171 Composition Operator on the Space of Functions

Acta Math. Univ. Comenianae 171 Vol. LXXXI, 2 (2012), pp. 171–183

COMPOSITION OPERATOR ON THE SPACE OF FUNCTIONS TRIEBEL-LIZORKIN AND BOUNDED VARIATION TYPE

M. MOUSSAI

Abstract. For a Borel-measurable function f : R → R satisfying f(0) = 0 and Z sup t−1 sup |f 0(x + h) − f 0(x)|p dx < +∞, (0 < p < +∞), t>0 R |h|≤t s n we study the composition operator Tf (g) := f◦g on Triebel-Lizorkin spaces Fp,q(R ) in the case 0 < s < 1 + (1/p).

1. Introduction and the main result

The study of the composition operator Tf : g → f ◦ g associated to a Borel- s n measurable function f : R → R on Triebel-Lizorkin spaces Fp,q(R ), consists in ﬁnding a characterization of the functions f such that s n s n (1.1) Tf (Fp,q(R )) ⊆ Fp,q(R ). The investigation to establish (1.1) was improved by several works, for example the papers of Adams and Frazier [1,2 ], Brezis and Mironescu [6], Maz’ya and Shaposnikova [9], Runst and Sickel [12] and [10]. There were obtained some necessary conditions on f; from which we recall the following results. For s > 0, 1 < p < +∞ and 1 ≤ q ≤ +∞ n s n s n • if Tf takes L∞(R ) ∩ Fp,q(R ) to Fp,q(R ), then f is locally Lipschitz continuous. n s n • if Tf takes the Schwartz space S(R ) to Fp,q(R ), then f belongs locally to s Fp,q(R). The ﬁrst assertion is proved in [3, Theorem 3.1]. The proof of the second assertion can be found in [12, Theorem 2, 5.3.1]. 1 Bourdaud and Kateb [4] introduced the functions class Up (R), the set of Lipschitz continuous functions f such that their derivatives, in the sense of distributions, satisfy Z 1/p 0 −1 0 0 p (1.2) Ap(f ) := sup t sup |f (x + h) − f (x)| dx < +∞, t>0 R |h|≤t

Received August 8, 2011; revised February 17, 2012. 2010 Mathematics Subject Classiﬁcation. Primary 46E35, 47H30. Key words and phrases. Triebel-Lizorkin spaces; Besov spaces; functions of bounded variation; composition operators. 172 M. MOUSSAI and are endowed with the seminorm

kfk 1 := inf(kgk + A (g)), Up (R) ∞ p where the infimum is taken over all functions g such that f is a primitive of g. In s n [4] the authors, proved the acting of the operator Tf on Besov space Bp,q(R ) for 1 1 ≤ p < +∞, 1 < s < 1 + (1/p) and f ∈ Up (R) with f(0) = 0. In [5] the same result holds for 0 < s < 1 + (1/p). s n In this work we will study the composition operator Tf on Fp,q(R ) for a func- 1 tion f which belongs to Up (R), then we will obtain a result of type (1.1). To do n n this, we introduce the set Vp(R ) of the functions g : R → R such that n Z 1/p X p 0 kgk n := kg 0 k 1 dx < +∞ Vp(R ) xj BV ( ) j n−1 p R j=1 R 1 where BVp (R) is the Wiener space of the primitives of functions of bounded p-variation (see Subsection 2.2 below for the definition) and n (1.3) g 0 (y) := g(x , . . . , x , y, x , . . . , x ), y ∈ , x ∈ . xj 1 j−1 j+1 n R R We will prove the following statement. Theorem 1.1. Let 0 < p, q < +∞ and 0 < s < 1 + (1/p). Then there exists a constant c > 0 such that the inequality (1.4) kf ◦ gk s n ≤ c kfk 1 kgk + kgk n Fp,q (R ) Up (R) p Vp(R ) n n 1 holds for all functions g ∈ Lp(R )∩Vp(R ) and all f ∈ Up (R) satisfying f(0) = 0. n n s n Moreover, for all such f, the operator Tf takes Lp(R ) ∩ Vp(R ) to Fp,q(R ). s n n s n n Remark. (i) Since Fp,q(R ) ,→ Lp(R ), then Tf maps from Fp,q(R ) ∩ Vp(R ) s n to Fp,q(R ) under the assumptions of Theorem 1.1. s n s n (ii) Since the Bessel potential spaces Hp (R ) = Fp,2(R ), 1 < p < ∞, Theorem 1.1 s n s n covers the results of composition operators in case Hp (R ) instead of Fp,q(R ). The paper is organized as follows. In Section2 we collect some properties of the s n 1 needed function spaces Fp,q(R ) and BVp (R). Section3 is devoted to the proof of the main result where in a first step we study the case of 1-dimensional which is the main tool when we prove Theorem 1.1. Also, our proof uses various Sobolev and Peetre embeddings, Fubini and Fatou properties, etc. In Section4 we give some corollaries and prove the sharp estimate of (1.4). Notation. We work with functions defined on the Euclidean space Rn. All n spaces and functions are assumed to be real-valued. We denote by Cb(R ) the Banach space of bounded continuous functions on Rn endowed with the supremum. Let D(Rn) (resp. S(Rn) and S0(Rn)) denotes the C∞-functions with compact support (resp. the Schwartz space of all C∞ rapidly decreasing functions and its topological dual). With k · kp we denote the Lp-norm. We define the differences n by ∆hf := f(· + h) − f for all h ∈ R . If E is a Banach function space on Rn, we denote by Eòc the collection of all functions f such that ϕf ∈ E for all COMPOSITION OPERATOR ON THE SPACE OF FUNCTIONS 173

