Equilibrium fluctuations for a driven tracer particle dynamics

Stochastic Processes and their Applications 85 (2000) 139–158www.elsevier.com/locate/spa

Equilibrium uctuations for a driven tracerparticle dynamics

Cl�audio Landima;b;∗, S�ergio B. VolchancaIMPA–Instituto de Matem�atica Pura e Aplicada, Estrada Dona Castorina 110, Jardim Botanico,

CEP 22460, Rio de Janeiro, BrazilbCNRS UPRES-A 6085, Universit�e de Rouen, 76.128 Mont Saint Aignan Cedex, France

cDepartamento de Matem�atica, PUC-Rio-Pontif ��cia Universidade Cat�olica do Rio de Janeiro,Rua Marques de Sao Vicente, G�avea, CEP 22453-900, Rio de Janeiro, Brazil

Received 11 January 1999; accepted 22 June 1999

Abstract

We study the equilibrium uctuations of a tagged particle driven by an external constant forcein an in�nite system of particles evolving in a one-dimensional lattice according to symmetricrandom walks with exclusion. We prove that when the system is initially in the equilibriumstate, the �nite-dimensional distributions of the di�usively rescaled position

√�X (�−2t) of the

tagged particle converges, as �→0, to the �nite-dimensional distributions of a mean zero Gaussianprocess whose covariance can be expressed in terms of a di�usion process. c© 2000 ElsevierScience B.V. All rights reserved.

MSC: Primary 60K35; Secondary 82A05

Keywords: Tagged particle; Equilibrium uctuations; Exclusion processes; Hydrodynamic limit

0. Introduction

In this paper we consider the motion of a tagged particle evolving in an in�nitelattice gas and which is driven by an external force. The motion occurs along thesites of the one-dimensional lattice through a local stochastic interaction. Particles per-form symmetric random walks except for the special tagged particle which, due to theaction of the external force, performs an asymmetric random walk, serving in this wayas a probe of the system’s global behavior. Interaction among particles results fromimposing an exclusion (or hard-core) condition, that is, particles are allowed to jumpto nearest-neighbor sites only, but the jump is suppressed in case the target site isalready occupied. We are interested in the long-time behavior of the tagged particle’strajectory Xt . This system can be thought of as a caricature model of the microscopicdynamics of particles of a gas or uid and has originated in attempts to derive a

∗ Corresponding author. IMPA–Instituto de Matem�atica Pura e Aplicada, Estrada Dona Castorina 110,Jardim Botanico, CEP 22460, Rio de Janeiro, Brazil.E-mail addresses: [email protected] (C. Landim), [email protected] (S.B. Volchan)

0304-4149/00/$ - see front matter c© 2000 Elsevier Science B.V. All rights reserved.PII: S0304 -4149(99)00069 -1

140 C. Landim, S.B. Volchan / Stochastic Processes and their Applications 85 (2000) 139–158

rigorous mechanical explanation of Brownian Motion. Of course, in a more realis-tic setting particles should evolve deterministically according to Hamilton’s equations.This has proven to be one of the hardest problems of classical statistical mechanics(cf. D�urr et al., 1985, 1986; Harris, 1965; Pellegrinotti et al., 1996; Soloveitchick, 1996;Spitzer, 1968). In general, progress in this path has been achieved by adding somewhatarti�cial simpli�cations. On the other hand, as past experience suggests, various aspectsof the macroscopic behavior of such many-particle systems are universal, in the senseof not depending too much on the underlying microscopic dynamics. Therefore, thestudy of highly simpli�ed models is not devoid of interest and can help us improveour understanding of such systems (Giacomin et al., 1996). Moreover, simpli�ed mod-els like the lattice gas can be highly non-trivial while sometimes yielding to a detailedand rigorous mathematical treatment.In a previous work (Landim et al., 1998) we proved a Law of Large Numbers for

the tagged particle’s position subject to a suitable di�usive rescaling of space and time.That is, for a large class of initial distributions of the random environment, we provedthat

limt→∞

Xt − X0√t

= v

in probability, where v is a real number depending only on the macroscopic densitypro�le of the initial state through the solution of a non-linear parabolic partial dif-ferential equation with boundary conditions. This result was based in establishing thehydrodynamic limit for an associated zero-range model from which the tagged particle’sposition behavior could be derived.As usual in probabilistic and statistical mechanical contexts, after having proved

a Law of Large Numbers the next step should consist in �nding a correspondingCentral Limit Theorem describing the uctuations of the system around the limitingvalue obtained previously. This requires an analysis on a �ner scale. In fact, the mainresult of this paper states that, in equilibrium,

limN→∞

XtN 2 − X0√N

= Zt (0.1)

for all �nite-dimensional distributions and where Z is a mean zero Gaussian process.The result can be interpreted as indicating that the tagged particle tends to be trappedbetween its neighbors and the uctuations of its position will depend directly on the uctuations of the density of particles around it. We observe that for one-dimensionalnearest-neighbor symmetric simple exclusion models Arratia (1983) proved (0.1) withconvergence in law.Our result is based on the study of the equilibrium uctuations for an associated

zero-range model, as before. As in the proof of the Law of Large Numbers, we neededto adapt the standard techniques of hydrodynamic limit theory of interacting particlesystems to our situation in which the system is in in�nite volume and the invariantmeasure has non-uniform density re ecting the presence of boundary conditions.

C. Landim, S.B. Volchan / Stochastic Processes and their Applications 85 (2000) 139–158 141

1. Notation and results

We consider a family of indistinguishable particles moving according to continuous-time, symmetric, nearest-neighbor random walks on Z with an exclusion rule thatprevents more than one particle per site. Added to this system is a tagged particle thatmoves according to an asymmetric random walk, having probability p ( 12 ¡p¡ 1) ofjumping to the right and probability q=1−p to the left and respecting the exclusionrule. The con�guration of the system is given by the pair (X; �), where X ∈ Z is theposition of the tagged particle, and � ∈ {0; 1}Z is the con�guration of all other particles(which is the tagged particle’s environment). Note that � is indexed by the sites of Z,so that �(z) is 0 or 1 according with the site z being occupied or empty. In particular,�(X ) = 0 as the site X is already occupied by the tagged particle. The system justdescribed is a Markov process whose in�nitesimal generator, acting on local functionsF :Z× {0; 1}Z → R, is given by

