Potential Theory

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For other titles in this series, go to www.springer.com/series/223

Lester L. Helms

Potential Theory

123

Lester L. Helms University of Illinois Department of Mathematics 1409 W. Green St. Urbana IL 61801, USA [email protected]

Editorial board: Sheldon Axler, San Francisco State University Vincenzo Capasso, Università degli Studi di Milano Carles Casacuberta, Universitat de Barcelona Angus MacIntyre, Queen Mary, University of London Kenneth Ribert, University of California, Berkeley Claude Sabbah, CNRS, École Polytechnique Endre Süli, University of Oxford Wojbor Woyczy´nski, Case Western Reserve University

ISBN 978-1-84882-318-1 e-ISBN 978-1-84882-319-8 DOI 10.1007/978-1-84882-319-8 Springer Dordrecht Heidelberg London New York British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library Library of Congress Control Number: 2009926475 Mathematics Subject Classification (2000): 33E30, 35C15, 35J05, 35J15, 35J40, 35J55, 35J67 c Springer-Verlag London Limited 2009 ° Apart from any fair dealing for the purposes of research or private study, or criticism or review, as permitted under the Copyright, Designs and Patents Act 1988, this publication may only be reproduced, stored or transmitted, in any form or by any means, with the prior permission in writing of the publishers, or in the case of reprographic reproduction in accordance with the terms of licenses issued by the Copyright Licensing Agency. Enquiries concerning reproduction outside those terms should be sent to the publishers. The use of registered names, trademarks, etc., in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant laws and regulations and therefore free for general use. The publisher makes no representation, express or implied, with regard to the accuracy of the information contained in this book and cannot accept any legal responsibility or liability for any errors or omissions that may be made. Printed on acid-free paper Springer is part of Springer Science+Business Media (www.springer.com)

Dedicated to my wife Dorothea Helms

Preface

The ﬁrst six chapters of this book are revised versions of the same chapters in the author’s 1969 book, Introduction to Potential Theory. At the time of the writing of that book, I had access to excellent articles, books, and lecture notes by M. Brelot. The clarity of these works made the task of collating them into a single body much easier. Unfortunately, there is not a similar collection relevant to more recent developments in potential theory. A newcomer to the subject will ﬁnd the journal literature to be a maze of excellent papers and papers that never should have been published as presented. In the Opinion Column of the August, 2008, issue of the Notices of the American Mathematical Society, M. Nathanson of Lehman College (CUNY) and (CUNY) Graduate Center said it best “...When I read a journal article, I often ﬁnd mistakes. Whether I can ﬁx them is irrelevant. The literature is unreliable.” From time to time, someone must try to ﬁnd a path through the maze. In planning this book, it became apparent that a deﬁciency in the 1969 book would have to be corrected to include a discussion of the Neumann problem, not only in preparation for a discussion of the oblique derivative boundary value problem but also to improve the basic part of the subject matter for the end users, engineers, physicists, etc. Generally speaking, the choice of topics was intended to make the book more pragmatic and less esoteric while exposing the reader to the major accomplishments by some of the most prominent mathematicians of the eighteenth through twentieth centuries. Most of these accomplishments had their origin in practical matters as, for example, Green’s assumption that there is a function, which he called a potential function, which could be used to solve problems related to electromagnetic ﬁelds. This book is targeted primarily at students with a background in a senior or graduate level course in real analysis which includes basic material on topology, measure theory, and Banach spaces. I have tried to present a clear path from the calculus to classic potential theory and then to recent work

vii

viii

Preface

on elliptic partial diﬀerential equations using potential theory methods. The goal has been to move the reader into a fertile area of mathematical research as quickly as possible. The author is indebted to L. H¨ ormander and K. Miller for their prompt responses to queries about the details of some of their proofs. The author also thanks Karen Borthwick of Springer, UK, for guiding the author through the publication process. Urbana, Illinois September, 2008

Lester L. Helms

Contents

0

Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.2 Useful Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 4

1

Laplace’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Fundamental Harmonic Function . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 The Mean Value Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Poisson Integral Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Gauss’ Averaging Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 The Dirichlet Problem for a Ball . . . . . . . . . . . . . . . . . . . . . . . . . 1.8 Kelvin Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.9 Poisson Integral for Half-space . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.10 Neumann Problem for a Disk . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.11 Neumann Problem for the Ball . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.12 Spherical Harmonics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7 7 8 9 10 14 21 24 32 33 39 42 49

2

The Dirichlet Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Sequences of Harmonic Functions . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Superharmonic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Properties of Superharmonic Functions . . . . . . . . . . . . . . . . . . . . 2.5 Approximation of Superharmonic Functions . . . . . . . . . . . . . . . . 2.6 Perron-Wiener Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 The Radial Limit Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8 Nontangential Boundary Limit Theorem . . . . . . . . . . . . . . . . . . . 2.9 Harmonic Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

53 53 54 59 65 71 75 88 92 100

ix

x

Contents

3

Green Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Green Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Symmetry of the Green Function . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Green Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Riesz Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Properties of Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

107 107 107 115 122 135 144

4

Negligible Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Superharmonic Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Reduction of Superharmonic Functions . . . . . . . . . . . . . . . . . . . . 4.4 Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Boundary Behavior of the Green Function . . . . . . . . . . . . . . . . . 4.6 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7 Sweeping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

149 149 149 158 163 178 182 189

5

Dirichlet Problem for Unbounded Regions . . . . . . . . . . . . . . . . 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Exterior Dirichlet Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 PWB Method for Unbounded Regions . . . . . . . . . . . . . . . . . . . . 5.4 Boundary Behavior . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Intrinsic Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Thin Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7 Thinness and Regularity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

197 197 197 204 210 223 224 228

6

Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Energy Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Mutual Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Projections of Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Wiener’s Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

241 241 241 249 257 260

7

Interpolation and Monotonicity . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 H¨ older Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Global Interpolation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Interpolation of Weighted Norms . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Inner Norms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6 Monotonicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

267 267 268 273 280 284 288

8

Newtonian Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Subnewtonian Kernels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Poisson’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 H¨ older Continuity of Second Derivatives . . . . . . . . . . . . . . . . . . . 8.5 The Reﬂection Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

303 303 303 316 319 326

Contents

9

xi

Elliptic Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Linear Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Constant Coeﬃcients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4 Schauder Interior Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5 Maximum Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.6 The Dirichlet Problem for a Ball . . . . . . . . . . . . . . . . . . . . . . . . . 9.7 Dirichlet Problem for Bounded Domains . . . . . . . . . . . . . . . . . . . 9.8 Barriers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

333 333 334 335 338 343 347 353 358

10 Apriori Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Green Function for a Half-space . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Mixed Boundary Conditions for Laplacian . . . . . . . . . . . . . . . . . 10.4 Nonconstant Coeﬃcients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

371 371 372 376 385

11 Oblique Derivative Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Boundary Maximum Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Curved Boundaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.4 Superfunctions for Elliptic Operators . . . . . . . . . . . . . . . . . . . . . 11.5 Regularity of Boundary Points . . . . . . . . . . . . . . . . . . . . . . . . . . .

391 391 391 405 410 420

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 439

Chapter 0

Preliminaries

0.1 Notation An element of the n-dimensional Euclidean space Rn , n ≥ 2, will n be denoted by x = (x1 , . . . , xn ) and its length |x| is deﬁned by |x| = ( i=1 x2i )1/2 . If . . , xn ) and y = (y1 , . . . , yn ) ∈ Rn , the inner product x · y is deﬁned x = (x1 , . n by x·y = i=1 xi yi , and the distance between x and y is deﬁned to be |x−y|. The angle between two nonzero vectors x and y is deﬁned to be the angle φ such that 0 ≤ φ ≤ π and x·y . cos φ = |x||y| The zero vector of Rn is denoted by 0 = (0, . . . , 0). The ball Bx,ρ with center x and radius ρ is deﬁned by Bx,ρ = {y; |x − y| < ρ}; the sphere with center x and radius ρ is deﬁned by Sx,ρ = ∂Bx,ρ = {y; |x−y| = ρ}. The closure of a set F ⊂ Rn is denoted by F − or by cl F , its interior by int F , and its complement by ∼ F . The standard basis for Rn will be denoted by {ε1 , . . . , εn } where εi has a 1 in the ith position and 0 in all other positions. The spherical coordinates of a point x = (x1 , . . . , xn ) = 0 in Rn are deﬁned as follows: if r = |x|, then θ=

x

1

r

,...,

xn r

is a point of S0,1 = ∂B0,1 , the unit sphere with center at 0. The r and θ of the ordered pair (r, θ) are called the spherical coordinates of the point x. This transformation from rectangular coordinates to spherical coordinates is essentially the mapping (x1 , . . . , xn ) → (θ1 , . . . , θn−1 , r) where

L.L. Helms, Potential Theory, Universitext, c Springer-Verlag London Limited 2009

1

2

0 Preliminaries

x1 r x2 θ2 = r .. . xn−1 θn−1 = r r = (x21 + · · · + x2n )1/2 . θ1 =

Letting θn = xn /r, θn is the cosine of the angle between x and the vector (0, . . . , 0, 1); that is, the angle between x and the xn -axis. The Jacobian of the map is easily calculated and its absolute value is given by ∂(x1 , . . . , xn ) rn−1 rn−1 = . = ∂(θ1 , . . . , θn−1 , r) (1 − θ2 − · · · − θ2 )1/2 |θn | 1 n−1 Surface area on a sphere Sx,ρ is a ﬁnite measure on the Borel subsets of Sx,ρ . The integral of a Borel measurable function f on Sx,ρ relative to surface area will be denoted by f dσ or f (x) dσ(x). Sx,ρ

Sx,ρ

The volume of Bx,ρ ⊂ Rn and surface area of Sx,ρ will be denoted by νn (ρ) and σn (ρ), respectively; when ρ = 1 these quantities will be denoted by νn and σn , respectively. The νn (ρ) and σn (ρ) are related by the equations νn (ρ) = ρn νn , σn (ρ) = ρn−1 σn , and νn = σn /n. Derivation of these equations can be found in Helms [28]. The σn are given by ⎧ πn/2 n if n is even , ⎪ ⎨ (n/2)! (0.1) σn = ⎪ ⎩ 2(n+1)/2 π(n−1)/2 if n is odd and > 1. 1·3·5···(n−2) Consider an extended real-valued function f deﬁned on a subset of Rn . When f is considered as a function of the spherical coordinates (r, θ) of x, f (x) will be denoted by f (r, θ). Suppose f is integrable on a ball B0,ρ . Then the integral of f over B0,ρ can be evaluated using spherical coordinates as in ρ n−1 f (x) dx = r f (r, θ) dσ(θ) dr. B0,ρ

0

|θ|=1

If β1 , . . . , βn are nonnegative n integers, β = (β1 , . . . , βn ) is called a multiindex, |β| is deﬁned by |β| = i=1 βi , and β! is deﬁned by β! = β1 ×β2 ×. . .× βn . If the real-valued function u = u(x1 , . . . , xn ) is deﬁned on a neighborhood of a point x ∈ Rn , put Di u = Dxi u = ∂u/∂xi , Dij u = Dxi xj u = ∂ 2 u/∂xi ∂xj ; more generally, if β = (β1 , . . . , βn ) is a multi-index, put

0.1 Notation

3

Dβ u(x) =

∂ |β|u ∂xβ1 1

· · · ∂xβnn

(x)

provided the indicated partial derivatives are deﬁned. The gradient ∇u of u at x is deﬁned by ∇u(x) = (D1 u(x), . . . , Dn u(x)). If the function u is a function of x, y ∈ Rn , then derivatives with respect to β u. just one of the variables, say x, will be denoted by D(x) n If Ω is an open subset of R and f is a bounded function on Ω, the norm of f is deﬁned by f 0,Ω = supx∈Ω |f (x)|. The set of continuous functions on Ω will be denoted by C 0 (Ω), the set of bounded continuous functions on Ω will be denoted by Cb0 (Ω), and C 0 (Ω − ) will denote the set of functions on Ω having continuous extensions to Ω − . If f ∈ C 0 (Ω) has a continuous extension to Ω − , the same symbol f will be used for its extension. If f ∈ C 0 (Ω) and limy→x,y∈Ω f (y) exists in R for all x ∈ ∂Ω, then f has a continuous extension to Ω − . If k is a positive integer or ∞, C k (Ω) will denote the set functions having continuous derivatives of all orders less than or equal to k on Ω; C k (Ω − ) will denote the set of functions on Ω with derivatives of order less than or equal to k having continuous extensions to Ω − . If k is a positive integer, Dk u 0,Ω will denote sup|β|=k Dβ u 0,Ω . A subscript 0 on C k (Ω) will signify the subset C0k (Ω) of functions in C k (Ω) having compact support in Ω. A measure on the Borel subsets of Ω which is ﬁnite on compact subsets of Ω is called a Borel measure and is determined by its values on the compact subsets of Ω. If u is an extended real-valued function with domain Ω ⊂ Rn , x ∈ Ω, and u(x) = +∞ (u(x) = −∞), then u is continuous in the extended sense at x if limy→x,y∈Ω u(y) = +∞ (limy→x,y∈Ω (y) = −∞); u is continuous in the extended sense on Ω if u is either continuous or continuous in the extended sense at each point of Ω. Let u be an extended real-valued function with domain Ω ⊂ Rn . For each y ∈ Rn , let N (y) be the collection of neighborhoods of y. If Ω0 ⊂ Ω and x is any point of Ω0− , deﬁne

lim inf u(y) = sup u(y) inf y→x,y∈Ω0

N ∈N (x)

y∈(N ∩Ω0 )∼{x}

lim sup u(y) = y→x,y∈Ω0

inf

N ∈N (x)

sup

u(y) ;

y∈(N ∩Ω0 )∼{x}

if Ω0 = Ω, the notation lim inf y→x u(y) and lim supy→x u(y) will be used. The function u is lower semicontinuous (l.s.c.) at x ∈ Ω if u(x) ≤ lim inf y→x u(y); upper semicontinuous (u.s.c.) at x ∈ Ω if u(x) ≥ lim supy→x u(y). If u is l.s.c. on Ω, then cu is l.s.c. or u.s.c. on Ω according as c ≥ 0 or c ≤ 0. If u is l.s.c. on Ω, then −u is u.s.c. on Ω. The function u is l.s.c. on Ω if and only if the set {x ∈ Ω; u(x) > c} is relatively open

4

0 Preliminaries

in Ω for every real number c. If {ua ; a ∈ A} is a family of l.s.c. functions with common domain Ω, then u = supa∈A ua is l.s.c. on Ω. If u and v are l.s.c. functions with a common domain Ω, then min (u, v) is l.s.c. as is u + v if deﬁned on Ω. If u is any extended real-valued function on Ω ⊂ Rn , the lower ˆ(x) = lim inf y→x u(y) is l.s.c. regularization uˆ of u deﬁned for x ∈ Ω − by u on Ω − . If u is l.s.c. on the compact set Ω ⊂ Rn , then it attains its minimum value on Ω. Moreover, if u is l.s.c. on Ω ⊂ Rn and there is a continuous function f on Rn such that u ≥ f on Ω, then there is an increasing sequence of continuous functions {fj } on Rn such that limj→∞ fj = u on Ω.

0.2 Useful Theorems In this section, several results used repeatedly in subsequent chapters will be stated without proofs. Proofs can be found in many intermediate real analysis texts; e.g., Apostol [1]. The following theorem is easily proved using the onedimensional version by introducing a function g(t) = u(tx + (1 − t)y), 0 ≤ t ≤ 1. Theorem 0.2.1 (Mean Value Theorem) If Ω is an open subset of Rn , u has continuous ﬁrst partial derivatives on Ω, x and y are two points of Ω such that the line segment {tx + (1 − t)y; 0 ≤ t ≤ 1} ⊂ Ω, then there is point z on the line segment joining x and y such that u(y) − u(x) = ∇u(z) · (y − x). The relationship of the limit of a convergent sequence of functions to the limit of the corresponding sequence of derivatives is the content of the next theorem. Theorem 0.2.2 Let {fj } be a sequence of real-valued functions on an open interval (a,b) having real-valued derivatives at each point of (a, b) such that a sequence {fj (x0 )} converges for some x0 ∈ (a, b). If the sequence {fj } converges uniformly to a function g on (a, b), then (i) the sequence {fj } converges uniformly to a function f on (a, b), and (ii) for each x ∈ (a, b), f (x) exists and f (x) = g(x). Recall that a family F of functions f on Ω ⊂ Rn is equicontinuous at x ∈ Ω if for each > 0 there is a δ > 0 such that |f (x) − f (y)| < for all f ∈ F and all y ∈ Bx,δ ∩ Ω, equicontinuous on Ω if equicontinuous at each point of Ω, and uniformly equicontinuous if the above δ does not depend upon x. Theorem 0.2.3 (Arzel` a-Ascoli Theorem) Let {fj } be a sequence of uniformly bounded and equicontinuous functions on a compact set K ⊂ Rn . Then there is a subsequence {fji } which converges uniformly on K to a continuous function f .

0.2 Useful Theorems

5

The Arzel`a-Ascoli theorem is valid for generalized sequences that are deﬁned as follows. Recall that a nonempty set A and a relation constitute a directed set if is reﬂexive and transitive, and for each pair α, β ∈ A there is a γ ∈ A such that γ α and γ β. A net {xα ; α ∈ A} is a function from the directed set A into a set X. If X is a metric space with metric d, the net {xα ; α ∈ A} converges to x ∈ X, written limA xα = x, if for every > 0 there is an α ∈ A such that d(xα , x) < for all α α . Theorem 0.2.4 (Arzel` a-Ascoli Theorem) Let {fα ; α ∈ A} be a net of uniformly bounded and equicontinuous functions on a compact set K ⊂ Rn . Then there is a subnet {fαβ } which converges uniformly on K to a continuous function f . If (X, M) is a measurable space, a signed measure μ on M is an extended real-valued, countably additive set function on M with μ(∅) = 0 and taking on only one of the two values +∞, −∞. A signed measure μ on a measurable space (X, M) has a decomposition μ = μ+ − μ− into a positive part μ+ and a negative part μ− . The total variation |μ| of μ is deﬁned by |μ| = μ+ + μ− . If both μ+ and μ− are ﬁnite measures, μ is said to be of bounded variation on X, in which case the norm of μ is deﬁned by μ = |μ|(X). If X is a locally compact Hausdorﬀ space, the smallest σ-algebra of subsets of X containing the closed subsets of X is called the class of Borel sets of X and is denoted by B(X). A measure μ on B(X) is a Borel measure if ﬁnite on each compact set. All Borel measures will be assumed to be regular Borel measures; that is, μ(Λ) = sup {μ(Γ ); Γ ⊂ Λ, Γ compact }, Λ ∈ B(X). If {μj } is a sequence of signed measures on B(X) of bounded variation, the sequence converges in the w∗ - topology to a signed measure μ of bounded variation if lim f dμj = f dμ j→∞

Cb0 (X).

for all f ∈ Lebesgue measure on Rn will be denoted by μn . The function u on the open set Ω ⊂ Rn is locally integrable on Ω if integrable with respect to Lebesgue measure on each compact subset of Ω. The following theorem is a useful tool for extracting a convergent subsequence from a sequence of signed measures. Theorem 0.2.5 Let {μj } be a sequence of signed measures on the Borel subsets of a compact metric space X with μj ≤ M for all j ≥ 1. Then there is a signed measure μ on the Borel subsets of X with u ≤ M and a subsequence {μji } that converges to μ in the w∗ -topology ; that is, for all f ∈ C 0 (X). f dμji = f dμ lim i→∞

Sketch of Proof: Since X is a compact metric space, C 0 (X) is separable. Let {fk } be countable dense subset of C 0 (X). For each k ≥ 1, the

6

0 Preliminaries

(k) fk d μj is bounded, and there is a subsequence {μj } of (1) (k−1) (k) {μj }, . . . {μn } such that limj→∞ fk dμj exists. The

sequence

the se-

quences Cantor diagonalization procedure then can be invoked to assert the existence of a subsequence that works for all fk . Corollary 0.2.6 If {fj } is a sequence in C 0 (X) that converges uniformly to f ∈ C 0 (X) and {μj } is a sequence of signed measures with μj ≤ M < +∞ for all j ≥ 1 that converges to μ in the w∗ - topology, then lim fj dμj = f dμ. j→∞

Sketch of Proof: Given > 0, there is an j( ) such that | f dμj − f dμ| < /2 and |fj − f | < /2M for all j > j( ). Apply triangle inequality. A linear space X is a normed linear space if to each x ∈ X there is associated a real number x with the following properties: (i) x ≥ 0 and x = 0 if and only if x = 0, (ii) αx = |α| x for all α ∈ R, x ∈ X, and (iii) x + y ≤ x + y for all x, y ∈ X. The function · is called a norm on X. If, in addition, (X, d) is a complete metric space where d(x, y) = x−y , then X is called a Banach space under the norm · . If it is necessary to consider more than one Banach space, the norm will be denoted by · X . If only the ﬁrst part of (i) holds, then ·

is called a seminorm on X. A linear operator on X is a linear map whose domain D(A) is a subspace of X and range R(A) is a subset of X. If A is a linear operator on the Banach space X with domain D(A) = X and there is a constant M such that Ax ≤ M x for all x ∈ X, then A is called a bounded linear operator, in which case the norm of A is deﬁned by

A = sup x =0

A linear operator need not be bounded.

Ax

.

x

Chapter 1

Laplace’s Equation

1.1 Introduction Potential theory has its origins in gravitational theory and electromagnetic theory. The common element of these two is the inverse square law governing the interaction of two bodies. The concept of potential function arose as a result oﬀ the work done in moving a unit charge from one point of space to another in the presence of another charged body. A basic potential function 1/r, the reciprocal distance function, has the important property that it satisﬁes Laplace’s equation except when r = 0. Such a function is called a harmonic function. Since the potential energy at a point due to a distribution of charge can be regarded as the sum of a large number of potential energies due to point charges, the corresponding potential function should also satisfy Laplace’s equation. Applications to electromagnetic theory led to the problem of determining a harmonic function on a region having prescribed values on the boundary of the region. This problem came to be known as the Dirichlet problem. A similar problem, connected with steadystate heat distribution, asks for a harmonic function with prescribed ﬂux or normal derivative at each point of the boundary. This problem is known as the Neumann problem. Another problem, Robin’s problem, asks for a harmonic function satisfying a condition at points of the boundary which is a linear combination of prescribed values and prescribed ﬂux. In this chapter, explicit formulas will be developed for solving the Dirichlet problem for a ball in n-space, uniqueness of the solution will be demonstrated, and the solution will be proved to have the right “boundary values.” Not nearly as much is possible for the Neumann problem. Explicit formulas for solving the Neumann problem for a ball are known only for the n = 2 and n = 3 cases.

L.L. Helms, Potential Theory, Universitext, c Springer-Verlag London Limited 2009

7

8

1 Laplace’s Equation

1.2 Green’s Theorem In this section, only real-valued functions on a bounded open set Ω ⊂ Rn will be considered. The Ω will be required to have a smooth boundary, this being the case when Ω is a ball in Rn or the interior of the region between two spheres, of which one is inside the other. Let v = (v1 , . . . , vn ) be a vector-valued function on Ω whose components vj have continuous extensions to Ω − and continuous ﬁrst partials on Ω. The divergence of v is deﬁned by div v =

n ∂vj . ∂xj j=1

The outer unit normal to the surface ∂Ω at a point x ∈ ∂Ω will be denoted by n(x). The starting point for this chapter is the following theorem which can be found in any advanced calculus textbook (e.g., [43]). Theorem 1.2.1 (Gauss’ Divergence Theorem) If v ∈ C 0 (Ω − )∩C 1 (Ω), then div v dx = (v · n) dσ Ω

∂Ω

where dσ denotes integration with respect to surface area. By convention, whenever “n” appears in an integral over a smooth surface it is understood to be the outer unit normal to the surface. Suppose now that u is a function deﬁned on a neighborhood of Ω − and has continuous second partials on that neighborhood. The Laplacian of u, Δu, is deﬁned by n ∂2u Δu = . ∂x2j j=1 If u is also a function of variables other than x and it necessary to clarify the meaning of the Laplacian, Δ(x) will signify that the Laplacian is relative to the coordinates of x. Let v be a second such function. Then u∇v = (uD1 v, . . . , uDn v) and div(u∇v) = uΔv + (∇u · ∇v). It follows from the divergence theorem that uΔv dx + (∇u · ∇v) dx = (u∇v · n) dσ Ω Ω ∂Ω = uDn v dσ ∂Ω

1.3 Fundamental Harmonic Function

9

since (u∇v, n) = u(∇v, n) and the latter inner product is just the directional derivative Dn v of v in the direction n. The following important identity is obtained by interchanging u and v and subtracting. Theorem 1.2.2 (Green’s Identity) If u, v ∈ C 2 (Ω − ), then (uΔv − vΔu) dx = (uDn v − vDn u) dσ. Ω

∂Ω

1.3 Fundamental Harmonic Function A real-valued function u on Rn having continuous second partials is called a harmonic function if Δu = 0 on Rn . Example 1.3.1 If f is an analytic function of a complex variable z, then the real and imaginary parts of f are harmonic functions. This fact follows from the Cauchy-Riemann equations. Thus, u(x, y) = ex cos y and v(x, y) = ex sin y are harmonic functions since they are the real and imaginary parts, respectively, of the analytic function f (z) = ez . Deﬁnition 1.3.2 The function u on an open set Ω ⊂ Rn is harmonic on Ω if u ∈ C 2 (Ω) and Δu = 0 on Ω. Remark 1.3.3 Note that harmonicity is preserved under rigid motions of Rn ; that is, if τ : Rn → Rn is an orthogonal transformation followed by a translation, τ (Ω) = {τ x; x ∈ Ω}, and uτ (y) = u(τ −1 y), y ∈ τ (Ω), then uτ is harmonic on Ω τ whenever u is harmonic on Ω. This is easy to verify if τ is a translation given by τ x = x + a, x ∈ Rn , for some ﬁxed a ∈ Rn . Suppose τ is deﬁned by the equation y = Ax where A is an orthogonal matrix. Since A−1 = AT , ∂u(AT y) ∂yj ∂u(AT y) ∂u(x) (x) = = aji ∂xi ∂yj ∂xi ∂yj j=1 j=1 n

and

Since

n

∂ 2u ∂ 2 u(AT y) (x) = a a . ji ki ∂x2i ∂yj ∂yk j=1 n

n i=1

n

k=1

aji aki is 1 or 0 according as j = k or j = k, respectively, Δ(x) u(x) = Δ(y) u(AT y) = Δ(y) u(A−1 y) = Δ(y) uτ (y).

Therefore, Δ(y) uτ (y) = 0 whenever y = Ax and Δ(x) u(x) = 0.

10

1 Laplace’s Equation

Remark 1.3.4 Note that directional derivatives are preserved under translation since this is true of ordinary derivatives. Using properties of orthogonal transformations, as in the above remark, it is easy to see that directional derivatives are preserved under orthogonal transformations and, in fact, that inner products are also preserved. Suppose u is harmonic on the open set Ω and y ∈ Ω. Then u(x) can be regarded as a function of the spherical coordinates (r, θ) relative to y where r = |x − y| and θ is the point of intersection of the line segment joining x and y and a unit sphere with center at y. Suppose that u is a function of r only. Then Δu, as a function of spherical coordinates, is easily seen to be given by Δu =

d2 u (n − 1) du , + dr2 r dr

r = 0.

The only functions that are harmonic on Rn ∼ {y} and a function of r only are those that satisfy the equation d2 u (n − 1) du =0 + dr2 r dr on Rn ∼ {y}. If n = 2, the general solution of this equation is A log r + B, where A and B are arbitrary constants. The particular solution u(r) = − log r is harmonic on Rn ∼ {y} and is called the fundamental harmonic function for R2 with pole y. If n ≥ 3, the general solution is Ar−n+2 + B; the particular solution u(r) = r−n+2 is called the fundamental harmonic function for Rn with pole y. The notations uy (x) and u(|x − y|) will be used interchangeably.

1.4 The Mean Value Property Most of the theorems about harmonic functions have their origin in the works of Gauss and Green concerning electromagnetic and gravitational forces. Theorem 1.4.1 (Gauss’ Integral Theorem) If u is harmonic on the ball B and u ∈ C 1 (B − ), then Dn u dσ = 0. ∂B

Proof: Put v = 1 in Green’s Identity, Theorem 1.2.2. Much more can be obtained from Green’s Identity. Theorem 1.4.2 (Green’s Representation Theorem) If u has continu− , then ous second partial derivatives on a closed ball B − = By,ρ

1.4 The Mean Value Property

11

(i) for n = 2 and x ∈ B, 1 u(x) = ((− log r)Dn u − uDn (− log r)) dσ(z) 2π ∂B 1 − Δu(− log r) dz; 2π B (ii) for n ≥ 3 and x ∈ B, −n+2 1 u(x) = r Dn u − uDn r−n+2 dσ(z) σn (n − 2) ∂B 1 − Δu · r−n+2 dz, σn (n − 2) B where r = |x − z|, z ∈ B − . Proof of (ii): Consider a ﬁxed x ∈ B, and let v(z) = |x − z|−n+2 = r−n+2 for z = x. Then v is harmonic on Rn ∼ {x}. Consider δ > 0 for which − − ⊂ By,ρ , and let Ω be the open set By,ρ ∼ Bx,δ . By Green’s Identity, Bx,δ

(uΔv − vΔu) dz =

(uDn v − vDn u) dσ(z).

Ω

∂Ω

Since v is harmonic on Ω and ∂Ω = ∂By,ρ ∪ ∂Bx,δ , vΔu dz = (uDn v − vDn u) dσ(z) − Ω

∂By,ρ

−

(uDn v − vDn u) dσ(z).

(1.1)

∂Bx,δ

The minus sign precedes the second term on the right because the outer unit normal to ∂Ω at a point of ∂Bx,δ is the negative of the outer unit normal to ∂Bx,δ . The next step is to let δ → 0 in Equation (1.1). If it can be shown that v is integrable over By,ρ , it would follow from the boundedness of Δu that vΔu is integrable and that lim vΔu dz = vΔu dz (1.2) δ→0

Ω

By,ρ

by the Lebesgue dominated convergence theorem, applied sequentially. Since v is bounded on Ω, it need only be shown that v is integrable on Bx,δ . By transforming to spherical coordinates with pole x,

|v| dz = σn

Bx,δ

0

δ

r−n+2 rn−1 dr = σn δ 2 /2.

12

1 Laplace’s Equation

Therefore, v is integrable on Bx,δ and Equation (1.2) holds. Consider the right side of Equation (1.1). The ﬁrst integral does not depend upon δ and only the second must be considered. Since n |Dn u| = |∇u · n| ≤ |n||∇u| = |∇u| = ( u2xi )1/2 , i=1 − and the ﬁrst partials of u are bounded on By,ρ , |Dn u| ≤ m on ∂Bx,δ for some constant m. Thus, vDn u dσ(z) ≤ m r−n+2 dσ(z) = mσn δ ∂Bx,δ ∂Bx,δ

and therefore,

lim

δ→0

vDn u dσ(z) = 0. ∂Bx,δ

Now consider

uDn v dσ(z).

lim

δ→0

∂Bx,δ

Since Dn v = Dr v = (−n + 2)δ −n+1 at points of ∂Bx,δ ,

1 uDn v dσ(z) = −σn (n − 2) u dσ(z) . σn δ n−1 ∂Bx.δ ∂Bx,δ The factor in parentheses is just the average of the continuous function u on ∂Bx,δ , and therefore has the limit u(x) as δ → 0. This shows that uDn v dσ(z) = −(n − 2)σn u(x). lim δ→0

∂Bx,δ

Taking the limit as δ → 0 in Equation (1.1), − vΔu dz = (uDn v − vDn u) dσ(z) + (n − 2)σn u(x), B

∂B

which is the equation in (ii). The proof of (i) is basically the same. According to the following theorem, the value of a harmonic function at a point y is equal to the average of its values over a sphere centered at y. Theorem 1.4.3 (Mean Value Property) If u is harmonic on a neighbor− , then hood of the closed ball By,ρ 1 u(y) = u(z) dσ(z). σn ρn−1 ∂By,ρ

1.4 The Mean Value Property

13

Proof: Only the n ≥ 3 case will be proved since the n = 2 case is similar. By Theorem 1.4.2 and the fact that Δu = 0 on B = By,ρ , 1 u(y) = |y − z|−n+2 Dn u − uDn |y − z|−n+2 dσ(z). σn (n − 2) ∂B For z ∈ ∂B, |y − z|−n+2 = ρ−n+2 and Dn |y − z|−n+2 = Dr r−n+2 |r=ρ = −(n − 2)ρ−n+1 . After substituting in the above integral, ρ−n+2 1 u(y) = Dn u dσ(z) + u dσ(z). σn (n − 2) ∂B σn ρn−1 ∂B The ﬁrst integral on the right is zero by Gauss’ Integral Theorem. Averaging over a sphere in this theorem can be replaced by averaging over a solid ball. Theorem 1.4.4 If u is harmonic on the ball B = By,ρ , then 1 u(z) dz. u(y) = νn ρn B Proof: Using spherical coordinates (r, θ) relative to the pole y, ρ 1 1 n−1 u(z) dz = r u(r, θ) dσ(θ) dr. νn ρn B νn ρn 0 |θ|=1 The integral within the parentheses is just the integal over a sphere of radius r relative to a uniformly distributed measure of total mass σn and is equal to σn u(y) by the mean value property. Therefore, ρ 1 1 σn u(z) dz = rn−1 σn u(y) dr = u(y). n n νn ρ B νn ρ 0 nνn The result follows from the fact that νn = σn /n. The content of the two preceding theorems is summarized by saying that the harmonic functions are mean valued or satisfy the averaging principle. Averages occur so frequently that a notation for them will be introduced. If u is a function integrable relative to surface area on the boundary ∂B of B = By,ρ , let 1 u dσ. L(u : y, ρ) = σn ρn−1 ∂B

14

1 Laplace’s Equation

If u is integrable on B relative to Lebesgue measure, let 1 A(u : y, ρ) = u dx. νn ρn B Using this notation, if h is harmonic on the open set R, then h(y) = L(h : y, ρ) = A(h : y, ρ) − whenever By,ρ ⊂ R. A partial converse will be proved later.

1.5 Poisson Integral Formula In the preceding section, it was shown that the value of a harmonic function at a point y is equal to the average of its values over a sphere centered at y. It is natural to ask if a similar result is true of any point inside the sphere. If − , then Green’s u is harmonic on a neighborhood of the closed ball B − = By,ρ representation theorem in the n ≥ 3 cases reduces to 1 u(x) = σn (n − 2)

(r−n+2 Dn u − uDn r−n+2 ) dσ(z)

∂B

where r = |x − z|, x ∈ B, z ∈ ∂B. This equation suggests the possibility of representing the harmonic function u in terms of the restriction of u to the boundary of B. For instance, if for each x ∈ B a harmonic function vx could be found such that r−n+2 + vx (z) vanishes on ∂B, then replacing r−n+2 by this sum would result in a representation of u in terms of just u|∂B and would not involve Dn u on ∂B. This is what will be done in the proof of the next theorem for the n ≥ 3 case; the proof in the n = 2 case is similar. The proof requires a mapping that allows passage back and forth between the interior and exterior of a sphere, called an inversion, which was ﬁrst introduced by Lord Kelvin [59]. Fix B = By,ρ and x ∈ B, and consider the radial line joining y to x. For x = y, choose x∗ on this radial line so that |x − y||x∗ − y| = ρ2 . Then x∗ = y +

ρ2 (x − y) |y − x|2

(1.3)

(1.4)

and is called the inverse of x relative to the sphere ∂B. Consider any z ∈ ∂B, and let φ be the angle between z − y and x − y. Then

1.5 Poisson Integral Formula

15

|z − x∗ |2 = ρ2 + |x∗ − y|2 − 2ρ|x∗ − y| cos φ |z − x|2 = ρ2 + |x − y|2 − 2ρ|x − y| cos φ. Replacing |x∗ − y| by ρ2 /|x − y| in the ﬁrst equation, solving for cos φ in the second, and substituting for cos φ in the ﬁrst, |z − x∗ |2 = ρ2

|z − x|2 . |y − x|2

Thus, for x ∈ B, x = y, z ∈ ∂B, and x∗ the inverse of x relative to ∂B, |y − x| |z − x∗ | = 1. ρ |z − x|

(1.5)

Theorem 1.5.1 If u is harmonic on a neighborhood of the closure of the ball B = By,ρ , x ∈ B, and x = y, then (i) for n = 2, u(x) = −

1 2π

|y − x| |z − x∗ | uDn log dσ(z), ρ |z − x| ∂B

(ii) for n ≥ 3, u(x)= −

1 σn (n − 2)

uDn ∂B

1 ρn−2 1 − n−2 |z − x| |y − x|n−2 |z − x∗ |n−2

dσ(z)

where x∗ is the inverse of x relative to ∂By,ρ . Proof of (ii): By Green’s representation theorem,

1 1 1 D u − uD u(x) = dσ(z) (1.6) n n σn (n − 2) ∂B |z − x|n−2 |z − x|n−2 since Δu = 0 on B. Since x∗ ∈ B − , |z − x∗ |−n+2 is harmonic on a neighborhood of B − and it follows from Green’s Identity that

1 1 1 D u − uD 0= dσ(z). (1.7) n n σn (n − 2) ∂B |z − x∗ |n−2 |z − x∗ |n−2 Multiplying both sides of this equation by α and subtracting from Equation (1.6), 1 1 1 −α u(x) = Dn u σn (n − 2) ∂B |z − x|n−2 |z − x∗ |n−2

1 1 dσ(z). −α −uDn n−2 |z − x| |z − x∗ |n−2

16

1 Laplace’s Equation

Letting α = (ρ/|x − y|)n−2 and using Equation (1.5), α=

ρn−2 |z − x∗ |n−2 = , |y − x|n−2 |z − x|n−2

z ∈ ∂B,

and the integral of the ﬁrst term on the right is zero. The functions within the parentheses in (i) and (ii) of the preceding theorem are called Green functions for the ball B. Deﬁnition 1.5.2 If n = 2, the Green given by ⎧ |z−x∗ | ⎪ log |y−x| ⎪ ρ |z−x| ⎪ ⎨ ρ GB (x, z) = log |z−x| ⎪ ⎪ ⎪ ⎩ +∞

function for the ball B = By,ρ is for z ∈ B ∼ {x}, x = y for z ∈ B ∼ {x}, x = y for z = x

where x∗ is the inverse of x with respect to ∂B. If n ≥ 3, the Green function is given by ⎧ ρn−2 1 1 ⎪ ⎪ |z−x|n−2 − |x−y|n−2 |z−x∗ |n−2 for z ∈ B ∼ {x}, x = y ⎪ ⎨ 1 1 GB (x, z) = |z−x| for z ∈ B ∼ {x}, x = y n−2 − ρn−2 ⎪ ⎪ ⎪ ⎩ +∞ for z = x. Using Equation (1.5), it is easy to see that limz→z0 GB (x, z) = 0 for all z0 ∈ ∂B and ﬁxed x ∈ B. Note also that GB (x, z) has been deﬁned by continuity for x = y since x∗ is not deﬁned when x = y. The appropriate value of GB (x, z) when x = y can be obtained as follows in the n = 2 case. By Equations (1.4) and (1.5), ρ2 − y) − ∗ (x − y) |y − x| (z |y−x|2 ρ |y − x| |z − x | = → ρ |z − x| ρ|z − x| |z − y| as x → y, x = y; that is, limx→y GB (x, z) = GB (y, z), z ∈ B ∼ {y}. If u is harmonic on a neighborhood of the closure of B = By,ρ , then the equations in (i) and (ii) of the preceding theorem can be written, respectively, as follows. For n = 2, 1 u(x) = − uDn GB (x, z) dσ(z), x ∈ B; 2π ∂B for n ≥ 3, u(x) = −

1 σn (n − 2)

uDn GB (x, z) dσ(z). ∂B

(1.8)

1.5 Poisson Integral Formula

17

The proof of the following theorem is the same as that of Theorem 1.5.1 except for the inclusion of the term involving Δu in Green’s representation theorem. Theorem 1.5.3 If u has continuous second partials on closure of the ball B = By,ρ , then (i) for each x ∈ B and n = 2, 1 1 u(x) = − u(z)Dn GB (x, z) dσ(z) − GB (x, z)Δu(z) dz; 2π ∂B 2π B (ii) for each x ∈ B and n ≥ 3, 1 u(x) = − u(z)Dn GB (x, z) dσ(z) σn (n − 2) ∂B 1 GB (x, z)Δu(z) dz. − σn (n − 2) B By evaluating the normal derivatives of the Green functions in these equations, a representation of the harmonic function u can be obtained that does not involve the inverse x∗ . The following theorem will be proved only in the n = 3 case; the proof of the n = 2 case is similar. Theorem 1.5.4 (Poisson Integral Formula) If u is harmonic on the ball B = By,ρ , u ∈ C 0 (B − ), and x ∈ B, then u(x) =

1 σn ρ

∂B

ρ2 − |y − x|2 u(z) dσ(z). |z − x|n

Proof: Suppose ﬁrst that u is harmonic on a neighborhood of B − . The gradient of GB (x, ·) is easily seen to be

1 ρn−2 1 ∗ ∇(z) GB (x, z) = −(n − 2) (z − x) − (z − x ) . |z − x|n |x − y|n−2 |z − x∗ |n Using the deﬁnition of x∗ and Equation (1.5),

(n − 2) |y − x|2 ∇(z) GB (x, z) = − (z − y) . (z − y) − |z − x|n ρ2 Since the outer unit normal to ∂B at z ∈ ∂B is (z − y)/ρ, Dn GB (x, z) = ∇GB (x, z) · =−

z−y ρ

(n − 2) 2 (ρ − |y − x|2 ). ρ|z − x|n

(1.9)

18

1 Laplace’s Equation

Substituting for Dn GB (x, z) in Equation (1.8), 1 ρ2 − |y − x|2 u(x) = u(z) dσ(z). σn ρ ∂B |z − x|n Suppose now that u is only harmonic on B and continuous on B − . Fix x ∈ By,ρ and let {ρk } be a sequence of positive numbers increasing to ρ with x ∈ By,ρk for all k ≥ 1. Since u is harmonic on a neighborhood of each By,ρk , 1 ρ2 − |y − x|2 u(x) = u(z) k dσ(z). σn ρk ∂By,ρ |z − x|2 k

Letting z = ρk θ, 1 u(x) = σn ρk

|θ|=1

u(ρk θ)

ρ2k − |y − x|2 n−1 ρ dσ(θ). |ρk θ − x|2 k

By uniform boundedness of the integrands and continuity, 1 ρ2 − |y − x|2 n−1 u(x) = u(ρθ) ρ dσ(θ) σn ρ |θ|=1 |ρθ − x|2 1 ρ2 − |y − x|2 = u(z) dσ(z). σn ρ ∂B |z − x|2 The deﬁnition of harmonicity of a function u on an open set Ω required that u ∈ C 2 (Ω). Much more is true as a consequence. Lemma 1.5.5 Let U and C be open and compact subsets of Rn , respectively, and let u = u(x, y) be a real-valued function on U × C with the property that ∂u/∂xj is continuous on U × C for each j = 1, . . . , n. Then ∂ ∂u u(x, y) dμ(y) = (x, y) dμ(y) x ∈ U, j = 1, . . . , n, ∂xj C C ∂xj for any ﬁnite measure μ on the Borel subsets of C. Proof: Let ν be a unit vector. It will be shown now that Dν u(x, y) dμ(y) = Dν u(x, y) dμ(y). C

C

Consider a ﬁxed x ∈ U . The left side of this equation is u(x + λν, y) − u(x, y) dμ(y). lim λ→0+ C λ Let B be a ball with center x such that B − ⊂ U , and consider only those λ for which x + λν ∈ B. By the mean value theorem of the calculus, for each such λ and each y ∈ C, there is a point ξλ,y on the line segment joining x to x + λν such that

1.5 Poisson Integral Formula

C

19

u(x + λν) − u(x, y) dμ(y) = λ

C

=

1 (∇u(ξλ,y ) · λν) dμ(y) λ Dν u(ξλ,y , y) dμ(y).

C

Since Dν u is continuous on B − × C, the integrand on the right is uniformly bounded for such λ and y ∈ C. Since μ(C) < +∞, the Lebesgue dominated convergence theorem can be applied as λ → 0+, sequentially, to obtain

Dν

u(x, y) dμ(y) =

Dν u(x, y) dμ(y).

C

C

Theorem 1.5.6 If u is harmonic on the open set Ω, then all partial derivatives of u are harmonic on Ω. − ⊂ Ω, and x ∈ B, then Proof: If y ∈ Ω, B = By,ρ ⊂ By,ρ

u(x) =

1 σn ρ

∂B

ρ2 − |y − x|2 u(z) dσ(z). |z − x|n

By the preceding lemma, ∂u 1 (x) = ∂xj σn ρ

∂B

∂ ∂xj

ρ2 − |y − x|2 |z − x|n

u(z) dσ(z).

By calculating the partial derivative under the integral sign and using the Lebesgue dominated convergence theorem, it can be shown that ∂u/∂xj is continuous on B. In the same way, it can be shown that partial derivatives of all orders are continuous on B. Since B can be any ball with B − ⊂ Ω, partial derivatives of all orders of u are continuous on Ω. Since n ∂2u i=1

∂x2i

= 0 on Ω,

n ∂ ∂2u = 0 on Ω ∂xj ∂x2i i=1

for each j = 1, . . . , n. Using continuity of the third partial derivatives, the order of diﬀerentiation can be interchanged to obtain n ∂ 2 ∂u = 0, ∂x2i ∂xj i=1

j = 1, . . . , n.

20

1 Laplace’s Equation

Since ∂u/∂xj has continuous second partials on Ω, ∂u/∂xj is harmonic on Ω, as well as all higher-order partial derivatives. Theorem 1.5.7 (Picard) If u is harmonic on Rn and either bounded above or bounded below, then u is a constant function. Proof: Since −u is harmonic if u is harmonic, it can be assumed that u is bounded below. Since the sum of a harmonic function and a constant function is harmonic, it can be assumed that u ≥ 0. Let x and y be distinct points, and consider balls Bx,δ and By, with By, ⊃ Bx,δ . By Theorem 1.4.4, u(z) dz ≤ u(z) dz = νn n u(y) νn δ n u(x) = Bx,δ

By,

and u(x) ≤ ( /δ)n u(y). Now let , δ → +∞ in such a way that /δ → 1 to obtain u(x) ≤ u(y). But since x and y are arbitrary points, u(y) ≤ u(x) and u is a constant function. According to the Poisson integral formula, if a harmonic function is zero on the boundary of a ball, then it is zero on the ball. This fact suggests the following principle, known as the maximum principle. Deﬁnition 1.5.8 1 A function u deﬁned on an open connected set Ω obeys the maximum principle if supx∈Ω u(x) is not attained on Ω unless u is constant on Ω; the minimum principle if inf x∈Ω u(x) is not attained on Ω unless u is constant on Ω. Theorem 1.5.9 If u is continuous on the open connected set Ω and for each x ∈ Ω there is a δx > 0 such that 1 u(y) dσ(y) u(x) = σn δ n−1 ∂Bx,δ whenever δ < δx , then u obeys the maximum and minimum principles. Proof: Suppose u attains its maximum value at a point of Ω and let M = {x ∈ Ω; u(x) = supy∈Ω u(y)} = ∅. Since u is continuous, M is relatively closed in Ω. It will be shown now that M is open in Ω. For x ∈ M , 1 u(x) = u(y) dσ(y) σn δ n−1 ∂Bx,δ whenever δ < δx . Consider any y ∈ Bx,δx and let δ0 = |y − x| < δx . Since 1 u(x) dσ(y), u(x) = σn δ0n−1 ∂Bx,δ0 1

This version of the maximum principle will be referred to later as the strong maximum principle.

1.6 Gauss’ Averaging Principle

21

(u(x) − u(y)) dσ(y) = 0. ∂Bx,δ0

Since x ∈ M, u(x) − u ≥ 0 on ∂Bx,δ0 and therefore u(x) − u = 0 a.e. on ∂Bx,δ0 . By continuity, u = u(x) on ∂Bx,δ0 and, in particular, u(y) = u(x). This shows that y ∈ M , that Bx,δx ⊂ M, and that M is both relatively closed and open in Ω. By the connectedness of Ω, either M = ∅ or M = Ω. Since the ﬁrst case has been excluded, M = Ω so that u is constant on Ω. Corollary 1.5.10 If u is harmonic on the open connected set Ω, then u satisﬁes both the maximum and minimum principles.

1.6 Gauss’ Averaging Principle Among the many contributions of Gauss to potential theory, one of the best known is the assertion that the gravitational potential at a point in space due to a homogeneous spherical body is the same as if the entire mass were concentrated at the center of the body. A classical version of this property will be stated ﬁrst. Lemma 1.6.1 If Bx,δ ⊂ Rn , then (i) for n = 2 and y ∈ R2 , 1 2πδ

1 dσ(z) = log |y − z| ∂Bx,δ

1 log |y−x|

if

|y − x| > δ

log 1δ

if

|y − x| ≤ δ

(ii) for n ≥ 3 and y ∈ Rn , 1 σn δ n−1

∂Bx,δ

1 dσ(z) = |y − z|n−2

1 |y−x|n−2

if

|y − x| > δ

1 δ n−2

if

|y − x| ≤ δ.

Proof: Only (i) will be proved, (ii) being the easier of the two. Fix x ∈ R2 and δ > 0, and let uy (z) = − log |y − z|, z ∈ R2 . Three cases will be considered − − . Suppose ﬁrst that y ∈ Bx,δ . according as y ∈ Bx,δ , y ∈ ∂Bx,δ , or y ∈ Bx,δ − Then uy is harmonic on a neighborhood of Bx,δ and the result follows from the mean value property. Consider now the case y ∈ ∂Bx,δ . For each n ≥ 1, let yn = x+(1+ n1 )(y −x). Then |yn −x| = (1+ n1 )|y −x| = (1+ n1 )δ, |yn −z| ≤ 3δ for all z ∈ ∂Bx,δ , and log

1 1 ≥ log , |yn − z| 3δ

z ∈ ∂Bx,δ .

22

1 Laplace’s Equation

It follows that the sequence of functions on the left is lower bounded on ∂Bx,δ . By the preceding case and Fatou’s Lemma, 1 1 1 1 dσ(z) ≤ lim inf dσ(z) log log n→∞ 2πδ ∂B 2πδ ∂Bx,δ |y − z| |y − z| n x,δ

1 = lim inf log n→∞ |yn − x| 1 . = log |y − x| On the other hand, since |yn − z| ≥ |y − z| for z ∈ ∂Bx,δ , log

1 1 = lim log |y − x| n→∞ |yn − x| 1 1 = lim dσ(z) log n→∞ 2πδ ∂B |y n − z| x,δ 1 1 ≤ dσ(z), log 2πδ ∂Bx,δ |y − z|

and the assertion is true for y ∈ ∂Bx,δ . Lastly, suppose y ∈ Bx,δ . Letting ux,δ denote the left side of the equation in (i) and applying Lemma 1.5.5 to it twice, it is easily seen that ux,δ is harmonic on Bx,δ . Letting γ denote the angle between the line segment joining x to y and x to z, 2π 1 1 dγ log ux,δ (y) = 2 2 2πδ 0 |y − x| + δ − 2δ|y − x| cos γ and it is easily seen that ux,δ is a function of r = |y − x| only and, as in Section 1.3, must be of the form α log r + β. Since 1 1 1 dσ(z) = log , ux,δ (x) = log 2πδ ∂Bx,δ |x − z| δ α = 0 and β = log (1/δ); that is, ux,δ (y) = log (1/δ), y ∈ Bx,δ . The preceding lemma can be interpreted as a statement about the average value of log (1/|y − z|) or 1/|y − z|n−2 for a unit mass concentrated on the point y. Arbitrary mass distributions will be considered now. Some preliminary calculations will be carried out for the n = 2 and n ≥ 3 cases separately. If μ is a measure on the Borel subsets of R2 with compact support Sμ , let Uμ (y) =

log

1 dμ(z) = |y − z|

log Sμ

1 dμ(z), |y − z|

y ∈ R2 .

1.6 Gauss’ Averaging Principle

23

For a ﬁxed y ∈ R2 , log (1/|y − z|) is lower bounded on Sμ and the above integral is deﬁned as a real number or +∞. Consider a ﬁxed ball Bx,δ . Since log (1/|y − z|) is lower bounded on ∂Bx,δ × Sμ , Tonelli’s theorem (c.f. [18]) implies that 1 1 1 dμ(z)dσ(y) Uμ (y) dσ(y) = log 2πδ ∂Bx,δ 2πδ ∂Bx,δ Sμ |y − z| 1 1 = dσ(y) dμ(z). log 2πδ ∂Bx,δ |y − z| Sμ If μ is a measure on the Borel subsets of Rn , n ≥ 3, with compact support Sμ , let 1 1 Uμ (y) = dμ(z) = dμ(z). |y − z|n−2 |y − z|n−2 Sμ The same argument implies that 1 σn δ n−1

μ

U (y) dσ(y) = ∂Bx,δ

Sμ

1 σn δ n−1

∂Bx,δ

1 dσ(y) dμ(z). |y − z|n−2

Only the n = 2 case of the following theorem will be proved, the n ≥ 3 case being essentially the same. Theorem 1.6.2 (Gauss’ Averaging Principle [23]) Let μ be a measure on the Borel subsets of Rn with compact support Sμ and let B=Bx,δ . If Sμ ∩ B − =∅, then the average of Uμ over ∂B is Uμ (x); if Sμ ⊂ B − , the average of Uμ over ∂B depends only upon the total mass of μ and is equal to μ(Sμ ) log (1/δ) in the n = 2 case and is equal to μ(Sμ )/δ n−2 in the n ≥ 3 case. Proof: (n = 2) From the above discussion, 1 1 1 μ dσ(y) dμ(z). U (y) dσ(y) = log 2πδ ∂Bx,δ 2πδ ∂Bx,δ |y − z| Sμ If Sμ ⊂ B − , it follows from Lemma 1.6.1 that 1 1 μ U (y) dσ(y) = log dμ(z) 2πδ ∂Bx,δ δ Sμ = μ(Sμ ) log (1/δ); if Sμ ∩ B − = ∅, then log (1/|y − z|) is a harmonic function of y and 1 1 dμ(z) = U μ (x) Uμ (y) dσ(y) = log 2πδ ∂Bx,δ |x − z| Sμ by the mean value property.

24

1 Laplace’s Equation

1.7 The Dirichlet Problem for a Ball If Ω is a nonempty open subset of Rn with compact closure and f is a real-valued function on ∂Ω, the Dirichlet problem is that of ﬁnding a harmonic function u on Ω such that limy→x,y∈Ω u(y) = f (x) for all x ∈ ∂Ω. As was noted in Section 0.1, this limiting behavior of u at the boundary of Ω implies that f is continuous on ∂Ω. Generally speaking, the Dirichlet problem does not have a solution even when Ω is a ball. Theorem 1.7.1 The solution of the Dirichlet problem for a nonempty, open connected set Ω with compact closure and a continuous boundary function f is unique if it exists. Proof: Let u1 and u2 be two solutions. Suppose there is a point z ∈ Ω such that u1 (z) > u2 (z). Then limy→x,y∈Ω [u1 (y) − u2 (y)] = 0 for all x ∈ ∂Ω. Let on Ω u1 − u2 w= 0 on ∂Ω. Thus, w is harmonic on Ω, continuous on Ω − , and zero on ∂Ω. Since w(z) > 0 and w is continuous on Ω − , w must attain a positive supremum at some point of Ω. By the maximum principle, Corollary 1.5.10, w must be constant on Ω. Since w = 0 on ∂Ω, w = 0 on Ω − , a contradiction. Therefore, u1 ≤ u2 on Ω. Interchanging u1 and u2 , u2 ≤ u1 , and the two are equal. According to the Poisson integral formula, the value of a harmonic function u at an interior point of a ball B is determined by values of u on ∂B, assuming that u has a continuous extension to ∂B. It is natural to ask if a function f on ∂B determines a function u harmonic on B that has a continuous extension to B − agreeing with f on ∂B. More generally, it might be asked if a measure on the boundary of B determines a harmonic function on B. Theorem 1.7.2 (Herglotz [29]) If μ is a signed measure of bounded variation on the Borel subsets of ∂By,ρ , then 1 u(x) = σn ρ

∂By,ρ

ρ2 − |y − x|2 dμ(z), |z − x|n

x ∈ By,ρ ,

is harmonic on By,ρ . Proof: Using Lemma 1.5.5, it can be shown that u has continuous second partials and that 1 ρ2 − |y − x|2 Δu(x) = Δ(x) dμ(z). σn ρ ∂By,ρ |z − x|n A tedious, but straightforward, diﬀerentiation shows that the integrand is zero for x ∈ By,ρ .

1.7 The Dirichlet Problem for a Ball

25

Corollary 1.7.3 If f is a Borel measurable function on ∂By,ρ and integrable relative to surface area, then ρ2 − |y − x|2 1 u(x) = f (z) dσ(z) σn ρ ∂By,ρ |z − x|n is harmonic on By,ρ . Since the Poisson integral formula will be referred to repeatedly, let ρ2 − |y − x|2 1 dμ(z), PI(μ : B)(x) = σn ρ ∂B |z − x|n where B = By,ρ and μ is a signed measure of bounded variation on ∂B. If B = By,ρ , the dependence on the parameters y and ρ will be exhibited by using the notation PI(μ : y, ρ). If μ is absolutely continuous relative to surface area on ∂B, then for each Borel set M ∈ ∂B μ(M ) = f (z) dσ(z) M

for some integrable function f on ∂B. In this case, let PI(μ : B) = PI(f : B). This notation will also be used if the domain of f contains ∂B. Note that PI(1 : B) = 1, that PI(μ : B) is linear in μ, that PI(μ : B) ≥ 0 if μ is a measure, and that PI(f : B) ≥ 0 if f is nonnegative. According to Corollary 1.7.3, an integrable boundary function f determines a harmonic function u on a ball B. In what way is u related to f ? For example, is it true that limy→x,y∈B u(y) = f (x) for x ∈ ∂B? The following three lemmas answer this question for a ball B = By,ρ . Lemma 1.7.4 If f is Borel measurable on ∂B, integrable relative to surface area on ∂B, u = PI(f : B) on B, and there is a constant k such that f ≤ k a.e.(σ) on a neighborhood of x0 ∈ ∂B, then lim supx→x0 ,x∈B u(x) ≤ k. Proof: It can be assumed that k ≥ 0 for if not, replace f by f − k. Choose > 0 such that f (z) ≤ k a.e.(σ) for z ∈ Bx0 , ∩ ∂B. Denoting the indicator function of Bx0 , by gx0 , , for x ∈ B u(x) = PI(gx0 , f : B) + PI((1 − gx0 , )f : B). Since f (z) ≤ k a.e.(σ) for z ∈ Bx0 , ∩ ∂B, PI(gx0 , f : B)(x) ≤ PI(k : B)(x) = kPI(1 : B)(x) = k. Suppose x ∈ Bx0 , /2 and z ∈ ∂B. Then |z − x| > /2 when |z − x0 | > for otherwise, |z − x0 | ≤ |z − x| + |x − x0 | ≤ . Thus,

26

1 Laplace’s Equation

|PI((1 − gx0 , )f : B)(x)| ≤

1 σn ρ

∼Bx0 , ∩∂B

2

ρ − |y − x|2 ≤ σn ρ( /2)n

ρ2 − |y − x|2 |f (z)| dσ(z) ( /2)n |f (z)| dσ(z).

∂B

Since |y − x| → ρ as x → x0 , PI((1 − gx0 , )f : B)(x) → 0 as x → x0 . Therefore, lim sup u(x) ≤ lim sup PI(gx0 , f : B)(x) x→x0

x→x0

+ lim sup PI((1 − gx0 , )f : B)(x) ≤ k. x→x0

Lemma 1.7.5 If f is Borel measurable on ∂B, integrable relative to surface area on ∂B, and u = PI(f : B) on B, then for x0 ∈ ∂B lim sup u(x) ≤ lim sup f (x). x→x0 ,x∈B

x→x0 ,x∈∂B

Proof: It can be assumed that the right side is ﬁnite for otherwise there is nothing to prove. If k is any number greater than the right member of the last inequality, then f (x) < k for all x ∈ ∂B in a neighborhood of x0 . By the preceding lemma, lim sup u(x) ≤ k; x→x0 ,x∈B

but since k is any number greater than lim supx→x0 ,x∈∂B f (x), the lemma is proved. Lemma 1.7.6 If f is Borel measurable on ∂B, integrable relative to surface area on ∂B, continuous at x0 ∈ ∂B, and u = PI(f : B) on B, then lim

x→x0 ,x∈B

u(x) = f (x0 ).

Proof: By the preceding lemma, lim supx→x0 ,x∈B u(x) ≤ f (x0 ). Since PI(−f :B) = −PI(f : B), lim sup −u(x) = lim sup PI(−f : B) ≤ −f (x0 ),

x→x0 ,x∈B

x→x0 ,x∈B

or lim inf x→x0 ,x∈B u(x) ≥ f (x0 ) and the result follows. Theorem 1.7.7 (Schwarz [56]) The Dirichlet problem is uniquely solvable for a ball B and a continuous boundary function f . The solution is given by PI(f : B). Proof: Uniqueness was proven in Theorem 1.7.1.

1.7 The Dirichlet Problem for a Ball

27

Remark 1.7.8 Put another way, if f ∈ C 0 (∂B), then u = PI(f : B) is harmonic on B, has a continuous extension to B − , and agrees with f on ∂B. According to Theorem 1.4.3, if u is harmonic on a neighborhood of a ball, then u has the mean value property. This property characterizes harmonic functions. Theorem 1.7.9 A function u on the open set Ω ⊂ Rn is harmonic if and only if u ∈ C 0 (Ω) and 1 u(x) = u(z) dσ(z) σn δ n−1 ∂Bx,δ − for every ball Bx,δ ⊂ Bx,δ ⊂ Ω.

Proof: The necessity follows from the continuity of harmonic functions and Theorem 1.4.3. Consider a function u ∈ C 0 (Ω) that satisﬁes the above − equation for every Bx,δ ⊂ Bx,δ ⊂ Ω. Fix such a ball. By Lemma 1.7.6, 0 − there is a function v ∈ C (Ω ) that agrees with u on ∂Bx,δ and is harmonic on Bx,δ . The diﬀerence u − v is then zero on ∂Bx,δ , and since it satisﬁes the hypothesis of Theorem 1.5.9, u − v satisﬁes both the maximum and minimum principles on Bx,δ . It follows that u = v on Bx,δ and u is harmonic on Bx,δ . − ⊂ Ω, u is harmonic on Ω. Since Bx,δ is an arbitrary ball with Bx,δ It is not necessary that the equation of the preceding theorem hold for − every ball Bx,δ ⊂ Bx,δ ⊂ Ω. Corollary 1.7.10 A function u on the open set Ω ⊂ Rn is harmonic if and only if u ∈ C 0 (Ω) and for each x ∈ Ω, there is a δx > 0 such that 1 u(x) = u(z) dσ(z) (1.10) σn δ n−1 ∂Bx,δ for all 0 < δ < δx , in which case 1 1 u(x) = u(z) dσ(z) = u(z) dz σn δ n−1 ∂Bx,δ νn δ n Bx,δ − ⊂ Ω. whenever Bx,δ

Proof: The necessity follows from the preceding theorem. Suppose u ∈ C 0 (Ω) and for each x ∈ Ω, there is a δx > 0 such that Equation (1.10) holds for − ⊂ Ω, then u|∂B is continuous and the all 0 < δ < δx . If B = Bx,ρ ⊂ Bx,ρ function PI(u : x, ρ)(y), y ∈ B, v(y) = u(y), y ∈ ∂B, is continuous on B − by Lemma 1.7.6. By deﬁnition, u − v = 0 on ∂B. Since v is harmonic on B, for each y ∈ B

28

1 Laplace’s Equation

u(y) − v(y) =

1 σn δ n−1

(u(z) − v(z)) dσ(z) ∂By,δ

for all suﬃciently small δ > 0. By Theorem 1.5.9, u − v satisﬁes the minimum and maximum principles on B. If u − v attains its maximum at some point of B, then it must be constant and therefore u − v = 0 on B; if u − v does not attain its maximum at a point of B, then u − v ≤ 0 on B. In either case, u ≤ v on B. Since u − v also satisﬁes the minimum principle on B, u = v on B and u is harmonic on B. Since B is arbitrary, u is harmonic on Ω. Lastly, since νn n = σn 1 νn δ n

1 u(z) dz = n ν nδ Bx,δ σn = νn δ n σn νn δ n = u(x).

δ

u(x + ρθ)ρn−1 dθ dρ

0

|θ|=1

δ

0

1 σn

u(x + ρθ) dθ

ρn−1 dρ

|θ|=1

δ

u(x)ρn−1 dρ

=

0

It is possible to relax the requirement of continuity in the preceding theorem at the expense of replacing surface averages by solid ball averages. In the proof of the following theorem, the symmetric diﬀerence of two sets A and B is denoted by AB and is deﬁned by the equation AB = (A ∼ B) ∪ (B ∼ A). Lemma 1.7.11 If u is locally integrable on the open connected set Ω and 1 u(y) dy u(x) = νn δ n Bx,δ whenever Bx,δ ⊂ Ω, then u is continuous and obeys the maximum and minimum principles on Ω. Proof: Note ﬁrst that u is real-valued. It will be shown now that the hypotheses imply that u is continuous on Ω. To see this, consider any x ∈ Ω − − ⊂ Ω. For any y ∈ Bx,δ , By,δ ⊂ Bx,2δ . Since and any δ > 0 such that Bx,2δ − − Bx,δ By,δ ⊂ Bx,2δ and u is integrable on Bx,2δ , |u(x) − u(y)| ≤

1 νn δ n

Bx,δ By,δ

|u(z)| dz → 0

as y → x by the absolute continuity of the Lebesgue integral, which proves that u is continuous at x. Suppose there is a point x0 ∈ Ω such that u(x0 ) = inf Ω u. Letting Σ = {y; u(y) = inf Ω u}, Σ is a relatively closed subset of Ω by continuity of u. For any y ∈ Σ and δ > 0 with By,δ ⊂ Ω,

1.7 The Dirichlet Problem for a Ball

1 νn δ n

29

(u(z) − u(y)) dz = 0. By,δ

Since y ∈ Σ, u − u(y) ≥ 0 on By,δ and it follows that u − u(y) = 0 a.e. in By,δ ; thus, u = u(y) on By,δ by continuity of u. This shows that By,δ ⊂ Σ and that Σ is an open subset of Ω. By the connectedness of Ω, Σ = ∅ or Σ = Ω. If Σ = Ω, then u is constant on Ω; otherwise, u does not attain its minimum at an interior point of Ω. Since −u satisﬁes the same hypotheses, u satisﬁes the maximum principle on Ω. Theorem 1.7.12 The function u is harmonic on the open set Ω if and only if u is locally integrable on Ω and 1 u(x) = u(y) dy νn δ n Bx,δ whenever Bx,δ ⊂ Ω. Proof: The necessity follows from the continuity of harmonic functions and Theorem 1.7.9. As to the suﬃciency, let u be locally integrable on Ω and satisfy the above equation. Since it suﬃces to prove that u is harmonic on each component of Ω, it will be assumed that Ω is connected. Let B = By,ρ be any ball with B − ⊂ Ω. It was shown in the preceding proof that u is continuous on Ω. Consider the harmonic function h = PI(u|∂B : y, ρ). By Lemma 1.7.6, limz→x,z∈B (u(z)−h(z)) = 0 for all x ∈ ∂B. Since u−h satisﬁes both the minimum and maximum principles on B by Lemma 1.7.11, u = h on B and u is harmonic on B. Since B is an arbitrary ball with B − ⊂ Ω, u is harmonic on Ω. The requirement in the preceding theorem that u satisfy a global solid ball averaging condition cannot be relaxed. This can be seen by examining the function u on R2 deﬁned by ⎧ if x < 0 ⎨ −1 u(x, y) = 0 if x = 0 ⎩ 1 if x > 0. This function is locally integrable and satisﬁes a local solid ball averaging principle but is not harmonic. If u1 , . . . , up are harmonic functions on the open set Ω and α1 , . . . , αp are real numbers, p then it is clear from the original deﬁnition of harmonic function that u = i=1 αi ui is harmonic on Ω. The following theorem extends this result to integrals. Theorem 1.7.13 Let U and V be open subsets of Rn , let μ be a measure on U , and let H be a nonnegative function on U ×V . If (i) for each y ∈ V, H(·, y) is continuous on U , (ii) for each x ∈ U, H(x, ·) is harmonic on V , and (iii) h(y) = U H(x, y) dμ(x) < +∞ for each y ∈ V , then h is harmonic on V .

30

1 Laplace’s Equation

Proof: By hypothesis, H(x, y) is continuous in each variable separately. This implies that H(x, y) is jointly measurable on U ×V . To see this, for each k ≥ 1 let ji ji + 1 Ik (j1 , . . . , jn ) = {(x1 , . . . , xn ); k ≤ xi < , i = 1, . . . , n} 2 2k where each ji ∈ Z, the set of integers. Deﬁne a map ψk : U → V by letting ψk (x) be any ﬁxed point of V ∩Ik (j1 , . . . , jn ) whenever x ∈ V ∩Ik (j1 , . . . , jn ). Note that limk→∞ ψk (x) = x for all x ∈ U and that limk→∞ H(ψk (x), y) = H(x, y) for each x ∈ U and y ∈ V . Since each function H(ψk (x), y) is jointly measurable on U × V , H(x, y) is also. Therefore, H(x, y) is a nonnegative jointly measurable function to which Tonelli’s theorem can be applied. Sup− ⊂ V . Then pose By,δ

H(x, ·) dμ(x) : y, δ

A(h, y, δ) = A =

U

H(x, y) dμ(x) U

= h(y) < +∞. This shows that h is locally integrable and satisﬁes the hypotheses of Theorem 1.7.12. Thus, h is harmonic on V . It would appear from Theorem 1.7.2 that the class of functions harmonic on a ball is much more extensive than the class obtained by solving the Dirichlet problem for a Borel measurable boundary function. For example, consider a point z0 ∈ ∂B0,ρ and a unit measure μ concentrated on z0 . Then for 0 < λ < 1 1 ρ2 − |λz0 |2 dμ(z) u(λz0 ) = PI(μ : B)(λz0 ) = σn ρ ∂B0,ρ |z − λz0 |n =

1 − λ2 1 → +∞ n−1 σn ρ (1 − λ)n

as λ → 1−; that is, PI(μ : B)(x) → +∞ as x approaches z0 along a radial line. On the other hand, if the measure μ is concentrated on some point z1 other than z0 , then u(λz0 ) = PI(μ : B)(λz0 ) =

1 ρ2 (1 − λ2 ) →0 σn ρ |z1 − λz0 |n

as λ → 1−. The function u is not determined by the boundary function f that is +∞ at z0 and is 0 at all other points of ∂B0,ρ since f = 0 a.e. (σ) and PI(f : B) = 0 on B.

1.7 The Dirichlet Problem for a Ball

31

Lemma 1.7.14 (Herglotz [29]) If u is harmonic on B = By,ρ and 1 |u| dσ ≤ k < +∞ for all δ < ρ, σn δ n−1 ∂By,δ then there is a signed measure μ of bounded variation on ∂B such that u = PI(μ : B) on B. Proof: If δ < ρ and x ∈ By,δ , then u(x) =

1 σn δ

∂By,δ

δ 2 − |x − y|2 u(z) dσ(z). |z − x|n

− If M is a Borel subset of By,ρ and δ < ρ, deﬁne μδ (M ) = u(z) dσ(z). M∩∂By,δ

Then μδ is concentrated on ∂By,δ and |u(z)| dσ(z) ≤ kσn δ n−1 ≤ kσn ρn−1

μδ = ∂By,δ

for all δ < ρ. Let {δj } be a sequence of positive numbers such that δj ↑ ρ. By Theorem 0.2.5, there is a subsequence of the sequence {μδj } that converges to a signed measure μ in the w∗ -topology with μ ≤ kσn ρn−1 . It can be assumed that the sequence {μδj } converges to μ in the w∗ -topology. Since the signed measure μδj is concentrated on ∂By,δj and δj ↑ ρ, μ is concentrated on ∂By,ρ . Consider any x ∈ B. By dropping a ﬁnite number of terms, if necessary, it can be assumed that |y − x| < δj < ρ for all j ≥ 1. Then u(x) =

1 σn δj

∂By,δj

δj2 − |y − x|2 dμδj (z). |z − x|n

Since the signed measures μδj are concentrated on the spherical shell {z; δ1 ≤ |z − x| ≤ ρ} and the sequence of integrands in the last equation converges uniformly to (ρ2 − |y − x|2 )/|z − x|n on this shell, by Corollary 0.2.6, ρ2 − |y − x|2 1 dμ(z). u(x) = σn ρ ∂By,ρ |z − x|n Theorem 1.7.15 (Herglotz) The harmonic function u on B = By,ρ is a diﬀerence of two nonnegative harmonic functions if and only if there is a signed measure μ of bounded variation on ∂B such that u = PI(μ : B). Proof: Suppose u = u1 − u2 , where u1 and u2 are nonnegative harmonic functions on B. If 0 < δ < ρ, then

32

1 Laplace’s Equation

1 σn δ n−1

1 |u| dσ ≤ n−1 σ δ n ∂By,δ

1 |u1 | dσ + n−1 σ δ n ∂By,δ

|u2 | dσ ∂By,δ

= u1 (y) + u2 (y) < +∞ and the necessity follows from Lemma 1.7.14. If u = PI(μ : B), where μ is a signed measure of bounded variation on ∂B, then μ = μ+ − μ− , where μ+ and μ− are ﬁnite measures on ∂B, and so u = PI(μ+ : B) − PI(μ− : B). The suﬃciency follows from Theorem 1.7.2.

1.8 Kelvin Transformation Consider a ball By,ρ ⊂ Rn . The transformation x → x∗ deﬁned by x∗ = y +

ρ2 (x − y), |y − x|2

x = y,

(1.11)

was used in the derivation of the Poisson integral formula and is called the Kelvin transformation or inversion with respect to ∂By,ρ . The point x∗ , lying on the radial line joining y to x, is called the inverse of x relative to ∂By,ρ . Aside from the derivation of the integral formula, the Kelvin transformation is useful for solving other problems. The following symmetry property of the Kelvin transformation will be − , then used later in the chapter. If x, y ∈ Bz,ρ |x − z||x∗ − y| = |y − z||x − y ∗ |.

(1.12)

This equation will be derived assuming that z = 0. Let γ denote the angle between the line segments joining 0 to x and 0 to y. Since x∗ = (ρ2 /|x|2 )x and y ∗ = (ρ2 /|y|2 )y, ρ4 ρ2 2 |y| cos γ + |y| − 2 |x|2 |x|

|y|2 ρ4 ρ2 2 = + |x| − 2 |x| cos γ |x|2 |y|2 |y| |y|2 ∗ = |y − x|2 . |x|2

|x∗ − y|2 =

Thus, |x||x∗ − y| = |y||y ∗ − x|. As with any other transformation, the eﬀect of the transformation on various geometric regions can be examined. Note ﬁrst that the transformation maps planes or spheres into planes or spheres, but not necessarily respectively. The calculations will be much simpler if it is assumed that y = 0, in which case

1.9 Poisson Integral for Half-space

x∗ =

33

ρ2 x |x|2

and

x=

ρ2 ∗ x . |x∗ |2

Starting with the general equation of a plane or sphere in Rn a

n

x2i +

i=1

n

bi xi + c = 0,

i=1

and making the substitution xi = (ρ2 /|x∗ |2 )x∗i , the equation aρ4 + ρ2

n i=1

bi x∗i + c

n

x∗2 i =0

i=1

is obtained, which is again the general equation of a plane or sphere. It is easily seen that a plane outside the sphere ∂B0,ρ will map into a sphere inside the sphere ∂B0,ρ which passes through the origin, and conversely, a sphere inside the sphere ∂B0,ρ , which passes through the origin, will map onto a plane outside the sphere ∂B0,ρ . The eﬀect of the Kelvin transformation on functions will also be examined. Let Ω be an open subset of Rn ∼ {y} and let Ω ∗ be the image of Ω under the map x → x∗ . If f ∗ is a function on Ω ∗ , the equation f (x) =

ρn−2 ∗ ρ2 f (y + (x − y), rn−2 r2

where r = |x − y| and rn−2 = 1 if n = 2, deﬁnes a function on Ω. The mapping f ∗ → f deﬁned in this way will be called the Kelvin operator. Theorem 1.8.1 The Kelvin operator preserves positivity and harmonicity. Proof: That positivity is preserved is obvious from the deﬁnition. The proof that f is harmonic on Ω whenever f ∗ is harmonic on Ω ∗ is accomplished by computing the Laplacian Δ(x) of ρn−2 ∗ ρ2 f (y + 2 (x − y)); n−2 r r the computation is straightforward but tedious.

1.9 Poisson Integral for Half-space It was shown in the preceding section that a nonnegative harmonic function u on a ball can be represented as the Poisson integral of a measure on the boundary of the ball. A similar result holds for half-spaces. Throughout this section Ω will be the half-space {(x1 , . . . , xn ); xn > 0}.

34

1 Laplace’s Equation

Theorem 1.9.1 If u is a nonnegative harmonic function on the open halfspace Ω = {(x1 , . . . , xn ); xn > 0}, then there is a nonnegative constant c and a Borel measure μ on ∂Ω such that u(x) = cxn +

2xn σn

∂Ω

1 dμ(z) |z − x|n

for all x ∈ Ω.

Proof: Consider an inversion relative to ∂By,1 where y = (0, . . . , 0, −1). The image of Ω under this map is the open ball Ω ∗ = {z ∗ ; |z ∗ − x∗0 | < 1/2} where x∗0 = (0, . . . , 0, − 21 ). Let u∗ be the image of u under the inversion. Then u∗ is a nonnegative, harmonic function on the ball Ω ∗ . By Theorem 1.7.15, there is a Borel measure μ∗ on ∂Ω ∗ such that u∗ (x∗ ) =

2 σn

∂Ω ∗

1 4

− |x∗ − x∗0 |2 ∗ ∗ dμ (z ) |z ∗ − x∗ |n

for all x∗ ∈ Ω ∗ .

Let c = (2/σn )μ∗ ({y}) ≥ 0 and let μ∗1 = μ∗ |∂Ω ∗ ∼{y} . Then 1

u∗ (x∗ ) = c 4

− |x∗ − x∗0 |2 2 + |y − x∗ |n σn

∂Ω ∗

1 4

− |x∗ − x∗0 |2 ∗ ∗ dμ1 (z ). |z ∗ − x∗ |n

Letting θ denote the angle between the line segment joining x∗ to y and the xn -axis, 1 − |x∗ − x∗0 |2 = −|x∗ − y|2 + |x∗ − y| cos θ 4 by the law of cosines. Since |x∗ − y||x − y| = 1 by deﬁnition of x∗ , 1 1 − |x∗ − x∗0 |2 = (−1 + |x − y| cos θ). 4 |x − y|2 Since |x − y| cos θ = xn + 1, where xn is the nth component of x, xn 1 . − |x∗ − x∗0 |2 = 4 |x − y|2 Letting φ be the angle between the line segment joining z ∗ to y, where z ∈ ∂Ω, and the line segment joining x∗ to y, |z ∗ − x∗ |2 = |z ∗ − y|2 + |x∗ − y|2 − 2|z ∗ − y||x∗ − y| cos φ 1 2 cos φ 1 + − = |z − y|2 |x − y|2 |z − y||x − y| 1 (|x − y|2 + |z − y|2 − 2|z − y||x − y| cos φ) = |z − y|2 |x − y|2 |z − x|2 = . |z − y|2 |x − y|2

1.9 Poisson Integral for Half-space

35

Therefore, ∗

∗

u (x ) = cxn |x − y| or

n−2

2xn + |x − y|n−2 σn

∂Ω ∗

|z − y|n ∗ ∗ dμ (z ) |z − x|n 1

1 |z − y|n ∗ ∗ 2xn (x − y) ∗ u + dμ (z ). y + = cx n |x − y|n−2 |x − y|2 σn ∂Ω ∗ |z − x|n 1

Note that the left side of this equation is just u(x). Denoting the map z ∗ → z ∗ ∗ induces a measure μ1 on ∂Ω by μ1 = μ∗1 T −1 . by T , the measure μ1 on ∂Ω n Deﬁning μ(E) = E |z − y| dμ1 (z) for any Borel set E, the last equation can be written 1 2xn u(x) = cxn + dμ(z). (1.13) σn ∂Ω |z − x|n Note: There is no reason to believe that the measure in the last equation is ﬁnite. The right side of Equation (1.13) will be denoted by PI(c, μ, Ω)(x) and is called the Poisson integral of the pair (c, μ) for the half-space Ω. The number c can be any real number and μ any signed measure. As before, if μ is the indeﬁnite integral of a measurable function f on ∂Ω relative to Lebesgue measure, let PI(c, f, Ω) = PI(c, μ, Ω); that is, 2xn f (z) dz, (1.14) PI(c, f, Ω)(x) = cxn + σn ∂Ω |z − x|n where x = (x1 , . . . , xn ) with xn > 0. A partial converse of the preceding theorem for the half-space Ω will be taken up now. It will be shown ﬁrst that 1 dz < +∞ (1.15) n ∂Ω |z − x| whenever n ≥ 2 and x ∈ Ω. To see this, let x be the projection of x = (x1 , . . . , xn ) onto ∂Ω and let z ∈ ∂Ω. Then |z − x|2 = x2n + |z − x|2 , and 1 1 1 dz ≤ dz + dz. n n |z − x| x |z − x|n ∂Ω ∂Ω∩{|z−x|<1} n ∂Ω∩{|z−x|≥1} Since the ﬁrst integral on the right is ﬁnite, only the second need be considered. Transforming to spherical coordinates in the (n − 1)-dimensional space ∂Ω relative to the pole x, ∂Ω∩{|z−x|≥1}

1 dz = |z − x|n

1

∞

|θ|=1

1 n−2 r dθdr = σn−1 < +∞. rn

36

1 Laplace’s Equation

It will be shown now that PI(0, 1, Ω) = 1 on Ω. By Equation (1.15), PI(0, 1, Ω)(x) =

2xn σn

∂Ω

1 dz |z − x|n

is ﬁnite for each x ∈ Ω. Equation (1.15) can also be used to show that PI(0, 1, Ω) is harmonic on Ω. The argument is straightforward, but tedious, and involves justiﬁcation of diﬀerentiation under the integral. Note that the above integral, as a function of x, is unaﬀected by adding to x a vector in ∂Ω. Such an operation is a translation of x in a direction normal to the xn -axis. In other words, ∂Ω |z − x|−n dz is a function of xn only. Let 1 2xn g(xn ) = PI(0, 1, Ω)(x) = dz, xn > 0. σn ∂Ω |z − x|n Since g is harmonic on Ω, d2 g/x2n = 0 and g(xn ) = axn + b. It will be shown now that a = 0. Since 2xn 1 dz g(xn ) = 2 σn ∂Ω (|z| + x2n )n/2

z 2 1 = d 2 n/2 σn ∂Ω (|z/xn | + 1) xn 2 1 = dz, σn ∂Ω (|z|2 + 1)n/2 g(xn ) is a constant function and a = 0. By transforming to spherical coordinates relative to the origin in ∂Ω, using the formula for the σn in Equation (0.1), and using standard integration techniques, 2 1 b = g(1) = dz σn ∂Ω (|z|2 + 1)n/2 +∞ 2 rn−2 = dσ(θ)dr 2 n/2 σn 0 |θ|=1 (r + 1) 2σn−1 +∞ rn−2 = dr 2 σn (r + 1)n/2 0 = 1. Therefore, g(xn ) = 1 for all xn > 0; that is, 2xn 1 PI(0, 1, Ω) = dz = 1 σn ∂Ω |z − x|n

for all x ∈ Ω.

(1.16)

The relationship of the harmonic function determined by a boundary function to the boundary function itself will be considered now.

1.9 Poisson Integral for Half-space

37

Theorem 1.9.2 If f is a bounded measurable function on ∂Ω and c is any real number, then u = PI(c, f, Ω) is harmonic on Ω. If f is continuous at z0 ∈ ∂Ω, then limx→z0 u(x) = f (z0 ); moreover, limxn →+∞ u(x)/xn = c. Proof: The latter statement will be proved ﬁrst. It follows from Equation (1.16) that 2 1 lim dz = 0. xn →+∞ σn ∂Ω |z − x|n Returning to Equation (1.14) and using the fact that f is bounded, u(x) 2 f (z) lim = c + lim dz = c. xn →+∞ xn xn →+∞ σn ∂Ω |z − x|n Suppose now that f is continuous at z0 ∈ ∂Ω and that |f (z)| ≤ M on ∂Ω. To show that limx→z0 u(x) = f (z0 ), it suﬃces to prove as in Lemma 1.7.4 that f (z) ≤ k in a neighborhood of z0 implies that lim supx→z0 u(x) ≤ k for all z, where k > 0. Suppose f (z) ≤ k for all z ∈ ∂Ω ∩ Bz0 , . Since the term cxn in Equation (1.14) has the limit zero as x → z0 , it suﬃces to consider just the integral term. The integral will be written as a sum of two integrals by splitting the boundary ∂Ω into the two regions ∂Ω ∩ {|z − z0 | < } and ∂Ω ∩ {|z − z0 | ≥ }. Since f (z) ≤ k on ∂Ω ∩ {|z − z0 | < }, f (z) 2xn dz ≤ kPI(0, 1, Ω) = k. σn ∂Ω∩{|z−z0 |< } |z − x|n Letting x denote the projection of x onto ∂Ω, f (z) 1 2M xn 2xn dz ≤ dz. σn ∂Ω∩{|z−z0 |≥ } |z − x|n σn |z − x|n n ∂Ω∩{|z−x| ≥ } Note that for z ∈ ∂Ω ∼ Bz0 , and x ∈ Bz0 , /2 , |z − x| ≥ /2. Using spherical coordinates relative to x, for x ∈ Bz0 , /2 ∂Ω∩{|z−z0 |≥ }

1 dz ≤ |z − x|n ≤

∂Ω∩{|z−x|≥ /2} +∞

|θ|=1

/2

1 dz |z − x|n

1 n−2 r dr dθ rn

2σn−1 . = Therefore, lim sup x→z0

∂Ω∩{|z−z0 |≥ }

f (z) 4M xn σn−1 = 0. dz ≤ lim sup |z − x|n σn x→z0

38

1 Laplace’s Equation

Hence, lim sup u(x) ≤ k + lim sup x→z0

x→z0

∂Ω∩{|z−x≥ }

f (z) dz ≤ k. |z − x|n

The last theorem illustrates the diﬃculty with uniqueness of the Dirichlet problem for unbounded regions. The solution for a bounded function on the boundary of a half-space is not unique since the choice of c in PI(c, f, Ω) is arbitrary. In some cases, a harmonic function can be extended harmonically across the boundary of a region. This will be shown to be the case when the region is a half-space under appropriate conditions. If y = (y1 , . . . , yn ) ∈ Rn , y r will denote the reﬂection of y across the xn = 0 hyperplane; that is, y r = (y1 , . . . , yn−1 , −yn ). If Λ ⊂ Rn , Λr will denote the set {y r ; y ∈ Λ}. Lemma 1.9.3 Let Ω = {(x1 , . . . , xn ) ∈ Rn ; xn > 0}, let Γ be a compact subset of Ω, and let u be a function on Ω − that is continuous on Ω − ∼ Γ , harmonic on Ω ∼ Γ , and equal to 0 on ∂Ω. Then the function u˜ deﬁned on Rn ∼ (Γ ∪ Γ r ) by ⎧ if x ∈ Ω ∼ Γ ⎨ u(x) 0 if x ∈ ∂Ω u ˜(x) = ⎩ if x ∈ (Ω ∼ Γ )r −u(xr ) is harmonic on Rn ∼ (Γ ∪ Γ r ). Proof: Since the function u˜ is clearly continuous on Rn ∼ (Γ ∪ Γ r ), the result follows from Corollary 1.7.10. This lemma will be used to prove a less trivial result. In the course of doing so, the fact that the Kelvin transformation is idempotent will be used; that is, if f is a real-valued function on an open set Ω, not containing the reference point of the inversion on which the transformation is based, then the Kelvin transform of f ∗ on Ω ∗ is just f on Ω. This property follows immediately from the deﬁnition of the transformation. − − ∼ Bx,δ , harmonic Theorem 1.9.4 Let 0 < δ < ρ. If u is continuous on Bx,ρ − on Bx,ρ ∼ Bx,δ , and equal to 0 on ∂Bx,ρ , then there is an > 0 such that u − . has a harmonic extension to Bx,ρ+ ∼ Bx,δ

Proof: Let y be any point of ∂Bx,ρ and consider an inversion relative to ∂By,2ρ . Under this inversion, the ball Bx,ρ maps onto a half-space Ω, which can be taken to be {(x1 , . . . , xn }; xn > 0} by choice of coordinates, with the hyperplane ∂Ω externally tangent to Bx,ρ , B = Bx,δ maps onto a ball B ∗ − with closure B ∗ − ⊂ Ω, and Bx,ρ ∼ Bx,δ maps onto Ω ∼ B ∗ − . The Kelvin transform u∗ of u is continuous on Ω − ∼ B ∗ , harmonic on Ω ∼ B ∗ − , and equal to 0 on ∂Ω. Now restrict u∗ to Ω − ∼ B − . By the preceding lemma,

1.10 Neumann Problem for a Disk

39

u∗ has a harmonic extension to Rn ∼ (B ∗ − ∪ (B ∗ − )r ), where (B ∗ − )r is the reﬂection of B ∗ − across ∂Ω. By simple geometry, it can be seen that a suf− ﬁciently small > 0 can be chosen so that Bx,ρ+ ∼ Bx,δ maps under the inversion onto a neighborhood of ∂Ω which also contains Ω ∼ B ∗ − . Restricting u∗ to this neighborhood, u∗ will be harmonic thereon and its transform − will be harmonic on Bx,ρ+ ∼ Bx,ρ , thereby extending u harmonically across ∂Bx,ρ .

1.10 Neumann Problem for a Disk Consider a nonempty open set Ω ⊂ Rn having compact closure and a smooth boundary. Given a real-valued function g on ∂Ω, the Neumann problem is that of ﬁnding a harmonic function u on Ω such that Dn u(x) = g(x), x ∈ ∂Ω. There is an obvious diﬃculty with uniqueness of the solution if u satisﬁes the above conditions and c is any constant, then u + c satisﬁes the same conditions. Also, not every function g can serve as a boundary function for the Neumann problem. If u ∈ C 2 (Ω − ) solves the Neumann problem for the boundary function g, then by taking v = 1 in Green’s Identity 0= Dn u(z) dσ(z) = g(z) dσ(z); ∂Ω

∂Ω

and it follows that a necessary condition for the solvability of the Neumann problem is that the latter integral be zero. Before getting into the details, the meaning of the statement Dn u(x) = g(x), x ∈ ∂Ω, should be clariﬁed since the solution u of the Neumann problem may be deﬁned only on Ω. If n(x) is the outer normal to ∂Ω at x, by deﬁnition Dn u(x) = limt→1− Dn(x) u(tx), where Dn(x) u denotes the derivative of u in the direction n(x). The Poisson integral solved the Dirichlet problem for a disk of any dimension n ≥ 2. An analogous integral for the Neumann problem is not available for all n ≥ 2. The n = 2 case will be considered in this section and the n ≥ 3 case in the next section. Consider the n = 2 case, using polar coordinates (r, θ) rather than rectangular coordinates. Let g be a real-valued function on the boundary of the disk B = By,ρ which satisﬁes the condition

2π

g(ρ, θ) dθ = 0.

(1.17)

0

Fourier series will be used to construct a harmonic function u on B satisfying the Neumann condition Dn u = g on ∂B. The use of Fourier series will not only provide a method of approximating the solution, but will also lead to

40

1 Laplace’s Equation

an integral representation analogous to the Poisson integral. Suppose the function g(ρ, θ) has the Fourier series expansion ∞

a0 + (an cos nθ + bn sin nθ), g(ρ, θ) = 2 n=1 where an =

1 π

bn =

1 π

2π

g(ρ, θ) cos nθ dθ

n≥0

g(ρ, θ) sin nθ dθ

n ≥ 1.

0

2π

0

(1.18)

1 2π g(ρ, θ) dθ = 0. π 0 Noting that the functions rn cos nθ, rn sin nθ, n ≥ 0, are harmonic on R2 (as the real and imaginary parts of z n ), it is conceivable that

By Equation (1.17),

a0 =

u(r, θ) = u0 +

∞

(αn rn cos nθ + βn rn sin nθ)

n=1

is a harmonic function. Since Dn u = ∂u/∂r for a disk, Dn u(ρ, θ) =

∞

(nαn ρn−1 cosnθ + nβn ρn−1 sin nθ),

n=1

formally at least. Choosing the αn and βn so that αn =

an nρn−1

βn =

bn , nρn−1

n ≥ 1,

it would appear that the Neumann condition Dn u = g on ∂By,ρ is satisﬁed. Substituting these values of αn and βn in the deﬁnition of u(r, θ), u(r, θ) = u0 +

∞

rn (an cos nθ + bn sin nθ). nρn−1 n=1

Since 1 an cos nθ + bn sin nθ = π 1 + π 1 = π

2π

cos nφ cos nθ g(ρ, φ) dφ

0 2π

sin nφ sin nθ g(ρ, φ) dφ 0

0

2π

cos n(φ − θ)g(ρ, φ) dφ,

(1.19)

1.10 Neumann Problem for a Disk

41

u(r, θ)

2π ∞ 1 rn = u0 + (ein(φ−θ) + e−in(φ−θ) )g(ρ, φ) dφ 2π n=1 nρn−1 0

n

n 2π ∞ ∞ 1 rei(φ−θ) 1 re−i(φ−θ) ρ + g(ρ, φ) dφ = u0 + 2π 0 n ρ n ρ n=1 n=1

for 0 ≤ r < ρ, 0 ≤ θ ≤ 2π. Using the fact that log (1 − t) = − |t| < 1, u(r, θ) = u0 +

ρ 2π

2π

log 0

∞ n=1

tn /n for

ρ2 g(ρ, φ) dφ. ρ2 − 2ρr cos (φ − θ) + r2

The ρ2 in the numerator can be dropped because of Equation (1.17) resulting in Dini’s formula 2π ρ u(r, θ) = u0 − log (ρ2 + r2 − 2ρr cos(φ − θ))g(ρ, φ) dφ. (1.20) 2π 0 Having found a formula for a solution to the Neumann problem for a disk, the Fourier series method could be dispensed and the formula examined independently of its origin. But in light of the fact that the Fourier series method also provides approximate solutions, the method will be retained for the next theorem. Theorem 1.10.1 If g(ρ, θ) is continuous and of bounded variation on [0, 2π] and ∂By,ρ g(z) dσ(z) = 0, then the function u(r, θ) = u0 −

ρ 2π

2π

0

log (ρ2 + r2 − 2ρr cos (θ − φ))g(ρ, φ) dφ

− belongs to C 0 (By,ρ ) ∩ C 2 (By,ρ ) and solves the Neumann problem for the boundary function g.

Proof: Under the conditions on g, the Dirichlet-Jordan test (c.f. [65]) implies that the Fourier series representation [Equation (1.18)] of g(ρ, θ) is valid for each θ ∈ [0, 2π]. Deﬁne u(r, θ) for 0 ≤ r ≤ ρ as in Equation (1.19). Applying Theorem 0.2.2 twice to the series in (1.19) via the Weierstrass M-test, it is easy to show that u(r, θ) is harmonic on By,ρ . Since the Fourier coeﬃcients an and bn of g are O( n1 ) (c.f. [65] )and r ≤ ρ, the series deﬁning u(r, θ) in − − Equation (1.19) converges uniformly on By,ρ , and therefore u ∈ C 0 (By,ρ ). Theorem 0.2.2 can be applied again to show that Dr u(r, θ) =

∞ rn−1 (an cos nθ + bn sin nθ). ρn−1 n=1

42

1 Laplace’s Equation

By Abel’s limit theorem (c.f. [1]), lim Dr u(r, θ) = g(r, θ),

r→ρ−

0 ≤ θ ≤ 2π.

The requirement that g be of bounded variation on ∂By,ρ in the preceding theorem is unnecessary for the conclusion. Theorem 1.10.2 If g is continuous on ∂By,ρ and ∂By,ρ g(z) dσ(z) = 0, then the function u deﬁned by ρ u(x) = u0 − (log |x − z|)g(z) dσ(z) x ∈ By,ρ 2π ∂By,ρ solves the Neumann problem for the boundary function g. Proof: By Equation (1.20) and Lemma 1.5.5, ur (r, θ) =

ρ 2π

2π

0

−2r + 2ρ cos (φ − θ) g(ρ, φ) dφ ρ2 + r2 − 2ρr cos (φ − θ)

2π 2π ρ ρ ρ2 − r 2 = g(ρ, φ) dφ − g(ρ, φ) dφ 2πr 0 ρ2 + r2 − 2ρr cos (φ − θ) 2πr 0 ρ = PI(g, (r, θ)). r By Lemma 1.7.6 Dn u(ρ, θ) = lim ur (r, θ) = g(ρ, θ). r→ρ−

Converting Equation (1.20) to rectangular coordinates, 1 1 log g(z) dσ(z), x ∈ By,ρ . u(x) = u0 + 2π ∂By,ρ |z − x|2 Putting x = y, the log factor becomes a constant and it is clear that the constant u0 is the value of the solution at the center of the disk. Remark 1.10.3 Integral representations of solutions to the Neumann problem for half-spaces, quadrants, circular annuli, etc., can be found in [15, 44].

1.11 Neumann Problem for the Ball In constructing the Green function GB (x, z) and the Poisson integral formula for a ball B = By,ρ ⊂ Rn , the representation

1.11 Neumann Problem for the Ball

u(x) =

1 σn (n − 2)

∂B

43

1 1 Dn u − uDn n−2 |z − x| |z − x|n−2

dσ(z)

of a harmonic function was modiﬁed by adding an appropriate harmonic function vx to 1/|z − x|n−2 in order to eliminate the integral of the ﬁrst term. It might be possible to eliminate the integral of the the second term by adding a harmonic function vx to 1/|z − x|n−2 so that its normal derivative vanishes on ∂B. This goal is not quite achievable, but the procedure can be used to produce a constant normal derivative. The resulting function KB (x, z) should also be a harmonic function of x for each z ∈ ∂B. The procedure will produce the following representation of a harmonic function u in terms of its boundary normal derivative Dn u: 1 u(x) = σn (n − 2)

KB (x, z)Dn u(z) dσ(z) + c ∂B

u(z) dσ(z),

x ∈ B.

∂B

(1.21) Finding an explicit function vx accomplishing the above can be done only in the n = 3 case. In this case, a function KB (x, z) and a constant c will be exhibited such that Δ(x) KB (x, z) = 0 Dn(z) KB (x, z) = c

for each z ∈ ∂B −

for each x ∈ B , z ∈ ∂B, x = z,

(1.22) (1.23)

where Dn(z) denotes the outer unit normal derivative at z ∈ ∂B. Before read√ ing on, the reader should verify that the function log (±w + u2 + v 2 + w2 ), in the usual calculus notation, is harmonic on its domain. In order to simplify the notation, it will be assumed that the ball B is centered at the origin; the general case will then follow by a translation. Returning to the discussion at the beginning of this section, a solution KB (x, z) of Equations (1.22) and (1.23) of the form 1 + k(x, z), 4πr will be constructed where r = |x−z|. This will be done for a special case ﬁrst. Consider a ﬁxed point x0 = (0, 0, t), 0 < t ≤ ρ, and the inverse x∗0 = (0, 0, ρ2 /t) of x0 relative to ∂B0,ρ . The procedure used in deriving the Poisson integral formula will be mimicked by choosing k(x0 , z) to be a multiple of 1/4πr1 where r1 = |x∗0 − z|. Rather than leaving the multiplier undetermined, for the sake of brevity it will be incorporated here by taking KB (x, z) =

k(x0 , z) = Letting z = (u, v, w) and r = |x0 − z|,

ρ 1 . t 4πr1

44

1 Laplace’s Equation

1 1 z Dn = ∇(z) · r ∂B |x0 − z| ρ ∂B

1 v w−t u = · (u, v, w) − 3,− 3,− 3 ρ r r r ∂B 2 tw − ρ = . ρr3 ∂B

Replacing x0 by

x∗0 ,

Dn

1 t∗ w − ρ2 = , r1 ∂B ρr13 ∂B

where t∗ = ρ2 /t. By Equation (1.12), tr1 = ρr on ∂B and

1 t3 (ρ2 /t)w − ρ2 = 3 Dn r1 ∂B ρ ρr3 ∂B t2 w − t = 2 3 , ρ r ∂B

and so Dn

1 ρ 1 2tw − ρ2 − t2 + = . r t r1 ∂B ρr3 ∂B

Since r2 = |x0 − z|2 = u2 + v 2 + (t − w)2 = ρ2 − 2tw + t2 on ∂B,

ρ 1 1 1 + = − . Dn 4πr t 4πr1 ∂B 4πρr ∂B The addition of ρ/4πtr1 to 1/4πρr does not result in a function KB (x0 , z) satisfying Equation (1.23). Fortunately, there is another function, namely h(x0 , z) = −

1 log (t∗ − w + r1 ), 4πρ

which when added to k(x0 , z) will come closer to satisfying Equation (1.23). The gradient of h(x0 , z) is given by ∇(z) h(x0 , z) = −

4πρr1

(t∗

1 (u, v, −r1 + w − t∗ ). − w + r1 )

Thus, Dn h(x0 , z)|∂B = ∇(z) h(x0 , z) · z ∂B

1 (r12 − (w − t∗ )2 ) + w(−r1 + w − t∗ ) = − 4πρ2 r1 (t∗ − w + r1 ) ∂B

1.11 Neumann Problem for the Ball

45

1 r1 − t∗ = − 4πρ2 r1 ∂B

1 t∗ = − 1 − 4πρ2 r1 ∂B 1 1 − = . 4πρr 4πρ2 ∂B Therefore,

Dn

1 ρ 1 1 1 ∗ + log (t − w + r1 ) = − − , 4πr t 4πr1 4πρ 4πρ2 ∂B

a constant. Adding −(1/4πρ2 ) log t to the function just constructed,

1 ρ 1 1 1 1 + log (t∗ − w + r1 ) − log t = − − . Dn 4πr t 4πr1 4πρ 4πρ 4πρ2 ∂B Deﬁne KB (x0 , z) =

ρ 1 1 1 1 + log (t∗ − w + r1 ) − log t. − 4πr t 4πr1 4πρ 4πρ

Recall that the above discussion pertains to the special case x0 = (0, 0, t), 0 < t≤ ρ. Consider now a ﬁxed point x = (x1 , x2 , x3 ) ∈ B − . Letting s = x21 + x22 and t = |x|, the point x0 = (0, 0, t) will be mapped onto the point x = (x1 , x2 , x3 ) by means of the orthogonal transformation ⎡

x2 s

⎢ x 1 A=⎢ ⎣− s 0

x1 x3 x1 st t x2 x3 x2 st t

− st

⎤ ⎥ ⎥. ⎦

x3 t

Letting y = Az for z ∈ B − , r = |x0 − z| = |x − y|, and r1 = |x∗0 − z| = |x∗ − y| since an orthogonal transformation preserves distances. Since t∗ = ρ2 /t = ρ2 /|x| and w = (x · y)/t, the function constructed above is mapped into the function ρ 1 1 + 4π|x − y| |x| 4π|x∗ − y|

2 ρ x·y 1 1 ∗ log − + |x − y| − log |x|. − 4πρ |x| |x| 4πρ

KB (x, y) =

By Equation (1.12), |x||x∗ − y| = |y||y ∗ − x| and it follows that the function

2 ρ x·y 2 ∗ ∗ − + |x − y| + log |x| log (ρ − x · y + |x||x − y|) = log |x| |x|

46

1 Laplace’s Equation

is a symmetric function of x and y; since it is a harmonic function of y for each x, it is a harmonic function of x for each y. Likewise, 1/|x||x∗ − y| is a symmetric function of x and y and is a harmonic function of x for y ∈ B − . Thus, Δ(x) KB (x, y) = 0 on B for each y ∈ ∂B. Moreover, by Remark 1.3.4, Dn KB (x, y)|∂B = Dn KB (x0 , z)|∂B = −(1/4πρ2 ) for each x ∈ B. Since y ∗ = y for y ∈ ∂B, KB (x, y) =

1 2 − log (ρ2 − x · y + ρ|x − y|), 4π|x − y| 4πρ

y ∈ ∂B.

Since adding a constant to KB (x, y) will not aﬀect the validity of Equations (1.22) and (1.23), KB (x, y) can be adjusted so that

2ρ2 1 1 KB (x, y) = + log . 2π|x − y| 4πρ ρ2 − x · y + ρ|x − y| The function KB (x, z) is called the Green function for the Neumann problem on the ball B or the Green function of the second kind for the ball B. The proof of the following theorem is essentially the same as the proof of Green’s representation theorem, Theorem 1.4.2. Theorem 1.11.1 If u is harmonic on a neighborhood of B − , then for each x ∈ B,

2ρ2 2 1 1 + log u(x) = Dn u(y) dσ(y) 4π ∂B |x − y| ρ ρ2 − x · y + ρ|x − y| 1 u(y) dσ(y). + 4πρ2 ∂B The following theorem is stated as an exercise by Kellogg in [35]. Theorem 1.11.2 If g ∈ C 0 (∂B) and ∂B g(z) dσ(z) = 0, then the function u(x) = KB (x, z)g(z) dσ(z) ∂B

solves the Neumann problem for the boundary function g. Proof: (Sobolev [58]) If x ∈ ∂B and 0 < t < 1, the function KB (tx, z) is bounded on ∂B and the above integral deﬁning u(tx) is ﬁnite. Letting r = |tx − z|, ∂KB (txi − zi )t 1 trzi − ρ(txi − zi )t . (tx, z) = − + ∂xi 2πr3 4πρ r(ρ2 − t(x · z) + ρr) By Lemma 1.5.5, ∂u (tx) = ∂xi

∂B

∂KB (tx, z)g(z) dσ(z), ∂xi

1.11 Neumann Problem for the Ball

and therefore

47

Dn(x) u(tx) =

Dn(x) KB (tx, z)g(z) dσ(z). ∂B

Fix x ∈ ∂B. In order to show that Dn(x) u(x) = limt→1− Dn(x) u(tx) = g(x), it will ﬁrst be shown that it suﬃces to prove the result for the special case g(x) = 0. Since at least one of the components of x = (x1 , x2 , x3 ) is diﬀerent from zero, it can be assumed that x1 = 0, say. Consider the linear function (u, v, w) = (ρg(x)/x1 )u for which Dn(x) (x) = (ρg(x)/x1 , 0, 0) · (x1 /ρ, x2 /ρ, x3 /ρ) = g(x). By Theorem 1.11.1 and the deﬁnition of u, (y) − u(y) = KB (y, z)(Dn (z) − g(z)) dσ(z) ∂B

with Dn (x) − g(x) = 0. If it can be shown that Dn(x) ( − u)(x) = 0, it would follow that Dn(x) u(x) = Dn(x) (x) = g(x). Henceforth, assume that g(x) = 0. To show that Dn(x) u(x) = 0, the special case x = (0, 0, ρ) will be proven, with the general case following from an orthogonal transformation. Consider x0 = (0, 0, t) where ρ/2 < t ≤ ρ. Then Dn(x) u(x) = limt→ρ− Dt u(x0 ), where Dt u(x0 ) = Dt KB (x0 , z)g(z) dσ(z). ∂B

Fix > 0 and choose γ0 ∈ (0, π/4) such that |g(z)| < whenever the angle γ between the line segments joining 0 to x and 0 to z is less than γ0 . Letting z = (u, v, w) and r = |x0 − z|, Dt KB (x0 , z) =

1 w−t 1 rw + ρw − ρt . + 2π r3 4πρr ρ2 − tw + ρr

The above integral over ∂B will be split into a sum of four integrals I1 , I2 , I3 , and I4 where I1 is the integral of the ﬁrst term of this function over ∂B ∩(γ < γ0 ) ∩ (t < w), I2 is the integral of the ﬁrst term over ∂B ∩ (γ < γ0 ) ∩ (w ≤ t), I3 is the integral of the second term over ∂B ∩ (γ < γ0 ), and I4 is the integral of both terms of the function over ∂B ∩ (γ ≥ γ0 ). Since t > ρ/2, 1 w−t g(z) dσ(z) |I1 | = ∂B∩(γ<γ0 )∩(t<w) 2π r3 ≤

2π

∂B∩(γ<γ0 )∩(t<w)

ρ−t dσ(z) r3

48

1 Laplace’s Equation

ρ 2 − t2 1 dσ(z) 2π ∂B r3 ρ + t ≤ 2 PI(1 : B)(x0 ) ≤

= 2 . Next consider I2 . For z ∈ ∂B ∩ (γ < γ0 ) ∩ (w ≤ t), r2 = u2 + v 2 + (w − t)2 = (ρ2 − tw) + t(t − w) ≥ t(t − w) ≥ (ρ/2)(t − w), and therefore |t − w| ≤ (2/ρ)r2 . It follows that 2 |t − w| |I2 | ≤ |g(z)| dσ(z) r3 ∂B∩(γ<γ0 )∩(w≤t) 4π 1 dσ(z). ≤ πρ ∂B∩(γ<γ0 )∩(w≤t) r By Theorem 1.6.1, (1/4πρ2 )

1 ∂B r

dσ(z) =

1 ρ

so that

|I2 | ≤ 4 . Turning to I3 , since tw ≤ ρ2 and r = u2 + v 2 + (w − t)2 ≥ |w − t| on ∂B ∩ (γ < γ0 ), another application of Theorem 1.6.1 shows that 1 rw + ρ(w − t) |I3 | = g(z) dσ(z) 2 4πρ ∂B∩(γ<γ0 ) r(ρ − tw + ρr) 1 ≤ dσ(z) 2πρ ∂B∩(γ<γ0 ) r = 2 . Lastly, since

∂B

g(z) dσ(z) = 0 and |

∂B∩(γ<γ0 )

g(z) dσ(z)| < 4πρ2 ,

g(z) dσ(z) < 4πρ2 , ∂B∩(γ≥γ0 )

1 g(z) dσ(z) − < . ∂B∩(γ≥γ0 ) 4πρ2

or

Thus, |I4 | ≤ |g|

∂B∩(γ≥γ0 )

2 w−t 1 1 rw + ρw − ρt 4π r3 + 4πρ r(ρ2 − tw + ρr) − − 4πρ2 dσ(z)+ .

Since r = |x0 − z| = t2 + ρ2 − 2ρt cos γ ≥ t2 + ρ2 − 2ρt cos γ0 > 0 for γ ≥ γ0 , the integrand on the right tends to zero boundedly in z as t → ρ−. Thus,

1.12 Spherical Harmonics

49

lim sup t→ρ−

Dt KB (x0 , z)g(z) dσ(z)≤9 . ∂B

Since > 0 is arbitrary, Dn(x) u(x) = limt→ρ− Dt u(x0 ) = 0.

1.12 Spherical Harmonics Solving the Neumann problem for a ball B = B0,ρ ⊂ Rn is less satisfactory for the n ≥ 4 case than for the n = 2 and n = 3 cases since the Fourier series method is not applicable, nor is there an explicit formula as in the n = 3 case. A series solution is possible using functions called spherical harmonics. This method of solution will be sketched only brieﬂy. For further details, see [16, 35, 44, 62]. Consider two points x, y ∈ B0,ρ ⊂ Rn , y = 0, n ≥ 3, and the fundamental harmonic function r−n+2 = |x − y|−n+2 with pole y. It is useful to think of y as a parameter in what follows. Letting γ denote the angle between the line segments joining 0 to x and 0 to y, 1 rn−2

1 + − 2|x||y| cos γ)(n−2)/2 1 1 = n−2 . |y| ((|x|/|y|)2 + 1 − 2(|x|/|y|) cos γ)(n−2)/2 =

(|x|2

|y|2

(1.24)

Letting u = cos γ and v = |x|/|y|, 1 1 1 = n−2 . rn−2 |y| (1 − 2uv + v 2 )(n−2)/2 Recalling the binomial coeﬃcients

α = α(α − 1) × · · · × (α − n + 1)/n!, n deﬁned for real α and n ≥ 1, and applying the generalized binomial theorem ∞ n−2 1 − 2 =1+ (2uv − v 2 )m 2 (n−2)/2 m (1 − 2uv + v ) m=1

√ provided |2uv − v 2 | < 1. Suppose v < 2 − 1. Since |u| ≤ 1, |2uv − v 2 | ≤ 2|u|v + v 2 ≤ 2v + v√2 = (v + 1)2 − 1 < 1 and the above series is absolutely convergent for v < 2 − 1. In fact, absolute convergence holds for v < 1 (c.f. [35]). Applying the binomial theorem to the binomials (2uv − v 2 )m ,

50

1 Laplace’s Equation

∞ m n−2 1 m − 2 = 1 + (−1)k 2m−k um−k v m+k . m k (1 − 2uv + v 2 )(n−2)/2 m=1 k=0

Making the substitution = m + k with k ﬁxed, ∞ [/2] − n−2 − k 1 2 = 1 + (−1)k 2−2k u−2k v , k −k (1 − 2uv + v 2 )(n−2)/2 =1 k=0 (1.25) where [/2] is the usual greatest integer function. The coeﬃcient of v m in this series will be denoted by Pn,m (u) and is given by [m/2]

Pn,m (u) =

k=0

− n−2 2 m−k

m−k k

(−1)k 2m−2k um−2k ,

and, in the n = 3 case, is called a Legendre polynomial. Consider the factor

m−2k m x·y |x| , 0 ≤ k ≤ m/2 |x||y| |y| 1 = 2(m−k) (x · y)m−2k (|x|2 )k |y|

um−2k v m =

in the general term of the series in Equation (1.25). It is clear that the latter expression is a polynomial in x1 , x2 , . . . , xn of degree m. Recalling that a function f (x) is homogeneous of degree m if f (λx) = λm f (x), this factor is also homogeneous of degree m. Thus, 1/rn−2 has the representation 1 rn−2

=

∞ 1 |y|n−2 m=0

ak1 ,...,kn xk11 xk22 · · · xknn ,

(1.26)

k1 +···+kn =m

where the k1 , . . . , kn are nonnegative integers. It will be shown now that each term of this series is a harmonic function. Let x = (|x1 |, . . . , |xn |), y = = (|y1 |, . . . , |yn |), u = (x · y)/|x||y| = (x · y)/|x||y|. Note that v = |x|/|y| √ |x|/|y| = v. As indicated above, |2uv + v 2 | ≤ 2uv + v 2 < 1 for v < 2 − 1. Thus, the binomial series 1+

∞ − n−2 2 (2uv + v 2 )m m

(1.27)

m=1

converges absolutely for v < 1+

∞

√ 2 − 1. Thus, the positive series

m

(−1)

m=1

− n−2 2 m

(2uv + v 2 )m

(1.28)

1.12 Spherical Harmonics

51

√ converges for v < 2 − 1. Expanding the binomials as above, this series becomes the following power series in |x1 |, . . . , |xn | ∞

bk1 ,...,kn |x1 |k1 × · · · × |xn |kn .

(1.29)

m=0 k1 +···+kn =m

Since the latter two series are positive series and√their partial sums are intertwined, the latter series converges for v < 2 − 1; that is, the series Equation √ and can be diﬀerentiated term by term √ (1.26) converges absolutely for v < 2 − 1. Thus, for |x| < ( 2 − 1)|y|, 0 = Δ(x)

1 rn−2

=

∞ 1 |y|n−2 m=0

ak1 ,...,kn Δ(x) (xk11 × · · · × xknn ).

k1 +···+kn =m

Since the series on the right is a power series in x1 , . . . , xn and none of the coeﬃcients vanish, each of the homogeneous polynomials in Equation (1.26) is a harmonic function. Example 1.12.1 If n = 3, m = 2, x = (x1 , . . . , xn ), y = (y1 , . . . , yn ), then P3,2 (u) = (3/2)(u2 − 13 ) and

x · y |x|2 P3,2 |x||y| |y|3

3 (x1 y1 + x2 y2 + x3 y3 )2 1 |x|2 = − 2 |x|2 |y|2 3 |y|3 1 3(x1 y1 + x2 y2 + x3 y3 )2 − (x21 + x22 + x23 )(y12 + y22 + y32 ) = 2|y|5 1 (y 2 (2x21 − x22 − x23 ) + y22 (2x22 − x21 − x23 ) + y32 (2x23 − x21 − x22 ) = 2|y|5 1 + 6y1 y2 (x1 x2 ) + 6y1 y3 (x1 x3 ) + 6y2 y3 (x2 x3 ).

2 |x| x·y Recalling that y = (y1 , y2 , y3 ) is regarded as a parameter, P3,2 |x||y| |y|3 is a linear combination of the following harmonic homogeneous polynomials in x1 , x2 , x3 of degree 2: 2x21 − x22 − x23 , 2x22 − x21 − x23 , 2x23 − x21 − x22 , x1 x2 , x1 x3 , x2 x3 . Among these six polynomials, ﬁve are linearly independent. The ﬁrst can be expressed as the negative of the sum of the second and third. Given a harmonic homogeneous polynomial Pn,m (x) in x = (x1 , . . . , xn ) of degree m, putting x = |x|θ, θ ∈ ∂B0,1 it can be written Pn,m (x) = |x|m Pn,m (θ) = |x|m Yn,m (θ),

52

1 Laplace’s Equation

where Yn,m (θ) = Pn,m (θ) is a function deﬁned on ∂B0,1 , called an ndimensional spherical harmonic function of degree m. For each m ≥ 1, there are kn,m = (2m + n − 2)(n + m − 3)!/(n − 2)!m! linearly independent n-dimensional spherical harmonic functions of degree m, denoted by (k) (k) Yn,m , k = 1, . . . , kn,m (c.f. [62] ). Yn,m is the commonly used notation for the spherical harmonic functions. These functions can be normalized so that (k) the family {Yn,m ; k = 1, . . . , kn,m , m ≥ 0} is a complete orthonormal system (1) relative to surface area on ∂B0,1 with Yn,0 equal to a constant. Now let g be a continuous function on ∂B0,ρ that satisﬁes the boundary condition g(z) dσ(z) = 0. ∂B0,ρ

Then g has the expansion n,m ∞ k

g(ρθ) =

(k) bn,m,k Yn,m (θ),

m=1 k=1

where bn,m,k =

(k) g(ρθ)Yn,m (θ) dσ(θ),

∂B0,1

with the term corresponding to m = 0 vanishing by the preceding equation (1) and the fact that Yn,0 is a constant. The solution u of the Neumann problem for g is then given by u(x) = u0 +

n,m ∞ k

m=1 k=1

bn,m,k

|x|m Y (k) (θ), mρn−1 n,m

where u0 is an arbitrary constant (for further details see [44]). For additional information about spherical harmonics, see the book by Axler, Bourdon, and Ramey [2]. This book contains information on how to access free software for generating spherical harmonic functions.

Chapter 2

The Dirichlet Problem

2.1 Introduction Just as the Riemann integral of a function on a ﬁnite interval can be deﬁned using lower and upper approximating sums, a solution to the Dirichlet problem can be constructed by using lower and upper functions called subharmonic and superharmonic functions, respectively. A simple example of a subharmonic function is a real-valued, continuous, convex function f on an open interval (a, b) ⊂ R; meaning, that if x, y ∈ (a, b) and 0 ≤ λ ≤ 1, then f (λx + (1 − λ)y) ≤ λf (x) + (1 − λ)f (y). In particular, if x = x0 − δ, y = x0 + δ and λ = 1/2, then f (x0 ) ≤

1 [f (x0 − δ) + f (x0 + δ)]; 2

that is, the value of f at the center of the interval (x0 − δ, x0 + δ) ⊂ (a, b) is less than or equal to the average of the values of f over the boundary of (x0 − δ, x0 + δ). Another simple example of a subharmonic function is the function f (x, y) = x2 + y 2 . Relaxing the requirement of continuity in the usual sense, for each y ∈ R3 the fundamental harmonic function uy (x) = |x − y|−1 is an example of a superharmonic function. More generally, if {y1 , . . . , yk } is a ﬁnite k collection of points in R3 , then u(x) = i=1 |yi − x|−1 is a superharmonic function. Collections of subharmonic and superharmonic functions will be used to prove the existence of solutions of the Dirichlet problem for rather general subsets of Rn , n ≥ 2 and to prove that the solution has the right boundary values. This chapter will also pave the way for development of a similar procedure for the Neumann problem.

