Random Homogenization of Fractional Obstacle Problems

We use a characterization of the fractional Laplacian as a Dirichlet to Neumann operator for an appropriate differential equation to study its obstacle problem in perforated domains.


Introduction
Given a smooth function ϕ : R n → R n and a subset T ε of R n , we consider v ε (x) solution of the following obstacle problem: (−∆) s v ε = 0 for x ∈ R n \ T ε and for x ∈ T ε if v ε (x) > ϕ(x). (1) The operator (−∆) s denotes the fractional Laplace operator of order s, where s is a real number between 0 and 1. It can be defined using Fourier transform, by F((−∆) s f )(ξ) = |ξ| 2s f (ξ). In particular, (1) can be seen as the Euler-Lagrange equation for the minimization of the . H s norm ||f || .
H s = || f (ξ)|ξ| s || L 2 with the constrain that f ≥ ϕ on T ε . We will see that this system of equations can also be stated as a boundary obstacle problem for elliptic degenerate equations.
In (1), the domain R n is perforated and the obstacle ϕ(x) is viewed by v ε (x) only on the subset T ε . A typical example of T ε is given by: with a ε ≪ ε. The goal of this paper is to study the asymptotic behavior of v ε as ε → 0. When T ε is given by (2), the effective equation satisfied by the limit of v ε strongly depends on the radius a ε : If a ε is large enough, the limit turns out to be an obstacle problem with obstacle ϕ(x). On the other hand, if a ε is small then the limiting problem is a simple elliptic equation without any obstacle condition. It is well known in the case of the regular Laplace operator (s = 1) that there is a critical size for a ε for which interesting behavior arises.
In the case of the regular Laplace operator, this problem was first studied for periodic T ε by L. Carbone and F. Colombini [CC80] and then in a more general framework by E. De Giorgi, G. Dal Maso and P. Longo [DGDML80] and G. Dal Maso and P. Longo [DML81], G. Dal Maso [DM81]. Our main reference will be the papers of D. Cioranescu and F. Murat [CM82a,CM82b], in which the case of a periodic distribution of balls is studied. More precisely, they prove that when s = 1 and if T ε is given by (2) with a ε = r 0 ε n n−2 , then the function v = lim ε→0 v ε solves where µ is a real number (depending on r 0 ) and w − = max(−w, 0). The obstacle condition thus disappears when ε goes to zero, but it gives rise to a new term µ(v − ϕ) − in the equation.
In [CM07], we generalize this result (still with s = 1) to sets T ε that are the union of small sets S ε (k) ⊂ R n still periodically distributed, but with random sizes and shapes. More precisely, we introduce a probability space (Ω, F, P) and we assume that for every ω ∈ Ω and every ε > 0 we are given some subsets S ε (k, ω) such that S ε (k, ω) ⊂ B ε (εk).
The only assumptions necessary to generalize the result of D. Cioranescu and F. Murat [CM82a]- [CM82b] is that each set S ε (k, ω) is of capacity of order ε n : cap(S ε (k, ω)) = ε n γ(k, ω) (this is where the critical exponent ε n n−2 comes from) and that the γ(k, ω) have some averaging properties (stationary ergodicity).
In the present paper, we extend the result of [CM07] to the case of fractional Laplace operators s ∈ (0, 1). We will show that under appropriate assumptions on the size of the sets S ε (k, ω), the function v(x) = lim ε→0 v ε (x) solves In the particular case of sets T ε of the form (2), the critical size is now given by a ε = r 0 ε n n−2s (the critical exponent n n−2s is related to the s-capacity of the sets S ε (k, ω)).
In the remainder of this section, we briefly motivate the problem and we introduce the extension problem for the fractional Laplace operators, which allows us to rewrite (1) as a boundary obstacle problem for a local (degenerate) elliptic operator. The precise hypothesis on T ε (ω) will be detailed in the following section in which the precise statement of the main theorem is also given. The remainder of the paper is devoted to the proof of our main statement.
It can be of interest to state equation (3) in a bounded domain D ⊂ R n+1 + : Introducing Σ = D ∩ {y = 0} and Γ = ∂D ∩ {y > 0}, we can consider the following boundary obstacle problem: with the boundary condition Equation (4) arises, for instance, in the modeling of diffusion through semi-permeable membranes (such as the membrane of a cell): The membrane is modeled by the surface Σ. The outside concentration of molecules is given by ϕ(x), and the transport of molecules through the membrane and in the direction of the concentration gradient is possible only across some given channels (represented by the set T ε ) and only from the outside of the cell ({y < 0}) toward the inside of the cell D. At equilibrium, the concentration inside the cell is then given by the solution u(x, y) of (4).