LF(X; �) = (12 )∑

z 6=X−1;X[F(X; �z;z+1�)− F(X; �)]

+p(1− �(X + 1))[F(X + 1; �)− F(X; �)]

+ q(1− �(X − 1))[F(X − 1; �)− F(X; �)];

where �z;z+1� is the con�guration obtained from �, exchanging the occupation variables�(z) and �(z + 1):

(�z;z+1�)(y) =

�(y) if y 6= z; z + 1;

�(z) if y = z + 1;

�(z + 1) if y = z:

For simplicity, we assume X0 = 0, where Xt stands for the tagged particle’s positionat time t. For a positive integer N and a density pro�le �0 : R→ [0; 1], let �N

�0(·) bethe Bernoulli product measure associated to �0:

�N�0(·){�; �(x) = 1}= �0(x=N )

which we take as the initial distribution for the random environment �0. Denote byP�N

�0·the probability measure on the path space D(R+;Z × {0; 1}Z) induced by the

Markov process with generator L and initial measure �0 × �N�0(·). Here �0 stands for

the Dirac measure concentrated on the origin. In Landim et al. (1998) we proved thatfor each t¿0 and for a large class of initial density pro�les �0, as N↑∞, XtN 2 =Nconverges in probability to a real number vt that depends only on �0. In the presentarticle we prove that for t¿0, and for the equilibrium initial distribution of the en-vironment, as N↑∞, the �nite-dimensional distributions of XtN 2 =

√N converge to the

�nite-dimensional distributions of a mean zero Gaussian process.The strategy begins by exploiting a connexion of the present model with a zero-range

model that we now describe. Start by labeling all particles. The tagged particle islabeled 0 and for j¿1, we label the jth particle at the right (resp. left) of the taggedparticle by j (resp. −j). Let then �(x), for x ∈ Z, be the number of holes (emptysites) between the particle labeled x and the particle labeled x + 1. This establishes


a transformation T:Z × {0; 1}Z → NZ taking each con�guration of {0; 1}Z with aparticle in some site X into a con�guration �= {�(x); x ∈ Z} ∈ NZ.The dynamics of the process (Xt; �t) induces a dynamics for the process �t that can

be described informally as follows. For all x 6= −1, if there is at least one particleat site x, at rate 1

2 one of them jumps to site x + 1 and, symmetrically, if there isat least one particle at site x + 1, at rate 1

2 one of them jumps to site x. The picturechanges between sites −1 and 0 due to the asymmetric behavior of the tagged particle.A particle jumps at rate q from site −1 to site 0 if there is at least a particle at −1and a particle jumps from site 0 to site −1 at rate p if there is at least a particle atthe origin.This process is a so called zero range process with an asymmetry at the origin. Note

that the position at time t of the tagged particle corresponds in the zero-range modelto the total number of jumps in [0; t] from 0 to −1 minus the total number of jumpsin the same time interval from −1 to 0, that is

Xt =∑x¿0

[�0(x)− �t(x)]: (1.1)

As stated, this is a formal sum whose precise sense we will explain later in connectionwith the notion of density �elds (cf. Lemma 3.1). Since in our zero-range processthe jumps of particles over all bonds, except bond {−1; 0}, are symmetric, we expectthe process to have a di�usive hydrodynamic behavior. This suggests considering theprocess accelerated by N 2 and to study the uctuation behavior from equilibrium of

XtN 2√N=

1√N

∑x¿0

[�0(x)− �tN 2 (x)]

as N↑∞. This “scaling” explains the factor t1=4 mentioned previously.Formally, we are dealing with a Markov process on Z whose generator, acting on

cylindrical functions f, is given by

(Lf)(�) =∑

x∈Z; |y|=1g(�x)[f(�x;x+y)− f(�)]px(y);

where �x;y is the con�guration obtained from � by letting a particle jump from x to y:

(�x;y)(z) =

�(x)− 1 if z = x;�(y) + 1 if z = y;�(z) otherwise;

px(y) =

p; x = 0; y =−1;q; x =−1; y = 1;1=2 otherwise

and g :N → R+ is a �xed jump rate given by g(k) = 1{k¿1}. For each ’, with06’¡p−1 this process has an invariant reversible product measure �’ with marginalsgiven by

�’{�: �(x) = k}= (1− ’x)’kx ; (1.2)

where

’x ={

p’; x6−1;q’; x¿0:


For N¿1, let �t = �Nt be the Markov process on NZ with generator N 2L: �t = �tN 2 .

Let �N = �N (�) be the empirical measure de�ned as the positive Radon measure on Robtained by assigning mass N−1 to each particle:

�N = N−1∑z∈Z

�(z)�z=N :

Set �Nt = �N (�t) and let M+ be the space of positive Radon measures on R. We

proved in Landim et al. (1998) that starting from a sequence of probability measures{�N ; N¿1}, associated to an initial density pro�le �0 :R→ R+ in the sense that

limN→∞

�N{∣∣∣∣∣ 1N

∑x∈Z

H (x=N )�(x)−∫RH (u)�0(u) du

∣∣∣∣∣¿�

}= 0

for all continuous functions H with compact support and all �¿ 0, then for all t¿0�Nt converges in probability to the absolutely continuous trajectory �(t; du) = �(t; u) duwhose density �(· ; ·) is the weak solution of the following non-linear parabolic partialdi�erential equation with boundary conditions at the origin:

@t�= (12)��(�);

p�(�(t; 0+)) = q�(�(t; 0−));@u�(�(t; 0+)) = @u�(�(t; 0−));�(0; ·) = �0(·);

(1.3)

where �(�) = �=1 + �.Fix once and for all 0¡’¡p−1. Denote by �+ (resp. �−) the density at the

right (resp. left) of the origin for �’: �± = E�’ [�(±1)]. To investigate the equilibrium uctuations of �N

t , we focus on the density �eld YNt (·) which acts on smooth functions

H as

YNt (H) = N−1=2∑

x∈ZH (x=N )[�t(x)− �x];

where �x = E�’ [�(x)]. Our aim is then to prove that YN converges to a stationarygeneralized Ornstein–Uhlenbeck process Y which is solution of the formal stochasticdi�erential equation

dYt = (12)�′(�)�Yt dt +

√�(�)=2∇ dWt;

where Wt is a generalized Brownian motion. In other words, Yt is a solution of thelinearized hydrodynamic equation (1.3) with a white noise added. To derive the lin-earization of Eq. (1.3), recall the de�nition of �± and de�ne