L.L. Helms, Potential Theory, Universitext, c Springer-Verlag London Limited 2009

53

54

2 The Dirichlet Problem

2.2 Sequences of Harmonic Functions In order to solve the Dirichlet and Neumann problems for regions more general than balls, it is necessary to consider successive approximations to solutions. This means that sequences of harmonic functions must be considered. The inequality of the next theorem is a fundamental tool of potential theory. Theorem 2.2.1 If By,ρ ⊂ Rn , 0 < δ < ρ, and h is any positive harmonic function on By,ρ , then ρ2 h(x) ≤ h(x ) ρ2 − δ 2

ρ+δ ρ−δ

n

for all x, x ∈ By,δ .

− , then ρ − δ ≤ |z − x| ≤ ρ + δ and Proof: If z ∈ ∂By,ρ and x ∈ By,δ

ρ2 − |x − y|2 ρ2 − |x − y|2 ρ2 − |x − y|2 ≤ ≤ . (ρ + δ)n |z − x|n (ρ − δ)n Replacing x by x and taking the ratio of the middle members,

n 2 ρ2 ρ+δ ρ − |x − y|2 ρ2 − |x − y|2 ≤ . (z − x)n ρ2 − δ 2 ρ − δ |z − x |n By Theorem 1.7.15, there is a measure μ on ∂By,ρ such that h = PI(μ, By,ρ ). Integrating both sides of the above inequality with respect to μ, keeping x, x ﬁxed, and multiplying both sides by (σn ρ)−1 , the inequality is obtained. Theorem 2.2.2 (Harnack’s Inequality) If K is a compact subset of the open connected set Ω, then there is a constant M , depending only upon K and Ω, such that h(x)/h(x ) ≤ M for all x, x ∈ K and all positive harmonic functions h on Ω. Proof: Suppose there is no such constant. Then for every positive integer k there is a positive harmonic function hk on Ω and points xk , yk ∈ K such that hk (xk ) ≥ k. hk (yk ) Since K is compact, the sequences {xk } and {yk } have subsequences with limits x and y, respectively, in K. Since Ω is connected, x and y can be joined by a polygonal path lying entirely in Ω. There is also a chain of balls B1 , . . . , Bp with closures in Ω which covers the path, x ∈ B1 , y ∈ Bp , Bi ∩ Bi+1 = ∅, i = 1, . . . , p − 1. If for each i, Theorem 2.2.1 is applied to Bi− and a slightly larger concentric ball, then there is a constant Mi such that

2.2 Sequences of Harmonic Functions

55

h(x )/h(y ) ≤ Mi whenever h is a positive harmonic function on Ω and x , y ∈ Bi− . For each i, let zi ∈ Bi ∩ Bi+1 , i = 1, . . . , p − 1. Then for any x ∈ B1 , y ∈ Bp , h(x ) h(x ) h(z1 ) h(zp−1 ) = × ···× ≤ M1 × · · · × Mp = M ∗ h(y ) h(z1 ) h(z2 ) h(y )

(2.1)

for any positive harmonic function h on Ω. Now choose k0 such that xk0 ∈ B1 , yk0 ∈ Bp and hk0 (xk0 ) ≥ k0 > M ∗ . hk0 (yk0 ) But this contradicts Inequality (2.1). The assumption that there is no such constant leads to a contradiction. It should not come as a surprise that the uniform limit of a sequence of harmonic functions is again harmonic. The same is true for generalized sequences or nets; all that counts is that the convergence is uniform. Lemma 2.2.3 (Harnack) If {hα ; α ∈ A} is a net of harmonic functions on the open set Ω ⊂ Rn which converges uniformly to h on Ω, then h is harmonic on Ω. Proof: Since the hα are continuous and the convergence is uniform, h is − ⊂ Ω. Using uniform convergence, continuous. Suppose x ∈ Ω and Bx,δ 1 hα (z) dσ(z) A σn δ n−1 ∂B x,δ 1 h(z) dσ(z). = σn δ n−1 ∂Bx,δ

h(x) = lim hα (x) = lim A

By Theorem 1.7.9, h is harmonic on Ω. Compactness properties of sets of harmonic functions will now be examined. In the proof of the following theorem, the symmetric diﬀerence A B of two sets A and B is deﬁned by A B = (A ∼ B) ∪ (B ∼ A). Theorem 2.2.4 If K is a compact subset of the open set Ω and F is a collection of harmonic functions on Ω which is uniformly bounded on a neighborhood of each point of Ω, then F is uniformly equicontinuous on K; moreover, each net {hα ; α ∈ A} in F has a subnet that converges uniformly on K. If {hα ; α ∈ A} is a convergent net of uniformly bounded harmonic functions on Ω, then h = limA hα is harmonic on Ω. Proof: Since K is compact, the family F is uniformly bounded on some − ⊂ U for all neighborhood U of K, and there is a ρ > 0 such that Bx,ρ x ∈ K. Let m = sup {|h(x)|; x ∈ U, h ∈ F }. For any h ∈ F and x, y ∈ K, by Theorem 1.4.4

56

2 The Dirichlet Problem

1 1 |h(x) − h(y)| = h(z) dz − h(z) dz νn ρn Bx,ρ νn ρn By,ρ m ≤ dz. νn ρn Bx,δ By,ρ Since the volume of Bx,ρ By,ρ can be made arbitrarily small by taking |x−y| small, given > 0, there is a δ > 0 such that vol (Bx,ρ By,ρ ) < (νn ρn /m) whenever |x−y| < δ. Thus, |h(x)−h(y)| < whenever h ∈ F, |x−y| < δ. This shows that the family F is uniformly equicontinuous on K and, in particular, equicontinuous on K. Consider now any net {hα ; α ∈ A} in F . Since the net {hα ; α ∈ A} is uniformly bounded and equicontinuous on K, there is a subnet that converges uniformly on K by the Arzel` a-Ascoli theorem, Theorem 0.2.4. Now let {hα ; α ∈ A} be a convergent net of uniformly bounded harmonic functions on Ω and let h = limA hα . Let U be an open subset of Ω with compact closure U − ⊂ Ω. Then a subnet {hαβ } converges uniformly on U − to h. By Lemma 2.2.3, h is harmonic on U . Since U is arbitrary, h is harmonic on Ω. Theorem 2.2.5 If {hk } is a monotone increasing (decreasing) sequence of harmonic functions on an open connected set Ω ⊂ Rn , then h = limk→∞ hk is either identically +∞(−∞) or harmonic on Ω. Proof: Suppose there is a point x ∈ Ω such that h(x) = limk→∞ hk (x) < +∞. Let U be any open set containing x with compact closure U − ⊂ Ω. For each k ≥ 1, the nonnegative harmonic function hk − h1 is either strictly positive or identically zero on Ω by the minimum principle, Theorem 1.5.10. By Harnack’s Inequality, Theorem 2.2.2, there is a positive constant M such that hk (y) − h1 (y) ≤ M (hk (x) − h1 (x)) ≤ M (h(x) − h1 (x)) for all y ∈ U − provided hk − h1 is strictly positive; since this inequality is trivially true if hk − h1 is identically zero on Ω, it holds for all k ≥ 1 and all y ∈ U − . This implies that the sequence {hk } is uniformly bounded on U . By Theorem 2.2.4, h is harmonic on U and therefore on Ω. Henceforth, “function” will mean “extended real-valued function.” Harmonic functions are always real-valued. For each of the following lemmas stated for l.s.c. (u.s.c.) functions, there is a corresponding result for u.s.c. (l.s.c.) functions. Deﬁnition 2.2.6 The family F of functions deﬁned on Ω is right-directed (left-directed) if for each pair u, v ∈ F there is a w ∈ F such that u ≤ w and v ≤ w (w ≤ u and w ≤ v). Lemma 2.2.7 If H = {hi ; i ∈ I} is a right-directed (left-directed) family of harmonic functions on an open connected set Ω ⊂ Rn , then h = supi∈I hi (h = inf i∈I hi ) is either identically +∞(−∞) or harmonic on Ω.

2.2 Sequences of Harmonic Functions

57

Proof: If i0 ∈ I, the hypothesis implies that h = sup {hi ; hi ≥ hi0 , i ∈ I}. It therefore will be assumed that H has a smallest element hi0 . Suppose there is a point x ∈ Ω such that h(x) = supi∈I hi (x) < +∞. Let U be any open set containing x having compact closure U − ⊂ Ω. As in the proof of Theorem 2.2.5, the family H is uniformly bounded on a neighborhood of each point of Ω. By Theorem 2.2.4, the net {hi ; i ∈ I} has a subnet that converges on U − to h. It follows from the same theorem that h is harmonic on U , and since U is any open set containing x, h is harmonic on Ω. Lemma 2.2.8 (Choquet) Let {fi ; i ∈ I} be a family of functions on an open set Ω ⊂ Rn . For J ⊂ I, let fJ (x) = inf i∈J fi (x), x ∈ Ω. Then there is a countable set I0 ⊂ I such that for every l.s.c. function g on Ω, g ≤ fI whenever g ≤ fI0 . Proof: It can be assumed that the fi take on values in [−π/2, π/2] since only order properties of functions are concerned. This can be accomplished by replacing each fi by arctan fi and noting that the arctan function is an order-preserving map of the extended real-number system. Let {Ωk } be a countable base for the topology of Ω. Consider αk = inf {fI (y); y ∈ Ωk }. Since there is a y ∈ Ωk such that fI (y ) is close to αk and an i ∈ I such that fi (y ) is close to fI (y ), for each k there is an ik ∈ I such that inf fik (y) < inf fI (y) +

y∈Ωk

y∈Ωk

1 . k

(2.2)

Let I0 = {ik ; k ≥ 1}. Suppose g is l.s.c. on Ω, g ≤ fI0 , > 0, and x ∈ Ω. Then g(y) > g(x) − /2 for y in some neighborhood of x. It follows that there is an Ωm such that (i) x ∈ Ωm , (ii) g(y) > g(x) − ( /2) for all y ∈ Ωm , and (iii) 1/m < /2. Then inf y∈Ωm g(y) ≥ g(x) − ( /2) or g(x) − inf g(y) < y∈Ωm

. 2

Since g ≤ fI0 ≤ fim , inf g(y) − inf fim (y) ≤ 0.

y∈Ωm

y∈Ωm

Rewriting Inequality (2.2) with k = m, inf fim (y) − inf fI (y) <

y∈Ωm

y∈Ωm

1 < . m 2

Adding corresponding members of the last three inequalities, g(x) − inf fI (y) < . y∈Ωm

58

2 The Dirichlet Problem

Since x ∈ Ωm ,

g(x) < inf fI (y) + ≤ fI (x) + . y∈Ωm

Since > 0 is arbitrary, g(x) ≤ fI (x) for all x ∈ Ω and g ≤ fI . The next result is a generalization of the classic theorem of Dini which states that if a monotone sequence of continuous functions on a compact space converges to a continuous function, then the convergence is uniform. Lemma 2.2.9 (Dini-Cartan) Let {fi ; i ∈ I} be a left-directed family of u.s.c. nonnegative functions deﬁned on a compact space X. If inf i∈I fi = 0, then for each > 0 there is an i0 ∈ I such that fi0 < . Proof: For each x ∈ X, there is an ix such that fix (x) < . Since fix is u.s.c., there is a neighborhood Ux of x such that fix (y) < for all y ∈ Ux . Since {Ux ; x ∈ X} is an open covering of X, there are ﬁnitely many points x1 , . . . , xp such that X ⊂ ∪pj=1 Uxj . Since the family {fi ; i ∈ I} is left-directed, there is a function fi0 ≤ fixj , j = 1, . . . , p. Consider any x ∈ X. Then x ∈ Uxj for some j and fi0 (x) ≤ fixj (x) < . The following generalization of the Lebesgue monotone convergence theorem will be needed later. Lemma 2.2.10 Let μ be a measure on the Borel subsets of a compact metric space X and let {fi ; i ∈ I} be a left-directed family of ﬁnite-valued u.s.c. functions on X. Then inf fi dμ = inf fi dμ. i∈I

i∈I

Proof: Since the inﬁmum of any family of u.s.c. functions is again u.s.c., the integral on the right side is deﬁned. If inf i∈I fi = 0, then the integral on the right is zero. By the preceding lemma, given > 0, there is an i0 ∈ I such that fi0 < . Thus, 0 ≤ inf fi dμ ≤ fi0 dμ ≤ μ(X). i∈I

Since is arbitrary, inf i∈I fi dμ = 0 and the lemma is true whenever inf i∈I fi = 0. Now suppose that inf i∈I fi = 0. Since g = inf i∈I fi is u.s.c. and bounded above, there is a sequence {φj } of continuous functions such that φj ↓ g and φj dμ ↓ g dμ. If λ > g dμ, there is a continuous function φ ≥ g such that φ dμ ≤ λ. Since φ is continuous, (fi − φ) is u.s.c. for each i and (fi − φ)+ = max (fi − φ, 0) is u.s.c.. Clearly, inf i∈I (fi − φ)+ = 0. From the ﬁrst part of the proof, inf (fi − φ) dμ ≤ inf (fi − φ)+ dμ = inf (fi − φ)+ dμ = 0. i∈I

i∈I

i∈I

2.3 Superharmonic Functions

Therefore,

59

inf

i∈I

fi dμ ≤

φ dμ ≤ λ;

but since λ is any number greater than g dμ, inf fi dμ ≤ inf fi dμ. i∈I

i∈I

The assertion follows since the opposite inequality is trivially true.

2.3 Superharmonic Functions Throughout this section, Ω will denote an open subset of Rn , n ≥ 2. The following result will serve as motivation for the deﬁnition of a superharmonic function. − ⊂ Ω, and Δu ≤ 0 on Ω, then Lemma 2.3.1 If u ∈ C 2 (Ω), B − = Bx,δ

u(x) ≥

1 σn δ n−1

u(z) dσ(z). ∂B

Proof: If q(y) = |y|2 , y ∈ Rn , then Δq = 2n and Δ(u − q) < 0 on Ω for every > 0. Let u − q on ∂B h= PI(u − q, B) on B. Then h is continuous on B − , agrees with u − q on ∂B, and is harmonic on B by Remark 1.7.8. Letting w = u − q − h on B − , Δw < 0 on B. If w attains its minimum value at x0 ∈ B, then ∂ 2 w ≥ 0, i = 1, 2, . . . , n ∂x2i x0 and Δw(x0 ) ≥ 0, a contradiction. Therefore, x0 ∈ ∂B and w = u − q − h ≥ 0 on B − ; that is, u − q ≥ h on B − . Thus, u(x) − q(x) ≥ h(x) = PI(u − q, B)(x). But since x is the center of the ball B, the latter is just the average of u − q over ∂B; that is, 1 (u(z) − q(z)) dσ(z). u(x) − q(x) ≥ σn δ n−1 ∂B The result follows by letting → 0+. If f is a nonnegative Borel function on ∂Bx,δ , then ∂Bx,δ f (z) dσ(z) is always deﬁned but may be +∞; more generally, if f = f + − f − is a Borel

60

2 The Dirichlet Problem

function on ∂Bx,δ , then ∂Bx,δ f (z) dσ(z) is deﬁned if at least one of the two integrals ∂Bx,δ f + (z) dσ(z), ∂Bx,δ f − (z) dσ(z) is ﬁnite. If f is l.s.c. on − − Bx,δ , then f is bounded below on Bx,δ and ∂Bx,δ f (z) dσ(z) is deﬁned as an extended real number. The operation of averaging a function over a sphere or ball occurs so frequently that it is worthwhile to introducea notation for such an operation. If f is a Borel function on ∂Bx,δ for which ∂Bx,δ f (z) dσ(z) is deﬁned, let L(f : x, δ) =

1 σn δ n−1

f (z) dσ(z); ∂Bx,δ

similarly, if f is a Borel function on Bx,δ , let 1 f (z) dz. A(f : x, δ) = νn δ n Bx,δ Deﬁnition 2.3.2 The extended real-valued, Borel measurable function u on Ω is − (i) super-mean-valued at x ∈ Ω if u(x) ≥ L(u : x, δ) whenever Bx,δ ⊂ Ω; (ii) super-mean-valued on Ω if super-mean-valued at each point of Ω; (iii) locally super-mean-valued at x ∈ Ω if there is a δx > 0 such that u(x) ≥ L(u : x, δ) for all δ ≤ δx ; (iv) locally super-mean-valued on Ω if locally super-mean-valued at each point of Ω.

There are analogous “sub-mean-valued” deﬁnitions obtained by reversing the inequalities. A function that is both sub-mean-valued and super-mean-valued in one of the above senses is said to be mean-valued in that sense. Deﬁnition 2.3.3 An extended real-valued function u on Ω is superharmonic on Ω if (i) u is not identically +∞ on any component of Ω, (ii) u > −∞ on Ω, (iii) u is l.s.c. on Ω, and (iv)u is super-mean-valued on Ω; u is subharmonic on Ω if −u is superharmonic on Ω. The set of functions satisfying (i), (ii), and (iii) of this deﬁnition will be denoted by L(Ω), the set of superharmonic functions by S(Ω), and the set of nonnegative superharmonic functions by S + (Ω). The phrase “on Ω” will be omitted whenever Ω = Rn . Alternative deﬁnitions of superharmonic functions will be considered. The following lemma will be proved only for the n ≥ 3 case, the n = 2 case being essentially the same except for the constant preceding integrals.

2.3 Superharmonic Functions

61

Lemma 2.3.4 If u ∈ C 2 (Ω), then u is superharmonic on Ω if and only if Δu ≤ 0 on Ω. Proof: The suﬃciency follows from the above deﬁnition and Lemma 2.3.1. Suppose there is a point x ∈ Ω where Δu(x) > 0. By continuity, there is a − − ⊂ Ω such that Δu(z) ≥ m > 0 on Bx,δ . By Theorem 1.5.3 closed ball Bx,δ and Equation (1.9), 1 u(z)Dn GB (x, z) dσ(z) u(x) = − σn (n − 2) ∂Bx,δ 1 GB (x, z)Δu(z) dz − σn (n − 2) Bx,δ 1 u(z)Dn GB (x, z)dσ(z) <− σn (n − 2) ∂Bx,δ 1 δ2 = u(z) dσ(z) σn δ ∂Bx,δ |z − x|n 1 u(z) dσ(z) = σn δ n−1 ∂Bx,δ = L(u : x, δ). Thus, Δu > 0 at some point of Ω implies that u cannot be superharmonic on Ω. Theorem 2.3.5 If Ω is a open connected subset of Rn , u ∈ L(Ω), and for each x ∈ Ω there is a δx > 0 such that Bx,δx ⊂ Ω and u(x) ≥ L(u : x, δ) (or u(x) ≥ A(u : x, δ)) for all δ < δx , then u satisﬁes the minimum principle on Ω. Proof: Suppose there is a point x0 ∈ Ω such that u(x0 ) = inf Ω u. Since u ∈ L(Ω), u is ﬁnite at some point of Ω and −∞ < u(x0 ) = inf Ω < +∞. Let K = {x : u(x) = inf Ω u}, which is a relatively closed subset of Ω by l.s.c. of u. It will be shown that K is also open. For any y ∈ K, there is a δy > 0 such that By,δy ⊂ Ω and u(y) ≥ L(u : y, δ) whenever δ < δy . Suppose there is a point z ∈ By,δy ∼ K. Let ρ = |y − z| < δy . Since y ∈ K, u(y) = L(u : y, ρ) or (u(z ) − u(y)) dσ(z ) = 0. ∂By,ρ

Since u − u(y) ≥ 0 on ∂By,ρ , u − u(y) = 0 a.e.(σ) on ∂By,ρ . But since u(z) > u(y), there is an α such that u(z) > α > u(y). By l.s.c. of u, there is a neighborhood Uz of z ∈ ∂By,ρ such that u > α > u(y) on Uz ∩ ∂By,ρ . Since the latter set has positive surface area, u − u(y) > 0 on a set of positive surface area, a contradiction. Thus, By,δy ⊂ K and K is open. Therefore, K = ∅ or K = Ω by the connectedness of Ω with K = ∅ excluded from the

62

2 The Dirichlet Problem

outset. It follows that K = Ω, and that u is constant if it attains its minimum value at a point of Ω. The proof using volume averages is similar. Corollary 2.3.6 If Ω is an open connected subset of Rn and u ∈ S(Ω), then u satisﬁes the minimum principle on Ω. If Ω is a bounded open set, u ∈ S(Ω), and lim inf z→x u(z) ≥ 0 for all x ∈ ∂Ω, then u ≥ 0 on Ω. Proof: That u satisﬁes the minimum principle follows from the preceding theorem. To prove the second assertion, let B be a ball containing Ω − . Extend u to be 0 on B − ∼ Ω, denoting the extension by the same symbol. Then u is l.s.c. on B − and attains its minimum value k at some point x0 ∈ B − . If k < 0, then x0 ∈ Ω. Consider any ball Bx0 ,δ with Bx−0 ,δ ⊂ Ω. Since u is superharmonic on Bx0 ,δ , it satisﬁes the minimum principle thereon by the ﬁrst part of the corollary. Since u attains its minimum at an interior point of Bx0 ,δ , it must be constant on Bx0 ,δ and so u = k on Bx0 ,δ . This shows that the set U = {x ∈ B − ; u(x) = k} is a nonempty open set. Moreover, V = {x ∈ B − ; u > k} is a nonempty open set also. Since B − = U ∩ V , the assumption that k < 0 contradicts the connectedness of B − . Therefore, k ≥ 0 on B − , and therefore u ≥ 0 on Ω. If u ∈ S(Ω) and h is harmonic on Ω, then it is easily seen that u+h ∈ S(Ω). Theorem 2.3.7 Let u ∈ S(Ω) and let U be an open subset of Ω with compact closure U − ⊂ Ω. If h is continuous on U − , harmonic on U , and u ≥ h on ∂U , then u ≥ h on U . Proof: Consider the function u − h on U − . Since u is l.s.c. on Ω and h is continuous on U − , lim inf (u(z) − h(z)) = lim inf u(z) −

z→x,z∈U

z→x,z∈U

lim

z→x,z∈U

h(z) ≥ u(x) − h(x) ≥ 0

for x ∈ ∂U . Therefore, u ≥ h on U by the preceding corollary. Alternative Deﬁnition of Superharmonic Function The function u ∈ L(Ω) is superharmonic on the open set Ω if for every open subset U of Ω with compact closure U − ⊂ Ω, the following statement is true: (P) h ∈ C 0 (U − ), h harmonic on U, u ≥ h on ∂U implies u ≥ h on U . Let S (Ω) be the class of functions u ∈ L(Ω) satisfying (P). By the preceding theorem, S(Ω) ⊂ S (Ω). It will be shown now that S (Ω) ⊂ S(Ω). Consider − any u ∈ S (Ω) and any closed ball Bx,δ ⊂ Ω. Since u is l.s.c. on ∂Bx,δ , there is a sequence {φj } of continuous functions on ∂Bx,δ such that φj ↑ u on ∂Bx,δ . Let φj on ∂Bx,δ hj = PI(φj : Bx,δ ) on Bx,δ . − , harmonic on Bx,δ , and u ≥ φj = hj on ∂Bx,δ . Then hj is continuous on Bx,δ Since u ∈ S (Ω), u ≥ hj on Bx,δ so that

2.3 Superharmonic Functions

63

u(x) ≥ hj (x) = PI(φj : Bx,δ )(x) = L(φj : x, δ). Since φj ↑ u on ∂Bx,δ , u(x) ≥ L(u : x, δ) by the Lebesgue monotone convergence theorem. It follows that u is super-mean-valued on Ω and S (Ω) = S(Ω). The following theorem is a precursor of yet another deﬁnition of superharmonic function. Theorem 2.3.8 If Ω is an open subset of Rn , u ∈ L(Ω), and u is locally super-mean-valued on Ω, then u ∈ S(Ω). Proof: It will be shown that u is super-mean-valued on Ω and therefore is superharmonic according to the original deﬁnition. Let Bx,δ be any ball with − ⊂ Ω. Consider any φ ∈ C 0 (∂Bx,δ ) with φ ≤ u on ∂Bx,δ and PI(φ : x, δ). Bx,δ Using the fact that u is l.s.c. on Ω and Lemma 1.7.6, lim inf (u(z) − PI(φ : x, δ)) ≥ u(x) − φ(x) ≥ 0 z→y

on ∂Bx,δ . Since Bx,δ is connected, u − PI(φ : x, δ) satisﬁes the minimum principle on Bx,δ by Theorem 2.3.5. Thus, u − PI(φ : x, δ) cannot attain its minimum value at a point of Bx,δ unless it is a constant thereon. This implies that u − PI(φ : x, δ) ≥ 0 on Bx,δ . Letting {φj } be a sequence in C 0 (∂Bx,δ ) such that φj ↑ u on ∂Bx,δ , u − PI(u : x, δ) ≥ 0 on Bx,δ . In particular, u(x) ≥ PI(u : x, δ)(x) = L(u : x, δ). Since Bx,δ can be any ball with closure in Ω, u is super-mean-valued on Ω. Yet Another Deﬁnition of Superharmonic Function An extended realvalued function u on the open set Ω is superharmonic if u ∈ L(Ω) and u is locally super-mean-valued on Ω. Let S (Ω) be the class of functions u ∈ L(Ω) which are locally super-meanvalued on Ω. By the preceding theorem, S (Ω) ⊂ S(Ω). Trivially, S(Ω) ⊂ S (Ω) and the two are equal. The last deﬁnition is usually the easiest to apply for verifying that a function is superharmonic. As lower semicontinuity is a local property of a function, superharmonicity is a local property; that is, it is only necessary to show that a function in L(Ω) is superharmonic on a neighborhood of each point in Ω to show that it is superharmonic globally. Example 2.3.9 The fundamental harmonic function uy with pole y is superharmonic on Rn . It is known that uy is harmonic on Rn ∼ {y} and therefore locally super-mean-valued on Rn ∼ {y}; uy is super-mean-valued at y since +∞ = uy (y) ≥ L(u : x, δ) for any δ > 0. Since uy ∈ L(Rn ), uy ∈ S(Rn ). Example 2.3.10 If u is harmonic on the open set Ω, then |u| is subharmonic on Ω and −|u| is superharmonic on Ω. Since u is ﬁnite-valued on Ω,

64

2 The Dirichlet Problem

− continuous on Ω, and |u(x)| = |L(u : x, δ)| ≤ L(|u| : x, δ) whenever Bx,δ ⊂ Ω, |u| is sub-mean-valued on Ω and is therefore subharmonic on Ω.

Example 2.3.11 If u is harmonic on the open set Ω, then u+ = max (u, 0) is subharmonic on Ω and u− = max (−u, 0) is subharmonic on Ω. Finiteness and continuity of u+ and u− follow from that of u. Since u(x) = L(u : x, δ) ≤ − − L(u+ : x, δ) whenever Bx,δ ⊂ Ω, u+ (x) ≤ L(u+ : x, δ) whenever Bx,δ ⊂ Ω. + Thus, u is sub-mean-valued on Ω and therefore subharmonic on Ω. The last example is a special case of a more general statement that a convex function of a harmonic function is a subharmonic function. Recall that a real-valued function φ on a ﬁnite or inﬁnite interval (a, b) is convex if φ(λx + (1 − λ)y) ≤ λφ(x) + (1 − λ)φ(y) whenever a < x < y < b and 0 < λ < 1. Such functions are continuous on (a, b) and have right- and left-hand derivatives at each point of (a, b). See Royden [54] for such facts. Let μ be a unit measure on the Borel subsets B of the open set Ω, and let f be a real-valued function taking on values in the domain (a, b) of the convex function φ. The inequality

f dμ ≤ φ(f ) dμ φ Ω

Ω

is known as Jensen’s Inequality. If φ is convex on (−∞, +∞), if f is integrable relative to μ, and if f takes on the values +∞ and −∞, then Jensen’s Inequality holds if φ is arbitrarily deﬁned at +∞ and −∞. This can be seen by applying Jensen’s Inequality to ∼F f dμ where F = {x ∈ Ω; |f (x)| = +∞}. Theorem 2.3.12 If u is harmonic on the open set Ω and φ is a convex function on an open interval containing the range of u, then φ(u) is subharmonic on Ω. Proof: φ(u) is ﬁnite-valued and continuous since both φ and u have these − ⊂ Ω, then φ(u(x)) = φ(L(u : x, δ)) ≤ L(φ(u) : x, δ) by properties. If Bx,δ Jensen’s Inequality; that is, φ(u) is sub-mean-valued on Ω and is therefore subharmonic on Ω. Example 2.3.10 is a special case of this theorem with φ(x) = |x|, x ∈ R. Theorem 2.3.13 If u is subharmonic on the open set Ω and φ is an increasing convex function on an open interval containing the range of u, then φ(u) is subharmonic on Ω. Proof: Since φ is real-valued, φ(u) is real-valued. Since a convex function is continuous, φ(u) is u.s.c. By Jensen’s Inequality and the fact that φ is

2.4 Properties of Superharmonic Functions

65

− increasing, φ(u(x)) ≤ φ(L(u : x, δ)) ≤ L(φ(u) : x, δ) whenever Bx,δ ⊂ Ω; that is, φ(u) is sub-mean-valued. Thus, φ(u) is subharmonic on Ω.

Example 2.3.14 If u is harmonic on the open set Ω and p > 1, then |u|p is subharmonic on Ω. It is known that |u| is subharmonic on Ω. Since 0 for x < 0 φ(x) = xp for x ≥ 0 is an increasing, convex function on R, |u|p is subharmonic on Ω.

2.4 Properties of Superharmonic Functions If u and v are superharmonic on the open set Ω ⊂ Rn , it is not apparent that u + v is superharmonic because of the requirement that it not be identically +∞ on any component of Ω. To show that u + v ∈ S(Ω), it must be shown that v cannot be identically +∞ on the set of points where u is ﬁnite. The following theorem will accomplish this and will also show that volume averages could have been used to deﬁne superharmonic functions. As usual, Ω will denote an open subset of Rn . Theorem 2.4.1 If u ∈ L(Ω) and for each x ∈ Ω there is a δx > 0 such that Bx,δx ⊂ Ω and u(x) ≥ A(u : x, δ) whenever δ < δx , then u is superharmonic on Ω. Moreover, if u is superharmonic on Ω and x ∈ Ω, then u(x) ≥ A(u : − ⊂ Ω. x, δ) whenever Bx,δ Proof: As superharmonicity is a local property, it suﬃces to show that u is superharmonic on each open, connected subset U of Ω with closure U − ⊂ Ω. Let u satisfy the hypotheses of the ﬁrst statement, let h be continuous on U − , harmonic on U , and u ≥ h on ∂U . By Theorem 2.3.5, u − h satisﬁes the minimum principle on U . Since u − h is l.s.c. on U − and u − h ≥ 0 on ∂U , u − h ≥ 0 on U by the minimum principle. It follows that u is superharmonic on U by the alternative deﬁnition of superharmonic function. Suppose now − ⊂ Ω. It will be proven now that u is superharmonic on Ω, x ∈ Ω, and Bx,δ that u(x) ≥ A(u : x, δ). If u(x) = +∞, the inequality is trivially true. Assume that u(x) < +∞. Since u(x) ≥ L(u : x, ρ) for 0 < ρ < δ, σn ρn−1 u(x) ≥ u(z) dσ(z). ∂Bx,ρ

Integrating over (0, δ), σn δ n u(x) ≥ n

δ

u(z) dσ(z)

0

∂Bx,ρ

dρ =

u(z) dz. Bx,δ

− ⊂ Ω. Since νn = σn /n, u(x) ≥ A(u : x, δ) whenever Bx,δ

66

2 The Dirichlet Problem

The fact that a superharmonic function u must be ﬁnite at some point of each component of Ω and that the average of u over a ball about each such point is also ﬁnite suggests that u must be ﬁnite a.e. on Ω; that is, except possibly on a subset of Ω of Lebesgue measure zero. Theorem 2.4.2 If u is superharmonic on the open set Ω, then u is ﬁnite a.e. on Ω relative to Lebesgue measure and Lebesgue integrable on each compact set K ⊂ Ω. Proof: Since it suﬃces to show that u is ﬁnite a.e. on each component of Ω, it might as well be assumed that Ω is connected. Since u is not identically +∞ on Ω, there is at least one point x ∈ Ω where u is ﬁnite. By the preceding theorem, there is a δ > 0 such that u is ﬁnite a.e. on Bx,δ . Thus, M = {x ∈ Ω; u is ﬁnite a.e. on Bx,δ for some δ > 0} = ∅. It will be shown now that M is open. Suppose x ∈ M . Then u is ﬁnite a.e. on some ball Bx,δ ⊂ Ω. Consider any y ∈ Bx,δ , and let 2ρ = min {|y − x|, δ − |y − x|}. Then By,ρ ⊂ Bx,δ , u is ﬁnite a.e. on By,ρ , and y ∈ M ; that is, Bx,δ ⊂ M . This shows that M is open. It will be shown next that M is relatively closed in Ω. Let {xi } be a sequence in M with limi→∞ xi = x ∈ Ω. Since Ω is open, there is a ball Bx, ⊂ Ω. Choose i so that xi ∈ Bx, /2 . Since xi ∈ M , there is a δi > 0 such that u is ﬁnite a.e. on Bxi ,δi . In particular, u is ﬁnite a.e. on Bxi ,δi ∩ Bx, /2 , which has positive Lebesgue measure. Thus, there is a point z ∈ Bxi ,δi ∩ Bx, /2 such that u(z) < +∞. By the preceding theorem, u is ﬁnite a.e. on Bz, /2 . Since x ∈ Bz, /2 , there is a ball Bx,ρ ⊂ Bz, /2 on which u is ﬁnite a.e.; that is, x ∈ M and M is relatively closed in Ω. Since M = ∅, M = Ω by the connectedness of Ω. By deﬁnition of M , to each x ∈ M there corresponds a ball Bx,δx ⊂ Ω on which u is ﬁnite a.e. Since such open sets cover Ω, a countable subcollection covers Ω. Since u is ﬁnite a.e. on each element of the countable covering, u is ﬁnite a.e. on Ω. Now let K be a compact subset of Ω. Cover K by a ﬁnite number of balls with centers xi and radii δi such that Bx−i ,δi ⊂ Ω, i = 1, . . . , p. Since u is ﬁnite a.e. on arbitrarily small balls containing each xi , it can be assumed that u(xi ) < +∞. This may require replacing xi by a nearby xi and increasing δi slightly. Then +∞ > u(xi ) ≥ A(u : xi , δi ) for i = 1, . . . , p. Since u is bounded below on pi=1 Bx−i ,δi , it can be assumed that u ≥ 0 on this set. Then p p −∞ < u(z) dz ≤ u(z) dz ≤ νn δin u(xi ) < +∞ K

and u is integrable on K.

i=1

Bxi ,δi

i=1

2.4 Properties of Superharmonic Functions

67

Theorem 2.4.3 If u is superharmonic on the open set Ω and B = Bx,δ is a ball with B − ⊂ Ω, then u is integrable relative to surface area on ∂B, PI(u : B) is harmonic on B, and u ≥ PI(u : B) on B. Proof: Since u is l.s.c. on ∂B, it can be assumed the u ≥ 0 on ∂B. Moreover, there is a sequence {φj } of nonnegative continuous functions on ∂B such that φj ↑ u on ∂B. Let PI(φj : B) on B vj = φj on ∂B. Since u ≥ vj = φj on ∂B, vj is harmonic on B, and vj is continuous on B − , u ≥ vj on B by the alternative deﬁnition of a superharmonic function. Because {vj } is an increasing sequence of harmonic functions on B, v = limj→∞ vj is either identically +∞ or harmonic on B by Theorem 2.2.5. Since u is ﬁnite a.e. on B, v is harmonic on B. It follows from the Lebesgue monotone convergence theorem that u ≥ v = PI(u : B) on B. Since +∞ > v(x) = PI(u : B)(x) = L(u : x, δ), u is integrable on ∂B. Spherical and volume averages of a superharmonic function as a function of the radius will be examined now. Lemma 2.4.4 If u is superharmonic on the open set Ω, x ∈ Ω, δx = d(x, ∼ Ω), then L(u : x, δ) and A(u : x, δ) are decreasing functions of δ on [0, δx ) which are right continuous at δ = 0 where L(u : x, 0) = A(u : x, 0) = u(x) by deﬁnition; moreover, A(u : x, δ) is continuous on [0, δx ). If u and v are superharmonic functions with u ≤ v a.e. on Ω or u = v a.e. on Ω, then u ≤ v on Ω or u = v on Ω, respectively. Proof: Suppose 0 ≤ δ < δ1 < δx . Deﬁne a function h on Bx,δ1 by h = PI(u : − Bx,δ1 ). Then u ≥ h on Bx,δ1 by Theorem 2.4.3. Since Bx,δ ⊂ Bx,δ1 and h is harmonic on the latter ball, L(u : x, δ) ≥ L(h : x, δ) = h(x) = L(u : x, δ1 ). This shows that L(u : x, δ) is monotone decreasing on [0, δx ). Since u is l.s.c. at x, {y : u(y) > u(x) − } is a neighborhood of x for each > 0 and L(u : x, δ) ≥ L(u(x) − : x, δ) = u(x) − for all suﬃciently small δ. This shows that L(u : x, δ) ↑ u(x) as δ ↓ 0. For any δ ≤ δ1 < δx , A(u : x, δ) is ﬁnite by Theorem 2.4.2 and 1 A(u : x, δ) = u(z) dz νn δ n Bx,δ δ 1 = ρn−1 u(x + ρθ) dσ(θ) dρ νn δ n 0

θ =1 δ σn = ρn−1 L(u : x, ρ) dρ. νn δ n 0

68

2 The Dirichlet Problem

This shows, among other things, that ρn−1 L(u : x, ρ) is integrable on [0, δ1 ]. Since the indeﬁnite integral of an integrable function is absolutely continuous, A(u : x, δ) is continuous on (0, δ1 ). The above equation also gives a representation of A(u : x, δ) on any closed interval [a, b] ⊂ (0, δx ) as a product of two bounded, absolutely continuous functions of δ on [a, b] and it follows directly from the deﬁnition of such functions that A(u : x, δ) is diﬀerentiable a.e. on (0, δx ). Recall also that the derivative of an indeﬁnite integral exists and is equal to the integrand a.e. relative to Lebesgue measure. Then for almost all δ ∈ (0, δ1 ), δ nσn σn L(u : x, δ) − ρn−1 L(u : x, ρ) dρ νn δ νn δ n+1 0 n = [L(u : x, δ) − A(u : x, δ)], δ

Dδ A(u : x, δ) =

since νn = σn /n. In view of the fact that L(u : x, ρ) is monotone decreasing on (0, δ), σn A(u : x, δ) = νn δ n ≥

σn νn δ n

δ

ρn−1 L(u : x, ρ) dρ 0

δ

ρn−1 dρ L(u : x, δ) 0

= L(u : x, δ),

(2.3)

and Dδ A(u : x, δ) ≤ 0 for almost all δ ∈ (0, δx ). If 0 < a < δ, then

δ

A(u : x, δ) − A(u : x, a) =

Dρ A(u : x, ρ) dρ, a

and A(u : x, δ) is a monotone decreasing function of δ on (0, δx ). Since A(u : x, 0) = u(x) ≥ A(u : x, δ) ≥ L(u : x, δ) for any δ ∈ (0, δx ), A(u : x, δ) is monotone decreasing on [0, δx ] and right continuous at 0. The fact that a superharmonic function is ﬁnite a.e. places a limitation on the set of inﬁnities of a superharmonic function. Theorem 2.4.5 If u and v are superharmonic on an open set Ω and c > 0, then cu, u + v, min (u, v) are superharmonic on Ω. Proof: That cu is superharmonic on Ω is obvious from the deﬁnition. As to u + v, note that u + v cannot be identically +∞ on any component U of Ω since each component has positive Lebesgue measure and both u and v are ﬁnite a.e. on U . Moreover, u + v > −∞ on Ω and u + v is l.s.c. on − ⊂ Ω, then u(x) ≥ L(u : x, δ) Ω since both have these properties. If Bx,δ and v(x) ≥ L(v : x, δ) so that u(x) + v(x) ≥ L(u + v : x, δ). Thus, u + v is superharmonic on Ω. It will be shown that min (u, v) is super-mean-valued

2.4 Properties of Superharmonic Functions

69

on Ω, the other properties of a superharmonic function being easy to verify. − ⊂ Ω, then u(x) ≥ L(u : x, δ) ≥ L(min (u, v) : x, δ) with the same If Bx,δ holding for v. Thus, min (u, v)(x) ≥ L(min (u, v) : x, δ). Theorem 2.4.6 If u is both subharmonic and superharmonic on an open set Ω, then u is harmonic on Ω. Theorem 2.4.7 If u is superharmonic on the open set Ω, x ∈ Ω, and Λ is a subset of Ω of Lebesgue measure zero, then u(x) = lim inf y→x,y ∈Λ∪{x} u(y); in particular, u(x) = lim inf y→x,y =x u(y). Proof: By l.s.c., u(x) ≤ lim inf u(y) ≤ y→x,y =x

lim inf

y→x,y ∈Λ∪{x}

u(y).

− If Bx,δ is any ball with Bx,δ ⊂ Ω, then

u(x) ≥ A(u : x, δ) ≥ inf{u(y); y ∈ Bx,δ , y ∈ Λ ∪ {x}}. Thus, u(x) ≥ lim inf y→x,y ∈Λ∪{x} u(y) and the two are equal. Convergence properties of monotone sequences of superharmonic functions will be considered; more generally, right-directed families of such functions will be considered. Theorem 2.4.8 If {ui ; i ∈ I} is a right-directed family of superharmonic functions on the open set Ω, then on each component of Ω the function u = supi∈I ui is either superharmonic or identically +∞. Proof: Since each ui is l.s.c. and the supremum of any collection of l.s.c. functions is again l.s.c., u is l.s.c. on Ω. For any i0 ∈ I, u ≥ ui0 > −∞ on Ω. − ⊂ Ω, then by Lemma 2.2.10 If Bx,δ u(x) = sup ui (x) ≥ sup L(ui : x, δ) = L(sup ui : x, δ) = L(u : x, δ). i∈I

i∈I

i∈I

Thus, u is super-mean-valued and is superharmonic on each component of Ω provided it is not +∞ thereon. The preceding theorem applies to monotone increasing sequences of superharmonic functions, but the analogous theorem for monotone decreasing sequences is not true. Consider, for example, the superharmonic function uy on R3 with pole y. For j ≥ 1, deﬁne uj = uy /j. Then {uj } is a monotone decreasing sequence of superharmonic functions. Letting u = inf j≥1 uj , u = 0 except at y where it is equal to +∞. Clearly, u is not superharmonic since it is not l.s.c. at y. Not only is the lack of semicontinuity of the limit a diﬃculty, but the requirement that the limit not take on the value −∞ is another problem.

70

2 The Dirichlet Problem

Theorem 2.4.9 If {ui ; i ∈ I} is a left-directed family of superharmonic functions on the open set Ω which is locally bounded below, then the lower regularization u ˆ of u = inf i∈I ui is superharmonic on Ω. Proof: The fact that the family is locally bounded below implies that u ˆ cannot take on the value −∞. For any i ∈ I, ui ≥ u and ui (x) = ˆ(x), x ∈ Ω, by Theorem 2.4.7 lim inf y→x,y =x ui (y) ≥ lim inf y→x,y =x u(y) = u and uˆ is not identically +∞ on any component of Ω. It will be shown that u ˆ is superharmonic using the deﬁnition of superharmonic function following Theorem 2.3.8. Let U be an open subset of Ω with compact closure U − ⊂ Ω. Let h be harmonic on U , continuous on U − , and u ˆ ≥ h on ∂U . Then ui ≥ h on ∂U for each i ∈ I. Since each ui is superharmonic on Ω, ui ≥ h on U ˆ = h on U . Therefore, u for each i ∈ I. Then u ≥ h on U and u ˆ ≥ h ˆ is superharmonic on Ω. It is apparent that the maximum of a convex function and a linear function is again a convex function that is partially linear. There is an analogous operation on superharmonic functions deﬁned as follows. Deﬁnition 2.4.10 If u is superharmonic on the open set Ω and B is a ball with B − ⊂ Ω, let PI(u : B) on B uB = u on Ω ∼ B. If u is subharmonic on Ω, uB is deﬁned similarly. The function uB (uB ) is called the lowering (lifting) of u over B. Lemma 2.4.11 If u is superharmonic on the open set Ω and B is a ball with B − ⊂ Ω, then uB ≤ u on Ω, harmonic on B, and superharmonic on Ω. Moreover, if v is a superharmonic function on B with v ≥ uB on B, then the function min (u, v) on B w= u on Ω ∼ B is superharmonic on Ω. Proof: By Theorem 2.4.3, uB is deﬁned, harmonic on B, and uB ≤ u on Ω. The fact that uB is superharmonic is a special case of the second assertion by taking v = uB on B, in which case w = uB . Let v be any superharmonic function on B satisfying v ≥ uB on B. Then w ≥ uB on Ω with w = uB = u on Ω ∼ B. Clearly, w is superharmonic on Ω ∼ ∂B so that it suﬃces to show that w is l.s.c. and super-mean-valued at points of ∂B. Consider any x ∈ ∂B. Then lim inf w(y) = lim inf u(y) ≥ lim inf u(y) ≥ u(x) = w(x).

y→x,y∈∼B

y→x,y∈∼B

y→x,y∈Ω

2.5 Approximation of Superharmonic Functions

71

Since u is l.s.c. on ∂B, lim inf w(y) ≥ lim inf uB (y) ≥ lim inf u(y) ≥ u(x) = w(x)

y→x,y∈B

y→x,y∈B

y→x,y∈∂B

by Lemma 1.7.5. This shows that w is l.s.c. at points of ∂B and therefore l.s.c. on Ω. Consider again x ∈ ∂B. Since w ≤ u on Ω, for any closed ball − Bx,δ ⊂Ω w(x) = u(x) ≥ L(u : x, δ) ≥ L(w : x, δ), which shows that w is super-mean-valued at points of ∂B and therefore locally super-mean-valued on Ω.

2.5 Approximation of Superharmonic Functions Consider an extended real-valued function u on an open set Ω ⊂ Rn . In order to smoothen u, at the expense of reducing its domain, a molliﬁer m : Rn → R deﬁned by ce

m(x) =

1 − 1−|x| 2

0

if |x| ≤ 1 if |x| > 1,

(2.4)

will be employed, where c is a constant chosen so that m is a probability density. It is known that m ∈ C0∞ (Rn ). Let u be a locally integrable function on Ω and let h > 0. For x ∈ Ωh = {y ∈ Ω; d(y, ∂Ω) > h}, deﬁne 1 Jh u(x) = n h

m Ω

y−x h

1 u(y) dy = n h

m Bx,h

y−x h

u(y) dy. (2.5)

It is easily seen that the operator Jh has the following properties: (i) u ≥ 0 on Ω implies that Jh u ≥ 0 on Ωh . (ii) Jh u ∈ C ∞ (Ωh ). (iii) Jh 1 = 1 on Ωh . (iv) u linear on Ω implies that Jh u = u on Ωh . The limit limh→0 Jh u(x), x ∈ Ω, is understood to be over those h for which x ∈ Ωh . Lemma 2.5.1 If u is superharmonic on the open set Ω and h > 0, then Jh u is a C ∞ (Ωh ) superharmonic function on Ωh , Jh u ≤ Jk u ≤ u on Ωh whenever 0 < k ≤ h, and limh→0 Jh u(x) = u(x) for each x ∈ Ω; if, in addition, u is harmonic on Ω, then Jh u = u on Ωh .

72

2 The Dirichlet Problem

Proof: That Jh u ∈ C ∞ (Ωh ) follows from the fact that m ∈ C0∞ (Rn ). It will be shown now that u is locally super-mean-valued on Ωh . Consider any ball − Bx,δ ⊂ Bx,δ ⊂ Ωh . Then A(Jh u : x, δ) =

1 νn δ n

=

m(z)u(y + hz) dz dy

Bx,δ

|z|≤1

m(z) |z|≤1

≤

1 νn δ n

u(y + hz) dy

dz

Bx,δ

m(z)u(x + hz) dz |z|≤1

= Jh u(x), and Jh u is locally super-mean-valued on Ωh . Using spherical coordinates (ρ, θ) relative to the pole x ∈ Ωh , the fact that m(ρθ/h) is independent of θ, and the superharmonicity of u,

h 1 ρθ Jh u(x) = n m u(x + ρθ)ρn−1 dθ dρ h 0 |θ|=1 h h ρθ 1 n−1 m u(x + ρθ) dθ dρ = n ρ h 0 h |θ|=1 h ρθ 1 m ≤ n ρn−1 σn u(x) dρ h 0 h = u(x)Jh 1(x) = u(x). Thus, Jh u ≤ u on Ωh . Consider any x ∈ Ω. Since u(x) = lim inf y→x,y =x u(y), given > 0 there is a neighborhood N of x such that inf y∈(N ∩Ω)∼{x} u(y) > u(x) − . Consider any h > 0 for which x ∈ Ωh and Bx,h ⊂ N . Then

y−x 1 m Jh u(x) = n u(y) dy h Bx,h h ≥ (u(x) − )Jh 1(x) = u(x) − . Thus for each x ∈ Ω, limh→0 Jh u(x) = u(x). If u is harmonic on Ω, then the above argument is applicable to both u and −u so that Jh u ≤ u and Jh u ≥ u on Ωh and the two are equal. It remains only to show that 0 < k < h implies Jh u ≤ Jk u on Ωh . Consider any x ∈ Ωh and k < h. Letting ρθ = (y − x)/h where θ is a point on the unit sphere with center at 0,

y−x 1 m Jh u(x) = n u(y) dy h Bx,h h

2.5 Approximation of Superharmonic Functions

1

73

m(ρθ)u(x + ρhθ)ρn−1 dθ dρ.

= 0

|θ|=1

Since m(ρθ) does not depend upon θ, by Theorem 2.4.4 Jh u(x) =

0

≤

0

1

1

m(ρθ)ρn−1 σn L(u : x, ρh) dρ m(ρθ)ρn−1 σn L(u : x, ρk) dρ

= Jk u(x) and Jh u ≤ Jk u ≤ u on Ωh whenever 0 ≤ k < h. Theorem 2.5.2 If u is superharmonic on the open set Ω ⊂ Rn and harmonic on Ω ∼ Γ where Γ is a compact subset of Ω, then there is an increasing sequence of superharmonic functions {uj } on Ω such that (i) uj ∈ C ∞ (Ω) for all j ≥ 1, (ii) limj→∞ uj = u on Ω, and (iii) if U is any neighborhood of Γ with compact closure U − ⊂ Ω, then uj = u on Ω ∼ U − for suﬃciently large j. Proof: Let U be a neighborhood of Γ with compact closure U − ⊂ Ω. Since {Ωh ; h > 0} is an open covering of the compact set U − , there is an h0 > 0 such that Γ ⊂ U ⊂ U − ⊂ Ωh0 and h0 < d(Γ, ∼ U ). Let {hj } be a sequence in (0, h0 ) that decreases to 0 and let Jhj u(x) if x ∈ Ωhj uj (x) = u if x ∈ Ω ∼ Ωhj . Since the Ωhj increase to Ω, the sequence {uj } increases to u on Ω. Since uj is superharmonic on Ωhj ⊃ U − , uj is superharmonic on U ; moreover, uj ∈ C ∞ (Ωhj ) implies that uj ∈ C ∞ (U ). It will be shown now that uj = u on a neighborhood of Ω ∼ U , thus implying that uj ∈ C ∞ (Ω) and is harmonic on Ω ∼ U − . By deﬁnition, uj = u on Ω ∼ Ωhj . Consider any x ∈ Ωhj ∼ U . Since x ∈ Ωhj and hj < d(Γ, ∼ U ), δj > 0 can be chosen so that Bx,δj ⊂ Ωhj and hj + δj < d(Γ, ∼ U ). For y ∈ Bx,δj , d(y, Γ ) ≥ d(x, Γ ) − d(x, y) > d(Γ, ∼ U ) − δj > hj so that By,δj ⊂ Ω ∼ Γ . Thus, uj (y) = Jhj u(y) = u(y) for all y ∈ Bx,δj . This proves that uj = u on a neighborhood of Ω ∼ U . It was shown in Lemma 2.4.4 that L(u : x, δ) is a decreasing function of δ if u is superharmonic. Much more is true. Recall that a function φ is concave on an interval (a, b) if −φ is convex on the interval; that is, φ(λx + (1 − λy) ≥ λφ(x) + (1 − λ)φ(y) whenever a < x < y < b and 0 < λ < 1.

74

2 The Dirichlet Problem

Theorem 2.5.3 Let u be superharmonic on the open set Ω and let Bx,ρ be a ball with ∂Bx,ρ ⊂ Ω for r1 ≤ ρ ≤ r2 . Then L(u : x, ρ) is a concave function of − log ρ if n = 2 and a concave function of ρ−n+2 if n ≥ 3 on (r1 , r2 ); in particular, L(u : x, ρ) is a continuous function of ρ on (r1 , r2 ). If, in addition, u is harmonic on Ω, then 1 Dn u(z) dσ(z) log ρ + const. (2.6) L(u : x, ρ) = 2π ∂Bx,r1 on (r1 , r2 ) if n = 2, and L(u : x, ρ) =

1 − σn (n − 2)

Dn u(z) dσ(z) ρ−n+2 + const.

(2.7)

∂Bx,r1

on (r1 , r2 ) if n ≥ 3. Proof: Assume ﬁrst that u ∈ C 2 (Ω). Letting Aρ denote the closed annulus − Bx,ρ ∼ Bx,r1 , r1 < ρ ≤ r2 , u has a bounded normal derivative on ∂Aρ . Since u is superharmonic on Ω, Δu ≤ 0 on Ω by Lemma 2.3.4. Applying Green’s theorem to the region Aρ ,

0≥

Dn u(z) dσ(z) −

Δu dz = Aρ

Dn u(z) dσ(z)

∂Bx,ρ

∂Bx,r1

for r1 < ρ ≤ r2 . Using the fact that Dn u is bounded on ∂Aρ , Dn u(z) dσ(z) = ρn−1 Dr (x + rθ)|r=ρ , dσ(θ) |θ|=1

∂Bx,ρ

u(x + (ρ + η)θ) − u(x + ρθ) dσ(θ) η |θ|=1 η→0 1 n−1 =ρ lim u(x + (ρ + η)θ) dσ(θ) η→0 η |θ|=1

= ρn−1

−

lim

u(x + ρθ) dσ(θ) |θ|=1

= σn ρn−1 Dρ L(u : x, ρ). Therefore, σn ρn−1 Dρ L(u : x, ρ) =

Δu(z) dz +

Aρ

Dn u(z) dσ(z). ∂Bx,r1

2.6 Perron-Wiener Method

75

If u is harmonic on Ω, then the ﬁrst term on the right is zero, the second term is a constant, and an integration with respect to ρ results in Equations (2.6) and (2.7). Suppose now that n ≥ 3. Letting ζ = 1/ρn−2 , ρ(ζ) the inverse function, and k the second term on the right of the last equation, −σn (n − 2)Dζ L(u : x, ρ(ζ)) = Δu(z) dz + k. Aρ(ζ)

As ρ increases,

Aρ

Δu(z) dz decreases; thus, Dζ L(u : x, ρ(ζ)) increases as

ζ decreases on (ζ1 , ζ2 ) where ζ1 = 1/r2n−2 , ζ2 = 1/r1n−2 . Therefore, L(u : x, ρ(ζ)) is a concave function of ζ = 1/ρn−2 ; that is, L(u : x, ρ) is a concave function of 1/ρn−2 . The proof of the n = 2 case is accomplished in the same way by letting ζ = − log ρ. Consider now an arbitrary superharmonic function u and the n = 3 case. By Lemma 2.5.1, there is a C ∞ (Ω) sequence {uj } of − superharmonic functions on a neighborhood of Bx,r ∼ Bx,r1 with uj ↑ u. 2 Note that the sequence {L(uj : x, ρ)} is increasing and uniformly bounded on (r1 , r2 ) since L(uj : x, ρ) ≤ L(uj : x, r1 ) ≤ L(u : x, r1 ) < +∞, the latter inequality holding because u is bounded on ∂Bx,r1 . For each j ≥ 1, there is a concave function ψj deﬁned on (ζ1 , ζ2 ) such that L(uj : x, ρ) = ψj (ρ−n+2 ) = ψj (ζ). Note that the sequence {ψj } is increasing and uniformly bounded on (ζ1 , ζ2 ). It is easily seen that ψ = limj→∞ ψj is also a concave function and that L(u : x, ρ) = ψ(ρ−n+2 ) for ρ ∈ (r1 , r2 ). The same argument applies to the n = 2 case. Continuity of L(u : x, ρ) on (r1 , r2 ) follows from the fact that a concave function on an open interval is continuous.

2.6 Perron-Wiener Method Let Ω be an open subset of Rn with nonempty boundary, and let f be an extended real-valued function on ∂Ω. The generalized Dirichlet problem is that of ﬁnding a harmonic function h corresponding to the boundary function f . Having shown that such a function exists, there is then the problem of relating the limiting behavior of h at boundary points of Ω to the values of f at such points. A standard method of proving the existence of an h is to approximate from above and below, and then show that the diﬀerence between the two can be made arbitrarily small. Deﬁnition 2.6.1 A family F of superharmonic functions on an open set Ω is saturated if (i) u, v ∈ F implies min (u, v) ∈ F (ii) u ∈ F and B a ball with B − ⊂ Ω implies uB ∈ F where

76

2 The Dirichlet Problem

uB =

PI(u : B) u

on B on Ω ∼ B.

Theorem 2.6.2 If F is a saturated family of superharmonic functions on Ω, then inf u∈F u is either identically −∞ or harmonic on each component of Ω. Proof: Let w = inf u∈F u and consider any closed ball B − ⊂ Ω. Since the assertion of the theorem pertains to the components of Ω, it can be assumed that Ω is connected. For u ∈ F, uB ≤ u, harmonic on B, and superharmonic on Ω by Lemma 2.4.11. Moreover, w = inf{uB ; u ∈ F }. Consider any u, v ∈ F and the corresponding uB , vB . Since F is left-directed, min (u, v) ∈ F so that (min (u, v)B ≤ min (uB , vB ). It follows that the family {uB ; u ∈ F } is left-directed. By Theorem 2.4.8, w = inf{uB ; u ∈ F } is identically −∞ or superharmonic on Ω. Since the functions uB are harmonic on B, w is identically −∞ or harmonic on B by Theorem 2.2.7. If w is not identically −∞ on Ω, then it is superharmonic on Ω and therefore harmonic on each ball B since superharmonic functions are ﬁnite a.e. on compact subsets of Ω by Theorem 2.4.2. Given a boundary function f , classes of functions that approximate the solution of the Dirichlet problem from above and from below will be considered. Deﬁnition 2.6.3 An extended real-valued function u on Ω is hyperharmonic (hypoharmonic) if it is superharmonic (subharmonic) or identically +∞ (−∞) on each component of Ω. Deﬁnition 2.6.4 The upper class Uf of functions determined by the function f on ∂Ω is given by Uf = {u; u hyperharmonic on Ω, lim inf u(y) ≥ f (x) y→x,y∈Ω

for all x ∈ ∂Ω, u bounded below on Ω}. The lower class Lf of functions determined by the function f on ∂Ω is given by Lf = {u; u hypoharmonic on Ω, lim sup u(y) ≤ f (x) y→x,y∈Ω

for all x ∈ ∂Ω, u bounded above on Ω}. Note that Uf contains the function that is identically +∞ on Ω, and Lf contains the function that is identically −∞ on Ω. Also note that Uf (Lf ) may contain only the identically +∞ (−∞) function, and also that the requirement that elements of Uf be bounded below may not be fulﬁlled by any nontrivial hyperharmonic function.

2.6 Perron-Wiener Method

77

Deﬁnition 2.6.5 The Perron upper solution for the boundary function f is given by H f = inf {u; u ∈ Uf }. The Perron lower solution is given by H f = sup {u; u ∈ Lf }. Whenever it is necessary to compare upper classes and upper solutions for two diﬀerent regions Ω and Σ, the upper classes will be denoted by UΩ f Ω

Σ

and UΣ g and the upper solutions by H f and H g , respectively, with a similar convention for lower classes and lower solutions. As a result of the following lemma, it can be assumed usually that Ω is connected. Lemma 2.6.6 If f is a function on ∂Ω and Λ is a component of Ω, then Λ Ω Ω H f |∂Λ = H f |Λ and H Λ f |∂Λ = H f |Λ . Proof: Let UΛ f |∂Λ be the upper class for Λ and f |∂Λ . It will be shown ﬁrst Ω Λ ∗ ⊂ UΩ that UΛ f |Λ = {u|Λ ; u ∈ Uf }. If u ∈ Uf |∂Λ , let u = u on Λ and f |∂Λ Λ Ω u∗ = +∞ on Ω ∼ Λ. Then u∗ ∈ UΩ f and Uf |∂Λ ⊂ Uf |Λ . Suppose now that u ∈ UΩ f and x ∈ ∂Λ. Then x ∈ ∂Ω and lim inf u(y) ≥ lim inf u(y) ≥ f (x),

y→x,y∈Λ

y→x,y∈Ω

x ∈ ∂Λ. Λ

Ω Λ Λ Ω Therefore, u|Λ ∈ UΛ f |∂Λ and Uf |Λ ⊂ Uf |Λ . Thus, Uf |∂Λ = Uf |Λ and H f |∂Λ = Ω

H f |Λ . Lemma 2.6.7 H f (H f ) is identically −∞ (+∞) or harmonic on each component of Ω. Proof: It can be assumed that Ω is connected. If Uf contains only the identically +∞ function, then H f = +∞ and the assertion is true. Suppose Uf contains a hyperharmonic function that is not identically +∞ on Ω. Then H f = inf {u; u ∈ Uf , u superharmonic on Ω}. If it can be shown that the family of superharmonic functions in Uf is a saturated family, it would then follow from Theorem 2.6.2 that H f is either identically −∞ or harmonic on Ω. Accordingly, let u1 and u2 be two superharmonic members of Uf . If x ∈ ∂Ω, then lim inf y→x,y∈Ω ui (y) ≥ f (x) and it is easy to show that lim inf min (u1 , u2 )(y) ≥ f (x).

y→x,y∈Ω

Since u1 and u2 are each bounded below on Ω, the same is true of the superharmonic function min (u1 , u2 ) and it has been shown that min (u1 , u2 ) ∈ Uf . If u is a superharmonic function in Uf and B is a ball with B − ⊂ Ω, then uB is superharmonic on Ω by Lemma 2.4.11, lim inf uB (y) = lim inf u(y) ≥ f (x)

y→x,y∈Ω

y→x,y∈Ω

78

2 The Dirichlet Problem

for all x ∈ ∂Ω, and uB is bounded below by the same constant bounding u; that is, uB is a superharmonic member of Uf . Therefore, the family of superharmonic functions in Uf is a saturated family. Deﬁnition 2.6.8 If H f = H f and both are harmonic, then f is called a resolutive boundary function ; in this case, Hf = H f = H f is called the Dirichlet solution for f . The solution of the Dirichlet problem need not be unique. In fact, given a n = {(x1 , . . . , xn ; xn > continuous boundary function f on the half-space R+ n 0} ⊂ R , according to Theorem 1.9.1 there is a one-parameter family of Dirichlet solutions corresponding to f . The diﬃculty here is related to the n fact that the region R+ is unbounded. This diﬃculty will be avoided by assuming that Ω is bounded for the remainder of this section. It is also possible that the Perron-Wiener method will result in the same harmonic function for two diﬀerent, but not too diﬀerent, boundary functions. Let Ω = B0,1 ∼ {0} ⊂ R3 , let B = B0,1 , and let 1 at 0 f= 0 on ∂B. Consider the function u(x) = 1/|x|, x ∈ Ω, the restriction of the fundamental harmonic function u0 to Ω. For every j ≥ 1, u/j ∈ Uf , so that H f = 0 on Ω. Since the zero function belongs to Lf , 0 ≤ H f ≤ H f = 0 and so Hf = H f = H f = 0 on Ω. On the other hand, if g = 0 on ∂Ω, then Hg = 0. Thus, f and g determine the same harmonic function although 0 = g(0) = f (0) = 1. In this case, the set of points at which f and g diﬀer does not aﬀect the Dirichlet solution. Such sets will be studied in greater detail later. What follows in this section is dependent upon Corollary 2.3.6 in which it is assumed that Ω is bounded. Since the conclusions of the corollary will be extended to unbounded Ω in Theorem 5.2.5, the results of the remainder of this section are valid for unbounded sets. Lemma 2.6.9 Let Ω be a bounded open subset of Rn . If u ∈ Lf and v ∈ Uf , then u ≤ v and H f ≤ H f on Ω. Proof: It can be assumed that Ω is connected. If either v is not superharmonic or u is not subharmonic, then u ≤ v trivially. It suﬃces to consider the case where v − u is superharmonic. If x ∈ ∂Ω and f (x) is ﬁnite, then lim inf (v − u)(y) ≥ lim inf v(y) − lim sup u(y) ≥ f (x) − f (x) = 0;

y→x,y∈Ω

y→x,y∈Ω

y→x,y∈Ω

if f (x) = +∞, then lim inf y→x,y∈Ω v(y) = +∞ and lim inf y→x,y∈Ω (v − u)(y) ≥ 0 by virtue of the fact that u ∈ Lf is bounded above on Ω and, similarly, if f (x) = −∞. Therefore, lim inf y→x,y∈Ω (v − u)(y) ≥ 0 for all

2.6 Perron-Wiener Method

79

x ∈ ∂Ω, and v ≥ u on Ω by Corollary 2.3.6. Using the fact that v ≥ u for all v ∈ Uf and each u ∈ Lf , H f ≥ u for all u ∈ Lf and therefore H f ≥ H f . If there is any merit to the Perron-Wiener method, it should give solutions that are consistent with certain natural solutions. Theorem 2.6.10 Let Ω be a bounded open subset of Rn . If f is bounded on ∂Ω and there is a harmonic function h on Ω such that limy→x,y∈Ω h(y) = f (x), x ∈ ∂Ω, then f is resolutive and Hf = h. Proof: As was noted in Section 0.1, h has a continuous extension to Ω − and is bounded on Ω. Therefore, h belongs to both Lf and Uf and H f ≤ h ≤ H f . Thus, H = H f = h, and since h is a harmonic function, f is a resolutive boundary function and h = Hf . In particular, if h is harmonic on a neighborhood of a compact set Ω and the Perron-Wiener method is applied to h|∂Ω , the solution is just what it should be. Eventually necessary and suﬃcient conditions will be determined in order that a boundary function be resolutive. For the time being, it is more convenient to study the class of resolutive functions. Each of the assertions of the following lemma has been proved already or is easily proved from the deﬁnitions. Lemma 2.6.11 Let Ω be a bounded open subset of Rn , let f and g be extended real-valued functions on ∂Ω, and let c be any real number. (i) If f = c on ∂Ω, then f is resolutive and Hf = c on Ω. (ii) H f +c = H f + c and H f +c = H f + c. If f is resolutive, then f + c is resolutive and Hf +c = Hf + c. (iii) If c > 0, then H cf = cH f and H cf = cH f . If f is resolutive, then cf is resolutive and Hcf = cHf for c ≥ 0. (iv) If f ≤ g, then H f ≤ H g and H f ≤ H g . (v) H −f = −H f . If f is resolutive, then −f is resolutive with H−f = −Hf . (vi) H f +g ≤ H f + H g and H f +g ≥ H f + H g whenever the sums are deﬁned. If f and g are resolutive and f + g is deﬁned, then f + g is resolutive with Hf +g = Hf + Hg . Lemma 2.6.12 Let Ω be a bounded open subset of Rn . The resolutive boundary functions form a linear class. Moreover, if {fj } is a sequence of such functions that converges uniformly to f , then f is resolutive and the sequence {Hfj } converges uniformly to Hf . Proof: If |fj − f | < , then fj − < f < fj + , H f ≤ H fj + = H fj + , and H f ≥ H fj − = H fj − . Therefore, |H f − Hfj | = |H f − H fj | < , and the sequence {Hfj } converges uniformly to H f . In the same way, the sequence {Hfj } converges uniformly to H f . Thus, H f = H f and since both are ﬁnite by the above inequalities, f is resolutive.

80

2 The Dirichlet Problem

Lemma 2.6.13 Let Ω be a bounded open subset of Rn . If {fj } is an increasing sequence of boundary functions, f = limj→∞ fj , and H f1 > −∞ on Ω, then H f = limj→∞ H fj ; if, in addition, the fj are resolutive, then H f = H f and f is resolutive if at least one of H f or H f is ﬁnite on each component of Ω. Proof: It can be assumed that Ω is connected. Since H fj ≤ H f , there is nothing to prove if limj→∞ H fj = +∞ on Ω. Therefore, it can be assumed that limj→∞ H fj (x0 ) < +∞ for some x0 ∈ Ω. Then −∞ < H f1 (x0 ) ≤ H fj (x0 ) ≤ limj→∞ H fj (x0 ) < +∞ for all j ≥ 1 and each of the functions H fj is harmonic by Lemma 2.6.7. Since the limit of an increasing sequence of harmonic functions is either harmonic or identically +∞ and limj→∞ H fj (x0 ) < +∞, limj→∞ H fj is harmonic on Ω. Suppose > 0. For each j ≥ 1, choose vj ∈ Ufj such that vj (x0 ) < H fj (x0 ) + 2−j , and deﬁne v = lim H fi + i→∞

∞

(vj − H fj ).

j=1

For each j ≥ 1, vj − H fj is superharmonic on Ω, nonnegative, and less than 2−j at x0 . From the inequality v ≥ lim H fi + (vj − H fj ) = ( lim H fi − H fj ) + vj i→∞

i→∞

and the fact that limi→∞ H fi ≥ H fj , v ≥ vj for each j ≥ 1. Since vj is bounded below, the same is true of v. By Theorem 2.4.8, v is superharmonic on Ω. Moreover, for each x ∈ ∂Ω and each j ≥ 1, lim inf v(y) ≥ lim inf vj (y) ≥ fj (x).

y→x,y∈Ω

y→x,y∈Ω

Therefore, lim inf y→x,y∈Ω v(y) ≥ f (x) for all x ∈ ∂Ω and consequently v ∈ Uf . This means that v(x0 ) ≥ H f (x0 ). Since lim H fj (x0 ) ≤ H f (x0 ) ≤ v(x0 ) ≤ lim H fj (x0 ) +

j→∞

j→∞

and is arbitrary, H f (x0 ) = limj→∞ H fj (x0 ). Since the latter quantity is ﬁnite at x0 by assumption, H f (x0 ) is ﬁnite and H f is therefore harmonic. Now H f −limj→∞ H fj is a nonnegative harmonic function that vanishes at x0 and therefore must be the constant zero function by the minimum principle. This proves the ﬁrst assertion that H f = limj→∞ H fj . Suppose now that the fj are resolutive. Then H fj = H fj and H f = lim H fj = lim Hfj = lim H fj ≤ H f ; j→∞

j→∞

j→∞

2.6 Perron-Wiener Method

81

it follows that H f = H f and that f is resolutive if at least one of H f and H f is ﬁnite on each component of Ω. Lemma 2.6.14 If u is a bounded superharmonic function on the bounded open set Ω such that f (x) = limy→x,y∈Ω u(y) exists for all x ∈ ∂Ω, then f is a resolutive boundary function. Proof: Note that f, H f , and H f are all bounded and that the latter two are harmonic functions. Clearly, u ∈ Uf and u ≥ H f . Then lim supy→x,y∈Ω H f ≤ lim supy→x,y∈Ω u(y) = f (x) for all x ∈ ∂Ω and therefore H f ∈ Lf . It follows that H f ≤ H f ; since the opposite inequality always holds, the two are equal and f is resolutive. Lemma 2.6.15 Let K be a compact subset of Rn , and let f be a continuous function on K. Given > 0 and a ball B ⊃ K, there is a function u that is the diﬀerence of two continuous superharmonic functions deﬁned on B such that supx∈K |u(x) − f (x)| < . Proof: By the Stone-Weierstrass theorem, there is a polynomial function u of the n coordinate variables such that supx∈K |f (x) − u(x)| < . Let B be any ball containing K, and let v(y) = −|y|2 . Then Δv = −2n, v is superharmonic by Lemma 2.3.4, and λv is a continuous superharmonic function on B for λ ≥ 0. Choose λ0 such that Δ(u + λ0 v) ≤ 0 on B. Then u = (u + λ0 v) − λ0 v with u + λ0 v and λ0 v continuous superharmonic functions on B. The existence of a harmonic function associated with each continuous boundary function will be established now. Theorem 2.6.16 (Wiener) If f is a continuous function on the boundary ∂Ω of a bounded open set Ω, then f is resolutive. Proof: Since ∂Ω is compact, the preceding lemma is applicable. Let B be a ball containing ∂Ω. Then there is a function of the form u = v − w, where v and w are continuous superharmonic functions on B, which approximates f uniformly on ∂Ω. Since v|∂Ω and w|∂Ω are resolutive by Lemma 2.6.14, u|∂Ω = v|∂Ω − w|∂Ω is resolutive. Since u|∂Ω approximates f uniformly on ∂Ω, f is resolutive by Lemma 2.6.12. Although Wiener’s theorem associates a harmonic function with each continuous boundary function, it leaves open the connection between the boundary behavior of the harmonic function and the boundary function. Deﬁnition 2.6.17 A point x ∈ ∂Ω is a regular boundary point if lim

y→x,y∈Ω

Hf (y) = f (x)

for all f ∈ C(∂Ω). Otherwise, x is an irregular boundary point. Ω is a regular region or Dirichlet region if every point of ∂Ω is a regular boundary point.

82

2 The Dirichlet Problem

Example 2.6.18 Let Ω = B0,1 ∼ {0} ⊂ R2 . Then 0 is an irregular boundary point. This follows from the fact that Hf = 0 on Ω for the continuous boundary function 1 if x = 0 f (x) = 0 if |x| = 1. In this case, limy→0,y∈Ω Hf (y) = 0 = 1 = f (0). The irregular boundary point of this example is an isolated point of the boundary. It will be seen later that this is true of all isolated boundary points. Necessary and suﬃcient conditions for a boundary point to be a regular boundary point will be taken up now. Deﬁnition 2.6.19 A function w is a local barrier at x ∈ ∂Ω if w is deﬁned on Λ ∩ Ω for some neighborhood Λ of x and (i) w is superharmonic on Λ ∩ Ω, (ii) w > 0 on Λ ∩ Ω, and (iii) limy→x,y∈Λ∩Ω w(y) = 0. The function w is a barrier at x ∈ ∂Ω on Ω if (i), (ii), and (iii) are satisﬁed with Λ = Ω, and a strong barrier at x ∈ ∂Ω if, in addition, inf {w(z); z ∈ Ω ∼ Λ} > 0 for each neighborhood Λ of x. Theorem 2.6.20 (Bouligand) Let Ω be a bounded open subset of Rn . If there is a local barrier at x ∈ ∂Ω, then there is a harmonic strong barrier h at x on Ω satisfying for all x ∈ ∂Ω ∼ {x}, and (i) lim inf y→x ,y∈Ω h(y) > 0 (ii) inf {h(y); y ∈ Ω ∼ Λx } > 0 for each neighborhood Λx of x. Proof: Let w be a local barrier at x ∈ ∂Ω. Deﬁne a function m on ∂Ω by putting m(y) = |y − x|, y ∈ ∂Ω. It will be shown that the function h = Hm has the required properties. Since the function |y − x| on Ω is subharmonic by Lemma 2.3.4 and belongs to the lower class Lm , |y − x| ≤ Hm (y) on Ω. Moreover, lim inf Hm (y) ≥ lim inf |y − x| = |x − x| > 0

y→x ,y∈Ω

y→x ,y∈Ω

for all x ∈ ∂Ω, x = x. It remains only to show that limy→x,y∈Ω Hm (y) = 0. Since w is a local barrier at x, there is a ball Bx,r such that w > 0 on Bx,r ∩Ω, superharmonic on Bx,r ∩Ω and limy→x,y∈Bx,r ∩Ω w(y) = 0. Consider any ρ for which 0 < ρ < r and let B = Bx,ρ . Two cases will be considered depending upon whether or not ∂B ∩ Ω is empty. Assume ﬁrst that ∂B ∩ Ω = ∅, in which case ∂(B ∩ Ω) ⊂ B − ∩ ∂Ω. Since m ≥ 0 on ∂Ω, it can be assumed that Hm = sup {u; u ∈ Lm , u ≥ 0}. For any such u and y ∈ ∂(B ∩ Ω) ⊂ B − ∩ ∂Ω, lim sup u(z) ≤ lim sup u(z) ≤ m(y) ≤ ρ.

z→y,z∈B∩Ω

z→y,z∈Ω

By Corollary 2.3.6, u ≤ ρ on B ∩ Ω, and since this is true of all such u ∈ Lm , Hm ≤ ρ on B ∩ Ω. Now assume that ∂B ∩ Ω = ∅, and choose

2.6 Perron-Wiener Method

83

a closed set F ⊂ ∂B ∩ Ω such that σ((∂B ∩ Ω) ∼ F ) < ρ/M where M = sup {|y − x|; y ∈ ∂Ω}. Note that m ≤ M on ∂Ω and let k = inf {w(y); y ∈ F } > 0. Deﬁne a function f on ∂B by putting M if y ∈ (∂B ∩ Ω) ∼ F f (y) = 0 otherwise . For u ∈ Lm , u ≥ 0, consider the function v = u − ρ − (M/k)w − PI(f : B) which is subharmonic on B ∩ Ω. It will be shown that lim supz→y,z∈B∩Ω v(z) ≤ 0 for all y ∈ ∂(B ∩ Ω). Note ﬁrst the following facts pertaining to such y: (i)

Since w is l.s.c. at points of F , −

lim inf

z→y,z∈B∩Ω

w(z) ≤ −k

for y ∈ F ;

otherwise, −

lim inf

z→y,z∈B∩Ω

w(z) ≤ 0.

(ii) Since m ≤ M and lim supz→y,z∈B∩Ω u(z) ≤ M for y ∈ ∂Ω, u ≤ M on Ω by Corollary 2.3.6 and lim sup u(z) ≤ M

for y ∈ ∂B ∩ Ω;

z→y,z∈B∩Ω

also, lim sup u(z) ≤ m(y) ≤ ρ

for y ∈ B ∩ ∂Ω.

z→y,z∈B∩Ω

(iii) On the set (∂B ∩ Ω) ∼ F, f is a continuous function and lim

z→y,z∈B∩Ω

PI(f : B) = f (y) = M

for y ∈ (∂B ∩ Ω) ∼ F ;

otherwise, −

lim inf

z→y,z∈B∩Ω

PI(f : B) ≤ 0.

Note also that ∂(B ∩ Ω) ⊂ ((∂B ∩ Ω) ∼ F ) ∪ F ∪ (∂Ω ∩ B − ). By considering the three cases y ∈ ((∂B ∩ Ω) ∼ F ), y ∈ F , and y ∈ (∂Ω ∩ B − ), it is easily seen that lim supz→y,z∈B∩Ω v(z) ≤ 0 for y ∈ ∂(B ∩ Ω) so that v ≤ 0 on B ∩ Ω by Corollary 2.3.6. Thus, u ≤ ρ + (M/k)w + PI(f : B)

on B ∩ Ω.

But since u is an arbitrary element of Lm , Hm ≤ ρ + (M/k)w + PI(f : B)

on B ∩ Ω.

84

2 The Dirichlet Problem

Since the average value of f is ρ/M and PI(f : B)(x) = L(f : x, B) ≤ M (ρ/M ) = ρ and PI(f : B) is continuous at x, lim supy→x,y∈B∩Ω PI(f : B)(y) ≤ ρ. Since limy→x,y∈Ω w(y) = 0, 0 ≤ lim sup Hm (y) = y→x,y∈Ω

lim sup Hm (y) ≤ 2ρ

y→x,y∈B∩Ω

in the ∂B ∩ Ω = ∅ case. In either case, lim supy→x,y∈Ω Hm (y) = 0 and (i) is satisﬁed. The function h = Hm clearly satisﬁes (ii). Simply put, if there is a local barrier at x ∈ ∂Ω for the bounded open set Ω, then the function H|y−x| is a strong barrier at x. It should also be noted that the existence of a barrier at a boundary point is a local property of Ω; that is, the existence of a barrier at x ∈ Ω depends only upon the relationship between x and the part of Ω near x. Since the assumption that there is a local barrier at a boundary point can be replaced by the assumption of a barrier or a strong barrier, the shorter term will be used in the statement of theorems. Lemma 2.6.21 Let Ω be a bounded open set. If f is bounded above on ∂Ω and there is a barrier at x ∈ ∂Ω, then lim sup H f (y) ≤ lim sup f (y);

y→x,y∈Ω

y→x,y∈∂Ω

or if f is bounded below on ∂Ω and there is a barrier at x ∈ ∂Ω, then lim inf H f (y) ≥ lim inf f (y).

y→x,y∈Ω

y→x,y∈∂Ω

Proof: By the preceding theorem, there is a barrier w at x deﬁned on Ω such that inf {w(y); y ∈ Ω ∼ Λx } > 0 for each neighborhood Λx of x. If L = lim supy→x,y∈∂Ω f (y) and > 0, then there is a neighborhood V of x such that f (y) < L + for all y ∈ ∂Ω ∩ V . Choose c > 0 such that

L+ +c inf w(y) > sup f (y) y∈Ω∼V

y∈∂Ω

and let u = L+ +cw. Then u is superharmonic on Ω and bounded below since w > 0 on Ω. Moreover, lim inf y→x ,y∈Ω u(y) ≥ L + + c lim inf y→x ,y∈Ω w(y) for x ∈ ∂Ω. If x ∈ ∂Ω ∼ V , then lim inf y→x ,y∈Ω w(y) ≥ inf y∈Ω∼V w(y) and

inf w(y) > sup f (y) ≥ f (x ). lim inf u(y) ≥ L + + c y→x ,y∈Ω

y∈Ω∼V

y∈∂Ω

On the other hand, if x ∈ ∂Ω ∩ V , then L + > f (x ) and lim inf u(y) ≥ L + > f (x ).

y→x ,y∈Ω

2.6 Perron-Wiener Method

85

This shows that lim inf y→x ,y∈Ω u(y) ≥ f (x ) for all x ∈ ∂Ω and that u ∈ Uf . Therefore, H f ≤ u on Ω and lim sup H f (y) ≤ lim sup u(y) ≤ L + + c lim sup w(y) = L + . y→x,y∈Ω

y→x,y∈Ω

y→x,y∈Ω

Since is arbitrary, lim supy→x,y∈Ω H f (y) ≤ L = lim supy→x,y∈∂Ω f (y). Corollary 2.6.22 Let Ω be a bounded open set with a barrier at x ∈ ∂Ω. If f is bounded on ∂Ω and continuous at x, then lim

y→x,y∈Ω

H f (y) =

lim

y→x,y∈Ω

H f (y) = f (x).

Proof: By the preceding lemma, continuity of f at x, and the fact that Hf ≤ Hf , lim sup H f (y) ≤ f (x) ≤ lim inf H f (y) ≤ lim inf H f (y).

y→x,y∈Ω

y→x,y∈Ω

y→x,y∈Ω

Thus, limy→x,y∈Ω H f (y) = f (x). Similarly, limy→x,y∈Ω H f (y) = f (x). Theorem 2.6.23 Let Ω be a bounded open set. A point x ∈ ∂Ω is a regular boundary point if and only if there is a barrier at x. Proof: The suﬃciency is immediate from the deﬁnition of a regular boundary point, the preceding corollary and Theorem 2.6.16. As to the necessity, assume that x is a regular boundary point, and let m(y) = |y − x|, y ∈ ∂Ω. As in the proof of Theorem 2.6.20, Hm (y) ≥ |y − x| > 0, y ∈ Ω. Since x is a regular boundary point and m is continuous on ∂Ω, limy→x,y∈Ω Hm (y) = m(x) = 0. This shows that Hm is a barrier at x. Theorem 2.6.24 If x is a regular boundary point for the bounded open set Ω and Ω0 is a subset of Ω for which x ∈ ∂Ω0 , then x is regular for Ω0 . Proof: A barrier at x relative to Ω is a barrier at x relative to Ω0 when restricted to the latter. Theorem 2.6.25 (Poincar´ e) If Ω is a bounded open set, if x ∈ ∂Ω, and there is a ball B with B ∩ Ω = ∅ and x ∈ ∂B, then x is a regular boundary point for Ω. Proof: By taking a ball internally tangent to B at x, it can be assumed that ∂B ∩ ∂Ω = {x}. If B = By,ρ , a constant c can be added to the fundamental harmonic function uy having pole y so that uy + c vanishes on ∂B. The function w = uy + c, restricted to Ω, is then a barrier at x for Ω. It follows that x is regular by Theorem 2.6.23.

86

2 The Dirichlet Problem

Example 2.6.26 The boundary points of a disk, square, and annulus in R2 are regular boundary points. The boundary points of a bounded convex subset of Rn are regular boundary points. In the one example of an irregular boundary point that has been considered, namely the point 0 of the boundary of Ω = B0,1 ∼ {0} ⊂ R2 , the region Ω is not simply connected. Generally speaking, irregular boundary points are more diﬃcult to illustrate in R2 than in Rn , n ≥ 3. Theorem 2.6.27 If x is a boundary point of the bounded open set Ω ⊂ R2 having the property that there is a line segment I ⊂∼ Ω with x as an end point, then x is a regular boundary point. Proof: Points of R2 will be regarded as complex numbers. Let B = Bx,δ be a ball such that I ∩ ∂B = ∅ and δ < 1. The function log (z − x) can be deﬁned to be continuous on B ∼ I by taking the branch cut along the ray starting at x and in the direction I. Letting

1 log |z − x| , =− w(z) = −Re log (z − x) | log (z − x)|2 w is harmonic on Ω ∩ B since it is the real part of a function that is analytic on Ω ∩ B. Moreover, w > 0 on Ω ∩ B since |z − x| < 1 on Ω ∩ B. It is easily seen that w is a barrier at x. More is true in the n = 2 case than is stated in this theorem. Recall that a continuum is a set containing more than one point which is closed and connected. For the sake of completeness, the following theorem is included here but the proof is deferred until Section 5.7. Theorem 2.6.28 (Lebesgue) If x is a boundary point of the bounded open set Ω ⊂ R2 having the property that there is a continuum ⊂∼ Ω with x ∈ , then x is a regular boundary point. The phrase “right circular cone” in the statement of the following theorem excludes a cone with zero half-angle. Theorem 2.6.29 (Zaremba) If x is a boundary point of the bounded open set Ω ⊂ Rn , n ≥ 2, and there is a truncated right circular cone in ∼ Ω with vertex at x, then x is a regular boundary point. Proof: By reducing the half-angle of the cone, it can be assumed that the truncated cone intersects ∂Ω only in {x}. Let Cx be a closed right circular cone with vertex x such that Cx ∩ Bx,ρ ∩ Ω = ∅ for some ρ > 0, and let Ω0 = Ω ∪ (Bx,ρ ∼ Cx ). Then x is a boundary point for Ω0 . If it can be shown that x is a regular boundary point for Ω0 , then it would follow from Theorem 2.6.24 that x is a regular boundary point for Ω. Since a barrier at x on Bx,ρ ∼ Cx is a local barrier at x on Ω ∩ (Bx,ρ ∼ Cx ), it is enough to

2.6 Perron-Wiener Method

87

show that there is a barrier at x on Bx,ρ ∼ Cx ; that is, there is a positive superharmonic function w on Bx,ρ ∼ Cx for which limy→x,y∈Bx,ρ ∼Cx w(y) = 0. In other words, it suﬃces to prove that x is a regular boundary point for Bx,ρ ∼ Cx . Henceforth, it will be assumed that Ω = Bx,ρ ∼ Cx . According to Theorem 2.6.25, all boundary points of Ω, except possibly for x, are regular boundary points. Let Ω2 be a magniﬁcation of Ω by a factor of 2; that is, Ω2 = {x + 2(y − x) : y ∈ Ω}. Also let r(y) = |y − x|, y ∈ ∂Ω2 . Consider the Dirichlet solution Hr on Ω2 corresponding to the boundary function r on ∂Ω2 . Since Hr is harmonic on Ω2 and, as in the proof of Theorem 2.6.20, Hr (z) ≥ |z − x| > 0 for z ∈ Ω2 , it remains only to show that limz→x,z∈ Ω Hr (z) = 0. Deﬁne a function u on Ω2− ∼ {x} by putting Hr (y) y ∈ Ω2 u(y) = r(y) y ∈ ∂Ω2 ∼ {x} and a second function v on Ω − ∼ {x} by putting v(y) = u(x + 2(y − x)). The functions u and v are continuous on Ω2− ∼ {x} and Ω − ∼ {x}, respectively. It will be shown next that u < v on ∂Ω ∩∂Bx,ρ . Since r ≤ 2ρ on ∂Ω2 , 2ρ ∈ Ur so that u = Hr ≤ 2ρ on Ω2 . Since u is harmonic on Ω2 , it satisﬁes the maximum principle on Ω2 by Corollary 1.5.10, and since it is clearly not constant on Ω2 , it cannot attain its maximum value at a point of Ω2 . Thus, u < 2ρ on Ω2 and on Ω2 ∩ ∂Ω ∩ ∂Bx,ρ in particular. But since v = 2ρ on Ω2 ∩ ∂Ω ∩ ∂Bx,ρ , u < v on the same set. Since points y ∈ ∂Ω2 ∩ ∂Ω are regular boundary points for Ω2 , limz→y,z∈∂Ω∩Ω2 u(z) = ρ; for such points, u(y) = ρ < 2ρ = v(y). Thus, u < v on ∂Ω ∩ ∂Bx,ρ . Since both functions are continuous on the compact set ∂Ω ∩ ∂Bx,ρ , there is a constant α with 1/2 < α < 1 such that u < αv on ∂Ω ∩ ∂Bx,ρ . On the remainder of ∂Ω ∼ {x}, u = (1/2)v. Therefore, u < αv on ∂Ω ∼ {x}. Since u and v are continuous on Ω − ∼ {x}, limz→y,z∈Ω (αv(z) − u(z)) ≥ 0 for y ∈ ∂Ω ∼ {x}. Consider the function αv − u + (ux + c) where > 0 and c has been chosen so that ux + c ≥ 0 on Ω − . Then 0 if y ∈ ∂Ω ∼ {x} lim inf (αv − u + (ux + c)) ≥ +∞ if y = x. z→y,z∈Ω − By Theorem 2.3.6, αv − u + (ux + c) ≥ 0 on Ω. For any closed ball By,δ ⊂ Ω,

αA(v : y, δ) − A(u : y, δ) + (A(ux : y, δ) + c) ≥ 0. Fixing δ > 0 and letting → 0, and then letting δ → 0, αv(y) − u(y) ≥ 0 for arbitrary y ∈ Ω. Therefore, lim sup u(z) ≤ α lim sup v(z) = α lim sup u(x + 2(z − x)). z→x,z∈Ω

z→x,z∈Ω

z→x,z∈Ω

88

2 The Dirichlet Problem

Since the two extreme limits are the same, lim sup Hr (z) = lim sup u(z) = 0. z→x,z∈Ω

z→x,z∈Ω

Remark 2.6.30 It is worth noting an interesting fact in the preceding proof. Namely, the fact that limz→y,z∈Ω (αv(z) − u(z)) ≥ 0 for all y ∈ ∂Ω except possibly for a single point was enough to prove that αv − u ≥ 0 on Ω, as though a single point is negligible. The proof was accomplished by exploiting the fact that ux = +∞ at x. The same proof would have worked for more than one point if a superharmonic function u could be found that takes on the value +∞ on the set of points where limz→y,z∈Ω (αv(z) − u(z)) ≥ 0 fails to hold. This suggests that sets of points where a superharmonic function takes on the value +∞ are negligible. Such sets are called polar sets and will be taken up later. Corollary 2.6.31 If Ω is an open subset of Rn , n ≥ 2, then there is an increasing sequence {Ωj } of regular bounded open sets with closures in Ω such that Ω = ∪Ωj . Proof: Let {Γj } be an increasing sequence of compact sets such that Ω = (1) (1) ∪Γj . Consider Γ1 and a ﬁnite covering of Γ1 by balls B1 , . . . , Bk1 having closures in Ω. By slightly increasing the radii of some of the balls, it can (1) 1 be assumed that each boundary point of Ω1 = ∪kj=1 Bj is the vertex of a truncated closed solid cone lying outside Ω1 ; that is, it can be assumed that Ω1 is a regular region containing Γ1 . Apply the same procedure to Ω1− ∪ Γ2 , etc.

2.7 The Radial Limit Theorem If h = PI(μ : B) on the ball B and μ is the indeﬁnite integral of a continuous boundary function f relative to surface area σ on ∂B, then limz→x,z∈B h(z) = f (x) for all x ∈ ∂B. Since the Radon-Nikodym derivative of μ with respect to σ is equal to f a.e.(σ), the boundary behavior of h is related to the derivative dμ/dσ. Since there are several ways of deﬁning a derivate, diﬀerent boundary limit theorems may be obtained depending upon the deﬁnition of the derivate used. Generally speaking, a deﬁnition that places stringent conditions on μ for the existence of the derivate will yield the strongest results. In this section, a weak derivate will be used ﬁrst to describe the behavior of h along radial approaches to the boundary of B. As a matter of notational convenience, let B = B0,1 . It is not essential the 0 is the center of the ball or that the radius is equal to 1. If z ∈ ∂B and 0 ≤ α ≤ π/2, Cz,α will denote a closed cone with vertex 0, axis coincident with the vector z, and half-angle α; that is,

2.7 The Radial Limit Theorem

Cz,α = {y; cos α ≤

89

(y, z) ≤ 1}. |y||z|

Each such cone determines a closed polar cap Kz,α = Cz,α ∩∂B having center z and spherical angle α. Let μ be a signed measure of bounded variation on the Borel subsets of ∂B. Deﬁnition 2.7.1 If limσ(K)→0 μ(K)/σ(K) exists, where K is any closed polar cap centered at x, the limit is called the symmetric derivate of μ with respect to σ at x and is denoted by Ds μ(x). The symmetric derivate is a weak derivate and results in the following weak boundary limit theorem. Although the theorem is stated for harmonic functions, versions of the theorem are true for superharmonic functions. Littlewood [41] proved an a.e. radial limit theorem for potentials GB μ in the n = 2 case, F. Riesz [53] extended Littlewood’s theorem to positive superharmonic functions in the n = 2 case, and Privalov [51, 52] extended the latter to the n ≥ 3 case. Theorem 2.7.2 Let μ be a signed measure of bounded variation on the Borel subsets of ∂B and let h = PI(μ : B). If Ds μ exists at x ∈ ∂B, then limλ→1− h(λx) = Ds μ(x). Proof: (B = B0,1 ) It will be shown ﬁrst that a = Ds μ(x), x ﬁxed, can be assumed to be zero. Consider the signed measure μ − aσ which is also of bounded variation. For this measure, Ds (μ − aσ)(x) = Ds μ(x) − aDs σ(x) = 0 and PI(μ− aσ : B) = PI(μ : B)− aPI(σ : B) = h− a. If it can be shown that limλ→1− (h(λx) − a) = 0, it would follow that limλ→1− h(λx) = a. Therefore, by replacing μ by μ − aσ, it can be assumed that a = 0. For x ∈ ∂B, 1 − |λx|2 1 h(λx) = dμ(z) σn ∂B |z − λx|n 1 1 − λ2 = dμ(z). σn ∂B |z − λx|n If γ denotes the angle between λx and the point z ∈ ∂B, then 1 1 − λ2 h(λx) = dμ(z) σn ∂B (1 + λ2 − 2λ cos γ)n/2 1 1 − λ2 = dμ(z). σn ∂B ((1 − λ)2 + 4λ sin2 (γ/2))n/2 With λ ﬁxed, the integrand depends only upon γ and a “zonal region” method can be used to evaluate the integral. Letting α(γ) = μ(Kx,γ ), 0 ≤ γ ≤ π, α is of bounded variation on [0, π] and

90

2 The Dirichlet Problem

π

h(λx) =

P (λ, γ) dα(γ) 0

where P (λ, γ) =

1 − λ2 1 , σn ((1 − λ)2 + 4λ sin2 (γ/2))n/2

0 < λ < 1, 0 ≤ γ ≤ π.

Since limγ→0 μ(Kx,γ )/σ(Kx,γ ) = 0, α(0) = μ(Kx,0 ) = 0. For each λ ∈ (0, 1), P (λ, ·) is positive, monotone decreasing on [0, π], and has a continuous derivative Pγ (λ, ·) on [0, π]. Consider also the function β(γ) = σ(Kx,γ ), 0 ≤ γ ≤ π, for which μ(Kx,γ |α(γ)| = |Ds μ(x)| = 0. = lim lim γ→0 β(γ) γ→0 σ(Kx,γ ) Expressing the surface area of Kx,γ as an integral and transforming to spherical coordinates (r, θ), β(γ) = σ(Kx,γ ) = σn−1

sin γ

0

rn−2 σn−1 sinn−1 γ dr ≥ 2 1/2 n−1 (1 − r )

and so γ n−1 n−1 ≤ lim sup lim sup β(γ) σn−1 γ→0 γ→0 Choose > 0 such that lim sup γ→0

γ sin γ

n−1

n−1 < +∞. σn−1

=

1 γ n−1 < β(γ)

and then choose φ ∈ (0, π) such that |α(γ)| ≤ β(γ) and γ n−1 ≤ β(γ) for all γ ∈ [0, φ]. Integrating by parts and noting that Pγ (λ, γ) ≤ 0, 0

φ

P (λ, γ) dα(γ) = α(φ)P (λ, φ) −

φ

≤ α(φ)P (λ, φ) −

α(γ)Pγ (λ, γ) dγ

0 φ

|α(γ)|Pγ (λ, γ) dγ

0

≤ α(φ)P (λ, φ) −

φ

0

β(γ)Pγ (λ, γ) dγ

= P (λ, φ)(α(φ) − β(φ)) +

φ

P (λ, γ) dβ(γ) 0

≤ P (λ, φ)(|α(φ)| − β(φ)) + ≤

φ

P (λ, γ) dβ(γ). 0

φ

P (λ, γ) dβ(γ) 0

2.7 The Radial Limit Theorem

91

Since

1 = PI(σ : B)(λx) =

π

P (λ, γ) dσ(z) =

P (λ, γ) dβ(γ), 0

∂B

integrating by parts 1≥

φ

P (λ, γ) dβ(γ) 0

= β(φ)P (λ, φ) −

φ

β(γ)Pγ (λ, γ) dγ

0

≥ β(φ)P (λ, φ) −

0

φ

γ n−1 Pγ (λ, γ) dγ

= P (λ, φ)(β(φ) − φn−1 ) + (n − 1) ≥

φ

P (λ, γ)γ n−2 dγ 0

φ

P (λ, γ)γ n−2 dγ. 0

Hence,

φ

π

P (λ, γ) dα(γ) +

h(λx) = 0

P (λ, γ) dα(γ) φ

φ

π P (λ, γ) β(γ) + φ P (λ, γ) dα(γ) ≤ φ 0 P (λ, γ) dβ(γ) π P (λ, γ) dα(γ) φ . ≤ + φ 0 P (λ, γ)γ n−2 dγ

0

Letting Vα denote the total variation of α, π

φ P (λ, γ) dVα (γ) |h(λx)| ≤ + φ 0 P (λ, γ)γ n−2 dγ

π P (λ, φ) φ dVα (γ) ≤ + φ 0 P (λ, γ)γ n−2 dγ ≤ + φ 0

|μ| P (λ,γ)γ n−2 P (λ,φ)

. dγ

92

2 The Dirichlet Problem

By Fatou’s Lemma, applied sequentially, lim inf λ→1−

0

φ

P (λ, γ)γ n−2 dγ ≥ P (λ, φ)

φ

lim inf λ→1−

0

P (λ, γ)γ n−2 dγ P (λ, φ)

φ

sinn (φ/2)

= 0

γ n−2 dγ sinn (γ/2)

= +∞, and therefore lim supλ→1− |h(λx)| ≤ . Since can be taken arbitrarily small, limλ→1− h(λx) = 0. Example 2.7.3 Consider B = B0,1 and the function 1 (x, y) ∈ ∂B, y ≥ 0 f (x, y) = 0 (x, y) ∈ ∂B, y < 0. Let μ(M ) = M f dσ for each Borel set M ⊂ ∂B. Then Ds μ exists for (x, y) ∈ ∂B and is equal to 1, 0, or 1/2 according to y > 0, y < 0, or y = 0, respectively. Consider the harmonic function h = PI(μ : B). Then limλ→1− h(λx, λy) = 1, 0, or 1/2 for (x, y) ∈ ∂B according to y > 0, y < 0, or y = 0, respectively.

2.8 Nontangential Boundary Limit Theorem Obtaining stronger results concerning the boundary behavior of harmonic functions necessitates replacement of the symmetric derivate by a more restrictive derviate. Deﬁnition 2.8.1 If limσ(K)→0 μ(K)/σ(K) exists, where K is any closed polar cap containing x, the limit is called the derivate of μ with respect to σ at x and is denoted by Dμ(x). Proof of the existence of a derivate requires a discussion of the Vitali covering theorem. Let X be a metric space with metric d. If F ⊂ X, the diameter of F is deﬁned to be d(F ) = sup {d(x, y); x, y ∈ F }. If F ⊂ X and x ∈ X, the distance from F to x is denoted by d(F, x) = inf {d(y, x); y ∈ F }. An -neighborhood of F ⊂ X is denoted by N (F, ) = {x; d(F, x) < }. Deﬁnition 2.8.2 Let μ be a measure deﬁned on the Borel subsets of a compact metric space X. A set A ⊂ X is said to be covered in the sense of Vitali by a family F of closed sets if each F ∈ F has positive μ measure and there is a constant α such that each point of A is contained in elements F ∈ F of arbitrarily small positive diameter for which μ(N (F, 3d(F )))/μ(F ) ≤ α. Note that if A is covered in the sense of Vitali by F , then any subset of A is also. A proof of the following theorem can be found in [18].

2.8 Nontangential Boundary Limit Theorem

93

Theorem 2.8.3 (Vitali Covering Theorem) Let μ be the completion of a ﬁnite measure on the σ-algebra of Borel subsets of a compact metric space X. If A ⊂ X is covered in the sense of Vitali by a family F of closed sets, then there is a sequence {Fj } of disjoint sets in F such that μ(A ∼ ∪Fj ) = 0. The Vitali covering theorem will be applied in the following context. For X take ∂B where B = B0,ρ ⊂ Rn . The family F will consist of all nondegenerate closed polar caps K ⊂ ∂B. To show that F covers ∂B in the sense of Vitali, it is necessary to estimate the surface area of polar caps. Let K = Kx,α , 0 < α ≤ π/2, be a closed polar cap on ∂B. Making free use of the fact that surface area is invariant under rotations of the sphere by assuming that x = (0, 0. . . . , 0, ρ), ρ σ(K) = · · · dy1 . . . dyn−1 . 2 − y2 − · · · − y2 1/2 2 2 2 (ρ 2 y1 +···+yn−1 ≤ρ sin α 1 n−1 ) Introducing the spherical coordinates (r, θ) for (y1 , . . . , yn−1 ), σ(K) = ρσn−1

0

ρ sin α

(ρ2

rn−2 dr. − r2 )1/2

It is easy to see that d(K) = 2ρ sin α. There is also an α0 with 0 < α0 < π/2 such that for α ≤ α0 7ρ sin α rn−2 σ(N (K, 3d(K)) ≤ ρσn−1 dr 2 (ρ − r2 )1/2 0 ρ sin α rn−2 n−1 dr. = ρσn−1 7 (ρ2 − 49r2 )1/2 0 Therefore, ρ sin α n−2 2 7n−1 0 (r /(ρ − 49r2 )1/2 ) dr σ(N (K, 3d(K)) ≤ . ρ sin α σ(K) (rn−2 /(ρ2 − r2 )1/2 ) dr 0

Considering the integrals as functions of their upper limits and applying the second mean value theorem of the diﬀerential calculus, there is a point ξ with 0 < ξ < ρ sin α such that the quotient on the right is equal to 7n−1 (ξ n−2 /(ρ2 − 49ξ 2 )1/2 ) . ξ n−2 /(ρ2 − ξ 2 )1/2 Therefore,

σ(N (K, 3d(K))) (ρ2 − ξ 2 )1/2 ≤ 7n−1 2 . σ(K) (ρ − 49ξ 2 )1/2

94

2 The Dirichlet Problem

As α → 0, ξ → 0 and the right member has the limit 7n−1 . This shows that there is a constant α0 with 0 < α0 < π/2 such that σ(N (k, 3d(K))) ≤ constant σ(K) whenever K = Kx,α is a closed polar cap with spherical angle α < α0 . Therefore, the family F of closed polar caps covers ∂B in the sense of Vitali. Unless stated otherwise, K will denote a closed polar cap in ∂B where B = B0,ρ . Lemma 2.8.4 Let μ be a measure on the Borel subsets of ∂B, and let 0 < r < +∞. (i)

If for each x in a set A ⊂ ∂B lim inf

σ(K)→0,x∈K

μ(K) < r, σ(K)

then each neighborhood of A contains an open set O such that σ(A ∼ O) = 0 and μ(O) < rσ(O). (ii) If for each x in a set A ⊂ ∂B lim sup σ(K)→0,x∈K

μ(K) > r, σ(K)

then each neighborhood of A contains a Borel set A0 such that σ(A ∼ A0 ) = 0 and μ(A0 ) > rσ(A0 ). Proof: To prove (i), let U be any open set with A ⊂ U ⊂ ∂B, and let F be the family of all closed polar caps K ⊂ U satisfying μ(K) < rσ(K). Then F covers A in the sense of Vitali. By the Vitali covering theorem, there is a sequence {Kj } of disjoint closed polar caps in F such that σ(A ∼ ∪Kj ) = 0. Let Kj◦ be the interior, relative to ∂B, of Kj . Since σ(Kj ∼ Kj◦ ) = 0, σ(A ∼ Kj◦ ) = 0 and μ(Kj◦ ) ≤ μ(Kj ) < rσ(Kj ) = rσ(Kj◦ ). Letting O = Kj◦ , O is open, O ⊂ U , and σ(A ∼ O) = 0. Moreover, μ(O) =

∞ j=1

μ(Kj◦ ) < r

∞

σ(Kj◦ ) = rσ(O).

j=1

To prove (ii), let U be an open set such that A ⊂ U ⊂ ∂B. By hypothesis, the family F of closed polar caps K ⊂ U satisfying μ(K) > rσ(K) covers A in the sense of Vitali. By the Vitali covering theorem, there is a sequence of disjoint closed polar caps in F such that σ(A ∼ Kj ) = 0. Let {Kj } A0 = Kj . Since

2.8 Nontangential Boundary Limit Theorem

μ(A0 ) =

∞

μ(Kj ) > r

j=1

∞

95

σ(Kj ) = rσ(A0 ),

j=1

A0 satisﬁes the requirements of (ii). It is assumed throughout the remainder of this section that σ is a complete measure. Lemma 2.8.5 Let μ be a measure on the Borel subsets of ∂B such that Dμ(x) = limσ(K)→0,x∈K μ(K)/σ(K) exists for almost all (σ) points x ∈ ∂B, and deﬁne Dμ arbitrarily at points where the limit does not exist. Then Dμ is Lebesgue measurable on ∂B. Proof: For each positive integer m, let {km1 , . . . , kmjm } be a ﬁnite collection of closed polar caps which covers ∂B and σ(Kmj ) < 1/m, j = 1, . . . , jm . Deﬁne a sequence of functions {fm } on ∂B as follows. Let Km0 = ∅. If x ∈ Kmj ∼ ∪j−1 i=1 Kmi , j = 1, . . . , jm , let fm (x) = μ(Kmj )/σ(Kmj ). Then each fm is measurable and the sequence {fm } converges a.e.(σ) to Dμ. Theorem 2.8.6 (Lebesgue) Let μ be a signed measure of bounded variation on the Borel subsets of ∂B. Then Dμ(x) exists for almost all (σ) points x ∈ ∂B and is Lebesgue measurable. Proof: By decomposing μ into its positive and negative variations, it can be assumed that μ is a measure. Let A be the set of points x ∈ ∂B for which lim sup σ(K)→0,x∈K

μ(K) μ(K) > lim inf . σ(K) σ(K)→0,x∈K σ(K)

If i is a nonnegative integer and j is a positive integer, let Aij be the set of x ∈ ∂B where lim sup σ(K)→0,x∈K

i+1 i μ(K) μ(K) > > > lim inf . σ(K) j j σ(K)→0,x∈K σ(K)

∗ Suppose > 0 where σ ∗ is the outer measure on ∂B induced by σ. Since σ (A) ∗ A = Aij , σ (Aij ) > 0 for some i and j. Given > 0, there is an open set U such that Aij ⊂ U and σ(U ) < σ ∗ (Aij )+ . Applying (i) of Lemma 2.8.4 with r replaced by i/j, there is an open set O such that O ⊂ U, σ(Aij ∼ O) = 0, and i i μ(O) < σ(O) < (σ ∗ (Aij ) + ). j j

Applying (ii) of Lemma 2.8.4 to the set Aij ∩ O, there is a Borel set M ⊂ O such that σ((Aij ∩ O) ∼ M ) = 0 and μ(M ) > ((i + 1)/j)σ(M ). Since Aij ⊂ M ∪ (Aij ∼ O) ∪ ((Aij ∩ O) ∼ M )

96

2 The Dirichlet Problem

and σ(Aij ∼ O) = 0 = σ((Aij ∩ O) ∼ M ), ∗

σ(M ) ≥ σ (Aij ). Thus, i ∗ i+1 ∗ i+1 (σ (Aij ) + ) > μ(O) ≥ μ(M ) > σ(M ) ≥ σ (Aij ). j j j Letting → 0, (i/j)σ ∗ (Aij ) ≥ ((i + 1)/j)σ ∗ (Aij ), a contradiction. Therefore, σ(A) = 0 and Dμ(x) = limσ(K)→0,x∈K μ(K)/σ(K) exists as a ﬁnite or inﬁnite limit for all x ∈ A. Let Z = {x ∈∼ A; Dμ(x) = +∞}. By (ii) of Lemma 2.8.4, for each positive integer j there is a Borel set Zj ⊂ ∂B such that σ(Z ∼ Zj ) = 0 and μ(Zj ) > jσ(Zj ). Since σ ∗ (Z) ≤ σ ∗ (Z ∩ Zj ) + σ ∗ (Z ∼ Zj ) = σ ∗ (Z ∩ Zj ) ≤ σ ∗ (Zj ), +∞ > μ(∂B) ≥ μ(Zj ) > jσ(Zj ) ≥ jσ ∗ (Z) for all positive integers j. Therefore, σ ∗ (Z) = 0 and Dμ < +∞ a.e.(σ) on ∼ A. The measurability of Dμ follows from the preceding lemma. The proof of the principal boundary limit theorem requires the exclusion of null sets of boundary points in addition to those where Dμ is not deﬁned. Theorem 2.8.7 (Lebesgue) Let μ be a signed measure of bounded variation on the Borel subsets of ∂B. (i) If f is an integrable function on ∂B and μ(M ) = M f dμ for each Borel set M ⊂ ∂B, then Dμ = f a.e.(σ). (ii) If μ is singular relative to σ, then Dμ = 0 a.e.(σ). Proof: Suppose μ(M ) = M f dσ with f integrable. It can be assumed that f ≥ 0 by decomposing f into its positive and negative parts, if necessary. Let A = {x ∈ ∂B; f (x) > Dμ(x)} and assume that σ(A) > 0. There are then positive numbers α and β such that Aαβ = {x ∈ ∂B; f (x) > α > β > Dμ(x)} and σ(Aαβ ) > 0. Since σ is a regular Borel measure, there is a compact set Γ ⊂ Aαβ with σ(Γ ) > 0. Thus, f dσ ≥ ασ(Γ ). (2.8) μ(Γ ) = Γ

Since Dμ(x) = limσ(K)→0,x∈K μ(K)/σ(K) < β for all x ∈ Γ , each neighborhood of Γ contains an open set O such that μ(O) < βσ(O) and σ(Γ ∼ O) = 0 by part (i) of Lemma 2.8.4. Since σ(Γ ∼ O) = 0 and μ is the indeﬁnite integral of an integrable function, a small open set can be adjoined to O so that Γ ⊂ O and μ(O) < βσ(O). Since Γ is the intersection of a sequence of such open sets O, μ(Γ ) ≤ βσ(Γ ) < ασ(Γ ) ≤ μ(Γ ), the latter inequality holding by Inequality (2.8), a contradiction. Therefore, σ(A) = 0. In a similar manner, using part (ii) of Lemma 2.8.4, it can be shown that σ({x ∈ ∂B; f (x) < Dμ(x)}) = 0. Therefore, Dμ = f a.e.(σ). Suppose now

2.8 Nontangential Boundary Limit Theorem

97

that μ is singular relative to σ. Since μ = μ+ − μ− and the positive and negative variations of μ are singular relative to σ, it can be assumed that μ is a measure and Dμ ≥ 0. There is then a measurable set M ⊂ ∂B such that σ(M ) = 0 and μ(∂B ∼ M ) = 0. Let P = {x ∈ ∂B; Dμ(x) > 0} and assume that σ(P ) > 0. Since Dμ is measurable and σ(M ) = 0, there is an > 0 and a compact set Γ ⊂ P ∼ M such that σ(Γ ) > 0 and Dμ(x) > for x ∈ Γ . By (ii) of Lemma 2.8.4, each neighborhood Γ has μ measure greater than σ(Γ ). Since Γ is the intersection of such neighborhoods, μ(Γ ) ≥ σ(Γ ) > 0; but this contradicts the fact that 0 ≤ μ(Γ ) ≤ μ(P ∼ M ) ≤ μ(∂B ∼ M ) = 0. Therefore, σ(P ) = 0 and Dμ = 0 a.e.(σ). Points x ∈ ∂B satisfying the equation of the following theorem are called Lebesgue points. Theorem 2.8.8 (Lebesgue) Let f be a measurable function on ∂B which is integrable relative to σ, and let μ(M ) = M f dσ for each Borel set M ⊂ ∂B. Then 1 |f (y) − f (x)| dσ(y) = 0 lim σ(K)→0,x∈K σ(K) K a.e.(σ) on ∂B. Proof: Let {ri } be an enumeration of the rationals. By the preceding theorem, for each i ≥ 1 there is a set Zi ⊂ ∂B with σ(Zi ) = 0 such that for x ∈ Zi , 1 lim |f (y) − ri | dσ(y) = |f (x) − ri |. σ(K)→0.x∈K σ(K) K Then Z = Zi has surface area zero. Let x ∈ ∂B ∼ Z, let > 0, and choose rj such that |f (x) − rj | < /2. Then 1 |f (y) − f (x)| dσ(y) lim σ(K)→0,x∈K σ(K) K 1 ≤ lim sup (|f (y) − rj | + |rj − f (x)|) dσ(y) σ(K)→0,x∈K σ(K) K 1 ≤ lim sup |f (y) − rj | dσ(y) + 2 σ(K)→0,x∈K σ(K) K = |f (x) − rj | + < . 2 It is not diﬃcult to construct examples of harmonic functions h on ∂B ⊂ R2 that behave badly in a neighborhood of a point z ∈ ∂B. In fact, suppose h = PI(δz : B) on B = B0,ρ , where δz is a unit measure supported by the singleton {z}. For x ∈ B, h(x) =

1 ρ2 − |x|2 . 2πρ |z − x|2

98

2 The Dirichlet Problem

Suppose x = λz and λ ↑ 1. Then lim h(λz) = λ↑1

ρ2 − λ2 ρ2 1+λ 1 1 lim lim = +∞. = 2πρ λ↑1 (1 − λ)2 ρ2 2πρ λ↑1 1 − λ

Thus, if x ∈ B approaches z along a radial line to z, then h(x) → +∞. Suppose now that x → z in the following manner. Taking the radial line through z as a polar axis, let (r, θ) be the polar coordinates of x where r = |x| and θ is the angle between the x and z vectors. Now let x → z subject to the condition r = ρ cos θ. In this case, lim

x→z,|x|=ρ cos θ

h(x) =

1 ρ2 (1 − cos2 θ) 1 lim 2 = . 2πρ θ→0 ρ (1 − cos2 θ) 2πρ

The values +∞ and 1/2πρ are not the only possible limiting values; for example, if x → z subject to the condition r = ρ cos2 θ, then the limiting value is 1/πρ. In the latter two cases, x approaches z along a path that is internally tangent to ∂B. If anything is to be said about the limiting behavior of h at z, then such tangential approaches to z must be excluded. For the remainder of this section, B will denote a ﬁxed ball with center at 0 of radius ρ. Deﬁnition 2.8.9 (i) A Stolz domain with vertex z ∈ ∂B and half-angle θ, denoted by Sz,θ , is deﬁned to be an open circular cone with vertex z, with the radial line to z as axis, and half-angle θ < π/2. (ii) A function u on B is said to have nontangential limit at z ∈ ∂B, denoted by θ − limx→z u(x), if limx→z,x∈Sz,θ ∩B u(x) = for all Stolz domains Sz,θ . Lemma 2.8.10 Let z ∈ ∂B, Sz,θ a Stolz domain, 0 < δ < 1, and Bδz,r a ball with center δz which is internally tangent to ∂Sz,θ . Then there is a constant m, depending only upon θ, such that if x, x ∈ Bδz,r and h is any positive harmonic function on B, then h(x)/h(x ) ≤ m; the constant m is given by

n 1 + sin θ 1 . m= cos2 θ 1 − sin θ Proof: Consider any > 0 for that θ + < π/2 and a ball Bδz,r that is internally tangent to Cz,θ+ . The ball Bδz,r has radius r = (r/ sin θ) sin (θ + ). Applying Theorem 2.2.1 to Bδz,r ⊂ Bδz,r , h(x) (r/ sin θ)2 sin2 (θ + ) ≤ h(x ) (r/ sin θ)2 sin2 (θ + ) − r2

(r/ sin θ) sin (θ + ) + r (r/sinθ) sin (θ + ) − r

n

for all positive harmonic functions h on B and all x, x ∈ Bδz,r . The constant m is obtained by letting θ + → π/2.

2.8 Nontangential Boundary Limit Theorem

99

Theorem 2.8.11 (Fatou [20]) Let μ be a signed measure of bounded variation on the Borel subsets of the boundary of B, and let h = PI(μ : B) on B. Then θ − limx→z h(x) exists and is equal to Dμ(z) a.e.(σ). Proof: It will be shown ﬁrst that it suﬃces to prove that ν is a measure on ∂B, Dν(z) = 0 ⇒ θ − lim PI(ν : B)(x) = 0 a.e.(σ). (2.9) x→z

Assume that this is the case and consider the decomposition μ = μa + μs of μ relative to σ, where μa and μs are the absolutely continuous and singular − + − parts of μ, respectively. Since μs = μ+ s −μs , where μs and μs are the positive + − and negative parts of μs , respectively, Dμs = Dμs − Dμs = 0 a.e.(σ) by (ii) of Theorem 2.8.7 and it follows that θ − lim PI(μs : B)(x) = 0 a.e.(σ). x→z

(2.10)

Now consider μa which can be written μa (M ) =

f dσ M

for each Borel set M ⊂ ∂B and some integrable function f on ∂B. By (i) of Theorem 2.8.7, Dμa = f a.e.(σ). For each Borel set M ⊂ ∂B and z ∈ ∂B, (μa − f (z)σ)(M ) = (f (y) − f (z)) dσ(y) M

and |μa − f (z)σ|(M ) =

|f (y) − f (z)| dσ(y). M

By Theorem 2.8.8, D|μa − f (z)σ|(z) = 0 for almost all (σ) points z ∈ ∂B. Let Za be the exceptional set of σ-measure zero. For z ∈ Za , D|μa −f (z)σ|(z) = 0 and by (2.9) θ − lim PI(|μa − f (z)σ| : B)(x) = 0. x→z

Since |PI(μa − f (z)σ : B)(x)| ≤ PI(|μa − f (z)σ| : B)(x), θ − lim (PI(μa : B)(x) − f (z)) = 0 x→z

for z ∈ Za . Include in the set Za the null set {z ∈ ∂B; Dμa (z) = f (z)}. Then for z ∈ Za , θ − lim PI(μa : B)(x) = f (z) = Dμa (z). x→z

In other words, θ − limx→z PI(μa : B)(x) = Dμa (z) a.e.(σ). It follows from this result and Equation (2.10) that θ − lim h(x) = θ − lim PI(μ : B)(x) = θ − lim (PI(μa : B)(x) = Dμa = Dμ x→z

x→z

x→z

100

2 The Dirichlet Problem

for almost all (σ) points z ∈ ∂B. This shows that it suﬃces to prove (2.9). Suppose ν is a measure on ∂B and Dν(z) = 0. Letting h = PI(ν : B) on B, limλ→1− h(λz) = 0 by Theorem 2.7.2. If Sz,θ is a Stolz domain, by the preceding lemma there is a constant m, depending only upon θ, such that h(x)/h(x ) ≤ m whenever x is suﬃciently close to z and x is the projection of x onto the radial line through z. Then h(x) |h(x ) − h(x)| = h(x ) 1 − ≤ h(x )(1 + m). h(x ) Since h(x ) → 0 as x → z, h(x) → 0 as x → z in Sz,θ .

