Geometric Analysis for the Metropolis Algorithm on Lipschitz Domains

Persi Diaconis,∗ Gilles Lebeau,†‡ Laurent Michel§

Departments of Statistics and Mathematics, Stanford University and Département de Mathématiques, Universitéde Nice Sophia-Antipolis

Abstract This paper gives geometric tools: comparison, Nash and Sobolev inequalities for pieces of the relevent Markov operators, that give useful bounds on rates of convergence for the Metropolis algorithm. As an example, we treat the random placement of N hard discs in the unit square, the original application of the Metropolis algorithm.

1 Introduction and Results

d Let Ω be a bounded, connected open subset of R . We assume that its boundary, ∂Ω, has Lipschitz d 1 R regularity. Let B1 be the unit ball of and ϕ(z) = 1 (z) so that ϕ(z)dz = 1. Let ρ(x) R vol(B1) B1 R be a measurable positive bounded function on Ω such that Ω ρ(x)dx = 1. For h ∈]0, 1], set x − y ρ(y) K (x, y) = h−dϕ min , 1 , (1.1) h,ρ h ρ(x) and let Th,ρ be the Metropolis operator associated with these data, that is, Z Th,ρ(u)(x) = mh,ρ(x)u(x) + Kh,ρ(x, y)u(y)dy, Ω Z (1.2) mh,ρ(x) = 1 − Kh,ρ(x, y)dy ≥ 0. Ω

Then the Metropolis kernel Th,ρ(x, dy) = mh,ρ(x)δx=y + Kh,ρ(x, y)dy is a Markov kernel, the 2 operator Th,ρ is self-adjoint on L (Ω, ρ(x)dx), and thus the probability measure ρ(x)dx on Ω is

∗Supported in part by the National Science Foundation, DMS 0505673. †Supported in part by ANR equa-disp Blan07-3-188618 ‡Corresponding author: Parc Valrose 06108 Nice Cedex 02, France; [email protected] §Supported in part by ANR equa-disp Blan07-3-188618

1 n n stationary. For n ≥ 1, we denote by Th,ρ(x, dy) the kernel of the iterated operator (Th,ρ) . For n any x ∈ Ω, Th,ρ(x, dy) is a probability measure on Ω, and our main goal is to get some estimates n on the rate of convergence, when n → +∞, of the probability Th,ρ(x, dy) toward the stationary probability ρ(y)dy. A good example to keep in mind is the random placement of N non-overlapping discs of radius ε > 0 in the unit square. This was the original motivation for the work of Metropolis et al. [MRR+53]. One version of their algorithm goes as follows: from a feasable configuration, pick a disc (uniformly at random) and a point within distance h of the center of the chosen disc (uniformly at random). If recentering the chosen disc at the chosen point results in a feasable configuration, the change is made. Otherwise, the configuration is kept as it started. If N is fixed and ε and h are small, this gives a Markov chain with a uniform stationary distribution over all feasable configurations. The state space consists of the N centers corresponding to feasible configurations. It is a bounded domain with a Lipschitz boundary when N is small (see Section 4, Proposition 4.1). The scientific motivation for the study of random packing of hard discs as a way of understanding the apparent existence of a liquid/solid phase transition for arbitrarily large temperatures (for suitably large pressure) is clearly described in Uhlenbeck [Uhl68, Sect. 5, p. 18]. An overview of the large literature is in Lowen [Löw00].Entry to the zoo of modern algorithms to do the simulation (particularly in the dense case) with many examples is in Krauth [Kra06]. Further discussion, showing that the problem is still of current interest, is in Radin [Rad08]. We shall denote by g(h, ρ) the spectral gap of the Metropolis operator Th,ρ. It is defined as the largest constant such that the following inequality holds true for all u ∈ L2(ρ) = L2(Ω, ρ(x)dx).

2 2 1 kuk − hu, 1i ≤ hu − T u, ui 2 , (1.3) L2(ρ) L2(ρ) g(h, ρ) h,ρ L (ρ) or equivalently, Z Z 2 1 2 |u(x) − u(y)| ρ(x)ρ(y)dxdy ≤ Kh,ρ(x, y)|u(x) − u(y)| ρ(x)dxdy. (1.4) Ω×Ω g(h, ρ) Ω×Ω

d Deﬁnition 1. We say that an open set Ω ⊂ R is Lipschitz if it is bounded and for all a ∈ ∂Ω d 0 there exists an orthonormal basis Ra of R , an open set V = V ×] − α, α[ and a Lipschitz map 0 η : V →] − α, α[ such that in the coordinates of Ra, we have

0 0 0 0 V ∩ Ω = y , yd < η(y ) , (y , yd) ∈ V ×] − α, α[ (1.5) V ∩ ∂Ω = y0, η(y0) , y0 ∈ V 0 .

Our ﬁrst result is the following:

d Theorem 1.1. Let Ω be an open, connected, bounded, Lipschitz subset of R . Let 0 < m ≤ M < ∞ be given numbers. There exists h0 > 0, δ0 ∈]0, 1/2[ and constants Ci > 0 such that for any h ∈]0, h0], and any probability density ρ on Ω which satisﬁes for all x, m ≤ ρ(x) ≤ M, the following holds true.

i) The spectrum of Th,ρ is a subset of [−1+δ0, 1], 1 is a simple eigenvalue of Th,ρ, and Spec(Th,ρ)∩ −2 [1 − δ0, 1] is discrete. Moreover, for any 0 ≤ λ ≤ δ0h , the number of eigenvalues of Th,ρ in 2 d/2 [1 − h λ, 1] (with multiplicity) is bounded by C1(1 + λ) . ii) The spectral gap g(h, ρ) satisﬁes

2 2 C2h ≤ g(h, ρ) ≤ C3h (1.6)

2 and the following estimates hold true for all integer n:

n n −ng(h,ρ) (1 − g(h, ρ)) ≤ sup kTh,ρ(x, dy) − ρ(y)dykTV ≤ C4e . (1.7) x∈Ω

The above results have to be understood as results for small h, other parameters of the problem being fixed. In particular, our estimates are certainly not sharp with respect to the dimension d of the space. For instance, a carefull look at the proof of estimate (1.7) shows that the constant C4 depends badly on the dimension d (if one tracks the dependance with respect to d, the bound d obtained by the Nash estimates can not be better than d , [DSC96]). Proving estimate on C4 with respect to the dimension would be of great interest. The next result will give some more information on the behavior of the spectral gap g(h, ρ) when h → 0. To state this result, let Z 1 Z 1 α = ϕ(z)z2dz = ϕ(z)|z|2dz = (1.8) d 1 d d + 2 and let us define ν(ρ) as the largest constant such that the following inequality holds true for all u in the Sobolev space H1(Ω): Z 2 2 1 αd 2 kukL2(ρ) − hu, 1iL2(ρ) ≤ |∇u| (x)ρ(x)dx, (1.9) ν(ρ) 2 Ω or equivalently, Z α Z |u(x) − u(y)|2ρ(x)ρ(y)dxdy ≤ d |∇u|2(x)ρ(x)dx. (1.10) Ω×Ω ν(ρ) Ω Observe that for a Lipschitz domain Ω, the constant ν(ρ) is well-defined thanks to Sobolev embedding. For a smooth density ρ, this number ν(ρ) > 0 is closely related to the unbounded operator 2 Lρ acting on on L (ρ).

−αd ∇ρ Lρ(u) = (4u + .∇u) 2 ρ (1.11) 1 2 D(Lρ) = u ∈ H (Ω), −∆u ∈ L (Ω), ∂nu|∂Ω = 0

We now justify and explain the choice of domain in (1.11). Background for the following discussion and tools for working in Lipschitz domains is in [AF03]. When Ω has smooth boundary, standard elliptic regularity results show that for any u ∈ H1(Ω) 2 −→ such that −∆u ∈ L (Ω), the normal derivative of u at the boundary, ∂nu = n (x).∇u|∂Ω is well defined and belongs to the Sobolev space H−1/2(∂Ω). Here, we denote by −→n (x) the incoming unit normal vector to ∂Ω at a point x. In the case where ∂Ω has only Lipschitz regularity, the s Sobolev spaces H (∂Ω) are well defined for all s ∈ [−1, 1]. The trace operator, γ0(u) = u|∂Ω maps 1 1/2 1 H (Ω) onto H (∂Ω) = Ran(γ0), and its kernel is Ker(γ0) = H0 (Ω). Equipped with the norm 1/2 ∗ kukH1/2 = inf{kvkH1 , γ0(v) = u} it is an Hilbert space. Then, for any ϕ ∈ H (∂Ω) , there exists −1/2 R 1/2 a unique v ∈ H (∂Ω) such that ϕ(u) = ∂Ω vudσ for all u ∈ H (∂Ω) (where σ is the measure induced on the boundary). For v ∈ H−1/2(∂Ω), the support of v can be defined in a standard way. The trace operator acting on vector fields u ∈ (L2)d with div(u) ∈ L2,

n 2 d 2 o −1/2 γ1 : u ∈ (L (Ω)) , div(u) ∈ L (Ω) → H (∂Ω), (1.12)

3 is then defined by the formula Z Z Z 1 div(u)(x)v(x)dx = − u(x).∇v(x)dx − γ1(u)v|∂Ωdσ(x), ∀v ∈ H (Ω). (1.13) Ω Ω ∂Ω 1 2 In particular, for u ∈ H (Ω) satisfying ∆u = div∇u ∈ L (Ω) we can define ∂nu|∂Ω = γ1(∇u) ∈ −1/2 1 H (∂Ω) and the set D(Lρ) is well defined. From (1.13) we deduce that for any u ∈ H (Ω) with ∆u ∈ L2 and any v ∈ H1(Ω) we have

αd hL u, vi 2 = h∇u, ∇vi 2 + h∂ u, ρvi −1/2 1/2 . (1.14) ρ L (ρ) 2 L (ρ) n H (∂Ω),H (∂Ω)

Then, it is standard that Lρ is the self-adjoint realization of the Dirichlet form α Z d |∇u(x)|2ρ(x)dx. (1.15) 2 Ω

A standard argument [RS78, Sects. 13, 14] using Sobolev embedding show that Lρ has a compact resolvant. Denote its spectrum by ν0 = 0 < ν1 < ν2 < . . . and by mj the multiplicity of νj. In particular, ν(ρ) = ν1. Observe also that m0 = 1 since KerL is spanned by the constant function equal to 1. To state our theorem, we need a basic definition: d Definition 2. Let Ω be a Lipschitz open set of R . We say that ∂Ω is quasi-regular if ∂Ω = Γreg ∪ Γsing, Γreg ∩ Γsing = ∅ with Γreg a finite union of smooth hypersurfaces, relatively open in ∂Ω, d and Γsing a closed subset of R such that −1/2 v ∈ H (∂Ω) and supp(v) ⊂ Γsing =⇒ v = 0. (1.16) Observe that (1.16) is obviously satisfied if ∂Ω is smooth, since in that case one can take Γsing = ∅. More generally, the boundary is quasi-regular if it is ‘piece-wise smooth’ in the following d sense: suppose Ω is a Lipschitz open set of R such that ∂Ω = Γreg ∪ Γsing, Γreg ∩ Γsing = ∅, where d d Γreg is a smooth hypersurface of R , relatively open in ∂Ω, and Γsing a closed subset of R such d that Γsing = ∪j≥2Sj where the Sj are smooth disjoint submanifolds of R such that

codim d S ≥ j, ∪ S = S , (1.17) R j k≥j k j −1/2 then Ω is quasi-regular, since in that case, if v ∈ H (∂Ω) is such that near a point x0, the d support of v is contained in a submanifold S of codimension ≥ 2 in R , then v = 0 near x0. This d −1 d follows from the fact that the distribution hu, φi = hv, φ|∂Ωi on R belongs to H (R ), and if 0 d −1 d u ∈ D (R ) is such that u ∈ H (R ) and supp(u) ⊂ {x1 = x2 = 0}, then u = 0. As an example, a d cube in R is quasi-regular. This ‘piece-wise smooth’ condition (often called “stratified”) is easy to visualize. In our applications (Section 4) it was hard to work with products of stratified sets. The definition we give works easily with products and is exactly what is needed in the proof. d Theorem 1.2. Let Ω be an open, connected, bounded and Lipschitz subset of R , such that ∂Ω is quasi-regular. Assume that the positive density ρ is continuous on Ω. Then lim h−2g(h, ρ) = ν(ρ). (1.18) h→0

Moreover, if the density ρ is smooth on Ω, then for any R > 0 and ε > 0 such that νj+1 − νj > 2ε for νj+2 < R, there exists h1 > 0 such that one has for all h ∈]0, h1], 1 − T Spec h,ρ ∩]0,R] ⊂ ∪ [ν − ε, ν + ε], (1.19) h2 j≥1 j j

1−Th,ρ and the number of eigenvalues of h2 in the interval [νj − ε, νj + ε] is equal to mj.

