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SOCIEDADE BRASILEIRA DE MATEMÁTICA ENSAIOS MATEMÁTICOS2015, Volume 29, 1–89

Lectures on singular stochasticPDEs

Massimiliano GubinelliNicolas Perkowski

Abstract. These are the notes for a course at the 18th Brazilian Schoolof Probability held from August 3rd to 9th, 2014 in Mambucaba. Theaim of the course is to introduce the basic problems of non–linear PDEswith stochastic and irregular terms. We explain how it is possible tohandle them using two main techniques: the notion of energy solutions in[Gonçalves and Jara, Arch. Ration. Mech. Anal., 2014] and [Gubinelli andJara, Stoch. Partial Di�. Equations: Analysis and Computations, 2013],and that of paracontrolled distributions, recently introduced in [Gubinelli,Imkeller, and Perkowski, Forum Math. Pi, 2015]. In order to maintaina link with physical intuitions, we motivate such singular SPDEs via ahomogenization result for a di�usion in a random potential.

2010 Mathematics Subject Classification: 60H15, 60G15, 35S50.

Contents

1 Introduction 5

2 Energy solutions 9

2.1 Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2 The Stochastic Burgers equation . . . . . . . . . . . . . . . 112.3 The Ornstein–Uhlenbeck process . . . . . . . . . . . . . . . 122.4 Gaussian computations . . . . . . . . . . . . . . . . . . . . . 192.5 The Itô trick . . . . . . . . . . . . . . . . . . . . . . . . . . 272.6 Controlled distributions . . . . . . . . . . . . . . . . . . . . 332.7 Existence of solutions . . . . . . . . . . . . . . . . . . . . . 34

3 Besov spaces 37

4 Di�usion in a random environment 44

4.1 The 2d generalized parabolic Anderson model . . . . . . . . 504.2 More singular problems . . . . . . . . . . . . . . . . . . . . 524.3 Hairer’s regularity structures . . . . . . . . . . . . . . . . . 54

5 The paracontrolled PAM 55

5.1 The paraproduct and the resonant term . . . . . . . . . . . 565.2 Commutator estimates and paralinearization . . . . . . . . 595.3 Paracontrolled distributions . . . . . . . . . . . . . . . . . . 665.4 Fixpoint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 705.5 Renormalization . . . . . . . . . . . . . . . . . . . . . . . . 715.6 Construction of the extended data . . . . . . . . . . . . . . 76

6 The stochastic Burgers equation 81

6.1 Structure of the solution . . . . . . . . . . . . . . . . . . . . 826.2 Paracontrolled solution . . . . . . . . . . . . . . . . . . . . . 83

Bibliography 87

3

Chapter 1

Introduction

The aim of these lectures is to explain how to apply controlled pathideas [12] to solve basic problems in singular stochastic parabolic equations.The hope is that the insight gained by doing so can inspire new applicationsor the construction of other more powerful tools to analyze a wider classof problems.

To understand the origin of such singular equations, we have chosen topresent the example of a homogenization problem for a singular potentialin a linear parabolic equation. This point of view has the added benefitthat it allows us to track back the renormalization needed to handle thesingularities as e�ects living on other scales than those of interest. Thebasic problem is that of having to handle e�ects of the microscopic scalesand their interaction through non–linearities on the macroscopic behaviourof the solution.

Mathematically, this problem translates into the attempt of makingSchwartz’s theory of distributions coexist with non–linear operations whichare notoriously not continuous in the usual topologies on distributions.This is a very old problem of analysis and has been widely studied. Theadditional input which is not present in the usual approaches is that thesingularities which force us to treat the problem in the setting of Schwartz’sdistributions are of a stochastic nature. So we dispose of two handles onthe problem: the analytical one and the probabilistic one. The right mixof the two will provide an e�ective solution to a wide class of problems.

A first and deep understanding of these problems has been obtainedstarting from the late ’90s by T. Lyons [25], who introduced a theory ofrough paths in order to settle the conflict of topology and non–linearityin the context of driven di�erential equations, or more generally in thecontext of the non–linear analysis of time–varying signals. Nowadays thereare many expositions of this theory [27, 9, 26, 8] and we refer the readerto the literature for more details.

5

6 M. Gubinelli and N. Perkowski

In [12, 13], the notion of controlled paths has been introduced in orderto extend the applicability of the rough path ideas to a larger class ofproblems that are not necessarily related to the integration of ODEs butwhich still retain the one–dimensional nature of the directions in which theirregularity manifests itself. The controlled path approach has been usedto make sense of the evolution of irregular objects such as vortex filamentsand certain SPDEs. Later Hairer understood how to apply these ideas tothe long standing problem of the Kardar–Parisi–Zhang equation [18], andhis insights prompted the researchers to try more ambitious approaches toextend rough paths to a multidimensional setting.

In [14], in collaboration with P. Imkeller, we introduced a notion ofparacontrolled distributions which is suitable to handle a wide class ofSPDEs which were well out of reach with previously known methods.Paracontrolled distributions can be understood as an extension ofcontrolled paths to a multidimensional setting, and they are based onnew combinations of basic tools from harmonic analysis.

At the same time, Hairer managed to devise a vast generalization ofthe basic construction of controlled rough paths in the multidimensionaland distributional setting, which he called the theory of regularitystructures [19] and which subsumes standard analysis based on Hölderspaces and controlled rough path theory but goes well beyond that. Justa few days after the lectures in Mambucaba took place, it was announcedthat Martin Hairer was awarded a Fields Medal for his work on SPDEsand in particular for his theory of regularity structures [19] as a tool fordealing with singular SPDEs. This prize witnesses the exciting period weare experiencing: we now understand sound lines of attack to long standingproblems, and there are countless opportunities to apply similar ideas tonew problems.

The plan of the lectures is the following. We start by discussing energysolutions [10, 11, 15] of the stationary stochastic Burgers equation (oneof the avatars of the Kardar–Parisi–Zhang equation).1 Energy solutionshave the advantage of being relatively easy to handle and of being based ontools that are familiar to probabilists. On the other side, they only applyin the specific example of the stochastic Burgers equation in equilibrium,and here we will only focus on the existence but not on the uniquenessof energy solutions. Starting our lectures in this way will allow us tointroduce the reader to SPDEs in a progressive manner, and also tointroduce Gaussian tools on the way (Wick products, hypercontractivity)and to present some of the basic phenomena that appear when dealing withsingular SPDEs. Next we set up the analytical tools we need in the restof the lectures: Besov spaces and some basic harmonic analysis based on

1The paper [11] is the revised published version of [10]. We would like to cite themtogether to acknowledge that the notion of energy solutions historically predates thatof Hairer in [18].

Chapter 1. Introduction 7

the Littlewood–Paley decomposition of distributions. In order to motivatethe reader and to provide a physical ground for the intuition to standon, we then discuss a homogenization problem for the linear heat equationwith random potential which describes di�usion in a random environment.This will allow us to derive the need for the weak topologies we shalluse and for irregular objects like the white noise from first principles and“concrete” applications. The homogenization problem also allows us tosee that there are naturally appearing renormalization e�ects and to keeptrack of their mathematical meaning. Starting from these problems weintroduce the two–dimensional parabolic Anderson model, the simplestSPDE in which most of the features of more di�cult problems are alreadypresent, and we explain how to use paraproducts and the paracontrolledansatz in order to keep the non–linear e�ect of the singular data undercontrol. Then we return to the stochastic Burgers equation and showhow to apply paracontrolled distributions in order to obtain the existenceand uniqueness of solutions also in the non–stationary case. e�ect of thesingular data under control. Then we return to the stochastic Burgersequation and show how to apply paracontrolled distributions in order toobtain the existence and uniqueness of solutions also in the non–stationarycase. e�ect of the singular data under control. Then we return tothe stochastic Burgers equation and show how to apply paracontrolleddistributions in order to obtain the existence and uniqueness of solutionsalso in the non–stationary case.

Acknowledgements. The authors would like to thank the twoanonymous referees for the careful reading and the manifold suggestionswhich helped up to greatly improve the manuscript. We would also liketo thank the organizers of the Brazilian Summer Schools in Probabilityfor the invitation and the researchers who attended the meeting for thewonderful atmosphere.

The main part of the research was carried out while N. P. wasemployed by Université Paris Dauphine. N. P. was supported by theFondation Sciences Mathématiques de Paris (FSMP) and by a public grantoverseen by the French National Research Agency (ANR) as part of the“Investissements d’Avenir” program (reference: ANR-10-LABX-0098).

Conventions and notations. We write a . b if there exists a constantC > 0, independent of the variables under consideration, such that a 6 Cb.Similarly we define &. We write a ƒ b if a . b and b . a. If wewant to emphasize the dependence of C on the variable x, then we writea(x) .x b(x).

If a is a complex number, we write aú for its complex conjugate.If i and j are index variables of Littlewood–Paley blocks (to be defined

below), then i . j is to be interpreted as 2i . 2j , and similarly for ƒ and


.. In other words, i . j means i 6 j + N for some fixed N œ N that doesnot depend on i or j.

We use standard multi-index notation: for µ œ Nd0

we write |µ| =µ

1

+ . . . + µd and ˆµ = ˆ|µ|/ˆµ1x1 . . . ˆµ

d

xd

, as well as xµ = xµ11

· . . . · xµd

d

for x œ Rd.For – > 0 we write C–

b for the bounded functions F : R æ R which areÂ–Ê times continuously di�erentiable with bounded and (– ≠ Â–Ê)–Höldercontinuous derivatives of order Â–Ê, equipped with the norm

ÎFÎC–

b

= supµ:0Æ|µ|ÆÂ–Ê

ÎˆµFÎLŒ + supµ:|µ|=Â–Ê

supx”=y

|ˆµF (x) ≠ ˆµF (y)||x ≠ y|–≠Â–Ê .

If we write u œ C –≠, then that means that u is in C –≠Á for all Á > 0.The C – spaces will be defined below.

If X is a Banach space with norm Î · ÎX and if T > 0, then we define CXand CTX as the spaces of continuous functions from [0, Œ) respectively[0, T ] to X, and CTX is equipped with the supremum norm Î · ÎC

T

X. If– œ (0, 1) then we write C–X for the functions in CX that are ––Höldercontinuous on every interval [0, T ], and we write

ÎfÎC–

T

X = sup06s<t6T

Îf(t) ≠ f(s)Î|t ≠ s|– .

Chapter 2

Energy solutions

The first issue one encounters when dealing with singular SPDEs is theill–posed character of the equation, even in a weak sense. Typically, theequation features some non–linearity that does not make sense in thenatural spaces where solutions live and one has to provide a suitablesmaller space in which it is possible to give an appropriate interpretationto “ambiguous quantities” that appear in the equation.

Energy solutions [11, 15] are a relatively simple tool in order to come upwith well–defined non–linearities. Moreover, proving existence of energysolutions or even convergence to energy solutions is usually a quite simpleproblem, at least compared to the other approaches like paracontrolledsolutions or regularity structures, where already existence requires quitea large amount of computations but where uniqueness can be establishedquite easily afterwards. The main drawback is that we lack of generaluniqueness results for energy solutions. Only very recently, after thecompletion of these notes, we were able to prove that energy solutionsfor the stationary stochastic Burgers equation are unique. This topic willnot be touched upon here. The interested reader can find the details inthe preprint [17].

2.1 Distributions

We will need to use distributions defined on the d-dimensional torus Td

where T = R/(2fiZ). We collect here some basic results and definitions.The space of distributions S Õ = S Õ(Td) is the set of linear maps f fromS = CŒ(Td,C) to C, such that there exist k œ N and C > 0 with

|Èf, ÏÍ| := |f(Ï)| 6 C sup|µ|6k

ÎˆµÏÎLŒ(Td

)

9


for all Ï œ S .Example 1. Clearly Lp = Lp(Td) µ S Õ for all p > 1, and more generallythe space of finite signed measures on (Td, B(Td)) is contained in S Õ.Another example of a distribution is Ï ‘æ ˆµÏ(x) for µ œ Nd

0

and x œ T.In particular, the Fourier transform Ff : Zd æ C,

Ff(k) = f(k) = Èf, ekÍ,

with ek = e≠iÈk,·Í/(2fi)d/2, is defined for all f œ S Õ, and it satisfies|Ff(k)| 6 |P (k)| for a suitable polynomial P . Conversely, if (g(k))kœZd isat most of polynomial growth, then its inverse Fourier transform

F ≠1g =ÿ

kœZd

g(k)eúk

defines a distribution (here eúk = eiÈk,·Í/(2fi)d/2 is the complex conjugate

of ek).Exercise 1. Show that the Fourier transform of Ï œ S decays faster thanany rational function (we say that it is of rapid decay). Combine this withthe fact that F defines a bijection from L2(Td) to ¸2(Zd) with inverse F ≠1

to show that F ≠1Ff = f for all f œ S Õ and FF ≠1g = g for all g ofpolynomial growth. Extend the Parseval formula

Èf, ÏúÍL2(Td

)

=⁄

Td

f(x)Ï(x)údx =ÿ

k

f(k)Ï(k)ú

from f, Ï œ L2(Td) to f œ S Õ and Ï œ S .Exercise 2. Fix a complete probability space (�, F ,P). On that space let› be a spatial white noise on Td, i.e. › is a centered Gaussian processindexed by L2(Td), with covariance

E[›(f)›(g)] =⁄

Td

f(x)g(x)dx.

Show that there exists › with P(›(f) = ›(f)) = 1 for all f œ L2, such that›(Ê) œ S Õ for all Ê œ �.

Hint: Show that E[q

kœZd

exp(⁄|›(ek)|2)/(1 + |k|d+1)] < Œ for somesuitable ⁄ > 0.

Linear maps on S Õ can be defined by duality: if A : S æ Sis such that for all k œ N there exists n œ N and C > 0 withsup|µ|6k Îˆµ(AÏ)ÎLŒ 6 C sup|µ|6n ÎˆµÏÎLŒ , then we set ÈtAf, ÏÍ =Èf, AÏÍ. Di�erential operators are defined by Èˆµf, ÏÍ = (≠1)|µ|Èf, ˆµÏÍ.If Ï : Zd æ C grows at most polynomially, then it defines a Fouriermultiplier

Ï(D) : S Õ æ S Õ, Ï(D)f = F ≠1(ÏFf).

Chapter 2. Energy solutions 11

Exercise 3. Use the Fourier inversion formula of Exercise 1 to show thatfor f œ S Õ, Ï œ S and for u, v : Zd æ C with u of polynomial growth andv of rapid decay

F (fÏ)(k) = (2fi)≠d/2

ÿ

¸

f(k ≠ ¸)Ï(¸)

andF ≠1(uv)(x) = (2fi)d/2ÈF ≠1u, (F ≠1v)(x ≠ ·)Í.

2.2 The Stochastic Burgers equationOur aim here is to motivate the ideas at the base of the notion of energysolutions. We will not insist on a detailed formulation of all the availableresults. The reader can always refer to the original paper [15] for missingdetails. Applications to the large scale behavior of particle systems arestudied in [11].

We will study the case of the stochastic Burgers equation on the torusT. The solution of the stochastic Burgers equation is the derivative of thesolution of the Kardar–Parisi–Zhang equation, a universal model for thefluctuations in random interface growth which has been at the center ofseveral spectacular results of the past years. Excellent surveys on the KPZequation and related areas are [6, 28, 29].

The unknown u : R+

◊ T æ R should satisfy

ˆtu = �u + ˆxu2 + ˆx›,

where › : R+

◊ T æ R is a space–time white noise defined on a givenprobability space (�, F ,P) fixed once and for all. That is, › is a centeredGaussian process indexed by L2(R

+

◊ T) with covariance

E[›(f)›(g)] =⁄

R+◊Tf(t, x)g(t, x)dtdx.

The equation has to be understood as a relation for processes whichare distributions in space with su�ciently regular time dependence. Inparticular, if we test the above relation with Ï œ S := S (T) := CŒ(T),denote with ut(Ï) the pairing of the distribution u(t, ·) with Ï, andintegrate in time over the interval [0, t], we formally get

ut(Ï) = u0

(Ï) +⁄ t

0

us(�Ï)ds ≠⁄ t

0

Èu2

s, ˆxÏÍds ≠⁄ t

0

›s(ˆxÏ)ds.


Let us discuss the various terms in this equation. In order to make senseof ut(Ï) and

s t

0

us(�Ï)ds, it is enough to assume that for all Ï œ Sthe mapping (t, Ê) ‘æ ut(Ï)(Ê) is a stochastic process with continuoustrajectories. Next, if we denote Mt(Ï) = ≠

s t

0

›s(ˆxÏ)ds then, at least bya formal computation, we have that (Mt(Ï))t>0,ÏœS is a Gaussian randomfield with covariance

E[Mt(Ï)Ms(Â)] = (t · s)ÈˆxÏ, ˆxÂÍL2(T)

.

In particular, for every Ï œ S the stochastic process (Mt(Ï))t>0

is aBrownian motion with covariance

ÎÏÎ2

H1(T)

:= ÈˆxÏ, ˆxÏÍL2(T)

.

We will use this fact to have a rigorous interpretation of the white noise› appearing in the equation. Here we used the notation M in order tostress the fact that Mt(Ï) is a martingale in its natural filtration andmore generally in the filtration Ft = ‡(Ms(Ï) : s 6 t, Ï œ H1(T)), t > 0.

The most di�cult term is of course the nonlinear one:s t

0

Èu2

s, ˆxÏÍds.In order to define it, we need to square the distribution ut, an operationwhich in general can be quite dangerous. A natural approach would be todefine it as the limit of some regularizations. For example, if fl : R æ R

+

is a compactly supported CŒ function such thatsR fl(x)dx = 1, and we

set flÁ(·) = fl(·/Á)/Á, then we can set Nt,Á(u)(x) =s t

0

((flÁ ú us)(x))2ds anddefine Nt(u) = limÁæ0

Nt,Á(u) whenever the limit exists in S Õ := S Õ(T),the space of distributions on T. Then the question arises which propertiesu should have for this convergence to occur.

2.3 The Ornstein–Uhlenbeck process

Let us simplify the problem and start by studying the linearized equationobtained by neglecting the non–linear term. Let X be a solution to

Xt(Ï) = X0

(Ï) +⁄ t

0

Xs(�Ï)ds + Mt(Ï) (2.1)

for all t > 0 and Ï œ S . This equation has at most one solution (forfixed X

0

). Indeed, the di�erence D between two solutions should satisfyDt(Ï) =

s t

0

Ds(�Ï)ds, which means that D is a distributional solution tothe heat equation. Taking Ï(x) = ek(x), where

ek(x) := exp(≠ikx)/Ô

2fi, k œ Z,


we get Dt(ek) = ≠k2

s t

0

Ds(ek)ds and then by Gronwall’s inequalityDt(ek) = 0 for all t > 0. This easily implies that Dt = 0 in S Õ forall t > 0.

To obtain the existence of a solution, observe that

Xt(ek) = X0

(ek) ≠ k2

⁄ t

0

Xs(ek)ds + Mt(ek)

and that Mt(e0

) = 0, while for all k ”= 0 the process —t(k) = Mt(ek)/(≠ik)is a complex valued Brownian motion (i.e. real and imaginary part areindependent Brownian motions with the same variance). The covarianceof — is given by

E[—t(k)—s(m)] = (t · s)”k+m=0

and moreover —t(k)ú = —t(≠k) for all k ”= 0 (where ·ú denotes complexconjugation), as well as —t(0) = 0. In other words, (Xt(ek)) is a complex–valued Ornstein–Uhlenbeck process ([23], Example 5.6.8) which solves alinear one–dimensional SDE and has an explicit representation given bythe variation of constants formula

Xt(ek) = e≠k2tX0

(ek) ≠ ik

⁄ t

0

e≠k2(t≠s)ds—s(k).

This is enough to determine Xt(Ï) for all t > 0 and Ï œ S .

Exercise 4. Show that (Xt(ek) : t œ R+

, k œ Z) is a complex Gaussianrandom field, that is for all n œ N, for all t

1

, . . . , tn œ R+

, k1

, . . . , kn œ Z,the vector

(Re(Xt1(ek1)), . . . , Re(Xtn

(ekn

)), Im(Xt1(ek1)), . . . , Im(Xtn

(ekn

)))

is multivariate Gaussian. Show that X has mean E[Xt(ek)] = e≠k2tX0

(ek)and covariance

E[(Xt

(ek

)≠E[Xt

(ek

)])(Xs

(em

)≠E[Xs

(em

)])] = k2”k+m=0

⁄t·s

0e≠k

2(t≠r)≠k

2(s≠r)dr

as well as

E[(Xt

(ek

)≠E[Xt

(ek

)])(Xs

(em

)≠E[Xs

(em

)])ú] = k2”k=m

⁄t·s

0e≠k

2(t≠r)≠k

2(s≠r)dr.

In particular,

E[|Xt(ek) ≠ E[Xt(ek)]|2] = 1 ≠ e≠2k2t

2 .

Next we examine the Sobolev regularity of X. For this purpose, we needthe following definition.


Definition 1. Let – œ R. Then the Sobolev space H– is defined as

H– := H–(T) :=I

fl œ S Õ : ÎflÎ2

H–

:=ÿ

kœZ(1 + |k|2)–|fl(ek)|2 < Œ

J.

We also write CH– for the space of continuous functions from R+

to H–.Lemma 1. Let “ 6 ≠1/2 and assume that X

0

œ H“ . Then almost surelyX œ CH“≠.Proof. Let – = “ ≠ Á and consider

ÎXt ≠ XsÎ2

H–

=ÿ

kœZ(1 + |k|2)–|Xt(ek) ≠ Xs(ek)|2.

Let us estimate the L2p(�) norm of this quantity for p œ N by writing

EÎXt ≠ XsÎ2pH–

=ÿ

k1,...,kp

œZ

pŸ

i=1

(1 + |ki|2)–EpŸ

i=1

|Xt(eki

) ≠ Xs(eki

)|2.

By Hölder’s inequality, we get


.ÿ

k1,...,kp

œZ

pŸ

i=1

(1 + |ki|2)–

pŸ

i=1

(E|Xt(eki

) ≠ Xs(eki

)|2p)1/p.

Note now that Xt(eki

) ≠ Xs(eki

) is a Gaussian random variable, so thatthere exists a universal constant Cp for which

E|Xt(eki

) ≠ Xs(eki

)|2p 6 Cp(E|Xt(eki

) ≠ Xs(eki

)|2)p.

Moreover,

Xt(ek) ≠ Xs(ek) = (e≠k2(t≠s) ≠ 1)Xs(ek) + ik

⁄ t

s

e≠k2(t≠r)dr—r(k),

leading to

E|Xt(ek) ≠ Xs(ek)|2

=(e≠k2(t≠s) ≠ 1)2E|Xs(ek)|2 + k2

⁄ t

s

e≠2k2(t≠r)dr

=(e≠k2(t≠s) ≠ 1)2e≠2k2s|X

0

(ek)|2 + (e≠k2(t≠s) ≠ 1)2k2

⁄ s

0

e≠2k2(s≠r)dr

+ k2

⁄ t

s

e≠2k2(t≠r)dr

=(e≠k2t ≠ e≠k2s)2|X0

(ek)|2 + 12(e≠k2

(t≠s) ≠ 1)2(1 ≠ e≠2k2s)

+ 12(1 ≠ e≠2k2

(t≠s)).


For any Ÿ œ [0, 1] and k ”= 0, we thus have

E|Xt(ek) ≠ Xs(ek)|2 . (k2(t ≠ s))Ÿ(|X0

(ek)|2 + 1),

while for k = 0 we have E|Xt(e0

) ≠ Xs(e0

)|2 = 0. Let us introduce thenotation Z

0

= Z \ {0}. Therefore,


.ÿ

k1,...,kp

œZ0

pŸ

i=1

(1 + |ki|2)–

pŸ

i=1

E|Xt(eki

) ≠ Xs(eki

)|2

. (t ≠ s)Ÿpÿ

k1,...,kp

œZ0

pŸ

i=1

(1 + |ki|2)–(k2

i )Ÿ(|X0

(eki

)|2 + 1)

. (t ≠ s)ŸpË ÿ

kœZ0

(1 + |k|2)–(k2)Ÿ(|X0

(ek)|2 + 1)Èp

. (t ≠ s)Ÿp1

ÎX0

Î2pH–+Ÿ

(T)

+Ë ÿ

kœZ0

(1 + |k|2)–(k2)ŸÈp2

.

