Composition Operators on Function Spaces with Fractional Order of Smoothness
G´erardBourdaud & Winfried Sickel
October 15, 2010
Abstract
In this paper we will give an overview concerning properties of composition oper-
ators Tf (g) := f ◦ g in the framework of Besov-Lizorkin-Triebel spaces. Boundedness and continuity will be discussed in a certain detail. In addition we also give a list of open problems.
2000 Mathematics Subject Classification: 46E35, 47H30. Keywords: composition of functions, composition operator, Sobolev spaces, Besov spaces, Lizorkin-Triebel spaces, Slobodeckij spaces, Bessel potential spaces, functions of bounded variation, Wiener classes, optimal inequalities.
1 Introduction
Let E denote a normed space of functions. Composition operators Tf : g 7→ f ◦g, g ∈ E, are simple examples of nonlinear mappings. It is a little bit surprising that the knowledge about
these operators is rather limited. One reason is, of course, that the properties of Tf strongly depend on f and E. Here in this paper we are concerned with E being either a Besov or a Lizorkin-Triebel space (for definitions of these classes we refer to the appendix at the end m n of this article). These scales of spaces generalize Sobolev spaces Wp (R ), Bessel potential s n s n s n spaces Hp (R ), Slobodeckij spaces Wp (R ) as well as H¨olderspaces C (R ) in view of the identities
m n m n • Wp (R ) = Fp,2(R ), 1 < p < ∞, m ∈ N;
s n s n • Hp (R ) = Fp,2(R ), 1 < p < ∞, s ∈ R;
1 s n s n s n • Wp (R ) = Fp,p(R ) = Bp,p(R ), 1 ≤ p < ∞, s > 0, s 6∈ N;
s n s n • C (R ) = B∞,∞(R ), s > 0, s 6∈ N.
In our opinion there is a very interesting interplay between integrability and regularity properties of f, g and of the composition f ◦ g. It is our aim to describe this in detail and to give a survey on the state of the art. Many times we will not give proofs but discuss illustrating examples. The theory is far from being complete. However we believe that a discussion of these operators in spaces of fractional order of smoothness s > 0 is appropriate and leads to a better understanding of the various phenomenons which occur. Convention: If there is no need for a distinction between Lizorkin-Triebel spaces and Besov s n s n s n spaces we will simply write Ep,q(R ) instead of Fp,q(R ) and Bp,q(R ), respectively. In the s n s n same spirit, Ep,q(R ) can denote Wp (R ) in case s ∈ N and p = 1 or +∞, although those spaces are not Lizorkin-Triebel nor Besov spaces.
An illustrating example
Let us have a look at the following boundary value problem:
∆u(x) + f(u(x)) = h(x) , x ∈ Ω , u = 0 , x ∈ ∂Ω .
Here Ω is an open and bounded subset of Rn with smooth boundary. Let L denote the solution operator for the Dirichlet problem for the Laplacian with respect to Ω. Then our boundary value problem can be reformulated as a fixed point problem
u = L(h − f(u)) . (1)
˚s s By Ep,q(Ω) we denote the collection of all functions in Ep,q(Ω) having vanishing boundary s−2 ˚s values (this makes sense if s > 1/p). Since L : Ep,q (Ω) → Ep,q(Ω) is an isomorphism a discussion of (1) requires s s−2 Tf (Ep,q(Ω)) ⊂ Ep,q (Ω)
including some estimates which relate the norms k T (u) k s−2 and k u k s . f Ep,q (Ω) Ep,q(Ω)
Remark 1 There are many papers dealing with problems as in our illustrating example. We refer e.g. to [28], [34], [49], where boundedness and continuity of composition operators are treated in connection with the Schr¨odingerequation.
2 The main problem
From a little bit more abstract point of view the above example indicates that we have to study the following problem.
Problem 1:
s0,`oc n s1,`oc n Suppose Ep0,q0 (R ) ⊂ Ep1,q1 (R ). Find necessary and sufficient conditions on f : R → R s.t.
s0 n s1 n Tf (Ep0,q0 (R )) ⊂ Ep1,q1 (R ) .
Remark 2 (i) For a given space of functions E on Rn, the associated space E`oc is the collection of all functions g s.t. u · g ∈ E, for all u ∈ D(Rn). s0,`oc n s1,`oc n (ii) Our assumption Ep0,q0 (R ) ⊂ Ep1,q1 (R ) indicates that there is no hope for an increase s0 n of the local regularity of the whole set Tf (Ep0,q0 (R )). (iii) The theory of the operators Tf , as we know it at this moment, does not depend very n much on the underlying domain Ω. So we discuss Tf on function spaces defined on R . (iv) Of course, it would make sense to replace Tf by more general mappings like
n N(g1, . . . , gd)(x) := f(x, g1(x), . . . , gd(x)) , x ∈ R , where f : Rn+d → R. In this generality these mappings N are called Nemytskij operators. Much less is known for these general mappings. In our survey only very few remarks will be made concerning this general situation.
There are hundred’s of references dealing with Problem 1 and its generalizations. However, only in very few cases, e.g. if f(t) := tm, m ∈ N, one knows the final answer. We will add a few remarks later on. For the moment we will turn to a simplified problem.
Problem 2: Find necessary and sufficient conditions on f : R → R s.t.
s n s n Tf (Ep,q(R )) ⊂ Ep,q(R ) . (2)
We will call the property (2) the acting property. In applications one needs more than the acting condition. In general one also needs boundedness and continuity. This justifies to consider the following modified problems.
3 Problem 2’: Find necessary and sufficient conditions on f : R → R s.t.
s n s n Tf : Ep,q(R ) → Ep,q(R ) is bounded.
