El Maati Ouhabaz

A "milder" version of Calderón's inverse problem for anisotropic conductivities and partial data

Keywords: AMS Subject Classification: 35P05, 35R30, 47D05, 47G30

Given a general symmetric elliptic operator

we define the associated Dirichlet-to-Neumann (D-t-N) map with partial data, i.e., data supported in a part of the boundary. We prove positivity, L p -estimates and domination properties for the semigroup associated with this D-t-N operator. Given La and L b of the previous type with bounded measurable coefficients a = {a kj , a k , a0} and b = {b kj , b k , b0}, we prove that if their partial D-t-N operators (with a0 and b0 replaced by a0 -λ and b0 -λ) coincide for all λ, then the operators La and L b , endowed with Dirichlet, mixed or Robin boundary conditions are unitarily equivalent. In the case of the Dirichlet boundary conditions, this result was proved recently by Behrndt and Rohleder [6]. We provide a different proof, based on spectral theory, which works for other boundary conditions.

Introduction

Let Ω be a bounded Lipschitz domain of R d with boundary ∂Ω. Let Γ 0 be a closed subset of ∂Ω with Γ 0 = ∂Ω and Γ 1 its complement in ∂Ω. We consider the symmetric elliptic operator on L 2 (Ω) given by the formal expression:

L a (λ) := d k,j=1 ∂ k (a kj ∂ j ) + d k=1 a k ∂ k -∂ k (a k .) + a 0 -λ
where a kj = a jk , a k , a 0 = a 0 ∈ L ∞ (Ω) and λ is a constant. We define the associated Dirichlet-to-Neumann (D-t-N) operator, N Γ1,a (λ), with partial data as follows: for ϕ ∈ H Let us consider first the case where a kj = σ(x)δ kj , a k = 0, k = 0, 1 . . . d, where σ ∈ L ∞ (Ω) is bounded from below (by a positive constant). A. Calderón's well known inverse problem asks whether one could determine solely the conductivity σ(x) from boundary measurements, i.e., from N Γ1 (0). For the global boundary measurements, i.e., Γ 1 = ∂Ω, the first global uniqueness result was proved by Sylvester and Uhlmann [START_REF] Sylvester | A global uniqueness theorem for an inverse boundary value problem[END_REF] for a C 2 -smooth conductivity when d ≥ 3. This results was extended to C 1+ -smooth conductivity by Greenleaf, Lassas and Uhlmann [START_REF] Greenleaf | The Calderon problem for conormal potentials, I: Global uniqueness and reconstruction[END_REF] and then by Haberman and Tataru [START_REF] Haberman | Uniqueness in Calderón's problem with Lipschitz conductivities[END_REF] to C 1 and Lipschitz conductivity close to the identity. Haberman [START_REF] Haberman | Uniqueness in Calderón's problem for conductivities with unbounded gradient[END_REF] proved the uniqueness for Lipschitz conductivity when d = 3, 4 and this was extended to all d ≥ 3 by Caro and Rogers [START_REF] Caro | Global uniqueness for the Calderón problem with Lipschitz conductivities[END_REF]. In the two-dimension case with C 2 -smooth conductivity, the global uniqueness was proved by Nachman [START_REF] Nachman | Global uniqueness for a two-dimensional inverse boundary value problem[END_REF]. This regularity assumption was completely removed by Astala and Päivärinta [START_REF] Astala | Calderón's inverse conductivity problem in the plane[END_REF] dealing with σ ∈ L ∞ (Ω). The inverse problem with partial data consists in proving uniqueness (either for the isotropic conductivity or for the potential) when the measurement is made only on a part of the boundary. This means that the trace of the solution u in (1.1) is supported on a set Γ D and the D-t-N operator is known on Γ N for some parts Γ D and Γ N of the boundary. This problem has been studied and there are some geometric conditions on Γ D and Γ N under which uniqueness is proved. We refer to Isakov [START_REF] Isakov | On uniqueness in the inverse conductivity problem with local data[END_REF], Kenig, Sjöstrand and Uhlmann [START_REF] Kenig | The Calderón problem with partial data[END_REF], Dos Santos et al. [START_REF] Santos Ferreira | Limiting Carleman weights and anisotropic inverse problems[END_REF], Imanuvilov, Uhlmann and Yamamoto [START_REF] Imanuvilov | The Calderón problem with partial data in two dimensions[END_REF] and the review paper [START_REF] Kenig | Recent progress in the Calderón problem with partial data[END_REF] by Kenig and Salo for more references and recent developments. Now we move to the anisotropic case. This corresponds to the general case where the conductivity is given by a general matrix a kj . As pointed out by Lee and Uhlmann in [START_REF] Lee | Determining anisotropic real-analytic conductivities by boundary measurements[END_REF], it is not difficult to see that a change of variables given by a diffeomorphism of Ω which is the identity on ∂Ω leads to different coefficients b kj without changing the D-t-N operator on the boundary. Therefore the single coefficients a kj are not uniquely determined in general. In [START_REF] Lee | Determining anisotropic real-analytic conductivities by boundary measurements[END_REF], Lee and Uhlmann proved that for real-analytic coefficients the uniqueness up to a diffeomorphism holds when the dimension d is ≥ 3. The same result was proved by Astala, Lassas and Päivärinta [START_REF] Astala | Calderón's inverse problem for anisotropic conductivity in the plane[END_REF] for the case d = 2 and L ∞ -coefficients.

In [START_REF] Behrndt | An inverse problem of Calderón type with partial data[END_REF], Behrndt and Rohleder considered general elliptic expressions L a and L b as above and prove that if the corresponding D-t-N operators N Γ1,a (λ) and N Γ1,b (λ) coincide for all λ in a set having an accumulation point in ρ(L D a )∩ρ(L D b ) then the operators L D a and L D b are unitarily equivalent. Here L D a is the elliptic operator L a with Dirichlet boundary conditions. This can be seen as a milder version of the uniqueness problem discussed above. The proof is based on the theory of extensions of symmetric operators and unique continuation results. It is assumed in [START_REF] Behrndt | An inverse problem of Calderón type with partial data[END_REF] that the coefficients are Lipschitz continuous on Ω. We give a different proof of this result which also works for other boundary conditions. Our main result is the following. Note that unlike [START_REF] Behrndt | An inverse problem of Calderón type with partial data[END_REF] we do not assume regularity of the coefficients when d = 2.

We shall restate this theorem in a more precise way after introducing some necessary material and notation. The proof is given in Section 4. It is based on spectral theory and differs from the one in [START_REF] Behrndt | An inverse problem of Calderón type with partial data[END_REF]. Our strategy is to use a relationship between eigenvalues of the D-t-N operator N Γ1,a (λ) and eigenvalues of the elliptic operator with Robin boundary conditions L µ a on Ω where µ is a parameter. One of the main ingredients in the proof is that each eigenvalue of the latter operator is a strictly decreasing map with respect to the parameter µ. Next, the equality of N Γ1,a (λ) and N Γ1,b (λ) allows us to prove that the spectra of L µ a and L µ b are the same and the eigenvalues have the same multiplicity. The similarity of the two elliptic operators with Dirichlet boundary conditions is obtained from the similarity of L µ a and L µ b by letting the parameter µ tend to -∞. During the proof we use some ideas from the papers of Arendt and Mazzeo [START_REF] Arendt | Spectral properties of the Dirichlet-to-Neumann operator on Lipschitz domains[END_REF] and [START_REF] Arendt | Friedlander's eigenvalue inequalities and the Dirichlet-to-Neumann semigroup[END_REF] which deal with a different subject, namely the Friendlander inequality for the eigenvalues of the Dirichlet and Neumann Laplacian on a Lipschitz domain. The ideas which we borrow from [START_REF] Arendt | Spectral properties of the Dirichlet-to-Neumann operator on Lipschitz domains[END_REF] and [START_REF] Arendt | Friedlander's eigenvalue inequalities and the Dirichlet-to-Neumann semigroup[END_REF] are then adapted and extended to our general case of D-t-N operators with variable coefficients and partial data. In Section 2 we define the D-t-N operator with partial data using the method of sesquilinear forms. In particular, for symmetric coefficients it is a self-adjoint operator on L 2 (Γ 1 ). It can be seen as an operator on L 2 (∂Ω) with a non-dense domain and which we extend by 0 to L 2 (Γ 0 ). Therefore one can associate with this D-t-N operator a semigroup (T Γ1 t ) t≥0 acting on L 2 (∂Ω). In Section 3 we prove positivity, sub-Markovian and domination properties for such semigroups. In particular, (T Γ1 t ) t≥0 extends to a contraction semigroup on L p (∂Ω) for all p ∈ [1, ∞). Hence, for ϕ 0 ∈ L p (Γ 1 ), one obtains existence and uniqueness of the solution in L p (∂Ω) to the evolution problem

∂ t ϕ + N Γ1,a (λ)ϕ = 0, ϕ(0) = ϕ 0 .
The results of Section 3 are of independent interest and are not used in the proof of the theorem stated above.

The partial D-t-N operator

Let Ω be a bounded open set of R d with Lipschitz boundary ∂Ω. The boundary is endowed with the (d -1)-dimensional Hausdorff measure dσ. Let a kj , a k , ãk , a 0 : Ω → C be bounded measurable for 1 ≤ k, j ≤ d and such that there exists a constant η > 0 for which

Re d k,j=1 a kj (x)ξ k ξ j ≥ η|ξ| 2 (2.1) for all ξ = (ξ 1 , • • • , ξ d ) ∈ C d and a.e. x ∈ Ω.
Let Γ 0 be an closed subset of ∂Ω and Γ 1 its complement in ∂Ω.

Elliptic operators on Ω.

We consider the space

V = {u ∈ W 1,2 (Ω), Tr(u) = 0 on Γ 0 = 0}, (2.2)
where Tr denotes the trace operator. We define the sesquilinear form

a : V × V → C by the expression a(u, v) = d k,j=1 Ω a kj ∂ k u∂ j v dx + d k=1 Ω a k ∂ k uv + ãk u∂ k v dx + a 0 uv dx (2.3)
for all u, v ∈ V . Here we use the notation ∂ j for the partial derivative ∂ ∂xj . It follows easily from the ellipticity assumption (2.1) that the form a is quasi-accretive, i.e., there exists a constant w such that Re a(u, u) + w u 2 2 ≥ 0 ∀u ∈ V. In addition, since V is a closed subspace of W 1,2 (Ω) the form a is closed. Therefore there exists an operator L a associated with a. It is defined by

D(L a ) = {u ∈ V, ∃v ∈ L 2 (Ω) : a(u, φ) = Ω vφ dx ∀φ ∈ V }, L a u := v.
Formally, L a is given by the expression

L a u = - d k,j=1 ∂ k (a kj ∂ j u) + d k=1 a k ∂ k u -∂ k ( ãk u) + a 0 u.
(2.4)

In addition, following [START_REF] Arendt | Spectral properties of the Dirichlet-to-Neumann operator on Lipschitz domains[END_REF] or [START_REF] Arendt | Friedlander's eigenvalue inequalities and the Dirichlet-to-Neumann semigroup[END_REF] we define the conormal derivative ∂ ∂ν in the weak sense (i.e. in H -1/2 (∂Ω) the dual space of H 1/2 (∂Ω) = Tr(W 1,2 (Ω))), then L a is subject to the boundary conditions

   Tr(u) = 0 on Γ 0 ∂u ∂ν = 0 on Γ 1 .
(2.5)

The conormal derivative in our case is usually interpreted as

d j=1 d k=1 a kj ∂ k u + ãj u ν j , where ν = (ν 1 , • • • , ν d )
is the outer unit normal to the boundary of Ω. For all this see [START_REF] Ouhabaz | Analysis of Heat Equations on Domains[END_REF], Chapter 4.

The condition (2.5) is a mixed boundary condition which consists in taking Dirichlet on Γ 0 and Neumann type boundary condition on Γ 1 . For this reason we denote this operator by L M a . The subscript a refers to the fact that the coefficients of the operator are given by a = {a kj , a k , ãk , a 0 } and M refers to mixed boundary conditions.

We also define the elliptic operator with Dirichlet boundary condition Tr(u) = 0 on ∂Ω. It is the operator associated with the form given by the expression (2.3) with domain D(a) = W 1,2 0 (Ω). It is a quasi-accretive and closed form and its associated operator L D a has the same expression as in (2.4) and subject to the Dirichlet boundary condition Tr(u) = 0 on ∂Ω.