If – < ≠1/2 ≠ Ÿ, the sum on the right hand side is finite and we obtain anestimation for the modulus of continuity of t ‘æ Xt in L2p(�; H–):


. (t ≠ s)Ÿp[1 + ÎX0

Î2pH–+Ÿ

].

Now Kolmogorov’s continuity criterion allows us to conclude that almostsurely X œ CH– whenever X

0

œ H–+Ÿ. ⇤Now note that the regularity of the Ornstein–Uhlenbeck process does

not allow us to form the quantity X2

t point–wise in time since by Fourierinversion Xt =

qk Xt(ek)eú

k, and therefore we should have

X2

t (ek) = (2fi)≠1/2

ÿ

¸+m=k

Xt(e¸)Xt(em).

Of course, at the moment this expression is purely formal since we cannotguarantee that the infinite sum converges. A reasonable thing to try is toapproximate the square by regularizing the distribution, taking the square,and then trying to remove the regularization. Let �N be the projector ofa distribution onto a finite number of Fourier modes:

(�N fl)(x) =ÿ

|k|6N

fl(ek)eúk(x).

Then �N Xt(x) is a smooth function of x and we can consider (�N Xt)2

which satisfies

(�N Xt)2(ek) = (2fi)≠1/2

ÿ

¸+m=k

I|¸|6N,|m|6N Xt(e¸)Xt(em).


We would then like to take the limit N æ +Œ. For convenience, we willperform the computations below in the limit N = +Œ, but one has tocome back to the case of finite N in order to make it rigorous.

ThenE[X2

t (ek)] =(2fi)≠1/2”k=0

ÿ

mœZ0

E[Xt(e≠m)Xt(em)]

=(2fi)≠1/2”k=0

ÿ

mœZ0

e≠2m2t|X0

(em)|2

+ (2fi)≠1/2”k=0

ÿ

mœZ0

m2

⁄ t

0

e≠2m2(t≠s)ds

andÿ

mœZ0

m2

⁄ t

0

e≠2m2(t≠s)ds = 1

2ÿ

mœZ0

(1 ≠ e≠2m2t) = +Œ.

This is not really a problem since in Burgers’ equation only components ofu2

t (ek) with k ”= 0 appear (due to the presence of the derivative). However,X2

t (ek) is not even a well–defined random variable. For the remainder ofthis subsection let us assume that X

0

= 0, which will slightly simplify thecomputation. If k ”= 0, we haveE[|X2

t (ek)|2] =E[X2

t (ek)X2

t (e≠k)]

=(2fi)≠1

ÿ

¸+m=k

ÿ

¸Õ+mÕ

=≠k

E[Xt(e¸)Xt(em)Xt(e¸Õ)Xt(emÕ)].

By Wick’s theorem (see [22], Theorem 1.28), the expectation can becomputed in terms of the covariances of all possible pairings of the fourGaussian random variables (3 possible combinations):E[Xt(e¸)Xt(em)Xt(e¸Õ)Xt(emÕ)] = E[Xt(e¸)Xt(em)]E[Xt(e¸Õ)Xt(emÕ)]

+ E[Xt(e¸)Xt(e¸Õ)]E[Xt(em)Xt(emÕ)]+ E[Xt(e¸)Xt(emÕ)]E[Xt(em)Xt(e¸Õ)].

Since k ”= 0, we have ¸ + m ”= 0 and ¸Õ + mÕ ”= 0 which allows us to neglectthe first term since it is zero. By symmetry of the summations, the twoother terms give the same contribution and we remain with

E[|X2

t (ek)|2] = 1fi

ÿ

¸+m=k

ÿ

¸Õ+mÕ

=≠k

E[Xt(e¸)Xt(e¸Õ)]E[Xt(em)Xt(emÕ)]

(2.2)

= 1fi

ÿ

¸+m=k

E[Xt(e¸)Xt(e≠¸)]E[Xt(em)Xt(e≠m)]

= 14fi

ÿ

¸+m=k

(1 ≠ e≠2¸2t)(1 ≠ e≠2m2t) = +Œ.


This shows that even when tested against smooth test functions, X2

t is notin L2(�). This indicates that there are problems with X2

t and indeed onecan show that X2

t (ek) does not make sense as a random variable.

To understand this better, observe that the Ornstein–Uhlenbeck processcan be decomposed as

Xt(ek) = ik

⁄ t

≠Œe≠k2

(t≠s)d—s(k) ≠ ike≠k2t

⁄0

≠Œek2sd—s(k),

where we extended the Brownian motions (—s(k))s>0

to two sided complexBrownian motions by considering independent copies. The interest in thisdecomposition is in the fact that it is not di�cult to show that the secondterm gives rise to a smooth function if t > 0, so all the irregularity of Xt

is described by the first term which we call Yt(ek) and which is stationaryin time. Note that Yt(ek) ≥ NC(0, 1/2) for all k œ Z

0

and t œ R, where wewrite

U ≥ NC(0, ‡2)

if U = V + iW , where V and W are independent random variableswith distribution N (0, ‡2/2). The random distribution Yt then satisfiesYt(Ï) ≥ N (0, ÎÏÎ2

L2(T)

/2), and moreover it is (1/Ô

2 times) the white noiseon T. It is also possible to deduce that the white noise on T is indeed theinvariant measure of the Ornstein–Uhlenbeck process, that it is the onlyone, and that it is approached quite fast [23].

So we should expect that, at fixed time, the regularity of the Ornstein–Uhlenbeck process is like that of the space white noise and this is a wayof understanding our di�culties in defining X2

t since this will be, modulosmooth terms, the square of the space white noise.

A di�erent matter is to make sense of the time–integral of ˆxX2

t . Letus give it a name and call it Jt(Ï) =

s t

0

ˆxX2

s (Ï)ds. For Jt(ek), thecomputation of its variance gives a quite di�erent result.

Lemma 2. Almost surely, J œ C1/2≠H≠1/2≠.

Proof. Proceeding as in (2.2), we have now

E[|Jt

(ek

)|2] = 1fi

k2⁄

t

0

⁄t

0

ÿ

¸+m=k

E[Xs

(e¸

)Xs

Õ (e≠¸

)]E[Xs

(em

)Xs

Õ (e≠m

)]dsdsÕ.

If s > sÕ, we have

E[Xs(e¸)XsÕ(e≠¸)] = 12e≠¸2

(s≠sÕ)(1 ≠ e≠2¸2sÕ

),


and therefore

E[|Jt

(ek

)|2] = k2

4fi

⁄t

0

⁄t

0

ÿ

¸+m=k

e≠(¸

2+m

2)|s≠s

Õ|(1 ≠ e≠2¸

2(s

Õ·s))(1 ≠ e≠2m

2(s

Õ·s))dsdsÕ

6 k2

4fi

⁄t

0

⁄t

0

ÿ

¸+m=k

e≠(¸

2+m

2)|s≠s

Õ|dsdsÕ

6 12fi

k2tÿ

¸+m=k

⁄ Œ

0e≠(¸

2+m

2)rdr

= 12fi

k2tÿ

¸+m=k

1¸2 + m2 .

Now for k ”= 0ÿ

¸+m=k

1¸2 + m2

.⁄

R

dx

x2 + (k ≠ x)2

. 1|k| .

So finally E[|Jt(ek)|2] . |k|t. From which is easy to conclude that at fixedt the random field Jt belongs almost surely to H≠1/2≠. Redoing a similarcomputation in the case Jt(ek) ≠ Js(ek), we obtain E[|Jt(ek) ≠ Js(ek)|2] .|k| ◊ |t ≠ s|. To go from this estimate to a path–wise regularity result ofthe distribution (Jt)t, following the line of reasoning of Lemma 1, we needto estimate the p-th moment of Jt(ek) ≠ Js(ek). We already used in theproof of Lemma 1 that all moments of a Gaussian random variable arecomparable. By Gaussian hypercontractivity (see Theorem 3.50 of [22])this also holds for polynomials of Gaussian random variables, so that

E[|Jt(ek) ≠ Js(ek)|2p] .p (E[|Jt(ek) ≠ Js(ek)|2])p.

From here we easily derive that almost surely J œ C1/2≠H≠1/2≠ which isthe space of 1/2≠Hölder continuous functions with values in H≠1/2≠. ⇤

This shows that ˆxX2

t exists as a space–time distribution but notas a continuous function of time with values in distributions in space.The key point in the proof of Lemma 2 is the fact that the correlationE[Xs(e¸)XsÕ(e≠¸)] of the Ornstein–Uhlenbeck process decays quite rapidlyin time.

The construction of the process J does not solve our problem ofconstructing

s t

0

ˆxu2

sds since we need similar properties for the full solutionu of the non–linear dynamics (or for some approximations thereof), andall we have done so far relies on explicit computations and the specificGaussian features of the Ornstein–Uhlenbeck process. But at least thisgive us a hint that indeed there could exist a way of making sense of theterm ˆxu(t, x)2, even if only as a space–time distribution, and that in doingso we should exploit some decorrelation properties of the dynamics.


So when dealing with the full solution u, we need a replacement for theGaussian computations based on the explicit distribution of X that weused above. This will be provided, in the current setting, by stochasticcalculus along the time direction. Indeed, note that for each Ï œ S theprocess (Xt(Ï))t>0

is a semimartingale in the filtration (Ft)t>0

.

Before proceeding with these computations, we need to develop sometools to describe the Itô formula for functions of the Ornstein–Uhlenbeckprocess. This will also serve us as an opportunity to set up some analysison Gaussian spaces.

2.4 Gaussian computationsFor cylindrical functions F : S Õ æ R of the form F (fl) =f(fl(Ï

1

), . . . , fl(Ïn)) with Ï1

, . . . , Ïn œ S and f : Rn æ R at least C2

b ,we have by Itô’s formula

dtF (Xt) =nÿ

i=1

Fi(Xt)dXt(Ïi) + 12

nÿ

i,j=1

Fi,j(Xt)dÈX(Ïi), X(Ïj)Ít,

where ÈÍt denotes the quadratic covariation of two continuoussemimartingales and where Fi(fl) = îf(fl(Ï

1

), . . . , fl(Ïn)) and Fi,j(fl) =ˆ2

i,jf(fl(Ï1

), . . . , fl(Ïn)), with î denoting the derivative with respect tothe i-th argument. Now recall that dXt(Ïi) = Xt(�Ïi)dt + dMt(Ïi) is acontinuous semimartingale, and therefore

dÈX(Ïi), X(Ïj)Ít = dÈM(Ïi), M(Ïj)Ít = ÈˆxÏi, ˆxÏjÍL2(T)

dt,

and then

dtF (Xt) =nÿ

i=1

Fi(Xt)dMt(Ïi) + L0

F (Xt)dt,

where L0

is the second–order di�erential operator defined on cylindricalfunctions F as

L0

F (fl) =nÿ

i=1

Fi(fl)fl(�Ïi) +nÿ

i,j=1

12Fi,j(fl)ÈˆxÏi, ˆxÏjÍL2

(T)

. (2.3)

Another way to describe the generator L0

is to give its value on thefunctions fl ‘æ exp(fl(Â)) for Â œ S , which is

L0

efl(Â) = efl(Â)(fl(�Â) ≠ 12 ÈÂ, �ÂÍL2

(T)

).


If F, G are two cylindrical functions (which we can take of the formF (fl) = f(fl(Ï

1

), . . . , fl(Ïn)) and G(fl) = g(fl(Ï1

), . . . , fl(Ïn)) for the sameÏ

1

, . . . , Ïn œ S ), we can check thatL

0

(FG) = (L0

F )G + F (L0

G) + E(F, G), (2.4)where the quadratic form E is given by

E(F, G)(fl) =ÿ

i,j

Fi(fl)Gj(fl)ÈˆxÏi, ˆxÏjÍL2(T)

. (2.5)

In particular, the quadratic variation of the martingale obtained in the Itôformula for F is given by

de ⁄ ·

0

nÿ

i=1

Fi(Xs)dMs(Ïi)f

t= E(F, F )(Xt)dt.

Lemma 3. (Gaussian integration by parts) Let (Zi)i=1,...,M bean M -dimensional Gaussian vector with zero mean and covariance(Ci,j)i,j=1,...,M . Then for all g œ C1

b (RM ) we have

E[Zkg(Z)] =ÿ

¸

Ck,¸E5

ˆg(Z)ˆZ¸

6.

Proof. Use that E[eiÈZ,⁄Í] = e≠È⁄,C⁄Í/2 and moreover that

E[ZkeiÈZ,⁄Í] = (≠i) ˆ

ˆ⁄kE[eiÈZ,⁄Í] = (≠i) ˆ

ˆ⁄ke≠È⁄,C⁄Í/2 = i(C⁄)ke≠È⁄,C⁄Í/2

= iÿ

¸

Ck,¸⁄¸E[eiÈZ,⁄Í] =ÿ

¸

Ck,¸E[ ˆ

ˆZ¸eiÈZ,⁄Í].

The relation is true for trigonometric functions and taking Fouriertransforms we see that it holds for all g œ S . Is then a matter of takinglimits to show that we can extend it to any g œ C1

b (RM ). ⇤As a first application of this formula let us show that E[L

0

F (÷)] = 0 forevery cylindrical function, where ÷ is a space white noise with mean zero,i.e. ÷(Ï) ≥ N (0, ÎÏÎ2

L2(T)

/2) for all Ï œ L2

0

(T), and ÷(1) = 0. Here wewrite L2

0

(T) for the subspace of all Ï œ L2(T) withsT Ïdx = 0. Indeed,

note that by polarization E[÷(Ïi)÷(�Ïj)] = 1

2

ÈÏi, �ÏjÍL2(T)

, leading to

Enÿ

i,j=1

12Fi,j(÷)ÈˆxÏi, ˆxÏjÍL2

(T)

= ≠Enÿ

i,j=1

12Fi,j(÷)ÈÏi, �ÏjÍL2

(T)

= ≠12

nÿ

i,j=1

ÈÏi, �ÏjÍL2(T)

E ˆ

ˆ÷(Ïi)Fj(÷)

= ≠nÿ

j=1

E[÷(�Ïj)Fj(÷)],


so that E[L0

F (÷)] = 0 (here we interpreted ˆjf as a function of n + 1variables, with trivial dependence on the (n + 1)-th one). In combinationwith Itô’s formula, this indicates that the white noise law should indeedbe a stationary distribution for X (convince yourself of it!). From now onwe fix the initial distribution X

0

≥ ÷, which means that Xt ≥ ÷ for allt > 0.

As another application of the Gaussian integration by parts formula, weget

12E[E(F, G)(÷)] = ≠1

2ÿ

i,j

E[Fi(÷)Gj(÷)]ÈÏi, �ÏjÍL2(T)

.

= ≠12

ÿ

i,j

E[(F (÷)Gj(÷))i]ÈÏi, �ÏjÍL2(T)

+ 12

ÿ

i,j

E[F (÷)Gij(÷)]ÈÏi, �ÏjÍL2(T)

= ≠ÿ

j

E[F (÷)Gj(÷)÷(�Ïj)]

+ 12

ÿ

i,j

E[F (÷)Gij(÷)]ÈÏi, �ÏjÍL2(T)

= ≠E[(FL0

G)(÷)].

Combining this with (2.4) and with E[L0

(FG)(÷)] = 0, we obtainE[(FL

0

G)(÷)] = E[(GL0

F )(÷)]. That is, L0

is a symmetric operator withrespect to the law of ÷.

Consider now the operator D, defined on cylindrical functions F by

DF (fl) =ÿ

i

Fi(fl)Ïi (2.6)

so that DF takes values in S Õ, the continuous linear functionals on S .Exercise 5. Show that D is independent of the specific representation ofF , that is if

F (fl) = f(fl(Ï1

), . . . , fl(Ïn)) = g(fl(Â1

), . . . , fl(Âm))

for all fl œ S Õ, thenÿ

i

îf(fl(Ï1

), . . . , fl(Ïn))Ïi =ÿ

j

ˆjg(fl(Â1

), . . . , fl(Âm))Âm.

Hint: One possible strategy is to show that for all ◊ œ S ,

ÈDF (fl), ◊Í = ddÁ

F (fl + Á◊)|Á=0

.


By Gaussian integration by parts we get

E[F (÷)ÈÂ, DG(÷)Í] + E[G(÷)ÈÂ, DF (÷)Í] =ÿ

i

E[(FG)i(÷)ÈÂ, ÏiÍ]

=2E[÷(Â)(FG)(÷)],

and therefore

E[F (÷)ÈÂ, DG(÷)Í] = E[G(÷)ÈÂ, ≠DF (÷) + 2flF (÷)Í].

So if we consider the space L2(law(÷)) with inner product E[F (÷)G(÷)],then the adjoint of D is given by DúF (fl) = ≠DF (fl) + 2flF (fl). LetDÂF (fl) = ÈÂ, DF (fl)Í and similarly for Dú

ÂF (fl) = ≠DÂF (fl)+2fl(Â)F (fl).

Exercise 6. Let (en)n>1

be an orthonormal basis of L2(T). Show that

L0

= 12

ÿ

n

Dúe

n

D�e

n

.

Recall that the commutator between two operators A and B is definedas [A, B] := AB ≠ BA. In our case we have

[D◊, DúÂ]F (fl) = (D◊Dú

Â ≠ DúÂD◊)F (fl) = 2ÈÂ, ◊ÍL2

(T)

F (fl),

whereas [Dú◊, Dú

Â] = 0. Therefore,

[L0

, DúÂ] =1

2ÿ

n

[Dúe

n

D�e

n

, DúÂ]

=12

ÿ

n

Dúe

n

[D�e

n

, DúÂ] + 1

2ÿ

n

[Dúe

n

, DúÂ]D

�en

=ÿ

n

Dúe

n

ÈÂ, �enÍL2(T)

= Dú�Â.

So if Â is an eigenvector of � with eigenvalue ⁄, then [L0

, DúÂ] = ⁄Dú

Â.Let now (Ân)nœN be an orthonormal eigenbasis for � with eigenvalues�Ân = ⁄nÂn and consider the functions

H(Âi1 , . . . , Âin

) : S Õ æ R, H(Âi1 , . . . , Âin

)(fl) = (DúÂ

i1· · · Dú

Âi

n

1)(fl).

Then

L0

H(Âi1 , . . . , Âin

) = L0

DúÂ

i1· · · Dú

Âi

n

1= Dú

Âi1

L0

DúÂ

i2· · · Dú

Âi

n

1 + ⁄i1DúÂ

i1· · · Dú

Âi

n

1 (2.7)= · · · = (⁄i1 + · · · + ⁄i

n

)H(Âi1 , . . . , Âin

),


where we used that L0

1 = 0. So these functions are eigenfunctions for L0

and the eigenvalues are all the possible combinations of ⁄i1 + · · · + ⁄in

fori1

, . . . , in œ N. We have immediately that for di�erent n these functionsare orthogonal in L2(law(÷)). They are actually orthogonal as soon asthe indices i di�er since in that case there is an index j which is in onebut not in the other and using the fact that Dú

Âj

is adjoint to DÂj

andthat DÂ

j

G = 0 if G does not depend on Âj we get the orthogonality.The functions H(Âi1 , . . . , Âi

n

) are polynomials and they are called Wickpolynomials.

Lemma 4. For all Â œ S , almost surely

(eD

úÂ 1)(÷) = e2÷(Â)≠ÎÂÎ2

.

Proof. If F is a cylindrical function of the form F (fl) =f(fl(Ï

1

), . . . , fl(Ïm)) with f œ S (Rm), then

E[F (÷)(eD

úÂ 1)(÷)] = E[eD

Â F (÷)] = E[F (÷ + Â)] = E[F (÷)e2÷(Â)≠ÎÂÎ2],

where the second step follows from the fact that if we note �t(÷) =F (÷ + tÂ) (note that every Â œ S can be interpreted as an element of S Õ)we have ˆt�t(÷) = DÂ�t(÷) and �

0

(÷) = F (÷) so that �t(÷) = (etD

Â F )(÷)for all t > 0 and in particular for t = 1. The last step is simply a Gaussianchange of variables. Indeed if we take Ï

1

= Â and Ïk‹Â for k > 2 we have

E[F (÷ + Â)] = E[f(÷(Â) + ÈÂ, ÂÍ, ÷(Ï2

), . . . , ÷(Ïm))]

since (÷+Â)(Ïk) = ÷(Ïk) for k > 2. Now observe that ÷(Â) is independentof (÷(Ï

2

), . . . , ÷(Ïm)) so that

E[f(÷(Â) + ÈÂ, ÂÍ, ÷(Ï2

), . . . , ÷(Ïm))]

=⁄

R

e≠z2/ÎÂÎ2

fiÎÂÎ2

E[f(z + ÈÂ, ÂÍ, ÷(Ï2

), . . . , ÷(Ïm))]

=⁄

R

e≠z2/ÎÂÎ2

fiÎÂÎ2

e2z≠ÎÂÎ2E[f(z, ÷(Ï

2

), . . . , ÷(Ïm))]

=E[F (÷)e2÷(Â)≠ÎÂÎ2].

To conclude the proof, it su�ces to note that E[F (÷)(eD

úÂ 1)(÷)] =

E[F (÷)e2÷(Â)≠ÎÂÎ2 ] for all cylindrical functions F implies that (eD

úÂ 1)(÷) =

e2÷(Â)≠ÎÂÎ2 . ⇤

Theorem 1. The Wick polynomials {H(Âi1 , . . . , Âin

)(÷) : n >0, i

1

, . . . , in œ N} form an orthogonal basis of L2(law(÷)).


Proof. Taking Â =q

i ‡iÂi in Lemma 4, we get

e2

qi

‡i

÷(Âi

)≠q

i

‡2i

ÎÂi

Î2= (eD

úÂ 1)(÷) =

ÿ

n>0

((DúÂ)n1)(÷)

n!

=ÿ

n>0

ÿ

i1,...,in

‡i1 · · · ‡in

n! H(Âi1 , . . . , Âin¸ ˚˙ ˝

n times

)(÷),

which is enough to show that any random variable in L2 can be expandedin a series of Wick polynomials showing that the Wick polynomials are anorthogonal basis of L2(law(÷)) (but they are still not normalized). Indeedassume that Z œ L2(law(÷)) but Z‹H(Âi1 , . . . , Âi

n

)(÷) for all n > 0,i1

, . . . , in œ N, then

0 =eq

i

‡2i

ÎÂi

Î2E[Z(eD

úÂ 1)(÷)]

=eq

i

‡2i

ÎÂi

Î2E[Ze2

qi

‡i

÷(Âi

)≠q

i

‡2i

ÎÂi

Î2]

=E[Ze2

qi

‡i

÷(Âi

)].

Since the ‡i are arbitrary, this means that Z is orthogonal to anypolynomial in ÷ (consider the derivatives in ‡ © 0) and then that it isorthogonal also to exp(i

qi ‡i÷(Âi)). So let f œ S (RM ) and ‡i = 0 for

i > m, and observe that

0 =(2fi)≠m/2

⁄d‡

1

· · · d‡mFf(‡1

, . . . , ‡m)E[Zeiq

i

‡i

÷(Âi

)]

=E[Zf(÷(Â1

), . . . , ÷(ÂM ))],

which means that Z is orthogonal to all the random variables in L2 whichare measurable with respect to the ‡–field generated by (÷(Ân))n>0

. Thisimplies Z = 0. That is, Wick polynomials form a basis for L2(law(÷)). ⇤

Example 2. The first few (un–normalized) Wick polynomials are

H(Âi)(fl) = DúÂ

i

1(fl) = 2fl(Âi),

H(Âi, Âj)(fl) = DúÂ

i

DúÂ

j

1 = 2DúÂ

i

fl(Âj) = ≠2”i=j + 4fl(Âi)fl(Âj),

and

H(Âi, Âj , Âk)(fl) = DúÂ

i

(≠2”j=k + 4fl(Âj)fl(Âk))= ≠4”j=kfl(Âi) ≠ 4”i=jfl(Âk) ≠ 4”i=kfl(Âj)

+ 8fl(Âi)fl(Âj)fl(Âk).