Remark 3 (i) Generally speaking, a mapping T of a metric space E to itself is said to be bounded if T (A) is bounded for all bounded set A ⊂ E. (ii) Coming back to our illustrating example it makes sense to ask for optimal inequalities describing the boundedness of Tf . We will take care of this problem as well in our survey. (iii) We do not know whether Problems 2 and 2’ have the same solution. Indeed, we do not have a counterexample in the framework of composition operators showing that the acting condition can occur without boundedness. Furthermore, let us mention that it is classically known that the acting condition implies at least a weak form of boundedness, see e.g. [39, pp. 275-276], [55, Lem. 5.2.4, 5.3.1/1] and [14, Section 4.2]. (iv) If we slightly generalize the class of operators by considering N(g)(x) := f(x, g(x)), x ∈ Rn, i.e. Nemytskij operators, then it is well-known that acting conditions and boundedness conditions may be different. For simplicity we only consider n = 1 and R replaced by [0, 1]. Then the classical example is given by ( 0 if u ≤ xs/2 , f (x, u) := s 1 x s/2 u2/s − u4/s if u > x ,
s where 0 < s < 1. The associated Nemytskij operator maps B∞,∞[0, 1] into itself, but is not bounded. We refer to [5].
Problem 2”: Find necessary and sufficient conditions on f : R → R s.t.
s n s n Tf : Ep,q(R ) → Ep,q(R ) is continuous.
Remark 4 Problem 2 ( Problem 2’) and Problem 2” have different answers in general. Below we will show that for some Lizorkin-Triebel spaces there exist noncontinuous bounded composition operators, see Corollary 1.
4 In our opinion also the following problem is of interest.
Problem 3: Characterize all function spaces E, where the following assertions are equivalent
• Tf (E) ⊂ E,
• Tf : E → E is bounded,
• Tf : E → E is continuous,
whatever be f : R → R.
At the end of our paper, see Section 7, we will give a list of spaces for which the acting property is equivalent to the boundedness and to the continuity. We will also produce a negative list, it means, we will collect also those spaces, for which such equivalences do not hold. Both authors have written surveys with respect to this topic in earlier times, see [11], [55, Chapt. 5] and [15]. To increase the readibility of this survey we allow some overlap with those articles. However, we mainly concentrate on the progress since 1995. In addition, a number of open problems is formulated within this text. This survey is organized as follows. After the Introduction we collect a number of necessary conditions for the acting property to hold. This will make the splitting in the discussion of sufficient conditions more transparent. We start with this discussion in Section 3 by con- centrating on Sobolev spaces. In Section 4 we will discuss the case of fractional order of smoothness. The results of this section motivate to study composition operators on intersec- tions which we will do in Section 5. Some generalizations to the vector-valued situation are considered in Section 6 (but here we concentrate on Sobolev spaces). As mentioned above, there will be a section devoted to Problem 3, namely Section 7. In the very short final section we make some concluding remarks.
Notation
As usual, N denotes the natural numbers, Z the integers and R the real numbers. If E and F are two Banach spaces, then the symbol E,→ F indicates that the embedding is continuous. The symbol c denotes a positive constant which depends only on the fixed parameters n, s, p, q and probably on auxiliary functions, unless otherwise stated; its value < > may vary from line to line. Sometimes we will use the symbols “ ∼ ” and “ ∼ ” instead
5 < of “≤” and “≥”, respectively. The meaning of A ∼ B is given by: there exists a constant > c > 0 such that A ≤ c B. Similarly ∼ is defined. The symbol A B will be used as an < < abbreviation of A ∼ B ∼ A.
m n n We denote by Cb (R ) the Banach space of functions on R which are continuous and n bounded, together with their derivatives up to order m, and by Cub(R ) the Banach space of bounded and uniformly continuous functions on Rn. The classical Sobolev spaces are m n ˙ m n denoted by Wp (R ), and their homogeneous counterparts by Wp (R ), for m ∈ N and s n 1 ≤ p ≤ ∞. Inhomogeneous Besov and Lizorkin-Triebel spaces are denoted by Bp,q(R ) s n ˙ s n ˙ s n and Fp,q(R ), respectively. We use the notation Fp,q(R ) and Bp,q(R ) for the homogeneous Lizorkin-Triebel and Besov spaces. For their definition, we refer to Section 9. General information about these function spaces, as well as the Wiener classes BVp, can be found s n ˙ s n e.g. in [55, 63, 64] (Fp,q(R )), [33, 37, 63] (Fp,q(R )), and [20, 21, 67] (BVp). Let us mention s n s n that we exclude the spaces F∞,q(R ), even in case when we write Ep,q(R ), 1 ≤ p, q ≤ ∞. If an equivalence class [f], for the a.e. equality, contains a continuous representative, then we call the class continuous and speak of values of f at any point (by taking the values of the continuous representative). n n If f is a function defined on R , and if h ∈ R , we put ∆hf(x) := f(x+h)−f(x). Throughout the paper ψ ∈ D(Rn) denotes a specific cut-off function, i.e., ψ(x) = 1 if |x| ≤ 1 and ψ(x) = 0 if |x| ≥ 3/2.
2 Necessary conditions
We begin with a collection of necessary conditions for the acting property. They are all found essentially before 1990. However, they give a good idea concerning the expectable solution of Problem 2.
Proposition 1 Let s > 0. Let f : R → R be a Borel measurable function. The acting n s n s,`oc property Tf (D(R )) ⊂ Ep,q(R ) implies f ∈ Ep,q (R).
n The proof follows easily by testing Tf on functions u ∈ D(R ) such that u(x) = x1 on some ball of Rn, see e.g. [4, Thm. 3.5] or [55, Thm. 5.3.1/2].
Proposition 2 Let s > 0. Let f : R → R be a Borel measurable function. s n s n 1,`oc (i) The acting property Tf (Ep,q ∩ L∞(R )) ⊂ Bp,∞(R ) implies f ∈ W∞ (R). s n n s n s n (ii) In case Ep,q(R ) 6⊂ L∞(R ), the acting property Tf (Ep,q(R )) ⊂ Bp,∞(R ) implies f ∈ ˙ 1 W∞(R).
