Pierre Bras
email: pierre.bras@sorbonne-universite.fr

Convergence rates of Gibbs measures with degenerate minimum

Introduction

Gibbs measures and their convergence properties are often used in stochastic optimization to minimize a function dened on R d . That is, let f : R d → R be a measurable function and let x ∈ R d be such that f admits a global minimum at x . It is well known [START_REF] Hwang | Laplace's method revisited: Weak convergence of probability measures[END_REF] that under standard assumptions, the associated Gibbs measure with density proportional to e -f (x)/t for t > 0, converges weakly to the Dirac mass at x , δ x , when t → 0. The Langevin equation dX s = -∇f (X s )ds + σdW s consists in a gradient descent with Gaussian noise. For σ = √ 2t, its invariant measure has a density proportional to e -f (x)/t (see for example [START_REF] Khasminskii | Stochastic stability of dierential equations[END_REF], Lemma 4.16), so for small t we can expect it to converge to argmin(f ) [START_REF] Arnak | Theoretical guarantees for approximate sampling from smooth and log-concave densities[END_REF] [START_REF] Barrera | Limit behavior of the invariant measure for Langevin dynamics[END_REF]. The simulated annealing algorithm [START_REF] Peter | Simulated Annealing: Theory and Applications[END_REF] builds a Markov chain from the Gibbs measure where the parameter t converges to zero over the iterations. This idea is also used in [START_REF] Saul | Recursive stochastic algorithms for global optimization in R d[END_REF], giving a stochastic gradient descent algorithm where the noise is gradually decreased to zero. Adding a small noise to the gradient descent allows to explore the space and to escape from traps such as local minima and saddle points which appear in non-convex optimization problems [START_REF] Lazarev | Convergence of Stochastic-Approximation procedures in the case of a Regression Equation with Several Roots[END_REF] [DPG + 14]. Such methods have been recently brought up to light again with SGLD (Stochastic Gradient Langevin Dynamics) algorithms [START_REF] Welling | Bayesian Learning via Stochastic Gradient Langevin Dynamics[END_REF] [LCCC15], especially for Machine Learning and calibration of articial neural networks, which is a high-dimensional non-convex optimization problem.

The rates of convergence of Gibbs measures have been studied in [START_REF] Hwang | Laplace's method revisited: Weak convergence of probability measures[END_REF], [START_REF] Hwang | A Generalization of Laplaces's Method[END_REF] and [START_REF] Athreya | Gibbs measures asymptotics[END_REF] under dierentiability assumptions on f . It turns out to be of order t 1/2 as soon as the Hessian matrix ∇ 2 f (x ) is positive denite. Furthermore, in the multiple well case i.e. if the minimum of f is attained at nitely many points x 1 , . . ., x m , [START_REF] Hwang | Laplace's method revisited: Weak convergence of probability measures[END_REF] proves that the limit distribution is a sum of Dirac masses δ x i with coecients proportional to det(∇ 2 f (x i )) -1/2 as soon as all the Hessian matrices are positive denite. If such is not the case, we can conjecture that the limit distribution is concentrated around the x i where the degeneracy is of the highest order.

The aim of this paper is to provide a rate of convergence in this degenerate setting, i.e. when x is still a strict global minimum but ∇ 2 f (x ) is no longer denite, which extends the range of applications of Gibbs measure-based algorithms where positive deniteness is generally assumed. A general framework is given in [START_REF] Athreya | Gibbs measures asymptotics[END_REF], which provides rates of convergence based on dominated convergence. However a strong and rather technical assumption on f is needed and checking it seems, to some extent, more demanding than proving the result. To be more precise, the assumption reads as follows: there exists a function g : R d → R with e -g ∈ L 1 (R d ) and α 1 , . . . , α d ∈ (0, +∞) such that

∀h ∈ R d , 1 t [f (x + (t α 1 h 1 , . . . , t α d h d )) -f (x )] -→ t→0 g(h 1 , . . . , h d ). (1)
Our objective is to give conditions on f such that (1) is fullled and then to elucidate the expression of g depending on f and its derivatives by studying the behaviour of f at x in every direction. Doing so we can apply the results from [START_REF] Athreya | Gibbs measures asymptotics[END_REF] yielding the convergence rate of the corresponding Gibbs measures. The orders α 1 , . . ., α d must be chosen carefully and not too big, as the function g needs to depend on every of its variables h 1 , . . ., h d , which is a necessary condition for e -g to be integrable. We also extend our results to the multiple well case.

We generally assume f to be coercive, i.e. f (x) → +∞ as ||x|| → +∞, C 2p in a neighbourhood of x for some p ∈ N and we assume that the minimum is polynomial strict, i.e. the function f is bounded below in a neighbourhood of x by some non-negative polynomial function, null only at x . Thus we can apply a multi-dimensional Taylor expansion to f at x , where the successive derivatives of f : R d → R are seen as symmetric tensors of R d . The idea is then to consider the successive subspaces where the derivatives of f are null up to some order ; using that the Taylor expansion of f (x + h) -f (x ) is non-negative, some cross derivative terms are null. However a diculty arises at orders 6 and higher, as the set where the derivatives of f are null up to some order is no longer a vector subspace in general. This diculty is linked with Hilbert's 17 th problem [START_REF] Hilbert | Ueber die Darstellung deniter Formen als Summe von Formenquadraten[END_REF], stating that a non-negative multivariate polynomial cannot be written as the sum of squares of polynomials in general. We thus need to change the denition of the subspaces we consider. Following this, we give a recursive algorithm yielding an adapted decomposition of R d into vector subspaces and a function g satisfying (1) up to a change of basis, giving a canonical higher order nested decomposition of f at x in degenerate cases. An interesting fact is that the case where the polynomial order of x is 10 or higher fundamentally diers from those of orders 2, 4, 6 and 8, owing to the presence of even cross terms which may be not null. The algorithm we provide works at the orders 10 or higher only under the assumption that all such even cross terms are null. In general, it is more dicult to get a general expression of g for the orders 10 and higher. We then apply our results to [START_REF] Athreya | Gibbs measures asymptotics[END_REF], where we give conditions such that the hypotheses of [START_REF] Athreya | Gibbs measures asymptotics[END_REF], especially (1), are satised so as to infer rates of convergence of Gibbs measures in the degenerate case where ∇ 2 f (x ) is not necessarily positive denite. The function g given by our algorithm is a non-negative polynomial function and non-constant in any of its variables, however it needs to be assumed to be coercive to be applied to [START_REF] Athreya | Gibbs measures asymptotics[END_REF]. We study the case where g is not coercive and give a method to deal with simple generic non-coercive cases, where our algorithm seems to be a rst step to a more general procedure. However, we do not give a general method in this case.

Our results are applied to Gibbs measures but they can also be applied to more general contexts, as we give a canonical higher order nested expansion of f at a minimum, in the case where some derivatives are degenerate.

For general properties of symmetric tensors we refer to [START_REF] Comon | Symmetric tensor and symmetric tensor rank[END_REF]. In the framework of stochastic approximation, [FP99] Section 3.1 introduced the notion of strict polynomial local extremum and investigated their properties as higher order "noisy traps".

The paper is organized as follows. In Section 3, we recall convergence properties of Gibbs measures and revisit the main theorem from [START_REF] Athreya | Gibbs measures asymptotics[END_REF]. This theorem requires, as an hypothesis, to nd an expansion of f at its global minimum ; we properly state this problem in Section 3.2 under the assumption of strict polynomial minimum. In Section 3.3, we state our main result for both single well and multiple well cases, as well as our algorithm. In Section 4, we detail the expansion of f at its minimum for each order and provide the proof. We give the general expression of the canonical higher order nested expansion at any order in Section 4.1, where we distinguish the orders 10 and higher from the lower ones. We then provide the proof for each order 2, 4, 6 and 8 in Sections 4.3, 4.4, 4.6 and 4.7 respectively. We need to prove that, with the exponents α 1 , . . ., α d we specify, the convergence in (1) holds ; we do so by proving that, using the non-negativity of the Taylor expansion, some cross derivative terms are zero. Because of Hilbert's 17 th problem, we need to distinguish the orders 6 and 8 from the orders 2 and 4, as emphasized in Section 4.5. For orders 10 and higher, such terms are not necessarily zero and must then be assumed to be zero. In Section 4.8, we give a counter-example if this assumption is not satised before proving the result. In Section 4.9, we prove that for every order the resulting function g is constant in none of its variables and that the convergence in (1) is uniform on every compact set. In Section 4.10, we study the case where the function g is not coercive and give a method to deal with the simple generic case. In Section 5, we prove our main theorems stated in Section 3.3 using the expansion of f established in Section 4. Finally, in Section 6, we deal with a "at" example where all the derivatives in the local minimum are zero and where we cannot apply our main theorems.

Denitions and notations

We give a brief list of notations that are used throughout the paper.

