Pierre Cornilleau

Luc Robbiano

Exponential stabilization of waves for the Zaremba boundary condition

Keywords: {ϕ, {ϕ, p}} {ϕ, {ψ, p}} {ψ, {ϕ, p}} {ψ, {ψ, p}} {{ϕ, p}, ϕ} {{ϕ, p}, ψ} 0 {{ϕ, p}, {ψ, p}} {{ψ, p}, ϕ} {{ψ, p}, ψ} {{ψ, p}, {ϕ, p}} 0 p}} {ϕ, {ψ, p}} {ψ, {ϕ, p}} {ψ, {ψ, p}}

In this article we prove, under some geometrical condition on geodesic flow, exponential stabilization of wave equation with Zaremba boundary condition. We prove an estimate on the resolvent of semigroup associated with wave equation on the imaginary axis and we deduce the stabilization result. To prove this estimate we apply semiclassical measure technics. The main difficulties are to prove that support of measure is in characteristic set in a neighborhood of the jump in the boundary condition and to prove results of propagation in a neighborhood of a boundary point where Neumann boundary condition is imposed. In fact if a lot of results applied here are proved in previous articles, these two points are new.

1 Introduction and results

Framework

In this article we are interested by stabilization of wave equation with Zaremba boundary condition.

To be precise we have to introduce some notation. Let Ω be a bounded open set in R d , with C ∞ boundary. Let ∂Ω D and ∂Ω N two open sets in ∂Ω such that ∂Ω D ∩ ∂Ω N = ∅ and ∂Ω D ∩ ∂Ω N = Γ, where Γ is a smooth manifold of dimension d-2. The manifold Γ is not necessary a connected set. Let P be a second order differential operator. We have P = 1≤j,k≤d D xj p jk (x)D x k + 1≤j≤d p j (x)D xj + p 0 (x), where p jk (x), p j (x) are in C ∞ (V ) where V is a neighborhood of Ω. The matrix (p jk (x)) jk is assumed positive definite for every x ∈ V . We assume that the operator defined by P u for u ∈ D = {u ∈ H 1 (Ω), P u ∈ L 2 (Ω), u |∂ΩD = 0, (∂ ν u) |∂ΩN = 0}, is self-adjoint and non negative. Here ∂ ν is the exterior normal derivative. Let a ∈ C ∞ (V ) be such that a(x) ≥ 0 for every x ∈ Ω. We associate with P the following wave equation

         ∂ 2 t u + P u + a(x)∂ t u = 0 in Ω × (0, ∞) (u, ∂ t u) |t=0 = (u 0 , u 1 ) ∈ H 1 (Ω) ⊕ L 2 (Ω) u = 0 on ∂Ω D × (0, ∞) ∂ ν u = 0 on ∂Ω N × (0, ∞) (1)
We associated the energy that is E(t, u 0 , u 1 ) = (P u|u) L 2 (Ω) + Ω |∂ t u| 2 dx, where (v|w) L 2 (Ω) =

We can find an introduction to this subject in Zworski [START_REF] Zworski | Semiclassical analysis[END_REF]. Problems for the Zaremba boundary condition was studied by several authors. In particular, for elliptic problem, Shamir [START_REF] Shamir | Regularization of mixed second order elliptic problems[END_REF] and Savaré [START_REF] Savaré | Regularity and perturbation results for mixed second order elliptic problems[END_REF] proved that the regularity of solution is not as for the Dirichlet boundary condition, there is a lack of regularity, s = 3/2 is critical in the Sobolev spaces H s if the datum is in L 2 . The problem is related with boundary problem in non smooth domain, with corner for instance. There is a large literature on this subject. For damping wave equation with Zaremba Boundary condition, the problem was studied by Bey and al. [START_REF] Bey | Singularities of the solution of a mixed problem for a general second order elliptic equation and boundary stabilization of the wave equation[END_REF], Cornilleau and al. [START_REF] Cornilleau | Nonlinear Neumann boundary stabilization of the wave equation using rotated multipliers[END_REF] where they prove exponential decay with multiplier method, and in [START_REF] Cornilleau | Carleman estimates for the Zaremba Boundary Condition and Stabilization of Waves[END_REF] where we only prove logarithmic decay but without geometric condition on the support of the damping. Same kind of result was proven by Fu [START_REF] Fu | Stabilization of hyperbolic equations with mixed boundary conditions[END_REF] for mixed boundary condition of Robin type.

In the following we described the geometry in section 1.2. This allows to give the precise assumption and the result in section 1.3. At the end of this section we give a description of proofs.

Geometry

Here we give the geometrical notion we use in this article. This framework comes from Melrose and Sjöstrand [START_REF] Melrose | Singularities of boundary value problems I[END_REF][START_REF] Melrose | Singularities of boundary value problems II[END_REF] and the reader may also find in Hörmander [START_REF] Hörmander | The analysis of linear partial differential operators[END_REF]Chapter 24] more informations and proofs. The characterisation of symplectic sub-manifold is probably classical and more details can be found in Grigis [START_REF] Grigis | Hypoellipticité et paramétrix pour des opérateurs pseudodifférentiels à caractéristiques doubles[END_REF].

Assumption on the symbol. We define the symbol of P by p(x, ξ) = 1≤j,k≤d

p jk (x)ξ j ξ k -1, (2)
Where p jk are C ∞ (Ω). Locally in a neighborhood of the boundary we can define Ω by ϕ > 0 with dϕ = 0. We can also choose coordinates (i.e. normal geodesic coordinates) such that ϕ(x) = x d and p(x, ξ) = ξ 2 d + R(x, ξ ′ ) -1 where x = (x ′ , x d ) and ξ = (ξ ′ , ξ d ).

Symplectic sub-manifold Σ. We can define a symplectic manifold Σ, contained into T * R d ∩{ϕ = 0}. We set Σ = {(x, ξ), ϕ(x) = 0 and {ϕ, p}(x, ξ) = 0}. The set Σ is a symplectic manifold as {ϕ, {ϕ, p}} = 0. In coordinates (x ′ , x d ), we have Σ = {x d = 0, ξ d = 0}, this manifold is isomorphic to T * ∂Ω and described by coordinates (x ′ , ξ ′ ). The Hamiltonian vector field H p is not a vector field on Σ, but for all X a vector field on T * R d , we can find unique fonctions α and β such X + αH ϕ + βH {ϕ,p} is a vector field on Σ. For H p we denote the associated vector H ′ p and an elementary computation leads to

H ′ p = H p + {p, {p, ϕ} {ϕ, {ϕ, p} H ϕ .
In coordinates (x ′ , x d ), H ′ p only depends on R and we have

H ′ p = H ′ R = d-1 j=1 (∂ ξj R(x ′ , 0, ξ ′ )∂ xj -∂ xj R(x ′ , 0, ξ ′ )∂ ξj ).
In particular the integral curves associated with H ′ p starting from a point into Σ stay into Σ. In coordinates (x ′ , x d ), we denote the integral curve starting from (x ′ , ξ ′ ), either γ g (s; x ′ , ξ ′ ), either γ g (x ′ , ξ ′ ), if s is implicit or γ g (s), if (x ′ , ξ ′ ) is implicit.

Symplectic sub-manifold Σ ′ . The manifold Γ can be locally defined by {ϕ = 0, ψ = 0}, where dϕ ∧ dψ = 0. We define the manifold Σ ′ by Σ ′ = {(x, ξ), ϕ(x) = ψ(x) = {ϕ, p}(x, ξ) = {ψ, p}(x, ξ) = 0}.

In these local coordinates, the manifold Σ ′ = {x 1 = x d = ξ 1 = ξ d = 0} which is isomorphic to T * Γ, and described by the coordinates (x ′′ , ξ ′′ ).

For all vector field X defined on T * R d , we can find unique functions α, β, γ, ζ such that X + αH ϕ + βH ψ + γH {ϕ,p} + ζH {ψ,p} is a vector field on Σ ′ . For X = H p , we denote the associated vector field H ′′ p and we have H ′′ p = H p + αH ϕ + βH ψ , as H p ϕ = H p ψ = 0 on Σ ′ and H ψ ϕ = 0. We can compute α and β but the precise values are not useful for general functions ϕ and ψ. In coordinates (x 1 , x ′′ , x d ), we have H ′′ p = H p + α∂ ξ1 + β∂ ξ d . The equations H ′′ p ξ 1 = H ′′ p ξ d = 0, on Σ ′ , give α = -∂ x1 R(0, x ′′ , 0, 0, ξ ′′ ) and β = -∂ x d R 1 (0, x ′′ , ξ ′′ ). In particular H ′′ p only depends on R 1 , and we have

H ′′ p = H ′′ R = H ′′ R1 = d-1 j=2 (∂ ξj R 1 (0, x ′′ , ξ ′′ )∂ xj -(∂ xj R 1 (0, x ′′ , ξ ′′ )∂ ξj ).
The integral curves starting from Σ ′ stay on Σ ′ . We denote the curves starting from (x ′′ , ξ ′′ ), γ sing (s; x ′′ , ξ ′′ ), γ sing (x ′′ , ξ ′′ ), if s is implicit and γ sing (s), if (x ′′ , ξ ′′ ) is implicit.

Description and topology of T *

b Ω. Let T * b Ω = T * ∂Ω ∪ T * Ω, this set is equipped with the following topology.

First if ρ ∈ T * Ω, a set V is a neighborhood of ρ if V contains an open set W of T * Ω such that ρ ∈ W .

Second if ρ = (x ′ 0 , ξ ′ 0 ) ∈ T * ∂Ω, a set V is a neighborhood of ρ if V contains a set

{(x ′ , ξ ′ ) ∈ T * ∂Ω, |x ′ 0 -x ′ | + |ξ ′ 0 -ξ ′ | ≤ ε} ∪ {(x, ξ) ∈ T * Ω, |x ′ 0 -x ′ | + |ξ ′ 0 -ξ ′ | ≤ ε and (x d , ξ d ) ∈ U ∩ {x d > 0}},
where ε > 0 and U is a neighborhood of

{(x d , ξ d ) ∈ R 2 , x d = 0} in R 2 .
In local coordinates where Ω is define by x d > 0, we define j : T * Ω → T * b Ω by j(x, ξ) = (x, ξ) if x ∈ Ω, and j(x, ξ) = (x ′ , ξ ′ ) if x d = 0. The map j is continuous for the topology given above. We can define more intrinsically j with the previous notation where Ω is given by ϕ(x) > 0. For (x, ξ) ∈ T * Ω, j(x, ξ) = (x, ξ) if x ∈ Ω and j(x, ξ) = (x, ξdx -({p, ϕ}/H 2 ϕ p)dϕ), if ϕ(x) = 0. We verify, in this last case that j(x, ξ) ∈ {ϕ = {p, ϕ} = 0}, as {p, ϕ}(x, dϕ) = {ϕ, {ϕ, p}}.

As usually we define the map π : T * b Ω → Ω, in local coordinates, as π(x, ξ) = x, if (x, ξ) ∈ T * Ω and π(x ′ , ξ ′ ) = x ′ , if (x ′ , ξ ′ ) ∈ T * ∂Ω.

Bicharacteristic and generalized flow. For (x, ξ) ∈ T * R d , we denote by γ(s; x, ξ) the integral curve of H p starting from (x, ξ). We use the same short notations γ(s) and γ(x, ξ) as above. Now we define the generalized bicharacteristic denoted by Γ(s, ρ) for ρ ∈ T * b Ω. To describe this curve in a neighborhood of the boundary we use the coordinates (x ′ , x d , ξ ′ , ξ d ) and we identify Σ ′ and T * ∂Ω and locally Ω = {x ∈ R d , x d > 0}. Moreover, we assume ρ ∈ char(P ) = {(x, ξ) ∈ T * Ω, p(x, ξ) = 0} ∪ {(x ′ , ξ ′ ) ∈ T * ∂Ω, R(x ′ , 0, ξ ′ ) -1 ≤ 0}.

Statement of Theorems

In the following we give the assumptions on the flows and these assumptions depend on the starting points. The assumptions also depend on the damping a and we assume a(x) ≥ 0 for every x ∈ Ω. We denote by ω = {x ∈ Ω, a(x) > 0}.

Definition 1.2. We say that P , a(x) and Ω satisfy the modified Geometric Control Condition (mGCC) if the bicharacteristic only has finite contact with the boundary (Definition 1.1) and the following assumptions are verified. Let ρ 0 ∈ char(P ).

• If π(ρ 0 ) / ∈ Γ we assume there exist s 0 ∈ R be such that πΓ(s 0 , ρ 0 ) ∈ ω and for every s ∈ [0, s 0 ], if πΓ(s, ρ 0 ) ∈ Γ then Γ(s, ρ 0 ) ∈ H.

• If π(ρ 0 ) ∈ Γ we assume there exist s 0 ∈ R be such that πγ sing (s 0 , ρ 0 ) ∈ ω.

Remark 2. This definition of mGCC is different from the usual GCC. We are not able to prove propagation of support of measure for generalized bicharacteristic hitting Γ except for hyperbolic points. It is possible that singularities can be create at Γ but we do not know what can happen. For points on Γ we are only able to prove propagation on Γ for integral curve of H ′′ R if we already know that measure is supported in the fiber above a point of Γ.

We recall the assumptions on P , a second order differential operator. We have

P = 1≤j,k≤d D xj p jk (x)D x k + 1≤j≤d p j (x)D xj + p 0 (x), (4)
where p jk (x), p j (x) are real valued and in C ∞ (V ) where V is a neighborhood of Ω. We assume that P is formally self-adjoint. The domain of P is given by D(P ) = {u ∈ H 1 (Ω), P u ∈ L 2 (Ω), u |∂ΩD = 0, (∂ ν u) |∂ΩN = 0}, where ∂ ν is the exterior normal derivative. With this domain P is self-adjoint, and

(P u|v) L 2 (Ω) = 1≤j,k≤d (p jk (x)D x k u|D xj v) L 2 (Ω) + ( 1≤j≤d p j (x)D xj u + p 0 (x)u|v) L 2 (Ω) , (5)
where u and v are in D(P). Moreover we assume P positive definite, there is δ > 0 such that (P u|u) L 2 (Ω) ≥ δ u 2 H 1 (Ω) , for every u ∈ D(P ).

To give a precise formulation of the wave equation we introduce H = H 1 (Ω) ⊕ L 2 (Ω), we denote by U = (u 0 , u 1 ) an element of H and the operator A is given by

A = 0 1 -P -a(x) , (7)
associated with the domain

D(A) = {U = (u 0 , u 1 ) ∈ H, P u 0 ∈ L 2 (Ω), u 1 ∈ H 1 (Ω), u 0 = 0 on ∂Ω D , ∂ ν u 0 = 0 on ∂Ω N }.
Let U be the solution of ∂ t U = AU satisfying U (0) = (u 0 , u 1 ) ∈ H, we have U (t) = (u(t), ∂ t u(t)) = e tA (u 0 , u 1 ), where e tA is the semigroup associated with A. Then u satisfies the wave equation

         ∂ 2
t u(x) + P u(x) + a(x)∂ t u(x) = 0 in Ω, u = 0 on ∂Ω D , ∂ ν u = 0 on ∂Ω N , (u(0), ∂ t u(0)) = (u 0 , u 1 ).

To U (t) = (u 0 (t), u 1 (t)), we associate the energy E(t, u 0 , u 1 ) = (P u 0 (t)|u 0 (t)) L 2 (Ω) + Ω |u 1 (t)| 2 dx.

We have ∂ t E(t, u 0 , u 1 ) = -2(au 1 |u 1 ) L 2 (Ω) ≤ 0. This implies E(t, u 0 , u 1 ) ≤ E(0, u 0 , u 1 ).

Remark 3. We have assumed that P is positive definite for simplicity but if P is non negative we can introduce H = H/ ker P , working on H instead of H we can obtain same results. For instance see [START_REF] Lebeau | Stabilisation de l'équation des ondes par le bord[END_REF] where this reduction is used.

The main result of the paper is the following Theorem 1. We assume that P , a(x) and Ω satisfy the modified Geometric Control Condition given in Definition 1.2. We assume that P has the form given in (4) and P is self-adjoint positive definite.

Let A be defined by [START_REF] Bey | Singularities of the solution of a mixed problem for a general second order elliptic equation and boundary stabilization of the wave equation[END_REF], we have

1) ∃M > 0, e tA L(H) ≤ M , 2) A -iµI, is invertible for all µ ∈ R, 3) ∃M > 0, (A -iµI) -1 L(H) ≤ M .
Remark 4. There are several results when the third item is replaced by other estimates as e C|µ| , |µ| α in these cases the energy decay with a speed depending of the estimate on the resolvent. For this kind of results we refer to Batty-Duyckaerts [START_REF] Batty | Non-uniform stability for bounded semi-groups on Banach spaces[END_REF], Borichev-Tomilov [START_REF] Borichev | Optimal polynomial decay of functions and operator semigroups[END_REF], Burq [START_REF] Burq | Mesures semi-classiques et mesures de défaut[END_REF], Lebeau [START_REF] Lebeau | Équation des ondes amorties[END_REF].

The first item is a consequence of the energy decay. The second is given by unique continuation theorem and also by the result given in [START_REF] Cornilleau | Carleman estimates for the Zaremba Boundary Condition and Stabilization of Waves[END_REF]Proposition 1.1]. The goal of this article is to prove the third item. From the Gearhart-Huang-Prüss test for the exponential stability (see [START_REF] Gearhart | Spectral theory for contraction semigroups on Hilbert space[END_REF], [START_REF] Prüss | On the spectrum of C 0 -semigroups[END_REF], [START_REF] Huang | Characteristic conditions for exponential stability of linear dynamical systems in Hilbert spaces[END_REF]Theorem 3], [START_REF] Engel | One-parameter semigroups for linear evolution equations[END_REF]), the three items of the previous theorem imply that the semigroup generated by A is exponentially stable and this implies the following theorem.

Theorem 2. We assume that P , a(x) and Ω satisfy the modified Geometric Control Condition given in Definition 1.2. We assume that P has the form given in (4) and P is self-adjoint positive definite. Let A defined by [START_REF] Bey | Singularities of the solution of a mixed problem for a general second order elliptic equation and boundary stabilization of the wave equation[END_REF], there exist C, c > 0 such that

e tA L(H) ≤ Ce -ct .
The outline of the proof is the following. In Section 2 we recall some tools on semiclassical pseudodifferential calculus (Section 2.1), we reduce the third item of Theorem 1 to a semiclassical estimate (Section 2.2 and Proposition 2.2), and we prove a basic estimate on the trace at the boundary (Section 2.3 and Proposition 2.6).

In Section 3 we construct a semiclassical measure and we prove that this measure is supported on the characteristic set. To do that at the boundary -for interior point the result is classical-we have to distinguish three kinds of points, hyperbolic points (see Section 3.1.2 and Proposition 3.2) and glancing points (see Section 3.1.3 and Proposition 3.6). Only for elliptic points (see Section 3.1.4) we need to consider the boundary conditions. In a neighborhood of boundary where we impose Dirichlet or Neumann boundary condition, we prove Proposition 3.11 and we deduce Proposition 3.12. The proof is delicate in a neighborhood of Γ (see Proposition 3.16). These estimates on trace allow us to prove Proposition 3.1 in Section 3.2. In Section 3.3 we prove that semiclassical measure is not identically zero and in Section 3.4 we prove that semiclassical measure is null on support of the damping.

We shall reach a contradiction if we also prove that the measure is identically null. This is done in the next sections.

In Section 4 we prove some properties of semiclassical measure. In Section 4.1, we obtain the action of Hamiltonian vector field on the semiclassical measure up to the boundary. The interior result is stated in Proposition 4.1. Propositions 4.7 and 4.8 are analogous results at boundary. In Section 4.2, we deduce from that a decomposition of the semiclassical measure in two measure, the measure restricted in interior and a boundary measure (Lemma 4.9). The action of Hamiltonian vector field allows us to deduce some properties of these measures. Lemma 4.10 describes the action of Hamiltonian on the interior measure, Lemma 4.11 gives precisions in neighborhood of hyperbolic points, Lemma 4.12 and Lemma 4.13 give properties of boundary measure in a neighborhood of Dirichlet and Neumann boundary, and Lemma 4.14 is the analogous in neighborhood of jump between Dirichlet and Neumann boundary conditions.

In Section 5 we prove the propagation of support of semiclassical measure. We have to distinguish the different cases, if a bicharacteristic hits boundary transversally or tangentially, ∂Ω D ∪ ∂Ω N or Γ. This allows to prove the main theorem.

In Appendix A we prove some regularity measure needed to prove some estimate in a neighborhood or Γ.

In Appendix B we prove some estimates on boundary trace in the case of Neumann boundary condition. This is useful to prove properties of semiclassical measure in a neighborhood of a diffractive point in a neighborhood of ∂Ω N . Appendix C is devoted to prove some technical results stated in the previous section.

This work is based on previous results, mention particularly, the course given by Patrick Gérard at IHP in 2015, the articles of Burq and Lebeau [START_REF] Burq | Mesures de défaut de compacité, application au système de Lamé[END_REF] and Luc Miller [START_REF] Miller | Refraction of high-frequency waves density by sharp interfaces and semiclassical measures at the boundary[END_REF]. We thank Claude Zuily for the first step in this kind of problem (see [START_REF] Robbiano | The Kato smoothing effect for Schrödinger equations with unbounded potentials in exterior domains[END_REF]), Belhassen Dehman, Matthieu Léautaud and Jérôme Le Rousseau for the working group where we have together studied this subject, Nicolas Burq to draw our attention on the Tataru paper [START_REF] Tataru | On the regularity of boundary traces for the wave equation[END_REF]. That allowed us to achieve the propagation of measure at boundary in the case of Neumann boundary condition.

Semiclassical formulation

Notations and pseudo-differential calculus

Here we summarize some result on pseudo-differential calculus. More details, results and extension are given in the Hörmander book [START_REF] Hörmander | The analysis of linear partial differential operators[END_REF]Chapter 18], Martinez [START_REF] Martinez | An introduction to semiclassical and microlocal analysis[END_REF] and Le Rousseau-Lebeau [START_REF] Rousseau | Applications to unique continuation and control of parabolic equations[END_REF]. Essentially we follow here this last article.

To a smooth function a(x, ξ), and h ∈ (0, 1) (a may depend on h but the constants, in the estimates given below, does not depend on h), we associate an operator by the following formula

Op sc (a)u = (2π) -d R d e ixξ a(x, hξ)û(ξ)dξ, where û(ξ) = R d e -ixξ u(x)dx.
This formula make sense under some assumption on a and u. In this paper we mainly use symbols in S k . We say that a ∈ S k if for every α, β ∈ N d there exists C = C α,β such that

|∂ α x ∂ β ξ a(x, ξ)| ≤ C ξ k-β , where ξ = (1 + |ξ| 2 ) 1/2 .
For a ∈ S k , Op sc (a)u has a sense for u ∈ S (R d ), and can be extended for u ∈ H s (R d ) for every s ∈ R.

In a neighborhood of ∂Ω it is useful to use a tangential calculus. For a smooth function a(x, ξ ′ ) we associate a tangential operator by the following formula

op sc (a)u = (2π) -d+1 R d-1 e ix ′ ξ ′ a(x, hξ ′ )ũ(ξ ′ , x d )dξ ′ , where ũ(ξ ′ , x d ) = R d-1 e -ix ′ ξ ′ u(x ′ , x d )dx ′ . This formula make sense if a ∈ S k tan , that is, for every α ∈ N d , β ∈ N d-1 , there exists C = C α,β such that |∂ α x ∂ β ξ ′ a(x, ξ ′ )| ≤ C ξ ′ k-β , where ξ ′ = (1 + |ξ ′ | 2 ) 1/2
. We also use this notation for pseudo-differential operator on the boundary x d = 0. In this case, a and u does not depend on x d .

For technical reason we also have to use other classes of symbols. In these cases we use the Hörmander's notations, for instance, S( ξ m , (dx) 2 + ξ ′ -2 (dξ ′ ) 2 ), S( ξ ′ m , (dx) 2 + (dξ ′ ) 2 ). In this case we keep the notations Op sc (a), if the symbol depends on ξ d and op sc (a), if the symbol does not depend on ξ d . We also use the notation op sc (b) = b(x, hD ′ ), in particular when we restrict a function on x d = 0, this allows to distinguish b(x, hD ′ ) and b(x ′ , 0, hD ′ ).

The main interest of pseudo-differential operators are the calculus of products, commutators, adjoints. We have for a ∈ S The asymptotic expansions may be extended to all power in h. Analogous formulas exist for op sc a.

Associated with this semiclassical calculus, we introduce the semiclassical Sobolev spaces. For

u ∈ S ′ (R d ), we define u H s sc = Op sc ( ξ s )u L 2 (R d ) , if Op sc ( ξ s )u ∈ L 2 (R d ). On the boundary we define for u ∈ S ′ (R d-1 ), |u| H s sc (x d =0) = | op sc ( ξ ′ s )u| L 2 (R d-1 ) , if op sc ( ξ ′ s )u ∈ L 2 (R d-1
). On the boundary we define (u|v

) 0 = R d-1 u(x ′ )v(x ′ )dx ′ .
We keep the same notation for a general Ω, namely (u|v

) 0 = ∂Ω u(x ′ )v(x ′ )dσ(x ′ )
, where σ is the superficial measure on ∂Ω.

Pseudo-differential operators act on Sobolev spaces. For a ∈ S k , there exists C > 0 such that

Op sc (a)u H s-k sc ≤ C u H s sc , for every u ∈ H s sc .
For a ∈ S k tan , there exists C > 0 such that

| op sc (a)u| H s-k sc ≤ C|u| H s sc (x d =0) , for every u ∈ H s sc (x d = 0).
When we consider Sobolev spaces on x d > 0, it is useful to consider op sc and distinguish variable

x d ∈ (0, ∞) and variables x ′ ∈ R d-1 . Let L 2 ((0, ∞), H s sc ) be the space such that u ∈ L 2 ((0, ∞), H s sc ) if R d-1 ∞ 0 | op sc ( ξ ′ s )u(x ′ , x d )| 2 dx d dx ′ = u 2 L 2 ((0,∞),H s sc ) < ∞.
We have the following estimate, let a ∈ S k tan , there exists C > 0, such that

op sc (a)u L 2 ((0,∞),H s-k sc ) ≤ C u L 2 ((0,∞),H s sc ) , , for every u ∈ L 2 ((0, ∞), H s sc ).
In the context of semiclassical Sobolev spaces we have the following trace formula. Let s > 0, there exists C > 0, such that

|u |x d =0 | H s sc ≤ Ch -1/2 u H s+1/2 sc , for every u ∈ H s+1/2 sc . (8)
We recall the Gårding inequality for semiclassical Sobolev spaces. Let a ∈ S 0 be such that a(x, ξ) ≥ 0, there exists C > 0, such that

Re(Op sc (a)u|u) L 2 (R d ) + Ch u 2 L 2 (R d ) ≥ 0. ( 9
)
Here and in what follows (w|v) L 2 (K) means the inner product in K.

For tangential symbol we have the analogous result. Let a ∈ S 0 tan be such that a(x, ξ ′ ) ≥ 0, there exists C > 0, such that

Re(op sc (a)u|u) L 2 ((0,∞)×R d-1 ) + Ch u 2 L 2 ((0,∞)×R d-1 ) ≥ 0. ( 10
)
We use consequences of this result. Let a ∈ S 0 tan , such that there exists

K > 0 such that |a(x, ξ ′ )| ≤ K, then op sc (a)u L 2 ((0,∞)×R d-1 ) ≤ 2K u L 2 ((0,∞)×R d-1 ) + Ch u L 2 ((0,∞)×R d-1 ) , (11)
where C > 0, depends on a finite number of seminorm of a. We have the same estimate at the boundary, if a ∈ S(1, (dx ′ ) 2 + (dξ ′ ) 2 ), and |a(x ′ , ξ ′ )| ≤ K, we have

| op sc (a)u| L 2 (R d-1 ) ≤ 2K|u| L 2 (R d-1 ) + Ch|u| L 2 (R d-1 ) , (12)
where C > 0, depends on a finite number of seminorm of a. In particular we use this estimate if a depends on a parameter but K is uniform with respect this parameter. In this case, in the previous estimate C depends on the parameter. For w ∈ L 2 (Ω), we extends w for x ∈ R d \ Ω by 0, and we use the following notations

w(x) = 1 Ω w(x) = w(x) if x ∈ Ω, 0 if x ∈ R d \ Ω. ( 13
)
If w ∈ L 2 (R d-1 × (0, ∞)), we extend w by 0 for x d < 0 and we use the notation w(x) = 1 x d >0 w(x).

