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北京十一选五开奖结果 www.frdg.net Finite range Decomposition of Gaussian Processes

arXiv:math-ph/0303013v3 28 Aug 2003

David C. Brydges? The University of British Columbia 1984 Mathematics Road Vancouver, B.C., Canada V6T 1Z2

G. Guadagni Department of Mathematics University of Virginia Charlottesville VA 22904-4137

P. K. Mitter? Laboratoire de Physique Math?matique e Universit? Montpellier 2 e Place E. Bataillon, Case 070 34095 Montpellier Cedex 05 France In Honour of G. Jona-Lasinio February 7, 2008

Abstract Let ? be the ?nite di?erence Laplacian associated to the lattice Zd . For dimension d ≥ 3, a ≥ 0 and L a su?ciently large positive dyadic integer, we prove that the integral kernel of the resolvent Ga := (a ? ?)?1 can be decomposed as an in?nite sum of positive semi-de?nite functions Vn of ?nite range, Vn (x?y) = 0 for |x?y| ≥ O(L)n . Equivalently, the Gaussian process on the lattice with covariance Ga admits a decomposition into independent Gaussian processes with ?nite range covariances. For a = 0, Vn has a limiting scaling form L?n(d?2) Γc,? x?y Ln as n → ∞. As a corollary, such decompositions also exist for fractional powers (??)?α/2 , 0 < α ≤ 2. The results of this paper give an alternative to the block spin renormalization group on the lattice.

keywords: gaussian processes, ?nite range decomposition, lattice, renormalization group, L?vy e processes.

1

Introduction

A smooth Gaussian process ζ(x) on Rd with the property that the expectation Eζ(x)ζ(y) = 0 when |x ? y| ≥ L will be said to have ?nite range L. What is the class of Gaussian processes φ that can be expressed as a sum φ = j ζj of independent ?nite range processes with ranges ? Lj for some L? Let us call such processes ?nite range decomposable. We can reformulate this in terms of the covariance: a Gaussian process φ is ?nite range decomposable if the covariance C(x, y) := Eφ(x)φ(y) can be written as a sum C = j Vj where each

? ?

Supported by NSERC Laboratoire Associ? au CNRS, UMR 5825 e

1

Vj (x, y) is positive semi-de?nite and has ?nite range ? Lj . In this form the question has already received a partial answer in the study of ground states for many-body Hamiltonians. In particular, in [HS02] Hainzl and Seiringer discuss this background and consider the decomposition

∞

V (x) =

0

dr g(r)χr/2 ? χr/2 (x)

(1.1)

of a radial function V (x) as a weighted integral of tent functions χr/2 ? χr/2 (x), where χr/2 is the indicator function of the ball of radius r/2. An explicit formula for g in terms of V is derived. For example, in three dimensions, 2 g(r) = ? (V ′′ (r)/r)′ π so necessary and su?cient conditions for g ≥ 0 in terms of V are readily formulated. In particular Coulomb and Yukawa potentials in three dimensions have decompositions with nonnegative g. This is relevant to our question because the tent function is positive semi-de?nite and therefore, when g(r) ≥ 0 and I is an interval [a, b), VI (x) :=

I

dr g(r)χr/2 ? χr/2 (x)

is also positive semi-de?nite. By breaking up the range of the r integration in (1.1) into a disjoint union of intervals, Ij := [Lj , Lj+1 ), j ∈ Z, we have V = VIj and there is a corresponding ?nite range decomposition φ = j ζj when φ is the Gaussian process with covariance V (x ? y) with g(r) ≥ 0 and ζj has covariance VIj . These decompositions are not the ?nal answer to our question, because we are also interested in kernels de?ned on the lattice Zd and furthermore one may get a wider class by not insisting on decompositions based on tent functions. For lattices or the continuum we have preliminary results that suggest that resolvents of quite general elliptic operators and fractional inverse powers of elliptic operators are candidates for such decompositions. Our interest in this question is rooted in the Renormalization Group (RG). In quantum ?eld theory and other contexts the RG is a method to calculate the expectation EZ of a functional Z = Z(φ) of a Gaussian ?eld φ. One decomposes φ = j≥1 ζj as a sum of independent Gaussian ?elds ζj and integrates out each ζj one at a time. Let Ej be the expectation that integrates out ζj . Then the RG is the sequence of maps Zj → Zj+1 := Ej+1 Zj . EZ is obtained from EZ = lim Zj , starting with Z0 := Z. [These ideas are explained further in Section 4]. The point is to choose the decomposition to have special properties so that each expectation Ej is more amenable to analysis than the whole expectation E and furthermore so that the map Zj → Zj+1 can be analysed within the context of dynamical systems. In particular the RG is very informative when the limiting map is autonomous, up to a scaling. This is possible when the covariances Vj of ζj are becoming self-similar. This means that there should exist a dimension [φ] such that Vj (x, y) =: L?2j[φ]Γj (L?j x, L?j y) de?nes a scaled covariance Γj which tends to a limit as j → ∞. When the covariances Vj are ?nite range, the map Zj → Zj+1 can be studied by using the independence of ζj (x) and ζj (y) for |x ? y| ≥ Lj . We amplify on this remark at the end of this introduction. In some ways, the use of these ?nite range covariances gives the simplest framework that goes beyond the hierarchical models. This program is also close to the technique that was invented by Fr¨hlich and Spencer in [FS81] to study the Kosterlitz-Thouless transition, but it is in o principle more precise and more robust. 2

Let ? be the ?nite di?erence Laplacian associated to the lattice Zd . We will consider decompositions for the kernel Ga (x ? y) of the resolvent (a ? ?)?1 and we will also consider the Green’s function of a stable L?vy process which is the kernel C(x ? y) of (??)?α/2 where 0 < α < 2. The e mass parameter a ∈ [0, ∞). We state our results for dimensions d ≥ 3 because lower dimensions require extra discussions for the case a = 0, but the basic construction is valid in lower dimensions as well. L is a parameter of the form 2p . p can be any su?ciently large integer. There are many technical results in this paper so we have summarised the main points in the following theorem which is a combination of results from the theorems in the rest of the paper. Theorem 1.1. For n = 0, 1, 2 . . . , and all a ≥ 0, there are positive semi-de?nite functions Γa (x) n de?ned for x ∈ (L?n Z)d such that 1. Ga (x ? y) = L?2n[φ] ΓL n

2n a

x?y Ln

with [φ] = (d ? 2)/2

2. Γa (x) = 0 for |x| ≥ 6L n ? 3. |Γa (p)| ≤ ck,L (1 + |p|)?k for p ∈ [?Ln π, Ln π]d n ? c,? ?n 4. Γa (p) := limn→∞ Γa (p) exists pointwise in p 5. Fix a positive integer l and let ? = L?l . Then Γa = limn→∞ Γa exists in L∞ ((?Z)d ). Furtherc,? n more the L∞ limit of any multiple lattice derivative of Γa also exists and is the corresponding n continuum derivative of Γa c,? 6. Analogous statements hold for C(x ? y), but with [φ] = (d ? α)/2. The multiscale expansion in (1) above is obtained in (3.31) of section 3, and the ?nite range property (2) is given in Lemmas 3.1, 3.2. The bound in (3) follows from Theorem 5.5 of section 5. (4) above is obtained in section 6 in the course of proving Theorem 6.1, see (6.24). (5) above is part of Theorem 6.1. Turning to (6) above, the ?nite range decomposition of C is obtained in section 4 ( see (4.2), (4.3) et seq.). A uniform bound on L?vy ?uctuation covariances is supplied e in section 5 ( see Corollary 5.6 ). The statement analogous to (5) above is part of Corollary 6.2 of section 6. Let us call the Gaussian ?elds in the decomposition ?uctuation ?elds. Other (wavelet) decompositions were developed in the context of the Block Spin Renormalization Group of Kadano? and Wilson by Gawedzki and Kupiainen, [GK80, GK83] as well as, in related work, by Balaban, [Bal82a, Bal82b]. Although the Gawedzki-Kupiainen ?uctuation ?elds are not ?nite range, they have their own advantages: notably they are independent lattice ?elds, determined by random variables de?ned on increasingly coarse lattices of spacing Ln . Our decomposition achieves this only in the weaker sense that, with high probability, ζn (x) ≈ ζn (y) for |x ? y| ? Ln , but ζn retains low probability variations on all smaller scales. This is a price paid for retaining translation invariance on small scales. In section 3 the aforementioned ?nite range decompositions are obtained. The rescaled ?uctuation covariances live on ?ner and ?ner lattices, but all have the same ?nite range. In section 4 probabilistic aspects of our construction are discussed and the ?nite range decomposition of the L?vy Greens function is obtained. We are interested in the L?vy Greens function (??)?α/2 bee e cause varying the parameter α a?ects the scaling of the associated ?eld and gives insight into the 3