2.9 Harmonic Measure Wiener’s theorem settles the problem of showing that each continuous function on the boundary of a bounded open set Ω ⊂ Rn determines a harmonic function. The same problem can be considered for more general boundary functions. By Wiener’s theorem, Theorem 2.6.16, each f ∈ C 0 (∂Ω) and x ∈ Ω determines a real number Hf (x). Consider the map Lx : C 0 (∂Ω) → R deﬁned by Lx (f ) = Hf (x). Using Lemma 2.6.11, it is easily seen that Lx is positive linear functional on C 0 (∂Ω). It follows from the Riesz representation theorem that there is a unique unit measure μx on the Borel subsets of ∂Ω such that Hf (x) = Lx (f ) = f dμx for all x ∈ Ω, f ∈ C 0 (∂Ω). Showing that two measures are equal can be a tedious process involving several steps, the ﬁrst of which is to show that the two measures agree on certain elementary sets. The following digression into measure theory will simplify this process. Let F0 be a collection of subsets of ∂Ω that contains ∂Ω and is closed under ﬁnite intersections. A system G of subsets of ∂Ω is called a Dynkin system or d-system if (i) ∂Ω ∈ G, (ii) E, F ∈ G with E ⊂ F implies that F ∼ E ∈ G, and (iii) if {Ej } is an increasing sequence in G, then E = ∪Ej ∈ G. Let d(F0 ) denote the smallest d-system containing F0 ; that is, d(F0 ) is the intersection of all d-systems containing F0 and is itself a d-system. Then d(F0 ) ⊃ F0 . Also, let σ(F0 ) denote the smallest σ-algebra containing F0 . Since σ(F0 ) is a d-system containing F0 , d(F0 ) ⊂ σ(F0 ). It will be shown now that d(F0 ) = σ(F0 ). Since a d-system is closed under complementation, it is closed under unions if closed under intersections. It will be shown ﬁrst that d(F0 ) is closed under ﬁnite intersections. Let H1 = {E ∈ d(F0 ); E ∩ F ∈ d(F0 ) for all F ∈ F0 } ⊂ d(F0 ).

2.9 Harmonic Measure

101

Since F0 is closed under ﬁnite intersections, F0 ⊂ H1 . It is easily checked that H1 is a d-system, and therefore that d(F0 ) ⊂ H1 . Since H1 ⊂ d(F0 ), the two are equal. Now let H2 = {E ∈ d(F0 ); E ∩ F ∈ d(F0 ) for all F ∈ d(F0 )} ⊂ d(F0 ). Again, H2 is a d-system. If F ∈ F0 , then E ∩ F ∈ d(F0 ) for all E ∈ H1 = d(F0 ) by deﬁnition of H1 and consequently F0 ⊂ H2 . Therefore, d(F0 ) ⊂ H2 and the two are equal. But this shows that d(F0 ) is closed under ﬁnite intersections, and therefore under ﬁnite unions. Since d(F0 ) is closed under monotone increasing limits and ﬁnite unions, d(F0 ) is closed under countable unions; that is, d(F0 ) is a σ-algebra and d(F0 ) = σ(F0 ). The advantage of using d-systems will be apparent in the proof of the following theorem. Theorem 2.9.1 If f is a lower bounded (or upper bounded) Borel measurable function on ∂Ω, then H f (x) = H f (x) = f dμx for all x ∈ Ω; if, in addition, f is bounded, then f is resolutive and Hf (x) = f dμx for all x ∈ Ω. Proof: Let F0 be the collection of compact (=closed) subsets of ∂Ω. If C ∈ F0 , there is a sequence {fj } in C 0 (∂Ω) such that fj ↓ χC , the indicator function of C, with 0 ≤ fj ≤ 1 for all j ≥ 1. It follows from Lemma 2.6.13 that χC is a resolutive boundary function and that HχC (x) = χC dμx , x ∈ Ω. Let G be the collection of sets E ⊂ ∂Ω such that χE is resolutive and HχE (x) = χE dμx , x ∈ Ω. Then F0 ⊂ G. Note that ∂Ω ∈ G since χ∂Ω = 1 and the constant functions are resolutive. If E, F ∈ G with E ⊂ F , then χF ∼E = χF − χE . Since χF and χE are resolutive, χF ∼E is resolutive by Lemma 2.6.11 and HχF ∼E (x) = HχF (x) − HχE (x) = χF dμx − χE dμx = χF ∼E dμx for all x ∈ Ω. This shows that G is closed under relative complements. Suppose {Ej } is an increasing sequence in G and E = ∪Ej . By Lemma 2.6.13, χE is resolutive and HχE (x) = lim HχEj (x) = lim χEj dμx = χE dμx , x ∈ Ω. j→∞

j→∞

This shows that G is closed under unions of increasing sequences and therefore that G is a d-system containing F0 . Thus, G = d(G) = σ(G) ⊃ σ(F0 ) so that G contains all Borel subsets of ∂Ω; that is, the indicator function of any Borel subset of ∂Ω is resolutive. It follows that any simple Borel measurable function χ on ∂Ω is resolutive with Hχ (x) = χ dμx , x ∈ Ω. If f is a nonnegative Borel measurable function on ∂Ω, then there is a sequence {fj } of simple Borel measurable functions such that 0 ≤ fj ≤ f and fj ↑ f . By Lemma 2.6.13,

102

2 The Dirichlet Problem

H f (x) = H f (x) = lim Hfj (x) = lim j→∞

j→∞

fj dμx =

f dμx , x ∈ Ω,

by the Lebesgue monotone convergence theorem. If f is only bounded below by α on ∂Ω, the result can be applied to f − α. If f is bounded, then H f and H f are ﬁnite on Ω by Lemma 2.6.11 and harmonic by Lemma 2.6.7. Theorem 2.9.2 If Λ is a component of the bounded open set Ω, the collection of Borel subsets of ∂Ω of μx measure zero is independent of x ∈ Λ. Proof: If E is a Borel subset of ∂Ω, then χE is resolutive and HχE (x) = χE dμx = μx (E), x ∈ Ω. If μx0 (E) = 0 for some x0 ∈ Λ, then the nonnegative harmonic function HχE = μ (E) must be identically zero on Λ by the minimum principle, Corollary 1.5.10. Let f be a bounded Borel measurable function on ∂Ω and let Λ be a component of Ω. Using the notation of Lemma 2.6.6, HfΩ = HfΛ|∂Λ on Λ, so that the value of Hf = HfΩ at a point x ∈ Λ is independent of the values of f on ∂Ω ∼ ∂Λ. Therefore, μx (∂Ω ∼ ∂Λ) = 0 for all x ∈ Λ; that is, each μx has support on the boundary of the component containing x. This argument shows that μΛ x is just the restriction of μx to ∂Λ. If E denotes any Borel subset of ∂Ω and Z denotes any subset of a Borel subset of ∂Ω of μx measure zero, the class Fx of sets of the form (E ∼ Z) ∪ (Z ∼ E) is a σ-algebra of subsets of ∂Ω. Letting F = ∩x∈Ω Fx , F is a σ-algebra of subsets of ∂Ω which includes the Borel subsets of ∂Ω. Since each μx can be uniquely extended to Fx , the same is true of F . The extension of μx to F will be denoted by the same symbol. Deﬁnition 2.9.3 Sets in F are called μ -measurable sets and extended real-valued functions measurable relative to F are called μ -measurable. The measure μx on F is called the harmonic measure relative to Ω and x. A μ -measurable function f on ∂Ω is μ -integrable if it is integrable relative to each μx , x ∈ Ω. Lemma 2.9.4 If f is a nonnegative μ -measurable function on ∂Ω, then H f (x) = H f (x) = f dμx for all x ∈ Ω; if f is any μ -measurable function on ∂Ω which is μy -integrable for somey in each component of Ω, then f is μ -integrable, resolutive, and Hf (x) = f dμx , x ∈ Ω. Proof: Consider any F ∈ F. For each x ∈ Ω, F ∈ Fx , and F has the form (E ∼ Z) ∪ (Z ∼ E) where E is a Borel subset of ∂Ω and Z is a subset of a Borel set A ⊂ ∂Ω with μx (A) = 0 (E and Z may depend upon x). Note that χF = χE +χZ −2χE∩Z and that χE is resolutive by Theorem 2.9.1. Since 0 ≤ H χZ (x) ≤ H χZ (x) ≤ H χA (x) and H χA (x) = μx (A) = 0 by Theorem 2.9.1, the nonnegative functions H χZ and H χZ are harmonic on the component of Ω containing x. Since H χZ (x) = H χZ (x) = 0, H χZ = H χZ = 0 on the

2.9 Harmonic Measure

103

component of Ω containing x by the minimum principle, Corollary 1.5.10. Likewise, H χE∩Z = H χE∩Z = 0 on the component of Ω containing x. By Lemma 2.6.11, 0 ≤ H χF ≤ H χF ≤ H χE + H χZ − 2H χE∩Z = HχE on the component of Ω containing x. Likewise, H χF ≥ H χF ≥ H χE + H χZ − 2H χE∩Z = HχE on the component of Ω containing x so that H χF = H χF = HχE . Thus, χF is resolutive and HχF (x) = χF dμx , , x ∈ Ω, by Theorem 2.9.1. Suppose now that f is a simple μ -measurable function. Then f is resolutive by Lemma 2.6.11 and H f (x) = H f (x) = f dμx , x ∈ Ω, by the preceding step. Now let f be a nonnegative μ -measurable function on ∂Ω. Then there is a sequence {fj } of μ -measurable simple functions such that 0 ≤ fj ≤ f and fj ↑ f on ∂Ω. By Lemma 2.6.13, x ∈ Ω. fj dμx = f dμx , H f (x) = H f (x) = lim Hfj (x) = lim j→∞

j→∞

If, in addition, f is μy -integrable for some y in each component of Ω, then this equation shows that H f and H f are ﬁnite at some point of each component of Ω and therefore are harmonic on Ω. It follows that f is resolutive and Hf (x) = f dμx , x ∈ Ω. Lastly, if f is any μ -measurable function on ∂Ω that is μy -integrable for some y in each component of Ω, then the above argument can be applied to f + and f − , the positive and negative parts of f . Theorem 2.9.5 (Brelot [6]) Let Ω be a bounded open set. A function f on ∂Ω is resolutive if and only if it is μ -integrable; in which case, Hf (x) = f dμx for all x ∈ Ω. Proof: The suﬃciency is just the preceding lemma. Assume that f is resolutive. If it can be shown that f is measurable relative to each Fx , x ∈ Ω, then f is measurable relative to F = ∩x∈Ω Fx . Fix x ∈ Ω and let Λ be the component of Ω containing x. In order to show that f is Fx -measurable, it suﬃces to show that f |∂Λ is equal a.e. (μx ) to a Borel measurable function on ∂Λ since μ(∂Ω ∼ ∂Λ) = 0 implies that every subset of ∂Ω ∼ ∂Λ

104

2 The Dirichlet Problem

belongs to Fx . Note also that the ﬁniteness of f dμx is independent of the values of f on ∂Ω ∼ ∂Λ. Therefore it can be assumed that Ω is connected. Let x0 be a ﬁxed point of Ω and let {vj } be a sequence in Uf such that limj→∞ vj (x0 ) = Hf (x0 ). It can be assumed that the sequence {vj } is decreasing. Consider a ﬁxed j ≥ 1. Since vj ∈ Uf , vj is bounded below by some αj as is the Borel measurable function fj (x) = lim inf y→x,y∈Ω vj (y). Since vj ∈ Uf , vj ≥ H fj ≥ H fj ≥ αj . It follows that the functions H fj and H fj are harmonic on Ω. By Theorem 2.9.1, each fj is resolutive. Since {vj } is a decreasing sequence, the sequence {fj } is also decreasing and f ∗ = limj→∞ fj is deﬁned. Using the fact that vj ∈ Uf , fj (x) = lim inf y→x,y∈Ω vj (y) ≥ f (x) and f ∗ ≥ f . Since the fj ’s are Borel measurable, the same is true of f ∗ . Since vj ≥ Hfj ≥ H f ∗ ≥ H f ∗ ≥ Hf and limj→∞ vj (x0 ) = Hf (x0 ), H f ∗ (x0 ) = H f ∗ (x0 ) = Hf (x0 ). It follows that H f ∗ and H f ∗ are harmonic and equal to Hf ; that is, f ∗ is resolutive with Hf ∗ = Hf . Similarly, there is an increasing sequence {gj } of Borel measurable functions on ∂Ω such that limj→∞ gj = f∗ ≤ f and Hf∗ = Hf . Since fj ≥ f ∗ ≥ f ≥ f∗ ≥ gj for each j and fj , gj are μx -integrable for each x ∈ Ω, f ∗ and f∗ are μx -integrable for each x ∈ Ω. In view of the fact that Hf ∗ (x) − Hf∗ (x) = (f ∗ − f∗ ) dμx and f ∗ − f∗ ≥ 0, f ∗ = f∗ a.e.(μx ). Therefore, f ∗ = f∗ = f a.e.(μx ) for each x ∈ Ω and f is Fx -measurable for each x ∈ Ω. Thus, f is F -measurable. Since f ∗ is μx -integrable for each x ∈ Ω, the same is true of f and so f is μ - integrable. Lastly, ∗ ∗ Hf (x) = Hf (x) = f dμx = f dμx , x ∈ Ω. Harmonic measures will now be examined from the viewpoint of the classic Lp spaces, p ≥ 1. Let Ω be a connected, bounded, open subset of Rn . According to Lemma 2.9.4, the class of μx -integrable boundary functions is independent of x ∈ Ω. This class constitutes the Banach space L1 (μx ) consisting of μx -integrable functions f on ∂Ω with norm deﬁned by f 1,μx = |f | dμx , under the usual proviso that functions that are equal a.e. relative to μx are identiﬁed. Since the class of Borel subsets of ∂Ω of μx measure zero is independent of x, L1 (μx ) is independent of x ∈ Ω; but it need not be true that

· 1,μx is independent of x. For any p > 1, Lp (μx ) consists of equivalence classes of μx -measurable boundary functions f for which 1/p

f p,μx =

|f | dμx p

< +∞.

Since f pp,μx = H|f |p (x) and the latter function is harmonic, the Banach space Lp (μx ) is independent of x ∈ Ω; but again the norm f p,μx may depend upon x. Consider any p ≥ 1. There are two notions of convergence in Lp (μx ), weak and strong convergence, which could possibly depend upon x. Consider any two points x, y ∈ Ω. If f p,μx = 0, then H|f |p (x) = 0 and H|f |p = 0 on Ω; in particular, f pp,μy = H|f |p (y) = 0. This shows

2.9 Harmonic Measure

105

that f p,μx = 0 if and only if f p,μy = 0. Let K = {x, y}. By Harnack’s inequality, Theorem 2.2.2, there is a constant k such that h(x )/h(x ) ≤ k for x , x ∈ K whenever h is a positive harmonic function on Ω. If f p,μx > 0, then f p,μ > 0 on Ω and

f pp,μx H|f |p (x) ≤ k; = p

f p,μy H|f |p (y) that is, f p,μx ≤ k 1/p f p,μy , with this inequality obviously holding if

f p,μx = 0. This shows that the norms · p,μx and · p,μy are equivalent norms on Lp (μ ) = Lp (μx ) = Lp (μy ). Thus, strong convergence in Lp (μx ) is independent of x ∈ Ω. Since the norms · p,μx and · p,μy are equivalent, the adjoint space L∗p (μx ) = Lq (μx ), (1/p + 1/q = 1 if p > 1; q = ∞ if p = 1) is independent of x and the weak topology on Lp (μx ) is independent of x. Thus, weak convergence in Lp (μx ) is independent of x ∈ Ω. Moreover, if {xα ; α ∈ A} is a convergent net in Ω having the limit x ∈ Ω, the fact that Hf is harmonic whenever f ∈ C 0 (∂Ω) implies that Hf (xα ) = f dμxα → f dμx = Hf (x), w∗

and therefore that μxα → μx . Lastly, note that the measures {μx ; x ∈ Ω} are mutually absolutely continuous; if μx (A) = 0 for some μ -measurable A ∈ ∂Ω, then μy (A) = 0 for any y ∈ Ω since the class of zero sets is independent of x.

Chapter 3

Green Functions

3.1 Introduction In 1828, Green introduced a function, which he called a potential, for calculating the distribution of a charge on a surface bounding a region in Rn in the presence of external electromagnetic forces. The argument that led to the function GB introduced in Section 1.5 is precisely the argument made by Green in [26]. The function introduced by Green is now called the Green function. After Green, the most general existence theorem for Green functions was proven by Osgood in 1900 for simply connected regions in the plane (see [49]). In this chapter it will be shown that a Green function can be deﬁned for some sets, but not all, which serve the same purpose as GB . The sets for which this is true are called Greenian sets. This chapter will culminate in an important theorem of F. Riesz which states that a nonnegative superharmonic on a Greenian set has a decomposition into the sum of a Green potential and a harmonic function.

3.2 Green Functions In Chapter 1, the Green function for a ball led to a solution of the Dirichlet problem for a measurable boundary function by way of the Poisson integral formula. The Green function for the ball B = By,ρ will be reexamined to establish certain properties. Lemma 3.2.1 GB is symmetric on B×B; moreover, for each x ∈ B, GB (x, ·) is stricly positive on B, superharmonic on B, and limz→z0 GB (x, z) = 0 for all z0 ∈ ∂B.

L.L. Helms, Potential Theory, Universitext, c Springer-Verlag London Limited 2009

107

108

3 Green Functions

Proof: It will be shown ﬁrst that GB is symmetric on B × B in the n ≥ 3 case, the n = 2 case being essentially the same. If x = y = z, then GB (x, z) = GB (z, x) = +∞. If x = y and z ∈ B ∼ {x}, then GB (x, z) =

1 1 − n−2 . |z − x|n−2 ρ

Since |y − z||y − z ∗ | = ρ2 by deﬁnition of z ∗ , 1 ρn−2 − |x − z|n−2 |z − y|n−2 |x − z ∗ |n−2 1 ρn−2 = − |x − z|n−2 (|y − z||y − z ∗ |)n−2 1 ρn−2 = − |x − z|n−2 (ρ2 )n−2 = GB (x, z).

GB (z, x) =

Lastly, suppose that x = y, z = y, and z ∈ B ∼ {x}. Then GB (x, z) =

1 ρn−2 − , n−2 n−2 |z − x| |x − y| |z − x∗ |n−2

and it suﬃces to show that the second term on the right is a symmetric function of x and z. To see this, let φ be the angle between the line segment joining x to y and z to y. Then by deﬁnition of x∗ , |z − x∗ |2 = |z − y|2 + |y − x∗ |2 − 2|z − y||y − x∗ | cos φ ρ4 ρ2 = |z − y|2 + − 2|z − y| cos φ |y − x|2 |y − x| and so |y − x|2 |z − x∗ |2 = |z − y|2 |y − x|2 + ρ4 − 2ρ2 |z − y||y − x| cos φ. Similarly, |y − z|2 |x − z ∗ |2 = |x − y|2 |y − z|2 + ρ4 − 2ρ2 |x − y||y − z| cos φ and it follows that |y − x||z − x∗ | = |y − z||x − z ∗ |.

(3.1)

Thus, GB (x, z) = GB (z, x) whenever x = y, z = y, and z ∈ B ∼ {x}. It is clear from the deﬁnition that GB (x, ·) is superharmonic on B. Using Equation (1.5) and the deﬁnition of GB (x, z), it is easy to see that limz→z0 GB (x, z) = 0 for all z0 ∈ ∂B. It follows from Corollary 2.3.6 that GB (x, ·) ≥ 0 on B. If GB (x, ·) attains its minimum value at a point of B, it

3.2 Green Functions

109

would have to be a constant function in contradiction to the fact that it is arbitrarily large near x. Therefore, GB (x, ·) is strictly positive on B. According to the following theorem, if a nonnegative function is harmonic on a neighborhood of a point, then it either has an inﬁnite pole at the point or has a harmonic extension to all of the neighborhood. Theorem 3.2.2 (Bˆ ocher [3]) Let Ω be a neighborhood of x0 ∈ Rn , and let u be deﬁned, harmonic, and nonnegative on Ω ∼ {x0 }. Then either (i) u can be deﬁned at x0 so as to be harmonic on Ω, or (ii) there is a c > 0 such that u = cux0 + v where v is harmonic on Ω. Proof: Consider a closed ball B − = Bx−0 ,δ ⊂ Ω. Deﬁne w(x) =

+∞ u(x) + ux0 (x)

if x = x0 if x ∈ Ω ∼ {x0 }.

Clearly, w is superharmonic on Ω. By Theorem 2.5.2, there is a sequence of superharmonic functions {wj } in C ∞ (Ω) such that wj ↑ w on Ω, with the additional property that outside of a small neighborhood of x0 , wj = w and is harmonic for suﬃciently large j. The rest of the proof will be carried out for the n ≥ 3 case, the n = 2 case only requiring that σn (n − 2) be replaced by 2π. By Theorem 1.5.3, for each x ∈ B 1 wj (x) = − wj (z)Dn GB (x, z) dσ(z) σn (n − 2) ∂B 1 GB (x, z)Δwj (z) dz. − σn (n − 2) B It can be seen from Equation (1.9) that −Dn GB (x, z) ≥ 0 on B. By the Lebesgue monotone convergence theorem, for each x ∈ B 1 w(x) = − w(z)Dn GB (x, z) dσ(z) σn (n − 2) ∂B 1 lim GB (x, z)Δwj (z) dz. (3.2) − σn (n − 2) j→∞ B For each j ≥ 1 and Borel set M ⊂ B − , deﬁne 1 ωj (M ) = − Δwj (z) dz. σn (n − 2) M By Lemma 2.3.4, Δwj ≤ 0 on Ω and ωj is a measure on the Borel subsets of B − . Moreover, if M ⊂ B − ∼ Bx0 ,ρ for some ρ > 0, then ωj (M ) = 0 for suﬃciently large j and limj→∞ ωj (M ) = 0. Now consider any y ∈ B, y = x0 , and a neighborhood Σ of x0 whose closure Σ − does not contain y and Σ − ⊂ B. Since w(y) < +∞, it follows from Equation (3.2) that

110

3 Green Functions

lim

j→∞

GB (y, z) dωj (z) < +∞. B

Since ωj (B − ∼ Σ − ) = 0 for suﬃciently large j, GB (y, z) dωj (z) < +∞. lim j→+∞

Σ−

By Lemma 3.2.1, GB (y, ·) > 0 on Σ − . By the continuity of GB (y, ·) on Σ − , there is a constant m > 0 such that GB (y, z) ≥ m for z ∈ Σ − . Therefore, GB (y, z) dωj (z) ≥ mωj (Σ − ), Σ−

and it follows that supj≥1 ωj (Σ − ) < +∞. Since ωj = ωj (B − ) = ωj (Σ − ) + ωj (B − ∼ Σ − ) = ωj (Σ − ), for suﬃciently large j there is a constant k > 0 such that ωj ≤ k for all j ≥ 1. By Lemma 0.2.5, there is a measure ω and w∗ w∗ a subsequence {ωjk } such that ωj → ω. It can be assumed that ωj → ω by replacing the sequence with the subsequence if necessary. Since ω is zero on the complement of each small neighborhood of x0 , ω is a nonnegative mass concentrated on {x0 }. If x = x0 , then GB (x, ·) is continuous at x0 and lim GB (x, z) dωj (z) = ω({x0 })GB (x, x0 ) j→∞

B

and Equation (3.2) becomes w(x) = PI(w : B)(x) + ω({x0 })GB (x, x0 ). It follows from the deﬁnition of w that u(x) = PI(w : B)(x) + ω({x0 })GB (x, x0 ) − ux0 for x ∈ B ∼ {x0 }. Since GB (x, x0 ) = GB (x0 , x) by Lemma 3.2.1, GB (·, x0 ) can be expressed as a sum of ux0 and a harmonic function. Therefore, for x ∈ B ∼ {x0 }, u(x) = v(x) + (ω({x0 }) − 1)ux0 where v is harmonic on B. If ω({x0 }) − 1 = 0, then u can be deﬁned at x0 to be harmonic on B by putting u(x0 ) = v(x0 ). It is not possible to have ω({x0 }) − 1 < 0 for in this case it would be true that u(x) → −∞ as x → x0 , contradicting the fact that u ≥ 0 on Ω ∼ {x0 }. If ω({x0 }) − 1 > 0, then the assertion is true with c = ω({x0 }) − 1. Historically, Green functions for general regions were constructed ﬁrst assuming smooth boundaries. It is possible to go directly to arbitrary regions. Deﬁnition 3.2.3 A Green Function for an open set Ω ⊂ Rn , if it exists, is an extended real-valued function GΩ on Ω × Ω with the following properties:

3.2 Green Functions

111

(i) GΩ (x, ·) = ux + hx where for each x ∈ Ω, hx is harmonic on Ω. (ii) GΩ ≥ 0. (iii) If for each x ∈ Ω, vx is a nonnegative superharmonic function on Ω and is the sum of ux and a superharmonic function, then vx ≥ GΩ (x, ·) on Ω. In other words, for each x ∈ Ω, GΩ (x, ·) is minimal in the class of functions described in (iii). The Green function for Rn , n ≥ 3, will be denoted simply by G; it will be seen later that this convention does not apply to the n = 2 case since R2 does not have a Green function. An open subset Ω of Rn having a Green function will be called a Greenian set. It will now be shown that the Green function GB for a ball B as deﬁned in Chapter 1 satisﬁes (i), (ii), and (iii) of this deﬁnition. It is apparent from the deﬁnition of GB (x, ·) that (i) is satisﬁed. By Lemma 3.2.1, (ii) is satisﬁed. Suppose now that for each x ∈ B, vx = ux + wx where vx ≥ 0 and wx is superharmonic on B. Moreover, let GB (x, ·) = ux + hx where hx is harmonic on B. Since both ux and hx have limits at points z0 ∈ ∂B, by Lemma 3.2.1 lim inf (wx (z) − hx (z))

z→z0 ,z∈B

= lim inf (ux (z) + wx (z) − (ux (z) + hx (z))) z→z0 ,z∈B

= lim inf (ux (z) + wx (z)) − z→z0 ,z∈B

lim

z→z0 ,z∈B

GB (x, z)

= lim inf (ux (z) + wx (z)) z→z0 ,z∈B

= lim inf vx (z) ≥ 0. z→z0 .z∈B

Since wx − hx is superharmonic on B, wx − hx ≥ 0 on B by Corollary 2.3.6; that is, vx −GB (x, ·) = (ux +wx )−(ux +hx ) = wx −hx ≥ 0 and GB (x, ·) ≤ vx on B. Thus, GB (x, ·) satisﬁes (iii) and can be justiﬁably called the Green function for the ball B. Theorem 3.2.4 The Green function for the open set Ω is unique if it exists. Proof: Let gΩ be a second Green function for Ω according to the above deﬁnition. By (iii) of the above deﬁnition, GΩ (x, ·) ≥ gΩ (x, ·) and gΩ (x, ·) ≥ GΩ (x, ·) for each x ∈ Ω. Thus, GΩ (x, ·) = gΩ (x, ·) for each x ∈ Ω. Lemma 3.2.5 If the open set Ω has a Green function, then inf y∈Ω GΩ (x, y)= 0 for each x∈Ω. Proof: Suppose c = inf y∈Ω GΩ (x, y) > 0 for some x ∈ Ω. Then GΩ (x, ·)−c ≥ 0 on Ω. Moreover, GΩ (x, ·) − c is nonnegative superharmonic, and representable as the sum of ux and a harmonic function. Since GΩ (x, ·) − c < GΩ (x, ·), the minimality of GΩ (x, ·) is contradicted. Lemma 3.2.6 If the open set Ω has a Green function and B is a ball with x ∈ B ⊂ B − ⊂ Ω, then GΩ (x, ·) is bounded outside B.

112

3 Green Functions

Proof: GΩ (x, ·) is continuous and nonnegative on Ω ∼ B. It must be strictly positive on the component of Ω containing x, for if GΩ (x, y) = 0 for some point y in this component, then GΩ (x, ·) would be identically zero thereon by the minimum principle, Corollary 2.3.6, in contradiction to the fact that GΩ (x, x) = +∞. Since B − ⊂ Ω, B − is a subset of the component containing x and GΩ (x, ·) is strictly positive on B − . Let m = supy∈∂B GΩ (x, y) > 0. Deﬁne GΩ (x, y), y ∈ B, g(x, y) = y ∈ Ω ∼ B. min (GΩ (x, y), 2m), Then g(x, ·) is superharmonic on B since it agrees with GΩ (x, ·) on B. Since g(x, ·) is the minimum of two superharmonic functions on Ω ∼ B − , it is superharmonic on Ω ∼ B − . Since GΩ (x, ·) ≤ m < 2m on ∂B and GΩ (x, ·) is continuous at points of ∂B, g(x, ·) agrees with GΩ (x, ·) on a neighborhood of ∂B and is superharmonic thereon. It follows that g(x, ·) is superharmonic on Ω. Since g(x, ·) − ux = GΩ (x, ·) − ux on B ∼ {x} and GΩ (x, ·) − ux is harmonic on B, g(x, ·) − ux can be deﬁned at x to be superharmonic on Ω. Denoting the extension by [g(x, ·) − ux ], 0 ≤ g(x, ·) = ux + [g(x, ·) − ux ], where the bracket function is superharmonic on Ω and GΩ (x, ·) ≤ g(x, ·) by the deﬁnition of the Green function. But since g(x, ·) ≤ 2m on Ω ∼ B, GΩ (x, ·) ≤ 2m on Ω ∼ B. The existence of a Green function for an open set Ω can be established by showing that Ω supports suﬃciently many nonnegative superharmonic functions. Deﬁnition 3.2.7 If Ω is an open subset of Rn and x ∈ Ω, let Bx = {vx ; vx ≥ 0, vx = ux + wx , wx ∈ S(Ω)}. The class Bx will generally depend upon Ω and x. It may well be that Bx = ∅ for some x ∈ Ω. Lemma 3.2.8 Bx is a saturated family of superharmonic functions on Ω ∼ {x}. Proof: Since there is nothing to prove if Bx = ∅, it can be assumed that Bx = ∅. If v1 , v2 ∈ Bx , then vi ≥ 0 and vi = ux + wi , i = 1, 2, where w1 and w2 are superharmonic on Ω. Since min (v1 , v2 ) = ux + min (w1 , w2 ) and min (w1 , w2 ) is superharmonic on Ω, min (v1 , v2 ) is nonnegative and superharmonic on Ω; that is, the minimum of two elements of Bx belongs to Bx . Suppose now that v = ux + w ∈ Bx where w is superharmonic on Ω. Let B be any ball with B − ⊂ Ω ∼ {x}. By Lemma 2.4.11, PI(v : B) on B vB = v on Ω ∼ B

3.2 Green Functions

113

is superharmonic on Ω. Since ux is harmonic on a neighborhood of B − , PI(v : B) = PI(ux +w : B) = ux +PI(w : B). Thus, vB = ux +wB is superharmonic on Ω and vB ∈ Bx . This completes the proof that Bx is saturated on Ω ∼ {x}. Lemma 3.2.9 If Bx = ∅ for each x ∈ Ω, then Ω has a Green function GΩ (x, ·) = inf {vx ; vx ∈ Bx }. Proof: Consider a ﬁxed x ∈ Ω and let wx = inf {vx ; vx ∈ Bx }. Since Bx is a saturated family of superharmonic functions on Ω ∼ {x} and wx ≥ 0, wx is harmonic on Ω ∼ {x} by Theorem 2.6.2. According to Bˆ ocher’s theorem, Theorem 3.2.2, there is a constant c ≥ 0 such that wx = cux +hx on Ω ∼ {x}, where hx is harmonic on Ω. It will be shown now that c = 1. Consider a ball B = Bx,δ with B − ⊂ Ω and the Green function GB (x, ·). It follows from (iii) of Deﬁnition 3.2.3 and the persuant discussion that vx ≥ GB (x, ·) on B for all vx ∈ Bx , and therefore that wx ≥ GB (x, ·) on B and that c = 0. Since cux + hx = wx ≥ GB (x, ·) = ux + h∗x on B where h∗x is harmonic on B, (h∗x − hx ) ≤ (c − 1)ux

on B ∼ {x}.

If c < 1, then the harmonic function h∗x − hx would have a limit −∞ at x, a contradiction. Therefore, c ≥ 1 and wx /c = ux + Hx on Ω where Hx is harmonic on Ω. Therefore, wx /c ∈ Bx and wx ≤ wx /c ≤ wx so that c = 1. Therefore, GΩ (x, ·) = wx (·) is the Green function for Ω. Theorem 3.2.10 Let Ω be an open subset of Rn . If (i) n = 2 and Ω is bounded, (ii) n = 2 and Ω is a subset of a set having a Green function, or (iii) n ≥ 3, then Ω has a Green function. Proof: (i) If n = 2 and Ω is bounded, then for each x ∈ Ω there is a constant cx such that ux + cx ∈ Bx . (ii) If n = 2 and Ω ⊂ Σ, and Σ has a Green function GΣ , then GΣ (x, ·)|Ω ∈ Bx for all x ∈ Ω. (iii) If n ≥ 3, ux ∈ Bx for all x ∈ Ω. Theorem 3.2.11 G(x, y) = ux (y) is the Green function for Rn , n ≥ 3. Proof: By the preceding theorem, the Green function G for Rn , n ≥ 3, exists. Fix x ∈ Rn . Then G(x, ·) = ux + hx where hx is harmonic on Rn . Since ux ∈ Bx , ux + hx ≤ ux and hx ≤ 0. By Picard’s theorem, Theorem 1.5.7, hx = c with c ≤ 0. Suppose c < 0. Since ux (y) → 0 as |y| → +∞, G(x, y) < 0 for large |y|. But this contradicts the fact that inf y∈Rn G(x, y) = 0 by Lemma 3.2.5. Therefore, c = 0 and hx = 0. Theorem 3.2.12 If Ω1 , Ω2 are Greenian subsets of Rn with Ω1 ⊂ Ω2 , then GΩ1 (x, ·) ≤ GΩ2 (x, ·) for each x ∈ Ω1 .

114

3 Green Functions

Proof: Let Bx (Ω1 ) be the Bx class relative to Ω1 . Thus, GΩ2 (x, ·)|Ω1 ∈ Bx (Ω1 ) and GΩ1 (x, ·) ≤ GΩ2 (x, ·) for each x ∈ Ω1 . Theorem 3.2.13 If the open set Ω is not everywhere dense in R2 , then it is Greenian. − ⊂∼ Ω. Proof: By hypothesis, there is a point x ∈ R2 and a closed ball Bx,δ − 2 Suppose, for example, that Ω ∩ B0,1 = ∅. The points of R will be interpreted as complex numbers in what follows. For any x, y ∈ Ω, x−1 , y −1 ∈ B0,1 and 1 − 1 < 2. x y

Then

2 2|xy| = log log > 0. 1 1 |x − y| x − y

Consider 2 = log 2|x| + log |y| − log |x − y|. vx (y) = log x1 − y1 It has just been shown that vx ≥ 0 on Ω for each x ∈ Ω. For ﬁxed x ∈ Ω, the ﬁrst term on the right is a constant on Ω, the second term is the fundamental harmonic function with pole at 0 and is harmonic on Ω, and the third term is the fundamental harmonic function with pole at x. Hence vx (y) = hx (y) + log

1 |x − y|

on Ω

where hx is harmonic on Ω. It follows that Bx = ∅ for each x ∈ Ω and the Green function for Ω exists by Theorem 3.2.9 Theorem 3.2.14 R2 does not have a Green function. Proof: Assume that R2 does have a Green function G(x, y) = − log |x − y| + hx (y) where hx is harmonic on R2 for each x ∈ R2 . Consider a ﬁxed x. Since − log |x − y| → −∞ as |y| → +∞, hx (y) → +∞ as |y| → +∞. If m is any positive integer, then there is a ball with its center at 0 such that hx ≥ m on the boundary of the ball. By the minimum principle, Corollary 1.5.10, hx ≥ m on the ball. Therefore, hx (0) = +∞, contradicting the fact that a harmonic function is real-valued. Thus, R2 does not have a Green function.

3.3 Symmetry of the Green Function

115

Theorem 3.2.15 If Λ ⊂ R2 is a nonempty closed set consisting only of isolated points, then Ω = R2 ∼ Λ does not have a Green function. Proof: Consider any x ∈ Ω. Since Ω is open, there is a neighborhood U ⊂ Ω of x outside of which GΩ (x, ·) is bounded by Lemma 3.2.6. Consider any y ∈ Ω. There is a neighborhood V of y that does not contain x or any of the other points of Λ such that GΩ (x, ·) is bounded and harmonic on V ∼ {y}. By Theorem 3.2.2, GΩ (x, ·) can be deﬁned at y so as to be harmonic on V . Since this is true of each y ∈ Ω, GΩ (x, ·) can be extended so that it is harmonic on Ω ∼ {x}. Writing GΩ (x, ·) = ux + hx , where hx is harmonic on Ω, this means that hx has a harmonic extension to all of R2 . The class Bx (R2 ) relative to R2 is then nonempty. The same can be seen to be true of Bx0 (R2 ) for any x0 ∈ R2 by means of a translation taking x to x0 , since harmonic functions map into harmonic functions and ux goes into ux0 under such a translation. It follows that R2 has a Green function, contrary to the preceding theorem.

3.3 Symmetry of the Green Function The principal result of this section is that the Green function is a symmetric function of its arguments. This result is achieved by introducing the concept of harmonic minorant and a smoothing operation known as balayage or sweeping out, the latter having its origin in the work of Poincar´e [50]. Let Ω be an open subset of Rn , let Λ be an open subset of Ω, and let u be a superharmonic function on Ω. Let {Bj } be a sequence of balls such that (i) Bj− ⊂ Λ for all j ≥ 1, (ii) Λ = ∪∞ j=1 Bj , and (iii) for each j ≥ 1, Bj occurs inﬁnitely often in the sequence. Deﬁne

u1 = uB1 =

PI(u : B1 ) u

on B1 on Ω ∼ B1

and, inductively, for j ≥ 2 uj = (uj−1 )Bj =

PI(uj−1 : Bj ) uj−1

on Bj on Ω ∼ Bj .

By Theorem 2.4.3, uj ≤ uj−1 on Ω, uj is harmonic on Bj , and uj is superharmonic on Ω. Since the sequence {uj } is monotone decreasing, u∞,Λ = limj→∞ uj is deﬁned on Ω.

116

3 Green Functions

Deﬁnition 3.3.1 u∞,Λ is called the reduction of u over Λ relative to Ω. The second subscript Λ will be omitted when Λ = Ω. It will be shown shortly that u∞,Λ is independent of the sequence {Bj } whenever u is bounded below by a harmonic function. Theorem 3.3.2 If u is superharmonic on the open set Ω and Λ is an open subset of Ω, then u∞,Λ is either harmonic or identically −∞ on each component of Λ and u∞,Λ = u on Ω ∼ Λ. Proof: By considering components, it can be assumed that Λ is connected. Let {Bj } be the sequence of balls and {uj } the corresponding sequence of superharmonic functions deﬁning u∞,Λ . Consider a ﬁxed j ≥ 1. Since Bj occurs inﬁnitely often in the sequence of balls, there is a subsequence {Bjk } such that Bj = Bjk for all k ≥ 1. On Bj , u∞,Λ = limk→∞ ujk . Since ujk is harmonic on Bjk = Bj , u∞,Λ is either harmonic or identically −∞ on Bj by Theorem 2.2.5. It is easy to see using a connectedness argument that u∞,Λ is either harmonic or identically −∞ on Λ. Since uj = u on Ω ∼ Λ for each j ≥ 1, u∞,Λ = u on Ω ∼ Λ. Even if u is not identically −∞ on any component of Λ, it cannot be concluded that u∞,Λ is superharmonic on Ω, since the limit of a decreasing sequence of l.s.c. functions need not be l.s.c. Lemma 3.3.3 If u is superharmonic on the open set Ω, Λ is an open subset of Ω, v is subharmonic on Λ, and v ≤ u on Λ, then v ≤ u∞,Λ ≤ u on Λ. Proof: Let {Bj } be the sequence of balls deﬁning u∞,Λ . On B1 , v ≤ PI(v : B1 ) ≤ PI(u : B1 ) = u1 . Moreover, v ≤ u = u1 on Λ ∼ B1 . Thus, v ≤ u1 on Λ and, by induction, v ≤ uj on Λ for all j ≥ 1. Deﬁnition 3.3.4 If u is superharmonic on the open set Ω, h is harmonic on Ω, and h ≤ u on Ω, then h is called a harmonic minorant of u. The function h is the greatest harmonic minorant of u, denoted by ghmΩ h, if h is a harmonic minorant and v ≤ h whenever v is a harmonic minorant of u. Theorem 3.3.5 If u is superharmonic on the open set Ω and u has a harmonic minorant on Ω, then u has a unique greatest harmonic minorant, namely u∞ . Proof: Let h be a harmonic minorant of u. By the preceding lemma, h ≤ u∞ ≤ u on Ω. Since h is harmonic on Ω, u∞ cannot take on the value −∞ on Ω and is therefore harmonic on Ω by Theorem 3.3.2. Since h can be any harmonic minorant of u, it follows that u∞ is the greatest harmonic minorant of u. Uniqueness is obvious. In deﬁning u∞,Λ , it was not shown that u∞,Λ is independent of the sequence of balls {Bj }. If u has a harmonic minorant on Λ, then u∞,Λ is independent of the sequence used, since it is the greatest harmonic minorant of u

3.3 Symmetry of the Green Function

117

over Λ, which is unique. Since u∞,Λ will be considered only when a harmonic minorant exists, further discussion is not warranted. Lemma 3.3.6 If u1 , u2 are superharmonic on the open set Ω and have harmonic minorants on Ω, then the greatest harmonic minorant of u1 + u2 exists and is the sum of the greatest harmonic minorants of u1 and u2 . Proof: Let {Bj } be a sequence of balls and let {uj } be the corresponding sequence of superharmonic functions deﬁning u∞ . Since the Poisson integral is linear in the boundary function, (u1 + u2 )j = (u1 )j + (u2 )j . Therefore, (u1 + u2 )∞ = (u1 )∞ + (u2 )∞ and the lemma follows from Theorem 3.3.5. Theorem 3.3.7 If the open set Ω is Greenian, then the greatest harmonic minorant of GΩ (x, ·), x ∈ Ω, is the zero function. Proof: It is known that GΩ (x, ·) ≥ 0 on Ω and that GΩ (x, ·) is superharmonic on Ω. By Theorem 3.3.5, GΩ (x, ·) has a greatest harmonic minorant, since the zero function is a harmonic function. Let hx be a harmonic minorant of GΩ (x, ·). Then GΩ (x, ·) − hx ≥ 0 and it can be written as the sum of ux and a harmonic function; that is, GΩ (x, ·) − hx ∈ Bx . Therefore, GΩ (x, ·) ≤ GΩ (x, ·) − hx by the minimality of GΩ (x, ·). This shows that hx ≤ 0 and that the zero function is the greatest harmonic minorant of GΩ (x, ·). The following theorem is a classic version of the domination principle (c.f.Theorem 4.4.8). Theorem 3.3.8 Let Ω be a Greenian subset of Rn . If x ∈ Ω, Λ is a neighborhood of x with compact closure in Ω, and u is a positive superharmonic function deﬁned on a neighborhood of Ω ∼ Λ with u ≥ GΩ (x, ·) on a neighborhood of ∂Λ, then u ≥ GΩ (x, ·) on Ω ∼ Λ. Proof: Letting

w=

GΩ (x, ·) on Λ min (GΩ (x, ·), u) on Ω ∼ Λ,

w is superharmonic on Ω. If GΩ (x, ·) − w is deﬁned to be 0 at x, then it is a subharmonic minorant of GΩ (x, ·) on Ω. By Lemma 3.3.3 and the fact that the greatest harmonic minorant of GΩ (x, ·) is the zero function, GΩ (x, ·) − w ≤ 0 so that u ≥ GΩ (x, ·) on Ω ∼ Λ. The method used to deﬁne u∞,Λ will be used to prove that the Green function is a symmetric function of its arguments. The proof will require a ∞ modiﬁcation of the sequence of balls {Bj } so that j=1 ∂Bj is closed relative to Ω. Such a sequence can be constructed as follows. Let {Ωk } be a sequence of open sets with compact closures such that Ωk ↑ Ω and Ωk− ⊂ Ω, k ≥ 1. It is easily seen using compactness arguments that there is a sequence of balls {Bj } such that only a ﬁnite number of balls intersect any Ωk− . If the sequence is modiﬁed so that each ball occurs inﬁnitely often in the sequence, ∞ then this will still be true. Suppose now that {xi } is a sequence in j=1 ∂Bj with

118

3 Green Functions

xi → x ∈ Ω. Since Ωk ↑ Ω, x ∈ Ωj0 for some j0 . Since only a ﬁnite number k0 of balls intersect Ωj−0 , there is some k0 such that xk ∈ j=1 ∂Bj for large k, and it follows that x ∈ ∂Bj for some j with 1 ≤ j ≤ k0 . Thus, x ∈ ∞ j=1 ∂Bj ∞ and j=1 ∂Bj is closed relative to Ω. It therefore can be assumed that there is a sequence of balls satisfying ∞ (i) Ω = j=1 Bj , (ii) Bj− ⊂ Ω for all j ≥ 1, ∞ (iii) j=1 ∂Bj is closed in Ω, and (iv) each Bj occurs inﬁnitely often in the sequence. Theorem 3.3.9 If the open set Ω has a Green function, then G(x, y) = G(y, x) for all x, y ∈ Ω; in particular, G(·, y) is harmonic on Ω ∼ {y} for each y ∈ Ω. Proof: Let {Bj } be a sequence of balls satisfying the above four properties. For each x ∈ Ω, G(x, ·) = ux + hx where hx is harmonic on Ω. To simplify the notation, ux (y) and hx (y) will be written u(x, y) and h(x, y), respectively. For each x ∈ Ω, Gj (x, ·) = uj (x, ·) + h(x, ·) where Gj (x, ·) and uj (x, ·) denote the jth result of applying the sweeping out method to functions G(x, ·) and u(x, ·), respectively. The second term on the right does not depend upon j because harmonic functions are invariant under the sweeping out process. As j → ∞, Gj (x, ·) approaches the greatest harmonic minorant of G(x, ·), which is the zero function by Theorem 3.3.7; likewise, uj (x, ·) → u∞ (x, ·). Therefore, 0 = u∞ (x, ·) + h(x, ·) and it follows that G(x, ·) = u(x, ·) − u∞ (x, ·).

(3.3)

It will be shown next that u∞ (·, y) is harmonic on Ω for each y ∈ Ω by reexamining the sweeping out process deﬁning u∞ (·, y). Consider any ﬁxed y ∈ Ω. If y ∈ B1 , then u1 (x, y) = u(x, y) for all x ∈ Ω and u1 (·, y) is harmonic on Ω ∼ {y}. If y ∈ B1 , then u1 (·, y) is harmonic on Ω ∼ B1− since it agrees with u(·, y) on Ω ∼ B1− and is harmonic on B1 since u1 (·, y) = PI(u(·, y) : B1 ) on B1 . Combining the two cases, u1 (·, y) is harmonic on Ω ∼ ({y} ∪ ∂B1 ). Now consider u2 (·, y). Since u2 (·, y) agrees with u1 (·, y) on Ω ∼ B2 , it is harmonic on (Ω ∼ ({y} ∪ ∂B1 )) ∼ B2− . Since u2 (·, y) is harmonic on B2 and therefore on (Ω ∼ ({y} ∪ ∂B1 )) ∩ B2 , it follows that u2 (·, y) is harmonic on Ω ∼ ({y} ∪ ∂B1 ∪ ∂B2 ). By an induction argument, uj (·, y) is harmonic on j Ω ∼ ({y} ∪ i=1 ∂Bi ). Since each ball occurs inﬁnitely often in the sequence {Bj }, there is a subsequence {Bjk } such that y ∈ Bjk , k ≥ 1. Since ujk (·, y) jk ∞ is harmonic on Ω ∼ i=1 ∂Bi ⊃ Ω ∼ i=1 ∂Bi and the sequence ∞{ujk (·, y)} decreases to u∞ (·, y) on each component of the open set Ω ∼ i=1 ∂Bi and

3.3 Symmetry of the Green Function

119

each ujk (·, y) ≥ 0, u∞ is harmonic on each component by Lemma 2.2.5 and therefore is harmonic on Ω ∼ ∞ i=1 ∂Bi . Since u∞ (·, y) is independent of the sequence of balls {Bj }, u∞ (·, y) is harmonic on Ω for each y ∈ Ω. Since GΩ (x, y) = ux (y) − u∞ (x, y) = uy (x) − u∞ (x, y), for each y ∈ Ω (i) u(·, y) − u∞ (·, y) ≥ 0 on Ω, and (ii) u(·, y) − u∞ (·, y) is the sum of uy and a harmonic function. By the minimality of the Green function GΩ (y, x) ≤ u(x, y) − u∞ (x, y) = GΩ (x, y). Interchanging x and y, GΩ (x, y) = GΩ (y, x). Corollary 3.3.10 Let Ω be a Greenian set, and let Ω0 be a component of Ω. If x ∈ Ω0 , then GΩ (x, y) = 0 for y ∈ Ω ∼ Ω0 and GΩ (x, ·) > 0 on Ω0 . Moreover, the Green function for Ω0 is given by GΩ0 = GΩ |Ω0 ×Ω0 . Proof: Consider a ﬁxed x ∈ Ω0 . Since GΩ (x, ·) = ux + hx ≥ 0, ux has a harmonic minorant on Ω, namely, −hx . It therefore has a greatest harmonic minorant h0 on Ω0 . Letting h0 on Ω0 h∗x = ux on Ω ∼ Ω0 , h∗x is harmonic on Ω and ux ≥ h∗x on Ω. By Equation (3.3), −hx is the greatest harmonic minorant of ux and therefore −hx ≥ h∗x . Thus, GΩ (x, ·) = ux + hx ≤ ux − h∗x = 0 on Ω ∼ Ω0 . Since it is always true that GΩ (x, ·) ≥ 0, GΩ = 0 on Ω ∼ Ω0 . If GΩ (x, y) = 0 for some point y ∈ Ω0 , then GΩ (x, ·) = 0 on Ω0 by the minimum principle, contradicting the fact that GΩ (x, x) = +∞. As to the second assertion, let {Bjk } be a subsequence of the sequence {Bj } of balls deﬁning the reduction of ux (·) = u(x, ·) over Ω such that Bj−k ⊂ Ω0 . Using such a sequence, it is easy to see that u∞,Ω0 (x, ·) = u∞ (x, ·)

on Ω0 .

By Equation (3.3), for x ∈ Ω0 GΩ0 (x, ·) = u(x, ·) − u∞,Ω0 (x, ·) = u(x, ·) − u∞ (x, ·) = GΩ (x, ·) on Ω0 . Hence, GΩ0 = GΩ |Ω0 ×Ω0 . Theorem 3.3.11 If {Ωj } is an increasing sequence of Greenian sets with Ω = Ωj , then either (i) Ω is Greenian and limj→∞ GΩj = GΩ on Ω × Ω, or (ii) Ω is not Greenian and limj→∞ GΩj = +∞ on Ω × Ω. Proof: Since GΩj (x, ·) and GΩ (x, ·), if deﬁned, vanish on components of Ω not containing x, it can be assumed that Ω is connected. For j ≥ 1 and (j) (j) x ∈ Ωj , by Equation (3.3) GΩj (x, ·) = ux − hx where hx is the greatest

120

3 Green Functions

harmonic minorant of ux on Ωj . Since the GΩj (x, ·) are eventually deﬁned and (j) increase at each point of Ω by Theorem 3.2.12, the hx are eventually deﬁned (j) and decrease at each point of Ω, and hx = limj→∞ hx is deﬁned on Ω. By Lemma 2.2.7, hx is either identically −∞ or harmonic on Ω. If hx is harmonic (j) (j) on Ω for some x, then hy is harmonic on Ω for all y ∈ Ω since hy (x) = hx (y) by the symmetry of the Green function, and so hy (x) = hx (y); in this case, vx = ux − hx ∈ Bx for each x ∈ Ω and the Green function GΩ exists by (j) Lemma 3.2.9. If h∗x is any harmonic minorant of ux on Ω, then h∗x ≤ hx on Ωj , and therefore h∗x ≤ hx on Ω; that is, hx is the greatest harmonic minorant of ux on Ω, and therefore GΩ (x, ·) = ux − hx = limj→∞ GΩj (x, ·) on Ω. On the other hand, if hx is identically −∞ for some x ∈ Ω, then hx is identically −∞ for all x ∈ Ω; if Ω were Greenian, the fact that GΩj (x, ·) ≤ GΩ (x, ·) would imply that GΩ (x, ·) = +∞ on Ω, a contradiction. Theorem 3.3.12 If the open set Ω is Greenian, then GΩ is continuous on Ω × Ω, in the extended sense at points of the diagonal of Ω × Ω. Moreover, − if Bx,δ ⊂ Ω, then L(GΩ (z, ·) : x, δ) is a continuous function of z ∈ Ω. Proof: GΩ is obviously continuous in the extended sense at points of the diagonal of Ω × Ω. If (x0 , y0 ) ∈ Ω × Ω, x0 = y0 , let Bx0 and By0 be balls with centers x0 and y0 , respectively, with disjoint closures in Ω. If x0 and y0 are in diﬀerent components of Ω, then GΩ = 0 on Bx0 × By0 and GΩ is trivially continuous at (x0 , y0 ). It therefore can be assumed that x0 and y0 are in the same component of Ω. By Lemma 3.2.6, there is a constant m > 0 such that GΩ (x0 , ·) ≤ m on By0 . Since GΩ (·, y) is positive and harmonic on Bx0 for each y ∈ By0 , there is a constant k > 0 such that GΩ (x, y)/GΩ (x0 , y) ≤ k for all x ∈ Bx0 and y ∈ By0 by Theorem 2.2.2. Therefore, GΩ (x, y) = (GΩ (x, y)/GΩ (x0 , y))GΩ (x0 , y) ≤ km for all x ∈ Bx0 and y ∈ By0 . This shows that the family {GΩ (x, ·); x ∈ Bx0 } is uniformly bounded on By0 . Since GΩ (x, y) = ux (y) + hx (y) and ux (y) is bounded on Bx0 × By0 , the family of harmonic functions {hx ; x ∈ Bx0 } is uniformly bounded on By0 and equicontinuous on any compact subset of By0 according to Theorem 2.2.4. Since hx (y) is a symmetric function, |GΩ (x, y) − GΩ (x0 , y0 )| ≤ |ux (y) − ux0 (y0 )| + |hx (y) − hx0 (y0 )| ≤ |ux (y) − ux0 (y0 )| + |hx (y) − hx (y0 )| + |hy0 (x) − hy0 (x0 )|. The ﬁrst term on the right can be made arbitrarily small by the joint continuity of ux (y) at (x0 , y0 ), the second term can be made arbitrarily small by the equicontinuity of the family {hx ; x ∈ Bx0 }, and the last term can be made arbitrarily small by the continuity of hy0 at x0 . Thus, GΩ (x, y) is continuous at − ⊂ Ω, and let νx,δ be a unit mass uniformly distributed (x0 , y0 ). Now let Bx,δ on ∂Bx,δ . Consider any z0 ∈ Ω ∼ ∂Bx,δ . If z is not in the component of Ω containing Bx,δ , then GΩ (z, ·) = 0 on ∂Bx,δ for all z near z0 and L(GΩ (z, ·) : x, δ)

3.3 Symmetry of the Green Function

121

is obviously continuous at z0 . Suppose now that z0 and Bx,δ belong to the same component Ω0 of Ω. Choose ρ > 0 such that Bz−0 ,ρ ∩ ∂Bx,δ = ∅. By the preceding corollary, the functions GΩ (z, ·), z ∈ Bz0 ,ρ , are harmonic and positive on Ω0 ∼ Bz−0 ,ρ . By Harnack’s inequality, Theorem 2.2.2, there is a constant K such that GΩ (z, y) ≤ KGΩ (z0 , y), y ∈ ∂Bx,δ and z ∈ Bz0 ,ρ . Since GΩ (z0 , ·) is bounded on ∂Bx,δ , the functions GΩ (z, ·) are dominated on ∂Bx,δ by the integrable function GΩ (z0 , ·). By the Lebesgue dominated convergence theorem, L(GΩ (z, ·) : x, δ) = GΩ (z, y) dνx,δ is continuous at z0 . Finally, suppose that z0 ∈ ∂Bx,δ . Since the functions L(GΩ (z, ·) : x, δ + r), r > 0 are continuous at z0 and increase to L(GΩ (z, ·) : x, δ) as r ↓ 0, L(GΩ (z, ·) : x, δ) is l.s.c. at z0 ; likewise, the functions L(GΩ (z, ·) : x, δ − r) are continuous at z0 and decrease to L(GΩ (z·) : x, δ) as r ↓ 0, so that L(GΩ (z·) : x, δ) is u.s.c. at z0 . Thus, L(GΩ (z, ·) : x, δ) is continuous at z0 . According to Theorem 3.2.10, every open subset Ω of Rn , n ≥ 3, has a Green function. The following theorem will help to narrow the class of open subsets of R2 without Green functions. Theorem 3.3.13 If Ω ⊂ R2 is a nonconnected open set, then Ω is Greenian. Proof: Let {Ωi } be the components of Ω, of which there are at least two. Suppose x ∈ Ω1 . By Theorem 3.2.13, Ω1 has a Green function; that is, there is a nonnegative superharmonic function vx on Ω1 of the form vx = ux + hx where hx is harmonic on Ω1 . Note that ux is harmonic on the remaining components of Ω. Extend hx to Ω by putting hx = −ux on Ω ∼ Ω1 . Then ux + hx ∈ Bx and Bx = ∅. Since this can be done for each component of Ω, Bx = ∅ for all x ∈ Ω and Ω has a Green function. Combining this theorem with Theorem 3.2.13, only the connected dense subsets of R2 present a problem in so far as the existence of a Green function is concerned. The boundary behavior of the Green function can be related to regular boundary points for the Dirichlet problem. Lemma 3.3.14 If Ω is a bounded open set, then GΩ (x, ·) = ux − Hux . Proof: Since ux is in the upper class deﬁning Hux , ux − Hux ≥ 0 on Ω. By Lemma 3.2.9, ux − Hux ≥ GΩ (x, ·) on Ω. Also from the deﬁnition of the Green function, GΩ (x, ·) = ux + hx , where hx is harmonic on Ω. Since GΩ (x, ·) ≥ 0, ux ≥ −hx on Ω and −hx belongs to the lower class deﬁning Hux ; that is, −hx ≤ Hux . Moreover, ux − Hux ≥ ux + hx implies that −hx ≥ Hux . This shows that −hx = Hux and that GΩ (x·) = ux − Hux . Theorem 3.3.15 Let Ω be a bounded open set. Then x ∈ ∂Ω is a regular boundary point for Ω if and only if limy→x,y∈Ω GΩ (z, y) = 0 for all z ∈ Ω.

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3 Green Functions

Proof: Suppose x is a regular boundary point for the Dirichlet problem. Consider any z ∈ Ω. Since uz is continuous at x, limy→x,y∈Ω GΩ (z, y) = uz (x) − limy→x,y∈Ω Huz = uz (x)− uz (x) = 0. Now suppose that limy→x,y∈Ω GΩ (z, y) = 0 for all z ∈ Ω. Let {Ωj } be the set of components of Ω andlet zj be a ﬁxed point of Ωj , j ≥ 1. It is easy to see that the function w = j GΩ (zj , ·) is a barrier at x.

3.4 Green Potentials Physically, a potential function represents the potential energy of a unit mass (charge) at a point of Rn due to a mass (charge) distribution on Rn . As such, potential functions are of interest in their own right. Deﬁnition 3.4.1 If μ is a signed measure on the Greenian set Ω, let GΩ μ(x) = GΩ (x, y) dμ(y), Ω

if deﬁned for all x ∈ Ω. If μ is a measure on Ω and GΩ μ is superharmonic on Ω, then GΩ μ is called the potential of μ. If μ = μ+ − μ− is a signed measure and both GΩ μ+ and GΩ μ− are superharmonic functions, then GΩ μ is called a Green potential. Under suitable conditions on μ, GΩ μ is deﬁned and superharmonic on Ω or a diﬀerence of two such functions. Lemma 3.4.2 If μ is a Borel measure on the Greenian set Ω, then GΩ μ is either superharmonic or identically +∞ on each component of Ω. sequence of open subsets of Ω with compact Proof: Let{Ωj } be an increasing closures Ωj− ⊂ Ω such that Ω = ∞ j=1 Ωj . For each x ∈ Ω, deﬁne mj (x, ·) = min (GΩ (x, ·), j) and uj (x) = mj (x, y) dμ(y). Ωj

Since μ is a Borel measure, μ(Ωj ) ≤ μ(Ωj− ) < +∞ for j ≥ 1. Since GΩ is continuous in the extended sense on Ω × Ω, GΩ (xi , y) → GΩ (x, y) as xi → x and the same is true of mj . Since mj is bounded by j, mj (xi , y) dμ(y) → mj (x, y) dμ(y) = uj (x) uj (xi ) = Ωj

Ωj

as xi → x by the Lebesgue dominated convergence theorem. Thus, each uj is continuous on Ω. Since Ωj ↑ Ω and mj (x, ·) ↑ GΩ (x, ·) as j → ∞, uj ↑ GΩ μ

3.4 Green Potentials

123

as j → ∞ by the Lebesgue monotone convergence theorem. As the limit of an increasing sequence of continuous functions, GΩ μ is l.s.c. on Ω. For any − Bx,δ ⊂ Ω, L(GΩ μ : x, δ) =

1 σn δ n−1

∂Bx,δ

GΩ (z, y) dμ(y) dσ(z).

Ω

Using Tonelli’s theorem and the fact that GΩ (·, y) = GΩ (y, ·), is superharmonic on Ω, 1 GΩ (z, y) dσ(z) dμ(y) L(GΩ μ : x, δ) = n−1 Ω σn δ ∂Bx,δ L(GΩ (·, y) : x, δ) dμ(y) = Ω ≤ GΩ (x, y) dμ(y) = GΩ μ(x). Ω

It follows that GΩ μ is nonnegative on Ω, l.s.c. on Ω, super-mean-valued on Ω, and therefore superharmonic on any component on which it is not identically +∞. Lemma 3.4.3 If μ is a measure on the Greenian set Ω such that μ(Ω) < +∞ and μ(B) = 0 for some closed ball B − ⊂ Ω, then GΩ μ is superharmonic on the component of Ω containing B. Proof: If B = Bx,δ , then GΩ (x, ·) is bounded outside B by Lemma 3.2.6. Since μ(B) = 0, GΩ μ(x) = GΩ (x, y) dμ(y) = GΩ (x, y) dμ(y) < +∞. Ω

Ω∼B

Therefore, GΩ μ is not identically +∞ on the component of Ω containing B and is superharmonic on that component by the preceding lemma. Theorem 3.4.4 If μ is a measure on the Greenian set Ω and μ(Ω) < +∞, then GΩ μ is a potential. Proof: Assume ﬁrst that Ω is connected. Consider any x ∈ Ω and a closed − ⊂ Ω. Then μ = μ|B + μ|Ω∼B where μ|B and μ|Ω∼B ball B − = Bx,δ are measures vanishing on Ω ∼ B and B, respectively. Clearly, GΩ μ = GΩ μ|B +GΩ μ|Ω∼B . Since μ|Ω∼B (B) = 0 and μ|Ω∼B (Ω) < +∞, GΩ μ|Ω∼B is superharmonic on Ω by the preceding lemma. Since B − ⊂ Ω, there is a closed ball B1− ⊂ Ω ∼ B − such that μ|B (B1 ) = 0. Since μ|B (Ω) < +∞, GΩ μ|B is superharmonic on Ω by the preceding lemma. It follows from Theorem 2.4.5 that GΩ μ = GΩ μ|B + GΩ μ|Ω∼B is superharmonic on Ω, assuming that Ω is connected. Suppose now that Ω is not connected, and let Ω = j Ωj ,

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3 Green Functions

where {Ωj } is the countable collection of components of Ω. Let μj = μ|Ωj . Then μj (Ωj ) < +∞. By Corollary 3.3.10, GΩj = GΩ |Ωj ×Ωj . Since GΩ (x, ·) vanishes outside of the component of Ω containing x by the same corollary GΩ μ(x) = GΩ (x, y) dμ(y) = GΩj (x, y) dμ(y) = GΩj μj (x) Ω

Ωj

for all x ∈ Ωj . By the ﬁrst part of the proof, GΩj μj is superharmonic on Ωj and therefore GΩ μ is superharmonic on Ω. Theorem 3.4.5 Let Ω be a Greenian set and let GΩ μ be the potential of a measure μ. Then the greatest harmonic minorant of GΩ μ is the zero function. Proof: Let u = GΩ μ and let {Bj } be a sequence of balls satisfying the requirements for deﬁning the reduction u∞ of the superharmonic function u over Ω. By Theorem 3.3.5, u∞ is the greatest harmonic minorant of u so that the assertion requires proving that u∞ = 0 on Ω. Let {uj } and {GΩ,j (·, y)} be the sequences of functions deﬁning u∞ and GΩ,∞ (·, y), respectively. By Theorem 3.3.7, the sequence {GΩ,j (·, y)} decreases to zero on Ω. The rest of the proof is based on the following result which will be proved using mathematical induction. uj (x) = GΩ,j (x, y) dμ(y), x ∈ Ω, j ≥ 1. (3.4) Ω

The ﬁrst step in the induction proof will be omitted because of its similarity with the general case. By deﬁnition, PI(uj−1 : Bj )(x) if x ∈ Bj uj (x) = uj−1 (x) if x ∈ Ω ∼ Bj , with a similar deﬁnition of GΩ,j (x, y) for ﬁxed y. Assuming that Equation (3.4) holds for j − 1, two cases will be considered. Suppose ﬁrst that x ∈ Bj . Then uj (x) = uj−1 (x), GΩ,j (x, y) = GΩ,j−1 (x, y), and GΩ,j−1 (x, y) dμ(y) = GΩ,j (x, y) dμ(y) uj (x) = uj−1 (x) = Ω

Ω

for x ∈ Bj , j ≥ 1. Now suppose that x ∈ Bj . By Tonelli’s theorem, uj (x) = PI(uj−1 : Bj )(x)

GΩ,j−1 (·, y) dμ(y) : Bj (x) = PI Ω = PI(GΩ,j−1 (·, y) : Bj )(x) dμ(y) Ω = GΩ,j (x, y) dμ(y), x ∈ Bj , j ≥ 1. Ω

3.4 Green Potentials

125

Thus, Equation (3.4) is true for all j ≥ 1. Since GΩ μ is superharmonic on Ω, it is ﬁnite a.e. on Ω. Choose any point x0 ∈ Ω such that GΩ μ(x0 ) < +∞. Then GΩ μ(x0 ) = GΩ (x0 , y) dμ(y) < +∞; Ω

that is, GΩ (x0 , ·) is integrable relative to μ. Since 0 ≤ GΩ,j (x0 , ·) ≤ GΩ (x0 , ·) and limj→∞ GΩ,j (x0 , ·) = 0, GΩ,j (x0 , y) dμ(y) = 0 u∞ (x0 ) = lim uj (x0 ) = lim j→∞

j→∞

Ω

by the Lebesgue dominated convergence theorem. Since u∞ (x0 ) ≥ 0 and is harmonic on Ω, u∞ = 0 on Ω by the minimum principle. Corollary 3.4.6 If Ω is a Greenian set, μ is a measure on Ω, and GΩ μ is harmonic on Ω, then μ is the zero measure. Proof: Let h = GΩ μ be harmonic on Ω. Since h is a harmonic minorant of itself, h = 0 by the preceding theorem. Thus, for any x ∈ Ω, Ω GΩ (x, y) dμ(y) = 0. Since GΩ (x, ·) > 0 on the component of Ω containing x, each component of Ω has μ-measure zero. It is possible for a potential to be harmonic on a proper subset of Ω. Theorem 3.4.7 If μ is a measure on the Greenian set Ω such that GΩ μ is a potential on Ω, then GΩ μ is harmonic on any open set of μ-measure zero. Proof: Let Λ be a nonempty open subset of Ω with μ(Λ) = 0. For any x ∈ Ω, GΩ μ(x) = GΩ (x, y) dμ(y). Ω∼Λ − ⊂ Λ. By Tonelli’s theorem, Suppose Bx,δ

A(GΩ μ : x, δ) = A =

GΩ (·, y) dμ(y) : x, δ

Ω∼Λ

A(GΩ (·, y) : x, δ) dμ(y).

Ω∼Λ − For y ∈ Ω ∼ Λ, GΩ (·, y) is harmonic on Λ. Since Bx,δ ⊂ Λ, A(GΩ (·, y) : x, δ) = GΩ (x, y) and A(GΩ μ : x, δ) = A(GΩ (·, y) : x, δ) dμ(y) Ω∼Λ = GΩ (x, y) dμ(y) Ω∼Λ

= GΩ μ(x).

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3 Green Functions

Since GΩ μ is superharmonic on Ω, it is locally integrable on Ω by Theorem 2.4.2 and, as in the proof of Theorem 1.7.11, continuous on Ω. It follows from Theorem 2.4.1 that GΩ μ is harmonic on Λ. It is possible to compare the Green potential for a region with that of another region, especially when one is a subset of the other. Theorem 3.4.8 Let Ω be a Greenian set, let Λ be an open subset of Ω, and let μ be a measure on Ω such that μ(Ω ∼ Λ) = 0 and GΩ μ is a potential. Then there is a nonnegative harmonic function h on Λ such that GΩ μ = GΛ μ|Λ +h on Λ. Proof: Since Λ ⊂ Ω, GΛ ≤ GΩ on Λ by Theorem 3.2.12. Thus, for each x ∈ Λ, GΛ μ|Λ (x) = GΛ (x, y) dμ(y) ≤ GΩ (x, y) dμ(y) = GΩ μ(x). Λ

Λ

This proves that GΛ μ|Λ is a potential. Since GΩ μ is superharmonic, it is ﬁnite-valued a.e. on Ω. Except possibly for x in a subset of Λ of Lebesgue measure zero, GΩ μ(x) − GΛ μ|Λ (x) = (GΩ (x, y) − GΛ (x, y)) dμ(y). Λ

Since GΩ (x, y) = ux (y) + hx (y) and GΛ (x, y) = ux (y) + h∗x (y), where hx and h∗x are harmonic on Ω and Λ, respectively, GΩ (x, y) − GΛ (x, y) = Hy (x) where Hy (x) is the symmetric function hx (y) − h∗x (y) on Λ × Λ. For each y ∈ Λ, Hy is harmonic on Λ, and for each x ∈ Λ, Hy (x) is harmonic on Λ. − Suppose Bx,δ ⊂ Λ. By Tonelli’s theorem, for x ∈ Λ as described above +∞ > A(GΩ μ − GΛ μ|Λ : x, δ) = A(GΩ (·, y) − GΛ (·, y)) : x, δ) dμ(y) Λ = A(Hy : x, δ) dμ(y) Λ = Hy (x) dμ(y). Λ

The last integral is harmonic on Λ by Theorem 1.7.13. Writing the last equation as A(GΩ μ : x, δ) = A(GΛ μ|Λ : x, δ) + Hy (x) dμ(y) Λ

and noting that the solid ball averages increase to GΩ μ(x) and GΛ μ|Λ (x), respectively, as δ ↓ 0 by Lemma 2.4.4, Hy (x) dμ(y) GΩ μ(x) = GΛ μ|Λ (x) + Λ

on Λ. The above theorem can be used to prove a converse to Theorem 3.4.7.

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127

Theorem 3.4.9 Let Ω be a Greenian set. If Gμ is the potential of a measure μ and is harmonic on the open set Λ ⊂ Ω, then μ(Λ) = 0. Proof: Since GΩ μ = GΩ μ|Λ + GΩ μ|Ω∼Λ and GΩ μ|Ω∼Λ is harmonic on Λ by Theorem 3.4.7, GΩ μ|Λ is harmonic on Λ. By the preceding theorem, GΩ μ|Λ = GΛ μ|Λ + h where h is harmonic on Λ. This implies that GΛ μ|Λ is harmonic on Λ, and by Corollary 3.4.6, μ|Λ (Λ) = 0 = μ(Λ). Historically, the Green potential was preceded by the logarithmic potential in the n = 2 case and the Newtonian potential in the n ≥ 3 case. The logarithmic potential of a measure μ on R2 having compact support is deﬁned by μ U (x) = − log |x − y| dμ(y), x ∈ R2 ; the Newtonian potential Uμ of a measure on Rn , n ≥ 3, is deﬁned by Uμ (x) =

1 dμ(y), |x − y|n−2

x ∈ Rn .

Since the Newtonian potential Uμ of a measure is just the Green potential Gμ, where G is the Green function for Rn , the preceding results apply directly to Uμ in the n ≥ 3 case; for example, if μ(R3 ) < +∞, then Uμ is superharmonic on R3 . A similar argument for the logarithmic potential cannot be made since R2 does not have a Green function. In order to show that Uμ is superharmonic in the n = 2 case when μ has compact support, let Bx0 ,δ be any ball and let C be the support of μ. Since − log |x − y| is continuous in the extended sense on Bx−0 ,δ × C, there is a constant c such that − log |x − y| + c ≥ 0 for x ∈ Bx−0 ,δ , y ∈ C. For x ∈ Bx−0 ,δ , (− log |x − y| + c) dμ(y).

Uμ (x) + cμ(C) =

As in the proof of Lemma 3.4.2, Uμ is l.s.c. on R2 . Calculating spherical averages and using Tonelli’s theorem, μ

L(U : x0 , δ) + cμ(C) = ≤

L(− log |x − y| + c : x0 , δ) dμ(y) (− log |x0 − y| + c) dμ(y)

= Uμ (x0 ) + cμ(C). Therefore, Uμ (x0 ) ≥ L(Uμ : x0 , δ) for any ball Bx0 ,δ and Uμ is super-meanvalued. Using the compactness of the support of μ, it can be seen that Uμ cannot take on the value −∞. If x0 ∈ C, then − log |x0 − y| is bounded on C and Uμ (x0 ) < +∞. This shows that Uμ is not identically +∞ on R2 and

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3 Green Functions

therefore that Uμ is superharmonic on R2 whenever μ has compact support. The potentials Uμ and GΩ μ can be compared in a sense. Theorem 3.4.10 If Ω ⊂ Rn , n ≥ 2, is Greenian and μ is a measure with compact support, then Uμ is superharmonic and diﬀers from GΩ μ on Ω by a harmonic function. Proof: Letting GΩ (x, ·) = ux + hx , GΩ μ(x) = ux (y) dμ(y) + hx (y) dμ(y),

x ∈ R2 .

Since hx is bounded on the support of μ, hx (y) dμ(y) < +∞ and the second term of the above equation is harmonic on Ω by Theorem 1.7.13. Thus, GΩ μ = Uμ + h with h harmonic on Ω. As an application of the results of this section, a region Ω ⊂ R3 having an irregular boundary point will be constructed using a potential function. Example 3.4.11 (Lebesgue Spine[37, 38]) Consider a measure concentrated on the line segment {(x, 0, 0); 0 ≤ x ≤ 1} and having a density f (x, 0, 0) = x, 0 ≤ x ≤ 1. The Newtonian potential u(x, y, z) of this measure at a point not on the x-axis is given by

1

u(x, y, z) = 0

=

0

1

ξ dξ ((ξ − x)2 + y 2 + z 2 )1/2 (ξ − x) dξ + x ((ξ − x)2 + y 2 + z 2 )1/2

1 0

1 dξ. ((ξ − x)2 + y 2 + z 2 )1/2

Using elementary calculus, the integrals can be evaluated to obtain u(x, y, z) = ((1 − x)2 + y 2 + z 2 )1/2 − (x2 + y 2 + z 2 )1/2 +x log (1 − x + ((1 − x)2 + y 2 + z 2 )1/2 ) −x log (−x + (x2 + y 2 + z 2 )1/2 ) = ((1 − x)2 + y 2 + z 2 )1/2 − (x2 + y 2 + z 2 )1/2 +x log (1 − x + ((1 − x)2 + y 2 + z 2 )1/2 ) +x log (x + (x2 + y 2 + z 2 )1/2 ) − x log (y 2 + z 2 ), the last term arising as a result of multiplying and dividing the expression 1 1 −x + (x2 + y 2 + z 2 ) 2 by x + (x2 + y 2 + z 2 ) 2 . It is easily seen that u has the limit 1 as (x, y, z) → (0, 0, 0) subject to the condition that y 2 + z 2 > 0, has the limit 0 as |(x, y, z)| → +∞, and has the limit +∞ as (x, y, z) approaches any point on the line segment {(x, 0, 0); 0 < x < 1}. Consider the open set Ω = {(x, y, z); < u(x, y, z) < 1 + c0 },

3.4 Green Potentials

129

where 0 < < 1 and c0 > 1. The set {(x, y, z); u(x, y, z) > } is a neighborhood of (0, 0, 0) and the set {(x, y, z); u(x, y, z) < 1 + c0 } is an open set which excludes the line segment {(x, 0, 0); 0 < x < 1}. It is easily seen that lim

|(x,y,z)|→0

u(x, y, z) = 1 −

lim

|(x,y,z)|→0

x log ρ2

where ρ2 = y 2 + z 2 . If (x, y, z) → (0, 0, 0) along a path for which x > 0, y 2 + z 2 = 0 and xk = ρ2 for some positive integer k, then u(x, y, z) → 1; note also that such points will be in Ω if suﬃciently close to (0, 0, 0). If, however, (x, y, z) → (0, 0, 0) along a path for which x > 0, y 2 + z 2 = 0 and ρ = exp (−c/2x) with c > c0 , then u(x, y, z) → 1 + c; note also that such points will be in ∼ Ω if suﬃciently close to (0, 0, 0). It follows that (0, 0, 0) is a boundary point for Ω. It remains only to show that (0, 0, 0) is an irregular boundary point for the region Ω. The function u, as the Newtonian potential of a measure with compact support, is harmonic on the open set Ω. As such, it is the solution of the Dirichlet problem for the boundary function which is equal to on the boundary of the region u > and equal to 1 + c0 on the boundary of the region u < 1 + c0 . For x < 0, 1 ξ dξ = 1 + x log (1 − x) − x log |x| u(x, 0, 0) = ξ − x 0 so that limx→0− u(x, 0, 0) = 1 < 1 + c0 . In general, little can be said about the continuity of a Green potential GΩ μ unless the measure μ is absolutely continuous with respect to Lebesgue measure. If Ω is a Greenian set and f is a measurable function on Ω, let GΩ f (x) = GΩ (x, y)f (y) dy Ω

provided the integral is deﬁned for all x ∈ Ω. Theorem 3.4.12 If Ω is a Greenian set and f is a bounded Lebesgue integrable function on Ω, then GΩ f is continuous on Ω. Proof: Since the positive and negative parts f + and f − are integrable over Ω and GΩ f = GΩ f + − GΩ f − , it can be assumed that 0 ≤ f ≤ m for some constant m. By Theorem 3.4.4, GΩ f is a potential. Let B = By,ρ be a ball with B − ⊂ Ω of radius less that 1/2. Then GΩ f = GΩ f χB + GΩ f χΩ∼B . Since the density f χΩ∼B assigns zero measure to B, GΩ f χΩ∼B is harmonic on B by Theorem 3.4.7 and therefore continuous on B. Consider now GΩ f χB . If it can be shown that GΩ f χB is continuous on B ∼ {y}, it would follow that GΩ f is continuous on B ∼ {y}; but since B = By,ρ is an arbitrary ball with closure in Ω, this would imply that GΩ f is continuous on Ω. Since GΩ f χB diﬀers from GB f by a function harmonic on B by Theorem 3.4.8, it suﬃces to prove that GB f is continuous on B ∼ {y}. It is necessary to consider the

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3 Green Functions

n = 2 and n ≥ 3 cases separately for the rest of the proof. Suppose n = 2 and x0 ∈ B ∼ {y} and consider any ball Bx0 ,r with closure in B. Since GB (x, z) = log

|y − x| |z − x∗ | ρ |z − x|

for z ∈ B ∼ {x}, where x∗ is the inverse of x relative to ∂B, |GB f (x) − GB f (x0 )| |y − x| |z − x∗ | |z − x0 | ≤ f (z) dz log ∗| B∼Bx ,r |y − x | |z − x| |z − x 0 0 0 + GB (x, z)f (z) dz + GB (x0 , z)f (z) dz. Bx0 ,r

Bx0 ,r

Recall that GB (x, z) is minimal in the class of nonnegative functions of the form − log |x − z| + wx (z) where wx is harmonic on B. Since |x − z| < 1 for x, z ∈ B, − log |x − z| is positive and majorizes GB (x, z). Thus for any x ∈ B, GB (x, z)f (z) dz ≤ m − log |x − z| dz. Bx0 ,r

Bx0 ,r

The integral on the right becomes larger if taken over Bx,r rather than Bx0 ,r . Thus, GB (x, z)f (z) dz ≤ m − log |x − z| dz Bx0 ,r

Bx,r r

= 2πm 0

−ρ log ρ dρ.

Since the latter integral approaches zero as r → 0, given > 0, r0 can be chosen so that r0 < |x0 − y| and GB (x, z)f (z) dz + GB (x0 , z)f (z) dz < Bx0 ,r0

Bx0 ,r0

for any x ∈ B. It follows that |GB f (x) − GB f (x0 )| |y − x| |z − x∗ | |z − x0 | f (z) dz + . ≤ log ∗ B∼Bx0 ,r0 |y − x0 | |z − x| |z − x0 | Consider only those x ∈ Bx0 ,r0 /2 . Since the logarithmic factor in the integral approaches zero boundedly as x → x0 , by the Lebesgue dominated convergence theorem,

3.4 Green Potentials

131

lim sup |GB f (x) − GB f (x0 )| ≤ .

x→x0 ,x∈B

Since is arbitrary, limx→x0 ,x∈B GB f (x) = GB f (x0 ) and GB f is continuous on B ∼ {y}. The proof of the n ≥ 3 case is just the same except for the fact that GB (x, z) ≤ |x − z|−n+2 and the estimate r GB (x, z)f (z) dz ≤ σn ρ dρ. Bx0 ,r

0

A digression into approximation theory is necessary in order to simplify later proofs. Lemma 3.4.13 If φ is a ﬁnite-valued, continuous function on the Greenian set Ω with compact support in a closed ball B − ⊂ Ω and B1 is a ball with B − ⊂ B1 ⊂ B1− ⊂ Ω, then there is a sequence of diﬀerences {uj − vj } of ﬁnite-valued, continuous potentials on Ω supported by measures in B1 such that uj = vj on Ω ∼ B1 and limj→∞ (uj − vj ) = φ uniformly on Ω. Proof: (n ≥ 3) Put φ(x) = 0 for x ∈ Rn ∼ Ω, B = Bx,δ and B1 = Bx,δ1 . Consider any j0 ≥ 1 for which 1/j0 < (δ1 − δ)/4 and the ball Bx,δ where δ = (δ1 + δ)/2. For j ≥ j0 , deﬁne J1/j φ to be zero on Ω ∼ B1 . Then {J1/j φ}j≥j0 is a sequence in C0∞ (Rn ) which converges uniformly to φ on Ω. By Theorem 1.5.3, for x ∈ B1 1 J1/j φ(x) = − GB1 (x, z)(ΔJ1/j φ)(z) dz σn (n − 2) B1 1 = GB1 (x, z)(ΔJ1/j φ)− (z) dz σn (n − 2) B1 1 GB1 (x, z)(ΔJ1/j φ)+ (z) dz − σn (n − 2) B1 1 1 GB1 (ΔJ1/j φ)− − GB1 (ΔJ1/j φ)+ . = σn (n − 2) σn (n − 2) The functions on the right are potentials of measures having supports in Bx,δ and are continuous by Theorem 3.4.12. By Theorem 3.4.8, J1/j φ =

1 1 GΩ (ΔJ1/j φ)− − GΩ (ΔJ1/j φ)+ + hj σn (n − 2) σn (n − 2)

(3.5)

where hj is harmonic on B1 . Since J1/j φ = 0 on Ω ∼ Bx,δ and GΩ (ΔJ1/j φ)± are harmonic on Ω ∼ Bx,δ , hj has a harmonic extension to Ω. Letting uj =

1 1 GΩ (ΔJ1/j φ)− and vj = GΩ (ΔJ1/j φ)+ , σn (n − 2) σn (n − 2)

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3 Green Functions

J1/j φ = uj − vj + hj . Since J1/j φ = 0 on Ω ∼ Bx,δ , vj − uj = hj on Ω ∼ Bx,δ . Thus, vj = uj +hj ≥ hj on Ω ∼ Bx,δ and, in particular, vj ≥ hj on ∂Bx,δ . By the minimum principle, vj ≥ hj on Bx,δ and therefore vj ≥ hj on Ω. Since the greatest harmonic minorant of vj on Ω is the zero function, hj ≤ 0 on Ω. A similar argument applied to uj shows that hj ≥ 0. Thus, hj = 0 on Ω and J1/j φ = uj − vj . Since the sequence {J1/j φ} converges uniformly to φ, the same is true of the sequence {uj − vj }. Finally, it follows from Equation (3.5) that uj = vj on Ω ∼ B1 since J1/j φ = 0 on Ω ∼ B1 and hj = 0 on Ω. The result of the following theorem is known as a partition of unity. The proof is taken from [48]. Theorem 3.4.14 (L. Schwartz [55]) Let Γ be a compact subset of Rn and let {U1 , . . . , Um } be an open covering of Γ . For j = 1, . . . , m, there is a φj ∈ C0∞ (Uj ) with 0 ≤ φj ≤ 1 such that x∈Γ ⇒

m

φj (x) = 1.

j=1

Proof: For j = 1, . . . , m, let Vj be an open subset of Uj with compact closure Vj− ⊂ Uj such that {V1 , . . . , Vm } is an open covering of Γ . Then dist(Γ, ∼ ∪m j=1 Vj ) > 0, − and for each j = 1, . . . , m, dist(Vj− , ∂Uj ) > 0. Let Vm+1 =∼ ∪m j=1 Vj and choose δ > 0 such that

dist(Γ, ∼ ∪m j=1 Vj ) > δ dist(Vj− , ∂Uj ) > δ,

j = 1, . . . m.

Each Vj , j = 1, . . . , m+1, can be covered by balls with centers yji ∈ Vj− , i ≥ 1, of radius δ so that at most ﬁnitely many of the balls intersect any compact set. Choosing the constant c of Section 2.5 so that m(0) = 1, let x − yji ψj (x) = m , j = 1, . . . , m + 1. δ i Then each ψj ∈ C0∞ (Uj ) and x∈Γ ⇒

m

ψj (x) = 0, ψm+1 (x) = 0.

j=1

Letting ψj (x) , φj (x) = m j=1 ψj (x)

x ∈ Rn , j = 1, . . . , m,

the functions φ1 , . . . , φm satisfy the above requirements.

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133

Theorem 3.4.15 If φ is a ﬁnite-valued, continuous function on Rn having compact support Γ in the Greenian set Ω and O is a neighborhood of Γ with compact closure O− ⊂ Ω, then there is a sequence of diﬀerences {uj − vj } of ﬁnite-valued, continuous potentials of measures having support in O such that uj = vj on Ω ∼ O− and limj→∞ (uj − vj ) = φ uniformly on Ω. − Proof: Let B1 , . . . , Bm be a ﬁnite collection of balls covering Γ with ∪m j=1 Bj ∞ n ⊂ O. Consider a partition of unity ψ1 , . . . , ψm where each ψj ∈ C0 (R ) has m m compact support in Bj and j=1 ψj (x) = 1 for x ∈ Γ . Then φ = j=1 φψj where each φψj is ﬁnite-valued and continuous with compact support in Bj . For j = 1, . . . , m, there is a sequence of diﬀerences {uij − vji } of ﬁnite-valued, continuous potentials such that uij = vji on Ω ∼ Bj ⊃ Ω ∼ ∪m j=1 Bj ⊃ Ω ∼ O− , and limi→∞ (uij − vji ) = φψj uniformly on Ω. Thus, ⎛ ⎞ m m uij − vji ⎠ = φ lim ⎝ i→∞

m

j=1

j=1

m

uniformly on Ω and j=1 uij = j=1 vji on Ω ∼ O− . Generally speaking, the boundary behavior of a potential on a region having a Green function reﬂects the boundary behavior of the Green function. Theorem 3.4.16 If μ is a measure with compact support Γ in an open ball B, then limx→x0 ,x∈B GB μ(x) = 0 for x0 ∈ ∂B. Proof: Let B = By,ρ and let z0 be a ﬁxed point of Γ . Also let Λ be a connected neighborhood of Γ such that Λ− ⊂ B. For x ∈ B ∼ Λ− , the functions GB (x, ·) are strictly positive harmonic functions on Λ. By Harnack’s inequality, Theorem 2.2.2, there is a constant k (depending upon Λ and Γ ) such that GB (x, z) ≤ kGB (x, z0 ) for all x ∈ B ∼ Λ− , z ∈ Γ . Since GB (x, z0 ) → 0 as x → x0 ∈ ∂B by Lemma 3.2.1, GB (x, z) → 0 as x → x0 ∈ ∂B uniformly for z ∈ Γ . Since the integrand in the equation GB μ(x) = Γ GB (x, z) dμ(z) can be made uniformly small as x → x0 , limx→x0 GB μ(x) = 0. The requirement that the support of μ be compact can be relaxed provided it has a density. Lemma 3.4.17 Let Ω be a Greenian connected set with ﬁnite Lebesgue measure if n ≥ 3 and bounded if n = 2, and let x0 ∈ ∂Ω. If f is a bounded measurable function on Ω and limx→x0 ,x∈Ω GΩ (x, z0 ) = 0 for some z0 ∈ Ω, then limx→x0 ,x∈Ω GΩ f (x) = 0. Proof: Suppose n ≥ 3. By consideration of the positive and negative parts of f , it can be assumed that 0 ≤ f ≤ m. Since f is integrable on Ω, GΩ f is a potential by Theorem 3.4.4. Now let {Γj } be an increasing sequence of compact subsets of Ω such that Ω = ∪j Γj and let GΩ (x, z)f (z) dz. φj (x) = Γj

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3 Green Functions

Then φj ↑ GΩ f by the Lebesgue monotone convergence theorem. Given > 0, choose j0 such that the Lebesgue measure λ0 of Ω ∼ Γj0 is less than and let α denote the Lebesgue measure of Ω. Since GΩ (x, z) is majorized by |x − z|−n+2 , |φj0 +j (x) − φj0 (x)| ≤

Γj0 +j ∼Γj

≤m Ω∼Γj0

1 f (z) dz |x − z|n−2 1 dz. |x − z|n−2

Now choose ρ such that the volume of Bx,ρ is λ0 ; that is, λ0 = νn ρn or ρ = (λ0 /νn )1/n . Then |φj0 +j (x) − φj0 (x)| ≤ m Ω∼Γj0

≤m

Bx,ρ

1 dz |x − z|n−2

1 dz |x − z|n−2

1 λ0 1 1 = σn mρ2 = σn m( )2/n ≤ σn m( )2/n 2 2 νn 2 νn uniformly with respect to x ∈ Ω and j ≥ 1. As in the preceding proof, limx→x0 ,x∈Ω GΩ (x, z) = 0 uniformly with respect to z ∈ Γj0 and there is a δ > 0 such that GΩ (x, z) < for all z ∈ Γj0 and all x satisfying |x − x0 | < δ; for such x, |φj0 (x)| ≤ GΩ (x, z)f (z) dz ≤ mα. Γj0

Thus, if |x − x0 | < δ and j ≥ 1, then |φj0 +j (x)| ≤ |φj0 +j (x) − φj0 (x)| + |φj0 (x)| 1 ≤ σn m( )2/n + mα. 2 νn Letting j → ∞, |GΩ f (x)| ≤

1 σn m( )2/n + mα; 2 νn

but since > 0 is arbitrary and the constants on the right do not depend upon and x, GΩ f (x) can be made arbitrarily small by taking x suﬃciently close to x0 . The proof in the n = 2 case is precisely the same except that GΩ (x, z) need not be majorized by − log |x − z|. Assuming that Ω is bounded, there is a constant c such that log (c/|x − z|) is positive on Ω and therefore majorizes GΩ (x, z).

3.5 Riesz Decomposition

135

3.5 Riesz Decomposition It will be shown in this section that a nonnegative superharmonic function can be represented as the sum of a potential and a harmonic function. A potential theory version of Tonelli’s theorem will be proved ﬁrst. Theorem 3.5.1 (Reciprocity Theorem) Let Ω be a Greenian subset of Rn , and let μ, ν be two measures on the Borel subsets of Ω. Then Gμ dν = Gν dμ. Ω

Ω

Proof: By Tonelli’s theorem and the symmetry of the Green function, Gμ(x) dν(x) = G(x, y) dμ(y) dν(x) Ω Ω Ω G(x, y) dν(x) dμ(y) = Ω Ω = G(y, x) dν(x) dμ(y) Ω Ω = Gν(y) dμ(y). Ω

Note that the reciprocity theorem holds even if Gμ is not integrable relative to ν, in which case the same is true of Gν relative to μ. A digression is necessary in order to develop some tools for showing that two measures are equal if they produce the same potential function. K + will denote the subset of C 0 (Ω) consisting of nonnegative, continuous functions with compact support. Deﬁnition 3.5.2 A subset K0+ ⊂ K + is called total if for each v ∈ K + with compact support Γ , each neighborhood Λ of Γ , and each > 0 there p corresponds a ﬁnite linear combination f = j=1 αj fj of elements of K0+ with αj > 0, fj ∈ K0+ , and fj having compact support in Λ for each j such that

v − f = sup |v(x) − f (x)| < . x∈Ω

If it is necessary to consider more than one region, the notation K + (Ω) will be used. Theorem 3.5.3 (Cartan [10]) Let K0+ be a subset of K + with the following properties: (i)

If f ∈ K0+ , then every translate of f is in K0+ .

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3 Green Functions

(ii) If x ∈ Ω and δ > 0, then there is an f ∈ K0+ that vanishes outside Bx,δ but does not vanish identically. Then K0+ is a total subset of K + . Proof: Without loss of generality, it can be assumed that 0 ∈ Ω. Let v be an element of K + with compact support Γ . If v is identically zero and > 0, then a scalar multiple of any function satisfying (ii) will approximate v uniformly on Ω within . It therefore can be assumed that v is not identically zero. Let Λ be a neighborhood of Γ with Λ− ⊂ Ω, let λ = d(Γ, Ω ∼ Λ) > 0, and let > 0. Since v is uniformly continuous on Ω, there is a δ < λ/2 such that |v(x) − v(y)| < /2 whenever |x − y| < δ. Then d(Γ, Ω ∼ Λ) > 2δ. By (ii), there is an fδ ∈ K0+ that vanishes outside B0,δ , does not vanish identically, and fδ (z) dz = 1. Since fδ is uniformly continuous, there is a ρ > 0 such that |fδ (x) − fδ (y)| < 2 v(z) dz whenever |x − y| < ρ. It can be assumed that δ + ρ < λ. Moreover, there is a sequence {Mj } of disjoint measurable sets such that Rn = ∪j Mj , diam Mj < ρ, and thus δ+ diam Mj < δ + ρ < λ. It can be assumed that only ﬁnitely many of the Mj have a nonempty intersection with Γ . Such a sequence of Mj ’s can be constructed by covering Rn by a sequence of balls each of diameter less than ρ with only a ﬁnite number intersecting Γ and then forming a disjoint sequence of measurable sets from the sequence of balls. Deﬁne the real-valued function h by h(x) = fδ (x − y)v(y) dy, x ∈ Rn . Then

|h(x) − v(x)| ≤ fδ (y)|v(y) − v(x)| dy ≤ 2 Bx,δ

fδ (y) dy = Bx,δ

2

for all x ∈ Rn ; that is, h approximates v uniformly within /2. It will now be shown that h can be approximated uniformly within /2 by a ﬁnite linear combination of translates of fδ . For each j ≥ 1, choose yj ∈ Mj . Then for any x ∈ Rn , h(x) − fδ (x − yj ) v(y) dy j

Mj

= fδ (x − y)v(y) dy − fδ (x − yj ) v(y) dy Mj j (fδ (x − y) − fδ (x − yj )v(y) dy . = j

Mj

3.5 Riesz Decomposition

137

Since diam Mj < ρ, |(x − y) − (x − yj )| = |yj − y| < ρ for y ∈ Mj and the ﬁrst factor in the integral is bounded in absolute value by /(2 v(y) dy). Therefore, h(x) − fδ (x − yj ) v(y) dy ≤ v(y) dy = . 2 2 v(y) dy Mj Mj j j This shows that h is approximated uniformly within /2 by the function f (x) = fδ (x − yj ) v(y) dy Mj

j

so that f approximates v uniformly within . Note that the sum in the last equation is a ﬁnite sum since only a ﬁnite number of the Mj have a nonempty intersection with the support of v. It remains only to show that each function gj (x) = fδ (x − yj ) v(y) dy Mj

has support in Λ provided it is not identically zero. Suppose Mj ∩ Γ = ∅ and fδ (x−yj ) = 0. Then |x−yj | < δ and d(x, Γ ) ≤ d(x, yj )+d(yj , Γ ) < δ +ρ < λ. Since λ = d(Γ, Ω ∼ Λ), x ∈ Ω ∼ Λ. This means that the support of those gj that are not identically zero is contained in Λ. Theorem 3.5.4 If Ω is an open subset of Rn, μ1 and μ2 are measures on + + Ω, K0 is a total subset of K , and Ω f dμ1 = Ω f dμ2 for all f ∈ K0+ , then μ1 = μ2 . Proof: Since any g ∈ K + can be approximated uniformly by a ﬁnite linear combination of elements of K0+ , the above equation holds for all f ∈ K + . Let Γ be a compact subset of Ω and let χΓ be its indicator function. Then there is a sequence {φj } in K + such that φj ↓ χΓ with 0 ≤ φj ≤ 1; for example, let φj (x) = max (2e−jd(x,Γ ) − 1, 0). Since Ω φj dμ1 = Ω φj dμ2 , j ≥ 1, μ1 (Γ ) = χΓ dμ1 = χΓ dμ2 = μ2 (Γ ) Ω

Ω

by the Lebesgue dominated convergence theorem. Hence, μ1 = μ2 since both are regular Borel measures. The following examples will be needed to construct total sets of continuous functions. − ⊂ Ω, δ > 0, and Example 3.5.5 (n=2) Let Ω be a Greenian set, let Bx,δ let νx,δ be a unit measure uniformly distributed on ∂Bx.δ . Letting 1 dνx,δ (z), log y ∈ R2 , ux,δ = |y − z| ∂Bx,δ

138

3 Green Functions

according to Lemma 1.6.1 ux,δ (y) =

log 1δ

if |y − x| ≤ δ

1 log |x−y|

if |y − x| > δ.

Putting τx,δ = νx,δ/2 − νx,δ , GΩ τx,δ is a continuous potential with ⎧ on Ω ⎨≥ 0 on Ω ∼ Bx,δ GΩ τx,δ (y) = 0 ⎩ ≡ 0.

(3.6)

To see this, consider ux,δ/2 − ux,δ . Then (i) for y ∈ Bx,δ/2 , ux,δ/2 (y) − ux,δ (y) = log

1 1 − log = log 2 > 0; (δ/2) δ

− (ii) for y ∈ Bx,δ ∼ Bx,δ/2 ,

ux,δ/2 (y) − ux,δ (y) = log

1 |x − y| 1 − log = − log ≥ 0; |x − y| δ δ

− (iii) for y ∈ Bx,δ ,

ux,δ/2 (y) − ux,δ (y) = log

1 1 − log = 0. |x − y| |x − y|

Since GΩ (y, ·) = uy + hy , where hy is harmonic on Ω, (uy (z) + hy (z)) dνx,δ/2 − (uy (z) + hy (z)) dνx,δ (z) GΩ τx,δ (y) = Ω Ω = ux,δ/2 (y) − ux.δ (y) + hy (z) dνx,δ/2 (z) − hy (z) dνx,δ (z). Ω

Ω

Since the last two integrals are just averages of the harmonic function hy , their diﬀerence is equal to hy (x) − hy (x) = 0. Thus, GΩ τx,δ (y) = ux,δ/2 (y) − ux,δ (y),

y∈Ω

and GΩ τx,δ has the properties listed in (3.6). Since it is easily seen that a translate of GΩ τx,δ is again a potential of the same kind, the collection K0+ − of potentials {GΩ τx,δ ; Bx,δ ⊂ Ω} is a total subset of K + . − Example 3.5.6 (n ≥ 3) Let Ω be a Greenian set, let Bx,δ ⊂ Ω, δ > 0, and let νx,δ/2 and νx,δ be deﬁned as in the preceding example. If τx,δ = νx,δ/2 − νx,δ , then the GΩ τx,δ have the same properties as in the preceding example.

3.5 Riesz Decomposition

139

Theorem 3.5.7 (Cartan [10],[11],[12]) If Ω is a Greenian set, then the − ⊂ Ω} is a total subset of collection K0+ of Green potentials {GΩ τx,δ ; Bx,δ + K . Proof: There is nothing to prove in view of the two preceding examples and Theorem 3.5.3. Theorem 3.5.8 Let μ and ν be measures on the Greenian set Ω such that GΩ μ and GΩ ν are potentials. If GΩ μ = GΩ ν at points where both are ﬁnite, then μ = ν. More generally, let Λ be an open subset of Ω such that GΩ μ = GΩ ν + h on the subset of Λ where both are ﬁnite and where h is harmonic on Λ. Then μ and ν are identical on the Borel subsets of Λ. Proof: Let Σ be the set of points in Λ where GΩ μ and GΩ ν are both ﬁnite. For x ∈ Σ, GΩ μ(x) = GΩ ν(x) + h(x). Since GΩ μ = GΩ μ|Λ + GΩ μ|Ω∼Λ and GΩ μ|Ω∼Λ is harmonic on Λ, GΩ μ(x) = GΩ μ|Λ (x) + h (x), x ∈ Σ, where h is harmonic on Λ. Applying the same argument to GΩ ν, GΩ ν(x) = ˜ = μ|Λ , GΩ ν|Λ (x) + h (x), x ∈ Σ, where h is harmonic on Λ. Letting μ ν˜ = ν|Λ , GΩ μ ˜(x) = GΩ ν˜(x) + h∗ (x), x ∈ Σ, where h∗ is harmonic on Λ. Since Λ ∼ Σ has Lebesgue measure zero, A(GΩ μ ˜ : x, δ) = A(GΩ ν˜ : x, δ) + h∗ (x) − whenever Bx,δ ⊂ Λ. Letting δ ↓ 0 and applying Lemma 2.4.4, GΩ μ ˜(x) = ∗ GΩ ν˜(x) + h (x) for all x ∈ Λ. Since Λ is an open subset of Ω, it has a Green function GΛ . Applying Theorem 3.4.8 to GΩ μ ˜ and GΩ ν˜, there is a harmonic − function h∗ on Λ such that GΛ μ ˜ = GΛ ν˜ + h∗ on Λ. For Bx,δ ⊂ Λ,

˜ : x, δ) = L(GΛ ν˜ : x, δ) + h∗ (x). L(GΛ μ Recalling that GΛ (x, y) = ux (y) + hx (y), the symmetry of the latter two functions, and applying Tonelli’s theorem L(GΛ μ ˜ : x, δ) = L((u· (y) + h· (y)) : x, δ) d˜ μ(y) Λ = L((uy (·) + hy (·)) : x, δ) d˜ μ(y) Λ = (ux,δ (y) + hx (y)) d˜ μ(y) Λ = ux,δ (y) d˜ μ(y) + hx (y) d˜ μ(y). Λ

Λ

Applying the same argument to GΛ ν˜, ux,δ (y) d˜ μ(y)+ hx (y) d˜ μ(y) = ux,δ (y) d˜ ν (y)+ hx (y) d˜ ν (y)+ h∗ (x). Λ

Λ

Λ

Λ

140

3 Green Functions

Replacing δ by δ/2 and subtracting, (ux,δ (y) − ux,δ/2 (y)) d˜ μ(y) = (ux,δ (y) − ux,δ/2 (y)) ν˜(y). Λ

Λ

˜ = ν˜ by Since the functions ux,δ − ux,δ/2 form a total subset of K + (Λ), μ Theorem 3.5.4. It follows that μ and ν agree on the Borel subsets of Λ. Lemma 3.5.9 Let u be superharmonic on the open set Ω, and let B be a ball with B − ⊂ Ω. Then there is a unique measure μ on B such that u = GB μ+h where h is the greatest harmonic minorant of u on B. Proof: The proof will be carried out only in the n ≥ 3 case, the n = 2 diﬀering only by certain constants appearing before integrals. Since the conclusion concerns only the representation of u on B, it can be assumed that Ω is a ball on which u is bounded below. It also can be assumed that u is harmonic on Ω ∼ B − for the following reason. Let u∞ be the reduction of u over Ω ∼ B − . Since u is bounded below on Ω, u∞ is harmonic on Ω ∼ B − and u∞ = u on B by Theorem 3.3.2. The function u∞ need not be superharmonic on Ω but the lower regularization v of u∞ is superharmonic on Ω by Theorem 2.4.9 since u∞ is the limit of a decreasing sequence of superharmonic functions. The lower regularization v diﬀers from u∞ only on ∂B. Therefore, v is superharmonic on Ω, v = u on B, and v is harmonic on Ω ∼ B − . In so far as the representation of u over B is concerned, it can be assumed that u has the same properties as v; that is, u is harmonic on Ω ∼ B − . Let B0 be a ball such that B − ⊂ B0 ⊂ B0− ⊂ Ω and let Λ be a neighborhood of B − such that Λ− ⊂ B0 . By Theorem 2.5.2 there is a sequence {uj } of superharmonic functions in C ∞ (Ω) such that (i) uj ↑ u as j → ∞, (ii) uj = u on Ω ∼ Λ− for all suﬃciently large j, and (iii) uj is harmonic on Ω ∼ Λ− for all suﬃciently large j. By discarding a ﬁnite number of the uj ’s, it can be assumed that (ii) and (iii) hold for all j. By Theorem 1.5.3, for x ∈ B0 1 uj (x) = − uj (z)Dn GB0 (x, z) dσ(z) σn (n − 2) ∂B0 1 − GB0 (x, z)Δuj (z) dz. σn (n − 2) B0 Since uj = u on ∂B0 , the ﬁrst integral does not depend upon j and is, in fact, PI(u : B0 ) which will be denoted by h0 . If M is a Borel subset of B0− , let 1 νj (M ) = − Δuj (z) dz. σn (n − 2) M

3.5 Riesz Decomposition

141

Since uj is superharmonic on Ω, Δuj ≤ 0 and νj is a measure on the Borel subsets of B0− with νj (∂B0 ) = 0. The above representation of uj on B0 then can be written uj = GB0 νj +h0 on B0 . Since uj ↑ u on Ω, w = limj→∞ GB0 νj is deﬁned on B0 . It will now be shown that w is the potential of a measure on B0 . Consider νj = μj (B0 ). By Green’s identity, Theorem 1.2.2, 1 1 Δuj (z) dz = − Dn uj (z) dσ(z); νj (B0 ) = − σn (n − 2) B0 σn (n − 2) ∂B0 but since uj = u outside Λ− , Dn uj (z) = Dn u(z) on ∂B0 and the latter integral is independent of j. Therefore, νj is independent of j and there is a subsequence of the sequence {νj } that converges to a measure ν on B0− in the w∗ -topology by Theorem 0.2.5. It can be assumed that the subsequence is the sequence itself. Letting j → ∞ in the equation uj (x) = GB0 (x, z) dνj (z) + h0 (z), x ∈ B0 , B0

requires justiﬁcation since the integrand GB0 (x, z) is not a bounded contin− uous function on B0− . Suppose Bx,δ ⊂ B0 . By Tonelli’s theorem, L(GB0 νj : x, δ) =

L(GB0 (·, z) : x, δ) dνj (z). B0

By Theorem 3.3.12, L(GB0 (·, z) : x, δ) is a continuous function of z ∈ B0 and, in fact, has a continuous extension to B0− since L(GB0 (·, z) : x, δ) = GB0 (x, z) − . Since for z ∈ B0 ∼ Bx,δ L(uj : x, δ) =

L(GB0 (·, z) : x, δ) dνj (z) + h0 (x), B0

w∗

νj → ν, and uj ↑ u, L(GB0 (·, z) : x, δ) dν(z) + h0 (x).

L(u : x, δ) = B0

Since u and GB0 (·, z) are superharmonic functions, the surface averages increase to u(x) and GB0 (x, z), respectively, as δ ↓ 0 and u = GB0 ν + h0 on B0 . Since GB0 ν = GB0 ν|B + GB0 ν|B0 ∼B and GB0 ν|B0 ∼B is harmonic on B, u = GB0 ν|B + h1 on B where h1 is harmonic on B. By Theorem 3.4.8, GB0 ν|B diﬀers from GB ν|B by a function that is harmonic on B. Letting μ = ν|B , it follows that u = GB μ + h where h is harmonic on B. Since the greatest harmonic minorant of GB μ on B is the zero function, the greatest harmonic minorant of u on B is h by Lemma 3.3.6. Let u = GB μ ˜ + h∗ be a second such representation. Since the greatest harmonic minorant of u on B

142

3 Green Functions

is unique, h = h∗ and so GB μ = GB μ ˜. It follows from the preceding theorem that μ = μ ˜ and that the representation of u is unique. Lemma 3.5.10 Let u be superharmonic on the open set Ω, and let Λ be an open subset of Ω with compact closure Λ− ⊂ Ω. Then there is a unique measure μ on Λ such that u = GΛ μ + h, where h is the greatest harmonic minorant of u on Λ. Proof: Suppose ﬁrst that Λ is a subset of a ball B with B − ⊂ Ω. By the preceding lemma, there is a measure ν on B such that u = GB ν+h0 , where h0 is the greatest harmonic minorant of u on B. Since GB ν = GB ν|Λ +GB ν|B∼Λ and the latter term is harmonic on Λ, u = Gν|Λ + h, where h is harmonic on Λ. Letting μ = ν|Λ , the assertion is true for this particular case. Returning to the general case, suppose only that Λ− is a compact subset of Ω. Then Λ = ∪pi=1 Λi , where each Λi is an open subset of a ball with closure in Ω. It was just shown that the assertion is true if p = 1. Since the proof is by induction on p, it suﬃces to consider the case Λ = Λ1 ∪ Λ2 , where Λ1 and Λ2 are open sets with compact closures in Ω for which the assertion is true. Then there are meausres μi on Λi and harmonic functions hi on Λi , i = 1, 2, such that u = GΛ1 μ1 + h1 u = GΛ2 μ2 + h2

on Λ1 on Λ2 .

(3.7)

Let Λ0 = Λ1 ∩ Λ2 . Comparing potentials, there are harmonic functions h∗i on Λi such that u = GΛ0 μi |Λ0 + h∗i , i = 1, 2. It follows from Theorem 3.5.8 that μ1 |Λ0 = μ2 |Λ0 . Thus, there is a measure μ on the Borel subsets of Λ = Λ1 ∪Λ2 such that μi = μ|Λi , i = 1, 2. Since GΛ μ = GΛ μ|Λi +GΛ μ|Λ∼Λi and the latter is harmonic on Λi , GΛ μ = GΛ μi + h∗i on Λi with h∗i harmonic on Λi , i = 1, 2. Comparing GΛ μi and GΛi μi , there is a harmonic function h∗i on Λi such that GΛ μ = GΛi μi + h∗i on Λi , i = 1, 2. It follows from Equations (3.7) that u = GΛ μ + H1

on Λ1

u = GΛ μ + H2

on Λ2 ,

where Hi is harmonic on Λi , i = 1, 2. It follows that H1 = H2 on Λ1 ∩ Λ2 and a harmonic function h can be deﬁned on Λ = Λ1 ∪ Λ2 such that h|Λi = Hi , i = 1, 2. The above equations then become simply u = GΛ μ + h. As in the preceding proof, h is the greatest harmonic minorant of u on Λ and μ is unique. The measure ν of the following theorem is known as the Riesz measure associated with the superharmonic function u and the region Ω. The result will be referred to as the Riesz Decomposition Theorem to distinguish it from the Riesz representation theorem concerning integral representations of linear functionals on spaces of continuous functions.

3.5 Riesz Decomposition

143

Theorem 3.5.11 (F. Riesz [53]) Let Ω be a Greenian set and let u be superharmonic on Ω. Then there is a unique measure ν on Ω such that if Λ is any open subset of Ω with compact closure Λ− ⊂ Ω, then u = GΛ ν|Λ + hΛ , where hΛ is the greatest harmonic minorant of u on Λ; if, in addition, u ≥ 0 on Ω, then u = GΩ ν+h where h is the greatest harmonic minorant of u on Ω. Proof: Let {Ωj } be an increasing sequence of open sets with compact closures in Ω such that Ω = ∪j Ωj . Let νj and hj be the measure and harmonic function on Ωj , respectively, of the preceding lemma. Then u = GΩj νj + hj on Ωj . By Theorem 3.5.8, νj and νj+k must agree on the Borel subsets of Ωj for every k ≥ 1. If Λ is any Borel subset of Ω, then {νj (Ωj ∩ Λ)} is a nondecreasing sequence and ν(Λ) = limj→∞ νj (Ωj ∩ Λ) is deﬁned. If νj is extended to be zero on Ω ∼ Ωj , then ν is the limit of an increasing sequence of measures and therefore is a measure. In particular, if Λ is a Borel subset of Ωj , then νj (Ωj ∩ Λ) = νj (Λ) = νj+k (Λ) = νj+k (Ωj+k ∩ Λ) for all k ≥ 1 so that νj (Λ) = limk→∞ νj+k (Ωj+k ∩ Λ) = ν(Λ); that is, νj = ν|Ωj . Now let Λ be any open set with compact closure Λ− ⊂ Ω and let u = GΛ νΛ + hΛ be the decompostion of the preceding lemma. Choose j0 such that Λ ⊂ Ωj for all j ≥ j0 and consider only such j. Since u = GΩj νj + hj on Ωj with hj harmonic on Ωj and GΩj νj = GΩj νj |Λ + GΩj νj |Ωj ∼Λ with the latter term harmonic on Λ, u = GΩj νj |Λ + h∗j with h∗j harmonic on Λ. Since GΩj νj |Λ and GΛ νj |Λ diﬀer by a harmonic function on Λ, u = GΛ νj |Λ + h∗j where h∗j is harmonic on Λ. It follows from the uniqueness of the representation of u on Λ that νΛ = νj |Λ = ν|Λ . This shows that u = GΛ ν|Λ + h∗j on Λ. It follows from this equation that h∗j is independent of j on Λ, and therefore u = GΛ νΛ + hΛ where hΛ is harmonic on Λ. Suppose u ≥ 0 on Ω. Then u = GΩj ν|Ωj + hj on Ωj where hj is harmonic on Ωj . Since GΩj ≤ GΩj+1 on Ωj × Ωj , for x ∈ Ωj , GΩj (x, y) dν(y) GΩj ν|Ωj (x) =

Ωj

Ωj

≤

GΩj+1 (x, y) dν(y)

≤

GΩj+1 (x, y) dν(y) Ωj+1

= GΩj+1 ν|Ωj+1 (x), showing that the sequence {GΩj+k ν|Ωj+k }k≥1 increases on Ωj and that the sequence {hj+k }k≥1 decreases on Ωj . In fact, for x ∈ Ωj , # ⏐ G(x, y) dν(y) = GΩ ν(x) GΩJ+k ν|Ωj+k (x) = χΩj+k GΩj+k (x, y) dν(y) by Theorem 3.3.11. Since 0 ≤ u = GΩj ν|Ωj + hj on Ωj , it follows from Theorem 3.3.6 that hj ≥ 0 for each j ≥ 1. Thus, the sequence {hj+k }k≥1 is a decreasing sequence of nonnegative harmonic functions on Ωj which has a

144

3 Green Functions

limit that is harmonic on Ωj by Theorem 2.2.7. Thus, a harmonic function h = limk→∞ hj+k can be deﬁned on Ωj which does not depend upon j. It follows that u = GΩ ν + h on Ω. Using Theorem 3.3.6 and the fact that the greatest harmonic function of GΩ ν is the zero function, h is the greatest harmonic minorant of u. Corollary 3.5.12 A positive superharmonic function on a Greenian set is the potential of a measure if and only if its greatest harmonic minorant is the zero function. Proof: The necessity is just Theorem 3.3.7. The suﬃciency follows from the preceding theorem.

3.6 Properties of Potentials In general, little can be said about the continuity properties of potentials GΩ μ without assuming that the measure μ has a density. The discontinuities of a potential of a measure can be removed by giving up a small part of the measure. This is the point of the second theorem below, but the ﬁrst theorem is of interest in its own right. Theorem 3.6.1 (Evans [19] and Vasilesco [60]) Let Ω be a Greenian set, let μ be a measure with compact support Γ ⊂ Ω, and let u = GΩ μ. If u|Γ is continuous at x0 ∈ Γ , then u is continuous at x0 . Proof: Only the n ≥ 3 case will be proved. If u(x0 ) = +∞, the result is trivially true by the l.s.c. of u. It therefore can be assumed that u(x0 ) < ∞. According to Theorem 3.4.2, the Newtonian potential Uμ = Gμ of the measure μ is superharmonic. By Theorem 3.4.10, u and Uμ diﬀer by a harmonic function on Ω. Since u|Γ is continuous at x0 , the same is true of Uμ |Γ . If Uμ is continuous at x0 , then the same will be true of u so that it suﬃces to show that Uμ is continuous at x0 . A map from Ω to Γ is deﬁned as follows. If x ∈ Ω, choose a point zx in Γ nearest x. Since x0 ∈ Γ , |zx − x0 | ≤ |zx − x| + |x − x0 | ≤ 2|x − x0 | for any x ∈ Ω so that zx → x0 as x → x0 . For y ∈ Γ , |x − zx | ≤ |x − y| and |zx − y| ≤ |zx − x| + |x − y| ≤ 2|x − y|. Therefore, 1 1 ≤ 2n−2 . |x − y|n−2 |zx − y|n−2

3.6 Properties of Potentials

145

If ν is any measure with support in Γ , by integrating 1 1 n−2 Uν (x) = dν(y) ≤ 2 dν(y) = 2n−2 Uν (zx ). |x − y|n−2 |zx − y|n−2 Since Uμ (x0 ) < +∞, given > 0 there is a ball B ⊂ Ω with center at x0 such that 1 dμ(y) < n−2 B |x0 − y| by absolute continuity of the integral with respect to μ. Let μ = μ|B and μ = μ|Ω∼B . By Theorem 3.4.7, Uμ is harmonic on B and, in particular, continuous at x0 . Since Uμ = Uμ + Uμ ,

|Uμ (x) − Uμ (x0 )| ≤ |Uμ (x) − Uμ (x0 )| + |Uμ (x) − Uμ (x0 )|

≤ |Uμ (x) − Uμ (x0 )| + 2n−2 Uμ (zx ) + .

