Rémi Peyre

A probabilistic approach to Carne's bound

Carne's bound is a sharp inequality controlling the transition probabilities for a discrete reversible Markov chain ( § 1). Its ordinary proof ( § 2) uses spectral techniques which look as efficient as miraculous. Here we present a new proof, comparing a "drift" for ways "out" and "back", to get the gaussian part of the bound ( § 3), and using a conditioning technique to get the flight factor ( § 5). Moreover we show how our proof is more "supple" than Carne's one and may generalize ( § 4.2).

Introduction

The Markov chain

Let V be a finite or countable set of points. Let us consider an irreducible Markov chain (X t ) t∈N on V , with transition kernel (p(x, y)) x,y∈V , and whose law is denoted by P x when starting at x. That chain is supposed to be reversible, i.e. we suppose that there exists a measure µ on V such that, for all x ∈ V 0 < µ(x) < ∞, and ∀x, y ∈ V µ(x)p(x, y) = µ(y)p(y, x).

(1.1)

By irreducibility, µ is then uniquely determined up to a multiplicative factor; in the sequel, we shall suppose it fixed. Note that we do not demand µ to be finite.

Then one may associate to the kernel a (non-oriented) graph (V, E) with vertices set V by defining the set of edges through {x, y} ∈ E ⇔ p(x, y) = 0.

(1.2)

(A priori that definition should determine an oriented graph, but actually p(x, y) = 0 ⇔ p(y, x) = 0 by reversibility). As usual, we shall write z ∼ z to mean that {z, z } ∈ E [START_REF] Th | Long range estimates for Markov chains[END_REF] . The graph distance, denoted by d, will stand for [START_REF] Th | Long range estimates for Markov chains[END_REF] Look out for the fact that ∼ is not an equivalence relation.

the length of the shortest path(s) in E joining two points. Speaking in terms of probability, one has: d(x, y) = inf t ∈ N ; p t (x, y) = 0 , (

where p t denotes the t-th convolution power of the kernel p.

This paper aims at explaining by probabilistic arguments an inequality due to Carne to sharply bound p t (x, y) above when d(x, y) √ t. Indeed to the best of our knowledge, all the methods developed so far to get that kind of bounds used spectral analysis techniques [1, 2]. We shall also show how our probabilistic approach allows us to generalize Carne-Varopoulos type bounds for more "flexible" distances than the graph distance.

Carne's bound and its history

In 1985, N. Th. Varopoulos [1] was the first to give a concentration result bounding p t (x, y) above for a reversible Markov chain, whose leading term was exp -d(x,y) 2

Ct

, C > 0 being an explicit constant depending on the transition kernel p. His method introduced a time-continuous Markov process on the cabled graph associated to (V, E), and studied the spectral properties of that process in an L 2 space. Moreover that proof required extra assumptions about the transition kernel.

The same year, T. K. Carne [2], by a simpler spectral method, got a finer result under the general assumptions stated in § 1.1: Theoreme 1.1 (Carne 1985). Suppose the hypotheses of § 1.1 are satisfied. Denote by P the L 2 (µ)-operator associated to the transition kernel p and let |P | stand for its norm, which is always 1 (see a more precise definition in § 2.1). Then:

p t (x, y) 2 µ(y) µ(x) 1/2 |P | t exp -d(x, y) 2 2t . (1.4)
My work was motivated by two goals: first, find a proof of theorem 1.1 which would be more natural than the original proof of Carne, then, adapt Carne-Varopoulos type bounds to distances which depend continuously on the transition kernel (see § 4.2).

Carne's proof

We give here the proof of [2] as it was exposed in [3].

Norm of the transition kernel

Let us first give a precise definition of P : Definition 2.1. P is the operator induced by P on L 2 (µ) through:

P f (x) = E x [f (X 1 )] = y∼x p(x, y)f (y), (2.1)
Then we define |P | as the operator norm of P in L 2 (µ), i.e. |P | = sup f L 2 (µ) =1 P f L 2 (µ) . Note that P is self-adjoint by reversibility of µ, and |P | 1 by Jensen's inequality.

A more intrinsic defintion of |P | is given by the following classical

Lemma 2.2 ([3, chap. 5-2]). For any x ∈ V , |P | = lim sup t -→ ∞ p t (x, x) 1/t = sup t 1 p t (x, x) 1/t . (2.2)

Chebychev's polynomials

Since P n f (x) = E x [f (X n )], one can write p t (x, y) = δ x µ(x) , P t δ y L 2 (µ) = |P | t µ(x) δ x , P |P | t δ y L 2 (µ) . (2.3)
The trick then consists in decomposing the polynomial Z t in the basis of Chebychev's polynomials. The following results are classical:

Lemma 2.3. For any k ∈ Z, there exists a unique polynomial

Q k (Z) satisfying ∀θ ∈ C Q k (cos θ) = cos(kθ), (2.4)
called the k-th (first type) Chebychev polynomial. It satisfies:

1. deg Q k = |k|; 2. |x| 1 ⇒ |Q(x)| 1; 3. ∀t ∈ N Z t = 1 2 t k∈Z t (t+k)/2 Q k (Z)
, where by convention t p = 0 whenever p / ∈ {0, 1, . . . , t}.

By property 3 in Lemma 2.3, formula (2.3) gives

p t (x, y) = |P | t 2 t µ(x) k∈Z t (t + k)/2 δ x , Q k P |P | δ y L 2 (µ)
.

(2.5)

The linear operator P |P | on L 2 (µ) is self-adjoint and its norm is 1 by construction; so it decomposes onto a countable orthonormal basis of eigenvectors as

P |P |   λ∈Spec(P/|P |) a λ v λ   = λ∈Spec(P/|P |) λa λ v λ , (2.6)
where v λ is the eigenvector associated to the eigenvalue λ, the eigenvalues being counted with multiplicity. By definition of |P | we have Spec

P |P | ⊂ [-1, 1]. So Q k P |P |   λ∈Spec(P/|P |) a λ v λ   = λ∈Spec(P/|P |) Q k (λ)a λ v λ , (2.7)
where the Q k (λ) are all of absolute value less than or equal to 1 by lemma 2.3, property 2. Then, the operator norm of

Q k P |P |
on L 2 (µ) is at most 1, so to write:

δ x , Q k P |P | δ y L 2 (µ) δ x L 2 (µ) • δ y L 2 (µ) = µ(x)µ(y). ( 2
δ x , Q k P |P | δ y L 2 (µ)
= 0. In the end,

p t (x, y) = |P | t 2 t µ(x) |k| d(x,y) t (t + k)/2 δ x , Q k P |P | δ y L 2 (µ) (2.8) |P | t 2 t µ(x) µ(x)µ(y) |k| d(x,y) t (t + k)/2 2|P | t µ(y) µ(x)

The Gaussian factor

As told before, this article presents a new, probabilistic proof of Carne's bound. In this section, only the Gaussian part of the bound will be considered. The fundamental estimate is the Theoreme 3.1. Let P be a Markov chain as described in § 1.1; let t 2; let x = y ∈ V ; then

p t (x, y) µ(y) µ(x) 1/2 exp - (d(x, y) -1) 2 2(t -1) . (3.1)
The following immediate corollary yields a more pleasant formula:

p t (x, y) p t (y, x) exp - (d -1) 2 t -1 . (3.4)
Now, rather than reasoning on the graph, which is a "complicated" object, we shall introduce a function ξ : V -→ R which measures how much the random walk X is closer to x or to y. ξ must satisfy:

Assumption 3.4. • ξ(x) = 0; ξ(y) = d; • ξ is 1-Lipschitz, i.e. for z ∼ v we have |ξ(v) -ξ(z)| 1.
Such a map ξ always exists since, for instance, the map d(x, •) always satisfies assumption 3.4. Each point of V tends to make ξ increase or decrease, depending on the values of the transition kernel. Let us denote by m(z) the expected value for the variation of ξ after the particle having visited z, i.e.

m(z) = E z [ξ(X 1 )] -ξ(z). (3.5)
Now, let (M u ) u 1 be the process defined by:

M u = ξ(X u ) -ξ(X 1 ) - u-1 s=1 m(X s ) = u-1 s=1 (ξ(X s+1 ) -ξ(X s ) -m(X s )) ; (3.6)
obviously M is a martingale starting at 0. Let us look at the chain starting at x. On the event

{X t = y}, one trivially has ξ(X t ) -ξ(X 1 ) d -1, hence E x [M t | X t = y] d -1 -E x t-1 u=1 m(X u ) X t = y . (3.7)
One may carry out the same reasoning starting at y, which gives:

E y [M t |X t = x] -(d -1) -E y t-1 u=1 m(X u ) X t = x . (3.8) What can we see? If the terms E x t-1 u=1 m(X u ) X t = y , resp. E y t-1 u=1 m(X u ) X t = x were not present in (3.7) and (3.8), these formulae would reduce to E x [M t |X t = y] d -1, resp. E y [M t |X t = x] -(d -1)
, so that we would observe a large deviation phenomenon on martingales, which would yield a control respectively on p t (x, y) and p t (y, x). Unfortunately, that phenomenon seems to be wiped out because of the terms m(X s ). The key idea then consists in noticing that, by reversibility, these m(X s ) are the same for the "way out" as for the "way back"; subsequently, if the m(X s ) tend to make the right hand side of (3.7) diminish (which would damp the large deviation phenomenon), then they tend to make the right hand side of (3.8) increase, which this time translates into a strengthening of the large deviation phenomenon. So, p t (x, y) and p t (y, x) cannot be large simultaneously, which will lead us to (3.4).

So, we consider X x , X y two independent chains with respective laws P x and P y ; let P x⊗y be their joint law. The respective realizations of (M u ) u 1 for the paths starting at x and at y are denoted by

(M x u ) u 1 and (M y u ) u 1 . By reversibility, ∀u ∈ {1, . . . , t -1} E x m(X u ) X t = y = E y m(X t-u ) X t = x .
(3.9)

Hence by combining (3.7) and (3.8),

E x⊗y M x t -M y t X x t = y and X y t = x 2(d -1). (3.10)
It remains to control the deviations of M x t -M y t . We remark that this random variable may be interpreted as the final value of a 2(t -1) steps martingale, whose steps satisfy the assumptions of the following lemma: Lemma 3.5. Let (F t ) t∈N be a filtration; let (X t ) t 1 be an adapted real-valued process with E[X t+1 |F t ] = 0 [START_REF] Carne | A transmutation formula for Markov chains[END_REF] . We suppose that, for all t ∈ N, Law(X t+1 |F t ) is [START_REF] Carne | A transmutation formula for Markov chains[END_REF] In other words, the X t 's are the increments of a martingale.

supported by an interval of length 2 almost surely. Then, letting u 0 be a fixed time, we have for all λ ∈ R: That point being taken for granted, we prove lemma 3.5 by induction on u:

E exp λ u t=1 X t exp u λ 2 2 . ( 3
• For u = 0 the result is trivial.

• Let u 1; suppose the result to be true for u -1. Let λ ∈ R; we write

E exp λ u t=1 X t = E exp λ u-1 t=1 X t E e λXu |F u-1 E exp λ u-1 t=1
X t e (u-1)λ 2 /2 by induction

• E e λXu |F u-1 ∞ e λ 2 /2 by (3.12)

e uλ 2 /2 . (3.13)
To conclude, it only remains to us to prove the following measure concentration lemma: Lemma 3.7. Let X be a centered real-valued random variable satisfying for some k > 0:

∀λ ∈ R E[e λX ] e kλ 2 /2 . (3.14) If A is an event such that E[X|A] C (3.15)
for some C 0, then Then it suffices to take λ = C/k to get the announced result.

P(A) exp - C 2 2k . ( 3
Now we conclude the proof of Theorem 3.1. Lemma 3.5 permits us to control the Laplace transform of M x t -M y t under P x⊗y :

∀λ ∈ R E x⊗y e λ(M x t -M y t ) e (t-1)λ 2 , (3.21)
and Formula (3.10) then gives, via Lemma 3.7:

P x⊗y (X x t = y and X y t = x) exp - (d -1) 2 t -1 , (3.22)
i.e. (3.4).

Generalization to a larger class of distances 4.1 Statement of the generalized theorem

Now we will show that the reasoning made above can in fact adapt to a whole class of distances. So let us consider a new distance on V , which we will also call d -to avoid confusions, the graph distance that we had defined by (1.3) will be denoted d G from now on. We have:

Theoreme 4.1. Suppose that d is built so that, if ξ : V -→ R is any 1-Lipschitz function with respect to d, one has, for all x ∈ V :

• E x [ξ(X 1 )] -ξ(x) B ; (4.1) • ∀λ 0 E x e λ{ξ(X 1 )-Ex[ξ(X 1 )]} e Aλ 2 , (4.2)
for some constants A and B independent of ξ. Then, for all x, y ∈ V , one has

p t (x, y) µ(y) µ(x) 1/2 exp - (d(x, y) -B) + 2 4At e B/2A µ(y) µ(x) 1/2 exp - d(x, y) 2 4At . (4.3)
Remark 4.2. We can already point out that, in the case when the distance is d G , it is impossible to get anything better than A = 1/2 and B = 1. Subsequently, the result will be worsened by a √ e factor compared to (3.2) -which remains negligible compared to the exponential part of the bound-: as we will see later, it is due to the difference in treating the first steps, since the specific argument that we have used for d G in the proof of Theorem 3.1 may not generalize.