Similarly, we define L N a to be the elliptic operator with Neumann type boundary conditions ∂u ∂ν = 0 on ∂Ω.

It is the operator associated with the form given by the expression (2.3) with domain

D(a) = W 1,2 (Ω). Note that L D a coincides with L M a if Γ 0 = ∂Ω and L N a coincides with L M a if Γ 0 = ∅.
Finally we define elliptic operators with Robin boundary conditions. Let µ ∈ R be a constant and define

a µ (u, v) = d k,j=1 Ω a kj ∂ k u∂ j v dx + d k=1 Ω a k ∂ k uv + ãk u∂ k v dx + a 0 uv dx -µ ∂Ω Tr(u)Tr(v)dσ (2.6)
for all u, v ∈ D(a µ ) := V . Again, Tr denotes the trace operator. Using the standard inequality (see [START_REF] Arendt | Spectral properties of the Dirichlet-to-Neumann operator on Lipschitz domains[END_REF] or [START_REF] Arendt | Friedlander's eigenvalue inequalities and the Dirichlet-to-Neumann semigroup[END_REF]),

∂Ω |Tr(u)| 2 ≤ ε u 2 W 1,2 (Ω) + c ε Ω |u| 2
which is valid for all ε > 0 (c ε is a constant depending on ε) one obtains that for some positive constants w and δ

Re a µ (u, u) + w Ω |u| 2 ≥ δ u 2 W 1,2 (Ω) .
From this it follows that a µ is a quasi-accretive and closed sesquilinear form. One can associate with a µ an operator L µ a . This operator has the same expression (2.4) and it is subject to the Robin boundary conditions

   Tr(u) = 0 on Γ 0 ∂u ∂ν = µ Tr(u) on Γ 1 .
(2.7)

Actually, the boundary conditions (2.7) are mixed Robin boundary conditions in the sense that we have the Dirichlet condition on Γ 0 and the Robin one on Γ 1 . For simplicity we ignore the word "mixed" and refer to (2.7) as the Robin boundary conditions. According to our previous notation, if µ = 0, then a 0 = a and L 0 a = L M a . Note that we may choose here µ to be a bounded measurable function on the boundary rather than just a constant.

The partial Dirichlet-to-Neumann operator on ∂Ω.

Suppose as before that a = {a kj , a k , ãk , a 0 } are bounded measurable and satisfy the ellipticity condition (2.1). Let Γ 0 , Γ 1 , V be as above and a is the sesquilinear form defined by (2.3). We define the space

V H := {u ∈ V, a(u, g) = 0 for all g ∈ W 1,2 0 (Ω)}. (2.8)
Then V H is a closed subspace of V . It is interpreted as the space of harmonic functions for the operator L a (given by (2.4)) with the additional property that Tr(u) = 0 on Γ 0 . We start with the following simple lemma.

Lemma 2.1. Suppose that 0 / ∈ σ(L D a ). Then V = V H ⊕ W 1,2 0 (Ω).
(2.9)

Proof. We argue as in [START_REF] Ter Elst | Analysis of the heat kernel of the Dirichlet-to-Neumann operator[END_REF], Section 2 or [START_REF] Arendt | Spectral properties of the Dirichlet-to-Neumann operator on Lipschitz domains[END_REF]. Let us denote by a D the form associated with L D a , that is, a D is given by (2.3) with D(a D ) = W 1,2 0 (Ω). There exists an operator L D a : W 1,2 0 (Ω) → W -1,2 (Ω) := W 1,2 0 (Ω) (the anti-dual of W 1,2 0 (Ω)) associated with a D in the sense

L D a h, g = a D (h, g) for all h, g ∈ W 1,2 0 (Ω). The notation •, • denotes the duality W 1,2 0 (Ω) - W 1,2 0 (Ω). Since 0 / ∈ σ(L D a )
, then L D a is invertible. Therefore L D a , seen as operator on W 1,2 0 (Ω) with domain W 1,2 0 (Ω), is also invertible on W 1,2 0 (Ω) since the two operators L D a and L D a have the same spectrum (see e.g., [START_REF] Arendt | Vector-Valued Laplace Transforms and Cauchy Problems[END_REF], Proposition 3.10.3). Now we fix u ∈ V and consider the (anti-)linear functional

F : v → a(u, v).
Clearly, F ∈ W 1,2 0 (Ω) and hence there exists a unique

u 0 ∈ W 1,2 0 (Ω) such that L D a u 0 = F , i.e., L D a u 0 , g = F (g) for all g ∈ W 1,2 0 (Ω). This means that a(u -u 0 , g) = 0 for all g ∈ W 1,2 0 (Ω) and hence u -u 0 ∈ V H . Thus, u = u -u 0 + u 0 ∈ V H + W 1,2 0 (Ω). Finally, if u ∈ V H ∩ W 1,2 0 (Ω) then a(u, g) = 0 for all g ∈ W 1,2 0 (Ω). This means that u ∈ D(L D a ) with L D a u = 0. Since L D a is invertible we conclude that u = 0.
As a consequence of Lemma 2.1, the trace operator Tr :

V H → L 2 (∂Ω) is injective and
Tr(V H ) = Tr(V ).

(2.10)

In the rest of this section we assume that 0 / ∈ σ(L D a ). We define on L 2 (∂Ω, dσ) the sesquilinear form b(ϕ, ψ) := a(u, v) (2.11) where u, v ∈ V H are such that ϕ = Tr(u) and ψ = Tr(v). This means that D(b) = Tr(V H ) and by (2.10)

D(b) = Tr(V H ) = Tr(V ).
(2.12)

Lemma 2.2.

There exist positive constants w, δ and M such that

Re b(ϕ, ϕ) + w ∂Ω |ϕ| 2 ≥ δ u 2 W 1,2 (Ω) (2.13)
and

|b(ϕ, ψ)| ≤ M Re b(ϕ, ϕ) + w ∂Ω |ϕ| 2 1/2 Re b(ψ, ψ) + w ∂Ω |ψ| 2 1/2 (2.14) for all ϕ, ψ ∈ D(b). In the first inequality, u ∈ V H is such that Tr(u) = ϕ. Proof. It is well known that Tr : W 1,2 (Ω) → L 2 (∂Ω) is a compact operator and since Tr : V H → L 2 (∂Ω) is injective it follows that for every > 0 there exists a constant c > 0 such that Ω |u| 2 ≤ u 2 W 1,2 + c ∂Ω |Tr(u)| 2 (2.15)
for all u ∈ V H (see, e.g., [START_REF] Arendt | Spectral properties of the Dirichlet-to-Neumann operator on Lipschitz domains[END_REF]). In particular,

Ω |u| 2 ≤ 1 -Ω |∇u| 2 + c 1 -∂Ω |ϕ| 2 . (2.16)
Now, let ϕ ∈ D(b) = Tr(V H ) and u ∈ V H such that ϕ = Tr(u). It follows from the ellipticity assumption (2.1) and the boundedness of the coefficients that for some constant

c 0 > 0 Re a(u, u) ≥ η 2 Ω |∇ u| 2 -c 0 Ω |u| 2 .
Therefore, using (2.16) and the definition of b we obtain

Re b(ϕ, ϕ) = Re a(u, u) ≥ ( η 2 - c 0 1 - ) Ω |∇ u| 2 - cc 0 1 -∂Ω |ϕ| 2 .
Taking > 0 small enough we obtain (2.13).

In order to prove the second inequality, we use the definition of b and again the boundedness of the coefficients to see that

|b(ϕ, ψ)| = |a(u, v)| ≤ C u W 1,2 v W 1,2 .
Thus, (2.14) follows from (2.13). Note that the domain Tr(V H ) of b may not be dense in L 2 (∂Ω) since functions in this domain vanish on Γ 0 . Indeed,

ϕ b := Re b(ϕ, ϕ) + w ∂Ω |ϕ| 2 1/2 in which w is as in (2.13). If (ϕ n ) is a Cauchy sequence for • b then by (2.13) the corresponding (u n ) ∈ V H with Tr(u n ) = ϕ n is a Cauchy sequence in V H . Since V H is a closed subspace of V it follows that u n is convergent to some u in V H .
H := D(b) L 2 (∂Ω) = L 2 (Γ 1 ) ⊕ {0}.
(2.17)

The direct inclusion follows from the fact that if

ϕ n ∈ D(b) converges in L 2 (∂Ω)
then after extracting a subsequence we have a.e. convergence. Since ϕ n = 0 on Γ 0 we obtain that the limit ϕ = 0 on Γ 0 . The reverse inclusion can be proved as follows. Let Γ 2 be a closed subset of R d with Γ 2 ⊂ Γ 1 and consider the space

E = {u |Γ2 : u ∈ W 1,∞ (R d ), u |Γ0 = 0}. Then E ⊂ C(Γ 2
) and an easy application of the Stone-Weierstrass theorem shows that E is dense in C(Γ 2 ). Now given

ϕ ∈ C c (Γ 1 ) and > 0 we find Γ 2 such that 1 Γ1\Γ2 2 < and u |Γ2 ∈ E such that u |Γ2 -ϕ C(Γ2) < . Finally we take χ ∈ C ∞ c (R d ) such that χ = 1 on Γ 2 . Then (uχ) |Ω ∈ V and uχ -ϕ L 2 (Γ1) ≤ u -ϕ L 2 (Γ2) + χ L 2 (Γ1\Γ2) ≤ |Γ 2 | + χ ∞ .
Here |Γ 2 | denotes the measure of Γ 2 . These inequalities together with the fact that C c (Γ 1 ) is dense in L 2 (Γ 1 ) imply (2.17).

We return to the form b defined above. We associate with b an operator N Γ1 . It is defined by

D(N Γ1 ) := {ϕ ∈ D(b), ∃ψ ∈ H : b(ϕ, ξ) = Γ1 ψξ ∀ξ ∈ D(b)}, N Γ1 ϕ = ψ.
The operator N Γ1 can be interpreted as an operator on L 2 (∂Ω) defined as follows: if ϕ ∈ D(N Γ1 ) then there exists a unique u ∈ V H such that ϕ = Tr (u) and

ϕ |Γ0 = 0, N Γ1 (ϕ) = ∂u ∂ν on Γ 1 . (2.18)
Again ∂u ∂ν is interpreted in the weak sense as the conormal derivative that is

d j=1 d k=1 a kj ∂ k u + ãj ϕ ν j .
In the particular case where a kj = δ kj and a 1 = • • • = a d = 0 the right hand side of (2.18) is seen as the normal derivative on the boundary. All this can be made precise by applying the Green formula if the boundary and the coefficients are smooth enough. We call N Γ1 the partial Dirichlet-to-Neumann operator on L 2 (∂Ω) or the Dirichletto-Neumann operator with partial data. The term partial refers to the fact that N Γ1 is known only on the part Γ 1 of the boundary ∂Ω. It follows from the general theory of forms that -N Γ1 generates a holomorphic semigroup e -tN Γ 1 on H. We define T Γ1 t on L 2 (∂Ω) by

T Γ1 t ϕ = e -tN Γ 1 (ϕ1 ) ⊕ 0.
We shall refer to (T Γ1 t ) t≥0 as the "semigroup" generated by -N Γ1 on L 2 (∂Ω). It is clear that

T Γ1 t L(L 2 (∂Ω)) ≤ e -w0t , t ≥ 0, (2.19)
for some constant w 0 . Note that if the form a is symmetric, then b is also symmetric and hence N Γ1 is self-adjoint. In this case, (2.19) holds with w 0 = inf σ(N Γ1 ) which also coincides with the first eigenvalue of N Γ1 . For all this, see e.g. [START_REF] Ouhabaz | Analysis of Heat Equations on Domains[END_REF], Chapter 1.

Positivity and domination

In this section we study some properties of the semigroup (T Γ1 t ) t≥0 . We assume throughout this section that a jk = a kj , ãk = a k , a 0 ∈ L ∞ (Ω, R).

