aa r X i v : . [ m a t h . P R ] F e b UNIVERSALITY OF DETERMINISTIC KPZ
SOURAV CHATTERJEEA
BSTRACT . The Kardar–Parisi–Zhang (KPZ) equation is believed to be the uni-versal limit of a large class of growing random surfaces. This intuition has beenmade mathematically precise to some extent for one-dimensional surfaces. Be-yond dimension one, there is no rigorous formulation of KPZ universality. Sur-prisingly, even in the simplest case where the growth is fully deterministic, thereseems to be no result in the literature that establishes the deterministic versionof the KPZ equation as a universal scaling limit. The aim of this article is to fillthis gap. Consider a deterministically growing surface of any dimension, wherethe growth at a point is an arbitrary nonlinear function of the heights at that pointand its neighboring points. Assuming that this nonlinear function is monotone,invariant under the symmetries of the lattice, equivariant under constant shifts,and twice continuously differentiable, it is shown that any such growing surfaceapproaches a solution of the deterministic KPZ equation in a suitable space-timescaling limit. In forthcoming papers, some aspects of the theory will be extendedto systems involving randomness. C ONTENTS
1. Result 22. Examples 43. Related literature 64. Forthcoming work 75. A priori control on first-order roughness 86. A random walk representation of the surface 107. A priori control on second-order roughness 118. Plan of the proof 129. Random walk transition probabilities 1210. Smoothness of transition probabilities 1511. Smoothness of the discrete surface 2112. Existence of subsequential limits 3013. Differentiability of subsequential limits 3214. Origin of the gradient squared field 3715. Duhamel representation for subsequential limits 3916. Uniqueness of subsequential limit 4417. The Cole–Hopf solution 46
Mathematics Subject Classification.
Key words and phrases.
KPZ equation, random surface, scaling limit, universality.Research partially supported by NSF grant DMS-1855484.
18. The case α = 0 ESULT
Let d be a positive integer. A d -dimensional discrete surface is a function from Z d into R , where the value of the function at a point denotes the height of thesurface at that point. We will consider discrete surfaces that evolve over time ac-cording to some deterministic local rule, to be made precise below. Let e , . . . , e d be the standard basis vectors of R d . Let A denote the set { , ± e , ± e , . . . , ± e d } ,consisting of the origin and its d nearest neighbors in Z d . Let B := A \ { } . Thesets A and B will be fixed throughout this paper. Let φ : R A → R be a function.We will say that the evolution of a deterministically growing d -dimensional surface f : Z ≥ × Z d → R is driven by the function φ if for each t ∈ Z ≥ and x ∈ Z d , f ( t + 1 , x ) = φ (( f ( t, x + a )) a ∈ A ) . (1.1)We assume that φ has the following set of properties, all of which are quite naturalfrom a physical point of view: • Equivariance under constant shifts.
For u ∈ R A and c ∈ R , let u + c denote the vector obtained by adding c to each coordinate of u . We assumethat φ ( u + c ) = φ ( u ) + c for each u and c . In other words, if the wholesurface at time t is raised by a fixed constant c , then the surface at time t + 1 is also raised by the same constant amount. • Monotonicity.
We assume that φ is monotone increasing. That is, if u dominates v in each coordinate, then φ ( u ) ≥ φ ( v ) . Thus, a surface that ishigher than another surface everywhere at time t will continue to remainso at time t + 1 . • Invariance under lattice symmetries.
We assume that φ is invariant underthe following set of symmetries. Let u ∈ R A , and let v be obtained byswapping u e i with u e j and u − e i with u − e j for some i and j . Then weassume that φ ( u ) = φ ( v ) . Also, if u is obtained by swapping u e i with u − e i for some i , we assume that φ ( u ) = φ ( v ) . This means that the surfacedoes not prefer to grow differently in different lattice directions. • Twice continuous differentiability.
We assume that φ is twice continuouslydifferentiable. This regularity condition is important for obtaining a KPZscaling limit. There are examples where the previous three conditions hold,but the limit is not KPZ because φ is not C [45].Incidentally, the equivariance and monotonicity conditions have appeared previ-ously in a famous work of Barles and Souganidis [4] on convergence of approxi-mation schemes for nonlinear second order equations.In the next section we will see various examples of φ that satisfy the aboveconditions. For now, let us assume that we have a function φ as above. Let g : NIVERSALITY OF DETERMINISTIC KPZ 3 R d → R be a Lipschitz function. For each ε > , define g ε : Z d → R as g ε ( x ) := g ( εx ) . Let f ε : Z ≥ × Z d → R be the function obtained using the evolution (1.1) withinitial condition f ε (0 , x ) = g ε ( x ) for all x . Our goal is to obtain a scaling limit of f ε as ε → . To obtain the scaling limit, we need to rescale time by ε − and spaceby ε − (that is, parabolic scaling), and also subtract off a large time-dependent‘renormalization term’. For t ∈ R , let [ t ] denote the greatest integer ≤ t . Fora point x = ( x , . . . , x d ) ∈ R d , let [ x ] denote the vector ([ x ] , . . . , [ x d ]) . For t ∈ R ≥ , x ∈ R d , and ε > , define the rescaled values t ε := [ ε − t ] , x ε := [ ε − x ] . Finally, define f ( ε ) : R ≥ × R d → R as f ( ε ) ( t, x ) := f ε ( t ε , x ε ) − t ε φ (0) . (1.2)To state the theorem, we need to define two quantities related to φ . For a ∈ A ,let ∂ a φ ( u ) denote the partial derivative of φ with respect to u a . By the invarianceunder lattice symmetries, ∂ b φ (0) is the same for all b ∈ B . Let this quantity bedenoted as β . Similarly, the second order derivatives ∂ b φ (0) and ∂ b ∂ − b φ (0) do notdepend on the choice b ∈ B . Define γ := ∂ b φ (0) − ∂ b ∂ − b φ (0) . The quantities β and γ will remain fixed throughout the rest of the paper. Note that β is nonnegative due to the monotonicity of φ . The following theorem is the mainresult of this paper. Theorem 1.1 (Universality of deterministic KPZ) . Let all notations and assump-tions be as above. First, suppose that β and γ are both nonzero. Then f ( ε ) ( t, x ) converges pointwise on R ≥ × R d to a function f as ε → , where f (0 , x ) = g ( x ) for all x , and for t > , f ( t, x ) = βγ log Z R d K ( t, x − y ) e γg ( y ) /β dy, (1.3) where K ( t, x ) = (4 πβt ) − d/ e −| x | / βt . The function f is continuous on R ≥ × R d ,is infinitely differentiable on R > × R d , and solves the deterministic KPZ equation ∂ t f = β ∆ f + γ |∇ f | (1.4) with initial condition f (0 , x ) = g ( x ) . Next, suppose that β = 0 and γ = 0 . Then f ( ε ) converges pointwise to the limit f given by f (0 , x ) = g ( x ) and f ( t, x ) = Z R d K ( t, x − y ) g ( y ) dy for t > . The function f is continuous on R ≥ × R d , is infinitely differentiableon R > × R d , and solves the heat equation ∂ t f = β ∆ f with initial condition f (0 , x ) = g ( x ) . Finally, if β = 0 , then γ must also be zero, and in this case f ( ε ) ( t, x ) converges to g ( x ) as ε → for any t and x . SOURAV CHATTERJEE
In the next section, we will discuss some examples of φ that satisfy the fourconditions stated above. A review of the related literature is in Section 3. Someforthcoming papers are discussed in Section 4. The remaining sections are devotedto the proof of Theorem 1.1. 2. E XAMPLES
Various classes of functions satisfy the four properties listed in the previoussection. Two special examples where the recursion (1.1) can be explicitly solvedare φ ( u ) = 12 d X b ∈ B u b and φ ( u ) = 1 θ log X b ∈ B e θu b , where θ is an arbitrary positive real number. The first example gives a determin-istic version of Edwards–Wilkinson growth [28], and the second one comes fromdirected polymers [18].Explicit solution of (1.1) seems to be impossible in almost any other situation.For instance, it is easy to see that any function of the form φ ( u ) = u + 12 d X b ∈ B q ( u b − u ) , (2.1)where q is a twice continuously differentiable function with ≤ q ′ ≤ everywhereon R , satisfies the given conditions. More generally, the conditions are satisfied byany function of the form φ ( u ) = u + cF (( u b − u ) b ∈ B ) , where F is a nondecreasing symmetric C function on R B with bounded deriva-tives, and c is a sufficiently small constant, so that the derivative of φ with respectto u is everywhere positive. These give very large classes of examples, none ofwhich seem to be explicitly solvable.The class of examples becomes even broader if we replace the C assumptionby the assumption that φ is a Lipschitz function. For example, then it includes thefunction φ ( u ) = max a ∈ A u a (2.2)which generates a deterministic version of directed last-passage percolation [43],as well as the function φ ( u ) = max { u + 1 , max b ∈ B u b } , (2.3)which gives a deterministic version of ballistic deposition [63]. The C assumptioncan, however, be easily reintroduced by convolving these functions with smooth NIVERSALITY OF DETERMINISTIC KPZ 5 kernels. For example, if φ satisfies the equivariance, monotonicity and invarianceassumptions, and is a Lipschitz function, then for any δ > , the convolution φ δ ( u ) := Z R A e −| u − v | / δ (2 πδ ) d/ φ ( v ) dv (2.4)also satisfies those three conditions, and moreover, is infinitely differentiable. Thus,we can find arbitrarily close approximations to the functions displayed in equa-tions (2.2) and (2.3) that satisfy all four conditions. It is known that without the C assumption — for example for the function displayed in (2.2) — the scaling limitmay not be the deterministic KPZ equation [45].A natural class of examples come from Glauber dynamics of gradient Gibbsmeasures with convex potentials. For instance, the growing surface generated bythe Glauber dynamics of the SOS model belongs to this class [11] (see also [42]).A gradient Gibbs measure is a measure on random surfaces formally represented as Z − exp (cid:18) − X x,y ∈ Z d , | x − y | =1 V ( f ( x ) − f ( y )) (cid:19) Y x ∈ Z d dλ ( f ( x )) , where λ is Lebesgue measure on R , V : R → R is an even function called the po-tential, and Z is the (formal) normalizing constant. The Glauber dynamics (Gibbssampler) for generating from a such a model proceeds by regenerating f ( x ) fromthe conditional distribution given ( f ( x + b )) b ∈ B for a randomly chosen vertex x ,and repeating ad infinitum. A natural deterministic version of this would be toupdate f ( x ) as the conditional expected value of f ( x ) given ( f ( x + b )) b ∈ B . Ex-plicitly, this gives a growth mechanism of the type displayed in equation (1.1),with φ ( u ) = R ∞−∞ t exp (cid:0) − P b ∈ B V ( u b − t ) (cid:1) dt R ∞−∞ exp (cid:0) − P b ∈ B V ( u b − t ) (cid:1) dt . (2.5)Suppose that V is convex, even, twice continuously differentiable, and sufficientlynicely behaved to allow derivatives with respect to u to be moved inside the aboveintegrals. Then φ is twice continuously differentiable. A simple calculation showsthat φ equivariant under constant shifts. Invariance under lattice symmetries isobvious from the definition. To see monotonicity, note that φ ( u ) has no dependenceon u , and for any b ∈ B , ∂φ∂u b = − Z ∞−∞ tV ′ ( u b − t ) ρ u ( t ) dt + Z ∞−∞ tρ u ( t ) dt Z ∞−∞ V ′ ( u b − t ) ρ u ( t ) dt. where ρ u is the probability density function proportional to e − P a ∈ B V ( u a − t ) . Since t
7→ − V ′ ( u b − t ) is a nondecreasing function of t , the FKG–Harris inequalityimplies that the above expression is nonnegative, and hence φ is monotone. Eventhough V is not C in many models of interest, one can consider approximations SOURAV CHATTERJEE by smoothing the potential. For example, in a continuous variant of the restrictedsolid-on-solid (RSOS) model, V is given by V ( x ) = ( if | x | ≤ , ∞ if | x | > . With this V in (2.5), an easy computation shows that φ ( u ) = 12 (max b ∈ B u b + min b ∈ B u b ) . This satisfies the equivariance, monotonicity and invariance conditions, but is nota C function. As before, this can be easily remedied by convolving with a smoothkernel as in (2.4). If V is not C , it is unlikely that the scaling limit is determin-istic KPZ, for the same reason why the φ displayed in (2.2) does not lead to adeterministic KPZ scaling limit.3. R ELATED LITERATURE
The KPZ equation was introduced by Kardar, Parisi and Zhang [44] to describethe growth of a generic randomly growing surface. Formally, the KPZ equation isgiven by ∂ t f = β ∆ f + γ |∇ f | + κξ, where ξ is a random field known as space-time white noise, and β , γ and κ arereal-valued parameters.The validity of the KPZ heuristic has been a topic of intense investigation in thelast three decades. A great deal of progress has been made in the one-dimensionalcase. A foundational challenge is to give a rigorous meaning to the KPZ equation.In dimension one, there are now many approaches to solving this problem, such asthe Cole–Hopf solution [6], regularity structures [38, 39], paracontrolled distribu-tions [34, 36], energy solutions [30–32, 37] and renormalization group [46].Many one-dimensional discrete processes have been shown to have the KPZscaling limit or properties indicative of a KPZ limit, such as directed polymers inthe intermediate disorder regime [1, 2, 26], polynuclear growth [53], the weaklyasymmetric exclusion processes [25, 56], log-gamma polymers [9] and Macdonaldprocesses [8]. Many of these results are based on exact formulas derived in priorwork, such as [3, 22, 43, 50, 57, 60–62]. Recently, more exotic objects related tothe KPZ equation have been rigorously treated, such as the KPZ fixed point [49],the KPZ line ensemble [21], and the Brownian landscape [24]. All of this is onlya small sample of the enormous literature that has grown around rigorous one-dimensional KPZ in the last ten years. For surveys, see [20, 54, 55].Beyond dimension one much less is known. Very few exact formulas are avail-able [10, 17, 52, 59]. Even the meaning of the KPZ equation is unclear in d ≥ .There has been some recent progress in making sense of the equation in d ≥ byregularizing the white noise term and then making the coefficient of |∇ f | tendto zero as the regularization is taken away [12, 15, 19, 27, 33, 47, 48]. However, NIVERSALITY OF DETERMINISTIC KPZ 7 it turns out that this limit is actually a solution of the Edwards–Wilkinson equa-tion [28], and is therefore not ‘true KPZ’. No discrete model has been shown toconverge to a true KPZ limit in d ≥ .Going beyond exactly solvable models, there have been some recent efforts to-wards understanding the universality of the KPZ equation in dimension one. Asignificant progress was made by Hairer and Quastel [40], who showed that if the |∇ f | term is replaced by a polynomial function of ∇ f , and the white noise is re-placed by the mollification of itself, the |∇ f | term reappears in a scaling limit asthe mollification is removed. A simpler proof of the result was given by Gubinelliand Perkowski [35], and extensions were recently obtained by Hairer and Xu [41]and Yang [66]. In a different approach to establishing universality, non-integrablemodels converging to one-dimensional KPZ were exhibited by Dembo and Tsai[25] and further developed by Yang [64, 65].The investigation of the deterministic KPZ equation and its variants as scalinglimits of discrete growth models was initiated by Krug and Spohn [45]. This workinspired a large body of follow-up work in physics, especially in modeling de-terministic traffic flows, such as the model of Biham, Middleton and Levine [7].However, not much was done on the rigorous side. The mathematical propertiesof the deterministic KPZ and related equations have been studied [5, 29], but con-vergence to the deterministic KPZ equation has not been rigorously established forany discrete model. 4. F ORTHCOMING WORK
The forthcoming paper [13] will consider growth mechanisms of the type (1.1),but with the additional feature that the growth is also allowed to involve the externalrandomness. That is, we will consider evolutions of the form f ( t + 1 , x ) = φ (( f ( t, x + a )) a ∈ A , z t +1 ,x ) , where z t,x are i.i.d. random variables. The function φ will be assumed to havethe equivariance, monotonicity and invariance properties, but the C assumptionwill be dropped, allowing for a much larger class of models. The goal of [13] willbe to investigate the fluctuations of the surface, and especially, conditions underwhich the surface is superconcentrated. Superconcentration, also called sublinearvariance in this context, means that Var( f ( t, x )) = o ( t ) as t → ∞ . Nontrivialsurface growth models are usually conjectured to be superconcentrated (with veryspecific conjectures about fluctuation exponents from the physics literature), butproofs are available only in a few special cases in dimension one.To devise a general technique for proving superconcentration in growing sur-faces, a new concept called subroughness will be introduced. Subroughness willmean that there exists two distinct points x and y such that E [( f ( t, x ) − f ( t, y )) ] = o ( t ) as t → ∞ . The main result of [13] will be that superconcentration is equiv-alent to subroughness. The result will be used to establish superconcentration invariants of the restricted solid-on-solid (RSOS) model and the ballistic depositionmodel, by first proving subroughness and then using the equivalence theorem to SOURAV CHATTERJEE deduce superconcentration. These will give the first nontrivial upper bounds onfluctuations in such models.In a second forthcoming paper [14], it will be shown that deterministic KPZ canalso arise as the scaling limit of random surfaces. Specifically, it will be shownthat for d ≥ , the surface generated by the ( d + 1) -dimensional directed polymermodel at sufficiently high temperature, when smoothed by taking microscopic localaverages, converges to the deterministic KPZ equation in a suitable scaling limit.After this work was completed, Panagiotis Souganidis (personal communica-tion) observed that using the Crandall and Lions theory of viscosity solutions [23,58] and the general method of Barles and Souganidis [4], it is possible to providea different and more flexible proof of Theorem 1.1. Such a proof will allow forconsiderable simplification of the arguments by not requiring any knowledge ofthe regularity properties of the Green’s function and the higher order estimates thatare used in the proof in this paper. It will also allow to deal with non-smooth φ ’sand other extensions. This work will appear in a forthcoming paper of Chatterjeeand Souganidis [16].The remainder of this paper is devoted to the proof of Theorem 1.1. After prov-ing some basic lemmas, a sketch of the proof is given in Section 8. The details areworked out in subsequent sections.5. A PRIORI CONTROL ON FIRST - ORDER ROUGHNESS
Let L denote the Lipschitz constant of the initial data g . Throughout the remain-der of this paper, C, C , C , . . . will denote constants that depend only on d , φ and L . The values of these constants may change from line to line, or even within aline. We will assume without loss of generality that ε ∈ (0 , .First, let us show that there is no loss of generality in assuming that φ (0) = 0 .This will simplify matters a bit by taking off the renormalization term in (1.2).Define e φ ( u ) := φ ( u ) − φ (0) , and let e f ε be obtained by evolving the surface with e φ instead of φ , but with the same initial data g ε . Lemma 5.1.