Proof. We follow again the proof of theorem 3.1: denote d = d(x, y), let ξ satisfy assumption 3.4, define m by (3.5) and let P x⊗y be the joint law of two independent chains of respective laws P x and P y ; we want to bound P x⊗y [X x t = y and X y t = x] above to conclude by formula (3.3). The first difference lies in the defintion of M : now, the martingale starts at time 0. So it is defined by:

M u = ξ(X u ) -ξ(X 0 ) - u-1 s=0 m(X s ) = u-1 s=0 ξ(X s+1 ) -ξ(X s ) -m(X s ) . (4.4)
Then we get:

   E x [M t | X t = y] = d -E x t-1 u=0 m(X u ) X t = y , E y [M t | X t = x] = -d -E y t-1 u=0 m(X u ) X t = x (4.5)
When we want to combine these two formulae as we did in (3.10), we observe that all the terms E[m(X u )] will cancel pairwise, except the terms corresponding to the first steps, i.e. to u = 0 in the two respective formulae. But we know exactly what these terms are, since under P x , we have X 0 = x a.s. (by definition!), resp. X 0 = y a.s. under P y . Thus

E x⊗y [M x t -M y t | X x t = y and X y t = x] = 2d -m(x) + m(y) 2d -2B. (4.6)
Taking into account assumption (4.2) -which plays here the role played before by Hoeffding's inequality-, we copy off the proof of Lemma 3.5 to get:

∀λ 0 E x⊗y e λ(M x t +M y t ) e 2Atλ 2 , (4.7)
and it only remains to conclude by Lemma 3.7 [START_REF] Lyons | Probabilities on trees and networks[END_REF] .

More flexible distances

Now we will show how one may build distances statisfying Theorem 4.1, such that the metric structure of V continuously depends on the transition kernel. The method developed below is certainly neither the best nor the most elegant, but it has the advantage to be of relative pedagogical simplicity.

(3) If it occurs that B d, Lemma 3.7 cannot be applied and we may only bound p t (x, y)p t (y, x) above by 1; this explains the positive part appearing in (4.3).

We keep on the principle of putting a length to each edge, but that time all the edges will not have the same size: indeed we will put a larger length to the edges that are more difficult to visit, in order to ensure that the metric structure of the graph will not be too much perturbed when we add a very "unlikely" edge.

Let α > 0 be an arbitrary parameter. To each couple (x, y) ∈ V × V , we associate a length (x, y) such that: ∀x ∈ V ∀a 0 P x (|ξ(X 1 ) -ξ(x)| 1 + a) e -a 2 /α . (4.9)

∀a 0 P x ( (x, X 1 ) 1 + a) e -a 2 /α , ( 4
Now we give a formula for (x, y) satisfying (4.8). First we define what we will call the β-entropy of a probability law: Definition 4.4. Let β ∈ (0, 1]; let p be a probability measure on a discrete space X . We call β-entropy of p the (possibly infinite) number:

H β (p) = x∈X p(x) 1-β (4) . (4.10)
A transition kernel p on V being given, we will also denote, for x ∈ V , H β (x) = H β (p(x, •)).

The β-entropy permits us to control the probability that the observed event is rare: Claim 4.5. Let β ∈ (0, 1]; let P be a probability law on a discrete state space X . We suppose H β (p) < ∞. Then, for all ∈ (0, 1], one has:

P (p(x) ) H β (p) β . (4.11) Proof. Use the identity H β (p) = E[p(x) -β
] and the fact that the map → -β is decreasing, then apply Markov's inequality.

So, one may choose the following expression for (x, y) to satisfy (4.8), where we set that, for a < 0, a 1/2 = 0:

(x, y) = 1 + √ α β ln(p(x, y) -1 ) -ln H β (x) 1/2 . (4.12)
Now, we want to show that (4.9) permits us to get (4.1) and (4.2) indeed. Let us begin with an easy observation: [START_REF] Serre | Matrices: Theory and Applications[END_REF] For β = 1 we set by continuity H β (p) = #{x ∈ X ; p(x) > 0}. Remark 4.8. The bound (4.15) is of quite poor quality close to 0, as it may be particularily striking in the case Y ≡ 1, where we get the bound e 2|λ| -2|λ|, while we know (Hoeffding's lemma 3.6) that e λ 2 /2 would work. In fact, we can compute that in a neighborhood of 0, the right hand side of (4.15) takes the form:

1 + E[Y 2 ] + 3E[Y ] 2 λ 2 2 + o(λ 2 ), (4.16)
while a variance calculation proves that in fact,

X(λ) 1 + E[Y 2 ] λ 2 2 + o(λ 2 ). ( 4

.17)

Proof. We may restrict ourselves to the case λ 0, the case λ 0 being then treated by using the result for -X.

We write

X(λ) = E e λ X -λ X . (4.18) But, since |X| Y , we have |X| Y , hence X ∈ [-Y -Y , Y + Y ].
And since, on an interval of the form [-a, a], the map x → e λx -λx takes its maximum at a, it follows that

X(λ) E e λ(Y +Y ) -λ(Y + Y ) = e λY Y (λ) -2λY . (4.19)
Now we have the following control on the Laplace transform of the random variable Y defined by (4.13): Proposition 4.9. If Y is a random variable whose law is given by (4.13), then for λ 0 one has:

E e λY 1 + √ πα 2 λ + α 2 4 λ 2 e λ+α 2 λ 2 /4 . (4.20)
Moreover, that bound is sharp close to 0, by which we mean that, if we rewrite (4.20) under the form

E[e λY ] f (λ), we have E[Y ] = d dλ λ=0 E[e λY ] = f (0) = 1 + √ π 2 α. Proof.
We begin with noticing that we can write Y = 1 + αZ, where Z is a random variable with law

dP(Z = z) = 1 z 0 2ze -z 2 dz. (4.21)
Then it suffices to prove that

∀λ 0 E e λZ 1 + √ π 2 λ + λ 2 4 e λ 2 /4 . (4.22)
To do that, we write:

E e λZ = e λ 2 /4 ∞ 0 2ze -(z-λ/2) 2 dz = t=z-λ/2 e λ 2 /4 ∞ -λ/2 (2t + λ)e -t 2 dt I(λ) , (4.23)
where I(0) = 0 and, by the theorem of differenciation under the integral,

I (λ) = ∞ -λ/2 e -t 2 dt √ π 2 + λ 2 , (4.24)
whence

I(λ) 1 + √ π 2 λ + λ 2 4
, and (4.20). From that we deduce the existence of a suitable value for A: Proposition 4.10. For all α > 0, there exists a constant A(α) < ∞ such that, if d satisfies condition (4.9) for the value α, then (4.2) is satisfied for the value A(α).

Proof. For Y with law (4.13), denoting

Y = E[Y ] = 1 + √ π
2 α, Proposition 4.9 shows that ln e λY E[e λY ] -2λY /λ 2 is bounded for λ -→ 0 and λ -→ ∞. By continuity, this function is thus bounded on the whole half-line [0, +∞). Lemma 4.7 then gives the existence of A.

Remark 4.11. We have not found any simple bound for A(α), but, for a given value of α, it is easy to compute numerically the maximum of the map λ → ln(e λY E[e λY ] -2λY )/λ 2 , which gives a suitable value for A.

A concrete example

We shall illustrate the preceding subsection by showing how our generalization gives some results in cases when the usual Carne bound is irrelevant.

Here we consider V to be the set of vertices of an inifinite 3-tree. The tree distance is denoted by d A , which will not be the same as the graph distance d G . Let L be an integer devised to become large and ε > 0 a real number devised to become small. We consider as the Markov chain on V the process which, from point x, jumps on each neighbor of x with probability (1 -ε)/3, and, with probability ε, chooses uniformly an arrival point in the (closed) ball centered on x with radius L. As one may check immediately, that chain is reversible and its invariant measure is the counting measure. Let us sum up: our process looks much like the simple random walk on a 3-tree, but sometimes the mobile may jump by roughly L units. We would like to say that, even if L is large, it suffices for ε to be small enough to get an exponential bound where L does not appear.

If we naively apply formula (1.4) to this transition kernel, we will not get anything interesting: indeed, small as ε might be, the graph distance is the same:

∀x, y ∈ G d G (x, y) = d A (x, y) L . (4.25)
Then, bounding below d G by d A /L, and merely bounding |P | by 1 [START_REF] Lyons | A crossing estimate for the canonical process on a Dirichlet space and a tightness result[END_REF] , Carne's bound yields

p t (x, y) 2 exp - d A (x, y) 2tL 2 . ( 4.26)
Concretely, if L = 17 and ε = 1/2 2 30 , it will give a bound with a d A (x, y)/578t in the exponential, which is strongly worse than the d A (x, y)/2t of the case ε = 0.

Yet it is obvious that the influence of large jumps should be nearly zero: the bound (4.26) thus must be improvable! So we will apply the techniques exposed in § 4.2. Here we have chosen arbitrarily α = 1 and β = 1/2. We suppose that L is large enough; in fact our computations will be valid as soon as L 2. Let us denote by N = 3 • 2 L -2 the cardinality of a ball of radius L. We have:

H 1/2 (x) = 3 1 -ε 3 + ε N 1/2 + (N -3) ε N 1/2 √ 3 + √ N ε 1/2 , (4.27)
hence H 1/2 (x) 2 for ε 1/(42 • 2 L ). So, if we bound over H 1/2 (x) by 2, we get that for 2 d A (x, y) L, one has:

(x, y) 1 + ln(N/ε)/2 -ln 2 1/2 , (4.28)
in particular (x, y) 1 + ln(ε -1 )/2.

( [START_REF] Lyons | A crossing estimate for the canonical process on a Dirichlet space and a tightness result[END_REF] Actually, here we could do better, but in this subsection we are only interested in the Gaussian part of the bound.

Our observation is that, if ε is small enough, d coincides with d A : indeed one has (x, y) = 1 for x ∼ y, (x, y) = ∞ if d A (x, y) > L, and (x, y) L for 2 d A (x, y) L, as soon as:

ε e -2L 2 (4.29)
-which is however a quite strong condition. (Note that for L large enough, condition (4.29) implies that ε 1/(42 • 2 L ) indeed.) Numerical computations for α = 1 give A = 8.09 . . ., resp. B = 1.88 . . .; so we have for ε e -2L 2 p t (x, y)

9 8 exp - d A (x, y) 2 33t . (4.30)
Although that bound undoubtedly improves Carne's bound (4.26) in "extreme" cases like that mentioned above, and even if it is certainly possible to get some better results by a more subtle choice of α and β, I find that bound rather disappointing in the sense that we still remain far from Carne's bound for the smallest values of ε. Anyway, Theorem 4.1 is theoretically interesting and may have better applications; in particular Lemma 4.7 can certainlty be improved.

The flight factor

Frame of the proof

The Gaussian bound (3.2) has got a disadvantage with respect to Carne's bound (1.4): in the case when |P | < 1, it does not show the exponential decreasing of p t (x, y) in the variable t. In fact, Lemma 2.2 implies p t (x, y)p t (y, x) p 2t (x, x) |P | 2t , whence

p t (x, y) µ(y) µ(x) 1/2 |P | t , (5.1)
but that is not enough to get again (1.4). The present section precisely aims at doing this. Here we will exclusively focus on the case when d = d G , cf. Remark 5.8 below. Let x, y ∈ V . For u a time devised to go to infinity, denote by

R u = {∃s u ; X s = x} (5.2)
the event which tells that the particle comes back at x at least once after time u. The strategy of our proof then consists in looking at our Markov chain conditioned to the event R u . Why this? Well, the fact that |P | < 1 expresses a possibility for the particle to "flee to infinity". That flight is responsible for the exponential decay with respect to t of the quantity p t (x, y)p t (y, x) introduced in (3.4), which measures the probability that the particle, starting at x, goes to y at time t and then comes back to x at time 2t. Conditioning with respect to R u then aims, in a way, at preventing the particle from going to infinity, which will give us a Markov chain for which |P | = 1, where the bound (3.2) will be relevant. Then it will remain to show that this conditioning selects sufficiently well the cases when the particle makes a return trip to get back a factor |P | t in (3.2). Our proof will use a kind of density argument: in a first step we will add some more assumptions on our Markov chain to carry out the reasoning, then in a second step we will prove that we can get rid of these extra assumptions by slightly perturbing the original Markov chain.

Proof under extra assumptions

We will use the following notation: Definition 5.1. We denote by τ x the hitting time of x by a walk on V , i.e. τ x = inf{t 0 ; X t = x}. For all z ∈ V , we denote:

R(z) = E z 1 τx<∞ |P | -τx . (5.3)
In this first part of the proof, we add to the assumptions of § 1.1 the following conditions: Assumption 5.2.

V is finite;

2. There exists a cemetery point ∂ ∈ V such that p(∂, •) = δ ∂ . We will denote V for V {∂};

3. The chain P is aperiodic on V .

Remark 5.3. Under the assumption 5.2, the chain will just be required to be irreducible and reversible on V ; moreover the definition of |P | will be that given by the formula (2.2) of lemma 2.2, for arbitrary x ∈ V .

