Uniqueness results in the inverse spectral Steklov problem
Germain Gendron

	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	Uniqueness results in the inverse spectral Steklov problem
    Germain Gendron 

		  Laboratoire de Mathématiques Jean Leray, UMR CNRS 6629, 
 2 Rue de la Houssinière BP 92208, F-44322 Nantes Cedex 03. 

		Email: germain.gendron@univ-nantes.fr 
    December 30, 2023
=========================================================================================================================================================================================


	
	
	
		
		This paper is devoted to an inverse Steklov problem for a particular class of n-dimensional manifolds having
		the topology of a hollow sphere and equipped with a warped product metric. We prove that the knowledge of the Steklov spectrum determines uniquely
		the associated warping function up to a natural invariance.
		
		
		
		
		

		
		Keywords. Inverse Calderón problem, Steklov spectrum, Weyl-Titchmarsh functions, Nevanlinna theorem, local Borg-Marchenko theorem.
		
		
		
		
		
	
		
				

	
	

	

	
	
	

§ INTRODUCTION

	
	

 §.§ The Calderón and Steklov problems.

	Let (M,g) be a smooth compact manifold of dimension n ≥ 2 with smooth boundary ∂ M. We consider the Dirichlet problem
	
    {    -Δ_g u=λ u   in  M
           u=ψ  on  ∂ M ,
    	.

	where ψ∈ H^1/2(∂ M) and λ∈ℝ is assumed to lie outside the Dirichlet spectrum σ(-Δ_g) of the Laplace-Beltrami operator -Δ_g.
	In a local coordinate system (x^i)_i=1, ...n, -Δ_g has the expression
	
	
    -Δ_g=-∑_1≤ i,j≤ n1/√(|g|)∂_i(√(|g|)g^ij∂_j),

	
	where we have set |g|=(g_ij) and (g^ij)=(g_ij)^-1.
	
	
	
	If λ∉σ(-Δ_g), the Dirichlet problem (<ref>) has a unique solution  u ∈ H^1(M), and we can define the so-called
	Dirichlet-to-Neumann (DN) operator as the map
	
    [        Λ_g(λ)             :    H^1/2(∂ M)             →   H^-1/2(∂ M);                                         ψ             ↦ ∂ u/∂ν|_∂ M ,;               ]

	
	where ∂_ν is the unit normal derivative with respect to the unit outer normal vector on ∂ M. This normal derivative has to be understood in the weak sense by :
	
    ∀ (ψ,ϕ)∈ H^1/2(∂ M)^2  : ⟨Λ_g(λ)ψ,ϕ⟩=∫_M⟨ du,dv⟩_g dVol_g+λ∫_Muv dVol_g,

	where u is the unique weak solution of the Dirichlet problem (<ref>), and where v is any element of H^1(M) such that v_|∂ M = ϕ. When ψ is sufficiently smooth,
	this definition coincides with the usual one in local coordinates, that is
        
    ∂_ν u = ν^i ∂_i u.

	The anisotropic Calderón problem can be initially stated as: does the knowledge of the DN map Λ_g(λ) at a fixed frequency λ determine uniquely
	the metric g ?
	
	Due to a number of gauge invariances, it is well-known that the answer to the above question is negative in general.
	An observation made by Luc Tartar (<cit.>, p.2) leads to the equality :
	
    Λ_g(λ)=Λ_ψ^*g(λ),

	
	where ψ:M→ M is any smooth diffeomorphism which is equal to the identity on the boundary, (here ψ^*g is the pullback of g by ψ). Moreover, in dimension n=2 and for λ=0, there is one more gauge invariance. Indeed, thanks to the conformal invariance of the Laplacian, for every positive function c, we have
	
    Δ_cg=1/cΔ_g.

	
	Consequently, the solutions of the Dirichlet problem (<ref>) associated to the metrics g and cg are the same when λ=0. Moreover, if we assume that c≡ 1 on the boundary, the unit outer normal vectors on ∂ M are also the same for both metrics. Therefore,
	
    Λ_cg(0)=Λ_g(0),

	and it is not possible to determine uniquely the metric from the DN map.
		
	Hence, the appropriate question to adress is :
	
	

		Assume n≥ 3 (resp. n=2 and λ0). If Λ_g(λ)=Λ_g̃(λ), is there a smooth diffeomorphism ψ:M→ M with ψ|_∂ M=id and ψ^*g=g̃ ?
		


	This problem is still largely open. Some special cases have been answered positively (see <cit.> and <cit.> for
	a presentation of the latest developments) but the general case seems to be very difficult to tackle. However, the question becomes simpler if we assume that (M,g) and (M,g̃)
	belong to the same conformal class. Precisely :
	
	

		For c∈ C^∞(M), assume Λ_g(λ)=Λ_cg(λ). Then, is there a smooth diffeomorphism ψ:M→ M such that  ψ|_∂ M=id
		and ψ^*(g)=cg  ?
	




Actually, we can precise the above question thanks to the following result due to  Lionheart (<cit.>) : any diffeomorphism ψ :M→ M  which satisfies ψ^*(cg)=g and
 ψ|_∂ M=id must be the identity. Then, the anisotropic Calderón inverse problem within the same conformal class can be replaced by

 

		If Λ_g(λ)=Λ_cg(λ), is it true that c=1  ?
	




Some deep results have been obtained in <cit.> for conformally transversally anisotropic manifolds (M,g) of dimension n ≥3, i.e for manifolds

    M⊂⊂ℝ× K,   g=c(x,y_K)(dx^2+g_K)




where (K,g_K) is a smooth compact manifold of dimension n-1. Under some geometrical conditions on K, such as simplicity
(a compact manifold K is said simple if any two points of K are connected by a unique geodesic and if its boundary is strictly convex) the conformal
factor c is entirely determined by the DN map at frequency λ=0.

In the same way, for a class of manifolds M= [0,1]× K which have the topology of a cylinder, various results of uniqueness (or non-uniqueness when the Dirichlet data and the Neumann data are measured on disjoint sets)
have been obtained  in (<cit.>). The proofs make use of separation of variables and introduces a connection between the DN map and the Weyl-Titchmarsh functions
associated to a separated ordinary differential equation corresponding to the horizontal variable of the cylinder. This connection allows to use some nice results from complex analysis.

In this paper, we are interested in a relative inverse problem within the conformal class of certain warped product manifold M=[0,1]×𝕊^n-1 (where 𝕊^n-1 is the (n-1)-sphere) but under a weaker assumption :



does the spectrum of the DN map characterize the conformal factor ?





We recall that Λ_g(λ) is an elliptic pseudodifferential operator of order 1 and is self-adjoint on L^2(∂ M, dS_g) where dS_g is the metric induced by g on the boundary ∂ M. Thus, the DN operator Λ_g(λ)  has a discrete spectrum denoted σ(Λ_g(λ)) accumulating at infinity

    σ(Λ_g(λ))={λ_0≤λ_1≤λ_2≤...≤λ_k→+∞}.




This spectrum is called the Steklov spectrum (see <cit.>, p.2). In the particular case where λ=0, using the Green's formula, we see that the DN operator is positive and we have λ_0 =0. The properties of the Steklov spectrum are highy sensitive to the smoothness of the boundary ∂ M. For example (see <cit.>): if ∂ M is smooth, the eigenvalues satisfy the Weyl formula for the Steklov spectrum:

    λ_j=2π(j/Vol(𝔹^n-1)Vol(∂ M))^1/n-1+O(1)

A much more refined asymptotic holds if M is a smooth surface, involving the lengths of the connected components of M, with a rate of decay of O(j^-∞) (see <cit.>), but this formula fails for polygons (<cit.>). In some particular cases, when the boundary ∂ M is just C^1, it is also possible to find a one term asymptotic of the Steklov spectrum counted with multiplicity (see <cit.>). For more specific domains with Lipschitz boundary, a recent paper (<cit.>) proves a two-term asymptotic formula for cuboids, i.e domains defined, for n≥ 3, as

    M=(-a_1,a_1)× ... × (-a_n,a_n)∈ℝ^n.

The formula for the counting function N(λ) of Steklov eigenvalues is the following (C_1,C_2∈ℝ):

    N(λ)=C_1Vol_n-1(∂ M)λ^n-1+C_2Vol_n-2(∂^2 M)λ^n-2+O(λ^η),

where ∂^2 M denotes the union of all the n-2 dimensional facets of M, η=2/3 if n=3 and η=n-2-1/n-1 if n≥ 4. As a corollary, the authors deduce also that, for a rectangle, the Steklov spectrum determines its side lengths: in other words they are Steklov spectrum invariants. Moreover, it is well-known that some other geometrical quantities of the boundary of a Riemannian manifold surface are Steklov spectrum invariants. Let us mention for instance (see <cit.> for details and reference therein):



  * the dimension of the manifold and the volume of its boundary.


  * When M≥ 3, the integral of the mean curvature on ∂ M.


  * When M=2, the number and the lengths of the connected components of ∂ M.



	
	

 §.§ The main result

	

	
	
	
In this section, we give the main results of the paper. Let M=[0,1]×𝕊^n-1 be a manifold equipped with the metric
 
    g=f(x)(dx^2+g_𝕊),

where 𝕊^n-1 is the unit sphere in ℝ^n, g_𝕊 is the  metric induced by the euclidean metric on ℝ^n, and where
the conformal factor f is a smooth positive function of the variable x ∈ [0,1] only. Note that the manifold M has two boundaries Γ_0= {0}×𝕊^n-1 and Γ_1 = {1}×𝕊^n-1. Thus, the DN map will be shown to have the structure of a matrix operator defined on H^1/2(𝕊^n-1)⊕ H^1/2(𝕊^n-1).

In this setting, we want to answer the following question : does the Steklov spectrum determine uniquely the warping function f(x)? This question has been answered positively in <cit.> when K=(0,1]×𝕊^n-1 equipped with the metric (<ref>). 
The difference between this case and the one we are studying here is that the boundary of K is connected whereas that of M is made of two connected components. Due to a natural gauge invariance, we emphasize it is hopeless to recover the metric from the spectrum data. Indeed, let ψ:M→ M be a smooth diffeomorphism. We have (<cit.>):


    Λ_ψ^*g(λ)=φ^*∘Λ_g(λ)∘φ^*^-1 ,




where φ:=ψ|_∂ M and where  φ^* : C^∞(∂ M)→ C^∞(∂ M) is the application defined by φ^*h:=h∘φ. As a by-product, one has:

    σ(Λ_ψ^*g(λ))=σ(Λ_g(λ)).




