RENORMALIZATION GROUP AND STOCHASTIC PDE'S

RENORMALIZATION GROUP AND STOCHASTIC PDE’S

arXiv:1410.3094v1 [math.PR] 12 Oct 2014

ANTTI KUPIAINEN1,2

Abstract. We develop a Renormalization Group (RG) approach to the study of existence and uniqueness of solutions to stochastic partial differential equations driven by space-time white noise. As an example we prove well-posedness and independence of regularization for the φ4 model in three dimensions recently studied by Hairer. Our method is ”Wilsonian”: the RG allows to construct effective equations on successive space time scales. Renormalization is needed to control the parameters in these equations. In particular no theory of multiplication of distributions enters our approach.

1. Introduction Nonlinear parabolic PDE’s driven by a space time decorrelated noise are ubiquitous in physics. Examples are thermal noise in fluid flow, random deposition in surface growth and stochastic dynamics for spin systems and field theories. These equations are of the form ∂t u = ∆u + F (u) + Ξ (1) where u(t, x) is defined on Λ ⊂ Rd , F (u) is a function of u and possibly its derivatives which can also be non-local and Ξ is white noise on R × Λ, formally E Ξ(t′ , x′ )Ξ(t, x) = δ(t′ − t)δ(x′ − x).

(2)

Usually in these problems one is interested in the behavior of solutions in large time and/or long distances in space. In particular one is interested in stationary states and their scaling properties. These can be studied with regularized versions of the equations where the noise is replaced by a mollified version that is smooth in small scales. Often one expects the large scale behavior is insensitive to such regularization. From the mathematical point of view and sometimes also from the physical one it is of interest to inquire the short time short distance properties i.e. the well-posedness of the equations without regularizations. Then one is encountering the problem that the solutions are expected to have very weak regularity, they are distributions, and it is not clear how to set up the solution theory for the nonlinear equations in distribution spaces. Recently this problem was adressed by Martin Hairer [1] who set up a solution theory for a class of such equations, including the KPZ equation in one spatial dimension and the nonlinear heat equation with qubic nonlinearity in three spatial dimensions. The class of equations discussed by Hairer are subcritical in the sense that the nonlinearity vanishes in small scales in the scaling that preserves the linear and noise terms in the equation. In physics terminology these equations are superrenormalizable. This means the following. Let Ξǫ be a mollified noise with short scale cutoff ǫ. One can write a formal series solution to the mollified version of eq. (1) by starting with the solution Date: October 14, 2014. 1

2

A. KUPIAINEN

ηǫ (t) = ηǫ (t, ·) of the linear (F = 0) equation and iterating: Z t e(t−s)∆ F (ηǫ (s))ds + . . . uǫ (t) = ηǫ (t) +

(3)

0

Typically, the random fields (apart from η) occurring in this expansion have no limits as ǫ → 0: even when tested by smooth functions their variances blow up. These divergencies are familiar from quantum field theory (QFT). Indeed, the correlation functions of uǫ have expressions in terms of Feynman diagrams and as in QFT the divergencies can be cancelled in this formal expansion by a adding to F extra ǫ-dependent terms, so-called counter terms. In QFT there is a well defined algorithm for doing this and in the superrenormalizable case rendering the first few terms in the expansion finite cures the divergences in the whole expansion. Hairer’s work can be seen as reformulating this perturbative renormalization theory as a rigorous solution theory for the subcritical equations. It should be stressed that [1] goes further by treating also rough non-random forces. In QFT there is another approach to renormalization pioneered by K. Wilson in the 60’s [2]. In Wilson’s approach adapted to the SPDE one would not try to solve equation (1), call it E, directly but rather go scale by scale starting from the scale ǫ and deriving effective equations En for larger scales 2n ǫ := ǫn , n = 1, 2, . . . . Going from scale ǫn to ǫn+1 is a problem with O(1) cutoff when transformed to dimensionless variables. This problem can be studied by a standard Banach fixed point method. The possible singularities of the original problem are present in the large n behavior of the corresponding effective equation. One views n → En as a dynamical system and attempts to find an initial condition at n = 0 i.e. modify E so that if we fix the scale ǫn = ǫ′ and then let ǫ → 0 (and as a consequence n → ∞) the effective equation at scale ǫ′ has a limit. It turns out that controlling this limit for the effective equations allows one then to control the solution to the original equation (1). In this paper we carry out Wilson’s renormalization group analysis to the qubic nonlinear heat equation in three dimensions. This equation is a good test case since its renormalization is non trivial in the sense that a simple Wick ordering of the nonlinearity is not sufficient. Our analysis is robust in the sense that it works for other subcritical cases like the KPZ equation. We prove almost sure local well-posedness for the mild (integral equation) version of (1) thereby recoverng the results in [1]. Our renormalization group method is a combination of the one developed in [3] for parabolic PDE’s and the one in [4] used for KAM theory. The content of the paper is as follows. In section 2 we define the model and state the result. The RG formalism is set up in a heuristic fashion in Sections 3 and 4. Sections 5 and 6 discuss the leading perturbative solution and set up the fixed point problem for the remainder. Section 7 states the estimates for the perturbative noise contributions and in Section 8 the functional spaces for RG are defined and the fixed point problem solved. The main result is proved in Section 9. Finally in Sections 10 and 11 estimates for the covariances of the various noise contributions are proved. 2. The ϕ43 model Let Ξ(t, x) be space time white noise on x ∈ T3 i.e. Ξ = β˙ with β(t, x) Brownian in time and white noise in space. Given a realization of the noise Ξ we want to make sense and solve the equation ∂t ϕ = ∆ϕ − ϕ3 − rϕ + Ξ,

ϕ(0) = ϕ0

(4)


3

on some time interval [0, τ ] and show τ > 0 almost surely. Due to the nonlinearity the equation (4) is not well defined. We need to define it through regularization. To do this we first formally write it in its integral equation form ϕ = G(−ϕ3 − rϕ + Ξ) + et∆ ϕ0 (5) where

(Gf )(t) =

Z

t

e(t−s)∆ f (s)ds.

0 R t (t−s)∆ dβ(s) 0 e

(for f = Ξ this stands for (Gξ)(t) = ). Next, introduce a regularization parameter ǫ > 0 and define Z t (1 − χ((t − s)/ǫ2 ))e(t−s)∆ f (s)ds. (6) (Gǫ f )(t) = 0

where χ ≥ 0 is a smooth bump, χ(t) = 1 for t ∈ [0, 1] and χ(t) = 0 for t ∈ [2, ∞). The regularization of (5) with ϕ0 = 0 is then defined to be ϕ = Gǫ (−ϕ3 − rǫ ϕ + Ξ).

(7)

We look for rǫ such that (7) has a unique solution ϕ(ǫ) which converges as ǫ → 0 to a non trivial limit. Note that since only t − s ≥ ǫ2 contribute in (6) Gǫ Ξ is a.s. smooth. Our main result is Theorem 1. There exits rǫ s.t. the following holds. For almost all realizations of the white noise Ξ there exists t(Ξ) > 0 such that the equation (7) has for all ǫ > 0 a unique smooth solution ϕ(ǫ) (t, x), t ∈ [0, t(Ξ)] and there exists ϕ ∈ D ′ ([0, t(Ξ)] × T3 ) such that ϕ(ǫ) → ϕ in D ′ ([0, t(Ξ)] × T3 ). The limit ϕ is independent of the regularization χ. Remark 2. We will find that the renormalization parameter is given by rǫ = r + m1 ǫ−1 + m2 log ǫ + m3

(8)

where the constants m1 and m3 depend on χ whereas the m2 is universal i.e. independent on χ. They of course agree with the mass renormalization needed to make sense of the formal stationary measure of (4) 1

µ(dφ) = e− 4

R

T3

φ(x)4

ν(dφ)

(9)

where ν is Gaussian measure with covariance (−∆ + r)−1 . Remark 3. This result can be extended a large class of initial conditions, deterministic or random. As an example we consider the random case where φ0 = η0 where η0 is the gaussian random field on T3 with covariance − 21 ∆−1 , independent of Ξ (this is the stationary state of the linear equation). Then Theorem 1 holds a.s. in the initial condition and Ξ, see Remark 6. 3. Effective equation Consider the cutoff problem ϕ = Gǫ (V (ϕ) + Ξ) for ϕ(t, x) on (t, x) ∈ [0, τ ] ×

T3

with

V (ϕ)(t, x) = −ϕ3 (t, x) − rǫ ϕ(t, x).

(10)

4

A. KUPIAINEN

Let us attempt increasing the cutoff ǫ to ǫ′ > ǫ by solving the equation (10) for scales between ǫ and ǫ′ . To do this split Gǫ = Gǫ′ + Γǫ,ǫ′ with (Γǫ,ǫ′ f )(t) =

Z

0

t

2

(χ((t − s)/ǫ′ ) − χ((t − s)/ǫ2 ))e(t−s)∆ f (s)ds.

Thus Γǫ,ǫ′ involves temporal scales between ǫ2 and ǫ′ 2 (and due to the heat kernel spatial scales between ǫ and ǫ′ ). Next, write ϕ = ϕ′ + Z and determine Z =

Z(ϕ′ )

as a function of

ϕ′

(11)

by solving the small scale equation

Z = Γǫ,ǫ′ (V (ϕ′ + Z) + Ξ). Eq. (10) will then hold provided

ϕ′

(12)

is a solution to the ”renormalized” equation

ϕ′ = Gǫ′ (V ′ (ϕ′ ) + Ξ)

(13)

where V ′ (φ′ ) = V (ϕ′ + Z(φ′ )). Eq. (13) is of the same form as (10) except that the cutoff has increased and V is replaced by V ′ . Combining (12) and (13) we see that the new V ′ can be obtained by solving a fixed point equation V ′ = V (· + Γǫ,ǫ′ (V ′ + Ξ)).

(14)

Finally, the solution of (10) is gotten from (11) as ϕ = ϕ′ + Γǫ,ǫ′ (V ′ (ϕ′ ) + Ξ)

(15) ′

ǫ′

ǫ′

Our aim is to study the flow of the effective equation V at scale as increases from 1 ǫ to τ 2 where [0, τ ] is the time interval where we try to solve the original equation. It will be convenient to do this step by step. We fix a number λ < 1 (taken to be small in the proof to kill numerical constants) and take the cutoff scale ǫ = λN .

(16)

V (N ) (ϕ) = −ϕ3 − rλN ϕ.

(17)

(N )

(18)

The corresponding V is denoted as (N )

Let Vn be the solution of (14) with ǫ′ = λn . We will construct these functions iteratively in n i.e. derive the effective equation on scale λn−1 from that of λn : (N )

Vn−1 = Vn(N ) (· + Γλn ,λn−1 (Vn−1 + Ξ)). The solution of (10) is then also constructed iteratively: let (N )

FN (ϕ) := ϕ and define

(N )

(N )

Fn−1 = Fn(N ) (· + Γλn ,λn−1 (Vn−1 + Ξ)). Then the solution of (10) is ϕ = Fn(N ) (ϕn ). where ϕn solves ϕn = Gλn (Vn(N ) (ϕn ) + Ξ).

(19)

(20)


5

Finally we want to control the limit N → ∞ where the regularization is removed i.e. (N ) (N ) construct the limits Vn = limN →∞ Vn and Fn = limN →∞ Fn which are then shown 2n to describe the solution of (4) on time interval [0, λ ]. How long this time interval will be i.e. how small n we can reach in the iteration depends on the realization of the noise. In informal terms, let Am be the event that these limits exist for n ≥ m. We will show P(Am ) ≥ 1 − O(λRm ).

