicz space ,(v) and is rearrangement-invariant with upper index .

Theorem 28. Let be open. Suppose is a Banach function space such that (AC) holds. Assume further that the function is in . If for some , the following inequality holds for some , then there exists with such that for all .
In particular the result holds if is(i)the Lebesgue space for some ,(ii)the Lorentz space , where and ,(iii)the Lorentz-Karamata space , where is slowly varying, , and ,(iv)the Orlicz space for some -function satisfying where is the density function of .

To help with orientation, let us give an overview of the remaining material in this section. With respect to Theorem 27, note that we carried out the proof of Corollary 14 in a high level of abstraction. Hence, we are left to verify that the spaces under question are normed spaces (we will show that they are even Banach function spaces) and the boundedness of the maximal operator.

Concerning the proof Theorem 28, we proved the corresponding part of Theorem 2 under the assumptions that the spaces under question are Banach function spaces and further satisfy the requirements listed in Properties 19. The existence of is classical (Proposition 29) as is the fact that (AC) implies (A) and (NC) (Proposition 30). Hence, we only need to point to the literature where (AC) was verified.

Proposition 29. Given a Banach function space, there is a space such that properties (B), (HI), (C), and (D) are satisfied.

Proof. We refer the reader to Sections 1.1 and 1.2 in [13] or to Section 3.1 in [14].

The definition of absolutely continuous norm can be found as Definition 3.1.11 in [14] and Definition 1.3.1 in [13].

Proposition 30. A Banach function space that has absolutely continuous norm satisfies (A) and (NC).

Proof. The first statement follows from Theorem 1.3.13 in [13] and the second from Proposition 1.3.6 in the same source.

5.1. General Statements

Proof of Theorem 27 (general statement). We basically have already proven the general statement in Corollary 14.

Proof of Theorem 28 (general statement). We gave the proof of the Lebesgue space case in such generality that the general statement follows from the proof of Theorem 2 and Propositions 29 and 30.

5.2. Lorentz-Karamata Spaces

Lorentz-Karamata spaces are detailed in [14], mainly Section 3.4.3.

Proof of Theorem 27, (ii) and (iii). The part that is left to prove is the boundedness of the maximal operator, which is guaranteed by Remark 3.5.17 in [14].

Proof of Theorem 28, (ii) and (iii). By Theorem 3.4.41 in [14], we see that is a rearrangement-invariant Banach space. A combination of Lemma 3.4.39 in [14] together with Lemma 3.4.43 in [14] gives the absolute continuity (AC) of the norm of .

5.3. Orlicz Spaces

The setting is the one of [17].

Lemma 31. If is an -function satisfying the -condition, then the Orlicz space has absolutely continuous norm.

Proof. Let . We verify first that Let . By the -condition, there exists a constant such that for all . Thus Suppose that is a measurable set. Then Let us fix . Again using the -condition, we have . Hence Example 1.3.3 in [13] tells us that has absolutely continuous norm. Thus if converges -almost everywhere to , then converges to zero as well. Thus, we have for large enough that implying that . Since was arbitrary, the absolute continuity of follows.

Proof of Theorem 27 (iv). We conclude from Gallardo’s result, Theorem 2.1 in [17], that it is sufficient that the complementary -function satisfies the -condition in . Again resorting to Gallardo, this time to Proposition 1.4 in [17], we obtain the conclusion.

Proof of Theorem 28 (iv). That is a Banach function space follows from Theorem 3.4.16 in [14]. To prove that has absolutely continuous norm, we note that inequality (90) together with Proposition 1.4 in [17] ensures that the complementary -function of satisfies the -condition. Lemma 31 verifies that has absolutely continuous norm.

5.4. Rearrangement-Invariant Spaces

The necessary definitions can be found in [13].

Proof of Theorem 27, (v). We are only left to show the boundedness of the maximal operator. However, this is the content of a result by Lorentz and Shimogaki, for example, stated as Theorem 3.5.17 in [13].

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors would like to thank Pilar Silvestre for her suggestions of how to prove the existence of intermediate derivatives. The research is supported by the Academy of Finland. The third author thanks the Department of Mathematics and Systems Analysis at Aalto University for its hospitality.