Thus, if we can find a diffeomorphism ψ preserving the warped product structure of the manifold M given by (<ref>), we are able to find a counterexample to uniqueness from the knowledge of the Steklov spectrum. For instance, consider the map


    ψ : (x,y)∈[0,1]×𝕊^n-1↦ (1-x,y)∈[0,1]×𝕊^n-1.




A straightforward computation gives ψ^*g=f(1-x)(dx^2+g_𝕊). Thus, the above discussion shows that Λ_g(λ)
and Λ_ψ^*g(λ) have the same Steklov spectrum. Now, we can reformulate more precisely our initial question. Let g and g̃ be two Riemannian metrics given by  (<ref>) with conformal factor f(x), (resp. f̃(x)). Assume that Λ_g(λ) and Λ_g̃(λ) are Steklov isospectral. Then, is it true that :



	f(x) = f̃(x) or f(x) = f̃(1-x)? 





In what follows, we answer positively this question in dimension n=2 with λ 0, and in dimension n≥3  for any frequency λ with an additional hypothesis on the metrics on the boundary ∂ M.



We choose the sphere 𝕊^n-1 as the transversal manifold of our cylinder for a purely technical reason. Indeed, we need a precise (in fact an exact) asymptotic of the eigenvalues of the Laplace Beltrami operator Δ_g_𝕊 on (𝕊^n-1,g_𝕊)  in order to get our uniqueness on the conformal factor f. Moreover, it is important to understand that we fix the transversal metric g_𝕊 in our inverse problem. Otherwise, it is known that we could find two non isometric transversal metrics such that the associated Riemannian manifolds are isospectral. This would lead to Steklov isospectral cylinders. Similarly, when dim(M)≥ 3, it is known (see <cit.>) that we can find Steklov isospectral manifolds M=[0,L]× K and M̃=[0,L]×K̃ (with K and K̃ not connected) such that the areas of the connected components of M and M̃ are not the same.
 
Our main result is the following:
	

Let M=[0,1]×𝕊^n-1 be a smooth Riemanniann manifold equipped with the metric
		

		g=f(x)(dx^2+g_𝕊)  , (resp. g̃=f̃(x)(dx^2+g_𝕊)),
		

and let λ be a frequency not belonging to the Dirichlet spectrum of  -Δ_g and -Δ_g̃ on M. Then,

	 
  * For n=2 and λ 0,
	    
    (σ(Λ_g(λ)) = σ(Λ_g̃(λ)))	⇔(f=f̃  or   f=f̃∘η)

		where η(x)=1-x for all x ∈ [0,1].
		
	
  * For n≥ 3, and if moreover
		
    f,f̃∈𝒞_b := {f∈ C^∞([0,1]), |f'(k)/f(k)|≤1/n-2, k=0  and  1 },
			
	
    (σ(Λ_g(λ)) = σ(Λ_g̃(λ)))	⇔(f=f̃  or   f=f̃∘η)



	
	
Let us explain briefly the outline of the proof. In both cases n=2 or n ≥ 3, the proof consists in four steps. We emphasize that in the first three steps, we do not use explicitly that the transversal manifold is the unit sphere 𝕊^n-1.




Step 1 : we follow the same approach as in <cit.>. Since the manifold M has the topology of a cylinder and is equipped with a warped product metric, we can use separation of variables and write the solution u of the Dirichlet problem (<ref>) as

    u(x,y)=∑_m=0^+∞u_m(x)Y_m(y),

reducing this problem to a countable family of Sturm-Liouville equations with boundary conditions :


    {    -v^”_m+qv_m=-μ_mv_m,  on  ]0,1[
       v_m(0)=f^n-2/4(0)ψ_m^0,  v_m(1)=f^n-2/4(1)ψ_m^1, .    ∀ m∈ℕ

where 

	q=(f^n-2/4)”/f^n-2/4-λ f and v_m=f^n-2/4u_m, ∀ m∈ℕ.



The sequence (Y_m) refers to an orthonormal basis of eigenvectors of the Laplace-Beltrami operator -Δ_g_𝕊 on the unit sphere.



Step 2 : we use the above decomposition to write the DN operator as an infinite matrix which is block diagonal. More precisely, let us introduce the basis  𝔅=({e^1_m,e^2_m})_m ∈ℕ, where e^1_m=(Y_m,0) and e^2_m=(0,Y_m). For each m∈ℕ, we denote Λ_g^m(λ) the restricted operator of Λ_g(λ) on the subspace spanned by {e^1_m,e^2_m}. We can write Λ_g(λ) in the basis 𝔅 as the infinite matrix:




    [Λ_g]_𝔅=[          Λ_g^1(λ)        0        0        ⋯;                 0 Λ_g^2(λ)        0        ⋯;                 0        0 Λ_g^3(λ)        ⋯;                 ⋮        ⋮        ⋮        ⋱ ]




Each (2,2) matrix Λ_g^m(λ) has a simple interpretation involving the so-called Weyl-Titchmarsh theory associated to the Sturm-Liouville equation (<ref>). More precisely, if we denote

    σ(-Δ_g_𝕊)={0=μ_0<μ_1≤μ_2≤...≤μ_m≤...→+∞},

the spectrum of -Δ_g_𝕊 and h:=f^n-2, we shall see that :

    Λ_g^m(λ)=[ -M(μ_m)/√(f(0))+1/4√(f(0))h'(0)/h(0)  -1/√(f(0))h^1/4(1)/h^1/4(0)1/Δ(μ_m);  -1/√(f(1))h^1/4(0)/h^1/4(1)1/Δ(μ_m) -N(μ_m)/√(f(1))-1/4√(f(1))h'(1)/h(1) ].


where M(z), N(z) are the Weyl-Titchmarsh functions and Δ(z) is the characteristic function associated to (<ref>). In particular, the trace (respectively the determinant) of these operators Λ_g^m(λ) are  meromorphic functions evaluated in μ_m. Moreover, one can prove that the Steklov Spectrum is made of two subsequences (λ^-(μ_m))_m∈ℕ and (λ^+(μ_m))_m∈ℕ satisfying the following asymptotic expansion 

    {   λ^-(μ_m)=-N(μ_m)/√(f(1))-1/4√(f(1))h'(1)/h(1)+O(√(μ_m)e^-√(μ_m)) 
       λ^+(μ_m)=-M(μ_m)/√(f(0))+1/4√(f(0))h'(0)/h(0)+O(√(μ_m)e^-√(μ_m))
    	.






Step 3 : we prove that the knowledge of the trace and the determinant of all restricted operators Λ_g^m(λ), for m large enough, characterizes f up to an involution: there is m_0∈ℕ such that :



	(Tr Λ_g^m(λ)= Tr Λ_g̃^m(λ) and Λ_g^m(λ)=Λ_g̃^m(λ) for all m ≥ m_0 ) 
	
	⇔ (f=f̃ or f=f̃∘η.)



Note that the third step requires to know the asymptotic expansion of the Steklov spectrum.



Step 4 : in this step, we use explicitely that the transversal manifold is the unit sphere 𝕊^n-1 equipped with the metric induced by the euclidean metric on ℝ^n. From the equality between the sets σ(Λ_g(λ))=(λ^±(μ_m))_m∈ℕ and σ(Λ_g̃(λ))=(λ̃^±(μ_m))_m∈ℕ, we want to deduce the equalities

    {   λ^-(μ_m)= λ̃^-(μ_m)
       λ^-(μ_m)= λ̃^+(μ_m)
    	.  or  {   λ^-(μ_m)= λ̃^+(μ_m)
       λ^+(μ_m)= λ̃^-(μ_m),
    	.

for integers m belonging to a set ℒ satisfying the Müntz conditions ∑_m∈ℒ1/m=+∞. Clearly, for every λ^±(μ_m) in σ(Λ_g(λ)), there is λ̃^-(μ_ℓ) or λ̃^+(μ_ℓ) in σ(Λ_g̃(λ)) such that 


	λ^±(μ_m)=λ̃^-(μ_ℓ) or λ^±(μ_m)=λ̃^+(μ_ℓ).


The assumption |f'(k)/f(k)|≤1/n-2 for k∈{0,1} is used here to ensure that m=ℓ if is m is large enough. This step leads to distinguish the cases f(0)=f(1) and f(0) f(1). Then we prove that Tr Λ_g^m(λ)= Tr Λ_g̃^m (λ) and Λ_g^m(λ)= Λ_g̃^m(λ) for all m large enough. The result follows from the Step 3.




§ REDUCTION TO ORDINARY DIFFERENTIAL EQUATIONS


In this section, (K,g_K) is an arbitrary closed manifold of dimension n-1 and M=[0,1]× K is equipped with the metric g=f(x)(dx^2+g_K).



 §.§ The separation of variables




The boundary ∂ M of the manifold M has two distinct connected components


	Γ_0={0}× K and Γ_1={1}× K,


so we can decompose H^1(∂ M) as the direct sum :


	H^1/2(∂ M)=H^1/2(Γ_0) ⊕ H^1/2(Γ_1).





Each  element ψ of H^1/2(∂ M) can be written as


ψ=[ ψ^0; ψ^1 ], ψ^0∈ H^1/2(Γ_0) and ψ^1∈ H^1/2(Γ_1).





The Laplacian -Δ_g_K is a self-adjoint operator on L^2(K) and has pure point spectrum (μ_m), with μ_0=0<μ_1≤μ_2≤... μ_m→+∞. We denote (Y_m)_m∈ℕ  the associated orthonormal Hilbert basis of the eigenvectors.

Now, we decompose ψ^0 and ψ^1 as:




    ψ^0=∑_m∈ℕψ_m^0 Y_m,    ψ^1=∑_m∈ℕψ_m^1 Y_m.




Then, we are looking for the unique solution u(x,y) of the Dirichlet problem (<ref>) in the form:

    u(x,y)=∑_m=0^+∞u_m(x)Y_m(y).





	
The equation (<ref>) is equivalent to the following countable system of Sturm-Liouville equations :


    {    -v^”_m+qv_m=-μ_mv_m,  on  ]0,1[
       v_m(0)=f^n-2/4(0)ψ_m^0,  v_m(1)=f^n-2/4(1)ψ_m^1, .    ∀ m∈ℕ,

where 

	q=(f^n-2/4)”/f^n-2/4-λ f and v_m=f^n-2/4u_m, ∀ m∈ℕ.