For any t ∈ Z ≥ and x ∈ Z d , e f ε ( t, x ) = f ε ( t, x ) − tφ (0) . Proof.
We will prove the identity by induction on t . It is obviously true when t = 0 , since both sides are equal to g ε ( x ) . Suppose that it holds for some t . Thenby equivariance under constant shifts, e f ε ( t + 1 , x ) = e φ (( e f ε ( t, x + a )) a ∈ A )= φ (( e f ε ( t, x + a )) a ∈ A ) − φ (0)= φ (( f ε ( t, x + a )) a ∈ A − tφ (0)) − φ (0)= φ (( f ε ( t, x + a )) a ∈ A ) − ( t + 1) φ (0)= f ε ( t + 1 , x ) − ( t + 1) φ (0) . This completes the proof of the lemma. (cid:3)
NIVERSALITY OF DETERMINISTIC KPZ 9
The above lemma shows that we can work with e φ and e f ε instead of φ and f ε .In other words, we can assume, without loss of generality, that φ (0) = 0 . We willwork under this assumption henceforth. We begin with two key observations aboutthe function φ . Lemma 5.2.
For each u ∈ R A and a ∈ A , ∂ a φ ( u ) ≥ , and P a ∂ a φ ( u ) = 1 .Proof. The first claim follows from monotonicity, and the second from equivari-ance. (cid:3)
Lemma 5.3.
For any u, v ∈ R A , | φ ( u ) − φ ( v ) | ≤ max a ∈ A | u a − v a | .Proof. Note that φ ( u ) − φ ( v ) = Z ∇ φ ( tu + (1 − t ) v ) · ( u − v ) dt. But by Lemma 5.2, |∇ φ ( tu + (1 − t ) v ) · ( u − v ) | = (cid:12)(cid:12)(cid:12)(cid:12)X a ∈ A ∂ a φ ( tu + (1 − t ) v )( u a − v a ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ X a ∈ A ∂ a φ ( tu + (1 − t ) v ) | u a − v a |≤ max a ∈ A | u a − v a | . This completes the proof of the lemma. (cid:3)
Lemma 5.3 has the following important consequence, which is crucial for allof the subsequent estimates. It says for any t , f ε ( t, · ) is Lipschitz on Z d underthe ℓ metric with Lipschitz constant Lε . In other words, the roughness of f ε ( t, · ) does not grow with t . The monotonicity and equivariance assumptions are mostlyneeded for this step. Without a control such as this, it is difficult to ensure that f ε has a smooth scaling limit as ε → . Lemma 5.4.
For any t and any two neighboring points x and y , | f ε ( t, x ) − f ε ( t, y ) | ≤ Lε.
Proof.
Note that | f ε (0 , x ) − f ε (0 , y ) | = | g ε ( x ) − g ε ( y ) | = | g ( εx ) − g ( εy ) | ≤ Lε | x − y | . This proves the claim for t = 0 . Suppose that it holds for some t . Then byLemma 5.3, for any two neighboring points x and y , | f ε ( t + 1 , x ) − f ε ( t + 1 , y ) | = | φ (( f ε ( t, x + a )) a ∈ A ) − φ (( f ε ( t, y + a )) a ∈ A ) |≤ max a ∈ A | f ε ( t, x + a ) − f ε ( t, y + a ) | . But x + a is a neighbor of y + a for each a . Thus, | f ε ( t, x + a ) − f ε ( t, y + a ) | ≤ Lε for each a . This completes the proof. (cid:3)
6. A
RANDOM WALK REPRESENTATION OF THE SURFACE
The main result of this section, stated below, represents the solution of (1.1)as a formula involving random walk transition probabilities. This solution is notexplicit, as it will involve terms that depend on f ε . We will use it later to obtain anintegral equation for the scaling limit, which will then lead to the solution displayedin Theorem 1.1.In addition to the quantities β and γ that have already been defined, let us alsodefine α := ∂ φ (0) . Note that by Lemma 5.2, α and β are nonnegative, and α + 2 dβ = 1 . Consider a random walk on Z d , starting at the origin at time , andjumping from x to x + a , a ∈ A , with probability α if a = 0 and β if a = 0 . Let p ( t, x ) be the probability that this random walk is at the point x at time t . Thennote that for any t ∈ Z ≥ and x ∈ Z d , p ( t, x ) = αp ( t − , x ) + β X b ∈ B p ( t − , x − b ) . (6.1)For each x ∈ Z d and t ∈ Z ≥ , let h ε ( t, x ) := f ε ( t, x ) − αf ε ( t − , x ) − β X b ∈ B f ε ( t − , x + b ) . (6.2)The following result represents the function f ε in terms of the transition probabili-ties of the random walk defined above and the functions h ε and g ε . Proposition 6.1.
For any t ∈ Z ≥ and x ∈ Z d , f ε ( t, x ) = X y ∈ Z d p ( t, x − y ) g ε ( y ) + X ≤ s ≤ t − X y ∈ Z d p ( s, x − y ) h ε ( t − s, y ) . Proof.
The proof is by induction on t . First, note that p (0 ,
0) = 1 , p (1 ,
0) = α and p (1 , b ) = β for all b ∈ B . Thus, by (6.2), X y ∈ Z d p (1 , x − y ) g ε ( y ) + X y ∈ Z d p (0 , x − y ) h ε (1 , y )= β X b ∈ B f ε (0 , x + b ) + αf ε (0 , x ) + h ε (1 , x ) = f ε (1 , x ) . This proves the claim for t = 1 . Now suppose that the claim holds up to time t − for some t ≥ . Then again by (6.2), f ε ( t, x ) = αf ε ( t − , x ) + β X b ∈ B f ε ( t − , x + b ) + h ε ( t, x )= α X y ∈ Z d p ( t − , x − y ) g ε ( y ) + α X ≤ s ≤ t − X y ∈ Z d p ( s, x − y ) h ε ( t − − s, y )+ β X y ∈ Z d X b ∈ B p ( t − , x + b − y ) g ε ( y )+ β X ≤ s ≤ t − X y ∈ Z d X b ∈ B p ( s, x + b − y ) h ε ( t − − s, y ) + h ε ( t, x ) . (6.3) NIVERSALITY OF DETERMINISTIC KPZ 11
Notice that by the relation (6.1), αp ( t − , x − y ) + β X b ∈ Z d p ( t − , x + b − y ) = p ( t, x − y ) , and similarly, αp ( s, x − y ) + β X b ∈ B p ( s, x + b − y ) = p ( s + 1 , x − y ) . Plugging these into (6.3) gives f ε ( t, x ) = X y ∈ Z d p ( t, x − y ) g ε ( y ) + X ≤ s ≤ t − X y ∈ Z d p ( s + 1 , x − y ) h ε ( t − − s, y )+ h ε ( t, x )= X y ∈ Z d p ( t, x − y ) g ε ( y ) + X ≤ s ≤ t − X y ∈ Z d p ( s, x − y ) h ε ( t − s, y )+ h ε ( t, x ) . Since p (0 ,
0) = 1 , completes the induction step. (cid:3)
7. A
PRIORI CONTROL ON SECOND - ORDER ROUGHNESS
We now prove a second-order refinement of Lemma 5.4. This lemma shows that | h ε ( t, x ) | is uniformly bounded by a constant times ε . We will see later that thiseffectively gets a control on the second-order smoothness of f ε , just as Lemma 5.4gives a control on the first-order smoothness. Lemma 7.1.
For each t ∈ Z ≥ and x ∈ Z d , | h ε ( t, x ) | ≤ Cε .Proof. Let u ∈ R A be the vector whose coordinates are u = 0 and u b = f ε ( t − , x + b ) − f ε ( t − , x ) for b ∈ B . Then note that by equivariance, f ε ( t, x ) = φ (( f ε ( t − , x + a )) a ∈ A )= f ε ( t − , x ) + φ ( u ) . (7.1)By Lemma 5.4 and the assumption that ε < , | u b | is uniformly bounded by Lε andhence by a constant that does not depend on ε . Therefore, by the twice continuousdifferentiability of φ , the assumption that φ (0) = 0 , and Taylor expansion, (cid:12)(cid:12)(cid:12)(cid:12) φ ( u ) − X b ∈ B u b ∂ b φ (0) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C X b ∈ B u b . Since | u b | ≤ Lε and ∂ b φ (0) = β , this gives (cid:12)(cid:12)(cid:12)(cid:12) φ ( u ) − β X b ∈ B u b (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cε . (7.2) Finally, note that since α = 1 − dβ , f ε ( t − , x ) + β X b ∈ B u b = (1 − dβ ) f ε ( t − , x ) + β X b ∈ B f ε ( t − , x + b )= αf ε ( t − , x ) + β X b ∈ B f ε ( t − , x + b ) . (7.3)Combining (7.3) with the formula (6.2) for h ε ( t, x ) , we get β X b ∈ B u b = f ε ( t, x ) − f ε ( t − , x ) − h ε ( t, x ) . Plugging this into (7.2) gives | f ε ( t, x ) − f ε ( t − , x ) − h ε ( t, x ) − φ ( u ) | ≤ Cε . But by (7.1), f ε ( t, x ) − f ε ( t − , x ) − φ ( u ) = 0 . This completes the proof of thelemma. (cid:3)
8. P
LAN OF THE PROOF
Having stated and proved Lemma 5.4, Proposition 6.1, and Lemma 7.1, we arenow in a position to discuss the general strategy for the proof of Theorem 1.1.The first step is to use the recursive formula for f ε ( t, x ) given in Proposition 6.1,together with the bound on h ε given by Lemma 7.1 and a number of delicate es-timates for first- and second-order discrete derivatives of random walk transitionprobabilities to show that f ε has subsequential scaling limits as ε → , all ofwhich are differentiable.The next step is to show that for any subsequential scaling limit f , the rescaledversion of h ε converges to γ |∇ f | . This uses the formula (6.2) of h ε ( t, x ) , to-gether with Taylor expansion and the a priori controls given by Lemma 5.4 andLemma 7.1. This step uses the invariance of φ under lattice symmetries and thefact that φ is twice continuously differentiable. Once this is established, Propo-sition 6.1 will again be used, together with a local central limit theorem for thetransition probability p ( t, x ) , to show that f satisfies the integral equation f ( t, x ) = Z R d K ( t, x − y ) g ( y ) dy + γ Z t Z R d K ( s, x − y ) |∇ f | ( t − s, y ) dyds, where K is the Gaussian kernel from the statement of Theorem 1.1. We will thenshow that this integral equation has a unique solution, given by the Cole–Hopfformula (1.3) displayed in Theorem 1.1.9. R ANDOM WALK TRANSITION PROBABILITIES
In this section we begin to work out the necessary estimates for random walktransition probabilities that are crucial for the carrying out the plan of the proofsketched out in Section 8. We will henceforth assume that α = 0 . The case α = 0 needs special treatment due to parity issues, which will be dealt with in Section 18. NIVERSALITY OF DETERMINISTIC KPZ 13
Let ξ be a random vector which takes value b with probability β for each b ∈ B ,and with probability α . Then the characteristic function ψ of ξ is given by ψ ( θ ) = E ( e i θ · ξ ) = α + 2 β d X i =1 cos θ i for θ = ( θ , . . . , θ d ) ∈ R d . We are going to use the Fourier inversion formula forexpressing the transition probability p ( t, x ) in terms of the characteristic function ψ . This makes it necessary to derive two important estimates for ψ , as done below. Lemma 9.1.
There is a positive constant C depending only on α , β , and d , suchthat for all θ ∈ [ − π, π ] d , | ψ ( θ ) | ≤ e − C | θ | .Proof. Since α > , β > , and α + 2 βd = 1 , it is easy to see that the map x
7→ | α + 2 βd cos x | equals at x = 0 and is strictly less than everywhere elsein the closed interval [ − π, π ] . From this, and the behavior of cosine near zero, itfollows that the function w ( x ) := ( (1 − | α + 2 βd cos x | ) /x if x ∈ [ − π, π ] \ { } ,βd if x = 0 is continuous and nonzero everywhere on [ − π, π ] . Thus, there is a positive constant C , depending only on α , β and d , such that w ( x ) ≥ C for all x ∈ [ − π, π ] . Inother words, | α + 2 βd cos x | ≤ − Cx for all x ∈ [ − π, π ] . Therefore, for any θ ∈ [ − π, π ] d , | ψ ( θ ) | ≤ d d X i =1 | α + 2 βd cos θ i |≤ d d X i =1 (1 − Cθ i ) = 1 − C | θ | d . The proof is completed by applying the inequality − x ≤ e − x . (cid:3) Lemma 9.2.
For any θ ∈ R d and t ∈ Z ≥ , | ψ ( θ ) t − e − βt | θ | | ≤ Ct | θ | . Proof.
By Taylor expansion, (cid:12)(cid:12)(cid:12)(cid:12) cos x − x (cid:12)(cid:12)(cid:12)(cid:12) ≤ x for all x ∈ R . By this, and the identity α + 2 βd = 1 , we have | ψ ( θ ) − β | θ | | = (cid:12)(cid:12)(cid:12)(cid:12) β d X i =1 (cid:18) cos θ i − θ i (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ β d X i =1 (cid:12)(cid:12)(cid:12)(cid:12) cos θ i − θ i (cid:12)(cid:12)(cid:12)(cid:12) ≤ β P di =1 θ i ≤ C | θ | . On the other hand, again by Taylor expansion, | e − x − x | ≤ x for all x ≥ . Combining this with the preceding display, we get | ψ ( θ ) − e − β | θ | | ≤ C | θ | . Finally, note that the absolute values of both ψ ( θ ) and e − β | θ | are both boundedabove by . Since | a − b | t ≤ t | a − b | for any two complex numbers a and b in theunit disk and any positive integer t , the claim follows. (cid:3) Now note that by Fourier inversion, we have p ( t, x ) = (2 π ) − d Z [ − π,π ] d e − i θ · x ψ ( θ ) t dθ = (2 π ) − d t − d/ Z [ − π √ t,π √ t ] d e − i t − / η · x ( ψ ( t − / η )) t dη, (9.1)where the second line was obtained by the change of variable η = √ tθ . We willnow simplify this formula to get successively simpler approximations p , p and p for p , and bound the approximation errors using Lemmas 9.1 and 9.2. First, let p ( t, x ) := (2 π ) − d t − d/ Z | η |≤ log t e − i t − / η · x ( ψ ( t − / η )) t dη. Then we have the following bound on the difference between p and p . Lemma 9.3.
For any t ∈ Z ≥ and x ∈ Z d , | p ( t, x ) − p ( t, x ) | ≤ C e − C (log t ) .Proof. Take any η with absolute value bigger than log t . Then by Lemma 9.1, | ψ ( t − / η ) | ≤ e − Ct − | η | ≤ e − Ct − (log t ) . Thus, | p ( t, x ) − p ( t, x ) | ≤ Ct − d/ Z log t< | η |≤ π √ td | ψ ( t − / η ) | t dη ≤ Ct − d/ Z log t< | η |≤ π √ td e − C (log t ) dη ≤ C e − C (log t ) , NIVERSALITY OF DETERMINISTIC KPZ 15 where in the last line we used the fact that any polynomial in t is rendered irrelevantby the presence of e − C (log t ) . This completes the proof of the lemma. (cid:3) Next, define p ( t, x ) := (2 π ) − d t − d/ Z | η |≤ log t e − i t − / η · x − β | η | dη, and finally, let p ( t, x ) := (2 π ) − d t − d/ Z R d e − i t − / η · x − β | η | dη = (4 πβt ) − d/ e −| x | / βt . (9.2)The following lemma shows that p is close to p . Lemma 9.4.
For any t ∈ Z ≥ and x ∈ Z d , | p ( t, x ) − p ( t, x ) | ≤ C e − C (log t ) .Proof. Using polar coordinates, we have | p ( t, x ) − p ( t, x ) | ≤ Ct − d/ Z | η | > log t e − β | η | dη = Ct − d/ Z ∞ log t r d − e − βr dr. It is not hard to see that the last expression is bounded by C e − C (log t ) . (cid:3) Note that we have not yet established the closeness of p and p . This will bedone in the next section.10. S MOOTHNESS OF TRANSITION PROBABILITIES
The primary goal of this section is to get upper bounds on the first- and second-order discrete derivatives of p . The plan is to show that these can be approximatedby the first- and second-order derivatives of p . The controls given by Lemma 9.3and Lemma 9.4 are sufficient to relate the derivatives of p with those of p , and thederivatives of p with those of p . Approximating the derivatives of p with thoseof p require more delicate control, which we will now obtain.For any function w : Z d → R , and any x, y, ∈ Z d , define δ y w ( x ) := w ( x + y ) − w ( x ) . For any y and z , the discrete derivative operators δ y and δ z commute, since δ y δ z w ( x ) = δ z w ( x + y ) − δ z w ( x )= w ( x + y + z ) − w ( x + y ) − ( w ( x + z ) − w ( x ))= w ( x + y + z ) − w ( x + z ) − ( w ( x + y ) − w ( x )) = δ z δ y w ( x ) . The following lemma shows that p and p are close, as are their first- and second-order discrete derivatives. The exact orders of the error bounds are important forsubsequent estimates. Lemma 10.1.