dynamical system Zj → Zj+1 . Moreover it is of intrinsic interest in various problems in probability theory. For example we may wish to study critical properties of self avoiding L?vy walks by renore malization group methods. We also discuss in this section renormalization group transformations based on the above ?nite range decompositions. Section 5 is devoted to bounds. Here our main result is Theorem 5.5. This theorem states that every member of the sequence of rescaled ?uctuation covariances is uniformly bounded in lattice Sobolev norms of arbitrarily high degree. The bound is independent of the lattice spacing. Finally in Section 6 we prove that the sequence converges in Sobolev norms in a precise sense to its continuum limit which is appropriately identi?ed. The continuum limit is smooth. This is the content of our main Theorem 6.1. We conclude with a brief indication of the role of the ?nite range property in the analysis of Zj → Zj+1 . Consider Z0 → Z1 when Z0 = Z0 (Λ, φ0 ) := e?λφ0 (x)

x∈Λ

4

where Λ ? Zd is a large box shaped subset of lattice points which is a disjoint union of some standard cube shaped subsets of lattice points ? ? Zd of side greater than the range of the ?rst ?eld ζ1 in a ?nite range decomposition φ0 = j≥1 ζj . Let φ1 := j≥2 ζj , X ? Zd and ? Z1 (X) = ? Note that Z1 (X) is independent of ζ1 . Then Z1 (Λ) = E1

??Λ

e?λφ1 (x)

x∈X

4

Z0 (?) = E1

??Λ

? ? Z0 (?) ? Z1 (?) + Z1 (?)

? Write δZ0 (?) := Z0 (?) ? Z1 (?). The product expands into a sum over X of terms

??X

? Z1 (?)

δZ0 (?)

??X

? In other words, X labels the factors where δZ0 is selected. We can partition X into disjoint connected components X1 , . . . XM , where Xi is connected when the cubes ? in Xi are such that one can pass between any pair of cubes by a path whose steps are nearest neighbour cubes in Xi . Let K0 (Xi ) := δZ0 (?)

??Xi

Then Z1 (Λ) =

1 M!

X1 ,...,XM

? Z1 (X0 )E1

K0 (Xi )

i

? where the connected sets Xj are disjoint and X0 := Λ \ ∪i≥1 Xi . Notice that Z1 can be and has been moved outside the expectation since it is independent of ζ1 . Now comes the key point: The sets Xj are connected unions of nearest neighbour cubes of side length greater than the range of 4

ζ1 . Since they are disjoint they are separated by a distance greater than the range of ζ1 . Therefore, by the ?nite range property, Z1 (Λ) = 1 M! ? Z1 (X0 ) E1 K0 (Xi )

i

(1.2)

X1 ,...,XM

The perturbation on a large volume Λ has been reduced to local calculations E1 K0 (Xi ) and the standard but heavy machinery of cluster expansions is being replaced by independence and geometry. Unlike Z0 (Λ) the image functional Z1 (Λ) no longer factors into contributions from boxes, which is the great simpli?cation of hierarchical models, but there is still a large part X0 of Λ where this property is retained. For the next RG map one proves that the more general form (1.2) is stable in the sense that Z2 can also be written in the same form but with a di?erent K1 and with larger cubes L?. The program is then to prove that Ej Kj (X) gives very little weight to connected sets ? X which are unions of many boxes Lj ?. This can be facilitated by making a better choice of Z1 , ?1 . In particular, one can replace λ in Z1 by some ? since the derivation is valid for other choices of Z other value λ1 chosen to minimise E1 K0 . This idea leads to a ?ow of the coupling constant λ → λ1 . The ?rst use of ?nite range covariances was in [MS00]. [BMS] is another appearance of ?nite range covariances.

2

Preliminaries

Throughout we will assume that d ≥ 3. Let L be a large integer power of 2. De?ne εn = L?n . We will be working on a sequence of lattices (εn Z)d ? Rd , (εn Z)d ? (εn+1 Z)d , with n = 0, 1, 2, ... and eventually passing to Rd . (εn Z)d is equipped with the discrete topology. The measurable sets are subsets of points and the measure dz on (εn Z)d is de?ned by

(εn Z)d

dz f (z) = εd n

z∈(εn Z)d

f (z)

(2.1)

We endow Rd with the distance function |x ? y| = max1≤j≤d |xj ? yj | R R U (R) = ? , 2 2 be an open cube of edge length R. De?ne Let

d

(2.2) (2.3)

? Rd

Uεn (R) = U (R) ∩ (εn Z)d and its boundary

(2.4)

?Uεn = {y ∈ Uεn : |x ? y| = εn , some x ∈ Uεn } (2.5) ? The distance function |.| is that induced from Rd . We denote by Uεn = Uεn ∪ ?Uεn the closure of Uεn . The lattice Laplacian ?εn is de?ned by the quadratic form (f, ??εn f )L2 ((εn Z)d ) = εd n 5

<x,y>

ε?2 |f (x) ? f (y)|2 n

(2.6)

where the sum runs over the nearest neighbour points in (εn Z)d . Let S = {?1 , ..., ed } be the set e ? d . For any such lattice unit vector e ∈ S de?ne forward and of standard unit basis vectors in Z backward lattice derivatives ?±e of a function in the direction e by (?±e f )(x) = ε?1 (f (x ± εn e) ? f (x)). n (2.7)

The backward derivative is de?ned so as to be the adjoint of the forward derivative. Then the de?nition (2.6) for the lattice Laplacian can be written as (f, ??εn f )L2 ((εn Z)d ) = εd n The corresponding resolvent Gan with ε a ≥ 0 is (2.9) |(?e f )(x)|2 (2.8)

x,e∈S

Gan (x ? y) = (??εn + a)?1 (x ? y) ε =

[?π/εn ,π/εn ]d

where ? ?εn (p) = 2ε?2 n

dd p eip.(x?y) ? (2π)d a ? ?εn (p)

d

(2.10)

?=1

cos(εn p? ) ? 1

(2.11)

3

Multiscale Decomposition of the Resolvent

We say that a function f (x, y) has ?nite range R if f (x, y) = 0 for |x ? y| ≥ R. Consider the resolvent in Zd Ga (x ? y) = (?? + a)?1 (x ? y) (3.1)

with a ≥ 0. We will ?rst develop a multiscale decomposition for the resolvent Ga , a ≥ 0 into smooth ?nite range positive semi-de?nite functions. As in (2.3) and (2.4), U (L) ? Rd is an open cube of edge length L in Rd and Uε (L) = U (L) ∩ (εZ)d the induced cube in (εZ)d with ?Uε (L) its boundary. Assume that the cube is centered at the origin. We will suppress the argument L when there is no risk of confusion. On the lattice there is no need to distinguish functions from measures. Nevertheless we use measures in cases where the associated continuum object is a measure. A case in point is the lattice a Poisson kernel PUε (x, du), which by de?nition is the measure supported on ?Uε such that

a h(x) = PUε (x, f ) := a PUε (x, du)f (u)

(3.2)

is the unique solution to the boundary value problem (??ε + a)h(x) = 0 : x ∈ Uε 6 (3.3)

h(x) = f (x) : x ∈ ?Uε

(3.4)

where f : ?Uε → R. Existence and uniqueness are easily proved since h solves a ?nite dimensional set of linear equations. Note that because a ≥ 0 a solution h(x) satis?es the weak maximum a principle. In Section 4 we will see an explicit construction of which shows that PUε is a defective probability measure. Defective means that the mass is at most one. (For a = 0 and f = 1, h ≡ 1 a which implies PUε (x, 1) is a probability measure). We will say h is a-harmonic in Xε if h solves (??ε + a)h(x) = 0 in X ∩ (εZ)d . Let g(x) be a rotationally invariant non-negative C ∞ (Rd ) function of compact support such that L (3.5) g(x) = 0 : |x| ≥ 4 with the normalization dx g(x) = 1 (3.6)

