Global compactness results for nonlocal problems

We obtain a Struwe type global compactness result for a class of nonlinear nonlocal problems involving the fractional $p-$Laplacian operator and nonlinearities at critical growth.

1. Introduction 1.1. Overview. In the seminal paper [21], M. Struwe obtained a very useful global compactness result for Palais-Smale sequences of the energy functional where Ω ⊂ R N is a smooth open and bounded set, N ≥ 3, λ ∈ R, and the space D 1,2 0 (Ω) is defined by The functional above is naturally associated with the semi-linear elliptic problem with critical nonlinearity in the sense that critical points of I are weak solutions of (1.1). Due to the presence of the term with critical growth in its definition, the functional I does not satisfy the Palais-Smale condition. In other words, sequences {u n } n∈N ⊂ D 1,2 0 (Ω) of "almost" critical points of I with bounded energy are not necessarily precompact in D 1,2 0 (Ω). Struwe's result gives a precise description of what happens when compactness fails at an energy level c. Roughly speaking, in this case there exists a (possibly trivial) solution v 0 to (1.1) and k profiles v k solving solving the purely critical problem on the whole space such that the sequence {u n } n∈N can be "almost" written as a superposition of v 0 , . . . , v k . More precisely, there exist {z i n } n∈N ⊂ R N and {λ i n } n∈N ⊂ R + converging to 0 as n → ∞, with where I ∞ is the energy functional associated with equation (1.2), i.e.
This kind of result is very useful to study the existence of ground states for nonlinear Schrödinger equations, Yamabe-type equations or various classes of minimization problems. Since then, several extensions of Struwe's result appeared in the literature for semi-linear elliptic problems. We refer the reader to [12,Lemma 5] for the case of the bilaplacian operator ∆ 2 with both Navier or Dirichlet boundary conditions and to [18,Theorem 1.1] for nonlocal problems involving the fractional Laplacian (−∆) s for s ∈ (0, 1). However, the linearity of the operator does not seem essential in the derivation of this type of results. In fact in [16,Theorem 1.2] (see also [1,25]) a similar result was obtained for signed Palais-Smale sequences of the functional associated with the problem −∆ p u + a |u| p−2 u = µ |u| p * −2 u in Ω u = 0 on ∂Ω, where a ∈ L N/p (Ω), µ > 0, ∆ p is the p−Laplacian operator and p * = N p/(N − p).
Applications of these results are provided to constrained minimization problems, to Brézis-Nirenberg type problems (see [16]) and to Bahri-Coron type problems (see [15]), namely the existence of positive solutions to the purely critical problem in Ω, when the domain Ω has a nontrivial topology. For the aforementioned results in the semi-linear case p = 2, we also refer to the monograph [24].

1.2.
Main results. Let 1 < p < ∞ and s ∈ (0, 1). The aim of this paper is to obtain a global compactness result for Palais-Smale sequences of the C 1 nonlocal energy functional I : D s,p 0 (Ω) → R defined by |u(x) − u(y)| p−2 (u(x) − u(y)) |x − y| N +s p dy, x ∈ R N .
A function u ∈ D s,p 0 (Ω) is a weak solution of (1.5) if |u(x) − u(y)| p−2 (u(x) − u(y)) (v(x) − v(y)) |x − y| N +s p dx dy +ˆR Our main result is the following Theorem 1.1. We assume hypothesis (NA). Let 1 < p < ∞ and s ∈ (0, 1) be such that s p < N . Let Ω ⊂ R N be an open bounded set with smooth boundary. Let {u n } n∈N ⊂ D s,p 0 (Ω) be a Palais-Smale sequence at level c for the functional I defined in (1.4).
Remark 1.4 (Radial case for N = 1). The previous results guarantees that, under the standing assumptions, a radial Palais-Smale sequence can concentrate only at the origin. This is due to the fact that functions in D s,p 0,rad verify some extra compactness properties on annular regions A R 0 ,R 1 = {x : R 0 < |x| < R 1 }, which go up to the exponent p * s (and even beyond). More precisely, we have compactness of the embeddings where p # s is the critical Sobolev exponent in dimension N = 1. As for N = 1 we have p * s = p # s , compactness ceases to be true for s p < N = 1 (see Proposition 4.1 and Remark 4.2). In the one-dimensional case, in Theorem 1.3 one would need (NA) as above.