4 Theorem 1.1 is proved in Section 2. This is done from the spectrum of the operator by comparison with a ‘ball walk’ on a big box B containing Ω. One novelty is the use of ‘normal extensions’ of functions from Ω to B allowing comparison of the two Dirichlet forms. When the Dirichlet forms and stationary distributions for random walk on a compact group are comparable, the rates of convergence are comparable as well [DSC93, Lemma 5]. Here, the Metropolis Markov chain is far from a random walk on a group. Indeed, because of the holding implicit in the Metropolis algorithm, the operator does not have any smoothing properties. The transfer of information is carried out by a Sobolev inequality for a spectrally-truncated part of the operator. This is transfered to a Nash inequality and then an inductive argument is used to obtain decay bounds on iterates of the kernel. A further technique is the use of crude Weyl type estimates to get bounds on the number of eigenvalues close to 1. All of these enter the proof of the total variation estimate (1.7). All of these techniques seem broadly applicable. Theorem 1.2 is proved in Section 3. It gives rigorous underpinnings to a general picture of the spectrum of the Metropolis algorithm based on small steps. This was observed and proved in special cases [DL09], [LM10]. The picture is this: because of the holding (or presence of the multiplier mh,ρ in (1.2)) in the Metropolis algorithm, the operator always has continuous spectrum. This is well isolated from 1 and can be neglected in bounding rates of convergence. The spectrum near 1 is discrete and for h small, merges with the spectrum of an associated Neumann problem. This is an analytic version of the weak convergence of the discrete time Metropolis chain to the Langevin diﬀusion with generator (1.11). In Section 4, we return to the hard disc problem showing that a suitable power of the operators and domains involved satisﬁes our hypothesis. Precisely, in Theorem 4.6 we shall prove that the results of Theorem 1.1 and Theorem 1.2 hold true in this case.

2 A Proof of Theorem 1.1

Let us recall that x − y ρ(y) K (x, y) = h−dϕ min , 1 , (2.1) h,ρ h ρ(x) so that Th,ρ(u) = u − Qh,ρ(u), Z Qh,ρ(u)(x) = Kh,ρ(x, y)(u(x) − u(y)) dy, Ω (2.2) ZZ 1 2 h(1 − Th,ρ)u, uiL2(ρ) = |u(x) − u(y)| Kh,ρ(x, y)ρ(x)dxdy. 2 Ω×Ω

Observe that since Ω is Lipschitz, there exists δ0 > 0 such that for any density ρ with 0 < m ≤ ρ(x) ≤ M one has supx∈Ωmh,ρ(x) ≤ 1 − 2δ0 for all h ∈]0, 1]. Indeed, it follows from (1.2) and (2.1) that mh−d Z 1 − mh,ρ(x) ≥ 1|x−y|

For x ∈ Ω such that dist(x, ∂Ω) > 2h, it follows that 1 − mh,ρ(x) ≥ 1 − m/M. If dist(x, ∂Ω) < 2h, using a Lipschitz local parametrization of the boundary, one shows easily that there exists a constant R d c > 0, independent of h such that Ω 1|x−y|

5 is constant. Therefore, 1 is a simple eigenvalue of Th,ρ. In particular, for any h > 0, the spectral gap satisﬁes g(h, ρ) > 0 (2.4) For the proof of Theorem 1.1, we will not really care about the precise choice of the density ρ. In fact, if ρ1, ρ2 are two densities such that m ≤ ρi(x) ≤ M for all x, then kρ − ρ k ρ (x) ≤ ρ (x) 1 + 1 2 ∞ , 2 1 m (2.5) kρ − ρ k K (x, y)ρ (x) ≤ K (x, y)ρ (x) 1 + 1 2 ∞ , h,ρ1 1 h,ρ2 2 m and this implies, using the deﬁnition (1.4) of the spectral gap and of ν(ρ),

g(h, ρ ) kρ − ρ k 3 1 ≤ 1 + 1 2 ∞ , g(h, ρ ) m 2 (2.6) ν(ρ ) kρ − ρ k 3 1 ≤ 1 + 1 2 ∞ . ν(ρ2) m In particular, it is suﬃcient to prove (1.6) for a constant density. The proof that for some δ0 > 0, independent of ρ, one has Spec(Th,ρ) ⊂ [−1 + δ0, 1] for all h ∈]0, h0] is the following: one has Z 1 2 hu + Th,ρu, uiL2(ρ) = Kh,ρ(x, y)|u(x) + u(y)| ρ(x)dxdy + 2hmh,ρu, uiL2(ρ). (2.7) 2 Ω×Ω

Therefore, it is suﬃcient to prove that there exists h0,C0 > 0 such that the following inequality 2 holds true for all h ∈]0, h0] and all u ∈ L (Ω): Z −d x − y 2 2 h ϕ |u(x) + u(y)| dxdy ≥ C0kukL2(Ω). (2.8) Ω×Ω h

Let ωj ⊂ Ω, ∪jωj = Ω be a covering of Ω such that diam(ωj) < h and for some Ci > 0 independent d of h, vol(ωj) ≥ C1h , and for any j, the number of k such that ωj ∩ ωk 6= ∅ is less than C2. Such a covering exists as Ω is Lipschitz. Then

2C1 2 ≥ kukL2(Ω). |B1| From (2.9), we get that (2.8) holds true. For the proof of (1.6) we need a suitable covering of Ω. Given > 0 small enough, there exists d 2 N some open sets Ω0,..., ΩN such that {x ∈ R , dist(x, Ω) ≤ } ⊂ ∪j=0Ωj, where the Ωj’s have the following properties:

6 2 1. Ω0 = {x ∈ Ω, d(x, ∂Ω) > }.

d 2. For j = 1,...,N, there exists rj > 0, an aﬃne isometry Rj of R and a Lipschitz map d−1 ˜ 0 0 0 ˜ ϕj : R → R such that, denoting φj(x , xd) = (x , xd + ϕj(x )) and φj = Rj ◦ φj, we have

φj is injective on B(0, 2rj)×] − 2, 2[

Ωj = φj(B(0, rj)×] − , [) (2.10) Ωj ∩ Ω = φj(B(0, rj)×]0, [)

φj (B(0, 2rj)×]0, 2[) ⊂ Ω

We put our open set Ω in a large box B =]−A/2, A/2[d and define an extension map E : L2(Ω) → 2 ∞ P 2 L (B). For j = 0,...,N we let χj ∈ C0 (Ωj) be such that j χj(x) = 1 for dist(x, Ω) ≤ . For any 2 −1 function u ∈ L (Ω), let uj, j = 0,...,N be defined in a neighborhood of Ωj by uj = u ◦ φj ◦ S ◦ φj , 0 0 0 0 where S(x , xd) = (x , −xd) if xd < 0 and S(x , xd) = (x , xd) if xd ≥ 0. For x ∈ Ω ∩ Ωj, one has uj(x) = u(x) and we define N X E(u)(x) = χj(x)uj(x). (2.11) j=0 ˜−1 0 0 We observe that φj (x) = (x , xd − ϕj(x )). Consequently, as ϕj is Lipschitz-continuous, then −1 φj and φj are also Lipschitz-continuous. Hence, formula (2.11), gives us an extension map from L2(Ω) into L2(B), which is also bounded from H1(Ω) into H1(B). For u ∈ L2(Ω), v ∈ L2(B), set

Eh,ρ(u) = h(1 − Th,ρ)u, uiL2(ρ), ZZ −d 2 (2.12) Eh(v) = h |v(x) − v(y)| dxdy. B×B,|x−y|≤h Since for A large, E(u) vanishes near the boundary of B, we can extend v = E(u) as an A- d P 2iπkx/A periodic function on , and write its Fourier series v(x) = E(u)(x) = d c (v)e with R k∈Z k −d R −2iπkx/A ck(v) = A B e v(x)dx. Then

2 d X 2 kE(u)kL2(B) = A |ck| , k (2.13) 2 d X 2 2 2 2 kE(u)kH1(B) = A (1 + 4π k /A )|ck| . k

1 1 2 d P 2 2 and since E is bounded from H (Ω) into H (B), then kukL2(Ω) ' A k |ck| and kukH1(Ω) ' d P 2 2 2 2 A k(1 + 4π k /A )|ck| , where f(u) ' g(u) means that there exists some constant C > 0 1 independent of u, such that C f(u) ≤ g(u) ≤ Cf(u). Moreover, one gets d X 2 Eh(v) = A |ck| θ(hk), k Z (2.14) θ(ξ) = |e2iπξz/A − 1|2dz. |z|≤1 Observe that the function θ is nonnegative, quadratic near 0. Moreover, it vanishes if and only if ξ = 0 and it has a positive lower bound for |ξ| ≥ 1. The next two lemmas show that the Dirichlet forms for u ∈ L2(Ω) and its extension to L2(B) are comparable.

7 Lemma 2.1. For all α > 1, there exists C > 0 and h0 > 0 such that 2 Eαh,ρ(u) ≤ CEh,ρ(u) ∀u ∈ L (Ω), ∀h ∈]0, h0]. (2.15) Proof. Using (2.2) and (2.5), we observe that it suffices to prove the lemma in the case where ρ(x) = ρ is constant, and we first show the result when Ω is convex. In that case, since |u(x)−u(y)| ≤ x+y x+y |u(x) − u( 2 )| + |u( 2 ) − u(y)|, one has −d Z Z (hα) 2 Eαh,ρ(u) = 1|x−y|≤αh|u(x) − u(y)| ρdxdy 2vol(B1) Ω Ω −d Z Z 2(hα) x + y 2 ≤ 1|x−y|≤αh|u(x) − u( )| ρdxdy (2.16) vol(B1) Ω Ω 2 −d Z 2(hα/2) 2 = 1|x−y|≤ αh |u(x) − u(y)| ρdxdy, vol(B1) φ(Ω×Ω) 2 x+y where φ(x, y) = (x, ). As Ω is convex φ(Ω × Ω) ⊂ Ω × Ω and we get Eαh,ρ(u) ≤ 4E αh (u). 2 2 ,ρ Iterating this process we obtain the anounced result for convex domains. + In the general case, we use the local covering introduced in (2.10). Let Ωi = Ωi ∩Ω (respectively − d + + Ωi = Ωi ∩ (R \ Ω)) and Ui(h) = {(x, y) ∈ Ωi × Ω, |x − y| ≤ αh}. Since by (2.2), Ω ⊂ ∪iΩi , we PN i have Eαh,ρ(u) ≤ i=0 Eαh,ρ(u) with −d Z i (αh) 2 Eαh,ρ(u) = 1|x−y|≤αh|u(x) − u(y)| ρdxdy. (2.17) 2vol(B1) Ui(h) 0 2 Let us estimate Eαh,ρ(u). For h ∈]0, /α[ and (x, y) ∈ U0(h), we have [x, y] ⊂ Ω. Therefore, the x+y change of variable φ(x, y) = (x, 2 ) maps U0(h) into Ω0 × Ω and we get as above −d Z 0 2(αh) x + y 2 Eαh,ρ(u) ≤ 1|x−y|≤αh|u(x) − u( )| ρdxdy ≤ 4E αh (u). (2.18) 2 ,ρ vol(B1) U0(h) 2 ˜ + ˜ + ˜ ± For i 6= 0 and h > 0 small enough, we remark that Ui(h) ⊂ Ωi ×Ωi , where Ωi = φi(B(0, 2ri)×{0 < ±xd < 2}). Denoting Qi = B(0, ri)×]0, [, Q˜i = B(0, 2ri)×]0, 2[, we can use the Lipschitz- ˜ ˜ + continuous change of variable φi : Qi → Ωi ⊂ Ω to get −d Z Z i (αh) 2 Eαh,ρ(u) ≤ Jφi (x)Jφi (y)1|φi(x)−φi(y)|≤αh|u ◦ φi(x) − u ◦ φi(y)| ρdxdy (2.19) 2vol(B1) Q˜i Q˜i −1 where the Jacobian Jφi of φi is a bounded function defined almost everywhere. As both φi, φi are Lipschitz-continuous, there exists Mi, mi > 0 such that for all x, y ∈ Q˜i we have mi|x − y| ≤ |φi(x) − φi(y)| ≤ Mi|x − y|. Therefore, Z Z i −d 2 Eαh,ρ(u) ≤ Ch 1 αh |u ◦ φi(x) − u ◦ φi(y)| ρdxdy, (2.20) |x−y|≤ m Q˜i Q˜i i where C denotes a positive constant changing from line to line. As Q˜i is convex, it follows from the study of the convex case that Z Z i −d 2 Eαh,ρ(u) ≤ Ch 1 h |u ◦ φi(x) − u ◦ φi(y)| ρdxdy |x−y|≤ M Q˜i Q˜i i Z Z −d 2 ≤ Ch 1|φi(x)−φi(y)|≤h|u ◦ φi(x) − u ◦ φi(y)| ρdxdy (2.21) Q˜i Q˜i Z Z −d 2 ≤ Ch 1|x−y|≤h|u(x) − u(y)| ρdxdy ≤ CiEh,ρ(u), ˜ + ˜ + Ωi Ωi

8 and the proof is complete.