An extension problem for fractional obstacle problems
Following L. Caffarelli, S. Salsa and L. Silvestre [CSS07], we can actually rewrite (1) as a boundary obstacle problem for all fractional powers s ∈ (0, 1). We rely for this on the following extension formula established by L. Caffarelli and L. Silvestre [CS06]: For a given function f (x) defined in R n , if we define u(x, y) by −div (y a ∇u) = 0 for (x, y) ∈ R n+1 We can thus rewrite the fractional obstacle problem (1) as follows: where a = 1 − 2s (note that a ∈ (−1, 1)). Our main result will concern problems such as (6) with possibly bounded domain D instead of R n+1 + . In the sequel, the theory of degenerate elliptic equations in weighted Sobolev spaces will play an important role. We refer to [FKS82] for many results that will be used in this paper.

Variational formulation
The system of equations (6) can also be written as a minimization problem.
For a given open subset D of R n+1 + , we denote by L 2 (D, |y| a ) the weighted L 2 space with weight |y| a and by W 1,2 (D, |y| a ) the corresponding Sobolev's space. We have We then introduce the energy functional: 2 |y| a |∇u| 2 dx dy and the set It is readily seen that (6) is the Euler-Lagrange equation associated to the minimization problem: (Note that since K ε is closed, convex and not empty, (7) has a unique solution u ε ∈ K ε ).
Finally, we notice (see [CS06]) that if u(x, y) is the extension of a function f (x) as in (5), then In particular, the minimization problem (7) is equivalent to the variational formulation of problem (1).
In this paper, we study the asymptotic behavior of the solutions of (7) for any open subset D of R n+1 + . The assumptions and the main result are made precise in the next section. The proof of the main theorem, which is details in Section 3, relies on the construction of an appropriate corrector. This construction is detailed in Sections 4 and 5.
2 Assumptions and Main result 2.1 The set T ε We consider a probability space (Ω, F, P). For all ω ∈ Ω, the set T ε (ω) is given by: where the sets S ε (k, ω) ⊂ R n satisfy the following assumptions: where cap s (A) denotes the s-capacity of subset A of R n+1 (defined below). Moreover, we assume that for some large constant M , and that there exists a constant γ > 0 such that γ(k, ω) ≤ γ for all k ∈ Z n and a.e. ω ∈ Ω.
This first assumption defines the critical size of the set T ε . It will guarantee that cap s (T ε ) remains finite as ε goes to zero. A natural definition for s-capacity of a subset A of R n is the following: Using the extension problem for the fractional Laplce operator (see [CS06] for details), an equivalent definition (up to a multiplicative constant) is given by We will use this second definition in this paper. If B n r is a n-dimensional ball, then its s-capacity in R n+1 is given by cap s (B n r ) = c n+1−a r n−1+a = c n+2s r n−2s for some constant c k . Assumption 1 is thus satisfied in particular if the sets S ε (k, ω) are balls centered on εZ n with radius r(k, ω)ε n n−2s .
Assumption 2: The process γ : Z n × Ω → [0, ∞) is stationary ergodic: There exists a family of measure-preserving transformations τ k : Ω → Ω satisfying γ(k + k ′ , ω) = γ(k, τ k ′ ω) for all k, k ′ ∈ Z n and ω ∈ Ω, and such that if A ⊂ Ω and τ k A = A for all k ∈ Z n , then P (A) = 0 or P (A) = 1 (the only invariant set of positive measure is the whole set).
This second assumption is necessary to ensure that some averaging process occur as ε goes to zero (the hypothesis of stationarity is the most general extension of the notions of periodicity and almost periodicity for a function to have some self-averaging behavior).

