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The Wadsworth Mathematics Series Series Editors

Raoul H. Bott, Harvard University David Eisenbud, Brandeis University Hugh L. Montgomery, University of Michigan Paul J. Sally, Jr., University of Chicago Barry Simon, California Institute of Technology Richard P. Stanley, Massachusetts Institute of Technology W. Beckner, A. Calderon, R. Fefferman, P. Jones, Conference on Harmonic Analysis in Honor of Antoni Zygmund M. Behzad, G. Chartrand, L. Lesniak-Foster, Graphs and Digraphs J. Cochran, Applied Mathematics: Principles, Techniques, and Applications W. Derrick, Complex Analysis and Applications, Second Edition R. Durrett, Brownian Motion and Martingales in Analysis A. Garsia, Topics in Almost Everywhere Convergence K. Stromberg, An Introduction to Classical Real Analysis R. Salem, Algebraic Numbers and Fourier Analysis, and L. Carleson, Selected Problems on Exceptional Sets

Brownian Motion and Martingales in Analysis

Richard Durrett University of California, Los Angeles

Wadsworth Advanced Books & Software Belmont, California A Division of Wadsworth, Inc.

Acquisitions t?ditor:.lohn Kimmel Production Editor: Marta Kongsle Designer: Janet Bollow Copy Editor: Mary Roybal Technical Illustrator: Brown & Sullivan

U 1984 by Wadsworth, Inc. All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transcribed, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher, Wadsworth Advanced Books & Software, Belmont, California 94002, a division of Wadsworth, Inc. Printed in the United States of America

1 2 3 4 5 6 7 8 9 10-88 87 86 85 84

ISBN 0-534-03065-3

Library of Congress Cataloging in Publication Data

Durrett, Richard, 1951Brownian motion and martingales in analysis. (The Wadsworth mathematics series) Bibliography: p. 313 Includes index. 1. Brownian motion processes. 2. Martingales (Mathematics) 1. Title. II. Series. QA274.75.D87 1984 515 84-7230 ISBN 0-534-03065-3

Preface

In the years that have passed since the pioneering work of Kakutani, Kac, and Doob, it has been shown that Brownian motion can be used to prove many results in classical analysis, primarily concerning the behavior of harmonic and analytic functions and the solutions of certain partial differential equations. In spite of the many pages that have been written on this subject, the results in this area are not widely known, primarily because they appear in articles that are scattered throughout the literature and are written in a style appropriate for technical journals. The purpose of this book, then, is to bring some of these results together and to explain them as simply and as clearly as we can. In Chapters 1 and 2, we introduce the two objects that will be the cornerstones for our later developments-Brownian motion and the stochastic integral. This material is necessary for all that follows, but after digesting Chapters 1 and 2, readers can turn to their favorite applications. The remaining seven chapters fall into four almost independent groups. In Chapters 3 and 4, we will study the boundary limits of functions that are harmonic in the upper half space H = {x a R": Xd > 0}. Chapter 3 is devoted to developing the relevant probabilistic machinery, that is, we define conditioned Brownian motions (or h-transforms) and give their basic properties. In Chapter 4, we apply these results to the study of harmonic functions; to be precise, we show that (modulo null sets) nontangential convergence, nontangential

boundedness, and finiteness of the "area function" are equivalent, and we investigate the relationship between these notions and their probabilistic counterparts. In Chapter 5, we turn our attention to analytic functions and use Brownian motion to prove results about their boundary limits and mapping properties (e.g., Picard's theorem). The first results are, I think, some of the most striking applications of Brownian motion. By observing that a complex Brownian motion never visits 0 at.any (positive) time, we can make Privalov's theorem "obvious," and we can Xb%mbiguously along Brownian paths to prove that functions in the vanlinnT4Ass have nontangential limits, without relying on factorization tl*lbremsto rei *'e the zeros. V

VI

Preface

In Chapters 6 and 7, we use Brownian motion to study the classical Hardy spaces HP, p > 0, that is, the set of functions that are analytic in D = {z: IzI < 1 } and have sup Jf(reb91'dO < oc. r 2, Kelvin's Transformations 78 Change of Measure, Girsanov's Formula 82 Martingales Adapted to Brownian Filtrations 85 Notes on Chapter 2 89 A Word about the Notes 90 ix

2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13 2.14

1

11

44

X

('onk'nhM

3

Conditioned Brownian Motions

3.1

3.2 3.3 3.4 3.5

Warm-Up: Conditioned Random Walks 91 Brownian Motion Conditioned to Exit H = Rd-1 x (0, oo) at 0 Other Conditioned Processes in H 97 Inversion in d > 3, B, Conditioned to Converge to 0 as t -* oo A Zero-One Law for Conditioned Processes 102

4

Boundary Limits of Harmonic Functions

91

4.1

105 Probabilistic Analogues of the Theorems of Privalov and

4.2 4.3

105 Spencer 108 Probability Is Less Stringent than Analysis Equivalence of Brownian and Nontangential Convergence in

4.4 4.5

d=2

94 100

113

Burkholder and Gundy's Counterexample (d = 3)

116

With a Little Help from Analysis, Probability Works in d > 3: Brossard's Proof of Calderon's Theorem

119

5

Complex Brownian Motion and Analytic Functions

5.1

Conformal Invariance, Applications to Brownian Motion 123 Nontangential Convergence in D 126 Boundary Limits of Functions in the Nevanlinna Class N 128 Two Special Properties of Boundary Limits of Analytic Functions 132 Winding of Brownian Motion in C - {0} (Spitzer's Theorem) 134 Tangling of Brownian Motion in C - { -1, 11 (Picard's Theorem) 139

5.2 5.3 5.4 5.5 5.6

123

6

Hardy Spaces and Related Spaces of Martingales

6.1

6.8 6.9

Definition of HP, an Important Example 144 First Definition of .11", Differences Between p > 1 and p = 1 146 A Second Definition of #P 152 Equivalence of H" to a Subspace of &P 155 Boundary Limits and Representation of Functions in HP 158 Martingale Transforms 162 J anson' s Ch aracter izat i on o f A' 166 Inequalities for Conjugate Harmonic Functions 170 Conjugate Functions of Indicators and Singular Measures 180

7

H' and BMO,

7.1

The Duality Theorem for .,#' 184 A Second Proof of (Jiu)* = .V.#0 188

6.2 6.3 6.4 6.5 6.6 6. 7

7.2 7.3 7.4

I' and J1(

144

184

Equivalence of BMO to a Subspace of M. && 192 The Duality Theorem for H 1, Fefferman-Stein Decomposition

199

('onlenN

xl

7.5 7.6 7.7 7.8

Examples of Martingales in -4.,# 205 The John-Nirenberg Inequality 208 The Garnett-Jones Theorem 211 A Disappointing Look at (.,ff°)* When p < 1

8

PDE's That Can Be Solved by Running a Brownian Motion 219

A 8.1

Parabolic Equations 219 The Heat Equation 220 The Inhomogeneous Equation 223 The Feynman-Kac Formula 229 The Cameron-Martin Transformation

8.2 8.3 8.4 B 8.5 8.6 8.7 8.8

215

234

Elliptic Equations 245 The Dirichlet Problem 246 Poisson's Equation 251 The Schrodinger Equation 255 Eigenvalues of A + c 263

9

Stochastic Differential Equations

9.1

PDE's That Can Be Solved by Running an SDE 271 Existence of Solutions to SDE's with Continuous Coefficients Uniqueness of Solutions to SDE's with Lipschitz Coefficients Some Examples 283 Solutions Weak and Strong, Uniqueness Questions 286 Markov and Feller Properties 288 Conditions for Smoothness 290 293 Notes on Chapter 9

9.2 9.3 9.4 9.5 9.6 9.7

Appendix A.1

A.2 A.3 A.4 A.5 A.6 A.7 A.8

271

A Primer of Probability Theory

294

Some Differences in the Language 294 Independence and Laws of Large Numbers 296 Conditional Expectation 300 Martingales 302 Gambling Systems and the Martingale Convergence Theorem Doob's Inequality, Convergence in L", p > 1 306 Uniform Integrability and Convergence in L' 307 Optional Stopping Theorems 309

References

313

Index of Notation Subject Index

327

325

274 278

303

1

Brownian Motion

1.1

Definition and Construction A d-dimensional Brownian motion is a process B t > 0, taking values in Rd, that has the following properties:

(i) if to < tl
0, then

P(B(s + t) - B(s) EA) = f

(2it)-1/2e-lxl2/21

A

(iii) with probability 1, t - B, is continuous. (i) says that B, has independent increments. (ii) says that the increment B,+, - Bs

has a d-dimensional normal distribution with mean 0 and covariance tl, that is, the coordinates B., - Br are independent and each one has a normal distribution with mean zero and variance t. (iii) is self-explanatory. The first question that must be confronted in any discussion of Brownian motion is, "Is there a process with these three properties?" The answer to this question is yes, of course. There are dozens of books about Brownian motion, and there are at least four or five essentially different constructions (one of which we will give below), so there can be no doubt that the process exists. For the moment, however, I want to try to shake your faith in this fact by pointing out that there are two things to worry about: (a) Are assumptions (i) and (ii) consistent? (b) If we specify the distribution of Bo, then (i) and (ii) determine the distribution of (B(t1), . . . for any finite set of times. Is (iii) consistent with these distributions?

To build the suspense a little bit, let's fix our attention on the case d = 1 and see what happens if we change (ii) to 1

2

I

Brownien Motion

(ii') P(8,+1 - Bs a A) =

dx JA

where f is a family of probability densities, that is, each ft z 0 and f f, = 1. If u > 0, then (i) implies I .f (x).f.(Y - x) dx,

(*)

or introducing the characteristic function (a.k.a. Fourier transform)

.r(e) = Je6f(x)dx (") f+.(B) =.f (0)f (e). It is easy to show that if t -+f ,(O) is continuous, then (") implies f (9) = exp(cet) (and if you work hard you can show this is true when X(0) is bounded and measurable), so the distributionsf(x) are far from arbitrary. When

f(x) =

(2nt)-1/2e-x2/21

f (O) = exp(-102/2)

so (*) holds, but shifting our attention now to question (b), this is not the only possibility. The Cauchy density with parameter t 1r(x) = lr(t 2 + x2)

has

f(9) = exp(-t!OI), so these f's are another possible choice, and the process that results is called the Cauchy process, C, t >- 0. It is easy to see from the formulas for the normal and Cauchy densities that (1)

Bt ° t 1/2B1

and

C1

tC1,

so as t-+ 0, P(IBI > e) -0, P(I CI > e) - 0. On this basis one might naively expect that both processes can be defined in such a way that the paths are continuous. This is true for Brownian motion but false for the Cauchy process, and, in fact, we will see in Section 1.9 that with probability 1 the set of discontinuities of C, is dense in [0, oc). The preceding discussion has hopefully convinced you that the fact that Brownian paths are continuous is not obvious, so we turn now to the somewhat

tedious details of the construction of Brownian motion. For pedagogical reasons, we will first pursue an approach that leads us to a dead end, and then we will retreat a little to rectify the difficulty.

Fix x c- R" and for each 0 < t l < ... < t define a measure µ11,.,,,t on (R")° by

I

I)rflnltlon end Construction

1.1

µ,,...,,,,(A, x ... x An) =

f',

dx1 ... j dx" ll n

where xo = x, to = 0, and

p,(x,y) = (2µt)

12e-Iv-x12/2r

In this notation, (+) says

µt,,u(R x A) = µt+"(A) This is the first step in showing that the family of µ's is a consistent set of finite

dimensional distributions (or, for short, f.d.d.'s), that is, if {s1, ... , sn-1 } c

... ,tn} and t;O{s1, ...

{t1,

sn-1}, then

µs,,....sn-1(A1 x ... x An-1)

=

x ... x Aj-1 x R x Aj x ... x An-1).

It is easy to check (details are left to the reader) that the measures given above are a consistent set of f.d.d.'s, so we can use Kolmogorov's extension theorem to give our first construction of Brownian motion. (2)

Let fl, = {functions w : [0, oo) - R} and .moo = a-algebra generated by the finite dimensional sets to): w (t;) E A; for 1 < i < n}, where each A; E 1d, the set of Borel subsets of Rd. Given a consistent set of f.d.d.'s µ,,....,,", there is a unique probability measure µ on (00,.Fo) so that for all the finite dimensional sets,

p({w:(o,.EAi,l 1, and if we pick S small enough, then r > 0 and E2-" < oc, so the Borel Cantelli lemma implies the following: (4)

For almost every w there is an N (which depends on co) so that for all n >- N,

I B(j2-") - B(12-")I < ((j - i)2-")y

for all 0 E(Am " - t2-n)2 = 2ntz2 z"E(Bi - 1)z m=1

M=1

With this established, the rest is easy. Using Chebyshev's inequality gives 7/ z"

PI

z

Y- Am " - t

> E < E Itz2 "E(Bi - 1)z,

\\m=1

and then the desired conclusion follows from the Borel Cantelli lemma.

(7) says that the quadratic variation of Bs, 0 < s < t is almost surely t. This shows that (a) locally the Brownian path looks like t 112 and (b) B, is irregular in a very regular way. The next result is more evidence for (a). Exercise 3 Generalize the proof of (5) to show that if y > 1/2, then with

probability 1, t

B, is not Holder continuous of order y at any point. (The

borderline case y = 1/2 will be treated in Section 1.3.)

1.2 The Markov Property Since Brownian motion has independent increments, it is easy (I hope) to believe that (1)

If we let s > 0, then B5 , - Bs, t > 0, is a Brownian motion that is independent of what happened before time s.

The last sentence is, I think, the clearest expression of the Markov property of Brownian motion. This section is devoted to formulating and proving a result that makes this precise, but reduces (1) to a rather cryptic formula that requires several definitions to even explain what it means. Why do I want to do this? It is not for sport or for the love of secret code,

but because the "cryptic form" of the Markov property is the most useful for doing computations and proving theorems. It is easy (I am told by my students) to get lost in the measure theoretic details, so before we plunge into

them, I will do two sample applications to convince you that the Markov property is easy to understand and use: If you want to compute PX(Bt = 0 for some t e [a, b]) where 0 < a < b, then it seems reasonable to break things down according to the value Example 1

$

I

Brownian Motion

of B. and use the fact that Brownian motion has "independent increments" to conclude that PX(B, = 0 for some t e [a, b])

= JPa(X,Y)1y(Bt = 0 for some t< b - a) dy where p (x, y) =

(27rt)-a12e-IX-yl2/2r

Example 2 The next level of complication is to compute P (B, = 0 for some t e [0, a] and some t e [a, b]). This time, when we break things up according to Ba the first factor in the answer changes to pa (x, y) = P; (B. = y, B, = 0 for some s c- [0, a] ) Pa (x, y),

but the second factor stays the same, that is, PX(B, = 0 for some t e [0, a] and some t e [a, b])

= JPa(XY)P(Br = 0 for some t:5 b - a) dy. Intuitively the last equality holds, because if we condition on the value of Ba, then the behavior of B, for t e [a, b] is independent of whether or not Brownian motion hit 0 in the interval [0, a]. The Markov property below will allow us to prove that this formula is correct. The first step in giving a precise statement of (1) is to explain what we mean by "what happened before time s." The first thing that comes to mind is

.5°=a(B,:r<s) where the right-hand side denotes the smallest a-field, which makes all the B r:5 s measurable. Ultimately, for technical reasons, we will trade this in for some a-fields that are a little large, but for the moment these (organic) a-fields are fine and we can state and prove our first version of the Markov property. (2)

If we let s > 0, then for any t > 0, B5+r - B5 is independent of what happened before time s. To be precise,

(2)

Ifs > 0, t > 0, and f is bounded, then for all x e R°, E.(.f(B,+S - B5)!. °) =

EE.f(B,+5 -

B5).

Proof This is almost an immediate consequence of the definition of Brownian motion. If sl < s2 < < s < s and A = {w : B5.((o) a C;, 1 < i:5 n} where the C, e. (the Borel subsets of Rd), then definition of Brownian motion implies 1A and f(B, - B5) are independent, so in this case (*) Ex(f(B, - B5);A) = Px(A)EX.f(B, - B5).

1.2

9

'rhr Murkov Property

To prove (2) we need to show that (+) holds for all A

To do this, we

will use an extension theorem, commonly called Dynkin's it - A theorem, which was tailor-made for situations like this. (3)

Let sad be a collection of subsets of Q that contains S2 and is closed under intersection. Let I be a collection of sets that satisfy

(i) if A, Be l and A

B, then A - Beg

(ii) if A E 9 and A,, T A, then A E 9.

If d c 9, then the a-field generated by .4, a(d) c 9. The proof of this result is not hard, but as you will discover if you try to prove it for yourself, it is not trivial either. Since the ideas involved in giving an efficient proof of this are not needed in the developments that follow, we will call this a result from measure theory and refer you to page 5 of Blumenthal and Getoor (1968) or page 34 of Billingsley (1979) for a proof. The formulation above is from Billingsley. With (3) in hand, the proof of (2) is trivial. Let sa?' = the sets of the form < s < s and let 9 = the collection {B,. E C;, 1 < i < n} where s1 < s2 < of A for which (*) holds, and observe that I clearly satisfies (i) and (ii), so 9 Q(sd) = .F°. (The reader should also notice that it is not so easy to show that if A, B E then A fl B E 9, but the proof of (i) is trivial.) Our next step toward the Markov property is (4)

If we let s z 0, then for any t >- 0, B,+r - Bs is independent of what happened before time s and has the same distribution as B, - B0. To be precise,

(4)

Ifs >- 0, t > 0, and f is bounded, then for all x E R°, EE(f(B,+s) I: °) = EB(,)f(BB)

where the right-hand side is the function p(x) = Es f(B,) evaluated at x = Bs.

Proof To prove (4), we will prove a slightly more general result (let g(x,y) _

f(x + M. (5)

If s >_ 0, t >- 0, and g is bounded, then Ex(9(Bs, Br+s - Bs) I.°fs°) = Ta(BS)

where (Pg(x) =

J(x, y)

(2nt)-12e-I yl2r21 dy.

To prove (5), we observe that if g(x, y) = 91(x)g2(y), then

(Pa(x) =91(x)J and

92(Y)(2itt)-12e-IrI2I2tdY

10

I

Brownlan Motion

Ex(9(Bs, B,+s - BS)I fF) = 91(B3)Ex(92(B,

-

91(BS)Es(92(81+5 - BS))

by (2), so the equality holds in this case.

To extend the last result to all bounded measurable functions, we need an extension theorem for functions. The one we will use is a "monotone class theorem." (6)

Let sad be a collection of subsets of 92 that contains Q and is closed under be a vector space of real-valued functions on S2 satisfying

intersection. Let (i) if A C-,4, 1A E

(ii) if f e ,Y are nonnegative and increase to a bounded function f, then f e .Y. contains all bounded functions on S2 that are measurable with respect Then to a(d) = the o-field generated by sad. As before, we will refer the reader to Blumenthal and Getoor (1968) for the proof (page 6 this time) and we will content ourselves with applying (6) to complete the proof of (5). Let ./ = the set of all rectangles (A x B where A, B E Md and Af = the set of all bounded g for which (5) holds. Taking g, = lA and g2 = 1B and applying our first result shows that (i) holds. It is clear that ., satisfies (ii), so applying (6) now proves (5) and hence (4). With (4) proved, the last step is to extend the reasoning to a general measurable "function of the path B,+s, t > 0." To describe this class of functions

we need some notation. Recall that our Brownian motions are a family P.,, x c- Rd of measures on (C, W). For s > 0, define the shift transformation O : C C by setting (Ow)(t) = co(s + t) for t > 0. In words, we cut off the part of the path before time s and then shift the time scale so that time s becomes time 0.

With this notation introduced, it is clear how to define a "function of the path B,+S, t > 0." It is simply Yo Os where Y: C R is some cC measurable function. We can now finally state the "cryptic form" of the Markov property. (7)

Ifs > 0 and Y is bounded and W measurable, then for all x c Rd, O) = EBtsi Y

EX(Yo 0.1

where the right-hand side is the function 9 (x) = EE Y evaluated at x = B.

Proof To prove (7) we start by retracing the steps in the proof of (4). First let s = t° < t1 < ... < t and Ai = B(ti) - B(ti_1). From the proof of (2), we see that if f1, . , f are bounded, .

.

E. (II f(oi) ; A) = E =1

f(A1)) Px(A), (fl j-1

so repeating the proof of (4) shows that if g is a bounded measurable function on (Rd)n+1

Ex(9(Bs, A1, ... , An)I°) = O(BS)

1.3

The Right Contlnuoue VIllrrllon, Blumenthal's 0-1 Law

Ii

where n

(Y) =

Jg(Y,z1, ... ,z

(z) dz,

proving (7) in the case where Y depends on finitely many coordinates, and applying the monotone class theorem again proves (7).

(7) is a very important formula which will be used many times below. You should try a few of the following exercises to see how (7) is used in computations. Exercise 1

Let B, be a one-dimensional Brownian motion. Let S = inf{t > 1 B, = 0} and T = inf{t > 0: B, = 0}. Show that S = To 0, and use the Markov property to conclude that if t 0, :

Px(S > 1 + t) = JPi(xY)P(T>t)dY. Let B, and T be as in Exercise 1 and let R = sup{t < 1 : B, = 0}. Use the Markov property to conclude that if 0 < t 5 1,

Exercise 2

Px(R < t) = JPt(xY)P(T> 1 - t) dy.

The last two formulas can be used to compute the distributions of R and S after we find Py(T > s) in Section 1.5. The next three examples will be important in Chapter 8. Exercise 3 Let u(t, x) = EX f(B,) where f is bounded. Use the Markov property

to conclude that if 0 < s < t, u(t - s, Bs).

E.(.f(B1)I.

Let u(t, x) = EE f o g(B,) dr where g is bounded. Use the Markov property to conclude that if 0 < s < t, Exercise 4

Ex

(f"g(B,)/dr

.

°l =

Jsg(B,)dr+u(t-s,B.). o

Exercise 5 Let c, = f o c(B,) dr and u(t, x) = Ex exp(c,) where c is bounded. Use the Markov property to conclude that if 0 < s < t, EX(exp(cr)I. °) = exp(c.,)u(t - s,B.).

1.3

The Right Continuous Filtration, Blumenthal's 0-1 Law The first part of this section is devoted to a technical matter. For reasons that will become apparent in Section 1.4, it is convenient to replace the fields J°

12

1

Brownian Motion

defined in the last section by the (slightly larger) fields

+= u>t

which are nicer because they are right continuous, that is,

\

(1) t>s

t>s

u

u>t

=n ,o= u

s

u>s

In words, the+ allow us "an infinitesimal peek at the future." Intuitively, this should not give us any information that is useful for predicting the value of Bt+s - B. In this section we will show that this intuition is correct, and furthermore that + and F° are equal (modulo null sets). The first step in proving the last result is to prove that the simplest form of the Markov property proved in Section 1.2 holds when 3 ° is replaced by

s

(2)

If f is bounded, then for all s, t > 0 and x e Rd, Ex(f(B,+s)jffl's+) = EB(s)f(B,)

where the right-hand side is the function 9(x) = Ex f(B,) evaluated at x = B(s).

Proof The result is trivial if t = 0, so we will suppose t > 0. Let r e (s, s + t). Applying (4) from Section 1.2 gives Ex(f(B,+s) I

°) = EB(r)f(Bt+s-r),

so if we let cp(x, u) = Ex

and integrate over A, we get

Ex(.f(B,+s) ; A) = Ex((P (Br, t + s - r) ; A).

Now if f is bounded and continuous, W (x, t) =

f(27[t)-d/2 e-I v-xj'12f(Y) dY

is a bounded continuous function of (x, t) ; so letting r j s, we conclude that Ex(f(Bt+s); A) = Ex((P(Bs, t); A)

for all bounded continuous functions. To extend this to all bounded functions, let A' = the set of all bounded functions for which the equality holds, let sad = the set of all rectangles (al, bl) x x (ad, bd) where for each i - oo < a; < b< < oo..*' is clearly a vector space that satisfies the hypotheses of our monotone class theorem ((6) in Section 1.2).

As for d, if A e.4, then 1A is an increasing limit of bounded nonnegative continuous f,,, so lA E.°.

With (2) established, it is routine to prove that the Markov property holds in the general form: (3)

Ifs >- 0 and Y is bounded and ' measurable, then for all x e Rd,

1.3

The Right Contlnuoui Filtration, Blumenthel's 0-I law

13

EE(Yo©SI.FS+) = EB(S)Y

where the right-hand side is the function cp(x) = E Y evaluated at x = Bs.

Proof Rereading the proof of (7) in Section 1.2 reveals that all we did was use induction and the monotone class theorem to turn the special case Yo BS = f(B,+S) into the general case. To help the reader check this, we will repeat the proof with the numbers changed below. To prove (3), we start by retracing the steps in the proof of (2). First let s = t° < tl < < t and 0; = B(ti) - B(ti_1). From the proof of (2), we see that iff1, ... , fn are bounded and continuous,

f(0i);A) =ExI

Ex \I

\t-1

t-1

f(4)I PX(A),

/

so repeating the proof of (2) shows that if g is a bounded measurable function on (Ra)n+1 , EX(g(BS, O1 , ... , 0n) I

+) _)I/(BS)

where n

I'

W (y) =

Jg(y,z1, ... , Zn)

Pt,-,,-1(Zi) dz, !=1

proving (3) in the case where Y depends on finitely many coordinates, and applying the monotone class theorem now proves the desired result. Combining (3) above with (7) from Section 1.2 gives EE(YoBSI #;+) = EB(S)Y= EX(YoO51°).

The last equality shows that conditional expectations of functions of the future are the same for and . °. Since it is trivial that E(X I.°rs+) = E(X I S°) for c F5+, we are led immediately to conclude that (4)

If Z e ' is bounded, then for all s > 0 and x e R", EE(Z

+) = Ex(Z

I3S°).

Proof By the monotone class theorem it suffices to prove the result when Z = II"-1 f (B,.) where the f are bounded and measurable, but in this case Z = X(Yo 9,) where Xe.F° and Yeq, so Ex(Z I

+) = XEE(Yo BSI3 S+) = XEB(S) YC FS°.

Since FO c F5+, it follows (from the definition of conditional expectation) that EX(Z Ex(Z I3S°). To steal a line from Chung (1974), page 341, "The reader is urged to ponder

over the intuitive meaning of this result and judge for himself whether it is obvious or incredible." Above we have treated the result as a technical necessity (i.e., an obvious result). The rest of this section is devoted to the other viewpoint.

14

I

Brownlmn Motion

The fun starts when we take s = 0 in (4) to get (5)

Blumenthal's 0-1 Law. If A e .moo, then for all x e Rd, Px(A) a {0,1 }.

Proof Since A e .Fo , 'A =

The Markov property (3) implies EB(o)lA = Ex 'A,

Px a.s.,

so PX(A) = E. 'A= lA(X) e {O, 11, proving the desired result.

In words, Blumenthal's 0-1 law says that the "germ field",o = nE,O. o is trivial. This result is very useful in studying the local behavior of Brownian paths. Let B, be a one-dimensional Brownian motion. (6)

Letr=inf{t _ 0:B,>0}. ThenP0(r=0)= 1. Proof P0(i < t) > P0(B, > 0) = 2

since the normal distribution is symmetric about 0. Letting t 10, we conclude P0(i=0)=li1mP0(r> t

I2'

so it follows from (5) that P0(r = 0) = 1. Once Brownian motion must hit (0, oo) immediately starting from 0, it must also hit (- oc, 0) immediately, and this forces it to cross 0 infinitely many times in [0, E] for any E > 0. (7)

Let t be a sequence of numbers that decreases to 0. Then PO almost surely, we have B,n > 0 for infinitely many n and B, < 0 for infinitely many n. Proof Let A _ {B,n > 0}. Then

P0(limsupAn) = lim P0I U A N-ao

n=N

/

lim sup PO(AN) = N-ao

2'

so it follows from the 0-1 law that P0(limsupAn) = 1. (7) is a prime example of how the Blumenthal 0-1 law forces the Brownian path to behave erratically. The following is a more humorous example: (8)

If you run a two-dimensional Brownian motion for a positive amount of time, it will write your name an infinite number of times, that is, if g : [0, 1] -. R2 is a continuous function, E > 0 and to 10, then P0 almost surely, we will have

1.3

The Right continuous Filtration, Blumenthal's 0-1 Law

B(Otn)

sup

15

_ g(O) < 1.

to

0 x) < 2Po(BT > x).

J

tES

Hint: It suffices to prove the result when S is finite. In this case the result is a consequence of independent increments and the fact that PO(B, > 0) = 1/2 for all t > 0. In Section 1.5 we will show that the inequality is an equality when

S = [0, T]. Exercise 4 e-Z2/2

ze-:2/2

dz < 1 X

x

=

J

e-x2/2

x 1

and

fe_z2/2dz

1

-e-x22

as x -+ oo.

x

Here and in what follows, f(x) - g(x) as x - oc meansf(x)/g(x) - 1 as x

oo.

Exercise 5 Law of the Iterated Logarithm. Let ap(t) = (2t log log(1/t))1/2. Then lim sup Bt/cp(t) = 1,

40

PO a.s.

16

1

Brownian Motion

(a) The first step is to explain why this is the right order of magnitude. To do this we observe P(B, > (ct log log(l /t) )1/2)

=

P(

> (clog log(1/t))1/21 ~

(cloglog(1/t))1/2

``

f

so if we let t" = a" where a < 1,

JJ

P(B(t") > (ct"loglog(l/t"))112)

- (clog(-nlog

a))_1/2(-n log a)cJ2'

so the sum is finite if c > 2 and infinite if c < 2. (b) To prove lim sup B,/(p(t) < 1, we observe that since t -+ (p (t) is increasing, using the result of Exercise 3 shows BS

PO C max

to+15s

.,

"`(1 + E)(P(tn+1))

by scaling, so if a is close to 1, then I(x-(l + E) > 1, and the result follows from the Borel Cantelli lemma. (c) To prove liminfBtl(p(t) > 1, observe that Po(B," - B,n+1 > (1 - E)(P(t")) = Po(Btn(1-a) > (1 - E)(P(tn)) = Pa (B,"

> (11 -Ea (P(t")) ,

so if a is close to 0, then (1 - E)/ l - a < 1, and the other Borel Cantelli lemma implies

P0(B," - B,n+1 > (1 - E)(p(tn) i.o.) = 1. An easy argument (see Chung (1974), Theorem 9.5.2, for details) now improves the last result to

PO(B," - B,"+1 > (1 - E)(,(4,) and B,"+1 > 0 i.o.) = 1,

which proves the desired result.

The results above concern the behavior of B, as t -' 0. By using a trick, we can use this result to get information about the behavior as t - oc. Exercise 6 If B, is a Brownian motion starting at 0, then so is X, = tB(1/t),

t>0.

Proof It is easy to check that EX, = 0, EX,2 = t, and EX,X, = S A t. Since < tn, (X(t1), ... , X(t")) has a multivariate normal disfor each t1 < t2
1. Combining the last result with Blumenthal's 0-1 law leads to a very useful result.

Let ffl;' = a(B5 : s > t). Let .sad =n

.'.

.sd is called the asymptotic a-field, since it concerns the asymptotic behavior of B, as t - oc. As was the case for _FO', one can think of a lot of events that are in .sad, but they are all trivial. (9)

If A e d, then either PX(A) = 0 or PX(A) = 1.

Proof Since the asymptotic a-field for B is the same as the germ a-field for X, we see that P0(A)e{0,1}. To improve this to the conclusion given, we so 1A can be written as lA = 1B o 01. Applying the Markov observe that A e property gives IPA) = EX(lB001) = EXEX(lBO011 -'Vi) = EXEB(1) lB

= J'(2x)2

dy. e-I X-vl2I2P,(B)

Taking x = 0, we see that if P0(A) = 0, then PP(B) = 0 for a.e. y and hence P,,(A) = 0 for every x. Repeating the last argument shows that if P0(A) = 1, then PX(A) = 1 for every x, and proves the desired result.

Remark: The reader should note that the conclusion above is stronger than that in Blumenthal's 0-1 law. In that case, for example, if A = {Bo e C}, then P(A) may depend on x.

1.4 Stopping Times In this section and the next we will develop and explore the most important property of Brownian motion-the strong Markov property. The intuitive idea is simple, but since the rigorous 1 rmulation requires a number of definitions that somewhat obscure the intuitive content, we will begin with a simple example.

Let B, be a one-dimensional Brownian motion, let a > 0, and let T. = inf{t > 0: B, = a} be the first time Brownian motion hits a. If we know that T. = t, then this says something about the past (BS 0 a for s < t) and the

IS

I

Brownian Motion

present (B, = a), but nothing about B for u > To, so it seems clear that B(T. + t), t 0, will have the same distribution as a Brownian motion starting at a and that this process will be independent of T.. The argument in the last paragraph may be convincing, but it is not rigorous. In' Section 1.5 we will prove a result that captures the essence of the reasoning used above and makes it precise. To prepare for this, we need to define and discuss a class of random times S that "say nothing about the behavior of the process after time S." A random variable S taking values in [0, oo] is said to be a stopping time if for all t >- 0, IS < t} If you think of our Brownian motion B, as giving the value of a stock or some other commodity, and think of S as the time at which you sell the quantity in question, then the condition IS < t} E.F,+ has a simple interpretaE+.

tion: The decision to sell at a time < t must be based on the information available at time t, that is, based on n,.0 a(BS : s < t + E). From the definition given above, it should be painfully obvious that we have made a choice between IS < t} and IS < t} and also between E.+ and E S °. The quantity we have defined above is what is officially known (see Dellacherie and Meyer (1978), page 115, (49.2)) as a "wide sense stopping time." This name comes from the fact that if t - 9, is an increasing family of o-fields, then {S < t} e 9, implies that

IS T. for all n, it follows that T = co and hence Tn T T.

(ii) If Tn T t < oo, then by the argument in (i), T > t. On the other hand, as B,. Since B(T,,) E G for all n, it follows that B E n Gn = K TT T t, B(T,,) and T!5; t, completing the proof.

One of the reasons we have written out the last proof in such great detail is to show its dependence on the fact that K is closed. (2) implies that if T = inf{ t > 0 : B, E An} is a stopping time and An T A, then T = inf {t > 0 : B, E A} is also ; but to go the other way, that is, let An A and conclude Tn T T, we need to know that A is closed. The last annoying fact makes it difficult to show that the hitting time of a Borel set is a stopping time, and, in fact, this is not true unless the a-fields are completed in a suitable way. To remedy this difficulty

and to bring ourselves into line with the "usual conditions" (les conditiones habituelles) of the general theory of Markov processes, we will now pause to complete our a-fields. Let X = {A : Pz(A) = 0 for all x e Rd}. A set A E AV is said to be a null set. Since our Brownian motion is a collection of measures P on our probability space (C, '), a set A can be safely ignored only if PX(A) = 0 for all x E Rd.

Let .W, = .F,+ v N' (where here, and if we ever use this notation below, sat v .4 means the smallest a-field containing sad and .4). {.F t >_ 0} is called

the augmented filtration and is the one which we will use throughout the rest of the book. It is trivial to see that the Markov property (formula (3) in Section 1.3) remains valid when . + is replaced by 3 (the conditional expectation is unchanged because we have only added null sets), so it seems reasonable to

use the larger filtration F, it allows more things to be measurable and still retains the Markov property. As we mentioned above, the real reason for wanting to use the completed filtration is that it is needed to make TA = inf{t > 0 : BLEAT measurable for every Borel set. Hunt (1957-8) was the first to prove this. The reader can find

20

1

Brownian Motion

a discussion of this result in Section 10 of Chapter 1 of Blumenthal and Getoor (1968) or in Chapter 3 of Dellacherie and Meyer (1978). We will not prove it here because the results we have given above imply the following: (5)

If A is a countable union of closed sets and T = inf{t > 0 : B,EA}, then T is a stopping time.

and this result is sufficient for our applications.

In Section 1.5 we will state and prove the strong Markov property. If one does not worry about details, it can be described in one sentence: If S is a stopping time, then formula (3) in Section 1.3 holds if we replace the fixed

time s by the stopping time S. To make the resulting formula meaningful, we have to define Os and S. It is clear how we should define the random shift: (6)

(Bs(0)(t) _

co(S((o) + t) on IS < oo} f undefined on {S = oo}. is a little more subtle. We defined F _ t >_ 0), so, by analogy, we should set .Fs = n>o Q(BtA(S+E)

The second quantity nE>o

t > 0). The definition we will give is less transparent than the last one, but easier to work with. (7)

.mss ={A:An {S01. Intuitively, a set A &A, if it depends only upon what happened before time S. To see why (7) says this, let A = {maxo<s<sIBSI > 3}. A obviously depends only on what happened before time S and is in .Fs, since for any t >_ 0

An{S3,S0. The next martingale is not as obvious as the first, but, as we will see in Chapter 2, it is just as important. (2)

B, - t is a martingale. The proof is a simple computation: EX(B, IBS) = EX(B., + 2BS(B5 - B5) + (B, - B5)2I# ) = B. + 2BSEX(B, - B.,13 s) + EX((B, - BS)21.fs)

=Bs +0+(t- s), since B, - Bs is independent of .mss and has mean 0 and variance t - s.

(1) and (2) generalize immediately to B,, the ith component of a d-dimensional Brownian motion. Repeating the proof of (2) shows (3)

If i

j, B,B/ is a martingale.

Proof EX(BtB/ I F)

= BsBs + BBEX(B! - Bs

BSEX(B,` -

BslJs)

+ EX((Br - BS) (B/ - Bj)I.` s)

Since B,` - Bs, B/ - Bs, and . are independent, the last three terms are 0. (1), (2), and (3) are special cases of the following result: (4)

If f E C 2 (i.e., the second-order partial derivatives of f are continuous) and f and Of are bounded, then

f(B,) - f 2 Af(B.,) ds is a martingale. 0

We will prove this result in Chapter 2, but unfortunately we want to use it in the next section and we do not want to give a direct proof now. Stuck in this position, we will engage in the somewhat undesirable approach of using (4), or more precisely the following corollary of (4), before we prove it. (5)

Let G be a bounded open set and i = inf{ t : B, 0 G}. If f E C 2 and Of = 0 in G, then f(B,,,z) is a martingale.

If the reader wants to insist on a strictly logical development, then he should read Chapter 2 and then come back to read the last part of Chapter 1. I think this is unnecessary, though. It is easy to see that the results in Sections 1.7-1.10 are not dependent on the proof of Ito's formula (which will be given in Section 2.8).