(21)

with large R i.e. the set of noise s.t. the equation (4) is well posed on time interval [0, τ ] has probability 1 − O(τ R ). 4. Renormalization group The equation (18) deals with scales between λn and λn−1 . Instead of letting the scale of the equations vary it will be be more convenient to rescale everything to fixed scale (of order unity) after which we need to iterate a O(1)-scale problem. This is the ”Wilsonian” approach to Renormalization Group. Let us define the space time scaling sµ by 1

(sµ f )(t, x) = µ 2 f (µ2 t, µx). The Green function of the heat equation and the space time white noise transform in a simple way under this scaling: 2 sµ Gs−1 µ = µ G,

2 sµ Gǫ s−1 µ = µ Gǫ/µ ,

d

sµ Ξ = µ−2 Ξµ

as one can easily verify by a simple changes of variables. Here Ξµ is space time white noise on R × µ−1 T3 . Set (22) φn = sλn ϕn where ϕn solves (20). Note that φn (t, x) is defined on x ∈ Tn with Tn := λ−n T3

(23)

Since ϕn was interpreted as a field involving spatial scales on [λn , 1] φn may be thought as a field involving scales on [1, λ−n ]. In these new variables eq. (20) becomes φn = G1 (vn(N ) (φn ) + ξn )

(24)

provided we define vn(N ) := λ2n sλn ◦ Vn(N ) ◦ s−1 λn

and

ξn := λ2n sλn Ξ. ξn is distributed as a space time white noise on R × Tn . Defining fn(N ) := sλn ◦ Fn(N ) ◦ s−1 λn

the solution of (10) is given by (N ) ϕ = s−1 λn fn (φn ).

(25)

The iterative equations (18) and (19) have their unit scale counterparts: recalling (16), denoting s := sλ and using sλn+1 = s ◦ sλn equation (18) becomes (N )

(N )

vn−1 (φ) = λ−2 s−1 vn(N ) (s(φ + Γ(vn−1 (φ) + ξn−1 )))

(26)

6

A. KUPIAINEN

where (Γf )(t) =

Z

t

0

(χ(t − s) − χ((t − s)/λ2 ))e(t−s)∆ f (s)ds.

(27)

Note the integrand is supported on the interval [λ2 , 2]. Eq. (19) in turn becomes (N )

(N )

fn−1 (φ) = s−1 fn(N ) (s(φ + Γ(vn−1 (φ) + ξn−1 ))).

(28)

(N )

Our task then is to solve the equation (26) for vn−1 to obtain the RG map (N )

vn−1 = Rn vn(N )

(29)

and then iterate this and (28) starting with (N )

vN (φ) = −λN φ3 − λ2N rλN φ = −λN φ3 − (λN m1 + λ2N (m2 log λN + m3 ))φ(30) (N )

fN (φ) = φ

(31) (N )

(N )

If the noise is in the set Am (to be defined) we then show the functions vn and fn have limits as N → ∞ and n ≥ m and allow to construct the solution to our original equation on time interval [0, λ2m ]. Equations (26), (28) and (24) involve the operators Γ and G1 respectively. These operators are infinitely smoothing and their kernels are local in space time. In particular the noise ζ = Γξn entering equations (26) and (28) has a smooth covariance which is short range in time Eζ(t, x)ζ(s, y) = 0 if |t − s| > 2λ−2 , (32) and it has gaussian decay in space. Hence the fixed point problem (26) turns out to be quite easy. As usual in RG studies one needs to keep track of the leading ”relevant” terms of (N ) vn which are revealed by a first and second order perturbative study of (26) to which we turn now. 5. Linearized renormalization group We will now study the fixed point equation (26) to first order in v. Define the map (Ln v)(φ) := λ−2 s−1 v(s(φ + Γξn−1 ))

(33)

Then (26) can be written as (N )

(N )

vn−1 (φ) = (Ln vn(N ) )(φ + Γvn−1 (φ))

(34)

and we see that Ln = DRn . The linear flow vn−1 = Ln vn from scale N to scale n is easy to solve by just replacing λ by λN −n : where

vn = (Ln+1 Ln+2 . . . LN vN )(φ) = λ−2(N −n) sn−N vN (sN −n (φ + ηn(N ) ))

(35) ηn(N ) = ΓN n ξn . −n 3 N Here ξn is space time white noise on Tn = λ T and Γn is given by (27) with λ replaced by λN −n i.e. its integral kernel is where we denoted

ΓN n (t, s, x, y) = χN −n (t − s)Hn (t − s, x − y) Hn (t, x − y) = et∆ (x, y)

(36)


7

the heat kernel on Tn and χN −n (s) := χ(s) − χ(λ−2(N −n) s)

(37)

[λ2(N −n) , 2].

is a smooth indicator of the interval The linearized flow is especially simple for a local v as the one we start with (30). For vN = φk we get vn = λn(k−5)/2 (φ + ηn(N ) )k and so we have ”eigenfunctions” (N )

Ln (φ + ηn(N ) )k = λ(k−5)/2 (φ + ηn−1 )k .

For k < 5 these are ”relevant”, for k > 5 they are ”irrelevant” and for k = 5 ”marginal”. (N ) The covariance of ηn is readily obtained from (36) (let t′ ≥ t): Z t (N ) ′ ′ (N ) Hn (t′ − t + 2s, x′ − x)χN −n (t′ − t + s)χN −n (s)ds Eηn (t , x )ηn (t, x) = :=

0 Cn(N ) (t′ , t, x′ , x)

In particular we have Eηn(N ) (t, x)2

=

Z

t

(38)

Hn (2s, 0)χN −n (s)2 ds.

(39)

0

This integral diverges as N − n → ∞ and is the source of the first renormalization constant in (30). We need to study the N and χ dependence of the solution to (7) . Since these ′(N ) dependencies are very similar we deal with them together. Thus let Γn the operator ′ (36) where the lower cutoff in (37) is modified to another bump χ : χ′N −n (s) = χ(s) − χ′ (λ−2(N −n) s).

(40)

Varying χ′ allows to study cutoff dependence of our scheme. Taking χ′ (s) = χ(λ−2 s) ′(N ) (N +1) in turn implies Γn = Γn and this allows us to study the N dependence and the ¸convergence as N → ∞. The following lemma controls these dependences for (39): Lemma 4. Let ρ :=

Z

0

∞

(8πs)−3/2 (1 − χ(s)2 )ds.

(41)

Then, Eηn(N ) (t, x)2 = λ−(N −n) ρ + δn(N ) (t) with and

1

|δn(N ) (t)| ≤ C(1 + t− 2 ) 1

(42)

|δn(N ) (t) − δn′(N ) (t)| ≤ C(t− 2 1[0,2λ2(N−n) ] (t) + e−cλ

−2N

)kχ − χ′ k∞

(43)

The Lemma is proved in Section 10. We will fix in (30) the first renormalization constant Defining

a = −3ρ

ρk = λ−k ρ, the first order solution to our problem is ) u(N := −λn ((φ + ηn(N ) )3 + 3ρN −n (φ + ηn(N ) )). n

(44) (45) (46)

8

A. KUPIAINEN (N )

Remark 5. Note that since we have the factor λn in un the counting of what terms (N ) are relevant, marginal or irrelevant depends on the order in λn . Thus λn (φ + ηn )k is is (N ) relevant for k < 3, marginal for k = 3 and irrelevant for k > 3. Similarly, λ2n (φ+ηn )k is marginal for k = 1 which is the source of the renormalisation constant b in (8). The terms of order λ3n are all irrelevant. For a precise statement, see Proposition 9. Remark 6. Consider the random initial condition discussed in Remark 3. We realize it in terms of the white noise on (−∞, 0] × Tn . In a regularized form we replace et∆ ϕ0 in(5) by Z 0 (1 − χ((t − s)/ǫ2 ))e(t−s)∆ Ξ(s)ds. (47) −∞

(N )

This initial condition can be absorbed to ηn . Indeed, the covariance (38) is just replaced by the stationary one Z t (N ) ′ ′ (N ) Eηn (t , x )ηn (t, x) = Hn (t′ − t + 2s, x′ − x)χN −n (t′ − t + s)χN −n (s)ds (48) −∞

(N )

and in particular δn

= 0 which makes the analysis in Section 11 actually less messy. 6. Second order calculation (N )

We will solve equation (26) by a fixed point argument in a suitable space of vn . Before going to that we need to spell out explicitly the leading relevant (in the RG (N ) sense) terms. By Remark 5 this requires looking at the second order terms in vn . (N ) (N ) To avoid too heavy notation we will drop the superscript (N ) in vn , ηn and other expressions unless needed for clarity. Let us write first separate in (26) the linear part: vn−1 = Ln vn + Gn (vn , vn−1 )

where we defined

Gn (v, v¯)(φ) = (Ln v)(φ + Γ¯ v (φ)) − (Ln v)(φ) Recalling the first order expression un in (46) we write

(49) (50)

vn = un + wn so that wn satisfies wn−1 = Ln wn + Gn (un + wn , un−1 + wn−1 ),

(51)

wN (φ) = −λ2N (m2 log λN + m3 )φ.

(52)

with the initial condition

O(λn )

O(λ2n ).

The reader should think about un as and wn as Next, we separate from the Gn -term in (51) the O(λ2n ) contribution. Since Ln un = un−1 we have Gn (un , un−1 )(φ) = un−1 (φ + Γun−1 (φ)) − un−1 (φ) (53) n which to O(λ ) equals Dun−1 Γun−1 where Hence

Dun−1 = −3λn−1 ((φ + ηn−1 )2 − ρN −n+1 ).

wn−1 = Ln wn + Dun−1 Γun−1 + Fn (wn−1 )

(54)


9

where Fn (wn−1 ) = Gn (un + wn , un−1 + wn−1 ) − Dun−1 Γun−1 (55) Fn is and will turn out to be irrelevant under the RG (i.e. it will contract in a suitable norm under the linear RG map Ln ). It is useful to solve (54) without the Fn term: let Un satisfy O(λ3n )

Un−1 = Ln Un + Dun−1 Γun−1 ,

The solution is

UN = −λ2N (m2 log λN + m3 )φ

2n N Un = Dun ΓN (56) n un − λ (m2 log λ + m3 )(φ + ηn ) N where Γn is defined in (37). This is seen by just doing one step RG with λ replaced by λN −n . Now write wn = Un + νn (57) so that νn−1 satisfies the equation

νn−1 = Ln νn + Fn (Un−1 + νn−1 ),

νN = 0

(58)

Eq. (58) is a fixed point equation that we will solve by contraction in a suitable space. 7. Noise estimates The time of existence for the solution depends on the size of the noise. In this Section we state probabilistic estimates for this size. The noise enters the fixed point equation (58) in the form Γξn−1 that enters the definition of Ln in (33) and in the polynomials un (46) and Un (56) of the random field ηn (35). According to Remark 5 in the second order term Un only the constant in φ term should be relevant under the linear RG and the linear in φ term should be marginal (neutral), the rest being irrelevant (contracting). We’ll see this indeed is the case and accordingly write Un (φ) = Un (0) + DUn (0)φ + Vn (φ) (59) where explicitly 3 2n N 2n Un (0) = 3λ2n (ηn2 − ρN −n )ΓN n (ηn − 3ρN −n ηn ) − λ (m2 log λ + m3 )ηn := λ ωn (60)

and

2n

(DUn (0)φ)(t, x) = λ zn (t, x)φ(t, x) + λ

2n

Z

zn (t, x, s, y)φ(s, y)dsdy

(61)

where and

3 zn = 6ηn ΓN n (ηn − 3ρN −n ηn )

2 N zn = 9(ηn2 − ρN −n )ΓN n (ηn − ρN −n ) − m2 log λ − m3 (here ηn2 − ρN −n is viewed as a multiplication operator). The random fields whose size we need to constrain probabilistically are then

ηn , ηn2 − ρN −n , ηn3 − 3ρN −n ηn , ωn zn , zn

(62) (63)

(64)

They belong to the Wiener chaos of white noise of bounded order and their size and regularity are controlled by studying their covariances. For finite cutoff parameter N these noise fields are a.s. smooth but in the limit N → ∞ they become distribution valued. These fields enter in the RG iteration (34) in the combination Γvn−1 i.e. they are always acted upon by the operator Γ which is infinitely smoothing. Therefore we estimate their size in suitable (negative index) Sobolev type norms which we now define. In addition to the fields (64) we also need to constrain the Gaussian field Γξn .