The transformation law for the Laplacian operator under conformal change of metric gives

    -Δ_g=f^-n+2/4(-Δ_g_0+q_f)f^n-2/4 ,

where g_0=dx^2+g_K and q_f=:(f^n-2/4)”/f^n-2/4. Thus, we have :

    -Δ_g u(x,y)=λ u(x,y)   ⇔ f^-n+2/4(-∂^2/∂ x^2-Δ_g_K+q_f)f^n-2/4u(x,y)=λ u(x,y)
       ⇔(-∂^2/∂ x^2-Δ_g_K+q_f)v(x,y)=λ f(x)v(x,y)


thanks to the change of variable v=f^n-2/4u. Writing v as


    v(x,y)=∑_m=0^+∞v_m(x)Y_m(y),

we get :

    -Δ_g u(x,y)=λ u(x,y)   ⇔∑_m=0^+∞(-∂^2/∂ x^2-Δ_g_K+q_f)v_m(x)Y_m=∑_m=0^+∞λ f(x)v_m(x)Y_m
       ⇔∑_m=0^+∞(-v”_m(x)+μ_mv_m(x)+q_f(x)v_m(x))Y_m
                                       =∑_m=0^+∞λ f(x)v_m(x)Y_m
       ⇔∀ m∈ℕ, 
     -v^”_m(x)+q(x)v_m(x)=-μ_mv_m(x),

where q=:q_f-λ f. Finally, v_m satisfies the above boundary conditions.




It turns out that this family of Sturm-Liouville equations fits into the so-called Weyl-Titchmarsh theory which we recall in the next section for the convenience of the reader.





 §.§ The Weyl-Titchmarsh functions




Consider the differential equation:

    -u”+qu=-zu,   z∈ℂ.

Let {c_0,s_0} and {c_1,s_1} be the two fundamental systems of solutions of (<ref>) with boundary conditions

    {    c_0(0)=1, c_0'(0)=0
        c_1(1)=1, c_1'(1)=0 .  and  {    s_0(0)=0, s_0'(0)=1
        s_1(1)=0, s_1'(1)=1. .




Since the wronskian W(f,g) = fg'-f'g of two solutions f and g of (<ref>) depends only  on the parameter z, we can define the following holomorphic functions:




The characteristic function Δ(z) of the equation (<ref>) is defined for every z∈ℂ by :

    Δ(z):=W(s_0,s_1)=s_0(1)=-s_1(0).

We also set  D(z):=W(c_0,s_1)=c_0(1) and E(z):=W(c_1,s_0)=c_1(0).


Now, we recall the three following results, (see for instance <cit.> and <cit.> for details).




	
Set Π^+={z∈ℂ, (z)>0} the half-right plane of the complex plane. The functions Δ, D and E are analytic on ℂ and have on Π^+ the following asymptotics as |z|→ +∞:



Δ(z)=sinh(√(z))/√(z) + O(e^|(√(z))|/z),  D(z)=cosh(√(z)) + O(e^|(√(z))|/√(z)),
   E(z)=cosh(√(z))+O(e^|(√(z))|/√(z)).


where √(z) is the principal square root of z. In particular, Δ(z), D(z) and E(z) are entire functions of order 1/2.





	
	The roots (α_j) of the function z↦Δ(z) are real and simple. they are the opposite of the eigenvalues of the operator -d^2/dx^2+q=:H on L^2((0,1),dx) with Dirichlet boundary conditions.







	
For all z∈ℂ, Δ(z) can be written as an infinite product. There is a constant C∈ℝ such that :


    Δ(z)=C∏_k=0^∞(1-z/α_k).






As a consequence of Proposition <ref> and Hadamard's factorization theorem, we can write Δ(z) as the infinite product :

    Δ(z)=Cz^p∏_k=0^∞(1-z/α_k),


with p∈{0,1}. In order to prove that z=0 is not a root of Δ, we use the same argument as in <cit.> (Remark 3.1 p.19). If Δ(0)=0, it follows from Proposition <ref>, that there is an eigenfunction u_0 associated to the eigenvalue 0 for the operator H= -d^2/dx^2+q. But, from Proposition <ref>, the function u:=u_0Y_0 is then a nontrivial solution of the Dirichlet problem :

    {    -Δ_g u=λ u   in  M
        u=0  on  ∂ M, .




which is not possible since  λ∉σ(-Δ_g).





As a by-product, we see that Δ (z) is uniquely determined by its (simple) roots (up to a multiplicative constant).




Now, consider the Weyl-Titchmarsh solutions ψ and ϕ of (<ref>) having the form :

    ψ(x)=c_0(x)+M(z)s_0(x),  ϕ(x)=c_1(x)-N(z)s_1(x)

and satisfying the Dirichlet boundary condition at x=1 and x=0 respectively. M(z), (resp. N(z)) are called the Weyl-Titchmarsh functions associated to (<ref>) and, using Wronskian identities, we have:


The Weyl-Titchmarsh functions M and N can be written as :

    ∀ z∈ℂ,  M(z)=-D(z)/Δ(z),   N(z)=-E(z)/Δ(z).




Since ψ(1)=0 and ϕ(0)=0, we have :



M(z)=-c_0(1,z)/s_0(1,z)=-D(z)/Δ(z), N(z)=c_1(0,z)/s_1(0,z)=-E(z)/Δ(z).




The four previous propositions are the key points to solve our uniqueness result. We will see that the Steklov spectrum can be expressed in terms of the Weyl-Titchmarsh and characteristic functions defined above. We will take advantage of the holomorphic properties of Δ(z), D(z) and E(z)  and we will use to the Nevanlinna theorem, (see the next section for details).












 §.§ Link between the DN map and the Weyl-Titchmarsh functions


First, we remark that, thanks to separation of variables, Λ_g(λ) leaves invariant each subspace spanned by {(Y_m,0),(0,Y_m)}. Indeed, if u is the solution of (<ref>), we have for each ψ∈ H^1/2(∂ M):

    Λ_g(λ)ψ=Λ_g(λ)[ ψ^0; ψ^1 ]=[ (∂_ν u)_|Γ_0; (∂_ν u)_|Γ_1 ]=[ -1/√(f(0))(∂_x u)_|x=0;  1/√(f(1))(∂_x u)_|x=1 ]

Consequently, for every m∈ℕ :

    Λ_g(λ)[ ψ^0_m; ψ^1_m ]⊗ Y_m=[ -1/√(f(0))u_m'(0);  1/√(f(1))u'_m(1) ]⊗ Y_m.




Its restriction on each space spanned by (1,0)⊗ Y_m and (0,1)⊗ Y_m is denoted Λ_g^m(λ). We can write Λ_g^m(λ)
the 2×2 matrix


Set :

    [   L^m(λ) T^m_R(λ); T^m_L(λ)   R^m(λ) ]

and we have by definition:


    [   L^m(λ) T^m_R(λ); T^m_L(λ)  R^m(_λ) ][ ψ_m^0; ψ_m^1 ]=[ -1/√(f(0))u_m'(0);  1/√(f(1))u'_m(1) ].






The full Steklov spectrum is then equal to the union of the eigenvalues of each operator Λ_g^m(λ). In the next Proposition, we express the restricted
operator Λ_g^m(λ) in terms of the Weyl-Titchmarsh functions.






  For all m∈ℕ, we have :
	
	
    Λ_g^m(λ)=[  -M(μ_m)/√(f(0))+ln'(h)(0)/4√(f(0)) -1/√(f(0))h^1/4(1)/h^1/4(0)1/Δ(μ_m); -1/√(f(1))h^1/4(0)/h^1/4(1)1/Δ(μ_m)  -N(μ_m)/√(f(1))-ln'(h)(1)/4√(f(1)) ]

	where h=:f^n-2






Recall that v_m is the m-th Fourier coefficient of v=h^1/4u with respect to Y_m. We have :


    Λ_g^m(λ)[ ψ_m^0; ψ_m^1 ]   =[ -1/√(f(0))u'_m(0);  1/√(f(1))u'_m(1) ]
       =[ -v_m'(0)/h^1/4(0)√(f(0))+1/4√(f(0))h'(0)/h(0)u_m(0);  v_m'(1)/h^1/4(1)√(f(1))-1/4√(f(1))h'(1)/h(1)u_m(1) ]
       =[ -v_m'(0)/h^1/4(0)√(f(0));  v_m'(1)/h^1/4(1)√(f(1)) ]+[  1/4√(f(0))h'(0)/h(0)ψ_m^0; -1/4√(f(1))h'(1)/h(1)ψ_m^1 ]




 It remains to find a simple expression of the first term of the (RHS). Since v_m is a solution of (<ref>), it can be written as a linear combination of the fundamental solutions {c_0,s_0} or {c_1,s_1} defined in (<ref>), i.e there exists (α,β,γ,δ)∈ℂ^4 such that :

    v_m=α c_0+β s_0=γ c_1+δ s_1.




Consequently:

    [ v_m(0); v_m(1) ]=[ α; γ ]=[ γ c_1(0)+δ s_1(0); α c_0(1)+β s_0(1) ]




The second equality can be rewritten as:

    [ α-γ c_1(0); γ-α c_0(1) ]=[ δ s_1(0); β s_0(1) ],

which is equivalent to :

    [       1 -c_1(0); -c_0(1)       1 ][ α; γ ]=[ δ s_1(0); β s_0(1) ].






As

    [ α; γ ]=[ h^1/4(0)u_m(0); h^1/4(1)u_m(1) ]=[ h^1/4(0)ψ_m^0; h^1/4(1)ψ_m^1 ]


we obtain:




    [        h^1/4(0)/s_1(0) -h^1/4(1)c_1(0)/s_1(0); -h^1/4(0)c_0(1)/s_0(1)        h^1/4(1)/s_0(1) ][ ψ_m^0; ψ_m^1 ]=[ δ; β ].






But δ=v'_m(1) and β=v'_m(0). Thus:




    Λ_g^m(λ)[ ψ_m^0; ψ_m^1 ]   =[ -v_m'(0)/h^1/4(0)√(f(0));  v_m'(1)/h^1/4(1)√(f(1)) ]+[  1/4√(f(0))h'(0)/h(0)ψ_m^0; -1/4√(f(1))h'(1)/h(1)ψ_m^1 ]
       =[            1/√(f(0))c_0(1)/s_0(1) -1/√(f(0))h^1/4(1)/h^1/4(0)s_0(1);  1/√(f(1))h^1/4(0)/h^1/4(1)s_1(0)           -1/√(f(1))c_1(0)/s_1(0) ][ ψ_m^0; ψ_m^1 ]+[  1/4√(f(0))h'(0)/h(0)ψ_m^0; -1/4√(f(1))h'(1)/h(1)ψ_m^1 ]
       =[  1/√(f(0))c_0(1)/s_0(1)+1/4√(f(0))h'(0)/h(0)            -1/√(f(0))h^1/4(1)/h^1/4(0)s_0(1);             1/√(f(1))h^1/4(0)/h^1/4(1)s_1(0) -1/√(f(1))c_1(0)/s_1(0)-1/4√(f(1))h'(1)/h(1) ][ ψ_m^0; ψ_m^1 ]


Recalling that:


    M(μ_m)=-c_0(1)/s_0(1),   N(μ_m)=c_1(0)/s_1(0)  et  Δ(μ_m)=-s_1(0)=s_0(1),


we get finally:


    Λ_g^m(λ)=[ -M(μ_m)/√(f(0))+1/4√(f(0))h'(0)/h(0)  -1/√(f(0))h^1/4(1)/h^1/4(0)1/Δ(μ_m);  -1/√(f(1))h^1/4(0)/h^1/4(1)1/Δ(μ_m) -N(μ_m)/√(f(1))-1/4√(f(1))h'(1)/h(1) ].






