LOCALIZATION TRANSITION FOR A POLYMER NEAR AN

263_61024404516.pdf

The Annals of Probability

1997, Vol

.25, No.3, 1334?1366

LOCALIZATION TRANSITION FOR A POLYMER

NEAR AN INTERFACE

1

BYERWINBOLTHAUSEN ANDFRANK DENHOLLANDER

University of Zurich and University of Nijmegen¨ ž.Consider the directed processi,Swhere the second componentii?0 ž.is simple random walk on?S?0.Define a transformed path measure 0 ?ž.ž.?by weighting eachn-step path with a factor exp?Ý??hsignS.

1?i?ni i

ž.Here,

?is an i.i.d.sequence of random variables taking values?1 ii?1 ž.?.with probability 1?2 acting as a random medium , while??0,?and ?.h?0,1 are parameters.The weight factor has a tendency to pull the path towards the horizontal, because it favors the combinationsS?0,i ???1 andS?0,???1.The transformed path measure describes a iii heteropolymer, consisting of hydrophylic and hydrophobic monomers, near an oil?water interface. We study the free energy of this model asn??and show that there ž. is a critical curve ??h ?where a phase transition occurs between c ž.localized and delocalized behavior in the vertical direction.We derive several properties of this curve, in particular, its behavior for ??0.To obtain this behavior, we prove that as ?,h?0 the free energy scales to its

Brownian motion analogue.

0. Introduction and main results.In this paper we solve a problem

ž.that was posed by Garel, Huse, Leibler and Orland 1989 and studied by ž.Sinai 1993.It involves a two-dimensional directed random polymer interact- ing with two solvents separated by an interface.Depending on the interac- ž.tion, the polymer either stays near the interface localization or wanders ž.away from it delocalization.The main problem is to determine the phase transition curve.

0.1.A random walk model.To define the model we need two ingredients;

ž.

1.S?S: a simple random walk on?starting at the origin, whereP,Eii?0

denote its probability law and expectation.

ž.2.

???:ani.i.d.sequence of random variables taking values?1 with ii?1

probability 1?2, where?,?denote its probability law and expectation.Received March 1996; revised June 1996.

1 Partially supported by Swiss National Science Foundation Contract 20-41925.94. AMS1991subject classifications.60F10, 60J15, 82B26. Key words and phrases.Random walk, Brownian motion, random medium, large deviations, phase transition. 1334

A POLYMER NEAR AN INTERFACE1335

?.?.Fix??0,?andh?0,1.Given?, define a transformed probability law Q ?,h,? onn-step paths by setting n ?,h,? n dQ1 nn

0.1S?exp???h?,ž. ž. ž .ž.Ý

iiii?0?,h,? dPZ ni?1 where signS,ifS?0, ž. ii

0.2??ž.

i ½ signS,ifS?0ž. i?1i ?,h,? ?ž.andZis the normalizing constant or partition sum.In 0.2 we could n ?put??0ifS?0.This would be a site rather than a bond model. ii ?,h,? ž. n We viewQas modelling the following situation.Think ofi,Sas nii?0 a directed polymer on? 2 , consisting ofnmonomers represented by the bonds in the path.The lower half plane is ''water,'' the upper half plane is ''oil.'' The monomers are of two different types, occurring in a random order indexed by ?.Namely,???1 means that monomeri''prefers water,''???1 means ii that it ''prefers oil.'' Since???1 when monomerilies in the water and i ž.???1 when it lies in the oil, we see that the weight factor in 0.1 i ''encourages matches and discourages mismatches.'' Forh?0 both types of monomers interact equally strongly with the water and with the oil, being ž.attracted by one and repelled by the other.However, forh?0,1 the monomers preferring oil have a stronger interaction with both the solvents than the monomers preferring water.The parameter ?is the overall interac- tion strength and plays the role of inverse temperature. ž. R EMARK.In0.1 we could put theh-dependence in the probability law of

ž.ž.

?, for instance, by picking????1?1?h?2 and writing?Ý??in iiii the exponent.This would describe a polymer where the two types of monomers occur with different densities but interact equally strongly with the solvents. žAlternatively we could make a mix of the two types ofh-dependence or even .allow for more general ?-sequences with exponential moments.For the proofs in this paper it is a slight advantage thathenters into the exponent. Nevertheless, all results carry over with only minor changes in the proofs. The way in which the polymer behaves near the interface is the result of a žcompetition between energy and entropy.The energy is minimal i.e., the .weight is maximal when all the monomers are placed in their preferred solvent, but this strategy has low entropy.On the other hand, the entropy is maximal when the polymer makes large excursions away from the interface, ž.but this strategy typically has high energy i.e., the weight is small.What do we expect will happen underQ ?,h,? asn??? n

1.??0.The vertical motion of the polymer is free simple random walk.

Since this is a null recurrent process, the polymer will not stay near the interface; that is, we have delocalization.

E.BOLTHAUSEN AND F.DEN HOLLANDER1336

2.??0,h?0.The polymer will want to stay close to the interface, so that

it can place as many monomers as possible in their preferred solvent and produce low energy.Indeed, wandering away from the interface would result in a misplacing of about half the monomers.The polymer can reduce this fraction by crossing the interface at a positive frequency.This lowers the entropy, but only by a small amount if the crossing frequency is small. ž.The estimates in Sinai 1993 show that for this strategy the gain exceeds the loss; that is, we have localization. 3. ??0,h?1.Now wandering away is again the winning strategy, simply because the monomers preferring water barely interact with either the water or the oil.By moving away in the upward direction the polymer can match all the monomers that prefer oil, thereby producing almost the minimal energy and almost the maximal entropy; that is, we have delocal- ization. The above intuitive picture seems to suggest that there is a critical curve

ž.in the

?,h-plane separating the localized from the delocalized phase.It is the goal of the present paper to prove the existence of this critical curve and to derive some of its properties. In order to give a precise definition of the two phases, we need the ž.following preliminary result proved in Section 1. ?.?.T

HEOREM1.For every??0,?and h?0,1 ,

1 ?,h,?

0.3 lim logZ???,hž. ž .

n nn?? exists?-a.s.and is nonrandom.

The function

?is the specific free energy of the polymer.It is immediate from

ž. ž. ž .0.1 and 0.3 that

??,his continuous, nondecreasing and convex in both ?variables.Note that our model makes perfect sense for ?,h??.Obviously,

ž.in this larger parameter space,

??,his everywhere finite, is symmetric and convex in both variables and hence is also continuous and unimodal in both ?variables.Moreover, it is easy to show that 0.4 ??,h??h.ž. ž . ž. 1?2 ž.Indeed, sinceP???1 for 1?i?n?C?nn??, it follows that i n ?,h,?

Z?Eexp???h?ž.

Ý nii Ž/ i?1 n ?exp???h?Olognž.ž. Ý i i?1

0.5ž.

?exp?hn?on,?-a.s.,ž. where in the last step we use the strong law of large numbers for?.Thus we ž.see that the lower bound in 0.4 corresponds to the strategy where the

A POLYMER NEAR AN INTERFACE1337

polymer wanders away in the upward direction.This leads us to the following definition. D

EFINITION1. We say that the polymer is:

ž. ž .

a localized if ??,h??h,

ž. ž .b delocalized if

??,h??h. ž. In case a the polymer is able to beat on an exponential scale the trivial strategy of moving upward.It is intuitively clear that this is only possible by crossing the interface at a positive frequency, which means that the path ž.measure localizes near the interface in a strong sense.In case b , on the other hand, the polymer is not able to beat the trivial strategy on an exponential scale.In principle it could still do better on a smaller scale, but ?we do not expect this at least not in the interior of the region described by ž.?b.We shall not derive any properties of the path measure, but just stick to ž.the above definition.See Section 0.4 for a further discussion.

Our first main theorem reads as follows.

ž. ž.ž.

T HEOREM2.For every??0,?there exists h??0,1such that the c polymer is: localized if0?h?h ?,ž. c

0.6ž.

delocalized if h?h?.ž. c

Moreover,

ž.?.

??h?is continuous and nondecreasing on0,?, c

0.7ž.

ž. ž.limh??1, limh??0.

???c??0c The proof of Theorem 2 is given in Section 2.It will also provide upper and

ž.lower bounds onh

?, namely: c 1 i limsuph??1,ž. ž . c ? ??0 1

0.8ž.ii liminfh??0,ž. ž.

c ???0 13 iii lim?1?h??log2, log2.ž. ž.ž. c 22???

0.2.A Brownian motion model.As??0, the reward to stay close to the

interface gets smaller and so the excursions of the polymer away from the interface will get longer.Therefore, intuitively we may expect to see a scaling behavior where bothSand ?can be approximated by Brownian motions.To make this more precise, we first define and describe the continuous analogue of the discrete model.As we shall see in Section 0.3, the scaling happens in a way that leads to a Brownian motion model.This model retains the full complexity of the random walk model, except that the Brownian scaling property gives rise to a simpler form of the phase separation curve.

E.BOLTHAUSEN AND F.DEN HOLLANDER1338

The two ingredients of the continuous model are two standard Brownian motions on?, denoted by:

ž. ž .

1B?B, tt?0

ž. ž .2

???, tt?0

˜˜ ˜˜

both starting at the origin.We writeP,E, respectively,?,?, to denote their ž. ž.probability law and expectation.Similarly as in 0.1 and 0.2 , the trans- ˜ ?,h,? formed probability lawQon paths of lengtht, given?, is defined by t ?,h,? ˜ dQ1 t t

0.9B?exp??d??hds.ž. ž. ž .ž.H

sss0?s?t ?,h,?