Since Uμ |Γ is continuous at x0 and Uμ |Γ is continuous at x0 , Uμ |Γ is continuous at x0 . Since zx ∈ Γ and zx → x0 as x → x0 , limx→x0 Uμ (zx ) = Uμ (x0 ) < . Thus, for x suﬃciently close to x0 ,

|Uμ (x) − Uμ (x0 )| ≤ |Uμ (x) − Uμ (x0 )| + 2n−2 + .

By continuity of Uμ at x0 , the ﬁrst term can be made arbitrarily small by taking x suﬃciently close to x0 ; that is, Uμ is continuous at x0 . Since Uμ diﬀers from u by a harmonic function on Ω, u is continuous at x0 . Theorem 3.6.2 Let Ω be a Greenian set, and let μ be a measure with compact support Γ ⊂ Ω such that GΩ μ < +∞ on Γ . Then given any > 0, there is a compact set Γ0 ⊂ Γ such that (i) μ(Γ ∼ Γ0 ) < , (ii) GΩ (μ|Γ0 ) < +∞ on Γ0 , and (iii) GΩ (μ|Γ0 ) is continuous on Ω. Proof: By Lusin’s theorem, given > 0 there is a compact set Γ0 ⊂ Γ such that μ(Γ ∼ Γ0 ) < and GΩ μ is continuous on Γ0 . Restrict all functions to Γ0 for the time being. Since GΩ (μ|Γ0 ) = GΩ μ − GΩ (μ|Γ ∼Γ0 ) on Γ0 , GΩ μ is continuous on Γ0 , and GΩ (μ|Γ ∼Γ0 ) is l.s.c. on Γ0 , GΩ (μ|Γ0 ) is u.s.c. on Γ0 . Since the function GΩ (μ|Γ0 ) is also l.s.c. on Γ0 , GΩ (μ|Γ0 )|Γ0 is continuous on Γ0 . By the preceding theorem, GΩ (μ|Γ0 ) is continuous on Γ0 . Since GΩ (μ|Γ0 ) is harmonic on Ω ∼ Γ0 , it is continuous on Ω. The proof of the preceding theorem can be modiﬁed to obtain an approximation of a potential by continuous potentials. Theorem 3.6.3 Let Ω be a Greenian set, and let μ be a measure on Ω with the property that there is a Borel set Σ ⊂ Ω with μ(Ω ∼ Σ) = 0 and GΩ μ < +∞ on Σ. Then there is a sequence ∞ of measures {μ n∞} with disjoint compact supports in Σ such that μ = n=1 μn , GΩ μ = n=1 GΩ μn , and for each n ≥ 1, GΩ μn is ﬁnite-valued and continuous on Ω.

146

3 Green Functions

Proof: Assume ﬁrst that μ is a ﬁnite measure. Applying Lusin’s theorem to the set Σ, there is a disjoint sequence of compact sets {Γm } in Σ such that μ(Ω ∼ ∪∞ n=1 Γn ) = 0 and GΩ μ|Γn is ﬁnite-valued and continuous on Γn . Let μn be the restriction of μ to Γn . As in the preceding proof, GΩ μn is ﬁnite-valued and continuous on Ω. The general case can be reduced to the ﬁnite measure case using the fact that μ is σ-ﬁnite. There is not much to be said about convergence properties of sequences of potentials. Theorem 3.6.4 Let {μj } be a sequence of measures on the Greenian set Ω that converges in the w∗ -topology to the measure μ. Then GΩ μ ≤ lim inf GΩ μj . j→∞

Proof: Consider any x ∈ Ω. Then GΩ (x, ·) is nonnegative and l.s.c. on Ω, and there is an increasing sequence {φj } in C 0 (Ω) such that φj ↑ GΩ (x, ·). Thus, φi dμ = lim φi dμj ≤ lim inf GΩ (x, ·)dμj . j→∞

j→∞

Letting i → ∞, GΩ μ(x) ≤ lim inf j→∞ GΩ μj (x). More stringent conditions are needed to improve on this result. Theorem 3.6.5 If Ω is a Greenian subset of Rn , {μj } is a sequence of meaw∗ sures with supports in the compact set Γ ⊂ Ω, μj → μ, and {GΩ μj } is an increasing sequence on Ω, then GΩ μ = limj→∞ GΩ μj and is superharmonic on Ω. Proof: Since μj (Γ ) → μ(Γ ), the latter is ﬁnite and GΩ μ is a potential by Theorem 3.4.4. By Theorem 2.4.8, the function u = limj→∞ GΩ μj is either superharmonic or identically +∞ on each component of Ω. It can be assumed that Ω is connected. Suppose x ∈ Ω ∼ Γ and Bx,ρ ⊂ Ω ∼ Γ . Since GΩ (x, ·) is bounded outside Bx,ρ by a constant Mx,ρ , GΩ μj (x) =

GΩ (x, y) dμj (y) =

∼Bx,ρ

GΩ (x, y) dμj (y) ≤ Mx,ρ μj (Γ ).

Thus, u(x) = limj→∞ GΩ μj (x) ≤ Mx,ρ μ(Γ ) < +∞ and it follows that u is − superharmonic on Ω. Consider any x ∈ Ω and any ball Bx,ρ with Bx,ρ ⊂ Ω. According to Theorem 3.3.12, L(GΩ (y, ·) : x, r) is a continuous function of y ∈ Ω. By Tonelli’s theorem, L(u : x, r) = lim L(GΩ μj : x, r) j→∞

3.6 Properties of Potentials

147

= lim L(GΩ (·, z) : x, r) dμj (z) j→∞ = L(GΩ (·, z) : x, r) dμ(z) = L(GΩ μ : x, r). Letting r ↓ 0, u(x) = GΩ μ(x) by Lemma 2.4.4.

Chapter 4

Negligible Sets

4.1 Introduction In the early development of potential theory, the singularities of superharmonic functions gave rise to the concept of a polar set. At the same time, the classic notion of capacity of a conductor in a grounded sphere became an object of interest to mathematicians. Both of these concepts are fully developed in this chapter and will culminate in the characterization of polar sets as sets of capacity zero. An essential part of this development is Choquet’s theory of capacities which has important applications to stochastic processes as well as potential theory. These concepts are used to settle questions pertaining to equilibrium distribution of charges, the existence of Green function for regions in Rn , and the boundary behavior of Green functions. It will be shown that an open subset of R2 has a Green function if and only if the region supports a positive superharmonic function.

4.2 Superharmonic Extensions In the proof of Zaremba’s theorem, Theorem 2.6.29, it was possible to conclude that a harmonic function u was nonnegative on Ω knowing that lim inf u(y) ≥ 0,

y→x,y∈Ω

x ∈ ∂Ω,

except possibly for a single point in ∂Ω. This suggests that a single point is in some sense negligible. By examining the proof more carefully, it can be seen that the same proof would work if there were a nonnegative superharmonic function taking on the value +∞ at those points x ∈ ∂Ω where lim inf y→x,y∈Ω u(y) ≥ 0 fails to hold. This suggests that the set of points

L.L. Helms, Potential Theory, Universitext, c Springer-Verlag London Limited 2009

149

150

4 Negligible Sets

where a superharmonic function takes on the value +∞ constitutes a negligible set as far as boundary behavior of potentials is concerned. Deﬁnition 4.2.1 A set Z ∈ Rn is a polar set if there is an open set Λ ⊃ Z and a superharmonic function u on Λ such that u = +∞ on Z (and perhaps elsewhere); Z is an inner polar set if each compact subset of Z is a polar set. Deﬁnition 4.2.2 A statement concerning points of Rn is said to hold quasi everywhere or q.e. if the set of points where the statement fails to hold is a polar set; the statement is said to hold inner quasi everywhere or inner q.e. if the set of points where it fails to hold is an inner polar set. Example 4.2.3 Let x0 be a ﬁxed point of Rn . Then Z = {x0 } is a polar set since ux0 is superharmonic on Rn and ux0 (x0 ) = +∞. Also, ux0 = 0 q.e. Example 4.2.4 Let {xj } be a sequence of distinct points in R3 . Then Z = {x1 , x2 , . . .} is a polar set. To see this, let y be a ﬁxed point in ∼ Z. Choose constants cj > 0 such that j

cj < +∞. |y − xj |

∞

−1 Deﬁning u(x) = , u is superharmonic on Rn j=1 cj |x − xj | by Theorem 2.4.8 since it is the limit of an increasing sequence of superharmonic functions and u(y) < +∞. Clearly, u = +∞ on Z.

Theorem 4.2.5 If Z is a polar set, then it is a subset of a Gδ polar set. Moreover, any subset of a polar set is a polar set. Proof: Let u be a superharmonic function on a neighborhood Λ of Z which is +∞ on Z. Since {x; u(x) = +∞} = ∩j ({x; u(x) > j} and each of the latter sets is open by the l.s.c. of u, {x; u(x) = +∞} is a Gδ set. The second statement is immediate from the deﬁnition of polar set. Theorem 4.2.6 If Z ⊂ Rn is a polar set, then Z has n-dimensional Lebesgue measure zero. Proof: Let u be superharmonic on the open set Λ ⊃ Z and +∞ on Z. Since u is ﬁnite a.e. by Theorem 2.4.2, Z has Lebesgue measure zero. Example 4.2.7 A line segment I ⊂ R2 of positive length is not a polar set. To see this, assume that there is a superharmonic function u on a neighborhood Λ of I with u = +∞ on I. It follows from the averaging principle that superharmonicity is invariant under rotations about a point. If x is an interior point of I, choose ρ > 0 such that Bx,ρ ⊂ Λ and choose a subinterval

4.2 Superharmonic Extensions

151

√ I1 of I with an end point at x of length less that ρ/ 2. Rotating I1 about x, an interval I2 can be constructed that is perpendicular to I1 and has an end point at x; a superharmonic function u2 can be constructed on the ball containing I1 and I2 which is +∞ on I2 by means of the rotation about x. Continuing in this way, a square and four superharmonic functions can be constructed on a neighborhood of the square with at least one of the four functions taking on the value +∞ on each side of the square. The sum v of the four functions is then superharmonic on a neighborhood of the square and is equal to +∞ on the boundary of the square. By the minimum principle, v must be +∞ on the square, contradicting the fact that a superharmonic function is ﬁnite a.e. on its domain. Thus, I is not polar. Example 4.2.8 A line segment in R3 is a polar set. Consider, for example, the line segment I joining (0, 0, 0) to (1, 0, 0), one-dimensional Lebesgue measure μ on I, and the Newtonian potential 1 dμ(z), x ∈ R3 Uμ (x) = |x − z| I 1 1 = dz. 2 + x2 + x2 )1/2 ((x − z) 1 0 2 3 By Theorem 3.4.4, Uμ is superharmonic. If x = (x1 , 0, 0) ∈ I, then μ

U (x) = 0

1

1 dz = +∞ |x1 − z|

so that Uμ = +∞ on I. This shows that I is a polar subset of R3 . The deﬁnition of a polar set Z requires that a superharmonic function u be deﬁned only on a neighborhood of Z and that it be inﬁnite on Z. On the one hand, this deﬁnition makes it easy to conﬁrm that some sets are polar, but on the other hand, makes it diﬃcult to show that the union of two disjoint polar sets is again polar. A global alternative is essential. Theorem 4.2.9 If Z ⊂ R2 is a polar set, then there is a superharmonic function u such that u = +∞ on Z. Proof: Let Λ be a neighborhood of Z on which there is deﬁned a superharmonic function u such that u = +∞ on Z. It can be assumed that u > 0 on Λ by replacing Λ with the open set Λ ∩ {y; u(y) > 0}, if necessary. Letting Λj = Λ ∩ B0,j , j ≥ 1, Z = ∪j (Λj ∩ Z). By a change of notation, if necessary, it can be assumed that each Λj is nonempty. Since Λj is bounded, it has a Green function GΛj by Theorem 3.2.10. By the Riesz decomposition theorem,Theorem 3.5.11, there is a μj associated with u such that u = GΛj μj + hj , where hj is harmonic on Λj . Since u = +∞ on Λj ∩ Z and GΛj μj ≤ GB0,j μj , the latter is +∞ on Λj ∩ Z. Deﬁning

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4 Negligible Sets

wj (x) =

log B0,j

2j dμj (y), |x − y|

x ∈ R2 ,

wj diﬀers from the logarithmic potential of μj only by a constant and is superharmonic by Theorem 3.4.10. Referring to Deﬁnition 1.5.2, log

2j ≥ GB0,j (x, y), |x − y|

x, y ∈ B0,j ,

so that wj = +∞ on Λj ∩ Z, since GB0,j μj = +∞ on Λj ∩ Z. In summary, (i) wj is superharmonic, (ii) wj ≥ 0 on B0,j , and (iii) wj = +∞ on Λj ∩ Z. Since each wj is ﬁnite a.e. on B0,1 , there is a point x0 ∈ B0,1 such that 0 < wj (x0 ) < +∞ for all j ≥ 1. A sequence of positive numbers {bj } therefore can be chosen so that the series j bj wj (x0 ) converges. Deﬁning v = j bj wj , v = +∞ on Z. It remains only to show that v is a superharmonic function. By Theorem 2.4.8, v is either superharmonic or identically +∞ on R2 . The former is true since v(x0 ) < +∞. Theorem 4.2.10 If Z ⊂ Rn , n ≥ 3, is a polar set, then there is a Borel measure ν on Rn such that the potential Gν = +∞ on Z. Proof: By hypothesis, there is an open set Λ ⊃ Z and a superharmonic function u on Λ with u = +∞ on Z. As in the preceding proof, it can be assumed that u > 0 on Λ. Let μ be the measure associated with u and Λ by the Riesz decomposition theorem. Extend μ to Rn by putting μ(∼ Λ) = 0. Then u = GΛ μ + h, where h is harmonic on Λ. Let {Λj } be a sequence of nonempty open sets having compact closures such that Rn = ∪j Λj . If μj = μ|Λj , then μj (Λj ) = μ(Λj ) < +∞ for each j ≥ 1. Also extend μj to Rn by putting μj (∼ Λj ) = 0. Letting uj = Gμj , uj ≥ 0 and is superharmonic since μj (Rn ) = μj (Λj ) < +∞. Since u = GΛ μ + h = GΛ μj + GΛ μ|Λ∼Λj + h on Λ and μ|Λ∼Λj (Λj ) = 0, u = GΛ μj + hj

on Λ ∩ Λj

where hj is harmonic on Λj by Theorem 3.4.7. Since u = +∞ on Z, GΛ μj = +∞ on Z ∩ Λj . By Theorem 3.4.8, ˜ j on Λ ∩ Λj GΛ μj = GΛ∩Λj μj + h ˜ j is harmonic on Λ∩Λj and it follows that GΛ∩Λ μj = +∞ on Z ∩Λj . where h j Applying Theorem 3.4.8 again, GΛj μj = GΛ∩Λj μj + h∗j on Λ ∩ Λj

4.2 Superharmonic Extensions

153

where h∗j is harmonic on Λ ∩ Λj and it follows that GΛj μj = +∞ on Z ∩ Λj . Since uj = Gμj ≥ GΛj μj = +∞ on Z ∩ Λj , uj = +∞ on Z ∩ Λj . Let y be a ﬁxed point of Λ1 such that uj (y) < +∞ for all j ≥ 1, and let {cj } be a sequence of positive numbers such that cj uj (y) converges. Put v = j cj uj . Since the uj ’s are nonnegative superharmonic functions and v(y) < +∞, v is a nonnegative superharmonic function on Rn by Theorem 2.4.8 and v = +∞ on Z. By the Riesz decomposition theorem, there is a Borel measure ν on Rn and a harmonic function H such that v = Gν + H. Since v = +∞ on Z, Gν = +∞ on Z. Since superharmonic functions are ﬁnite a.e. according to Theorem 2.4.2, polar sets have Lebesgue measure zero. The following theorem describes polar sets as small sets in another way. Theorem 4.2.11 If Z is a polar set, its intersection with any sphere has surface area zero. Proof: By the two preceding theorems, there is a superharmonic function u such that u = +∞ on Z. If B = Bx,ρ is any ball, then L(u : x, ρ) is ﬁnite by Theorem 2.4.3. It follows that Z ∩ ∂B has surface area zero. Theorem 4.2.12 If {Zj } is a sequence of polar sets in Rn , then ∪j Zj is a polar set. Proof: For each j ≥ 1, let uj be superharmonic and uj = +∞ on Zj . Since the uj ’s are ﬁnite a.e., there is a point y ∈ B0,1 such that uj (y) < +∞ for all j ≥ 1. Since uj is l.s.c., inf |x|≤j uj (x) is ﬁnite. Thus, a sequence of positive numbers {bj } can be chosen such that bj (uj (y) − inf uj (x)) |x|≤j

j

converges. Let u=

j

bj (uj − inf uj (x)), |x|≤j

and consider a ﬁxed ball B0,k . All terms except the ﬁrst k − 1 of the series deﬁning u are nonnegative on B0,k and the series ∞ j=k

bj (uj − inf uj (x)) |x|≤j

is superharmonic on B0,k by Theorem 2.4.8. Since the sum of the ﬁrst k − 1 terms is superharmonic on B0,k , u is superharmonic on B0,k for every positive integer k and therefore superharmonic on Rn . Since uj = +∞ on Zj , u = +∞ on Z. In the following discussion, {Bxi ,ri ; i ≥ 1} will be a countable base for the Euclidean topology on Rn consisting of balls.

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4 Negligible Sets

Remark 4.2.13 Polarity is a local property; that is, if for each x ∈ Z there is a ball Bx containing x such that Z ∩ Bx is polar, then Z is polar. This can be shown as follows. For each x ∈ Z, there is a ball Bxj ,rj such that x ∈ Bxj ,rj ⊂ Bx with Z ∩ Bxj ,rj polar since it is a subset of the polar set Z ∩ Bx . It therefore can be assumed that each Bx belongs to the countable base {Bxi ,ri ; i ≥ 1}. Since Z ⊂ ∪(Z ∩ Bx ) and the latter is a countable union of polar sets, Z is a subset of a polar set by the preceding theorem and is therefore polar. Similarly, inner polarity is a local property; that is, if for each x ∈ Z there is a ball Bx containing x such that Z ∩ Bx is inner polar, then Z is inner polar. To see this, let Γ be a compact subset of Z. As above, it can be assumed that each Bx is a member of the countable base {Bxi ,ri ; i ≥ 1}. Suppose Bx = Bxj ,rj . Since Γ ∩Bx is inner polar and each Γ ∩Bx−j ,rj −1/m , m ≥ − 1, is a compact subset of Γ ∩ Bx , Γ ∩ Bx = ∪∞ m=1 Γ ∩ Bx−j,rj −1/m is a polar set. Since Γ ⊂ (Γ ∩ Bx ), Γ is polar and therefore Z is inner polar. The preceding results provide a quantitative measure of the smallness of polar sets. The results will be applied to the problem of extending superharmonic functions across polar sets and to the existence of Green functions in the two-dimensional case. Lemma 4.2.14 Let Ω be an open set, and let Z be a relatively closed polar subset of Ω. If x ∈ Z and {xj } is a sequence of distinct points in Ω ∼ Z such that x = limj→∞ xj , then there is a superharmonic function v on Ω such that v = +∞ on Z and v(xj ) < +∞ for all j ≥ 1. Proof: For each j ≥ 1, a number δj > 0 can be chosen so that Bx−j ,δj ⊂ Ω ∼ Z and Bx−i ,δi ∩ Bx−j ,δj = ∅ for i = j. It is easy to see that limj→∞ δj = 0. Since Z is polar, there is a superharmonic function v such that v = +∞ on Z. By Lemma 2.4.11, the function PI(v : B1 ) on B1 v1 = v on Ω ∼ B1 , where B1 = Bx1 ,δ1 , is superharmonic on Ω. Then v = v1 = +∞ on Z and v1 (x1 ) < +∞. Inductively deﬁne a sequence of superharmonic functions {vj } by putting PI(vj−1 : Bj ) on Bj vj = vj−1 on Ω ∼ Bj , where Bj = Bxj ,δj . The sequence {vj } is then a decreasing sequence of superharmonic functions on Ω, vj = +∞ on Z, and vj (xj ) < +∞ for all j ≥ 1. Consider the function v˜ = limj→∞ vj . Given any neighborhood of x, the vj ’s agree outside the neighborhood except possibly for a ﬁnite number of terms of the sequence. This means that v˜ is superharmonic on Ω ∼ {x}. To show that v˜ is superharmonic on Ω, it suﬃces to show that v˜ is l.s.c. at x, v˜ being locally super-mean-valued at x since v˜(x) = +∞. By the l.s.c. of v at x and

4.2 Superharmonic Extensions

155

the fact that v(x) = +∞, for each integer k ≥ 1 there is a neighborhood Λk of x such that v > k on Λk . Moreover, it can be assumed that there is an integer jk such that Bj− ⊂ Λk for all j ≥ jk . It follows that vj ≥ k on Λk for all j ≥ jk , and therefore that v˜ ≥ k on Λk . Thus, +∞ = v˜(x) = limy→x v˜(y); that is, v˜ is continuous (in the extended sense) at x. The following theorem was ﬁrst proved by Schwarz [56] for a bounded harmonic function on a deleted neighborhood of a point, then by Bouligand [5] for bounded harmonic functions and Z compact polar, and then by Vasilesco [61] for Z a compact polar set. Theorem 4.2.15 (Brelot [7]) Let Ω be an open set and let Z be a relatively closed polar subset of Ω. If u is superharmonic on Ω ∼ Z and locally bounded below on Ω, then it has a unique superharmonic extension to Ω. Proof: Note ﬁrst that Ω ∼ Z is dense in Ω, for if this were not the case, then Z would have a nonempty interior and therefore positive Lebesgue measure, a contradiction. Since Ω ∼ Z is dense in Ω, a function u˜ can be deﬁned on Ω by putting u˜ = lim inf y→x,y∈Ω∼Z u(y) for x ∈ Ω. Since u is locally bounded below on Ω, u ˜ cannot take on the value −∞. Clearly, u ˜ is l.s.c. on Ω. Since u ˜ = u on the open set Ω ∼ Z, u ˜ is superharmonic on Ω ∼ Z and therefore not identically +∞ on any component of Ω. To show that u ˜ is superharmonic on Ω, it remains only to show that u ˜ is locally super-mean-valued on Ω. This is true at points of the open set Ω ∼ Z since u ˜ = u on this set. Consider any x ∈ Z and let {xj } be a sequence of distinct points in Ω ∼ Z such that limj→∞ xj = x and limj→∞ u ˜(xj ) = u ˜(x). Since Z is a polar set, there is a superharmonic function v such that v = +∞ on Z. By the preceding lemma, ˜ cannot take on it can be assumed that v(xj ) < +∞ for each j ≥ 1. Since u the value −∞ and is l.s.c. on Ω, u˜ + v is deﬁned and l.s.c. on Ω for every − ⊂ Ω, +∞ = (˜ u + v)(y) ≥ L(˜ u + v : y, ρ). This > 0. For y ∈ Z and By,ρ shows that u ˜ + v is superharmonic on Ω. Since v(xj ) < +∞ and u : xj , δ) + A(v : xj , δ), u ˜(xj ) + v(xj ) ≥ A(˜ letting → 0 u˜(xj ) ≥ A(˜ u : xj , δ) =

1 νn δ n

u(y) dy. χBxj ,δ (y)˜

Since u ˜ is bounded below on the compact closure of a neighborhood of x, a δ can be chosen so that all but a ﬁnite number of the integrands on the right are bounded below. By Fatou’s Lemma, u ˜(x) = lim u ˜(xj ) ≥ A(˜ u : x, δ) j→∞

for all suﬃciently small δ. This concludes the proof that u ˜ is superharmonic on Ω. To prove uniqueness, let u be a second such extension of u. Since Z

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has Lebesgue measure zero, u˜ = u a.e. on Ω and A(˜ u : x, δ) = A(u : x, δ) ˜(x) = u(x) for all x ∈ Ω by Lemma 2.4.4. whenever Bx,δ ⊂ Ω. Letting δ ↓ 0, u

Corollary 4.2.16 Let Ω be an open set, Z a relatively closed polar subset of Ω, and u a function that is harmonic on Ω ∼ Z and locally bounded on Ω. Then u has a unique harmonic extension. Proof: By the preceding theorem, u has a superharmonic extension u on Ω. Applying the same result to −u, it has a subharmonic extension u on Ω with u = u on Ω ∼ Z. Since Z has Lebesgue measure zero, A(u : x, δ) = A(u : x, δ) whenever Bx,δ ⊂ Ω. Letting δ ↓ 0, u (x) = u (x) for all x ∈ Ω and u is therefore harmonic on Ω. Lemma 4.2.17 If Ω is an open set, then ∂Ω is polar if and only if ∼ Ω is polar. Proof: Since ∂Ω ⊂∼ Ω, ∼ Ω polar implies that ∂Ω is polar and the suﬃciency is proved. Assume that ∂Ω is polar. By Theorems 4.2.9 and 4.2.10, there is a superharmonic function u such that u = +∞ on ∂Ω. If Λ is any component of int(∼ Ω), then u = +∞ on ∂Λ ⊂ ∂(int(∼ Ω)) = ∂Ω and therefore u = +∞ on Λ by the minimum principle, Corollary 2.3.6. Since Λ is any component of int(∼ Ω), u = +∞ on int(∼ Ω) and the same is true on its boundary. Thus, u = +∞ on ∼ Ω, proving that ∼ Ω is polar. The concept of polar set can shed some light on the existence of a Green function for a subset of R2 . Theorem 4.2.18 If Ω is an open subset of R2 having a Green function, then ∼ Ω is not a polar set (equivalently, ∂Ω is not a polar set). Proof: Assume that ∼ Ω is polar and ﬁx x ∈ Ω. Since GΩ (x, ·) ≥ 0 on Ω, by Theorem 4.2.15 it has a nonnegative superharmonic extension on R2 , denoted by the same symbol. By Theorem 2.5.3, L(GΩ (x, ·) : x, δ) is a concave function φ of − log δ. If (ξ) = aξ + b is any supporting line for the convex set {(ξ, η); η ≤ φ(ξ), −∞ < ξ < +∞}, then L(GΩ (x, ·) : x, δ) = φ(− log δ) ≤ −a log δ + b. Since L(GΩ (x, ·) : x, δ) ↑ +∞ as δ ↓ 0, a must be nonnegative; on the other hand if a > 0, then −a log δ + b is negative for large δ and therefore a = 0; that is, every supporting line for the above convex set is horizontal. This implies that φ is a constant function. Thus, L(GΩ (x, ·) : x, δ) is a constant function on (0, +∞), a contradiction since L(GΩ (x, δ) : x, δ) ↑ +∞ as δ ↓ 0. Therefore, ∼ Ω cannot be polar. In deﬁning the upper class Uf for a function f on the boundary ∂Ω, it was required of a function u ∈ Uf that it satisfy the condition lim inf y→x,y∈Ω u(y) ≥ f (x) for all x ∈ ∂Ω. This requirement can be relaxed so that the inequality

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157

holds only quasi everywhere on ∂Ω. Let Upf and Lpf be deﬁned in the same way as Uf and Lf , respectively, except that the boundary condition is required to hold only quasi everywhere. Theorem 4.2.19 If f is a boundary function for the bounded open set Ω, then H f = inf {u; u ∈ Upf } and H f = sup {u; u ∈ Lpf }. Proof: Consider any u ∈ Upf . Then u is superharmonic or identically +∞ on each component of Ω, u is bounded below on Ω, and lim inf y→x,y∈Ω u(y) ≥ f (x) for all x ∈ ∂Ω ∼ Z, where Z is a polar subset of ∂Ω. Let v be a superharmonic function such that v = +∞ on Z. Since Ω is bounded, it can be assumed that v ≥ 0 on Ω. Then for any > 0, lim inf y→x,y∈Ω (u + v)(y) ≥ f (x) for all x ∈ ∂Ω. Therefore, u + v ≥ H f on Ω. If H f is identically −∞ on a component of Ω, then u ≥ H f on that component; if H f is identically +∞ on a component of Ω, then u ≥ H f a.e. on that component and, consequently, u ≥ H f on that component; if H f is harmonic on a component of Ω and By,δ is a ball in that component, then A(u : y, δ) + A(v : y, δ) ≥ H f (y) and it follows that u ≥ H f on that component by letting → 0 ﬁrst and then letting δ ↓ 0. Therefore, inf {u; u ∈ Upf } ≥ H f . Since inf {u; u ∈ Upf } ≤ inf {u; u ∈ Uf } = H f , H f = inf {u; u ∈ Upf }. Generally speaking, the values of a boundary function can be changed on a polar subset of the boundary without aﬀecting the Dirichlet solution. Corollary 4.2.20 Let Ω be a bounded open set. If f and g are functions on ∂Ω such that f = g on ∂Ω except possibly on a polar subset of ∂Ω, then H f = H g and H f = H g ; if, in addition, f is resolutive, then g is also and Hf = Hg on Ω. Proof: If u ∈ Uf , then lim inf y→x,y∈Ω u(y) ≥ f (x) = g(x) quasi everywhere on ∂Ω and u ∈ Upg ; that is, Uf ⊂ Upg so that H g ≤ H f . Interchanging f and g, H f = H g . The following theorem exhibits an essential diﬀerence between the n = 2 and n ≥ 3 cases. In the n = 2 case, the point at inﬁnity is a negligible point; but in the n ≥ 3 case, the point at inﬁnity cannot be ignored. Theorem 4.2.21 Let Ω be a Greenian set, let u be superharmonic and bounded below on Ω, and suppose there is a constant α such that lim inf u(y) ≥ α

y→x,y∈Ω

for all x ∈ ∂Ω except possibly for an inner polar set. If (i)n = 2, or (ii)n ≥ 3 and Ω is bounded, or (iii)n ≥ 3, Ω is unbounded, and lim inf |y|→+∞,y∈Ω u(y) ≥ α, then u ≥ α on Ω.

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Proof: It will be shown ﬁrst that the exceptional set is a polar set. Letting Z denote the exceptional set, Z = {x ∈ ∂Ω; lim inf u(y) < α} y→x,y∈Ω

& ∞ % 1 = x ∈ ∂Ω; |x| ≤ n, lim inf u(y) ≤ α − y→x,y∈Ω n n=1 Since each of the sets of the countable union is a compact subset of Z, each is polar and so Z is polar. It therefore can be assumed that the exceptional set is a polar set. If Ω is bounded, then u ≥ H α = α by Theorem 4.2.19. Suppose now that Ω is unbounded. The rest of the proof requires separate consideration of the n = 2 and n ≥ 3 cases. Consider ﬁrst the n = 2 case. Since Ω has a Green function, ∂Ω is not a polar set by Theorem 4.2.18. It follows that there is a point x0 ∈ ∂Ω such that lim inf y→x0 ,y∈Ω u(y) ≥ α. If η is any positive number, then there is a ball B = Bx0 ,δ such that u(y) > α − η for y ∈ Ω ∩ B − . By suitable choice of a constant c, the function v(y) = log |y − x0 | + c can be made harmonic and nonnegative on Λ = Ω ∼ B − and will satisfy the condition lim|y|→+∞,y∈Λ v(y) = +∞. For each > 0, the function u + v has the properties that lim inf y→x,y∈Λ (u + v)(y) ≥ α − η for x ∈ ∂Λ except possibly for a polar set and lim|y|→+∞,y∈Λ (u + v)(y) = +∞. Thus, for all suﬃciently large integers j, (u + v) ≥ α − η on ∂Bx0 ,j ∩ Λ, and therefore lim inf y→x,y∈Λ∩Bx0 ,j (u + v) ≥ α − η for all x ∈ ∂(Λ ∩ Bx0 ,j ) except possibly for a polar set. It follows from the bounded case that u + v ≥ α − η on Λ ∩ Bx0 ,j for all large j. Therefore, u + v ≥ α − η on Λ. Letting → 0, u ≥ α − η on Λ. But since u > α − η on Ω ∩ B − , u ≥ α − η on Ω. Since η is an arbitrary positive number, u ≥ α on Ω. Consider the n ≥ 3 case with lim inf |y|→+∞,y∈Ω u(y) ≥ α. For each η > 0, there is a positive integer j0 such that u(y) ≥ α − η for all y ∈ Ω ∼ B0,j , j ≥ j0 . Then lim inf y→x,y∈Ω∩B0,j u(y) ≥ α − η for x ∈ ∂(Ω ∩ B0,j ) except possibly for a polar set. By the ﬁrst part of the proof for bounded sets, u ≥ α − η on Ω ∩ B0,j for every j ≥ j0 . It follows that u ≥ α − η on Ω and that u ≥ α on Ω since η is an arbitrary positive number.

4.3 Reduction of Superharmonic Functions In this section, the notion of “balayage,” due to Poincar´e, will be developed. As conceived by Poincar´e, “balayage” or “sweeping out” is just the smoothing process used to deﬁne the reduction u∞,Λ of the superharmonic function u. Let Ω be a Greenian subset of Rn , let u be a nonnegative superharmonic function on Ω, and if Λ is any subset of Ω let ΦuΛ = {v ∈ S(Ω); v ≥ 0 on Ω, v ≥ u on Λ}.

4.3 Reduction of Superharmonic Functions

159

Deﬁnition 4.3.1 RuΛ = inf {v; v ∈ ΦuΛ } is called the reduction of u over Λ relative to Ω. In using the notation RuΛ , it will be understood that u is a superharmonic function. Clearly, if u and v are nonnegative superharmonic functions on Ω which agree on Λ, then RuΛ = RvΛ . RuΛ , as the inﬁmum of a family of superharmonic functions, need not be superharmonic. Consider, for example, the fundamental harmonic function u0 for R3 with pole 0. Let Ω = R3 and Λ = {0}. Then RuΛ = +∞ on Λ, zero elsewhere, and therefore not l.s.c. In the general case, RuΛ fails to be superharmonic on Ω only because of the l.s.c. requirement. Since u ∈ ΦuΛ and elements of ΦuΛ are nonnegative, 0 ≤ RuΛ ≤ u. Thus, RuΛ > −∞ and RuΛ cannot be identically +∞ on any component of Ω since this is true of u. RuΛ can be seen to be super-meanvalued on Ω as follows. If Bx,δ ⊂ Ω and v ∈ ΦuΛ , then v(x) ≥ L(v : x, δ) ≥ L(RuΛ : x, δ). Thus, RuΛ (x) ≥ L(RuΛ : x, δ) (assuming that RuΛ is measurable). This shows that only the l.s.c. requirement stands in the way of RuΛ being superharmonic on Ω. ˆ u of Ru is superharmonic on Ω. Lemma 4.3.2 The lower regularization R Λ Λ Moreover, ˆ u ≤ Ru ≤ u on Ω, (i) 0 ≤ R Λ Λ (ii) u = RuΛ on Λ, ˆ u on the interior of Λ, and (iii) u = RuΛ = R Λ ˆ u on Ω ∼ Λ− and both are harmonic on Ω ∼ Λ− . (iv) RuΛ = R Λ Proof: The ﬁrst three assertions are obvious. To prove (iv), it is only necessary to show that RuΛ is harmonic on Ω ∼ Λ− . As in the proof of Theorem 2.6.2, ΦuΛ is a saturated family of superharmonic functions on Ω ∼ Λ− and RuΛ is therefore harmonic on Ω ∼ Λ− by the same theorem. ˆ u diﬀers from Ru only on the boundAccording to the preceding lemma, R Λ Λ ˆ u < Ru . If u0 is the fundaary of Λ. At such points, it may be that R Λ Λ mental harmonic function on R3 with pole 0, Ω = R3 , and Λ = {0}, then ˆ u (0) < Ru (0) = +∞. It will be shown later that, in general, R ˆ u diﬀers 0=R Λ Λ Λ u from RΛ only on a polar set. Remark 4.3.3 If Ω has components {Ωj } and Λ ⊂ Ω, it is easy to see that ˆu = R ˆu R on Oj , j ≥ 1. Λ

Λ∩Oj

ˆ u is called the regularized reduction of u over Λ relDeﬁnition 4.3.4 R Λ ative to Ω. Theorem 4.3.5 If u and v are nonnegative superharmonic functions on a Greenian set Ω, then

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4 Negligible Sets

ˆu ≤ R ˆu , (i) Λ ⊂ Σ ⊂ Ω ⇒ R Λ Σ u v ˆ ˆ (ii) u ≤ v ⇒ RΛ ≤ RΛ , ˆ λu = λR ˆu, (iii) λ > 0 ⇒ R Λ Λ ˆu + R ˆv , ˆ u+v ≤ R (iv) R Λ Λ Λ ˆ u is positive or identically zero on each component of Ω, (v) R Λ ˆ u = Ru on Ω, and (vi) Λ open subset of Ω ⇒ R Λ Λ ˆ u is a potential. (vii) Γ a compact subset of Ω ⇒ R Γ Proof: If Λ ⊂ Σ, then ΦuΣ ⊂ ΦuΛ and RuΛ ≤ RuΣ . Assertion (i) follows by taking the lower regularization of each side of this inequality. Assertions (ii) and (iii) are proved in the same way. If f ∈ ΦuΛ and g ∈ ΦvΛ , then f +g ∈ Φu+v Λ ˆ u+v ≤ Ru+v ≤ f + g, where R ˆ u+v , f, and g are superharmonic on Ω. and R Λ Λ Λ − For the time being, ﬁx g. If Bx,δ ⊂ Ω, then ˆ u+v : x, δ) ≤ A(f : x, δ) + A(g : x, δ) A(R Λ ≤ f (x) + A(g : x, δ). Taking the inﬁmum over f ∈ ΦuΛ , ˆ u+v : x, δ) ≤ RuΛ (x) + A(g : x, δ). A(R Λ Since superharmonic functions are integrable on compact subsets of Ω by Theorem 2.4.2, the two averages are continuous on Ωδ = {y ∈ Ω; d(y, ∼ Ω) > δ}. Thus, ˆ u+v : x, δ) ≤ R ˆ uΛ (x) + A(g : x, δ), A(R Λ Letting δ ↓ 0,

ˆ u+v (x) ≤ R ˆ uΛ (x) + g(x), R Λ

on Ωδ .

on Ω.

Repeating this argument, ˆ u+v ≤ R ˆ uΛ + R ˆ vΛ , R Λ

on Ω.

Assertion (v) follows from the minimum principle. Consider assertion (vi). By ˆ u = Ru = u on Λ. Thus, R ˆ u ∈ Φu and Ru ≤ R ˆ u on the preceding lemma, R Λ Λ Λ Λ Λ Λ u u ˆ Ω. Since the opposite inequality is always true, RΛ = RΛ on Ω. To show that ˆ u is a potential, it suﬃces to show that the greatest harmonic minorant of R Γ ˆ u is the zero function by Corollary 3.5.12. Assume ﬁrst that u is bounded R Γ on Γ . Let Ω0 be any component of Ω and let x0 ∈ Ω0 . By Theorem 3.3.7 and Corollary 3.3.10, the greatest harmonic minorant of GΩ (x0 , ·)|Ω0 is the zero function. Since GΩ (x0 , ·) is superharmonic and strictly positive on Ω0 , it has a strictly positive inﬁmum on Ω0 ∩ Γ . For some λ > 0, λGΩ (x0 , ·) majorizes u on Ω0 ∩ Γ and the function

4.3 Reduction of Superharmonic Functions

u=

λGΩ (x0 , ·) u

161

on Ω0 on Ω ∼ Ω0

ˆ u ≤ Ru ≤ λGΩ (x0 , ·) on Ω0 and it follows belongs to ΦuΓ ; therefore, 0 ≤ R Γ Γ ˆ u on Ω0 is the zero function. Since that the greatest harmonic minorant of R Γ ˆ u on Ω is Ω0 is any component of Ω, the greatest harmonic minorant of R Γ u ˆ is a potential by Riesz’s representation theorem, the zero function and R Γ Theorem 3.5.11. Suppose now that u is unbounded on Γ . By the Riesz representation theorem, u = v + h where v is a potential and h is harmonic on ˆ v is a potential. Since h ˆ v ≤ Rv ≤ v and v is a potential, R Ω. Since 0 ≤ R Γ Γ Γ h ˆ is a potential on Ω by the ﬁrst part of the proof. By is bounded on Γ, R Γ ˆv + R ˆ h . Since the sum of two potentials is again a potential, ˆu ≤ R (iv), R Γ Γ Γ u ˆ RΓ is a potential. Corollary 4.3.6 If u is a nonnegative superharmonic function on the Greenian set Ω, then there is an increasing sequence of potentials {uj } having compact supports such that limj→∞ uj = u. Proof: Let Ωj } be an increasing sequence of open sets with compact closures ˆ u − is a potential that Ωj− ⊂ Ω such that Ω = ∪Ωj . For each j ≥ 1, uj = R Ω j

agrees with u on Ωj and uj ↑ u.

Corollary 4.3.7 If u and v are nonnegative, bounded, superharmonic functions on Ω and Λ ⊂ Ω, then ˆ uΛ (x) − R ˆ vΛ (x)| ≤ sup |u(x) − v(x)|. sup |R

x∈Ω

x∈Ω

ˆv ≤ R ˆ u+c ≤ R ˆu + Proof: Let c = supx∈Ω |u(x) − v(x)|. Then v − u ≤ c, R Λ Λ Λ c u ˆ ≤R ˆ +c. Interchange u and v and combine the two inequalities to obtain R Λ Λ the result. Theorem 4.3.8 If Γ is a compact subset of a connected Greenian set Ω, ˆ 1 is identically zero on Ω if and only if Γ is a polar set. then R Γ ˆ 1 is identically zero on Ω. Since R ˆ 1 = R1 on Ω ∼ Γ, R1 Proof: Suppose R Γ Γ Γ Γ is identically zero on Ω ∼ Γ . Let x0 be any point of Ω ∼ Γ . For each ∞ n ≥ 1, choose un ∈ Φ1Γ such that un (x0 ) < 1/2n , and let u = n=1 un . By Theorem 2.4.8, u is either superharmonic or identically +∞ on Ω. Since ∞ u(x0 ) = n=1 un (x0 ) ≤ 1, u is superharmonic on Ω. Since un ≥ 1 on Γ, u = +∞ on Γ and Γ is therefore polar. Suppose now that Γ is a compact ˆ 1 on Ω ∼ Γ . Since u is harmonic on Ω ∼ Γ , it polar subset of Ω. Let u = R Γ ˆ1 ; has a unique superharmonic extension to Ω by Theorem 4.2.15, namely R Γ 1 ˆ = GΩ μ, μ by Corollary 4.2.16, the extension is harmonic on Ω. Letting R Γ ˆ 1 is identically zero on Ω. must be the zero measure by Corollary 3.4.6 and R Γ

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Theorem 4.3.9 If Γ is a compact nonpolar subset of a connected Greenian set Ω having a Green function, then there is a measure μ with support in Γ such that GΩ μ is positive, continuous, and bounded on Ω. ˆ 1 is positive at some point of Ω by Proof: Since Γ is nonpolar, the potential R Γ the preceding lemma and therefore positive on Ω by the minimum principle. ˆ 1 = GΩ ν for some measure ν with support in Γ , by Theorem 3.6.2 Since R Γ there is a measure μ with support in Γ such that GΩ μ is positive and conˆ 1 ≤ 1, GΩ μ is bounded on Ω. tinuous on Ω. Since GΩ μ ≤ GΩ ν = R Γ Lemma 4.3.10 If u is superharmonic on By,ρ and 0 < r < ρ, then there is a superharmonic function u˜ that agrees with u on By,r and is ﬁnite on ∼ By,r . Proof: By reducing ρ slightly it can be assumed that u is bounded below on − and superharmonic on a neighborhood of By,ρ . It also can be assumed By,ρ − − ˆ u , the regularized reduction that u ≥ 0 on By,ρ . Let Γ = By,r and consider R Γ ˆ u = u on By,r and is harmonic on of u over Γ relative to By,ρ . Now R Γ − ˆ u is the potential of a measure By,ρ ∼ By,r . By the preceding theorem, R Γ − ˆ u (z) = 0 for having support in By,r ⊂ By,ρ . By Theorem 3.4.16, limz→z0 R Γ all z0 ∈ ∂By,ρ . Deﬁne u ˆ (z) R if z ∈ By,ρ Γ w(z) = 0 if z ∈ ∂By,ρ . By Theorem 1.9.4, w can be continued harmonically across ∂By,ρ ; that is, − there is a number δ < ρ − r and a function w ˜ deﬁned on By,ρ+δ such that − w ˜ agrees with w on ∂By,ρ and is harmonic on By,ρ+δ ∼ By,ρ−δ . Now choose ˜ on ∂By,ρ and αuy + β ≤ w ˜ on α and β such that αuy + β agrees with w By,ρ+δ/2 ∼ B y,ρ . Then u˜ =

w ˜ αuy + β

− on By,ρ − on ∼ By,ρ

has the desired properties. Theorem 4.3.11 If Z ⊂ Rn is a polar set and x0 ∈ Z, then there is a superharmonic function u such that u = +∞ on Z and u(x0 ) < +∞; moreover, if Z is a polar subset of the Greenian set Ω and x0 ∈ Ω ∼ Z, then there is a potential u on Ω with these properties. Proof: Let v be a superharmonic function taking on the value +∞ on Z. For each y ∈ Z, let By,ρy be a ball such that x0 ∈ By,ρy . A countable number of balls corresponding to a sequence {yj } in Z suﬃces to cover Z. Let Bj be the ball corresponding to yj . By the preceding lemma, there is a superharmonic function vj such that vj = +∞ on Bj ∩ Z and is ﬁnite

4.4 Capacity

163

outside Bj− . Since vj is bounded below on Bj− , there is a constant αj such that vj + αj ≥ 0 on Bj− . Now let {βj } be a sequence of positive numbers such that j βj (vj (x0 ) + αj ) converges. By Theorem 2.4.8, the function u = j βj (vj + αj ) has the desired properties. As to the second assertion, let μ be the measure associated with the function u by the Riesz representation theorem, Theorem 3.5.11. Consider the sequence of balls {Bj0 } where Bj0 = Bx0 ,j , j ≥ 1. Since u(x0 ) is ﬁnite and u(x0 ) = GBj0 μ|Bj0 (x0 ) + hj (x0 ) by the same theorem, where hj is harmonic on Bj0 , GBj0 μ|Bj0 (x0 ) is ﬁnite for each ˜ j where h ˜ j is harmonic on B ) by j ≥ 1. Since GΩ μ|B 0 (x0 ) = GB 0 μ|B 0 (x0 )+ h j

j

j

j

Theorem 3.4.8, GΩ μ|Bj0 (x0 ) is also ﬁnite for each j ≥ 1. Thus, for each j ≥ 1 a cj > 0 can be chosen so that cj μ|Bj0 (Ω) < 1/2j and cj GΩ μ|Bj0 (x0 ) < 1/2j . ∞ Since ∞ j≥1 cj μ|Bj0 is a ﬁnite measure on Ω, GΩ j=1 cj μ|Bj0 is a potential on Ω by Theorem 3.4.4. This fact along with the inequality k j=1

∞ GΩ (cj μ|Bj0 ) ≤ GΩ ( cj μ|Bj0 ) j=1

implies that the sequence of superharmonic functions ∞ on the left increases to a potential function v on Ω. Note that v(x0 ) = j=1 cj GΩ μ|Bj0 (x0 ) ≤ 1. Since u = GBj0 μ|Bj0 + hj where hj is harmonic on Bj , GBj0 μ|Bj0 = +∞ on ˜ j where h ˜ j is harmonic on Bj0 ∩Z for each j ≥ 1. Since GΩ μ|Bj0 = GBj0 μ|Bj0 + h ∞ 0 0 Bj , GΩ μ|Bj0 = +∞ on Bj ∩ Z for each j ≥ 1. Thus, v = j=1 cj GΩ μ|Bj0 = +∞ on Z. Corollary 4.3.12 There are discontinuous ﬁnite-valued superharmonic functions. Proof: Let Z be a nonclosed polar set and let x be a limit point of Z that is not in Z. By the preceding theorem, there is a superharmonic function v such that v(x) < +∞ and v = +∞ on Z. Letting u = min (v, v(x) + 1), u is a ﬁnite-valued superharmonic function that takes on the value v(x)+1 on Z.

4.4 Capacity If an electrical charge is placed on a conductor in the interior of a grounded sphere, the charge will distribute itself on the surface of the conductor in such a way that the potential within the conductor is constant. The ratio of the total charge to the value of this constant is known as the capacity of the conductor. As the capacity depends only upon the ratio, the capacity is equal to the total charge if the potential is equal to 1 within the conductor.

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4 Negligible Sets

In the terminology of the preceding section, if the conductor is represented ˆ 1 = GB μ, where μ is by a compact subset Γ of a ball B, the potential is R Γ some measure on ∂Γ , and the capacity of Γ is just μ(∂Γ ). The mathematical concept of capacity was deﬁned ﬁrst by Wiener [63] for compact sets in the n ≥ 3 case in the following way. Let Γ be a compact set and let u be a harmonic function on ∼ Γ corresponding to the boundary function 1 on ∂Γ and satisfying lim|x|→∞ u(x) = 0. Then Wiener deﬁned the capacity C(Γ ) of Γ by the equation 1 Dn u dσ C(Γ ) = − 4π Σ where Σ is a smooth surface encompassing Γ . Throughout this section, Ω will be a Greenian subset of Rn and Γ , with or without subscripts, will denote a compact subset of Ω. The collection of compact subsets of Ω will be denoted by K(Ω) and the collection of open subsets by O(Ω). When u = 1 on Ω, the superscript 1 will be omitted from ˆ 1 . As a function of Γ , the maps Γ → RΓ and Γ → R ˆ Γ will be R1Γ and R Γ ˆ denoted by R and R , respectively. ˆ Γ is called the capacitary poDeﬁnition 4.4.1 For each Γ ∈ K(Ω), R ˆ Γ = GΩ μΓ is called the tential of Γ . The unique measure μΓ for which R capacitary distribution for Γ . The capacity of Γ relative to Ω, denoted by C(Γ ), is deﬁned by the equation C(Γ ) = μΓ (Γ ) with C(∅) = 0. The deﬁnition of capacity of a compact set Γ agrees with Wiener’s deﬁnition for the case Ω = Rn , n ≥ 3. Example 4.4.2 If Γ is a compact subset of Ω = Rn , n ≥ 3, and B is any ball containing Γ , then 1 ˆ Γ (x) dσ(x) C(Γ ) = − Dn(x) R σn (n − 2) ∂B ˆ Γ is relative to Ω. This formula for calculating C(Γ ) can be derived where R as follows. In this case, 1 ˆ Γ (x) = R μΓ (dy). |x − y|n−2 Γ Using Tonelli’s theorem, 1 ˆ Γ (x) dσ(x) − Dn(x) R σn (n − 2) ∂B 1 1 Dn(x) μΓ (dy) dσ(x) =− σn (n − 2) ∂B |x − y|n−2 Γ 1 1 Dn(x) dσ(x)μΓ (dy). =− σn (n − 2) Γ ∂B |x − y|n−2

4.4 Capacity

165

− Letting By,δ ⊂ By,δ ⊂ B and applying Green’s identity, Theorem 1.2.2, to − the region B ∼ By,δ , the last expression is equal to

−

1 σn (n − 2)

Dn(x) Γ

∂By,δ

1 dσ(x)μΓ (dy) = μΓ (Γ ) = C(Γ ). |x − y|n−2

− Example 4.4.3 If D = Bx,δ ⊂ Ω = Rn , n ≥ 3, then

ˆ D (y) = R so that 1 C(D) = − σn (n − 2)

1 δ n−2 |x−y|n−2

if y ∈ D if y ∈ D

∂Bx,2δ

ˆ D (y) dσ(y) = δ n−2 . Dn(y) R

Thus, the capacity of a closed ball in R3 is its radius. − ∼ Bx,ρ , 0 < ρ < δ, is an annulus in Ω = Example 4.4.4 If A = Bx,δ n ˆ ˆ D of the preceding example so that R , n ≥ 3, then RA is the same as the R n−2 . This is in keeping with the intuitive notion that an electrical C(A) = δ charge placed on the annulus will gravitate to the outer sphere.

As Lebesgue measure theory commences with the area of simple regions, it is possible to extend the concept of capacity commencing with compact sets in an analogous way. Deﬁnition 4.4.5 If E is a subset of Ω, the inner capacity of E, denoted by C∗ (E), is deﬁned by C∗ (E) = sup{C(K); K ∈ K(Ω), K ⊂ E}. Clearly, C∗ (K) = C(K) if K ∈ K(Ω), and C∗ is a nonnegative extended realvalued set function on the class of all subsets of Ω with C(∅) = C∗ (∅) = 0. Deﬁnition 4.4.6 If E is a subset of the Greenian set Ω, the outer capacity of E, denoted by C ∗ (E), is deﬁned by C ∗ (E) = inf{C∗ (O) : O ∈ O(Ω), O ⊃ E}. The outer capacity C ∗ is a nonnegative extended real-valued set function on the class of all subsets of Ω.

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4 Negligible Sets

Deﬁnition 4.4.7 A subset E of Ω is capacitable if C∗ (E) = C ∗ (E), in which case the capacity C(E) is deﬁned to be the common value. As is the case for Lebesgue measure, every Borel subset of Ω is capacitable. The proof of this fact requires several steps and will be completed only in the next section. The initial steps are limited to Greenian sets Ω ⊂ Rn having a certain property, known at this stage to be valid only for balls. The theorems will be stated, however, for any Greenian set as it will be shown in the next section that every Greenian set has this property. Property B: The open set Ω has a Green function and limy→x,y∈Ω GΩ μ(y) = 0 quasi everywhere on ∂Ω for all measures μ having compact support in Ω. If Ω is an unbounded open subset of Rn , then lim|y|→+∞,y∈Ω GΩ μ(y) = 0 since GΩ is dominated by the Green function for Rn , namely x − y −n+2 . According to Theorem 3.4.16, balls are known to have Property B. It will be shown in the next section that every open set having a Green function has Property B. Some of the proofs of the following results do not require that this property hold. If this property is required, it will be indicated at the beginning of the proof. The following theorem is known as the domination principle. Theorem 4.4.8 (Maria [42], Frostman [22]) Let Ω be a Greenian set, μ a measure on Ω, and v a positive superharmonic function on Ω. If GΩ μ < +∞ a.e.(μ) on the support of μ and GΩ μ ≤ v inner q.e. on the support of μ, then GΩ μ ≤ v on Ω. Proof: (Assuming Property B) According to Theorem 3.6.3, there is a sequence of measures {μn } such that each μn in the restriction of μ to a compact subset of the support of μ, GΩ μn is ﬁnite-valued and continuous on Ω, and GΩ μn ↑ GΩ μ. Since GΩ μn ≤ GΩ μ < +∞ a.e.(μ), GΩ μn < +∞ a.e.(μn ) and GΩ μ ≤ v inner quasi everywhere on the support of μn . If it can be shown that GΩ μn ≤ v on Ω, it would then follow that GΩ μ ≤ v on Ω. It therefore suﬃces to prove the result assuming that μ has compact support Γ ⊂ Ω and GΩ μ is ﬁnite-valued and continuous on Ω. If x is a point of Γ for which GΩ μ(x) ≤ v(x), then lim inf v(y) ≥ v(x) ≥ GΩ μ(x) =

y→x,y∈Ω

lim

y→x,y∈Ω

GΩ μ(y).

Thus, lim inf (v − GΩ μ)(y) ≥ 0

y→x,y∈Ω

for x ∈ Γ except possibly for an inner polar set. This inequality also holds for x ∈ ∂Ω except possibly for a polar set and, in case Ω is an unbounded subset of Rn , n ≥ 3, x = ∞. By Theorem 4.2.21, v − GΩ μ ≥ 0 on Ω.

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167

Remark 4.4.9 Let F = {ui ; i ∈ I} be a left-directed family of locally uniformly bounded below superharmonic functions on Ω, and let uˆ be the lower regularization of u = inf i∈I ui . Then there is a decreasing sequence {wj } in F such that u ˆ is the lower regularization of inf j≥1 wj . This can be seen as follows. By Theorem 2.2.8, there is a countable set I0 ⊂ I such that if g is l.s.c. on Ω and g ≤ inf j≥1 uij , then g ≤ inf i∈I ui . Letting w1 = ui1 and inductively replacing uij by a wj ∈ F satisfying wj ≤ min (uij , wj−1 ), the decreasing sequence {wj } in F has the property that inf i∈I ui ≤ inf j≥1 wj ≤ inf j≥1 uij . Letting w = inf j≥1 wj , u ˆ ≤ w; ˆ but since w ˆ is l.s.c. and w ˆ ≤ inf j≥1 wj ≤ ˆ ≤ inf i∈I ui . Therefore, w ˆ≤u ˆ and the two are equal; that is, u ˆ inf j≥1 uij , w is the lower regularization of a decreasing sequence in F . It is necessary to prove the following theorem in two steps, the ﬁrst establishing the conclusion inner quasi everywhere. The proof will be revisited. Theorem 4.4.10 (Cartan [11]) Let F = {ui ; i ∈ I} be a family of locally uniformly bounded below superharmonic functions on an open set Ω, and let ˆ = u q.e. on Ω. u = inf i∈I ui . Then u Partial proof:(ˆ u = u inner quasi everywhere) Since adjoining the minima of all ﬁnite subsets of F to F has no eﬀect on u or u ˆ, it can be assumed that F is left-directed. Since the conclusion is a local property, it can be assumed that Ω is a ball and that all ui are nonnegative on Ω. By the preceding remark, it can be assumed that F is a decreasing sequence {uj } of nonnegative superharmonic functions with u = limj→∞ uj . Furthermore, it suﬃces to prove that u = u ˆ inner quasi everywhere on a slightly smaller concentric ˆ uj , the regularized reduction ball B ⊂ Ω. Also, each uj can be replaced by R B u ˆ j on B. Since R ˆ uj is a potential on of uj on B relative to Ω, since uj = R B B ˆ uj = GΩ μj for some measure μj with support in B − . It therefore can be Ω, R B assumed that uj = GB μj , j ≥ 1. By taking a concentric ball slightly larger than B and the reduction of 1 over it, a measure λ can be found so that GB λ = 1 on B − . By the reciprocity theorem, for j ≥ 1, − μj (B ) = 1 dμj = GΩ λ dμj = GΩ μj dλ ≤ Gμ1 dλ; by reversing the steps, the latter integral is equal to μ1 (B − ), and therefore, μj (B − ) ≤ μ1 (B − ) < +∞ for all j ≥ 1. Thus, the set {μj (B − ) : j ≥ 1} is bounded, and there is a subsequence of the sequence {μj } that converges in the w∗ - topology to some measure μ with support in B − . It can be assumed that the sequence itself has this property. If m is any positive number, then u = lim uj ≥ lim min (GΩ (·, y), m) dμj (y) = min (GΩ (·, y), m) dμ(y). j→∞

j→∞

Letting m ↑ ∞, u ≥ GΩ μ on Ω. Consider the set M = {u > u ˆ}, let Γ be any compact subset of M , and assume there is a measure ν with support in Γ

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4 Negligible Sets

such that GΩ ν is bounded and continuous on Ω. By the reciprocity theorem, Theorem 3.5.1, GΩ μ dν =

GΩ ν dμ < +∞.

Since GΩ μj ≤ GΩ μ1 , j ≥ 1, u dν = lim GΩ μj dν = lim GΩ ν dμj = GΩ ν dμ = GΩ μ dν; j→∞

j→∞

that is, (u − GΩ μ) dν = 0. Since u ≥ GΩ μ on Ω, u = GΩ μ a.e.(ν), u ≥ uˆ ≥ ˆ a.e.(ν). Thus, ν(Γ ) = ν(M ) = 0. It follows from GΩ μ and therefore u = u Theorem 4.3.9 that Γ is a polar set and that u = u ˆ inner quasi everywhere. Lemma 4.4.11 If Γ ∈ K(Ω) and u is a ﬁnite-valued, nonnegative superˆ u = Ru except possibly at points of ∂Γ , harmonic function on Ω, then R Γ Γ u u u ˆ ˆ RΓ = RΓ = u q.e. on Γ , and RΓ = RuΓ q.e. on Ω. Proof: (Assuming Property B) The ﬁrst assertion follows from Lemma 4.3.2. ˆ u inner quasi everywhere on By the proved part of Theorem 4.4.10, RuΓ = R Γ ˆ u = u inner quasi everywhere on Γ . Since Ω. Since RuΓ = u on Γ, R Γ ˆ uΓ < u} = Γ ∩ {R

∞ % ˆ uΓ ≤ (1 − 1 )u} Γ ∩ {R j j=1

and each term of the union is polar, the countable union is polar by Theorem 4.2.12. Lemma 4.4.12 If u is a ﬁnite-valued, positive, superharmonic function on Ω, then RuΓ1 ∪Γ2 + RuΓ1 ∩Γ2 ≤ RuΓ1 + RuΓ2 (4.1) and ˆ uΓ ∪Γ + R ˆ uΓ ∩Γ ≤ R ˆ uΓ + R ˆ uΓ R 1 2 1 2 1 2

(4.2)

for all Γ1 , Γ2 ∈ K(Ω). Proof: (Assuming Property B) If x ∈ Γ1 ∩ Γ2 , then (4.1) is true since all terms are equal to 1. If, for example, x ∈ Γ1 ∼ Γ2 , then RuΓ1 ∪Γ2 (x)+RuΓ1 ∩Γ2 (x) = u(x)+RuΓ1 ∩Γ2 (x) ≤ u(x)+RuΓ2 (x)=RuΓ1 (x)+RuΓ2 (x). u u u ˆu ˆu ˆu By the preceding lemma, R Γ1 ∪Γ2 = RΓ1 ∪Γ2 , RΓ1 ∩Γ2 = RΓ1 ∩Γ2 , RΓ1 = RΓ1 , ˆ u = Ru quasi everywhere on Γ1 ∪ Γ2 . Thus, (4.2) of the lemma and R Γ2 Γ2 holds quasi everywhere on Γ1 ∪ Γ2 . By Lemma 3.3.6, the greatest harmonic minorant of the left side of the latter inequality is the zero function and is therefore the potential of a measure with support in Γ1 ∪ Γ2 . It follows from the domination principle that

4.4 Capacity

169

ˆ uΓ ∪Γ + R ˆ uΓ ∩Γ ≤ R ˆ uΓ + R ˆ uΓ R 1 2 1 2 1 2

on Ω.

u ˆu Since RuΓ1 ∪Γ2 is harmonic on ∼ (Γ1 ∪ Γ2 ), R Γ1 ∪Γ2 = RΓ1 ∪Γ2 on ∼ (Γ1 ∪ Γ2 ), etc. Thus, (4.1) holds on ∼ (Γ1 ∪ Γ2 ), and since it already has been seen to hold on Γ1 ∪ Γ2 , it holds on Ω.

Lemma 4.4.13 If u is a ﬁnite-valued, nonnegative superharmonic function ˆ u is the greatest potential of measures μ with on Ω and Γ ∈ K(Ω), then R Γ support in Γ such that GΩ μ ≤ u. Proof: (Assuming Property B) Let μ be any measure with compact support in Γ satisfying GΩ μ ≤ u on Ω. Since u is ﬁnite-valued on Ω, GΩ μ < +∞ on Ω. If v ∈ ΦuΓ , then GΩ μ ≤ u ≤ v on Γ so that GΩ μ ≤ v on Ω by the domination principle Theorem 4.4.8. Thus, RuΓ ≥ GΩ μ on Ω and since GΩ μ ˆ u ≥ GΩ μ. Since R ˆ u is the potential of a measure ν with is l.s.c. on Ω, R Γ Γ u ˆ = GΩ ν ≥ GΩ μ for all measures μ with support in Γ . support in Γ, R Γ Lemma 4.4.14 Let u be a ﬁnite-valued, continuous, positive superharmonic function on Ω and let {Γj } be a monotone sequence in K(Ω) with limit Γ ∈ K(Ω). If {Γj } is an increasing sequence, then limj→∞ RuΓj = RuΓ and ˆu = R ˆ u on Ω. limj→∞ R Γj Γ (ii) If {Γj } is a decreasing sequence, then limj→∞ RuΓj = RuΓ on Ω.

(i)

Proof: (Assuming Property B) Suppose ﬁrst that {Γj } is an increasing sequence. Since RuΓj = u on Γj and RuΓ = u on Γ, limj→∞ RuΓj = RuΓ on Γ . Since RuΓj ≤ RuΓ on Ω, limj→∞ RuΓj ≤ RuΓ on Ω ∼ Γ and limj→∞ RuΓj ≤ RuΓ ˆ u = Ru q.e. on Ω, R ˆ u = Ru q.e. on Ω, and the union of on Ω. Since R Γj

Γj

Γ

Γ

ˆu ≤ R ˆ u on Ω. Since countably many polar sets is again polar, limj→∞ R Γj Γ u u ˆ is superharmonic on Ω by Theorem 2.4.8, R ˆu ˆ = limj→∞ R limj→∞ R Γj Γ Γj ˆ u is a potential, R ˆ u ≤ limj→∞ R ˆ u on Ω by the domiq.e. on Γ , and R Γ

Γ

Γj

ˆu = R ˆ u on Ω. Since nation principle, Theorem 4.4.8. Therefore, limj→∞ R Γj Γ ˆ u = Ru and R ˆ u = Ru on Ω ∼ Γ, limj→∞ Ru = Ru on Ω. Now let {Γj } R Γj Γj Γ Γ Γj Γ be decreasing sequence of compact subsets of Ω. Deﬁning w = limj→∞ RuΓj on Ω and applying Theorem 2.4.9, the lower regularization w ˆ of w is superharmonic on Ω. By the Riesz decomposition theorem, w ˆ = GΩ μ + h where μ is a measure on Ω and h is the greatest harmonic minorant of ˆ u on Ω. Since R ˆ u is a pow. ˆ Since 0 ≤ w ≤ RuΓ1 on Ω, 0 ≤ w ˆ ≤ R Γ1 Γ1 tential, the greatest harmonic minorant of w ˆ is the zero function; that is, h = 0 and w ˆ = GΩ μ. For ﬁxed j0 ≥ 1, consider the functions RuΓj , j ≥ j0 , which are harmonic on Ω ∼ Γj0 . By Theorem 2.2.5, w is harmonic and therefore equal to w ˆ on Ω ∼ Γj0 . Since j0 ≥ 1 is arbitrary, w ˆ is harmonic on Ω ∼ Γ . By Theorem 3.4.6, the support of μ is contained in Γ .

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4 Negligible Sets

ˆ u by the preceding lemma. Since Since w ˆ = GΩ μ ≤ u on Ω, w ˆ ≤ R Γ u u u ˆ ˆ u . Taking the lower regularization RΓj ≥ RΓ ≥ RΓ for all j ≥ 1, w ≥ R Γ ˆ u and consequently w ˆ u . Since w is harmonic on of both sides, w ˆ≥R ˆ =R Γ Γ u u u ˆ Ω ∼ Γ, w ˆ = w = limj→∞ RΓj ≥ RΓ ≥ RΓ and RuΓ = limj→∞ RuΓj on Ω ∼ Γ . Since RuΓj = u on Γ and RuΓ = u on Γ , RuΓ = limj→∞ RuΓj on Ω. Example 4.4.15 Part (ii) of the preceding lemma is not true for regularized reduced functions. For example, if u = 1 and Γj = {x; |x| ≤ 1/j}, then ˆ 1 (0) = 0 = R ˆ 1 (0). 1 = limj→∞ R Γj {0} ˆ are functions of two variables x It should be remembered that R and R and Γ . Although the variable x will be suppressed in the following, it should be kept in mind that x is ﬁxed. Except for the last item in the following list, ˆ Γ were proved in the preceding section. the following properties of RΓ and R ˆ Γ ≥ 0 on Ω. (i) 1 ≥ RΓ ≥ R (ii) RΓ = 1 on Γ . ˆ Γ on the interior of Γ . (iii) 1 = RΓ = R ˆ (iv) RΓ = RΓ on Ω ∼ Γ and both are harmonic on Ω ∼ Γ . ˆ Γ is the potential of a measure with support in Γ . (v) R The last fact follows from Theorems 3.4.4 and 3.4.9. It has been seen that it ˆ Γ (x) < RΓ (x) for some x. It was shown in Lemma 4.4.11 is possible to have R that the set for which this is true is polar, assuming Property B. It will take several steps to show that the set of such points is a negligible set for Greenian sets in general. Deﬁnition 4.4.16 A nonnegative, extended real-valued set function φ deﬁned on a class of sets F is strongly subadditive on F if F is closed under ﬁnite unions and ﬁnite intersections and A, B ∈ F ⇒ φ(A ∪ B) + φ(A ∩ B) ≤ φ(A) + φ(B), and (ii) countably strongly subadditive on F if F is closed under countable unions and countable intersections and (i)

φ(∪∞ j=1 Bj ) +

∞ j=1

φ(Aj ) ≤

∞

φ(Bj ) + φ(∪∞ j=1 Aj )

j=l

whenever {Aj } and {Bj } are sequences in F satisfying Aj ⊂ Bj , j ≥ 1. Lemma 4.4.17 The nonnegative, ﬁnite-valued set function C(Γ ) deﬁned on K(Ω) has the following properties. (i) (Monotonicity) C(∅) = 0 and Γ1 ⊂ Γ2 ⇒ C(Γ1 ) ≤ C(Γ2 ). (ii) If {Γj } is a monotone sequence in K(Ω) with limit Γ ∈ K(Ω), then limj→∞ C(Γj ) = C(Γ ). (iii) (Strong subadditivity) C(Γ1 ∪ Γ2 ) + C(Γ1 ∩ Γ2 ) ≤ C(Γ1 ) + C(Γ2 ).

4.4 Capacity

171

Proof: (Assuming Property B) Only (ii) will be proved. The same method can be used to prove (i) and (iii) using Theorem 4.3.5 and Lemma 4.4.12, respectively. Note that RΓj → RΓ by the preceding lemma. Let Λ be an open ˆ Λ− = GΩ ν set having compact closure Λ− with ∪Γj ⊂ Λ ⊂ Λ− ⊂ Ω, and let R ˆ where ν is a measure with compact support in ∂Λ. Thus, RΓj = RΓj and ˆ Γ = RΓ a.e.(ν) since these functions diﬀer only at points in Λ. By the R reciprocity theorem, Theorem 3.5.1, C(Γj ) = Γj 1 dμΓj = GΩ ν dμΓj = ˆ Γj dν = RΓj dν → RΓ dν. Reversing the steps, the GΩ μΓj dν = R latter integral is just C(Γ ) and so C(Γj ) → C(Γ ) as j → ∞. It is apparent that C∗ (E) is an increasing function of E ⊂ Ω and that C∗ (Γ ) = C(Γ ) for all compact sets Γ ⊂ Ω. At the beginning of this section, a subset E of the Greenian set Ω was deﬁned to be capacitable if C∗ (E) = C ∗ (E) and the capacity C(E) was deﬁned to be the common value. It is easy to see that C∗ (E) ≤ C ∗ (E) for all E ⊂ Ω and that the open sets are capacitable. Remark 4.4.18 The part of (ii) of the preceding lemma pertaining to monotone decreasing sequences is equivalent to the following assertion. (ii ) (Right-continuity) If Γ ∈ K(Ω) Ω and > 0, then there is a neighborhood Λ of Γ such that C(Σ) − C(Γ ) < for all Σ ∈ K(Ω) satisfying Γ ⊂ Σ ⊂ Λ. To see that (ii) implies (ii ), let {Λj } be a decreasing sequence of open sets such that Γ ⊂ Λj , j ≥ 1, and Γ = ∩Λ− j . By (ii), given > 0 there is a ) < C(Γ ) + . For any compact set Σ with Γ ⊂ Σ ⊂ j0 ≥ 1 such that C(Λ− j0 − Λj0 , C(Σ) ≤ C(Λj0 ) < C(Γ ) + and C(Σ) − C(Γ ) < . Assuming (ii ), let {Γj } be a decreasing sequence of compact subsets of Ω with Γ = ∩Γj . Given > 0, let Λ be a neighborhood of Γ such that C(Σ) − C(Γ ) < whenever Σ is a compact set with Γ ⊂ Σ ⊂ Λ. There is then a j0 ≥ 1 such that Γ ⊂ Γj ⊂ Λ for all j ≥ j0 so that C(Γ ) ≤ C(Γj ) < C(Γ ) + for all j ≥ j0 ; that is, limj→∞ C(Γj ) = C(Γ ). It was noted previously that all open sets are capacitable and that C(Γ ) = C∗ (Γ ) for all Γ ∈ K(Ω). Note that if O ∈ O(Ω) and Γ is a compact subset of O, then C∗ (O) = sup {C(K); K ∈ K(Ω), Γ ⊂ K ⊂ O}. Lemma 4.4.19 Compact and open subsets of Ω are capacitable. Proof: (Assuming Property B) By (ii ) of the preceding remark, given > 0 there is an open set Λ such that C(Σ) − C(Γ ) < for any compact set Σ satisfying Γ ⊂ Σ ⊂ Λ. Thus, C∗ (Γ ) ≤ C∗ (Λ) = sup {C(Σ); Σ ∈ K(Ω), Γ ⊂ Σ ⊂ Λ} ≤ C(Γ ) + .

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4 Negligible Sets

If O is any open set satisfying Γ ⊂ O ⊂ Λ, then C∗ (Γ ) ≤ C∗ (O) ≤ C∗ (Λ) and C(Γ ) = C∗ (Γ ) ≤ inf {C∗ (O); O ∈ O(Ω), Γ ⊂ O} = inf {C∗ (O); O ∈ O(Ω), Γ ⊂ O ⊂ Λ} ≤ C∗ (Λ) < C(Γ ) + . But since the ﬁrst inﬁmum is just C ∗ (Γ ) and is arbitrary, C(Γ ) = C∗ (Γ ) = C ∗ (Γ ). Extending the conclusion of this lemma to Borel sets requires several steps. Lemma 4.4.20 If A1 , . . . , Am , B1 , . . . , Bm ∈ K(Ω) with Aj ⊂ Bj , j = 1, . . . , m, then m m m (C(Bj ) − C(Aj )). C(∪j=1 Bj ) − C(∪j=1 Aj ) ≤ j=1

Proof: (Assuming Property B) Consider the m = 2 case ﬁrst. Taking Γ1 = B1 and Γ2 = A1 ∪ B2 in (iii) and using (i) of Lemma 4.4.17, C(B1 ∪ B2 ) + C(A1 ) ≤ C(B1 ∪ B2 ) + C(A1 ∪ (B1 ∩ B2 )) ≤ C(B1 ) + C(A1 ∪ B2 ); also taking Γ1 = A1 ∪ A2 and Γ2 = B2 , C(A1 ∪ B2 ) + C(A2 ) ≤ C(A1 ∪ B2 ) + C((A1 ∪ B2 ) ∪ A2 ) ≤ C(A1 ∪ A2 ) + C(B2 ). Thus, C(B1 ∪ B2 ) + C(A1 ) + C(A2 ) ≤ C(B1 ) + C(A1 ∪ B2 ) + C(A2 ) ≤ C(B1 ) + C(A1 ∪ A2 ) + C(B2 ) and the assertion is true for m = 2. Assume the assertion is true for m − 1. By the m = 2 case, m−1 m−1 m C(∪m j=1 Bj ) − C(∪j=1 Aj ) = C(∪j=1 Bj ∪ Bm ) − C(∪j=1 Aj ∪ Am ) m−1 ≤ (C(∪m−1 j=1 Bj ) − C(∪j−1 Aj )) + (C(Bm ) − C(Am ))

≤ =

m−1

(C(Bj ) − C(Aj )) + (C(Bm ) − C(Am ))

j=1 m

(C(Bj ) − C(Aj ))

j=1

and the assertion is true for m whenever it is true for m − 1. The conclusion follows by induction. Theorem 4.4.21 The outer capacity has the following properties. (i)

A, B ⊂ Ω, A ⊂ B ⇒ C ∗ (A) ≤ C ∗ (B).

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173

(ii) Aj ⊂ Ω, j ≥ 1, Aj ↑ A ⇒ C ∗ (Aj ) → C ∗ (A). (iii) Aj ∈ K(Ω), j ≥ 1, Aj ↓ A ⇒ C ∗ (Aj ) → C ∗ (A). (iv) Aj , Bj ⊂ Ω, Aj ⊂ Bj , j ≥ 1 ⇒ C ∗ (∪∞ j=1 Bj ) +

∞

C ∗ (Aj ) ≤

j=1

∞

C ∗ (Bj ) + C ∗ (∪∞ j=1 Aj ).

j=1

Proof: (Assuming Property B) Assertion (i) is trivial and assertion (iii) is the same as (ii) of Lemma 4.4.17. The other assertions will require several steps. Step 1. Oj ∈ O(Ω), j ≥ 1, Oj ↑ O ⇒ C(Oj ) → C(O). For each j ≥ 1, let Oj = ∪∞ k=1 Γjk with Γjk ∈ K(Ω), let Σj = ∪i,k≤j Γik , j ≥ 1, and let Γ be any compact subset of O. Thus, each Σj is compact, Σj ↑ O, and Σj ∩ Γ ↑ Γ . Since ∞ C(Oj ) = C ∪∞ k=1 Γjk = C ∪i≤j ∪k=1 Γik ≥ C ∪i,k≤j Γik ∩ Γ = C(Σj ∩ Γ ), by (ii) of Lemma 4.4.17 lim C(Oj ) ≥ lim C(Σj ∩ Γ ) = C(Γ ).

j→∞

j→∞

Since Γ is any compact subset of O, lim C(Oj ) ≥ C∗ (O) = C(O).

j→∞

As the opposite inequality is always true, limj→∞ C(Oj ) = C(O). Step 2. A1 , . . . , Am , B1 , . . . , Bm ⊂ Ω, Aj ⊂ Bj , j = 1, . . . , m ⇒ C ∗ (∪m j=1 Bj ) +

m j=1

C ∗ (Aj ) ≤

m

C ∗ (Bj ) + C ∗ (∪m j=1 Aj ).

(4.3)

j=1

This inequality will be proved for the m = 2 and open sets case ﬁrst. Let U1 , U2 , V1 , V2 be open sets with Uj ⊂ Vj , j = 1, 2. Let Γ1 , Γ2 , and Σ be compact subsets with Γ1 ⊂ U1 , Γ2 ⊂ U2 , and Σ ⊂ V1 ∪ V2 . The sets Σ ∼ V1 and Σ ∼ V2 are disjoint closed sets with Σ ∼ V1 ⊂ V2 and Σ ∼ V2 ⊂ V1 . There are disjoint open sets O1 and O2 such that Σ ∼ V1 ⊂ O1 and Σ ∼ V2 ⊂ O2 . Letting Σ1 = Σ ∼ O1 and Σ2 = Σ ∼ O2 , Σ = Σ1 ∪ Σ2 where Σ1 and Σ2 are compact subsets of V1 and V2 , respectively. By Lemma 4.4.20, C((Σ1 ∪ Γ1 ) ∪ (Σ2 ∪ Γ2 )) + C(Γ1 ) + C(Γ2 ) ≤ C(Σ1 ∪ Γ1 ) + C(Σ2 ∪ Γ2 ) + C(Γ1 ∪ Γ2 ).

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4 Negligible Sets

Since Σ ⊂ (Σ1 ∪ Γ1 ) ∪ (Σ2 ∪ Γ2 ) and Σ ∼ Oi ⊂ Vi , i = 1, 2, C(Σ) + C(Γ1 ) + C(Γ2 ) ≤ C(V1 ) + C(V2 ) + C(U1 ∪ U2 ). Taking the supremum over Γ1 , Γ2 , and Σ, C(V1 ∪ V2 ) + C(U1 ) + C(U2 ) ≤ C(V1 ) + C(V2 ) + C(U1 ∪ U2 ) so that Inequality (4.3) is true for m = 2 and open sets. As in the proof of Lemma 4.4.20, Inequality (4.3) is true for all open sets and m ≥ 2. Consider now arbitrary sets Aj , Bj , j = 1, . . . m. If some C ∗ (Bj ) = +∞ or C ∗ (∪m j=1 Aj ) = +∞, then Inequality (4.3) is trivially true. It therefore can be assumed that all terms on the right side of the inequality are ﬁnite. If > 0, there is an open set O ⊃ ∪m j=1 Aj and open sets Oj ⊃ Bj , j = 1, . . . , m, such that m m C(Oj ) + C(O) ≤ C ∗ (Bj ) + C ∗ (∪m j=1 Aj ) + . j=1

j=1

Letting Uj = Oj ∩ O ⊂ Oj , j = 1, . . . , m, C(∪m j=1 Oj )

+

m

C(Uj ) ≤

C(∪m j=1 Uj )

+

j=1

m

C(Oj ).

j=1

Since ∪m j=1 Uj ⊂ O, C(∪m j=1 Oj ) +

m j=1

C(Uj ) ≤ C(O) +

m

C(Oj )

j=1

≤ C ∗ (∪m j=1 Aj ) +

m

C ∗ (Bj ) + .

j=1 m Since ∪m j=1 Bj ⊂ ∪j=1 Oj , the ﬁrst term on the left is greater than or equal ∗ m to C (∪j=1 Bj ); since Aj ⊂ Bj ⊂ Oj and Aj ⊂ O, Aj ⊂ Oj ∩ O = Uj , j = 1, . . . , m, and the second term on the left is greater than or equal ∗ to m j=1 C (Aj ). Therefore,

C ∗ (∪m j=1 Bj ) +

m

C ∗ (Aj ) ≤ C ∗ (∪m j=1 Aj ) +

j=1

m

C ∗ (Bj ) + .

j=1

Since is arbitrary, the assertion is true for m ≥ 2 and arbitrary subsets of Ω. Step 3. Aj ⊂ Ω, j ≥ 1, Aj ↑ A ⇒ C ∗ (Aj ) → C ∗ (A). This result is trivial if limj→∞ C ∗ (Aj ) = +∞. It therefore can be assumed that the limit is ﬁnite. If > 0, for each j ≥ 1 an open set Oj ⊃ Aj can be chosen so that C(Oj ) < C ∗ (Aj ) + /2j . By the preceding step of the proof

4.4 Capacity

175 ∗ m C(∪m j=1 Oj ) ≤ C (∪j=1 Aj ) +

m

(C(Oj ) − C ∗ (Aj ))

j=1 ∗

< C (Am ) + . ∞ By the ﬁrst step of this proof, C(∪m j=1 Oj ) → C(∪j=1 Oj ) and ∞ ∗ C ∗ (A) = C ∗ (∪∞ j=1 Aj ) ≤ C(∪j=1 Oj ) ≤ lim C (Am ) + . m→∞

Since is arbitrary and limm→∞ C ∗ (Am ) ≤ C ∗ (A), C ∗ (A) = limm→∞ C ∗ (Am ). This completes the proof of (ii). Step 4. Assertion (iv) follows from Inequality (4.3) and the preceding step by letting m → ∞. In order to show that the class of capacitable sets includes the Borel sets, an excursion into set theory is required. After this excursion, Borel sets will be shown to be capacitable along with certain sets called analytic sets. Consider a pair (X, F ) where X is some nonempty set and F is a collection of subsets of X that includes the empty set. F is called a paving of X and the pair (X, F ) is called a paved space. Let N ∞ = N × N × · · · be the set of all inﬁnite sequences of positive integers. Elements of N ∞ will be denoted by n = (n1 , n2 , . . .). Deﬁnition 4.4.22 Let (X, F ) be a paved space. A Suslin scheme is a map (n1 , . . . , nk ) → Xn1 ...nk ∈ F from the set of all ﬁnite sequences of positive integers into F . (ii) The nucleus of the Suslin scheme is the uncountable union % (Xn1 ∩ Xn1 n2 ∩ Xn1 n2 n3 ∩ · · · );

(i)

n∈N ∞

this set is said to be analytic over F . (iii) The class of sets analytic over F will be denoted by A(F ). It is easy to see that F ⊂ A(F ) as follows. If A ∈ F, for each ﬁnite sequence (n1 , . . . , nk ) of positive integers let Xn1 ...nk = A. The map (n1 , . . . , nk ) → Xn1 ...nk is then a Suslin system and % A= (Xn1 ∩ Xn1 n2 ∩ · · · ) n∈N ∞

so that A ∈ A(F ). To show that a subset B of X belongs to A(F ), it must be shown that there is a Suslin system (n1 , . . . , nk ) → Xn1 ···nk so that % B= (Xn1 ∩ Xn1 n2 ∩ · · · ). n∈N ∞

176

4 Negligible Sets

Lemma 4.4.23 A(F ) is closed under countable unions and countable intersections. Proof: Let {Aj } be a sequence in A(F ), and let A = ∪Aj . For each j ≥ 1, (j) let (n1 , . . . , nk ) → Xn1 ···nk be a Suslin scheme with % Aj = (Xn(j) ∩ Xn(j) ∩ Xn(j) ∩ · · · ). 1 1 n2 1 n2 n3 n∈N ∞

Consider a one-to-one and onto map α : N → N × N . If (m1 · · · mk ) is a (j) ﬁnite sequence of positive integers, deﬁne Xm1 ···mk = Xn1 ···nk if α(m1 ) = (j, n1 ), mi = ni , i = 2, . . . k. Thus, % A= (Xm1 ∩ Xm1 m2 ∩ Xm1 m2 m3 ∩ · · · ) n∈N ∞

and A ∈ A(F ). Therefore, A(F ) is closed under countable unions. Now let {Bj } be sequence in A(F ), and let B = ∩Bj . For each j ≥ 1, let (j) (n1 , . . . , nk ) → Xn1 ···nk be a Suslin system with % (Xn(j) ∩ Xn(j) ∩ Xn(j) ∩ ···) Bj = j1 j1 nj2 j1 nj2 nj3 nj ∈N ∞

where nj = (nj1 , nj2 , . . .). Consider any x ∈ ∩Bj . To each such x there corresponds a sequence of sequences {nj } such that ' x∈ (Xn(j) ∩ Xn(j) ∩ Xn(j) ∩ · · · ). j1 j1 nj2 j1 nj2 nj3 j≥1

Arrange the sequence of sequences {nj } into a single sequence m = {mj } by applying a diagonal procedure so that m = (n11 , n21 , n12 , n31 , n22 , n13 , n41 , . . .). Given the latter sequence, the sequence of sequences {nj } can be reconstructed. Now let Xm1 = Xn(1) , Xm1 m2 = Xn(2) , Xm1 m2 m3 = Xn(1) , Xm1 m2 m3 m4 = Xn(3) ,.... 11 21 11 n12 31 Then B⊂

%

(Xm1 ∩ Xm1 m2 ∩ Xm1 m2 m3 ∩ · · · ).

m∈N ∞

Conversely, any x in the latter set can be seen to belong to B by constructing a sequence of sequences {nj } from the {mj } sequence as indicated above. Thus, B ∈ A(F ) and the latter is closed under countable intersections.

4.4 Capacity

177

If X is a metric space and F is the collection of closed subsets of X, then A(F ) contains the open sets as well as the closed sets. If, in addition, X is σ-compact and F is the collection of compact subsets of X, then A(F ) contains all open sets. Although closed under countable unions and countable intersections in both cases, A(F ) is not necessarily a σ-algebra since it may not be closed under complementation. Theorem 4.4.24 If X is a σ-compact metric space and F is the class of compact subsets of X, then A(F ) contains all Borel subsets of X. Proof: Let A0 (F ) be the collection of all ﬁnite unions of ﬁnite intersections of elements of F and complements of such sets. Then A0 (F ) is an algebra and F ⊂ A0 (F ) ⊂ A(F ). Since A(F ) is a monotone class (that is, closed under countable unions and countable intersections), by a standard theorem of measure theory it must contain the smallest σ-algebra containing A0 (F ) (c.f., Halmos [27] ), since the latter contains all compact sets, A(F ) ⊃ σ(F ) = class of Borel subsets of X. Lemma 4.4.25 (Sierpinski) Let (X, F ) be a paved space, let A be the nucleus of a Suslin scheme (n1 , . . . , nk ) → Xn1 ···nk ∈ F, and let {bj } be a sequence in N . Deﬁne % (Xn1 ∩ Xn1 n2 ∩ · · · ) Aj = {n:ni ≤bi ,i≤j}

Bj =

%

(Xn1 ∩ · · · ∩ Xn1 ···nj )

{n:ni ≤bi ,i≤j}

B = ∩j≥1 Bj . Then Aj ⊂ A, Aj ⊂ Bj , Bj+1 ⊂ Bj , j ≥ 1, and B ⊂ A. Proof: All of the conclusions are easy to verify except for the last, namely that B ⊂ A. Consider any x ∈ B. For each k ≥ 1, there is a k-tuple (nk1 , . . . nkk ) with nki ≤ bi , i ≤ k such that x ∈ Xnk1 ∩ Xnk1 nk2 ∩ · · · ∩ Xnk1 ···nkk . Consider a ﬁxed sequence of such k-tuples corresponding to x. Since nk1 ∈ {1, 2, . . . b1 }, k ≥ 1, there is an integer m1 ∈ {1, . . . , b1 } such that nk1 = m1 for inﬁnitely many k and x ∈ Xm1 = Xk1 for all such k. Considering only those k-tuples having m1 as the ﬁrst integer, there is an m2 ∈ {1, . . . , b2 } such that m2 is the second integer for inﬁnitely many k-tuples having m1 as the ﬁrst integer so that x ∈ Xm1 ∩ Xm1 m2 = Xm1 ∩ Xm1 nk2 for all such k. By an induction argument, there is a sequence {mj }, with mj ≤ bj , j ≥ 1, such that x ∈ Xm1 ∩ Xm1 m2 ∩ Xm1 m2 m3 ∩ · · · ⊂ A.

178

4 Negligible Sets

Note that if X is a σ-compact metric space and F consists of compact sets, then the B and Bj of Sierpinski’s lemma are also compact. Theorem 4.4.26 (Choquet [13]) If Ω is a Greenian set with paving K(Ω), then the analytic sets are capacitable; in particular, the Borel subsets of Ω are capacitable. Proof: (Assuming Property B) Consider A ∈ A(K(Ω)). If C ∗ (A) = 0, then C∗ (A) = 0 and A is capacitable. It therefore can be assumed that C ∗ (A) > 0. Let 0 ≤ α < C ∗ (A), and let A be the nucleus of the Suslin scheme (n1 , . . . , nk ) → Xn1 ···nk ∈ K(Ω). Then % A= (Xn1 ∩ Xn1 n2 ∩ · · · ). n∈N ∞

Since A1 =

%

(Xn1 ∩ Xn1 n2 ∩ · · · ) ↑ A as b1 ↑ +∞,

{n:n1 ≤b1 }

b1 can be chosen so that C ∗ (A1 ) > α by (ii) of Theorem 4.4.21. Suppose b1 , . . . , bj−1 have been chosen so that C ∗ (Aj−1 ) > α. Since % Aj = (Xn1 ∩ Xn1 n2 ∩ · · · ) ↑ Aj−1 , {n:ni ≤bi ,i≤j}

bj can be chosen so that C ∗ (Aj ) > α. The sequence {bj } in Sierpinski’s Lemma is thus deﬁned inductively. Since Aj ⊂ Bj , C ∗ (Bj ) ≥ C ∗ (Aj ) > α, j ≥ 1. Since B ∈ K(Ω) and the Bj ∈ K(Ω), C(Bj ) > α, j ≥ 1, and C(B) > α by (iii) of Theorem 4.4.21. Since α is any number less than C ∗ (A), C ∗ (A) = sup {C(B) : B ∈ K(Ω), B ⊂ A} = C∗ (A) and A is capacitable.

4.5 Boundary Behavior of the Green Function The proofs of the results of the preceding section, but not the results themselves, were conditional upon showing that the Greenian set Ω has Property B. It is known that each ball B has Property B. The validity of the results of the preceding section will be established here by showing that all Greenian sets have Property B. Lemma 4.5.1 If Ω is a bounded open set and x ∈ Ω, then GΩ (x, ·) = ux − Hux on Ω. Proof: Since ux ∈ Uux , ux − Hux ≥ 0 on Ω. From the deﬁnition of the Green function, ux − Hux ≥ GΩ (x, ·) on Ω. Also from the deﬁnition, GΩ (x, ·) = ux + hx where hx is harmonic on Ω. Since GΩ (x, ·) ≥ 0, ux ≥ −hx on Ω, and

4.5 Boundary Behavior of the Green Function

179

therefore −hx ∈ Lux ; that is, −hx ≤ Hux . Moreover, ux − Hux ≥ ux + hx implies that −hx ≥ Hux . Thus, −hx = Hux and GΩ (x, ·) = ux − Hux . Theorem 4.5.2 (Cartan [11]) A subset Z of a Greenian set Ω is polar if and only if it has capacity zero. Proof: (Assuming Property B) Let Z ⊂ Ω be a polar set. It follows from (ii) of Theorem 4.4.21 that C ∗ (Z) = limj→∞ C ∗ (Z ∩ Λj ) whenever {Λj } is an increasing sequence of open sets with Ω = ∪Λj and with compact closures Λ− j ⊂ Ω, j ≥ 1. It therefore can be assumed that Z ⊂ Λ where Λ is open with compact closure Λ− ⊂ Ω. By Theorems 4.2.9 and 4.2.10, there is a superharmonic function u = +∞ on Z. By the Riesz decomposition theorem, Theorem 3.5.11, there is a measure μ on Λ such that u = GΛ μ + h where h is harmonic on Λ. Since u = +∞ on Z, GΛ μ = +∞ on Z. Since GΛ ≤ GΩ on Λ × Λ, GΩ μ = +∞ on Z. Since GΩ μ is l.s.c., Or = {x ∈ Ω; GΩ μ(x) > r}, r > 0, is an open subset of Ω. Consider any compact set Γ ⊂ Or and the capacitary distribution μΓ corresponding to Γ and Ω. By the reciprocity ˆ Γ = GΩ μΓ ≤ 1 on Ω, theorem, Theorem 3.5.1, and the fact that R 1 1 1 C(Γ ) = dμΓ ≤ GΩ μ dμΓ = GΩ μΓ dμ ≤ μ(Λ− ). r r r It follows that C(Or ) ≤ μ(Λ− )/r. Since C ∗ (Z) ≤ inf r>0 C(Or ), C ∗ (Z) = 0 and Z has capacity zero. As to the converse, assume that C ∗ (Z) = 0. Since a countable union of polar sets is polar, it suﬃces to show that the intersection of Z with any component of Ω is polar. Since the polar property is a local property, it suﬃces to prove that the part of Z in a ball B with compact closure B − ⊂ Ω is polar. It can be assumed that Z ⊂ B. Let O be a nonempty open subset of B with O− ⊂ B. Then both Z ∩ O and Z ∩ (B ∼ O) have outer capacity zero by (i) of Theorem 4.4.21. If it can be shown that each of these two sets is polar, then it would follow that Z is polar. For each of these two sets, there is a point not in the set and neighborhoods separating the two. Assume that Z itself has this property; that is, there is a point y ∈ B ∼ Z and open sets O1 and O2 such that (i)y ∈ O1 ⊂ B, (ii)Z ⊂ O2 ⊂ B, and (iii) O1 ∩ O2 = ∅. Since C ∗ (Z) = 0, for each j ≥ 1 there is an open set Λj with Z ⊂ Λj ⊂ O2 and C(Λj ) < 1/j 2 . Also for each j ≥ 1, there is an increasing sequence of open sets {Wj,k } with compact closures − − ⊂ Wj,k+1 ⊂ Λj and Λj = ∪k≥1 Wj,k . For each j, k ≥ 1, let Γjk = Wj,k Wj,k ˆ Γjk relative to Ω which and consider the corresponding capacitary potential R is equal to 1 on Wj,k . Since GΩ (y, ·) is bounded outside any ball containing y, supx∈O2 GΩ (y, x) ≤ m for some m. Since the support of μΓjk is contained − in Γjk = Wj,k ⊂ Λj ⊂ O2 , ˆ Γjk (y) = R

GΩ (y, x) dμΓjk (x) ≤ mμΓjk (Γjk ) ≤ mC(Γjk ) ≤ m/j 2 .

180

4 Negligible Sets

ˆ Γjk is deﬁned ˆ Γjk }k≥1 is an increasing sequence, wj = limk→∞ R Since {R 2 ˆ Γjk }k≥1 is uniformly bounded with wj (y) ≤ m/j . Since the sequence {R by 1, wj is a nonnegative superharmonic function on Ω by Theorem 2.4.8. ˆ Γjk = 1 on Wj,k , wj = 1 on Λj ⊃ Z. Letting w = ∞ wj , w(y) = Since R j=1 ∞ ∞ 2 w (y) ≤ m/j < +∞. Since the partial sums of the series deﬁnj j=1 j=1 ing w are nonnegative superharmonic functions, w is superharmonic on B by Theorem 2.4.8. Since each wj = 1 on Z, w = +∞ on Z. This shows that Z is polar. Note that there is no mention of Property B in the statement of the following theorem. All of the preceding results are applicable since the proof is reduced to subsets of a ball that is known to have Property B. Theorem 4.5.3 Analytic inner polar sets are polar. Proof: Let Z be an analytic inner polar set. Since Z = ∪j≥1 (Z ∩ B0,j ), if it can be shown that each term is polar, then it would follow that Z is polar, since a countable union of polar sets is polar by Theorem 4.2.12. It is easily seen that each term is an analytic inner polar set. Thus, it can be assumed that Z is an analytic inner polar subset of a ball B. Since each compact subset Γ of Z is polar and C(Γ ) = 0 by the preceding theorem, C∗ (Z) = sup {C(Γ ); Γ ∈ K(B), Γ ⊂ Z} = 0. Since analytic sets are capacitable by Theorem 4.4.26, C(Z) = 0. By the preceding theorem, Z is polar.

Completion of proof of Theorem 4.4.10 It has been shown that the set of points in Ω where u ˆ = u is an inner polar set. By the previous theorem, the exceptional set is a polar set; that is, u ˆ = u q.e. The deﬁnition of a polar set was independent of any set Ω. This is also true of sets of capacity zero. If Z is a subset of a Greenian set Ω1 of capacity zero relative to Ω1 , then it is a polar set; if it is also a subset of another Greenian set Ω2 , then it is a polar subset, and therefore of capacity zero relative to Ω2 . Theorem 4.5.4 If Ω is a Greenian set and x ∈ Ω, then limy→z,y∈Ω GΩ (x, y) = 0 q.e. on ∂Ω. Proof: By Corollary 2.6.31, there is an increasing sequence {Ωj } of regular bounded open sets with compact closures in Ω such that Ω = ∪Ωj . Fix ˜ Ω (x, ·) be equal to GΩ (x, ·) x ∈ Ω. It can be assumed that x ∈ Ω1 . Let G ˜ Ωj to be GΩj (x, ·) on on Ω and equal to zero on ∼ Ω; similarly, deﬁne G ˜ Ωj (x, ·) ≤ G ˜ Ω (x, ·). By Theorem 3.3.11, Ωj and zero on ∼ Ωj . Then 0 ≤ G − ˜ ˜ GΩj (x, ·) ↑ GΩ (x, ·) on Ω. Consider a ball B = Bx,ρ with Bx,ρ ⊂ Ω. By ˜ Lemma 3.2.6, GΩ (x, ·) is bounded on ∼ B and it follows that the sequence ˜ Ωj (x, ·)} is uniformly bounded on ∼ B. Since each G ˜ Ωj (x, ·) is harmonic {G

4.5 Boundary Behavior of the Green Function

181

˜ Ωj (x, y) = 0 for z ∈ ∂Ωj by Theorem 3.3.15, on Ωj ∼ B and limy→z,y∈Ωj G ˜ Ωj (x, ·) is subharmonic on ∼ B. Let g = limj→∞ G ˜ Ωj (x, ·) on ∼ B, and let G g(z) = lim supy→z g(y), z ∈ Rn , denote the upper regulariztion of g. Note that g = 0 on ∂Ω. By Theorem 4.4.10, g = g quasi everywhere. Since g = GΩ (x, ·) on Ω ∼ B, for z ∈ ∂Ω g(z) = lim sup g(y) = lim sup g(y) = lim sup GΩ (x, y). y→z

y→z,y∈Ω

y→z,y∈Ω

Since g = 0 on ∂Ω, lim supy→z,y∈Ω GΩ (x, y) = 0 quasi everywhere on ∂Ω. Corollary 4.5.5 The set of irregular boundary points of a Greenian set Ω is a polar set. Moreover, if h is a bounded harmonic function on Ω and lim

y→x,y∈Ω

h(y) = 0

for each regular boundary point x ∈ ∂Ω, then h = 0 on Ω. Proof: The ﬁrst assertion follows from Theorem 3.3.15. Let f (x) = lim inf h(y), y→x,y∈Ω

x ∈ ∂Ω.

By the ﬁrst assertion, f = 0 except possibly for a polar subset of ∂Ω. By Corollary 4.2.20, Hf = H0 = 0 on Ω. Since h ∈ Uf , h ≥ 0 on Ω. Similarly, h ≤ 0 on Ω. According to the following theorem, all Greenian sets have Property B so that all the results of this section and the preceding section apply to all Greenian sets. Theorem 4.5.6 Let Ω be a Greenian set. If μ is a measure with compact support Γ ⊂ Ω, then limy→z,y∈Ω GΩ μ(y) = 0 q.e. on ∂Ω. If Ω has ﬁnite Lebesgue measure when n ≥ 3 or is bounded when n = 2, then limy→z,y∈Ω Gf (y) = 0 q.e. whenever f is a bounded measurable function on Ω. If Ω is regular, then these statements hold at every point of ∂Ω. Proof: Let Γ be any compact subset of Ω, and let {Ωj } be the sequence of components of Ω. Then Γ ⊂ ∪pj=1 Ωj for some p ≥ 1. For 1 ≤ j ≤ p, let xj be a ﬁxed point in Γ ∩ Ωj . By the preceding theorem, for each such j there is a polar set Zj ⊂ ∂Ω such that limy→z,y∈Ω GΩ (xj , y) = 0 for all z ∈ ∂Ω ∼ Zj . Letting Z = ∪Zj , limy→z,y∈Ω GΩ (xj , y) = 0, z ∈ ∂Ω ∼ Z, 1 ≤ j ≤ p. For 1 ≤ j ≤ p, let Oj be a neighborhood of Γ ∩ Ωj with Oj− ⊂ Ωj . By Harnack’s inequality, Theorem 2.2.2, for such j there is a constant kj such that GΩ (x, y) ≤ kj GΩ (xj , y),

x ∈ Γ ∩ Ωj , y ∈ Ωj ∼ Oj ;

182

4 Negligible Sets

this inequality is trivially satisﬁed if y ∈ Ωj . Thus, GΩ (x, y) ≤ kj GΩ (xj , y),

x ∈ Γ ∩ Ωj , y ∈ Ω ∼ ∪pj=1 Oj ,

and therefore GΩ (x, y) ≤ max (kj GΩ (xj , y)), 1≤j≤p

x ∈ Γ, y ∈ Ω ∼ ∪pj=1 Oj .

It follows that limy→z,y∈Ω GΩ (·, y) = 0 uniformly on Γ for each z ∈ ∂Ω ∼ Z. In particular, if the measure μ has support in Γ , then lim Gμ(y) = lim z ∈ ∂Ω ∼ Z. GΩ (x, y) dμ(x) = 0, y→z,y∈Ω

y→z,y∈Ω

The proof of the second assertion is the same as that of Lemma 3.4.17 using, in the notation of that lemma, the fact that limx→x0 ,x∈Ω GΩ (x, z) = 0 uniformly with respect to z ∈ Γj0 provided x0 ∈ ∂Ω does not belong to a polar subset of ∂Ω.

4.6 Applications It is now possible to show that a nonempty open subset of R2 is Greenian if and only if it supports a nonconstant, positive superharmonic function. Theorem 4.6.1 (Myrberg [47]) If Ω is a nonempty open subset of R2 , then the following three conditions are equivalent: (i) ∼ Ω is not a polar set. (ii) There is a nonconstant, positive superharmonic function on Ω. (iii) Ω is Greenian. Proof: The assertion (iii) ⇒ (i) is just Theorem 4.2.18. To show that (i) ⇒ (ii), assume that ∼ Ω is not polar. By Lemma 4.2.17, this assumption is equivalent to assuming that ∂Ω is not polar. Let {Bj } be a sequence of balls covering R2 . By Theorem 4.2.12, there is a j0 ≥ 1 such that Γ = ∂Ω ∩ Bj−0 is not polar. Letting B be any ball such that Bj−0 ⊂ B, Γ is a compact ˆ Γ , relative to B, which is strictly positive on B by subset of B. Consider R Theorem 4.3.5 and harmonic on B ∼ Γ . Since Γ is not polar, by Lemma 4.4.11 ˆ Γ (x) = RΓ (x) = 1. Thus, there are points there is a point x ∈ Γ such that R ˆ Γ (y) = ˆ z ∈ Ω such that RΓ (z) is arbitrarily close to 1. Since limy→z,y∈B R limy→z,y∈B GB μΓ (y) = 0 for all z ∈ ∂B by Theorem 3.4.16, the function ˆ Γ (y) R y ∈ Ω ∩ (B ∼ Γ ) u(y) = 0 y∈Ω∼B is subharmonic on Ω, takes on the value 0, and takes on values arbitrarily close to 1, but not the value 1 by the maximum principle. The function

4.6 Applications

183

v = 1 − u on Ω is a nonconstant, positive superharmonic function on Ω. It only remains to show that (ii) ⇒ (iii). Suppose there is a nonconstant, positive superharmonic function u on Ω. For each j ≥ 1, Ωj = Ω ∩ B0,j is a Greenian set by Theorem 3.2.10 and the sequence {Ωj } increases to Ω. By the Riesz representation theorem, Theorem 3.5.11, for each j ≥ 1 there is a measure μj and a harmonic function hj ≥ 0 on Ωj such that u = GΩj μj + hj on Ωj . As in the proof of Theorem 3.5.11, there is a measure μ on Ω such that μj = μ|Ωj . Extending GΩj (x, ·) to R2 by putting GΩj (x, y) = 0 for y ∈ Ωj , u(x) = GΩj μ(x) + hj (x) ≥ GΩj (x, y) dμ(y). If Ω were not Greenian, then GΩj ↑ +∞ on Ω × Ω by Theorem 3.3.11, and it would follow that u = +∞ on Ω, a contradiction. Thus, Ω is Greenian. Now that it is known that all Greenian sets have Property B, it can be ˆ u (x) = Ru (x) is a negligible set. shown that the set of points where R Λ Λ Corollary 4.6.2 If u is a nonnegative superharmonic function on the Greenˆ u = Ru q.e. on Ω. ian set Ω and Λ ⊂ Ω, then R Λ Λ Proof: Theorem 4.4.10. Theorem 4.6.3 If u is a nonnegative superharmonic function on the Greenˆ u = Ru on Ω ∼ Λ. ian set Ω and Λ ⊂ Ω, then R Λ Λ Proof: It is known from (ii) of Lemma 4.3.2 that RuΛ = u on Λ, and from ˆ u = Ru quasi everywhere on Ω. Let Z be the the preceding corollary that R Λ Λ exceptional polar set, and let x0 be any point of Ω ∼ Λ. By Theorem 4.3.11, there is a positive superharmonic function w on Ω such that w = +∞ on ˆ u + w ≥ u on Λ for every > 0, and it Λ ∩ Z and w(x0 ) < +∞. Then R Λ u ˆ u + w ≥ Ru . Since w(x0 ) < +∞ follows from the deﬁnition of RΛ that R Λ Λ u u ˆ and is arbitrary, RΛ (x0 ) ≥ RΛ (x0 ). Since the opposite inequality is always ˆ u = Ru on Ω ∼ Λ. true, there is equality at x0 . Thus, R Λ Λ Corollary 4.6.4 If Λ ⊂ Ω, u and v are nonnegative superharmonic functions on the Greenian set Ω with u = v q.e. on Λ, and Z is a polar subset of ˆu = R ˆu ˆu ˆv Λ, then R Λ Λ∼Z and RΛ = RΛ . ˆu ˆu Proof: By Theorem 4.3.5, R Λ∼Z ≤ RZ . Consider any nonnegative superharmonic function w on Ω that majorizes u on Λ ∼ Z. By Theorem 4.3.11, if x is any point of Λ ∼ Z, there is a positive superharmonic function v on Ω such that v(x) < +∞ and v = +∞ on Z. Then w + v majorizes u on Λ for every ˆ u (x) ≤ w(x) ˆ u (x) ≤ w(x) + v(x), and letting ↓ 0, R > 0. Therefore, R Λ Λ ˆ u ≤ Ru for all x ∈ Λ ∼ Z. Therefore, R Λ Λ∼Z on Ω, and by taking the lower ˆu ≤ R ˆu regularization of both sides, R Λ Λ∼Z and the two are equal. The second equation follows from the ﬁrst.

184

4 Negligible Sets

According to Lemma 4.4.14, if {Γj } is an increasing sequence of compact ˆ u for any ˆu ↑ R subsets of the Greenian set Ω with compact limit Γ , then R Γj Γ ﬁnite-value, nonnegative, continuous superharmonic function u on Ω. This result can be improved. Theorem 4.6.5 If u is a nonnegative superharmonic function on the Greenian set Ω, {Λj } is an increasing sequence of subsets of Ω, and Λ = ∪Λj , then ˆu = R ˆu. limj→∞ R Λj Λ ˆ u } is increasing and is majorized by the suProof: Since the sequence {R Λj ˆ u , the function v = limj→∞ R ˆ u is superharmonic perharmonic function R Λ Λj ˆ u . By Corollary 4.6.2, R ˆ u diﬀers from u on Λj on at most a and v ≤ R Λ Λj polar subset of Λj . Therefore, v diﬀers from u on Λ on at most a polar subset Z of Λ. Consider any x ∈ Λ ∼ Z. There is then a positive superharmonic function w on Ω such that w(x) < +∞ and w = +∞ on Z. Then for every ˆ u on Ω. If B − ⊂ Ω, > 0, v + w ≥ u on Λ, and therefore v + w ≥ RuΛ ≥ R Λ x,δ then ˆ uΛ : x, δ); A(v : x, δ) + A(w : x, δ) ≥ A(R ˆ u : x, δ) whenever B − ⊂ Ω. By Lemma 2.4.4, letting ↓ 0, A(v : x, δ) ≥ A(R Λ x,δ ˆ u (x) for all x ∈ Ω; that is, v = limj→∞ R ˆu ≥ R ˆ u . It follows that v(x) ≥ R Λ Λj Λ ˆu = R ˆu. v = limj→∞ R Λj

Λ

Lemma 4.6.6 If Λ is an open subset of the Greenian set Ω and f is a ∂Ω resolutive boundary function, then the function Ω Hf on Ω ∩ ∂Λ g= f on ∂Ω ∩ ∂Λ is resolutive for Λ and HfΩ = HgΛ on Λ. Ω Proof: Consider any u ∈ UΩ f . Then u ≥ Hf on Ω. For x ∈ Ω ∩ ∂Λ,

lim inf u(z) ≥ lim inf HfΩ (z) = HfΩ (x);

z→x,z∈Λ

z→x,z∈Λ

while for x ∈ ∂Ω ∩ ∂Λ, lim inf z→x,z∈Λ u(z) ≥ lim inf z→x,z∈Ω u(z) ≥ f (x). Ω Λ Ω Λ Therefore, u ∈ UΛ g and H f ≥ H g on Λ. Similarly, Hg ≥ H f and it follows Λ

Ω

Ω that 0 ≤ H g − H Λ g ≤ H f − H f . Since f is ∂Ω resolutive, the latter diﬀerence is equal to zero and so g is ∂Λ resolutive. Using the fact that f is ∂Ω Λ Ω Ω Λ resolutive, HfΩ ≤ H Λ f ≤ H g ≤ Hf on Λ, and therefore Hf = Hg . The operation on the positive superharmonic function u described below is an extension of the lowering operation of Deﬁnition 2.4.10.

Lemma 4.6.7 If Λ is an open subset of the Greenian set Ω having compact closure Λ− ⊂ Ω and u is a positive superharmonic function on Ω, then the lower regularization vˆ of the function

4.6 Applications

185

v=

u HuΛ

on Ω ∼ Λ on Λ

is superharmonic on Ω. Λ

Λ Proof: Since u ≥ H u ≥ H Λ u ≥ 0 on Λ, Hu is deﬁned by Theorem 2.9.1. Let − B be a ball with B ⊂ Ω. It will be shown ﬁrst that v ≥ HvB on B. Since HuΛ ≤ u on Λ, v ≤ u on Ω and HvB ≤ HuB ≤ u = v on B ∼ Λ. Now let v on Ω ∼ B w= HvB on B.

Note that w = HvB ≤ HuB ≤ u = v on ∂Λ ∩ B; that is, w ≤ v on ∂(B ∩ Λ). It follows that HwB∩Λ ≤ HvB∩Λ on B ∩ Λ. Since HvB = HwB∩Λ and HvB∩Λ = HuΛ on B ∩ Λ by the preceding lemma, it follows that HvB ≤ HuΛ on B ∩ Λ. Since v = HuΛ on Λ, HvB ≤ v on B. Noting that vˆ ≤ v, HvˆB ≤ HvB ≤ v on B. Since HvˆB is continuous on B, HvˆB ≤ vˆ on B for any ball B with B − ⊂ Ω and vˆ is superharmonic on Ω. Theorem 4.6.8 If u is a positive superharmonic function on the Greenian set Ω and Λ is an open subset of Ω with compact closure Λ− ⊂ Ω, then Λ ˆu R Ω∼Λ = Hu on Λ; moreover, if {Λj } is an increasing sequence of such sets Λ with Ω = ∪Λj , then u is a potential if and only if limj→∞ Hu j = 0 on Ω.