6 Remark 5 (i) The necessity of (local) Lipschitz continuity for the acting condition in the framework of function spaces with fractional order of smoothness has been observed for the first time by Igari [36] (p = q = 2, 0 < s < 1, s 6= 1/2). Then the result has been extended to Besov and Lizorkin-Triebel spaces by the first named author and some co-authors. We refer to [12] for more detailed references and for the proof of Prop. 2. Let us also mention the recent publication [4, Thm. 3.1 and 3.2], where a proof in a more general context is given. s n s n s n (ii) Observe that Ep,q(R ) ,→ Bp,∞(R ). Hence, the Nikol’skij-Besov space Bp,∞(R ) is the s n largest space within the family Ep,q(R ).
s n Proposition 3 Suppose 1 + 1/p < s < n/p. Then the acting property Tf (Ep,q(R )) ⊂ s n Bp,∞(R ) implies f(t) = c t for some constant c. The phenomenon described in Proposition 3 is well known since Dahlberg [30] had published m n m n his short note in 1979. He had proved that the implication Tf (Wp (R )) ⊂ Wp (R ) requires f(t) = c t for some constant c. Extensions to Besov and Triebel-Lizorkin spaces have been given by Bourdaud [8, 11]. Extensions to values p, q < 1 can be found in Runst [54] and s n in [59]. Runst has also been the first who had investigated implications of Tf (Ep,q(R )) ⊂ s n Bp,∞(R ). A proof of Prop. 3 can be also found in [55, 5.3.1] and [4, Thm. 3.3]. As many times in the theory of Besov-Lizorkin-Triebel spaces in limiting situations the third index q has some influence.
Proposition 4 Suppose 1 + 1/p = s < n/p. s n s n (i) Let q > 1. Then the acting property Tf (Bp,q(R )) ⊂ Bp,q(R ) implies f(t) = c t for some constant c. s n s n (ii) Let p > 1. Then the acting property Tf (Fp,q(R )) ⊂ Fp,q(R ) implies f(t) = c t for some constant c.
This has been proved by the first named author in [10] and [11], see also [55, Lem. 5.3.1/2] and 1+(1/p) n [4, Thm. 3.3]. The existence of nontrivial composition operators on Bp,1 (R ), n > p + 1, 2 n and on F1,q(R ), n > 2, is an open problem. Remark 6 The degeneracy is connected to the existence of unbounded functions in s n s n Ep,q(R ). Instead, if we consider Ep,q ∩ L∞(R ), the Dahlberg phenomenon disappears. s n Indeed Ep,q ∩ L∞(R ) is known to be a Banach algebra for the pointwise product. Hence, all entire functions, vanishing at 0, act on it. For this problem, we refer also to Section 5.
1,`oc s,`oc Remark 7 Notice that the embedding W∞ (R) ,→ Ep,q (R) holds if 0 < s < 1, while the reverse embedding holds for s > 1 + (1/p). In view of the four preceding propositions, the values s = 1, s = 1 + (1/p) and s = n/p appear as the critical ones for Problem 2.
7 3 Sobolev spaces
For an easier reading of the paper we first discuss Problems 2, 2’, 2” for Sobolev spaces. Ac- cording to Remark 3, we present boundedness in the same section than the acting condition.
3.1 The acting condition on Sobolev spaces
In view of Section 2, in particular Remarks 6 and 7, it makes sense to separate the discussion into the cases
• m = 1;
m n n • m ≥ 2 and Wp (R ) ,→ L∞(R );
m n n • m ≥ 2 and Wp (R ) 6⊂ L∞(R ).
Theorem 1 Suppose 1 ≤ p ≤ ∞. Let f : R → R be a Borel measurable function s.t. f(0) = 0. Then it holds
f 0 ∈ L`oc( ) p > n ∞ R 1 n 1 n or p = 1 = n Tf (Wp (R )) ⊂ Wp (R ) ⇐⇒ 0 f ∈ L∞(R) otherwise. In either case we have
0 k f ◦ g k 1 n ≤ k f k k g k 1 n . (3) Wp (R ) L∞(R) Wp (R )
Remark 8 (i) For a proof we refer to Marcus and Mizel [42], see also [6, Chapt. 9]. (ii) If f and g are C1 functions, and if f(0) = 0, then Z Z Z |f(g(x))|p dx = |f(g(x)) − f(0)|p dx ≤ k f 0 kp |g(x)|p dx , (4) L∞( ) n n R n R R R Z ∂g p Z ∂g p f 0(g(x)) (x) dx ≤ k f 0 kp (x) dx , i = 1, . . . , n . L∞( ) n ∂x R n ∂x R i R i Hence, (3) follows under these extra conditions. By (4) also the role of the condition f(0) = 0 becomes clear. The extension of these estimates to the general case is more complicated.
m n n Theorem 2 Let m = 2, 3,..., and let 1 ≤ p < ∞. We suppose that Wp (R ) ,→ L∞(R ). Let f : R → R be a Borel measurable function s.t. f(0) = 0.
8 m n m,`oc (i) Then the composition operator Tf maps Wp (R ) into itself if, and only if, f ∈ Wp (R). m,`oc (ii) For f ∈ Wp (R) we have
m k f ◦ g kW m( n) ≤ C(f, g) k g kW m( n) + k g k m n , (5) p R p R Wp (R ) where
(m) C(f, g) := c k f k m−1 + k f k W∞ (Ig) Lp(Ig)
n Ig := {t ∈ R : |t| ≤ kgkL∞(R )} , and c is independent of f and g.
Remark 9 (i) For a proof we refer to [9], see also [55, 5.2.4]. s n (ii) The estimate (5) will be typical for the supercritical case, i.e., the case Ep,q(R ) ,→ n L∞(R ). Recall the chain rule
|γ| γ X X (`) α1 α` D (f ◦ g) = cγ,`,α1,...,α` (f ◦ g) D g ··· D g , (6) `=1 α1+···+α`=γ |αi|6=0
i i i where γ := (γ1, . . . , γn), α := (α1, . . . , αn), i = 1, . . . , `, are multi-indices and cγ,`,α1,...,α` are certain combinatorical constants. Let γ = (m, 0,..., 0). Then the sum on the right-hand side contains the terms
m m (m) ∂g 0 ∂ g (f ◦ g) and (f ◦ g) m . ∂x1 ∂x1 From this point of view an estimate as given in (5) looks natural.