˜˜

dP Z0 t Here, signB,ifB?0, ž. ss

0.10??ž.

s ½

0, ifB?0,

s ?.the first integral is an Ito integral, and the parameters ?,hare both in 0,?.ˆ ž.The analogue of Theorem 1 proved in Section 3 reads as follows. ?.T

HEOREM3.For every?,h?0,?,

1 ?,h,?

˜˜

0.11 lim logZ???,hž. ž.

t tt?? ˜ exists?-a.s.and is nonrandom. ˜ž.The function?has the same qualitative properties as?in 0.3 , including ž.the lower bound in 0.4.Therefore we can maintain the same distinction between phases as in Definition 1.

The Brownian scaling property tells us that

0.12B,

??aB2,a?2for alla?0,ž. ž .ž. ss D s?as?as?0s?0 where?means equality in distribution.This implies that, for fixed?,h D and as a random variable in?, ˜ ?,h,? ˜ a?,ah,?

20.13Z?Zfor allt?0 anda?0.ž.

tDt?a Hence 1

˜˜

0.14??,h??a?,ahfor alla?0.ž. ž. ž .

2 a

˜ž.It immediately follows from 0.14 that

?has the followingscaling form: ˜ 2 ž.ž.?.ž.??,K??SSK?forK?0,?, withK?SSKcon-0.15 ž. ž.tinuous, nondecreasing and convex, satisfyingSSK?K. ž. The analogue of Theorem 2 proved in Section 3 now reads as follows.

A POLYMER NEAR AN INTERFACE1339

ž?THEOREM4.There exists K?0,1such that

c

SSK?KifK?K,ž.

c

SSK?Kif0?K?K.ž.

c

0.16ž.

˜˜ž. ž. ž.By 0.15 , Theorem 4 implies that ??,h??hforh?K?and??,h??h c forh?K?; that is, the phase separation curve is the straight line??K?. cc Although the picture here looks fairly simple, the complexity of the model is hidden in the constantK, which seems to be a very ungainly and complex c object.We have rough bounds onK, but nothing like a sequence of bounds c that could be expected to converge toK. c

0.3.Weak interaction limit.We are now ready to formulate our main

results concerning the weak interaction limit of the random walk model and its relation to the Brownian motion model. ?.T

HEOREM5.For every?,h?0,?,

1 ˜

0.17 lim?a?,ah???,h.ž. ž . ž.

2 aa?0 ž.Although 0.17 is intuitively plausible, the estimates needed for its proof are quite delicate.The reason is that our paths carry exponential weight factors, which are very sensitive to fluctuations.One should keep in mind that, at least in the localized region, the path exhibits a behavior that has an exponentially small probability under the free path measure.It is therefore clear that the result cannot be proved by a routine application of invariance principles. We shall not prove Theorem 5 separately, as it is a consequence of the more powerful but more technical Theorem 6 below.A proof of Theorem 5 ž.would be simpler and more transparent than that of Theorem 6 given in Section 4.However, the unfortunate fact is that Theorem 5 alone does not lead to a determination of the tangent at ??0 of the phase separation curve in the discrete model.In fact, it only yields 1

0.18 liminfh??K.ž. ž.

cc ???0

ž.ž.Indeed, pickK?K.Then, by 0.15?0.17 ,

c 1 ˜

0.19 lim?a,aK??1,K?K.ž. ž . ž.

2 aa?0 ž. 2

ž.This implies

?a,aK?Kaand henceha?aKfor small enougha, c ž.which proves 0.18 after lettinga?0 followed byK?K.It is clear that a c

ž.statement like 0.17 does not yield

1

0.20 suph??K,ž. ž.

cc ? ??0

E.BOLTHAUSEN AND F.DEN HOLLANDER1340

˜ 2

ž. ž .simply because

?1,K?KforK?Kdoes not imply that?a,aK?aK c for small enougha. In order to remedy this situation, we introduce the 'excess' free energies ??,h???,h??h,ž. ž.

0.21ž.

˜˜

??,h???,h??h,ž. ž. ˜ so that the delocalized region is characterized by??0, respectively,??0. Our main result for the weak interaction limit is the following. ? ž. ?

THEOREM6.Fix??0.Let h?0,h?0and??0satisfy1??h?h.

Then 1 ? ˜ ?a?,ah?1????,h,ž.ž.ž. 2 a0.22 ž. 1 ? ˜ ??,h?1???a?,ahž.ž . ž . 2 a for small enough a. ˜ Theorem 6 and the continuity of?and?obviously imply Theorem 5. Theorem 6 is also sufficiently strong to give us the following corollary. C

OROLLARY1.

1

0.23 limh??K.ž. ž.

cc ???0

ž. ž.

? To get 0.20 from the first line in 0.22 , pickh?K,??0 and??1, c

˜ž. ž. žž..h?1?2

?K.Since?1,K?0, it follows that?a,a1?2?K?0 cc c ž. ž .and henceha?a1?2?Kfor small enougha.Now leta?0 and??0. cc The idea behind Theorem 6 is that by slightly varyinghwe can dominate the errors that arise in the approximation of the random walk by the

Brownian motion.

R EMARK. Theorem 6 can be shown to carry over to the version of the model where theh-dependence sits in the probability law of ?.For the Brownian motion model there is no distinction between the two versions. Apparently, the weak interaction limit is largely independent of the details of the model.This is essentially a stability result.Stability is crucial for our understanding of the localization problem, and typically hard to prove for path measures with exponential weight factors.

0.4.Open problems.Our distinction between the localized and the delo-

calized phase, as given in Definition 1, is in terms of the specific free energy rather than the path measure itself.We would like to show that in the ž.localized phase ''Struly localizes,'' in the sense that it stays close to i0?i?n the horizontal, while in the delocalized phase it does not.For instance, consider the following two questions.

A POLYMER NEAR AN INTERFACE1341

?,h,? ž.1.For fixedi, doesQS??converge to a nondegenerate limit law as ni n??? ž. ?,h,?

ž??4?2.Is there ad?d

?,h?0 such that limQ1?i?n:S?0?n n??ni ??.?d? ?,d???1 for all??0? No doubt the answer is ''yes'' in the localized phase and ''no'' in the delocal- ized phase, but this remains to be proven.Other interesting questions are: How does the free energy behave close to the critical curve? How large are the excursions of the path away from the horizontal?

ž.Sinai 1993 proved that if

??0,h?0, then the path localizes in the

ž.following sense: there exist numbers

??0,???0 and random variables

ž. ž.n

?,k?such that 00 ?,0,? ?? ??ž?.k supQS?k?ež. ni ?? logn?i?n?logn

0.24ž.

fork?k?,n?n?,?-a.s.ž. ž. 00 We expect that Sinai's arguments can be extended to cover the whole localized region. ž. ž.One could hope to make some progress on problems 1 and 2 above by looking at the times when the path intersects the interface.In the localized ž.region these times admit a Gibbsian description in the limit asn??. However, this leads to a Gibbs measure with a random long-range potential having both signs, which is a notoriously difficult object.Nevertheless, we expect that a limiting measure exists and that it has exponentially decaying correlations. Even the delocalized region is not trivial.It seems intuitively clear that, at ?ž.?least in the interior of this region i.e., forh?h ?, the path just behaves as c simple random walk conditioned to stay positive, which is well known to have Brownian scaling with the so-called Brownian meander as limiting measure ?ž.?see Bolthausen 1976.However, it appears to be difficult to exclude the possibility of rare returns to the interface. ž.Grosberg, Izrailev and Nechaev 1994 obtain localization for the case where ?is periodic instead of random. ž. ?,0,? Albeverio and Zhou 1996 prove that if??0,h?0, then logZ n ž.satisfies a LLN and a CLT as a random variable in?.However, there is no description of the mean and the variance.They further show that ?,0,?

ž.HQ?d?-a.s.both

n maxj?i:S?S?0,S?0 fori?k?j,?4 ij k

0?i?j?n

0.25ž.

??maxS i 0?i?n are of order lognasn??, which is typical for a localized path. ž. ž.Grosberg, Izrailev and Nechaev 1994 and Sinai and Spohn 1996 study an annealed version of the model in whichZ ?,h,? is averaged w.r.t.?.The n free energy and the critical curve can in this case be computed exactly.

E.BOLTHAUSEN AND F.DEN HOLLANDER1342

However, the quenched version described in the present paper is qualita- tively very different and considerably more complex.

1. Proof of Theorem 1.The proof consists of two parts.In Lemma 1 we

prove that the claim holds when the random walk is constrained to return to the origin at time 2n.In Lemma 2 we show how to remove this constraint. Fix ?andh.Define 2n ??, ?41.1Z?Eexp???h?1S?0,ž. ž . Ý

2nii2n

Ž/ i?1 where we recall the notation introduced in Section 0.1. ž. ?,? LEMMA1.The limitlim 1?2nlogZ exists and is constant?-a.s. n??2n

PROOF. We need the following three properties.

?,??,?T 2m ?,?

ž.I.Z?ZZfor all 0?m?n, withTthe left-shiftT

??