10

A. KUPIAINEN

Let K1 be the operator (−∂t2 + 1)−1 on L2 (R) i.e. it has the integral kernel 1

K1 (t, s) = 2 e−|t−t | ′

(65)

Let K2 = (−∆ + 1)−2 on L2 (Tn ) which has a continuous kernel K2 (x, y) = K2 (x − y) satisfying K2 (x) ≤ Ce−|x| . (66) Set K := K1 K2 . (67) ∞ Define Vn to be be the completion of C0 (R+ × Tn ) with the norm kvkVn = sup kKvkL2 (ci )

(68)

i

where ci is the unit cube centered at i ∈ Z × (Z3 ∩ Tn ). To deal with the bi-local field zn in (63) we define for z(t, x, s, y) in C0∞ (R+ × Tn × R+ × Tn ) X kK ⊗ KzkL2 (ci ×cj ) (69) kzkVn = sup i

j

Now we can specify the admissible set of noise. Let γ > 0 and define events Am , m > 0 (N ) in the probability space of the space time white noise Ξ as follows. Let ζn denote any (N ) one of the fields (64). We want to constrain the size of ζn on the time interval [0, τn−m ] where we will denote τn−m = λ−2(n−m) . To do this choose a smooth bump h on R with h(t) = 1 for t ≤ −λ2 and h(t) = 0 for t ≥ − 21 λ2 and set hk (t) = h(t−τk ) so that hk (t) = 1 for t ≤ τk − λ2 and hk (t) = 0 for t ≥ τk − 21 λ2 (the reason for these strange choices will become clear in Section 8). The first condition on Am is that for all N ≥ n ≥ m the following hold: khn−m ζn(N ) kVn ≤ λ−γn (70) (N )

We need also to control the N and χ dependence of the noise fields ζn . Recall that (N ) we can study both by varying the lower cutoff in the operator Γn in (36). We denote ′(N ) by ζn any of the resulting noise fields. Our second condition on Am is that for all N ≥ n ≥ m and all cutoff functions χ, χ′ with bounded C 1 norm khn−m (ζn′(N ) − ζn(N ) )kVn ≤ λγ(N −n) λ−γn .

(71)

The final condition concerns the fields Γξn entering the RG iteration (26). Note that these fields are N independent and smooth and we impose them a smoothness condition given in (72). We have: Proposition 7. There exist renormalization constants m2 and m3 such that for some γ > 0 almost surely Am holds for some m < ∞. On Am we will control the RG iteration for scales n ≥ m. This will enable us to solve the equation (4) on the time interval [0, λ2m ]. 8. Fixed point problem We will now fix the noise Ξ ∈ Am for some m > 0 and set up a suitable space of functions νn (φ) for the fixed point problems (58) for n ≥ m. Since the noise contributions take values in Vn we let νn take values there as well. The noise enters in (26) in the argument of vn in the combination Γ(vn−1 + ξn−1 ). Since Γ is infinitely smoothing this means that we may take the domain of νn (φ) to consist of suitably smooth functions φ. Since the noise contributions become distributions in


11

the limit N → ∞ and enter multiplicatively with φ e.g. in (46) we need to match the smoothness condition for φ with that of the noise. Finally, since (24) implies φn ≡ 0 on [0, 1] we let the φ be defined on [1, τn−m ]. With these motivations we take the domain Φn of vn to consist of φ : [1, τn−m ]× Tn → R which are C 2 in t and C 4 in x with ∂ti φ(1, x) = 0 for i ≤ 2 and all x ∈ Tn . We equip Φn with the sup norm X kφkΦn := k∂ti ∂xα φk∞ . i≤2,|α|≤4

We will now set up the RG map (34) in a suitable space of vn , vn−1 defined on Φn and Φn−1 respectively. 1 First, note that for φ ∈ Φn−1 , sφ(t, x) = λ 2 φ(λ2 t, λx) is defined on [λ−2 , τn−m ] × Tn . We extend it to [1, τn−m ] × Tn by setting sφ(t, x) = 0 for 1 ≤ t ≤ λ−2 . Next, Γξn−1 vanishes (a.s.) on [0, λ2 ] and thus sΓξn−1 vanishes on [0, 1]. We can now state the final condition for the set Am : for all n > m we demand ksΓξn−1 kΦn ≤ λ−γn .

(72)

λ−2γn

Let Bn ⊂ Φn be the open ball centered at origin of radius rn = and Wn (Bn ) be the space of analytic functions from Bn to Vn equipped with the supremum norm which we denote by k · kBn . We will solve the fixed point problem (34) in this space. We collect some elementary properties of these norms in the following lemma, proven in Section 10: Lemma 8. (a) sΓ : Vn−1 → Φn and hn−1−m Γ : Vn−1 → Vn−1 are bounded operators with norms bounded by C(λ). Moreover sΓhn−1−m v = sΓv as elements of Φn . (b) s : Φn−1 → Φn and s−1 : Vn → Vn−1 are bounded with 1

ksk ≤ λ 2 ,

1

ks−1 k ≤ Cλ− 2 .

(c) Let φ ∈ C 2 (R) and v ∈ Vn . Then φv ∈ Vn and kφvkVn ≤ CkφkC 2 kvkVn . The linear RG (33) is controlled by Proposition 9. Given λ < 1, γ > 0 there exists n(γ, λ) s.t. for n ≥ n(γ, λ) Ln maps 1 5 Wn (Bn ) into Wn−1 (λ− 2 Bn−1 ) with norm kLn k ≤ Cλ− 2 . 1

Proof. Let v ∈ Wn (Bn ) and φ ∈ λ− 2 Bn−1 . By (72) and Lemma 8 (a,b) ks(φ + Γξn−1 )kΦn ≤ λ−2γ(n−1) + λ−γn ≤ λ−2γn

1

for n ≥ n(γ, λ). Hence vn (s(φ + Γξn−1 )) is defined and analytic in φ ∈ λ− 2 Bn−1 i.e. Ln 1 maps Wn (Bn ) into Wn−1 (λ− 2 Bn−1 ) and by Lemma 8(b) kLn vk

5

1 −2

λ

Bn−1

≤ Cλ− 2 kvkBn

As a corollary of Lemma 8(c) and (70) we obtain for n ≥ m and N ≥ n: ) (1−6γ)n khn−m u(N n kBn ≤ Cλ

(73)

khn−m (Un(N ) (0) + DUn(N ) (0)φ)kBn ≤ Cλ(2−3γ)n .

(74)

and Consider next

(N ) Vn :

12

A. KUPIAINEN

Lemma 10. There exist λ0 > 0, γ0 > 0 so that for λ < λ0 , γ < γ0 and m > m(γ, λ) if Ξ ∈ Am then for all N ≥ n ≥ m khn−m Vn(N ) kBn ≤ Aλγ(N −n) λ(2−10γ)n .

(75)

Proof. Inductively assume (75). The n = N − 1 case follows easily from the definition (56) and (73) since ΓN N −1 = Γ. For the inductive step note first that by (33) and hn−1−m get

(λ2 ·)

hn−m−1 Ln = Ln hn−m−1 (λ2 ·)

(76)

is supported on [0, τn−m − ]. Since hn−m = 1 on [0, τn−m − 1 2

λ2 ]

hn−m−1 (λ2 t) = hn−m−1 (λ2 t)hn−m (t)

so that

we

(77)

(N )

hn−1−m Vn−1 = hn−1−m (Ln hn−m Vn(N ) )

(78)

By construction (N )

Vn−1 (φ) = (Ln Vn(N ) )(φ) − (Ln Vn(N ) )(0) − D(Ln Vn(N ) )(0)φ.

Combining these with Proposition 9 and Lemma 8(b) we get by a Cauchy estimate 5

(N )

3

1

khn−1−m Vn−1 kBn−1 ≤ hλ− 2 (λ− 2 rn−1 /rn −1)−2 khn−m Vn(N ) kBn ≤ 2hλ− 2 λ−4γ khn−m Vn(N ) kBn (79) 1 −15γ 2γ− 21 2 > 2. We denoted h = khkC 2 . The claim follows provided 2hλ < 3. if λ (N )

Next we will rewrite the fixed point equation (34) in a localized form. Define vñ (N ) hn−m vn . By (78) and Lemma 8(a) we may write (34) as (N )

(N )

vñ−1 (φ) = hn−1−m (Ln vñ(N ) )(φ + Γ˜ vn−1 (φ)). Thus the ν fixed point problem (58) becomes ñ−1 + νñ−1 )), νñ−1 = hn−1−m (Ln νñ + Fñ (U

=

(80)

νÑ = 0

(81)

where Fñ is as in (55) i.e. Fñ (w) = Gn (˜ un + w ñ , u ñ−1 + w) − D˜ un−1 Γ˜ un−1

(82)

Thus with only a slight abuse of notation we will drop the tildes and h factors in the norms in the following: Proposition 11. There exist λ0 > 0, γ0 > 0 so that for λ < λ0 , γ < γ0 and m > m(γ, λ) if Ξ ∈ Am then then for all N ≥ n − 1 ≥ m the equation (81) has a unique solution (N ) νn−1 ∈ Wn (Bn−1 ). These solutions satisfy 1

(N )

kνn(N ) kBn ≤ λ(3− 4 )n

(83)

and νn converge in Wn (Bn ) to a limit νn ∈ Wn (Bn ) as N → ∞. νn is independent on the small scale cutoff: νn = νn′ . 1

Proof. We solve (81) by Banach fixed point theorem in the ball kνn−1 kBn−1 ≤ λ(3− 4 )(n−1) . By Proposition 9 we have 5

1

1

1

kLn νn kBn−1 ≤ Cλ− 2 λ(3− 4 )n = Cλ 4 λ(3− 4 )(n−1)

(84)

Next we estimate the Fn term in (81). Fn is given in (82) so we need to start with Gn given in (53). Let f (z) := Gn (v, z¯ v )(φ). For φ ∈ Bn−1 we have by Lemma 8(a) and (72) 1

v (φ)kVn−1 ks(φ + Γξn−1 + zΓ¯ v (φ))kΦn ≤ λ 2 λ−2γ(n−1) + λ−γn + |z|C(λ)k¯


13

Hence f (z) is analytic in the ball |z| ≤ c(λ)λ−2γn k¯ v (φ)k−1 Vn−1 We use this first to estimate gn := G(un , un−1 ) − Dun−1 Γun−1 . We have gn = f (z) − f (0) − zf ′ (0) with v = un and v¯ = un−1 . By a Cauchy estimate and (73) kgn kBn−1 ≤ C(λ)λ4γn kun kBn kun−1 k2Bn−1 ≤ C(λ)λ(3−14γ)n .

(85)

Next we write Gn (un + wn , un−1 + wn−1 ) − G(un , un−1 ) = Gn (un , un−1 + wn−1 ) − G(un , un−1 ) + Gn (wn , un−1 + wn−1 ) and use Cauchy estimates to get kGn (un , un−1 + wn−1 ) − G(un , un−1 )kBn−1 kGn (wn , un−1 + wn−1 )|Bn−1

≤ C(λ)λ2γn kun kBn kwn−1 kBn−1 ≤ C(λ)kwn kBn k(kun−1 kBn−1 + kwn−1 kBn−1 ).

Combining this with (73), (74), (75) and (85) we then conclude kFn (Un−1 + νn−1 )kBn−1 ≤ C(λ)(λ(3−14γ)n + λ(1−4γ)n kνn−1 kBn−1 ).