n ϕ ∈ D(R ). As usual, constants c, c1,... are strictly positive and depend only on the ﬁxed parameters n, s, p, q; their values may vary from line to line.

2. Function spaces

2.1. Triebel-Lizorkin spaces

Let 0 < a ≤ ∞. For all measurable functions f on Rn, we set Z Z a Z q/u p/q 1/p s,u,a −sq 1 u dt Mp,q (f) := t n |∆hf(x)| dh dx . n t t R 0 |h|≤t Definition 2.1. Let 0 < p < +∞ and 0 < q ≤ +∞. Let s be a real satisfying 1 1 1 < s < 2 and s > n max − 1, − 1 . p q n s n Then, a function f ∈ Lp(R ) belongs to Fp,q(R ) if n X s−1,1,∞ kfk s n := kfk + M (∂ f) < +∞. Fp,q (R ) p p,q j j=1 s n The set Fp,q(R ) is a quasi Banach space for the quasi-norm defined above. For the equivalence of the above definition with other characterizations we refer to [15, Theorem 3.5.3] from which we recall the following statement. Proposition 2.2. Let 0 < p < +∞ and 0 < q, u ≤ +∞. Let s be a real satisfying 1 1 1 1 1 < s < 2 and s > n max − , − . p u q u n s n Then, a function f ∈ Lp(R ) belongs to Fp,q(R ) if and only if s,u,∞ (2.1) kfkp + Mp,q (f) < +∞ s n and the expression (2.1) is an equivalent quasi-norm in Fp,q(R ). Moreover, this s,u,∞ s,u,a assertion remains true if one replaces Mp,q by Mp,q for any fixed a > 0. The argument of the equivalence of above quasi-norms that we can replace the integration for t ∈]0, +∞[ by t ≤ a for a fixed positive number a is the part of the integral for which t > a can be easily estimated by the Lp-norm. Embeddings. Triebel-Lizorkin spaces are spaces of equivalence classes w.r.t. al- most everywhere equality. However, if such an equivalence class contains a continuous representative, then usually we work with this representative and call also the equivalence class a continuous function. Later on we need the following continuous embeddings: s n (i) The spaces Fp,q(R ) are monotone with respect to s and q, more exactly s n t n t n Fp,∞(R ) ,→ Fp,q(R ) ,→ Fp,∞(R ) if t < s and 0 < q ≤ ∞. s n s n s n (ii) With Besov spaces, we have Bp,1(R ) ,→ Fp,q(R ) ,→ Bp,∞(R ). s n n (iii) If either s > n/p or s = n/p and 0 < p ≤ 1, then Fp,q(R ) ,→ Cb(R ). For various further embeddings we refer to [14, 2.3.2, 2.7.1] or [12, 2.2.2, 2.2.3]. 174 M. MOUSSAI