We need a further class of functions. If E is a normed space of functions on Rn, then the `oc space Eunif is the collection of all g ∈ E s.t.
k g kEunif := sup k g ψ( · − a) kE < ∞ , n a∈R where ψ is the cut-off function from our list of conventions.
m n n Theorem 3 Let m = 2, 3,..., and let 1 ≤ p < ∞. We suppose that Wp (R ) 6⊂ L∞(R ). Let f : R → R be a Borel measurable function s.t. f(0) = 0. m n m n 0 m−1 (i) If m = n/p ≥ 2, then Tf (Wp (R )) ⊂ Wp (R ) holds if, and only if, f ∈ Wp,unif (R). 0 m−1 (ii) Let m = n/p ≥ 2 and f ∈ Wp,unif (R). Then
m k f ◦ g kW m( n) ≤ Cf k g kW m( n) + k g k m n (7) p R p R Wp (R )
9 where Z a+1 1/p (m) p Cf := c k f k ˙ 1 + sup |f (t)| dt (8) W∞(R) a∈R a and c is independent of f and g. m n (iii) Let either 1 < p < ∞ and 2 ≤ m < n/p, or p = 1 and 3 ≤ m < n. Then Tf (Wp (R )) ⊂ m n Wp (R ) holds if, and only if, G(t) = c t, t ∈ R, for some c ∈ R. 2 n 2 n 00 (iv) If n ≥ 3, then Tf (W1 (R )) ⊂ W1 (R ) holds if, and only if, f ∈ L1(R). 00 (v) Let f ∈ L1(R). Then
0 00 k f ◦ g k 2 n ≤ c ( k f k + k f k ) k g k 2 n (9) W1 (R ) L∞(R) L1(R) W1 (R ) and c is independent of f and g.
Remark 10 (i) For a proof we refer to [9], see also [55, 5.2.4] and [14]. The crucial idea in the proofs of Thm. 2 and Thm. 3 (except part (iii)) consists in an integration by parts involving the norm of the Sobolev space. Part (iii) in Thm. 3 is a particular case of Proposition 3, see some further comments in Subsection 4.3. m (ii) The class Wp,unif (R) plays a crucial role in the composition problem, not only in the critical case m = n/p. This becomes clear if we consider instead of the usual Sobolev space m n m ˙ 1 n Wp (R ) the so-called Adams-Frazier space Wp ∩ Wmp(R ), see Theorem 25. We turn back to the problem touched in Remark 9(ii). Here is an improvement of the estimate (5) in the supercritical case.
m n n Theorem 4 Let m = 2, 3,..., and let 1 ≤ p < ∞. We suppose that Wp (R ) ,→ L∞(R ). m Let f : R → R be in Wp (R) s.t. f(0) = 0. Then there exists a constant c s.t.
m−1/p k f ◦ g kW m( n) ≤ c k f kW m( ) k g kW m( n) + k g k m n . (10) p R p R p R Wp (R ) m n Here c is independent of f and g ∈ Wp (R ). Remark 11 A proof of (10) has been given in [14]. The exponent m − 1/p is the optimal m,`oc one, see Proposition 6 in Section 4. However, the result does not extend to all f in Wp (R). A counterexample is given by f(t) = sin t, t ∈ R, see [14].
3.2 Continuity of composition operators on Sobolev spaces
1 n Theorem 5 (i) Let 1 ≤ p < ∞. Every composition operator Tf , which maps Wp (R ) into itself, is continuous.
(ii) Let 1 < p < ∞, m ∈ N and m > n/p. Then every composition operator Tf , which maps m n Wp (R ) into itself, is continuous.
10 m n m n Proof We concentrate on (ii). Under the given restrictions we have Wp (R ) = Fp,2(R ) in the sense of equivalent norms. From Thm. 2 we derive the equivalence of the acting m n m n m,`oc condition Tf (Wp (R )) ⊂ Wp (R ) and f ∈ Wp (R), f(0) = 0. Furthermore, the bound- edness of Tf follows from the estimate (5). In this situation we can apply Proposition 7, see paragraph 4.2.3, and obtain that Tf must be continuous as well.
Remark 12 Part (i) is a famous result of Marcus and Mizel [43]. Part (ii) seems to be a novelty.
4 Spaces of fractional order of smoothness
According to Section 2, it makes sense to separate the discussion into the cases
• 0 < s < 1;
• 1 < s < 1 + 1/p;
• 1 + 1/p < s < n/p;
• max(1 + 1/p, n/p) < s.
4.1 The case of low smoothness
In case 0 < s < 1, the only known necessary condition, see Remark 7, turns out to be also sufficient.
Theorem 6 Let 1 ≤ p, q ≤ ∞ and 0 < s < 1. Let f : R → R be a Borel measurable function s.t. f(0) = 0. (i) The following assertions are equivalent:
s n s n (a) Tf (Ep,q(R )) ⊂ Ep,q(R );
s n s n (b) Tf : Ep,q(R ) → Ep,q(R ) is bounded;
0 s n n 0 `oc s n n (c) Either f ∈ L∞(R) if Ep,q(R ) 6⊂ L∞(R ) or f ∈ L∞ (R) if Ep,q(R ) ,→ L∞(R ).
(ii) Let f 0 be as in (c). Then
0 k f ◦ g k s n ≤ k f k k g k s n Ep,q(R ) L∞(Ig) Ep,q(R )
s n holds for all g ∈ Ep,q(R ) and Ig is defined as in Thm. 2.
11 s n Proof The space Ep,q(R ) is defined by first order differences, see Definition 2 in Section 0 9. For those differences we have the obvious inequality |∆h(f ◦ g)| ≤ kf k∞ |∆hg|. For the
estimate of the Lp-term we refer to (4). Both inequalities together prove (ii) and at the same time the implications (c) ⇒ (b), (a). The nontrivial implication (a) ⇒ (c) follows by Proposition 2.
By using real interpolation for Lipschitz-continuous operators, see [52], one can add conti- nuity to the list in part (i) of the theorem.
Theorem 7 Let 1 ≤ p, q < ∞ and 0 < s < 1. Let f : R → R be a continuous function 0 s n n 0 `oc s n s.t. f(0) = 0 and either f ∈ L∞(R) if Bp,q(R ) 6⊂ L∞(R ) or f ∈ L∞ (R) if Bp,q(R ) ,→ n s n s n L∞(R ). Then Tf : Bp,q(R ) → Bp,q(R ) is continuous.
The proof in [55, Thm. 5.5.2/3] uses the real interpolation formula
n 1 n s n Lp(R ),Wp (R ) = Bp,q(R ) , 0 < s < 1 , 1 ≤ q ≤ ∞ , s,q
in connection with a result of Maligranda [41] concerning the continuity of a nonlinear operator T with respect to real interpolation, see [55, Sect. 2.5].