2n2m2n?2mi

?. i?1

ž.ž

?,? .II.n?1?2n?logZis bounded from above. 2n

ž.ž.III.?T

????????. ž. ?4 Property I follows from 1.1 by inserting an extra indicator 1S?0 and 2m using the Markov property ofSat time 2m.Property II holds because ?logZ ?,? ?log?Z ?,?

ž.ž.

2n2n 2n 2n ?4?logEcosh?exp?h?1S?0ž. Ý i2n Ž/ i?1

1.2ž.

?2nlogcosh???h.ž. ž ?,? .Property III is trivial.Thus,??logZis a superadditive process.It 2nn?0 ?ž.therefore follows from the superadditive ergodic theorem Kingman 1973 , ?ž. ?,? Theorem 1 that lim 1?2nlogZconverges?-a.s.and in mean, and is n??2n measurable w.r.t.the tail?-field of?.Since the latter is trivial, the limit is constant?-a.s.?

Our original partition sum was

2n ?

1.3Z?Eexp???h?,ž. ž .

Ý 2nii Ž/ i?1 ž.which is 1.1 but without the indicator.Thus, in order to prove Theorem 1 we ?ž ?? .?must show that this indicator is harmless asn??.Since logZ?Z?

2n2n?1

ž. ?1?h, it will suffice to considerneven. L

EMMA2.There exists C?0such that Z

?,? ?Z ? ?CnZ ?,? for all n

2n2n2n

and?.

A POLYMER NEAR AN INTERFACE1343

PROOF. The lower bound is obvious.The upper bound is proved as follows. By conditioning on the last hitting time of 0 prior to time 2n, we may write n2n ????,?,

Z?Z?ZEexp???h?ž.

ÝÝ

2n2n2n?2kii

Ž k?1i?2n?2k?1 ?? ?1A?AS?0?4 n,kn,k2n?2k /

1.4ž.

n2n a k?? ?,?, ?Z?ZEexp???h?ž.

ÝÝ

2n2n?2kii

Žb kk?1i?2n?2k?1 ?? ?1B?BS?0.?4 n,kn,k2n?2k /

Here we abbreviate the events

? ?4A?S?0 for 2n?2k?1?i?2n, n,ki ? ?4B?S?0 for 2n?2k?1?i?2n,S?0 n,ki2n

1.5ž.

and similarly forA ? ,B ? , and their probabilities n,kn,k ? ? ? ?a?PA S?0?PA S?0,

ž.ž.

kn,k2n?2kn,k2n?2k

1.6ž.

? ? ? ?b?PB S?0?PB S?0

ž.ž.

kn,k2n?2kn,k2n?2k ž. ž.both independent ofn.The reason for the second equality in 1.4 is that ???1 for all 2n?2k?1?i?2non the eventsA ? ,B ? and???1 in,kn,ki ??

ž.for all 2n?2k?1?i?2non the eventsA,B

?is fixed. n,kn,k

Next, there existC,C?0 such thata?C?k

1?2 andb?C?k 3?2 for

12k1k2

ž.allk?1.Moreover, without the factora?bthe last sum in 1.4 is precisely kk Z ?,? .Hence 2n C 1? ??,

1.7Z?1?nZ.?ž.

2n2n Ž/ C 2

Lemmas 1 and 2 complete the proof of Theorem 1.

2. Proof of Theorem 2.The proof proceeds in a sequence of five steps,

?ž.?organized as Sections 2.1 and 2.2.Define recall 0.21 2.1 ??,h???,h??h.ž. ž. ž. Let

2.2DD?

?,h:??,h?0?4ž. ž.ž. ž.be the region of delocalization see Definition 1. ž.

2.1.Existence,continuity and monotonicity of h

?. c ž. ž .STEP1. If?,h?DD, then???,h???DDfor all?,??0 satisfying ž. ???1?h??.

E.BOLTHAUSEN AND F.DEN HOLLANDER1344

n

ž. ž.P

ROOF. Since?Ý??h??hn?on?-a.s., we have the following i?1i ?ž.?equivalence recall that??0by 0.4: ??,h?0ž. n 1 ?lim logEexp???h??1?0,?-a.s.ž.ž. Ý ii Ž/ nn??i?1

2.3ž.

ž.Thus, to prove the claim we must show that if the r.h.s.of 2.3 holds for

ž. ž .

?,h, then it also holds for???,h??.To see this, write n ?????h????1ž.ž .ž . Ý ii i?1 n ????h??1ž.ž. Ý ii i?1

2.4ž.

n ????h????????1.ž. ž. Ý ii i?1 Since??1 and???1, the last sum is less than or equal to 0 when ii ž. ??1?h????0.? ?.For ??0,?define ??2.5h??infh?0,1 :?,h?DD.?4ž. ž. ž . c žž..By continuity of?, we have?,h??DD.It therefore follows from Step 1 c ž. ž.that?,h?DDfor allh?h?, so that the localized and the delocalized c ž.phase are separated by a single critical curve:??h?. c ž. ž .?.STEP2. i??h?is continuous and nondecreasing on 0,?. c

ž. ž ž..?.ii

???1?h?is continuous and nondecreasing on 0,?. c ž. ž .PROOF. i We know that??,h?0 is convex in?with boundary value

ž. ž. ž .

?0,h?0.Therefore, if?,h?DDthen also???,h?DDfor all??0.

ž.Hence

??h?is nondecreasing.Step 1 shows that its slope at the point? c žž..is bounded from above by 1?h???.Since this is finite for??0, we get c ž. ž.continuity on 0,?.Continuity at??0 follows from Step 3 i.

ž.ii This is easily deduced from Step 1.?

ž.

2.2.Bounds on h

?. c ž. ž . ž .STEP3.h??1?2?logcosh 2?.Consequently: c

ž. ž . ž .i limsup 1?

?h??1, ??0c 1

ž. ž ž..ii liminf?1?h??log2.

???c2

ž.P

ROOF. The claim will follow once we prove that?,h?DDfor allh?

ž. ž.1?2

?logcosh 2?.This will be done by checking the property in the r.h.s.of

ž.2.3.

A POLYMER NEAR AN INTERFACE1345

ž.Estimate??,hfrom above as follows:

n 1 ??,h?lim?logEexp???h??1ž. ž .ž . Ý ii

Ž/Ž/

nn??i?1 n 1 ?liminf logE?exp???h??1ž.ž. Ý ii

Ž/Ž/

nn??i?1

2.6ž.

n 111
?2?ž1?h.?2?ž?1?h. ?liminf logE1e?e. Ł ????14 i Ž/ n22n??i?1 žThe first equality is a direct consequence of the superadditivity see Section .1.The r.h.s.is less than or equal to 0 as soon as the term between square brackets is less than or equal to 1.?

ž.ž.

S

TEP4. liminf 1??h??0.

??0c PROOF. The idea is to find a strategy of the polymer for which the

ž.contribution to the free energy exceeds

?hsee Definition 1.The computa- tions below are easy but a bit lengthy, due to a necessary fine-tuning of constants.The proof comes in three parts. ž. i As was shown in Section 1, 1 ?

2.7??,h?lim?logZž. ž .ž.

n nn?? ? ?ž.? ? withZour partition sum see 1.3.We begin by rewritingZin terms of nn the excursions ofSaway from the origin.To that end, define ??0,??infi??:S?0,j?0,?4

0j?1ji

??maxj?0:??n?4 nj

2.8ž.

and 2.9 ?x?logcoshx.ž. ž. Let ? n

2.10HS,???? ??h??? ??h.ž.ž. ž. ž.

ÝÝ Ý

nii

Ž/Ž/

j?1ž?ž?i??,?i??,n j?1j? n Then, using the up?down symmetry ofSfor each excursion, we can write ?

2.11Z?EexpHS,?.ž. ž.ž.

nn ž.ii The length of a typical free excursion has distributionfgiven by l2 "

2.12zf l?1?1?z,ž. ž.

Ý l which is the generating function for the probability of first return to the ž.origin of simple random walk.In order to bound 2.11 from below, we shall

E.BOLTHAUSEN AND F.DEN HOLLANDER1346

be looking for a strategy of the path in which the excursions have distribution 1l ?2 "2.13fl?fl1??,l?1.ž . ž. ž. Ž/ 1?? žThis corresponds to a random walk with drift?towards the origin i.e., 1

ž?ž.?..S?S??1 with probability 1?

??signSfori?0.Here 0? i?1ii2 ??1 is a parameter we shall optimize over.

The following lemma is an intermezzo.Abbreviate

??Ý?. Ii?Ii ?.?.LEMMA3.For all??0,?and h?0,1 ? ? ,h?supfl??????hlž. ž.ž.ž.Ý

ž0,l?½

1??

0???1l

2.14ž.

1??1??

?log 1???log 1??.ž. ž. 5 2?2?

PROOF. LetP

0 andP ? denote the laws of simple random walk, respec- nn

ž.tively, random walk with drift

?, restricted ton-step paths.Then from 2.13 , ? ? ?? nÝflž. dP fl?n?? ?nn n

S????ž. ž .ž.Łijj?1i?00

fÝfldPž. j?1l?n??n2.15ž.? n n?2? ??12 n ?1??1??.ž.ž. ž. ž.Using Jensen's inequality, we get from 2.11 and 2.15 that ? dP n ?? logZ?logEexpHS,??logž. nnn0 Ž/ dP n dP ? n ?? ?EHS,??Elogž.ž. nn n0 Ž/ dP n

2.16ž.

n ??2 ?EHS,??E??1 log 1???log 1??.ž. ž .ž .ž.ž. nn nn 2 ?4 Now,??1?minj?0:??nis a stopping time.Moreover, a straightfor- nj ? ž. ž .ward calculation yieldsE??1????.Therefore, the optional sampling n1 theorem gives us 1 ? ?