Some other properties of Wick polynomials can be derived using thecommutation relation between D and Dú. By linearity Dú

Ï+Â = DúÏ + Dú

Â,so that using the symmetry of H we get

Hn(Ï + Â) := H (Ï + Â, . . . , Ï + Â)¸ ˚˙ ˝n

=ÿ

06k6n

3n

k

4H(Ï, . . . , Ï¸ ˚˙ ˝

k

, Â, . . . , Â¸ ˚˙ ˝n≠k

).

Then note that by Lemma 4 we have

(eD

úÏ1)(÷)(eD

úÂ 1)(÷) = e2÷(Ï)≠ÎÏÎ2

e2÷(Â)≠ÎÂÎ2= e2÷(Ï+Â)≠ÎÏ+ÂÎ2

+2ÈÏ,ÂÍ

= (eD

úÏ+Â 1)(÷)e2ÈÏ,ÂÍ.

Expanding the exponentials,

ÿ

m,n

Hm(Ï)m!

Hn(Â)n! =

ÿ

r,¸

Hr(Ï + Â)r!

(2ÈÏ, ÂÍ)¸

¸!

=ÿ

p,q,¸

H(p˙ ˝¸ ˚

Ï, . . . , Ï,

q˙ ˝¸ ˚Â, . . . , Â)

p!q!(2ÈÏ, ÂÍ)¸

¸! ,

and identifying the terms of the same homogeneity in Ï and Â respectivelywe get

Hm(Ï)Hn(Â) =ÿ

p+¸=m

ÿ

q+¸=n

m!n!p!q!¸!H(

p˙ ˝¸ ˚Ï, . . . , Ï,

q˙ ˝¸ ˚Â, . . . , Â) (2ÈÏ, ÂÍ)¸

.

(2.8)This gives a general formula for such products. By polarization of thismultilinear form, we can also get a general formula for the productsof general Wick polynomials. Indeed taking Ï =

qmi=1

ŸiÏi and Â =qnj=1

⁄jÂj for arbitrary real coe�cients Ÿ1

, . . . , Ÿm and ⁄1

, . . . , ⁄n, wehave

Hm(mÿ

i=1

ŸiÏi)Hn(nÿ

j=1

⁄jÂj)

=ÿ

i1,...,im

ÿ

j1,...,jn

Ÿi1 · · · Ÿim

⁄j1 · · · ⁄jm

H(Ïi1 , . . . , Ïim

)H(Âj1 , . . . , Âjn

).

Deriving this with respect to all the Ÿ, ⁄ parameters and setting them tozero, we single out the term

ÿ

‡œSm

,ÊœSn

H(Ï‡(1)

, . . . , Ï‡(m)

)H(ÂÊ(1)

, . . . , ÂÊ(n)

)

= m!n!H(Ï1

, . . . , Ïm)H(Â1

, . . . , Ân),


where Sk denotes the symmetric group on {1, . . . , k}, and where we usedthe symmetry of the Wick polynomials. Doing the same also for the righthand side of (2.8) we get

H(Ï1

, . . . , Ïm)H(Â1

, . . . , Ân)

=ÿ

p+¸=m

ÿ

q+¸=n

1p!q!¸!

ÿ

i,j

H(p˙ ˝¸ ˚

Ïi1 , . . . , Ïip

,

q˙ ˝¸ ˚Âj1 , . . . , Âj

q

)Ÿ

r=1

(2ÈÏip+r

, Âjq+r

Í),

where the sum over i, j runs over i1

, . . . , im permutation of 1, . . . , m andsimilarly for j

1

, . . . , jn. Since H(Ïi1 , . . . , Ïip

, Âj1 , . . . , Âjq

)(÷) is orthogonalto 1 whenever p + q > 0, we obtain in particular

E[H(Â1

, . . . , Ân)(÷)H(Â1

, . . . , Ân)(÷)] = 1n!

ÿ

i,j

nŸ

r=1

(2ÈÂir

, Âjr

Í)

=ÿ

‡œSn

nŸ

r=1

(2ÈÂr, Â‡(r)

Í).

In conclusion, we have shown that the family

Ó1 ÿ

‡œSn

nŸ

r=1

(2ÈÂr, Â‡(r)

Í)2≠1/2

H(Âi1 , . . . , Âin

)(÷) : n > 0, i1

, . . . , in œ NÔ

is an orthonormal basis of L2(law(÷)).

Remark 1. In our problem it will be convenient to take the Fourierbasis as basis in the above computations. Let ek(x) = exp(ikx)/

Ô2fi =

ak(x)+ibk(x) where (Ô

2ak)kœN and (Ô

2bk)kœN form together a real valuedorthonormal basis for L2(T). Then fl(ek)ú = fl(e≠k) whenever fl is realvalued, and we will denote Dk = De

k

= Dak

+ iDbk

and similarly forDú

k = Dúa

k

≠ iDúb

k

= ≠D≠k + 2fl(e≠k). In this way, Dúk is the adjoint

of Dk with respect to the Hermitian scalar product on L2(�;C) and theOrnstein–Uhlenbeck generator takes the form

L0

=ÿ

kœN(Dú

ˆx

ak

Dˆx

ak

+ Dúˆ

x

bk

Dˆx

bk

) = 12

ÿ

kœZk2Dú

kDk (2.9)

(convince yourself of the last identity by observing that DúkDk +Dú

≠kD≠k =2(Dú

ak

Dak

+ Dúb

k

Dbk

)!). Similarly,

E(F, G) =ÿ

kœZk2(DkF )ú(DkG). (2.10)


2.5 The Itô trickWe are ready now to start our computations. Recall that we want toanalyse Jt(Ï) =

s t

0

ˆxX2

s (Ï)ds using Itô calculus with respect to theOrnstein–Uhlenbeck process. We want to understand Jt as a correctionterm in Itô’s formula: if we can find a function G such that L

0

G(Xt) =ˆxX2

t , then we get from Itô’s formula⁄ t

0

ˆxX2

s ds = G(Xt) ≠ G(X0

) ≠ MG,t,

where MG is a martingale depending on G. Of course, G will not bea cylindrical function but we only defined L

0

on cylindrical functions.So to make the following calculations rigorous we would again have toreplace ˆxX2

t by ˆx�nX2

t and then pass to the limit, see the paper [15]for details. As before we will perform the calculations already in thelimit N = +Œ, in order to simplify the computations and not to obscurethe ideas through technicalities. The next problem is that the pointwiseevaluation

s t

0

ˆxX2

s (x)ds does not make any sense because the integral willonly be defined as a space distribution. So we will consider

G : S Õ æ S Õ

instead of G : S Õ æ C. Note however that we can reduce every such G toa function from S Õ to C by considering fl ‘æ G(fl)(ek) for all k.

Now for a fixed k, we have

ˆxX2

t (ek) = ikÔ2

ÿ

¸+m=k

Xt(e¸)Xt(em) = ikÔ2

ÿ

¸+m=k

H¸,m(Xt), (2.11)

where H¸,m(fl) = 1

4

(Dú≠¸Dú

≠m1)(fl) = fl(e¸)fl(em) ≠ 1

2

”¸+m=0

is a secondorder Wick polynomial so that L

0

H¸,m = ≠(¸2 + m2)H¸,m by (2.7).Therefore, it is enough to take

G(Xt)(ek) = ≠ikÿ

¸+m=k

H¸,m(Xt)¸2 + m2

. (2.12)

This corresponds to the distribution G(Xt)(Ï) = ≠s Œ

0

ˆx(es�Xt)2(Ï)ds(check it!). Then

G(Xt)(Ï) = G(X0

)(Ï) + MG,t(Ï) + Jt(Ï),

where MG,t(Ï) is a martingale with quadratic variation

dÈMG,ú(Ï), MG,ú(Ï)Ít = E(G(ú)(Ï), G(ú)(Ï))(Xt)dt.


We can estimate

E[|Jt

(Ï) ≠ Js

(Ï)|2p] .p

E[|MG,t

(Ï) ≠ MG,s

(Ï)|2p] +E[|G(Xt

)(Ï) ≠ G(Xs

)(Ï)|2p].

To bound the martingale expectation, we will use the following Burkholderinequality:

Lemma 5. Let m be a continuous local martingale with m0

= 0. Thenfor all T > 0 and p > 1,

E[supt6T

|mt|2p] 6 CpE[ÈmÍpT ].

Proof. Start by assuming that m and ÈmÍ are bounded. Itô’s formulayields

d|mt|2p = (2p)|mt|2p≠1dmt + 12(2p)(2p ≠ 1)|mt|2p≠2dÈmÍt,

and therefore

E[|mT |2p] = CpEË ⁄ T

0

|ms|2p≠2dÈmÍs

È6 CpE[sup

t6T|mt|2p≠2ÈmÍT ].

By the Cauchy–Schwarz inequality we get

E[|mT |2p] 6 CpE[supt6T

|mt|2p](2p≠2)/2pE[ÈmÍpT ]1/p.

But now Doob’s Lp inequality yields E[supt6T |mt|2p] 6 C ÕpE[|mT |2p], and

this implies the claim in the bounded case. The unbounded case can betreated with a localization argument. ⇤

Applying Burkholder’s inequality, we obtain

E[|Jt(Ï) ≠ Js(Ï)|2p] .p EË---

⁄ t

s

E(G(ú)(Ï), G(ú)(Ï))(Xr)dr---pÈ

+ E[|G(Xt)(Ï) ≠ G(Xs)(Ï)|2p]

6 (t ≠ s)p≠1

⁄ t

s

E[|E(G(ú)(Ï), G(ú)(Ï))(Xr)|p]dr

+ E[|G(Xt)(Ï) ≠ G(Xs)(Ï)|2p]= (t ≠ s)pE[|E(G(ú)(Ï), G(ú)(Ï))(÷)|p]

+ E[|G(Xt)(Ï) ≠ G(Xs)(Ï)|2p],

using that Xr ≥ ÷. Now

DmG(fl)(ek) = ≠2ikfl(ek≠m)

(k ≠ m)2 + m2

,


and therefore

E(G(ú)(ek), G(ú)(ek))(fl) =ÿ

m

m2D≠mG(fl)(e≠k)DmG(fl)(ek)

= 4k2

ÿ

¸+m=k

m2

|fl(e¸)|2(¸2 + m2)2

. k2

ÿ

¸+m=k

|fl(e¸)|2¸2 + m2

,

which implies that

E[|E(G(ú)(ek), G(ú)(ek))(÷)|] . k2Eÿ

¸+m=k

|÷(e¸)|2¸2 + m2

. k2

ÿ

¸+m=k

1¸2 + m2

. |k|.

A similar computation gives also that

E[|E(G(ú)(ek), G(ú)(ek))(÷)|p] . |k|p.

Further, we have

E[|G(Xt)(ek) ≠ G(Xs)(ek)|2] . k2

ÿ

¸+m=k

EË |H¸,m(Xt) ≠ H¸,m(Xs))2

(¸2 + m2|2È

. k2|t ≠ s|ÿ

¸+m=k

m2

(¸2 + m2)2

. |k||t ≠ s|.

And finally, since G is a second order polynomial of a Gaussian process wecan apply once more Gaussian hypercontractivity to obtain

E[|Jt(ek) ≠ Js(ek)|2p] .p (t ≠ s)p|k|p.

The advantage of the Itô trick with respect to the explicit Gaussiancomputation is that it goes over to the non–Gaussian case. Indeed notethat while the boundary term G(Xt)(Ï) ≠ G(Xs)(Ï) has been estimatedusing a lot of the Gaussian information about X, we used only the law ata fixed time to handle the term

s t

sE(G(ú)(Ï), G(ú)(Ï))(Xr)dr.

In order to carry over these computation to the solution of the non–lineardynamics u, we need to replace the generator of X with that of u and tohave a way to handle the boundary terms. The idea is now to reverse theMarkov process u in time, which will allow us to kill the antisymmetric


part of the generator and at the same time kill the boundary terms. Indeedobserve that if u solves the stochastic Burgers equation, then formally wehave the Itô formula

dtF (ut) =nÿ

i=1

Fi(ut)dMt(Ïi) + LF (ut)dt,

where L is now the full generator of the non–linear dynamics, given by

LF (fl) = L0

F (fl) +ÿ

i

Fi(fl)Èˆxfl2, ÏiÍ = L0

F (fl) + BF (fl),

whereBF (fl) =

ÿ

k

(ˆxfl2)(ek)DkF (fl).

Formally, the non–linear term is antisymmetric with respect to theinvariant measure of L

0

. Indeed since B is a first order operator

E[(BF (÷))G(÷)] = E[(B(FG)(÷))] ≠ E[F (÷)(BG(÷))] = ≠E[F (÷)(BG(÷))](2.13)

provided E[BF (÷)] = 0 for any cylinder function F . Let us show this. Wehave

E[BF (÷)] =ÿ

k

E[(ˆx÷2)(ek)DkF (÷)]

= ≠ÿ

k

E[(Dk(ˆx÷2)(ek))F (÷)] +ÿ

k

E[Dk[(ˆx÷2)(ek)F (÷)]].

But now we get from (2.11)

Dk(ˆx÷2)(ek) =Ô

2ik÷(e0

) = fi≠1/2ikÈ÷, 1Í = 0,

where we used that È÷, 1Í = 0. Gaussian integration by parts then formallygives

E[BF (÷)] =ÿ

k

E[Dk[(ˆx÷2)(ek)F (÷)]] =ÿ

k

E[÷(ek)(ˆx÷2)(ek)F (÷)]

= E[È÷, ˆx÷2ÍF (÷)] = 13E[È1, ˆx÷3ÍF (÷)] = 0

since È1, ˆx÷3Í = ≠Èˆx1, ÷3Í = 0 (but of course È÷, ˆx÷2Í is not welldefined).

The dynamics of u backwards in time has a Markovian description whichis the subject of the next exercise.


Exercise 7. Let (yt)t>0

be a stationary Markov process on a Polish space,with semigroup (Pt)t>0

and stationary distribution µ. Show that if P út

is the adjoint of Pt in L2(µ), then (P út ) is a semigroup of operators on

L2(µ) (that is P ú0

= id and P ús+t = P ú

s P út as operators on L2(µ)). Show

that if y0

≥ µ, then for all T > 0 the process yt = yT ≠t, t œ [0, T ], isalso Markov, with semigroup (P ú

t )tœ[0,T ]

, and that µ is also an invariantdistribution for (P ú

t ). Show also that if (Pt) has generator L then (P út )

has generator Lú which is the adjoint of L with respect to L2(µ).

Now if we reverse the process in time letting ut = uT ≠t, we have bystationarity

E[F (ut)G(u0

)] = E[F (uT ≠t)G(uT )] = E[F (u0

)G(ut)].

So if we denote by L the generator of u:

E[LF (u0

)G(u0

)] = ddt

----t=0

E[F (ut)G(u0

)]

= ddt

----t=0

E[F (u0

)G(ut)]

=E[LG(u0

)F (u0

)],

which means that L is the adjoint of L, that is

LF (fl) = L0

F (fl) ≠ BF (fl) = L0

F (fl) ≠ÿ

k

(ˆxfl2)(ek)DkF (fl).

In other words, the reversed process solves

ut(Ï) = u0

(Ï) +⁄ t

0

us(�Ï)ds +⁄ t

0

Èu2

s, ˆxÏÍds ≠⁄ t

0

›s(ˆxÏ)ds

for a di�erent space-time white noise ›. Then Itô’s formula for u gives

dtF (ut) =nÿ

i=1

Fi(ut)dMt(Ïi) + LF (ut)dt,

where for all test functions Ï, the process M(Ï) is a martingale in thefiltration of u with covariance

dÈM(Ï), M(Â)Ít = ÈˆxÏ, ˆxÂÍL2(T)

dt.

Combining the Itô formulas for u and u, we get

F (uT )(Ï) = F (u0

)(Ï) + MF,T (Ï) +⁄ T

0

LF (us)(Ï)ds


and

F (u0

)(Ï) = F (uT )(Ï) = F (u0

)(Ï) + MF,T (Ï) +⁄ T

0

LF (us)(Ï)ds

= F (uT )(Ï) + MF,T (Ï) +⁄ T

0

LF (us)(Ï)ds,

and summing up these two equalities gives

0 = MF,T (Ï) + MF,T (Ï) +⁄ T

0

(L + L)F (us)(Ï)ds,

that is

2⁄ T

0

L0

F (us)(Ï)ds = ≠MF,T (Ï) ≠ MF,T (Ï).

An added benefit of this forward–backward representation is that the onlyterm which required quite a lot of informations about X, that is theboundary term F (Xt)(Ï) ≠ F (Xs)(Ï) does not appear at all now. Asabove if 2L

0

F (fl) = ˆxfl2, we end up with

⁄ T

0

ˆxu2

s(Ï)ds = ≠MF,T (Ï) ≠ MF,T (Ï). (2.14)

Exercise 8. Perform a similar formal calculation as in (2.13) to see thatE[LF (÷)] = 0 for all cylindrical functions F , so that ÷ should also beinvariant for the stochastic Burgers equation. Combine this with (2.14) toshow that setting N N

t (Ï) =s t

0

ˆx(�N us)2(Ï)ds we have

E[|N Nt (ek) ≠ N N

s (ek)|2p] .p (t ≠ s)p|k|p

and letting N N,Mt = N N

t ≠ N Mt we get

E[|N N,Mt (ek) ≠ N N,M

s (ek)|2p] .p (|k|/N)Áp(t ≠ s)p|k|p

for all 1 6 N 6 M . Use this to derive that

(E[ÎN N,Mt ≠ N N,M

s Î2pH–

])1/2p .p,– N≠Á/2(t ≠ s)1/2

for all – < ≠1 ≠ Á, and realize that this estimate allows you to provecompactness of the approximations N N and then convergence to a limit Nin L2p(�; C1/2≠H≠1≠).


2.6 Controlled distributionsLet us cook up a definition which will allow us to rigorously perform theformal computations above in a general setting.

Definition 2. Let u, A : R+

◊ T æ S Õ(T) be a couple of generalized (i.e.distribution-valued) processes such that

i. For all Ï œ S (T) the process t ‘æ ut(Ï) is a continuoussemimartingale satisfying

ut(Ï) = u0

(Ï) +⁄ t

0

us(�Ï)ds + At(Ï) + Mt(Ï),

where t ‘æ Mt(Ï) is a martingale with quadratic variationÈM(Ï), M(Â)Ít = ÈˆxÏ, ˆxÂÍL2

(T)

t and t ‘æ At(Ï) is a finite variationprocess with A

0

(Ï) = 0.

ii. For all t > 0 the random distribution Ï ‘æ ut(Ï) is a zero mean spacewhite noise with variance ÎÏÎ2

L20/2.

iii. For any T > 0 the reversed process ut = uT ≠t has again propertiesi, ii with martingale M and finite variation part A such that At(Ï) =≠(AT (Ï) ≠ AT ≠t(Ï)).

Any pair of processes (u, A) satisfying these condition will be calledcontrolled by the Ornstein–Uhlenbeck process and we will denote the setof all such processes with Q

ou

.

Theorem 2 ([15], Lemma 1). Assume that (u, A) œ Qou

and for anyN > 1, t > 0, Ï œ S let

N Nt (Ï) =

⁄ t

0

ˆx(�N us)2(Ï)ds

Then for any p > 1 (N N )N>1

converges in Lp(�) to a space–timedistribution N œ C1/2≠H≠1≠.

We are now at a point where we can give a meaning to our originalequation.

Definition 3. A pair of random distribution (u, A) œ Qou

is an energysolution to the stochastic Burgers equation if it satisfies

ut(Ï) = u0

(Ï) +⁄ t

0

us(�Ï)ds + Nt(Ï) + Mt(Ï)

for all t > 0 and Ï œ S . That is if A = N .


Now we are in a relatively standard setting of needing to prove existenceand uniqueness of such energy solutions. Note that in general the solutionsare pairs of processes (u, A).

Remark 2. The notion of energy solution has been introduced (in a slightlydi�erent way) in the work of Gonçalves and Jara [11] on macroscopicuniversal fluctuations of weakly asymmetric interacting particle systems.

2.7 Existence of solutionsFor the existence, the way to proceed is quite standard. We approximatethe equation, construct approximate solutions and then try to have enoughcompactness to have limiting points which then naturally will satisfy therequirements for energy solutions. For any N > 1 consider solutions uN

toˆtu

N = �uN + ˆx�N (�N uN )2 + ˆx›

These are generalized functions such that

duNt (ek) = ≠k2uN

t (ek)dt + [ˆx�N (�N uN )2](ek)dt + ikd—t(k)

for k œ Z and t > 0. We take u0

to be the white noise with covarianceu

0

(Ï) ≥ N (0, ÎÏÎ2/2). The point of our choice of the non–linearity isthat this (infinite–dimensional) system of equations decomposes into afinite dimensional system for (vN (k) = �N uN (ek))k:|k|6N and an infinitenumber of one–dimensional equations for each uN (ek) with |k| > N .Indeed if |k| > N we have [ˆx�N (�N uN )2](ek) = 0 so ut(ek) = Xt(ek) theOrnstein–Uhlenbeck process with initial condition X

0

(ek) = u0

(ek) whichrenders it stationary in time (check it). The equation for (vN (k))|k|6N

reads

dvNt (k) = ≠k2vN

t (k)dt + bk(vNt )dt + ikd—t(k), |k| 6 N, t > 0

wherebk(vN

t ) = ikÿ

¸+m=k

I|¸|,|k|,|m|6N vNt (¸)vN

t (m).

This is a standard finite–dimensional ODE having global solutions for allinitial conditions which gives rise to a nice Markov process. The fact thatsolutions do not blow up even if the interaction is quadratic can be seenby computing the evolution of the norm

At =ÿ

|k|6N

|vNt (k)|2


and by showing that

dAt =2ÿ

|k|6N

vNt (≠k)dvN

t (k)

= ≠ 2ÿ

|k|ÆN

k2|vNt (k)|2dt + 2

ÿ

|k|6N

vNt (≠k)bk(vN

t )dt

+ 2ikÿ

|k|6N

vNt (≠k)d—t(k).

Since A is nonnegative, we increase its absolute value by omitting the firstcontribution. But nowÿ

|k|6N

vNt (≠k)bk(vN

t ) = 2iÿ

k,¸,m:¸+m=k

I|¸|,|k|,|m|6N kvNt (¸)vN

t (m)vNt (≠k)

= ≠2iÿ

k,¸,m:¸+m+k=0

I|¸|,|k|,|m|6N (k)vNt (¸)vN

t (m)vNt (k)

and by symmetry of this expression it is equal to

≠23 i

ÿ

k,¸,m:¸+m+k=0

I|¸|,|k|,|m|6N (k + ¸ + m)vNt (¸)vN

t (m)vNt (k) = 0,

so |At| Æ |A0

+ Mt| where dMt = 2q

|k|6N I|k|6N (ik)vNt (≠k)d—t(k). Now

E[M2

T ] .⁄ T

0

ÿ

|k|6N

k2|vNt (k)|2dt . N2

⁄ T

0

Atdt

and then by martingales inequalities

E[ suptœ[0,T ]

(At)2] 6 2E[A2

0

] + 2E[ suptœ[0,T ]

(Mt)2] 6 2E[A2

0

] + 8E[M2

T ]

6 2E[A2

0

] + CN2

⁄ T

0

E(At)dt.