Assumption 5.2 permits to obtain sharp results about the recurrence behaviour of the chain: Lemma 5.4. Suppose we have an irreducible and reversible Markov chain satisfying Assumption 5.2. Then:

1. There exist two constants 0 < c 1 c 2 < ∞ such that ∀t 0 c 1 |P | t P x (R t ) c 2 |P | t .
(5.4)

2. P x (R t+1 )/P x (R t ) -→ t -→ ∞ |P |. 3. For all z ∈ V , R(z) < ∞.
thye whole of Assumption 5.2 as soon as ε > 0. Thus, for x, y ∈ V , t > 0 and ε > 0, we have

p t ε (x, y) √ e µ(y) µ(x) 1/2 |P ε | t exp - d(x, y) 2 2t . (5.23)
To conclude, we just have to notice that p t ε (x, y), resp. P ε , are functions of ε continuous at 0. Indeed, the finite-sized matrix M ε varies continuously with ε, thus its spectral radius |P ε | also varies continuously, as well as p t ε (x, y) which is the coefficient number (y, x) of M t ε .

Infinite graphs

Now we turn to the general case, i.e. we consider a chain that merely satisfies the assumptions of § 1.1. Let us give a mark ν(z) > 0 to each vertice z of V , in such a way that for all ε > 0, #{z ∈ V ; ν(z) > 0} is finite. Let us fix x, y ∈ V , and let us take ε > 0 arbitrarily small (we shall always suppose ε < ν(x), ν(y) to avoid certain problems). We define a finite set V ε equipped with a transition kernel (p ε (z, v)) z,v∈Vε through the following way: Definition 5.10. V ε is obtained by identifying all the points with ν-mass less than ε to a cemetery point ∂:

V ε = {z ∈ V ; ν(z) ε} ∪ {∂}.
(5.24)

From now on we will denote the points of V in the same way as their images on V ε . Then p ε is the kernel p projected on V ε , with the requirement that ∂ is a cemetery point:

∀z, v ∈ V ε p ε (z, v) =          p(z, v) if z, v ∈ V ε ; 0 if z = ∂ and v ∈ V ε ; 1 if z = v = ∂; ν(w)<ε p(z, w) if z ∈ V ε and w = ∂.
(5.25)

Then the chain P ε satisfies points 1 and 2 of Assumption 5.2, and it is reversible with measure µ Vε . This chain may not be irreducible, but we can suppose that such is the case by keeping only the irreducible component of V ε containing x.

So the relation (5.21) is satisfied for V ε equipped with P ε ; it only remains to prove that p

t ε (x, y) -→ ε -→ 0 p t (x, y), resp. |P ε | -→ ε -→ 0 |P |.
Let us deal immediately with the operator norm. The very construction of p ε ensures that for all z, v ∈ V ε , we have p t ε (z, v) p t (z, v). Taking z = v = x, the characterization (2.2) of |P | immediately gives that |P ε | |P |, which is enough for us (but convergence when ε -→ 0 is also true). Now, we observe that the law of the t first steps of the chain generated by p ε converges to the law of the initial chain in the sense of total variation. Indeed, given the way how V ε and p ε are constructed respectively from V and p, we have a canonical map which associates a walk on V (ε) to a walk on V , so that the law P x maps into the law P ε

x . That map is defined as follows: the points of the walk on V are sent onto their projections on V ε until the image walk hits ∂, and from that time on the walk stays at ∂. In particular, if a realization of the original chain stays in V ε up to time t, its image by our map is kept safe on {0, . . . , t}, and so In particular, p t ε (x, y) -→ ε -→ 0 p t (x, y), and so (5.21) is satisfied for V equipped with p, QED.

A Appendix: Finite sub-Markov chains

This appendix aims at proving Lemma 5.4. Let us recall that in that lemma, we consider a sub-Markov chain on a finite graph V , given by a kernel p, which is irreducible and aperiodic (the fact that the chain is reversible is not used in the proof of lemma 5.4). Denote by n the cardinality of V .

The study of the chain may be expressed into matricial terms: we introduce the matrix M = p(v, z) z,v∈ V . (A.1)

Then the aperiodicity condition translates into the existence of a time t 0 such that, for all t t 0 , M t 0 has strictly positive coefficients (actually t 0 = n 2 would always do). On the other hand, Lemma 5.9 above permits us to consider |P | as the spectral radius of M . So we have in hand all the assumptions to apply the strongest form of the Perron-Frobenius theorem, whose general statement and proof the reader can find in [4, chap. 5]: (A.2)

  .8) and we define the length of the edge [xy] by |[xy]| = min{ (x, y), (y, x)}. Then, for any 1-Lipschitz function ξ on V , we have:

Claim 4 . 6 .

 46 Let ξ be a 1-Lipchitz function on V for a distance d built as above. Then there exists a random variable Y whose repartition map satisfies ∀a 0 P(Y 1 + a) = e -a 2 /α , i.e. Y has a law with densitydP(Y = y) = 1 y 1 2(y -1) α e -(y-1) 2 /α dy,(4.13)such that one has:P x -p.s. |ξ(X 1 ) -ξ(x)| Y. (4.14)So, we easily find that we can takeB = E[Y ] = 1 + √ π2 α into (4.1). To get a formula for A in (4.2), things are a bit more complicated. The tool which we will use is the Lemma 4.7. Let Y be a positive random variable whose Laplace tranform Y (λ) = E e λY is supposed to be finite for all λ 0, and let us denote Y = E[Y ]. Let X be a real-valued random variable satisfying |X| Y a.s.; let us denote X = E[X] and X = X -E[X]. Then the Laplace transform of X satisfies: ∀λ ∈ R X(λ) e |λ|Y Y (|λ|) -2|λ|Y . (4.15)

-

  P x V {0,...,t} TV P x ∃u ∈ {0, . . . , t} X u / ∈ V ε t u=0 P x (ν(X u ) < ε)

Proposition A. 1 (

 1 Perron-Frobenius). |P | is a simple eigenvalue of M , and all the other eigenvalues of M have an absolute value strictly less than |P |. Moreover, the eigenvector v associated to the eigenvalue |P | has all its entries strictly positive. Now, let us begin with proving point 1 of Lemma 5.4, and in a first step let us prove the second inequality. Markov's strong property gives us the overmuliplicativity relation: ∀t, u 0 P x (R t+u ) P x (R t )P x (R u ).

The proof of this lemma, which is rather technical, is postponed to the Appendix. Now we are armed to prove Carne's bound for a Markov chain satisfying Assumption 5.2. Let us fix t 0; we have the key proposition: Proposition 5.5. The law of (X s ) 0 s 2t under P x (•|R u ) converges when u -→ ∞ (for the total variation norm on V {0,...,2t} ) to the law P x of the Markov chain on V starting at x, with transition probabilities:

w∼z R(w)p(z, w) [START_REF] Lunt | Integrability of functionals of Dirichlet processes, probabilistic representations of semigroups, and estimates of heat kernels[END_REF] .

(5.5)

Proof. It is true in a general framework that P(•|R u ) is a time-inhomogeneous Markov chain with P X s+1 = v X s = z and R u = S s+1,u (v)p(z, v) w∼z S s+1,u (w)p(z, w) , (5.6) where we let S s+1,u (z) = P(R u |X s+1 = z). Our attack will consist in proving that, for all s ∈ {0, . . . , 2t -1}, for all z ∈ V , we have

To begin with, let us notice that S s+1,u (z) = P z (R u-(s+1) ), which we shall denote by S u-(s+1) (z); then what we want to prove can be written:

(5.7)

The idea consists in splitting the probability space according to the value of τ x , thanks to the strong Markov property:

whence S u (z)

.

(5.9)

Let us fix an arbitrarily small ε > 0. Since s P z (τ x = s)|P | -s converges (cf. Lemma 5.4-3), we may introduce a time u 0 for which one has s>u 0 P z (τ x = s) ε. Then, for u u 0 one has on the one hand,

.10) [START_REF] Lunt | Integrability of functionals of Dirichlet processes, probabilistic representations of semigroups, and estimates of heat kernels[END_REF] That defines a Markov chain on V indeed because R(∂) = 0.

on the other hand,

) hence (5.7) by letting ε -→ 0.

Now we want to look at the chain P . First, P is clearly irreducible. Then, one has: Proposition 5.6. The chain P is reversible, and its invariant measure is:

(5.13) where R + (x) is defined by: Definition 5.7. We denote by τ +

x the return time to x, i.e.:

Then R + (x) is defined by:

(5.15)

Proof. Let z, v ∈ V with z = x. We can lighten the expression of p (z, v), since by Markov's property,

thus (5.5) can be rewritten as

(5.17)

In the case when z = x, the same argument leads to

So, it only remains to use (5.17), (5.18) and the reversibility of µ under P to get the reversibility of µ under P . (7) The careful reader may have noticed that R(x) = 1; we let that factor appear for ease of understanding. Now we are ready to end the proof. We observe that, by Markov's property, p t (x, y)p t (y, x) = P x (X t = y and X 2t = x), which we will denote by P x (A). For u 2t, Bayes' formula gives:

hence, letting u go to infinity:

(5.20)

Now, the Markov chain P satisfies the assumptions of theorem 3.1, hence

, and finally we get the desired formula:

(5.21)

Remark 5.8. The reasoning carried out above cannot apply to other distances than d G : indeed, the distance which appears in (5.21) in fact comes as the distance associated to the process P . When one works with the graph distance, that distance is the same for P and for P , but this is no more true if d depends in a more subtle way on the transition kernel.

The density argument

Now, we want to get rid of Assumption 5.2. We will proceed in two steps: first we will just relax Assumption 5.2-3, then we will deal with the general case.

Relaxing the aperiodicity condition

We consider a finite set V with a transition law (p(x, y)) x,y∈V , such that there exists a cemetery point ∂ ∈ V satisfying Assumption 5.2-2. We suppose that the Markov chain defined by p is irreducible and reversible on V = V {∂}, with a reversible measure µ. We denote by n the cardinality of V , and by M the matrix p(y, x) x,y∈ V . The following lemma gives an algebraic characterization of the value |P | defined in (2.2): Lemma 5.9 ([3, chap. 5-2]). |P | is the spectral radius of M .

For ε ∈ [0, 1[, let p ε be the transition kernel defined by:

(5.22)

The Markov chain P ε generated by p ε is an irreducible reversible chain whose graph and reversible measure are the same as for p = p 0 , and which satisfies

We deduce that, for all t 1, one has P x (R t ) 1/t lim sup u -→ ∞ P x (R u ) 1/u . Moreover, if |P | < 1, we have, for all t 1,

hence lim sup u -→ ∞ P x (R u ) 1/u |P |, that last relation being also trivially true in the case when |P | = 1. Finally, the second inequality of (5.4) is satisfied for t 1 with c 2 = 1, the case t = 0 being trivial.

For the lower bound, we will only show that there exists a constant c 3 > 0 such that one has, for t large enough:

the first inequality of (5.4) then will follow for t large enough, and the case when t is small then will be dealt with by finiteness, thanks to noticing that irreducibility ensures that P x (R t ) > 0 for all t ∈ N. To prove (A.4), let us consider the eigenvector v = (v i ) i∈ V associated to the eigenvalue |P |. We shall keep in mind that, by Proposition A.1,

then gives for all t 0:

Let now t 0 be such that one has ∀z, v ∈ V p t 0 (z, v) > 0 (such a t 0 exists by aperiodicity, as we noticed it supra), and let us denote η = min z,v∈ V p t 0 (z, v) > 0. For t t 0 , by (A.6) we can fix z 1 such that p t-t 0 (z 1 , x)

hence (A.4) with c 3 = ηv x /nv|P | t 0 . Now, let us look at the fine behaviour of the sequence P x (R t ) when t -→ ∞. By Markov's property,

subsequently to prove property 2 in Lemma 5.4, we just have to show that we have

|P | for all z ∈ V . More precisely, we will show that there exists a constant c 4 (z) > 0 such that

p t (x, z) can be rewritten in matricial terms as T δ z M t δ x . Now, by A.1, if Ṁ stands for the matrix of the projection on Rv relatively to the sum of the characteristic spaces for the eigenvalues of M other than |P |:

(A.9) Subsequently, p t (x, z)/|P | t tends to the value c 4 (z) = T δ y Ṁ δ x when t tends to infinity. The non-nullity of c 4 (z) then is a consequence of point 1 which we have proved a few lines above: indeed, taking again the notations t 0 and η used above, we have by Markov's property: ∀t t 0 p t (x, z) p t-t 0 (x, x)p t 0 (x, z) c 2 |P | t-t 0 η, (A.10) hence c 4 c 2 > 0. Point 3 may be the most subtle. In fact we will prove that for all z in V , P z (τ x = t) decreases exponentially with an exponential decay factor strictly less than |P |. More precisely, we will estimate the decay factor of P z (τ x t and τ ∂ t) = P z (∀u ∈ {0, . . . , t -1})(X u = x, ∂) . In other words, we have to look at the decay speed of the sub-Markov chain associated to the transition kernel p, but restricted to V {x}. In matricial words, it is the spectral radius of the matrix M * , which is the matrix M where the x-th line has been replaced by zeroes. Let us denote by |P * | its spectral radius. The weak form of Perron-Frobenius theorem (cf. [4]) claims that there exists a |P * |-eigenvector v * with positive or zero entries for M * . Each entry of M * is less than or equal to the corresponding entry of M , and moreover M * = M ; since M is the matrix of an irreducible aperiodic chain, it follows that, for t sufficiently large, each entry of (M * ) t is strictly less than the correponding entry of M t . So, for t sufficiently large:

Further readings and acknowledgements

The use of the martingales introduced in § 3 can actually be seen as a discrete variant of the forward/backward martingale decomposition technique of Lyons and Weian [5], as was pointed out to me by Laurent Saloff-Coste. In fact, paper [6] gives a bound for continuous diffusions whose spirit is quite close to that of (3.1). The present work was launched by informal discussions with my colleagues Yann Ollivier and Vincent Beffara, who usefully encouraged and helped me when necessary.


