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ELECTRONIC COMMUNICATIONS in PROBABILITY

GEOMETRIC INTERPRETATION OF HALF-PLANE

CAPACITY

STEVEN LALLEY1 Department of Statistics University of Chicago

email: [email protected]

GREGORY LAWLER2 Department of Mathematics University of Chicago

email: [email protected]

HARIHARAN NARAYANAN

Laboratory for Information and Decision Systems MIT

email: [email protected]

SubmittedAugust 30, 2009, accepted in final formOctober 8, 2009 AMS 2000 Subject classification: 60J67, 97I80

Keywords: Brownian motion, Conformal Invariance, Schramm-Loewner Evolution

Abstract

Schramm-Loewner Evolution describes the scaling limits of interfaces in certain statistical me-chanical systems. These interfaces are geometric objects that are not equipped with a canonical parametrization. The standard parametrization of SLE is via half-plane capacity, which is a confor-mal measure of the size of a set in the reference upper half-plane. This has useful harmonic and complex analytic properties and makes SLE a time-homogeneous Markov process on conformal maps. In this note, we show that the half-plane capacity of a hullAis comparable up to multiplica-tive constants to more geometric quantities, namely the area of the union of all balls centered in Atangent toR, and the (Euclidean) area of a 1-neighborhood ofAwith respect to the hyperbolic

metric.

1

Introduction

SupposeAis a bounded, relatively closed subset of the upper half planeH. We callAa compactH

-hull ifAis bounded andH\Ais simply connected. Thehalf-plane capacityofA, hcap(A), is defined

in a number of equivalent ways (see[1], especially Chapter 3). IfgAdenotes the unique conformal 1RESEARCH SUPPORTED BY NATIONAL SCIENCE FOUNDATION GRANT DMS-0805755.

2RESEARCH SUPPORTED BY NATIONAL SCIENCE FOUNDATION GRANT DMS-0734151.

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transformation ofH\AontoHwithg

A(z) =z+o(1)asz→ ∞, thengAhas the expansion gA(z) =z+hcap(A)

z +O(|z| −2), z

→ ∞.

Equivalently, ifBt is a standard complex Brownian motion andτA=inf{t≥0 :Bt6∈H\A}, hcap(A) = lim

y→∞y

Ei y”Im(Bτ

A)

— .

Let Im[A] =sup{Im(z):zA}. Then if yIm[A], we can also write

hcap(A) = 1

π

Z∞

−∞

Ex+i y”Im(Bτ

A)

— d x.

These last two definitions do not requireH\Ato be simply connected, and the latter definition

does not requireAto be bounded but only that Im[A]<.

ForH-hulls (that is, for relatively closedAfor whichH\Ais simply connected), the half-plane

capacity is comparable to a more geometric quantity that we define. This is not new (the second author learned it from Oded Schramm in oral communication), but we do not know of a proof in the literature3. In this note, we prove the fact giving (nonoptimal) bounds on the constant. We start with the definition of the geometric quantity.

Definition 1. For anH-hull A, lethsiz(A)be the2-dimensional Lebesgue measure of the union of all balls centered at points in A that are tangent to the real line. In other words

hsiz(A) =area 

 

[

x+i yA

B(x+i y,y)

 ,

whereB(z,ε)denotes the disk of radiusεabout z. In this paper, we prove the following.

Theorem 1. For everyH-hull A,

1

66hsiz(A)<hcap(A)< 7

2πhsiz(A).

2

Proof of Theorem 1

It suffices to prove this for weakly boundedH-hulls, by which we meanH-hullsAwith Im(A)<

and such that for eachε >0, the set{x+i y:y> ε}is bounded. Indeed, forH-hulls that are not

weakly bounded, it is easy to verify that hsiz(A) =hcap(A) =.

We start with a simple inequality that is implied but not explicitly stated in[1]. Equality is achieved whenAis a vertical line segment.

Lemma 1. If A is anH-hull, then

hcap(A) Im[A] 2

2 . (1)

3After submitting this article, we learned that a similar result was recently proved by Carto Wong as part of his Ph.D.

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Proof. Due to the continuity of hcap with respect to the Hausdorff metric onH-hulls, it suffices to

prove the result forH-hulls that are path-connected. For twoH-hullsA

1⊆A2, it can be seen using the Optional stopping theorem that hcap(A1)hcap(A2). Therefore without loss of generality,A can be assumed to be of the formη(0,T]whereηis a simple curve withη(0+)R, parameterized

so that hcap[η(0,t]) = 2t. In particular, T = hcap(A)/2. If gt = (0,t], then gt satisfies the Loewner equation

tgt(z) = 2 gt(z)−Ut

, g0(z) =z, (2)

whereU :[0,T]R is continuous. Suppose Im(z)2>2 hcap(A)and letYt =Im[gt(z)]. Then

(2) gives

tYt2≤ 4Yt

|gt(z)Ut|2≤4, which implies

YT2Y024T>0. This implies thatz6∈A, and hence Im[A]22 hcap(A).

The next lemma is a variant of the Vitali covering lemma. Ifc>0 andz=x+i y∈H, let I(z,c) = (xc y,x+c y),

R(z,c) =I(z,c)×(0,y] ={x′+i y′:|x′−x|<c y, 0<y′≤y}.