(3.1)

We recall that L D a is the elliptic operator with Dirichlet boundary conditions defined in the previous section. Its associated symmetric form a D is given by (2.3) and has domain W 1,2 0 (Ω). We shall need the accretivity assumption of a D (or equivalently the self-adjoint operator L D a is non-negative) which means that a D (u, u) ≥ 0 for all u ∈ W 1,2 0 (Ω).

e., it maps non-negative functions of L 2 (∂Ω) into non-negative functions). b) Suppose in addition that

a 0 ≥ 0 and a k = 0 for all k ∈ {1, • • • , d}. Then (T Γ1 t ) t≥0 is a sub-Markovian semigroup.
Recall that the sub-Markovian property means that for ϕ ∈ L 2 (∂Ω) and t ≥ 0 0 ≤ ϕ ≤ 1 ⇒ 0 ≤ T Γ1 t ϕ ≤ 1. This property implies in particular that (T Γ1 t ) t≥0 extends from L 2 (∂Ω) to L p (∂Ω) for all p ∈ [2, ∞[. Since a is symmetric then so is b and one obtains by duality that (T Γ1 t ) t≥0 extends also to L p (∂Ω) for p ∈ [START_REF] Arendt | Vector-Valued Laplace Transforms and Cauchy Problems[END_REF][START_REF] Arendt | Spectral properties of the Dirichlet-to-Neumann operator on Lipschitz domains[END_REF].

Proof. The proof follows exactly the same lines as for Theorem 2.3 in [START_REF] Ter Elst | Analysis of the heat kernel of the Dirichlet-to-Neumann operator[END_REF]. a) By the well known Beurling-Deny criteria (see [START_REF] Davies | Heat Kernel and Spectral Theory[END_REF], Section 1.3 or [START_REF] Ouhabaz | Analysis of Heat Equations on Domains[END_REF], Theorem 2.6), it suffices to prove that ϕ + ∈ D(b) and b(ϕ + , ϕ -) ≤ 0 for all realvalued ϕ ∈ D(b). Note that the fact that D(b) is not densely defined does not affect the the statements of the Beurling-Deny criteria. Let ϕ ∈ D(b) be real-valued. There exists a real-valued u ∈ H V such that ϕ = Tr(u). Then ϕ + = Tr(u + ) ∈ Tr(V ) = TrH V = D(b). This follows from the fact that v + ∈ V for all v ∈ V (see [START_REF] Ouhabaz | Analysis of Heat Equations on Domains[END_REF], Section 4.2). By Lemma 2.1 we can write u + = u 0 +u 1 and u

-= v 0 +v 1 with u 0 , v 0 ∈ W 1,2 0 (Ω) and u 1 , v 1 ∈ H V . Hence, u = u + -u -= (u 0 -v 0 ) + (u 1 -v 1 ). Since u, u 1 -v 1 ∈ H V it follows that u 0 = v 0 . Therefore, b(ϕ + , ϕ -) = a(u 1 , v 1 ) = a(u 1 , v 0 + v 1 ) = a(u 0 + u 1 , v 0 + v 1 ) -a(u 0 , v 0 + v 1 ) = a(u + , u -) -a(u 0 , v 0 ) = -a(u 0 , v 0 ) = -a(u 0 , u 0 ) = -a D (u 0 , u 0 ).
Here we use the fact that

a(u + , u -) = d k,j=1 Ω a kj ∂ k (u + )∂ j (u -) + d k=1 Ω a k ∂ k u + u -+ a k u + ∂ k u - + Ω a 0 u + u -= 0.
By assumption (3.2) we have a D (u 0 , u 0 ) ≥ 0 and we obtain b(ϕ + , ϕ -) ≤ 0. This proves the positivity of (T Γ1 t ) t≥0 on L 2 (∂Ω). b) By [START_REF] Ouhabaz | Invariance of closed convex sets and domination criteria for semigroups[END_REF] or [START_REF] Ouhabaz | Analysis of Heat Equations on Domains[END_REF], Corollary 2.17 it suffices to prove that 1 ∧ ϕ := inf(1, ϕ) ∈ D(b) and b(1 ∧ ϕ, (ϕ -1) + ) ≥ 0 for all ϕ ∈ D(b) with ϕ ≥ 0. Let ϕ ∈ D(b) and suppose that ϕ ≥ 0. Let u ∈ H V be real-valued such that ϕ = Tr(u). Note that 1∧u ∈ V (see [START_REF] Ouhabaz | Analysis of Heat Equations on Domains[END_REF], Section 4.3). We decompose 1∧u

= u 0 +u 1 ∈ W 1,2 0 (Ω)⊕H V . Then (u -1) + = u -1 ∧ u = (-u 0 ) + (u -u 1 ) ∈ W 1,2 0 (Ω) ⊕ H V . Therefore, b(1 ∧ ϕ, (ϕ -1) + ) = a(u 1 , u -u 1 ) = a(u 0 + u 1 , u -u 1 ) = a(u 0 + u 1 , -u 0 + u -u 1 ) + a(u 0 + u 1 , u 0 ) = a(u 0 + u 1 , -u 0 + u -u 1 ) + a(u 0 , u 0 ) = d k,j=1 Ω a kj ∂ k (1 ∧ u)∂ j ((u -1) + ) + Ω a 0 (1 ∧ u)(u -1) + + a D (u 0 , u 0 ) = Ω a 0 (u -1) + + a D (u 0 , u 0 ) ≥ 0.
This proves that b(1 ∧ ϕ, (ϕ -1) + ) ≥ 0.

Next we have the following domination property. Theorem 3.2. Suppose that a kj , a k , ãk and a 0 satisfy (3.1). Suppose also that L D a is accretive with 0 / ∈ σ(L D a ). Let Γ 0 and Γ0 be two closed subsets of the boundary such that Γ 0 ⊆ Γ0 . Then for every 0 The next result shows monotonicity with respect to the potential a 0 . This was already proved in [START_REF] Ter Elst | Analysis of the heat kernel of the Dirichlet-to-Neumann operator[END_REF] Theorem 2.4, in the case where L D a = -∆ + a 0 . The proof given there works also in the general framework of the present paper.

≤ ϕ ∈ L 2 (∂Ω) 0 ≤ T Γ1 t ϕ ≤ T Γ1 t ϕ.
As above let a kj , a k and a 0 be real-valued and let (T Γ1,a0 t ) t≥0 denote the semigroup (T Γ1 t ) t≥0 defined above. Suppose that b 0 is a real-valued function and denote by (T Γ1,b0

Proof of the main result

In this section we prove Theorem 1.1. We recall briefly the operators introduced in Section 2.

For µ ∈ R and recall the operator L µ a associated with the form a µ given by (2.6) with domain D(a µ ) := V and V is again given by (2.2). The operator associated with a µ is L µ a . It is given by the formal expression (2.4) and it is subject to mixed and Robin boundary conditions (2.7). We also recall that L D a is the operator subject to the Dirichlet boundary conditions and L M a is subject to mixed boundary conditions. Fix λ / ∈ σ(L D a ). We denote by N Γ1,a (λ) the partial D-t-N operator with the coefficients {a kj , a k , a 0 -λ}. It is the operator associated with the form

b(ϕ, ψ) := d k,j Ω a kj ∂ k u∂ j v dx + d k=1 Ω a k ∂ k uv + a k u∂ k v dx + (a 0 -λ)uv dx where u, v ∈ V H (λ) with Tr(u) = ϕ, Tr(v) = ψ and V H (λ) := {u ∈ V, a(u, g) = λ Ω ug for all g ∈ W 1,2 0 (Ω)}, (4.1)
This space is the same as in (2.8) but now with a 0 replaced by a 0 -λ. We restate the main theorem using the notation introduced in Section 2.

∈ σ(L µ a ) = σ(L µ b ) with λ / ∈ σ(L D a ) = σ(L D b ), the sets {Tr(u), u ∈ Ker(λI -L µ a )} and {Tr(v), v ∈ Ker(λI -L µ b )} coincide.
The same property holds for the operators L M a and L M b . We shall need several preparatory results. We start with the following theorem which was proved in [START_REF] Arendt | Spectral properties of the Dirichlet-to-Neumann operator on Lipschitz domains[END_REF] and [START_REF] Arendt | Friedlander's eigenvalue inequalities and the Dirichlet-to-Neumann semigroup[END_REF] in the case where a kj = δ kj , a k = 0, a 0 is a constant and Γ 1 = ∂Ω. Theorem 4.2. Let a = {a kj = a jk , a k = ãk , a 0 = a 0 } be bounded measurable functions on Ω such that a kj satisfy the ellipticity condition

(2.1). Let µ, λ ∈ R and λ / ∈ σ(L D a ). Then: 1) µ ∈ σ(N Γ1,a (λ)) ⇔ λ ∈ σ(L µ a ). In addition, if u ∈ Ker(λ -L µ a ), u = 0 then ϕ := Tr(u) ∈ Ker(µ -N Γ1,a (λ)) and ϕ = 0. Conversely, if ϕ ∈ Ker(µ - N Γ1,a (λ)), ϕ = 0, then there exists u ∈ Ker(λ -L µ a ), u = 0 such that ϕ = Tr(u). 2) dim Ker(µ -N Γ1,a (λ)) = dim Ker(λ -L µ a )
. Proof. We follow a similar idea as in [START_REF] Arendt | Spectral properties of the Dirichlet-to-Neumann operator on Lipschitz domains[END_REF] and [START_REF] Arendt | Friedlander's eigenvalue inequalities and the Dirichlet-to-Neumann semigroup[END_REF]. It is enough to prove that the mapping S : Ker(λ -L µ a ) → Ker(µ -N Γ1,a (λ)), u → Tr(u) is an isomorphism. First, we prove that S is well defined. Let u ∈ Ker(λ -L µ a ). Then u ∈ D(L µ a ) and L µ a u = λu. By the definition of L µ a we have u ∈ V and for all

v ∈ V d k,j=1 Ω a kj ∂ k u∂ j v + d k=1 Ω a k ∂ k uv + a k u∂ k v + Ω a 0 uv -λ Ω uv = µ ∂Ω Tr(u)Tr(v). (4.2)
Taking v ∈ W 1,2 0 (Ω) yields u ∈ V H (λ). Note that (4.2) also holds for v ∈ V H (λ). Hence it follows from the definition of N Γ1,a (λ) that

ϕ := Tr(u) ∈ D(N Γ1,a (λ)) and N Γ1,a (λ)ϕ = µϕ. This means that S(u) ∈ Ker(µ -N Γ1,a (λ)). Suppose now that u ∈ Ker(λ-L µ a ) with u = 0. If S(u) = 0 then u ∈ W 1,2 0 (Ω). Therefore, it follows from (4.2) that for all v ∈ V d k,j=1 Ω a kj ∂ k u∂ j v + d k=1 Ω a k ∂ k uv + a k u∂ k v + Ω (a 0 -λ)uv = 0. ( 4.3)
This implies that u ∈ V H (λ). We conclude by Lemma 2.1 that u = 0. Thus S is injective. We prove that S is surjective. Let ϕ ∈ Ker(µ-N Γ1,a (λ)). Then by the definition of N Γ1,a (λ), there exists u ∈ V H (λ) such that ϕ = Tr(u) and u satisfies (4.2) for all

v ∈ V H (λ). If v ∈ V we write v = v 0 + v 1 ∈ W 1,2 0 (Ω) ⊕ V H (λ)
and see that (4.2) holds for u and v. This means that u ∈ D(L µ a ) and L µ a u = λu.

Lemma 4.3. For λ ∈ R large enough, (λ + L µ a ) -1 converges in L(L 2 (Ω)) to (λ + L D a ) -1 as µ → -∞. This is Proposition 2.6 in [START_REF] Arendt | Spectral properties of the Dirichlet-to-Neumann operator on Lipschitz domains[END_REF] when a kj = δ kj , a k = a 0 = 0. The proof given in [START_REF] Arendt | Spectral properties of the Dirichlet-to-Neumann operator on Lipschitz domains[END_REF] remains valid in our setting. Note that the idea of proving the uniform convergence here is based on a criterion from [START_REF] Daners | Dirichlet problems on varying domains[END_REF] (see Appendix B) which states that it is enough to check that for all (f n ), f ∈ L 2 (Ω)

f n f ⇒ (λ + L µn a ) -1 f n → (λ + L D a ) -1 f, ( 4.4)
for every sequence µ n → -∞. The first convergence is in the weak sense in L 2 (Ω) and the second one is the strong convergence. It is not difficult to check (4.4).