For any t ∈ Z ≥ and x, y, z ∈ Z d , | p ( t, x ) − p ( t, x ) | ≤ C t − ( d +2) / (log t ) C , | δ y p ( t, x ) − δ y p ( t, x ) | ≤ C | y | t − ( d +3) / (log t ) C , | δ y δ z p ( t, x ) − δ y δ z p ( t, x ) | ≤ C | y || z | t − ( d +4) / (log t ) C . Proof.
By Lemma 9.2, | ( ψ ( t − / η )) t − e − β | η | | ≤ Ct − | η | . Thus, | p ( t, x ) − p ( t, x ) | ≤ Ct − d/ Z | η |≤ log t | ( ψ ( t − / η )) t − e − β | η | | dη ≤ Ct − d/ − Z | η |≤ log t | η | dη ≤ Ct − d/ − (log t ) d +4 . This proves the first inequality in the statement of the lemma. Next, note that δ y p ( t, x ) = (2 π ) − d t − d/ Z | η |≤ log t ( e − i t − / η · y − e − i t − / η · x ( ψ ( t − / η )) t dη and δ y p ( t, x ) = (2 π ) − d t − d/ Z | η |≤ log t ( e − i t − / η · y − e − i t − / η · x − β | η | dη. By the inequality | e − i a − | ≤ | a | for a ∈ R , this gives | δ y p ( t, x ) − δ y p ( t, x ) | ≤ t − ( d +1) / | y | Z | η |≤ log t | η || ( ψ ( t − / η )) t − e − β | η | | dη. Now proceeding as before, we get the second inequality. Similarly, noting that δ y δ z p ( t, x )= (2 π ) − d t − d/ Z | η |≤ log t ( e − i t − / η · y − e − i t − / η · z − · e − i t − / η · x ( ψ ( t − / η )) t dη and δ y δ z p ( t, x )= (2 π ) − d t − d/ Z | η |≤ log t ( e − i t − / η · y − e − i t − / η · z − e − i t − / η · x − β | η | dη, and proceeding just as above, we get the third inequality in the statement of thelemma. (cid:3) Now we have all the necessary ingredients for getting bounds on the derivativesof p using bounds on the derivatives of p . The following lemma gives the boundsfor p . NIVERSALITY OF DETERMINISTIC KPZ 17
Lemma 10.2.
Take any t ∈ Z ≥ and x, y, z ∈ Z d . Let a := min {| x | , | x + y |} ,b := min {| x | , | x + y | , | x + z | , | x + y + z |} . Then | δ y p ( t, x ) | ≤ C e − C a /t t − ( d +1) / | y | , | δ y δ z p ( t, x ) | ≤ C e − C b /t t − ( d +1) / min {| y | , | z | , t − / | y || z |} . Proof.
Let u ( q ) := e −| q | for q ∈ R d , so that p ( t, x ) = (4 πβt ) − d/ u ((4 βt ) − / x ) . Let v be any derivative of u , of any order. Clearly, v ( q ) is some polynomial in q times u ( q ) , and therefore | v ( q ) | ≤ C e − C | q | for all q ∈ R d , where C and C are some positive constants depending on v . In particular, the maximum value of | v | on a line joining two points q and q ′ is bounded above by C e − C min {| q | , | q ′ | } .This gives | δ y p ( t, x ) | = (4 πβt ) − d/ | u ((4 βt ) − / ( x + y )) − u ((4 βt ) − / x ) |≤ C t − ( d +1) / | y | e − C a /t . This proves the first claim of the lemma. For the second claim, let us assumewithout loss of generality that | y | ≤ | z | . Note that by the above inequality, | δ y δ z p ( t, x ) | ≤ | δ y p ( t, x + z ) | + | δ y p ( t, x ) |≤ C e − C b /t t − ( d +1) / | y | . (10.1)On the other hand, by Taylor approximation and the above observation about thederivatives of u , | δ y p ( t, x ) − (4 πβt ) − d/ (4 βt ) − / y · ∇ u ((4 βt ) − / x ) |≤ C t − ( d +2) / | y | e − C min {| x + y | , | x | } /t , and the same inequality holds with x replaced by x + z on both sides. This gives | δ y δ z p ( t, x ) | = | δ y p ( t, x + z ) − δ y p ( t, x ) |≤ Ct − ( d +1) / | y ||∇ u ((4 βt ) − / ( x + z )) − ∇ u ((4 βt ) − / x ) | + C e − C b /t t − ( d +2) / | y | ≤ C e − C b /t t − ( d +2) / ( | y || z | + | y | ) ≤ C e − C b /t t − ( d +2) / | y || z | , (10.2)where we used | y | ≤ | z | in the last step. Combining (10.1) and (10.2) completesthe proof of the second claim. (cid:3) Finally, we need the following crude bound on p ( t, x ) to deal with values of x whose norms are ‘too large’. Lemma 10.3.
For any t ∈ Z ≥ and x ∈ Z d , p ( t, x ) ≤ C e − C | x | /t . Proof.
Take any t ∈ Z ≥ and x = ( x , . . . , x d ) ∈ Z d . Without loss of generality,suppose that | x | ≥ | x i | for all i . Consider the evolution of the first coordinateof our random walk. At each time step, the first coordinate increases by one withprobability β , decreases by one with probability β , and remains the same withprobability − β . Thus, it evolves like a sum of i.i.d. {− , , } -valued randomvariables. It is now easy to see by Hoeffding’s inequality for sums of i.i.d. boundedrandom variables that at time t , the probability that the first coordinate equals x is at most C e − C x /t . Since | x | ≥ | x i | for all i , we have x ≥ | x | /d . Thiscompletes the proof of the lemma. (cid:3) We are now ready to obtain the required estimates for the first- and second-order discrete derivatives of p . Proposition 10.4 gives the bound on first-orderderivatives, and following that, Proposition 10.5 gives the bound on second orderderivatives. Proposition 10.4.
Take any t ∈ Z ≥ and x, y ∈ Z d . Let a := min {| x | , | x + y |} .Then we have | δ y p ( t, x ) | ≤ C e − C a /t t − ( d +1) / | y | + C t − ( d +3) / (log t ) C | y | . Moreover, we also have | δ y p ( t, x ) | ≤ C e − C a /t .Proof. The bounds hold trivially if y = 0 . So let us assume that y = 0 . Again, thebounds hold trivially if t = 1 . So let us also assume that t ≥ . By Lemma 9.3 andthe triangle inequality, | δ y p ( t, x ) − δ y p ( t, x ) |≤ | p ( t, x + y ) − p ( t, x + y ) | + | p ( t, x ) − p ( t, x ) |≤ C e − C (log t ) . Similarly, by Lemma 9.4, | δ y p ( t, x ) − δ y p ( t, x ) | ≤ C e − C (log t ) . Next, by Lemma 10.1, | δ y p ( t, x ) − δ y p ( t, x ) | ≤ C | y | t − ( d +3) / (log t ) C . Combining these with the bound on | δ y p ( t, x ) | from Lemma 10.2, and observ-ing that C | y | t − ( d +3) / (log t ) C dominates C e − C (log t ) since y = 0 and t ≥ ,proves the first claim of the proposition. The second claim follows from Lemma 10.3and the triangle inequality. (cid:3) Proposition 10.5.
Take any t ∈ Z ≥ and x, y, z ∈ Z d . Let b := min {| x | , | x + y | , | x + z | , | x + y + z |} . NIVERSALITY OF DETERMINISTIC KPZ 19
Then we have | δ y δ z p ( t, x ) | ≤ C e − C b /t t − ( d +1) / min {| y | , | z | , t − / | y || z |} + C | y || z | t − ( d +4) / (log t ) C . Moreover, we also have | δ y δ z p ( t, x ) | ≤ C e − C b /t .Proof. The bound holds trivially if either y = 0 or z = 0 , because in that case, δ y δ z p ( t, x ) = 0 . So let us assume that y and z are both nonzero. Then, the boundholds trivially if t = 1 . So let us further assume that t ≥ . By Lemma 9.3 and thetriangle inequality, | δ y δ z p ( t, x ) − δ y δ z p ( t, x ) |≤ | p ( t, x + y + z ) − p ( t, x + y + z ) | + | p ( t, x + y ) − p ( t, x + y ) | + | p ( t, x + z ) − p ( t, x + z ) | + | p ( t, x ) − p ( t, x ) |≤ C e − C (log t ) . Similarly, by Lemma 9.4, | δ y δ z p ( t, x ) − δ y δ z p ( t, x ) | ≤ C e − C (log t ) . Next, by Lemma 10.1, | δ y δ z p ( t, x ) − δ y δ z p ( t, x ) | ≤ C | y || z | t − ( d +4) / (log t ) C . Combining these with the bound on | δ y δ z p ( t, x ) | from Lemma 10.2, and the ob-servation that C | y || z | t − ( d +4) / (log t ) C dominates C e − C (log t ) since y and z are nonzero and t ≥ , proves the first claim of the proposition. The second one isa consequence of Lemma 10.3 and the triangle inequality. (cid:3) We now have the necessary bounds for spatial derivatives of p . We will alsoneed bounds for first-order derivatives with respect to t , and space-times mixedderivatives. For r ∈ Z ≥ and a function w : Z ≥ × Z d → R , define the temporalderivative δ ′ r w ( t, x ) := w ( t + r, x ) − w ( t, x ) . Our goal is to get bounds on quantities like δ ′ r p ( t, x ) and δ ′ r δ y p ( t, x ) . The first step,as before, is to get the corresponding bounds for p instead of p . Lemma 10.6.
For any two integers t ≥ r ≥ , and any x ∈ Z d , | δ ′ r p ( t, x ) | ≤ C e − C | x | /t rt − ( d +2) / . Proof.
Treating t as a continuous variable in the formula (9.2) for p , let p ′ denoteits derivative with respect to t . Then note that | p ′ ( t, x ) | ≤ C t − ( d +2) / (1 + t − | x | ) e − C | x | /t ≤ C t − ( d +2) / e − C | x | /t . Since r ≤ t , the maximum value of this derivative in the time interval [ t, t + r ] isbounded by C t − ( d +2) / e − C | x | /t . By the mean value theorem, this completes theproof of the lemma. (cid:3) Lemma 10.7.
Take any two integers t ≥ r ≥ , and any x, y ∈ Z d . Let a :=min {| x | , | x + y |} . Then | δ ′ r δ y p ( t, x ) | ≤ C e − C a /t | y | rt − ( d +3) / . Proof.
In this proof we will use K as shorthand for C e − C a /t . Fixing x and y ,let w ( t ) := δ y p ( t, x ) and let w ′ denote its derivative with respect to t , considering t as a continuousparameter. Let u ( q ) := e −| q | and v ( q ) := q · ∇ u ( q ) for q ∈ R d . Then w ′ ( t ) = − d πβ ) − d/ t − ( d +2) / ( u ((4 βt ) − / ( x + y )) − u ((4 βt ) − / x )) −
12 (4 πβt ) − d/ t − ( v ((4 βt ) − / ( x + y )) − v ((4 βt ) − / x )) . It is not hard to see that |∇ v ( q ) | ≤ C e − C | q | for all q . Thus, | v ((4 βt ) − / ( x + y )) − v ((4 βt ) − / x ) | ≤ Kt − / | y | . Similarly, | u ((4 βt ) − / ( x + y )) − u ((4 βt ) − / x ) | ≤ Kt − / | y | . Combining, we get | w ′ ( t ) | ≤ Kt − ( d +3) / | y | . Since | w ( t + r ) − w ( t ) | ≤ r max t ≤ s ≤ t + r | w ′ ( s ) | and r ≤ t , this shows that | δ ′ r δ y p ( t, x ) | ≤ Kt − ( d +3) / | y | r. This completes the proof of the lemma. (cid:3)
Now that we have the required bounds for the derivatives of p , we can use themto deduce the corresponding bounds for the derivatives of p . Proposition 10.8.
For any two integers t ≥ r ≥ , and any x ∈ Z d , | δ ′ r p ( t, x ) | ≤ C e − C | x | /t rt − ( d +2) / + C t − ( d +2) / (log t ) C . Moreover, we also have | δ ′ r p ( t, x ) | ≤ C e − C | x | /t .Proof. Since ≤ r ≤ t , it is easy to see that the case t = 1 is trivial. So let usassume that t ≥ . First, note that by Lemma 9.3, | δ ′ r p ( t, x ) − δ ′ r p ( t, x ) | ≤ | p ( t + r, x ) − p ( t + r, x ) | + | p ( t, x ) − p ( t, x ) |≤ C e − C (log t ) . Similarly, by Lemma 9.4, | δ ′ r p ( t, x ) − δ ′ r p ( t, x ) | ≤ C e − C (log t ) . NIVERSALITY OF DETERMINISTIC KPZ 21
Next, by Lemma 10.1, | δ ′ r p ( t, x ) − δ ′ r p ( t, x ) |≤ | p ( t + r, x ) − p ( t + r, x ) | + | p ( t, x ) − p ( t, x ) |≤ C t − ( d +2) / (log t ) C . Combining the above inequalities with the bound on | δ ′ r p ( t, x ) | from Lemma 10.6,and the observation that C t − ( d +2) / (log t ) C dominates C e − C (log t ) since t ≥ , we get the first inequality in the statement of the lemma. The second one followsfrom Lemma 10.3 and the triangle inequality. (cid:3) Proposition 10.9.
Take any two integers t ≥ r ≥ , and any x, y ∈ Z d . Let a := min {| x | , | x + y |} . Then | δ ′ r δ y p ( t, x ) | ≤ C e − C a /t t − ( d +3) / r | y | + C | y | t − ( d +3) / (log t ) C . Moreover, we also have | δ ′ r δ y p ( t, x ) | ≤ C e − C a /t .Proof. The bounds hold trivially if y = 0 . So let us assume that y = 0 . Thenagain, the bounds are trivial if t = 1 . So let us also assume that t ≥ . First, notethat by Lemma 9.3, | δ ′ r δ y p ( t, x ) − δ ′ r δ y p ( t, x ) |≤ | p ( t + r, x + y ) − p ( t + r, x + y ) | + | p ( t, x + y ) − p ( t, x + y ) | + | p ( t + r, x ) − p ( t + r, x ) | + | p ( t, x ) − p ( t, x ) |≤ C e − C (log t ) . Similarly, by Lemma 9.4, | δ ′ r δ y p ( t, x ) − δ ′ r δ y p ( t, x ) | ≤ C e − C (log t ) . Next, by Lemma 10.1, | δ ′ r δ y p ( t, x ) − δ ′ r δ y p ( t, x ) |≤ | δ y p ( t + r, x ) − δ y p ( t + r, x ) | + | δ y p ( t, x ) − δ y p ( t, x ) |≤ C | y | t − ( d +3) / (log t ) C . Combining the above inequalities with the bound on | δ ′ r δ y p ( t, x ) | from Lemma10.7, and the observation that C | y | t − ( d +3) / (log t ) C dominates C e − C (log t ) since y = 0 and t ≥ , we get the first inequality in the statement of the lemma.The second one follows from Lemma 10.3 and the triangle inequality. (cid:3)
11. S
MOOTHNESS OF THE DISCRETE SURFACE
In this section we will derive bounds on the first- and second-order discretederivatives of f ε using the bounds on the derivatives of p obtained in the previoussection, the recursive formula from Proposition 6.1, and the uniform bound on h ε from Lemma 7.1. Our first result gives a bound on second-order spatial derivativesof f ε . Proposition 11.1.
Take any t ∈ Z ≥ and any x, y, z ∈ Z d with | y | ≤ | z | . Then | δ y δ z f ε ( t, x ) |≤ C ε | y || z | log(2 + t/ | z | )+ C (1 + ε | x | + ε | y | + ε | z | + εt / ) t − / | y | min { , t − / | z |} + C (1 + ε | x | + ε | y | + ε | z | + εt / log t ) | y || z | t − (log t ) C . Proof. If y = 0 or z = 0 , the left side is zero and hence the bound holds trivially.So let us assume that y and z are nonzero. By Proposition 6.1, we get δ y δ z f ε ( t, x ) = X w ∈ Z d δ y δ z p ( t, x − w ) g ε ( w )+ X ≤ s ≤ t − X w ∈ Z d δ y δ z p ( s, x − w ) h ε ( t − s, w ) . (11.1)Take any ≤ s ≤ t − (which exists only if t ≥ ; otherwise, we do not have toworry about such s ). By Lemma 7.1, | h ε ( t − s, w ) | ≤ Cε for all w . Thus, (cid:12)(cid:12)(cid:12)(cid:12) X w ∈ Z d δ y δ z p ( s, x − w ) h ε ( t − s, w ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cε X w ∈ Z d | δ y δ z p ( s, x − w ) | . (11.2)Let D be the set of all w such that at least one of the quantities | x − w | , | x − w + y | , | x − w + z | and | x − w + y + z | is less than √ s log s . Then by the second boundfrom Proposition 10.5, we get X w / ∈ D | δ y δ z p ( s, x − w ) | ≤ C X | v |≥√ s log s e − C | v | /s ≤ C e − C (log s ) . (11.3)Fixing x , y and z , define q ( w ) := min {| x − w | , | x − w + y | , | x − w + z | , | x − w + y + z |} for w ∈ Z d . Now, it is not hard to see that for any C , X v ∈ Z d e − C | v | /s ≤ C ′ s d/ (11.4)where C ′ depends on C and d . This implies that X w ∈ Z d e − Cq ( w ) /s ≤ X w ∈ Z d ( e − C | x − w | /s + e − C | x − w + y | /s + e − C | x − w + z | /s + e − C | x − w + y + z | /s )= 4 X w ∈ Z d e − C | w | /s ≤ C ′ s d/ , NIVERSALITY OF DETERMINISTIC KPZ 23 where C ′ depends on C and d . Also, note that | D | ≤ Cs d/ (log s ) d . Therefore,the first bound from Proposition 10.5 gives X w ∈ D | δ y δ z p ( s, x − w ) | ≤ C s − ( d +1) / | y | min { , s − / | z |} X w ∈ D e − C q ( w ) /s + C | y || z | s − ( d +4) / (log s ) C | D |≤ C s − / | y | min { , s − / | z |} + C | y || z | s − (log s ) C . (11.5)Combining (11.2), (11.3) and (11.5), and the assumption that y and z are nonzero(which allows the bound from (11.5) to dominate the bound from (11.3)), we have (cid:12)(cid:12)(cid:12)(cid:12) X w ∈ Z d δ y δ z p ( s, x − w ) h ε ( t − s, w ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C ε | y | s − / min { , s − / | z |} + C ε | y || z | s − (log s ) C . Consequently, (cid:12)(cid:12)(cid:12)(cid:12) X ≤ s ≤ t − X w ∈ Z d δ y δ z p ( s, x − w ) h ε ( t − s, w ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C ε | y | X ≤ s ≤ t − s − / min { , s − / | z |} + C ε | y || z | . Now note that X ≤ s ≤ t − s − / min { , s − / | z |} ≤ X ≤ s ≤| z | s − / + X | z |
For any two integers t ≥ r ≥ and any x ∈ Z d , | δ ′ r f ε ( t, x ) | ≤ C ε r (1 + log( t/r )) + C ε (log t ) C + C rt − (1 + ε | x | + εt / )+ C (1 + ε | x | + εt / log t ) t − (log t ) C . NIVERSALITY OF DETERMINISTIC KPZ 25
Proof.