One of the nice things about martingales is that they allow us to compute various quantities associated with Brownian motion. We will see a number of

1.7

Hitting Probabilities, Recurrence, and Transience

27

instances of this later, especially in the next section. The exercises below give a number of other applications. Exercise 1

Let B, be a d-dimensional Brownian motion and let T =

I B, I = r). Use the fact that if Ixl < r, EXT = (r2 - Ixlz)/d. inf { t :

I B, I z

- td is a martingale to conclude that

If u is harmonic (i.e., Au = 0) and bounded, then (4) implies u(B,) is a martingale. Combine this observation with the martingale convergence theorem and (9) of Section 1.3 to prove Liouville's theorem : Any bounded harmonic function is constant. Exercise 2

Exercise 3 The Exponential Martingale. Let B, be a one-dimensional Brownian motion.

(i) Use the fact that E0(exp(OB,)) = exp(ezt/2) to prove that exp(0B, - ezt/2) is a martingale.

(ii) Let a>0and T=inf{t:B,=a+bt}.Ifb0and

0 >- 2b, then the martingale in (i) is bounded, so the optional stopping theorem can be applied to conclude ez

1=Eal expIOBT -

2T ;T- 0}. We start with the one-dimensional case. Let a < x < b and T = inf {t : B, 0 (a, b) }. Since B, is a martingale, it is easy to guess the distribution of BT: (1)

PX(BT=a)=b-a PX(BT=b)

a =bx - a.

(This is the only probability distribution with support {a, b} and mean x.)

28

I

Brownlsn Motion

Proof The first step is to show that T < oo, Px a.s. Observe that if y e (a, b),

Py(T> 1) 0,

f"(Y) = 2 dPY) From the last equation we see f'(y) = Cy-d'2 guarantees A(p = 0 for x

0,

so we can let

cp(x)=loglxl d=2 (p(x) = 1x12-d

d > 3.

We are now ready to imitate the proof of (1) in d > 2. Let Sr = inf{t : 1 B,l = r} and r < R. Since qp is bounded and has A(p = 0 in {x: r < lxl < R}, applying the optional stopping theorem at T = S, A SR gives cp(x) = Excp(BT) = cp(r)P(S,. < SR) + cp(R)(1 - P(S, < SR)),

and solving gives (4)

Px(S, < SR) -

g,(R) - gq(x) (G(R) - (P(r)

In d = 2, the last formula says (5)

Px(S, < SR) =

logR - loglxl log R - l og r oo in (5), the right-hand side goes to 1; so Px(S, < 00) = 1

If we fix r and let R

for x, r > 0, and repeating the proof of (3) shows that two-dimensional Brownian motion is recurrent in the sense that if G is any open set, then Ps(B, e G i.o.) = 1. If we fix R, let r -- 0 in (5), and let So = inf{t > 0: B, = 0}, then for x

0

Px(S0 <SR) 3

-'log(jx-yl)

d=2

-I Ix - yj

d= 1.

The reader should note that in each case, G(x,y) = Ccp(I x - yI) where p is the

harmonic function we used in Section 1.7. This is, of course, no accident. x - G(x, 0) is obviously spherically symmetric and, as we will see in Section 8.6, satisfies AG(x, 0) = 0 for x 0, so the results above imply G (x, 0) = A + Bop (I x I ).

The formulas above correspond to A = 0, which is nice and simple. But what about the weird looking B's? What is special about them? The answer is simple: They are chosen to make AG(x, 0) _ -So (a point mass at 0) in the distributional sense. It is easy to see that this happens in d = 1: tp'(x) =

1

x>0 x n} and let Hs = H Since H" E A2, we know how to define H" B. On the other

hand, if m < n, H'-B- H" B = (Hm - H") B = 0 for t < Tm, so we can define H B by setting (H B)s = (H" B), for s < T".

48

2

Stochastic Integration

Remark: We will show in Section 2.11 that this collection of integrands is essentially unimprovable, that is, if we let

T=inf{t: JH52ds=oo then on

{JT

0

Hs ds = o0

lim tfT

oo .

2.2 Integration w.r.t. Discrete Martingales Our second step toward the general definition of the stochastic integral is to discuss integration w.r.t discrete time martingales. The definition of the integral in this case is trivial, but by looking at the developments in the right way, we will jump to an important conclusion-if we want our stochastic integrals to be martingales, then the integrands should be "predictable" (a notion we will describe in this section) rather than merely optional. Let n > 0, be a martingale w.r.t.. . If H,,, n > 1, is any process, we can define n

F,

H.(Xm-Xm J

M=1

By analogy with results in Section 2.1, you might expect that if H e .F and each H. is bounded, then (H X). is a martingale. A simple example shows that this is false. Let S. be the symmetric simple random walk, that is, S. _ 1 + +

-1) = 1/2. Let where the ; are independent and have P(c , = 1) = .F. = but if we let H = S. is a martingale (H- S),, = Y bm(`Sm -'Sm-1) _ Y m = n, m=1

m=1

which is clearly not a martingale. If we think of cn as the net amount of money we would win for each dollar

bet at time n and let H,, be the amount of money we bet at time n, then the "problem" with the last example becomes clear: We should require that H e 5,,_1 for n > 1 (and let Fo = {0, 0}), that is, our decision on how much to bet at time n must be based on the previous outcomes 1, ... ,i_1 and not on the outcome we are betting on! A process H that has H effl _1 for all n 1 is said to be predictable since its value at time n can be predicted (with certainty) at time n - 1. The next

-

result shows that this is the right class of integrands for discrete time martingales.

2.2

(I)

49

Integration w.r.t. INecrete Martingales

Let X" be a martingale. If H is predictable and each H" is bounded, then (H X)" is a martingale.

Proof The boundedness of the H. implies El (H X)" I < oo for each n. With this established, we can compute conditional expectations to conclude E((H. X)"+1

(H' X)" + E(H"+I (X"+, - X")I (HX)" + H"+1E(X"+1 - X.13") = (H' X)",

since Hn+1 c- .f" and E(Xn+1 -

0.

The definition and proof given above relied very heavily on the fact that the time set was {0, 1, 2, . . . }. The distinction between predictable and optional becomes very subtle in continuous time, so we will begin by considering a simple example : Let (S2, P) be a probability space on which there is defined

a random variable T with P(T < t) = t for 0 < t < 1 and an independent random variable with P( = 1) = P( = - 1) = 1/2. Let X. _

0 tT

and let . = o (XS : s < t). X is a martingale with respect to .°t-;, but f o X S dXs = Z = 1, so Y, = J0 Xs dX5 is not, and hence if we want our stochastic integrals which are martingales we must impose some condition that rules out X. The problem with the last example is the same as in the first case, and again

there is a gambling interpretation that illustrates what is wrong. Consider the game of roulette. After the wheel is spun and the ball is rolled, people can bet at any time before (-). One way of requiring that our bet be made strictly before T is to require that the amount

of money we have bet at time t is left continuous, that is, we cannot react instantaneously to take advantage of a jump in the process we are betting on. Weakening the last requirement we can, by analogy with the optional a-field, state the following definition.

Let H be the a-field of subsets of [0, oo) x fl that is generated by the left continuous adapted processes. A process H is said to be predictable if H(s,co)ell. The definition of H, like the definition of A, makes it easy to verify that something is predictable, but it does not tell us what sets in H look like. This time, however, it is easy to describe the a-field precisely: (2)

II = a((a, b] x A : A e Fo).

Proof Clearly, the right-hand side c H. Let H(s, co) be adapted and left continuous and let H"(s, co) = H(m2-", (o) for m2-" < s < (m + 1)2-". Since H is adapted, H"(s, co) E a((a, b] x A : A c.ya), and since H is left continuous, H"(s, w) - H(s, co) as n - oo.

50

2

Stochutlc Integration

Remark: Based on the result above, you might guess that A = a([a, b) x A : A e.F.), but you would be wrong. I would like to thank Bruce Atkinson for catching this mistake in an earlier version of this chapter.

then H is the Exercise 1 Show that if H(s, co) = l(a,b](s)WA(W) where A limit of a sequence of optional processes; therefore, H is optional and II c A.

2.3 The Basic Ingredients for Our Stochastic Integral To define a stochastic integral we need four ingredients:

a probability space (fl, F, P)

a filtration F = {. t > 01 a process X, t > 0, that is adapted to F a class of integrands H, t >_ 0. In Section 2.2 we described the class of integrands that we will consider: the predictable processes. In this section, we will describe the assumptions that we will make concerning the process X and the filtration IF. This will require a number of definitions and explanations. (1)

X is said to be a local martingale (w.r.t. F) if there are stopping times T. T 00 t >_ 0}). The stopping times T. are so that X,,,Ta is a martingale (w.r.t. said to reduce X. In the same way, we can define local submartingale, locally bounded, locally of bounded variation, and so on. Remark :

You should think of a local martingale as something that would be a martingale if it had EIXI < oc. There are several reasons for working with local martingales rather than with martingales :

(i) It frees us from worrying about integrability. For example, if X, is a martingale and 9 is a convex function, then tp(X,) is always a local submartingale, but we can conclude that p(X,) is a submartingale only if we know EI (p(X,)I < oo, a fact that may be either difficult to check or false in some applications. (ii) Often we will deal with processes defined on a random time interval [0, T). If T < oo, then the concept of martingale is meaningless, but it is trivial to define a local martingale : If there are stopping times T T T so that.... (iii) Since most of our theorems will be proved by introducing stopping times T. to reduce the problem to a question about nice martingales, the proofs are no harder for local martingales defined on a random time interval than for martingales. Reason (iii) is more than just a feeling. There is a construction that makes

2.3

The Bask Ingredients for Our Stochastic Integral

51

it almost a theorem. Let X be a local martingale defined on [0, r) and let T T r be a sequence of stopping times that reduces X. Let t T1

0_ 1

A,,=

E(Y,, I

Y,.-1

From the definition, it is immediate that A,, c a submartingale), and

E(Y

-

E(Y,,I.Fn-1)

is increasing (since Y is

- A.

= Y,.-1 - A,,-1,

so A has the desired properties. To see that A is unique, observe that if B is another process with the desired property, then A,, - B,, is a martingale and Therefore An and it follows by induction that A,, - B = A 0 - Bo = 0. The key to the uniqueness may be summarized as "any predictable discrete time martingale is constant." The last statement fails miserably if predictable is replaced by optional, and with a little work (exercise) one can construct a submartingale Y for which there are many optional increasing processes with AO = 0 that make Y - A. a martingale (e.g., Y,, = b1 + + where the ; are independent and have Ec1 > 0). The last paragraph explains why predictability is needed in discrete time. The same arguments apply in continuous time, that is, the requirement that A, be predictable and increasing is needed to rule out the possibility of producing another process Ai by adding a predictable martingale (e.g., Brownian motion) to A,. The only change is that the uniqueness statement must be formulated more carefully. (3)

Any local martingale that is the difference of two predictable increasing processes is constant.

Remark: This is implicit in the proof of the Doob-Meyer decomposition, but to make it clear that (2) is the only result we are taking for granted, we will show that (2) (3).

54

2

stochastic interation

Proof If X, = A, - A; is a local martingale and A A; are predictable and increasing, then Y, = A, is a local submartingale that has two Doob-Meyer

decompositions, since Y, - (A, - Ao) = Ao is a local martingale and Y, (A; - Ao) = X, - Ao is also. Since the Doob-Meyer decomposition is unique, it follows that A, - Ao = A; - A;, that is, A, - A' = A 0 - Ao. With a little care we can improve (3) to (4)

Any (continuous) local martingale that is predictable and locally of bounded variation is constant. Remark: The result is true in general, but for simplicity we prove it here only for continuous processes.

Proof In light of (3), all we have to do is show that a predictable process that is locally of bounded variation can be written as a difference of two predictable

increasing processes. To do this we open up any real analysis book (e.g., Royden (1968)) and observe that if X, is optional, continuous, and locally of

bounded variation, then the decomposition given there for a function of bounded variation expresses X, as a difference of two optional continuous increasing processes, proving (4).

The variance process is important for defining and using the stochastic integral. Since we will spend a lot of time considering this process and discussing

its properties below, we will drop the subject for the moment and turn to the definition of the covariance of two local martingales. (5)

If X and Y are two local martingales, we let

<X, Y>, = a(<X + Y>, - <X - Y>,). If X and Y are random variables with mean zero,

cov(X, Y) = EXY = 4(E(X + Y)2 - E(X - Y)2) = 4(var(X + Y) - var(X - Y)), so it is natural, I think, to call <X, Y), the covariance of X and Y. The following result is useful for computing <X, Y>,. (6)

<X, Y>, is the unique predictable process A, that is locally of bounded variation, has A o = 0, and makes X, Y, - A, a local martingale.

Proof From the definition, it is easy to see that

X, Y,-<X, >,=a[(X,+Y,)2-<X+Y>,-(X,-Y,)2+<X-Y>,] is a local martingale. To prove the converse, observe that if A, and A; are two processes with the desired property, then A, - A; = (X, Y, - A') - (X, Y, - A,)

is a predictable local martingale that is locally of bounded variation and hence ° 0. For some of the developments below, the following result will be important.

2.5

Integretlon w.r.t. Continuous Local Martlngala

Exercise 1

55

If S :!g Tare stopping times and <X>s = <X>,., then X is constant

on [S, T]. Sketch of Proof It suffices to prove the result when X and (T - S)1(S5. J0

Proof Under the assumptions above, E((H X),(K Y),) = E

,

so it suffices to prove the first result. Let H = l(a,b]C and K = 1(c 41D. Since the first formula is linear in H and in K, we can assume without loss that (a, b] and (c, d] are either disjoint or equal.

Case 1. b < c. In this case, f o HSKS d <X, Y% = 0, so we need to show that

, = 0, in other words, (H X), (K Y), is a martingale. To prove this we observe that

(H - X),(K X), =

0

0 < oo a.s. for all t >- 0} is also easy. Let T. = inf{t : f o H.,2 d<X> > n}, let Hs = Hs 1(s n}, it suffices to show that if X is a bounded martingale and H e 112 (X), then (H X), is a martingale. Let H" be a sequence of elements in b111 that converges to H in 112 (X). It follows from (3) above that (H"- X), is a martingale. In other words, ifs < t, E((H"' X)11. ) = (H" - X)S.

To complete the proof, it suffices to show that this equality persists when n -* oo. The first step is to show that (H" X)5 - (H X)5 in L2. To do this, we observe that Doob's inequality and the isometry property of H -. H X imply E(sup I (H" - X), - (H' X), 12) -< 4IIH" X - H XII2 t

To deal with the conditional expectation, we observe that from Jensen's inequality X),)2

-+ 0. E[E((H" . X ) ,1 .) - E((H . X), I. )] 2 < E((H" . X), - (H. ) - E((H X ), 1,F,) in L2, and it follows that Therefore E((H" X ) !

E((H- X)tlfs) = (H' X)s For several applications that follow (e.g., in Chapters 8 and 9), it is important to make a trivial extension of the integral we have defined above. S is said to be a continuous semimartingale if St can be written as X, + A,

2.6

59

The Kunits-Watanabe Ioequallty

where X, is a continuous local martingale and A, is a continuous adapted process that is locally of bounded variation. A nice feature of continuous semimartingales that is almost unheard of for their more general counterparts is (10)

Let S, be a semimartingale. If X, and A, are chosen so that AO = 0, then the decomposition S, = X, + A, is unique. Proof If X, + A, is another decomposition, then A, -A; is a continuous local martingale and locally b.v., so A, - A; is constant and hence - 0. Given a unique decomposition, it is easy to extend our definition of the

stochastic integral to continuous semimartingales. If H e (bII = the set of locally bounded predictable processes, we can define (H' A), (o)) =

t H.,(o)) dA,(w)

J0

as a Lebesgue-Stieltjes integral (which exists for a.e. co), and since t bH

II3(X),

we can define (H X), and (since by the uniqueness of the decomposition this is an unambiguous definition. From the definitions above, it follows immediately that we have (11)

If X is a semimartingale and He (bII, then (H X), is a semimartingale.

2.6 The Kunita-Watanabe Inequality In this section we will prove a Cauchy-Schwarz inequality due to Kunita and Watanabe and apply this result to extend the formula given in Section 2.5 for the covariance of two stochastic integrals. (1)

If X and Y are local martingales and H and K are two measurable processes, then almost surely 'o

Jo

IH5K5I Id<X,Y>51

l1/2

f o

HS2d<X>S/

'0

)1/2

(fo

where ld<X, Y>51 stands for dV5 where Vs is the variation of r - <X, Y>, on [0,s].

Remark: This result is from Meyer (1976). He attributes the proof to P. Priouret. Notice that H and K are not assumed to be predictable. We assume only that H(s, co) and K(s, (o) are measurable with respect to 9 x .F where . 9 is the Borel subsets of R. The reason we can attain this level of generality is that the notion of martingale does not enter into the proof after the first line.

60

2


Proof

Observe that if s 5 t, <X+ AY, X + AY>, >- <X+ AY, X + A,Y>S. If we let <M, N>' = <M, N>t - <M, N>S, then Step ] :

0:!5; <X+AY,X+A,Y>t- <X+A,Y,X+AY>S _ <X, X>.,+ 22 <X, y>S + )2 < Y, y>s

for all s, t, and A. Now a quadratic ax2 + bx + c that is nonnegative at all the rationals and not identically 0 has at most one real root (i.e., b2 - 4ac < 0), so we have that (<X,

y>.,1)2 < <X, X>5< y' y>1

Step 2: Let 0 = to < t 1 < < t be an increasing sequence of times, let hi, ki, 1 < i:!5; n, be random variables, and define simple measurable processes n

H(s, (o) = Y hi (o)) I

(S)

i=1 n

K(s,w) _ Yki(w)Iui 1'k>(s) i=1

From the definition of the integral, the result of Step 1, and the CauchySchwarz inequality, it follows that n

f HKKd<X,

Y>S

Jo

< Y- Ihikil I <X, Y>ti-1 i=1

i=1

Ihil1<x'X>ri-1S

I
s)1/2 (fo.

0

Step 3: Let M be a large number and let T = inf{t : <X>t or t > M}. By the monotone convergence theorem, it suffices to prove (1) when H = K = 0 for s >- T and I HS I, l KS l < M for s :!g T. Having restricted our attention to [0, T], <X> and are finite measures; so using the bounded convergence theorem, we see that (2) holds for bounded measurable processes. To improve (2) to (1) (and complete the proof), let JS be a measurable process taking values in { -1,1 } such that

61

The Kunlts-W.tenube Inequ.Ilty

2.6

f ld<X, Y>sl = J JSd<X, Y>, 0

0

and apply (2) to H, = IHSI and Ks = JIK,I With (1) established, we are now ready to generalize (4) of Section 2.5 from

HebH, to HelI3(X). (3)

If X and Y are local martingales and H e II3 (X), Ke II3 (Y), then t = J HS K, d <X, Y>s 0

Proof What we need to show is that

(*) (H' X)t(K- Y)t - J H., Ksd<X, Y>s 0

is a local martingale. By stopping at T. = inf{t : IX,1, Y1, f o IH,I2 d<X>., or f 'O 1K.,12 ds > n}, it suffices to show that if X and Yare bounded, He II2(X), K e II2 (Y), and Zt is the quantity defined in (*), then Zt is a martingale.

Let H" and K" be sequences of elements of bII, that converge to H and K in r12 (X) and 112 (Y), respectively, and let Zt" be the quantity that results when H" and K" replace H and K in (*). By results in Section 2.5, Zt" is a martingale, that is, if s < t, E(ZZ" I.F,) = Z,". To complete the proof it suffices to show that this equality persists when n --p oo. The first step is to show Zs --+ Zs in L'. The triangle inequality implies that

E supI(H"' X),(K Y)t - (H. X)r(K- Y)rl t

<E supI((H" - H)' X),(K"' t

Y)tl

+ El sup l (H - X ), ((K" - K)- Y), j

To estimate the first term, we observe that from the inequalities of CauchySchwarz and Doob, it is

< {E(supI((H" - H)' X)tI2)E(supI(K"'

Y)t12)11/2

t

t

< 411(H" - H) X112 IIK

Y112 = 411H" - HIIxIIK"IIY , 0

as n -+ oo, since II H" - HII x -- 0 and IIK"IIY -' IIKIly < oo. A similar estimate shows El suptl(H X)t((K" - K) Y)t11 -> 0, so we have shown (H"- X)s(K" . X)s

- (H X),(K. Y)5 in L1. Repeating the last argument and using the Kunita-Watanabe inequality instead of Cauchy-Schwarz, we see that

JtHSK. d<X,Y>., - (HHK,d<X,Y>S o

0

f'JHs -H31IKslld<X,Y>SI+ f IH5IIKs -KslId<X,Y>,I, 0

0

62

2

Stochosde Integretlon

and the first term 0 < (JtIx"-H312d<X>sll/Z(JtI

"I2dsJl/2

Therefore J H"Ks d <X, Y>, -' 0

Hs Ks d <X, Y>s fo

in L1, and it follows that Z, - Z, in L1. To deal with E(Z" we observe that from Jensen's inequality EIE(Z,"l9S) - E(Z1 S )1 5. 0

Remark: If A, is a continuous process that is locally of bounded variation, then

f(A,) - f(A0) = JJ'(AS)dAS. 0

As the reader will see in the proof, the second term comes from the fact that local martingale paths have quadratic variation <X>,, and the 1/2 in front of it comes- from expanding f in a Taylor series.

Proof By stopping at TM = inf{t : IX I or <X> >- M}, it suffices to prove the result when IXj and <X> < M. From calculus we know that if a < b, there is a c(a, b) a [a, b] such that (2)

f(b) -f(a) = (b - a)f'(a) + Z(b - a)2f"(c(a,b)). Let t be a fixed positive number. For each S > 0, define a random partition of [0, t] by

or

<X> -<X>.>S}.

(Note that this partition depends on S even though we have not recorded the dependence in the notation.) From (2) it follows that

(3)

65

Chunee of Vorlobin, It6'u Formul

2.8

f(X) -f(X0) = Y-f(X1i+1) -f(Xn) i

= Y_f'(Xri)(Xri+1 - Xii) + i

where gi(w)

I

gi(w)(X,i+1 -

X,i)2

=f"(c(X.,X=i+1)).

Comparing (3) with (1), it becomes clear that we want to show

f'(X'i) (X'i+1 - Xii) -

(4a)

(4b)

f'(Xs) dXS Jo

1 2E9i(w)(Xti+1

f"(Xjd<X>s.

- Xi)2 -' o

The proof of (4a) is easy. If we let H." =f (X,,) when se [ti, ti+1) and = 0 otherwise, then (H".

(Xi)(X,i+1 - Xti) =

X)t

Let H, = f'(XX) when s < t and = 0 otherwise. Since fis uniformly continuous on [-M, M], we see that as n --> oo, supIHs - HI -> 0. Since <X>. :!5; M, it follows that H" - H in 112 (X) and hence H" X - H X in .,412. To prove (4b), we start by proving the result when f" = 1, that is, (5)

As S

0,

Y_ (X1i+1

- X i)2 - <X )t

in probability.

-

Proof Let Di = (X,i+1 X.)2 - (<X>,i+1 - <X)t.) E(A1I. ) = 0, and if i < j, then Die . , so

and

observe

E(A1A) = EE(AiOtl. )= E(AiE(AAI. ,)) = 0. and it follows that E(Y Ai)2 = Y_ EDiA, = E EO?. i

iJ

i

Using the trivial inequality (x - y)2 < (2x)2 + (2y)2 now gives

A < 4(X,i+1 - Xi)4 + 4(<X )[i+1

- <X >,i)2.

To estimate the right-hand side, we observe that by the definition of the ti,

Y(<X )ri+1

- (X >,i)2 < b j (<X )ri+1 - <X >,i) = 6<X>, (Xii+1 - Xi)4 (Xi+l - Xi)2' i i

62

that

66

2

Stochiude Inteandon

and since martingale increments are orthogonal, z

E ((x+i-X,)Z Combining the estimates above shows

EA < 4(SM + 52M2), and it follows that Y-(Xt,+i - X f"(X At) uniformly in s, so to complete the proof all we have to do is justify taking the limit inside the integral (6)

f

dAs

fof(Xs)d<X>.,o

-+

o

o

To do this, we observe that by taking subsequences we can suppose that with probability 1, A",,, converges weakly to <X>SA,, in other words, if we fix

w and regards

A,,, and s - <X>s,, as distribution functions, then the

associated measures converge weakly. Having done this, we can fix w and deduce (6) from the following simple result: (7)

If (i) measures µ on [0, t] converge weakly to µm, a finite measure, and (ii) g is a sequence of functions with Ig,, I < M that have the property that whenever s a [0, t] - s we have g(s) then as n - oc J gn dµn

f g dµ..

Proof By letting µ;,(A) = t]), we can assume that all the P. are probability measures. A standard construction (see Exercise 1 below) shows that there is a sequence of random variables X,, such that X has distribution µ g(XW), and as n - oo, X,, - X,. a.s. The convergence of g to g implies so the result follows from the bounded convergence theorem. (7) is the last piece in the proof of (1). Tracing back through the proof, we

2.9

67

N xtennlon to I unctloer of Several tiemlmsrtlnpeles

see that (7) implies (6), which in turn completes the proof of (4b), so adding (4a) and using (3) gives that for each t1,

Jf'(X)dXs +'

f(X,) -f(X0) = 0

Jt

-(XS)d<X>s

a.s.

0

Since each side of the formula is a continuous function oft, it follows that with probability 1 the equality holds for all t >- 0, the statement made in (1). Exercise 1

to a limit j

Let y. be a sequence of probability measures that converges weakly

and let

oc,x]), 1 < n < oo, be the corresponding

sequence of distribution functions. Let U be a random variable with P(0 < U < u) = u for all u e (0, 1), and let X = F,-1(U) where F1(y) _ sup{x:

y}. Then X,, has distribution

and X, --+ X. a.s.

Remark: Although (1) is by far the most important formula in this section, (5) is also useful because it says that local martingale paths have quadratic variation <X>,. This result has the following useful consequences: Exercise 2 If S < T are stopping times and X is constant on [S, T], then <X>S = <X>TExercise 3

If X is a continuous martingale with bounded variation, then

<X> - 0 and hence X is constant. The reader should note that the proof of the last result is somewhat circular, since the definition of <X> relies on the uniqueness of the Doob-Meyer decomposition, which already implies the result above (see (4) in Section 2.4).

2.9 Extension to Functions of Several Semimartingales In this section, we use the version of Ito's formula that was proved in the last section to prove a much more powerful form of the result. The extension is based on the following simple application of the formula derived in the last section so it is almost an independent proof. If we letf(x) = x2 in (1) in the last section, then we get (1)

X,Z - Xo =

f 2X5dXs+<X>, 0

so ifXo=0,

J2XdX. = XX2 - <X >, in contrast to So 2s ds = t 2. From (1), we get the following useful result:

68

2

(2)

Integration by Parts.


X, Y, - XoYo=

f

r YsdX., +J'XSdYc+<X,Y>1.

0

0

Proof Applying (1) to X, + Y, and X, - Y gives (X, + Y,)2 - (Xo + Yo)2 =

f

2Xs d(X + Y)., + f t 2 Y d(X + Y)., + <X + Y>,

0

(X, - Y)2 - (Xo - Yo)2 =

f

0

2X, d(X - Y)S -

f 2Y d(X - Y)5 + <X - Y>,. 0

0

The linearity of (H X) in X allows us to break the stochastic integrals into two pieces, using the distributive law: (3)

H- (X+Y)=(H Multiplying the two equations above by 1/4 and subtracting gives (2). If AS is of bounded variation and Y is continuous, then it follows from the theory of Riemann-Stieltjes integration that A,Y-AO YO = f YdA5+ J, A,dY. 0

o

Adding this to (2), we see that if R, = X, + A, is a continuous semimartingale and Y is a continuous local martingale, then R, Y, - RO YO = f , Y dRs +

f Rs dYs + <X, Y>,.

Repeating the last argument and interchanging the roles of X and Y, we see that if S, = Y + A' is another continuous semimartingale, then (4)

R, S, - Ro So = f , Ss dRs + fo, Rs dSs + <X, Y>,. Jo

so (2) holds for semimartingales if we define the covariance of two semimartingales to be the covariance of their "martingale parts." (This is unambiguous, since we have a unique decomposition.) The last observation and induction allow us to prove the following generalization of Ito's formula: (5)

... , X,d are continuous semimartingales and f : R° - R has continuous second-order partial derivatives, then If X,1,

.f(Xc) -.f(Xo) _

fo Dif(XS) d4s + 1

Y JDf(A;)d<XiXJ>s. 2i, o

Remark: At the end of the proof, we will state a version of this result that requires less differentiability on f, and we will prove the more general result by saying "the same proof works." To prepare for this, the reader should observe that f appears only in the first paragraph of the proof.

2.9

69

Extenelon to FunctlonN of Several Scmlmrrtingxlefi

Proof By stopping, it suffices to prove the result when I X,' l < M for all i, t. Since any continuous function on [ - M, M]" can be approximated by polynomials g^ in such a way that gn, Dig", and Dijg^ converge to f, Di f, and Dij f uniformly, it suffices to prove the result when f is a polynomial and, by linearity, xk^ where k1, k2, .... k" E {1, . . . , d}. (The when f is a monomial xklxk2

reader should note that k1..... k" are superscripts, not powers, e.g., our monomial might be x1x4xlx1x2.) If n = 1 and k1 = k, then (5) says that

Xk -Xo= f

1dXs,

which is trivially true. To prove the result for a general monomial, we use induction. Let Y = IIm=, Xk (m) be a monomial for which (5) holds, and let Zt = Xk("+1) Applying (4) gives

YZ,= JZ5dY.+ f Y SdZZ+,. 0

0

Applying (5) to Y gives Yt

-

n X.,k(m)

t

iwe , observe that by the formula for the covariance of stochastic integrals, t

,J

jt

[ =

XS(m)

. 1 d<Xk(i)

Xk(n+1)>s

0m

i= J4Srdr, 0

74

2


so the infinitesimal drift is d (independent of position) and the infinitesimal variance is four times the value of S (which is 1x12 if the Brownian motion is at x). Exercise 5 Let d _> 2 and let

loglxl d=2 (P(x) _ IIX12` d > 3. Use Ito's formula to show that q,(R,) is a local martingale. Exercise 6 Suppose

X, - J

b(XX) ds is a local martingale

0

and

<X >, =

f r a(XS) ds

where a > 0 is continuous and b is bounded on compact intervals. Use Ito's formula to show 9(X,) is a local martingale if and only if 1]f (p "(x)a(x)

+ (p'(x)b(x) = 0,

so if we normalize cp to have q(0) = 0 and 9'(0) = 1, then

9W =

f z exp

(-

dy.

Jo a(z) dZ/

o

In Chapter 9 we will show how to construct processes X that satisfy the two conditions above. We will call such an X "a diffusion process with infinitesimal drift b(x) and infinitesimal variance a(x)," and we will call cP "the natural scale for this process." One of the reasons for interest in the natural scale is that it allows us to generalize the results in Section 1.7. Exercise 7 Let X be the process described in Exercise 6 and let T. = inf{t : X, = a}. Then

ps(T. n}), integrate over A, and let n - oo, we get B2

Jr

j(t) - P(A) = 0 - 2

j(s) ds, wherej(s) = E(ei°Xs ; A). 0

Since we know a priori that I j(s)I < 1, it follows that j is continuous, j has a continuous derivative ... Differentiating the equation gives 02

j'(t) =

(t),

which, together with the initial condition j(0) = P(A), shows that j(t) _ P(A)e-0242. Since j(t) = E(eiOXt ; A), this shows that E(e`°XtI.4--o) = e-02y2, that is, Xr is independent of FQ and has a normal distribution with mean 0 and variance t. An immediate consequence of (1) is: (2)

Every continuous local martingale with <X> = oo is a time change of Brownian motion.

76

2


Let y(u) = inf{t : <X>, > u} and let Y. = X(U). Since y(u), u >- 0, is an increasing family of stopping times, the optional stopping theorem implies that (3)

YU, .FY(u), u > 0, is a local martingale.

Proof By stopping at T = inf{t : IX,I > n}, it suffices to show that if X is a bounded martingale, then Y is a martingale. To do this we observe that if ul < u2, then y(ul) < y(u2) are stopping times, so we have E(XV(U2)I'1Y(u,))

= XY(u,) = Y(ui),

proving (3). Repeating the argument above shows (4)

YY - u, JFY(u), u >- 0, is a local martingale.

Exercise 1 of Section 2.4 implies that u -+ Y is continuous, so combining (3) and (4) with Levy's characterization shows that Y is a Brownian motion.

The result above can be easily extended to local martingales with P(<X> < oo) > 0 (and we will see below that this is an important extension). Let T = <X>.. The ,proofs of (3) and (4) show that Y and YY - u are local martingales on [0, T), and repeating the proof of (1) shows that YU, U < T, is a Brownian motion run for a random amount of time. Since Brownian paths are continuous, it follows immediately from the last observation that (5)

lim X exists on {<X>, < oo}. tTOD

In Chapter 1 we showed that Brownian motion has (6)

lirn sup B, = oo, lim inf B,

oo.

Using this observation, we can sharpen (5) to conclude that (7)

The following sets are equal almost surely: C = {lim X exists} t- 00

B= {sup JXJ 3,f(x) = Ax where A is an orthogonal matrix. The first step in deriving these results is to prove the following generalization of Levy's theorem ((1) in the last section) (1)

If X,', ... , X" are continuous local martingales with Xo = 0 and <X`, X'>, _ b;,t, then X, = (X,', . . . , X°) is an n-dimensional Brownian motion. Proof If Y = Y, a(X,` where Y ; a? = 1, then it follows from the proof of the one-dimensional result that ifs < t, then

rs)I. ) = e-(r-s)e2/z or, in terms of the X`, E(epEj(eaj)(xi-xs)) . ) = e-(r-s)e2/2

Since any v c R" can be written as 9a where Y_l a; = 1 and 0 e [0, oo), this shows that X, has independent increments and X, - XS has a d-dimensional normal distribution with mean 0 and covariance (t - s)I, proving (1). From (1), it follows immediately that a vector (X,', . . . , X°) of continuous local martingales is a time change of Brownian motion if and only if there is an increasing process A, such that <X`, Xj>, = b;,A,. The most important instance of martingales of this type occurs when we compose an analytic function with a complex Brownian motion. To explain this we need some notation. Let C, be a complex Brownian motion, that is, C, = B' + iB, where B,', B, are independent Brownian motions. Let f be analytic in C; in other words, if we write z = x + iy and f = u + iv, then f satisfies the Cauchy-Riemann equations au

av

au

-all

OX

ay

ay

ax

Differentiating again gives a2u axe

_

a2y

a2u

axay

aye

azv ayax'

and since v has partial derivatives of all orders,

2.12

Conformal Invariance In d

Au=-+-

a2u

32u

ax2

2, Kelvin's Transformations

a2v

aZU

axay

ay2

ayax

79

=o

that is, u is harmonic. A similar argument shows that v is harmonic, so it follows from results in Section 2.10 that U, = u(C,) and V, = v(C,) are martingales, and J rAt

r =

(2)

0 rhi

Vv(Cs) 2 ds

< 0

where Vu =

au au lay av ay) and Vv = (ax' ax' ay

From the- Cauchy-Riemann equations, it follows that Vu(z) 12 = Vv(z) 12 for all z, so , = - 0} be the filtration generated by a one-dimensional Brownian

motion B. In this section we will show (1) all local martingales adapted to B are continuous and (2) every martingale XE.,#2(B) can be written as X0 +

86

2

stochastic Inteustion

f o Hs dBg where He 1Z2(B). The generalizations of these results to d dimensions will be obvious once the reader has seen the proofs below. (1)

All local martingales adapted to B are continuous.

Proof Let X, be a local martingale adapted to B and let T. be a sequence of is stopping times that reduces X. It suffices to show that for each n, continuous, or in other words, it is enough to show that the result holds for martingales of the form Y, = E(YI -4,) where Ye .,,. We build up to this level of generality in three steps.

f is a bounded continuous function. If t >- n, Let Y = so t - Y, is trivially continuous for t > n. If t < n, the Markov property of Brownian motion implies that Step 1:

then Y =

E(YI -4,) = h(n - t, B,) where

h(s x) =

f

1

2res

e-(y-x)2asf(y) dy

It is easy to see that h(s,x) is a continuous function on (0, oo) x R, so Y is continuous for t < n. To check continuity at t = n, observe that changing variables y = x + (z/,) gives h (s, x) = Jy=l=_e:z2/2f(x

?) dz,

+ ssf

so the dominated convergence theorem implies that as t T n, h(n - t, B,) f(B,,) Step 2: Let Y = f1(B,1) f2 (B,2) where t 1 < t2 < n and f1 f2 are bounded and continuous. If t >- t1, then Y =.fi(B,,)E(f2(Br)I °. ),

so the argument from Step 1 implies that Y is continuous on [t1, oo). On the other hand, if t < t1, then Y = E(Y1I -4,) and Y, =.f1(B,,)E(.f2(B,2)I-4r1) f, (B1,)g(Br1)

where

9(x) =

1

e cy X11 V2 `1y 2(y)dy

2n(t2 - t1)

is a continuous function, so

Y,= E(fi(B,)g(B,)j.

)

for

t 0, there is a random variable X` of the form considered in Step 2 that has EI X` - Y' < e. Now

IE(XÌ_V,)-E(YIM,)I <E(IX`- YIIM,), and if we let Z, = E(I X` - YI IM,), it follows from Doob's inequality that

2P(supZ,>2)<EZn=EIX`- YI<e. t _- 0, that is, X, Y, is a martingale.

The first definition corresponds to E<X, Y> = 0, so it is much less restrictive than the second. In some circumstances, however, if Y is orthogonal to all the martingales in a subspace .K, then it is also uncorrelated with them. A subspace .K of . #2 is said to be stable if it is

(a) closed in the #2 topology (b) closed under stopping : if X e .N' and T is a stopping time, XT = XrA, E .N'. (3)

Let N' be a stable subspace. If X e .N' and He 112(X ), then H X e .N'.

Proof The key to the proof is the following: (4)

Let -4/' be a stable subspace and .N'' = { Y : (X, Y) = 0 for all X E X) be the orthogonal complement of A. If X e .N' and Ye A', then X, Y, is a martingale.

88

2


Proof Let X e ./V'. If T is a stopping time, then X T e .N', so if Ye .N'', then 0 = (XT, Y) = E(XTYom) = E(XTYT),

(4)

since martingale increments are orthogonal. (31 now follows from (5)

If for all bounded stopping times T we have EZT = EZo, then Z, is a martingale.

Proof Lets < t, A E ., and let T = son A and T = ton A`. By hypothesis, EZT = EZo = EZ, so subtracting EZ, lAc from both sides, we find EZS 'A = EZ, IA. Since this holds for all A e Fs, then ZS = E(Z, I, ), and Z, is a martingale.