�(u) = �+1{u¿0} + �−1{u¡0}: (1.4)

It is clear that � is a stationary solution of the hydrodynamic equation (1.3). Let ��(·)be an initial data which is close to that solution in the sense that �−1(�� − �) → �as � ↓ 0. That is, �(·) is the “deviation” from equilibrium in �rst (or linear) orderin �. We �rst �nd the boundary conditions for �. As �(·) and ��(· ; ·) are solutionsof the hydrodynamic equation, they satisfy the boundary condition p�(��(t; 0+)) =q�(��(t; 0−)) and p�(�+) = q�(�−), from which follows, subtracting term by term


and expanding in �, that p�′(�+)�(t; 0+)= q�′(�−)�(t; 0−). On the other hand, fromthe second boundary condition, we have that @u�(��(t; 0+))=@u�(��(t; 0−)). A simpleTaylor expansion then gives that �′(�+)@u�(t; 0+) = �′(�−)@u�(t; 0−). We can now�nd the di�erential equation for �(·). We know that ��(· ; ·) satis�es∫

Rdu ��

t (u)H (u) =∫Rdu ��

0(u)H (u)− ( 12 )∫ t

0ds∫Rdu @uH (u) @u�(��

s)

for smooth functions H (·). Expanding in � we obtain that � satis�es the equation∫Rdu �t(u)H (u)−

∫Rdu �0(u)H (u) =−( 12 )

∫ t

0ds∫Rdu @uH (u)�′(�) @u�s(u):

Notice that in this formula �′(�(u)) =�′(�+) (resp. �′(�−)) for u¿ 0 (resp. u¡ 0).Hence, � is the solution of the linear second-order equation with boundary conditionsat the origin:

@t�= (12)�′(�)��;

p�′(�+)�(t; 0+) = q�′(�−)�(t; 0−);�′(�+)@u�(t; 0+) = �′(�−)@u�(t; 0−);�(0; ·) = �0(·):

(1.5)

In this formula, � is the stationary solution (1.4) of the original equation. The aim ofthis article is thus to prove that the density �eld YN

t converges to the solution Yt of(1.5) with a white noise added.Denote by C2K (R) the space of continuous functions on R, with compact support

and that are twice continuously di�erentiable on R∗ = (−∞; 0) ∪ (0;+∞). LetC2;∗K = {H ∈ C2K ; qH ′(0+) = pH ′(0−); �′(�+)H ′′(0+) = �′(�−)H ′′(0−)}:

Notice that for H ∈ C2;∗K ,∫Rdu �t(u)H (u)−

∫Rdu �0(u)H (u) = (12 )

∫ t

0ds∫Rdu @2uH (u)�

′(�)�s(u)

= (12 )∫ t

0ds∫RduH (u)�′(�)@2u�s(u)

because of the boundary conditions of � and H at the origin.On C2;∗K de�ne the operator A:

A= (12)�′(�)�=

{(1=2)�′(�+)� in R+;(1=2)�′(�−)� in R−:

(1.6)

De�ne m+ and m− by

m− = q�′(�−) and m+ = p�′(�+)

and let m(u)=m−1{u¡0}+m+1{u¿0}. It is not di�cult to check that −A is a symmetricand non-negative operator on L2(m(u) du). Denote by D(A) its domain in L2(m(u) du).In view of the Hille–Yosida theorem, to show that A is the generator of a Markovsemigroup Tt , we only need to check that the Poisson equation

f −Af = g (1.7)


has a solution f∈D(A) for every g ∈ L2(m(u) du) and every ¿ 0. Fix such a functiong and ¿ 0. Let f be de�ned by

f(u) =

a−eb−u +∫ 0

−∞

12b−�−

e−b−|u−v|g(v) dv; u¡ 0;

a+e−b+u +∫ +∞

0

12b+�+

e−b+|u−v|g(v) dv; u¿ 0:

Here �± = (12)�′(�±); b± =

√ =�± and a+; a− are chosen so that f(0+) = f(0−);

qf′(0+) = pf′(0−). It is straightforward to check that f(·) is a solution of (1.7).Denote by Tt (resp. Wt) the semigroup (resp. di�usion process) associated to thegenerator A.To state the main theorems of this paper, we need some more notation and concepts.

Let L be the linear operator L = (u2=2�′(�))−A. Denote by 〈·; ·〉 (resp. 〈·; ·〉m) theinner product on L2(du) (resp. L2(m(u) du)). For k¿0 let Hk be the Sobolev spaceobtained by completion with respect to the inner product

〈f; g〉m;k = 〈f;L kg〉m:In the Appendix we construct an orthonormal basis {Hj; j¿0} of L2(m(u) du) so that,

Hk =

f ∈ L2(m(u) du);

∑j¿0

(2j + 1)k〈f;Hj〉2m ¡∞ :

Denote by H−k the dual of Hk with respect to L2(m(u) du). We consider the density�eld YN

t (·) as taking values in the Sobolev space H−k for some su�ciently large k. Fixa time T ¿ 0, a positive integer k0 and let D([0; T ];H−k0 ) be the space of H−k0 -valuedfunctions which are right-continuous with left limits. Denote by PN the probabilitymeasure on D([0; T ];H−k0 ) induced by the process YN and the product measure �’de�ned in (1:8). Let P�’ be the probability measure on D([0; T ];NZ) induced by theprobability measure �’ and the process �t and let EPN [·] and E�’ [·] denote the respectiveexpectations.

Theorem 1.1. Fix a positive integer k0¿3. Let P be the probability measure concen-trated on C([0; T ];H−k0 ) corresponding to the generalized Ornstein–Uhlenbeck pro-cess Yt with mean 0 and covariance

EP[Yt(H)Ys(G)] =∫R�(u)(T|t−s|H)(u)G(u) du

for all H;G ∈ Hk0 . Here �(·) is the variance of �(·) : �(u)=�+1{u¿0}+�−1{u¡0} and�± = E�’ [�(±1)2] − �2±. Then the sequence PN converges weakly to the probabilitymeasure P.

Fix 0¡�¡ 1 and denote by �� the product measure on {0; 1}Z∗ with marginalsgiven by

��{�; �(x) = 1}={

�; x¡ 0;1− (q=p)(1− �); x¿ 0:


It is easy to check that �� is a reversible invariant measure for the environment asseen from the tagged particle. Moreover, �� is sent to �’ with ’ = p−1(1 − �) bythe transformation T de�ned in the beginning of this section. For this �xed value of’=’(�), recall that Wt stands for the di�usion process with generator A (that dependson ’= ’(�)).