(86)

Combining (84) and (86) and using Lemma 8(c) to bound the hn−1−m factor in (81) we 1 conclude that for γ < 1/56 the ball kνn−1 kBn−1 ≤ λ(3− 4 )(n−1) is mapped by the RHS of (81) to itself. The map is also a contraction if n ≥ n(λ) since by (84) L is and kFn (Un−1 + ν1 ) − Fn (Un−1 + ν2 )kBn−1 ≤ C(λ)λ(1−6γ)n kν1 − ν2 kBn−1 . (N )

Let us address the convergence as N → ∞ and cutoff dependence of νn . Recall ′(N ) we can deal with both questions together with νn . The estimate (71) together with Lemma 10 implies that kun(N ) − un′(N ) kBn

kUn(N ) − Un′(N ) kBn

≤ Cλγ(N −n) λ(1−6γ)n

≤ Cλγ(N −n) λ(2−3γ)n

These estimates lead to as in (86) (N )

′(N )

kFn(N ) − Fn′(N ) kBn−1 ≤ C(λ)(λγ(N −n) λ(3−14γ)n + λ(1−4γ)n kνn−1 − νn−1 kBn−1 ). As in (84) we get 5

kLn (νn(N ) − νn′(N ) )kBn−1 ≤ Cλγ(N −n) λ− 2 kνn(N ) − νn′(N ) kBn

(87)

Hence for small γ we obtain inductively for m ≤ n ≤ N 1

kνn(N ) − νn′(N ) kBn ≤ Cλγ(N −n) λ(3− 4 )n . (N )

This establishhes the convergence of νn time cutoff.

to a limit that is independent on the short

14

A. KUPIAINEN

9. Proof of Theorem 1 We are now ready to construct the solution φ(ǫ) of the ǫ cutoff equation (7). Recall that formally φ(ǫ) is given on time interval [0, λ2m ] by eq. (25) (with n = m) with φm given as the solution of eq. (24) on time interval [0, 1]. Hence we first need to study the f iteration eq. (28). This is very similar to the v iteration (26) except there is no fixed (N ) point problem to be solved and there is no multiplicative λ−2 factor. As in (80) for vn we study instead of (28) the localized iteration (N ) (N ) fñ−1 (φ) = hn−1−m s−1 fñ(N ) (s(φ + Γ(˜ vn−1 (φ) + ξn−1 )))

(88)

(N ) (N ) for fñ = hn−m fn . The following Proposition is immediate : (N )

Proposition 12. Let νñ ∈ Wn (Bn ), m ≤ n ≤ N be as in Proposition 11. Then for (N ) m ≤ n ≤ N fñ ∈ Wn (Bn ) and fn(N ) (φ) = φ + ηn(N ) + gn(N ) (φ).

with

(89)

3

k˜ gn(N ) kBn ≤ λ 4 n .

(N ) gn

(90)

and converge in Wn (Bn ) as N → ∞ to a limit gñ ∈ Wn (Bn ) which is independent on the short time cutoff. Proof. We have (N )

(N )

(N )

gñ−1 (φ) = hn−1−m (Γ˜ vn−1 (φ) + s−1 gñ(N ) (s(φ + Γ(˜ vn−1 (φ) + ξn−1 ))) (N )

Since k˜ vn−1 kBn−1 ≤ Cλ(1−3γ)(n−1) Lemma 8(b) implies (N )

1

3

3

k˜ gn−1 kBn−1 ≤ C(λ)λ(1−3γ)(n−1) + Cλ− 2 λ 4 n ≤ λ 4 (n−1) (N )

The convergence and cutoff independence follows from that of vn .

We need the following lemma: Lemma 13. G1 is a bounded operator from Vn to Φn and G1 (hn−1−m (λ2 ·)v) = G1 v. Proof of Theorem 1 . We claim that if Ξ ∈ Am the solution ϕ(N ) of equation (7) with ǫ = λN is given by (recall (25)) (N ) ϕ(N ) = s−m f˜m (0)

on the time interval [0, n>m

1 2

λ−2m ].

(91)

Let φn ∈ Φn be defined inductively by φm = 0 and for (N )

φn = s(φn−1 + Γ(˜ vn−1 (φn−1 ) + ξn−1 )). We claim that for all m ≤ n ≤ N φn ∈ Bn and φn = G1 (˜ vn(N ) (φn ) + ξn ).

(92) (93)

Indeed, this holds trivially for n = m since the RHS vanishes identically on [0, 1]. Suppose φn−1 ∈ Bn−1 satisfies (N )

φn−1 = G1 (˜ vn−1 (φn−1 ) + ξn−1 ). Then, first by by Lemma 8(b) 1

kφn kΦn ≤ λ 2 kφn−1 kΦn−1 + C(λ)λ−γn ≤ λ−2γn

(94)


so that φn ∈ Bn . Second, we have by (94) and (92) (N )

15

(N )

φn = s((G1 + Γ)(˜ vn−1 (φn−1 ) + ξn−1 )) = G1 λ2 s(˜ vn−1 (φn−1 ) + ξn−1 ) = G1 (hn−1−m (λ2 ·)˜ vn(N ) (φn ) + ξn ) = G1 (˜ vn(N ) (φn ) + ξn )

(95)

where in the third equality we used the RG iteration (80) and in the last equality Lemma 13. From (88) we have since φm = 0 (N ) (N ) (N ) (N ) (0) = h0 s−1 f˜m+1 (φm+1 ) = h0 h1 (·/λ2 )s−2 f˜m+2 (φm+2 ) = h0 s−2 f˜m+2 (φm+2 ) f˜m

where we used (77). Iterating we get (N ) (N ) f˜m (0) = h0 s−(N −m) fÑ = h0 hN −m (·/λ2(N −m) )s−(N −m) φN = h0 s−(N −m) φN (N )

(N )

(N )

again by (77). Now φN ∈ BN solves (93) with vÑ (φ) = hN −m vN (φ) with vN by (30). Since hN −m = 1 on [0, τN −m − λ2 ] we obtain

(96) given

φN = G1 (˜ vn(N ) (φn ) + ξn ) = G1 (vn(N ) (φn ) + ξn ).

and thus ϕ(N ) = s−N φN solves (91) on the time interval [0, λ−2m ]. (96) then gives (N ) (0) h0 (λ−2m ·)ϕ(N ) = s−m f˜m

so that (91) holds on the time interval [0, 21 λ−2m ]. (N )

By Proposition 12 fm (0) converges in Vm to a limit ψm which is independent on the short distance cutoff. Convergence in Vm implies convergence in D ′ ([0, 1] × Tm ). The claim follows from continuity of s−m : D ′ ([0, 1] × Tm ) → D ′ ([0, λ2m ] × T1 ). 10. Kernel estimates In this Section we prove Lemmas 8, 13 and 4 and give bounds for the various kernels entering the proof of Proposition 7. 10.1. Proof of Lemma 8 and 13. Lemma 8 (a) and Lemma 13 Let v ∈ C0∞ (R+ × Tn−1 ). Then Z ∞ k(λ2 t − s)e(t−s)∆ v(s)ds sΓv(t) = 0

1

where k(τ ) = λ 2 (χ(τ ) − χ(τ /λ2 )) vanishes for τ ≤ λ2 . Hence sΓv ∈ C0∞ ([1, ∞) × Tn ). Next, write v = (−∂t2 + 1)(−∆ + 1)2 Kv so that setting w = Kv we have Z 2 k(λ2 t − s)(−∆ + 1)2 e(λ t−s)∆ (−∂s2 + 1)w(s)ds (97) sΓv(t) = R

Integrating by parts we get Z 2 sΓv(t) = ((−∂s2 + 1)k(λ2 t − s)(−∆ + 1)2 e(λ t−s)∆ )w(s)ds. R

∂ta (k(λ2 t

2

The kernels − s)∂xα e(λ t−s)∆ (x − y)) are smooth, exponentially decreasing in |x − y| and and supported on λ2 t − s ∈ [λ2 , 2] for all a and α. Hence the corresponding operators Oaα satisfy |(Oaα 1ci w)(t, x)| ≤ C(λ)e−cd(i,(t,x)) kwkL2 (ci ) which in turn yields our claim ksΓvkΦn ≤ C(λ)kvkVn−1 . Hence sΓ extends to a bounded operator from Vn−1 to Φn .

16

A. KUPIAINEN

G1 v is given by (97) with k is replaced by 1 − χ(t − s). The time integral is confined to [0, τn−m ] so that kG1 vkΦn ≤ C(λ, n)kvkVn . 1 Lemma 8 (b). Recall (sφ)(t, x) = λ 2 φ(λ2 t, λx) on [λ−2 , τn−1−m ] × Tn−1 . Since λ < 1 1 we then get ksφkΦn ≤ λ 2 kφkΦn−1 . Next, let v ∈ C0∞ (R+ × Tn ) and w = Kv. First write Ks−1 v = Ks−1 (−∂t2 + 1)(−∆ + 1)2 w = K1 (−λ4 ∂t2 + 1)K2 (−λ2 ∆ + 1)2 s−1 w.

Next, K1 (−λ4 ∂t2 + 1) = λ4 + (1 − λ4 )K1 and

K2 (−λ2 ∆ + 1)2 = λ4 + 2λ2 (1 − λ2 )(−∆ + 1)−1 + (1 − λ2 )2 K2 . 1

Thus we need to show the operators Ki , (−∆ + 1)−1 and λ 2 s−1 are bounded in the norm supi k · kL2 (ci ) uniformly in λ. For Ki this follows from the bounds (65) and (66) and for the third one from (−∆ + 1)−1 (x, y) ≤ Ce−c|x−y| |x − y]−1 . Finally, let ci /λ := {(t/λ2 , x/λ)|(t, x) ∈ ci } Z X 1 kλ 2 s−1 wk2L2 (ci ) = λ5 kwk2L2 (c) ≤ C |w|2 ≤ λ5 ci /λ

c∩(ci /λ)6=∅

Lemma 8 (c) Let φ = φ(t) be in and v ∈ C0∞ (R+ × Tn ) and set again w = Kv so that φv = φ(−∂t2 + 1)(−∆ + 1)2 w Using φ∂t2 f = ∂t2 (φf ) − 2∂t (∂t φf ) + ∂t2 φf we get C 2 (R)

K(φv) = φw + 2∂t K1 (∂t φw) − K1 (∂t2 φw).

Since the operators ∂t K1 and K1 have bounded exponentially decaying kernels and thus are bounded in the norm supi k · kL2 (ci ) we get kφvkVn ≤ CkφkC 2 kvkVn .

−x2 /4t

10.2. Proof of Lemma 4. Let H(t, x) = et∆ (0, x) = (4πs)−3/2 e kernel on R3 . Then X Hn (t, x) = H(t, x + λ−n i).

be the heat (98)

i∈Z3

Denoting ǫ = λ2(N −n) and separating the i = 0 term we have Z t (N ) 2 (8πs)−3/2 (χ(s)2 − χ(s/ǫ2 )2 )ds + α(t) Eηn (t, x) =

(99)

0

where |α(t)| ≤

XZ i6=0

0

2

(8πs)−3/2 e−i

2 /(4sλ2n )

≤ Ce−cλ

−2n

.

Let α′ (t) have the lower cutoff replaced by χ′ . Then Z −2n −2N kχ − χ′ k∞ . |α(t) − α′ (t)k ≤ C s−3/2 |χ(s/ǫ′2 ) − χ(s/ǫ2 )2 |e−λ /4s ds ≤ Ce−cλ Denote the first term in (99) by β(t, ǫ) and β ′ (t, ǫ) where the lower cutoff is χ′ . Then Z ∞ ′ −1 β (∞, ǫ) = ǫ (8πs)−3/2 (χ(ǫ2 s)2 − χ′ (s)2 )ds = ǫ−1 ρ′ − ρ 0

So δn(N ) (t) = α(t) + ρ − γ(t, ǫ)


with γ(t, ǫ) = β(∞, ǫ) − β(t, ǫ) = Moreover ′

|γ(t, ǫ) − γ (t, ǫ)| =

Z

∞ t

Z

∞

t

17

1

(8πs)−3/2 (χ(s)2 − χ(s/ǫ2 )2 )ds ≤ Ct− 2 . 1

(8πs)−3/2 |χ(s/ǫ2 )2 − χ′ (s/ǫ2 )2 |ds ≤ Ct− 2 kχ − χ′ k∞ 1[0,ǫ] (t).