With (4) established, it is easy to prove (3): If Ye.K', then <X, Y> = 0, so = f H., d <X, Y>s = 0 0

and E((H X)c Y.) = Ec = 0. This shows that H X is in the orthogonal complement of .N', that is, H X e.41". We have proved (3). Since {H X : He 112(X)} is a stable subspace, it follows from (3) that we have (6)

The smallest stable subspace containing X is {H X : He 112(X)}. Having established some properties of stable subspaces, we are now ready to prove the second result we stated at the beginning of this section.

(2)

If Xe .4'2(B) and X0 = 0, then there is an He 112(B) such that

X,= f

H, dB.,. 0

Proof Let ,' = {H B : He n2(B) }.. ( is a stable subspace of A'2. Let Y be a martingale in the orthogonal complement .K'. We will show that Y =_ 0. The key to doing this is the following formula: E(eiO t-B,)

)o2/2.

Ye-et-S

(7)

As we observed in Section 2.11 when we proved a similar formula, it suffices to prove the result when s = 0. Ito's formula shows that eteat

= 1 + iB

t eboas

dBs - e2 f t e ieas ds.. 2 Jo

o

Let A e.Fo and multiply both sides by 'A Y. If we take expectations and let j(t) = E(e'eBt lA Y), then we get

j(t) = E(1A Y) + ME (Y f r 1Ae'eas) \

o

/

- 22

f,

ds.

Jo

Since Y, is a martingale and A e moo, E(lAY) = E(lA Yo). To deal with the second term on the right, we observe that Z, = j O 1Ae`BBS dBs e .K, so it follows from (3)

that = 0 and hence that E(YZZ) = 0. Combining the last two observa-

Notes on Chapter 2

89

tions with the formula above shows that e2

j(t) = E(1AYo) -

2

Jt

j(s)ds. 0

As we argued in Section 2.11, this result implies that

Pt) =

82

j(t) j(0) = E(1AYo),

so we have j(t) = e-e2,12E(IAY0), proving (7).

The next step in the proof is to extend (7) to a statement about an arbitrary finite set of increments of Bt. Let 0 = to < t1
1. The simple random walk is the discrete analogue of Brownian motionit has independent increments and paths that are as continuous as they can be. 0) Let N = inf{n >- 0: S. = 0}. The random walk analogue of is SN = 0), but there is a big difference. If the starting point So = xeH, then PX(SN = 0) > 0, so the process can be defined by elementary conditional probabilities. In order to understand how (SnANI SN = 0) behaves, we start by considering what happens at the first step. If So = x e H, then by the definition of conditional probability and the Markov property, (1)

PX(S, = YJSN = 0) =

PX(S1 = Y,SN = 0) PX(SN = O)

- PX(S1

= Y)Py(SN = 0) PX(SN = O)

It is easier to see what is going on if we write the right-hand side in more that is, abstract notation. Let p(x, y) be the transition probability for

11/2dif xeH, Ix - yj = 1

p(x,y)=

1

ifxoH,x=y

0

otherwise.

Let h(x) = PX(SN = 0), and let g(x, y) = PX(S, = yISN = 0). By considering the two cases x e H and x = 0, we see that for all x with h(x) > 0, 91

92

(2)

3

Conditioned Brownlrn Motion

q(x,y) =

h(x)-1p(x,y)h(y)

The special relationship between q and p causes some nice cancellation to occur when we consider the joint distribution of the first two steps. If x, y e H, then a simple extension of the computation in (1) shows that

Px('S1=y,S2=ZISN=0)=PX(S1=y,S2=Z,SN=O)/Px(SN=0) = p (x, y)p(y, Z)h (Z)/h (x) = (h(x)-1p(x,y)h(y))(h(y)-'p(y,Z)h(Z))

= q(x,y)q(y,z)

(and similar computations show that the result above is true whenever y or x = 0), so the joint distribution of the first two steps is the product of the onestep probabilities, a trait indicative of the Markov property. The reader can easily confirm that the pattern above persists for any finite dimensional distribution and hence that (S,,,,NI SN = 0) is a Markov chain with state space HU {0} and transition probability q given by (2). The construction above can be generalized considerably. Let p(x, y) be the transition probability for a Markov chain on a countable set S, that is, p(x, y) >- 0 and Y,,p(x, y) = 1. Let h(x) >- 0 be a harmonic function for p, that is,

h(x) = Yp(x,y)h(y) Y

Given these two ingredients, we can define a transition probability by setting

q(x,y) =

h(x)-1p(x,y)h(y)

for all x with h(x) > 0. The q defined in the last paragraph is commonly called an h-transform of p. These processes were introduced by Doob in a paper titled "Conditioned Brownian Motion and the Boundary Limits of Harmonic Functions," so the reader should not be surprised that these processes will appear several times below. The first occurrence will be in the next section : by choosing the right harmonic function, we will get an h-transform of Brownian motion with Bt = 0 a.s. The answer will be obvious when you see it, so you should spend a moment thinking about what the h should be. It is one of the functions in Chapter. 1. In Section 3.3 we will encounter another h-transform : Brownian motion conditioned to never leave H (again you should try to guess the h), so for the rest of this section we will consider the analogous problem for simple random walk. Since the conditioning affects only the last component, it suffices to consider the one-dimensional case. Imitating the proof of (1) in Section 1.7 shows that if Tx = inf{n > 0 : S = x} and a < x < b, then (3)

b-x Px(T° TM), compute its transition probability qM(x, y), and let M -, oo. From

formula (2) above, we see that if 0 < x < M and Ix - yl =1, then gM(x,y) _

I P(To> TM) r -.

2 P.(To > TM) ly/M y 2 x/M 2x

-

A remarkable aspect of the last formula is that M does not appear on the righthand side (except in the requirement that x < M), so if x > 0 and Ix -- yl = 1, then (4)

q (x, y) = Mim qM(x, y) = 2x

is the transition probability for random walk conditioned to have To = oo. The reader should observe that q. is itself an h-transform with h(x) = x, and the generalization to Brownian motion should then be obvious. The last sentence tells you one of the things I wanted you to guess. For one last hint about the other, observe that in the situations described above, h = 0 on the forbidden set and this keeps the h-transform from going there. Exercise 1

It is interesting to see what happens when we condition an asym-

metric random walk, that is, let P(X = 1) = p, P(X,, _ -1) = q = I - p, and let S. = X, + + X with X,, X2 ... independent. (a) Define a function qp by setting cp(O) = 0 and p(x) = cp(x - 1) + (q/p)X, x 0 0, and check that is a martingale. (b) Repeating the argument for (3) above, we see that PX(Tb S} and let Ta =inf{t : Y e Ga}.

P(Ta < T") = Ez(ho(B:n); T < T")/ho(z)

Since ho(B,n) is a bounded martingale, using the optional stopping theorem at time Ta shows that P(T6 < T") = Ez(ho(B(Ta)); TT < T")lho(z) < sup{ho(w) : we Ga}/ho(z)

T, and our construction is almost complete. The last detail is to move the measure back to (C, W). This is no problem, however: co -+ Y(w) maps 0 into C, so we simply let P° be the image of P under this mapping.

The construction of P° above can be extended by translation to define measures PZB, 0 E Rd 1, that are "Brownian motion starting at z and conditioned to exit H at (0, 0)." The next result justifies the description in quotation marks. (3)

If

then PZ (A) = lim PZ (A I I BT - (0, 0) I < e).

Proof Let D. = {>li : 19 - 01 < e} and let h`(z) = PZ(I BT - (0, 0)I < e)

=

Jh(z)d/. De

Since he(B,A,,,) is a bounded martingale, we have for each A EA that PZ(AIIBL - (0,0)l < e)

A)lhe(z) = PB(A)

ase-+0. With (3) established, it is easy both to believe and to prove: (4)

Let A e .S and z c H. If we let g (0) = P .-'(A), then g(Bz) = PZ(AI Bt)

Proof By the bounded convergence theorem, it suffices to prove the result when A n > 0. To prove the result in this case, we observe that from the definition of conditional expectation and the strong Markov property it follows that if I is a Borel subset of 8H, then PZ(B. EI,A) =

EZ(hr(B(;)); A) where hl(z) = the probability of exiting H in I starting from z. Now hr(z) =

I he(z) d0,

,J

so applying Fubini's theorem (everything is > 0) and recalling the relevant definitions, we find that

3.3

Other Conditioned Proceases In 11

97

PP(B,aI,A) = JEz(ho(B);A)dO

_

ho(z)PB(A) d0 I

= JIho(z)g(0)d9 I

= Ez(g(BT); Bz E I).

Since I is an arbitrary Borel set, this shows that g(Bt) is a version of P2(AIBT) and completes the proof.

Note: The approach we have taken to the definition of PB follows Appendix II of Brossard (1975) with some minor modifications. This approach is slightly different from the one used in Section 3.1 (see the definition given in (2)), but it is easy to make the connection. Exercise 1

If ffis bounded and f = 0 on all, then

Exf(B,) = 4,'(x, y)f(y) dv, where

9B(x,Y) = he(x)-p'(x,y)ho(Y)

and pH is the transition probability for Brownian motion killed when it leaves H (see Section 1.9 for a description of this process).

3.3 Other Conditioned Processes in H As we promised in Section 3.1, in this section we will define Brownian motion

conditioned to never leave H. In light of the discussion above, there is not much to say: It is an h-transform with h(z) = Zd, and it is defined almost exactly like the processes constructed in the last section. Let G. = Rd-1 x (0, 2"), let T. = inf{t : B,0G"}, and let .F (T"). If A e .F., we let (1)

PP°(A) = Ez(h.(B,);A)/h,(z) where h., (z) = Zd. As before, it is easy to see that these definitions are consistent,

so if we let Y"(co) = B( A T"), then we can construct (Y1, Y2, ) on some probability space with Y"(t) = Yn+1(t) for t :!g T", and we can define a process on [0, lim T") by letting Y, = Y"(t) for t < T". To see that lim T. = oo a.s., we 0 (since he = 0 on 8H), so the strong Markov observe that property of Brownian motion and an obvious scaling imply that (T,.,, - T")/2", n = 1 , 2, ... , are indpendent and identically distributed, which is more than enough for the desired conclusion.

98

3

CondltbnW Brownlnn Modons

With two examples of conditioned Brownian motion constructed, it is natural to ask if there are any other interesting examples, or what is the same : Can we describe the set of all nonnegative functions that are harmonic

in H? To do this we will start by considering the analogous problem in D = {z:Izl- 0 is harmonic in D if and only if there is a finite measure µ on 8D such that

u(z) = JkY(z)diu(Y) where ky(z)

= Iz-_

Izlz

1

. yid

Proof Now ky(z) is the probability density (with respect to the surface prob-

ability measure ir) of exiting D at y when we start at z, so as we showed in Section 1.10, each of the functions ky is harmonic. It is routine to show, using the result on differentiating under the integral sign given in Section 1.10, that A

Jk(z) du(y) = JAk(z) du(y) = 0

(details are left to the reader), so it follows that all the functions defined above are harmonic.

To prove the converse requires a little more thought, but less tedious computations. For 0 < r < 1, let u, be the measure on 8D that has density tp(y) = u(ry) with respect to n. u, has total mass Ju(rY)dlr(Y) = u(0).

Since all these measures are concentrated on a compact set and have the same

total mass, the Helly selection theorem implies that there is a subsequence p,, that converges to a limit u. It follows from the definition of µ, that J

ky(z) d,(Y) = Jky(z)u(rY) dir(t) = u(rz).

Letting r = r and n --), oo, we get

u(z) = lim u(zr,,) = J'k(z) dy(y),

since p,, converges weakly top and y - ky(z) is continuous. To translate (2) into a result about H, we will use one of Kelvin's transforma-

tions (discussed in Section 2.12) to map D one-to-one onto H. To figure out which transformation we want, we start by calculating what the inversion J(z) = z/IzI2 does to D(ed, 1), the ball of radius 1 centered at ed = (0, ... ,0, 1).

3.3

Other Conditioned Procewe In //

99

If zeaD(ed, 1) _ {z : Iz - ed1 = 1}, then Z1 + ... + Zd-1 + (Zd - 1)2 = 1.

The last component of J is :

JAZ) = Zd/(Z1 + ... + zd) The first equality above implies that

zl+...

Z2

2

so Jd(z) = 1/2 for all ze3D(ed,1). A little thought shows that J maps D(ed, 1) one-to-one onto the half space H' = {z : Zd > 1/2}, so K(z) = J(z + ed) - (ed/2) maps D one-to-one onto H. From (6) in Section 2.12, it follows that there is also a one-to-one correspondence between positive harmonic functions in D and in H, that is, if u is harmonic in D, then I xI2-du(K(x)) is harmonic in H, and, conversely, if u is harmonic in H, then IxI2-du(K(x)) is harmonic in D. Combining the last observation with (2) gives an integral representation for positive harmonic functions in H. It is easy to see without computation what functions will appear in the decomposition. The ky(z) with y -ed get mapped to Zd

Cylz

- K(y)Id'

that is, the probability density of exiting D at y gets mapped to a constant (that depends upon y) times the probability density of exiting at K(y), and when y = -ed, ky gets mapped to Czd. To check the second claim, observe that this ky is mapped to a function harmonic in H that vanishes on 8H, and that mapping things back to D and using (2) shows that there is (up to a constant multiple) only one such function). Combining the observations above with (2) gives: (3)

If u > 0 is harmonic in H, then there is a constant C and a measure µ on 8H such that u(z) = CZd +

f

he(z) dµ(9)

aH

Remark: The reader should note that the measure µ is not necessarily finite and that this is only an "only if" statement. To assert that such a u is harmonic, we need a condition such as he(ed) dµ(9) < oo 'aH

to assert that u < oo in H. It turns out that the condition just mentioned is necessary and sufficient for u * oo and harmonic in D, but we will not pursue this here.

100

3

Conditioned Brownien Motions

(3) gives us the answer to the question asked at the beginning of this section. It shows that all nonnegative functions that are harmonic in H are linear combinations of zd and the he's, so the processes we have constructed in the last two

sections are the only interesting examples of conditioned processes in H. Nontrivial linear combinations being uninteresting, since, for instance, if p is a probability measure and h(z) = Jho(z)diu(O),

then t he corresponding h-transform is just Brownian motion conditioned to have B

a

f

3.4 Inversion in d >_ 3, Bt Conditioned to Converge to 0

ast -goo Let J(x) = x/IxI2. If B, is a Brownian motion in d >_ 3, then (as we observed in Section 2.12) J(B,) is not a Brownian motion, since J(B,) -+ 0 as t - oo. What it is, of course, is an h-transform of Brownian motion. To prove this and IJ(x)I2-d. Now to discover the h, let u be a harmonic function and let g(x) =

J(J(x)) = x, g(J(x)) =

IxI2-du(J(x)) is harmonic, so

IxI2-d, and

B,I2-du(J(B,))I Bo = J(x)) = I J(x)I2-du(J(J(x)))'= g(x)u(x)

E(g(J(B,))u(J(B,))I J(Bo) = x) = E(I

Since u is harmonic, we can rewrite the last expression as

g(x)u(x) = g(x) fP,(x,y) u(y) dy where p,(x,Y) = P(B, = YI Bo = x) =

let P,(x, y) = If weJPt(xY)(Y)u(Y)dY

(2nt)-d/2e-Ix-y12/2t.

yI J(B0) = x), then we have shown that = E(g(J(B,))u(J(B,))I J(Bo) = x)

= fg(x)&(:!, Y) g(y)u(y)dy g(Y)

for all harmonic functions u. It follows that Pt(x,Y) = g(x)Pt(x,Y)g(Y)-1,

that is, J(B,) is an h-transform of Brownian motion with h(x) = 1/g(x) =

XI2-d,

a constant multiple of the potential kernel defined in Section 1.8. After the fact, it should be obvious that this is the right h. It is nonnegative, is harmonic when x 0 0, and converges to 0 as IxI -> oo.

Inversion In d a 1, B, Conditioned to Converge to 0 ss i -+ w

3.4

101

Using the last result, we can compute the occupation time density for Brownian motion killed when it leaves D = {z; I z I < 11. Let f be bounded and

=0onD`,andletT=inf{t:B,0D}. Ind> 3, o

w(x) = E. f f(B,) ds < oo, 0

and the strong Markov property implies that w(x) = Ex f f(BS) ds + EXw(B.). 0

so we have

Ex J If(BS) ds = w(x) - Exw(B,). 0

To compute the second term on the right, we observe that since J(x) = x on 8D, it follows that (for x 0) Exw(B.) = Exw(J(B,)) = h(J(x))-1EJ(x)w(B,)h(B.)

where h(x) =

IxI2-d. Since h

= 1 on 8D, it follows that

E,,w(B.) = h(J(x))-1E,(x)w(Bc) = Ix12-dE!(x)

tf(B5)ds

= X 2-d E!(x)

f(B5) ds, J0W

since f=0 on D`. Combining bining the results of the last two paragraphs shows that (for x

ds = w(x) -

Ex

0) :

IxI2-dw(J(x)),

so we have (1)

E. J f(B.,)ds = JGD(xY)f(Y)dY 0

where

GD(x,Y) = G(x,y) -

IxI2-dG(J(x),Y)

It is trivial to extend the last formula to x = 0. To do this, we observe that

asx-+0, x12-dG(J(x),Y) I

=

CIxl2-dlxIxI-2

__

yl 2-d

converges to C(= F(d/2)/lyd-2). If we set Go(0, y) = G(0, y) - C, then (1) also holds for d >_ 3.

102

3

('onditlonod Brownlan Motions

3.5 A Zero-One Law for Conditioned Processes This section is devoted to proving a zero-one law, which is an important tool both for establishing properties of the conditioned processes and for deducing analytical results about boundary limits of harmonic functions from their probabilistic counterparts. Since our only applications will be to the second topic, we will prove the zero-one law only for Brownian motion conditioned to exit H at (0, 0). At the end of the proof, however, it will be obvious how to prove the result for other conditioned processes. To state our result, we need several definitions. (1)

An event Ae. lA o 0T = 1A

is said to be shift invariant if for all stopping times T < T P= a.s. for all z e H

where 0T is the (random) shift operator defined in Section 1.5 by (BTw)(t) = w(T(w) + t).

In words, we can determine whether or not a shift invariant event occurs by looking at B T :!g t < T, that is, these events concern the behavior of Bt as t T T. The following are typical examples of shift invariant events: (i) {lim u(B1) exists} t1 r

(ii) {lim sup I u(B,) I < oo } tTr

(iii) {B, e A infinitely often as t T T} {to: for all s > 0 there is a u e (T - E, T) with B. e A}.

Let 5 be the collection of unvariant events. Although the examples above suggest that there are a large number of events in 5, the next result shows that

this impression is wrong. These events are all 0 or Q (give or take a set of measure 0). (2)

If A e f, then z -+ PB(A) is constant and the constant is either 0 or 1. The key to the proof is the following lemma, which is a strong Markov property

for P°: (3)

If A e Fz and T < T is a stopping time, then EB(1A0OTI FT) = P07T)(A).

Proof We will prove the result first for T < T,,. By the bounded convergence theorem, it suffices to prove the result when A eJFm(-.F(Tm)) and m > n. If Be .° T, then by the definition of/ P f, EO L lA o 0T ; B] = Ez [hO (Bs ) (lmA o BT) ; B]/h9 (z)

= E. [(he(B=m) 1A) ° 0T ; B]/ho(z),

since Btm(w) = BTm(0TW). Applying the strong Markov property and using the

A r*ro-one Law for Co.awoned rrocea ea

3.5

103

definition of P2 with z = B(T), we find that the above = EZ[EB(T)(he(B=) lA); B]lhe(z) = EZ[he(BT)PB(T)(A); B]lhe(z)

To convert this to the desired form, we observe that the optional stopping theorem implies that if Ce.FT, then EE(he(BT)lc) = EjhO(B(TJ) IC), so the definition of Pe for Ce.FT may be rewritten as P!(C) = EZ(he(BT) lc)/he(z),

and it follows that if Ye.FT is bounded, EB(Y) = EE(he(BT)Y)lhe(z)

Applying this result with Y = P, (T) (A)1B and tracing back through the chain of equalities above, we get: EB [lA o OT; B] = EB [Pa(T)(A) ; B]

for all Be.IFT, which is the desired result in the case T:!!9 r,,. To remove this assumption, apply the result above to T A T,,, let B = B n {T< and let n - oo to conclude that the last equality holds in general.

Proof of (2) Let A E.f and let (p(z) = PB(A). From (3), it follows that if T < 'r is a stopping time, then (p(z) = P°(A) = Pe(A o BT) = Ee(PB(A0OTI Ee(Pa(T)(A)) = EB((p(BT)) = EZ(he(BT)(p(BT))lhe(z)

Applying this result at T = inf{t : B,OD(z, S)}, we see that g(z) = he(z)(p(z) has the averaging property g(z) =

J aD(z, a)

g(Y) d7r(Y),

so it follows from (7) in Section 2.10, that g is C °° and has Ag = 0. Since g > 0, it follows from (3) of Section 3.3 that g(z) = Cza +

Jh(z)dz(Y) e

and, since 0 < (p(z) < 1, g(z) = he(z)(p(z) < he(z). Combining the last two observations, we conclude that u(101') = 0 and C = 0, so (p is constant. With then this result in hand, we can easily complete the proof. If Be.J _

0 OA

= Ez(E'('A00,,1°n);B) = (A) ; B)

104

3

Conditioned Brownien Motion

It follows from the definition of cp and the fact that qp is constant that the above

=

B)

= cp(z)PB(B)

= Pe(A)PB(B),

and we have shown that PB(A (1 B) = P°(A)PB(B)

for all B e

and hence for all B e

Letting B = A in the last equation gives

PB(A) _ (P°(A))2,

so P!(A) = 0 or 1, and the proof of (2) is complete. Remark :

As we promised at the beginning of this section, it is easy to gener-

alize this result to other conditioned processes, since the keys to the proof above are (a) the strong Markov property of the conditioned process and (b) the fact that if g is harmonic and 0 < g(x) < ha(x), then g(x) = che(x). In the terminology of the theory of convex sets, (b) says that he is an extreme point of the cone of nonnegative functions that are harmonic in H, and our general zero-one law may be phrased as follow: If h is an extreme point, then the zero-one law holds for the corresponding h-transform. Looking at the representations given in Section 3.3, we can easily see that in H or D the converse is also true. If h is a nontrivial linear combination, then the asymptotic random variable limit. B, has a nontrivial distribution.

Note: The proof of the zero-one law in this section, like the definition given in Section 3.2, is from Appendix II of Brossard (1975).

4 Boundary Limits of Harmonic Functions

4.1

Probabilistic Analogues of the Theorems of Privalov and Spencer Let u be a function that is harmonic in H =

x (0, cc). As the section titles in this chapter indicate, we are concerned with the existence of the limit of u(z) when z - (0, 0) e 8H in two special ways: Rd-1

(a) when z -+ (0, 0) along Brownian paths (i.e., u(B,a) -> a limit a.s. where BB is a Brownian motion conditioned to exit H at (0, 0)) (b) when z - (0, 0) nontangentially. To explain notion (b), we need a few definitions. For each a > 0, let Va be a cone of height 1 with opening a and peak at 0, that is,

Va={(x,y)cH:Ix -0I 3

Privalov(1916) Marcinkiewicz and Zygmund (1938) Spencer (1943)

2' _ .N' Calderon (1950a) .N' c sat .N D sat

Calderon (1950b) Stein (1961)

4.1

Probabilistic Analogues of the Theorems of Prlvslov and Spencer

107

From the order in which the results were proved, the reader can see the relative difficulties of the three results.

So much for history. The main task in this chapter is to prove that if u

is harmonic in H, then 2' = At = d a.e. The first step in doing this is to define the analogous probabilistic sets 2*, V*, and s.4* using Brownian motion and to show that 2* = }( * = d* a.e. The generalization of nontangential convergence or boundedness to the Brownian setting is easy. We ask for convergence or boundedness along almost every Brownian path, and to get a subset of Rd-1 we let {O: lim u(B,) exists PO a.s. }

ttr

)

U* = sup I u(B)I t 0. If Pa(o = r) > 0 and U* = supr 0, and it follows from the zero-one law (theorem (2) in Section 3.5) that P°(U* < oo) = 1 for all z, that is, 0 e X*. To prove (1), we need only to prove the innocent-looking fact stated in (2). This result would be easy to prove if were always a finite union of intervalsthe only exceptional points would be the end points of the intervals-but unfortunately, the set -Y./' can be very ugly. A typical nightmare is u(z) = YBcehe(z) where the sum is over all rational 0, he(z) is the probability density of exiting H

110

4

Boundary Limits of Harmonic Function

x+ 2 Figure 4.2

at 0, and the ce > 0 are chosen so that the sum converges and is harmonic in H. In this case, A' = 8H - UD(0, re) where the union is over rational 0 and re > 0,

so )r is dense and it is hard to tell which 0 c.*' will be "good" (that is, have PB(a = T) > 0 for some z). We will eventually derive a criterion (due to Marcinkiewicz) for a 0 E A' to be good, but the reader would not see the reason for the definition if we gave it now, so we will first see what kind of estimates we can get on PB(r = T) and then introduce the criterion when it will appear more natural. Proof of (2) We start by observing that, in order to prove that PB(a = T) > 0 for some z, all we have to do is look at 8S2 near 0. Let I, J be intervals centered at 0 with radii r and 2r, respectively. Let

Al = {(x,y)e092:xeI,y < 1) A2=8S2-A1 a; = inf{t: B,EA4} i = 1, 2. It is easy to estimate Pe(a2 < T). By definition, Pe(a2 < T) = EZ(he(B(a2)); a2 < T)lhe(z) Now C1 = sup {ho(z) : z c A2 } < oo, so it follows that (3)

PB(a2 < T) < C1/h0(z),

and if we let z -> (0, 0) in such a way that he(z) co, then Pa(o 2 < T) - 0. This occurs, for instance, if z --> (0, 0) in Vb for any b < oo, so to prove (2) we need to estimate PB(a1 < T) for zE V6. As in the case of a2i we start with the equality

Peal < T) = EZ(he(B(a1)); a, < T)/he(z), but this time we have to work harder to estimate the right-hand side. We begin by looking at the geometry of 8S2 and how it relates to the Poisson kernel.

If (x, y)eA1, then y = d(x,.f)/a where d(x,. t) is the distance from x to the closest point in A'. . Let x' c- (x - ay12, x + ay/2) and y' = d(x', -' )Ia. Since y = d(x, )F)la, then x' E.Yi and, consulting the definition of S2, we see that (x', y') is the point on 8S) that is above x' (see Figure 4.2). As the reader can easily check, we have

2

2he(x, y), we have 2

x+(ay/2)

.-(ay/2)

x+(ay/2)

hx,(x,Y) dx'

g(x')h.,(x,Y) dx' ? 9 he(x,Y)

x-(ay/2)

and x+(ay/2)

y12

2

hx (x,Y)dx' > ay (ay/2)2 = a'

x-(ay/2)

so if we let C2 = 9a/4, it follows that if (x, y) e A 1, Cx+(ay/2)

g(x')hx.(x,Y) dx'

he(x,Y) d(x, .-(), so we can replace the integral over .'f by an integral over 0 E [x - d(x, Jfl, x + d(x, If we make the replacement and change variables toy = 0 - x, it follows that n for z E E,,, then it will follow from (1) that N1 u = oo for all 0 E [0, 1]'. This takes care of [0,1]2 c .N'`. The next part of the construction is designed to keep [0,1]2 c -4/*. Let x [0, 2b"] A. = [ - b,,, bn] x [ - b,,, B"= {z:Iz - WI n + 2 so that P(O, o, 2) (Bt E A. - B,, for all 0 < t < T) >- 1 - 2" .

118

4

Boundary Llmite of Harmonic Functions

Since Y1/2" < oc, it follows from the last result that with probability I there is

an N (that will depend on co) such that Bt E AN - BN for all 0:5 t < r, and hence we can construct u in such a way that it is bounded on each set A. - B", it will then follow that {sup«LIu(B)I < oo} has P(0.0,2) probability 1, and from (4) of Section 3.2 that the event has P(o, o, 2) probability 1 for a.e., 0. At this point our mission is clear : We want to construct a harmonic function

that is large on E. and small on A. - B". To do this, we use the following result of J. L. Walsh (1929).

Let K be a compact set in Rd such that R" - K is connected, and suppose u is harmonic on an open set containing K. Then u can be uniformly approximated by harmonic polynomials on K. RUNGE'S THEOREM FOR R°

K,, = (A" - Bn) U E,, is compact, and R3 - K,, is connected. Let U and V be disjoint open sets with A. - B" c U and D. c V. Let wn(z) equal 0 on U and equal 2,, on V. Then w" is harmonic on U U V, and Runge's theorem guarantees that we can find a harmonic polynomial u,, such that (2)

for zeA, - B,, Iu,,(z)I < 1/2" Iu,, (z) - 2,I < 1/2" for zeE". From (2) it follows immediately that we have

(3)

X I u"(z) I converges uniformly on compact subsets of H. n=1

Proof If K is compact, then K c A" - B" for all n sufficiently large. Since each u" is harmonic, that is, if z e H and D(z, r) c H, then (4)

un(y) dit(y),

un(Z) = y eJoD(x,r)

so it follows from (3) that u(z) = Y', u,,(z) also satisfies (4) and hence is harmonic in H. From the construction above, we see that u is bounded on each set A,, - B", so it remains only to pick the A. such that u(z) > n for z e E". Let Al = 2. E1 = {(l/2, 1/2, 1)} and (1/2, /2,1/2EA" - B. for all n 2, so we have (recall e" < 2-") °° 1\"

u(1/2,1/2,1)= X u"(1/2,1/2,1)>2- Y_

2

n=2 2

n=1

> 1.

Suppose now that A1, ... , An_1 have been chosen so that u(z) > m for all z c Em, m < n - 1. If we pick 2,, such that n-1

inf Y um (z) + 2n > n + l , zEEO m=1

it follows that if z E E", then

u(z)>n+l-

ao

m=n+1

m

(,)

>n.

4.5

Broewrd's Proof of Calderon's Theorem

119

4.5 With a Little Help from Analysis, Probability Works in d >_ 3: Brossard's Proof of Calderon's Theorem In the last section we saw that when d > 3, A^ * - .N' may have positive measure. This result makes morally (if not logically) certain that it is impossible to show

that 2' = .N' by using purely probabilistic methods. In this section, we will show that if we borrow a simple result from analysis, namely, (1)

If u is harmonic and l ul < M in H, then u has a nontangential limit at a.e. point of 8H, then we can use probability to prove the local version of this result :

(2)

If u is harmonic and Nau < M for all 0 e S, then u has a nontangential limit at a.e. point of S. (2) implies that .Na c 2', so when we combine this with the trivial contain-

ments 2 c Y. c .Na, we get Calderon's theorem: 2' = K. (1) is a basic fact about harmonic functions and can be found in Chapter 2 of Stein and Weiss (1971), but it is a simple result and the analytical proof has some interesting probabilistic aspects, so I will give the details here. The first step in proving (1) is to prove: (3)

If l u(x) l < M for all x e H, then there is a function f with l f(x) l < M so that u(z) = EZ f(BL) where z = inf{t : B, 0 H}, that is, u(z) = Jho(z)f(0)dO.

Remark: This is Theorem 2.5 of Stein and Weiss (1971). It gives the Poisson integral representation for bounded harmonic functions. Since u + M > 0 in H, this theorem is, except for the assertion about f, a special case of (3) in Section 3.3.

Notation: Since functions of the form given in (3) will appear many times below, we let Yf(z) = I he(z)f(0) d0.

Proof of (3) We begin by proving the following semigroup property for u(-,y): (4)

If u is bounded in every HE = {(x, y) : y > E}, then for all r, s > 0

u(x, r + s) = f he(x, r)u(O, s) d0.

120

4

Boundary Limits of Harmonic Functions

Proof Let T = inf{t : B O H,}. Since

is a bounded martingale,

u(x, r + s) = E(x.,+s)U(BT)

=

f ho(x, r)u(0, s) d9.

Remark: The stopping time argument used above is a substitute for (and proof of!) the maximum principle used in the analytic proof (see Stein and Weiss (1971), pages 52-53).

With (4) established, the rest of the proof of (3) is soft analysis. Since I u I < M in H, there is a sequence yk 10 such that u( , yk) converges weakly to a limit f, that is, for all g e L' Ju(OYk)o(O) dO

Jf(O)o(O) de.

and so we have in particular that

f he(x,Y)f(O)d9.

Jho(x y)u(O, Yk) dO

-, By (4), the left-hand side is u(x, y + yk). Since u is continuous in H, u(x, y + yk) - u(x, y) as yk 10, proving (3). With the Poisson integral representation established, the proof of (1) is reduced to the task of showing that functions of the form u = gf have nontangential limits a.e. Since the value of u comes from integrating f with respect to a spherically symmetric kernel, it does not take too much inspiration to guess that the Lebesgue points, that is, the points where 1

r"

If(x+O)-f(B)Idx->0

are going to be good. The next result confirms this guess and, since a.e. point is a Lebesgue point, proves (1). (5)

If 00 is a Lebesgue point for f, then u = 9/f has nontangential limit f(00) at (001 0).

Remark: This result and its proof are classics (see Stein and Weiss (1971), pages 62-63).

Proof Since he(x, y) >- 0 and f he(x, y)d9 = 1, I u(x,Y) -.f(eo)I = Jho(xY)(f(O) -f(00))dO J he(x,Y)I.f(0) -.f(Bo)I d9.

4.3

Broeuurd'N Proof of C.Ideron'M'11eorem

121

Now if (x, y) e Va(00), then h9(x,Y) h0(00,Y)

0)2

(/(00 + Y2)d/2 ((X - 0)2 + y2)d/2 if 100 - 01

(a2 + 1)d/2 K2a2+1

d/2

(K - 1)2a2 + 1)

< ay

if 100 - 01 = Kay where K > 1.

The last expression - 1 as K -+ oc, so the ratio above is bounded, and it follows that 1 u(x,Y) -f(Oo)1 < C I ho(Oo,Y)I f(0) -f(0o)I d0 - 0

if 00 is a Lebesgue point. With the proof of (1) completed, we turn now to the main business of this

section, namely, the proof of (2). We begin with our usual construction. Let .7' _ {0 : Na(0) < n}, let 0 = UBEi- Va(O), and let a be the exit time from Q. I u1 < n on fl, so as t t o, u(B,) -* a limit that we call u(B0), and the optional stopping theorem implies that u(z) = Ezu(Ba)

Since {a = T} c {Bt e . r }, we can write u(Z) = uo(z) + u1(z) + u2(z) where u2(z) = Ez(u(B0) : a < T)

2ul(z)

uo(z) = Ez(u(Bj: Be.*) = -Ez(u(B,) : Bte.*, a < T). Now u0 is a bounded harmonic function, so by (1) it has nontangential limits a.e., and, furthermore, the limit of u0 is f 1, To handle ul and u2, we observe that each is bounded by nPz(a < T), so it suffices to show that for a.e. 0 c _* , Pz(a < T) - 0 when z (0, 0) in Vb(0). (This is similar to the result for PO proved in Section 4.2, but it is simpler.) To prove this result, we use an idea due to Calderon: it suffices to show that Pz(a < T) < CPZ(B,O_C) for the right-hand side is a bounded harmonic

function that, by (3) and (5) above, at a.e. point of V. To prove the inequality, observe that if (x, y) e H - ), then either y > 1 or y < d(x, it")la. If y > 1 (the trivial case), then P(x.Y)(B, 0

) >-

P(x.Y)(BTe[-n, n]) > el.

If y < d(x, 1')/a, then D(x, ay) c . ''° and P(x.Y)(BT0'

) < P(x.Y)(B,eD(x,ay)) > e2(a) > 0,

122

4

Boundary Umlt of Harmonic Functions

so there is an e > 0 (that depends on a) such that inf P=(B,O.,r) >_ E.

z c H-11

From the Markov property, P.(B,

P. (BT0.7E'',o'EPZ(a_ 0, is a time

change of Brownian motion. Since 0 is not in the range of exp, it follows that Pl (So < oc) =

0 for some u >_ 0) = 0. 123

124

5


6

4L 27ri

0

- 27ri ez

Figure 5.1

(2)

Let S1 = inf{t >- 0 : IBI 5 1}. If Ix! > 1, then

PX(S1 u}, then is a Brownian motion starting at x. Let T1 = inf {t > 0 : I Cr I > 1). It is trivial to show that P1IX(T1 < oo) = 1 (see Exercise I in Section 1.7 for a more

general result). At time t = T1, IC,I = 1, so at time u = o(T1) (which is - 0 : exp(CT) = F/(Y(aT)) = B(QT) = B,

that is, exp maps the exit distributions from G to corresponding exit distributions from H. With the last identity established, it is trivial to compute the exit distribution from G. PO (Ins CT = it/2, Re CT >- a) = P1(Im Bz > e °)

=

rte e°

x2 + 1 i

dx

by (2) from Section 1.9. Changing variables x = tan y, dx = seczy dy converts the above to

f

'

1 TC

dy=

1

2

- - tan - (e), 1

it

and differentiating gives (recall that (tan-lx)' _ (1 + (4)

P0(Im CT = n/2, Re CT = a) =

1

x2)-')

e°

it e2°+1

= (2n cosh a)-1 The last example is just one of many that can be done with this method. The technique used above can, in principle at least, be used to compute the exit distribution for any simple connected region, since the Riemann mapping theorem implies that any such region (that is not the whole plane) can be mapped one-to-

one onto the disk. Carrying out the details of this computation, however, is usually very difficult even in simple examples such as polygons where the Schwarz-Christoffel formula gives an almost explicit formula for the mapping.

Note: Ever since Levy discovered the conformal invariance, many people have used it to prove results about two-dimensional Brownian motion; see Ito and

126

5


McKean (1964), McKean (1969), Davis (1979a, 1979b), and Lyons and McKean (1980). Formulas (1), (2), and (3) each appear several times in these references.