4.2 The case of high smoothness
In view of the results presented in Section 2, there is a natural conjecture.
Conjecture 1 Let s > 1 + (1/p). The composition operator Tf , associated to a Borel s n s n s,`oc measurable function f : R → R, maps Ep,q ∩L∞(R ) to Ep,q(R ) if, and only if, f ∈ Ep,q (R) and f(0) = 0.
m n The validity of the above Conjecture is known in case of Sobolev spaces (recall Wp (R ) = m n Fp,2(R ), 1 < p < ∞, m ∈ N), see Thm. 1 and [14, Thm. 2]. For n = 1, it is also valid for any Lizorkin-Triebel space and for some Besov spaces, see Subsection 4.2.1. The extension to dimensions n > 1 is an open question.
4.2.1 The acting condition in the one-dimensional situation
To begin with we deal with a simplified situation where we give a sketch of the proof. However, the used arguments are also typical for the more general results which will be mentioned below.
12 We need some more classes of functions. For a function g : R → R, we denote by kgkBVp the supremum of numbers N !1/p X p |g(bk) − g(ak)| , k=1 taken over all finite sets {]ak, bk[; k = 1,...,N} of pairwise disjoint open intervals. A function g is said to be of bounded p-variation if kgkBVp < +∞. The collection of all such functions is called the Wiener class and denoted by BVp. We refer to [20, 21, 67] for a discussion of these classes. Their importance in the composition problem is related to the Peetre’s embedding 1/p Bp,1 (R) ,→ BVp(R) , (11) see e.g. [53] and [20, Thm. 5].
0 s−1 Proposition 5 Let 1 ≤ p < ∞, p ≤ q ≤ ∞ and 1 + 1/p < s < 2. Let f ∈ Bp,q (R) s.t. f(0) = 0. Then there exists a constant c > 0 such that the inequality
0 0 s−(1/p) k f ◦ g kBs ( ) ≤ c k f k s−1 k g kBs ( ) + k g k (12) p,q R Bp,q (R) p,q R BVsp−1
s holds for all real analytic functions g in Bp,q(R).
Remark 13 By (11) and by assumption s > 1 + (1/p), a weaker version of (12) is given by
0 0 s−(1/p) k f ◦ g kBs ( ) ≤ c k f kBs−1( ) k g kBs ( ) + k g k s−1 . (13) p,q R p,q R p,q R Bp,q (R) Proof Because of
0 1 Bs−1( ) ,→ Bs−1−1/p( ) ,→ Cs ( ) ,→ L ( ) , 0 < s0 < s − 1 − , p,q R ∞,∞ R ub R ∞ R p the function f 0 is a continuous. Hence we may apply the chain rule and obtain
(f ◦ g)0 = (f 0 ◦ g) g0 in the pointwise sense. Furthermore, we will use that
0 k u kBs ( ) k u kL ( ) + k u k s−1 (14) p,q R p R Bp,q (R) for all distribution u if s > 0, see [63, 2.3.8].
Step 1. Since the estimate of the Lp-term is as in (4), we only have to deal with the homogeneous part of the Besov norm. For brevity we put Z 1/p p ωp(f, h) := |∆hf(x)| dx , h ∈ R . R 13 Using (14) we have to estimate
Z 1 0 q 1/q ωp((f ◦ g) , h) dh s−1 . −1 |h| |h| Since 0 0 0 0 ωp((f ◦ g) g , h) ≤ k f k∞ ωp(g , h) + U(h) , where Z 1/p 0 p 0 p U(h) := |∆h(f ◦ g)(x)| |g (x)| dx , R we are reduced to prove that Z 1 U(h) q dh1/q s−1 −1 |h| |h| can be estimated by the right-hand side of (12). Step 2. Without loss of generality we may assume h > 0. The set of zeros of g0 is discrete,
and its complement in R is the union of a family (Il)l of nonempty open disjoint intervals. 0 For any h > 0 we denote by Il the (possibly empty) set of x ∈ Il whose distance to the right
endpoint of Il is greater than h, and we set
00 0 0 Il := Il \ Il , al := sup |g | . Il
By gl we mean the restriction of g to Il, hence a strictly monotone smooth function. If 0 Il 6= ∅, then we have
−1 0 |g(gl (y) + h) − y| ≤ al h for y ∈ gl(Il ) , (15)
−1 where gl denotes the inverse function of gl. Substep 2.1. Let Z !1/p p Ωp(f, t) := sup |∆hf(x)| dx , t > 0 . (16) R |h|≤t By (15) and by a change of variable we find Z 0 p 0 p p−1 p 0 |∆h(f ◦ g)(x)| |g (x)| dx ≤ al Ωp(f , alh) . (17) 0 Il
By the Minkowski inequality w.r.t. Lq/p, and by Proposition 8 in Section 9, we obtain
14 Z ∞ 1 X q/p dh1/q ap−1 Ωp(f 0, a h) h(s−1)p l p l h 0 l 1/p q/p !p/q X Z ∞ 1 dh ≤ ap−1 Ωp(f 0, a h) hp(s−1) l p l h l 0
1/p Z ∞ 0 q 1/q X p−1+p(s−1) Ωp(f , t) dt = a l ts−1 t l 0 sp−1 1/p 0 X 0 ≤ c k f k s−1 sup |g | . Bp,q ( ) R I l l Now we follow [20, proof of Thm. 7]. By definition, the function g0 vanishes at the endpoints of Il. Let βl be one of these endpoints. Furthermore, there is at least one point ξl ∈ Il such that 0 0 |g (ξl)| = sup |g | . Il Hence X X sup |g0|sp−1 = |g0(ξ ) − g0(β )|sp−1 ≤ kg0ksp−1 . l l BVsp−1 I l l l By (17), we conclude that
Z ∞ Z q/p 1/q 1 X 0 p 0 p dh (s−1)p |∆h(f ◦ g)(x)| |g (x)| dx h 0 h 0 l Il 0 0 s−(1/p) ≤ c k f k s−1 k g k . (18) Bp,q (R) BVsp−1
0 00 Substep 2.2. Since g vanishes at the right endpoint of Il , it holds
|g0(x)| ≤ sup |g0(x) − g0(x + v)| |v|≤h
00 for all x ∈ Il . Thus we obtain Z X 0 p 0 p 0 p 0 p |∆h(f ◦ g)(x)| |g (x)| dx ≤ (2kf k∞) Ωp(g , h) . (19) 00 l Il By Proposition 8, we conclude that
Z ∞ Z q/p 1/q 1 X 0 p 0 p dh (s−1)p |∆h(f ◦ g)(x)| |g (x)| dx h 00 h 0 l Il 0 ≤ c k f k kgk s . (20) ∞ Bp,q(R)