Rd

We restrict g(x) to the lattice (εZ)d and choose the normalization constant cε such that dx cε g(x) = 1 (3.7)

(εZ)d

Since g(x) is a continuous function of compact support Riemann sums converge. Hence cε is a continuous function of ε on the compact set 0 ≤ ε ≤ 1 and thus uniformly bounded. Moreover cε → 1 as ε → 0. Now comes the main idea. The point of the function g is to avoid needing detailed knowledge of the Poisson kernel for the lattice. We are about to use the Poisson kernel to de?ne an averaging operator that leaves a-harmonic functions unchanged. This property leads to the ?nite range property in Lemma 3.1. It is relatively easy to prove that our averaging operator is smoothing (uniformly in lattice spacing) because when checking di?erentiability, derivatives either fall on g a which is smooth by choice or on the x argument of PUε (x, du) with x forced to be away from the boundary ?Uε so that the easy part of standard elliptic techniques is su?cient to prove smoothness uniformly in the lattice spacing. Given a function f : (εZ)d → R we de?ne the averaging map : f → Aa (L)f ε where (Aa (L)f )(x) = ε

(εZ)d

dz cε g(z ? x)

a PUε (L) (x ? z, du)f (u + z)

(3.8)

Note that this can also be written as (Aa (L)f )(x) = ε

a dz cε g(z ? x)PUε (L,z) (x, f )

(3.9)

(εZ)d

where Uε (L, z) is the translate of the cube Uε (L) so that its center is now z . In summing over the translates we have put in the smooth function g , and not the delta function, because we will need to take derivatives with respect to x and this is hard to do if x is the center of the cube. 7

?From (3.8) we see that for every ?xed x, (Aa (L)f )(x) de?nes a bounded, positive linear functional ε on C0 ((εZ)d ), the space of functions of compact support on (εZ)d . We have |(Aa (L)f )(x)| ≤ f ε

∞

Now this integration over all translates makes Aa (L) translation invariant. Translation invariε ance plays an essential role in the proof of positive semi-de?niteness of the ?uctuation covariance constructed below, (see Lemma 3.1). Proof: For b ∈ (εZ)d , let fb (x) = f (x ? b). By change of variables z → z ? b in (3.8) (Aa (L)f )(x ? b) = (Aa (L)fb )(x). (3.10) ε ε

so that the norm of this linear functional is ≤ 1. This gives a family of defective probability measures Aa (L)(x, du) on (εZ)d : ε (Aa (L)f )(x) = ε The Fourier transform of this measure ? Aa (p) = ε satis?es ? |Aa (p)| ≤ ε De?ne the ?uctuation covariance Γa (x ? y) = Ga (x ? y) ? (Aa (L)Ga Aa (L)? )(x ? y) ε ε ε ε ε where by de?nition (Aa (L)Ga Aa (L)? )(x ? y) = ε ε ε Aa (L)(x, du)Ga (u ? v)Aa (L)(y, dv) ε ε (3.14) (3.13)

(εZ)d (εZ)d (εZ)d

Aa (L)(x, du)f (u) ε

(3.11)

du Aa (L)(0, du)e?ip.u ε

du Aa (L)(0, du) ≤ 1 ε

(3.12)

The latter is the analogue of the block spin covariance in statistical mechanics, [GK86]. Lemma 3.1. Γa and Aa (L)Ga Aa (L)? are positive semi-de?nite. Γa has ?nite range, ε ε ε ε ε Γa (x ? y) = 0 : |x ? y| ≥ 3L ε ?ε and the Fourier transform Γa (p) is continuous in p including at p = 0, uniformly in ε. Proof. First we prove the ?nite range property. By the de?nition of the Poisson kernel, if f is a a-harmonic in x + Uε , then PUε (x, f ) = f (x). Since cε g in the de?nition of Aa (L) was chosen to be ε a probability density with support in Uε (L/4), for f a-harmonic in x + Uε (5L/4), (Aa (L)f )(x) = ε

(εZ)d a dz cε g(z ? x)PUε (L,z) (x, f ) = (εZ)d

(3.15)

dz cε g(z ? x)f (x) = f (x)

When |x ? y| ≥ 3L, x + Uε (5L/4) and y + Uε (5L/4) are disjoint. Therefore Ga (u ? v) is a-harmonic ε in each argument in the appropriate region and therefore (Aa (L)Ga Aa (L)? )(x ? y) = Ga (x ? y) ε ε ε ε 8

which proves (3.15). Now we prove positive de?niteness. By translation invariance, we can take the fourier transform of Γa (x ? y) to get ε ?ε ?ε ?ε Γa (p) = (1 ? |Aa (p)|2 )Ga (p) (3.16) ? ? ? Now Ga (p) ≥ 0 and |Aa (p)| ≤ 1 where we have used (3.12). Hence Γa (p) ≥ 0. This proves positive ε ε ε a . The positive de?niteness of Aa Ga Aa ? is obvious. de?niteness of Γε ε ε ε ? Continuity of Γa (p): By (2.10),

ε

?ε Ga (p) =

1 ? εn (p) a??

?ε which holds because p derivatives of Aa (p) are moments for Aa and this is a probability measure ε of compact support which therefore has moments of all orders. Choose ε = εn?1 for n ≥ 1 and write (3.13) in a rescaled form. It is easy to check that for u, v ∈ (εn?1 Z)d we have 2 u?v (3.17) Gan?1 (u ? v) = L?(d?2) GLn a ε ε L

a PUε

?ε is continuous if a > 0 or p = 0. If a = 0 then (3.16) shows that the p?2 singularity of Γa (p) is cancelled by ?ε ?ε ?ε 1 ? Aa (p) = Aa (0) ? Aa (p) = o(p2 )

(x, du) n?1 (R,z)

L = PU

R z εn ( L , L )

2a

x du , L L

(3.18)

De?ne the sequence of functions gn on Rd by gn (z) = Lnd g(Ln z) (3.19)

where g is the function introduced earlier (see (3.5), (3.6)) and observe that because of the normalization (3.6), the function gn is also normalized : dz gn (z) = 1 (3.20)

Rd

Moreover, from the support property of g, we have that gn (x) = 0 : |x| ≥ Let Rm = L?(m?1) As in (3.11) we have in (εn Z)d for n ≥ m ≥ 0 the measure Aan ,m (Rm )(x, du) given by ε

(εn Z)d

1 4Ln?1

(3.21)

(3.22)

du Aan ,m (Rm )(x, du) f (u) = ε

(εn Z)d

a dz cεn?m gm (x ? z)PUε (Rm ,z)(x, f )

(3.23)

Note that Aan ,0 = Aan ε ε 9 (3.24)

as de?ned earlier. Observe that

(εn Z)d

dz cεn?m gm (z) =

(εn?m Z)d

dz cεn?m g(z) = 1

(3.25)

from the de?nition of the constants cεn in (3.7). The de?nition (3.19) together with (3.23) and (3.18) imply the scaling relation x du 2a Aan?1 ,m?1 (Rm?1 )(x, du) = ALn ,m (Rm )( , ) ε ε L L Applying this to the righthand side of (3.13) we get for n ≥ 1

a a Gan?1 (x ? y) = Γan?1 (x ? y) + L?(d?2) ALn ,1 (1) GLn a ALn ,1 (1)? ε ε ε ε ε

2 2 2

(3.26)

x?y L

(3.27)

We can now iterate (3.27) starting with n = 1, n-times using the same principle. De?ne for n ≥ 1

n

Aa = n For n = 0 we set We also de?ne on the (εn Z)d lattice

Aan ,n+1?j (L?(n?j) ) ε

j=1

(3.28)