From now on, we denote by (λ µ a,n ) n≥1 the eigenvalues of L µ a , repeated according to their multiplicities. We have for each µ ∈ R

λ µ a,1 ≤ λ µ a,2 ≤ • • • → +∞.
Similarly for the eigenvalues (λ D a,n ) n≥1 of L D a . These eigenvalues satisfy the standard min-max principle since the operators L µ a and L D a are self-adjoint by our assumptions.

A well known consequence of the previous lemma is that the spectrum of L µ a converges to the spectrum of L D a . More precisely, for all k, λ µ a,k → λ D a,k as µ → -∞. (4.5)

In addition, we have the following lemma which will play a fundamental role.

Lemma 4.4. Let a = {a kj = a jk , a k = ãk , a 0 = a 0 } be bounded measurable functions on Ω such that a kj satisfy the ellipticity condition (2.1). If d ≥ 3 we assume in addition that the coefficients a kj and a k are Lipschitz continuous on Ω. Then for each k, µ → λ µ a,k is strictly decreasing on R and λ a,k → -∞ as µ → +∞.

Proof. Firstly, by the min-max principle λ µ a,k ≤ λ D a,k and the function µ → λ µ a,k is non-increasing. Fix k ≥ 0 and suppose that µ → λ µ a,k is constant on [α, β] for some α < β. For each µ we take a normalized eigenvector u µ such that Tr(u µ+h ) → Tr(u µ ) in L 2 (∂Ω) as h → 0 (or as h n → 0 for some sequence h n ). Indeed, due to regularity properties µ → λ µ a,k is continuous (see [START_REF] Kato | Perturbation Theory for Linear Operators[END_REF], Chapter VII) and hence (λ µ+h a,k ) h is bounded for small h. The equality a µ+h (u µ+h , u µ+h ) = λ µ+h a,k implies that a µ+h (u µ+h , u µ+h ) is bounded w.r.t. h (for small h). This latter property and ellipticity easily imply that (u µ+h ) h is bounded in V . After extracting a sequence we may assume that (u µ+h ) h converges weakly in V to some u as h → 0. The compactness embedding of V in L 2 (Ω) as well as the compactness of the trace operator show that (u µ+h ) h converges to u in L 2 (Ω) and Tr(u µ+h ) converges to Tr(u) in L 2 (∂Ω). On the other hand for every v ∈ V , the equality

a µ+h (u µ+h , v) = λ µ+h a,k Ω u µ+h v dx
shows that the limit u is a normalized eigenvector of L µ a for the eigenvalue λ µ a,k . We take u µ := u and obtain the claim stated above.

Observe that

Γ1

Tr(u µ+h )Tr(u µ ) dσ = 0 (4.6)

for all h = 0 and µ, µ + h ∈ [α, β]. Indeed, using the definition of the form a µ (see (2.6)) we have

λ Ω u µ+h u µ dx = a µ+h (u µ+h , u µ ) = a µ (u µ+h , u µ ) -h Γ1 Tr(u µ+h )Tr(u µ ) dσ = λ Ω u µ+h u µ dx -h Γ1 Tr(u µ+h )Tr(u µ ) dσ.
This gives (4.6). Now, letting h → 0 we obtain from (4.6) and the fact that Tr(u µ+h ) converges to Tr(u µ ) as h → 0 that Tr(u µ ) = 0 on Γ 1 for all µ ∈ [α, β].

Hence Tr(u µ ) = 0 on ∂Ω since u µ ∈ V . Hence, L µ has an eigenfunction u µ ∈ W 1,2 0 (Ω). Note that if d = 2 or if d ≥ 3 and the coefficients a kj and a k are Lipschitz continuous on Ω, then the operator L a has the unique continuation property (see [START_REF] Schulz | On the unique continuation property of elliptic divergence form equations in the plane[END_REF] for the case d = 2 and [START_REF] Wolff | Recent work on sharp estimates in second-order elliptic unique continuation problems[END_REF] for d ≥ 3). If d ≥ 3 and hence the coefficients are Lipschitz on Ω, we apply Proposition 2.5 in [START_REF] Behrndt | An inverse problem of Calderón type with partial data[END_REF] to conclude that u µ = 0, but this is not possible since u µ 2 = 1. If d = 2 we argue in a similar way. Indeed, let Ω be an open subset of R 2 containing Ω and such that Γ 0 ⊂ ∂ Ω and Ω \ Ω contains an open ball. We extend all the coefficients to bounded measurable function ãkj , ãk and ã0 on Ω. In addition, ãkj = ãjk on Ω and satisfy the ellipticity condition. We extend u µ to ũµ ∈ W 1,2 0 ( Ω) by 0 outside Ω. We define in Ω the elliptic operator L ã as previously. For v ∈ C ∞ c ( Ω) we note that v |Ω ∈ V and hence

Ω L ã( ũµ )vdx = a µ (u µ , v |Ω ) = λ Ω u µ v |Ω = λ Ω ũµ v.
The term Ω L ã( ũµ )v is of course interpreted in the sense of the associated sesquilinear form and the first equality uses the fact that ũµ is 0 on Ω \ Ω and u µ ∈ W 1,2 0 (Ω). Hence, ũµ satisfies

(L ã -λ)( ũµ ) = 0
in the weak sense on Ω. We conclude by the unique continuation property ( [START_REF] Schulz | On the unique continuation property of elliptic divergence form equations in the plane[END_REF]) that ũµ = 0 on Ω since it is 0 on an open ball contained in Ω \ Ω . We arrive as above to a contradiction. Hence, µ → λ µ a,k is strictly decreasing on R.

It remains to prove that for any k, λ µ a,k → -∞ as µ → +∞. By the min-max principle

λ µ 1 ≤ d k,j=1 Ω a kj ∂ k u∂ j u + 2 Re d k=1 Ω a k ∂ k uu + Ω a 0 |u| 2 -µ Γ1 |Tr(u)| 2
for every normalized u ∈ V . Taking u such that Tr(u) = 0 shows that λ µ a,1 → -∞ as µ → +∞. Suppose now that λ µ a,k > w for some w ∈ R, k > 1 and all µ ∈ R. Taking the smallest possible k we have λ µ a,j → -∞ as µ → +∞ for j = 1, • • • , k-1. Of course, λ µ a,j > w for all j ≥ k and we may choose w / ∈ σ(L D a ). Remember also that µ → λ µ a,j is strictly decreasing for j = 1, • • • , k -1. On the other hand, by Theorem 4.2 we have σ(N Γ1,a (w)) ⊂ {µ ∈ R, λ µ a,j = w, j = 1, • • • , k -1}. Using the fact that λ µ a,j → -∞ as µ → +∞ and µ → λ µ a,j is strictly decreasing for j = 1, • • • , k -1 we see that we can choose w such that the set {µ ∈ R, λ µ a,j = w, j = 1, • • • , k -1} is finite and hence σ(N Γ1,a (w)) is finite which is not possible since L 2 (Γ 1 ) has infinite dimension. Related results to Lemma 4.4 can be found in [START_REF] Arendt | Friedlander's eigenvalue inequalities and the Dirichlet-to-Neumann semigroup[END_REF] (see Proposition 3) and [START_REF] Rohleder | Strict inequality of Robin eigenvalues for elliptic differential operators on Lipschitz domains[END_REF]. In both papers the proofs use the unique continuation property.

We shall also need the following lemma. This result is easy to prove, see Lemma 2.4 in [START_REF] Behrndt | An inverse problem of Calderón type with partial data[END_REF]. have the same multiplicity. In addition, j = k. Indeed, if k < j then

λ µ b,1 ≤ λ µ b,2 ≤ • • • ≤ λ µ b,k ≤ • • • ≤ λ µ b,j = λ µ a,k .
Each λ µ b,m coincides with an eigenvalue of L µ a (with the same multiplicity) and hence λ µ a,k is (at least) the j-th eigenvalue of L µ a with j > k which is not possible. The same argument works if j < k. Using Lemma 4.4 we see that for any k there exists a discrete set J ⊂ R such that λ µ a,k = λ µ b,k for every µ ∈ R \ J. By continuity of µ → λ µ a,k and µ → λ µ b,k these two functions coincide on R. This proves (4.7) and also that the multiplicities of the eigenvalues λ µ a,k and λ µ b,k are the same.

The similarity property follows by a classical argument. Recall that L µ a and L µ b are self-adjoint operators with compact resolvents. It follows that here exist orthonormal bases Φ n and Ψ n of L 2 (Ω) which are eigenfunctions of L µ a and L µ b , respectively. Define the mapping

U : L 2 (Ω) → L 2 (Ω), Φ n → Ψ n . Thus for f = n (f, Φ n )Φ n ∈ L 2 (Ω), U(f ) = n (f, Φ n )Ψ n .
The notation (f, Φ n ) is the scalar product in L 2 (Ω). Clearly,

U(f ) 2 2 = n |(f, Φ n )| 2 = f 2 2 .
The mapping U is an isomorphism. In addition, if L µ a Φ n = λ µ a,n Φ n then for f ∈ D(L µ b )

UL µ a U -1 (f ) = UL µ a U -1 n (f, Ψ n )Ψ n = UL µ a n (f, Ψ n )Φ n = U n (f, Ψ n )λ µ a,n Φ n = n (f, λ µ b,n Ψ n )Ψ n
= L µ b (f ). Thus, L µ a and L µ b are unitarily equivalent. This proves assertion i).

ii) Choose µ = 0 in the previous assertion.

1 2 (

 2 ∂Ω) with ϕ = 0 on Γ 0 , one solves the Dirichlet problem L a (λ)u = 0 weakly in W 1,2 (Ω) with u = ϕ on ∂Ω, (1.1) and defines (in the weak sense) N Γ1,a (λ)ϕ := d j=1 d k=1 a kj ∂ k u + a j ϕ ν j on Γ 1 . (1.2) 1 Here ν = (ν 1 , • • • , ν d ) is the outer unit normal to the boundary of Ω. The operator N Γ1,a (λ) is interpreted as the conormal derivative on the boundary. It is an operator acting on L 2 (∂Ω). See Section 2 for more details.

Theorem 1 . 1 .

 11 Suppose that Ω is a bounded Lipchitz domain of R d with d ≥ 2. Let Γ 0 be a closed subset of ∂Ω, Γ 0 = ∂Ω and Γ 1 its complement. Let a = {a kj , a k , a 0 } and b = {b kj , b k , b 0 } be bounded functions on Ω such that a kj and b kj satisfy the usual ellipticity condition. If d ≥ 3 we assume in addition that the coefficients a kj , b kj , a k and b k are Lipschitz continuous on Ω. Suppose that N Γ1,a (λ) = N Γ1,b (λ) for all λ in a set having an accumulation point in ρ(L D a ) ∩ ρ(L D b ). Then: i) The operators L a and L b endowed with Robin boundary conditions are unitarily equivalent. ii) The operators L a and L b endowed with mixed boundary conditions (Dirichlet on Γ 0 and Neumann type on Γ 1 ) are unitarily equivalent. iii) The operators L a and L b endowed with Dirichlet boundary conditions are unitarily equivalent. In addition, for Robin or mixed boundary conditions, the eigenfunctions associated to the same eigenvalue λ / ∈ σ(L D a ) = σ(L D b ) coincide on the boundary of Ω.

Corollary 2 . 3 .

 23 The form b is continuous, quasi-accretive and closed. Proof. Continuity of b is exactly (2.14). Quasi-accretivity means that Re b(ϕ, ϕ) + w ∂Ω |ϕ| 2 ≥ 0 for some w and all ϕ ∈ D(b). This follows from (2.13). Now we prove that b is closed which means that D(b) is complete for the norm

  Set ϕ := Tr(u). We have ϕ ∈ D(b) and the definition of b together with continuity of Tr as an operator from W 1,2 (Ω) to L 2 (∂Ω) show that ϕ n converges to ϕ for the norm • b . This means that b is a closed form.

(3. 2 ) 3 . 1 .

 231 Theorem Suppose that 0 / ∈ σ(L D a ), (3.1) and that L D a is accretive. a) The semigroup (T Γ1 t ) t≥0 is positive (i.

Proof.