Main result
We are now ready to state our main result: and let T ε (ω) be a subset of Σ satisfying Assumptions 1 and 2 above. There exists a constant α 0 ≥ 0 such that for any ϕ(x, y) ∈ C 1,1 (D) the solution u ε (x, y, ω) of converges W 1,2 (D, |y| a )-weak and almost surely ω ∈ Ω to a function u(x, y) solution of the following minimization problem where w − = max(0, −w). If, moreover, there exists γ > 0 such that γ(k, ω) ≥ γ for all k ∈ Z n and a.e. ω ∈ Ω, then α 0 > 0.
We stated Theorem 2.1 in its most general form. It contains the semipermeable membrane problem, as well as our original problem (1) with the fractional operator. More precisely, if we have D = R n+1 + and if we consider the trace v(x) = u(x, 0) in Theorem 2.1 we get: Corollary 2.3 Let T ε be a subset of R n (n ≥ 2) satisfying Assumptions 1 and 2 above. There exists α 0 ≥ 0 such that for any ϕ(x) ∈ C 1,1 (R n ), the solution v ε (x, ω) of (1) converges, as ε → 0, H s (D)-weak and almost surely to a function v(x) solution of As in Cioranescu -Murat [CM82a,CM82b] and Cafarelli-Mellet [CM07], the proof of Theorem 2.1 relies on the construction of an appropriate corrector. More precisely, we use the following result: Proposition 2.4 Let T ε (ω) be a subset of R n satisfying Assumptions 1 and 2 above. There exists a non-negative constant α 0 such that for every bounded subset D of R n+1 + , there is a function w ε 0 (x, y, ω) defined in D and satisfying and s. and for any φ ∈ D(D) such that φ ≥ 0, we have: The proof of Proposition 2.4 will occupy most of this paper. We stress the fact that Assumptions 1 and 2 are sufficient but by no mean necessary to the proof of this Proposition. Any set T ε (ω) such that Proposition 2.4 holds would be admissible for Theorem 2.1.
The condition (13) may seem rather obscure and the next Lemma will suggest a nicer (but stronger) condition to replace it. However (13) is the condition that appears naturally in the proof of Theorem 2.1.
This lemma also gives an indication of how to construct w ε (x, y, ω): We will look for a constant α 0 such that the solution of (14) converges to zero in Then, we have: In order to pass to the limit in (15), we note that we have the following convergences: Hence the first term in the right hand side of (15) goes to zero. Moreover

Related problems
Before turning to the proof of Theorem 2.1, we briefly mention other results that follow from Proposition 2.4: If we consider energy functionals of the form , then a proof similar to that of Theorem 2.1 shows that the homogenization of the following equation More interestingly, we can replace the constrain v ε ≥ ϕ on T ε by a Dirichlet condition of the form v ε = 0 on T ε . This amounts to minimizing J (v) in the convex set The corresponding Euler equation is We can then show that the solution

Proof of Theorem 2.1
In this section, we prove that Theorem 2.1 follows from Proposition 2.4. For the sake of simplicity, we assume that D is a bounded domain in R n+1 + . This allows us to take the corrector w ε (x, y, ω) given by Proposition 2.4 and corresponding to the domain D. When D is unbounded, we note that every integral involving w ε is computed with a compactly supported test function φ. We can thus use, for each of them, the corrector w ε corresponding to the domain supp φ. The final result is of course independent of w ε .
The maximum principle and the natural energy estimate easily give that u ε is bounded in L ∞ (D) ∩ W 1,2 0 (D, |y| a ) almost surely. In particular, there exists a function u(x, u, ω) such that In order to prove Theorem 2.1, we have to show that where J α is the energy associated to the limiting problem, given by: Equality (16) will be a consequence of the following lemmas: Proof of Theorem 2.1: For any v ∈ D(D), we consider the function v + (v − ϕ) − w ε (note that this function satisfies the obstacle constrain). Its energy is given by: Lemma 3.1 and the weak convergence of w ε to 0 in W 1,2 (D, |y| a ) thus implies Morever, it is readily seen that the function v and therefore On the other hand, Lemma 3.2 gives and so Equality (16) follows by a density argument.
Using (12), we obtain: Property (13), together with the facts that u ε ∈ L ∞ (D) and It follows that for any test function z ∈ D(D) we have: We can now take a sequence z n that converges to u strongly in W 1,2 (D, |y| a ) and such that z n (·, 0) converges to u(·, 0) strongly in L 2 (Σ, |y| a ). Using the which concludes the proof.