2.17 limE??1?.ž. ž .

nn n1??n??

ž. ž.ž

? .In order to bound ??,h?lim 1?n?logZ, it therefore remains to n??n consider ?E ?

HS,??E

? ?HS,?ž. ž.ž.ž.ž.ž. nn n n ? n ? ?E??? ??h.ž.

ÝÝ

ni Ž/

Ž/Ž/

j?1ž?i??,? j?1j

2.18ž.

A POLYMER NEAR AN INTERFACE1347

By stationarity of the?-sequence, the summands are functions of??? jj?1 only.Applying the optional sampling theorem again we get ? n 1 ? limE??? ??hž.

ÝÝ

ni Ž/

Ž/nn??Ž/

j?1ž?i??,? j?1j

2.19ž.

? 1 ? ? ?E??? ??h.ž. Ý ni

Ž/Ž/Ž/1??

j?1

žTo handle the last excursion

???, note that?is linearly bounded and ??1? nn ? .that the excursion times have an exponential moment underP.Putting the n estimates together we obtain the claim.? ž. iii The proof of Step 4 can now be complete as follows.Because ??0 and ?is convex, we have ? ?????hlž.ž.

ž0,l?

????0??????hl???0ž.ž .ž.

ž0,l?ž0,1?ž0,l?

2.20ž.

1 ????????0??hl.ž.ž.

ž0,l?ž0,l?2

ž. 1?2 Next, note that there existsA?0 such that?????0?Alfor

ž0,l?ž0,l?

ž. ž.alll?1.Now pickh???and????in Lemma 3, insert 2.13 and 2.20 ,

ž.?ž.

l ? 3?2

ž.and use thatfl?1??1B?ll??, to obtain

1 ?

2.21 liminf??,???BI A,?,???,ž. ž . ž .

2

2?????0

where ?dx1 2 "

2.22IA,?,??exp??x?Ax??x.ž. ž .ž.H

3?2 Ž/ 2x 0 ž.The constants?,?can still be optimized.PickM?2?BI A,0,0 and put ž. ??M?.Then, as??0, the r.h.s.of 2.21 converges to a number greater ž. 2 than 1.Therefore we have proved that??,?????for?,?sufficiently ž.small.This proves the claim in Step 4 recall Definition 1.? 3

žž..STEP5. lim?1?h??log2.

???c2

ž.P

ROOF. Recall Step 2 ii.The claim is proved as follows.As???, the path will tend to make short excursions.Therefore we bound the partition sum from below by requiring all excursions to have length 2: 2n ? ?4Z?Eexp???h?1S?0 for 0?m?nž. Ý

2nii2m

Ž/ i?1

2.23ž.

n n 1 ???cosh?????2h.ž.ž.Ł

2m?12m2

m?1

E.BOLTHAUSEN AND F.DEN HOLLANDER1348

ž.?Use the up?down symmetry ofSfor each excursion.It follows that recall

ž.?2.9:

1 ? ,h?lim?logZž.ž. 2n 2nn?? 11 ????log2????????2hž.ž. 12

222.24

ž. 1 ??log22

11 1 1

??2?1?h??2?1?h??2?h.ž. ž. ž.ž.ž. ½5

24 4 2

ž. ž

?2x .ž .Next, insert ?x?x?log2?Oe x??.Pickh?1?M??with

M?0 arbitrary.Then for

???,

MM13 1

??,1???1??M?log2??2Mž. ½5

Ž/Ž/

??48 82.25 ž. ?Oe ?4? .ž. 3 As soon asM?log2, the term between braces is greater than 0, implying 2 ž.?ž. ž.?that?,1?M???DDfor?sufficiently large cf.2.2 and 2.3.But then

ž.?ž.?žž..h

??1?M??for?sufficiently large cf.2.6 , that is,?1?h??M. cc ?

ž. ž . ž .

Steps 2?5 prove Theorem 2 as well as Properties i?iii in 0.8.

3. Proofs of Theorems 3 and 4.Essentially the same arguments as in

the proofs of Theorems 1 and 2 carry over to the continuous case.We only indicate which points need modification. ?4ž.3.1.Proof of Theorem3.We cannot insert 1B?0 , sincePB?0?0 tt ?ž.?compare with 1.1.However, this problem is easily handled through a comparison argument.Recall the notation introduced in Section 0.2.

Define

t ??,

˜˜??3.1Z?infEexp

??d??hds1B?1B?x.?4ž. ž .

Htsst0

Ž/ ??x?10 Then: ˜ ?,? ˜ ?,? ˜ T u ?,?uu

ž.I.Z?ZZfor all 0?u?t, withTthe left-shiftT

?? tut?us ???. u?su

˜˜

?,?

ž.ž .II.t?1?t?logZis bounded from above.

t ˜ u

˜ž.ž.III.?T????????for allu?0.

Properties I and III are obvious.Property II holds because

˜˜

?,?

˜˜

?,? ?logZ?log?Z

ž.ž.

tt t ˜˜??3.2?logE?exp??d??hds1B?1B?0?4ž. ž . H ss t0

Ž/Ž/

0

A POLYMER NEAR AN INTERFACE1349

t 12

˜???logEexp?t??h?ds1B?1B?0?4

H st02 Ž/ 0 12 ?t???h, ž. 2 where the equality follows from the martingale property tt 122
˜ ?3.3?expfsd??expfsds,f?L0,t.ž . ž. ž..ž.HH s2 Ž/ 00 ˜ ?,?

ž.Thus,

??logZis a superadditive process. tt?0 In order to apply the superadditive ergodic theorem, we need an additional ?regularity condition that is absent in the discrete time setting, namely see ž.?Kingman 1973 , Theorem 4 the following property. IV.

˜˜

?,? ??3.4?sup logZ??for allT??, ž. s,t Ž/

0?s?t?T

˜ ?,? ?.whereZis the partition sum over the time intervals,t; that is, s,t t ??,

˜˜??3.5Z?infEexp

??d??hdu1B?1B?x.?4ž. ž .

Hs,tuuts

Ž/ ??x?1s To prove Property IV, we first note that, for all?, ˜ ?,?

˜??3.6 infZ?inf infPB?1?B?x?0 for allT??.

ž.ž.

s,tts

0?s?t?T0?s?t?T??x?1

˜ž.Use Jensen's inequality together withE??0.Hence it suffices to prove u ž.3.4 without the absolute value signs.But this we may estimate as follows:

˜˜

?,? ?sup logZ s,t Ž/

0?s?t?T

t

˜˜

?logE?sup exp??d??hduž. Huu

Ž/Ž

s

0?s?t?T

3.7ž.

 ?1?B??1B?0. ts /

ž. ž .

t The exponent in 3.7 is bounded from above by?ht?s??H?d?. su u ˜ Moreover, we note that under the law?the last integral is just Brownian 2 ˜ motion, since??1 almost everywhereP-a.s.Thus we obtain u ??,

˜˜ ˜

3.8?sup logZ??hT?log?sup exp????.ž. ž .

s,ttsŽ/Ž/

0?s?t?T0?s?t?T

??But the last integral is finite, because 2 ?sup?has an exponential

0?u?Tu

moment.This proves Property IV. Properties I?IV guarantee that the superadditive ergodic theorem applies: ˜ ?,? ˜ž.lim 1?tlogZconverges?-a.s.and in mean, and is t??t

3.9ž.

˜ constant?-a.s.

E.BOLTHAUSEN AND F.DEN HOLLANDER1350

ž.Thus we have the LLN for the quantity defined in 3.1.In order to get it for ???4our original partition sum, it remains to remove 1B?1 and inf from t?x??1

ž.3.1.This will be done in two pieces.

Define

t ??,

˜˜??Zx?Eexp

??d??hds1B?1B?x,?4ž. ž . H tsst0 Ž/ 0 t ?

˜˜

Zx?Eexp??d??hds B?x.ž. ž .

H tss0 Ž/ 0

3.10ž.

˜ ?,? ˜ ?,? ž. ž.In 3.9 we have the LLN forZ?infZx.The key estimates are t?x??1t now ˜ ?,? ˜ ?,? ˜ ?,? iZ?Z0?C?Zfor alltand?,ž. ž . ž . tt t

3.11ž.

˜ ?,? ˜ ? ˜ ?,? iiZ0?Z0?CtZ0 for alltand?.ž. ž. ž. ž. tt t ž.The lower bounds are trivial.The upper bound in ii is obtained from an ž.almost literal transcription of the proof of Lemma 2.The upper bound in i follows from a coupling argument.Indeed, since two Brownian motions starting at 0, respectively,x, hit each other after a finite time a.s., we have ˜ ?,? ˜ ?,? ˜?ž. ž .?ž. ž.supZ0?Zx?C?withC???,?-a.s. ?x??1tt ˜ ? ˜ ? ž. ž.The conclusion of 3.11 is that our original partition sumZ?Z0 has tt

˜˜

?,? ž. ž.the same?-a.s.constant growth rate asZin 3.1. t ž??3.2.Proof of Theorem4.All we have to do is show thatK?0,1 since c ž.?ž. ž .the rest follows from 0.15.As this inclusion follows from 0.8 and 0.23 , strictly speaking there is no need to give a proof here.Still, we indicate a direct proof of the lower bound forKbecause it is instructive. c

Fix?,h.In Section 3.1 we saw that

1 ?