We endow R d with its canonical basis (e 1 , . . . , e d ) and the Euclidean norm denoted by || • ||. For x ∈ R d and r > 0 we denote by B(x, r) the Euclidean ball of R d of center x and radius r. For E a vector subspace of R d , we denote by p E : R d → E the orthogonal projection on E. For a decomposition of R d into orthogonal subspaces, For v 1 , . . ., v k vectors in R d and T a tensor of order k of R d , we denote the tensor product

R d = E 1 ⊕ • • • ⊕ E p , we say that an orthogonal transformation B ∈ O d (R) is adapted to this decomposition if for all j ∈ {1, . . . , p}, ∀i ∈ {dim(E 1 ) + • • • + dim(E j-1 ) + 1, . . . , dim(E 1 ) + • • • + dim(E j )}, B • e i ∈ E j .
T • (v 1 ⊗ • • • ⊗ v k ) = i 1 ,...,i k ∈{1,...,d} T i 1 •••i k v 1 i 1 . . . v k i k . More generally, if j ≤ k and v 1 , . . . , v j are j vectors in R d , then T • (v 1 ⊗ • • • ⊗ v j ) is a tensor of order k -j such that: T • (v 1 ⊗ • • • ⊗ v j ) i j+1 ...i k = i 1 ,...,i j ∈{1,...,d} T i 1 ...i k v 1 i 1 . . . v j i j .
For h ∈ R d , h ⊗k denotes the tensor of order k such that

h ⊗k = (h i 1 . . . h i k ) i 1 ,...,i k ∈{1,...,d} . For a function f ∈ C p R d , R , we denote ∇ k f (x) the dierential of order k ≤ p of f at x, as ∇ k f (x)
is the tensor of order k dened by:

∇ k f (x) = ∂ k f (x) ∂x i 1 • • • ∂x i k i 1 ,i 2 ,...,i k ∈{1,...,d}
.

By Schwarz's theorem, this tensor is symmetric, i.e. for all permutation σ ∈ S k ,

∂ k f (x) ∂x i σ(1) • • • ∂x i σ(k) = ∂ k f (x) ∂x i 1 • • • ∂x i k .
We recall the Taylor-Young formula in any dimension, and the Newton multinomial formula.

Theorem 1 (Taylor-Young formula). Let f : R d → R be C p and let x ∈ R d . Then:

f (x + h) = h→0 p k=0 1 k! ∇ k f (x) • h ⊗k + ||h|| p o(1).
We denote by k i 1 ,...,ip the p-nomial coecient, dened as:

k i 1 , . . . , i p = k! i 1 ! . . . i p ! .
Theorem 2 (Newton multinomial formula). Let h 1 , . . . , h p ∈ R d , then

(h 1 + h 2 + • • • + h p ) ⊗k = i 1 ,...,ip∈{0,...,k} i 1 +•••+ip=k k i 1 , . . . , i p h ⊗i 1 1 ⊗ • • • ⊗ h ⊗ip p . (2)
For T a tensor of order k, we say that T is non-negative (resp. positive) if

∀h ∈ R d , T • h ⊗k ≥ 0 (resp. T • h ⊗k > 0). (3)
We denote L 1 (R d ) the set of measurable functions f : R d → R that are integrable with respect to the Lebesgue measure on R d . We denote by λ d the Lebesgue measure on R d . For f : R d → R such that e -f ∈ L 1 (R d ), we dene for t > 0, C t := R d e -f /t -1 and π t the Gibbs measure π t (x)dx := C t e -f (x)/t dx.

For a family of random variables (Y t ) t∈(0,1] and Y a random variable, we write

Y t L -→ t→0 Y meaning that (Y t ) weakly converges to Y .
We give the following denition of a strict polynomial local minimum of f : Denition 1. Let f : R d → R be C 2p for p ∈ N and let x be a local minimum of f . We say that f has a strict polynomial local minimum at x of order 2p if p is the smallest integer such that:

∃r > 0, ∀h ∈ B(x , r) \ {0}, 2p k=2 1 k! ∇ k f (x ) • h ⊗k > 0. (4)
Remarks :

1. A local minimum x of f is not necessarily strictly polynomial, for example, f : x → e -||x|| -2 and x = 0.

If

x is polynomial strict, then the order is necessarily even, because if x is not polynomial strict of order 2l for some l ∈ N, then we have h n → 0 such that the Taylor expansion in h n up to order 2l is zero ; by the minimum condition, the Taylor expansion in h n up to order 2l + 1 must be non-negative, so we also have

∇ 2l+1 f (x ) • h ⊗2l+1 n = 0.
For f : R d → R such that min R d (f ) exists, we denote by argmin(f ) the arguments of the minima of f , i.e.

argmin(f ) = x ∈ R d : f (x) = min R d (f ) .
Without ambiguity, we write "minimum" or "local minimum" to designate f (x ) as well as x . Finally, we dene, for x ∈ R d and p ∈ N: R) : f admits a strict polynomial local minimum at x of order 2p .

A p (x ) := f ∈ C 2p (R d , R) : f admits a local minimum at x . A p (x ) := f ∈ C 2p (R d ,
3 Convergence of Gibbs measures

Properties of Gibbs measures

Let us consider a Borel function f : R d → R with e -f ∈ L 1 (R d ). We study the asymptotic behaviour of the probability measures of density for t ∈ (0, ∞):

π t (x)dx = C t e -f (x)
t dx when t → 0. When t is small, the measure π t tends to the set argmin(f ). The following proposition makes this statement precise.

Proposition 1. Let f : R d → R be a Borel function such that

f := essinf(f ) = inf{y : λ d {f ≤ y} > 0} > -∞,
and

e -f ∈ L 1 (R d ). Then ∀ε > 0, π t ({f ≥ f + ε}) -→ t→0 0.
Proof. As f > -∞, we may assume without loss of generality that f = 0 by replacing f by f -f .

Let ε > 0. It follows from the assumptions that f ≥ 0 λ d -a.e. and λ d {f ≤ ε} > 0 for every ε > 0. As e -f ∈ L 1 (R d ), we have

λ d {f ≤ ε/3} ≤ e ε/3 R d e -f dλ d < +∞.
Moreover by dominated convergence, it is clear that

C -1 t ↓ λ d {f = 0} < +∞.
We have

C t ≤ f ≤ε/3 e -f (x) t dx -1 ≤     e -ε 3t λ d {f ≤ ε 3 } >0     -1
.

Then

π t {f ≥ ε} = C t f ≥ε e -f (x) t dx ≤ e ε/3t f ≥ε e -f (x)/t dx λ d {f ≤ ε 3 } ≤ e -ε/3t C -1 3t λ d {f ≤ ε 3 } -→ t→0 0, because if f (x) ≥ ε, then e -f (x) t ≤ e -2ε 3t e -f (x) 3t
, and where we used that

C -1 3t ≤ C -1 1 if t ≤ 1/3 Now, let us assume that f : R d → R is continuous, e -f ∈ L 1 (R d
) and f admits a unique global minimum at x so that argmin(f ) = {x }. In [START_REF] Athreya | Gibbs measures asymptotics[END_REF] is proved the weak convergence of π t to δ x and a rate of convergence depending on the behaviour of f (x + h) -f (x ) for small enough h. Let us recall this result in detail ; we may assume without loss of generality that x = 0 and f (x ) = 0.

Theorem 3 [START_REF] Athreya | Gibbs measures asymptotics[END_REF]. Let f : R d → [0, ∞) be a Borel function such that :

1. e -f ∈ L 1 (R d ).
2. For all δ > 0, inf{f (x), ||x|| > δ} > 0.

There exist

α 1 , . . . , α d > 0 such that for all (h 1 , . . . , h d ) ∈ R d , 1 t f (t α 1 h 1 , . . . , t α d h d ) -→ t→0 g(h 1 , . . . , h d ) ∈ R. 4. R d sup 0<t<1 e -f (t α 1 h 1 ,...,t α d h d ) t dh 1 . . . dh d < ∞.
For 0 < t < 1, let X t be a random vector with distribution π t . Then e -g ∈ L 1 (R d ) and

(X t ) 1 t α 1 , . . . , (X t ) d t α d L -→ X as t → 0 (5)
where the distribution of X has a density proportional to e -g(x 1 ,...,x d ) .

Remark: Hypothesis 2. is veried as soon as f is continuous, coercive (i.e. f (x) -→ +∞ when ||x|| → +∞) and that argmin(f ) = {0}.

To study the rate of convergence of the measure π t when t → 0 using Theorem 3, we need to identify α 1 , . . . , α d and g such that the condition (5) holds, up to a possible change of basis. Since x is a local minimum, the Hessian

∇ 2 f (x ) is positive semi-denite. Moreover, if ∇ 2 f (x ) is positive denite, then choosing α 1 = • • • = α d = 1
2 , we have:

1 t f (t 1/2 h) -→ t→0 1 2 h T • ∇ 2 f (x ) • h := g(x).
And using an orthogonal change of variable:

R d e -g(x) dx = R d e -1 2 d i=1 β i y 2 i dy 1 . . . dy d < ∞,
where the eigenvalues β i are positive. However, if ∇ 2 f (x ) is not positive denite, then some of the β i are zero and the integral does not converge.

Statement of the problem

We still consider the function f : R d → R and assume that f ∈ A p (x ) for some x ∈ R d and some integer p ≥ 1. Then our objective is to nd

α 1 ≥ • • • ≥ α d ∈ (0, +∞) and an orthogonal transformation B ∈ O d (R) such that: ∀h ∈ R d , 1 t [f (x + B • (t α * h)) -f (x )] -→ t→0 g(h 1 , . . . , h d ), (6)
where t α denotes the vector (t α 1 , . . . , t α d ) and where g : R d → R is a measurable function which is not constant in any h 1 , . . . , h d , i.e. for all i ∈ {1, . . . , d}, there exist h 1 , . . . ,

h i-1 , h i+1 , . . . , h d ∈ R d such that h i → g(h 1 , . . . , h d ) is not constant. ( 7
)
Then we say that α 1 , . . . , α d , B and g are a solution of the problem (6). The hypothesis that g is not constant in any of its variables is important ; otherwise, we could simply take

α 1 = • • • = α d = 1
and obtain, by the rst order condition:

1 t [f (x + t(h 1 , . . . , h d )) -f (x )] -→ t→0 0.