Aa = 1 0 Γa = Aa Γan Aa ? n n ε n (3.29) (3.30)

and

a Gn = Aa Gan Aa ? n n ε

Then we have the multiscale decomposition for the resolvent

n?1

G (x ? y) =

a

L?j(d?2) ΓL j

j=0

2j a

x?y Lj

L + L?n(d?2) Gn

2n a

x?y Ln

(3.31)

which is valid for a ≥ 0. The special case a = 0 gives the multiscale decomposition for the massless Green’s function

n?1

G0 (x ? y) = Lemma 3.2. For all n ≥ 0

L?j(d?2) Γ0 j

j=0

x?y Lj

0 + L?n(d?2) Gn

x?y Ln

(3.32)

Γa (x ? y) = 0 : |x ? y| ≥ 6L n

Proof. Γa is a multiple convolution, so the range of Γa is the sum of the ranges of the convolved funcn n a tions. ?From the de?nition (3.23) of Aan ,m (R), the support property (3.21) of gm and PUεn (R)(x, u) ε 1 vanishes if |u ? z| > R, we ?nd that the range of Aan ,m (R) is R + 4Lm?1 . From the de?nition (3.28) ε n a the range of Aa is ?(n?j) (1 + 1 ) which is less than 3 for L large. By construction of An n j=1 L 4L a is less than 3L by Lemma 3.1. The Lemma follows. the range of Γεn 10

4

Probabilistic Aspects

Multiscale decomposition for the L?vy Green’s function. e e Let xt , 0 < α < 2, be the stable L?vy process in Z d , (stable in the sense that its scaling limit (α) is stable). Note that Ex (|xt |) < ∞, provided α > 1. However in the following we will allow for the full admissable range 0 < α < 2. The L?vy Green’s function C is given by e C(x ? y) = (??)?α/2 (x ? y) = const

∞ 0 (α)

da a?α/2 Ga (x ? y)

(4.1)

with 0 < α < 2. In this range it is easy to verify that the integral representation converges. C has a ?nite range multiscale decomposition, which we obtain by inserting the multiscale decomposition (3.31) for Ga into the integral representation (4.1) of the L?vy Green’s function C e to get

n?1

C(x ? y) =

L?j(d?2) const

j=0 ∞ 0 0

∞

da a?α/2 ΓL j

2n a

2j a

x?y + Lj

+L?n(d?2) const After rescaling in a in each term we get

n?1

L da a?α/2 Gn

x?y Ln

C(x ? y) = where

L?2j[φ] Γj

j=0

x?y Lj

+ L?2n[φ] Cn

x?y Ln

(4.2)

∞

Γj = const

0 ∞

da a?α/2 Γa j

a da a?α/2 Gn

Cn = const

0

[φ] =

d?α 2

(4.3)

Note that [φ] as de?ned above is the canonical dimension of the the scalar Gaussian ?eld φ distributed with covariance C. We have [φ] > 0, since d ≥ 3 and 0 < α < 2. Γj and Cn are well de?ned because of the bounds provided below (see section 5, Corollary 5.6). Moreover by Lemma 3.2 and (4.3) Γn (x ? y) = 0 : |x ? y| ≥ 6L ?From Lemma 3.2 and (4.3), Cn and Γn are positive semi-de?nite and thus qualify as covariances of Gaussian measures denoted ?Cn , ?Γn . The multiscale decomposition (4.2) now gives rise to renormalization group transformations. From (4.2) we get for x, y ∈ (εn Z)d Cn (x ? y) = Γn (x ? y) + L?2[φ] Cn+1 ( 11 x?y ) L (4.4)

and hence we have a sequence of RG transformations zn+1 (φ) = where and d?Cn (φ) zn (φ) = with C0 = C given by (4.1). Poisson Kernel. Let xt , t ≥ 0 be continuous time simple random walk with right continuous paths and state space (εn Z)d . The characteristic function is E(eip.xt ) = et?εn (p) and the generator of the Markov process xt is the lattice Laplacian ?εn . Note that the semigroup et?εn is a contraction on L∞ ((εn Z)d ). In the discrete topology the latter coincides with the space (n) of bounded continuous functions. Hence et?εn is a Feller semigoup so that xt is strong Markov with respect to stopping times τ . Now let xε be the above process in (εZ)d . Let Px the probability measure for the process t conditioned to start at x. As in (2.3) and (2.4), U (R) is an open cube of radius R in Rd , Uε (R) the induced cube in (εZ)d , and the boundary ?Uε (R) is de?ned as in (2.5). Let τUε be the ?rst exit time from Uε . Then τUε is also the ?rst hitting time of ?Uε from the interior. Ex (τUε ) < ∞ because Uε (R) is bounded, and hence τUε < ∞, Px a.s.. Let A be a subset of points in ?Uε . It is a standard result in probability that the measure a (x, dy) de?ned on ?U by PUε ε

a PUε (x, A) = Ex (e?aτUε 1xτU

ε (n)

d?Γn (ζ) zn (ζ + φL?1 )

(4.5)

x φL?1 (x) = L?[φ] φ( ) L d?Cn+1 (φ) zn+1 (φ) (4.6)

(n)

?

(n)

?A )

(4.7)

is the Poisson kernel we de?ned in (3.2). Also, since Px (xτUε ∈ ?U ) = 1 the total mass is

a PUε (x, ?Uε ) = Ex (e?aτUε ) ≤ 1

(4.8)

The same construction works in Rd with xε replaced by standard Brownian motion xt , Uε (R) t replaced by U (R) and A taken to be any Borel subset of ?U (R). Let now R = Rm = L?(m?1) , with 0 ≤ m ≤ n. In Section 6 we will need an estimate for a 1 ? PUεn (x, ?Uεn ) where it is understood Uεn = Uεn (Rm ). Observe that 1 ? e?aτ ≤ aτ so

a 0 ≤ 1 ? PUεn (x, ?Uεn ) ≤ aEx (τUε ).

We can estimate the mean exit time as follows. Let f be a smooth function in Rd bounded in a neighbourhood of U (R). Then by the strong Markov property, for any x ∈ Uεn (Rm ), Ex

0 τUε

dt(?εn f )(xt ) = Ex (f (xτUε )) ? f (x) 12

3 We choose x = 0 and f (x) = |x|2 in U ( 2 Rm ). Then by a simple computation in U (Rm ) we obtain ?εn f (x) = 2d. Moreover as shown in Section 6, (6.14), ?Uεn (Rm ) ? ?U (R). Hence we get the bound 1 R2 Ex (τUε ) ≤ Ex (x2Uε ) ≤ m . τ 2d 2 Thus we have proved the following

Lemma 4.1. Let Rm = L?(m?1) , 0 ≤ m ≤ n Then we have the bound

a 0 ≤ 1 ? PUεn (Rm ) (x, ?Uεn (Rm )) ≤ a 2 Rm 2

Now recall the de?nition of the measure Aan ,m (Rm )(0, du) on (εn Z)d given in (3.23). cεn?m gm (z) ε is a probability density in (εZ)d . Hence from (Lemma 4.1) we get Corollary 4.2. 0≤1?

(εn Z)d 2 Rm 2

Aan ,m (Rm )(0, du) ≤ a ε

5

Bounds

We will give uniform bounds on Fourier transforms and Sobolev norms of arbitrary high index. In Section 6 we will prove in the latter norms the convergence of the sequences Cn , Γn . The limiting covariances will thus turn out to be in C p , ?p ≥ 0. e e We recall the de?nition of the lattice derivative ?e in (2.7). In particular S = {?1 , ...?d } is the standard basis of unit vectors. For e ∈ S, ?e is the forward partial derivative and for ?e ∈ S, ?e is the backward partial derivative. We de?ne the n th lattice derivative ?n1 ,...,en = ?e1 .....?en e Let X be a connected open set in Rd . We de?ne Xε = X ∩ (εZ)d We now de?ne the lattice Sobolev norm

k

.