Rémi Peyre

Comparison between W 2 distance and Ḣ-1 norm, and Localization of Wasserstein distance

Keywords: Wasserstein distance, homogeneous Sobolev norm, localization

It is well known that the quadratic Wasserstein distance W 2 (•, •) is formally equivalent, for infinitesimally small perturbations, to some weighted H -1 homogeneous Sobolev norm. In this article I show that this equivalence can be integrated to get non-asymptotic comparison results between these distances. Then I give an application of these results to prove that the W 2 distance exhibits some localization phenomenon: if µ and ν are measures on R n and ϕ : R n → R + is some bump function with compact support, then under mild hypotheses, you can bound above the Wasserstein distance between ϕ • µ and ϕ • ν by an explicit multiple of W 2 (µ, ν).

1 Non-asymptotic equivalence between W 2 distance and Ḣ-1 norm

Introduction

In all this section, M denotes a connected Riemannian manifold endowed with its distance dist(•, •) and its standard measure λ provided by the volume form (so, in the case M = R n , λ is the Lebesgue measure). Let us give a few standard definitions which will be at the core of our work:

• For µ, ν two positive measures on M , denoting by Π (µ, ν) the set of (positive) measures on M × M whose respective marginals are µ and ν, for π ∈ Π (µ, ν) one defines

I(π) := ˆM×M dist(x, y) 2 π(dx, dy) (1)
and then

W 2 (µ, ν) := inf{I(π) | π ∈ Π (µ, ν)} 1/2 . ( 2
)
W 2 is a (possibly infinite) distance, called the quadratic Wasserstein distance [13, § 7.1]. Note that this distance is finite only between measures having the same total mass.

• On the other hand, for µ a (positive) measure on M , if f is a C 1 real function on M , one denotes

f Ḣ1 (µ) := ˆM |∇f (x)| 2 µ(dx) 1/2 , (3)
which defines a semi-norm; for ν a signed measure on M , one then denotes

ν Ḣ-1 (µ) := sup{| f, ν | | f Ḣ1 (µ) 1}, (4)
where the duality product f, ν denotes the integral of the function f against the measure ν. [ * ] We observe that • Ḣ-1 (µ) defines a (possibly infinite) norm, which we will call the Ḣ-1 (µ) weighted homogeneous Sobolev norm. Note that this norm is finite only for measures having zero total mass. In the case µ is the standard measure, we will merely write " Ḣ-1 " for " Ḣ-1 (λ)".

The W 2 Wasserstein distance is an important object in analysis; but it is nonlinear, which makes it harder to study. For infinitesimal perturbations however, the linearized behaviour of W 2 is well known: if µ is a positive measure on M and dµ is an infinitesimally small perturbation of this measure, [ †] one has formally (see [13, § 7.6] or [9, § 7]) W 2 (µ, µ + dµ) = dµ Ḣ-1 (µ) + o(dµ). [ ‡] (5)

More precisely, one has the following equality, known as the Benamou-Brenier formula ([2, Prop. 1.1] for M = R n ; see [START_REF] Pratelli | Equivalence between some definitions for the optimal mass transport problem and for the transport density on manifolds[END_REF] for general Riemannian manifolds): for two positive measures µ, ν on M ,

W 2 (µ, ν) = inf ˆ1 0 dµ t Ḣ-1 (µt) µ 0 = µ, µ 1 = ν . (6)
Then, a natural question is the following: are there non-asymptotic comparisons between the W 2 distance and the Ḣ-1 norm? Concretely, we are looking for inequalities like

C a µ -ν Ḣ-1 (µ) W 2 (µ, ν) C b µ -ν Ḣ-1 (µ) (7)
for constants 0 < C a C b < ∞, under mild assumptions on µ and ν.

[ * ] The rationale behind the use of duality notation in this article is that we cannot use the notation "dµ" to refer to the measure of a small volume: see indeed Footnote [ †] below.

[ †] Beware that here dµ denotes a small measure on M , not the value of µ on a small volume. [ ‡] This formula has to be understood in the sense that, for every measure ν, one has W2(µ, µ

+ εν) ε→0 = |ε| ν Ḣ-1 (µ) + o(ε).
As explained in the references cited, some regularity assumptions on ν shall be required for that property to hold rigorously: in particular, one must have ν µ with a bounded and smooth enough density.

 W 2 distance and Ḣ-1 norm Rémi Peyre 1.2 Controlling W 2 by Ḣ-1 Theorem 1. For any positive measures µ, ν on M , W 2 (µ, ν) 2 µ -ν Ḣ-1 (µ) . (8)
Proof. We suppose that µ -ν Ḣ-1 (µ) < ∞, otherwise there is nothing to prove. For t ∈ [0, 1], let

µ t := (1 -t)µ + tν, (9)
so that µ 0 = µ, µ 1 = ν and dµ t = (µ -ν)dt. Then, by the Benamou-Brenier formula (6):

W 2 (µ, ν) ˆ1 0 µ -ν Ḣ-1 (µt) dt. (10)
Now, we use the following key lemma, whose proof is postponed:

Lemma 2.
If µ, µ are two measures such that µ ρµ for some ρ > 0, then

• Ḣ-1 (µ ) ρ -1/2 • Ḣ-1 (µ) . [ §]
Here obviously µ t (1 -t)µ, so

W 2 (µ, ν) ˆ1 0 (1 -t) -1/2 µ -ν Ḣ-1 (µ) dt = 2 µ -ν Ḣ-1 (µ) , (11) qed.
Corollary 3. If µ ρλ for some ρ > 0, then

W 2 (µ, ν) 2ρ -1/2 µ -ν Ḣ-1 . (12)
Proof. Just use that

• Ḣ-1 (µ) ρ -1/2 • Ḣ-1 by Lemma 2.
Proof of Lemma 2. Take µ ρµ and let ν be a signed measure on M such that µ + ν is positive; then µ + ρν is also positive. For m a measure on M , we denote by diag(m) the measure on M × M supported by the diagonal whose marginals (which are equal) are m, i.e.:

diag(m) (A × B) := m(A ∩ B); (13)
with that notation,

π ∈ Π (µ, µ + ν) ⇒ ρπ + diag(µ -ρµ) ∈ Π (µ , µ + ρν), (14)
and

I ρπ + diag(µ -ρµ) = ρI(π). (15)
Therefore, taking infima, 16) [ §] Beware that here '•' stands for a measure, not for a function: otherwise the formula would be false.-When f is a function, f Ḣ-1 (µ) stands for the Ḣ-1 (µ) norm of the measure having density f w.r.t. µ.

W 2 (µ , µ + ρν) 2 = inf{I(π ) | π ∈ Γ (µ , µ + ρν)} inf I ρπ + diag(µ -ρµ) π ∈ Γ (µ, µ + ν) = ρ inf{I(π) | π ∈ Γ (µ, µ + ν)} = ρW 2 (µ, µ + ν) 2 . (



W 2 distance and Ḣ-1 norm Rémi Peyre For infinitesimally small ν, [ ¶] it follows by Equation ( 5) that ρν

2 Ḣ-1 (µ ) ρ ν 2 Ḣ-1 (µ) , hence ν Ḣ-1 (µ ) ρ -1/2 ν Ḣ-1 (µ)
. This relation remains true even for non-infinitesimal ν by linearity, which ends the proof. Remark 4. Lemma 2 could also be proved very quickly by using the definition (3)-( 4) of the Ḣ-1 (µ) norm. The proof above, however, has the advantage that it does not need the precise expression of • Ḣ-1 (µ) , but only the fact that it is the linearized W 2 distance.

1.3 Controlling Ḣ-1 by W 2 Theorem 5. Assume M has nonnegative Ricci curvature. Then for any positive measures µ, ν on M such that µ ρ 0 λ and ν ρ 1 λ,

µ -ν Ḣ-1 2(ρ 1/2 0 -ρ 1/2 1 ) ln(ρ 0 / ρ 1 ) W 2 (µ, ν). ( 17
)
(For ρ 1 = ρ 0 , the right-hand side of (17) is to be taken as ρ • Ḣ-1 (µt) by Lemma 2. Then, by the integral triangle inequality for normed vector spaces,

µ -ν Ḣ-1 = ˆ1 0 dµ t Ḣ-1 ˆ1 0 dµ t Ḣ-1 ˆ1 0 ρ (1-t)/2 0 ρ t/2 1 dµ t Ḣ-1 (µt) = ˆ1 0 ρ (1-t)/2 0 ρ t/2 1 dt W 2 (µ, ν) = 2(ρ 1/2 0 -ρ 1/2 1 ) ln(ρ 0 / ρ 1 ) W 2 (µ, ν), (18)
qed.

Remark 7. Taking into account the dimension n of the manifold M , the bound on µ t ∞ could be refined into

µ t ∞ (1 -t) µ 0 -1/n ∞ + t µ 1 -1/n ∞ -n (19)
[ ¶] To make rigorous the formal argument of taking an infinitestimally small ν, according to Footnote [ ‡] above, one would have to replace ν by εν1, where ν1 is a regular enough measure, and to let ε tend to 0; then the regularity assumption on ν1 would be relaxed by a classical approximation argument. Anyway, Lemma 2 can also be proved easily and rigorously without referring to optimal transportation at all, cf. Remark 4 below.



W 2 distance and Ḣ-1 norm Rémi Peyre (cf. [START_REF] Mccann | A convexity principle for interacting gases[END_REF]Theorem 2.3]), which would yield a slightly sharper bound in Equation (17), namely:

µ -ν Ḣ-1 ˆ1 0 (1 -t)ρ -1/n 0 + tρ -1/n 1 -n/2 dt W 2 (µ, ν) =      ρ 1/2-1/n 0 -ρ 1/2-1/n 1 (n/2-1)(ρ -1/n 1 -ρ -1/n 0 ) W 2 (µ, ν) n 2; log(ρ 1 /ρ 0 ) 2(ρ -1/2 0 -ρ -1/2 1 ) W 2 (µ, ν) n = 2. ( 20
)
For n = 1 it turns out that one can let tend ρ 1 → ∞ in (20) without making the integral diverge; which leads to a much more powerful result:

Theorem 8. When M is an interval of R, then under the sole assumption that µ ρ 0 λ, one has for all positive measures ν on M :

µ -ν Ḣ-1 2ρ 1/2 0 W 2 (µ, ν). ( 21
)
Remark 9. For n 2 there is no hope to get a bound valid for all ν, because then it can occur that W 2 (µ, ν) < ∞ but µ -ν Ḣ-1 = ∞: for instance, take µ to be the uniform measure on the 2-dimensional sphere and ν a Dirac mass.

2 Application to localization of Wasserstein distance

Introduction

In all this section, we work in the Euclidian space R n , whose norm is denoted by |•|. dist(x, A) := inf{|x -y| | y ∈ A} denotes the distance between a point x and a set A; A c denotes the complement of A; λ denotes the Lebesgue measure. We will use the following notation to handle measures:

• For µ a measure on R n and f : R n → R n a measurable map, f * µ denotes the pushforward of µ by f , that is,

f * µ (A) := µ(f -1 (A)).
• For µ a measure on R n and ϕ : R n → R + a nonnegative measurable function, ϕ • µ denotes the measure such that ϕ • µ (dx) := ϕ(x)µ(dx).

We will also use the following norms on measures:

• µ Ḣ-1 (ν) has the same definition as in § 1;

•

µ 1 := ´Rn|µ(dx)| is the total variation norm of µ; [ ]
• For ν a positive measure with µ ν, we define

µ L 2 (ν) := ˆsupp ν µ(dx) ν(dx) 2 ν(dx) 1/2 . [ * * ] (22)
For A ⊂ R n , we also denote

• L 2 (A) for • L 2 (1 A •λ) .
[ ] Note that in the case µ is a positive measure on R n , then µ 1 is nothing but µ(R n ). [ * * ] What we denote here by µ(dx)/ν(dx) here is what is commonly called dµ/dν (x): indeed, as we already told, in this article we reserve the use of "dµ" to denote a mass distribution of infinitesimally small magnitude, rather than for the mass of an infinitely small volume.

 W 2 distance and Ḣ-1 norm Rémi Peyre
The goal of this section is to give an application of Theorem 1 to the problem of localization of the quadratic Wasserstein distance. Morally, the question is the following: take two measures µ, ν on R n being close to each other in the sense of W 2 distance; is it true that µ and ν remain close when you consider their restrictions to a subset of R n ? Concretely, if ϕ is a non-negative real function on R n with compact support (plus some technical assumptions to be specified later), we want to bound above W 2 (aϕ•µ, ϕ•ν) by some multiple of W 2 (µ, ν)-where, in the former expression, a is a constant factor ensuring that aϕ • µ and ϕ • ν have the same mass (for otherwise the distance between ϕ • µ and ϕ • ν is generically infinite).