Lemma 2.2. There exist C0, h0 > 0 such that the following holds true for any h ∈]0, h0] and any u ∈ L2(ρ). 2 2 Eh,ρ(u)/C0 ≤ Eh (E(u)) ≤ C0 Eh,ρ(u) + h kukL2 . (2.22) 2 As a byproduct, there exists C1 such that for all h ∈]0, h0] and any function u ∈ L (ρ) such that

2 −2 kukL2(ρ) + h h(1 − Th,ρ)u, uiL2(ρ) ≤ 1,

1 the function u admits a decomposition u = uL +uH with uL ∈ H (Ω), kuLkH1 ≤ C1, and kuH kL2 ≤ C1h. Proof. Using the second line of (2.5), we may assume that the density ρ is constant equal to 1. The proof of the left inequality in (2.22) is obvious. For the upper bound, we remark that there exists PN j,1 j,2 C > 0 such that Eh(E(u)) ≤ C j=0(Eh + Eh ) with Z j,1 −d 2 2 Eh = h 1|x−y|≤h|χj(x) − χj(y)| |uj(x)| dxdy (2.23) B×B and Z j,2 −d 2 2 Eh = h 1|x−y|≤h|χj(y)| |uj(x) − uj(y)| dxdy. (2.24) B×B ∞ As the functions χj are regular, there exist someχ ˜j ∈ C0 (B) equal to 1 near the support of χj such that Z Z j,1 −d 2 2 2 2 Eh ≤ Ch χ˜j(x)|uj(x)| 1|x−y|≤h|x − y| dy dx ≤ Ch kukL2(Ω). (2.25) B B

j,2 In order to estimate Eh one has to estimate the contribution of the points x ∈ Ω, y∈ / Ω and x∈ / Ω, y∈ / Ω. All the terms are treated in the same way and we only examine Z j,3 −d 2 2 Eh = h 1|x−y|≤h|χj(y)| |uj(x) − uj(y)| dxdy Ω×(B\Ω) Z (2.26) −d 2 −1 2 = h 1|x−y|≤h|χj(y)| |u(x) − u ◦ φj ◦ S ◦ φj (y)| dxdy, ˜ + − Ωj ×Ωj

d d with S deﬁned below (2.10). Let σ : R → R be the symmetry with respect to {yd = 0}, so + that Sσ = Id on {yd < 0}. We use the Lipschitz-continuous change of variable ψj : y ∈ Ωj 7→ −1 − φj ◦ σ ◦ φj (y) ∈ Ωj to get Z j,3 −d 2 2 Eh ≤ Ch 1|x−ψj (y)|≤h|χj ◦ ψj(y)| |u(x) − u(y)| dxdy. (2.27) ˜ + + Ωj ×Ωj We claim that there exists β > 0 such that

−1 ˜ + + |ψj(y) − x| ≥ β |x − y| ∀(x, y) ∈ Ωj × Ωj . (2.28)

−1 Indeed, as both φj and φj are Lipschitz-continuous, (2.28) is equivalent to ﬁnding β > 0 such that

−1 −1 ˜ + + |σ(y) − x| ≥ β |x − y| ∀(x, y) ∈ φj (Ωj × Ωj ), (2.29)

9 which is obvious with β = 1. From (2.28) it follows that for some α > 1, one has Z j,3 −d 2 Eh ≤ Ch 1|x−y|≤αh|u(x) − u(y)| dxdy ≤ CEαh,ρ(u), (2.30) ˜ + + Ωj ×Ωj and the upper bound is then a straightforward consequence of Lemma 2.1. 2 2 The proof of the decomposition of u goes as follows. Start from u ∈ L (ρ) such that kukL2(ρ) + −2 h h(1 − Th,ρ)u, uiL2(ρ) ≤ 1 and let v = E(u). It follows from (2.14) and (2.22) that

X 2 0 2 −d 2 |ck| θ(hk) ≤ C0h = A C0h . (2.31) d k∈Z

P 2iπkx/A Let vL = |hk|≤1 cke and vH = v − vL. Since θ is quadratic near 0 and only vanishes at the origin, there exists some constant C > 0 such that θ(ξ) ≥ C|ξ|2 and hence

2 X 2 2 0 2 Ch |ck| |k| ≤ C0h (2.32) |hk|≤1

2 2 0 1 Dividing by h , this shows that kvLkH1(B) ≤ C0/C and hence, uL = (vL)|∂Ω is bounded in H (Ω) uniformly with respect to h. Similarly, since θ is bounded from below by some C > 0 on |ξ| ≥ 1, it follows from (2.31) that

2 X 2 0 2 kvH kL2(B) ≤ C |ck| ≤ C0h . (2.33) h|k|>1

Setting uH = (vH )|∂Ω, this shows that kuH kL2(Ω) = O(h) and the proof of Lemma 2.2 is complete.

We are in position to prove the estimate (1.6) on the spectral gap. Once again, using the second line of (2.5), we may assume that the density ρ is constant equal to 1. To show the right inequality, it suffices to plug a function u ∈ C∞(Ω) with support contained in a ball Q ⊂ Ω and such that R 0 Ω u(x)ρ(x)dx = 0 into (1.3). As Q is convex, it follows from Taylor’s formula that for such u, we 2 have hu − Thu, ui = O(h ). To show the left inequality in (1.6), we first observe that it is clearly satisfied when Ω is convex. Indeed, given u ∈ L2(Ω) we have by Cauchy-Schwarz

Z K(h)−1 Z 2 −1 X 2 |u(x) − u(y)| dxdy ≤ Ch |u (x + k~(y − x)) − u (x + (k + 1)~(y − x)) | dxdy, Ω×Ω k=0 Ω×Ω (2.34) −1 0 where K(h) is the greatest integer ≤ h and ~ = 1/K(h). With the new variables x = x+k~(y−x), 0 0 0 d y = x + (k + 1)~(y − x), one has dx dy = ~ dxdy and we get Z Z 2 −d−1 0 0 2 0 0 |u(x) − u(y)| dxdy ≤ Ch K(h) 1 0 0 |u(x ) − u(y )| dx dy , (2.35) |x −y |<~diam(Ω) Ω×Ω Ω×Ω By lemma 2.1, this proves the left inequality in (1.6) in the case where Ω is convex. + ˜ + In the general case, we can ﬁnd some open sets contained in Ω, ωj ⊂⊂ Ωj ⊂⊂ Ωj , j = + ˜ + + 1,...,N+M such that for j = 1,...,N,Ωj , Ωj are given in the proof of Lemma 2.1, (Ωj )j=N+1,...,N+M

10 M+N + N+M Ω are convex Ω0 ⊂ ∪j=N+1Ωj ,Ω ⊂ ∪j=1 ωj, and where A ⊂⊂ B means that A ⊂ B. Hence for h > 0 small enough,

N+M Z X −d 2 Eh,ρ(u) ≥ C h 1|x−y| 0 independent of h such that the second sum in (2.36) is bounded from below by

N+M N+M X Z X Z Ch2 (u(x) − u(y))2 dxdy ≥ Ch2 (u(x) − u(y))2 dxdy. (2.37) + j=N+1 ωj ×Ωj j=N+1 ωj ×Ω,|x−y|

On the other hand, thanks to the fact that φj is a Lipschitz diﬀeomorphism, there exists α > 0 such that 1|x−y|

N X Z Ch2 (u(x) − u(y))2 dxdy. (2.38) j=1 ωj ×Ω,|x−y|

Combining (2.36), (2.37) and (2.38), we get Z 2 2 Eh,ρ(u) ≥ Ch (u(x) − u(y)) dxdy (2.39) Ω×Ω,|x−y| 0 independent of h. Since by (2.4) we have g(a, ρ) > 0, we get Z 2 2 Eh,ρ(u) ≥ Ch (u(x) − u(y)) dxdy (2.40) Ω×Ω The proof of (1.6) is complete.

Lemma 2.3. There exists δ0 ∈]0, 1/2[ and some constants Ci independent on λ and h such that 2 Spec(Th,ρ) ∩ [1 − δ0, 1] is discrete, and for any 0 ≤ λ ≤ δ0/h , the number of eigenvalues of Th,ρ in 2 d/2 [1−h λ, 1] (with multiplicity) is bounded by C1(1+λ) . Moreover, any eigenfuntion Th,ρ(u) = λu with λ ∈ [1 − δ0, 1] satisﬁes the bound

−d/2 kukL∞ ≤ C2h kukL2 . (2.41)

Proof. To get (2.41), we just write that since λ is not in the range of mh, one has

1 Z x − y ρ(y) u(x) = h−dϕ min , 1 u(y)dy, λ − mh(x) Ω h ρ(x)

11 and we apply Cauchy–Schwarz. The important point here is the estimate on the number of eigenvalues in [1 − h2λ, 1] by a power of λ. This is obtained by the min-max and uses (2.22). The min-max gives: if for some closed subspace F of L2(ρ) with codim(F ) = N one has for all u ∈ F , −2 2 2 h h(1 − Th,ρ)u, uiL2(ρ) ≥ λkukL2(ρ), then the number of eigenvalues of Th in [1 − h λ, 1] (with multiplicity) is bounded by codim(F ) = N. Then, we ﬁx c > 0 small enough, and we choose for F the subspace of functions u such that their extension v = E(u) is such that the Fourier coeﬃcients satisfy ck(E(u)) = 0 for |k| ≤ D with hD ≤ c. The codimension of this space F is exactly the num- d ∗ ber of k ∈ Z such that |k| ≤ D, since if p is a trigonometric polynomial such that E (p) = 0, we R will have Ω p(x)u(x)dx = 0 for any function u with compact support in Ω and such that E(u) = u, and this implies p = 0. Thus codim(F ) ' (1 + D)d. On the other hand, the right inequality in −2 2 2 (2.22) gives for u ∈ F , h h(1 − Th,ρ)u, uiL2(ρ) ≥ C0(D − C1)kukL2(ρ) for universal C0,C1, since by (2.14), there exists C > 0 such that one has θ(hk)h−2 ≥ CD2 for all D ≤ c/h and all |k| > D. The proof of our lemma is complete.

We are now ready to prove the total variation estimate (1.7). We use the notation Th = Th,ρ. 2 Let Π0 be the orthogonal projector in L (ρ) on the space of constant functions Z Π0(u)(x) = 1Ω(x) u(y)ρ(y)dy. (2.42) Ω Then n n 2 sup kTh (x0, dy) − ρ(y)dykTV = kTh − Π0kL∞→L∞ . (2.43) x0∈Ω

To show the left hand side, it suﬃces to consider an eigenfunction eh associated to the eigenvalue ∞ (1 − g(h, ρ)) and normalized in L (Ω). Since eh is orthogonal to constant functions, we get n n n ∞ immediatly 2 supx0∈Ω kTh (x0, dy) − ρ(y)dykTV ≥ kTh eh − Π0ehkL = (1 − g(h, ρ)) . Let us now prove the right inequality. We have to show that there exist C0, h0, such that for any n and any h ∈]0, h0], one has

n −ngh,ρ kTh − Π0kL∞→L∞ ≤ C0e . (2.44)

Observe that since we know that for h0 small, the estimate (1.6) holds true for any ρ, we may −2 assume n ≥ Ch . In order to prove (2.44), we split Th into three pieces, using spectral theory. −2 Let 0 < λ1,h ≤ · · · ≤ λj,h ≤ λj+1,h ≤ · · · ≤ h δ0 be such that the eigenvalues of Th in the 2 interval [1 − δ0, 1[ are the 1 − h λj,h, with associated orthonormal eigenfunctions ej,h,

2 Th(ej,h) = (1 − h λj,h)ej,h, (ej,h|ek,h)L2(ρ) = δj,k. (2.45)

Then we write Th − Π0 = T1,h + T2,h + T3,h with

X 2 T1,h(x, y) = (1 − h λj,h)ej,h(x)ej,h(y), −α λ1,h≤λj,h≤h X 2 T2,h(x, y) = (1 − h λj,h)ej,h(x)ej,h(y), (2.46) −α −2 h <λj,h≤h δ0

T3,h = Th − Π0 − T1,h − T2,h.

n n n n Here α > 0 is a small constant that will be chosen later. One has Th − Π0 = T1,h + T2,h + T3,h, and we will get the bound (2.44) for each of the three terms. We start by very rough bounds. Since

12 −d there are at most Ch eigenvalues λj,h and using the bound (2.41), we get that there exists C independent of n ≥ 1 and h such that

n n −3d/2 kT1,hkL∞→L∞ + kT2,hkL∞→L∞ . ≤ Ch (2.47) n ∞ n n n n Since Th is bounded by 1 on L , we get from Th − Π0 = T1,h + T2,h + T3,h

n −3d/2 kT3,hkL∞→L∞ ≤ Ch (2.48)

Next we use (1.2) to write Th = mh + Rh with

kmhkL∞→L∞ ≤ γ < 1, (2.49) −d/2 kRhkL2→L∞ ≤ C0h . p From this, we deduce that for any p = 1, 2,... , one has Th = Ap,h+Bp,h, with A1,h = mh,B1,h = Rh p p and the recurrence relation Ap+1,h = mhAp,h,Bp+1,h = mhBp,h + RhTh . Thus one gets, since Th is bounded by 1 on L2, p kAp,hkL∞→L∞ ≤ γ , (2.50) −d/2 p −d/2 kBp,hkL2→L∞ ≤ C0h (1 + γ + ··· + γ ) ≤ C0h (1 − γ).