5.2 Nontangential Convergence in D In the next two sections, we will be concerned with the nontangential limits of analytic functions. For some of the developments, we will need to know that several of the theorems proved in Chapter 4 for the upper half space H are also valid in D. In this section, we will use a combination of proof by analogy and by conformal mapping to prove the results we need. The first thing we have to do is define nontangential convergence in our new setting. Let Sa(6) be the convex hull of the disk {lzl < a} and the point e'°. (The S is for Stolz domain.) A function f, analytic or not, is said to have nontangential c whenever z e Sa(O) and z,, - e'°. limit cat e'0 if for all a < 1, The first step in investigating nontangential convergence in D probabilistically is to define Brownian motion conditioned to exit D at e'9. It is easy to see that this can be done by imitating the approach used in Section 3.2. We let we let T. = inf{t : B,l = 1 - 1/n}, and, if A (1)

PB(A) =

A)lke(z)

where

1lz - etelz 1212

ke(z)

is the probability density of exiting D at e'° starting from z. Since ke >- 0 is harmonic, repeating the arguments of Section 3.2 shows that we can paste these processes together using Kolmogorov's extension theorem, to get a process Y defined for t < T. The last problem, then, is to show that Y -+ ei° as t -+,r by modifying the proof of (2) from Section 3.2. Filling in the details in the sketch above is left as an exercise for the reader. The first result we want to generalize from H to D is that (2)

Y*=_V*=Y=.N', where these quantities are defined in the obvious way by analogy with the definitions given in Section 4.1. It should not be hard to believe that we could prove this result by patiently working our way through Sections 4.1 to 4.3 and noticing that the proof given there works, with minor modifications, in our new setting. Fortunately, there is another alternative.

Proof g(z) = (z - 1)/(z + 1) maps H' = {Rez > 0} one-to-one onto D. Since g is analytic, if we let a = inf{t : B, H'}, then Levy's theorem (and a little common sense) shows that

is a time change of a Brownian motion start-

5.2

Nontangentld Convergence In t)

127

ing at g(B0) and run until it exits D. The last result makes it easy to believe that "g maps B, conditioned to exit H at y to a time change of B, conditioned to exit D at g(y)," and, in fact, using (3) of Section 3.2, the reader can easily convert this belief into a proof. With the statement in quotation marks established, the desired result fol-

lows immediately, since the last conclusion implies that g(2') = 2D and g(.N'f;) _ .4,D (where the subscript indicates the domain under consideration), and it follows from the definitions of nontangential convergence and boundedness that g(54) = 22D and g(.KH) = VD. The argument above can obviously be used to obtain many other results about conditioned processes or harmonic functions in D from facts about the corresponding objects in H. For the arguments in the next two sections, we will also need to generalize results (3) and (5) of Section 4.5 from H to D. (3)

If Iu(z)I < M for all zeD, then there is a function f with I f(x) I < M so that u(z) = E.. f(B,) and, furthermore, u has nontangential limit f at a.e. point of 8D.

Proof Let g, H', and a be defined as in the proof of (2). Since g is analytic, a simple computation (or an application of Levy's theorem) shows that v(z) = u(g(z)) is harmonic in H' = {Rez > 0}. Since Iv(z)I < M, it follows from (3) and (5) in Section 4.5 that there is a function fo defined on 8H' so that v(z) _ EZ fo(B,) and v has nontangential limit fo at a.e. point of 8H'. To take these two results back to D, letf(z) = fo(g(z)) for z e 8D - {11 and let h(z) = (1 + z)/(1 - z) be the inverse of g. Since h(B,), t < u, is a time change of Brownian motion run until it exits D by unscrambling the definitions and applying Levy's theorem we get u(z) = Ez f(BL). The second conclusion follows from the second paragraph of the proof of (2). Exercise 1 Let u be a measurable function defined in D and define the two corresponding maximal functions UU* = sup{Iu(B,)I : t < i} Nau(O) = sup{Iu(z)I : zeS,(O)}.

Use the reasoning we called Calderon's argument in Section 4.5 to show that there is a constant C (whose value depends only on a) such that for all , > 0, A) All. A similar result holds in the upper half space : fdx

1)(UL* > A) A) 1.

In view of the counterexample in Section 4.4, it is much more difficult to show results in the other direction.

128

5

Complex Brownien Motion and Analytic Functions

5.3 Boundary Limits of Functions in the Nevanlinna Class N A function f analytic in D = {z : Izl < 1) is said to be in the Nevanlinna class N if 2n

sup , 0 be chosen so that D(0, 6) fl Zf = 0. Since f(0) = 1, there is a unique analytic function g that has g(O)=O and exp(g(z)) = f(z) for all z E D(0, S), so for t < Ta =inf {t : B OD(0, S)}, L, = g(B,). The last observation shows that fort < Ta, L, is an analytic function of Brownian motion. By iterating this result and using the strong Markov property, we get : (3)

L, t < T, is a complex local martingale.

Proof We begin with some notation: A` = U D(Y, E) yEA

T' = inf{t : B, E (Zf U

T. =

T is the sequence of times that we will use to reduce L. See Figure 5.2 for a picture of (Zf U D)1I The first thing to check is that T T T as n - oo. Since L, is continuous, it is immediate that T2 T T. To see that T T, we observe that PO (B, e Z f for some

t> 0) = 0, and since the Brownian path is continuous, there is a lower bound on the distance between Bt, 0 < t < T, and the closed set Zf fl D(0, r).

5.3

Boundary limits of Function In the Nevanllnna Clan N

129

Figure 5.2

To prove that L(t A T.) is a martingale, we have to further subdivide time. The strategy is simple : For each z e D - Z1, we will choose a ball D(z, SZ) r D - Z1, define the associated exit times a = inf {t : B, 0 D (z, SZ) }, and then exhaust T. by iterating these exit times. To do this requires a little patience and a lot of notation. Let

SE(z) _ (1 - i)sup{S : D(z, S) c D - Zf} R.(z) = D(z, 6.(z)) u ,(z) = inf (t: Bt RE (z) }

So= 0,Zo=0 and form

- 1,

S. = inf{t > Sm_1 : B,0Riin(Zm_1)} A T. Z. = B(Sm)

M = inf{m : S. = The next result explains the reason for our choice of 6,(z). (4)

For each E > 0, there is a constant p (E) > 0 such that if z c- D - Z1, then Pz(u.(z) E (Zf U D`)`) > p(E)

Proof Since sup {6 : D(z, S) c D - Zf} < 1, there is at least one point, say zo ED` U Zf, that is within e/2 of the boundary of R,(z). Since D(zo, E) C (Zf U D`)`, looking at Figure 5.2 and experimenting with a few cases, one finds that

inf P-(a.(E) E D(zo, E)) > 0,

ZED-Zf

proving (4).

110

5

Complex Brownian Motion and Analytic Function

From (4) and the strong Markov property, it follows immediately that

P(M > m) < (1 - p(IIn))m, so P(M < oo) = 1. Since SM = T", all that remains is to put the pieces together to prove (3). The argument preceding the statement of (3) shows that L(t A S1) is a martingale. Combining this result with the strong Markov property shows that for any m, L(t A Sm) is a martingale, that is, E(L(t A S.)IJ¢"'S) = L(s A Sm).

Since I LI I < n for t< T,,2 and T.2 > Sm, it follows from the bounded convergence theorem that L(t A T") is a martingale.

Remark: In Section 6.4, we need to use the fact that iff is an analytic function with f(O) = 1 and we define G, to be the unique continuous process that has Go = 1 and G," = f(Bj, then G, t < r, is a local martingale. The reader should look back over the last proof now to verify that (a) the proof above (with trivial modifications) proves this fact and (b) in the proof above we never use the fact that f e N. To use the condition f c- N to conclude something about the convergence of L, as OT, T, we have to replace the stopping times T" by i, = inf{t : IB,I > r}. To do this we will discard Im L,, because we have very little control over its size (and we know that there is a trick for deducing the convergence of Im L, from that of Re L,). Even Re L, is not so nice on {z : IzI < r}. The function f is bounded, so Re L, = log I f(Bt) I is bounded above, but not below, if f has zeros. To get around

this problem and to make the connection with the condition feN, we truncate below. (5)

If a > - oo and U° = (Re L,) v a, then Ua(t n r,) is a submartingale. Proof In the proof of (3) we showed that L(t A T") is a martingale, so it follows that U°(t A T") is a submartingale, and the optional stopping theorem gives

E(U°(t n r, n T")I,Fs) > U°(s A i, n T").

Since U,", t < T is uniformly bounded, letting n - oo and using the bounded convergence theorem proves (5).

Remark: Continuing to prepare for Section 6.4, we observe that IG,I =IFI1/" is bounded for t < T so the argument above shows that G(t n r,) is a complex martingale. Up to this point, we have not used the fact that f e N. The next result ends this generality. (6)

If f e N, then lim Re L, exists a. s. `T`

5.3

Boundary Limits of Functions In the Nevsnllnn. Cl.. N

131

Proof Since f e N, J2n

EU°(T,) < Ial +

log+ I f(re`B)I dn(O) < C < oo. 0

Let T = inf{t : IB,I >_ 1 - 1/n} and let y : [0, oo) -+ [0, T) be the stretching func-

tion defined in Section 2.3. y is a predictable increasing function that has y(n) < T, so U°(y,), t z 0, is an L1-bounded submartingale, and it follows that Y° as as t - oe, U" (y,) converges to a limit Y° a.s. If Y° > a, we have Re L, t -- T. On the other hand, if Y° < a for all a > - oo, then lim sup Re L, < inf Y° tt=

oo,

a

so the limit exists in this case also, and we have proved (6). Combining (6) with the results of Section 2.11, we can early conclude that (7)

If f E N, then lim Im L, exists a.s. et=

Proof Let U, = Re L V = Im L,. Since L, is locally an analytic function of B the variance processes , and , are equal. It follows from a result in Section 2.11 that {lim U exists} _ {, < oo}

={, oo, observing that Re L(t A T,) is bounded above, and using Fatou's

132

S


lemma (upside down), we see that E Re L(-r,),-!f 0, so E Re L(T,)- < E Re L(t,)+, and it follows from another application of Fatou's lemma that El logIF,II < cc. Exercise 1

Jensen's Formula.

(a) Show that iff(z) is analytic and free from zeros in IzI < r, then d9 logl f(0)I = Jloglf(rei0)p.

(b) If f has zeros a1, ... , a in IzI < r (multiple zeros being repeated according to their multiplicity), then apply (a) to

rz - a,z

F(z) =f(z) i=1 r(z - a;) which is free from zeros and has IF(z)I = I f(z)I on IzI = r, and conclude that J logl f(re`B)I

d9

= logIF(0)I

= logl f(0)I - `Y log (L_) What is the probabilistic interpretation of the correction term? Note:

As the date on the Nevanlinnas' paper suggests, these results are class-

ical. For an analytic treatment, see Duren (1970), pages 16-17. (1) is his Theorem 2.1; (2) and (7), when combined, are his Theorem 2.2. The proofs given here, like many of the results in the book, are "new" but can be traced directly to results in the literature. The idea of taking logs along the Brownian path is a natural generalization of a proof in Section 6.4 (where we take nth roots), a tactic that in turn was suggested by a remark in the introduction of Getoor and Sharpe (1972).

5.4 Two Special Properties of Boundary Limits of Analytic Functions In this section, we will prove two results that show that the possibilities for the boundary behavior of a function analytic in D are much more limited than those of a function harmonic in D. The first result is a very strong uniqueness theorem (that fails miserably for harmonic functions). (1)

If an analytic function f has nontangential limit 0 on a set of positive measure, then f =_ 0.

Remark: This result is due to Privalov (1924). IffeN, (1) is a consequence of (8) in Section 5.3, but the f here is not assumed to satisfy any growth condition.

5.4 Two Special I'ropcrtlee of Boundary l.lmlte of Analytic Functions

133

Proof If f has nontangential limit 0 on a set of positive measure, then it follows from (2) in Section 5.2 that P(f(B,) - 0 as t T T) > 0. If f is not = 0, then .f is nonconstant, and it follows from Levy's theorem that f(B,), t < T, is a time change of Brownian motion run for an amount of time

f

I

The last conclusion, however, contradicts the first, since if W is a twodimensional Brownian motion, then

Px(W=0for some t>0)=0 and

P0

,(W-). 0ast-.oo)=0.

Remark: The proof above is taken from Burkholder (1976). As he says on page 147: The advantage of a good probabilistic proof is not that the technical details become easier, although this is sometimes the case, but that the underlying

ideas become more transparent. For example, the truth of Privalov's theorem becomes evident once we see that it is a question of whether or not Brownian motion hits a particular point in the complex plane with positive probability. By using the fact that two-dimensional Brownian motion is recurrent, we can get another result about the boundary behavior of analytic functions that is due to Plessner (1928). As in Chapter 4, it is easy to prove the probabilistic analogue. (2)

With probability 1, either (i) lim f(B,) exists rTT

or (ii) for all e > 0, { f(Br) : t e [T - E, T) } is dense in C.

Proof If we let U, = Ref(B,), then the first event is a.s. equal to { < 00} and the second is a.s. { (U>t = oo }. In order to translate this statement into analytical terms, we need a lemma. Let S2,(O) be the Stolz domain defined in Section 5.2, and if A is a Borel set, let S.(A) = UeEASS(O) (3)

For each a < 1 and Borel set A, as e ---> 0, P(BT e A and B, e SS(A) for all t e [T - e, T)) --> P(B, e A).

Proof Let h(z) = PZ(B, E A). This result is an immediate consequence of two observations:

1.14

S

Complex Brownian Motion and Analytic Function.

(a) A simple generalization of Calderon's argument in Section 4.5 shows that

there is an e > 0 (that depends only on a) such that if z 0 S2(A), then h (z) < 1 - E. (b) It follows from results in Section 5.2 that h(B1) -> lA(BT) as t T T.

With (3) established, we are ready to prove Plessner's theorem. (4)

Except for a set of 0 of Lebesgue measure zero, either

(i) f has nontangential limit at e`° or (ii) for all a > 0j (S,,(0)) is dense in C.

Proof Let A be a set of positive measure such that for all 0 e A, limf(z) does not exist as z - e'° in S2(0). If we let Po be the law of Brownian motion conditioned to exit D at eie, it follows from results in Section 5.2 that if 0 e A, then Po (1 m f(B,) exists) = 0, Ift and it follows from (4) in Section 3.2 that PO Clim f(B1) exists, BT E Al = 0.

Combining the last result with (2) above shows that for all e > 0, P0(BtaA, {f(B1) : tE[T - E, T)} is not dense in C) = 0, which in view of (3) implies that f(S,,(A)) is dense in C. Let A,, = {0 : lim f (z) does

not exist as z --> e`° within S2(0)}. Since the last conclusion holds for any set A c A. that has positive measure, a simple argument (which is left to the reader) shows that for a.e. 0EAa, S,, (0) is dense in C. Note: The main ideas in this section, including the fact that boundary values of analytic functions can be studied via Levy's theorem, are due to Doob (1961). As usual, however, we are not using the original source. Our proof of Privalov's theorem follows Burkholder (1976). Our proof of Plessner's theorem is from Davis (1979a) ; specifically, (2), (3), and (4) are his Theorems 4.2, 4.3, and 4.6. For classical proofs of these results, see Theorems 1.9 and 1.10 in Chapter 14 of Zygmund (1959).

5.5 Winding of Brownian Motion in C - {0} (Spitzer's Theorem) Let B, and B' be two independent Brownian motions with Bo = 1 and Bo = 0, and let C, = B, + iB' be a complex Brownian motion. Since Ct almost surely never hits 0, we can define the total angle swept out up to time t to be the unique

5.5

Winding of Brownlen Motion In C- 10} (Spltzer's Theorem)

135

process 0, with continuous paths that has 00 = 0 and sin(B,) = B,21I('I for all t > 0. In words, the process 0, records the angle and keeps track of the number of times the path has wound around 0, counting clockwise loops - 2n and counterclockwise loops +2n. Spitzer (1958) proved the following limit theorem for 0,: (1)

As t - oc, P(20,/log t< y) --> Jy

dx

+xZ

l

1

n*

Spitzer's proof is ingenious but requires a lot of computation (see Ito and McKean (1964), pages 270-271, for a succinct version). In this section, we use the conformal invariance of Brownian motion to give a simple proof of this result. The proof given below is based on Durrett (1982) but incorporates some

improvements that I learned from Messulam and Yor (1982) and an earlier work of D. Williams (unpublished).

Let D, = A, + iB, be a complex Brownian motion with Do = 0, and let F = exp(D,). By Levy's theorem, if we let

a,= J0 Iexp(D.)12ds and

yu=inf{t: a, >u}, then C. = F(y"), u > 0, is a complex Brownian motion with Co = 1. The first advantage of constructing C, as F(y,) is that we can easily write down the angle process 0, = B(y,). The second, and more crucial, observation is that (2)

a, = f Iexp(DS)I2ds = f exp(2A5)ds, 0

so y and B are independent. Let S = y(e2u). Time e2" in process C corresponds to time S in process D. The main idea of the proof is to show that 9(e2u) - B(y(e2u)) = B(S") is approximately B(Tu), where T. = inf{t : A, > u} (our guess is motivated by (2) above). Once we show this, (1) follows immediately, because a simple scaling gives a

(3)

B(Tu)/u

B(T1),

and the right-hand side, which is the hitting distribution of {(x, y) : x = 1}, is (by results in Section 1.9) a Cauchy distribution with parameter 1. To show that B(SS) is approximately B(Tu), we start by showing that S. is

approximately T. The following estimate is crude, but it is sufficient for proving (1). (4)

Ife>0,then asu-+oo P(TJ(l_o < S. < Tuo +E)) -> I.

1.36

5

Complex Brownian Motion end Analytic Function

Proof To get the lower bound, observe that P(SU u(1 + e/2)}. From (2), it follows that a(To(1+E)) ? LE(u) exp(2u(1 + E/2)),

so we have (5b)

P(S,, > Tu(1+e))
Le(u)

> 0),

a

which 0 as u -+ oo since LE(u) uLE(1). With (4) established, completing the proof of (1) is routine. If we let

DE(u) = sup{IB(t) - B(T)I/u: te[T.(1_E), T(1+e)]} and recall that O(e2u) = B(SS), then (4) implies that for fixed E > 0,

P(

B(TT)

O(e2u)

U

U

as u -+ oo. Now the distribution of A(u) is independent of u and tends to 0 as 6 - 0, and (3) says that B(T,)/u d B(T1), so it follows from a routine computation that 0(e2u) = B(T1) u

as u --* oo

stands for convergence in distribution). With a little effort, (1) can be improved to a limit theorem that describes when the winding occurs. Using the notation introduced in the proof of (1), we can state this result as the following : (where

(6)

As u -> oo, the finite-dimensional distributions of O(e2ò)/u, t > 0, converge to those of B(T,), t >- 0 (a Cauchy process with parameter 1).

To understand what (6) says, look at Figure 5.3 and observe that if u = 1 - E and E is small, then B(T1_E) is almost B(T1). Therefore, if N = ego, then most of

5.5

Winding of Brownion Motion in ('

-

{0} (Spitzer's Theorem)

137

time S. zT

u/2

u

1

Figure 5.3

the winding has occurred by time e2u(1-`) = N1 ` = o(N). To see when in [0,N'-'] most of this winding occurred, observe that if t1 < t2 < < t.(a) are the times of the jumps > 6 in magnitude, the winding at time N is almost a sum of contributions from times proportional to N`1, N`2, ... , N`n(b).

Technical Remark: Some probabilists may have expected (6) to contain a claim of weak convergence as a sequence of random elements of D (the space

of functions that are right continuous and have left limits-see Billingsley (1968) for details). We have refrained from doing this because the claim is false : C = the set of continuous functions that is closed in the usual (Skorokhod J1) topology, so we cannot have weak convergence. Refinement 1: Rate of Convergence In his unpublished paper, Williams said, "The reader will find it a useful exercise to see how good a `rate of convergence' ... he can get by refining our method." We will give a partial solution below

and leave the rest of the details to the reader. Looking back at the proof, we see that the difference between O(e2u)/u and B(T")/u = B(T1) is determined by the size of S" - T", which in turn was controlled by the estimates (5a)

P(Su < T"(1-E)) _< P(ez"E < T

(5b)

P(SS > T.(,,,) r, f(B,), 0 < t < T, is not homotopic to 0 in C. For then we can let a = inf {t > i : C, e 0}. There are then a number of ways of

proving (3). Davis gave one proof (1974) and later suggested another one ((1979a), page 920). The proof given below is our version of his second proof. Most of the effort is spent introducing a lot of notation for keeping track of the tangling around the two points. We start by introducing six sets (see Figure 5.4),

Ao={X+iy:-1 <X0}

A4={x+iy:lxl=l,y 0, let Ti+i = inf{t > 7 : f(B,) c A - A;,,} where the sequence of indices i is defined by f(BT) a Ai. for n > 0 (recall that we have assumed f(0) = 0, so io = 0). It is clear that if we are given the sequence of indices, then we can compute the homotopy class of the path, but this sequence contains a lot of irrelevant information, so we define a reduced sequence inductively by the following procedure: (a) At time 0, we write 0. (b) At time T,,, if there is only one number in the string, we add i to the end of the string. (c) At time T,,, if there are k > 1 numbers in the string and the new number i is the same as the next to last, we erase the last number. Otherwise we add i to the end of the string.

Convention (c) allows us to delete irrelevant loops from the sequence of numbers. (To check your understanding of the procedure, you should try the example drawn in Figure 5.4. Its reduced sequence is 03240.) Besides cutting down on the length of the sequence, the erasing also drastically reduces the number of possibilities for the sequence. The reader can easily check that between two successive occurrences of 0, the sequence must be one of four types :

5.6

Tangling of Brownian Motion In C- { - 1, I } (Picard's Theorem)

141

(i) 03240, 0324130, 032413240, .. . (ii) 03140, 0314230, 031423140, .. . (iii) 04230, 0423140, 042314230, .. . (iv) 04130, 0413240, 041324130, ... .

From the observations above, it is clear that we can compute the homotopy class of the path from the reduced sequence and that to prove the theorem it suffices to show that the length of the reduced sequence -> oo almost surely. The

fact that the reduced sequence - oo can be made obvious by the observation that if the last two digits of the string are 03 or 04, then the probability we will add a number is greater than 1/2, while in any other case the probability we will add a number equals 1/2. To make the last argument precise, we would need to obtain an estimate on

how frequently i = 0. Rather than trying to do this, we will adopt a slightly different approach to avoid this difficulty. Let No = 0, and for k >- 1 let Nk = inf{n > Nk_1 : i = 0 and the reduced sequence corresponding to iNk-1 ... i is not 0}. At time Rk = TNk, the number of zeros in the reduced string will either be the same or differ by one from the number of zeros at time Rk_1. To describe the conditions that result in the three possible outcomes, we need some definitions. Any of the sequences that appear in the lists (i), (ii), (iii), and (iv) above is called a zero-block. A zero-block b2 is said to be the inverse of the zero-block b1 if b2 is obtained by reading b1 backwards, that is, if b1 = 0324130, then b2 = 0314230. The reason for the name inverse is explained by the next result. Let ak be the last zero-block in the reduced string at time Rk, and let bk be the zero-block that If bk = the number of zeros at time Rk is obtained by reducing ... minus the number of zeros at time Rk_1, then

1.I

-1 _ k

,'Nk.

if bk is the inverse of ak

if the first nonzero digit of bk (which must be 3 or 4) is different from the last nonzero digit of ak 0 otherwise.

+1

At this point we have almost, but not quite, avoided the difficulty referred to above. It follows from the strong Markov property and symmetry that 1I,

(Rk)) = 2

PO(Yk+1 = - 1 I.F(Rk)) = h(ak,f(BRk)),

where h is a function that has h(a, x) < 1/2 for all a, x. It is somewhat unfortunate that sup h (a, x) = 1 /2 (contrary to what we claimed in an earlier version of this proof), but this difficulty is easy to remedy. The culprits are the zero-blocks of length five : 03240, 03140, 04230, and 04130. In this first or second case we have lim h (a, x) = 1/2, .1 xt1

142

s

Complex Brownlin Motion and Analytic Functions

since the next winding will, with a probability approaching 1, be around 1, and symmetry implies that the two possibilities have equal probability. A similar argument shows that in the third or fourth case

lim h(a,x) = 1/2,

x1-1

but the other zero-blocks, luckily, are our friends. It is easy to check that if a is not one of the four zero-blocks mentioned above, then sup h (a, x) = C(a) < 1 /2

XE(-1,1)

because each such a involves winding about both -1 and 1, and hence has lim h (a, x) = lim h (a, x) = 0. x11

xl-1

Combining this with the observation that if a is a sequence of zero-blocks with length co and x e (-1, 1) then h(a,,, xa) - 0, it follows that if we exclude the four bad zero-blocks then sup h (a, x) < 1/2. a, X

To complete the proof now, it suffices (as we will show below) to ignore the

bad zeros and count only the good ones. Let No = 0 and for k >- 1 let Nk = inf{n > Nk_1 : i = 0 and the reduced sequence which corresponds to LNk_1 ... L contains a zero-block that has length greater than 5}. If we call a zero that is

at the right end of a zero-block of length greater than 5 a "good zero," then arguments given above show that at time Rk = TNk the number of good zeros in the reduced string will be the same or differ by one from the number of zeros at time Rk_1, and, furthermore, if we let a', b;,, and k be defined in the obvious way then we can repeat the arguments above with "primes" attached. Having done this we come at last to a point where (at last!?) we can use the strong Markov property and symmetry to conclude that Pork+1 = 1 J.F(Rk)) = 1/2

{(BR,))

where h' is a function that has h'(a', x) < 1/2 for all x and a'. The last observavation makes it clear that Sk = ' + + k -> oc a.s. To prove this, let rik be a sequence of random variables that have nk =k on 0} and are defined on {lk = 0} in such a way that

Po(ilk = -1

I-IF(Rk-1)) = q

If we let S,' = nl +

+ qk, then S is a sum of independent random variables,

so

kk2-q>0a.s. and it follows that Sk > Sk -* oo a.s.

S.6

Tangling of Brownian Motion In C- { - I, I } (Pkard'. Theorem)

143

Problem I Having shown that the length of the reduced string that describes the homotopy class -+ co a.s., it is natural to inquire about the rate ot'growth. 11' Lt denotes the length at time t, then it is trivial that

Lt1?4min(

10"

where O is the winding around a defined in Section 5.5. This suggests (to me at least) that I Ltl should grow faster than log t. What is the right rate of growth? Someone who understands Jacobi's modular function well can probably solve this problem by looking carefully at how Brownian motion exits H (see McKean (1969)). Once one proves Picard's little theorem using Brownian motion, it is natural to try to prove his "great theorem" as well. This theorem states that if f has an essential singularity at z = a, then in each neighborhood of a, f assumes

each complex number (with one possible exception) an infinite number of times. See Conway (1978), page 300, for the relevant definitions and an analytic

proof. Davis (1979a) has given a probabilistic proof, but since his proof still requires a fair amount of complex analysis, we refer the reader to pages 928-930 of his paper for the details. A second type of extension of the results above is to ask if Brownian motion in C - {-1, 1} gets tangled in other senses as well. McKean (1969) claimed on

page 112 that "the plane Brownian path undoes itself i.o. as t T 00 from the point of view of homology with integral coefficients." This is incorrect. Lyons and McKean (1980) "corrects this error and takes the matter further." Note: I would like to thank Jim Pitman for pointing out an error in the original version of the proof given above. He is not to be blamed, however, if my "first aid" fails to save the victim.

6 Hardy Spaces and Related Spaces of Martingales

6.1

Definition of H", an Important Example Let D = {z : I z I < 1} be the unit disk in the complex plane. A function u that is harmonic in D is said to be in h", 0 < p < oo, if

I u(re`BI d(0)
0. To find u = Rep and v = Im p, we write

l+z 1-z P(z)1-z 1-z

1+z-z-IzI2

I1-z12

so

u(z) = 1 - IZI2 I1-z12 144

v(z) - 2Imz I1-zi2*

6.1

145

Definition of NP, an Important Example

We have seen u(z) many times above. It is the probability density (w.r.t. n) that a Brownian motion starting at z will exit D at 1. From this interpretation of u and the strong Markov property it follows that 2n

u(re`B) dn(O) = u(0) = 1 0

for all r < 1 and also that if 0

0, then as r -+ 1, u(re`8) -+ 0. Combining the last two observations shows that if p > 1, fo2n

ju(reò)JPdn(9) -' oo

as r - 1 (if not, then u - O in L 1), so u e h 1 but u O h P for any p>

1.

At this point, we might hope that p e H1. In Section 6.5, we will see that the limiting behavior of u precludes this, but for the moment we do not know this, so we will show that p 0 H' by computing f02 n

I v(reà) I dn(O) = 2 fon

2r sin 9

dO

1 - 2r cos O + r2 2n it

_ log(1 - 2r cos 9 + r2) io

=2logl I+rl-.oo

as

It doesn't miss by much since lim

2r sin O

sin O

,t, 1 + r2 - 2r cos 9

O

1 - cos 9

0

'

and as 0 -+ 0,

-

sing 0+0(03) 1-cos0 Z+0(94)

2

0'

so we have lim v(re`B) e LP

rt1

for all p < 1.

As our experience with u indicates, this result is not, in general, enough to conclude that v e hP (in that case limit 1 u(re`B) = 0 e L°°). This time, however,

nothing bad happens. For r > 0, 1 + r2 > 2r z 2r cos 0, so I2r sin 01 < 12r sin 0l 1 + r2 - 2r cos O - 2r - 2r cos 9

-

I sin o l

1 - cos 9

and if p < 1, it follows that dd(v)
I and p - 1

147

Let B, be a Brownian motion starting at 0 (defined for convenience on our

special probability space (C, ')) and let F be the filtration generated by IS,. Let X* = sup, IXI and let . #P = {X: X is a local martingale w.r.t. 9 and EIX*IP < oo}. If p >- 1, then EIX*I < oo, so X is uniformly integrable, and a standard martingale convergence theorem implies that

(i) as t oo, X, (ii) X, = E(Xao I

X.. a.s.

)

It is clear that El X. I P < EIX * I P. When p > 1, there is a converse inequality. (1)

If Xc

P with p > 1, then

EIX,IP < EIX*IP
2)<E(X,;X,>A) without any restriction on X0 or A, since in this case E(X,I .T) >- XT and

XT>A on {TA}. To obtain (1) from (2), we integrate and use Fubini's theorem (everything

148

6

Hardy Spaces and Related Spaces of Mortingsloa

is nonnegative) as follows:

EX°=

j

p21P(X,>A)dl

0

Jp2i (A1 `

o

=L f Xr+ P

P- l

J

XdP I d) x r> x}

rp2p-2

(

\o

dA dP

fn X,+XP-1 dP.

If we let q = pl(p - 1) be the exponent conjugate to p and apply Holder's inequality, we see that the above q(EI X,+ IP)1/P(EIX[I P) 1/9

At this point, we would like to divide both sides of the inequality above by (EI XrI p)1/9 to get

(*)

(EX/') 1/P < q(EI X,+I P)'/P. of

Unfortunately, the laws arithmetic do not allow us to divide by something that may be oo. To remedy this difficulty, we observe that P(X, n n > 1) < P(X, > 2), so repeating the proof above shows that (E(Xr A n)P)1/P < q(EI X,+I P)1/P

and letting n -+ oo proves (*).

The last step in proving (1) is to let t -+ oo and use the monotone and dominated convergence theorems to conclude that if we let X = sup, X,, then we have P

(1')

EIXIP
- 1, . IIP defines a norm on .. #P. From formula (1) and the discussion preceding it, we see that X -> X a(B, : t > 0) = W. maps dl" one-to-one into a subspace of where Since we only allow continuous martingales in . P, the image of .1/" under this map is, in general, L"(Jm), however, we also only consider Brownian filtrations, so it follows from (1) and results in Section 2.14 that if p > 1, X -+ X"" are maps .,12p one-to-one onto L"(F.), and furthermore that dl" and equivalent as metric spaces. When p = 1, the last two claims are false. Rather than giving an explicit counterexample, I will prove the following result, which II

I learned from Gundy (1980a). The theorem originally appeared in Gundy (1969). (3)

If X > 0 is a continuous martingale with Xo = 1, then

6.2

First Definition of .N°, Differences Between p > I snd p - I

149

EX* Z 1 + EX log' X., where X,. = lim X, (which exists, since X >_ 0).

Proof Let TA = inf{t : X, > 2}. If 2 > 1, then X(TA) = 2 on {Tz < co {X* > Al, so it follows from (2) that 2P(Tz < t) = E(X l(T

< )).

Now as t -+ co, Xt --> X. and 1(TZ 1 + Jf 2 'E(X.,) 1(xm>z)) d2 1

= 1 + E (X. f

)

-11(x1O>z) d2

= 1 + E(X. (log X.)+). Let M1 be the image of , " under the mapping X - Xc. The last result shows that if Y is a nonnegative random variable with EY = 1 and EY(log Y) +

= co, then Y e L 1 - M 1. The next result shows that these are the only nonnegative random variables in L' - M1. (4)

If X > 0 is a continuous submartingale and X7 = sup { Xsl : s < t}, then EX,* < 2(1 + EX, log+ c).

Proof To simplify the computation, we will first transform the estimate obtained in (2). (5)

Under the hypotheses of (2) or (2)', if I > 0, then 2P(X, > 22) < E(X,; X, > 2).

Proof By assumption, we have

22F(X, >21)<E(X,;X, >22) <E(X,;X, >2)+E(X,;Xt 0 a.s., so X, - E(X,,,) I.9;) and X0 21.

6.2

Flat Definition of .,A", Differences Between p > I and p

151

I

Note: As the story unfolds, the reader will see that there is a close contlccl ion between this example and Rep, where p(z) = (1 + z)/(1 - z) is the function we studied in Section 6.1. If we wanted to, we could define.*' P and 2P for p > 1 in the obvious ways, .)VP = {martingales X with supElX,IP < oo} Y P = {X, = E(X I. ) with X e LP},

but there is no reason to do this. Standard results from martingale theory imply that V-P = IfP = . P for p > 1. Because of this fact, we will not consider .%° and IfP, p > 1, below, and we will simplify our notation by defining IIXILr = SupEIXI

IIXII,=EIXXI if XX=E(XWI.'). Another thing that might occur to you is to define A log . ( = {martingales X with sup EI X I log' X < oo }. The proof of (6) given above shows that .7t' log _V' c dl', but it is easy to show that the inclusion is strict. Let Z >_ 0 be a random variable with EZ < co and EIZI log' IZI = oo. If we let B, be a Brownian motion that is independent of Z,

letT=inf{t:B, [-Z,Z]}, and let

X* = Z, so Xede', but

urn EIXI log IXI = EIZI log' IZI = oo, t-OD

so X O X' log X . Generalizing this example, we see that . # 1 cannot be characterized as {martingales X with sup Et/i (X,) < oc } for any 0. Two formulas sum up this section :

.''1

If1 Z) .,/f1

AT log JY

.'P=YP=_#P for p> 1. Exercise 1

Let

If°° _ {X: X, = E(XI,,), XeL°°} ..lf°° = {martingales X : X * e L°° }.

Show that 2°° = W°°. Remark: It is interesting to note that although the proof of (1) is very simple, the constant is the best possible. For a proof of this assertion and a lot more, see Dubins and Gilat (1978). They conjecture that there is no martingale that

is * 0 and has EIX*IP=

p-1 PEIX

IP

1:52

6

Hardy Spacer and Related Spaces of Martingaka

Pitman (1979) has proved this for the case p = 2 and has shown that, although the inequality is sharp, it can be improved to

E sup XX+ - inf X,

(

t

I

2

/

< 4EX.2.

6.3 A Second Definition of #p In this section, we will introduce a second equivalent norm for the .. #P spaces. This norm is the analogue of the area function in analysis and, as we shall see below, it is much easier to work with in many circumstances. In Section 2.1,

we associated with each continuous local martingale X, t > 0, a predictable increasing process <X>,, with <X>0 = 0, that makes Xr2 - <X>, a local martingale. Since t <X>, is increasing, lim,. <X>t exists. If we let <X>,, denote the limit, and if p > 1, we can define a norm by letting 112IIP>

where II YII c denotes the L° norm (EI YIP)1IP (the 1/2 is necessary because , = c2<X>,)

One rationale for this definition can be found in Section 2.11. If we let yt = sup{s : <X%< t}, then X(y,) is a Brownian motion run for an amount of time <X>,,, so the Brownian scaling relationship B, = t112B, suggests that X* and <X> 0I2 should be about the same size. It is a delicate problem to formulate a result which makes this precise. If Xt = then <X>,. = 1, but since P(X * > A) > 0 for all 2, there can be no inequality of the form P(X * > 2) <AP(<X>1/2 > 2/B). The other direction is just as bad. If T = inf{t : IB,I = 1} 00 and Xt = BT, then X* - 1, but A) > 0 for all A. To get around these problems, we change the left-hand side of the desired inequalities. (1)

Let /3 > 1 and S > 0. If T is a stopping time for Brownian motion, then (la) P(B* > PA, T112 < 62)
A) z

(Ib) P(r1/2 > #A, B* < 62) < pea

1

P(T1/2 > 2).

Remark: As we will see in the proof of (2), the key point here is that for some /3 < oo, we have an inequality of the form P(f > #A, g < 6A) < C0P(f > A),

where C,, -+0as6-.0. Proof It is enough to prove the result for T bounded, for if the result holds for T A n for all n > 1, it also holds for T. Let S1 = inf{t : I B(T A t)I > Al. Let S2 = inf{t : IB(T A t) I > fit}. Let T - (S2)2.

6.3

153

A Second Definition of ..M'

It is easy to check that if B* = sups,, IB3I, then P(B* > $1, T1/2 < 61)

(fi-I)2) < (/3 _ 1)-2.1-2E(B(T A S2 A T) - B(T A S1 A T))2. Now if R1 A),

proving the first inequality. To prove the other inequality Let S1 = inf{t : (T A t)112 > .1}. Let S2 = inf{t : (T A t) 112 >

Let T = inf{t : IB(T A t)I > 6.1}.

Again, it is easy to check that P(T 1/2 > #A, B* < 62) < P(T A S2 A

T - T A S1 A T> $2)2 - 22)

A) < C$,KP(g > A) for all A that satisfy P(g > A) < KP(g > $2), $, K > 1. The next result shows why we are interested in good A inequalities. First, we need a definition. A function cp is said to be moderately increasing if /P is a nondecreasing function with 9(0) = 0 and if there is a constant K such that

p(21)0. Examples

(i) T(x)=x",00. Proof To prove the result, it suffices to show: (3)

If X, Y > 0 satisfy

P(X>22, Y2) forallb>0 and 9 is a moderately increasing function, then there is a constant C (that depends only on the growth rate K) such that

Etp(X) < CE9(Y). Proof It is enough to prove the result for bounded gyp, for if the result holds for cp A n for all n > 1, it also holds for gyp. T is the distribution function, a measure on [0, oo) that has

T(h) = foodtp(2) = J

1(h>A)d(p(2). 0

If Z is a nonnegative random variable, taking expectations and using Fubini's theorem gives E4p(Z) = E f

l(Z>A) dgp(A) =

JP(Z> A) d4 (2).

