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HAL Id: hal-03145320

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Moderate Deviations for the SSEP with a Slow Bond

Xiaofeng Xue, Linjie Zhao

To cite this version:

Xiaofeng Xue, Linjie Zhao. Moderate Deviations for the SSEP with a Slow Bond. 2021. �hal-03145320�

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Moderate Deviations for the SSEP with a Slow Bond

Xiaofeng Xue

and Linjie Zhao

Abstract

We consider the one dimensional symmetric simple exclusion process with a slow bond. In this model, particles cross each bond at rate N

2

, except one particular bond, the slow bond, where the rate is N. Above, N is the scaling parameter. This model has been considered in the context of hydrodynamic limits, fluctuations and large deviations. We investigate moderate deviations from hydrodynamics and obtain a moderate deviation principle.

Keywords: exclusion process, slow bond, moderate deviation, exponential martingale.

1 Introduction

The symmetric simple exclusion process (SSEP) with a slow bond was introduced in [6] by Franco, Gonçalves and Neumann to consider the macroscopic effect of the slow bond on the hydrodynamic profile. They derived from this microscopic system PDEs with boundary condi- tions, which has become a popular topic recently [1, 9, 12]. The process evolves on the discrete ring with N sites, where N is the scaling parameter. There is at most one particle per site.

Particles cross each bond at rate N

2

except one particular bond, where the rate is N .

The hydrodynamic limit of the SSEP with a slow bond has been well understood [6, 8]. The hydrodynamic equation turns out to be the heat equation with Robin’s boundary conditions:

 

t

ρ (t, u) =

u2

ρ (t, u), t > 0, u T\{ 0 } ,

u

ρ (t, 0

+

) =

u

ρ (t, 0

) = ρ (t, 0

+

) ρ (t, 0

) , t > 0,

ρ(0, u) = γ (u), u T ,

(1.1)

where T is the continuous ring, 0

+

and 0

denote respectively the right limit and left limit at site 0, and γ( · ) is the initial density profile. Then it is natural to consider the equilibrium fluctuations and large deviations from the hydrodynamic limit. Equilibrium fluctuations have been studied in [7] and large deviations in [11] by Franco, Gonçalves and Neumann.

To better understand the SSEP with a slow bond, we consider the moderate deviations from the hydrodynamic limit, which gives asymptotic behavior of the model between the central limit theorem and the large deviation. As far as we know, the only paper concerned about moderate deviations from hydrodynamics is [14] authored by Gao and Quastel, where the classic SSEP was considered. For literatures about theories of moderate deviations, see References [2–4,13,18–20]

and so on.

E-mail: [email protected] Address: School of Science, Beijing Jiaotong University, Beijing 100044, China.

E-mail: [email protected] Address: Inria Lille-Nord Europe, France.

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A main physical motivation to investigate the moderate deviation theory is due to its appli- cation in statistical inference. Generally speaking, assuming that θ is a parameter of a model in statistical physics while { ϑ

n

}

n≥1

is a series of stochastic elements arising from the random sample path of the model that can be observed by the researchers, if one can show that ϑ

n

converges weakly to θ under a moderate deviation principle with rate function I(ϵ), then for any ϵ > 0,

P

n a

n

n

θ) > ϵ

e

a

2n n I(ϵ)

for sufficiently large n and positive sequence { a

n

}

n≥1

satisfying

ann

0 and

an2n

+ as

n + . As a result, h

ϑ

n

a

n

ϵ

n , ϑ

n

+ a

n

ϵ n

i

is a confidence interval of θ with the advantage that the length of the interval converges to 0 while the confidence level of the interval converges to 1 exponentially as n + .

Next, we introduce the SSEP with a slow bond and main results. The process evolves on T

N

= { 0, 1, . . . , N 1 } the ring with N sites, with the convention N 0. Therefore, the state space is { 0, 1 }

TN

. For each configuration η ∈ { 0, 1 }

TN

, η(x) = 1 means site x is occupied by a particle, and η(x) = 0 means site x is vacant. The infinitesimal generator L

N

of the process is

L

N

f (η) = N [f(η

1,0

) f (η)] + N

2

X

x∈TN , x̸=−1

[f (η

x,x+1

) f (η)],

where

η

x,y

(u) =

 

 

η(u) if u ̸ = x, y, η(y) if u = x, η(x) if u = y

for any x ̸ = y. Denote by { η

t

}

t≥0

the process with generator L

N

. We suppress the dependence of the process { η

t

}

t≥0

on N for short.