Main results : rate of convergence of Gibbs measures

Theorem 4 (Single well case). Let f : R d → R be C 2p with p ∈ N and such that:

1. f is coercive, i.e. f (x) -→ +∞ when ||x|| → +∞.

argmin(f

) = 0. 3. f ∈ A p (0) and f (0) = 0. 4. e -f ∈ L 1 (R d ).
Let (E k ) k , (α i ) i , B and g to be dened as in Algorithm 1 stated right after, so that for all h ∈ R d ,

1 t [f (x + B • (t α * h)) -f (x )] -→ t→0 g(h),
and where g is not constant in any of its variables. Moreover, assume that g is coercive and the following technical hypothesis if p ≥ 5:

∀h ∈ R d , ∀(i 1 , . . . , i p ) ∈ {0, 2, • • • , 2p} p , (8) i 1 2 + • • • + i p 2p < 1 =⇒ ∇ i 1 +•••+ip f (x ) • p E 1 (h) ⊗i 1 ⊗ • • • ⊗ p Ep (h) ⊗ip = 0.
Then the conclusion of Theorem 3 holds, with:

1 t α 1 , . . . , 1 t α d * (B -1 • X t ) L -→ X as t → 0,
where X has a density proportional to e -g(x) .

Algorithm 1. Let f ∈ A p (x ) for p ∈ N.

1. Dene (F k ) 0≤k≤p-1 recursively as:

F 0 = R d F k = {h ∈ F k-1 : ∀h ∈ F k-1 , ∇ 2k f (x ) • h ⊗ h ⊗2k-1 = 0}. 2. For 1 ≤ k ≤ p -1, dene the subspace E k as the orthogonal complement of F k in F k-1 . By abuse of notation, dene E p := F p-1 . 3. Dene B ∈ O d (R)
as an orthogonal transformation adapted to the decomposition

R d = E 1 ⊕ • • • ⊕ E p .

Dene for

1 ≤ i ≤ d, α i := 1 2j for i ∈ {dim(E 1 ) + • • • + dim(E j-1 ) + 1, . . . , dim(E 1 ) + • • • + dim(E j )}. (9)
5. Dene g : R d → R as

g(h) = 2p k=2 1 k! i 1 ,...,ip∈{0,...,k} i 1 +•••+ip=k i 1 2 +•••+ ip 2p =1 k i 1 , . . . , i p ∇ k f (x ) • p E 1 (B • h) ⊗i 1 ⊗ • • • ⊗ p Ep (B • h) ⊗ip . (10)
Remarks :

1. The function g is not unique, as we can choose any base B adapted to the decomposition

R d = E 1 ⊕ • • • ⊕ E p .
2. The case p ≥ 5 is fundamentally dierent from the case p ≤ 4, since Algorithm 1 may fail to provide such (E k ) k , (α i ) i , B and g if the technical hypothesis (8) is not fullled, as explained in Section 4.8. This yields fewer results for the case p ≥ 5.

3. For p ∈ {1, 2, 3, 4}, the detail the expression of g in ( 17), ( 18), ( 20) and ( 23) respectively.

4. The function g has the following general properties : g is a non-negative polynomial of order 2p; g(0) = 0 and ∇g(0) = 0.

5. The condition on g to be coercive may seem not natural. We give more details about the case where g is not coercive in Section 4.10 and give a way to deal with the simple generic case of non-coercivity. However dealing with the general case where g is not coercive goes beyond the scope of our work.

6. The hypothesis that g is coercive is a necessary condition for e -g ∈ L 1 (R d ). We actually prove in Proposition 3 that it is a sucient condition.

Still following [START_REF] Athreya | Gibbs measures asymptotics[END_REF], we study the multiple well case, i.e. the global minimum is attained in a nite number of points in R d , say {x 1 , . . . , x m } for some m ∈ N. In this case, the limiting measure of π t will have its support in {x 1 , . . . , x m }, with dierent weights.

Theorem 5 [START_REF] Athreya | Gibbs measures asymptotics[END_REF]. Let f : R d → [0, ∞) measurable such that:

1.

e -f ∈ L 1 (R d ).
2. For all δ > 0, inf{f (x), ||x -

x i || > δ, 1 ≤ i ≤ m} > 0.
3. There exist

(α ij ) 1≤i≤m 1≤j≤d
such that for all i, j, α ij ≥ 0 and for all i:

1 t f (x i + (t α i1 h 1 , . . . , t α id h d )) -→ t→0 g i (h 1 , . . . , h d ) ∈ [0, ∞).
4. For all i ∈ {1, . . . , m},

R d sup 0<t<1 e -f (x i +(t α i1 h 1 ,...,t α id h d )) t dh 1 . . . dh d < ∞.
Then, let α := min 1≤i≤m d j=1 α ij and let J := i ∈ {1, . . . , m} : d j=1 α ij = α . For 0 < t < 1, let X t be a random vector with distribution π t . Then:

X t L -→ t→0 1 j∈J R d e -g j (x) dx i∈J R d e -g i (x) dx • δ x i .
Theorem 6 (Multiple well case). Let f : R d → R be C 2p for p ∈ N and such that:

1. f is coercive i.e. f (x) -→ +∞ when ||x|| → +∞.
2. argmin(f ) = {x 1 , . . . , x m } and for all i, f (x i ) = 0.

For all

i ∈ {1, . . . , m}, f ∈ A p i (x i ) for some p i ≤ p. 4. e -f ∈ L 1 (R d ).
Then, for every i ∈ {1, . . . , m}, we consider (E ik ) k , (α ij ) j , B i and g i as dened in Algorithm 1, where we consider f to be in A p i (x i ), so that for every h ∈ R d :

1 t f (x i + B i • (t α i * h)) -→ t→0 g i (h 1 , . . . , h d ) ∈ [0, ∞),
where t α i is the vector (t α i1 , . . . , t α id ) and where g i is not constant in any of its variables. Furthermore, we assume that for all i, g i is coercive and the following technical hypothesis for every i such that

p i ≥ 5: ∀h ∈ R d , ∀(i 1 , . . . , i p i ) ∈ {0, 2, . . . , 2p i } p i , i 1 2 + • • • + i p i 2p < 1 =⇒ ∇ i 1 +•••+ip i f (x i ) • p E i1 (h) ⊗i 1 ⊗ • • • ⊗ p E ip i (h) ⊗ip i = 0.
Let α := min 1≤i≤m d j=1 α ij and let J := i ∈ {1, . . . , m} : d j=1 α ij = α . Then:

X t -→ t→0 1 j∈J R d e -g j (x) dx i∈J R d e -g i (x) dx • δ x i .
Moreover, let δ > 0 be small enough so that the balls B(x i , δ) are disjoint, and dene the random vector X it to have the law of X t conditionally to the event ||X t -x i || < δ. Then:

1 t α i1 , . . . , 1 t α id * (B -1 i • X it ) L -→ X i as t → 0,
where X i has a density proportional to e -g i (x) .

4 Expansion of f at a local minimum with degenerate derivatives

In this section, we aim at answering to the problem stated in (6) in order to devise conditions to apply Theorem 3. This problem can also be considered in a more general setting, independently of the study of the convergence of Gibbs measures. It provides a non degenerate higher order nested expansion of f at a local minimum when some of the derivatives of f are degenerate. Note here that we only need x to be a local minimum instead of a global minimum, since we only give local properties. For k ≤ p, we dene the tensor of order k, T k := ∇ k f (x ).

Expansion of f for any order p

In this section, we state our result in a synthetic form. The proofs of the cases p = 1, 2, 3, 4 are individually detailled in Sections 4.3, 4.4, 4.6 and 4.7 respectively.

Theorem 7. Let f : R d → R be C 2p for some p ∈ N and assume that f ∈ A p (x ) for some x ∈ R d .

1. If p ∈ {1, 2, 3, 4}, then there exists orthogonal subspaces of R d , E 1 , . . . , E p such that

R d = E 1 ⊕ • • • ⊕ E p ,
and satisfying for every h ∈ R d :

1 t f x + t 1/2 p E 1 (h) + • • • + t 1/(2p) p Ep (h) -f (x ) (11) -→ t→0 2p k=2 1 k! i 1 ,...,ip∈{0,••• ,k} i 1 +•••+ip=k i 1 2 +•••+ ip 2p =1 k i 1 , . . . , i p T k • p E 1 (h) ⊗i 1 ⊗ • • • ⊗ p Ep (h) ⊗ip . ( 12
)
The convergence is uniform with respect to h on every compact set. Moreover, let B ∈ O d (R) be an orthogonal transformation adapted to the decomposition

E 1 ⊕ • • • ⊕ E p , then 1 t [f (x + B • (t α * h)) -f (x )] -→ t→0 g(h), (13)
where

g(h) = 2p k=2 1 k! i 1 ,...,ip∈{0,...,k} i 1 +•••+ip=k i 1 2 +•••+ ip 2p =1 k i 1 , . . . , i p T k • p E 1 (B • h) ⊗i 1 ⊗ • • • ⊗ p Ep (B • h) ⊗ip (14)
is not constant in any of its variables h 1 , . . . , h d and

α i := 1 2j for i ∈ {dim(E 1 ) + • • • + dim(E j-1 ) + 1, . . . , dim(E 1 ) + • • • + dim(E j )}. (15)
2. If p ≥ 5 and if there exist orthogonal subspaces of R d , E 1 , . . . , E p such that

R d = E 1 ⊕ • • • ⊕ E p
and satisfying the following additional assumption

∀h ∈ R d , ∀(i 1 , . . . , i p ) ∈ {0, 2, . . . , 2p} p , (16) i 1 2 + • • • + i p 2p < 1 =⇒ T i 1 +•••+ip • p E 1 (h) ⊗i 1 ⊗ • • • ⊗ p Ep (h) ⊗ip = 0,
then (12) stills holds true, as well as the uniform convergence on every compact set. Moreover, if B ∈ O d (R) is an orthogonal transformation adapted to the previous decomposition, then (13) still hold true. However, depending on the function f , such subspaces do not necessarily exist.