Now Gronwall’s inequality gives

E[ suptœ[0,T ]

(At)2] . eCN2TE[A2

0

],

from where we can deduce (by a continuation argument) that almost surelythere is no blowup at finite time for the dynamics. The generator LN forthe Galerkin dynamics is given by

LN F (fl) = L0

F (fl) + BN F (fl),


whereBN F (fl) =

ÿ

k

I|k|6N (ˆxfl2)(ek)DkF (fl).

And again the non–linear drift BN is antisymmetric with respect to theinvariant measure of L

0

by a computation similar to that for the full driftB. Next, using Echeverría’s criterion [7] we can obtain the invariance ofthe white noise from its infinitesimal invariance which can be checked atthe level of the generator LN . Finally it is also possible to rigorously showthat the reversed process is a Markov process with generator

LN F (fl) = L0

F (fl) ≠ BN F (fl),

thus proving that the reversed non-linear drift is the opposite of theforward one. Taking

ANt (ek) =

⁄ t

0

bk(vNs )ds

we obtain that (vN , AN ) œ Qou

. Note that this result depends on the factthat we kept the full linear part L

0

of the generator. A more standardGalerkin truncation would have lead us to a process which is controlled bythe Galerkin–truncated OU process. Estimates would have resulted in asimilar way but our setup is simpler.

Given that (vN , AN ) is controlled by the OU process, the Itô trickapplied to AN provides enough compactness in order to pass to the limit asN æ Œ and build an energy solution to the Stochastic Burgers equation.See [15] for additional details on the limiting procedure and [30] for detailson how to implement the Itô trick on the level of di�usions.

Remark 3. There is however one small catch: For a controlleddistribution (u, A) we required A(Ï) to be of finite variation for every testfunction Ï. The solution (vN , AN ) to the truncated equation will satisfythis, but in the limit A(Ï) will only have vanishing quadratic variationand it will not be of finite variation (in other words u(Ï) is a Dirichletprocess and not a semimartingale). Luckily in this setting it is still possibleto derive an Itô formula and everything goes through as described above,see [15] for details.

Chapter 3

Besov spaces

Here we collect some classical results from harmonic analysis which we willneed in the following. We concentrate on distributions and SPDEs on thetorus, but everything in this Section applies mutatis mutandis on the fullspace Rd, see [14]. The only problem is that then the stochastic terms willno longer be in the Besov spaces C – which we encounter below but ratherin weighted Besov spaces. Handling SPDEs in weighted function spaces ismore delicate and we prefer here to concentrate on the simpler situationof the torus.

We will use Littlewood–Paley blocks to obtain a decomposition ofdistributions into an infinite series of smooth functions. Of course, we havealready such a decomposition at our disposal: f =

qk f(k)eú

k. But it turnsout to be convenient not to consider each Fourier coe�cient separately, butto work with projections on dyadic Fourier blocks.

Definition 4. A dyadic partition of unity (‰, fl) consists of twononnegative radial functions ‰, fl œ CŒ(Rd,R), where ‰ is supported ina ball B = {|x| 6 c} and fl is supported in an annulus A = {a 6 |x| 6 b}for suitable a, b, c > 0, such that

1. ‰ +q

j>0

fl(2≠j ·) © 1 and

2. supp(‰) fl supp(fl(2≠j ·)) = ÿ for j > 1 and supp(fl(2≠i·)) flsupp(fl(2≠j ·)) = ÿ for all i, j > 0 with |i ≠ j| > 1.

We will often write fl≠1

= ‰ and flj = fl(2≠j ·) for j > 0.

Dyadic partitions of unity exist, see [1]. From now on we fix a dyadicpartition of unity (‰, fl) and define the dyadic blocks

�jf = flj(D)f = F ≠1(flj f), j > ≠1,

37


where here and in the following we use that every function on Rd can benaturally interpreted as a function on Zd. We also use the notation

Sjf =ÿ

i6j≠1

�if

as well as Ki = (2fi)d/2F ≠1fli so that

Ki ú f = F ≠1(fliFf) = �if.

From this representation we can also see the reason for considering smoothpartitions rather than indicator functions: From Young’s inequality weget only ÎI

[2

j ,2j+1)

(|D|)fÎLŒ Æ ÎF ≠1I[2

j ,2j+1)

ÎL1ÎfÎLŒ . jÎfÎLŒ forf œ LŒ, whereas Îflj(D)fÎLŒ . ÎfÎLŒ uniformly in j.

Every dyadic block has a compactly supported Fourier transform and istherefore in S . It is easy to see that f =

qj>≠1

�jf = limjæŒ Sjf forall f œ S Õ.

For – œ R, the Hölder–Besov space C – is given by C – = B–Œ,Œ(Td,R),

where for p, q œ [1, Œ] we define

B–p,q = B–

p,q(Td,R) =Ó

f œ S Õ : ÎfÎB–

p,q

=1 ÿ

j>≠1

(2j–Î�jfÎLp)q2

1/q

< ŒÔ

,

with the usual interpretation as ¸Œ norm if q = Œ. Then B–p,q is a Banach

space and while the norm Î·ÎB–

p,q

depends on (‰, fl), the space B–p,q does not

and any other dyadic partition of unity corresponds to an equivalent norm(for (p, q) = (Œ, Œ) this follows from Lemma 10 below, for the generalcase see [1], Lemma 2.69). We write Î·Î– instead of Î·ÎB–

Œ,Œ .

Exercise 9. Let ”0

denote the Dirac delta in 0. Show that ”0

œ C ≠d.If – œ (0, Œ) \ N, then C – is the space of Â–Ê times di�erentiable

functions whose partial derivatives of order Â–Ê are (– ≠ Â–Ê)–Höldercontinuous (see page 99 of [1]). Note however, that for k œ N the space C k

is strictly larger than Ck, the space of k times continuously di�erentiablefunctions. Below we will give the proof for – œ (0, 1), but before we stillneed some tools.

Recall that Schwartz functions on Rd are functions f œ CŒ(Rd) suchthat for every multiindex µ and all n Ø 0 we have

supxœRd

(1 + |x|)n|ˆµf(x)| < Œ.

Lemma 6. (Poisson summation) Let Ï : Rd æ C be a Schwartz function.Then

F ≠1Ï(x) =ÿ

kœZd

F ≠1

Rd

Ï(x + 2fik),

for all x œ Td, where F ≠1

Rd

Ï(x) = (2fi)≠d/2

sRd

Ï(y)eiÈx,yÍdy.

Chapter 3. Besov spaces 39

Proof. Let g(x) =q

kœZd

F ≠1

Rd

Ï(x + 2fik). The function F ≠1

Rd

Ï is ofrapid decay since Ï œ S so the sum converges absolutely and defines acontinuous function g : Rd æ R which is periodic of period 2fi in everydirection. The Fourier transform over the torus Td of this function is

Fg(y) =⁄

Td

e≠iÈx,yÍg(x) dx

(2fi)d/2

=⁄

Td

ÿ

kœZd

F ≠1

Rd

Ï(x + 2fik)e≠iÈx+2fik,yÍ dx

(2fi)d/2

since e≠iÈ2fik,yÍ = 1 for all y œ Zd. By dominated convergence the sumand the integral can be combined in an overall integration over Rd:

Fg(y) =⁄

Rd

F ≠1

Rd

Ï(x)e≠iÈx,yÍ dx

(2fi)d/2

= FRdF ≠1

Rd

Ï(y) = Ï(y),

where FRdf(x) = F ≠1

Rd

f(≠x). So we deduce that g(x) = F ≠1Ï(x). ⇤

Exercise 10. Show that Î·Î– 6 Î·Î— for – 6 —, that Î·ÎLŒ . Î·Î– for– > 0, that Î·Î– . Î·ÎLŒ for – 6 0, and that ÎSj · ÎLŒ . 2j–Î · Î– for– < 0. These inequalities will be very important for us in the followingand we will often use them without mentioning it specifically.

Hint: When proving Î·Î– . Î·ÎLŒ for – 6 0, you might need Poisson’ssummation formula.

The following Bernstein inequality is extremely useful when dealing withfunctions with compactly supported Fourier transform.

Lemma 7. (Bernstein inequality) Let B be a ball and k œ N0

. For any⁄ > 1, 1 6 p 6 q 6 Œ, and f œ Lp with supp(Ff) ™ ⁄B we have

maxµœNd

:|µ|=kÎˆµfÎLq .k,B ⁄k+d(

1p

≠ 1q

)ÎfÎLp .

Proof. Let Â be a compactly supported CŒ function on Rd such thatÂ © 1 on B and write Â⁄(x) = Â(⁄≠1x). Then

ˆµf(x) = ˆµF ≠1(Â⁄Ff)(x) = (2fi)d/2Èf, ˆµ(F ≠1Â⁄)(x ≠ ·)Í= (2fi)d/2(f ú ˆµ(F ≠1Â⁄))(x).

By Young’s inequality, we get

ÎˆµfÎLq . ÎfÎLpÎˆµ(F ≠1Â⁄)ÎLr ,

where 1 + 1/q = 1/p + 1/r. Now it is a short exercise to verify


Î · ÎLr 6 Î · Î1/rL1 Î · Î1≠1/r

LŒ , and

..ˆµ!F ≠1Â⁄

"..L1 =

⁄

Td

---ÿ

k

ˆµ!F ≠1

Rd

Â⁄

"(x + 2fik)

---dx

6⁄

Rd

|ˆµ(F ≠1

Rd

Â⁄)(x)|dx

= ⁄|µ|⁄

Rd

⁄d|(ˆµF ≠1

Rd

Â)(⁄x)|dx

ƒ ⁄|µ|,

whereas

supxœTd

---ÿ

k

ˆµ(F ≠1

Rd

Â⁄)(x + 2fik)--- = ⁄d+|µ| sup

xœTd

---ÿ

k

(ˆµF ≠1

Rd

Â)(⁄(x + 2fik))---

. ⁄d+|µ| supxœTd

ÿ

k

(1 + ⁄|x + 2fik|)≠2d

. ⁄d+|µ| supxœTd

ÿ

k

(1 + |x + 2fik|)≠2d

. ⁄d+|µ|.

We end up with

ÎˆµfÎLq . ÎfÎLpÎˆµ(F ≠1Â⁄)ÎLr

. ÎfÎLp⁄|µ|/r⁄(d+|µ|)(1≠1/r)

= ÎfÎLp⁄|µ|+d(1/p≠1/q).

⇤It then follows immediately that for – œ R, f œ C –, µ œ Nd

0

, we haveˆµf œ C –≠|µ|. Another simple application of the Bernstein inequalities isthe Besov embedding theorem, the proof of which we leave as an exercise.

Lemma 8. (Besov embedding) Let 1 6 p1

6 p2

6 Œ and 1 6 q1

6q

2

6 Œ, and let – œ R. Then B–p1,q1 is continuously embedded into

B–≠d(1/p1≠1/p2)

p2,q2 .

Exercise 11. In the setting of Exercise 2, use Besov embedding to showthat E[Î›Îp

≠d/2≠Á] < Œ for all p > 1 and Á > 0 (in particular › œ C ≠d/2≠

almost surely).Hint: Estimate E[Î›Î2p

B–

2p,2p

] using Gaussian hypercontractivity(equivalence of moments).

As another application of the Bernstein inequality, let us show thatC – = C– for – œ (0, 1).


Lemma 9. For – œ (0, 1) we have C – = C–, the space of –≠Höldercontinuous functions, and

ÎfÎ– ƒ ÎfÎC– = ÎfÎLŒ + supx”=y

|f(x) ≠ f(y)|dTd(x, y)–

,

where dTd(x, y) denotes the canonical distance on Td.

Proof. Start by noting that for f œ C – we have ÎfÎLŒ 6 qj Î�jfÎLŒ 6q

j 2≠j–ÎfÎ– . ÎfÎ–. Let now x ”= y œ Td and choose j0

with2≠j0 ƒ dTd(x, y). For j 6 j

0

we use Bernstein’s inequality to obtain

|�jf(x) ≠ �jf(y)| . ÎD�jfÎLŒdTd(x, y). 2jÎ�jfÎLŒdTd(x, y)6 2j(1≠–)ÎfÎ–dTd(x, y),

whereas for j > j0

we simply estimate

|�jf(x) ≠ �jf(y)| . Î�jfÎLŒ . 2≠j–ÎfÎ–.

Summing over j, we get

|f(x) ≠ f(y)| 6ÿ

j6j0

2j(1≠–)ÎfÎ–dTd(x, y) +ÿ

j>j0

2≠j–ÎfÎ–

ƒ ÎfÎ–(2j0(1≠–)dTd(x, y) + 2≠j0–) ƒ ÎfÎ–dTd(x, y)–.

Conversely, if f œ C–, then we estimate Î�≠1

fÎLŒ . ÎfÎLŒ . For j > 0,the function flj satisfies

s(F ≠1flj)(x)dx = 0, and therefore

|�jf(x)| =---⁄

Td

F ≠1flj(x ≠ y)(f(y) ≠ f(x))dy---

=---⁄

Td

ÿ

k

F ≠1

Rd

flj(x ≠ y + 2fik)(f(y) ≠ f(x))dy---

=---⁄

Rd

F ≠1

Rd

flj(x ≠ y)(f(y) ≠ f(x))dy---.

Now |f(y) ≠ f(x)| 6 ÎfÎC–dTd(x, y)– 6 ÎfÎC– |x ≠ y|–, and thus we endup with

|�jf(x)| 6 ÎfÎC–

---2jd

⁄

Rd

|(F ≠1

Rd

fl)(2j(x ≠ y))||x ≠ y|–dy---

= ÎfÎC–2≠j–---2jd

⁄

Rd

|(F ≠1

Rd

fl)(2j(x ≠ y))||2j(x ≠ y)|–dy---

. ÎfÎC–2≠j–.

⇤


The following lemma, a characterization of Besov regularity for functionsthat can be decomposed into pieces which are localized in Fourier space,will be immensely useful in what follows.

Lemma 10.

1. Let A be an annulus, let – œ R, and let (uj) be a sequence ofsmooth functions such that Fuj has its support in 2jA , and such thatÎujÎLŒ . 2≠j– for all j. Then

u =ÿ

j>≠1

uj œ C – and ÎuÎ– . supj>≠1

{2j–ÎujÎLŒ}.

2. Let B be a ball, let – > 0, and let (uj) be a sequence of smooth functionssuch that Fuj has its support in 2jB, and such that ÎujÎLŒ . 2≠j–

for all j. Then

u =ÿ

j>≠1

uj œ C – and ÎuÎ– . supj>≠1

{2j–ÎujÎLŒ}.

Proof. If Fuj is supported in 2jA , then �iuj ”= 0 only for i ≥ j. Hence,we obtain

Î�iuÎLŒ 6ÿ

j:j≥i

Î�iujÎLŒ

6 supk>≠1

{2k–ÎukÎLŒ}ÿ

j:j≥i

2≠j–

ƒ supk>≠1

{2k–ÎukÎLŒ}2≠i–.

If Fuj is supported in 2jB, then �iuj ”= 0 only for i . j. Therefore,

Î�iuÎLŒ 6ÿ

j:j&i

Î�iujÎLŒ

6 supk>≠1

{2k–ÎukÎLŒ}ÿ

j:j&i

2≠j–

. supk>≠1

{2k–ÎukÎLŒ}2≠i–,

using – > 0 in the last step. ⇤When solving SPDEs, we will need the smoothing properties of the heat

semigroup. We define L – = CC – fl C–/2LŒ for – œ (0, 2). For T > 0 weset L –

T = CT C – fl C–/2

T LŒ and we equip L –T with the norm

Î · ÎL –

T

= max{Î · ÎCT

C – , Î · ÎC

–/2T

LŒ}.


The notation L – is chosen to be reminiscent of the operator L = ˆt ≠ �and indeed the parabolic spaces L – are adapted to L in the sense thatthe temporal regularity “counts twice”, which is due to the fact that Lcontains a first order temporal but a second order spatial derivative. If wewould replace � by a fractional Laplacian ≠(≠�)‡, then we would haveto consider the space CC – fl C–/(2‡)LŒ instead of L –.

We have the following Schauder estimate on the scale of (L –)– spaces:

Lemma 11. Let – œ (0, 2) and let (Pt)t>0

be the semigroup generated bythe periodic Laplacian, F (Ptf)(k) = e≠t|k|2

Ff(k). For f œ CC –≠2 defineJf(t) =

s t

0

Pt≠sfsds. Then Jf is the solution to L Jf = f , Jf(0) = 0,and we have

ÎJfÎL –

T

. (1 + T )ÎfÎCT

C –≠2

for all T > 0. If u œ C –, then t ‘æ Ptu is the solution to L P·u = 0,P

0

u = u, and we have

Ît ‘æ PtuÎL –

T

. ÎuÎ–.

Bibliographic notes. For a gentle introduction to Littlewood–Paleytheory and Besov spaces see the recent monograph [1], where most of ourresults are taken from. There the case of tempered distributions on Rd isconsidered. The theory on the torus is developed in [31]. The Schauderestimates for the heat semigroup are classical and can be found in [14, 16].

Chapter 4

Di�usion in a randomenvironment

Let us consider the following d-dimensional homogenization problem. FixÁ > 0 and let uÁ : R

+

◊ Td æ R be the solution to the Cauchy problem

ˆtuÁ(t, x) = �uÁ(t, x) + Á≠–V (x/Á)uÁ(t, x), uÁ(0) = u

0

, (4.1)

where V : TdÁ æ R is a random field defined on the rescaled torus

TdÁ = (R/(2fiÁ≠1Z))d. This model describes the di�usion of particles in

a random medium (replacing ˆt by iˆt gives the Schrödinger equation of aquantum particle evolving in a random potential). For a review of relatedresults the reader can give a look at the recent paper of Bal and Gu [2].The limit Á æ 0 corresponds to looking at the large scale behavior of themodel since (4.1) can be understood as the equation for the macroscopicdensity uÁ(t, x) = u(t/Á2, x/Á) which corresponds to a microscopic densityu : R

+

◊ TdÁ æ R evolving according to the parabolic equation

ˆtu(t, x) = �u(t, x) + Á2≠–V (x)u(t, x), u(0, ·) = u0

(Á·).

Slightly abusing notation, we do not index u or V by Á despite the factthat they of course depend on it. We assume that V : Td

Á æ R is Gaussianand has mean zero and homogeneous correlation function CÁ given by

CÁ(x ≠ y) = E[V (x)V (y)] = (Á/Ô

2fi)dÿ

kœÁZd

eiÈx≠y,kÍR(k).

On R : Rd æ R+

we make the following hypothesis: for some — œ (0, d]we have R(k) = |k|—≠dR(k) where R œ S (Rd) is a smooth radial functionof rapid decay. For — < d it would be equivalent to require that spatialcorrelations (in the limit Á æ 0) decay as |x|≠— . For — = d this hypothesis

44

Chapter 4. Di�usion in a random environment 45

means that spatial correlations are of rapid decay. Indeed by dominatedconvergence

limÁæ0

CÁ(x) =⁄

Rd

dk

(2fi)d/2

eiÈx,kÍR(k) =⁄

Rd

dk

(2fi)d/2

eiÈx,kÍ|k|—≠dR(k)

= (2fi)d/2

!F ≠1

Rd

(| · |—≠d) ú F ≠1

Rd

(R)"

(x).

Here we applied the formula of Exercise 3, which also holds for the Fouriertransform on Rd. Now F ≠1

Rd

(R) œ S (Rd) and F ≠1

Rd

(| · |—≠d)(x) ƒ |x|≠— if0 < — < d (see for example Proposition 1.29 of [1]), so limÁæ0

|CÁ(x)| .|x|≠— for |x| æ +Œ.

Let us write VÁ(x) = Á≠–V (x/Á) so that (4.1) can be rewritten asˆtu

Á = �uÁ + VÁuÁ, and let us compute the variance of the Littlewood–Paley blocks of VÁ.

In order to perform more easily some computations we can introduce afamily of centered complex Gaussian random variables {g(k)}kœÁZ0 suchthat g(k)ú = g(≠k) and E[g(k)g(kÕ)] = ”k+kÕ

=0

and represent VÁ(x) as

VÁ(x) = Ád/2≠–

(Ô

2fi)d/2

ÿ

kœÁZd

eiÈx,k/ÁÍ

R(k)g(k).

Lemma 12. Assume — ≠ 2– > 0.We have for any Á > 0 and i > 0 andany 0 6 Ÿ 6 — ≠ 2–:

E[|�iVÁ(x)|2] . 2(2–+Ÿ)iÁŸ.

This estimate implies that if — > 2–, then for all ” > 0 we have VÁ æ 0in L2(�; B≠–≠”

2,2 (Td)) as Á æ 0.

Proof. A spectral computation gives

�iVÁ(x) = Ád/2≠–

(Ô

2fi)d/2

ÿ

kœÁZd

eiÈx,k/ÁÍfli(k/Á)

R(k)g(k)

so

E[|�iVÁ(x)|2] = Ád(Ô

2fi)≠dÁ≠2–q

kœÁZd

fli(k/Á)2R(k)= (

Ô2fi)≠dÁd≠2–

qkœÁZd

fl(k/(Á2i))2R(k). Ád≠2–2id supkœÁ2

iA R(k),(4.2)

where A is the annulus in which fl is supported. Now recall that — Æ dso that (Á2i)—≠d Ø 1 whenever Á2i 6 1, which leads to E[|�iVÁ(x)|2] .2idÁd≠2–(Á2i)—≠d = Á—≠2–2i— in that case. The assumption — ≠ 2– > 0


then implies E[|�iVÁ(x)|2] . 2(2–+Ÿ)iÁŸ for any 0 6 Ÿ 6 — ≠ 2–. In thecase Á2i > 1 we use that

sRd

R(k)dk < +Œ to estimate

Ádÿ

kœÁZd

fl(k/(Á2i))2R(k) 6 Ádÿ

kœZd

R(Ák) .⁄

Rd

R(k)dk < +Œ,

and then E[|�iVÁ(x)|2] . Á≠2– . 22–i(Á2i)Ÿ for any small Ÿ > 0. ⇤

Remark 4. Using Gaussian hypercontractivity, we get from Lemma 12that

E[|�iVÁ(x)|2p] . E[|�iVÁ(x)|2]p . 2(2–+Ÿ)piÁŸp

whenever p Ø 1, and therefore

limÁæ0

E[ÎVÁÎ2p

B≠–≠”

2p,2p

] = limÁæ0

ÿ

iØ≠1

2i(≠–≠”)2p

⁄

TE[|�iVÁ(x)|2p]dx = 0

whenever ” > 0. By the Besov embedding theorem, this shows that for allp, ” > 0

limÁæ0

E[ÎVÁÎpC ≠–≠”

] = 0.

Slightly improving the computation carried out in equation (4.2) wecan also see that if — ≠ 2– < 0, then essentially VÁ does not convergein any reasonable sense since the variance of the Littlewood–Paley blocksexplodes.

Remark 5. The same calculation as in (4.2) shows that

E[�iVÁ(x)�jVÁ(x)] = 0

whenever |i ≠ j| > 1, because in that case fliflj © 0.

The previous analysis shows that it is reasonable to take – 6 —/2 inorder to have some hope of obtaining a well defined limit as Á æ 0. In thiscase VÁ stays bounded in probability (at least) in spaces of distributionsof regularity ≠–≠. This brings us to the problem of obtaining estimatesfor the parabolic PDE

L uÁ(t, x) = (ˆt ≠ �)uÁ(t, x) = VÁ(x)uÁ(t, x), (t, x) œ [0, T ] ◊ Td,

depending only on negative regularity norms of VÁ. On one side theregularity of uÁ is then limited by the regularity of the right hand sidewhich cannot be better than that of VÁ. On the other side the productof VÁ with uÁ can cause problems since we try to multiply an (a priori)irregular object with one of limited regularity.