Let θ = 1 − δ0 < 1 so that kT3,hkL2→L2 ≤ θ. Then one has n n n kT3,hkL∞→L2 ≤ kT3,hkL2→L2 ≤ θ , and for n ≥ 1, p ≥ 1, one gets, using (2.50) and (2.48),

p+n p n kT3,h kL∞→L∞ = kTh T3,hkL∞→L∞ n n ≤ kAp,hT3,hkL∞→L∞ + kBp,hT3,hkL∞→L∞ (2.51) −3d/2 p −d/2 n ≤ Ch γ + C0h θ (1 − γ). 2n −α ˜n ˜ Taking p = n, we get kT3,hkL∞→L∞ ≤ Ch θ for some α > 0 and 0 < θ < 1. This shows that there exists some constants C > 0, µ > 0, such that

n −µn kT3,hkL∞→L∞ ≤ Ce , ∀h, ∀n ≥ 1/h, (2.52) n and thus the contribution of T3,h is far smaller than the bound we have to prove in (2.44). n −d Next, for the contribution of T2,h, we just write, since there are at most Ch eigenvalues λj,h and using the bound (2.41),

n X 2 n T2,h(x, y) = (1 − h λj,h) ej,h(x)ej,h(y), −α −2 h <λj,h≤h δ0 (2.53) n −3d/2 2−α n kT2,hkL∞→L∞ ≤ Ch (1 − h ) .

Thus we get for some Cα > 0,

2−α n − nh −2+α/2 kT2,hkL∞→L∞ ≤ Cαe 2 , ∀h, ∀n ≥ h , (2.54) and thus this contribution is still neglectible for h ∈]0, h0] for h0 small. It remains to study the n contribution of T1,h. 2 Let Eα be the (ﬁnite dimensional) subspace of L (ρ) spanned by the eigenvectors ej,h, λj,h ≤ −α −dα/2 h . By Lemma 2.3, one has dim(Eα) ≤ Ch . We next prove a Sobolev-type inequality for the form Eh,ρ. For background on Sobolev and the following Nash inequality, see [DSC96], [SC97].

13 Lemma 2.4. There exist α > 0, p > 2 and C independent of h such that for all u ∈ Eα, the following inequality holds true:

2 −2 2 2 kukLp ≤ Ch Eh,ρ(u) + h kukL2 . (2.55) P Proof. Clearly, one has for u = −α a e ∈ E , λ1,h≤λj,h≤h j j,h α

2 2 X 2 2 Eh,ρ(u) + h kukL2 = h (1 + λj,h)|aj| . −α λ1,h≤λj,h≤h

−2 2 2 −2 Take u ∈ Eα such that h (Eh,ρ(u) + h kukL2 ) ≤ 1. Then by (2.22), one has h Eh(E(u)) ≤ C0. ∞ P 2iπkx/A Let ψ(t) ∈ C ( ) be equal to 1 near t = 0, and for v(x) = d c (v)e , set 0 R k∈Z k X 2iπkx/A v = vL + vH , vL(x) = ψ(h|k|)ck(v)e . (2.56) d k∈Z

Then v = vL + vH is a decomposition of the extension v = E(u) in low frequencies (vL) and high R −d x−y frequencies (vH ). One has vL(x) = d h θ( h )v(y)dy, where θ is the function in the Schwartz ˆ R space deﬁned by θ(2πz/A) = ψ(|z|). Hence, the map v 7→ vL is bounded uniformly in h on all the spaces Lq for 1 ≤ q ≤ ∞. Using the deﬁnition of ψ, it follows from from (2.32) that

kvLkH1(B) ≤ C. (2.57)

2d Thus, with uL = vL|Ω and uH = vH |Ω, we get kuLkH1(Ω) ≤ C so by Sobolev for p < d−2 ,

kuLkLp ≤ C. (2.58) One the other hand, one has also by (2.22),

−2 h Eh (E(ej,h)) ≤ C0(1 + λj,h), (2.59) and this implies, by (2.33),

−2 2 −α h kE(ej,h)H kL2 ≤ C0(1 + λj,h) ≤ C0(1 + h ). (2.60) 1/2 −d/2 Thus for α ≤ 1, we get kE(ej,h)H kL2 ≤ Ch . On the other hand, since kej,hkL∞ ≤ Ch , using the deﬁnition of the low frequency cut-oﬀ we get

−d/2 kE(ej,h)H kL∞ ≤ kE(ej,h)kL∞ + kE(ej,h)LkL∞ ≤ CkE(ej,h)kL∞ ≤ Ch . By interpolation we can ﬁnd some p > 2 such that

1/4 kE(ej,h)H kLp ≤ C0h . (2.61)

P −2 2 2 Thus one gets, for u = −α a e ∈ E with h (E (u) + h kuk 2 ) ≤ 1, λ1,h≤λj,h≤h j j,h α h,ρ L X kuH kLp ≤ |aj|kE(ej,h)H kLp −α λ1,h≤λj,h≤h (2.62) 1/4 1/2 1/4 −dα/4 ≤ C0h dim(Eα) kukL2 ≤ Ch h . Our lemma follows from (2.58) and (2.62) if one takes α small. Observe that here, the estimate on the number of eigenvalues (i.e., the estimation of the dimension of Eα) is crucial. The proof of Lemma 2.4 is complete.

14 p p−2 2 p−1 p−1 From Lemma 2.4, using the interpolation inequality kukL2 ≤ kukLp kukL1 , we deduce the Nash inequality, with 1/D = 2 − 4/p > 0,

2+1/D −2 2 2 1/D kukL2 ≤ Ch Eh,ρ(u) + h kukL2 kukL1 , ∀u ∈ Eα. (2.63)

−α 2 2 2 For λj,h ≤ h , one has h λj,h ≤ 1, and thus for any u ∈ Eα, one gets Eh,ρ(u) ≤ kukL2 −kThukL2 and thus we get, from (2.63),

2+1/D −2 2 2 2 2 1/D kukL2 ≤ Ch kukL2 − kThukL2 + h kukL2 kukL1 , ∀u ∈ Eα. (2.64)

n n n n For 0 < α ≤ 2, we deduce from (2.52),(2.54), and Th − Π0 = T1,h + T2,h + T3,h, that there exists C2 such that n −2+α/2 kT1,hkL∞→L∞ ≤ C2, ∀h, ∀n ≥ h , (2.65) 2 and thus since T1,h is self adjoint on L ,

n −2+α/2 kT1,hkL1→L1 ≤ C2, ∀h, ∀n ≥ h . (2.66)

−2+α/2 2 Fix p ' h . Take g ∈ L such that kgkL1 ≤ 1 and consider the sequence cn, n ≥ 0,

n+p 2 cn = kT1,h gkL2 . (2.67)

Then 0 ≤ cn+1 ≤ cn, and from (2.64) and (2.66), we get

1 1+ 2D −2 2 n+p 1/D cn ≤ Ch cn − cn+1 + h cn kT gk 1 1,h L (2.68) 1/D −2 2 ≤ CC2 h cn − cn+1 + h cn .

2 1 2D From this inequality, we deduce that there exist A ' CC2 sup0≤n≤h−2 (2 + n)(1 + h − (1 − n+2 ) ) −2 Ah−2 2D which depends only on C,C2,D, such that for all 0 ≤ n ≤ h , one has cn ≤ ( 1+n ) , and thus −2 there exist C0 which depends only on C,C2,D, such that for N ' h , one has cN ≤ C0. This implies N+p kT1,h gkL2 ≤ C0kgkL1 , (2.69) and thus taking adjoints, N+p kT1,h gkL∞ ≤ C0kgkL2 , (2.70) and so we get, for any n and with N + p ' h−2,

N+p+n 2 n kT1,h gkL∞ ≤ C0(1 − h λ1,h) kgkL2 . (2.71)

And thus for n ≥ h−2,

−2 2 n −(n−h )h λ1,h λ1,h −ng(h,ρ) −2 kT1,hkL∞→L∞ ≤ C0e = C0e e , ∀h, ∀n ≥ h 2 (2.72) −nC2h ≤ C1e for some C1,C2 > 0 idependant on h, by (1.6). This concludes the proof of Theorem 1.1. Remark 1. We believe that (2.41) is true with a power of Λ instead of a power of h with λ = 1−h2Λ. We have no proof for this which is why we use a Nash inequality for T1,h.

15 Remark 2. The above proof seems to apply for a more general choice of the elementary Markov −d x−y kernel h ϕ( h ). Replace ϕ by a positive symmetric measure of total mass 1 with support in the unit ball, and let Th be the Metropolis algorithm with this data. Assume that one is able to prove that for some δ0 > 0 one has Spec(Th) ⊂ [−1 + δ0, 1] for all h ≤ h0, and that for some power M, one has for some C, c > 0,

M −d Th (x, dy) = µh(x, dy) + Ch 1|x−y|≤chρ(y)dy, µh(x, dy) ≥ 0.

Then there exists γ < 1 such that kµhkL∞ ≤ γ. Moreover, the right inequality in (2.22) and (2.41) M M M are still valid for Th . Also, the spectral gap of Th is given by formula (1.4) with Th (x, dy) in place of Kh,ρ(x, y)dy, and therefore the left inequality in (1.6) holds true, and the right one is true, ∞ 2 since if ρ is constant, for any θ ∈ C0 (Ω), one has u − Thu ∈ O(h ). We shall use these remarks in the study of the hard disc problem, in Section 4.

3 A Proof of Theorem 1.2

In this section, we suppose additionally that Ω is quasi-regular (Deﬁnition 2). For a given continuous ∞ density ρ, using (2.6) and an approximation of ρ in L by a sequence of smooth densities ρk on Ω, one sees that the ﬁrst assertion (1.18) of Theorem 1.2 is a consequence of the second one (1.19). Assume now that ρ is smooth.

∞ Lemma 3.1. Let θ ∈ C (Ω) be such that supp(θ) ∩ Γsing = ∅ and ∂nθ|Γreg = 0. Then, with Qh,ρ deﬁned in (2.2), Lρ deﬁned in (1.11),

2 5/2 Qh,ρ(θ) = h Lρ(θ) + r, krkL2 = O(h ). (3.1)

Proof. For θ ∈ C∞(Ω) and x ∈ Ω, we can use the Taylor formula to get Z 1 ∇ρ(x) 2 2 Qh,ρ(θ)(x) = min 1 + h .z + O(h |z| ), 1 vol(B1) A(x,h) ρ(x)   (3.2) h2 X −h∇θ(x).z − z z ∂ ∂ θ(x) + O(h3|z|3) dz,  2 i j xi xj  i,j

d + − ± with A(x, h) = {z ∈ R , |z| < 1, x + hz ∈ Ω}. As A(x, h) = A (x, h) ∪ A (x, h), with A (x, h) = {z ∈ A(x, h), ±(ρ(x + hz) − ρ(x)) ≥ 0}, it follows by an easy computation that

d h Z h2 X Z Q (θ)(x) = − ∇θ(x). zdz − ∂ ∂ θ(x) z z dz h,ρ vol(B ) 2vol(B ) xi xj i j 1 A(x,h) 1 i,j=1 A(x,h) h2 Z ∇ρ(x) (3.3) − .z∇θ(x).zdz + r(x) vol(B1) A−(x,h) ρ(x)

=f1(x) + f2(x) + f3(x) + r(x),

3 with krkL∞(Ω) = O(h ). Let χ = 1d(x,∂Ω)<2h, then for j = 2, 3,

5/2 kχfjkL2(Ω) ≤ kχkL2(Ω)kfjkL∞ (Ω) = O(h ), (3.4)

16 thanks to the support properties of χ. Moreover, for x ∈ supp(1 − χ), A(x, h) = {|z| < 1} and the αd 2 change of variable z 7→ −z shows that (1 − χ)f2 = −(1 − χ) 2 h ∆θ(x) thanks to (1.8). Hence, α f (x) = − d h2∆θ(x) + r(x), (3.5) 2 2

5/2 with krkL2 = O(h ). 2 To compute f3(x) for x ∈ supp(1 − χ), we first observe that |f3(x)| ≤ Ch |∇ρ(x)||∇θ(x)|. 5/2 1/2 We thus get k1|∇ρ|≤h1/2 f3kL∞ ≤ Ch k∇θkL∞ . At a point x where |∇ρ(x)| ≥ h , we may ∇ρ(x) ⊥ z.∇ρ(x) ⊥ − write z = t |∇ρ(x)| + z , t = |∇ρ(x)| and z .∇ρ(x) = 0. In these coordinates, one has A (x, h) = {|z| < 1, (t, z⊥), t|∇ρ(x)| + O(h(t2 + |z⊥|2)) ≤ 0}. From |∇ρ(x)| ≥ h1/2 we get that the symmetric difference R between A−(x, h) and {t ≤ 0} satisfies meas(R) = O(h1/2) (the symmetric difference of two sets A, B is A ∪ B \ A ∩ B). Therefore Z 2 (1 − χ)(x) ∇ρ(x) 1|∇ρ|≥h1/2 (1−χ)f3(x) = −h 1|∇ρ|≥h1/2 .z∇θ(x).zdz+r(x), (3.6) vol(B1) {|z|<1,∇ρ(x).z≤0} ρ(x)