Theorem 1.2. Let �0 × �� be the initial measure of the process (Xt; �t). The �nite-dimensional distributions of N−1=2XtN 2 converge to the �nite-dimensional distributionsof the mean zero Gaussian process Zt with covariance given by

E[ZtZs] =∫ ∞

0Pu[Wt60]�(u) du+

∫ ∞

0Pu[Ws60]�(u) du

−∫ ∞

0Pu[W|t−s|60]�(u) du (1.9)

and such that Z0 = 0.

2. The density �eld

In this section we prove Theorem 1.1. We follow the approach presented in Chang(1994) and we refer the reader to Spohn (1991), Kipnis and Landim (1999) for refer-ences and details omitted here.The proof of Theorem 1.1 is divided in three steps. We �rst prove that the sequence

PN is tight and that all limit points are concentrated on continuous trajectories. Then,we show that all limit points P of the sequence PN solve the following martingaleproblem. For each t¿0, denote by Ft the �-algebra generated by Ys(H) for 06s6tand H in Hk0 . For each H in Hk0

MA;Ht = Yt(H)− Y0(H)−

∫ t

0Ys(AH) ds;

KA;Ht = (MA;H

t )2 − ‖BH‖22t(2.1)

are (P;Ft)-martingales and Y0 is a mean zero Gaussian �eld with covariance given by

EP[Y0(H)Y0(G)] =∫R�(u)H (u)G(u) du (2.2)

for H; G in Hk0 . HereA is the di�erential operator de�ned in (1.6), BH=√

�(�)=2∇Hand ‖G‖22=〈G;G〉. We conclude the proof of Theorem 1.1 invoking Holley and Stroock(1978) theory on generalized Ornstein–Uhlenbeck processes that states that for eachmeasure q on Hk0 , there is a unique probability measure on C([0; T ];H−k0 ) that solvesthe martingale problem (2.1) and whose marginal at time 0 is q.We postpone the proof of the tightness to Section 4. We prove here that all limit

points of the sequence PN solve the martingale problem (2.1), (2.2). Fix a limit pointP and assume without loss of generality that PN converges to P. Recall the de�nitionof the �ltration Ft . We �rst prove that the restriction of P to F0 is a Gaussian �eldwith covariance matrix given by (2.2).


Lemma 2.1. For every continuous function H with compact support and every t ¿ 0;

limN→∞

log E�’ [exp{iYt(H)}] =−( 12 )〈�H;H 〉:

Proof. Recall that the initial state is the product invariant measure �’. Hence, for each�xed continuous function with compact support,

log E�’ [exp{iYt(H)}] =∑x∈Z

logE�’ [exp{iN−1=2H (x=N )[�(x)− �x]}]:

By Taylor expansion, the last expectation is equal to

1− ( 12 )N−1�(x=N )H (x=N )2 + O(N−3=2):

Therefore, summing over x and using that log(1 + x) = x +O(x2) we obtain that

log E�’ [exp{iYt(H)}] =−( 12 )N−1∑x∈Z

�(x=N )H (x=N )2 + O(N−1=2)

which converges, as N↑∞, to −(1=2)〈�H;H 〉.

From this lemma and the linearity of Yt it is not di�cult (cf. Kipnis and Landim,1999, Chapter 11) to prove that P restricted to F0 is a Gaussian �eld:

Corollary 2.2. Restricted to F0; P is a Gaussian �eld with covariance given by

EP[Y0(G)Y0(H)] = 〈�H;G〉:

To prove that P solves the martingale problem (2.1), for each H in C2;∗K de�ne themartingales MN;H

t , KN;Ht by

MN;Ht = YN

t (H)− YN0 (H)−

∫ t

0N 2LYN

s (H) ds;

KN;Ht = (MN;H

t )2 − N 2∫ t

0{L(YN

s (H))2 − 2YN

s (H)LYNs (H)} ds:

Denote by ∇N and �N the discrete derivative and the discrete Laplacian: (∇Nf)(x)=N [f(x + 1) − f(x)] and (�Nf)(x) = N 2[f(x + 1) + f(x − 1) − 2f(x)]. Note thatN 2L�(x) = (1=2)�Ng(�(x)) for x 6= −1; 0. A simple computation taking into accountthe boundary terms shows that N 2LYN (H) is equal to

( 12 )N−1=2 ∑

x 6=−1;0H (x=N )�Ng(�(x))

+N 3=2H (−1=N ){(1=2)g(�(−2)) + pg(�(0))− [(1=2) + q]g(�(−1))}

+N 3=2H (0){(1=2)g(�(1)) + qg(�(−1))− [(1=2) + p]g(�(0))}:Let �x=E�’ [g(�(x))] and notice that ��x=0 for x 6= −1; 0. We may therefore replace�Ng(�(x)) by �N [g(�(x)) − �x] in the �rst expression. A double summation by parts


in this term gives that N 2LYN (H) is equal to

( 12 )N−1=2 ∑

x 6=−1;0(�NH)(x=N )(gx − �x) + (12 )N

3=2(g−1 − �−1)

×{H (−2=N )− H (−1=N )}+ (12)N 3=2(g0 − �0){H (1=N )− H (0)}+N 3=2[pg0 − qg−1]{H (−1=N )− H (0)}:

Here we denoted g(�(x)) by gx and we used the fact that �−1 = �−2; �0 = �1. Recallthat H belongs to C2;∗K so that pH ′(0−)=qH ′(0+). On the other hand, an elementarycomputation shows that q�−1 = qE�’ [g(�(−1))] = qp’ which is also equal to p�0. Inparticular, expanding the last three terms in the previous formula, we obtain that theyare equal to

√Np−1[q+ (1=2)]H ′(0+)(pg0 − qg−1) + O(N−1=2):

For x in Z denote by Wx;x+1 the instantaneous current over the bond {x; x + 1},i.e., the rate at which a particle jumps from x to x + 1 minus the rate at which aparticle jumps from x+1 to x. In particular, for each x ∈ Z; L�(x)=Wx−1; x −Wx;x+1,Wx;x+1 = (12 )[g(�(x)) − g(�(x + 1))] for x 6= −1 and W−1;0 = qg(�(−1)) − pg(�(0)).Therefore, we end up with

MN;Ht = YN

t (H)− YN0 (H)−

∫ t

0( 12 )N

−1=2 ∑x 6=−1;0

(�NH)(x=N )[g(�s(x))− �x] ds

+∫ t

0�NW−1;0(s) ds+O(N−1=2); (2.3)

where �N =√Np−1[q+ (12)]H

′(0+).Observe that there is an homogeneous part involving di�erences of type g(�(x)))−�x,

for all sites x 6= −1; 0, and a boundary term W−1;0 involving only the sites x =0;−1. In the next lemma we show that the boundary term converges to zero and inthe proposition which follows we state the Boltzmann–Gibbs principle that allows toexpress the homogeneous part in terms of the density �eld YN

s (H), closing the equationof the martingale MN;H

t in terms of the density �eld.