The Fatou property. Well-known the Triebel-Lizorkin space has the Fatou prop- s n erty, cf. [8]. We will brieﬂy recall it. Any f ∈ Fp,q(R ) can be approximated (in 0 n the weak sense in S (R )) by a sequence (fj)j≥0 such that any fj is an entire function of exponential type s n f ∈ F ( ) and lim sup k f k s n ≤ c k fk s n j p,q R j Fp,q (R ) Fp,q (R ) j→+∞ with a positive constant c independent of f. Vice versa, if for a tempered distri- 0 n bution f ∈ S (R ), there exists a sequence (fj)j≥0 such that s n f ∈ F ( ) and A := lim sup k f k s n < +∞ , j p,q R j Fp,q (R ) j→+∞ s n and limj→+∞ fj = f in the sense of distributions, then f belongs to Fp,q(R ) and there exists a constant c > 0 independent of f such that kfk s n ≤ cA. Fp,q (R ) 2.2. Functions of bounded variation

For a function g : R → R, we set

N 1/p X p (2.2) νp(g) := sup |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 νp(g) < +∞. Clearly, by considering a ﬁnite sequence with only two terms, we obtain |g(x) − g(y)| ≤ νp(g), for all x, y ∈ R, hence g is a bounded function. The set of (generalized) primitives 1 of functions of bounded p-variation is denoted by BVp (R) and endowed with the seminorm

kfk 1 := inf ν (g), BVp (R) p where the inﬁmum is taken over all functions g whose f is the primitive. For more details about this space we refer to [11] or [5]. However, we need to recall some embeddings 1 1 BVp (R) ,→ Up (R)(2.3) (equality in case p = 1), see [5, Theorem 5] for the proof which is given for 1 < p < +∞ and can be easily extended to 0 < p ≤ 1, see also [7, Theorem 9.3]. The Peetre embedding theorem ˙ 1+(1/p) 1 ˙ 1+(1/p) (2.4) Bp,1 (R) ,→ BVp (R) ,→ Bp,∞ (R) , (1 ≤ p < +∞), where the dotted space is the homogeneous Besov space.

Example. Let α ∈ R. We put uα(x) := |x + α| − |α| for all x ∈ R, and

fα(x, y) := uα(x)χ[0,1](y) + uα(y)χ[0,1](x), ∀x, y ∈ R, 0 where χ[0,1] denotes the indicatrix function of [0, 1]. Clearly that νp(uα) = 2 and 2 n kχ k 1 = 0. Then it holds f ∈ V ( ) with kf k 2 = 4. The V ( ) [0,1] BVp (R) α p R α Vp(R ) p R space is deﬁned in Section1. COMPOSITION OPERATOR ON THE SPACE OF FUNCTIONS 175

3. Proof of the result

Theorem 1.1 can be obtained from the following statement. Proposition 3.1. Let 0 < p, q < +∞, 0 < u < min(p, q) and 0 < s < 1/p. Then there exists a constant c > 0 such that the inequality s,u,∞ 0 (3.1) M ((f ◦ g) ) ≤ ckfk 1 kgk 1 p,q Up (R) BVp (R) 1 1 1 holds for all f ∈ Up (R) ∩ C (R) and all real analytic functions g in BVp (R). Proof. For a better readability we split our proof in two steps. Step 1. Let us prove s,u,a 0 (1/p)−s (3.2) M ((f ◦ g) ) ≤ c a kfk 1 kgk 1 p,q Up (R) BVp (R) 1 1 1 for all a > 0 and all f ∈ Up (R)∩C (R) and all real analytic functions g in BVp (R). Assume ﬁrst a = 1. By the assumptions on f and g it holds (f ◦g)0 = (f 0 ◦g) g0. 0 0 0 We have k(f ◦ g) k∞ ≤ kf k∞ kg k∞ and 0 0 0 0 0 0 |∆h((f ◦ g)g )(x)| ≤ kf k∞ |∆hg (x)| + |g (x)| |∆h(f ◦ g)(x)|. Hence s,u,1 0 0 s,u,1 0 Mp,q ((f ◦ g) ) ≤ kf k∞ Mp,q (g ) + V (f; g), where V (f; g) (3.3) Z Z 1 1 Z t q/u dtp/q 1/p := t−sq |∆ (f 0 ◦ g)(x)|u|g0(x)|udh dx . t h t R 0 −t −j s,u,1 0 R 1 P∞ R 2 Estimate of Mp,q (g ). By writing 0 ··· = j=0 2−j−1 ··· and by an elementary computation, we have −j Z 1 Z t q/u ∞ Z 2 −sq1 0 u dt X −sq 0 q dt t |∆hg (x)| dh ≤ c1 t sup |∆hg (x)| t t −j−1 t 0 −t j=0 2 |h|≤t ∞ X jsq 0 q ≤ c2 2 sup |∆hg (x)| . −j j=0 |h|≤2