˜˜˜

3.12??,h?lim?logZ.ž. ž.ž.

t tt?? We begin by expressing our partition sum in terms of the excursions ofB ?4?.away from the origin.LetNN?s?0:B?0.Then 0,??NN??Iis a sjj ?countable union of disjoint open intervals having full measure Revuz and

ž.?Yor 1991 , Chapter XII.Let

?43.13J?j:I?0,t?0,ž. .?4 tj where we reserve the index 0 for the interval betweentand the last hitting time of the origin prior to timet.Then, using the up?down symmetry ofBfor each excursion, we can write ?

˜˜??3.14Z?Eexp?????hI.ž.

Ýž.tIj

j Ž/ j?J t

A POLYMER NEAR AN INTERFACE1351

Here,?denotes the increment of?over the setI, and?was defined in I ž. ž. ž.2.9.The representation in 3.14 is the continuous analogue of 2.10 and

ž.2.11.

˜ ? ˜ ? Fix??0.LetP,Edenote the probability law and expectation of Brown- ian motion with drift ?towards the origin.Then it follows from the ?ž.?Cameron?Martin formula Chung and Williams 1990 , Theorem 9.10 , re- ?ž.?spectively, the Tanaka formula Revuz and Yor 1991 , Theorem VI.1.2 , that ˜ ? dP

Bž.ž.

s0?s?t0 ˜ dP 1 tt 2 ?exp??signBdB???signBds?4?4ž. ž. HH ss s 2 00

3.15ž.

1 2 ???exp ?L?B??t,ž. tt 2 ?.whereLis the local time at the origin in the time interval 0,t.Next, t 11???

˜˜ ˜ž. ž?

according to Tanaka's formula underP, we haveEL???EB1B ttt221

4.ž. ž.ž.?0?

??O1.Therefore, substituting 3.15 into 3.14 and using 2

Jensen's inequality, we obtain

11 2?

˜˜˜??3.16

??,h????limsup?E?????hI.ž.ž.

Ýž.Ij

jŽ/ Ž/ 2t t??j?J t ž.It remains to compute the r.h.s.of 3.16.This is essentially parallel to ž.ž.2.18?2.22.In order to be able to properly count excursions, one first has to cut away the excursions that have length smaller than ?and then let??0.

We leave this to the reader.

4. Proof of Theorem 6.Recall the notation introduced in Sections 0.1

and 0.2.Define for the random walk model, ??1, i????14 i ??t 1 ??,h??logEexp?2????h,ž. ž . Ý tii

Ž/Ž/t

i?1

4.1ž.

??,h?lim??,hž. ž. t t?? and for the Brownian motion model, ??1, s????14 s 1 t

˜˜˜

??,h??logEexp?2??d??hds,ž. ž . Htss

Ž/Ž/

t 0

4.2ž.

˜˜

??,h?lim??,h.ž. ž. t t??

By the law of large numbers for?, respectively,?,

??,h???,h??h,ž. ž.

4.3ž.

˜˜

??,h???,h??h.ž. ž.

It suffices to consider the case??1.

E.BOLTHAUSEN AND F.DEN HOLLANDER1352

4.1.Outline of the proof of Theorem6.Theorem 6 is proved by a series of

approximation steps.Our approximations will depend on two auxiliary pa- rameters ?and?, where 0????.Later on, we shall lett??,a?0,??0, ž. 2 ??0 in this order.There will be no danger in assuming thatt?a,t??, ??a 2 ,???are all integers, which we shall do in order to avoid a plethora of brackets. Below we shall make a number of quite similar comparisons.In order to write these in a compact form, we introduce the following notation.

ž. ž.

D EFINITION2. Letfa,handga,hbe real-valued functions. t,?,?t,?,? ? ž. ? We writef?gif for any 0?h?h,??0 satisfying 1??h?hthe

ž.following is true: there exists

?such that for 0????there exists?? 000

ž.such that for 0?

???there existsa?,?such that 00 ?

222limsupfa,h?1??ga1??,h?0ž.ž. ž.ž.

t,?,?tž1??.,?ž1??.,?ž1??. t??

4.4ž.

for 0?a?a. 0

Here?,?,amay depend onh,h

? ,?.We writef?giff?gandg?f. 000 Note that?is a transitive relation and therefore?is an equivalence relation. The function for which we shall make such comparisons will be of the form 1

4.5fa,h??logEexp?2aH a,h,ž. ž . ž .ž.ž.

t,?,?t,?,? t ž. where the HamiltonianHa,his a random variable defined on the t,?,? žproduct space of the random walk and the random medium having as .probability measure the product ofPand?.Similar functions will be considered for the Brownian motion and medium. ?? ž.Now suppose that we want to provef?f, wherefa,hhas the t,?,? ? ž.HamiltonianHa,h.We can do this in the following way: t,?,?

1.SplitHinto two parts

4.6H?H

žI.

?H

žII.

.ž.

2.Apply Holder, Jensen and Fubini to get, for??0,¨

1

žI.

fa,h??logEexp?2a1??Hž. ž .ž.ž. t,?,? t1??ž. 1 ?1žII. ?logE?exp?2a1??H.ž.ž.ž. ?1 t1??ž.

4.7ž.

A POLYMER NEAR AN INTERFACE1353

ž. ?

3.The crucial point will be, for given 1??h?h, to choose the splitting in

such a way that 4.8H

žI.

?H

žI.

a,h ? ?H222 ? a1??,h ?

ž. ž . ž .ž.

t,?,?tž1??.,?ž1??.,?ž1??.

žII.žII.

ž ? .and thatH?Ha,h,hsatisfies t,?,? 1 ??1žII.

4.9 limsup logE?exp?2a1??Ha,h,h?0ž. ž .ž.ž.ž.

t,?,? t t??

ž.with

?,?,achosen appropriately in the sense of Definition 2.

ž.ž.

?

Clearly, 4.6?4.9 implyf?f.

Before we proceed, let us agree on some conventions about constants:A, B,Care generic positive constants, not necessarily the same at different occurrences.They may depend onh,h ? ,?, but not on the running parameters t,a, ?,?.

ž.Return to 4.1?4.3.Let

1

2?a,h??a,ah,ž. ž .

t,?,?t?a2 a4.10 ž.

˜˜

?a,h??1,hž. ž. t,?,?t

ž.which in fact do not depend on

?,?, respectively,?,?,a.What we finally ˜ want to prove is???, since by Definition 2 this implies Theorem 6.In order to achieve this, we shall introduce three intermediate quantities i

ž.ž .Fa,hi?1,2,3 and prove that

t,?,? 123
˜

4.11??F?F?F??.ž.

ž.The proof of 4.11 comes in four steps, organized as Sections 4.2?4.5.In order not to overburden notations, we shall often not explicitly express dependen- cies ona, ?,?. One of the crucial aspects of the proof is that the statement of Theorem 6 žž .. ž .does not allow for error factors of the form exp ?a,?,?twith?a,?,? tending to zero asa,?,??0.The reader should keep this in mind.

4.2.Coarse graining of the RW.We start by definingF

1 .Divide time into intervals of length ??a 2 : 22

4.12I?j?1??a,j??a,j?1.ž. ž .ž

j

Put??0 and

0

4.13??infj??????:S?0 for somei?I,k?1.ž. ž.?4

kk?1ij That is,?,?,...number the intervals in which the walk returns to the 12 ž.origin leaving gaps of at least????1 in the numbering.Define 2

4.14I?I?0,t?a,k?1,ž.ž?kj

Ž/ ??j?? k?1k ?4? 42and putm?maxk:I???mink:??t??. t?ak k

E.BOLTHAUSEN AND F.DEN HOLLANDER1354

For 1?k?m2, we sets?1 if the random walk is negative just prior t?ak to its first zero inI, ands?0 otherwise.Fork?m2, on the other ?kt?a k hand, we sets?1 if the random walk is negative att?a 2 , ands?0 kk otherwise.Let 4.15Z ???.ž. ž. Ý ki i?I k We can now define our first intermediate quantity: 1 11

Fa,h??logEexp?2aH a,h,ž. ž.ž.ž.

t,?,?t,?,? t m2 t?a 1 ??Ha,h?sZ??ah I.ž. ž.?4Ý t,?,?kk k k?1

4.16ž.

STEP1.??F

1 . P

ROOF. The proof comes in six parts.

ž.?ž. ž .?i We have recall 4.1 and 4.10 : 1 ?a,h??logEexp?2aH a,h,ž. ž.ž.ž. t,?,?t,?,? t 2 m2 t?at?a

Ha,h????ah????ah.ž. ž. ž.

ÝÝÝ

t,?,?ii ii i?1k?1i?I k

4.17ž.

?ž.ž.?Remark that, by a trivial rescaling of the parameters see 4.12?4.14 , we have

4.18Ha,

?h?H222?a,hfor any??0,ž. ž . ž . t,?,??t,??,?? and the same forH 1 .Furthermore, for anyh,h?0, 12

Ha,h?H

1 a,hž. ž. t,?,?1t,?,?2 m2m2 t?at?a ?ah?h??ah????s.ž. ž .ž.