Lemma 2.3. Let {�N ; N¿1} be a sequence of numbers such that lim supN→∞ �N =N=0. Then; for each t¿0

lim supN→∞

E�’

[(∫ t

0�NW−1;0(s) ds

)2]= 0:

Proof. Let V (�s)=�NW−1;0(s). Since �’ is reversible, by Proposition A.1.6.1 in Kipnisand Landim (1999), there exists a �nite constant C0 such that

E�’

[(∫ t

0V (�s) ds

)2]6 C0t〈V; (−N 2L)−1V 〉

:=C0t suph{2〈V; h〉 − N 2〈h;Lh〉}:

In this formula 〈·; ·〉 stands for the inner product in L2(�’) and the supremum is takenover all bounded cylinder functions h. For x ∈ Z, denote by dx the con�guration with


just one particle at x. Also, the summation of two con�gurations is to be understoodcomponentwise. Fix a bounded cylinder function h. Making the change of variables�= �− d0; �′ = �− d−1 we obtain that∫

W−1;0h d�’ = pq’∫[h(�+ d−1)− h(�+ d0)] d�’: (2.4)

The same change of variables gives that

〈h;−Lh〉¿pq’∫[h(�+ d−1)− h(�+ d0)]2 d�’:

Therefore,

2�N 〈W−1;0; h〉 − N 2〈h;Lh〉6qp’

∫{2�N [h(�+ d0)− h(�+ d−1)]− N 2[h(�+ d−1)− h(�+ d0)]2} d�’

6qp’�2NN 2 :

This concludes the proof of the lemma.

We now close the equation for the martingale MN;Ht in terms of the density �eld

YNt . This is the content of the Boltzmann–Gibbs Principle stated below. The proofpresented in Kipnis and Landim (1999, Chapter 11) is easily adapted to our context.This result was �rst proved by Brox and Rost (1984) in the context of homogeneouszero-range processes.

Proposition 2.4. For all continuous functions G(·) on R∗ of compact support andevery t ¿ 0,

limN→∞

E�’

∫ t

0N−1=2 ∑

x 6=0;−1G(x=N ) {g(�s(x))− �x − �′(�x)[�s(x)− �x]} ds

2

=0:

Note that �x = �(�x). It follows from this proposition applied to G = �NH andLemma 2.3 that MN;H

t is equal to

YNt (H)− YN

0 (H)−∫ t

0ds (1=2)N−1=2 ∑

x 6=−1;0(�NH)(x=N )�′(�x)[�s(x)− �x] + AN ;

where AN is a remainder term that converges to 0 in L2(P�’). Since H belongs toC2;∗K , we may replace the discrete Laplacian by the continuous one. In this case, since�′(�)�H is continuous, we may replace summation over x 6= −1; 0 by a summationover x ∈ Z to obtain that

MN;Ht = YN

t (H)− YN0 (H)−

∫ t

0YNs (AH) ds+ AN : (2.5)

The expression of AN changed but still converges to 0 in L2(P�’). It is not di�cult todeduce from the previous identity that MA;H

t de�ned in (2.1) is a (P;Ft)-martingalefor each H in Hk0 . This is the content of the next proposition.


Proposition 2.5. For each H in Hk0 ; MA;Ht is a (P;Ft)-martingale.

Proof. Fix �rst H in C2;∗K ∩ Hk0 . By de�nition, MA;Ht is measurable with respect to

the �ltration Ft . In order to prove that MA;Ht is a martingale it su�ces to check that

EP[MA;Ht U ] = EP[MA;H

s U ]

for all 06s6t6T and all U = 1{Ysi (Hi)∈Ai; 16i6n}, where n is a positive integer,

06s16 · · ·6sn6s and, for each 16i6n; Hi belongs to C2;∗K and Ai is some measur-able subset of R. Since MN;H

t is a martingale, E�’ [MN;Ht U ] = E�’ [MN;H

s U ]. Hence, weonly need to show that these expectations converge, respectively, to EP[M

A;Ht U ] and

EP[MA;Hs U ].

By (2.5), since U is bounded,

limN→∞

E�’ [MN;Ht U ] = lim

N→∞EPN

[{Yt(H)− Y0(H)−

∫ t

0ds Ys(AH)

}U]:

A simple computation shows that the term inside braces is square integrable, uni-formly in N . In particular, since U is bounded and PN converges weakly to P andsince, by Proposition 4.1 below, P is concentrated on continuous paths, EPN [M

N;Ht U ]

converges to EP[MA;Ht U ] as N↑∞. Since this statement remains in force with s in

place of t; MA;Ht is a (P;Ft)-martingale for each H∈C2;∗K ∩ Hk0 . Fix now H in Hk0 .

With a sequence Hn of functions in C2;∗K ∩ Hk0 such that Hn, �′(�)�Hn convergesto H; �′(�)�H in L2(P), it is easy to show that MA;H

t is a (P;Ft)-martingale. Thisconcludes the proof of the lemma.

We now turn to the martingale KN;Ht . An elementary computation gives that

N 2LYN (H)2 −2N 2YN (H)LYN (H) is equal to

( 12 )N−1 ∑

x 6=0;−1g(�(x))[(∇NH)(x=N )]2

+N−1pg−1[(∇NH)(−1=N )]2 + N−1qg0[(∇NH)(0)]2:

Lemma 2.6. For every continuous function G on R∗ of compact support

limN→∞

E�’

∫ t

0N−1 ∑

x 6=0;−1G(x=N )[g(�s(x))− �x]

2= 0:

The proof is elementary because �’ is a product stationary measure. It follows fromthis result applied to G =∇NH and from the previous computation that

KN;Ht = (MN;H

t )2 − t2N

∑x 6=−1;0

[(∇NH)(x=N )]2�x + AN ;

where AN converges to 0 in L2(P�’). Here again we may replace the discrete gradientby the continuous one and extend the summation to Z. Repeating the arguments pre-sented in the proof of Proposition 2.5, we show that KA;H

t is a (P;Ft)-martingale foreach H ∈ Hk0 .