§ A CHARACTERISATION BY THE TRACE AND THE DETERMINANT


In this section, (K,g_K) is still an arbitrary closed manifold of dimension n-1 and M=[0,1]× K is equipped with the metric g=f(x)(dx^2+g_K). We have the following result:




	
	Assume that, for every m∈ℕ, we have :
	
	

	(Λ_g^m(λ))=(Λ^m_g̃(λ)) and Tr(Λ_g^m(λ))=Tr(Λ^m_g̃(λ)).
	
Then  :
	
    f=f̃  or   f=f̃∘η

	
	where, for all x∈ [0,1], η(x)=1-x.





	
	This Proposition is still true if the equalities about the trace and the determinant of Λ_g^m(λ) are are satisfied for m≥ m_0, with m_0∈ℕ.




In order to prove this proposition, let us calculate the eigenvalues of the operator Λ_g^m(λ) and their asymptotics.




	
	Λ_g^m(λ) has two eigenvalues λ^-(μ_m) and λ^+(μ_m) whose asymptotics are given by :
	
    {   λ^-(μ_m)m→∞=√(μ_m)/√(f(1))-ln(h)'(1)/4√(f(1))+O(1/√(μ_m))	 
       λ^+(μ_m)m→∞=√(μ_m)/√(f(0))+ln(h)'(0)/4√(f(0))+O(1/√(μ_m)).
    	.







	If f(0)<f(1), we have λ^-(μ_m)<λ^+(μ_m). This assumption will be made, without loss of generality, each time that f(0) f(1). The notations λ^-(μ_m) and λ^+(μ_m) refer to that choice.





	The Weyl's law for the eigenvalues of the Laplace-Beltrami operator gives the following asymptotic :
	
    μ_m = 4π^2 (Vol(𝔹^n-1)Vol(K))^-2/n-1m^2/n-1 + O(1),
 
so, by replacing it in Lemma <ref>, one has :

    λ^-(μ_m)   m→∞=√(μ_m)/√(f(1))+O(1)
       =2π/√(f(1))(Vol(𝔹^n-1)Vol(K))^-1/n-1m^1/n-1+O(1).

The boundary component Γ_1 consists in copy of K equipped with the metric γ_1=f(1)g_K. It follows that we have Vol(Γ_1)=∫_Γ_1 dVol_γ_1=f(1)^n-1/2Vol(K), hence 1/√(f(1))=Vol(K)^1/n-1/Vol(Γ_1)^1/n-1. Consequently :

    λ^-(μ_m)= 2π(m/Vol(𝔹^n-1)Vol(Γ_1))^1/n-1+O(1).

In other words, an asymptotic of λ^-(μ_m) is exactly given by the Weyl's law restricted to the connected component boundary Γ_1. In the same way, one can prove :

    λ^+(μ_m)= 2π(m/Vol(𝔹^n-1)Vol(Γ_0))^1/n-1+O(1).

We recognize again the Weyl's law, restricted to the connected component boundary Γ_0.





The equalities proved in Lemma <ref> highlight the link that exists between the Steklov spectrum and the spectrum of -Δ_g_𝕊. Let us denote the eigenvalues of the Laplace-Beltrami operator on (𝕊,f(0)g_𝕊) by
	
    μ_0^(0)≤μ_1^(0)≤μ_2^(0)≤ ... → +∞

with μ_m^(0)=μ_m/f(0) for m∈ℕ. Lemma <ref> implies in particular that there is a constant C^(0)_f>0 only depending on the conformal factor f at x=0 such that

    |λ^+(μ_m)-√(μ_m^(0))|≤ C_f^(0)

This can be related to results obtained in <cit.> (Theorem 1.7 p.2) where it is proved that, for a bounded domain ∂Ω⊂ℝ^n with boundary of class C^2 which has only one boundary component, there is a bound C_Ω>0 depending on Ω such that

    |λ_m-√(μ_m)|≤ C_Ω  ∀ m∈ℕ.

where λ_m and μ_m are respectively the m^th eigenvalue of the DN map and the the m^th eigenvalue of the Laplace-Beltrami operator on the boundary. One can notice that, in our case, C_f^(0) only depends on the metric on the boundary component (𝕊^n-1,f(0)g_𝕊). This fact can be compared to a recent result (see <cit.>) where it is proved that the previous bound C_Ω can be chosen uniformly with respect to a class of manifolds ℳ satisfying some geometrical conditions only in a neighborhood of the boundary (Theorem 3, p.3). 



In the same way, Lemma <ref> implies also the existence of C^(1)_f>0 only depending on the conformal factor f at x=1 such that

    |λ^-(μ_m)-√(μ_m^(1))|≤ C^(1)_f

where μ_m^(1)=μ_m/f(1) is the m^th eigenvalue of the Laplace-Beltrami operator on (𝕊,f(1)g_𝕊).  



Let us prove Lemma <ref>.



 We distinguish two cases :
	
	
		
		
  ∙ Assume f(0) f(1) (for instance f(0)<f(1)).
	
	
	The characteristic polynomial P(X) of Λ_g^m(λ) is : 						
	
	
    P(X)=X^2-Tr(Λ_g^m(λ))X+(Λ_g^m(λ)).

	
	
	
	To simplify the notation, we set :
	
	

	C_0=ln(h)'(0)/4√(f(0)), C_1=ln(h)'(1)/4√(f(1)).



Thanks to Propositions <ref> and <ref>, for m large enough,  Tr(Λ_g^m(λ)) and (Λ_g^m(λ)) satisfy:
	
	
    {   Tr(Λ_g^m(λ))=-M(μ_m)/√(f(0))-N(μ_m)/√(f(1))+C_0-C_1.	 
       (Λ_g^m)=(-M(μ_m)/√(f(0))+C_0)(-N(μ_m)/√(f(1))-C_1)+O(μ_m e^-2√(μ_m)).
    	.

	
	
	
	The asymptotics of the discriminant δ of P(X) depending on M(μ_m) and N(μ_m) can thus be written :
	
    δ   = (-M(μ_m)/√(f(0))+C_0-N(μ_m)/√(f(1))-C_1)^2-4(-M(μ_m)/√(f(0))+C_0)(-N(μ_m)/√(f(1))-C_1)
                                                                       +O(μ_m e^-2√(μ_m)).
       =(-M(μ_m)/√(f(0))+C_0+N(μ_m)/√(f(1))+C_1)^2+O(μ_m e^-2√(μ_m)).

	
	Now, let us recall the result obtained by Simon in <cit.> :
	
	
		
		M(z^2) has the following asymptotic expansion :
		
    ∀ A∈ℕ,  -M(z^2)z →∞=z+∑_j=0^Aβ_j(0)/z^j+1+o(1/z^A+1)

		where, for every x∈[0,1], β_j(x) is defined by : {   β_0(x)=1/2q(x)
   β_j+1(x)=1/2β'_j(x)+1/2∑_l=0^jβ_l(x)β_j-l(x).
		.
	
	
	Of course, by symmetry, one has immediately:
	
	
		
		N(z^2) has the following asymptotic expansion :
		
    ∀ A∈ℕ,  -N(z^2)_z →∞=z+∑_j=0^Aγ_j(0)/z^j+1+o(1/z^A+1)

		where, for all x∈[0,1], γ_j(x) is defined by : {   γ_0(x)=1/2q(1-x)
   γ_j+1(x)=1/2γ'_j(x)+1/2∑_l=0^jγ_l(x)γ_j-l(x).
		.
	
	
	





We deduce from Theorem <ref> and Corollary <ref>:
	
	
    -M(μ_m)/√(f(0))+N(μ_m)/√(f(1))=(1/√(f(0))-1/√(f(1)))_> 0√(μ_m)+O(1/√(μ_m)).


Thus, recalling that

    δ   =(-M(μ_m)/√(f(0))+C_0+N(μ_m)/√(f(1))+C_1)^2+O(μ_m e^-2√(μ_m)).


we obtain:


    √(δ)   =(N(μ_m)/√(f(1))-M(μ_m)/√(f(0))+C_0+C_1)√(1+O(e^-2√(μ_m)))
       =N(μ_m)/√(f(1))-M(μ_m)/√(f(0))+C_0+C_1+O(√(μ_m)e^-2√(μ_m)).




Hence :
	

    {   λ^-(μ_m)=1/2[(-M(μ_m)/√(f(0))-N(μ_m)/√(f(1))+C_0-C_1)-√(δ)] 
       λ^+(μ_m)=1/2[(-M(μ_m)/√(f(0))-N(μ_m)/√(f(1))+C_0-C_1)+√(δ)],
    .

	
	and therefore, substituting C_1 and C_2 by their values and M(μ_m) and N(μ_m) by their asymptotics, we get :
	
	
    {   λ^-(μ_m)=√(μ_m)/√(f(1))-ln(h)'(1)/4√(f(1))+O(1/√(μ_m))	 
       λ^+(μ_m)=√(μ_m)/√(f(0))+ln(h)'(0)/4√(f(0))+O(1/√(μ_m)).
    	.

	
	
	
  ∙ Assume now f(0)=f(1). In this case, the restricted DN map
			
	
	
    Λ_g^m(λ)=[ -M(μ_m)/√(f(0))+ln(h)'(0)/4√(f(0))                 -1/√(f(0))1/Δ(μ_m);                 -1/√(f(0))1/Δ(μ_m) -N(μ_m)/√(f(1))-ln(h)'(1)/4√(f(1)) ]




is a symmetric matrix and we can use the well-known result:
		
		
	
Let H be a Hilbert space, A∈ℒ(H) be a selfadjoint operator. Let ϵ>0. Assume there exists λ_0∈ℝ and u_0∈ H a unit vector such that (A-λ_0 Id)u_0≤ϵ. Then there exists an element λ in the spectrum of A such that |λ-λ_0|≤ϵ.
	
We apply this theorem with A=Λ_g^m(λ), λ_0=-M(μ_m)/√(f(0))+ln(h)'(0)/4√(f(0)) and U_0=[ 1; 0 ].



