# Higher Order Sobolev-Type Spaces on the Real Line

Authors: Bogdan Bojarski; Juha Kinnunen; Thomas Zürcher
Journal: Journal of Function Spaces
 (2014)

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## 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.


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## Body

## 1. Introduction

The general opinion in the literature on Sobolev spacesWm,p(Rn) and their generalizations is that the one-dimensional case n=1 may not be interesting. In the first order case m=1, the theory is essentially contained in the fundamental theorem of calculus
(1)f(x)-f(y)=∫xy‍f′(t)dt,x,y∈R,
and the higher order case m>1 reduces recursively to the first order case according to the following definition.Definition 1 (Sobolev spaces).
Assume thatΩ⊂R is an open interval with finite Lebesgue measure in R and let X be a vector space of functions. One says that f∈WXm if there exists F∈X that agrees with f almost everywhere and has weak derivatives of order m in X and the derivatives of order r, 0≤r≤m-1, 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 thatΩ⊂R is an open interval with finite measure. A real valued (continuous) function f:Ω→R is in the Sobolev space Wm,p(Ω):=WLpm(Ω)  (m≥1, p>1) if and only if for each x and y∈R the mth difference of the function f at the point x with step h=(y-x)/m(2)Δmf(x;y):=Δhmf(x):=∑j=0m‍-1m-jmjfx+jh
satisfies the inequality
(3)Δmfx;y≤x-ymafx+afy
for some function af∈Lp(Ω).Note that this result is false for domains with infinite measure as nonzero constant functions satisfy (3) with af=0 but are not in   Lp(Ω) (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 m=1 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 Theorem2 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 (x-y)m-1 and of terms like (x-y)+m-1. The so-calledHermite-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 af are multiples of the Hardy-Littlewood maximal function of the corresponding derivatives. Reference [8] studies the possibility to find af 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 f∈L1m[a,b] to express that the function f:[a,b]→R is (m-1)-times differentiable and that the derivative of order m-1 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 bydivided differences.Definition 3 (divided difference).
Lety1,…,yr+1 be points in a domain Ω⊂R. One assumes that if yi has multiplicity n, then f:Ω→R is (n-1)-times differentiable in yi. First one considers the case where y1≤y2≤⋯≤yr+1 are ordered. One defines for r≥1(4)y1f:=fy1,y1,…,yr+1f:=y2,…,yr+1f-y1,…,yrfyr+1-y1,ify1≠yr+1,fry1r!otherwise.
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 omittedy1 in the nominator of the minuend and yr+1 in the nominator of the subtrahend and then divided by yr+1-yr. We claim that we obtain the same result if we replace y1 by yk and yr+1 by ym as long as yk≠ym. 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
(5)[y^1]-[y^m]ym-y1=y^1,y^m-y^1,y^r+1yr+1-ym-y^1,y^m-y^m,y^r+1yr+1-y1×ym-y1-1=(yr+1-y1)[y^1,y^m]-(yr+1-y1)[y^1,y^r+1]-(yr+1-ym)[y^1,y^m]+(yr+1-ym)[y^m,y^r+1]×yr+1-ymym-y1yr+1-y1-1.
We collect the first and third term and expand the second one:
(6)[y^1]-[y^m]ym-y1=(ym-y1)[y^1,y^m]-(ym-y1)[y^1,y^r+1]-yr+1-ymy^1,y^r+1+yr+1-ymy^m,y^r+1×yr+1-ymym-y1yr+1-y1-1.
Collecting the first two terms and the second two, cancelling, and applying the induction hypothesis give
(7)[y^1]-[y^m]ym-y1=[y^1]yr+1-y1-[y^r+1]yr+1-y1=[y1,…,yr+1].The following formula is a useful tool to compute divided differences.Lemma 5 (Hermite-Genocchi formula).
Lety1,…,yr+1 be points in a domain Ω⊂R. One supposes that f:Ω→R is such that f(n-1) is absolutely continuous if not all yi coincide and that additionally f(n)(y1) exists if all points are the same. Then
(8)[y0,…,yn]f=∫01‍dt1∫0t1‍dt2⋯∫0tn-1‍fny0+y1-y0t1+⋯+yn-yn-1tndtn.Proof (see page 122 in [10] as well).
Ifn=1, we have that
(9)∫01‍f′y0+y1-y0t1dt1=f(y1)-f(y0)y1-y0,y0≠y1,f′(y0),y0=y1,=[y0,y1]f=[y1,y0]f,y0≠y1,[y0,y0]f,y0=y1.
Assume now that the claim is true for some n. We assume here that yn+1≠yn and leave the case when yn and yn+1 coincide to the reader (computation (14) might be helpful). Then, evaluating the right most integral,