Equivalently, we can define the process in the following way. For each i ̸ = 1, let { Y

i

(t) }

t≥0

be a Poisson process with rate N

2

and { Y

1

(t) }

t≥0

be a Poisson process with rate N . Assume that all these Poisson processes are independent. Then at any event moment of Y

i

( · ), η(i) and η(i + 1) exchange their values.

The SSEP with a slow bond has a family of invariant measures indexed by the particle density. To be precise, let ν

ρ

, ρ [0, 1], be the product measure on T

N

with marginals given by

ν

ρ

: η(x) = 1} = ρ, ∀x T

N

.

Then, it can be checked easily that ν

ρ

, ρ [0, 1], are reversible measures for the process { η

t

}

t≥0

. To define the empirical density and rate functions, we need to introduce some definitions and notations and then discuss some topological issues. We identify T with [0, 1), and thus 0

+

with 0 and 0

with 1. By the boundary conditions imposed on the hydrodynamic equation (1.1), it is reasonable to consider test functions G C

1

[0, 1] with the property

G

(0) = G

(1) = G(0) G(1). (1.2)

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The result of this paper relies heavily on the above kind of functions, especially trigonometric functions satisfying (1.2). Define G

0

as

G

0

:= span sin

k

n

x 1 2

n≥1

[ cos 2nπx

n0

! ,

where k

n

is the unique solution to the equation

x2

= tan

x2

in (2n 1)π, (2n + 1)π

for each n 1. It can be checked easily that any G G

0

satisfies (1.2). According to [10, Theorem 1]

given by Franco and Landim, we can prove the set of the above trigonometric functions is a basis in L

2

[0, 1], which is crucial to construct the topology of this paper.

Lemma 1.1. The set

sin k

n

(x 1/2)

n1

S

cos 2nπx

n0

is an orthogonal basis of L

2

[0, 1].

We put the proof of Lemma 1.1 in the appendix.

Let M be the space of linear (not necessarily bounded) functionals on G

0

endowed with the following topology: for any A

n

M , n 1, and A M ,

n→

lim

+

A

n

= A in M if and only if lim

n→+

A

n

k

) = A

k

) for all integers k, where θ

n

(x) = sin k

n

(x 1/2)

for n 1 and θ

n

(x) = cos 2nπx

for n 0. The above topology is metrizable and the metric d ( · , · ) is given by

d A

1

, A

2

= X

−∞<n<+∞

1 2

|n|

|A

1

n

) A

2

n

) |

1 + |A

1

n

) A

2

n

)| , A

1

, A

2

M .

It can be checked directly that the space M is complete and separable under the above metric.

Note that a bounded signed measure µ on [0, 1] can be identified with an element in M in the sense that µ(f ) = R

[0,1]

f (x)µ(dx) for any f G

0

.

Remark 1.2. We construct the above topology for technical reasons. Mainly, we cannot show the uniqueness or existence of the weak solution to a PDE arising from hydrodynamics of the SSEP with a slow bond under a Girsanov’s transformed measure. However, if we do not distinguish two measures µ

1

and µ

2

satisfying µ

1

n

) = µ

2

n

) for all n, the above PDE can be reduced to an ODE on M , the existence and uniqueness of the solution to which can be rigorously proved.

For mathematical details, see Section 4 and appendix. We also underline that µ

1

n

) = µ

2

n

) for all n does not mean µ

1

= µ

2

under the usual weak topology, since the product of functions on G

0

may not be in G

0

when applying the Stone-Weierstrass Theorem [5, Theorem 7.5.3]. Indeed, the readers can check directly that the function sin(k

1

(x 1/2)) sin(k

2

(x 1/2)) does not belong to G

0

since the function does not satisfy (1.2).

In the following, we will fix a horizon time T > 0. Let D [0, T ], M

be the space of càdlàg functions from [0, T ] to M endowed with the Skorohod topology. Define the rescaled central empirical density µ

Nt

(du) as

µ

Nt

(du) := 1 a

N

X

x∈TN

t

(x) ρ)δ

x/N

(du),

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where

N a

N

N , i.e.,

lim sup

N→∞

N

a

N

= lim sup

N→∞

a

N

N = 0.

We will regard µ

N

:= { µ

Nt

}

0≤t≤T

as a random element taking values in D [0, T ], M .

Let G be the family of functions G : [0, T ] × [0, 1] R with the following forms: there exist M N and b

m

(t) C

1

([0, T ]), M m M such that

G(t, u) = X

M m=−M

b

m

(t) θ

m

(u), (t, u) [0, T ] × [0, 1].

