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Divergence of a stationary random vector field can be always positive (a Weiss' phenomenon)

a r X i v :0709.1270v 2 [m a t h .P R ] 11 S e p 2007

Divergence of a stationary random vector ?eld can be always positive (a Weiss’phenomenon)

Boris Tsirelson

Abstract

The divergence of a stationary random vector ?eld at a given point is usually a centered (that is,zero mean)random variable.Strangely enough,it can be equal to 1almost surely.This fact is another form of a phenomenon disclosed by B.Weiss in 1997.

Introduction

If a random vector ?eld is stationary (that is,shift-invariant in distribu-tion),then the expectation of its divergence must vanish,since it is equal to the divergence of the expectation,thus,the divergence of a constant vector ?eld.This simple argument is conclusive provided that the expectations are well-de?ned (especially,for Gaussian processes).Waiving existence of ?rst moments we cannot ask about the expected divergence,but we still may ask,whether the divergence can be always (strictly)positive,or not.The answer appears to be negative in dimension one but a?rmative in dimension two.The former is evident,while the latter is demonstrated by a non-evident counterexample constructed below following an idea of B.Weiss [1].

The phenomena under consideration manifest themselves equally well in two forms,discrete and continuous.Starting with the discrete setup we treat two lattices Z and Z 2as two (in?nite)graphs.

?10

12

Z

Z 2

By a vector ?eld me mean a real-valued function on oriented edges such that its sum over the two orientations of an edge vanishes.By the divergence of a vector ?eld we mean the following real-valued function on vertices:given a vertex,we sum over the (two or four)outgoing edges the values of the vector ?eld.

?8

5

1

a vector ?eld its divergence

dimension one

For a vector ?eld v =(v x,x +1)x ∈Z on Z ,its divergence is given by

(div v )x =v x,x +1?v x ?1,x

for x ∈Z .

If a random vector ?eld v on Z is stationary then the two random variables v ?1,0and v 0,1are identically distributed.If also (div v )0≥0a.s.,then v ?1,0≤v 0,1a.s.,therefore P v ?1,0≤a

a stationary random vector ?eld v on Z cannot satisfy the condi-tion P (div v )0>0 =1.

In dimension two the situation is di?erent.

Theorem 1.There exists a stationary random vector ?eld v on Z 2whose divergence is equal to 1everywhere,almost sure.

(For the proof see

Sect.1.)The corresponding result in the continuous setup follows by smoothing,namely,convolution with the indicator function of the square (?0.5,0.5)×(?0.5,0.5).In other words:if,say,v =1on the edge ((0,0),(0,1))and v =0on other edges of Z 2,then the ?rst (horizontal)component of the smoothed vector ?eld on R 2at (x,y )is equal to 1?|x ?0.5|if |x ?0.5|<1,|y |<0.5(and 0otherwise).

vector ?eld

discrete

2

The divergence of the smoothed?eld is equal to+1on the square(?0.5,0.5)×(?0.5,0.5)and?1on(0.5,1.5)×(?0.5,0.5);just the smoothed div v.

divergence

discrete

+

+

+

+

?

?

?

?

Having v such that div v=1everywhere on Z2we get the smoothed diver-gence equal to1almost everywhere on R2(and in the distributional sense). Some additional smoothing gives a smooth vector?eld of divergence1ev-erywhere.

1The construction and the proof

In order to keep the matter as discrete as possible,from now on we consider only integer-valued vector?elds on Z2.

Lemma1.If there exist stationary random vector?elds v1,v2,...such that

(a)P (div v n)x=1 →1as n→∞,for every vertex x of the graph Z2,

(b)sup n P |(v n)y|>C →0as C→∞,for every edge y of the graph Z2, then there exists a stationary random vector?eld v such that

(A)P (div v)x=1 =1for every vertex x of the graph Z2,

(B)P |v y|>C ≤sup n P |(v n)y|>C for every C and every edge y of the graph Z2.

Proof.The distributionμn of v n is a probability measure on the space Z E of all maps E→Z;here E is the(countable)set of all edges of https://www.sodocs.net/doc/c23090205.html,ing the one-point compacti?cation

Z E.All probability measures on

Random vector ?elds v n will be constructed out of non-random ?nite fragments.For example,n =2:

the fragment

The fragment of size 3×3(whose construction will be explained later)is repeated,forming a (double-)periodic vector ?eld (non-random).Shifting this periodic ?eld we get 3·3=9periodic ?elds.These 9vector ?elds are the possible values of the random vector ?eld v 2;they have equal probabilities (1/9),by de?nition (of v 2).Thus,v 2is stationary.

Note that the divergence of v 2at a given point (say,the origin)takes on two values,+1(with probability 8/9)and ?8(with probability 1/9).Also,the value of v 2on a given horizontal edge takes on three values ?3,0,3with probabilities 1/9,7/9,1/9respectively.

The fragment of size 3×3,used above,is the second element of a sequence,whose n -th element is of size (2n ?1)×(2n ?1).The sequence is constructed recursively.Its ?rst element,of size

1×1,is trivial:just a single vertex,no edges.The (n +1)-th fragment contains four copies of the n -th fragment (two of them being turned upside down)connected as

follows:

=

4

Here are the?rst three fragments:

n=1

n=2

n=3

For each fragment,the vector?eld conforms to the given oriented graph,and its divergence is equal to+1at all vertices except for one vertex.Such vector ?eld exists and is unique,since the graph is a tree;the divergence at the root

is equal to? (2n?1)2?1 =?(22n?2n+1

).Note the consistency:the four

copies of the n-th fragment occur in the(n+1)-th fragment as vector?elds (not only graphs).

For each n we construct v n out of the n-th fragment in the same way as we did it for n=2.Clearly,v n is an integer-valued stationary random vector ?eld on Z2,and

P (div v n)x=1 =1?1

C)as C→∞,for every edge y of the graph Z2.

Proof.First,the maximal possible value of|(v n)y|is equal to22n?2n+1(this is the value on the edge that enters the root of the tree).Second,

P |(v n+1)y|≤22n?2n+1 ≥4·(2n?1)2

(2n+i?1)2

5

for i=1,2,...Therefore

P |(v n+i)y|≤22n ≥ 2n+i?2i2n+i?1 2≥(1?2?n)2≥

≥1?2·2?n;

P |(v n)y|>22k ≤2·2?k;

sup

n

the lemma follows.

2Remarks and questions

Remark1.It is easy to see that the constructed sequence(v n)n converges in distribution.No need to choose a subsequence(as in the proof of Lemma1). Question1.Is the condition E

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