By Proposition 6.1, δ ′ r f ε ( t, x ) = X w ∈ Z d δ ′ r p ( t, x − w ) g ε ( w )+ X ≤ s ≤ t − X w ∈ Z d δ ′ r p ( s, x − w ) h ε ( t − s, w )+ X ≤ s ≤ r − X w ∈ Z d p ( s, x − w ) h ε ( t + r − s, w ) . (11.12)Recall that by Lemma 7.1, | h ε ( s, w ) | ≤ Cε for all s and w . Therefore, (cid:12)(cid:12)(cid:12)(cid:12) X ≤ s ≤ r − X w ∈ Z d p ( s, x − w ) h ε ( t + r − s, w ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cε X ≤ s ≤ r − X w ∈ Z d p ( s, x − w ) = Cε r. (11.13)Similarly, we have (cid:12)(cid:12)(cid:12)(cid:12) X ≤ s ≤ r X w ∈ Z d δ ′ r p ( s, x − w ) h ε ( t − s, w ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cε X ≤ s ≤ r X w ∈ Z d ( p ( s + r, x − w ) + p ( s, x − w )) ≤ Cε r. (11.14)Finally, take any r + 1 ≤ s ≤ t − . (If t ≤ r + 1 , then no such s exists and we donot have to worry about this case.) Let D := { w : | x − w | ≤ √ s log s } . Then bythe second bound from Lemma 10.8, (cid:12)(cid:12)(cid:12)(cid:12) X w / ∈ D δ ′ r p ( s, x − w ) h ε ( t − s, w ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C ε X w / ∈ D e − C | x − w | /s ≤ C ε e − C (log s ) . But, by the first bound from Proposition 10.8 and the inequality (11.4), (cid:12)(cid:12)(cid:12)(cid:12) X w ∈ D δ ′ r p ( s, x − w ) h ε ( t − s, w ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C ε X w ∈ D rs − ( d +2) / e − C | x − w | /s + C ε | D | s − ( d +2) / (log s ) C ≤ C ε rs − + C ε s − (log s ) C . Adding the last two displays, we get that for any r + 1 ≤ s ≤ t − , (cid:12)(cid:12)(cid:12)(cid:12) X w ∈ Z d δ ′ r p ( s, x − w ) h ε ( t − s, w ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C ε rs − + C ε s − (log s ) C . (11.15) Combining (11.14) and (11.15), we get (cid:12)(cid:12)(cid:12)(cid:12) X ≤ s ≤ t − X w ∈ Z d δ ′ r p ( s, x − w ) h ε ( t − s, w ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C ε r (1 + log( t/r )) + C ε (log t ) C . (11.16)Next, let D := { w : | x − w | ≤ √ t log t } . Then by the second bound fromProposition 10.8 and the facts that r ≤ t and | g ε ( w ) | ≤ C (1 + ε | w | ) , we get (cid:12)(cid:12)(cid:12)(cid:12) X w / ∈ D δ ′ r p ( t, x − w ) g ε ( w ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C X w / ∈ D (1 + ε | w | ) e − C | x − w | /t ≤ C (1 + ε | x | ) e − C (log t ) . Similarly, by the first bound from Proposition 10.8, (cid:12)(cid:12)(cid:12)(cid:12) X w ∈ D δ ′ r p ( t, x − w ) g ε ( w ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C X w ∈ D (1 + ε | w | ) rt − ( d +2) / e − C | x − w | /t + C t − ( d +2) / (log t ) C X w ∈ D (1 + ε | w | ) ≤ C rt − (1 + ε | x | + εt / )+ C (1 + ε | x | + εt / log t ) t − (log t ) C . Since t − (log t ) C dominates e − C (log t ) , combining the last two displays gives (cid:12)(cid:12)(cid:12)(cid:12) X w ∈ Z d δ ′ r p ( t, x − w ) g ε ( w ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C rt − (1 + ε | x | + εt / )+ C (1 + ε | x | + εt / log t ) t − (log t ) C . (11.17)Combining (11.12), (11.13), (11.16) and (11.17) completes the proof. (cid:3) The next result gives bounds for mixed space-time derivatives of f ε . Again,the proof uses Proposition 6.1 and the bounds on mixed derivatives of p from theprevious section. Proposition 11.3.
For any two integers t ≥ r ≥ and any x, y ∈ Z d , | δ ′ r δ y f ε ( t, x ) | ≤ C ε √ r | y | + C ε | y | r − / (log t ) C + C rt − / | y | (1 + ε | x | + ε | y | + εt / )+ C | y | (1 + ε | x | + ε | y | + εt / log t ) t − / (log t ) C . NIVERSALITY OF DETERMINISTIC KPZ 27
Proof.
The bound holds trivially if y = 0 . So let us assume that y = 0 . ByProposition 6.1, δ ′ r δ y f ε ( t, x ) = X w ∈ Z d δ ′ r δ y p ( t, x − w ) g ε ( w )+ X ≤ s ≤ t − X w ∈ Z d δ ′ r δ y p ( s, x − w ) h ε ( t − s, w )+ X ≤ s ≤ r − X w ∈ Z d δ y p ( s, x − w ) h ε ( t + r − s, w ) . (11.18)By Lemma 7.1, (cid:12)(cid:12)(cid:12)(cid:12) X ≤ s ≤ r − X w ∈ Z d δ y p ( s, x − w ) h ε ( t + r − s, w ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cε X ≤ s ≤ r − X w ∈ Z d | δ y p ( s, x − w ) | . Take any ≤ s ≤ r − . Let D be the set of all w such the minimum of | x − w | and | x + y − w | is less than √ s log s . Then by the second bound from Proposition 10.4,we get X w / ∈ D | δ y p ( s, x − w ) | ≤ C e − C (log s ) . On the other hand, by the first bound from Proposition 10.4 and the inequality(11.4), we have X w ∈ D | δ y p ( s, x − w ) | ≤ C s − ( d +1) / | y | X w ∈ D ( e − C | x − w | /s + e − C | x + y − w | /s )+ C | y | s − ( d +3) / (log s ) C | D |≤ C s − / | y | + C | y | s − / (log s ) C . Combining the last three displays, we get (cid:12)(cid:12)(cid:12)(cid:12) X ≤ s ≤ r − X w ∈ Z d δ y p ( s, x − w ) h ε ( t + r − s, w ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cε √ r | y | . Since the contributions from the s = 0 and s = 1 terms amount to at most Cε , wehave (cid:12)(cid:12)(cid:12)(cid:12) X ≤ s ≤ r − X w ∈ Z d δ y p ( s, x − w ) h ε ( t + r − s, w ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cε √ r | y | . (11.19) Similarly, (cid:12)(cid:12)(cid:12)(cid:12) X ≤ s ≤ r X w ∈ Z d δ ′ r δ y p ( s, x − w ) h ε ( t − s, w ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cε X ≤ s ≤ r X w ∈ Z d ( | δ y p ( s + r, x − w ) | + | δ y p ( s, x − w ) | ) ≤ Cε √ r | y | . (11.20)Finally, take any r + 1 ≤ s ≤ t − . As before, let D be the set of all w such theminimum of | x − w | and | x + y − w | is less than √ s log s . Then by Lemma 7.1and the second bound from Proposition 10.9, (cid:12)(cid:12)(cid:12)(cid:12) X w / ∈ D δ ′ r δ y p ( s, x − w ) h ε ( t − s, w ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C ε e − C (log s ) . On the other hand, by the first bound from Proposition 10.9 and the inequality(11.4), (cid:12)(cid:12)(cid:12)(cid:12) X w ∈ D δ ′ r δ y p ( s, x − w ) h ε ( t − s, w ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C ε | y | X w ∈ D rs − ( d +3) / ( e − C | x − w | /s + e − C | x + y − w | /s )+ C ε | y || D | s − ( d +3) / (log s ) C ≤ C ε r | y | s − / + C ε | y | s − / (log s ) C . Adding the last two displays and using the assumption that y = 0 , we get that forany r + 1 ≤ s ≤ t − , (cid:12)(cid:12)(cid:12)(cid:12) X w ∈ Z d δ ′ r δ y p ( s, x − w ) h ε ( t − s, w ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C ε r | y | s − / + C ε | y | s − / (log s ) C . (11.21)Combining (11.20) and (11.21), we get (cid:12)(cid:12)(cid:12)(cid:12) X ≤ s ≤ t − X w ∈ Z d δ ′ r δ y p ( s, x − w ) h ε ( t − s, w ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C ε √ r | y | + C ε | y | r − / (log t ) C . (11.22)Next, let D be the set of all w such the minimum of | x − w | and | x + y − w | is lessthan √ t log t . Then by the second bound from Proposition 10.9 and the facts that NIVERSALITY OF DETERMINISTIC KPZ 29 r ≤ t and | g ε ( w ) | ≤ C (1 + ε | w | ) for all w , we get (cid:12)(cid:12)(cid:12)(cid:12) X w / ∈ D δ ′ r δ y p ( t, x − w ) g ε ( w ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C X w / ∈ D (1 + ε | w | )( e − C | x − w | /t + e − C | x + y − w | /t ) ≤ C (1 + ε | x | + ε | y | ) e − C (log t ) . Similarly, by the first bound from Proposition 10.9, (cid:12)(cid:12)(cid:12)(cid:12) X w ∈ D δ ′ r δ y p ( t, x − w ) g ε ( w ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C | y | X w ∈ D (1 + ε | w | ) rt − ( d +3) / ( e − C | x − w | /t + e − C | x + y − w | /t )+ C | y | t − ( d +3) / (log t ) C X w ∈ D (1 + ε | w | ) ≤ C rt − / | y | (1 + ε | x | + ε | y | + εt / )+ C | y | (1 + ε | x | + ε | y | + εt / log t ) t − / (log t ) C . Combining the last two displays, and observing that | y | t − / dominates e − C (log t ) ,we get (cid:12)(cid:12)(cid:12)(cid:12) X w ∈ Z d δ ′ r δ y p ( t, x − w ) g ε ( w ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C rt − / | y | (1 + ε | x | + ε | y | + εt / )+ C | y | (1 + ε | x | + ε | y | + εt / log t ) t − / (log t ) C . (11.23)Combining (11.18), (11.19), (11.22) and (11.23) completes the proof. (cid:3) We need one more estimate, to deal with situations where t is small. In this case, f ε ( t, x ) is expected to be close to g ε ( x ) . The following result makes this precise. Proposition 11.4.
For any t ∈ Z ≥ and any x ∈ Z d , | f ε ( t, x ) − g ε ( x ) | ≤ Cε √ t + Cε t. Proof.
By Proposition 6.1, | f ε ( t, x ) − g ε ( x ) | ≤ X y ∈ Z d p ( t, x − y ) | g ε ( y ) − g ε ( x ) | + X ≤ s ≤ t − X y ∈ Z d p ( s, x − y ) | h ε ( t − s, y ) | . By the Lipschitz property of g and simple properties of random walks, X y ∈ Z d p ( t, x − y ) | g ε ( y ) − g ε ( x ) | ≤ Cε X y ∈ Z d p ( t, x − y ) | x − y |≤ Cε √ t. On the other hand, by Lemma 7.1, X ≤ s ≤ t − X y ∈ Z d p ( s, x − y ) | h ε ( t − s, y ) | ≤ Cε X ≤ s ≤ t − X y ∈ Z d p ( s, x − y )= Cε t. Combining the last two displays completes the proof. (cid:3)
12. E
XISTENCE OF SUBSEQUENTIAL LIMITS
We will now use the estimates from the previous section to show that the rescaledand renormalized function f ( ε ) converges pointwise to a limit function along a sub-sequence. In fact, any subsequence will have a further subsequence that converges.Moreover, we will show that the convergence is uniform on compact sets. In thenext section, we will show that the convergence happens also at a finer scale, en-suring convergence of discrete spatial derivatives to the derivatives of the limit.In this section and in all subsequent sections, we will use the following conven-tions that were introduced in Section 1. For t ∈ R , let [ t ] denote the greatest integer ≤ t . For a point x = ( x , . . . , x d ) ∈ R d , let [ x ] denote the vector ([ x ] , . . . , [ x d ]) .For t ∈ R ≥ , x ∈ R d , and ε > , let t ε := [ ε − t ] , x ε := [ ε − x ] . Since we are working under the assumption that φ (0) = 0 , we have f ( ε ) ( t, x ) = f ε ( t ε , x ε ) . The key idea, unsurprisingly, is to show that the family { f ( ε ) } ε> isequicontinuous. This consists of the following two lemmas. Lemma 12.1.
Take any t ∈ R ≥ and x, y ∈ R d . Then lim sup ε → | f ( ε ) (0 , x ) − f ( ε ) ( t, y ) | ≤ C | x − y | + Ct.
Proof.
By Lemma 5.4, | f ( ε ) (0 , x ) − f ( ε ) (0 , y ) | = | f ε (0 , x ε ) − f ε (0 , y ε ) |≤ Cε | x ε − y ε |≤ C | x − y | . (12.1)On the other hand, by Proposition 11.4, | f ( ε ) (0 , y ) − f ( ε ) ( t, y ) | = | f ε (0 , y ε ) − f ε ( t ε , y ε ) | = | g ε ( y ε ) − f ε ( t ε , y ε ) |≤ Cε √ t ε + Cε t ε ≤ Ct. (12.2)The proof is completed by adding the two inequalities. (cid:3)
NIVERSALITY OF DETERMINISTIC KPZ 31
Lemma 12.2.
Take any real numbers s > and t ∈ [ s, s ) , and any x, y ∈ R d .Then lim sup ε → | f ( ε ) ( s, y ) − f ( ε ) ( t, x ) |≤ C | x − y | + C ( t − s ) (cid:18) st − s + 1 + | x | + √ ss (cid:19) , where the second term is interpreted as zero if s = t .Proof. By Lemma 5.4, | f ( ε ) ( s, y ) − f ( ε ) ( s, x ) | = | f ε ( s ε , y ε ) − f ε ( s ε , x ε ) |≤ Cε | y ε − x ε |≤ C | x − y | . (12.3)If s = t , this completes the proof. So let us assume that s < t . Let r ε := t ε − s ε . Since < t − s < s , r ε must be less than s ε and bigger than when ε is smallenough. Therefore by Proposition 11.2, | f ( ε ) ( s, x ) − f ( ε ) ( t, x ) | = | δ r ε f ε ( s ε , x ε ) |≤ C ε r ε (1 + log( s ε /r ε )) + C ε (log t ε ) C + C r ε s − ε (1 + ε | x ε | + εs / ε )+ C (1 + ε | x ε | + εs / ε log s ε ) s − ε (log s ε ) C . (12.4)Note that ε r ε → t − s , ε s ε → s , and ε | x ε | → | x | as ε → . From this, it is easyto compute the limit of the right side in the above display as ε → . This completesthe proof. (cid:3) We are now ready to prove the main result of this section.
Proposition 12.3.
Given any sequence ε n → , there is a subsequence alongwhich f ( ε ) converges pointwise everywhere on R ≥ × R d . Moreover, the conver-gence is guaranteed to be uniform on compact subsets of R ≥ × R d . Any such limit f is continuous and satisfies f (0 , x ) = g ( x ) for all x .Proof. Take any t ∈ R ≥ and x ∈ R d . By Proposition 11.4 and the Lipschitzproperty of g , | f ( ε ) ( t, x ) | = | f ε ( t ε , x ε ) |≤ | g ε ( x ε ) | + Cε √ t ε + Cε t ε ≤ C (1 + | x | + √ t + t ) . (12.5)In particular, | f ( ε ) ( t, x ) | is bounded by a number that does not depend on ε . Thus,given any sequence ε n → , we can find (by a diagonal argument) a subsequence ε n k along which f ( ε nk ) ( t, x ) converges to a limit f ( t, x ) for every t ∈ Q ≥ and x ∈ Q d , where Q is the set of rational numbers. Now take any x ∈ R d and any δ > . Find y ∈ Q d such that | x − y | < δ . Thenby Lemma 12.1, lim sup k →∞ | f ( ε nk ) (0 , x ) − f (0 , y ) | ≤ Cδ.