(10)∫01‍dt1∫0t1‍dt2⋯∫0tn‍fn+1y0+∑k=1n‍yk-yk-1tk+yn+1-yntn+1∑k=1n‍dtn+1=∫01‍dt1∫0t1‍dt2⋯∫0tn-1‍fny0+∑k=1n-1‍yk-yk-1tk+yn+1-yn-1tn∑k=1n-1‍-fny0+∑k=1n‍yk-yk-1tkdtn+1·1yn+1-yn=1yn+1-yn([y0,…,yn-1,yn+1]f-[y0,…,yn]f)=[y0,…,yn+1]f.We use the Hermite-Genocchi formula to compute finite differences of polynomials.Lemma 6.
Leta be a point in R and m∈N0. We set
(11)pm,a(y):=y-am.
Then
(12)y0,…,ynpm,a=0,0≤m<n,1,m=n.Proof.
Aspm,a is smooth, we can use the Hermite-Genocchi formula, Lemma 5. If 0≤m<n, then pm,a(n)=0; hence the Hermite-Genocchi formula gives the claim. If m=n, then pm,a(m)=m!; we compute the integral in the Hermite-Genocchi formula via induction over m.
We let first for fixedt0≥0(13)y0,…,ymt0:=∫0t0‍dt1∫0t1‍dt2⋯∫0tm-1‍1dtm.
We claim that
(14)y0,…,ymt0=t0mm!,m∈N.
For m=1, this can easily be seen. Assume that the claim holds for some m. Then
(15)y0,…,ym+1t0=∫0t0‍dt1y0,…,ymt1=∫0t0‍t1mm!dt1=t0m+1(m+1)!.
The Hermite-Genocchi formula tells us y0,…,ympm,a=y0,…,ym1·m!, giving the proof.The following notation is on page 15 in [11].Definition 7.
Set(16)x-y+0=1,x≥y,0,x<y,
and for m>1(17)x-y+m-1=x-ym-1,x≥y,0,x<y.
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 x does not differ from at least two yi. However, we are only interested in cases where Qim lives under an integral, and we agree here to set Qim(x)=0 in points where we do not have enough regularity.Definition 8 (B-spline).
Let⋯≤y-1≤y0≤y1≤y2≤⋯ be a sequence of real numbers. Given integers i and m>0, one defines (note that x is fixed so that the finite difference is with respect to y)
(18)Qim(x)=-1m[yi,…,yi+m]x-y+m-1,ifyi<yi+m,0otherwise.
We call Qim the mth order B-spline associated with the knots yi,…,yi+m.
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).
Letf∈L1m[a,b]. Then for all a≤y≤b(19)fy=∑j=0m-1‍-1jDjf(b)b-yjj!+∫ab‍x-y+m-1-1mDmf(x)(m-1)!dx.Proof.
The proof uses induction and integration by parts. Note that, by replacinga by y in the integral, we may delete the + in x-y+. The case m=1 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 m and fix f∈L1m+1[a,b]. Then
(20)∫yb‍x-ym-1m+1Dm+1f(x)m!dx=b-ym-1m+1Dmf(b)m!-∫yb‍x-ym-1-1m+1Dmf(x)(m-1)!dx=-b-ym-1mDmf(b)m!-∑j=0m-1‍-1jDjf(b)b-yjj!+f(y).We now cite a part of Theorem 4.23 in [11]. Note that m is a positive integer and D+j means that we take the right hand derivative of order j. We only need and prove the equality for j=0. We also assume that not all points coincide.Theorem 10 (Peano representation).
Fix0≤j≤m-n, where n is the maximum multiplicity of yi,…,yi+m. Then
(21)[yi,…,yi+m]f=∫yiyi+m‍-1jD+jQimxDm-jfx(m-1)!dx
for all f∈L1m-jmin⁡iyi,max⁡iyi.Proof of the Peano representation forj=0.
We take the dual Taylor expansion from Theorem9. Then, we apply [yi,…,yi+m] 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 factorD+Qim in the integrand.Lemma 11.
Suppose thatyi≤⋯≤yi+m and yi<yi+m. Then there is a constant C such that
(22)D+Qim(x)=[yi,…,yi+m]x-y+m-1≤Cyi+m-yi.Proof.
We want to estimate(23)yi,…,yi+mx-y+m-1=yi+1,…,yi+mx-y+m-1-yi,…,yi+m-1x-y+m-1yi+m-yi.
Assume k<m-1. We claim that
(24)dkdykx-y+m-1=-1k(m-1)⋯(m-k)x-y+m-1-k.
This claim is easily seen to be true for y≠x. That it is true also for y=x follows by induction. We also note that
(25)dm-2dym-2x-y+m-1
is absolutely continuous on bounded intervals and that
(26)-1(m-1)m-1!x-y+0
is a weak derivative that equals the derivative whenever y≠x. We first assume that not all yj,…,yj+m-1 coincide. An application of the Hermite-Genocchi formula, Lemma 5, gives
(27)[yj,…,yj+m-1]x-y+m-1=∫01‍dt1∫0t1‍dt2⋯∫0tm-2‍-1m-1(m-1)!·x-yj+∑k=1m-1‍yj+k-yj+k-1tk+0dtm-1.
We use now the fact that the integrand is bounded from above by (m-1)! and the computation (14) to obtain
(28)[yj,…,yj+m-1]x-y+m-1≤1.
Suppose now that yj=⋯=yj+m-1≠x. Then [xj,…,xj+m-1]x-y+m-1 is bounded from above by 1 if x>yi and vanishes if yi<x. In the end, we find a constant C such that
(29)D+[yi,…,yi+m]x-y+m-1≤Cyi+m-yi.In the next result, we use the notation(30)Mg(x)=sup⁡z⨍xzg
for the famous Hardy-Littlewood maximal function of g introduced in the seminal paper [12] of Hardy-Littlewood in Acta Math. (1930).Theorem 12.
Supposey0≤⋯≤ym, y0<ym and f is in L1m[y0,ym] for some m∈N. Then there is some constant C such that
(31)[y0,…,ym]f≤CMDmfy0+MDmfym.Proof.