5/2 ⊥ with krkL∞ = O(h ). Using the change of variable z 7→ z − 2z , we get

2 αd ∇ρ(x) 1 1/2 (1 − χ)f (x) = −h 1 1/2 (1 − χ)(x) .∇θ(x) + r(x), (3.7) |∇ρ|≥h 3 |∇ρ|≥h 2 ρ(x) and therefore, using (3.4), we get

α ∇ρ(x) f (x) = −h2 d .∇θ(x) + r(x), (3.8) 3 2 ρ(x)

5/2 5/2 with krkL2 = O(h ). It remains to show that kf1kL2(Ω) = O(h ). Using the change of variable 0 R z 7→ −z we easily obtain (1−χ)f1 = 0. Hence, it suﬃces to show that f1(x, h) = χ A(x,h) z.∇θ(x)dz 0 satisﬁes kf1kL∞(Ω) = O(h). As Γsing is compact and supp(θ) ∩ Γsing = ∅, this is a local problem near any point x0 of the regular part Γreg of the boundary. Let ψ be a smooth function such that 0 near x0 = (0, 0) one has Ω = {xd > ψ(x )}. For x close to x0 one has

n d 0 0 o A(x, h) = z ∈ R , |z| < 1, xd + hzd > ψ(x + hz ) . (3.9)

Set n d 0 0 0o A1(x, h) = z ∈ R , |z| < 1, xd + hzd > ψ(x ) + h∇ψ(x )z , (3.10) then the symmetric diﬀerence R between A(x, h) and A1(x, h) satisﬁes meas(R) = O(h) uniformly in x close to x0. This yields Z 0 f1(x, h) = ∇θ(x).v(x, h) + r(x), v(x, h) = zdz, (3.11) A1(x,h)

0 with krkL∞ = O(h). Let ν(x) be the vector ﬁeld deﬁned by ν(x) = (−∇ψ(x ), 1). Observe that 0 ψ(x )−xd ν(x) R v(x, h) = φ( h|ν(x)| ) |ν(x)| with φ(a) = |z|<1,z >a z1dz, vanishes for dist(x, ∂Ω) > Ch and that for 1 −→ x ∈ ∂Ω, ν(x) is collinear to the unit normal to the boundary n (x). Since ∂nθ|Γreg = 0, we thus get 0 kf1kL∞ = O(h). The proof of our lemma is complete.

17 Let us recall that we denote 1 = ν0 < ν1 < ··· < νj < . . . the eigenvalues of Lρ and mj the associated multiplicities. We introduce the bilinear form

αd a (u, v) = h∇u, ∇vi 2 + hu, vi 2 . (3.12) ρ 2 L (ρ) L (ρ)

1 This deﬁnes an Hilbertian structure on H (Ω) which is equivalent to the usual one. We write k.k 1 Hρ for the norm induced by aρ. We denote

∞ D0 = θ ∈ C (Ω), θ = 0 near Γsing, ∂nθ|Γreg = 0 . (3.13)

1 Lemma 3.2. D0 is dense in H (Ω).

1 Proof. Let f ∈ H (Ω) be orthogonal to D0 for the inner product aρ deﬁned above. Then, it is ∞ 2 orthogonal to C0 (Ω) so that (Lρ+1)f = 0 in the sense of distributions. In particular −∆f ∈ L (Ω). Hence we can use the Green formula (1.14) to get for any θ ∈ D0, since aρ(f, θ) = 0,

h∂nf, ρθiH−1/2,H1/2 = 0. (3.14)

∞ ˜ ˜ For any ψ ∈ C0 (Γreg), using smooth local coordinates we can find ψ in D0 such that ψ|∂Ω = ψ. Consequently, ˜ h∂nf, ρψiH−1/2,H1/2 = h∂nf, ρψiH−1/2,H1/2 = 0. (3.15) −1/2 Hence, ∂nf|Γreg = 0. This shows that ∂nf|∂Ω ∈ H is supported in Γsing. From (1.16) this implies ∂nf|∂Ω = 0. This shows that f ∈ D(Lρ). As the operator Lρ + 1 is strictly positive, this implies f = 0. The proof of our lemma is complete. We are now in position to complete the proof of Theorem 1.2. Let K ∈ N be fixed (independent 2 of h). We first observe that if νh ∈ [0,M] and ψk,h ∈ L (ρ), k = 1,...,K satisfy kψk,hkL2 = 1, −2 h Qh,ρψk,h = νhψk,h and ψi,h orthogonal to ψj,h for any j 6= i, then thanks to Lemma 2.2 the 2 family (ψk,h)h∈]0,1] is relatively compact in L (ρ) for any k, so that we can suppose (extracting a 2 subsequence hn) that νh → ν, ψk,h → ψk in L (ρ), kψkkL2 = 1 and ψi orthogonal to ψj for any 1 j 6= i. Moreover by Lemma 2.2, the limit ψk belongs to H (ρ). Given θ ∈ D0, it follows from self-ajointness of Qh,ρ and Lemma 3.1 that

−2 1/2 0 = (h Qh,ρ − νh)ψk,h, θ L2(ρ) = hψk,h, (Lρ − νh)θiL2(ρ) + O(h ). (3.16)

Making h → 0 we obtain hψk, (Lρ − ν)θiL2(ρ) = 0 for all θ ∈ D0. It follows that (Lρ − ν)ψk = 0 1 in the distribution sense, and integrating by parts that ∂nψk vanish on Γreg. Since ψk ∈ H (ρ), we get as above using (1.16) that ∂nψk = 0, and it follows that ψk ∈ D(Lρ). This shows that ν is an eigenvalue of Lρ, and thus (1.19) is satisﬁed. Moreover, the inequality dim(Ker(Lρ − ν)) ≥ K, shows that that for any > 0 small enough, there exists h > 0 small enough such that

−2 ]Spec(h Qh,ρ) ∩ [νj − , νj + ] ≤ mj, (3.17) for h ∈]0, h]. It remains to show that there is equality in (3.17), and we shall proceed by induction on j. Let > 0, small, be given such that for 0 ≤ νj ≤ M + 1, the intervals Ij = [νj − , νj + ] are −2 PN disjoint. Let (µj(h))j≥0 be the increasing sequence of eigenvalues of h Qh,ρ, σN = j=1 mj and (ek)k≥0 the eigenfunctions of Lρ such that for all k ∈ {1+σN , . . . , σN+1}, one has (Lρ−νN+1)ek = 0. As 0 is a simple eigenvalue of both Lρ and Qh,ρ, we have clearly ν0 = µ0 = 0 and m0 = 1 = −2 ]Spec(h Qh,ρ) ∩ [ν0 − , ν0 + ].

18 −2 Suppose that for all n ≤ N, mn = ]Spec(h Qh,ρ) ∩ [νn − , νn + ]. Then, one has by (1.19), for h ≤ hε,

µ1+σN (h) ≥ νN+1 − . (3.18) 1 By the min-max principle, if G is a ﬁnite dimensional subspace of H with dim(G) = 1 + σN+1, one has −2 µσN+1 (h) ≤ sup hh Qh,ρψ, ψiL2(ρ) (3.19) ψ∈G,kψk=1

Thanks to Lemma 3.2, for all ek, 0 ≤ k ≤ σN+1 and all α > 0, there exists ek,α ∈ D0 such that ke − e k 1 ≤ α. Let G be the vector space spanned by the e , 0 ≤ k ≤ σ . For α small k k,α Hρ α k,α N+1 enough, one has dim(Gα) = 1 + σN+1. From Lemma 3.1, one has

−2 1/2 hh Qh,ρek,α, ek0,αiL2(ρ) = hLρek,α, ek0,αiL2(ρ) + Oα(h ). (3.20)

αd Since e ∈ D , one has hL e , e 0 i 2 = h∇e , ∇e 0 i 2 and h∇e , ∇e 0 i 2 = h∇e , k,α 0 ρ k,α k ,α L (ρ) 2 k,α k ,α Lρ k,α k ,α Lρ k ∇e 0 i 2 + O(α). Therefore, for ψ ∈ G , kψk = 1, we get k Lρ α

−2 1/2 hh Qh,ρψ, ψiL2(ρ) ≤ νN+1 + Cα + Oα(h ). (3.21)

Taking α > 0 small enough and h < hα << 1, we obtain from (3.19) and (3.21), µσN+1 (h) ≤ νN+1 + −2 . Combining this with (3.18) and (3.17), we get mN+1 = ]Spec(h Qh,ρ) ∩ [νN+1 − , νN+1 + ]. The proof of Theorem 1.2 is complete.

4 Application to Random Placement of Non-Overlapping Balls

In this section, we suppose that Ω is a bounded, Lipschitz, quasi-regular, connected, open subset d of R with d ≥ 2. Let N ∈ N,N ≥ 2 and > 0 be given. Let ON, be the open bounded subset of Nd R , N ON, = x = (x1, . . . , xN ) ∈ Ω , ∀ 1 ≤ i < j ≤ N, |xi − xj| > d 1 and recall that B1 denotes the unit ball in and ϕ(z) = 1 (z). We introduce the kernel R vol(B1) B1

N 1 X xj − yj K (x, dy) = δ ⊗ · · · ⊗ δ ⊗ h−dϕ dy ⊗ δ ⊗ · · · ⊗ δ , (4.1) h N x1 xj−1 h j xj+1 xN j=1

2 and the associated Metropolis operator on L (ON,) Z Th(u)(x) = mh(x)u(x) + u(y)Kh(x, dy), (4.2) ON, with Z mh(x) = 1 − Kh(x, dy). (4.3) ON, 2 The operator Th is Markov and self -adjoint on L (ON,). The conﬁguration space ON, is the set d of N disjoint closed balls of radius /2 in R , with centers at the xj ∈ Ω. The topology of this set, and the geometry of its boundary is generally hard to understand, but since d ≥ 2, ON, is clearly non-void and connected for a given N if is small enough. The Metropolis kernel Th is associated to the following algorithm: at each step, we choose uniformly at random a ball, and we move its

19 d center uniformly at random in R in a ball of radius h. If the new configuration is in ON,, the change is made. Otherwise, the configuration is kept as it started. In order to study the random walk associated to Th, we will assume that N and are such that N is small enough. Under this condition, we prove in Proposition 4.1 that the open set ON, is connected, Lipschitz and quasi-regular, and in Proposition 4.4 we prove that the kernel of the M iterated operator Th (with M large, but independent of h) admits a suitable lower bound, so that we will be able to use Remark 2 at the end of Section 2. The main results are collected together in Theorem 4.6 below. We define Γreg and Γsing the set of regular and singular points of ∂ON, as follows. Denote NN = {1,...,N}. For x ∈ ON, set

R(x) = {i ∈ NN , xi ∈ ∂Ω},

S(x) = {τ = (τ1, τ2) ∈ NN , τ1 < τ2 and |xτ1 − xτ2 | = } , (4.4) r(x) = ]R(x), s(x) = ]S(x).

The functions r and s are lower semi-continuous and any x ∈ ON, belongs to ∂ON, iﬀ r(x)+s(x) ≥ 1. Deﬁne Γreg = x ∈ ON,, s(x) = 1 and r(x) = 0 (4.5) ∪ x ∈ ON,, s(x) = 0,R(x) = {j0} and xj0 ∈ ∂Ωreg and Γsing = ∂ON, \ Γreg. Then Γsing is clearly closed, and the Γreg is the union of smooth disjoint Nd hypersurfaces in R .

Proposition 4.1. There exists α > 0 such that for N ≤ α, the set ON, is arcwise connected, Lipschitz and quasi-regular. Remark 3. Observe that in the above Proposition, the smallness condition on is N ≤ α where α > 0 depends only on Ω. The condition Nd ≤ c, which says that the density of the balls is suﬃciently small, does not imply that the set ON, has Lipschitz regularity. As an example, if 2 Ω =]0, 1[ is the unit square in the plane, then x = (x1, . . . , xN ), xj = ((j − 1), 0), j = 1,...,N, 1 with = N−1 is a conﬁguration point in the boundary ∂ON,. However, ∂ON, is not Lipschitz at x: otherwise, there would exist νj = (aj, bj) such that (x1 + tν1, . . . , xN + tνN ) ∈ ON, for t > 0 small enough, and this implies a1 > 0, aj+1 > aj and aN < 0 which is impossible. Proof. For ν ∈ Sp−1, p ≥ 1 and δ ∈]0, 1[, denote

p Γ±(ν, δ) = {ξ ∈ R , ±hξ, νi > (1 − δ)|ξ|, |hξ, νi| < δ} . (4.6) p We remark [AF03] that an open set O ⊂ R is Lipschitz if and only if it satisﬁes the cone property: p−1 ∀a ∈ ∂O, ∃δ > 0, ∃νa ∈ S , ∀b ∈ B(a, δ) ∩ ∂O we have

p b + Γ+(νa, δ) ⊂ O and b + Γ−(νa, δ) ⊂ R \ O. (4.7)

Let us ﬁrst show that ON, is connected for N small. For x ∈ ON, deﬁne

I(x) = inf |xi − xj|. (4.8) i6=j Then I(x) > and we have the following lemma.