The auxiliary corrrector 4.1 Notations and scheme of the proof
We recall that R n+1 and we fix a bounded domain D ⊂ R n+1 + . For any x 0 ∈ R n and y 0 > 0, we introduce the following notation for the Euclidian balls:

The fundamental solution
We recall (see [CSS07] for details) that the function where δ(x) denotes the Dirac distribution centered at 0 in R n . We also have div (y a ∇h) = −µ n,a δ(x, y) in R n+1 where δ(x, y) denotes the Dirac distribution centered at 0 in R n+1 and for some constant µ n,a .

An auxiliary corrector
One of the key point in the proof of Proposition 2.4 is to see that away from εk, the set S ε (k, ω) is equivalent to a (n+1)-dimensional ball. More precisely, we introduce the capacitary potential ϕ ε k (x, y, ω) associated to the set S ε (k, ω). It is defined by the following minimization problem: It is readily seen that, almost surely in ω, ϕ ε k (x, y, ω) satisfies and by definition of the capacity as seen in the introduction, Assumption 1 yields Moreover, we have the following lemma (the proof of which is presented in Appendix A): for all (x, y) such that |(x − εk, y)| ≥ ε n n−1+a R δ and for all ε > 0. Moreover, R δ depends only on the constant M appearing in Assumption 1 (in particular, R δ is independent of k and ω).
This Lemma will play a fundamental role in the proof of Proposition 2.4 (see Section 5). It suggests that at distance ε n n−1+a R away from εk, the corrector w ε should behave like the function For later use, we introduce the notation The first step in the proof, and the main goal of this section is to construct a function w ε that would be a good approximation of w ε away from εk and that behaves like h ε k at distance a ε R from εk For that purpose, we introduce where r(k, ω) is chosen in such a way that h ε k (x, y) = 1 on ∂B + r(k,ω)a ε (εk), i.e. r(k, ω) = 2ν n+1+a µ n,a γ(k, ω) 1/(n−1+a) .
We will prove the following proposition: Proposition 4.2 There exist a non-negative real number α 0 (independent of the choice of D) and a function w ε (x, y, ω) satisfying for almost all ω ∈ Ω, such that s. ω ∈ Ω (21) Moreover, we have: The goal of this section is to establish Proposition 4.2. The main advantage of w ε over w ε is that the former only depends on the capacity of S ε (k, ω). This explain why no assumptions are needed on the shape of S ε (k, ω). In the last section of the paper (Section 5), we will see how to use both the functions ϕ ε k (near εk) and the corrector w ε (at distance a ε R of εk) in order to prove Proposition 2.4.

