Superposition operators and functions of boundedp-variation II

Gerard Bourdaud, Massimo Lanza de Cristoforis, Winfried Sickel

31/08/04

AbstractWe continue the study of the superposition operator Tf : g 7→ f ◦ g, on

the space BV 1p (I) of primitives of real-valued functions of bounded p-variation

on an interval I. We give first a characterization of the functions f such thatTf takes BV 1

p (I) to itself. Then we characterize functions f for which Tf iscontinuous, uniformly continuous, and differentiable, as a mapping of BV 1

p (I)to itself, respectively. By exploiting the Peetre’s Imbedding Theorem and theFubini property, we derive partial results on continuity of Tf in Besov spacesBsp,q(Rn), for a smoothness parameter s satisfying 0 < s ≤ 1 + (1/p).

2000 Mathematics Subject Classification: 46E35, 47H30.Keywords: Functions of bounded p-variation, homogeneous and inhomogeneous Besovspaces, Peetre’s Imbedding Theorem, continuity and differentiability of superpositionoperators.

1 Introduction

The present paper is a continuation of the analysis of [9]. In [9], we have characterizedthe set of all functions f : R → R such that the associated superposition (composition)operator Tf : g 7→ f ◦ g acts in BV 1

p (R). Here BV 1p (R) denotes the set of primitives of

functions of bounded p-variation in the sense of Wiener (see Subsection 3.1). It turnsout that whenever the acting property

Tf (BV1p (R)) ⊆ BV 1

p (R)

holds, then the operator Tf is also bounded. Here we continue with a detailed studyof the continuity and differentiability properties of Tf .For an extensive treatment of superposition operators and function spaces, we referto the monographs of Appell and Zabrejko [1], and of [19, Chapter 5], and to thesurvey papers [3, 4]. Our specific interest in BV 1

p has been motivated in [9]. Theclasses BV 1

p represent one of the possible limiting cases for sublinear estimates ofsuperposition operators in function spaces with positive smoothness, see [9] for furtherdetails. What also contributes to make the classes BV 1

p interesting is that one canapply the interpolation theorem of Peetre for nonlinear operators to deduce propertiesof Tf in Besov spaces Bs

p,q with 0 < s ≤ 1 + (1/p) and 1 ≤ q ≤ ∞ from correspondingproperties of Tf in BV 1

p .

1

2 The main results

In our investigations, we have realized a substantial difference between the propertiesof Tf in function spaces defined on bounded intervals, and the properties of Tf in func-tion spaces defined on unbounded intervals. Accordingly, we first prove the followingTheorem, which clarifies the behaviour of Tf in the space BV 1

p on bounded intervalsand which can be considered as a complement to a corresponding result of [9].

In all statements of this paper, we assume p to be an a priori fixed number in [1,+∞[,and I to be an a priori fixed nonempty interval of R,

unless otherwise specified (as in Propositions 2, 3, 11, 13, 21.)

Theorem 1 Let f : R → R be a Borel measurable function. Then the followingstatements hold.

(i) The superposition operator Tf maps BV 1p (I)∩Lp(I) or BV 1

p (I)∩L∞(I) to BV 1p (I)

if and only if f ∈ BV 1p,`oc(R).

(ii) Let I be unbounded. Then Tf maps BV 1p (I) to itself if and only if f ∈ BV 1

p (R).

To describe the continuity properties of Tf we need to introduce some other classes offunctions. Let Ep and E1

p denote the closure of BVp(R) ∩ C∞(R) in BVp(R) and theclosure of BV 1

p (R) ∩ C∞(R) in BV 1p (R), respectively.

Theorem 2 (i) Let f ∈ BV 1p,`oc(R). Then Tf is a continuous mapping of BV 1

p (I)∩Lp(I) or of BV 1

p (I) ∩ L∞(I) to BV 1p (I) if and only if f ∈ E1

p,`oc.

(ii) Let f ∈ BV 1p,`oc(R), f(0) = 0. Then Tf is a continuous mapping of BV 1

p (I)∩Lp(I)to itself or of BV 1

p (I) ∩ L∞(I) to itself if and only if f ∈ E1p,`oc.

(iii) Let I be unbounded and f ∈ BV 1p (R). Then Tf is a continuous mapping of

BV 1p (I) to itself if and only if f ∈ E1

p .

As usual, in Theorem 2 and in the sequel, we assume that the intersection of twonormed spaces is endowed with the sum of the norms of the two spaces. Clearly,if I is bounded, then BV 1

p (I) = BV 1p (I) ∩ L∞(I) and statement (i) applies. Since

E1p,`oc and E1

p are proper subclasses of BV 1p,`oc(R) and BV 1

p (R), respectively, we do nothave equivalence of the acting property, boundedness and continuity of superpositionoperators as in case of the first order Sobolev spaces W 1,p (cf. Marcus and Mizel [15].)

The sufficiency part of Theorem 2 is a consequence of a stronger result. Namely, thefollowing holds.

Theorem 3 The composition operator T : E1p ×BV 1

p (R) → BV 1p (R) defined by

T (f, g) := f ◦ g

is continuous.

2

As is well-known for superposition operators, Lipschitz continuity is a rare property(cf. Appell and Zabrejko [1].) As a matter of fact, we can even prove that the uniformcontinuity of Tf can hold only if f is affine.

Theorem 4 If f : R → R is a function such that Tf is an uniformly continuousmapping of BV 1

p (I) ∩ Lp(I) to BV 1p (I), then f is an affine function.

Actually, in Section 8 we prove a stronger result by replacing the target space BV 1p (I)

by a Banach space having a weaker norm. Finally, we analyze the differentiability ofTf . The differentiability of Tf is one of the properties indicated as those for which theboundedness of I really makes a difference. Indeed, we have the following.

Theorem 5 Let f : R → R be a Borel measurable function. Then the following twostatements hold.

(i) If I is unbounded and Tf : BV 1p (I) → BV 1

p (I) is differentiable, then f is anaffine function.

(ii) The following properties are equivalent:

(a) Tf : BV 1p (I) ∩ Lp(I) → BV 1

p (I) ∩ Lp(I) is of class C1.

(b) Tf : BV 1p (I) ∩ Lp(I) → BV 1

p (I) is of class C1.

(c) f is differentiable and f ′ ∈ E1p,`oc.

As a supplement to our investigations into superposition operators on BV 1p , we dis-

cuss the behaviour of superposition operators on the Besov spaces B1+(1/p)p,1 (Rn). The

exponent 1 + (1/p) is in a certain sense critical in the framework of Besov spaces, andthus of particular interest (see comments in [9] and references therein.) To analyze thecontinuity of Tf in Besov spaces, we need to introduce another class of functions, whichhave been first considered by Bourdaud and Kateb in connection with the acting con-ditions of Tf in the Besov spaces Bs

p,q with the ‘subcritical’ exponent 0 < s < 1+ (1/p)(cf. [7].) We denote by Up(R) the set of measurable functions f : R → R such that

‖ f ‖Up := supt>0

(1

t

∫ ∞

−∞sup|h|<t

|f(x+ h)− f(x)|p dx)1/p

<∞ .

We associate to Up(R) a space of distributions Up(R) by taking the set of functionswhich are almost everywhere equal to at least an element of Up(R), and then by takingthe quotient modulo equality almost everywhere. Then we endow Up(R) with theseminorm

‖f‖Up(R) := inf{‖g‖Up(R) : g ∈ Up(R), g = f a.e.} ∀f ∈ Up(R) .

Then we denote by U1p (R) the set of Lipschitz continuous functions f : R → R such

that f ′ has a bounded measurable representative of class Up(R), and we set

‖f‖U1p (R) := |f(0)|+ ‖f ′‖∞ + ‖f ′‖Up(R) ∀f ∈ U1

p (R) .

3

We shall also need the following notation. If E is a Banach space contained in thespace of distributions D′(Rn), then we set

C∞[E] :={g ∈ C∞(Rn) : g(α) ∈ E for all α ∈ Nn

}(1)

and we denote by clE C∞[E] the closure of C∞[E] in E. In [9], we have proved the

chain of continuous imbeddings

BVp(R) ↪→ Up(R) ↪→ B1/pp,∞(R) ,

and the boundedness of Tf as a mapping of B1+(1/p)p,1 (Rn) to B

1+(1/p)p,∞ (Rn) for f ∈ U1

p (R),

f(0) = 0. Here B1/pp,∞(R) denotes the homogeneous version of B

1/pp,∞(R). Then we have

the following result for Besov spaces in Rn.

Theorem 6 Let f : R → R be a Borel measurable function. Then the followingstatements hold.

(i) If Tf : B1+(1/p)p,1 (Rn) → B

1+(1/p)p,∞ (Rn) is continuous, then the function f belongs

locally to clB

1+(1/p)p,∞ (R)

C∞[B1+(1/p)p,∞ (R)].

(ii) Let f(0) = 0. If f belongs to clU1p (R)C

∞[U1p (R)], then the mapping

Tf : B1+(1/p)p,1 (Rn) → B1+(1/p)

p,∞ (Rn)

is continuous.

(iii) Let 0 < s < 1 + (1/p) and let 1 ≤ q < ∞. Under the same restrictions on f asin (ii), the operator Tf maps Bs

p,q(Rn) to Bsp,q(Rn) continuously.

2.1 An example: the absolute value

The most interesting example in this context is given by the function f(t) := |t|, andwe now summarize the properties of Tf .

(i) The function f belongs to BV 1p (R) \ E1

p,loc.

(ii) The operator Tf maps BV 1p (I) to BV 1

p (I) and there exists a constant c such that

‖ |g| ‖BV 1p (I) ≤ c ‖ g ‖BV 1

p (I) ∀g ∈ BV 1p (I) . (2)

(iii) The operator Tf maps B1+(1/p)p,1 (Rn) to B

1+(1/p)p,∞ (Rn) and there exists a constant

c such that

‖ |g| ‖B

1+(1/p)p,∞ (Rn)

≤ c ‖ g ‖B

1+(1/p)p,1 (Rn)

∀g ∈ B1+(1/p)p,1 (Rn) . (3)

(iv) The operator Tf is not continuous from B1+(1/p)p,1 (Rn) to B

1+(1/p)p,∞ (Rn).

4

(v) Let 0 < s < 1 + (1/p) and 1 ≤ q ≤ ∞. The operator Tf maps Bsp,q(Rn) to

Bsp,q(Rn) and there exists a constant c such that

‖ |g| ‖Bsp,q(Rn) ≤ c ‖ g ‖Bs

p,q(Rn) ∀g ∈ Bsp,q(Rn) . (4)

(vi) Let 0 < s < 1 and 1 ≤ q < ∞. The operator Tf considered as a mapping ofBsp,q(Rn) to Bs

p,q(Rn) is continuous.