  Let Γ1 be the complement of Γ0 in ∂Ω. Denote by b and b the sesquilinear forms associated with N Γ1 and N Γ1 , respectively. Clearly, b is a restriction of b and hence it is enough to prove that D( b) is an ideal of D(b) and apply [22] or [23], Theorem 2.24. For this, let 0 ≤ ϕ ≤ ψ with ϕ ∈ D(b) and ψ ∈ D( b). This means that ϕ and ψ are respectively the traces on ∂Ω of u, v ∈ W 1,2 (Ω) such that ϕ = Tr(u) = 0 on Γ 0 and ψ = Tr(v) = 0 on Γ0 . Since 0 ≤ ϕ ≤ ψ we have ϕ = 0 on Γ0 . This equality gives ϕ ∈ D( b) and this shows that D( b) is an ideal of D(b).

t) 3 . 3 .

 33 t≥0 be the semigroup of N Γ1 with coefficients a kj , a k and b 0 (i.e. a 0 is replaced by b 0 ). Then we have Theorem Suppose that a kj , a k , ãk and a 0 satisfy (3.1). Suppose again that 0 / ∈ σ(L D a ) and L D a is accretive. If a 0 ≤ b 0 then 0 ≤ T Γ1,b0 t ϕ ≤ T Γ1,a0 t ϕ for all 0 ≤ ϕ ∈ L 2 (∂Ω) and t ≥ 0.

Theorem 4 . 1 .

 41 Suppose that Ω is a bounded Lipchitz domain of R d with d ≥ 2. Let Γ 0 be a closed subset of ∂Ω, Γ 0 = ∂Ω and Γ 1 = ∂Ω \ Γ 0 . Let a = {a kj = a jk , a k = ãk , a 0 = a 0 } and b = {b kj = b jk , b k = bk , b 0 = b 0 } be bounded measurable functions on Ω such that a kj and b kj satisfy the ellipticity condition (2.1). If d ≥ 3 we assume in addition that the coefficients a kj , b kj , a k and b k are Lipschitz continuous on Ω. Suppose that N Γ1,a (λ) = N Γ1,b (λ) for all λ in a set having an accumulation point in ρ(L D a ) ∩ ρ(L D b ). Then: i) The operators L µ a and L µ b are unitarily equivalent for all µ ∈ R. ii) The operators L M a and L M b are unitarily equivalent. iii) The operators L D a and L D b are unitarily equivalent. Moreover, for every λ

Lemma 4 . 5 .

 45 For every ϕ, ψ ∈ Tr(V ), the mappingλ → N Γ1,a (λ)ϕ, ψ is holomorphic on C \ σ(L D a ).

Proof of Theorem 1 . 1 .

 11 As above, we denote by (λ µ b,n ) n≥1 and (λ D b,n ) n≥1 the eigenvalues of the self-adjoint operators L µ b and L D b , respectively. It follows from Lemma 4.5 and the assumptions thatN Γ1,a (λ) = N Γ1,b (λ) for all λ ∈ C \ (σ(L D a ) ∪ σ(L D b )). i) We show that for all µ ∈ R σ(L µ a ) = σ(L µ b ),(4.7)and the eigenvalues have the same multiplicity. Fix µ ∈ R and suppose that λ= λ µ a,k ∈ σ(L µ a ) \ (σ(L D a ) ∪ σ(L D b )). By Theorem 4.2, µ ∈ σ(N Γ1,a (λ)) = σ(N Γ1,b (λ)) and hence λ ∈ σ(L µ b ). Thus, λ = λ µ a,k = λ µ b,jfor some j ≥ 1. The second assertion of Theorem 4.2 shows that λ µ a,k and λ µ b,j

iii)

  As mentioned above, by Lemma 4.3 we have (4.5). The same property holds for L µ b , that is, λ µ b,k → λ D b,k as µ → -∞. It follows from assertion (i) that λ D a,k = λ D b,k for all k ≥ 1 and have the same multiplicity. We conclude as above that L D a and L D b are unitarily equivalent. Finally, another application of Theorem 4.2 shows that Tr(Ker(λ -L µ a )) = Tr(Ker(λ -L µ b )) for λ / ∈ σ(L D a ) = σ(L D b ).

Acknowledgements. The author wishes to thank Jonathan Rohleder for very helpful remarks and discussions on the proof of Lemma 4.4. He wishes also to thank the referee for his/her very careful reading of the manuscript.


Jean-Baptiste Boyer
email: jeaboyer@math.cnrs.fr

ON THE REFLECTED RANDOM WALK ON R +

Keywords: Markov Chains, Poisson's equation, Gordin's method, renewal theorem, random

ou non, émanant des établissements d'enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

Introduction 2 1. Induced Markov chains 3 1.1. Definitions 3 1.2. Application to the study of invariant measures 7 2. Induction and the renewal theorem 9 3. Application to the relfected random walk on R + 14 References 21

Introduction

Let ρ be a borelian probability measure on R having a moment of order 1 and a drift λ = R ydρ(y) < 0.

Consider the random walk on R + starting at x ∈ R + and defined for any n ∈ N by

X 0 = x X n+1 = |X n + Y n+1 |
where (Y n ) is an iid sequence of law ρ.

We note P the Markov operator associated to this random walk. This is the operator defined for any borelian and bounded function f on R + and any x ∈ R + by

P f (x) = R f (|x + y|)dρ(y)
The aim of this article is to study this Markov chain and to do so, we will use a standard technique (known as Gordin's method) which consists in finding a solution g to the "Poisson equation" g -P g = f for f in a certain Banach space.

Using results of Glynn and Meyn, we will prove that under a regularity assumption on the measure ρ, there always is a solution to this equation if f is borelian and bounded but this solution will not be bounded in general and this prevents us from studying the large deviation principle and complicates the study of the central limit theorem and of the law of the iterated logarithm.

However, we will see in section 1 that the solution to Poisson's equation satisfies some equations (see proposition 1.3) and using a "stopped" renewal theorem that we will state in section 2 (see corollary 2.6) we will prove the following Corollary (3.7). Let ρ be an absolutely continuous probability measure on R having a moment of order 1, a negative drift λ = R ydρ(y) < 0 and such that ρ(R * + ) > 0. Let (X n ) be the reflected random walk on R + defined by ρ.

Let ν be the unique stationary probability measure on R + (it exists according to [START_REF] Leguesdron | Marche aléatoire sur le semi-groupe des contractions de R d . Cas de la marche aléatoire sur R+ avec choc élastique en zéro[END_REF] or [START_REF] Peigné | On recurrence of reflected random walk on the half-line. With an appendix on results of Martin Benda[END_REF]).

Then, for any directly Riemann-integrable function f on R + such that f dν = 0, there is a bounded and a.e. continuous function g on R + such that f = g -P g and lim x→+∞ g(x) = 0 This will allow us to prove the following Proposition (3.8 and 3.11). Let ρ be an absolutely continuous probability measure on R having a moment of order 1, a negative drift λ = R ydρ(y) < 0 and such that ρ(R * + ) > 0. Let (X n ) be the reflected random walk on R + defined by ρ. For any directly Riemann-integrable function f on R + such that f dν = 0, there are constants C 1 , C 2 ∈ R * + such that for any ε ∈]0, 1], any x ∈ R + and any n ∈ N * ,

P x 1 n n-1 k=0 f (X k ) ε C 1 e -C 2 ε 2 n
In particular, for any x ∈ R + ,

1 n n-1 k=0
f (X k ) -→ 0 P x -a.e. and in L 1 (P x )

Moreover, is ρ as a moment of order 2 + ε for some ε ∈ R * + , then, for any directly Riemann-integrable function f on R + with f dν = 0 and any x ∈ R + ,

1 √ n n-1 k=0 f (X k ) L -→ N 0, σ 2 (f )
Where we noted N (0, 0) the Dirac mass at 0 and

σ 2 (f ) = R + g 2 -(P g) 2 dν
with g the bounded function given by corollary 3.7 and such that f = g -P g. Moreover, if σ 2 (f ) = 0, then

lim sup n-1 k=0 f (X k ) 2nσ 2 (f ) ln ln(n) = 1 a.e. and lim inf n-1 k=0 f (X k ) 2nσ 2 (f ) ln ln(n) = -1 a.e.
1. Induced Markov chains The aim of this section is to study the process of induction of Markov chains by stopping times and to link the induced chain to the original one. We study it in a general case as it doesn't use any particular property of the reflected random walk.

1.1. Definitions. Let (X n ) be a Markov chain on a standard Borel space X. We define a Markov operator on X setting, for a borelian function f and x ∈ X,

P f (x) = E[f (X 1 )|X 0 = x]
Given a stopping time τ , we can study the Markov chain (X τ n ) n∈N where τ n is defined by

τ 0 ((X n )) = 0 τ k+1 ((X n )) = τ k ((X n )) + τ (θ τ k ((Xn)) (X n ))
where θ stands for the shift on X N .

We note Q the sub-Markov operator associated to (X τ n ), that is, for a borelian function g on X and x ∈ X,

Q(g)(x) = {τ <+∞} g(X τ )dP x ((X n )) If, for any x ∈ X, P x ({τ < +∞}) = 1, then Q is a Markov operator.
Finally, we define two other operator on X setting, for a borelian non negative function f and x ∈ X,

Sf (x) = {τ =1} f (X 1 )dP x ((X n )) (1.1) Rf (x) = {τ <+∞} f (X 0 ) + • • • + f (X τ -1 )dP x ((X n )) (1.2)
Definition 1.1 (θ-compatible stopping times). We say that a stopping time τ is θ-compatible if for all x ∈ X, P x ({τ = 0}) = 0 and for P

x -a.e. (X n ) ∈ X N , τ ((X n )) 2 implies that τ (θ(X n )) = τ ((X n )) -1.
Example 1.2. Let Y be a borelian subset of X and τ Y the time of first return in Y:

τ Y ((X n )) = inf{n ∈ N * ; X n ∈ Y} Then, τ Y is θ-compatible.
Moreover, τ n Y as we defined it coresponds to the time of n-th return to Y. For x ∈ X, we set u(x) = E x τ Y and we call Y strongly Harris-recurrent if u is finite on X. This imply in particular that for any x in X, τ Y is P x -a.e. finite.

Indeed for any borelian non negative function f and any x ∈ X, we have that

Qf (x) = {τ <+∞} f (X τ )dP x = +∞ n=1 E x f (X n )1 {τ =n} = +∞ n=1 E x f (X n )1 Y c (X 1 ) . . . 1 Y c (X n-1 )1 Y (X n ) = +∞ n=1 (P 1 Y c ) n-1 P (f 1 Y ) = +∞ n=0 (P 1 Y c ) n P (f 1 Y ) Rf (x) = {τ <+∞} f (X 0 ) + • • • + f (X τ -1 )dP x = +∞ n=0 E x f (X n )1 {τ n+1} = f (x) + +∞ n=1 E x f (X n )1 Y c (X 1 ) . . . 1 Y c (X n ) = f (x) + +∞ n=1 (P 1 Y c ) n (f )(x) = +∞ n=0 (P 1 Y c ) n (f )(x) Sf (x) = {τ =1} f (X 1 )dP x = f (X 1 )1 Y (X 1 )dP x = P (f 1 Y ) Thus, we have that (R + Q)f = (I d + RP )f , RSf = Qf , (P -S)Qf = P (1 Y c Qf ) = Qf -Sf and (P -S)Rf = P (1 Y c Rf ) = Rf -f .
Note that P, Q, R, S, P -S, Q -S and R -I d are positive operators and so the computations we made make sense for any non negative borelian function f . Next lemma generalizes those relations for any θ-compatible stopping time.

Proposition 1.3. Let τ be a θ-compatible stopping time such that for any x ∈ X, τ is P x -a.e. finite.

For any non negative borelian function f on X, we have :

(R + Q)f = (I d + RP )f (I d + P R)f = (I d + S)Rf (I d + S)Qf = (S + P Q)f RSf = Qf
Proof. Let f be a borelian non negative function on X and x ∈ X.