Then for any G G ,

u

G(t, 0) =

u

G(t, 1) = G(t, 0) G(t, 1), t [0, T ]. (1.3) We sometimes write G

t

(u) for G(t, u). For G G , define the extended Laplacian ∆ ˜ as

∆G ˜

t

(u) = (

u2

G

t

(u) if u ̸ = 0,

u2

G

t

(0

+

) if u = 0.

Fix a density ρ (0, 1). Denote by Q

Nρ

the law of { µ

Nt

}

0≤t≤T

with initial distribution ν

ρ

. Let P

Nρ

be the law of the process { η

t

}

0≤t≤T

with initial distribution ν

ρ

, and E

Nρ

the corresponding expectation. Let E

νρ

be the expectation with respect to ν

ρ

. For µ D([0, T ], M ), define

I(µ) := I

ini

0

) + I

dyn

(µ), I

ini

0

) := sup

γ∈G0

µ

0

(γ) ρ(1 ρ) 2

Z

1

0

γ

2

(u) du

,

I

dyn

(µ) := sup

G∈G

T

(µ, G) ρ(1 ρ) Z

T

0

(G

t

(0) G

t

(1))

2

dt

ρ(1 ρ) Z

T

0

Z

1

0

(∂

u

G

t

(u))

2

du dt

,

(1.4)

where

T

(µ, G) := µ

T

(G

T

) µ

0

(G

0

) Z

T

0

µ

t

(∂

t

+ ˜ ∆)G

t

dt. (1.5)

Now we are ready to state the main result of the paper.

Theorem 1.3. For any closed set C of D ([0, T ], M ), lim sup

N→∞

N

a

2N

log Q

Nρ

[C] ≤ − inf

µ∈C

I (µ), (1.6)

and for any open set O of D ([0, T ], M ), lim inf

N→∞

N

a

2N

log Q

Nρ

[O] ≥ − inf

µ∈O

I(µ). (1.7)

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Remark 1.4. We recall the large deviation principle of the SSEP with a slow bond established in [11] by Franco and Neumann for a comparison. Note that definitions and notations in this remark are not utilized elsewhere. Let

π

tN

(du) = 1 N

X

x∈TN

η

t

(x)δ

x/N

(du), π

N

= { π

Nt

}

0≤t≤T

, then it was shown in [11] that, roughly speaking,

P

N

π) exp {− N J(π) }

assuming uniqueness for the weak solution to hydrodynamic equation associated to the perturbed process, where

J (π) = sup

H∈C1,2([0,T]×[0,1])

{ b

H

(π) Φ

H

(π)}

with b

H

(π) given by

b

H

(π) = ρ

T

, H

T

⟩ − ⟨ ρ

0

, H

0

⟩ − Z

T

0

ρ

t

, (∂

t

+ ∆) H

t

dt

Z

T

0

{ ρ

t

(0)∂

u

H

t

(0) ρ

t

(1)∂

u

H

t

(1) } dt + Z

T

0

t

(0) ρ

t

(1)) δH

t

(0) dt and Φ

H

(π) given by

Φ

H

(π) = Z

T

0

⟨χ(ρ

t

), (∂

u

H

t

)

2

dt + Z

T

0

ρ

t

(1) (1 ρ

t

(0)) ψ (δH

t

(0)) dt +

Z

T

0

ρ

t

(0) (1 ρ

t

(1)) ψ ( δH

t

(0)) dt,

where ρ

t

is the Radon-Nikodym derivative of π

t

with respect to the Lebesgue measure, ψ(x) = e

x

x 1, χ(ρ) = ρ(1 ρ) and δH

t

(0) = H

t

(0) H

t

(1). Since e

x

x 1 =

x22

+ o(x

2

) as | x | decreases to 0 and b

H

(π) equals

T

(π, H) when H satisfies (1.3), the rate function I

dyn

can be intuitively considered as the quadratic part of J about its minimum, which is a common relationship between large and moderate deviations for many models in statistical physics.

Notation. For deterministic positive sequences { b

n

}

n≥1

, { c

n

}

n≥1

and random sequence { X

n

}

n≥1

, we write b

n

= o(c

n

) if lim sup

n→∞

b

n

/c

n

= 0 and b

n

= O (c

n

) if lim sup

n→∞

b

n

/c

n

< C for some constant C independent of n. We also write b

n

= O

G

(c

n

) to stress the dependence on some parameter G of the constant C. We write X

n

= o

p

(c

n

) if X

n

/c

n

0 in probability as n → ∞ , and X

n

= o

exp

(c

n

) if

lim sup

n→+

1 c

n

log P | X

n

| > ϵ

= −∞ , ϵ > 0.