Proof of Theorem 1.1. By Proposition 4.1 the sequence PN is tight and all its limitpoints are concentrated on continuous paths. Fix a limit point P of the sequence PN .We proved in this section that P solves the martingale problem (2.1), (2.2). The proofof Theorem (1.4) in Holley and Stroock (1978) can be adapted to show that thereexists at most one solution of (2.1), (2.2). This concludes the proof of the theorem.

3. Fluctuation of the asymmetric tagged particle

In this section we use the results proven for the zero-range model to deduce a CentralLimit Theorem for the asymmetric tagged particle in the nearest-neighbor symmetricsimple exclusion process. We follow the approach presented by Rost and Vares (1985)for the case of a symmetric tagged particle.Recall from Section 2 that (Xt; �t) is a one-dimensional nearest-neighbor symmetric

exclusion process with an asymmetric tagged particle that sits at the origin at t=0. Wede�ned in Section 1 a correspondence between this exclusion process and a symmetriczero-range process with an asymmetry at the bond {−1; 0}. In this correspondence theposition of the tagged particle at time t is given by the total number of jumps from 0to −1 minus the total number of jumps from −1 to 0 in the time interval [0; t] .Fix 06�61 and recall the de�nition of the product measure �� given in Section 1.

Let ’=p−1(1− �). Since �t = �Nt is the zero-range process with generator N

2L, it isaccelerated by N 2, in contrast with the process (Xt; �t) which is not. Denote by SN

+(t)(resp. SN

−(t)) the total number of jumps from −1 to 0 (resp. 0 to −1) in the timeinterval [0; t] for the process �. The position of the tagged particle is such that

XtN 2 = SN−(t)− SN

+(t):

For n¿1, let Gn(u) = [1 − (u=n)]+1{u¿0}. The next result gives a precise meaningto identity (1.1).

Lemma 3.1. There exists a �nite constant c such that

supN¿1

E�’

(XtN 2√

N− 1√

N

∑x¿0

Gn(x=N )[�0(x)− �t(x)]

)26ctn

for all n¿1 and t ∈ [0; T ].

Proof. First observe that

SN+(t)− qN 2

∫ t

0g(�s(−1)) ds; SN

−(t)− pN 2∫ t

0g(�s(0)) ds

are P�’ -martingales. Hence, so is Mt = XtN 2 + N 2∫ t0 W−1;0(s) ds. For n¿1, let Y n

t =∑x¿0 Gn(x=N )(�t(x)− �x) and Vn

t = Y n0 − Y n

t . Then

Mnt = Vn

t +∫ t

0N 2∑x¿0

Gn(x=N )L�s(x) ds


is a P�’ -martingale. A straightforward computation shows that the integral term of theprevious expression is equal to

N 2∫ t

0

{W−1;0(s) +

1nN

nN−1∑x=0

Wx;x+1(s)

}ds:

Thus, XtN 2 − Vnt is equal to

Mt −Mnt +

Nn

∫ t

0

nN−1∑x=0

Wx;x+1(s) ds:

We claim that this expression converges to 0 in L2(P�’) as n ↑ ∞. Notice that theexpression appearing in the statement of Lemma 2.3 is just the current W−1;0. Repeatingthe arguments presented there, we deduce that

E�’

(∫ t

0

Nn

nN−1∑x=0

Wx;x+1(s) ds

)2 ds6CtNn

for some �nite constant C.For each x denote by J x;x+1

t the total number of jumps from x to x + 1 minus thetotal number of jumps from x + 1 to x in the time interval [0; t] for the process �.Thus, jx;x+1t = J x;x+1

t − N 2∫ t0 Wx;x+1(s) ds is a martingale for each x in Z and Mt =

−j−1;0t ; Mnt =−j−1;0t + (nN )−1

∑nN−1x=0 jx;x+1t . In particular,

E�’ [(Mnt −Mt)2] = E�’

( 1

nN

nN−1∑x=0

jx;x+1t

)2

=1

(nN )2

nN−1∑x=0

tN 2(�x + �x+1) =2N�0t

n:

Here we used the fact that the martingales { jx;x+1t ; x ∈ Z} are mutually orthogonalbecause they do not have common jumps and that E�’ [(j

x;x+1t )2] = tN 2(�x + �x+1).

It follows from this lemma that for each �xed t the sequence {YN0 (Gn) − YN

t (Gn);n¿1} is Cauchy in L2(P�’), uniformly over N¿1. In consequence, the sequenceY0(Gn)−Yt(Gn) is Cauchy in L2(P) (Since Gn−Gm does not belong to Hk0 ; Yt(Gn−Gm)is understood as the limit of Yt(H

m;nk ), where Hm;n

k belongs to Hk0 and converges inL2(du) to Gn − Gm).Let Zt be the L2(P) limit of Y0(Gn)−Yt(Gn). Recall that Wt is the di�usion process

with generator A. For 06s6t, the covariance EP[ZsZt] is given by (1.9). Indeed,since Zt is the L2(P) limit of Y0(Gn)− Yt(Gn),

EP[ZsZt] = limn→∞ EP[{Y0(Gn)− Ys(Gn)}{Y0(Gn)− Yt(Gn)}]:


Recall that Tt stands for the semigroup associated to the generator A. By Theorem 1.1,for each �xed n, the previous expectation is equal to∫

RGn(u)[Gn(u)− TtGn(u)]�(u) du+

∫RGn(u)[Gn(u)− TsGn(u)]�(u) du

−∫RGn(u)[Gn(u)− T|t−s|Gn(u)]�(u) du:

Since∫R Gn(u)[Gn(u) − TtGn(u)] du converges, as n ↑ ∞, to ∫∞

0 Pu[Wt60] du, (1.9)is proven.We are now in a position to prove Theorem 1.2.

Proof of Theorem 1.2. Since YN0 (Gn) − YN

t (Gn) converges, as n ↑ ∞, to N−1=2XtN 2

in L2(PN ) uniformly in N and since Y0(Gn) − Yt(Gn) converges, as n ↑ ∞, to Zt inL2(P), it is not hard to show that

E�’ [f(N−1=2Xt1N 2 ; : : : ; N−1=2XtkN 2 )]

converges to EP[f(Zt1 ; : : : ; Ztk )] for every bounded Lipschitz-continuous function f andevery 06t1¡ · · ·¡tk6T .