We mention that (iv) is new. Statement (iii) instead has been proved by Savare [20]for n = 1 and in [9] for n > 1. Furthermore, (v) represents for us the starting pointof our interest into those questions, and has been proved independently by Bourdaudand Meyer [10] and Oswald [16]. In addition, we mention that Triebel [22, Prop. 25.4,p. 361] has given a third proof of statement (v), and has proved that the operator Tfas in statement (v) is not uniformly continuous.We prove the above properties in Subsection 10.4.

2.2 A second example: logarithmic perturbations of the ab-solute value

It is interesting to compare the behaviour of g 7→ |g| with that of g 7→ ψ ◦ g, whereψ : R → R is defined by

ψ(t) := |t| ρ(t)log |t|

if t 6= 0, ψ(0) = 0 ,

with ρ ∈ D(R) and supp ρ ⊆ [−1/2, 1/2], and ρ(t) = 1 on [−1/e, 1/e]. Then the func-tion ψ and the operator Tψ have the following properties, which we prove in subsection10.5.

(a) The function ψ belongs to E1p .

(b) The operator Tψ maps BV 1p (I) to BV 1

p (I) and there exists a constant c such that

‖ψ ◦ g‖BV 1p (I) ≤ c ‖ g ‖BV 1

p (I) ∀g ∈ BV 1p (I) . (5)

(c) The operator Tψ maps B1+(1/p)p,1 (Rn) to B

1+(1/p)p,∞ (Rn) and there exists a constant

c such that

‖ψ ◦ g‖B

1+(1/p)p,∞ (Rn)

≤ c ‖g‖B

1+(1/p)p,1 (Rn)

∀g ∈ B1+(1/p)p,1 (Rn) . (6)

(d) The operator Tψ is continuous from B1+(1/p)p,1 (Rn) to B

1+(1/p)p,∞ (Rn).

(e) Let 0 < s < 1 + (1/p), 1 ≤ q ≤ ∞. The operator Tψ maps Bsp,q(Rn) to Bs

p,q(Rn)and there exists a constant c such that

‖ψ ◦ g‖Bsp,q(Rn) ≤ c ‖g‖Bs

p,q(Rn) ∀g ∈ Bsp,q(Rn) . (7)

(f) If q ∈ [1,∞[, then the operator in (e) is continuous.

The above properties of the superposition operator do not change if log |t| is replacedby iterated logarithms like log | log |t|| or log | log | log |t||. The above properties of Tψdo not involve p. In order to observe a dependence of Tψ on p, we need to introducean oscillatory term around t = 0 in the sense of the following subsection.

5

2.3 A third example: perturbations of the absolute value in-volving oscillations

Since the classes BV 1p increase with p, it is interesting to exhibit functions which belong

to BV 1p for p > p0 but not for p ≤ p0. To do so, we consider the family of functions

ψα,β : R → R defined by

ψα,β(t) := |t|α+1 ρ(t) sin(|t|−β

)if t 6= 0, ψα,β(0) := 0 , (8)

with ρ as above and 0 < β < α. Then the function ψα,β and the operator Tψα,βhave

the following properties, which we prove in subsection 10.6.

(a) The function ψα,β belongs to BV 1p (R) if and only if

1

p<α

β− 1 (9)

(a’) The function ψα,β belongs to E1p if and only if (9) is satisfied.

(b) The operator Tψα,βmaps BV 1

p (I) to BV 1p (I) if and only if (9) is satisfied.

Furthermore, if (9) holds, then the operator Tψ satisfies the properties (c)-(f) of sub-section 2.2, with ψ replaced by ψα,β.

2.4 Some open problems

The theory of nonlinear superposition operators acting in Besov or in Triebel-Lizorkinspaces with a positive ‘smoothness’ parameter s is still more or less at its beginning,and there is a huge collection of open problems. We now mention some of those whichinvolve function spaces with smoothness parameter less than or equal to 1 + (1/p).

Besov spaces

• Establish whether the mapping g 7→ |g| is continuous in Besov spaces with 1 ≤s < 1 + (1/p) (compare with property (vi) in Subsection 2.1).

• Characterize all functions f such that Tf maps B1+(1/p)p,1 (Rn) to B

1+(1/p)p,∞ (Rn) (see

comments preceding Theorem 6 and [9]).

• Characterize all functions f such that Tf maps B1+(1/p)p,1 (Rn) to B

1+(1/p)p,∞ (Rn) and

is continuous (see Theorem 6 (i), (ii)).

• Characterize all functions f such that Tf maps Bsp,q(Rn) to Bs

p,q(Rn) with 1 ≤s < 1 + (1/p).

Concerning this point, the best known result is the following sufficient conditionof Kateb [11]. If f ∈ B

1+(1/p)p,∞ (R), and f ′ ∈ L∞(R), and f(0) = 0, then Tf maps

Bsp,q(Rn) to Bs

p,q(Rn).

• Characterize all functions f such that Tf maps Bsp,q(Rn) to Bs

p,q(Rn) with 1 ≤s < 1 + (1/p) and is continuous.

6

Lizorkin-Triebel spaces

There is an other scale of function spaces, in some sense parallel to that of Besovspaces, the so-called Lizorkin-Triebel spaces F s

p,q(Rn). The classes F sp,q(Rn) are certain

generalizations of Sobolev spaces of fractional order (Bessel potential spaces), cf. e.g.Triebel [21]. They are close to the corresponding Besov spaces in the sense that thecontinuous imbedding

Bsp,1(Rn) ↪→ F s

p,q(Rn) ↪→ Bsp,∞(Rn)

holds for all 1 ≤ q ≤ ∞. We now formulate a few open problems for the classesF sp,q(Rn). We mention that there is no counterpart of Theorem 6 in the context of

Lizorkin-Triebel spaces. More precisely, if 1 < p <∞, and 1 + (1/p) < n/p, and if the

composition operator Tf maps F1+(1/p)p,1 (Rn) to B

1+(1/p)p,∞ (Rn), then f must be an affine

function (cf. [5] and [19, Remark 5.3.1/11].)

• Characterize all functions f such that Tf maps F sp,q(Rn) to F s

p,q(Rn) with 1 ≤ s <1 + (1/p).

• Characterize all functions f such that Tf maps F sp,q(Rn) to F s

p,q(Rn) with 1 ≤ s <1 + (1/p) and is continuous.

The paper is organized as follows. In Section 3, we collect some basic propertiesof functions of bounded p-variation. In Section 4, we investigate the space of theprimitives of the functions of bounded p-variation, namely the space BV 1

p (R). Section5 is devoted to the study of the classes Ep and E1

p . Then we prove Theorem 1 inSection 6. The proofs of Theorems 2 and 3 are given in Section 7. The nonexistence ofnontrivial uniformly continuous superposition operators in BV 1

p is shown in Section 8.The differentiability properties are investigated in Section 9. In Section 10 we considerthe case of n variables and we prove Theorem 6. In section 10, we also consider theexamples of Subsections 2.1-2.3. Finally, we include in an Appendix some technicalstatements required in the paper, and for which we claim no credit.

In the sequel, all functions are assumed to be real valued.

3 Basic properties of Vp(I) and BVp(I)

3.1 The definition of Vp(I). Restriction and extensions

Following Wiener [23], a function f : I → R is said to be of bounded p-variation, ifthere exists c > 0 such that

N∑k=1

|f(tk)− f(tk−1)|p ≤ cp ,

for all finite sequences t0 < t1 < · · · < tN in I. The infimum of such constants c isdenoted by νp(f, I). We denote by Vp(I) the set of functions f : I → R of bounded

7

p-variation. We shall use the abbreviations Vp := Vp(R) and νp(f) := νp(f,R). Byconsidering a sequence with only two terms, we obtain

|f(x)− f(y)| ≤ νp(f, I) ∀x, y ∈ I . (10)

Hence, every function of Vp(I) is bounded. One can easily prove that Vp(I) becomes aBanach space if endowed with the following norm:

‖f‖Vp(I) := supx∈I

|f(x)|+ νp(f, I) ∀f ∈ Vp(I) .

We state in the following Proposition an elementary, but fundamental property offunctions of bounded p-variation.

Proposition 1 Let a ∈◦I and I1 :=]−∞, a[∩I, I2 := [a,∞[∩I.

1. If f ∈ Vp(I), thenνpp(f, I1) + νpp(f, I2) ≤ νpp(f, I) . (11)

2. If fj ∈ Vp(Ij) for j = 1, 2, then the function f : I → R such that f |Ij = fj forj = 1, 2, belongs to Vp(I); moreover

νp(f, I) ≤ νp(f1, I1) + νp(f2, I2) + supI1

|f1|+ supI2

|f2| . (12)

Proof The first inequality is obvious. To establish the second one, we take a sequencex0 < x1 < · · · < xN in I with x0 < a < xN , and defineM by inequality xM−1 < a ≤ xM .Then

N∑j=1

|f(xj)− f(xj−1)|p

≤M−1∑j=1

|f(xj)− f(xj−1)|p + |f(xM)− f(xM−1)|p +N∑

j=M+1

|f(xj)− f(xj−1)|p

≤ νp(f1, I1)p + νp(f2, I2)

p +

(supI1

|f1|+ supI2

|f2|)p

.

Hence, inequality (12) follows.

Remark 1 According to Proposition 1, if J ⊆ I then we have a bounded restrictionoperator from Vp(I) to Vp(J) and many different bounded extension operators fromVp(J) to Vp(I). Another consequence of Proposition 1 is the following. Let I be aclosed interval of R and f : R → R a function with support in I. If f |I ∈ Vp(I), thenf ∈ Vp(R) and

‖f‖Vp(R) ≤ 2‖f |I‖Vp(I) .

8

3.2 Limiting properties

The existence of left and/or right limits for a function f ∈ Vp is a classical property(see Wiener [23]). We now prove it in a form, which will be useful in the sequel.

Proposition 2 Let I be an interval of R such that b := sup I /∈ I. If f0 ∈ Vp(I), then

limVp(I)×I3(f,c)→(f0,b)

νp(f, [c, b[) = 0 .

Proof By definition of the p-variation of f0, we have

νp(f0, I) = limc→b−

νp

(f0, I∩]−∞, c[

).

By inequality (11), we deduce that limc→b− νp(f0, [c, b[) = 0. For any f ∈ Vp(I), wehave

νp(f, [c, b[) ≤ νp(f − f0, I) + νp(f0, [c, b[) ,

and thus the proof is complete.

By exploiting inequality (10) with I = [c, b[, and Proposition 2, and by letting c→ b−,we obtain the following Proposition.

Proposition 3 Let I be an interval of R such that b := sup I /∈ I. Let f0 ∈ Vp(I).Then h := limx→b− f0(x) exists in R. Moreover,

limVp(I)×I3(f,x)→(f0,b)

f(x) = h .