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We remark on these last points that the constant throughout the paper may be different from line to line.

The rest of the paper is devoted to the proof of Theorem 1.3. In Section 2 we give several super-exponential estimates that are necessary in the proof of upper and lower bounds as a preparation. Moderate upper bounds are proved in Section 3. Our proof follows a strategy similar with that introduced in [14], except for some details modified due to technical reasons caused by the slow bond. First, as introduced above, we have to choose a proper topology and to consider the empirical density as a random element taking values in the linear functional space M , instead of the dual of Schwartz functions. Second, an extra super-exponential estimate (Lemma 2.1) is needed. Third, because of the topology constructed, we have to use a different version of Minimax Theorem (Theorem 3.2) from the one in [14]. Moderate lower bounds are proved in Section 4. A crucial step in the proof is the utilizing of a generalized Girsanov’s theorem to give the hydrodynamic equation of the model under a transformed measure.

2 Super-exponential Decay

In this section, we mainly present three super-exponential estimates that are critical when making some replacements and proving exponential tightness.

Lemma 2.1. For any continuous function G : [0, T ] R and any δ, t > 0, lim sup

N→∞

1

a

N

log P

Nρ

Z

t

0

s

(0)(1 η

s

( 1)) ρ(1 ρ)) G

s

ds > δ

= −∞ . (2.1) The same result holds with η

s

(0)(1 η

s

( 1)) replaced by η

s

( 1)(1 η

s

(0)).

Proof. We only present the proof of (2.1) since the rest is the same. For any integer M > 0 and x T

N

, define η

M,R

(x) (resp. η

M,L

(x)) as the average density over the box of size M to the right (resp. left) of site x ,

η

M,R

(x) = 1 M

x+M

X

1 y=x

η(y), η

M,L

(x) = 1 M

X

x y=x−M+1

η(y).

Note that for every integer M > 0,

η(0)(1 η(−1)) ρ(1 ρ) = η(0) η

M,R

(0)

(1 η(−1)) + η

M,R

(0)(η

M,L

(−1) η(−1)) + η

M,R

(0) ρ

1 η

M,L

(−1) + ρ ρ η

M,L

(−1)

.

Since for any positive sequences { b

N

}

N≥1

and { c

N

}

N≥1

, lim sup

N→∞

1

a

N

log(b

N

+ c

N

) max

lim sup

N→∞

1

a

N

log b

N

, lim sup

N→∞

1

a

N

log c

N

,

to prove (2.1), we only need to prove for any δ > 0, lim sup

N→∞

1 a

N

log P

Nρ

Z

t

0

η

s

(0) η

M,Rs

(0)

(1 η

s

(−1)) G

s

ds > δ

= −∞, (2.2)

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lim sup

N→∞

1 a

N

log P

Nρ

Z

t

0

η

M,Rs

(0)(η

sM,L

(−1) η

s

(−1))G

s

ds > δ

= −∞, lim sup

N→∞

1 a

N

log P

Nρ

Z

t

0

η

sM,R

(0) ρ

1 η

sM,L

( 1) G

s

ds

> δ

= −∞ , (2.3) and

lim sup

N→∞

1

a

N

log P

Nρ

Z

t

0

ρ ρ η

M,Ls

( 1) G

s

ds

> δ

= −∞ . We only prove (2.2) and (2.3), since the remaining two terms are similar.

For any A > 0, by Chebyshev’s inequality, the formula on the left-hand side of (2.2) is bounded from above by

a

N

+ 1 a

N

log E

Nρ

exp

A Z

t

0

η

s

(0) η

M,Rs

(0)

(1 η

s

(−1)) G

s

ds

. (2.4)

Since e

|x|

e

x

+ e

x

, we can remove the modulus in the expectation above. By the Feynman- Kac formula (see [15, Lemma A.1.7.2] by Kipnis and Landim for example), the second term in (2.4) is bounded by

1 a

N

Z

t

0

ds sup

fdensity

AG

s

Z

η(0) η

M,R

(0)

(1 η(−1)) f (η)

ρ

− D

N

(f ; ν

ρ

)

,

where D

N

(f ; ν

ρ

) is the Dirichlet form of f associated with ν

ρ

given by D

N

(f ; ν

ρ

) := Dp

f , (−L

N

) p f

E

νρ

= N hp

f(η

1,0

) p f (η)

i

2

+ N

2

X

x∈TN , x̸=1

hp f

x,x+1

) p f (η)

i

2

.