4. Tightness

We prove in this section that the sequence PN de�ned in Section 1 is tight.

Proposition 4.1. The sequence {PN ; N¿1} is tight. Moreover; all its limit points areconcentrated on continuous paths.

We adapt the arguments exposed in Kipnis and Landim (1999, Chapter 11). Weprove in this section that the sequence is tight with respect to the uniform topology.This in turn implies that the limit points are concentrated on continuous trajectories.Recall the de�nition of the operator L . Let {Hj; j¿0} be the L2(m(u) du) ortho-

normal basis of L-eigenfunctions obtained in the Appendix. Notice that

Lf =∑j¿0

〈f;Hj〉m(2j + 1)Hj

for all f in D(A). Recall that for k¿1; Hk stands for the Hilbert space obtained bycompletion of D(L k) with respect to the inner product 〈·; ·〉m;k de�ned by 〈f; g〉m;k =〈f;L kg〉m for f; g in D(L k). It can be shown that Hk={f ∈ L2;

∑j¿0 (2j+1)

k〈f;Hj〉2m¡∞}.Let H−k be the dual of Hk with respect to L2(m(u) du). It is again easy to check that

H−k consists of all sequences {〈f;Hj〉m; j¿0} such that∑j¿0 (2j+1)−k〈f;Hj〉2m ¡∞.

In particular,

· · · ⊇H−k ⊇ · · ·⊇H0⊇H1⊇ · · ·⊇Hk ⊇ · · · :

Fix k¿1 and denote by w�(Y ) the modulus of continuity de�ned by

w�(Y ) = sup|s−t|6�06s; t6T

‖Yt − Ys‖−k :


A family {PN ; N¿0} of probability measures on D([0; T ];H−k) is tight if

limA→∞

lim supN→∞

PN

[sup

06t6T‖Yt‖−k¿A

]= 0;

lim�→0

lim supN→∞

PN [w�(Yt)¿�] = 0; ∀�¿ 0:(4.1)

Moreover, in this case all limit points are concentrated on continuous paths.The next result provides an estimate that allows us to prove (4.1).

Lemma 4.2. For each T ¿ 0; there exists a �nite constant C(T ) such that

E�’[sup

06t6T|Yt(Hj)|2

]6C(T )

{1N

∑x∈Z

Hj(x=N )2 +1N

∑x∈Z

[(∇Hj)(x=N )]2}

for all N¿1.

Proof. Fix j¿0 and denote by Mt(j) =MNt (j) the martingale de�ned by

Mt(j) = YNt (Hj)− YN

0 (Hj)−∫ t

0�1; j(s) ds;

where �1; j(s) = N 2LYNs (Hj). This identity permits us to write YN

t (Hj) in terms ofYN0 (Hj), the martingale Mt(j) and the time integral of �1; j(·). We shall estimate thesethree terms separately.A simple computation shows that

E�’ [(Y0(Hj))2] =1N

∑x∈Z

Hj(x=N )2�(x=N ):

On the other hand, since Mt(j) is a martingale, by Doob’s inequality,

E�’[sup

06t6T|Mt(j)|2

]64E�’ [M 2

T (j)]:

The computations performed just before Lemma 2.6 show that this expectation isbounded by CTN−1∑

x∈Z [∇NHj(x=N )]2 for some �nite constant C.It remains to consider the integral term

∫ t0 �1; j(s) ds. Fix a0¿ 0. Let g(t) =∫ t

0 �1; j(s) ds and (u) = ea0Nu − 1. By the Garsia–Rodemich–Rumsey inequality,

sup06t6T

|g(t)|64∫ T

0−1(4B=u2)u−1=2 du;

where B =∫ T0 ds

∫ T0 dt (|g(t) − g(s)|=√t − s). With our choice of ; −1(u) =

(a0N )−1log(1 + u). By an integration by parts,∫ T

0u−1=2 log

{1 +

4Bu2

}du62

√T{log(1 +

4BT 2

)+ 4}

because B=(4B+ u2)6 14 . Therefore, by the two previous estimates,

E�’[exp

{(a0N=8

√T ) sup

06t6T|g(t)|

}]6e4

{1 +

4T 2E�’ [B]

}: (4.2)


Recall the de�nition of B. By Lemma 4.4 below, with a=a0N=8√T ; E�’ [B] is bounded

above by 2T 2 exp{a20N∑

x∈Z (∇NHj(x=N ))2} − T 2. From this estimate and (4.2) weobtain (cf. Kipnis and Landim, 1999, Section 11.4) that

E�’[exp

{(a0N=8

√T ) sup

06t6T|g(t)|

}]68e4 exp

{a20N

∑x∈Z

(∇NHj(x=N ))2}

:

Maximizing over a0, taking logarithms in both sides and applying Jensen’s inequality,we deduce that

E�’[sup

06t6T|g(t)|2

]6c0TN−1∑

x∈Z(∇NHj(x=N ))2

for some universal constant c0. This concludes the proof of the lemma.

Corollary 4.3. For k¿3;

lim supN→∞

E�’[sup

06t6T‖Yt‖2−k

]¡∞;

limn→∞ lim sup

N→∞E�’

[sup

06t6T

∑j¿n

(2j + 1)−k〈Yt; Hj〉2m]= 0:

Proof. By de�nition of ‖Yt‖2−k ,

‖Yt‖2−k =∑j¿0

(2j + 1)−k〈Yt; Hj〉2m:

In particular, by the previous lemma,

limN→∞

E�’[sup

06t6T‖Yt‖2−k

]6C(T )

∑j¿0

(2j + 1)−k{〈Hj; Hj〉m + 〈H ′j ; H

′j 〉m}:

Here and below we used the fact that there exists a positive constant A such thatA〈H;H 〉m6〈H;H 〉6A−1〈H;H 〉m. Since {Hj; j¿0} is an orthonormal basis, 〈Hj; Hj〉m=1 for all j¿0. On the other hand, by an integration by parts, 〈H ′

j ; H′j 〉m6c〈−AHj; Hj〉m

for some �nite constant c. Since −A6L and LHj =(2j+1)Hj; 〈H ′j ; H

′j 〉 is bounded

by c(2j + 1). The right-hand side of the previous expression is thus �nite for k¿3.To prove the second statement of the corollary, observe that the expectation is

bounded above by

∑j¿n

(2j + 1)−kE�’[sup

06t6T〈Yt; Hj〉2m

]:

By the estimates obtained in the previous lemma for �xed N , this expression issummable for k¿3 and thus converges to 0 as n ↑ ∞.