10.3. Covariance and response function bounds. We prove now bounds for the covariance and response kernels (36) and (38) that are needed for the probabilistic estimates in Section 11. These kernels are translation invariant in the spatial variable (N ) (N ) and we will denote Cn (t′ , t, x′ , x) simply by Cn (t′ , t, x′ − x) and similarly for the other kernels. As before primed kernels and fields have the lower cutoff χ′ . We need to introduce the mixed covariance Let us define

Cn′(N ) (t′ , t, x′ − x) := Eηn′(N ) (t′ , x′ )ηn(N ) (t, x) Cn (τ, x) = CnN (τ, x) = Gn (t, x) =

(100)

sup sup Cn′(N ) (t′ , t, x)

|t′ −t|=τ N ≥n

sup |Cn′(N ) (t′ , t, x) − Cn(N ) (t′ , t, x)|

|t′ −t|=τ

sup ΓN n (t, x)

N ≥n

N GnN (t, x) = |Γ′N n (t, x) − Γn (t, x)|

The regularity of these kernels is summarized in

Lemma 14. Cn ∈ Lp (R × Tn ) uniformly in n for p < 5 and for γp > 0 for p < 5. Gn ∈ for γp′ > 0 for p < 5/3.

kCnN kp ≤ Cλγp (N −n) kχ − χ′ k∞ .

Lp (R

(101)

× Tn ) uniformly in n for p < 5/3 and ′

kGnN kpp ≤ Cλγp (N −n) kχ − χ′ k∞ .

(102)

Proof. As in (38) we have, for t′ ≥ t: Z t ′(N ) ′ ′ Hn (|t′ − t| + 2s, x′ − x)χ1N −n (t′ − t + s)χ2N −n (s)ds Cn (t , t, x − x) =

(103)

0

where χ1 = χ′ , χ2 = χ or vice versa depending on t′ > t or t′ < t. The heat kernel Hn is pointwise positive. Using χ1N −n (t′ − t + s)χ2N −n (s) ≤ 1[0,2] (s)1[0,2] (t′ − t) we may bound where

Cn′(N ) (t′ , t, x) ≤ Cn (t′ − t, x)1[0,2] (t′ − t) Cn (τ, x) =

Z

(104)

2

Hn (τ + 2s, x)ds

0

From (98) we get Hn (τ, x) =

X

m∈Z3

(4πt)−3/2 e−

(x+λ−n m)2 4t

.

(105)

18

A. KUPIAINEN

Therefore Cn (τ, x) =

X

c(τ, x + λ−n m)

(106)

m∈Z3

where c(τ, x) =

Z

2

2

− 4(τx+2s)

(4π(τ +2s))−3/2 e

0

1

2

ds ≤ C(e−cx (x2 +τ )− 2 1τ ∈[0,2] +τ −3/2 e−cx

2 /τ

1τ >2 )). (107)

Combining with (106) and (104) (with

χ′

= χ) 1

2

Cn (τ, x) ≤ Ce−cx (x2 + τ )− 2 1[0,2] (τ )

(108)

and so Cn ∈ Lp if p < 5. To get (102) use

|χ1k (t′ − t + s)χ2k (s) − χ1k (t′ − t + s)χ2k (s)| ≤ 1[0,2λ2k ] (s)1[0,2] (t′ − t)kχ − χ′ k∞ .

Hence CnN (τ, x) ≤ where cM (τ, x) :=

Z

2λ2M

X

m∈Z3

cN −n (τ, x + λ−n m)1[0,2] (τ )kχ − χ′ k∞ 2

− 4(τx+2s)

(4π(τ + 2s))−3/2 e

ds = λ−M c0 (λ−2M τ, λ−M x).

(109)

0

Hence using (107) kcM 1[0,2] kpp = λ(5−p)M kc0 1[0,2λ−2M ] kpp ≤ λ(5−p)M (1 +

Z

3

τ 2 (1−p) 1[2,2λ−2M ] dt) ≤ CλγM

with γ > 0 for p < 5. This yields (102). In the same way 2 Gn (t, x) ≤ Ct−3/2 e−cx /t 1[0,2] (t) which is in

Lp

for p < 5/3. (102) follows then as above.

(110)

We will later also need the properties of the kernel ′ Sn(N ) (t′ , t, x) := Cn(N ) (t′ , t, x)ΓN n (t , t, x).

(111)

Sn (τ, x) := sup sup Sn(N ) (t′ , t, x)

(112)

Set |t′ −t|=τ N ≥n

We get from (108) and (110) Sn (t, x) ≤ Ct−2 e−cx

2 /t

1[0,2] (t) ∈ Lp ,

p < 5/4.

Finally let Sn′N have χ′ in all the lower cutoffs in CnN and ΓN n and set Sn′(N ) (τ, x) = sup |Sn′(N ) (t′ , t, x) − Sn(N ) (t′ , t, x)|. |t′ −t|=τ

Then ′′

kSnN kpp ≤ Cλγp (N −n) kχ − χ′ k∞ .

(113)

for some γp′′ > 0 for p < 5/4.

Remark 15. Recall from Remark 6 that or the initial condition (47) the covariance of (N ) ηn is the stationary one (48). Hence Lemma 14 holds for it as well.


19

11. Proof of Proposition 7 In this section we prove Proposition 7. The strategy is straightforward. We need to compute the covariances of the various fields in (64) and establish enough regularity for them. Covariance estimate is all we need since the probabilistic bounds are readily derived from it. The covariances have of course expressions in terms of Feynman diagrams only one of which is diverging as N → ∞. The renormalization constant b is needed to cancel that divergence. We don’t introduce the terminology of diagrams since the ones that enter are simple enough to be expressed without that notational device. 11.1. Covariance bound. We will deduce Proposition 7 from a covariance bound for (N ) (N ) the fields in (64). Let ζn (t, x) or ζn (t, x, s, y) be any of the fields in (64). Let ˜ ′ , t, x) = e 12 dist(t′ ,I) K(t′ − t, x)hn−m (t) K(t

(114)

where I = [0, λ−2(n−m) ] and define

) ˜ n(N ) ρ(N = Kζ n

) ˜ ⊗ Kζ ˜ n(N ) . ρ(N =K n

or

Then

1

) kK ζñ(N ) kL2 (ci ) ≤ Ce− 2 dist(i0 ,I) kρ(N n kL2 (ci ) .

(115)

where i0 is the time component of i and similarly for the bi-local case. We bound the (N ) covariance of ρn : Proposition 16. There exist renormalization constants m1 , m2 , m3 and γ > 0 s.t. for all 0 ≤ n ≤ N < ∞ ) 2 Eρ(N ≤ C n (t, x)

(N )

(116)

) 2 E(ρ′ n (t, x) − ρ(N ≤ Cλγ(N −n) kχ − χ′ k∞ n (t, x)) (N ) E(ρ′ n (t, x, s, y)

) 2 Eρ(N n (t, x, s, y) ) 2 − ρ(N n (t, x, s, y))

(117)

−c(|t−s|+|x−y|)

≤ Ce

≤ Cλ

γ(N −n) −c(|t−s|+|x−y|)

e

(118) ′

kχ − χ k∞ (119)

(N )

Proof of Proposition 7. ρn (t, x)2 belongs to the inhomogenous Wiener Chaos of bounded order m (in fact m ≤ 10). Thus we get for all p > 1 ) 2p ) 2 p Eρ(N ≤ (2p − 1)pm (Eρ(N n (t, x) n (t, x) )

(see [6], page 62). Using Hölder, (120) and (116) in turn we deduce Z (N ) 2p (N ) 2p ) 2p E(kρn kL2 (ci ) ) ≤ E(kρn kL2p (ci ) ) = E(ρ(N n (t, x)) ci Z (N ) 2 p ≤ Cp ((Eρn (t, x)) ) ≤ Cp

(120)

(121)

ci

and thus by (115) so that and finally

E(kK ζñ(N ) k2p ) ≤ Cp e−pdist(i0 ,In ) L2 (ci ) 1

P(kK ζñ(N ) kL2 (ci ) ≥ R) ≤ Cp (R−1 e− 2 dist(i0 ,In ) )2p

P(kζñ(N ) kVn ≥ λ−γn ) ≤

X i

1

Cp (λγn e− 2 dist(i0 ,In ) )2p ≤ Cp λ2γpn λ2m λ−5n .

(122)

20

A. KUPIAINEN

For the bi-local fields we proceed in the same way. First as in (121) we deduce Z ) 2p ((Eρn(N ) (t, x, s, y)2 )p ≤ Cp (Ce−cdist(ci ,cj ) )2p E(kρ(N k ) ≤ C p n L2 (ci ×cj ) ci ×cj

and then use exponential decay to control X P(kζñ(N ) kVn ≥ λ−γn ) ≤ Cp (λγn e−c(dist(i0 ,In)+dist(ci ,cj )) )2p ≤ Cp λ2pγn λ2m λ−5n . i,j

Next we turn to (71) which we recall we want to hold for all cutoff functions χ, χ′ with bounded C 1 norm. We proceed by a standard Kolmogorov continuity argument. (N ) Let f (χ) := K ζñ . Without loss we may consider χ in the ball Br of radius r at origin in C 1 [0, 1]. As above we conclude from (117) E(kf (χ) − f (χ′ )k2p ) ≤ Cp (ǫkχ − χ′ k∞ e−dist(i0 ,In) )p L2 (ci )

(123)

with ǫ = λγ(N −n) . Let χn , n = 1, 2, . . . be the Fourier coefficients of χ in the basis 1, sin 2πx, cos 2πx, sin 4πx, cos 4πx, . . . . Hence |χn | ≤ Crn−1 . Let χm n = χn for n ≤ m ′ and χm n = χn for n > m. Let QN be the dyadic rationals in [0, 1]. Then, for β ∈ (0, 1) ∞ X

′

kf (χ) − f (χ )kL2 (ci ) ≤

m=1

m

kf (χ ) − f (χ

m+1

)kL2 (ci ) ≤

∞ X

m=1

χ′m |β

|χm −

∞ X

2βN ∆N,m

N =0

where ∆N,m =

sup t,t′ ∈Q

and gm (t) =

N

,|t−t′ |=2−N

f (χ1 , . . . , χm−1 , t, χ′m+1 , . . . ).

and then, taking β
R) ≤ C2N (R−2 ǫ2−N e−dist(i0 ,In ) )p 1 2

and 2βp > 1

P( sup kf (χ) − f (χ′ )kL2 (ci ) > R) ≤ χ,χ′ ∈Br

≤

X

m,N

X

m,N

P(∆N,m > R|χm − χ′m |−β 2−βN )

C2N (R−2 ǫ2−N (1−2β) r β e−dist(i0 ,In ) m−2β )p ≤ C(R−2 ǫr β e−dist(i0 ,In ) )p

since |χm − χ′m | ≤ Crm−1 . We conclude then (N )

P( sup kζ˜′ n χ,χ′ ∈Br

1

− ζñ(N ) kVn ≥ λ 2 γ(N −n) λ−γn ) ≤ Cp r βp λpγ(N −n) λ(2γp−5)n λ2m .