3.3 A multiplication property

Proposition 4 The Banach space Vp(I) is a Banach algebra with respect to the point-wise multiplication of functions. Moreover

‖fg‖Vp(I) ≤ ‖f‖Vp(I) ‖g‖Vp(I) ∀f, g ∈ Vp(I) . (13)

Proof Let x0 < x1 < · · · < xN be a finite sequence in I and f , g in Vp(I). Then wehave( N∑

j=1

|fg(xj)− fg(xj−1)|p)1/p

≤

(N∑j=1

|f(xj)(g(xj)− g(xj−1))|p)1/p

+

(N∑j=1

|g(xj−1)(f(xj)− f(xj−1))|p)1/p

≤ ( supI|f |) νp(g, I) + ( sup

I|g|) νp(f, I) .

Hence, ‖fg‖Vp(I) ≤ ‖f‖Vp(I) ‖g‖Vp(I).

9

3.4 The definition of BVp(I)

Definition 1 Let f : I → R be a function having only discontinuities of the first kind.Then f is said to be normalized if

f(x) =1

2(f(x+) + f(x−)) ∀x ∈

◦I ,

andf(x) = lim

y→x,y∈◦I

f(y) ∀x ∈ I ∩ ∂I .

Now we have the following three Propositions, whose verification is straightforward.

Proposition 5 Let f be a normalized function on I. If f is vanishes almost every-where, then f vanishes everywhere. Moreover f ∈ L∞(I) if and only if f is bounded,and

‖f‖∞ = supx∈I

|f(x)| .

Proposition 6 If f is a function in Vp(I), then the function f defined by

f(x) :=1

2(f(x+) + f(x−)) ∀x ∈

◦I ,

andf(x) := lim

y→x,y∈◦I

f(y) ∀x ∈ I ∩ ∂I ,

is normalized, and belongs to Vp(I), and satisfies the following inequalities:

νp(f , I) ≤ νp(f, I), supI|f | ≤ sup

I|f | .

Proposition 7 Let f : I → R be a measurable function. Then there exists at mostone normalized function f : I → R, such that f = f almost everywhere in I. If fexists and if g : I → R is a measurable function equal to f almost everywhere, then gexists and is equal to f .

We define BVp(I) by taking the functions which are almost everywhere equal to anelement of Vp(I), and then by taking the quotient modulo equality almost everywhere.By the above Propositions, each f ∈ BVp(I) contains a unique normalized f ∈ Vp(I)such that

νp(f , I) ≤ νp(g, I)

for any function g ∈ Vp(I) belonging to the class f , and such that

supI|f | = ‖f‖∞ .

BVp(I) becomes a Banach space if endowed with the following norm

‖f‖BVp(I) := νp(f) + supI|f | ∀f ∈ BVp(I) .

The space BVp(I) inherits many properties of Vp(I), such as the extension and restric-tion properties, and it is a Banach algebra.

10

4 Basic properties of BV 1p (I)

4.1 Definition and basic properties

To define the space BV 1p (I) on the interval I, we choose a point x0 ∈ I. If I = R or

[0,+∞[, we take x0 = 0. Then we have the following.

Definition 2 We shall say that a function f : I → R, belongs to BV 1p (I) if there exist

α ∈ R and g ∈ BVp(I) such that

f(x) = α+

∫ x

x0

g(t) dt ∀x ∈ I . (14)

If (14) holds, then f is Lipschitz continuous, the derivative f ′ exists almost everywherein I, and f ′(x) = g(x) a.e.. According to Proposition 7, there exists a unique normalizedfunction g ∈ Vp(I) such that (14) holds. Each time when we consider a functionf ∈ BV 1

p (I), we will denote by f ′ the normalized representative of f ′ in Vp(I). Weendow BV 1

p (I) with the norm

‖f‖BV 1p (I) := |f(x0)|+ ‖f ′‖BVp(I) ,

for which BV 1p (I) becomes a Banach space. It is immediate to verify that choosing

another point x0 ∈ I, we obtain an equivalent norm in BV 1p (I).

Proposition 8 If f ∈ BV 1p (I), then f can be extended in a unique way to a function

in BV 1p (I) with the same norm of f in BV 1

p (I).

Proof If f ∈ BV 1p (I), then f is uniformly continuous. Hence, the existence and

uniqueness of the extension of f to I follows. Assume for instance that I = [a, b[, withb <∞. We extend f ′ as a function g on [a, b] by taking g(b) = limx→b− f

′(x). Clearly,g ∈ Vp([a, b]) with the same norm of f ′, and the function

x 7→ f(x0) +

∫ x

x0

g(t)dt , ∀x ∈ [a, b] ,

coincides with the extension of f to I.

Remark 2 By the above Proposition, and by exploiting obvious affine transforma-tions, the study of BV 1

p (I) spaces can be reduced to that of the following three cases:I = R, I = [0,+∞[, I = [0, 1].

4.2 The behaviour near infinity

Proposition 9 Let f0 ∈ BV 1p (R) and limx→+∞ f ′0(x) 6= 0. Then there exists a neigh-

borhood V of f0 in BV 1p (R) such that f(x) → ±∞ as x → +∞, uniformly on f ∈ V ,

where ± stands for the sign of limx→+∞ f ′0(x).

11

Proof Assume that H := limx→+∞ f ′0(x) > 0. Let R > 0 be such that

f ′0 ([R,+∞[) ⊆ [2H/3,+∞[ .

If f ∈ BV 1p (R) and ‖f − f0‖BV 1

p (R) ≤ H/3, then we have

f(x) ≥ f(0) +

∫ R

0

f ′(t) dt+H

3(x−R) ≥ H

3x+ f0(R)− H

3(1 + 2R) ,

for every x ≥ R. Hence, the proof is complete.

4.3 Nonhomogeneous BV 1p spaces

The homogeneity property

‖f(λ(.))′‖BVp = λ‖f ′‖BVp ∀λ > 0

holds for every function f in BV 1p (R) or in BV 1

p ([0,+∞[). Thus the spaces BV 1p (R)

and BV 1p ([0,+∞[) can be seen as analoguous to a homogeneous Sobolev space, for

instance to the spaceW 1,p(R) := {f : f ′ ∈ Lp(R)} ,

endowed with the seminorm ‖f ′‖p. Exactly as for W 1,p — whose nonhomogeneouscounterpart is the usual Sobolev space W 1,p(R) := W 1,p(R) ∩ Lp(R) — we can considerthe space BV 1

p (I) ∩ Lp(I) and endow it with the norm

‖f‖BV 1p (I) + ‖f‖Lp(I) .

If I is bounded, then BV 1p (I) ∩ Lp(I) = BV 1

p (I) with equivalent norms.

If I is unbounded, then the set BV 1p (I) is not contained in BVp(I) (for instance the

function f(x) = x belongs to BV 1p (I) but does not belong to BVp(I), since it is un-

bounded). This difficulty disappears in the nonhomogeneous space BV 1p (I) ∩ Lp(I).

Indeed, the following holds.

Proposition 10 BV 1p (I) ∩ Lp(I) is continuously imbedded into BVp(I).

Proof We divide our proof into three steps.Step 1. We first assume that I is bounded. If f ∈ BV 1

p (I), then f is Lipschitzcontinuous. Hence,(

N∑j=1

|f(xj)− f(xj−1)|p)1/p

≤N∑j=1

|f(xj)− f(xj−1)| ≤ ‖f ′‖∞ |I| ,

for each finite increasing sequence {xj}Nj=0 in I. Thus we have νp(f, I) ≤ |I| ‖f‖BV 1p (I).

Step 2. Assume I = R. By Peetre’s theorem ([9, Thm. 5], Peetre [17]) we have thefollowing continuous imbeddings:

BV 1p (R) ∩ Lp(R) ↪→ B1+(1/p)

p,∞ (R) ↪→ B1/pp,1 (R) ↪→ BVp(R) .

Step 3. In case I = [0,+∞[, we use an extension to R — see Subsection 4.5.

12

4.4 Multiplication properties

By combining the Leibniz formula and Propositions 4 and 10, we obtain the following.

Theorem 7 BV 1p (I) ∩ Lp(I) is a Banach algebra.

Remark 3 If I is bounded, then Theorem 7 implies that BV 1p (I) is a Banach algebra.

If I is unbounded, BV 1p (I) is not an algebra. Indeed, the function x 7→ x belongs to

BV 1p (I), but the function x 7→ x2 does not. However, a partial result holds in BV 1

p (R).

Proposition 11 Let I be a compact interval of R. If f, g ∈ BV 1p (R) and g has support

in I, then fg ∈ BV 1p (R). Furthermore, there exists c > 0 such that

‖fg‖BV 1p (R) ≤ c‖f‖BV 1

p (R) ‖g‖BV 1p (R) ∀f, g ∈ BV 1

p (R), supp g ⊆ I .

Proof Since the support of g is compact, we have g ∈ Lp(R). Then by Propositions 4and 10 there exists c′ > 0 such that

‖f ′g‖BVp(R) ≤ ‖f ′‖BVp(R)‖g‖BVp(R) ≤ c′‖f ′‖BVp(R)(‖g‖BV 1p (R) + ‖g‖Lp(R)) ,

for all f, g ∈ BV 1p (R), with supp g ⊆ I. Moreover, we have

‖g‖Lp(R) ≤ |I|1/p(|g(0)|+ |I|‖g′‖L∞(R)) ≤ |I|1/p‖g‖BV 1p (R)(1 + |I|) .

Since fg′ vanishes outside of I, Remark 1, Propositions 1, 4 and 10 imply that thereexist c1, c2 > 0 depending on I such that,

‖fg′‖BVp(R) ≤ 2‖fg′‖BVp(I) ≤ 2‖f‖BVp(I)‖g′‖BVp(I) ,

‖f‖BVp(I) ≤ c1

(‖f‖BV 1

p (I) + ‖f‖Lp(I)

)≤ c2‖f‖BV 1

p (R)

for all functions f and g as in the statement. Thus the proof is complete.

4.5 Restriction and extensions

Let J be a subinterval of I. The restriction mapping f 7→ f |J is easily seen to bebounded and linear from BV 1

p (I) to BV 1p (J).

There exist various extension operators from BV 1p (J) to BV 1

p (I). They can be obtainedby the following Proposition, which is an immediate consequence of Proposition 1.

Proposition 12 Let a ∈◦I and I1 :=]−∞, a] ∩ I, I2 := [a,+∞[∩I. Let

Φ := {(g1, g2) ∈ BV 1p (I1)×BV 1

p (I2) : g1(a) = g2(a)} .

Φ is a Banach subspace of the space BV 1p (I1)×BV 1

p (I2), and the mapping

U : g 7→ (g|I1 , g|I2)

is a Banach space isomorphism of BV 1p (I) onto Φ.