This question, which was my initial motivation for the results of § 1, was asked to me by Xavier Tolsa, who needed such a result for his paper [START_REF] Tolsa | Mass transport and uniform rectifiability[END_REF] on characterizing uniform rectifiability in terms of mass transport. Actually Xavier managed to devise a proof of his own [12, Theorem 1.1], but it was quite long (about thirty pages) and involved arguments of multi-scale analysis. With Theorem 1 at hand, however, the reasoning becomes far more direct; moreover we will be able to relax some of the assumptions of Xavier's theorem.

Statement of the theorem

Theorem 10. Let µ, ν be (positive) measures on R n having the same total mass; let B be a ball of R n (whose radius will be denoted by R when needed). Assume that on B, the density of µ w.r.t. the Lebesgue measure is bounded above and below:

∃ 0 < m 1 m 2 < ∞ ∀x ∈ B m 1 λ(dx) µ(dx) m 2 λ(dx). (23)
Let ϕ : R n → R + be a function such that:

(i) ϕ is zero outside B;

(ii) There exist 0 < c 1 c 2 < ∞ such that for all x ∈ B, c

1 dist(x, B c ) 2 ϕ(x) c 2 dist(x, B c ) 2 . (iii) ϕ is k-Lipschitz for some k < ∞.
Then, denoting a

:= ϕ • ν 1 / ϕ • µ 1 , W 2 (aϕ • µ, ϕ • ν) C(n) c 3/2 2 m 3/2 2 c 3/2 1 m 3/2 1 kc -1/2 1 W 2 (µ, ν), (24)
for C(n) < ∞ some absolute constant only depending on n. Moreover, one can bound explicitly

C(n) in such a way that C(n) = O(n 1/2 ) when n → ∞. [ † †]
Remark 11. Theorem 10 relaxes the assumptions of Theorem 1.1 of [START_REF] Tolsa | Mass transport and uniform rectifiability[END_REF] on the following points: first, Tolsa's theorem required that |∇ϕ| was bounded by a multiple of dist(•, B c ), while ours does not impose any specific control on |∇ϕ| near the boundary of B; second, Tolsa's theorem worked only for radially symmetric ϕ. Also, contrary to [START_REF] Tolsa | Mass transport and uniform rectifiability[END_REF], our conclusions state explicitly how the bound on W 2 (aϕ • µ, ϕ • ν) depends on the constants k, c 1 , c 2 , m 1 , m 2 and on the dimension n.

[ † †] For instance, with the estimates of this article, one finds that C(n) := 47n 1/2 fits-though this may be strongly suboptimal.

 W 2 distance and Ḣ-1 norm Rémi Peyre Remark 12. Actually the constraint that the support of ϕ is a ball is of little importance: we could assume as well that it would be a cube, a simplex, or many other shapes, as the corollary below shows:

Corollary 13. Make the same assumptions as in Theorem 10, except that B need not be a ball: instead, we only assume that, denoting by B • the (true) ball having the same volume as B, there exists a bijection Φ : B ↔ B • mapping the uniform measure on B onto the uniform measure on B • (i.e. such that Φ * (1

B • λ) = 1 B• • λ)
such that Φ is bi-Lipschitz (i.e. such that both Φ and Φ -1 are Lipschitz). Denote by

Φ Lip and Φ -1
Lip the optimal Lipschitz constants for resp. Φ and Φ -1 . Then, the conclusion of Theorem 10 remains true, except that now you have to replace the factor C(n) by Lip and c 2 Φ -1 2 Lip . Therefore, applying (24):

( Φ Lip Φ -1 Lip ) 5 C(n). (25
W 2 (aϕ • • µ • , ϕ • • ν • ) C(n) Φ 4 Lip Φ -1 4 Lip c 3/2 2 m 3/2 2 c 3/2 1 m 3/2 1 W 2 (µ • , ν • ). (26)
But the optimal transportation plan from µ to ν, with cost W 2 (µ, ν) 2 , can be pushed forward by Φ into a (not optimal in general) transportation plan from µ • to ν • , whose cost will then be

Φ 2 Lip W 2 (µ, ν) 2 ; so W 2 (µ • , ν • ) Φ Lip W 2 (µ, ν). Similarly W 2 (aϕ • µ, ϕ • ν) Φ -1 Lip W 2 (aϕ • • µ • , ϕ • • ν • ).
The announced result follows.

Proof of the main theorem

In the sequel we will shorthand W 2 (µ, ν) =: w, and also ϕ • µ =: μ, resp. ϕ • ν =: ν. Let g =: Id + S be a map achieving optimal transportation from ν to µ, i.e. such that µ = g * ν with ´Rn|S(y)| 2 ν(dy) = w 2 . [ ‡ ‡] Our strategy will consist in transforming ν into aμ according to the following procedure:

x We apply the transportation plan g to ν; this transforms ν into some measure μ * . The measure μ * is not supported by B a priori, so we split it into μ * [ ‡ ‡] Actually such an g does not always exist, as it can occur that the optimal transportation plan from ν to µ "splits points" if ν is not regular enough. However it would suffice to use the general formalism of transportation plans to handle that case: we do not do it here to keep notation light, but this is straightforward. Also note that it is not obvious that the infimum in ( 2) is attained: again, that is not a real problem as our proof still works by considering a sequence of transportation plans approaching optimality.

B + μ * c := 1 B • μ * + 1 B c • μ * .
[ * ] Observe that aB + ac = a.

 W 2 distance and Ḣ-1 norm

Rémi Peyre

Then, denoting by W 2 (x), W 2 (y), W 2 (z) the respective Wasserstein distances of these steps, we shall have

W 2 (ν, aμ) W 2 (x) + (W 2 (y) 2 + W 2 (z) 2 ) 1/2 .
Let us begin with bounding the cost of Step x. The squared cost of this step is

W 2 (x) 2 = ˆ|S(y)| 2 ν(dy) = ˆ|S(y)| 2 ϕ(y)ν(dy) sup ϕ × ˆ|S(y)| 2 ν(dy) = sup ϕ × w 2 c 2 R 2 w 2 , ( 27
) whence W 2 (x) c 1/2
2 Rw. Now consider Step y. As a c μ is supported by B, one has obviously

W 2 (y) 2 ˆ dist(x, B) + 2R 2 μ * c (dx) = ˆBc dist(x, B) + 2R 2 μ * (dx). (28)
From that we deduce that W 2 (y) 2c

1/2 2 Rw by the following computation:

ˆBc dist(x, B) + 2R 2 μ * (dx) = ˆg(y)/ ∈B dist(g(y), B) + 2R 2 ϕ(y)ν(dy) c 2 ˆy∈B g(y) / ∈B dist(g(y), B) + 2R 2 dist(y, B c ) 2 ν(dy) c 2 ˆy∈B g(y) / ∈B R dist(g(y), B) + 2R dist(y, B c ) 2 ν(dy) 4c 2 R 2 ˆy∈B g(y) / ∈B dist(g(y), B) + dist(y, B c ) 2 ν(dy) 4c 2 R 2 ˆ|y -g(y)| 2 ν(dy) = 4c 2 R 2 w 2 . ( 29
)
Step z is the difficult one. We begin with observing that it is easy to bound the L 2 (B) distance between μ * B and μ: indeed, denoting by f =: Id + T the inverse map of g [ †] ,

μ * B -μ 2 L 2 (1 B •µ) = ˆB μ * (dx) -ϕ(x)µ(dx) µ(dx) 2 µ(dx) = ˆB ϕ(f (x)) -ϕ(x) 2 µ(dx) k 2 ˆRn |x -f (x)| 2 µ(dx) = k 2 ˆ|T (x)| 2 µ(dx) = k 2 w 2 , ( 30
)
(where we used that μ * (dx

) = ν d(f (x)) = ϕ(f (x))ν d(f (x)) = ϕ(f (x))µ(dx)), so that μ * B -μ 2 L 2 (B) k 2 m 2 w 2 . [ ‡] (31)
[ †] For f to exist, g should be bijective, which is not always true stricto sensu; but we can safely carry out the reasoning with pretending so, by the same argument as in Footnote [ ‡ ‡] on page 7.

[ ‡] Remember that when ν stands for a measure, ν L 2 (µ) means what is more commonly denoted by dν/dµ L 2 (µ) , so that the relation µ mλ implies that ν 2

L 2 (λ)
m ν 2 L 2 (µ) -while on the other hand, when f stands for a function, one has f 2

L 2 (µ) m f 2 L 2 (λ) .
 W 2 distance and Ḣ-1 norm

Rémi Peyre

Now we have to link

• L 2 (B) with • Ḣ-1 (µ)
. This is achieved by the following lemma, whose proof is postponed: Lemma 14. Define λ to be the measure on B such that λ(dx) := dist(x, B c ) 2 λ(dx). Then, for any signed measure m on B having total mass zero:

m Ḣ-1 ( λ) C 1 (n) 1/2 m L 2 (B) , (32)
where C 1 (n) is some absolute constant only depending on n. Moreover, taking C 1 (n) := (2e + 1)n -1 ∨ 8e fits.

Thanks to Theorem 1 and Lemma 14, we have that

W 2 (z) 2 a B μ -μ * B Ḣ-1 (a B μ) 2(a B c 1 m 1 ) -1/2 a B μ -μ * B Ḣ-1 ( λ) 2C 1 (n) 1/2 (a B c 1 m 1 ) -1/2 a B μ -μ * B L 2 (B) . ( 33
)
Next, we compute

a B μ -μ * B L 2 (B) = μ * B 1 μ 1 μ -μ * B L 2 (B) | μ * B 1 -μ 1 | μ 1 μ L 2 (B) + μ * B -μ L 2 (B) μ L 2 (B) μ 1 μ * B -μ 1 + μ * B -μ L 2 (B) μ L 2 (B) μ 1 λ(B) 1/2 + 1 μ * B -μ L 2 (B) c 2 m 2 c 1 m 1 λ(B) 1/2 λ L 2 (B) λ 1 + 1 μ * B -μ L 2 (B) [ §] √ 6 c 2 m 2 c 1 m 1 + 1 μ * B -μ L 2 (B) (31)
( √ 6 + 1) c 2 m 2 c 1 m 1 km 1/2 2 w, (34)
so that, combining (33) and (34), we have got:

W 2 (z) (2 √ 6 + 2)C 1 (n) 1/2 a -1/2 B c 2 m 3/2 2 c 1 m 3/2 1 k c 1/2 1 w. ( 35
)
Equation ( 35) is the kind of bound we were looking for, provided a B 1. Though this will be the case in practice (since we are mainly interested in cases where ν is close to µ and thus μ * is close to μ), this is not quite satisfactory yet. So, what can we do when a B 1, that is, when μ * B 1 μ 1 ? In fact that case is easier, because transportation between small measures has low cost, while w has to be large to make μ * B very different from μ. The computations are the following. First, it is obvious that

W 2 (z) = W 2 (μ * B , a B μ) 2R μ * B 1/2 1 . ( 36
)
[ §] This step comes from the computation λ(B)

1/2 λ L 2 (B) / λ 1 = ( ´1 0 r n-1 dr) 1/2 × ´1 0 (1 -r) 4 r n-1 dr 1/2 / ´1 0 (1 -r) 2 r n-1 dr = 6(1 + n)(2 + n) / (3 + n)(4 + n) 1/2 √ 6 ∀n.  W 2 distance and Ḣ-1 norm Rémi Peyre Next, observing that ϕ(f (x)) c 1 c 2 ϕ(x) -2c 1 dist(x, B c )|T (x)|, [ ¶] we compute that μ * B 1 = ˆB ϕ(f (x))µ(dx) ˆB c 1 c 2 ϕ(x) -2c 1 dist(x, B c )|T (x)| µ(dx) c 1 c 2 μ 1 -2c 1 ˆB dist(x, B c ) 2 µ(dx) 1/2 ˆB|T (x)| 2 µ(dx) 1/2 = c 1 c 2 μ 1 -2c 1 dist(•, B c ) 2 • µ 1/2 1 w c 1 c 2 μ 1 -2c 1 m 1/2 2 λ 1/2 1 w, (38)
whence

w c 1 c 2 μ 1 -μ * B 1 + 2c 1 m 1/2 2 λ 1/2 1 = c 1 c 2 -a B + μ 1 2c 1 m 1/2 2 λ 1/2 1 m 1/2 1 2c 1/2 1 m 1/2 2 c 1 c 2 -a B + μ 1/2 1 . (39) So, W 2 (z) 2R μ * B 1/2 1 = 2Ra 1/2 B μ 1/2 1 4Rc 1/2 1 m 1/2 2 m 1/2 1 a 1/2 B ( c 1 c 2 -a B ) + w. (40)
In the end, choosing either (35) if a B c 1 / 2c 2 or (40) if c 1 / 2c 2 , and observing that c 1 kR -1 , one has always:

W 2 (z) (4 √ 3 + 2 √ 2)C 1 (n) 1/2 ∨ 4 √ 2 c 3/2 2 m 3/2 2 c 3/2 1 m 3/2 1 k c 1/2 1 w. (41)
Remark 15. To bound W 2 (z) in the situation where a B 1, we could also have started from "ϕ(f

(x)) ϕ(x) -k|T (x)|" (instead of "ϕ(f (x)) c 1 c 2 ϕ(x) - 2c 1 dist(x, B c )|T (x)|"
) to get another bound analogous to (38). Following such an approach, the factor (c 2 / c 1 ) 3/2 in (40) would be improved into (c 2 / c 1 ) in the analogous formula; however the dimensional factor would behave in O(n) rather than in O(n 1/2 ).