We use the Peano representation, Theorem10, with i=j=0. Then, using the bound of D+Qim(x) found in Lemma 11,
(32)y0,…,ymf≤1(m-1)!∫y0ym‍D+Qim(x)Dmf(x)dx≤Cym-y0∫y0ym‍Dmf(x)dx.
For y∈[y0,ym], we have
(33)[y0,…,ym]f≤Cym-y0∫y0y‍Dmf(x)dx+∫yym‍Dmf(x)dx
and further
(34)[y0,…,ym]f≤C⨍y0yy-y0ym-y0Dmf(x)dx+⨍yymym-yym-y0Dmf(x)dx.
The first factors in both integrals are bounded from above by 1. 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.
Letx,y be two points in a domain Ω⊂R. Let m∈N and xk=x0+hk, where h=(y-x)/m. Then each f:Ω→R satisfies
(35)[x0,…,xm]f=1hmm!Δmf(x;y)(x=x0,xm=y).Proof.
We use induction. Form=1, we have
(36)[x0,x1]f=fx1-fx0x1-x0.
As h=x1-x0, x=x0, y=x1, and m=1, we obtain the claim. Suppose that the claim is true for some m∈N. Then
(37)Δhm+1fx=∑j=0m+1‍-1m+1-jm+1jfx+jh=∑j=0m+1‍-1m+1-jmj-1+mjf(x+jh)=∑j=0m+1‍-1m-jmjfx+j+1h-∑j=0m+1‍-1m-jmjf(x+jh)=∑j=0m‍-1m-jmjfx1+jh-∑j=0m‍-1m-jmjf(x+jh)=hmm!x1,…,xm+1f-x0,…,xmf=hmm!hm+1x0,…,xm+1f.We can now formulate Theorem12 in terms of finite differences.Corollary 14 (Corollary of Theorem12).
LetΩ⊂R be a domain. Suppose f∈L1m(Ω) for some m∈N; then there is a constant C=C(m) such that
(38)Δmf(x;y)≤Cx-ymMDmfx+MDmfy,vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvx,y∈Ω.
In particular, we have the following special case. Suppose X and Y are normed spaces of functions such that the maximal operator M:X→Y is bounded. If f∈WXm(Ω) for some m≥1, then
(39)Δmfx;y≤x-ymafx+afy
for some function af∈Y depending on f.Proof.
We first use Lemma13 and then Theorem 12 with step h=(y-x)/m. There is some constant C=C(m) such that
(40)Δmfx;y=hmm![x0,…,xm]f≤ChmMDmfx+MDmfy≤x-ymCMDmfx+CMDmfy.Proof of Theorem2 (Sobolev functions satisfy pointwise inequality).
We combine Corollary14 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.
LetΩ⊂R be a domain and φ∈C0n(Ω); that is, φ has compact support in Ω and is n-times continuously differentiable. Then
(41)[x0,…,xn]φ-φ(n)(y)n!⟶0
uniformly in Ω as x0,…,xn→y.Proof.
By Taylor’s formula, forx∈conv{x0,…,xn}, there is ξx∈conv{x0,…,xn} such that
(42)φx=φ(y)+φ′(y)(x-y)+⋯+φ(n-1)(y)x-yn-1(n-1)!+1n!φ(n)(ξx)x-yn.
We apply [x0,…,xn] on both sides (with variable x) and use the computation of the divided differences of polynomials carried out in Lemma 6 to obtain
(43)[x0,…,xn]φ(x)=1n!φ(n)(ξx).
As φ(n) is uniformly continuous on compact intervals, the claim follows.Lemma 16.
AssumeΩ⊂R is open and K⊂Ω is compact. Then there is a neighborhood of K that is a subset of Ω.Proof.
For eachx∈K, we find a ball B(x,rx)⊂Ω. Suppose the statement of the lemma is false. Then for every n∈N there is zn∈Ω∖K with dist⁡(zn,K)<1/n. Thus, for each n∈N, there is yn∈K with d(zn,yn)<1/n. Since K is compact, there is a subsequence (ynk)k of (yn)n that converges to some y∈K. Now, y lies in one of the balls B(x,rx)⊂Ω. For k large enough, we have that ynk and znk are in B(x,rx)⊂Ω. This is a contradiction.Lemma 17.
SupposeΩ⊂R is open and K⊂Ω compact. Then there exists ε>0 such that
(44)(Ω-h)∩K=Ω∩K
for h<ε.Proof.
We first show that any member of the left hand side is contained in the right hand side; this is true for anyh∈R. If z is a member of the left hand side, then it is in K, and since K is a subset of Ω, it is also a member of the right hand side. To prove the other inclusion, note that, by Lemma 16, there is ε>0 such that the ε-neighborhood Kε of K lies in Ω. Now suppose that z is in the right hand side and h<ε. Then for x:=z+h, we have x∈Kε⊂Ω implying that z=x-h∈Ω-h. That z∈K 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 M+ denotes the cone of measurable functions whose values lie in [0,∞].Definition 18 (Banach function norm; Banach function space).
Assume thatR⊂R. ABanach function norm on R is a map ρ from M+(R) to [0,∞] such that, for all f, g, and fn (n∈N) in M+(R), all scalars λ≥0, and all measurable subsets E of R, the following are true, where μ is the Lebesgue measure:(P1)
ρ(f)=0 if, and only if, f=0  μ-almost everywhere, ρ(λf)=λρ(f), ρ(f+g)≤ρ(f)+ρ(g);(P2)
if0≤g≤f  μ-almost everywhere, then ρ(g)≤ρ(f);(P3)
if0≤fn↑f  μ-almost everywhere, we have ρ(fn)↑ρ(f);(P4)
ifμ(E)<∞, then ρ(χE)<∞;(P5)
ifμ(E)<∞, then there is a constant C=C(E,ρ)<∞ such that ∫E‍fdx≤Cρ(f).Given such a function norm ρ, the set X=X(ρ) of all extended measurable functions f (identifying functions equal μ-almost everywhere) such that ρ(f)<∞ is called aBanach function space (BFS for short), and we define