We point out that, contrary to [16,24], on the weight function a we merely assume it to be in L N/sp (Ω), avoiding an additional coercivity assumption (see [24, condition (B), p.125]) which was used in [16,24] to get the boundedness of the Palais-Smale sequence {u n } n∈N .
The proof by Struwe in [21] is essentially based upon iterated rescaling arguments, jointly with an extension procedure to show the non-triviality of the weak limits. The latter seems hard to adapt to the nonlocal cases, namely when s > 0 is not integer. Thus we prove Theorem 1.1 by basically following the scheme of Clapp's paper [7]. A delicate point will be proving that the weak limits appearing in the construction are non-trivial. As a main ingredient, we use a Caccioppoli inequality for solutions of (−∆) s p u = f (see Proposition 2.9 below). Remark 1.5 (The case p = 2). In the Hilbertian setting, namely for p = 2 and 0 < s < N/2, Theorem 1.1 has been recently proved in [18] by appealing to the so-called profile decomposition of Gerard, see [13]. The latter is a general result describing the compactness defects of general bounded sequences in D s,2 0 (R N ), which are not necessarily Palais-Smale sequences of some energy functional. See also [17,Theorem 1.4], where some improved fractional Sobolev embeddings are obtained. We point out that for p = 2 such an approach does not seem feasible. Indeed, the paper [14] suggests that the decomposition (1.8) should not be expected for a generic bounded sequence in D s,p 0 (R N ) (see [14, page 387]). We also observe that some form of the global compactness result of [18] was also derived in [20] in the study of Coron-type results in the fractional case. Remark 1.6. We also consider a version of the above theorem stated for Palais-Smale sequences with sign, namely Palais-Smale sequences {u n } n∈N with the additional property that the negative parts {(u n ) − } n∈N converges to zero in L p * s . This is particularly interesting if c is a minimax type level (i.e. with mountain pass, saddle point or linking geometry). Indeed, in this case it is often possible to obtain a Palais-Smale sequence with sign at level c via deformation arguments of Critical Point Theory, see [24, Theorem 2.8].
1.3. Notations. For 1 < p < ∞ we consider the monotone function J p : R N → R N defined by We recall that this satisfies We denote by B r (x 0 ) the N −dimensional open ball of radius r, centered at a point x 0 ∈ R N . The symbol · L p (Ω) stands for the standard norm for the L p (Ω) space. For a measurable function u : be its Gagliardo seminorm. For s p < N , we consider the space The previous result implies the following splitting properties.
Lemma 2.2. Let {u n } n∈N ⊂ D s,p 0 (R N ) be such that u n ⇀ u in D s,p 0 (R N ) and u n → u almost everywhere, as n → ∞. Then: Proof. Statement (i 1 ) follows by Lemma 2.1 by choosing With the same choices, we can also obtain (i 3 ) from (2.1). Statement (i 2 ) directly follows from (2.1) with the choices once we recalled that a weakly convergent sequence in D s,p 0 (R N ) weakly converges in L p * s (R N ) as well, thanks to Sobolev inequality. This concludes the proof.
Let I and I ∞ be the functionals defined by (1.4) and (1.6). We recall that I ∈ C 1 (D s,p 0 (Ω)), I ∞ ∈ C 1 (D s,p 0 (H)) and In the following, we repeatedly use the inclusion D s,p 0 (Ω) ֒→ D s,p 0 (R N ). Lemma 2.3. Let a ∈ L N/sp (Ω), assume that {u n } n∈N is bounded in L p * s (Ω) and that u n → u almost everywhere in Ω. Then Proof. Let us set . It is not difficult to see that {φ n } n∈N is bounded in L σ/(σ−1) (Ω) and converges to 0 almost everywhere in Ω, thanks to the assumptions on {u n } n∈N . Thus we obtain and the last limit is zero. Indeed, by Young inequality and Fatou Lemma for every 0 < τ ≪ 1, This proves and by the arbitrariness of τ > 0, we get the conclusion.