Let α := min(1, p/q). By using the monotonicity of the `r-norms (i.e. `1 ,→ `1/α) and by the Minkowski inequality w.r.t Lp/(αq), since q < +∞, we obtain

Z ∞ p/(αq) 1/p s,u,1 0 X jsαq 0 αq Mp,q (g ) ≤ c1 2 sup |∆hg (x)| dx −j R j=0 |h|≤2 ∞ Z (αq)/p 1/(αq) X jsαq 0 p ≤ c2 2 sup |∆hg (x)| dx −j j=0 R |h|≤2 ∞ 1/(αq) X j(s−(1/p))αq ≤ c 2 kgk 1 . 3 Up (R) j=0 From the embedding (2.3) and the assumption on s, the desired estimate holds. 176 M. MOUSSAI

Estimate of V (f; g). In (3.3) the integral with respect to h can be limited to the interval [0, t] denoting the corresponding expression by V+(f; g). Let us notice that the estimate with respect to [−t, 0] will be completely similar. Again, by applying the Minkowski inequality twice, it holds

V+(f; g) Z Z 1 Z 1 dtu/q p/u 1/p ≤ t−(s+(1/u))q |∆ (f 0 ◦ g)(x)|q |g0(x)|q dh dx h t R 0 h Z 1 Z u/p Z ∞ dtu/q 1/u ≤ |∆ (f 0 ◦ g)(x)|p |g0(x)|p dx t−(s+(1/u))q dh h t 0 R h Z 1 Z u/p 1/u −(su+1) 0 p 0 p ≤ c h |∆h(f ◦ g)(x)| |g (x)| dx dh . 0 R Case 1: Assume that g0 does not vanish on R. By the Mean Value Theorem and by the change of variable y = g(x), we ﬁnd

V+(f; g) Z 1 Z u/p 1/u 0 1−(1/p) −(su+1) 0 0 p ≤ c1kg k∞ h sup |f (v + y) − f (y)| dy dh ≤ 0 0 R |v| hkg k∞ Z 1 1/u 0 u((1/p)−s)−1 ≤ c kfk 1 kg k h dh 2 Up (R) ∞ 0 ≤ c kfk 1 kgk 1 . 3 Up (R) BVp (R) Case 2: Assume that the set of zeros of g0 is nonempty. Then it is a discrete set whose complement in R is the union of a family (Il)l of open disjoint intervals. 0 For any h > 0, we denote by Il,h the set of x ∈ Il whose distance to the boundary of Il is greater than h. We set

00 0 I := Il \ I and gl := g . l,h l,h |Il

0 Clearly the function gl is a diﬀeomorphism of Il onto g(Il). Let us notice that Il,h is an open interval, possibly empty. In case it is not empty, we have

−1 0 0 (3.4) | g(gl (y) + h) − y | ≤ h sup |g |, ∀y ∈ gl(Il,h). Il

00 The set Il,h is an interval of length at most 2h or the union of two such intervals, and g0 vanishes at one of the endpoints of these or those intervals. We write V+(f; g) ≤ V1(f; g) + V2(f; g), where

Z 1 Z u/p 1/u −(su+1) X 0 p 0 p V1(f; g) := h |∆h(f ◦ g)(x)| |g (x)| dx dh 0 0 l Il,h

0 00 and V2(f; g) is deﬁned in the same way by replacing Il,h by Il,h. COMPOSITION OPERATOR ON THE SPACE OF FUNCTIONS 177

Indeed, by the assumption on g, for any Il there exists ξl ∈ Il such that 0 0 |g (ξl)| = sup |g (t)|. t∈Il

Furthermore, set βl the right endpoint of Il. The open intervals { ]ξl, βl[ }l are pairwise disjoint. Then the assertion (3.5) follows from

X 0 p X 0 0 p 0 p sup |g (t)| = |g (ξl) − g (βl)| ≤ νp(g ) . t∈I l l l

(See (2.2) for the deﬁnition of νp).