15 Putting (4), (17), (18) and (20) together we obtain (12). The proof is complete.
s To derive the acting property with respect to Bp,q(R) one uses a specific density argument (to get rid of the restriction to real analytic g), the Fatou property, see, e.g., [22] and [32]. s To obtain the acting property for Bp,q(R) for higher values of s, m + 1/p < s < m + 1, m ≥ 2, one can use an induction argument in combination with
(m) k u kBs ( ) k u kL ( ) + k u k s−m (21) p,q R p R Bp,q (R) if s > 0, see [63, 2.3.8]. However, to get an optimal inequality for those values of s and to deal with m ≤ s ≤ m + 1/p, m ≥ 2, more effort is needed. To describe this we have to introduce a further space of functions. Since from now on we do not give proofs and the arguments in the Lizorkin-Triebel case are similar (however, a bit more sophisticated) we turn to the general notation. The outcome is the following, see [23], [24], [25] and [48] for all details.
Theorem 8 Let 1 < p < ∞, 1 ≤ q ≤ ∞ (p ≤ q if E = B), and s > 1 + (1/p). The
composition operator Tf , associated to a Borel measurable function f : R → R, acts on s s,`oc Ep,q(R) if, and only if, f(0) = 0 and f ∈ Ep,q (R). Remark 14 We believe that the restriction p ≤ q in the B-case is connected with the method of proof but not with the problem itself.
Acting property and boundedness are equivalent in this situation. To describe this, we use ˙s the smooth cut-off function ψ and define ψt(x) := ψ(x/t), t > 0. We denote by Ep,q(R) the space of all functions in L∞(R) which belong to the homogeneous Besov-Lizorkin-Triebel ˙ s space Ep,q(R), and endow it with the natural norm
kfk ˙s := kfk ˙ s + k f kL ( ) . Ep,q(R) Ep,q(R) ∞ R Theorem 9 Let 1 < p < ∞, 1 ≤ q ≤ ∞ (p ≤ q if E = B), and s > 1 + (1/p). s,`oc (i) Let f ∈ Ep,q (R) and f(0) = 0. Then there exists a constant c > 0 such that the inequality
0 s−(1/p) k f ◦ g kEs ( ) ≤ c k (f ψk g k ) k ˙s−1 kgkEs ( ) + kgk s (22) p,q R L∞(R) Ep,q (R) p,q R Ep,q(R) s holds for all such functions f and all g ∈ Ep,q(R). (ii) There exists a constant c > 0 such that the inequality
s−(1/p) k f ◦ g kEs ( ) ≤ c k f kEs ( ) kgkEs ( ) + kgk s (23) p,q R p,q R p,q R Ep,q(R) s s holds for all functions f ∈ Ep,q(R), f(0) = 0, and all functions g ∈ Ep,q(R).
16 4.2.2 Optimal inequalities (I)
We would like to discuss the quality of the estimate (23). However, for doing that there is no need to concentrate on n = 1. The following simple arguments work for any n.
Proposition 6 Let 1 ≤ p, q ≤ ∞ and s > 1+1/p. Let h : [0, ∞[→ [0, ∞[ be a nondecreasing s−(1/p) n function satisfying h(x) = o x for x → +∞. Let N be any semi-norm on D(R ). If f is a continuous function such that, for some constant c > 0, the inequality k f ◦ g k s n ≤ c N(g) + h (N(g)) (24) Bp,∞(R ) holds for all g ∈ D(Rn), then f is a polynomial of degree ≤ s.
Proof We take the natural number m ≥ 1 such that m−1 ≤ s < m. Let ϕ be a function n n in D(R ) s.t. ϕ(x) = x1 for all x ∈ [−1, 1] . Then it holds (f ◦ aϕ)(x) = f(ax1) for all x ∈ [−1, 1]n and all a > 0. Hence
m m n ∆te1 (f ◦ aϕ)(x) = ∆atf(ax1) , ∀x ∈ [−1/2, 1/2] , ∀t ∈ [0, 1/2m] , ∀a > 0 .
Using this identity and (24), we deduce
Z a/2 1/p m p s (1/p)+1−s (1/p)−s |∆t f(x)| dx ≤ c t a N(ϕ) + h(a N(ϕ)) a −a/2 for all t ∈]0, 1] and a > 2m. By taking a to +∞, and by applying the assumption on h, we deduce m ∆t f(x) = 0 a.e., ∀t ∈]0, 1] . Then by a standard argument, we deduce that f is a polynomial of degree at most m − 1.
Remark 15 Proposition 6 yields that the exponent s−1/p in (13), (22), and (23) is optimal. It also implies the optimality of the estimate (10).
4.2.3 Continuity of composition operators in the one-dimensional situation
There is a general continuity theorem in [18, Cor. 2] which can be applied for all n, if Conjecture 1 is valid.
s n Proposition 7 Let s > 1 + 1/p. If Conjecture 1 is valid for Ep,q(R ), then a composition operator Tf is continuous if, and only if, f(0) = 0 and f belongs to the closure of the smooth s,`oc functions in Ep,q (R).
17 In consequence of Thm. 8 and this proposition we immediately get the following.
Corollary 1 Let 1 < p < ∞ and s > 1 + (1/p). Let f : R → R be Borel measurable. (i) Let 1 ≤ q < ∞ (p ≤ q if E = B). Then the following assertions are equivalent :
s s • Tf satisfies the acting condition Tf (Ep,q(R)) ⊂ Ep,q(R).
s • Tf is a bounded mapping of Ep,q(R) to itself.
s s • Tf : Ep,q(R) → Ep,q(R) is continuous.
s,`oc • f ∈ Ep,q (R) and f(0) = 0. (ii) The following assertions are equivalent :
s s • Tf satisfies the acting condition Tf (Ep,∞(R)) ⊂ Ep,∞(R).
s • Tf is a bounded mapping of Ep,∞(R) to itself.
s,`oc • f ∈ Ep,∞ (R) and f(0) = 0.
s s s (iii) There exist functions f in Ep,∞(R), f(0) = 0, such that Tf : Ep,∞(R) → Ep,∞(R) is not continuous.