(45)fX=ρ(f),f∈X.We leave it to the reader to verify that Lebesgue spaces satisfy the following properties (X′ can be chosen as the dual space).Properties 19.
Given a Banach function spaceX, we require the existence of a Banach function space X′ such that (B)
X contains each essentially bounded function whose support has finite measure,(HI)
Hölder’s inequality:∫‍fgdx≤fXgX′,(C)
X′ is complete,(D)
gX=sup⁡∫Ω‍gh:h∈M+,hX′≤1,(AC)
absolutely continuous norm: if(χEn)n is a sequence of characteristic functions converging pointwise almost everywhere to 0, then fχEnX′ converges to zero for all f∈X′,(A)
each functionf∈X′ can be approximated by simple functions (in X′-norm),(NC)
iff∈X′ and hn are step functions with 0≤hn↑f, then
(46)f-hnsgn⁡fX′⟶0asn⟶∞.Lemma 20.
SupposeΩ⊂R is open and X is a Banach function space fulfilling the items listed in Properties 19. Assume that f:Ω→R is in X. We let λf:X′∩C0r+1(Ω)→R be defined as
(47)λf(φ)=∫Ω‍fφ(r)dx.
If
(48)Δrf(x,y)≤x-yrafx+afy,foralmosteveryx,y∈Ω
for some af∈X, then λf has a continuous extension to X′ with norm bounded from above by 2rrafX.Proof.
Our goal is to show(49)∫Ω‍f(x)φ(r)(x)dx≤2rrafXφX′,
because λf extends then by the theorem of Hahn-Banach to a bounded functional on X′ with norm bounded from above by 2rrafX. Without loss of generality, we may assume that af is nonnegative. If f were be smooth, we could just use integration by parts and use as bound the limiting case of the pointwise inequality (48). As f is not necessarily smooth, we have to operate on the level of finite differences.
Let us chooseφ∈C0(Ω) and denote its support by K. Lemma 17 ensures the existence of some positive ε>0 such that (Ω-jh)∩K=Ω∩K for 0≤j≤r as long as h<ε/r. We assume in the following that h has been chosen accordingly.
The following integrals exist by Hölder’s inequality becauseφ and thus Δ-hrφ(x) are bounded with compact support and f is integrable. Consider
(50)∫Ω‍f(x)φ(r)(x)dx=∫Ω‍f(x)-h-rΔ-hrφ(x)dx+∫Ω‍fxφrx--h-rΔ-hrφxdx.
We compute the following integral:
(51)∫Ω‍f(x)Δ-hrφ(x)dx=∫Ω‍f(x)∑j=0r‍-1r-jrjφ(x-jh)dx=∑j=0r‍-1r-jrj∫Ω‍f(x)φ(x-jh)dx=∑j=0r‍-1r-jrj∫Ω-jh‍f(y+jh)φ(y)dy=∑j=0r‍-1r-jrj∫(Ω-jh)∩K‍f(y+jh)φ(y)dy.
We have chosen ε>0 such that (Ω-jh)∩K=Ω∩K. We continue
(52)∫Ω‍f(x)Δ-hrφ(x)dx=∑j=0r‍-1r-jrj∫Ω∩K‍f(y+jh)φ(y)dy=∑j=0r‍-1r-jrj∫Ω‍f(y+jh)φ(y)dy=∫Ω‍∑j=0r‍-1r-jrjf(y+jh)φ(y)dy=∫Ω‍Δhrf(x)φ(x)dx.
With the aid of (52), we can now rewrite (50) as
(53)∫Ω‍f(x)φ(r)(x)dx=∫Ω‍-h-rΔhrf(x)φ(x)dx+∫Ω‍fxφrx--h-rΔ-hrφxdx.
Inequality (48) leads to
(54)∫Ω‍fxφrxdx≤rr∫Ω‍afx+afx+rhφxdx+∫Ω‍fxφrx--h-rΔ-hrφxdx.
Using Hölder’s inequality (HI) for the first summand, we are left to show that the second summand tends to 0 as h approaches 0. 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 (L1,L∞)-Hölder inequality and the uniform convergence of the divided difference to φ(r)(x)/r! established in Lemmas 13 and 15.After having verified thatλf has a continuous extension to X′, we want to show that this extension has a representation as integral.Lemma 21.
AssumeΩ⊂R is open. Suppose the properties listed in Properties 19 are met. Let λf be defined as in (47) and bounded on X′. Then there exists g∈X with gX≤λf such that
(55)λf(φ)=-1r∫Ω‍g(x)φ(x)dx
for all φ∈X′, especially for all φ∈C0∞(Ω).Proof.
The proof is heavily influenced by the proof of Theorem 1.4.1 in [13]. We first want to find a signed measure ν such that
(56)λf(φ)=∫Ω‍φdν
for each φ∈X′. Then, we want to apply the Lebesgue-Radon-Nikodym theorem to replace dν by (-1)rg(x)dx for some integrable function g for which we then show that it is contained in X.
We start by assuming that the measure ofΩ is finite. For each μ-measurable set A, we set
(57)ν(A):=λfχA.
We start by proving the σ-additivity of ν (for the μ-measurable sets). In the first step, we just show the additivity. By induction, it is enough to consider only two sets.
LetA1 and A2 be two disjoint μ-measurable sets. Then
(58)νA1∪A2=λfχA1∪A2=λfχA1+χA2=λf(χA1)+λfχA2=ν(A1)+νA2.
We move to the verification of theσ-additivity. Let A1,A2,… be countably many pairwise disjoint μ-measurable sets. We first want to show that ∑i‍χAi converges in X′.