Proof: If v=

u HuΛ

on Ω ∼ Λ on Λ,

then vˆ is superharmonic on Ω. Since u is l.s.c. on ∂Λ, there is a sequence {fj } in C 0 (∂Λ) such that fj ↑ u on ∂Λ. Since lim inf v(z) = lim inf HuΛ (z) ≥ lim inf HfΛj (z) = fj (y)

z→y,z∈Λ

z→y,z∈Λ

z→y,z∈Λ

for all y ∈ ∂Λ that are regular boundary points for Λ, lim inf z→y,z∈Λ v(z) ≥ u(y) for all y ∈ ∂Λ except possibly for a polar set according to Corollary 4.5.5. Since v = u on Ω ∼ Λ, lim inf z→y,z∈Ω∼Λ v(z) = lim inf z→y,z∈Ω∼Λ u(z) ≥ u(y), and therefore vˆ(y) = lim inf z→y,z∈Ω v(z) ≥ u(y) for all y ∈ ∂Λ except possibly for a polar set. Since vˆ = v = u on Ω ∼ Λ− , vˆ ≥ u on Ω ∼ Λ except ˆu . possibly for a polar set. It follows from Corollary 4.6.4 that vˆ ≥ R Ω∼Λ If w is any nonnegative superharmonic function on Ω that majorizes u on Ω ∼ Λ, then w|Λ ∈ UΛ u and w ≥ v on Λ; therefore, w ≥ v on Ω. This ˆu shows that RuΩ∼Λ ≥ v on Ω, and consequently that R ˆ on Ω. As the Ω∼Λ ≥ v u ˆ opposite inequality was just proved, RΩ∼Λ = vˆ = v = HuΛ on Λ. Now let {Λj } be an increasing sequence of open subsets of Ω with compact closures Λj ˆu Λ− j ⊂ Ω such that Λj ↑ Ω. It was just shown that Hu = RΩ∼Λj on Λj . Since Λ Λ j+1 j u u ˆ ˆ R ≤R by Theorem 4.3.5, Hu ≤ Hu on Λj for each j ≥ 1. Ω∼Λj+1

Ω∼Λj

186

4 Negligible Sets Λ

The sequence {Hu j } is then a decreasing sequence of nonnegative harmonic Λ functions (with due regard given to domains), and h = limj→∞ Hu j deﬁnes a ˜ is any nonnegative nonnegative harmonic function on Ω with u ≥ h ≥ 0. If h Λj Λj ˜ ˜ harmonic minorant of u on Ω, then Hu ≥ Hh˜ = h on Λj , j ≥ 1, and h ≥ h; that is, h is the greatest harmonic minorant of u on Ω, and the result follows from Corollary 3.5.12. Theorem 4.6.9 (G. Mokobodzki and D. Sibony [46]) If u, u1 , and u2 are nonnegative superharmonic functions on the Greenian set Ω with u ≤ u1 + u2 , then there are unique nonnegative superharmonic functions u ˜1 , u ˜2 on Ω for which u ˜ 1 ≤ u1 , u ˜2 ≤ u2 , and u = u ˜1 + u ˜2 . Proof: Letting u ˜1 be the lower regularization of inf {v ∈ S(Ω); v ≥ 0, u ≤ v + u2 }, u ≤ u˜1 + u2 q.e. on Ω by Theorem 4.4.10. Since polar sets have Lebesgue ˜1 + u2 on measure zero by Theorem 4.2.6, u ≤ u ˜1 + u2 a.e. on Ω and u ≤ u Ω by Lemma 2.4.4; moreover, u˜1 is superharmonic on Ω by Theorem 2.4.9. Letting u ˜2 be the lower regularization of inf {v ∈ S(Ω); v ≥ 0, u ≤ u ˜1 + v}, u≤u ˜1 + u ˜2 a.e. on Ω and therefore u ≤ u ˜1 + u ˜2 on Ω as above; moreover, u ˜2 is superharmonic on Ω as above. Since u is a member of each of the sets deﬁning u ˜1 and u ˜2 , u ≥ u ˜1 a.e. on Ω and u ≥ u˜2 a.e. on Ω so that u ≥ u ˜1 on Ω and u ≥ u ˜2 on Ω. It will be shown now that u ≥ u ˜1 + u˜2 on Ω. Since ˜2 at x so that u(x) − u ˜2 (x) may be indeterminant, redeﬁne u − u u(x) − u ˜2 (x) if u ˜2 (x) < +∞ (u − u ˜2 )(x) = lim inf y→x,˜u2 (y)<+∞ (u(y) − u ˜2 (y)) if u ˜2 (x) = +∞. Then u − u ˜2 is l.s.c. at points x for which u ˜2 (x) = +∞. It will be shown now that u − u˜2 is l.s.c. at points x for which u ˜2 (x) < +∞. Consider any ˜iB = (˜ ui )B , i = 1, 2 and using the ball B = Bx,δ with B − ⊂ Ω. Letting u inequality u ≤ u ˜1 + u˜2 , u = uB + (u − uB ) ≤ u˜1B + (˜ u2B + u − uB ) ≤ u ˜1 + (˜ u2B + u − uB ). ˜2B on B. The function v = u˜2B + u − uB is superharmonic on B with v ≥ u Consider the function min (˜ u2 , v) on B w= u ˜2 on Ω ∼ B. Since v ≥ u ˜2B and u ˜2 ≥ u˜2B on B, w ≥ u ˜2B on B so that w is a nonnegative superharmonic function on Ω by Lemma 2.4.11. Since u ≤ u ˜1 + u ˜2 and

4.6 Applications

187

u≤u ˜1 + v, u ≤ u ˜1 + min (˜ u2 , v) = u ˜1 + w on B and u ≤ u ˜1 + u ˜2 = u ˜1 + w on Ω ∼ B, u ≤ u˜1 + w on Ω. Since w is a nonnegative superharmonic function, w ≥ u˜2 and v ≥ w ≥ u˜2 on B, it follows that uB − u ˜2B ≤ u − u ˜2

(4.4)

on B. Note that the function on the left is continuous on B. Therefore, ˜2B (x) ≤ lim inf (u − u˜2 )(y). uB (x) − u y→x,y∈Ω

Since superharmonic functions are integrable on spheres, ˜2B (x) = (u − u ˜2 )B (x) = L(u : x, δ) − L(˜ u2B : x, δ) → u(x) − u ˜2 (x) uB (x) − u as δ → 0. It follows from Inequality (4.4) that u(x) − u˜2 (x) ≤ lim inf (u − u ˜2 )(y) y→x,y∈Ω

and u − u ˜2 is l.s.c. at x whenever u ˜2 (x) < +∞ and therefore l.s.c. on Ω. From Inequality (4.4), ˜2 )B (x) = PI(u − u˜2 , B)(x) = L(u − u ˜2 : x, δ), u(x) − u ˜2 (x) ≥ (u − u so that u − u ˜2 satisﬁes the averaging principle and is therefore a nonnegative ˜2 , u − u ˜2 ≥ u˜1 a.e. on Ω. superharmonic function on Ω. Since u = (u − u ˜2 ) + u Since the opposite inequality was established above, u = u˜1 + u ˜2 on Ω. To prove uniqueness, assume that there are nonnegative superharmonic functions v˜1 and v˜2 such that u = v˜1 + v˜2 , v˜1 ≤ u1 and v˜2 ≤ u2 . Then u ≤ v˜1 + u2 and therefore u ˜1 ≤ v˜1 a.e. on Ω by deﬁnition of u ˜1 and consequently on all of Ω. ˜1 + u ˜2 = v˜1 + v˜2 , u ˜i = v˜i , i = 1, 2 on Ω. Similarly, u ˜2 ≤ v˜2 on Ω. Since u Theorem 4.6.10 If {uj } is an increasing sequence of nonnegative superharmonic functions on the Greenian set Ω with superharmonic limit u and ˆ uj . ˆ u = limj→∞ R Λ ⊂ Ω, then R Λ Λ ˆ uj ≤ R ˆ uj ≤ R ˆ u , j ≥ 1, w = limj→∞ R ˆ u and is superharmonic Proof: Since R Λ Λ Λ Λ u u j j ˆ = R = uj q.e. on Λ, w = u q.e. on Λ; on Ω by Theorem 2.4.8. Since R Λ Λ that is, there is a polar set Z ⊂ Λ such that w = u on Λ ∼ Z. Since u ˆu ˆu ˆu ˆu = R R Λ Λ∼Z by Corollary 4.6.4 and RΛ∼Z ≤ RΛ∼Z by Lemma 4.3.2, RΛ = uj u u w ˆ ˆ RΛ∼Z ≤ RΛ∼Z = RΛ∼Z ≤ w = limj→∞ RΛ . Lemma 4.6.11 If u is a nonnegative superharmonic function on the open set Ω, then there is an increasing sequence {uj } of nonnegative, ﬁnite-valued, continuous superharmonic functions on Ω such that u = limj→∞ uj on Ω. Proof: Let {Ωj } be an increasing sequence of open subsets of Ω with compact ˆ u = u on Ωj and is harmonic on closures in Ω such that Ω = ∪Ωj . Then R Ωj − u u ˆ ˆ Ω ∼ Ω . Note that R ≤R on Ω for all j ≥ 1. For each j ≥ 1, let j

Ωj

Ωj+1

δj = min (d(Ωj , ∼ Ω), 1/j)) and let

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4 Negligible Sets

uj (x) =

ˆ u )(x) Jδj /3 (R Ωj u ˆ (x) R Ωj

if d(x, Ωj ) ≤ δj /3, x ∈ Ω if d(x, Ωj ) > δj /3, x ∈ Ω

where the operator J is deﬁned by Equation (2.5.1). The sequence {uj } has the required properties. Lemma 4.6.12 If Ω is a Greenian subset of Rn , Λ ⊂ Ω, and v is a nonnegative, ﬁnite-valued superharmonic function that is continuous on Λ, then RvΛ = inf {RvO ; O ∈ O(Ω), O ⊃ Λ}. Proof: It can be assumed that Ω is connected and that v > 0 on Ω. Let u be any positive superharmonic function on Ω that majorizes v on Λ. Let > 0, ﬁx x ∈ Ω, and choose n ≥ 1 such that (1/n)RvΛ (x) < . Then (1+ n1 )u−v > 0 on an open set On ⊃ Λ, and therefore (1 +

1 )u ≥ RvOn ≥ inf {RvO ; O ∈ O(Ω), O ⊃ Λ}. n

Thus, RvΛ (x) + ≥ inf {RvO (x); O ∈ O, O ⊃ Λ}. The result follows by letting → 0 and using the fact that the resulting opposite inequality is always true. Theorem 4.6.13 If {uj } is a sequence of nonnegative superharmonic func∞ tions on the Greenian set Ω such that u = j=1 uj is superharmonic on Ω ˆ uj . ˆu = ∞ R and Λ ⊂ Ω, then R Λ Λ j=1 Proof: Suppose ﬁrst that the sequence is of length two. By Theorem 4.3.5, ˆ u1 + R ˆ u2 so that it suﬃces to prove the opposite inequality. ˆ u1 +u2 ≤ R R Λ Λ Λ According to Theorem 4.6.9, there are nonnegative superharmonic functions ˆ u1 +u2 = v1 + v2 with vi ≤ R ˆ ui , i = 1, 2. Assume that v1 and v2 such that R Λ Λ u u u +u 1 2 1 2 ˆ = u2 , and R ˆ ˆ = u1 , R = u1 + u2 on Λ. Since Λ is open. Then R Λ Λ Λ uj ˆ vj ≤ RΛ ≤ uj , j = 1, 2, and v1 + v2 = u1 + u2 on Λ, vj = uj , j = 1, 2 on Λ. ˆ uj , j = 1, 2, that R ˆ u1 +u2 = v1 + v2 ≥ R ˆ u1 + R ˆ u2 , and It follows that vj ≥ R Λ Λ Λ Λ u +u u u 1 2 1 2 ˆ ˆ +R ˆ whenever Λ is open. Suppose now that therefore that R =R Λ Λ Λ u1 and u2 are nonnegative, ﬁnite-valued, continuous superharmonic functions on Ω and Λ is an arbitrary subset of Ω. For any O ∈ O(Ω) with O ⊃ ˆ u1 + R ˆ u2 . By the preceding Λ, RuO1 +u2 = RuO1 + RuO2 ≥ RuΛ1 + RuΛ2 ≥ R Λ Λ u1 +u2 u u ˆ 1 +R ˆ 2 . Taking the lower regularization of both sides, lemma, RΛ ≥R Λ Λ ˆ u1 +u2 ≥ R ˆ u1 + R ˆ u2 and the two are equal. Lastly, suppose that u1 and u2 R Λ Λ Λ are any nonnegative superharmonic functions on Ω and Λ is any subset of Ω. According to Theorem 4.6.11, there are increasing sequences {u1j } and {u2j } of nonnegative, ﬁnite-valued, continuous superharmonic functions such that ˆ u1j +u2j = R ˆ u1j + R ˆ u2j by the preceding ui = limj→∞ uij , i = 1, 2. Then, R Λ Λ Λ

4.7 Sweeping

189

ˆ u1 +u2 = R ˆ u1 + R ˆ u2 and the theorem is true for step. By Theorem 4.6.10, R Λ Λ Λ sequences of length two. This result has a trivial extension to ﬁnite sequences. Suppose now that {uj } is an inﬁnite sequence. Then for each k ≥ 1, k

ˆ j=1 uj = R Λ

k

ˆ uj . R Λ

j=1

ˆu = By Theorem 4.6.10, R Λ

∞ j=1

ˆ uj . R Λ

4.7 Sweeping If u = GΩ μ is the potential of a measure μ on the Greenian set Ω with ˆu μ(Ω) < +∞ and Λ is an open subset of Ω, then R Ω∼Λ is again a potential that is harmonic on Λ. By the Riesz decomposition theorem, Theorem 3.5.11, ˆu R Ω∼Λ = GΩ μΩ∼Λ where μΩ∼Λ (Λ) = 0. The measure μΩ∼Λ is obtained from μ by “sweeping” the measure in Λ out of Λ. The operation taking μ into μΩ∼Λ is called the sweeping of μ, and μΩ∼Λ is called the swept measure. An iteration of the sweeping operation leads to a cumbersome notation. ˆ Λ (u, ·) will be used interchangeably. If ˆ u and R For this reason, the notations R Λ x ∈ Ω, the unit measure concentrated on {x} will be denoted by δΩ (x, ·). Deﬁnition 4.7.1 If Λ is a subset of the Greenian set Ω, deﬁne GΛ Ω (x, ·) and Λ a unique measure δΩ (x, ·) by the equation Λ ˆ GΛ (x, y) = R (G (x, ·), y) = GΩ (y, z) δΩ (x, dz); Λ Ω Ω Ω

if μ is a measure on Ω, GΛ Ω μ is deﬁned by the equation GΛ μ(x) = GΛ Ω Ω (x, y) μ(dy). Λ (x, ·) is called the sweeping of δΩ (x, ·) onto Λ. That such a The measure δΩ measure exists is a consequence of Theorem 3.5.11.

Lemma 4.7.2 If Λ is any subset of the Greenian set Ω, then for any x ∈ Ω Λ (x, Ω) ≤ 1, (i) δΩ Λ Λ (ii) δΩ (x, ·) = δΩ (x, ·) and δΩ (x, {x}) = 1 if x is an interior point of Λ, Λ (iii) δΩ (x, ·) has support in ∂Λ whenever x ∈ Ω ∼ int Λ.

Proof: (i) If V is any open subset of Ω with compact closure V − ⊂ Ω, 1 ˆ = GΩ μV . Since 1 ≥ u = GΩ μV = GΩ (x, y) dμV (y) and let u = R V ˆ Λ (GΩ (x, ·), y), by Tonelli’s theorem GΩ (x, y) ≥ R

190

4 Negligible Sets

1≥

ˆ Λ (GΩ (x, ·), y) dμV (y) R Λ = GΩ (y, z) δΩ (x, dz) dμV (y) Λ (x, dz) = GΩ (z, y) dμV (y) δΩ Λ = u(z) δΩ (x, dz) Λ Λ ≥ u(z) δΩ (x, dz) = δΩ (x, V ). V

Λ Letting V ↑ Ω, δΩ (x, Ω) ≤ 1. (ii) If x is an interior point of Λ, then Λ ˆ GΩ (y, z) δΩ (x, dz) = RΛ (GΩ (x, ·), y) = GΩ (x, y) = GΩ (y, z) δΩ (x, dz) Λ Λ so that δΩ (x, ·) = δΩ (x, ·) by the uniqueness of δΩ (x, ·). (iii) If x is not an ˆ interior point of Λ, then RΛ (GΩ (x, ·), ·) is harmonic on the interior of Λ and Λ on Ω ∼ Λ− so that δΩ (x, ·) has its support in ∂Λ by Theorem 3.4.9. Λ Λ By (iii) above, δΩ (x, {x}) = 0 if x ∈ Ω ∼ Λ− so that δΩ (x, {x}) is 0 or 1 except possibly for x ∈ ∂Λ. It will be shown later in this section that the same is true of points in ∂Λ. Λ (·, M ) is a Borel measurable function Lemma 4.7.3 For each Borel set M, δΩ on Ω.

Proof: Let P be the set of bounded, continuous potentials u = GΩ μ on Ω for which μ has compact support in Ω, and let F be the set of all diﬀerences ˆ u as an f = u − v, u, v ∈ P having compact supports in Ω. Considering R Λ operator, it can be extended to F by putting ˆf = R ˆ uΛ − R ˆ vΛ R Λ whenever f = u − v ∈ F. If the function f ∈ F has two representations f = u − v and f = u − v , then u + v = u + v, ˆ uΛ + R ˆ vΛ = R ˆ u+v = R ˆ u +v = R ˆ uΛ + R ˆ vΛ R Λ Λ ˆu − R ˆv = R ˆ u − R ˆ v . Thus, R ˆ f is wellby Theorem 4.6.13, and therefore R Λ Λ Λ Λ Λ ˆ f (x) is easily seen to be deﬁned. For each x ∈ Ω, the map f → Lx (f ) = R Λ linear on F . Letting F be the collection of uniform limits of sequences in F , F contains all continuous functions with compact support in Ω according to Theorem 3.4.15. It will be shown next that Lx has an extension to F as a bounded linear functional. Let {fj } be a sequence in F that converges uniformly to f . By Corollary 4.3.7,

4.7 Sweeping

191

ˆ fi (x) − R ˆ fj (x)| ≤ |fi − fj | sup |Lx (fi ) − Lx (fj )| ≤ sup |R Λ Λ

x∈Ω

x∈Ω

from which it follows that the limit Lx (f ) = limj→∞ Lx (fj ) exists. It is easily seen using a similar inequality that Lx (f ) does not depend upon the sequence {fj }. Consider any nonnegative f ∈ F and a sequence {fj } in F with fj = uj − vj which converges uniformly to f . If j = − inf x∈Ω (uj (x) − vj (x)), then uj + j ≥ vj , ˆ uj + j ≥ R ˆ vj , ˆ uj + j ≥ R R Λ Λ Λ and therefore u

v

ˆ j −R ˆ j ) ≥ lim (− j ) = 0. Lx (f ) = lim Lx (fj ) = lim (R Λ Λ j→∞

j→∞

j→∞

This shows that Lx is positive and linear on F . Since ˆ uj − R ˆ vj | ≤ lim |uj − vj | = lim |fj | = |f |, |Lx (f )| = lim |Lx (fj )| = lim |R Λ Λ j→∞

j→∞

j→∞

j→∞

Lx is a bounded, positive linear functional on F . By the Riesz representation theorem, for each x ∈ Ω there is a measure ρΛ Ω (x, ·) on the Borel subsets of Ω with ρΛ (x, Ω) ≤ 1 such that Ω Lx (f ) = f (y) ρΛ f ∈ F. Ω (x, dy), ˆ u (x) − R ˆ v (x), Lx (f ) is a Borel measurable function for Since Lx (f ) = R Λ Λ all f ∈ F and all f ∈ F by approximation. Since the latter contains all continuous functions with compact support in Ω, ρΛ Ω (x, M ) is a Borel function of x for each Borel set M ⊂ Ω. The conclusion of the theorem will follow if it Λ (x, ·) = ρΛ can be shown that δΩ Ω (x, ·). Suppose u = GΩ μ is the potential of a measure with compact support Γ ⊂ Ω. Let {Ωj } be an increasing sequence of open subsets of Ω with compact closures such that Ω = ∪Ωj , Γ ⊂ Ω1 , Ωj− ⊂ Ωj+1 for all j ≥ 1. By Theorem 2.5.2, there is an increasing sequence {ui } of continuous potentials for which ui = u on Ω ∼ Ω1 and limi→∞ ui = u on Ω. Moreover, for each i ≥ 1, there is an increasing sequence {uik } of continuous ˆu ˆu potentials such that limk→∞ uik = R Ω∼Ωi and uik = RΩ∼Ωi on Ω ∼ Ωi+1 for all k ≥ 1. Then ui − uii ∈ F and ui uii ˆ ˆ RΛ (z) − RΛ (z) = (ui (y) − uii (y)) ρΛ (4.5) Ω (z, dy). ˆu ˆu Since {R Ω∼Ωi } is a decreasing sequence of functions on Ω and each RΩ∼Ωi ˆu is harmonic on Ωi , limi→∞ R Ω∼Ωi is a nonnegative harmonic function that is dominated by the potential u. Since the greatest harmonic minorant of u is the zero function, ˆ uΩ∼Ω (z) = 0, z ∈ Ω. lim R i i→∞

192

4 Negligible Sets

Therefore, ˆ uii (z) ≤ lim sup uii (z) ≤ lim R ˆ uΩ∼Ω (z) = 0. lim sup R Λ i i→∞

i→∞

i→∞

ˆ uii (z) = uii (y) ρΛ Since R Ω (z, dy), it follows from Equation (4.5) and Λ Theorem 4.6.10 that u ˆ RΛ (z) = u(y) ρΛ Ω (z, dy). Taking u(x) = GΩ (x, ·), ˆ Λ (GΩ (x, ·), z) = R

GΩ (x, y) ρΛ Ω (z, dy).

Λ Thus, ρΛ Ω (x, ·) satisﬁes the deﬁnition of δΩ (x, ·) and the two are equal.

Theorem 4.7.4 If u is a nonnegative superharmonic function on the Greenian set Ω and Λ ⊂ Ω, then u Λ ˆ RΛ (x) = u(y) δΩ (x, dy), x ∈ Ω. Proof: In the course of the preceding proof, it was shown that the result is true if u is the potential of a measure having compact support in Ω. By Corollary 4.3.6, there is a sequence of such potentials that increases to u. The result follows from the Lebesgue monotone convergence theorem and Theorem 4.6.10. Lemma 4.7.5 If u is a nonnegative superharmonic function on the Greenian ˆ B (R ˆ u , ·) = R ˆ A (R ˆ u , ·) = R ˆu. set Ω and A ⊂ B ⊂ Ω, then R A B A ˆ u ≤ u, Proof: Since A ⊂ B and R B ˆ A (R ˆ uA , ·) ≤ R ˆ A (R ˆ uB , ·) ≤ R ˆ A (u, ·) R and ˆ A (R ˆ uA , ·) ≤ R ˆ B (R ˆ uA , ·) ≤ R ˆ A (u, ·). R ˆ A (R ˆ A (u, ·). Since R ˆ u = u q.e. on A by ˆ A (u, ·) = R It suﬃces to show that R A ˆ A (u, ·) = R ˆ A (R ˆ u , ·) according to Corollary 4.6.4. Corollary 4.6.2, R A Theorem 4.7.6 If u is a nonnegative superharmonic function on the Greeˆu,R ˆ u ), then nian set Ω, A, B ⊂ Ω, and u = min (R A B

RuA∪B + RuA∪B = RuA + RuB

4.7 Sweeping

and

193 ˆ uA∪B ˆ uA + R ˆ uB . ˆ uA∪B + R =R R

Proof: On A ∼ B, RuA = u ≥ RuB so that u + u = u + RuB = RuA + RuB . Interchanging A and B, u+u = RuA +RuB on B ∼ A. On A∩B, RuA = u = RuB ˆu + R ˆ u q.e. on so that u + u = u + u = RuA + RuB . Therefore, u + u = R A B A ∪ B. By the preceding lemma and Corollary 4.6.4, ˆ uA∪B + R ˆ uA∪B ˆ A∪B (u + u , ·) R =R ˆ A∪B (R ˆ uA + R ˆ uB , ·) =R ˆ uA , ·) + R ˆ A∪B (R ˆ uB , ·) ˆ A∪B (R =R

ˆ uA + R ˆ uB =R on A ∪ B. By Theorem 4.6.3, on Ω ∼ (A ∪ B),

RuA∪B + RuA∪B = RuA + RuB . Since this equation holds on A ∪ B, it holds on Ω. Deﬁnition 4.7.7 If u is a Borel measurable function on the Greenian set Ω, Λ deﬁne the function δΩ (·, u) by putting Λ δΩ (x, u) =

Λ u(y) δΩ (x, dy)

provided the integral is deﬁned for all x ∈ Ω. If μ is a Borel measure on Ω, Λ deﬁne the measure δΩ (μ, ·) on the Borel subsets M of Ω by putting Λ δΩ (μ, M ) =

Λ δΩ (x, M ) μ(dx)

provided the integral is deﬁned for all M . Lemma 4.7.8 For each x and each Borel subset M of the Greenian set · (x, M ) is subadditive; that is, Ω, δΩ A∪B A B (x, M ) ≤ δΩ (x, M ) + δΩ (x, M ). δΩ

Proof: Using the notation in the proof of the Lemma 4.7.3, let f = u1 − u2 ≥ 0 where u1 and u2 are bounded, continuous potentials of measures with compact support and let ui = min (RuAi , RuBi ), i = 1, 2. By Theorem 4.7.6 and Theorem 4.7.4, A∪B A∪B A B (x, ui ) + δΩ (x, ui ) = δΩ (x, ui ) + δΩ (x, ui ), δΩ

i = 1, 2.

194

4 Negligible Sets

Since u1 ≥ u2 , u1 ≥ u2 and A∪B A B (x, u1 − u2 ) ≤ δΩ (x, u1 − u2 ) + δΩ (x, u1 − u2 ). δΩ

Thus, A∪B A B δΩ (x, f ) ≤ δΩ (x, f ) + δΩ (x, f )

for all nonnegative f ∈ F having compact support. Since any continuous function on Ω with compact support can be approximated uniformly by a sequence of such functions according to Theorem 3.4.15, the last inequality holds for all continuous functions with compact support. Using the usual methods of measure theory to pass from continuous functions to Borel sets, the result holds for all Borel sets. Lemma 4.7.9 (Zero-one Law) If Ω is a Greenian set, x ∈ Ω, Λ ⊂ Ω, and u is a nonnegative superharmonic function on Ω, then Λ (x, Z) = 0 for all polar sets Z not containing x, (i) δΩ Λ (x, {x}) = 0 or 1, (ii) δΩ Λ Λ∩U (iii) if δΩ (x, {x}) = 0, then δΩ (x, {x}) = 0 for all neighborhoods U of x, and ˆu limU↓{x} R Λ∩U = 0 on Ω

ˆ Λ (R ˆ u , x) = 0 limU↓{x} R U whenever u(x) < +∞ , and Λ Λ∩U ˆ u (x) = u(x) for all (x, {x}) = 1, then δΩ (x, {x}) = 1 and R (iv) if δΩ Λ∩U neighborhoods U of x. Proof: (i) Let Z be a polar set not containing x. By Theorem 4.3.11, there is a nonnegative superharmonic function v with v(x) < +∞ such that v = +∞ on Z. Since v Λ Λ ˆ v(y) δΩ (x, dy), +∞ > v(x) ≥ RΛ (x) = v(y) δΩ (x, dy) ≥ Z Λ Λ∩U (x, Z) = 0. (ii) Consider δΩ (x, {x}) as a function of the neighborhood δΩ U of x. If U1 and U2 are any neighborhoods of x with U1 ⊂ U2 , then by the preceding lemma Λ∩(U2 ∼U1 )

Λ∩U2 Λ∩U1 (x, {x}) ≤ δΩ (x, {x}) + δΩ δΩ

(x, {x}).

Λ∩(U ∼U )

Since the support of δΩ 2 1 (x, ·) is in the closure of U2 ∼ U1 , the second Λ∩U term on the right is zero. Thus, as a function of U , δΩ (x, {x}) increases as U decreases. By Lemma 4.7.5, ˆ Λ∩U (R ˆ Λ (GΩ (x, ·)) ˆ Λ∩U (GΩ (x, ·)) = R R

4.7 Sweeping

195

so that

Λ∩U GΩ (y, z) δΩ (x, dz)

Λ Λ∩U GΩ (y, z) δΩ (w, dz) δΩ (x, dw) Λ Λ∩U = GΩ (y, z) δΩ (w, dz) δΩ (x, dw).

=

Since the function on the left side of this equation is a potential, by the uniqueness assertion of the Riesz decomposition theorem, Theorem 3.5.11, Λ∩U δΩ (x, ·) =

Λ Λ∩U δΩ (w, ·) δΩ (x, dw).

Λ∩U Since δΩ (x, ·) vanishes on polar sets not containing x by (i), Λ∩U Λ Λ∩U δΩ (x, {x}) = δΩ (x, {x})δΩ (x, {x}).

(4.6)

Λ (x, {x}) = 0 or 1, proving (ii). If Taking U = Ω, it follows that δΩ Λ Λ∩U δΩ (x, {x}) = 0, then Equation (4.6) implies that δΩ (x, {x}) = 0 for all Λ Λ∩U neighborhoods U of x; if δΩ (x, {x}) = 1, then δΩ (x, {x}) = 1 for all neighΛ∩U borhoods of x since δΩ (x, {x}) decreases as U increases. It remains only Λ (x, {x}) = 1, to prove the assertions concerning the reduced function. If δΩ then Λ∩U ˆ uΛ∩U (x) = u(y) δΩ R (x, dy) = u(x). Λ∩U This proves (iv). Lastly, suppose δΩ (x, {x}) = 0 and u(x) < +∞. To show u ˆ that limU↓{x} R = 0 on Ω it suﬃces to consider a decreasing sequence of Λ∩U ˆu balls {Bj } with centers at x, with closures in Ω, and ∩Bj = {x}. Then R Λ∩Bj ˆu is a decreasing sequence of potentials on Ω with limit w = limj→∞ R Λ∩Bj

which is harmonic on Ω ∼ {x}. Restricting w to Ω ∼ {x}, by Bˆ ocher’s theorem, Theorem 3.2.2, it has a nonnegative superharmonic extension w ˜ to Ω of the form cGΩ (x, ·) + h, where h is harmonic on Ω. Since cGΩ (x, ·) + ˆu h ≤ R Λ∩Bj on Ω ∼ {x} and both functions are superharmonic, the same inequality holds on Ω. By the Riesz decomposition theorem, Theorem 3.5.11, h is the greatest harmonic minorant of w ˜ and is therefore nonnegative. Since ˆu ˜ = cGΩ (x, ·) ≤ h≤R Λ∩Bj and the latter is a potential, h = 0 on Ω. Thus, w ˆu R ≤ u on Ω. Since u(x) < +∞, c = 0 and w ˜ = w = 0 on Ω ∼ {x}. Λ∩Bj Λ u ˆ (x) ≤ u(x) < +∞, the sequence {R ˆu Since u(y) δΩ (x, dy) = R Λ Λ∩Bj } is Λ dominated by the δΩ (x, ·) integrable function u so that by Lemma 4.7.5 and the result just proved

196

4 Negligible Sets

ˆ uΛ∩B (x) = lim R ˆ Λ (R ˆ uΛ∩B )(x) lim R j j j→∞ Λ ˆ uΛ∩B (z) δΩ = lim (x, dz) R j j→∞ Λ ˆ uΛ∩B (z) δΩ = lim R (x, dz) j

j→∞

j→∞

Ω∼{x}

= 0. ˆ u } to show that The same argument can be applied to the sequence {R Bj ˆ uB = 0 on Ω ∼ {x} lim R j

j→∞

Λ (x, ·) integrable function u and that the sequence is dominated by the δΩ so that Λ ˆ Λ (R ˆ uB , x) = lim ˆ uB (y) δΩ lim R (x, dy) = 0. R j j j→∞

j→∞

Chapter 5

Dirichlet Problem for Unbounded Regions

5.1 Introduction The principle problem associated with unbounded regions is the lack of uniqueness of the solution to the Dirichlet problem. To achieve uniqueness, the point at inﬁnity ∞ will be adjoined to Rn with the enlarged space den noted by R∞ . This will require redeﬁnition of harmonic and superharmonic functions. The Dirichlet problem for the exterior of a ball will be solved by a Poisson type integral. Using this result, it will be shown that the PerronWiener-Brelot method can be used to solve the Dirichlet problem for unbounded regions. Poincar´e’s exterior ball condition and Zaremba’s exterior cone condition are suﬃcient conditions for a ﬁnite boundary point to be a regular boundary point for the Dirichlet problem. Both conditions preclude the boundary point from being “too surrounded” by the region. On the other hand, the Lebesgue spine is an example of a region that does “surround” a boundary point too much; in some sense, the complement of the region is “thin” at the boundary point. A concept of thinness will be explored and related to a topology on n ﬁner than the metric topology which is more natural from the potential R∞ theoretic point of view. The words “open,” “neighborhood,” “continuous,” etc., will be preﬁxed by “ﬁne” or “ﬁnely” when used in this context.

5.2 Exterior Dirichlet Problem n Consider the one-point compactiﬁcation R∞ = Rn ∪{∞} of Rn . The topology n for R∞ is obtained by adjoining to a base for the Euclidean topology of Rn the class of sets of the form Rn ∼ Γ , where Γ is a compact subset of Rn in n that the usual topology. In this chapter, Ω will denote an open subset of R∞ might include the point at inﬁnity. As usual, if Ω is an open subset of Rn ,

L.L. Helms, Potential Theory, Universitext, c Springer-Verlag London Limited 2009

197

198

5 Dirichlet Problem for Unbounded Regions

the Euclidean boundary of Ω will be denoted by ∂Ω, whereas the boundary n will be denoted by ∂∞ Ω. of Ω relative to the one-pont compactiﬁcation R∞ The following deﬁnition is consistent with the earlier deﬁnition. Deﬁnition 5.2.1 The extended real-valued function u on the open set Ω is superharmonic on Ω if it is superharmonic on Ω ∼ {∞} in the usual sense, and if ∞ ∈ Ω, then (i) u is l.s.c. at ∞ and (ii) u(∞) ≥ L(u : x, δ) for all x ∈ Rn and δ > 0 such that ∼ Bx,δ ⊂ Ω; the function u is subharmonic on Ω if −u is superharmonic on Ω; if both u and −u are superharmonic on Ω, then u is harmonic on Ω. n . Consider the funcThe constant functions are obviously harmonic on R∞ −n+2 n tion ux (y) = |x − y| deﬁned on R∞ , n ≥ 3, by putting ux (∞) = 0; although this function is superharmonic on Rn , it is not superharmonic on n R∞ since 0 = ux (∞) < L(u : x, δ) for any x ∈ Rn and δ > 0. It is, however, n subharmonic on R∞ ∼ {x}. It will be shown shortly that the only functions n are the constant functions. that are superharmonic on R∞ A Poisson integral type representation of functions harmonic on a neighborhood of ∞ is possible. − Lemma 5.2.2 If u is nonnegative and harmonic on Rn ∼ By,ρ , continuous n on R ∼ By,ρ , and α = limx→∞ u(x) exists in R, then − (i) for x ∈ R2 ∼ By,ρ

u(x) =

1 2πρ

∂By,ρ

|y − x|2 − ρ2 u(z) dσ(z) |z − x|2

and α = L(u : y, r) for all r ≥ ρ; − , n ≥ 3, (ii) for x ∈ Rn ∼ By,ρ 1 u(x) = σn ρ

∂By,ρ

|y − x|2 − ρ2 αρn−2 u(z) dσ(z) − + α; |z − x|n |x − y|n−2

moreover, if α = L(u : y, ρ), then L(u : y, r) = α for all r > ρ. Proof: If x ∈ Rn ∼ By,ρ , let x∗ denote the inverse of x relative to ∂By,ρ as deﬁned in Equation (1.11). Under this inversion map, the region Rn ∼ By,ρ − − maps onto the region By,ρ ∼ {y} and induces a function u∗ on By,ρ ∼ {y} by means of the equation |x∗ − y|n−2 ∗ ∗ u (x ) = u(x), ρn−2

− x∗ ∈ By,ρ ∼ {y}

(5.1)

(u∗ (x∗ ) = u(x) in the n = 2 case). The function u∗ is harmonic on By,ρ ∼ {y} − and continuous on By,ρ ∼ {y} by Theorem 1.8.1. By Theorem 3.2.2, there is

5.2 Exterior Dirichlet Problem

199

a constant c ≥ 0 and a harmonic function v on By,ρ such that u∗ = cuy + v − so that the on By,ρ ∼ {y}. Note that v has a continuous extension to By,ρ − preceding equation holds on By,ρ ∼ {y}. The n = 2 and n ≥ 3 cases will be treated separately for part of the proof. (i) (n = 2) Using the fact that |x − y||x∗ − y| = ρ2 , u(x) = u∗ (x∗ ) = −c log |x∗ − y| + v(x∗ ) = −c log

ρ2 + v(x∗ ). |x − y|

Since α = limx→∞ u(x) exists in R, c = 0, and u(x) = u∗ (x∗ ) = v(x∗ ). Applying the Poisson integral formula to v, using Equation (1.5), the fact that |x∗ − y||x − y| = ρ2 , and the fact that u = v on ∂By,ρ , 1 u(x) = 2πρ =

1 2πρ

∂By,ρ

ρ2 − |x∗ − y|2 v(z) dσ(z) |z − x∗ |2

∂By,ρ

|y − x|2 − ρ2 u(z) dσ(z). |z − x|2

− Note that α = limx→∞ u(x) = L(u : y, ρ). Since u∗ = v on By,ρ ∼ {y}, for r ≥ ρ, 1 L(u : y, r) = u(y + rθ) dσ(θ) 2π |θ|=1 1 = u∗ (y + (ρ2 /r)θ) dσ(θ) 2π |θ|=1 1 = v(y + (ρ2 /r)θ) dσ(θ) 2π |θ|=1

= v(y). Therefore, α = L(u : y, ρ) = L(u : y, r) for all r > ρ. (ii) (n ≥ 3) Since |x∗ − y|n−2 ∗ ∗ |x∗ − y|n−2 u(x) = u (x ) = ρn−2 ρn−2

c ∗ + v(x ) |x∗ − y|n−2

and v is harmonic on By,ρ , α = lim u(x) = x→∞

c ; ρn−2

that is, u∗ = αρn−2 uy + v on By,ρ ∼ {y}. Since uy = ρ−n+2 on ∂By,ρ , u = u∗ = v + α on ∂By,ρ and v = u − α on ∂By,ρ . Returning to the equation u∗ = αρn−2 uy + v and applying the Poisson integral representation to v

200

5 Dirichlet Problem for Unbounded Regions

u∗ (x∗ ) =

αρn−2 1 −α+ ∗ n−2 |x − y| σn ρ

∂By,ρ

ρ2 − |x∗ − y|2 u(z) dσ(z) |x∗ − z|n

(5.2)

for x∗ ∈ By,ρ ∼ {y}. Using the fact that |x − y||x∗ − y| = ρ2 and Equation (1.5), 1 ρ2 − |x∗ − y|2 u(z) dσ(z) σn ρ ∂By,ρ |z − x∗ |n |x − y|2 − ρ2 |x − y|n−2 1 = u(z) dσ(z). ρn−2 σn ρ ∂By,ρ |z − x|n By Equation (5.2), |x∗ − y|n−2 ∗ ∗ u (x ) ρn−2 αρn−2 |x − y|2 − ρ2 1 =α− + u(z) dσ(z) n−2 |x − y| σn ρ ∂By,ρ |z − x|n

u(x) =

− for x ∈ Rn ∼ By,ρ . To prove the second assertion of (ii), a relation between the spherical averages of u and u∗ will be established. If δ ≥ ρ, then 1 L(u : y, δ) = u(y + δθ) dσ(θ) σn |θ|=1 ρn−2 |δθ|n−2 = u(y + δθ) dσ(θ). σn δ n−2 |θ|=1 ρn−2

Since the integrand is just u∗ (y + (ρ2 /δ)θ), ρn−2 1 ρ2 L(u : y, δ) = n−2 u∗ (y + θ) dσ(θ) δ σn |θ|=1 δ =

ρn−2 L(u∗ : y, ρ2 /δ). δ n−2

(5.3)

Returning to the equation u∗ = αρn−2 uy + v, L(u∗ : y, ρ2 /δ) = αρn−2 L(uy : y, ρ2 /δ) + L(v : y, ρ2 /δ). Since uy = δ n−2 /ρ2(n−2) on ∂By,ρ2 /δ , L(u∗ : y, ρ2 /δ) = α

δ n−2 + v(y). ρn−2

In particular, L(u∗ : y, ρ) = α + v(y). Using the fact that u = u∗ on ∂By,ρ , α = L(u : y, ρ) = L(u∗ : y, ρ) = α + v(y) so that v(y) = 0 and that

5.2 Exterior Dirichlet Problem

L(u : y, ρ) =

201

ρn−2 L(u∗ : y, ρ) = α δ n−2

whenever δ ≥ ρ. The requirement that α = L(u : y, ρ) in (ii) of the above lemma need not be satisﬁed. For example, if ux = 1/|x − y|n−2 , n ≥ 3, α = limy→∞ ux (y) = 0 < L(ux : x, ρ) = ρ−n+2 . Note that the equation in (ii) of the lemma can be put into a simpler form under the additional hypothesis that u(z) 1 α = L(u : y, ρ) = dσ(z). σn ρ ∂By,ρ ρn−2 In this case, u(x) =

1 σn ρ

∂By,ρ

|y − x|2 − ρ2 1 1 − + n−2 n n−2 |z − x| |x − y| ρ

u(z) dσ(z)

− . for x ∈ Rn ∼ By,ρ

Deﬁnition 5.2.3 If f is a Borel measurable function on ∂By,ρ that is integrable relative to surface area, the Poisson Integral of f relative to n n − R∞ ∼ By,ρ is deﬁned for x ∈ R∞ ∼ By,ρ by PI(f :∼ By,ρ )(x)

|y − x|2 − ρ2 1 1 1 = + − f (z) dσ(z) σn ρ ∂By,ρ |z − x|n ρn−2 |x − y|n−2 when n ≥ 3 and by 1 PI(f :∼ By,ρ )(x) = 2πρ

∂By,ρ

|y − x|2 − ρ2 f (z) dσ(z) |z − x|2

when n = 2. The integrand in the n ≥ 3 case is deﬁned to be ρ−n+2 when x = ∞. Note that the Poisson integral has been deﬁned so that PI(f :∼ By,ρ ) is harmonic on ∼ By,ρ . Theorem 5.2.4 (Brelot [8]) If f is a Borel measurable function on ∂By,ρ that is integrable relative to surface area, then u = PI(f :∼ By,ρ ) is harn − − u(z) ≤ lim sup monic on R∞ ∼ By,ρ and lim supz→x,z∈∼By,ρ z→x,z∈∂By,ρ f (z) for all x ∈ ∂By,ρ ; if, in addition, f is continuous at x ∈ ∂By,ρ , then − u(z) = f (x). limz→x,z∈∼By,ρ

202

5 Dirichlet Problem for Unbounded Regions

Proof: The n ≥ 3 case will be proved with only minor modiﬁcations required for the n = 2 case. Applying the Laplacian Δ(x) to both sides of the equations deﬁning the Poisson integral, it is easy to justify taking the Laplacian under − the integral; since the integrand is harmonic, u is harmonic on Rn ∼ By,ρ . It is also easily seen that 1 lim u(x) = f (z) dσ(z) = u(∞); x→∞ σn ρn−1 ∂By,ρ that is, u is continuous at ∞ and u(∞) = L(f : y, ρ). It remains only to show − that u(∞) = L(u : x, δ) whenever ∼ Bx,δ ⊂ Rn ∼ By,ρ . This will be done in two steps, the ﬁrst of which speciﬁes that x = y and δ > ρ. Applying the inversion relative to ∂By,ρ , ρn−2 L(f : y, ρ) − L(f : y, ρ) |x∗ − y|n−2 ρ2 − |x∗ − y|2 1 f (z) dσ(z). + σn ρ ∂By,ρ |z − x∗ |n

u∗ (x∗ ) =

(5.4)

Since the last term on the right is PI(f : By,ρ ) and PI(f : By,ρ )(y) = L(f : y, ρ), δ n−2 − L(f : y, ρ) + L(f : y, ρ) ρn−2 δ n−2 = L(f : y, ρ) n−2 ρ

L(u∗ : y, ρ2 /δ) = L(f : y, ρ)

whenever δ > ρ. From the proof of the preceding lemma, L(u : y, δ) =

ρn−2 L(u∗ : y, ρ2 /δ), δ n−2

so that L(u : y, δ) = L(f : y, ρ) = u(∞) whenever δ > ρ. Consider now − . Choose δ0 > ρ such that any closed ball Bx−0 ,δ with ∼ Bx0 ,δ ⊂ Rn ∼ By,ρ − − n − By,δ0 ⊂ Bx0 ,δ . Then Bx0 ,δ ∼ By,δ0 ⊂ R ∼ By,ρ . Since u is harmonic on the latter region, u is harmonic on a neighborhood of Bx−0 ,δ ∼ By,δ0 and by Green’s identity, Theorem 1.2.2, Dn u(z) dσ(z) = Dn u(z) dσ(z). ∂By,δ0

∂Bx0 ,δ

Since L(u : y, δ) = L(u : y, δ0 ) for all δ > δ0 as was just shown, the integral on the left is zero by Theorem 2.5.3 and so both integrals are zero. By the same theorem, L(u : x0 , δ ) is a constant for δ > δ. Since u is continuous at ∞, its spherical average over ∂Bx0 ,δ converges to u(∞) = L(f : y, ρ) as δ → +∞. Thus, L(u : x0 , δ ) = u(∞) = L(f : y, ρ) for all δ > δ. Since this is true

5.2 Exterior Dirichlet Problem

203

− for all δ for which ∼ Bx0 ,δ ⊂ Rn ∼ By,ρ , it also holds when δ = δ. Thus, − . Now consider any point u(∞) = L(u : x0 , δ) whenever ∼ Bx,δ ⊂ Rn ∼ By,ρ x ∈ ∂By,ρ . By Equation (5.1), Lemma 1.7.5, and Equation (5.4),

lim sup

u(z) =

− z→x,z∈∼By,ρ

= ≤

lim sup

|z ∗ − y|n−2 ∗ ∗ u (z ) ρn−2

lim sup

PI(f : By,ρ )(z ∗ )

z ∗ →x,z ∗ ∈By,ρ z ∗ →x,z ∗ ∈By,ρ

lim sup

f (z).

z→x,z∈∂By,ρ

The last assertion follows from Lemma 1.7.6. n Theorem 5.2.5 If u is superharmonic on an open connected set Ω ⊂ R∞ , n then u satisﬁes the minimum principle on Ω. If Ω is any open subset of R∞ , u is superharmonic on Ω, and lim inf z→x,z∈Ω u(z) ≥ 0 for all x ∈ ∂∞ Ω, then u ≥ 0 on Ω.

Proof: If ∞ ∈ Ω, the ﬁrst assertion is just Corollary 2.3.6. If ∞ ∈ Ω, let m = inf {u(y); y ∈ Ω} and let M = {y ∈ Ω; u(y) = m}. Then M is relatively closed by the l.s.c. of u. If y ∈ M ∩ Rn , then u must be equal to m on a neighborhood of y. Suppose ∞ ∈ M . Since ∞ ∈ Ω, there is an x0 and δ > 0 such that ∂Bx0 ,ρ ⊂ Ω whenever ρ ≥ δ. Since ∞ ∈ M, u(∞) = L(u : x0 , ρ) for all ρ ≥ δ and therefore u = m on ∼ Bx0 ,δ . This shows that u = m on a neighborhood of ∞. Hence, M is both open and relatively closed. Since Ω is connected, M = Ω or M = ∅; in the ﬁrst case, u is constant on Ω and in the second case u does not attain its minimum on Ω. To prove the second assertion, deﬁne a function u ˆ on Ω − as follows: u(y) if y ∈ Ω u ˆ(y) = if y ∈ ∂∞ Ω. lim inf z→y,z∈Ω u(z) Then uˆ is l.s.c. on Ω − and uˆ ≥ 0 on ∂∞ Ω. Consider any component Λ of Ω. n is compact, Λ− is compact Then ∂∞ Λ ⊂ ∂∞ Ω and uˆ ≥ 0 on ∂∞ Λ. Since R∞ − and u ˆ attains its minimum on Λ . If uˆ attains a negative minimum at an interior point of Λ, then the minimum principle is contradicted. Thus, u ˆ≥0 on each component Λ of Ω and therefore uˆ ≥ 0 on Ω. n Corollary 5.2.6 If u is superharmonic on R∞ , then u is a constant function. n n is compact and u is l.s.c. on R∞ , u attains its minimum Proof: Since R∞ n on R∞ and is therefore constant by the minimum principle.

Example 5.2.7 (Green Function for a Half-space) Let n Ω = R+ = {(x1 , . . . , xn ) ∈ Rn ; xn > 0}, n ≥ 2,

204

5 Dirichlet Problem for Unbounded Regions

and consider u(x, y) = ux (y), the fundamental harmonic function with pole at x ∈ Ω. For x = (x1 , . . . , xn ) ∈ Rn , let x = (x , xn ), where x = (x1 , . . . , xn−1 ) ∈ Rn−1 , and let xr = (x1 , . . . , xn−1 , −xn ) be the reﬂection of x across the hyperplane xn = 0. Note that u((x , 0), y) = u((x , 0), yr ) since |(x , 0) − y| = |(x , 0) − yr |. It now will be shown that u(x, y) − u(x, yr ) is the Green function for the half-space Ω by showing that the diﬀerence satisﬁes (i), (ii), and (iii) of Deﬁnition 3.2.3. Using the fact that |x − y| ≤ |x − yr |, u(x, y) − u(x, yr ) ≥ 0 for x, y ∈ Ω. Since u(x, yr ) = u(xr , y) for x, y ∈ Ω, it is clear that u(x, yr ) is a harmonic function of y so that u(x, y) − u(x, yr ) has the form of a sum of ux and a harmonic function on Ω. Fix x ∈ Ω and let vx be a nonnegative superharmonic function on Ω that is the sum of ux and a superharmonic function on Ω. Consider the function vx (y) − u(x, y) + u(x, yr ) = vx (y) − u(x, y) + u(xr , y) and any y0 ∈ R0n = {(y1 , . . . , yn ) ∈ Rn ; yn = 0}. Then lim inf (vx (y) − u(x, y) + u(xr , y)) = lim inf vx (y) ≥ 0

y→y0 ,y∈Ω

y→y0 ,y∈Ω

since vx ≥ 0 on Ω. Moreover, lim inf

|y|→+∞,y∈Ω

(vx (y) − u(x, y) + u(xr , y)) ≥ +

lim inf

vx (y)

lim inf

(u(xr , y) − u(x, y)).

|y|→+∞,y∈Ω |y|→+∞,y∈Ω

The ﬁrst term on the right is nonnegative since vx ≥ 0 on Ω, and it is easily seen that lim|y|→+∞,y∈Ω (u(xr , y) − u(x, y)) = 0. Thus, lim inf (vx (y) − u(x, y) + u(x, yr )) ≥ 0

y→y0 ,y∈Ω

for all y0 ∈ ∂∞ Ω.

By the preceding theorem, vx (y) ≥ u(x, y) − u(x, yr ) on Ω and (iii) of Deﬁnition 3.2.3 is satisﬁed. Thus, GΩ (x, y) = u(x, y) − u(x, yr ).

5.3 PWB Method for Unbounded Regions The ﬁrst few results of this section have been studied before and do not warrant elaboration. The following theorem follows easily from the deﬁnitions. n and Theorem 5.3.1 If u and v are superharmonic on the open set Ω ⊂ R∞ c > 0, then

(i) cu is superharmonic on Ω, (ii) u + v is superharmonic on Ω, and (iii) min (u, v) is superharmonic on Ω.

5.3 PWB Method for Unbounded Regions

205

n Lemma 5.3.2 Let u be superharmonic on the open set Ω ⊂ R∞ . If Λ is an − − open subset of Ω with Λ ⊂ Ω, h is continuous on Λ and harmonic on Λ, and u ≥ h on ∂∞ Λ, then u ≥ h on Λ.

Proof: Since lim inf z→x,z∈Λ (u − h)(z) ≥ 0 on ∂∞ Λ, the result follows from Theorem 5.2.5. As in Lemma 2.4.11, the part of a superharmonic function over the complement of a ball can be lowered to obtain a harmonic patch. n n with R∞ ∼ B ⊂ Ω for some Lemma 5.3.3 If Ω is an open subset of R∞ ball B = Bx,δ and u is superharmonic on Ω, the function u on B − ∩ Ω u∼B = n ∼ B− PI(u :∼ B) on R∞ n is superharmonic on Ω, harmonic on R∞ ∼ B − , and u ≥ u∼B on Ω.

Proof: Let {φj } be an increasing sequence of continuous functions on ∂∞ (∼ B) such that φj ↑ u and let φj on ∂∞ (∼ B) hj = on Ω ∼ B − . PI(φj :∼ B) n ∼ B and harmonic on By Theorem 5.2.4, each hj is continuous on R∞ n − n R∞ ∼ B . Since u ≥ hj on ∂∞ (∼ B), u ≥ hj on R∞ ∼ B by the preceding n ∼ B − . Since φj ↑ u on ∂∞ (∼ B), lemma; that is, u ≥ PI(φj :∼ B) on R∞

u ≥ PI(u :∼ B) = u∼B

n on R∞ ∼ B−.

Obviously, u ≥ u∼B on Ω. The proof that u∼B is superharmonic on Ω is the same as in the proof of Lemma 2.4.11. Lemma 5.3.4 If {ui ; i ∈ I} is a right-directed family of harmonic functions n on the open set Ω ⊂ R∞ , then u = supi∈I ui is either identically +∞ or harmonic on each component of Ω. Proof: Since the assertion concerns components, it can be assumed that Ω is connected; it might as well be assumed that ∞ ∈ Ω, for otherwise the assertion is just Lemma 2.2.7. By the same lemma, u is either identically +∞ or harmonic on Ω ∼ {∞}. If ∼ Bx,δ ⊂ Ω, then ui = PI(ui :∼ Bx,δ ) − − n n ∼ Bx,δ for each i ∈ I and u = PI(u :∼ Bx.δ ) on R∞ ∼ Bx,δ by on R∞ Lemma 2.2.10. If u is harmonic on Ω ∼ {∞}, it is integrable on ∂Bx,δ and − n therefore harmonic on R∞ ∼ Bx,δ . If u is identically +∞ on Ω ∼ {∞}, then u(∞) = L(u : x, δ) = +∞ and u is identically +∞ on Ω. Deﬁnition 5.3.5 A family F of superharmonic functions on the open set n Ω ⊂ R∞ is saturated over Ω if it is saturated over Ω ∼ {∞} in the usual sense and if ∞ ∈ Ω, then

206

5 Dirichlet Problem for Unbounded Regions

(i) u, v ∈ F implies min (u, v) ∈ F, and (ii) for every ball B ⊂ Rn with B − ⊂ Ω and u ∈ F, the function u on B − u∼B = PI(u :∼ B) on Ω ∼ B − belongs to F . Theorem 5.3.6 If F is a saturated family of superharmonic functions on n the open set Ω ⊂ R∞ , then the function v = inf u∈F u is either identically −∞ or harmonic on each component of Ω. Proof: It can be assumed that Ω is connected and that ∞ ∈ Ω. By Theorem 2.6.2, v is identically −∞ or harmonic on Ω ∼ {∞}. Consider any ball B ⊂ Rn with B − ⊂ Ω. If u ∈ F, then u∼B ≤ u, u∼B is harmonic on Ω ∼ B − , and v = inf u∈F u∼B on Ω ∼ B − . As in the proof of Theorem 2.6.2, the family {u∼B : u ∈ F } is left-directed. By Lemma 5.3.4, v is identically −∞ or harmonic on Ω ∼ B − . Since Ω ∼ B − and Ω ∼ {∞} overlap, v is identically −∞ or harmonic on Ω. The concept of polar set has an extension to the one point compactiﬁcation of Rn . n Deﬁnition 5.3.7 A set Z ⊂ R∞ is a polar set if to each x ∈ Z there corresponds a neighborhood Λ of x and a superharmonic function u on Λ such that u = +∞ on Z ∩ Λ. 2 n Theorem 5.3.8 {∞} is a polar subset of R∞ but not of any R∞ , n ≥ 3. 2 Proof: First consider R∞ . The function u(x) = log |x| is harmonic on R2 ∼ − B0,1 and has the limit +∞ as |x| → +∞. Deﬁning u(∞) = +∞, u is easily − 2 ∼ B0,1 . This shows that {∞} is a polar seen to be superharmonic on R∞ 2 n , n ≥ 3, and let u subset of R∞ . Now let Λ be any neighborhood of ∞ in R∞ be superharmonic on Λ. Then ∼ Bx,δ ⊂ Λ for some x and δ > 0. Let δ1 > δ. By Theorem 2.5.3, L(u : x, ρ) is a concave function of ρ−n+2 for ρ ≥ δ1 . This implies that there are constants a and b such that L(u : x, ρ) ≤ aρ−n+2 + b for ρ ≥ δ1 . Therefore, lim supρ→+∞ L(u : x, ρ) ≤ b < +∞. If u(∞) = +∞, then L(u : x, ρ) would have to approach +∞ as ρ → +∞ by the l.s.c. of u at ∞. Therefore, u(∞) < +∞ for any such u and ∞ is not a polar subset of n , n ≥ 3. R∞ n A polar subset of R∞ , n ≥ 3, is just a polar subset of Rn as previously 2 may include the point at inﬁnity. Note deﬁned; but a polar subset of R∞ n that Theorems 4.2.9 and 4.2.10 cannot be generalized in the context of R∞ in view of Corollary 5.2.6. n is said to be hyperharmonic As in Chapter 2, a function u on Ω ⊂ R∞ (hypoharmonic) on Ω if on each component of Ω the function u is superharmonic (subharmonic) or identically +∞ (−∞). If f is an extended real-valued function on ∂∞ Ω, the upper and lower classes associated with Ω and f are deﬁned by

5.3 PWB Method for Unbounded Regions

207

Uf = {u; lim inf u(z) ≥ f (x) for all x ∈ ∂∞ Ω, u hyperharmonic z→x,z∈Ω

and bounded below on Ω}, Lf = {u; lim sup u(z) ≤ f (x) for all x ∈ ∂∞ Ω, u hypoharmonic z→x,z∈Ω

and bounded above on Ω}, respectively; the corresponding upper and lower solutions are deﬁned as H f = inf{u; u ∈ Uf } and H f = sup{u; u ∈ Lf }. Noting that the assertions of Lemma 2.6.11 follow from Corollary 2.3.6 and the order properties of functions, the same assertions are valid in the present context with Corollary 2.3.6 replaced by Theorem 5.2.5. The boundary function f is said to be resolutive for Ω if H f = H f and both are harmonic functions. Since the collection of functions in Uf , which are superharmonic on a given component of Ω, forms a suaturated family over that component, on each component of Ω the function H f is either identically +∞, identically −∞, or harmonic. The same is true of H f . Several of the following results will require that the complement of an open n be nonpolar. In the n = 2 case, this means that R2 ∼ Ω must subset Ω of R∞ be nonpolar so that Ω ∩ R2 is Greenian; but Ω may or may not contain ∞. This requirement is of no consequence in the n ≥ 3 case for the following n , n ≥ 3, and ∼ Ω is polar. In this case, ∞ ∈ Ω and reason. Suppose Ω ⊂ R∞ ∂∞ Ω ⊂ Rn . If h were any bounded harmonic function on Ω, then it could n by Corollary 4.2.16; but such be extended to a harmonic function on R∞ functions are constant functions by Corollary 5.2.6. Thus, the PWB method applied to a bounded Borel measurable boundary function will yield only constant functions. n with ∼ Ω nonpolar and {fj } Lemma 5.3.9 If Ω is an open subset of R∞ is a sequence of resolutive boundary functions that converge uniformly to f , then f is resolutive and the sequence {Hfj } converges uniformly to Hf .

Proof: The proof is the same as the proof of Lemma 2.6.12. n Lemma 5.3.10 Let Ω be an open subset of R∞ with ∼ Ω nonpolar. If u is a bounded superharmonic function on Ω such that f (x) = limz→x,z∈Ω u(z) exists for all x ∈ ∂Ω, then f is a resolutive boundary function.

Proof: The proof is the same as the proof of Lemma 2.6.14. Remark 5.3.11 Recalling that the Kelvin transformation preserves positivity and harmonicity, it will now be shown that it preserves superharmonicity as well. Suppose Ω ⊂ Rn ∼ {y} and u is superharmonic on Ω. Let Ω ∗ ⊂ Rn be the inversion of Ω relative to ∂By,ρ , and let u∗ be the Kelvin transform of u. Note that the Kelvin transformation preserves the order properties of functions. An open subset of Ω ∗ with compact closure in Ω ∗ is the inversion

208

5 Dirichlet Problem for Unbounded Regions

Λ∗ of a set Λ ⊂ Ω with the same properties; similarly, any function that is continuous on the closure of Λ∗ and harmonic on Λ∗ is the Kelvin transform h∗ of a function h on Λ with the same properties. If u∗ ≥ h∗ on ∂Λ∗ for any such function h∗ , then u ≥ h on ∂Λ and u ≥ h on Λ since u is superharmonic on Ω. Thus, u∗ ≥ h∗ on Λ∗ and u∗ is superharmonic on Ω ∗ by the alternative deﬁnition of a superharmonic function discussed in Section 2.3. Remark 5.3.12 Another facet of the Kelvin transformation is important in the discussion of barriers and regular boundary points. Suppose Ω ⊂ n , x ∈ ∂∞ Ω, u is a positive superharmonic function on Ω such that R∞ − limz→x,z∈Ω u(z) = 0, and there is a closed ball By,ρ ⊂∼ Ω − . If x = ∞ ∗ and u is the Kelvin transform of u relative to ∂By,ρ , then lim ∗ ∗

z ∗ →x ,z ∈Ω

u∗ (z ∗ ) = ∗

lim

z→x,z∈Ω

|z − y|n−2 u(z) = 0; ρn−2

that is, a barrier u at x relative to Ω is transformed into a barrier u∗ at x∗ relative to Ω ∗ provided x = ∞. If x = ∞ and n ≥ 3, this result does not necessarily hold because of the indeterminancy of (|z − y|n−2 /ρn−2 )u(z) as z → ∞; if, however, x = ∞ and n = 2, then a positive superharmonic function that approaches zero at ∞ has a Kelvin transform that is positive and approaches zero at y. The same is true in the n = 2 case if Ω ⊂ By,ρ , y ∈ ∂∞ Ω, and u is a positive superharmonic function that approaches zero at y; that is, the Kelvin transform of u is a positive superharmonic function on Ω ∗ that approaches zero at ∞. n by mapAn inversion relative to a sphere ∂By,ρ can be extended to R∞ ping y onto ∞ and ∞ onto y. This extension, however, may not preserve harmonicity.

Lemma 5.3.13 If u is superharmonic on By,ρ , harmonic on a neighborhood n − of y, and u(y) = 0, then the function u∗ deﬁned on R∞ ∼ By,ρ by ∗

∗

u (x ) =

ρn−2 |x∗ −y|n−2 u(x),

0

− x∗ ∈ Rn ∼ By,ρ

x∗ = ∞

n − is superharmonic on R∞ ∼ By,ρ . n − ∼ By,ρ , Proof: Consider the n ≥ 3 case ﬁrst. It is clear that u∗ is l.s.c. on R∞ ∗ n − and from the above discussion that u is superharmonic on R ∼ By,ρ . It − . remains only to show that L(u∗ : x, δ) ≤ 0 = u∗ (∞) whenever Bx,δ ⊃ By,ρ Since u is harmonic on a neighborhood of y, 0 = u(y) = L(u : y, r) for all suﬃciently small r. Since L(u : y, r) = (ρn−2 /rn−2 )L(u∗ : y, ρ2 /r) by Equation (5.3), L(u∗ : y, r) = 0 for all suﬃciently large r. By Theorem 2.5.3 and Green’s identity, Theorem 1.2.2, L(u∗ : x, r) = 0 for suﬃciently large r. Suppose r1 < r2 and L(u∗ : x, r1 ) = L(u∗ : x, r2 ) = 0. For any ρ < r < r1 ,

5.3 PWB Method for Unbounded Regions

209

1/r1n−2 is between 1/rn−2 and 1/r2n−2 which implies that there is a λ ∈ (0, 1) such that 1/r1n−2 = λ(1/rn−2 ) + (1 − λ)(1 − r2n−2 ). By Theorem 2.5.3, there is a concave function φ on (ρ, +∞) such that L(u∗ : x, δ) = φ(1/δ n−2 ) whenever δ > ρ. By the concavity of φ, φ(1/r1n−2 ) ≥ λφ(1/rn−2 ) + (1 − λ)φ(1/r2n−2 ). That is, 0 = L(u∗ : x, r1 ) ≥ λL(u∗ : x, r) + (1 − λ)L(u∗ : x, r2 ) = λL(u∗ : x, r). − . This shows that u∗ is Thus, L(u∗ : x, r) ≤ 0 = u∗ (∞) whenever Bx,r ⊃ By,ρ n − superharmonic on R∞ ∼ By,ρ . The proof of the n = 2 case is basically the same using the fact that L(u∗ : x, r) is a concave function of − log r. n Lemma 5.3.14 Let Ω be an open subset of R∞ such that ∼ Ω − = ∅. If 0 f ∈ C (∂Ω) and > 0, then there are continuous superharmonic functions u and v deﬁned on a neighborhood of Ω − such that supx∈∂Ω |f (x) − (u(x) − v(x))| < .

Proof: If ∞ ∈ Ω − , the assertion is just that of Lemma 2.6.15. The two cases ∞ ∈ Ω and ∞ ∈ ∂∞ Ω will be considered separately. Suppose ﬁrst that − ∞ ∈ Ω. Since ∼ Ω − = ∅, there is a ball By,ρ with By,ρ ⊂∼ Ω − . An inversion ∗ relative to ∂By,ρ maps Ω into Ω ⊂ By,ρ with y an interior point of Ω ∗ , and the corresponding Kelvin transformation induces a continuous function f ∗ on ∂Ω ∗ . Applying Lemma 2.6.15 to the compact set ∂Ω ∗ ∪ {y} and the continuous function that agrees with f ∗ on ∂Ω ∗ and is equal to zero at y, there are continuous superharmonic functions u∗0 and v0∗ on By,ρ such that sup |f ∗ (x∗ ) − (u∗0 (x∗ ) − v0∗ (x∗ ))| <

x∗ ∈∂Ω ∗

and

2

. 2 Since L(u∗0 : y, r) increases to u∗0 (y) as r↓0 and L(u∗0 : y, r)=PI(u∗0 , By,r )(y), PI(u∗0 , By,r )(y) increases to u∗0 (y) as r ↓ 0. Similarly for v0∗ . Thus, r can be chosen small enough so that the continuous superharmonic functions u∗0 on By,ρ ∼ By,r ∗ u1 = PI(u∗0 , By,r ) on By,r |u∗0 (y) − v0∗ (y)| <

v1∗

=

v0∗ ∗ PI(v0 , By,r )

on By,ρ ∼ By,r on By,r

satisfy sup |f ∗ (x∗ ) − (u∗1 (x∗ ) − v1∗ (x∗ ))| <

x∗ ∈∂Ω ∗

2

210

and

5 Dirichlet Problem for Unbounded Regions

. 2 The functions u∗1 and v1∗ are then harmonic on a neighborhood of y. Let u∗ = u∗1 − u∗1 (y) and v ∗ = v1∗ − v1∗ (y). Then u∗ and v ∗ are superharmonic on By,ρ , are harmonic on a neighborhood of y, and vanish at y. Moreover, |u∗1 (y) − v1∗ (y)| <

sup |f ∗ (x∗ ) − (u∗ (x∗ ) − v ∗ (x∗ ))| ≤

x∗ ∈∂Ω ∗

sup |f ∗ (x∗ ) − (u∗1 (x∗ ) − v1∗ (x∗ ))|

x∗ ∈∂Ω ∗

+ |u∗1 (y) − v1∗ (y)| < .

By the preceding lemma, the Kelvin transforms u and v of u∗ and v ∗ , respectively, are superharmonic on Ω, provided both are deﬁned to be zero at ∞. It is clear that u and v are continuous superharmonic functions on a neighborhood of Ω − . Since |x∗ − y|n−2 ∗ ∗ |f (x ) − (u∗ (x∗ ) − v ∗ (x∗ ))| ρn−2 x∗ ∈∂Ω ∗ ≤ sup |f ∗ (x∗ ) − (u∗ (x∗ ) − v ∗ (x∗ )| < ,

sup |f (x) − (u(x) − v(x))| = sup x∈∂Ω

x∗ ∈∂Ω ∗

the assertion is proved in the ∞ ∈ Ω case. Suppose now that ∞ ∈ ∂∞ Ω and that f is continuous on ∂∞ Ω. Since f can be approximated uniformly by a function that is constant on a neighborhood of ∞, it suﬃces to prove the result for such functions. It also suﬃces to prove the result for functions that are zero in a neighborhood of ∞, for then an appropriate constant can be added to u or v. It therefore can be assumed that f is continuous on ∂∞ Ω and equal to zero on a neighborhood of ∞. The Kelvin transform f ∗ of f relative to a ball By,ρ ⊂∼ Ω − is then continuous on ∂Ω ∗ and zero on a neighborhood of y ∈ ∂Ω ∗ . The remainder of the proof is the same as the proof of the preceding case. n Lemma 5.3.15 If Ω is an open subset of R∞ with ∼ Ω − = ∅ and f ∈ 0 C (∂∞ Ω), then f is resolutive.

Proof: Since f can be approximated on ∂∞ Ω by the diﬀerence of two continuous superharmonic functions deﬁned on a neighborhood of ∂Ω by the preceding lemma and the restriction of such functions to ∂∞ Ω are resolutive by Lemma 5.3.10, the assertion follows from Lemma 5.3.9.

5.4 Boundary Behavior In order to replace the condition that ∼ Ω − = ∅ in the last result of the preceding section by the condition that ∼ Ω is not polar, it is necessary to extend the concepts of regular boundary point and barrier.

5.4 Boundary Behavior

211

n Let Ω be any open subset of R∞ . A point x ∈ ∂∞ Ω is a regular boundary point if limz→x Hf (z) = f (x) for all f ∈ C 0 (∂∞ Ω); a function u is a local barrier at x ∈ ∂∞ Ω if u is a positive superharmonic function deﬁned on Λ ∩ Ω for some neighborhood Λ of x with limz→x,z∈Λ∩Ω u(z) = 0; if, in addition, Λ = Ω, then u is called a barrier at x ∈ ∂∞ Ω. n Lemma 5.4.1 Let Ω be an open subset of R∞ such that ∼ Ω − = ∅. If there is a local barrier at x ∈ ∂∞ Ω, then there is a barrier w at x deﬁned on Ω with the additional property that inf{w(z); z ∈ Ω ∼ Λx } > 0 for all neighborhoods Λx of x.

Proof: If ∞ ∈ Ω − , the assertion is the same as that of Theorem 2.6.20. It therefore can be assumed that ∞ ∈ Ω − . Suppose ﬁrst that ∞ ∈ ∂∞ Ω. If x = ∞ and n ≥ 3, there is always a barrier at x with the additional property, namely, the fundamental harmonic function |z − y|−n+2 with ﬁxed − pole y ∈∼ Ω − . It therefore can be assumed that x ∈ ∂Ω. Let By,ρ ⊂∼ Ω − , ∗ let Ω be the inversion of Ω relative to ∂By,ρ , and let v be a local barrier at x. Then the Kelvin transform v ∗ (z ∗ ) =

|z − y|n−2 v(z), ρn−2

z∈Ω

is a local barrier at x∗ ∈ ∂Ω ∗ . By Theorem 2.6.20 there is a barrier w∗ at x∗ deﬁned on Ω ∗ such that inf{w∗ (z ∗ ); z ∗ ∈ Ω ∗ ∼ Vx∗∗ } > 0 for all neighborhoods Vx∗∗ of x∗ . It is easy to see that the Kelvin transform w of w∗ is a barrier at x deﬁned on Ω with the additional property. Consider now the case ∞ ∈ Ω. In this case, ∂Ω is a compact subset of Rn . Suppose there is a local barrier at x ∈ ∂Ω. The boundary function m(z) = |z − x| is then continuous on ∂Ω and resolutive by Lemma 5.3.15. The proof that limz→x,z∈Ω Hm (z) = 0 is precisely the same as in the proof of Theorem 2.6.20. This time, however, it is not true that the function |z − x| is in the lower class Lm since it is not subharmonic at ∞ ∈ Ω. If it can be shown that there is a number α > 0 such that Hm (z) ≥ α|z − x| on Ω in a neighborhood of x, then the minimum principle can be used to show that Hm has the required additional property. By considering the Dirichlet problem for Ω ∗ and the Kelvin transform m∗ of m, it can be seen that Hm is strictly positive on Ω since it is true of Hm∗ . Consider any ball Bx,ρ1 ⊃ ∂Ω, and let k = inf{Hm (z); z ∈ ∂Bx,ρ1 } > 0. Choosing α < 1 such that αρ1 < k/4, α|z − x| < k/4 for z ∈ ∂Bx,ρ1 . Moreover, there is a function u ∈ Lm such that u(z) > k/2 for z ∈ ∂Bx,ρ1 . The function − max (u(z), α|z − x|) z ∈ Ω ∩ Bx,ρ 1 w(z) = u(z) z ∈ Ω ∼ Bx,ρ1 is a member of Lm . This shows that Hm (z) ≥ α|x − z| for all z ∈ Ω ∩ Bx,ρ1 .

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5 Dirichlet Problem for Unbounded Regions

The condition that ∼ Ω − = ∅ in the following theorem can be replaced by the less restrictive condition that ∼ Ω is nonpolar. See Remark 5.4.11. n such that ∼ Ω − = ∅. If Theorem 5.4.2 Let Ω be an open subset of R∞ the function f is bounded above on ∂∞ Ω and there is a barrier at x ∈ ∂Ω, then lim supz→x,z∈Ω H f (z) ≤ lim supz→x,z∈∂∞ Ω f (z). A ﬁnite point x ∈ ∂Ω is regular if and only if there is a barrier at x. The point ∞ is always a regular boundary point if n ≥ 3. A ﬁnite point x ∈ ∂∞ Ω is a regular boundary point if and only if x∗ is a regular boundary point for Ω ∗ , the inversion of Ω relative − ⊂∼ Ω − ; if ∞ ∈ ∂Ω and n = 2, then ∞ is a regular boundary to ∂By,ρ ⊂ By,ρ point if and only if y is a regular boundary point for Ω ∗ .

Proof: The proof of the ﬁrst assertion is basically the same as the proof of Lemma 2.6.21. It follows from the ﬁrst assertion that a boundary point x is regular if there is a barrier at x. As to the converse, suppose x is a ﬁnite regular boundary point for Ω. If g(z) = min (|z − x|, 1), z ∈ ∂Ω, then Hg is a barrier at x. If x = ∞ and n ≥ 3, there is always a barrier at x, namely, the fundamental harmonic function with a pole in ∼ Ω − . Since a boundary point x is regular if and only there is a barrier at x, the last two statements follow from Remark 5.3.12. n Lemma 5.4.3 If Ω is an open subset of R∞ such that ∼ Ω − = ∅, then the set I of irregular boundary points of Ω is a polar set. − − n − ⊂ By,δ ⊂ By,δ ⊂∼ Ω − = ∅, I ⊂ R∞ ∼ By,ρ . The Proof: Letting By,ρ ∗ inversion I of the set I relative to ∂By,ρ is contained in the set of irregular boundary points of Ω ∗ by the preceding theorem. (Note that y may be an irregular boundary point for Ω ∗ , whereas y ∗ = ∞ is not an irregular boundary point for Ω in the n ≥ 3 case.) Since I ∗ is a polar subset of By,ρ , there is a superharmonic function v ∗ on By,ρ such that v ∗ = +∞ on I ∗ . Consider − . Since v = +∞ on the Kelvin transform v of v ∗ , deﬁned only on Rn ∼ By,ρ I ∼ {∞}, I ∼ {∞} is a polar set. If n = 2 and ∞ ∈ I, then I is a polar 2 by Theorem 5.3.8; if n ≥ 3, then ∞ set since {∞} is a polar subset of R∞ cannot belong to I since ∞ is always a regular boundary point when n ≥ 3, n and therefore I is a polar subset of R∞ , n ≥ 3. n Lemma 5.4.4 If Ω is an open subset of R∞ with ∼ Ω − = ∅ and Z is a polar subset of ∂∞ Ω, then there is a positive superharmonic function v deﬁned on a neighborhood of Ω − such that limz→x,z∈Ω v(z) = +∞ for all x ∈ Z. − − − n ⊂ By,δ ⊂ By,δ ⊂∼ Ω − = ∅, Z ⊂ R∞ ∼ By,δ . If x ∈ Proof: Letting By,ρ − n Z ∼ {∞}, there is a neighborhood W ⊂ R ∼ By,δ of x and a superharmonic function w = +∞ on W ∩Z. Then the Kelvin transform of w relative to ∂By,ρ is +∞ on W ∗ ∩ Z ∗ . It follows that the image of Z ∼ {∞} is a polar subset of By,ρ . Since adjoining a single point of Rn to a polar set results in a polar set, Z ∗ is a polar subset of By,ρ . Now let j0 be the smallest integer greater than δ,

5.4 Boundary Behavior

213

− − let Zj0 = Z ∩ (By,j ∼ By,δ ), and let Zj = Z ∩ (By,j ∼ By,j−1 ), j ≥ j0 . Then 0 ∗ Z ∼ {∞} = ∪j≥j0 Zj and each Zj is a polar subset of By,ρ . It follows that there is a superharmonic function vj∗ on By,ρ such that vj∗ = +∞ on Zj∗ . By replacing vj∗ on By,ρ2 /j by its Poisson integral as in Lemma 2.4.11, it can be assumed that vj∗ is harmonic on a neighborhood of y without aﬀecting its values on Zj∗ . By adding a constant, it can be assumed that vj∗ (y) = 0. Then the Kelvin transform vj of vj∗ , with vj (∞) deﬁned to be zero, is superharmonic − n on R∞ ∼ By,δ by Lemma 5.3.13 and vj = +∞ on Zj . Since vj is l.s.c. on the n compact set R∞ ∼ By,δ , a constant can be added to vj to make it positive thereon without aﬀecting itsinﬁnities. Now let {αj }j≥j0 be a sequence of positive numbers such that j≥j0 αj vj (∞) < +∞ on Z ∼ {∞}. Since ∞ cannot be in the polar set Z in the n ≥ 3 case, the proof is complete in this case; if ∞ ∈ Z in the n = 2 case, replace v(z) by v(z) + log (|z − y|/δ). n Lemma 5.4.5 Let Ω be an open subset of R∞ with ∼ Ω − = ∅. If f and g are functions on ∂∞ Ω such that f = g on ∂∞ Ω except possibly for a polar subset of ∂∞ Ω, then H f = H g and H f = H g ; if, in addition, f is resolutive, then g is also and Hf = Hg .

Proof: Using the preceding lemma, the result follows as in the proofs of Theorem 4.2.19 and Corollary 4.2.20. The method used to prove the following lemma is due to Schwarz [57] and is known as the alternating method. The method was originally used to construct the Dirichlet solution on a region that is the union of two regions known to be regular (c.f. Kellogg [35]). n such that ∼ Ω is not polar and Lemma 5.4.6 If Ω is an open subset of R∞ Z is a polar subset of ∂∞ Ω, then there is a positive superharmonic function v on Ω such that limy→z,y∈Ω v(y) = +∞ for all z ∈ Z.

Proof: If ∼ Ω − = ∅, the assertion is just that of the preceding lemma. It therefore can be assumed that ∼ Ω = ∂∞ Ω and that ∂∞ Ω is not po− − lar. Fix By,ρ ⊂ Ω. Then Z is a polar subset of the boundary of Ω ∼ By,ρ . − Consider ∂∞ (Ω ∼ By,ρ ) = ∂∞ Ω ∪ ∂By,ρ . Since the points of ∂By,ρ are reg− ular boundary points for Ω ∼ By,ρ and ∂∞ Ω is not polar, some point of − by Lemma 5.4.3. By ∂∞ Ω must be a regular boundary point for Ω ∼ By,ρ − Lemma 5.4.4, there is a positive superharmonic function w on Ω ∼ By,ρ such that limy→z,y∈Ω w(y) = +∞ for all z ∈ Z ⊂ ∂∞ Ω. Moreover, if B is a ball − with B − ⊂ Ω, then it can be assumed that w is harmonic containing By,ρ − ˆ w − relon B ∼ By,ρ ; for if this is not the case, w can be replaced by R Ω∼B − ative to Ω ∼ By,ρ without changing its boundary behavior at points of ∂Ω − ⊂ B. Then w according to Lemma 4.3.2. Choose ρ2 > ρ1 > ρ such that By,ρ 2 − is harmonic on a neighborhood of By,ρ2 ∼ By,ρ1 . Two sequences of functions {uj } and {vj } on By,ρ2 and Ω ∼ By,ρ1 , respectively, will be constructed as follows. Deﬁning

214

5 Dirichlet Problem for Unbounded Regions

u1 =

PI(w : By,ρ2 ) w

on By,ρ2 on ∂By,ρ2 ,

u1 is continuous on By,ρ2 and harmonic on the interior. Let 0 on ∂∞ Ω f1 = on ∂By,ρ1 u1 − w

and deﬁne v1 =

Hf1 u1 − w

− on Ω ∼ By,ρ 1 on ∂By,ρ1 ,

where Hf1 is the Dirichlet solution corresponding to f1 and the region Ω ∼ − − . Since all points of ∂By,ρ1 are regular boundary points for Ω ∼ By,ρ , v1 By,ρ 1 1 is continuous on Ω ∼ By,ρ1 and harmonic on the interior. Now let PI(w + v1 : By,ρ2 ) on By,ρ2 u2 = w + v1 on ∂By,ρ2 . − and harmonic on the interior. Again letting Then u2 is continuous on By,ρ 2

f2 = and deﬁning

v2 =

0 u2 − w

Hf2 u2 − w

on ∂∞ Ω on ∂By,ρ1 , − on Ω ∼ By,ρ 1 on ∂By,ρ1 ,

v2 is continuous on Ω ∼ By,ρ1 and harmonic on the interior. The sequences {uj } and {vj } are so deﬁned inductively. On ∂By,ρ1 , vj = uj − w and so vj+1 − vj = uj+1 − uj

on ∂By,ρ1

for j ≥ 1; whereas on ∂By,ρ2 , uj+1 = w + vj and so uj+1 − uj = vj − vj−1

on ∂By,ρ2

for j ≥ 2. Let αj =

sup

|vj+1 (z) − vj (z)| =

z∈∂By,ρ1

sup

|uj+1 (z) − uj (z)|

z∈∂By,ρ1

and let βj =

sup z∈∂By,ρ2

|uj+1 (z) − uj (z)| =

sup

|vj (z) − vj−1 (z)|

z∈∂By,ρ2

for j ≥ 2. Consider the boundary function g that is equal to 1 on ∂By,ρ1 and 0 on ∂∞ Ω and the associated Dirichlet solution Hg . Since all points of ∂By,ρ1

5.4 Boundary Behavior

215

− − and some points of ∂∞ Ω are regular for Ω ∼ By,ρ , 0 < Hg < 1 on Ω ∼ By,ρ . 1 1 It follows that there is a constant q < 1 such that Hg < q on ∂By,ρ2 . If g˜ − is any continuous function on the boundary of Ω ∼ By,ρ such that |˜ g | ≤ g, 1 then |Hg˜ | < q on ∂By,ρ2 . Since (vj+1 − vj )/αj = H(fj+1 −fj )/αj on Ω ∼ − , |(fj+1 − fj )/αj | = |(uj+1 − uj )/αj | ≤ 1 on ∂By,ρ1 , and |fj+1 − fj | = 0 By,ρ 1 on ∂Ω, |(vj+1 − vj )/αj | = |H(fj+1 −fj )/αj | < q on ∂By,ρ2 .

Thus, |vj+1 − vj | ≤ qαj

on ∂By,ρ2 .

Therefore, βj+1 ≤ qαj , j ≥ 1. Since uj+1 − uj is harmonic on By,ρ2 , |uj+1 − uj | ≤

|uj+1 (z) − uj (z)| = βj

sup z∈∂By,ρ2

on ∂By,ρ1 and it follows that α j ≤ βj , j ≥ 1. Hence, αj+1 ≤ βj+1 ≤ qαj for j ≥ 1, and the positive series αj is dominated by a convergent geometric series. This shows that the series v1 +

∞

(vj+1 − vj )

j=1 − . Similarly, βj+1 ≤ qβj , and the positive seconverges uniformly on Ω ∼ By,ρ 1 ries βj is dominated by a convergent geometric series, thereby establishing that ∞ u = lim uj = u1 + (uj+1 − uj ) j→∞

j=1

deﬁnes a function that is harmonic on By,ρ2 . Since vj = uj −w on ∂By,ρ1 , v = u − w on ∂By,ρ1 . Also, uj+1 = w + vj on ∂By,ρ2 implies that w = u − v on ∂By,ρ2 ; that is, u = v + w on both ∂By,ρ1 and ∂By,ρ2 , but since both u and − − v + w are harmonic on By,ρ2 ∼ By,ρ , u = v + w on By,ρ2 ∼ By,ρ . It follows 1 1 that the function u on By,ρ2 w ˜= − v+w on Ω ∼ By,ρ 1 has the desired properties. n Corollary 5.4.7 (Bouligand [4]) If Ω is an open subset of R∞ such that ∼ Ω is not polar and h is a bounded harmonic function on Ω such that limy→z,y∈Ω h(y) = 0 for all z ∈ ∂∞ Ω except possibly for a polar set, then h = 0 on Ω.

Proof: Let Z be the exceptional subset of ∂∞ Ω. By the preceding lemma, there is a positive superharmonic function w on Ω such that limy→z,y∈Ω w(y) = +∞ for all z ∈ Z. Then for every > 0, h + w is superharmonic on Ω and lim inf y→z,y∈Ω (h+ w)(y) ≥ 0 for all z ∈ ∂∞ Ω. By Theorem 5.2.5, h+ w ≥ 0

216

5 Dirichlet Problem for Unbounded Regions

on Ω. Since w is ﬁnite a.e. on Ω ∩ Rn , h ≥ 0 a.e.on Ω ∩ Rn , and it follows from the continuity of h on Ω that h ≥ 0 on Ω. Applying the same argument to −h, h = 0 on Ω. n Lemma 5.4.8 Let Ω be an open subset of R∞ such that ∼ Ω is not polar − n and let By,ρ ⊂ Ω ∩ R . If f is a bounded function on ∂Ω, then

H f = inf{u; u hyperharmonic on Ω, u bounded below onΩ, and lim inf u(y) ≥ f (z) for all z ∈ ∂Ω that are regular boundary

y→z,y∈Ω

− }. points for Ω ∼ By,ρ

Proof: Since the right side of the equation is less than or equal to H f , it suﬃces to show that u ≥ H f for every u in the indicated set. This inequality is obviously true on any component of Ω on which u is identically +∞. It therefore can be assumed that u is superharmonic on Ω. Note ﬁrst that the − is a polar subset of ∂∞ Ω set Z of irregular boundary points of Ω ∼ By,ρ by Lemma 5.4.3. By the preceding lemma, there is a positive superharmonic function w on Ω such that limy→z,y∈Ω w(y) = +∞ for all z ∈ Z. For every > 0, u + w ∈ Uf and therefore u + w ≥ H f . Since w is ﬁnite a.e. on Ω ∩ Rn , u ≥ H f a.e. on Ω ∩ Rn . Thus, for x ∈ Ω ∩ Rn , u(x) ≥ limδ→0 A(u : x, δ) ≥ limδ→0 A(H f : x, δ) = H f (x). This shows that u ≥ H f on Ω ∩ Rn ; if ∞ ∈ Ω, it follows from Deﬁnition 5.2.1 that for any x ∈ Rn and δ > 0 for which ∼ Bx,δ ⊂ Ω, u(∞) ≥ L(u : x, δ) ≥ L(H f : x, δ) = H f (∞), showing that u ≥ H f on Ω. n Theorem 5.4.9 Let Ω be an open subset of R∞ such that ∼ Ω is not polar. Then each f ∈ C 0 (∂∞ Ω) is resolutive.

Proof: Since f is bounded on the compact set ∂∞ Ω, the same is true of H f − and H f and both are harmonic. Let m = supz∈∂Ω |f (z)| and let By,ρ ⊂ Ω. − The Dirichlet solution Hg corresponding to Ω ∼ By,ρ and the boundary function m+1 on ∂By,ρ g= f on ∂∞ Ω − − Hg (z) = f (x) whenever x ∈ is harmonic on Ω ∼ By,ρ and limz→x,z∈Ω∼By,ρ − ∂∞ Ω is a regular boundary point for Ω ∼ By,ρ . Moreover, the function

u=

m+1 Hg

− on By,ρ − on Ω ∼ By,ρ

is superharmonic on Ω and limz→x,z∈Ω u(z) = f (x) whenever x ∈ ∂∞ Ω is − . By the preceding lemma, u ≥ H f . a regular boundary point for Ω ∼ By,ρ Similarly, there is a subharmonic function v on Ω such that v ≤ H f and limz→x,z∈Ω v(z) = f (x) whenever x ∈ ∂∞ Ω is a regular boundary point for

5.4 Boundary Behavior

217

− Ω ∼ By,ρ . It follows that limz→x,z∈Ω (H f − H f )(z) = 0 for all x ∈ ∂∞ Ω that − ; but since the irregular boundary are regular boundary points for Ω ∼ By,ρ − points of Ω ∼ By,ρ constitute a polar set by Lemma 5.4.3, H f − H f = 0 by the preceding corollary. By the preceding theorem, each f ∈ C 0 (∂∞ Ω) determines a harmonic function Hf on Ω. For each x ∈ Ω, the map x → Hf (x) is a linear functional on C 0 (∂∞ Ω) to which the Riesz representation theorem may be applied to obtain the integral representation Hf (x) = f dμx

where μx is a Borel measure on ∂∞ Ω with μx (∂∞ Ω) = 1 since H1 (x) = 1. If {fj } is an increasing sequence of boundary functions such that H f1 > −∞ on Ω and f = limj→∞ fj , then the proof of Lemma 2.6.13 is valid in the present context and results in H f = limj→∞ H fj ; if the fj are also resolutive, then H f (x) = H f (x) =

f dμx

(5.5)

on Ω, where the three quantities are all +∞ if any one is +∞. This result can be used to show that the indicator functions of compact subsets of ∂∞ Ω are resolutive. Using the minimum principle of Theorem 5.2.5 as in Theorem 2.9.2, the class of μx null sets is independent of x on any component of Ω. If for each x ∈ Ω, Fx is the completed Borel σ-algebra of subsets of ∂∞ Ω relative to μx , then F = ∩x∈Ω Fx is called the σ-algebra ofμ -measurable sets; those F -measurable functions f on ∂∞ Ω for which |f | dμx < +∞ for some x in each component of Ω are called μ -integrable functions. As before, μx , extended to F , will be called harmonic measure relative to x and Ω. If necessary, the dependence of harmonic measure on the set Ω will be noted by using the notation μΩ x. For the remainder of this section, Ω is an open subset of Rn having a n is allowed to be a boundary point of Ω Green function. The point ∞ ∈ R∞ n but not an interior point. As a subset of R∞ , ∞ ∈∼ Ω and ∼ Ω is not polar in the n ≥ 3 case; in the n = 2 case, Rn ∼ Ω is not polar by Theorem 4.6.1 so n ∼ Ω is not polar. The results of this chapter are therefore applicable that R∞ n to Ω as a subset of R∞ . If F is a relatively closed subset of Ω, HfΩ∼F will denote the Dirichlet solution corresponding to f and Ω ∼ F , regarded as a n . subset of R∞ Theorem 5.4.10 Let Ω be a Greenian subset of Rn such that ∼ Ω is nonpolar. If Z is a polar subset of ∂∞ Ω, then μx (Z) = 0 for all x ∈ Ω. Moreover, if f is a resolutive function on ∂∞ Ω and g = f q.e. on ∂∞ Ω, then g is resolutive on ∂∞ Ω and Hg = Hf on Ω.

218

5 Dirichlet Problem for Unbounded Regions

Proof: Fix x ∈ Ω. By Lemma 5.4.6, there is a superharmonic function u on Ω with u(x) < +∞ such that limy→z,y∈Ω u(y) = +∞ for all z ∈ Z. Since (1/n)u ∈ UΩ χZ , 0 ≤ HχZ (x) ≤ (1/n)u(x) for all n ≥ 1 and therefore μx (Z) = χZ dμx = HχZ (x) = 0. The second assertion follows from the ﬁrst and Equation (5.5). Remark 5.4.11 Let Ω be an open subset of Rn such that ∼ Ω is not polar. The results of Theorem 5.4.2 then hold for such Ω. To see this, consider a − ⊂ Ω. If f ∈ C 0 (∂∞ Ω), let closed ball B − = Bx,δ g=

f HfΩ

on ∂∞ Ω on ∂Bx,δ .

By considering upper and lower PWB solutions, it is easily seen that Ω∼B −

−

Ω∼B ≤ Hg HΩ f ≤ Hg

Ω

≤ Hf

−

on Ω ∼ B − . Since f is resolutive, HfΩ = HgΩ∼B on Ω ∼ B − . Thus, HfΩ has −

the same boundary properties as HgΩ∼B at points of ∂∞ Ω. Since the results of Theorem 5.4.2 apply to Ω ∼ B − , they also apply to Ω. Note also that if y ∈ ∂Ω is a regular boundary point for Ω, then it is also a regular boundary point for Ω ∼ B − and there is a barrier u at y on Ω ∼ B − having the additional property that inf {u(z); z ∈ (Ω ∼ B − ) ∼ Λ} > 0 for each neighborhood Λ of y. It is easy to see that the barrier u has a superharmonic extension on Ω satisfying inf {u(z); z ∈ Ω ∼ Λ} > 0 for each neighborhood Λ of y. Theorem 5.4.12 Let Ω be a subset of Rn such that ∼ Ω is nonpolar and let x ∈ Ω. If y0 is a ﬁnite, regular boundary point, then limy→y0 ,y∈Ω GΩ (x, y)= 0. If n ≥ 3 and ∞ ∈ ∂∞ Ω, then limy→∞ GΩ (x, y) = 0. If n = 2 and ∞ is a regular boundary point for Ω, then limy→∞,y∈Ω GΩ (x, y) = 0. Proof: Suppose ﬁrst that n ≥ 3 and ∞ ∈ ∂∞ Ω. Since GΩ (x, ·) ≤ G(x, ·) by Theorem 3.2.12, limy→∞ GΩ (x, y) = 0. Suppose now that y0 is a ﬁnite, regular boundary point for Ω. By Theorem 5.4.2 and the preceding remark, there is a barrier u at y0 on Ω such that inf {u(z); z ∈ Ω ∼ Λ} > 0 for all neighborhoods Λ of y0 . Choose α > 0 such that Λα = {z ∈ Ω; GΩ (x, z) > α} has compact closure in Ω. Choosing n ≥ 1 so that nu > α on ∂Λα , nu ≥ GΩ (x, ·) on ∂Λα and therefore nu ≥ GΩ (x, ·) on Ω ∼ Λα by Theorem 3.3.8. It follows that limy→y0 ,y∈Ω GΩ (x, y) = 0. Λ It is possible to relate the swept measures δΩ (x, ·) to the harmonic meaΛ sures μx by means of the following theorem. Theorem 5.4.13 If u is a nonnegative superharmonic function on the Greenian set Ω ⊂ Rn , Λ is an open subset of Ω, and

5.4 Boundary Behavior

219

f=

u on Ω ∩ ∂∞ Λ , 0 on ∂∞ Ω ∩ ∂∞ Λ,

ˆu then f is resolutive and HfΛ = R Ω∼Λ on Λ. Proof: Let v be a nonnegative superharmonic function on Ω with v ≥ u on Ω ∼ Λ. Then lim inf y→x,y∈Λ v(y) ≥ f (x) for x ∈ ∂∞ Λ ∩ Ω and lim inf y→x,y∈Λ v(y) ≥ 0 = f (x) for x ∈ ∂∞ Ω ∩ ∂∞ Λ. Thus, v ∈ UΛ f . It Λ

u follows that RuΩ∼Λ ≥ H f ≥ H Λ f ≥ 0 on Λ. Since RΩ∼Λ ≤ u and the latter is ﬁnite a.e. on Λ, f is resolutive relative to Λ, HfΛ ≤ RuΩ∼Λ , and therefore ˆ u . Now consider any w ∈ UΛ . Since u|Λ ∈ UΛ , min (w, u|Λ ) ∈ UΛ . HfΛ ≤ R Ω∼Λ f f f Deﬁne w ˜ on Ω by on Λ w ˜ = min (w, u|Λ ) u on Ω ∼ Λ.

Then w ˜ is superharmonic on Ω, majorizes u on Ω ∼ Λ, and therefore w ˜≥ ˆu on Ω. Since w ≥ w ˜ on Λ, RuΩ∼Λ ≥ R Ω∼Λ Λ ˆ uΩ∼Λ H f = HfΛ ≥ R

on Λ

ˆu and it follows that HfΛ = R Ω∼Λ on Λ. Remark 5.4.14 If Λ− ⊂ Ω, then by Theorem 4.7.4 Λ Ω∼Λ (x, dy), x ∈ Λ. u dμx = u(y) δΩ Ω∼Λ (x, ·) whenever Applying Lemma 5.3.14 to Ω ∼ Λ− , it follows that μΛ x = δΩ − Λ ⊂ Ω.

Let Ω be a Greenian subset of Rn . Consider the following extension gΩ (x, ·) n of GΩ (x, ·) to R∞ for x ∈ Ω. ⎧ if y ∈ Ω GΩ (x, y) ⎨ 0 if y ∈ Rn ∼ Ω − gΩ (x, y) = ⎩ lim supz→y,z∈Ω GΩ (x, z) if y ∈ ∂∞ Ω. In the case of unbounded Ω, the deﬁnition of gΩ (x, ∞) is consistent with the behavior of GΩ (x, ·) at ∞. According to Theorem 4.5.4, gΩ (x, ·) = 0 q.e. on ∂Ω. If Ω is unbounded, gΩ (x, ∞) = 0 in the n ≥ 3 case; in the n = 2 case, ∞ is a polar subset of ∂∞ Ω according to Theorem 5.3.8 so that qΩ (x, ·) = 0 q.e. on ∂∞ Ω. The function gΩ (x, ·) has the following properties: (i) gΩ (x, ·) = 0 q.e. on ∂∞ Ω , (ii) gΩ (x, ·) is subharmonic on Rn ∼ {x}, and (iii) if n ≥ 3 or n = 2 and ∞ is a regular boundary point for Ω, then gΩ (x, ∞) = 0.

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5 Dirichlet Problem for Unbounded Regions

Theorem 5.4.15 If Λ is an open subset of the Greenian set Ω ⊂ Rn , then for each x ∈ Ω, the function gΩ (x, ·)|∂∞ Λ is a resolutive boundary function, and (ii) gΛ (x, y) = GΩ (x, y) − HgΛΩ (y,·) (x), x ∈ Λ, y ∈ Ω . (i)

Proof: For x ∈ Ω, let fx (y) =

GΩ (x, y) if y ∈ Ω ∩ ∂∞ Λ 0 if y ∈ ∂∞ Ω ∩ ∂∞ Λ.

By the preceding theorem, Theorem 5.4.13, the function fx is resolutive on ∂∞ Λ. Since gΩ = 0 q.e. on ∂∞ Ω, fx = gΩ (x, ·) q.e. on ∂∞ Λ so that gΩ (x, ·) is resolutive by Theorem 5.4.10 and (i) is true. The proof of (ii) will be split into two cases. Λ Case (i): (y ∈ Ω ∼ ∂∞ Λ). Since GΩ (y, ·)|Λ ∈ UΛ fy , GΩ (y, ·)|Λ ≥ Hfy on Λ. Λ Since Hfy is a harmonic minorant of GΩ (y, ·) on Λ, ghmΛ GΩ (y, ·) ≥ HfΛy on Λ. Consider any v ∈ LΛ fy . Then for z ∈ ∂∞ Λ

lim sup ghmΛ GΩ (y, x) ≥ lim sup v(x) ≥ fy (x),

x→z,x∈Λ

x→z,x∈Λ

Λ ghmΛ GΩ (y, ·) ∈ LΛ fy and therefore ghmΛ GΩ (y, ·) ≤ Hfy on Λ. Thus,

ghmΛ GΩ (y, ·) = HfΛy on Λ. Since fy = gΩ (y, ·) q.e. on ∂∞ Λ, ghmΛ GΩ (y, ·) = HgΛΩ (y,·) on Λ. By Theorem 3.4.8, for x ∈ Λ, GΩ (y, x) = GΛ (y, x) + h where h is a nonnegative harmonic function on Λ. Since ghm GΛ (y, ·) = 0 on Λ, ghm GΩ (y, ·) = h on Λ by Lemma 3.3.6. Therefore, for x ∈ Λ, gΛ (x, y) = GΛ (x, y) = GΛ (y, x) = GΩ (y, x) − ghmΛ GΩ (y, x) = GΩ (x, y) − HgΩ (y,·) (x) and (ii) holds in the y ∈ Ω ∼ ∂∞ Λ case. Before discussing the next case, it will be shown that ˆ Ω∼Λ (GΩ (y, ·), x) = R ˆ Ω∼Λ (GΩ (x, ·), y), R

x ∈ Λ, y ∈ Ω.

(5.6)

For x, y ∈ Λ, it follows from Case (i) that GΛ (x, y) = GΩ (x, y) − HgΩ (y,·) (x). Since GΛ and GΩ are symmetric functions, HgΩ (y,·) (x) is a symmetric function on Λ × Λ. Since gΩ (y, ·) = GΩ (y, ·) on Ω ∩ ∂∞ Λ and gΩ (y, ·) = 0 quasi everywhere on ∂∞ Ω ∩ ∂∞ Λ, for x, y ∈ Λ

5.4 Boundary Behavior

221

Λ ˆ Ω∼Λ (GΩ (y, ·), x) = H Λ ˆ R gΩ (y,·) (x) = HgΩ (x,·) (y) = RΩ∼Λ (GΩ (x, ·), y) (5.7)

by Theorem 5.4.13. Suppose now that x ∈ Λ and y ∈ Ω ∼ Λ. Letting Λn = Λ ∪ By,1/n and using the preceding equation, ˆ Ω∼Λn (GΩ (x, ·), y). ˆ Ω∼Λn (GΩ (y, ·), x) = R R By Theorem 4.6.5, ˆ Ω∼(Λ∪{y}) (GΩ (y, ·), x) = R ˆ Ω∼(Λ∪{y}) (GΩ (x, ·), y). R Since Ω ∼ (Λ∪{y}) diﬀers from Ω ∼ Λ only by a polar set, by Corollary 4.6.4 the preceding equation holds with Ω ∼ (Λ ∪ {y}) replaced by Ω ∼ Λ. Thus, Equation (5.6) has been established. Case (ii): (y ∈ Ω ∩ ∂∞ Λ). Suppose ﬁrst that y is a regular boundary point ˆ Ω∼Λ (GΩ (y, ·), ·) = RΩ∼Λ (GΩ (y, ·), ·) q.e. on Ω and polar sets for Λ. Since R have Lebesgue measure zero, there is a set Z ⊂ Ω of Lebesgue measure ˆ Ω∼Λ (GΩ (y, ·), x) = RΩ∼Λ (GΩ (y, ·), x) for x ∈ Ω ∼ Z. By zero such that R Theorem 2.4.7, ˆ Ω∼Λ (GΩ (x, ·), y) = R

lim inf

z→y,z ∈Z∪{y}

RΩ∼Λ (GΩ (x, ·), z).

The lim inf on the right can be calculated by taking the minimum of the lim inf on Λ ∼ (Z ∪ {y}) and the lim inf on (Ω ∼ Λ) ∼ (Z ∪ {y}). Since y is a regular boundary point for Λ, lim inf z→y,z∈Λ∼(Z∪{y})

RΩ∼Λ (GΩ (x, ·), z) =

lim z→y,z∈Λ∼(Z∪{y}

HgΛΩ (x,·) (z) = GΩ (x, y)

by continuity of G( x, ·) at y. Moreover, since RΩ∼Λ (GΩ (x, ·), ·) = GΩ (x, ·) on Ω ∼ Λ, lim inf

z→y,z∈(Ω∼Λ)∼(Z∪{y})

RΩ∼Λ (GΩ (x, ·), z) =

lim inf

z→y,z∈(Ω∼Λ)∼(Z∪{y})

GΩ (x, z)

= GΩ (x, y).

ˆ Ω∼Λ (GΩ (x, ·), y) = GΩ (x, y), x ∈ Λ, whenever y is a regular Therefore, R boundary point for Λ. By Theorem 5.4.13, Equation (5.6), Equation (5.7), and the above equation, ˆ Ω∼Λ (GΩ (y, ·), x) GΩ (x, y) − HgΛΩ (y,·) (x) = GΩ (x, y) − R ˆ Ω∼Λ (GΩ (x, ·), y) = GΩ (x, y) − R = GΩ (x, y) − GΩ (x, y) = 0

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5 Dirichlet Problem for Unbounded Regions

for x ∈ Λ and y ∈ Ω ∩ ∂∞ Λ a regular boundary point for Λ; since GΩ is a symmetric function, HgΛΩ (y,·) (x) = HgΛΩ (x,·) (y) for such x and y. For x, z ∈ Λ, by Case (i), gΛ (x, z) = GΩ (x, z) − HgΛΩ (z,·) (x) = GΩ (x, z) − HgΛΩ (x,·) (z). Letting z → y and using the fact that y is a regular boundary point for Λ, gΛ (x, y) = GΩ (x, y) − HgΛΩ (x,·) (y) = GΩ (x, y) − HgΛΩ (y,·) (x). Thus for x ∈ Λ,

gΛ (x, y) = GΩ (x, y) − HgΛΩ (y,·) (x),

except possibly for a polar subset of Ω ∩ ∂∞ Λ. Combining the two cases, the above equation holds for all y ∈ Ω except possibly for a polar subset. Since gΛ (x, ·) is subharmonic on Rn ∼ {x} and GΩ (x, ·) is harmonic on Ω ∼ {x}, both sides are subharmonic on Ω ∼ {x} and the equation holds for all y ∈ Ω by Theorem 2.4.7. Theorem 5.4.16 If Ω is a Greenian subset of Rn and Λ ⊂ Ω, then Λ ˆ ˆ GΛ Ω (x, y) = RΛ (GΩ (x, ·), y) = RΛ (GΩ (y, ·), x) = GΩ (y, x),

x, y ∈ Ω. (5.8)

Proof: Suppose ﬁrst that Λ is closed. Then Ω ∼ Λ is open and ˆ Λ (GΩ (x, ·), y) = R ˆ Λ (GΩ (y, ·), x), R

x, y ∈ Ω ∼ Λ

by Equation (5.6). If either x ∈ Λ or y ∈ Λ, replace Λ by Λn = Λ ∼ ˆ Λn (GΩ (x, ·), y) = R ˆ Λn (GΩ (y, ·), x); letting n → ∞, (Bx,1/n ∪By,1/n ) so that R ˆ ˆ RΛ∼{x,y} (GΩ (x, ·), y) = RΛ∼{x,y} (GΩ (y, ·), x) by Theorem 4.6.5. Since Λ differs from Λ ∼ {x, y} by a polar set, Equation (5.8) holds whenever Λ is a closed set. It then follows from Theorem 4.6.5 that the equation holds for Fσ sets and, in particular, for open sets. Lastly, consider any Λ ⊂ Ω and any open set O ⊂ Ω, O ∈ O(Ω). By Equation (5.8), (vi) of Theorem 4.3.5, and Lemma 4.6.12, RΛ (GΩ (x, ·), y) = RΛ (GΩ (y, ·), x),

x, y ∈ Ω.

ˆ Λ (GΩ (x, ·), y) and RΛ (GΩ (y, ·), x) = R ˆ Λ (GΩ Since RΛ (GΩ (x, ·), y) = R (y, ·), x) for x, y ∈ Ω ∼ Λ by Theorem 4.6.3, ˆ Λ (GΩ (x, ·), y) = R ˆ Λ (GΩ (y, ·), x), R

x, y ∈ Ω ∼ Λ.

5.5 Intrinsic Topology

223

If x ∈ Λ or y ∈ Λ, apply this result to Λ ∼ {x, y} and use the fact that this set diﬀers from Λ by a polar set to arrive at Equation (5.8). Theorem 5.4.17 If u is the potential of a measure μ on the Greenian set Ω and Λ ⊂ Ω, then Λ Λ ˆ uΛ = GΩ δΩ R (μ, ·) = GΛ Ω μ = δΩ (·, u).

Proof: The ﬁrst and fourth terms are equal by Theorem 4.7.4. By Tonelli’s theorem, Λ Λ Λ GΩ μ(x) = GΩ (x, y) μ(dy) = GΩ (y, z) δΩ (x, dz) μ(dy) Λ ˆ uΛ (x) = u(z) δΩ (x, dz) = R so that the ﬁrst and third terms are equal. By the symmetry of GΛ Ω (x, y) established in the preceding theorem, Λ Λ Λ GΩ δΩ (μ, ·)(x) = GΩ (x, y) δΩ (μ, dy) = GΩ (x, y) δΩ (z, dy) μ(dz) Λ = GΩ (x, y) δΩ (z, dy) μ(dz) = GΛ Ω (z, x) μ(dz) Λ = GΛ Ω (x, z) μ(dz) = GΩ μ(x). This shows that the second and third terms are equal. Lastly, Λ Λ GΛ μ(x) = G (x, y) μ(dy) = GΩ (y, z) δΩ (x, dz) μ(dy) Ω Ω Λ ˆ uΛ (x) = u(z) δΩ (x, dz) = R so that the ﬁrst and third terms are equal.

5.5 Intrinsic Topology In view of the central role of superharmonic functions in potential theory, a topology deﬁned by such functions would be better suited to potential theory problems than the metric topology. Deﬁnition 5.5.1 The ﬁne topology is the smallest topology on Rn for which all superharmonic functions are continuous in the extended sense. Since each fundamental harmonic function ux is superharmonic on Rn and {y ∈ Rn ; ux (y) > α} is a ball, balls with arbitrary centers and radii are

224

5 Dirichlet Problem for Unbounded Regions

ﬁnely open. It follows that the metric topology is coarser than the ﬁne topology. In particular, open sets are ﬁnely open, continuous functions are ﬁnely continuous, etc. Since it was shown in Corollary 4.3.12 that there are ﬁnitevalued discontinuous superharmonic functions, the ﬁne topology is strictly ﬁner than the metric topology. If u is any superharmonic function on Rn and α, β are any real numbers, sets of the type {x; u(x) > α} and {x; u(x) < β} must be ﬁnely open; sets of the ﬁrst type are open but sets of the second type must be adjoined to the metric topology. The collection of such sets can be taken as a subbase for the ﬁne topology. If u is superharmonic on the open set Ω ⊂ Rn , then u is ﬁnely continuous on Ω. To see this, let x ∈ Ω and let B be any ball containing x with B − ⊂ Ω. By the Riesz representation theorem, u diﬀers from a Green potential GB ν by a (continuous) harmonic function on B; since GB ν diﬀers from the superharmonic Newtonian potential Uν by a harmonic function on B, u diﬀers from a superharmonic function on Rn by a harmonic function on B. Thus, u is ﬁnely continuous at x. The ﬁne closure of a set A ⊂ Rn will be denoted by clf A, the ﬁne interior by intf A, the ﬁne limit by f-lim, the ﬁne limsup by f-limsup, etc. If O is ﬁnely open, the notation Of will be used as a reminder that is ﬁnely open; if Γ is ﬁnely closed, Γ f will be used as a reminder of this fact. A set Λ will be called a ﬁne neighborhood of a point x ∈ Λ if it is a superset of a ﬁnely open set containing x. Theorem 5.5.2 If Z is a polar subset of Rn , then Z has no ﬁne limit points; in particular, the points of Z are ﬁnely isolated points. Proof: Assume that x ∈ Rn is a ﬁne limit point of Z. Then x is a limit point of Z. Since this is true if and only if x is a limit point of Z ∼ {x} and the latter is polar, it can be assumed that x ∈ Z. By Theorem 4.3.11, there is a superharmonic function u such that u(x) < lim inf y→x,y∈Z u(y) = r < +∞ for any r > 0. Thus, there is a neighborhood O of x such that u(y) > 1 2 (r + u(x)) for y ∈ O ∩ Z; but since u is ﬁnely continuous at x, there is a ﬁne neighborhood Of of x such that u(y) < 12 (r + u(x)) for y ∈ Of . Therefore, (O ∩ Z) ∩ Of = (O ∩ Of ) ∩ Z = ∅. Since O ∩ Of is a ﬁne neighborhood of x that does not intersect Z, x is not a ﬁne limit point of Z, a contradiction. Example 5.5.3 Consider the sequence of points {xj } in R2 with xj = (1/j, 0), j ≥ 1. There is a ﬁne neighborhood of (0, 0) that contains no points of the sequence.

5.6 Thin Sets Consider an open set Ω and a point x ∈ Ω. Then x is a regular boundary point for the Dirichlet problem if there is a cone in ∼ Ω having x as its vertex. For a boundary point to be irregular, ∼ Ω must somehow be thinner than a cone at x. The concept of “thinness” of a set at a point expresses this more precisely.

5.6 Thin Sets

225

Deﬁnition 5.6.1 A set Λ is thin at x if x is not a ﬁne limit point of Λ. Theorem 5.6.2 A polar set Z is thin at every point of Rn . Proof: By Theorem 5.5.2, no point of Rn is a ﬁne limit point of Z. In particular, a line segment in R3 is thin at each of its points since it is polar by Example 4.2.8. Also note that Λ is thin at x whenever x is not a limit point of Λ. Theorem 5.6.3 A set Λ is thin at a limit point x of Λ if and only if there is a superharmonic function u on a neighborhood U of x such that u(x) <

lim inf

y→x,y∈Λ∩U∼{x}

u(y).

Proof: It can be assumed that x ∈ Λ. Suppose that Λ is thin at x. There is then a ﬁne neighborhood Of of x such that Of ∩ Λ = ∅. Since x is a limit point of Λ, Of is not a metric neighborhood of x. Since the class of sets of the form {y; v(y) < α} and {y; v(y) > β} for arbitrary α, β and arbitrary superharmonic functions v constitute a subbase for the ﬁne topology, it can be assumed that Of has the form Of = ∩ji=1 {y; ui (y) < αi } ∩ U where U is a neighborhood of x and the ui are superharmonic. Since x ∈ Of , ui (x) < αi for i = 1, . . . , j and > 0 can be chosen so that 0 < < min1≤i≤j (αi − ui (x)). With this choice of , it can be assumed that Of = ∩ji=1 {y; ui (y) < ui (x) + j } ∩ U . Now let u = i=1 ui which is superharmonic. It remains only to show that u(x) < lim inf y→x,y∈Λ∼{x} u(y). Since the ui are l.s.c. at x, there is a neighborhood V ⊂ U of x such that ui (y) > ui (x) − ( /j) for 1 ≤ i ≤ j and y ∈ V . Consider any y ∈ Λ ∩ V . Since Of ∩ Λ = ∅, y ∈ Of and so uk (y) ≥ uk (x) + for some k. Thus, u(y) =

j

ui (y) >

i=1

j

ui (x) −

i=1

j−1 + = u(x) + ; j j

that is, u(y) > u(x)+ /j for all y ∈ Λ∩V . Therefore, lim inf y→x,y∈Λ∼{x} u(y) > u(x), as was to be proved. Conversely, suppose there is a superharmonic function u on a neighborhood U of x such that u(x) <

lim inf

y→x,y∈Λ∩U∼{x}

u(y) = r.

It can be assumed that r < +∞ since u can be replaced by min (u, u(x) + 1). Consider Of = {y; u(y) < 12 (u(x)+r). By choice of r, there is a neighborhood V ⊂ U of x such that u(y) > 12 (u(x) + r) for y ∈ Λ ∩ V . Then Of ∩ (Λ ∩ V ) = (Of ∩ V ) ∩ Λ = ∅. Since Of ∩ V is a ﬁne neighborhood of x, x is not a ﬁne limit point of Λ and Λ is thin at x. Note that the function u constructed in the proof of the necessity is superharmonic on Rn .

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5 Dirichlet Problem for Unbounded Regions

Theorem 5.6.4 A set Λ is thin at a limit point x of Λ if and only if there is a superharmonic function u such that u(x) <

u(y) = +∞.

lim

(5.9)

y→x,y∈Λ∼{x}

If x belongs to the Greenian set Ω, then there is a potential u = GΩ μ satisfying (5.9) for which μ has compact support in Ω. Proof: The suﬃciency was proved in the preceding theorem. To prove the necessity, suppose Λ is thin at x. As usual, it can be assumed that x ∈ Λ. By the preceding comment, there is a superharmonic function v such that v(x) < limy→x,y∈Λ∼{x} v(y). If the right side is +∞, there is nothing to prove. Assume that the right side is ﬁnite. Let B = Bx,r be any ball. By the Riesz decomposition theorem, Theorem 3.5.11, there is a measure ν on B and a harmonic function h on B such that v = GB ν + h on B. Since h is continuous at x, it can be assumed that v = GB ν on B. Let { j } be a decreasing sequence of positive numbers. Since

GB (x, y) dν(y) +

v(x) = Bx,j

GB (x, y) dν(y) < +∞ B∼Bx,j

and the second integral on the right increases to v(x),

lim

j→∞

GB (x, y) dν(y) = lim

j→∞

Bx,j

GB (x, y) dνj (y) = 0

⊂ B. By passing to where νj (M ) = ν(M ∩ Bx, j ) for each Borel set M a subsequence, if necessary, it can be assumed that j GB νj (x) < +∞. The function u = j GB νj is then superharmonic on B and ﬁnite at x. Let δ = lim inf y→x,y∈Λ (v(y) − v(x)) > 0 and let vj = GB νj . Then v = vj + hj where hj is harmonic on Bx, j , and it follows that lim inf (vj (y) − vj (x))

y→x,y∈Λ

≥ lim inf (vj (y) + hj (y)) + lim inf (−hj (y) − vj (x)) y→x,y∈Λ

y→x,y∈Λ

= lim inf (v(y) − v(x)) ≥ δ y→x,y∈Λ

for all j ≥ 1 and that ⎛ lim inf ⎝

y→x,y∈Λ

p j=1

vj (y) −

p j=1

⎞ vj (x)⎠ ≥ pδ

5.6 Thin Sets

227

for all p ≥ 1. Therefore, lim inf

p

y→x,y∈Λ

vj (y) ≥

j=1

p

vj (x) + pδ ≥ pδ

j=1

p for all p ≥ 1. Since u = j vj ≥ j=1 vj , lim inf y→x,y∈Λ u(y) ≥ pδ for all p ≥ 1. Therefore, lim inf y→x,y∈Λ u(y) = +∞. Now extend u to Rn using Lemma 4.3.10. Suppose now that x belongs to the Greenian set Ω, and let B = Bx,r be a ball with B − ⊂ Ω. Choose a decreasing sequence { j } of r. By Tonelli’s theorem, u = positive numbers as above for which 1 < G ν = G ( ν ) on B. Letting μ = B j B j j j j νj on B, GΩ μ has the same properties as u = GB μ since the two diﬀer by a harmonic function on B. Corollary 5.6.5 If Λ is thin at a limit point x of Λ, then lim

r→0

σ(∂Bx,r ∩ Λ) = 0. σ(Bx,r )

Proof: By the preceding theorem, there is a superharmonic function u such that u(x) < limy→x,y∈Λ∼{x} u(y) = +∞. It can be assumed that u is nonnegative on a neighborhood of x. For suﬃciently small r > 0, 1 u(x) ≥ L(u : x, r) = u(y) dσ(y) σn rn−1 ∂Bx,r 1 ≥ u(y) dσ(y) σn rn−1 ∂Bx,r ∩Λ

σ(∂Bx,r ∩ Λ) ≥ inf u(y) . y∈Bx,r ∩Λ σ(∂Bx,r ) Since u(x) < +∞ and the last factor tends to +∞ as r → 0, lim

r→0

σ(∂Bx,r ∩ Λ) = 0. σ(∂Bx,r )

This section will be concluded by showing that thinness is invariant under certain transformations. A map τ : Rn → Rn is measurable if τ −1 (M ) is a Borel subset of Rn for each Borel set M ⊂ Rn ; if, in addition, |τ x − τ y| ≤ |x − y| for all x, y ∈ Rn , then τ is called a contraction map. Theorem 5.6.6 If Λ is thin at x and τ : Rn → Rn is a contraction mapping such that |τ x − τ y| = |x − y| for all y ∈ Rn , then τ (Λ) is thin at τ x. Proof: If x is not a limit point of Λ, it is easy to see that τ x is not a limit point of τ (Λ) and thinness is preserved. It therefore can be assumed that x is a limit point of Λ and that x ∈ Λ. By Theorem 5.6.4, there is a superharmonic

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5 Dirichlet Problem for Unbounded Regions

function u such that u(x) < lim inf y→x,y∈Λ u(y) = +∞. It can be assumed that u is nonnegative on a ball B containing x and, in fact, that u = GB μ where μ is a measure on B. Since u = GB μ diﬀers from the Newtonian (logarithmic) potential Uμ by a harmonic function that is continuous at x, it can be assumed that μ u(y) = U (y) = u(y, z) dμ(z) where u(y, z) = uy (z). Extend μ to Rn by putting μ(Rn ∼ B) = 0. The map τ induces a measure μτ −1 on the Borel subsets M ⊂ Rn by the equation (μτ −1 )(M ) = μ(τ −1 (M )). Consider −1 Uμτ (y) = u(y, z) d(μτ −1 )(z). Applying the transformation τ , μτ −1

U

(y) =

Therefore, Uμτ

−1

u(y, τ z) dμ(z).

(τ y) =

u(τ y, τ z) dμ(z). −1

Since |τ y − τ z| ≤ |y − z|, u(τ y, τ z) ≥ u(y, z) and Uμτ (τ y) ≥ Uμ (y). Note −1 that Uμτ is superharmonic with μτ −1 U (τ x) = u(τ x, τ z) dμ(z) = u(x, z) dμ(z) = Uμ (x) < +∞. Since |τ x − τ y| = |x − y| for all y, lim

z→τ x,z∈τ (Λ)

Uμτ

−1

(z) =

lim

τ y→τ x,y∈Λ

Uμτ

−1

Thus, Uμτ (τ x) < limz→x,z∈τ (Λ) Uμτ Theorem 5.6.4.

−1

−1

(τ y) ≥

lim

y→x,y∈Λ

Uμ (y) = +∞.

(z) and τ (Λ) is thin at τ x by

5.7 Thinness and Regularity Regularity of boundary points relative to the Dirichlet problem can be characterized in terms of thinness. The latter will be characterized in terms of regularized reduced functions ﬁrst.