In particular, lim sup k →∞ f ( ε nk ) (0 , x ) − lim inf k →∞ f ( ε nk ) (0 , x ) ≤ Cδ.
Since this holds for any δ , we see that f ( ε nk ) (0 , x ) converges to a limit as k → ∞ .Moreover, by (12.5), the sequence is bounded. Hence the limit is finite. Let us callit f (0 , x ) .Next, take any t > and x ∈ R d . Take any s ∈ ( t/ , t ] ∩ Q and y ∈ Q d . Thenby Lemma 12.2, lim sup k →∞ | f ( s, y ) − f ( ε nk ) ( t, x ) |≤ C | x − y | + C ( t − s ) (cid:18) st − s + 1 + | x | + √ ss (cid:19) . The right side can be made arbitrarily small by bringing y close to x and s close to t . Then by the same argument as before, we conclude that f ( t, x ) := lim k →∞ f ( ε nk ) ( t, x ) exists and is finite. Continuity of f follows from Lemmas 12.1 and 12.2. Proposi-tion 11.4 shows that f (0 , x ) = g ( x ) . Lastly, it is easy to deduce using the inequal-ities (12.3) and (12.4) from the proof of Lemma 12.2 that if ( t n k , x n k ) → ( t, x ) ∈ R > × R d , then f ( ε nk ) ( t n k , x n k ) → f ( t, x ) . If ( t n k , x n k ) → (0 , x ) ∈ { } × R d ,the same deduction can be made using the inequalities (12.1) and (12.2) from theproof of Lemma 12.1. It is a standard exercise to deduce uniform convergence oncompact sets from these facts. (cid:3)
13. D
IFFERENTIABILITY OF SUBSEQUENTIAL LIMITS
The goal of this section is to show that if f ( ε ) converges to f along a subse-quence, then f is differentiable in x (for any fixed t ), and the first-order discretederivatives of f ( ε ) converge pointwise to the corresponding derivatives of f . More-over, the derivatives are continuous in t and x . As in the previous section, theresults of this section are also based on the estimates from Section 11.Let e , . . . , e d be the standard basis vectors of R d . For t ∈ Z ≥ and x ∈ Z d ,define the discrete derivative D i f ε ( t, x ) := f ε ( t, x + e i ) − f ε ( t, x ) ε . Also, for t ∈ R ≥ and x ∈ R d , define D i f ( ε ) ( t, x ) := D i f ε ( t ε , x ε ) . NIVERSALITY OF DETERMINISTIC KPZ 33
Let Df ( ε ) ( t, x ) denote the vector ( D i f ( ε ) ( t, x )) ≤ i ≤ d . We need several lemmasabout the behaviors of these discrete derivatives. The first lemma shows that theyare uniformly bounded. Lemma 13.1.
For each ≤ i ≤ d , | D i f ε | and | D i f ( ε ) | are bounded by L every-where.Proof. This is just a restatement of Lemma 5.4. (cid:3)
The next lemma shows that the discrete derivatives have a certain degree ofsmoothness.
Lemma 13.2.
For any ≤ i, j ≤ d , t ∈ Z ≥ , x ∈ Z d , and k ∈ Z \ { } , we have | D i f ε ( t, x + ke j ) − D i f ε ( t, x ) | ≤ J ( ε, k, t, x ) , where J ( ε, k, t, x ):= C ε | k | log(2 + t/k )+ C ε − (1 + ε | x | + ε | k | + εt / ) t − / min { , t − / | k |} + C ε − (1 + ε | x | + ε | k | + εt / log t ) | k | t − (log t ) C for some suitable constants C and C that depend only on φ , L and d .Proof. Note that D i f ε ( t, x + ke j ) − D i f ε ( t, x ) = ε − ( δ e i f ε ( t, x + ke j ) − δ e i f ε ( t, x ))= ε − δ e i δ ke j f ε ( t, x ) . The desired bound now follows from Lemma 11.1. (cid:3)
The function J defined in the above lemma describes an important remainderterm. We need the following fact about it. Lemma 13.3.
For any positive integers k and t , any ε ∈ (0 , , and any x ∈ Z d , k X j =1 J ( ε, j, t, x ) ≤ kJ ( ε, k, t, x ) + Cεk . Proof.
Let C be the constant displayed in the definition of J . From the definitionof J , it is immediate that for any ε , t and x , the function J ( ε, k, t, x ) + 2 C ε | k | log | k | is increasing in | k | . Thus, for any positive integers k ≥ j ≥ , J ( ε, j, t, x ) ≤ J ( ε, k, t, x ) + 2 C εk log( k/j ) . The proof is completed by observing that P kj =1 log( k/j ) ≤ Ck for some universalconstant C . (cid:3) Lemma 13.4.
For any ≤ i ≤ d , t ∈ Z ≥ , x ∈ Z d , and k ∈ Z \ { } , | f ε ( t, x + ke i ) − f ε ( t, x ) − kεD i f ε ( t, x ) | ≤ ε | k | J ( ε, k, t, x ) + Cε k . Proof.
Suppose that k > . Then note that f ε ( t, x + ke i ) − f ε ( t, x ) − kεD i f ε ( t, x )= ε k − X j =1 ( D i f ε ( t, x + je i ) − D i f ε ( t, x )) . Bounding the absolute value of each term in the above sum using Lemma 13.2, weget | f ε ( t, x + ke i ) − f ε ( t, x ) − kεD i f ε ( t, x ) | ≤ ε k − X j =1 J ( ε, j, t, x ) . By Lemma 13.3, this completes the proof when k > . The proof for k < is similar, since Lemma 13.2 works for both positive and negative k in the sameway. (cid:3) Lemma 13.5.
For any ≤ i ≤ d , t ∈ R > , x ∈ R d , and a ∈ R , lim sup ε → (cid:12)(cid:12)(cid:12)(cid:12) f ( ε ) ( t, x + ae i ) − f ( ε ) ( t, x ) a − D i f ( ε ) ( t, x ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ Q ( a, t, x ) , where Q ( a, t, x ) := C | a | log(2 + t/a )+ C (1 + | x | + | a | + t / ) t − / min { , t − / | a |} for some suitable constant C that depends only on φ , L and d .Proof. Let k ε be the integer such that [ ε − ( x + ae i )] = x ε + k ε e i . (13.1)(Note that there is such an integer because [ ε − ( x + ae i )] and x ε agree on allcoordinates other than i .) Note also that f ( ε ) ( t, x + ae i ) − f ( ε ) ( t, x ) − aD i f ( ε ) ( t, x )= f ε ( t ε , x ε + k ε e i ) − f ε ( t ε , x ε ) − aD i f ε ( t ε , x ε ) . Let p be the value of coordinate i of the vector x ε , and let q be the value of coordi-nate i of the vector [ ε − ( x + ae i )] . Then on the one hand, | q − ε − ( x i + a ) | ≤ .On the other hand, by (13.1), q = p + k ε . But | p − ε − x i | ≤ . Combining thesethree observations, we get | k ε − ε − a | ≤ . (13.2)Combining this with Lemma 13.1, we get lim ε → ( a − k ε ε ) D i f ε ( t ε , x ε ) = 0 . NIVERSALITY OF DETERMINISTIC KPZ 35
Thus, lim sup ε → | f ( ε ) ( t, x + ae i ) − f ( ε ) ( t, x ) − aD i f ( ε ) ( t, x ) | = lim sup ε → | f ε ( t ε , x ε + k ε e i ) − f ε ( t ε , x ε ) − k ε εD i f ε ( t ε , x ε ) | . By Lemma 13.4, | f ε ( t ε , x ε + k ε e i ) − f ε ( t ε , x ε ) − k ε εD i f ε ( t ε , x ε ) |≤ ε | k ε | J ( ε, k ε , t ε , x ε ) + Cε k ε . Note that as ε → , εx ε → x and ε t ε → t . Also, by (13.2), εk ε → a . Pluggingthese limits into the right side of the above display, we get the required result. (cid:3) Let us now fix a sequence ε n → such that f ( ε n ) converges pointwise (and uni-formly on compact sets) to a limit f . Such a sequence exists by Proposition 12.3.Moreover, the function f is continuous and satisfies f (0 , x ) = g ( x ) for all x . Wewill now show that f is differentiable in x , the derivatives are continuous in x and t , and the discrete first-order derivatives of f ( ε n ) converge to the correspondingderivatives of f . In the next two lemmas, Q denotes the function appearing as theerror bound in Lemma 13.5. Lemma 13.6.
For any ≤ i ≤ d , t ∈ R > , x ∈ R d , and a, b ∈ R \ { } , (cid:12)(cid:12)(cid:12)(cid:12) f ( t, x + ae i ) − f ( t, x ) a − f ( t, x + be i ) − f ( t, x ) b (cid:12)(cid:12)(cid:12)(cid:12) ≤ Q ( a, t, x ) + Q ( b, t, x ) . Proof.
Applying Lemma 13.5 and dividing both sides by a , and then doing thesame with a replaced by b , and finally, applying the triangle inequality, we arriveat lim sup ε → (cid:12)(cid:12)(cid:12)(cid:12) f ( ε ) ( t, x + ae i ) − f ( ε ) ( t, x ) a − f ( ε ) ( t, x + be i ) − f ( ε ) ( t, x ) b (cid:12)(cid:12)(cid:12)(cid:12) ≤ Q ( a, t, x ) + Q ( b, t, x ) . The proof is now completed by replacing ε by ε n and lim sup ε → by lim n →∞ . (cid:3) Note that Q ( a, t, x ) → as a → . Therefore, Lemma 13.6 implies that as a → , the numbers ( f ( t, x + ae i ) − f ( t, x )) /a have the Cauchy property, andhence the limit lim a → f ( t, x + ae i ) − f ( t, x ) a exists and is finite. Let us call this limit ∂ i f ( t, x ) , and let us denote the vector ( ∂ i f ( t, x )) ≤ i ≤ d by ∇ f ( t, x ) . We will show below that f is differentiable in x with gradient ∇ f , and also that Df ( ε n ) converges to ∇ f . The following lemma isthe key step in the proof. Lemma 13.7.
For any t ∈ R > and x ∈ R d , and any ≤ i ≤ d , lim n →∞ D i f ( ε n ) ( t, x ) = ∂ i f ( t, x ) . Moreover, for any a = 0 , (cid:12)(cid:12)(cid:12)(cid:12) f ( t, x + ae i ) − f ( t, x ) a − ∂ i f ( t, x ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ Q ( a, t, x ) . Proof.
By Lemma 13.5, for any a = 0 , lim sup n →∞ (cid:12)(cid:12)(cid:12)(cid:12) f ( t, x + ae i ) − f ( t, x ) a − D i f ( ε n ) ( t, x ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ Q ( a, t, x ) . (13.3)Consequently, f ( t, x + ae i ) − f ( t, x ) a − Q ( a, t, x ) ≤ lim inf n →∞ D i f ( ε n ) ( t, x ) ≤ lim sup n →∞ D i f ( ε n ) ( t, x ) ≤ f ( t, x + ae i ) − f ( t, x ) a + Q ( a, t, x ) . Since ∂ i f ( t, x ) = lim a → ( f ( t, x + ae i ) − f ( t, x )) /a and Q ( a, t, x ) → as a → ,this proves the first claim. The second claim follows by combining the first claimwith the inequality (13.3). (cid:3) We now arrive at the main result of this section.
Proposition 13.8.
The function ∇ f : R > × R d → R d is continuous. Moreover,the function Df ( ε n ) converges to ∇ f uniformly on any compact subset of R > × R d as n → ∞ . In particular, for any sequence of points ( t n , x n ) converging to a point ( t, x ) ∈ R > × R d , we have Df ( ε n ) ( t n , x n ) → ∇ f ( t, x ) .Proof. Fix t > and ≤ i ≤ d . Take any x ∈ R d , z ∈ R d \ { } , and ε > . Let z ′ ε := ( x + z ) ε − x ε . Then note that D i f ( ε ) ( t, x + z ) − D i f ( ε ) ( t, x ) = ε − ( δ e i f ε ( t ε , x ε + z ′ ε ) − δ e i f ε ( t ε , x ε ))= ε − δ e i δ z ′ ε f ( t ε , x ε ) . Therefore, by Lemma 11.1, | D i f ( ε ) ( t, x + z ) − D i f ( ε ) ( t, x ) |≤ C ε | z ′ ε | log(2 + t ε / | z ′ ε | )+ C ε − (1 + ε | x ε | + ε | z ′ ε | + εt / ε ) t − / ε min { , t − / ε | z ′ ε |} + C ε − (1 + ε | x ε | + ε | z ′ ε | + εt / ε log t ε ) | z ′ ε | t − ε (log t ε ) C . (13.4)Now note that as ε → , ε t ε → t , εx ε → x and εz ′ ε → z . Thus, replacing ε by ε n in the above display, taking n → ∞ , and using Lemma 13.7, we get | ∂ i f ( t, x + z ) − ∂ i f ( t, x ) |≤ C | z | log(2 + t/ | z | ) + C (1 + | x | + | z | + t / ) t − / min { , t − / | z |} . NIVERSALITY OF DETERMINISTIC KPZ 37
Since the right side tends to zero as z → , this proves the continuity of ∂ i f ( t, x ) in x . Next, take some r ∈ (0 , t ) , and let r ′ ε := ( t + r ) ε − t ε . Then D i f ( ε ) ( t + r, x ) − D i f ( ε ) ( t, x ) = ε − ( δ e i f ε ( t ε + r ′ ε , x ε ) − δ e i f ε ( t ε , x ε ))= ε − δ ′ r ′ ε δ e i f ( t ε , x ε ) . Therefore, by Proposition 11.3, | D i f ( ε ) ( t + r, x ) − D i f ( ε ) ( t, x ) |≤ C ε p r ′ ε + C εr ′ ε − / (log t ε ) C + C ε − r ′ ε t − / ε (1 + ε | x ε | + ε + εt / ε )+ C ε − (1 + ε | x ε | + ε + εt / ε log t ε ) t − / ε (log t ε ) C . (13.5)Replacing ε by ε n and letting n → ∞ , we get | ∂ i f ( t + r, x ) − ∂ i f ( t, x ) | ≤ C √ r + Crt − / (1 + | x | + √ t ) . This proves the continuity of ∂ i f in t .Next, by the bounds (13.4) and (13.5) and the first assertion of Lemma 13.7, itis easy to see that if ( t n , x n ) is a sequence of points converging to a point ( t, x ) ∈ R > × R d , then Df ( ε n ) ( t n , x n ) → ∇ f ( t, x ) . It is a standard fact that this isequivalent to uniform convergence on compact sets. (cid:3) The following result is an immediate corollary of Proposition 13.8.
Corollary 13.9.
For any t > , f ( t, x ) is continuously differentiable in x , withderivative ∇ f defined above. Moreover, |∇ f | is uniformly bounded by L √ d .Proof. By definition, ∂ i f is the partial derivative of f in direction i . Since thesepartial derivatives are continuous by Proposition 13.8, it follows that f ( t, · ) is con-tinuously differentiable for any fixed t , and ∇ f is its gradient. Lemma 13.1 showsthat | Df ( ε ) | is uniformly bounded by L √ d everywhere, for any ε . By Lemma 13.7,this shows that |∇ f | is also bounded by the same quantity. (cid:3)
14. O
RIGIN OF THE GRADIENT SQUARED FIELD
As in the previous section, let us fix a sequence ε n → such that f ( ε n ) convergesto a limit f . In this section, we will show that a suitably rescaled version of h ε n converges to a scalar multiple of |∇ f | . Using Proposition 6.1, this will allow uslater to derive an integral equation for f .Take any b, b ′ ∈ B such that b = b ′ and b = − b ′ . Let γ := ∂ b φ (0) , γ := ∂ b ∂ − b φ (0) , γ := ∂ b ∂ b ′ φ (0) . By the invariance of φ under lattice symmetries, these numbers do not depend onthe choices of b and b ′ . Note that the quantity γ in the statement of Theorem 1.1 equals γ − γ . Define H ε ( t, x ) := γ X b ∈ B ( δ b f ε ( t − , x )) + γ X b ∈ B δ b f ε ( t − , x ) δ − b f ε ( t − , x )+ γ X b,b ′ ∈ Bb = b ′ ,b = − b ′ δ b f ε ( t − , x ) δ b ′ f ε ( t − , x ) . The following lemma shows that h ε is close to H ε . Lemma 14.1.
For any ε > , t ∈ Z ≥ and x ∈ Z d , | h ε ( t, x ) − H ε ( t, x ) | ≤ ε F ( ε ) , where F is a function determined solely by φ (and not depending on t or x ), suchthat F ( ε ) → as ε → .Proof. Let u ∈ R A be the vector whose coordinates are u = 0 and u b = δ b f ε ( t − , x ) for b ∈ B . Then recall the identity (7.1), which says that f ε ( t, x ) = f ε ( t − , x ) + φ ( u ) . By Lemma 5.4, | u b | ≤ Cε for all b . Since φ is a C function and φ (0) = 0 , thisimplies that (cid:12)(cid:12)(cid:12)(cid:12) f ε ( t, x ) − f ε ( t − , x ) − X b ∈ B u b ∂ b φ (0) − X b,b ′ ∈ B u b u b ′ ∂ b ∂ b ′ φ (0) (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cε F ( ε ) , where F is determined solely by φ , and F ( ε ) → as ε → . It is easy to checkthat the quantity on the left equals | h ε ( t, x ) − H ε ( t, x ) | . (cid:3) For t ∈ R > and x ∈ R d , define h ( ε ) ( t, x ) := ε − h ε ( t ε , x ε ) ,H ( ε ) ( t, x ) := ε − H ε ( t ε , x ε ) . The following proposition is the main result of this section. It shows that h ( ε n ) converges pointwise to | γ |∇ f | . The first step in the proof is to show that thisholds for H ( ε n ) , and then use Lemma 14.1 to deduce that it also holds for h ( ε n ) . Proposition 14.2.