We have :

    A-λ_0I_2=[                                      0                     -1/√(f(0))1/Δ(μ_m);                     -1/√(f(0))1/Δ(μ_m) -N(μ_m)/√(f(1))-ln(h)'(1)/4√(f(1))-λ_0 ]




Hence :

    (A-λ_0 I_2)U_0=-1/√(f(0))[        0; 1/Δ(μ_m) ]= O(√(μ_m)e^-√(μ_m)).




Lemma <ref> gives λ^+_m∈σ(Λ_g^m(λ)) such that : 
    λ^+_m   =-M(μ_m)/√(f(0))+ln(h)'(0)/4√(f(0))+O(√(μ_m)e^-√(μ_m))
       =√(μ_m)/√(f(0))+ln(h)'(0)/4√(f(0))+O(1/√(μ_m))




from Theorem <ref>. The second eigenvalue λ_m^2 (which is distinct from λ^1_m otherwhise Λ_g^m(λ) would be a homothety) can be deduced from the first one :

    λ_m^-   =Tr(Λ_g^m(λ))-λ_m^+
       =-N(μ_m)/√(f(1))-ln(h)'(1)/4√(f(1))+O(√(μ_m)e^-√(μ_m))
       =√(μ_m)/√(f(1))-ln(h)'(1)/4√(f(1))+O(1/√(μ_m)).







	Under the hypothesis of Proposition <ref>, we have the following alternative :
	

	{   f(0)=f̃(0)	 
   f(1)=f̃(1)
	.  or  {   f(0)=f̃(1)	 
   f(1)=f̃(0).
	.
	

	




We begin with:

    Tr(Λ_g^m(λ))/√(μ_m)=	Tr(Λ_g̃^m(λ))/√(μ_m),  ∀ m∈ℕ.

Thus, it follows from Lemma <ref> that

    1/√(f(0))+1/√(f(1))=1/√(f̃(0))+1/√(f̃(1)).


In the same way, thanks to the relations:

    det(Λ_g^m(λ))/μ_m=	det(Λ_g̃^m(λ))/μ_m  ∀ m∈ℕ,

we get :

    1/√(f(0)f(1))=1/√(f̃(0)f̃(1)),

and the proof is complete.




Case 1 :  f(0)=f̃(0) et f(1)=f̃(1).



 Under the hypotheses of Proposition <ref>, we have:

    Δ(z)=Δ̃(z),  ∀ z∈ℂ.





 We write the equality Tr(Λ_g^m(λ))=Tr(Λ^m_g̃(λ)) as follows : for all μ_m∈σ(-Δ_g_K),


    (-M(μ_m)/√(f(0))+ln(h)'(0)/4√(f(0)))+(-N(μ_m)/√(f(1))-ln(h)'(1)/4√(f(1)))=
    (   -M̃(μ_m)/√(f(0))+ln(h̃)'(0)/4√(f(0)))
       +(-Ñ(μ_m)/√(f(1))-ln(h̃)'(1)/4√(f(1))).



Thus, using Theorem <ref>, one gets when m → +∞,

    ln(h)'(0)/4√(f(0))-ln(h)'(1)/4√(f(1))=ln(h̃)'(0)/4√(f(0))-ln(h̃)'(1)/4√(f(1)).

It follows that

    M(μ_m)/√(f(0))+N(μ_m)/√(f(1))=M̃(μ_m)/√(f(0))+Ñ(μ_m)/√(f(1)),

or equivalently

    Δ̃(μ_m)(D(μ_m)/√(f(0))+E(μ_m)/√(f(1)))=Δ(μ_m)(D̃(μ_m)/√(f(0))+Ẽ(μ_m)/√(f(1))).




In the same way, using det(Λ_g^m(λ))=det(Λ^m_g̃(λ)), we have:

    (M(μ_m)/√(f(0))-   ln'(h)(0)/4√(f(0)))(N(μ_m)/√(f(1))+ln'(h)(1)/4√(f(1)))-1/√(f(0)f(1))Δ^2(μ_m)
       =(M̃(μ_m)/√(f(0))-ln(h̃)'(0)/4√(f(0)))×(Ñ(μ_m)/√(f(1))+ln'(h̃)(1)/4√(f(1)))-1/√(f(0)f(1))Δ̃^2(μ_m).




Multiplying on both sides by Δ^2(μ_m)Δ̃^2(μ_m), we obtain:


    Δ̃^2(μ_m)[(-D(μ_m)/√(f(0))-Δ(μ_m)ln'(h)(0)/4√(f(0)))(-E(μ_m)/√(f(1))
    +Δ(μ_m)ln'(h)(1)/4√(f(1)))-1/√(f(0)f(1))]
       =Δ^2(μ_m)[(-D̃(μ_m)/√(f(0))-Δ̃(μ_m)
    ln(h̃)'(0)/4√(f(0)))(-Ẽ(μ_m)/√(f(1))+Δ̃(μ_m)ln'(h̃)(1)/4√(f(1)))-1/√(f(0)f(1))]




Now, let us show that the equalities (<ref>) and (<ref>) can be analytically extended with respect to μ_m in the half-right complex plane.
First, let us recall the definition of the so-called Nevanlinna class:




	Set Π^+={z∈ℂ, (z)>0} the half-right plane of the complex plane. The Nevanlinna class 𝒩(Π^+) is the set of analytic functions f on Π^+ such that
	
    0<r<1sup∫_-π^πln^+|f(1-re^iθ/1+re^iθ)|dθ <∞,

	with:
	
    ln^+(x)={   ln x  if  ln(x)≥ 0
       0  if  ln(x)< 0.
    	.





We have the following result <cit.>:




	Let h∈ H(Π^+) an analytic function on Π^+, A and C two constants. Assume :
	
    |h(z)|≤ Ce^A(z),  ∀ z∈Π^+.

	Then h∈𝒩(Π^+).



Thus, thanks to the asymptotics of Proposition <ref>, the holomorphic functions defined by δ:z↦Δ(z^2), d:z↦ D(z^2) and e:z↦ E(z^2)
belong to 𝒩(Π^+). Let us recall now a useful uniqueness theorem involving functions in the Nevanlinna class (see <cit.> for instance):	


	Let h∈𝒩(Π^+) and ℒ⊂ℝ_+ be a countable set such that
	
	

		∑_μ∈ℒ1/μ=+∞.
	
 Then :
	
    (h(μ)=0, ∀μ∈ℒ)⇒ h≡ 0   on Π^+.






Now, by the Weyl law,  (cf <cit.>), we get:


    μ_mm→ +∞=c(K)m^2/n-1+O(1)

where c(K)=(2π)^2/(ω_1vol(K))^2/n-1 and ω_1 is the volume of the unit ball in ℝ^n-1. Thus, for a fixed T∈ℕ, we have :

    μ_(mT)^n-1/c(K)T^2=m^2+O(1)/c(K)T^2.


As a consequence, for m and T large enough, the real numbers μ_(mT)^n-1 are always distinct. Now, we set :

    ℒ={√(μ_(mT)^n-1),  m∈ℕ}.

Using √(μ_(mT)^n-1)m→+∞∼√(c(K))Tm, one has:


    ∑_μ∈ℒ1/μ=+∞.

Thus, thanks to Theorem <ref>, the relations (<ref>) and (<ref>) are still true if one replaces μ_m by z^2 ∈ℂ, then by z.






Now, let us prove that  Δ(z)=Δ̃(z) for any z∈ℂ.
We recall that these functions are entire of order 1/2 and their roots are simple. So, using Hadamard's factorization theorem (see Proposition <ref>), we deduce that these functions
are entirely described by their roots (up to a multiplicative constant). Consequently, in order to prove that Δ=Δ̃, it is enough to show that their roots are the same.

Set 𝒫={α_j∈ℂ, Δ(α_j)=0}. Let α_k be in 𝒫 and let us show that Δ̃(α_k)=0. By definition, -α_k is an eigenvalue of the Sturm Liouville operator H = -d^2/dx^2+q with Dirichlet
boundary conditions. Thus, from Proposition <ref>, α_k is real and since the potential q is real, the quantities D(α_k) and E(α_k) are also real. Using (<ref>)
and (<ref>) with μ_m replaced with α_k, we obtain the following system:



    {   (D(α_k)/√(f(0))+E(α_k)/√(f(1)))Δ̃(α_k)=0
       (D(α_k)E(α_k)-1/√(f(0)f(1)))Δ̃(α_k)^2=0.
    .

To finish the proof, we distinguish two cases:




	
  ∙ If D(α_k)/√(f(0)) +E(α_k)/√(f(1)) 0 then Δ̃(α_k)=0.
	
  ∙ Otherwise D(α_k)/√(f(0))=-E(α_k)/√(f(1)). Then, substituting in the second equality, we have:
	
    (D(α_k)^2/f(0)+1/√(f(0)f(1)))Δ̃(α_k)^2=0


Since D(α_k)∈ℝ, we conclude that Δ̃(α_k)=0. By a standard symmetric argument, it follows Δ and Δ̃ have the same zeros. So, there exists C such that
Δ=CΔ̃. But,  as Δ(z) and Δ̃(z) have the same asymptotics :

    Δ(z),Δ̃(z)∼sinh(√(z))/√(z),

we deduce that C=1 and the proof is complete.







	It is interesting to note that Δ(z)=Δ̃(z) for all z∈ℂ implies that (in fact is equivalent to) q and q̃ are isospectral.







As a consequence, we can simplify (<ref>) (with μ_m replaced by z), and we get on ℂ :


    {   (M(z)/√(f(0))-ln'(h)(0)/4√(f(0)))+(N(z)/√(f(1))+
    ln'(h)(1)/4√(f(1)))=(M̃(z)/√(f(0))-ln(h̃)'(0)/4√(f(0)))
                                                           +
    (Ñ(z)/√(f(1))+ln'(h̃)(1)/4√(f(1)))
       (M(z)/√(f(0))-ln'(h)(0)/4√(f(0)))(N(z)/√(f(1))+ln'(h)(1)/4√(f(1)))=(M̃(z)/√(f(0))-
    ln(h̃)'(0)/4√(f(0)))
                                                           ×(Ñ(z)/√(f(1))+ln'(h̃)(1)/4√(f(1))).
    .

Thus, for each z∈ℂ, we have the following alternative:




    {   M(z)/√(f(0))-ln'(h)(0)/4√(f(0))=M̃(z)/√(f(0))-ln(h̃)'(0)/4√(f(0))
       N(z)/√(f(1))+ln'(h)(1)/4√(f(1))=Ñ(z)/√(f(1))+ln'(h̃)(1)/4√(f(1)),
    	.

or 
    {   M(z)/√(f(0))-ln'(h)(0)/4√(f(0))=Ñ(z)/√(f(1))+ln'(h̃)(1)/4√(f(1))
       N(z)/√(f(1))+ln'(h)(1)/4√(f(1))=M̃(z)/√(f(0))-ln(h̃)'(0)/4√(f(0)).
    	.