We first write η(0) η

M,R

(0) as a telescope sum, η(0) η

M,R

(0) = 1

M

M

X

1 x=0

x−1

X

y=0

(η(y) η(y + 1)) .

Making the transformations η η

y,y+1

, by Cauchy-Schwarz inequality, we obtain that there

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exists a constant C only depending on G such that for any B > 0, AG

s

Z

η(0) η

M,R

(0)

(1 η(−1)) f (η)

ρ

= AG

s

2M

M

X

1 x=0

x−1

X

y=0

Z

(η(y) η(y + 1)) (1 η( 1)) f (η) f

y,y+1

)

ρ

AB||G||

4M

M

X

1 x=0

x−1

X

y=0

Z p f (η) p

f

y,y+1

)

2

ρ

+ A || G ||

4BM

M

X

1 x=0

x−1

X

y=0

Z p f(η) + p

f

y,y+1

)

2

ρ

C AB

N

2

D

N

(f; ν

ρ

) + AM B

.

Taking B = N

2

A

1

C

1

, we bound (2.4) by

A>0

inf

a

N

+ A

2

C

2

tM N

2

a

N

= δ

2

N

2

4C

2

tM a

N

. We prove (2.2) by choosing M = [N/2].

As above, for any A > 0, the formula on the left-hand side of (2.3) is bounded by

a

N

+ 1

a

N

log E

Nρ

exp

A

Z

t

0

η

M,Rs

(0) ρ

1 η

M,Ls

( 1) G

s

ds

.

As before, we can first remove the modulus. By Jensen’s inequality and the invariance of the measure ν

ρ

, we bound the above formula by

a

N

+ 1 a

N

log 1

t Z

t

0

ds E

νρ

exp

AtG

s

η

M,R

(0) ρ

1 η

M,L

( 1) . (2.5) By Taylor’s expansion, the expectation in the above formula is less than or equal to

X

k≥0

A

2k

t

2k

|| G ||

2k

(2k)! E

νρ

h

η

M,R

(0) ρ

2k

i

+ X

k≥0

A

2k+1

t

2k+1

|| G ||

2k+1

(2k + 1)! E

νρ

h η

M,R

(0) ρ

2k+1

i

(1 + At || G ||

) X

k≥0

A

2k

t

2k

|| G ||

2k

(2k)! E

νρ

h

η

M,R

(0) ρ

2k

i .

We claim that there exists a constant C(ρ) such that E

νρ

h

η

M,R

(0) ρ

2k

i

C(ρ)

k

k!

M

k

. (2.6)

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Since 2

k

(k!)

2

(2k)!, the expectation in (2.5) is bounded by CA exp { CA

2

/M } for some constant C = C(t, G, ρ). Therefore, we bound (2.5) by

a

N

+ CA

2

M a

N

+ log C + log A a

N

.

Recall we have set M = [N/2]. We prove (2.3) by taking A = M δ/(2C).

At last, we only need to check Equation (2.6) to complete this proof. By Fubini’s theorem, E

νρ

h

η

M,R

(0) ρ

2k

i

= Z

+

0

2kt

2k1

P

νρ

η

M,R

(0) ρ t dt.

For 0 t < 1 ρ and θ 0, by Chebyshev’s inequality, P

νρ

η

M,R

(0) ρ t

e

tM θ

E

νρ

h

e

θM

(

ηM,R(0)−ρ

) i

=

e

θt

E

νρ

h

e

θ(η(0)ρ)

i

M

=

e

θ(t+ρ)

h

e

θ

ρ + 1 ρ i

M

. Let θ = log

(t+ρ)(1ρ(1tρ)ρ)

, then

P

νρ

η

M,R

(0) ρ t

e

MI(t)

(2.7)

for 0 t < 1 ρ, where

I (t) = (1 t ρ) log(1 ρ) + (t + ρ) log(t + ρ) + (1 t ρ) log(1 ρ t) (t + ρ) log ρ.

We define I (1 ρ) = log ρ. Note that lim

t1ρ

I (t) = log ρ and P

νρ

η

M,R

(0) ρ 1 ρ

= P

νρ

(η(0) = 1)

M

= ρ

M

,

hence Equation (2.7) holds for 0 t 1 ρ and I is continuous in [0, 1 ρ]. It is easy to check that I (0) = 0 and I (t) > 0 for t (0, 1 ρ]. By L’Hospital’s rule,

lim

t↓0

I (t)

t

2

= 1 2ρ(1 ρ) .