In this article we use the symbol : A B means, there exists C > 0, A ≤ CB, where C is independent of parameters.

We denote z s = exp(s log(z)), where log z is the principal value of the logarithm, where z ∈ C\R -.

Evolution equation and resolvent estimate

We begin the proof of the third item of Theorem 1. We may assume |µ| ≥ 1 as A -iµI is invertible for all µ ∈ R, by second item and µ → (A -iµI) -1 is continuous from R to L(H, H).

Let F = (f 0 , f 1 ) ∈ H and let U = (u 0 , u 1 ) ∈ D(A) be such that AU -iµU = F , we have

u 1 -iµu 0 = f 0 -P u 0 -au 1 -iµu 1 = f 1 . ( 14
)
Lemma 2.1. Assume that there exists C 1 > 0 such that

|µ| u 0 L 2 (Ω) + ∇u 0 L 2 (Ω) ≤ C 1 ( f 0 H 1 (Ω) + f 1 L 2 (Ω) ) for all (f 0 , f 1 ) ∈ H, (15)
where (u 0 , u 1 ) are the solution of [START_REF] Cornilleau | Carleman estimates for the Zaremba Boundary Condition and Stabilization of Waves[END_REF]. Then there exists C 2 > 0 such that

U H ≤ C F H , for all F ∈ H,
where AU -iµU = F .

Proof. By [START_REF] Cornilleau | Carleman estimates for the Zaremba Boundary Condition and Stabilization of Waves[END_REF], u 1 = iµu 0 + f 0 , so that

u 1 L 2 (Ω) ≤ f 0 L 2 (Ω) + |µ| u 0 L 2 (Ω) f 0 H 1 (Ω) + f 1 L 2 (Ω) ,
and [START_REF] Davies | Spectral theory and differential operators[END_REF] gives

u 0 H 1 (Ω) f 0 H 1 (Ω) + f 1 L 2 (Ω)
, for |µ| ≥ 1. This gives the result.

Formula ( 14) implies the following equation on u 0

-P u 0 + µ 2 u 0 -iµau 0 = af 0 + iµf 0 + f 1 .
To use semiclassical tools, we set h = 1/µ, we multiply (15) by h 2 , we obtain the following equation on u 0 -h

2 P u 0 + u 0 -ihau 0 = ah 2 f 0 + ihf 0 + h 2 f 1 ,
and ( 15) is equivalent to

u 0 L 2 (Ω) + h∇u 0 L 2 (Ω) ≤ Ch( f 0 H 1 (Ω) + f 1 L 2 (Ω) ). ( 16
)
We shall prove this inequality by contradiction. If ( 16) is false, up to a normalization, there exist a sequence h n → 0 as n → ∞ denoted for sake of simplicity by h,

(u h ) h ∈ H 1 (Ω) and (f h 0 , f h 1 ) h ∈ H satisfying -h 2 P u h + u h -ihau h = ah 2 f h 0 + ihf h 0 + h 2 f h 1 u h L 2 (Ω) + h∇u h L 2 (Ω) = 1 h( f h 0 H 1 (Ω) + f h 1 L 2 (Ω) ) → 0 as h → 0. ( 17
)
Let g h 0 = ahf h 0 + hf h 1 and g h 1 = ihf h 0 , ( 17) is equivalent to

-h 2 P u h + u h -ihau h = hg h 0 + g h 1 u h L 2 (Ω) + h∇u h L 2 (Ω) = 1 g h 0 L 2 (Ω) + g h 1 H 1 (Ω) → 0 as h → 0. ( 18
)
Proposition 2.2. There exist β > α > 0, there exists θ ∈ C ∞ 0 (R), supported in [α, β], there exists

(ũ h ) h , satisfying ũh L 2 (Ω) + h∇ũ h L 2 (Ω) ≤ C, for some C > 0, such that -h 2 P v h + v h -ihav h = hq h , v h L 2 (Ω) = 1 and h∇v h L 2 (Ω) ≤ 2, q h L 2 (Ω) → 0 as h → 0, ( 19
)
where

v h = θ(h 2 P )ũ h . Proof. Let ψ ∈ C ∞ 0 (R), 0 ≤ ψ ≤ 1 such that ψ(s) = 1 if s ≤ 1 0 if s ≥ 2.
Let A > 0 be sufficiently large to be fixed below. We have

1 = lim n→∞ ψ(2 -n A -1 s) = ψ(sA -1 ) + ∞ k=1 ψ(2 -k A -1 s) -ψ(2 -k+1 A -1 s) .
Setting φ(s) = ψ(s)ψ(2s), we have

1 = ψ(sA -1 ) + ∞ k=1 φ(2 -k A -1 s) and φ is supported in [1/2, 2]. ( 20
)
By functional calculus for auto-adjoint operators, we have

I = ψ(A -1 h 2 P ) -ψ(Ah 2 P ) + ψ(Ah 2 P ) + ∞ k=1 φ(2 -k A -1 h 2 P ). ( 21
)
Lemma 2.3. There exists C > 0 such that

ψ(Ah 2 P )u h L 2 (Ω) ≤ C A -1 + h + h g h 0 L 2 (Ω) + g h 1 L 2 (Ω) , (22)
∞ k=1 φ(2 -k A -1 h 2 P )u h L 2 (Ω) ≤ CA -1 1 + h g h 0 L 2 (Ω) + g h 1 L 2 (Ω) . (23)
Proof. We apply ψ(Ah 2 P ) to equation [START_REF] Engel | One-parameter semigroups for linear evolution equations[END_REF], we obtain

-h 2 P ψ(Ah 2 P )u h + ψ(Ah 2 P )u h -ihψ(Ah 2 P )(au h ) = hψ(Ah 2 P )g h 0 + ψ(Ah 2 P )g h 1 .
Let ψ(s) = sψ(s) we have

ψ(Ah 2 P )u h = A -1 ψ(Ah 2 P )u h + ihψ(Ah 2 P )(au h ) + hψ(Ah 2 P )g h 0 + ψ(Ah 2 P )g h 1 .
As |ψ(s)| ≤ 1 and | ψ(s)| ≤ C, we obtain [START_REF] Gérard | Microlocal defect measures[END_REF].

To prove [START_REF] Gérard | Ergodic properties of eigenfunctions for the Dirichlet problem Duke[END_REF] first we estimate φ(2 -k A -1 h 2 P )u h . Let φ ∈ C ∞ 0 ((0, ∞)) to be fixed below. We apply φ(2 -k A -1 h 2 P ) to equation [START_REF] Engel | One-parameter semigroups for linear evolution equations[END_REF], we obtain

-h 2 P φ(2 -k A -1 h 2 P )u h = -φ(2 -k A -1 h 2 P )u h + ih φ(2 -k A -1 h 2 P )(au h ) + h φ(2 -k A -1 h 2 P )g h 0 + φ(2 -k A -1 h 2 P )g h 1 .
Let φ(s) = -s -1 φ(s) be supported in [1/2, 2]. We obtain

2 k Aφ(2 -k A -1 h 2 P )u h = -φ(2 -k A -1 h 2 P )u h + ih φ(2 -k A -1 h 2 P )(au h ) + h φ(2 -k A -1 h 2 P )g h 0 + φ(2 -k A -1 h 2 P )g h 1 .
This yields

2 k A φ(2 -k A -1 h 2 P )u h L 2 (Ω) ≤ C 1 + h g h 0 L 2 (Ω) + g h 1 L 2 (Ω)
. Summing over k we obtain [START_REF] Gérard | Ergodic properties of eigenfunctions for the Dirichlet problem Duke[END_REF]. Let θ(s) = ψ(A -1 s)ψ(As), by Lemma 2.3 and (21) choosing A sufficiently large and h ∈ (0, h 0 ] for h 0 > 0 sufficiently small, we can have u hθ(h 2 P )u h L 2 (Ω) as small as we want. From equation [START_REF] Engel | One-parameter semigroups for linear evolution equations[END_REF] multiplying by u h and integrating by parts, we obtain

-(P u h |u h ) L 2 (Ω) + Ω |u h | 2 dx = ih(au h |u h ) L 2 (Ω) + h(g h 0 |u h ) L 2 (Ω) + (g h 1 |u h ) L 2 (Ω) .
Taking h 0 sufficiently small, we have

h 2 (P u h |u h ) L 2 (Ω) = Ω |u h | 2 dx + ε h where ε h → 0 as h → 0. ( 24
)
We now observe that h 2 (P u h |u h ) L 2 (Ω) is equivalent to H 1 sc (Ω)-norm, uniformly with respect to h ∈ (0, 1). From [START_REF] Engel | One-parameter semigroups for linear evolution equations[END_REF], the assumption on norms consequently gives u h L 2 (Ω) ≥ C 0 + εh , where C 0 > 0 and εh → 0 as h → 0.

Let

v h = θ(h 2 P )u h / θ(h 2 P )u h L 2 (Ω)
. Now we prove that v h satisfies the equation. We apply θ(h 2 P ) to (18), we have

h 2 P θ(h 2 P )u h + θ(h 2 P )u h -ihθ(h 2 P )(au h ) = hθ(h 2 P )g h 0 + θ(h 2 P )g h 1
which is equivalent to

h 2 P v h + v h -ihav h = θ(h 2 P )u h -1 L 2 (Ω) ih[θ(h 2 P ), a]u h + hθ(h 2 P )g h 0 + θ(h 2 P )g h 1 = hq h . ( 25
)
To obtain the estimate on q h we have to prove

[θ(h 2 P ), a]u h L 2 (Ω) → 0 as h → 0, (26)
h -1 θ(h 2 P )g h 1 L 2 (Ω) → 0 as h → 0. ( 27
)
To do that we need the following result proved below.

Lemma 2.4. Let φ ∈ C ∞ 0 (0, ∞), and a ∈ C ∞ 0 (Ω), there exists C > 0 such that

[φ(h 2 P ), a]w L 2 (Ω) ≤ Ch w L 2 (Ω) , [∂ xj , φ(h 2 P )]w L 2 (Ω) ≤ C w L 2 (Ω) , ∂ xj φ(h 2 P )w L 2 (Ω) ≤ C w H 1 (Ω) , for j = 1, . . . , d.
Estimate ( 26) is a direct consequence of Lemma 2.4. To prove [START_REF] Huang | Characteristic conditions for exponential stability of linear dynamical systems in Hilbert spaces[END_REF], let θ(s

) = s -1 θ(s) ∈ C ∞ 0 (0, ∞), we compute h -2 θ(h 2 P )g h 1 2 L 2 (Ω) = h -2 (θ(h 2 P )g h 1 |θ(h 2 P )g h 1 ) L 2 (Ω) , = h -2 (h 2 P θ(h 2 P )g h 1 |θ(h 2 P )g h 1 ) L 2 (Ω) , = 1≤j,k≤d (p jk ∂ xj θ(h 2 P )g h 1 |∂ x k θ(h 2 P )g h 1 ) L 2 (Ω) + 1≤j≤d (-ip j ∂ xj θ(h 2 P )g h 1 |θ(h 2 P )g h 1 ) L 2 (Ω) + (p 0 θ(h 2 P )g h 1 |θ(h 2 P )g h 1 ) L 2 (Ω)
as θ(h 2 P )g h 1 and θ(h 2 P )g h 1 are in domain of P we can apply [START_REF] Andersson | The propagation of singularities along gliding rays[END_REF]. By Lemma 2.4, we obtain

h -2 θ(h 2 P )g h 1 2 L 2 (Ω) ≤ C g h 1 2
H 1 (Ω) . This completes the proof of Proposition 2.2 because arguing as we did to obtain Formula (24), we get v h L 2 (Ω) = 1 so that hDv h L 2 (Ω) = 1 + ε h with ε h → 0 as h → 0.

Proof. Lemma 2.4. We start with the following lemma which will be proved below.

Lemma 2.5. Let C 1 > 0. There exists C > 0 such that

(-h 2 P + z) -1 f L 2 (Ω) ≤ C| Im z| -1 f L 2 (Ω) , h∂ xj (-h 2 P + z) -1 f L 2 (Ω) ≤ C| Im z| -1 f L 2 (Ω) , (-h 2 P + z) -1 h∂ xj f L 2 (Ω) ≤ C| Im z| -1 f L 2 (Ω) , for all |z| ≤ C 1 and f ∈ L 2 (Ω).
To prove Lemma 2.4 we use the Helffer-Sjöstrand formula,

φ(h 2 P ) = - 1 π ∂z φ(x, y)(-h 2 P + z) -1 dxdy,
where z = x + iy ∈ C and φ is an almost analytic extension of φ (see [START_REF] Helffer | Équation de Schrödinger avec champ magnétique et équation de Harper[END_REF]Proposition 7.2] and [START_REF] Davies | Spectral theory and differential operators[END_REF]): φ is compactly supported and satisfies φ(x, 0) = φ(x),

| ∂z φ(x, y)| ≤ C N |y| N , for every N.
We recall that ∂z = (1/2)(∂ x + i∂ y ). The Helffer-Sjöstrand formula gives

[φ(h 2 P ), a] = - 1 π ∂z φ(x, y)[(-h 2 P + z) -1 , a]dxdy = 1 π ∂z φ(x, y)(-h 2 P + z) -1 [-h 2 P, a](-h 2 P + z) -1 dxdy.
As [-h 2 P, a] is a sum of terms of following type bh 2 ∂ xj and ch 2 , where b and c are into C ∞ (Ω), the first estimate of the lemma is given by the two following estimates

(-h 2 P + z) -1 ch 2 (-h 2 P + z) -1 w L 2 (Ω) h 2 | Im z| -1 (-h 2 P + z) -1 w L 2 (Ω) h 2 | Im z| -2 w L 2 (Ω) , (28)
by the first estimate of Lemma 2.5.

(-

h 2 P + z) -1 bh 2 ∂ xj (-h 2 P + z) -1 w L 2 (Ω) h| Im z| -1 h∂ xj (-h 2 P + z) -1 w L 2 (Ω) h| Im z| -2 w L 2 (Ω) , (29)
by the two first estimates of Lemma 2.5.

For the second estimate we have by the Helffer-Sjöstrand formula

[∂ xj , φ(h 2 P )] = - 1 π ∂z φ(x, y)[∂ xj , (-h 2 P + z) -1 ]dxdy = 1 π ∂z φ(x, y)(-h 2 P + z) -1 [∂ xj , -h 2 P + z](-h 2 P + z) -
(-h 2 P + z) -1 h 2 ∂ xj b∂ x k (-h 2 P + z) -1 w L 2 (Ω) | Im z| -1 h∂ xj (-h 2 P + z) -1 w L 2 (Ω) , | Im z| -2 w L 2 (Ω) ,
By the second and third estimate of Lemma 2.5. This gives the second estimate of Lemma 2.4.

To prove the third estimate of Lemma 2.4 we write

∂ xj φ(h 2 P ) = φ(h 2 P )∂ xj + [∂ xj , φ(h 2 P )].
The first term is clearly estimated by H1 -norm and second term is estimated by the second inequality of Lemma 2.4.

Proof. Lemma 2.5. Let u = (-h2 P + z) -1 f , we have u ∈ H 1 (Ω) and u satisfies (-h 2 P + z)u = f and the Zaremba boundary condition. Multiplying the equation by u, integrating over Ω and an performing integration by parts, we get

-(h 2 P u|u) L 2 (Ω) + z u 2 L 2 (Ω) = (f |u). ( 30
)
Taking the imaginary part of equation we have

| Im z| u 2 L 2 (Ω) ≤ f L 2 (Ω) u L 2 (Ω)
, which gives the first estimate.

Taking the real part of (30) and from [START_REF] Batty | Non-uniform stability for bounded semi-groups on Banach spaces[END_REF] we have ∇u 2

L 2 (Ω) (P u|u) L 2 (Ω)
, we obtain

h∇u 2 L 2 (Ω) f L 2 (Ω) u L 2 (Ω) + u 2 L 2 (Ω) .
This gives the second estimate with the previous result.

It is sufficient to prove the third estimate of Lemma 2.5 for

f ∈ C ∞ 0 (Ω) as C ∞ 0 (Ω) is dense in L 2 (Ω). We have (-h 2 P + z) -1 h∂ xj f 2 L 2 (Ω) = ((-h 2 P + z) -1 h∂ xj f |(-h 2 P + z) -1 h∂ xj f ) = -(f |h∂ xj (-h 2 P + z) -1 (-h 2 P + z) -1 h∂ xj f ) ≤ C| Im z| -1 f L 2 (Ω) (-h 2 P + z) -1 h∂ xj f L 2 (Ω) ,
by the previous result. This gives the third estimate of Lemma 2.5.

A priori estimate on traces

In this section we assume that Ω is locally given near a point of the boundary by x d > 0 and we denote the variables by x = (x ′ , x d ) where x ′ ∈ R d-1 , and we set

R d + = {(x ′ , x d ) ∈ R d , x d > 0}. In these local coordinates 1 we have h 2 D 2 x d v h + (R(x, hD ′ ) -1 + iha)v h = hq h .
When there are no ambiguity we only write R instead of R(x, hD ′ ). Proposition 2.6. Let v h given in Proposition 2.2. Then, there exists C > 0 such that

|(v h ) |x d =0 | H 1/2 sc ≤ Ch -1/2 , (31)
|(hD x d v h ) |x d =0 | H -1/2 sc ≤ Ch -1/2 . ( 32
)
Proof. As v h ∈ H 1 sc (Ω), the trace formula [START_REF] Bardos | Sharp sufficient conditions for the observation, control and stabilization of waves from the boundary[END_REF] gives the first estimate.

Let χ ∈ C ∞ (R) be such that χ(x d ) = 1 if x d ≤ δ, χ(x d ) = 0 if x d ≥ 2δ and 0 ≤ χ ≤ 1. Firstly h| op sc ( ξ ′ -1/2 )hD x d v h (x ′ , 0)| 2 L 2 = -i ∞ 0 hD x d χ(x d )| op sc ( ξ ′ -1/2 )hD x d v h (x ′ , x d )| 2 dx d = - ∞ 0 hχ ′ (x d )| op sc ( ξ ′ -1/2 )hD x d v h (x ′ , x d )| 2 dx d -2i ∞ 0 χ(x d ) Re op sc ( ξ ′ -1/2 )h 2 D 2 x d v h (x ′ , x d )| op sc ( ξ ′ -1/2 )hD x d v h (x ′ , x d ) dx d = - ∞ 0 hχ ′ (x d )| op sc ( ξ ′ -1/2 )hD x d v h (x ′ , x d )| 2 dx d -2i ∞ 0 χ(x d ) Re op sc ( ξ ′ -1/2 )(-R + 1)v h (x ′ , x d )| op sc ( ξ ′ -1/2 )hD x d v h (x ′ , x d ) dx d -2i ∞ 0 χ(x d ) Re op sc ( ξ ′ -1/2 )h(q h (x ′ , x d ) -i(av h )(x ′ , x d ))| op sc ( ξ ′ -1/2 )hD x d v h (x ′ , x d ) dx d = I 1 + I 2 + I 3 .
Secondly, we have

|I 1 | h hD x d v h 2 L 2 (Ω) ≤ C, |I 3 | h( q h L 2 (Ω) + v h L 2 (Ω) ) hD x d v h L 2 (Ω) ≤ C.
Finally, we obtain

I 2 = -2i ∞ 0 χ(x d ) Re op sc ( ξ ′ -1 )(-R + 1)v h (x ′ , x d )|hD x d v h (x ′ , x d ) dx d . By tangential semiclassical pseudo-differential calculus, op sc ( ξ ′ -1 )(-R + 1) is of order 1, then |I 2 | op sc ( ξ ′ )v h L 2 (Ω) hD x d v h L 2 (Ω) ≤ C.
This achieves the proof of the proposition.

3 Semiclassical measure and the characteristic set

Support of the semiclassical measure

We now define a semiclassical measure associated with

(v h ) h . It is classical, as (v h ) h is bounded in L 2 (Ω), that there exists µ a measure on T * (R d ) such that, up to extraction of subsequence of (v h ) h , we have for all b(x, ξ) ∈ C ∞ 0 (R 2d ), (Op sc (b)v h |v h ) L 2 (R d ) → µ, b as h → 0. ( 33
)
For first expositions on microlocal defect measure or H-measure see [START_REF] Gérard | Microlocal defect measures[END_REF][START_REF] Tartar | H-measures, a new approach for studying homogenisation, oscillations and concentration effects in partial differential equations[END_REF]. For semiclassical measure or Wigner measure see [START_REF] Burq | Mesures semi-classiques et mesures de défaut[END_REF][START_REF] Gérard | Mesures semi-classiques et ondes de Bloch Séminaire sur les Équations aux Dérivées Partielles[END_REF][START_REF] Miller | Refraction of high-frequency waves density by sharp interfaces and semiclassical measures at the boundary[END_REF][START_REF] Robbiano | The Kato smoothing effect for Schrödinger equations with unbounded potentials in exterior domains[END_REF]. The goal of this section is to prove the following result.

Proposition 3.1. The measure µ is supported in Ω × R d and pµ = 0, where p is the semiclassical symbol of h 2 P -1 (see (2)).

To prove this proposition we consider four sets in T * (R d ), exterior or interior points (i.e. x ∈ R d \Ω or x ∈ Ω), hyperbolic points (i.e., x ∈ ∂Ω and R(x, ξ ′ ) -1 < 0), glancing points (i.e., x ∈ ∂Ω and R(x, ξ ′ ) -1 = 0), and elliptic points (i.e., x ∈ ∂Ω and R(x, ξ ′ ) -1 > 0).

The proofs for exterior and interior points are classical, we give the proofs for the sake of completeness. The proofs for hyperbolic and glancing points are similar to the proofs given by Burq and Lebeau [START_REF] Burq | Mesures de défaut de compacité, application au système de Lamé[END_REF] in the context of defect measures. The proof for elliptic points is specific to Zaremba boundary condition. Of course in ∂Ω D ∩ ∂Ω N the proof is also classical.

Interior and exterior points

Clearly, for χ

(x) ∈ C ∞ 0 (R d \ Ω) and φ(ξ) ∈ C ∞ 0 (R d ), (Op sc (χ(x)φ(ξ))v h |v h ) → 0 as h → 0. Then µ is supported in Ω × R d . Let χ(x) ∈ C ∞ 0 (Ω), φ(ξ) ∈ C ∞ 0 (R d ) and χ 1 (x) ∈ C ∞ 0 (Ω) be such that χ 1 χ = χ, we have, by symbol calculus as the symbol p(x, ξ) -1 is in S 2 (R d × R d ) and χ(x)φ(ξ) ∈ S -∞ (R d × R d ), Op sc (χ(x)φ(ξ))(h 2 P -1) = Op sc (χ(x)φ(ξ)p(x, ξ)) + h Op sc (r 0 ) = Op sc (χ(x)φ(ξ))χ 1 (x)(h 2 P -1) + h Op sc (r ′ 0 ), where r 0 , r0 , r ′ 0 ∈ S 0 (R d × R d ). As χ 1 is compactly supported in Ω and P is a local operator χ 1 (x)(h 2 P -1)v h = χ 1 (x)(h 2 P -1)v h , we have (Op sc (χ(x)φ(ξ))(h 2 P -1)v h |v h ) = (Op sc (χ(x)φ(ξ))χ 1 (x)(h 2 P -1)v h |v h ) + hO( v h 2 L 2 (Ω) ) = (Op sc (χ(x)φ(ξ))χ 1 (x)(-iha -hq h )v h |v h ) + hO( v h 2 L 2 (Ω) ) → 0 as h → 0. ( 34
)
We also have

(Op sc (χ(x)φ(ξ))(h 2 P -1)v h |v h ) = (Op sc (χ(x)φ(ξ)p(x, ξ))v h |v h ) + hO( v h 2 L 2 (Ω) ) → µ, p(x, ξ)χ(x)φ(ξ) as h → 0.
This and [START_REF] Martinez | An introduction to semiclassical and microlocal analysis[END_REF] give that µ, p(x, ξ)χ(x)φ(ξ) = 0. This proves Proposition 3.1 for the interior points, as the space spanned by functions χ

(x)φ(ξ) is dense in C ∞ 0 (Ω × R d ).
Before proving Proposition 3.1 in a neighborhood of the boundary, we have to prove estimates more precise than Proposition 2.6. We have to distinguish microlocally hyperbolic, glancing and elliptic points. For hyperbolic and glancing points the boundary condition play no role. For elliptic points we have to distinguish points in ∂Ω D ∪ ∂Ω N , and points in Γ. The results are stated in Proposition 3.12 and in Proposition 3.16.

Hyperbolic points

Let δ > 0 be sufficiently small, let

χ = χ ε ∈ C ∞ (R 2d ) be such that χ(x, ξ ′ ) = 1 if R(x, ξ ′ ) -1 ≤ -ε and x d ≤ δ 0 if R(x, ξ ′ ) -1 ≥ -ε/2 or x d ≥ 2δ, (35)
and 0 ≤ χ ≤ 1. Observe that χ ∈ S -∞ tan as supported for |ξ ′ | ≤ C, where C depends on R. Proposition 3.2. For any ε > 0, there exists

C ε > 0 such that | op sc (χ |x d =0 )(v h ) |x d =0 | H 1 sc ≤ C ε | op sc (χ |x d =0 )(hD x d v h ) |x d =0 | L 2 ≤ C ε , (36)
for all h ∈ (0, 1].

Remark 5. The estimate on traces, in the hyperbolic region, are better than the one proved in Proposition 2.6.

We begin the proof by giving a localization result which is useful in each region defined in what follows.

Lemma 3.3. Let χ(x, ξ ′ ), χ 1 (x, ξ ′ ) ∈ S 0 tan , be such that χ 1 (x, ξ ′ ) = 1, for (x, ξ ′ ) in the support of χ. We assume 0 ≤ χ ≤ 1 and 0 ≤ χ 1 ≤ 1. Let w h = op sc (χ)v h , there exists q h 2 such that h 2 D 2 x d + op sc (χ 2 1 (R(x, ξ ′ ) -1)) w h = hq h 2 , where q h 2 L 2 (Ω) ≤ C, (37)
for some C > 0 depending on semi-norms of χ and χ 1 .

Proof. We have, for

x d > 0, (h 2 D 2 x d + R(x, hD ′ ) -1)w h = h op sc (χ)(q h -iav h ) + [R(x, hD ′ ), op sc (χ)]v h + [h 2 D 2 x d , op sc (χ)] = hq h 1 .
The symbol of [R(x, hD ′ ), op sc (χ)] is in hS 1 tan and by exact symbol calculus with D x d , we have

[h 2 D 2 x d , op sc (χ)] = -ih 2 D x d op sc (∂ x d χ) -ih op sc (∂ x d χ)hD x d = -2ih op sc (∂ x d χ)hD x d + h 2 op sc (r 0 ),
where r 0 ∈ S 0 tan . By the properties of v h given in Proposition 2.2 and the previous formulas, there exists C > 0 such that

q h 1 L 2 ≤ C. We have R(x, hD ′ ) -1 = op sc (χ 2 1 (R(x, ξ ′ ) -1)) + op sc ((1 -χ 2 1 )(R(x, ξ ′ ) -1)), thus h 2 D 2 x d + op sc (χ 2 1 (R(x, ξ ′ ) -1)) w h = hq h 1 -op sc ((1 -χ 2 1 )(R(x, ξ ′ ) -1)) op sc (χ)v h = hq h 2 .
As (1χ 2 1 )χ = 0 by assumption, we deduce from symbol calculus that

op sc ((1 -χ 2 1 )(R(x, ξ ′ ) -1)) op sc (χ)v h L 2 (Ω) ≤ Ch.
This gives [START_REF] Miller | Refraction of high-frequency waves density by sharp interfaces and semiclassical measures at the boundary[END_REF].

Proof of Proposition 3.2. Let χ 1 ∈ C ∞ (R 2d ) be such that χ 1 (x, ξ ′ ) = 1 if R(x, ξ ′ ) -1 ≤ -ε/2 and x d ≤ 2δ 0 if R(x, ξ ′ ) -1 ≥ -ε/4 or x d ≥ 3δ, in particular χ 1 is 1 on the support of χ. Let b(x, ξ ′ ) = χ 1 (x, ξ ′ ) 1 -R(x, ξ ′ ) 1/2 , observe that 1 -R(x, ξ ′ ) > 0 on the support of χ 1 .
We denote w h = op sc (χ)v h , where χ, is defined by [START_REF] Melrose | Singularities of boundary value problems I[END_REF].

