A sharpness result for powers of Besov functions

A recent result of Kateb asserts that f∈Bp,qs(ℝn) implies |f|μ∈Bp,qs(ℝn) as soon as the following three conditions hold: (1) 0≺s≺μ

We denote here by B s p,q (R n ) the Besov space -see for instance [9] for the definition. Throughout the paper, we assume that p, q ∈ [1, +∞] and s, µ > 0. In the three following theorems, we shall see that the above conditions are the best possible.
Theorem 1 Assume that µ is not an even integer. Then there exists f ∈ D(R n ) such that |f | µ / ∈ B µ+(1/p) p,q (R n ), for every q < ∞.
Theorem 2 Let µ > 1. Assume that s < n/p or that s = n/p and q > 1. Then there exists a positive unbounded function f such that f ∈ B s p,q (R n ) but f µ / ∈ B s p,q (R n ).
Now we turn to the sharpness of the condition µ > 1. If µ < 1, the function t → |t| µ is not locally Lipschitz continuous; then by a general result of Bourdaud [1], there exist functions f in B s p,q (R n ) such that |f | µ / ∈ B s p,q (R n ). Here we want to go further: the same phenomenon can appear even if the function f is bounded and positive.
In the second section, we construct some specific test functions in Besov spaces. In the third one, we give the proofs of the three theorems. We denote by c, c 1 , c 2 , . . . various strictly positive constants.

Some functions in Besov spaces
Let ρ be a C ∞ function on R such that ρ(x) = 1 for x ≤ e −3 and ρ(x) = 0 for x ≥ e −2 . We denote by ∆ the Laplace operator on R n .
is indefinitely differentiable on R n and satisfies as |ξ| → ∞, for any k ∈ N. In case α = 0, the above estimation can be improved as follows: Proof. This is probably a known result. We outline the proof, following the method of S. Wainger [10, thm. 3, p. 27]. We need a few notations: • J µ is the classical Bessel function (see e.g. [11]) ; • h 0 (t) := | log t| α (log | log t|) −σ , for 0 < t < 1/e ; An easy computation yields the formula with b α,σ,0,m = α(α − 1) · · · (α − m + 1). Then we have First of all, we establish the following alternative formula for u k : for any integer ν ≥ k + 1, where the summation is extended to all the triples (m, j, l) ∈ N 3 such that m + j + l = ν and such that Proof of (3). It is essentially the lemma 8 of Wainger [10]. We first prove (3), for any ν ≥ 1 and without the restriction (4), by using repeated integrations by parts and the identity (see [11, p. 45]). If we take ν = k, the formula becomes In the above formula, we consider the term corresponding to m = j = 0, i.e. l = k, and we perform a further integration by parts; by (5), we obtain By this way, we obtain the desired formula for ν = k + 1 and with condition (4). To obtain the general case ν > k + 1, we pursue the integrations by parts.
It remains to estimate the various terms of (3). We choose ν > (n/2) + 2k + (1/2) and we recall the classical properties of Bessel functions: In the formula (3), we first consider the terms such that j = 0. From the formula (2), we deduce the following estimations: where σ can be replaced by σ + 1 in case α = 0. By using the conditions and the estimations (6), we obtain for r sufficiently large. If we turn to the terms such that j > 0, we have the following trivial estimation: That ends up the proof of proposition 1.
Proposition 2 Let f be the function defined by (1).
Proof. To each couple (α, σ), we associate the sequence (ε j ) j≥2 defined by which belongs to l q if and only if (α, σ) ∈ U q .
Step 1. Let us assume (α, σ) ∈ U q . We are going to prove that f ∈ B n 1,q (R n ), which the smallest relevant Besov space. We use the Littlewood-Paley setting, so we consider a function ψ ∈ D(R n ), with support in the annulus 1 ≤ |ξ| ≤ 3, such that j∈Z ψ(2 j ξ) = 1 (∀ξ = 0) , and we define the operators L j by Since f is clearly integrable, it suffices to prove the following: To do so, we consider the functions g j defined by g j (ξ) := ψ(ξ) f (2 j ξ) .
Then θ ∈ D(R n ) and θ(x) dx = 0. If f would belong to B 0 ∞,q (R n ), then, by a theorem of Peetre [5, thm. 4, p.164] (see also [2, prop. 19]), the sequence defined by f j (x) := 2 nj θ(2 j (x − y))f (y)dy , ∀j ∈ N , would satisfy the following property: Now we are going to prove that a property which contradicts (10). If ω denotes the volume of the unit sphere in R n , we have The second above term plays no role, since it is a O(2 −j ). On the other hand, by formula (2) we have for sufficiently large j's . That ends up the proof of (11) and of proposition 2.
Proof. Step 1. Case σ < 1. Of course we may assume that γ < 1. Thanks to known properties of power functions, we are reduced to estimate we deduce By assumptions, we have αp − γp(β + 1) < −n and αp > −n. Hence we have the wished estimation.
Step 2. Case σ ≥ 1. Now we have to estimate We shall use the following identity : By combining this identity with (13) and with As in Step 1, the two above last terms are estimated by |h| αp+n β+1 . Moreover a short discussion shows that |h| 2p for some r > σ.
Step 3. Now we prove the estimation from below in case of g(t) = sin 2 t 2 . We give the proof for σ ≥ 1 (the case σ < 1 is similar and easier). Assume |h| ≤ c 1 |x| β+1 , with c 1 sufficiently small. Then by (13) we have By using again the identity, we see that J(h) is greater than It is easily seen that ε≤|x|≤1 |x| a | cos(|x| −β )| p dx ≈ ε a+n (ε → 0) , for any a < −n. Then J(h) ≥ c 5 |h| αp+n β+1 , for |h| sufficiently small.

Proof of Theorem 1
This theorem is a classical result. We outline the proof, for the sake of completeness. The function t → ρ(|t|) µ |t| µ does not belong to B µ+(1/p) p,q (R), for q < +∞ (cf. for instance [6, 2.3.1, p. 44]). Hence the function defined by f (x) := x 1 ρ(|x|), for x ∈ R n , belongs to D(R n ), but |f | µ does not belong to B in a neighborhood of 0, the second part of proposition 3 yields f µ / ∈ B s p,q (R).