Proof of Lemma 14

It still remains to prove Lemma 14, whose statement we recall to be:

Lemma. Denoting λ := dist(•, B c ) 2 • λ,
one has, for any signed measure m on B having total mass zero:

m Ḣ-1 ( λ) (2e + 1)n -1 ∨ 8e 1/2 m L 2 (B) . (42)
-In the sequel, " (2e + 1)n -1 ∨ 8e" will be shorthanded into "C 1 (n)".

Remark 16. The bound (42) is within a constant factor of being optimal, uniformly in n, as one sees by taking a linear function f in (45).

[ ¶] This follows from the computation:

ϕ(f (x)) c1 dist(f (x), B c ) 2 c1 dist(x, B c ) -|T (x)| 2 + c1 dist(x, B c ) 2 -2c1 dist(x, B c )|T (x)| c1 c2 ϕ(x) -2c1 dist(x, B c )|T (x)|. (37)  W 2 distance and Ḣ-1 norm Rémi Peyre
Proof of the lemma. We begin with translating the lemma into a functional analysis statement by a duality argument. Recall the duality definition of m Ḣ-1 ( λ) from § 1:

m Ḣ-1 ( λ) := sup{| f, m | | f Ḣ1 ( λ) 1}. (43)
There is a similar duality formula for m L 2 (B) :

m L 2 (B) = sup{| f, m | | f L 2 (B) 1}, (44)
where, for f a function, f L 2 (B) has its usual meaning, namely

f L 2 (B) := ´B f (x) 2 λ(dx) 1/2
. Since m is assumed to have total mass zero, | f, m | does not change when one adds a constant to f . On the other hand, when f describes the set

{ f 0 + a | a ∈ R}, f L 2 (B)
is minimal when a is such that f has zero mean on B, while the value of f Ḣ1 ( λ) remains constant. [ ] As a consequence, we can restrict the supremum in ( 43) and (44) to those f having zero mean on B. Thus, the lemma will be implied [ * * ] by proving that

f, 1 B • λ = 0 ⇒ f L 2 (B) C 1 (n) 1/2 f Ḣ1 ( λ) . ( 45
)
Going back to the definitions of • Ḣ-1 ( λ) and • L 2 (B) , relaxing the condition on f to be centred by projecting it orthogonally in L 2 (B) onto the subspace of centred functions, and denoting by P the uniform probability measure on B, Equation (45) turns into:

∀f Var P (f ) C 1 (n) ˆdist(x, B c ) 2 |∇f (x)| 2 P (dx), (46)
which we recognize to be a so-called "improved Poincaré inequality" [START_REF] Boas | Integral inequalities of Hardy and Poincaré type[END_REF][START_REF] Hurri-Syrjänen | An improved Poincaré inequality[END_REF]. In general, Poincaré inequalities, bounding the variance of f by a quadratic integral of its first derivative, are linked with the exponential convergence of a certain diffusion Markov process towards equilibrium (cf. [1, Chap. 2]): that probabilistic vision initially guided me to tackle Equation (46), although this will not be apparent in the sequel.

To prove (46), the first key idea (inspired by [START_REF] Bobkov | Spectral gap and concentration for some spherically symmetric probability measures[END_REF]) is to separate radial and spherical coordinates. This is, considering the bijection

ϕ : (0, R) × S n-1 → B {0} (r, θ) → rθ (47)
(the origin of space being set at the center of B), we introduce the measure P := ϕ -1 * P , which is obviously the product measure Pr ⊗ Pθ , where Pr is the probability measure on (0, R) such that Pr (dr) := nR -n r n-1 dr, resp. Pθ is the uniform measure on the sphere S n-1 . With this notation, we perform can a change of variables to see that ( 46) is equivalent to proving that, for all g ∈ L 2 ( P ):

C 1 (n) -1 Var P (g) ˆR 0 ˆSn-1 (R -r) 2 |∇ r g(r, θ)| 2 + r -2 |∇ θ g(r, θ)| 2 Pr (dr) Pθ (dθ), (48)
where ∇ r and ∇ θ denote the gradient along resp. the r coordinate and the θ coordinate. [ † †] We will denote the right-hand side of (48) by E(g, g).

[ ] Here we implicitly assume that ´B|f (x)|λ(dx) < ∞, which is legitimate since an approximation argument allows to restrict the suprema in ( 43) and (44) to those f having a C ∞ continuation on cl(B).

[ * * ] Actually there is even equivalence.

[ † †] In the latter case, we have to use the Riemannian definition of the gradient on S n-1 .



W 2 distance and Ḣ-1 norm

Rémi Peyre

Because P = Pr ⊗ Pθ , we know that L 2 ( P ) can be seen as (the closure of) the tensor product of L 2 ( Pr ) and L 2 ( Pθ ):

L 2 ( P ) = cl(L 2 ( Pr ) ⊥ ⊗ L 2 ( Pθ )), (49)
where the symbol ' ⊥ ⊗' means that the Hilbertian structure of L 2 ( P ) is compatible with the Hilbertian structures of L 2 ( Pr ) and L 2 ( Pθ )-i.e., that h

a ⊗ u a , h b ⊗ u b L 2 ( P ) = h a , h b L 2 ( Pr) × u a , u b L 2 ( Pθ )
. Now consider the spherical harmonics Y 0 , Y 1 , . . ., which by definition are an orthonormal basis, in L 2 ( Pθ ), of eigenfunctions of the Laplace-Beltrami operator Δ on S n-1 ; and call 0 , 1 , . . . the associated eigenvalues, which are known to be such that (up to permuting indices) Y 0 ≡ 1 with 0 = 0, and i -(n -1) ∀i = 0 (see for instance [START_REF] Seeley | Spherical harmonics[END_REF]). By construction,

L 2 ( Pθ ) = cl ⊥ i∈N (R • Y i ) ; therefore, one has that L 2 ( P ) = cl ⊥ i∈N L 2 ( Pr ) • Y i : (50)
in other words, the functions of L 2 ( P ) are those of the form

g(r, θ) = i∈N h i (r)Y i (θ), (51)
with i h i 2 L 2 ( Pr) < ∞, and the correspondence is bijective. An interesting point is that, then, one has:

Var P (g) = Var Pr (h 0 ) + i =0 h i 2 L 2 ( Pr) . ( 52
)
On the other hand, one has

E(g, g) = -Lg, g L 2 ( P ) , (53)
where

Lg (r, θ) := (R-r) 2 Δ r g+ (n-1) (R -r) 2 r -2(R-r) e r •∇ r g+ (R -r) 2 r 2 Δ θ g. (54)
From (54) we see that, since the Y i are eigenfunctions of Δ θ , all the L 2 ( Pr ) • Y i are invariant by L, and that one has:

E(g, g) = i∈N ˆR 0 (R -r) 2 |∇h i (r)| 2 -i (R -r) 2 r 2 h i (r) 2 Pr (dr). (55)
So, proving (48) becomes equivalent to proving that both following formulas hold for all h ∈ L 2 ( Pr ):

Var Pr (h) C 1 (n) ˆR 0 (R -r) 2 |∇h(r)| 2 Pr (dr); (56)
h 2 L 2 ( Pr) C 1 (n) ˆR 0 (R -r) 2 |∇h(r)| 2 + (n -1) (R -r) 2 r 2 h(r) 2 Pr (dr). (57)
Let us start with (56). In all the sequel of the proof, we introduce

b := 1 -n -1 . ( 58
)
By the Cauchy-Schwarz inequality, one has, for all r ∈ (bR, R):

h(r)-h(bR) 2 = ˆr bR h (s)ds 2 ˆr bR (R-s) -3/2 ds × ˆr bR (R-s) 3/2 |∇h(s)| 2 ds 2 (R -r) -1/2 -(R -bR) -1/2 ˆr bR (R -s) 3/2 |∇h(s)| 2 ds 2(R -r) -1/2 ˆr bR (R -s) 3/2 |∇h(s)| 2 ds. ( 59
)
Integrating and using Fubini's formula, it follows that

ˆR bR h(r) -h(bR) 2 Pr (dr) 2 ˆR s=bR ˆR r=s nR -n (R -r) -1/2 r n-1 dr (R -s) 3/2 |∇h(s)| 2 ds 2 ˆR s=bR ˆR r=s nR -n (b -1 s) n-1 (R -r) -1/2 dr (R -s) 3/2 |∇h(s)| 2 ds = 2b -(n-1) ˆR s=bR ˆR r=s (R -r) -1/2 dr (R -s) 3/2 |∇h(s)| 2 Pr (ds) = 4b -(n-1) ˆR s=bR (R -s) 2 |∇h(s)| 2 ds. ( 60
)
One can apply the same line of reasoning for r ∈ (0, bR): the (unweighted this time) Cauchy-Schwarz inequality then yields h(r) -h(bR) ˆr br (R -s) The second term of the right-hand side of (65) is itself bounded by 4b -(n-1) ˆR s=b 2 R ˆR r=s (R -r) -1/2 dr (R -s)

Combining (66), ( 67) and (69), we finally get the wanted bound (57).

Remark 17. At the time I wrote that proof I was not aware of the already existing results on improved Poincaré inequalities, in particular [6, Theorem 1.3], which Equation ( 46) is actually a particular case of; nor were the people whom I had asked about such inequalities. Compared to the result of [START_REF] Hurri-Syrjänen | An improved Poincaré inequality[END_REF] however, my Equation (46) states an explicit value for the constant in the inequality, which moreover is within a constant factor of being optimal, uniformly in the dimension n; also, it uses a quite different proof, which may be interesting

)

  Proof. Consider the measures µ • := Φ * µ and ν • := Φ * ν, and the bump function ϕ • := ϕ • Φ -1 ; then, µ • , ν • and ϕ • satisfy the original assumptions of Theorem 10, the roles of 'm 1 ' and 'm 2 ' (in the ball situation) being held by m 1 and m 2 (in the general situation) themselves, the role of 'k' being held by Φ -1 Lip k, and the roles of 'c 1 ' and 'c 2 ' being held by c 1 / Φ 2

y

  Denoting a c := μ * c 1 / μ 1 , we then transform μ * c into a c μ according to an arbitrary transference plan. z Finally, denoting a B := μ * B 1 / μ 1 , [ * ] we transform μ * B into a B μ according to the optimal transference plan: the cost of this operation is W 2 (μ * B , a B μ), which we bound above by 2 μ * B -a B μ Ḣ-1 (a B μ) thanks to Theorem 1.

2 ( 2 ˆbR 0 |∇h(s)| 2

 2202 bR -r) × ´bR r |∇h(s)| 2 ds, whence: ˆbR 0 h(r) -h(bR) 2 Pr (dr) ˆbR s=0 ˆs r=0 nR -n (bR -r)r n-1 dr |∇h(s)| 2 ds bR -(n-1)ˆbR s=0 ˆs r=0 nr n-1 dr |∇h(s)| 2 ds = bR ˆbR 0 |∇h(s)| 2 s n ds bn -1 R Pr (ds) b(1 -b) -2 n -1 ˆbR 0 (R -s) 2 |∇h(s)| 2 Pr (ds).(61)Summing (60) and (61), we get thatˆR 0 h(r) -h(bR) 2 Pr (dr) 4b -(n-1) ∨ b(1 -b) -2 n -1 ˆs 0 (R -s) 2 |∇h(s)| 2 Pr (ds),(62) where 4b -(n-1) ∨ b(1 -b) -2 n -1 can itself be bounded by (n -1) ∨ 4e . The left-hand-side of (62) being an upper bound for Var Pr (h), this proves (56). Now we turn to (57). For r ∈ (bR, R) we have, similarly to (59), that h(r) -h(br) 2 2(R -r) -1/2

  3/2 |∇h(s)| 2 ds, 2 2h(br) 2 + 4(R -r) -1/2 ˆr br (R -s) 3/2 |∇h(s)| 2 ds. (R -r) -1/2 dr (R -s) 3/2 |∇h(s)| 2 ds. (65)By change of variables, the first term of the right-hand side of (65) is equal to 2b -n ´bR b 2 R h(s) 2 Pr (ds), which we can bound by

	W 2 distance and Ḣ-1 norm			Rémi Peyre
	so that
	h(r) (64)
	Then, integrating and applying Fubini's formula:
	ˆR bR nR -n r n-1 2b -(n-2) (1 -b) -2 h(r) 2 Pr (dr) 2 h(br) 2 Pr (dr) + ˆR bR 4 ˆb-1 s∧R ˆR s=b 2 R r=s∨bR n -1 ˆbR b 2 R (n -1) (R -r) 2 h(s) 2 r 2 2ne ˆR 0 Pr (ds) (n -1)	(R -r) 2 r 2	h(s) 2	Pr (ds). (66)
				(63)
				

  3/2 |∇h(s)| 2 Pr (ds) 8e ˆR 0 (R -s) 2 |∇h(s)| 2 Pr (ds).(67)This way, we have bounded ´R bR h(r) 2 Pr (dr). On the other hand, it is trivial that, for r bR,

		h(r) 2	b 2 (n -1)(1 -b) 2 × (n -1)	(R -r) 2 r 2	h(r) 2 ,	(68)
	whence:
	ˆbR 0	h(r) 2	Pr (dr) (n -1)	ˆR 0	(n -1)	(R -r) 2 r 2	h(r) 2	Pr (dr).