Hk (Xε )

of a function f by dx|?j 1 ,...,ej f (x)|2 e (5.1)

f

2 Hk (Xε )

=

j=0

2?j

±e1 ,...,±ej ∈S Xε

Lemma 5.1. Let g be a C ∞ (Rd ) function. Then for every k ≥ 0 there exists a constant Ck independent of ε such that fg

Hk (Xε )

≤ Ck g

C k (Rd )

f

Hk (Xε )

For every k ≥ 0, and any L1 function g, f ?g

Hk ((εZ)d ))

≤ g

L1 ((εZ)d )

f

Hk ((εZ)d ))

13

Proof. To prove this we take the square of the norm on the left hand side and then use the (εZ)d lattice modi?cation of the Leibniz rule : ?ε,e (f g) = (?ε,e f )g + f ?ε,eg + ε?ε,e f ?ε,eg (5.2)

Derivatives on g are bounded in the C k (Rd ) norm and all ε dependent constants can be majorised by setting ε = 1. This proves the ?rst inequality. The second inequality, which is a form of Young’s convolution inequality, is proved exactly as in the continuum. In the following we will exploit a lattice version of elliptic regularity. Let f be a bounded function in (εZ)d . Let U (R) ? Rd be an open cube centered at the origin and of edge length R. 1 Let Uε (R) = U (R) ∩ (εZ)d . Let ? ? U ( 1 R) be an open connected set. Then ?ε ? Uε ( 4 R). De?ne 4

a ha (x) = PUε (x, f )

Recall from Section 2 (see (3.3), (3.4)) et seq.) that ha (x) is the solution of the Dirichlet problem (??ε + a)ha (x) = 0 : x ∈ Uε ha (u) = f (u) : u ∈ ?Uε (5.3)

and that the maximum principle holds, because a ≥ 0. We have Proposition 5.2 (lattice elliptic regularity). With ?ε de?ned as above, and ? k ≥ 0 ha

Hk (?ε )

≤ CR (1 + a)? 2 f

1

L∞ ((εZ)d ) .

Remark : This is well known in the continuum. For completeness we give a proof of the lattice version in Appendix A. We choose ε = εn and apply the proposition to Aan ,m=1 (L0 )f (x′ ) := ε

(εn Z)d a dz cεn?1 g1 (x′ ? z)PUn ε

n (1,z)

(x′ , f )

which is the j = n term in the product (3.28) de?ning Aa . n Corollary 5.3. Let ?εn be as in Proposition 5.2 with R = 1. Then for every k ≥ 0 and every n ≥ 1 there exists a constant Ck,L independent of εn such that Aan ,m=1 (L0 )f ε

Hk (?εn )

≤ Ck,L (1 + a)? 2 f

1

L∞ ((εn Z)d ) .

1

(5.4) (5.5)

?j 1 ,...,ek Aan ,m=1 (L0 )f ε e

L∞ ((εn Z)d )

≤ Ck,L (1 + a)? 2 f

L∞ ((εn Z)d ) .

Proof. The integral over z in (5.4,3.9) can be restricted to Uεn (2) because the range of g1 is 1/4. Therefore Aan ,m=1 (L0 )(·, f ) ε dz cεn?1 g1

Uεn (2) C k (Rd )

1

Hk (?εn )

n (1,z)

≤

Hk (?εn )

a PUn ε

(·, f )

≤ ck,L (1 + a)? 2 f 14

L∞ ((εn Z)d )

We have used the ?rst inequality of Lemma 5.1, absorbing the C k (Rd ) norm of g1 in the constant since g has been ?xed once for all , and then using Proposition 5.2. To prove (5.5): By the embedding of high degree Sobolev space into L∞ , (Lemma B.1), reviewed in Appendix B, we pass from (5.4) to ?j 1 ,...,ek Aan ,m=1 (L0 )f ε e

L∞ ((εn Z)d )

≤ Ck,L (1 + a)? 2 f

1

L∞ ((εn Z)d ) .

(5.6)

noting that (5.4) applies to any translate x + ?εn of ?εn . Lemma 5.4. For every integer k ≥ 1, and every n ≥ 1, ? a constant ck,L independent of εn such that , 1 ?k ? ?ε (5.7) Aan ,m=1 (L0 )(p) ≤ ck,L (1 + a)? 2 ??εn (p) + 1 Proof. This is the essentially the standard proof that the Fourier transform of a smooth function of compact support has rapid decay. Let h(x, p) =

(εn Z)n

Aan ,m=1 (L0 )(x, du)e?ipu e

k

Then ? ??εn (p) + 1 h(x, p) =

k (εn Z)n

Aan ,m=1 (L0 )(x, du) ??u,εn + 1 e?ipu e

where the u subscript on ?u,εn indicates the variable it di?erentiates. Since ?u,εn is a linear combination of lattice translations under which the lattice is invariant, the exact analogue of integration by parts is valid and we continue with =

(εn Z)n

??u,εn + 1 Aan ,m=1 (L0 )(x, du)e?ipu ε

k

By the translation invariance (see (3.10)) of Aan ,m=1 (L0 )(x, du) we can change the derivatives to x ε =

(εn Z)n

??x,εn + 1 Aan ,m=1 (L0 )(x, du)e?ipu ε

k

By (5.5) in Corollary 5.3 with f (u) = exp(ip.u),

(εn Z)n

??x,εn + 1 Aan ,m=1 (L0 )(x, du)e?ipu ≤ ck,L (1 + a)? 2 ε

k

1

Collecting these relations we have ? ??εn (p) + 1 h(x, p) ≤ ck,L (1 + a)? 2 By setting x = 0 we ?nish the proof. Fourier transforms are naturally de?ned on the Brillouin zone Bε = [?π/ε, π/ε]d There is a constant c independent of ε such that ? p2 ≥ ??ε (p) ≥ cp2 for p ∈ Bε (5.9) (5.8)

k

1

which follows from t2 /2 ≥ 1 ? cos t ≥ ct2 on [?π, π]. 15

Theorem 5.5. ?n ≥ 0 and ?k ≥ 0, ? a constant ck,L independent of n such that ?n |Γa (p)| ≤ ck,L (1 + a)?1 (1 + p2 )?2k for p ∈ Bεn Γa n Proof. ?From (2.10) and (5.9), ? ?ε 0 ≤ Gan (p) ≤ (??εn )?1 ≤ ?ε Γan (p) ≤ cL (1 + p2 )?1 c p2

Hk ((εn Z)d )

(5.10) (5.11)

≤ ck,L (1 + a)?1

Combining this with (3.16) and the continuity assertion of Lemma 3.1 we have

?ε Case n = 0: It is su?cient to prove that Γa0 (p) is bounded by C(1 + a)?1 with C uniform in p because B?0 is bounded. Referring to (3.16,2.10) we ?nd that ?ε |Γa0 (p)| ≤ 2 ? ? ? ? |1 ? Aa0 (0)| |Aa (0) ? Aa0 (p)| |1 ? Aa0 (p)| ε ε ε ≤2 + 2 ε0 ? ε (p) ? ε (p) a a?? 0 a?? 0

The ?rst term is continuous at a = 0 by Corollary 4.2. Therefore it is bounded by C(1 + a)?1 . The second term is bounded by C(1 + a)?1 using the same argument (existence of moments of the Poissson measure) as in the proof of Lemma 3.1. ?ε Case n ≥ 1: by Lemma 5.4, and the bound |Aan ,m (p)| ≤ 1 which we use for m ≥ 2 ?n |Aa (p)|2 ≤ Ck,L (1 + a)?1 p2 + 1 Using these estimates in (3.29) we obtain (5.10). Proof of (5.11). We have the easily established bound ? |p2 + ?εn (p)| ≤ O(1)ε2 |p|4 n so that and hence for any m ≥ 0 ? 0 ≤ 1 ? ?εn (p) ≤ (1 + p2 )(1 + O(1)ε2 p2 ) n

?k

? ?n ?n (1 ? ?εn (p))m |Γa (p)|2 ≤ O(1)(1 + p2 )2m |Γa (p)|2 ≤ ck,q,L (1 + a)?1 (1 + p2 )?q Γa n

2 Hm ((εn Z)d )

for any q ≥ 0 by choosing k in (5.10) su?ciently large. Taking q > d proves the theorem because ≤ dd p

[? επ , επ ]d n n

(2π)

d

? ?n (1 ? ?εn (p))m |Γa (p)|

2

Now turn to the L?vy ?uctuation covariance given in (4.3). Using the bounds provided in e Theorem 5.5 we get Corollary 5.6. For 0 < α < 2, all k = 0, 1, . . . , and all n ≥ 0, Γn

Hk ((εn Z)d )