For any sequence ( t n , x n ) converging to a point ( t, x ) ∈ R > × R d , we have lim n →∞ h ( ε n ) ( t n , x n ) = γ |∇ f ( t, x ) | . Proof.
For any t ∈ Z ≥ and x ∈ Z d , note that δ e i f ε ( t − , x ) = εD i f ε ( t − , x )= εD i f ( ε ) ( ε ( t − , εx ) . NIVERSALITY OF DETERMINISTIC KPZ 39
Similarly, δ − e i f ε ( t − , x ) = − εD i f ε ( t − , x − e i )= − εD i f ( ε ) ( ε ( t − , ε ( x − e i )) . Next, changing notation, let ( t n , x n ) and ( t, x ) be as in the statement of the propo-sition. Let s n := [ ε − n t n ] and y n := [ ε − n x n ] . Then by the above identities, H ( ε n ) ( t n , x n ) = ε − n H ε n ( s n , y n )= γ d X i =1 (cid:20) ( D i f ( ε n ) ( ε n ( s n − , ε n y n )) + ( D i f ( ε n ) ( ε n ( s n − , ε n ( y n − e i ))) (cid:21) − γ d X i =1 D i f ( ε n ) ( ε n ( s n − , ε n y n ) D i f ( ε n ) ( ε n ( s n − , ε n ( y n − e i ))+ γ X ≤ i = j ≤ d D i f ( ε n ) ( ε n ( s n − , ε n y n ) D j f ( ε n ) ( ε n ( s n − , ε n y n ) − γ X ≤ i = j ≤ d D i f ( ε n ) ( ε n ( s n − , ε n y n ) D j f ( ε n ) ( ε n ( s n − , ε n ( y n − e j )) . Now note that ε n s n → t and ε n y n → x as n → ∞ . Therefore by Proposition 13.8,we have the following limits, for each i : lim n →∞ D i f ( ε n ) ( ε n ( s n − , ε n y n ) = ∂ i f ( t, x ) , lim n →∞ D i f ( ε n ) ( ε n ( s n − , ε n ( y n − e i )) = ∂ i f ( t, x ) . Plugging these into the previous display, we get lim n →∞ H ( ε n ) ( t n , x n ) = ( γ − γ ) |∇ f ( t, x ) | = γ |∇ f ( t, x ) | . But by Lemma 14.1, the limit of h ( ε n ) ( t n , x n ) must also be the same. (cid:3)
15. D
UHAMEL REPRESENTATION FOR SUBSEQUENTIAL LIMITS
As in the previous two sections, fix a sequence ε n → such that f ( ε n ) con-verges to a limit f . In this section we will derive an integral equation for f . Theidea is to use Proposition 14.2 to show that the recursive equation displayed inProposition 6.1 yields an integral equation for f in the scaling limit.Recall the transition probability p ( t, x ) defined in Section 6. For ( t, x ) ∈ R ≥ × R d and ε > , define p ( ε ) ( t, x ) := ε − d p ( t ε , x ε ) . Also, recall the Gaussian kernel K ( t, x ) = (4 πβt ) − d/ e −| x | / βt defined in thestatement of Theorem 1.1. The following result shows that p ( ε ) converges to K uniformly on compact sets as ε → . Lemma 15.1.
For any sequence ( t n , x n ) converging to a point ( t, x ) ∈ R > × R d and any sequence ε n → , lim n →∞ p ( ε n ) ( t n , x n ) = K ( t, x ) . Proof.
Recall the functions p , p , and p from Section 9. For ( t, x ) ∈ R > × R d ,define p ( ε )3 ( t, x ) := ε − d p ( t ε , x ε ) . Note that ( t n ) ε n ∼ ε − n t as n → ∞ , since t n → t > . Thus, by Lemmas 9.3, 9.4and 10.1, we have that as n → ∞ , p ( ε n ) ( t n , x n ) − p ( ε n )1 ( t n , x n ) = O ( ε − dn e − C (log ε n ) ) ,p ( ε n )1 ( t n , x n ) − p ( ε n )2 ( t n , x n ) = O ( ε − dn ε d +2 n | log ε n | C ) ,p ( ε n )2 ( t n , x n ) − p ( ε n )3 ( t n , x n ) = O ( ε − dn e − C (log ε n ) ) . All of the above tend to zero as n → ∞ . Thus, lim n →∞ ( p ( ε n ) ( t n , x n ) − p ( ε n )3 ( t n , x n )) = 0 . On the other hand, from the explicit expression (9.2) for p , we have that lim n →∞ p ( ε n )3 ( t n , x n ) = K ( t, x ) . This completes the proof of the lemma. (cid:3)
Define g ( ε ) : R d → R as g ( ε ) ( x ) := g ε ( x ε ) = g ( ε [ ε − x ]) . Recall the definition of h ( ε ) from the previous section. The following lemma ex-presses the recursive formula from Proposition 6.1 as an integral involving thefunctions p ( ε ) , g ( ε ) and h ( ε ) . Lemma 15.2.
For any ( t, x ) ∈ R > × R d and ε ∈ (0 , √ t ) , f ( ε ) ( t, x )= Z R d p ( ε ) ( ε t ε , ε ( x ε − y ε )) g ( ε ) ( εy ε ) dy + Z ε t ε Z R d p ( ε ) ( ε s ε , ε ( x ε − y ε )) h ( ε ) ( ε ( t ε − s ε ) , εy ε ) dyds. Proof.
The map x [ x ] takes cubes of unit volume in R d to elements of Z d . Themap is surjective, and these cubes cover the whole of R d . This implies that for anyabsolutely summable w : Z d → R , X x ∈ Z d w ( x ) = Z R d w ([ x ]) dx. (15.1) NIVERSALITY OF DETERMINISTIC KPZ 41
Similarly, for any absolutely summable w : Z → R , and any two integers a < b , b − X t = a w ( t ) = Z ba w ([ t ]) dt. The condition ε < √ t implies that t ε ≥ . So, using the above identities andProposition 6.1, we get f ( ε ) ( t, x ) = f ε ( t ε , x ε )= X y ∈ Z d p ( t ε , x ε − y ) g ε ( y ) + X ≤ s ≤ t ε − X y ∈ Z d p ( s, x ε − y ) h ε ( t ε − s, y )= Z R d p ( t ε , x ε − [ y ]) g ε ([ y ]) dy + Z t ε Z R d p ([ s ] , x ε − [ y ]) h ε ( t ε − [ s ] , [ y ]) dyds. Applying the changes of variable z = εy and u = ε s in the above integrals, weget f ( ε ) ( t, x )= ε − d Z R d p ( t ε , x ε − z ε ) g ε ( z ε ) dz + ε − ( d +2) Z ε t ε Z R d p ( u ε , x ε − z ε ) h ε ( t ε − u ε , z ε ) dzdu = Z R d p ( ε ) ( ε t ε , ε ( x ε − z ε )) g ( ε ) ( εz ε ) dz + Z ε t ε Z R d p ( ε ) ( ε u ε , ε ( x ε − z ε )) h ( ε ) ( ε ( t ε − u ε ) , εz ε ) dzdu. This completes the proof of the lemma. (cid:3)
Our goal is to take ε to zero in the integral equation from Lemma 15.2. Forthat, we need to apply the dominated convergence theorem. The next two lemmasprepare the ground for that. Lemma 15.3.
Take any δ > . There are constants C and C , depending only on d and δ , such that for any t ∈ R >δ , x, y ∈ R d , and ε ∈ (0 , √ δ ) , p ( ε ) ( ε t ε , ε ( x ε − y ε )) ≤ C e − C | x − y | /t . Proof.
In this proof, C , C , . . . will denote constants that may depend only on d and δ . Note that t ε ≥ . So, by Lemmas 9.3, 9.4 and 10.1, p ( ε ) ( ε t ε , ε ( x ε − y ε )) = ε − d p ( t ε , x ε − y ε ) ≤ ε − d p ( t ε , x ε − y ε ) + C ε − d t − ( d +2) / ε (log t ε ) C = ε − d (4 πβt ε ) − d/ e −| x ε − y ε | / βt ε + C ε − d t − ( d +2) / ε (log t ε ) C . (15.2) Now note that since t > δ > ε , we have ε − t ≤ t ε ≤ ε − t. (15.3)Also, since | ε − x − x ε | ≤ √ d and | ε − y − y ε | ≤ √ d , we have ε − | x − y | ≤ | ε − x − x ε | + | x ε − y ε | + | ε − y − y ε | ) ≤ C + C | x ε − y ε | . (15.4)Using (15.3) and (15.4) in (15.2), we get p ( ε ) ( ε t ε , ε ( x ε − y ε )) ≤ C t − d/ e − C | x − y | /t + C ε t − ( d +2) / (log( ε − t )) C . Similarly, by (15.3), (15.4), and Lemma 10.3, p ( ε ) ( ε t ε , ε ( x ε − y ε )) ≤ C ε − d e − C | x − y | /t . Combining the last two displays, we get p ( ε ) ( ε t ε , ε ( x ε − y ε )) ≤ C t − d/ e − C | x − y | /t + C min { ε t − ( d +2) / (log( ε − t )) C , ε − d e − C | x − y | /t } . (15.5)Using the inequality min { a, b } ≤ a ( d +1) / ( d +2) b / ( d +2) , we get min { ε t − ( d +2) / (log( ε − t )) C , ε − d e − C | x − y | /t }≤ εt − ( d +1) / e − C | x − y | /t (log( ε − t )) C ≤ C e − C | x − y | /t , (15.6)where the last inequality holds because εt − ( d +1) / (log( ε − t )) C ≤ C (log( ε − t )) C ( ε − t ) / ≤ C . Plugging the bound from (15.6) into the right side of (15.5) completes the proof ofthe lemma. (cid:3)
Lemma 15.4.
For any t ∈ R ≥ , x ∈ R d , and ε ∈ (0 , , Z R d p ( ε ) ( ε t ε , ε ( x ε − y ε )) dy = 1 . Proof.
Note that by (15.1), Z R d p ( ε ) ( ε t ε , ε ( x ε − y ε )) dy = ε − d Z R d p ( t ε , x ε − y ε ) dy = Z R d p ( t ε , x ε − [ z ]) dz = X z ∈ Z d p ( t ε , x ε − z ) = 1 , where the last equality holds because p ( t ε , · ) is a probability mass function. (cid:3) NIVERSALITY OF DETERMINISTIC KPZ 43
Finally, we are ready to state and prove the main result of this section.
Proposition 15.5.
For any ( t, x ) ∈ R > × R d , f ( t, x ) = Z R d K ( t, x − y ) g ( y ) dy + γ Z t Z R d K ( s, x − y ) |∇ f | ( t − s, y ) dyds. Proof.
Fix t and x . Take any δ ∈ (0 , t/ . Define f ( ε ) δ ( t, x ):= Z R d p ( ε ) ( ε t ε , ε ( x ε − y ε )) g ( ε ) ( εy ε ) dy + Z tδ Z R d p ( ε ) ( ε s ε , ε ( x ε − y ε )) h ( ε ) ( ε ( t ε − s ε ) , εy ε ) dyds. Then by Lemma 15.2 and the facts that ε t ε ≤ t , and s ε ≤ t ε whenever s ≤ t , weget | f ( ε ) ( t, x ) − f ( ε ) δ ( t, x ) |≤ (cid:12)(cid:12)(cid:12)(cid:12)Z δ Z R d p ( ε ) ( ε s ε , ε ( x ε − y ε )) h ( ε ) ( ε ( t ε − s ε ) , εy ε ) dyds (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)Z tε t ε Z R d p ( ε ) ( ε s ε , ε ( x ε − y ε )) h ( ε ) ( ε ( t ε − s ε ) , εy ε ) dyds (cid:12)(cid:12)(cid:12)(cid:12) . By Lemma 7.1, | h ( ε ) | is uniformly bounded by C . Therefore using the identityfrom Lemma 15.4 and the above bound, we get | f ( ε ) ( t, x ) − f ( ε ) δ ( t, x ) | ≤ Cδ + Cε . (15.7)Now, by Lemma 15.1, lim ε → p ( ε ) ( ε s ε , ε ( x ε − y ε )) = K ( s, x − y ) for any s ∈ R > and x, y ∈ R d . By the continuity of g , we have g ( ε ) ( εy ε ) → g ( y ) as ε → . Moreover, by the Lipschitz property of g , we have that | g ( ε ) ( εy ε ) | ≤ C + C | y | where C and C do not depend on ε . Therefore by Lemma 15.3 andthe dominated convergence theorem, we get lim ε → Z R d p ( ε ) ( ε t ε , ε ( x ε − y ε )) g ( ε ) ( εy ε ) dy = Z R d K ( t, x − y ) g ( y ) dy. (15.8)Next, by Proposition 14.2, we have that for any t > s > and y ∈ R d , lim n →∞ h ( ε n ) ( ε n ( t ε n − s ε n ) , ε n y ε n ) = γ |∇ f | ( t − s, y ) . Moreover, by Lemma 7.1, h ( ε ) is uniformly bounded by a constant that does notdepend on ε . Therefore again by Lemma 15.3 and the dominated convergence theorem, lim n →∞ Z tδ Z R d p ( ε n ) ( ε n s ε n , ε n ( x ε n − y ε n )) h ( ε n ) ( ε n ( t ε n − s ε n ) , ε n y ε n ) dyds = γ Z tδ Z R d K ( s, x − y ) |∇ f | ( t − s, y ) dyds. (15.9)Combining (15.7), (15.8) and (15.9), we get (cid:12)(cid:12)(cid:12)(cid:12) f ( t, x ) − Z R d K ( t, x − y ) g ( y ) dy − γ Z tδ Z R d K ( s, x − y ) |∇ f | ( t − s, y ) dyds (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cδ. (15.10)Next, note that by Corollary 13.9, |∇ f | is uniformly bounded by a constant. Also,for any s and x , Z R d K ( s, x − y ) dy = 1 . This gives (cid:12)(cid:12)(cid:12)(cid:12)Z δ Z R d K ( s, x − y ) |∇ f | ( t − s, y ) dyds (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cδ.
Combining this with (15.10) gives (cid:12)(cid:12)(cid:12)(cid:12) f ( t, x ) − Z R d K ( t, x − y ) g ( y ) dy − γ Z t Z R d K ( s, x − y ) |∇ f | ( t − s, y ) dyds (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cδ.
Since δ is arbitrary, this completes the proof. (cid:3)
16. U
NIQUENESS OF SUBSEQUENTIAL LIMIT
We will now show that the integral equation displayed in Proposition 15.5 hasat most one solution among all possible subsequential limits of f ( ε ) . Proposition 16.1.
There can be at most one function f : R > × R d → R thatsatisfies all of the following conditions:(1) For each t > , f is differentiable in x .(2) The gradient of f with respect to x , denoted by ∇ f , is uniformly boundedover R > × R d .(3) For every ( t, x ) ∈ R > × R d , f ( t, x ) = Z R d K ( t, x − y ) g ( y ) dy + γ Z t Z R d K ( s, x − y ) |∇ f | ( t − s, y ) dyds. NIVERSALITY OF DETERMINISTIC KPZ 45
Proof.