Remarks:

1. The limit (12) can be rewritten as:

2p k=2 i 1 ,...,ip∈{0,••• ,k} i 1 +•••+ip=k i 1 2 +•••+ ip 2p =1 T k • p E 1 (h) ⊗i 1 i 1 ! ⊗ • • • ⊗ p Ep (h) ⊗ip i p ! .
2. For p ∈ {1, 2, 3, 4}, we explicitly give the expression of the sum (12) and the p-tuples (i 1 , . . . , i p ) 18), ( 20) and ( 23) respectively. 3. For p ∈ {1, 2, 3, 4}, we give in Algorithm 1 an explicit construction of the orthogonal subspaces E 1 , . . . , E p as complementaries of annulation sets of some derivatives of f . 4. The case p ≥ 5 is fundamentally dierent from the case p ∈ {1, 2, 3, 4}. The strategy of proof developed for p ∈ {1, 2, 3, 4} fails if the assumption (16) is not satised. In 4.8 a counter-example is detailed. The case p ≥ 5 yields fewer results than for p ≤ 4, as the assumption (16) is strong.

such that i 1 2 + • • • + ip 2p = 1, in (17), (
5. For p ≥ 5, such subspaces E 1 , . . ., E p may also be obtained from Algorithm 1, however (16) is not necessarily true in this case.

The proof of Theorem 7 is given rst individually for each p ∈ {1, 2, 3, 4}, in Sections 4.3, 4.4, 4.6, 4.7 respectively. The proof for p ≥ 5 is given in Section 4.8. The proof of the uniform convergence and of the fact that g is not constant is given in Section 4.9.

Review of the one dimensional case

We review the case d = 1, as it guides us for the proof in the case d ≥ 2. The strategy is to nd the rst derivative f (m) (x ) which is non zero and then to choose α 1 = 1/m. Proposition 2. Let f : R → R be C p for some p ∈ N and let x be a strict polynomial local minimum of f . Then :

1. The order of the local minimum m is an even number and f (m) (x ) > 0.

f (

x + h) = h→0 f (x ) + f (m) (x ) m! h p + o(h m )
Then α 1 := 1/m is the solution of (6) and

1 t (f (x + t 1/m h) -f (x )) -→ t→0 f (m) (x ) m! h m
which is a non-constant function of h, since f (m) (x ) = 0. The direct proof using the Taylor formula is left to the reader.

Proof of Theorem 7 for

p = 1 Let f ∈ A 1 (x ).
The assumption that x is a strict polynomial local minimum at order 2 implies that ∇ 2 f (x ) is positive denite. Let us denote (β i ) 1≤i≤d its positive eigenvalues. By the spectral theorem, let us write ∇ 2 f (x ) = BDiag(β 1:d )B T for some B ∈ O d (R). Then:

1 t (f (x + t 1/2 B • h) -f (x )) -→ t→0 1 2 d i=1 β i h 2 i . (17)
Thus, a solution of ( 6) is

α 1 = • • • = α d = 1 2 , B, and
g(h 1 , . . . , h d ) = 1 2 d i=1 β i h 2 i
, which is a nonconstant function of every h 1 , . . . , h d , since for all i, β i is positive.

In the following, our objective is to establish a similar result when ∇ 2 f (x ) is not necessarily positive denite.

Proof of Theorem 7 for p = 2

Theorem 8. Let f ∈ A 2 (x ). Then there exist orthogonal subspaces E and F such that R d = E ⊕ F , and that for all h ∈ R d :

1 t f (x + t 1/2 p E (h) + t 1/4 p F (h)) -f (x ) -→ t→0 1 2 ∇ 2 f (x ) • p E (h) ⊗2 + 1 2 ∇ 3 f (x ) • p E (h) ⊗ p F (h) ⊗2 + 1 4! ∇ 4 f (x ) • p F (h) ⊗4 . (18)
Moreover, if f ∈ A 2 (x ), then this is a solution to the problem (6), with E 1 = E, E 2 = F , α dened in (15), B adapted to the previous decomposition and g dened in (14).

Remark: The set of 2-tuples (i 1 , i 2 ) such that i 1 2 + i 2 4 = 1, are (2, 0), (1, 2) and (0, 4), which gives the terms appearing in the sum in (12).

Proof. Let

F := {h ∈ R d : ∇ 2 f (x ) • h ⊗2 = 0}. By the spectral theorem and since ∇ 2 f (x ) is positive semi-denite, F = {h ∈ R d : ∇ 2 f (x ) • h = 0 ⊗1 } is a vector subspace of R d . Let E be the orthogonal complement of F in R d .
For h ∈ R d we expand the left term of (18) using the Taylor formula up to order 4 and the multinomial formula (2), giving

4 k=2 1 k! i 1 ,i 2 ∈{0,...,k} i 1 +i 2 =k k i 1 , i 2 t i 1 2 + i 2 4 -1 T k • p E (h) ⊗i 1 ⊗ p F (h) ⊗i 2 + o(1).
The terms with coecient t a , a > 0, are o(1) as t → 0. By denition of F we have ∇ 2 f (x )•p F (h) = 0 ⊗1 , so we also have

∇ 3 f (x ) • p F (h) ⊗3 = 0
by the local minimum condition. This yields the convergence stated in (18). Moreover, if x is a local minimum of polynomial order 4, then by the local minimum condition, ∇ 4 f (x ) > 0 on F in the sense of (3). Moreover, since ∇ 2 f (x ) > 0 on E, then the limit is not constant in any h 1 , . . . , h d .

Remark:

The cross odd term is not necessarily null. For example, consider

f : R 2 -→ R (x, y) -→ x 2 + y 4 + xy 2 .
Then f admits a global minimum at x = 0 since |xy 2 | ≤ 1 2 (x 2 + y 4 ). We have E 1 = R(1, 0), E 2 = R(0, 1) and for all (x, y) ∈ R 2 , T 3 • (xe 1 ) ⊗ (ye 2 ) ⊗2 = 2xy 2 is not identically null.

4.5 Diculties beyond the 4th order and Hilbert's 17 th problem If we do not assume as in the previous section that ∇ 4 f (x ) is not positive on F , then we carry on the development of f (x +h) up to higher orders. A rst idea is to consider F 2 := {h ∈ F : ∇ 4 f (x )•h ⊗4 = 0} ⊆ F and E 2 a complement subspace of F 2 in F , and to continue this process by induction as in Section 4.4. However, F 2 is not necessarily a subspace of F .

Indeed, let T be a symmetric tensor dened on R d of order 2k with k ∈ N. As T is symmetric, there exist vectors v 1 , . . . , v q ∈ R d , and scalars λ 1 , . . . , λ q ∈ R such that T = i λ i (v i ) ⊗2k (see [START_REF] Comon | Symmetric tensor and symmetric tensor rank[END_REF], Lemma 4.2.), so

∀h ∈ R d , T • h ⊗2k = q i=1 λ i (v i ) ⊗2k • h ⊗2k = q i=1 λ i v i , h 2k . For k = 2 and T = ∇ 2k f (x ) | F , since x is a local minimum, we have, identifying F and R d , ∀h ∈ R d , T • h ⊗2k ≥ 0
Then, we could think it implies that for all i, λ i ≥ 0, and then

T • h ⊗2k = 0 =⇒ ∀i, v i , h = 0
which would give a linear caracterization of {h ∈ R d : T • h ⊗2k = 0} and in this case, F 2 would be a subspace of F . However this reasoning is not correct in general as we do not have necessarily that for all i, λ i ≥ 0.

We can build counter-examples as follows. Since T is a non-negative symmetric tensor, T can be seen as a non-negative homogeneous polynomial of degree 2k with d variables. A counter-example at order 2k = 4 is T (X, Y, Z) = ((X -Y )(X -Z)) 2 , which is a non-negative polynomial of order 4, but {T = 0} = {X = Y or X = Z}, which is not a vector space.