≤ ck,L

(5.12)

where the constant on the right hand side is independent of n. 16

6

Convergence

Theorem 5.5 and Corollary 5.6 provide uniform bounds in Sobolev norms for ?uctuation and block covariances. In particular they are uniform in the lattice spacing εn . We will now prove that these sequences converge to their formal continuum limits. Continuum objects have the subscript c in place of ε. Thus, as in (2.9), Ga (x ? y) = c dd p ip.(x?y) ? a e Gc (p) (2π)d (6.1) (6.2)

Rd

?c Ga (p) = (a + p2 )?1 Recall from Section 3 that

R R d U (R) ≡ Uc (R) = ? , ? Rd 2 2 represents an open cube of edge length R. In analogy to (3.23) with cc := 1 we de?ne the continuum average Aa (Rm )(x, , du) by c,m du Aa (Rm )(x, du) f (u) = c,m

a dz cc gm (x ? z)Pc,U (Rm ,z) (x, f )

(6.3)

Rd

Rd

where

a PUc (R,z) (x, f ) = ?Uc (R,z) a du PUc (R,z) (x, u)f (u)

is the solution

ha (x) c

to the continuum Dirichlet problem (??c + a)ha (x) = 0 : x ∈ Uc (R) c ha (x) = f (x) : x ∈ ?Uc (R) c (6.4)

With these notations, the Fourier transform of the continuum analogue of (3.13) is ? ? ? Γa (p) = Ga (p) ? |Aa (R0 )(p)|2 Ga (p) c c c,0 c and that of (3.29) is ? c,n Γa =

m=1 n

(6.5)

?c,m ?c |Aa (Rm )|2 Γa

(6.6)

The Lemmas, Propositions, Theorems and their Corollaries of sections 3,4 and 5 remain true in the continuum with the following caveat : in the continuum the uniform Sobolev bounds of Theorem 5.5 and Corollary 5.6 hold only for n ≥ 1. Aside from this caveat their proofs are identical and need no repetition. When referring to them for the continuum objects we shall simply mention them as the continuum analogues of the relevant results for the lattice. Recall that Bεn is the Brillouin zone, de?ned in (5.8). The main result is Theorem 6.1. For every integer k ≥ 0, Γa → Γa c,n c,? 17 (6.7)

in Hk (Rd ). Moreover, for every ?xed lattice (εl Z)d , 0 ≤ l ≤ n the restriction of Γa (x) to (εl Z)d n converges to the continuum Γa (x) restricted to (εl Z)d in the Sobolev norm c,? Γa ? Γa c,? n

Hk ((εl Z)d )

→ 0 as n → ∞

(6.8)

Moreover multiple lattice derivatives of Γa converge to the corresponding continuum derivatives of n Γa in the L∞ ((εl Z)d ) norm. c,? For the Levy ?nite range decomposition (4.2) we apply the last theorem to (4.3) and obtain Corollary 6.2. For all k ≥ 0, Γc,? ? Γn

Hk ((εl Z)d )

→ 0 as n → ∞

(6.9)

Moreover multiple lattice derivatives of Γn converge to the corresponding continuum derivatives of Γc,? in the L∞ ((εl Z)d ) norm. We now give some Lemmas which will be employed in the proof of Theorem 6.1. In the following lemmas we consider continuum functions f : Rd → R and use the same symbol for the lattice function f : Zd → R de?ned by restriction. Continuum and lattice integration are to be distinguished by the domain of integration. Lemma 6.3. f (. + h) ? f (.) f (. + h) ? f (.)

Hk (Rd )

≤ c|h|. f ≤ c|h|. f

Hk+1 (Rd )

where |h| is the norm in Zd ? Rd . Moreover if |h| ≤ R we have

Hk (Uc (R)) Hk+1 (Uc (2R))

(6.10)

Proof. See Theorem 3.3 on page 42 of [Agm65]. Lemma 6.4. De?ne Q = ??ε ? (??c ) Then we have for every k ≥ 0, Qf

Hk (Rd )

≤ cε f

Hk+3 (Rd )

(6.11)

where the constant c is independent of ε. Proof. ?e,ε is the forward lattice derivative in (εZ)d , and ?? , its L2 ((εZ)d ) adjoint, the backward e,ε lattice derivative. Forward and backward derivatives commute. ?e,c is the continuum derivative in direction e and the adjoint is ?? = ??e,c . A calculation shows that e,c ?? ?e,ε f (x) = e,ε Therefore Qf (x) = ?

e 0 1 1 1

dt

0 1 0

ds (??2 )f x + [t ? s]εe e,c

(6.12)

dt

0

ds

?2 f x + [t ? s]εe ? ?2 f (x) e,c e,c 18

since ??ε =

e

?? ?e,ε e,ε

the lemma is proved by taking norms under the integrals and using Lemma 6.3 (which gives the factor ε). We can now describe the main idea. Let f : ?Uc (R) → R be a continuum function f : Rd → R restricted to the continuum boundary of Uc (R). We need to estimate the di?erence between the solution ha to the continuum Dirichlet problem and the solution han to the lattice Dirichlet problem. ε c This will be done by restricting ha to the lattice. The restriction also solves a lattice Dirichlet c problem, but with a non-zero right hand side involving Qha , which by Lemma 6.4 is O(εn ). Thus c we consider R R (6.13) han (x) ? ha (x) : x ∈ Uεn ( ) ? Uc ( ) c ε 4 4 han (x) ? ha (x) = 0 : x ∈ ?Uεn (R) c ε in Sobolev norms. The di?erence satis?es zero boundary conditions because we will arrange that the lattice boundary points ?Uεn (R) all lie on the continuum boundary ?Uc (R) and both solutions have boundary values f restricted to ?Uεn (R). The lattice cube Uε (R) = U (R) ∩ (εZ)d has as its boundary ?Uε (R) = {y ∈ Uε (R) : |x ? y| = ε, some x ∈ Uε (R)} ?From Section 2 we have εn = L?n , L = 2p , p ≥ 1. In addition we now choose , in accord with (3.22),R = Rm = L?(m?1) : 0 ≤ m ≤ n. Then not only do we have Uεn (R) ? Uc (R) := U (R) but also ?U εn (R) ? ?Uc (R) (6.14) 2?np

Rm 2

This last statement follows from the observation that εn = and = so that for Rm 0 ≤ m ≤ n we have 2 ∈ εn Z ? R. This means that for d = 1 the boundary points of Uεn (Rm ) coincide with the boundary points of U (Rm ). For d > 1 we are in cubes and the above reasoning easily generalises to (6.14). Lemma 6.5. Let f : Rd → R, Then for R = L?(m?1) : 0 ≤ m ≤ n we have ?k1 ,...,ek han ? ha c ε e

L∞ (Uεn ( R )) 4

2?mp+p?1

≤ cL,R,k εn f

L∞ (?Uc (R))

where the constant cL,R,k is independent of n. 19

Proof. See Appendix A. To proceed further we need a formula. First de?ne a new ?nite di?erence derivative that acts on continuum functions by : ? ?e,ε f (z) = Then we have Lemma 6.6.

d?1 Rd 1 0

dt (1 ? t)?e,c f (z + tε)

dz f (z) ?

dz f (z) = ε

(εZ)d j=0 (εZ)d?j ×Rj

? dz ?ej+1 ,c f (z)

(6.15)

Proof. We obtain case d = 1 by dz f (z) ? dz ε?1

εZ [z,z+ε]

dz f (z) =

εZ z∈εZ [z,z+ε]

R

dx f (x) ?