Let f and f be two functions satisfying all of the given conditions. Let u := f − f and v := |∇ f | − |∇ f | , so that u ( t, x ) = γ Z t Z R d K ( s, x − y ) v ( t − s, y ) dyds. The first step is to show that when t > , the right side of the above equation can bedifferentiated with respect to x and the derivative can be moved inside the integralto give ∇ u ( t, x ) = γ Z t Z R d ∇ K ( s, x − y ) v ( t − s, y ) dyds. (16.1)To prove this, define for each δ ∈ (0 , t ) , u δ ( t, x ) := γ Z tδ Z R d K ( s, x − y ) v ( t − s, y ) dyds. Since v is a uniformly bounded function and s is uniformly bounded away fromzero in the above integral, it is not difficult to show by the dominated convergencetheorem that u δ is differentiable and ∇ u δ ( t, x ) = γ Z tδ Z R d ∇ K ( s, x − y ) v ( t − s, y ) dyds. Again, since v is uniformly bounded, this shows that for any δ > δ ′ > , |∇ u δ ( t, x ) − ∇ u δ ′ ( t, x ) | ≤ C Z δδ ′ Z R d |∇ K ( s, x − y ) | dyds ≤ C Z δδ ′ Z R d s − | x − y | K ( s, x − y ) dyds ≤ C √ δ. Thus, ∇ u δ ( t, x ) converges uniformly as δ → . By a slight modification of theabove display, it is easy to see that the limit equals the right side of (16.1). More-over, by the boundedness of v , u δ ( t, x ) → u ( t, x ) as δ → . From these observa-tions and the standard condition for convergence of derivatives, we get (16.1).Next, for each s , define a ( s ) := sup y ∈ R d |∇ u ( s, y ) | , b ( s ) := sup y ∈ R d | v ( s, y ) | . Note that these quantities are finite since ∇ f and ∇ f are uniformly bounded.Since ∇ K ( s, x − y ) = (2 βs ) − K ( s, x − y )( x − y ) , the identity (16.1) gives us a ( t ) ≤ C Z t Z R d s − | x − y | K ( s, x − y ) b ( t − s ) dyds ≤ C Z t s − / b ( t − s ) ds = C Z t ( t − s ) − / b ( s ) ds. Now, since ∇ f and ∇ f are uniformly bounded, | v ( s, y ) | = ||∇ f ( s, y ) | − |∇ f ( s, y ) | |≤ C ||∇ f ( s, y ) | − |∇ f ( s, y ) ||≤ C |∇ f ( s, y ) − ∇ f ( s, y ) | = C |∇ u ( s, y ) | . Thus, a ( t ) ≤ C Z t ( t − s ) − / a ( s ) ds. Fix some
T > . Then for any ≤ t ≤ T , H ¨older’s inequality gives a ( t ) ≤ C (cid:18)Z t ( t − s ) − / ds (cid:19) / (cid:18)Z t a ( s ) ds (cid:19) / ≤ CT / (cid:18)Z t a ( s ) ds (cid:19) / . Thus, for every t ≤ T , a ( t ) ≤ CT / Z t a ( s ) ds. Moreover, the function a is uniformly bounded. So by Gr¨onwall’s lemma, it nowfollows that a ( t ) = 0 for all t ∈ [0 , T ] . Since T is arbitrary, this completes theproof. (cid:3)
17. T HE C OLE –H OPF SOLUTION
We will now complete the proof of Theorem 1.1 under the condition that α = 0 .First, suppose that β and γ are both nonzero. Let b := γ/β . For t > and x ∈ R d ,define u ( t, x ) := Z R d K ( t, x − y ) e bg ( y ) dy = Z R d K ( t, y ) e bg ( x − y ) dy. Then u is strictly positive everywhere. Using the Lipschitzness of g and the domi-nated convergence theorem, it is not difficult to verify u is infinitely differentiablein R > × R d , continuous on the closure of this domain, and solves the heat equation ∂ t u = β ∆ u with initial condition u (0 , · ) = e bg ( · ) . Define v ( t, x ) := 1 b log u ( t, x ) . Then v is also infinitely differentiable in R > × R d , continuous in R ≥ × R d , andsolves the equation ∂ t v = ∂ t ubu = β ∆ ubu (17.1) NIVERSALITY OF DETERMINISTIC KPZ 47 with initial condition v (0 , · ) = g ( · ) . Moreover, if g is differentiable (in addition tobeing Lipschitz), then ∇ v ( t, x ) = R R d K ( t, y ) e bg ( x − y ) ∇ g ( x − y ) dy R R d K ( t, y ) e bg ( x − y ) dy , which shows that |∇ v | is uniformly bounded by L (the Lipschitz constant of g ). If g is only Lipschitz, and not differentiable, then it is not hard to show that |∇ v | isstill uniformly bounded by L , by approximating g with a sequence of differentiableLipschitz maps.Now, from the definition of v , an easy calculation gives ∆ v = ∆ ubu − |∇ u | bu = ∆ ubu − b |∇ v | . (17.2)Combining (17.1) and (17.1), we get ∂ t v = β ∆ v + βb |∇ v | = β ∆ v + γ |∇ v | . Next, for ( t, x ) ∈ R > × R d , define w ( t, x ) := Z R d K ( t, x − y ) g ( y ) dy,w ( t, x ) := γ Z t Z R d K ( t − s, x − y ) |∇ v | ( s, y ) dyds,w ( t, x ) := v ( t, x ) − w ( t, x ) − w ( t, x ) . Since K is a heat kernel and g is sufficiently well-behaved, it is easy to see that w satisfies the heat equation ∂ t w = β ∆ w . Similarly, it is not hard to show using standard arguments, and the boundednessand continuity of |∇ v | , that w solves ∂ t w ( t, x ) = γ |∇ v | ( t, x ) + β ∆ w ( t, x ) . Thus, ∂ t w = ∂ t v − ∂ t w − ∂ t w = ( β ∆ v + γ |∇ v | ) − β ∆ w − ( γ |∇ v | + β ∆ w )= β ∆ w. Now, w extends to a continuous function on R ≥ × R d by defining w (0 , x ) = g ( x ) . By the boundedness of |∇ v | , w extends to a continuous function on R ≥ × R d by defining w (0 , x ) = 0 . Lastly, as observed before, v also extends contin-uously, with v (0 , x ) = x . Thus, w extends continuously to the boundary with w (0 , x ) = 0 . From this, and the fact that w solves the heat equation displayedabove and is sufficiently well-behaved due to the Lipschitzness of g and the bound-edness of |∇ v | , it is now a standard exercise to show that w ( t, x ) = 0 for all t and x . This is equivalent to saying that v satisfies the integral equation of Propo-sition 16.1. We have already observed that v satisfies the other two conditions of Proposition 16.1. Thus, by Proposition 15.5, every subsequential limit of f ( ε ) must equal v . It is now easy to argue using Proposition 12.3 that f ( ε ) convergespointwise to v . This completes the proof of Theorem 1.1 when α , β and γ are allnonzero.Next, suppose that only α and β are nonzero, but γ = 0 . Then define v ( t, x ) := Z R d K ( t, x − y ) g ( y ) dy. Then immediately by Propositions 15.5 and 16.1, we see that any subsequentiallimit of f ( ε ) must be equal to v , and thus conclude that f ( ε ) converges pointwiseto v .Finally, suppose that β = 0 . We claim that γ , γ and γ must all be zero in thiscase. Fix some b ∈ B . For any δ ∈ R , let u δ ∈ R A be the vector defined as u δa = 0 for a ∈ A \ { b } and u δb = δ . Since φ (0) = 0 and β = 0 , Taylor expansion gives φ ( u δ ) = 12 γ δ + o ( δ ) as δ → . But this shows that φ ( u δ ) is not a monotone function of δ in a smallenough neighborhood of if γ = 0 . This violates the monotonicity of φ . Thus, γ must be zero. Next, define v δ ∈ R A as v δb = v δ − b = δ and v δa = 0 for a ∈ A \ { b, − b } . Since φ (0) = β = γ = 0 , Taylor expansion gives φ ( v δ ) = γ δ + o ( δ ) as δ → . Since v δ is monotone increasing in δ , the above expression violates themonotonicity of φ when δ is in a small enough neighborhood of zero if γ = 0 .Thus, γ = 0 . Finally, take any b, b ′ ∈ B such that b = b ′ and b = − b ′ . Define w δ ∈ R A as w δb = w δb ′ = δ and w δa = 0 for a ∈ A \ { b, b ′ } . Since φ (0) = β = γ = 0 , Taylor expansion gives φ ( w δ ) = γ δ + o ( δ ) as δ → . Proceeding as before, this yields γ = 0 .Since β = γ = γ = γ = 0 , we get that for any u ∈ R A with u = 0 , φ ( u ) = o ( | u | ) as | u | → . Since Lemma 5.4 is still valid, this shows that | f ε ( t, x ) − f ε ( t − , x ) | = | φ (( f ( t − , x + a ) − f ( t − , x )) a ∈ A ) |≤ ε F ( ε ) , where F ( ε ) is a function of ε that tends to zero as ε → . Thus, for any ( t, x ) ∈ R > × R d , | f ( ε ) ( t, x ) − g ( x ) | = | f ε ( t ε , x ε ) − f ε (0 , x ε ) |≤ t ε X s =1 | f ε ( s, x ε ) − f ε ( s − , x ε ) |≤ t ε ε F ( ε ) . Since t ε ε → t and F ( ε ) → as ε → , this shows that f ( ε ) ( t, x ) → g ( x ) as ε → . This wraps up the proof of Theorem 1.1 when α = 0 . NIVERSALITY OF DETERMINISTIC KPZ 49
18. T
HE CASE α = 0 We will now deal with the case α = 0 . The main issue is that the randomwalk described in Section 6 is now periodic, which necessitates several changes invarious steps of the proof. The results of Sections 5, 6 and 7 continue to remainvalid. A number of changes need to be made to the results of Section 9. First,Lemma 9.1 needs to be modified. Note that when α = 0 , we have β = 1 / d , andhence ψ ( θ ) = E ( e i θ · ξ ) = 1 d d X i =1 cos θ i . Let π be the element of R d whose coordinates are all equal to π . For θ ∈ R d , define q ( θ ) := min {| θ | , | θ − π |} . The following lemma is the new version of Lemma 9.1.
Lemma 18.1.
There is a positive constant C depending only d , such that for all θ ∈ [ − π/ , π/ d , | ψ ( θ ) | ≤ e − Cq ( θ ) .Proof. Let U denote the interval [ − π/ , π/ . Let δ be the smallest positive so-lution of cos x = 1 − / d . Let I := [ − δ, δ ] , I = [ π − δ, π + δ ] and I := I ∪ I .Note that | cos x | ≤ − / d for all x ∈ U \ I . Take any θ = ( θ , . . . , θ d ) ∈ U d \ I d .Then θ i / ∈ I for at least one i , and hence | ψ ( θ ) | ≤ d (cid:18) d − − d (cid:19) = 1 − d . Since q is uniformly bounded in U d , the above inequality shows that there is a smallenough positive constant C such that | ψ ( θ ) | ≤ e − C q ( θ ) for all θ ∈ U d \ I d .Next, take θ ∈ I d . Let J be the set of all j such that θ j ∈ I , and let J be theset of all j such that θ j ∈ I . Suppose that J and J are both nonempty. Then,since cos x ≤ for all x ∈ I and cos x ≤ − / d for all x ∈ I , we have ψ ( θ ) ≤ | J | d + (cid:18) − d (cid:19) | J | d = | J | − | J | d + | J | d ≤ d − d + d − d = 1 − d + 12 d , and similarly, since cos x ≥ − / d for all x ∈ I and cos x ≥ − for all x ∈ I , ψ ( θ ) ≥ (cid:18) − d (cid:19) | J | d − | J | d = | J | − | J | d − | J | d ≥ − dd − d − d = − d + 12 d . Thus, there is a positive constant C such that if the sets J and J are bothnonempty, then | ψ ( θ ) | ≤ e − C q ( θ ) .Next, suppose that J is empty. Then θ i ∈ I for each i . There is a positiveconstant C such that | cos x | ≤ − C x for all x ∈ I . Therefore, in this case, | ψ ( θ ) | ≤ d d X i =1 | cos θ i |≤ d d X i =1 (1 − C θ i )= 1 − C | θ | ≤ e − C | θ | ≤ e − C q ( θ ) . Finally, suppose that J is empty, so that θ i ∈ I for each i . Observe that | cos x | = | cos( x − π ) | ≤ − C ( x − π ) for all x ∈ I . Thus, as above, we have | ψ ( θ ) | ≤ e − C | θ − π | ≤ e − C q ( θ ) . So, if C := min { C , C , C } , then | ψ ( θ ) | ≤ e − Cq ( θ ) for all θ ∈ U d . (cid:3) Lemma 9.2 also needs to be modified. The following is the new version ofLemma 9.2.
Lemma 18.2.
For any θ ∈ R d and t ∈ Z ≥ , we have | ψ ( θ ) t − e − βt | θ | | ≤ Ct | θ | , | ψ ( π + θ ) t − ( − t e − βt | θ | | ≤ Ct | θ | . Proof.
The proof of the first inequality is just the same as the proof of Lemma 9.2.For the second, notice that ψ ( π + θ ) = 1 d d X i =1 cos( π + θ i ) = − d d X i =1 cos θ i = − ψ ( θ ) . So, by the first inequality applied with t = 1 , we have | ψ ( π + θ ) + e − β | θ | | = | − ψ ( θ ) + e − β | θ | | ≤ C | θ | . But recall that if a and b are complex numbers in the unit disk, and t is a positiveinteger, then | a t − b t | ≤ t | a − b | . Applying this in the above inequality, with a = ψ ( π + θ ) and b = − e − β | θ | , we get the desired result. (cid:3) Next, we introduce a different form of the Fourier inversion formula for transi-tion probabilities: p ( t, x ) = (2 π ) − d Z [ − π/ , π/ d e − i θ · x ψ ( θ ) t dθ = (2 π ) − d t − d/ Z [ − π √ t/ , π √ t/ d e − i t − / η · x ( ψ ( t − / η )) t dη, (18.1) NIVERSALITY OF DETERMINISTIC KPZ 51
The formula for p also needs to be changed. We define p ( t, x ) := (2 π ) − d t − d/ Z | η |≤ log t e − i t − / η · x ( ψ ( t − / η )) t dη + (2 π ) − d t − d/ Z | η −√ tπ |≤ log t e − i t − / η · x ( ψ ( t − / η )) t dη. The following is the new version of Lemma 9.3.
Lemma 18.3.
For any t ∈ Z ≥ and x ∈ Z d , | p ( t, x ) − p ( t, x ) | ≤ C e − C (log t ) .Proof. Without loss, let us assume that t is so large that the regions | η | ≤ log t and | η − √ tπ | ≤ log t do not intersect. Let us take η outside the union of these tworegions. Then note that q ( t − / η ) = min {| t − / η | , | t − / η − π |} ≥ t − / | log t | . Therefore, by Lemma 18.1, | ψ ( t − / η ) | ≤ e − Cq ( t − / η ) ≤ e − Ct − (log t ) . Now proceeding as in the proof of Lemma 9.3, we get the required bound. (cid:3)
For x = ( x , . . . , x d ) ∈ R d , define s ( x ) := P di =1 x i . We will say that x is evenif s ( x ) is even, and x is odd if s ( x ) is odd. We define p ( t, x ) and p ( t, x ) to betwice the values defined in Section 9 if t and x have the same parity, and equal tozero if not. With these new definitions, Lemma 9.4 clearly remains valid.Next, we modify the results of Section 10. The following is the new version ofLemma 10.1. Lemma 18.4.
For any t ≥ Z ≥ and x, y, z ∈ Z d such that x and t have the sameparity, and y and z are even, we have | p ( t, x ) − p ( t, x ) | ≤ C t − ( d +2) / (log t ) C , | δ y p ( t, x ) − δ y p ( t, x ) | ≤ C | y | t − ( d +3) / (log t ) C , | δ y δ z p ( t, x ) − δ y δ z p ( t, x ) | ≤ C | y || z | t − ( d +4) / (log t ) C . Proof.