Next we produce a Palais-Smale sequence for I ∞ from a Palais-Smale sequence for I.
(Ω) be a Palais-Smale sequence for I at the level c. Assume that Then, passing if necessary to a subsequence, {v n } n∈N := {u n −u} n∈N ⊂ D s,p 0 (Ω) is a Palais-Smale sequence for the functional I ∞ at the level c − I(u). Moreover, we have Proof. We first observe that (2.2) readily gives that I ′ (u) = 0, i.e. u is a critical point of I. By definition and hypothesis (2.2), we have that {|v n | p } n∈N is bounded in L p * s /p (Ω) and v n → 0 a.e. on Ω. Thus it follows that |v n | p converges weakly in L p * s /p (Ω) to 0. Since a ∈ L (p * s /p) ′ (Ω), we can infer lim n→∞ˆΩ a |v n | p dx = 0.
A similar argument, shows that Ω a |u n | p dx =ˆΩ a |u| p dx + o n (1).
. By using the three previous displays and Lemma 2.1 for L p * s (Ω), we have Finally, by virtue Lemma 2.3 applied to the sequence u n − u, we have , whereô n (1) denotes a sequence going to zero in D −s,p ′ (Ω). By using assertions (i 2 ), (i 3 ) and Lemma 2.3 we further get andô n (1) still denotes a sequence going to zero in D −s,p ′ (Ω). This concludes the proof.

2.2.
Scaling invariance and related facts. The following result follows from a direct computation, we leave the verification to the reader. Lemma 2.5 (Scaling invariance). For z ∈ Ω and λ > 0, we set Then, the following facts hold: Next, we transform a Palais-Smale sequence for I ∞ into a new one via rescaling and localization. Assume that {u n } n∈N ⊂ D s,p 0 (Ω) is a Palais-Smale sequence for I ∞ at level c and that the rescaled sequence is a Palais-Smale sequence for I ∞ at level c − I ∞ (v) and such that Proof. Let us assume (2.4), under this assumption the sets Ω n converges to R N . Thus, for every ϕ ∈ C ∞ 0 (R N ) with compact support, we can assume that Ω n contain the support of ϕ for n sufficiently large. From Lemma 2.5 and the hypothesis on {u n } n∈N , it readily follows By arbitrariness of ϕ ∈ C ∞ 0 (R N ), we get the desired conclusion. Before going on, we observe that since v is a critical point of I ∞ , from Lemma B.1 we get For the second part of the statement, we first observe that w n ∈ D s,p 0 (Ω) thanks to Lemma A.1. Thanks to (2.6) we can apply Lemma A.2: by using this and (i 1 ) of Lemma 2.2, we have thanks to the fact that λ n /̺ n converges to 0, by assumption. From the scaling properties of Lemma 2.5, this yields which proves (2.5). Similarly to (2.7), we also have By scaling, (2.7) and (2.8) we get It is only left to show that {w n } n∈N is a Palais-Smale sequence. For any ϕ ∈ D s,p where o n (1) is independent of ϕ. Indeed, by using the compact notations we have We focus on the nonlocal term, the other being easier. By Hölder inequality this is estimated by Let us suppose for simplicity that 1 p > 2. Then we use (1.12) and Hölder inequality with exponents By recalling the definitions of Z n and V n , we get that the first term is uniformly bounded, while the second one coincides with , which converges to 0 thanks to Lemma A.2. This proves (2.9) and by using it in conjunction with Lemma 2.2, we get where o n (1) is independent of ϕ. We now use that {u n } n∈N is a Palais-Smale sequence and that I ′ ∞ (v), ϕ n = 0 by the first part of the proof. This allows us to conclude.
Proof. Under the assumption (2.10), the proof is the same as in the first part of Lemma 2.6, we only have to observe that in this case the sets Ω n converge to a half-space H.
Next we prove that nonsingular scalings of weakly vanishing sequences are weakly vanishing. Then Proof. Take any continuous functional F ∈ D −s,p ′ (R N ). Then, there exists a function ϕ ∈ We have, by a change of variables, On the other hand, introducing the functions of Ψ n , Ψ ∈ L p ′ (R 2N ) by setting and Ψ n → Ψ strongly in L p ′ (R 2N ) as n → ∞, since λ n → λ 0 > 0 and z n → z 0 .