We letEk:=∪i=1kAi and E:=∪i=1∞Ai. Then, since (Ek)k is increasing,
(59)μ(E)=μ⋃k=1∞Ek=lim⁡k→∞μEk.
As E has finite measure, given ε>0, there exists k0∈N such that
(60)1-εμE<μEk0.
By the disjointness of the sets Ai, we see that
(61)μE=μEk0+μ⋃i=k0+1∞Ai.
It follows that
(62)μ⋃i=k0+1∞Ai=μ(E)-μ(Ek0)<μ(E)-(1-ε)μ(E)=εμ(E).
We have for m<n(63)∑i=1n‍χAi-∑i=1mχAi‍X′=∑i=m+1n‍χAiX′≤χ∪i=m+1∞AiX′.
By the considerations before and the fact that ·X′ is absolutely continuous as stipulated in (AC), we see that ∑i=1n‍χAi is a Cauchy sequence. Now, as X′ is complete by (C), the Cauchy sequence converges in X′. We have, by the continuity of λf,
(64)ν∪iAi=λfχ∪iAi=λf∑i=1∞‍χAi=lim⁡n→∞λf∑i=1n‍χAi=lim⁡n→∞∑i=1n‍λfχAi=lim⁡n→∞∑i=1n‍νAi=∑i=1∞‍νAi.
Thus ν is σ-additive.
In order to apply the theorem of Lebesgue-Radon-Nikodym, see, for example, Section 6.9 in [15], and concluding the existence of g^∈L1(Ω) such that
(65)ν(A)=∫A‍g^dx,Aμ-measurable,
we need to show that ν is absolutely continuous with respect to μ. To do so, assume that μ(N)=0. Then ν(N)=λf(χN). By the absolute continuity of the norm of X′ required in (AC), we see that χNX′=0, and therefore ν(N)=0. Replacing g^ by -g^ if necessary, we obtain
(66)λf(χA)=-1r∫Ω‍g^(x)χA(x)dx.
Now, we assume that Ω does not necessarily have finite measure. We write Ω=∪nΩn as union of open sets with finite measure such that Ωn⊂Ωn+1. By the considerations above, we find for each n∈N a gn such that
(67)λf(χA)=-1r∫Ω‍gn(x)χA(x)dx
for each measurable set A⊂Ωn. It is easily seen that gn and gn+1 agree almost everywhere on Ωn. This gives rise to a function g satisfying
(68)λfχA=-1r∫Ω‍gxχAxdx,A⊂⋃n=1NΩnforsomeN.
By linearity, we have for each simple function s:
(69)λf(s)=-1r∫Ω‍g(x)s(x)dx.
To show that g∈X, we want to use property (D).
Leth be a nonnegative simple function with support in some Ωn. Then hsgn⁡(g) is also a simple function with support in Ωn. It is a finite linear combination of characteristic functions of sets in Ωn. Hence
(70)∫‍hgdx=∫‍h·sgn⁡(g)gdx=λfhsgn⁡g.
Using the boundedness of λf, we see that
(71)∫‍hgdx≤λfhX′.
Ifh is an arbitrary function in X′, then we may construct a sequence (hn)n=1∞ of simple functions, each hn having support in Ωn, such that 0≤hn↑h  μ-almost everywhere. By the monotone convergence theorem, we have by (71) that
(72)∫‍hgdx=lim⁡n→∞∫‍hngdx≤limsup⁡n→∞λfhnX′.
By (P3) in Definition 18, we see that
(73)∫‍hgdx≤λfhX′
and Property (D) gives that g∈X. In the last step, we want to upgrade the integral representation (69) to hold for arbitrary functions in φ∈X′. As before, we approximate φ by simple functions hn with support contained in Ωn and 0≤hn↑φ in μ-almost everywhere point. Define sn:=hnsgn⁡(φ). By (NC), sn converges to φ in X′ and as both λf and “its integral representation” are continuous and agree on the set of step functions, they agree on X′.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 Proposition22 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 Theorem2 (pointwise inequality implies membership in Sobolev space).
We combine Lemmas20 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.
LetΩ⊂R be a domain with finite Lebesgue measure in R. Suppose X is a Banach function space satisfying (B) and (D). Assume that m∈N. Then the following sets coincide: (a)
A={f∈X: there exists F∈X, where F agrees almost everywhere with f and has weak derivatives up to order m in X}, where the derivatives of order r, 0≤r≤m-1, are absolutely continuous,(b)
B={f∈X:f(k)∈X,k∈{0,…,m}}, where f(k) are the weak derivatives of f,(c)
C={f∈X:f(m)∈X}, where f(m) is the weak derivative of f of order m.We see thatA⊂B⊂C. We now show that C⊂A.Let us give an outline of the proof. Giveng:Ω→R (we will choose g=f(m)), by integrating, we construct in Lemma 23 an absolutely continuous function Gm together with its weak derivatives G0,…,Gm-1, which are absolutely continuous, and Gm(m)=g. However, Gm and f 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 P such that Gm+P agrees with F almost everywhere and shares with Gm the other required properties. We conclude the proof of Theorem 26 by settling the membership questions with the aid of Proposition 25.Lemma 23.
AssumeΩ⊂R is a domain. We further assume that g:Ω→R is in Lloc⁡1(Ω). One fixes x*∈Ω and defines