Assume that VÁ converges to zero in C “≠2 for “ > 0. It is thenreasonable to assume that also VÁuÁ œ CT C “≠2, uniformly in Á > 0, and


that uÁ œ CT C “ as a consequence of the regularizing e�ect of the heatoperator (Lemma 11). We will see in Section 5.1 below that the productVÁuÁ is under control only if “ + “ ≠ 2 > 0, that is if “ > 1. If VÁ æ 0 inC ≠1+, it is not di�cult to show that uÁ converges as Á æ 0 to the solutionu of the linear equation L u = 0 (for example this will follow from ouranalysis below, but in fact it is much simpler to show). In this case therandom potential will not have any e�ect in the limit.

The interesting situation then is when “ 6 1. To understand what couldhappen in this case let us use a simple transformation of the solution. WriteuÁ = exp(XÁ)vÁ where XÁ satisfies the equation L XÁ = VÁ with initialcondition XÁ(0, ·) = 0. Then

L uÁ = exp(XÁ)!vÁL XÁ + L vÁ ≠ vÁ(ˆxXÁ)2 ≠ 2ÈˆxXÁ, ˆxvÁÍRd

"

= exp(XÁ)vÁVÁ.

Since exp(XÁ) > 0 on [0, T ] ◊ Td, this implies that vÁ satisfies

L vÁ ≠ vÁ|ˆxXÁ|2 ≠ 2ÈˆxXÁ, ˆxvÁÍRd = 0, (t, x) œ [0, T ] ◊ Td.

Our Schauder estimates imply that XÁ = JVÁ œ CT C “ with uniformbounds in Á > 0, so that the problematic term is |ˆxXÁ|2 for which thisestimate does not guarantee existence.

Note that J(eiÈ·,kÍ)(t, x) = eiÈx,kÍ(1 ≠ e≠t|k|2)/|k|2, which yields

ˆxXÁ(t, x) = Ád/2≠–

(Ô

2fi)d/2

ÿ

kœÁZd

0

eiÈx,k/ÁÍGÁ(t, k)g(k) (4.3)

where Zd0

= Zd\{0} and where

GÁ(t, k) = ik

Á

[1 ≠ e≠t|k/Á|2 ]|k/Á|2

R(k).

Lemma 13. Assume that

‡2 = (Ô

2fi)d

⁄

Rd

R(k)k2

dk < +Œ.

Then if – = 1 and t > 0 we have

limÁæ0

E[|ˆxXÁ|2(t, x)] = ‡2,

and if – < 1 and t > 0

limÁæ0

E[(|ˆxXÁ|)2(t, x)] = 0.

Moreover

Var[�q(|ˆxXÁ|2)(t, x)] . Á4≠4– min(‡4, (Á2q)—≠2ÎRÎŒ‡2).


Proof. A computation similar to that leading to equation (4.2) gives

E[|ˆxXÁ|2(t, x)] = Ád(Ô

2fi)dÁ≠2–ÿ

kœÁZd

0

|k/Á|2Ë ⁄ t

0

e≠(t≠s)|k/Á|2ds

È2

R(k)

= Ád(Ô

2fi)dÁ2≠2–ÿ

kœÁZd

0

[1 ≠ e≠t(k/Á)

2 ]2k2

R(k),

which for Á æ 0, t > 0, and – Æ 1 tends to

limÁæ0

E[|ˆxXÁ|2(t, x)] = I–=1

(Ô

2fi)d

⁄

Rd

R(k)k2

dk = I–=1

‡2.

Let us now study the variance of |ˆxXÁ|2(t, x). Using equation (4.3) wehave

�q(|ˆxXÁ|2)(t, x)

= Ád≠2–

(2fi)d/2

ÿ

k1,k2œÁZd

0

eiÈk1+k2,x/ÁÍflq((k1

+ k2

)/Á)GÁ(t, k1

)GÁ(t, k2

)g(k1

)g(k2

).

By Wick’s theorem ([22], Theorem 1.28)

Cov(g(k1

)g(k2

), g(kÕ1

)g(kÕ2

)) =E[g(k1

)g(kÕ1

)]E[g(k2

)g(kÕ2

)]+ E[g(k

1

)g(kÕ2

)]E[g(k2

)g(kÕ1

)]=Ik1+kÕ

1=k2+kÕ2=0

+ Ik1+kÕ2=k2+kÕ

1=0

,

which implies

Var[�q

(|ˆx

XÁ|2)(t, x)] = 2Á2d≠4–

(2fi)d

ÿ

k1,k2œÁZd

0

(flq

((k1+k2)/Á))2|GÁ

(t, k1)|2|GÁ

(t, k2)|2.

For any q > 0 (the case q = ≠1 is left to the reader), the variables k1

andk

2

are bounded away from 0 and we have

Var[�q(|ˆxXÁ|2)(t, x)] . Á2d+4≠4–ÿ

k1,k2œÁZd

0

(flq((k1

+k2

)/Á))2

|R(k1

)||R(k2

)||k

1

|2|k2

|2 .

A first estimate is obtained by just dropping the factor flq((k1

+k2

)/Á) andresults in the bound

Var[�q(|ˆxXÁ|2)(t, x)] . Á2d+4≠4–ÿ

k1,k2œÁZd

0

|R(k1

)||R(k2

)||k

1

|2|k2

|2 . Á4≠4–‡4.


Another estimate proceeds by taking into account the constraint given bythe support of flq((k

1

+ k2

)/Á). In order to satisfy k1

+ k2

≥ Á2q we musthave k

2

. k1

≥ Á2q or Á2q . k1

≥ k2

. In the first case

Á2d+4≠4–ÿ

k1,k2œÁZd

0

Ik2.k1≥Á2

q

|R(k1

)||R(k2

)||k

1

|2|k2

|2

.2q(—≠2)Ád+—+2≠4–ÎRÎŒÿ

k2œÁZd

0

Ik2.Á2

q

|R(k2

)||k

2

|2

.(Á2q)—≠2ÎRÎŒ

⁄dk

|R(k)||k|2

.(Á2q)—≠2Á4≠4–ÎRÎŒ‡2

since |R(k1

)|/|k1

|2 . ÎRÎŒ(Á2q)—≠d≠2. If Á2q . k1

≥ k2

we similarly have

Á2d+4≠4–ÿ

k1,k2œÁZd

0

IÁ2

q.k1≥k2

|R(k1

)||R(k2

)||k

1

|2|k2

|2

.2q(—≠2)Ád+—+2≠4–ÎRÎŒÿ

k2œÁZd

0

IÁ2

q.k2

|R(k2

)||k

2

|2

.(Á2q)—≠2Á4≠4–ÎRÎŒ

⁄dk

|R(k)||k|2

.(Á2q)—≠2Á4≠4–ÎRÎŒ‡2.

⇤This lemma shows that the interesting situation is – = 1. Then,

provided ‡2 < +Œ and — > 2 we have |ˆxXÁ|2(t) æ ‡2 in L2(�; C 0≠) forall t > 0, and in fact the convergence is uniform for t œ [c, C] whenever0 < c < C. Since all the operations that appear in the equation for vÁ arecontinuous, it is then easy to see that vÁ converges to the solution of thePDE

L v = ‡2v (4.4)and since XÁ is a continuous linear functional of VÁ, we have XÁ æ 0 inCT C “ and thus we finally obtain the convergence of (uÁ)Á>0

to the samev.

Thus, we have (modulo technical details) shown the following theorem:Theorem 3. Let — œ (0, d] and let R = | · |—≠dR, where R œ S (Rd)is a smooth radial function of rapid decay, and assume that ‡2 =(Ô

2fi)dsRd

R(k)/k2dk < Œ. Let V : TdÁ æ R be a continuous Gaussian

function with mean zero and correlation

E[V (x)V (y)] = CÁ(x ≠ y) = (Á/Ô

2fi)dÿ

kœÁZd

eiÈx≠y,kÍR(k).


Consider the solution uÁ : R+

◊ Td æ R to the Cauchy problem

ˆtuÁ(t, x) = �uÁ(t, x) + Á≠–V (x/Á)uÁ(t, x), uÁ(0) = u

0

,

where u0

œ CŒ(Td). If – œ (0, 1 · —/2), then uÁ converges to the solutionu of

ˆtu(t, x) = �u(t, x), u(0) = u0

.

However, if 1 = – < —/2, then uÁ converges to the solution v of

ˆtv(t, x) = �v(t, x) + ‡2v(t, x), v(0) = u0

.

4.1 The 2d generalized parabolic Andersonmodel

The case – = 1 and — = 2 remains open in the previous analysis. When— = 2 we cannot expect ‡2 to be finite and moreover from the abovecomputations we see that the variance of |ˆxXÁ|2 remains finite and doesnot go to zero, so the limiting object should satisfy a stochastic PDE ratherthan a deterministic one. If we let ‡2

Á(t) = E[|ˆxXÁ|2(t, x)] (which dependson time but which is easily shown to be independent of x œ T2), then weexpect that solving the renormalized equation

L uÁ = VÁuÁ ≠ ‡2

Á uÁ

should give rise in the limit to a well defined random field u satisfyingu = eX v, where

L v = v’ + 2ÈˆxX, ˆxvÍRd

and where X is the limit of XÁ as Á æ 0 while ’ is the limit of(ˆxXÁ)2 ≠ ‡2

Á . The relation of uÁ with uÁ is uÁ(t, x) = e≠

st

0‡2

Á

(s)dsuÁ(t, x).

The renormalization procedure is therefore equivalent to a time–dependentrescaling of the solution to the initial problem. Without renormalization,the solution will simply drift of to +Œ, so in order to see a nontrivialbehavior, we have to put ourselves in a di�erent reference frame bymultiplying with e

≠s

t

0‡2

Á

(s)ds. One familiar situation where such a needfor renormalization arises is in the central limit theorem: If (Yn) is asequence of i.i.d. random variables with unit variance and mean µ > 0,then (n≠1/2

qnk=1

Yk) diverges to +Œ, but once we subtract the divergingconstants n1/2µ we get that (n≠1/2

qnk=1

Yk ≠ n1/2µ) converges weakly toa standard Gaussian distribution.

We will study the renormalization and convergence problem for a moregeneral equation of the form

L uÁ = F (uÁ)VÁ, (4.5)


where F : R æ R is a su�ciently smooth function, in general non–linear.One possible motivation is that if zÁ solves the linear PDE L zÁ = zÁVÁ

and we set uÁ = Ï(zÁ) for some invertible Ï : R æ R such that ÏÕ > 0,then

L uÁ = ÏÕ(zÁ)L zÁ ≠ÏÕÕ(zÁ)|ˆxzÁ|2 = ÏÕ(zÁ)zÁVÁ ≠ÏÕÕ(zÁ)(ÏÕ(zÁ))≠2|ˆxuÁ|2

and thus uÁ satisfies the PDE

L uÁ = F1

(uÁ)VÁ + F2

(uÁ)(ˆxuÁ)2

where

F1

(x) = ÏÕ(Ï≠1(x))Ï≠1(x) and F2

(x) = ≠ÏÕÕ(Ï≠1(x))(ÏÕ(Ï≠1(x)))≠2.

In the situation we are interested in, the second term in the right hand sideis simpler to treat than the first term. So, for the time being, we will dropit and we will concentrate on the equation (4.5) in d = 2 with – = 1 andshort ranged (— = d) potential V which we refer to as generalized parabolicAnderson model (gpam).

Under these conditions, VÁ converges to the white noise in space whichwe usually denote with › and our aim will be to set up a theory in whichthe operations involved in the definition of the dynamics of the gpam arewell defined, including the possibility of the renormalization which alreadyappears in the linear case as hinted above.

While the reader should always have in mind a limiting procedure from awell defined model like the ones we were considering so far, in the followingwe will mostly discuss the limiting equation. The specific phenomenaappearing when trying to track the oscillations of the term F (uÁ)VÁ asÁ æ 0 will be described by a renormalized product F (u) ù › and so wewrite the gpam as

L u(t, x) = F (u(t, x)) ù ›(x), u(0) = u0

. (4.6)

In the linear case F (u) = u, the problem of the renormalization can besolved along the lines suggested above. Another possible line of attackcomes from the theory of Gaussian spaces and in particular from Wickproducts, see for example [21]. However, the definition of the Wick productrelies on the concrete chaos expansion of its factors, and since nonlinearfunctions change the chaos expansion in a complicated way, there is littlehope of directly extending the Wick product approach to the nonlinearcase and moreover using these non–local (in the probability space) objectscan deliver solutions which are not physically acceptable [5].

Equation (4.6) is structurally very similar to the stochastic di�erentialequation

ˆtv(t) = F (v(t))ˆtBH(t), v(0) = v

0

, (4.7)


where BH denotes a fractional Brownian motion with Hurst index H œ(0, 1). There are many ways to solve (4.7) in the Brownian case. Since weare interested in a way that might extend to (4.6) where the irregularityappears along the two–dimensional spatial variable x, we should excludeall approaches based on information, filtrations, and a direction of time; inparticular, any approach that works for H ”= 1/2 might seem promising.But Lyons’ theory of rough paths [25] equips us exactly with the techniqueswe need to solve (4.7) for general H. More precisely, if for H > 1/3 we aregiven

s ·0

BHs dBH

s , then we can use the controlled rough path integral [12] tomake sense of

s ·0

fsdBHs for any f which “looks like” BH , and this allows us

to solve (4.7). So the main ingredients required for controlled rough pathsare the integral

s ·0

BHs dBH

s for the reference path BH , and the fact thatwe can describe paths which look like BH . It is worthwhile to note thatwhile we need probability theory to construct

s ·0

BHs dBH

s , the constructionof

s ·0

fsdBHs is achieved using pathwise arguments and it is given as a

continuous map of f and (BH ,s ·

0

BHs dBH

s ). As a consequence, the solutionto the SDE (4.7) depends pathwise continuously on (BH ,

s ·0

BHs dBH).

By the structural similarity of (4.6) and (4.7), we might hope to extendthe rough path approach to (4.6). The equivalent of BH is given by thesolution Ë to L Ë = ›, Ë(0) = 0, and the equivalent of

s ·0

BHs dBH

s turnsout to be the renormalized product Ë ù ›. Then we might hope that givenË ù › we are able to define f ù › for all f that “look like Ë”, however this isto be interpreted. Of course, rough paths can only be applied to functionsof a one–dimensional index variable, while for (4.6) the problem lies in theirregularity of › in the spatial variable x œ T2.

In the following we combine the ideas from controlled rough paths withBony’s paraproduct, a tool from functional analysis that allows us toextend rough paths to functions of a multidimensional parameter. Usingthe paraproduct, we are able to make precise in a simple way what wemean by “distributions looking like a reference distribution”. We can thendefine products of suitable distributions and solve (4.6) as well as manyother interesting singular SPDEs.

4.2 More singular problemsKeeping the homogenization problem as leitmotiv for these lectures,we could consider also space–time varying environments VÁ(t, x) =Á≠–V (t/Á2, x/Á). The scaling of the temporal variable is chosen so thatit is compatible with the di�usive scaling from a microscopic description,where V (t, x) has typical variation in space and time in scales of order1. Assume that d = 1, then when the random field V is Gaussian, zeromean, and with short–range space–time correlations, the natural choicefor the magnitude of the macroscopic fluctuations is – = 3/2. In this


case VÁ converges as Á æ 0 to a space–time white noise ›. Understandingthe limit dynamics as Á æ 0 of the solution uÁ to the linear equationL uÁ = VÁuÁ represents now a more di�cult problem than in the timeindependent situation. A Gaussian computation shows that the randomfield XÁ, solution to L XÁ = VÁ (e.g. with zero initial condition), staysbounded in CT C 1/2≠ as Á æ 0. Since L is a second order operator (if weuse an appropriate parabolic weighting of the time and space regularities),› is expected to live in a space of distributions of regularity ≠3/2≠. Thisis to be compared with the ≠1≠ of the space white noise which had tobe dealt with in the gpam. Renormalization e�ects are then expected tobe stronger in this setting and the limiting object, which we denote withw, should satisfy a (suitably renormalized) linear stochastic heat equationwith multiplicative noise (she)

L w(t, x) = w(t, x) ù ›(t, x), w(0) = w0

. (4.8)

As indicated by the computations in the more regular case, it is usefulto consider the change of variables w = eh which is called Cole–Hopftransformation. Here h : [0, Œ) ◊T æ R is a new unknown which satisfiesnow the Kardar–Parisi–Zhang (kpz) equation:

L h(t, x) = (ˆxh(t, x)) ù 2 + ›(t, x), h(0) = h0

(4.9)

where the di�culty comes now from the squaring of the derivative butwhich has the nice feature to be additively perturbed by the space–timewhite noise, a feature which simplifies many considerations. Anotherrelevant model in applications is obtained by taking the space derivativeof kpz and letting u(t, x) = ˆxh(t, x) in order to obtain the stochasticconservation law

L u(t, x) = ˆx(u(t, x)) ù 2 + ˆx›(t, x), u(0) = u0

, (4.10)

which we will refer to as the stochastic Burgers equation (sbe). In all thesecases, ù denotes a suitably renormalized product.

The kpz equation was derived by Kardar–Parisi–Zhang in 1986 as auniversal model for the random growth of an interface [24]. For a longtime it could not be solved due to the fact that there was no way to makesense of the nonlinearity (ˆxh) ù 2 in (4.9). The only way to make senseof kpz was to apply the Cole-Hopf transform [3]: solve she (4.8) (whichis accessible to Itô integration) and set h = log w. But there was nointrinsic interpretation of what it means to solve (4.9). Finally, in 2011Hairer [18] used rough paths to give a meaning to the equation and toobtain solutions directly at the kpz level. In Section 6 we will sketchhow to recover his solution in the paracontrolled setting. Applicationsof the techniques used by Hairer to solve the kpz problem to a moregeneral homogenization problem with ergodic potentials (not necessarilyGaussian) have been studied in [20].


4.3 Hairer’s regularity structuresIn [19], Hairer introduces a theory of regularity structures which can alsobe considered a generalization of the theory of controlled rough pathsto functions of a multidimensional index variable. Hairer fundamentallyrethinks the notion of regularity. Usually a function is called smooth ifit can be approximated around every point by a polynomial of a givendegree (the Taylor polynomial). Naturally, the solution to an SPDEdriven by –say– Gaussian space-time white noise is not smooth in thatsense. So in Hairer’s theory, a function is called smooth if locally it can beapproximated by the noise (and higher order terms constructed from thenoise). This induces a natural topology in which the solutions to semilinearSPDEs depend continuously on the driving signal.

At this date it seems that the theory of regularity structures has a widerrange of applicability than the paracontrolled approach described in [14],but also at the expense of a very deep conceptual sophistication. Thereare problems (like the one–dimensional heat equation with multiplicativenoise and general nonlinearity) that cannot be solved using paracontrolleddistributions, but these problems seem also quite di�cult (even if doableand there is work in progress) to tackle with regularity structures.Moreover, equations of a more general kind, say dispersive equationsor wave equations, are still poorly (or not at all) understood in bothapproaches.

Chapter 5

The paracontrolled PAM

As we have tried to motivate in the previous sections we are looking for atheory for pam which describes the possible limits of the equation

L u = F (u)÷ (5.1)

driven by su�ciently regular ÷ but as ÷ is converging to the space whitenoise ›. From this point of view we are looking for a priori estimates onthe solution u to (5.1) which depend only on distributional norms of ÷.So in the following we will assume that we have at hand only a uniformcontrol of ÷ in CT C “≠2 for some “ > 0. For the application to the 2dspace white noise we could take “ = 1≠, but we will not use this specificinformation in order to probe the range of applicability of our approachand we will only assume that the exponent “ is such that 3“ ≠ 2 > 0.

Assume for a moment that we are in the simpler situation “ > 1 andu

0

œ C “ and let us try to solve equation (5.1) via Picard iterations (un)n>0

starting from u0 © u0

. Since F preserves the CC “-regularity (which canbe seen by identifying CC “ with the classical space of bounded Hölder–continuous functions of space), the product F (u0(t))÷ is well defined as anelement of C “≠2 for all t > 0 since 2“ ≠ 2 > 0 and we are in conditionto apply Corollary 1 below on the product of elements in Hölder–Besovspaces. Now by Lemma 11, the heat semigroup generated by the Laplaciangains two degrees of regularity so that the solution u1 to L u1 = F (u0)÷,u1(0) = u

0

, is in CC “ . From here we obtain a contraction on CT C “ forsome small T > 0 whose value does not depend on u

0

, which gives usglobal in time existence and uniqueness of solutions. Note that in onedimension the space white noise has regularity C ≠1/2≠ (see Exercise 11)so taking “ = 3/2≠ we have determined that the one–dimensional pamcan be solved globally in time with standard techniques.

55


When the condition 2“ ≠ 2 > 0 is not satisfied we still have thatif ÷ œ CT C “≠2 then u œ L “ = CT C “≠2 fl C

“/2

T LŒ by the standardparabolic estimates of Lemma 11. However with the regularities at hand wecannot use Corollary 1 anymore to guarantee the continuity of the operator(u, ÷) ‘æ F (u)÷. Moreover, as already seen in the simpler homogenizationproblems of Theorem 3 above, this is not a technical di�culty but a realissue of the regime “ 6 1. We expect that controlling the model in thisregime can be quite tricky since limits exists when ÷ æ 0 but the limitingsolution still feels residual order one e�ects from the vanishing drivingsignal ÷. This situation cannot be improved from the point of view ofstandard analytic considerations. What is needed is a finer control of thesolution u which allows to analyse in more detail the possible resonancesbetween the fluctuations of u and those of ÷.

Before going on we will revise the problem of multiplication ofdistributions in the scale of Hölder–Besov spaces, introducing the basictool of our general analysis: Bony’s paraproduct.

5.1 The paraproduct and the resonant term

Paraproducts are bilinear operations introduced by Bony [4] to linearizea class of nonlinear hyperbolic PDEs in order to analyse the regularity oftheir solutions. In terms of Littlewood–Paley blocks, a general product fgof two distributions can be (at least formally) decomposed as

fg =ÿ

j>≠1

ÿ

i>≠1

�if�jg = f ª g + f º g + f ¶ g.

Here f ª g is the part of the double sum with i < j ≠ 1, f º g is the partwith i > j + 1, and f ¶ g is the “diagonal” part, where |i ≠ j| 6 1. Moreprecisely, we define

f ª g = g º f =ÿ

j>≠1

j≠2ÿ

i=≠1

�if�jg and f ¶ g =ÿ

|i≠j|61

�if�jg.

Of course, the decomposition depends on the dyadic partition of unity usedto define the blocks �j , and also on the particular choice of the pairs (i, j)in the diagonal part. The choice of taking all (i, j) with |i ≠ j| 6 1 intothe diagonal part corresponds to the fact that the partition of unity canbe chosen such that supp F (�if�jg) ™ 2jA if i < j ≠ 1, where A is asuitable annulus. If |i≠j| 6 1, the only apriori information on the spectralsupport of the various term in the double sum is supp F (�if�jg) ™ 2jB,

Chapter 5. The paracontrolled PAM 57

that is they are supported in balls and in particular they can have non–zero contributions to very low wave vectors. We call f ª g and f º gparaproducts, and f ¶ g the resonant term.

Bony’s crucial observation is that f ª g (and thus f º g) is always a well-defined distribution. Heuristically, f ª g behaves at large frequencies likeg (and thus retains the same regularity), and f provides only a frequencymodulation of g. The only di�culty in constructing fg for arbitrarydistributions lies in handling the diagonal term f ¶ g. The basic resultabout these bilinear operations is given by the following estimates.

Theorem 4. (Paraproduct estimates) For any — œ R and f, g œ S Õ wehave

Îf ª gÎ— .— ÎfÎLŒÎgÎ— , (5.2)

and for – < 0 furthermore

Îf ª gÎ–+— .–,— ÎfÎ–ÎgÎ— . (5.3)

For – + — > 0 we have

Îf ¶ gÎ–+— .–,— ÎfÎ–ÎgÎ— . (5.4)

Proof. There exists an annulus A such that Sj≠1

f�jg has Fouriertransform supported in 2jA , and for f œ LŒ we have

ÎSj≠1

f�jgÎLŒ 6 ÎSj≠1

fÎLŒÎ�jgÎLŒ . ÎfÎLŒ2≠j—ÎgÎ— .