0 Estimate of V2(f; g). Using both the elementary inequality |∆h(f ◦ g)(x)| ≤ 0 00 2kf k∞ and the properties of Il,h, it holds 1/p Z 1 1/u 0 X 0 p u((1/p)−s)−1 V2(f; g) ≤ c1 kf k∞ sup |g | h dh I l l 0

≤ c kfk 1 kgk 1 . 2 Up (R) BVp (R)

Hence we obtain (3.2) with a = 1. We put gλ(x) := g(λx) for all x ∈ R and all λ > 0. Then (3.2) can be obtained for all a > 0 since kg k 1 = akgk 1 a BVp (R) BVp (R) and s,u,a 0 (1/p)−s−1 s,u,1 0 Mp,q ((f ◦ g) ) = a Mp,q ((f ◦ ga) ).

Step 2: Proof of (3.1). Let a > 0. Let f and g be as in Proposition 3.1. By Proposition 2.2 it holds s,u,∞ 0 0 0 s,u,a 0 M ((f ◦ g) ) ≤ k(f ◦ g) k s = k(f ◦ g) k + M ((f ◦ g) ). p,q Fp,q (R) p p,q Applying (3.2), we obtain s,u,∞ 0 0 0 (1/p)−s (3.6) M ((f ◦ g) ) ≤ kf k kg k + c a kfk 1 kgk 1 p,q ∞ p 1 Up (R) BVp (R) 178 M. MOUSSAI with a positive constant c1 depending only on s, p and q (see the end of Step 1). Now, by replacing g by gλ in (3.6), (gλ is deﬁned in Step 1), and by using the equality s,u,∞ 0 s+1−(1/p) s,u,∞ 0 Mp,q (f ◦ gλ) = λ Mp,q ((f ◦ g) ), we deduce s,u,∞ 0 Mp,q ((f ◦ g) ) (3.7) −s 0 0 (1/p)−s (1/p)−s ≤ λ kf k kg k + c a λ kfk 1 kgk 1 ∞ p 1 Up (R) BVp (R) for all a, λ > 0. Taking a = 1/λ. Now letting λ → +∞ in (3.7), we obtain the desired result. Remark. Proposition 3.1 is also valid in the n-dimensional case. The inequality (3.1) becomes s−1,u,∞ M (∂ (f ◦ g)) ≤ ckfk 1 kgk n , (j = 1, . . . , n) p,q j Up (R) Vp(R ) 1 1 n for all f ∈ Up (R) ∩ C (R) and all real analytic functions g in Vp(R ). Proof of Theorem 1.1. Step 1. Observe that the conditions f(0) = 0 and f 0 ∈ L∞(R) imply 0 kf ◦ gkp ≤ kf k∞ kgkp n which is suﬃcient for the estimate Tf (g) with respect to Lp(R )-norm.

Step 2: The case 1 < s < 1 + (1/p) and n = 1. We ﬁrst consider a function 1 1 1 f ∈ Up (R), of class C and a function g real analytic in Lp(R) ∩ BVp (R). By Proposition 3.1, it holds (3.8) kf ◦ gk s ≤ c kfk 1 kgk + kgk 1 . Fp,q (R) Up (R) p BVp (R) 1 1 Now we prove (3.8) in the general case. Let g ∈ Lp(R) ∩ BVp (R) and f ∈ Up (R). We introduce a function ρ ∈ D(R) satisfying ρ(0) = 1, and we set ϕj(x) := 2jnF −1ρ(2jx) for all x ∈ R and all j ∈ N; here F −1ρ denotes the inverse Fourier transform of ρ. We set also

fj := ϕj ∗ f − ϕj ∗ f(0) and gj := ϕj ∗ g.

Then the function gj is real analytic and gj → g in Lp(R). We have also

(3.9) kg k 1 ≤ c kgk 1 , ∀j ∈ . j BVp (R) BVp (R) N

To prove (3.9), let {]ak, bk[, k = 1,...,N} be a set of pairwise disjoint intervals. By the Minkowski inequality, it holds

N Z p1/p X 0 0 ϕj(y) g (bk − y) − g (ak − y) dy k=1 R Z N p1/p X 0 0 ≤ |ϕj(y)| g (bk − y) − g (ak − y) dy. R k=1 COMPOSITION OPERATOR ON THE SPACE OF FUNCTIONS 179