4.2.4 Acting conditions in the general n-dimensional situation
In dimension n > 1, Conjecture 1 has not been proved up to now. We give here two sufficient acting conditions. In the first one, we try to approach the minimal assumptions on f. In the second one, we obtain an optimal estimate, but with stronger regularity assumptions on f.
Theorem 10 Let 1 ≤ p ≤ ∞, 1 ≤ q ≤ ∞, max(n/p, 1) < s < µ. We suppose f ∈ Cµ(R) s n s n and f(0) = 0. Then Tf maps Ep,q(R ) into Ep,q(R ).
µ s,`oc Remark 16 (i) Of course, we have C (R) ⊂ Ep,q (R). For a proof of Thm. 10 we refer to [55, Sect. 5.3.6, Thm. 2]. (ii) Let E = B. Under some extra conditions on s, it is proved in [13] that one can replace µ s+ε,`oc C (R) by Bp,q (R) with ε > 0 arbitrary.
Theorem 11 Let 1 ≤ p ≤ ∞, 1 ≤ q ≤ ∞, m ∈ N and max(m, n/p) < s < m + 1. We suppose f ∈ Cm+1(R) and f(0) = 0. Then there exists a constant c s.t.
s k f ◦ g kEs ( n) ≤ c k f k m+1 (kgkEs ( n) + kgk s n ) (25) p,q R Cb (R) p,q R Ep,q(R )
s n holds for all such f and all g ∈ Ep,q(R ).
18 Proof The theorem has been proved essentially in [55, 5.3.7, Thm. 1 and 2]. Only one further remark is needed. Because of s > max(1, n/p) we have n n n s − = s 1 − > 1 − p ps ps and therefore, the continuous embeddings
s n 1 n s n 1 n Bp,q(R ) ,→ Bps,v(R ) and Fp,q(R ) ,→ Fps,v(R ) (26)
for arbitrary v ∈]0, ∞].
Remark 17 (i) Probably the exponent s in (25) can be improved to s−1/p under additional assumptions on f, see Proposition 6 and Thm. 4. (ii) The estimation of the norm of kf ◦ gk in Theorem 11 is better than in Theorem 10. In the detailed statement of Theorem 10, the norm kf ◦ gk is controlled by kgkµ instead of kgks for kgk ≥ 1. (iii) Both theorems have a long list of forerunners. Let us mention at least a few: Moser 1960 (Sobolev spaces) [47], Mizohata 1965 (Bessel potential spaces with p = 2) [46], Peetre 1970 (Besov spaces, nonlinear interpolation) [52], Adams 1976 (Bessel potential spaces) [1], Meyer 1981 (Bessel potential spaces) [45], Runst 1986 (Besov-Lizorkin-Triebel spaces) [54], Adams & Frazier 1992 (Bessel potential spaces) [2, 3].
4.3 The intermediate case (I)
We consider the case 1 + 1/p < s < n/p. By Proposition 3, the composition operator necessarily lowers the regularity. The study of Tf for f(t) := sin t, t ∈ R, is particularly enlightening in this respect. Let us define
−α n gα(x) = ψ(x) |x| , x ∈ R , α > 0. (27)
Then it is well known (and not very complicated to prove) that in case s > 0 n g ∈ F s ( n) ⇐⇒ 0 < α < − s (28) α p,q R p and n n g ∈ Bs ( n) ⇐⇒ either 0 < α < − s or α = − s and q = ∞ (29) α p,q R p p
19 hold, cf. [55, Lem. 2.3.1/1]. Then
m ∂ (m) −α −α−2 m m sin(gα(x)) sin (|x| )(|x| x1) + lower order terms , ∂x1 in a certain neighborhood of the origin. Compare this with the m-th order derivative of the original function. It turns out that
m ∂ −α−2m m m gα(x) |x| x1 + lower order terms (|x| → 0) . ∂x1 Since we have (α + 1)m > α + m if m ≥ 2 this shows that the local singularity of the composition becomes stronger than the singularity of the original function gα. Going back to the general situation, a natural question consists in asking for the best possible image space, hence we turn to Problem 1 for the rest of this subsection. The following strange number will play a certain role n + 1 n − s ∗ ∗ p p p % = % (s, p, n) := n . (30) p − s + 1
Theorem 12 Suppose 1 + 1/p < s < n/p. Let f be a non-polynomial Borel measurable ∗ function. Then for every r > % (s, p, n) there exists a compactly supported function gr ∈ s n r n Ep,q(R ) such that the composition f ◦ gr does not belong to Bp,∞(R ).
Remark 18 (i) A proof in case of smooth f can be found in [58, 59, 55]. However, by the same arguments one can deal with non-smooth functions. (ii) If 1 + 1/p < s < n/p, then 1 + 1/p < %∗ < s. This indicates a certain loss of smoothness. s n %∗ n Whether there exists a nonlinear function f such that Tf (Ep,q(R )) ⊂ Bp,∞(R ) is still an open question. (iii) Also the following observation is of some interest. We study the difference d(s) := s−%∗(s, p, n) for fixed n and p (n > p+1). Obviously, d(1+1/p) = d(n/p) = 0 and d(s) > 0 if 1 + 1/p < s < n/p. Moreover, the function d(s) is concave on this interval, hence, it
attains a maximal value d(s0) there. We have
2 n rn − 1 rn − 1 p − 1 s0 := + 1 − and d(s0) = − 1 + p p p pp(n − 1)
This shows that d(s) has a bound depending on p and n, but it does not have an a priori bound for fixed p and independent of n.