Lemma 4.2. There exists α0 > 0 such that for any N ∈ N, > 0 with N ≤ α0, there exists δN, > 0 such that for any x ∈ ON, with I(x) < α0/N, there exists a continuous path γ : [0, 1] → ON, such that γ(0) = x and I(γ(1)) ≥ I(x) + δN,.

20 Proof. As Ω is bounded and Lipschitz, a compactness argument shows that there exists δ0 > 0, r0 > 0 such that

d−1 ∀x0 ∈ Ω, ∃ν ∈ S , ∀x ∈ B(x0, r0) ∩ Ω, x + Γ+(ν, δ0) ⊂ Ω (4.9) d−1 d ∀x0 ∈ ∂Ω, ∃ν ∈ S , ∀x ∈ B(x0, r0) ∩ ∂Ω, x + Γ−(ν, δ0) ⊂ R \ Ω.

∗ 3 2 Let 0 < α0 < min(δ0, r0)/100. For K ∈ N denote δK = α0/K , ρK = 10α0/K . Observe that it suﬃces to show the following statement:

∗ ∀K ∈ N , ∀ ∈]0, α0/K], ∀N ∈ NK , ∀x ∈ ON, s.t. I(x) < α0/K, ∃γ ∈ C([0, 1], ON,), s.t. (4.10)

γ(0) = x, I(γ(1)) ≥ I(x) + δK and ∀t ∈ [0, 1], |x − γ(t)|∞ ≤ NρK

Let K ≥ 1 and 0 < < α0/K. We proceed by induction on N ∈ NK . (Recall that NK = {0, 1,...,K}.) In the case N = 1, there is nothing to show. Suppose that the above property holds true at rank N − 1 and let x ∈ ON, be such that I(x) < α0/K (this is possible since < α0/K). Introduce the equivalence relation on NN defined by i 'x j iff xi and xj can be connected by a path lying in ∪k∈NN B(xk, 40α0/K) and denote by c(x) the number of equivalence class. Suppose that c(x) ≥ 2. Then there exists a partition NN = I ∪ J, such that NI = ]I ≥ 1, NJ = ]J ≥ 1 and for all i ∈ I, j ∈ J, |xi − xj| > 40α0/K. By induction, there exists a path N γI : [0, 1] → Ω I ∩{(xi)i∈I , ∀i 6= j, |xi −xj| > } such that γI (0) = (xi)i∈I , I(γI (1)) ≥ I(γI (0))+δK and |γI (0) − γI (t)|∞ < NI ρK . The same construction for the set J provides a path γJ with the same properties. Define the pathγ ˜ on [0, 1] by (˜γ(t))i = (γI (t))i for i ∈ I and (˜γ(t))j = (γJ (t))j for j ∈ J. Since 40α0/K − (NI + NJ )ρK > α0/K + δK > ,γ ˜ has values in ON, and we have I(˜γ(1)) ≥ I(x) + δK as well as

|x − γ(t)|∞ < max(NI ,NJ )ρK ≤ (N − 1)ρK . (4.11)

Suppose now that there is only one equivalence class. Then for all k ∈ NN , |x1 − xk| ≤ d−1 40α0N/K ≤ 40α0 < r0, where r0 is deﬁned in (4.9). In particular, there exists ν ∈ S such that for all y ∈ B(x1, 40α0) ∩ Ω, y + Γ+(ν, δ0) ⊂ Ω. On the other hand, we can suppose without loss of generality that hx1, νi ≤ · · · ≤ hxN , νi. (4.12)

For j ∈ {1, ..., N} set aj = jρK and

γ(t) = (x1 + ta1ν, ..., xN + taN ν), t ∈ [0, 1] (4.13)

Then, one has |x − γ(t)|∞ ≤ sup aj = NρK , xj + tajν ∈ Ω since NρK ≤ δ0, and for i < j

Thus one has 2 2 2 2 I(γ(1)) ≥ I(x) + ρK ≥ (I(x) + δK ) (4.15) The proof of lemma 4.2 is complete.

Using this lemma, it is easy to show that ON, is arcwise connected for N small. For x ∈ ON,, deﬁne Ix = {y ∈ ON,, ∃γ ∈ C ([0, 1], ON,) , γ(0) = x, γ(1) = y} . (4.16)

21 We ﬁrst show easily that there exists y ∈ Ix such that I(y) ≥ α0/N if N < α0. Let M = maxy∈Ix I(y). As I is a bounded function, M is ﬁnite and given γ ∈]0, δN,/2[, there exists y1 ∈ Ix such that I(y1) ≥ M − γ. If I(y1) < α0/N, Lemma 4.2 shows that there exists y2 ∈ Ix such that I(y2) ≥ I(y1) + δN, > M which is impossible. This shows that there exists y ∈ Ix such that d−1 I(y) ≥ α0/N. Now by 4.9, for any x ∈ Ω, there exists νx ∈ S such that x+tνx ∈ Ω for t ∈ [0, δ0] and dist(x + tνx, ∂Ω) ≥ t sin θ0 with cos θ0 = 1 − δ0. Let α1 = α0 sin θ0/20. Then for Nε ≤ α1, and

I(y) ≥ α0/N, γ(t) = (y1 + t(α0/4N)νy1 , ..., yN + t(α0/4N)νyN ), t ∈ [0, 1] is a path in ON, and one 0 0 0 0 0 has with γ(1) = y = (y1, ..., yN ), I(y ) ≥ α0/2N, and dist(yj, ∂Ω) > 3ε for all j. Let CN, be the set of x ∈ ON, such that I(x) ≥ α0/2N and dist(xj, ∂Ω) > 3ε for all j. It remains to show that for any x, y ∈ CN, there exists a continous path γ from x to y, with values d in ON, for N ≤ α1. Decreasing α0 we may assume 6cdα0 < vol(Ω) with cd = vol(B(0, 1)). Decreasing α1, we get that for any x, y ∈ CN, with N ≤ α1, there exists z ∈ CN, such that

|xp − zq| ≥ α0/2N and |yp − zq| ≥ α0/2N ∀p, q ∈ NN . (4.17)

One can easily choose the zj by induction, since for any x, y ∈ ON, and any z1, ..., zl ∈ Ω with N N l d −d 0 ≤ l ≤ N − 1 we have vol(∪j=1B(xj, α0/N) ∪j=1 B(yj, α0/N) ∪j=1 B(zj, α0/N)) ≤ 3Ncdα0N < vol(Ω)/2 < vol({x ∈ Ω, dist(x, ∂Ω)} > 3). Thus we are reduce to show that if y, z ∈ CN, satisfy (4.17), there exists a continuous path γ from y to z, with values in ON, if N ≤ α1. We look for a path γ of the form γ = γN ◦ · · · ◦ γ1, where the path γj moves only the jth ball from yj to zj. Let us explain how to choose γj. As Ω is connected, there exists an analytic pathγ ˜1 which connect y1 to z1 in Ω. We have to modify the pathγ ˜1 into a new path γ1 in order that

|γ1(t) − yj| > ∀j ∈ {2,...,N}. (4.18)

Let K = {t ∈ [0, 1], ∃j0 ∈ {2,...,N}, |γ˜1(t) − yj0 | ≤ 2}. If K is empty, we set γ1 =γ ˜1. If K is non empty, since the pathγ ˜1 is analytic and I(y) ≥ α0/2N > 4, K is a disjoint union of intervals, K = [a1, b1] ∪ ... ∪ [aL, bL] and for any l ∈ {1, ..., L} there exists a unique jl such that

|γ˜1(t) − yjl | ≤ 2 for t ∈ [al, bl]. For t∈ / K we set γ1(t) =γ ˜1(t) and for t ∈ [al, bl] we replaceγ ˜1 by a continuous path γ1 connectingγ ˜1(al) toγ ˜1(bl) on the sphere |x − yjl | = 2 which is contained in Ω. Then γ1(t) is continuous. Moreover, as I(y) > 4, for any j ∈ {2, ..., N} and t ∈ [0, 1] we have |γ1(t) − yj| ≥ 2. In particular, the path t ∈ [0, 1] 7→ (γ1(t), y2, . . . , yN ) has values in ON, and 0 connects y andy ˜ := (z1, y ). From (4.17) it is clear thaty ˜ ∈ CN, and that (4.17) holds true with y replace byy ˜. This permits iterating the construction to build a continuous path from y to z. Thus ON, is connected for N < α1.

Let us now prove that ∂ON, has Lipschitz regularity for N ≤ r0/2, where r0 is is such that (4.9) holds. For a given , we will prove this fact by induction on N ∈ [1, r0/2]. The case N = 1 is obvious since ∂Ω is Lipschitz. Let x ∈ ∂ON,. The equivalence relation i ' j iﬀ xi and xj can be connected by a path lying in the union of closed balls of radius /2, gives us a partition r {1,...,N} = ∪k=1Fk such that

|xi − xj| > ∀k 6= l, ∀i ∈ Fk, ∀j ∈ Fl; (4.19) |xnl − xnl+1 | = ∀k, ∀i 6= j ∈ Fk, ∃(nl) ∈ Fk, 1 ≤ l ≤ m, n1 = i, nm = j.

d The Cartesien product O1 × O2 of two bounded Lipschitz open subsets Oi ⊂ R i has Lipschitz regularity. Thus, if r ≥ 2, the induction hypothesis on N shows that ∂ON, has Lipschitz regularity near x. Thus we may assume r = 1, and therefore, for all i, j one has |xi − xj| ≤ (N − 1) ≤ r0/2.

22 Thus there exists x0 ∈ Ω such that xi ∈ B(x0, r0/2), and (4.9) gives us a unit vector ν and δ0 > 0. We set ξi = αxi + ν (4.20) with α > 0 small such that tξi ∈ Γ+(ν, δ0/2) for t > 0 small. We choose β > 0, ρ > 0, t0 > 0 such 2 2 2 2 that β << α , β << δ0, ρ << α , ρ << r0, t0|ξi| << α , t0 << δ0. d Let x ∈ ∂ON, such that |xj − xj| ≤ ρ and θi ∈ R such that |θi| ≤ β. Let ξi = ξi + θi, and ξ = (ξ1, ··· , ξN ). One has tξi ∈ Γ+(ν, δ0) for t ∈]0, t0] and tξi ∈ Γ−(ν, δ0) for t ∈ [−t0, 0[. From 2 hxi − xj, ξi − ξji = hxi − xj, ξi − ξji + O(β) = α|xi − xj| + O(β + ρ) (4.21) and 2 2 2 2 |(xi + tξi) − (xj + tξj)| = |xi − xj| + 2thxi − xj, ξi − ξji + t |ξi − ξj| (4.22) 2 we get that the function t ∈ [−t0, t0] 7→ gi,j(t) = |(xi + tξi) − (xj + tξj)| is strictly increasing. Since by (4.9) we have xi + tξi ∈ Ω for t ∈]0, t0], we get x + tξ ∈ ON, for t ∈]0, t0]. It remains to show x+tξ∈ / ON, for t ∈ [−t0, 0[. If there exists two indices i, j such that |xi −xj| = , this follows from 2 gi,j(t) < for t < 0. If there exists one indice i such that xi ∈ ∂Ω, this follows from tξi ∈ Γ−(ν, δ0) and the second line of (4.9) which implies xi + tξi ∈/ Ω for t ∈ [−t0, 0[. Thus ∂ON, is Lipschitz.

−1/2 Let us ﬁnally prove that ON, is quasi-regular. Let u ∈ H (∂ON,) be supported in Γsing. We have to show that u is identically zero. This is a local problem near any point x ∈ Γsing.

Let x be such that s(x) = 0, R(x) = {j0} (say j0=1) and xj0 ∈ ∂Ωsing. Denote DN, = {x ∈ d N (R ) , |xi − xj| > , ∀1 ≤ i < j ≤ N}. Let χ be a cut-oﬀ function supported near x such that d N−1 ∞ N−1 supp(χ) ⊂ (R × Ω ) ∩ DN,. Then, for any ψ ∈ C0 (Ω ) the linear form uψ deﬁned on H1/2(∂Ω) by huψ, fi = hχu, f(x1)ψ(x2, . . . , xN )i (4.23) is continuous and supported in ∂Ωsing. As ∂Ω is quasi-regular, it follows that uψ is equal to zero for all ψ and hence, χu = 0. Therefore, we can suppose that u is supported in the set {r(x)+s(x) ≥ 2}. Nd Let v be the distribution on R

hv, ϕi = hu, ϕ|∂ON, i (4.24) −1 Nd −1 Then v ∈ H (R ) and its support is equal to supp(u). The Sobolev space H is preserved by bi-Lipschitz maps. Therefore, if there exists a bi-Lipschitz map Φ deﬁned near x such that locally one has Φ(supp(u)) ⊂ {y1 = y2 = 0}, then u is identically 0 near x. For n ∈ N, n ≥ 2, introduce the following property:

(Pn) : for any x ∈ Γsing with r(x) + s(x) = n, we have u = 0 near x. (4.25) This property is proved by induction on n. By lower semicontinuity of the functions r and s, we may assume in the proof that for x ∈ supp(u) closed to x, one has r(x) = r(x) and s(x) = s(x) and hence R(x) = R(x) and S(x) = S(x). Therefore, we are reduced to prove that for x ∈ Γsing −1/2 with r(x) + s(x) ≥ 2 and u ∈ H (∂ON,) supported in R(x) = R(x) and S(x) = S(x), we have u = 0 near x .