Hence

It(t)2

is continuous and strictly positive on [0, 1 ρ]. Let J

1

(ρ) = inf

0t1ρIt(t)2

, which is strictly positive, then

P

νρ

η

M,R

(0) ρ t

e

M J1(ρ)t2

(2.8)

for all M 1 and any t [0, 1 ρ] by Equation (2.7). Note that P

νρ

η

M,R

(0) ρ t

= 0 for t > 1 ρ, hence Equation (2.8) holds for all M 1 and t 0. A similar argument proves that there exists J

2

(ρ) > 0 such that

P

νρ

η

M,R

(0) ρ ≤ −t

e

M J2(ρ)t2

for all M 1 and t 0. Let J(ρ) = inf { J

1

(ρ), J

2

(ρ) } , then

E

νρ

h

η

M,R

(0) ρ

2k

i

= Z

+

0

2kt

2k1

P

νρ

η

M,R

(0) ρ t dt

Z

+

0

4kt

2k1

e

M J(ρ)t2

dt = 2k!

M

k

(J(ρ))

k

.

Equation (2.6) follows by taking C(ρ) =

J(ρ)2

. This completes the proof.

(11)

Lemma 2.2. For any G C ([0, T ] × [0, 1]) and any δ, t > 0,

lim sup

N→∞

1

a

N

log P

Nρ

 

Z

t

0

1 N

X

x∈TN , x̸=−1

s

(x) η

s

(x + 1))

2

2ρ(1 ρ) G

s

x N

ds

> δ

  = −∞ .

Lemma 2.3. Let G C[0, 1]. Then for any t > 0, lim sup

A→∞

lim sup

N→∞

N

a

2N

log P

Nρ

sup

0⩽tT

Z

t

0

µ

Ns

, G ds > A

= −∞ , (2.9)

and for any ϵ, t > 0, lim sup

δ→0

lim sup

N→∞

N

a

2N

log P

Nρ

"

sup

|t−s|≤δ

Z

t

s

µ

Nu

, G du

> ϵ

#

= −∞ . (2.10)

The proof of [14, Lemmas 2.1 and 2.2] also applies to the above two lemmas. The main ingredients are the invariance of the Bernoulli product measure ν

ρ

. For that reason we omit the proof.

3 Upper Bound

In this section, we prove (1.6) the moderate deviations upper bound. The strategy is first proving upper bound over compact sets, and then extending to closed sets, which follows from the exponential tightness.

Fix G G . By Feynman-Kac formula (see [15, A.1.7]), M

tN

(G) := f (t, η

t

)

f (0, η

0

) exp

Z

t

0

s

f + L

N

f

f (s, η

s

) ds

(3.1) is a positive mean-one martingale, where

f (t, η) := f

G

(t, η) := exp

 

a

N

N X

x∈TN

(η(x) ρ) G

t

(x/N )

 

. Notice that

f (t, η

t

) = exp a

2N

N

µ

Nt

, G

t

. A simple calculation yields that

(∂

s

f + L

N

f)(s, η

s

) = f (s, η

s

) a

2N

N µ

Ns

, ∂

s

G

s

+ N

exp

a

N

N

s

( 1) η

s

(0))

G

s

0

N

G

s

1

N

1

+ X

x∈TN , x̸=1

N

2

exp a

N

N

s

(x) η

s

(x + 1))

G

s

x + 1 N

G

s

x

N

1 

  .

(12)

By Taylor’s expansion, N

exp

a

N

N

s

( 1) η

s

(0))

G

s

0

N

G

s

1

N

1

= a

N

s

( 1) η

s

(0))

G

s

0

N

G

s

1

N

+ a

2N

2N (η

s

(−1) η

s

(0))

2

G

s

0 N

G

s

−1 N

2

+ O

G

a

3N

N

2

,

and for x ̸= −1, N

2

exp

a

N

N

s

(x) η

s

(x + 1))

G

s

x + 1 N

G

s

x

N

1

= N a

N

s

(x) η

s

(x + 1))

G

s

x + 1 N

G

s

x N

+ a

2N

2 (η

s

(x) η

s

(x + 1))

2

G

s

x + 1 N

G

s

x

N

2

+ O

G

a

3N

N

4

.