0

From our assumption it follows that

P(X > 22) = P(X > 22, Y:5 52) + P(Y > b2) < 62P(X > A) + P(Y > 62), or integrating dq(2), E4 (2-1X) =

f

P(2-1X < 2) dq (2)

0

< 62E9 (X) + Egp(b-1Y).

Pick 62K < 1 and then pick N > 0 so that 2' > b-1. From the growth condition and the monotonicity of gyp, it follows that

6A Equivalence of Ho too Subspace of .No

115

E9(6-'Y)!5 KNEcO(Y)

Combining this with the last inequality and using the growth condition gives E(p(X) < KEcp(2-1X) < K62Etp(X) + KN+1Eco(f

Solving for E.p(X) now gives KN+1

Ecp(X)
1, it follows from Doob's inequality that 12p2p

1

)2"EIGI2P

Taking a supremum over r and applying the monotone convergence theorem gives

EI F*I P = EIG*(r)I2p
1/n by taking nth roots, with the result that p

EIFF*Ip < np pT1dp(f) Letting n - oo, the last inequality gives El F* I p < edp(f ),

proving (2) with C = e. The result in (2) shows that the Hp norm of f and the .,#,! norm of f(B,), t < r, are equivalent. As we mentioned above, we want to go one step further and consider Hp as a space of real local martingales by using the mapping f - Re f(B,), t < T. Since every harmonic function u has a conjugate harmonic function u defined by the requirements that 4(0) = 0 and f = u + iu is analytic in D, the mapping is one-to-one on He = { fe Hp : f(0) = 0}. The next result shows that if we restrict our attention to this subspace, then the Hp norm off and the .A'' norms of Ref(B,), t < r, are equivalent (i.e., (1) holds). (5)

There is a constant C e (0, oo) (that depends only on p) such that for all f e Ho, EIU*Ip < EIF*Ip < CEIU*IP.

Proof Since l a l < (a2 + b2)1/2 for all a, b, we have I u (z) l < I f (z) I for all z, proving the inequality on the left. To prove the other inequality, we observe that if we let U = u(B,), t < r, then it follows from results in Sections 2.11 and 6.4 that E(U*)p < CEpl2 All,

where C depends only on a, and integrating gives zn

Eo(UU*)P < C

l NQul P(0) d7r(0) 0

(a result that holds for any measurable u). To prove the other inequality, we use (2) from Section 4.3 (which generalizes easily from H to D). This result implies that there is an E > 0 such that if z e S8(0), then

Po(B,, 0 < t < r, makes a loop around z) > E, so it follows from the maximum principle that Pa(U* > Nau(0)/2) >_ E,

and hence that E0(U*)P > E f lNaull(0)dn(0). Combining this with the other inequality and (1) above gives the maximal function characterization of HP cdd(f) 1/p and consider g'1" to reduce the result to the easy case where p > 1. The reader should note that taking nth roots to reduce to the trivial case p > 1 was also the key to the proof of (3) in Section 6.2, but in that proof, we

used the fact that Brownian motion did not hit zero to construct a pathwise nth root, so we did not have to factor out the zeros.

6.6 Martingale Transforms Martingale transforms are a natural generalization of the following: Example 1 Let f = u + iv be an analytic function with f(0) = 0, and let B, be a complex Brownian motion starting at 0. Ito's formula implies that u(B,) = f Vu(BS) dB. 0

v(B,) = JVv(Bs).dBs, 0

and the Cauchy-Riemann equations say that Vv =

(_0 0)Vu,

so if we let

H,=Vu(B,) A= (-0

11 0

'

then we can write

u(B) =

JJJ.dB , 0

v(B) = JAI-I.dB3. 0

The last equation obviously makes sense if B is a d-dimensional Brownian motion, H is a locally bounded (R'-valued) predictable process, and A is any d x d matrix. To define the transform of a general local martingale X in this setting, we now recall that since we have assumed that our filtration is generated by a Brownian motion, it follows from results in Section 2.14 that

Xt=Xo+ f 0

Mmrtingde TnMformw

6.6

163

We can therefore define the transform of X by A as

(A*X)t= (whenever this makes sense, e.g., if H is locally bounded). In this section, we will study properties of martingale transforms as mappings between the V' spaces. The results we prove here are analogues of classical results about conjugate functions that we will prove by probabilistic methods in Section 6.7. The first and most basic result is: (1)

If p > 0, X -. A * X is a bounded linear transformation from #P to M". Proof If Xt = 0

then

(A*X)t= f 0

so

E(sup G,)'IP = E(sup F) > E(sup IXI ), so X E .4f 1. If we keep track of the constants, we get

E(sup I X I) < E(sup G,)' '

. 1

m

2 i=o

o

The first term on the right-hand side is a local martingale, so to prove the result we need to show that we can pick p < 1 so that the second term plus the third term is 0. If X= f Hsd Bs, then

<X= JAHJ2ds o

and

<X`, X'>' = J(AiJ1AJ1)ds. o

To complete our proof, we need to show: (4)

22 - 1) Yi; 4xtx;(At4P, A;(P) + (1 + 1x12) > 1Aj4,12 > 0 1

(P

for all x e R" 1 and cp e Rd. To do this, we observe that if B,; is the angle between xjA;cp and x;A;cp, then Y(x;A;cp,x;A;(p) _ Y1x;Aicp1lx;A;cpl cosBtj 1, this equation defines a norm on hP. Our first result, due to M. Riesz (1927), shows that if p > 1, u -+ u is a bounded linear map from hP to hP. (1)

If p > 1, then there is a constant C (that depends only on p) such that Ilullp < CIIuIIP.

Proof Let U = u(B,), t < T, and let U = u(B,), t < T. If llullP = 1, then Doob's inequality implies that E(U*) < (p/(p - 1))P, and it follows from results in Section 6.4 that Ilullp < El

U*IP < KEI U*IP,

so (1) holds with C = K11P(pl(p - 1)). To be fair, we should observe that (1) is easy to prove analytically-there is a simple argument using Green's theorem that is due to P. Stein (1933) and that gives a much better value for Cp, namely,

Cp=2(Pl(P- 1))lh' 1 0} = {re'B : r > 0, 0 < n/2}, f,, maps D one-to-one onto the cone IF,, = {reÒ : r > 0,

0 < fln/2}. From the last observation (and Levy's theorem), it follows that j,,(Bt), t < 7, is a time change of a Brownian motion C, that starts at 1 and runs until it leaves F''. Drawing a picture (see Figure 6.2) reveals that if up _ Re f,, and vp = Im f,, , then

limIvP(B) I = tanfl imlu,,(BJ)I. tT=

If 1
A}) - 2/n2 as A -+ oc.

The next result, due to Kolmogorov (1925), shows that (up to a constant multiple) this is the worst behavior we can have for ueh1. (2)

There is a constant C such that if ueh1, then n({e: lu(eà)I >_ 2}) < Ca.-11IuII1.

Remark: In the jargon, this is called a weak type (1, 1) inequality. We will give Davis's (1974) proof, because it has the advantage of identifying the best

constant C = 1.347... . Proof Let ZZ = Xt + iY be a complex Brownian motion, and let a = inf{t : I Y I = 11. Since t - I X, I is a submartingale and the stopping time a depends

only upon the Y component, it is easy to see that if x and - 1 < y < 1 are real numbers, (i) EoIXaI >EIYIXal (ii) Es+i,IXal > EirI X1I.

The next step is to show :

(iii) Let x be a real number and fi a stopping time for Z. Then EXIX API > EOIX.IPX(, ? a).

Proof By (ii) and the triangle inequality, EOIXXI < E.IXal < EXIX..pl + EXI Xa - X.A$I.

By the strong Markov property, Exl X8 -

Xa

-

and we have, using (i),

a> (the centering makes it look like starting at 0, but I Y,,, until a). Combining the results above, we have EOIXQI <ExIXZAPI +EoIX2IPP(«> fi),

proving (iii).

> 0, so there is less time

174

6

Hardy Spaces and Related Spaces of Martlnaalee

We are now ready to prove Kolmogorov's inequality. Let f = u + iu. By Levy's theorem, if we let a(t) = f O I f'(B5)IZ ds fort < T, then Z,(,) = f(B,), t < T,

defines a Brownian motion run for an amount of time y = Q(T). If we write Z, = X, + iY and let T, = inf{t : IB,I > r}, then IX",(tr)I = iu(BtAT,)I is a bounded submartingale, so EI X.Aa(t,)I 1})=P(a }) < 2IIu111

7r(l

+y2),

JA

and this constant is the best possible.

Hint: The function that shows that this inequality is sharp is

f(z) = it + (1 -

z)'

i+

which maps D one-to-one onto {x + iy : x > O, y < 2} and has f(0) = 1. In this case, the reflection principle shows that

n({O:u(e`")>2})=2 f Z

dy + y2). n(1

Applying (2) to u(rz) shows that if 1lull1 = 1, then 7C({O :

I u(re")I > A}) < CA-',

soifp.l})dA

Jp1 C` d= 1

P,

and we have proved another inequality due to Kolmogorov: (3)

If u e h1 and p < 1, there is a constant C (that depends only on p) such that dd(u) < CIIull f.

The argument above proves (3), but since it is not very interesting and does not give a very good value of the constant, we will give two more proofs of (3). The first is a purely probabilistic one, and because of this, it gives a

crude value for the constant. The second is an analytic translation of the probabilistic argument and gives the best constant for positive u. Proof 2 Again without loss of generality, we can suppose that u > 0 in D and IIuhl1 = 1. In this case, u(0) = 1 and u(0) = 0, so (u(B,), u(B,)), t < 'r, is a time change of a two-dimensional Brownian motion starting from (1,0) and running for an amount of time y. Since u >- 0, y must be smaller than 71, the time it takes a Brownian motion starting at 1 to hit 0, and it follows that

176

n

1lardy Spacra and Related Spacca of M.rtlnialra

dp(u) < /;(U*)p < CE 200< -CETv1Z.

In Chapter 1, we found that the probability density of T1 is (tit) - 3/2 t- 3/2 e- 1/2t ,

l

so ifp < 1, ETf2 =

(2n)-3/2

f aD

t(p- 3)/2 e - 1/21 dt.

0

To evaluate the integral, let u = 1/2t to obtain (2ir)-3/22(1-p)12

r u-(1+p)l2e-"du J0

and observe that the value of the integral is I'((1 - p)/2) if p < 1. The reader should observe that if u(x) = Re p(x), then y = T1. This suggests that for nonnegative functions Ilullp/Ilullp should be largest for the Poisson mapping. The proof above cannot be used to prove this fact, since we have used the clumsy estimates dp(u) < EI U*Ip < CEc/2. This inaccuracy can be avoided if we abandon the correspondence and translate the proof into analytical terms. Proof 3 The idea for this proof is due to Littlewood (1926). Again without loss of generality, we can suppose that u > 0 in D and II U 111 = 1. A function f analytic in D is said to be subordinate to g if there is an analytic function co with Ico(z)I < Izl such thatf(z) = g(w(z)).

One reason we are interested in this concept is that

(i) If u > 0 and u(0) = 1, thenf = u + iu is subordinate to p(z) _ (1 + z)/(1 - z).

Proof Let w(z) = p-1(f(z)); co is analytic and maps D into D and 0 into p-1(1) = 0. Let a(z) = w(z)lz, z 0, and a(z) = w'(0) at z = 0; o is analytic in D and has I u(z) I < 1 on 8D, so it follows from the maximum principle that Io(z)I < 1 in D, that is, Iw(z)I 5 Izl. Another reason for our interest in subordination is (ii) If f is subordinate to g, then f02

I f(reà)I 'd7r(0) < I

I g(reà)I pd7r(0)

for all r < 1.

0

Proof Let B, be a complex Brownian motion and 'rr = inf{t : IBI > r}. By hypothesis, EI J (Br)I p = El g(w(B1r))I p

6.8

Inequalities for Conjugate Harmonic Functions

177

If we subject w(B,), t < z to Levy's time change, we get a Browniun mot ion B,' run for an amount of time y,. Since Iw(z)I < Izi, it follows that y, r; inf{t : IB;I > r}. Combining this with the fact that Ig(B',Tr)IPisa submurtingulc gives

EIg((o(B,,))IP=EIg(BY.)IP<EIg(Bt)IP and proves (ii). Combining (i) and (ii) proves (3) with CP = dP(q), where q(z) = Imp (z).

To compute the value of the constant, write p(z) = R(z)e"(z), where R(z) = I p (z) I and cb(z) E (- n/2, 7r/2). Since there is no confusion about which root to take, F(z) = R(z)Pe'P(z) is analytic in D. By the mean-value theorem,

1 = Re F(0) = f

Re F(reie) d7r(O)

,J n n

= f

R(re'B)Pcos(p(D(re`B))d7r(O). n

As r

1, cb(re'B) -+ sgn(9)n/2 and R(re'B) -> Iq(e's)I. Since gehP and Icos(p(D(reie))I < 1, it follows from the dominated convergence theorem that

1= f

cosP2 lim f rT1

,1

Iq(reie)IPdm(0), n

so dd(q) = sec(pn/2).

Remark: If at the beginning of the computation of the constant we take an arbitrary f with u = Ref > 0 and use the inequality I ((re') I < n/2, we get a purely analytical proof with the constant given above. This argument is due to Hardy (1928). Proof 3 shows that (3a)

If u > 0, dd(u) < sec (12) IIuIIl

(3b)

If uEh1, dP(u) < 2 sec 12 I IIuIIl

.

The Poisson kernel shows that (3a) is a sharp result. Since (3b) is obtained from (3a) by using the triangle inequality, we should expect that (3b) is not sharp, and indeed it is not. Burgess Davis (1976) solved the problem of finding

the optimal constant in inequality (3b). He showed that the smallest value for C, is IIull,,, where u = (k1 + k_1)/2. In this case, the corresponding analytic function is g(z)

1 (I +z 1-z =2 1-z+1+z

2z (1-zZ),

178

6

Hardy Spaces and Related Spaces of Mortinplea

which maps D one-to-one onto S = C - {x + iy : x = 0, IyI >- 11. To prove his theorem, Davis uses Levy's theorem to reduce the result to an optimal stopping problem for Brownian motion, which is solved by considering related discrete time problems. Since the argument is rather lengthy and the improvement on (3b) is rather slight, the reader is referred to Davis's paper for the details or

Davis (1979b) for a sketch of the proof. A. Baernstein (1978) has given a purely analytical proof of this result. Up to this point, we have only discussed u e h° for p > 1. For 1 < p < oo, the class hP was preserved under conjugation. For p = 1, this was false, but it

was almost true: uehP for all p < 1. When p < 1, things fall apart-there is an analytic function f = u + iv such that uehP for all p < 1, and yet foN (hence v 0 hP for any p > 0). The example is a randomly chosen function f(Z, (0) =

(4)

n=1

Sn(w)g(ZZn)

where g(z) = z/(1 - z2) and fin, n > 1, are independent random variables with

-1)= 1/2 (for analysts, let S = [0, 1] and n((o) = sgn(cos(2nw)), the nth Rademacher function). I claim that (4a)

For every w, Ref (. , co) a hP for all p < 1 and

(4b)

With probability 1, Yo = (0: lim,fl f(re`B, (o) exists} has Lebesgue measure 0.

The proof of (4a) is a straightforward, but somewhat lengthy, calculation and is therefore left to the reader (see Duren (1970), page 66). We will proceed, then, with the more interesting claim (4b), which is implied by the following result. (5)

Let gn(z), n > 1, be complex-valued and continuous in IzI < 1 except at a finite number of points z with IzI = 1.

If (i) Y' O, I g (z) I < oo and for all r < 1 the convergence is uniform on D, {z : IzI < r}, and (ii) for each N, as r T 1 we have oo

gn(re`B)I2. oo

uniformly in 0,

n=N

then with probability 1, f(z, (o) = Y',

(w)gn(z) has a radial limit almost

nowhere.

Proof Let E = {(0, (o) : lim,.1 f(re`B, co) exists}. It suffices to show that for almost every 0 the section Ee = {w : (0, co) e E} has probability 0, for then the desired conclusion follows from Fubini's theorem.

6.8

179

Inequalities for Conjugate Harmonic Functions

Suppose 0 is such that eie is not a discontinuity point of any of the an, i,ntl P(EB) = a > 0. We are going to construct a decreasing sequence of events An. n > 0, with AO = EB and do this in such a way that limn-,. P(A,) Z a/2. I'or

k,n> 1, we let B,(k) = {coeAn_1 : I f(re`B,w)I < k for all r < 1}.

B,(k) T An_1 as k T oo, so if we have already constructed An_1 with P(A,,_1) z

(1/2 + 1/(n + 1))a, then we can pick kn so that P(B,(kn)) > (1/2 + 1/(n + 2))a and let An = B,(k,J. Let B =n,, 1 An. By construction, P(B) > a/2 and, for all n, z

00

i 9m(re') m(w)

1B < k2P(B).

m=n

The functions bi4, 0 < j < k < oo, are an (incomplete) orthonormal set in L2(S2), so

Y

(E1B

m)2 _< II1B 2.

n

0<m y) = PP(A) (Im CT > y).

To compute the value of the right-hand side, we observe that z --, exp(in(x - 1/2)) maps the strip {0 < Rez < 1} one-to-one onto the half space {Rez > 0} (see Figure 6.3) and sends 1

-.i

1 + i t -> ie-z' 2

a -- exp(in(a - 1/2)),

so if we let a = n(A), b = n(a - 1/2), and c = -2n/2, then it follows from results in Section 1.9 that _fec

CT > y) =

1

cos b

n (cos b)2 + (y - sin b)2

dy,

and changing variables x cos b = y - sin b shows the last integral

=1

((tan-, e`

sin cos b b)

- tan'

(-e-

- e cos b

in b)l

Remark: To be fair to the analysts, we should say that the proof above is essentially due to Calderon (1966) and was later rediscovered by Davis (1973b),

who wrote the proof in probabilistic language. We leave it to the reader to

1$2

6

Hardy Spaces and Related Spaces of Martlnf{ales

a 0

1

exp(iir(z - 1))

-le-ami:

exp(i17(a -

Figure 6.3

show that (1) is consistent with the result given by Stein and Weiss (1959) (see (4.3) on page 273) that sinh Y + i sin

/

expl tl{O:IcP(e)I >Y}I) = 2

2

A2

sinh 2 - i sin 12 I

At this point, the reader can probably guess how we are going to prove the result for u = 9aµ when u is singular. Let f = u + iu and observe that

(a) f(B,), t < t, is a time change of a Brownian motion Cr starting at f(O) = µ(8D) and running for an amount of time

y= Jf(B)I2ds. (b) If 0 < µ(8D) < oc, then 0 < u < oo in D and u(B) a (0, oo) for all t < T. oo)}. This implies that y Y)

This time it is trivial to compute the right-hand side. If we let a = µ(8D), then

Pa(Im C(T') > y) = l JJ

a 1t

a2

x2

dx.

The proof above is due to B. Davis (1973).

7 H1 and BMO, Jf 1 and Rd(

7.1

The Duality Theorem for # 1 In Chapter 6, we saw that ifp > 1, X -. X.,, maps .,!!P one-to-one onto in such a way that the., #P norm of X is equivalent to the LP norm of X.. From this observation, it follows immediately that every continuous linear functional qP on MP can be written as cp(X) = E(X,, Y), where YEL9, q = p/(p - 1). When p < 1, the equivalence of MP and LP breaks down and the reasoning above fails to identify the dual space (,#P)*. In this section, we will consider the problem of describing (.,lfl)*. The first step in the solution is to introduce a decomposition due to Bernard and Maisoneuve (1977), which expresses a general XE.,!!1 as a .,#1. As will be the case many times below, (a) sum of very simple elements of the probabilistic definition was developed after and imitates the definition in-

vented by analysts (see Coifman (1974), Latter (1977), Coifman and Weiss (1977)) and, more embarrassingly, (b) we will often assume (e.g., in the proof of (4)) that our martingales start at 0 at time 0 and forget to mention this. (1)

A martingale A E.,kl is said to be an atom if there is a stopping time T such that

(i) A,=Oif tc"A"E.,ff' and 11X111 < >1c"1-Clearly, di1. The (3) allows us to construct many examples of martingales in next result shows that every XE if 1 can be built up in this way and furthermore

thatY-lc"1 c,,A, is a telescoping series, so we have N

X, - Y c"Ai = X(t) - X(t A TN+1) + X(t A T-N). n= -N

The last term on the right is oo. Up to this point, the values of the cn's and the precise form of the stopping times have not entered into the proof. We must now choose the c"'s to make the A"'s atoms. To do this, we observe that

X(t A Tn+l) - X(t A T")I < I2n+1 _ (-2")I = 3 ' 2", so if we want IAt I < P(Tn < co)-', we must pick cn = 3 - 2"P(Tn < oo).

Having done this, we find that

EIcnl=Y3'2"P(X*>2") n 2"

2n

P(X* > 2")dy

= n

:!5; 6

J

f2"-

P(X*>Y)dy=611XII1.

0

Remark 1: It is important to observe that we use 2", n e Z, and not just n > 0. We do this (that is, use n < 0) and use a sequence that grows geometrically, so

that the picture remains the same, if we multiply by 2'. The last feature is

1N6

7

H' wd BMO..t' snd 44&J

crucial if we are going to prove an estimate like (4), which is unaffected if the quantities under consideration are multiplied by a constant. Remark 2: While it is important to let T" = inf{t : I X,I > a"} and to pick c" = (a + 1)a"P(T" < oo), the actual choice of a is not crucial. If we repeat the last computation above in this generality, we find that Y_ICI
0, then cP(T' < oc) Z E(I YO -YT E(E(I Y. - YTII JET) 1(T' x,))

> cP(T' < oo), a contradiction, so P(T' < oc) = 0.

7.2 A Second Proof of (,&')* = R,& O In this section, we will give a second proof of the duality theorem for -#' following Meyer (1976). This approach starts with a somewhat different definition of -4,V9.

Let Xe.,#2 with X0 = 0. We say that Xe-"02 if there is a constant c such that, for all stopping times T, (1)

E(XQ - XT)2 < c2P(T < oo).

The infimum of the constants with this property is called the _q,#02 norm of X and is denoted by > *.

7.2

A Secood Proof of (. 4')* - 4Y4N

189

Remark 1: From Jensen's inequality for conditional expectations, it follows

that if Xe. .,lf02, then Xe-V..lf0 and > Z IIXII*. An inequality in the other direction, > 5 CIIXII*, is also true and is a consequence oI' the two proofs of the duality theorem. We will also give a direct proof of the second inequality in Section 7.6. XT Remark 2: Since E((X, the definition of -4.,k02 can be written as

(1)

IJIT)

= E(<X )m -

E(<X> - <X )r) : c2P(T < oo) or, in view of (9) in Section 7.1, as

E(<X> -

c2

a.s.

The first step in our second proof of the duality theorem is the same as in the first proof. We show that every continuous linear functional on .,K1 comes from a Ye 9.,0 02. (2)

If 9 is a continuous linear functional on A1, then there is a YE-V.,#02 such that for all Xe.df2 cp (X) = EX. Y..

Proof It suffices to prove the result when 1 = sup { I (p(X) I : > 1 5 1 } = 1. Jensen's inequality implies that «X))2 = (E<X)W)112 z E(<X>10012) _ >,, so tp induces a continuous linear functional on .,#2, and it follows as in Section 7.1 that there is a Ye.112 such that 9(X) = EX, Y. for all Xe A'2.

To show that Ye"02, let T be a stopping time and let X, = Y, - YT,,,. X is the stochastic integral J Y where J = 1[T,.,), so using our formula for the covariance of two stochastic integrals ((3) of Section 2.6), we have <X, Y> = JJd3 o

<X,X)r =

J,2d1"2 dr.

EJ

Since <X>, has bounded variation, ordinary (Riemann-Stieljes) integration gives <X>-112

d <X )t = 2<X> a1o/2,

J0 so

E

f

<X>t 1/2d<X>, = 2>1

0

To estimate the second integral, we fix co and integrate by parts to obtain E J oo <X>112 dt = E<X> 0

=E

r

J0

2

-

fOD

d<X>i12

0

- td<X>tl2

At this point, it is very easy to complete the proof if we leave out one detail. I claim that since <X> is adapted to F, the last expression

7.2

A Second Proof of (.t') - :M.A'

= E f " E(< Y>.

191

- ,I A)d<X>9 2

If you accept this, then there is nothing left to show, for definition (I") of 5U. WC) 2

implies that the above is < KY>)*E(<X>1/2) 00 = sI < (2>1)1J2(t

which completes both the proof of (3) and the proof of the duality theorem. As we mentioned in Section 7.1, one advantage of the new proof is that it gives

a formula for the linear functional that is valid on the whole space and not just on a dense subset.

7.3 Equivalence of BMO to a Subspace of L# Our next goal is to prove the duality theorem (H1)* = BMO. One-third of the work for the proof of this result was done in Sections 6.4 and 6.5, when we showed:

(a) f - Re f(B,), t < r, maps Ho one-to-one into .,ktl and, furthermore, the //T' H' norm off is equivalent to the Xe..,R#', norm of its image (b) if we let X, = Ref(BfAT), then so if we let M: f-+ Ref(BEAT), then the results in (a) hold when the r is erased.

It follows from (b) that M(H0) is a closed subspace of .#1 and that all continuous linear functionals on Ho have the form A(Mf), where A e (.,#1)*. Since (.,ff 1)* = _4.,#0 (the second third of the work), it then remains to identify M(H, )* c -4.,#(9 and to show that we can map BMO one-to-one onto M(Ho )*

in such a way that the BMO norm of a function is equivalent to the .4,f(9 norm of its image. The answer to the first question can be found by very naive reasoning :

M(HH) = {XE.,ffl : X0 = 0, X, = h(B,), t r}, a space we call lf,; , so if L.,L(O,, is defined in the obvious way, we should have

When we prove that (H1)* = BMO in Section 7.4, we will show that this reasoning is correct. To prepare for that result, this section is devoted to determining which harmonic functions h give rise to martingales in (.,ff,; )* =

R.,1f0,, (the last third of the work). The first baby step in doing this is to observe : (1)

If we let ('

Yf= J ke(z).f(e`B) dn(0), then h = Yip where 9 e LZ(8D, n).

7.3

Equivalence of BMO to a Subspace of A WO

193

Proof c .,llti , so the result follows from (2) in Section 7.3. The last result is not much but, in view of the remarks at the end of Scct ion 6.5, it allows us to think of h as being a function (p e LZ (8D, n) and to conclude that t < T

XJh(Bt) t

- gq(Bz)

t > T

is a continuous martingale e.,ll2. A more substantive conclusion results if we use the definition of M.,KO: for all stopping times T, E((Xao - XT)ZI FT) S C2.

So it follows from the strong Markov property that

E((XX -

`YT)2I_f

T) = W(BTnz)

where w(z) = 0 on OD and w(z) = Jko(z)(co(ei) - YTV(z))'21r(O)

when z e D. So in terms of w, the condition for x to be in M.,lfO is (2)

For all z e D, w (z) < c2.

The aim of this section is to show that the last definition is equivalent to the following notion in analysis. (3)

T is in BMO if there is a constant c such that for all intervals I, 19

- J(q(eio) - #(p(z))2ke(z) dir(e),

for this would imply, by the argument above, that i(z) z w(z). The proof of (6) was easy, because all we had to do was show that some multiple of ke(z) was >- je(z). This statement is false if we interchange the roles of je and k9, so we will have to work harder to prove the result we want-we have to add up a large number of functions of the form cl[a,b] to make something z ke. Because of this difficulty, we will prove a slightly weaker result than (*) that is still sufficient to prove that the norms are equivalent. (7)

There is a constant B such that for all z e D, B (sup

)?

(z)

J(cp(ei0

- #(p(z))2ke(z)d7t(9).

Proof As we mentioned above, we prove this by adding up multiples of lla bi to make something > ke(z). The first step is to introduce the intervals. Let

Io = [-(1 - r)n,(1 - r)ir] and for n > 1, I,, = 2"I0, let N be the largest integer such that 2N(1 - r) < 1, and write

F. =

10 + n 1 + fl--,-]-IN For a picture of the decomposition, see Figure 7.1.

19b

7

H' ud sM0, .,#' and iM.,re)

FT

Figure 7.1

k8(.9) = 19/(1 + 180(1 - cosO)), Io =

N= 3

Estimating the integral over Io is easy. If 0 e Io, then (4) implies that ( ao)

1-r2 < 1-r2 ke((1-r)2+2r(1-cos0) r) = (1-r)2 = i + r < 2je(r),

so we have (co(eie)

(bo) 20(r) ? J

- Pl(o(r))2ke(r) dn(0).

to

(The reason for this strange numbering will become apparent as the proof goes on.) Estimating the rest of the integrals requires more work. The first step in estimating the integral over I" - In_, is to observe

(a") There is a constant C such that if r > 1/2 and n < N, (1 - r)ke(r) < C4-" for all 0eI,, - In_1.

Proof From (4), it follows that

(1 - r)ka(r) = (l + r) 1 + 2r(1 - cos 0)1 (1 - r) Now if n < N and 0 E I" -

then

1 - cos0 > I - cos(2"-ln(1 - r)),

and we have from calculus that

inf

XE(0,l)

1 - Zosx-e>0, x

7.3

Equivalence of BMO to a SubNpace of :N. w V

197

so

(1 - r)ke(r) < 2(1 + E(2"-17,)2)-1, proving (an). The estimate in (a") takes care of ke(r) on In - In_1. To estimate the rest of the integral fin (7), we let 1/2

IIfIII

=(

If 12 dO) JI"

and

a" =

f

J9 jel

In

F

.

Since I I I I n2 is a norm, II(p

- a0II2

IIp - anII2 + E Ilak - ale-1II2 k=1

The first term 1/2

((p

-a")2dO)

1 /2 and 0 e [ - n, n] - IN, then

(1 - r)ke(r) < 2(1 +

(2Nn)2)-1

and repeating the arguments used to prove

proves :

(q - Ycp)'ko(z)dO < C'Il(pjI*NZ/2-N

(bN+t)

Adding up the estimates (bo) + (bl) +

+ (bN+l) and recalling that II0I* _

sup, w(z), proves (7).

With (7) proved, it follows immediately (for reasons given after the proof of (6)) that we have (8)

For all z e D, BII(pII*

? w(z).

Combining this result with (6) shows that if we let «(V>>* = sup w(z) Z

(= the M.,#C norm of the associated martingale, Ytp(BtAt)), then A IIq'II* <

* in a slightly different way. Let tp e L2(8D, n) and let u = ° p. Since u(z) = E,cp(B,), then w(z) = E,(u(Bj) - u(z))2 = E,u(B)2 - u(z)Z. Since U, = u(B,,,) e .,tl Z, it follows that

w(z) = EZ J0

,

7.4

The Durllty,rheorem for 11', F'effermrn-Stein DecompoNltlon

20.1

so we have m

E<X, Y>c = Y E<X,AT * Y`). t=o

and, since this equality holds for all X e .,#' ,

Y= I AT * Y` i=O

Now in the arguments above we started with an arbitrary Ye. -V.#0 so it follows from the last equation that we have : (7)

If A1, ..., A. do not have a common eigenvector in R', then any YeR,#O can be written as

Y°+

m

AT*Y`,

with Yèfor i = 0, 1, ... , in, and furthermore this can be done in such a way that II YÌI.

CII Y11*-

Since the proof of (7) uses the Hahn-Banach theorem, it does not tell us how to find the martingales Y°, ... , Ym. In simple cases, this can be done "by hand." Let d = 2,

Example 1

A = (_ 0

\

1

11

0 '

and let Xt = B,'AL where

i=

inf{t : I B, 01 = 11. If we let Y, = BfAt 0 and pick a, b so that (1, 0) = a0 + bAO,

then

Xt=af

tAT

( A S

A0 dB,

,Jo

,10

= (a Y,) + A * (b YY).

Finding a decomposition in general, however, seems to be quite a difficult problem. A. Uchiyama succeeded in doing this recently for d-adic martingales (1982a) and has generalized the construction to BMO functions (1982b), but

I do not know how to prove the result for RMO, and I invite the reader to consider this problem.

The key to deducing the duality theorem for Hp from its probabilistic analogue was the fact that the mapping M (mnemonic for Martingale) defined by M: p

Y p(B,j

t > 0

has I M(p 11 * < C II (p I * . For some applications, we will need to know that the inverse N, defined by I

N:X-+cp(e'e)=E(XXIBT=e`°), is also continuous.

204

7

H' and BMU, .N' and s ,#O

(8)

There is a constant C such that, for all X c- -4,#0,

INXII*
- 1. Let N be a stopping time

for .F(T"), that is, {N < n} E.F(T"), and suppose that for all n > 1, P(N > (Tn_1)) < 0 < 1 on {N > n - 1}. If we let X, = B(t A TN), then X E R.,W0 and as the reader can easily show

nI.

IIXII* < A(1 + (1- 0)-1). The four examples above should give you an idea of what type of martingales are in -4.Af(9 and, hopefully, make the results in the next two sections more obvious. To lead you in the direction of the John-Nirenberg inequality ((1) in the next section), we will now compute P(X* > A) for the three unbounded examples. Example 2.

P(sup X, > ..) = 2P(B1 > i±) =

P(X * > 2) < 2P(sup, Xt > 2) =

4

2

e k2 /2 r

e-v212 dy.

exze X212 dx

21T

4

e x2/22-1. As 2 -+ o o.

2n

Example 3.

If n is a positive integer, then

P(X*>n)_(l/2)".

2011

7 Wand IMO.. N' and :4..WE)

E x a m p l e 4.

I f n is a positive integer, then

P(X * > nA) < P(N > n) < (1 - 0)". Exercise 1

Compute the -4.,lfC norm of the martingale in Example 3.

(a) By the strong Markov property and independent increments of Brownian motion, it suffices to consider the case XT = a e (- 1, 1) a.s. on IT < co}. (b) When XT = a, the distribution of IXX - XTI is given by the following table :

X.

I XX -XTI

-1

a+1

Probability (1-a) 2

( a +1 ) 2 /I)n+i

2

n

n-a

(a 2 1)

I\

Summing over the possibilities, we get

E(IX.-XTI)- (a+1)(2-a)/2 ifa>-0 P(T Al I < Ce-".

II

II

Proof (*) implies that II U II * < C, so 110: Iq(eè)l > All < 21tP(U* > A).-5 C'e-'Ice.

It is easy to improve (3) to the usual result. (4)

Let C, y be the constants in (3). If II'II* < 1, then for all intervals I, {O E I :

I q(eè) - cprl > All
V (eie) _ cp(exp(i(a + Oh))) - gyp is in BMO with I IV/ II * < 1 and 10 = 0, so (4) follows from applying (3) to 0. The conclusion of (1) can be improved in a similar way.

210

7 W .ndBMo,.M' snd91_

(5)

Let C be the constant in (1). If IIXII* S 1, then for all stopping times T, P (sup I tzT

- XTI > A) :9 Ce--"P(T < oc). /l

Proof This could be proved in the same manner as (4), but it seems easier to observe that repeating the proof of (1) with Ro = T proves (5). From (5), we see that the definition of -4,#C as sup E(I

XT I

I T < oo) < oo

T

implies the much stronger conclusion that (6)

For all a < (eIIXII*)-1, supE(ealxw -xTIIT < oo) < o0 T

(in both cases, the supremum is over all stopping times). From (6), it follows that ...,df0 can be defined as the set of all martingales with

supE(IXc,, -XTI"IT x) : IIXII* < 1}.

By stopping at the first time, XXI > x, we see that a(x) < 1/x and a(x)a(y) > a(x + y). Taking logs gives log a(x) + log a(y) > log a(x + y), and one easily concludes from this that lim sup 1 log a(n) = inf 1 log a(m) < 0, R- D

n

m>1 m

proving the result with a constant that is not explicit, but is trivially the best possible.

7.7 The Garnett-Jones Theorem In this section, we will determine which martingales in -4.lf(9 are almost bounded, or, to be precise, we will find the closure of .,#°° in

The solution

to this problem is again the probabilistic analogue of a previously proven analytical result, and the analytical result can be recovered from the probabilistic one. To state the result, we first need some notation. Let ao(X) be the supremum of the set of all a so that supE(ealXw_XTIIT < oc) < oc

T

where the sup is taken over all stopping times (by (6) in Section 7.6, ao(X) > (eII X II*) ' > 0). Intuitively, ao(X) is the exponential rate at which P(I XX - XTI > .) goes to zero for the worst choice of T. Given this interpretation, it should not be surprising that ao(X) can be used to measure how well an X in M MO can be approximated by a Ye.#°°. (1)

There are constants c, Ce(0, oo) such that c

ao(X)

< inf IIX - YII* < Y.M-

C ao(X)

and, consequently, the closure of .,lf°° in R.&(9 is {X : ao(X) = oc }.

212

7

H' and BMO.. #' and 4U.wtV

Remark: This result and the proof we will give below are due to Varopoulos (1979).

Proof The inequality on the left is an easy consequence of the John-Nirenberg

inequality. If YeA' and IIX - Y I* = a, it follows from (5) in Section 7.6 that if Z = X - Y and a > 1/e, then E(e°ho_ZTlla) < C,,P(T < oc), so

P(I Za, -ZTI > 2) < Let Ao = II YII, If K > 2 and A > K.,, Then

P(IX(-XTI>2)A-22o)
(l K)2) C

e-Qa(K-2)/KaP(T

< Ca

< oc).

At this point, we have shown that ao(X) > a(K - 2)/Ka. Letting a 1/e and K-> oo gives that ao(X) > Ilea, so the left-hand inequality in (1) holds with c = 1/e (the constant is inherited from the John-Nirenberg inequality). To prove the other inequality is more difficult. Given a large a and a martin-

gale Xe.A.J1 with ao(X) > a, we need to construct Ze.#' with IIx - Z Ih < C/a. This construction is accomplished in two stages. First, we construct an approximating martingale of the form described in Example 4 of Section 7.5, but we are forced to take 2 large to make 0 < 1, and we end up with a M'#6 norm that is too large. Then we use an ingenious construction, due to Varopoulos, to introduce a sequence of stopping times that smooths the transitions

between the times constructed in our first attempt and reduces the M.# C norm to the right size. The first part of the construction is straightforward. We let 2 be fixed and define R. inductively by letting Ro = 0 and Rn = inf{t > Rn_1 : I X(t) - X(Rn_1)I

> Al for n >- 1. If we let n = X(R,,) - X(Rn-1) on {Rn < oo} and Z =

Y-, S. 1(Rn m1(Rm2"/P} A" _ (X(t A Tn+1) - X(t A T"))/c" cn = (21/P + 1)2' "P(T < oc)1/P.