εZ

dz f (z)

εZ

=

dx f (x) ? f (z) = ε

? dz ?e,ε f (z)

and then the general case is obtained by applying this formula iteratively. The Fourier transform of Aan ,m (Rm )(0, du) is given by ε ?ε Aan ,m (Rm )(p) = cεn?m where han ,m (z, p) = ε

?Uεn (Rm a PUεn (Rm ) (z, du)eip.u

(εn

Z)d

dz gm (z)e?ip.z han ,m (z, p) ε

? Likewise, there is the continuum Fourier transform Aa (Rm )(p) de?ned by the same formulas with c,m εn , εn?m replaced by c and with cc := 1. ?a ? We wish to estimate the di?erence Aa c,m+1 (Rm+1 )(p) ? Aεn ,m+1 (Rm+1 )(p). This is provided by the following Lemma : Lemma 6.7. For all integers k ≥ 0, and n ≥ 1, ? a constant ck,L,m independent of n such that ?a ? Aa c,m+1 (Rm+1 )(p) ? Aεn ,m+1 (Rm+1 )(p) ≤ ck,L,m εn Proof. It is easy to see using Lemma 6.6 that we can write ?c,m+1 (Rm+1 )(p) ? Aa ,m+1 (Rm+1 )(p) = R1 (p) + R2 (p) + R3 (p) ?ε Aa n where R1 (p) = (1 ? cεn?1?m ) R2 (p) = cεn?1?m

(εn Z)d (εn Z)d

(6.16)

dz e?ip.z gm+1 (z)ha c,m+1 (z, p)

a dz e?ip.z gm+1 (z) ha c,m+1 (z, p) ? hεn ,m+1 (z, p)

20

d?1

R3 (p) = εn

j=0

(εn Z)d?j ×Rj

? dz e?ip.z ?ej+1 ,c gm+1 (z)ha c,m+1 (z, p)

We observe that as follows from the de?nition (3.7), Lemma 6.6, and the fact that g is a smooth function in Rd of compact support. The integral is bounded by O(1). For the term involving R2 we use Lemma 6.5 with f (u) = exp(ip.u) which produces the small factor εn in the bound. Finally in R3 the O(εn ) factor is already there. In the integrand ha c,m+1 (z, p) and derivatives are L∞ bounded on the support of gm+1 by (the continuum versions of) Proposition 5.2 and Lemma B.1, with f (u) = exp(ip.u). Proof. (Theorem 6.1) Let ? > 0 and ?x any p = 0. Fix any a ≥ 0. We will ?rst prove that ? c,? ?n |Γa (p) ? Γa (p)| < ? for all su?ciently large n. ?ε ?ε Recall that Aan ,m (p) and Aan ,m (p) are Fourier transforms of defective probability measures supported in a cube of side Rm . Now ?ε |1 ? Aan ,m (p)| ≤ 1 ? Aan ,m (0, du) + ε Aan ,m (0, du)|1 ? exp(?ip.x)|. ε |1 ? cεn?1?m | ≤ cL εn

Therefore, by |1 ? exp(?ip.x)| ≤ c|p.x|, and Corollary 4.2,

2 ?ε |1 ? Aan ,m (p)| ≤ cRm |p| + caRm

(6.17)

Note that the same bound holds in the continuum, because Corollary 4.2 remains true in the continuum. ? c,n ?From the de?nition of Γa we get ? c,n+1 ? Γa = |Aa ? c,n ?c,n+1 (Rn+1 )|2 ? 1 Γa ? c,n Γa whence, using the continuum analogue of (6.17),

2 ?a ? ?a |Γa c,n+1 ? Γc,n | ≤ 2cRn+1 |p| + 2caRn+1 |Γc,n |

? Now Γa sati?es the uniform bound of Theorem 5.5, Rn+1 decreases geometrically with increasing c,n ? c,n n. Therefore we see from the previous inequality that Γa form a Cauchy sequence. This proves ? c,? the existence of the limit Γa satisfying the bound of Theorem 5.5 and the ?rst part of Theorem 6.1 has been proved. Choose n su?ciently large so that ε ? c,? ? c,n |Γa (p) ? Γa (p)| < 4 21 (6.18)

?ε Γan (p) is uniformly continuous in p by Lemma 3.1. Therefore, as shown in the course of proving Theorem 5.5 , there exists a constant γ independent of εn such that ?ε |Γan (p)| ≤ γ Now (6.17) implies that

2 ?ε ?c,m |Aan ,m | ? |Aa | ≤ 2cRm |p| + 2caRm

(6.19)

??,m Since Rm → 0 geometrically fast and |Aa (p)| ≤ 1, we can choose N depending on ? such that for all n > N , n n ? ?ε ?a |2 ? |Aan ,m |2 < (6.20) |Ac,m 4γ

m>N m>N

By Lemma 6.7 there is a constant CN such that

N m=1 N

? |Aa |2 ? c,m

m=1

? |Aan ,m |2 < CN εn < ε

? 4γ

(6.21)

? (6.22) 4 by the de?nitions (3.13) , the explicit Fourier transforms 2.10 6.2) and Lemma 6.7. ?From the de?nition (3.29) we see that the four inequalities (6.19, 6.20, 6.21, 6.22) imply that ?c ?ε |Γan ? Γa | < ? c,n ? n |Γa ? Γa | < (6.23) and (6.18) establish the pointwise convergence ? ? Γa (p) → Γa (p) n c,? (6.24) 3? 4 (6.23)

for all su?ciently large n. Finally,

By the dominated convergence theorem using Theorem 5.5 for domination, we have, for any ?xed compact set X ? Rd in momentum space and any k

X

?n ? c,? dp |Γa (p) ? Γa (p)|2 (1 + p2 )k → 0

as n → ∞. This proves (6.8) of the theorem, because we can choose X to be a ?xed Brillouin zone in the dual of (εl Z)d . The convergence in the L∞ ((εl Z)d ) norm follows by Sobolev embedding (see Appendix B).

7

Acknowledgements

We thank Gianni Jona-Lasinio for his hospitality in Rome and for many stimulating conversations over the years. PKM thanks G?rard Menessier and Andr? Neveu for fruitful discussions. We thank e e an anonymous referee for ?ngering an incorrect Lemma.

22

A

Lattice Elliptic Regularity

(a ? ?ε )h = g (A.1)

Suppose that h solves on (εZ)d and ? has compact support. Then, on (εZ)d ,

?h(a ? ?ε )h = ?hg Integrate over (εZ)d . By the de?nition of ??ε this can be rewritten as a

(εZ)d

dz ?h2 +

(εZ)d

dz

e∈S

(?e ?h)(?e h) =

dz ?hg

(εZ)d

(A.2)

Surprisingly, calculation shows that this can be rewritten as a

(εZ)d

dz ?h2 +

1 2

±e∈S

(εZ)d

dz ?(?e h)2 =

1 dz ?hg + (?ε ?)h2 2 (εZ)d

(A.3)

In other words with forward and backward derivatives the lattice gives the same formula as the continuum without corrections that go to zero with ε. For a ≥ 0 and for ? ≥ 0, (A.3) implies 1 2

±e∈S (εZ)d

dz ?(?e h)2 ≤ 1 a

1 dz ?hg + (?ε ?)h2 2 (εZ)d

(A.4)

(εZ)d

dz ?h2 ≤

1 dz ?hg + (?ε ?)h2 2 (εZ)d dz |?j 1 ,...,ej h|2 ? e

(A.5)

Let [h]2 = 2?j ?,j

±e1 ,...,±ej ∈S (εZ)d

By applying j ?nite di?erence derivatives to (A.1) we ?nd that (A.4) is also true for h and g replaced by derivatives of h and g. By the Cauchy-Schwartz inequality on the ?rst term in (A.4),

2 [h]2 ?,j+1 ≤ [h]?,j [g]?,j + [h] 1 ?ε ?,j

2

Simplify the ?rst term using the inequality ab ≤ 1 (a2 + b2 ) and use the resulting inequality to 2 iteratively reduce the order of the top derivative in the Sobolev norm,

k

h We obtain,

2 ?,k

:=

j=0

2?j

±e1 ,...,±ej ∈S (εZ)d

dz |?j 1 ,...,ej h|2 ? e

(A.6)

h

2 ?k ,k

≤ h

2 ?0 ,0

+ g

2 ?0 ,k?1

(A.7)

where ?0 is the ?nal member of a sequence ?j of non-negative functions chosen such that ?j?1 ≥ 1 1 ?j + ?ε ?j , ?j?1 ≥ ?j 2 2 23 (A.8)

Proof. (Proposition 5.2). We are given that h solves (A.1) in Uε (R) with g = 0. We can estimate the L2 norm in Uε (R/2) in two di?erent ways. Firstly, For all a ≥ 0 we can use the maximum principle |h| ≤ f L∞ (?Uε (R)) h

L2 (Uε (R/2))

≤ cRd/2 f

L∞ (?Uε (R))

Secondly, for a ≥ 1, we can choose ? in (A.5) to be one on U (R/2) and zero outside U (R) to obtain h