Using the summation by parts formula, M

tN

(G) = exp a

2N

N

µ

Nt

, G

t

⟩ − ⟨ µ

N0

, G

0

⟩ − Z

t

0

µ

Ns

, (∂

s

+ ˜ ∆)G

s

ds (3.2)

Z

t

0

N

a

N

s

( 1) η

s

(0))

G

s

0 N

G

s

−1 N

ds (3.3)

Z

t

0

1

2 (η

s

( 1) η

s

(0))

2

G

s

0 N

G

s

1 N

2

ds (3.4)

Z

t

0

N a

N

s

(0) ρ)

N

G

s

0

N

s

( 1) ρ)

N

G

s

2

N

ds (3.5)

Z

t

0

1 2N

X

x∈TN , x̸=−1

s

(x) η

s

(x + 1))

2

N

G

s

x N

2

ds (3.6)

+ O

G

a

N

N

+ O

G

1 a

N

, (3.7)

where

N

is the discrete space derivative,

N

G

s

(x/N ) := N [G

s

((x + 1)/N ) G

s

(x/N)]. By the boundary condition (1.3) imposed on G, the sum of (3.3) and (3.5) is of order O

G

a

N1

. Therefore,

M

tN

(G) = exp a

2N

N

(

t

N

) Z

t

0

1

2 (η

s

( 1) η

s

(0))

2

G

s

0 N

G

s

1 N

2

ds

Z

t

0

1 2N

X

x∈TN , x̸=−1

s

(x) η

s

(x + 1))

2

N

G

s

x N

2

ds + O

G

1 a

N

 

  .

(13)

Lemma 3.1 (Upper bounds over compact sets). For any compact set K D ([0, T ], M ), lim sup

N→∞

N

a

2N

log Q

Nρ

[K] ≤ − inf

µ∈K

I (µ). (3.8)

Proof. For any δ > 0 and any G G , let

B

N,δ

=

 

 

Z

T

0

X

x∈TN , x̸=1

1

2N (η

t

(x) η

t

(x + 1))

2

N

G

t

x

N

2

dt Z

T

0

Z

1

0

ρ(1 ρ) ( G

t

(u))

2

dt du

< δ

 

 

\ (

Z

T

0

1

2 (η

t

( 1) η

t

(0))

2

ρ(1 ρ) G

t

0

N

G

t

1

N

2

dt < δ

) .

By Lemmas 2.1, 2.2 and the assumption a

N

N , lim sup

N→∞

N

a

2N

log P

Nρ

[B

N,δc

] = −∞ . Therefore, for any γ G

0

,

lim sup

N→∞

N

a

2N

log P

Nρ

N

K] lim sup

N→∞

N

a

2N

log P

Nρ

µ

N

K B

N,δ

= lim sup

N→∞

N

a

2N

log E

Nρ

h

M

TN

(G)

1

M

TN

(G)1

{µN∈K}T BN,δ

i

sup

µ∈K

T

(µ) + ρ(1 ρ) Z

T

0

(G

t

(0) G

t

(1))

2

dt +ρ(1 ρ)

Z

T

0

Z

1

0

(∂

u

G

t

(u))

2

du dt µ

0

(γ)

+ O (δ) + lim sup

N→∞

N

a

2N

log E

Nρ

M

TN

(G) exp a

2N

N µ

N0

, γ

.

Because { M

tN

(G) } is a mean one martingale and ν

ρ

is a product measure, direct calculations yield that

lim sup

N→∞

N

a

2N

log E

Nρ

M

TN

(G) exp a

2N

N µ

N0

, γ

= ρ(1 ρ) 2

Z

1

0

γ(u)

2

du.

Letting δ 0, and then minimizing over G G , γ G

0

, lim sup

N→∞

N

a

2N

log P

Nρ

N

K] inf

G∈G, γ∈G0

sup

µ∈K

T

(µ) + ρ(1 ρ) Z

T

0

(G

t

(0) G

t

(1))

2

dt +ρ(1 ρ)

Z

T

0

Z

1

0

(∂

u

G

t

(u))

2

du dt µ

0

(γ ) + ρ(1 ρ) 2

Z

1

0

γ (u)

2

du

.

In order to exchange the supremum and infimum above, we use the following version of

Minimax Theorem proved by Nikaidô.

(14)

Theorem 3.2 (Minimax Theorem, [16, Theorem 1]). Let X be a linear space endowed with separative topology and Y a linear space. Moreover, assume X is compact. Let f : X × Y R satisfy that f (x, y) is convex in y for each fixed x, and concave in x for each fixed y. Furthermore, f (x, y) is continuous in x for each fixed y. If sup

xX

inf

yY

f (x, y) is finite, then

sup

x∈X

y

inf

Y

f (x, y) = inf

y∈Y

sup

x∈X

f (x, y).

We finish the proof by taking X = K D ([0, T ], M ) , Y = G × G

0

and f (µ, (G, γ)) =

T

(µ) + ρ(1 ρ)

Z

T

0

(G

t

(0) G

t

(1))

2

dt + ρ(1 ρ)

Z

T

0

Z

1

0

(∂

u

G

t

(u))

2

du dt µ

0

(γ) + ρ(1 ρ) 2

Z

1

0

γ(u)

2

du for any µ X and (G, γ) Y.