Lemma 3.4. There exist C ε > 0, q h 3 and q h 4 such that

hD x d -op sc (b) hD x d + op sc (b) w h = hq h 3 hD x d + op sc (b) hD x d -op sc (b) w h = hq h 4 ,
where q h j L 2 (Ω) ≤ C ε for j = 3, 4. As the semi-norms of χ, χ 1 depend on ε the estimates depend on ε and to keep in mind that, we denote these constants by C ε .

Proof. Let k = 1, 2. We have

hD x d -(-1) k op sc (b) hD x d + (-1) k op sc (b) = h 2 D 2 x d + (-1) k hD x d op sc (b) -(-1) k op sc (b)hD x d -op sc (b) 2 = h 2 D 2 x d + op sc (χ 2 1 (R(x, ξ ′ ) -1)) + ±h op sc (r 1 ), as op sc (b) 2 = op sc (b 2 ) + h op sc (r 1
) and exact symbol calculus, we have [START_REF] Miller | Refraction of high-frequency waves density by sharp interfaces and semiclassical measures at the boundary[END_REF], this proves the lemma.

r 1 = (-1) k D x d b + r1 . As χ 1 is compactly supported, we have r 1 , r1 ∈ S 0 tan . Then op sc (r 1 )w h L 2 (Ω) ≤ C ε op sc (χ)v h L 2 (Ω) ≤ C ε . With
Let z h k = hD x d -(-1) k op sc (b) w h , by Lemma 3.4, we have, for k = 1, 2, hD x d + (-1) k op sc (b) z h k = hq h k ,
where qh 1 = q h 3 and qh 2 = q h 4 . By Proposition 2.2 and Lemma 3.4, we have

z h k L 2 (Ω) ≤ C ε and qh k L 2 (Ω) ≤ C ε for k = 1, 2. Lemma 3.5. There exists C ε > 0 such that |(z h k ) |x d =0 | L 2 ≤ C ε . Proof. We have h∂ x d R d-1 |z h k (x ′ , x d ) | 2 dx ′ = 2 Re R d-1 ihD x d z h k (x ′ , x d ) z h k (x ′ , x d )dx ′ = 2 Re R d-1 i(-1) k+1 op sc (b) z h k (x ′ , x d ) z h k (x ′ , x d )dx ′ + 2 Re R d-1 ihq h k z h k (x ′ , x d )dx ′ .
Integrating with respect to x d from 0 to ∞, we obtain

h R d-1 |z h k (x ′ , 0)| 2 dx ′ Re(i op sc (b)z h k |z h k ) L 2 (Ω) + h qh k L 2 (Ω) z h k L 2 (Ω) . (38)
As

(i op sc (b)z h k |z h k ) L 2 (Ω) = (z h k | -i op sc (b) * z h k ) L 2 (Ω) and op sc (b) * = op sc (b) + h op sc (r 0 ) where r 0 ∈ S 0 tan , we have 2 Re(i op sc (b)z h k |z h k ) L 2 (Ω) ≤ C ε h z h k 2 L 2 (Ω) . Then (38) implies Lemma 3.5.
The definition of z h k , implies

(hD x d w h ) |x d =0 + op sc (b 0 )(w h ) |x d =0 = (z h 1 ) |x d =0 (hD x d w h ) |x d =0 -op sc (b 0 )(w h ) |x d =0 = (z h 2 ) |x d =0 , (39)
where

b 0 = b |x d =0 . Then we have 2(hD x d w h ) |x d =0 = (z h 1 ) |x d =0 + (z h 2 ) |x d =0 . As (hD x d op sc (χ)v h ) |x d =0 = (op sc (χ) |x d =0 (hD x d v h ) |x d =0 + h op sc (D x d χ)) |x d =0 (v h ) |x d =0 ,
and from [START_REF] Lebeau | Équation des ondes amorties[END_REF] we deduce that |(hD

x d w h ) |x d =0 | L 2 ≤ C ε by Lemma 3.5.
This gives the second estimate of [START_REF] Melrose | Singularities of boundary value problems II[END_REF].

From [START_REF] Robbiano | Spectral analysis of the interior transmission eigenvalue problem[END_REF] we also have

| op sc (b 0 )(w h ) |x d =0 | L 2 ≤ C ε . ( 40
) Let χ 2 ∈ C ∞ 0 (R d × R d-1
), be such that (χ 2 ) |x d =0 = 1, on the support of χ |x d =0 and supp χ 2 ⊂ {χ 1 = 1}. By symbol calculus we have

op sc χ 2 (1 -R(x, ξ ′ )) -1/2 |x d =0 op sc (b 0 ) = op sc (χ 2 ) |x d =0 + h op sc (r 0 ), where r 0 ∈ S -1
tan , and

op sc (χ 2 ) |x d =0 op sc (χ |x d =0 ) = op sc (χ |x d =0 ) + h op sc (r 1 ),
where r 1 ∈ S -1 tan . We can write 40) and [START_REF] Lebeau | Équation des ondes amorties[END_REF]. This achieves the proof of Proposition 3.2.

(w h ) |x d =0 = op sc (χ) |x d =0 (v h ) |x d =0 = op sc (χ 2 ) |x d =0 (w h ) |x d =0 -h op sc (r 1 )(w h ) |x d =0 = op sc χ 2 (1 -R(x, ξ ′ )) -1/2 |x d =0 op sc (b 0 )(w h ) |x d =0 -h op sc (r 0 )(w h ) |x d =0 -h op sc (r 1 )(w h ) |x d =0 . As χ 2 is compactly supported, op sc ( ξ ′ ) op sc χ 2 (1 -R(x, ξ ′ )) -1/2 |x d =0 has a symbol in S 0 tan and op sc ( ξ ′ ) op sc (r k ), k = 0, 1 have symbols in S 0 tan , then | op sc (χ |x d =0 )(v h ) |x d =0 | H 1 sc ≤ C ε | op sc (b 0 )(w h ) |x d =0 | L 2 + h| op sc (χ |x d =0 )(v h ) |x d =0 | L 2 ≤ C ε , applying (

Glancing points

Let χ = χ ε ∈ C ∞ (R 2d ) be such that χ(x, ξ ′ ) = 1 if |R(x, ξ ′ ) -1| ≤ 2ε and x d ≤ 2δ 0 if |R(x, ξ ′ ) -1| ≥ 3ε or x d ≥ 3δ, (41)
and 0 ≤ χ ≤ 1. Observe that χ ∈ S -∞ tan as supported for ξ ′ in a compact set, as R(x, ξ ′ ) is bounded. Let w h = op sc (χ)v h . We have the following estimate on the traces on w h . Proposition 3.6. There exists C > 0 such that, for any ε > 0,

| op sc (χ |x d =0 )(v h ) |x d =0 | H 1 sc ≤ Cε 1/4 h -1/2 + C ε h -3/8 (42) | op sc (χ |x d =0 )(hD x d v h ) |x d =0 | L 2 ≤ Cε 3/4 h -1/2 + C ε h -3/8 , (43)
for C ε > 0 .

Remark 6. Compared with the estimates stated in Proposition 2.6, we have the same power of h but with a power of ε in front of h -1/2 . The term

C ε h -3/8 = C ε h -1/2 h 1/
4 is a remainder. This gives a gain in this microlocal region.

Proof. In this proof C is a constant independent of ε and we denote by

C ε a constant depending on ε. Let χ 1 ∈ C ∞ (R 2d ) be such that χ 1 (x, ξ ′ ) = 1 if |R(x, ξ ′ ) -1| ≤ ε and x d ≤ δ 0 if |R(x, ξ ′ ) -1| ≥ -ε/2 or x d ≥ δ,
and 0 ≤ χ 1 ≤ 1. The symbol χ 1 ∈ S -∞ tan since it is supported in |ξ ′ | ≤ 2.
We have by Lemma 3.3

(h 2 D 2 x d + op sc (χ 2 1 (R(x, ξ ′ ) -1)))w h = hq h 2 , where q h 2 L 2 (Ω) ≤ C ε ,
for some C ε > 0.

Then we have

R d + |h 2 D 2 x d w h | 2 dx ≤ 2 R d + | op sc (χ 2 1 (R(x, ξ ′ ) -1))w h | 2 dx + 2 R d + |hq h 2 | 2 dx. (44)
First, by symbol calculus, we have

op sc (χ 2 1 (R(x, ξ ′ ) -1))w h = op sc (χ 2 1 (R(x, ξ ′ ) -1)) op sc (χ)v h = op sc (χ(R(x, ξ ′ ) -1))v h + h op sc (r 0 )v h ,
where r 0 ∈ S 0 tan . Observe that the semi-norms of r 0 depend on ε. This gives

op sc (χ 2 1 (R(x, ξ ′ ) -1))w h L 2 (Ω) ≤ op sc (χ(R(x, ξ ′ ) -1))v h L 2 (Ω) + C ε h v h L 2 (Ω) . (45)
On the support of χ, we have

|R(x, ξ ′ ) -1| ≤ 2ε then 4ε 2 -χ 2 (x, ξ ′ )(R(x, ξ ′ ) -1) 2 ≥ 0 and χ 2 (x, ξ ′ )(R(x, ξ ′ ) -1) 2 ∈ S 0
tan as χ is compactly supported. By Gårding inequality [START_REF] Borichev | Optimal polynomial decay of functions and operator semigroups[END_REF] and by symbol calculus, we have

4ε 2 v h 2 L 2 (x d >0) -op sc (χ(R(x, ξ ′ ) -1))v h 2 L 2 (x d >0) ≥ -C ε h v h 2 L 2 (x d >0) .
We deduce from this equation, ( 44) and ( 45)

h 2 D 2 x d w h L 2 (x d >0) ≤ (Cε + C ε h 1/2 ) v h L 2 (x d >0) + Ch q h 2 L 2 (x d >0) .
By estimates on v h and q h 2 , we obtain

h 2 D 2 x d w h L 2 (x d >0) ≤ Cε + C ε h 1/2 . ( 46
) Lemma 3.7. Let g ∈ L 2 (x d > 0) be supported in R d-1 × [0, 1]. We assume D x d g ∈ L 2 (x d > 0) then h|g |x d =0 | 2 L 2 ≤ 2 hD x d g L 2 (x d >0) g L 2 (x d >0) . (47) Proof. Since h R d-1 |g(x ′ , 0)| 2 dx ′ = -i R d-1 ∞ 0 hD x d |g(x ′ , x d )| 2 dx ′ dx d = -2i Re R d-1 ∞ 0 hD x d g(x ′ , x d )g(x ′ , x d )dx ′ dx d ,
we obtain the lemma by Cauchy-Schwarz inequality.

Lemma 3.8. There exists C > 0, such that for all

g ∈ L 2 (x d > 0) be supported in R d-1 × [0, 1]. Moreover we assume D x d g ∈ L 2 (x d > 0) and D 2 x d g ∈ L 2 (x d > 0), we have hD x d g 2 L 2 (x d >0) ≤ C h 2 D 2 x d g L 2 (x d >0) g L 2 (x d >0) (48)
Proof. We have

R d-1 ∞ 0 |hD x d g(x ′ , x d )| 2 dx ′ dx d = R d-1 ∞ 0 hD x d hD x d g(x ′ , x d )g(x ′ , x d ) dx ′ dx d - R d-1 ∞ 0 h 2 D 2 x d g(x ′ , x d )g(x ′ , x d )dx ′ dx d ,
we yields

hD x d g 2 L 2 (x d >0) ≤ h|hD x d g(x ′ , 0)| L 2 |g(x ′ , 0)| L 2 + h 2 D 2 x d g L 2 (x d >0) g L 2 (x d >0) . (49)
As g ∈ H 2 (x d > 0), we can apply Lemma 3.7 to hD x d g to obtain

h|(hD x d g) |x d =0 | 2 L 2 ≤ 2 h 2 D 2 x d g L 2 (x d >0) hD x d g L 2 (x d >0) . (50)
This estimate and (47) yield

h|g |x d =0 | L 2 |(hD x d g) |x d =0 | L 2 ≤ 2 h 2 D 2 x d g 1/2 L 2 (x d >0) hD x d g L 2 (x d >0) g 1/2 L 2 (x d >0)
. From this estimate and (49) we obtain (48). Lemma 3.9. There exists C > 0 such that, for any g ∈ H

2 (x d > 0) supported in R d-1 × [0, 1], we have h|g |x d =0 | 2 L 2 ≤ C h 2 D 2 x d g 1/2 L 2 (x d >0) g 3/2 L 2 (x d >0) (51) h|(hD x d g) |x d =0 | 2 L 2 ≤ C h 2 D 2 x d g 3/2 L 2 (x d >0) g 1/2 L 2 (x d >0) (52)
Proof. The first estimate is obtained from (47) and (48). The second estimate is obtained from (50) and (48).

Before applying the previous lemma to w h , we have to estimate uniformly this function. As χ depend on ε the norm of op sc (χ) as an operator on L 2 depends on ε. Nevertheless applying (11) a consequence of Gårding inequality and as |χ| ≤ 1 we have

w h L 2 (x d >0) = op sc (χ)v h L 2 (x d >0) ≤ C v h L 2 (x d >0) + C ε h 1/2 v h L 2 (x d >0) ≤ C + C ε h 1/2 .
From ( 46) and (51), we moreover have

h|(w h ) |x d =0 | 2 L 2 ≤ (Cε + C ε h 1/2 ) 1/2 (C + C ε h 1/2 ) 3/2 ≤ Cε 1/2 + C ε h 1/4
, which gives [START_REF] Shamir | Regularization of mixed second order elliptic problems[END_REF].

From ( 46) and (52), we also have

h|(hD x d w h ) |x d =0 | 2 L 2 ≤ (Cε + C ε h 1/2 ) 3/2 (C + C ε h 1/2 ) 1/2 ≤ Cε 3/2 + C ε h 1/4 , which gives (43) as (hD x d op sc (χ)v h ) |x d =0 = op sc (χ |x d =0 )(hD x d v h ) |x d =0 + h op sc (hD x d χ) |x d =0 (v h ) |x d =0 ,
and the last term can be estimated by [START_REF] Lebeau | Équation des ondes amorties[END_REF].

Elliptic points

We start with the notation introduced in Proposition 2.2. Let χ = χ ε ∈ C ∞ (R 2d-1 ), be such that

χ(x, ξ ′ ) = 1 if R(x, ξ ′ ) -1 ≥ 2ε and x d ≤ δ 0 if R(x, ξ ′ ) -1 ≤ ε or x d ≥ 2δ, (53)
and 0 ≤ χ ≤ 1. We have χ ∈ S 0 tan . In this region the support of χ is not bounded, we have to take care of the symbol classes we use.

Let

ρ(x, ξ ′ ) = R(x, ξ ′ ) -1 1/2 if (x, ξ ′ ) satisfies R(x, ξ ′ ) -1 > 0. Let χ 1 (x, ξ ′ ) ∈ S 0
tan be such that, χ 1 = 1 on the support of χ, and supp

χ 1 ⊂ {R(x, ξ ′ ) -1 ≥ ε/2} ∪ {x d ≤ 3δ}. Observe that χ 1 ρ ∈ S 1
tan . We have

hD x d + i op sc (χ 1 ρ) hD x d -i op sc (χ 1 ρ) = h 2 D 2 x d -ihD x d op sc (χ 1 ρ) + i op sc (χ 1 ρ)hD x d + op sc (χ 1 ρ) 2 = h 2 D 2 x d -i[hD x d , op sc (χ 1 ρ)] + op sc (χ 1 ρ) 2 ,
and

[hD x d , op sc (χ 1 ρ)] = h op sc D x d (χ 1 ρ) , where D x d (χ 1 ρ) ∈ S 1 tan , op sc (χ 1 ρ) 2 = op sc χ 2 1 (R(x, ξ ′ ) -1) + h op sc (r 1 )
, where r 1 ∈ S 1 tan .

Then

hD x d + i op sc (χ 1 ρ) hD x d -i op sc (χ 1 ρ) = h 2 D 2 x d + op sc χ 2 1 (R(x, ξ ′ ) -1) + h op sc (r 1 ),
where r1 ∈ S 1 tan . Applying Lemma 3.3 we obtain

hD x d + i op sc (χ 1 ρ) hD x d -i op sc (χ 1 ρ) op sc (χ)v h = hq h 2 , ( 54
)
where

q h 2 L 2 (x d >0) ≤ C ε . Let z = hD x d -i op sc (χ 1 ρ) op sc (χ)v h ; z
depends on h but for the sake of simplicity we prefer to denote it z. From (54) we have

hD x d + i op sc (χ 1 ρ) z = hq h 2 , in x d > 0. We then have 2 Re (hD x d + i op sc (χ 1 ρ))z|i op sc (χ 1 ρ)z L 2 (x d >0) h q h 2 L 2 (x d >0) op sc (χ 1 ρ)z L 2 (x d >0) . (55)
We recall the integration by parts formula in semiclassical context,

(u|hD x d w) L 2 (x d >0) = (hD x d u|w) L 2 (x d >0) -ih(u |x d =0 |w |x d =0 ) 0 , (56)
for u and w sufficiently smooth. Taking w = i op sc (χ 1 ρ)z and u = z we have

(z|ihD x d op sc (χ 1 ρ)z) L 2 (x d >0) = (hD x d z|i op sc (χ 1 ρ)z) L 2 (x d >0) -ih(z |x d =0 |i op sc (χ 1 ρ)z |x d =0 ) 0 . As ihD x d op sc (χ 1 ρ) = i op sc (χ 1 ρ)hD x d + h op sc (∂ x d (χ 1 ρ))
, and op sc (χ 1 ρ) = op sc (χ 1 ρ) * + h op sc (r 0 ), where r 0 ∈ S 0 tan we obtain 2 Re(hD

x d z|i op sc (χ 1 ρ)z) L 2 (x d >0) = h Re(z |x d =0 | op sc (χ 1 ρ)z |x d =0 ) 0 + hK, (57)
where

|K| hDz L 2 (x d >0) z L 2 (x d >0) + z 2 L 2 (x d >0) .
From equation satisfied by z, we have

hD x d z L 2 (x d >0) op sc (χ 1 ρ)z L 2 (x d >0) + h q h 2 L 2 (x d >0) op sc ( ξ ′ )z L 2 (x d >0) + h q h 2 L 2 (x d >0) .
Then

|K| op sc ( ξ ′ )z L 2 (x d >0) z L 2 (x d >0) + h 2 q h 2 2 L 2 (x d >0) . (58)
From ( 55), ( 57) and (58) we yield

op sc (χ 1 ρ)z 2 L 2 (x d >0) + h Re(z |x d =0 | op sc (χ 1 ρ)z |x d =0 ) 0 h op sc ( ξ ′ )z L 2 (x d >0) z L 2 (x d >0) + h 2 q h 2 2 L 2 (x d >0) , (59)
as we can estimate

h q h 2 L 2 (x d >0) op sc (χ 1 ρ)z L 2 (x d >0) ≤ α op sc (χ 1 ρ)z 2 L 2 (x d >0) + C α h 2 q h 2 2 L 2 (x d >0) ,
and absorb the term op sc (χ

1 ρ)z 2 L 2 (x d >0)
, by the left hand side of (59) if α is sufficiently small. Lemma 3.10.

op sc ( ξ ′ )z L 2 (x d >0) op sc (χ 1 ρ)z L 2 (x d >0) + h hDv h L 2 (x d >0) + h v h L 2 (x d >0) .
Proof. Let χ 2 (x, ξ ′ ) ∈ S 0 tan be such that, χ 2 = 1 on the support of χ 1 , and supp

χ 1 ⊂ {R(x, ξ ′ ) -1 ≥ ε/4} ∪ {x d ≤ 4δ}. We have by symbol calculus op sc (χ 2 ρ -1 ) op sc (χ 1 ρ) = op sc (χ 1 ) + h op sc (r 1 ), where r 1 ∈ S -1
tan . Thus we obtain

op sc ( ξ ′ ) op sc (χ 1 )z L 2 (x d >0) op sc (χ 1 ρ)z L 2 (x d >0) + h z L 2 (x d >0) .
From this estimate we obtain

op sc ( ξ ′ )z L 2 (x d >0) op sc ( ξ ′ ) op sc (1 -χ 1 )z L 2 (x d >0) + op sc (χ 1 ρ)z L 2 (x d >0) + h z L 2 (x d >0) . (60)
From definition of z, we can write z = op sc (χ) hD

x d + op sc (r ′ 1 ) v h , where r ′ 1 ∈ S 1 tan .
From symbol calculus and support properties of χ and χ 1 the operator op sc ( ξ ′ ) op sc (1χ 1 ) op sc (χ) is bounded on L 2 by Ch. From (60) we thus have

op sc ( ξ ′ )z L 2 (x d >0) h hDv h L 2 (x d >0) + h v h L 2 (x d >0) + op sc (χ 1 ρ)z L 2 (x d >0) + h z L 2 (x d >0) .
We obtain the statement as we can absorb h z L 2 (x d >0) by the left hand side.

From (59) and Lemma 3.10 we deduce

op sc (χ 1 ρ)z 2 L 2 (x d >0) + h Re(z |x d =0 | op sc (χ 1 ρ)z |x d =0 ) 0 h 2 hDv h 2 L 2 (x d >0) + h 2 v h 2 L 2 (x d >0) + h 2 q h 2 2 L 2 (x d >0) , as z L 2 (x d >0) ≤ op sc ( ξ ′ )z L 2 (x d >0) .
From Lemma 3.10 we obtain

op sc ( ξ ′ )z L 2 (x d >0) h hDv h 2 L 2 (x d >0) + h v h L 2 (x d >0) + h q h 2 L 2 (x d >0) . (61)
From equation satisfied by z we have

hD x d z L 2 (x d >0) op sc ( ξ ′ )z L 2 (x d >0) + h q h 2 L 2 (x d >0) .
From this estimate, (61) and trace formula (8), we deduce

|z |x d =0 | H 1/2 sc ≤ C ε h 1/2 . ( 62
)
From definition of z, we have, for x d > 0 and by symbol calculus,

z = op sc (χ)hD x d v h -i op sc (χρ)v h + hz 1 ,
where z 1 = op sc (r 0 )v h and r 0 ∈ S 0 tan . In particular

z 1 L 2 (x d >0) + hDz 1 L 2 (x d >0) ≤ C ε . Let u 0 = h(D x d v h ) |x d =0 , u 1 = (v h ) |x d =0 , χ 0 = χ |x d =0 and ρ 0 = ρ |x d =0 . From (62), we have op sc (χ 0 )u 0 -i op sc (χ 0 ρ 0 )u 1 = h 1/2 z 4 , where |z 4 | L 2 ≤ C ε . Let Φ ∈ C ∞ (∂Ω), we have Φ op sc (χ 0 ) = op sc (χ 0 )Φ + h op sc (r 0 ) and Φ op sc (χ 0 ρ 0 ) = op sc (χ 0 ρ 0 )Φ + h op sc (r 0 ),
where r 0 , r0 ∈ S 0 tan by symbol calculus. From Proposition 2.6, we have

op sc (χ 0 )Φu 0 -i op sc (χ 0 ρ 0 )Φu 1 = h 1/2 z 5 , where |z 5 | L 2 ≤ C ε . ( 63
)
With this equation we can obtain trace estimates into ∂Ω D and ∂Ω N .

Proposition 3.11. Let Φ ∈ C ∞ (∂Ω). If Φ is supported on ∂Ω D . Then | op sc (χ 0 )Φu 0 | L 2 ≤ C ε h 1/2 .
If Φ is supported on ∂Ω N . Then

| op sc (χ 0 )Φu 1 | H 1 sc ≤ C ε h 1/2 . Proof. If Φ is supported on ∂Ω D , then Φu 1 = 0, (63) gives the first result. If Φ is supported on ∂Ω N , then Φu 0 = 0. Let χ 1 ∈ C ∞ (R d-1 × R d-1 ) be such that χ 1 = 1 on the support of χ 0 , supp χ 1 ⊂ {R(x ′ , 0, ξ ′ ) -1 ≥ ε/2}, χ 1 ∈ S 0
tan , in particular we have χ 0 χ 1 = χ 0 and ρ 0 = 0 on support of χ 1 . We have op sc ξ ′ op sc (χ 1 ρ -1 0 ) op sc (χ 0 ρ 0 ) = op sc ξ ′ op sc (χ 0 ) + h op sc (r 0 ), where r 0 ∈ S 0 tan , by symbol calculus. From Proposition 2.6 and (63), we have

| op sc (χ 0 )Φu 1 | H 1 sc ≤ | op sc ξ ′ op sc (χ 1 ρ -1 0 ) op sc (χ 0 ρ 0 )Φu 1 | L 2 + h| op sc (r 0 )Φu 1 | L 2 ≤ h 1/2 | op sc ξ ′ op sc (χ 1 ρ -1 0 )z 5 | L 2 + h| op sc (r 0 )Φu 1 | L 2 ≤ C ε h 1/2 .
This gives the second estimates. Proposition 3.12. Let Φ ∈ C ∞ (∂Ω). We assume that Φ is supported either on ∂Ω D or on ∂Ω N .

h 1/2 |(v h ) |x d =0 | H 1 sc → 0 as h → 0, h 1/2 |(hD x d v h ) |x d =0 | L 2 → 0 as h → 0.
Proof. Let ε > 0, we can find χ H , χ G and χ E satisfying respectively the assumption of Propositions 3.2, 3.6, 3.11 and furthermore the relation χ H + χ G + χ E = 1. Applying the results of Propositions in each region, we deduce the proposition.

In what follows we shall prove estimates in a neighborhood of ∂Ω N ∩ ∂Ω D . The result is more delicate and less classical than these obtained for Dirichlet or Neumann boundary condition.

Now we assume that Φ is supported in a neighborhood of a point of ∂Ω N ∩ ∂Ω D . We can assume that locally this set is given by x 1 = 0 and the support of Φ is contained into a fixed domain in x ′ and into {|x 1 | ≤ µε/2} where µ > 0 will be fixed below sufficiently small. Here and in what follows ε is the one used to define hyperbolic, glancing, elliptic regions (see respectively [START_REF] Melrose | Singularities of boundary value problems I[END_REF], [START_REF] Savaré | Regularity and perturbation results for mixed second order elliptic problems[END_REF] and ( 53)). We assume that supp u 0 ⊂ {x 1 ≤ 0} and supp u 1 ⊂ {x 1 ≥ 0}. We can choose the local coordinates such that

R(x ′ , 0, ξ ′ ) = ξ 2 1 + R 0 (x ′′ , ξ ′′ ) + x 1 r 2 (x ′ , ξ ′ ), where x ′ = (x 1 , x ′′ ) and ξ ′ = (ξ 1 , ξ ′′ ), (64)
R 0 ∈ S( ξ ′′ 2 , (dx ′′ ) 2 + ξ ′′ -2 (dξ ′′ ) 2 ) and r 2 ∈ S 2 tan . Indeed in normal geodesic coordinates we have R(x ′ , 0, ξ ′ ) = ξ 2 1 + R 1 (x ′ , ξ ′′ ) = ξ 2 1 + R 0 (x ′′ , ξ ′′ ) + x 1 r 2 (x ′ , ξ ′
) and in fact r 2 (x ′ , ξ ′ ) does not depend on ξ 1 but we do not use this property in what follows.

Let

α(x ′′ , ξ ′ ) = ξ 2 1 + R 0 (x ′′ , ξ ′′ ) -1 + iε 1/2
, and let

β(x ′′ , ξ ′′ ) = R 0 (x ′′ , ξ ′′ ) -1 + iε 1/2 , ( 65
) be such that Im β(x ′′ , ξ ′′ ) > 0, for all (x ′′ , ξ ′′ ) ∈ R d-2 × R d-2 . We have α(x ′′ , ξ ′ ) = ξ 1 + iβ(x ′′ , ξ ′′ ) 1/2 ξ 1 -iβ(x ′′ , ξ ′′ ) 1/2 ,
Observe that Re β(x ′′ , ξ ′′ ) > 0, we deduce that

ξ 1 → ξ 1 + iβ(x ′′ , ξ ′′ ) ±1/2 are holomorphic functions in {Im ξ 1 > 0}, ξ 1 → ξ 1 -iβ(x ′′ , ξ ′′ ) ±1/2 are holomorphic functions on {Im ξ 1 < 0}. Let v 1 = op sc ξ 1 -iβ(x ′′ , ξ ′′ ) 1/2 Φu 1 .
The operator op sc ξ 1 -iβ(x ′′ , ξ ′′ ) 1/2 is a convolution operator with respect x 1 and its kernel is supported in x 1 ≥ 0. As u 1 is supported in x 1 ≥ 0, this implies that v 1 is supported in x 1 ≥ 0. Let v 0 = op sc (ξ 1 + iβ(x ′′ , ξ ′′ )) -1/2 Φu 0 . As u 0 is supported in x 1 ≤ 0 and (ξ 1 + iβ(x ′′ , ξ ′′ )) -1/2 is a holomorphic function in {Im ξ 1 > 0}, v 0 is supported in x 1 ≤ 0. We first prove the following lemma.