ÝÝ ÝÝ12i2ii k

k?1k?1i?Ii?I kk

4.19ž.

In order to prove??F

1 , we splitH?H

žI.

?H

žII.

with 4.20H

žI.

?H 1 a,1??h ? ?H222 1 a1??,h ? ,ž. ž. ž.ž. ž. t,?,?t,?,?tž1??.,?ž1??.,?ž1??.

ž. ž .

?žII. and take the r.h.s.of 4.19 withh?h,h?1??hasH.On the other 12 hand, in order to proveF 1 ??, we splitH 1 ?H ?žI. ?H ?žII. with 4.21H ?žI. ?Ha,1??h ? ?H222a1??,h ? ,ž. ž. ž.ž. ž. t,?,?t,?,?tž1??.,?ž1??.,?ž1??.

ž. ž .

??žII. and take minus the r.h.s.of 4.19 withh?1??h,h?hasH.We 12 žshall prove that if we choosea,?,?small enough in this order because of .ž.Definition 2 , then also the requirement in 4.9 is met: 1 ??1žII.

4.22 limsup logE?exp?2a1??Ha,h,h?0,ž. ž .ž.ž.ž.t,?,?

t t?? and the same withH ?žII. instead ofH

žII.

.This will prove the claim in Step 1.

A POLYMER NEAR AN INTERFACE1355

ž. ž .ii To prove 4.22 , we first carry out the expectation over?: ?1žII. ?exp?2a1??Ha,h,h?ž.ž.ž. t,?,? 2m t?a ?2?1 ?exp?2a1??h?1??h?ž.ž.ž.ÝÝ i k?1i?I k 2m t?a ?2?1 ?exp?2a1??1??h??sž. ž .ž.ÝÝ ik k?1i?I k

4.23ž.

2m t?a ?1 ?exp logcosh 2a1????sž.?4ž.ÝÝ ik k?1i?I k 22mm
t?at?a 22
???expAa ??s?Ba?

ÝÝ ÝÝik i

k?1k?1i?Ii?I kk ž ? .for some constantsA,B?0 which depend onh,h, ?but not ont,a,?,?. The crucial point is that the second summand in the exponent is able to kill the first summand for arbitraryA,B?0, provided the parametersa, ?,? are chosen appropriately.Thus, to complete the proof of??F 1 , it remains to show that 22mm
t?at?a 1 22
??4.24 limsup logEexpAa ??s?Ba??0.ž.

ÝÝ ÝÝ

ik i tŽ/ t??k?1k?1i?Ii?I kk This is a problem about simple random walk and its zeroes.The only difference betweenH

žII.

andH ?žII. is that the second summand on the r.h.s. comes with a minus andh,hinterchanged.However, this obviously leads 12 ž. ž.to the same type of estimate as 4.23.Therefore 4.24 proves Step 1 completely. ž. ž .iii To prove 4.24 , we introduce the standard return times of the random walk: ?4T?0,T?infi?T:S?0,l?1,

0ll?1i

l2?minl:T?t?a 2 ?4 t?al

4.25ž.

and the excursion times 4.26 ??T?T,1?l?l2,??t?a 2 ?T.ž.ž. lll?1t?al l22?1t?at?a We further define??1 if the sign of thelth excursion is negative, and l ??0 otherwise.Then, obviously, we can write the second summand in the l

ž.r.h.s.of 4.24 as

m2l2 t?at?a

4.27????.ž.

ÝÝ Ýill

k?1l?1i?I k

E.BOLTHAUSEN AND F.DEN HOLLANDER1356

m2 t?a ??Next we estimate the first summandÝÝ ??sin terms of the same k?1i?Ii k k quantities.Putt?0, and lettbe the first zero of the random walk in the 0k ž. 2 ž?2intervalI1?k?m, andt?t?a.On the time intervalt,t ?t?am k?1k2kt?a the random walk makes a number of excursions, andsjust depends on the k sign of the last one; that is,s?1 if and only if this is negative.By k construction, only this last excursion can have length greater than or equal to

ž.ž

2 . 2 ?ž. ž.? ?????a???asee 4.12 and 4.13.It follows that ifiis not in an excursion of length less than ??a 2 andidoes not belong to one of the intervalsI, then ? k

4.28??sfor thekwithi?I.ž.

ik k

ž??

2 .From these considerations we obtain recall thatI? ??a ? k m2l2 t?at?a ?? ??24.29??s??1???m.ž.

ÝÝ Ý

ik ll t?a22 ½5 aa k?1l?1i?I k ž. ž. ž.Combining 4.27 and 4.29 we see that, in order to prove 4.24 , it now suffices to show that 2l t?a 1? 2 limsup logEexpAa?1?? Ý ll2 ½5

Žta

t??l?1

4.30ž.

2l t?a 2

2?A?m?Ba???0

Ý t?all / l?1 for appropriatea,?,?.

ž. živ As the

?'s are independent of the?'s 0 or 1 with probability 1?2 ll . 2 each , we can integrate out the former and replace?BaÝ??in the r.h.s.of lll 112
ž. ž ž ..4.30 byÝlog?exp?Ba?.We next claim that ll22 l2 t?a 1112

2A?m?1?log?exp?Ba??0ž.ž.ž.Ý

t?al222

4.31ž.

l?1 for 0?????.ž. 0 ž?ž.2To see why, pick any of the intervalst,t1?k?m.If any of the k?1kt?a ž? 2 excursions ont,thas length greater than or equal to??a, then for the k?1k lindexing this excursion we have

11 112

4.32 log?exp?Ba??log?exp?B?ž. ž .ž.ž.ž.

l22 22 and hence 1112

4.33A??log?exp?Ba??0 for 0?????.ž. ž.ž.ž.

l0222

Therefore

ž.k

1112

4.34A??log?exp?Ba??0 for 0?????,ž. ž.ž.ž.Ý

l0222 l

A POLYMER NEAR AN INTERFACE1357

žk.

ž?whereÝmeans summing over all the excursions ont,t.On the other k?1k ž? 2 hand, if all excursions ont,thave length less than??a, then for all k?1k thelindexing these excursions we have

11 122

4.35 log?exp?Ba???Ba?for 0????ž.ž.ž.

ll022 4 and so

ž. ž.kk

111 122

A??log?exp?Ba??A??Ba?

ž.ž.ÝÝ

ll222 8

4.36ž.

ll 12 ?A??Ba t?t.ž. kk?18 ?ž.?ž 2 .ž . 2

By construction, however,t?t?????1??a?????afor 1

kk?1 ž.2?k?mand so the r.h.s.of 4.36 is less than or equal to 0 for 0??? t?a

ž. ž. ž. ž.

??.Combining 4.34 with 4.36 and summing onk, we get 4.31.Thus, in 0 ž.order to prove 4.30 , it now remains to show that 2l t?a 1? 2 limsup logEexpAa?1?? Ý ll2 ½5

Žta

t??l?1

4.37ž.

2l t?a 111
2 ?log?exp?Ba??0.

ž.Ý

l Ž/ /222 l?1

ž.v Observe next that

l2l2 t?at?a ?111 22

Aa?1???log?exp?Ba?

ž.ÝÝll l2

½5 Ž/ 222a
l?1l?1 l2l2 t?at?a ?111 ?? ?22 ?Aa?1???log?exp?Ba?

ž.ÝÝll l2

½5 Ž/ 222a
l?1l?1

4.38ž.

1 ?A??log2,2 ? ž. ? ?ž.?2where???1?l?lbut??T?Tcompare with 4.26. ll t?alll222?1t?at?at?a 1 ž.Clearly,A??log2 is negligible after taking thet??limit in 4.37.By 2 the optional sampling theorem, it therefore suffices to prove that ?111 ?? ?22

4.39EexpAa?1???log?exp?Ba??1ž.ž.

11 12 ½5

Ž/Ž/

222a
for appropriatea,?.

E.BOLTHAUSEN AND F.DEN HOLLANDER1358

ž.vi For fixed??0, a Riemann approximation together with the asymp- ž ? . 3?2

ž.totic formulaP

??k?C?kk??even yields 1 1? ??2 limEexpAa?1?? 112

½5½

Ž aaa?0 111
?2 ?log?exp?Ba??1 ž. 1

5Ž//

222

4.40ž.

?dx111 ?4?CexpAx1x???log?exp?Bx?1.ž. H

3?2½5Ž/

222x
0 ž. ž.Clearly, the r.h.s.of 4.40 is?0 when 0? ???.This proves 4.37 and 0 completes the proof of Step 1.?

4.3.From discrete to continuous medium.We next replace the i.i.d.

Bernoulli random variables

?by i.i.d.standard normal random variables?.ˆ ii ?Therefore, we define our second intermediate quantity as compare with

ž.?4.16

1 22
ˆ

Fa,h??logEexp?2aH a,h,ž. ž.ž.ž.

t,?,?t,?,? t m2 t?a 2 ??Ha,h?sZ ??ah I,ž. ž.?4ˆÝ t,?,?kk k k?1

4.41ž.

ˆ where?is expectation w.r.t.?.ˆ

STEP2.F

1 ?F 2 . P

ROOF. The proof comes in three parts.