It follows from computations in Section 7.1 that (i) holds. To check (ii), we observe that dP(>cnA"J < ICnIP

_ (21/P + I)PY2"P(T < oc)

"

_

(21/P

+ 1)P Y 2"P((X *)P > 2") n

< 2(21/P + 1)E(X*)P.

Remark: In the argument above, the fact that p < 1 is important. If p > 1, the triangle inequality gives that cnA"I m-

P

2"))1/P,

and the right-hand side > (21/P + 1) (yn 2"P((X *)P > 2"))1/P. Since the spaces .diP, p > 1, are easy to handle directly, we leave it to the reader to figure out how to define atoms for p > 1.

7.8

217,

A Dlwppolnting Look of (A")* Whenp < I

With the atomic decomposition established, it is easy to find the droll of .#P. From results in Section 7.2, if 6p is a continuous linear functional on . NP,

then there is a Ye.,lf2 such that if Xe.,N2, (p(X) = EX. Y.. To see what properties Y must have, we follow (7) of Section 7.1. Let T be a stopping time,

let Z. = sgn(Y. - YT), and let Z, =

Since Z,i 5 1, X, = (Z, ZTAr)/2P(T < oo)l/P is an atom. By a computation in Section 7.1, E I Y. - Y1.I =

E(Z.(Y. - YT)) = E((Z,0 - ZT) Y.), SO

EIY - YTI

E(X,

2P(T < 00)11P'

Taking the supremum over all stopping times, we see that there must be a constant c such that for all T, (5)

EI Y - YTI < cP(T < oo)'IP.

If p = 1, this inequality reduces to (6) of Section 7.1, so on the surface everything is fine. We have a family of spaces, which we might call MP, 0 < p < 1, that generalize R..#D and have (.,NP)* c MP. Unfortunately condition (5) is too strong when p < 1: (6)

If p < 1, then the only martingales Ye 4'2 that satisfy (5) are constant.

Proof The idea of the proof is simple. Suppose that C = 1. If we can find a T such that 0

and if we can pick an > 0 and define another stopping time

T' =

(T 00

if E(I Y.,, - YTI I FT) > E

otherwise

with P(T' < oo) = S, a small number chosen to have PIP-' < E, then we are done, since if (5) holds with c = 1, we have

P(T' < oo)1/P >_ E(I Y. - YT.I) > eP(T' < 00), which implies that 6'IP-1 > e, a contradiction. The technical problems we now face are (a) to find a stopping time with E(I Y. - YTII JFT) * 0 (easy) and (b) to shrink P(T < oo) to an appropriate size. To solve these problems, we observe that since every continuous local martingale is a time change of Brownian motion, it suffices to prove the result when Y, = B,A, and z is a stopping time. Let T. = inf{t : IB,I > a). Since Ye.,N2 and I Y. - Y,I < 2Y*, it follows from Theorem 9.5.4 in Chung (1974) that as E(I Y. - Y(Ta)II _F(Ta)) -'E(IYYII n a>o

.

(Ta)).

The right-hand side is E I Y. I (exercise), but we do not need this. The expectation

218

7

H' ud DM0, .N' rnd A,#&

of the limit is I Y,,,1, which is > 0 if Y * 0 (recall that YY = E(YY IA,)), so it must be positive with positive probability, and hence, if a is small enough and T = Ta, then E(I Y. - YTII.`FT) # 0, and we have accomplished (a). Since T' has a continuous distribution, the solution of (b) is trivial. We let I

T, - IT 100

if E(I Y. - YTII FT) otherwise

and 0 < T < A

and vary A to make P(T' < oo) the right size.

Remark: The last part of the proof obviously breaks down for dyadic martingales. In that setting, if you want a fixed value for the stopping time, say T = 1, then the probability of taking on that value cannot be arbitrarily small. It is this curiosity that allows nontrivial examples on the dyadic filtration.

8 PDE's That Can Be Solved by Running a Brownian Motion

A Parabolic Equations In the first half of this chapter, we will show how Brownian motion can be used to construct (classical) solutions of the following equations: U, = 20u

ut=2Au+g u,=ZOu+cu in (0, oo) x R" subject to the boundary condition: u is continuous in [0, oo) x Rd and u(0, x) = f(x) for x e Rd. The solutions to these equations are (under suitable assumptions) given by Ex(f(B,))

J(t E(f(B)+-s,BS)ds) \

E. (f(B) exp (f o

c

(BS) ds)

-

b(BS) dB, fo I b(BS) 2 dso /I \J 2 In words, the solutions may be described as follows : E. .f(Br) eXp

(i) To solve the heat equation, run a Brownian motion and let u(t, x) _ Ex.f (B,)

(ii) To solve the inhomogeneous equation u, - ZAu = g, add the integral of g along the path. 219

221)

8

PDE's That Can Be Solved by Running a Brownian Motion

(iii) To introduce cu, multiply f(B,) by exp(f o c(B5) ds) before taking expected values. In more picturesque terms, we think of the Brownian particle as having mass 1 at time 0 and changing size according to m' = c(B,) m, and when we take expected values, we take the particle's weight into account. (iv) To introduce b Vu, we multiply f(B,) by what may now seem to be a very strange-looking factor. By the end of Section 8.4, this factor will look very natural, that is, it will be clear that this is the only factor that can do the job. In the first four sections of this chapter, we will say more about why the expressions we have written above solve the indicated equations. In order to bring out the similarities and differences in these equations, we have adopted a

rather robotic style. Formulas (2) through (6) and their proofs have been developed in parallel in the four sections, and at the end of each section we discuss what happens when something becomes unbounded.

8.1

The Heat Equation In this section, we will consider the following equation:

(1)

(a) u, = 'Au in (0, oo) x Rd (b) u is continuous in [0, oo) x Rd and u(0, x) = f(x). The equation derives its name from the fact that if the units of measurement are chosen suitably and if we let u(t, x) be the temperature at the point x c Rd at time t when the temperatures at time 0 were given by f(x), then u satisfies (1).

The first step in solving (1), as it will be many times below, is to prove : (2)

If u satisfies (a), then M, = u(t - s, B.) is a local martingale on [0, t).

Proof Applying Ito's formula gives s

- u,(t - r, B,) dr

u(t - s, BS) - u(t, Bo) = 0

+ JVu(t - r,B,)dB, 0

Du(t - r, B) dr,

+2 0

which proves (2), since - u, + i Au = 0 and the second term is a local martingale.

If we now assume that u is bounded, then M, 0 < s < t, is a bounded martingale. The martingale convergence theorem implies that as s T t, M, converges to a limit. If u satisfies (b), this limit must bef(B), and since MS is uniformly integrable, it follows that

Ms = E,,(f(B,)I3 ) Taking s = 0 in the last equation gives us a uniqueness theorem.

8.1

(3)

221

The Nest Equation

If there is a solution of (1) that is bounded, it must be v(t, x) = Exf(B,).

Now that (3) has told us what the solution must be, the next logical step is to find conditions under which v is a solution. It is (and always will be) casy to show that v is a "generalized solution," that is, we have (4)

Suppose f is bounded. If v is smooth (i.e., it has enough continuous derivatives so that we can apply Ito's formula in the form given at the end of Section 2.9), then it satisfies (a).

Proof /The Markov property implies that Ex(f(B,)IJ1s) = EB(S)(f(Bt-s)) = v(t - s, B.).

Since the left-hand side is a martingale, v(t - s, BS) is also. If v is smooth, then repeating the calculation in the proof of (2) shows that

v(t-s,B.)-v(t,B0)= fo (-vt+ZAv) (t- r, B,) dr 5

+ a local martingale, so it follows that the integral on the right-hand side is a local martingale. Since this process is continuous and locally of bounded variation, it must be = 0, and hence, - vt + zAv = 0 in (0, oo) x Rd (vt and Ov are continuous, so if - vt + Av 0 at some point (t, x), then it is 0 on an open neighborhood of that point, and, hence, with positive probability the integral is * 0, a contradiction). It is easy to give conditions that imply that v satisfies (b). In order to keep the exposition simple, we first consider the situation when f is bounded. In this situation, the following condition is necessary and sufficient: (5)

If f is bounded and continuous, then v satisfies (b).

Proof (Bt - Bo) d= t1"2N, where N has a normal distribution with mean 0 and variance 1, so if t 0 and x -+ x, the bounded convergence theorem implies that

v(t,

Ef(x + t, 2N) -f(x). The last step in showing that v is a solution is to find conditions that guarantee that it is smooth. In this case, the computations are not very difficult. (6)

If f is bounded, then v e C°° and hence satisfies (a).

Proof We will show only that v c C Z, since that is all we need to apply Ito's formula. By definition, v(t, x) = Exf(B) =

$(2xt)_42e_1x_Y1212tf(Y)dY.

222

8

PDE'. flit ('an Be Solved by Running a Brownian Motion

A little calculus gives Die-Ix-yl2/2t =

-(xi -

Yi)e-Ix-y12/2tt

- Yi)2 Dve-Ix-yl2/2r = (xi - Yi)(x, Dite-Ix-yl2/2r = ((xi

t)e-Ix-yl2/2tt2 Y,)e-Ix-yl2/2rt2

i 5 j.

If f is bounded, then it is easy to see that for a = i or ij,

f

IDe-Ix-yl2/2tl

I f(Y)I dy < o0

and is continuous in R°, so the result follows from our result on differentiating under the integral sign (an exercise at the end of Section 1.10). For some applications, the assumption that f is bounded is too restrictive. To see what type of unbounded f we can allow, we observe that, at the bare minimum, we need Ex I f(B) I < oo for all t. Since I E.If(Br)I = J (2nt)

wee-Ix-yl2/2tIf(y)Idy,

a condition that guarantees this is (*)

IxI 2log+If(x)I -.0

as x- oo.

By repeating the proofs of (5) and (6) above and doing the estimates more carefully, it is not hard to show: (7)

If f is continuous and satisfies (*), then v satisfies (1).

Note: All we have done in this section is rewrite well-known results in a different language. An analyst (see, for example, Folland (1976), Section 4A) would write the first equation as

Btu-Du=0 u(0, x) = f(x)

and take Fourier transforms to get

t) -

t) = 0 =f(e)e-4,M21CI2r

t)

Now KK(x) = (47rt) -d/2 e- 1-"1'/4' has e-4n21C12t,

so it follows that u(t, x) = JKt(x - Y)f(Y) dY,

and we have derived the result without reference to Brownian motion.

8.2

The Inhomogeneous Equstton

223

Given the simplicity of the derivation above, we would be foolish to chiim that Brownian motion is the best way to study the heat equation in (0, cx)) x R. The situation changes, however, if we turn our attention to (0, oo) x G, where G is an open set (e.g., G = {z : IzI < 1}), and try to solve (1')

(a) u, = 'Au in (0, oo) x G (b) u is continuous in [0, oc) x G and u(0, x) = f(x) x e G u(t, x) = 0 t > 0, x e 8G. In this context, the analyst (for example, Folland (1976), Sections 4B and 7E) must look for solutions of the form f (x) exp(A,jt), "separation of variables," and show that the initial condition can be written as

f(x) _

a; f (x).

Proving this even in the special case G = {z: Izi < 1} requires a lot more work than when G = Rd, but for Brownian motion the amount of work is almost the same in both cases. We let T = inf{t : B, t G } and let

v(t, x) = EX(f(B); t < T). Repeating the proofs above shows

(2)

If u satisfies (a), then MS = u(t - s, Bs) is a local martingale on [0, T A t).

(3')

If there is a solution of (1) that is bounded, it must be v(t, x).

(4)

If v is smooth, then it satisfies (a).

(5')

If f is bounded and continuous, then u is continuous in [0, oo) x G and u(0, x)

f(x). If the reader is patient, he or she can also show that (6')

If f is bounded, then v, Div, and Djv, 1 < i, j < d, all exist and are continuous, so v satisfies (a).

Note: v will not necessarily satisfy the other boundary condition u(t, y) = 0 for y e G. We will discuss this point when we consider the Dirichlet problem in Section 8.5.

8.2 The Inhomogeneous Equation In this section, we will consider what happens when we add a function g(t, x) to the equation we considered in the last section, that is, we will study (1)

(a) u, = 'Au + g in (0, oo) x Rd (b) u is continuous in [0, oo) x Rd and u(0, x) = f(x).

224

8 PDE'e Thai ('en Be Sowed by RunIa a Brownlen Motion

The first step is to observe that we know how to solve the equation when g = 0, so we can restrict our attention to the case f 0. Having made this simplification, we will now solve the equation above by blindly following the procedure used in the last section. The first step is to prove (2)

If u satisfies (a), then

M, = u(t - s, Bs) + f sg(t - r, B,) dr 0

is a local martingale on [0, t).

Proof Applying Ito's formula gives

u(t - s, Bs) - u(t, Bo) = fos (-ut +

ZAu)(t - r, B,) dr + fos Vu(t

- r, B,) dB,,

wh ich proves (2), since - ur + Au = -g and the second term is a local martingale.

If g is bounded and u is bounded on [0, t] x R" and satisfies (a), then M 0 S s < t, is a bounded martingale. By the argument in the last section, if u satisfies (b), then

lim M, = fo g(t - s, B,) ds ,Tr

and

g(t - s, Bs) dsl).

M, = Ex o

Taking s = 0 gives Suppose g is bounded. If there is a solution of (1) that is bounded on [0, t] x Rd, it must be //

v(t,x) = ExI

g(t - s,Bs)ds). r,B,)drl.°Fs/

\ fo Again, it is easy to show (4)

Suppose g is bounded. If v is smooth, then it satisfies (a) ax, in (0, oo) x Rd.

Proof The Markov property implies that rg(t -

Ex\J

s

t-s

J(t-r,B,)dr+EB(.,)(fo

l

225

&2 7% I.homoQsn.ou. Equatlon

Since the left-hand side is a martingale, it follows that J3

v(t - s, B) +

g(t - r, B,) dr 0

is also. If v is smooth, then repeating the calculation in the proof of (2) shows that s

g(t - r, B,) dr

v(t - s, Bs) - v(t, Bo) + J0

= f (-vt+ZAv+g)(t-r,B,)dr 0

+ a local martingale,

Again, we conclude that the integral on the right-hand side is a local martingale and, hence, must be = 0, so we have (- vt + 4v + g) = 0 a.e. The next step is to give a condition that guarantees that v satisfies (b). As in the last section, we will begin by considering what happens when everything is bounded. (5)

If g is bounded, then v satisfies (b).

Proof If I g I < M, then rt

g(t - s, Bs) ds

E.

<Mt- 0.

J0

The last step in showing that v is a solution is to check that it is smooth enough. In this case,

v(t,x) = J'dr J(2xs)_2e__2/2sg(t - s,Y)dy, 0

and the normal density (2irs)-d/2e-Iv-XI2/2s

PS(x,Y) _

- oo if x = y and s -+ 0, so things are not as simple as they were in the last section. The expression we have written for v above is what Friedman (1964) would call a volume potential and would write as V(x, t) =

f f

Z(x, t ; , T)

i) d di.

J To D

To translate between notations, set To = 0, D = Rd, Z(x, t ; , T) = Pt-,(x, ),

and change variables s = t - r, y = . Because of their importance for the

226

$

PDE'e That (:nn Be Solved by Ruuing . Brownlen Motion

parametrix method, the differentiability properties of volume potentials are well known. Since the calculations necessary to establish these properties are quite tedious, we will content ourselves to state what the results are and indicate why they are true. The results we will state are just Theorems 2 to 5 of Friedman (1964), so the reader who is interested in knowing the whole story can find the missing details there. (6a)

If g is a bounded measurable function, then v(t, x) is continuous on (0, oo) x R°.

Proof This result follows easily from the bounded convergence theorem. (6b)

If g is bounded and measurable, then the partial derivatives Div = avlax; are continuous, and Div = J' JDP(xY)(t - s,Y)dyds.

Proof The right-hand side is

-

f f ,J

0

(2ns)-d/z (Xi - Yi) e- Ix-vl2/2sg(t S

,J

- s,Y) dY ds

= - JEx[(xi oS

-

B

s/g(t - S, B)]

Although the last formula looks suspicious because we are integrating s-1 near 0, everything is really all right. If lgl < M, then Exl(x1 - B5)g(t - s, BS)l < MEEI x; - BsI = CMs1'2, so

- B8)g(t - S, BS)l < 00

f

0

and it follows from the exercise in Section 1.10 that the partial derivatives Div exist and are continuous. The computations above are not hard to make rigorous, but you should save your strength. Things get very nasty when we take second derivatives. (6c)

Suppose that g is a bounded continuous function and that for any N < oo there are constants C, a e (0, oc) such that l g(t, x) - g(t, y)l < CJx - yla whenever

lxl, lyl, and t < N. Then the partial derivatives Dijv = a2v1ax;axx are continuous, and

Dv =

JDiiPs(xY)(t - s, y) dy ds. Ef

Proof Suppose for simplicity that i = j. In this case, the right-hand side is

227

8.2 The Inhomo$sneoa. Equ.tlo.

ft

f

(2ns)-az

((xi -Z Y,Z - S

)

`

Jo,J

e- i=-yi2/zsg(t - s,Y) dy ds

=

fo'

-sg(t-.s,B.)]tIS,

ExL(xi-BZ)

S

but this time, however, ExI(xt - Bs )2 - SI = sEoI(Bi)2 - 1I so

(x;-B:)2-S

= 00.

S2

We can overcome this problem if f is Holder continuous at x, because we can write [((x;-s2t2- sg(t

E.

- sB)

S

s

=Exl(x`-B2) S

s(g(t-s,Bj -g(t-s,x)))f

The second expression < Cs-1+a, so its integral from s = 0 to t converges absolutely, and with a little work (6c) follows. (See Friedman (1964), pages 10-12, for more details.) The last detail now is : (6d)

Let g be as in (6c). Then 8v/8t exists, and

x) + at (t, x) = g(t,

f'drf

A-,(x,Y)g(r,Y) dY

o

Proof To take the derivative w.r.t. t, we rewrite v as

v(t,x) = JJ'Ptr(xY)frY)'1Y. 0

Differentiating the left-hand side w.r.t. t gives two terms. Differentiating the upper limit of the integral gives g(t, x). Differentiating the integrand gives

fai--

pt-.(X y)g(r, y) dy dr

0

tr lJ

2d(2n(t

-

2

r))-(d+2)12

exp (

c

+

fj

r

+ IX (2n(t - r))-dl2 12(t

2(t

-) I g(r,Y) dy dr

X-Y

- r)I2) exp ( 2I(t -

2d :tdrr -Exg(r,Bt_r)dr+

'

dr

) 2) g(r,Y) dy dr

2N

PDE's'Ib.t Can Ile Solved by Rwtnlni a Brownimn Motion

In the second integral, we can use the fact that Ex(Ix - Bt_,I2) = C(t - r) to cancel one of the t - is and make the second expression like the first, but even if we do this,

' dr

J0t-r This is the difficulty that we experienced in the proof of (6c), and the remedy is the same : We can save the day if g is locally uniformly Holder continuous in x. For further details, see pages 12-13 of Friedman (1964). As in the last section we can generalize our results to unbounded g's. To see what type of unbounded g's can be allowed, we will restrict our attention to the homogeneous case g(t, x) = f(x). At the bare minimum, we need f(B,) ds < oo,

Ex J0

and if we want (b) to hold, we need to know that if t 10 and x

x, then

f0,"

f(B5) ds -* 0.

Ex

If we strengthen the last result to uniform convergence for x c R", then we get a definition that is essentially due to Kato (1973). A function f is said to be in Kd if (*) lim sup Ex (fot I f(B5) I ds) = 0. t4 o

x

By Fubini's theorem, we can write the above as t m sup J kt(x, y) I .f(y) I dy = 0,

where J(27s)_d/2e_k_Y12/2sds.

kt(x,y) = 0

By considering the asymptotic behavior of kt(x, y) as t -+ 0 and we can cast this condition in a more analytical form as

(**) limsupJ a. o

x

w(Ix-yl).f(y)dy=0, Ix-Y13 d=2

1

d= 1.

IZI

(P(z) _

(d z)

This is Theorem 4.5 of Aizenman and Simon (1982). The details of the proof,

8.3

The Feynman-Koc Formula

229

though simple, are a little tedious, so they are left to the reader. We will have more to say about these spaces at the end of the next section. For the developments there, we will also need the space K,"', which is defined in the obvious way: fcKK°° if for every R < co, fl(lxl _ 0 be a function on Rd with

a = sup Ex (f,f(BS) dsl < 1. J

x

T hen sup E. exp

(f f(BS) dsl < (1 - a)-1.

x

Proof The Markov property and nonnegativity off imply that sup Ex X

f ... j' dsl ...

f(Bsn) < an,

so the desired result follows by noticing that

J...J

Jo...

nl

Jo

and summing on n. From the last result, it should be clear why assuming c c Kd is natural in this context. This condition guarantees that sup E. ( Jt I c (B.) Ids) X

/

o

-0

and, hence, that sup Ex exp x

\ fo, I c (

BS) I

ds) J

- 1.

232

s

PDE'r Tbat Can B. 9olwd by Ruing Brownian Motion

With these two results in hand, we can proceed with developing the theory much as we did in the case of bounded coefficients. Since Simon (1982) has written a lengthy and very readable account of how to develop the theory in this generality, we will content ourselves just to briefly describe a few of the results given in his paper. Part of our motivation for doing this is to establish the connection

between the notation we use and the way in which mathematical physicists write things. To make it easy for the reader to find the results in Simon's paper, we have used his theorem numbers below.

Let Ho = - A and V = - c z

H=Ho+V=-A+V, and define a linear operator e- 1H by setting

(e-`Hf) (x) = E. (exp

(_

, V (BS) ds).f(B,) I .

J

THEOREM B.1.1 Let V- E Kd and V E KK'0c. Then for every t> 0 and p < q < co, e-`H is bounded from L° to L.

Proof Since the proof relies on some things that we have not explained above, we simply sketch the proof and refer the reader to Simon (1982) for details. Let Ile " 111,9 denote the norm of e-`H as a map from L" to L. Step 1: p = oo, q = eo. The Feynman-Kac formula shows that if t is small, then

e 1H A. < Cll.fll., so the semigroup property e-(s+t)H = e-sHe-H imples that lle-t"11.0 m < CeAt,

where A = T -'In C. Step 2: p = 2, q = co. Using the Cauchy-Schwarz inequality in the FeynmanKac formula gives e-tH fl < (e-t(Ho+2V)1)1/2(e-uHol fI2)1/2.

Applying Step 1 to Ho + 2 V gives e-t(Ho+2v)Ih.0 < C'e' `,

and an easy estimate shows that e-tH09lh = sup J(2xt)_2pt(x,y)g(y)dy z

< (2nt)-112II9ll1,

8.3

233

The Feynmra-Kac Formula

so

IIe-tH fhlm < C"IIf112

Step 3: p = 1, q = 2. Since 0, e-tH f has a distributional gradient in L'10; THEOREM B.3.4

To get more smoothness, one has to assume more boundedness. Suppose for simplicity that d > 2 and, for 0 < a < 2, let Kd be the set of all functions that satisfy :

(i) sup JIX X

< oo if a # 1

1)

Ix -

(ii) limsup r4o

yI-(d-2+a)I f(y)I dy

x

D(x,r)

yI-(d-2+a)I f(y)I dy

= 0 if a = 1.

234

$

PDE'e 111.t ('Mn Be Solved by Running is Brownian Motion

Let a < 2. Let V E Kd, V+ E Kd'°°. Suppose that the restric-

THEOREM 18.3.5

tion of' V to some hounded open set G lies in Kd. If feL°°, then for each t > 0, e- "`feC°(G) = the set of functions whose derivatives of order [a] are Holder continuous of order a - [a].

Remarks: The fact that V is not supposed to be smooth restricts the last

result to a < 2. The reader should also observe that by writing e-` = e-`12 e-`12 and applying Theorem B.1.1, we can conclude that the results above hold if f E L°° is replaced by f E UP1 L°.

Inspired by the work of Feynman (1948), Kac (1949) proved the first version of what is now known as the Feynman-Kac formula. He proved his result in d = 1 for potentials V = - c, which are bounded below, by discretizing time, passing to the limit, and ignoring a few details along the way. Rosenblatt (1951) extended Kac's work to d>- 2 and filled in the missing details (e.g., Note :

Holder continuity is needed if one wants the solution to be C 2). Since that time,

there have been a number of papers extending the result to more general processes and potentials. The results we have mentioned above are only a small sample of what is known. If the reader would like to see more examples of how

probability can be used to study these problems, he should look at McKean (1977), Berthier and Gaveau (1978), and at recent work of Carmona and Simon. Perhaps the best place to begin is Simon's (1982) survey paper.

8.4 The Cameron-Martin Transformation In this section, we will consider what happens when we add b . Vu to the righthand side of the equation considered in Section 8.1, that is, we will study: (1)

(a) u

x Rd and u(0, x) = f(x).

in

Physically, the extra term corresponds to a force field. In this section, we will see that the probabilistic effect is to add an infinitesimal drift b to our Brownian motion. The first step in solving (1) is to prove: (2)

If u satisfies (a), then MS = u(t - s, BS) exp

\Jo (f

S

1

b(B,) dB, - 2 JI1'(8r)I2th)

is a local martingale on [0, t).

Proof/ Let ZS =f o b(B,) dBr - i f o I b(Br) 12 dr. Applying Ito's formula to J (x0 , x 1 , ... , xd+ 1) = U (t - x0 , X1, ... , xd) exp (xd+1)

X°=s,X. =Bsfor1 n}. The exponenis a local martingale. So tial formula implies that exp(ZZ) - exp(Xt i<X>) observing that <X>, z 0 and stopping at s A T. gives

Eexp(ZSAT) = Eexp(Z0) = 1.

Letting n - oc now and using Fatou's lemma shows Eexp(Z5) < 1, that is, if we let Y = exp(Z5), then Y is an Ll bounded martingale. Applying the last result with b replaced by 2b gives 1

>Eexp(2 f Sb(B,)-dB,-4.1 f lb(B,)I2dr) \\

/

2 ,J o

,10

> exp(-sb*)EY2, where b* = suplb(x)I, and we can use the martingale convergence theorem to conclude that if u satisfies (1), then lim M, = f(B) exp(ZZ) stt

and

M. = EE(f(B)eXp(Z,)I.), so taking s = 0 proves (3). There are many other ways to prove (3). Let

Exercise

Xs= 0

and observe that <X>s = J S I b (B,)12 dr < Cs, 0

so it follows from Levy's theorem (see Section 2.11) that

E (sup exp (Xs)) < E (exp ( sup B.) o<s = fo Ib(B,)I2 ds, it follows from the exponential formula that exp(Z,) is a local martingale/Ps. Let T = inf{t : IZ,I > n}. By stopping at T. A t and letting n -+ oo, we see that Since X,

Q. (C) = E,, exp Z, < 1.

In some situations (e.g., b bounded), we will have Q.,(C) = 1. The next result shows that when this occurs, Q,, makes the coordinate maps behave like a Brownian motion plus a drift.

2,3$

B

(6)

If QS(C) = I, then under Qx

PDE's That ('en Be Solved by Running a Brownian Motion

W, = B, - Jb(B5)th 0

is a Brownian motion starting at x.

Proof Let W' be the jth component of W, Our first goal is to prove that W' is a one-dimensional Brownian motion starting at x;. The first step in doing this is to observe that it suffices to show that if for all 0, U = exp(iO W' + 02t/2) is a local martingale under QX, because then EQX(exp(i0(W' + 02t/2))I.) = exp(iOW' + 02s/2), and

exp(-02(t - s)/2).

EQx(exp(i0(W' -

In other words, W' - WS' is independent of J and has a normal distribution with mean 0 and variance t - s. By (2) from Section 2.13, U is a local martingale/Qx if and only if U exp(Z,) is a local martingale/Px. Unscrambling the definitions gives

U exp(Z,) = exp (i0 (B?

- f" b'(BS) ds) + 02 t 1 o

,1

- 12

exp\fotb(B.) dBs

J

fo

where

C, = ioBi + f b(B.,) dB,, 0

and

D, =

- 02t

2

+ i0

b'(BS) ds + 0

1

2

`

I b(BS)I2 ds.

fo

Now, if we let X, = fo b(BS) dB, then

, = Jbi(Bs)ds ,

and <X >=

' t 0

,

so D, = i, and it follows from the exponential formula that U exp(Z,) is a local martingale.

8.4

239

The Cameron-Martin' 1'ranaformatlon

The computation above shows that each component WJ is u local martingale. To complete the proof, we observe that (5) in Section 2.8 implics that <W', W'5, is the same under Q as it was under P,,, so <W`, WJ>, = t1Ut. The desired result follows from the characterization of multidimensional 1rownian motion given in Section 2.12. (6) shows that under Qx, the coordinate maps, which we will now call X,((o) = co, satisfy

X,=W+ f

b(Xs)ds

,I0

where W is a Brownian motion, an equation that we can write formally as (*) dX, = dB, + b(X,) dt

(here, to facilitate comparison with the results above, we have replaced the W (for Wiener process) with the letter B, which we usually use to denote Brownian motion). There is an obvious connection between (*) and (1): (2')

If u satisfies (a) and X satisfies (*), then M, = u(t - s, XS) is a local martingale on [0, t).

Proof Applying Ito's formula gives s

s

-u,(t - r, X,) dr +

u(t - s, Xs) - u(0, Xo) =

Vu(t - r, X,) dX, J0

0 s

+2Y_

D;ju(t-r,X,)d<X`,X'>,.

ii fo

Since <X`, X'>, = 6,,r, it follows that the right-hand side

f

- u, (t - r, X,) dr + a local martingale

o S

d

+Y f i=1

D,u(t - r, X,)b`(X,) dr

0

+ 2 J Lu(t - r, X,) dr, 0

which proves (2') since - u, + b Vu + 'Au = 0. From the discussion above, it should be clear how to approach the problem of solving (*) and (1) when b is only locally bounded. Let b" = bl(jxjs"). Since b" is bounded, we can solve (*) and get a process X" that satisfies the original equation for t < T" = inf{t : IXI > n}. The measures µ" on (C, W) that give rise to X" have the property that if m < n, then it,, and It agree on .F (T.), so we can let n co and construct a process that solves the original equation for t < T. =

240

8

PDE'. flit ('en Be Solved by Running

Browden Motion

lim T". When 17:,, (x) } has positive probability, we say that the process explodes. The next result gives a simple condition that rules out explosion. (7)

If x b(x) < C(1 + lxl2), then the process does not explode starting from any

xeR°. Proof Let b", X", and T" be as above. Let g(x) = 1 + lx12 and, to ease the notation, let Y = X,". Applying Ito's formula ula gives

9(Y)-9(Yo)=Y J2YsdYs+1Y2d 2 and let G = D - {0}, where D = {x : Ixj < 1}. If we let To = inf{t > 0: B, = 01, then PO(To = oo) = 1, so 0 is not a regular point of OR Example 1 (trivial)

Example 2 Lebesgue's Thorn. Let d>3 3 and let G= (-1, 1)d - U-n=1 [2-", 2-n-1] x [-an, (See Figure 8.1 for a look at G (1 {x : x3 = ... = Xd = 0}. Younger readers will notice that G is a cubistic cousin of Pac-Man with infinitely many very small teeth.) I claim that if a" j 0 sufficiently fast, then 0 is an]d-1

not a regular point of 8G. To prove this result, we observe that since threedimensional Brownian motion is transient and P0((B, , Bi) = (0, 0) for some t > 0) = 1, then with probability 1, a Brownian motion B, starting at 0 will not hit In = {x : x1 e [2-", 2-n-1], X2 = X3 = = Xd = 01, and furthermore, for a.e. w the distance between {B.: 0 < s < oo } and In is positive. From the last observation, it follows immediately that if we let Tn = inf{t : B, c- [2-", 2-n-1] x

[a.,an]d-1} and pick an small enough, then P0(Tn < oo) < 3-". Now Y' l 3-"= 3-1(3/2) = 1/2, so if we let i =inf{t > 0 : BOG} and u =inf{t > 0 : B, (-1, 1)d}, then we have PO(T< a) :5

PO

n=1

(7'n PO(T = o-) >

2

and 0 is an irregular point. The last two examples show that if G` is too small near y, then y may be

8.5

The Dirlchlet Problem

249

Figure 8.1

irregular. The next result shows that if G` is not too small near y, then y is regular. (5c)

Poincare's Cone Condition. If there is a cone V having vertex y such that v fl D (y, r) c G `, then y is a regular point.

Proof The first thing we have to do is explain what we mean by a cone. In Section 4.1, we defined Va = {(x,y)c- H: I x - 01 < ay,y,

< 1},

an object that might be called a cone with opening a, vertex (0, 0), and direction ed = (0, ... , 0, 1). Generalizing this definition, we define a cone with opening a, vertex z 1, and direction z2 as follows :

Va(zl,z2)={z:z=z1+y(z2+w)where w 122 andllwll 0, where a is a constant that depends only on a, so an easy argument shows that if VV(z, z') fl D(z, r) c G` for some r > 0, then

2011

8

PDE'a That Can Be Solved by Running a Brownian Motion

lim inf PZ(B, E G`) >_ e.

40

Combining the last conclusion with the trivial inequality P.(-r < t) > PZ(B, e G`) shows that

PZ(T=0)=limPZ(Te, tdo

and it follows from Blumenthal's zero-one law that PZ(r = 0) = 1.

The last result, called Poincare's cone condition, is sufficient for most examples (e.g., if G is a region with a smooth boundary). The ultimate result on

regularity is a necessary and sufficient condition due to Wiener (1924). To describe Wiener's test, we would have to define and explain the notion of capacity, so we will content ourselves to state what Wiener's test says about Example 2 above and refer the reader to Ito and McKean (1964), page 259, or Port and Stone (1978), page 68, for details. (5d)

In d = 3, Po(T = 0) = 0 if and only if - oo < Y log (2"a") < oo. n=1

In d > 4, Po(T = 0) = 0 if and only if Y (2"an)d- 3 < oo. n=1

In contrast, Poincare's cone condition implies that Po(T = 0) = 1 if lim inf 2"a" > 0. n_ao

The last result completes our discussion of the boundary condition, so we now turn our attention to determining when v is smooth. As in Section 8.1, this is true under minimal assumptions on f. (6)

Let G be any open set. If f is bounded, then v is smooth and, hence, satisfies (a).

Proof Let x e G and pick 6 > 0 so that D(x, 6) c G. If we let a = inf {t : B, D(x, 6)}, then the strong Markov property implies that (*) v(x) = EE.f(B,) = EX[EB(Q)f(Bt)] v(Y) dn(Y), D(x,a)

where it is surface measure on D(x, 6) normalized to be a probability measure, so it follows from (7) in Section 2.10 that v c- C°°. As in the last four sections, our last topic is to discuss what happens when something becomes unbounded. This time we will focus on G and ignore f. By repeating the arguments above, we can easily show the following:

8.6

(7a)

Polison's Equation

251

Suppose that f is bounded and continuous and that each point of 8G is regular. If for all xeG, Px(T < oo) = 1, then v satisfies (1) and is the unique solution. Conversely, we have

(7b)

Suppose that f is bounded and continuous and that each point of 8G is regular. If for some x e G, PX(T < oc) < 1, then the solution of (1) is not unique. Proof of (7b) Since h(x) = PX(T = oo) has the averaging property given in (*), it is C°° and has Ah = 0. Since each point of 8G is regular, P,(T = oo) < Px(T > 1)

- 0 as x - y e 8G. The last two observations show that h is a solution of (1) with f - 0, proving (7b). By working a little harder, one can show that adding aPP(T = oo) is the only way to produce new bounded solutions. (7c)

Suppose that f is bounded and continuous. If u is bounded and satisfies (1), then there is a constant C such that u(x) = EXf(BT) + CPP(T = ao).

Proof See Port and Stone (1978), Theorem 4.2.12.

8.6 Poisson's Equation In this section, we will see what happens when we add a function of x to the equation considered in the last section, that is, we will study (1)

(a) -Au = -gin G

_

(b) u is continuous in G and u = 0 on 8G.

If G = R" and the boundary condition is ignored, then any solution of (1) is called a potential of the charge distribution, because (if the units are chosen correctly) the gradient of the solution gives the force field associated with electrical charges distributed according to g. As in the first five sections of this chapter, the first step in solving (1) is to show (2)

Let T = inf{t : B, 0 G }. If u satisfies (a), then

M= u(B) + J(Bs)ds 0

is a local martingale on [0, T).

Proof Applying Ito's formula as we did in the last section gives u(B,) - u(B0) = fo Vu(BS) dBs + J Du(BS) ds, 2

0

252

8

PDB's not ('rn N. Solved by Rannin a Brownlan Motion

N and the first term on the right-hand side is a local martinpalc on 0, r). If G is hounded, then /s.,r (x) for all xeG (see Exercise 1 in Section 1.7), so if g is hounded and u is hounded and satisfies (a), then for s < T which proven (2). sinec JtAu -

5 1fu11z. + tIIgII.,

so M,, N < T, is a uniformly integrable martingale, and if u satisfies (b), then lim Al, = fo g(B,) dt

.fr

M,

Ex\Jog(B,)dtIFs//

.

Taking s = 0 now gives (3)

Suppose that G and g are bounded. If there is a solution of (1) that is bounded, it must be

v(x) = E. (fo g(B,) dt.

/

Again, it is easy to show (4)

Suppose that G and g are bounded. If v is smooth, then it satisfies (a) a.e. in G.

Proof The Markov property implies that on {T > s}, Ex\Jog(B,)dtl#s/

=

J g(B)di + En(s)(J(Bf)dt).

Since the left-hand side is a local martingale on [0, T), it follows that g(B,)dt + v(B.,)

is also. If v is smooth, then repeating the calculation in the proof of (2) shows that for s e [0, T), s

s

v(B,) - v(B0) + f g(B,) dr = f (ZOu + g) (B,) dr + a local martingale, 0

0

so again, we conclude that the integral on the right-hand side is a local martingale and, hence, must be - 0, so we have 'Au + g = 0 a.e. in D. After the discussion in the last section, the conditions needed to guarantee that the boundary conditions hold should come as no surprise. (5)

Suppose that G and g are bounded. Let y be a regular point of G. If x,, e G and y, then 0.

x

8.6

253

PolMon'. Equation

Proof We begin by observing: (i) In the last section we showed that if r - 0. P,,.(r > E) -, 0, and (ii) if G is bounded, then we have (see Exercise I of Section 1.7) C = sup Exr < oo. X

Combining the last two observations with the Markov property shows that for any r; > 0, Iv(x0I < EIIgII. +

E),

which proves (5). Last, but not least, we come to the question of smoothness. We begin with the case d > 3, because in this case,

dt < oo,

w(x) = E. f Joo00

so the strong Markov property implies that

w(x) = Ex f

dt + Exw(BL),

o

and we have and

(*) v(x) = w(x) - Exw(BT).