13

We now present two applications of Proposition 12.

• If a, b ∈ R, a < b, f ∈ BV 1p ([a, b]), then we can extend f to R by setting

E1f(x) := f(a) for x < a , E1f(x) := f(b) for x > b .

• If a ∈ R, a convenient way to extend f ∈ BV 1p ([a,+∞[) to the real line consists

in using the symmetry with respect to a, that is to define

E2f(x) := f(2a− x) for x < a .

By Proposition 12 it follows that E1 and E2 are bounded as operators from BV 1p ([a, b])

and BV 1p ([a,+∞[) to BV 1

p (R), respectively.

We observe that E2 is also a bounded operator from BV 1p ([a,+∞[) ∩ Lp([a,+∞[) to

BV 1p (R) ∩ Lp(R), and that the same applies if we replace Lp by L∞. The use of E2

allows to prove Proposition 10 in case of [0,+∞[, a point left open in the proof ofProposition 10. Finally, we can define a bounded extension operator

E3 : BV 1p ([a, b]) → BV 1

p (R) ∩ Lp(R)

by taking ϕ ∈ D(R) such that ϕ(x) = 1 on [a, b] and by setting E3f := (E1f)ϕ.Boundedness of E3 follows by that of E1 and by Proposition 11.

Remark 4 If I is a compact interval of R, then there exists c > 0 such that

‖f‖BV 1p (R) ≤ c ‖f |I‖BV 1

p (I)

for all continuous functions f of R to itself such that supp f ⊆ I and f |I ∈ BV 1p (I).

Indeed, f = E1 (f |I).

5 Regular functions in BVp(R) and BV 1p (R)

We now characterize the classes Ep and E1p defined in Section 2 (just before Theorem

2.) Let f : R → R be a function, ε ∈]0,+∞]. We denote by wp,ε(f) the supremum ofthe sums ∑

k

|f(bk)− f(ak)|p ,

for all finite families of disjoint open intervals ]ak, bk[, such that bk − ak < ε for everyk. Then wp,∞(f) = νpp(f) and the function ε 7→ wp,ε(f) is obviously increasing. We set

wp(f) := limε→0

wp,ε(f) = infε>0

wp,ε(f) .

It is straightforward to verify that w1/pp is a continuous seminorm on Vp. The functional

wp has been introduced by Wiener [23] with the notation V (p), which we prefer to avoidhere.

Proposition 13 Let p ∈]1,+∞[, f ∈ BVp(R), and f the normalized representative off . Then the following four properties are equivalent:

14

(i) f belongs to Ep.

(ii) wp(f) = 0.

(iii) limx→0 ‖τxf − f‖BVp(R) = 0.

(iv) f ∈ clBVp(R)C∞[BVp(R)].

Proof We divide our proof into three steps.Step 1. We first assume that f ∈ Ep and prove that (ii) holds. By continuity of the

seminorm w1/pp on Vp, it suffices to prove that wp(f) = 0 for all f ∈ C∞(R) ∩ Vp. Let

ε > 0. By Proposition 2, there exist a, b ∈ R, a < b, such that

νpp(f, ]−∞, a]) + νpp(f, [b,+∞[) ≤ ε/2 . (15)

Let M be an upper bound for |f ′| on [a− 1, b+ 1]. Let η ∈]0, 1] be such that

Mpηp−1(b− a+ 2) ≤ ε/2 . (16)

Let x0 < x1 < · · · < xN be a finite sequence such that xk − xk−1 < η for all k. Let

J := {k ∈ {1, . . . , N} : {xk−1, xk} ∩ [a, b] 6= ∅} .

Clearly, if k ∈ J , then {xk−1, xk} ⊆ [a− 1, b+ 1] and we have

N∑k=1

|f(xk)− f(xk−1)|p ≤∑k∈J

|f(xk)− f(xk−1)|p +ε

2≤ ε ,

by (15) and (16). Thus we obtain wp(f) = 0.

Step 2. We now assume that wp(f) = 0, and we prove that (iii) holds. Let ε > 0.By assumption, there exists η > 0 such that wp,η(f) ≤ ε. Let 0 < y < η and letx0 < x1 < · · · < xN be any finite sequence in R. We set

A := {k : xk − xk−1 ≤ y} , B := {k : xk − xk−1 > y} .

Then∑k∈A

|f(xk − y)− f(xk−1 − y)− f(xk) + f(xk−1)|p

≤ 2p−1

(∑k∈A

|f(xk − y)− f(xk−1 − y)|p +∑k∈A

|f(xk)− f(xk−1)|p).

Since the intervals ]xk−1 − y, xk − y[ (k ∈ A) are disjoint and have length less than η,the above expression is at most 2pε. On the other hand∑k∈B

|f(xk − y)− f(xk−1 − y)− f(xk) + f(xk−1)|p

≤ 2p−1

(∑k∈B

|f(xk − y)− f(xk)|p +∑k∈B

|f(xk−1 − y)− f(xk−1)|p).

15

By the definition of B, any two distinct intervals in the family {]xk−y, xk[}k∈B, or in thefamily {]xk−1 − y, xk−1[}k∈B are disjoint. Since the length of each interval in such twofamilies is less than η, then the above expression is at most 2pε. Thus we have provedthat lim

y→0+νp(τyf − f) = 0. Arguing similarly, we obtain lim

y→0−νp(τyf − f) = 0. Since

equality wp(f) = 0 implies the uniform continuity of f , we have limy→0

supR|τyf − f | = 0.

We conclude thatlimy→0

‖τyf − f‖Vp = 0 .

Step 3. By Proposition 18 of the Appendix, (iv) follows by (iii). Finally, (i) is anobvious consequence of (iv).

Remark 5 Proposition 13 remains true for p = 1, if we replace condition (ii) bycondition(ii’) the distributional derivative of f is in L1(R).Thus E1 turns out to be the set of distributions f such that f ′ ∈ L1(R).

Now the characterization of Ep by means of condition (iii) of Proposition 13 has acounterpart in case of E1

p . Namely, we have the following.

Proposition 14 A function f ∈ BV 1p (R) belongs to E1

p if and only if

limx→0

‖τxf − f‖BV 1p (R) = 0 . (17)

Proof We divide our proof into two steps.Step 1. We first assume that (17) holds. Let ε > 0. Since limx→0 ‖τxf ′ − f ′‖BVp(R) =0, then Proposition 13 implies that there exists g ∈ BVp(R) ∩ C∞(R) such that‖f ′ − g‖BVp(R) ≤ ε. Now we set

h(x) := f(0) +

∫ x

0

g(t) dt ∀x ∈ R .

Then h ∈ BV 1p (R) ∩ C∞(R) and

‖h− f‖BV 1p (R) = ‖g − f ′‖BVp(R) ≤ ε .

Step 2. We now assume that f = limk→∞ fk in BV 1p (R), with fk ∈ BV 1

p (R) ∩ C∞(R).Then f ′ = limk→∞ f ′k in BVp(R). Since f ′k is C∞, Proposition 13 implies that

‖τxf ′ − f ′‖BVp(R) → 0 .

Moreover, the continuity of f at 0 implies that |τxf(0) − f(0)| = |f(−x) − f(0)| → 0as x→ 0.

Remark 6 According to Remark 5, E11 is the set of functions whose second distribu-

tional derivative belongs to L1(R).

16

We now introduce the local versions of Ep and E1p . If F is any distribution space on Rn,

we say that f ∈ Floc if fϕ ∈ F for every ϕ ∈ D(Rn). We first mention the followingresult, which is of immediate verification (see also Propositions 4 and 11.)

Proposition 15 A function f belongs to BVp,loc(R) and to BV 1p,loc(R) if and only if

f |I belongs to BVp(I) and to BV 1p (I) for every compact interval I of R, respectively.

Then we have the following.

Proposition 16 Let f ∈ BVp,loc(R) (resp. BV 1p,loc(R)). Then f ∈ Ep,loc (resp. E1

p,loc)if and only if τxf → f in BVp(I) (resp. BV 1

p (I)) as x→ 0, for every compact intervalI of R.

Proof We divide our proof into four steps.Step 1. We first assume that f ∈ Ep,loc. Let a > 0. Let ϕ ∈ D(R) be such that ϕ(x) = 1for |x| ≤ a+ 1. By assumption and by Proposition 13, we have

‖τx(fϕ)− fϕ‖BVp(R) → 0 as x→ 0 .

Hence,‖τx(fϕ)− fϕ‖BVp([−a,a]) → 0 as x→ 0 .

Now we observe that τx(fϕ)(y) = τxf(y), for all y ∈ [−a, a] and |x| ≤ 1. We concludethat

‖τxf − f‖BVp([−a,a]) → 0 as x→ 0 . (18)

Step 2. We now assume that (18) holds for every a > 0 and we take ϕ ∈ D(R). Thenwe choose a > 0 such that supp τxϕ ⊆ [−a, a] for all x ∈ [−1, 1]. By inequality (13)and by Remark 1, there exists c > 0 such that

‖τx(fϕ)− fϕ‖BVp(R) ≤ ‖(τxf − f)τxϕ‖BVp(R) + ‖(τxϕ− ϕ)f‖BVp(R)

≤ c(‖τxf − f‖BVp([−a,a])‖ϕ‖BVp(R) + ‖f‖BVp([−a,a])‖τxϕ− ϕ‖BVp(R)

),

for all x ∈ [−1, 1]. By Proposition 13 we have

‖τxϕ− ϕ‖BVp(R) → 0 ,

and accordingly fϕ ∈ Ep.

Step 3. We now assume that f ∈ E1p,loc. Arguing as in Step 1, we obtain

‖τxf − f‖BV 1p ([−a,a]) → 0 as x→ 0 . (19)

for all a > 0.

Step 4. We now assume that (19) holds for every a > 0. We argue as in Step 2. ByRemarks 3 and 4, there exists c > 0 such that

‖τx(fϕ)− fϕ‖BV 1p (R)

≤ c(‖ϕ‖BV 1

p (R)‖τxf − f‖BV 1p ([−a,a]) + ‖f‖BV 1

p ([−a,a])‖τxϕ− ϕ‖BV 1p (R)

),

for all x ∈ [−1, 1]. Thus we conclude that fϕ ∈ E1p by Proposition 14.

17

6 Acting conditions

We now prove Theorem 1.

Sufficiency of the conditions follows by [9, Thm. 4] (see also Proposition 10 and Sub-section 4.5.) Necessity in case of BV 1

p (R) is immediate, since we can test Tf on thefunction idR, which belongs to BV 1

p (R).