Note that p ( t, x ) = (2 π ) − d t − d/ Z | η |≤ log t e − i t − / η · x ( ψ ( t − / η )) t dη + (2 π ) − d t − d/ Z | η |≤ log t e − i( π + t − / η ) · x ( ψ ( π + t − / η )) t dη = (2 π ) − d t − d/ Z | η |≤ log t e − i t − / η · x ( ψ ( t − / η )) t dη + (2 π ) − d t − d/ ( − s ( x ) Z | η |≤ log t e − i t − / η · x ( ψ ( π + t − / η )) t dη By Lemma 18.2, | ( ψ ( t − / η )) t − e − β | η | | ≤ Ct − | η | , | ( ψ ( π + t − / η )) t − ( − t e − β | η | | ≤ Ct − | η | . Lastly, note that p ( t, x ) = (1 + ( − s ( x )+ t )(2 π ) − d t − d/ Z | η |≤ log t e − i t − / η · x − β | η | dη Combining the last three displays, we get | p ( t, x ) − p ( t, x ) | ≤ Ct − d/ Z | η |≤ log t | ( ψ ( t − / η )) t − e − β | η | | dη + Ct − d/ Z | η |≤ log t | ( ψ ( t − / η )) t − ( − t e − β | η | | dη ≤ Ct − d/ − Z | η |≤ log t | η | dη ≤ Ct − d/ − (log t ) d +4 . This proves the first inequality in the statement of the lemma. Next, note that since y is even, we have e − i π · y = 1 . This gives δ y p ( t, x )= (2 π ) − d t − d/ Z | η |≤ log t ( e − i t − / η · y − e − i t − / η · x ( ψ ( t − / η )) t dη + (2 π ) − d t − d/ ( − s ( x ) Z | η |≤ log t ( e − i t − / η · y − e − i t − / η · x · ( ψ ( π + t − / η )) t dη. Moreover, the evenness of y ensures that t and x have the same parity if and onlyif t and x + y have the same parity. Thus, δ y p ( t, x )= (1 + ( − s ( x )+ t )(2 π ) − d t − d/ Z | η |≤ log t ( e − i t − / η · y − e − i t − / η · x − β | η | dη. It is now easy to complete the rest of the proof following the same steps as in theproof of Lemma 18.4 and implementing the same modifications as above. (cid:3)
As a result of Lemma 18.4, the remaining results of Sections 10 and 11 remainvalid after the following modifications: • In Lemma 10.2, we have to insert the additional condition that x and t havethe same parity, and y and z are even. • Lemma 10.3 can remain as it is, with the extra observation that p ( t, x ) = 0 when t and x do not have the same parity. NIVERSALITY OF DETERMINISTIC KPZ 53 • Proposition 10.4 remains valid if y is even. We do not need to add thecondition and that x and t have the same parity, since δ y p ( t, x ) = 0 if thatis not the case (and y is even). • Similarly, Proposition 10.5 remains valid if y and z are even. • In Lemma 10.6, we need r to be even, and in Lemma 10.7, we need both r and y to be even. • Similarly, in Proposition 10.8, we need r to be even, and in Proposi-tion 10.9, we need both r and y to be even. • Because of the above changes, we need y and z to be even in Proposi-tion 11.1. Similarly, we need r to be even in Proposition 11.2, and both r and y to be even in Proposition 11.3. • Proposition 11.4 remains valid without any changes.From Section 12 onwards, we have to be cautious about the definition of f ( ε ) . Wewill work with two different definitions, and show that they both converge to thesame limit. From that, we will deduce that the original f ( ε ) also converges to thatlimit.First, let us define for a real number a , [ a ] := 2[ a/ . This map takes the interval [2 k, k + 2) to the number k , for any k ∈ Z . Next, let [ a ] := [ a ] + 1 . This map takes [2 k, k + 2) to k + 1 . For x = ( x , . . . , x d ) ∈ R d , define [ x ] := ( ([ x ] , [ x ] , [ x ] , . . . , [ x d ]) if [ x ] + · · · + [ x d ] is even, ([ x ] , [ x ] , [ x ] , . . . , [ x d ]) otherwise.Similarly, define [ x ] := ( ([ x ] , [ x ] , [ x ] , . . . , [ x d ]) if [ x ] + · · · + [ x d ] is even, ([ x ] , [ x ] , [ x ] , . . . , [ x d ]) otherwise.Note that the definitions are meant to ensure that [ x ] is always even and [ x ] isalways odd. Note also that for any x , [ x ] is either equal to [ x ] or equal to [ x ] .This observation will be useful in the following construction. Define two newversions of f ( ε ) , called f ( ε, and f ( ε, , as follows. Let f ( ε, ( t, x ) := ( f ε ([ ε − t ] , [ ε − x ] ) if [ ε − t ] is even, f ε ([ ε − t ] , [ ε − x ] ) if [ ε − t ] is odd,and let f ( ε, ( t, x ) := ( f ε ([ ε − t ] , [ ε − x ] ) if [ ε − t ] is even, f ε ([ ε − t ] , [ ε − x ] ) if [ ε − t ] is odd.We will show that both f ( ε, and f ( ε, converge pointwise to the same f as ε → .Now note that by the observation from the previous paragraph, we have that for any t , x , and ε , f ( ε ) ( t, x ) has to be equal to either f ( ε, ( t, x ) or f ( ε, ( t, x ) . Thus, itwill follow that f ( ε ) ( t, x ) must also be converging to f ( t, x ) as ε → .Let us now prove the pointwise convergence of f ( ε, to f . The proof for f ( ε, is similar.First, note that Lemma 12.1 remains valid with f ( ε, instead of f ( ε ) . Sameis true for Lemma 12.2, but special care has to be taken because the originalLemma 12.2 uses Proposition 11.2, and in the modified version of Proposition 11.2,we need r to be even. The proof of Lemma 12.2 goes through as before when r ε is even. However, r ε is not guaranteed to be even. If for some ε , r ε turns out tobe odd, we then proceed as follows. Suppose that s ε is even and t ε is odd. Let x ε := [ ε − x ] and x ′ ε := [ ε − x ] . Then f ( ε ) ( s, x ) − f ( ε ) ( t, x ) = f ε ( s ε , x ε ) − f ε ( t ε , x ′ ε ) . By Lemma 7.1, (cid:12)(cid:12)(cid:12)(cid:12) f ε ( t ε , x ′ ε ) − d X b ∈ B f ε ( t ε − , x ′ ε + b ) (cid:12)(cid:12)(cid:12)(cid:12) = | h ε ( t ε , x ′ ε ) | ≤ Cε . Since t ε − − s ε is even, we can now use Proposition 11.2 and Lemma 5.4 to boundthe differences | f ε ( t ε − , x ′ ε + b ) − f ε ( s ε , x ε ) | . Together with the above inequality,this suffices to complete the proof of Lemma 12.2 for f ( ε, . Proposition 12.3 for f ( ε, follows from this.In Section 13, we need to change the definitions of D i f ε and D i f ( ε ) . We define D i f ε ( t, x ) := f ε ( t, x + 2 e i ) − f ε ( t, x )2 ε , and for t ∈ R ≥ and x ∈ R d , let D i f ( ε, ( t, x ) := ( D i f ε ([ ε − t ] , [ ε − x ] ) if [ ε − t ] is even, D i f ε ([ ε − t ] , [ ε − x ] ) otherwise.The results of Section 13 remain valid after the following modifications (adoptingthe new definition of D i f ε , and replacing D i f ( ε ) by D i f ( ε, everywhere): • Lemmas 13.1 and 13.3 remain as is. • Lemmas 13.2 and 13.4 need the additional condition that k is even. • Lemma 13.5 remains valid as is, because the number k ε in its proof is nowguaranteed to be even, so that the preceding two lemmas can be applied.(In the proof, we have to change x ε to denote [ ε − x ] if t ε is even, and [ ε − x ] if t ε is odd. Similarly, k ε has to be defined as the integer suchthat [ ε − ( x + ae i )] = x ε + k ε e i if t ε is even and the integer such that [ ε − ( x + ae i )] = x ε + k ε e i if t ε is odd.) • Consequently, Lemmas 13.6, 13.7, Proposition 13.8 and Corollary 13.9remain valid. In the proof of Proposition 13.8, we again run into the slightdifficulty — as in the proof of the new version of Lemma 12.2 — that r ε need not be even, so Proposition 11.3 may not be directly applicable. Thedifficulty is overcome easily by the same trick as we used for Lemma 12.2. NIVERSALITY OF DETERMINISTIC KPZ 55
We need some further caution in Section 14. First, for a ∈ A and ( t, x ) ∈ Z ≥ × Z d ,define u a,ε ( t, x ) := f ε ( t, x + a ) − d X b ∈ B f ε ( t, x + b ) . Let u ε ( t, x ) denote the vector ( u a,ε ( t, x )) a ∈ A . Then redefine H ε as H ε ( t, x ) := γ X b ∈ B ( u b,ε ( t − , x )) + γ X b ∈ B u b,ε ( t − , x ) u − b,ε ( t − , x )+ γ X b,b ′ ∈ Bb = b ′ ,b = − b ′ u b,ε ( t − , x ) u b ′ ,ε ( t − , x ) . We will show that Lemma 14.1 remains valid with this new H ε (but the same old h ε ). For that, we need some preparation. Lemma 18.5. If α = 0 , then we must have that ∂ ∂ a φ (0) = 0 for all a ∈ A .Proof. Take any a ∈ A . First, suppose that a = 0 . Take any δ ∈ R . Let u δ be thevector in R A with u δ = δ and u δb = 0 for all b = 0 . Then by Taylor expansion andthe facts that φ (0) = 0 and ∂ φ (0) = α = 0 , we have φ ( u δ ) := δ ∂ φ (0) + o ( δ ) as δ → . If ∂ φ (0) = 0 , this is not a monotone function of δ in a small enoughneighborhood of zero. This contradicts the monotonicity of φ . Thus, ∂ φ (0) = 0 .Next, suppose that a = 0 . Given δ, δ ′ ∈ R , let u δ,δ ′ ∈ R A be the vector with u δ,δ ′ = δ , u δ,δ ′ a = δ ′ , and u δ,δ ′ b = 0 if b = 0 and b = a . Then, since we alreadyshowed that ∂ φ (0) = 0 , and we are given that φ (0) = 0 and ∂ φ (0) = 0 , Taylorexpansion gives φ ( u δ,δ ′ ) = δ ′ ∂ a φ (0) + δ ′ ∂ a φ (0) + δδ ′ ∂ ∂ a φ (0) + o (max { δ , δ ′ } ) as δ, δ ′ → . Thus, φ ( u δ,δ ′ ) − φ ( u δ,δ ′ ) = δδ ′ ∂ ∂ a φ (0) + o (max { δ , δ ′ } ) as δ, δ ′ → . Now choose δ ′ = − δ sign( ∂ ∂ a φ (0)) . With this choice of δ ′ , we get φ ( u δ,δ ′ ) − φ ( u δ,δ ′ ) = − δ | ∂ ∂ a φ (0) | + o ( δ ) as δ → . Take δ decreasing to zero. Then the left side is always nonnegative,by the monotone increasing nature of φ ; and the right side is negative for smallenough δ , unless ∂ ∂ a φ (0) = 0 . Thus, ∂ ∂ a φ (0) must be zero. (cid:3) The following result is our modified version of Lemma 14.1, that works for themodified H ε defined above. Lemma 18.6.
For any ε > , t ∈ Z ≥ and x ∈ Z d , | h ε ( t, x ) − H ε ( t, x ) | ≤ ε F ( ε ) , where F is a function determined by φ (and not depending on t or x ), such that F ( ε ) → as ε → .Proof. Note that h ε ( t, x ) = f ε ( t, x ) − d X b ∈ B f ε ( t − , x + b )= φ ( u ε ( t − , x )) . By Lemma 5.4, | u a,ε ( t, x ) | ≤ Cε for all a , t and x . We know that φ (0) = 0 , ∂ φ (0) = 0 , ∂ b φ (0) = 1 / d for all b ∈ B , and by Lemma 18.5, ∂ ∂ a φ (0) = 0 forall a ∈ A . Thus, by Taylor expansion, (cid:12)(cid:12)(cid:12)(cid:12) h ε ( t, x ) − d X b ∈ B u b,ε ( t − , x ) − X b,b ′ ∈ B ∂ b ∂ b ′ φ (0) u b,ε ( t − , x ) u b ′ ,ε ( t − , x ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cε F ( ε ) , where F is determined solely by φ , and F ( ε ) → as ε → . Now, d X b ∈ B u b,ε ( t − , x ) = 12 d X b ∈ B (cid:18) f ε ( t − , x + b ) − d X b ′ ∈ B f ε ( t − , x + b ′ ) (cid:19) = 0 . On the other hand, X b,b ′ ∈ B ∂ b ∂ b ′ φ (0) u b,ε ( t − , x ) u b ′ ,ε ( t − , x ) = H ε ( t, x ) . Combining the last three displays completes the proof. (cid:3)
The next goal is to prove a modification of Proposition 14.2. Again, some prepa-ration is needed. We need the following enhancement of Lemma 13.4.
Lemma 18.7.
For any t ∈ Z ≥ , x, y ∈ Z d with y even, and any k ∈ Z \ { } , | f ε ( t, x + ky ) − f ε ( t, x ) − kδ y f ε ( t, x ) | ≤ C ( y ) ε | k | J ( ε, k, t, x ) + C ( y ) ε k , where J is as in Lemma , and C ( y ) is a constant that depends only on φ , L , d and y .Proof. Suppose that k > . Then note that f ε ( t, x + ky ) − f ε ( t, x ) − kδ y f ε ( x )= k X j =1 ( δ y f ε ( x + jy ) − δ y f ε ( x ))= k X j =1 δ y δ jy f ε ( x ) . NIVERSALITY OF DETERMINISTIC KPZ 57
Observe that y and jy are even. It is now easy to complete the proof using a suitablemodification of Lemma 13.2, and Lemma 13.3. The proof for k < is similar. (cid:3) Lemma 18.8.
Fix some even y ∈ Z d . Take any sequence ( t n , x n ) in Z > × Z d such that t n has the same parity as x n for each n , and ( ε n t n , ε n x n ) → ( t, x ) ∈ R > × R d . Then lim n →∞ ε − n δ y f ε n ( t n , x n ) = y · ∇ f ( t, x ) . Proof.
Take any a > . Let k n := [ a/ε n ] . By Lemma 18.7, (cid:12)(cid:12)(cid:12)(cid:12) f ε n ( t n , x n + k n y ) − f ε n ( t n , x n ) k n ε n − ε − n δ y f ε n ( t n , x n ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C ( y ) J ( ε n , k n , t n , x n ) + C ( y ) ε n k n . Now notice that by the uniform convergence assertion of the version of Proposi-tion 12.3 for f ( ε, , we have lim n →∞ f ε n ( t n , x n ) = lim n →∞ f ( ε n , ( ε n t n , ε n x n ) = f ( t, x ) , and similarly, lim n →∞ f ε n ( t n , x n + k n y ) = f ( t, x + ay ) . Also, k n ε n → a . Finally, note that lim n →∞ J ( ε n , k n , t n , x n )= C a log(2 + t/a ) + C (1 + | x | + | a | + t / ) t − / min { , t − / | a |} . Note that this is the quantity Q ( a, t, x ) defined in Lemma 13.5. Thus, we get lim sup n →∞ (cid:12)(cid:12)(cid:12)(cid:12) f ( t, x + ay ) − f ( t, x ) a − ε − n δ y f ε n ( t n , x n ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C ( y ) Q ( a, t, x ) . Since Q ( a, t, x ) → and a − ( f ( t, x + ay ) − f ( t, x )) → y · ∇ f ( t, x ) as a → ,this completes the proof of the lemma. (cid:3) Lemma 18.9.
Take any b ∈ B , and any sequence ( t n , x n ) in Z > × Z d such that t n and x n have opposite parities for each n , and ( ε n t n , ε n x n ) → ( t, x ) ∈ R > × R d .Then lim n →∞ ε − n u b,ε n ( t n , x n ) = b · ∇ f ( t, x ) Proof.
Note that for any ε , t , x and b , u b,ε ( t, x ) = f ε ( t, x + b ) − d X b ′ ∈ B f ε ( t, x + b ′ )= 12 d X b ′ ∈ B ( f ε ( t, x + b ) − f ε ( t, x + b ′ ))= 12 d X b ′ ∈ B δ b − b ′ f ε ( t, x + b ′ ) . Thus, by Lemma 18.8, we have lim n →∞ ε − n u b,ε n ( t, x ) = 12 d X b ′ ∈ B lim n →∞ ε − n δ b − b ′ f ε n ( t n , x n + b ′ )= 12 d X b ′ ∈ B ( b − b ′ ) · ∇ f ( t, x )= b · ∇ f ( t, x ) . This completes the proof of the lemma. (cid:3)
Lemma 18.10.
Take any sequence ( t n , x n ) in Z > × Z d such that t n has the sameparity as x n for each n , and ( ε n t n , ε n x n ) → ( t, x ) ∈ R > × R d . Then lim n →∞ ε − n H ε n ( t n , x n ) = γ |∇ f ( t, x ) | . Proof.
Note that ε − n H ε n ( t n , x n ):= γ X b ∈ B ( ε − n u b,ε n ( t n − , x n )) + γ X b ∈ B ( ε − n u b,ε n ( t n − , x n ))( ε − n u − b,ε n ( t n − , x n ))+ γ X b,b ′ ∈ Bb = b ′ ,b = − b ′ ( ε − n u b,ε n ( t n − , x n ))( ε − n u b ′ ,ε n ( t n − , x n )) . Applying Lemma 18.9 to each term, we get lim n →∞ ε − n H ε n ( t n , x n ) = γ X b ∈ B ( b · ∇ f ( t, x )) + γ X b,b ′ ∈ Bb = b ′ ,b = − b ′ ( b · ∇ f ( t, x ))( b ′ · ∇ f ( t, x )) . Now note that for any b ∈ B , the sum of all b ′ ∈ B \ { b, − b } is zero. Also, notethat X b ∈ B ( b · ∇ f ( t, x )) = 2 |∇ f ( t, x ) | . This completes the proof of the lemma. (cid:3)
Next, we turn to modifying the results of Section 15. For each even x ∈ Z d ,there is a region of volume in R d that maps to x under the map y [ y ] . Theseregions form a partition of R d . From this, it follows that for any w : Z d → R , X x ∈ Z d ,x even w ( x ) = 12 Z R d w ([ x ] ) dx. NIVERSALITY OF DETERMINISTIC KPZ 59
Similarly, X x ∈ Z d ,x odd w ( x ) = 12 Z R d w ([ x ] ) dx, whenever the sums are absolutely convergent. For ( t, x ) ∈ R ≥ × R d , define p ( ε, ( t, x ) := ( ε − d p ([ ε − t ] , [ ε − x ] ) if [ ε − t ] is even, ε − d p ([ ε − t ] , [ ε − x ] ) if [ ε − t ] is odd.Also, let h ( ε, ( t, x ) := ( ε − h ε ([ ε − t ] , [ ε − x ] ) if [ ε − t ] is even, ε − h ε ([ ε − t ] , [ ε − x ] ) if [ ε − t ] is odd. Proposition 18.11.
Take any sequence ( t n , x n ) → ( t, x ) ∈ R > × R d . Then lim n →∞ h ( ε n ) ( t n , x n ) = γ |∇ f ( t, x ) | . Proof.
This follows immediately from Lemma 18.6 and Lemma 18.10. (cid:3)
Take any ( t, x ) ∈ R > × R d . Define t ε := [ ε − t ] . If t ε is even, let x ε := [ ε − x ] .Otherwise, let x ε := [ ε − x ] . Then note that t ε and x ε have the same parity, and f ( ε, ( t, x ) = f ε ( t ε , x ε ) . Thus, by Proposition 6.1, f ( ε ) ( t, x ) = f ε ( t ε , x ε )= X y ∈ Z d p ( t ε , x ε − y ) g ε ( y ) + X ≤ s ≤ t ε − X y ∈ Z d p ( s, x ε − y ) h ε ( t ε − s, y ) . Suppose that t ε is even. Then x ε is also even, and hence p ( s, x ε − y ) = 0 if s and y do not have the same parity. Thus, if s is even, then X y ∈ Z d p ( s, x ε − y ) h ε ( t ε − s, y )= X y ∈ Z d ,y even p ( s, x ε − y ) h ε ( t ε − s, y )= 12 Z R d p ( s, x ε − [ y ] ) h ε ( t ε − s, [ y ] ) dy = ε d +2 Z R d p ( ε, ( ε s, ε ( x ε − [ y ] )) h ( ε, ( ε ( t ε − s ) , ε [ y ] ) dy. One gets a similar expression for odd s , with [ y ] instead of [ y ] . Thus, if we define χ ( s ) := 0 if [ s ] is even and χ ( s ) := 1 if [ s ] is odd, then X ≤ s ≤ t ε − X y ∈ Z d p ( s, x ε − y )= ε d +2 Z t ε Z R d p ( ε, ( ε [ s ] , ε ( x ε − [ y ] χ ( s ) )) h ( ε, ( ε ( t ε − [ s ]) , ε [ y ] χ ( s ) ) dyds = Z ε t ε Z R d p ( ε, ( ε [ ε − u ] , ε ( x ε − [ ε − z ] χ ( ε − u ) )) · h ( ε, ( ε ( t ε − [ ε − u ]) , ε [ ε − z ] χ ( ε − u ) ) dzdu. Now proceeding as in the proof of Proposition 15.5, using Lemma 18.6 and Propo-sition 18.11 instead of Lemma 14.1 and Proposition 14.2, and suitable modifica-tions of Lemma 15.3 and Lemma 15.4, we complete the proof of Proposition 15.5for any subsequential limit of f ( ε, . The rest of the proof of Theorem 1.1 nowproceeds as before. A CKNOWLEDGEMENTS
I thank Ivan Corwin, Persi Diaconis, Panagiotis Souganidis, and Lexing Yingfor helpful comments and references.R
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