∙ As a first step, assume that the first system is satisfied for all z in ℂ. Then, using the asymptotics of M(z), (resp.  M̃(z)), we get:


    ln'(h)(0)/4√(f(0))=ln(h̃)'(0)/4√(f(0))

and therefore :

    M(z)=M̃(z)  for  all  z ∈ℂ\ℝ.




Thus, thanks to the Borg-Marchenko's theorem (see <cit.> for instance), we deduce:


    q=q̃  sur  [0,1].

Recall that :

    q=(h^1/4)”/h^1/4-λ f

where we have set h=f^n-2 (respectively q̃ and h̃=f̃^n-2). In particular, in the 2 dimensional case, we get immediateley f=f̃ since λ 0.
In dimension n ≥ 3, we see that f and f̃ same verify the same ODE with f(0)=f̃(0). Moreover, the relation


    ln'(h)(0)/4√(f(0))=ln(h̃)'(0)/4√(f̃(0)),

implies :

    f'(0)=f̃'(0).

Thus, the Cauchy-Lipschitz's theorem says that f=f̃..






∙ Secondly, assume there exists z_0∈ℂ satisfying :


    M(z_0)/√(f(0))-ln'(h)(0)/4√(f(0))M̃(z_0)/√(f(0))-ln(h̃)'(0)/4√(f̃(0)).




By a standard continuity argument, there is a ball B of center z_0 such that :


    M(z)/√(f(0))-ln'(h)(0)/4√(f(0))M̃(z)/√(f(0))-ln(h̃)'(0)/4√(f̃(0)),  ∀ z∈ B.
 Then, necessarily, we have:

    M(z)/√(f(0))-ln'(h)(0)/4√(f(0))=Ñ(z)/√(f(1))+ln(h̃)'(1)/4√(f̃(1)),  ∀ z∈ B.

Then, using the analytic continuation principle, the previous equality is true for every z∈ℂ\𝒫, where 𝒫 is the set of roots of Δ(z). Thanks to the asymptotics of M(z) and Ñ(z), one gets :

    f(0)=f(1)  (=f̃(1))  and  ln'(h)(0)/4√(f(0))=-ln(h̃)'(1)/4√(f̃(1)).

Hence, simplifying in (<ref>), we obtain
 
    M(z)=Ñ(z)  for  all   z ∈ℂ\ℝ.

By symmetry, Ñ has the same role as M̃ for the potential x↦q̃(1-x). Now, it follows, from the Borg-Marchenko's theorem that:

    q(x)=q̃(1-x)  ∀ x∈ [0,1],

and as previously, one gets:

    f=f̃∘η.




Case 2 : f(0)=f̃(1) and f(1)=f̃(0).



The proof is identical interchanging the roles of M and N.






§ UNIQUENESS RESULTS ON THE TRACE AND THE DETERMINANT


In this section, we assume that (K,g_K)=(𝕊^n-1,g_𝕊), and our main result is the following:



Assume that σ(Λ_g(λ))=σ(Λ_g̃(λ)). Then, there is m_0∈ℕ such that :


	∀ m∈ℕ, m≥ m_0⇒ (Λ_g^m(λ))=(Λ_g̃^m(λ)) and Tr(Λ_g^m(λ))=Tr(Λ_g̃^m(λ)).







Before giving the proof of this proposition, let us begin by the following lemma:




	Under the hypothesis σ(Λ_g(λ))=σ(Λ_g̃(λ)), we have the alternative :
	

		{   f(0)=f̃(0)	 
   f(1)=f̃(1)
	.  or  {   f(0)=f̃(1)	 
   f(1)=f̃(0).
	.
	


	

 First, let us define  the set:
	
    Σ(Λ_g(λ))={λ^±(κ_m),   m∈ℕ}

where κ_m is the m-th eigenvalue of the usual Laplace-Beltrami operator on the sphere 𝕊^n-1 and counted without multiplicity. Thanks to our hypothesis, one has obviously	

    Σ(Λ_g(λ))=Σ(Λ_g̃(λ)).

When K=𝕊^n-1, we have an explicit formula for κ_m (see for instance <cit.>) :
	
    κ_m=m(m+n-2),  ∀ m∈ℕ.

	
	
	
	The proof involves two steps.
	
	

		Step 1. First we have :
		
    f(0)^n-1/2+f(1)^n-1/2=f̃(0)^n-1/2+f̃(1)^n-1/2.
 
	
	
	
	Indeed, it is known that the Steklov spectrum determines the volume of the boundary of M : this is an immediate consequence of the Weyl's law (<ref>) for Steklov eigenvalues (see <cit.>).
	We have ∂ M=Γ_0∪Γ_1 where, for i∈{0,1}, Γ_i is the sphere 𝕊^n-1 equipped with the metric γ_i=f(i)g_𝕊. Hence :
		
    Vol(∂ M)=Vol(∂M̃)   ⇔∫_Γ_0 dVol_γ_0 + ∫_Γ_1 dVol_γ_1 = ∫_Γ_0 dVol_γ̃_0+∫_Γ_1 dVol_γ̃_1
       ⇔(f(0)^n-1/2+f(1)^n-1/2)Vol(𝕊^n-1)=(f̃(0)^n-1/2+f̃(1)^n-1/2)Vol(𝕊^n-1),

	
	and this proves the claim.
	
	
	
 Step 2. We show that : f(0)∈{f̃(0),f̃(1)}.




Assume this statement is false. Without loss of generality, assume that f(0)<min{f̃(̃0̃)̃,f̃(1)}. Then the equality (<ref>) implies : f(1)>max{f̃(0),f̃(1)}.



Our strategy is the following : we prove that one of the elements of Σ(Λ_g(λ)) is not in Σ(Λ_g̃(λ)) and this shall give a contradiction.



Let ε and A two positive numbers. Set :


    α_A=inf{|λ^+(κ_n)-λ^-(κ_m)| |  m,n≥ A}.




We claim that α_A≤1/2√(f(1))+ε for a sufficiently large A. 





Indeed, let n∈ℕ and set 

	m=max{j∈ℕ  |  λ^-(κ_j)<λ^+(κ_n)-1/2√(f(1))-ε/2}




By definition of m, λ^-(κ_m+1)≥λ^+(κ_n)-1/2√(f(1))-ε. But :

    λ^-(κ_m+1)   =λ^-(κ_m)+1/√(f(1))+O(1/m)
       < λ^+(κ_n)-1/2√(f(1))-ε/2+1/√(f(1))+O(1/m)
       <λ^+(κ_n)+1/2√(f(1))+ε/2+O(1/m)


Hence 


	α_A≤ |λ^+(κ_n)-λ^-(κ_m+1)|≤1/2√(f(1))+ε, for m,n≥ A sufficiently large.






From the above, we can now choose (m,n)∈ℕ^2 such that :


    -1/2√(f(1))-ε≤λ^+(κ_n)-λ^-(κ_m)≤1/2√(f(1))+ε.


The equality of the sets Σ(Λ_g(λ)) and Σ(Λ_g̃(λ)) gives two elements λ̃_1 and λ̃_2 of Σ(Λ_g̃(λ)) such that {λ^-(κ_m),λ^+(κ_n)}={λ̃_1,λ̃_2}. Remark that if λ^-(κ_m)λ^+(κ_n) then the elements λ̃_1 and λ̃_2 can not both belong to the same sequence (λ̃^-(κ_ℓ)) or (λ̃^+(κ_p)) because, for indexes m and n large enough, one would get :

    |λ̃_1-λ̃_2|≥min{1/√(f̃(0)),1/√(f̃(1))}-ε

and, by choosing ε small enough, this would contradict (<ref>). Consequently, we have the existence of (ℓ,p)∈ℕ^2 such that

    {λ^-(κ_m),λ^+(κ_n)}={λ̃^-(κ_ℓ),λ̃^+(κ_p)}.


Let us assume first that

    {   λ^-(κ_m)= λ̃^-(κ_ℓ) 
       λ^+(κ_n)= λ̃^+(κ_p).  .




	
  * Case 1 : λ^-(κ_m)≤λ^+(κ_n). Then λ^-(κ_m+1) is not equal to any element of Σ(Λ_g̃(λ)). Indeed we have on one hand :
	
	
    λ^-(κ_m+1)   =λ^-(κ_m)+1/√(f(1))+O(1/m)
       =λ̃^-(κ_ℓ)+1/√(f(1))+O(1/m)
       =λ̃^-(κ_ℓ+1)+[1/√(f(1))-1/√(f̃(1))_<0]+O(1/m)+O(1/ℓ)

	
	Hence, for m and ℓ large enough : λ̃^-(κ_ℓ)<λ^-(κ_m+1)<λ̃^-(κ_ℓ+1). On the other hand :
	
	
    λ^-(κ_m+1)   = λ^-(κ_m)+1/√(f(1))+O(1/m)
       ≤λ^+(κ_n)+1/√(f(1))+O(1/m)
       = λ̃^+(κ_p)+1/√(f(1))+O(1/m)
       =λ̃^+(κ_p+1)+[1/√(f(1))-1/√(f̃(0))_<0]+O(1/m)+O(1/p).

	Moreover, from (<ref>) and (<ref>) : 
    λ^-(κ_m+1)   =λ^-(κ_m)+1/√(f(1))+O(1/m)
       =λ^-(κ_m)+1/2√(f(1))+ε+O(1/m)+1/2√(f(1))-ε
       ≥λ̃^+(κ_p)+1/2√(f(1))-2ε

	
	
	
	
	As a result, for m and p large enough : λ̃^+(κ_p)<λ^-(κ_m+1)<λ̃^+(κ_p+1)
	
	The sequences (λ̃^-(κ_m)) and (λ̃^+(κ_m)) being strictly increasing (at least for m large enough), none of the elements of Σ(Λ_g̃(λ)) can be equal to λ^-(κ_m+1). This refutes Σ(Λ_g̃(λ))=Σ(Λ_g(λ)).
	
	
	
	
  * Case 2 : λ^-(κ_m)> λ^+(κ_n). With similar arguments, one can prove :
	
    max(λ̃^-(κ_ℓ-1), λ̃^+(κ_p-1))< λ^-(κ_m-1) < min(λ̃^-(κ_ℓ), λ̃^+(κ_p)).

	


Now, if we assume that

    {   λ^-(κ_m)= λ̃^+(κ_p) 
       λ^+(κ_n)= λ̃^-(κ_ℓ),  .


one can also prove, by interchanging the roles of f̃(0) and f̃(1), that λ^-(κ_m+1) or λ^-(κ_m-1) does not belong to the set Σ(Λ_g̃(λ)).