Another counterexample given in [START_REF] Motzkin | The arithmetic-geometric inequality[END_REF] at order 2k = 6 is the following. We dene

T (X, Y, Z) = Z 6 + X 4 Y 2 + X 2 Y 4 -3X 2 Y 2 Z 2
By the arithmetic-geometric mean inequality and its equality case, T is non-negative and T (x, y, z) = 0 if and only if z 6 = x 4 y 2 = x 2 y 4 , so that

{T = 0} = R   1 1 1   ∪ R   -1 1 1   ∪ R   1 -1 1   ∪ R   1 1 -1   .
Hence, {T = 0} is not a subspace of R 3 . In particular T cannot be written as i λ i (v i ) ⊗2k with λ i ≥ 0.

In fact, this problem is linked with the Hilbert's seventeenth problem that we recall below.

Problem 1 (Hilbert's seventeeth problem). Let P be a non-negative polynomial with d variables, homogeneous of even degree 2k. Find polynomials P 1 , . . . , P r with d variables, homogeneous of degree k, such that P = r i=1 P 2 i Hilbert proved in 1888 [START_REF] Hilbert | Ueber die Darstellung deniter Formen als Summe von Formenquadraten[END_REF] that there does not always exist a solution. In general {T = 0} is not even a submanifold of R d . Indeed, taking T :

h → ∇ 2k f (x ) • h ⊗2k , we have ∂ h T • h = 2k∇ 2k f (x ) • h ⊗2k-1
is not surjective in h = 0, so the surjectivity condition for {T = 0} to be a submanifold is not fullled.

Proof of Theorem 7 for p = 3

We slightly change our strategy of proof developed in Section 4.4. For k ≥ 2, we dene F k recursively as

F k := {h ∈ F k-1 : ∀h ∈ F k-1 , ∇ 2k f (x ) • h ⊗ h ⊗2k-1 = 0}, (19)
instead of {h ∈ F k-1 : ∇ 2k f (x ) • h ⊗2k = 0}.
Then, by construction, F k is a vector subspace of R d .

Theorem 9. Let f ∈ A 3 (x ). Then there exist orthogonal subspaces of R d , E 1 , E 2 and F 2 , such that

R d = E 1 ⊕ E 2 ⊕ F 2 ,
and such that for all h ∈ R d ,

1 t f (x + t 1/2 p E 1 (h) + t 1/4 p E 2 (h) + t 1/6 p F 2 (h)) -f (x ) (20) -→ t→0 1 2 ∇ 2 f (x ) • p E 1 (h) ⊗2 + 1 2 ∇ 3 f (x ) • p E 1 (h) ⊗ p E 2 (h) ⊗2 + 1 4! ∇ 4 f (x ) • p E 2 (h) ⊗4 + 4 4! ∇ 4 f (x ) • p E 1 (h) ⊗ p F 2 (h) ⊗3 + 10 5! ∇ 5 f (x ) • p E 2 (h) ⊗2 ⊗ p F 2 (h) ⊗3 + 1 6! ∇ 6 f (x ) • p F 2 (h) ⊗6 .
Moreover, if f ∈ A 3 (x ), then this is a solution to the problem (6), with E 3 = F 2 , α dened in (15), B adapted to the previous decomposition and g dened in (14).

Remark: The set of 3-tuples (i 1 , i 2 , i 3 ) such that i 1 2 + i 2 4 + i 3 6 = 1, are (2, 0, 0), (1, 2, 0), (0, 4, 0), (1, 0, 3), (0, 2, 3), (0, 0, 6), which gives the terms appearing in (12).

Proof. We consider the subspace

F 1 := {h ∈ R d : T 2 • h ⊗2 = 0} = {h ∈ R d : T 2 • h = 0 ⊗1 },
since T 2 ≥ 0. Then, let E 1 be the orthogonal complement of F 1 in R d and consider the vector subspace of F 1 dened by

F 2 = {h ∈ F 1 : ∀h ∈ F 1 , T 4 • h ⊗ h ⊗3 = 0}.
Let E 2 be the orthogonal complement of F 2 in F 1 . Then we have

R d = E 1 ⊕ F 1 = E 1 ⊕ E 2 ⊕ F 2 .
For h ∈ R d we expand the left term of (20) using the Taylor formula up to order 6 and the multinomial formula (2), giving

6 k=2 1 k! i 1 ,i 2 ,i 3 ∈{0,...,k} i 1 +i 2 +i 3 =k k i 1 , i 2 , i 3 t i 1 2 + i 2 4 + i 3 6 -1 T k • p E 1 (h) ⊗i 1 ⊗ p E 2 (h) ⊗i 2 ⊗ p F 2 (h) ⊗i 3 + o(1),
and we prove the convergence stated in (20).

All the terms with coecient t a where a > 0 are o(1) as t → 0.

Order 2: we have

T 2 •p E 2 (h) = 0 ⊗1 and T 2 •p F 2 (h) = 0 ⊗1 so the only term for k = 2 is 1 2 T 2 •p E 1 (h) ⊗2 .
Order 3: Since x is a local minimum and T 2 • p F 1 (h) ⊗2 = 0, we have T 3 • p F 1 (h) ⊗3 = 0. Then, using property Proposition 6, if the factor p E 1 (h) does not appear as an argument in T 3 , then the corresponding term is zero.

Let us prove that

T 3 • p E 1 (h) ⊗ p F 2 (h) ⊗2 = 0.
Using Theorem 8 with

E = E 1 , F = E 2 ⊕ F 2 , we have in particular that for all h ∈ R d , 1 2 T 2 • p E (h) ⊗2 + 1 2 T 3 • p E (h) ⊗ p F (h) ⊗2 + 1 4! T 4 • p F (h) ⊗4 ≥ 0. ( 21
)
Then taking h ∈ E 1 ⊕ F 2 so that h = p E 1 (h) + p F 2 (h) and with

T 4 • p F 2 (h) |F 1 ≡ 0 ⊗3 , (22)
we may rewrite (21) as

1 2 T 2 • p E 1 (h) ⊗2 + 1 2 T 3 • p E 1 (h) ⊗ p F 2 (h) ⊗2 ≥ 0.
Now, considering h = λh, we have that for all λ ∈ R,

λ 2 1 2 T 2 • p E 1 (h) ⊗2 + λ 2 T 3 • p E 1 (h) ⊗ p F 2 (h) ⊗2 ≥ 0, so that necessarily T 3 • p E 1 (h) ⊗ p F 2 (h) ⊗2 = 0.
Let us prove that

T 3 • p E 1 (h) ⊗ p E 2 (h) ⊗ p F 2 (h) = 0.
We use again (21), with p

F (h) = p E 2 (h) + p F 2 (h), so that 1 2 T 2 • p E 1 (h) ⊗2 + 1 2 T 3 • p E 1 (h) ⊗ p E 2 (h) + p F 2 (h) ⊗2 + 1 4! T 4 • p E 2 (h) + p F 2 (h) ⊗4 ≥ 0.
But using ( 22) and that

T 3 • p E 1 (h) ⊗ p F 2 (h) ⊗2 = 0, we obtain 1 2 T 2 • p E 1 (h) ⊗2 + 1 2 T 3 • p E 1 (h) ⊗ p E 2 (h) ⊗2 + T 3 • p E 1 (h) ⊗ p E 2 (h) ⊗ p F 2 (h) + 1 4! T 4 • p E 2 (h) ⊗4 ≥ 0. Now, considering h = p E 1 (h) + p E 2 (h) + λp F 2 (h), we have that for all λ ∈ R, 1 2 T 2 • p E 1 (h) ⊗2 + 1 2 T 3 • p E 1 (h) ⊗ p E 2 (h) ⊗2 + λT 3 • p E 1 (h) ⊗ p E 2 (h) ⊗ p F 2 (h) + 1 4! T 4 • p E 2 (h) ⊗4 ≥ 0, so necessarily T 3 • p E 1 (h) ⊗ p E 2 (h) ⊗ p F 2 (h) = 0. The last remaining term for k = 3 is 1 2 T 3 • p E 1 (h) ⊗ p E 2 (h) ⊗2 .
Order 4: If the factor p E 1 (h) does not appear and if the factor p F 2 (h) appears at least once, then using (22) the corresponding term is zero. If p E 1 (h) appears, the only term with a non-positive

exponent of t is 4 4! T 4 • p E 1 (h) ⊗ p F 2 (h) ⊗3 . So the only terms for k = 4 are 1 4! T 4 • p E 2 (h) ⊗4 and 4 4! T 4 • p E 1 (h) ⊗ p F 2 (h) ⊗3 .
Order 5: The terms where p E 1 (h) appears at least once have a coecient t a with a > 0 so are o(1) when t → 0. We have

T 2 • p F 2 (h) ⊗2 = 0, T 3 • p F 2 (h) ⊗3 = 0, T 4 • p F 2 (h) ⊗4 = 0 and since x is a local minimum, we have T 5 • p F 2 (h) ⊗5 = 0.
Let us prove that

T 5 • p E 2 (h) ⊗ p F 2 (h) ⊗4 = 0. Let h ∈ R d . We have 1 t 11/12 f (x + t 1/4 p E 2 (h) + t 1/6 p F 2 (h)) -f (x ) -→ t→0 1 4! T 5 • p E 2 (h) ⊗ p F 2 (h) ⊗4 ≥ 0.
Hence, considering h = λh, we have for every λ ∈ R,

λ 5 T 5 • p E 2 (h) ⊗ p F 2 (h) ⊗4 ≥ 0,
which yields the desired result.