Now we prove the necessity of condition f ∈ BV 1p,loc(R). By an affine transformation, we

can assume that [−1, 1] ⊆ I. By using extension from BV 1p ([−1, 1]) to BV 1

p (I)∩Lp(I),we obtain that Tf takes BV 1

p ([−1, 1]) to itself. Let a > 0. By testing Tf on thefunction x 7→ ax, which belongs to BV 1

p ([−1, 1]), we see that x 7→ f(ax) belongs toBV 1

p ([−1, 1]). By a dilation argument, we conclude that the restriction of f belongs toBV 1

p ([−a, a]).

The necessity of condition f ∈ BV 1p (R), in case of BV 1

p ([0,+∞[), is a consequence ofthe following lemma.

Lemma 1 If Tf takes BV 1p ([0,+∞[) to itself, then Tf takes BV 1

p (R) to itself.

Proof By a symmetry argument, Tf takes BV 1p (] −∞, 0]) to itself. Then we apply

Proposition 12 to intervals ] − ∞, 0] and [0,+∞[. If g ∈ BV 1p (R), then we have

(g1, g2) := U(g) ∈ Φ and accordingly (f ◦ g1, f ◦ g2) ∈ Φ. Hence,

f ◦ g = U−1(f ◦ g1, f ◦ g2) ∈ BV 1p (R) ,

and thus the proof is complete.

7 Continuity conditions

7.1 A preliminary continuity property

Sufficient conditions for continuity of Tf rely upon the following preliminary result.

Proposition 17 If ϕ ∈ C3(R) ∩ BV 1p (R), then the superposition operator Tϕ is con-

tinuous from BV 1p (R) to itself.

Proof Let g0 ∈ BV 1p (R). We first prove the continuity at g0 of the mapping g 7→

(ϕ′ ◦ g)g′ from the normed space BV 1p (R) to Vp(R), endowed with the seminorm νp.

Then the result is a consequence of the following three elementary remarks.

(i) supR|g| ≤ |g(0)|+ νp(g,R), for all g ∈ Vp(R),

(ii) g 7→ ϕ′(g(0))g′(0) is a continuous map of BV 1p (R) to R,

(iii) g 7→ ϕ(g(0)) is a continuous map of BV 1p (R) to R.

18

Step 1. We first prove that if I is a compact interval, then there exists a constant cdepending only on I, g0, ϕ, such that the inequality

νp ((ϕ′ ◦ g)g′ − (ϕ′ ◦ g0)g′0, I) ≤ c‖g − g0‖BV 1

p (R) , (20)

holds for all g ∈ BV 1p (R) such that ‖g − g0‖BV 1

p (R) ≤ 1. Since BV 1p (I) is continuously

imbedded in C(I), there is a compact interval J such that g(I) ⊆ J whenever

‖g − g0‖BV 1p (R) ≤ 1 .

Since ϕ′ ◦ g0 is Lipschitz continuous on I, then ϕ′ ◦ g0 ∈ Vp(I). Hence, Proposition 4implies that

νp ((ϕ′ ◦ g)g′ − (ϕ′ ◦ g0)g′0, I) (21)

≤ ‖ϕ′ ◦ g − ϕ′ ◦ g0‖Vp(I) ‖g′‖Vp(I) + ‖ϕ′ ◦ g0‖Vp(I) ‖g′ − g′0‖Vp(I) .

Then it suffices to estimate ‖ϕ′ ◦ g − ϕ′ ◦ g0‖Vp(I). Clearly, we have

(ϕ′ ◦ g)(x)− (ϕ′ ◦ g0)(x) =

∫ 1

0

ϕ′′(g0(x) + t(g(x)− g0(x)))(g(x)− g0(x)) dt ∀x ∈ R .

By the definition of p-variation and by the Minkowski inequality, we obtain

νp (ϕ′ ◦ g − ϕ′ ◦ g0, I) ≤∫ 1

0

νp (ϕ′′ ◦ (g0 + t(g − g0))(g − g0), I) dt . (22)

Then by Proposition 4, we conclude that

νp (ϕ′′ ◦ (g0 + t(g − g0))(g − g0), I) ≤ ‖ϕ′′′|J‖∞(1 + ‖g0‖Vp(I))‖g − g0‖Vp(I) , (23)

for all t ∈ [0, 1]. Then by Proposition 10, and by inequalities (21), (22), (23), andby the obvious inequality ‖ϕ′ ◦ g − ϕ′ ◦ g0‖L∞(I) ≤ ‖ϕ′′|J‖∞‖g − g0‖L∞(I), we obtaininequality (20).

Step 2. Now we prove that

νp ((ϕ′ ◦ g)g′, [R,+∞[) + νp ((ϕ′ ◦ g)g′, ]−∞,−R]) → 0 , (24)

as the couple (g,R) tends to (g0,+∞) in BV 1p (R)× [0,+∞].

We consider for instance νp ((ϕ′ ◦ g)g′, [R,+∞[). By [9, Thm. 3], we have

νp ((ϕ′ ◦ g)g′, [R,+∞[) (25)

≤ νp (ϕ′, g([R,+∞[)) ( sup[R,+∞[

|g′|) + 21/p‖ϕ‖BV 1p (R) νp (g′, [R,+∞[) .

By Proposition 2, we know that νp (g′, [R,+∞[) tends to zero as g → g0 in BV 1p (R)

and R→ +∞. We now turn to consider the first term in the right hand side of (25).

1- If limx→+∞ g′0(x) = 0, then we have

νp (ϕ′, g([R,+∞[)) ( sup[R,+∞[

|g′|) ≤ ‖ϕ‖BV 1p (R) ( sup

[R,+∞[

|g′|) .

19

Then by Proposition 3, we know that sup[R,+∞[ |g′| tends to zero as (g,R) tends to(g0,+∞).

2- Now let limx→+∞ g′0(x) 6= 0. We consider for example the case in which such limitis > 0. By Proposition 9, we consider a ball B of center g0 and radius ≤ 1, such thatg(x) → +∞ as x → +∞, uniformly for g ∈ B. Let ε > 0. By Proposition 2, thereexists a number M such that νp (ϕ′, [M,+∞[) ≤ ε. By the above choice of B, thereexists a number R such that g([R,+∞[) ⊆ [M,+∞[ for any g ∈ B. For such functionsg, we have

νp (ϕ′, g([R,+∞[)) ( sup[R,+∞[

|g′|) ≤ (supR|g′|) νp (ϕ′, g([R,+∞[)) ≤

(1 + (sup

R|g′0|)

)ε .

Thus we have proved property (24). By inequality (20) and by the limiting relation(24), we deduce the continuity of (ϕ′ ◦ g)g′ in the variable g, and thus the proof ofProposition 17 is complete.

7.2 Proof of Theorem 3

By exploiting [9, Thm. 4] and Proposition 17 we are now ready to prove our mainsufficient condition for the continuity of Tf . We mention that we exploit ideas of [13,p. 928].For simplicity, we denote by ‖ − ‖ the norm in BV 1

p (R). Let (f ], g]) ∈ E1p × BV 1

p (R).Let ε > 0 be given arbitrarily.By assumption, there exists ϕ ∈ C∞(R) ∩ BV 1

p (R) such that ‖f ] − ϕ‖ ≤ ε. ByProposition 17, the map Tϕ is continuous from BV 1

p (R) to itself. Then there existsδ ∈]0, 1] such that ∥∥ϕ ◦ g − ϕ ◦ g]

∥∥ ≤ ε if∥∥g − g]

∥∥ ≤ δ .

Hence, if (f, g) ∈ BV 1p (R) × BV 1

p (R) with∥∥g − g]

∥∥ ≤ δ and∥∥f − f ]

∥∥ ≤ ε, then [9,Thm. 4] implies that

‖f ◦ g − f ] ◦ g]‖≤ ‖f − ϕ‖

(1 + 21/p‖g‖

)+∥∥ϕ ◦ g − ϕ ◦ g]

∥∥+ ‖ϕ− f ]‖(1 + 21/p‖g]‖

)≤ 2ε

(1 + 21/p + 21/p‖g]‖

)+ ε+ ε

(1 + 21/p‖g]‖

),

and thus the proof is complete.

7.3 Proof of Theorem 2

We are now able to prove a characterization of all continuous superposition operatorson BV 1

p spaces. We divide our argument into seven steps.

Step 1. Let f ∈ E1p . By Theorem 3, Tf is continuous from BV 1

p (R) to itself. Byexploiting extension and restriction operators, Tf can be shown to be continuous fromBV 1

p (I) to itself.

20

Step 2. We now assume that f ∈ E1p,loc and prove that Tf is continuous from BV 1

p (I)∩L∞(I) to BV 1

p (I). Of course it suffices to prove continuity of Tf on any ball B ofBV 1

p (I)∩L∞(I). We observe that there exists a > 0 such that g(I) ⊆ [−a, a] for everyg ∈ B. Let ϕ ∈ D(R) be such that ϕ(x) = 1 on [−a, a]. Then f ◦ g = (fϕ) ◦ g for everyg ∈ B and we are reduced to Step 1. By Proposition 10, BV 1

p (I)∩Lp(I) is continuouslyimbedded in BV 1

p (I) ∩ L∞(I). Then condition f ∈ E1p,loc also implies continuity of Tf

on BV 1p (I) ∩ Lp(I).

Step 3. We now assume that f ∈ E1p,loc, f(0) = 0 and prove that Tf is continuous

from BV 1p (I) ∩ Lp(I) to itself. The case of BV 1

p (I) ∩ L∞(I) can be treated similarly.By Step 2 it suffices to show the continuity of Tf from BV 1

p (I) ∩ Lp(I) to Lp(I). Letg0 ∈ BV 1

p (I). By Proposition 10, BV 1p (I) ∩ Lp(I) is continuously imbedded in L∞(I).

Thus there exists a compact interval J such that g(I) ⊆ J for all g ∈ BV 1p (I) ∩ Lp(I)

such that ‖g − g0‖ ≤ 1. Since f is Lipschitz continuous on J , we have∫J

|f ◦ g − f ◦ g0|p dx ≤ Lipp(f |J)∫J

|g − g0|p dx ∀g ∈ BV 1p (I) ∩ Lp(I), g(I) ⊆ J .

Hence, Tf is continuous at g0.

Step 4. We now assume that Tf is continuous from BV 1p (R) to itself. In particular Tf

is continuous at the identity function in R, which means that

limy→0

τyf = f in BV 1p (R) . (26)

Hence f ∈ E1p by Proposition 14.

Step 5. We now assume that Tf is continuous from BV 1p ([0,∞[) to itself. By Proposi-

tion 12, we see that Tf is continuous also from BV 1p (R) to itself. Thus we are reduced

to the previous step, and we can conclude that f ∈ E1p (see also Remark 2.)