2 L2 (Uε (R/2))

and then use the maximum principle to bound the right hand side by the L∞ (?Uε (R)) norm. Therefore, for any smooth ?0 supported in Uε (R/2) h

2 ?0 ,0

≤ CR a?1 h

2 L2 (Uε (R))

(A.9)

≤ CR (1 + a)?1 f

2 L∞ (?Uε (R))

(A.10)

Let ?k , ?k?1 , . . . , ?0 ≥ 0 be C ∞ continuum functions with compact support in U (R/2) such that ?k = 1 on ? and (A.8) holds. Apply (A.7) and (A.10) to obtain Proposition 5.2. √ Remark on Exponential Decay: The correct a dependence is exp(?O( aR)). We outline how to do this using a method suggested by [Agm82]. (A.5), for g = 0, can be rewritten as dz wh2 ≤ 0

(εZ)d

1 w := (a ? ?ε )? 2 Consider the choice ? = exp(?u) with u(x) = 1√ a 1 + x2 2

Let ?0 be the continuum Laplacian. The ?nite di?erence Laplacian ?ε ? can be written as an integral over ?0 ?: for example, in one dimension

1 1 0

?ε ?(x) =

0

ds dt ?0 ?(x + [t ? s]ε)

√ Using this we ?nd ?ε u ≥ 0 and w ≥ c(a, ε)? with c(a, ε) ≈ exp(? aε). Now replace exp(?u) by ? = exp(?u)ψ where ψ is a smooth, positive, monotonic decreasing function such that ψ = 1 on U ( 2 R) and vanishes outside U (R). Then w ≥ c(a, ε)? ≥ 0 on U ( 2 R) so by taking the part of the 3 3 integral where w ≥ 0 to the right hand side of the bound and discarding part of the integral where w ≥ 0 we get c(a, ε)

Uε (R/2)

dz ?h2 ≤

Uε (R)\Uε ( 2 R) 3

dz (?w)h2 ≤ f

2 L∞ (?Uε (R))

Uε (R)\Uε ( 2 R) 3

dz |w|

which gives decay because √ ?|U (R/2) ≥ exp(O( aR))?|U (R)\U ( 2 R)

3

24

Preparation for proof of Lemma 6.5: Suppose (A.1) holds in a domain Uε (R) and h vanishes on ?Uε (R). Then, with ? = 1, a ≥ 0, (A.2) becomes,

e Uε (R)

dz (?e h)(?e h) ≤

dz hg

Uε (R)

(A.11)

We estimate the right hand side by the Cauchy-Schwartz inequality and substitute the result into the Poincar? inequality, which is e h

2 L2 (U? (R))

≤ CR

dz (?e h)(?e h)

e Uε

(A.12)

where CR is independant of ε. Then h

L2 (U? (R))

≤ CR g

L2 (U? (R))

(A.13)

Returning to (A.7) and using (A.13) we obtain Lemma A.1. Let ? ≥ 0 be a C ∞ continuum function with compact support in U (R). By restriction it de?nes functions on all lattices, denoted by the same letter. Then for ? su?ciently small, there exists a constant CR,? such that the solution to (A.1), with zero boundary conditions on ?Uε (R), satis?es h ?,k ≤ CR,φ g Hk?1 (U? (R)) where the constant CR,φ is independant of ?. Proof. (Lemma 6.5) Write as in lemma 4.3, ??εn+1 = ??εn + Q. Let h = han ? han+1 . This has ε ε zero boundary conditions on ?U εn (R) and (??εn + a)h(x) = g(x) : x ∈ Uεn (R) where g = Qhan+1 . Now apply Lemma A.1 and estimate g by Lemma 6.4 followed by Propoε sition 5.2. The proof is completed by Sobolev embedding, see Lemma B.1 below, taking k in Lemma A.1 su?ciently large.

B

Sobolev Spaces on the Lattice

d d Lemma B.1. Let I d := [0, 1]d , Iε = I d ∩ (εZ)d and X be any subset of Iε . Then

u Proof.

L∞ (X)

≤ Cd u

d Hd (Iε )

A smooth continuum function u(x), where x = (x1 , . . . , xd ), satis?es

x1 xd

x1 . . . xd u(x) =

0

dy1 . . .

0

dyd ?1 . . . ?d (y1 . . . yd u(y1 , . . . , yd ))

and therefore, for x ∈ I d := [0, 1]d , |u(x)| ≤ Cd u |x1 . . . xd | 25

Hd (I d )

The same proof adapts to the lattice (εZ)d with integrals and derivatives being replaced by sums d and ?nite di?erences so that for u : Iε → R,

x∈Jε

sup |u(x)| ≤ Cd u

d Hd (Iε )

where Jε := [1/2, 1]d . For d = 2, I 2 is the union of [1/2, 1]2 , [0, 1/2]2 , [1/2, 1] × [0, 1/2] and ε [0, 1/2] × [1/2, 1] and by symmetry the same bound holds with Jε replaced by any of these boxes. The same argument applies for all dimensions d, which implies the lemma.

d Lemma B.2 (Poincare Inequality). Let u : Iε → R be any function vanishing on the boundary d . There exists C independent of ε such that of Iε

d Iε

dx|u|2 ≤ C ?u

2 d L2 (Iε )

Proof. Let ui (x) be the ?nite di?erence partial derivative with respect to component xi of x. Then, for i = 1, . . . , d, |u(x1 , . . . , xd )| ≤

Iε

dxi |ui (x1 , . . . , xd )|

The right hand side is a function of all components of x except xi . Take the product over i followed by 2/d root or power. |u(x)| ≤

2

2 d

i

Iε

dxi |ui |

≤

i

Iε

dxi |ui |

2

1 d

d Integrate both sides over (x1 , . . . , xd ) ∈ Iε and use the H¨lder inequality on the right hand side. o

d Iε

dx|u| ≤

2

dx

i

d Iε

Iε

dxi |ui |

2

1 d

≤

dx

d Iε

i

|ui |2

because the extra integral integrates to unity.

References

[Agm65] Shmuel Agmon. Lectures on elliptic boundary value problems. Prepared for publication by B. Frank Jones, Jr. with the assistance of George W. Batten, Jr. Van Nostrand Mathematical Studies, No. 2. D. Van Nostrand Co., Inc., Princeton, N.J.-Toronto-London, 1965. [Agm82] Shmuel Agmon. Lectures on exponential decay of solutions of second-order elliptic equations: bounds on eigenfunctions of N -body Schr¨dinger operators, volume 29 of Matheo matical Notes. Princeton University Press, Princeton, NJ, 1982. [Bal82a] Tadeusz Balaban. (Higgs)2,3 quantum ?elds in a ?nite volume. I. A lower bound. Commun. Math. Phys., 85(4):603–626, 1982.

26

[Bal82b] Tadeusz Balaban. (Higgs)2,3 quantum ?elds in a ?nite volume II. An upper bound. Commun. Math. Phys., 86:555–594, 1982. [BMS] [FS81] [GK80] [GK83] [GK86] D.C Brydges, P.K. Mitter, and B. Scoppola. Critical (Φ4 )3, ? . hep-th/0206040, mp-arc 02-273, Commun. Math. Phys., (in press). J. Fr¨hlich and T. Spencer. Kosterlitz-Thouless transition in the two dimensional Abelian o spin systems and Coulomb gas. Comm. Math. Phys. , 81:527, 1981. K. Gawedzki and A. Kupiainen. A rigorous block spin approach to massless lattice theories. Comm. Math. Phys. , 77:31–64, 1980. K. Gawedzki and A. Kupiainen. Block spin renormalization group for dipole gas and (▽φ)4 . Ann. Phys., 147:198, 1983. K. Gawedzki and A. Kupiainen. Asymptotic freedom beyond perturbation theory. In K. Osterwalder and R. Stora, editors, Critical Phenomena, Random Systems, Gauge Theories. Les Houches, North Holland, 1986. Christian Hainzl and Robert Seiringer. General decomposition of radial functions on Rn and applications to N -body quantum systems. Lett. Math. Phys., 61(1):75–84, 2002. P. K. Mitter and B. Scoppola. Renormalization group approach to interacting polymerised manifolds. Comm. Math. Phys., 209(1):207–261, 2000.

[HS02] [MS00]

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