Lemma 3.13. There exists C > 0, such that for all ε ∈ (0, 1) we have, for every h ∈ (0, 1)

|v 1 | L 2 ≤ Cε 1/5 h -1/2 + C ε h -3/8 , |v 0 | L 2 ≤ Cε 1/5 h -1/2 + C ε h -3/8 ,
where C ε > 0 depends on ε.

Proof. Observe that (ξ 1 -iβ(x ′′ , ξ ′′ )) ±1/2 ∈ S( ξ ′ ±1/2 , (dx ′ ) 2 + (dξ ′ ) 2 ) then op sc (ξ 1 -iβ(x ′′ , ξ ′′ )) -1/2 op sc (ξ 1 -iβ(x ′′ , ξ ′′ )) 1/2 = Id + h op sc (r 0 ), where r 0 ∈ S(1, (dx ′ ) 2 + (dξ ′ ) 2 ).
From the definition of v 1 we have

op sc (ξ 1 -iβ(x ′′ , ξ ′′ )) -1/2 v 1 = Φu 1 + h op sc (r 0 )u 1 . (66)
As op sc (χ 0 ρ 0 ) op sc (r 0 ) has a symbol in S( ξ ′ , (dx ′ ) 2 + (dξ ′ ) 2 ), we obtain

| op sc (χ 0 ρ 0 ) op sc (r 0 )v 1 | H -1/2 sc ≤ C ε |u 1 | H 1/2 sc ≤ C ε h -1/2
by Proposition 2.6. Then from (63) we obtain

op sc (χ 0 )Φu 0 -i op sc (χ 0 ρ 0 ) op sc ξ 1 -iβ(x ′′ , ξ ′′ ) -1/2 v 1 = h 1/2 z 6 , where |z 6 | H -1/2 sc ≤ C ε . (67)
To determine u 0 and u 1 we have to use the support properties of these functions. To do that, we have to modify the operators acting on these functions, note that op sc (χ 0 ρ 0 ) does not preserve the support of u 0 . We introduce three cutoff functions χ H , χ G and χ E , be such that

χ H + χ G + χ E = 1 and supp χ H ⊂ {R(x ′ , 0, ξ ′ ) -1 ≤ -ε/2}, 0 ≤ χ H ≤ 1 supp χ G ⊂ {|R(x ′ , 0, , ξ ′ ) -1| ≤ ε}, 0 ≤ χ G ≤ 1 supp χ E ⊂ {R(x ′ , 0, ξ ′ ) -1 ≥ ε/2}, 0 ≤ χ E ≤ 1.
We then have χ H , χ G and χ E , the χ |x d =0 , defined respectively in the hyperbolic, glancing and elliptic regions (after multiplying ε by a fix factor). We have Φu 0 = op sc (χ H )Φu 0 + op sc (χ G )Φu 0 + op sc (χ E )Φu 0 . From hyperbolic estimate given in Proposition 3.2, Proposition 2.6 and symbol calculus, we have

| op sc (χ H )(Φu 0 )| L 2 ≤ C ε .
From glancing estimate given in Proposition 3.6, Proposition 2.6 and symbol calculus, we have

| op sc (χ G )(Φu 0 )| L 2 ≤ Cε 3/4 h -1/2 + C ε h -3/8 .
We deduce that

|Φu 0 -op sc (χ E )(Φu 0 )| L 2 ≤ Cε 3/4 h -1/2 + C ε h -3/8 . ( 68
)
Lemma 3.14. With the previously defined notations, we have

| op sc (χ H α) op sc ξ 1 -iβ(x ′′ , ξ ′′ ) -1/2 v 1 | H -1/2 sc ≤ C ε (69) | op sc (χ G α) op sc ξ 1 -iβ(x ′′ , ξ ′′ ) -1/2 v 1 | H -1/2 sc ≤ Cε 3/4 h -1/2 + C ε h -3/8 , (70)
where

α(x ′′ , ξ ′ ) = ξ 2 1 + R 0 (x ′′ , ξ ′′ ) -1 + iε 1/2 .
Proof. Let χH be such that χH = 1 on the support of χ H and χH = 0 if R 0 (x ′ , ξ ′ ) -1 ≥ -ε/4. Let χG be such that χG = 1 on the support of χ G and χH = 0 if |R 0 (x ′ , ξ ′ ) -1| ≥ 2ε. Let J = G or H. By symbol calculus, we have op sc (αχ J ) = op sc (αχ J ) op sc ( χJ ) + h op sc (r 0 ), where r 0 ∈ S 0 tan . From (66), we deduce

| op sc (χ J α) op sc ξ 1 -iβ(x ′′ , ξ ′′ ) -1/2 v 1 | H -1/2 sc ≤ | op sc (χ J α)Φu 1 | H -1/2 sc + h| op sc (χ J α) op sc (r 0 )u 1 | H -1/2 sc ≤ | op sc (χ J α) op sc ( χJ )u 1 | H -1/2 sc + C ε h|u 1 | H 1/2 sc . If J = H we have | op sc (χ H α) op sc ξ 1 -iβ(x ′′ , ξ ′′ ) -1/2 v 1 | H -1/2 sc ≤ C ε | op sc ( χH )u 1 | H 1/2 sc + C ε h|u 1 | H 1/2 sc .
we obtain (69) from Propositions 2.6 and 3.2.

If J = G, using that |x 1 | ≤ µε/2, on the support of χ G , we have |α| ≤ Cε 1/2 . As | op sc (χ J α) op sc ( χJ )u 1 | H -1/2 sc ≤ C| op sc (χ J α) op sc ( χJ )u 1 | L 2 ,
we apply Gårding inequality [START_REF] Burq | Mesures de défaut de compacité, application au système de Lamé[END_REF] and we obtain

| op sc (χ G α) op sc ξ 1 -iβ(x ′′ , ξ ′′ ) -1/2 v 1 | H -1/2 sc ≤ Cε 1/2 | op sc ( χG )u 1 | H 1/2 sc + C ε h|u 1 | H 1/2 sc ,
where, at the right hand side, we have estimated the L 2 -norm by the H 1/2 -norm. We obtain (70) from Propositions 2.6 and 3.6.

Following (67) (with χ 0 = χ E ), (68) and Lemma 3.14, we have

Φu 0 -i op sc α(χ H + χ G ) op sc (ξ 1 -iβ(x ′′ , ξ ′′ )) -1/2 v 1 -i op sc (χ E ρ 0 ) op sc (ξ 1 -iβ(x ′′ , ξ ′′ )) -1/2 v 1 = z 7 ,
where

|z 7 | H -1/2 sc ≤ Cε 3/4 h -1/2 + C ε h -3/8
. Applying op sc (ξ 1 + iβ(x ′′ , ξ ′′ )) -1/2 to this equation, we obtain

v 0 -i op sc (ξ 1 + iβ(x ′′ , ξ ′′ )) -1/2 op sc α(χ H + χ G ) op sc (ξ 1 -iβ(x ′′ , ξ ′′ )) -1/2 v 1 -i op sc (ξ 1 + iβ(x ′′ , ξ ′′ )) -1/2 op sc (χ E ρ 0 ) op sc (ξ 1 -iβ(x ′′ , ξ ′′ )) -1/2 v 1 = op sc (ξ 1 + iβ(x ′′ , ξ ′′ )) -1/2 z 7 . (71)
We have to precisely estimate z 8 = op sc (ξ 1 + iβ(x ′′ , ξ ′′ )) -1/2 z 7 . By the definition of β(x ′′ , ξ ′′ ) there exists

C 0 > 0 such that Re β(x ′′ , ξ ′′ ) ≥ C 0 ε, as R 0 (x ′′ , ξ ′′ ) -1 ≥ -1 moreover if |ξ ′′ | ≥ 2, then Re β(x ′′ , ξ ′′ ) ≥ C 0 ξ ′′ . We deduce that |ξ 1 + iβ(x ′′ , ξ ′′ )| ≥ C 0 ε ξ ′ . This implies that |(ξ 1 + iβ(x ′′ , ξ ′′ )) -1/2 | ≤ C 0 ε -1/2 ξ ′ -1/2
, where C 0 > 0. We have by symbol calculus

|z 8 | L 2 ≤ | op sc (ξ 1 + iβ(x ′′ , ξ ′′ )) -1/2 ξ ′ 1/2 op sc ( ξ ′ -1/2 )z 7 | L 2 + C ε h| op sc ( ξ ′ -1/2 )z 7 | L 2 ,
and by Gårding inequality [START_REF] Burq | Mesures de défaut de compacité, application au système de Lamé[END_REF] and the estimate on z 7 , we have

|z 8 | L 2 ≤ Cε -1/2 | op sc ( ξ ′ -1/2 )z 7 | L 2 + C ε h| op sc ( ξ ′ -1/2 )z 7 | L 2 , ≤ Cε 1/4 h -1/2 + C ε h -3/8 .

By symbol calculus, as (ξ

1 + iβ(x ′′ , ξ ′′ )) -1/2 α(ξ 1 -iβ(x ′′ , ξ ′′ )) -1/2 = 1, we have op sc (ξ 1 + iβ(x ′′ , ξ ′′ )) -1/2 op sc α(χ H + χ G ) op sc (ξ 1 -iβ(x ′′ , ξ ′′ )) -1/2 = op sc (χ H + χ G ) + h op sc (r 0 ), ( 72
)
where r 0 ∈ S(1, (dx ′ ) 2 + (dξ ′ ) 2 ).

By the same argument, we have

op sc (ξ 1 + iβ(x ′′ , ξ ′′ )) -1/2 op sc (χ E ρ 0 ) op sc (ξ 1 -iβ(x ′′ , ξ ′′ )) -1/2 = op sc (χ E ρ 0 α -1 ) + h op sc (r 0 ), ( 73
)
where r 0 ∈ S(1,

(dx ′ ) 2 + (dξ ′ ) 2 ). Indeed (ξ 1 ± iβ(x ′′ , ξ ′′ )) -1/2 ∈ S( ξ ′ -1/2 , (dx ′ ) 2 + (dξ ′ ) 2 ) and χ E ρ 0 ∈ S( ξ ′ , (dx ′ ) 2 + (dξ ′ ) 2 ).
The following lemma gives a precise estimate on χ E (ρ 0 α -1 -1). We shall exploit that α and ρ 0 are close. Lemma 3.15. We have

op sc (χ E ρ 0 α -1 )v 1 -op sc (χ E )v 1 L 2 ≤ Cε 1/5 h -1/2 + C ε h -3/8 . Proof. Let Φ(x 1 ) supported in {|x 1 | ≤ µε} and Φ = 1 on the support of Φ. Let b be either the symbol χ E ρα -1 or χ E , we have b ∈ S(1, (dx ′ ) 2 + (dξ ′ ) 2 ).
By symbol calculus we have op

sc (b(1 -Φ))v 1 = h op(r)u 1 , where r ∈ S( ξ ′ 1/2 , (dx ′ ) 2 + (dξ ′ ) 2 ). By Proposition 2.6 we have | op sc (b(1 -Φ))v 1 | L 2 ≤ C ε h|u 1 | H 1/2 sc ≤ C ε h 1/2 .
Then we can considerate op sc ( Φχ E ρα -1 ) and op sc ( Φχ E ), instead of respectively op sc (χ E ρα -1 ) and op sc (χ E ).

We introduce three cutoff functions

χ j ∈ C ∞ (R d-2 × R d-1
) be such that

χ 1 = 1 if ξ 2 1 + R 0 (x ′′ , ξ ′′ ) -1 ≤ ε 4/5 , 0 if ξ 2 1 + R 0 (x ′′ , ξ ′′ ) -1 ≥ 3ε 4/5 , χ 2 = 1 if 1 ≥ ξ 2 1 + R 0 (x ′′ , ξ ′′ ) -1 ≥ 2ε 4/5 , 0 if ξ 2 1 + R 0 (x ′′ , ξ ′′ ) -1 ≤ ε 4/5 and |ξ ′ | 2 -1 ≥ 3, χ 3 = 1 if ξ 2 1 + R 0 (x ′′ , ξ ′′ ) -1 ≥ 2, 0 if ξ 2 1 + R 0 (x ′′ , ξ ′′ ) -1 ≤ 1, χ 1 + χ 2 + χ 3 = 1 and 0 ≤ χ j ≤ 1 for j = 1, 2, 3.
Estimation on the support of χ E χ 1 .

On the support of χ E χ 1 we have ξ 2 1 + R 0 (x ′′ , ξ ′′ ) -1 ≤ 3ε 4/5 and R(x ′ , 0, ξ ′ ) -1 ≥ ε/2, in particular |ξ ′ | is bounded. We compute on this domain

ρ 0 (x ′ , ξ ′ ) -α(x ′′ , ξ ′ ) = D(x ′ , ξ ′ ) x 1 r 2 (x ′ , ξ ′ ) -iε , (74)
where

D(x ′ , ξ ′ ) = (R(x ′ , 0, ξ ′ ) -1) 1/2 + (ξ 2 1 + R 0 (x ′′ , ξ ′′ ) -1 + iε) 1/2 -1
. Observe that we have

Re(ξ 2 1 + R 0 (x ′′ , ξ ′′ ) -1 + iε) 1/2 > 0 and (R(x ′ , 0, ξ ′′ ) -1) 1/2 ≥ 2 -1/2 ε 1/2 , then |D(x ′ , ξ ′ )| ≤ C 0 ε -1/2 . We deduce that |ρ 0 -α| ≤ C 0 ε 1/2 if |x 1 | ≤ µε
where µ was introduced in the definition of Φ. As

|ξ 2 1 + R 0 (x ′′ , ξ ′′ ) -1 -(R(x ′ , 0, ξ ′ ) -1)| ≤ C|x 1 | ≤ Cµε,
if µ is chosen sufficiently small, on the support of χ E , we have

ξ 2 1 + R 0 (x ′′ , ξ ′′ ) -1 ≥ C 0 ε, for C 0 > 0. We deduce that |α(x ′′ , ξ ′ )| ≥ C 1 ε 1/2 , for C 1 > 0.
This implies that

|(ρ 0 -α)α -1 | ≤ C 2 on the support of χ E χ 1 , for C 2 > 0 and |x 1 | ≤ µε. ( 75
) Let χ1 (x ′ ξ ′ ) = 1 on the support of χ 1 , 0 if ξ 2 1 + R 0 (x ′′ , ξ ′′ ) -1 ≥ 4ε 4/5 or R(x ′ , 0, ξ ′ ) -1 ≤ ε/4.
By symbol calculus in classes S( ξ ′ s , (dx ′ ) 2 + (dξ ′ ) 2 ), we have

op sc ( Φχ E χ 1 ρ 0 α -1 ) -op sc ( Φχ E χ 1 ) v 1 = op sc ( Φχ E χ 1 (ρ 0 -α)α -1 ) op sc ( χ1 )v 1 + h op sc (r 0 )v 1 ,
where r 0 ∈ S(1, (dx ′ ) 2 + (dξ ′ ) 2 ). By Gårding inequality ( 12) and (75) we have

op sc ( Φχ E χ 1 ρ 0 α -1 ) -op sc ( Φχ E χ 1 ) v 1 L 2 ≤ C| op sc ( χ1 )v 1 | L 2 + C ε h|v 1 | L 2 . ( 76
)
Observe that

op sc ( χ1 )v 1 = op sc ξ 1 -iβ(x ′′ , ξ ′′ ) 1/2 Φ op sc ( χ1 )u 1 + h op sc (r 1/2 )u 1 ,
by semiclassical symbol calculus in S( ξ ′ s , (dx ′ ) 2 + (dξ ′ ) 2 ), where r 1/2 ∈ S( ξ ′ 1/2 , (dx ′ ) 2 + (dξ ′ ) 2 ). As χ1 is supported in |R(x ′ , 0, ξ ′ ) -1| ≤ Cε 4/5 , we can apply Proposition 3.6 with ε 4/5 instead of ε. From (76), Proposition 2.6 and as

|v 1 | L 2 ≤ C ε |u 1 | H 1/2 sc , we have op sc (χ E χ 1 ρ 0 α -1 ) -op sc (χ E χ 1 ) v 1 H 1/2 sc ≤ C| op sc ( χ1 )u 1 | H 1 sc + C ε h|u 1 | H 1/2 sc . ≤ Cε 1/5 h -1/2 + C ε h -3/8 . ( 77
)
Estimation on the support of χ E χ 2 .

Equation ( 74) is valid in the supports of χ E χ 2 and Φ. As ξ 2 1 +R 0 (x ′′ , ξ ′′ )-1 ≥ ε 4/5 , on this domain, we have Re(ξ

2 1 +R 0 (x ′′ , ξ ′′ )-1+iε) 1/2 ≥ C 0 ε 2/5 , for C 0 > 0. We deduce that |D(x ′ , ξ ′ )| ≤ Cε -2/5 and |α -1 (x ′′ , ξ ′ )| ≤ Cε -2/5 , then |(ρ 0 -α)α -1 | ≤ Cε 1/5
. We conclude by semiclassical symbol calculus in S( ξ ′ s , (dx ′ ) 2 + (dξ ′ ) 2 ), Gårding inequality (12) and Proposition 2.6 that

op sc ( Φχ E χ 2 ρ 0 α -1 ) -op sc ( Φχ E χ 2 )v 1 L 2 = op sc ( Φχ E χ 2 (ρ 0 -α)α -1 ) op sc ξ 1 -iβ(x ′′ , ξ ′′ ) 1/2 Φu 1 L 2 ≤ Cε 1/5 h -1/2 + C ε . ( 78
)
Estimation on the support of χ E χ 3 .

On the supports of χ E χ 3 and Φ, we have Re(ξ

2 1 + R 0 (x ′′ , ξ ′′ ) -1 + iε) ≥ C 0 ξ ′ 2 for C 0 > 0. We deduce that Re(ξ 2 1 + R 0 (x ′′ , ξ ′′ ) -1 + iε) 1/2 ≥ C 0 ξ ′ for C 0 > 0 and |(ρ 0 -α)α -1 | ≤ Cε. By semiclassicalsymbol calculus in S( ξ ′ s , (dx ′ ) 2 + (dξ ′ ) 2
), Gårding inequality and Proposition 2.6 we have

op sc ( Φχ E χ 3 ρ 0 α -1 ) -op sc ( Φχ E χ 3 )v 1 L 2 = op sc ( Φχ E χ 3 (ρ 0 -α)α -1 ) op sc ξ 1 -iβ(x ′′ , ξ ′′ ) 1/2 Φu 1 L 2 ≤ Cεh -1/2 + C ε . (79)
As χ 1 + χ 2 + χ 3 = 1, Formulas (77), ( 78) and (79) give the conclusion of Lemma 3.15.

From (71)-( 73) and Lemma 3.15 we have

v 0 -iv 1 = z 9 , where |z 9 | L 2 ≤ Cε 1/5 h -1/2 + C ε h -3/8 . ( 80
)
As v 0 is supported in x 1 ≤ 0 and v 1 is supported in x 1 ≥ 0 , if we restrict (80) on x 1 > 0, we obtain

|v 1 | L 2 (x1>0) = |v 1 | L 2 ≤ Cε 1/5 h -1/2 + C ε h -3/8 ,
and we deduce

|v 0 | L 2 ≤ Cε 1/5 h -1/2 + C ε h -3/8
. This proves Lemma 3.13.

Proposition 3.16. With the previously defined notation, we have

h 1/2 |(v h ) |x d =0 | H 1/2 sc = h 1/2 |u 1 | H 1/2 sc → 0 as h → 0, h 1/2 |(hD x d v h ) |x d =0 | H -1/2 sc = h 1/2 |u 0 | H -1/2 sc → 0 as h → 0.
Proof. We have to introduce another small parameter ν > 0 chosen below such that

ν >> ε. Let χ H , χ G and χ E in C ∞ (R d-1 × R d-1 ) such that χ H (x ′ , ξ ′ ) is supported in R(x ′ , 0, ξ ′ ) -1 ≤ -ν, χ G (x ′ , ξ ′ ) is supported in |R(x ′ , 0, ξ ′ ) -1| ≤ 2ν, χ E (x ′ , ξ ′ ) is supported in R(x ′ , 0, ξ ′ ) -1 ≥ ν, χ H + χ G + χ E = 1. Let ψ D , ψ Z and ψ N in C ∞ (∂Ω) such that ψ D is supported in x 1 ≤ -εµ/4 ψ Z is supported in |x 1 | ≤ εµ/2 ψ N is supported in x 1 ≥ εµ/4 ψ D + ψ Z + ψ N = 1.
To be clear, the ε is the one used in elliptic region. We recall the estimates obtained in previous sections. From Proposition 3.2 we have

| op sc (χ H )u 1 | H 1 sc ≤ C ν | op sc (χ H )u 0 | L 2 ≤ C ν . (81)
From Proposition 3.6 we have

| op sc (χ G )u 1 | H 1 sc ≤ Cν 1/4 h -1/2 + C ν h -3/8 |(op sc (χ G )u 0 ) |x d =0 | L 2 sc ≤ Cν 3/4 h -1/2 + C ν h -3/8 . (82)
In the elliptic region, we estimate v j and we have to estimate u j for j = 0, 1. To be precise,

v 0 = op sc (ξ 1 + iβ(x ′′ , ξ ′′ )) -1/2 ψ Z u 0 and v 1 = op sc ξ 1 -iβ(x ′′ , ξ ′′ ) 1/2 ψ Z u 1
, where β is defined by formula (65) and Φ = ψ Z . By Lemma 3.13, we have

|v 1 | L 2 ≤ Cε 1/5 h -1/2 + C ε h -3/8 , |v 0 | L 2 ≤ Cε 1/5 h -1/2 + C ε h -3/8 . Thus op sc (χ E (ξ 1 -iβ(x ′′ , ξ ′′ )) -1/2 )v 1 = op sc (χ E )ψ Z u 1 +h op sc (r 0 )u 1 , where r 0 ∈ S(1, (dx ′ ) 2 +(dξ ′ ) 2 ).
Lemma 3.17. On the support of χ E , we have

|(ξ 1 -iβ(x ′′ , ξ ′′ )) -1/2 | ≤ Cν -1/4 ξ ′ -1/2 = C ν ξ ′ -1/2
, where C ν does not depend on ε.

Proof. We have to consider different cases.

If |ξ ′′ | ≥ C, where C sufficiently large to have R 0 (x ′′ , ξ ′′ ) ≥ 2, we have Re

β(x ′′ , ξ ′′ ) ≥ C 1 ξ ′′ , for C 1 > 0. Then |ξ 1 -iβ(x ′′ , ξ ′′ )| ≥ C 2 ξ ′ , for C 2 > 0. If |ξ ′′ | ≤ C and |ξ 1 | sufficiently large, we have |ξ 1 -iβ(x ′′ , ξ ′′ )| ≥ C 3 ξ 1 ≥ C 4 ξ ′ . If |ξ ′ | bounded, on the support of χ E , if ε is sufficiently small with respect ν, we have ξ 2 1 + R 0 (x ′′ , ξ ′′ ) -1 ≥ ν/2. If R 0 (x ′′ , ξ ′′ ) -1 ≥ δ 1 ν, for δ 1 > 0, then Re β(x ′′ , ξ ′′ ) ≥ δ 2 ν 1/2 , for δ 2 > 0. If δ 1 is sufficiently small and R 0 (x ′′ , ξ ′′ ) -1 ≤ δ 1 ν, then ξ 2 1 ≥ ν/4 and |ξ 1 -iβ(x ′′ , ξ ′′ )| ≥ δ 3 ν 1/2
. In all cases, we get that |ξ 1iβ(x ′′ , ξ ′′ )| ≥ δ 3 ν 1/2 ξ ′ . This implies the result.

By symbol calculus, we have

| op sc (χ E )ψ Z u 1 | H 1/2 sc ≤ | op sc ( ξ ′ 1/2 ) op sc (χ E (ξ 1 -iβ(x ′′ , ξ ′′ )) -1/2 )v 1 | L 2 + C ν,ε h|u 1 | H 1/2 sc ≤ | op sc ( ξ ′ 1/2 χ E (ξ 1 -iβ(x ′′ , ξ ′′ )) -1/2 )v 1 | L 2 + C ν,ε h|u 1 | H 1/2 sc ,
and by Gårding inequality [START_REF] Burq | Mesures de défaut de compacité, application au système de Lamé[END_REF] and Lemma 3.13 we have

| op sc (χ E )ψ Z u 1 | H 1/2 sc ≤ C ν |v 1 | L 2 + C ν,ε h 1/2 + C ν,ε h|u 1 | H 1/2 sc ≤ C ν ε 1/5 h -1/2 + C ν,ε h -3/8 + C ν,ε h|u 1 | H 1/2 sc . ( 83
)
For v 0 we have

op sc (χ E (ξ 1 + iβ(x ′′ , ξ ′′ )) 1/2 )v 0 = op sc (χ E )ψ Z u 0 + h op sc (r 0 )u 0 , where r 0 ∈ S(1, (dx ′ ) 2 + (dξ ′ ) 2 ).
A proof analogous to the one of Lemma 3.17 gives

|(ξ + iβ(x ′′ , ξ ′′ )) 1/2 | ≤ C ν ξ ′ 1/2
, we have by Gårding inequality and symbol calculus

| op sc (χ E )ψ Z u 0 | H -1/2 ≤ C ν |v 0 | L 2 + C ν,ε h 1/2 + C ν,ε h|u 0 | H -1/2 sc ≤ C ν ε 1/5 h -1/2 + C ν,ε h -3/8 + C ν,ε h|u 0 | H -1/2 sc . (84)
From Proposition 3.11, we have

| op sc (χ E )ψ D u 0 | L 2 ≤ C ν,ε h 1/2 and ψ D u 1 = 0 | op sc (χ E )ψ N u 1 | H 1 sc ≤ C ν,ε h 1/2 and ψ N u 0 = 0. ( 85
)
As u j = op sc (χ H )u j + op sc (χ G )u j + op sc (χ E )ψ N u j + op sc (χ E )ψ Z u j + op sc (χ E )ψ D u j , we have, by ( 81)-(85),

h 1/2 |u j | H -1/2+j sc ≤ Cν (3-2j)/4 + C ν ε 1/5 + C ν,ε h 1/8 + C ν,ε h 3/2 |u j | H -1/2+j sc .
Choosing first ν sufficiently small, second ε sufficiently small, we can absorb the right hand side term

C ν,ε h 3/2 |u j | H -1/2+j
sc by the left hand side term taking h sufficiently small. The limit superior with respect to h of the left hand side can be estimated by any positive number. This proves Proposition 3.16.

Support of semiclassical measure in a neighborhood of boundary

We can now prove Proposition 3.1, that is, the measure µ is supported on p = 0, in a neighborhood of x 0 ∈ ∂Ω.

Proof. The proof is based on the results obtained by Proposition 2.2 and Proposition 3.16. We recall that in local coordinates p(x, ξ) = ξ 2 d + R(x, ξ ′ ) -1. We have with the notation ( 13)

[h 2 D x d + R(x, hD x ′ ) -1]v h = hq h -ih(hD x d v h ) |x d =0 ⊗ δ x d =0 -ih(v h ) |x d =0 ⊗ hD x d δ x d =0 . (86) Let ϕ ∈ C ∞ 0 (R ξ d ) and χ ∈ C ∞ 0 (R x d ) be such that χ is supported in a neighborhood of 0. Let ℓ ∈ C ∞ 0 (R d-1 x ′ × R d-1 ξ ′ )
. By symbol calculus we have

Op sc χ(x d )ϕ(ξ d )ℓ(x ′ , ξ ′ ) [h 2 D x d + R(x, hD x ′ ) -1] = Op sc χ(x d )ϕ(ξ d )ℓ(x ′ , ξ ′ )(ξ 2 d + R(x, ξ ′ ) -1) + h Op sc (r 0 ),
where r 0 ∈ S 0 . Then

I := Op sc χ(x d )ϕ(ξ d )ℓ(x ′ , ξ ′ ) [h 2 D x d + R(x, hD x ′ ) -1]v h |v h = Op sc χ(x d )ϕ(ξ d )ℓ(x ′ , ξ ′ )(ξ 2 d + R(x, ξ ′ ) -1) v h |v h =:A + h(Op sc (r 0 )v h |v h ) =:B
.