2.3.
Estimates for solutions. Next we prove a Caccioppoli inequality, which will turn out to be the main technical tool in order to handle Step 3 in the proof of Theorem 1.1.
Proof. The proof is the same as that of Caccioppoli inequality [5,Proposition 3.5]. The only differences are that here F is not necessarily (represented by) a function and that the test function ψ can cross the boundary ∂Ω. We insert the test function 2 ϕ = ψ p u, where ψ ∈ C ∞ 0 (Ω) is as in the statement. Then we get We now split the double integral in three parts: |x − y| N +s p u(x) ψ(x) p dx dy, and |x − y| N +s p u(y) ψ(y) p dx dy The first integral I 1 can be estimated exactly as in [5,Proposition 3.5], with the choices there. This gives (2.12) For the estimate of I 2 we proceed similarly to [5], by observing that the positivity assumption on u can be dropped. Namely, we simply observe that by monotonicity of τ → J p (τ ), for x ∈ Ω ′ we have if u(x) < 0. Thus in both cases we get Then we obtain The third integral can be estimated in a similar fashion. By inserting the above estimates in (2.11), we get the conclusion.
Let us set S p,s := inf which is nothing but the sharp constant in the Sobolev inequality for D s,p 0 (R N ), namely (2.14) for all u ∈ D s,p 0 (R N ). 2 Observe that this is a legitimate test function, since ψ p u ∈ D s,p 0 (R N ) by Lemma A.1 and ψ p u ≡ 0 outside Ω.
It is useful to remark that if u ∈ D s,p 0 (E) weakly solves in some open set E ⊂ R N (E = R N is allowed) and for some µ > 0, then we get . Combining this with (2.14) yields the following universal lower bounds for the norms of the nontrivial solutions of problem (2.15), that is This in turn entails the following universal estimate for the energy of solutions This lower bound can be improved, if we consider sign-changing solutions. This is the content of the next useul result.
Lemma 2.10 (Energy doubling). Assume that u ∈ D s,p 0 (E) is a sign-changing weak solution to (2.15) where µ > 0 and E is a (possibly unbounded) domain in R N . Then Proof. For p = 2, see [20,Lemma 2.5]. In the general case, the heuristic idea is to exploit the fact that u ± := max{±u, 0} ∈ D s,p 0 (E) \ {0} are both positive subsolutions of (2.15). Thus the above universal estimates hold for both of them separately. More precisely, it is readily seen that for a.e. (x, y) ∈ R 2N the following inequalities hold Then, testing equation (2.15) by u + (respectively −u − ) yields As before, we can combine these equalities with S p,s u ± By summing up these two inequalities, we get the first estimate in (2.17). The second one is then obtained by observing that from the equation we have Finally, for the third estimate in (2.17) we observe that from the previous identity , which completes the proof.
3. Proof of Theorem 1.1 We divide the proof into five steps.
Step 1. We first observe that the Palais-Smale sequence {u n } n∈N is bounded in D s,p 0 (Ω). In fact, by hypothesis we have as n → ∞, which yields In turn, by Hölder inequality and (3.2), with simple manipulations it follows where C > 0 depends on N, s, p, µ, c and the norm of a, but not on n. Whence, from Step 2. If u 1 Since this sequence is bounded in D s,p 0 (Ω), this yields that [u 1 n ] D s,p (R N ) → 0 as n goes to ∞, thus completing the proof. Let us now suppose that {u 1 n } n∈N does not converge to 0 in L p * s (R N ). Then, up to a subsequence, we have inf n∈NˆRN |u 1 n | p * s dx := δ 0 > 0.