First assume r(x) = s(x) = 1. Then, we can suppose without losing generality, that u is N−1 supported near x in G = (∂Ω × Ω ) ∩ {|xi − x2| = } for some i ∈ {1, 3,...,N}. Denoting xi = (xi,1, . . . , xi,d), we may assume that near x, G is given by two equations, 0 0 x1,1 = α(x1), x1 = (x1,2, ..., x1,d) 0 0 (4.26) x2,k = β(x2, xi), x2 = (x2,1, ..., x2,k−1, x2,k+1, ..., x2,d).

23 0 0 0 0 with α Lipschitz and β smooth. Then, ν(x) = (x1,1 − α(x1), x2,k − β(x2, xi), x1, x2, x3, . . . , xN ) Nd 2 Nd−2 deﬁnes a local bi-Lipschitz homeomorphism of R such that ν ◦ G ⊂ {0} × R . Therefore, ν∗(v) vanishes identically near ν(x) and hence u is null near x. We may thus assume that s(x) ≥ 2 or r(x) ≥ 2. In the case s(x) ≥ 2, the support of u near x is contained in a set A of the form |x1 − x2| = |x2 − x3| = or |x1 − x2| = |x3 − x4| = . Since A is Nd a subvariety of R of codimension 2, we get as above that u is null near x. In the case r(x) ≥ 2, (N−2)d the support of u near x is contained in a set B of the form ∂Ω × ∂Ω × R which is near x Nd−2 bi-Lipschitz homeomorphic to (y1 = y2 = 0) × R , and thus u is null near x. The proof of Proposition 4.1 is complete.

Nd Nd d Nd Deﬁne for j ∈ NN the two functions πj from R to R and σj from R to R by

πj(x1, . . . , xj, . . . , xN ) = (x1,..., 0, . . . , xN ), (4.27) σj(y) = (0,...,y,..., 0), so that x = πj(x) + σj(xj). The following geometric lemma will be the main ingredient of the proof of Proposition 4.4.

Lemma 4.3. Let α0 = r0/10 with r0 given by (4.9). For all N ∈ N and ∈]0, α0/N], there exists d−1 δN, > 0 and a ﬁnite covering (Ul)l of ON, such that for all l, there exists j and ν ∈ S such that x + σj(Γ+(ν, δN,)) ⊂ ON, ∀x ∈ Ul ∩ ON,, (4.28) where Γ+ is deﬁned in (4.6)

0 Proof. Since ON, is compact, we have to prove that for any given x ∈ ON,, there exist r > 0, d−1 0 δ = δN, > 0, j and ν ∈ S such that (4.28) holds true for x ∈ ON, ∩ B(x , r). This means that we can select one ball, and that moving only this ball by a vector in Γ+(ν, δ) while keeping the other balls ﬁxed, results in an admissible conﬁguration. We shall proceed by induction on N ≥ 1. For N = 1, this is true since Ω is Lipschitz. Let N ≥ 2. If one can write {1,...,N} as the disjoint union I ∪ J with ]I ≥ 1, ]J ≥ 1, and

0 0 |xi − xj | ≥ 5 ∀i ∈ I, ∀j ∈ J, (4.29) then, by the induction hypothesis, the result is true for some δN, ∈]0, 4[. Thus, using the deﬁnition 0 0 of α0, we may assume that all the xi are in a small neighborhood of a given point y ∈ Ω and 0 0 supk|xk − y | ≤ r0/2. By (4.9) there exist ν, δ0 > 0, r0 > 0 such that

0 y ∈ Ω and |y − y | ≤ r0 =⇒ y + Γ+(ν, δ0) ∈ Ω (4.30)

0 0 0 0 0 It remains to show that there exist j, r0 ∈]0, r0[, and ν , δ0 > 0, with Γ+(ν , δ0) ⊂ Γ+(ν, δ0), such 0 0 0 0 that for all x = (x1, . . . , xN ) ∈ ON, with dist(x, x ) ≤ r0, and all z ∈ xj + Γ+(ν , δ0), one has |z − xk| > for all k 6= j. This will be a consequence of the following property:

0 d−1 0 0 0 0 ∀β > 0, ∃j, ∃ν ∈ S s.t. |ν − ν| ≤ β and ν .(xj − xk) > 0 ∀k 6= j. (4.31)

In fact, if (4.31) holds true, ﬁrst take β small enough, such that for all ν0 ∈ Sd−1 with |ν0 − 0 0 0 0 ν| ≤ β there exists δ0 > 0 with Γ+(ν , δ0) ⊂ Γ+(ν, δ0); then (4.31) gives us a pair ν , j such 0 0 0 0 0 0 0 that ν .(xj − xk) > 0 ∀k 6= j. For r0 > 0, δ0 > 0 small enough, we get for all ξ ∈ Γ+(ν , δ0) 0 0 0 and all x ∈ ON,, dist(x, x ) ≤ r0, that infk6=jξ.(xj − xk) ≥ δ0|ξ|, and thus there exists t0 such

24 2 that for t ∈ [0, t0] and k 6= j, the function t 7→ |xk − (xj + tξ)| is strictly increasing for all 0 0 0 0 x ∈ ON,, dist(x, x ) ≤ r0 and all ξ ∈ Γ+(ν , δ0). 0 Let us show that (4.31) holds true. If j 7→ ν.xj achieves its maximum at a single j, then (4.31) 0 0 d−1 0 0 0 is obvious with ν = ν . Otherwise, the set A = {ν ∈ S , ∃j 6= k, ν .(xj − xk) = 0} is contained in a ﬁnite union of equators in the sphere Sd−1, with ν ∈ A, and thus (4.31) is still obvious by taking ν0 ∈ Sd−1 \ A close to ν. The proof of Lemma 4.3 is complete.

Proposition 4.4. Let N, be such that Lemma 4.3 holds true. There exists h0 > 0, c0, c1 > 0 and ∗ M ∈ N such that for all h ∈]0, h0], one has M −Nd x − y Th (x, dy) = µh(x, dy) + c0h ϕ dy, (4.32) c1h where for all x ∈ ON,, µh(x, dy) is a positive Borel measure.

∗ k 1 Proof. For k ∈ denote B = B k (0, 1) the unit Euclidean ball and ϕ (z) = 1 (z). For N R k vol(Bk) Bk x, y ∈ ON,, we set dist(x, y) = sup1≤i≤N |xi − yi|. For N ≥ 1, denote by Kh,N the kernel given in ∗ (4.1). It is suﬃcient to prove the following: there exists h0 > 0, c0, c1 > 0 and M(N) ∈ N such that for all h ∈]0, h0], one has for all nonnegative function f, Z M(N) −Nd Kh,N (f)(x) ≥ c0h f(y)dy. (4.33) y∈ON,,dist(y,x)≤c1h

0 0 First note that it is suﬃcient to prove the weaker version: for all x ∈ ON,, there exist M(N, x ), r = 0 r(x ) > 0, c0 = c0(x0) > 0, c1 = c1(x0) > 0, h0 = h0(x0) > 0 such that for all h ∈]0, h0], all x ∈ ON, and all nonnegative function f Z 0 M(N,x0) −Nd dist(x, x ) ≤ 2r =⇒ Kh,N (f)(x) ≥ c0h f(y)dy. (4.34) y∈ON,,dist(y,x)≤c1h

Let us verify that (4.34) implies (4.33). Decreasing r(x0) if necessary, we may assume that any set 0 {dist(x, x ) ≤ 2r(x0)} is contained in one of the open set Ul of Lemma 4.3. There exists a ﬁnite set F 0 0 such that ON, ⊂ ∪x0∈F {dist(x, x ) ≤ r(x0)}. Let M(N) = supx0∈F M(N, x0), ci = minx0∈F ci(x0) 0 0 0 0 and h0 = minx0∈F h0(x0). One has to check that for any x ∈ F and any x with dist(x, x ) ≤ r(x ), the right inequality in (4.34) holds true with M(N) = M(N, x0) + n in place of M(N, x0) for some 0 constants c0, c1, h0. Let l be such that dist(x, x ) ≤ r(x0) implies x ∈ Ul. Let j and Γ+(ν, δ) be given by Lemma 4.3. Clearly, if f is nonnegative, one has Z M(N,x0)+1 1 −d M(N,x0) Kh,N (f)(x) ≥ h ϕ(z/h)Kh,N (f)(x + σj(z))dz (4.35) N x+σj (z)∈ON,

0 0 0 0 0 0 For dist(x, x ) ≤ 2r(x ) − c1h/2, and |z| ≤ c1h/2, z ∈ Γ+(ν, δ), one has dist(x + σj(z), x ) ≤ 2r(x ) 0 0 and by (4.28), x + σj(z) ∈ ON,. Moreover, dist(y, x) ≤ c1h/2 =⇒ dist(y, x + σj(z)) ≤ c1h. From (4.35) and (4.34) we thus get, with a constant Cδ depending only on the δ given by Lemma 4.3, 0 and for h ≤ h0, 0 0 0 M(N,x0)+1 dist(x, x ) ≤ 2r(x ) − c1h/2 =⇒ Kh,N (f)(x) Z (4.36) Cδ 0 −Nd ≥ c0h f(y)dy. N 0 y∈ON,,dist(y,x)≤c1h/2

25 By induction on n, we thus get

0 0 0 M(N,x0)+n dist(x, x ) ≤ 2r(x ) − c1h =⇒ Kh,N (f)(x) n Z (4.37) Cδ 0 −Nd ≥ c0h f(y)dy. N 0 h y∈ON,,dist(y,x)≤c1 2n 0 0 0 Since n is bounded, we get the desired result with h0 = min(minx0∈F r(x )/c1, h0). To complete the proof, let us show (4.34) by induction on N. The case N = 1 is obvious. 0 Suppose that (4.34) holds for N − 1 discs. Let x ∈ ON, be ﬁxed. Thanks to Lemma 4.3, we can suppose that there exists an open neighborhood U of x0 a direction ν ∈ Sd−1 and δ > 0 such that 0 (4.28) holds with j = 1. Let us denote x = (x1, x ) and

Kh,N = Kh,N,1 + Kh,N,> (4.38) with −d Z h x1 − y1 0 Kh,N,1f(x) = ϕ f y1, x dy1. (4.39) 0 N (y1,x )∈ON, h δ We also denote G(ν, δ) = {x1 ∈ Γ+(ν, δ), |x1| > 2 }. Then, we have the following: 0 Lemma 4.5. For any δ ∈]0, δ/2], there exists C > 0, α > 0, h0 > 0 and r0 > 0 such that 0 N−1 0 ∀r ∈]0, r0], ∀h ∈]0, h0], ∀x ∈ U ∩ ON,, ∀x˜ ∈ x + h(G(ν, δ ) × B(0, r) ) with x˜ ∈ ON−1,, we have x˜ ∈ ON, and 0 Kh,N,>f(˜x) ≥ CKαh,N−1 (f(˜x1,.)) x˜ , (4.40) ∗ for any nonnegative function f. In particular, for all M ∈ N , there exists C, r0, h0, α as above 0 N−1 such that ∀x ∈ U ∩ ON, and ∀x˜ ∈ x + h(G(ν, δ ) × B(0, r) ), we have M M 0 Kh,N,>f(˜x) ≥ CKαh,N−1 (f(˜x1,.)) x˜ . (4.41) Proof. Inequality (4.41) is obtained easily from (4.40) by induction on M. To prove (4.40), observe that for nonnegative f and α ∈]0, 1[ we have

N h−d X Z K f(˜x) ≥ f(˜x , . . . , y ,..., x˜ )dy , (4.42) h,N,> N 1 j N j j=2 Aj,α,h(˜x) with Aj,α,h(˜x) = {z ∈ Ω, |x˜j −z| < αh and ∀k 6= j, |x˜k −z| > }. Let Bj,α,h(˜x) = {z ∈ Ω, |x˜j −z| < αh and ∀k 6= 1, j, |x˜k − z| > }. Then Aj,α,h ⊂ Bj,α,h and we claim that for α, r > 0 small enough 0 N−1 0 andx ˜ ∈ x + h(G(ν, δ ) × B(0, r) ) withx ˜ ∈ ON−1,, we have Bj,α,h(˜x) = Aj,α,h(˜x). Indeed, 0 0 letx ˜1 = x1 + hu1 with u1 ∈ G(ν, δ ) andx ˜ ∈ ON−1, be such that |x˜j − xj| < hr. Then for z ∈ Bj,α,h(˜x), |x˜1 − z| = |x1 − xj + hv1|, (4.43)

xj −x˜j x˜j −z with v1 = u1 + h + h . Taking α, r small enough (w.r.t. δ) it follows that v1 ∈ Γ+(ν, δ). Consequently, Lemma 4.3 shows that |x˜1 − z| > and hence z ∈ Aj,α,h(˜x) (the same argument shows thatx ˜ ∈ ON,). Therefore,

N h−d X Z K f(˜x) ≥ f(˜x , . . . , y ,..., x˜ )dy h,N,> N 1 j N j j=2 Bj,α,h(˜x) (4.44) (N − 1)vol(Bd) = K (f(˜x ,.)) x˜0 , N αh,N−1 1 and the proof of Lemma 4.5 is complete.