Step 6. We now assume that Tf is continuous from BV 1p (I)∩Lp(I) to BV 1

p (I). Arguingas in the proof of Theorem 1, we can assume that Tf is continuous from BV 1

p ([−1, 1])to itself. Let a > 0 and ga(x) := ax for |x| ≤ 1. Since ga + y → ga in BV 1

p ([−1, 1]) asy → 0, we have

f ◦ (ga + y) → f ◦ ga in BV 1p ([−1, 1]) .

By a dilation argument, we conclude that τyf → f in BV 1p ([−a, a]). Then by Propo-

sition 16, we have f ∈ E1p,`oc. The case of BV 1

p (I) ∩ L∞(I) can be treated similarly.

Step 7. If Tf is continuous from BV 1p (I) ∩ Lp(I) to itself, then Tf is continuous from

BV 1p (I)∩Lp(I) to BV 1

p (I) and we can conclude by Step 6. The case of BV 1p (I)∩L∞(I)

can be treated similarly.

21

8 A degeneracy result for uniform continuity

We shall prove that — except for affine functions f — the superposition operator Tfis not uniformly continuous from BV 1

p (I) ∩ Lp(I) to BV 1p (I) (cf. Theorem 4.) In fact,

a stronger result holds. To formulate it, we introduce the generalized Holder space offunctions having a given modulus of continuity.Let ω be a strictly increasing continuous function on [0,+∞[, such that ω(0) = 0. Wedefine Cω(I) to be the set of real valued functions f defined on I such that

‖f‖Cω(I) := supx,y∈I,x<y

|f(x)− f(y)|ω(y − x)

<∞ .

We denote by Cω,loc(I) the set of functions f , defined on I, such that f |J ∈ Cω(J),for any compact subinterval J of I. We endow Cω,loc(I) with the family of seminorms‖f |J‖Cω(J), for all compact subintervals J of I.

Theorem 8 Let f : R → R be a function such that Tf is an uniformly continuousmapping from BV 1

p (I) ∩ Lp(I) to Cω,loc(I). Then f is an affine function.

Proof We can assume that [0, 1] ⊆ I. By exploiting restriction and extension oper-ators, we conclude that Tf is an uniformly continuous mapping from BV 1

p ([0, 1]) toCω([0, 1]). By evaluating Tf on affine functions, we see that f is continuous. Withoutloss of generality, we can assume f(0) = 0. By assumption there exists η > 0 such that

g, h ∈ BV 1p ([0, 1]), ‖g − h‖BV 1

p ([0,1]) ≤ η ⇒ ‖f ◦ g − f ◦ h‖Cω([0,1]) ≤ 1 .

By taking h(x) := ax and g(x) := ax+ b for x ∈ [0, 1], we obtain

|f(ax+ b)− f(ax)− f(ay + b) + f(ay)| ≤ ω(x− y) , (27)

for all 0 ≤ y < x ≤ 1, |b| ≤ η and a ∈ R.Let u be a nonzero real number. Let a ∈ R \ {0} be such that x := u/a ∈ [0, 1]. Theninequality (27) implies that

|f(u+ b)− f(u)− f(b)| ≤ ω(u/a) ∀b ∈ [−η, η] .

Taking the limit as |a| → +∞, we obtain

f(u+ b) = f(u) + f(b) ∀u ∈ R \ {0} , ∀b ∈ [−η, η] .

Then a standard argument implies that f(x) = f(1)x for all x ∈ R.

9 Differentiability

This section will be devoted to the proof of Theorem 5. We first prove statement (i).Arguing as in the proof of Lemma 1, we can assume that I = R. Our first goal is tocompute dTf (0).v for all v ∈ BV 1

p (R). Since convergence in BV 1p (R) implies pointwise

convergence in R, we have

limt→0

f(tv)− f(0)

t= dTf (0).v ∀v ∈ BV 1

p (R) , (28)

22

pointwise in R. Then the existence of f ′(0) follows by taking v = 1 in (28). Then againby (28) we deduce that

dTf (0).v = f ′(0)v ∀v ∈ BV 1p (R) . (29)

We now set F (x) := f(x)− f(0)− f ′(0)x. Since TF is differentiable at 0, equality (29)applied to F implies that

dTF (0).v = 0 ∀v ∈ BV 1p (R) .

Then by equality F (0) = 0, we obtain

limv→0

‖F ◦ v‖BV 1p (R)

‖v‖BV 1p (R)

= 0.

Now we choose va(x) := ax, with a ∈ R. Clearly, ‖va‖BV 1p (R) = |a|. Then we have

lima→0

νp (a(F ′ ◦ va))|a|

= 0 . (30)

However, it is immediate to verify that

νp(F′ ◦ va) = νp(F

′) ∀a ∈ R \ {0} .By (30), we deduce that νp(F

′) = 0, and thus F ′ must be constant.

We now prove statement (ii). To begin with, we consider the implication (c) ⇒ (a).For simplicity, we denote by ‖ − ‖ the norm of BV 1

p (I) ∩ Lp(I). By condition (c), f isof class C1. Hence, we have the following identity:

f(x+ y) = f(x) + yf ′(x) +

∫ 1

0

(f ′(x+ ty)− f ′(x)) y dt ∀x, y ∈ R .

Thus

f ◦ (g + h) = f ◦ g + h(f ′ ◦ g) +

∫ 1

0

(f ′ ◦ (g + th)− f ′ ◦ g) h dt , (31)

for every g, h ∈ BV 1p (I) ∩ Lp(I). By Theorems 2 and 7, the function

t 7→ (f ′ ◦ (g + th))h− (f ′ ◦ g)his continuous from [0, 1] to BV 1

p (I) ∩ Lp(I), and thus we can interpret the integral inthe right hand side of (31) as a vector valued Riemann integral and obtain

‖f ◦ (g + h)− f ◦ g − h(f ′ ◦ g)‖ ≤ c

(supt∈[0,1]

‖f ′ ◦ (g + th)− f ′ ◦ g‖

)‖h‖ .

Since Tf ′−f ′(0) is a continuous mapping from BV 1p (I) ∩ Lp(I) to itself, the above in-

equality shows that Tf is of class C1 of BV 1p (I) ∩ Lp(I) to itself, and that

dTf (g).h = (f ′ ◦ g)h . (32)

Implication (a) ⇒ (b) is obvious. Finally, we prove the implication (b) ⇒ (c). We usean argument of [12]. By exploiting extension and restriction operators, we can assumethat Tf is a differentiable mapping from BV 1

p ([0, 1]) to itself. Arguing as in the proofof (29), we obtain formula (32). Taking h = 1 in formula (32), we obtain that Tf ′ iscontinuous from BV 1

p ([0, 1]) to itself, and thus f ′ ∈ E1p,loc by Theorem 2 (i).

23

10 Superposition operators and Besov spaces

10.1 Besov spaces

To introduce Besov spaces, we use the characterization by differences. We set

∆hf(x) := f(x+ h)− f(x) ∀h, x ∈ Rn . (33)

Let p, q ∈ [1,+∞], M ∈ N \ {0}. If 0 < s < M , then the Besov space Bsp,q(Rn) is the

set of all functions f ∈ Lp(Rn) such that

‖f‖Bsp,q(Rn) := ‖ f ‖p +

(∫Rn

(1

|h|s

(∫Rn

|∆Mh f(x) |pdx

)1/p)q

dh

|h|n

)1/q

< +∞ ,

with the usual modifications in cases p = ∞ or q = ∞. Formally, this definition dependsonM . However, it is well-known that for differentM > s one obtains equivalent norms.For the theory of Besov spaces, we refer the reader to the monographs of Peetre [17]and of Triebel [21].

10.2 Preliminaries

To begin with, we prove a necessary condition for the continuity of a superpositionoperator in a general context. For each f : R → R, we set

f [(x) := f(x1) ∀x := (x1, x2, . . . , xn) ∈ Rn . (34)

Then the following holds.

Theorem 9 Let E be a translation invariant Banach Distribution Space (see Defini-tion 3 in the Appendix) satisfying the following condition.

(J) ϕf ∈ E for all ϕ ∈ D(Rn) and f ∈ E.

Let f : R → R be a Borel function such that Tf : D(Rn) → E is continuous. Then

limy→0

‖ τy(ϕf [)− ϕf [ ‖E = 0 (35)

for all ϕ ∈ D(Rn).

Proof We first observe that condition (J) and the Closed Graph Theorem imply that,for any compact subset K of Rn, there exist C > 0 and m ∈ N such that

‖ϕf ‖E ≤ C sup|α|≤m

‖ϕ(α) ‖∞ ‖ f ‖E (36)

for all ϕ ∈ D(Rn) with support contained in K and all f ∈ E.

Now we divide our proof into two steps.

24

Step 1. We show that f [ ∈ E`oc. Let ϕ ∈ D(Rn). Let g ∈ D(Rn) be such that g(x) = x1

on the support of ϕ. By exploiting inclusion Tf [D(Rn)] ⊆ E, we obtain f ◦ g ∈ E.Then condition (J) implies that the product (f ◦ g)ϕ = ϕf [ belongs to E.

Step 2. We now prove (35). Let ϕ ∈ D(Rn). Let

K :={x ∈ Rn : dist (x, suppϕ) ≤ 1

}and let g ∈ D(Rn) be such that g(x) = x1 on K. Then we have

(τy(f[)− f [)ϕ = (f ◦ τyg − f ◦ g)ϕ ,

for |y| ≤ 1. Let ψ ∈ D(Rn) be such that ψ(x) = 1 if dist (x,K) ≤ 1. By translationinvariance of E, and by (36), there exist C > 0 and m ∈ N such that

‖ τy(ϕf [)− ϕf [ ‖E ≤ ‖ (τyϕ− ϕ)τyf[ ‖E + ‖ (τyf

[ − f [)ϕ ‖E= ‖ (τyϕ− ϕ)τy(f

[ ψ) ‖E + ‖ (τyf[ − f [)ϕ ‖E

≤ C(

sup|α|≤m

‖ τyϕ(α) − ϕ(α) ‖∞ ‖ f [ ψ ‖E + ‖ f ◦ τyg − f ◦ g ‖E sup|α|≤m

‖ϕ(α) ‖∞),

for |y| ≤ 1. Hence, the continuity of Tf and the uniform continuity of ϕ(α) imply (35).

We now turn to the proof of Theorem 6.

10.3 Proof of Theorem 6

We first prove the sufficient conditions of the statement of Theorem 6, and we startwith statement (ii). For case n = 1 we actually have the following stronger statement.

Theorem 10 If f : R → R is a Borel measurable function of class(clU1

p (R)C∞[U1

p (R)])`oc

such that f(0) = 0, then Tf is continuous from B1+(1/p)p,1 (R) to B

1+(1/p)p,∞ (R).