Now, for all y ∈ R, the intervals ]ak −y, bk −y[(k = 1,...,N) are pairwise disjoint. Then N 1/p X 0 0 p −1 0 |gj(bk) − gj(ak)| ≤ kF ρk1νp(g ) , ∀j ∈ N. k=1 Hence we obtain (3.9). ∞ The functions fj are C such that fj(0) = 0 and satisfy

(3.10) kf k 1 ≤ c kfk 1 , ∀j ∈ . j Up (R) Up (R) N To prove (3.10), for all t > 0 and all h ∈ [−t, t] we trivially have Z 0 0 0 0 |ϕj ∗ f (x + h) − ϕj ∗ f (x)| ≤ |ϕj(y)| sup |f (x − y + z) − f (x − y)|dy. R |z|≤t By the Minkowski inequality, we have Z 0 0 p sup |ϕj ∗ f (x + h) − ϕj ∗ f (x)| dx R |h|≤t Z Z 1/p p 0 0 p ≤ |ϕj(y)| sup |f (x − y + z) − f (x − y)| dx dy R R |z|≤t −1 p 0 p ≤ tkF ρk1 Ap(f ) , (see (1.2) for the deﬁnition of Ap). Consequently, 0 0 −1 0 0 Ap(fj) + kfjk∞ ≤ kF ρk1(Ap(f ) + kf k∞) and we obtain the desired result. On the other hand, we have

(3.11) lim kfj − fk∞ = 0. j→+∞

To prove (3.11), since limj→+∞ ϕj ∗ f(0) = f(0) = 0, the Lipschitz continuous of f yields Z 0 |fj(x) − f(x)| ≤ kf k∞ |x − y||ϕj(x − y)|dy + |ϕj ∗ f(0)| R −j 0 ≤ c 2 kf k∞ + |ϕj ∗ f(0)|. Then the desired result holds. By the same argument, we obtain −j 0 (3.12) kgj − gk∞ ≤ c 2 kg k∞.

Now we apply (3.8) to fj and gj. Then by (3.9) and (3.10), we obtain (3.13) kf ◦ g k s ≤ c kfk 1 kgk + kgk 1 . j j Fp,q (R) Up (R) p BVp (R) The elementary inequality 0 kf ◦ g − fj ◦ gjk∞ ≤ kf k∞kg − gjk∞ + kf − fjk∞ complemented by (3.11)–(3.12) yields the convergence of the sequence {fj ◦gj}j∈N to f ◦ g in L∞(R). Since |hfj ◦ gj − f ◦ g, ψi| ≤ kfj ◦ gj − f ◦ gk∞kψk1, ∀ψ ∈ D(R), 180 M. MOUSSAI thus we conclude that limj→+∞ fj ◦ gj = f ◦ g in the sense of distributions. Hence, s by the Fatou property of Fp,q(R), see Subsection 2.1, we deduce (3.8). Step 3: The case 1 < s < 1 + (1/p) and n ≥ 2. We use the notation (1.3). Since Triebel-Lizorkin space has the Fubini property (see [12, p. 70]), by (3.1) it holds

n Z 1/p X p 0 kf ◦ gk s n ≤ c kf ◦ g 0 k s dx Fp,q (R ) 1 xj F ( ) j n−1 p,q R j=1 R n Z 1/p X p p 0 ≤ c kfk 1 kg 0 k + kg 0 k 1 dx 2 Up (R) xj p xj BV ( ) j n−1 p R j=1 R ≤ c kfk 1 kgk + kgk n . 3 Up (R) p Vp(R ) Step 4: The case 0 < s ≤ 1. Due to the monotonicity of the Triebel-Lizorkin scale with respect to the smoothness parameter s, the result holds. Indeed, let 1 < t < 1 + (1/p). From Step 3, we have (1.4) with kf ◦ gk t n instead of Fp,q (R ) t n s n kf ◦ gk s n . Now we apply the continuous embedding F ( ) ,→ F ( ). Fp,q (R ) p,q R p,q R This completes the proof. Remark. In case n = 1 and 1 ≤ p, q < +∞ the inequality (1.4) becomes kf ◦ gk s ≤ ckfk 1 kgk s + kgk 1 Fp,q (R) Up (R) Fp,q (R) BVp (R) 1 s 1 1 for all g ∈ Lp(R) ∩ BVp (R), since Fp,q(R) ∩ BVp (R) = Lp(R) ∩ BVp (R) if s < ˙ 1+(1/p) 1+(1/p) 1 + (1/p). To prove this equality, we have Bp,∞ (R) ∩ Lp(R) = Bp,∞ (R) 1+(1/p) s (see [12, 2.6.2, p. 95]). Then by (2.4) and by both Bp,∞ (R) ,→ Bp,1(R) and s n s n 1 s Bp,1(R ) ,→ Fp,q(R ), it holds Lp(R) ∩ BVp (R) ,→ Fp,q(R).