Order 2

(2, 0, 0, 0) Order 3

(2, 1, 0, 0) Order 4 (0, 4, 0, 0), (1, 1, 0, 2), (1, 0, 3, 0) Order 5 (1, 0, 0, 4), (0, 2, 3, 0), (0, 3, 0, 2) Order 6 (0, 1, 3, 2), (0, 2, 0, 4), (0, 0, 6, 0) Order 7 (0, 1, 0, 6), (0, 0, 3, 4) Order 8 (0, 0, 0, 8)

Table 1: Terms expressed as 4-tuples in the development (23)

The only remaining term for p = 5 is

10 5! T 5 • p E 2 (h) ⊗2 ⊗ p F 2 (h) ⊗3 .
Order 6: The only term for k = 6 is 1 6! T 6 • p F 2 (h) ⊗6 ; the other terms have a coecient t a with a > 0, so are o(1) when t → 0.

Remark : As in Theorem 8 and the remark that follows, the remaining odd cross-terms cannot be proved to be zero using the same method of proof, and may be actually not zero. For example, consider:

f : R 2 -→ R (x, y) -→ x 4 + y 6 + x 2 y 3 , which satises h → ∇ 5 f (x ) • p E 2 (h) ⊗2 ⊗ p F 2 (h) ⊗3 ≡ 0.

Proof of Theorem 7 for p = 4

Theorem 10. Let f ∈ A 4 (x ). Then there exist orthogonal subspaces of R d , E 1 , E 2 , E 3 and F 3 such that

R d = E 1 ⊕ E 2 ⊕ E 3 ⊕ F 3 ,
and for all h ∈ R d ,

1 t f (x + t 1/2 p E 1 (h) + t 1/4 p E 2 (h) + t 1/6 p E 3 (h) + t 1/8 p F 3 (h)) -f (x ) -→ t→0 8 k=2 1 k! i 1 ,...,i 4 ∈{0,...,k} i 1 +•••+i 4 =k k i 1 , . . . , i 4 T k • p E 1 (h) ⊗i 1 ⊗ p E 2 (h) ⊗i 2 ⊗ p E 3 (h) ⊗i 3 ⊗ p F 3 (h) ⊗i 4 . ( 23
)
These terms are summarized as tuples (i 1 , . . . , i 4 ) in Table 1. Moreover, if f ∈ A 4 (x ), then this is a solution to (6), with E 4 = F 3 , α dened in (15), B adapted to the previous decomposition and g dened in (14).

Proof. As before, we dene the subspaces F 0 := R d and by induction:

F k = h ∈ F k-1 : ∀h ∈ F k-1 , T 2k • h ⊗ h ⊗3 = 0 for k = 1, 2, 3. We dene E k as the orthogonal complement of F k in F k-1 for k = 1, 2, 3, so that R d = E 1 ⊕ E 2 ⊕ E 3 ⊕ F 3 . E 1 F 1 T 2 ≥ 0 T 2 = 0 E 2 F 2 T 4 ≥ 0 T 4 = 0 E 3 F 3 T 6 ≥ 0 T 6 = 0
Table 2: Illustration of the subspaces Then we apply a Taylor expansion up to order 8 to the left side of (23) and the multinomial formula (2), which reads

8 k=2 1 k! i 1 ,...,i 4 ∈{0,...,k} i 1 +•••+i 4 =k k i 1 , . . . , i 4 t i 1 2 +•••+ i 4 8 -1 T k • p E 1 (h) ⊗i 1 ⊗ p E 2 (h) ⊗i 2 ⊗ p E 3 (h) ⊗i 3 ⊗ p F 3 (h) ⊗i 4 + o(1)
.

If i 1 2 + • • • + i 4 8 > 1 then the corresponding term is in o(1) when t → 0. If i 1 2 + • • • + i 4
8 < 1 then the corresponding term diverges when t → 0, so we need to prove that actually

T k • p E 1 (h) ⊗i 1 ⊗ p E 2 (h) ⊗i 2 ⊗ p E 3 (h) ⊗i 3 ⊗ p F 3 (h) ⊗i 4 = 0. ( 24
) If i 1 2 + i 2 4 + i 3 6 + i 4 8 < 1 but if we also have i 1 2 + i 2 4 + i 3 6 + i 4
6 < 1, then by applying the property at the order 6 (Theorem 9) with the 3-tuple (i 1 , i 2 , i 3 + i 4 ), we get (24).

So we only need to consider 4-tuples

such that i 1 2 + i 2 4 + i 3 6 + i 4 8 < 1 and i 1 2 + i 2 4 + i 3 6 + i 4 6 ≥ 1.
We can remove all the terms which are null by the denitions of the subspaces E 1 , E 2 , E 3 , F 3 . The remaining terms are:

For k = 4:

t 21/24 6 T 4 • p E 1 (h) ⊗ p F 3 (h) ⊗3 , t 11/12 2 T 4 • p E 1 (h) ⊗ p E 3 (h) ⊗ p F 3 (h) ⊗2 , t 23/24 2 T 4 • p E 1 (h) ⊗ p E 3 (h) ⊗2 ⊗ p F 3 (h). For k = 5 : t 21/24 12 T 5 •p E 2 (h) ⊗2 ⊗p F 3 (h) ⊗3 , t 11/12 4 T 5 •p E 2 (h) ⊗2 ⊗p E 3 (h)⊗p F 3 (h) ⊗2 , t 23/24 4 T 5 •p E 2 (h) ⊗2 ⊗ p E 3 (h) ⊗2 ⊗ p F 3 (h). For k = 6 : t 21/24 5! T 6 • p E 2 (h) ⊗ p F 3 (h) ⊗5 , t 11/12 4! T 6 • p E 2 (h) ⊗ p E 3 (h) ⊗ p F 3 (h) ⊗4 , t 23/24 12 T 6 • p E 2 (h) ⊗ p E 3 (h) ⊗2 ⊗ p F 3 (h) ⊗3 .
First, we note that

1 t 21/24 f (x + t 1/2 p E 1 (h) + t 1/4 p E 2 (h) + t 1/6 p E 3 (h) + t 1/8 p F 3 (h)) -f (x ) -→ t→0 1 6 T 4 • p E 1 (h) ⊗ p F 3 (h) ⊗3 + 1 12 T 5 • p E 2 (h) ⊗2 ⊗ p F 3 (h) ⊗3 + 1 5! T 6 • p E 2 (h) ⊗ p F 3 (h) ⊗5 ≥ 0. Then, considering h = λp E 1 (h) + p E 2 (h) + p E 3 (h) + p F 3 (h), we have that for all λ ∈ R, λ 6 T 4 • p E 1 (h) ⊗ p F 3 (h) ⊗3 + 1 12 T 5 • p E 2 (h) ⊗2 ⊗ p F 3 (h) ⊗3 + 1 5! T 6 • p E 2 (h) ⊗ p F 3 (h) ⊗5 ≥ 0, so necessarily T 4 • p E 1 (h) ⊗ p F 3 (h) ⊗3 = 0.
Then, considering h = p E 2 (h) + λp F 3 (h) for λ ∈ R, we get successively that the two other terms are null.

Likewise, we prove successively that the terms in t 11/12 are null, and then that the terms in t 23/24 are null. This yields the convergence stated in (23).

4.8 Counter-example and proof of Theorem 7 with p ≥ 5 under the hypothesis (16) Algorithm 1 may fail to yield such expansion of f for orders no lower than 10 if the hypothesis (16) is not fullled. Indeed for p ≥ 5, there exist p-tuples (i 1 , . . . , i p ) such that i 1 2 + • • • + ip 2p < 1 and i 1 , . . ., i p are all even. Such tuples do not appear at orders 8 and lower, but they do appear at orders 10 and higher, for example (0, 2, 0, 0, 4) for k = 6. In such a case, we cannot use the positiveness argument to prove that the corresponding term

T k • p E 1 (h) ⊗i 1 ⊗ • • • ⊗ p Ep (h) ⊗ip
is zero, and in fact, it may be not zero.

Let us give a counter example. Consider

f : R 2 -→ R (x, y) -→ x 4 + y 10 + x 2 y 4 . Then f ∈ A 5 (0) and we have E 1 = {0}, E 2 = R • (1, 0), E 3 = {0}, E 4 = {0}, F 4 = R • (0, 1). But 1 t f (t 1/4 , t 1/10 ) = 1 t t + t + t 9/10
goes to +∞ when t → 0.

Now, let us give the proof of Theorem 7 for p ≥ 5. In this proof, we assume that the subspaces E 1 , . . . , E p given in Algorithm 1 satisfy the hypothesis (16).