By Lemma 10, we thus obtain (5.2). The proof of (5.3) and (5.4) works inthe same way, where for estimating f ¶ g we need – + — > 0 because theterms of the series are supported in a ball and not in an annulus. ⇤

In combination with Exercise 10 above, we deduce the following simplecorollary:

Corollary 1. Let f œ C – and g œ C — with – + — > 0, then the product(f, g) ‘æ fg is a bounded bilinear map from C – ◊ C — to C –·—. Whilef ª g, f º g, and f ¶ g depend on the specific dyadic partition of unity, theproduct fg does not.

The independence of the product from the dyadic partition of unityeasily follows by taking smooth approximations.

The ill–posedness of f ¶ g for –+— 6 0 can be interpreted as a resonancee�ect since f ¶ g contains exactly those part of the double series where fand g are in the same frequency range. The paraproduct f ª g can beinterpreted as frequency modulation of g, which should become more clearin the following example.


Example 3. In Figure 5.1 we see a slowly oscillating positive function u,while Figure 5.2 depicts a fast sine curve v. The product uv, which hereequals the paraproduct u ª v since u has no rapidly oscillating components,is shown in Figure 5.3. We see that the local fluctuations of uv are due tov, and that uv is essentially oscillating with the same speed as v.

Figure 5.1: The function u Figure 5.2: The function v

Figure 5.3: The function u ª v

Example 4. If f œ C “(T) and g œ C ”(T) with “ + ” > 1, then wecan define

sfdg :=

s(fˆtg), which is well defined since ˆtg œ C ”≠1 and

“ +” ≠1 > 0, and since integration is a linear map. In this way we recoverthe Young integral [32].

Example 5. Let BH be a fractional Brownian bridge on T (or simply afractional Brownian motion on [0, fi], reflected on [fi, 2fi]) and assume thatH > 1/2. We have Ï(BH) œ C H≠ for all Lipschitz continuous Ï, andˆtB

H œ C (H≠1)≠, and in particular Ï(BH)ˆtBH is well-defined. This can

be used to solve SDEs driven by BH in a pathwise sense.

The condition – + — > 0 is essentially sharp, at least at this level ofgenerality, see [32] for counterexamples. It excludes of course the Browniancase: if B is a Brownian motion, then almost surely B œ C –

loc

for all– < 1/2 (meaning that ÏB œ C – whenever Ï is a smooth compactlysupported function), so that ˆtB œ C –≠1

loc

and thus B ¶ ˆtB fails to be welldefined. See also [26], Proposition 1.29 for an instructive example whichshows that this is not a shortcoming of our description of regularity, butthat it is indeed impossible to define the product BˆtB as a continuousbilinear operation on distribution spaces.

Other counterexamples are given by our discussion of the homogeniza-tion problem in Theorem 3 above. More simply, one can consider thefollowing situation.

Example 6. Consider the sequence of functions fn : T æ C givenbyfn(x) = ein2x/n. Then it is easy to show that ÎfnÎ“ æ 0 for all “ < 1/2.


However let

gn(x) = Re fn(x) Im ˆxfn(x) = (cos(n2x))2 = cos(2n2x) + 12

Then gn æ 1/2 in C 0≠ which shows that the map f ‘æ (Re f)(ˆx Im f)cannot be continuous in C “ if “ < 1/2. Pictorially the situation issummarized in Figure 5.4, where we sketched the three dimensional curvegiven by x ‘æ (Re fn(x), Im fn(x),

s x

0

gn(y)dy) for various values of n andin the limit.

Figure 5.4: Resonances give macroscopic e�ects

5.2 Commutator estimates and paralin-earization

The product F (u)÷ appearing in the right hand side of pam can bedecomposed via the paraproduct ª as a sum of three terms

F (u)÷ = F (u) ª ÷ + F (u) ¶ ÷ + F (u) º ÷.

The first and the last of these terms are continuous in any topology we willchoose for F (u) and ÷. The resonant term F (u) ¶ ÷ however is problematic.It gathers the products of the oscillations of F (u) and ÷ on comparabledyadic scales and these products can contribute to all larger scales insuch a way that microscopic oscillations might build up to a macroscopice�ect which does not disappear in the limit (as we have already seen inTheorem 3). If the function F is smooth enough, then we expect theresonances between F (u) and ÷ to correspond to the resonances betweenu and ÷, and as we will see this is justified.

The expected regularity of the di�erent terms is

F (u) ª ÷¸ ˚˙ ˝“≠2

+ F (u) ¶ ÷¸ ˚˙ ˝2“≠2

+ F (u) º ÷¸ ˚˙ ˝2“≠2

, (5.5)

but unless 2“ ≠ 2 > 0 the resonant term F (u) ¶ ÷ cannot be controlledusing only the CC “–norm of u and the CC “≠2–norm of ÷. If F is at least


C2, we can use a paralinearization result (stated precisely in Lemma 16below) to rewrite this term as

F (u) ¶ ÷ = F Õ(u)(u ¶ ÷) + �F (u, ÷), (5.6)

with a remainder �F (u, ÷) œ C 2“≠2 provided 3“ ≠ 2 > 0. The di�culty isnow localized in the linearized resonant product u ¶ ÷. In order to controlthis term, we would like to exploit the fact that the function u is not ageneric element of CC “ but that it has a specific structure, since L u hasto match the paraproduct decomposition given in (5.5) where the leastregular term is expected to be F (u) ª ÷ œ CC “≠2.

In order to do so, we postulate that the solution u is given by thefollowing paracontrolled ansatz:

u = uX ª X + u˘,

for functions uX , X, u˘ such that uX , X œ CC “ and the remainderu˘ œ CC 2“ . This decomposition allows for a finer analysis of the resonantterm u ¶ ÷: indeed, we have

u ¶ ÷ = (uX ª X) ¶ ÷ + u˘ ¶ ÷ = uX(X ¶ ÷) + C(uX , X, ÷) + u˘ ¶ ÷, (5.7)

where the commutator is defined by C(uX , X, ÷) = (uX ª X) ¶ ÷ ≠uX(X ¶ ÷). Observe now that the term u˘ ¶ ÷ does not pose any furtherproblem, as it can be controlled in CC 3“≠2. The key point is now that thecommutator is a bounded multilinear function of its arguments as long asthe sum of their regularities is strictly positive, see Lemma 14 below. Byassumption, we have 3“ ≠ 2 > 0, and therefore C(uX , X, ÷) œ CC 2“≠2.

The only problematic term which remains to be handled is thus thebilinear functional of the noise given by X ¶ ÷. Here we need to makethe assumption that X ¶ ÷ œ CC 2“≠2 in order for the product uX(X ¶ ÷)to be well defined. This assumption is not guaranteed by the analyticalestimates at hand, and it has to be added as a further requirement to ourconstruction.

Granting this last step, we have obtained that the right handside of equation (5.1) is well defined and a continuous function of(u, uX , u˘, X, ÷, X ¶ ÷) œ CC “ ◊ CC “ ◊ CC 2“ ◊ CC “ ◊ CC “≠2 ◊ CC 2“≠2.

It remains to check that the paracontrolled ansatz is coherent with theequation satisfied by solutions to pam. Let us first consider the linearexample F (u) = u. Here we saw that the solution is of the form u = eXvwith

L v = v|ˆxX|2 + 2Èˆxv, ˆxXÍR2 ,

where |ˆxX|2 œ CC 2“≠2 by Lemma 13 and ˆxX œ CC “≠1 and thereforev œ CC 2“ by the Bony and Schauder estimates. Note that here we have a


clash of notation, because a priori the X that we defined in Section 4 doesnot have to be equal to the paracontrolling distribution X. But of course,as the notation suggests, we will see momentarily that we can choose themto be the same. In the setting of Section 4, we have in particular

u = eXv = v ª eX + CC 2“ = v ª (eX ª X) + CC 2“ ,

where the notation u = v ª eX + CC 2“ means that u ≠ v ª eX œ CC 2“ ,and where we used a paralinearization result in last step (see Lemma 15below). Now the double paraproduct f ª (g ª h) satisfies

Îf ª (g ª h) ≠ (fg) ª hÎ–+— . ÎfÎ–ÎgÎ–ÎhÎ— ,

see [4], and therefore u = (veX) ª X +CC 2“ = u ª X +CC 2“ which showsthat the paracontrolled ansatz is at least justified in the linear case andindeed we can choose the paracontrolling distribution to be X.

In the nonlinear case, the paracontrolled ansatz and the Leibniz rule forthe paraproduct imply that (5.1) can be rewritten as

L u = L (uX ª X + u˘)= uX ª L X + [L , uX ª ]X + L u˘

= F (u) ª ÷ + F (u) ¶ ÷ + F (u) º ÷,

where we recall that [L , uX ª ]X = L (uX ª X) ≠ uX ª L X denotes thecommutator. If we choose X such that L X = ÷ and we set uX = F (u),then we can use (5.6) and (5.7) to obtain the following equation for theremainder u˘:

L u˘ = F Õ(u)F (u)(X ¶ ÷) + F (u) º ÷ ≠ [L , F (u) ª ]X+F Õ(u)C(F (u), X, ÷) + F Õ(u)(u˘ ¶ ÷) + �F (u, ÷). (5.8)

Lemma 18 below ensures that J [L , F (u) ª ]X œ CC 2“ wheneverF (u) œ L “ (which easily follows from u œ L “ by using the incrementcharacterization of C “ regularity), and combining the paraproductestimates with the estimates for C and �F that we discussed above,we see that all the other terms on the right hand side are in CC 2“≠2.So the Schauder estimate Lemma 11 allows us to control u˘ in CC 2“ .Together with u = F (u) ª X + u˘, equation (5.8) gives an equivalentdescription of the solution, because we only rewrote the original problem.This allows us to obtain a priori estimates on u and u˘ in terms of(u

0

, Î÷Î“≠2

, ÎX ¶ ÷Î2“≠2

), see Chapter 5 of [14] for details. It is nowstraightforward to show that if F œ C3

b , then u depends continuouslyon the data (u

0

, ÷, X ¶ ÷), so that we have a robust strategy to pass to thelimit in (4.5) and to make sense of the solution to (5.1) also for irregular÷ œ CC “≠2 as long as “ > 2/3.


In the remainder of this section we will prove the results(paralinearization and various key commutators) which we used in thediscussion above, before going on to gather the consequences of our analysisin the next section. When the time dependence does not play any role westate the results for distributions depending only on the space variable asthe extension to time varying functions will not add further di�culty.

Lemma 14. Assume that –, —, “ œ R are such that – + — + “ > 0 and— + “ ”= 0. Then for f, g, h œ CŒ the trilinear operator

C(f, g, h) = ((f ª g) ¶ h) ≠ f(g ¶ h)

satisfiesÎC(f, g, h)Î—+“ . ÎfÎ–ÎgÎ—ÎhÎ“ , (5.9)

and can thus be uniquely extended to a bounded trilinear operator fromC –◊C — ◊C – to C —+“ .

Proof. For — + “ > 0 this follows from the paraproduct estimates, so let— + “ < 0. By definition

C(f, g, h) =ÿ

i,j,k,¸

�i(�jf�kg)�¸h(Ij<k≠1

I|i≠¸|61

≠ I|k≠¸|61

)

=ÿ

i,j,k,¸

�i(�jf�kg)�¸h(Ij<k≠1

I|i≠¸|61

I|k≠¸|6N ≠ I|k≠¸|61

),

where we used that F (Sk≠1

f�kg) has support in an annulus 2kA , sothat �i(Sk≠1

f�kg) ”= 0 only if |i ≠ k| 6 N ≠ 1 for some fixed N œ N,which in combination with |i ≠ ¸| 6 1 yields |k ≠ ¸| 6 N . Now theassumptions on our partition of unity guarantee that for fixed k, theterm

q¸ I26|k≠¸|6N �kg�¸h is spectrally supported in an annulus 2kA ,

so thatq

k,¸ I26|k≠¸|6N �kg�¸h œ C —+“ and we may add and subtractf

qk,¸ I26|k≠¸|6N �kg�¸h to C(f, g, h) while maintaining the bound (5.9).

It remains to treatÿ

i,j,k,¸

�i(�jf�kg)�¸hI|k≠¸|6N (Ij<k≠1

I|i≠¸|61

≠ 1)

= ≠ÿ

i,j,k,¸

�i(�jf�kg)�¸hI|k≠¸|6N (Ij>k≠1

+ Ij<k≠1

I|i≠¸|>1

). (5.10)

We estimate both terms on the right hand side separately. For m > ≠1we have (recall that for indices of Littlewood–Paley blocks, i . j is to be


read as 2i . 2j , that is i Æ j + c for some fixed c):...�m

1 ÿ

i,j,k,¸

�i(�jf�kg)�¸hI|k≠¸|6N Ij>k≠1

2...LŒ

6ÿ

j,k,¸

I|k≠¸|6N Ij>k≠1

Î�m(�jf�kg�¸h)ÎLŒ

.ÿ

j&m

ÿ

k.j

2≠j–ÎfÎ–2≠k—ÎgÎ—2≠k“ÎhÎ“

.ÿ

j&m

2≠j(–+—+“)ÎfÎ–ÎgÎ—ÎhÎ“ . 2≠m(–+—+“)ÎfÎ–ÎgÎ—ÎhÎ“ ,

using — + “ < 0 to getq

k.j 2k(—+“) . 2j(–+—). It remains to estimate thesecond term in (5.10). For |i ≠ ¸| > 1 and i ≥ k ≥ ¸, any term of the form�i(·)�¸(·) is spectrally supported in an annulus 2¸A , and therefore

...�m

1 ÿ

i,j,k,¸

�i(�jf�kg)�¸hI|k≠¸|6N Ij<k≠1

I|i≠¸|>1

2...LŒ

.ÿ

i,j,k,¸

Ij<k≠1

Ii≥k≥¸≥mÎ�i(�jf�kg)�¸hÎLŒ

.ÿ

j.m

2≠j–ÎfÎ–2≠m—ÎgÎ—2≠m“ÎhÎ“ . 2≠m(—+“)ÎfÎ–ÎgÎ—ÎhÎ“ .

⇤

Remark 6. For — + “ = 0 we can apply the commutator estimate with“Õ < “, as long as – + — + “Õ > 0.

Our next result is a simple paralinearization lemma for non–linearoperators.

Lemma 15 (see also [1], Theorem 2.92). Let – œ (0, 1), — œ (0, –],and let F œ C

1+—/–b . There exists a locally bounded map RF : C – æ C –+—

such thatF (f) = F Õ(f) ª f + RF (f) (5.11)

for all f œ C –. More precisely, we have

ÎRF (f)Î–+— . ÎFÎC

1+—/–

b

(1 + ÎfÎ1+—/–– ).

If F œ C2+—/–b , then RF is locally Lipschitz continuous:

ÎRF (f) ≠ RF (g)Î–+— . ÎFÎC

2+—/–

b

(1 + ÎfÎ– + ÎgÎ–)1+—/–Îf ≠ gÎ–.


Remark 7. Since every element of C – is bounded, the result immediatelyextends to unbounded F œ C1+—/–: Simply replace F by an element ofC

1+—/–b which agrees with F on the image of f .

Proof. [Proof of Lemma 15] The di�erence F (f) ≠ F Õ(f) ª f is given by

RF (f) = F (f) ≠ F Õ(f) ª f =ÿ

i>≠1

[�iF (f) ≠ Si≠1

F Õ(f)�if ] =ÿ

i>≠1

ui,

and every ui is spectrally supported in a ball 2iB. For i < 1, we simplyestimate ÎuiÎLŒ . ÎFÎC1

b

(1 + ÎfÎ–). For i > 1 we use the fact thatf is a bounded function to write the Littlewood–Paley projections asconvolutions and obtain

ui

(x)

=⁄

Ki

(x ≠ y)K<i≠1(x ≠ z)[F (f(y)) ≠ F Õ(f(z))f(y)]dydz

=⁄

Ki

(x ≠ y)K<i≠1(x ≠ z)[F (f(y)) ≠ F (f(z)) ≠ F Õ(f(z))(f(y) ≠ f(z))]dydz,

where Ki = F ≠1fli, K<i≠1

=q

j<i≠1

Kj , and where we used thatsKi(y)dy = fli(0) = 0 for i > 0 and

sK<i≠1

(z)dz = 1 for i > 1. Nowwe can apply a first order Taylor expansion to F and use the —/––Höldercontinuity of F Õ in combination with the ––Hölder continuity of f , todeduce

|ui

(x)|

.ÎF ÎC

1+—/–

b

ÎfÎ1+—/–

–

⁄|K

i

(x ≠ y)K<0(x ≠ z)| ◊ |z ≠ y|–+—dydz

=ÎF ÎC

1+—/–

b

ÎfÎ1+—/–

–

2≠(i≠1)(–+—)

◊⁄

|2(i≠1)dK1(2i≠1(x ≠ y))2(i≠1)dK<0(2i≠1(x ≠ z))| ◊ |2i≠1(z ≠ y)|–+—dydz

.ÎF ÎC

1+—/–

b

ÎfÎ1+—/–

–

2≠i(–+—).

Therefore, the estimate for RF (f) follows from Lemma 10. The estimatefor RF (f) ≠ RF (g) is shown in the same way. ⇤

Let g be a distribution belonging to C — for some — < 0. Then the mapf ‘æ f ¶ g behaves, modulo smoother correction terms, like a derivativeoperator:

Lemma 16. Let – œ (0, 1), — œ (0, –], “ œ R be such that – + — + “ > 0and – + “ ”= 0. Let F œ C

1+—/–b . Then there exists a locally bounded map

�F : C – ◊ C “ æ C –+“ such that

F (f) ¶ g = F Õ(f)(f ¶ g) + �F (f, g) (5.12)


for all f œ C – and all smooth g. More precisely, we have

Î�F (f, g)Î–+“ . ÎFÎC

1+—/–

b

(1 + ÎfÎ1+—/–– )ÎgÎ“ .

If F œ C2+—/–b , then �F is locally Lipschitz continuous:

Î�F (f, g) ≠ �F (u, v)Î–+“

.ÎFÎC

2+—/–

b

(1 + ÎfÎ– + ÎuÎ–)1+—/–(1 + ÎvÎ“)(Îf ≠ uÎ– + Îg ≠ vÎ“).

Proof. Use the paralinearization and commutator lemmas above todeduce that

�F (f, g) = F (f) ¶ g ≠ F Õ(f)(f ¶ g)= RF (f) ¶ g + (F Õ(f) ª f) ¶ g ≠ F Õ(f)(f ¶ g)= RF (f) ¶ g + C(F Õ(f), f, g),

so that the claimed bounds easily follow from Lemma 14 and Lemma 15.⇤

Besides this sort of chain rule, we also have a Leibniz rule for f ‘æ f ¶ g:

Lemma 17. Let – œ (0, 1) and “ < 0 be such that 2–+“ > 0 and –+“ ”= 0.Then there exists a bounded trilinear operator �◊ : C –◊C –◊C “ æ C –+“ ,such that

(fu) ¶ g = f(u ¶ g) + u(f ¶ g) + �◊(f, u, g)for all f, u œ C –(R) and all smooth g.

Proof. It su�ces to note that fu = f ª u + f º u + f ¶ u, which leads to

�◊(f, u, g) = (fu) ¶ g≠f(u ¶ g)≠u(f ¶ g) = C(f, u, g)+C(u, f, g)+(f ¶ u) ¶ g.

⇤Lemma 18. Let — < 1, – œ R, and let f œ L — and G œ CC – withL G œ CC –≠2. There exists H = H(f, G) such that L H = [L , f ª ]Gand H(0) = 0. Moreover H œ CC –+— fl C(–·—)/2LŒ and for all T > 0

ÎHÎC

(–·—)/2T

LŒ + ÎHÎCT

C –+— . ÎfÎL —

T

(ÎGÎCT

C – + ÎL GÎCT

C –≠2).

Proof. Let T > 0 and let fÁ be a time mollification of f such thatÎˆtfÁÎC

T

LŒ . Á—/2≠1ÎfÎL —

T

and ÎfÁ ≠fÎCT

LŒ . Á—/2ÎfÎL — for all Á > 0.For example we can take fÁ = flÁ ú f with flÁ(t) = fl(t/Á)/Á and fl : R æ Rcompactly supported, smooth, and of unit integral. For i > ≠1 we have

L �i

H = �i

L H

= �i

[L ((f ≠ fÁ

) ª G) ≠ (f ≠ fÁ

) ª L G] + �i

[L (fÁ

ª G) ≠ fÁ

ª L G] ,


so that

L �i(H ≠ (f ≠ fÁ) ª G) = ≠�i [(f ≠ fÁ) ª L G] + �i [L (fÁ ª G) ≠ fÁ ª L G]= �i [(fÁ ≠ f) ª L G] + �i [L fÁ ª G ≠ 2ˆxfÁ ª ˆxG] ,

with initial condition �i(H ≠ (f ≠ fÁ) ª G)(0) = ≠(�i(f ≠ fÁ) ª G)(0).The Schauder estimates for L (Lemma 11) give

Î�i(H + (f ≠ fÁ) ª G)ÎL –+—

T

. Î�i [(f ≠ fÁ) ª L G] + �i [(L fÁ) ª G ≠ 2ˆxfÁ ª ˆxG]ÎCT

C –+—≠2

+ Î(�i(f ≠ fÁ) ª G)(0)Î–+— .

Choosing Á = 2≠2i, we have

Î�i((f ≠ fÁ) ª G)ÎCT

C –+— . 2—iÎ�i((f ≠ fÁ) ª G)ÎCT

C –

. 2—iÎf ≠ fÁÎCT

LŒÎGÎCT

C –

. ÎfÎL —

T

ÎGÎCT

C –

and exactly the same argument also gives

Î�i [(f ≠ fÁ) ª L G]ÎCT

C –+—≠2 . ÎfÎL —

T

ÎL GÎCT

C –≠2 .

Since — < 1, we further get

Î�i [L fÁ ª G + ˆxfÁ ª ˆxG]ÎCT

C –+—≠2

.2i(—≠2)ÎˆtfÁÎCT

LŒÎGÎCT

C – + ÎfÁÎCT

C — ÎGÎCT

C –

.ÎfÎL —

T

ÎGÎCT

C – + ÎfÎCT

C — ÎGÎCT

C – .

Combining everything, we end up with

Î�iHÎCT

C –+— . ÎfÎL —

T

(ÎGÎCT

C – + ÎL GÎCT

C –≠2),

which gives the estimate for the space regularity of H since Î�iHÎCT

LŒ .2≠(–+—)iÎ�iHÎC

T

C –+— . The time regularity of H can be controlledsimilarly by noting that (f ≠ fÁ) ª G œ C

(–·—)/2

T LŒ, uniformly in Á. ⇤

5.3 Paracontrolled distributionsHere we build a calculus of distributions satisfying a paracontrolled ansatz.We start by defining a suitable space of such objects.

Definition 5. Let – > 0 and — œ (0, –] be such that – + — œ (0, 2),and let u œ L –. A pair of distributions (f, fu) œ L – ◊ L — is called


paracontrolled by u if f ˘ = f ≠ fu ª u œ CC –+— fl L —. In that case wewrite f œ D— = D—(u), and for all T > 0 we define the norm

ÎfÎD—

T

= ÎfÎC

–/2T

+ ÎfuÎL —

T

+ Îf ˘ÎCT

C –+— + Îf ˘ÎC

—/2T

LŒ .

If u œ L – and (f , f u) œ D—(u), then we also write

dD—

T

(f, f) = Îfu ≠ f uÎL —

T

+ Îf ˘ ≠ f ˘ÎCT

C –+— + Îf ˘ ≠ f ˘ÎC

—/2T

LŒ .

Note that in general f and f do not live on the same space, so dD—

T

is nota distance.

Of course we should really write (f, fu) œ D— since given f and u, thederivative fu is usually not uniquely determined. But in the applicationsthere will always be an obvious candidate for the derivative, and noconfusion will arise.