By definition of the semiclassical measure, the term A converges to µ|χ(

x d )ϕ(ξ d )ℓ(x ′ , ξ ′ )(ξ 2 d + R(x, ξ ′ ) -1) as h to 0. The term B is estimated by Ch v h 2 L 2 (Ω)
and this converges to 0 as h to 0 by Proposition 2.2.

By (86) we also have

I = h Op sc χ(x d )ϕ(ξ d )ℓ(x ′ , ξ ′ ) q h |v h -ih Op sc χ(x d )ϕ(ξ d )ℓ(x ′ , ξ ′ ) (hD x d v h ) |x d =0 ⊗ δ x d =0 |v h -ih Op sc χ(x d )ϕ(ξ d )ℓ(x ′ , ξ ′ ) (v h ) |x d =0 ⊗ hD x d δ x d =0 |v h = I 1 + I 2 + I 3 .
Obviously we have

|I 1 | ≤ Ch q h L 2 (Ω v h L 2 (Ω , as χ(x d )ϕ(ξ d )ℓ(x ′ , ξ ′ ) ∈ S 0 .
Then I 1 → 0 as h → 0 by [START_REF] Fu | Stabilization of hyperbolic equations with mixed boundary conditions[END_REF]. By exact calculus, we have

Op sc χ(x d )ϕ(ξ d )ℓ(x ′ , ξ ′ ) = χ(x d ) op sc ℓ(x ′ , ξ ′ ) Op sc (ϕ(ξ d )).
Let w j = ((hD x d ) 1-j v h ) |x d =0 , we have for j = 0, 1

h| Op sc χ(x d )ϕ(ξ d )ℓ(x ′ , ξ ′ ) w j ⊗ (hD x d ) j δ x d =0 |v h | = h| χ(x d ) op sc ℓ(x ′ , ξ ′ ) w j ⊗ Op sc ϕ(ξ d ) (hD x d ) j δ x d =0 |v h | ≤ h|w j | H j-1/2 Op sc ϕ(ξ d ) (hD x d ) j δ x d =0 L 2 (R) v h L 2 (Ω) , (87)
where we have used, to estimate w 0 , that ξ ′ 1/2 ℓ(x ′ , ξ ′ ) is bounded on L 2 , as ℓ is compactly supported. A direct computation in Fourier variable gives that Op sc ϕ(ξ d ) (hD

x d ) j δ x d =0 L 2 (R) h -1/2
. From (87), we obtain

|I 2 | + |I 3 | ≤ C(|(v h ) |x d =0 | H 1/2 + |(D x d v h ) |x d =0 | H -1/2 )h 1/2 → 0 as h → 0, by Proposition 3.16. We conclude that µ|χ(x d )ϕ(ξ d )ℓ(x ′ , ξ ′ )(ξ 2 d + R(x, ξ ′ ) -1 = 0 and by density of functions spanned by χ(x d )ϕ(ξ d )ℓ(x ′ , ξ ′ ) in C ∞ 0 (R d × R d ), we have that p(x, ξ)µ|φ(x, ξ) = 0, for all φ ∈ C ∞ 0 (R d × R d ).
This gives the conclusion of Proposition 3.1.

The semiclassical measure is not identically null

Proposition 3.18. The measure µ constructed at the beginning of Section 3.1 for the sequence (v h ) h satisfying (19) is not identically 0, i.e. µ ≡ 0.

Proof. Let φ ∈ C ∞ 0 (R) be such that φ = 1 in a neighborhood of 0. Let s ∈ (0, 1/2). Let φ R (ξ) = φ(|ξ|/R), we have Op sc (1 -φ R )v h L 2 (R d ) ≤ CR -s Op sc ( ξ s )v h L 2 (R d ) ≤ CR -s v h H s sc (Ω) ≤ CR -s v h H 1 sc (Ω) ≤ CR -s , as w H s sc (R d ) is equivalent to w H s sc (Ω) (uniformly with respect to h ∈ (0, 1)) if w is supported in Ω. Then for R sufficiently large, Op sc (1 -φ(|ξ|/R)v h L 2 (R d ) ≤ 1/2. We thus have (Op sc (φ R )v h |v h ) L 2 (R d ) = v h 2 L 2 (R d ) -(Op sc (1 -φ R )v h |v h ) L 2 (R d ) ≥ 1 -Op sc (1 -φ(|ξ|/R)v h L 2 (R d ) ≥ 1/2. Let χ ∈ C ∞ 0 (R d ) be such that χ(x) = 1 for x ∈ Ω and χ ≥ 0, we have (Op sc (φ R )v h |v h ) L 2 (R d ) = (Op sc (φ R (ξ)χ(x))v h |v h ) L 2 (R d ) → µ, χ(x)φ R (ξ) as h → 0,
we obtain µ, χ(x)φ R (ξ) ≥ 1/2. Then µ is not identically null.

The semiclassical measure is null on the support of a

Before proving the result we need to extend the space of test functions acting on µ. We have the following lemma.

Let

b(x, ξ ′ ) ∈ C ∞ (R d × R d-1
), we can give a sense to the expression µ, b(x, ξ ′ )ξ j d for all j ∈ N.

Lemma 3.19. Let Φ ∈ C ∞ 0 (R), be such that Φ(σ) = 1, for |σ| ≤ 1. Let j ∈ N, then the quantity µ, b(x, ξ ′ )ξ j d Φ(|ξ|/R) does not depend on R for sufficiently large R. By definition, we denote µ, b(x, ξ ′ )ξ j d = lim R→∞ µ, b(x, ξ ′ )ξ j d Φ(|ξ|/R) . Proof. As pµ = 0, µ is supported in |ξ ′ | 2 + ξ 2 d ≤ C 0 , for C 0 > 0, sufficiently large. If R is suf- ficiently large and R ′ > R, Φ(|ξ|/R) -Φ(|ξ|/R ′ ) = 0, if |ξ| ≤ R, in particular if R 2 > C 0 . Then b(x, ξ ′ ) Φ(|ξ|/R) -Φ(|ξ|/R ′ ) = 0, on the support of µ. This proves that µ, b(x, ξ ′ )ξ j d Φ(|ξ|/R) , does not depend on R if R is sufficiently large. Proposition 3.20. We have aµ = 0. Proof. From Proposition 2.2 we have -h 2 P v h + v h -ihav h = hq h . The inner product with v h lead to (-h 2 P v h +v h -ihav h |v h ) = h(q h |v h ).
Taking the imaginary part of this equation, as (P v h |v h ) is real, we have -(av h |v h ) = Im(q h |v h ). As | Im(q h |v h )| ≤ q h v h → 0, as h → 0, we have (av h |v h ) → 0, as h → 0. Let Φ ∈ C ∞ 0 (R), be such that Φ = 1 in a neighborhood of 0, and 0 ≤ Φ ≤ 1. By Lemma 3.19, the limit when h goes to 0 of (Op sc (a(x)Φ(|ξ|/R))v h |v h ) does not depend on R for R large enough. By Gårding inequality [START_REF] Bony | Calcul symbolique et propagation des singularités pour les équations aux dérivées partielles non linéaires[END_REF], as a ≥ 0, (Op sc (a(x)Φ(|ξ|/R))v h |v h ) and (Op sc (a(x)(1 -Φ(|ξ|/R)))v h |v h ) are non negative modulo O(h).

Consequently, we have lim h→0 (Op sc (a(x)Φ(|ξ|/R))v h |v h ) ≥ 0, lim h→0 (av

h |v h ) = 0 moreover (Op sc (a(x)(1 -Φ(|ξ|/R)))v h |v h ) has a limit and lim h→0 (Op sc (a(x)(1 -Φ(|ξ|/R)))v h |v h ) ≥ 0. Since lim h→0 (Op sc (a(x)Φ(|ξ|/R))v h |v h ) + lim h→0 (Op sc (a(x)(1 -Φ(|ξ|/R)))v h |v h ) = 0,
we deduce that lim h→0 (Op sc (a(x)Φ(|ξ|/R))v h |v h ) = 0. This implies that µ, a = 0, and as a and µ are non negative, we deduce aµ = 0.

Measure properties 4.1 Action of Hamiltonian

We first recall the main results proved in previous sections. There exists a sequence (v h ) h satisfying the following properties

h 2 P v h -v h + ihav h = hq h , v h L 2 (Ω) = 1 and h∇v h L 2 (Ω) ≤ 2, q h L 2 (Ω) → 0 as h → 0, h 1/2 |(v h ) |x d =0 | H 1/2 sc → 0 as h → 0, h 1/2 |(hD x d v h ) |x d =0 | H -1/2 sc → 0 as h → 0, pµ = 0. ( 88
)
We also proved that aµ = 0 but in the following we do not systematically use this property. These results was stated in Proposition 2.2, Proposition 3.1 and Proposition 3.16.

Interior formula

We begin by stated that the measure µ is propagated along the H p flow in interior of domain. This property is classical but the proof is simpler in this case than in a neighborhood of boundary, even if the main ideas are used.

Proposition 4.1. Let b ∈ C ∞ 0 (Ω × R d ). We have H p µ -2aµ, b = 0.
Proof. We consider the following quantity

A = ih -1 b(x, hD) h 2 P -1 + iha v h |v h L 2 (Ω) -ih -1 b(x, hD)v h | h 2 P -1 + iha v h L 2 (Ω) = ih -1 b(x, hD)hq h |v h L 2 (Ω) -ih -1 b(x, hD)v h |hq h L 2 (Ω) .
We have |A| v h L 2 (Ω) q h L 2 (Ω) , then A goes to 0 as h. As b is supported far away the boundary of Ω, we have b

(x, hD)v h | h 2 P -1 + iha v h L 2 (Ω) = h 2 P -1 -iha b(x, hD)v h |v h L 2 (Ω) . As the principal symbol of b(x, hD)a -ab(x, hD) is O(h), we have A = ih -1 b(x, hD), h 2 P -1 v h |v h L 2 (Ω) -2 ab(x, hD)v h |v h L 2 (Ω) + O(h) v h 2 L 2 (Ω) ,
and by symbol calculus the principal symbol of b(x, hD), h 2 P -1 is -ih{b, p}. Then

A = (Op sc ({b, p}) -2a(x)b(x, hD))v h |v h L 2 (Ω) + O(h) v h 2 L 2 (Ω) ,
then A → µ, {b, p} -2ab as h → 0, which gives the result.

Limit computations

In the following section quantities as (op sc (a)h j D j

x d v h |v h ) for j = 0, 1, 2 appear. We need to evaluate their limits in term of the measure µ. To do so we shall now state some technical results.

Proposition 4.2. Let b 0 (x, ξ ′ ) ∈ S(1, (dx) 2 + (dξ ′ ) 2 ), and b 1 (x, ξ ′′ ), b 2 (x, ξ ′′ ) ∈ C ∞ 0 (R d × R d-2 ). Let b(x, ξ ′ ) = b 0 (x, ξ ′ )+ b 1 (x, ξ ′′ )ξ 1 + b 2 (x, ξ ′′ )ξ 2 1 . We have op sc (b)v h |v h L 2 (x d >0) → µ, b as h → 0.
For the proof we need the following lemma.

Lemma 4.3. Let Φ ∈ C ∞ 0 (R), be such that Φ(σ) = 1, for σ in a neighborhood of 0. For all s ∈ (0, 1/2), there exists C > 0, such that

1 -Op sc Φ(ξ d /R) 1 x d >0 v L 2 (R d ) ≤ CR -s v H s sc (x d >0) 1 -Op sc Φ(ξ d /R)Φ(|ξ ′ |/R) 1 x d >0 v L 2 (R d ) ≤ CR -s v H s sc (x d >0) ,
for all v ∈ H s (x d > 0), for all h ∈ (0, 1) and all R > 1.

We recall that

H s sc (x d > 0) = {u ∈ D ′ (x d > 0) : ∃w ∈ H s sc (R d ), w |x d >0 = u},
and for u ∈ H s sc (x d > 0), we define u H s sc (x d >0) = inf{ w H s sc (R d ) , w |x d >0 = u}. We recall that for s ∈ [0, 1/2) and u ∈ H s sc (x d > 0), one has 1 x d >0 u ∈ H s sc (R d ) and u H s sc (x d >0) and 1 x d >0 u H s sc (R d )
define two equivalent norms (uniformly with respect to h ∈ (0, 1)).

Proof. Let w = 1 x d >0 v. We have

1 -Op sc Φ(ξ d /R) 1 x d >0 v 2 L 2 (R d ) 1 -Φ(hξ d /R) 2 | ŵ(ξ ′ , ξ d )| 2 dξ 1 -Φ(hξ d /R) /(|hξ d | s ) 2 hξ d 2s | ŵ(ξ ′ , ξ d )| 2 dξ. But 1 -Φ(hξ d /R) /(|hξ d | s ) ≤ R -s 1 -Φ(hξ d /R) /(|hξ d /R| s ) R -s ,
then, as s < 1/2, we obtain that

1 -Op sc Φ(ξ d /R) 1 x d >0 v 2 L 2 (R d ) R -2s hξ 2s | ŵ(ξ ′ , ξ d )| 2 dξ R -2s v 2 H s (x d >0) , (89)
which is the first estimate of statement. By the same method, we prove that (for any s > 0)

1 -Op sc Φ(ξ ′ /R) 1 x d >0 v 2 L 2 (R d ) R -2s hξ ′ 2s | ŵ(ξ ′ , ξ d )| 2 dξ R -2s v 2 H s (x d >0) , ( 90
) As 1 -Φ(ξ d /R)Φ(|ξ ′ |/R) = 1 -Φ(ξ d /R) + Φ(ξ d /R) 1 -Φ(|ξ ′ |/R) , we have 1 -Op sc Φ(ξ d /R)Φ(|ξ ′ |/R) 1 x d >0 v L 2 (R d ) ≤ 1 -Op sc Φ(ξ d /R) 1 x d >0 v L 2 (R d ) + 1 -op sc Φ(|ξ ′ |/R) 1 x d >0 v L 2 (R d ) ,
as Op sc Φ(ξ d /R) is bounded by 1 on L 2 . From ( 89) and (90), we obtain the second estimate of statement.

Proof of Proposition 4.2. Let Φ ∈ C ∞ 0 (R), be such that Φ(σ) = 1, for σ in a neighborhood of 0. We treat the terme b 0 . We have as op sc (b 0 ) is a tangential operator

op sc (b 0 )v h |v h L 2 (x d >0) = op sc (b 0 )1 x d >0 v h |1 x d >0 v h L 2 (R d ) = op sc (b 0 ) Op sc Φ(ξ d /R)Φ(|ξ ′ |/R) 1 x d >0 v h |1 x d >0 v h L 2 (R d ) + op sc (b 0 ) Op sc (1 -Φ(|ξ ′ |/R))Φ(ξ d /R) 1 x d >0 v h |1 x d >0 v h L 2 (R d ) + op sc (b 0 ) Op sc 1 -Φ(ξ d /R)) 1 x d >0 v h |1 x d >0 v h L 2 (R d ) = A 1 + A 2 + A 3 .
By definition of semiclassical measure and from Lemma 3.19 we have

A 1 → µ, b 0 (x, ξ ′ )Φ(ξ d /R)Φ(|ξ ′ |/R) = µ, b 0 as h → 0.
We have

|A 2 | 1 x d >0 Op sc (1 -Φ(|ξ ′ |/R))v h L 2 (R d ) 1 x d >0 v h L 2 (R d ) R -1 v h L 2 (R,H 1 sc (R d-1 )) v h L 2 (x d >0)
, and, for s ∈ (0, 1/2) we have

|A 3 | Op sc (1 -Φ(ξ d /R))1 x d >0 v h L 2 (R d ) 1 x d >0 v h L 2 (R d ) R -s v H s sc (x d >0) v L 2 (x d >0) ,
as op sc (b 0 ) is bounded on L 2 (R d ) and from Lemma 4.3. As A 1 + A 2 + A 3 does not depend on R we obtain the result for b 0 .

In the following we only consider the term b 2 ξ 2 1 , the term b 1 ξ 1 can be managed as the previous with some minor modifications.

op sc (b 2 ξ 2 1 )v h |v h L 2 (x d >0) = op sc (b 2 ξ 2 1 )1 x d >0 v h |1 x d >0 v h L 2 (R d ) = op sc (b 2 ξ 2 1 ) Op sc Φ(ξ d /R)Φ(|ξ 1 |/R) 1 x d >0 v h |1 x d >0 v h L 2 (R d ) + op sc (b 2 ξ 2 1 ) Op sc (1 -Φ(ξ d /R))Φ(|ξ 1 |/R) 1 x d >0 v h |1 x d >0 v h L 2 (R d ) + op sc (b 2 ξ 2 1 ) Op sc (1 -Φ(|ξ 1 |/R) 1 x d >0 v h |1 x d >0 v h L 2 (R d ) = B 1 + B 2 + B 3 As previously B 1 → µ, b 2 ξ 2 1 Φ(ξ d /R)Φ(|ξ 1 |/R) = µ, b 2 ξ 2 1 as h → 0.
We need to prove a regularity result on v h given by the following lemma which is proven in Appendix A.

Lemma 4.4 (Zaremba regularity result). Let s ∈ (0, 1/2). There exists C > 0, such that for any v h satisfying (88), we have v h H 1+s sc (Ω) ≤ C. We have after an integration by parts and symbol calculus, for s ∈ (0, 1/2)

|B 2 | Op sc (1 -Φ(|ξ d |/R) 1 x d >0 hD x1 v h L 2 (x d >0) v h L 2 (x d >0,H 1 sc (R d-1 )) + h v h 2 H 1 sc (x d >0) R -s hD x1 v h H s (x d >0) v h L 2 (x d >0,H 1 sc (R d-1 ) + h v h 2 H 1 sc (x d >0)
and

|B 3 | (1 -Φ(|ξ 1 |/R) v h L 2 (x d >0,H 1 sc (R d-1 )) v h L 2 (x d >0,H 1 sc (R d-1 )) + h v h 2 H 1 sc (x d >0) R -s v h L 2 (x d >0,H 1+s sc (R d-1 )) v h L 2 (x d >0,H 1 sc (R d-1 )) + h v h 2 H 1 sc (x d >0)
where we have applied Lemma 4.4. Then we can conclude as for the term b 0 .

In the following lemma we consider the quantity op sc (b)

h 2 D 2 x d v h |v h L 2 (x d >0)
, which is not clearly well defined. But as v h satisfied (88), we can prove that h 2 D 2

x d v h ∈ L 2 (x d > 0, H -1 sc (R d-1
)). The inner product in tangential variables need to be interpreted as a duality product

H -1 , H 1 . Proposition 4.5. Let b(x, ξ ′ ) ∈ C ∞ 0 (R d × R d-1
), we have

op sc (b)h 2 D 2 x d v h |v h L 2 (x d >0) → µ, bξ 2 d as h → 0.
Proof. We use the equation satisfied by v h (see (88)).

op sc (b)h 2 D 2 x d v h |v h L 2 (x d >0) = op sc (b)h(q h -iav h )|v h L 2 (x d >0) -op sc (b)(R(x, hD x ′ ) -1)v h |v h L 2 (x d >0) = A + B.
Clearly

|A| h( q h L 2 (x d >0) + v h L 2 (x d >0) ) v h L 2 (x d >0) → 0 as h → 0. By symbol calculus, op sc (b)(R(x, hD x ′ ) -1) = op sc b(x, ξ ′ )(R(x, ξ ′ ) -1) + h op sc (r 0 )
, where r 0 ∈ S 0 tan . Then, by Proposition 4.2, we have B →µ, b(R -1) . Let Φ be as given in Lemma 3.19, we have for λ sufficiently large

-µ, b(x, ξ ′ )(R(x, ξ ′ ) -1) = -µ, b(x, ξ ′ )(R(x, ξ ′ ) -1)Φ(ξ d /λ) = -µ, b(x, ξ ′ )(ξ 2 d + R(x, ξ ′ ) -1)Φ(ξ d /λ) + µ, b(x, ξ ′ )ξ 2 d Φ(ξ d /λ) = µ, b(x, ξ ′ )ξ 2 d Φ(ξ d /λ) = µ, b(x, ξ ′ )ξ 2 d ,
as pµ = 0. Which gives the lemma.

Proposition 4.6. Let b(x, ξ ′ ) ∈ S(1, (dx) 2 + (dξ ′ ) 2 ), we have

op sc (b)hD x d v h |v h L 2 (x d >0) → µ, bξ d as h → 0.
Proof. Let Φ be as given in Lemma 3.19, we have for λ > 0,

op sc (b)hD x d v h |v h L 2 (x d >0) = op sc (b)1 x d >0 hD x d v h |1 x d >0 v h L 2 (R d ) = op sc (b) Op sc Φ(ξ d /λ) 1 x d >0 hD x d v h |1 x d >0 v h L 2 (R d ) + op sc (b) Op sc 1 -Φ(ξ d /λ) 1 x d >0 hD x d v h |1 x d >0 v h L 2 (R d ) = A + B.
By symbol calculus in S(1, (dx) 2 + (dξ) 2 ), we have

op sc (b) Op sc 1 -Φ(ξ d /λ) = Op sc 1 -Φ(ξ d /λ) * op sc (b) + h Op sc (r 0 ), where r 0 ∈ S(1, (dx) 2 + (dξ) 2 ). Then |B| op sc (b)1 x d >0 hD x d v h | Op sc 1 -Φ(ξ d /λ) 1 x d >0 v h L 2 (R d ) + h hD x d v h L 2 (R ) v h L 2 (R d ) λ -s + h,
by Lemma 4.3 and a priori estimates (88).

Next we treat the term A. We have 1

x d >0 hD x d v h = hD x d 1 x d >0 v h + ih(v h ) |x d =0 ⊗ δ x d =0 . Then we have A = op sc (b) Op sc Φ(ξ d /λ) hD x d 1 x d >0 v h |1 x d >0 v h L 2 (R d ) + ih op sc (b) Op sc Φ(ξ d /λ) (v h ) |x d =0 ⊗ δ x d =0 |1 x d >0 v h L 2 (R d ) = A 1 + A 2 .
We have Op sc Φ(ξ d /λ) hD x d = Op sc ξ d Φ(ξ d /λ) and by symbol calculus in S(1, (dx

) 2 + (dξ) 2 ), we have op sc (b) Op sc ξ d Φ(ξ d /λ) = Op sc ξ d Φ(ξ d /λ)b + h Op sc (r 0 ), where r 0 ∈ S(1, (dx) 2 + (dξ) 2 ). Then A 1 = op sc (b) Op sc Φ(ξ d /λ) hD x d 1 x d >0 v h |1 x d >0 v h L 2 (R d ) → µ, ξ d Φ(ξ d /λ)b = µ, ξ d b , if λ is sufficiently large.
Let Φ be such that F Φ = Φ, where F is the Fourier transform. We have Op sc Φ(ξ d /λ) δ

x d =0 = h -1 λ Φ(λx d /h). Then |A 2 | op sc (b) (v h ) |x d =0 λ Φ(λx d /h) |1 x d >0 v h L 2 (R d ) λ|(v h ) |x d =0 | L 2 (R d-1 ) | Φ(λx d /h)| L 2 (R) v h L 2 (x d >0) λ 1/2 h 1/2 |(v h ) |x d =0 | L 2 (R d-1 ) , as | Φ(λx d /h)| L 2 (R) = Ch 1/2 λ -1/2
. From (88), we have A 2 → 0, as h → 0. From estimates on A 1 , A 2 and B we obtain the result.

Boundary formulas

The two next propositions are the analogous of Proposition 4.1 at the boundary.

Proposition 4.7. Let b ∈ C ∞ 0 (R d × R d-1 ) (resp. b ∈ C ∞ 0 (R d × R d-2
) ) be real valued functions. We have the following formula

H p µ -2aµ, b = µ, {b, p} -2ab = lim h→0 2 Re b(x ′ , 0, hD ′ )(v h ) |x d =0 |(hD x d v h ) |x d =0 0 . ( resp. lim h→0 2 Re b(x ′ , 0, hD ′′ )(v h ) |x d =0 |(hD x d v h ) |x d =0 0 .) (91) In particular b(x ′ , 0, hD ′ )(v h ) |x d =0 |(hD x d v h ) |x d =0 0 (resp. b(x ′ , 0, hD ′′ )(v h ) |x d =0 |(hD x d v h ) |x d =0 0 ), have limits as h → 0. Remark 7.
In what follow we write b(x ′ , 0, hD ′ ) even if b only depends on variables (x, ξ ′′ ). With Proposition 3.16 we can only proof that b

(x ′ , 0, hD ′ )(v h ) |x d =0 |(hD x d v h ) |x d =0 0 , is a o(h -1
). In the proof below we show that the quantity has a limit and converges to the left hand side of (91). We are not able to prove that

(v h ) |x d =0 or (hD x d v h ) x d =0 are bounded.
Proof. We recall (56) the integration by parts formula in semiclassical context.

(u|hD x d w) L 2 (x d >0) = (hD x d u|w) L 2 (x d >0) -ih(u |x d =0 |w |x d =0 ) 0 ,
for u and w sufficiently smooth.

To proof the lemma we compute the following quantity by two different manners.

A = ih -1 b(x, hD ′ ) h 2 D 2 x d + R(x, hD ′ ) -1 + iha v h |v h L 2 (x d >0) -ih -1 b(x, hD ′ )v h | h 2 D 2 x d + R(x, hD ′ ) -1 + iha v h L 2 (x d >0) = ih -1 b(x, hD ′ )hq h |v h L 2 (x d >0) -ih -1 b(x, hD ′ )v h |hq h L 2 (x d >0) . Then |A| q h L 2 (x d >0) v h L 2 (x d >0) → 0 as h → 0 from (88).
To compute A we now integrate by parts. We have

b(x, hD ′ )v h | R(x, hD ′ ) -1 v h L 2 (x d >0) = R(x, hD ′ ) -1 b(x, hD ′ )v h |v h L 2 (x d >0) , ihb(x, hD ′ )av h |v h L 2 (x d >0) -b(x, hD ′ )v h |ihav h L 2 (x d >0) = 2ih op sc (b(x, ξ ′ )a(x))v h |v h L 2 (x d >0) + O(h 2 ),
as first R(x, hD ′ ) is self-adjoint and does not contain derivative with respect x d and second by symbol

calculus b(x, hD ′ )a = ab(x, hD ′ ) = op sc (a(x)b(x, ξ ′ )) up to O(h). We have b(x, hD ′ )v h |h 2 D 2 x d v h L 2 (x d >0) = hD x d b(x, hD ′ )v h |hD x d v h L 2 (x d >0) -ih b(x ′ , 0, hD ′ )(v h ) |x d =0 |(hD x d v h ) |x d =0 0 = h 2 D 2 x d b(x, hD ′ )v h |v h L 2 (x d >0) -ih b(x ′ , 0, hD ′ )(hD x d v h ) |x d =0 |(v h ) |x d =0 0 -ih b(x ′ , 0, hD ′ )(v h ) |x d =0 |(hD x d v h ) |x d =0 0 .

Then we have

A = ih -1 [b(x, hD ′ ), h 2 D 2 x d + R(x, hD ′ ) -1 ]v h |v h L 2 (x d >0) -2 op sc (b(x, ξ ′ )a(x))v h |v h L 2 (x d >0) -b(x ′ , 0, hD ′ )(hD x d v h ) |x d =0 |(v h ) |x d =0 0 -b(x ′ , 0, hD ′ )(v h ) |x d =0 |(hD x d v h ) |x d =0 0 + O(h).
From the structure of b we claim that

[b(x, hD ′ ), h 2 D 2 x d + R(x, hD ′ ) -1 ] = -ih Op sc {b, p} + h 2 op sc (r 0 )
where

r 0 ∈ S(1, (dx) 2 + (dξ ′ ) 2 ). Indeed, the assumptions imply that b(x, ξ ′ ) ∈ S(1, (dx) 2 + (dξ ′ ) 2 ), and
h 2 D 2 x d + R is a sum of terms c(x)h 2 D xj D x k . By exact symbol calculus we have [b(x, hD ′ ), c(x)h 2 D xj D x k ] = [b(x, hD ′ ), c(x)]h 2 D xj D x k + c(x)[b(x, hD ′ ), hD xj ]hD x k + c(x)hD xj [b(x, hD ′ ), hD x k ] = [b(x, hD ′ ), c(x)]h 2 D xj D x k + c(x)[b(x, hD ′ ), hD xj ]hD x k + c(x)[b(x, hD ′ ), hD x k ]hD xj + c(x)[hD xj , [b(x, hD ′ ), hD x k ]] = -ih Op sc ({b, cξ j ξ k }) + h 2 op sc (r 0 ), where r0 is in S(1, (dx) 2 + (dξ ′ ) 2 ).
First {b, p} = -2ξ d ∂ x d b + {b, R}, we can apply Propositions 4.2 and 4.6 as {b, R} is the sum of terms in S(1, (dx) 2 + (dξ ′ ) 2 ) or of the form q(x, ξ ′′ ), q(x, ξ ′′ )ξ 1 and q(x, ξ ′′ )ξ 2 1 , where

q ∈ C ∞ 0 (R d × R d-2 ). Second |(op sc (r 0 )v h |v h )| v h 2 L 2 (x d >0) we can conclude that ih -1 [b(x, hD ′ ), h 2 D 2 x d + R(x, hD ′ ) -1 ]v h |v h L 2 (x d >0) -2 op sc (b(x, ξ ′ )a(x))v h |v h L 2 (x d >0) → µ, {b, p} -2ba as h → 0.
By symbol calculus we have op sc (b(x ′ , 0, ξ ′ )) * = op sc (b(x ′ , 0, ξ ′ ))+ h op sc (r 0 ), where r 0 ∈ S(1, (dx

) 2 + (dξ ′ ) 2 ). Then b(x ′ , 0, hD ′ )(hD x d v h ) |x d =0 |(v h ) |x d =0 0 = (hD x d v h ) |x d =0 |b(x ′ , 0, hD ′ )(v h ) |x d =0 0 + B, where |B| ≤ h|(v h ) |x d =0 | H 1/2 |(hD x d v h ) |x d =0 | H -1/2 →
0 as h → 0 by (88). This gives the conclusion of Lemma.