5.7 Thinness and Regularity

229

Deﬁnition 5.7.1 An extended real-valued function u deﬁned on the open set Ω peaks at x ∈ Ω if u(x) > supy∈Ω∼O u(y) for every neighborhood O of x. If Ω is Greenian, there is no problem in ﬁnding a positive superharmonic function u that peaks at x ∈ Ω, namely, u = GΩ (x, ·). Theorem 5.7.2 Let Ω be a Greenian set and let u be a positive superharmonic function that peaks at x ∈ Ω. A set Λ ⊂ Ω is thin at x if and only if ˆ u (x) < u(x). R Λ Proof: Note that Λ is thin at x if and only if Λ ∼ {x} is thin at x. By ˆu ˆu Corollary 4.6.4, R Λ∼{x} = RΛ . It therefore can be assumed that x ∈ Λ, in u u ˆ (x) = R (x). It will be shown now that Ru (x) < u(x) implies which case R Λ Λ Λ that Λ is thin at x. Since Λ is thin at x if x is not a limit point of Λ, it can be assumed that x is a limit point of Λ. Since RuΛ (x) < u(x), there is a positive superharmonic function v on Ω that majorizes u on Λ such that v(x) < u(x). Therefore, lim inf v(y) ≥ lim inf u(y) ≥ u(x) > v(x) y→x,y∈Λ

y→x,y∈Λ

and Λ is thin at x by Theorem 5.6.4. Suppose now that Λ is thin at x. If x is not a limit point of Λ, it is easily seen that RuΛ (x) < u(x) since u peaks at x. Assuming that x is limit point of Λ, by Theorem 5.6.4 there is a potential v on Ω such that v(x) < lim inf y→x,y∈Λ v(y). It can be assumed that v is bounded above on Ω by replacing v by min (v, v(x) + 1) if necessary. Let α be such that v(x) < α < lim inf y→x,y∈Λ v(y). For λ > 0, let wλ = u(x)+λ(v−α). There is a neighborhood O of x such that v(y) > α for y ∈ O∩Λ. Then wλ ≥ u(x) ≥ u(y) for y ∈ O∩Λ since u peaks at x. Moreover, u(x)−u(y) ≥ γ > 0 for y ∈ Ω ∼ O since u peaks at x. Since v is bounded, a positive number λ0 can be chosen so that λ0 |v(y)−α| ≤ γ ≤ u(x)−u(y) for y ∈ Ω ∼ O. Then λ0 (v(y)−α) ≥ u(y)− u(x) for y ∈ Ω ∼ O and wλ0 (y) = u(x) + λ0 (v(y) − α) ≥ u(y) for y ∈ Ω ∼ O. Since wλo (y) ≥ u(y) for y ∈ O ∩ Λ, wλ0 ≥ u on Λ, and therefore wλ0 ≥ RuΛ . ˆ u (x). By choice of α, u(x) > u(x) + λ0 (v(x) − α) = wλ0 (x) ≥ RuΛ (x) ≥ R Λ Lemma 5.7.3 If B is a ball of radius 12 , then there is a positive, continuous, superharmonic function u on B such that for each x ∈ B the function u has a decomposition into a sum of two positive, continuous, superharmonic functions of which one peaks at x. Proof: Only the n ≥ 3 case will be discussed, the n = 2 case being essentially the same. Deﬁne 1 dz, y ∈ B. u(y) = |y − z|n−2 B It follows from Theorem 3.4.12 that u is a positive, continuous, superharmonic − function on B. Consider any x ∈ B and choose r > 0 such that Bx,r ⊂ B. Letting

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5 Dirichlet Problem for Unbounded Regions

u1 (y) = Bx,r

1 dz, |y − z|n−2

u2 (y) = B∼Bx,r

1 dz, |y − z|n−2

y∈B y ∈ B,

u = u1 + u2 . Using Gauss’ Averaging Principle, Theorem 1.6.2, to evaluate the integrals, it can be shown that u1 peaks at x. Theorem 5.7.4 The set of points of Λ ⊂ Rn where Λ is thin is a polar set. Proof: Since thinness is a local property and a countable union of polar sets is again polar, it suﬃces to consider the case for which Λ is a subset of a ball of radius 12 . Suppose Λ is thin at x ∈ Λ. Let w = w1 + w2 be the superharmonic function described in the preceding lemma. Taking reduced functions relative ˆ w1 (x) < w1 (x) by Theorem 5.7.2. By Theorem 4.3.5, to B, R Λ w ˆw ˆ w1 ˆ w2 R Λ (x) ≤ RΛ + RΛ (x) < w1 (x) + w2 (x) = w(x) = RΛ (x)

ˆ w (y) < Rw (y)}. By Corollary 4.6.2, the latter set is polar and and x ∈ {y; R Λ Λ it follows that the set of points in Λ where Λ is thin is a polar set. Corollary 5.7.5 A set Z ⊂ Rn is polar if and only if it is thin at each of its points. Proof: If Z is thin at each of its points, then Z is polar by the preceding theorem. By Theorem 5.6.2, a polar set is thin at each of its points. Regularity of a boundary point of a Greenian region Ω can be related to thinness of ∼ Ω at the point. Lemma 5.7.6 Let F be a closed subset of Rn . Then F is not thin at the ﬁnite point x ∈ ∂F if and only if there is a positive superharmonic function w deﬁned on some neighborhood O of x such that limy→x,y∈O∼F w(y) = 0. Proof: Suppose F is not thin at the ﬁnite point x ∈ ∂F . It can be assumed that F ⊂ B = Bx,δ . Also let u(y) = δ − |y − x| for y ∈ B. Then u is a positive ˆ u relative to B. superharmonic function on B that peaks at x. Consider R F u ˆ (x) = u(x) by Theorem 5.7.2 and Then R F lim inf

y→x,y∈B∼F

ˆ uF (y) ≥ lim inf R ˆ uF (y) ≥ R ˆ uF (x) = u(x). R y→x,y∈B

ˆ u on B ∼ F . If w vanConsider the superharmonic function w = u − R F ishes at some point of B ∼ F , it must be identically zero on B ∼ F by the minimum principle; this would imply that u is harmonic on B ∼ F since ˆ u is harmonic there, a contradiction since the Laplacian of u is strictly R F negative. Thus, w > 0 on B ∼ F . Moreover, lim supy→x,y∈B∼F w(y) ≤ ˆ u (y) ≤ 0. Therefore, limy→x,y∈B∼F w(y) = 0. u(x) − lim inf y→x,B∼F R F

5.7 Thinness and Regularity

231

This proves the necessity with O = B. To prove the converse, let w be a positive, superharmonic function on O ∼ F for some neighborhood O of x such that limy→x,y∈O∼F w(y) = 0, and let B = Bx,δ ⊂ O. It can be assumed that O = B. Assume that F is thin at x; that is, there is a subharmonic function v on B such that v(x) = 1 and v(z) ≤ −1 on B ∩ (F ∼ {x}) according to Theorem 5.6.4. Let B1 be a second ball with center x with B1− ⊂ B. Since v ≤ −1 on F ∩ ∂B1 , v < 0 on a neighborhood W of F ∩ ∂B1 , and for any positive number λ, v − λw ≤ 0 on W ∩ (∂B1 ∼ F ). Since w > 0 on ∂B1 ∼ W, inf z∈∂B1 ∼W w(z) > 0. Along with the fact that v is bounded above on ∂B1 ∼ W , this implies that λ0 > 0 can be chosen so that v − λ0 w ≤ 0 on ∂B1 ∼ W . Since v − λ0 w ≤ 0 on W ∩ (∂B1 ∼ F ) also, lim sup (v(z) − λ0 w(z) − ux (z)) ≤ 0

z→y,z∈B1 ∼F

for any > 0 whenever y ∈ ∂B1 ∼ F . This inequality also holds at points y ∈ B1− ∩ (∂F ∼ {x}) for in a neighborhood of such a point v is strictly negative; it also holds at x, since in this case the left side is −∞. By Corollary 2.3.6, v − λ0 w − ux ≤ 0 on B1 ∼ F for every > 0. Therefore, v ≤ λ0 w on B1 ∼ F and lim supy→x,y∈B1 ∼F v(y) ≤ 0. Since lim supy→x,y∈F ∼{x} v(y) ≤ −1, v(x) ≤ lim supy→x,y =x v(y) ≤ 0. But by Theorem 2.4.7, 1 = v(x) = lim supy→x,y∈B v(y), a contradiction. Theorem 5.7.7 A ﬁnite point x ∈ ∂Ω is a regular boundary point for the Greenian set Ω if and only if ∼ Ω is not thin at x. Proof: By the preceding lemma, ∼ Ω is not thin at the ﬁnite point x ∈ ∂Ω if and only if there is a barrier at x. The existence of a barrier is necessary and suﬃcient for regularity by Remark 5.4.11 and Theorem 5.4.2. Corollary 5.7.8 A line segment I ⊂ R2 is not thin at its end points. Proof: It will be shown ﬁrst that I cannot be thin at an interior point x. Since thinness is a local property, any closed subinterval of I containing x as an interior point will be thin at x if I is thin at x. It suﬃces to show that a closed line segment I cannot be thin at an interior point. Let B be a ball containing I in its interior. Regarding x as a boundary point B ∼ I, x is a regular boundary point for B ∼ I by Theorem 2.6.27 and I is not thin at x by the preceding theorem. Suppose now that I is thin at an end point x. Then it follows from Theorem 5.6.4 that there is a superharmonic function w such that w(x) < limy→x,y∈I∼{x} w(y) = +∞. Since superharmonicity is invariant under a rotation about x, a line segment I can be constructed having x as an interior point and a superharmonic function w (as a sum of w and two rotations of w about x) can be constructed such that w (x) < limy→x,y∈I ∼{x} w (y); that is, I is thin at an interior point, but this is not possible by the ﬁrst part of the proof.

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5 Dirichlet Problem for Unbounded Regions

Theorem 5.7.9 If Ω ⊂ R2 and ∼ Ω is thin at x, then there are arbitrarily small ρ > 0 such that (∼ Ω) ∩ ∂Bx,ρ = ∅. Proof: If x is not a limit point of ∼ Ω, the assertion is trivially true. Let x be a limit point of ∼ Ω, and let (r, θ) be the polar coordinates of a point in the plane relative to the pole x and a ray having x as an end point. Consider the map τ : R2 → taking the point (r, θ) into the point (r, 0). It is easily seen that |τ x − τ y| = |x − y| for all y ∈ R2 . Suppose there is a ρ0 > 0 such that (∼ Ω) ∩ Bx,ρ = ∅ for all ρ ≤ ρ0 . Then τ (∼ Ω) contains the line segment ˜ = {(ρ, 0) : 0 < ρ ≤ ρ0 }. Since τ (∼ Ω) is thin at x by Theorem 5.6.6, ˜ is thin at x, but this contradicts the preceding corollary and there is no such ρ.

Remark 5.7.10 (Proof of Theorem 2.6.28) Let Ω be an open subset of R2 , let x ∈ ∂Ω, and let be a continuum in ∼ Ω containing the point x. By Theorem 5.7.7, x is a regular boundary point for Ω if and only if ∼ Ω is not thin at x. Assume that ∼ Ω is thin at x. By the above theorem, there is a ρ > 0 − ∩ ), such that (∼ Ω) ∩ ∂Bx,ρ = ∅. Since ∩ ∂Bx,ρ = ∅, = (Bx,ρ ∩ ) ∪ (∼ Bx,ρ contradicting the connectedness of . Thus, Ω is not thin at x and x is a regular boundary point for Ω. Theorem 5.7.11 (Brelot [9]) Let Ω be a Greenian subset of Rn , x ∈ Ω, Λ ⊂ Ω, and u any nonnegative superharmonic function on Ω. Then Λ ˆu (x, {x}) = 1, GΛ (i) if x ∈ clf Λ, then δΩ Ω (x, ·) = GΩ (x, ·), and RΛ∩U (x) = u(x) for all neighborhoods U of x, and Λ ˆu (x, {x}) = 0, GΛ (ii) if x ∈ clf Λ, then δΩ Ω (x, ·) = GΩ (x, ·), and limU↓{x} RΛ∩U = 0 on Ω whenever u(x) < +∞ ,

Proof: (i) If x ∈ clf Λ and U is any neighborhood of x, then x is a ﬁne limit point of Λ and therefore a ﬁne limit point of Λ ∩ U . By Corollary 4.6.2, ˆu R Λ∩U = u quasi everywhere on Λ ∩ U . Thus, there is a polar set Z ⊂ Λ ∩ U ˆu such that R Λ∩U = u on (Λ ∩ U ) ∼ Z. Since Z does not have any ﬁne limit points by Theorem 5.5.2, there is a ﬁne neighborhood Of of x such f ˆu that Z ∩ Of = ∅. Thus, R Λ∩U = u on (Λ ∩ U ) ∩ O . Since both funcu ˆ tions are ﬁnely continuous, RΛ∩U (x) = u(x). Taking u = GΩ (x, ·) and ˆ U = Ω, GΛ Ω (x, x) = RΛ (GΩ (x, ·), x) = GΩ (x, x) by the preceding step. From Λ Λ the deﬁnition of δΩ , δΩ (x, {x}) = δΩ (x, {x}) = 1. (ii) Suppose x ∈ clf Λ. By adjoining a countable set to Λ, it can be assumed that x is a limit point of Λ. This modiﬁcation will have no eﬀect on reductions since the set of augmented points is a polar set. Since x is not a ﬁne limit point of Λ, the latter is thin at x. By Theorem 5.6.4, there is a superharmonic function v such that v(x) < +∞ and limy→x,y∈Λ∼{x} u(y) = +∞. If > 0, choose a ball B with center at x such that u > 1/ on Λ ∩ (B ∼ {x}). By Theorem 4.7.4 and Lemma 4.3.5, Λ∩B Λ∩B ˆ 1 (x) ≤ R ˆ u (x) ≤ u(x); that is, δΩ (x, {x}) ≤ δΩ (x, Ω) = R Λ∩B Λ∩B Λ∩B Λ (x, {x}) = 0 limB↓{x} δΩ (x, {x}) = 0. It follows from Lemma 4.7.9 that δΩ

5.7 Thinness and Regularity

233

Λ Λ∩U for if δΩ (x, {x}) = 1, then δΩ (x, {x}) = 1 for all neighborhoods U of x. ˆ The assertion that limU↓{x} RΛ∩U = 0 on Ω follows from the same lemma.

Lemma 5.7.12 If Ω is a Greenian subset of Rn , then there is a bounded, continuous, nonnegative superharmonic function w satisfying ˆw clf Λ ∩ Ω = {x; R Λ (x) = w(x)} for every Λ ⊂ Ω. Proof: Let {Bj } be a sequence of balls with closures in Ω that forms a basis for the topology of Ω, and let w=

∞

ˆ 1B 2−j R j

j=1

ˆ 1 ≤ 2−j and R ˆ 1 is continuous on Ω, w is a bounded, on Ω. Since 2−j R Bj Bj continuous, superharmonic function on Ω by the Weierstrass M-test and Theorem 2.4.8. By Theorem 4.6.13, for any Λ ⊂ Ω ˆw R Λ =

∞

ˆ Λ (R ˆ 1B , ·). 2−j R j

j=1

ˆ w (x) = w(x) by Theorem 5.7.11. Consider any If x ∈ clf Λ ∩ Ω, then R Λ ˆ Λ (R ˆ 1 , x) can be x ∈ Ω ∼ clf Λ and those j for which x ∈ Bj . Since R Bj made arbitrarily small by making j suﬃciently large according to (iii) of ˆ w (x). ˆ 1 (x) = 1 for such j, w(x) > R Lemma 4.7.9 and R Bj Λ Theorem 5.7.13 If u is the potential of a measure μ on the Greenian set ˆ u = GΩ μΛ , then μΛ is supported by clf Λ; in particular, if u Ω, Λ ⊂ Ω, and R Λ is a nonnegative superharmonic function on Ω, the Riesz measure associated ˆ u is supported by clf Λ. with R Λ Λ (x, ·) is supported by clf Λ; that is, Proof: It will be shown ﬁrst that δΩ Λ δΩ (x, Ω ∼ clf Λ) = 0. Let w be a superharmonic function as described ˆw = R ˆ (R ˆ w , ·) by Lemma 4.7.5, R ˆw = in the preceding lemma. Since R Λ Λ w Λ ΛΛ Λ ˆ ˆ ˆ w(y) δΩ (·, dy), and RΛ (RΛ , ·) = RΛ (y) δΩ (·, dy), Λ ˆw (w(y) − R x ∈ Ω. Λ (y)) δΩ (x, dy) = 0,

ˆ w , {y; w(y) > R ˆ w (y)} = Ω ∼ clf Λ has δ Λ (x, ·) measure zero; Since w ≥ R Λ Λ Ω Λ that is, δΩ (x, ·) is supported by clf Λ. Now let u = GΩ μ be the potential of a measure μ on Ω, and let Λ Λ μ = δΩ (y, ·) μ(dy).

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5 Dirichlet Problem for Unbounded Regions

Clearly, μΛ is also supported by clf Λ. Since Λ GΩ μΛ (x) = GΩ (x, z) δΩ (y, dz) μ(dy) Λ = GΩ (x, z)δΩ (y, dz) μ(dy) = GΛ Ω (y, x) μ(dy) = GΛ Ω (x, y) μ(dy) = GΛ Ω μ(x) ˆ u = GΛ μ by Theorem 5.4.17, R ˆ u = GΩ μΛ . Thus, by Theorem 5.4.16 and R Λ Ω Λ Λ μΛ = μ and μΛ is supported by clf Λ. Lastly, suppose u is a nonnegative superharmonic function on Ω, let νΛ be the Riesz measure associated with ˆ u and let vΛ = GΩ νΛ . Using the notation just introduced, νΛ = ν Λ and νΛ R Λ Λ is therefore supported by clf Λ. Theorem 5.7.14 (Brelot [9]) Let u be a nonnegative superharmonic function on the Greenian set Ω and let μ be the Riesz measure associated with u. (i)

If x ∈ Ω, then f-limy→x,y∈Ω

u(y) u(y) = μ({x}) = inf . GΩ (x, y) y∈Ω∼{x} GΩ (x, y)

(5.10)

(ii) If x ∈ Ω is a limit point of Λ ⊂ Ω and Λ is thin at x, then there is a potential u on Ω for which lim

y→x,y∈Λ

u(y) = +∞. GΩ (x, y)

(5.11)

Conversely, if there is a nonnegative superharmonic function u on Ω ∼ {x} satisfying Equation (5.11), then Λ is thin at x. Proof: (i) Let c denote the third expression in Equation (5.10). Since u(y) = GΩ (z, y) dμ(z) ≥ μ({x})GΩ (x, y) for y ∈ Ω ∼ {x}, c ≥ μ({x}). On the other hand, u − cGΩ (x, ·) ≥ 0 on Ω ∼ {x} and is superharmonic thereon. By Theorem 4.2.15, the diﬀerence has a superharmonic extension v on Ω. Thus, u = cGΩ (x, ·) + v so that u = GΩ (z, ·) d(cδ{x} + ν)(z) + h

5.7 Thinness and Regularity

235

where ν is the Riesz measure and h is the harmonic function associated with v. Since the Riesz decomposition is unique according to Theorem 3.5.11, μ({x}) ≥ c. Thus, μ({x}) = c and the right equation in (i) is proved. It can be assumed that both sides of the second equation in (i) vanish, for if not, replace u by the nonnegative superharmonic function u − μ({x})GΩ (x, ·), extended across {x} as above. Assume that f-limsupy→x,y∈Ω

u(y) > 0. GΩ (x, y)

Then there is a b > 0 and a ﬁne neighborhood U f of x such that u(y) ≥b GΩ (x, y)

for y ∈ U f .

Since x ∈ U f ⊂ clf U f , by Theorem 5.7.11 f ˆ u f ≥ bGU u≥R Ω (x, ·) = bGΩ (x, ·) U

so that

u ≥ b > 0, GΩ (x, ·)

a contradiction. (ii) The second assertion of (ii) will be proved ﬁrst. If u is a nonnegative superharmonic function on Ω ∼ {x} for which Equation (5.11) holds, then u has a nonnegative superharmonic extension to Ω by Theorem 4.2.15. According to Equation (5.10), there is a ﬁne neighborhood Of of x on which u/GΩ (x, ·) is bounded above on Of ∼ {x} by some constant M < +∞. On the other hand, by Equation (5.11) there is an ordinary neighborhood U of x such that u/GΩ (x, ·) > M on (U ∩ Λ) ∼ {x}. Therefore, (Of ∩ U ) ∼ {x}) ∩ Λ = ∅. Thus, x is not a ﬁne limit point of Λ and Λ is therefore thin at x. To prove the ﬁrst assertion of (ii), let x be a limit point of Λ ⊂ Ω which is thin at x. It suﬃces to show that there is a potential u on the component of Ω containing x satisfying Equation (5.11). It therefore can be assumed that Ω is connected. Let y be a ﬁxed point of ∈ Ω ∼ {x}. Since GΩ (x, y) < +∞, by (ii) of Theorem 5.7.11 there is a decreasing sequence {Un } of neighborhoods of x with y ∈∼ U1− such that ˆ Λ∩Un (GΩ (x, y))(y) < 2−n , R

n ≥ 1.

ˆ Λ∩Un (GΩ (x, ·))(y) = RΛ∩Un (GΩ (x, ·))(y), there is a positive, superSince R harmonic function un majorizing GΩ (x, ·) on Ω ∩ Un such that un (y) < 2−n . By replacing un by min n , GΩ (x, ·)), it can be assumed that each un is a (u ∞ potential. Letting u = n=1 un , u is superharmonic on Ω since u(y) < +∞.

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5 Dirichlet Problem for Unbounded Regions

Since u ≥ nGΩ (x, ·) on Λ ∩ Un for each n ≥ 1, Equation (5.11) holds. By considering the Riesz decomposition of u, it can be assumed that u is a potential on Ω. Theorem 5.7.15 Let Λ ⊂ Rn and let φ be any inversion with respect to a sphere. (i)

Either φ(Λ) is thin at φ(∞) for every inversion φ, or φ(Λ) is not thin at φ(∞) for any inversion φ. (ii) If Λ is unbounded, then φ(Λ) is thin at φ(∞) for every inversion φ if and only if there is a nonnegative superharmonic function u on a deleted neighborhood of ∞ such that u(y) = +∞ log (y) lim u(y) = +∞

lim

if n = 2

y→∞,y∈Λ

if n ≥ 3;

y→∞,y∈Λ

in the n ≥ 3 case, there is a nonnegative superharmonic function u on Rn satisfying the last equation. (iii) If x and φ(x) are ﬁnite, φ(Λ) is thin at φ(x) if and only if Λ is thin at x. Proof: Since (i) and (ii) apply only to unbounded sets, Λ will be assumed to be unbounded. (i)(n = 2) Consider an arbitrary inversion y → y ∗ relative to a sphere ∂B where B = Bx,δ . Suppose Λ∗ is thin at x. By replacing Λ by Λ ∩ (∼ B − ), if necessary, it can be assumed that Λ∗ ⊂ B. Since x is a limit point of Λ∗ , according to Theorem 5.7.14 there is a potential v on B such that v(y) = +∞. lim y→x,y∈Λ∗ GB (x, y) Since GB (x, y) = − log |x − y| + hx where hx is harmonic on B, lim

y→x,y∈Λ∗

−

v(y) = +∞. log |x − y|

Letting v ∗ denote the Kelvin transform of v, lim ∗

y ∗ →∞,y

v ∗ (y ∗ ) = ∈Λ log |y ∗ − x| =

lim

v(y) log δ 2 − log |y − x|

lim

−

y→x,y∈Λ∗

y→x,y∈Λ∗

= +∞.

v(y) log |y − x|

5.7 Thinness and Regularity

237

Since limy∗ →∞,y∗ ∈Λ (log |y ∗ − x|/ log |y ∗ |) = 1, lim ∗

y ∗ →∞,y

v ∗ (y ∗ ) = +∞. ∈Λ log |y ∗ |

Letting u = v ∗ , limy→∞,y∈Λ u(y)/ log |y| = +∞. This proves (ii) in the n = 2 case. Since the preceding steps are reversible, the existence of such a function u on a neighborhood of x suﬃces to prove that Λ∗ is thin at x for every inversion with respect to a sphere. (ii)(n ≥ 3) Again consider an arbitrary inversion with respect to a sphere ∂B where B = Bx,δ , and suppose Λ∗ is thin at x. By Theorem 5.7.14, there is a potential v on Rn such that lim

y→x,y∈Λ∗

v(y) = +∞. GΩ (x, y)

Since it is easily seen that lim

y→x,y∈Λ∗

and G(x, y) =

GΩ (x, y)/G(x, y) = 1

1 |x − y ∗ |n−2 = , n−2 |x − y| δ 2n−4

δ 2n−4 v(y) v(y) = . G(x, y) |x − y ∗ |n−2 Since the expression on the right is just v ∗ (y ∗ ), the Kelvin transform of v, lim

y ∗ →∞,y ∗ ∈Λ

v ∗ (y ∗ ) = +∞.

Since v ∗ is a nonnegative superharmonic function on Rn ∼ {x}, it has a superharmonic extension to Rn by Theorem 4.2.15 which will be denoted by the same symbol. Letting u(y) = v ∗ (y ∗ ), limy→∞,y∈Λ u(y) = +∞. As before, the preceding steps are reversible. Assertion (iii) follows from Remark 5.3.11. n Deﬁnition 5.7.16 The ﬁne topology on R∞ is the smallest topology conn for which taining the set {∞}, ﬁnely open subsets of Rn , and sets Λ ⊂ R∞ φ(Λ ∼ {∞}) is a deleted ﬁne neighborhood of φ(∞) for every inversion φ with respect to a sphere.

Since a set Λ is thin at x ∈ Rn if and only if ∼ Λ is a ﬁne neighborhood of x, the requirement that φ(Λ ∼ {∞}) is a deleted ﬁne neighborhood of φ(∞) holds for all inversions if it holds for one. The fact that {∞} is ﬁnely open in n means that ∞ is ﬁnely isolated and cannot be a ﬁne limit point of any R∞ n . subset of R∞ n Deﬁnition 5.7.17 A set Λ ⊂ R∞ is thin at ∞ if φ(Λ ∼ {∞}) is thin at φ(∞) for every inversion φ with respect to a sphere.

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5 Dirichlet Problem for Unbounded Regions

Let Ω be a Greenian subset of Rn , possibly unbounded. Regularity of ﬁnite boundary points has been characterized in terms of thinness of ∼ Ω at x, and regularity of ∞ in the n ≥ 3 case has been established. It remains only to characterize regularity of ∞ in the n = 2 case. Theorem 5.7.18 Let Ω be an unbounded Greenian subset of R2 . Then ∞ is a regular boundary point for Ω if and only if ∼ Ω is not thin at ∞. Proof: Consider any inversion y → y ∗ with respect to a sphere ∂Bx,δ . By Remark 5.4.11 and Theorem 5.4.2, ∞ is a regular boundary point for Ω if and only if x is a regular boundary point for Ω ∗ . By the same results, the latter is true if and only if R2 ∼ Ω ∗ is not thin at x, and this is true if and only if R2 ∼ Ω is not thin at ∞. The ﬁne topology on Rn seems to be the natural topology for studying the boundary behavior of superharmonic functions. In some cases, the ﬁne limit of a superharmonic function can be identiﬁed. Example 5.7.19 Let x ∈ Rn , let O be a neighborhood of x, and let u be a nonnegative superharmonic function on O ∼ {x}. Then f-limy→x,y∈O u(y) = lim inf u(y). y→x,y∈O

This follows from the fact that u has a superharmonic extension to O, denoted by u, and that both are then equal to u(x). Example 5.7.20 If u is a nonnegative superharmonic function on Ω = Rn ∼ − , then Bx,δ f-limy→∞,y∈Ω u(y) = lim inf u(y) if n = 2 y→x,y∈Ω

and f-limy→∞,y∈Ω

u(y) u(y) = lim inf n−2 y→∞,y∈Ω |y|n−2 |y|

if n ≥ 3.

Consider an inversion with respect to ∂Bx,δ . Since u∗ (y ∗ ) = u(y) in the n = 2 case, f-limy→∞,y∈Ω u(y) = f-limy∗ →x,y∗ ∈Ω ∗ u∗ (y ∗ ) =

lim inf

y ∗ →x,y ∗ ∈Ω ∗

u∗ (y ∗ ) = lim inf u(y) y→∞,y∈Ω

by the preceding example. The same argument can be applied to the second equation. Theorem 5.7.21 (Cartan) If x is a ﬁne limit point of the set Λ and g is an extended real-valued function having a ﬁne limit λ at x, then there is a ﬁnely open set V f of x such that limy→x,y∈Λ∩V f g(y) = λ.

5.7 Thinness and Regularity

239

Proof: It can be assumed that λ is ﬁnite for otherwise g can be replaced by arctan g. It also can be assumed that Λ is a subset of a ball B with center x. If {Ij } is a decreasing sequence of open intervals such that Ij ↓ {λ}, then for each j ≥ 1 there is a ﬁne neighborhood Ujf of x such that g(y) ∈ Ij for all y ∈ (Λ ∼ {x}) ∩ Ujf . If x is an interior point of inﬁnitely many of the Ujf , then it suﬃces to take V f = B. It therefore can be assumed that x is an interior point of at most ﬁnitely many of the Ujf and, in fact, that x is a limit point of ∼ Ujf for every j ≥ 1. Since Ujf is a ﬁne neighborhood of x, ∼ Ujf is thin at x and there is a positive superharmonic function uj such that uj (x) <

lim

y→x,y∈B∼Ujf

uj (y) = +∞.

(5.12)

Let {αj } be a sequence of positive numbers such that j αj uj (x) < +∞. Then u = j αj uj is superharmonic and is ﬁnite at x. Now let Fj = B ∩ (∼ Ujf ) ∩ {y; αj uj (y) > j}, and let F = ∪j Fj . Then x ∈ F . It will now be shown that F is thin at x. This is trivially true if x ∈ F − and it therefore can be assumed that x is a limit pointof F . If p is a positive integer, αj uj (y) > p for y ∈ Fj , j ≥ p and u(y) = j αj uj (y) ≥ p. For each j < p, there is a neighborhood Vj of x p for y ∈ Fj ∩ Vj by Inequality (5.12). Then y ∈ ∩p−1 such that αj uj (y) ≥ j=1 Vj implies that u(y) = j αj uj (y) ≥ p. Since p is an arbitrary positive integer, limy→x,y∈F u(y) = +∞ and that F is thin at x. Letting V f be the ﬁne interior of ∼ F, V f is then a ﬁne neighborhood of x. If Wj = B ∩ {y : αj uj (y) > j}, then Wj ∩ (∼ Ujf ) = Fj ⊂ Wj ∩ F and Ujf ⊃ Wj ∩ Ujf ⊃ Wj ∩ (∼ F ) ⊃ Wj ∩ V f . If y ∈ Wj ∩ V f ∩ (Λ ∼ {x}) ⊂ Ujf ∩ (Λ ∼ {x}), then g(y) ∈ Ij . This proves that limy→x,y∈V f ∩Λ g(y) = λ. Theorem 5.7.22 (Brelot) If F is a closed subset of Rn that is thin at x and v is a positive superharmonic function on U ∼ F for some neighborhood U of x, then v has a ﬁne limit at x. Proof: If x is not a limit point of F , then there is nothing to prove since in this case v is ﬁnely continuous at x. It therefore can be assumed that x is a limit point of F . There is then a superharmonic function u such that u(x) < limy→x,y∈F u(y) = +∞. Since u is ﬁnely continuous at x, u(x) = f-limy→x,y∈∼F u(y) < +∞.

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5 Dirichlet Problem for Unbounded Regions

If lim inf y→x,y∈∼F (v+u)(y) = +∞, then limy→x,y∈∼F (v+u)(y) exists and, in particular, f-limy→x,y∈∼F (v + u)(y) = +∞. It follows that f-limy→x,y∈∼F v(y) exists and is equal to +∞. Suppose now that λ = lim inf y→x,y∈∼F (v+u)(y) < +∞ and let p be a positive integer with p > λ. Since limy→x,y∈F u(y) = +∞, there is a neighborhood O of x having compact closure in U such that u(y) > p for y ∈ O− ∩ (F ∼ {x}) ⊃ O ∩ (F ∼ {x}). Since F is closed, lim inf z→y,z∈∼F (v + u)(z) ≥ lim inf z→y,z∈∼F u(z) ≥ u(y) > p for y ∈ O ∩ (∂F ∼ {x}). Let min (u + v), p) on (O ∼ F ) ∼ {x} w= p on (O ∩ F ) ∼ {x}. At each point y ∈ O ∩ (∂F ∼ {x}), w is equal to p on a neighborhood of y. Thus, w is superharmonic on O ∼ {x} and since O has compact closure in U , it is bounded below on O. By Theorem 4.2.15, w has a superharmonic extension to O. Noting that lim inf y→x,y∈∼F (v + u)(y) = λ < p, w agrees with v + u on ∼ F in a neighborhood of x. Since w is ﬁnely continuous at x, f-limy→x,y∈∼F (v + u)(y) exists. Lastly, f-limy→x,y∈∼F v(y) exists since f-limy→x,y∈∼F u(y) exists.

Chapter 6

Energy

6.1 Introduction In 1828, Green used a physical argument to introduce a function, which he called a potential function, to calculate charge distributions on conducting bodies. The lack of mathematical rigor led Gauss in a 1840 paper to propose a procedure for ﬁnding equilibrium distributions based on the fact that such a distribution should have minimal potential energy. This led to the study of the functional (G − 2f )σ dS where σ is a density function and dS denotes integration with respect to surface area on the boundary of a conducting body. Gauss assumed the existence of a distribution minimizing the functional. Frostman proved the existence of such a minimizing distribution in a 1935 paper. After deﬁning the energy of a measure, properties of energy will be related to capacity and equilibrium distributions as developed in Section 4.4. The chapter will conclude with Wiener’s necessary and suﬃcient condition for regularity of a boundary point for the Dirichlet problem.

6.2 Energy Principle Throughout this section, Ω will denote a connected Greenian subset of Rn with Green function GΩ . Deﬁnition 6.2.1 If μ, ν are Borel measures on Ω, the mutual energy [μ, ν]e of μ and ν is deﬁned by [μ, ν]e = GΩ μ dν and the energy |μ|2e is deﬁned by |μ|2e = [μ, μ]e . The set of Borel measures μ of ﬁnite energy |μ|2e will be denoted by E + . L.L. Helms, Potential Theory, Universitext, c Springer-Verlag London Limited 2009

241

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6 Energy

Note that the integral deﬁning [μ, ν] is always deﬁned, even though it may be inﬁnite, and that 0 ≤ [μ, ν]e = GΩ μ dν = GΩ ν dμ = [ν, μ]e ≤ +∞, by the reciprocity theorem, Theorem 3.5.1. It follows that [μ, ν]e is a symmetric, bilinear form. If GΩ μ is bounded by m and μ has compact support Γ ⊂ Ω, then μ has ﬁnite energy and 2 |μ|e = GΩ μ dμ ≤ mμ(Γ ). An example will be given later to show that GΩ μ can take on the value +∞ even though μ has ﬁnite energy. Example 6.2.2 Let Ω = Bx,ρ ⊂ Rn , let B = Bx,δ where 0 < δ < ρ, and let ∂B ∂B μ = δΩ (x, ·). By deﬁnition of δΩ (x, ·), |μ|2e = GΩ μ dμ ∂B ∂B = GΩ (y, z) δΩ (x, dz) δΩ (x, dy) ∂B ˆ ∂B (GΩ (x, ·))(y) δΩ (x, dy). = R In the n = 2 case, GΩ (x, z) = log (ρ/|x − z|) = log (ρ/δ) for z ∈ ∂B, and in the n ≥ 3 case, GΩ (x, z) = (1/|x − z|n−2 ) − (1/ρn−2 ). Since ˆ ∂B (GΩ (x, ·), z) = GΩ (x, z) for z ∈ ∂B, δ ∂B (x, ·) has support in ∂B, and R Ω ∂B (x, ∂B) = 1, δΩ log ρδ if n = 2 2 |μ|e = 1 1 if n ≥ 3 δ n−2 − ρn−2 ∂B It is easily seen that δΩ (x, ·) is a unit uniform measure on ∂B.

Theorem 6.2.3 An analytic set Z is polar if and only if μ(Z) = 0 for all μ ∈ E + with support in Z. Proof: It is easily seen that the assertion is true if it is true for any analytic subset of B0,n , n ≥ 1. It therefore can be assumed that Z is a subset of a Greenian set Ω. Assume that Z is an analytic nonpolar set. Then Z has positive capacity relative to Ω by Theorem 4.5.2. It follows that there is a compact set Γ ⊂ Z with C(Γ ) > 0. Consider the capacitary distribution ˆ 1 ≤ 1. Thus, μΓ for which 0 < C(Γ ) = μΓ (Γ ) < +∞ and GΩ μΓ = R Γ + μΓ ∈ E and μΓ (Z) > 0. This proves the suﬃciency. Suppose now that Z is an analytic polar set and that μ ∈ E + has support in Z. By Theorem 4.3.11,

6.2 Energy Principle

243

there is a nonnegative superharmonic function w on Ω such that w = +∞ on Z. Since μ has ﬁnite energy, GΩ μ < +∞ a.e.(μ). Let Λ = {x; w(x) = +∞} and Λj = Λ ∩ {x; GΩ μ(x) ≤ j}, j ≥ 1. Since Λ ∼ ∪j Λj = Λ ∩ {x; GΩ μ(x) = +∞}, μ(Λ ∼ ∪j Λj ) = 0. Fix j ≥ 1, let Γ be any compact subset of Λj , and let ν = μ|Γ . Since GΩ ν ≤ GΩ μ, GΩ ν ≤ j on Γ . Since ν has support in Γ and w = +∞ on Γ, GΩ ν ≤ w on the support of ν for every > 0. By the Maria-Frostman domination principle, Theorem 4.4.8, GΩ ν ≤ w on Ω for every > 0 and it follows that GΩ ν = 0. Thus, ν is the zero measure and ν(Γ ) = μ(Γ ) = 0 for all compact Γ ⊂ Λj . Since μ(Λj ) =

sup

μ(Γ ),

Γ ⊂Λj ,Γ ∈K(Ω)

μ(Λj ) = 0 for all j ≥ 1. Since ∪Λj ⊂ Λ and μ(Λ ∼ ∪Λj ) = 0, μ(Λ) = 0 and μ(Z) = 0. Theorem 6.2.4 Let Ω be a Greenian subset of Rn and let {μj }, {νj } be sequences of Borel measures on Ω. (i) If GΩ μj ≤ GΩ νj , j = 1, 2, then [μ1 , μ2 ]e ≤ [ν1 , ν2 ]e . (ii) If μ, ν are Borel measures on Ω with GΩ μ ≤ GΩ ν, then |μ|e ≤ |ν|e ; moreover, if |μ|e = |ν|e < +∞, then μ = ν. (iii) If {GΩ μj } and {GΩ νj } are increasing sequences of potentials on Ω with potentials GΩ μ and GΩ ν as limits, respectively, then lim [μj , νj ]e = [μ, ν]e .

j→∞

(iv) If {GΩ μj } and {GΩ νj } are decreasing sequences of potentials, then the lower regularizations of their limits are potentials GΩ μ and GΩ ν, respectively, and lim [μj , νj ]e = [μ, ν]e j→∞

provided |μ1 |e , |ν1 |e , and [μ1 , ν1 ]e are ﬁnite. Proof: (i) By the reciprocity theorem, Theorem 3.5.1, [μ1 , μ2 ]e = GΩ μ1 dμ2 ≤ GΩ ν1 dμ2 = GΩ μ2 dν1 ≤ GΩ ν2 dν1 = [ν1 , ν2 ]e . (ii) Take μi = μ, νi = ν, i = 1, 2, in (i) to obtain |μ|2e ≤ |ν|2e ; if |μ|e = |ν|e < +∞, then the above inequalities are equalities so that GΩ μ = GΩ ν a.e.(μ). Since GΩ ν < +∞ a.e.(ν), GΩ μ ≤ GΩ ν on Ω by the MariaFrostman domination principle, Theorem 4.4.8. Similarly, GΩ ν ≤ GΩ μ on Ω. Thus, GΩ μ = GΩ ν and μ = ν by Theorem 3.5.8.

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6 Energy

(iii) By (i), {[μj , νj ]e } is an increasing sequence with limj→∞ [μj , νj ]e ≤ [μ, ν]e . Since GΩ μi ≤ GΩ μj for j ≥ i, by the Lebesgue monotone convergence theorem and the reciprocity theorem lim [μj , νj ]e = lim GΩ μj dνj ≥ lim inf GΩ μi dνj j→∞ j→∞ j→∞ = lim GΩ νj dμi = GΩ ν dμi = GΩ μi dν. j→∞

Since the latter sequence of integrals increases to [μ, ν]e , limj→∞ [μj , νj ]e ≥ [μ, ν]e and the two are equal. (iv) By (i), [μj , νj ]e ≤ [μ1 , ν1 ]e for all j ≥ 1 and limj→∞ [μj , νj ]e is deﬁned and ﬁnite; moreover, each GΩ μj is νj - integrable and each GΩ νj is μj -integrable. Note also that limj→∞ GΩ μj = GΩ μ quasi everywhere and limj→∞ GΩ νj = GΩ ν quasi everywhere by Theorem 4.4.10. Since polar sets are null sets for measures of ﬁnite energy by the above theorem, lim [μj , νj ]e = lim GΩ μj dνj ≥ lim ( lim GΩ μi ) dνj j→∞ j→∞ j→∞ i→∞ = lim GΩ μ dνi = GΩ νi dμ j→∞ ≥ ( lim GΩ νi ) dμ = GΩ ν dμ = [μ, ν]e . i→∞

On the other hand, for each i ≥ 1, lim [μj , νj ]e ≤ lim inf GΩ μi dνj = lim inf GΩ νj dμi . j→∞

j→∞

j→∞

Since the sequence {GΩ νj }j≥i is dominated by the μi -integrable function GΩ νi , lim [μj , νj ]e ≤ ( lim GΩ νj ) dμi = GΩ ν dμi j→∞ j→∞ = GΩ μi dν ≥ ( lim GΩ μi ) dν i→∞ = GΩ μ dν = [μ, ν]e . Thus, limj→∞ [μj , νj ]e = [μ, ν]e . It was shown in Theorem 4.3.5 that for a nonnegative superharmonic funcˆ u is a potential. tion u on Ω and a compact set Γ ⊂ Ω, R Γ ˆ 1 is Deﬁnition 6.2.5 A subset Λ of the Greenian set Ω is energizable if R Λ the potential of a measure μΛ .

6.2 Energy Principle

245

It is clear that a subset of an energizable set is again energizable, that compact subsets of Ω are energizable, and that polar subsets of Ω are energizable. Even though μΛ is not deﬁned when Λ is not energizable, the formal expression |μΛ |e = +∞ will be used in this case so that |μΛ |e is deﬁned for all subsets of Ω. Lemma 6.2.6 The set E + of energizable subsets of Ω has the following properties: (i) If Λ ∈ E + , then there is an open set O such that Λ ⊂ O ∈ E + . (ii) If Λ ∈ E + , then |μΛ |2e = μΛ (clf Λ) = μΛ (Ω). (iii) |μΛ |2e is an increasing function of Λ ∈ E + . ˆ 1 = R1 = 1 ˆ 1 > 1/2}. Since R Proof: Consider the open set U = {x ∈ Ω; R Λ Λ Λ quasi everywhere on Λ, U covers Λ except possibly for a polar subset of Λ. By Theorem 4.3.11, there is a potential u that is identically +∞ on the polar set ˆ 1 +u Λ ∼ U . Letting O = U ∪{u > 1}, O is an open set containing Λ. Since 2R Λ 1 1 1 1 ˆ + u on O, R ˆ ≤ R ≤ 2R ˆ + u. Since the latter is a potential with 1 ≤ eR Λ O O Λ ˆ 1 is a potential so that O ∈ E + and O is an energizable open is a potential, R O ˆ 1 = 1 on clf Λ by Theorem 5.7.11 and μΛ is set containing Λ. (ii) Since R Λ supported by clf Λ by Theorem 5.7.13, ˆ 1Λ dμΛ = μΛ (clf Λ) = μΛ (Ω), |μΛ |2e = GΩ μΛ dμΛ = R (6.1) proving (ii). Consider Λ1 , Λ2 ∈ E + with Λ1 ⊂ Λ2 ⊂ Ω, then GΩ μΛ1 = ˆ1 ≤ R ˆ 1 = GΩ μΛ2 and |μΛ1 |2e = GΩ μΛ1 dμΛ1 ≤ GΩ μΛ2 dμΛ1 = R Λ2 Λ1 GΩ μΛ2 dμΛ2 = |μΛ2 |2e by the reciprocity theorem, GΩ μΛ1 dμΛ2 ≤ Theorem 3.5.1.

Theorem 6.2.7 If Λ is an energizable subset of the Greenian set Ω, then |μΛ |2e = inf {|μO |2e ; Λ ⊂ O ∈ O(Ω)}. Proof: Since there is nothing to prove if |μΛ |2e = +∞, it can be assumed that μΛ has ﬁnite energy. By the preceding lemma, there is an energizable open set U ⊃ Λ so that it suﬃces to consider only those open sets O with ˆ 1 is a potential. By Lemma 4.6.12, Λ ⊂ O ⊂ U . For such O, R O R1Λ = inf {R1O ; O ∈ O(Ω), Λ ⊂ O ⊂ U }. By Choquet’s Lemma, Lemma 2.2.8, there is a decreasing sequence {Oj } of ˆ 1 is equal to the lower regularizaopen sets with Λ ⊂ Oj ⊂ U such that R Λ tion of ˆ 1O . lim R1Oj = lim R j j→∞

j→∞

By part (iv) of Theorem 6.2.4, limj→∞ |μOj |2e = |μΛ |2e .

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6 Energy

Theorem 6.2.8 If Λ is an energizable subset of the the Greenian set Ω, then C∗ (Λ) ≤ μΛ (Ω) = μΛ (clf Λ) = C ∗ (Λ); if, in addition, Λ is capacitable, then μΛ (Ω) = |μΛ |2e = C(Λ). Proof: If Γ is any compact subset of Λ, then C(Γ ) = μΓ (Γ ) = μΓ (Ω) ≤ μΛ (Ω) by Lemma 6.2.6. Taking the supremum over such compact sets Γ, C∗ (Λ) ≤ μΛ (Ω). Now let O be any open set containing Λ and let {Γj } be an increasing sequence of compact subsets of O such that Γj ↑ O. Then |μΓj ∩Λ |2e = μΓj ∩Λ (Ω) ≤ μΓj (Ω) = C(Γj ) ≤ C(O) for all j ≥ 1. By ˆ1 ˆ1 Theorem 4.6.5, GΩ μΓj ∩Λ = R Γj ∩Λ ↑ RΛ = GΩ μΛ on Ω and it follows from Theorem 6.2.4 that |μΓj ∩Λ |2e ↑ |μΛ |2e . Therefore, μΛ (Ω) = |μΛ |2e ≤ C(O) for all open sets O ⊃ Λ. Taking the inﬁmum over such sets, μΛ (Ω) ≤ C ∗ (Λ). Thus, C∗ (Λ) ≤ |μΛ |2e = μΛ (Ω) ≤ C ∗ (Λ). It remains only to show that μΛ (Ω) = C ∗ (Λ). Since open sets O are capacitable, |μO |2e = μO (Ω) = C(O) = C∗ (O). By Theorem 6.2.7, μΛ (Ω) = |μΛ |2e = inf {C∗ (O); Λ ⊂ O ∈ O(Ω)} = C ∗ (Λ). The following fact will be used in the course of the next proof. Let {μj } be a sequence of measures on a compact Hausdorf space X which converges in the w∗ -topology to the measure μ. The sequence of product measures {μj × μj } on X × X then converges in the w∗ -topology to the product measure μ × μ. For continuous fucntions of the form f (x)g(y) on X × X where f, g ∈ C 0 (X),

f (x) dμj (x) g(y) dμj (y)

→ f (x) dμ(x) g(y) dμ(y) = f (x)g(y) dμ(x) dμ(y)

f (x)g(y) dμj (x) dμj (y) =

as j → ∞. This results extends to ﬁnite sums of such products. Since the class A of such functions is a subalgebra of C 0 (X ×X) which separates points and contains the constant functions, A is a dense subset of C 0 (X × X) by the Stone-Weierstrass theorem. The above convergence can be extended to any function in C 0 (X × X) by a simple approximation. The following theorem was proposed, but not justiﬁed, as a procedure for solving the Dirichlet problem by Gauss in 1840. Theorem 6.2.9 (Gauss-Frostman) Let Γ be a compact subset of the Greenian set Ω, let f ∈ C 0 (Γ ), and for each measure μ with support in Γ let

6.2 Energy Principle

247

Φf (μ) =

(GΩ μ − 2f ) dμ.

Then there is a measure μ ∈ E + with support in Γ that minimizes Φf . Moreover, GΩ μ ≤ f on the support of μ and GΩ μ ≥ f on Γ except possibly for a polar set. Proof: By (vii) of Theorem 4.3.5, there is a measure ν with support in Γ such ˆ 1 = GΩ ν on Ω; for this measure, |Φf (ν)| ≤ ν(Γ )(1 + 2 f 0,Γ ) < that 1 ≥ R Γ +∞. Replacing ν by ν for 0 < < 1, Φf ( ν) = 2 Gν dν − f dν ≤ Φf (ν). If Φf (ν) ≤ 0, then inf μ Φf (μ) ≤ 0; if Φf (ν) ≥ 0, then there are measures μ with Φf (μ) arbitrarily close to 0 so that inf μ Φf (μ) ≤ 0. Let α = inf Φf (μ) ≤ 0, β = inf {GΩ (x, y); x, y ∈ Γ }, and γ = sup{f (y); y ∈ Γ }. Then GΩ (x, y) dμ(x) dμ(y) − 2 f (x) dμ(x) ≥ βμ2 (Γ ) − 2γμ(Γ ). Φf (μ) = Since βm2 − 2γm has a minimum value of −γ 2 /β, α ≥ −γ 2 /β > −∞. Letting {μj } be a sequence of measures with supports in Γ such that Φf (μj ) → α as j → ∞, Φf (μj ) ≥ βμ2j (Γ ) − 2γμj (Γ ) = μj (Γ )(βμj (Γ ) − 2γ) and it follows that the sequence {μj (Γ )} is bounded. There is then a subsequence of the sequence {μj }, which can be assumed to be the sequence itself, w∗

and a measure μ with support in Γ such that μj → μ. Let k be any positive w∗

integer. Since μj × μj → μ × μ and min (k, GΩ (x, y)) is a bounded continuous function on Γ × Γ , min (k, GΩ (x, y)) dμ(x) dμ(y) GΩ (x, y) dμ(x) dμ(y) = lim k→∞ = lim lim min (k, GΩ (x, y)) dμj (x) dμj (y) k→∞ j→∞ ≤ lim inf GΩ (x, y) dμj (x) dμj (y). j→∞

Therefore,

α ≤ Φf (μ) =

GΩ (x, y) dμ(x) dμ(y) − 2 f (x) dμ(x) ≤ lim inf GΩ (x, y) dμj (x) dμj (y) − 2 lim f (x) dμj (x) j→∞

j→∞

= lim Φf (μj ) = α. j→∞

This shows that μ minimizes Φf and that μ ∈ E + . It will be shown now that GΩ μ ≤ f on the support of μ. Suppose x is in the support of μ and

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GΩ μ(x) > f (x). Then there is a neighborhood U of x such that GΩ μ > f on U ∩ Γ . Letting ν = μ|U and 0 < ≤ 1, μ − ν is a measure with support in Γ and Φf (μ) ≤ Φf (μ − ν). Since Φf (μ) = GΩ μ dμ − 2 f dμ ≤ Φf (μ − ν) = GΩ (μ − ν) d(μ − ν) − 2 f d(μ − ν) = GΩ μ dμ − 2 GΩ μ dν + 2 GΩ ν − 2 f dμ + 2 f dν,

(f − GΩ μ) dν + GΩ ν dν ≥ 0. 2 Since GΩ ν ≤ GΩ μ and μ ∈ E +, GΩ ν dν < +∞ by Theorem 6.2.4. Letting → 0 in the above equation, (f − GΩ μ) dν ≥ 0. Since f − GΩ μ < 0 on the support of ν, ν(U ) = μ(U ) = 0; that is, the support of μ is contained in {x : GΩ μ(x) ≤ f }. Lastly, consider the assertion that GΩ μ ≥ f on Γ except possibly for a polar set. For each n ≥ 1, let μΓn be the capacitary distribution corresponding to the compact set Γn = Γ ∩ {x : f (x) ≥ GΩ μ(x) + n1 }. If t > 0, μ + tμΓn is a measure with support in Γ and Φf (μ) ≤ Φf (μ + tμΓn ). As above, t (GΩ μ − f ) dμΓn + GΩ μΓn dμΓn ≥ 0. 2 Letting t → 0, 1 0 ≤ (GΩ μ − f ) dμΓn ≤ − μΓn (Γn ) n from which it follows that C(Γn ) = μΓn (Γn ) = 0. Thus, Γn is polar for every n ≥ 1 and GΩ μ ≥ f on Γ except possibly for a polar set. Putting f = 1 in the deﬁnition of Φf , Gauss’s integral Φ1 (μ) = (GΩ μ − 2) dμ is obtained. It will be shown that the measure μ which minimizes Φ1 (μ) is just the capacitary distribution μΓ corresponding to the compact set Γ . Theorem 6.2.10 (Energy Principle) If μ and ν are Borel measures on the Greenian set Ω, then 0 ≤ [μ, ν]e ≤ |μ|e |ν|e . Proof: By (iii) of Theorem 6.2.4, it suﬃces to prove the inequality when μ and ν have supports in a compact set Γ ⊂ Ω. Let f be a bounded, nonnegative continuous function on Ω such that f ≤ GΩ μ and let μ ∈ E + be the measure with support in Γ which minimizes Φf . Since μ ∈ E + and polar sets have μ -measure zero by Theorem 6.2.3, GΩ μ = f a.e.(μ ) on the support of μ . It follows that

6.3 Mutual Energy

249

and Φf (μ ) =

f dμ =

GΩ μ dμ − 2

GΩ μ dμ

f dμ = −

GΩ μ dμ = −|μ |2e .

For any α > 0, Φf (αν) ≥ Φf (μ ) = −|μ |2e ; that is, α2 |ν|2e − 2α f dν + |μ |2e ≥ 0 for every α > 0. Since the inequality is trivially true if α ≤ 0, it holds for all α and f dν ≤ |ν|e |μ |e . Since GΩ μ ≤ f ≤ GΩ μ on the support of μ and GΩ μ is harmonic oﬀ the support of μ , GΩ μ is ﬁnite-valued and GΩ μ ≤ GΩ μ on Ω by the Maria-Frostman domination principle, Theorem 4.4.8. By (ii) of Theorem 6.2.4, |μ |e ≤ |μ|e , and therefore f dν ≤ the supremum over all bounded, nonnegative continuous |ν|e |μ|e . Taking f ≤ GΩ μ, GΩ μ dν ≤ |μ|e |ν|e . It is now possible to give an example of a measure μ ∈ E + for which GΩ μ can take on the value +∞. Example 6.2.11 Let Z be a nonempty compact polar subset of the Greenian set Ω. Then Z has capacity zero. For each j ≥ 1, let Uj be an open set containing Z with compact closure Uj− ⊂ Ω such that C(Uj− ) < 2−2j , and let μj be the capacitary distribution of Uj− . Since GΩ μj ≤ 1, μj ∈ E + . ∞ Moreover, GΩ μj = 1 on Uj ⊃ Z. Letting μ = j=1 μj , μ is a measure for which GΩ μ = +∞ on Z. Since GΩ μ dμ = |μj |e |μk |e GΩ μj dμk ≤ j,k

≤

j,k

j,k

2−2j 2−2k =

2−2j

2

< 1,

j

μ ∈ E +.

6.3 Mutual Energy If μ and ν are two Borel measures on the Greenian set Ω, the Green potential GΩ (μ−ν) is certainly not deﬁned if both μ and ν are inﬁnite on the same set. To get around this, μ − ν will be regarded as a formal diﬀerence. Two such diﬀerences μ − ν and μ − ν will be regarded as the same if μ + ν = μ + ν

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and written μ− ν ∼ μ − ν . Strictly speaking, μ− ν should be regarded as the ordered pair (μ, ν) in the product space E + ×E + in which two pairs (μ, ν) and (μ , ν ) are regarded as the same if μ + ν = μ + ν, written (μ, ν) ∼ (μ , ν ). This relation is reﬂexive and transitive but not symmetric. In showing that (μ, ν) ∼ (μ , ν ) and (μ , ν ) ∼ (μ , ν ) ⇒ (μ, ν) ∼ (μ , ν ), the equation μ + μ + ν + ν = μ + μ + ν + ν is obtained; since all measures are assumed to be ﬁnite on compact subsets of Ω, μ+ν = μ +ν on compact subsets of Ω and therefore on the Borel subsets of Ω by the assumption that all measures are regular Borel measures. Deﬁnition 6.3.1 Let E = E + × E + = {(μ, ν); μ, ν ∈ E + }. If λi = (μi , νi ) with μi , νi ∈ E + , i = 1, 2, the mutual energy [λ1 , λ2 ] is deﬁned by [λ1 , λ2 ]e = [μ1 , μ2 ]e − [μ1 , ν2 ]e − [μ2 , ν1 ]e + [ν1 , ν2 ]e ; the energy |λ|2e of λ ∈ E is deﬁned by |λ|2e = [λ, λ]e . The latter part of this deﬁnition assumes that the energy of an element λ ∈ E is nonnegative. This follows from the energy principle, Theorem 6.2.10, for if λ = (λ1 , λ2 ), then [λ, λ]e = |λ1 |2e − 2[λ1 , λ2 ]e + |λ2 |2e ≥ |λ1 |2e − 2|λ1 |e |λ2 |e + |λ2 |2e = (|λ1 |e − |λ2 |e )2 ≥ 0. It can be seen also that the deﬁnition of the mutual energy [λ1 , λ2 ] is independent of the ordered pairs used to represent λ1 and λ2 as follows. If (μi , νi ) ∼ (μi , νi ), i = 1, 2, then μi + νi = μi + νi ,

i = 1, 2.

Using these equations, the above expression deﬁning [λ1 , λ2 ] can be converted into the expression that would be obtained if the (μi , νi ), i = 1, 2, were used instead. Addition and multiplication by real numbers is deﬁned on E by (μ, ν) + (μ , ν ) = (μ + μ , ν + ν ) (cμ, cν) if c ≥ 0 c(μ, ν) = (−cν, −cμ) if c < 0. The zero element of E is given by 0 = (0, 0). It is easily checked that E is a vector space over the reals. It is also easily seen that [λ1 , λ2 ]e = [λ2 , λ1 ]e for all λ1 , λ2 ∈ E.

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251

Theorem 6.3.2 If μ, ν ∈ E, then |[μ, ν]e | ≤ |μ|e |ν|e and |μ+ν|e ≤ |μ|e +|ν|e . Proof: Since the energy is nonnegative, [μ − αν, μ − αν]e ≥ 0 for all α ∈ R; that is, |μ|2e − 2α[μ, ν]e + α2 |ν|2e ≥ 0. Therefore,

(−2[μ, ν]e )2 ≤ 4|μ|2e |ν|2e

and so |[μ, ν]e | ≤ |μ|e |ν|e . By the preceding step, |μ + ν|2e = |μ|2e + 2[μ, ν]e + |ν|2e ≤ |μ|2e + 2|μ|e |ν|e + |ν|2e = (|μ|e + |ν|e )2 . Theorem 6.3.3 The energy of μ ∈ E is zero if and only if μ ∼ 0. Proof: According to the preceding theorem if |μ|e = 0, then [μ, ν]e = 0 for all ν ∈ E. If μ = (μ1 , μ2 ) with μ1 , μ2 ∈ E + , then [μ1 , ν]e = [μ2 , ν]e for all − ν ∈ E. According to Theorem 3.5.7, for each pair x, δ such that Bx,δ ⊂Ω there is a signed measure τx,δ of ﬁnite energy such that the class of functions {GΩ τx,δ } is a total subset of K + (Ω). Since GΩ τx,δ dμ1 = GΩ μ1 dτx,δ = [μ1 , τx,δ ]e = [μ2 , τx,δ ]e = GΩ τx,δ dμ2 and the set {GΩ τx,δ } is total in K + (Ω), μ1 = μ2 by Theorem 3.5.4 and μ = (μ1 , μ1 ) ∼ 0. Conversely, if μ ∼ 0, then the energy is zero. From the Hilbert space point of view, E is a pre-Hilbert space in that it is a vector space over the reals with [μ, ν]e serving as an inner product and |μ|e as a norm but lacks the completeness property as a metric space. The following example shows that E is not complete. Example 6.3.4 (Cartan) Let Ω = R3 , and for each j ≥ 1 let λj = μj − νj be a signed measure where μj is a unit measure uniformly distributed on a sphere with center at the origin of radius 1 − 4−j and νj is deﬁned similarly on a concentric sphere of radius 1. By Gauss’s averaging principle, Theorem 1.6.2, 1 . [λj , λj ]e = Gμj dμj − 2 Gμj dνj + Gνj dνj = j 4 −1 ∞ It follows that the series j=1 |λj |e converges. Since p p+k p+k p+k λj − λj = λj ≤ |λj |e → 0 j=1

j=1

e

j=p+1

e

j=p+1

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6 Energy

p as p → ∞ independently of k, the sequence of partial sums { j=1 λj } is a Cauchy sequence in the energy norm. Suppose there is a λ = (λ , λ ) in E such that p lim λj − λ = 0. (6.2) p→∞

By Theorem 6.3.2, lim

p (

p→∞

e

j=1

) λj , ν

j=1

e

= [λ, ν]e

for all ν ∈ E + ; that is, lim

p→∞

p λj ) = Gν dλ − Gν dλ . Gν d( j=1

In particular, this equation holds for all the potentials Gτx,δ of Theorem 3.5.7 − with Bx,δ ⊂ B = B0,1 . If K0+ denotes the set of all such potentials, then K0+ is a total subset of the class K + (B) of nonnegative continuous functions on B having compact support in B by the same theorem. Since the Gτx,δ = 0 − outside Bx,δ , for f ∈ K0+ (B) lim

p→∞

p p λj ) = lim μj ) = f λ − f dλ . f d( f d( p→∞

j=1

j=1

By deﬁnition of a total set, Deﬁnition 3.5.2, there is a nonnegative linear combination f of elements of K0+ (B) that majorizes 1 on the closure of the ball of radius 1 − 4−p with |f | ≤ 2. Then p

μj (B) ≤

p f d( μj )

j=1

and +∞ = lim

p→∞

p j=1

j=1

μj (B) ≤ lim

p→∞

p μj ) ≤ 2λ (B − ). f d( j=1

Thus, λ is not a Borel measure, a contradiction. Therefore, there is no λ ∈ E satisfying Equation (6.2). Consider a Borel measure μ and a superharmonic function u on the Greenian set Ω for which GΩ μ ≤ u a.e.(μ) on the support of μ. Application of the Maria-Frostman domination principle, Theorem 4.4.8, requires that GΩ μ < +∞ on Ω to conclude that GΩ μ ≤ u on Ω. The ﬁniteness condition can be eliminated if μ has ﬁnite energy.

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253

Theorem 6.3.5 (Cartan) If u is a superharmonic function on the Greenian set Ω, μ ∈ E + , and GΩ μ ≤ u a.e.(μ) on the support of μ, then GΩ μ ≤ u on Ω. Proof: The function v = min (u, GΩ μ) is a potential GΩ ν with ν ∈ E + since GΩ ν ≤ GΩ μ. Since (GΩ μ > u) = (GΩ μ>GΩ ν) and μ(GΩ μ>u)=0, (GΩ μ − GΩ ν) dμ = 0 ≤ (GΩ μ − GΩ ν) dν. It follows that |μ − ν|2e = |μ|2e − 2[μ, ν]e + |ν|2e ≤ 0. In view of the fact that energy is nonnegative, |μ − ν|e = 0 and μ ∼ ν by the preceding theorem. Since μ and ν are measures, this means that μ = ν; that is, u ≥ min (u, GΩ μ) = GΩ ν = GΩ μ. Given a sequence {μj } in E + , there are three modes of convergence of the sequence to a measure μ ∈ E + : (i) strong convergence in the energy norm; that is, limj→∞ |μj − μ|e = 0, (ii) weak convergence in the pre-Hilbert space E; that is, limj→∞ [μj , ν]e = [μ, ν]e for all ν ∈ E (equivalently, for all ν ∈ E + ), and (iii) vague convergence ; that is, limj→∞ f dμj = f dμ for all f ∈ C00 (Ω). Strong convergence implies weak convergence since |[μj , ν]e − [μ, ν]e | ≤ |μj − μ|e |ν|e for all ν ∈ E. Lemma 6.3.6 If {μj } is a sequence in E that converges weakly to μ ∈ E, then |μ|e ≤ lim supj→∞ |μj |e . Proof: Since |μ|2e = [μ, μ]e ≤ |[μ − μj , μ]e + |[μj , μ]|e ≤ |[μ − μj , μ]e | + |μj |e |μ|e , |μ|2e ≤ lim sup |[μj − μ, μ]e | + |μ|e lim sup |μj |e = |μ|e lim sup |μj |e . j→∞

j→∞

j→∞

Theorem 6.3.7 Weak convergence and strong convergence are equivalent for strong Cauchy sequences in E + . Proof: As noted above, strong convergence implies weak convergence. Let {μj } be a strong Cauchy sequence in E + . It need only be shown that weak convergence of the sequence {μj } to μ ∈ E + implies that limj→∞ |μj −μ|e = 0. Since the sequence {μj − μk } converges weakly to μ − μk for each k ≥ 1, |μ − μk |e ≤ lim sup |μj − μk |e j→∞

by the preceding lemma. Since the sequence {μj } is strong Cauchy, limk→∞ |μ − μk |e = 0.

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Theorem 6.3.8 If μ ∈ E + and GΩ μ is the limit of a monotone sequence of potentials {GΩ μj }, μj ∈ E + , j ≥ 1, on the Greenian set Ω, then the sequence {μj } converges strongly to μ. Proof: Suppose GΩ μj ≤ GΩ μk . Then |μk −

μj |2e

=

|μk |2e

−2

≤ |μk |2e − 2

GΩ μk dμj + |μj |2e GΩ μj dμj + |μj |2e

(6.3)

= |μk |2e − |μj |2e . This shows that the sequence {|μj |e } is monotone. If the sequence {GΩ μj } is increasing, the same is true of the sequence {|μj |e }, and 2 |μj |e = GΩ μj dμj ≤ GΩ μ dμj = GΩ μj dμ ≤ |μ|2e , and therefore |μj |e ≤ |μ|e , j ≥ 1. If the sequence {GΩ μj } is decreasing, the same is true of the sequence {|μj |e } and |μj |e ≤ |μ1 |e , j ≥ 1. In either case, the sequence {|μj |e } is a bounded Cauchy sequence. It follows from Inequality (6.3) that the sequence {μj } is a strong Cauchy sequence. By the preceding theorem, it suﬃces to prove that the sequence converges weakly to μ; that is, GΩ μj dν = GΩ μ dν lim j→∞

for all ν ∈ E + , but this follows from the Lebesgue dominated convergence theorem. The ordered pair λ = (μ, ν) ∈ E has compact support if both μ and ν have compact supports. In this case, GΩ μ and GΩ ν are both deﬁned, but GΩ (μ − ν) = GΩ μ − GΩ ν may not be deﬁned because of the inﬁnities of GΩ μ and GΩ ν. The notation GΩ (μ−ν) will be used only when the diﬀerence GΩ μ − GΩ ν is deﬁned everywhere on Ω. Theorem 6.3.9 The set of μ ∈ E having compact support with GΩ μ ∈ C00 (Ω) is dense in E. Proof: Consider any μ ∈ E + , let {Γj } be an increasing sequence of compact subsets of Ω such that Ω = ∪Γj , and let μj = μ|Γj , j ≥ 1. Since each μj has compact support, GΩ μj is a potential and {GΩ μj } is a sequence of potentials increasing to GΩ μ. By the preceding theorem, the sequence {μj } converges strongly to μ. This shows that the set E0+ of μ ∈ E + having compact support in Ω is strongly dense in E + . Now consider any μ ∈ E0+ . By taking volume averages of GΩ μ, a sequence {μj } in E0+ can be constructed so that GΩ μj ↑ GΩ μ with each GΩ μj ∈ C 0 (Ω). By the preceding theorem, the sequence {μj } converges strongly to μ. This shows that the set of μ ∈ E0+ with

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255

continuous GΩ μ is strongly dense in E + . Consider now an arbitrary element of E0+ with continuous u = GΩ μ. By Corollary 2.6.31, there is an increasing sequence {Ωj } of regular open sets with Ωj− ⊂ Ω such that Ω = ∪Ωj . It Ω

can be assumed that the support of μ is contained in Ω1 . Let Hu j be the harmonic function on Ωj corresponding to the boundary function u|∂Ωj . Now let Ωj Hu on Ωj uj = u on Ω ∼ Ωj . By Theorem 4.6.8, uj ↓ 0 on Ω as j → ∞. Letting uj = GΩ μj , GΩ μj ≤ GΩ μ implies that μj ∈ E + and, in fact, that μj ∈ E0+ since uj = u outside a neighborhood of the support of μ on which u is harmonic. Since uj = GΩ μj ↓ 0, the sequence {μj } converges strongly to the zero measure by the preceding theorem. It follows that the sequence {μ − μj } in E converges strongly to μ. Note that GΩ μ − GΩ μj = u − uj = 0 outside Ωj and therefore GΩ (μ − μj ) ∈ C00 (Ω). Therefore, any μ ∈ E + can be approximated strongly by a ν ∈ E having compact support with GΩ ν ∈ C00 (Ω). If μ = μ1 − μ2 ∈ E where μi ∈ E + , i = 1, 2, then each μi can be approximated strongly by a νi ∈ E having compact support with GΩ νi ∈ C00 (Ω). Theorem 6.3.10 Let {μj } be a sequence in E + . Strong convergence of the sequence implies weak convergence and norm boundedness. If the sequence {μj } is norm bounded, then weak convergence and vague convergence are equivalent. Proof: If the sequence {μj } converges strongly to μ ∈ E + , then norm boundedness follows from the inequality ||μj |e − |μ|e | ≤ |μj − μ|e . Suppose the sequence {μj } is norm bounded by m and converges weakly to μ ∈ E + . Consider the signed measures τx,δ of Theorem 3.5.3. Since each GΩ τx,δ is the − ⊂ Ω, Green potential of a signed measure of ﬁnite energy whenever Bx,δ

lim

j→∞

GΩ τx,δ , dμj =

GΩ τx,δ dμ;

that is, limj→∞ f dμj = f dμ for all f in a total subset K0+ of K + (Ω). Let Γ be any compact subset of Ω. It now will be shown that the sequence {μj (Γ )} is bounded. To do this, let f be a nonnegative linear combination of elements of K0+ that majorizes 1 on Γ . Since lim sup μj (Γ ) ≤ lim sup f dμj = f dμ < +∞, j→∞

j→∞

the sequence {μj (Γ )} is bounded. Consider any g ∈ K + (Ω). If > 0, choose f ∈ K0+ such that |g − f | < . Since f and g have compact supports in Ω, there is a compact set Γ containing the supports of f and g and a constant α such that μj (Γ ) ≤ α, j ≥ 1. Since

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6 Energy

g dμj − g dμ ≤ |g − f | dμj + f dμj − f dμ + |f − g| dμ ≤ α + f dμj − f dμ + μ(Γ ) and f ∈ K0+ , limj→∞ g dμj = g dμ; that is, the sequence {μj } converges to μ vaguely. Assume now that |μj |e ≤ m, j ≥ 1, and that the sequence {μj } converges to μ vaguely. As in the proof of Theorem 3.6.4, GΩ μ ≤ lim inf j→∞ GΩ μj . By Fatou’s Lemma, for each p ≥ 1 GΩ μ dμp ≤ (lim inf GΩ μj ) dμp ≤ lim inf GΩ μj dμp ≤ m2 . j→∞

j→∞

Therefore, GΩ μ dμ ≤ (lim inf GΩ μp ) dμ ≤ lim inf GΩ μ dμp ≤ m2 p→∞

p→∞

and μ ∈ E + . To show that limj→∞ [μj , ν]e = [μ, ν]e for all ν ∈ E, it suﬃces to prove this result for all ν in a dense subset of E. By the preceding theorem, the set of ν ∈ E having compact support in Ω with Gν ∈ C00 (Ω) is dense in E. For such ν, lim [μj , ν]e = lim GΩ ν dμj = GΩ ν dμ = [μ, ν]e j→∞

j→∞

by the vague convergence of the sequence {μj } to μ. Theorem 6.3.11 (Cartan [11]) E + is complete. Proof: Let {μj } be a sequence in E + that is strong Cauchy. For any ν in E, the inequality |[μj , ν]e − [μk , ν]|e ≤ |μj − μk |e |ν|e implies that the sequence {[μj , ν]e } is Cauchy in R. Taking ν = τx,δ , where − Bx,δ ⊂ Ω, L(f ) = lim f dμj j→∞

exists for each f in a total subset K0+ of K + . The linear functional L(f ) can be extended to K + as follows. First let Γ be any compact subset of Ω and let f ∈ K0+ majorize 1 on Γ . Since the sequence { f dμj } is bounded by some m > 0, μj (Γ ) ≤ f dμj ≤ m, j ≥ 1. This shows that the sequence {μj (Γ )} is bounded for any compact subset Γ of Ω. Now consider any g ∈ K + having compact support Γ and a neighborhood U of Γ having compact closure U − ⊂ Ω. Choose m > 0 so that μj (U − ) ≤ m for all j ≥ 1. For each k ≥ 1, let gk ∈ K0+ have compact support in U such that the sequence {gk } converges uniformly to g on Ω. Since

6.4 Projections of Measures

257

g dμi − g dμj ≤ |g − gk | dμi + gk dμi − gk dμj + |gk − g| dμj ≤ 2m g − gk 0 + gk dμi − gk dμj . Given > 0, choose k so that |g − gk | < /4m. Then g dμi − g dμj ≤ + gk dμi − gk dμj 2 from which it follows that the sequence { g dμj } is Cauchy and L(g) = lim g dμj j→∞

exists for all g ∈ K + . It follows that there is a measure μ such that lim g dμj = g dμ j→∞

for all g ∈ K + ; that is, the sequence {μj } converges vaguely to μ. Since the sequence is strongly bounded, the sequence converges weakly to μ by the preceding theorem and strong convergence follows from Theorem 6.3.7.

6.4 Projections of Measures The operation of sweeping a measure deﬁning a potential on a Greenian set can be formulated as an operator on the pre-Hilbert space E by projecting a measure μ ∈ E + onto a convex subset of E + . Recall that a subset F of E is convex if λμ + (1 − λ)ν ∈ F whenever μ, ν ∈ F and 0 < λ < 1. Lemma 6.4.1 If F is a nonempty, closed, convex subset of E + and μ ∈ E + , then there is a unique measure μF ∈ F such that |μ − μF |e ≤ |μ − ν|e for all ν ∈ F. Proof: Consider any two measures α, β ∈ F. Then γ = (α + β)/2 ∈ F since F is convex. Now |μ − γ|2e = |(μ − α) + (α − γ)|2e 1 = |(μ − α) + (α − β)|2e 2 1 2 = |μ − α|e + |α − β|2e + [μ − α, α − β]e . 4

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6 Energy

Similarly, 1 |μ − γ|2e = |μ − β|2e + |α − β|2e + [μ − β, β − α]e . 4 Therefore, 1 2|μ − γ|2e = |μ − α|2e + |μ − β|2e + |α − β|2e + [β − α, α − β]e 2 1 2 2 = |μ − α|e + |μ − β|e − |α − β|2e 2 so that

(6.4)

|α − β|2e = 2|μ − α|2e + 2|μ − β|2e − 4|μ − γ|2e .

Let m = inf ν∈F |μ − ν|e . If {νj } is any sequence in F , then |νj − νk |2e ≤ −4m2 + 2|μ − νj |2e + 2|μ − νk |2e . Choosing a sequence {νj } in F such that |μ − νj |e ↓ m, lim supj,k→∞ |νj − νk |e ≤ 0 and it follows that the sequence {νj } is strong Cauchy in F . Since E + is complete and F is closed, there is an α ∈ F such that |μ − α|e = m = inf |μ − ν|e . ν∈F

Uniqueness of a measure minimizing |μ − ν|e can be shown as follows. If α and β are two measures satisfying |μ − α|e = |μ − β|e = m, then 1 2m2 ≤ 2m2 − |α − β|2e 2 by Equation (6.4) and α = β. Deﬁnition 6.4.2 If F is a nonempty, closed, convex subset of E + and μ ∈ E + , the measure μF of the preceding lemma is called the projection of μ onto F . Lemma 6.4.3 Let F be a nonempty, closed, convex subset of E + and let μ ∈ E + . Then μ0 is the projection of μ on F if and only if [μ−μ0 , ν −μ0 ]e ≤ 0 for all ν ∈ F. Proof: If α ∈ F, then |μ − α|2e ≤ |μ − ν|2e for all ν ∈ F if and only if 2[μ − α, ν − α]e ≤ |α − ν|2e for all ν ∈ F. It now will be shown that the latter inequality is true if and only if 2[μ − α, ν − α]e ≤ 0 for all ν ∈ F. The suﬃciency of the latter statement is trivial. If ν ∈ F and 0 < λ < 1, then ν = (1−λ)α+λν = α+λ(ν−α) ∈ F and

6.4 Projections of Measures

259

2[μ − α, λ(ν − α)]e ≤ |λ(ν − α)|2e or

2[μ − α, ν − α]e ≤ λ|ν − α|2e .

Letting λ → 0, 2[μ − α, ν − α]e ≤ 0 for all ν ∈ F. Taking α = μ0 , |μ − μ0 |2 ≤ |μ − ν|2 for all ν ∈ F if and only if [μ − μ0 , ν − μ0 ] ≤ 0 for all ν ∈ F. Recall that a convex set F is a cone with vertex α if α + β ∈ F implies that α + λβ ∈ F for all λ ≥ 0. Lemma 6.4.4 Let F be a nonempty, closed, convex cone in E + with vertex 0 and let μ ∈ E + . Then μ0 is the projection of μ onto F if and only if (i) [μ − μ0 , ν]e ≤ 0 for all ν ∈ F and (ii) [μ − μ0 , μ0 ]e = 0. Proof: Assume that μ0 is the projection of μ onto F . By the preceding lemma, [μ − μ0 , ν − μ0 ]e ≤ 0 for all ν ∈ F. For any ν ∈ F, (μ0 + ν)/2 ∈ F, μ0 + ν ∈ F, and [μ − μ0 , ν]e ≤ 0 from the preceding inequality. This proves (i). In particular, [μ− μ0 , μ0 ]e ≤ 0. Putting ν = 0 in the above inequality, [μ − μ0 , μ0 ]e ≥ 0 and (ii) is proved. Conversely, assume that μ0 satisﬁes (i) and (ii). Then [μ − μ0 , ν − μ0 ]e = [μ − μ0 , ν]e − [μ − μ0 , μ0 ]e ≤ 0 for all ν ∈ F and μ0 is the projection of μ onto F by the preceding lemma. If Γ is a compact subset of Ω, F (Γ ) will denote the set of measures ν ∈ E + with support in Γ . F (Γ ) is easily seen to be a convex cone in E + . That F (Γ ) is strongly closed can be seen as follows. Consider a sequence {μj } in F (Γ ) that converges strongly to μ ∈ E + . This implies that the sequence is norm bounded and weakly convergent to μ. Since weak convergence and vague convergence are equivalent for norm-bounded sequences, the sequence {μj } converges vaguely to μ. Since the μj have support in Γ , the same is true of μ; that is, F (Γ ) is closed. It therefore makes sense to consider the projection μF (Γ ) of μ onto F (Γ ). Theorem 6.4.5 If Γ is a compact subset of the Greenian set Ω and μ ∈ E + , then μ0 ∈ F(Γ ) is the projection of μ onto F (Γ ) if and only if (i) GΩ μ0 ≤ GΩ μ on Ω and (ii) GΩ μ0 = GΩ μ on Γ except possibly for a polar set.

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6 Energy

Proof: Suppose μ0 is the projection of μ onto F (Γ ). By the preceding lemma, (i ) (GΩ μ − GΩ μ0 ) dν ≤ 0 for all ν ∈ F(Γ ) and (ii ) (GΩ μ − GΩ μ0 ) dμ0 = 0. Consider the set {x; GΩ μ(x) > GΩ μ0 (x)} and any ν ∈ F(Γ ). Since the restriction of ν to this set also belongs to F (Γ ), (GΩ μ − GΩ μ0 ) dν ≤ 0. {GΩ μ>GΩ μ0 }

This implies that ν({x; GΩ μ(x) > GΩ μ0 (x)}) = 0 for all ν ∈ F(Γ ). Taking ν = μ0 , GΩ μ ≤ GΩ μ0 a.e.(μ0 ). In conjunction with (ii ), this means that GΩ μ = GΩ μ0 a.e.(μ0 ) on the support of μ0 . By Theorem 6.3.5, GΩ μ0 ≤ GΩ μ on Ω. Thus, μ0 satisﬁes (i). By Theorem 6.2.3, GΩ μ = GΩ μ0 , except possibly for a polar set since μ0 ∈ E + . Conversely, suppose μ0 satisﬁes (i) and (ii). Then (ii) implies that GΩ μ0 = GΩ μ a.e.(ν) for all ν ∈ F(Γ ) ⊂ E + by the same theorem. Properties (i ) and (ii ) follow immediately from the preceding lemma. Corollary 6.4.6 If Γ is a compact subset of the Greenian set Ω and μ ∈ E + , ˆ Γ (GΩ μ, ·). then GΩ μF (Γ ) = R ˆ Γ (GΩ μ, ·), R ˆ Γ (GΩ μ, ·) is the potential of a measure Proof: Since GΩ μ ≥ R ˆ Γ (GΩ μ, ·) = μ0 ∈ E + by (ii) of Theorem 6.2.4. By Corollary 4.6.2, GΩ μ0 = R GΩ μ on Γ except possibly for a polar set. It follows from the preceding theˆ Γ (GΩ μ, ·) = GΩ μF (Γ ) . orem that μ0 = μF (Γ ) ; that is, R The Gauss-Frostman theorem, Theorem 6.2.9, can be generalized to potentials GΩ μ, μ ∈ E + , by deﬁning ΦGΩ μ (ν) = GΩ ν dν − 2 GΩ μ dν, ν ∈ E +. Theorem 6.4.7 Let F be a nonempty, closed convex subset of E + and let μ ∈ E + . Then μF is the unique measure that minimizes ΦGΩ μ over F . Proof: |μ − ν|2e = ΦGΩ μ (ν) + |μ|2e .

6.5 Wiener’s Test Wiener’s original theorem characterized irregularity of a boundary point of a region Ω in terms of the capacities of parts of ∼ Ω in neighborhoods of the boundary point. Since thinness of the complement and irregularity are equivalent, it suﬃces to relate thinness of ∼ Ω to the capacities of certain subsets of ∼ Ω.

6.5 Wiener’s Test

261

Let Ω be a Greenian subset of Rn , let x ∈ Ω, and let Λ be any subset of Ω. For α > 1, let Aj = {y ∈ Ω; αj ≤ u(x, y) ≤ αj+1 }, j j let Aj = ∪∞ i=j Ai , let Λj = Λ ∩ Aj , j ≥ 1, and let Λ = Λ ∩ A . Choose k ≥ 1 large enough so that (Ak )− ⊂ Ω, diam(Ak ) < 1, and Ak is a subset of the component of Ω containing x. Several lemmas will be required to prove the celebrated theorem of Wiener.

Theorem x if and only if (a) the ∞ 6.5.1 (Wiener’s Test [64]) Λ is thinat ∞ ˆ1 series j=0 αj C ∗ (Λj ) converges or (b) the series j=0 R Λj (x) converges. Several lemmas will be required to prove Wiener’s result. The proof that ∞ ˆ1 the series j=0 R Λj (x) converges as a consequence of the thinness at x will be accomplished by showing that the two series corresponding to odd numbered terms and even numbered terms converge. This amounts to considering alternate annuli A2j+1 and alternate annuli A2j , respectively. To avoid cumbersome notation, only the annuli with even subscripts will be considered; the treatment of odd subscripts being essentially the same. Lemma 6.5.2 For each j ≥ k, let Γj be a nonempty compact subset of Λj and let Γ = ∪∞ j=k Γ2j . If α > 1, there is a constant m > 0, depending only upon α and k, such that u(y, z) ≤m u(y, x)

z ∈ Γ2j , y ∈ Γ ∼ Γ2j , j ≥ k.

Proof: Consider the n = 2 case ﬁrst. If u(y, x) ≤ α2j−1 and α2j ≤ u(x, z) ≤ α2j+1 , then u(y, z) − log |y − z| = u(y, x) u(y, x) − log (|y − x| − |x − z|) ≤ u(y, x)

|x − z| ≤ 1 − α−2k+1 log 1 − . |x − y| Since

2j−1 e−u(x,z) |x − z| (α−1) = −u(x,y) ≤ e−α |x − y| e

and the last expression has the limit 0 as j → ∞ uniformly in y and z, it follows that u(y, z)/u(y, x) is bounded uniformly in j and y, z as described above. Suppose now that u(x, y) ≥ α2j+2 and α2j ≤ u(x, z) ≤ α2j+1 . In this case |x−z| log − 1 |x−y| u(y, z) ≤1− . u(y, x) u(y, x)

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Since

2j+1 |x − z| (α−1) ≥ eα , |x − y| 2j+1 (α−1) log eα −1 u(y, z) ≤1− , u(y, x) α2j+2

where the last term has the limit 1 + (1 − α)/α as j → ∞, and again it follows that u(y, z)/u(y, x) is bounded uniformly in j and y, z as described above. This proves the assertion in the n = 2 case. Consider now the n ≥ 3 case. Similarly, if u(x, y) ≤ α2j−1 and α2j ≤ u(x, z) ≤ α2j+1 , then u(y, z) ≤ u(y, x)

α1/(n−2) α1/(n−2) − 1

n−2

and u(y, z)/u(y, x) is uniformly bounded in j and y, z as described above; if u(x, y) ≥ α2j+2 and α2j ≤ u(x, z) ≤ α2j+1 , then u(y, z) ≤ u(y, x)

1

n−2

α1/(n−2) − 1

and u(y, z)/u(y, x) is uniformly bounded in j and y, z as described above. Lemma 6.5.3 There is a constant c > 0, depending only on k, such that 1 GΩ (x, ·) ≤ u(x, ·) ≤ cGΩ (x, ·) c

on (Ak )− .

Proof: Since GΩ (x, ·) = u(x, ·) − ghmΩ u(x, ·) on Ω, GΩ (x, ·) ≤ u(x, ·) on k (Ak )− . Note that Ak = Bx,δ ∼ {x} where δ = αk/(n−2) if n ≥ 3 and δ = e−α in the n = 2 case. Since GΩ (x, ·) and u(x, ·) are positive and continuous on ∂Bx,δ , there is a constant c > 1 such that cGΩ (x, ·) − u(x, ·) ≥ 0 on ∂Bx,δ so that lim inf (cGΩ (x, y) − u(x, y)) ≥ 0 y→z,y∈Bx,δ

for z ∈ ∂Bx,δ . Since lim inf (cGΩ (x, y) − u(x, y)) = lim inf ((c − 1)u(x, y) − c ghm u(x, y)) = +∞, y→x

y→x

Ω

lim inf y→x,y∈Bx,δ ∼{x} (cGΩ (x, y) − u(x, y)) ≥ 0 on ∂(Bx,δ ∼ {x}). By Corollary 2.3.6, cGΩ (x, ·) − u(x, y) ≥ 0 on Bx,δ ∼ {x} = Ak . Thus, u(x, ·) ≤ cGΩ (x, ·) on (Ak )− . Since c > 1, (1/c)GΩ (x, ·) ≤ GΩ (x, ·) ≤ u(x, ·) on (Ak )− . Lemma 6.5.4 If Γ is a compact subset of the Greenian set Ω, then lim

|x−y|→0,x,y∈Γ

GΩ (x, y) = 1. u(x, y)

6.5 Wiener’s Test

263

Proof: Since Γ is covered by ﬁnitely many components of Ω, it suﬃces to prove the assertion for the part of Γ in each of the ﬁnitely many components. Thus, it can be assumed that Ω is connected. Since GΩ (x, y) = u(x, y)−hx (y), where hx is a nonnegative harmonic function on Ω, GΩ (x, y)/u(x, y) = 1 − hx (y)/u(x, y) whenever |x − y| is suﬃciently small. It suﬃces to show that there is a constant m > 0 such that hx (y) ≤ m for all x, y ∈ Γ . By the minimum principle, each hx = 0 or hx > 0 on Ω. Consider x ∈ Γ for which the latter holds, any y ∈ Γ , and ﬁxed y0 ∈ Γ . By Harnack’s inequality, Theorem 2.2.2, there is a constant k > 0 such that hx (y) ≤ khx (y0 ). Since GΩ and u are symmetric functions, hx (y0 ) = hy0 (x) and hx (y) ≤ hy0 (x). Since hy0 is bounded on Γ , there is a constant m such that hx (y) ≤ m whenever x, y ∈ Γ with hx > 0 on Ω. If hx = 0 on Ω, this result is trivially true. Lemma 6.5.5 If Γ is a compact subset of the Greenian set Ω, there is a constant c > 0 such that GΩ (x, y) ≤ cu(x, y), x, y ∈ Γ . Proof: If x and y belong to diﬀerent components of Ω, GΩ (x, y) = 0 and the result is trivially true for any choice of c. Only the case with x, y in the same component need be considered. Since Γ is covered by ﬁnitely many components of Ω, it suﬃces to prove the result for the part of Γ in each of the components and then take c to be the maximum of a ﬁnite collection of constants. It therefore can be assumed that Ω is connected. The result follows from a simple compactness argument and the preceding lemma. Lemma 6.5.6 The series ∞ ˆ 1 (x) converges. R j=0

∞ j=0

αj C ∗ (Λj ) converges if and only if the series

Λj

ˆ1 Proof: Since Λ− j is compact, RΛj is a potential by (vii) of Theorem 4.3.5. ˆ 1 = GΩ μj , the measure μj is supported by clf Λj ⊂ Λ− by TheoLetting R Λj

j

rem 5.7.13. By Theorem 6.2.8, C ∗ (Λj ) = μj (Ω) and j ∗ j j 1 dμj α C (Λj ) = α μj (Ω) = α ≤ u(x, y) μj (dy) ≤ αj+1 μj (Ω) = αj+1 C ∗ (Λj ).

Thus,

αj C ∗ (Λj ) ≤ Uμj (x) ≤ αj+1 C ∗ (Λj ).

ˆ 1 (x) = GΩ μj (x) ≤ Uμj (x), the convergence of ∞ αj C ∗ (Λj ) imSince R Λj j=0 ∞ ˆ 1 (x). By Lemma 6.5.3, there is a constant plies the convergence of j=0 R Λj c > 0 such that u(x, ·) ≤ cGΩ (x, ·) on (Ak )− . Thus, ˆ 1Λ (x) αj C ∗ (Λj ) ≤ Uμj (x) ≤ cGΩ μj (x) = cR j

264

6 Energy

so that convergence of the series ∞ series j=0 αj C ∗ (Λj ).

∞ j=0

ˆ 1 (x) implies convergence of the R Λj

The following proof of Wiener’s theorem simpliﬁes early versions and can be found in [17]. Proof of Theorem 6.5.1: Suppose ﬁrst that Theorem 4.7.6, for j ≥ k ˆ 1∪m Λ (x) ≤ R i=j i

m

ˆ 1Λ (x) ≤ R i

i=j

∞

∞ j=0

ˆ 1 (x) < +∞. By R Λj

ˆ 1Λ (x). R i

i=j

ˆ 1 j (x) as m → ∞ by Theorem 4.6.5, Since the term on the left increases to R Λ ˆ 1 j (x) ≤ R Λ

∞

ˆ 1Λ (x). R i

i=j

ˆ 1 j (x) = 0. Since Since the latter term has the limit 0 as j → ∞, limj→∞ R Λ 1 1 j ˆ ˆ RΛj (x) = RΛj ∪{x} (x) by Corollary 4.6.4, Λ ∪ {x} is a neighborhood of x, ˆ1j and R Λ ∪{x} (x) < 1 for large j, Λ is thin at x according to Theorem 5.7.11. Assume now that Λ is thin at x. It can be assumed that x ∈ ∂Λ ∼ Λ for the following reasons. Since Λ is thin at x, x is not a ﬁne limit point of Λ and there is a ﬁne neighborhood Of of x such that Of ∩ Λ = ∅, which follow that means that x ∈ Λ; if it were true that x ∈∼ Λ− , it would then ∞ Λj = ∅ for large j so that C ∗ (Λj ) = 0 for large j and the series j=0 C ∗ (Λj ), ∞ ˆ 1 (x), would converge trivially. For each j, let {Γji } be equivalently j=0 R Λj ˆ1 ↑ R ˆ1 an increasing sequence of compact sets such that Γji ↑ Λj . Since R Γji Λj by Theorem 4.6.5, for each j ≥ k there is a compact set Γj ⊂ Λj such that ˆ 1 (x) > R ˆ 1 (x) converges, ˆ 1 (x) − (1/2j ). To show that the series ∞ R R Γj Λj Λj j=0 ∞ ∞ ˆ 1 (x) and ˆ1 it suﬃces to show that the two series R Γ2j j=k j=k RΓ2j+1 (x) converge. As indicated at the beginning of this section, only the ﬁrst series will be shown to be convergent, the argument for the second series being the same. Let Γ = ∪∞ j=k Γ2j . If the intersection of Γ with some neighborhood of ∞ ˆ 1 (x) x is polar, then at most a ﬁnite number of terms of the series j=0 R Γ2j are nonzero and the series converges. It therefore can be assumed that the intersection of Γ with each neighborhood of x is nonpolar. Since Γ ⊂ Λ and Λ is thin at x, Γ is thin at x. By Theorem 5.6.4, there is a potential u such ˆ u = Ru = u quasi that u(x) < +∞ and limy→x,y∈Γ ∼{x} u(y) = +∞. Since R Γ Γ u ˆ everywhere on Γ , RΓ has the limit +∞ at x on Γ except possibly for a polar ˆ u (x) ≤ u(x) < +∞, R ˆ u is ﬁnite at x. Since R ˆ u ≤ u and u is a set. Since R Γ Γ Γ ˆ u is a potential with R ˆ u = GΩ μ for some measure μ supported potential, R Γ Γ by the compact set Γ ∪ {x}. For j ≥ k, let

6.5 Wiener’s Test

265

μ2j = μ|Γ2j , μj = μ|Ω∼Γ2j so that μ = μ2j + μ2j . By Lemmas 6.5.2, 6.5.3, and 6.5.5 there is a constant c > 0, not depending upon j, such that GΩ (z, y) ≤ cGΩ (x, y),

y ∈ Γ ∼ Γ2j , z ∈ Γ2j , j ≥ k.

Using the fact that this inequality is trivially true when y = x, for j ≥ k ˆ uΓ (x) ≤ cu(x) < +∞ GΩ μ2j ≤ cGΩ μ2j (x) ≤ cGΩ μ(x) = cR

on Γ2j .

ˆ u = GΩ μ = GΩ μΓ2j + GΩ μ quasi everywhere on Γ, GΩ μΓ2j ≥ Since u = R 2j Γ u − cu(x) quasi everywhere on Γ, GΩ μΓ2j ≥ 1 quasi everywhere on Γ2j for suﬃciently large j ≥ k. Thus, ˆ 1Γ ≤ 1 ≤ GΩ μ2j R 2j ˆ 1 is a potential quasi everywhere on Γ2j for suﬃciently large j ≥ k. Since R Γ2j ˆ 1 ≤ GΩ μ2j on Ω by Theorem 6.3.5. of a measure with support in Γ2j , R Γ2j Thus, ˆ Γ2j (x) ≤ ˆ Γ (x) ≤ u(x) < +∞ R GΩ μ2j (x) ≤ GΩ μ(x) ≤ R j≥i

j≥i

for suﬃciently large i, proving that the series

∞ j=k

ˆ 1 (x) converges. R Γ2j

Example 6.5.7 (Zaremba’s Cone Condition) Let Λ be a subset of Rn , n ≥ 3, let x ∈ ∂Λ, and let K be a closed cone with vertex at x, nonempty interior, and K ⊂ Λ. Zaremba’s cone condition in Theorem 2.6.29 implies that x is a regular boundary point of ∼ Λ. Wiener’s theorem can be used to prove Λ is not thin at x as follows. Fix α > 1 and consider the series ∞ that j α C(Λ j ). Since Λj = Λ ∩Aj ⊃ Aj ∩K, it suﬃces to show that the series j=0 ∞ j n α C(A j ∩ K) diverges. Consider any orthogonal transformation φ of R j=0 ˆ φ(A ∩K) (y) = R ˆ Aj ∩K (φ−1 y) and using the equation about x. By symmetry, R j for calculating C(φ(Aj ∩ K)) in Example 4.4.2, C(φ(Aj ∩ K)) = C(Aj ∩ K). By the compactness of Aj , there a ﬁnite number of such transformations (j) (j) (j) φi , . . . , φm such that Aj ⊂ ∪m i=1 φi (Aj ∩ K). By Lemma 4.4.20, C(Aj ) ≤ m (j) i=1 C(φi (Aj ∩ K)) = mC(Aj ∩ K). As calculated in Example 4.4.4, j

C(Aj ) = α− n−2 . Therefore, ∞ j=0

∞

αj C(Aj ∩ K) ≥

∞

1 j 1 n−3 α C(Aj ) = α n−2 j = +∞. m j=0 m j=0

Chapter 7

Interpolation and Monotonicity

7.1 Introduction In order to take advantage of known facts about the Dirichlet and Neumann problems, it is necessary to study transformations from one region onto another and their eﬀects on smoothness properties of transformed functions. In later chapters, it will be shown that the solution of the Neumann problem for a spherical chip with a speciﬁed normal derivative on the ﬂat part of the boundary can be morphed into a solution of an elliptic equation with an oblique derivative boundary condition. The method for accomplishing this is called the continuity method and requires showing that the mapping from known solutions to potential solutions is a bounded transformation. The essential problem here is to establish inequalities known as apriori inequalities which are based on the assumption that there are solutions. The passage from spherical chips to regions Ω with curved boundaries requires relating the norms of functions and their transforms. The ﬁnal step involves the adaptation of the Perron-Wiener method and the extraction of convergent subsequences in certain Banach spaces called H¨ older spaces. Eventually, very strong conditions will have to be imposed on Ω. So strong, in fact, that the reader might reasonably conclude that only a spherical chip, or some topological equivalent, will satisfy all the conditions. As the ultimate application will involve spherical chips, the reader might beneﬁt from assuming that Ω is a spherical chip, at least up to the point of passing to regions with curved boundaries . But as some of the inequalities require less stringent conditions on Ω, the conditions on Ω will be spelled out at the beginning of each section. Since these conditions will not be repeated in the formal statements, it is important to refer back to the beginning of each section for a description of Ω.

L.L. Helms, Potential Theory, Universitext, c Springer-Verlag London Limited 2009

267

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7 Interpolation and Monotonicity

7.2 H¨ older Spaces As solutions of elliptic equations require at least that second-order partial derivatives exist and be continuous, the approximation methods used in proving the existence of solutions must involve slightly stronger conditions. It is not necessary to go so far as requiring the existence of third partials as some kind of smoothness between continuous second partials and the existence of third partials will suﬃce. Collections of such functions constitute the H¨ older spaces. The deﬁnitions of this section apply to any subset Ω of Rn . Deﬁnition 7.2.1 If k is a nonnegative integer, 0 < α ≤ 1, and u ∈ C k (Ω), let [u]k+0,Ω =

sup

|Dβ u(x)|,

x∈Ω,|β|=k

|u|k+0,Ω = u k,Ω =

k

[u]j+0,Ω ,

j=0

[u]k+α,Ω =

|Dβ u(x) − Dβ u(y)| , |x − y|α x,y∈Ω,|β|=k sup

|u|k+α,Ω = u k,Ω + [u]k+α,Ω . For k ≥ 1, note that |u|k+0,Ω is not the same as |u|k,Ω = |u|(k−1)+1,Ω . The collection of functions u on Ω for which |u|k+α,Ω < +∞ will be denoted by Hk+α (Ω). For u ∈ Hk+α (Ω) it is clear from the deﬁnition of [u]k+α,Ω that Dβ u, |β| = k, is uniformly continuous on Ω and has a continuous extension to Ω − which will be denoted by the same symbol. It is easily seen that the Hk+α (Ω) spaces are normed linear spaces with norm |u|k+α,Ω . These spaces are called H¨ older spaces. In general, [u]k+α,Ω is only a seminorm. The collection of functions u on Ω such that u ∈ Hk+α (Γ ) for all compact subsets Γ of Ω will be denoted by Hk+α,loc (Ω). In the remainder of this section, Ω will denote a bounded open subset of Rn . Granted that C 0 (Ω − ) is a Banach space, it will be shown that C k (Ω − ) is a Banach space for every k ≥ 1. Theorem 7.2.2 C k (Ω − ), k ≥ 1, is a Banach space under the norm u k,Ω . k − Proof: Let {uj }∞ j=1 be a Cauchy sequence in C (Ω ). Since

Dβ ui − Dβ uj 0,Ω ≤ ui − uj k,Ω 0 − for each β with |β| ≤ k, the sequence {Dβ uj }∞ j=1 is Cauchy in C (Ω ), and β 0 − β β thus there is a v ∈ C (Ω ) such that D uj → v as j → ∞ uniformly on

7.2 H¨ older Spaces

269

Ω − . If it can be shown that v β = Dβ u for |β| ≤ k where uj → u = v 0 as j → ∞ uniformly on Ω − , it would then follow that

Dβ uj − Dβ u 0,Ω → 0 as j → ∞

uj − u k,Ω = |β|≤k

and that uj → u in C k (Ω − ). This statement will be proved by a ﬁnite induction argument on |β| using Theorem 0.2.2. Suppose m is a multi-index with all 0 s except for a 1 in the mth position, |β + m | ≤ k, and Dβ uj → Dβ u as j → ∞ uniformly on Ω − . Since D m (Dβ uj ) = Dβ+ m uj → v β+ m as j → ∞ uniformly on Ω − , v β+ m = D m (Dβ u) = Dβ+ m u. Thus, uj → u in C k (Ω − ). Theorem 7.2.3 Hk+α (Ω − ) is a Banach space under the norm |u|k+α,Ω − . − Proof: If {uj }∞ j=1 is a Cauchy sequence in Hk+α (Ω ), then it is a Cauchy k − k − sequence in C (Ω ) and there is a u ∈ C (Ω ) such that uj − u k,Ω − → 0 as j → ∞. Thus, it suﬃces to show that [Dβ uj − Dβ u]α,Ω − → 0 as j → ∞ for each β with |β| = k. If |β| = k and > 0, there is an N ( ) ≥ 1 such that

[Dβ ui − Dβ uj ]α,Ω − = sup

x,y∈Ω −

x =y

|(Dβ ui (x) − Dβ uj (x)) − (Dβ ui (y) − Dβ uj (y))| |x − y|α

< for i, j ≥ N ( ). Thus, |(Dβ ui (x) − Dβ uj (x)) − (Dβ ui (y) − Dβ uj (y))| ≤ |x − y|α for x, y ∈ Ω − , i, j ≥ N ( ). Letting i → ∞, |(Dβ u(x) − Dβ uj (x)) − (Dβ u(y) − Dβ uj (y))| ≤ |x − y|α for x, y ∈ Ω − , j ≥ N ( ). Since the sequence {[Dβ uj ]α,Ω } is bounded in R, say by M > 0, it follows from this inequality that [Dβ u]α,Ω − ≤ [Dβ uj ]α,Ω − + ≤ M + , j ≥ N ( ), that u ∈ Hk+α (Ω − ), and that [Dβ u − Dβ uj ]α,Ω − ≤

for j ≥ N ( ).

Thus, [Dβ u − Dβ uj ]α,Ω − → 0 as j → ∞ and uj → u in Hk+α (Ω − ). A useful tool for proving the existence of solutions of an equation is to construct a sequence of functions in H2+α (Ω) and then extracting a convergent subsequence to approximate the solution. The principle embodied in the following theorem will be called the subsequence selection principle. The subsequence selection principle also applies to sequences in Hk+α (Ω), k ≥ 1. The theorem will be stated and proved for k = 2. To simplify the notation, D and D2 will serve as generic symbols for ﬁrst- and second-order partial derivatives, respectively.

270

7 Interpolation and Monotonicity

Theorem 7.2.4 If Ω is a convex, bounded, open subset of Rn and {uj } is a sequence in H2+α (Ω), 0 < α ≤ 1 for which there is an M > 0 such that |uj |2+α,Ω ≤ M, j ≥ 1, then the uj , Duj , and D2 uj have continuous extensions to Ω − and there is a function u on Ω − and a subsequence {ujk } of the {uj } sequence such that the sequences {ujk }, {Dujk }, and {D2 ujk } converge uniformly on Ω − to continuous functions u, Du, and D2 u, respectively. Proof: Note ﬁrst that the inequality |D2 uj (x) − D2 uj (y)| ≤ [uj ]2+α ≤ M, |x − y|α

x, y ∈ Ω

implies that the D2 uj are uniformly continuous on Ω and have continuous extensions to Ω − which will be denoted by the same symbol. The inequality also implies that the sequence {D2 uj } is uniformly equicontinuous on Ω − . Since |D2 uj |0,Ω = [uj ]2+0,Ω ≤ M, j ≥ 1, the sequence {D2 uj } is uniformly bounded on Ω − . It follows from the Arzel´a-Ascoli Theorem, Theorem 0.2.3, that there is a subsequence of the {D2 uj } sequence, which can be assumed to be the sequence itself by a change of notation if necessary, which converges uniformly on Ω − to a continuous function v (2) on Ω − . By the mean value theorem of the calculus, there is a zj on the line segment joining x to y such that √ √ |Duj (x) − Duj (y)| = |∇Duj (zj )| ≤ n[uj ]2+0,Ω ≤ nM. |x − y| This shows that the Duj are uniformly continuous on Ω, have continuous extensions to Ω − which will be denoted by the same symbol, and that the {Duj } are uniformly bounded and equicontinuous on Ω − . Again it follows from the Arzel´a-Ascoli Theorem that there is a subsequence of the {Duj } sequence, which can be assumed to be the sequence itself, which converges uniformly on Ω − to a continuous function v (1) on Ω − . Applying the same argument to the {uj } sequence, there is a subsequence of the {uj } sequence, which can be assumed to be the sequence itself, which converges uniformly on Ω − to a continuous function u on Ω − . By Theorem 0.2.2, Du = v (1) = limj→∞ Duj and D2 u = v (2) = limj→∞ D2 uj on Ω. Note that Du and D2 u have continuous extensions to Ω − . The roles played by a function u on Ω and a function g on a portion Σ of the boundary of Ω are distinctly diﬀerent and the norm of the latter will be deﬁned diﬀerently based on the idea of considering g to be the restriction of a function u to Σ as discussed in [25]. The following deﬁnition and subsequent discussion is also applicable if stated with the 1+α replaced by k+α for k ≥ 1. Deﬁnition 7.2.5 For g : Σ → R, let |g|1+α,Σ = inf {|u|1+α,Ω ; u ∈ H1+α (Ω), u|Σ = g}}, and let H1+α (Σ) be the set of all g for which |g|1+α,Σ < +∞.

7.2 H¨ older Spaces

271

According to the following theorem, it can be assumed that a boundary function g in H1+α (Σ) has an extension g˜ on Ω of the same norm. Theorem 7.2.6 If Ω is a convex, bounded, open subset of Rn , Σ is a relatively open subset of ∂Ω, and g : Σ → R with |g|1+α,Σ < +∞, then there is a u ∈ H1+α (Ω) such that u|Σ = g and |u|1+α,Ω = |g|1+α,Σ . Proof: Let {un } be a sequence in H1+α (Ω) for which |un |1+α,Ω ↓ |g|1+α,Σ and un |Σ = g. Since the sequence {|un |1+α,Ω } is bounded, by the subsequence selection principle there is a subsequence {unj } and a function u on Ω − such that the sequences {unj } and {Dunj } converge uniformly on Ω − to u and Du, respectively, as j → ∞. Since |unj |0,Ω − → |u|0,Ω − , |Dunj |0,Ω − → |Du|0Ω − , and lim inf j→∞ |Dunj |α,Ω − ≤ |g|1+α,Σ −|u|0,Ω − −|Du|0,Ω − , for any x, y ∈ Ω − lim inf j→∞

|Dunj (x) − Dunj (y)| ≤ lim inf [Dunj ]α,Ω − j→∞ |x − y|α ≤ |g|1+α,Σ − |u|0,Ω − − |Du|0,Ω − .

Since Dunj (x) → Du(x) and Dunj (y) → Du(y) as j → ∞, [Du]α,Ω − ≤ |g|1+α,Σ − |u|0,Ω − − |Du|0,Ω − so that |u|1+α,Ω − ≤ |g|1+α,Σ . Since u belongs to the class deﬁning |g|1+α,Σ , the two are equal. The proof of the following theorem is an easy exercise. Theorem 7.2.7 H1+α (Σ) is a normed linear space under the norm |g|1+α,Σ . The norms |u|k+α,Ω on Hk+α (Ω) are global in that the points of Ω are treated equally. An alternative is to deemphasize points of Ω near the boundary by using a weight function. Deﬁnition 7.2.8 If φ(x) is a positive, real-valued function on Ω, k is a nonnegative integer, φ(x, y) = min (φ(x), φ(y)), u ∈ C k (Ω − ), 0 < α ≤ 1, b ∈ R, and k + α + b ≥ 0, let (b)

[u; φ]k+0,Ω = b) |u; φ|k+0,Ω

sup

φk+b (x)|Dβ u(x)|

x∈Ω,|β|=k

= u; φ

(b)

k,Ω =

k

(b)

[u; φ]j+0,Ω

j=0 (b)

[u; φ]k+α,Ω = (b)

sup

φk+α+b (x, y)

x,y∈Ω,|β|=k (b)

(b)

|Dβ u(x) − Dβ u(y)| |x − y|α

|u; φ|k+α,Ω = | u; φ |k+0,Ω +[u; φ]k+α,Ω (0)

|u; φ|k+α,Ω = |u; φ|k+α,Ω .

272

7 Interpolation and Monotonicity (b)

(b)

(b)

(b)

H¨older classes H0 (Ω; φ), Hα (Ω; φ), Hk+0 (Ω; φ), and Hk+α (Ω; φ) are deﬁned as above and are easily seen to be normed linear spaces. These classes will be of particular interest for (i) φ(x) = c, a constant, (ii) φ(x) = dx = d(x) = dist(x, ∂Ω). (b)

(b)

(b)

(b)

When φ(x) = 1, [u; 1]k+0,Ω , u; 1 k,Ω , [u; 1]k+α,Ω , and |u; 1|k+α,Ω are the same as [u]k+0,Ω , u k,Ω , [u]k+α,Ω , and |u|k+α,Ω , respectively, as deﬁned (0) above. The superscript (b) is dropped when b = 0; that is, |u; φ|k+α,Ω is (b)

written simply |u; φ|k+α . Whenever the notation H2+α (Ω; φ) is used, it is understood that 2 + b + α ≥ 0. (b)

(b)

Theorem 7.2.9 Hk+α (Ω; d) is a Banach space under the norm |u; d|k+α,Ω . Proof: If δ ∈ (0, 1), let Ωδ = {y ∈ Ω; d(y) > δ}. There is then a constant C such that (b) (7.1) C|u|k+α,Ωδ ≤ |u; d|k+α,Ω where C = min {dj+b (Ω), dk+α+b (Ω), δ j+b , δ k+α+b , j = 0, . . . , k}, where d(Ω) is the diameter of Ω. Let {uj } be a Cauchy sequence in (b) Hk+α (Ω; d). By the above inequality, the sequence of restrictions of the uj to Ωδ− is Cauchy in Hk+α (Ωδ ) and has a limit u. The function u has an obvious (b) extension to Ω which is independent of δ. Since the sequence {|uj ; d|k+α,Ω } (b)

is Cauchy in R, there is a constant M such that |uj ; d|k+α,Ω ≤ M for all j ≥ 1. If β is a multi-index with |β| = ≤ k, then d+b (x)|Dβ uj (x)| ≤ M,

x ∈ Ω, j ≥ 1. (b)

Letting j → ∞, d+b (x)|Dβ u(x)| ≤ M so that [u; d]+0,Ω ≤ M . Similarly, (b)

(b)

[u; d]k+α,Ω ≤ M , proving that u ∈ Hk+α (Ω; d). If > 0 and |β| = ≤ k, there is an m such that (b)

d+b (x)|Dβ ui (x) − Dβ uj (x)| ≤ |(ui − uj ); d|k+α,Ω <

, k+1

i, j ≥ m.

Letting j → ∞ and taking the supremum over x ∈ Ω and β with |β| = , (b)

|ui − u; d|+0,Ω <

, k+1

i ≥ m;

(b)

(b)

Similarly, [ui − u; d]k+α,Ω < /(k + 1) and it follows that |ui − u; d|k+α,Ω < (b)

for i ≥ m; that is, the sequence {ui } converges to u in Hk+α (Ω; d).

7.3 Global Interpolation

273

7.3 Global Interpolation Given norms | · |s , | · |u , and | · |t on a normed linear space X with s < u < t, interpolation inequalities compare | · |u to some function of | · |s and | · |t , usually a linear function of the latter of the form |x|u ≤ a|x|s + b|x|t ,

x ∈ X.

Such inequalities can be derived from certain convexity properties of norms due to H¨ ormander. It will be assumed throughout this section that Ω is a bounded, convex, open subset of Rn . Since only norms of functions on Ω will be considered in this section, the subscript Ω will be omitted from the norm notation; that is, |u|k+α,Ω will appear as |u|k+α . Lemma 7.3.1 If k is a positive integer and k < b = m + σ, 0 < σ ≤ 1, then there is a constant C = C(b, d(Ω)) = nm/2 (max (d(Ω)σ , d(Ω), 1))m such that |u|k ≤ C|u|b . Proof: Let y0 be a ﬁxed point of Ω, let yτ v(y) = Dτ u(y0 ), τ! |τ |=m

and let w = u − v. Consider any multi-index β with |β| = m. Then Dβ v(y) = Dβ u(y0 ), y ∈ Ω, and Dβ w(y) = Dβ u(y)− Dβ v(y) = Dβ u(y)− Dβ u(y0 ). Note that Dβ w(y0 ) = 0 and that [w]b = [u]b since [w]b = sup [Dβ w]σ = sup [Dβ u − Dβ u(y0 )]σ |β|=m

|β|=m

= sup [Dβ u]σ = [u]b .

(7.2)

|β|=m

Now consider any multi-index β with |β| = m − 1. For 1 ≤ j ≤ n, let j be the vector having 1 as its jth component and 0 as its other components. By the mean value theorem, there is a point z on the line segment joining x to y such that |Dβ w(x) − Dβ w(y)| |x − y| = |∇Dβ w(z)| = |(Dβ+ 1 w(z), . . . , Dβ+ n w(z)| = Dβ+ 1 w(z) − Dβ+ 1 w(y0 ), . . . , Dβ+ n w(z) − Dβ+ n w(y0 ) n 12 |Dβ+ i w(z) − Dβ+ i w(y0 )| 2 2σ = |z − y0 | |z − y0 |σ i=1 √ √ ≤ nd(Ω)σ [w]m+σ = nd(Ω)σ [w]b .

274

7 Interpolation and Monotonicity

√ √ This shows that [w]m = [w](m−1)+1 ≤ nd(Ω)σ [w]b ≤ nd(Ω)σ [u]b by Equation (7.2). Consider [u]m = [w + v]m ≤ [w]m + [v]m and the last term in particular. By the mean value theorem, [v]m = =

sup

sup

|τ |=m−1 x,y∈Ω x =y

sup

|Dτ v(x) − Dτ v(y)| |x − y|

sup |∇Dτ v(z)|

|τ |=m−1 x,y∈Ω x =y

where z is a point on the line segment joining x to y. Since |∇Dτ v(z)| = |(Dτ + 1 v(z), . . . , Dτ + n v(z)| = |(Dτ + 1 u(y0 ), . . . , Dτ + n u(y0 )|, |∇Dτ v(z)| ≤

√

n[u]m+0 so that [v]m ≤

√ n[u]m+0 .

Therefore, [u]m ≤ [w]m + [v]m √ √ ≤ nd(Ω)σ [u]b + n[u]m+0 √ ≤ n max (d(Ω)σ , 1)([u]m+0 + [u]b ) where

√ n max (d(Ω)σ , 1) ≥ 1. Thus,

|u|m = |u|(m−1)+1 = u m−1 + [u]m ≤

m−1

[u]j+0 +

√ n max (d(Ω)σ , 1)([u]m+0 + [u]b )

j=0

√ ≤ n max (d(Ω)σ , 1)|u|b . Replacing b by m and m by m − 1, √ |u|m−1 ≤ n max (d(Ω), 1)|u|m ≤ n(max (d(Ω)σ , d(Ω), 1))2 |u|b . Repeating the last step m − k + 1 times, m−k+1 |u|k ≤ n(m−k+1)/2 max (d(Ω)σ , d(Ω), 1) |u|b . The conclusion follows with C = C(b, d(Ω)) = nm/2 (max (d(Ω)σ , d(Ω), 1))m . The growth of |u|a as a function of a will be considered now.

7.3 Global Interpolation

275

Theorem 7.3.2 If 0 ≤ a < b = m + σ, 0 < σ ≤ 1, then |u|a ≤ C|u|b with C = C(b, d(Ω)) = nm/2 max (1, d(Ω)σ , d(Ω)). Proof: Three cases will be considered according to a > m, a = m, a < m. Case (i) (a > m) In this case, a = m + ρ where 0 < ρ < 1, ρ < σ. Since [Dβ u]ρ ≤ d(Ω)σ−ρ [Dβ u]σ whenever |β| = m, [u]m+ρ ≤ d(Ω)σ−ρ [u]m+σ and |u|a = ≤

m j=0 m

[u]j+0 + [u]m+ρ [u]j+0 + d(Ω)σ−ρ [u]m+σ

j=0

≤ max (1, d(Ω)σ−ρ )|u|b . Since d(Ω)σ−ρ ≤ 1 if d(Ω) ≤ 1 and d(Ω)σ−ρ ≤ d(Ω)σ if d(Ω) > 1, |u|a ≤ max (1, d(Ω)σ )|u|b . In this case, the constant in the assertion can be taken as max (1, d(Ω))σ ). Case (ii) (a = m) This case is covered by the preceding lemma with the m constant nm/2 (max (1, d(Ω)σ , d(Ω)) . Case (iii) (a < m) Suppose a = + ρ < m < b = m + σ where 0 < ρ ≤ 1. Consider any multi-index β with |β| = . Since [Dβ u]ρ = sup x,y∈Ω

x =y

|Dβ u(x) − Dβ u(y)| |x − y|1−ρ |x − y|

≤ d(Ω)1−ρ sup |∇Dβ u(z)| x,y∈Ω

≤

√

x =y

1−ρ

nd(Ω)

[u]+1+0 ,

where z is a point on the line segment joining x to y, √ [u]+ρ ≤ nd(Ω)1−ρ [u]+1+0 and |u|a =

[u]j+0 + [u]+ρ

j=0

≤

[u]j+0 +

√ nd(Ω)1−ρ [u]+1+0

j=0

≤

+1 √ n max (1, d(Ω)1−ρ ) [u]j+0 . j=0

276

7 Interpolation and Monotonicity

Since + 1 ≤ m, the latter sum is less than or equal to |u|b . As above, d(Ω)1−ρ ≤ max (1, d(Ω)) and the constant of the assertion can be taken as √ n max (1, d(Ω)). The constant C(b, d(Ω)) = nm/2 max (1, d(Ω)σ , d(Ω)) will suﬃce for all three cases. A polynomial approximation is needed for the proof of the next lemma. Given a function f on a neighborhood U of the origin in Rn and a ﬁnite set Ξ ⊂ U , the polynomial p in n variables of degree k is said to interpolate f on Ξ if p = f on Ξ. The number of terms of such a polynomial is

k n+j−1 n+k , = j n j=0 the jth term of the sum being the number of terms having degree j (c.f., Feller [21], pp. 36, 62). It is shown in [14] that there is a subset Ξ = {ξ1 , . . . , ξM } of U having M = n+k points such that the polynomial n p(y) =

M

pj (y)f (ξj )

j=0

interpolates f on Ξ. The polynomial p is unique, the pj are of degree less than or equal to k and are products of linear factors, and the pj depend only upon n, k, Ξ and not on any f . Lemma 7.3.3 (H¨ omander [34]) Let 0 < a ≤ k < b where k is an integer. If [u]a ≤ 1, [u]b ≤ 1, and α is a multi-index with |α| = k, then there is a constant C = C(a, b)) and a polynomial function P of d(Ω) that increases with d(Ω) such that Dα u 0 ≤ C max (d(Ω)a , d(Ω)b )P (d(Ω)). Proof: Suppose the result is true when a ≤ 1. Consider any multi-index β with |β| = j. If a = j + σ, 0 < σ ≤ 1, then 0 < a − j ≤ k − j < b − j. Since [Dβ u]a−j ≤ [u]a ≤ 1, [Dβ u]b−j ≤ [u]b ≤ 1, and a − j ≤ 1, for any multi-index γ with |γ| = k − j, Dγ (Dβ u) 0 ≤ C. Thus, Dα u 0 ≤ C for any α with |α| = k. It therefore can be assumed that a ≤ 1. Since neither the hypothesis nor the conclusion is aﬀected by adding a constant to u, it can be assumed that there is a point x0 ∈ Ω such that u(x0 ) = 0. Since [u]a ≤ 1, |u(y)| = |u(y) − u(x0 | ≤ |y − x0 |a ≤ d(Ω)a and u 0 ≤ d(Ω)a . Let b = m + γ, 0 < γ ≤ 1. For each x ∈ Ω, consider the Taylor formula of order m − 1 with remainder u(y) =

m−1

Dα u(z) Dα u(x) (y − x)α + (y − x)α , α! α!

j=0 |α|=j

|α|=m

where z is a point on the line segment joining x to y. Letting Px (y) =

m Dα u(x) (y − x)α , α! j=0 |α|=j

y ∈ Ω,

y ∈ Ω,

7.3 Global Interpolation

277

Taylor’s formula can be written u(y) = Px (y) +

Dα u(z) − Dα u(x) (y − x)α . α!

|α|=m

Therefore, |u(y) − Px (y)| ≤

|Dα u(x) − Dα u(z)| |y − x||α| α!

|α|=m

≤ [u]b

1 |y − x|m+γ ≤ C(b)d(Ω)b . α!

|α|=m

Combining this inequality with the bound for u 0 , there is a constant C = C(a, b)) such that |Px (y)| ≤ C max (d(Ω)a , d(Ω)b ), y ∈ Ω. Let p be the unique polynomial of degree m that interpolates the function m Dα u(x) α z , α! j=0

z ∈ Rn

|α|=j

on the set Ξ = {ξ1 , . . . , ξM } discussed above. Since this function interpolates itself on Ξ and p is unique, Px (y) =

M

pj (y − x)Px (ξj ),

y ∈ Ω.

j=0

If α is any multi-index with |α| ≤ m, then α

D Px (y) =

M

Dα pj (y − x)Px (ξj )

j=0

so that |Dα Px (y)| ≤

M

|Dα pj (y − x)| |Px (ξj )|

j=0

≤ C max (d(Ω)a , d(Ω)b )

M

|Dα pj (y − x)|,

y ∈ Ω.

j=0

Note that the sum is bounded by a polynomial function P of d(Ω) that increases with d(Ω). Thus, there is a constant C = C(a, b) such that |Dα Px (y)| ≤ C max (d(Ω)a , d(Ω)b )P (d(Ω)) for |α| ≤ m, y ∈ Ω. Since Dα Px (x) = Dα u(x), the conclusion follows from the preceding inequality.