We now take 0 < δ < δ 0 , to be specified later on, and introduce the Levy concentration function Q n (r) := sup ξ∈R NˆB r (ξ) For all n ∈ N, the function r → Q n (r) is continuous on R + (see Lemma 3.1 below). This and the fact that Q n (0) = 0 and Q n (∞) > δ imply the existence of {λ 1 n } n∈N ⊂ R + such that Moreover, since |u n | p * s vanishes outside Ω, still by Lemma 3.1 we know that Before proceeding further, we record the following observation: since if λ 1 n ≥ diam(Ω), then we obtain that the sequence {λ 1 n } n∈N is bounded. This in turn implies that {z 1 n } n∈N is bounded as well, by construction. We consider now the sequence v 1 n : Ω n → R defined by In light of Lemma 2.5 the sequence {v 1 n } n∈N is bounded in D s,p 0 (R N ) (because so is {u 1 n } n∈N ) and thus we can assume that up to a subsequence. Observe also that and this in turn implies that Step 3. The argument that we exploit in this step is substantially different from the argument originally devised by Struwe in [21], requiring a delicate extension procedure on the sequence of approximate solutions. We rather follow a related argument contained in [7].
We now recall that for functions in D s,p 0 (B 3/2 (0)) the following Sobolev inequality holds (see [5,Proposition 2.3] with the choices r = 3/2 and R = 2 there) (3.7) for a constant T = T (N, s, p) > 0. By the Hölder inequality and (3.7), since h v 1 n ∈ D s,p 0 (B 3/2 (0)), it follows thatˆR for some positive constant T depending only on N, s, p. We now observe that by the very Then, by applying Proposition 2.9 for every n ∈ N with the choices (3.9) Observe that thanks to (3.4), we know that B 2 (0) ∩ Ω n in a non-empty open set. We proceed to estimate the terms on the right-hand side of (3.9). For the first term on the right-hand side, we haveˆB thanks to the local strong L p convergence to 0 of {v 1 n } n∈N . For the second term on the right-hand side of (3.9), we observe that for the same reason we havê while by Hölder inequality, for every y ∈ B 3/2 (0) we get which is uniformly bounded. For the third term, by using inequality (3.8), and recalling (3.4) and (3.6), we havê For the last term, since I ′ ∞ (u 1 n ) → 0, we learn from (a 2 ) of Lemma 2.5 that sup ϕ∈D s,p 0 (Ωn) . By introducing the previous estimates in (3.9), we thus get By using again the Sobolev inequality (3.7), this in turn implieŝ By arbitrariness of h ∈ C ∞ 0 (B 1 (z)), we obtain that {v 1 n } n∈N converges to zero in L p * s loc (B 1 (z)). Finally, taking into account the condition (3.6) and the arbitrariness of z ∈ B 1/2 (0), we obtain that {v 1 n } n∈N converges to zero in L p * s (B 1 (0)), which contradicts (3.4). Hence, v 1 = 0.
Step 4. We have already seen in Step 2 that the sequences {z 1 n } n∈N and {λ 1 n } n∈N are bounded, thus we may assume that z 1 n → z 1 0 ∈ R N and λ 1 n → λ 1 0 ≥ 0. If λ 1 0 > 0 then as a consequence of the fact that u 1 n ⇀ 0 in D s,p 0 (Ω), we have v 1 n ⇀ 0 in D s,p 0 (R N ) by Lemma 2.8 and this is impossible by the previous Step 3. Thus λ 1 n → 0 and by construction this implies lim n→∞ dist(z 1 n , ∂Ω) = 0 and z 1 0 ∈ Ω.
We now distinguish two cases: In the first case, by Lemma 2.6 we have Moreover, by recalling (3.5), we obtain that z 1 n ∈ {x ∈ Ω : dist(x, ∂Ω) ≥ λ 1 n }, for n sufficiently large. In the second case, by Lemma 2.7 we would have v 1 ∈ D s,p 0 (H) for a suitable half-space H and On account of Assumption (NA), this case is ruled out. We set ̺ 1 n = dist(z 1 n , ∂Ω)/2 and take ζ ∈ C ∞ 0 (B 2 (0)) a standard cut-off function, such that ζ ≡ 1 on B 1 (0). We consider the sequence by construction we have that λ 1 n /̺ 1 n converges to 0, as n goes to ∞. Thus Lemma 2.6 assures that {u 2 n } n∈N is a Palais-Smale sequence for I ∞ at the energy level Step 5. We can iterate the previous construction to cook-up a sequence {v k } k∈N of critical points of I ∞ and, for every k ∈ N, sequences {z k n } n∈N , {λ k n } n∈N , {̺ k n } n∈N and {u k n } n∈N ⊂ D s,p 0 (Ω) with where ζ is the same cut-off function as above. By construction, we have that {u k n } n∈N is a Palais-Smale sequence for I ∞ at the energy level and, furthermore, Observe that each v 1 , . . . , v k−1 is a critical point of I ∞ , thus from (2.16) we get which implies that this iterative construction must stop at some k 0 ∈ N. As at the beginning of Step 2, this means that [u k 0 n ] D s,p (R N ) → 0 as n goes to ∞. This in turn yields (1.8), (1.9) and (1.10), as desired.