Using the Markov property and τ being a θ-compatible stopping time, we have that for any n ∈ N * ,

E x f (X n )1 {τ n} = E x P f (X n-1 )1 {τ n} And so, (R + Q)f (x) = E x f (X 0 ) + • • • + f (X τ )dP x = E x +∞ n=0 f (X n )1 {τ n} = f (x) + +∞ n=1 E x P f (X n-1 )1 {τ n} = f (x) + RP f (x) Moreover, as τ is θ-compatible, {τ 2} Rf (X 1 )dP x ((X n )) = {τ 2} f (X 1 ) + • • • + f (X τ -1 )dP x ((X n )) Thus, f (x)+P Rf (x) = f (x) + {τ =1} Rf (X 1 )dP x ((X n )) + {τ 2} Rf (X 1 )dP x ((X n )) = f (x) + SRf (x) + {τ 2} f (X 1 ) + • • • + f (X τ -1 )dP x ((X n )) = SRf (x) + f (X 0 ) + • • • + f (X τ -1 )dP x ((X n )) = SRf (x) + Rf (x) Then, by definition of S, {τ =1} f (X 1 )dP x ((X n )) = {τ =1} f (X τ )dP x ((X n )), so, Sf (x) + P Qf (x) = 1 {τ =1} (f (X 1 ) + Qf (X 1 )) + 1 {τ 2} f (X τ )dP x ((X n )) = SQf (x) + Qf (x) Finally, for any n ∈ N * , E x Sf (X n-1 )1 {τ n} = {τ =n+1} f (X n+1 ) therefore, RSf (x) = E x +∞ n=1 Sf (X n-1 )1 {τ n} = +∞ n=1 E x Sf (X n-1 )1 {τ n} = ∞ n=1 E x f (X n+1 )1 {τ =n+1} = Qf (x)
And this finishes the proof of this proposition.

Lemma 1.4. Let (X n ) be a Markov chain on a standard borelian space X.

Let ν be a finite P -invariant measure on X and τ a θ-compatible stopping time such that for ν-a.e x ∈ X, lim n→+∞ P x (τ n) = 0.

Then, for any non negative borelian function f on X, we have

X f dν = X SRf dν Proof. According to proposition 1.3, f + P Rf = Rf + SRf . So, if Rf ∈ L 1 (X, ν), as ν is P -invariant, we get the lemma. If f ∈ L 1 (X, ν
), we will get the lemma by approximation. First, we assume that f is bounded. In general, Rf ∈ L 1 (X, ν) so, we approximate it with a sequence of integrable functions.

More precisely, for n ∈ N * , we note R n the operator defined like R but associated to the stopping time min(n, τ ) (which is not θ-compatible).

That is to say, for a borelian non negative function f and any x ∈ X,

R n f (x) = E x min(τ,n)-1 k=0 f (X k )
As {min(τ, n) = 1} = {τ = 1} for n 2, the operator S associated to min(τ, n) does not depend on n for n 2.

As min(τ, n) is not θ-compatible, we can't use proposition 1.3, but we have for n 2, that

P R n f (x) = E x min(τ •θ,n)-1 k=0 f (X k+1 ) = SR n f (x) + {τ 2} min(τ -1,n)-1 k=0 f (X k+1 )dP x = SR n f (x) + {τ 2} min(τ,n+1)-1 k=1 f (X k )dP x = SR n f + R n+1 f -f
And, as f is bounded, for any x ∈ X, |R n f (x)| n f ∞ and so R n f is integrable since µ is a finite measure and,

SR n f -f dν = P R n f -R n+1 f dν = R n f -R n+1 f dν = f (X n )1 {τ n} dP x ((X n ))dν(x) = P f (X n-1 )1 {τ n} dP x ((X n ))dν(x) So, SR n f -f dν f ∞ X P x ({τ n})dν(x) -→ 0 (by monotone convergence)
and using the monotone convergence theorem, we get the expected result for borelian bounded functions. If f is not bounded and non-negative, we take an increasing sequence (f n ) of bounded positive functions which converges to f and we get the expected result by monotone convergence.

Example 1.5. If τ is the return time to some strongly Harris-recurrent set Y, then Sf (x) = P (f 1 Y )(x). Moreover for every P -invariant measure ν and every

f ∈ L 1 (X, ν), such that Rf is ν-a.e. finite, X SRf dν = Y Rf dν.
In particular, with f = 1, we have that, Y Eτ dν = ν(X). This is Kac's lemma for dynamical systems.

1.2. Application to the study of invariant measures. In this subsection, X is a complete separable metric space endowed with it's Borel tribe and "measure" stands for "borelian measure". We assume that there exist (at least) a P -invariant probability measure on X.

We also fix a θ-compatible stopping time τ such that for any x in X, E x τ is finite.

Lemma 1.6. Let µ be a finite non-zero P -invariant borelian measure on X. Then, S * µ is a finite non-zero Q-invariant measure on X. Moreover, R * S * µ = µ and S * µ is absolutely continuous with respect to µ.

Proof. First, for all non negative f ∈ B(X) and all x ∈ X, Sf (x) P f (x). So, Sf dµ P f dµ = f dµ since µ is P -invariant and f is bounded. And this proves that S * µ is absolutely continuous with respect to µ. So, as Fubuni's theorem proves that it is σ-additive, S * µ is a finite measure on X.

Moreover, we saw in lemma 1.4 that for all non negative borelian function f on X, SRf dµ = f dµ and this proves that R * S * µ = µ.

Then, we need to prove that S * µ(X) > 0. But, for all x ∈ X,

P k S(1)(x) = E x S1(X k ) P x ({τ = k + 1}) So, n-1 k=0 P k S(1) P x ({τ n + 1}).
And, as µ is P -invariant, taking the integral on both sides, we get that,

nS * µ(X) x∈X P x ({τ n})dµ(x)
Finally, we use that for µ-a.e. x ∈ X, lim n P x (τ n) = 1 and the dominated convergence theorem, tells us that 0 < µ(X) lim nS * µ(X), so S * µ(X) > 0.

Lemma 1.7. Let ν be a non-zero Q-invariant borelian measure on X. Then, R * ν is a non zero P -invariant measure on X.

Moreover, S * R * ν = ν and ν is absolutely continuous with respect to R * ν.

Finally, if QR(1) is bounded on X, then R * ν is a finite measure if and only if ν is.

Remark 1.8. The technical assumption QR1 bounded on X is reasonable. More specifically, using the same notations as in remark 1.2, we call Y linearily recurrent if sup y∈Y E y τ Y is finite.

In this case,

R1(x) = E x τ Y and QR1(x) = E x R1(X τ Y ) sup y∈Y E y τ Y since for any x ∈ X, P x (X τ Y ∈ Y) = 1 be definition of τ Y .
Proof. To prove that R * µ is a measure, one just have to prove that it is σ-additive.

Let (A n ) be a sequence of pairwise disjoint borelian subsets of X and n ∈ N. As R is a linear operator, we have that R(1

∪ n k=0 A k )dν = n k=0 R1 A k dν, thus, R * ν
is finitely additive. But, according to the monotone convergence theorem, the left side of this equation converges to R(1 ∪An )dν and this finishes the proof that R * ν is σ-additive.

Moreover, for all non negative f ∈ B(X), f Rf , so ν(f ) ν(Rf ) and ν is absolutely continuous with respect to R * ν and R * ν(X) > 0.

Then, proposition 1.3 shows that for any positive borelian function f , Rf + Qf = f + RP f . Applying this to f = 1 A for some borelian set A, and taking the integral over ν, we get that

R1 A + Q1 A dν = 1 A + RP 1 A dν. But, ν is Q-invariant so if ν(A) is finite, we get that R1 A dν = RP 1 A dν. If ν(A) is infinite, the result still holds since in this case, R1 A dν = ν(A) = Q * ν(A) = RP 1 A dν = +∞. Thus, for any borelian set A, R * ν(A) = P * R * ν(A) that is to say, R * ν is P -invariant.
As RS = Q and ν is Q-invariant, we directly have that S * R * ν = ν.

For the last point, assume that QR(1) is a bounded function on X. If R * ν is finite, then so is ν since ν(X) R * ν(X). Assume that ν is finite. Then according to Chacon-Ornstein's ergodic theorem (see chapter 3 theorem 3.4 in [START_REF] Krengel | Ergodic theorems[END_REF]), there exist a Q-invariant non negative borelian function g * such that g * dν = R1dν and for ν-almost every x ∈ X,

1 n n-1 k=0 Q k R1(x) -→ g * (x)
And, since QR is bounded on X and

R1(x) = E x τ is finite, we get that g * (x) QR ∞ for ν-a.e x ∈ X. So, g ∈ L ∞ (X, ν) ⊂ L 1 (X, ν) since ν(X) < +∞ and R1dν QR ∞ ν(X) < +∞.
We saw in the previous lemmas that R ans S act on invariant measure. As they are linear operators and the set of invariant measures is convex, next proposition shows that they also preserve the ergodic measures (in some sense since they do not preserve probability measures).

Corollary 1.9. Let (X n ) be a Markov chain on a complete separable metric space X and τ a θ-compatible stopping time such that for any x ∈ X, E x τ is finite. Define P , Q, R and S as previously and assume that QR1 is bounded on X.

Then, S * and R * are reciproqual linear bijections between the P -invariant finite measures and the Q-invariant ones which preserve ergodicity.

Proof. We already saw in lemma 1.6 and 1.7 that S * (resp R * ) maps the P -invariant (resp. Q-invariant) finite non zero measures onto the Q-invariant (resp. P -invariant) ones and that they are reciproqual to each-other. Thus, it remains to prove that the image by S * or R * of an ergodic measure still is ergodic. To do so, we use the linearity of S * and R * and that ergodic probability measures are extreme points of the set of invariant probability measures for a Markov chain in a complete separable metric space.

Let µ be a P -ergodic finite non zero measure. We assume without any loss of generality that µ is a probability measure. We saw in lemma 1.6 that S * µ is a Q-invariant non zero finite measure.

Assume that S * µ = S * µ(X)(tν 1 + (1t)ν 2 ) where ν 1 and ν 2 are two Q-invariant probability measures and t ∈ [0, 1].

Then, we get that

µ = R * S * µ = S * µ(X)(tR * ν 1 + (1 -t)R * ν 2 ). But µ is ergodic, so 1 R * ν 1 (X) R * ν 1 = 1 R * ν 2 (X) R * ν 2 .
And applyting S * again, we obtain that

ν 1 = ν 2 , hence, S * µ is Q-ergodic.
The same proof holds to show that if ν is Q-ergodic, then R * ν is P -ergodic.

Induction and the renewal theorem

In this section, we use the renewal theorem on R to prove a "stopped renewal theorem" in corollary 2.6.

Let ρ be a borelian probability measure on R and define a random walk on R starting at x ∈ R by (2.1)

X 0 = x X n+1 = X n + Y n+1 where (Y n ) has law ρ ⊗N .
We assume that ρ has a moment of order 1 and a negative drift λ = R ydρ(y) < 0.

In particular, for ρ ⊗N -a.e (Y n ) ∈ R N , n k=0 Y k converges to -∞. We note P the Markov operator associated to ρ. This is the operator defined for any bounded borelian function f on R and any x ∈ R by

P f (x) = R f (x + y)dρ(y)
We note τ the time of first return to ] -∞, 0] :

τ ((X n )) = inf{n ∈ N * , X n ∈] -∞, 0]}
This is a ϑ-compatible stopping time and our assumption on ρ implies (see P1 in section 18 of [START_REF] Spitzer | Principles of random walk[END_REF]) that for any x ∈ R, E x τ < +∞ In this section, we are interested in the operator R defined as in section 1 for any non negative borelian function f on R and any x ∈ R by

Rf (x) := E x τ -1 k=0 f (X k )
The study of the operator is very close from renewal theory : indeed, if ρ(R + ) = 0 and f is null on R * -, then for any x ∈ R,

Rf (x) = +∞ n=0 P n f (x)
Therefore, we make the following definition that is usual in renewal theory :

Definition 2.1. Let f be a borelian function on R. We say that f is directly Riemannintegrable if

lim h→0 + h n∈Z inf x∈[nh,(n+1)h] |f (x)| = lim h→0 + h n∈Z sup x∈[nh,(n+1)h] |f (x)| < +∞
In the sequel, we will use the following characterisation Lemma 2.2 (Lebesgue's criterion for Riemann-integrability). Let f be a bounded function on R.

Then, f is directly Riemann integrable if and only if it is a.e. continuous and for some

h ∈ R * + , n∈Z sup x∈[nh,(n+1)h] |f (x)| < +∞
In the next three lemmas, we are going to prove that, noting R, S the operators defined as in section 1 and associated to τ , then for any directly Riemann-integrable function f , SRf is also directly Riemann-integrable.