To extend the moderate deviations upper bound to any closed set, it suffices to show the exponential tightness of the sequence { Q

N

}

N≥1

, which follows from Lemma 3.3 as in [14].

Lemma 3.3. For any G G

0

,

lim sup

A→∞

lim sup

N→∞

N

a

2N

log P

Nρ

sup

0⩽tT

µ

Nt

, G > A

= −∞ , (3.9)

and for any ϵ > 0, lim sup

δ→0

lim sup

N→∞

N

a

2N

log P

Nρ

sup

0<t⩽δ

µ

Nt

µ

N0

, G > ε

= −∞. (3.10)

We first explain why the above lemma implies exponential tightness. For any m N, k Z and any δ, A > 0, define

B

k,A

= (

sup

0≤t≤T

| µ

t

k

) | ≤ A )

, B

k,m,δ

= (

sup

0≤|t−s|≤δ

|

t

µ

s

)(θ

k

) | ≤ 1 m

) .

Then by Lemma 3.3, for any n > 0, there exist A = A(n, k) and δ = δ(m, k, n) such that sup

N≥1

Q

Nρ

B

k,Ac

< e

(a2N/N)nk

, sup

N≥1

Q

Nρ

B

ck,m,δ

< e

(a2N/N)nkm

.

Let

K

n

=

 

\

k≥1

B

k,A(n,k)

 

\ 

\

k, m≥1

B

k,m,δ(m,k,n)

 

.

It can be checked that K

n

is a compact set for each n 1. Moreover, Q

Nρ

[ K

nc

] is bounded by a

multiple of exp {− (a

2N

/N )n } . This proves the exponential tightness.

(15)

Proof of Lemma 3.3. We first prove (3.9). Since (3.4) and (3.6) are bounded, we only need to show that

lim sup

A→∞

lim sup

N→∞

N

a

2N

log P

Nρ

sup

0⩽tT

N

a

2N

log M

tN

(G) + µ

N0

, G + Z

t

0

µ

Ns

, ∆G ˜ ds > A

= −∞ , which is a consequence of

lim sup

A→∞

lim sup

N→∞

N

a

2N

log P

Nρ

sup

0⩽tT

N

a

2N

log M

tN

(G)

> A/3

= −∞ , (3.11)

lim sup

A→∞

lim sup

N→∞

N

a

2N

log P

Nρ

 1 a

N

X

x∈TN

0

(x) ρ) G (x/N )

> A/3

 = −∞ , (3.12)

and

lim sup

A→∞

lim sup

N→∞

N

a

2N

log P

Nρ

sup

0⩽tT

Z

t

0

µ

Ns

, ∆G ˜ ds > A/3

= −∞ . (3.13)

Notice that (3.13) follows from Lemma 2.3. To prove (3.11), without loss of generality, we first remove the modulus since otherwise we can replace G by G. Then

P

Nρ

sup

0⩽tT

N

a

2N

log M

tN

(G) > A/3

= P

Nρ

sup

0⩽tT

M

tN

(G) > exp Aa

2N

3N

4 exp

Aa

2N

3N

E

Nρ

h

M

TN

(G)

2

i

4 exp

C || G ||

2

+ || G

||

2

T Aa

2N

3N

.

This proves (3.11). For (3.12), removing the modulus inside the probability as before and then by Chebyshev’s inequality,

N

a

2N

log P

Nρ

 1 a

N

X

x∈TN

0

(x) ρ) G (x/N) > A/3

≤ − A/3 + N

a

2N

log E

Nρ

exp

 

a

N

N X

x∈TN

0

(x) ρ) G (x/N )

 

= −A/3 + N a

2N

X

x∈TN

log

1 + C(ρ)a

2N

N

2

G(x/N )

2

+ O

G

a

3N

N

3

≤ − A/3 + C(ρ) N

X

x∈TN

G(x/N)

2

+ O

G

a

N

N

.

This proves (3.12) by letting N → ∞ and then A → ∞ .

Next we prove (3.10). Fix A > 0, which will converge to infinity after δ 0, N → ∞ . From Equations (3.2)-(3.7) with G replaced by AG, we only need to prove (3.10) for the following four terms:

N

Aa

2N

log M

tN

(AG), Z

t

0

µ

Ns

, ∆G ˜ ds,

A Z

t

0

1

2 (η

s

( 1) η

s

(0))

2

G 0

N

G 1

N

2

ds,

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