In
Step 2 above we used the following result, which is well-known. We record its proof for the sake of completeness.
Lemma 3.1. Let f ∈ L 1 (R N ), then its Levy concentration function is a continuous function. If f ≡ 0 outside a bounded set K with smooth boundary, then for every r ≥ 0 the supremum in the definition of Q f (r) is actually a maximum. More precisely, we have Proof. The function Q f is monotone non decreasing. Observe that for every ξ ∈ R N , the function is continuous, then Q f is lower semicontinuous as a supremum of continuous functions. Let us suppose that there exists r 0 > 0 such that By monotonicity and lower semicontinuity of Q f , this means that ℓ + > ℓ − = Q f (r 0 ). Let us set ε = ℓ + − Q f (r 0 ), then for every r > r 0 we have By definition of Q f , we can then choose ξ 0 = ξ 0 (ε, r) ∈ R N such that Since the measure of the annulus B r (ξ 0 ) \ B r 0 (ξ 0 ) converges to 0 as r ց r 0 , this gives the desired contradiction. Let us now assume that f = 0 almost everywhere in R N \ K. For every r > 0 the function is continuous and it vanishes if B r (ξ) ⊂ R N \ K. This happens if dist(ξ, K) > r and we conclude the proof.
Remark 3.2. We observe that if the level c satisfies then k in Theorem 1.1 is either 0 (compactness holds) or k = 1 (compactness fails). In the second case, the unique function v 1 must have constant sign and be different from 0 almost everywhere. Indeed, let us assume (3.10) and observe that I(v 0 ) ≥ 0, since v 0 is a critical point of I. If we suppose that v 1 is sign-changing, from Lemma 2.10 and the decomposition (1.10) we would get , thus contradicting (3.10). This implies that v 1 has constant sign and we can conclude that v 1 = 0 almost everywhere, thanks to Proposition B.3.
We say that {u n } n∈N ⊂ D s,p 0 (Ω) is a Palais-Smale sequence with sign for I at level c if it is a Palais-Smale sequence and lim n→∞ (u n ) − L p * s (Ω) = 0.
With minor modifications in the proof of Theorem 1.1, we can get the following variant for Palais-Smale sequences with sign. We leave the details to the reader.
Theorem 3.3. We assume hypothesis (NA). Let 1 < p < ∞ and s ∈ (0, 1) be such that N > s p. Let Ω ⊂ R N be an open bounded set with smooth boundary. Let {u n } n∈N ⊂ D s,p 0 (Ω) be a Palais-Smale sequence with sign for the functional I defined in (1.4) at level c.
Then there exist: in Ω,  Then we have the compact embedding for every 1 ≤ q < p # s and every K ⋐ R N \ {0}.
Proof. Let us start with the case s p > 1. We remark that we already know that the embedding D s,p 0 (B R ) ֒→ L p loc (R N ) is compact (for example, see [3, Theorem 2.7]). A simple interpolation argument permits to infer the desired conclusion. Indeed, let us take q > p, a set K ⋐ R N \ {0}, for every u ∈ D s,p 0,rad (B R ) by using Lemma 4.3 and (4.1) we obtain Thanks to this we can get the desired conclusion.
As far as the case sp ≤ 1 is concerned, we still use that D s,p 0 (B R ) compactly embeds into L p (B R ) and then the assertion follows by Lemma 4.3 jointly with a standard interpolation argument in Lebesgue spaces.