Lemma 2.3. Let ρ be a borelian probability measure on R having a moment of order 1 and a negative drift λ = R ydρ(y) < 0.

Note τ the time of first return to ] -∞, 0] and R the associated operator defined as in section 1.

Then, for any directly Riemann-integrable function f on R, the function Rf is bounded on R.

Proof. To prove this proposition, we are going to use the classical renewal theorem. Indeed, for any x ∈ R we have that

|Rf (x)| R|f |(x) = E x τ -1 k=0 |f (X k )| E x +∞ n=0 |f (X n )| = +∞ n=0 P n |f |(x)
But, if the measure is non-lattice1 , according to the renewal theorem (see [Fel71]), we have that

lim x→-∞ +∞ n=0 P n |f |(x) = 0 and lim x→+∞ +∞ n=0 P n |f |(x) = -1 λ R |f |dν
and this proves our lemma in the non-lattice case. The same kind of arguments holds in the lattice case and allows us to conclude.

Lemma 2.4. Let ρ be a borelian probability measure on R having a moment of order 1 and a negative drift λ = R ydρ(y) < 0.

Then, for any directly-Riemann integrable function f on R that is null on R -, we have that

n∈Z sup x∈[n,(n+1)] |SRf (x)| < +∞
Proof. First, according to the lemma 2.3, Rf is bounded and for any x ∈ R,

|SRf (x)| Rf ∞ S1(x) = Rf ∞ ρ(] -∞, -x])
So, as the function (x → ρ(] -∞, -x])) is decreasing, we have that

n∈N sup x∈[n,(n+1)] |SRf (x)| Rf ∞ n∈N ρ(] -∞, -n])
Moreover, using that

ρ(] -∞, -n]) = +∞ m=n ρ(] -m -1, -m])
We have that We now have to control the sum over Z -. Since for any x ∈ R -, f (x) = 0, we have, using Markov's property, that for any x ∈ R,

SRf (x) = E x 1 {τ =1} τ 2 -1 k=1 f (X k ) = E x 1 {τ =1,τ 2 >2} τ 2 -1 k=2 f (X k ) = E x 1 {τ =1,τ 2 >2} Rf (X 2 ) So, |SRf (x)| E x 1 {τ =1}∩{τ 2 >2} |Rf (X 2 )| Rf ∞ P x (τ = 1, τ 2 > 2)
But,

P x (τ = 1, τ 2 > 2) = R 2
1 {x+y 1 0,x+y 1 +y 2 >0} dρ(y 1 )dρ(y 2 ) ρ * 2 (]x, +∞[) So, using the fact that the function (x → ρ * 2 (]x, +∞[)) is non-decreasing, we have that

0 n=-∞ sup x∈[n-1,n] |SRf (x)| Rf ∞ +∞ n=0 ρ * 2 (]n, +∞[) = Rf ∞ +∞ n=0 +∞ m=n ρ * 2 (]m, m + 1]) Rf ∞ +∞ m=0 (m + 1)ρ * 2 (]m, m + 1]) Rf ∞ 1 + +∞ 0 ydρ(y)
And this finishes the proof of the lemma.

Lemma 2.5. Let ρ be a borelian probability measure that is the sum of an absolutely continuous measure ρ 1 and a discrete measure ρ 2 . Then, for any bounded and a.e. continuous function f on R, Rf and Sf are also a.e. continuous.

Proof. For any x ∈ R, we have that

Sf (x) = R f (x + y)1 {x+y 0} dρ(y) = R f (x + y)1 {x+y 0} ϕ(y)dy + y∈supp ρ 2 f (x + y)1 {x+y 0} ρ 2 (y)
Where we noted ϕ the density of ρ 1 and we used the fact that ρ 2 is atomic.

But, x → f (x)1 {x 0} is bounded and ϕ is integrable so the function

x → R f (x + y)1 {x+y 0} ϕ(y)dy
is continuous on R (as a convolution product of an integrable and a bounded function). And, as supp ρ 2 is denumerable and f is a.e. continuous, the function

x → y∈supp ρ 2 f (x + y)1 {x+y 0} ρ 2 (y)
still is a.e. continuous. This proves that Sf is a.e. continuous.

To prove that Rf is a.e. continuous, note that for any x ∈ R, we can write

Rf (x) = +∞ n=0 E x f (X n )1 {τ >n}
Moreover, we have that

E x f (X n )1 {τ >n} f ∞ P x (τ > n)
so, using the fact that the function (x → P x (τ > n)) is non-decreasing, we have that for any x 0 ∈ R and any x ∈] -∞, x 0 ],

E x f (X n )1 {τ >n} f ∞ P x 0 (τ > n)
So the convergence in the series is uniform on every compact subset of R since we already saw that

n P x 0 (τ > n) = E x 0 τ < +∞
Therefore, the set of continuity points of Rf contains the intersection of the sets of continuity points of the functions x → E x f (X n )1 {τ >n} for n ∈ N. Moreover, for any x ∈ R and any n ∈ N,

E x f (X n )1 {τ >n} = R n f (x + y 1 + • • • + y n )1 ∀k∈[1,n], x+y 1 +•••+y k >0 dρ ⊗n ((y i ))
And we can see that this function is a.e. continuous by using the same kind of arguments than in the proof of the a.e. continuity of Sf .

Corollary 2.6. Let ρ be a borelian probability measure on R having a moment of order 1 and a negative drift λ = R ydρ(y) < 0. Assume that ρ is non lattice 2 and the sum of an absolutely continuous measure and of a discrete one.

Note τ the time of first return to ] -∞, 0] and note R and S the Markov operators defined for any borelian bounded function f on R and any x ∈ R by

Rf (x) = E x τ -1 k=0 f (X k ) and Sf (x) = {τ =1} f (X 1 )dP x ((X n ))
Then, for any directly Riemann-integrable function f on R,

lim x→+∞ Rf (x) = -1 λ R (I d -SR)f (u)du
Proof. By definition of the operator R, we shall assume without any loss of generality that f = 0 on R -.

According to the previous lemmas (2.2, 2.3, 2.4 and 2.5) we have that f -SRf is directly Riemann-integrable. Thus, we can apply the renewal theorem to get that

lim x→+∞ +∞ n=0 P n (I d -SR)f (x) = -1 λ R (I d -SR)f (u)du But, for any n ∈ N, n-1 k=0 P k (I d -SR)f (x) = n-1 k=0 P k (I d -P )Rf (x) = Rf (x) -P n Rf (x)
and, as f is null on R -and Rf is bounded according to lemma 2.3, we have that

lim n→+∞ P n Rf (x) = 0 2 For any α ∈ R, ρ(αZ) < 1.
Thus, for any x ∈ R,

Rf (x) = +∞ n=0 P n (I d -SR)f (x)
Which is what we intended to prove.

3. Application to the relfected random walk on R +

In this section, we use the previous results for the stopped renewal theorem to study the regularity at infinity of the solution of Poisson's equation for the reflected random walk on R + .

Let ρ be a probability measure on R such that R |y|ρ(dy) is finite, R yρ(dy) < 0 and ρ(R * + ) = 0. These last two asumptions means that for ρ ⊗N

-a.e (Y n ) ∈ R N , n k=0 Y k converges to -∞ but for any fixed M ∈ R, ρ ⊗N ({(Y n ) ∈ R N ; ∃n ∈ N n-1 k=0 Y k M }) > 0.
Let (Y n ) n∈N be an iid seqence of random variables of law ρ. We define the reflected random walk starting at x on R + by

X 0 = x for all n ∈ N, X n+1 = |X n + Y n+1 | Defined like this, (X n , Y n ) n∈N is a Markov chain on R + × R.
As Peigné and Woess in [START_REF] Peigné | On recurrence of reflected random walk on the half-line. With an appendix on results of Martin Benda[END_REF], we define a stopping time which we call the time of first reflection by

(3.1) τ ((X n , Y n )) = inf{n ∈ N * ; X n-1 + Y n < 0}
We see that τ is θ-compatible since it is the time of first return in R * -for the unreflected random walk on R driven by ρ.

Since yρ(y)dy < 0, n-1 k=0 Y k ---→ n→∞ -∞ a.e., and τ n x is well defined (finite almost everywhere for all positive real number x).

We define the operator P , Q, R and S as in section 1.

Remark 3.1. For a borelian non negative function f on R + × R we defined P (f )(x, y) = E (x,y) f (X 1 , Y 1 ). But, since X 1 and Y 1 are independant of Y 0 , we have that

P (f )(x, y) = E x f (X 1 , Y 1 ) = R f (|x + y|, y)ρ(dy).
In particular, if f itself does not depend of it's second argument, we have P (f )(x, y) = E (x,y) f (X 1 ) = E x f (X 1 ) and we find the usual Markov operator associated to (X n ). The same argument applies to Q, R and S (defined as in section 1) Those considerations are just made to prove that τ is θ-compatible so we can apply the results of the previous sections, but we can anyway "forget" about the second variable.

From now on, we identify functions on R + and functions on R + × R which don't depend on their second variable and we make the abuse of notations that come with this identification.

As we will need some regularity assumption on ρ, we make the following Definition 3.2 (Spread-out probability measure on R). We say that a probability measure ρ on R is spread-out if there exist m ∈ N * such that ρ * m is not singular with respect to Lebesgue's measure on R.

First, we have the following Theorem 3.3 (Leguesdron [START_REF] Leguesdron | Marche aléatoire sur le semi-groupe des contractions de R d . Cas de la marche aléatoire sur R+ avec choc élastique en zéro[END_REF], Peigné-Woess [START_REF] Peigné | On recurrence of reflected random walk on the half-line. With an appendix on results of Martin Benda[END_REF]). Let ρ be a spread-out probability measure on R having an moment of order 1 and a negative drift λ = R ydρ(y). Consider (X n ), the reflected random walk associated to ρ.

There exist a unique P -invariant probability measure ν on R + . Moreover, if ρ(R * + ) = 0, then supp ν = [0, M ], else supp ν = R + where M =inf supp (ρ) (which may be infinite).

Finally, the reflected random walk is topologically irreducible on supp ν.

To solve Poisson's equation, we are going to use the theory of petite sets developed by Glynn, Meyn and Tweedie (see [START_REF] Glynn | A Liapounov bound for solutions of the Poisson equation[END_REF] and [MT93]).

Definition 3.4. Let a be a probability measure on N. A set C ⊂ X is called ν-petite where ν is a non trivial measure on X if for any borelian subset A of X and any x ∈ C, n∈N

a n P n (1 A )(x) ν(A)
Proposition 3.5. Let ρ be a spread out probability measure on R having a moment of order 1 and such that ρ(R * + ) > 0. Then the reflected random walk on R + defined by ρ is irreducible and every compact set is petite.

Proof. As we already saw in theorem 3.3, if |y|ρ(dy) is finite, reflected random walk is open-set irreducible on R + .

Therefore, is we prove that (X n ) is a T-chain (see [MT93]), using the first point of Theorem 6.0.1 in [MT93] we will get that the chain is m-irreducible and then, using the second point of this theorem, we will get that every compact set is petite.

We need to find (a n ) ∈ [0, 1] N such that n a n = 1 and a substochastic transition kernel T such that ∀x ∈ X T (1)(x) > 0, for any borelian set A, T (1 A ) is lower semicontinuous and

∀x ∈ X ∀A ∈ B(X) +∞ n=0 a n P n (1 A )(x) T (1 A )(x)
We assume without any loss of generality that ρ is compactly supported and we note M ∈ R * + such that supp ρ ⊂ [-M, +∞[) Let m be such that ρ * m is not singular with respect to Lebesgue's measure. We note ψ it's Radon-Nikodym's derivative.

Let

ε ∈ R * + such that ρ([ε, +∞[) > 0 (such ε exists since ρ(R * + ) > 0). We note ρ ε the measure defined by ρ ε (A) = ρ(A ∩ [ε, +∞[) X ε
n the random walk assoicated to ρ ε and P ε the submarkov operator associated to ρ ε that is to say :

P ε (f )(x) = R f (|x+y|)dρ ε (y) = +∞ ε f (x + y)dρ(y).
The main idea of this proof is that, using ρ ε , we can "escape" any compact set, and in particular, if we walk for a long enough time N with ρ ε (N such that N ε mM ), we can be sure that the time of first reflection for the walk starting at X ε N is greater than m. And thus, we can use the hypothesis that ρ is spread out.