 

Thus f(0)∈{f̃(0),f̃(1)}. Associated to the equality (<ref>), this gives the wanted conclusion.







From now on, we assume that f(0)=f̃(0) and f(1)=f̃(1), since the case 
 f(0)=f̃(1) and f(1)=f̃(0) is obtained by substituting the roles of λ̃^-(κ_m) and λ̃^+(κ_m). First, let us begin by a simple case:



 §.§ The case f(0)= f(1)




Without loss of generality, we can assume that f(0)=f(1)=1. Thanks to Lemma <ref>, Λ_g^m(λ) has two eigenvalues λ^-(μ_m) and λ^+(μ_m) whose asymptotics are given by :

    {   λ^-(μ_m)=√(μ_m)-ln(h)'(1)/4√(f(1))+O(1/√(μ_m))	 
       λ^+(μ_m)=√(μ_m)+ln(h)'(0)/4√(f(0))+O(1/√(μ_m)).
    .

We recall that : 
    √(κ_m)=m+n-2/2+O(1/m),

and for m∈ℕ, we set V_m={λ^-(κ_m), λ^+(κ_m)} and Ṽ_m:={λ̃^-(κ_m), λ̃^+(κ_m)}.




∙ The two dimensional case




In this case, the eigenvalues λ^±(κ_m) have the following asymptotics :


    {   λ^-(κ_m)=m+O(1/m)	 
       λ^+(κ_m)=m+O(1/m).
    .


For m large enough, the sets V_m and Ṽ_m are both included in the interval [m-1/4,m+1/4]. In particular V_m∩Ṽ_m'=∅ if m m'. The equality Σ(Λ_g(λ))=Σ(Λ_g̃(λ)) leads to the equalities V_m=Ṽ_m if m is greater than some index m_0. Consequently, there is m_0 such that, for m≥ m_0 :

    {   λ^-(κ_m)+λ^+(κ_m)=λ̃^-(κ_m)+λ̃^+(κ_m) 
       λ^-(κ_m)λ^+(κ_m)=λ̃^-(κ_m)λ̃^+(κ_m).
    	.




Of course, the previous equalities are still true when κ_m is replaced by μ_m. Thus, we have proved:





	∀ m∈ℕ,  m≥ m_0⇒ Tr(Λ^m_g)=Tr(Λ^m_g̃) and (Λ^m_g)=(Λ^m_g̃).






∙ The n ≥ 3 dimensional case





In this case, we use the following asymptotics of the eigenvalues λ^±(κ_m):


    {   λ^-(κ_m)=m+n-2/2-h'(1)/4+O(1/m)	 
       λ^+(κ_m)=m+n-2/2+h'(0)/4+O(1/m).
    .




Contrary to the two dimensional case, we can not conclude that the sets V_m and Ṽ_m are equal. This is due to the presence of the constants h'(0) and h'(1) in these asymptotics. We have the following Proposition:



	
There is ℒ⊂ℕ such that ∑_m∈ℒ1/m=+∞ satisfying :
	
    ∀ m∈ℒ,  {   λ^-(κ_m)+λ^+(κ_m)=λ̃^-(κ_m)+λ̃^+(κ_m) 
       λ^-(κ_m)λ^+(κ_m)=λ̃^-(κ_m)λ̃^+(κ_m) .



 We start with the following Lemma:


	
There is ℒ_1⊂ℕ such that ∑_m∈ℒ_11/m=+∞ satisfying :

    (λ^-(κ_m)=λ̃^-(κ_m),  ∀ m∈ℒ_1)  or  (λ^-(κ_m)=λ̃^+(κ_m),  ∀ m∈ℒ_1)




Since Σ(Λ_g(λ))=Σ(Λ_g̃(λ)), we have the inclusion :


    {λ^-(κ_m), m∈ℕ}⊂{λ̃^-(κ_m), m∈ℕ}∪{λ̃^+(κ_m), m∈ℕ}.


Thus, there exists  a sequence of integers (a_m) such that ∑_m∈ℕ1/a_m=+∞ and :



{λ^-(κ_a_m), m∈ℕ}⊂{λ̃^-(κ_m), m∈ℕ} or {λ^-(κ_a_m), m∈ℕ}⊂{λ̃^+(κ_m), m∈ℕ}.





For instance, let us study  the first case (since the second case is similar). We can find another sequence of integers (ã_m) such that :

    λ^-(κ_a_m)=λ̃^-(κ_ã_m),  ∀ m∈ℕ.

Thanks to Lemma <ref>, one obtains :


    a_m-ã_m=h̃'(1)/4-h'(1)/4+O(1/a_m).


Therefore, the sequence of integers (a_m-ã_m) converges to the integer h̃'(1)/4-h'(1)/4=n-2/4f'(1)/f(1)-n-2/4f̃'(1)/f̃(1).

Recalling that f,f̃∈𝒞_b={f∈ C^∞([0,1]), |f'(i)/f(i)|≤1/n-2, i∈{0,1}}, we get:

    |n-2/4f'(1)/f(1)-n-2/4f̃'(1)/f̃(1)|≤1/2.

As this quantity must be an integer, we have proved that, for m≥ m_0 :

    a_m=ã_m.


We conclude the proof of the Lemma setting ℒ_1={a_m,  m≥ m_0}.





Now, we can finish the proof of the Proposition. For instance, assume that:

    λ^-(κ_m)=λ̃^-(κ_m),  ∀ m∈ℒ_1.

Repeating the previous argument with λ^+(κ_m), m ∈ℒ_1, we have the inclusion:


    {λ^+(κ_m), m∈ℒ_1}⊂{λ̃^-(κ_m), m∈ℕ}∪{λ̃^+(κ_m), m∈ℕ}


There is a sequence of integers, denoted (b_m)∈ℒ_1, such that ∑_m∈ℕ1/b_m=+∞ and :



	{λ^+(κ_b_m), m∈ℕ}⊂{λ̃^-(κ_m), m∈ℕ} or {λ^+(κ_b_m), m∈ℕ}⊂{λ̃^+(κ_m), m∈ℕ}.





  ∙ In the second case, there is a sequence (b̃_m) such that :

    λ^+(κ_b_m)=λ̃^+(κ_b̃_m)

which implies as previously that b_m=b̃_m. Then, setting ℒ={b_m,  m∈ℕ} we have for all m∈ℒ :

    {   λ^-(κ_m)=λ̃^-(κ_m)	 
       λ^+(κ_m)=λ̃^+(κ_m)
    .




  ∙ In the first case, we still could show that :

    λ^+(κ_b_m)=λ̃^-(κ_b_m)  (=λ^-(κ_b_m))

But this is not possible because Λ^b_m_g(λ) would then be an homothety, which contradicts the matrix expression of Λ^b_m_g(λ).

Hence the proof of Proposition <ref> is complete.









 
We recall that 𝒫 is the set of roots of Δ(z). For every z∈ℂ\𝒫, let us set 

    B(z)   =   -D(z)+E(z)/Δ(z)+C_0-C_1 
    
    C(z)   =   (-D(z)/Δ(z)+C_0)(-E(z)/Δ(z)
    -C_1)-1/√(f(0)f(1))Δ(z)^2,

where 


	C_0=ln(h)'(0)/4√(f(0)), C_1=ln(h)'(1)/4√(f(1)).



For every m in ℕ, we have (cf proof of Lemma <ref>):



	B(κ_m)=λ^-(κ_m)+λ^+(κ_m) and C(κ_m)=λ^-(κ_m)λ^+(κ_m).





We introduce the functions g_1 and g_2 on Π^+ as follows:

    g_1(z)   = Δ(z^2)Δ̃(z^2)[B(z^2)-B̃(z^2)]
    
    	g_2(z)   = Δ(z^2)^2Δ̃(z^2)^2[C(z^2)-C̃(z^2)].

We claim that g_1 and g_2 are identically zero. Indeed :


	
  ∙ g_1,g_2 are holomorphic on Π^+.
	
	
  ∙ g_1,g_2∈𝒩(Π^+) thanks to the estimates of Proposition <ref>.
	
	
  ∙ Thanks to Proposition <ref>, we have g_1(√(κ_m))=g_2(√(κ_m))=0 for every m∈ℒ. As √(κ_m)∼ m, one has ∑_m∈ℒ1/√(κ_m)=+∞. Thus, we can conclude, by Nevanlinna's theorem :
	
    g_1≡ g_2≡ 0   on  Π^+.



In particular, for every m in ℕ :


    {   λ^-(κ_m)+λ^+(κ_m)=λ̃^-(κ_m)+λ̃^+(κ_m)	 
       λ^-(κ_m)λ^+(κ_m)=λ̃^-(κ_m)λ̃^+(κ_m)
    .




and, by replacing κ_m by μ_m we have for every m in ℕ :





	Tr(Λ^m_g)=Tr(Λ^m_g̃) and (Λ^m_g)=(Λ^m_g̃).



	

 §.§ The case f(0) f(1)




Without loss of generality, assume that f(0)<f(1). Thanks to the asymptotics:


    √(κ_m)=m+n-2/2+O(1/m)
 and Lemma <ref>, we have :
	
    {   λ^-(κ_m)=m/√(f(1))+n-2/2√(f(1))-ln(h)'(1)/4√(f(1))+O(1/m)	 
       λ^+(κ_m)=m/√(f(0))+n-2/2√(f(0))+ln(h)'(0)/4√(f(0))+O(1/m).
    .




Since f(0)≠ f(1), the proof is a little more delicate. As previously, we have to find a subset ℒ⊂ℕ such that ∑_m∈ℒ1/m=+∞ and such that λ^-(κ_m)=λ̃^-(κ_m) or λ^+(κ_m)=λ̃^+(κ_m) for all m  ∈ℒ. We have the following Proposition:


	
	There is ℒ⊂ℕ satisfying ∑_m∈ℒ1/m=+∞ and such that :

    {λ^-(κ_m),   m∈ℒ}⊂{λ̃^-(κ_m),   m∈ℕ}



	

	
If the subset of the elements of {λ^-(κ_m),   m∈ℕ} belonging to {λ̃^+(κ_m),   m∈ℕ} is finite, the proposition is true. If not, there exist two strictly increasing sequences  φ,ψ :ℕ→ℕ such that:

    λ^-(κ_ψ(m))=λ̃^+(κ_φ(m)).

	
	We emphasize that the functions φ and ψ are built so that an integer m∈ℕ which is not in the image of ψ (respectively in that of φ) satisfies λ^-(κ_m)=λ̃^-(κ_n) for some n∈ℕ (respectively λ^+(κ_n)=λ̃^+(κ_m) for some n∈ℕ).