4. Concluding remarks

4.1. Some corollaries

In this section we ﬁx a smooth cut-oﬀ function ϕ ∈ D(R) such that ϕ(x) = 1 for −1 |x| ≤ 1. We put ϕt(x) := ϕ t x , ∀x ∈ R and for all t > 0. Also for brevity we s n s n n introduce the space Fp,q(R ) := Fp,q(R ) ∩ L∞(R ) endowed with the quasi-norm

kfk s n := kfk s n + kfk . Fp,q (R ) Fp,q (R ) ∞ Theorem 1.1 has a consequence for the case of functions f which are only locally 1 in Up (R). Corollary 4.1. Let s, p, q be real numbers as in Theorem 1.1. Then there exists a constant c > 0 such that the inequality (4.1) kf ◦ gk s n ≤ ckfϕ k 1 kgk s n + kgk n Fp,q (R ) kgk∞ Up (R) Fp,q (R ) Vp(R ) s n n 1,`oc holds for all functions g ∈ Fp,q(R ) ∩ Vp(R ) and all f ∈ Up (R) satisfying f(0) = 0. Moreover, for all such functions f, the composition operator Tf takes s n n s n Fp,q(R ) ∩ Vp(R ) to Fp,q(R ). COMPOSITION OPERATOR ON THE SPACE OF FUNCTIONS 181

Proof. Since f ◦ g = (fϕkgk∞ ) ◦ g and (fϕt)(0) = 0, the result follows from Theorem 1.1. There is consequence of Theorem 1.1 that we can obtain the equivalence of acting condition and boundedness. Corollary 4.2. Let s, p, q be real numbers as in Theorem 1.1. Let f be a func- 1,`oc tion in Up (R) satisfying f(0) = 0. Then the following assertions are equivalent: s n n s n (i) Tf satisﬁes the acting condition Tf (Fp,q(R ) ∩ Vp(R )) ⊆ Fp,q(R ). s n n s n (ii) Tf maps bounded sets in Fp,q(R ) ∩ Vp(R ) into bounded sets in Fp,q(R ). Proof. Let t > 0. By (4.1), it holds

(4.2) kf ◦ gk s n ≤ c tkfϕ k 1 Fp,q (R ) t Up (R) s n n for all g ∈ F ( ) ∩ V ( ) such that kgk s n + kgk n ≤ t. Now, from p,q R p R Fp,q (R ) Vp(R ) s n n (4.2), we conclude that Tf maps bounded sets in Fp,q(R ) ∩ Vp(R ) into bounded s n sets in Fp,q(R ). s n s n Remark. If n/p < s < 1 + (1/p), then we can replace Fp,q(R ) by Fp,q(R ) in s n n Corollaries 4.1–4.2, since Fp,q(R ) ,→ Cb(R ). We show that Theorem 1.1 can be extended to the case of the boundedness between Besov spaces and Triebel-Lizorkin spaces. Corollary 4.3. Let 1 ≤ p, q < +∞ and 0 < s < 1 + (1/p). Then there exists a constant c > 0 such that the inequality

kf ◦ gkF s ( n) ≤ c kfkU 1( ) kgk 1+(1/p) n p,q R p R Bp,1 (R )

1+(1/p) n 1 holds for all functions g ∈ Bp,1 (R ) and all f ∈ Up (R) satisfying f(0) = 0. 1+(1/p) n s n Moreover, for all such functions f, the operator Tf takes Bp,1 (R ) to Fp,q(R ). Proof. This is an easy consequence of Theorem 1.1 and the following continuous embedding