More precisely, if f is a non negative borelian function on R + and N ∈ N is such that N ε mM , then,

P m+N f (x) R m+N f (||x + y 1 | + • • • + y m+N |)dρ ⊗N ε ⊗ dρ ⊗m ((y i )) R m+N f (x + y 1 + • • • + y m+N )dρ ⊗N ε ⊗ dρ ⊗m ((y i )) R 2 f (x + v + u)dρ * N ε (v)dρ * m (u) R 2 f (x + v + u)dρ * N ε (v)ψ(u)du R f (x + u) R ψ(u -v)dρ * N ε (v)du = R f (x + u)ψ 1 (u)du
Where, for u ∈ R,

ψ 1 (u) = R ψ(u -v)ρ * N ε (dv). For n ∈ N, let ψ n 2 (u) = min(n, ψ 1 (u)) ∈ L 1 ∩ L ∞ .
By definition, ψ n 2 is a non decreasing sequence of positive functions and using the monotone convergence theorem, we have that

ψ n 2 (u)du -----→ n→+∞ ψ 1 du = ρ ε (R) N = ρ([ε, +∞[) N ∈ ]0, 1]
. So there exist n 0 ∈ N such that ψ n 0 2 (u)du = 0. We note ψ n 0 2 = ψ 2 and we resume our computations. By construction,

P 2m+2N f (x) R P m+N f (x + u)ψ 2 (u)du R 2 f (x + u + v)ψ 2 (v)ψ 2 (u)dudv R f (x + u) R ψ 2 (u -v)ψ 2 (v)dv du = R f (x + u)ψ 3 (u)du Where ψ 3 (u) = R ψ 2 (u -v)ψ 2 (v)dv.
But, ψ 3 is the convolution product is of an integrable and a bounded function so it is continuous and R ψ 3 (u)du = R ψ 2 (u)du 2 > 0.

Let ψ 4 be a non zero non negative continuous function on R such that ψ 4 ψ 3 and supp ψ 4 is compact.

We note T the operator defined for any borelian bounded function f on R and any x ∈ R by

T f (x) = R f (x + u)ψ 4 (u)du
Using the dominated convergence theorem we have that for any borelian and bounded function f , T f is continuous. Moreover, for all x ∈ R + , 1 T 1(x) = R ψ 4 dλ > 0.

And, we get that for every borelian non negative function f and every x in R + ,

P 2N +2m f (x) T f (x)
and this finishes the proof of the proposition.

Proposition 3.6. Let ρ be a spread out probability measure on R having a moment of order 1 and such that 0 < R + ydρ(y) < R -(-y)dρ(y) . Then, there is a constant C such that for any borelian and bounded fonction f on R + , there is a function g on R + such that f = g -P g + f dν And,

sup x∈R |g(x)| 1 + |x| C f -f dν ∞
Moreover, if ρ is the sum of an absolutely continuous measure and of a discrete one and if f is a.e. continuous, then so is g.

Proof.

For any x ∈ R + , let u(x) = 1 + x -λ where we noted λ = R ydρ(y) < 0. Then,

P u(x) = 1 + -x -∞ x + y λ dρ(y) + +∞ -x -x -y λ dρ(y) = 1 + x λ (ρ(] -∞, -x]) -ρ(] -x, +∞])) - 1 λ +∞ -x ydρ(y) - -x -∞ ydρ(y) = 1 + x -λ ) (1 -2ρ(] -∞, -x])) -1 + 2 λ -x -∞ ydρ(y) = u(x) -1 + 2 x λ ρ(] -∞, -x]) + 2 λ -x -∞ ydρ(y) Moreover, xρ(] -∞, -x]) = -x -∞ xdρ(y) -x -∞ |y|dρ(y)
So, as ρ has a moment of order 1,

lim x→+∞ P u(x) -u(x) = -1
Thus, there are x 0 ∈ R + and b ∈ R such that

P u u - 1 2 + b1 [0,x 0 ]
Therefore, using proposition 3.5 we can apply the theorem 2.3 in [START_REF] Glynn | A Liapounov bound for solutions of the Poisson equation[END_REF] and get a constant C such that for any borelian bounded function f on R + there is a borelian function g such that sup

x |g(x)| u(x) C f -f dν ∞ and f = g -P g + f dν
The fact that g is a.e. continuous when f is a.e. continuous is also proved since we have an explicit formula for the function g given in [START_REF] Glynn | A Liapounov bound for solutions of the Poisson equation[END_REF] and using the same ideas as in the proof of lemma 2.5.

From now on, the assumption on ρ being only spread-out is not enough (since the stopped renewal theorem we have in this case doesn't hold for these probability measures) so we are going to ask that it is absolutely continuous instead.

Corollary 3.7. Let ρ be an absolutely continuous probability measure on R having a moment of order 1, a negative drift λ = R ydρ(y) < 0 and that ρ(R * + ) = 0. Then, for any directly Riemman-integrable function f on R + such that f dν = 0, there is a bounded and a.e. continuous function g on R + such that f = g -P g and lim x→+∞ g(x) = 0 Proof. According to the previous lemma, there is an a.e. continuous function g on R + such that f = g -P g and sup

x∈R + |g(x)| 1 + |x|
We note τ the time of first reflection. This is the stopping time defined by

τ ((X n )) = inf{n ∈ N | X n+1 = -X n -Y n }
Moreover, we note R and S the Markov operators associated to τ and defined as in section 1.

Note that for x 0 and before the reflection, the walk is the same as the unreflected random walk. Therefore, as in section 2 we have that for any x ∈ R + ,

E x τ < +∞
The stopping time τ is ϑ-compatible so, we can use the relations of proposition 1.3 to get that for any x ∈ R + ,

Rf (x) = R(g -P g)(x) = g(x) -Qg(x)
Moreover, we also have that Qg(x) = RSg(x) so, we get that g

(x) = R(f + Sg)(x) Moreover, |Qg(x)| = |Eg(X τ )| CE x 1 + X τ
But, using that 0 X τ -Y τ , we have that

E x X τ -E x Y τ R |y|dρ(y)
So the function Qg is bounded on R + and as Rf is bounded on R + (according to lemma 2.3 that we can use for x 0 since the operator R for the reflected random walk and for the unreflected one are the same), this proves that g is bounded on R + . Thus, using lemma 2.4 and lemma 2.5, we get that Sg is directly Riemann-integrable. Therefore, we can apply corollary 2.6 to the function f + Sg to get that g = R(f + Sg) has a limit l at infinity. Noting g 1 = gl we have that g 1 is a.e. continuous, bounded,

g 1 -P g 1 = f and lim x→+∞ g 1 (x) = 0
And this is finally what we intended to prove. Proposition 3.8 (Large deviations inequality). Under the assumptions of corollary 3.7.

For any directly Riemann-integrable function f on R + such that f dν = 0, there are constant C 1 , C 2 ∈ R * + such that for any ε ∈]0, 1], any x ∈ R + and any n ∈ N * ,

P x 1 n n-1 k=0 f (X k ) ε C 1 e -C 2 ε 2 n
In particular, for any x ∈ R + ,

1 n n-1 k=0 f (X k ) -→ 0 P x -a.e.
Proof. To prove the result, we are going to use Asuma-Hoeffding's inequality. First, we write f = g -P g where g is the function given by corollary 3.7. Write, for any n ∈ N,

1 n n-1 k=0 f (X k ) = 1 n n-1 k=0 g(X k+1 ) -P g(X k ) + 1 n (g(X 0 ) -g(X n )) Thus, 1 n n-1 k=0 g(X k+1 ) -P g(X k ) 1 n n-1 k=0 f (X k ) - 2 n g ∞
and so, using Azuma-Hoeffding's inequality, if nε > 2 g ∞ , we have that

I n (x) : = P x 1 n n-1 k=0 f (X k ) > ε P x n-1 k=0 g(X k+1 ) -P g(X k ) nε -2 g ∞ 2 exp -(nε -2 g ∞ ) 2 2n(2 g ∞ ) 2
and this last inequality is what we intended to prove. The law of large numbers now comes from Borel-Cantelli's lemma.

To prove the central limit theorem and the law of the iterated logarithm, we will need a weaker version of the law of large numbers (to show that the variance converges). This will be the following Lemma 3.9. Let ρ be an absolutely continuous probability measure on R having an moment of order 2 + ε for some ε ∈ R * + and such that 0 < R + ydρ(y) < R -(-y)dρ(y). For any borelian and bounded function f on R + and any x ∈ R + ,

1 n n-1 k=0 f (X k ) -→ f dν P x -a.e.
Remark 3.10. In this lemma, we don't ask the function f to be directly Riemannintegrable but the price we have to pay is a stronger moment hypothesis on ρ.

Proof. Write f = g -P g + f dν with g the function given by 3.6 such that C(g) = sup 1 n s < +∞ and this proves, using the law of large numbers for martingales (see theorem 2.18 in [START_REF] Hall | Martingale limit theory and its application[END_REF]) that if ρ has a moment of order 2 + ε for some ε ∈ R * + , then for any x ∈ R,

1 n n-1 k=0
g(X k+1 ) -P g(X k ) -→ 0 P x -a.e. and in L 1 (P x ) Doing the same kind of computations, we also prove that 1 n g(X n ) converges to 0 and this proves the expected result.

Corollary 3.11. Let ρ be an absolutely continuous probability measure on R having an moment of order 2 + ε for some ε ∈ R * + and such that 0 < R + ydρ(y) < R -(-y)dρ(y). Then, for any directly Riemann-integrable function f on R + with f dν = 0 and any

x ∈ R + , 1 √ n n-1 k=0 f (X k ) L -→ N 0, σ 2 (f ) Where σ 2 (f ) = R + g 2 -(P g) 2 dν
with g the bounded function given by corollary 3.7 and such that f = g -P g. Moreover, if σ 2 (f ) = 0, then lim sup n-1 k=0 f (X k ) 2nσ 2 (f ) ln ln(n) = 1 a.e. and lim inf n-1 k=0 f (X k ) 2nσ 2 (f ) ln ln(n) = -1 a.e.

Proof. First, we use that

n-1 k=0 f (X k ) = g(X 0 ) -g(X n ) + n-1 k=0
g(X k+1 ) -P g(X k )

Let M n = n-1 k=0 g(X k+1 ) -P g(X k ). Then, M n is a martingale with bounded difference sequence.

Moreover, noting (F n ) the filtration associated to this martingale, we have that

E x (M n+1 -M n ) 2 |F n = E x g(X n+1
) 2g(X n+1 )P g(X n ) + (P g(X n )) 2 |F n = P (g 2 )(X n ) -(P g) 2 (X n )

And so,

1 n n-1 k=0 E x (M k+1 -M k ) 2 |F k = 1 n n-1 k=0 P g 2 (X k ) + 1 n n-1 k=0 (P g) 2 (X k )
But, the function g is bounded on R, so, according to lemma 3.9,

lim n→+∞ 1 n n-1 k=0 E x (M k+1 -M k ) 2
|F k = g 2 dν -(P g) 2 dν P xa.e. and in L 1 (P x ) Thus, using the central limit theorem and the law of the iterated logarithm for martingales with bounded increments (cf. [Bro71] and theorem 4.8 and corollary 4.2 in [HH80]), we get the expected result.

  -(m + 1), -m]) = m∈N (m + 1)ρ(] -(m + 1), -m]) = ρ(R -) + m∈N mρ(] -(m + 1), -m])

  k+1 ) -P g(X k ) Moreover, for any s ∈ [1, +∞[,E x |g(X k+1 ) -P g(X k )| s 2 s P k+1 |g| s (x) 2 s C(g) s P k+1 u s (x) with u s (x) = (1 + x) s .But, doing the same computations as in the proof of lemma 3.6, we see that if ρ has a moment of order s + 1, then there areC, B ∈ R * + such that u s C (u s+1 -P u s+1 + B) so, +∞ n=1 1 n s E x |g(X k+1 ) -P g(X k )| s CC(g) s 2 s +∞ n=1 1 n s P n+1 (u s+1 -P u s+1 + B) C2 s C(g) s +∞ n=1 P n+1 (I d -P )u s+1 n s C2 s C(g) s u s+1 + B +∞ n=1

For any α ∈ R, ρ(αZ) < 1