(74)G0(x0):=g(x0),Gk(xk):=∫x*xk‍Gk-1(xk-1)dxk-1,k∈N,
where the integrals are Lebesgue integrals, and we use the notation
(75)∫x*xk‍Gk-1(xk-1)dxk-1=∫[x*,xk]‍Gk-1(xk-1)dxk-1,x*≤xk,-∫[xk,x*]‍Gk-1(xk-1)dxk-1,xk<x*.
Then (a)
Gk is locally absolutely continuous for all k∈N, especially locally integrable,(b)
ifφ∈C0∞(Ω), and 0≤r≤k, one has
(76)∫Ω‍Gkφ(r)=-1r∫Ω‍Gk-rφ.Proof.
Let us prove (a) by induction. Assumek=1. Note that G1(x*)=0. Thus, we have for x*≤x1(77)G1(x1)-G1(x*)=G1(x1)=∫x*x1‍g(x0)dx0.
As g∈L1([x*,x1]), we see that G1 is absolutely continuous. The case x1<x* is treated similarly.
Assume now that the claim is true for somek. Now Gk+1(xk+1) 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 fork=0; thus we may assume that k≥1. We make an induction over r. The case r=0 is clear. Let us choose an interval [a,b]⊂Ω containing a neighborhood of the support of φ. As Gk and φ(r) are both absolutely continuous in [a,b], 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 φ(r)(a)=φ(r)(b)=0(78)∫Ω‍Gkφ(r+1)=∫[a,b]‍Gkφ(r+1)=-1∫a,b‍Gk′φ(r).
Noting that Gk′=Gk-1 almost everywhere, we see that
(79)∫Ω‍Gkφ(r+1)=(-1)∫[a,b]‍Gk-1φ(r)=(-1)∫Ω‍Gk-1φ(r).
Using the induction hypothesis, the claim follows.We cite Theorem 2.5.3 in [10] without giving its proof.Theorem 24.
LetA=[a,b], R, or R+ and r≥0. If the function f∈Lloc⁡1(A) satisfies
(80)∫A‍fφ(r)dx=0
for all φ∈C0∞(A), then there is a polynomial P of degree <r such that f=P almost everywhere on A.To show thatGk is in X, we need to verify that (if there is danger of confusion with respect to which variable we take the norm, we use notations as X(xk))
(81)∫x*xk‍Gk-1(xk-1)dxk-1X(xk)<∞
under the assumption that Gk-1 is in X. Actually, bounding Gk from above by its absolute value, this means that we have first to take the L1-norm and then the norm in X. We however would like to switch the norms.We adapt Proposition 2.1 in [16] to our setting.Proposition 25.
Letρ be a Banach function norm satisfying property (D). If f:Ω×Ω→R is a measurable function, and one has fx and fy functions on Ω such that fx(y)=f(x,y)=fy(x), then
(82)ρfx1≤ρfy1.The casesX∈{L1,L∞} are easy to prove. If X=Lp for some 1<p<∞, 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.
LetΩ⊂R be a domain, and let r≥1. Suppose X is a Banach function space satisfying (B) and (D). If f is locally in X and has a generalized rth derivative g locally in X, then there is F agreeing almost everywhere with f, having absolutely continuous intermediate weak derivatives, and the weak derivative of order r agrees almost everywhere with g. Furthermore, the derivatives of F are locally in X. If Ω 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 inX.Proof.
Choosex*∈Ω and let Gk be defined as in Lemma 23. Assume φ∈C0∞(Ω). Then, using the fact that g is a weak derivative of f of order r and (b) in Lemma 23(83)∫Ω‍f-Grφrdx=∫Ω‍fφ(r)dx-∫Ω‍Grφ(r)dx=-1r∫Ω‍gφdx--1r∫Ω‍gφdx=0.
Using Theorem 24, we see that f-Gr=P almost everywhere, where P is a polynomial of degree <r. Thus f=P+Gr almost everywhere and for φ∈C0∞(Ω) and 0≤m≤r(84)∫Ω‍fφ(m)dx=∫Ω‍P+Grφ(m)=-1m∫Ω‍Gr-m+Pmφdx.
Thus Gr-m+P(m) is the mth generalized derivative of f. As it is locally absolutely continuous (if m<r) it lies locally in X. In the case m=r the same conclusion holds by the assumption on g.
Moreover, there existsF, for example, F=P+Gr, that agrees with f almost everywhere, such that its generalized derivatives up to order r-1 are locally absolutely continuous and the rth generalized derivative equals g almost everywhere.
Asg is locally in X and the other derivatives are locally absolutely continuous, they are locally in L∞ and thus locally in X by (B).
Suppose now thatΩ has additionally finite measure. Assume that f and g are in X globally. We want to show that the intermediate derivatives are in X as well.
Note in our setting that, by (P5) in Definition18, membership in X implies membership in L1. We proceed by induction. Suppose that Gk-1∈X. We will apply Proposition 25. We set, denoting by convx*,xk the closed interval with endpoints x* and xk,
(85)f(xk-1,xk)=Gk-1(xk-1)χconv{x*,xk}(xk-1).
Then
(86)GkxkX(xk)=∫x*xk‍Gk-1(xk-1)dxk-1X(xk)=∫x*xk‍f(xk-1,xk)dxk-1X(xk)≤fxk-1,xkL1(xk-1)X(xk)≤fxk-1,xkXxkL1(xk-1)≤Gk-1xk-1χΩXxkL1xk-1<∞.