The last equation, allows us to verify that v is smooth by proving that w is, a task that is made simple by the fact that w(x) = c J I x - ylz-dg(Y) dy.

The first derivative is easy: (6a)

If g is bounded and has compact support, then w is C'. Proof As before, we will content ourselves to show that the expression we get by differentiating under the integral sign converges and leave it to the reader to apply Exercise 1 of Section 1.10 to make the argument rigorous.

Ix so

(x. _

ylz-d

C

\(z y)zJ1

/Z

2 - d,-Yi)z D,Ix_yIz-d=()((x 2

and we have

Diw(x) = c J (2 - d)iX`_ yig(Y)dY,

2(x,-Yi),

LOA

8

PD6's Thai ('en He Solved by Ruudni a Brownian Motion

the integral on the right-hand side being convergent, since Yld g(.v) I dy O)

As in Section 8.2, trouble starts when we consider second derivatives. If

i#j, then ylz-d = (2 - d)(-d)lx - YI-d-2(x1 - yi)(x, - Y) Dijl x In this case, the estimate used above leads to JDiilx

-y12_dl < Ix- yl-d,

which is (just barely) not locally integrable. As in Section 8.2, if g is Holder continuous of order a, we can get an extra Ix - yla to save the day. The details are tedious, so we will content ourselves to state the result : (6b)

If g is Holder continuous, then w is C2.

The reader can find a proof either in Port and Stone (1978), pages 116-117, or in Gilbarg and Trudinger (1977), pages 53-55. Combining (*) with (6a) and (6b) gives (6)

Suppose that G is bounded. If g is Holder continuous, then v is smooth and hence satisfies (a).

Proof (6b) implies that w is CZ. Since w is bounded, it follows from (6) in the last section that x -p Esw(BT) is C. The last result settles the question of smoothness in d > 3. To settle the question in d = 1 and d = 2, we need to find a substitute for (*). To do this, we let

w(x) = JG(xY)g(Y)dY where G is the potential kernel defined in Section 1.8, that is, I

n

G(x,y) _

log(lx - yl) d=2

- llx - yl

d = 1.

G was defined as

p, (x, y) - a, dt 0 foo

where the a, were chosen to make the integral converge, so if $ gdx = 0, we see that

T0

J G(x,y)9(Y)dy = li m Ex

9(Bt)dt.

N.7

255

The Schrlidinger Equation

Using this interpretation of w, we can easily show that (*) holds, so tlgnin our problem is reduced to proving that w is C 2, which is a problem in calculuN. (nice all the computations are done, we find that (6) holds in d:5 2 and that in d I ,

it is sufficient to assume that g is continuous. The reader can find dctuils in either of the sources given above. On the basis of what we have done in the last five sections, our next step should be to consider what happens when something becomes unbounded. For the sake of variety, however, we will not do this, but instead, we will show how (1) can be used to study Brownian motion. Example 1 Let d = 1, G = (-1, 1), and g = 1. In this case, formulas (3) through (6) imply that v(x) = EX2 is the unique solution of (1)

(a) ?u"(x) _ -1 in (- 1, 1) (b) u is continuous on [-1, 1] and u(- 1) = u(1) = 0, so u(x) = 1 - x2. Once you see the one-dimensional case, it is easy to do the general case. Exercise 2

Let d > 2, D = {x: I x < 11, T = inf {t : B, 0 D}. Then

Exr = d(1 - Ix12). Remark: This result can also be proved by observing that IB,I2 - dt is a martingale, so Ix12 = 1 - dExr.

8.7 The Schrodinger Equation In this section, we will consider what happens when we add cu to the left-hand side of the equation considered in Section 8.5, that is, we will study: (1)

(a) 2'Au+cu=0inG (b) u is continuous in G and u = f on aG. We will explain the physical significance of this equation in the next section. For the moment, you should consider it simply as the inevitable next step in the progression established in Sections 8.1, 8.2, 8.3, 8.5, and 8.6. As in the first six sections of this chapter, the first step in solving (1) is to show

(2)

Let i = inf{t : Bt G }. If u satisfies (a) then

Mt = u(B,) exp (I c(B) ds) ff

is a local martingale on [0, r).

26

8

PDE'e That ('em Be Solved by fling a Brownian Motion

Proof Let c, =

c(B,)dLs. Applying Ito's formula gives

u(B,) exp(c,) - u(B0) =

fr

exp(c,)Vu(B5) dB, +

u(BS) exp(cs) dos o

0 e

+

Au(B,) exp(c,) ds, 0

2

which proves (2), since dc, = c(B,) ds, IAu + cu = 0, and the first term on the right-hand side is a local martingale on [0, r). At this point, the reader might expect that the next step, as it has been six times before, is to assume that everything is bounded and conclude that if there is a solution of (1) that is bounded, it must be v(x) = EE(.f(Bt) exp(cz))

We will not do this, however, because the following simple example shows that this result is false. Example 1 Let d = 1, G = (-n/2, n/2), and c = 1/2. If u(x) = cos x, then u' (x) = - sin x and u"(x) _ - cos x, so z u" + cu = 0 and u = 0 on 0G. But

there is obviously another solution: u = 0. We will see below that the trouble with the last example is that c = 1/2 is too large or, to be precise, if we let t

w(x) = Exexp(

c(B,)ds ),

then w = oo. The rest of this section is devoted to showing that if w * oo, then "everything is fine." The development will require several stages. The first step is to show (3a)

If w # oc and G is connected, then w(x) < oo for all x e G.

Proof Let c* = suplc(x)I. By Exercise 1 in Section 1.7, we can pick ro so small that if T, = inf{t : JB, - Bol > r} and r:5 ro, then Exexp(c*T,) < 2 for all xeG. If D(x, r) c G, then the strong Markov property implies that w(x) = E.[exp(c(T,))w(B(T,))] < Ex[exp(c*T,)w(('B(T,))]

= EX[exp(c*T)] J

w(y)dir(y), 8D(z,r)

since the exit time T, and location are independent (here TC is surface measure on D(x, r) normalized to be a probability measure). If S < ro and D(x, S) C G, multiplying the last equality by rd-1 and integrating from 0 to b gives

w(x) < 2 S

f

w(z) dz

D(z,6)

where C is a constant (that depends only on d).

8.7

257

The Schrtldinger Equation

Repeating the argument above and using c(T,) > -c*T,, givca it lower bound of w(x) > 2-'-C d

f",(X,6) w(Y) dy. ,a)

Combining the last two bounds gives: (3b)

Let S < ro. If D(x, 2S) c G and y e D(x, S), then w(x) > 2-(d+2)w(Y)

Proof w(x) > 2-1(26)d fD(x,26) w(z) dz

2-1 (26) d

w(z) dz

J d

>2 1( S)d 2 1Cw(Y)_ 2-(d+2)w(y). From (3b), we see that if w(x) < co, 26 < ro, and D(x, 26) < G, then w < oo

on D(x, S). From this result, it follows that Go = {x: w(x) < oo} is an open subset of G. It is easy to see that Go is also closed (if x - y e G and D(y, 36) c G, then for n sufficiently large, D(x,,, 26) c G and y e D(x,,, S), so w(y) < oc). From the last results, it follows (if G is connected) that Go = G, so (3a) holds. With (3a) established, we are ready to prove our uniqueness result. (3)

Suppose that f and c are bounded and that w # oc. If there is a solution of (1) that is bounded, it must be

v(x) = E.(f(B,) exp(ct)). Proof If u satisfies (a), then (2) implies that X3 = u(BS,,,)exp(cSA,) is a local martingale on [0, T). If u, f, and c are bounded and u satisfies (b), then letting s T T gives

u(x) = Ex(f(BT) exp(cT) ; T < t) + Ex(u(B,) exp(c,); T > t).

Since f is bounded and w(x) = Ex exp(ct) < oo, the dominated convergence theorem implies that as t --> co, the first term converges to

Ex(f(Bj To show that the second term -0, we observe that EX[u(B,) exp(c,); T > t] = Ex[Ex(u(B,)

T > t]

= Ex[u(B,) eXp(c,)w(B,); T > t]

and use the trivial inequality

w(x) > exp(-c*)PP(T < 1)

2

8

PDE'u 7ba1 Van Be Solvod by Rrrta/ Brownian Motion

to conclude that inf w(x) = is > 0. XEo

Replacing w(B,) by e,

EE[u(B,) exp(c,); T > t] < e-'EE[u(B,) exp(c.); r > t] < E 1IIuIImEx[exp(ci); T > t] - 0 as t

oo, since w(x) = E., exp(cT) < oo.

This completes our consideration of uniqueness. The next stage in our program, fortunately, is as easy as it always has been. (4)

Suppose that f and c are bounded and that w * oo. If v is smooth, then it satisfies (a) a.e. in G.

Proof The Markov property implies that on {T > s}, Ex(exp(c)f(BT)IS') = exp(cs)Ea(s)(eXp(c,)f(BT)) Since the left-hand side is a local martingale on [0, T), it follows that exp(c,)v(B,)

is also. If v is smooth, then repeating the calculation in the proof of (2) shows that for s e [0, T),

v(B) exp(c,) - v(B0) =

f(f& + cv) (B,) exp(c,) dr + a local martingale,

so it follows that the integral on the right-hand side is a local martingale and, hence, must be = 0, so we have ZAv + cv = 0 a.e. in G. Having proved (4), the next step is to consider the boundary condition. As in the last two sections, we need the boundary to be regular. (5)

Suppose that f and c are bounded and that w # oo. If f is continuous, then v satisfies (b) at each regular point of 8G. Proof Let y be a regular point of 8G. We showed in (5a) and (5b) of Section 8.5 that if S > 0 and x - y, then Px,,(r < S) -> l and Px,,(BT a D(y, S)) - 1, so if cl < M, then Psn(e-"ra < exp(c3) < e"i") - 1 and, since f is bounded and continuous,

EXJexp(c)f(B.); T < S) -+f(x). To control the contribution from the rest of the space, we observe that

Exjexp(ct)f(Bj; T > S) < eMall f < eMaIIfIImjIwII.P..(r

r > 8) a)

0.

This brings us finally to the problem of determining when v is smooth enough to be a solution. To solve the problem in this case, we use the same

8.7

29

The Schr6dinger Equation

trick used in Section 8.3 to reduce our result to the previous case. We observe that exp

(f

T

T

c(BS) dsl

=1+

J

o

T

c(B) exp (

\

,)o

c(B,) dr) ds, s

/

so multiplying by f(BT) and taking expected values gives

v(x) = 1 +

Ex

(c(Bs)exP(f

c(B,) dr)1(S2, 2, G= {z: Izl < 11, and c=- -fl < 0. Again, v(x) = Ex exp(- fir) < 1, so v is the unique solution of (1). This time, however, it is not so trivial to guess the answer, so we just state the result: v(x) = CIXII-112Id/2

1(

where I, is the modified Bessel function m

m2

(Zv+ 2m

/(m!I'(v + m + 1))

M=0

(see Ciesielski and Taylor (1962) or Knight (1981), pages 88-89, for details). It is one of the great mysteries of life that the distribution of z starting from 0 is the same as the total amount of time a (d + 2)-dimensional Brownian motion spends in {z : I z I > 1 } (which can also be computed using the methods of this section). For more on this phenomenon, see Getoor and Sharpe (1979).

tis

Example 4 G = Rd, c(x) = -a - flk(x), where a, fl e(0, oc), and k(x) > 0. Since G is unbounded, this example is, strictly speaking, not covered by the results above. However,

E. (exp (- f f 'k (Bs) ds)) < 1 o

JJJ

and we have supposed that a > 0, so v(x) = fo dt e-'Ex

(exp (_fljk(Bs)ds)) f

nicely convergent, and it is not hard to show (details are left to the reader) that v is the bounded solution of (1), the boundary condition (b) being regarded as vacuous.

8.7

'the SchrtMtnaer Iqustlon

261

The most famous instance of this solution occurs when d = 1 and

k(x) =

I

t0

x>0 x 0} I < Bt) =

1

n

dr

r(1 - r)

= 2 aresin(O). it

The reader should note that t-' l {s E [0, t] : Bs > 0} I

1/2

as t oc. In fact, the distribution of this quantity is independent of t, and its density has a minimum at t = 1/2! Examples where (1) can be solved explicitly are rare. A second famous example, due to Cameron and Martin (1944b), is k(x) = x2. In this case, Eo (exp

(- f

f (B5)2 ds)) =

(sech((2p)1/2t))1i2.

ff

The reader is invited to try to derive this equation. The proof given on pages 10-11 of Kac (1949) is a beautiful example of Kac's computational ability. Up to now, we have focused our attention on the question of the existence of solutions to (1). The probabilistic formula for the solution can also be used to study properties of the solution. Perhaps the most basic result is (7)

Harnack's Inequality. Let u > 0 and satisfy 'Au + cu = 0 in D = {x : I xI < 1}. Then for any r < 1, there is a constant C (depending only on r, c) such that if x, y e D (0, r), then

u(x) < Cu(y).

Proof Pick ro so small that if Tr = inf { t : I B, - Bo I > r} and r < ro, then Exexp(c*Tr) < 2, where c* = supIc(x)I. Repeating the first computation in the proof of (3a), we see that if S < ro and D(x, S) c G, then

u(x) < 2 b° f D(x,S)

u(z) dz

762

8

PDE's 9181 Can Be Solved by Rpmly u Brownian Motion

Figure 8.2

and

u(x) > 2-'-C J

u(y) dy.

,(x,6) Therefore, since we have assumed that u >- 0, we can repeat the proof of (3b) to conclude that if S < ro, D(x, 26) c G, and y e D(x, S), then (*) u(x) 2-(d+2)u(y) The desired result now follows from a simple covering argument. Fix r < 1

and pick b < ro A (1 - r)/2. Given x, y e D(0, 1 - r), there is a sequence of points xo = x, x1, ... , xm = y (with m < 3/6), such that for all i, we have Ixi - xi_1I < 2(5/3 and D(xi, 26) c D (see Figure 8.2). It follows from (*) that the inequality holds with C = 2(a+2)316

Remarks: (a) The constant in the proof above grows like exp(C(1 - r)-1) ; by working harder, you can get (1 - r)-,' (exercise). (b) The result in (7) is true if D is replaced by any connected open set G and D(0, r) is replaced by K, a compact subset of G. The details of the covering argument are much more complicated in this case. See Chung (1982), Exercise 3 on page 205, for a proof, or spend an hour or two and find your own. (c) The result above is also true if we assume only that c e Kd°` (the space defined in Section 8.2). The original proof, due to Aizenman and Simon (1982), was given in this generality. The fact that c can be unbounded causes quite a bit of trouble, but by following the outline of the proof of Theorem B. 1.1 given in Section 8.3 and using a clever time reversal argument as a substitute for the

8.8

EIReev.l.m of A + c

263

self-adjointness used in Step 3, they succeed in proving the key eslimule: 11' S < 60, then u(x) < C

u(y)dn(y) Jan(z, a)

(their (1.13)), and once this result is established, things follow pretty much as before. The reader can find the details clearly explained in their paper. Exercise 1

Use the Poisson integral representation ((2) of Section 3.3) to show

that ifu>0andAu=0inD,then foranyr cpl, it follows that

cpo= in fl-0:wc+0 *oc in G} 5 cp1. Having proved that 2o < cpl, we are now in a situation where we can use results of Donsker and Varadhan (1974-1976). Our next step (following their (1976) paper) is to show that (4)

cpi

=inf sup c + Au u

x

where the infimum is taken over all u e C+ = {u a C°° : u > 0 on G }.

Proof If 1 > >/i2, there is a ueC°° that is > 0 on G and has

c+Au t1 0, (0 + (1 0)c)2 < 0 + (1 - 0)cz. This is true when c = 0. Differentiating the difference reveals that

aac = 2(1 - e)c - 2(1- 0)(0 + (1- 0)c) >0 ifc> 1

J v(t x) at v(t1

(t, x)

dµo(x)

x) (A + cl v(t, x) dµo(x)

J

_ f cdµo + J Av(t,x)dµo(x)

? Jcdµo+APO) =0i, so

f log(v(t, x)) dµo > 4i, t. Jensen's inequality implies that log Jdµo(x)v(t, x) > 0, t,

and we have

J dµo(x)v(t, x) > exp(4i, t), which proves (6). The last link in the chain is: (7)

4P 1 5 4.

Proof Let 1= q,1. We want to show that 1 is in the spectrum, that is, if we let R1f(x) = f

m

e-uEx(.f(B,)

exp(c)) dt,

0

then R, is not a bounded operator on C(G). If R, 1 is bounded, then we have 00 >

j'dzo(x)J e`v(t,x)dt o

f dte " f dµo(x)v(t,x), ,J

but it follows from (4), (5), and the proof of (6), that the last expression is

270

N

PDE's That Can Be Solved by Rumina

Brownian Motion

dt r "cxp(ot t) = oo.

Z 0

This is a contradiction, which proves (7) and completes the chain

A0;, A,j(x)D,j f(x) + Y, b,(x)D, f(x), where the A,j(x) and b,(x) are

(for the moment) arbitrary. In this chapter, we will consider the following equation : (1)

(a) u, = Lu in (0, oo) x Rd (b) u is continuous in [0, co) x Rd and u(0, x) = f(x).

In Section 8.4, we solved (1) in the special case A(x) - I by first solving the stochastic differential equations

dX"=dB,+b(XX)dt Xo =x and then running the resulting processes to solve (1): u(t, x) = Ef(X, ).

On the basis of the results for A - I, it seems reasonable to try to solve the general case by solving

Xo = x

(*) dXX = o (XX) dB, + b(Xx) dt

(where o is a d x d matrix) and letting u(t, x) = Ef(XX ). To see which a to pick, we apply Ito's formula to get s

u(t - s, X) - u(t, Xo) =

- u, (t - r, X,) dr 0

+I D,u(t-r,X,)dX, '

oS s

+2

ij

Jo

D;,u(t - r, X,) d<X', X'>r 271

272

9

Stochutlc I)Ifferentlol Equatloou

(here we have taken the liberty of dropping the superscript x to simplify the formulas). Now dXs = bi(XS) ds + Y aij(Xs) dBs ,

j

so it follows from the formula for the covariance of two stochastic integrals that <Xi1XjX= Y_

aik(X,)ajk(Xs)dS

k

and we have 5

u(t - s, X5) - u(t, X0) = f - u, (t - r, X,) dr 0

+Y

Diu(t - r, X,)bi(Xr) dr

i

+ a local martingale 5

+ 2Y

Diju(t - r,Xr)(aaT)ij(X,)dr,

so if CUT = A, then we have

u(t - s, XS) - u(t, Xo) =

(- ur + Lu) (t - r, X,) dr + a local martingale. 0

The condition A = auT obviously restricts the set of A's we can consider, for if A = auT, then for each xERd"1 (i.e., Rd viewed as d x 1 matrices), XTAx = XTQQTX = Ior TXIZ > 0,

that is, A is nonnegative definite. If we assume (as we can without loss of generality) that A is symmetric, then this condition is also sufficient, because results in linear algebra tell us that any nonnegative definite symmetric matrix can be written as UT DU, where U is an orthogonal matrix (i.e., UTU = I) and D is a diagonal matrix. This observation allows us to define a by setting or = UTCU, where C > 0 is the diagonal matrix that has CZ = D. With this choice of a, we have: (2)

If u satisfies (i) and X satisfies (*), then M., = u(t - s, Xs) is a local martingale on [0, t).

Proof By computations above,

u(t - s, Xs) - u(t, Xo) = f (- ut + Lu) (t - r, X,) dr + a local martingale. o

If If u satisfies (i), the first term is = 0.

Remark: Looking back at the proof of (2), we see that after we used Ito's formula to conclude that

9.1 M's That Can Be Solved by Running n SDE

273

u(t - s, Xs) - u(t, Xo) = f - U, (t - r, X,) dr 0

+1: f5 Diu(t-r,X,)dX, i

2

0

+ 1Y_

D;ju(t - r, X,)d<X`, XJ) o S

all we did was work out what dX, and d<X`, X'>, were and plug in their values, so we have the same conclusion if X satisfies : (**)

(i) For each i, XX - f o b;(X) ds is a local martingale (ii) For each i, j,

A;;(X)ds.

<X`,X'>, = J0

We will see in the next section that there is essentially no difference between (*) and (**).

As has been the case many times before in Chapter 8, the last result leads immediately to a uniqueness theorem. (3)

Let X be a solution of (*) (or (**)), with X0 = x. If there is a solution of (1) that is bounded, it must be v(t, x) = Ef(X).

Proof If u satisfies (1), then MS = u(t - s, X) is a bounded local martingale on [0, t) that converges to f(X) as s r t and satisfies M, = E(f(X,) j A), so taking s = 0 proves (3).

Remark: (3')

The reader should note that (3) is also a uniqueness result for (*):

Suppose there is a solution of (1) that is bounded. If X, and X7 are two solutions of (*) with X0 = Xo = x (constructed, perhaps, on different probability spaces using different Brownian motions), then

Ef(X:) = Ef(X') (2) and (3) may look the same as the corresponding steps in Sections 8.1 through 8.7, but when we start to consider the existence of solutions, things become very different. We first have to construct solutions of (*) and then run them to produce solutions of (1). The first task will be accomplished in Sections 9.2 through 9.5 (since I am a probabilist, a neophyte, and a pedagogue,

we will spend some time investigating the countryside on the way to our destination-constructing "weak" solutions of (*)). In Sections 9.6 and 9.7, we turn our attention to v(t, x) = Ef(XX) and prove the analogues of the results

274

9

Stochutk 1)Iffercntrl Equatlou

that we called (4), (5), and (6) in Sections 8.1 through 8.7. The first two results, which are easy consequences of the Markov and Feller properties, are dispensed with in Section 9.6. In Section 9.7, we confront (but do not conquer) the problem of proving (6).

9.2 Existence of Solutions to SDE's with Continuous Coefficients In this section, we will describe Skorohod's approach to constructing solutions of stochastic differential equations. In order to focus our attention on o and not on b (which we have already considered in Section 8.4), we assume that b - 0. The same method works when there is a (bounded) continuous b * 0, but there are more estimates to do and, as parenthetical qualification suggests, many of the complications involve issues (e.g., explosions if b is too big) that we have considered earlier. Skorohod's idea for solving stochastic differential equations was to discretize time to get an equation that can be solved by induction, and then pass

to the limit and extract subsequential limits to solve the original equation. Given this approach, it is natural (and almost necessary) to assume that each A,3 is a continuous function of x, and, for the moment, we will suppose that A is bounded, that is, IAij(x)I < M for all i, j, x. For each n, define X"(t) by setting X"(0) = x and, for m2-" < t < (m + 1)2-",

X"(t) = X"(m2-") + o,(X"(m2-"))(B, - B(m2-")), where the second term is the matrix Q(X"(m2-")) times the Brownian increment

(B, - B(m2-n)). Since X" is a stochastic integral with respect to Brownian motion, the formula for covariance of stochastic integrals implies

<X", xi>, = k k

J0

Aij(Xn([2"s]/2")) us,

and it follows that if s < t, then

<M(t-s), so we have (see Section 6.3)

E sup I X,i(u) - XX(s)I° < CEI<XX>, - <Xi>slp12 uc[s.r]

s, we observe that if t < T, then t

XXX? - JA()ds = 0

YY(t, YYit) - JAiJ(Y(s))ds. 0

9.2

Existence of Solutions to SDE's with Contlnuoue Coefficients

Changing variables s = o, in the integral above and observing that )'(n,) converts the right-hand side to

277

r

Ylt)

yr(nYv) -

fo

which is a local martingale, so we have .

JAj(Xs)ds.

<X' 1X'>1 =

0

As before, we can construct a Brownian motion B such that for t < T,

X, - X0 = J a(Xs) dB, If P(T = oo) = 1, then we have solved (*) for all times and we are done. If P(T < oo) > 0, then, as you might guess from the discussion in Section 8.4, we are also done, for the process has exploded, in other words, lim,1T Y = 00 a.s. on IT < oo }. It is easy to explain why this is true-for any R, there is a 6 such that if Ixl < R and S2R = inf{t : IXI > 2R}, then Px(S2R > S) > 6; therefore, with probability 1, X, cannot leave I x I < 2R and return to I x I < R infinitely

many times in a finite time interval. A full proof, however, requires the strong Markov property, which we have not established, and a number of unpleasant

details, so we leave the rest to the reader. Again, a complete proof can be found in Ikeda and Watanabe (1981), this time on pages 160-162. To get an idea of when explosions occur, consider the following: A;i(x) = (1 + Ix16)5;j. _ In these processes, if we let g(x) = (1 + lxl8)-1, then A(x) = I, so only the Example 1

second part of the construction is necessary: W

T= Jo ( 1+IBIb)-1ds. In d = 1, 2, Brownian motion is recurrent, so T = oo (and the process never explodes). In d > 3, Fubini's theorem implies that

ExT = E.

(1 + iBsl6)-1 ds 0

=C

lx

The integrand -

f

-

1

1

y1d-2 I + Iylb lyl2-a-d

dy.

as y -. oo, so if

r2-b-drd-1 dr < o0

(i.e ., S > 2), we have EXT < oo.

Conversely, we have (2)

If trace(A) < C(1 + lxl2), then QX(T = cc) = 1.

27K

9

Stochutlc Differential Equ.tlor

Proof This proof is the same as the proof of (7) in Section 8.4. Let p(x) _ 1 + 1x12. By Ito's formula, w(Xt)-(P

(x°)=Y

it,

i=1 Jo

o t

d

= a local martingale +

A;;(X5) ds. 0i=1

The last integral < C f o cp(X5) ds, so another application of Ito's formula shows

that e-ctcp(X) - tp(Xo) = a local martingale + Y f t e-cSAjj(Xs) ds 0

+ J(_C)eco(X)ds = a local supermartingale.

If we let T. = inf{t: IXI > n}, it follows from the optional stopping theorem that Note: Our treatment of the existence of solutions follows Section 4.2 of Ikeda and Watanabe (1981), who in turn got their proof from Skorohod (1965). The treatment of explosions here and in Section 8.4 is from Section 10.2 in Stroock and Varadhan (1979). They also give a more refined result due to Hasminskii.

9.3

Uniqueness of Solutions to SDE's with Lipschitz Coefficients The first and simplest existence and uniqueness result was proved by K. Ito (1946).

(1)

If for all i, j, x, and y we have I a (x) - aij(y) I < KI x - y I and I bi(x) - bi(y) I < KI x - y1, then the stochastic differential equation X= x+ fo t a(XS) dB., + fo t b(X) ds

has a unique solution.

Proof We construct the solution by successive approximation. Let X° = x and define :

X' = x + fo a(X: -1) dBs

+ J b(X: -1) ds for n > 1. 0

9.3

279

Uolqueo.M of Solution to SDE'. with Llp.chltz Codflclsnte

Let A,,(t) = E ( sup Xs - Xs -' j

.

I

Ossst

J

We estimate A. by induction. The first step is easy: XSl

= x + a(x)Bs + b(x)s,

so

IX91- X°I < Ia(x)Bsi + Ib(x)Is.

Squaring and using the fact that sup Ia(x)BSI ossst

a 1112

sup lo(x)BI,

o<s 2, we recall that Ia + b12 < 2a2 + 2b2 and observe that this implies that for n >- 1,

2E sup

ta( s) -

a(Xs-1)dBs I 2

o5t5T fo t

+ 2E sup

05tsT

2

fb(Xn) - b(Xs -1) ds 0

To bound the second term, we observe that the Cauchy-Schwarz inequality implies that e b'(Xs) - b`(Xs 1) dsl2
2-" i.o

0 0, so cp = 0 and, hence, X = Y. To remove the integrability condition on Y, observe that for any R < oe, we can modify a and b outside IxI > R in such a way that a and b still satisfy IQ;;(x) - a;;(y)I < KI x - yI and Ib;(x) - bt(y)I < KI x - yI, but

a and b are = 0 off some compact set. When we do this, any solution has E(supo<s 2, but in this range we can conclude that the solution is unique.

Although the condition required by Ito's approach is not sharp in this case, there is a good reason why Ito's approach, or any other approach that proves uniqueness, does not work for small 6-the solution of (*) is not unique

284

9

Stochutk Differential Equations

when 6 < 1. To prove this statement observe that by using the approach in the second part of Section 9.2, we can think of a(x) = 1/g(x) where g(x) = IxI-a and solve the equation by time-changing a Brownian motion. If 6 < 1, then

(a) Eo f IB,I ads= f s-612EoIBi ads 0

0

=

fos-112 dsl EoIB1I-6 < o0

(since the first factor is < oo for 6 < 2 and the second for 6 < 1), so we can construct a solution of dX, = I XSlaiz dB, that starts at 0 and does not stay there. Once we have two solutions starting from 0, it is easy to see that there are

two solutions starting from x 0 0. If x > 0

and

i = inf{t : B, = 0}, then by

results in n Section 1.9,

(b) E. J

xlyl-a(x AY)dy

TIBSI-a1(a,<x)ds= I z

Y1-ady < 00

(whenever 6 < 2), and we have I B,I a1(Bs>z)ds
_ 1 and you cannot reach 0 when 6 >- 2 (see Exercises 1 and 2 below for converses). On the basis of this conclusion,

you might guess that the solution is unique when 1 < 6 < 2 and that all solutions stop when they hit 0. This conjecture is indeed true. The first fact is a consequence of a theorem of Yamada and Watanabe (1971). (1)

Suppose that (i) there is a strictly increasing function p with la(x) - a(y)I < p(I x - yl) that has p(0) = 0 and

JP2(u)du = c0 for all a> 0

9.4

285

Some I xemplee

(ii) there is an increasing and concave function ) with I b(x) - b(y)I c A (I % that has A (O) = 0 and

i'I )

A-1(u)du= oo forallE>0. fo,

Then (*) has a unique solution.

Proof See pages 168-170 of Ikeda and Watanabe (1981). (1) implies that our equation has a unique solution when 6 > 1, so we now have a complete picture of Example 3. unique for 6 > 1 The solution of (*) is not unique for 6 < 1.

-

While (1) helps us with Example 3; Example 2, on the other hand, helps us to understand the reason for the difference between the assumptions about a and b in (1). When b(x) = x1 and a = 0, (1) implies that

unique for 6 > 1 I not unique for 6 < 1,

solution of (*) jl

which is the same as the conclusion for Example 3, except for the fact that it pertains to a(x) = Ixlaiz To justify my remark that "you cannot escape when 6 > 1," show that for all e > 0, Exercise 1

IBSI-1 ds = oc P. a.s. fo

Proof By scaling and monotonicity, it suffices to prove the result where a is replaced by T1 = inf{t : IB,I - 1}, and to do this, it suffices to show that T1

lim sup nom

fo

I B.I 11(IBSI 0 PO a.s.

To prove the last result, let R0 = 0, Sn = inf{t > Rn : IB,I = 2-"}, Rn+1 = inf{tr> Sn : B, = 0} for n -> 0, and N = sup{n : S. < T1}. Observe that N f S, N >Y 2"(Sm - Rm) 0Tl > Y M=O Rm

J

m=0

where EN > 2" and the S. - R. are i.i.d. with E(Sm - Rm) = C2 2n. A simple argument (compute variances) shows that the lim inf >- C a.s. Exercise 2 To justify my remark that "you cannot reach 0 when 6 > 2," show

that if r=inf{t:B,=0}, then for allx

JIBI_2ds= oo P., a.s. 0

0

286

9 Stochudc Dlfferontlil Equtlow

Proof The game is the same as in the last example, but the stopping times are different and don't work as well. Let S_1 = 0, R. = inf{t > Sn_1 : IB,I = 2"j, S. = inf{t > R" : IB,I = 2-"} for n >- 0, and N = sup{n : S" < T1}. Again, we have T, 0

N`

Sm

N

> Y 2Z"(S, - Rm),

L

m=0 Rm

m=0

but this time EN = 1 and E22"(Sm - Rm) = C, so an even simpler argument shows that the lim inf = oo a.s.

9.5 Solutions Weak and Strong, Uniqueness Questions Having solved our stochastic differential equation twice and seen some examples, our next step is to introduce some terminology that allows us to describe in technical terms what we have done. The solution constructed in Section 9.3 using Ito's method is called a strong solution. Given a Brownian motion A and an x e R°, we constructed a process X, on the same probability space in such a way that

(*) X = x +

f

A(X,) dB, +

0

f

b(XS) ds.

0

In contrast, the solution constructed in Section 9.2 by discretizing and taking limits is called a weak solution. Its weakness is that we first defined X, on some probability space and then constructed a Brownian motion such that (*) holds. With each concept of solution there is associated a concept of uniqueness. We say that pathwise uniqueness holds, or that there is a unique strong solution, if whenever B, is a Brownian motion (defined on some probability space (f2,.F, P)) and X and X are two strong solutions of (*), it follows that, with probability 1, X = Xt for all t >- 0.

We say that distributional uniqueness holds, or that there is a unique weak solution, if all solutions of (*) give rise to the same probability law on (C, ') when we map co e f2 --> X(w) E C.

Ito's theorem implies that pathwise uniqueness holds when the coefficients

are Lipschitz continuous. It is easy to show that in this case there is also distributional uniqueness. (1)

If the coefficients o and b are Lipschitz continuous, then distributional uniqueness holds.

Proof Let B, and B, be two Brownian motions, and let Xe and X, be the sequence of processes defined in the proof of Ito's theorem when X° = X° = x and the Brownian motions are B, and Bt, respectively. An easy induction shows that for each n, X" and X" have the same distribution, so letting n - oo proves M.

9.5

Solutlonn Week and Strong, Unlqueneaa Questions

287

With two notions of solution, it is natural, and almost inevitable, to risk about the relationship between the two concepts. A simple example due to Tanaka (see Yamada and Watanabe (1971)) shows that, contrary to the naive idea that it is easier to be weak than to be strong, you may have it unique weak solution but several strong solutions.

Let a(x) = 1 for x > 0 and = -1 for x < 0. Let W be a Brownian motion starting at 0 and let Example 1

B: = I a(W)dW Since B = e W is a local martingale with = t, B is a Brownian motion. The associative law implies that

a(W)-B=u(W)Z-W= W, so we have

W=

Ja()dBs.

Since a(- x)

c (x) for all x

0, we also have

-W = f a(-W)dBs. 0

The last two equations show that there is more than one strong solution of dXs = o (xs) dBs. To prove that there is a unique weak solution, we observe that if dXs = a(XX)dBs, then X is a local martingale with <X> = t and, hence, a Brownian motion.

In the other direction, we have the following result of Yamada and Watanabe (1971) : (2)

Pathwise uniqueness implies distributional uniqueness.

We will not prove this because we are lazy and this is not important for the developments below. The reader can find a discussion of this result in Williams

(1981) and a proof in either Ikeda and Watanabe (1981), pages 149-152, or Stroock and Varadhan (1979), Section 8.1. The last result and Tanaka's example are the basic facts about the differences between pathwise and distributional uniqueness. The best results about distributional uniqueness are due to Stroock and Varadhan (1969). To avoid the consideration of explosion, we will state their result only for bounded coefficients. (3)

Suppose that

(i) A is bounded, continuous, and positive definite at each point (ii) b is bounded and Borel measurable. Then there is a unique weak solution.

Much of Chapters 6 and 7 in Stroock and Varadhan (1979) is devoted to preparing for and proving (3), so we will not go into the details here. The key

20

9 Stochudc Differential Equador

step is to prove the result when b = 0 and I Aij(x) - 5,i1 5 s for all x E R°. The reader can find a nice exposition of this part in Ikeda and Watanabe (1981), pages 171-176.

In the material that follows, we will develop the theory of stochastic differential equations only for coefficients that are Lipschitz continuous. We do this not only because we have omitted the proof of (3), but also because in the developments in Section 9.6, the proofs of the Markov and Feller properties, we will use the fact that our processes are constructed by Ito's iteration scheme. You can also prove these results in the generality of (3) by knowing that there is a unique solution to the martingale problem, but for this you have to read Stroock and Varadhan (1979).

9.6 Markov and Feller Properties Having constructed solutions of (*) and having considered their nature and number at some length, we finally turn our attention to finding conditions that guarantee that v(t, x) = Ef(X") is a solution of (1) of Section 9.1. In Section 8.1, when we dealt with XX = B, a Brownian motion, the first step was to observe that if f is bounded, then the Markov property implies that

Ex(f(B)IJs) = v(t - s, B,), so v(t - s, B,) is a martingale. To generalize this proof to our new setting, we need to show that the X, have the Markov property, that is, (1)

If f is bounded and continuous and v(t, x) = Ef(Xx), then

E(f(Xt)I.)=v(t-s,X ). Proof Let X,.x(t) (the process starting at x at time s) be defined as the solution of

X, = x +

a(X,) dB,. + fS b(X,) dr g

for t >- s, and X,,x(t) = x for t < s. It follows from uniqueness that if s < I < u, then

X,,x(u) = X,,x,,x(,)(u)

a.s.

(recall that all the random variables X, x(u) are defined on the same pyobability space, (C, ', P)). From the last result, it follows immediately that if < s < t < u, if A 1, ... , A. are Borel sets, and if f is bounded, then 0< s1 < E(.f(XXx); Xx(s1)EA1, ... ,

X x(s1) E A1, ... , X x(s,,) c (recall that Xx = X0.x(t)). To prove (1), it is enough to prove the following:

9.6

289

Markov end Pallor Propertler

v(u - t, Xx).

(2)

To this end, observe that for any y e R°, Xr.v(u) a u(Br+s - Br, s Z 0)

and is, therefore, independent of ., so

E(f(Xr,v(u))I) = Ef(X,,,(u)) = v(u - t,y). Now if Y : S - R° is .yr measurable and takes on only a finite number of values yl, ... , y,,, then n

X"Y(u) _ Y l(r=vt)X y1(u)

a.s.

It follows from the last result that

E(f(XX,r(u))I.) = v(u - t, Y). To prove this equality, let A e., with A c { Y = y,} and observe that it follows from the first result that E(f(Xt,y(u)); A) = E(f(X1,v,(u)); A)

= v(u - t, y;)P(A) = E(v(u - t, Y); A). To extend our results to a general Ye., (and, hence, to prove (1')), pick a sequence Yn of random variables that take on only finitely many values and have Y. - Ya.s. and EI Yn - YI2 - 0 as n - oo. From the continuity result (e') in Section 9.3 (or, to be more precise, the remark afterwards), we have that XX,rn(u) - XX,r(u)

in L2

and, hence, E(f(Xr,r(u))I.°1t)

in L.