Remark 8. The space D— does not depend on the specific dyadicpartition of unity. Indeed, Bony [4] has shown that if ª is theparaproduct constructed from another partition of unity, then Îfu ª u ≠fu ª uÎC

T

C –+— . ÎfuÎCT

C — ÎuÎCT

C – .

Nonlinear operations As an immediate consequence of Lemma 14 wecan multiply any distribution that is paracontrolled by u with a given v,provided that we know how to multiply u with v (of course always undersuitable regularity assumptions):

Theorem 5 (also see Theorem 3.7 of [14]). Let –, — œ R, “ < 0, with–+— +“ > 0 and –+“ ”= 0. Let u œ CC –, v œ CC “ , and let ’ œ CC –+“ .Then

D—(u) – f ‘æ f · v := f ª v + f º v + f ˘ ¶ v + C(fu, u, v) + fu’ œ CC “

defines a bounded linear operator and for all T > 0 we have the bound

Î(fv)˘ÎCT

C –+“ := Îf · v ≠ f ª vÎCT

C –+“

. ÎfÎD—

T

(ÎvÎCT

C “ + ÎuÎCT

C –ÎvÎCT

C “ + Î’ÎCT

C –+“ ) .

If there exist sequences of smooth functions (un) and (vn) converging to uand v in CC – and CC “ respectively for which (un ¶ vn) converges to ’ inCC –+“ , then f · v does not depend on the dyadic partition of unity usedto construct it.

Furthermore, there exists a quadratic polynomial P so that if u, v, ’satisfy the same assumptions as u, v, ’ respectively, if f œ D—(u), and


if

M = maxÓ

ÎuÎCT

C – , ÎvÎCT

C “ , Î’ÎCT

C –+“ , ÎuÎCT

C – , ÎvÎCT

C “ ,

Î’ÎCT

C –+“ , ÎfÎD—

T

(u)

, ÎfÎD—

T

(u)

Ô,

then

Î(fv)˘ ≠ (f v)˘ÎCT

C –+“

6P (M)!dD— (f, f) + Îu ≠ uÎC

T

C – + Îv ≠ vÎCT

C “ + Î’ ≠ ’ÎCT

C –+“

".

Proof. Given Lemma 14 (and the paraproduct estimates Theorem 4),the proof is straightforward and we leave most of it as an exercise. Let usonly comment on the independence of the partition of unity: let (un, vn)be as announced and define fn := fu ª un + f ˘. Then

limnæŒ

fn

vn

= limnæŒ

!f

n

ª vn

+ fn

º vn

+ f ˘ ¶ vn

+ C(fu, un

, vn

) + fu(un

¶ vn

)"

= f ª v + f º v + f ˘ ¶ v + C(fu, u, v) + fu’ = f · v.

Since the pointwise product fnvn does not depend on the partition of unity,also the limit must be independent.

The bound on the di�erence is obtained by using the boundedness andmultilinearity of all operators involved. ⇤

From now on we will assume that there exist smooth functions (un) and(vn) converging to u and v respectively for which (un ¶ vn) converges to ’,so that the product does not depend on the partition of unity, and we willusually write fv rather than f ·v. Later we will see that the resonant term(un ¶ vn) must often be renormalized by subtracting a large constant, butthis will not a�ect the independence of the product from the partition ofunity.

To solve equations involving general nonlinear functions, we needto examine the stability of paracontrolled distributions under smoothfunctions.

Theorem 6. Let – œ (0, 1) and — œ (0, –]. Let u œ L –, f œ D–(u), andF œ C

1+—/–b . Then F (f) œ D— with derivative (F (f))u = F Õ(f)fu, and

for all T > 0

ÎF (f)ÎD—

T

. ÎFÎC

1+—/–

b

(1 + ÎfÎ2

D–

T

)(1 + ÎuÎ2

L –

T

).

Moreover, there exists a polynomial P which satisfies, for all F œ C2+—/–b ,

u œ L –, f œ D–(u), and

M := maxÓ

ÎuÎL –

T

, ÎuÎL –

T

, ÎfÎD–

T

(u)

, ÎfÎD–

T

(u)

Ô,


the bound

dD—

T

(F (f), F (f)) 6 P (M)ÎFÎC

2+—/–

T

(dD–

T

(f, f) + Îu ≠ uÎL –

T

).

The proof is not very complicated but rather lengthy, and we do notpresent it here. The reader can find it in [14].

Schauder estimate for paracontrolled distributions The Schauderestimate Lemma 11 is not quite su�cient: we also need to understand howthe heat kernel acts on the paracontrolled structure.

Theorem 7. Let – œ (0, 1) and — œ (0, –]. Let u œ CC –≠2 and L U = uwith U(0) = 0. Let fu œ L —, f ˘ œ CC –+—≠2, and g

0

œ C –+—. Then(g, fu) œ D—(U), where g solves

L g = fu ª u + f ˘, g(0) = g0

,

and we have the bound

ÎgÎD—

T

(U)

. Îg0

Î–+— + (1 + T )(ÎfuÎL —

T

(1 + ÎuÎCT

C –≠2) + Îf ˘ÎCT

C –+—≠2)

for all T > 0. If furthermore u, U , f u, f ˘, g0

, g satisfy thesame assumptions as u, U, fu, f ˘, g

0

, g respectively, and if M =max{ÎfuÎL —

T

, ÎuÎCT

C –≠2 , 1}, then

dD—

T

(g, g) .Îg0 ≠ g0Î–+—

+ (1 + T )M(Îfu ≠ f uÎL —

T

+ Îu ≠ uÎC

T

C–≠2 + Îf ˘ ≠ f ˘ÎC

T

C–+—≠2 ).

Proof. Let us derive an equation for the remainder g˘. We have

L g˘ = L g ≠ L (f Õ ª U)= [fu ª u + f ˘] ≠ fu ª L U ≠ [L (fu ª U) ≠ fu ª L U ]= f ˘ ≠ [L , fu ª ]U.

Since – · — = — we can now apply Lemma 18 to see that there existsH œ CC –+— fl C—/2LŒ such that L H = [L , fu ª ] U , so we can applythe standard Schauder estimates of Lemma 11 to L (g˘ + H) = f ˘ to get

Îg˘ÎCT

C –+— + Îg˘ÎC

(–·—)/2T

LŒ .ÎfuÎL —

T

(ÎUÎCT

C – + ÎL UÎCT

C –≠2)

+ Îf ˘ÎCT

C –+—≠2 .

The estimate for g˘ ≠ g˘ can be derived in the same way. ⇤


Bibliographic notes. Paraproducts were introduced in [4]. For a niceintroduction see [1]. The commutator estimate Lemma 14 is from [14],but the proof here is new and the statement is slightly di�erent. In [14],we require the additional assumption – œ (0, 1) under which C mapsC – ◊ C — ◊ C “ to C –+—+“ and not only to C —+“ . Theorem 6 is from [14].

Theorem 7 is new, but it is implicitly used in [14]. The estimatespresented here will only allow us to consider regular initial conditions.More general situations can be covered by working on spaces allowing fora singularity at 0, such as

Óf œ C ((0, Œ), C –) : sup

tœ(0,T ]

Ît≠“f(t)ÎC – < Œ for all T > 0Ô

and similar for the temporal regularity. This is also done in [14].Of course it is easily possible to replace the Laplacian by more general

pseudo-di�erential operators. We only used two properties of �: the factthat �(f Õ ª U) ≠ f Õ ª (�U) is relatively regular, and that the semigroupgenerated by � has a su�ciently strong regularization e�ect. This is alsotrue for the fractional Laplace operator and more generally for a widerange of pseudo-di�erential operators.

5.4 FixpointLet us now give the details for the solution to pam in the space ofparacontrolled distributions. Assume that F : R æ R is in C1+Á

b forsome Á > 0 such that (2 + Á)“ > 2.

Let Y œ CC “ and let u œ D“(Y ). We will see below how to chooseY , for the moment it is an arbitrary CC “ function. From Theorem 6 weknow that F (u) œ DÁ“(Y ):

D“(Y ) u ‘æF (u)≠≠≠≠≠æ DÁ“(Y ). (5.13)

Assume now that Y ¶ ÷ œ CC 2“≠2 is given – note that for the regularityassumptions we made, Y ¶ ÷ is not a continuous functional of Y and ÷but must be controlled using other means, say stochastic computations!Under this assumption, Theorem 5 applied with u = Y , v = ÷, and’ = Y ¶ ÷ shows that for all f œ DÁ“(Y ) we have f÷ = (f÷)˘ + f ª ÷ with(f÷)˘ œ CC 2“≠2 – it is here that we use (2 + Á)“ > 2. Integrating againstthe heat kernel and assuming that u

0

œ C 2“ , we obtain from Theorem 7(with u = ÷, fu = f , f ˘ = (f÷)˘) that the solution (J(f÷)(t) + Ptu0

)t>0

to L (J(f÷) + P·u0

) = f÷, J(f÷)(0) + P0

u0

= u0

, is in D“(X), where Xsolves L X = ÷ and X(0) = 0. In other words, we have a map

DÁ“(Y ) f ‘æP·u0+J(f÷)≠≠≠≠≠≠≠≠≠≠æ D“(X), (5.14)


and combining (5.13) and (5.14) we get

D“(Y ) u ‘æF (u)≠≠≠≠≠æ DÁ“(Y ) F (u) ‘æP·u0+J(F (u)÷)≠≠≠≠≠≠≠≠≠≠≠≠≠≠æ D“(X),

so that for all T > 0 we can define

�T : D“T (Y ) æ D“

T (X), �T (u) = (P·u0

+ J(F (u)÷))|[0,T ]

.

To set up a Picard iteration domain and image space should coincide whichmeans we should take Y = X. Refining the analysis, we obtain a scalingfactor T ” when estimating the D“

T (X)–norm of �T (u). This allows us toshow that for small T > 0, the map �T leaves suitable balls in D“

T (X)invariant, and therefore we obtain the (local in time) existence of solutionsto the equation under the assumption X ¶ ÷ œ CC 2“≠2.

To obtain uniqueness we need to suppose that F œ C2+Áb . In that

case Theorem 6 gives the local Lipschitz continuity of the map u ‘æ F (u)from D“

T (X) to DÁ“T (X), while Theorem 5 and Theorem 7 show that

f ‘æ u0

+ J(f÷) defines a Lipschitz continuous map from DÁ“T (X) to

D“T (X). Again we can obtain a scaling factor T ”, so that �T defines a

contraction on a suitable ball of D“T (X) for some small T > 0.

Even better, �T not only depends locally Lipschitz continuously on u,but also on the extended data (u

0

, ÷,X ¶ ÷), and therefore the solutionto (5.1) depends locally Lipschitz continuously on (u

0

, ÷, X ¶ ÷).

5.5 RenormalizationSo far we argued under the assumption that X ¶ ÷ exists and has a su�cientregularity. This should be understood via approximations as the existenceof a sequence of smooth functions (÷n) that converges to ÷, such that(Xn ¶ ÷n) converges to X ¶ ÷. However, as we will see below, this hypothesisis questionable and actually not satisfied at all in the problem we areinterested in. More concretely, recall that we would like to take ÷ = › tobe the two–dimensional space white noise. If then Ï is a Schwartz functionon R2 and if Ïn = nÏ(n·) and

÷n(x) = Ïn ú ›(x) =⁄

R2Ïn(x ≠ y)›(y)dy =

ÿ

kœZ2

È›, Ïn(x + 2fik ≠ ·)Í,

then we will see below that there exist constants (cn) with limn cn = Œ,such that (Xn ¶ ÷n ≠ cn) converges in CT C 2“≠2 for all T > 0.

This is not a problem with our specific approximation. Thehomogenization setting shows that even for ÷ æ 0 there are cases wherethe limiting equation is nontrivial. In the paracontrolled setting we have


a continuous dependence of the solution on the data (÷, X ¶ ÷), so thisnon–triviality of the limit can only mean that it is X ¶ ÷ which does notconverge to zero.

Another way to see that there is a problem is to consider the followingrepresentation of the resonant term: use L X = ÷ to write

X ¶ ÷ = X ¶ L X

= 12L (X ¶ X) + ˆxX ¶ ˆxX

= |ˆxX|2 + 12L (X ¶ X) ≠ 2ˆxX ª ˆxX.

Integrating this equation over the torus and over t œ [0, T ], we get⁄ T

0

⁄

T2X ¶ ÷dxdt =

⁄ T

0

⁄

T2|ˆxX|2dxdt + 1

2

⁄ T

0

⁄

T2L (X ¶ X)dx

≠ 2⁄ T

0

⁄

T2(ˆxX ª ˆxX)dxdt.

Writing L = ˆt ≠ � and using that X(0) = 0 andsT2 �Âdx = 0 for all Â

(which can be seen using integration by parts and pulling the operator �on the constant function 1), we thus get

⁄ T

0

⁄

T2X ¶ ÷dxdt =

⁄ T

0

⁄

T2|ˆxX|2dxdt + 1

2

⁄

T2(X(T ) ¶ X(T ))dx

≠ 2⁄ T

0

⁄

T2(ˆxX ª ˆxX)dxdt.

So if X ¶ ÷ œ CT C 2“≠2 and X œ CT C “ , then all the terms should bewell defined and finite (the integral over T2 corresponds to testing adistribution against to constant test function 1). This would mean thats T

0

sT2 |ˆxX|2dxdt < +Œ, but on the other side a direct computation

shows that ⁄

T2|ˆxX(t, ·)|2dx = +Œ

for any t > 0 almost surely if ÷ is the space white noise. Note also thatthe problematic term |ˆxX|2 is exactly the correction term appearing inthe analysis of the linear homogenization problem in Section 4.

In order to prove the convergence of the smooth solutions in general,we should introduce corrections to the equation to remove the divergentconstant cn. Let us see where the resonant product X ¶ ÷ appears. We


have

(F (u)÷)˘ = F (u) º ÷ + (F (u))˘ ¶ ÷ + C((F (u))X , X, ÷) + (F (u))X(X ¶ ÷).(5.15)

Now (F (u))X = F Õ(u)uX by Theorem 6, and if u solves the equationL u = F (u)÷ = F (u) ª ÷ + (F (u)÷)˘, then Theorem 7 with u = ÷, X = Ushows that uX = F (u). So we should really consider the renormalizedequation

L un = F (un) ù ÷n := F (un)›n ≠ F Õ(un)F (un)cn,

where we recall that (cn) are the diverging constants for which (Xn ¶ ÷n ≠cn) converges. In that case we have

L un =F (un) ª ÷n + F (un) º ÷n + (F (un))˘ ¶ ÷n

+ C(F Õ(un)F (un), Xn, ÷n) + F Õ(un)F (un)(Xn ¶ ÷n ≠ cn),

and now all the terms on the right hand side are under control and we cansafely pass to the limit, for which we obtain the equation

L u = F (u) ù ÷ := (F (u) ù ÷)˘ + F (u) ª ÷, (5.16)

where (F (u) ù ÷)˘ is calculated using X ù ÷ = limn(Xn ¶ ÷n ≠ cn) in theplace of X ¶ ÷ in (5.15). Formally, we also denote this product by

F (u) ù ÷ = F (u)÷ ≠ F Õ(u)F (u) · Œ,

so that the solution u will satisfy

L u = F (u) ≠ F Õ(u)F (u) · Œ.

Note that the correction term has exactly the same form as theItô/Stratonovich corrector for SDEs. For the reader familiar with roughpaths this will not come as a surprise: Changing the iterated integrals ofa rough path B from some given

s ·0

BsdBs tos ·

0

BsdBs + Ï introducesa correction term +F Õ(y)F (y)ˆtÏ in the ODE ˆty = F (y)ˆtB. In oursetting the resonant term takes the role of the iterated integrals, and sincethe structure of the ODE and gpam is very similar changing the resonantterm has a similar e�ect as changing the iterated integrals in the ODEexample.

Remark 9. The convergence properties of (Xn ¶ ÷n) are in stark contrastto the ODE setting: if we consider the equation ˆtu = F (u)’ rather thanpam, then we should replace X by Z with ˆtZ = ’. But then we have inone dimension Z ¶ ’ = 1/2ˆt(Z ¶ Z), so that the convergence of (Zn ¶ ’n)to Z ¶ ’ comes for free with the convergence of (Zn) to Z. Indeed, ˆt is a


bounded linear operator from C “ to C “≠1 whenever “ œ R, and Z ‘æ Z ¶Zis continuous from C “ to C 2“ whenever “ > 0. So if (Zn) converges toZ in a Hölder space of positive regularity, then (ˆt(Zn ¶ Zn)) convergesto ˆt(Z ¶ Z). This specific representation of Z ¶ ’ comes from the Leibnizrule for ˆt and it is the reason why rough path theory is trivial in onedimension, at least as long as one considers those rough paths which arelimit of smooth paths. Of course, the argument breaks down as soon as Zhas at least two components. As we have discussed, for the second orderdi�erential operator L we have di�erent rules and obtain

(X ¶ ÷) = (X ¶ L X) = 12L (X ¶ X) + (ˆxX ¶ ˆxX),

so that in our setting the nontrivial term is ˆxX ¶ ˆxX.

These considerations lead naturally to the following definition.

Definition 6. (pam–enhancement) Let “ œ (2/3, 1) and let

X “pam

™ C “≠2 ◊ CC 2“≠2

be the closure of the image of the map

�pam

: CŒ ◊ C([0, Œ),R) æ X “pam

,

given by�

pam

(◊, f) = (◊, � ù ◊) := (◊, � ¶ ◊ ≠ f), (5.17)where � = J◊, that is L � = ◊ and �(0) = 0. We will call �

pam

(◊, f) therenormalized pam–enhancement of the driving distribution ◊. For T > 0we define X “

pam

(T ) = X “pam

|[0,T ]

and we write ÎXÎX “

pam(T )

for the norm ofX œ X “

pam

(T ) in the Banach space C “≠2 ◊ CT C 2“≠2. Moreover, we definethe distance dX “

pam(T )

(X, X) = ÎX ≠ XÎX “

pam(T )

.

Remark 10. In the homogenization example of Section 4 we would take◊ = VÁ and � = XÁ.

Remark 11. It would be more elegant to renormalize � ¶ ◊ with a constantand not with a time-dependent function, as we discussed above. Indeed thisis possible, see Chapter 5 of [14]. But since here we chose �(0) = 0, wehave �(0) ¶ ◊ = 0 and therefore (�n(0) ¶ ◊n ≠cn) diverges for any divergingsequence of constants (cn). A simple way of avoiding this problem is toconsider the stationary version � given by

�(x) =⁄ Œ

0

Pt� ”=0

◊(x)dt,

where � ”=0

denotes the projection on the non-zero Fourier modes, � ”=0

u =u ≠ (2fi)≠d/2u(0). But then � does not depend on time and in particular


�(0) ”= 0, so that we have to consider irregular initial conditions in theparacontrolled approach which complicates the presentation. Alternatively,we could observe that in the white noise case there exist constants (cn) sothat (Xn(t) ¶ ›n ≠cn) converges for all t > 0, and while the limit (X(t) ù ›)diverges as t æ 0, it can be integrated against the heat kernel. Again,this would complicate the presentation and here we choose the simple (andcheap) solution of taking a time-dependent renormalization.Theorem 8. Let “ œ (2/3, 1) and Á > 0 be such that (2 + Á)“ > 2. LetX = (÷, X ù ÷) œ X “

pam

, F œ C2+Áb , and u

0

œ C 2“ . Then there exists aunique solution u œ D“(X) to the equation

L u = F (u) ù ÷, u(0) = u0

,

up to the (possibly finite) explosion time · = ·(u) = inf{t > 0 : ÎuÎD“

t

=Œ} > 0.

Moreover, u depends on (u0

,X) œ C 2“ ◊ X “pam

in a locally Lipschitzcontinuous way: if M, T > 0 are such that for all (u

0

,X) with Îu0

Î2“ ‚

ÎXÎX “

pam(T )

6 M , the solution u to the equation driven by (u0

,X) satisfies·(u) > T , and if (u

0

, X) is another set of data bounded in the above senseby M , then there exists C(F, M) > 0 for which

dD“

T

(u, u) 6 C(F, M)(Îu0

≠ u0

Î2“ + dX “

pam(T )

(X, X)).Proof. We only have to turn the formal discussion of Section 5.4 intorigorous mathematics. The small factor T ” on page 71 is obtained froma scaling argument and while this does not require any new insights it issomewhat lengthy and we refer to [14, 16] for details.

Let us just indicate how to iterate the construction to obtain theexistence of solutions up to the explosion time · . Let us assume thatwe constructed the paracontrolled solution (u, uX) (with uX = F (u)) on[0, T

0

] for some T0

> 0. Now we no longer have X(T0

) = 0, and also theinitial condition u(T

0

) is no longer in C 2“ . But we only used X(0) = 0to see that the initial condition for u˘ is u˘(0) = u

0

, and we only usedu

0

œ C 2“ to obtain a C 2“ initial condition for u˘. On the next interval,the initial condition for u˘ is u˘(T

0

) = u(T0

) ≠ F (u(T0

)) ª X(T0

) which isin C 2“ by construction, since we already know that u˘ œ C([0, T

0

], C 2“).As for the continuity in (u

0

,X), let (u0

, X) be another set of data alsobounded by M . Then the solutions u and u both are bounded in D“

T

by some constant c = c(F, M) > 0. So by the continuity properties ofthe paracontrolled product (and the other operations involved), we canestimate

dD“

T

(u, u) 6 P (c)1

Îu0

≠ u0

Î2“ + dX “

pam(T )

(X, X) + T ”dD“

T

(u, u)2

for a polynomial P . The local Lipschitz continuity on [0, T ] immediatelyfollows if we choose T > 0 small enough. This can be iterated to obtainthe local Lipschitz continuity on “macroscopic” intervals. ⇤


Remark 12. For the local in time existence it is not necessary to assumeF œ C2+Á

b . It su�ces to have F œ C2+Á. This can be seen by considering aball containing u

0

(x) for all x œ Td, a function F œ C2+Áb which coincides

with F on this ball, and by stopping u upon exiting the ball.In the linear case F (u) = u we have global in time solutions:

in general we only get local in time solutions because we pick up asuperlinear (polynomial) estimate when applying the paralinearizationresult Theorem 6. This step is not necessary if F is linear, and all theother estimates are linear in u.

5.6 Construction of the extended data

In order to apply Theorem 8 to equation (5.1) with white noiseperturbation, it remains to show that if › is a spatial white noise on T2,then (›, X ù ›) defines an element of X “

pam

whenever “ œ (2/3, 1). In otherwords, we need to construct X ù › and control its regularity.

Since Pt› is a smooth function for every t > 0, the resonant termPt› ¶ › is a smooth function, and therefore we could formally set X(t) ¶ › =s t

0

(Ps› ¶ ›)ds. But we will see that this expression does not make sense.Recall that (›(k))kœZ2 is a complex valued, centered Gaussian process

with covarianceE[›(k)›(kÕ)] = ”k+kÕ

=0

, (5.18)

and such that ›(k)ú = ›(≠k).

Lemma 19. For any x œ T2 and t > 0 we have

gt

= E[(Pt

›)(x)›(x)] = E[(Pt

› ¶ ›)(x)] = E[�≠1(Pt

› ¶ ›)(x)] = (2fi)≠2ÿ

kœZ2

e≠t|k|2.

In particular, gt does not depend on the partition of unity used to definethe ¶ operator, and

s t

0

gsds = Œ for all t > 0.

Proof. Let x œ T2, t > 0, and ¸ Ø ≠1. Then

E[�¸(Pt› ¶ ›)(x)] =ÿ

|i≠j|61

E[�¸(�i(Pt›)�j›)(x)],

where exchanging summation and expectation is justified because it canbe easily verified that the partial sums of �¸(Pt› ¶ ›)(x) are uniformlyLp–bounded for any p Ø 1. Now Pt = e≠t|·|2(D), and therefore we get


from (5.18)

E[�¸(�i(Pt›)�j›)(x)]

=(2fi)≠1

ÿ

k,kÕœZ2

eúk+kÕ(x)fl¸(k + kÕ)fli(k)e≠t|k|2

flj(kÕ)E[›(k)›(kÕ)]

=(2fi)≠2

ÿ

kœZ2

fl¸(0)fli(k)e≠t|k|2flj(k) = ”¸=≠1

(2fi)≠2

ÿ

kœZ2

fli(k)flj(k)e≠t|k|2.