278

7 Interpolation and Monotonicity

If c is a positive integer, the expression [u]c in the statement of the following theorem can be interpreted as [u]c+0 or [u](c−1)+1 since [u](c−1)+1 ≤ C[u]c+0 . If the theorem is proved for [u]c+0 , then it is also true for [u](c−1)+1 . Theorem 7.3.4 (H¨ ormander [34]) If 0 ≤ a < c < b, 0 < λ < 1, and c = λa + (1 − λ)b, then [u]c ≤ C[u]λa [u]1−λ (7.3) b and

|u|c ≤ C|u|λa |u|1−λ . b

(7.4)

where C = C(a, b, d(Ω)) is a constant. Proof: If [u]b ≤ [u]a , then by Theorem 7.3.2 [u]c ≤ C[u]b = C[u]λb [u]1−λ ≤ C[u]λa [u]1−λ , b b where C = C(b, d(Ω)). It therefore can be assumed that [u]a < [u]b for the remainder of the proof, which will be dealt with in several steps. Step 1 Consider the case that there is no integer between a and b. Suppose ﬁrst that 0 ≤ a < c < b ≤ 1. Then for x, y ∈ Ω, |u(x) − u(y)| = |x − y|c

|u(x) − u(y)| |x − y|a

λ

|u(x) − u(y)| |x − y|b

1−λ

so that [u]c ≤ [u]λa [u]1−λ when a > 0 and [u]c ≤ 2λ [u]λa [u]1−λ when a = 0. In b b 1−λ λ λ either case, [u]c ≤ 2[u]a [u]1−λ ≤ 2|u| |u| . Suppose now that k≤a

2 sup |α|=k

1−λ λ [Dα u]λa−k [Dα u]1−λ ≤ 2|u|λa |u|1−λ . b−k ≤ 2[u]a [u]b b

Step 2 Suppose now that c is an integer k ≥ 1. It can be assumed that |u|a > 0 for otherwise u = 0, a trivial case that can be excluded. For x ∈ Ω and 0 < < 1, let u x (y) = u((1 − )x + y),

y ∈ Ω.

It is easily seen that [u x ]a ≤ a [u]a , [u x ]b ≤ b [u]b , and Dβ u x (x) = k Dβ u(x) if |β| = k. Since 0 < [u]a < [u]b , 0 < < 1 can be chosen so that b [u]b = a [u]a . Since |u x / a [u]a |a ≤ 1 and |u x / a [u]a |b ≤ 1, it follows from Lemma 7.3.3 that β u x (x) ≤ C(b, d(Ω)) if |β| = k, D a [u]a

7.3 Global Interpolation

279

where C(b, d(Ω)). Thus, k |Dβ u(x)|=|Dβ u x (x)| ≤ C(b, d(Ω)) a [u]a =C(b, d(Ω))( a [u]a )λ ( b [u]b )1−λ . Since c = k = λa + (1 − λ)b, |Dβ u(x)| ≤ C(b, d(Ω))[u]λa [u]1−λ . Taking the b supremum over x ∈ Ω and β satisfying |β| = k, [u]c ≤ C(b, d(Ω))[u]λa [u]1−λ ≤ C(b, d(Ω))|u|λa |u|1−λ . b b

(7.5)

Step 3 Let c(λ) = λa + (1 − λ)b, 0 ≤ λ ≤ 1, and assume that the inequality [u]c(λ) ≤ C(a, b, d(Ω))|u|λa |u|1−λ b

(7.6)

holds when λ = λ1 and λ = λ2 . If 0 < μ1 < 1, μ2 = 1 − μ1 , and λ = λ1 μ1 + λ2 μ2 , then c(λ) = μ1 c(λ1 ) + μ2 c(λ2 ) = μ1 c(λ1 ) + (1 − μ1 )c(λ2 ). By Step 1, if there is no integer between c(λ1 ) and c(λ2 ), then [u]c(λ) = [u]μ1 c(λ1 )+(1−μ1 )c(λ2 ) 1 1 ≤ 2[u]μc(λ [u]1−μ c(λ2 ) 1)

μ1 μ2 1−λ2 λ2 1 ≤ C(a, b, d(Ω)) ([u]λa 1 [u]1−λ [u] [u] a b b

≤ C(a, b, d(Ω))|u|λa |u|1−λ ; b that is, Inequality (7.6) holds for c(λ) for all λ between λ1 and λ2 providing there is no integer between c(λ1 ) and c(λ2 ). Note that the inequality holds for λ = 1 and λ = 0. Step 4 Consider any c ∈ (a, b). If there is no integer in (a, b), then [u]c ≤ 2|u|λa |u|1−λ by Step 1. If c is an integer in (a, b), then [u]c ≤ |u|λa |u|1−λ by Step b b 2. Suppose c is not an integer and there are one or more integers k1 , . . . , kt in (a, b) ordered so that a < k1 < · · · < kt < b. By considering the three cases c ∈ (a, k1 ), c ∈ (ki−1 , ki ), and c ∈ (kt , b) and the fact that Inequality (7.5) holds for each ki , Inequality (7.5) holds for each c ∈ (a, b). This completes the proof of Inequality (7.3). Step 5 Suppose a = j + α and c = + γ, then |u|c =

j i=0

[u]i +

[u]i + [u]c .

i=j+1

For i = 0, . . . , j, [u]i ≤ |u|i ≤ C(a, b, d(Ω))|u|a ≤ C(a, b, d(Ω))|u|λa |u|1−λ by b Theorem 7.3.2. For i = j + 1, . . . , , a ≤ i < b so that [u]i ≤ C(a, b, d(Ω))|u|λa |u|1−λ b by Step 2. Lastly, [u]c ≤ C(a, b, d(Ω))|u|λa |u|1−λ by the preceding step. b

280

7 Interpolation and Monotonicity

In applying Step 3 to the interval (ki−1 , ki ), C = C(ki−1 , ki , d(Ω)) can be replaced by C(a, b, d(Ω)) since the kj s can be expressed as explicit functions of a and b. Theorem 7.3.5 If 0 ≤ a < c < b and > 0, there is a constant C = C(a, b, c, , d(Ω)) such that [u]c ≤ C[u]a + [u]b

(7.7)

|u|c ≤ C|u|a + |u|b .

(7.8)

and Proof: Letting c = λa + (1 − λ)b, 0 < λ < 1, the result follows from Inequality (7.3) when [u]b = 0. It therefore can be assumed that [u]b = 0. It suﬃces to prove the result when [u]b = 1 for then ( u ) ( u ) ≤C + [u]b c [u]b a implies that [u]c ≤ C[u]a + [u]b . In the case |u|b = 1, [u]c ≤ C[u]λa by Theorem 7.3.4, where C = C(a, b, d(Ω)). It is easily seen that there is a linear function Kx + /C such that xλ ≤ Kx + /C for x ≥ 0 where K depends upon , c, and C; in fact, K=λ

(λ−1)/λ C(1 − λ)

where λ is determined from c. Thus, [u]c ≤ C[u]λa ≤ C(K[u]a + /C) = CK[u]a + , where CK may depend upon c through λ. The second assertion follows in exactly the same way.

7.4 Interpolation of Weighted Norms In this section, Ω can be any bounded open subset of Rn . Convexity is not required. The section applies to any weight function φ > 0 satisfying the inequality |φ(x) − φ(y)| ≤ M |x − y|α , x, y ∈ Ω, for some ﬁxed α ∈ (0, 1]. As the results will be applied only to the weight function φ(x) = d(x) = d(x, ∂Ω), the results of this section will be stated only for d. In the proof of the following lemma, Dj u will denote a generic symbol for a jth order partial derivative of u. The lemma is a special case of a more general result that can be found in [25] allowing arbitrary j, k ≥ 0 . As the consideration of many cases is required in the proof, only a few cases will be dealt with to illustrate techniques.

7.4 Interpolation of Weighted Norms

281

Lemma 7.4.1 If j + α < 2 + β where j = 0, 1, 2, 0 ≤ α, β ≤ 1, and u ∈ H2+β (Ω; d), then for any > 0 there is a constant C = C( , j, α, β) such that [u; d]j+α,Ω ≤ C u 0,Ω + [u; d]2+β,Ω

(7.9)

|u; d|j+α,Ω ≤ C u 0,Ω + [u; d]2+β,Ω

(7.10)

Proof: Since [u; d]0+0,Ω = u 0,Ω , Inequality (7.9) is trivially satisﬁed in any case with j = 0, α = 0. The cases (j = 1, α = 1, β = 0), (j = 2, 0 ≤ α < 1, β = 0), and (j = 2, α = 1) are excluded by the requirement that j + α < 2 + β. Let x, y be two points of Ω labeled so that d(x) < d(y), let 0 < μ ≤ 12 , let δ = μd(x) and B = Bx,δ . For any y ∈ B, d(y) ≥ d(x)/2. Consider the j = 1, α = 0, β = 0 case. For a ﬁxed i = 1, . . . , n, let x and x be the points where a diameter of B parallel to the xi -axis intersects ∂B. For some point x = (x1 , . . . , xi−1 , xi , xi+1 , . . . , xn ) on the diameter, |Dxi u(x)| =

1 |u(x ) − u(x )| ≤ u 0,Ω 2δ δ

and |Dxi u(x)| ≤ |Dxi u(x| +

xi

xi

Dxi xi u(x1 , . . . , xi−1 , yi , xi+1 , . . . , xn ) dyi

1 ≤ u 0,Ω + δ sup |Dxi xi u(y)| δ y∈B 1 ≤ u 0,Ω + δ sup d−2 (y)d2 (y)|Dxi xi u(y)|. δ y∈B Thus, d(x)|Dxi u(x)| ≤

d(x) 4δd(x)

u 0,Ω + 2 sup d2 (y)|Dxi xi u(y)| δ d (x) y∈B

≤ μ−1 u 0,Ω + 4μ[u; d]2+0,Ω . Thus, [u; d]1+0,Ω ≤ μ−1 u 0,Ω + 4μ[u; d]2+0,Ω . Given > 0, choose μ < min (1/2, /4). Then [u; d]1+0,Ω ≤ C( ) u 0,Ω + [u; d]2+0,Ω

(7.11)

where C( ) = 1/μ. This proves Inequality (7.9) in the j = 1, α = β = 0 case. Suppose now that j = 0, 1, α ∈ (0, 1), β = 0. For some point z on the line segment joining x to y ∈ B,

282

7 Interpolation and Monotonicity

dj+α (x, y)

|Dj u(x) − Dj u(y)| = dj+α (x, y)|x − y|1−α |∇(Dj u)(z)| |x − y|α ≤ dj+α (x)δ 1−α d−j−1 (z)dj+1 (z)|∇(Dj u)(z)| √ ≤ 2j+1 dα−1 (x)δ 1−α n[u; d]j+1+0,Ω √ = 2j+1 nμ1−α [u; d]j+1+0,Ω .

On the other hand, if y ∈ B, then dj+α (x, y)

|Dj u(x) − Dj u(y)| dj+α (x, y) ≤ (|Dj u(x)| + |Dj u(y)|) α |x − y| δα ≤ μ−α (dj (x)|Dj u(x)| + dj u(y)|Dj u(y)|) ≤ 2μ−α [u; d]j+0,Ω .

Combining these two cases and taking the supremum over x, y ∈ Ω, √ [u; d]j+α,Ω ≤ 2μ−α [u; d]j+0,Ω + 2j+1 nμ1−α [u; d]j+1+0,Ω . √ Given > 0, choose μ such that 2j+1 nμ1−α < . Then for j = 0, 1, α ∈ (0, 1) and β = 0, (7.12) [u; d]j+α,Ω ≤ C( )[u; d]j+0,Ω + [u; d]j+1+0,Ω where C( ) = C(n, α, j, ) = 2μ−α . Consider now the j = 1 or 2, α = 0, and 0 < β ≤ 1 case. Deﬁning x , x , x as above, Dj u(x) =

1 |Dj−1 u(x ) − Dj−1 u(x )| ≤ sup |Dj−1 u(y)|. 2δ δ y∈B

Therefore, |Dj u(x)| ≤ |Dj u(x)| + |Dj u(x) − Dj u(x)| 1 ≤ sup d−j+1 (y)dj−1 (y)|Dj−1 u(y)| δ y∈B + δ β sup d−j−β (x, y)dj+β (x, y) y∈B

≤

|Dj u(x) − Dj u(y)| |x − y|β

2j−1 μβ 2j+β [u; d]j−1+0,Ω + j [u; d]j+β,Ω j−1 δd (x) d (x)

so that dj u(x)|Dj u(x)| ≤ 2j−1 μ−1 [u; d]j−1+0,Ω + 2j+β μβ [u; d]j+β . If > 0, by taking the supremum over x ∈ Ω and choosing μ such that 2j+β μβ < , there is a constant C( ) = C(j, , β) such that [u; d]j+0,Ω ≤ C( )[u; d]j−1+0,Ω + [u; d]j+β,Ω

(7.13)

7.4 Interpolation of Weighted Norms

283

whenever j = 1 or 2, α = 0, and 0 < β ≤ 1. Inequalities (7.11), (7.12), and (7.13) are not quite enough to prove Inequality (7.9). Consider the j = 2, α = 0, 0 < β ≤ 1 case. Given > 0, choose 1 > 0 such that 2 1 < . There is then a constant C1 ( 1 ) satisfying Inequality (7.13). Choosing 2 > 0 such that 1 − C1 ( 1 ) 2 ≥ 1/2, there is a constant C2 ( 2 ) satisfying Inequality (7.11). Thus, [u; d]2+0,Ω ≤ C1 ( 1 )[u; d]1+0,Ω + 1 [u; d]2+β,Ω ≤ C1 ( 1 )(C2 ( 2 ) u 0,Ω + 2 [u; d]2+0,Ω ) + 1 [u; d]2+β,Ω from which it follows that C1 ( 1 )C2 ( 2 ) 1

u 0,Ω + [u; d]2+β,Ω 1 − C1 ( 1 ) 2 1 − C( 1 ) 2 ≤ C( ) u 0,Ω + [u; d]2+β,Ω

[u; d]2+0,Ω ≤

(7.14)

where C( ) is the coeﬃcient of u 0 in the ﬁrst line. By successive applications of Inequalities (7.11), (7.12), (7.13), and (7.14), the following cases of Inequality (7.9) can be established: (i) j = 0, 1, 2 and α = 0 (ii) j = 0, 1 and α ∈ (0, 1). Cases (j = 0, α = 1), (j = 1, α = 1, 0 < β ≤ 1), and (j = 2, α ∈ (0, 1), 0 < β ≤ 1) require separate arguments. Consider ﬁrst the j = 0, α = 1 case. As in the argument leading to Inequality (7.12), √ [u; d]0+1,Ω ≤ 2μ−1 u 0,Ω + 2 n[u; d]1+0,Ω . Now apply Inequality (7.11) to the second term and then Inequality (7.14). Consider now the j = 1, α = 1, 0 < β ≤ 1 case. Using the fact that d(x, y) = d(x) ≤ 2d(z ) for any z ∈ B, for some point z on the line segment joining x and y, d(x, y)2

|Du(x) − Du(y)| |Du(x) − Du(y)| ≤ d(x, y)2 χB (y) |x − y| |x − y| |Du(x)| + |Du(y)| χ∼B (y) + d(x, y)2 δ 2d(x) [u]1+0,Ω ≤ d(x, y)2 |∇Du(z)|χB (y) + δ 2 ≤ 4[u; d]2+0,Ω + [u]1+0,Ω . μ

Choose μ = 12 and apply Inequalities (7.11) and (7.14) to the latter two terms. Lastly, consider the j = 2, α ∈ (0, 1) case. If y ∈ B, then

284

7 Interpolation and Monotonicity

d2+α (x, y)

|D2 u(x) − D2 u(y)| |x − y|α

= d2+α (x)|x − y|β−α d−2−β (x, y)d2+β (x, y)

|D2 u(x) − D2 u(y)| |x − y|β

≤ dα−β (x)δ β−α [u; d]2+β,Ω = μβ−α [u; d]2+β,Ω . On the other hand, if y ∈ B, then |D2 u(x) − D2 u(y)| |x − y|α 2+α (x, y) −2 d ≤ d (x, y)(d2 (x)|D2 u(x)| + d2 (y)|D2 u(y)|) ≤ 2μ−α [u; d]2+0,Ω . δα

d2+α (x, y)

Applying Inequality (7.14) to [u; d]2+0,Ω with = μ2α , [u; d]2+α,Ω ≤ μβ−α [u; d]2+β,Ω + 2μ−α (C1 (μ2α ) u 0 + μ2α [u; d]2+β,Ω ) = 2C1 (μ2α )μ−α u 0,Ω + (μβ−α + 2μα )[u; d]2+β,Ω . Given > 0, choose μ so that μβ−α + μα < to obtain Inequality (7.9). Lastly, Inequality (7.10) follows by applying (7.9) to each of the terms of j |u; d|j+α,Ω = i=0 [u; d]i+0,Ω + [u; d]2+α,Ω .

7.5 Inner Norms If Ω is a bounded open subset of Rn and Σ is a relatively open subset of ∂Ω * is deﬁned by the equation with ∂Ω ∼ Σ = ∅, d*x = d(x) * = dist(x, ∂Ω ∼ Σ), x ∈ Ω. d*x = d(x) * − Since ∂Ω ∼ Σ is closed, d* is positive and continuous on Ω; in fact, |d(x) * d(y)| ≤ |x − y| for x, y ∈ Ω. Recalling that a ∧ b denotes the minimum of * y) = d*xy = d*x ∧ d*y for x, y ∈ Ω. For the two real numbers a and b, let d(x, * * *δ is deﬁned for δ > 0, let Ωδ = {x ∈ Ω; d(x) > δ}, if nonempty. Thus, Ω * 0 < δ < sup {η > 0; Ωη = ∅} and is an open set. If Ω is a convex set, it does *δ = Ωδ and is *δ is convex; but if Σ = ∅, then Ω not necessarily follow that Ω *δ are convex. It will be assumed at the end of this section that Ω and the Ω bounded, convex, open subsets of Rn . In addition, it will be assumed also that Ω is stratiﬁable, a condition assuring that the annulus between *δ and Ω *2δ is of uniform width. Since the ﬁrst theorem does not require this Ω assumption, the deﬁnition of stratiﬁable appears after this theorem. All of these assumptions are fulﬁlled for a particular case of interest here; namely, Ω is a spherical chip and Σ is the interior of the ﬂat part of the boundary.

7.5 Inner Norms

285

Deﬁnition 7.5.1 If Ω is a bounded open subset of Rn , Σ is a relatively open subset of ∂Ω, 0 ≤ α ≤ 1, and k + b + α ≥ 0, let (b) *δ = ∅}, |u|k+α,Ω = sup {δ k+α+b |u|k+α,Ω* δ ; 0 < δ < 1, Ω

*δ = ∅}, [u]k+α,Ω = sup {δ k+α+b [u]k+α,Ω* δ ; 0 < δ < 1, Ω (b)

(b)

(b)

and let Hk+α (Ω) be the set of u such that |u|k+α,Ω < +∞. There is no restriction that Σ be a proper subset of ∂Ω in this deﬁnition. In *δ and Ωδ are taken as the same. the Σ = ∅ or Σ = ∂Ω cases, Ω Remark 7.5.2 The requirement that δ ∈ (0, 1)) in the deﬁnition is based on the fact that the behavior of |u|a,Ω* δ is of interest only for small δ. Any *δ = ∅ for all δ < δ0 could serve as well at the number δ0 ∈ (0, 1) for which Ω expense of increasing the bound by a factor that does not depend upon u; that is, if sup0<δ≤δ0 δ a+b |u|a,Ω* δ ≤ M , then *δ = ∅} ≤ sup {δ a+b |u|a,Ω* δ ; 0 < δ < 1, Ω

1 a+b +1 M. δ0

To see this, suppose the ﬁrst inequality is true. Consider those δ for which *δ = ∅ and let q be the smallest integer such that 1 ∧ d(Ω) ≤ δ0 ≤ δ < 1 and Ω *δ ⊂ Ω *δ/q , qδ0 . Since δ/q < (1)/q ≤ δ0 and Ω δ a+b q

|u|a,Ω* δ ≤

δ a+b q

|u|a,Ω* δ/q ≤ M

so that δ a+b |u|a,Ω* δ ≤ q a+b M, Since q <

1 δ0

δ0 ≤ δ < 1.

+ 1,

δ

a+b

|u|a,Ω* δ ≤

1 +1 δ0

a+b M,

δ0 ≤ δ < 1.

As this inequality is trivially true for 0 < δ < δ0 , sup {δ

a+b

*δ = |u|a,Ω* δ ; 0 < δ < 1, Ω ∅} ≤

1 +1 δ0

a+b M.

It is easily seen that for 0 < δ0 < 1, there is a constant C = C(δ0 ) such that 1 (b) *δ = ∅} ≤ C|u|(b) . |u| ≤ sup {δ a+b |u|a,Ω* δ ; 0 < δ < δ0 , Ω a,Ω C a,Ω

286

7 Interpolation and Monotonicity

Theorem 7.5.3 If k is a nonnegative integer, 0 < α ≤ 1, and k + b + α ≥ 0, (b) (b) then Hk+α (Ω) is a Banach space under the norm |u|k+α,Ω . (b)

Proof: It is easily checked that Hk+α (Ω) is a normed linear space. Let (b)

(b)

{un } be a Cauchy sequence in Hk+α (Ω). The sequence {|un |k+α,Ω } is then bounded by K, say. Let {δ(j)} be a sequence in (0, 1) decreasing to zero. (b) Since δ(j)k+b+α |un |k+α,Ω* δ(j) ≤ |un |k+α,Ω ≤ K, the sequence {|un |k+α,Ω* δ(j) } is bounded for each j ≥ 1. Using the subsequence selection principle, an in(0) duction argument, and letting un = un , the functions un , Dun , . . . , Dk un − * * − , and a have continuous extensions to Ωδ(j) , there is a function vj on Ω δ(j) (j)

(j−1)

(j)

(j)

subsequence {un } of the {un } sequence such that un → vj , Dun → − k (j) k * *− Dvj , . . . , D un → D vj on Ωδ(j) with vj = vj−1 on Ω δ(j−1) . Thus, there − * is a function v on Ω that agrees with vj on Ω , j ≥ 1. Consider the diδ(j)

(j)

agonal sequence {uj } that is a subsequence of all of the above sequences (j)

(j)

→ v, Duj

so that uj

(j)

→ Dv, . . . , Dk uj , → Dk v on Ω as j → ∞. It will (b)

(j) (b)

be shown now that v ∈ Hk+α (Ω). Note ﬁrst that |uj |k+α,Ω ≤ K for all *δ . Since δ k+b+α |u(j) (x)| * ≤ j ≥ 1. Consider any δ ∈ (0, 1) and x, y ∈ Ω j 0,Ωδ (j) (b) *δ , ≤ K < +∞, letting j → ∞, taking the supremum over x ∈ Ω |u | j

k+α,Ω

(b)

and then taking the supremum over δ, |v|0,Ω ≤ K < +∞. Similar arguments (j)

j)

(j)

(b)

apply to |Duj |0,Ω , . . . , |Dk uj |0,Ω , and [Dk uj ]α,Ω . Thus, |v|k+α < +∞ and (b)

(j)

(b)

v ∈ Hk+α (Ω). To show that uj → v in Hk+α (Ω), consider any δ ∈ (0, 1) *δ . Using the fact that the sequence is Cauchy, given > 0 there and x, y ∈ Ω (i) (j) (b) is an N ( ) ≥ 1 such that |ui − uj |k+α,Ω < /(k + 1) for all i, j ≥ N ( ). (i)

(j)

For any δ ∈ (0, 1), δ k+b+α |ui − uj |k+α,Ω* δ < /(k + 1). Expanding the (i)

(j)

norm according to its deﬁnition, the last term δ k+b+α [Dk ui − Dk uj ]αΩ* δ dominates (i)

δ

k+b+α

(j)

(i)

(j)

|Dk ui (x) − Dk uj (x) − Dk ui (y) − Dk uj (y)| . ≤ |x − y|α k+1

*δ , δ k+b+α [Dk u(i) − Letting j → ∞ and taking the supremum over x, y ∈ Ω i k D v]α,Ω* δ < /(k + 1). Similar arguments can be applied to the other (i)

terms, resulting in δ k+b+α |ui − v|k+α,Ω* δ < . Taking the supremum over (i)

(b)

δ ∈ (0, 1), |ui − v|k+α,Ω < for all i ≥ N ( ). This proves that the sequence (i)

(b)

{ui } converges to v ∈ Hk+α (Ω). Since the sequence {un } is Cauchy and a ()b)

subsequence converges to v ∈ Hk+α (Ω), the sequence {un } converges to v in ()b)

Hk+α (Ω).

7.5 Inner Norms

287

The above deﬁnition of inner norm cannot be applied to a boundary func* tion if Σ = ∂Ω, as an open set in its own right, because d(x) and Σδ are not deﬁned in this case. As in Deﬁnition 7.2.5, the inner norm of a boundary function can be based on the idea that g is the restriction of a function u on Ω − to Σ. Deﬁnition 7.5.4 For g : Σ → R, let (1+b)

(1+b)

(1+b)

|g|1+α,Σ = inf {|u|1+α,Ω ; u ∈ H1+α (Ω), u|Σ = g} (1+b)

(1+b)

and let H1+α (Σ) be the set of all g for which |g|1+α,Σ < +∞. (1+b)

Theorem 7.5.5 If Σ = ∅, 0 < α ≤ 1, 2 + b + α ≥ 0, H1+α (Σ) is a normed (1+b) linear space under the norm |g|1+α,Σ . (1+b)

Proof: As it is a routine exercise to show that |g|1+α,Σ is a seminorm, only the (1+b)

fact that |g|1+α,Σ = 0 implies that g = 0 on Σ will be proved. Assuming that (1+b)

(1+b)

(1+b)

|g|1+α,Σ = 0, there is a sequence {un } in H1+α (Ω) such that |un |1+α,Ω ↓ 0 (1+b)

as n → ∞. Since the sequence of norms is bounded, |un |1+α,Ω ≤ K, n ≥ 1, for some K > 0. For each δ ∈ (0, 1), δ 2+b+α |un |1+α,Ω* δ ≤ K, and therefore each sequence {|un |1+α,Ω* δ } is bounded. Let {δ(m)} be a sequence in (0, 1) that decreases to 0. For each m ≥ 1, {|un |1+α,Ω* δ(m) } is a bounded sequence in H1+α,Ω* δ(m) . As in the proof of Theorem 7.5.3 for each m ≥ 1, there is a (m) (m−1) *δ(m) subsequence {un } of the {un } sequence and a function vm on Ω (m) (m) * − and vm = vm−1 such that un → vm and Dun → Dvm uniformly on Ω δ(m) *δ(m−1) . Consider the diagonal sequence {u(m) on Ω m } which is a subsequence of all of the above sequences. It follows that there is a function v on Ω such (m) (m) that um → v and Dum → Dv on Ω as m → ∞. Given > 0, there is (1+b) an N ≥ 1 such that |un |1+α,Ω < for all n ≥ N . By looking at the terms (m) (1+b)

(1+b)

of |um |1+α,Ω pointwise and taking the limit as m → ∞, |v|1+α,Ω < for (1+b)

every > 0 so that |v|1+α,Ω = 0 and, in particular, that [v]0,Ω* δ = 0 for every (m)

δ ∈ (0, 1). Thus, v = 0 on Ω. Since the um therefore g = 0 on Σ.

= g on Σ, v = g on Σ, and

(1+b)

Theorem 7.5.6 If g : Σ → R and |g|1+α,Σ < +∞, then there is a u ∈

(1+b) H1+α (Ω)

such that u|Σ = g and

(1+b) |u|1+α,Ω (1+b)

(1+b)

= |g|1+α,Σ . (1+b)

(1+b)

Proof: Let {un } be a sequence in H1+α (Ω) for which |un |1+α,Ω ↓ |g|1+α,Σ (m)

and un |Σ = g. As in the preceding proof there is a subsequence {um } and

288

7 Interpolation and Monotonicity (m)

(m)

a function u on Ω such that um → u and Dum → Du on Ω as n → ∞. As (1+b) (1+b) (1+b) (1+b) before, |u|1+α,Ω ≤ |g|1+α,Σ and u|Σ = g. Since u ∈ H1+α (Ω), |u|1+α,Ω = (1+b)

|g|1+α,Σ .

7.6 Monotonicity Geometrically, a spherical chip Ω is the smaller of two regions determined by intersecting a solid ball with a hyperplane. It will be assumed in this section that a coordinate system is chosen so that Ω = {x ∈ Rn : |x − x0 | < ρ, xn > 0}, where x0 = 0 or x0 is a point on the negative xn -axis, and that Σ is a relatively open subset of ∂Ω ∩ {x : xn = 0}. When Ω is a ball in the statement of the theorem, Σ is the empty set in which case d* = d and *δ = Ωδ . Ω Theorem 7.6.1 If Ω is a spherical chip with Σ a relatively open subset of ∂Ω ∩ {x : xn = 0} or a ball in Rn with Σ = ∅, j, k = 1, 2, b > −1, 0 ≤ α, β ≤ 1, j + α < k + β, j + b + α ≥ 0, and > 0, then there is a constant C0 (b, α, β, , d(Ω)) such that (b) (b) [u]j+α,Ω ≤ C0 d*b u 0,Ω + [u]k+β,Ω

and (b) (b) |u|j+α,Ω ≤ C0 d*b u 0,Ω + [u]k+β,Ω .

Proof: Since δ < 1, |u|j+α,Ω* δ = u 0,Ω* δ +

j

[u]i+0,Ω* δ + [u]j+α,Ω* δ ,

i=1

*δ , and δ j−i+α < 1 for 1 ≤ i ≤ j, d* > δ on Ω δ j+b+α |u|j+α,Ω* δ ≤ δ j+b+α u 0,Ω* δ +

j

δ j+b+α [u]i+0,Ω* δ + δ j+b+α [u]j+α,Ω* δ

i=1

≤ d(Ω)j+α d*b u 0,Ω +

j

(b)

(b)

[u]i+0,Ω + [u]j+α,Ω

(7.15)

i=1

and the second assertion follows from the ﬁrst by repeated application of the ﬁrst to the terms on the right side. It also can be assumed that the right members of the above inequalities are ﬁnite, for otherwise there is nothing to prove. Consider the following four cases:

7.6 Monotonicity

289

(i) j = 1, k = 2, α = β = 0 (ii) j ≤ k, α = 0, β > 0 (iii) j ≤ k, α > 0, β = 0 (iv) j ≤ k, α > 0, β > 0. It will be shown ﬁrst that there is a constant C = C(b, α, β, ), not depending on Ω, such that (b) (b) (b) [u]2+0,Ω ≤ C[u]1+0,Ω + [u]2+β,Ω . (7.16) For this purpose, let x be a ﬁxed point of Ω, let μ ≤ 1/2 be a positive number that will be speciﬁed later, let δ = d*x /2, ρ = μδ, B = Bx,ρ , and B + = B ∩ {x : xn > 0}(B + = B if Ω is a ball). If y ∈ B + , then d*y ≥ *δ ; that is, B + ⊂ Ω *δ . d*x − |x − y| ≥ (3/4)d*x = (3/2)δ > δ so that y ∈ Ω + Similarly, if y ∈ B , then d*y ≤ d*x + |x − y| ≤ (5/4)d*x . Therefore, 3* 5 dx ≤ d*y ≤ d*x , 4 4 If r is any real number, then d*rx ≤ A(r)d*ry ,

y ∈ B+.

(7.17)

y ∈ B+

(7.18)

where A(r) = max ((4/5)r , (4/3)r ). For i = 1, . . . , n − 1, let xi , xi be the end points of the diameter of B parallel to the xi -axis; for i = n, let xn = x and xn = x + ρ n ; i.e., the latter is the point of intersection in Ω of ∂B and a line through x perpendicular to R0n . Moreover, x will denote a point on the line segment joining xi to xi resulting from an application of the mean value *δ is not required since the mean theorem. (Note that convexity of Ω or Ω value theorem is applied to the convex set B + .) Since 1 |Dj u(xi ) − Dj u(xi )| ≤ sup |Dj u(y)|, 2ρ ρ y∈B + and

i = 1, . . . , n − 1

2 |Dj u(xi ) − Dj u(xi )| ≤ sup |Dj u(y)|, ρ ρ y∈B +

i = n,

for some point x on the line segment joining xi to xi , |Dji u(x)| ≤

2 sup |Dj u(y)|, ρ y∈B +

i = 1, . . . , n

by the mean value theorem. Since |x − x| ≤ ρ = μδ, |Dji u(x)| ≤ |Dji u(x)| +

|Dji u(x) − Dji u(x)| |x − x|β |x − x|β

2 sup |Dj u(y)| + μβ δ β [u]2+β,Ω* δ μδ y∈B + 2 β β ≤ [u] * + μ δ [u]2+β,Ω *δ . μδ 1+0,Ωδ ≤

290

7 Interpolation and Monotonicity

Multiplying both sides by δ 2+b , 2 1+b δ [u]1+0,Ω* δ + μβ δ 2+b+β [u]2+β,Ω* δ μ 2 (b) (b) ≤ [u]1+0,Ω + μβ [u]2+β.Ω . μ

δ 2+b |Dji u(x)| ≤

* Since δ = d(x)/2, 3+b (b) (b) * 2+b |Dij u(x)| ≤ 2 [u]1+0,Ω + 22+b μβ [u]2+β.Ω . d(x) μ

*η , For x ∈ Ω η 2+b |Dij u(x)| ≤

23+b (b) (b) [u]1+0,Ω + 22+b μβ [u]2+β.Ω . μ

*η and then the supremum over 0 < η < 1, Taking the supremum over x ∈ Ω (b)

[u]2+0,Ω ≤

23+b (b) (b) [u]1+0,Ω + 22+b μβ [u]2+β,Ω . μ

Choosing μ such that 22+b μβ < , μ < 1/2 and letting C = 23+b /μ, the proof of Inequality (7.16) is complete. Case (i) j = 1, k = 2, α = β = 0. For some point x on the line segment joining xi to xi , |Di u(x| =

1 |u(xi ) − u(xi )| ≤ sup |u(y)|, 2ρ ρ y∈B +

and |Di u(x)| = so that

i = 1, . . . , n − 1

2 |u(xi ) − u(xi )| ≤ sup |u(y)|, ρ ρ y∈B +

|Di u(x)| = Di u(x) +

x

x

i=n

Dii u dxi

2 sup |u(y)| + ρ sup |Dii u(y)|. ≤ ρ y∈B + y∈B + Multiplying both sides by δ 1+b and using the fact that ρ = μδ ≤ d*x /4, δ 1+b |Di u(x)| ≤ ≤

21−b sup |d*b u(y)| + μδ 2+b [u]2+0,Ω* δ μ y∈B + x 21−b (b) A(b) d*b u 0,Ω + μ[u]2+0,Ω . μ

7.6 Monotonicity

291

*η , 0 < η < 1 Therefore, for all x ∈ Ω 4A(b) *b (b)

d u 0,Ω + μ21+b [u]2+0,Ω . d*1+b x |Di u(x)| ≤ μ *η and then the supremum over 0 < η < 1, Taking the supremum over x ∈ Ω choosing μ such that μ21+b < , and taking C = 4A(b)/μ, (b) (b) [u]1+0,Ω ≤ C d*b u 0,Ω + [u]2+0,Ω .

(7.19)

Case (ii) j ≤ k, α = 0, β > 0. Suppose ﬁrst that j = 1, k = 2. By Inequality (7.16), there is a constant C0 such that (b)

(b)

(b)

[u]2+0,Ω ≤ C0 [u]1+0,Ω + [u]2+β,Ω . Choose 1 > 0 such that 1 C0 < 1/2 and 1 < 1/2. Applying Inequality (7.19) with this choice of 1 < 1, there is a constant C1 such that (b) (b) [u]1+0,Ω ≤ C1 d*b u 0,Ω + 1 [u]2+0,Ω

≤ C1 d*b u 0,Ω + 1 C0 [u]1+0,Ω + 1 [u]2+β,Ω . (b)

(b)

Since 1 C0 < 1/2, the second term on the right can be transposed to the left resulting in (b) (b) [u]1+0,Ω ≤ 2C d*b u 0,Ω + [u]2+β,Ω (7.20) where C = 2C1 . Suppose now that j = 1 and k = 1. In this case, |Di u(x)| ≤ |Di u(x)| + ≤

|Di u(x) − Di u(x)| |x − x|β |x − x|β

2 sup |u(y)| + μβ δ β [u]1+β,Ω* δ . ρ y∈B +

Multiplying both sides by δ 1+b = (d*x /2)1+b , d*1+b 21−b (b) x sup |d*b u(y)| + μβ [u]1+β,Ω |D u(x)| ≤ i 21+b μ y∈B + x ≤

21−b A(b) *b (b)

d u 0,Ω + μβ [u]1+β,Ω . μ

*η Thus, for all x ∈ Ω 4A(b) *b (b)

d u 0,Ω + 21+b μβ [u]1+β,Ω . d*1+b x |Di u(x)| ≤ μ

292

7 Interpolation and Monotonicity

*η and then the supremum over 0 < η < 1, Taking the supremum over x ∈ Ω 1+b β choosing μ such that 2 μ < , μ < 1/2, and taking C = 4A(b)/μ, (b) (b) [u]1+0,Ω ≤ C d*b u 0,Ω + [u]1+β,Ω .

(7.21)

Suppose now that j = 2 and k = 2. By Inequality (7.16), there is a constant C0 such that (b) (b) (b) [u]2+0,Ω ≤ C0 [u]1+0,Ω + [u]2+β,Ω . 2 Choose 1 > 0 such that 1 C0 < /2. By Inequality (7.20), there is a constant C1 such that (b) (b) [u]1+0,Ω ≤ C1 d*b u 0,Ω + 1 [u]2+β,Ω . Thus, (b) (b) (b) [u]2+0,Ω ≤ C0 C1 d*b u 0,Ω + C0 1 [u]2+β,Ω + [u]2+β,Ω 2 (b) ≤ C d*b u 0,Ω + [u]2+β,Ω

(7.22)

where C = C0 C1 , completing the proof of Case (ii). Case (iii) j ≤ k, α > 0, β = 0. The hypothesis that j + α < k + β requires in this case that j = 1, k = 2, and α ∈ (0, 1). Fix δ > 0, let x, y be distinct *2δ , let μ < 1/2 be a constant to be speciﬁed later, let ρ = μδ, and points in Ω let B = Bx,ρ . If z ∈ B + , then d*z ≥ d*x − |x − z| > (2 − μ)δ ≥ 3δ/2 > δ, *δ . For some point x on the line segment joining x to y, showing that B + ⊂ Ω ρ1+b+α

|Di u(x) − Di u(y)| |Di u(x) − Di u(y)| |x − y|1−α χ|x−y|<ρ = ρ1+b+α |x − y|α |x − y| |Di u(x)| + |Di u(y)| + ρ1+b+α χ|x−y|≥ρ ρα ≤ ρ2+b |∇Di u(x)| + 2ρ1+b [u]1+0,Ω* 2δ .

Therefore, δ 1+b+α

√ |Di u(x) − Di u(y)| ≤ μ1−α nδ 2+b [u]2+0,Ω* δ + 2μ−α δ 1+b [u]1+0,Ω* 2δ α |x − y| √ (b) (b) ≤ nμ1−α [u]2+0,Ω + 2−b μ−α [u]1+0,Ω .

*2δ and then the supremum over 0 < δ < 1, Taking the supremum over x, y ∈ Ω (b)

[u]1+α,Ω ≤

√ 1−α (b) (b) nμ [u]2+0,Ω + 2−b μ−α [u]1+0,Ω .

Applying Inequality (7.19) with = μ, there is a constant C0 such that [u]1+0,Ω ≤ C0 d*b u 0,Ω + μ[u]2+0,Ω . (b)

(b)

7.6 Monotonicity

293

Therefore, (b)

[u]1+α,Ω ≤

√

(b) (b) nμ1−α [u]2+0,Ω + 2−b C0 μ−α d*b u 0,Ω + 2−b μ1−α [u]2+0,Ω .

√ Choosing μ such that ( n + 2−b )μ1−α < , μ < 1/2, and letting C = 2−b C0 μ−α , (b) (b) [u]1+α,Ω ≤ C d*b u 0,Ω + [u]2+0,Ω , (7.23) completing the proof of Case (iii). Case (iv) j ≤ k, α > 0, β > 0. The j = 1, k = 2 case follows by applying Inequality (7.23) and then Inequality (7.22). Only the j = k = 1, 2 cases need to be discussed in detail. Suppose ﬁrst that j = k = 1. Fix δ > 0, let x, y be *2δ , and let μ, ρ, B be deﬁned relative to x as above. Then distinct points in Ω ρ1+b+α

|Di u(x) − Di u(y)| |Di u(x) − Di u(y)| = ρ1+b+α |x − y|β−α χ|x−y|<ρ α |x − y| |x − y|β + ρ1+b (|Di u(x)| + |Di u(y)|)

so that (b)

(b)

(b)

μ1+b+α [u]1+α,Ω ≤ μ1+b+β [u]1+β,Ω + 2−b μ1+b [u]1+0,Ω . Therefore,

(b)

(b)

(b)

[u]1+α,Ω ≤ μβ−α [u]1+β,Ω + 2−b μ−α [u]1+0,Ω . Applying Inequality (7.21) with = μβ , there is a constant C0 such that [u]1+α,Ω ≤ μβ−α [u]1+β,Ω + 2−b C0 μ−α d*b u 0,Ω + 2−b μβ−α [u]1+β,Ω . (b)

(b)

(b)

Choosing μ so that (1 + 2−b )μβ−α < and taking C = 2−b C0 μ−α , [u]1+α,Ω ≤ C d*b u 0,Ω + [u]1+β,Ω . (b)

(b)

The j = k = 2 case is essentially the same. Corollary 7.6.2 If the conditions of the preceding theorem are satisﬁed and j = 1, 2, b > −1, 0 ≤ α ≤ 1, then there is a constant C0 = C0 (b, α, d(Ω)) such that (b) (b) . ≤ C0 d*b u 0,Ω + [u] |u| j+α,Ω

j+α,Ω

(b)

Proof: Apply Theorem 7.6.1 to the terms [u]i+0,Ω in Inequality (7.15). *δ are It will be assumed in the remainder of this section that Ω and the Ω bounded, convex, open subsets of Rn .

294

7 Interpolation and Monotonicity

Theorem 7.6.3 If 0 ≤ a1 < a2 , b1 ≤ b2 , a1 + b1 ≥ 0, 0 < λ < 1, and (a, b) = λ(a1 , b1 ) + (1 − λ)(a2 , b2 ), then (b ) λ (b2 ) 1−λ (b) |u|a,Ω ≤ C |u|a11,Ω |u|a2 ,Ω where C = C(a1 , a2 , b1 , b2 , d(Ω)). Proof: By Theorem 7.3.4, δ a+b |u|a,Ω* δ ≤ Cδ a+b |u|λa ,Ω* |u|1−λ *δ a2 ,Ω 1 δ λ 1−λ δ a2 +b2 |u|a2 ,Ω* δ = C δ a1 +b1 |u|a1 ,Ω* δ λ 1−λ (b ) (b ) ≤ C |u|a11,Ω . |u|a22,Ω The result follows by taking the supremum over δ ∈ (0, 1). In the course of the proof of the following and subsequent proofs, the symbol C will be used as a generic symbol for a constant that may very well change from line to line or even within a line. Lemma 7.6.4 If a > 1 and a + b ≥ 0, then there is a constant C = C(a, d(Ω)) such that (b) (b+1) (b+1) (b) |u|a,Ω ≤ C |Dα u|a−1,Ω and |Dα u|a−1,Ω ≤ C|u|a,Ω . (7.24) |α|≤1

|α|≤1

Proof: Suppose 1 ≤ k ≤ a < k + 1. Then |u|a,Ω* δ ≤ u 0,Ω* δ +

Dα u 0,Ω* δ + [Dα u]a−k,Ω* δ . 1≤|α|≤k

|α|=k

Recalling that j is a unit vector with 1 in the jth place, for 1 ≤ |α| ≤ k, α = α + j for some j with |α | ≤ k − 1. Thus,

|u|a,Ω* δ ≤ u 0,Ω* δ +

+

n

|α |≤k−1 j=1 n

|α |=k−1 j=1

≤ u 0,Ω* δ + C

Dα (D j u) 0,Ω* δ

[Dα (D j u)](a−1)−(k−1),Ω* δ

n

|D j u|a−1,Ω* δ .

j=1

Since u 0,Ω* δ ≤ |u|a−1,Ω* δ , |u|a,Ω* δ ≤ |u|a−1,Ω* δ + C

n j=1

|D j u|a−1,Ω* δ ≤ C

|α|≤1

|Dα u|a−1,Ω* δ .

7.6 Monotonicity

295

Multiplying both sides of this inequality by δ a+b and taking the supremum over δ ∈ (0, 1), the ﬁrst inequality of (7.24) is obtained. On the other hand, |Dα u|a−1,Ω* δ |α|≤1

= |u|a−1,Ω* δ +

|Dα u|a−1,Ω* δ

|α|=1

≤ |u|a−1,Ω* δ +

|α|=1

|β|≤k−1

+

β

α

[D (D u)](a−1)−(k−1),Ω* δ

|β|=k−1

= |u|a−1,Ω* δ +

Dβ (Dα u) 0,Ω* δ

1≤|β|≤k

Dβ u 0,Ω* δ +

[Dβ u]a−k,Ω* δ .

|β|=k

Replacing |u|a−1,Ω* δ by its deﬁnition,

|Dα u|a−1,Ω* δ ≤

|α|≤1

|β|≤k−1

Dβ u 0,Ω* δ +

+ ≤2

|β|≤k

+

D u 0,Ω* δ +

Dβ u 0,Ω* δ +

[Dβ u](a−1)−(k−1),Ω* δ

|β|=k−1

β

1≤|β|≤k

[Dβ u]a−k,Ω* δ

|β|=k

sup

* |β|=k−1 x,y∈Ωδ

|Dβ u(x) − Dβ u(y)| |x − y|a−k

[Dβ u]a−k,Ω* δ .

|β|=k

By the mean value theorem for derivatives and the fact that 1 − a + k ≥ 0, there is a constant K = K(a, d(Ω)) such that

|Dα u|a−1,Ω* δ

|α|≤1

≤2

|β|≤k

≤2

|β|≤k

≤C

Dβ u 0,Ω* δ +

|β|≤k

Dβ u 0,Ω* δ +

≤ C|u|a,Ω* δ .

∇Dβ u 0,Ω* δ |x − y|

* |β|=k−1 x,y∈Ωδ

Dβ u 0,Ω* δ + K

sup

|β|=k

|β|=k

Dβ u 0,Ω* δ

[Dβ u]a−k,Ω* δ

+

|x − y|a−k |β|=k β + [D u]a−k,Ω* δ

|β|=k

[Dβ u]a−k,Ω* δ

296

7 Interpolation and Monotonicity

For each α with |α| ≤ 1, |Dα u|a−1,Ω* δ ≤ C|u|a,Ω* δ . Multiplying both sides by (b+1)

(b)

δ a+b and taking the supremum over δ ∈ (0, 1), |Dα u|a−1,Ω ≤ C|u|a,Ω . By summing over |α| ≤ 1, the second inequality of (7.24) is obtained. *2δ0 = ∅ for Deﬁnition 7.6.5 If O is a bounded open subset of Rn with O some δ0 > 0, then O is stratiﬁable if there is a constant K > 0 with the *δ , there is a y ∈ O *2δ satisfying property that for 0 < δ ≤ δ0 and x ∈ O |x − y| ≤ Kδ. Example 7.6.6 Let ρ0 > 0, let y be any point in Rn with yn < 0 and n , and let Σ = By,ρ0 ∩ R0n . The latter is the |yn | < ρ0 , let Ω = By,ρ0 ∩ R+ n *δ = By,ρ0 −δ ∩ R+ for ﬂat face of a spherical chip. Taking δ0 = (ρ0 + yn )/4, Ω * 0 < δ ≤ δ0 . Note that each Ωδ is a bounded, convex, open subset of Ω. For *2δ with |x − y| ≤ Kδ where *δ , 0 < δ < δ0 , there is a y ∈ Ω each x ∈ Ω K=

ρ0 1

((ρ0 − 2δ0 )2 − yn2 ) 2

;

thus, spherical chips are stratiﬁable. (b)

Monotonicity of |u|a,Ω with respect to each of the parameters a and b will be considered now. It is easy to see that (b)

(b )

|u|a,Ω ≤ |u|a,Ω whenever b > b .

(7.25)

The following theorem and its proof are adapted from the paper cited below that deals with the Σ = ∅ case. Theorem 7.6.7 (Gilbarg and H¨ ormander[24]) If Ω is a stratiﬁable, *δ is convex, 0 ≤ a < a, a + b ≥ 0 bounded, convex open subset of R0n , each Ω and b is not a negative integer or 0, then there is a constant C = C(a, b, d(Ω)) such that (b) (b) (7.26) |u|a ,Ω ≤ C|u|a,Ω . Proof: Consider the a ≤ 1 case ﬁrst. Suppose it has been shown that (b)

(b)

|u|0,Ω ≤ C|u|a,Ω and

(b)

(b)

|u|−b,Ω ≤ C|u|a,Ω

if b > 0

(7.27)

if b < 0, a + b ≥ 0.

(7.28)

In the ﬁrst case, a = λ · 0 + (1 − λ)a for some λ ∈ (0, 1] and it follows from Theorem 7.6.3 that λ 1−λ (b) (b) (b) (b) |u|a ,Ω ≤ C |u|0,Ω ≤ C|u|a,Ω . |u|a,Ω

7.6 Monotonicity

297

In the second case, a = λ(−b) + (1 − λ)a for some λ ∈ (0, 1] and it follows from Theorem 7.6.3 that λ 1−λ (b) (b) (b) (b) |u|a,Ω |u|a ,Ω ≤ C0 |u|−b,Ω ≤ C|u|a,Ω . It therefore suﬃces to prove Inequalities (7.27) and (7.28) for the a ≤ 1 case. To prove the ﬁrst, assume that b > 0 and let δ0 and K be the constants (b) speciﬁed in the stratiﬁable deﬁnition. It can be assumed that |u|a,Ω is ﬁnite, for otherwise there is nothing to prove. For any δ ≤ δ0 , δ b+a sup

*δ x,y∈Ω

|u(x) − u(y)| (b) ≤ |u|a,Ω . |x − y|a

*δ , choose k ≥ 1 such Consider only those δ for which δ < δ0 /2. For any x ∈ Ω k *2δ such that that δ0 /2 < 2 δ ≤ δ0 . By the preceding lemma, there is a x1 ∈ Ω |x − x1 | ≤ Kδ. Thus, δ b+a

|u(x) − u(x1 )| (b) ≤ |u|a,Ω (Kδ)a

or (b)

|u(x) − u(x1 )| ≤ K a δ −b |u|a,Ω . Thus, |u(x)| ≤ |u(x1 )| + |u(x) − u(x1 )| (b)

≤ |u(x1 )| + K a δ −b |u|a,Ω . *2δ , there is a x2 ∈ Ω *4δ such that Applying the same argument to x1 ∈ Ω (b)

(b)

|u(x)| ≤ |u(x2 )| + K a (2δ)−b |u|a,Ω + K a δ −b |u|a,Ω . *2k δ ⊂ Continuing in this way, there is a ﬁnite sequence x1 . . . . , xk with xk ∈ Ω *δ /2 such that Ω 0 |u(x)| ≤ |u(xk )| + K a

k−1

(b) (2j δ)−b |u|a,Ω .

j=0

Since (δ0 /2)a+b |u|a,Ω* δ

0 /2

(b) *δ /2 , ≤ |u|a,Ω and xk ∈ Ω 0

|u(xk )| ≤ |u|a,Ω* δ

0 /2

≤

2a+b (b) |u|a,Ω . δ0a+b

298

7 Interpolation and Monotonicity

Therefore, 2a (b) (b) a −bj |u|a,Ω |u| + K 2 δ0a a,Ω j=0 ∞

δ b |u(x)| ≤ ≤

2a (b) (b) |u| + (2b K a /(2b − 1))|u|a,Ω δ0a a,Ω (b)

(7.29) (b)

for all x ∈ Ωδ , δ ≤ δ0 /2, and it follows that |u|0,Ω ≤ C|u|a,Ω where C = 2a /δ0a + (2b K a /(2b − 1)). This completes the proof of Inequality (7.27). *δ and let Turning to Inequality (7.28), assume that b < 0. Consider x, y ∈ Ω t be a ﬁxed positive number. Two cases will be considered. Case (i) (δ < t). In this case, choose k ≥ 1 such that t < 2k δ ≤ 2t. There *2δ , x2 ∈ Ω *4δ , . . . , xk ∈ Ω *2k δ such that |x − x1 | ≤ are then points x1 ∈ Ω k−1 Kδ, |x1 − x2 | ≤ K2δ, . . . , |xk−1 − xk | ≤ K2 δ. Letting x0 = x and xt = *2k δ ⊂ Ω *t , xk ∈ Ω |x − xt | ≤

k

|xi−1 − xi | ≤ Kδ

i=1

k

2i−1 ≤ Kδ2k .

i=1

Since a+b

|u(x ) − u(y )| (b) ≤ |u|a,Ω |x − y |a

* , 0 < < δ0 , for all x , y ∈ Ω

*t ⊂ Ω *δ , and xt ∈ Ω |u(x) − u(xt )| ≤

k

|u(xi−1 ) − u(xi )|

i=1 (b)

k

(b)

k

≤ |u|a,Ω

|xi−1 − xi |a (2i−1 δ)−a−b

i=1

≤ |u|a,Ω

(K2i−1 δ)a (2i−1 δ)−a−b

i=1

≤ K a t−b

−b k 1 2k−i i=1 (b)

≤ Ct−b |u|a,Ω . *t with |y − yt | ≤ Kδ2k such that *δ there is a yt ∈ Ω In the same way, for y ∈ Ω (b)

|u(y) − u(yt )| ≤ Ct−b |u|a,Ω .

7.6 Monotonicity

299

It follows that |u(x) − u(y)| ≤ |u(x) − u(xt )| + |u(xt ) − u(yt )| + |u(yt ) − u(y)| (b)

(b)

≤ 2Ct−b |u|a,Ω + |xt − yt |a (2k δ)−a−b |u|a,Ω (b)

(b)

≤ 2Ct−b |u|a,Ω + |xt − yt |a t−a−b |u|a,Ω . Since |xt −yt | ≤ 2Ct+|x−y|, putting t = |x−y|, |xt −yt | ≤ (2C +1)|x−y| and |u(x) − u(y)| (b) ≤ (2C + (2C + 1)a )|u|a,Ω , |x − y|−b

|x − y| > δ.

(7.30)

Case (ii) (δ ≥ t) In this case, (b)

|u(x) − u(y)| ≤ |x − y|a δ −a−b |u|a,Ω so that (b)

|u(x) − u(y)| ≤ |u|a,Ω |x − y|a

t a+b δ

(b)

t−a−b ≤ |u|a,Ω |x − y|a t−a−b .

Taking t = |x − y|, |u(x) − u(y)| (b) ≤ |u|a,Ω , |x − y|−b

|x − y| ≤ δ.

(7.31)

Combining this inequality with Inequality (7.30) and taking the supremum *δ , over x, y ∈ Ω (b)

(b)

|u|−b,Ω ≤ C|u|a,Ω . This completes the proof in the a ≤ 1 case. The a > 1 case will be dealt with by an induction argument. Assume that the lemma is true for a − 1 and b is neither a negative integer or 0. By Lemma 7.6.4, if a > 1, then (b)

|u|a ,Ω ≤ C

(b+1)

|Dα u|a −1,Ω ≤ C

|α|≤1

(b+1)

(b)

|Dα u|a−1,Ω ≤ C|u|a,Ω .

(7.32)

|α|≤1

The 0 ≤ a ≤ 1 < a case can be proved using the preceding results as follows. Note that (b)

(b)

|u|a ,Ω ≤ C|u|1,Ω by the “a ≤ 1” case. Suppose ﬁrst that 1 < a ≤ 2 so that a − 1 ≤ 1. Applying the mean value theorem,

300

7 Interpolation and Monotonicity

δ 1+b |u|1,Ω* δ = δ 1+b |u|0+1,Ω* δ = δ 1+b ( u 0,Ω* δ + [u]0+1,Ω* δ )

Dα u 0,Ω* δ ≤ δ 1+b u 0,Ω* δ + ≤

(1+b) |u|0,Ω

+

|α|=1

(1+b)

|Dα u|0,Ω .

|α|=1

Since a − 1 ≤ 1, by the above “a ≤ 1” case, (1+b)

|Dα u|0,Ω

(1+b)

(b)

≤ C|Dα u|a−1,Ω ≤ C sup δ a+b |u|a,Ω* δ = C|u|a,Ω ; δ>0

and by Theorem 7.3.2, (1+b)

|u|0,Ω

(1+b)

(b)

≤ |u|a−1,Ω = sup δ a+b |u|a−1,Ω* δ ≤ sup δ a+b |u|a,Ω* δ = |u|a,Ω . δ>0

Therefore,

(b)

δ>0

(b)

|u|1,Ω ≤ C|u|a,Ω , so that

(b)

(b)

(b)

|u|a ,Ω ≤ C|u|1,Ω ≤ C|u|a,Ω ,

1
The remaining cases are proved by induction using Inequality (7.32). Remark 7.6.8 The above proof does not apply when b = 0, as can be seen in the derivation of Inequality (7.29). The requirement that b not be a negative integer is exhibited in Inequality (7.32); for example, if b = −1, (−1) (0) |Dα u|a −1,Ω |u|a ,Ω = |α|≤1

and the induction process cannot be applied, since b = 0 in the terms in the sum. If a + b ≥ 0, a + b ≥ 0, b ≤ b, a ≥ a, by combining Inequalities (7.25) and (7.26) (b) (b ) |u|a,Ω ≤ C|u|a ,Ω (7.33) where C = C(a, b, a b , d(Ω)). We now will consider compactness properties (−b) of the Ha (Ω) spaces. Recall that b = m + σ with 0 < σ ≤ 1. Lemma 7.6.9 (Gilbarg and H¨ ormander [24]) If 0 < a < a, 0 < b < b, b ≤ a, b ≤ a , and neither b nor b is a nonnegative integer, then ev(−b) ery bounded sequence (uj ) in Ha (Ω) has a subsequence that converges in (−b ) Ha (Ω).

7.6 Monotonicity

301 (−b)

Proof: Suppose |uj |a,Ω ≤ M for all j ≥ 1. By Inequality (7.26), |uj |b,Ω = (−b)

(−b)

|uj |b,Ω ≤ C|uj |a,Ω ≤ M, j ≥ 1, from which it follows that each uj ∈ C m (Ω − ). It follows that the sequences {Dα uj }, 0 ≤ |α| ≤ m are bounded and equicontinuous on Ω − . By Theorems 0.2.2 and 0.2.3, there is a subsequence {ujk } and a function u such that Dα ujk − Dα u 0,Ω → 0 as k → ∞, |α| = 0, . . . , m. By letting k → ∞ in the deﬁnition of |ujk |b,Ω , |u|b,Ω ≤ M , and |ujk − u|b,Ω ≤ 2M, k ≥ 1. Consider any c ∈ (0, b). Then c = tb + (1 − t)0 for some 0 < t < 1. By Lemma 7.6.3, (−c)

|ujk − u|c,Ω = |ujk − u|c,Ω

(−b)

≤ C(|ujk − u|b,Ω )t (|ujk − u|0,Ω )1−t 1−t

≤ C(2M )t ( ujk − u 0,Ω )

and limk→∞ |ujk − u|c,Ω = 0. If a = b < b, this result is applicable with (−b ) c = a so that |ujk − u|a ,Ω → 0 as k → ∞. It therefore can be assumed that a > b . By comparing the slopes of the lines in R2 through (a, −b) and the points (0, 0), (a , −b ), the rest of the proof can be split into three cases. Case (i) a/b = a /b . In this case, the point (a , −b ) lies on the line joining (0, 0) to (a, −b) so that there is a 0 < t < 1 such that t 1−t (−b ) (−b) (0) |ujk − u|a ,Ω ≤ C |ujk − u|a,Ω |ujk − u|0,Ω 1−t

≤ C(2M )t ( ujk − u 0,Ω )

.

(−b )

Therefore, limk→∞ |ujk − u|a ,Ω = 0. Case (ii) a/b > a /b . In this case, the line through (a , −b ) and (a, −b) will intersect the line y = −x in a point (c, −c) with 0 < c < b. Thus, there is a 0 < t < 1 such that 1−t (−b ) (−c) |ujk − u|a ,Ω ≤ C(2M )t |ujk − u|c,Ω (−b )

and limk→∞ |ujk − u|a ,Ω = 0. Case (iii) a/b < a /b . In this case, choose c so that b < c < min (a , b). This is possible since b < a and b < b. Let (c, −d) be the point on the line through (a , −b ) and (a, −b). Since −d > −c, for some 0 < t < 1, 1−t (−b ) (−d) |ujk − u|a ,Ω ≤ C(2M )t |ujk − u|c,Ω 1−t (−c) ≤ C(2M )t |ujk − u|c,Ω (−b )

by Inequality (7.6.3). Since 0 < c < b, limk→∞ |ujk − u|a ,Ω = 0.

Chapter 8

Newtonian Potential

8.1 Introduction If Ω is an open subset of Rn , a function k : Ω × Ω → R is called a kernel and can be used to transform a measurable function g : Ω → R into a new function gˆ : Ω → R by putting gˆ(x) = k(x, y)g(y) dy, x ∈ Ω, Ω

provided the integral is deﬁned. Properties of k and g will be examined now that will ensure that gˆ has some desirable property; e.g., that gˆ is continuous on Ω. The most obvious criterion for continuity of gˆ is that g be integrable on Ω, that k be bounded on Ω × Ω, and that k(·, y) is continuous on Ω for each y ∈ Ω; for if {xn } is an arbitrary sequence in Ω with limit x, then lim gˆ(xn ) = lim k(xn , y)g(y) dy = k(x, y)g(y) dy = gˆ(x) n→∞

n→∞

Ω

Ω

by the Lebesgue dominated convergence theorem. If the integrability condition on g is dropped, additional conditions on k must be imposed. Rather than stating some preliminary results speciﬁc to the Newtonian, the results will be formulated in a more general context for use in later chapters.

8.2 Subnewtonian Kernels Throughout this section, Ω will be an open subset of Rn , Σ will be a relatively open subset of ∂Ω, and k will be a kernel on (Ω ∪ Σ) × (Ω ∪ Σ) that is continuous in the extended sense on (Ω ∪ Σ) × (Ω ∪ Σ) and continuous oﬀ the diagonal of (Ω ∪ Σ) × (Ω ∪ Σ); in addition, it will be assumed that for each L.L. Helms, Potential Theory, Universitext, c Springer-Verlag London Limited 2009

303

304

8 Newtonian Potential

y ∈ Ω the partial derivatives Dxi k(x, y), Dxi xj k(x, y), and Dxi xj xk k(x, y) are continuous on (Ω∪Σ) ∼ {y} and continuous in the extended sense on (Ω∪Σ), and have the following properties: |k(x, y)| ≤ C|x − y|−n+2 , |Dxi k(x, y)| ≤ C|x − y|−n+1 , −n

|Dxi xj k(x, y)| ≤ C|x − y|

,

−n−1

|Dxi xj xk k(x, y)| ≤ C|x − y|

,

1 ≤ i ≤ n,

(8.1) (8.2)

1 ≤ i, j ≤ n,

(8.3)

1 ≤ i, j, k ≤ n

(8.4)

for all x ∈ Ω ∪ Σ, y ∈ Ω, where C is a constant depending upon parameters n, d(Ω), etc. Such a kernel will be called a subnewtonian kernel. If Ω is a bounded open subset of R2 , the logarithmic potential − log |x − y| is a subnewtonian potential. See Section 8.3 for details. Throughout this section k(x, y) will denote a subnewtonian kernel. Most of the results of this section are adapted from [25]. Example 8.2.1 Consider α satisfying 2 < α < n and a bounded open region Ω. The function k(x, y) = |x−y|−n+α is called a Riesz potential of order α. It is easily seen that such potential functions are subnewtonian. An extensive treatment of Riesz potentials can be found in [36]. Lemma 8.2.2 If > 0 and |β| = 0, 1, then Cσn 2−|β| β , |D(x) k(x, y)| dy ≤ 2 − |β| Ω∩Bx, moreover,

lim

→0

x ∈ Ω ∪ Σ;

Ω∩Bx,

β |D(x) k(x, y)| dy = 0

uniformly for x ∈ Ω ∪ Σ.

Proof: The second assertion follows from the ﬁrst. By Inequalities (8.1) and (8.2), Ω∩Bx,

β |D(x) k(x, y)| dy

≤C Bx,

1 Cσn 2−|β| . dy = 2 − |β| |x − y|n−2+|β|

Lemma 8.2.3 If Γ is a compact subset of Ω and |β| = 0, 1 , then β |D(x) k(x, y)| dy < +∞ Γ

for all x ∈ Ω ∪ Σ. Proof: Fix x ∈ Ω ∪ Σ. For δ > 0, β β |D(x) k(x, y)| dy = |D(x) k(x, y)| dy + Γ

Γ ∩Bx,δ

Γ ∼Bx,δ

β |D(x) k(x, y)| dy.

8.2 Subnewtonian Kernels

305

By the preceding lemma, δ > 0 can be chosen so that the ﬁrst term is less β k(x, ·) is continuous and therefore bounded on the compact than 1. Since D(x) set {y; y ∈ Γ, |x − y| ≥ δ}, the second term is ﬁnite. Deﬁnition 8.2.4 An operator K corresponding to the kernel k is deﬁned as follows. If f is a measurable function on Ω and x ∈ Ω ∪ Σ, let k(x, y)f (y) dy Kf (x) = Ω

if ﬁnite. If Σ is a subset of the boundary of the open set Ω, C k (Ω ∪ Σ) will denote the set of functions on Ω whose derivatives of order less than or equal to k are continuous on Ω and have continuous extensions to Ω ∪ Σ. Theorem 8.2.5 If f is a bounded and integrable function on Ω, then Kf ∈ Cb0 (Ω ∪ Σ). Proof: It will be shown ﬁrst that Kf (x) is deﬁned for all x ∈ Ω ∪ Σ. Fix x ∈ Ω ∪ Σ and > 0. Then 1 |k(x, y)||f (y)| dy + C |f (y)| dy |Kf (x)| ≤ n−2 Ω∩Bx, Ω∼Bx, |x − y| C ≤ (sup |f |) |k(x, y)| dy + n−2 |f (y)| dy. Ω Ω∩Bx, Ω∼Bx, The ﬁrst integral is ﬁnite by Lemma 8.2.3 and the second is ﬁnite by the integrability of f . Now consider a function ψ ∈ C 2 (R) satisfying 0 ≤ ψ ≤ 1, 0 ≤ ψ ≤ 2, ψ(t) = 0 for t ≤ 1, and ψ(t) = 1 for t ≥ 2. For each > 0 and x ∈ Ω ∪ Σ, let k(x, y)ψ(|x − y|/ )f (y) dy. K f (x) = Ω

Since f is integrable on Ω, by Inequality (8.1) |K f (x)| ≤ |k(x, y)|ψ(|x − y|/ )|f (y)| dy Ω∼Bx,

≤C Ω∼Bx,

≤

C n−2

1 |f (y)| dy |x − y|n−2

(8.5)

|f (y)| dy < +∞ Ω

and K f (x) is deﬁned for all x ∈ Ω ∪ Σ. Let {xj } be a sequence in Ω ∪ Σ such that xj → x ∈ Ω ∪ Σ. Since |k(x, y)ψ(|x − y|/ )f (y)| ≤

C |f (y)|, n−2

y ∈ Ω ∼ Bx, ,

306

8 Newtonian Potential

and the latter function is integrable on Ω, limj→∞ K f (xj ) = K f (x) by the Lebesgue dominated convergence theorem. This shows that K f ∈ Cb0 (Ω∪Σ). By Lemma 8.2.2, |Kf (x) − K f (x)| ≤ |k(x, y)(1 − ψ(|x − y|/ ))f (y)| dy Ω ≤ (sup |f |) |k(x, y)| dy → 0 Ω

Ω∩Bx,2

uniformly for x ∈ Ω ∪ Σ as → 0. Thus, Kf is continuous on Ω ∪ Σ. As to the second assertion, by what has just been proved there is an 0 > 0 such that |Kf (x)| ≤ 1 + |K 0 f (x)| and the latter term is bounded uniformly for x ∈ Ω ∪ Σ by Inequality (8.5). Remark 8.2.6 Only Inequalities (8.1) and (8.2) are used in the proof of the following theorem. Theorem 8.2.7 If f is bounded and integrable on Ω, then Kf ∈ C 1 (Ω ∪ Σ), Dxi k(x, y)f (y) dy for all x ∈ Ω ∪ Σ, Dxi Kf (x) = Ω

and [Kf ]1+0,Ω < +∞. Proof: Fix 1 ≤ i ≤ n. Deﬁne g : Ω ∪ Σ → R by the equation Dxi k(x, y)f (y) dy x ∈ Ω ∪ Σ. g(x) = Ω

Let ψ be the function described in the preceding proof. For each > 0 and x ∈ Ω ∪ Σ, let g (x) = Dxi k(x, y) ψ(|x − y|/ )f (y) dy. Ω

Since | Dxi k(x, y) ψ(|x − y|/ )f (y)| ≤

C C ψ(|x − y|/ )|f (y)| ≤ n−1 |f (y)| |x − y|n−1

and the latter function is integrable on Ω, g is continuous on Ω ∪ Σ by the Lebesgue dominated convergence theorem. By Lemma 8.2.2, |g(x) − g (x)| = Dxi k(x, y) 1 − ψ(|x − y|/ ) f (y) dy Ω ≤ sup |f | |Dxi k(x, y)| dy → 0 Ω

Ω∩Bx,2

8.2 Subnewtonian Kernels

307

as → 0 uniformly for x ∈ Ω ∪ Σ. It follows that g is continuous on Ω ∪ Σ. Letting k(x, y)ψ(|x − y|/ )f (y) dy

K f (x) = Ω

and using the fact that f is integrable on Ω, |K f (x)| = k(x, y)ψ(|x − y|/ )f (y) dy Ω∼Bx, C ≤ |f (y)| dy |x − y|n−2 Ω∼Bx, C ≤ n−2 |f (y)| dy < +∞ Ω and K f is deﬁned for all x ∈ Ω ∪ Σ. It follows from Lemma 8.2.2 that |Kf (x) − K f (x)| ≤ |k(x, y)(1 − ψ(|x − y|/ ))f (y) dy Ω ≤ sup |f | |k(x, y)| dy → 0 Ω

Ω∩Bx,2

uniformly for x ∈ Ω ∪ Σ as → 0. Thus, lim →0 K f = Kf uniformly on Ω ∪ Σ. For ﬁxed > 0 and x, y ∈ Ω, 1 xi − yi Dxi k(x, y)ψ(|x − y|/ ) = k(x, y)ψ (|x − y|/ ) |x − y| + Dxi k(x, y) ψ(|x − y|/ ).

(8.6)

Since ψ (|x − y|/ ) = 0 in Bx, and |k(x, y)| ≤ C/ n−2 outside Bx, , the ﬁrst term on the right is dominated by 2C/ n−1 ; the second term on the right is dominated by C/ n−1 . Thus 2C |Dxi k(x, y)ψ(|x − y|/ ) | ≤ n−1

(8.7)

uniformly for x, y ∈ Ω. Letting i denote a unit vector having 1 as its ith coordinate, K f (x + λ i ) − Kf (x) λ k(x + λ i , y)ψ(|x + λ i − y|/ ) − k(x, y)ψ(|x − y|/ ) f (y) dy = λ Ω = Dxi k(ξλ , y)ψ(|ξλ − y|/ ) f (y) dy Ω

308

8 Newtonian Potential

where ξλ is a point on the line segment joining x to x+λ i . Since the ﬁrst factor in the last integral is bounded uniformly for x, y ∈ Ω by Inequality (8.7) and f is integrable on Ω, Dxi k(x, y)ψ(|x − y|/ ) f (y) dy. Dxi K f (x) = Ω

It is clear from Equation (8.6) that the derivative in the latter integral is continuous and bounded, and therefore K f ∈ C 1 (Ω). Using the fact that ψ(|x − y|/ ) = 1 whenever |x − y|/ ≥ 2, Dxi k(x, y)(1 − ψ(|x − y|/ ) f (y) dy g(x) − Dxi K f (x) = Ω = Dxi k(x, y)(1 − ψ(|x − y|/ )) f (y) dy, x ∈ Ω. Ω∩Bx,2

By Inequality (8.2), C Dxi k(x, y)(1 − ψ(|x − y|/ )) ≤ 2 |k(x, y)| + , |x − y|n−1 It follows from Lemma 8.2.2 that

|g(x) − Dxi K f (x)| ≤ sup |f |

Ω

Ω∩Bx,2

x, y ∈ Ω.

2 C |k(x, y)| + dy |x − y|n−1

2 |k(x, y)| dy + 2Cσn ≤ sup |f | Ω∩Bx,2 Ω

→0

uniformly for x ∈ Ω as → 0. Since K1/n f converges uniformly to Kf on Ω and Dxi K1/n f converges uniformly to g on Ω, Dxi Kf is deﬁned and Dxi Kf = g on Ω by Theorem 0.2.2; that is, Dxi Kf (x) = g(x) = Dxi k(x, y)f (y) dy, x ∈ Ω. Ω

Since it was shown above that g is continuous on Ω ∪ Σ, Dxi Kf has a continuous extension to Ω ∪ Σ. For x ∈ Ω ∪ Σ and > 0, |Dxi Kf (x)| ≤ |Dxi k(x, y)| |f (y)| dy + |Dxi k(x, y)| |f (y)| dy Ω∩Bx,

Ω∼Bx,

≤ sup |f | Ω

|Dxi k(x, y)| dy + Ω∩Bx,

C n−1

|f (y)| dy. Ω

The ﬁrst term on the right can be made less than 1 uniformly for x ∈ Ω ∪ Σ for some choice = 0 by Lemma 8.2.2 and the second term is then ﬁnite for this choice of by the integrability of f . Taking the supremum over x ∈ Ω ∪Σ and then over 1 ≤ i ≤ n, it follows that [Kf ]1+0,Ω < +∞.

8.2 Subnewtonian Kernels

309

To conclude that Kf is twice diﬀerentiable, hypotheses stronger than those of the preceding theorem are required. The appropriate class of functions f required for stronger results is the class of H¨ older continuous functions. Deﬁnition 8.2.8 (i) If f is a real-valued function deﬁned on a bounded set Ω ⊂ Rn , x0 ∈ Ω, and 0 < α ≤ 1, then f is H¨ older continuous at x0 with exponent α if there is an M (x0 ) > 0 such that |f (x) − f (x0 )| ≤ M (x0 )|x − x0 |α ,

for all x ∈ Ω.

(ii) f is uniformly H¨ older continuous on Ω with exponent α if there is an M (Ω) > 0 such that |f (x) − f (y)| ≤ M (Ω)|x − y|α ,

for all x, y ∈ Ω.

(iii) f is locally H¨ older continuous on Ω with exponent α, 0 < α ≤ 1, if f is uniformly H¨ older continuous with exponent α on each compact subset of Ω. If α = 1 in (i), then f is customarily said to be Lipschitz continuous at x0 . The class of functions satisfying (ii) and (iii) are the H¨ older spaces Hα (Ω) and Hα,loc (Ω), respectively, deﬁned in Section 7.2. Let Ω be a bounded open subset of Rn with a boundary made up of a ﬁnite number of smooth surfaces. Such a set will be said to have a piecewise smooth boundary. If { 1 , . . . , n } is a vector base for Rn , the inner product of j and n(x) at x ∈ ∂Ω will be denoted by γj (x) = j · n(x) whenever n(x) is deﬁned. The equation Dyj u(y) dy = u(y)γj (y) dσ(y), u ∈ C 2 (Ω − ), (8.8) Ω

∂Ω

is usually proved, as in [43], as a preliminary version of the Divergence Theorem. In the course of the proof of the following theorem, reference will be made to subsets of Ω of the form Ωδ = {x ∈ Ω; d(x, ∼ Ω) > δ} where δ > 0. Moreover, the following fact will be used in the proof. If g is a diﬀerentiable function on (0, ∞), then Dxi g(|x − y|) = −Dyi g(|x − y|),

(8.9)

Theorem 8.2.9 Let f be bounded and locally H¨ older continuous with exponent α, 0 < α ≤ 1, on the bounded open set Ω, and let Ω0 be any bounded open set containing Ω − with a piecewise smooth boundary. If k(x, y) = k(|x − y|) satisﬁes Inequalities (8.1) through (8.4)on Ω0− × Ω0− with Σ = ∂Ω0 and w(x) = Kf (x) = k(|x − y|)f (y) dy, Ω

310

8 Newtonian Potential

then w ∈ C 2 (Ω) and

Dxi xj k(|x − y|) f (y) − f (x) dy Ω0 Dxi k(|x − y|)γj (y) dσ(y), −f (x)

Dxi xj w(x) =

x ∈ Ω, (8.10)

∂Ω0

where f is extended to be 0 on Ω0 ∼ Ω. Proof: It suﬃces to show that w ∈ C 2 (Ωδ ) for every δ > 0 with Ωδ = ∅. Fix δ > 0 and consider any > 0 for which 4 < δ. Let |f (x)| ≤ M for all x ∈ Ω and |f (y) − f (x)| ≤ M |x − y|α for all x, y ∈ Ωδ/2 . Fix x ∈ Ωδ . Then Bx,2 ⊂ Ωδ/2 and by Inequality (8.3),

Dxi xj k(|x − y|) f (y) − f (x) dy

Ω0

Dxi xj k(|x − y|)f (y) − f (x) dy

≤ Bx,

+ Ω0 ∼Bx,

Dxi xj k(|x − y|)f (y) − f (x) dy

1 |x − y|α dy |x − y|n

≤ CM Bx,

+2CM Ω0 ∼Bx,

1 dy |x − y|n

rα−1 dθ dr +

= CM 0

|θ|=1

2CM μn (Ω0 ) n

2CM α = CM σn + n μn (Ω0 ) < +∞ α Since x ∈ Ωδ ⊂ Ω − ⊂ Ω0 , by Inequality (8.2),

∂Ω0

Dxi k(|x − y|)γj (y) dσ(y) ≤ C

∂Ω0

≤

C δ n−1

1 dσ(y) |x − y|n−1

σ(∂Ω0 ) < +∞.

Deﬁne ui,j (x) =

Dxj Dxi k(|x − y|) f (y) − f (x) dy Ω0 Dxi k(|x − y|)γj (y) dσ(y) −f (x) ∂Ω0

(8.11)

8.2 Subnewtonian Kernels

311

for x ∈ Ωδ , i, j = 1, . . . , n. Also let v = Dxi Kf on Ωδ and let Dxi k(|x − y|) ψ(|x − y|/ )f (y) dy v (x) = Ω Dxi k(|x − y|) ψ(|x − y|/ )f (y) dy, = Ω0

where ψ is the function in the proof of Theorem 8.2.5. The last equation holds since f = 0 on Ω0 ∼ Ω. Applying Theorem 8.2.7 to v, Dxi k(|x − y|) 1 − ψ(|x − y|/ ) f (y) dy |v(x) − v (x)| = Ω = Dxi k(|x − y|) 1 − ψ(|x − y|/ ) f (y) dy Bx,2 1 ≤ CM dy |x − y|n−1 Bx,2 = CM σn 2 → 0 as → 0. This shows that lim →0 v = v uniformly on Ωδ . Since the kernel used to deﬁne v satisﬁes Inequalities (8.1) and (8.2), Theorem 8.2.7 is applicable by virtue of Remark 8.2.6 so that v ∈ C 1 (Ω), and it follows from Theorem 8.2.7 that Dxj v (x) = Dxj Dxi k(|x − y|) ψ(|x − y|/ ) f (y) dy (8.12) Ω 0 Dxj Dxi k(|x − y|) ψ(|x − y|/ ) (f (y) − f (x)) dy = Ω0 Dxj Dxi k(|x − y|) ψ(|x − y|/ ) dy +f (x) Ω0 Dxj Dxi k(|x − y|) ψ(|x − y|/ ) (f (y) − f (x)) dy = Ω0 −f (x) Dxi k(|x − y|) ψ(|x − y|/ )γj (y) dσ(y), ∂Ω0

the last equation holding by Equations (8.8) and (8.9). For y ∈ ∂Ω0 , |x − y| > 2 and ψ(|x − y|/ ) = 1 so that the last integral is equal to Dxi k(|x − y|)γj (y) dσ(y). ∂Ω0

It follows from Equations (8.11) and (8.12) that |ui,j (x) − Dxj v (x)| Dxj Dxi k(|x − y|) 1 − ψ(|x − y|/ ) (f (y) − f (x)) dy . = Ω0

312

8 Newtonian Potential

Since ψ(|x − y|/ ) = 1 for |x − y| > 2 , |ui,j − Dxj v (x)| = Dxj Dxi k(|x − y|) 1 − ψ(|x − y|/ ) (f (y) − f (x)) dy Bx,2 α ≤M Dxj Dxi k(|x − y|) 1 − ψ(|x − y|/ ) |x − y| dy Bx,2

≤M

(1 − ψ(|x − y|/ ))Dxj xi k(|x − y|)

Bx,2

1 (xj − yj ) + Dxi k(|x − y|) ψ (|x − y|/ ) |x − y|α dy |x − y|

2 ≤M |Dxj xi k(|x − y|)| + |Dxi k(|x − y|)| |x − y|α dy Bx,2

1 2 1 |x − y|α dy + ≤ CM n n−1 |x − y| |x − y| Bx,2 ≤

(8.13)

CM σn 2 (2 )α+1 σn (2 )α + CM → 0 as → 0 α α+1

uniformly for x ∈ Ωδ . Since the v converge uniformly to v = Dxi Kf on Ωδ and the Dxj v converge uniformly to ui,j on Ωδ , by Theorem 0.2.2 Dxi xj w = Dxi xj Kf = ui,j on Ωδ , i, j = 1, . . . , n. Returning to Equation (8.12), Dxj v (x) =

Dx j

Dxi k(|x − y|) ψ(|x − y|/ ) f (y) dy.

Ω∼Bx,

It is easily seen that the integrand is dominated by an integrable function, namely, a constant, and that Dxi v is continuous on Ωδ . Since the Dxi v converge uniformly to ui,j = Dxi xj w on Ωδ , the latter function is continuous on Ωδ . Since this is true for each δ > 0, w ∈ C 2 (Ω). Recall that the symbol C is used as a generic symbol for a constant that may very well change from line to line or even within a line. Whenever possible, the dependence of C on certain parameters will be indicated by C = C(n, a, b), etc. It will be assumed in the remainder of this section that k(x, y) is a function of |x − y|; that is, k(x, y) = k(|x − y|). If x = (x1 , . . . , xn ) ∈ Rn , let x = (x , xn ) where x = (x1 , . . . , xn−1 ), and let n R+ = {(x , xn ) ∈ Rn ; xn > 0}

R0n = {(x , xn ) ∈ Rn ; xn = 0} n R− = {(x , xn ) ∈ Rn ; xn < 0};

8.2 Subnewtonian Kernels

313

n moreover, if A ⊂ Rn , let A± = A ∩ R± and A0 = A ∩ R0n . Let Ω be a n bounded open subset of R+ , x0 ∈ Ω, B1 = Bx0 ,ρ and B2 = Bx0 ,2ρ with B2+ ⊂ Ω. The balls B1 and B2 may or may not have a nonempty intersection with R0n . The function f on B2+ of the following lemma is understood to be zero outside B2+ . Recall that Dk u 0,Ω stands for the supremum of norms of derivatives of u of order k.

Lemma 8.2.10 If B1 = Bx0 ,ρ and B2 = Bx0 ,2ρ are concentric balls with center x0 in Ω, f ∈ Hα (B2+ ), α ∈ (0, 1), and w = Kf on Ω, then w ∈ H2+α (B1+ ) and [w]2+0,B + + ρα [w]2+α,B + ≤ C(n, α)( f 0,B + + ρα [f ]α,B + ). 1

1

2

2

Proof: Note that f is bounded and uniformly H¨ older continuous on B2+ . Suppose i = n or j = n, the latter, for example. If y ∈ ∂B2+ ∩ R0n , then γj (y) = 0, and it follows from Equation (8.10) that Dxi xj w(x) =

B2+

Dxi xj k(|x − z|)(f (z) − f (x)) dz

−f (x)

(∂B2 )+

Dxi k(|x − z|)γj (z) dσ(z)

for x ∈ B1+ . If i = n instead, interchange i and j in this equation. Using Inequalities (8.2) and (8.3) and the fact that B2+ ⊂ B2 ⊂ Bx0 ,3ρ , |Dxi xj w(x)| ≤

B2+

|Dxi xj k(|x − z|)||f (z) − f (x)| dz

+|f (x)| ≤ [f ]α,B + 2

(∂B2 )+

Bx,3ρ

+ f 0,B + 2

|Dxi k(|x − y|)| dσ(z)

C(n) |z − x|α dz |x − z|n

(∂B2

)+

C(n) dσ(z) |x − z|n−1

σn C(n)σn 3α α ρ [f ]α,B + + C(n) f 0,B + ≤ (2ρ)n−1 . 2 2 α (2ρ)n−1 Thus, there is a constant C(n, α) such that |Dxi xj w(x)| ≤ C(n, α)(ρα [f ]α,B + + f 0,B + ), 2

2

x ∈ B1+ .

(8.14)

+ Now let y be any other point in B1+ , t = |x − y|, ξ = 12 (x + y), and Bξ,t = n Bξ,t ∩ R+ . Then

314

8 Newtonian Potential

Dxi xj w(y) − Dxi xj w(x) = (Dxi k(|x − z|) − Dxi k(|y − z|)γj (z) dσ(z) f (x) (∂B2 )+ +(f (x) − f (y)) Dxi k(|y − z|)γj (z) dσ(z) (∂B2 )+ Dxi xj k(|x − z|)(f (x) − f (z)) dz + + Bξ,t ∩B2+

+

+ Bξ,t ∩B2+

Dxi xj k(|y − z|)(f (z) − f (y)) dz

+(f (x) − f (y))

+ B2+ ∼Bξ,t

+

+ B2+ ∼Bξ,t

Dxi xj k(|x − z|) dz

(Dxi xj k(|x − z|) − Dxi xj k(|y − z|)(f (y) − f (z)) dz.

ˆ Denote the terms on the right by I1 , . . . , I6 , respectively. For some point x on the line segment joining x to y, |I1 | ≤ f 0,B + |x − y| 2

(∂B2 )+

|∇(x) Dxi k(|ˆ x − z|)| dσ(z).

Using Inequality (8.3) and the fact that |ˆ x − z| ≥ ρ for z ∈ (∂B2 )+ , |I1 | ≤ f 0,B + |x − y| 2

(∂B2

≤ f 0,B + |x − y|C(n) 2

)+

C(n) dσ(z) |ˆ x − z|n

1 σn (2ρ)n−1 . ρn

Since |x − y| ≤ |x − y|α (2ρ)1−α , there is a constant C(n, α) such that |I1 | ≤ C(n, α) f 0,B + 2

|x − y|α . ρα

Similarly, |I2 | ≤ C(n)[f ]α,B + |x − y|α . Using Inequality (8.3) and the fact that 2 + ⊂ Bx,3t/2 , Bξ,t |I3 | ≤

+ Bξ,t ∩B2+

|Dxi xj k(|x − z|)(f (x) − f (z))| dz

≤ C(n)[f ]α,B + 2

|x − z|α−n dz Bx,3t/2

≤ C(n, α)[f ]α,B + |x − y|α . 2

8.2 Subnewtonian Kernels

315

Similarly, |I4 | ≤ C(n, α)[f ]α,B + |x − y|α . 2

By Equation (8.8), |I5 | ≤ [f ]α,B + |x − y| Dxi xj k(|x − z|) dz 2 + + B2 ∼Bξ,t Dxi k(|x − z|)γj (z) dσ(z). = [f ]α,B + |x − y|α 2 + + α

∂(B2 ∼Bξ,t )

Since |x − z| > ρ for z ∈ ∂B2 and |x − z| > t/2 for z ∈ ∂Bξ,t , C(n) Dxi k(|x − z|)γj (z) dσ(z) ≤ dσ(z) n−1 + + ∂(B2 ∼B ) ∂B2 |x − z| ξ,t C(n) dσ(z) + |x − z|n−1 ∂Bξ,t ≤ C(n)2n−1 σn + C(n)2n−1 σn = C(n)2n σn . Therefore, |I5 | ≤ [f ]α,B + |x − y|α C(n)2n σn . 2

Lastly, for some x ˆ on the line segment joining x to y, |I6 | ≤ |x − y|

+ B2+ ∼Bξ,t

|∇(x) Dxi xj k(|ˆ x − z|)||f (y) − f (z)| dz.

By Inequality (8.4), |I6 | ≤ C(n)|x − y|[f ]α,B + 2

+ B2+ ∼Bξ,t

|y − z|α dz. |ˆ x − z|n+1

Since |y − z| ≤ 32 |ξ − z| ≤ 3|ˆ x − z| for z ∈ B2 ∼ Bξ,t , t = |x − y|, and + ⊂∼ Bξ,t , B2+ ∼ Bξ,t |I6 | ≤ C(n, α)|x − y|[f ]α,B + tα−1 2

= C(n, α)[f ]α,B + |x − y|α . 2

Thus, there is a constant C(n, α) such that

|Dxi xj w(y) − Dxi xj w(x)| ≤ C(n, α)|x − y|α

1

f 0,B + + [f ]α,B + 2 2 ρα

316

8 Newtonian Potential

and it follows that

[Dxi xj w]α,B + ≤ C(n, α) 1

1 + + [f ] + .

f

0,B2 α,B2 ρα

(8.15)

Taking the supremum over pairs (i, j) in Inequalities (8.14) and (8.15) and then combining, [w]2+0,B + + ρα [w]2+α,B + ≤ C(n, α) f 0,B + + ρα [f ]α,B + . 1

1

2

2

Since f is bounded and integrable on Ω, |w|1+0,B + < +∞ by Theorem 8.2.5 1 and Theorem 8.2.7. This fact combined with the preceding inequality establishes that w ∈ H2+α (B1+ ). This proof fails in the α = 1 case because the bound on |I6 | involves a divergent integral. Corollary 8.2.11 If Ω is a bounded open subset of Rn , x0 ∈ Ω, B1 = Bx0 ,ρ , B2 = Bx0 ,2ρ with B2 ⊂ Ω, f ∈ Hα (B2 ), α ∈ (0, 1), and w = Kf on Ω, then w ∈ H2+α (B1 ) and [w]2+0,B1 + ρα [w]2+α,B1 ≤ C(n, α) f 0,B2 + ρα [f ]α,B2 . n Proof: If Ω ⊂ R+ , then B1+ = B1 and B2+ = B2 . If Ω is any subset of Rn , n . the preceding case can be applied by translating Ω onto R+ n Theorem 8.2.12 Let Ω be a bounded open subset of R+ and let Σ be a n relatively open subset of ∂Ω ∩ R0 . If α ∈ (0, 1), f ∈ Hα (Ω), and w = Kf on Ω, then w ∈ C 2 (Ω) ∩ C 1 (Ω ∪ Σ).

Proof: It follows from the preceding corollary that w ∈ C 2 (Ω). The assertion that w ∈ C 1 (Ω ∪ Σ) follows from Theorem 8.2.7.

8.3 Poisson’s Equation Taking the kernel of the preceding section to be the Newtonian kernel u(|x − y|), the results therein can be used to show that Poisson’s equation Δw = f has a solution and that w inherits smoothness properties of f . Recall that the notations u(x, y), ux (y), u(|x − y|), u(r) for the fundamental harmonic function are used interchangeably. If f : Rn → R is Lebesgue integrable and has compact support, the Newtonian potential of f is the function Uf : Rn → R deﬁned by Uf (x) = u(|x − y|)f (y) dy. Rn

8.3 Poisson’s Equation

317

The kernel u(x, y) is continuous in the extended sense on Rn × Rn and continuous oﬀ the diagonal of Rn ×Rn . It will be shown now that u(x, y) satisﬁes Inequalities (8.1), (8.2), (8.3), and (8.4). Letting δij = 0 or 1 according to j = i or j = i, respectively, the following derivatives and inequalities are easily veriﬁed: ⎧ x −y i i ⎨ − |x−y| if n = 2 2 (8.16) Dxi u(x, y) = ⎩ −(n − 2) xi −yin if n ≥ 3, |x−y| ⎧ (xi −yi )(xj −yj ) 1 ⎪ if n = 2 ⎨ − |x−y|2 δij + 2 |x−y|4 (8.17) Dxi xj u(x, y) = ⎪ j −yj ) ⎩ (n − 2) − 1 n δij + n (xi −yi )(xn+2 if n ≥ 3, |x−y| |x−y| ⎧ 1 ⎨ |x−y| if n = 2 (8.18) |Dxi u(x, y)| ≤ ⎩ n−2n−1 if n ≥ 3, |x−y| ⎧ 3 if n = 2 ⎨ |x−y|2 |Dxi xj u(x, y)| ≤ (8.19) ⎩ (n−2)(n+1) if n ≥ 3 n |x−y| where i, j = 1, 2, . . . , n. More generally, there is a constant C(n, β) such that |Dβ u| ≤ C(n, β)

1 |x − y|n−2+k

if |β| = k ≥ 1.

(8.20)

All of the results of the preceding section are applicable to u(x, y) on Ω×Ω for any bounded open subset Ω of Rn . The statement that u ∈ C 0 (Ω − ) ∩ C 2 (Ω) means that the restriction of u to Ω − is in C 0 (Ω − ) and that the restriction of u to Ω is in C 2 (Ω). Theorem 8.3.1 If the function f is bounded and locally H¨ older continuous with exponent α ∈ (0, 1], on the bounded open set Ω, then the function u=

1 Uf (2 − n)σn

belongs to C 0 (Ω − ) ∩ C 2 (Ω) and satisﬁes Δu = f on Ω; if, in addition, g is continuous on ∂Ω, then there is a unique u in C 0 (Ω − ) ∩ C 2 (Ω) satisfying Δu = f on Ω and limy→x u(y) = g(x) for each regular boundary point x ∈ ∂Ω. Proof: (n ≥ 3) Extending f to be 0 outside Ω, the function v = Uf is continuous on Rn by Theorem 8.2.5 and belongs to C 2 (Ω) by Theorem 8.2.9. If Ω0 is any bounded open set containing Ω with a piecewise smooth boundary, the second partials of v have the representation

318

8 Newtonian Potential

Dxi xj v(x) =

(Dxi xj u(x, y))(f (y) − f (x)) dy Dxi u(x, y)γj (y) dσ(y), −f (x) Ω0

x ∈ Ω,

∂Ω0

by the same theorem. Taking i = j and summing over i, Δv(x) = ((x) u(|x − y|))(f (y) − f (x)) dy Ω0

−f (x)

n

Dxi u(|x − y|)γi (y) dσ(y).

∂Ω0

i=1

Since Δ(x) u(|x − y|) = 0 for y = x, Δv(x) = −f (x)

n i=1

Dxi u(|x − y|)γi (y) dσ(y).

∂Ω0

Now let Ω0 = Bx,ρ , where ρ is chosen so that Ω − ⊂ Bx,ρ . By Equation (8.16), Δv(x) = −f (x)

n i=1

−(n − 2)

∂Bx,ρ

f (x)(n − 2) ρn i=1 n

=

(xi − yi ) γi (y) dσ(y) |x − y|n

(xi − yi )γi (y) dσ(y). ∂Bx,ρ

By Equation (8.8), n f (x)(n − 2) Dyi (xi − yi ) dy ρn i=1 Bx,ρ f (x)n(n − 2) ρ =− rn−1 dθ dr ρn 0 |θ|=1

Δv(x) =

= (2 − n)σn f (x). 1 The function vˆ = (2−n)σ v therefore satisﬁes Poisson’s equation Δˆ v = f n 0 − on Ω and belongs to C (Ω ) by Theorem 8.2.5. Consider the continuous boundary function h = g − vˆ|∂Ω and the corresponding harmonic function v + ΔHh = f on Ω, and Hh . If w = vˆ + Hh , then w ∈ C 2 (Ω), Δw = Δˆ limy→x w(y) = limy→x (ˆ v (y) + Hh (y)) = vˆ|∂Ω (x) + g(x) − vˆ|∂Ω (x) = g(x) for all regular boundary points x ∈ ∂Ω. Suppose now that w1 and w2 satisfy the conclusions of the theorem. Then w1 − w2 ∈ C 2 (Ω), Δ(w1 − w2 ) = 0 on Ω, and limy→x (w1 − w2 )(y) = 0 for all regular boundary points x ∈ ∂Ω. By the uniqueness of the solution to the Dirichlet problem for continuous boundary functions, w1 = w2 on Ω.

8.4 H¨ older Continuity of Second Derivatives

319

8.4 H¨ older Continuity of Second Derivatives It was shown in Theorem 8.3.1 that Poisson’s equation Δw = f has a solution w ∈ C 2 (Ω) whenever f is bounded and locally H¨ older continuous with exponent α ∈ (0, 1] on the bounded open set Ω. The purpose of the next few results is to show that w ∈ H2+α,loc (Ω) when α ∈ (0, 1). Theorem 8.4.1 If Γ is a compact subset of the bounded open set Ω ⊂ Rn , β = (β1 , . . . , βn ) is a multi-index, and u is a harmonic function on Ω, then

|β| n|β|

Dβ u 0,Γ ≤

u 0,Ω d where d = dist(Γ, ∂Ω). Proof: Consider any y ∈ Γ , let ρ = d/|β|, and consider the concentric balls B1 , . . . , B|β| where Bj = By,jρ , j = 1, . . . , |β|. Since all derivatives of u are harmonic on Ω, 1 Dxi u(y) = Dxi u(z) dz, i = 1, . . . , n, νn ρn B1 by the mean value property. By Equation (8.8), 1 Dxi u(y) = u(z)γi (z) dσ(z) νn ρn ∂B1 so that |Dxi u(y)| ≤

n sup |u(z)|. ρ z∈B − 1

Replacing u by Dxj u, j = 1, . . . , n,

n |Dxi xj u(y)| ≤ sup |Dxj u(z)|. ρ z∈B − 1

Applying the same argument to each z ∈ B1− and using the fact that Bz,ρ ⊂ B2 ,

n n |Dxj u(z)| ≤ sup |u(x)| ≤ sup |u(x)|. ρ x∈Bz,ρ − ρ x∈B − 2

Thus,

2 n |Dxj xj u(y)| ≤ sup |u(z)|. ρ z∈B − 2

320

8 Newtonian Potential

Continuing in this manner, |Dβ u(y)| ≤

|β|

|β| n n sup |u(z)| ≤

u 0,Ω . ρ ρ z∈B − |β|

Taking the supremum over y ∈ Γ and using the fact that ρ = d/|β|, Dβ u 0,Γ ≤ (n|β|/d)|β| u 0,Ω . Theorem 8.4.2 If B1 = Bx0 ,ρ and B2 = Bx0 ,2ρ are concentric balls with closures in the bounded open set Ω ⊂ Rn , f ∈ Hα,loc (Ω), α ∈ (0, 1), u ∈ C 2 (Ω), and Δu = f on Ω, then |u; ρ|2+α,B1 ≤ C(n, α) u 0,B2 + ρ2 |f ; ρ|α,B2 . Proof: Note that u 0,B2 < +∞ since u ∈ C 2 (Ω) and that f can be replaced by f χB2 where χB2 is the indicator function of B2 . Extend f to be 0 outside Ω. By Theorem 8.2.7, w = (1/(2 − n)σn )Uf ∈ C 1 (Rn ). Since Δw = f on B2 by Theorem 8.3.1, Δ(u − w) = 0 on B2 , v = u − w is harmonic on B2 , and v is the Dirichlet solution corresponding to the boundary function (u − w)|∂B2 . A bound for w 0,B2 can be found as follows, the n = 2 case being deferred to the end of the proof. If x ∈ B2 , then B2 ⊂ Bx,4ρ and 1 1 |w(x)| = f (y) dy n−2 (2 − n)σn B2 |x − y| 1 1 ≤

f 0,B2 dy (n − 2)σn |x − y|n−2 Bx,4ρ ≤ C(n)ρ2 f 0,B2 and

w 0,B2 ≤ C(n)ρ2 f 0,B2 .

(8.21)

Therefore

v 0,B1 ≤ v 0,B2 ≤ u 0,B2 + w 0,B2 ≤ C(n)( u 0,B2 + ρ2 f 0,B2 ). (8.22) By Theorem 8.4.1, for any multi-index β with |β| ≥ 1

D v 0,B1 β

|β| n|β| ≤

v 0,B2 ρ

|β| n|β| ≤ C(n) ( u 0,B2 + ρ2 f 0,B2 ). ρ

Hence, if |β| = j = 1 or 2, then

[v]j+0,B1 ≤ C(n)

nj ρ

j

( u 0,B2 + ρ2 f 0,B2 );

(8.23)

8.4 H¨ older Continuity of Second Derivatives

321

if |β| = 2 and x, y ∈ B1 , then for some point z on the line segment joining x to y |Dβ v(x) − Dβ v(y)| = |∇(x) Dβ v(z)||x − y|1−α |x − y|α

3 3n ≤ C(n)

v 0,B2 |x − y|1−α ρ

3 3n ≤ C(n) (2ρ)1−α ( u 0,B2 + ρ2 f 0,B2 ). ρ Thus, ρ2+α [v]2+α,B1 ≤ C(n)( u 0,B2 + ρ2 f 0,B2 ).

(8.24)

Combining Inequalities (8.22), (8.23), and (8.24), |v, ρ|2+α,B1 =

2

ρj [v]j+0,B1 + ρ2+α [v]2+α,B1

j=0

≤ C(n)( u 0,B2 + ρ2 f 0,B2 )

(8.25)

≤ C(n)( u 0,B2 + ρ2 |f ; ρ|α,B2 ). By Theorem 8.2.7, Dxi u(|x − y|)f (y) dy,

Dxi w(x) = B2

x ∈ B2− ,

and by Inequality (8.18), |Dxi w(x)| ≤ C(n) f 0,B2 B2

1 dy |x − y|n−1

x ∈ B2− .

Since B2 ⊂ Bx,3ρ for x ∈ B1 , |Dxi w(x)| ≤ C(n)ρ f 0,B2 . Therefore, [w]1+0,B1 ≤ C(n)ρ f 0,B2 . By Corollary 8.2.11, [w]2+0,B1 + ρα [w]2+α,B1 ≤ C(n, α)( f 0,B2 + ρα [f ]α,B2 ).

(8.26)

322

8 Newtonian Potential

Using Inequalities (8.21) and (8.26), 2 |w; ρ|2+α,B1 = ρj [w]j+0,B1 + ρ2+α [w]2+α,B1 j=0

≤ C(n, α)(ρ2 f 0,B2 + ρ2 f 0,B2 + ρ2 f 0,B2 + ρα [f ]α,B2 ) ≤ C(n, α)ρ2 |f ; ρ|α,B2 . Combining this inequality with Inequality (8.25), |u; ρ|2+α,B1 ≤ C(n, α)( u 0,B2 + ρ2 |f ; ρ|α,B2 ). Returning to the n = 2 case, the functions u(x, y) and f (x, y) can be regarded as functions of three variables without change to the hypotheses, the supnorms Dj u , f 0,B2 , or the seminorms [u; ρ]2+α , [f ]α . Corollary 8.4.3 If f ∈ Hα,loc (Ω), α ∈ (0, 1), and {uj } is a uniformly bounded sequence of functions in C 2 (Ω) satisfying Poisson’s equation Δuj = f on Ω, then there is a subsequence that converges uniformly on compact subsets of Ω to a function u ∈ C 2 (Ω) for which Δu = f on Ω. Proof: For each x ∈ Ω, let Bx,ρ , Bx,2ρ be balls with closures in Ω. Then {Bx,ρ ; x ∈ Ω} is an open covering of Ω and there is a countable subcovering {Bxi ,ρi ; i ≥ 1}. Note that uj ∈ C 2 (Ω) implies that uj restricted to Bx−i ,ρi belongs to C 2 (Bx−i ,ρi ). By hypothesis, there is a constant C0 such that

uj 0,Ω ≤ C0 for j ≥ 1. Consider the sequence {uj } on Bx1 ,ρ1 . Using the fact that f ∈ Hα (Bx−1 ,2ρ1 ) implies |f ; ρ1 |α,Bx1 ,2ρ1 ≤ (1 + ρα 1 )|f |α,B −

x1 ,2ρ1

and the preceding theorem, |uj ; ρ1 |2+α,Bx1 ,ρ1 ≤ C(n, α)( uj 0,Bx1 ,2ρ1 + ρ21 |f ; ρ1 |α,Bx1 ,2ρ1 ) ≤ C(n, α)(C0 + ρ21 |f ; ρ1 |α,Bx1 ,2ρ1 ) = M < +∞,

j ≥ 1.

Recalling that D and D2 are generic symbols for ﬁrst- and second-order partials, respectively, for all j ≥ 1, |uj |0,Bx1 ,ρ1 + ρ1 [Duj ]0,Bx1 ,ρ1 + ρ21 [D2 uj ]0,Bx1 ,ρ1 + ρ2+α [D2 uj ]α,Bx1 ,kρ1 ≤ M . 1 Therefore, |uj |2+α,Bx1 ,ρ1 ≤ M / min (1, ρ1 , ρ21 , ρ2+α ), and it follows from the 1 subsequence selection principle, Theorem 7.2.4, that there is a continuous u (1) on Bx1 ,ρ1 and a subsequence {uj } of the {uj } sequence such that the se(1)

(1)

(1)

quences {uj }, {Duj }, and {D2 uj } converge uniformly on Bx−1 ,ρ1 to u, Du,

8.4 H¨ older Continuity of Second Derivatives (1)

and D2 u, respectively. Since Δuj now the sequence

(1) {uj }

323

= f on Ω, Δu = f on Bx−1 ,ρ1 . Consider

on Bx2 ,ρ2 . The above argument applies to this case (2)

(1)

also and there is a subsequence {uj } of the {uj } sequence such that the (2)

(2)

(2)

sequences {uj }, {Duj }, and {D2 uj } converge uniformly on Bx−2 ,ρ2 to continuous functions v, Dv, and D2 v, respectively, on Bx−2 ,ρ2 with Δv = f on Bx−2 ,ρ2 . The functions u and v must agree on Bx−1 ,ρ1 ∩ Bx−2 ,ρ2 and v can be relabeled as u so that u is deﬁned on Bx−1 ,ρ1 ∪ Bx−2 ,ρ2 . By an induction argu(i)

(i−1)

ment, for each i ≥ 2, there is a subsequence {uj } of the {uj (i) (i) {uj }, {Duj },

} sequence

(i) {D2 uj } 2

and converge uniformly such that the sequences i on k=1 Bx−k ,ρk to continuous functions u, Du, and D u with Δu = f on i (j) − k=1 Bxk ,ρk . Finally, the diagonal sequence {uj } is a subsequence of every sequence discussed above and there is a function u ∈ C 2 (Ω) such that the (j) (j) (j) sequences {uj }, {Duj }, and {D2 uj } converge uniformly on each compact set Γ ⊂ Ω to the continuous functions u, Du, and D2 u with u ∈ C 2 (Ω) and Δu = f on Ω. The requirement that f be bounded in Theorem 8.3.1 can be relaxed in case Ω is a ball. The proof of the following theorem necessitates looking at the eﬀect of truncation of a function satisfying a H¨ older condition. If a, b, c are any real numbers, it is easily seen that | min (a, b) − min (a, c)| ≤ |b − c|. If f ∈ Hα (Ω), 0 < α < 1 and k is any real number, then | min (k, f (x)) − min (k, f (y))| ≤ |f (x) − f (y)| from which it follows that the function min (f, k) ∈ Hα (Ω). The same result is true of max (f, k). In particular, the functions fj in the following proof belong to Hα,loc (B). Theorem 8.4.4 If B is a ball in Rn , b ∈ (−1, 0), α ∈ (0, 1), and f ∈ (2+b) Hα,loc (B) satisﬁes [f ; d]0,B < +∞, there is then a unique function u ∈ C 0 (B − ) ∩ C 2 (B) satisfying Δu = f on B and u = 0 on ∂B. Moreover, (b)

(2+b)

[u; d]0,B ≤ C[f ; d]0,B

(8.27)

where C = C(b, ρ). (2+b)

Proof: Let B = Bx0 ,ρ , r = |x − x0 |, K = [f ; d]0,B , and v = (ρ2 − r2 )−b for r ≤ ρ. Using the polar form of the Laplacian as in Section 1.3 and using the fact that −n + 2(1 + b) < 0, Δv = 2b(ρ2 − r2 )−b−2 n(ρ2 − r2 ) + 2(1 + b)r2 ≤ 4b(1 + b)ρ2 (ρ2 − r2 )−b−2 ≤ 4b(1 + b)ρ−b (ρ − r)−b−2 . Since d(x)2+b |f (x)| ≤ K by hypothesis and d(x) = ρ − r, |f (x)| ≤ Kd(x)−2−b = K(ρ − r)−b−2 ≤ C0 KΔv

324

8 Newtonian Potential

where C0 = 1/4(b(1 + b)ρ−b ). Granted the ﬁrst assertion of the theorem, the second assertion will be proved now. Assume that Δu = f on B and u = 0 on ∂B. Then Δ(−C0 Kv ± u)(x) ≤ −|f (x)| ± f (x) ≤ 0 on B with −C0 Kv ± u = 0 on ∂B. By Corollary 2.3.6, −C0 Kv ± u ≥ 0 on B so that |u(x)| ≤ −C0 Kv = −C0 K(ρ2 − r2 )−b ≤ −C0 K(2ρ)−b d(x)−b . (2+b)

Letting C = −C0 (2ρ)−b , d(x)b |u(x)| ≤ CK = C[f ; d]0,B , completing the proof of the second assertion. Now let {Bj } be an increasing sequence of concentric balls with closures in B exhausting B. For each j ≥ 1, let mj = supx∈Bj |f (x)| and let ⎧ ⎨ −mj if f (x) ≤ −mj fj (x) = f (x) if |f (x)| < mj ⎩ mj if f (x) ≥ mj . By Theorem 8.3.1, for each j ≥ 1 there is a uj ∈ C 0 (B − ) ∩ C 2 (B) satisfying Δuj = fj on B and uj = 0 on ∂B. By Inequality (8.27), (2+b) (2+b) ρb |uj (x)| ≤ d(x)b |uj (x)| ≤ C[fj ; d]0,B ≤ C[f ; d]0,B on B, proving that the sequence {uj } is uniformly bounded on B. It follows from the above corollary that there is a function u ∈ C 2 (B) such that Δu = f on B. Since (2+b) d(x)b |u(x)| ≤ C[f ; d]0,B on B, limx→y u(x) = 0 for all y ∈ ∂B. Theorem 8.4.5 If Ω is a bounded open subset of Rn , u ∈ C 2 (Ω), f ∈ (2) Hα (Ω; d), α ∈ (0, 1), and Δu = f on Ω, then (2) |u; d|2+α,Ω ≤ C u 0,Ω + |f ; d|α,Ω where C = C(n, α). Proof: It can be assumed that u 0,Ω < +∞, for otherwise the inequality is trivial. For the time being, ﬁx x ∈ Ω and let δ = d(x)/3, B1 = Bx,δ , B2 = Bx,2δ . For i, j = 1, . . . n, d(x)|Dxi u(x)| + d2 (x)|Dxi xj u(x)| ≤ (3δ)[u]1+0,B1 + (3δ)2 [u]2+0,B1 . Since the sum on the right is included in the sum deﬁning |u; δ|2+α,B1 except for a constant, by Theorem 8.4.2 d(x)|Dxi u(x)| + d2 (x)|Dxi xj u(x)| ≤ C(n, α)( u 0,B2 + δ 2 |f ; δ|α,B2 ).

8.4 H¨ older Continuity of Second Derivatives

325

Since d(y) ≥ d(x)/3 = δ for y ∈ B2 , δ 2 |f ; δ|α,B2 ≤ sup d2 (y)|f (y)| + y∈B2

sup

y ,y ∈B2

d2+α (y , y )

|f (y ) − f (y )| |y − y |α

(2)

≤ C|f ; d|α,Ω .

(8.28)

Therefore, (2)

|u; d|2+0,Ω ≤ C(n, α)( u 0,Ω + |f ; d|α,Ω ).

(8.29)

Now consider any x, y ∈ Ω with d(x) ≤ d(y) and deﬁne δ, B1 , B2 as above. Since d(x, y) = d(x) = 3δ, d2+α (x, y)

|Dxi xj u(x) − Dxi xj u(y)| ≤ (3δ)2+α [u]2+α,B1 |x − y|α +3α 2(3δ)2 [u]2+0,Ω .

Since δ 2+α [u]2+α,B1 ≤ [u; δ]2+α,B1 ≤ |u; δ|2+α,B1 , by Theorem 8.4.2 and Inequality (8.28) δ 2+α [u]2+α,B1 ≤ C(n, α)( u 0,B2 + δ 2 |f ; δ|α,B2 ) (2)

≤ C(n, α)( u 0,Ω + |f ; d|α,Ω ). By Inequality (8.29) d2+α (x, y)

|Dxi xj u(x) − Dxi xj u(y)| (2) ≤ C(n, α)( u 0,Ω + |f ; d|α,Ω ). |x − y|α

This shows that (2)

[u; d]2+α,Ω ≤ C(n, α)( u 0,Ω + |f ; d|α,Ω ). (2)

Hence, |u; d|2+α,Ω = |u; d|2+0,Ω + [u; d]2+α,Ω ≤ C(n, α)( u 0,Ω + |f ; d|α,Ω ). Corollary 8.4.6 If Ω is a bounded open subset of Rn , u ∈ Cb0 (Ω)∩C 2 (Ω), f ∈ (2) Hα (Ω; d), α ∈ (0, 1), and Δu=f on Ω, then u∈H2+α,loc (Ω). Proof: Let Γ be a compact subset of Ω and let d0 = d(Γ, ∂Ω) > 0. By the preceding theorem, (2)

)|u|2+α,Γ ≤|u; d|2+α,Ω ≤ C(n, α)( u 0,Ω +|f ; d|α,Ω ) < +∞. min (1, d0 , d20 , d2+α 0 Thus, u ∈ H2+α,loc (Ω).

326

8 Newtonian Potential

8.5 The Reﬂection Principle The classic reﬂection principle, known as Schwarz’s Reﬂection Principle, can be described as follows. Let Ω be an open subset of Rn that is symmetric n with respect to R0n with Ω0 = Ω ∩ R0n = ∅, let Ω+ = Ω ∩ R+ , and let n 0 Ω− = Ω ∩ R− . If h ∈ C (Ω+ ∪ Ω0 ) is harmonic on Ω+ with h = 0 on Ω0 , then the function ⎧ if (x , xn ) ∈ Ω+ ⎨ h(x , xn ) ˜ 0 if (x , xn ) ∈ Ω0 h(x , xn ) = ⎩ −h(x , −xn ) if (x , xn ) ∈ Ω− is harmonic on Ω by Corollary 1.7.10, since it is continuous there and satisﬁes the local averaging principle. Throughout this section, unless noted otherwise, Ω will denote a spherical n n for some x0 ∈ R− ∪ R0n and δ > 0. It will be chip; that is, Ω = Bx0 ,δ ∩ R+ n assumed that Bx0 ,δ ∩ R0 = ∅. Σ will denote the interior of the ﬂat portion of the boundary of Ω. Recall that y r = (y , −yn ) if y = (y , yn ) ∈ Rn and that Ω r = {y r ; y ∈ Ω}. Each point of Σ is then an interior point of the convex region Ω = Ω ∪ Ω r ∪ Σ. Note that each point of the boundary of Ω is a regular boundary point for the Dirichlet problem, since the Zaremba cone condition is satisﬁed at each such point. Theorem 8.5.1 If h ∈ C 0 (∂Ω ∼ Σ), then there is a function u ∈ C 0 (Ω − ) ∩ C 2 (Ω ∪ Σ) satisfying Δu = 0 on Ω, Dn u = 0 on Σ, and limy→x,y∈Ω u(y) = h(x) for each x ∈ ∂Ω ∼ Σ. Proof: Deﬁne h on ∂Ω by putting h(x , xn ) on ∂Ω ∼ Σ h (x , xn ) = h(x , −xn ) on (∂Ω ∼ Σ)r . Note that h (x , xn ) = h (x , −xn ) on ∂Ω . Now consider u in the class Uh of lower bounded superharmonic functions with lim inf y→x u(y) ≥ h (x), x ∈ ∂Ω . Then u(x , −xn ) is a lower bounded superharmonic function on Ω . Since lim inf

(y ,yn )→(x ,xn )

u(y , −yn ) ≥ h (x , −xn ) = h (x , xn ),

u(x , −xn ) is in Uh . Letting u (x , xn ) = min [u(x , xn ), u(x , −xn )], u is a lower bounded superharmonic function on Ω , and it is easily seen that lim inf

(y ,yn )→(x ,xn )

u (y , yn ) ≥ h (x , xn ).

Note that u (x , xn ) = u (x , −xn ). Thus u ∈ Uh ⇒ u ∈ Uh and u ≤ u. Therefore, H h = inf {u : u ∈ Uh } and H h (x , xn ) = H h (x , −xn ) because this is true of each u . Similarly, H h (x , xn ) = H h (x , −xn ). In particular, if

8.5 The Reﬂection Principle

327

h is continuous on ∂Ω ∼ Σ, then H h = H h , and we put Hh = H h = H h . Now consider a point (x , 0) ∈ Σ, which is an interior point of Ω . Since Hh is harmonic on a neighborhood of (x , 0) and Hh (x , xn ) − Hh (x , −xn ) 2xn 1 Hh (x , xn ) − Hh (x , 0) 1 Hh (x , −xn ) − Hh (x , 0) = + , 2 xn 2 −xn

0=

Dxn Hh (x , 0) = 0. Letting u be the restriction of Hh to Ω, it follows that u ∈ C 0 (Ω − ) ∩ C 2 (Ω ∪ Σ), since Hh is harmonic on Ω and on a neighborhood of each point of Σ. Lastly, if x ∈ ∂Ω ∼ Σ, then limy→x,y∈Ω u(y) = limy→x,y∈Ω Hh (y) = h (x) = h(x). The following lemma results from combining Theorem 8.2.7 and Theorem 8.3.1. The set Ω need not satisfy the standing assumption stated at the beginning of this section. Lemma 8.5.2 Let f be bounded and locally H¨ older continuous with exponent α ∈ (0, 1] on the bounded open set Ω and let w(x) =

1 (2 − n)σn

u(|x − y|)f (y) dy,

x ∈ Rn .