Effective equation
The main idea to prove Proposition 4.2 (and in particular (21)) makes use of the fact that h ε k (x, y, ω) solves: Proposition 4.2 will thus be a consequence of the following proposition: satisfies: This proposition is the main step in the proof of Proposition 4.2 and its proof will occupy most of section.
Equation (22) then becomes: In order to find the critical α 0 for which the solution v ε 0 has the appropriate behavior near the lattice points k ∈ Z n , we follow the method developed by Caffarelli-Souganidis-Wang in [CSW05] and which was already the corner stone in [CM07]: We introduce the following obstacle problem, for every open set A ⊂ R n and for every real number α ∈ R: (26) We then define the smallest super-solution of the obstacle problem: It is readily seen that the function v α,A satisfies and Remark 4.4 The function It is radially symmetric around k and sup |x|=1, y>0 h α,k (x, y) ≤ r n−1+a . In particular, the maximum principle and (28) We now want to show that there exists a critical α 0 such that the followings hold: 1. The solution of the obstacle problem v α,A (x, y, ω) behaves like h α,k (x, y, ω) near any point k ∈ A ∩ Z n .
2. The solution of (25) is not far from v α,A .
For that purpose, we introduce the following quantity, which measures the size of the contact set along the boundary {y = 0}: where |A| denotes the Lebesgue measure of a set A in R n .
The starting point of the proof is the following lemma: has the same distribution for all k ∈ Z n .
Proof of Lemma 4.5: Assume that the finite family of sets (A i ) i∈I is such that and so which gives the subadditive property. Assumption 2 then yields which gives the last assertion of the lemma.
Since m α (A, ω) ≤ |A|, and thanks to the ergodicity of the transformations τ k , it follows from the subadditive ergodic theorem (see [DMM86]) that for each α, there exists a constant ℓ(α) such that where B t (0) denotes the ball centered at the origin with radius t. Note that the limit exists and is the same if instead of B t (0), we use cubes or balls centered at tx 0 for some x 0 . If we scale back and consider the function The next lemma summarizes the properties of ℓ(α): Lemma 4.6 (i) ℓ(α) is a nondecreasing functions of α.
The proof of this Lemma is rather technical and of little interest. It is presented in full details in Appendix B. Using Lemma 4.6, we can define α 0 = sup{α ; ℓ(α) = 0}.
We observe that α 0 is finite (Lemma 4.6 (iii)) and that α 0 is non negative (Lemma 4.6 (ii)). Moreover, α 0 is strictly positive if the γ(k, ω) are bounded from below almost surely by a positive constant.
We now fix a bounded subset A of R n and we denote by the solutions of (27) corresponding to ε −1 A. We also introduce the rescaled function w ε α (x, y, ω) = ε 1−a v ε α (x/ε, y/ε, ω).
In order to complete the proof of Proposition 4.3, we are first going to prove that w ε α satisfies inequality (23), and then that the solution w ε 0 of (22) behaves like w ε α . We recall the definition of h α,k : and we introduce the scaled function h ε α,k (x, y) := ε 1−a h α,k (x/ε, y/ε).
(ii) The proof of (ii) is more delicate and is split in several steps. Preliminary: First of all since A is bounded, we have A ⊂ B n R (x 0 ) for some R. Without loss of generality, we can always assume that B n R (x 0 ) = B n 1 (0). If we consider v ε α (x, y, ω) = v α,ε −1 B n 1 (x, y, ω), the solution of (27) corresponding to A = B n ε −1 (0), it is readily seen that v ε α (x, y, ω) ≤ v ε α (x, y, ω) for all (x, y) ∈ R n+1 + a.e. ω ∈ Ω.
It is thus enough to prove (ii) for v ε α .
In the sequel, we will need the following consequence of Lemma 4.5 (see [CSW05] for the proof): Lemma 4.9 For any ball B n r (x 0 ) ∈ B n 1 (0), the following limit holds, a.s. in ω: Step 1: We now start the proof: For any δ > 0, we can cover B n ε −1 by a finite number N (≤ Cδ −n ) of balls B n i = B n δε −1 (ε −1 x i ) with radius δε −1 and center ε −1 x i . Since α > α 0 , we have ℓ(α) > 0. By Lemma 4.9, we deduce that for every i, there exists ε i such that if ε ≤ ε i , then the n + 1 dimensional ball with same radius and same center as B n i , we now have to show that v ε α remains small in each B + i as long as we stay away from the lattice points k ∈ Z n . More precisely, we want to show that Step 2: Let η(x) be a nonnegative function defined in R n such that 0 ≤ η(x) ≤ 1 for all x, η(x) = 1 in B 1/8 and η = 0 in R n \B 1/4 . We then consider the function u = v ε α ⋆ x η where ⋆ x indicates the convolution in R n with respect to the x-variable. The function u(x, y) is nonnegative on 2B + i and satisfies div (y a ∇u) = 0 where C is a universal constant depending only on n, r and α. We deduce: Lemma 4.10 There exists a universal constant C such that Proof: We write u = u 1 + u 2 where u 1 and u 2 are two functions solution of div (y a ∇u i ) = 0 in 2B + i and satisfying lim y→0 and lim y→0 y a ∂ y u 2 (x, y) = 0 for x ∈ 2B n i , u 2 (x, y) = u(x, y) for (x, y) ∈ ∂(2B + i ) ∩ {y > 0}. The maximum principle and the fact that B i has radius δε −1 yield: On the other hand, boundary Harnack inequality for degenerate elliptic equation (see [FKS82]) implies The Lemma follows easily.
For the next step, we will need the following lemma: a universal constant and ω n+a = B 1 (x 0 ,0) |y| a dx dy.
Proceeding as in [CS06], we now reflect w about the plane {y = 0}. The function is now defined in the whole space R n+1 and it satisfies div (|y| a ∇w) ≤ 0 in B r (x 0 , 0).
We can thus use the mean value formula (see [CSS07]): Since α ≥ 0, we see that v ≤ w and so 2 ω n+a r n+a B + r (x 0 ,0) y a v(x, y) dx dy ≤ 1 ω n+a r n+a Br(x 0 ,0) |y| a w(x, y) dx dy hence the lemma.
Step 3: . Lemma 4.11 thus applies and yields: We want to deduce an upper bound on u in B i . Since u ≥ 0, we note that Then, using the definition of u (and the fact that η(x) = 0 outside B n 1/4 (x)), we deduce: Which, together with (34) yields: Using Lemma 4.10 we see that for every δ and for ε small enough, we have: (36) Step 4: We now want to use (36) to get an upper bound on v ε α . For that purpose, we note that lim y→0 y a ∂ y v ε α ≥ 0 in B i \ ∩ k∈Z n {k}, and so a proof similar to that of Lemma 4.11 yields v ε α (x, y) ≤ C n+a for all (x, y) ∈ B i \ ∩ k∈Z n B 1/4 (k). Inequality (37) and the definition of u(x, y) yield that for all (x, y) in B i \ ∩ k∈Z n B 1/4 (k), we have: Inequality (36) therefore implies sup (x,y)∈∪ k∈Z n B + 1 (k)\B Step 5: In order to complete the proof of the lemma, we only have to notice that since inf ∂B 1/2 h α,k (x, y) ≥ −Cα, (38) and the definition of v ε This conclude the proof of Lemma 4.7, and we are now in position to complete the proof of Propositions 4.3.
We recall that w ε 0 is solution of In order to prove Proposition 4.3, we have to establish (23). This is done in two steps using the properties of the function w ε α : 1. For every α > α 0 , we have div (y a ∇(w ε 0 − w ε α )) = 0 for (x, y) ∈ R n+1 and therefore sup (x,y)∈R N+1 In particular, we thus have and Lemma 4.7 (ii) (since α > α 0 ) yields: (Note that this argument shows the continuity of w ε α with respect to α).