Proposition 4.8. Let b ∈ C ∞ 0 (R d × R d-1
) be a real valued we have the following formula

-H p µ + 2aµ, bξ d = µ, {p, ξ d b} + 2abξ d = lim h→0 Re b(x ′ , 0, hD ′ ) R(x ′ , 0, hD ′ ) -1 (v h ) |x d =0 |(v h ) |x d =0 0 -b(x ′ , 0, hD ′ )(hD x d v h ) |x d =0 |(hD x d v h ) |x d =0 0 . (92)
In particular this means that

Re b(x ′ , 0, hD ′ ) R(x ′ , 0, hD ′ ) -1 (v h ) |x d =0 |(v h ) |x d =0 0 -b(x ′ , 0, hD ′ )(hD x d v h ) |x d =0 |(hD x d v h ) |x d =0 0 ,
has a limit as h → 0.

Remark 8. As in Proposition 4.7 the right hand side of (92) does not have a priori limit and we do not know if each term of the sum has a limit.

Proof. We begin by an observation on regularity of traces. From the definition of q h (see [START_REF] Helffer | Équation de Schrödinger avec champ magnétique et équation de Harper[END_REF]), the terms θ(h 2 P )g h j are in D(P ) for j = 1, 2. The term [θ(h 2 P ), a]u h = θ(h 2 P )(au h )aθ(h 2 P )u h is in D(P ), it is clear for θ(h 2 P )(au h ) and aθ(h 2 P )u h is in H 1 (Ω), a direct computation shows that P (aθ(h 2 P )u h ) is in L 2 (Ω) and aθ(h 2 P )u h satisfies the Zaremba trace condition as θ(h 2 P )u h satisfies it. This implies that q h ∈ H 1 (Ω). In particular we have

h 2 D 2 x d v h = hq h -(R(x, hD ′ ) -1)v h , then (op sc (b)h 2 D 2 x d v h ) |x d =0 = (h op sc (b)q h ) |x d =0 -(op sc (b)(R(x, hD ′ ) -1 + iha)v h ) |x d =0 ∈ L 2 (x d = 0
), for b compactly supported and using properties (88). In this analysis we do not estimate the size of the norm with respect h but this allows to give a sense to some terms appearing in what follows. We introduce the following quantity which is real

A = ih -1 (b(x, hD ′ )hD x d + hD x d b(x, hD ′ ) * )v h |(h 2 D 2 x d + R(x, hD ′ ) -1 + iha)v h L 2 (x d >0) -(h 2 D 2 x d + R(x, hD ′ ) -1 + iha)v h |(b(x, hD ′ )hD x d + hD x d b(x, hD ′ ) * )v h L 2 (x d >0) = (b(x, hD ′ )hD x d + hD x d b(x, hD ′ ) * )v h |q h L 2 (x d >0) -q h |(b(x, hD ′ )hD x d + hD x d b(x, hD ′ ) * )v h L 2 (x d >0) .
As hD x d b(x, hD ′ ) * = b(x, hD ′ ) * hD x d + op sc (r 0 ) where r 0 ∈ S 0 tan , we obtain

|A| q h L 2 (x d >0) ( v h L 2 (x d >0) + hD x d v h L 2 (x d >0) ) → 0 as h → 0, by (88)
. Let

B = (h 2 D 2 x d + R(x, hD ′ ) -1 + iha)v h |(b(x, hD ′ )hD x d + hD x d b(x, hD ′ ) * )v h L 2 (x d >0) . (93)
We have by integrations by parts

B = (b(x, hD ′ )hD x d + hD x d b(x, hD ′ ) * )(h 2 D 2 x d + R(x, hD ′ ) -1 + iha)v h |v h L 2 (x d >0) + (b(x, hD ′ )hD x d + hD x d b(x, hD ′ ) * )ihav h |v h L 2 (x d >0) -ih (b(x, hD ′ ) + b(x, hD ′ ) * )(h 2 D 2 x d + R(x, hD ′ ) -1)v h |x d =0 |(v h ) |x d =0 0 . ( 94
) Let C = (b(x, hD ′ )hD x d + hD x d b(x, hD ′ ) * )v h |h 2 D 2 x d v h L 2 (x d >0) .
We have by integration by parts

C = hD x d b(x, hD ′ )hD x d + hD x d b(x, hD ′ ) * v h |hD x d v h L 2 (x d >0) -ih (b(x, hD ′ )hD x d + hD x d b(x, hD ′ ) * )v h |x d =0 |(hD x d v h ) |x d =0 0 = h 2 D 2 x d (b(x, hD ′ )hD x d + hD x d b(x, hD ′ ) * )v h |v h L 2 (x d >0) -ih hD x d b(x, hD ′ )hD x d + hD x d b(x, hD ′ ) * v h |x d =0 |(v h ) |x d =0 0 -ih (b(x, hD ′ )hD x d + hD x d b(x, hD ′ ) * )v h |x d =0 |(hD x d v h ) |x d =0 0 . (95)
The terms with damping term a, coming from A and B, give a term

-4 ihab(x, hD ′ )hD x d v h |v h L 2 (x d >0) + h 2 op sc (r 0 )v h |v h L 2 (x d >0) + h 2 op sc (r 0 )hD x d v h |v h L 2 (x d >0) ,
where r 0 , r0 ∈ S 0 tan . From this, (94) and (95) we obtain

A = ih -1 h 2 D 2 x d + R(x, hD ′ ) -1, b(x, hD ′ )hD x d + hD x d b(x, hD ′ ) * v h |v h L 2 (x d >0) + 4 ab(x, hD ′ )hD x d v h |v h L 2 (x d >0) + hO( v h 2 L 2 (x d >0) + hDv h 2 L 2 (x d >0) ) -(b(x, hD ′ ) + b(x, hD ′ ) * )(h 2 D 2 x d + R(x, hD ′ ) -1)v h |x d =0 |(v h ) |x d =0 0 + hD x d b(x, hD ′ )hD x d + hD x d b(x, hD ′ ) * v h |x d =0 |(v h ) |x d =0 0 + (b(x, hD ′ )hD x d + hD x d b(x, hD ′ ) * )v h |x d =0 |(hD x d v h ) |x d =0 0 . (96)
By symbol calculus we have

ih -1 h 2 D 2 x d + R(x, hD ′ ) -1, b(x, hD ′ )hD x d + hD x d b(x, hD ′ ) * (97) = Op sc {ξ 2 d + R(x, ξ ′ ) -1, 2b(x, ξ ′ )ξ d } + h op sc (r 0 ) + h op sc (r 0 )hD x d , (98)
where r 0 , r0 ∈ S 0 tan . Propositions 4.2, 4.5 and 4.6 imply that Op sc {ξ

2 d + R(x, ξ ′ ) -1, 2b(x, ξ ′ )ξ d }v h |v h L 2 (x d >0) → µ, {p, 2b} as h → 0. (99)
By Proposition 4.6 the term 4 ab(x, hD ′ )hD

x d v h |v h L 2 (x d >0) → µ, 4baξ d as h → 0. And we have |(h op sc (r 0 )v h |v h ) L 2 (x d >0) | h v h 2 L 2 (x d >0) → 0 as h → 0, |(h op sc (r 0 )hD x d v h |v h ) L 2 (x d >0) | h hD x d v h L 2 (x d >0) v h L 2 (x d >0) → 0 as h → 0.
Then this and (99) imply that

ih -1 h 2 D 2 x d + R(x, hD ′ ) -1, b(x, hD ′ )hD x d + hD x d b(x, hD ′ ) * v h |v h L 2 (x d >0) + 4 ab(x, hD ′ )hD x d v h |v h L 2 (x d >0) → µ, {p, 2b} + 4abξ d as h → 0. ( 100
)
We now treat the boundary terms coming from (96). We have by symbol calculus

b(x, hD ′ ) + b(x, hD ′ ) * = 2b(x, hD ′ ) + h op sc (r 0 ), b(x, hD ′ )hD x d + hD x d b(x, hD ′ ) * = 2b(x, hD ′ )hD x d + h op sc (r 0 ), hD x d (b(x, hD ′ )hD x d + hD x d b(x, hD ′ ) * ) = (b(x, hD ′ ) + b(x, hD ′ ) * )h 2 D 2 x d + h op sc (r 0 ) + h op sc (r 0 )hD x d ,
where r 0 , r0 ∈ S 0 tan . Then boundary terms of A can be written as

-(2b(x, hD ′ ))(R(x, hD ′ ) -1)v h |x d =0 |(v h ) |x d =0 0 + 2b(x, hD ′ )hD x d v h |x d =0 |(hD x d v h ) |x d =0 0 , (101)
up to terms estimated by h| (op(r

0 )v h ) |x d =0 |(v h ) |x d =0 0 | + h| (op(r 0 )hD x d v h ) |x d =0 |(v h ) |x d =0 0 |
, and these terms converge to 0 from (88). Recalling that A → 0 as h → 0 and A real valued, we deduce Proposition 4.8 from ( 96), ( 100) and (101).

Properties and support of semiclassical measure

Here we decompose the measure into an interior measure and a measure supported at boundary. Moreover this last measure is supported on ξ d = 0.

Lemma 4.9. There exists a non negative Radon measure µ ∂ on

x d = ξ d = 0 such that µ = 1 x d >0 µ + µ ∂ ⊗ δ x d =0 ⊗ δ ξ d =0 . Furthermore µ ∂ is supported on R(x ′ , 0, ξ ′ ) -1 = 0.
Proof. We apply Proposition 4.8. We observe that

H p = 2ξ d ∂ x d -(∂ x d R)∂ ξ d + H ′ R where H ′ R = d-1 j=1 (∂ ξj R)∂ xj -(∂ xj R)∂ ξj . We have H p (ξ d b) = 2ξ 2 d (∂ x d b) -(∂ x d R)b + ξ d H ′ R b. Let b = b ε = εχ(x d /ε)ℓ(x, ξ ′ ), where χ ∈ C ∞ 0 (R), such that χ(0) = 0 and χ ′ (0) = 1, ℓ ∈ C ∞ 0 (R d × R d-1
). We have

H p (ξ d b ε ) + 2ab ε ξ d = 2ξ 2 d χ ′ (x d /ε)ℓ(x, ξ ′ ) + 2εχ(x d /ε)ξ 2 d ∂ x d ℓ -(∂ x d R)εχ(x d /ε)ℓ(x, ξ ′ ) + ξ d εχ(x d /ε)H ′ R ℓ + 2εaχ(x d /ε)ℓ(x, ξ ′ ).
Clearly H p (ξ d b ε ) + 2ab ε ξ d is uniformly bounded on the support of µ and

H p (ξ d b ε ) + 2ab ε ξ d → 2ξ 2 d χ ′ (0)ℓ1 x d =0
everywhere as ε → 0. Then by Lebesgue's dominated convergence theorem we have µ, H p (ξ d b ε ) + 2ab ε ξ d → µ, 2ξ 2 d ℓ1 x d =0 as ε → 0. As χ(0) = 0, the right hand side of (92) is 0 for every h. Then µ, 2ξ 2 d ℓ1 x d =0 = 0. This means that 1 x d =0 µ is supported on ξ d = 0. We denote µ ∂ the measure

1 ξ d =0 1 x d =0 µ. As µ = 1 x d >0 µ + 1 x d =0 µ, we have µ = 1 x d >0 µ + µ ∂ ⊗ δ x d =0 ⊗ δ ξ d =0 . We have by Proposition 3.1, (ξ 2 d + R(x, ξ ′ ) -1)1 x d =0 µ = 0, then (R(x ′ , 0, ξ ′ ) -1)µ ∂ = 0.
This gives the conclusion of Lemma.

The Hamiltonian of the interior measure is a priori a distribution of order one supported on x d = 0. The following lemma says that this quantity is a measure if the Hamiltonian vector field is transverse to the boundary. Lemma 4.10. We assume that aµ = 0. There exists a distribution (of order 1) µ 0 defined on x d = 0, such that H p (µ1 x d >0 ) = δ x d =0 ⊗ µ 0 . Moreover, in a neighborhood where H p x d > 0, µ 0 is a non negative Radon measure, and in a neighborhood where H p x d < 0, µ 0 is a non positive Radon measure.

Proof. The support of H p (µ1 x d >0 ) as a distribution is x d ≥ 0 and H p (µ) = 0 on x d > 0. Then H p (µ1 x d >0 ) is supported on x d = 0. This implies that there exist n ≥ 0 and µ j distributions on We recall that G d = {(x ′ , ξ ′ ), R(x ′ , 0, ξ ′ ) = 1 and ∂ x d R(x ′ , 0, ξ ′ ) < 0} is the set of diffractive points.

The following lemma states that the diffractive points whose projection belong to ∂Ω D are not in the support of the measure. Lemma 4.12. We assume aµ = 0. We have 1 G d ∩(∂ΩDN ×R d-1 ) µ ∂ = 0, where ∂Ω DN = ∂Ω D ∪ ∂Ω N .

Proof. We apply Proposition 4.8. We have to choose an adapted function b. Let χ ∈ C ∞ be such that χ(σ) = 0 if |s| ≥ 2, χ(σ) = 1 if |σ| ≤ 1. We apply Proposition 4.8 with b(x, ξ ′ ) = χ((1 -R(x, ξ ′ ))/ε)ℓ(x, ξ ′ )χ(x d /ε), where ε > 0 will be chosen in what follows and ℓ is supported in a neighborhood of a point of

∂Ω DN × R × R d-1 . We recall that H p = 2ξ d ∂ x d -(∂ x d R)∂ ξ d + H ′
R (see the proof of Lemma 4.9). We have

H p (ξ d b) = 2ξ 2 d -(∂ x d R)χ ′ ((1 -R(x, ξ ′ ))/ε)ℓ(x, ξ ′ )χ(x d /ε)/ε + χ((1 -R(x, ξ ′ ))/ε)(∂ x d ℓ(x, ξ ′ ))χ(x d /ε) + χ((1 -R(x, ξ ′ ))/ε)ℓ(x, ξ ′ )χ ′ (x d /ε)/ε -(∂ x d R)χ((1 -R(x, ξ ′ ))/ε)ℓ(x, ξ ′ )χ(x d /ε) + χ((1 -R(x, ξ ′ ))/ε)χ(x d /ε)ξ d H ′ R ℓ(x, ξ ′ ). ( 103
)
We claim

H p (bξ d ) is uniformly bounded on ξ 2 d + R(x, ξ ′ ) = 1, H p (bξ d ) → -(∂ x d R(x ′ , 0, ξ ′ ))1 R(x ′ ,0,ξ ′ )=1 1 x d =0 ℓ(x ′ , 0, ξ ′ ) as ε → 0 for all (x, ξ), such that ξ 2 d + R(x, ξ ′ ) = 1. ( 104
)
As µ is supported on

ξ 2 d + R(x, ξ ′ ) -1 = 0, then ξ 2 d /ε = (1 -R(x, ξ ′
))/ε on the support of µ, this implies that the three first terms in (103) are bounded. It is easy to prove that they converge to 0 as ε to 0. The fourth term is bounded and converges to -(∂

x d R(x ′ , 0, ξ ′ ))1 R(x ′ ,0,ξ ′ )=1 1 x d =0 ℓ(x ′ , 0, ξ ′ ).
In the last term as |R(x, ξ ′ ) -1|/ε is bounded, thus |ξ d | is bounded by C √ ε and then this term converges to 0 as ε → 0. This proves (104). From that we can conclude that µ, H p (ℓξ d ) converges to

µ, -(∂ x d R(x ′ , 0, ξ ′ ))1 R(x ′ ,0,ξ ′ )=1 1 x d =0 ℓ(x ′ , 0, ξ ′ ) = µ ∂ , -(∂ x d R(x ′ , 0, ξ ′ ))1 R(x ′ ,0,ξ ′ )=1 ℓ(x ′ , 0, ξ ′ ) , (105)
as ε → 0, which is a non negative term if l is non negative and supported in a neighborhood of a point of G d .

We now assume ℓ supported on a neighborhood of a point of ∂Ω D . Let

A ε = lim h→0 b(x ′ , 0, hD ′ ) R(x ′ , 0, hD ′ ) -1 (v h ) |x d =0 |(v h ) |x d =0 0 -b(x ′ , 0, hD ′ )(hD x d v h ) |x d =0 |(hD x d v h ) |x d =0 0 = -lim h→0 b(x ′ , 0, hD ′ )(hD x d v h ) |x d =0 |(hD x d v h ) |x d =0 0 , (106)
as v h satisfies the Dirichlet boundary condition. We want to prove that A ε ≤ 0. By Gårding inequality (9) (in fact used for R d-1 instead R d ) we have

-b(x ′ , 0, hD ′ )(hD x d v h ) |x d =0 |(hD x d v h ) |x d =0 0 ≤ C ε h|(hD x d v h ) |x d =0 | 2 L 2 .
Taking the limit as h → 0, Proposition 3.12 implies that A ε ≤ 0. As (105) is non negative, this prove that

µ ∂ , -(∂ x d R(x ′ , 0, ξ ′ ))1 R(x ′ ,0,ξ ′ )=1 ℓ(x ′ , 0, ξ ′ ) = 0. ( 107
) As χ ′ (x d , ε)χ(ξ d /ε)ξ 2 d /ε, χ(x d /ε)χ(ξ d /ε)ξ d /ε and ξ d χ(ξ d /ε)χ(x d /ε
) are uniformly bounded and goes to 0 as ε goes to 0, and χ(ξ d /ε)χ(x d /ε) goes to 1 x d =0,ξ d =0 and is uniformly bounded. We have µ, H p b goes to µ, -1 x d =0,ξ d =0 (∂ x d R(x ′ , 0, ξ ′ ))ℓ(x ′ , ξ ′ ) as ε goes to 0 and we have

µ, -1 x d =0,ξ d =0 ∂ x d R(x ′ , 0, ξ ′ )ℓ(x ′ , ξ ′ ) = -µ ∂ , (∂ x d R(x ′ , 0, ξ ′ ))ℓ(x ′ , ξ ′ ) . (111)
The term 2aµ ∂ , b(x ′ , 0, ξ ′ , 0) goes to 0 as ε, as b goes to 0 uniformly as ε goes to 0. Now we compute the limite as ε → 0 of μ1 , ∂ ξ d b(x ′ , 0, ξ ′ , 0) . We have

∂ ξ d b(x ′ , x d , ξ ′ , ξ d ) = ℓ(x ′ , ξ ′ )χ(ξ d /ε)χ(x d /ε) + ℓ(x ′ , ξ ′ )χ ′ (ξ d /ε)χ(x d /ε)ξ d /ε. Then ∂ ξ d b(x ′ , 0, ξ ′ , 0) = ℓ(x ′ , ξ ′ ) and μ1 -(∂ x d R(x ′ , 0, ξ ′ ))µ ∂ , ∂ ξ d b(x ′ , 0, ξ ′ , 0) = μ1 -(∂ x d R(x ′ , 0, ξ ′ ))µ ∂ , ℓ(x ′ , ξ ′ ) = -µ ∂ , (∂ x d R(x ′ , 0, ξ ′ ))ℓ(x ′ , ξ ′ )
from ( 110) and (111). We deduce that μ1 = 0 and the last result of the lemma from (109).

We have an analogous result to Lemma 4.13 in a neighborhood of Zaremba condition. To be precise we recall the notations defined in Formula (64), we have R(x ′ , 0, ξ ′ ) = ξ 2 1 + R 0 (x ′′ , ξ ′′ ) + x 1 r 2 (x, ξ ′ ), where x ′ = (x 1 , x ′′ ) and ξ ′ = (ξ 1 , ξ ′′ ), R 0 ∈ S( ξ ′′ 2 , (dx ′′ ) 2 + ξ ′′ -2 (dξ ′′ ) 2 ) and r 2 ∈ S 2 tan .

Lemma 4.14. We assume that (v h ) h satisfies the boundary Zaremba condition neighborhood of (0,

x ′′ 0 ). Let ℓ ∈ C ∞ 0 (R d-1 x ′ × R d-2 ξ ′′ ), where ξ ′ = (ξ 1 , ξ ′′ ). Then H ′ R µ ∂ -2aµ ∂ + µ 0 , ℓ = 0. In particular if µ ∂ is a measure supported on x 1 = ξ 1 = 0, this means that µ ∂ = μ∂ ⊗ δ x1=ξ1=0 and if µ 0 is supported on x 1 = ξ 1 = ξ d = 0,
this means that there exists μ0 , and μα,β distributions of order 1 on R d-1

x ′′ × R d-1 ξ ′′ , for α = (0, 0, 0) or (1, 0, 0) and β ∈ {(j, 0, k), j, k = 0 or 1} such that

µ 0 = μ0 ⊗ δ x1=ξ1=ξ d =0 + |α|+|β|=1 μα,β ⊗ ∂ α x ∂ β ξ δ x1=ξ1=ξ d =0 . Then H ′′ R0 μ∂ -2aμ ∂ + μ0 = 0, where H ′′ R0 = 2≤j≤d-1 ∂ ξj R 0 (x ′′ , ξ ′′ )∂ xj -∂ xj R 0 (x ′′ , ξ ′′ )∂ ξj . Proof. Let b ∈ C ∞ 0 (R d × R d-2
) be such that b |x d =0 = ℓ. We apply Proposition 4.7 to b(x, ξ ′′ ). As in the end of the proof of Lemma 4.11 we have ℓ(x ′ , hD ′′ )(v h ) |x d =0 |(hD x d v h ) |x d =0 0 = 0. Then we have H p µ-2aµ, b(x, ξ ′′ ) = 0. We follow the same ideas of the proof of Lemma 4.13. From Formula (108) we have

H ′ R µ ∂ -2aµ ∂ + µ 0 , ℓ -2 a1 x d >0 µ, b = 0.
Let χ ∈ C ∞ 0 (R) be such that χ(σ) = 1 in a neighborhood of 0. Taking b ε (x, ξ ′′ ) = χ(x d /ε)ℓ(x ′ , ξ ′′ ) and letting ε goes to 0, we obtain

H ′ R µ ∂ -2aµ ∂ + µ 0 , ℓ = 0. We have H ′ R = 2ξ 1 ∂ x1 + H ′′ R0 + H ′ x1r2 , and H ′ x1r2 ℓ(x ′ , ξ ′′ ) = x 1 H ′ r2 ℓ(x ′ , ξ ′′ ), as ℓ independent of ξ 1 . We deduce from the form of µ ∂ that H ′ R µ ∂ = H ′′ R0 μ∂ ⊗ δ x1=ξ1=0 . Taking ℓ(x ′ , ξ ′′ ) = χ(x 1 ) l(x ′′ , ξ ′′ ), we deduce from the form of µ 0 H ′ R µ ∂ -2aµ ∂ + µ 0 , ℓ = H ′′ R0 μ∂ -2aμ ∂ + μ0 , ℓ = 0.
This implies the result.

Support propagation results

In this section we prove propagation of support of semiclassical measure under the assumption mGCC, see Definition 1.2.

Proposition 5.1. We assume that P , a(x) and Ω satisfy mGCC. Let µ the semiclassical measure constructed from (v h ) h and satisfying (33), we have µ = 0.

We prove the propagation result, first in interior which is a classical result, second in a neighborhood of a point on the boundary with Dirichlet or Neumann conditions and third in a neighborhood of a point on Γ.

A Proof of Lemma 4.4, Zaremba regularity result

It is well known that the solution of elliptic equation of second order with Zaremba boundary condition is in H s with s < 3/2 for a data in L 2 , see for instance Shamir [START_REF] Shamir | Regularization of mixed second order elliptic problems[END_REF], Savaré [START_REF] Savaré | Regularity and perturbation results for mixed second order elliptic problems[END_REF] . Here we have to prove that the solution is in semiclassical Sobolev spaces.

We start from (88) and equation h 2 P v hv h + ihav h = hq h . We have h 2 P v h + v h = r h where r h = 2v hihav h + hq h and we deduce r h L 2 (Ω) ≤ C. We observe that h 2 P + 1 is a semiclassical elliptic operator. To prove the result we follow the method used in Section 3.1.4 with the advantage that the operator is globally elliptic and we keep more or less the same notations introduced in this section. In particular we do not have to use microlocal cutoff.