For |i ≠ j| > 1 we have fli(k)flj(k) = 0 and therefore

gt = E[(Pt› ¶ ›)(x)]= E[(Pt›)(x)›(x)]

= (2fi)≠2

ÿ

kœZ2

ÿ

i,j

fli(k)flj(k)e≠t|k|2

= (2fi)≠2

ÿ

kœZ2

e≠t|k|2,

while E[(Pt› ¶ ›)(x) ≠ �≠1

(Pt› ¶ ›))(x)] = 0. ⇤Exercise 12. Let Ï be a Schwartz function on R2 and set

›n(x) = ((n2Ï(n·)) ú ›)(x)

=⁄

R2n2Ï(n(x ≠ y))›(y)dy

=ÿ

kœZ2

È›, n2Ï(n(x + 2fik ≠ ·))Í

for x œ T2. Write FR2Ï(z) =sR2 e≠iÈz,xÍÏ(x)dx. Show that

E[(Pt›n ¶ ›n)(x)] = E[�≠1

(Pt›n ¶ ›n)(x)] = (2fi)≠2

ÿ

kœZ2

e≠t|k|2|FR2Ï(k/n)|2.

Hint: Use Poisson summation.

The diverging time integral motivates us to study the renormalizedproduct X ¶ › ≠

s ·0

gsds, wheres ·

0

gsds is an “infinite function”:

Lemma 20. Set

(X ù ›)(t) =⁄ t

0

(Ps› ¶ › ≠ gs)ds.

Then E[ÎX ù ›ÎpC

T

C 2“≠2(T2

)

] < Œ for all “ < 1, p Ø 1, T > 0. Moreover,if Ï is a Schwartz function on R2 with

sÏ(x)dx = 1, if ›n = Ïn ú › with

Ïn = n2Ï(n·) for n œ N, and Xn(t) =s t

0

Ps›nds, then

limnæŒ

E[ÎX ù › ≠ (Xn ¶ ›n ≠ fn)ÎpC

T

C 2“≠2(T2

)

] = 0


for all p Ø 1, where for all x œ T2

fn(t) = E[Xn(t, x)›n(x)] = E[(Xn(t) ¶ ›n)(x)]

= (2fi)≠2

ÿ

kœZ2\{0}

|FR2Ï(k/n)|2|k|2 (1 ≠ e≠t|k|2

) + (2fi)≠2t.

Proof. To lighten the notation, we will only show thatE[ÎX ù ›Îp

CT

C 2“≠2 ] < Œ. The convergence of (Xn ¶ ›n ≠ fn) to X ù ›is shown by applying dominated convergence, and we leave it as anexercise. Let t > 0 and define �t = Pt› ¶ › ≠ gt. Let us start byestimating E[|�¸�t(x)|2] for ¸ > ≠1 and x œ T2. Lemma 19 yields�¸gt = 0 = E[�¸(Pt› ¶ ›)(x)] for ¸ Ø 0 and x œ T2, and �≠1

gt = gt =E[�≠1

(Pt› ¶ ›)(x)], so that E[|�¸�t(x)|2] = Var(�¸(Pt› ¶ ›)(x)). But

�¸(Pt› ¶ ›)(x)

=ÿ

kœZ2

eúk(x)fl¸(k)F (Pt› ¶ ›))(k)

=(2fi)≠1

ÿ

k1,k2œZ2

ÿ

|i≠j|61

eúk1+k2(x)fl¸(k1

+ k2

)fli(k1

)e≠t|k1|2›(k

1

)flj(k2

)›(k2

),

and therefore

Var(�¸(Pt› ¶ ›)(x))

=(2fi)≠2

ÿ

k1,k2

ÿ

kÕ1,kÕ

2

ÿ

|i≠j|61

ÿ

|iÕ≠jÕ|61

eúk1+k2(x)fl¸(k1

+ k2

)fli(k1

)e≠t|k1|2flj(k

2

)

◊ eúkÕ

1+kÕ2(x)fl¸(kÕ

1

+ kÕ2

)fliÕ(kÕ1

)e≠t|kÕ1|2

fljÕ(kÕ2

) Cov(›(k1

)›(k2

), ›(kÕ1

)›(kÕ2

)),

where exchanging summation and expectation can be justified a posterioriby the uniform Lp–boundedness of the partial sums. Now Wick’s theorem([22], Theorem 1.28) gives

Cov(›(k1

)›(k2

), ›(kÕ1

)›(kÕ2

))=E[›(k

1

)›(k2

)›(kÕ1

)›(kÕ2

)] ≠ E[›(k1

)›(k2

)]E[›(kÕ1

)›(kÕ2

)]=E[›(k

1

)›(k2

)]E[›(kÕ1

)›(kÕ2

)] + E[›(k1

)›(kÕ1

)]E[›(k2

)›(kÕ2

)]+ E[›(k

1

)›(kÕ2

)]E[›(k2

)›(kÕ1

)] ≠ E[›(k1

)›(k2

)]E[›(kÕ1

)›(kÕ2

)]=(”k1+kÕ

1=0

”k2+kÕ2=0

+ ”k1+kÕ2=0

”k2+kÕ1=0

),

which leads to

Var(�¸

(Pt

› ¶ ›)(x)) =(2fi)≠4ÿ

k1,k2

ÿ

|i≠j|61

ÿ

|iÕ≠j

Õ|61

I¸.i

I¸.i

Õ fl2¸

(k1 + k2)fli

(k1)flj

(k2)

◊ [fli

Õ (k1)flj

Õ (k2)e≠2t|k1|2+ fl

i

Õ (k2)flj

Õ (k1)e≠t|k1|2≠t|k2|2].


Observe that there exists c > 0 such that e≠2t|k|2 . e≠tc2

2i for allk œ supp(fli) fi supp(flj) with i, j Ø ≠1 and |i ≠ j| 6 1. Thus

Var(�¸(Pt› ¶ ›)(x))

.ÿ

i,j,iÕ,jÕ

I¸.iIi≥j≥iÕ≥jÕ

ÿ

k1,k2

Isupp(fl

¸

)

(k1

+ k2

)Isupp(fl

i

)

(k1

)Isupp(fl

j

)

(k2

)e≠2tc2

2i

.ÿ

i:i&¸

22i22¸e≠tc2

2i . 22¸

t

ÿ

i:i&¸

e≠tcÕ2

2i . 22¸

te≠tcÕ

2

2¸

,

where in the third step we used that t22i . et(c≠cÕ)2

2i for all 0 < cÕ < c.Consider now X ù ›(t) =

s t

0

�sds. We have for all 0 6 s < t

E[ÎX ù ›(t) ≠ X ù ›(s)Î2p

B2“≠22p,2p

]

=ÿ

¸

22p¸(2“≠2)

⁄

T2E[|�¸(X ù ›(t) ≠ X ù ›(s))(x)|2p]dx.

Since the random variable �¸(X ù ›(t) ≠ X ù ›(s))(x) lives in the secondnon-homogeneous chaos generated by the Gaussian white noise ›, we mayuse Gaussian hypercontractivity ([22], Theorem 3.50) to bound

E[|�¸(X ù ›(t) ≠ X ù ›(s))(x)|2p] . E[|�¸(X ù ›(t) ≠ X ù ›(s))(x)|]2p

61 ⁄ t

s

E[|�¸�r(x)|]dr2

2p

.

But we just showed that

E[|�¸�r(x)|] 6 E[|�¸�r(x)|2]1/2 = (Var(�¸(Pr› ¶ ›)(x)))1/2

. r≠1/22¸e≠ 12 rcÕ

2

2¸

= r≠1/22¸e≠rcÕÕ2

2¸

for cÕÕ = cÕ/2 > 0, and therefore1E

ËÎX ù ›(t) ≠ X ù ›(s)Î2p

B2“≠22p,2p

È21/2p

.1 ÿ

¸

12¸(2“≠2)

⁄ t

s

r≠1/22¸e≠rcÕÕ2

2¸

dr2

2p21/2p

6ÿ

¸

2¸(2“≠1)

⁄ t

s

r≠1/2e≠rcÕÕ2

2¸

dr

.⁄ t

s

r≠1/2

⁄ Œ

≠1

(2x)2“≠1e≠rcÕÕ2

2x

dxdr.

The change of variable y =Ô

r2x leads to1E

ËÎX ù ›(t)≠X ù ›(s)Î2p

B

2“≠22p,2p

È21/2p

.⁄

t

s

r≠1/2r≠(2“≠1)/2⁄ Œ

0y2“≠2e≠c

ÕÕy

2dydr.


For “ > 1/2, the integral in y is finite and we end up with

1E

ËÎX ù ›(t) ≠ X ù ›(s)Î2p

B2“≠22p,2p

È21/2p

.⁄ t

s

r≠“dr . |t ≠ s|1≠“

provided that “ œ (1/2, 1). So for large enough p we can use Kolmogorov’scontinuity criterion to deduce that (modulo taking a modification of X ù›)we have E[ÎX ù ›Î2p

CT

B2“≠22p,2p

] < Œ for all T > 0. Since this holds for all“ < 1, the claim now follows from the Besov embedding theorem, Lemma 8.

⇤Combining Theorem 8 and Lemma 20, we are finally able to solve (5.1)

driven by a space white noise.

Corollary 2. Let Á > 0 and let F œ C2+Áb and assume that u

0

is a randomvariable that almost surely takes its values in C 2“ for some “ œ (2/3, 1)with (2 + Á)“ > 2. Let › be a spatial white noise on T2. Then there existsa unique solution u to

L u = F (u) ù ›, u(0) = u0

,


t

=Œ} which is almost surely strictly positive.

If (Ïn) and (›n) are as described in Lemma 20, and if (u0,n) converges

in probability in C 2“ to u0

, then u is the limit in probability of the solutionsun to

L un = F (un) ù ›n, un(0) = u0,n.

Remark 13. We even have a stronger result: We can fix a null set outsideof which X ù › is regular enough, and once we dispose of that null set wecan solve all equations for any regular enough u

0

and F simultaneously,without ever having to worry about null sets again. This is for exampleinteresting when studying stochastic flows or when studying equations withrandom u

0

and F .The pathwise continuous dependence on the signal is also powerful

in several other applications, for example support theorems and largedeviations. For examples in the theory of rough paths see [9].

Chapter 6

The stochastic Burgersequation

Let us now return to the stochastic Burgers equation sbe

L u = ˆxu2 + ˆx›, u(0) = u0

, (6.1)

where u : [0, Œ)◊T æ R, › is a space-time white noise, and ˆx denotes thespatial derivative. As we argued before, the solution u cannot be expectedto behave better than the Ornstein–Uhlenbeck process X, the solution ofthe linear equation L X = ˆx›, and as we saw in Section 2 X(t) is for allt > 0 a smooth function of the space variable plus a space white noise. ByExercise 11, the white noise in dimension 1 has regularity C ≠1/2≠. ThusX œ CC ≠1/2≠, and in particular u2 is the square of a distribution and apriori not well defined.

What raises some hope is that in Lemma 2 we were able to show thatˆxX2 exists as a space–time distribution. So as in the previous examplesthere are stochastic cancellations going into ˆxX2. The energy solutionapproach was designed to take those cancellations into account in the fullsolution u, but while it allowed us to work under rather weak assumptionswhich easily gave us existence of solutions, it did not give us su�cientcontrol to have uniqueness of solutions. On the other side, a suitableparacontrolled ansatz for the solution u will allow us to transfer thecancellation properties of X to u and it will allow us to construct ˆxu2 as acontinuous bilinear map, from where existence and uniqueness of solutionseasily follows.

81


6.1 Structure of the solutionIn this discussion we consider the case of zero initial condition and smoothnoise ›, and we analyze the structure of the solution. Let us expand uaround the Ornstein–Uhlenbeck process X with L X = ˆx›, X(0) = 0.Setting u = X + u>1, we have

L u>1 = ˆx(u2) = ˆx(X2) + 2ˆx(Xu>1) + ˆx((u>1)2).

Let us define the bilinear map

B(f, g) = Jˆx(fg) =⁄ ·

0

P·≠sˆx(f(s)g(s))ds.

Then we can proceed by performing a further change of variables in orderto remove the term ˆx(X2) from the equation by setting

u = X + B(X, X) + u>2. (6.2)

Now u>2 satisfiesL u>2 = 2ˆx(XB(X, X)) + ˆx(B(X, X)B(X, X))

+2ˆx(Xu>2) + 2ˆx(B(X, X)u>2) + ˆx((u>2)2). (6.3)

We can imagine to make a similar change of variables to get rid of theterm

2ˆx(XB(X, X)) = L 2B(X, B(X, X)).As we proceed in this inductive expansion, we generate a number of explicitterms, obtained by various combinations of X and B. Since we will haveto deal explicitly with at least some of these terms, it is convenient torepresent them with a compact notation involving binary trees. A binarytree · œ T is either the root • or the combination of two smaller binarytrees · = (·

1

·2

), where the two edges of the root of · are attached to ·1

and ·2

respectively. For example

(••) = , ( •) = , ( •) = , ( ) = , . . .

Then we define recursively

X• = X, X(·1·2) = B(X·1 , X·2),

giving

X = B(X, X), X = B(X , X), X = B(X , X), X = B(X , X ),

and so on. In this notation the expansion (6.2)–(6.3) reads

u = X + X + u>2, (6.4)

u>2 = 2X + X + 2B(X, u>2) + 2B(X , u>2) + B(u>2, u>2). (6.5)

Chapter 6. The stochastic Burgers equation 83

Remark 14. We observe that formally the solution u of sbe can beexpanded as an infinite sum of terms labelled by binary trees:

u =ÿ

·œTc(·)X· ,

where c(·) is a combinatorial factor counting the number of planartrees which are isomorphic (as graphs) to · . For example c(•) =

1, c( ) = 1, c( ) = 2, c( ) = 4, c( ) = 1 and in generalc(·) =

q·1,·2œT I

(·1·2)=· c(·1

)c(·2

). Alternatively, we may truncate thesummation at trees of degree at most n and set

u =ÿ

·œT ,d(·)<n

c(·)X· + u>n,

where we denote by d(·) œ N0

the degree of the tree · , given by d(•) = 0and then inductively d((·

1

·2

)) = 1 + d(·1

) + d(·2

). For example d( ) = 1,

d( ) = 2, d( ) = 3, d( ) = 3. We then obtain for the remainder

u>n =ÿ

·1, ·2 : d(·1) < n, d(·2) < nd((·1·2)) > n

c(·1

)c(·2

)X(·1·2)

+ÿ

· :d(·)<n

c(·)B(X· , u>n) + B(u>n, u>n). (6.6)

Our aim is to control the truncated expansion under the naturalregularity assumptions in the white noise case, X œ CC ≠1/2≠. Since (6.6)contains the term B(X, u>n) which in turn contains the paraproductJˆx(u>n ª X), the remainder u>n will be at best in CC 1/2≠. But then thesum of the regularities of X and u>n is negative, and the term B(X, u>n)is not well defined. We therefore continue the expansion up to the point(turning out to be u>3) where we can set up a paracontrolled ansatz forthe remainder, which will allow us to make sense of ˆx(X ¶ u>n) and thusof B(X, u>n).

6.2 Paracontrolled solutionInspired by the partial tree series expansion of u we set up a paracontrolledansatz of the form

u = X + X + 2X + uQ, uQ = uÕ ª Q + u˘, (6.7)

where the functions uÕ, Q and u˘ are for the moment arbitrary, butwe assume uÕ, Q œ L “ and u˘ œ L 2“ , where from now on we fix


“ œ (1/3, 1/2). For such u, the nonlinear term takes the form

ˆxu2 = ˆx(X2 + 2X X + (X )2 + 4X X) + 2ˆx(uQX)

+ 2ˆx(X (uQ + 2X )) + ˆx((uQ + 2X )2), (6.8)

which gives us an equation for uQ:

L uQ

=ˆx

((X )2 + 4X X) + 2ˆx

(uQX) + 2ˆx

(X (uQ + 2X )) + ˆx

((uQ + 2X )2)

=L X + 4L X + 2ˆx

(uQX) + 2ˆx

(X (uQ + 2X )) + ˆx

((uQ + 2X )2).(6.9)

In Lemma 1 we showed that X œ CH≠1/2≠. But now we understand Besovspaces and Gaussian hypercontractivity well enough so that we can returnto the proof and modify the argumentation in order to show that X œCC ≠1/2≠. If we then formally apply the paraproduct estimate Theorem 4(which is of course not possible since the regularity requirements forthe resonant term are not satisfied), we obtain X2 œ CC ≠1≠ and thenˆxX2 œ CC ≠2≠. Therefore, X = J(ˆxX2) should be in CC 0≠. Notethat Lemma 11 does not apply here, because ≠2≠ is not in (≠2, 0). Butwe only needed this requirement to control the temporal regularity in LŒ

of the image of J . For arbitrary – œ R we have Ju œ CC – wheneveru œ CC –≠2, see for example Lemma A.9 in [14]. Similarly we derivethe formal regularities of the remaining driving terms: X œ L 1/2≠,X œ L 1/2≠, and X œ L 1≠. In terms of “, we can encode this as

X œ CC “≠1, X œ CC 2“≠1, X œ L “ , X œ L “ , X œ L 2“ .

Under these regularity assumptions the term 2ˆx(X (uQ+X ))+ˆx((uQ+X )2) is well defined and the only problematic term in (6.9) is ˆx(uQX).Using the paracontrolled structure of uQ, we can make sense of ˆx(uQX)as a bounded operator provided that Q ¶ X œ CC 2“≠1 is given. Inother words, the right hand side of (6.9) is well defined for paracontrolleddistributions.

Next, we should specify how to choose Q and which form uÕ will takefor the solution uQ. We have formally

L uQ

=L X + 4L X + 2ˆx

(uQX) + 2ˆx

(X (uQ + 2X )) + ˆx

((uQ + 2X )2)

=4ˆx

(X X) + 2ˆx

(uQX) + CC 2“≠2

=4X ª ˆx

X + 2uQ ª ˆx

X + CC 2“≠2,

Chapter 6. The stochastic Burgers equation 85

where we assumed that not only L X œ CC “≠2, but that ˆx(X ¶ X) œCC 2“≠1 (which implies L X œ CC “≠2, but also the stronger statementL X ≠ X ª ˆxX œ CC 2“≠2). By Theorem 7, uQ is paracontrolled byJ(ˆxX), and in other words we should set Q = J(ˆxX). The derivative uÕ

of the solution uQ will then be given by uÕ = 4X + 2uQ.Unlike for pam, here we do not need to introduce a renormalization.

This is due to the fact that we di�erentiate after taking the square: toconstruct u2, we would have to subtract an infinite constant and formallyconsider u ù 2 = u2 ≠ Œ, or at the level of the approximation u2

n ≠ cn. Butthen

ˆxu ù 2 = limnæŒ

ˆx(u2

n ≠ cn) = limnæŒ

ˆxu2

n = ˆxu2.

So we obtain the following description of the driving data for the stochasticBurgers equation.

Definition 7. (sbe–enhancement) Let “ œ (1/3, 1/2) and let

Xsbe

™ CC “≠1 ◊ CC 2“≠1 ◊ L “ ◊ L 2“ ◊ CC 2“≠1 ◊ CC 2“≠1

be the closure of the image of the map �sbe

: C(R+

, CŒ(T)) æ Xsbe

givenby

�sbe

(◊) = (X(◊), X (◊), X (◊), X (◊), (X ¶ X)(◊), (Q ¶ X)(◊)),(6.10)

whereX(◊) = J(ˆx◊),

X (◊) = B(X(◊), X(◊)),X (◊) = B(X (◊), X(◊)),X (◊) = B(X (◊), X (◊)),

Q(◊) = J(ˆxX(◊)).

(6.11)

We will call �sbe

(◊) the sbe–enhancement of the driving distribution ◊.For T > 0 we define X

sbe

(T ) = Xsbe

|[0,T ]

and we write ÎXÎXsbe(T )

forthe norm of X in the Banach space CT C “≠1 ◊ CT C 2“≠1 ◊ L “

T ◊ L 2“T ◊

CT C 2“≠1 ◊ CT C 2“≠1. Moreover, we define the distance dXsbe(T )

(X, X) =ÎX ≠ XÎXsbe(T )

.

For every X œ Xsbe

, there is an associated space of paracontrolleddistributions:

Definition 8. Let X œ Xsbe

. Then the space of paracontrolled distributionsD“(X) is defined as the set of all (u, uÕ) œ CC “≠1 ◊ L “ with

u = X + X + 2X + uÕ ª Q + u˘,


where u˘ œ L 2“ . For T > 0 we define

ÎuÎD“

T

= ÎuÕÎL “

T

+ Îu˘ÎCT

C 2“ .

If X œ Xsbe

and (u, uÕ) œ D“(X), then we also write

dD“

T

(u, u) = ÎuÕ ≠ uÕÎL “

T

+ Îu˘ ≠ u˘ÎCT

C 2“

T

.

We now have everything in place to solve sbe driven by X œ Xsbe

.

Theorem 9. Let “ œ (1/3, 1/2). Let X œ Xsbe

, write ˆx◊ = L X, and letu

0

œ C 2“ . Then there exists a unique solution u œ D“(X) to the equation

L u = ˆxu2 + ˆx◊, u(0) = u0

, (6.12)


t

=Œ} > 0.

Moreover, u depends on (u0

,X) œ C 2“ ◊ Xsbe

in a locally Lipschitzcontinuous way: if M, T > 0 are such that for all (u

0

,X) with Îu0

Î2“ ‚

ÎXÎXsbe(T )

6 M , the solution u to the equation driven by (u0

,X) satisfies·(u) > T , and if (u

0

, X) is another set of data bounded in the above senseby M , then there exists C(M) > 0 for which

dD“

T

(u, u) 6 C(M)(Îu0

≠ u0

Î2“ + dXsbe(T )

(X, X)).

Proof. By definition of the term ˆxu2, the distribution u œ D“(X)solves (6.12) if and only if uQ = u ≠ X ≠ X ≠ 2X solves

L uQ = L X +4ˆx

(X X)+2ˆx

(uQX)+2ˆx

(X (uQ+2X ))+ˆx

((uQ+2X )2)

with initial condition uQ(0) = u0

. This equation is structurally verysimilar to pam (5.1) and can be solved using the same arguments, whichwe do not reproduce here. ⇤

For this result to be of any use we still have to show that if › is the space-time white noise, then there is almost surely an element of X

sbe

associatedto ˆx›. While for pam we needed to construct only one term, here we haveto construct five terms: X , X , X , X ¶ X, Q ¶ X. For details we referto [16]. Alternatively we can simply di�erentiate the extended data whichHairer constructed for the KPZ equation in Chapter 5 of [18].

The same approach allows us to solve the KPZ equation L h =(ˆxh) ù 2 + ›, and if we are careful how to interpret the product w ù ›,then also the linear heat equation L w = w ù ›. In both cases the solutiondepends continuously on some suitably extended data that is constructedfrom › in a similar way as described in Definition 7. Moreover, the formallinks between the three equations that we discussed in Section 4.2 can bemade rigorous. These results are included in [16].

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Massimiliano GubinelliCEREMADE & CNRS UMR 7534, Université Paris Dauphine and IUF,France.Hausdor� Center of Mathematics & Institute of Applied Mathematics,Universität Bonn, [email protected]://www.iam.uni-bonn.de/abteilung-gubinelli/home

Nicolas PerkowskiInstitut für MathematikHumboldt–Universität zu Berlin, [email protected]://www.math.hu-berlin.de/~perkowsk/

https://www.math.hu-berlin.de/~perkowsk/