Proof of Proposition 4.1. We have seen that in order to prove tightness, one needs tocheck (4.1). The �rst condition in (4.1) follows directly from the previous corollary.To prove the second one, in view of the previous corollary, we need to show that for


all �¿ 0 and n¿1

lim�→0

lim supN→∞

P�’

sup

|s−t|6�06s; t6T

∑j6n

〈Yt − Ys; Hj〉2(2j + 1)−k ¿ �

= 0:

We observe that YNt (Hj)−YN

s (Hj)=Mt(j)−Ms(j)+∫ ts �1; j(r) dr. The martingale part

is handled as in (Kipnis and Landim, 1999, Chapter 11). For the integral part one usesthe proof in Lu (1994) for nongradient systems. We leave the details to the reader.

We conclude this section proving an estimate used in Lemma 4.2.

Lemma 4.4. For all 06t6T and a¿ 0;

E�’[exp

{a∣∣∣∣∫ t

s�1; j(u) du

∣∣∣∣}]62 exp

{a2(t − s)N−1∑

x∈Z(∇NHj(x=N ))2

}:

Proof. Since e|x|6ex + ex, it is enough to prove that the left-hand side without theabsolute value is bounded by one half of the right-hand side. Since �’ is an invariantmeasure, by Feynman–Kac formula,

E�’[exp

{∫ t−s

0a�1; j(u) du

}]6exp

{(t − s) sup

f

{∫a�1; jf d�’ − N 2D(f)

}};

where the supremum is taken over all densities with respect to �’ and −D(f) =〈√f;L

√f〉�’ . Recall the de�nition of the current Wx;x+1 introduced in Section 2.

We have that

a�1; j = a√N∑x∈Z

(∇NHj)(x=N )Wx;x+1:

The integration by parts formula (2.4) for the current shows that 2a∫�1; jf d�’ is

bounded above by

8a2

N

∑x∈Z

[(∇NHj)(x=N )]2 + N 2D(f):

Therefore,

supf

{2∫

a�1; jf d�’ − N 2D(f)}6(8a2=N )

∑x∈Z

[(∇NHj)(x=N )]2

which concludes the proof of the lemma.

Acknowledgements

C.L. was partially supported by CNPq grant 300358=93-8. C.L. and S.V. acknowl-edge partial support of PRONEX 41.96.0923.00 “Fenomenos Cr��ticos em Probabilidadee Processos Estoc�asticos”.


Appendix

Consider the linear operator L = (u2=2�′(�)) −A. In this Appendix we constructan orthonormal basis of L-eigenfunctions {Hj; j¿0} for L2(m(u) du) such that

Hj ∈ D(A) and LHj = (2j + 1)Hj (5.1)

for each j¿0.Recall that �(u)=�−1{u¡0}+�+1{u¿0}, where �±=�′(�±). Let J0(u)=exp(−u2=2�)

and notice that it belongs to the space

C∞;∗0 = {H ∈ C∞

0 (R); qH ′(0+) = pH ′(0−); �+H ′′(0+) = �−H ′′(0−)};where C∞

0 (R) stands for the space of smooth functions on R∗ that are rapidly decayingand continuous on R. For n¿1, let 2n(u) = (u=

√�)2n; 2n−1(u) = b(u)(u=

√�)2n−1,

where b(u) =√

�+p1{u¿0} +√

�−q1{u¡0}. Set Jn(u) = n(u)e−u2=2� for n¿1. Noticethat Jn(·) belong to C∞;∗

0 for n¿1.We claim that the set {Jn(·); n¿0} is complete in L2(m(u) du) or, equivalently, that

the set { n(u); n¿0} is complete in L2(e−u2=2� m(u) du). So let f∈L2(e−u2=2� m(u) du)and suppose that 〈f; n〉� = 0 for all n¿0. Here 〈·; ·〉� stands for the inner product inL2(e−u2=2� m(u) du).Let f0(u)=f(u)m(u), which belongs to L2(e−u2=2� du). For the even terms, we have∫

R(u=

√�)2nf0(u)e−u2=2� du= 0

from which follows, after a change of variables, that∫R(t2n=2n!)u2nf0(u

√�)e−u2=2

√� du= 0

for each t ∈ R. Thus, summing over n¿0 we get∫Rcos(tu)A�(u) du= 0 (5.2)

for all t ∈ R, where A�(u) =√�f0(u

√�)e−u2=2.

Repeating the same argument for the odd terms, we get that∫Rsin(tu)b(u)A�(u) du= 0

for all t ∈ R.Decomposing A�(·) in its even and odd parts, denoted, respectively, by Ae(·) and

Ao(·), we get that∫RAe(u)sin(tu) du= 0 and

∫RAe(u)cos(tu) du= 0:

The second identity follows from (5.2) and the fact that cos(·) is an even function.Therefore, the Fourier transform of the Ae(·) vanishes in R so that Ae(u) = 0 almosteverywhere.Now, let B�(u) = b(u)A�(u) and denote by Bo(·) its odd part. A similar argument

shows that Bo(u) = 0 almost everywhere. Therefore, for almost all u¿ 0, b(u)A�(u)−b(−u)A�(−u) = 0 and A�(u) + A�(−u) = 0. From these two identities it follows


that A�(u) = 0 almost everywhere in R. This proves that {Jn; n¿0} is complete inL2(m(u)du).Recall that L=(u2=�)−A. A simple computation shows that LJ0=J0; LJ1=3J1 and,

for n¿2; LJn=(2n+1)Jn − n(n− 1)Jn−2. From these relations it is easy to constructan orthonormal basis {Hj; j¿0} of L-eigenfunctions for L2(m(u) du), satisfying therequirements (5.1).

References

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Equilibrium fluctuations for a driven tracer particle dynamics