We work in a neighborhood of the boundary in coordinate (x ′ , x d ) and Ω is given by x d > 0 and Γ by x 1 = 0. The symbol of the operator is given by ξ

2 d + R(x, ξ ′ ) + 1 and R(x ′ , 0, ξ ′ ) = ξ 2 1 + R 0 (x ′′ , ξ ′′ ) + x 1 r 2 (x ′ , ξ ′ ).
Let χ a cutoff in a neighborhood of a point of Γ, it will fix to 0 in what follows. We set w h = χδ v h where χδ (x) = χ(x/δ). Let χ another cutoff function such that χ(x) = 1 if x is contained in a neighborhood of supp χ and we set χ δ (x) = χ(x/δ). As v h is uniformly in H 1 sc (x d > 0) and r 2 is a differential operator, we have

h 2 D 2 x d w h + h 2 D 2 x1 w h + op sc (R 0 (0, x ′′ , x d , ξ ′′ ))w h + x 1 χ δ (x) op sc (r 2 (x, ξ ′ ))w h = r 0 h , where r 0 h L 2 (Ω) ≤ C. Let ρ(x, ξ ′ ) = (ξ 2 d + ξ 2 1 + R 0 (0, x ′′ , x d , ξ ′′ ) + x 1 χ δ (x)r 2 (x, ξ ′ ) + 1) 1/2 . By symbol calculus, we have in x d > 0 (hD x d + i op sc (ρ))(hD x d + i op sc (ρ))w h = r 1 h where r 1 h L 2 (Ω) ≤ C. (113)
Let z = (hD x d + i op sc (ρ))w h , we then have

(hD x d + i op sc (ρ))z = r 1 h .
2 Re((hD

x d + i op sc (ρ))z|i op sc (ρ)z) ≤ 2 r 1 h L 2 (Ω) op sc (ρ)z L 2 sc (x d >0) . (114)
Integrating by parts (see (56)) we have

(i op sc (ρ)z|hD x d z) = (ihD x d op sc (ρ)z|z) -ih(i op sc (ρ)z |x d =0 |z |x d =0 ) 0 .
We deduce 2 Re(hD

x d z|i op sc (ρ)z) = (i[hD x d , op sc (ρ)]z|z) + h(op sc (ρ)z |x d =0 |z |x d =0 ) 0 . As |(i[hD x d , op sc (ρ)]z|z)| h z 2 L 2 (0,+∞,H 1 sc ) , we deduce from (114) op sc (ρ)z 2 L 2 (x d >0) + h(op sc (ρ)z |x d =0 |z |x d =0 ) 0 r 1 h L 2 (Ω) op sc (ρ)z L 2 (x d >0) + h z 2 L 2 (0,+∞,H 1 sc ) .
As op sc (ρ -1 ) op sc (ρ) = Id + h op sc (r 1 ), where r 1 ∈ S -1 tan we have

z 2 L 2 (0,+∞,H 1 sc ) op sc (ρ)z 2 L 2 (x d >0) + h z L 2 . We deduce h(op sc (ρ)z |x d =0 |z |x d =0 ) 0 r 1 h 2 L 2 (Ω)
. And by tangential Gårding inequality [START_REF] Bony | Calcul symbolique et propagation des singularités pour les équations aux dérivées partielles non linéaires[END_REF]

(in R d-1 instead of R d ) applied to op sc ( ξ ′ -1/2 ) op sc (ρ) op sc ( ξ ′ -1/2 ) op sc ( ξ ′ 1/2 )z |x d =0 | op sc ( ξ ′ 1/2 )z |x d =0 0 we obtain, as ξ ′ -1 ρ ≥ C > 0, h|z |x d =0 | H 1/2 sc r 1 h 2 L 2 (Ω)
. By definition of z, we have

(hD x d w h ) |x d =0 + i op sc (ρ 0 )(w h ) |x d =0 = h -1/2 r 2 h , where |r 2 h | H 1/2 sc ≤ C, (115)
where

ρ 0 (x ′ , ξ ′ ) = (R(x ′ , 0, ξ ′ ) + 1) 1/2 . Let β(x ′′ , ξ ′′ ) = (R 0 (x ′′ , ξ ′′ ) + 1) 1/2 . We have ρ 0 (x ′ , ξ ′ ) = (ξ 2 1 + β 2 (x ′′ , ξ ′′ ) + x 1 χ δ r 2 (x ′ , ξ ′ )) 1/2 . Let u 0 = (hD x d w h ) |x d =0 and u 1 = (w h ) |x d =0 , we recall that supp u 0 ⊂ {x 1 ≤ 0} and supp u 1 ⊂ {x 1 ≥ 0}.
As in Section 3.1.4, we have

ξ 1 → ξ 1 + iβ(x ′′ , ξ ′′ )
±1/2 are holomorphic functions on Im ξ 1 > 0,

ξ 1 → ξ 1 -iβ(x ′′ , ξ ′′ ) ±1/2 are holomorphic functions on Im ξ 1 < 0. Let v 0 = op sc (ξ 1 + iβ) -1/2 u 0 . As u 0 is supported in x 1 ≤ 0 and (ξ 1 + iβ) -1/2 is a holomorphic function on Im ξ 1 > 0, v 0 is supported in x 1 ≤ 0. We have ξ 1 -iβ(x ′′ , ξ ′′ ) ∈ S( ξ ′ , (dx ′ ) 2 + (dξ ′ ) 2 ), then (ξ 1 -iβ(x ′′ , ξ ′′ )) ±1/2 ∈ S( ξ ′ ±1/2 , (dx ′ ) 2 + (dξ ′ ) 2
). This implies by symbol calculus that

op sc (ξ 1 -iβ) -1/2 op sc (ξ 1 -iβ) 1/2 = Id + h op sc (s 0 ), (116)
where

s 0 ∈ S(1, (dx ′ ) 2 +(dξ ′ ) 2 ). If h is sufficiently small, Id+h op sc (s 0 ) is invertible on H s sc for every s. Let v 1 = op sc ξ 1 -iβ(x ′′ , ξ ′′ ) 1/2 (Id+h op sc (s 0 )) -1 u 1 , thus we have op sc ξ 1 -iβ(x ′′ , ξ ′′ ) -1/2 v 1 = u 1 .
Moreover from (116), op sc (s 0 ) map distribution supported on x 1 ≥ 0 to distribution supported on x 1 ≥ 0, then by Neumann series, (Id + h op sc (s 0 )) -1 also satisfies this property. This implies that v 1 is supported on x 1 ≥ 0. From (115), we obtain

op sc (ξ 1 + iβ) -1/2 v 0 + i op sc (ξ 1 + iβ) -1/2 op sc (ρ 0 ) op sc (ξ 1 -iβ) -1/2 v 1 = h -1/2 r 3 h , (117)
where

|r 3 h | H 1 sc ≤ C. The principal symbol of op sc (ξ 1 +iβ) -1/2 op sc (ρ 0 ) op sc (ξ 1 -iβ) -1/2 is by a simple computation (ξ 1 + iβ) -1/2 ρ 0 (ξ 1 -iβ) -1/2 = 1 + x 1 χ δ r 2 (ξ 2 1 + β 2 ) 1/2 ρ 0 + (ξ 2 1 + β 2 ) 1/2 = 1 + x 1 χ δ r 3 ,
where r 3 ∈ S(1, (dx ′ ) 2 + (dξ ′ ) 2 ). Formula (117) reads

op sc (ξ 1 + iβ) -1/2 v 0 + iv 1 + ix 1 χ δ op sc (r 3 )v 1 = h -1/2 r 3 h .
We restrict this equation on x 1 > 0, as op sc (ξ 1 + iβ) -1/2 v 0 is supported on x 1 ≤ 0, we obtain

i(v 1 ) |x1>0 + ix 1 χ δ (op sc (r 3 )v 1 ) |x1>0 = h -1/2 (r 3 h ) |x1>0 . As |w |x1>0 | H s sc ≤ |w| H s sc , we obtain that |(v 1 ) |x1>0 | H s sc (x1>0) ≤ |x 1 χ δ op sc (r 3 )v 1 | H s sc + h -1/2 |r 3 h | H s sc . ( 118
) Lemma A.1. Let χ ∈ C ∞ 0 (R d ) and χ δ (x) = χ(x/δ)
, where δ > 0. Then there exist C > 0 such that for every δ > 0,

x 1 χ δ C α ≤ Cδ 1-α , for α ∈ (0, 1) and x 1 χ δ L ∞ ≤ Cδ. Proof. First, we have |x 1 χ(x/δ)| ≤ Cδ. Second, |(x 1 + y 1 )χ((x + y)/δ) -x 1 χ(x/δ)| ≤ 2Cδ and |∂ x (x 1 χ(x/δ))| ≤ C, then we have |(x 1 + y 1 )χ((x + y)/δ) -x 1 χ(x/δ)| ≤ C|y|.
For α ∈ (0, 1), interpolating both estimates, we have |(x 1 + y 1 )χ((x + y)/δ)x 1 χ(x/δ)| ≤ Cδ 1-α |y| α . Which gives the result.

Lemma A.2. Let s ∈ (0, 1) and 0 < s < α < 1, there exists C > 0, such that for every f ∈ C α sc (R d ), and g ∈ H s sc (R d ), f g ∈ H s sc (R d ) and we have

f g H s sc (R d ) ≤ C f C α sc (R d ) g H s sc (R d ) .
Here we say that f

∈ C α sc (R d ) if f is bounded and h α |f (x + y) -f (x)| ≤ C|y| α . The norm on C α sc (R d ) is f L ∞ (R d ) + sup x,y∈R d h α |f (x + y) -f (x)||y| -α .
Proof. We can follow the classical proof that the multiplication by C α functions are bounded operators on H s , using Littlewood-Paley theory and para-product in spirit of Bony [START_REF] Bony | Calcul symbolique et propagation des singularités pour les équations aux dérivées partielles non linéaires[END_REF]. We recall the Littlewood-Paley decomposition (see [START_REF] Gearhart | Spectral theory for contraction semigroups on Hilbert space[END_REF] for notations), we have w

= k≥-1 ∆ k w where F (∆ k (w))(ξ) = φ(2 -k hξ)F (w)(ξ) for k ≥ 0 and F (∆ -1 (w))(ξ) = ψ(hξ)F (w)(ξ). Let S k = -1≤j≤k ∆ j . By assumptions on f , we have S k f L ∞ (R d ) ≤ C f L ∞ (R d )
and by the usual proof of characterization of C α functions with Littlewood-Paley decomposition we have

∆ k f L ∞ (R d ) ≤ C2 -kα f C α sc (R d )
for k ≥ 0. By assumptions on g, we have ∆

k g L 2 (R d ) ≤ c k 2 -sk for k ≥ 0, where (c k ) ℓ 2 (N) ≤ C g H s sc (R d ) , ∆ -1 g L 2 (R d ) ≤ C f L 2 (R d ) and S k g L 2 (R d ) ≤ C g L 2 (R d ) . The product f g = k≥-1 S k (f )∆ k (g) + k≥0 S k-1 (g)∆ k (f ).
We estimate each term in previous formula. For j ≥ 2 we have

∆ j k≥-1 S k (f )∆ k (g) L 2 (R d ) ∆ j k≥j-2 S k (f )∆ k (g) L 2 (R d ) k≥j-2 S k (f ) L ∞ (R d ) ∆ k (g) L 2 (R d ) 2 -sj f L ∞ (R d ) k≥j-2 c k 2 -(k-j)s ,
and

d j = k≥j-2 c k 2 -(k-j)s ∈ ℓ 2 * ℓ 1 ⊂ ℓ 2 , where (d j ) ℓ 2 g H s sc (R d ) . ∆ j k≥0 S k-1 (g)∆ k (f ) L 2 (R d ) ≤ ∆ j k≥j-2 S k-1 (g)∆ k (f ) L 2 (R d ) ≤ k≥j-2 S k-1 (g) L 2 (R d ) ∆ k (f ) L ∞ (R d ) g L 2 (R d ) f C α sc (R d ) k≥j-2 2 -kα g L 2 (R d ) f C α sc (R d ) 2
-js 2 -j(α-s) . As (2 -j(α-s)) is in ℓ 2 , and as the result is obvious for j ≤ 1 we obtain the result.

Observe that

f C α sc (R d ) f C α (R d ) . Then by Lemmas A.1 and A.2 we have |x 1 χ δ w| H s sc ≤ Cδ 1-α |w| H s sc . From (118) we have |(v 1 ) |x1>0 | H s sc (x1>0) ≤ Cδ 1-α |v 1 | H s sc + h -1/2 |r 3 h | H s sc , for ≤ s < α < 1/2. As |v 1 | H s sc = |(v 1 ) |x1>0 | H s sc (x1>0) for s ∈ [0, 1/2), we have |(v 1 ) |x1>0 | H s sc (x1>0) ≤ Ch -1/2 |r 3 h | H s sc ≤ Ch -1/2
, for δ sufficiently small. We obtain

|u 1 | H s+1/2 (x1>0) ≤ | op sc ξ 1 -iβ(x ′′ , ξ ′′ ) -1/2 v 1 | H s+1/2 ≤ C|v 1 | H s ≤ Ch -1/2 .
From (115) we deduce that

|u 0 | H s-1/2 ≤ Ch -1/2 .
The solution w h of semiclassical elliptic problem with boundary condition satisfying

|(w h ) |x d =0 | H s+1/2 sc ≤ Ch -1/2 and |(hD x d w h ) |x d =0 | H s-1/2 ≤ Ch -1/2 ,
is in H 1+s sc (x d > 0) and w h H 1+s (x d >0) ≤ C. This result is well-known and it is a consequence of Formula (60) in [START_REF] Robbiano | Spectral analysis of the interior transmission eigenvalue problem[END_REF]. This achieves the proof of Lemma 4.4.

B A priori estimate for the trace of solution for Neumann boundary condition

We begin this section by recall some result on semiclassical Fourier Integral Operator.

Lemma B.1. Let (x ′ 0 , ξ ′ 0 ) be such that R(x ′ 0 , 0, ξ ′ 0 ) -1 = 0. For all x d in a neighborhood of 0, there exist a smooth symplectic transformation κ : U 0 → U 1 where U 0 and U 1 are some open set respectively of R This lemma is classical. We can find a proof in Hörmander [START_REF] Hörmander | The analysis of linear partial differential operators[END_REF]Theorem 21.1.6]. This means we can complete the coordinate R -1 in a symplectic manner.

d-1 x ′ × R d-1 ξ ′ and of R d-1 y ′ × R d-1 η ′ , satisfying (x ′ 0 , ξ ′ 0 ) ∈ U 0 , (0, 0) ∈ U 1 , κ(x ′ 0 , ξ ′ 0 ) = (0,
To avoid ambiguity even if x d = y d we denote x d when we work in (x, ξ ′ ) variables and y d otherwise.

We call a symbol of order 0 a symbol a ∈ S(

. ii) For all ã ∈ C ∞ 0 (U 0 ), F -1 op sc (ã)F = op sc (a)
, where a = κ * ã + hb where b is a symbol of order 0. In particular we have F -1 op sc (η 1 χ2 (y, η ′ ))F = op sc (χ 2 (R-1))+h op sc (b), where χ ∈ C ∞ 0 (U 0 ), χ ≥ 0 and b a symbol of order 0.

iii) there exist θ a symbol of order 0, B a bounded operator on L 2 such that op sc (θ) * = op sc (θ),

κ * χ = χ and (∂ x d F )F -1 = ih -1 op sc (θ) + hB.
iv) If the operators A and à are such that A = F -1 ÃF then

∂ x d A = F -1 ∂ y d à + ih -1 [ Ã, op sc (θ)] + h[ Ã, B] F
where B is the operator defined previously.

v) In particular we have κ * {η 1 , θ} = ∂ x d R in a neighborhood of (x ′ 0 , 0, ξ ′ 0 ).

Remark 10. Zworski states the result for Weyl quantification. It is clear that we can deduce the result for classical quantification. In the proof of Lemma B.2 we use Weyl quantification but in the rest of this section we shall use classical quantification to be coherent with notation used in this article.

A proof of Lemma B.2 is given in Section C.

Here we adapt, in the framework of semiclassical analysis, the results obtained by Tataru [START_REF] Tataru | On the regularity of boundary traces for the wave equation[END_REF] especially Lemma 4.3, Propositions 4.5 and 4.7. We essentially keep the notation used in that paper.

From now we shall use two semiclassical quantifications of symbol, one with parameter h and the other with parameter h 1/3 . To avoid ambiguity or confusion between both, we do not use the notation op sc but we use classical quantification. For instance, for a a symbol of order 0 we have op sc (a) = op(a(x, hξ ′ )) that is we keep the h or h 1/3 in the notation.

Let g = |dy| 2 + h 2/3 h 1/3 η 1 -2 |dη ′ | 2 , this metric gives symbol classes essentially as semiclassical symbol classes with h 1/3 for semiclassical parameter. We let to the reader to check that g is slowly varying and σ temperate. The "h" defined by Hörmander associated with g is h 1/3 h 1/3 η 1 -1 . It is the quantity we gain in the asymptotic expansion for the symbol calculus. In particular the function h 1/3 η 1 ν is a g continuous and σ, g temperate for every ν ∈ R. We refer to [26, Chapter 18, Sections 4 and 5] for definitions used freely here.

From (88) we have -h 2 ∂ 2

x d v h + op(R(x, hξ ′ ) -1)v h = hq h . Let v h = op(χ 1 (x, hξ ′ ))v h , where χ 1 is supported where χ = 1 and χ is the cutoff function define in Lemma B.2. By symbol calculus we have h 2 ∂ 2 x d v h + op(χ 2 (x, hξ ′ )(R(x, hξ ′ ) -1))v h = hq h (119) where q h = op(χ 1 (x, hξ ′ ))q hh -1 [h 2 D 2 x d , op(χ 1 (x, hξ ′ ))]v h + h -1op(χ 1 (x, hξ ′ )) op(R(x, hξ ′ ) -1) + op(χ 2 (x, hξ ′ )(R(x, hξ ′ ) -1)) op(χ 1 (x, hξ ′ )) is bounded on L 2 (x d > 0) from symbol calculus and support properties of χ and χ 1 . Moreover we have

|v h | L 2 (R d-1 ) |v h | L 2 (R d-1 ) .
We recall some properties of the Airy function which is denoted by Ai. It verifies the equation Ai ′′ (z)z Ai(z) = 0 for z ∈ C, Ai is real on the real axis and Ai(z) = Ai(z). Let ω = e 2iεπ/3 for ε = ±1.

For x ∈ R, let α(x) = -ω Ai ′ (ωx)/ Ai(ωx) ∈ C ∞ (R). As the zero of Ai are on the negative real axis, the function α is well defined for x ∈ R and smooth. The function α satisfies the following properties.

Lemma B.3. We have i) α(x) = -√ x + 1 4x + b 1 (x) for x > 0 ii) α(x) = εi √ -x + 1 4x + b 2 (x) for x < 0 iii) Re α(x) < 0 for all x ∈ R, iv) α satisfies the differential equation α ′ (x) = α 2 (x)x.

where b j ∈ S( x -5/2 , |dx| 2 ) for j = 1, 2.

The proof of lemma is given in Section C. Let rd be such that κ * rd = -∂ x d R. We assume that locally ∂ x d R < 0, this implies that rd > 0 in a neighborhood of (0, 0).

Let ã(y, η ′ ) = h 1/3 χ(y, hη ′ )r ). We assume that on the support of χ(y, hη ′ ), rd (y, hη ′ ) > 0. In what follows we denote ρ = (y, hη ′ ). We define à = op(ã), Ψ = h -1/3 op( h 1/3 η 1 -1/2 ), and let A = F -1 ÃF , Ψ = F -1 ΨF . We have ã ∈ S(h 1/3 h 1/3 η 1 1/2 , g) as h -1/3 h 1/3 η 1 -1/2 ∈ S(h -1/3 h 1/3 η 1 -1/2 , g) and from Lemma B.3 Proposition B.4. Let v h satisfying properties (119). There exist C 0 > 0 such that

|(h∂ x d v h -Av h ) |x d =0 | 2 L 2 (R d-1 ) + Ψ(h∂ x d v h -Av h ) 2 L 2 (x d >0) ≤ C 0 v h 2 H 1 sc (x d >0) + C 0 q h 2 L 2 (x d >0) .
This result is equivalent to the following. Let w h = F (h∂ x d v h -Av h ), there exist C 0 > 0 such that

|(w h ) |y d =0 | 2 L 2 (R d-1 ) + Ψw h 2 L 2 (y d >0) ≤ C 0 v h 2 H 1 sc (x d >0) + C 0 q h 2 L 2 (x d >0) .
Here and in this section we denote u H 1 sc (x d >0) = u L 2 (x d >0) + h∇u L 2 (x d >0) . Proof. To ease notation we write |u| instead of |u| L 2 (x d ) when there is no ambiguity on the fact that the L 2 norm is taken on variables x ′ or y ′ at point x d or y d . By the same abuse of notation we write the inner product (.|.) instead of (.|.) L 2 (R d-1 ) (x d ). We compute

1 2 ∂ x d |h∂ x d v h -Av h | 2 = Re ∂ x d (h∂ x d v h -Av h )|h∂ x d v h -Av h = Re h∂ 2 x d v h -(∂ x d A)v h -A∂ x d v h |h∂ x d v h -Av h = Re h -1 op(χ 2 (x, hξ ′ )(R(x, hξ ′ ) -1))v h -q h -(∂ x d A)v h -A∂ x d v h |h∂ x d v h -Av h ,
from (119). Then we have

1 2 ∂ x d |h∂ x d v h -Av h | 2 = I 1 + I 2 + I 3 + I 4 , (120)
where

I 1 = Re -h -1 A(h∂ x d v h -Av h )|h∂ x d v h -Av h I 2 = Re h -1 (op(χ 2 (x, hξ ′ )(R(x, hξ ′ ) -1)) -A 2 )v h |h∂ x d v h -Av h I 3 = Re -(∂ x d A)v h |h∂ x d v h -Av h I 4 = -Re q h |h∂ x d v h -Av h .
Using w h = F (h∂ x d v h -Av h ), we have

I 1 = Re -h -1 F AF -1 w h |w h = Re -h -1 Ãw h |w h .
From Lemma B.3 we have -h -1 Re ã(y, η) ≥ δh -2/3 χ2 (ρ) h 1/3 η 1 -1 , and h -1 Re ã(y, η) ∈ S(h -2/3 h 1/3 η 1 1/2 , g). Then from Fefferman-Phong inequality (see [START_REF] Hörmander | The analysis of linear partial differential operators[END_REF]Theorem 18.6.8]) and as the real part of symbol of op(h -1/3 χ(ρ) h 1/3 η 1 -1/2 ) * op(h -1/3 χ(ρ) h 1/3 η 1 -1/2 ) is h -2/3 χ2 (ρ) h 1/3 η 1 -1 modulo an operator bounded on L 2 , we have

I 1 ≥ δ|h -1/3 op( χ(ρ) h 1/3 η 1 -1/2 )w h | 2 -C|w h | 2 ,
for C > 0.

From Lemma C.1 we obtain

I 1 ≥ δ| Ψw h | 2 -C |w h | 2 + |v h | 2 + |h∂ x d v h | 2 . ( 121
)
We have The symbol of Ã2 is ã2 ∈ S(h 2/3 h 1/3 η 1 , g) modulo a term in S(h, g). From definition of ã and Lemma B.3 we have ã2 = h 2/3 χ2 (ρ)r We have h -1/3 χ2 (ρ)r Proof. We have

I 2 =
1 2 ∂ x d |C(h∂ x d v h -Av h )| 2 = Re ∂ x d (C(h∂ x d v h -Av h ))|C(h∂ x d v h -Av h ) = Re (∂ x d C)(h∂ x d v h -Av h ) + C(h∂ 2 x d v h -(∂ x d A)v h -A∂ x d v h )|C(h∂ x d v h -Av h ) = Re (∂ x d C)(h∂ x d v h -Av h ) + C(h -1 op(χ 2 (x, hξ ′ )(R(x, hξ ′ ) -1))v h -q h -(∂ x d A)v h -A∂ x d v h )|C(h∂ x d v h -Av h ) ,
from (119). Then we have

1 2 ∂ x d |C(h∂ x d v h -Av h )| 2 = J 1 + J 2 + J 3 + J 4 + J 5 = K 1 + K 2 + K 3 + K 4 + K 5 , (128)
where

J 1 = Re C * (∂ x d C)(h∂ x d v h -Av h )|h∂ x d v h -Av h J 2 = -Re (∂ x d A)v h |C * C(h∂ x d v h -Av h ) J 3 = -Re q h |C * C(h∂ x d v h -Av h ) J 4 = Re (-C * C)A∂ x d v h |h∂ x d v h -Av h J 5 = Re h -1 op(χ 2 (x, hξ ′ )(R(x, hξ ′ ) -1))v h |C * C(h∂ x d v h -Av h ) .
Taking as in the proof of Proposition B.4, w h = F (h∂ x d v h -Av h ) and from Lemma B.2 we write To estimate K 1 , observe that the symbol of C * (∂ y d C) is in ∈ S(h -1/3 h 1/3 η 1 -1/2 , g), then

K 1 = Re C * (∂ y d C)w h |w h K 2 = -Re (∂ y d Ã)F v h | C * Cw h -Re h[ Ã, B]F v h | C * Cw h -Re F op(b(x, hξ ′ ))v h | C * Cw h K 3 = Re F q h | C * Cw h
|K 1 | | Ψw h ||w h |. ( 129
)
From symbol calculus the symbol of (∂ y d Ã) * C * C is in S(1, g), thus this operator is bounded on L 2 . Clearly the terms [ Ã, B] coming from remainder term of ∂ x d A (see iv) Lemma B.2) and op(b(x, hξ ′ )) coming from remainder term of F op(χ 2 (x, hξ ′ )(R(x, hξ ′ ) -1))F -1 (see ii) Lemma B.2) are bounded on L 2 . As C * C has a symbol in S(h -1/3 h 1/3 η 1 -1/2 , g), we obtain

|K 2 | |w h ||v h | + |v h || Ψw h |. (130)
For the same argument we have To estimate the other terms of K 4 observe that the symbol of -h -1 C * C Ã + ih -1 C * [ C, op(θ)]) is -h -1 c2 ã + h -1 c{c, θ(ρ)} modulo a symbol in S(h -2/3 h 1/3 η 1 -1 , g) and this term can be estimate by | Ψw h | 2 . We compute The term h -1 ch -1/6 { χ2 (ρ), θ(ρ)}β(ζ) ∈ S(h -1/3 h 1/3 η 1 -1/2 , g), and the term of K 4 coming from this term can be estimate by | Ψw h ||w h |. For the other term we have h -1/3 {ζ, θ(ρ)} = {r

(ρ).

The term h -1 ch -1/6 χ 2 (ρ)β ′ (ζ)h 1/3 {r -2/3 d (ρ), θ(ρ)}η 1 ∈ S(h -1/3 h 1/3 η 1 -1/2 , g) and the term of K 4 coming from this term can be estimate by | Ψw h ||w h |.

Thus, modulo remainder terms, the symbol of -h -1 C * C Ã + ih -1 C * [ C, op(θ)]) is given by L =h -1 c2 ã + h -1 ch -1/6 χ2 (ρ)β ′ (ζ)h 1/3 {η 1 , θ(ρ)}r

The last term is K 5 . The symbol of h -1 Ã2 + ih -1 [ Ã, op(θ(ρ))]h -1 op(hη 1 χ2 (ρ)) is h -1 ã2 + h -1 {ã, θ(ρ)}η 1 χ2 (ρ) modulo a symbol in S(1, g). We have {ã, θ(ρ)} = h 1/3 α(ζ){( χr (133)

The proof of the claim is given below. With this claim, ( 126) and what we do above, the operator h -1 Ã2 + ih -1 [ Ã, op(θ(ρ))]h -1 op(hη 1 χ2 (ρ))) gives a term bounded by |v h |. As the symbol of C * C is in S(h -1/3 h 1/3 η 1 -1/2 , g), we obtain that

|K 5 | |v h || Ψw h |. (134)
From ( 127), ( 128), ( 129), ( 130), ( 131), ( 132) and (134) we obtain

1 2 ∂ x d |C(h∂ x d v h -Av h )| 2 -|q h | 2 + | Ψw h | 2 + |v h | 2 + |h∂ x d v h | 2 .
Integrating this inequality between 0 and σ > 0, we have, estimating the term coming from | Ψw h | by Proposition B.4,

|C(h∂

x d v h -Av h )| 2 (0) q h 2 + v h

  k and b ∈ S m , Op sc (a) Op sc (b) = Op sc (c), where c ∈ S m+k , and c admits an asymptotic expansion, c(x, ξ) = a(x, ξ)b(x, ξ) + hd(x, ξ), where d ∈ S m+k-1 . [Op sc (a), Op sc (b)] = h Op sc (c), where c ∈ S m+k-1 , and c admits an asymptotic expansion, c(x, ξ) = -i{a, b}(x, ξ) + hd(x, ξ), where d ∈ S m+k-2 , and {a, b}(x, ξ) = d j=1 ∂ ξj a(x, ξ)∂ xj b(x, ξ)-∂ xj a(x, ξ)∂ ξj b(x, ξ) is the Poisson bracket. At some point, it is useful to use that the commutator between an operator and a derivative admits an exact formula, we have [hD xj , Op sc a] = -ih Op sc (∂ xj a). For a ∈ S k , we have Op sc (a) * = Op sc (b) and b(x, ξ ′ ) = ā(x, ξ) + hc(x, ξ) where b ∈ S k , c ∈ S k-1 .

  0), and κ * (η 1 ) = R -1. Moreover x d acts as a parameter and κ is smooth with respect x d = y d .

1 / 3 d 3 d

 133 (y, hη ′ )α(ζ) where ζ = h 1/3 η 1 r-2/(y, hη ′

Re h - 1 F

 1 (op(χ 2 (x, hξ ′ )(R(x, hξ ′ ) -1) -A 2 )v h |w h = Re h -1 (op(hη 1 χ2 (ρ)) -Ã2 )F v h |w h + Re B 0 v h |w h ,where B 0 = op sc (b) is bounded on L 2 (see Lemma B.2).

2 / 3 d 3 d 3 d

 2333 (ρ)α 2 (ζ) = h 2/3 χ2 (ρ)r 2/(ρ) ζ + α ′ (ζ) = hη 1 χ2 (ρ) + h 2/3 χ2 (ρ)r 2/(ρ)α ′ (ζ).

2 / 3 d

 23 (ρ)α ′ (ζ) ∈ S(h -1/3 h 1/3 η 1 -1/2 , g), we then obtain |I 2 | |v h | | Ψw h | + |w h |(122)We haveI 3 = -Re F (∂ x d A)F -1 F v h |w h = -Re (op(∂ y d ã(ρ)) + ih -1 [op(ã(ρ)), op(θ(ρ))]F v h |w h -Re h[op(ã(ρ)), B]F v h |w h , from Lemma B.2.

K 4 =K 5 =

 45 Re (-h -1 C * C Ã + ih -1 C * [ C, op(θ(ρ))])w h |w h + Re h C * [ C, B]w h |w h -Re (h -1 Ã2 + ih -1 [ Ã, op(θ(ρ))]h -1 op(hη 1 χ2 (ρ)))F v h | C * Cw h .

|K 3 |

 3 |q h || Ψw h |. (131) To estimate the last term of K 4 we write C * [ C, B] = C * CB -C * B C, the first term gives a term estimated by | Ψw h ||w h | and the second is estimated | Cw h | 2 | Ψw h ||w h |.

h 1 / 6

 16 {c, θ(ρ)} = { χ2 (ρ), θ(ρ)}β(ζ) + {β(ζ), θ(ρ)}χ 2 (ρ) = { χ2 (ρ), θ(ρ)}β(ζ) + {ζ, θ(ρ)}χ 2 (ρ)β ′ (ζ).