(124)

We still need to deal with the last condition on Am in (72). By (27) ζ := Γξn−1 is a Gaussian field with covariance Z t ′ ′ Hn (t′ − t + 2s, x′ − x)χ(t′ − t + s)χ(s)ds Eζ(t , x )ζ(t, x) = 0

where χ is smooth with support in [λ2 , 2]. Eζ(t′ , x′ )ζ(t, y) is smooth, compactly supported in t′ − t and exponentially decaying in x − y. We get then by standard Gaussian estimates [5] for a ≤ 2, |α| ≤ 4 and thus

P(k∂ta ∂xα Γξn+1 kL∞ (ci ) > r) ≤ C(λ)e−c(λ)r

2

P(ksΓξn−1 kΦn > λ−2γn ) ≤ C(λ)λ2m λ−5n e−c(λ)λ

−4n

(125)


21

Combining (122), (124) and (125) with a Borel-Cantelli argument gives the claims (70) and (71). 11.2. Normal ordering. Since the fields (64) are polynomials in gaussian fields the computation of their covariances is straightforward albeit tedious. To organize the computation it is useful to express them in terms of ”normal ordered” expressions in the field ηn . This provides also a transparent way to see why the renormalization constant b is needed. We suppress again the superscript (N ) unless needed for clarity. Define the ”normal ordered” random fields : ηn := ηn , : ηn2 := ηn2 − Eηn2 , : ηn3 := ηn3 − 3ηn Eηn2

(126)

δn (t) := E(ηn(N ) (t, x))2 − 3ρN −n .

(127)

and and (recall Lemma 4) Then

ηn2 − ρN −n =: ηn2 : +δn , ηn3 − 3ρN −n ηn =: ηn3 : +3δn ηn In virtue of Lemma 4 δn terms will turn out to give negligible contribution. These normal ordered fields have zero mean and the following covariances: E : ηn2 : (t, x) : ηn2 : (s, y) = 2Cn (t, s, x, y)2

(128)

E : ηn3 : (t, x) : ηn3 : (s, y) = 6Cn (t, s, x, y)3 . where Cn is defined in (38). Next we process ωn , ζn and zn :

(129)

and

3 N N 3 2 N N ωn = 3 : ηn2 : ΓN n : ηn : +(m2 log λ + m3 )ηn + 3δn Γn : ηn : +9 : ηn : Γn δn ηn + 9δn : Γn δn ηn 3 N ζn = 6ηn ΓN n : ηn : +18ηn Γn δn ηn 2 N N 2 2 N zn = 9 : ηn2 : ΓN n : ηn : +(m2 log λ + m3 ) + 9δn Γn : ηn : +9 : ηn : Γn δn Consider first the terms not involving δn , call them ω ˜ n , ζñ and zñ . We normal order ω ñ and zñ : N N 2 2 N 3 ω ˜ n = (18Cn2 ΓN n − m2 log λ − m3 )ηn + 18 : ηn Cn Γn ηn : +3 : ηn Γn ηn :

zñ =

(18Cn2 ΓN n

N

− m2 log λ − m3 ) + 36 :

ηn Cn ΓN n ηn

: +9 :

2 ηn2 ΓN n ηn

:

(130) (131)

where the product means point wise multiplication of kernels, e.g. the operator Cn2 ΓN n has the kernel Bn (t, x, s, y) := Cn (t, x, s, y)2 ΓN (132) n (t, x, s, y). For Proposition 16 it suffices to bound separately the covariances of all the fields in (130) and (131) as well as the δn -dependent ones. 11.3. Regular fields. We will now prove the claims of Proposition 16 for the fields ζn not requiring renormalization. It will be convenient to denote the space time points (t, x) by the symbol z. We recall from (65) and (66) the bounds for the kernel K(z ′ − z) of the operator K which imply ˜ of (114): a similar bound for K ˜ ′ , z) ≤ Ce− 12 |z ′ −z| := K(z ′ − z). 0 ≤ K(z

We start with the local fields. Let ζn (z) be one of them and Eζn (z1 )ζn (z2 ) := Hn (z1 , z2 )

(133)

22

A. KUPIAINEN

Then (Hn is non negative, see below) Z 2 ˜ z1 )K(z, ˜ z2 )Hn (z1 , z2 )dz1 dz2 K(z, Eρn (z) = Z 1 ≤ C e− 2 (|z−z1 |+|z−z2|) Hn (z1 , z2 )dz1 dz2 ≤ CkHn k1

(134)

where Hn (z) :=

sup Hn (z1 , z2 )

z1 −z2 =z

Since Hn is given in terms of products of the covariances Cn and ΓN n we get an upper by the translation invariant upper bounds Cn and bound for H by replacing Cn and ΓN n N Gn defined and bounded in Lemma 14. Let us proceed case by case. (a) ζn =: ηni :. Using (128) and (129) we have

Hn (z) ≤ Cn (z)i

so by Lemma 14 kHn k1 ≤ kCn kii < ∞. In the same way we get (N ) E(ρ′n (z) − ρn (z))2 ≤ CkCn ki−1 i kCn ki ≤ Cλ

5−i (N −n) i

2

≤ Cλ 3 (N −n)

(N )

(b) ζn =: ηn Cn Γn ηn2 :. We have E : ηn (z1 )ηn (z2 )2 :: ηn (z3 )ηn (z4 )2 := 2Cn (z1 , z3 )Cn (z2 , z4 )2 +4Cn (z1 , z4 )Cn (z2 , z3 )Cn (z2 , z4 ) and so

Z

Eζn (z1 )ζn (z3 ) = ·

(2Cn (z1 , z3 )Cn (z2 , z4 )2 + 4Cn (z1 , z4 )Cn (z2 , z3 )Cn (z2 , z4 ))

S(z1 , z2 )S(z3 , z4 )dz2 dz4

(135)

where S(z1 , z2 ) := Cn (z1 , z2 )Γn (z1 , z2 ) ≤ Sn (z12 )

where we use the notation z12 = z1 − z2 and Sn is defined in (112). Thus Z 0 ≤ Eζn (z1 )ζn (z3 ) ≤ 2Cn (z13 ) Sn (z12 )Sn (z34 )Cn (z24 )2 dz2 dz4 + Z 4 Sn (z12 )Sn (z34 )Cn (z14 )Cn (z23 )Cn (z24 ))dz2 dz4 := A(z13 ) + B(z13 )

(136)

so that Since A = Cn (Sn ∗ Sn ∗

Cn2 )

Eρn (z)2 ≤ C(kAk1 + kBk1 ).

(137)

we get by Hölder and Young’s inequalities

kAk1 ≤ kCn kp kSn ∗ Sn ∗ Cn2 kq ≤ kCn kp kSn k2r kCn2 ks

where 2 + 1q = 2r + 1s . We can take e.g. p = 3, q = 3/2, r = 1 and s = 3/2 and by Lemma 14 this is finite. As for B, we write Z kBk1 = 4 Sn (z − z2 )Sn (z4 )Cn (z − z4 )Cn (z2 )Cn (z24 ))dzdz2 dz4 = (Sn ∗ (Cn (Sn ∗ Cn )) ∗ Cn )(0) ≤ kSn kr kCn (Sn ∗ Cn )kp kCn kq


23

by Young’s inequality with 2 = 1r + p1 + 1q . Since Cn ∈ La for a < 5 and Sn ∈ Lb for b < 5/4 we have Sn ∗ Cn ∈ Lc for c < ∞ and so Cn (Sn ∗ Cn ) ∈ Lp for p < 5. So we may take e.g. p = q = 2 and r = 1. In the same way, using (102) and (113) we obtain E(ρ′n (z) − ρn (z))2 ≤ Cλγ(N −n) kχ − χ′ k∞

(138)

for some γ > 0. 3 (c) ζn =: ηn2 ΓN n ηn :. Now

E : ηn (z1 )2 ηn (z2 )3 :: ηn (z3 )2 ηn (z4 )3 := 12Cn (z1 , z3 )2 Cn (z2 , z4 )3 2

(139)

2

+36Cn (z1 , z4 ) Cn (z2 , z3 ) Cn (z2 , z4 ) + 36Cn (z1 , z3 )Cn (z1 , z4 )Cn (z2 , z3 )Cn (z2 , z4 )2 . The first two terms have the same topology (as Feynman diagrams!) as the A and B above and we call their contributions with those names again. Thus N 3 2 N 2 3 kAk1 ≤ kCn2 kp kΓN n ∗ Γn ∗ Cn kq ≤ kCn kp kΓn kr kCn ks

r where 2 + 1q = 2r + 1s . Now ΓN n is in L for r < 5/3. So we may take for instance p = 2, 2 q = 2, s = 4/3, r = 8/7 so that kAk1 ≤ CkCn k54 kΓN n k8/7 . For B we get N 2 2 N N 2 2 kBk1 = 36(ΓN n ∗ (Cn (Γn ∗ Cn )) ∗ Cn )(0) ≤ 36kΓn kr kCn (Γn ∗ Cn )kp kCn kq

b with 2 = 1r + 1p + 1q . Since Cn ∈ La for a < 5 and ΓN n ∈ L for b < 5/4 we have by p c N 2 Young ΓN n ∗ Cn ∈ L for c < 5 and so Cn (Γn ∗ Cn ) ∈ L for p < 5/2. So we may take e.g. p = q = 2 and r = 1. Finally the last term in (139), call it D, is bounded by Z N 2 kDk1 ≤ 36 Cn (z)ΓN n (z − z2 )Γn (z4 )Cn (z − z4 )Cn (z2 )Cn (z24 ) dzdz2 dz4 Z := f (z2 , z4 )g(z2 , z4 )dz2 dz4

where f (z2 , z4 ) =

R

Cn (z)ΓN n (z − z2 )Cn (z − z4 )dz. We have Z f (z2 , z4 )dz2 = (Cn ∗ Cn )(z4 )

(140)

which is in L∞ since Cn ∈ Lp , p < 5. Hence Z kDk1 ≤ C g(z2 , z4 )dz2 dz4 = C(Cn ∗ Cn2 ∗ ΓN n )(0) < ∞ 1 2 ∞ since ΓN n ∈ L and Cn ∗ Cn ∈ L .

Now we turn to the bi-local fields ζ(z1 , z2 ) and set Eζ(z1 , z2 )ζ(z3 , z4 ) := Hn (z)

We proceed as in (134) ′

2

Eρn (z , z)

≤ C

Z

1

(R×Tn )4

Let Y := sup i

X j

′

′

e− 2 (|z−z1 |+|z−z3|+|z −z2 |+|z −z4 |) Hn (z)dz.

sup z ′ ∈ci ,z∈cj

Eρn (z ′ , z)2 ec|z−z | . ′

(141)

24

A. KUPIAINEN

(137) follows from Y < ∞. Let

1

Hn (z) := sup Hn (z1 + u, z2 + u, z3 + u, z4 + u)e 2 (|z1 −z2 |+|z3 −z3 |) u

We have Y

≤ C sup z

Z

1

(R×Tn )4

e− 2 (|z−z1 |+|z−z3|+|z −z2 |+|z −z4 |+|z1−z2 |+|z3−z3 |) ec|z−z | Hn (z)dz. ′

′

′

The integrand is actually independent on z as Hn is translation invariant: ˜ n (z14 , z24 , z34 ) Hn (z) = H

˜ n (z1 , z2 , z3 ) = H(z1 , z2 , z3 , 0). We can then conclude with H ˜ n k1 Y ≤ CkH

˜ n k1 < ∞. Let us again proceed i.e. we need to show for the various bi-local fields that kH by cases. (d) ζn =: ηn Cn ΓN n ηn :. We have 1 e 2 (|z12 |+|z34 |) Hn (z)) ≤ Sñ (z12 )Sñ (z34 )(Cn (z13 )Cn (z24 ) + Cn (z14 )Cn (z23 ))

where Sñ (z) = ec|z| Sn (z) so that

˜ n k1 ≤ 2kSñ ∗ Cn k22 < ∞ kH

since by Lemma 14 Sñ is in Lp , p < 5/4 and Cn is in Lp , p < 5 so that Sñ ∗ Cn ∈ Lp , p < ∞. (119) goes in the same way where at least one of the Cn or Sn is replaced by (N ) (N ) Cn or Sn .