By the result above for simple Y, the left-hand side is v(u - t, Yn). To complete the proof of (2) (and, hence, of (1)), it suffices to prove (3)

Suppose that f is bounded and continuous. Then for fixed t, x -> v(t, x) = Ef(X x) is continuous.

Proof The continuity result (e') in Section 9.3 implies that

E(sup IXS -XXIZSlx-yl2CT, ÒSsST

lJ

where CT is a constant whose value depends only on T. From this result, it follows immediately that if xn - x, X, n - Xt in probability and, hence, v(t,

Remark:

Ef(Xx^) -. Ef(Xx) = v(t, x).

The proof of (1) given above follows Stroock and Varadhan (1979), pages 128-130. 1 think it is a good example of the power of the idea from measure

290

9

Stochardc Dlfferoadal Equadooe

theory that to prove an equality for the general variable Y, it is enough to prove the result for an indicator function (for then you can extend by linearity and take limits to prove the result). With the Markov property established, it is now easy to prove (4)

Suppose that f is bounded. If v is smooth, then it satisfies part (a) of (1) in Section 9.1.

Proof The Markov property implies that E(.f(X, )I#;) = v(t - s, X. ,x).

Since the left-hand side is a martingale, v(t - s, Xs) is also a martingale. If v is smooth, then repeating the computation in the proof of (2) (at the beginning of this chapter) shows that s

v(t - s, XS) - v(t, x) =

(- v, + Lv) (t - r, X,) dr + a local martingale, 0

where

Lf= 1 YA`VD;;f+ Yb`Dif, 2 ii

so it follows that the integral on the right-hand side is a local martingale. Since

this process is continuous and locally of bounded variation, it must be = 0 and, hence, (- v, + Lv) = 0 in (0, ao) x Rd. With (4) established, the next step is to show the following: (5)

If f is bounded and continuous, then v satisfies part (b) of (1) of Section 9.1.

Proof From the continuity result used in the last proof, it follows that if

x - x and t - 0, then

x in probability and, hence,

v(tn, xn) = E.f(X',,(t,,)) - f(x)

9.7 Conditions for Smoothness In this section, we finally confront the problem of finding conditions that guarantee that v(t, x) = Ef(X) is smooth and, hence, satisfies (1)

(a) u, = Lu in (0, oo) x Rd (b) u is continuous in [0, oo) x Rd and u(0, x) = f(x).

In this category, the probabilistic approach has not been very successful. By purely analytical methods (i.e., the parametrix method; see Friedman (1964), Chapter 1), one can show (2)

Suppose that Aj(x) and b;(x) are bounded for each i and j and that

9.7

Condltlm for Smoottm u

291

(a) there is an a > 0 such that for all x, y e Rd, >Y,Aij(x)yj > alYl l

(b) there is a J > 0 and C < oo such that for all i, j, x, and y, l Aij(x) - A.,(Y)l 0, and write Xi = X+ to prove the result in general. (2)

If X1, .. . , X. are independent random variables, then the a-fields S i = a(Xi)

are independent, and, consequently, if fl, ... , f are Borel functions, then f1(Xi), . . . ,fn(XX) are independent.

298

Appendix

(3)

Generalize the proof of (2) to conclude that if 1 < n1 < nz < the f : S2 R"i-ni-1 are Borel measurable, then


A (X1 , ... ,

Jk(Xnk-1 +1 ,

.

< nk = n and

, Xnk)

and independent.

Hint: Start with f that are of the form H .1,1.(X.) and then use the approach described for (1) to work your way up to theJ general result. Sequences of independent random variables are very important in probability theory, because they are (hopefully) what is generated when an experiment is repeated or a survey is taken. Motivated by this example, one of the main problems of the subject is to give conditions under which the "sample mean" (X1 + + Xn)/n approaches the "true mean" as n - oo, and to estimate the rate at which this occurs. Much of the first quarter of a graduate probability course is devoted to developing these results, but since they are not essential for what we will do, we will just prove one sample result to illustrate some of the concepts in this section and then state two more results in order to give you a taste of the theory. (4)

Let X1, Xz, ... be a sequence of independent and identically distributed random

variables (i.e., P(X; < x) is independent of i) that have EXz < oc. As n -, oo, 1(X1 n

+

EX1 in probability.

+ Xn)

Proof Letµ=EX;, Y=X; -p. Now n(X1+...

+XX)_P=n(Y1+... +Y),

so it suffices to show that the right-hand side - 0 in probability. The key to doing this is to observe that

)2(n n =E

E

YY

Y

i=1j=1

i=1 n

_

n

n

EY Y = > t=1j=1

(since if i

nEY1

i=1

j, EY Yj = EYEY, = 0). If we let S. = Y1 +

then

ES = Cn,

i.e.,

S" z

E

n

= C n

and Chebyshev's inequality implies that

/ Sn

PI

//

1\z

>E <E-LEIS"I =CE

\n- -

n

-c

n

->O.

+ Yn and C = EYz,

A.2

299

todepeadence and Lows of Large Numbers

The result above is a weak form of the weak law of large numhcrs. The strongest form of the strong law is (5)

Let X1, X2, ... be a sequence of i.i.d. random variables (for a translation, see (4)) with EI Xj I < co. As n

n (X1 + ... + Xn) -, EX1

oc,

a.s.

Analysts may recognize this result as being a consequence of Birkhoff's ergodic theorem. It is easy to show (see Exercise 2 below) that EI X; I < oo is necessary + X,)/n to converge to a finite limit so the condition in (5) is for (X1 + "sharp." The weak law holds in a little greater generality: (6)

Let X1, X2, ... be a sequence of i.i.d. random variables. There is a sequence of constants an such that 1

n

+ Xn) - a - 0 in probability

(Xl +

if and only if nP(I XiI > n) - 0. In this case, we can take an = E(Xi ; I Xi 1 < n).

Proof We leave the proof as an exercise for the reader. Once someone WIN you to look at Xi,n = Xi I(IXil,,n) and use Chebyshev's inequality, the rest is not hard. See Feller (1971), page 235, for a solution. The next two exercises are much easier, but the first is much more important.

The Second Borel-Cantelli Lemma. If A1, A2, ... are independent and >P(An) = co, then P(An i.o.) = 1. Exercise 1

Hint:

/

PI

If M:5 N < oo, then

\

A I = [1 (1 - P(An)),

n n/ =M

and, for any M, the right-hand side - 0 as N

co.

Exercise 2 Let Xl, X2, ... be a sequence of i.i.d. random variables with EI X,I = oo. Then

P(ISn-S.+1I>ni.o.)1, so Sn/n cannot converge to a finite limit on a set of positive probability. Hint : 00

n=0

W

P(IX;I>n)> fo P(IXI>x)dx=El XI.

300

Appendix A Primer of Probability Theory

A.3 Conditional Expectation Given a probability space (52, .to, P), a a-field F c .moo, and a random variable

Xe Fo (i.e., FO measurable) with EI XI < oo, we define E(XI.F) to be any random variable Y that has (1) Ye.F

(ii) for all A e .F, f

dP = fA YdP.

A

Y is said to be a version of E(XI fl. Any two versions of E(XI -F) are equal almost surely.

Interpretation: We think of .t as describing the information we have at our disposal : For each A e F, we know whether or not A has occurred. E(X I.F) is, then, our "best guess" of the value of X given the information at our disposal. Some examples should help to clarify this. In each case, you should check that the answer we have given satisfies (i) and (ii).

Example l If X e .F, then E(X I.F) = X, that is, if we know X, then our "best guess" is X itself. In general, the only thing that can keep X from being E(XIF) is condition (i). Example 2

Suppose that 521, ... , n. is a partition of S2 into disjoint sets, each

of which has positive probability, and that F = a(521, ... , 52 ), the a-field generated by these sets. Then on 52;,

E(XI F) = E(X ; f1i) P(52i)

In words, the information in F tells us the element of the partition which contains our outcome, and given this information, our best guess for X is the average value of X over 52;.

Example 3 Suppose that X is independent of F, that is, P({X e Al fl B) =

P(XeA)P(B) for all Ae. and Be.F. In this case, E(XI,fl = EX, that is, the information in .F is of no help in guessing the value of X. Let X > 0 and let Q be the measure that has density X with respect to P, that is, Q (A) = JA XdP. Let P' and Q' be the restrictions of P and Q to .F. Then Q' 1, then E(XI _F) in LP

(f) dominated convergence if X -> X a. s. and Y with EY < co, then E(XXI.t) -), E(XI f) a.s.

These properties are not very hard to prove using the definition of conditional expectation. To prove (a), we simply check that the right-hand side a .F and has the same integral as aX + Y over all A e .F. To prove (b), we observe that if A e .F,

f fA XdP< SA and applying this result to A = {E(XIfl > JA

A

we conclude that A = 0,

that is, E(XI.F) < E(YI). With these two examples as a guide, you should be able to prove (c) through

(e), but, for the moment, I want to discourage you from doing this. It is more important to understand the following properties of conditional expectation, which have no analogue for ordinary expectation, and I leave the proofs of these properties as recommended exercises. E(E(Xj

)) = E(X).

If F c .F2, then (i) E(E(XI.°,Fi) I Fz) = E(X I.°wi) (ii) E(E(XI _O--z)I.Fj) = E(XI.F1).

In words, the smaller a-field always wins out. (3)

IfAe4andEIYI , then X. is said to be a supermartingale or submartingale, respectively.

To give an example of a martingale, consider successive tosses of a fair coin, and let = 1 if the nth toss is heads and = -1 if the nth toss is tails.

Let S. = 1 +

+ . S is called the (symmetric) simple random walk. It

represents the amount of money a gambler has after the nth toss if each time the coin is tossed he bets one dollar on the coin coming up heads. Let # = As we mentioned before, we think of J as giving the information we have at time n, which in this case is the outcomes of the first n tosses. S To prove this, we observe that S. = 1 + is a martingale with respect to oo, and, by (3) in Section A.2, n+1 is independent of + we have

/ + Ee(Sn+1 I .F.) = E(S

KK

bn+1 l

E(SS I Win) +

I .F.)

If the successive tosses have P( = 1) < 1/2, then the computation above shows that

S corresponds to betting on an unfavorS is able game, we see that there is nothing "super" about a supermartingale. The

name comes, instead, from the fact that superharmonic functions, when composed with Brownian motion, give rise to supermartingales. For what follows, we will need a few simple facts about martingales, the proofs of which we leave as exercises for the reader.

A.5 Gambling Systems and the Martingele Convergence Theorem

303

(1)

If X is a martingale and m < n, then

(2)

If X is a martingale and T is a convex function with El (p(X,)l < rfj for nil n, then qp(X,) is a submartingale.

(3)

If X,, is a submartingale and qp is an increasing convex function with E(p(X,) < CZ)

for all n, then (p(X,) is a submartingale. (4)

Orthogonality of Martingale Increments. If X. is a martingale with EX,2 < 00 for all n and V!9 m 5 n, then

E((Xn-Xm)Xi)=0. Since the proof of (4) is a classic example of manipulating conditional expectations, we will give the proof and let the reader justify the steps

E((X, - Xm)Xi) = EE((X, - Xm)XiI&m = E[X,E(X, - XmI `fm)] = E[Xi(E(Xnl

Xm)]

=0. From the proof above, it is immediate that we have (5)

Under the hypotheses of (4), E((Xn - Xm)ZIFm) = E(X,

X.2.

A.5 Gambling Systems and the Martingale Convergence Theorem Let ? be an increasing sequence of o-fields. H. is said to be predictable if for all n > 1. In words, the value of the process at time n may be predicted (with certainty)

from the information available at time n - 1. You should think of H. as the amount of money a gambler bets at time n. This amount can be based on the outcomes at times 1, ... , n - 1, but it cannot depend on the outcome at time n. Once we start thinking of H. as a gambling system, it is natural to ask how much money we would win if we used it. For concreteness, let us suppose that the game consists of flipping a coin and that for each dollar bet we win one dollar when the coin comes up heads and lose one dollar when the coin comes up tails (most games in casinos reduce to this situation when you ignore all the ritual). Let S,, be the net amount of money we would have won at time n if we had bet one dollar each time. If we bet according to a gambling system H, then our net winnings at time n would be

304

Appendix


n

(H. S)n = Y Hm(Sm - Sm-1), m=1

since S. - Sm_1 = + I or -1 when the mth toss results in a win or loss, respectively.

-

The next result is the most basic fact about gambling systems and is, apparently, little known. It says that there is no system for beating an unfavorable game. (1)

Let Xn be a supermartingale. If H,, >- 0 is predictable and each H is bounded, then (H- X)n is a supermartingale.

Proof (H- X). + E(Hn+1(X,,+1 - Xn)I` n)

E((H.

= (H' X),, + Hn+1E(Xn+1 (1-1 -

since E(Xn+1 -

0 and Hn+, > 0.

Remark: The same result is obviously also valid for submartingales and for martingales (and in the second case without the restriction H. > 0). To keep from being repetitious, we will state our results for only one type of process and leave it to the reader to translate the result to the other two. In my remark preceding (1), I did not mean to dismiss gambling systems as worthless. There is one system that allows us to prove the martingale convergence theorem. Let h > 0, let No = 0, and for k >- 1 let N2k_1 = inf{m > N2k_2 : X. < a}

N2k=inf{m> N2,_1:Xm?a+h} if N2k_1 < m - 1 < N2k for some k otherwise U. = sup{k: N2k < n}.

Hm

11 0

Since X(N2k-1) < a and X(N2k) > a+ h, between times N2k_1 and N2k, X crosses from < a to > a + h. H. is a gambling system that tries to take advantage

of the "upcrossings." In stock-market terms, we buy when Xm < a and sell when Xm > a + h. In this way, every time an upcrossing is completed, we make

a profit

h. Last but not least, U. is the number of upcrossings completed

by time n. (2)

The Upcrossing Inequality. If X. is a submartingale, then

hEU, < E(X - a) + - E(Xo - a) Proof Let Y. = (Xn - a)+. Since Y is a submartingale that upcrosses [0,h] the same number of times that X,, upcrosses [a, a + h], it suffices to prove the result when a = 0 and Xn > 0. In this case, we have that

A.5 Gambling Systems and the Martingele Convergence Theorem

35

hU" S (H X)",

since a final incomplete upcrossing (if there is one) makes a nonncgolivc contribution to the right-hand side. Let Km = 1 - Hm.

and it follows from (1) that E(K X)" > E(K X)o = 0, so E(H X)" S E(X" Xo), proving (2) in the special case and, hence, in general.

Remark: We have proved the result above in its classical form even though this approach is a little misleading. The key fact is that E(K X)" > 0, that is, even by buying high and selling low, we cannot lose money on a submartingale,

or in other words, it is the reluctance of submartingales to go from above a + h to below a that limits the number of upcrossings. From the upcrossing inequality, we easily obtain (3)

The Martingale Convergence Theorem. If X. is a submartingale with sup,, EX < oo, then as n -> oo, X. converges almost surely to a limit X with El XI < oo.

Proof Since (X - a)+ < X + + IaI, (2) implies that EU" 5 (IaI + EIX"I)/h, so if we let U = lim U,, be the number of upcrossings of [a, a + h] by the whole sequence, then EU < co and, hence, U < oo a.s. Since this result holds for all rational a and h,

U {lim inf X" < a < a + h < lim sup X" a,hEQ

rm

n-m

has probability 0, and lim sup X" = lim inf X"

n-

n,ao

a.s.,

which implies X = limX" exists a.s. Fatou's lemma guarantees that EX+ S

lim inf EX < oo, so X < oc a.s. To see that X > - oo, we observe that EX,- = EX - EX" < EX,, - EXo (since X,, is a submartingale), and another application of Fatou's lemma shows that EX - < lim inf EX,,- < oo. From (3), it follows immediately that we have (4)

If X" < 0 is a submartingale, then as n -+ oo, X"

X a.s.

The last two results are easy to rationalize. Submartingales are like increasing sequences of real numbers-if they are bounded above, they must converge almost surely. The next example shows that they need not converge in L1.

Example 1 (Double or nothing) Suppose that we are betting on a symmetric simple random walk and we use the following system: H" _

2n-1

on S,,-, = n - 1

0

otherwise.

3%

Appendix

A Primer of ProbablUty Theory

In words, we start by betting one dollar on heads. If we win, we add our winnings to our original bet and bet everything again. When we lose, we lose everything

and quit playing. Let X" = 1 + (H- S)n. From (1), it follows that X is a martingale. The definition implies that X. > 0, so EI XnI = EX. = EXo = 1, but it is easy to see that P(X" > 0) = 2-", so X" -. 0 a.s. as n - oo. This is a very important example to keep in mind as you read the next three sections.

A.6

Doob's Inequality, Convergence in L", p > 1 A random variable N is said to be a stopping time if {N = n} e .F" for all n < oo.

If you think of N as the time a gambler stops gambling, then the condition above says that the decision to stop at time n must be measurable with respect to the information available at that time. The following is an important property of stopping times : (1)

If X. is a submartingale and N is a stopping time, then XfAN is a submartingale.

Proof Let Hn = 1(N> n). Since {N > n } = {N< n - 1 }` E and it follows from (1) of the last section that (H - X). = (2)

Hn is predictable, is a submartingale.

If X. is a submartingale and N is a stopping time with P(N < k) = 1, then EXo < EXN < EXk .

Proof Since XNnn is a submartingale, it follows that EXo = :!g EXNnk = EXN. To prove the right-hand inequality, let H. = 1(N<m). Now {N < m} = {N < m - 11 so (H X)n = Xn - XN,n is a submartingale and EXk EXN = E(H X)k > E(H X )O = 0. Remark: Let Xn be the martingale described in the last section and let N = inf In: X,, = 0}. Then EX, = 1 > 0 = EXN, so the first inequality need not hold for unbounded stopping times. In Section A.8, we will consider conditions that guarantee that EXo < EXN for unbounded N. From (2), we immediately get (3)

Doob's Inequality. If X. is a submartingale and A = 10<M Al , then

).P(A) < EX. lA < EX. .

Proof LetN=inf{m:Xm>Aorm>n}.XN>AonAandN=nonA`,so it follows from (2) that ).P(A) < E(XN 1A) < EX. lA

(observe that XN = Xn on A`). The second inequality in (3) is trivial.

A.7

Uniform IntearebWty and Convergence In L'

307

Integrating the inequality in (3) gives (4)

If X,, = max Xm, then for p > 1, OSmSn

p

E(X,°)
M)I =0.

\ieI

M- O

J

Uniformly integrable families can be very large. (1)

If X e L', then {E(XI.fi) } is uniformly integrable.

0, then the dominated converProof If A. is a sequence of sets with gence theorem implies that E(I XI ; Aj 0. From the last result, it follows that if e > 0, we can pick S > 0 such that if P(A) < S, then E(I XI ; A) < e.

Pick M large enough so that EI XIIM < S. Jensen's inequality and the definition of conditional expectation imply that

E(JE(XI.F) JE(XI.F) > M) < E(I XI ; E(IXI I.F)

> M),

and we have that P(E(I XI I _F) > M)

< E(E(IMI I.f )) = EIXI < s,

so for this choice of M,

suupE(IE(XI.9)I ; E(XI.f)I > M) < e, and since a was arbitrary, the collection is uniformly integrable. Another common example is Exercise 1

Let (p be any function with 9(x)lx

oo as x - oo, for example,

qp(x)=x" or cp(x)=xlogx. If Eq(IXi1) 0 and M is large, then the second term < e.

A.8

:9

Optional Stopping Theorems

To bound the third term, we observe that uniform integrability implieN that that EIXI < sup. EIXI < oo, M the

that

sup El X

E.

- XI < 2e, proving (b).

the

(b)=(c): EIXI - EIXU :!9 El IX.1-IXI <EIX -XI. (c) =>(a): Let /IM(x) = x on [0, M - 1], OM = 0 on [M, oo), and let dIM he

linear on [M - 1, M]. If M is large, EIXI - Eom(I XI) < e/2. The bounded convergence theorem implies that E//M( 1 X I) - EI//M(X I ), so using (c) we get that if n > no, E(IXXI ; IXI > M):!5; EI X1 - E,M(Xfl) < E.

By choosing M larger, we can make E(I XI ; I XI > M) < r: for 0 < n < no, so X is uniformly integrable.

We are now ready to state the main theorems of this section. Since we have already done all the work, the proofs are short. (3)

For a submartingale, the following statements are equivalent:

(a) it is uniformly integrable (b) it converges in L'.

Proof (a) => (b): Uniform integrability implies that supElXl < ou, which by (3) of Section A.5 implies that X -> Xa.s., which by (2) above implies that X -> X in L'. The converse, (b) (a), is a corollary of (2). (4)

For a martingale, the following statements are equivalent:

(a) it is uniformly integrable (b) it converges in L' (c) there is an integrable random variable X such that X = E(XI..). Proof (a) (b) => (c) :

(b) : This result follows from the proof given in (3). Let X = lim X,,. If m > n, then E(Xm I-F) = X,,, so if A e .F,,,

E(X ; A) = E(Xm ; A). As m - oo, Xm IA --> X1A in L1, so we have that E(X ; A) _

E(X, A) for all A e 3, and it follows that X. = E(Xl S ). (c) (a): This result follows from (1) above.

A.8 Optional Stopping Theorems In this section, we will prove a number of results that allow us to conclude that if X. is a submartingale and M :!g N are stopping times, then EXM < EXN. The first step is to show (1)

If X. is a uniformly integrable submartingale, then for any stopping time N, XNAf is uniformly integrable.

310

Appendix A Primer of Probability Theory

Proof We begin by observing that XNnn is a submartingale with EXlvnn < EX

(since X is a submartingale), so sup EX,',,, < sup EX.' < oo n

(since/Xn - X. in L1), and it follows from the martingale convergence theorem that XNAn -+ XN a.s. and EI XNI < oo. With this result established, the rest is easy, since M) < E(I XNI ; I XNI > M) + E(IXnI ; IXnI > M)

E(IXNnnI ;

(2)

and Xn is uniformly integrable. From (1) it follows immediately that we have: If Xn is a uniformly integrable submartingale, then for any stopping time N< oo,

EXo < EXN < EX..

Proof (2) in Section A.6 implies that EXa < EXNnn 5 EX.. Letting n - oo and observing that (1) above and results in Section A.7 imply that XNAf - XN in L' and X. - X. in L1, gives the desired result. From (2), we get the following useful corollary: (3)

The Optional Stopping Theorem. If XNAf is a uniformly integrable submartingale,

then for any stopping time M < N, EXM < EXN.

We have given (3) a name since it is the basic result in this section and it is usually the result we are referring to when we use the words "optional stopping theorem" in the text. In applying (3), the following fact is useful: (4)

If Xn is a submartingale and N < co is a stopping time with (a) EIXNI < oo (b) E(I XXI ; N > n) - 0, then XNnn is uniformly integrable.

Proof E(IXNnnI; IXNnnI > M) M) + E(IXXI;N> n, IXnI > M).

Let e > 0. If we pick no large enough, then E(IXnI ; N > n) < e/2 for all n > no. Having done this, we can pick M large enough so that E(I XNI ; I XNI > M) < E/2

and, for 0:!9 n :!g no, E(IXnI; IXnI > M) < e/2, so it follows from the first inequality that E(I XNnnI ; IXNnnI

> M) < E

for all n, and hence XN.,, is uniformly integrable.

Finally, there is one stopping theorem that does not require uniform integrability :

A.8

(5)

Options[ Stopping Theorem.

311

If Xn is a nonnegative supermartingale and N.-!5; oc is a stopping time, then EXo Z EXN, where X. = lim Xn (which exists by (4) of Section A.5). Proof EXo > EXNAf. The monotone convergence theorem implies that E(XN, N < oo) = lim E(XN ; N< n), n- w

and Fatou's lemma implies that E(XX ; N = oo) n), n-m

so adding the last two lines gives the desired result.

References

In the list of references below, we have for convenience shortened the standard abbreviations for the sources referred to most often : AMS BAMS

CMP CPAM JFA LMS

LNM MAA PAMS

PJM Sem.

TAMS TPA

ZfW

American Mathematical Society Bulletin of the American Mathematical Society Communications on Mathematical Physics Communications on Pure and Applied Mathematics Journal of Functional Analysis London Mathematical Society Lecture Notes in Mathematics, Springer-Verlag, New York Mathematical Association of America Proceedings of the American Mathematical Society Pacific Journal of Mathematics Seminar de Probabilites (Strasbourg) Transactions of the American Mathematical Society Theory of Probability and Its Applications Zeitschrift fur Wahrscheinlichkeitstheorie and Verwandte Gebiete

Agmon, S. (1965). Lectures on elliptic boundary value problems. Van Nostrand, New York. (1982). Lectures on exponential decay of solutions of second-order elliptic equaliona, Mathematical Notes. Princeton Univ. Press, Princeton. , Ahlfors, L. (1966). Complex analysis. 2d ed. McGraw-Hill, New York.

Aizenman, M., and B. Simon (1982). Brownian motion and a Harnack inequality for Schrodinger operators. CPAM 35:209-273. Ash, J. M., ed. (1976). Studies in harmonic analysis. MAA Studies in Mathematics, vol, 13, MAA, Washington, D.C.

Austin, D. G. (1966). A sample function property of martingales. Ann. Math, Slat, 37:1396-1397. Bachelier, L. (1900). Theorie de las speculation. Ann. Sci. Ecole Norm. Sup. 17:21-86, Baernstein, A., II (1978). Some sharp inequalities for conjugate functions. Indiana Math, J, 27:833-852. 313

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Bary, N. K. (1964). A treatise on trigonometric series. Pergamon Press, New York. Bernard, A., and B. Maisoneuve (1977). Decomposition atomique de martingales de la classe H1. In Sem. XII, pp. 303-323. Springer LNM 649. Bers, L., F. John, and M. Schechter (1964). Partial differential equations. AMS, Providence, RI. Berthier, A., and B. Gaveau (1978). Critere de convergence des fonctionelles de Kac et applications en mechanique et en geometrie. JFA 29:416-424. Billingsley, P. (1968). Convergence of probability measures. John Wiley, New York. Blumenthal, R. M., and R. K. Getoor (1968). Markov processes and their potential theory. Academic Press, New York. Breiman, L. (1968). Probability. Addison-Wesley, New York. Brelot, M., and J. L. Doob (1963). Limites angulaires et limites fines. Ann. Inst. Fourier (Grenoble) 13:395-415. Brossard, J. (1975). These de troisieme cycle. (1976). Comportement nontangentiel et comportement Brownien des fonctions

harmonique dans un demi-espace: Demonstration probabiliste d'un theoreme de Calderon et Stein. In Sem. XII, pp. 378-397. Springer LNM 649. Burkholder, D. L. (1962). Successive conditional expectation of an integrable function. Ann. Math. Stat. 33:887-893. (1964). Maximal inequalities as necessary conditions for almost everywhere convergence. ZfW 3:75-88. (1966). Martingale transforms. Ann. Math. Stat. 37:1494-1505. (1970). Martingale inequalities. In Martingales, ed. H. Dinges, pp. 1-8. Springer LNM 190. (1973). Distribution function inequalities for martingales. Ann. Prob. 1: 19-42. (1975). One-sided maximal functions and H°. JFA 18:429-454. (1976). Harmonic analysis and probability. In Studies in harmonic analysis, ed. Ash (1976), pp. 136-149. (1977a). Brownian motion and classical analysis. Proc. Symp. Pure Math. 31:5-14. (1977b). Exit times of Brownian motion, harmonic majorization and Hardy spaces. Adv. in Math. 26:182-205. (1978). Boundary value estimation of the range of an analytic function. Michigan Math. J. 25:197-211. (1979a). Martingale theory and harmonic analysis in Euclidean spaces. In Harmonic analysis in Euclidian space, ed. Weiss and Wainger (1979). (1979b). A sharp inequality for martingale transforms. Ann. Prob. 7:858-863.

(1981). A geometrical characterization of Banach spaces in which martingale difference sequences are unconditional. Ann. Prob. 9:997-1011. (1982). A nonlinear partial differential equation and the unconditional constant of the Haar system in L. BAMS (New Series) 7:591-595. (in press). Boundary value problems and sharp inequalities for martingale transforms. Ann. Prob.

Burkholder, D. L., B. Davis, and R. F. Gundy (1972). Integral inequalities for convex functions of operators and martingales. In Proceedings of the Sixth Berkeley Symposium, vol. II, pp. 223-240. Burkholder, D. L., and R. F. Gundy (1970). Extrapolation and interpolation of quasilinear operators on martingales. Acta Math. 124:249-304.

(1972). Distribution function inequalities for the area integral. Studia Math. 44:527-544.

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(1979b). Regularity projects of Schrodinger and Dirichlet semigroups. JI"A 33:259-296. Carmona, R., and B. Simon (1981). Pointwise bounds on eigenfunctions and wave packots in n-body quantum systems V. CMP 80:59-98. Chao, J. A., and M. H. Taibleson (1973). A subregularity inequality for conjugate systems on local fields. Studia Math. 46:249-257. Chung, K. L. (1974). A course in probability theory. 2d ed. Academic Press, New York. (1976). Excursions in Brownian motion. Archiv fur Math. 14:155-177. (1979). On stopped Feynman-Kac functionals. In Sem. XIV, pp. 347-356. Springer LNM 784. (1981). Feynman-Kac functional and the Schrodinger equation. In Seminar on Stochastic Processes I, ed. E. Cinlar, K. L. Chung, and R. Getoor. Birkhauser, Boston. (1982). Lectures from Markov processes to Brownian motion. Springer-Verlag, New York. Chung, K. L., R. Durrett, and Z. Zhongxin (1983). Extension of domains with finite gauge. Math. Annalen 264:78-79. Chung, K. L., and P. Li (1983). Comparison of probability and eigenvalue methods for the Schrodinger equation. Adv. in Math. Chung, K. L., and K. M. Rao (1980). Sur la theorie du potentiel avec la fonctionelle de

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Feynman-Kac. C. R. Acad. Sci. Paris 29OA: 629-63 1. (1981). Potential theory with the Feynman-Kac functional. Prob. Math. Stat. 1. Chung, K. L., ahd S. R. S. Varadhan (1979). Positive solutions of the Schrodinger equation in one dimension. Studia Math. 68:249-260. Ciesielski, Z., and S. J. Taylor (1962). First passage times and sojurn times for Brownian motion in space and exact Hausdorff measure. TAMS 103: 434-450. Coifman, R. R. (1974). A real variable characterization of H". Studia Math. 51:269-274. Coifman, R. R., and C. Fefferman (1974). Weighted norm inequalities for maximal functions and singular integrals. Studia Math. 51:241-250.

Coifman, R. R., and R. Rochberg (1980). Another characterization of BMO. PAMS 79:249-254. Coifman, R. R., and G. Weiss (1977). Extensions of Hardy spaces and their use in analysis. BAMS 83:569-645. Conway, J. B. (1978). Functions of one complex variable. 2d ed. Springer-Verlag, New York. Courant, R., and D. Hilbert (1953, 1962). Methods of mathematical physics. Vols. I, II. Interscience, New York. Davis, B. (1968). Comparison tests for martingale convergence. Ann. Math. Stat. 39:21412144.

(1969). A comparison test for martingale inequalities. Ann. Math. Stat. 40:505-508.

(1970). On the integrability of the martingale square function. Israel J. Math. 8:187-190. (1973a). On the distributions of conjugate functions of nonnegative measures. Duke Math. J. 40:695-700. (1973b). An inequality for the distribution of the Brownian gradient function. PAMS 37:189-194. (1974). On the weak type (1, 1) inequality for conjugate functions. PAMS 44:307311.

(1975). Picard's theorem and Brownian motion. TAMS 213:353-362. (1976). On Kolmogorov's inequalities 11a < Coll f 111, 0 < p < 1. TAMS 222:179192.

(1979a). Applications of the conformal invariance of Brownian motion. In Harmonic analysis in Euclidean space, ed. Weiss and Wainger (1979). (1979b). Brownian motion and analytic functions. Ann. Prob. 7:913-932. (1980). Hardy spaces and rearrangements. TAMS 261:211-233. Deift, P., W. Hunziker, B. Simon, and E. Vock (1978). Pointwise bounds on eigenfunctions

and wave packets in n-body quantum systems IV. CMP 64:1-34. Dellacherie, C. (1980). Un survol de la theorie de l'integrale stochastique. Stoch. Proc. Appl. 10:115-144. Dellacherie, C., and P. A. Meyer (1978). Probabilities and potential. English translation of Probabilites et potentiel. North Holland, Amsterdam. Doleans-Dade, C. (1970). Quelques applications de la formule de changements de variables pour les semimartingales. ZJW 16:181-194.

Donsker, M. D., and M. Kac (1950). A sampling method for determining the lowest eigenvalue and the principal eigenfunction of Schrodinger's equation. J. Res. Nat. Bureau of Standards 44:551-557. Donsker, M. D., and S. R. S. Varadhan (1974). Asymptotic evaluation of certain Wiener integrals for large time. In Proceedings of the International Conference on Function Space Integration. Oxford Univ. Press. (1975a). Asymptotics for the Wiener sausage. CPAM 28:525-566; errata, p. 677.

Rd.resoes

317

(1975b). On a variational formula for the principal eigenvaluc fur upeuiturs with a maximum principle. Proc. Nat. Acad. Sci., USA 72:780-783. (1975-1976). Asymptotic evaluation of certain Markov process exhrrlntwns for large time. CPAM (I) 28:1-47; (II) 28:279-301; (III) 29: 389-461. (1976). On the principal eigenvalue of second-order elliptic differontlal operators. CPAM 29:595-622. Doob, J. L. (1953). Stochastic processes. John Wiley, New York. (1954). Semimartingales and subharmonic functions. TA MS 77:86 121 (1955a). Martingales and one-dimensional diffusion. TA MS 78:168 208. (1955b). A probability approach to the heat equation. TAMS 80:216 280.

(1956). Probability methods applied to the first boundary value problem. In Proceedings of the Third Berkeley Symposium, vol. II, pp. 49 NO.

(1957). Conditional Brownian motion and the boundary limits of harmonic functions. Bull. Soc. Math. France 85:431-458. (1958a). Boundary limit theorems for a half-space. J. Math, Pure-,% Appl. 37:385392.

(1958b). Probability theory and the first boundary value prohlciii Illinois J. Math.

2:19-36. (1959). A nonprobabilistic proof of the relative Fatou theorem ,4nn. Inst. FourIa (Grenoble) 9:293-300. (1960). Relative limit theorems in analysis. J. Analyse Math. 8 2H) 306. (1961). Conformally invariant cluster value theory. Illinois J. Moth. 5:521-549. (1962). Boundary properties of functions with finite Dirichict integrals. Ann. Inst, Fourier (Grenoble) 12: 573-622. (1963). One-sided cluster-value theorems. Proc. LMS 13:461 470. (1964). Some classical function theory theorems and their modern versions. Ann, Inst. Fourier (Grenoble) 15:113-136.

(1966). Remarks on the boundary limits of harmonic functions. J. SIAMNumerical Analysis 3:229-235. Dubins, L. E., and D. Gilat (1978). On the distribution of maxima of mariingales. PAMS 68:337-338. Dubins, L. E., and G. Schwarz (1965). On continuous martingales. Proc. Nat. Acad. Sci., USA 53:913-916.

Dunford, N., and J. T. Schwarz (1957). Linear operators, Part I: General theory. Interscience, New York. Duren, P. L. (1970). The theory of H" spaces. Academic Press, New York. Duren, P. L., B. W. Romberg, and A. L. Shields (1969). Linear functionals on HP spaces with 0 < p < 1. J. Reine Angew. Math. 238: 32-60.

Durrett, R. (1982). A new proof of Spitzer's result on the winding of two-dimensional Brownian motion. Ann. Prob. 10:244-246. Dvoretsky, A., P. Erdos, and S. Kakutani (1961). Nonincreasing everywhere of the Brownian motion process. In Proceedings of the Fourth Berkeley Symposium, vol. II, pp. 103-116. Dynkin, E. B. (1960). Markov processes and related problems of analysis. Russian Math. Surveys 15, no. 2, pp. 1-24. (1963). Markov processes and problems in analysis. In Proceedings of the International Congress, Stockholm, 1962. AMS Translations, series 2, vol. 31, pp. 1-24, (1981). Markov processes and related problems of analysis Selected papers. LMS Lecture Notes. Cambridge Univ. Press, Cambridge.

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Feller, W. (1971). An introduction to probability theory and its applications. Vol. II. John Wiley, New York. Feynman, R. J. (1948). Space-time approach to nonrelativistic quantum mechanics. Rev. Mod. Phys. 20:367-387. Folland, G. B. (1976). Introduction to partial differential equations. Princeton Univ. Press, Princeton. Freedman, D. (1970). Brownian motion and diffusion. Holden-Day, San Francisco. Friedman, A. (1964). Partial differential equations of parabolic type. Prentice-Hall, Englewood Cliffs, NJ. (1969). Partial differential equations. Holt, Rinehart & Winston, New York. (1975). Stochastic differential equations and applications. Academic Press, New York. Garnett, J. (1979). Two constructions in BMO. In Harmonic analysis in Euclidean space, ed. Weiss and Wainger (1979). (1980). Bounded analytic functions. Academic Press, New York. Garnett, J., and P. Jones (1978). The distance in BMO to L'. Ann. Math. 108:373-393. (1982). BMO from dyadic BMO. PJM99:351-371. Garsia, A. M. (1970). Topics in almost everywhere convergence. Markham, Chicago.

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(1974b). Hp spaces of martingales 0 l Ho (A*X),

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91#0

186

M N

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186

ao(X)

211

Q(X)

296

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188

Kd

228

E(X I.F)

300

Cole Mathematics Series)

Cole

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Brooks Cole Laboratory Series for Chemistry

Mortal Cole

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Halo: The Cole Protocol

Halo: The Cole Protocol

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Cole, Abigail & Megan

Cole for Christmas

Cole, Abigail & Megan

Cole Mathematics Series)

Cole

Cole

Cole

Martina Cole

Brooks Cole Laboratory Series for Chemistry

Mortal Cole

Cole Cameron's Revenge

Halo: The Cole Protocol

Halo: The Cole Protocol

La doctora Cole

Cole for Christmas

Halo The Cole Protocol

The Cole Protocol

Halo: The Cole Protocol

Halo: The Cole Protocol

Halo: The Cole Protocol

Scout 'n' Cole

Halo The Cole Protocol

Sunset Express: An Elvis Cole Novel (Elvis Cole Novels)

Brownian Motion and Martingales in Analysis (Wadsworth & Brooks Cole Mathematics Series)

Introduction to Fourier Analysis and Wavelets (Brooks Cole Series in Advanced Mathematics)

Fourier Analysis and Its Applications (The Wadsworth and Brooks Cole Mathematics Series)

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Fourier Analysis and Its Applications (The Wadsworth and Brooks Cole Mathematics Series)

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