ž. i We couple the random variables ?and?.Remark that these randomˆ ii variables enter intoF 1 andF 2 only via their partial sums over intervals of 2 ?ž.ž.?length??arecall 4.12?4.15.We can define?and?on a commonˆ ?probability space such that for anyj?1 see Komlos, Major and Tusnady

´´

ž.?1975, 1976 ,

? ??ck 3

4.42?????clog?k?ce,k?1,ž.ž.ˆÝii122

Ž/ a i?I j for some constantsc,c,c?0.Here?*denotes the coupling measure 123
obtained by independently repeating the KMT-coupling in each??a 2 -interval

I.It suffices to proveF?F.Namely, the

?and?enter symmetricallyˆ j12ii ž.into 4.42 , and therefore the proof ofF?Fwill be exactly the same upon 21
exchange of?and?.ˆ

Following our general scheme, we choose

m2 t?a 1 ??Ha,h?sZ ??ah Iž. ž.?4Ýt,?,?kk k

4.43ž.k?1

?H

žI.

a,h?H

žII.

a,hž. ž. t,?,?t,?,?

A POLYMER NEAR AN INTERFACE1359

with m2 t?a ?žI. ??Ha,h?sZ ??a1??hIž. ž.ž .?4ˆÝ t,?,?kk k k?1 m2 t?a

žII.

Ha,h?s???ž.ž.ˆÝÝ

t,?,?kii k?1i?I k

4.44ž.

m2 t?a ? ???ah?1? ?hsI.ž.ž.

Ýkk

k?1

With this choice, we have

4.45H

žI.

a,h?H222 2 a1??,h ? ,ž. ž. ž .ž. t,?,?tž1??.,?ž1??.,?ž1??. ž. ž.as required by 4.8 , and so we must show that 4.9 is met: 1 ?1žII.

4.46 limsup logE?*exp?2a1??Ha,h?0.ž. ž.ž.ž.ž.

t,?,? t t?? ž. ž .ii To prove 4.46 , we next claim that for arbitraryA,B?0, 2m t?a ?*expAa s???ž.ˆÝÝkii Ž/ k?1i?I k

4.47ž.

2m t?a 2 ???expBa s Ifor 0?a?a?.ž. Ý kk0 k?1 ž.To see why 4.47 is true, note that, by the independence of the coupling in disjoint ??a 2 -intervals, it suffices to prove that 2 ??4.48?*expAa????expBa I.ž.ž.ˆž.Ý ii1 Ž/ i?I 1

ž.But 4.42 gives

?*expAa???ž.ˆÝii Ž/ i?I 1 ?? ?expAa clog?expAa clog?k Ý 1122

Ž/ Ž /Ž/

aa k?1

4.49ž.

? ? ?*????clog?kž.ˆÝ ii12 Ž/ a i?I 1 ? ?expAa clog 1?cexp?kc?Aa.ž.ž.Ý

123½52Ž/

a k?1

ž. ž

2 ??.This is clearly less than or equal to expB ??expBa Iwhen 0?a? 1

ž.a

?. 0

ž. ž

?1 .ž ?1 .ž ž . ? .ž.iii PickingA?21??,B?21??h?1??hin 4.47 and ž. ž.recalling 4.44 , we get 4.46.This completes the proof of Step 2.?

E.BOLTHAUSEN AND F.DEN HOLLANDER1360

4.4.From discrete to continuous process.The next step consists in replac-

ing the random walk by a Brownian motion.For the random walk we have

ž. ž

2defined in 4.13 the random times?,...,?m?mfor short hence-

1mt?a .forth.For convenience we put ??t??.Write m mm 2 ??4.50asZ??ah I?sa??ahž. ž.?4ˆˆž.ÝÝÝÝ kk k k i k?1k?1??j??i?I k?1kj and note that 2

4.51a??ah?????h???,ž.ž.ˆž.ž.ÝÝ

iD??kk?1 kk?1k?1Ž/ ??j??i?I k?1kj k?1

ž.where

????are the scaled random times and?is a Brownian kk s0?s?t medium independent of the random walk.

LetQbe the distribution of

4.52??m;s,...,s;?,...,?,ž.ž.

1m1m žwhich of course depends on all the parameterst,a, ?,?Qis a probability .ž.ž.ž.distribution on a finite set.Then, in view of 4.41 , 4.50 and 4.51 , we may

ž.write with an obvious abuse of notation :

1 22
˜ Fa,h??logEexp?2aH a,h,ž. ž.ž.Ž/t,?,?Qt,?,? t m 1 2

Ha,h?s????h???,ž.ž.?4Ý

t,?,?k??kk?1 kk?1 a k?1

4.53ž.

˜ where?is the expectation over?.Remark now that?can be interpreted as a

ž ž .. ž

2 .functional on the space of continuous pathsfs, defined byfia? 0?s?t ž 2 .aS0?i?t?awith linear interpolation.Replacing the law of the random i ˜ž.walk by the law of a Brownian motionB, we get a distributionQof s0?s?t ˜ ?.Obviously,QandQare mutually absolutely continuous.We therefore define our third intermediate quantity as 1 32
˜ Fa,h??logEexp?2aH a,hž. ž.ž.˜Ž/t,?,?Qt,?,? t4.54 ž. 1 3 ˜ ??logEexp?2aH a,h,ž.ž.Ž/Qt,?,? t where ˜ 1dQ 32

4.55H?H?log.ž.

2adQ S

TEP3.F

2 ?F 3 . ž. ?

PROOF. We again use our splitting.If 1??h?h, then

4.56H 2 a,h?H

žI.

a,h ? ?H

žII.

a,h,h ? ,ž. ž. ž. ž . t,?,?t,?,?t,?,?

A POLYMER NEAR AN INTERFACE1361

ž.where, as required by 4.8,

H

žI.

a,h ? ?H 3 a,1??h ?

ž. ž .ž.

t,?,?t,?,? ?H222 3 a1??,h ? ,ž.ž. tž1??.,?ž1??.,?ž1??. ? m h?1??hž. ?žII.

Ha,h,h?s???ž.ž.Ý

t,?,?kk k?1 a k?1

4.57ž.

˜ 1dQ ?log.2adQ

žII.

ž ? .ž.Observe thatHa,h,hdoes not depend on ?.According to 4.9,in t,?,? order to proveF 2 ?F 3 we have to show that m ˜ 1dQ

4.58 limsup logEexp?As????Blog?0ž.ž.Ý

Qkkk?1

tdQŽ/ t??k?1 ž ?1 .ž ž . ? .ž ?1 .withA?21? ?h?1??h,B?1??and for?,?,aappropri- ate.This is, however, immediate from Lemma 4 below upon putting in the ˜ lower estimate fordQ?dQand integrating out thesafterward.Indeed, k sincesare 0 or 1 with probability 1?2 each, the summand can be replaced k by 11 11

4.59 log?exp?A????log?exp?A?.ž. ž .ž.ž.ž.ž.

kk?122 22

The proof ofF

3 ?F 2 is similar after putting in the upper estimate for ˜ dQ?dQ.? ž. L

EMMA4.There exists???a,?,??0satisfying

4.60 lim limsup

?a,?,??0for all??0ž. ž . ??0a?0 such that dQ mm

4.61 1?????1??.ž. ž . ž.ž .

˜ dQ P

ROOF. The proof comes in three parts.

ž. i Letk,lbe positive integers such thatk?lis even.Define

4.62qk,l?PS?0 fork?i?k?l,S?0?S?0.

ž. ž. ž .

ik?l0

ž. ž?.Let furtherpx?PS?xS?0 fork?xeven.

kk0 žAssume thatk,lare odd.We are faced here with the usual parity problems.The case wherek,lare even is handled by slight modification.We .neglect such trivial points in the following discussion.Then, via the reflec-

E.BOLTHAUSEN AND F.DEN HOLLANDER1362

tion principle, l qk,l?2pxPS?0 for 0?i?l,S?0?S?xž. ž.ž . Ý ki l0 x?1 l ?pxp x?1?px?1ž.ž. ž. Ý kl?1l?1 x?1

4.63ž.

l 2x ?pxpx.ž. ž. Ý kl l x?1 ž.Now, 0?2x?l?21?x?l, so using the Bernstein large deviation esti-

ž. ž.mates forpxandpxwe get

kl ž. 3?5 k?ll 2x2x

4.64pxpx?1?o1pxpx,ž . ž. ž. ž. ž. ž.ž.

ÝÝ

kl kl ll x?1x?1

ž.whereo1 refers tok,l??jointly.But fork??,

2 2x px?1?o1 exp?ž. ž.ž.( k Ž/ ?k2k4.65ž. ? 3?5

4uniformly inx?1,...,k.

ž. ž

Substitution into 4.63 yields a Riemann approximation only the oddx's .count , 2 ?12 1 1x11 qk,l?1?o1dx2xexp??ž. ž.ž. H

Ž/"

2?l2klkl

0

4.66ž.

 2k ?1?o1.ž.ž. " ?k?llž.

ž. ž

2 .ii We fix now ?,?,aas usual with???,??ainteger.For integers 2

ž. ž .j?2, 1?y?

??awe obtain from 4.66 asa?0 only half of thel's count : ??j?1?j?ž.

Pmini?:S?0??,S?0

iy2222 ½5

Ž/Ž

Ž/ aaaa j??a 2 ? ?q?y,l Ý 2 Ž/ a 2

ž.l?j?1??a?1

2 2j??a "??a?y12ž. ?1?o1ž.ž. Ý 2 "

2???a?y?llž.2

ž.l?j?1??a?1

4.67ž.

2j?j?1?ž.