Proof of Proposition 4.2
In order to complete the proof of Proposition 4.2, we construct a corrector w ε which is equal to 1 on the (n+1)-dimensional balls B + r(k,ω)a ε (εk). More precisely, we recall that D is a bounded subset of R n+1 + , and we introduce We then define a corrector w ε (x, y, ω) which will satisfy all the conditions of Proposition 2.4, with the set T ε instead of T ε . In particular, we will prove that w ε behaves like h ε k near the B + r(k,ω)a ε (εk). We consider the following obstacle problem: and we define: w ε (x, y, ω) = inf {w(x, y, ω) ; w solution of (39)} .
It is readily seen that w ε satisfies (20). So in order to complete the proof of Proposition 4.2, we only have to show that w ε is bounded uniformly in L ∞ (D) and that w ε −→ 0 in W 1,2 loc (D, |y| a )-weak as ε goes to zero.
Strong convergence in L 2 (D, |y| a ): First of all, since w ε = 1 = h ε α,k (x, y) + o(1) on T ε , (23) implies which in turn implies (using Proposition 4.3 again): In particular, we get: Moreover, a simple computation shows that and it is readily seen that (40) implies We deduce: and since #{Z n ∩ ε −1 Σ} ≤ Cε −n for all n, we have: In particular as ε goes to zero.
Bound in W 1,2 (D, |y| a ): Using the definition if w ε and an integration by parts, we get: The L ∞ bound thus yieds which completes the proof.