Proof We must show that if g0 ∈ B1+(1/p)p,1 (R), then Tf is continuous at g0. Since

B1+(1/p)p,1 (R) is imbedded into L∞(R) (cf. e.g. [19, 4.6.4, Thm. 1, pp. 221–222]), we can

argue as in the proof of Theorem 2 and conclude that it suffices to prove the continuityof Tϕf at g0 for all ϕ ∈ D(R). By assumption, we have ϕf ∈ clU1

p (R)C∞[U1

p (R)].We now divide our proof into two steps.Step 1. We first consider the specific case in which f ∈ C∞[U1

p (R)]. By assumption,ϕf belongs to D(R). Then by Theorem 2, Tϕf is continuous from BV 1

p (R) ∩ Lp(R) to

itself. By Peetre’s Imbedding Theorem (cf. e.g. [9, Thm. 5], Peetre [17]), B1+(1/p)p,1 (R)

is continuously imbedded in BV 1p (R) ∩ Lp(R), and BV 1

p (R) ∩ Lp(R) is continuously

imbedded in B1+(1/p)p,∞ (R). Thus we deduce the continuity of Tϕf . Then by the above

remark, we also have the continuity of Tf .

25

Step 2. We now consider the general case and we show continuity of Tϕf . We argueexactly as in the proof of Theorem 3. Let ψ be an element of C∞[U1

p (R)]. By [9,Thm. 9] there exists c > 0 such that

‖h◦g‖B

1+(1/p)p,∞ (R)

≤ c‖h‖U1p (R)‖g‖B1+(1/p)

p,1 (R)∀h ∈ U1

p (R), h(0) = 0, ∀g ∈ B1+(1/p)p,1 (R) .

(37)Then we have

‖(fϕ) ◦ g − (fϕ) ◦ g0‖B1+(1/p)p,∞ (R)

≤ c‖fϕ− ψ‖U1p (R)

(‖g‖

B1+(1/p)p,1 (R)

+ ‖g0‖B1+(1/p)p,1 (R)

)+ ‖ψ ◦ g − ψ ◦ g0‖B1+(1/p)

p,∞ (R)

By Step 1, and by assumption on f , the continuity of Tϕf at g0 follows.

Now we turn to consider the case n ≥ 2. To do so, we need the following two technicalLemmas.

Lemma 2 Let (X,µ) be a measure space. Let {fk}k∈N be a sequence of measurablefunctions of X to [0,∞[. If

limk→∞

∫X

fk dµ = 0 ,

then there exists a subsequence {fkj}j∈N such that

limj→∞

fkj= 0 a.e. in X, sup

j∈Nfkj

∈ L1(X,µ) .

For a proof of Lemma 2, we refer the reader to Rudin [18, Proof of Thm. 3.1]. Tointroduce the second Lemma, we need some notation. For all maps g of Rn to R,r = 1, . . . , n, x′ := (x1, . . . , xn−1) ∈ Rn−1, we define the map gr,x′ of R to R by settinggr,x′(y) = g(x1, . . . , xr−1, y, xr, . . . , xn−1) for all y ∈ R.Then we have the following.

Lemma 3 There exists a constant c > 0 such that(∫Rn−1

‖gr,x′‖pB

1+(1/p)p,1 (R)

dx′)1/p

≤ c‖g‖B

1+(1/p)p,1 (Rn)

∀g ∈ B1+(1/p)p,1 (Rn) ,

for all r ∈ {1, . . . , n}.

Proof We first assume that p > 1, and we denote by e1, . . . , en the canonical basis ofRn. By exploiting the Minkowski inequality and by choosing an equivalent norm onBesov spaces, we can assert that there exists a constant c > 0 such that(∫

Rn−1

‖gr,x′‖pB

1+(1/p)p,1 (R)

dx′)1/p

≤ c(∫

Rn−1

‖gr,x′‖pLp(R)dx′

+

∫Rn−1

(∫ 1

0

(∫R|∆2

h(gr,x′(y))|p dy)1/p dh

h2+(1/p)

)pdx′)1/p

≤ c[(∫

Rn

|g(x)|p dx)1/p

+

∫ 1

0

(∫Rn

|∆2herg(x)|p dx

)1/p dh

h2+(1/p)

]≤ c‖g‖

B1+(1/p)p,1 (Rn)

∀g ∈ B1+(1/p)p,1 (Rn) .

26

Case p = 1 can be treated similarly by replacing the differences of order 2 with differ-ences of order 3.

We are now ready for the following.

Proof of statement (ii) of Theorem 6 As in the previous proof, we assume thatp > 1. Indeed, case p = 1 can be treated similarly. It clearly suffices to show that if{gk}k∈N is a sequence in B

1+(1/p)p,1 (Rn) converging to some g ∈ B1+(1/p)

p,1 (Rn), then thereexists a subsequence {gkj

}j∈N such that

limj→∞

f ◦ gkj= f ◦ g in B1+(1/p)

p,∞ (Rn) .

Since f is Lipschitz continuous and f(0) = 0, we can argue as in Step 3 of the proof ofTheorem 2 and prove that the above limiting relation holds in Lp(Rn). Then it sufficesto prove that

limj→∞

sup0<h≤1

1

h1+(1/p)

(∫Rn

|∆2her

(f ◦ gkj− f ◦ g)(x)|p dx

)1/p

= 0 , (38)

for all r ∈ {1, . . . , n}. By Lemma 3 there exists c > 0 such that∫Rn−1

‖(gk)r,x′ − gr,x′‖pB

1+(1/p)p,1 (R)

dx′ ≤ cp‖gk − g‖pB

1+(1/p)p,1 (Rn)

∀k ∈ N.

By Lemma 2, there exists a subsequence {gkj}j∈N such that

limj→∞

‖(gkj)r,x′ − gr,x′‖p

B1+(1/p)p,1 (R)

= 0 for almost all x′ ∈ Rn−1 , (39)

and such that the function Gr of Rn−1 to R defined by

Gr(x′) := sup

j∈N‖(gkj

)r,x′ − gr,x′‖pB

1+(1/p)p,1 (R)

∀x′ ∈ Rn−1 ,

belongs to L1(Rn−1). By Theorem 10, we conclude that

limj→∞

‖f ◦ (gkj)r,x′ − f ◦ gr,x′‖p

B1+(1/p)p,∞ (R)

= 0 for almost all x′ ∈ Rn−1 . (40)

By inequality (37), there exists c > 0 such that

‖f ◦ (gkj)r,x′ − f ◦ gr,x′‖p

B1+(1/p)p,∞ (R)

≤ cp‖f‖pU1p (R)

(Gr(x

′) + ‖gr,x′‖pB

1+(1/p)p,1 (R)

)for almost all x′ ∈ Rn−1 .

By (39), and by Lemma 3, and by (40), and by the Dominated Convergence Theorem,we deduce that

limj→∞

(∫Rn−1

‖f ◦ (gkj)r,x′ − f ◦ gr,x′‖p

B1+(1/p)p,∞ (R)

dx′)1/p

= 0 .

27

Since the argument of the limiting relation in (38) is less or equal to(∫Rn−1

(sup

0<h≤1

1

h1+(1/p)

∫R|∆2

h

(f ◦ (gkj

)r,x′ − f ◦ gr,x′)(y)|p dy

)dx′)1/p

≤(∫

Rn−1

‖f ◦ (gkj)r,x′ − f ◦ gr,x′‖p

B1+(1/p)p,∞ (R)

dx′)1/p

,

we conclude that (38) holds.

We now turn to the proof of the remaining two statements.

Proof of statement (iii) of Theorem 6 We shall use nonlinear interpolation ofoperators following earlier works of Bona and Scott [2] and Maligranda [14]. Theseauthors have observed that real interpolation of nonlinear operators can be used totransfer continuity properties. Maligranda’s result [14, Thm. 2] can be stated asfollows. Let 0 < Θ < 1 and 1 ≤ q <∞. Let A0, A1, B0, B1 be Banach spaces such thatA1 ↪→ A0 and B1 ↪→ B0. Let T be an operator with the following properties: thereexists a constant c such that

‖Ta− Tb ‖B0 ≤ c ‖ a− b ‖A0 , ∀a, b ∈ A0 ;

‖Ta ‖B1 ≤ c ‖ a ‖A1 , ∀a ∈ A1 ;

and T : A1 → B1 is continuous. If in addition the pair (A0, A1) has an (Θ, q)-approximate identity, then T is continuous as a mapping from (A0, A1)Θ,q to (B0, B1)Θ,q.Hence, part (iii) follows from (i) and (ii), as long as we can show that the pairs

(A0, A1) := (Lp(Rn), B1+(1/p)p,1 (Rn)) have an (Θ, q)-approximate identity, and we take

Θ = s1+(1/p)

and (B0, B1) := (Lp(Rn), B1+(1/p)p,∞ (Rn)). For the existence of the approxi-

mate identity, we refer to [19, Lemma 2.5].

Proof of statement (i) of Theorem 6 Let n ≥ 2. Case n = 1 can be treated

similarly. We apply Theorem 9 with E = B1+(1/p)p,∞ (Rn). Let ϕ ∈ D(R) and ψ ∈

D(Rn−1), ϕ⊗ ψ the tensorproduct of ϕ and ψ. Then the continuity of Tf implies

limy→0

‖τy((ϕ⊗ ψ) f [

)− (ϕ⊗ ψ) f [ ‖

B1+(1/p)p,∞ (Rn)

= 0 ,

(cf. (34).) Let h = (h1, 0, . . . , 0) and y = (y1, 0, . . . , 0). Since

‖ τy((ϕ⊗ ψ) f [

)− (ϕ⊗ ψ) f [ ‖Lp(Rn) = ‖ψ ‖Lp(Rn−1) ‖ τy1(ϕf)− ϕf ‖Lp(R)

and ∫Rn

∣∣∣∆Mh

(τy

((ϕ⊗ ψ) f [

)− (ϕ⊗ ψ) f [

)∣∣∣p dx= ‖ψ ‖pLp(Rn−1)

∫R

∣∣∣∆Mh

(τy1(ϕf)− ϕf

)∣∣∣p dx1 ,

28

for M = 2 if p > 1, and with M = 3 if p = 1, we obtain

limy→0

‖τy(ϕf)− ϕf ‖B

1+(1/p)p,∞ (R)

= 0 .

Then statement (i) holds by Proposition 18 of the Appendix.

10.4 Proof of the properties of the absolute value

We now turn to prove the properties of the absolute value stated in (i), (ii), (iv) and(vi) of Section 2.1. Let f(t) := |t|.

We first prove statement (i). Let χ ∈ D(R), χ = 1 on [−1, 1]. Obviously,

‖τx(fχ)− fχ‖BV 1p (R) ≥ ‖τx(fχ)′ − (fχ)′‖BVp(R) ≥ 2 ∀x ∈ R \ {0} .

Hence, Proposition 14 implies that fχ 6∈ E1p .