		
	
	
Replacing  λ^+(κ_φ(m)) and λ̃^-(κ_ψ(m)) by their asymptotics in the equality (<ref>), we have :

    φ(m)/√(f(0))+ln(h)'(0)/4√(f(0))+n-2/2√(f(0))+O(1/φ(m))=ψ(m)/√(f(1))-ln(h)'(1)/4√(f(1))+n-2/2√(f(1))+O(1/ψ(m)).

Setting C=-ln(h)'(1)/4√(f(1))-ln(h)'(0)/4√(f(0))+n-2/2√(f(1))-n-2/2√(f(0)) and noticing that the previous equality implies that O(1/ψ(m))=O(1/φ(m)), one can write (<ref>) as follows:

    φ(m)/√(f(0))=ψ(m)/√(f(1))+C+O(1/φ(m)).




Now, we have the following Lemma:


		
		There exists m_0∈ℕ such that for all m≥ m_0,  ψ(m+1)≥ψ(m)+ 2.

	
	

		Set A=√(f(1))/√(f(0))>1 and C'=-√(f(1))C. It follows from (<ref>) that:
		
    ψ(m)=Aφ(m)+C'+O(1/φ(m)).

		Assume ψ(m+1)=ψ(m)+1. Then :
		
    ψ(m)+1=ψ(m+1)   =Aφ(m+1)+C'+O(1/φ(m))
       ≥ A(φ(m)+1)+C'+O(1/φ(m))
       =Aφ(m)+C'+A+O(1/φ(m))
       =ψ(m)+A+O(1/φ(m)).

Thus, we get:
		
    1≥ A+O(1/φ(m))

which is clearly false for m  large enough.	




Consequently, the range of ψ  doesn't contain two consecutive integers. We deduce from this the following Lemma :




	
Denote a_1,a_2,... the sequence of all integers that are not in the range of ψ. There exists C>0 such that this sequence satisfies :

    a_m≤ 2m+C.





We set p_0=ψ(m_0) and C:= a_p_0 -2p_0. Clearly, (<ref>) is true for m=p_0. Now, assume that (<ref>) is satisfied for a fixed m≥ p_0. Then :

	
  ∙ either a_m+1 is not in the range of ψ and so : a_m+1=a_m+1,
	
  ∙ or a_m+1 is in the range of ψ. Then a_m+2 is not in the image of ψ since it does not contain consecutive integers and consequently a_m+1=a_m+2.

In both cases, we get a_m+1≤ 2(m+1)+C. Then, the proof of the Lemma follows by a standard induction argument.



In particular, Lemma <ref> shows that :

    ∑_n∈ℕ1/a_n=+∞.




Thus, setting ℒ={a_n,   n∈ℕ}, we have :




    {λ^-(κ_m),   m∈ℒ}⊂{λ̃^-(κ_m),   m∈ℕ}

which concludes  the proof of the Proposition.




As previously, we have to treat differently the case of the dimension n=2 and n≥ 3.





  §.§.§ The two dimensional case:





	
	
		
	For all m ∈ℒ, one has:
	
    λ^-(κ_m)=λ̃^-(κ_m).

	



By construction, for each element m of ℒ, there exists ℓ(m)∈ℕ such that :

    λ^-(κ_m)=λ̃^-(κ_ℓ(m)).

Using Lemma <ref>, we get:

    m=ℓ(m)+O(1/m)+O(1/ℓ(m)),

which implies m=ℓ(m) for m large enough.





Let us consider again the functions B and C defined in (<ref>) and (<ref>). Setting

    R(z)=B(z)^2-4C(z),
 we get:

    λ^±(κ_m)=1/2(B(κ_m)±√(R(κ_m))).


We can define on Π^+ the function λ^-(z) by λ^-(z)= 1/2(B(z)-√(R(z))).



	
	For every t∈ℝ_+ large enough, we have the alternative :
	
	

		λ^-(t)=λ̃^-(t) or λ^+(t)=λ̃^+(t).
	





From Lemma <ref>, there is ℒ⊂ℕ satisfying ∑_m∈ℒ1/m=+∞ and such that


    ∀ m∈ℒ,     B(κ_m)-√(R(κ_m))=B̃(κ_m)-√(R̃(κ_m)).

One has
	
    (B(κ_m)-B̃(κ_m))^2=(√(R(κ_m))-√(R̃(κ_m)))^2,

and then

    [(B(κ_m)-B̃(κ_m))^2-R(κ_m)-R̃(κ_m)]^2-4R(κ_m)R̃(κ_m)_:=g_1(κ_m)=0.

For z ∈Π^+, let us define:

    g_2(z)=Δ(z^2)^4Δ̃(z^2)^4g_1(z^2).

	
	Then g_2 is identically zero. Indeed :
	
	
		
  ∙ g_2 is holomorphic on Π^+.
		
		
  ∙ g_2∈𝒩(Π^+) thanks to the estimates of Proposition <ref>.
		
		
  ∙ We have g_2(√(κ_m))=0 for every m∈ℒ, which enable us to conclude, by Nevanlinna's theorem, that
		
    g_2≡ 0   on  Π^+.

		
	
We have obtained:

    ∀ z∈Π^+,    [(B(z^2)-B̃(z^2))^2-R(z^2)-R̃(z^2)]^2-4R(z^2)R̃(z^2)=0.

It easily follows that for z∈Π^+, we have four alternatives:
	
    λ^-(z^2) = λ̃^-(z^2),  λ^+(z^2) = λ̃^-(z^2),  λ^-(z^2) =λ̃^+(z^2)  or  λ^+(z^2) = λ̃^+(z^2).


We can note that the asymptotics of Lemma <ref>, involving λ^±(κ_m), can be extended on ℝ_+ for λ^±(t) (the proof remains exactly the same). Now, if a real sequence of positive numbers (t_m) satisfies, for instance, λ^+(t_m^2) = λ̃^-(t_m^2) with t_m→ +∞, then, using Lemma <ref> with t_m^2 instead of κ_m we get f(0)=f(1), and that contradicts our hypothesis. We use the same argument to exclude the third case when z is a positive real number large enough. Finally, we have for t positive large enough :
	
    λ^-(t) = λ̃^-(t)  or  λ^+(t) = λ̃^+(t).


Case 1 : Assume now there exists x∈ℝ_+ as large as we want such that λ^-(x) λ̃^-(x). By a standard continuity argument, we can find an interval I centered in x such that :



∀ t∈ I, λ^-(t) λ̃^-(t),

 
which implies necessarily that ∀ t∈ I, λ^+(t) = λ̃^+(t). Moreover, there is L>0 such that the real function

    t↦λ^+(t)-λ̃^+(t)
  is analytic on the interval [L,+∞[. Thus, ∀ t≥ L, one has λ^+(t) = λ̃^+(t) by the analytic continuation principle. One deduces there exists m_0∈ℕ such that, for m∈ℒ and m≥ m_0:

    λ^+(κ_m) = λ̃^+(κ_m)    and   λ^-(κ_m) = λ̃^-(κ_m).

Without loss of generality, assume that minℒ is greater than m_0. As a by product, one gets:
	
    ∀ m∈ℒ,  {   λ^-(κ_m)+λ^+(κ_m)=λ̃^-(κ_m)+λ̃^+(κ_m) 
       λ^-(κ_m)λ^+(κ_m)=λ̃^-(κ_m)λ̃^+(κ_m) .

Hence, we have proved the same result as the one of Proposition <ref>. We can deduce similarly that both previous equalities are in fact true for every m∈ℕ, i.e 


	∀ m∈ℕ, Tr(Λ^m_g)=Tr(Λ^m_g̃) and (Λ^m_g)=(Λ^m_g̃).


We are thus brought back to the third section.



Case 2 : For all t∈ℝ, t large enough, we have λ^-(t)=λ̃^-(t). In particular, for all m∈ℕ large enough, λ^-(κ_m)=λ̃^-(κ_m). Hence, for m large enough, one also has λ^+(κ_m)=λ̃^+(κ_m). Indeed, let m∈ℕ. If λ^+(κ_m)∈(λ̃^+(κ_ℓ)), there is ℓ(m)∈ℕ such that λ^+(κ_m)=λ̃^+(κ_ℓ(m)). One can prove, as in the proof of Lemma <ref> that ℓ(m)=m. The same holds if we assume that λ̃^+(κ_m)∈(λ^+(κ_ℓ)). The only other option is that there are integers p and ℓ such that 

    {   λ^+(κ_m) = λ̃^-(κ_p)
       λ̃^+(κ_m)= λ^-(κ_ℓ)
    	.

Then λ^+(κ_m)-λ̃^+(κ_m)=λ̃^-(κ_p)-λ^-(κ_ℓ), which implies

    O(1/m)=p-ℓ/√(f(1))+O(1/p)+O(1/ℓ).

Hence p=ℓ for m,p,ℓ large enough and so λ^+(κ_m)=λ̃^+(κ_m). There is then m_0∈ℕ such that


	∀ m≥ m_0, Tr(Λ^m_g)=Tr(Λ^m_g̃) and (Λ^m_g)=(Λ^m_g̃).


We are brought back again to the third section (see Remark <ref> p. 13).



  §.§.§ The case of the  dimension n≥ 3.




The following arguments are more or less immediate adaptations of the two dimensional case. We just refer to it for details.




	There exists ℒ⊂ℕ such that  ∑_m∈ℒ1/m=+∞ and :
	
    λ̃^-(κ_m)=λ^-(κ_m),   ∀ m∈ℒ.




By definition of the sequence (a_n) defined in Lemma <ref>, there is another sequence of integers (ã_n) such that :
	
    λ^-(κ_a_n)=λ̃^-(κ_ã_n)

As in the proof of Lemma <ref>, we show that  a_m=ã_m and we set ℒ={a_m,  m∈ℒ}.



We have the following Lemma (the proof is identical to the two dimensional case).

For all t∈ℝ_+ large enough, we have the alternative :

    λ^-(t) = λ̃^-(t),    λ^+(t) = λ̃^+(t).




We deduce then, for all m∈ℒ :


λ^-(κ_m) = λ̃^-(κ_m) and λ^+(κ_m) = λ̃^+(κ_m).



It follows that :



	∀ m∈ℕ, Tr(Λ^m_g)=Tr(Λ^m_g̃) and (Λ^m_g)=(Λ^m_g̃).






Aknowledgements: The author would like to thank Thierry Daudé and François Nicoleau for their encouragements and helpful discussions. The author also thanks the referees for valuable remarks and comments that improved the first version of this paper. 
	









acm

	
   	

  
  
  
      
      
      
      
      
      
      
 
   
 
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
	  









  