Proof of proposition 2.4
This section is devoted to the proof of the main proposition. We recall that the sets S ε (k, ω) are subsets of R n with unspecified shapes and they satisfy cap s (S ε (k, ω)) = ε n γ(k, ω).
Lemma 4.1 gives the existence of a function ϕ ε k (x, y, ω) such that and we let α 0 and w ε (x, y, ω) be given by Proposition 4.2. We then have: 1. For a given δ > 0, Lemma 4.1 implies that for every k ∈ Z n and ω ∈ Ω there exists a constant R δ (k, ω) such that and for all ε > 0. It is readily seen that for any R there exists ε 1 (R) such that a ε R ≤ ε σ /4 for all ε ≤ ε 1 .
2. Inequality (21) in Proposition 4.2 implies that for given δ and R, there exists ε 2 (δ, R) < ε 1 (R) such that for all ε ≤ ε 2 (δ, R), we have Thanks to (43), Inequality (44) holds in particular in B + 2a ε R \B + a ε R (εk). The corrector will be constructed by gluing together the functions ϕ ε k (near the sets S ε (k)) and the function w ε (away from the sets S ε (k)). The gluing has to be done very carefully so that the corrector satisfies all the properties listed in Proposition 2.4: For a given ε, we define δ ε to be the smallest positive number such that (43) and (44) hold with δ = δ ε and R = R δε . From the remarks above, we see that δ ε is well defined as soon as ε is small enough (say smaller than ε 2 (1, R 1 )). Moreover, for any δ > 0, there exists ε 0 = ε 2 (δ, R δ ) such that δ ε ≤ δ for all ε ≤ ε 0 . In particular lim ε→0 δ ε = 0.
From now on, we write R ε = R δε .
In order to define w ε , we introduce the cut-off function η ε (x, y) defined on D and such that We can always choose η in such a way that . We now set: It satisfies To simplify the notations in the sequel, we denote The properties of w ε are summarized in the following lemma, which implies Proposition 2.4: Lemma 5.1 The function w ε satisfies the following properties: (i) w ε (x, 0) = 1 for x ∈ S ε and ||w ε || L ∞ (D) ≤ C.
(ii) w ε converges to zero in L 2 (D, |y| a )-strong as ε goes to zero.
(iii) Next, we want to show that w ε is bounded in W 1,2 (D, |y| a ). First, we note that outside ∪ k∈Z n B ε/2 (εk) we have ∇w ε = ∇ w ε which is bounded in W 1,2 (D, |y| a ). Next, we see that in B ε/2 (εk), we have: Since w ε and ϕ ε are both bounded in W 1,2 (D, |y| a ) (thanks to (18)), we see that in order to show that ∇w ε is bounded in L 2 (D, |y| a ), we only have to show that D y a |∇η ε ( w ε − ϕ ε )| 2 dx dy ≤ C.
For that purpose, we notice that (42) and (44) yield and so, using the definition of η ε (x, y), we deduce: where we used the fact that we can always assume that δ ε < 1 and R ε ≥ 1. For latter use, we note that we actually proved (iv) It remains to show that (13) holds. We only show the inequality (the equality follows easily). Let v ε be a sequence of functions satisfying: Then for any φ ∈ D(D), we have: where we used the fact that div (y a ∇ w ε ) = 0 on supp η ε and div (y a ∇ϕ ε ) = 0 on supp (1 − η ε ). The first term goes to zero thanks to (46) and the weak convergence of ∇v ε in L 2 (D, |y| a ), and the boundary terms satisfy Finally, the last two terms can be rewritten as: Using the weak convergence of ∇ w ε and ∇ϕ ε to zero, we see that in order to prove (13), it only remains to prove that Since v ε is bounded in L ∞ , it is enough to show that For that purpose, we recall that In particular, interior gradient estimates (see [CSS07]) implies

A Proof of Lemma 4.1
We now turn to the proof of Lemma 4.1. We take k = 0 and we recall that ϕ ε 0 is the capacity potential associated to S ε (0). It satisfies (17) and (18). We then introduce the function for all x, ξ and y. If y > 0, we deduce that for any function ϕ(x, y), we have: τ a ∂ τ G(x, ξ, y, τ )ϕ(ξ, 0) dξ = µ n,a ϕ(x, y).
In order to conclude, we recall that S ε (0) ⊂ B M a ε (0) and so we have |ξ| ≤ M a ε in the previous integral. If (x, y) is such that |(x, y)| ≥ Ra ε with R ≥ 8M , we deduce that for all ξ ∈ S ε (0), we have: We can thus write ϕ ε 0 (x, y) − 2 µ n,a ε n γ(0)h(x, y) h(x, y) where the right hand side is bounded by δ 2 µn,a ε n γ(0)h(x, y) if R is large enough.
B Proof of Lemma 4.6.
(i) For a given set A, it is readily seen from the definition of v α,A that if α ′ ≤ α, then v α ′ ,A is admissible for the obstacle problem with α: It follows that v α,A ≤ v α ′ ,A for any α, α ′ such that α ′ ≤ α and so α → m α (A, ω) is nondecreasing. The result follows from the definition of ℓ(α).
(ii) If α is negative, then we have lim y→0 y a ∂ y v α,tB (x, y) < 0 for x ∈ R n .
Since we can do this in any ball B n 1 (k), we must have m α (tB n , ω) = 0 for all t > 0, and so ℓ(α) = 0 for all α small enough.