2 (e) ζn =: ηn2 ΓN n ηn :. We get

1 ˜ N (z12 )Γ ˜ N (z23 )(Cn (z13 )2 Cn (z24 )2 + Cn (z14 )2 Cn (z23 )2 e 2 (|z12 |+|z34 |) Hn (z)) ≤ 4Γ n n + Cn (z13 )Cn (z24 )Cn (z14 )Cn (z23 ))

The first two terms on the RHS have the same topology as in (e): their contribution to ˜ N is in L1 and C 2 in L2 . ˜ n k1 is bounded by CkΓ ˜ N ∗ C 2 k2 which is finite since Γ kH n n n 2 n The third term is treated as the analogous one D in (d), let us call it D again. Again R its L1 norm is given by kDk1 = f (z2 , z3 )g(z2 , z3 )dz2 dz3 where f is as above and is in L∞ . g(z2 , z3 ) = Cn (z2 − z3 )Cn (z2 )ΓN n (z3 )and thus Z kDk1 ≤ C g(z2 , z4 )dz2 dz4 = C(Cn ∗ Cn ∗ ΓN n )(0) < ∞. Let us finally turn to the δn terms in ωn , ζn and zn . Starting with ωn , and the term 3 ζn = δn ΓN n : ηn : we have 3 N Hn (z1 , z2 ) ≤ δn (t1 )δn (t2 )(ΓN n ∗ Cn ∗ Γn )(z12 ).

p −c|z−zi | δ (t ) we N 3 The function gn = ΓN n i n ∗ Cn ∗ Γn is in L , p < 5/3. Letting fn (zi ) = e p see from Lemma 4 that fn ∈ L , p < 2 and so

Eρn (z)2 ≤ C(fn ∗ gn ∗ fn )(0)

is finite by Young. To get the bound (119) we use (42) to get for p < 2 kfn′ − fn kp ≤ Cλ

2−p (N −n) 2p

kχ − χ′ k∞ .


25

Consider next the case where δn is on the ”other side”: ζn =: ηn2 : ΓN n δn ηn . Replacing Cn by the translation invariant upper bound Cn we have Z Hn (z1 , z2 ) ≤ Gn (z1 , z2 , z3 , z4 )δn (t3 )δn (t4 )dz3 dz4 with N 2 Gn (z) = ΓN n (z13 )Γn (z24 )(2Cn (z12 ) Cn (z34 )+4Cn (z12 )Cn (z13 )Cn (z24 )+4Cn (z12 )Cn (z14 )Cn (z23 ))

˜ Ñ At the cost of replacing Cn and ΓN n by Cn and Γn we may replace δn by the f in the 2 upper bound for Eρn (z) : Z Z 2 ˜ Eρn (z) ≤ Gn (z1 , z2 , z3 , z4 )f (z3 )f (z4 )dz1 dz2 dz3 dz4 = gn (z3 − z4 )f (z3 )f (z4 )dz3 dz4 This is bounded if gn is in Lp , p > 1 which is now straightforward. As the last case consider an example of a bi-local field the second last term in zn : 2 ζn = δn ΓN n : ηn :. By the now familiar steps Z 2 Eρn (z) ≤ g(z1 − z2 )f (z1 )f (z2 )dz1 dz2 p Ñ ˜2 ˜ N with g = Γ n ∗ C ∗ Γn in L , p < 5/2.

11.4. Renormalization. We are left with the first terms in (130) and (131) that require fixing the renormalization constants m2 and m3 . Define (product is of kernels as usual) Bn(N ) = (Cn(N ) )2 ΓN n

(142)

Let B∞ denote the set of bounded operators L∞ (R × Tn ) → L∞ (R × R3 ) and B1 the ones L∞ (R) × L1 (Tn ) → L∞ (R × R3 ). We prove: Proposition 17. There exist constants βi s.t. the operator ˜ n(N ) − (β2 log λN −n + β3 ) id) Rn(N ) := K(B satisfies

kRn(N ) kBi ≤ C,

(N )

kR′ n

− Rn(N ) kBi ≤ Cλ(N −n) kχ − χ′ k∞ ,

i = 1, ∞

(143)

uniformly in 0 ≤ n ≤ N < ∞. β1 universal (i.e. independent on χ).

We fix the renormalization constants mi = 18βi . This means that e.g. the first term in (130) becomes 18Rn(N ) ηn + (m2 log λn + m3 )ηn . The second term once multiplied by λ2n fits into the bound (74). The i = ∞ case of Proposition 17 takes care of the deterministic term in (131) since khn−m (Bn(N ) − β1 log λN −n + β2 )φkVn ≤ kRn(N ) kB∞ kφk∞ .

and kφk∞ ≤ kφkΦn . The i = 1 case is needed for the first term in (130). Indeed, we have Z ) 2 Eρ(N (z) = (Rn(N ) (t, t1 ) ∗ Rn(N ) (t, t2 ) ∗ Cn(N ) (t1 , t2 ))(0)dt1 dt2 n where ∗ is spatial convolution. We have

sup kCn(N ) (t1 , t2 )k1 < ∞

t1 ,t2

26

A. KUPIAINEN

so that indeed Eρn(N ) (z)2 ≤ CkRn(N ) kB∞ kkRn(N ) kB1 . ˜ B to be bounded is Recall also that by (134) a sufficient condition for kKBk i sup |B(t + ·, t, ·)| ∈ L1 (R × Tn ).

(144)

t

Proof. Let us first remark that it suffices to work in R3 instead of Tn . Indeed, recall (N ) (N ) (N ) Bn = 18(Cn )2 Bn where the product is defined as pointwise multiplication of ker(N ) ) ˜ (N nels. Let Cñ and Γ be given by (38) and (36) where the Tn heat kernels Hn are n 3 replaced by the R heat kernel H and similarly let Let ˜ (N ) (t′ , t, x) δBn(N ) (t′ , t, x) := Bn(N ) (t′ , t, x) − B n (N ) (N ) (N ) (N ) From (106) and a similar representation for Γn we infer that Cn − Cñ and Γn − ) ˜ (N Γ are in Lp (R × Tn ) for all 1 ≤ p ≤ ∞. Proceeding as in Lemma 14 we then conclude n (N ) supt |δBn (t + τ, t, x)| is in Lp for p < 5/4 and the analogue of (113) holds. Hence (144) holds. For the rest of this Section we work in R × R3 and fix the UV cutoff λN −n := ǫ and denote the operators simply by C and Γ. Also, we set χǫ (s) := χ(s) − χ(s/ǫ2 ). With (N ) these preliminaries we will start to work towards extracting from the operator Bn (132) the divergent part responsible for the renormalization. First we’ll derive a version of the fluctuation-dissipation relation relating C and Γ: 1

∂t′ C(t′ , t) = − 2 Γ(t′ , t) + A(t′ , t).

(145)

where A(t′ , t) will give a non singular contribution to B. To derive (145) write (38) in operator form and differentiate in t′ : Z t ′ (t′ −t)∆ ∂t′ C(t , t) = e (∆ + ∂t′ )e2s∆ χǫ (t′ − t + s)χǫ (s)ds 0

Next, write ∆e2s∆ = 21 ∂s e2s∆ and integrate by parts to get Z t 1 1 (t′ +t)∆ ′ (t′ −t)∆ ′ χǫ (t )χǫ (t) + e e2s∆ (∂t − 2 ∂s )(χǫ (t′ − t + s)χǫ (s))ds ∂t C(t , t) = − 2 e 0

′

′

:= a0 (t , t) + a ˜(t , t) Now recall that χǫ (s) = χ(s) − χ(ǫ−2 s) and write denoting τ = t′ − t: 1

(∂t′ − 2 ∂s )(χǫ (t′ − t + s)χǫ (s)) = ρ1 (τ, s) + ρ2 (τ, s) + ρ3 (τ, s) with ρ1 (τ, s) =

1 (−χǫ (τ 2

+ s)χ′ (s) + χ′ (τ + s)χǫ (s))

1

ρ2 (τ, s) = − 2 χǫ (s)∂s χ(ǫ−2 (τ + s)) ρ3 (τ, s) =

1 χ (τ 2 ǫ

+ s)∂s χ(ǫ−2 s)

and correspondingly a ˜(t′ , t) = a1 (t′ , t) + a2 (t′ , t) + a3 (t′ , t).

(146)


27

ρ1 localizes s-integral to s > O(1) and gives a smooth contribution as ǫ → 0. dµ(s) := −∂s χ(ǫ−2 s)ds is a probability measure supported on [ǫ2 , 2ǫ2 ] so ρ2 localizes s and τ to O(ǫ2 ). The main term comes from ρ3 Z t 1 a3 (t′ , t) = − 2 eτ ∆ e2s∆ χǫ (τ + s)dµ(s) = where 1

a4 (t′ , t) = 2 eτ ∆

0 1 ′ − 2 Γ(t , t)

Z

t

+ a4 (t′ , t)

(147)

(χǫ (τ ) − e2s∆ χǫ (τ + s))dµ(s)

(148)

A(t′ , t) = a0 (t′ , t) + a1 (t′ , t) + a2 (t′ , t) + a4 (t′ , t).

(149)

0

(145) follows then with

Inserting (145) into (132) and using Z Z Z t Z Z t 3 3 ds∂t C(t, s, x−y) φ(s, y) = ∂t dy dsC(t, s, x−y) φ(s, y)− C(t, t, x−y)3 φ(t, y)dy dy 0

0

we obtain (Bφ)(t) = D(t) ∗ φ(t) + ∂t (Eφ)(t) + (F φ)(t)

with

(150)

D(t, x) = 12C(t, t, x)3 Z Z t (Eφ)(t, x) = −12∂t dy dsC(t, s, x − y)3 φ(s, y) 0 Z Z t dsA(t, s, x − y)C(t, s, x − y)2 φ(s, y). (F φ)(t, x) = −36 dy 0

We estimate these three operators in turn. (a) D. The Fourier transform of D in x is given by Z ˆ t, p + k)C(t, ˆ t, k + q)C(t, ˆ t, q)dkdq ˆ D(t, p) = C(t, where ˆ t, q) = C(t,

Z

t

2

e−sq χǫ (s)2 ds

(151)

(152)

0

The integral in (151) diverges at p = 0 logarithmically as ǫ → 0. Doing the gaussian integrals over k, q we have Z 3 Y 2 −3 ˆ p) = (4π) χǫ (si )2 dsi e−α(s)p d(s)−3/2 D(t, [0,t]3

i=1

s2 s3 where α(s) := − s1d(s) and d(s) := s1 s2 +s1 s3 +s2 s3 . Let us study the cutoff dependence ˆ of D. First by differentiating and changing variables

ˆ p) = − ǫ∂ǫ D(t,

3 32π 3

Z

[0,t/ǫ2 ]3

2

e−α(s)(ǫp) d(s)−3/2 s1 ∂s1 χ(s1 )2 ds1

3 Y (χ(ǫ2 si )2 − χ(si )2 )dsi . i=2

(153)

28

A. KUPIAINEN

Let 3 αǫ (t, p) := − 32π 3

Z

−α(s)(ǫp)2

e

−3/2

d(s)

[0,t/ǫ2 ]3

3 Y s1 ∂s1 χ(s1 ) ds1 (1 − χ(si )2 )dsi . (154) 2

i=2

ˆ p) − αǫ (t, p). Since the s1 integral is supported on [1, 2] and and set α ˜ ǫ (t, p) = ǫ∂ǫ D(t, the others on si ≥ 1 we have d(s) ≍ s2 s3 , α(s) ≍ 1 which leads to Z (s2 s3 )−3/2 1[1,2] (s1 )1[ǫ−2 ,∞) (s2 )1[1,∞) (s3 ) ≤ Cǫ. (155) |˜ αǫ (t, p)| ≤ C R3+