Remark 4.2 (The exponent p # s ). We observe that p # s coincides with the one-dimensional Sobolev exponent. In the case s p < 1 it is not possible to go beyond this exponent in Proposition 4.1. Indeed, for s p < 1 it is not difficult to construct a bounded sequence {u n } n∈N ⊂ D s,p 0,rad (B 1 (0)) such that for a suitable compact set Let us consider the spherical shells If we denote by 1 E the characteristic function of a set E, we observe that the functions u n = n 1 An belong to D s,p 0,rad (B 1 (0)). Indeed, if P (E) denotes the perimeter of a smooth set E ⊂ R N , we have where the last inequality is [3,Corollary 4.4]. It is not difficult to see that On the other hand, for q > p/(1 − s p) we have u n q L q (R N ) = n q |A n | ≃ n q r n = n q− p 1−s p , which diverges. We also point out that the very same example shows that in the limit case q = p # s the embedding is continuous, but not compact.
The previous result was based on the following Radial Lemma for fractional Sobolev spaces. We give the proof for the reader's convenience. For more general results valid in Besov and Triebel spaces, we refer the reader to [19] and [22,Chapter 6]. Lemma 4.3 (A nonlocal Radial Lemma). Let 1 < p < ∞ and s ∈ (0, 1). Let B R be the ball centered at the origin with radius R > 0. Then we have the continuous embeddings: Proof. We divide the proof in three cases.
Case s p > 1. Let 0 < ̺ < R, since u is a radial function we get We observe that the integral is well-defined, since u has a trace in L p (∂B ̺ ) thanks to the hypothesis s p > 1. We can now use the trace inequality for D s,p (B ̺ ) (see [23,Section 3.3.3]), so to obtain for some C = C(N, p, s) > 0. In order to get the desired estimate, it is now sufficient to use Poincaré inequality (which again needs s p > 1) This gives for some C = C(N, s, p) > 0. Observe that inequality (4.1) holds for |x| ≥ R as well, since u ≡ 0 on R N \ B R . We now take K ⋐ R N \ {0}. Then, there exists 0 < R 0 < R 1 such that which proves the desired embedding.
Case s p < 1. Let u ∈ D s,p 0,rad (R N ). We first show that for every 0 < R 0 < R 1 we have (with a slight abuse of notation) for some C = C(N, s, p, R 0 , R 1 ) > 0. Indeed, by arguing as in [2, Lemma B.2], we have For ̺ = r, we make the change of variables Then, the previous expression becomes For every 0 < R 0 < R 1 we thus obtain In order to estimate the last integral, we observe that for R 0 ≤ ̺ ≤ R 1 and R 0 ≤ r ≤ R 1 we have Thus, we proceed as follows (we assume for simplicity N ≥ 3) By spending this information into (4.3), we obtain (4.2). Observe that on the right-hand side of (4.2) we have the one-dimensional Gagliardo seminorm of the function u on the interval [R 0 , R 1 ]. By using Sobolev embedding in dimension 1, we know that for some S = S(s, p, R 0 , R 1 ) > 0. We now prove the claimed embedding. As above we take K ⋐ R N \ {0}. Then, there exists 0 < R 0 < R 1 such that For u ∈ D s,p 0,rad (R N ) we have for some C = C(N, s, p, R 1 ) > 0. We now use polar coordinates, take advantage of the fact that R 0 > 0 and use formula (4.2). Therefore, we have (the constant C may vary from line to line) In the last line we used (4.5). Finally, by using that R 1 < +∞, we get for some C = C(N, s, p, R 0 , R 1 ) > 0. This concludes the proof in the case s p < 1.
Case s p = 1. This is the same proof as before, we only need to observe that in this case, in place of (4.5), we have for every 1 ≤ t < ∞ (4.6)ˆR for some T = T (s, t, p, R 0 , R 1 ) > 0. Then we can proceed as above, we leave the details to the reader.

4.2.