?1?o1 arctan?arctan ,ž.ž. ((22

Ž/???ya??ya

ž. ž .

whereo1 refers toa?0, uniformly in 0? ????2 for a fixed??0, 2 ž.1?y???aandj?2.The uniformity injis of crucial importance.

A POLYMER NEAR AN INTERFACE1363

ž. ž.Equation 4.67 is also true forj?1, although 4.66 is obviously not correct for fixedland onlyk??.However, some rough estimate like "ž. ž. ž .qk,l?p0?C?k?lsuffices to show that the smalll's in 4.67 are k?l negligible. ž.iii By weak convergence of random walk to Brownian motion, we get

ž.from 4.67 that for 0?

????2, 0?y??andj?1:˜ ò ?4Pinfu??:B?0???j?1?,j??B?0ž.žž. uy˜

2j?j?1?ž.

?arctan?arctan ((

Ž/???y??y˜˜

4.68ž.

ž.which, of course, can also be proved directly.

Now define

?a,?,?;y,y,jž.˜ 2222
Pmini???a:S?0???a?j?1??a,j??a?S?0ž.?4žž.ž.4.69iyž. ? ˜ ?4Pinfu??:B?0???j?1?,j??B?0ž.žž. uy˜ and

4.70?a,?,??sup sup sup?a,?,?;y,y,j?1.ž. ž . ž .˜

2 j?10?y??˜1?y???a ˜ž.Then 4.61 follows immediately from the definition of ?,QandQ.Combining

ž.ž. ž.4.67?4.70 , we arrive at 4.60.?

4.5.Coarse graining of the BM.The final step must consist in getting rid

ž. 3

ž.of

?,?we have already said goodbye toa.The quantityFin 4.54 is 1 ž.similar toFin 4.16 , but all defined in terms of the Brownian motion and its zeroes in ?-intervals with gaps of size?.The point is to remove these

ž.restrictions by letting

??0,??0 in this order. 3 ˜

STEP4.F??.

P ROOF. This is quite parallel to Step 1 and we can therefore be brief.For ž. 3 žthe reader's and our own convenience, we stick to the proof ofF??the 3 .argument for??Fbeing similar.The proof comes in six parts.

ž. ž .

i Define the random function ?as follows.For 1?k?m, put s0?s?t ž? ??1 on the interval?,?if the Brownian motion is negative just prior sk?1k to its first zero in this interval, and??0 otherwise.On the last interval s ž? ?,tput??1ifB?0, and??0 otherwise.Then m?1st s

4.71??sfors??,?and 1?k?m,ž.ž

sk k?1k where thesare defined in terms of the Brownian motion. k

E.BOLTHAUSEN AND F.DEN HOLLANDER1364

ž.ii Our quantities no longer depend ona, so we need a slight modification of our general scheme.Put 33

Hh?aH a,hž. ž .

t,?,?t,?,? m ?s????h???ž.ž.½5Ý k??kk?1 kk?1 k?1 m ? k ??d??hds,ž. ÝH ss ? k?1k?1

4.72ž.

t ˜

Hh??d??hdsž. ž .

H tss 0 m ? k ??d??hds.ž. ÝH ss ? k?1k?1 Then 1 33

˜˜

Fh??logEexp?2Hh,ž. ž.

Ž/Ž/t,?,?t,?,?

t4.73 ž. 1

˜˜˜˜

?h??logEexp?2Hh.ž. ž.

Ž/Ž/tt

t

Remark next that, by Brownian rescaling,

1 33

222H1??h?Hh,ž. ž.ž.

t,?,?Dtž1??.,?ž1??.,?ž1??. 1?? 1

˜˜

2H1??h?Hh.ž. ž.ž.

tDtž1??. 1??

4.74ž.

Furthermore,

m ? k 3 ˜

Hh?Hh?h?h?dsž. ž. ž .

ÝH t,?,?1t212s ? k?1k?1 m ? k ????d??hds,ž.ž . ÝH ss s2 ? k?1k?1

4.75ž.

ž.which is completely analogous to 4.19.

ž.iii It should now be clear that the argument runs parallel to Step 1, so ?ž.?we have to show that compare with 4.24 m 1 t ˜??4.76 limsup logEexpA???ds?Bs????0ž.ž.ÝH ss kkk?1 Ž/ t 0 t??k?1 for?,?appropriate.The Brownian motion has at most a finite number of ž?excursions of length greater than or equal to ?in the interval 0,t.We ž?denote byJthe complement of these excursion intervals in 0,t.By the t,? definition of?, we have s t ????4.77 ???ds?J?m?.ž. H ss t,? 0

A POLYMER NEAR AN INTERFACE1365

?See the derivation of the corresponding estimate for the random walk in ž.?ž. ž. ž4.29.Substituting 4.77 into 4.76 and afterward integrating out thes0 k .or 1 with probability 1?2 each , we see that it suffices to prove 1

˜??limsup logEexpAJ?A

?m t,? Ž t t??

4.78ž.

m 11 ?log?exp?B????0.ž.ž.Ý kk?1 Ž/ / 22
k?1

ž.?ž.?As

?????1?k?m, we trivially have compare with 4.31 kk?1 m 111

A?m?1?log?exp?B????0ž.ž.ž.ž.Ý

kk?1222

4.79ž.k?1

for 0?????.ž. 0

Therefore it suffices to prove

1

˜??limsup logEexpAJ

t,? Ž t t??

4.80ž.

m 111
?log?exp?B????0ž.ž.Ý kk?1 Ž/ / 222
k?1 for appropriate?,?.

ž.iv The

?are in fact stopping times for the Brownian motion.They are k related to another sequence of stopping times:????0 and 00 ??inft????:B?0,?4 kk?1t ??j?if??j?1?,j?,k?1,ž.ž kk

4.81ž.

until the smallestmsuch that??t.By construction, it is clear that m ?ž?? ?,??J?2?for allk. k?1kt,? ž. ž .Next, remark that in 4.80 we may replace the last?which is justtby m 1 žž .?žj?if??j?1?,j?, provided we add log2 in the exponent which is m2 .ž.irrelevant in thet??limit.Therefore, we prove 4.80 in this form. ž. ž .v Clearly,?isFF-measurable, whereFFis the natural filtra- k?1?ss?0 k?1 tion of the Brownian motion.Furthermore, because??????? k?1k?1k?1 we have ?44.82????inft????:B?0????.ž. kk?1k?1tk?1 Therefore, givenFF, the conditional distribution of???dominates ?kk?1 k?1 ž.the conditional distribution of the r.h.s.of 4.82 , which is independent ofFF ? k?1 and just the distribution of???.By the optimal sampling theorem it 1 ?ž.?therefore suffices to prove compare with 4.39 111
?44.83EexpA?1????log?exp?B????1ž. ž .ž.ž.ž.

11 1222

ž.for 0????and 0?????.

00

ž.vi As

?does not depend on?, we can first let??0, and it therefore 1 suffices to prove 11

4.84E?exp?B??exp?A?for 0????."ž. ž . ž .

Ž/1022

E.BOLTHAUSEN AND F.DEN HOLLANDER1366

?ž.?But ?has an explicit density compare with 4.66 : 1 " 1?

4.85P????ds?ds,s?0.ž. ž .

1 " ???ssž.

Therefore

?11 2dv11 2

4.86E?exp?B???exp?B?v.ž. ž . ž .

((H 12 Ž/

22?221?v

0 Since 111 1
22

4.87 lim 1??exp?B?v?Bvž. ž .

( ½5 ?22 4??0 ?2 ž 2 .andHv?1?vdv??, it follows from Fatou that 0 111

4.88 lim 1?E?exp?B???.ž. ž .

( 1

½5Ž/

?22??0

ž.This implies 4.84.?

ž. Steps 1?4 combine to give 4.11 , proving Theorem 6. Acknowledgment.We thank Herbert Spohn for discussions.

REFERENCES

ž.ALBEVERIO, S. and ZHOU, X. Y. 1996 . Free energy and some sample path properties of a random walk with random potential.J.Statist.Phys.83573?622. ž. B OLTHAUSEN, E. 1976 . On a functional central limit theorem for random walks conditioned to stay positive.Ann.Probab.4480?485. ž. C HUNG, K. L. and WILLIAMS, R. J. 1990 .Introduction to Stochastic Integration, 2nd ed.

Birkhauser, Boston.

¨ ž.GAREL, T., HUSE, D. A., LEIBLER, S. and ORLAND, H. 1989 . Localization transition of random chains at interfaces.Europhys.Lett.89?13. ž. G ROSBERG, A., IZRAILEV, S. and NECHAEV, S. 1994 . Phase transition in a heteropolymer chain at a selective interface.Phys.Rev.E501912?1921. ž. K INGMAN, J. F. C. 1973 . Subadditive ergodic theory.Ann.Probab.6883?909.

ž.ž.

K OMLOS, J., MAJOR, P. and TUSNADY, G. 1975 1976 . An approximation of partial sums of´´ independent RV's and the sample DF.I and II.Z.Wahrsch.Verw.Gebiete32111?131,

3433?58.

ž. R EVUZ, D. and YOR, M. 1991 .Continuous Martingales and Brownian Motion.Springer, Berlin. ž. S INAI,YA. G. 1993 . A random walk with random potential.Theory Probab.Appl.38382?385. ž. S INAI,YA. G. and SPOHN, H. 1996 . Remarks on the delocalization transition for heteropolymers.

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