Furthermore, let α ˜ ′ǫ (t, p) be gotten by replacing the lower cutoffs χ(si ) by another one ′ χ (si ). Then |˜ α′ǫ (t, p) − α ˜ ǫ (t, p)| ≤ Cǫkχ − χ′ k∞ . (156) Rǫ ′ dǫ Since α ˜ 0 (t, p) = 0 we get that 0 α ˜ ǫ′ (t, p) ǫ′ satifies (155) and (156) as well. Thus all the divergences come from αǫ (t, p). Note that α0 (t, p) = α0 is independent on t and p. Set aǫ (t, p) = αǫ (t, p) − α0 . We get √ (157) |aǫ (t, p)| ≤ C((ǫ2 p2 + ǫ/ t) ∧ 1) We fix the renormalization constant β2 = α0 and define ˆ p) − α0 log ǫ. d(t, p) := D(t,

(158)

Combining above we get Z 1 √ dǫ′ |d(t, p)| = | (˜ αǫ′ (t, p) + aǫ′ (t, p)) ′ | ≤ C(1 + log(1 + p2 + 1/ t)). (159) ǫ ǫ Next we write Z 3 Y X 1 −3/2 −α(s)(ǫp)2 (1 − χ(si )2 )dsi (160) s ∂ αǫ (t, p) = − d(s) e i si 32π 3 [0,t/ǫ2 ]3 i=1 i Z ∞ Z 3 3 Y X 1 −3/2 −λα(s)(ǫp)2 (1 − χ(λsi )2 )dsi d(s) ∂ s − 1)e = − δ( dλ i λ 32π 3 0 [0,t/λǫ2 ]3 i=1

i=1

At ǫ = 0 this implies after an integration by parts Z 3 3 Y X 1 −3/2 α0 = − dsi . si − 1)d(s) δ( 32π 3 R3+ i=1

i=1

i.e. α0 is universal (i.e. independent of χ). Finally let us vary the cutoff. Replace χ by χσ = σχ + (1 − σ)χ′ . Since ∂σ aǫ = ∂σ (aǫ − a0 ) we get from (160) Z ∞ Z 2 α(s)(ǫp)2 e−λα(s)(ǫp) d(s)−3/2 dµ(s) dλ ∂σ αǫ (t, p) = 0

+

3 Z ∞ X i=1

0

[0,t/λǫ2 ]3

2 2

dλ(t/λ ǫ )

Z

2

[0,t/λǫ2 ]2

e−λα(s)(ǫp) d(s)−3/2 dµ(s)|si =t/λǫ2 (161)

where dµ(s) =

3 3 Y X 3 ′ (1 − χσ (λsi )2 ). s − 1)(χ (λs ) − χ (λs ))χ (λs ) δ( i σ 1 1 σ 1 σ 16π 3 i=1

i=2

Start with the first term in (161). Since λsi ≥ 1 the λ-integral is supported in λ ≥ 3. On the support of χσ (λs1 ) − χ′σ (λs1 ) s1 ∈ [λ−1 , 2λ−1 ]. By symmetry we may assume


29

s2 ≤ s3 and then s3 ≥ 1/6. Hence in the support of the χ’s α ≍ s1 and d ≍ s2 . We get the bound Z 1 Z ∞ Z 2/λ 2 2 −3/2 ds2 s2 ǫkχ − χ′ k∞ ≤ C(ǫp)2 e−(ǫp) kχ − χ′ k∞ . ds1 dλ C(ǫp)2 e−(ǫp) 1/λ

1/λ

1/3

For the second term, if i = 1 then s1 = t/λǫ2 ∈ [1/λ, 2/λ] implies t ∈ [ǫ2 , 2ǫ2 ]. Again α ≍ s1 , d ≍ s2 ≤ s3 and we end up with the bound C1(t ∈ [ǫ2 , 2ǫ2 ])kχ − χ′ k∞ .

If i 6= 1 the same bound results. We may summarize this discussion in |d(t, p) − d′ (t, p)| ≤ C(ǫ + 1(t ∈ [ǫ2 , 2ǫ2 ])kχ − χ′ k∞ .

(162)

−|t′ −t|

The operator K(t′ , t)d(t) acts as a Fourier multiplier with 12 e (p2 +1)−2 d(t, p). Since d(t, p) is analytic in a strip |Im p| ≤ c we get in x-space from the above bounds 1

|(K(t′ , t)d(t))(x)| ≤ Ce−|t −t|−c|x| (1 + log(1 + t− 2 )) ′

′

|(K(t′ , t)(d′ (t) − d(t)))(x)| ≤ Ce−|t −t|−c|x| (ǫ + 1(t ∈ [ǫ2 , 2ǫ2 ]))kχ − χ′ k∞ .

Hence

˜ |(Kdφ)(t , x)| ≤ ′

Z

1

1

′

e− 2 |t −t|−c|x−y| (1 + log(1 + t− 2 ))φ(t, y)dtdy

which is in L∞ for φ ∈ L∞ × L∞ and for φ ∈ L∞ × L1 as well. This gives the first bound in (143). The second is similar. (b) E. We have Eφ = ∂t C 3 φ where C 3 (t, s, x) := C(t, s, x)3 . Thus integrating by parts Khn−m Eφ = K∂t hn−m C 3 φ + ∂t Khn−m C 3 φ.

(163)

By (65) |∂t K(z)| = K(z) so we may use (133) for the second term as well to get ˜ n−m Eφk∞ ≤ CkK ∗ C 3 ∗ φk∞ . kKh

By Lemma 14 C 3 is in L1 (R×Tn) and hence K∗C 3 is in L1 (R×Tn ) and in L1 (R)×L∞ (Tn ) so that the first estimate of (143) follows. The second is similar. (c) F . By (149) F has four contributions, call them F0 , F1 , F2 , F4 . Start with F1 . Since ρ1 is supported in s ≥ 1 the kernel is bounded (in fact smooth) |a1 (t′ , t, x)| ≤ Ce−c|x] .

and so by (108) we get

|F1 (t + τ, t, x)| ≤ Ce−c|x](x2 + τ )−1 .

Hence (144) holds for F1 uniformly in ǫ and the first bound in (143) follows. For the second one we proceed as in Lemma 14 to get for f1 (τ, x) := supt |F1 (t + τ, t, x) − F1′ (t + τ, t, x)| that kf1 k1 ≤ Cǫγ for some γ > 0. Consider next F2 = −36a2 C 2 . We show: where

F2 (t′ , t, x) = ǫ−5 p((t′ − t)/ǫ2 , x/ǫ) + r(t′ , t, x) 2

|p(τ, x)| ≤ Ce−cx 1τ ≤2 and r satisfies (143). To derive (169) we note that by a change of variables Z t/ǫ2 1 a2 (t + τ, t, x) = − 2 H(τ + 2ǫ2 s, x)(1 − χ(s))∂s χ(τ /ǫ2 + s)ds 0

(164)

(165)

30

A. KUPIAINEN

where we also noted since ∂s χ is supported on [1, 2] χ(ǫ2 s) = 1. Using scaling property of the heat kernel H(τ + ǫ2 s, x) = ǫ−3 H(τ /ǫ2 + 2s, x/ǫ) we get then τ t x a2 (t + τ, t, x) = ǫ−3 α( 2 , 2 , ) ǫ ǫ ǫ with Z t

1

α(τ, t, x) = − 2

0

H(τ + s, x)(1 − χ(s))∂s χ(τ + s)ds.

Note that α depends on t only on t ≤ 2 and is bounded by 2

|α(τ, t, x)| ≤ Ce−cx 1τ ≤2 ,

(166)

2

|α(τ, t, x) − α(τ, ∞, x| ≤ Ce−cx 1τ ≤2 1t≤2 .

Comparing two lower cutoffs χ and χ′ we get

2

|α(τ, t, x) − α′ (τ, t, x| ≤ Ce−cx 1τ ≤2 kχ − χ′ k∞ . By similar manipulations we obtain C(t + τ, t, x)2 = ǫ−2 c( where c(τ, t, x; ǫ) =

Z

t

0

τ t x 2 , , ; ǫ) ǫ2 ǫ2 ǫ

H(τ + 2s, x)(χ(ǫ2 (τ + s)) − χ(τ + s))(χ(ǫ2 s) − χ(s))ds.

Since H(τ + 2s, x) ≤ C(1 + s)−3/2 on support of the integrand we get |c(τ, t, x; ǫ)| ≤ C(1 + |x|)−1

and

1

|c(τ, t, x; ǫ) − c(τ, ∞, x; 0)| ≤ C(ǫ(1 + ǫ|x|)−1 + (1 + |x| + t)− 2 )

with an extra kχ − χ′ k∞ factor if we compare two lower cutoffs. (169) follows with p(τ, x) = −36c(τ, x, ∞, 0)2 α(τ, ∞, x).

The error term satisfies 1

|r(t + τ, t, x)| ≤ Cǫ−5 (ǫ + (1 + t/ǫ2 )− 2 + 1t≤2ǫ2 )e−cx with an extra kχ −

χ′ k∞

′ ′ ˜ |(Krφ)(t , x )| ≤

Z

2 /ǫ2

1τ ≤2ǫ2

(167)

factor if we compare two lower cutoffs. Hence ′

′

e−c|t −τ −t|+|x −x| |r(t + τ, t, x − y)|φ(t, y)|dτ dx ≤ Cǫkφk

both in the norm L∞ × L∞ and in L∞ × L1 . This and similar statement with kχ − χ′ k∞ gives (143). LetR p˜ be the operator with the kernel ǫ−5 p((t′ − t)/ǫ2 , (x′ − x)/ǫ) − p0 δ(z ′ − z) where p0 = p(z)dz. Then Z (Khn−m p˜)(z + v, z) = (K(v − uǫ )h(z + uǫ ) − K(v)hn−m (z))p(u)du. (168) ˜ p˜φk∞ ≤ Cǫkφk in both norms. where uǫ = (ǫ2 u0 , ǫu). The bound (165) then implies kK ′ A similar statement holds with kχ − χ k∞ . p0 contributes the the renormalization constant m2 . The analysis of F4 = −36a4 C 2 parallels that of F2 so we are brief: F2 (t′ , t, x) = ǫ−5 q((t′ − t)/ǫ2 , x/ǫ) + s(t′ , t, x)

(169)


31

where s satisfies (143) and with a(τ, x) =

Z

q(τ, x) = −36c(τ, x, ∞, 0)2 a(τ, x) (H(τ, x)(1 − χ(τ )) − H(τ + 2s, x)(1 − χ(τ + s))∂s χ(s)ds.

Since ∂s χ is supported on [1, 2] and 1 − χ on τ ≥ 1 we get |a(τ, x)| ≤ (1 + τ )−5/2 e−cx Since c(τ, x, ∞, 0) ≤ C(1 + |x|)−1 we end up with |q(τ, x| ≤ C(1 + τ )−5/2 (1 + |x|)−2 e−cx

2 /τ

2 /τ

.

.

We may now proceed as in (168).

The analysis of the term F0 proceeds along similar lines and is omitted. References [1] M. Hairer, A theory of regularity structures. Invent. Math. (2014). doi:10.1007/s00222-014-0505-4. [2] K. Wilson, http://www.nobelprize.org/nobel prizes/physics/laureates/1982/wilson-lecture.pdf [3] J.Bricmont, A.Kupiainen, G.Lin Renormalization Group and Asymptotics of Solutions of Nonlinear Parabolic Equations, Commun. Pure.Appl.Math. 47, 893-922 (1994) [4] J.Bricmont, K.Gawedzki, A.Kupiainen KAM theorem and quantum field theory, Commun. Math Phys 201 (1999) 3, 699-727 [5] Vladimir I. Bogachev, Gaussian Measures (Mathematical Surveys and Monographs) American Mathematical Society1998 [6] D. Nualart, The Malliavin calculus and related topics. Probability and its Applications (New York). Springer-Verlag, Berlin, second ed., 2006 University of Helsinki, Department of Mathematics and Statistics, P.O. Box 68 , FIN00014 University of Helsinki, Finland E-mail address: [email protected]