Proof. We develop (11), which reads:

2p k=2 1 k! i 1 ,...,ip∈{0,...,k} i 1 +•••+ip=k k i 1 , . . . , i p t i 1 2 +•••+ ip 2p -1 T k • p E 1 (h) ⊗i 1 ⊗ • • • ⊗ p Ep (h) ⊗ip + o(1) =: S.
The terms such that i 1 2 + • • • + ip 2p < 1 may diverge when t → 0, so let us prove that they are in fact null. Let

α := inf                  i 1 2 + • • • + i p 2p : h -→ 2p k=2 1 k! i 1 ,...,ip∈{0,...,k} i 1 +•••+ip=k i 1 2 +•••+ ip 2p =α k i 1 , . . . , i p T k • p E 1 (h) ⊗i 1 ⊗ • • • ⊗ p Ep (h) ⊗ip ≡ 0                 
, and assume by contradiction that α < 1. Then we have for all h ∈ R d :

t 1-α S -→ t→0          2p k=2 1 k! i 1 ,...,ip∈{0,...,k} i 1 +•••+ip=k i 1 2 +•••+ ip 2p =α k i 1 , . . . , i p T k • p E 1 (h) ⊗i 1 ⊗ • • • ⊗ p Ep (h) ⊗ip          ≥ 0,
+•••+ip=k i 1 2 +•••+ ip 2p =α λ i 1 1 . . . λ ip p k i 1 , . . . , i p T k • p E 1 (h) ⊗i 1 ⊗ • • • ⊗ p Ep (h) ⊗ip ≥ 0. (25)
Now, we x h ∈ R d such that the polynomial in (25) in the variables λ 1 , . . . , λ p is not identically zero, and we consider k max its highest homogeneous degree, so that we have i 1 ,...,ip∈{0,...,kmax}

i 1 +•••+ip=kmax i 1 2 +•••+ ip 2p =α λ i 1 1 . . . λ ip p k max i 1 , . . . , i p T kmax • p E 1 (h) ⊗i 1 ⊗ • • • ⊗ p Ep (h) ⊗ip ≥ 0.
If k max is odd, this yields a contradiction, taking λ 1 = • • • = λ p =: λ → ±∞. If k max is even, we consider the index l 1 such that i l 1 =: a 1 is maximal and the coecients in the above sum with i l 1 = a 1 are not all zero. Then xing all the λ l for l = l 1 and taking λ l 1 → ∞, we have

i 1 ,...,ip∈{0,...,kmax} i 1 +•••+ip=kmax i 1 2 +•••+ ip 2p =α i l 1 =a 1 λ i 1 1 . . . λ ip p k max i 1 , • • • , i p T kmax • p E 1 (h) ⊗i 1 ⊗ • • • ⊗ p Ep (h) ⊗ip ≥ 0.
Thus, if a 1 is odd, this yields a contradiction. If a 1 is even, we carry on this process by induction : knowing l 1 , . . . , l r , we choose the index l r+1 such that l r+1 / ∈ {l 1 , . . . , l r }, the corresponding term

i 1 ,...,ip∈{0,...,kmax} i 1 +•••+ip=kmax i 1 2 +•••+ ip 2p =α i l 1 =a 1 ,...,i l r+1 =a r+1 λ i 1 1 . . . λ ip p k max i 1 , . . . , i p T kmax • p E 1 (h) ⊗i 1 ⊗ • • • ⊗ p Ep (h) ⊗ip
is not identically null and such that i l r+1 =: a r+1 is maximal. Necessarily, a r+1 is even. In the end we will nd a non-zero term whose exponents i are all even which contradicts assumption (16).

4.9 Proofs of the uniform convergence and of the non-constant property

In this section we prove the additional properties claimed in Theorem 7 : the uniform convergence with respect to h on every compact set and the fact that the function g is not constant in any of its variables h 1 , . . . , h d .

Proof. First, let us prove that the convergence is uniform with respect to h on every compact set. Let ε > 0 and let R > 0. By the Taylor formula at order 2p, there exists δ > 0 such that for ||h|| < δ,

f (x + h) -f (x ) - 2p k=2 1 k! i 1 +•••+ip=k k i 1 , . . . , i p T k • p E 1 (h) ⊗i 1 ⊗ • • • ⊗ p Ep (h) ⊗ip ≤ ε||h|| 2p .
Now, let us consider t → 0 and h ∈ R d with ||h|| ≤ R. Then we have:

∀t ≤ max 1, δ R 1/(2p) , ||t 1/2 p E 1 (h) + • • • + t 1/(2p) p Ep (h)|| ≤ δ, so that 1 t [ f (x + t 1/2 p E 1 (h) + • • • + t 1/(2p) p Ep (h)) -f (x ) - 2p k=2 1 k! i 1 +•••+ip=k k i 1 , . . . , i p •t i 1 2 +•••+ ip 2p -1 T k • p E 1 (h) ⊗i 1 ⊗ • • • ⊗ p Ep (h) ⊗ip ≤ ε t ||t 1/2 p E 1 (h) + • • • + t 1/(2p) p Ep (h)|| 2p .
We proved or assumed that the terms such that i 1 2 + • • • + ip 2p < 1 are zero. We denote by g 1 (h) the sum in the last equation with the terms such that i 1 2 + • • • + ip 2p = 1 and by g 2 (h) the sum with the terms such that i

1 2 + • • • + ip 2p > 1.
We also dene a as the smallest exponent of t appearing in g 2 (h):

a := min i 1 2 + • • • + i p 2p : i 1 , . . . , i p ∈ {0, . . . , 2p}, i 1 + • • • + i p ≤ 2p, i 1 2 + • • • + i p 2p > 1 > 1.
So that:

1 t f (x + t 1/2 p E 1 (h) + • • • + t 1/(2p) p Ep (h)) -f (x ) -g 1 (h) (26) ≤ t a-1 |g 2 (h)| + ε t ||t 1/2 p E 1 (h) + • • • + t 1/(2p) p Ep (h)|| 2p .
We remark that h → g 2 (h) is a polynomial function so is bounded on every compact set. We also have:

ε t ||t 1/2 p E 1 (h) + • • • + t 1/(2p) p Ep (h)|| 2p ≤ ε(t 1/(2p) ) 2p t ||h|| 2p = ε||h|| 2p .
So (26) converges to 0 as t → 0, uniformly with respect to h on every compact set. Now let us assume that f ∈ A p (x ) ; we prove that the function g dened in (10) is not constant in any of its variables in the sense of (7

). Let B ∈ O d (R) adapted to the decomposition R d = E 1 ⊕• • •⊕E p . We have: 1 t [f (x + B • (t α * h)) -f (x )] -→ t→0 g(h).
Let i ∈ {1, . . . , p} and k such that v i := B • e i ∈ E k . Let us assume by contradiction that g does not depend on the i th coordinate. Considering the expression of g in ( 10) and setting all the variables outside E k to 0, we have:

∀h ∈ E k , λ ∈ R → T 2k • (h + λv i ) ⊗2k
is constant. Then applying (2), we have:

∀h ∈ E k , T 2k • v i ⊗ h ⊗2k-1 = 0.
Moreover, for h ∈ F k-1 , let us write h = h + h where h ∈ E k and h ∈ F k , so that

T 2k • v i ⊗ h ⊗2k-1 = T 2k • v i ⊗ h ⊗2k-1 = 0,
where we used that

∀h (3) ∈ F k-1 , T 2k • h ⊗ h (3) ⊗2k-1 = 0
following (19), and Proposition 6. Considering the denition of E k as the orthogonal complement of F k , which is dened in (19), the last equation contradicts that v i ∈ E k .

Non coercive case

The function g we obtain in Algorithm 1 is a non-negative polynomial function which is constant in none of its variables. However, this does not always guarantee that e -g ∈ L 1 (R d ), or even that g is coercive. Indeed, g can be null on an unbounded continuous polynomial curve, while the polynomial degree of the minimum x of f is higher than the degree of g in these variables. For example, let us consider

f : R 2 → R (27) (x, y) → (x -y 2 ) 2 + x 6 .
Then f ∈ A 3 (0) and using Algorithm 1, we get g(x, y) = (x -y 2 ) 2 , which does not satisfy e -g ∈ L 1 (R d ). In fact this case is highly degenerate, as, with

f ε (x, y) := f (x, y) + εxy 2 = x 2 + y 4 -(2 -ε)xy 2 + x 6 ,
we have that g ε (x, y) = x 2 + y 4 -(2 -ε)xy 2 satises e -gε ∈ L 1 (R d ) for every ε ∈ (0, 4) and that x is not the global minimum of f ε for every ε ∈ (-∞, 0) ∪ (4, ∞).

We now prove that instead of assuming e -g ∈ L 1 (R d ), we can only assume that g is coercive, which is justied in the following proposition. More specic conditions for g to be coercive can be found in [START_REF] Bajbar | Coercive Polynomials and Their Newton Polytopes[END_REF] and [START_REF] Bajbar | Coercive polynomials: stability, order of growth, and Newton polytopes[END_REF].

Proposition 3. Let g : R d → R be the polynomial function obtained from Algorithm 1. If g is coercive, then e -g ∈ L 1 (R d ).