Proof of Theorem 1.3. The proof is the same as that of Theorem 1.1, we only need to modify Step 4 and Step 5 as follows. With the previous notations, as in the proof of Theorem 1.1 we already know that λ 1 n → 0 as n goes to ∞. We now show that this implies that (4.7) Indeed, if this was not the case, up to a subsequence, one would have |z 1 n | ≥ τ 0 eventually for some τ 0 > 0. Taking into account Proposition 4.1, observing that p * s < p # s for N ≥ 2 and recalling that u 1 n converges to 0 almost everywhere, we have u 1 n → 0 in L p * s (K) for every K ⋐ R N \ {0}. Then, for 0 < τ < τ 0 , we conclude since, eventually B λ 1 n (z 1 n ) ∩ B τ (0) = ∅, thanks to the convergence of λ 1 n to 0. The property (4.7) in turn implies that in Step 4 we are in the case covered by Lemma 2.6, i.e. lim n→∞ 1 λ 1 n dist(z 1 n , ∂B) = ∞, thus we do not need Assumption (NA) this time. Then, in order to prove (1.11), we need to remove the translations by z i n from (1.8). This is done by appealing to (4.7) and continuity of L p norms with respect to translations. Indeed, by triangle inequality we have .
By observing that the both norms converge to 0, we get the conclusion.

Appendix A. A truncation Lemma
The following result is proved in [9, Lemma 5.3] under the stronger assumption u ∈ D s,p (R N )∩ L p (R N ). We need to remove the last integrability assumption.
Lemma A.1. Let ψ be a Lipschitz function with compact support and u ∈ D s,p 0 (R N ). Then ψ u ∈ D s,p 0 (R N ) and we have the estimate Proof. We notice that With a simple change of variables, the last integral can be written aŝ By using Hölder inequality with exponents p * s /p and N/sp, Fubini Theorem and triangle inequality, the previous integral can be estimated bŷ |u(x)| p |ψ(x)| p |h| N +s p dh dx |u(x)| p |ψ(x + h)| p |h| N +s p dx dh.
For the first integral containing ψ, we observe that the function For the second integral containing ψ, by using that h → |h| N +s p is integrable at infinity, we simply havê For the last integral, we just observe that for every |h| > 1, the function ψ(· + h) is compactly supported. We thus havê By collecting all the estimates, we conclude the proof.
The following result has been curcially exploited in the proof of Theorem 1.1, in order to localize the rescaled sequences.
Third integral. We proceed similarly as before for the integral in x, we have for every |y| ≥ 4/µ n B 2/µn |ζ(µ n x)| p |x − y| N +s p dx ≤ C µ −N n ζ p L ∞ |y| −N −s p ≤ C ζ p L ∞ |y| −s p , while for 2/µ n ≤ |y| ≤ 4/µ n we can use the Lipschitz character of ζ and get In conclusion we get The first term tends to 0 as before. The second one vanishes since |v| p |y| −s p is integrable, thanks to Hardy inequality for D s,p (R N ) (see [11,Theorem 1.1]). where u − = max{−u, 0} and C = C(N, p, s) > 0 is a constant.
Proof. The proof is exactly the same of that of the Logarithmic Lemma for supersolutions in the case a ≡ 0, see [8,Lemma 1.3]. We take a test function φ ∈ C ∞ 0 (B 3/2 r (x 0 )) such that Then we insert the test function ϕ = φ p (δ + u) 1−p in the equation. By using that Ω a u p−1 φ p (δ + u) p−1 dx ≤ˆB Proposition B.3 (Minimum principle). Let 1 < p < ∞ and s ∈ (0, 1) be such that s p < N .
Let Ω ⊂ R N be an open bounded connected set, a ∈ L N/sp (Ω) and let u ∈ D s,p 0 (Ω) \ {0} be a non negative function such that (−∆) s p u + a u p−1 ≥ µ u q−1 , in Ω, for some µ ∈ R and p ≤ q ≤ p * s . Then we have u > 0 almost everywhere in Ω. Proof. The proof is the same of that for the case a + ≡ 0 and µ ≥ 0, which is contained in [2,Theorem A.1]. It is sufficient to replace the logarithmic estimate there with the one of Lemma B.2 and use that u ∈ L p * s (Ω). We leave the details to the reader.