20 We will use the following abbreviation:
Z w Z w 1/p −rp−1 m+1 p r k f ka˙ p([−w,w]) := |h| | ∆h f(y) | dy dh (31) 0 −w
r with m ∈ N and m + 1 > r > 0. The notation k f ka˙ p([−w,w]) reminds on a norm in a homogeneous Besov space. In some sense it is an incomplete one.
s n r n Theorem 13 Suppose Tf (Ep,q(R )) ⊂ Ep,p(R ) and 0 < r ≤ s < n/p. Let f ∈ L∞(R). Then it follows that −γ r sup w k f ka˙ p([−w,w]) < ∞ (32) w≥1 for all n 1 n n p + p p − s − r p − s + 1 γ > γ0(s, n, p, r) := n . (33) p − s Remark 19 (i) For the proof we refer to [61]. (ii) Let us comment on the condition (33). Fix n and p and consider s ↑ n/p. Then the lower bound for γ tends to infinity, which means that the necessary condition (32) becomes less restrictive. This is connected with the fact that local singularities in spaces with s ↑ n/p become weaker and weaker, cf. (27) and (28). If we fix also s and consider r ↓ 0, then the necessary condition (32) becomes again weaker (since the lower bound of γ increases). r n Clearly, that corresponds to the fact that the spaces Ep,p(R ) become larger. A similar observation gives a converse result if r ↑ s.
r (iii) A first essential conclusion of Theorem 13 is obtained by observing that k f ka˙ p([−w,w]) is a
r non-decreasing function in w. So, whenever f is not a polynomial of low degree, k f ka˙ p([−w,w]) is bounded from below by a positive constant. Then Theorem 13 says (in the case that f ∗ is bounded) that γ0 ≥ 0 if, and only if, r ≤ ρ , which also follows from Theorem 3. Hence, Theorem 13 represents an extension of Theorem 3.
r n r n r n r n Replacing the space Ep,p(R ) = Bp,p(R ) = Fp,p(R ) by the Sobolev space Wp (R ), r ∈ N, we have necessary and sufficient conditions. Let
m n X ` o Pm = g : g(t) = a` t , a` ∈ R, ` = 0, . . . , m , m ∈ N , `=0 and Z w 1/p −γ (m) p Aγ,p(f) := sup w | f (y) | dy , (34) w≥1 −w which replaces (31).
21 ∗ Theorem 14 Suppose 0 ≤ r ≤ s < n/p. Recall, ρ and γ0 are defined in (30) and (33), respectively. Let f : R → R be a Borel measurable function if r ≥ 1 and a continuous s n r n ∗ function if r = 0, but not an element of Pr−1. Then Tf (Ep,p(R )) ⊂ Wp (R ) implies r ≤ ρ , r,`oc f(0) = 0, f ∈ Wp (R) and Aγ0,p(f) < ∞.
Now we turn to the sufficient conditions. We concentrate on r ≥ 2 (for r = 0, 1 we refer to [60]).
Theorem 15 Suppose 1 < p < ∞, 2 ≤ r ≤ s < n/p and define again γ0 by (33). Let s n r n f(0) = 0. If Aγ0,p(f) < ∞, then Tf (Ep,p(R )) ⊂ Wp (R ) holds. Moreover, there exists some constant c such that
1 γ0+r− p k f ◦ g kW r( n) ≤ c Aγ ,p(f) k g kEs ( n) + k g k s n (35) p R 0 p,p R Ep,p(R ) holds with c independent of f and g.
Remark 20 Both theorems are proved in [60]. However, the used arguments are as in [30, 8, 11, 54] (Thm. 14) and [9] (Thm. 15).
Various examples are treated in [59, 60, 61]. Here we concentrate on smooth periodic func- tions. If f 6≡ 0 is periodic and smooth, then Z w | f (m)(y) |p dy w , w ≥ 1 . −w
Hence Aγ,p(f) < ∞ if, and only if, γ ≥ 1/p. In such a situation Theorems 14 and 15 yield final results. Here another strange number occurs. Let
n p % = %(s, n, p) := n . (36) p − s + 1 Obviously, % < %∗ if p < ∞.
Corollary 2 Let 1 < p < ∞ and let 2 ≤ r ≤ s < n/p. Suppose that f is periodic, ∞ s n r n f ∈ C (R), and f 6≡ 0. Then Tf (Ep,p(R )) ⊂ Wp (R ) holds if, and only if, f(0) = 0 and r ≤ %.
This has a fractional counterpart, see [55, 5.3.6] and [61].
Theorem 16 Let 1 ≤ p < ∞ and 1 < s < n/p. Suppose that f is periodic, f(0) = 0, f ∈ C∞(R), and f 6≡ 0. Then the following assertions are equivalent: s n r n s n r n (i) Tf (Fp,q(R )) ⊂ Bp,∞(R ). (ii) Tf (Fp,∞(R )) ⊂ Fp,q(R ). (iii) r ≤ %.
22 In [55, Thm. 5.3.6/3] we have been able also to prove the following associated inequality.
Theorem 17 Let f 0 ∈ C∞(R) and let f(0) = 0. Suppose 1 ≤ q, r ≤ ∞ and 1 < s < n/p. Let % be as in (36). Then there exists a constant c such that
% % k f ◦ g kF ( n) ≤ c (k g kF s ( n) + k g k s n ) (37) p,r R p,q R Fp,q(R )
s n holds for all g ∈ Fp,q(R ).
Remark 21 The proof of Thm. 17 relies on a specific estimate of products in Lizorkin- Triebel spaces, see [55, Thm. 4.6.2/5]. There is also counterpart for Besov spaces but less satisfactory.
4.4 The intermediate case (II)
1,`oc s,`oc s,`oc 1,`oc Now we assume that 1 < s < 1+1/p. Since W∞ (R) 6⊂ Ep,q (R) and Ep,q (R) 6⊂ W∞ (R), we have two independent necessary acting conditions, but we do not know if these two conditions are sufficient. Indeed, at this moment we do not have a conjecture how does the solution of Problem 2 looks like. The best sufficient condition obtained so far is connected with a new class of functions which has been introduced for the first time by Bourdaud and Kateb. 1 We define Up (R) as the set of Lipschitz continuous functions f on R such that
0 −1/p 0 kf kUp := sup t Ωp(f , t) < +∞ , (38) t>0 see (16) for the definition of Ωp.
1 Theorem 18 Let 1 ≤ p < +∞ and 0 < s < 1 + (1/p). If f ∈ Up (R) and f(0) = 0, then s n s n Tf (Bp,q(R )) ⊂ Bp,q(R ). Moreover, the inequality