Proof. Let

A k := Span (e i : i ∈ {dim(E 1 ) + • • • + dim(E k-1 ) + 1, . . . , dim(E 1 ) + • • • + dim(E k )})
for k ∈ {1, . . . , p}. By construction of g, note that for all t ∈ [0, +∞),

g p k=1 t 1/2k p A k (h) = tg(h).
Since g is coercive, there exists R ≥ 1 such that for every h with ||h|| ≥ R, g(h) ≥ 1. Then, for every h ∈ R d , we have:

g(h) = g p k=1 p A k (h) = g p k=1 ||h|| 1/2k R 1/2k p A k R 1/2k h ||h|| 1/2k = ||h|| R g p k=1 p A k R 1/2k h ||h|| 1/2k . Then, for ||h|| ≥ R, p k=1 p A k R 1/2k h ||h|| 1/2k 2 = p k=1 R 1/k ||h|| 1/k ||p A k (h)|| 2 ≥ R ||h|| ||h|| 2 = R||h|| ≥ R 2 ≥ R, so that g(h) ≥ ||h|| R which in turn implies e -g ∈ L 1 (R d ).
We now deal with the simplest conguration where the function g is not coercive, as described in (28), by dealing with the case where f is given by (27), which is an archetype of such conguration. However, dealing with the general case is more complicated and to give a general formula for the rate of convergence of the measure π t in this case is not our current objective. Proposition 4. Let the function f be given by (27). Then, if (X t , Y t ) ∼ C t e -f (x,y)/t dxdy, we have:

X t t 1/6 , Y 2 t -X t t 1/2 -→ t→0 C e -x 6 √ x e -y 2 √ π 1 x≥0 dxdy, where C = ∞ 0 e -x 6 √ x dx -1 .
Proof. First, let us consider the normalizing constant C t . We have :

C -1 t = R 2 e -(x-y 2 ) 2 +x 6 t dxdy = 2t 3/4 ∞ -∞ e -t 2 x 6 ∞ 0 e -(y 2 -x) 2 dy dx = t 3/4 ∞ -∞ e -t 2 x 6 ∞ -x e -u 2 √ u + x dy dx = t 7/12 ∞ -∞ e -x 6 ∞ -t -1/3 x e -u 2 t 1/3 u + x du dx ∼ t→0 t 7/12 ∞ 0 e -x 6 √ x ∞ -∞ e -u 2 du dx,
where the convergence is obtained by dominated convergence and where we performed the change of variables x = t -1/6 x and u = t -1/2 (y 2 -x). Then we consider, for a 1 < b 1 and a 2 < b 2 ,

P X t t 1/6 ∈ [a 1 , b 1 ], Y 2 -X t 1/2 ∈ [a 2 , b 2 ] .
Performing the same changes of variables and using the above equivalent of C t completes the proof.

More generally, if the function g is not coercive and if we can write, up to a change of basis,

g(h 1 , . . . , h d ) = Q 1 (h 1 , h 2 ) 2 + Q 2 (h 3 , h 4 ) 2 + • • • + Q r (h 2r-1 , h 2r ) 2 + g(h 2r+1 , . . . , h d ), (28)
where the Q i are polynomials with two variables null on an unbounded curve (for example,

Q i (x, y) = (x -y 2 ), Q i (x, y) = (x 2 -y 3 ), Q i (x, y) = x 2 y 2 )
, and where g is a non-negative coercive polynomial, then

a 1 ((X t ) 1 , (X t ) 2 , t) , . . . , a r ((X t ) 2r-1 , (X t ) 2r , t) , 1 t α 2r+1 , . . . , 1 t α d * B • ((X t ) 2r+1 , . . . , (X t ) d ) -→ t→0 b 1 (x 1 , x 2 ) . . . b r (x 2r-1 , x 2r )Ce -g(x 2r+1 ,...,x d ) dx 1 . . . dx 2r dx 2r+1 • • • dx d ,
where C is a normalization constant, B ∈ O d-2r-1 (R) is an orthogonal transformation and for all k = 1, . . . , r, a k : R 2 × (0, +∞) → R 2 and b k is a density on R 2 . Such a k and b k can be obtained by applying the same method as in Proposition 4. Algorithm 1 yields the rst change of variable for this method, given by the exponents (α i ) (in the proof of Proposition 4, the rst change of variable is t -1/2 x and t -1/4 y) and thus seems to be the rst step of a more general procedure in this case. However, we do not give a general formula as the general case is cumbersome. Moreover, we do not give a method where the non coercive polynomials Q i depend on more than two variables, like Q(x, y, z) = (x -y 2 ) 2 + (x -z 2 ) 2 .

The method sketched in Proposition 4 cannot be direclty applied to this case.

5 Proofs of Theorem 4 and Theorem 6 using Theorem 7

Single well case

We now prove Theorem 4.

Proof. Using Theorem 7, we have for all h ∈ R d :

1 t f (B • (t α * h)) -→ t→0 g(h).
To simplify the notations, assume that there is no need of a change of basis i.e. B = I d . We want to apply Theorem 3 to the function f . However the condition e -g(h) dh.

Since g is coercive, using Proposition 3 we have e -g ∈ L 1 (R d ) and it follows from Theorem 3 that if X t has density π t (x) := C t e -f (x)/t , then ( X t ) 1 t α 1 , . . . ,

( X t ) d t α d L -→ X as t → 0,
where X has density proportional to e -g(x) . Now, let us prove that if X t has density proportional to e -f (x)/t , then we also have

(X
I 1 + I 2 ,
where I 1 is the integral on the set B(0, δ) and I 2 on B(0, δ) c . We have then:

|I 2 | ≤ ||ϕ|| ∞ (π t (B(0, δ) c ) + π t (B(0, δ) c )) -→ t→0 0,
where we used Proposition 1. On the other hand, we have f = f on B(0, δ), so that

|I 1 | ≤ ||ϕ|| ∞ |C t -C t | B(0,δ) e -f (x) t dx ≤ ||ϕ|| ∞ 1 - C t C t .
And we have:

C t C t = e -f (x)
t dx e -f (x) t dx = B(0,δ) e -f (x) t dx + B(0,δ) c e -f (x) t dx B(0,δ) e -f (x) t dx + B(0,δ) c e -f (x) t dx .

By Proposition 1, we have when t → 0 B(0,δ) c e -f (x) t dx = o B(0,δ) e -f (x) t dx B(0,δ) c e -f (x) t dx = o B(0,δ) e -f (x) t dx , so that C t /C t → 1, so I 1 → 0, which then implies (29).

Multiple well case

We now prove Theorem 6.

Proof. The rst point is a direct application of Theorem 5. For the second point, we remark that X it has a density proportional to e -f i (x)/t , where

f i (x) := f (x) if x ∈ B(x i , δ) +∞ else.
We then consider f i as in Section 5.1:

f i (x) = f i (x) if x ∈ B(x i , δ) ||x|| 2 else.
and still as in Section 5.1, we apply Theorem 3 to f i and then prove that random variables with densities proportional to e -f i (x)/t and e -f i (x)/t respectively have the same limit in law.

Innitely at minimum

In this section, we deal with an example of innitely at global minimum, where we cannot use a Taylor expansion. for some a > 0. Furthermore, assume that f is coercive and e -f ∈ L 1 (R d ). Then, if X t has density π t ,

log 1/2 1 t • X t L -→ X as t → 0,
where X ∼ U(B(0, 1)).

Proof. Noting that ||x||>1 e -f (x)/t dx → 0 as t → 0 by dominated convergence, we have

C t ∼ t→0 B(0,1)
e -e , where the convergence of the integral is obtained by dominated convergence. Then we have, for -1 < a i < b i < 1 and i a 2 i < 1, i b 2 i < 1:

P log 1/2 1 t • X t ∈ d i=1 [a i , b i ] = C t log d/2 1 t (b i ) (a i ) e -t 1 |x| 2 -1 dx -→ t→0 d i=1 (b i -a i )
Vol(B(0, 1))

.

For

  a, b ∈ R d , we denote by a * b the element-wise product, i.e. ∀i ∈ {1, . . . , d}, (a * b) i = a i b i .

Proposition 5 .

 5 Let f : R d → R such that ∀x ∈ B(0, 1), f (x) = e 1), f (x) > a

  by the local minimum property. Then, considering h = λ 1 p E 1 (h) + • • • + λ p p Ep (h), we have, for all h ∈ R d and λ 1 , . . . , λ d ∈ R,

	2p	1
	k=2	k!	i 1 ,...,ip∈{0,...,k}
			i 1

  is not necessarily true. Instead, let ε > 0 and we apply Theorem 3 to f , where f is dened as:f (h) = f (h) if h ∈ B(0, δ) ||h|| 2 else,and where δ > 0 will be xed later. Then f satises the hypotheses of Theorem 3. The only difcult point to prove is the last condition of Theorem 3. If t ∈ (0, 1] and h ∈ R d are such that(t α 1 h 1 , . . . , t α d h d ) / ∈ B(0, δ), then f (t α 1 h 1 , . . . , t α d h d ) t = ||(t α 1 h 1 , . . . , t α d h d )|| 2 t ≥ ||h|| 2 ,because for all i, α i ≤ 1 2 . If t and h are such that (tα 1 h 1 , . . . , t α d h d ) ∈ B(0, δ), then choosing δ such that for all (t α 1 h 1 , . . . , t α d h d ) ∈ B(0, δ), f (t α 1 h 1 , . . . , t α d h d ) t -g(h) ≤ ε,which is possible because of the uniform convergence on every compact set (see Section 4.9), we derive that f (t α 1 h 1 , . . . , t α d h d ) t ≥ g(h) -ε.

	Hence
	sup
	R d

R d sup 0<t<1 e -f (t α 1 h 1 ,...,t α d h d ) t dh 1 . . . dh d < ∞ 0<t<1 e -f (t α 1 h 1 ,...,t α d h d ) t dh 1 . . . dh d ≤ R d e -||h|| 2 dh + e ε R d

Acknowledgements

I would like to thank Gilles Pagès for insightful discussions.

A Properties of tensors Proposition 6. Let T k be a symmetric tensor of order k in R d . Let E be a subspace of R d . Assume that ∀h ∈ E, T k • h ⊗k = 0.

Then we have

Proof. Using (2), we have for h 1 , . . . , h k ∈ E and λ 1 , . . . , λ k ∈ R,

which is an identically null polynomial in the variables λ 1 , . . . , λ k , so every coecient is null, in particular ∀h 1 , . . . ,