Acknowledgements. I warmly thank Franck Barthe for his much precious help in providing me with the technical tools for the proof of Lemma 14. I am also much grateful to the anonymous COCV referee for their very relevant remarks on my initial manuscript.

* Supported by the Austrian Science Fund (FWF) under grant P25815.


Staging the 'devoted' Mr Amanjit Singh in Upstairs, Downstairs (BBC1, 2010-2012)

Florence CABARET, ERIAC, Rouen University

The recent BBC TV series Upstairs, Downstairs, which is a reboot of a 1970s TV series taking place between 1903 and 1930, i starts in London's prestigious Belgravia in 1936, with the arrival at n°165 Eaton Place of a young couple of new owners, Sir and Lady Holland, who have just returned from the Unites States where Sir Hallam worked as a member of the British embassy. His wife Lady Agnes is soon shown as going to a domestic employment agency to recruit a set of six servants as the couple needs to entertain diplomats and other prominent people. Created by Call the Midwife screenwriter Heidi Thomas, Upstairs, Downstairs is very often compared to Downton Abbey (ITV, 2010(ITV, -2015)): ii indeed, along with films such as Gosford Park (2001), TV series like Downton Abbey and Upstairs, Downstairs are quite typical of a fairly recent body of British fictional filmic and TV production which tends to provide a greater visibility to "life below stairs". Indeed, if "characters from below" were already present in 18 th c. novels, they have come back to the fore as protagonists and narrators with Ishiguro's Remains of the Days (1989) and its butler narrator, iii or with Jo Baker's recent Longbourn (2013) which rewrites Pride and Prejudice from the point of view of Elizabeth Bennet's housemaid.

Nevertheless, in spite of a symmetrical title that may sound as if it were suggesting a form of quantitative equality in the representations of characters from both upstairs and downstairs, the TV series tends to focus more on the upstairs protagonists as they actually drive the narrative arc, which mainly relies on the interweaving of the national history of Great Britain on the eve of World War II and of the household stories related to Sir Hallam, who works for the British government and Foreign Affairs. Yet, we will see that the figure of Mr Amanjit, a personal secretary who belongs neither to upstairs nor to downstairs, is construed as a Gramscian "subaltern" by the TV series in so far as his characterization remains fragmentary and elliptic while he simultaneously turns out to participate in the national history of Great Britainso that he also happens to share the limelight with his masters on some occasions. Quite interestingly, it is the female masters' involvement with world politics that leads to the introduction of the character of Mr Amanjit Singh, who arrives at Eaton Place along with Lady Maud Holland, who has been living in India for over thirty years on account of the diplomatic activities of her recently deceased husband. Now a widow, Lady Maud returns to England to live with her son (Sir Hallam) and she arrives in the first episode with an urn, a monkey and a Sikh Indian personal secretary (Mr Amanjit), who is there to help her write her memoirs after being at her service in India for some years already.

In his case, the server-served relationship is therefore reduplicated by the colonizer-colonized relationship at a time when India still belonged to the British Empire and I shall bear in mind here how this specific status of his influences the kind of relationship he may have with the Holland family.

We shall first see that, however personal a secretary Mr Amanjit may be, no sign of any form of intimacy, in the sense of physical proximity or emotional closeness, is ever exchanged between him and his mistress, nor with the Holland family at large. In the TV series, this obvious physical remoteness of Mr Amanjit's presence is all the more foregrounded by the elliptic visual and narrative functioning of the show, a characteristic that appears to relegate the body of the Indian secretary off-screen, or in the background, or in the margins of many shots in the first half of season 1. Yet, what may be regarded as a form of physical distance on the parts of both masters and personal secretary is actually countered by a strong form of reciprocal loyalty, leading the two parties to engage with matters that entail an individual and collective dimension, or, to put it differently, an intimate and a national dimension, which is made visually perceptible on the screen as well. So that more than being a mere "rhetorical doubling" (Robbins 1986: x) of the master protagonists, Mr Amanjit triggers subplots that eventually become master plots as the masters themselves join him or support him in his enterprises, most of them turning to be political enterprises. Still, if the series promotes a form of growing social and public agency of the character of Mr Amanjit, it is obvious that his autonomy remains possible within the confines of the masters' supervision and paternal/maternal companionship -which can also be read in a broader perspective as a contemporary interrogation on would-be "model migrants" from the former Commonwealth as both self-governing and faithful, instead of dependent, or rebellious and potentially threatening.

Mr Amanjit is attached to the person of Lady Maud for his intellectual qualities, which ranks him higher than all the other servants who are circumscribed to physical work in the house, and which also exemplifies the degree of education and mastery of oral and written English a certain number of Indian people could display as descendents of "Macaulay's children". His relative superiority first sets Mr Amanjit apart from the rest of the servants and confines him to Lady Maud's study-room where he types the memoirs she dictates to him on a daily basis. One may note that memoirs point to the fairly public dimension of her narrative, which is not an autobiography, so that Mr Amanjit is never allowed into more than an account of a personal viewpoint on experiences that are never intimate. What's more, only one scene in the whole of season 1 actually stages him working with Lady Maud, while the few scenes when he is shown interacting with her leave him literally in the background as he takes care of Lady Maud's monkey or brings her objects she requires. So, at first, no conversation proper is ever shown between the two characters and the one and only working scene of season 1 in the study room confines Mr Amanjit to a silent role as he types Lady Maud's oral story without interfering in any way in her text. Ironically though, and proleptically, Mr Amanjit is later shown tuning in the wireless for the Hollands in the drawing room and playing the piano at one of their parties (S01E01), so that the series stages him as a discreet presence, whose own personal voice is not heard yet in discussions, but who is constantly heard as the physical provider of news information or as the final musical touch to a social gathering of political acquaintances. It may be tempting to describe such a mute role as that of the jewel in the crown of the Holland family, which it is undeniably -but only up to a certain point. Generally speaking, the TV series never clarifies the conditions of Mr Amanjit's recruitment, nor his legal status as a citizen of the Commonwealth working in Great Britain. We may assume that he is employed as a bonded servant to Lady Maud, but we are never explained how he is entitled to remain at the service of the young Holland couple when Lady Maud dies between season 1 and season 2. The fact that her death takes place in a narrative ellipsis may account for the reinforced vagueness of his subsequent official status. While the recruitment of the other servants is the object of a substantial number of scenes in the show, silence prevails as far Mr Amanjit's presence is concerned. This somehow instils the idea that he is part of Lady Maud's heritage --not as an object though, but rather as a sort of foster family member whose involvement in household matters will prove more and more central.

Still, "family member" remains an inappropriate term as Mr Amanjit never shares his meals with the Hollands, just like any of the other servants, with whom he is also forbidden to eat by Lady Maud who wants him to be served in her study room by one of the maids. His inbetween status in the house, as belonging neither completely to upstairs nor to downstairs, implicitly points to his ethnic specificity, even though once again this is never given as a reason of the physical distance that he feels he must maintain with his masters, but also with the servants. Interestingly, the fact that his Indian origins is an accepted fact that keeps him apart from the other servants is unveiled thanks to a conversation he has in the middle of season 1 with a new maid, Rachel, who reveals to him that, as a Jew, she had to resign from her position as a university professor in Frankfort so as to flee Germany with her daughter.

Her explicit speech about the racism she suffered from may be read as an indirect comment on his own situation as a character who is relatively marginalised on account of his ethnic background. So Rachel's own in-between status as a demoted ethnic character enables them to share an intimacy that would otherwise have been hindered. Thanks to her arrival at Eaton Place, we learn about private details concerning Mr Amanjit's past family life as well as geographical and social provenance. Quite tellingly, the first time Rachel appears on the screen illustrates this shift of focus visually operated by the camera (S01E02): Mr Amanjit is proof-reading Lady Maud's text and is shown on the left-hand side of the shot, in shallow focus though standing in the foreground. As he turns round to look at the newly-arrived Rachel, the focus on his face sharpens and attracts our attention to his growing relevance for the plot, considering that Lady Maud remains at the centre of the shot, even though blurred in the background and slightly pushed to the right of the shot. Yet, the circumstances that led to both his wife and child's deaths remain unknown to the spectators, pointing once again to the show's narrative elliptical functioning as far as servants are concerned, literally illustrating Antonio Gramsci's Prison Notebooks when he states that "The history of subaltern social group is necessarily fragmented and episodic" (Gramsci 1971: 54), and that the intertwining with the history of "civil society and thereby with the history of States and groups of States" is another typical trait of the subaltern's intervention in world affairs, as we shall see in a second time.

But, to come back to Mr Amanjit and his lack of obvious intimacy with his masters, one may underline that if no physical contact is ever shown between himself and the Holland family, iv a subtle emotional proximity is made obvious when, at the opening of season 2, Mr Amanjit is seen holding in his hands the urn encasing the ashes of his dead mistress which he brings back at Eaton Place, as a son could do. This moment also mirrors the situation of Lady Maud at the beginning of season 1, when she puts the urn of her late husband on the mantelpiece of her study room while referring to the Sikhs' conception of death as a passing though fire. The gradual blurring of boundaries between his own body and bodies from a higher social class is also discreetly hinted at in several scenes with Rachel, the former German teacher (when she hangs the laundry in the courtyard just like his mother used to, when she is about to faint in the study room, when he accompanies her in the crowd scene during the anti Oswald Mosley march, and when he eventually discovers Rachel's dead body in her bedroom, in S01E02), and later with Rachel's daughter Lotte (whom he appears to physically shelter when she is brought back at the Hollands' after her mother's death in S01E02 & E03), or again with Mrs Fuller, Lotte's headmistress (with whom he takes a photograph in the street with Lotte in between them -standing together as a close family group in S02E04). Thus, the TV series gives us to understand that Mr Amanjit is not limited to having physical caring contacts with animals, such as when he "protects" Lady Maud's pet monkey from the fright of the maids (S01E01) or when he tends to the little abandoned bird that he finds with young Johnny (S01E01). Actually, far from reducing him to the animal level, these two scenes also point to the range of his abilities, which are not limited to typing and proof-reading skills, and to some of his moral qualities such as self-control when confronted to panic and sympathy when confronted to the destitution of a weaker being. This moral dimension of his character is definitely the one that is most invested by the TV series, and the domain where his interactions with master characters are made most manifest through a combination of more frequent physical presence in the shots and gradual integration in the dialogues between himself and the master characters.

Mr Amanjit is thus staged as evolving from being a fairly invisible, self-effaced coloured individual to becoming a growingly dependable, self-asserting member of the household who reaches centre stage and becomes more visible on the screen. Indeed, even though he remains a secondary character like most of the servants, he is one of those who are most actively in touch with « upstairs » and one who tends to defend respect and loyalty to the masters more eagerly than the other servants. Oddly enough, while some spectators discussing the show on blogs and forums find Mr Amanjit too ill-defined so that the TV series could well have done without him, I think that this character, who is obviously not designed to attract the viewer's immediate sympathy -contrary to the glamorous chauffer, the boisterous cook, the faulty young Johnny or the hard-working butler-is conceived so as to embody a form of moral devotion to his masters, who are also promoted as acting loyally towards him. One may think again of the opening of season 2, which shows Mr Amanjit "coming back home" at Eaton Place with the urn, so that we infer Sir Hallam has kept him at their service with no further explanation, as if the TV series took for granted that Mr Amanjit's unswerving faithfulness to his mother's most whimsical and demanding attitudes were all to his credits and placed Sir Hallam in the moral obligation not to terminate his working contract. But it is true that this is left to the spectator's interpretation, which may be read as one of the show's drawbacks as far as the fleshing out of this master-servant relation is concerned, or which may also be read as reflecting a relation which the TV series constructs as based on implicitness, reserve and a form of reciprocal trust which goes without saying. Indeed, as the TV series progresses, Mr Amanjit is being entrusted with missions (rather than duties) testifying to the growing confidence he is attributed by Sir Hallam, and later by his aunt Blanche, but also by Rachel and by Mrs Fuller. Some of his missions are directly connected to domestic services, but each time they reveal a commitment that goes beyond Mr Amanjit's professional obligations and puts him on an equal moral footing than some master characters, and sometimes on a superior footing. v However, the TV series also offers a nuanced portrayal of a man who may not have the choice of his morality because of his inferior position as a servant. We may think of the moment when he replaces the missing chauffeur at Sir Hallam's request and secretly drives back Sir Hallam's step sister Persephone from prison. If this intervention is a means to avoid the shame of having a family member related to an infamous detention, it also illustrates a conflict of loyalties that Mr Amanjit cannot but submit to as Rachel was arrested for taking part in the march organised by Oswald Mosley's right-wing party whereas Mr Amanjit had joined the counter march with Rachel.