1+(1/p) n n (4.3) Bp,1 (R ) ,→ Vp(R ). To prove (4.3), we use the notation (1.3) and the equivalent norm in Besov space given by n Z 1 1/q X −sq 2 q dt kfkp + t k∆ fk , (0 < s < 2), tej p t j=1 0 n where {e1, . . . , en} denotes the canonical basis of R , see [15, p. 96]. 1+(1/p) n ˙ 1+(1/p) 1+(1/p) Let f ∈ Bp,1 (R ). Since Bp,1 (R) ∩ Lp(R) = Bp,1 (R) (in the sense of equivalent norms, see, e.g. [15]), then by (2.4), we get n Z 1/p X p 0 kfk n ≤ c kf 0 k dx . Vp(R ) xj 1+(1/p) j n−1 Bp,1 (R) j=1 R 182 M. MOUSSAI

n−1 Using the Minkowski inequality with respect to Lp(R ), it follows Z Z 1 p Z 1 p −(1+(1/p)) 2 dt 0 −(1+(1/p)) 2 dt t k∆ f 0 k dx ≤ t k∆ fk tek xj p j tek p n−1 t t R 0 0 for j, k ∈ {1, . . . , n}. Then we obtain the desired result. Remark. As in Corollary 4.1 we can see the case when the function f associated 1 to the composition operator Tf belongs locally to Up (R). Indeed, if 1 ≤ p, q < +∞ and 0 < s < 1 + (1/p), it holds that

s n 1 kf ◦ gkF ( ) ≤ c kfϕkgk∞ kU ( ) kgk 1+(1/p) n p,q R p R Bp,1 (R ) 1,`oc 1+(1/p) n n for all f ∈ Up (R) such that f(0) = 0 and all g ∈ Bp,1 (R ) ∩ L∞(R ). 4.2. Sharpness of estimate For simplicity we deﬁne

kgk := kgk s n + kgk n . Fp,q (R ) Vp(R ) According to Corollary 4.1, there is a substantial class of nonlinear functions f for which there exist constants cf = c(f) > 0 such that s n n k f ◦ g k s n ≤ c kgk, ∀g ∈ F ( ) ∩ V ( ). Fp,q (R ) f p,q R p R In this form the inequality is optimal if we avoid linear functions in the following sense. Proposition 4.4. Let Ω: [0, +∞) → [0, +∞) be a continuous function satisfying (4.4) lim t1/p Ω(t) = 0. t→+∞ If f is a function such that the inequality

kf ◦ gk s n ≤ Ω(kgk)(4.5) Fp,q (R ) s n n holds for all g ∈ Fp,q(R ) ∩ Vp(R ), then f is an aﬃne function (linear, if we assume that f(0) = 0). Proof. Let us deﬁne a smooth function ϕ ∈ D(Rn) such that ϕ(x) = 1 on the n 2 cube Q := [−1, 1] and ϕ(x) = 0 if x∈ / 2Q. We put ∆h := ∆h ◦ ∆h and 0 n−1 ga(x) := ax1ϕ(x) , (x = (x1, x ) ∈ R × R , a > 0).

We have kgak ∼ a and ∆2 (f ◦ g )(x) = ∆2 f(ax ), (∀x ∈ √1 Q, ∀h ∈ √1 Q, ∀a > 0). h a ah1 1 2( n) 4( n)

s n s n By the above inequality, the embedding Fp,q(R ) ,→ Bp,∞(R ) and the assumption (4.5), we obtain √ Z a/(2 n) 1/p |∆2 f(y)|p dy ≤ c |h|s a1/p Ω(kg k) √ ah1 1 a −a/(2 n) 1/p ≤ c2 a Ω(kgak), (∀h : |h| ≤ 1/4).

By setting u := ah1, we deduce that √ Z a/(2 n) 1/p 2 p 1/p √ |∆uf(y)| dy ≤ c1 a Ω(c2a), ∀a > 0, ∀u : |u| ≤ a. −a/(2 n) By applying the assumption (4.4) on Ω and taking a to +∞, we obtain Z +∞ p |f(y + 2u) − 2f(y + u) + f(y)| dy = 0, ∀u ∈ R. −∞ Hence f(y + 2u) − 2f(y + u) + f(y) = 0 a.e., ∀y, u ∈ R. Then f 0(y + 2u) − f 0(y + u) = 0, i.e., 0 0 0 it implies f (u) = f (0) (∀u ∈ R). We deduce that f is a constant. References

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M. Moussai, Department of Mathematics, University of M’Sila, P.O. Box 166, 28000 M’Sila, Algeria, e-mail: [email protected]