## 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 Theorem2.We list the results and prove them simultaneously by splitting the proofs into parts according to the function spaces.Theorem 27.
AssumeΩ⊂R is a domain, X is a normed space, and m∈N. If the maximal function M:X→X is bounded, then there is a constant C=C(X,m) such that for each continuous f∈WXm there is af∈X with af≤CfWXm and
(87)Δmf(x;y)≤x-ymafx+afy.
In particular such af exists if(i)
1<p≤∞ and X is the Lebesgue space Lp(Ω),(ii)
1<p<∞, 1≤q≤∞, and X is the Lorentz space Lp,q(Ω),(iii)
1<p<∞, 1≤q≤∞, b is slowly varying, and X is the Lorentz-Karamata space Lp,q;b(Ω),(iv)
ϕ is an N-function with inf⁡s>0sφ(s)/ϕ(s)>1, where φ is the density function of ϕ and X is the Orlicz space Lϕ*(Ω),(v)
Ω=R and X is rearrangement-invariant with upper index α¯X<1.Theorem 28.
LetΩ⊂R be open. Suppose X is a Banach function space such that (AC) holds. Assume further that the function f:Ω→R is in X. If for some r∈N0, the following inequality holds
(88)Δrfx,y≤x-yrafx+afy
for some af∈X, then there exists g∈X with gX≤2rrafX such that
(89)∫Ω‍f(x)φ(r)(x)dx=-1r∫Ω‍g(x)φ(x)dx
for all φ∈C0∞(Ω).
In particular the result holds ifX is(i)
the Lebesgue spaceLp(Ω) for some 1<p<∞,(ii)
the Lorentz spaceLp,q(Ω), where 1<p<∞ and 1≤q<∞,(iii)
the Lorentz-Karamata spaceLp,q;b(Ω), where b is slowly varying, 1<p<∞, and 1<q≤∞,(iv)
the Orlicz spaceLϕ*(Ω) for some N-function ϕ satisfying
(90)inf⁡s>0sφ(s)ϕ(s)>1,where φ is the density function of ϕ.To help with orientation, let us give an overview of the remaining material in this section. With respect to Theorem27, 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 Theorem28, 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 X′ 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 spaceX′ 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 Theorem27 (general statement).
We basically have already proven the general statement in Corollary14.Proof of Theorem28 (general statement).
We gave the proof of the Lebesgue space case in such generality that the general statement follows from the proof of Theorem2 and Propositions 29 and 30.