Statement (ii) follows by using restriction and extension operators and [9, Thm. 4].

Now we prove statement (iv). By Theorem 6, it suffices to prove that f does not

belong to(clB

1+(1/p)p,∞ (R)

C∞[B1+(1/p)p,∞ ]

)`oc

. By the continuous imbedding of B1+(1/p)p,∞ (R)

into B1∞,∞(R), it suffices to prove that fχ does not belong to the closure of C∞[B1

∞,∞]in B1

∞,∞(R). By [8, Subsection 2.3], we have

limh→0+

| g(x+ h) + g(x− h)− 2g(x) |h

= 0 , (41)

for all g in the closure of C∞[B1∞,∞] in B1

∞,∞(R). However, a straightforward compu-tation shows that fχ does not satisfy (41).

By Marcus and Mizel [15], Tf is continuous in W 1,p(Rn). Since Tf is Lipschitz contin-uous in Lp(Rn), we can exploit Maligranda’s result [14, Thm. 2] and argue as in theproof of statement (iii) of Theorem 6 to conclude that statement (vi) holds.

10.5 Proof of the statements of Subsection 2.2

We now prove properties (a)–(f) of the operator Tψ introduced in Subsection 2.2. Adirect computation shows that ψ ∈ W 2,1(R). Then by the density of D(R) in W 2,1(R),and by the continuous imbeddings

W 2,1(R) ↪→ BV 11 (R) ↪→ BV 1

p (R) ,

we conclude that ψ ∈ E1p . Then statement (b) follows by [9, Thm. 4], and by use of

restriction and extension operators. Statements (c) and (e) follow by [9, Thms. 5, 11].By Proposition 14 and by Proposition 18 of the Appendix, we have

ψ ∈ clBV 1p (R)C

∞[BV 1p (R)] .

Thus statements (d) and (f) follow by Theorem 6 and by the imbedding of BV 1p (R)

into U1p (R) (cf. [9, Thm. 5].)

29

10.6 Proof of the statements of Subsection 2.3

Step 1. Let p ∈]1,+∞[. By [6, Prop. 3] and by standard imbeddings between Besovspaces, we know that ψα,β ∈ Bσ

p,1(R) as soon as

σ < min

(2,

1

β + 1

(α+ 1 +

1

p

)).

In particular, if σ := 1 + 1p

satisfies such inequality, i.e., if (9) is satisfied and p > 1,

then we have ψα,β ∈ B1+ 1

p

p,1 (R) and thus ψα,β ∈ BV 1p (R).

Step 2. If (9) holds, with p = 1, then an easy computation shows that ψ′′α,β ∈ L1(R).Hence, ψα,β ∈ BV 1

1 (R).

Step 3. Now we assume that1

p=α

β− 1 .

Clearly,ψ′α,β(t) = (α+ 1)tα sin

(t−β)− βtα−β cos

(t−β),

for 0 < t < 1/e. We now define a sequence {tk}k∈N by setting

tk := (kπ)−1/β k ∈ N .

Let k0 ∈ N be such that tk < 1/e if k ≥ k0. By an easy computation, and by ourassumption on p, we obtain

|ψ′α,β(tk+1)− ψ′α,β(tk)| ≥ βπ(β−α)/β

(1

k

)1/p

,

for k ≥ k0, a condition which obviously implies ψα,β /∈ BV 1p (R). If 1

p> α

β− 1, then

we take q ∈ [1,+∞[ with 1p> 1

q= α

β− 1, and we have ψα,β /∈ BV 1

q (R). Hence,

ψα,β /∈ BV 1p (R).

Step 4. We now assume that (9) holds and prove that ψα,β ∈ E1p . In case p = 1,

the membership of ψα,β in E1p follows by property ψ′′α,β ∈ L1(R) (cf. Remark 6). In

case p > 1, we have ψα,β ∈ B1+(1/p)p,1 (R). By density of D(R) into B

1+(1/p)p,1 (R) (cf. e.g.,

Triebel [21, Thm. 2.3.3],) and by Peetre’s Imbedding Theorem, we can conclude thatψα,β ∈ E1

p .

Step 5. To prove properties (b)-(f), we can repeat the arguments from Subsection 10.5.The necessity in (b) follows by Theorem 1 and by statement (a’).

11 Appendix

We now discuss the relation between the conditions

30

limy→0

‖τyg − g‖E = 0 , (42)

andg ∈ clEC

∞[E] (43)

for an element g of a function space E. To do so, we introduce the following wellknown.

Definition 3 We define a Banach Distribution Space (a BDS) to be a vector sub-space E of D′(Rn) endowed with a norm which renders E a Banach space continuouslyimbedded in D′(Rn). We say that a BDS (E, ‖−‖) is translation invariant if properties

τyg ∈ E, ‖τyg‖E = ‖g‖E .

hold for all g ∈ E and y ∈ Rn.

Then we have the following.

Proposition 18 Let E be a translation invariant BDS. If g ∈ E satisfies (42), then gsatisfies also (43).

Proof Let g ∈ E satisfy (42). Then by translation invariance of E, the map whichtakes y ∈ Rn to τyg is continuous, and we can consider the E-valued Riemann integral∫

Rn ϕ(y)τyg dy for all ϕ ∈ D(Rn). We now verify that∫Rn

ϕ(y)τyg dy = ϕ ∗ g ∀ϕ ∈ D(Rn) , (44)

where the right hand side is the ordinary convolution of ϕ and g. Indeed, by exploitingstandard properties of the vector valued Riemann integral and the continuity of thepairing 〈·, ·〉 between D(Rn) and its dual, and the continuity of the imbedding of Einto D′(Rn), we deduce that⟨∫

Rn

ϕ(y)τyg dy, ψ

⟩=

∫Rn

ϕ(y)〈τyg, ψ〉 dy =

∫Rn

ϕ(y)〈g, τ−yψ〉 dy

=

⟨g,

∫Rn

ϕ(y)τ−yψ dy

⟩= 〈g, ϕ ∗ ψ〉 = 〈g ∗ ϕ, ψ〉 ∀ϕ, ψ ∈ D(Rn) ,

where ϕ(x) := ϕ(−x). Now we take ϕ ∈ D(Rn) with∫

Rn ϕ(x) dx = 1, ϕ ≥ 0, and weset

ϕλ(x) := λ−nϕ(λ−1x), gλ(x) := g ∗ ϕλ ∀λ > 0 .

Then gλ ∈ C∞(Rn) and g(α)λ = g ∗ ϕ(α)

λ for all α ∈ Nn. Then equality (44) ensures that

g(α)λ ∈ E for all α ∈ Nn. Moreover, by continuity of ϕλ(y)(τyg − g) in the variable y,

we have

‖gλ − g‖E ≤∫

Rn

ϕλ(y)‖τyg − g‖E dy ∀λ > 0 ,

and thus condition (42) implies condition (43).

Instead, it seems that condition (43) implies condition (42) only with some extra as-sumptions on E, which we now consider. First, we introduce the following.

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Remark 7 If E is a translation invariant BDS, and if (42) holds for all g in a linearsubspace F of E, then (42) holds for all elements of clEF .

Then we have the following.

Proposition 19 Let E be a translation invariant BDS such that D(Rn) is a densesubset of E. Then both (42) and (43) hold for all elements of E.

Proof By Remark 7 and by Proposition 18, it suffices to show that (42) holds for allg ∈ D(Rn). Now we fix g ∈ D(Rn), and we take a compact subset K of Rn such thatsupp τyg ⊆ K for each |y| ≤ 1. By uniform continuity of g and of its derivatives, wehave limy→0 τyg = g in D(K). Since D(K) is contained in E and E is continuouslyimbedded in D′(Rn), the Closed Graph Theorem implies that D(K) is continuouslyimbedded in E. Hence, (42) holds for g.

Now we prove a related statement for the strong dual E ′ of E. We note that if E is aBDS which contains D(Rn) as a dense subset, then we can identify the elements of E ′

with those distributions which are continuous on (D(Rn), ‖ − ‖E), i.e., by transposingthe injection of D(Rn) into E. Then we have the following.

Proposition 20 Let E be a translation invariant BDS such that D(Rn) is a densesubset of E. Then an element g of E ′ satisfies (42) if and only if it satisfies (43).

Proof By Remark 7 and by Proposition 18, it suffices to show that (42) holds for allg ∈ C∞[E ′]. As we can see below, we can actually prove that (42) holds under theweaker assumption that only g and its first order derivatives belong to E ′. Clearly,

〈g − τyg, ϕ〉 =n∑j=1

yj

∫ 1

0

〈∂jg, τ−tyϕ〉 dt ∀ϕ ∈ D(Rn) , (45)

where the argument of the integrals in the right hand side are real valued continuousfunctions on [0, 1]. Thus by continuity of g on (D(Rn), ‖ − ‖E), we obtain

|〈g − τyg, ϕ〉| ≤n∑j=1

|yj|‖∂jg‖E′ ,

for all ϕ ∈ D(Rn) such that ‖ϕ‖E=1. Hence,

‖g − τyg‖E′ ≤n∑j=1

|yj|‖∂jg‖E′ ,

and thus (42) holds for g.

Finally, we have the following statement for the case of Besov spaces.

Proposition 21 Let s ∈ R, p, q ∈ [1,+∞]. Then the following statements hold.

(i) If p, q ∈ [1,∞[, then all elements of Bsp,q(Rn) satisfy both conditions (42) and

(43).

32

(ii) If at least one between p and q equals +∞, then an element g ∈ Bsp,q(Rn) satisfies

condition (42) if and only if it satisfies condition (43).

Proof Statement (i) follows by Proposition 19 and by the density of D(Rn) in Bsp,q(Rn)

(cf. e.g., Triebel [21, Thm. 2.3.3].) Let bsp,q(Rn) denote the closure of D(Rn) in Bsp,q(Rn).

Let p′ denotes the conjugate exponent of p. Then the dual of bsp,q(Rn) can be identifiedwith B−s

p′,q′(Rn), s ∈ R, 1 ≤ p, q ≤ ∞, cf. Triebel [21, Remark 2 in 2.11.2] and referencestherein. Hence, statement (ii) follows from Proposition 20.

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Gerard Bourdaud: Institut de Mathematiques de Jussieu, Projet d’analyse fonction-nelle, Case 186, 4 place Jussieu, 75252 Paris Cedex 05, France.E-mail: bourdaud@ccr.jussieu.fr

Massimo Lanza de Cristoforis: Dipartimento di Matematica Pura ed Applicata, Uni-versita di Padova, Via Belzoni 7, 35131 Padova, Italia.E-mail: mldc@math.unipd.it

Winfried Sickel: Mathematisches Institut, FSU Jena, Ernst-Abbe-Platz 1-2, O7743Jena, Germany.E-mail: sickel@minet.uni-jena.de

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