Lemma 2. Suppose A is a weakly boundedH-hull and c>0. Then there exists a finite or countably

infinite sequence of points{z1=xi+i y1,z2=x2+i y2, , . . .} ⊂A such that:

y1y2y3≥ · · ·;

the intervalsI(x1,c),I(x2,c), . . .are disjoint;

A

∞ [

j=1

R(zj, 2c). (3)

Proof. We define the points recursively. LetA0=Aand given{z1, . . . ,zj}, let

Aj=A\

j [

k=1

R(zj, 2c)  .

IfAj=;we stop, and ifAj=6 ;,we choosezj+1=xj+1+i yj+1∈Awithyj+1=Im[Aj]. Note that if kj, then|xj+1xk| ≥2c ykc(yk+yj+1)and henceI(zj+1,c)∩ I(zk,c) =;. Using the weak boundedness ofA, we can see that yj0 and hence (3) holds.

Lemma 3. For every c>0, let

ρc:=2

p

2

π arctan

€

eθŠ, θ=θc= π 4c.

Then, for any c>0, if A is a weakly boundedH-hull and x

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Proof. By scaling and invariance under real translation, we may assume that Im[A] = y0 = 1 and x0 = 0. Let S = Sc be defined to be the set of all points z of the form x +iu y where x+i yA\ R(i, 2c)and 0<u1.

Clearly,SA=A\ R(i, 2c).

Using the capacity inequality[1, (3.10)]

hcap(A1A2)hcap(A2)hcap(A1)hcap(A1A2), (4) we see that

hcap(SA)hcap(S)hcap(A)hcap(SA). Hence, it suffices to show that

hcap(SA)hcap(S)ρ2 c.

Let f be the conformal map ofH\SontoHsuch thatzf(z) =o(1)asz→ ∞. LetS∗:=SA. By properties of halfplane capacity[1, (3.8)]and (1),

hcap(S∗)hcap(S) =hcap[f(S\S)] Im[f(i)] 2

2 .

Hence, it suffices to prove that

Im[f(i)]p2ρ= 4

π arctan

€

eθŠ. (5)

By construction,S∩ R(z, 2c) =;. LetV = (2c, 2c)×(0,) ={x+i y :|x|<2c,y>0}and let

τV be the first time that a Brownian motion leaves the domain. Then[1, (3.5)],

Im[f(i)] =1Ei”Im(B τS)

—

≥P¦B

τS∈[−2c, 2c]

©

≥P¦B

τV ∈[−2c, 2c]

© .

The mapΦ(z) =sin(θz)mapsV ontoHsending[2c, 2c]to[1, 1]andΦ(i) =isinhθ. Using conformal invariance of Brownian motion and the Poisson kernel inH, we see that

B

τV ∈[−2c, 2c]

© = 2

πarctan

1

sinhθ

= 4

π arctan

€ eθŠ. The second equality uses the double angle formula for the tangent.

Lemma 4. Suppose c>0and x1+i y1,x2+i y2, . . .are as in Lemma 2. Then

hsiz(A)[π+8c] ∞ X

j=1

y2j. (6)

If c≥1, then

π

∞ X

j=1

y2j hsiz(A). (7)

Proof. A simple geometry exercise shows that

area 

 

[

x+i y∈R(zj,2c)

B(x+i y,y)

(5)

Since

the upper bound in (6) follows. Sincec1, and the intervalsI(zj,c)are disjoint, so are the disks

B(zj,yj). Hence,

Combining this with the upper bound in (6) with anyc>0 gives hcap(A)

For the upper bound, choose a covering as in Lemma 2. Subadditivity and scaling give

hcap(A)

Combining this with the lower bound in (6) withc=1 gives hcap(A)

hsiz(A) ≤

hcap[R(i, 2)]

π .

Note thatR(i, 2)is the union of two real translates ofR(i, 1), hcap[R(i, 2)]2 hcap[R(i, 1)] whose intersection is the interval(0,i]. Using (4), we see that

hcap(R(i, 2))2 hcap(R(i, 1))hcap((0,i]) =2 hcap(R(i, 1))1 2.

ButR(i, 1)is strictly contained inA′:={zH:|z| ≤p2}, and hence

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The last equality can be seen by consideringh(z) =z+2z−1which mapsH\AontoH. Therefore,

hcap[R(i, 2)]<7

2,

and hence

hcap(A) hsiz(A) <

7 2π.

An equivalent form of this result can be stated4in terms of the area of the 1-neighborhood ofA (denoted hyp(A)) in the hyperbolic metric. The unit hyperbolic ball centered at a point x+ιy is the Euclidean ball with respect to which x+ιy/e)and x = ιy e are diametrically opposite boundary points. For anyc, choosing a covering as in Lemma 2,

hyp(A)<

e

2 ‹2

π+4ec

X

j=1 y2j.

So by (8),

hcap(A) hyp(A) > ρ

2 c

e

2 ‹2

π+4ec −1

.

Settingcto8 5,

hcap(A) hyp(A) >

1 100.

For anyc>ee−1

2 ,

hyp(A)π

‚ ee−1

2

Œ2 X

j=1 y2j. So by (9),

hcap(A) hyp(A) <

hcap[R(i, 3)]

πee−1

2

2 .

hcap(R(i, 3))hcap(R(i, 1)) +hcap(R(i, 2))hcap((0,i])5. Therefore,

1 100<

hcap(A) hyp(A) <

20

π(ee−1)2.

References

[1] G. Lawler, Conformally Invariant Processes in the Plane, American Mathematical Society, 2005. MR2129588

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