26 Using this lemma we can complete the proof of (4.34). In the following computations, C denotes a strictly positive constant independent of h, whose value may change from line to line. Let p ∈ N, α ∈]0, α0] and x ∈ ON,, then

p+1 p Kh,N f(x) ≥ Kh,N,1Kh,N,>f(x) −d Z h p 0 ≥ Kh,N,>f z1, x dz1 0 0 N (z1,x )∈ON,,z1∈x1+hG(ν,δ ) (4.45) −d Z h p 0 ≥ C Kαh,N−1 (f(z1,.)) x dz1, 0 0 N (z1,x )∈ON,,z1∈x1+hG(ν,δ ) thanks to Lemma 4.5. From the induction hypothesis we can choose p ∈ N so that Z Z p+1 −Nd 0 0 Kh,N f(x) ≥ Ch f z1, y dy dz1 (4.46) 0 0 0 0 0 (z1,x )∈ON,,z1∈x1+hG(ν,δ ) |x −y |<αh,y ∈ON−1, Hence, for any β ∈]0, 1] we get Z p+2 p+1 −Nd 0 0 Kh,N f(x) ≥ Kh,N Kh,N,1f(x) ≥ Ch f y1, y γh(x, y1)dy1dy , (4.47) Dα,β,h(x) with 0 0 Dα,β,h(x) = y ∈ ON,, |x − y | < αh, |x1 − y1| < βh (4.48) and Z −d γh(x, y1) = h 1|z1−y1|

We have to show that γh is bounded from below by a positive constant, uniformly with respect to 0 (x, y1) when |x1 − y1| < βh. For z1 ∈ x1 + hG(ν, δ ), one has |z1 − y1| ≤ |z1 − x1| + |x1 − y1| ≤ 0 0 hδ + hβ < h for β and δ small. Thus for |x1 − y1| < βh one has Z Z −d 0 γh(x, y1) = h dz1 = 1(x1+hu,x )∈ON, du. (4.50) 0 0 0 (z1,x )∈ON,,z1∈x1+hG(ν,δ ) u∈G(ν,δ )

Using Lemma 4.3 again, we get for |x1 − y1| < βh Z γh(x, y1) = du = C0 > 0. (4.51) u∈G(ν,δ0) Plugging this lower bound into (4.47), gives Z p+2 −Nd Kh,N ≥ Ch f(y)dy, (4.52) Dα,β,h(x) and the proof of (4.34) is complete. This completes the proof of Proposition 4.4.

By Proposition 4.1, we can consider the Neumann Laplacian |∆|N on ON, deﬁned by

αd |∆|N = − ∆, 2N (4.53) 1 2 D(|∆|N ) = u ∈ H (ON,), −∆u ∈ L (ON,), ∂nu|∂ON, = 0 .

We still denote 0 = ν0 < ν1 < ν2 < . . . the spectrum of |∆|N and mj the multiplicity of νj. Our main result is the following.

27 Theorem 4.6. Let N ≥ 2 be ﬁxed. Let > 0 be small enough such that Proposition 4.1 and Proposition 4.4 hold true. Let R > 0 be given and β > 0 such that the spectrum νj of the Neumann Laplacian (4.53) satisﬁes νj+1 − νj > 2β for all j such that νj+2 ≤ R. There exists h0 > 0, δ0 ∈]0, 1/2[ and constants Ci > 0 such that for any h ∈]0, h0], the following hold true:

i) The spectrum of Th is a subset of [−1 + δ0, 1], 1 is a simple eigenvalue of Th, and Spec(Th) ∩ [1 − δ0, 1] is discrete. Moreover, 1 − T Spec h ∩]0,R] ⊂ ∪ [ν − β, ν + β]; h2 j≥1 j j (4.54) 1 − T ]Spec h ∩ [ν − β, ν + β] = m ∀ν ≤ R; h2 j j j j

−2 2 and for any 0 ≤ λ ≤ δ0h , the number of eigenvalues of Th in [1 − h λ, 1] (with multiplicity) dN/2 is bounded by C1(1 + λ) . ii) The spectral gap g(h) satisﬁes −2 lim h g(h) = ν1 (4.55) h→0+ and the following estimate holds true for all integer n:

n dy −ng(h) sup kTh (x, dy) − kTV ≤ C4e . (4.56) x∈ON, vol(ON,)

The rest of this section is devoted to the proof of Theorem 4.6. R Let µh(x, dy) be given by (4.32) and µh(f)(x) = f(y)µh(x, dy). Thanks to the positivity ON, M of µh(x, dy), using the Markov property of Th and Lipschitz-continuity of the boundary, we get 0 for some δ0 > 0, independant of h > 0, small enough Z −Nd x − y 0 kµhkL∞,L∞ ≤ 1 − inf c0h ϕNd dy < 1 − δ0. (4.57) x∈O N, ON, c1h

2 Since by (4.32) µh is self-adjoint on L (ON,), we also get 0 kµhkL1,L1 ≤ 1 − δ0, (4.58) 0 and by interpolation it follows that kµhkL2,L2 ≤ 1 − δ0. In particular the essential spectrum of M 0 0 1/M Th is contained in [0, 1 − δ0] so that σess(Th) ⊂ [0, 1 − 2δ0] with 2δ0 = 1 − (1 − δ0) . Thus Spec(Th) ∩ [1 − δ0, 1] is discrete. Let us verify that, decreasing δ0 > 0, we may also assume

Spec(Th) ⊂ [−1 + δ0, 1]. (4.59)

M Thanks to the Markov property of Th , to prove this, it suﬃces to ﬁnd M ∈ 2N + 1 such that Z Z 2 M 2 (u(x) + u(y)) Th (x, dy)dx ≥ δ0kukL2 , (4.60) ON, ON,

2 for any u ∈ L (Ω). Thanks to the proof of Proposition 4.4, there exists M ∈ 2N + 1 such that Z Z Z 2 M −Nd 2 x − y (u(x) + u(y)) Th (x, dy)dx ≥ c0h (u(x) + u(y)) ϕNd dxdy. ON, ON, ON,×ON, c1h (4.61)

28 Hence, (4.59) follows from (4.61) and (2.8). Nd Following the strategy of Section 2 we put ON, in a large box B =] − A/2, A/2[ and, thanks 2 2 to Proposition 4.1, there is an extension map E : L (ON,) → L (B) which is also bounded from 1 1 H (ON,) into H (B). Deﬁne

k 2 Eh,k(u) = h(1 − Th )u, uiL (ON,), (4.62) and deﬁne Eh as in Section 2. Moreover, the identities (2.13) and (2.14) remain true with obvious modiﬁcations.

Lemma 4.7. There exist C0, h0 > 0 such that the following holds true for any h ∈]0, h0] and any 2 u ∈ L (ON,): 2 2 Eh (E(u)) ≤ C0 Eh,M (u) + h kukL2 . (4.63) Proof. Thanks to Lemma 2.2 we have Z ! 2 −Nd x − y 2 2 Eh (E(u)) ≤ C0 (u(x) − u(y)) c0h ϕNd dydx + h kukL (ON,) . (4.64) ON,×ON, c1h

Combined with (4.32), this shows that

Z ! 2 M 2 2 Eh (E(u)) ≤ C0 (u(x) − u(y)) Th (x, dy)dx + h kukL (ON,) , (4.65) ON,×ON, and the proof is complete.

2 2 Lemma 4.8. For any 0 ≤ λ ≤ δ0/h , the number of eigenvalues of Th in [1 − h λ, 1] (with Nd/2 multiplicity) is bounded by C1(1+λ) . Moreover, any eigenfunction Th(u) = λu with λ ∈]1−δ0, 1] satisﬁes the bound −Nd/2 kukL∞ ≤ C2h kukL2 . (4.66) M M Proof. Suppose that Th(u) = λu with λ ∈ [1 − δ0, 1], then Th u = λ u and thanks to (4.32), we get M −Nd/2 k(µh − λ )ukL∞ = O(h ). (4.67) k The estimate (4.66) follows from (4.57). Let ζk(λ, h) be the number of eigenvalues of Th in the 2 2 interval [1 − h λ, 1] for h λ < δ0. Thanks to Lemma 4.7, we can mimick the proof of Lemma 2.3 to get Nd/2 ζM (λ, h) ≤ C(1 + λ) . (4.68) Then from (4.59), one has 1 − (1 − h2λ)k ζ (λ, h) = ζ , h . (4.69) 1 k h2 Combining (4.68) and (4.69), we easily obtain the announced estimate. The proof of Lemma 4.8 is complete.

The rest of the proof of Theorem 4.6 follows the strategy of Sections 2 and 3. Using the spectral decomposition (2.45), (2.46) we get easily the estimates (2.52) and (2.54), and it remains to estimate n T1,h. Following the proof of Lemma 2.4, we can ﬁnd α > 0 small enough and C > 0 such that the following Nash inequality holds with 1/D = 2 − 4/p > 0:

2+1/D −2 2 2 1/D kukL2 ≤ Ch Eh,M (u) + h kukL2 kukL1 , ∀u ∈ Eα. (4.70)

29 From this inequality, we deduce that for k ≥ h−2, kM −kMg(h) kT1,h kL∞,L∞ ≤ Ce , (4.71) −2 kM kM and this implies for k ≥ h , since the contributions of T2,h ,T3,h are negligible, kM 0 −kMg(h) kTh kL∞,L∞ ≤ C e . (4.72) ∞ −2 As Th is bounded by 1 on L we can replace kM by n ≥ h in (4.72) and (4.56) is proved. Assertion (4.55) is an obvious consequence of (4.54). The proof of (4.54) is the same as the one of Theorem 1.2. Thus, the following lemma will end the proof of Theorem 4.6. ∞ Lemma 4.9. Let θ ∈ C (ON,) be such that supp(θ) ∩ Γsing = ∅ and ∂nθ|Γreg = 0. Then 2 5/2 (1 − Th)θ = h |∆|N θ + r, krkL2 = O(h ). (4.73) ∞ Proof. Let θ ∈ C (ON,) be such that supp(θ)∩Γsing = ∅ and ∂nθ|Γreg = 0 and denote Qh = 1−Th. 1 PN Then Qh = N j=1 Qj,h with h−d Z Qj,hθ(x) = 1|xj −y| (θ(x) − θ (πj(x) + σj(y))) dy. (4.74) vol(B1) Ω

Let χ0(x) = 1dist(x,∂ON,)<2h. The same proof as in Section 3 shows that

αd 2 2 3 (1 − χ )Q θ(x) = − h ∂ θ(x) + O ∞ (h ), (4.75) 0 j,h 2 j L so that 2 3 (1 − χ0)Qhθ(x) = h |∆|N θ(x) + OL2 (h ). (4.76) 1/2 2 We study χ0Qhθ. As kχ0kL2 = O(h ) it suﬃces to show that kχ0QhθkL∞ = O(h ). On the other hand, by Taylor expansion we have Z hχ0(x) 2 ∞ χ0Qj,hθ(x) = − Πk6=j1|xj +hz−xk|>1xj +hz∈Ωz.∂jθ(x)dz + OL (h ). (4.77) vol(B1) |z|<1 Hence, it suﬃces to show that

N X Z v(x) = χ0(x) Πk6=j1|xj +hz−xk|>1xj +hz∈Ωz.∂jθ(x)dz (4.78) j=1 |z|<1 satisﬁes kvkL∞ = O(h). Since dist(sup(θ), Γsing) > 0, there exists disjoint compact sets Fl ⊂ {s(x) = 0,R(x) = l}, and Fi,j ⊂ {r(x) = 0,S(x) = (i, j)} such that

sup(χ0θ) ⊂ ∪l {x, dist(x, Fl) ≤ 4h} ∪i,j {x, dist(x, Fi,j) ≤ 4h} .

If x ∈ supp(χ0θ) is in {x, dist(x, F1) ≤ 4h}, then the same parity arguments as in Section 3 show that Z v(x) = χ0(x) z.∂1θ(x)dz = O(h). (4.79) |z|<1,x1+hz∈Ω

If x ∈ supp(χ0θ) is in {x, dist(x, F1,2) ≤ 4h}, then Z

v(x) = χ0(x) z.(∂1θ(x)1|x1+hz−x2|> + ∂2θ(x)1|x2+hz−x1|>)dz (4.80) |z|<1 and the result follows from (x1 − x2).(∂1θ − ∂2θ)(x) = 0(h) for {x, dist(x, F1,2) ≤ 4h}, since ∂nθ vanishes on the boundary |x1 − x2| = . The proof of Lemma 4.9 is complete.

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