### 5.2. Lorentz-Karamata Spaces

Lorentz-Karamata spaces are detailed in [14], mainly Section 3.4.3.Proof of Theorem27, (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 Theorem28, (ii) and (iii).
By Theorem 3.4.41 in [14], we see that X 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 X′.


### 5.3. Orlicz Spaces

The setting is the one of [17].Lemma 31.
Ifϕ is an N-function satisfying the Δ2-condition, then the Orlicz space Lϕ* has absolutely continuous norm.Proof.
Letf∈Lϕ*(Ω). We verify first that
(91)∫Ω‍ϕ(f)dx<∞.
Let ε>0. By the Δ2-condition, there exists a constant κ=κ(f(ϕ)+ε) such that
(92)ϕfϕ+εs≤κϕ(s)
for all s≥0. Thus
(93)∫Ω‍ϕfdx=∫Ω‍ϕffϕ+εfϕ+εdx≤κ∫Ω‍ϕffϕ+εdx≤κ.
Suppose that E is a measurable set. Then
(94)fχE(ϕ)=inf⁡λ>0:∫Ω‍ϕ(λ-1fχE)dx≤1=inf⁡λ>0:∫Ω‍ϕ(λ-1f)χEdx≤1.
Let us fix λ>0. Again using the Δ2-condition, we have ϕ(λ-1s)≤κ(λ-1)ϕ(s). Hence
(95)∫Ω‍ϕ(λ-1f)χEdx≤κ∫Ω‍ϕ(f)χEdx.
Example 1.3.3 in [13] tells us that L1 has absolutely continuous norm. Thus if χEn converges μ-almost everywhere to 0, then ∫Ω‍ϕ(f)χEn converges to zero as well. Thus, we have for n large enough that
(96)κ∫Ω‍ϕ(f)χEndx≤1
implying that fχEn(ϕ)≤λ. Since λ>0 was arbitrary, the absolute continuity of f follows.Proof of Theorem27 (iv).
We conclude from Gallardo’s result, Theorem 2.1 in [17], that it is sufficient that the complementary N-function satisfies the Δ2-condition in [0,∞). Again resorting to Gallardo, this time to Proposition 1.4 in [17], we obtain the conclusion.Proof of Theorem28 (iv).
ThatX is a Banach function space follows from Theorem 3.4.16 in [14]. To prove that X′ has absolutely continuous norm, we note that inequality (90) together with Proposition 1.4 in [17] ensures that the complementary N-function of ϕ satisfies the Δ2-condition. Lemma 31 verifies that X′ has absolutely continuous norm.


### 5.4. Rearrangement-Invariant Spaces

The necessary definitions can be found in [13].Proof of Theorem27, (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].


## 5.1. General Statements

Proof of Theorem27 (general statement).
We basically have already proven the general statement in Corollary14.Proof of Theorem28 (general statement).
We gave the proof of the Lebesgue space case in such generality that the general statement follows from the proof of Theorem2 and Propositions 29 and 30.


## 5.2. Lorentz-Karamata Spaces

Lorentz-Karamata spaces are detailed in [14], mainly Section 3.4.3.Proof of Theorem27, (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 Theorem28, (ii) and (iii).
By Theorem 3.4.41 in [14], we see that X 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 X′.


## 5.3. Orlicz Spaces

The setting is the one of [17].Lemma 31.
Ifϕ is an N-function satisfying the Δ2-condition, then the Orlicz space Lϕ* has absolutely continuous norm.Proof.
Letf∈Lϕ*(Ω). We verify first that
(91)∫Ω‍ϕ(f)dx<∞.
Let ε>0. By the Δ2-condition, there exists a constant κ=κ(f(ϕ)+ε) such that
(92)ϕfϕ+εs≤κϕ(s)
for all s≥0. Thus
(93)∫Ω‍ϕfdx=∫Ω‍ϕffϕ+εfϕ+εdx≤κ∫Ω‍ϕffϕ+εdx≤κ.
Suppose that E is a measurable set. Then
(94)fχE(ϕ)=inf⁡λ>0:∫Ω‍ϕ(λ-1fχE)dx≤1=inf⁡λ>0:∫Ω‍ϕ(λ-1f)χEdx≤1.
Let us fix λ>0. Again using the Δ2-condition, we have ϕ(λ-1s)≤κ(λ-1)ϕ(s). Hence
(95)∫Ω‍ϕ(λ-1f)χEdx≤κ∫Ω‍ϕ(f)χEdx.
Example 1.3.3 in [13] tells us that L1 has absolutely continuous norm. Thus if χEn converges μ-almost everywhere to 0, then ∫Ω‍ϕ(f)χEn converges to zero as well. Thus, we have for n large enough that
(96)κ∫Ω‍ϕ(f)χEndx≤1
implying that fχEn(ϕ)≤λ. Since λ>0 was arbitrary, the absolute continuity of f follows.Proof of Theorem27 (iv).
We conclude from Gallardo’s result, Theorem 2.1 in [17], that it is sufficient that the complementary N-function satisfies the Δ2-condition in [0,∞). Again resorting to Gallardo, this time to Proposition 1.4 in [17], we obtain the conclusion.Proof of Theorem28 (iv).
ThatX is a Banach function space follows from Theorem 3.4.16 in [14]. To prove that X′ has absolutely continuous norm, we note that inequality (90) together with Proposition 1.4 in [17] ensures that the complementary N-function of ϕ satisfies the Δ2-condition. Lemma 31 verifies that X′ has absolutely continuous norm.


## 5.4. Rearrangement-Invariant Spaces

The necessary definitions can be found in [13].Proof of Theorem27, (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].



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