-2/ 3 d

 3 (ρ), θ(ρ)}η 1 + {η 1 , θ(ρ)}r-2/3 d

= 3 d 3 d 3 d

 333 -h -1 h -1/3 χ2 2 (ρ)β 2 (ζ)h 1/3 χ(ρ)r 1/(ρ)α(ζ) h -1 h -1/6 χ2 (ρ)β(ζ)h -1/6 χ2 (ρ)β ′ (ζ)h 1/3 rd r-2/(ρ) from v) of Lemma B.2.We thus obtainL = h -1 χ2 2 (ρ)β(ζ)r 1/(ρ)β(ζ) χ(ρ)α(ζ)β ′ (ζ) ∈ S(h -1 , g). As χ is equal 1 on the support of χ2 2 , β ≥ 0 and -β(ζ) Re α(ζ)β ′ (ζ) ≥ 0 we have Re L ≥ 0. We can apply sharp Gårding inequality (see [26, Theorem 18.6.7]), we yield, taking account remainder terms K 4 ≥ -C | Ψw h | 2 + | Ψw h ||w h | .

1 / 3 d 1 1/ 2 1 3 d 3 d 3 d 3 d(ρ)η 1 -h - 1 / 3 ( χr 2 / 3 d

 1312333311323 )(ρ), θ(ρ)} + h 2/3 ( χr1/3 d )(ρ)α ′ (ζ)η 1 {r -2/3 d (ρ), θ(ρ)} + h 2/3 ( χr -1/3 d )(ρ)α ′ (ζ){η 1 , θ(ρ)}.The first two terms give a term estimated by h 1/3 h 1/3 η as on the support of χ we have |η 1 | h -1 . Then both terms give associated operators bounded on L 2 . Modulo a bounded operator on L 2 we have to consider the symbol, taking account v) of Lemma B.2h -1/3 χ2 (ρ)r 2/(ρ)α 2 (ζ)h -1/3 ( χr 2/)(ρ)α ′ (ζ)η 1 χ2 (ρ) = h -1/3 χ2 (ρ)r 2/(ρ) α 2 (ζ)α ′ (ζ)h 1/3 r-2/)(ρ)(1χ(ρ))α ′ (ζ). The first term is null from differential equation satisfying by α and the value of ζ. We claim that | op h -1/3 ( χr 2/3 d )(ρ)(1χ(ρ))α ′ (ζ) F v h | |v h |.

2 H 1 sc

 21 (x d >0) + |C(h∂ x d v h -Av h )| 2 (σ). (135) As |C(h∂ x d v h -Av h )| 2 (σ) = ( C * Cw h , w h ),we have from (127)|C(h∂ x d v h -Av h )| 2 (σ) | Ψw h | 2 (σ) + |v h | 2 (σ) + |h∂ x d v h | 2 (σ)Integrating estimate (135) between two positive values of σ and estimating as above the term | Ψw h | 2 (σ), we obtain the conclusion of Proposition B.6.

  1 dxdy, and [∂ xj , -h 2 P + z] is a sum of term of type h 2 ∂ xj b∂ x k , ch 2 ∂ xj and dh 2 where b, c and d are in C ∞ (Ω). The terms with ch 2 ∂ xj and dh 2 were estimated in[START_REF] Rousseau | Applications to unique continuation and control of parabolic equations[END_REF] and[START_REF] Rousseau | Geometric control condition for the wave equation with a time-dependent observation domain[END_REF]. For the term h 2 ∂ xj b∂ x k , we have

  1, |dx| 2 + |dξ ′ | 2 ) or in S(1, |dy| 2 + |dη ′ | 2 ). In this section we only use tangential symbol, but as in what follows we have to use different classes of symbols, we prefer use everywhere the same kind of notation. Lemma B.2. Associated with κ, there exists F a semiclassical Fourier Integral Operator satisfying the following properties, i) F is a unitary operator uniformly with respect x d

Rigorously, the laplace-Beltrami operator has a term h

Dx d . We can eliminate this term after a conjugaison of operator by a function non null everywhere but this has no influence on the proof given here. For simplicity we choose to keep v h ) instead of the conjugated function.

x d = 0 such that H p (µ1

x d =0 ⊗ µ k . Suppose that n ≥ 1 and let χ ∈ C ∞ 0 (R), be such that χ (k) (0) = 0 for k = 0, . . . n -1 and χ (n)

), we have

We also have

which is bounded uniformly with respect ε and supported in a fixed compact set. Then if n ≥ 2, H p ε n χ(x d /ε)b(x, ξ ′ ) → 0 everywhere as ε → 0 and (102), Lebesgue's dominated convergence theorem imply that µ n = 0 for n ≥ 2. If n = 1, H p εχ(x d /ε)b(x, ξ ′ ) → 2ξ d 1 x d =0 b(x, ξ ′ ) everywhere as ε → 0. Lebesgue's dominated convergence theorem and (102) imply that

as µ1 x d >0 , 2ξ d 1 x d =0 b(x, ξ ′ ) = 0 we find that µ 1 = 0. Then we have H p (µ1 x d >0 ) = δ x d =0 ⊗ µ 0 , where µ 0 is a distribution of order 1.

If H p x d = 0, let (x(s; x ′ , ξ), ξ(s; x ′ , ξ)) the solution to ( ẋ, ξ) = H p satisfying the initial condition (x(0; x ′ , ξ), ξ(0; x ′ , ξ)) = (x ′ , 0, ξ). We verify that the map (s, x ′ , ξ) → (x(s; x ′ , ξ), ξ(s; x ′ , ξ)) locally is one to one and transforms ∂ s in H p . Moreover, s = 0 is transformed in x d = 0 and if H p x d > 0, s > 0 is transformed in x d > 0, if H p x d < 0, s < 0 is transformed in x d > 0. In coordinates (s, x ′ , ξ) the equation H p (µ1

, where we keep the notations µ, µ 0 in variables (s, x ′ , ξ) for the images of µ, µ 0 . If H p x d > 0, we have µ1 s>0 = (1 s>0 ds) ⊗ µ 0 and if H p x d < 0, we have µ1 s<0 = -(1 s<0 ds) ⊗ µ 0 . As µ is non negative, we obtain that µ 0 is a measure and its sign.

At the hyperbolic region the measure µ 0 has a particular structure given by this lemma. Lemma 4.11. We assume that aµ = 0. Let (x ′ 0 , ξ ′ 0 ) be a hyperbolic point (i.e. R(x ′ 0 , 0, ξ ′ 0 ) < 1). Locally in a neighborhood of (x ′ 0 , ξ ′ 0 ), there exist µ + and µ -non negative measures on R d-1

x ′ × R d-1 ξ ′ such that µ 0 = µ + ⊗δ ξ d = √ 1-R(x ′ ,0,ξ ′ ) -µ -⊗δ ξ d =- √ 1-R(x ′ ,0,ξ ′ ) . Moreover, if in a neighborhood of x ′ 0 , (v h ) h satisfies the Dirichlet boundary condition or the Neumann boundary condition then µ + = µ -.

In a neighborhood of Γ we have the following property, if µ + = 0 (resp µ -= 0) then µ -= 0 (resp µ + = 0).

Proof. As µ is supported on ξ 2 d + R(x, ξ ′ ) -1 = 0, this implies that δ x d =0 ⊗ µ 0 is supported on ξ 2 d + R(x, ξ ′ ) -1 = 0, then µ 0 is supported on

where X is a vector field tangent to

is supported on a part of the boundary such that (v h ) h satisfies the Dirichlet boundary condition (resp. Neumann boundary condition), then the right hand side of (91) is 0. Then we have

) positive. Here we use notation defined above Formula (64). As

by support properties and passing to the limit we have b

, as µ + is a non negative measure we obtain µ + = 0 in a neighborhood of ρ.

Then 1 G d µ ∂ = 0 on ∂Ω D as ℓ can be arbitrary chosen.

We now assume ℓ supported on a neighborhood of a point of ∂Ω N . With A ε defined in (106) we have, as (hD

From Proposition B.7, A ε goes to 0 as ε → 0. Then from (105) and Proposition 4.8 we obtain (107) in this case and 1

This lemma describes how the support of the boundary measure propagates along the boundary.

, where μ0 is a distribution and μ1 is a Radon measure. Then

as

As µ 0 is of order 1 and supported on ξ d = 0, we have

, where μj are distributions. To prove that μ1 is a Radon measure, we test

When ε → 0 the supremum on ϕ is estimated by supremum of ψ(x ′ , ξ ′ ). We also have µ 0 , ϕ converging to μ1 , ψ(x ′ , ξ ′ ) as ε → 0. We deduce that μ1 , ψ(x ′ , ξ ′ ) is estimated by the supremum of ψ(x ′ , ξ ′ ), this implies that μ1 is a Radon measure.

By Proposition 4.

) is supported in ∂Ω D ∪ ∂Ω N and in a neighborhood of (x ′ 0 , 0, ξ ′ 0 , 0), we have

), and χ ∈ C ∞ 0 (R), be such that χ(σ) = 1 in a neighborhood of 0. From (108) we obtain

) is uniformly bounded and bχ(x d /ε) converges to b1 x d =0 everywhere, we obtain

We deduce that H ′ R µ ∂ -2aµ ∂ + μ0 = 0, which gives the first conclusion of Lemma. We deduce from (108)

We then can write

). Now we choose an adapted b to apply (110). Let χ ∈ C ∞ 0 (R) be such that χ(σ) = 1 for σ in a neighborhood of 0. Let ℓ ∈ C ∞ 0 (x ′ , ξ ′ ) be supported in a neighborhood of (x ′ 0 , ξ ′ 0 ). We set b(x, ξ) = ξ d ℓ(x ′ , ξ ′ )χ(ξ d /ε)χ(x d /ε), where ε > 0. We have

Propagation in interior domain Ω

Here we use the fact that aµ = 0. From Proposition 4.1, we have H p µ = 0. It is then classical that µ in invariant by the flow of H p . More precisely, let ρ 0 ∈ T * Ω and we assume that γ(s, ρ 0

Propagation at boundary: hyperbolic points

The propagation results given in this section are classical for Dirichlet boundary condition. We prove a propagation result for Neumann boundary and in a neighborhood of Γ which is new in context of semiclassical measure. We use the geometry context defined in section 1.2, in particular j and the definition of the different flows.

We prove that the support of measure is locally empty in the future assuming that in the past the support of measure is locally empty but by symmetry we can deduce that the support of measure is locally empty in the past if we assume that the support of measure is locally empty in the future.

Recall that we choose coordinates such that p(x, ξ) = ξ 2 d + R(x, ξ ′ ) -1 and locally Ω = {x d > 0}. In this section we use that aµ = 0.

We recall that a point (

We apply Lemmas 4.9 and 4.11. We have µ = 1 x d >0 µ and

We call γ ± the integral curve of

If we assume that the support of µ = 1 x d >0 µ is empty in a neighborhood of γ -(s) for s < 0 and |s| sufficiently small, then µ -= 0. This implies µ + = 0 and H p µ = 0. As µ |x d <0 = 0 and γ + (s) is in x d < 0 for s < 0 and |s| sufficiently small, this implies that µ = 0 in a neighborhood of γ + (0). Remark 9. In a neighborhood of points in ∂Ω D ∪ ∂Ω N we can prove a propagation of measure because we have proved µ + = µ -but we do not know if this property is true for points in Γ.

Propagation at ∂Ω

Propagation at gliding points

We recall that a point (

Let γ be the integral curve of H p starting from (x ′ 0 , 0, ξ ′ 0 , 0). Then γ(s) into {x d < 0} for s = 0 and |s| sufficiently small. In a neighborhood of (x ′ 0 , ξ ′ 0 ) in T * ∂Ω all the point are either hyperbolic, or gliding. We assume that j -1 γ g (s 0 ; x ′ 0 , ξ ′ 0 ) ∩ supp µ = ∅ for s 0 < 0 where |s 0 | is sufficiently small. Here γ g (s;

, then all the point ρ in a neighborhood of j -1 γ g (s 0 ; x ′ 0 , ξ ′ 0 ) are not in the support of µ. By continuity of Γ the curve Γ(s; ρ) hit the boundary at ρ ′ in a neighborhood of (x ′ 0 , ξ ′ 0 ). If ρ ′ is an hyperbolic point, by the previous result the point j -1 Γ(s; ρ) are not in the support of µ. If ρ ′ is a gliding point, all the points Γ(s; ρ) are strictly gliding. In particular this implies that µ is supported on x d = 0, then 1 x d >0 µ = 0 and µ 0 = 0. We can apply Lemma 4.13 and

Propagation at diffractive points

We recall that a point (

We keep the previous notation for γ. For a point ρ in a neighborhood of (x ′ 0 , ξ ′ 0 ) ∈ T * ∂Ω ∪ T * Ω, there are three cases, first the integral curve passing through ρ hits x d = 0 at an hyperbolic point and by previous result the integral curve is not in support of µ, second it does not hit x d = 0 and the integral curve is not in the support of µ by propagation result in interior, third the integral curve hits x d = 0 at a diffractive point. Then the support of µ ∂ is in G g and the support of 1

If (x ′ 0 , 0) ∈ ∂Ω D ∪∂Ω N we can apply Lemma 4.12, then µ ∂ = 0 in a neighborhood of (x ′ 0 , 0, ξ ′ 0 ). As the integral curves hitting x d = 0 at an hyperbolic points are not in the support of 1 x d >0 µ, then µ 0 is supported on ξ d = 0. We can apply Lemma 4.13 to obtain µ 0 = 0. Then H p µ = H p (1 x d >0 µ) = 0 and as, by assumption, γ(s) is not in the support of µ for s < 0, |s| sufficiently small, we deduce that γ(0) is not in the support of µ.

Propagation at boundary: integral curves with high contact order

We recall that if

. By the assumption made (see Definition 1.1) there exist k ∈ N, k ≥ 3 and α = 0 such that

. For each k we assume that we have already proved that the integral curves hitting x d = 0 at a point in G j for j < k or H are not in the support of µ.

Case k even, α < 0 The integral curve of H p starting from a point belonging to T * Ω in a neighborhood of (x ′ 0 , ξ ′ 0 ) in T * ∂Ω ∪ T * Ω eventually hits x d = 0 at a point ρ ′ , in H or G j for j ≤ k (see Section 1.2 for definition of G j ). By assumptions and by induction this integral curve is not in the support of µ except if ρ ′ is in G k , but in this case this integral curve is in x d ≤ 0. This implies that 1 x d >0 µ = 0, then µ 0 = 0. By Lemma 4.13, we have H ′ R µ ∂ = 0 and as, by assumption, γ g (s) is not in the support of µ ∂ for s < 0, |s| sufficiently small, we deduce that γ g (0) is not in the support of µ ∂ .

Case k odd, α < 0 By the same argument as in previous case, the integral curve of H p starting from a point belonging to T * Ω in a neighborhood of (x ′ 0 , ξ ′ 0 ) in T * ∂Ω ∪ T * Ω hits G k or is not in the support of µ. Denote by ρ ′ the point of this integral curve hitting x d = 0. The generalized bicharacteristic starting from ρ ′ is on x d = 0 for s > 0 and in x d > 0 for s < 0, and for s > 0 all the points on the integral curve of H ′ R are in G g , if |s| is sufficiently small. As, by assumption the generalized bicharacteristic is not in support of µ in the past, this means that 1 x d >0 µ = 0 then µ 0 = 0. We can apply Lemma 4.13 then H ′ R µ ∂ = 0. But γ g (s) is not in support of µ = µ ∂ for s < 0 and |s| sufficiently small as γ g (s) ∈ G d , then γ g (0) is not in the support of µ ∂ .

Case k even, α > 0 By induction, only the generalized bicharacteristics with the same order of contact k and the same sign condition α > 0 can be in the support of µ. Applying Lemma 4.13 we have -∂

We deduce H p µ = 0. Then the propagation the support of µ is invariant by the flow of H p .

Case k odd, α > 0 By induction, only the generalized bicharacteristics with the same order of contact k and the same sign condition α > 0 can be in the support of µ. We can apply Lemma 4.13, and by the same argument used in the previous case, we have H p µ = 0 and as γ(s) is in x d < 0 for |s| < 0 sufficiently small, γ(s) is not in the support of µ for s < 0 and by propagation γ(0) is not in support of µ.

Proof of Proposition 5.1, first case. By assumption for a point ρ

we can apply the results obtained in this section to prove that j -1 Γ(s 1 , ρ) ∩ supp µ = ∅, and reach a contradiction.

Propagation on Γ

Now we prove Proposition 5.1 in the second case, i.e. π(ρ) ∈ Γ. We recall that we can change the coordinates such that locally in a neighborhood of Γ we have Ω = {x d > 0} and Γ =

. By the result obtained from the previous section, supp µ ⊂ {x 1 = 0,

We have to distinguish two cases, first if R 0 (x ′′ , ξ ′′ ) -1 < 0 and second R 0 (x ′′ , ξ ′′ ) = 1 even if the result is the same in both cases.

where

). We have

Then µ + = µ -as ξ 1 (x ′′ , ξ ′′ ) = 0 in a neighborhood of (x ′′ 0 , ξ ′′ 0 ). Now we take ℓ(x ′ , ξ ′′ ) = b(x ′′ , ξ ′′ )χ(x 1 ). Observe that (∂ ξ1 S(x ′ , ξ ′ )) = x 1 ∂ ξ1 r 2 (x ′ , ξ ′ ) is null on x 1 = 0, we deduce that

then, as µ + = µ -, H ′′ R0 µ + = H ′′ R0 µ -= 0 with the previous equation. Then the support of µ + , µ - and µ ∂ propagate along the integral curves of H ′′ R0 . By assumption mGCC (see Definition 1.2) all these curves hit the set a ≥ δ > 0, we obtain that µ ∂ = 0 in a neighborhood of such a point (x ′ 0 , ξ ′ 0 ).

Case

By the result obtained in previous section the measure µ ∂ is supported on x 1 = ξ 1 = 0, then we have

, where μ is a non negative Radon measure. We deduce from (112), as ∂ ξ1 S(x ′ , ξ ′ ) = 0 on x 1 = 0,

Taking ℓ(x ′ , ξ ′′ ) = b(x ′′ , ξ ′′ )χ(x 1 ), where χ ∈ C ∞ 0 (R) and χ(s) = 1 for s in a neighborhood of 0, by Lemma 4.14, and arguing as in the previous case, we have H ′′ R0 μ = 0, then the supports of μ and µ ∂ propagate along the integral curves of H ′′ R0 . As in the previous case we obtain µ ∂ = 0 in a neighborhood of such a point (x ′ 0 , ξ ′ 0 ).

Observe that ã(ρ) ∈ S(1, g) as h 1/3 h 1/3 η 1 1/2 is bounded on support of χ(ρ). Then op(∂ y d ã(ρ)) and h[op(ã(ρ)), B] are bounded operator on L 2 .

From properties of α and symbol calculus, the symbol of

Then we obtain

We have

From ( 120), ( 121), ( 122), ( 123) and ( 124) we have

Observe that

We also have

We deduce from (125) that

Integrating this inequality between 0 and σ > 0 we have

from (127). Integrating this inequality between two positive values of σ and as

) we obtain the result.

To state the next result we have to introduce another operator.

Lemma B.5.

There exist a function β ∈ C ∞ (R) satisfying the following properties

A proof is given in Section C. We recall the notation ã(y, η ′ ) = h 1/3 χ(ρ)r Let c(y, η ′ ) = h -1/6 χ2 (ρ)β(ζ), where χ2 is supported on { χ = 1} and χ2 = 1 on a neighborhood of (0, 0). We have ã ∈ S(h 1/3 h 1/3 η 1 1/2 , g) and c ∈ S(h -1/6 h 1/3 η 1 -1/4 , g). We define C = op(c) and C = F -1 CF .

Proposition B.6. Let v h satisfying properties (119). There exists C 0 > 0 such that

Proof of Claim (133). As κ * χ1 = χ 1 , from Lemma B.2 we thus have

where K is bounded on L 2 . We then have

The first term coming from op( χ1 (ρ)) gives an operator with null symbol. As

then the second term is also bounded by |v h |. This proves the claim.

be supported on a neighborhood of 0 and

) supported on {χ 2 = 1} for every x d , where κ * χ2 = χ 2 . We moreover assume

Proof. We can assume that the support of ℓ is contained in {χ 1 = 1}. We then have from symbol calculus

By symbol calculus and the support properties of χ and χ2

modulo an operator with symbol in S(h 1/2 h 1/3 η 1 -3/4 , g). From properties of traces, see (88), this remainder term goes to 0 as h to 0.

Let l be such that κ * l = ℓ and κ * χ 4 (η

In what follow, to be coherent with our notation we define χ4 = χ 4 and we use the notation χ4 when the function is defined in (y, η) variables.

From Lemma B.2 we have

where r 0 is of order 0. The term coming from r 0 goes to 0 as h to 0, from properties of traces. Then it is sufficient to prove that lim ε→0 lim h→0 op(hη

) and from support properties of l and χ2 , we have

, modulo an operator with symbol in S(h, g) then this last term involves a term going to 0 as h to 0. Then we obtain an estimation, modulo a term going to 0 as h to 0,

, and this gives the sought estimate. We have for k ≥ 1,

As hη 1 is bounded on the support of χ(k) 4 (hη 1 /ε) both terms are bounded by h k-1/3 . From estimate h 2/3 h 1/3 η 1 hη 1 we have

which proves the claim. Observe that the constant of estimation are not uniform with respect ε.

With the previous claim and as

As

, on the support of χ4 (hη 1 /ε), we deduce from Gårding inequality (see [START_REF] Hörmander | The analysis of linear partial differential operators[END_REF]Theorem 18.6.7]) that the operator norm from

where C is independent of ε and C ε may depend on ε. From that and (136) we deduce the result.

C Proof of Lemmas

Lemma C.1. Let χ3 ∈ C ∞ 0 (U 1 ) be such that (1 -χ3 ) χ1 = 0 where κ * χ1 = χ 1 . We have

Proof. We write Ψ = h -1/3 op( χ3 (ρ)

), it suffices to estimates terms coming from the second operator. Recall that w h = F (h∂

, g) and ã ∈ S(h 1/3 h 1/3 η 1 1/2 , g), we have

We have

The second term gives

As κ * χ1 = χ 1 , from Lemma B.2 we thus have F -1 op( χ1 (ρ))F = op(χ 1 (x, hξ ′ )) + hK, where K is bounded on L 2 . Then the first term at the right hand side of (137) gives

as the both terms are bounded on L 2 indeed the asymptotic expansion of the first symbol is null and the second is bounded by h 2/3 times an operator bounded on L 2 . This concludes the proof of the lemma.

Proof of Lemma B.2. We recall that Zworski use Weyl quantification. We give the proof in this context. From that, it is easy to obtain the result for classical quantification. We denote by op w sc (a) the operator associated with symbol a by the Weyl quantification. Items i) and ii) come for Zworski [START_REF] Zworski | Semiclassical analysis[END_REF]Theorem 11.5]. To prove the others Items we have to use the construction of F given by Zworski [START_REF] Zworski | Semiclassical analysis[END_REF]Section 11.1 and 11.2]. Let κ t a smooth family of symplectic transformations, t ∈ [0, 1], κ 0 = Id and κ 1 = κ and q t ∈ C ∞ 0 (U 0 ) be real valued, such that ∂ t κ t = (κ t ) * H qt (see [START_REF] Zworski | Semiclassical analysis[END_REF]Theorem 11.4]). Let Q(t) = op w sc (q t ) (here as we use the Weyl quantification, Q(t) is selfadjoint). Let F (t) the solution of

The Fourier Integral Operator we search, is F = F (1). We then have for

The Duhamel formula yields

Taking q0 (t) such that κ * t q0 (t) = ∂ x d q(t) which it is possible as κ t is a diffeomorphism, we have from Item ii), F -1 (σ) op w sc (q 0 (σ))F (σ) = ∂ x d Q(σ) + h op w sc (q 1 ). We can repeat the construction taking κ * t q1 (t) = q 1 (t) and we have

, where B(σ) is bounded on L 2 uniformly with respect σ. From that and taking t = 1 we deduce Item iii).

To prove Item iv) we have

Which gives Item iv) as there exist a symbol θ 1 such that op sc (θ 1 ) = op w sc (θ). Let χ 1 and χ 2 be C ∞ 0 functions such that χ 1 = 1 and χ 2 = 1 in a neighborhood of (x ′ 0 , 0, ξ ′ 0 ), and we assume χ 2 supported on {χ 1 = 1}. Applying iv) to A = (R -1)χ 1 , taking account ii), we obtain κ * ({η 1 , θ} χ2 (y, η ′ )) = χ 2 ∂ x d R where κ * χ2 = χ 2 . We deduce Item v) taking the previous formula on {χ 2 = 1}.

Proof of Lemma B.3. The asymptotic expansion of Ai is well-known (see [START_REF] Abramowitz | Handbook of mathematical functions with formulas, graphs, and mathematical tables[END_REF]Formula 10.4.59]. We recall that for z ∈ C with | arg z| < π we have

. From that we obtain

where f 0 = ℓ 0 = 1, f 1 = 1/6 and ℓ 1 = 1/4.

For x > 0, as ω 3/2 = -1, we have,

As this asymptotic expansion is also valid for derivative with respect x we deduce Item i).

For x < 0, we have ωx = -e 2iεπ/3 |x| = e -iεπ/3 |x| to have | arg(ωx)| < π. As -ωe -iεπ/6 = -εi we obtain

This gives Item ii) from properties of this asymptotic expansion.

Let F (z) = Ai ′ (z)/ Ai(z) we have F ′ (z) = z -A 2 (z) for z different of a zero of Ai which are on the negative real axis. As α(x) = -ωF (ωx) we have

This gives Item iv).

Item iii) is probably classical but we do not find this property in literature. Here we give a proof of that. Let α 1 (x) and α 2 (x) real valued be such that α(x) = α 1 (x) + iα 2 (x). We have

We also use a nice formula given in [45, Section 3]

where ν j 's are the zeros of Ai (observe that ν j < 0) and C 1 ∈ R is an explicit negative constant. We deduce

It is easy to prove that both series converge. We fix ε = 1, observe that α(x) is the α(x) defined with ε = -1. Observe that α 2 (x) > 0, indeed if α 2 (x 0 ) = 0 for some x 0 ∈ R, α 2 is identically null by uniqueness of System (139). Then α 2 (x) > 0 as it is true for x ≪ 0 from Item ii).

We have

We deduce that

Assuming x ≤ 0 we have α ′ 2 (x) < 0 and from (139), α 1 (x) < 0. Now for x ≥ 0 we compute (α 1 /α 2 ) ′ (x) = -α 2 1 (x)α 2 (x)α 3 2 (x)xα 2 (x) α 2 2 (x)

< 0.

This implies (α 1 /α 2 )(x) ≤ (α 1 /α 2 )(0) < 0 then α 1 (x) < 0. This concludes Item iii). .

From definition of γ 2 we have for x > 0, γ 2 > 0, γ ′ 2 < 0 and as Re α < 0 we have γ ′ 2 + γ 2 Re α < 0 . Let χ 0 ∈ C ∞ (R) be such that χ 0 (x) = 0 if x ≤ 0, χ 0 (x) = 1 if x ≥ 1 and we assume χ ′ 0 ≥ 0 on R. Let β = χ 0 γ 2 + (1χ 0 )γ 1 . Clearly β is a smooth function. We have β ′ = χ 0 γ ′ 2 + (1χ 0 )γ ′ 1 + χ ′ 0 (γ 2γ 1 ). As γ 1 > 0, if C 0 is chosen sufficiently small, γ 2γ 1 < 0 on the support of χ ′ 0 . As χ ′ 0 ≥ 0, χ ′ 0 (γ 2γ 1 ) ≤ 0. This and above properties imply i). We only have to prove ii) for x < 0. That is a consequence of asymptotic expansion of Airy function (138). Indeed for x < 0, ωx = |x|e -iεπ/3 , then (ωx) 3/2 = -εi|x| 3/2 . We then have

3 (ωx) 2/3 . Clearly the asymptotic expansion satisfies symbol estimates. This asymptotic expansion also gives estimates iii) for x < 0 with |x| sufficiently large. For x > 0 iii) is obvious, and by construction β > 0. This achieves the proof.