Conversely, another scene underlines the sense of adequation between personal and professional commitment, when Lady Agnes asks Mr Amanjit to replace the butler, Mr Pritchard. The latter has fallen back to alcoholism after being jilted by a lady friend once she discovers that he was a conscience objector during World War I, leaving others to sacrifice their lives on the battlefield while he was "only" an ambulance man. An earlier conversation between Mr Amanjit and Sir Hallam, which is one of the rare actual discussions they have together while alone, had enabled Mr Amanjit to disclose that he was a soldier in the Indian troops that fought along the allies in France during World War I. He then confides to Sir Hallam that he was in the Jallundur 59 th Brigade and confirms him that he was wounded at Ypres, where most of the Indian soldiers involved in this battle died or were injured. This moment of man-to-man exchange, and even manly intimate confidence on Mr Amanjit's part, is inserted just after Sir Hallam has decreed the personal secretary to be the only guardian of his mother's letters, journals and papers which his aunt was trying to go through on her own (S02E01). This first sign of great trust on Sir Hallam's part is immediately followed by another similar moment when Sir Hallam hands him the gun he used to carry when he was in the navy, so as to protect the household in the event of a German invasion taking place in his absence (S02E01). So the TV series chooses here to foreground Mr Amanjit as not only replacing male key figures downstairs, but also potentially replacing the master figure as moral and physical defender, both of the written heritage of his mother and of his entire household. This agency by proxy, which is a conventional motif of patronizing colonial history and fiction, is clearly a way of foregrounding the master's wisdom and sense in the prospect of circumstances that will radically alter the traditional order of things. But, as season 2 draws to a close, the character of Mr Amanjit is no longer only depicted as reliable, discreet and selfforgetful since a form of empowerment, triggered by his own self-confidence and his masters' respect and trust, makes him more enterprising and influential, and no less dependable. The moral and political sharing of common values thus appears to be the basis of an intimacy which is a subtle mixture of physical and intellectual trust as far as his masters are concerned, rekindling in this contemporary TV series the ideal of the colonized subject as both trustworthy and self-reliant, even in, or especially in, situations of crisis.

Still, the growing agency of the character of Mr Amanjit remains under control and operates from the heart of a household that has adopted him, from the centre of the British Empire and close to the heart of the British government. Thus, if two other scenes relegate Sir Hallam in the wings of the action in order to favour Mr Amanjit's initiatives, these initiatives are clearly backed by the master figure. Such is the case of Mr Amanjit's search for Lotte's German father, for which he writes many letters to try and find out the prison where he was sent to, even though this quest proves fruitless. The other example is that of Mr Amanjit's participation in the organisation of a network of refugee Jewish children along with the headmistress Mrs Fuller, as well as with Blanche with whom he reaches an understanding after their quarrel, and also with Lady Agnes who contributes to raise funds for the network while Sir Hallam deals with the required entrance visas (S02E02). Thus, on several occasions starting at the end of season 1, Mr Amanjit is staged in short moments of political conversations where his ideas are opposed, debated, or willingly accepted by the strongwilled and independent character of Blanche in particular, underlining his decisive participation in the group of master characters. Some shots visually enhance this newly acquired status, as when he sits side by side with other volunteers looking for host families, or finds himself elected by the camera among the British people waiting for the refugee children at the strain station (S02E02), or again when Lady Agnes is shown literally looking up to him as he advises her about the servants' new management (S02E03).

However, the equal position he gradually acquires with the masters of the house is toned down by two other very different episodes, and the portrayal of the character once again appears fairly balanced, however sketchy it remains. When Mr Amanjit's excess of zeal, as he himself calls it, throws him into a temper and an argument with the butler about what it means to refuse to wage war, he starts practising shooting in the back garden so that both Blanche and Lady Agnes are compelled to intervene and momentarily disarm him (S02E01). In this case, the female masters' agency can be interpreted as exposing Mr Amanjit's own agency as maybe too precocious and potentially dangerous (a replica of Great Britain's overall position about Indian nationalists' growing demand for self-government?). Yet, in the same episode, Mr Amanjit, acting as butler with Sir Hallam's guests, is bluntly rebuffed by Stanhope, the Foreign Affairs Secretary, who considers a servant (again, he does not mention the fact that he is Indian, but he probably does not need to) has no right to participate in a political conversation. Blanche later supports Mr Amanjit, underlining their respective status as "outsiders" and "misfits" who yet manage to achieve things together, but she does not dare say it in Stanhope's face (S02E01). This time, as her name conspicuously underscores, it is not so much Blanche's ethnic dimension which makes her side with Mr Amanjit but rather her inordinate situation as a middle-aged single working woman who had a homosexual affair with a married woman and mother of two. The frontal confrontation with a "white master" raises a question which shakes Mr Amanjit's tendency to defend loyalty to masters over servants' professional claims in a subsequent exchange with Mr Pritchard, the butler he replaces: Mr Amanjit: "It's not easy to bite your tongue when hearing such stupidity. [referring to Stanhope's erroneous worldview]" Mr Pritchard: "A butler hears nothing." Mr Amanjit: "I don't understand. (…) I'm a servant but before that I'm a man." Mr Pritchard: "So you have no place downstairs." (S02E01)

This passage is interesting in so far as it is colour-blind and appears to state that social division is stronger than racial hierarchy, that notions of social superiority and inferiority may be more accepted by servants themselves (also a well-known motif of integrated alienation), but it also points to the position of a white servant speaker who feels less threatened by his dependence in a white society than a coloured servant in this very same white society. A last example indeed reminds Mr Amanjit that the outside world is not as benevolent as the Holland masters, when he is asked by a waiter in a tea parlour to move from the front room to the back room as there is no place for him in the front room. It is one of the rare moments when the character of Mr Amanjit is so explicitly reminded (by a member of the domestic servants, who may act on his own accord and/or on account of the owner of the place) that he is Indian-born and that his presence visually spoils the place.

Thus Mr Amanjit gradually appears as a quasi flawless model of devoted colonial servant embodying the benefits and advantages of self-government and collaboration for the good of Great Britain. He is thus depicted as a servant who generally embraces the cause of Great Britain but remains oblivious to what was going on in India in the 1930s, at a time when frustrated claims of home-rule had given way to demands for total independence and partition, especially after the rejection of the 1935 Government of India Act. Indeed, Mr Amanjit appears to never question the option of Indians supporting the war effort or not.

Added to the fact that the TV series represents a Sikh Indian character rather than a Hindu or a Muslim character, this choice may be a way to overlook the internal divergences that opposed the Indian National Congress (with a Hindu majority) to the Muslim League as to the renewal of India's participation to a second world war, and to recycle the now well-known presence of a majority of Sikh soldiers in the British army during the two world wars. This reality contributed to fuel the stereotypical figure of the fierce Sikh warrior who does not betray, going back to the numerous Sikh soldiers who remained faithful to the British colonizers during the 1857 Sepoy Mutiny. On the eve of the 2014 commemorations of the commitment of Indian troops in World War I celebrated with cultural and historical events such as conferences, exhibitions and monuments, the TV series turns the character of Mr Amanjit into a complex fetish object/subject of mutual trust between colonised servant and master colonizer. What's more, in a post-9/11 and 7/7 context, the TV series also reminds its viewers of the crucial supportive role played by loyal colonial subjects in the defence of Great Britain and its allies during the two world conflicts -somehow inviting spectators to question contemporary suspicion as regard descendants of these idealized devoted servants and fighters now demonized by some as potential "enemies from within", whether they be Muslims, Hindus or Sikhs.

Upstairs, Downstairs can thus be viewed as retracing the first moments of a British Asian diaspora in the making, at a time when emigrating to Great Britain was still a singular enterprise -just before the Partition of India and Pakistan and the post-war reconstruction era in Great Britain initiated a wider movement of migrations of workers and technicians, who were later followed by their families. By reminding TV viewers that their recent national and European history was the result of numerous cooperations based on trust and common moral and political values between colonizers and colonized, between served and server, the show also provides positive images of an Indian servant as a steadfast supporter and protector of Great Britain, like many others in his time, and many others today. What's more, by turning Mr Amanjit into a dedicated and reliable typist, not departing from his mistress's text and the master text of the heydays of Great Britain as a nation that still counted in the balance of world power, Upstairs, Downstairs gives us to watch a character who paves the way for the emancipation of Indian writing from the master's dictation -as if Mr Amanjit stood as a precursor of those emancipated writers and artists whose own personal voices started to be heard and seen in British and world culture fifty years later. For that matter, it is relevant that Mr Amanjit should follow in the steps of an independent woman writing about her own perceptions of diplomatic affairs in India, which was still regarded as a male sphere of influence in spite of British women's growing involvement in political life. Quite tellingly, one shot of a page being typed by Mr Amanjit reveals a text in which Lady Maud appears to be quite aware of the intricacies of the domestic and the political when she writes that "'Domestic space' undoubtedly represented politicised space." (S01E02) -a fleeting but undeniable subtext for the TV series itself where the upstairs/downstairs domestic space and its inhabitants are both literally and metaphorically politicised.

As such a progressive image of the master-servant / coloniser-colonised relationship is clearly idealized on both the master's and the servant's parts by the series, it appears to me that the TV series tries -and partly fails-to write back to the 1980 British tradition of TV Raj revival films by recycling the "heritage film" fashion so characteristic of British fiction programs in the 1980s. Indeed, Mr Amanjit is depicted as the embodiment of a successful and fruitful outcome of the colonizer-colonized relationship that evolves into a partnership which remains unequal but which turns out to be beneficial for both parties. In that respect, Upstairs, Downstairs favours a positive representation of the consequences of both colonization and "deserved" gradual emancipation, which flatters a certain British view and nationalist Indian view of India's independence as "a gentlemen's agreement." Simultaneously, the TV series opts for a geographical shift of perspective, showing that before 1947, Indian people were no longer exotic foreigners leaving abroad (as in Raj revival films) but that they had started to settle in Great Britain and become close neighbours -reminding us that the history of Indo-Pakistani migration to Great Britain is a progressive one, initiated as much by the British as by the Indo-Pakistani people themselves. But this perspective was already introduced on British big and small screens by such script writers as Hanif Kureishi in My Beautiful Laundrette (1985) or The Buddha of Suburbia (BBC, 1993) -so that Upstairs Downstairs Gramsci, Antonio (1971). Selections from the Prison Notebooks, London: Lawrence and Wishart.

Robbins, Bruce (1986Bruce ( , 1993)). The Servant's Hand: English Fiction from Below, Duke University Press.

Sen, Indrani (2010 ). "Discourses of gendered 'loyalty'. Indian women in nineteenth century 'mutiny ' fiction" (pp. 111-127) in Biswamoi Pati, The Great Rebellion in India in 1857: Exploring Transgressions, Contests and Diversities, Routledge.

i Upstairs Downstairs (ITV, 1971(ITV, -1975) ) consists of 68 episodes divided into 5 series.

ii Downton Abbey happened to be launched a few weeks only after the first episode of Upstairs, Downstairs and, in spite of the great popularity of the original series broadcast on ITV between 1970 and 1975, Downton Abbey proved to be much more successful than Upstairs, Downstairs. If the latter plays a lower key than the glamorous Downton Abbey, it is my intention here to point out a particularity not chosen by the rival TV series, which is embodied by the choice to introduce a regular Indian character in the cast.

iii Later famously performed on the big screen by Anthony Hopkins and Emma Thomson as the former housekeeper of Darlington Hall in the 1993 film directed by James Ivory, and written by by Ruth Prawar Jhabwallah. iv Contrary to the chauffeur Spargo, who becomes at one stage the secret lover of Sir Hallam's step sister, and contrary to the butler Mr Pritchard, who delivers the first baby of Lady Agnes in an emergency situation. v When he sides with Rachel "against" Lady Persephone's adherence to British right-wing ideas and German national-socialist characters; when he remains faithful to the memory of his departed wife while Sir Hallam betrays Lady Agnes under their own roof with her own sister, who appears as an easy scapegoat responsible for Sir Hallam's lapse and who is gotten rid of at the end thanks to her convenient suicide.

  looks a bit outdated, or at least progressive in a retrograde and benign way. The casting of Art Malik in the role of Mr Amanjit as a middle-aged compliant man may also come as a disappointment for spectators who remember his central and sensational role as a young British Asian student in the mini-series The Jewel in the Crown(ITV, 1984) taking place inIndia between 1942 and 1946. In this story, a dashing Hari Kumar discovers India after living most of his life in England, and challenges British propriety abroad by falling in love with a young British white woman, which proves fatal for the two of them. Even though a Raj revival TV series, The Jewel in the Crown was audacious enough to cast the unknown Art Malik as a rebellious young man who felt more British than Indian and died from challenging the prejudices and violence of some British protagonists. Upstairs, Downstairs ironically gives us to see a more tamed version and interpretation of the young independent and rebellious mind he used to embody in The Jewel in the Crown, implicitly pointing to the lack of prominent roles this kind of programmes can offer today to an actor of his age and status. Bibliographical references: Crespino, Joseph (2011). "All Strom's Children: Gender, Race, and Memory in the 20th Century American South" (pp. 112-128) in Gyanendra Pandey, Subalternity and Difference: Investigations from the North and the South, Routledge.