Ω

Then w ∈ C 0 (Ω − ) ∩ C 1 (Ω ∪ Σ) ∩ C 2 (Ω) with Δw = f on Ω. The solution w of the equation Δw = f is not unique, for if w is replaced by w + h, where h is harmonic on Rn , then another solution is obtained. Lemma 8.5.3 If f ∈ H0+α (Ω), α ∈ (0, 1], there is a function w ∈ C 0 (Ω − )∩ C 1 (Ω ∪ Σ) ∩ C 2 (Ω) such that Δw = f on Ω and Dn w = 0 on Σ. Proof: Since f ∈ H0+α (Ω) implies that f is uniformly continuous on Ω, f has a continuous extension to Ω − , also denoted by f . Let f (x , xn ) if xn ≥ 0 f (x , xn ) = f (x , −xn ) if xn ≤ 0. It is easily seen that f ∈ H0+α (Ω ). Applying the preceding lemma to f , there is a bounded function w ∈ C 0 (Ω − ) ∩ C 2 (Ω ) such that Δw = f on Ω . For each y ∈ R0n , let Γy = {yn ; (y , yn ) ∈ Ω }. Using the fact that f (y , yn ) = f (y , −yn ), the fact that u(|(x , xn ) − (y , yn )|) = u(|(x , −xn ) − (y , −yn )|), and Fubini’s Theorem,

328

8 Newtonian Potential

(2 − n)σn w(x , xn ) u(|(x , xn ) − (y , yn )|)f (y , yn ) dyn dy = Rn 0

=

Rn 0

=

Rn 0

=

Rn 0

Γy

u(|(x , −xn ) − (y , −yn )|)f (y , yn ) dyn

dy

Γy

Γyr

u(|(x , −xn ) − (y , zn )|)f (y , −zn ) d(−zn )

dy

u(|(x , −xn ) − (y , zn )|)f (y , zn ) dzn dy

Γy

= (2 − n)σn w(x , −xn ). The change from d(−zn ) to dzn in going from the fourth line to the ﬁfth is justiﬁed since Γyr is replaced by Γy . It follows that w is symmetric with respect to R0n . By the argument used above to show that Dxn H(h )(x , 0) = 0, Dn w = 0 on Σ. The sets Ω and Σ of the following lemma need not satisfy the standing assumption stated at the beginning of this section. Lemma 8.5.4 (Boundary Point Lemma [33]) If Ω is a bounded open subset of Rn , Σ is a relatively open subset of ∂Ω, u ∈ C 1 (Ω ∪ Σ) ∩ C 2 (Ω) with Δu ≥ 0 on Ω, x0 ∈ Σ, u(x0 ) > u(x) for all x ∈ Ω, and there is a ball B ⊂ Ω internally tangent to Σ at x0 , then Dn u(x0 ) > 0. 2

2

Proof: Let B = By,ρ , x ∈ By,ρ , r = |x − y|, v(x) = e−kr − e−kρ , where k is a constant to be chosen, and 0 < δ < ρ. Then v = 0 on ∂By,ρ , Δv = 2 (4k 2 r2 − 2kn)e−kr , and k can be chosen so that Δv ≥ 0 on the annulus 2 2 Aρ,δ = By,ρ ∼ By,δ . Since u(x)−u(x0 ) < 0 on ∂By,δ and v = e−kδ −e−kρ on ∂By,δ , there is a constant > 0 such that u(x)−u(x0 )+ v ≤ 0 on ∂By,δ . This inequality also holds on ∂By,ρ where v = 0. Since Δ(u−u(x0 )+ v) ≥ Δv ≥ 0 on Aρ,δ and u − u(x0 ) + v ≤ 0 on ∂Aρ,δ , u − u(x0 ) + v ≤ 0 on Aρ,δ by the maximum principle. Thus, u(x) − u(x0 ) v(x) − v(x0 ) ≤ − , |x − x0 | |x − x0 |

x ∈ Aρ,δ .

As x → x0 in Aρ,δ normal to Σ, the limit of the left term is the inner normal derivative of u at x0 . Thus, the outer normal derivative of u at x0 is 2

Dn u(x0 ) ≥ − Dn v(x0 ) = − Dr v|r=ρ = 2 kρe−kρ > 0. Lemma 8.5.5 If f ∈ H0+α (Ω), α ∈ (0, 1], and g ∈ C 0 (∂Ω ∼ Σ), then there is a unique u ∈ C 0 (Ω − )∩C 1 (Ω∪Σ)∩C 2 (Ω) such that Δu = f on Ω, Dn u = 0 on Σ, and u = g on ∂Ω ∼ Σ.

8.5 The Reﬂection Principle

329

Proof: By Lemma 8.5.3, there is a w ∈ C 0 (Ω − ) ∩ C 1 (Ω ∪ Σ) ∩ C 2 (Ω) such n ∼ Ω, the that Δw = f on Ω and Dn w = 0 on Σ. Extending f to be 0 on R+ function w constructed by way of Lemma 8.5.2 can be regarded as an element n of Cb0 (R+ ∪ R0n ) by Theorem 8.2.5. In particular, w|∂Ω ∈ C 0 (∂Ω ∼ Σ). By Theorem 8.5.1, there is a v ∈ C 1 (Ω ∪ Σ) ∩ C 2 (Ω) such that Δv = 0 on Ω, Dn v = 0 on Σ, and lim

y→x,y∈Ω

v(y) = g(x) − w|∂Ω (x)

for all x ∈ ∂Ω ∼ Σ. Since v ∈ C 1 (Ω ∪ Σ), limy→x,y∈Ω v(y) exists in R for all x ∈ Σ. As noted in Section 0.1, the function v has a continuous extension on Ω − . Letting u = w + v, Δu = Δw + Δv = f on Ω, Dn u = Dn w + Dn v = 0 on Σ, and lim

y→x,y∈Ω

u(y) =

lim

y→x,y∈Ω

w(y) +

lim

y→x,y∈Ω

v(y)

= w|∂Ω (x) + g(x) − w|∂Ω (x) = g(x) for all x ∈ ∂Ω ∼ Σ. Since w is continuous on Ω − and v has a continuous extension on Ω − , u ∈ C 0 (Ω − ). To prove uniqueness, assume that both w1 and w2 satisfy the conclusions of the lemma. The diﬀerence w = w1 − w2 satisﬁes Δw = 0 on Ω, Dn w = 0 on Σ, and w = 0 on ∂Ω ∼ Σ. Since w ∈ C 0 (Ω − ), it attains it maximum value at some point x0 ∈ Ω − . If x0 is an interior point of Ω, then w is constant by the maximum principle for harmonic functions, and therefore w = 0 on Ω − because w = 0 on ∂Ω ∼ Σ. If w does not attain its maximum value at an interior point of Ω, then x0 ∈ ∂Ω and w(x) < w(x0 ) for all x ∈ Ω. By the preceding lemma, x0 ∈ Σ because Dn w(x) = 0 for x ∈ Σ. In this case, x0 ∈ ∂Ω ∼ Σ and w(x) ≤ w(x0 ) = 0 for all x ∈ Ω − . Since −w has the same properties as w, w ≥ 0 on Ω − and so w = w1 −w2 = 0 on Ω − . A standard method for passing from the homogeneous boundary condition on Σ to the nonhomogeneous condition is to incorporate the boundary function into the function f . In addition to the standing assumptions regarding the spherical chip, another condition will be imposed in the following lemn where B is ball with center at (0, −ρ0 ) and mas. Suppose Ω = B + = B ∩ R+ √ of radius ρ where 0 < ρ0 < ρ. If the condition ρ0 > ρ/ 2 is imposed and x = (x , xn ) ∈ Ω, then the ball in R0n with center x of radius xn will lie in Σ. This can be seen by considering the intersection of Ω with hyperplanes containing the xn -axis. If g : Σ → R is a boundary function, recall that |g|1+α,Σ = inf {|u|1+α,Ω ; u ∈ H1+α (Ω), u|Σ = g}. Lemma 8.5.6 If√Ω = B + where B is a ball with center at (0, −ρ0 ) and of radius ρ, ρ0 > ρ/ 2, α ∈ (0, 1], and g ∈ H1+α (Σ), then there is a function ψ ∈ H2+α (Ω) satisfying ψ = 0 on Σ, Dn ψ = g on Σ, |ψ|2+α,Ω ≤ C|g|1+α,Σ , and Δψ ∈ H0+α (Ω).

330

8 Newtonian Potential

Proof: According to Theorem 7.2.6, it can be assumed that g ∈ H1+α (Ω). Consider any η ∈ C 2 (R0n ) with support in B0,1 ⊂ R0n and Rn η(z) dz = 1. 0 Let ψ(x , xn ) = xn g(x − xn z, 0)η(z) dz. (8.30) Rn 0

√ Under the hypothesis that ρ0 > ρ/ 2, it can be seen from geometrical arguments that the point x − xn z ∈ Σ for all z ∈ B0,1 ⊂ R0n . For x = (x , xn ) ∈ Ω, ψ(x , 0) = 0, Dn ψ(x , 0) = g(x , 0). Therefore, [ψ|0,Ω ≤ [g]0,Σ ≤ C|g|1+α,Ω . For i = n, integrating by parts with respect to zi and making use of the fact that η(z) = 0 for z outside the ball with center at x of radius xn , Dxi ψ(x , xn ) = xn

=− = For i, j = n, Dxi xj ψ(x , xn ) =

Rn 0

Rn 0

Rn 0

Rn 0

Dxi g(x − xn z, 0)η(z) dz

Dzi g(x − xn z, 0)η(z) dz

g(x − xn z, 0)Dzi η(z) dz.

(8.31)

Dxj g(x − xn z, 0)Dzi η(z) dz.

(8.32)

Similarly, Dxn ψ(x , xn ) = xn

n−1

Rn 0 j=1

Dxj g(x − xn z, 0)(−zj )η(z) dz

+ = xn

=−

n−1

g(x − xn z, 0)η(z) dz

1 Dzj g(x − xn z, 0) − (−zj η(z)) dz n xn j=1 R0 + g(x − xn z, 0)η(z) dz

n−1

Rn 0

Rn 0

j=1

g(x − xn z, 0)Dzj (zj η(z))dz

+ =−

Rn 0

Rn 0

Rn 0

g(x − xn z, 0)η(z) dz

g(x − xn z, 0) ∇η(z) · z + (n − 2)η(z) dz.

(8.33)

8.5 The Reﬂection Principle

331

By Equations (8.31) and (8.33), [ψ]1+0,Ω ≤ C[g]0,Σ ≤ C|g|1+α,Ω . For i = n, Dxi g(x − xn z, 0) ∇η(z) · z + (n − 2)η(z) dz. (8.34) Dxn xi ψ(x , xn ) = − Rn 0

Similarly,

Dxn xn ψ(x , xn ) =

Rn 0

∇g(x −xn z, 0)·z ∇η(z)·z +(n−2)η(z) dz. (8.35)

As before, [ψ]2+0,Ω ≤ C[g]1+0,Σ ≤ C|g|1+α,Ω = C|g|1+α,Σ . Since |(x −xn z)− (y − yn z)| ≤ |x − y|(1 + |z|), for i = n |Dxi xj ψ(x , xn ) − Dxi xj ψ(y , yn )| |x − y|α |Dxj g(x − xn z, 0) − Dxj g(y − yn z), 0| ≤ (1 + |z|)α |Dzi η(z)| dz − x z) − (y − y z)|α |(x n n Rn 0 ≤ C[g]1+α,Σ ≤ C|g|1+α,Ω = C|g|1+α,Σ .

(8.36)

Taking the supremum over x, x ∈ Ω, [ψ]2+α,Ω ≤ C|g|1+α,Σ . Therefore, |ψ|2+α,Ω ≤ C|g|1+α,Σ . Theorem 8.5.7 If Ω satisﬁes the conditions of the preceding lemma and f ∈ H0+α (Ω), α ∈ (0, 1], g ∈ H1+α (Σ) and h ∈ C 0 (∂Ω ∼ Σ), then there is a unique w ∈ C 0 (Ω − ) ∩ C 1 (Ω ∪ Σ) ∩ C 2 (Ω) satisfying Δw = f on Ω, Dn w = g on Σ, and w = h on ∂Ω ∼ Σ. Proof: Consider the function ψ of the preceding lemma for which Dn ψ = g on Σ and Δψ ∈ H0+α (Ω) and the problem (i) Δv = f − Δψ on Ω, (ii) Dn v = 0 on Σ, and (iii) v = h − ψ on ∂Ω ∼ Σ. Since ψ ∈ H2+α (Ω), it is uniformly continuous on Ω and has a continuous extension to ∂Ω. Since f − Δψ ∈ H0+α (Ω) and h − ψ ∈ C 0 (∂Ω ∼ Σ), by Lemma 8.5.5 there is a v ∈ C 0 (Ω − ) ∩ C 1 (Ω ∪ Σ) ∩ C 2 (Ω) satisfying (i)-(iii). Letting w = v + ψ, Δw = Δv + Δψ = f on Ω, Dn w = Dn v + Dn ψ = g on Σ, and w = v + ψ = h − ψ + ψ = h on ∂Ω ∼ Σ. Since ψ ∈ H2+α (Ω) implies that ψ ∈ C 0 (Ω − ) ∩ C 1 (Ω ∪ Σ) ∩ C 2 (Ω), w ∈ C 0 (Ω − ) ∩ C 1 (Ω ∪ Σ) ∩ C 2 (Ω). Uniqueness of w follows as in the proof of Lemma 8.5.5.

Chapter 9

Elliptic Operators

9.1 Introduction Up to this point, all of the concepts pertain to the Dirichlet and Neumann problems for the Laplacian. In this and later chapters, the Laplacian will be replaced by an elliptic operator as described below and mixed boundary conditions will be allowed that specify the value of a function on part of the boundary and the value of its normal derivative on the remainder of the boundary as in Theorem 8.5.7. Let Ω be a bounded open subset of Rn with a smooth boundary ∂Ω. A precise smoothness requirement will be speciﬁed later. For i, j = 1, . . . , n, let aij , bi , c be real-valued functions on Ω with aij = aji . In addition, for i = 1, . . . , n, let βi , γ be real-valued functions on ∂Ω. An elliptic operator is an operator acting on real-valued functions u on Ω of the form Lu(x) =

n

∂2u ∂u (x) + bi (x) (x) + c(x)u(x), ∂xi ∂xj ∂xi i=1 n

aij (x)

i,j=1

x ∈ Ω,

with suitable conditions imposed on the coeﬃcients. In addition, a boundary condition can be speciﬁed by an operator M deﬁned by the equation Mu(x) =

n i=1

βi (x)

∂u (x) + γ(x)u(x) = 0, ∂xi

x ∈ ∂Ω.

In terms of the operators L and M, the Dirichlet problem is a special case of the problem Lu = f on Ω, u = g on ∂Ω, and the Neumann problem is a special case of the problem Lu = f on Ω, Mu = g on ∂Ω. The latter problem is known as the oblique derivative problem. L.L. Helms, Potential Theory, Universitext, c Springer-Verlag London Limited 2009

333

334

9 Elliptic Operators

Each of the results of this chapter are stated under minimal conditions on the coeﬃcients of L; all of these conditions are satisﬁed if (0) (1) (2)

aij 0,Ω +|aij ; d|α,Ω + bi 0,Ω +|bi ; d|α,Ω + c 0,Ω +|c; d|α,Ω < +∞. sup 1≤i,j≤n

9.2 Linear Spaces Results pertaining to the Dirichlet problem for the Laplacian operator can be carried over to a general elliptic operator by the method of continuity. This method requires the use of some theorems about mappings between normed linear spaces. A map T from a Banach space B into a normed linear space X is bounded if there is a constant C such that Tx X ≤ C x B . The smallest C for which this is true is denoted by T ; that is, T = supx∈B,x =0 T x X/ x B . A map T : B → B is a contraction map if there is a constant λ < 1 such that Tx − Ty B ≤ λ x − y B for all x, y ∈ B. The map T : B → X is a linear map if T(ax + by) = aTx + bTy for all a, b ∈ R and all x, y ∈ B. A map T : B → B is a bounded linear operator or simply a bounded operator on T if bounded and linear as deﬁned above. Theorem 9.2.1 If T is a contraction map on the Banach space B, then there is a unique x ∈ B such that Tx = x. Proof: Let x0 be an arbitrary point of B and let λ < 1 be a constant such that Tx ≤ λ x for all x ∈ B. It is easy to see that Tj x ≤ λj x for all j ≥ 1, x ∈ B. For each j ≥ 1, let xj = Tj x0 . If j ≥ i, then

xj − xi ≤

j k=i+1

≤

j k=i+1

xk − xk−1 =

j

Tk−1 x1 − Tk−1 x0

k=i+1

λk−1 x1 − x0 ≤

λi

x1 − x0 → 0 as i → ∞. 1−λ

This shows that the sequence {xj } is Cauchy in B and therefore converges to some x ∈ B. Since T is a continuous map, Tx = limj→∞ Txj = limj→∞ xj+1 = x. If Ty = y also, then x − y = Tx − Ty ≤ λ x − y

with λ < 1 and therefore x = y. Theorem 9.2.2 (Method of Continuity) Let B be a Banach space, let X be a normed linear space, and let T0 , T1 be bounded linear operators from B into X. For each t ∈ [0, 1], let Tt = (1 − t)T0 + tT1 . If there is a constant C, not depending upon on t, such that

x B ≤ C Tt x X , for all t ∈ [0, 1], x ∈ B, then T1 maps B onto X if and only if T0 maps B onto X.

(9.1)

9.3 Constant Coeﬃcients

335

Proof: Suppose Ts maps B onto X for some s ∈ [0, 1]. By Inequality (9.1), Ts is one-to-one and therefore has an inverse T−1 s : X → B. For ﬁxed t ∈ [0, 1] and y ∈ X, there is an x ∈ B satisfying the equation Tt x = y if and only if Ts x = y + (Ts − Tt )x = y + (t − s)T0 − (t − s)T1 . This equation holds if and only if −1 x = T−1 s y + (t − s)Ts (T0 − T1 )x;

that is, if and only if the map T : B → B deﬁned by −1 Tx = T−1 s y + (t − s)Ts (T0 − T1 )x

has a ﬁxed point. Since T−1 s y B ≤ C y X for all y ∈ X,

Tx1 − Tx2 X = (t − s)T−1 s (T0 − T1 )(x1 − x2 ) B ≤ C|t − s| T0 − T1

x1 − x2 B ≤ C|t − s|( T0 + T1 ) x1 − x2 B so that T is a contraction operator on B if |t − s| < δ =

1 . C( T0 + T1 )

By the preceding theorem, T has a ﬁxed point whenever |t − s| < δ; that is, if |t − s| < δ, then Tt maps B onto X whenever Ts maps B onto X. By subdividing the interval [0, 1] into subintervals of length less than δ, the mapping Tt is onto for all t ∈ [0, 1] provided it is onto for some s ∈ [0, 1], in particular if s = 0 or s = 1.

9.3 Constant Coeﬃcients It will be assumed in this section that α ∈ (0, 1) and that the coeﬃcients aij , i, j = 1, . . . , n, of the elliptic operator deﬁned in Section 9.1 are all constants, that aij = aji , i, j = 1, . . . , n, and that the bi , i = 1, . . . n, and c are all 0. In this case, the operator will be denoted by L0 u(x) =

n i,j=1

aij Dij u(x).

336

9 Elliptic Operators

It also will be assumed that there are positive constants m and M such that n aij xi xj ≤ M |x|2 . (9.2) m|x|2 ≤ i,j=1

This assumption implies that the matrix A = [aij ] is positive deﬁnite. The following facts from elementary matrix theory will be needed (c.f., Hohn [31]). There is an n × n orthogonal matrix O = [oij ] such that OAOt = diag[λ1 , . . . , λn ] where λ1 . . . , λn are the eigenvalues of A and the superscript t signiﬁes the transpose matrix. Regarding elements of Rn as column vectors for purposes of this discussion, if y = Ox, then t

x Ax =

n

aij xi xj =

i,j=1

n

λi yi2 .

i=1

It also is known from matrix theory that O preserves norms; that is, |Ox| = |x|, x ∈ Rn . It follows that m = min (λ1 , . . . , λn ) and M = max (λ1 , . . . , λn ). Now let −1 D = [λi 2 δij ] = diag[1/ λ1 , . . . , 1/ λn ] and deﬁne T : Rn → Rn by the equation y = T x = DOx. Since xt Ax =

n

λ2i (T x)2i ,

i=1

m|x|2 ≤

n

λ2i (T x)2i ≤ M |x|2 .

i=1

Thus,

m M |x|2 ≤ (T x)2i ≤ 2 |x|2 , 2 M m i=1 n

and there is a constant c = c(m, M ) such that 1 |x| ≤ |T x| ≤ c|x|. (9.3) c If Ω is an open subset of Rn , let Ω ∗ be the image of Ω under T . Recalling that d(y) = dist(y, ∼ Ω), the d notation will be augmented by adding the subscript Ω in order to deal with the two sets Ω and Ω ∗ simultaneously. It is easily seen using the above inequality that 1 dΩ (x) ≤ dΩ ∗ (y) ≤ cdΩ (x). (9.4) c Consider now any function u on Ω. The transformation T deﬁnes a function u∗ on the image Ω ∗ of Ω by the equation u∗ (y) = u(x) where y = T x. If u ∈ C 2 (Ω) or u ∈ Hα (Ω), then u∗ ∈ C 2 (Ω ∗ ) or u∗ ∈ Hα (Ω ∗ ), respectively. If u ∈ C 2 (Ω) and f ∈ Hα (Ω), the equation L0 u = f on Ω is transformed into the equation

9.3 Constant Coeﬃcients

337

Δu∗ (y) = f ∗ (y),

y ∈ Ω∗

with u∗ ∈ C 2 (Ω ∗ ) and f ∗ ∈ Hα (Ω ∗ ). This result provides a means of solving the equation L0 u = f on the bounded open set Ω by solving the equation Δu∗ = f ∗ on Ω ∗ . (b) The norms |u; dΩ |2+α,Ω and |f ; dΩ |2+α,Ω are related to |u∗ ; dΩ ∗ |2+α,Ω ∗ (b)

and |f ∗ ; dΩ ∗ |2+α,Ω ∗ , respectively, as in 1 |u; dΩ |2+α,Ω ≤ |u∗ ; dΩ ∗ |2+α,Ω ∗ ≤ c|u; dΩ |2+α,Ω , c

(9.5)

1 (b) (b) (b) |f ; dΩ |α,Ω ≤ |f ∗ ; dΩ ∗ |α,Ω ∗ ≤ c|f ; dΩ |α,Ω (9.6) c where c = C(n, m, M ). These inequalities follow from Inequalities (9.3) and (9.4). Just one of the inequalities will be proven as an illustration; namely, |u; dΩ |2+α,Ω ≤ c|u∗ ; dΩ ∗√ |2+α,Ω ∗ . Clearly, u 0,Ω = u∗ √0.Ω ∗ . Suppose y = n T x. Then yi = j=1 (1/ λi )oij xj so that ∂yi /∂xj = (1/ λi )oij . For each j, ∂u∗ ∂yi ∂u∗ 1 ∂u √ oij = = ∂xj ∂yi ∂xj ∂yi λi i=1 i=1 n

n

and therefore n

1 |dΩ ∗ (y)Di u∗ (y)| √ |oij | λi i=1 n 1 ∗ √ |oij | . ≤ c[u ; dΩ ∗ ]1+0,Ω ∗ λi i=1

|dΩ (x)Dj u(x)| ≤ c

With j ﬁxed, let zi = 1 if oij ≥ 0 and zi = −1 if oij < 0. Then |z| = (T z)i = Thus,

(9.7) √ n and

n n 1 1 √ oij zj = √ |oij |. λi λi i=1 i=1

n √ 1 √ |oij | = (T z)i ≤ |T z| ≤ c|z| = c n λi i=1

(9.8)

and it follows that [u; dΩ ]1+0,Ω ≤ C(n, m, M )[u∗ ; dΩ ∗ ]1+0,Ω ∗ . Similarly, [u; dΩ ]2+0,Ω ≤ C(n, m, M )[u∗ ; dΩ ∗ ]2+0,Ω ∗ and [u; dΩ ]2+α,Ω ≤ C(n, m, M )[u∗ ; dΩ ∗ ]2+α,Ω ∗ . Therefore, |u; dΩ |2+α,Ω =

2 j=0

[u; dΩ ]j+0,Ω + [u; dΩ ]2+α,Ω ≤ C(n, m, M )|u∗ ; dΩ ∗ |2+α,Ω ∗ .

338

9 Elliptic Operators

The following lemma is an immediate consequence of Theorem 8.4.5 and Inequalities (9.5) and (9.6). (2)

Lemma 9.3.1 If u ∈ C 2 (Ω), f ∈ Hα (Ω; d), α ∈ (0, 1), and L0 u = f on Ω, then (2) |u; d|2+α,Ω ≤ C(n, α, m, M )( u 0,Ω + |f ; d|α,Ω ).

9.4 Schauder Interior Estimates The following terminology regarding the operator L will be used. If there are positive functions m and M on the open set Ω ⊂ Rn such that m|λ|2 ≤

n

aij (x)λi λj ≤ M |λ|2 for all x ∈ Ω, λ ∈ Rn ,

(9.9)

i,j=1

the operator L is said to be elliptic on Ω; if m is a positive constant, L is said to be strictly elliptic on Ω. Granted that the solution of the equation Lu = f is bounded, a bound for the weighted norm |u; d|2+α,Ω can be obtained. Such estimates are called Schauder interior estimates. All of the results in this section can be found in [25]. Lemma 9.4.1 If α ∈ (0, 1), a + α ≥ 0, b + α ≥ 0, a + b + α ≥ 0, f ∈ (a) (b) (a+b) (Ω; φ) and Hα (Ω; φ), g ∈ Hα (Ω; φ), then f g ∈ Hα (a+b)

(a)

(b)

|f g; φ|α,Ω ≤ |f ; φ|α,Ω |g; φ|α,Ω .

(9.10)

Proof: By deﬁnition, (a+b)

|f g; φ|α,Ω = sup φa+b (x)|f (x)g(x)| x∈Ω

+ sup φa+b+α (x, y) x,y∈Ω (a)

|f (x)g(x) − f (y)g(y)| |x − y|α

(b)

≤ |f ; φ|0,Ω |g; φ|0,Ω + sup φa+b+α (x, y) x,y∈Ω (a)

(b)

|f (x)||g(x) − g(y)| + |g(y)||f (x) − f (y)| |x − y|α (a)

(b)

≤ |f ; φ|0,Ω |g; φ|0,Ω + [f ; φ]0,Ω [g; φ]α,Ω (b)

(a)

+|g; φ|0,Ω [f ; φ]α,Ω (a) (a) (b) (b) ≤ |f ; φ|0,Ω + [f ; φ]α,Ω |g; φ|0,Ω + [g; φ]α,Ω (a)

(b)

= |f ; φ|α,Ω |g; φ|α,Ω .

9.4 Schauder Interior Estimates

339 (2)

Lemma 9.4.2 If α ∈ (0, 1), g ∈ Hα (Ω, d), μ ≤ 1/2, x0 ∈ Ω, δ = μd(x0 ), and B = Bx0 ,δ , then (2)

(2)

(2)

(2)

|g; dB |α,B ≤ 4μ2 [g; d]0,Ω + 8μ2+α [g; d]α,Ω ≤ 8μ2 |g; d|α,Ω .

(9.11)

Proof: For x ∈ B, dB (x) = d(x, ∂B) < δ and d(x) ≥ (1 − μ)d(x0 ) so that (2)

|g; dB |α,B = sup d2B (x)|g(x)| + sup d2+α B (x, y) x∈B

x,y∈B

|g(x) − g(y)| |x − y|α

≤ δ 2 sup d−2 (x)d2 (x)|g(x)| x∈B

+ δ 2+α sup d−2−α (x, y)d2+α (x, y) x,y∈B

|g(x) − g(y)| |x − y|α

≤

δ2 sup d2 (x)|g(x)| (1 − μ)2 d2 (x0 ) x∈Ω

+

|g(x) − g(y)| δ 2+α sup d2+α (x, y) (1 − μ)2+α d2+α (x0 ) x,y∈Ω |x − y|α

≤

μ2 μ2+α (2) (2) [g; d]0,Ω + [g; d]α,Ω 2 (1 − μ) (1 − μ)2+α (2)

(2)

≤ 4μ2 [g; d]0,Ω + 8μ2+α [g; d]α,Ω (2)

≤ 8μ2 |g; d|α,Ω . Lemma 9.4.3 Let Ω be a bounded open subset of Rn and let L be an elliptic operator on Ω satisfying (0)

(1)

(2)

|aij ; d|α,Ω , |bi ; d|α,Ω , |c; d|α,Ω ≤ K, α ∈ (0, 1), i, j = 1, . . . , n.

(9.12)

(2)

If u ∈ H2+α (Ω; d), f ∈ Hα (Ω; d), α ∈ (0, 1), and Lu = f , then (2)

|u; d|2+α,Ω ≤ C( u 0,Ω + |f ; d|α,Ω ).

(9.13)

where C = C(m, n, α, K). Proof: By Inequality (7.10), it suﬃces to prove the result for [u; d]2+α,Ω . Let x0 , y0 be two points of Ω with x0 = y0 and labeled so that d(x0 ) ≤ d(y0 ), let μ ≤ 1/2 be a constant that will be further speciﬁed later, let δ = μd(x0 ), and let B = Bx0 ,δ . The equation Lu = f can be related to an elliptic operator with constant coeﬃcients by writing n

aij (x0 )Dij u(x) =

i,j=1

−

n i,j=1 n

(aij (x0 ) − aij (x))Dij u(x)

bi (x)Di u(x) − c(x)u(x) + f (x).

i=1

340

9 Elliptic Operators

Denote the left side of this equation by L0 u and the right side by f0 . Note that for y0 ∈ Bx0 ,δ/2 , d(y0 , ∂B) ≥ δ/2 so that by Lemma 9.3.1,

2+α δ |Dij u(x0 ) − Dij u(y0 )| (2) ≤ C u 0,B + |f0 ; dB |α,B (9.14) 2 |x0 − y0 |α where C = C(m, n, α, K). Since δ = μd(x0 ) = μd(x0 , y0 ), d2+α (x0 , y0 )

|Dij u(x0 ) − Dij u(y0 )| 1 (2) ≤ 2+α C u 0,B + |f0 ; dB |α,B . (9.15) α |x0 − y0 | μ

Each of the terms on the right side of the inequality (2)

|f0 ; dB |α,B ≤

n

(2)

|(aij (x0 ) − aij )Dij u; dB |α,B +

i,j=1

+

n

(2)

|bi Di u; dB |α,B

i=1

(2) |cu; dB |α,B

+

(2) |f ; dB |α,B

(9.16)

can be estimated as follows. The subscripts on aij will be omitted temporarily. Keep in mind that D2 is a generic symbol for a second-order partial derivative. By Inequalities (9.10) and (9.11), (2)

|(a(x0 ) − a)D2 u; dB |α,B (0)

(2)

≤ |a(x0 ) − a; dB |α,B |D2 u; dB |α,B (2) (2) ≤ |a(x0 ) − a; dB |α,B 4μ2 [D2 u; d]0,Ω + 8μ2+α [D2 u; d]α,Ω ≤ |a(x0 ) − a; dB |α,B 4μ2 [u; d]2+0,Ω + 8μ2+α [u; d]2+α,Ω . Using the fact that (1 − μ)d(x0 ) ≤ d(x), (1 − μ)d(x0 ) ≤ min (d(x0 ), d(x)) = d(x0 , x) for x ∈ B, |a(x0 ) − a(x)| |x − x0 |α |a(x0 ) − a(x)| ≤ μα d(x0 )α |x − x0 |α μα |a(x0 ) − a(x)| ≤ dα (x0 , x) (1 − μ)α |x − x0 |α α μ ≤ [a; d]α,Ω ≤ 2α μα [a; d]α,Ω . (1 − μ)α

|a(x0 ) − a(x)| = |x − x0 |α

so that a(x0 ) − a 0,B ≤ 2α μα [a; d]α,Ω . Moreover, [a(x0 ) − a; dB ]α,B = sup dα B (x, y) x,y∈B

|a(x) − a(y)| |x − y|α

≤ μα d(x0 )α sup d(x, y)−α d(x, y)α ≤

(1

x,y∈B α μ d(x0 )α [a; d]α,Ω − μ)α d(x0 )α

|a(x) − a(y)| |x − y|α

≤ 2α μα [a; d]α,Ω .

9.4 Schauder Interior Estimates

341

Combining these two inequalities, |a(x0 ) − a; dB |α,B ≤ 2α+1 μα [a; d]α,Ω . Since |a; d|α,Ω ≤ K by Inequality (9.12), (2) |(a(x0 )−a)D2 u; dB |α,B ≤ 2α+1 μα [a; d]α,Ω 4μ2 [u; d]2+0,Ω +8μ2+α [u; d]2+α,Ω ≤ 32Kμ2+α [u; d]2+0,Ω +μα [u; d]2+α,Ω . Applying Inequality (7.10) to [u; d]2+0,Ω with = μα , there is a constant C = C(μ, α) such that (2)

|(a(x0 ) − a)D2 u; dB |α,B ≤ 64Kμ2+α (C u 0,Ω + μα [u; d]2+α,Ω ).

(9.17)

By Inequalities (9.11) and (9.10), (2)

(2)

|bi Di u; dB |α,B ≤ 8μ2 |bi Di u; d|α,Ω (1)

(1)

≤ 8μ2 |bi ; d|α,Ω |Di u; d|α,Ω ≤ 8Kμ2 |u; d|1+α,Ω . Applying Inequality (7.10) to |u; d|1+α,Ω with = μ2α , there is a constant C = C(μ, α) such that (2) |bi Di u; dB |α,B ≤ 8Kμ2 C u 0,Ω + μ2α |u; d|2+α,Ω .

(9.18)

Similarly, there is a constant C = C(μ, α) such that (2)

(2)

(0)

|cu; dB |α,B ≤ 8μ2 |c; d|α,Ω |u; d|α,Ω ≤ 8Kμ2 (C u 0,Ω + μ2α |u; d|2+α,Ω ).

(9.19)

By Inequality (9.11), (2)

(2)

|f ; dB |α,B ≤ 8μ2 |f ; d|α,Ω .

(9.20) (2)

In combining the above inequalities to estimate |f0 ; dB |α,B , it is essential to keep track of the dependence of constants on the data; in particular, it is essential that the coeﬃcient of μ2+2α [u; d]2+α,Ω does not depend upon μ. As usual, C will be a generic symbol for constants depending upon the data but the symbol for the coeﬃcient of μ2+2α [u; d]2+α,Ω will exhibit the dependence explicitly. Returning to Inequality (9.16) and using Inequalities (9.17), (9.18), (9.19), and (9.20), (2)

(2)

|f0 ; d|α,B ≤ Cμ2 u 0,Ω + C(K)μ2+2α [u]2+α,Ω + Cμ2 [f ; d]α,Ω .

342

9 Elliptic Operators

Returning to Inequality (9.15), for y0 ∈ Bx0 ,δ/2 d(x0 , y0 )2+α

|Dij u(x0 ) − Dij u(y0 )| |x0 − y0 |α

(9.21)

(2)

|f ; d|α,Ω

u 0,Ω

u 0,Ω ≤ C 2+α + C + C(K)μα |u; d|2+α,Ω + C . α μ μ μα Using Inequality (7.10) with = μ2α /2α+1 , for y0 ∈ Bx0 ,δ/2 d(x0 , y0 )2+α

|Dij u(x0 ) − Dij u(y0 )| 2α+1 ≤ α d(x0 )α [u; d]2+0,Ω α |x0 − y0 | δ α+1 2 μ2α ≤ α (C u 0,Ω + α+1 [u; d]2+α,Ω ) μ 2 α ≤ C u 0,Ω + μ [u; d]2+α,Ω .

Combining this result with Inequality (9.21), (2)

[u; d]2+α,Ω ≤ C u 0,Ω + C(K)μ |u; d|2+α,Ω + C α

|f ; d|α,Ω μα

.

Adding |u; d|2+0,Ω to both sides and applying Inequality (7.10) with = μα and choosing μ < 1/2 so that (C(K) + 1)μα < 1/2, there is a constant C = C(K) such that (2) |u; d|2+α,Ω ≤ C u 0,Ω + |f ; d|α,Ω . Theorem 9.4.4 Let L be an elliptic operator satisfying (0)

(1)

(2)

|aij ; d|α,Ω , |bi ; d|α,Ω , |c; d|α,Ω ≤ K,

i, j = 1, . . . , n .

(9.22)

(2)

If f ∈ Hα (Ω; d), α ∈ (0, 1), b ∈ (−1, 0], and {uj } is a sequence of uniformly (b) bounded functions in H2+α (Ω; d) satisfying the equation Lui = f on Ω, then there is a subsequence that converges uniformly on compact subsets of Ω to a function u ∈ H2+α (Ω; d) ⊂ H2+α,loc (Ω) for which Lu = f on Ω. Proof: Suppose ﬁrst that b ∈ (−1, 0). Since d(x) ≤ d(Ω), d(x)b ≥ d(Ω)b , x ∈ (b) Ω, and d(Ω)b |ui ; d|2+α,Ω ≤ |ui ; d|2+α,Ω < +∞, i ≥ 1. By Lemma 9.4.3, (2) (2) |ui ; d|2+α,Ω ≤ C ui 0,Ω + |f ; d|α,Ω ≤ C K + |f ; d|α,Ω < +∞ (9.23) where K = supi≥1 ui 0,Ω . If b = 0, this inequality follows from the same lemma. Now let Γ be any compact, convex subset of Ω and choose 0 < δ < min (1, d(Γ, ∂Ω) so that Γ ⊂ Ωδ . Since |ui ; d|2+α,Ω* δ ≤ |u; d|2+α,Ω ≤ (2)

C(K + |f ; d|α,Ω < +∞, it follows from the subsequence selection principle,

9.5 Maximum Principles

343

Theorem 7.2.4, that the sequences {ui }, {Dui }, and {D2 ui } are uniformly bounded and equicontinuous on Γ . Let {Bj } be a sequence of balls with closures in Ω which exhausts Ω. Applying the above argument to each Bj− , (j)

there is a function u(j) on each Bj and a subsequence {ui } of the {ui } (j) (j) (j) sequence such that the sequences {ui }, {Dui }, and {D2 ui } converge to − (j) (j) 2 (j) u , Du , and D u , respectively, on Bj . Note that the u(j) agree on (i)

overlapping parts of the Bj . The sequence {ui } is a subsequence of each of the above sequences and there is a function u on Ω such that the sequences (i) (i) (i) {ui }, {Dui }, and {D2 ui } converge to u, Du, and D2 u, respectively, on Ω. (i) (i) Since Inequality (9.23) holds for each ui , each term of |ui |2+α,Ω is bounded (2) by C(K + |f ; d|α,Ω ); for example, d(x, y)2+α

|D2 ui (x) − D2 ui (y)| (2) , ≤ C K + |f ; d| α,Ω |x − y|α (i)

(i)

i ≥ 1, x, y ∈ Ω.

Letting i → ∞ and taking the supremum over x, y ∈ Ω and Djk , [u; d]2+α,Ω < +∞. Similarly, for the other terms of |u; d|2+α,Ω . Thus, u ∈ H2+α (Ω; d) ⊂ H2+α,loc (Ω).

9.5 Maximum Principles Although an existence theorem for the Dirichlet problem corresponding to an elliptic operator L has not been proven, a uniqueness result can be proved using a weak maximum principle. Before stating the principle, some results from basic analysis will be reviewed. Consider the quadratic function Q(y) =

n

cpq yp yq

p,q=1

where the n × n matrix [cpq ] is a real symmetric matrix. The quadratic form Q is said to be nonnegative deﬁnite if Q(y) ≥ 0 for all y ∈ Rn ; if, in addition, Q(y) = 0 only when y = 0, then Q is said to be positive deﬁnite. Lemma 9.5.1 If u ∈ C 2 (Ω), x0 ∈ Ω, and u(x0 ) = supx∈Ω (x), then the matrix [−Dxi xj (x0 )] is nonnegative deﬁnite. Proof: Let B = Bx0 ,δ ⊂ Ω and let x ∈ Rn . Since u attains its maximum value at x0 , Dxi u(x0 ) = 0, 1 ≤ i ≤ n. By Taylor’s formula with remainder (c.f. [1]), for 0 < t|x| < δ, 0 ≥ u(x0 + tx) − u(x0 ) =

n i,j=1

Dxi xj u(x0 + t x)txi txj

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9 Elliptic Operators

for some 0 < t < t. Thus, n

Dxi xj u(x0 + t x)xi xj ≤ 0.

i,j=1

Letting δ → 0,

n i.j=1

Dxi xj u(x0 )xi xj ≤ 0.

Lemma 9.5.2 Let Ω be a bounded open subset of Rn and let L be an elliptic operator with c = 0 and supx∈Ω,1≤i≤n (|bi (x)|/m(x)| < +∞. If u ∈ C 0 (Ω − ) ∩ C 2 (Ω) and Lu ≥ 0 on Ω, then u attains its maximum value on ∂Ω; moreover, if Lu > 0 on Ω, then u cannot attain its maximum value at an interior point of Ω. Proof: It will be shown ﬁrst that if Lu > 0, then u cannot attain its maximum value at an interior point of Ω. Suppose u attains its maximum value at an interior point x0 of Ω. At such a point, the matrix [−Dij u(x0 )] is real, symmetric, and nonnegative deﬁnite. Since the matrix A = [aij (x0 )] of coefﬁcients is positive deﬁnite, there is anonsingular matrix S = [sij ] such that n A = S t S (c.f.,[31]). Since aij (x0 ) = k=1 ski skj , Lu(x0 ) = − =− =−

n

aij (x0 )(−Dxi xj u(x0 )) i,j=1 n n

ski skj (−Dxi xj u(x0 ))

i,j=1 k=1 n n k=1

ski (−Dxi xj u(x0 ))skj .

i,j=1

Since the double sum is nonnegative for each k = 1, . . . , n, Lu(x0 ) ≤ 0, a contradiction. Therefore, u cannot attain its maximum value at an interior point of Ω if Lu > 0 on Ω. Suppose now that Lu ≥ 0 on Ω. By hypothesis, there is a constant M0 such that |bi |/m ≤ M0 , i = 1, . . . , n. Since a11 ≥ m, for any constant γ > M0 Leγx1 = (γ 2 a11 + γb1 )eγx1 ≥ m(x)(γ 2 − γM0 )eγx1 > 0. For every > 0, L(u + eγx1 ) ≥ Leγx1 > 0 on Ω. By the ﬁrst part of the proof, there is a z = z( ) ∈ ∂Ω such that u(z) + eγz1 ≥ u(x) + eγx1 for all x ∈ Ω. Thus, supz∈∂Ω u(z) + supz∈∂Ω eγz1 ≥ supz∈∂Ω (u(z) + eγz1 ) ≥ u(x) + eγx1 ≥ u(x) for all x ∈ Ω. Letting → 0, supz∈∂Ω u(z) ≥ u(x) for all x ∈ Ω. Thus, supz∈∂Ω u(z) ≥ supx∈Ω u(x). Recall that u+ = max (u, 0) and u− = max (−u, 0).

9.5 Maximum Principles

345

Corollary 9.5.3 (Weak Maximum Principle) Let Ω be a bounded open subset of Rn and let L be an elliptic operator on Ω with c ≤ 0 and sup

(|bi (x)|/m(x)) < +∞.

x∈Ω,1≤i≤n

If u ∈ C 0 (Ω − ) ∩ C 2 (Ω) and Lu ≥ 0 or Lu ≤ 0, then sup u ≤ sup u+ or inf u− ≤ inf u, Ω

∂Ω

∂Ω

Ω

respectively; if Lu = 0, then sup |u| ≤ sup |u|. Ω

∂Ω

Proof: Suppose Lu ≥ 0 on Ω and let L1 u = Lu − cu. Since Lu ≥ 0 on Ω, L1 u ≥ −cu on Ω. Letting Ωu+ = {x ∈ Ω; u(x) > 0}, L1 u ≥ −cu ≥ 0 on Ωu+ . If Ωu+ = ∅, then u ≤ 0 on Ω, u+ = 0, and supΩ u ≤ sup∂Ω u+ ; if Ωu+ = ∅, applying the previous lemma to L1 on Ωu+ , u attains its maximum value on ∂Ωu+ . But since u = 0 on ∂Ωu+ ∩Ω, u attains its maximum value on ∂Ω where u = u+ . Thus, supΩ u ≤ sup∂Ω u+ . Suppose now that Lu = 0 on Ω. For any x ∈ Ω, u(x) ≤ sup∂Ω u+ ≤ sup∂Ω |u|; since L(−u) = 0, −u(x) ≤ sup∂Ω |u|. Therefore, |u(x)| ≤ sup∂Ω |u| for all x ∈ Ω. A uniqueness theorem and comparison principle follow easily from the weak maximum principle. Theorem 9.5.4 Let Ω be a bounded open subset of Rn and let L be an elliptic operator with c ≤ 0 and supx∈Ω,1≤i≤n (|bi (x)|/m(x)) < +∞. If u, v ∈ C 0 (Ω − ) ∩ C 2 (Ω) satisfy Lu = Lv on Ω and u = v on ∂Ω, then u = v on Ω. If Lu ≥ Lv on Ω and u ≤ v on ∂Ω, then u ≤ v on Ω. The weak maximum principle does not exclude the possibility that a function u satisfying Lu ≥ 0 attains a maximum value at an interior point of Ω as was the case with the Laplacian for nontrivial functions. A stronger form of the weak maximum principle is needed to exclude this possibility. The open set Ω ⊂ Rn is said to satisfy an interior sphere condition at x0 ∈ ∂Ω if there is a ball B ⊂ Ω with x0 ∈ ∂B. The upper inner normal derivate of u at x0 ∈ ∂Ω is deﬁned by Dν+ u(x0 ) = lim sup t→0+

u(x0 + tν(x0 )) − u(x0 ) t

where ν = −n. Lemma 9.5.5 (Hopf [33], Boundary Point Lemma) Let L be a strictly elliptic operator on the open set Ω ⊂ Rn with c ≤ 0 and with |aij |, |bi |, |c| ≤ K, i, j = 1, . . . , n and let u ∈ C 2 (Ω) satisfy Lu ≥ 0 on Ω. If for x0 ∈ ∂Ω,

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9 Elliptic Operators

(i) u can be extended continuously to x0 , (ii) u(x0 ) > u(x) for all x ∈ Ω, and (iii) the interior sphere condition is satisﬁed at x0 , then the upper inner normal derivative at x0 satisﬁes Dν+ u(x0 ) < 0 whenever u(x0 ) ≥ 0; if, in addition, c = 0, then Dν+ (x0 ) < 0 regardless of the sign of u(x0 ). Proof: Let B = By,ρ0 be a ball in Ω with x0 ∈ ∂B. Fix 0 < ρ1 < ρ0 and − deﬁne a function w on the closed annulus A = By,ρ ∼ By,ρ1 by putting 0 2

2

w(x) = e−γr − e−γρ0 where ρ1 < r = |x − y| < ρ0 and γ is a constant to be speciﬁed below. By Schwarz’s inequality, Lw(x) = e−γr

2

n 4γ 2 aij (xi − yi )(xj − yj ) i,j=1

− 2γ

n

(aii + bi (xi − yi ) + cw

i=1

≥ e−γr

2

n aii − 2γ|b|r + c 4γ 2 mr2 − 2γ i=1

≥e

−γr 2

(4γ

2

mρ21

√ − 2γnK − 2γ nKρ0 − K).

It follows that γ can be chosen so that Lw ≥ 0 on A. Since u − u(x0 ) < 0 on ∂By,ρ1 , there is an > 0 such that u − u(x0 ) + w ≤ 0

on ∂By,ρ1 .

Since w = 0 on ∂By,ρ0 , u − u(x0 ) + w ≤ 0 on ∂By,ρ0 and therefore on ∂A. Since L(u − u(x0 ) + w) = Lu − cu(x0 ) + Lw ≥ −cu(x0 ) ≥ 0 whenever u(x0 ) ≥ 0, u − u(x0 ) + w ≤ 0 on A by Corollary 9.5.3. For x0 + tν ∈ A, u(x0 + tν(x0 )) − u(x0 ) w(x0 + tν(x0 )) w(x0 + tν(x0 )) − w(x0 ) ≤ − = − t t t 2

so that Dν+ u(x0 ) ≤ − Dν+ w(x0 ) = Dr w|r=ρ0 = −2γ ρ0e−γρ0 < 0. If c = 0, then the preceding steps are valid without any condition on u(x0 ). Theorem 9.5.6 (Hopf [32], Strong Maximum Principle) Let L be a strictly elliptic operator on the connected open set Ω ⊂ Rn with c ≤ 0 and |aij |, |bi |, |c| ≤ K, i, j = 1, . . . , n, and let u ∈ C 0 (Ω − ) ∩ C 2 (Ω) satisfy Lu ≥ 0 on Ω. Then u cannot attain a nonnegative maximum value at a point of Ω unless it is constant; if, in addition, c = 0 and u attains its maximum value at a point of Ω, then u is a constant function.

9.6 The Dirichlet Problem for a Ball

347

Proof: Assume that u is nonconstant and attains a nonnegative maximum value M at an interior point of Ω. Let ΩM = {x ∈ Ω; u(x) < M } = ∅. Under these assumptions, ∂ΩM ∩ Ω = ∅, for if not, then ∂ΩM ⊂ ∂Ω which − implies that Ω = ΩM ∪ (Ω ∼ ΩM ) = ΩM ∪ (Ω ∼ ΩM ), contradicting the fact that Ω is connected. Thus, there is a point x0 ∈ ΩM such that d(x0 , ∂ΩM ) < d(x0 , ∂Ω). Let B be the largest ball in ΩM having x0 as its center. Then u(y) = M for some y ∈ ∂B and u < M on B. Since ΩM satisﬁes the interior sphere condition at y, the inner normal derivative of u on B at y is strictly negative according to the preceding lemma. But this is a contradiction since any directional derivative of u at y must be zero. Thus, if u attains a nonnegative maximum value at an interior point of Ω, it must be a constant function. If, in addition, c = 0, the preceding lemma can be applied in the same way. Theorem 9.5.7 (Hopf [32]) Let Ω be a connected open subset of Rn , let L be a strictly elliptic operator with c ≤ 0 and locally bounded coeﬃcients. If u ∈ C 0 (Ω − ) ∩ C 2 (Ω) satisﬁes Lu ≥ 0 on Ω, then u cannot attain a nonnegative maximum value on Ω unless it is constant on Ω. If, in addition, c = 0, then u cannot attain its maximum value on Ω unless it is constant on Ω. Proof: Suppose M = supx∈Ω − u(x) ≥ 0 and let ΩM = {y ∈ Ω; u(y) = M }. Then ΩM is a relatively closed subset of Ω. Suppose x0 ∈ ΩM and B = Bx0 ,δ ⊂ Bx−0 ,δ ⊂ Ω. By the preceding theorem, u = M on B since u(x0 ) = M . Therefore, B ⊂ ΩM ; that is, ΩM is relatively closed and open in Ω. Since Ω is connected, ΩM = ∅ or ΩM = Ω.

9.6 The Dirichlet Problem for a Ball It will be assumed in this section that the coeﬃcients of L belong to Hα,loc (Ω); in particular, the coeﬃcients belong to Hα (B) when restricted to a ball B with closure in Ω. If it were possible to solve the Dirichlet problem corresponding to an elliptic operator L and a ball B, then the Perron-Wiener-Brelot method could be used to solve the Dirichlet problem for more general regions. The following lemma reduces to Theorem 9.4.3 when b = 0. Lemma 9.6.1 Let L be a strictly elliptic operator on the bounded open set Ω ⊂ Rn satisfying Inequalities (9.12). If u ∈ H2+α (Ω; d), α ∈ (0, 1), satisﬁes (b) (2+b) the equation Lu = f on Ω, |u; d|0,Ω < +∞, and |f ; d|α,Ω < +∞ for b ∈ (−1, 0), then (b)

(b)

(2+b)

|u; d|2+α,Ω ≤ C |u; d|0,Ω + |f ; d|α,Ω

where C = C(n, α, b).

348

9 Elliptic Operators

Proof: For the time being, ﬁx x ∈ Ω. Letting δ = d(x)/2, B = Bx,δ , and applying Theorem 9.4.3, d1+b (x)|Du(x)| + d2+b (x)|D2 u(x)| ≤ db (x)|u; d|2+α,B

b ≤ d (x)C sup |u(y)| + sup d2 (y)|f (y)| y∈B

y∈B

|f (y ) − f (y )| + sup d2+α (y , y ) |y − y |α y ,y ∈B

≤ C sup db (x)|u(y)| + sup db (x)d2 (y)|f (y)| y∈B

y∈B b

2+α

+ sup d (x)d y ,y ∈B

|f (y ) − f (y )| (y , y ) |y − y |α

.

Since d(y) < δ + d(x) = (3/2)d(x) for y ∈ B, d(x) > (2/3)d(y) and (b) (2+b) d1+b (x)|Du(x)| + d2+b (x)|D2 u(x)| ≤ C |u; d|0,B + |f ; d|α,B (b) (2+b) ≤ C |u; d|0,Ω + |f ; d|α,Ω . Taking the supremum over x ∈ Ω, (b) (b) (2+b) |u; d|2+0,Ω ≤ C |u; d|0,Ω + |f ; d|α,Ω .

(9.24)

Consider now two points x, y ∈ Ω labeled so that d(x) ≤ d(y) with x = y and deﬁne δ and B as before. As above, for |x − y| ≤ δ d2+α+b (x, y)

|D2 u(x) − D2 u(y)| ≤ db (x)|u; d|2+α,B |x − y|α (b) (2+b) ≤ C |u; d|0,Ω + |f ; d|α,Ω .

On the other hand, for |x − y| > δ d2+α+b (x, y)

|D2 u(x) − D2 u(y)| 2α 2+α+b (|D2 u(x)| + |D2 u(y)|) ≤ d (x) |x − y|α dα (x) ≤ 2α (d2+b (x)|D2 u(x)| + d2+b (y)|D2 u(y)|) (b)

≤ 4|u; d|2+0,Ω (b)

(2+b)

≤ 4C(|u; d|0,Ω + |f ; d|α,Ω ), the latter inequality following from Inequality (9.24). Taking the supremum over x, y ∈ Ω and combining the result with Inequality (9.24), (b)

(b)

(b)

(b)

(2+b)

|u; d|2+α,Ω = |u; d|2+0,Ω + [u; d]2+α,Ω ≤ C(|u; d|0,Ω + |f ; d|α,Ω ).

9.6 The Dirichlet Problem for a Ball

349

Lemma 9.6.2 Let L be a strictly elliptic operator with c ≤ 0 on a ball B = Bx0 ,ρ ⊂ Rn satisfying |aij |, |bi |, |c| ≤ K

on B, i, j = 1, . . . , n. (2+b)

If u ∈ C 0 (B − ) ∩ C 2 (B), α ∈ (0, 1), b ∈ (−1, 0), f ∈ Hα (B; d), Lu = f on (b) (2+b) B, and u = 0 on ∂B, then |u; d|0,B ≤ C|f ; d|0,B where C = C(n, b, ρ, m, K). (2+b)

Proof: Since f ∈ Hα

(B; d),

A = sup d2+b (x)|f (x)| < +∞. x∈B

The inequality is trivial if A = 0, for then f = 0 and u = 0 by the weak maximum principle, Corollary 9.5.3. It can be assumed that A > 0. Let w1 (x) = (ρ2 − r2 )−b where r = |x − x0 |. If x = (x1 , . . . , xn ) and x0 = (x01 , . . . , x0n ), then Lw1 (x) = +

+

n

aii (2b)(ρ2 − r2 )−b−1

i=1 n

aij 4b(b + 1)(ρ2 − r2 )−b−2 (xi − x0i )(xj − x0j )

i,j=1 n

bi (2b)(ρ2 − r2 )−b−1 (xi − x0i ) + c(x)(ρ2 − r2 )−b

i=1

2

2 −b−2

= b(ρ − r )

n

4(b + 1)aij (xi − x0i )(xj − x0j )

i,j=1

c(x) 2 2 + 2b(ρ − r ) (ρ − r ) (aii + bi (xi − x0i )) + 2b i=1 √ ≤ b(ρ2 − r2 )−b−2 4(1 + b)mr2 + 2(ρ2 − r2 )(mn − K nr) . 2

2 −b−1

n

Since the ﬁrst term within the parentheses has the limit 4(1 + b)mρ2 as r → ρ, there is 0 < ρ0 < ρ such that Lw1 (x) ≤ −k(ρ2 − r2 )−b−2 for some constant k and ρ0 ≤ r < ρ. Since (ρ2 − r2 )−b−2 = (ρ − r)−b−2 (ρ + r)−b−2 ≥ (ρ − r)−b−2 (2ρ)−b−2 , there are positive constants c1 and c2 such that c1 (ρ − r)−b−2 0 ≤ r < ρ0 (9.25) Lw1 (x) ≤ −c2 (ρ − r)−b−2 ρ0 ≤ r < ρ. Now let w2 (x) = eaρ − ea(x1 −x01 ) where a is a positive constant that will be speciﬁed later. Then Lw2 (x) = −a11 a2 ea(x1 −x01 ) − b1 aea(x1 −x01 ) + c(x)(eaρ − ea(x1 −x01 ) ).

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9 Elliptic Operators

Since c ≤ 0, eaρ − ea(x1 −x01 ) ≥ 0, and a11 ≥ m, Lw2 (x) ≤ ea(x1 −x01 ) a(−ma + |b1 |)

|b1 | a(x1 −x01 ) a a− = −me . m Choosing a so that a ≥ 1 + supB (|b1 |/m),

|b1 | |b1 | b1 b1 − a a− ≥ 1 + sup 1 + sup ≥1 m m B m B m and therefore Lw2 (x) ≤ −mea(x1 −x01 ) ≤ −me−aρ

on B.

Note that Lw2 (x) ≤ 0 on B. For 0 ≤ r < ρ0 , using the fact that b ∈ (−1, 0) Lw2 (x) ≤ −me−aρ (ρ − r)−b−2 (ρ − r)2+b ≤ −c3 (ρ − r)−b−2 where c3 = me−aρ (ρ − ρ0 )2+b . Therefore, −c3 (ρ − r)−b−2 Lw2 (x) ≤ 0

if 0 ≤ r < ρ0 if ρ0 ≤ r < ρ.

(9.26)

Consider the constants η1 = 1/c2 , η2 = (1 + cc12 )/c3 , and the function w = η1 w1 + η2 w2 . By Inequalities (9.25) and (9.26), Lw(x) ≤ −(ρ − r)−b−2 = −d−b−2 (x) ≤ −

|f (x)| . A

Therefore, L(Aw ± u)(x) ≤ −|f (x)| ± f (x) ≤ 0. When r = ρ, w = η2 (eaρ − ea(x1 −x01 ) ) ≥ 0. Since u = 0 on ∂B by hypothesis, Aw ± u ≥ 0 on ∂B. By the minimum principle, Corollary 9.5.3, Aw ± u ≥ 0 on B; that is, |u(x)| ≤ Aw(x)

on B.

An upper bound for the function w can be obtained as follows. First note that w1 (x) = (ρ2 − r2 )−b = (ρ − r)−b (ρ + r)−b ≤ ρ−b (ρ − r)−b

on B.

Since w2 (x) = eaρ − ea(x1 −x01 ) depends only upon the ﬁrst coordinate of x, it can be assumed that x is on the x1 -axis. When x1 − x01 = ρ, w2 (x) = 0. By the mean value theorem, |w2 (x)| = |∇(x) w2 (z)|(ρ − r) ≤ aeaρ (ρ − r)−b ρ1+b = c4 (ρ−r)−b where z lies on the line segment joining x to x0 +(ρ, 0, . . . , 0). Thus, |u(x)| ≤ AC(ρ − r)−b = ACd−b (x)

9.6 The Dirichlet Problem for a Ball

351

so that (b)

(2+b)

|u; d|0,B = sup db (x)|u(x)| ≤ AC = C sup d2+b (x)|f (x)| = C|f ; d|0,B . x∈B

x∈B

Theorem 9.6.3 If α ∈ (0, 1), b ∈ (−1, 0), B is a ball in Rn , and L is a strictly elliptic operator on Ω with coeﬃcients satisfying |aij |α,B , |bi |α,B , |c|α,B ≤ K < +∞,

i, j = 1, . . . , n.

(b)

(2+b)

then L is a bounded, linear map from H2+α (B; d) into Hα

(9.27)

(B; d).

Proof: The assertion follows if it can be shown that there is a constant C such that (2+b)

(2+b)

(2+b)

(b)

|aij Dij u; d|α,B , |bi Di u; d|α,B , |cu; d|α,B ≤ C|u; d|2+α,B (b)

for u ∈ H2+α (B; d). By Inequality (9.10), (2+b)

(0)

(2+b)

(2+b)

|aij Dij u; d|α,B ≤ |aij ; d|α,B |Dij u; d|α,B ≤ max (1, d(B)α )K|Dij u; d|α,B (2+b)

(0)

(2+b)

(2+b)

|bi Di u; d|α,B ≤ |bi ; d|α,B |Di u; d|α,B ≤ max (1, d(B)α )K|Di u; d|α,B (2+b)

(0)

(2+b)

(2+b)

|cu; d|α,B ≤ |c; d|α,B |u; d|α,B ≤ max (1, d(B)α )K|u; d|α,B

(2+b)

so that it suﬃces to show that there is a constant C such that |Dij u; d|α,B , (2+b)

(2+b)

(b)

|Di u; d|α,B , |u; d|α,B ≤ C|u; d|2+α . An upper bound for the second-order term follows from the inequality (2+b)

(b)

(b)

(b)

|Dij u; d|α,B ≤ [u; d]2+0,B + [u; d]2+α,B ≤ |u; d|2+α,B . Consider two points x, y ∈ B so labeled that d(x) ≤ d(y). A simple convexity argument shows that if z is a point on the line segment joining x to y, then d(z) ≥ d(x). Applying the mean value theorem, 2+b)

|Di u; d|α,B = sup d2+b (x)|Di u(x)| + sup d2+b+α (x, y) x∈B

x,y∈B

√ (b) (b) ≤ d(B)[u; d]1+0,B + d(B) n[u; d]2+0,B √ (b) ≤ d(B)(1 + n)|u; d|2+α,B .

|Di u(x) − Di u(y)| |x − y|α

Another application of the mean value theorem results in the inequality √ (2+b) (b) (b) |u; d|α,B ≤ d2 (B)[u; d]0,B + d2 (B) n[u; d]1+0,B √ (b) ≤ d2 (B)(1 + n)|u; d|2+α,B .

352

9 Elliptic Operators

This shows that (2+b)

(b)

(b)

|Lu; d|α,B ≤ C|u; d|2+α,B for all u ∈ H2+α (B; d). Lemma 9.6.4 If α ∈ (0, 1), b ∈ (−1, 0), B = Bx0 ,ρ is a ball in Rn , and (2+b) (b) (B; d), then there is a unique u ∈ C 0 (B − ) ∩ H2+α (B; d) such that f ∈ Hα Δu = f on B and u = 0 on ∂B. Proof: It is easily seen that f ∈ Hα,loc (B). By Theorem 8.4.4, there is a unique u ∈ C 0 (B − ) ∩ C 2 (B) such that Δu = f on B and u = 0 on ∂B. By Lemma 9.3.1, (2)

(2+b)

|u; d|2+α,B ≤ C( u 0,B + |f ; d|α,B ) ≤ C( u 0,B + ρ−b |f ; d|α,B ) < +∞ so that u ∈ H2+α (B; d). By Theorem 8.4.4, (b)

(2+b)

2+b)

[u; d]0,B ≤ [f ; d]0,B

≤ |f ; d|α,B < +∞. (b)

The hypotheses of Lemma 9.6.1 are satisﬁed and so u ∈ H2+α (B; d). Theorem 9.6.5 Let B be a ball in Rn , let α ∈ (0, 1), b ∈ (−1, 0), and let L be a strictly elliptic operator on B with c ≤ 0 satisfying Inequality (9.27). If (2+b) (b) |f ; d|α,B < +∞, then there is a unique u ∈ C 0 (B − ) ∩ H2+α (B; d) satisfying Lu = f on B, u = 0 on ∂B, and (b)

(b)

(2+b)

|u; d|2+α,B ≤ C(|u; d|0,B + |f ; d|α,B ) < +∞

(9.28)

where C = C(n, α, b, m, K). Proof: For 0 ≤ t ≤ 1, let Lt u = tLu + (1 − t)Δu. Since L and Δ are bounded linear operators from the Banach space B1 = (b) (2+b) H2+α (B; d) into the Banach space B2 = Hα (B; d), the same is true of each Lt with the m in Inequality (9.2) replaced by mt = min (1, m) and the K in Inequality (9.27) replaced by Kt = max (1, K). Consider the Dirichlet prob(b) lem Lt u = f on B and u = 0 on ∂B. Suppose there is a u ∈ H2+α (B; d) sat(b) (b) isfying the equation Lt u = f . Since d(x)(b) |u(x)| ≤ [u; d]0,B ≤ |u; d|2+α,B < +∞, the function u has a continuous extension, denoted by the same symbol, to B − with u = 0 on ∂B. By Lemma 9.6.2, (b)

(2+b)

|u; d|0,B ≤ C|f ; d|0,B

(2+b)

≤ C|f ; d|α,B < +∞.

9.7 Dirichlet Problem for Bounded Domains

353

Using the fact that d(B)b ≤ d(x)b for x ∈ B, it is easily seen that u ∈ (b) H2+α (B; d) implies u ∈ H2+α (B; d) and it follows from Lemma 9.6.1 that (b)

(b)

(2+b)

(2+b)

|u; d|2+α,B ≤ C1 (|u; d|0,B + |f ; d|α,B ) ≤ C|f ; d|α,B or, equivalently,

u B1 ≤ C Lt u B2

where C is independent of t. Since the Laplacian L0 = Δ is an invertible map from B1 onto B2 by Lemma 9.6.4, it follows from Theorem 9.2.2 that L is (2+b) (b) invertible. Thus, given f ∈ Hα (B; d) there is a u ∈ H2+α (B; d) such that Lu = f on B and u = 0 on ∂B. Uniqueness follows from Theorem 9.5.4. Theorem 9.6.6 Let B be a ball in Rn , let α ∈ (0, 1), b ∈ (−1, 0), and let L be a strictly elliptic operator with c ≤ 0 and coeﬃcients satisfying In(2+b) equalities (9.22). If |f ; d|α,B < +∞ and g ∈ C 0 (∂B), there is a unique u ∈ C 0 (B − ) ∩ H2+α,loc (B) that solves the problem Lu = f on B and u = g on ∂B. Proof: Consider any g ∈ C 3 (B − ) and the Dirichlet problem Lv = f − Lg on (2+b) B and v = 0 on ∂B. Since f − Lg ∈ Hα (B; d), by the preceding theorem (b) 0 − there is a unique v ∈ C (B ) ∩ H2+α (B; d) that solves this problem and satisﬁes Inequality (9.28). Letting u = v + g, Lu = f on B, u = g on ∂B, and (b) u ∈ C 0 (B − ) ∩ H2+α (B; d). Consider any g ∈ C 0 (∂B). It can be assumed that 0 − g ∈ C (B ) by the Tietze Extension Theorem. Now let {gj } be a sequence in C 3 (B − ) that converges uniformly to g on B − . For each j ≥ 1, let vj be the solution of the problem Lvj = f − Lgj on B and vj = gj on ∂B as in the preceding step. Letting uj = vj + gj , Luj = f on B, uj = gj on ∂B, (b) and uj ∈ C 0 (B − ) ∩ H2+α (B; d). Since L(ui − uj ) = 0 for i, j ≥ 1, by the weak maximum principle, Corollary 9.5.3, ui − uj 0,B − ≤ gi − gj 0,∂B , the sequence {uj } is Cauchy on B − , and therefore uniformly bounded on B − . By Corollary 9.4.4, there is a subsequence that converges uniformly on compact subsets of B to a function u ∈ H2+α,loc (B) for which Lu = f on B and u = g on ∂B. Since the sequence {uj } converges uniformly on B − , u also belongs to C 0 (B − ).

9.7 Dirichlet Problem for Bounded Domains Throughout this section, Ω will be a bounded open subset of Rn , L will be a strictly elliptic operator with c ≤ 0 and coeﬃcients aij , bi , c, i, j = 1, . . . n, in Hα,loc (Ω). Unless otherwise noted, f will be a ﬁxed function in Cb0 (Ω) ∩ (2+b) (Ω; d), α ∈ (0, 1), b ∈ (−1, 0). Hα

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9 Elliptic Operators

Deﬁnition 9.7.1 u ∈ C 0 (Ω − ) is a superfunction if for every ball B with closure in Ω and v ∈ C 0 (B − ) ∩ C 2 (B), Lv = f on B, u ≥ v on ∂B implies u ≥ v on B. Reversing the inequalities of this deﬁnition results in the deﬁnition of a subfunction. The fact that u is a superfunction does not mean that −u is a subfunction. Otherwise, superfunctions and subfunctions share most of the properties of superharmonic and subharmonic functions. (2+b)

Deﬁnition 9.7.2 If B = Bx,δ , α ∈ (0, 1), b ∈ (−1, 0), |f ; d|α,B < +∞, and g ∈ C 0 (B − ), LS(x, B, g) will denote the unique solution of the problem Lv = f on B,

v = g on ∂B.

Lemma 9.7.3 u ∈ C 0 (Ω − ) is a superfunction if and only if u ≥ LS(x, B, u) on B for every ball B = Bx,δ with closure in Ω. Proof: Suppose ﬁrst that u is a superfunction on Ω and let v = LS(x, B, u) on B. Then Lv = f on B and v = u on ∂B. Thus, u ≥ v = LS(x, B, u) on B. On the other hand, suppose u ≥ LS(x, B, u) on B. Consider any v ∈ C 0 (B − ) ∩ C 2 (B) satisfying Lv = f on B and u ≥ v on ∂B. Since L(LS(x, B, u) − LS(x, B, v)) = 0 on B and LS(x, B, u) ≥ LS(x, B, v) on ∂B, by Theorem 9.5.4 u ≥ LS(x, B, u) ≥ LS(x, B, v) = v on B, showing that u is a superfunction on Ω. Lemma 9.7.4 u ∈ C 0 (Ω − )∩C 2 (Ω) is a superfunction if and only if Lu ≤ f on Ω; moreover, if u ∈ C 0 (Ω − ) ∩ C 2 (Ω) is a superfunction and k ≥ 0, then u + k is a superfunction. Proof: Suppose ﬁrst that Lu ≤ f on Ω. Let B be a ball with closure in Ω and let v ∈ C 0 (B − ) ∩ C 2 (B) satisfy Lv = f on B and u ≥ v on ∂B. By Theorem 9.5.4, u ≥ v on B and u is a superfunction. Now let u be a superfunction and assume that Lu(x0 ) > f (x0 ) for some point x0 ∈ Ω. Then there is a ball B = Bx0 ,δ such that Lu > f on B. By Theorem 9.6.6, there is a function uB on B − such that LuB = f on B and uB = u on ∂B. Since L(u − uB ) > f − f = 0 on B, by the weak maximum principle, Corollary 9.5.3, u − uB ≤ sup∂B (u − uB )+ = 0 on B so that u ≤ uB on B. Since u is a superfunction, u ≥ uB on B, and therefore u = uB on B so that Lu = LuB on B. Thus, f (x0 ) = LuB (x0 ) = Lu(x0 ) > f (x0 ), a contradiction, and it follows that Lu ≤ f on Ω. The second assertion follows easily from the ﬁrst and the fact that Lk = ck ≤ 0.

9.7 Dirichlet Problem for Bounded Domains

355

Lemma 9.7.5 If v ∈ C 0 (Ω − ) is a superfunction, u ∈ C 0 (Ω − ) is a subfunction, and v ≥ u on ∂Ω, then v ≥ u on Ω. Proof: Let m = inf Ω − (v − u) and let Γ = {y ∈ Ω; v(y) − u(y) = m}. If Γ = ∅, then v − u attains its minimum value on ∂Ω, and therefore v − u ≥ 0 on Ω − since v − u ≥ 0 on ∂Ω. Assume Γ = ∅ and let z be a point of Γ such that d(z) = d(Γ, ∂Ω) > 0. Let B = Bz,δ be a ball with closure in Ω. Note that B ∩ (Ω ∼ Γ ) = ∅ since z ∈ ∂Γ . By Theorem 9.6.6, there is then a vB ∈ C 0 (B − ) ∩ C 2 (B) satisfying LvB = f on B and vB = v on ∂B. Thus, vB ≤ v on B since v is a superfunction. Likewise, there is a uB ∈ C 0 (B − ) ∩ C 2 (B) satisfying LuB = f on B, uB = u on ∂B, and uB ≥ u on B. Note that L(vB − uB ) = 0 on B and that m = v(z) − u(z) ≥ vB (z) − uB (z). Since vB − uB = v − u ≥ m on ∂B, vB − uB ≥ m on B by the weak maximum principle, Corollary 9.5.3. It follows that vB (z) − uB (z) = m so that vB − uB attains its minimum value at an interior point of B; if m ≤ 0, then vB − uB would be constant on B by the strong maximum principle, Theorem 9.5.6. But this contradicts the fact that B ∩ (Ω ∼ Γ ) = ∅ where vB − uB > m. Thus, m ≥ 0 and v ≥ u on Ω. Lemma 9.7.6 If u ∈ C 0 (Ω − ) is a superfunction, B is a ball with closure in Ω, Lv = f on B, v = u on ∂B, then uB =

v u

on B on Ω ∼ B

is a superfunction. If u and v are superfunctions on Ω and u ≥ v on Ω, then uB ≥ vB on Ω. Proof: Let B1 be an arbitrary ball with closure in Ω, and let w1 ∈ C 0 (B1− ) ∩ C 2 (B1 ) satisfy Lw1 = f on B1 , w1 = uB on ∂B1 . Since u is a superfunction, u ≥ uB on Ω and, in particular, u ≥ w1 on ∂B1 . Using the fact that u is a superfunction again, u ≥ w1 on B1 . Since uB = u on Ω ∼ B, uB = u ≥ w1 on B1 ∼ B. Noting that LuB = Lv = f on B, L(uB − w1 ) = 0 on B ∩ B1 . Since uB = u ≥ w1 on ∂B ∩ B1 and uB ≥ w1 on ∂B1 ∩ B, uB − w1 ≥ 0 on (∂B ∩B1 )∪(∂B1 ∩B). By continuity, uB −w1 ≥ 0 on ∂(B ∩B1 ) ⊂ (∂B ∩B1− )∪ (∂B1 ∩ B − ). It follows from the weak maximum principle, Corollary 9.5.3, that uB ≥ w1 on B ∩ B1 and therefore on B1 . This proves that uB is a superfunction. Let u and v be superfunctions on Ω with u ≥ v on Ω. Since L(uB − vB ) = 0 on B and uB − vB = u − v ≥ 0 on ∂B, uB − vB ≥ 0 by the weak maximum principle. The function uB of the preceding lemma will be called the lowering of u over B; the corresponding operation for subfunctions will be called the lifting of u over B and will be denoted by uB .

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9 Elliptic Operators

Lemma 9.7.7 If u1 , u2 ∈ C 0 (Ω − ) are superfunctions, then min (u1 , u2 ) is a superfunction. Proof: Let u = min (u1 , u2 ) and consider any ball B with closure in Ω. Suppose Lw = f on B and u ≥ w on ∂B. Then ui ≥ w on ∂B so that ui ≥ w on B, i = 1, 2. Therefore, u = min (u1 , u2 ) ≥ w on B and u is a superfunction. Deﬁnition 9.7.8 A function v ∈ C 0 (Ω − ) is a supersolution relative to the bounded function g on ∂Ω if v is a superfunction and v ≥ g on ∂Ω. The set of supersolutions relative to g will be denoted by U(f, g). There is a corresponding deﬁnition of subsolution relative to the bounded function g on ∂Ω; the set of such functions will be denoted by L(f, g). According to Lemma 9.7.5, each subsolution relative to g is less than or equal to each supersolution relative to g. The hypothesis of the following lemma is an exception to the requirement that f be a ﬁxed function. (2+b)

(Ω; d) with supf ∈F f 0,Ω ≤ M < Lemma 9.7.9 If F ⊂ Cb0 (Ω) ∩ Hα +∞, G is a collection of bounded functions on ∂Ω with supg∈G g 0,∂Ω ≤ N < +∞, then ' ' U(f, g) ∩ C 2 (Ω − ) = ∅ and L(f, g) ∩ C 2 (Ω − ) = ∅; f ∈F ,g∈G

moreover, the set

f ∈F ,g∈G

f ∈F ,g∈G U(f, g)

L(f, g) is uniformly bounded f ∈F ,g∈G

below (above) by a constant. Proof: Choose d > 0 so that Ω ⊂ {x; −d ≤ x1 ≤ d}. Let β=

sup

(|bi (x)|/m),

x∈Ω,1≤i≤n

let ρ > β + 1, and let L1 = L − c. Since β < ρ − 1, L1 eρ(d+x1 ) = a11 ρ2 eρ(d+x1 ) + b1 ρeρ(d+x1 ) ≥ mρ(ρ − β)eρ(d+x1 ) ≥ m. Letting v+ = N +

1 2ρd (e − eρ(d+x1 ) )M, m

Lv+ = L1 v+ + cv+ = L1 (−eρ(d+x1 ) )M/m + cv+ ≤ −M ≤ − f 0 ≤ f for all f ∈ F. Since v+ ∈ C 0 (Ω − ) ∩ C 2 (Ω) and v+ ≥ N ≥ g on ∂Ω, v+ is a supersolution relative to g for all g ∈ G and therefore ' U(f, g) ∩ C 2 (Ω − ) = ∅. f ∈F ,g∈G

9.7 Dirichlet Problem for Bounded Domains

Now let v− = −N −

357

1 2ρd (e − eρ(d+x1 ) )M. m

Then Lv− = L1 v− + cv− ≥ L1 (eρ(d+x1 ) )M/m ≥ M ≥ f 0 ≥ f. for all f ∈ F. Since v− ∈ C 0 (Ω − ) ∩ C 2 (Ω) and v− ≤ −N ≤ g on ∂Ω, v− is a subsolution relative to g for all g ∈ G and it follows that ' L(f, g) ∩ C 2 (Ω − ) = ∅. f ∈F ,g∈G

Consider now a ﬁxed f and g. Since each subsolution is less than or equal to each supersolution, U(f, g) is bounded below by each element of L(f, g) and L(f, g) is bounded above by each element of U(f, g). Since v+ ≤ N + e2ρd M/m, the elements of L(f, g) are uniformly bounded above by the constant N + e2ρd M/m. Similarly, the elements of U(f, g) are uniformly bounded below by the constant −N − e2ρd M/m. Deﬁnition 9.7.10 The Perron supersolution for the bounded function g + on ∂Ω is the function Hf,g = inf{v; v ∈ U(f, g)} and the Perron subsolu− tion for g is the function Hf,g = sup{v; v ∈ L(f, g)}. It can be assumed that U(f, g) is uniformly bounded on Ω by considering only those u for which u ≤ u0 for some u0 ∈ U(f, g) since + = inf {u ∧ u0 ; u ∈ U(f, g)}. Hf,g + − − It follows that Hf,g is bounded on Ω as is Hf,g . Clearly, Hf,g ≤ u for all − + u ∈ U(f, g) and consequently Hf,g ≤ Hf,g . The same assumption can be made for L(f, g). − + Deﬁnition 9.7.11 If Hf,g = Hf,g on Ω, then g is said to be L-resolutive and Hf,g is deﬁned to be the common value. + Lemma 9.7.12 If g is a bounded function on ∂Ω, then Hf,g ∈ H2+α (B; d) + − for every closed ball B with B ⊂ Ω and satisﬁes Poisson’s equation LHf,g = + 2 f on Ω; in particular, Hf,g ∈ C (Ω).

Proof: By Lemma 2.2.8, there is a countable set Ig ⊂ U(f, g) such that for + every l.s.c. function w on Ω, w ≤ Hf,g whenever w ≤ inf {v; v ∈ Ig }. By Lemma 9.7.7, it can be assumed that Ig is a decreasing sequence {uj } of uniformly bounded functions on Ω. Let u = limj→∞ uj and let B be a ﬁxed ball with B − ⊂ Ω. By replacing each uj by (uj )B , if necessary, it can be assumed that each uj satisﬁes the equation Luj = f on B. By Theorem 9.4.4, there is a subsequence {ujk } that converges uniformly on compact subsets of B to a function w ∈ H2+α (B; d) for which Lw = f on B. Since w = u

358

9 Elliptic Operators

on B, Lu = Lw = f on B and u ∈ H2+α (B; d). Since w is continuous on B, u is continuous on B, and since B is an arbitrary ball with closure in Ω, u is + since each ujk ∈ U(f, g). continuous on Ω and Lu = f on Ω. Clearly, u ≥ Hf,g Since u is continuous on Ω and u ≤ inf {v; v ∈ Ig }, u ≤ inf {v; v ∈ U(f, g)} = + + + on Ω. Therefore, u = Hf,g on Ω and LHf,g = Lu = f on Ω. Hf,g

9.8 Barriers Throughout this section, Ω will be a bounded open subset of Rn , L will be a strictly elliptic operator with c ≤ 0 and coeﬃcients aij , bi , c, i, j = 1, . . . n, in Hα (Ω). Unless noted otherwise, f will be a ﬁxed bounded function in (2+b) Hα (Ω; d), α ∈ (0, 1), b ∈ (−1, 0). ± Under the hypothesis that g is a bounded function on ∂Ω, Hf,g satisfy the ± equations LHf,g = f on Ω. In the case of the Laplacian with f = 0 on Ω, it was possible to show that every continuous function g on ∂Ω is Δ-resolutive. This is the statement of Wiener’s theorem, Theorem 2.6.16. Deﬁnition 9.8.1 If g is a bounded function on ∂Ω that is continuous at x0 ∈ ∂Ω, a sequence of functions {wj+ } in C 0 (Ω − ) is an approximate upper barrier at x0 for g if (i) each wj+ is a supersolution relative to g and (ii) limj→∞ wj+ (x0 ) = g(x0 ). An approximate lower barrier {wj− } is deﬁned by using subsolution in place of supersolution. If both an approximate upper barrier and an approximate lower barrier at x0 for g exist, the pair {wj± } is called an approximate barrier at x0 for g. Lemma 9.8.2 Let g be a bounded function on ∂Ω that is continuous at x0 ∈ ∂Ω. If there is an approximate barrier {wj± } at x0 for g, then ± limy→x0 ,y∈Ω Hf,g (y) = g(x0 ). Proof: Since every subsolution for g is less than or equal to every supersolution for g, + wj− ≤ Hf,g ≤ wj+

so that + (x) − g(x0 ) wj− (x) − wj− (x0 ) + wj− (x0 ) − g(x0 ) ≤ Hf,g

≤ wj+ (x) − wj+ (x0 ) + wj+ (x0 ) − g(x0 )

9.8 Barriers

359

for x ∈ Ω. Given > 0, choose j0 ≥ 1 such that wj+0 (x0 ) − g(x0 ) < /2 and wj−0 (x0 ) − g(x0 ) > − /2. Then wj−0 (x) − wj−0 (x0 ) −

+ ≤ Hf,g (x) − g(x0 ) ≤ wj+0 (x) − wj+0 (x0 ) + 2 2

and since wj±0 ∈ C 0 (Ω − ), + + − /2 ≤ lim inf (Hf,g (x) − g(x0 )) ≤ lim sup (Hf,g (x) − g(x0 )) ≤ /2 x→x0 ,x∈Ω

x→x0 ,x∈Ω

+ and so limx→x0 ,x∈Ω Hf,g (x) = g(x0 ).

Lemma 9.8.3 If x0 ∈ ∂Ω, g is a bounded function on ∂Ω that is continuous at x0 , and w ∈ C 0 (Ω − ) ∩ C 2 (Ω) satisﬁes (i) Lw ≤ −1 on Ω, (ii) w(x0 ) = 0, and (iii) w > 0 on ∂Ω ∼ {x0 }, then for each j ≥ 1 there is a constant γj > 0 such that the functions wj+ = g(x0 ) + (1/j) + γj w and wj− = g(x0 ) − (1/j) − γj w are supersolutions and subsolutions, respectively, and {wj± } is an approximate barrier at x0 for g. Proof: Using the continuity of g at x0 , for each j ≥ 1 there is a neighborhood U of x0 in ∂Ω such that g(x0 ) − (1/j) ≤ g(x) ≤ g(x0 ) + (1/j) for x ∈ U . On ∂Ω ∼ U, w is bounded below by a positive constant and it follows that there is a constant βj > 0 such that g(x0 ) −

1 1 − βj w(x) ≤ g(x) ≤ g(x0 ) + + βj w(x) j j

for x ∈ ∂Ω.

For each j ≥ 1, let γj = max (βj , f − c(g(x0 ) + 1j ) 0 ) and let wj± = g(x0 ) ±

1 ± γj w j

on Ω − .

Clearly, wj− (x) ≤ g(x) ≤ wj+ (x), x ∈ ∂Ω. Since Lw ≤ −1 on Ω and c ≤ 0, Lwj+ = c(g(x0 ) + (1/j)) + γj Lw ≤ cg(x0 ) + (c/j) − γj ≤ f . This shows that each wj+ is a superfunction. Since wj+ (x0 ) = g(x0 ) + (1/j) → g(x0 ) as j → ∞, {wj+ } is an upper barrier at x0 for g. Similarly, {wj− } is a lower barrier at x0 for g. By analogy with the classic case, a point x0 ∈ ∂Ω is called a L-regular ± boundary point if limy→x0 ,y∈Ω Hf,g (y) = g(x0 ) whenever g is bounded on ∂Ω and continuous at x0 . The region Ω is called L-regular if each boundary point is L-regular. It will be shown below that Poincar´e’s exterior sphere

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9 Elliptic Operators

condition at a point x0 ∈ ∂Ω suﬃces to show that x0 is a L-regular boundary point; that is, the existence of a ball B with Ω − ∩B − = {x0 } suﬃces to prove L-regularity at x0 . Assuming that Ω has a smooth boundary and that the aij , bi , c, and f are locally Lipschitz continuous on Ω, Herv´e [30] has shown that a boundary point x0 ∈ ∂Ω is L-regular if and only if it is Δ-regular. The following lemma in conjunction with Lemma 9.8.2 shows that lim

y→x0 ,y∈Ω

Hg± (y) = g(x0 )

if Ω satisﬁes an exterior sphere condition at x0 ∈ ∂Ω. It will be shown below that the exterior sphere condition can be replaced by a Zaremba cone condition. Lemma 9.8.4 (Poincar´ e [50]) If the bounded open set Ω satisﬁes an exterior sphere condition at x0 ∈ ∂Ω, the coeﬃcients of L are bounded, and g is a bounded function on ∂Ω that is continuous at x0 , then there is an approximate barrier at x0 for g. Proof: Let B = By,δ satisfy the exterior sphere condition at x0 , let r = |x−y|, and let vβ (x) = δ −β − r−β , β > 0. It can be assumed that δ < 1. Since r−β < δ −β for r > δ, Lvβ (x) = −β(β + 2)r−β−4

n

aij (xi − yi )(xj − yj )

i,j=1

+ βr−β−2

n aii + bi (xi − yi ) + c(x) δ −β − r−β i=1

≤ βr−β−2 (−β − 2)m +

n aii + bi (xi − yi ) ,

x ∈ Ω.

i=1

Since r−β−2 < δ −β−2 for x ∈ Ω and the coeﬃcients of L are bounded, there is a constant k > 0 such that x ∈ Ω. Lvβ (x) ≤ βδ −β−2 (−β − 2)m + k , Since the right side of this inequality has the limit −∞ as β → +∞, there is a positive constant β0 such that Lvβ0 (x) ≤ −1 for all x ∈ Ω. Letting w(x) = vβ0 (x), Lw(x) ≤ −1 for x ∈ Ω, w(x0 ) = 0, and w > 0 on ∂Ω ∼ {x0 }. Thus, there is an approximate barrier at x0 for g by the preceding lemma. It was shown in Theorem 2.6.29 that the existence of a cone C in ∼ Ω with vertex at x0 ∈ ∂Ω implies that x0 is a Δ-regular boundary point for Ω. By making use of a polar representation of certain functions on Ω, it will be shown that this same condition can be used to prove the existence of an approximate barrier at x0 for the Dirichlet problem for elliptic operators.

9.8 Barriers

361

The following results from the calculus will be needed to derive the polar representation. If (ρ, φ) are the polar coordinates of the point (x, y) ∈ R2 and u is a function deﬁned on a neighborhood of (x, y), under suitable conditions the partial derivatives ux and uy can be converted to polar coordinates as follows: 1 ∂u = uρ cos φ − uφ sin φ ∂x ρ 1 ∂u = uρ sin φ + uφ cos φ ∂y ρ

(9.29) (9.30)

The second partials uxx , uxy , and uyy also can be converted to polar coordinates as follows: 2 sin2 φ ∂ 2 u sin φ cos φ ∂ 2 u ∂2u 2 ∂ u + = cos φ − 2 ∂x2 ∂ρ2 ρ ∂ρ∂φ ρ2 ∂φ2 2 sin φ cos φ ∂u sin φ ∂u + +2 ρ2 ∂φ ρ ∂ρ

1 − 2 sin2 φ ∂ 2 u ∂ 2u ∂2u sin φ cos φ ∂ 2 u = sin φ cos φ 2 + − ∂x∂y ∂ρ ρ ∂φ∂ρ ρ2 ∂φ2

2 −1 + sin φ ∂u sin φ cos φ ∂u + − ρ ∂ρ ρ2 ∂φ 2 ∂2u sin φ cos φ ∂ 2 u u ∂u ∂ sin φ cos φ +2 = sin2 φ 2 − 2 2 2 ∂y ∂ρ ρ ∂φ ρ ∂φ∂ρ cos2 φ ∂u cos2 φ ∂ 2 u + . + ρ ∂ρ ρ2 ∂φ2

In particular, when φ = 0 ux = uρ , uy = uxx = uρρ ,

uxy =

1 uφ ρ

1 1 uρφ − 2 uφ , ρ ρ

uyy =

(9.31) 1 1 uρ + 2 uφφ . ρ ρ

(9.32)

For notational convenience, the derivation of a polar representation of a function at a point x0 ∈ ∂Ω will be carried out by assuming that x0 = 0. For x ∈ Rn , let r(x) = |x| and let θ(x) = cos−1 (xn /r(x)) for x = 0. Functions of the form u(r, θ) will be considered in order to obtain a polar representation of Lu on a cone having the xn -axis as its axis of symmetry and of half-angle θ0 ; more precisely, if x is a point of the cone, then a polar representation of Lu(x), dependent upon x, will be derived. For this purpose, it is convenient to introduce a second coordinate system dependent upon the ﬁxed point x, which will be referred to as the y-coordinate system. It is easier to describe the construction of the y-coordinate system in

362

9 Elliptic Operators y1

xn x

R

Sn−1

y3 µ y2

µ

r

x0

Fig. 9.1 Polar Representation

geometrical terms. Consider the sphere r = r(x) and the tangent hyperplane πn−1 (x) to this sphere at x. The unit vector η1 = x/|x| is then normal to πn−1 (x) at x. Now let η2 be the unit vector at x tangent to the circle obtained by intersecting the sphere r = r(x) with the plane spanned by the point x and the xn -axis, oriented in the direction of increasing θ. Now let η3 , . . . , ηn be an orthogonal basis for the orthogonal complement in πn−1 (x) of the linear space spanned by η2 . Then the vectors η2 , . . . , ηn constitute an northogonal basis for πn−1 (x). Then any point y ∈ Rn can be written y = i=1 yi ηi for suitable choice of the yi and (y1 , . . . , yn ) are the coordinates of y relative to the y-coordinate system centered at x. After a translation and orthogonal transformation, a function f in x1 , . . . , xn coordinates induces a function f˜ in y1 , . . . , yn coordinates by the equation f˜(y) = f (x + Oy), where O = [oij ] is an orthogonal matrix; in particular, f˜(0) = f (x). If S is any sphere in Rn and u is a constant function on S, then the rate of change of u at any point y of S in the direction of a tangent line at y is zero. In particular, if (y1 , y2 , 0, . . . , 0) is any point in the y1 y2 -plane and S is a sphere in Rn having center at y = 0 and passing through the point (y1 , y2 , 0, . . . , 0), and the function u is constant on S, then for i ≥ 3, uyi = 0 at (y1 , y2 , 0, . . . , 0) since the yi axis is orthogonal to the y1 y2 -plane and therefore a tangent line. Transforming the operator Lu =

n

aij Dxi xj u +

i,j=1

n

bi Dxi u + cu

i=1

in the x-coordinate system to the y-coordinate system, Lu =

n i,j=1

a ˜ij Dyi yj u +

n

˜bi Dy u + c˜u i

i=1

n where A˜ = [˜ aij ] = Ot AO, ˜bj = i=1 bi oij , and c˜(y) = c(x + Oy).

(9.33)

9.8 Barriers

363

Letting Sn−1 denote the sphere obtained by intersecting the cone having the xn -axis as axis of symmetry and half-angle θ = θ(x) with the sphere of radius r = r(x), the function u is constant on Sn−1 and equal to u(r, θ) there. As remarked above, uyi = 0, i ≥ 3, at any point of the y1 y2 -plane. Therefore, (uyi )y1 (x) = u(yi )y2 (x) = 0,

i ≥ 3.

Consider the matrix D = [uyi yj ] that has the form D2 0 D= , 0 Dn−2 where D2 and Dn−2 are real symmetric matrices of size 2×2 and (n−2)×(n− 2), respectively. Thus, there is an orthogonal matrix Q such that QDn−2 Qt is a diagonal matrix of which the entries on the diagonal are the eigenvalues of Dn−2 . If variables z3 , . . . zn are deﬁned by the equation ⎡ ⎤ ⎡ ⎤ y3 z3 ⎢ .. ⎥ ⎢ .. ⎥ ⎣ . ⎦ = Q⎣ . ⎦, zn

yn

then QDn−2 QT has the form indicated in the matrix ⎡ u y1 y1 u y1 y2 0 0 . . . 0 ⎢ u y2 y1 u y2 y2 0 0 . . . 0 ⎢ ⎢ 0 u z3 z3 0 . . . 0 D=⎢ 0 ⎢ .. .. .. .. . . ⎣ . . . . . 0 0

0

0

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

0 . . . u zn zn

Consider any line through x that is orthogonal to the y1 y2 -plane. The hyperplane through parallel to the y1 -axis will intersect Sn−1 in a circle. The geometrical relationship of the line to the circle is independent of the line; that is, by symmetry, any second derivative of u at x in any direction orthogonal to the y1 y2 -plane is independent of the direction. Hence, uzi zi , i ≥ 3, are all equal. Since these second partials are the eigenvalues of the Dn−2 , Dn−2 must be scalar multiple of the identity matrix (c.f. [31], Lemma 9.5.3, and Exercise 26, p. 386). Since z3 = nj=3 q3j yj is orthogonal to the y1 y2 -plane, uz3 z3 = uy3 y3 and so uzj zj = uz3 z3 = uy3 y3 for j ≥ 3. Therefore, ⎡

u y1 y1 u y1 y2 0 ⎢ u y2 y1 u y2 y2 0 ⎢ ⎢ 0 u y3 y3 D=⎢ 0 ⎢ .. .. .. ⎣ . . . 0 0 0

0 0 0 .. .

... ... ... .. .

0 0 0

0 0 . . . u y3 y3

⎤ ⎥ ⎥ ⎥ ⎥. ⎥ ⎦

(9.34)

364

9 Elliptic Operators

Consider now a function u of the form u(r, θ) = rλ f (θ)). At any point (y1 , y2 ) in the y1 y2 -plane with polar coordinates (ρ, θ + φ) with pole at 0, the value of u is u(ρ, θ + φ) = ρλ f (θ + φ). By the ﬁrst equation in (9.32), when φ=0 uy1 y1 = uρρ = λ(λ − 1)ρλ−2 f (θ). At x, ρ = r(x) and so uy1 y1 (x) = λ(λ − 1)rλ−2 f (θ).

(9.35)

Similarly, using the second equation in (9.32), uy1 y2 (x) = uy2 y1 (x) = rλ−2 (λ − 1)f (θ)

(9.36)

and using the third equation in (9.32), uy2 y2 = rλ−2 (λf (θ) + f (θ)).

(9.37)

Expressing uy3 y3 in terms of r, λ, and f (θ) is more complicated. When y1 = y2 = 0, u is a constant on Sn−1 and has the value u(r, θ) there. For such points, r = (x2n + R2 )1/2 , where R = r sin θ so that u is a function of R only on Sn−1 . Letting (R, ψ) denote polar coordinates in the plane orthogonal to the xn -axis and passing through the point x, by the third equation of (9.32) 1 uR . R Since uR is the directional derivative of u in any direction radiating from the conical axis in the x = xn plane, it can be calculated in the y1 y2 -plane. Returning to polar coordinates (ρ, φ) in the y1 y2 -plane, u = u(ρ, θ + φ) there. Note that uρ = ur at x and uφ = uθ when φ = 0. Since θ is the angle between the y2 -axis and the R-direction as depicted in Figure 9.1, u y3 y3 =

π π − θ) + uy2 sin ( − θ) 2 2 = uy1 sin θ + uy2 cos θ.

uR = uy1 cos (

By translating the y1 , y2 coordinate system to 0, it is easy to see from Equations (9.29) and (9.30) that 1 uy1 = uρ cos φ − uφ sin φ ρ 1 uy2 = uρ sin φ + uφ cos φ. ρ When φ = 0, uy1 = uρ and uy2 = (1/ρ)uφ . At x, uy1 = ur and uy2 = Therefore, 1 uR = ur sin θ + uθ cos θ r

1 r uθ .

9.8 Barriers

365

and u y3 y3 =

1 uR 1 = ur + 2 uθ cot θ = rλ−2 (λf + f cot θ). R r r

(9.38)

The above calculations of the uyi yj lead to a polar representation of L(rλ f (θ)). Theorem 9.8.5 (K. Miller [45]) If λ > 0 and u = rλ f (θ) ∈ C 2 ([0, ∞) × [0, π]), then ˜11 λ(λ − 1)f + 2˜ L rλ f (θ) = rλ−2 a a12 (λ − 1)f + a ˜22 (f + λf ) (9.39) ˜ − 2)(λf + f cot θ) + rλ−1 (˜b1 λf + ˜b2 f + c˜rf ) + d(n where d˜ =

1 n−2

n i=3

aii ∈ [m, M ] and the eigenvalues of the matrix

˜12 a ˜11 a a ˜21 a ˜22

(9.40)

are in [m, M ]. Proof: The above equation follows immediately from Equations (9.35) to (9.38). Since the eigenvalues of A and its diagonal entries are in the interval [m, M ], the same is true of A˜ = OT AO. Since the above matrix is a diagonal ˜ its eigenvalues are in [m, M ]. submatrix of A, The construction of an approximate barrier at x0 ∈ ∂Ω depends upon showing that there is a function w deﬁned on Ω in a neighborhood of x0 satisfying the properties listed in Lemma 9.8.3; in particular, the function w must satisfy the inequality Lw ≤ −1 on Ω. The function w = rλ f (θ) will be used in the construction of an approximate barrier at x0 by choosing f so that L(rλ f (θ))/rλ−2 < 0 on Ω in a neighborhood of x0 and then modifying w. Suppose that every cone C with vertex at x0 , no matter how sharp, contains a point of Ω arbitrarily close to x0 . It would then follow that there is a sequence (rn , θn ) in Ω with limit x0 such that the term f (θ) cot θn in Equation (9.39) could become inﬁnite. By postulating that there is a cone C ⊂∼ Ω with vertex at x0 , the term f (θ) cot θ can be controlled on Ω. Consider any > 0, a positive constant K, and a function f on [0, 2π] satisfying the diﬀerential equation mf − Kf = −3 subject to the following conditions: (i) f (0) = 1, f > 0 on [0, 2π] and f ∈ C 2 ([0, 2π]) (ii) f (0) = 0 and f < 0 on (0, 2π] (iii) f < 0 on [0, 2π].

(9.41)

366

9 Elliptic Operators

The function

3 m m K 3 θ − emθ + 1 + 2 K K K is easily seen to satisfy Equation (9.41) and all of the above conditions save possibly for the requirement that f (θ) > 0 for θ ∈ [0, 2π]. Since the sum on the right side has the limit 1 uniformly on [0, 2π] as → 0, there is an > 0 such that the function f satisﬁes all of the above conditions. The condition that f (0) = 1 in (i) is only a normalizing constant and any other positive constant would do as well. f (θ) =

Lemma 9.8.6 If 0 < θ0 < π, there are constants 0 < r0 < 1, 0 < λ0 < 1 and a function f ∈ C 2 ([0, π]) such that L(rλ f (θ)) ≤ −1, rλ−2

θ ∈ [0, θ0 ]), r < r0 , |λ| < λ0 .

(9.42)

Proof: It can be assumed that r < 1 and |λ| < 1. Fix θ0 ∈ (0, π) and let f be any function as described above. Since f (θ) cot θ ≤ f (θ) cot θ0 and a ˜22 f (θ) ≤ mf (θ) on (0, θ0 ], it follows from Equation (9.39) that ˜11 λ(λ − 1)f + 2˜ L(rλ f (θ)) ≤ rλ−2 a a12 (λ − 1)f + mf + a ˜22 λf ˜ − 2)(λf + f cot θ0 ) + rλ−1 (˜b1 f + ˜b2 f + c˜rf ). + d(n Using the fact that the eigenvalues and diagonal entries of the matrix (9.40) are in [n, M ], it is easily seen that |a12 | = |a21 | ≤ M . Moreover, there are constants A, B > 0 such that L(rλ f (θ)) ≤ rλ−2 (M |λ| |λ − 1| + M |λ| + M (n − 2)|λ|)f + mf − (2M |λ − 1| + M (n − 2)| cot θ0 | + A)f + Brλ−1 f. Now let f be a solution of Equation (9.41) satisfying (i), (ii), and (iii) with K = 4M +M (n−2)| cot θ0 |+A. Letting ελ (θ) = M (|λ| |λ−1|+|λ|+(n−2)|λ|) and noting that f ≤ 1, L(rλ f (θ)) ≤ rλ−2 (mf − Kf + ελ (θ)) + Brλ−1 f where ελ (θ) → 0 uniformly on [0, θ0 ] as λ → 0. By choice of f , L(rλ f (θ)) ≤ −3 + ελ (θ) + Brf. rλ−2 Now choose r0 > 0 and 0 < λ0 < 1 such that ελ (θ) < for θ ∈ (0, θ0 ], |λ| < λ0 , and Brf ≤ Br < for r < r0 and then replace f / by f .

9.8 Barriers

367

Corollary 9.8.7 If x0 ∈ ∂Ω and there is a truncated right circular cone C ⊂∼ Ω of half-angle 0 < ψ < π with vertex at x0 , then there is 0 < λ0 < 1, a neighborhood U of x0 , and a function f ∈ C 2 ([0, π]) such that L(rλ f (θ)) < −1,

|λ| < λ0 , (r, θ) ∈ U ∩ Ω.

Proof: Consider ∼ C ⊃ Ω which is a right circular cone of half-angle θ0 = π − ψ with vertex at x0 . Applying Lemma 9.8.6, there are constants 0 < r0 < 1, 0 < λ0 < 1 and a function f ∈ C 2 (([0, π]) such that L(rλ f (θ)) ≤ −1, rλ−2

θ ∈ [0, θ0 ], r < r0 , |λ| < λ0 .

Since rλ−2 → +∞ as r → 0, there is a neighborhood U of x0 such that L(rλ f (θ)) < −1,

|λ| < λ0 , (r, θ) ∈ U ∩ Ω.

The condition of the above corollary is known as Zaremba’s cone condition. (2+b)

(Ω; d) with supf ∈F f 0,Ω ≤ M < Lemma 9.8.8 If F ⊂ Cb0 (Ω) ∩ Hα +∞, x0 ∈ ∂Ω, g is bounded on ∂Ω and continuous at x0 , and there is a right circular cone C ⊂∼ Ω of half-angle 0 < ψ < π with vertex at x0 , then lim

y→x0 ,y∈Ω

± Hf,g (y) = g(x0 ) uniformly for f ∈ F.

Proof: If {wj± } is an approximate barrier at x0 for g simultaneously for all + f ∈ F, then wj− ≤ Hf,g ≤ wj+ and the conclusion follows as in Lemma 9.8.2. It therefore suﬃces to prove that there is an approximate barrier {wj± } at x0 which applies simultaneously to all f ∈ F. Consider any u in the set ' (U(f, g) ∩ C 2 (Ω − )) f ∈F

which is nonempty by Lemma 9.7.9. It can be assumed that g(x0 ) + u is a supersolution simultaneously for all f ∈ F and u(x0 ) ≥ 0, for if not, replace u by |g|0 +|u|0 +u since g(x0 )+|g|0 +|u|0 +u is a superfunction by Lemma 9.7.4. Let Uj be a neighborhood of x0 such that (i) g(x) < g(x0 ) + 1j , x ∈ ∂Ω ∩ Uj λ (ii) L(r f0 (θ)) < −1 on Uj ∩ Ω where w = rλ f0 (θ) is a function of the preceding corollary with λ > 0. Now choose k(j) ≥ 1 such that (iii) k(j) > M + |c|0 g(x0 ) + 1j ≥ f − L g(x0 ) + 1j for all f ∈ F (iv)

1 j

+ k(j)w > u on ∂Uj ∩ Ω.

0

368

9 Elliptic Operators

On Uj ∩ Ω,

1 L g(x0 ) + + k(j)w ≤ L g(x0 ) + j

≤ L g(x0 ) +

≤ L g(x0 ) +

1 − k(j) j

1 1 − f − L g(x0 ) + j j 0

1 1 + f − L g(x0 ) + j j

= f. It follows that the function g(x0 ) + (1/j) + k(j)w is a superfunction on Uj ∩ Ω for all f ∈ F and g by Lemma 9.7.4. Let min {g(x0 ) + 1j + k(j)w, g(x0 ) + u} on Uj ∩ Ω wj+ = g(x0 ) + u on Ω − ∼ Uj . To show that wj+ is a superfunction on Ω, according to Lemma 9.7.4 it suﬃces to show that Lwj+ ≤ f for all f ∈ F at each point of Ω since wj+ ∈ C 0 (Ω − ) ∩ C 2 (Ω). By Lemma 9.7.7, wj+ is a superfunction on Uj ∩ Ω and on Ω ∼ Uj− . Thus, Lwj+ ≤ f on each of these two sets by Lemma 9.7.4. It remains only to show that Lwj+ ≤ f at each point z ∈ ∂Uj ∩ Ω. Since g(x0 ) + (1/j) + k(j)w(z) > g(x0 ) + u(z), there is a neighborhood V of z such that g(x0 ) + (1/j) + k(j)w > g(x0 ) + u on V . Since wj+ = g(x0 ) + u on V and the latter is a superfunction on Ω, Lwj+ ≤ f on V and, in particular, Lwj+ (z) ≤ f . This completes the proof that wj+ is a superfunction on Ω. On ∂Ω ∩ Uj , g(x0 ) + (1/j) + k(j)w ≥ g(x0 ) + (1/j) > g so that wj+ ≥ g on ∂Ω ∩ Uj . On ∂Ω ∼ Uj , wj+ = g(x0 ) + u ≥ g since g(x0 ) + u is assumed to be a supersolution relative to g. Thus, wj+ ≥ g on ∂Ω and wj+ is a supersolution on Ω for all f ∈ F relative to g. At x0 , 1 wj+ (x0 ) = min {g(x0 ) + , g(x0 ) + u(x0 )} j and

lim wj+ (x0 ) = min {g(x0 ), g(x0 ) + u(x0 )} = g(x0 ).

j→∞

This completes the proof that {wj+ } is an upper approximate barrier at x0 simultaneously for all f ∈ F. Only minor modiﬁcations of the above argument are needed to construct a lower approximate barrier {wj− } at x0 . Theorem 9.8.9 Let Ω be a L-regular bounded open subset of Rn and let L be a strictly elliptic operator with coeﬃcients in Hα (Ω), c ≤ 0, and let (2+b) (Ω; d), α ∈ (0, 1), b ∈ (−1, 0). If g ∈ C 0 (∂Ω), the Dirichlet f ∈ Cb0 (Ω)∩Hα problem Lu = f on Ω, u = g on ∂Ω has a unique solution u ∈ H2+α,loc (Ω).

9.8 Barriers

369

Proof: The requirement of regularity means that at each point x of ∂Ω, ± − (y) = g(x) for all x ∈ ∂Ω. Redeﬁning Hf,g , if necessary, to limy→x,y∈Ω Hf,g − + − be equal to g on ∂Ω, Hf,g ∈ U(f, g) so that Hf,g ≤ Hf,g and the two are + − equal. Letting Hf,g = Hf,g = Hf,g , LHf,g = f on Ω and Hf,g = g on ∂Ω. Uniqueness follows from Theorem 9.5.7.

Chapter 10

Apriori Bounds

10.1 Introduction As is well known, the mere fact that a function of a complex variable is diﬀerentiable on a domain in the complex plane implies that the function has many other properties; for example, the real and imaginary parts satisfy the Cauchy-Riemann equations, the function is inﬁnitely diﬀerentiable on the domain, etc. In addition, if a function on a domain Ω in Rn has continuous second partials and satisﬁes Laplace’s equation thereon, then other properties follow as a consequence, such as the averaging property, etc. It will be shown (b) in this chapter that there are limitations on the size of the H2+α (Ω) norm of a solution of the oblique boundary derivative problem for elliptic equations. These inequalities are known as apriori inequalities since it is assumed that a function is a solution without actually knowing that there are solutions. The establishment of such inequalities will pave the way for proving the existence of solutions in the next chapter. Eventually, very strong conditions will have to be imposed on Ω. So strong, in fact, that the reader might reasonably conclude that only a spherical chip, or some topological equivalent, will satisfy all the conditions. As the ultimate application will involve spherical chips, the reader might beneﬁt from assuming that Ω is a spherical chip. But as some of the inequalities require less stringent conditions on Ω, the conditions on Ω will be spelled out at the beginning of each section. Since these conditions will not be repeated in the formal statements, it is important to refer back to the beginning of each section for a description of Ω.

L.L. Helms, Potential Theory, Universitext, c Springer-Verlag London Limited 2009

371

372

10 Apriori Bounds

10.2 Green Function for a Half-space n Let Ω = R+ and let γ, β1 , . . . , βn be constants with γ ≤ 0 in the n ≥ 3 case, γ < 0 in the n = 2 case, βn > 0, and |β| = 1 where β = (β1 , . . . , βn ). For n x ∈ R+ , deﬁne a boundary operator M0 by the equation

M0 u(x) = γu(x) +

n

βi Di u(x) = γu(x) + β · ∇u(x),

i=1

which can be written M0 u = γu + Dβ u, where Dβ u is the directional derivative in the β direction. Although M0 is primarily of interest as an operator on functions on R0n , it can be regarded as an operator on functions with domain Ω. To solve the oblique derivative problem Δu = f

on Ω,

M0 u = 0

on R0n ,

n n × R+ will be constructed so that the a Green function G1/2 (x, y) on R+ solution is given by u(x) = G1/2 (x, y)f (y) dy. Rn +

The function G1/2 (x, y) is deﬁned by the equation G1/2 (x, y) = u(|x − y|) − u(|x − y r |) − 2βn

0

∞

eγs

∂ u(|x − y r + sβ|) ds, ∂xn

n where u(|x − y|) is the fundamental harmonic function with pole at x ∈ R+ . It will be shown below that the same equation can be used to deﬁne G1/2 n− n n− n on R+ × R+ , recalling that R+ denotes the closure of R+ . The following calculations will be carried out in the n ≥ 3 case, the n = 2 case being essentially the same assuming that γ < 0. n− × The integral part of G1/2 (x, y) suggests deﬁning a kernel h(x, y) on R+ n R+ by the equation

∞

∂ eγs u(|x − y r + sβ|) ds ∂x n 0 ∞ eγs |x − y r + sβ|−n (xn + yn + sβn ) ds. = 2βn (n − 2)

h(x, y) = −2βn

(10.1)

0

n− n The integral can be seen to be ﬁnite for all (x, y) ∈ R+ × R+ as follows. r r Letting ξ = (x − y )/|x − y |,

10.2 Green Function for a Half-space

373

∞

eγs |x − y r + sβ|−n+1 ds ∞ r r −n+2 = 2βn (n − 2)|x − y | eγ|x−y |s |ξ + sβ|−n+1 ds 0 ∞ r −n+2 |ξ + sβ|−n+1 ds. ≤ 2βn (n − 2)|x − y |

|h(x, y)| ≤ 2βn (n − 2)

0

0

Putting βt = (β1 , . . . , βn−1 , 0) and βn= (0, . . . , 0, βn ), ξ · β = ξ · βt + ξ · βn n− n = ξ · βt + ξn βn ≥ ξ · βt ≥ −|βt | = − 1 − βn2 for all x ∈ R+ , y ∈ R+ , and so |ξ + sβ|2 = 1 + 2sξ · β + s2 ≥ 1 − 2s 1 − βn2 + s2 . Since the latter function has the minimum value βn2 , |ξ + sβ|−n+1 ≤ 1/βnn−1 . Noting that |ξ + sβ| ≥ ||sβ| − |ξ|| = |s − 1|, |ξ + sβ|−n+1 ≤ 1/|s − 1|n−1 . The integrand of the above integral is therefore dominated on [0, +∞) by the integrable function min (1/βnn−1 , 1/|s − 1|n−1 ). This shows that h(x, y) is n− n , y ∈ R+ . Moreover, there is a constant C(n, βn ) such ﬁnite for all x ∈ R+ that |h(x, y)| ≤ C(n, βn )|x − y r |−n+2

n− n x ∈ R+ , y ∈ R+ .

(10.2)

n− n n− n If (xn , yn ) is a sequence in R+ × R+ that converges to (x, y) ∈ R+ × R+ , it follows in the same way from the Lebesgue dominated convergence theorem n− n × R+ . It is easy that limn→∞ h(xn , yn ) = h(x, y); i.e., h is continuous on R+ r n to see by a simple geometrical argument that |x − y| ≤ |x − y |, x, y ∈ R+ . Therefore,

|h(x, y)| ≤ C(n, βn )

1 , |x − y|n−2

n− n x ∈ R+ , y ∈ R+ .

(10.3)

n It will be shown now that M0 G1/2 (·, y) = 0 on (xn = 0) for each y ∈ R+ . n For ﬁxed y ∈ R+ ,

∂ u(|x − y|) − u(|x − y r |) ∂xi i=1 = βn (−n + 2) |x − y|−n (xn − yn ) − |x − y r |−n (xn + yn )

n

βi

so that when xn = 0, Dβ u(|x − y|) − u(|x − y r |) = 2(n − 2)|x − y|−n βn yn .

(10.4)

374

10 Apriori Bounds

Moreover, n

∂ βi ∂x i i=1

− 2βn

+ 2βn2 (n − 2)

e 0

= −2nβn (n − 2)

∞

∞

0

γs

∂ u(|x − y r + sβ|) ds ∂xn

eγs |x − y r + sβ|−n−2

n

βi (xi − (y r )i + sβi )

i=1

× (xn + yn + sβn ) ds ∞

0

eγs |x − y r + sβ|−n ds.

Using the fact that n d (|x − y r + sβ|−n ) = −n|x − y r + sβ|−n−2 βi (xi − (y r )i + sβi ), ds i=1

integrating by parts, and simplifying, ∞ ∂ ∂ − 2βn βi eγs u(|x − y r + sβ|) ds ∂xi ∂xn 0 i=1 ∞ d eγs (|x − y r + sβ|−n )(xn + yn + sβn ) ds = 2βn (n − 2) ds 0∞ 2 eγs |x − y r + sβ|−n ds + 2βn (n − 2)

n

0

= −2βn (n − 2)|x − y r |−n (xn + yn ) ∞ − 2βn γ(n − 2) eγs |x − y r + sβ|−n (xn + yn + sβn ) ds 0

= −2βn (n − 2)|x − y r |−n (xn + yn ) ∞ ∂ + 2γβn eγs |x − y r + sβ|−n+2 ds ∂x n 0 = −2βn (n − 2)|x − y r |−n (xn + yn ) − γh(x, y). Putting xn = 0 in this equation, using Equation (10.4), and using the fact that G1/2 (x, y) = h(x, y) when xn = 0, M0 G1/2 (x, y) = γG1/2 (x, y) + 2(n − 2)|x − y|−n βn yn − 2(n − 2)|x − y|−n βn yn − γG1/2 (x, y) = 0. Thus, M0 G1/2 (·, y) = 0 on R0n .

10.2 Green Function for a Half-space

375

Lemma 10.2.1 There is a constant C(n, βn ) such that 1 ≤ i ≤ n − 1, (i) Dxi h(x, y) = −Dyi h(x, y), (ii) Dxn h(x, y) = Dyn h(x, y), β h(x, y)| ≤ C(n, βn )|x − y|−n+2−k if |β| = k, 0 ≤ k ≤ 3, (iii) |D(x) n− n for all x ∈ R+ , y ∈ R+ .

Proof: The ﬁrst two assertions are straightforward and the k = 0 case of (iii) was proved above. For 1 ≤ i ≤ n − 1, ∞ eγs |x − y r + sβ|−n−2 |xi − yi + sβi | |Dxi h(x, y)| ≤ 2βn n(n − 2) 0

≤ 2βn n(n − 2)

0

∞

×|xn + yn + sβn | ds eγs |x − y r + sβ|−n ds.

Letting ξ = (x − y r )/|x − y r |, |Dxi h(x, y)| ≤ 2βn n(n − 2)|x − y r |−n+1

∞ 0

eγ|x−y

r

|s

|ξ + sβ|−n ds.

As above, the integrand of the above integral is dominated on (0, +∞) by the integrable function min (1/βnn , 1/|s − 1|n ). Thus, ∞ |Dxi h(x, y)| ≤ 2βn n(n − 2)|x − y r |−n+1 min (1/βnn , 1/|s − 1|n ) ds 0

and there is a constant C = C(n, βn ) such that n− n |Dxi h(x, y)| ≤ C(n, βn )|x − y r |−n+1 , 1 ≤ i ≤ n − 1, x ∈ R+ , y ∈ R+ .

When i = n, Dxn h(x, y) = 2βn (n − 2)

0

∞

eγs − n|x − y r + sβ|−n−2 (xn + yn + sβn )2 + |x − y r + sβ|−n ds

so that

|Dxn h(x, y)| ≤ 4βn n(n − 2)

0

∞

eγs |x − y r + sβ|−n ds.

The integral is the same as in the preceding step so that the same bound is obtained except for a factor of 2 which can be absorbed into the constant C(n, βn ). Estimates for |Dβ h(x, y)|, |β| = k, follow in exactly the same way using the integrable function min (1/β n−1+k , 1/|s − 1|n−1+k ). This establishes (iii) with |x − y r | in place of |x − y|. Since |x − y| ≤ |x − y r | for all n , |x − y r |−n+2−k ≤ |x − y|−n+2−k and (iii) holds. x, y ∈ R+

376

10 Apriori Bounds

n n This lemma shows that the kernel h(x, y) on R+ × R+ satisﬁes Inequalities (8.1), (8.2), (8.3), and (8.4) of Section 8.2. All of the results of that section therefore apply to the kernel h(x, y). It was shown in Section 8.3 that the same results apply to the Newtonian kernel u(|x− y|). Since |x− y| ≤ |x− y r |, the kernel u(|x−y r |) also satisﬁes the conditions of Section 8.2. It follows that G1/2 (x, y) satisﬁes the same conditions and all of the results of Section 8.2 are applicable to G1/2 (x, y).

Deﬁnition 10.2.2 If f is a bounded measurable function with compact supn port in R+ , deﬁne G1/2 f corresponding to the kernel G1/2 by the equation 1 n G1/2 f (x) = x ∈ R+ . G1/2 (x, y)f (y) dy, (2 − n)σn Theorem 10.2.3 If f is a bounded measurable function with compact supn , f is locally H¨ older continuous with exponent α ∈ (0, 1], and port in R+ n n n u = G1/2 f , then u ∈ C 2 (R+ ) ∩ C 1 (R+ ∪ R0n ), Δu = f on R+ , and M0 u = 0 n on R0 . n Proof: To show that Δu = f on R+ , consider 1 1 G1/2 f (x) = u(|x − y|)f (y) dy − u(|x − y r |)f (y) dy (2 − n)σn (2 − n)σn ∞ ∂ 2βn − eγs u(|x − y r + sβ|)f (y) dy. (2 − n)σn 0 ∂xn n By Theorem 8.3.1, the ﬁrst term is in C 2 (R+ ) and its Laplacian is just f . It therefore suﬃces to show that the second and third terms are harmonic n functions on Dδ = {x ∈ R+ ; xn > δ} for each δ > 0. Apply Theorem 1.7.13 to n . The assertion show that these functions are harmonic. Thus, Δu = f on R+ n that u ∈ C 1 (R+ ∪ R0n ) follows from Theorem 8.2.7. The fact that M0 u = 0 on R0n follows from the fact that MG1/2 (·, y) = 0 on R0n .

Remark 10.2.4 This proof includes the fact that G1/2 f has the representation 1 G1/2 f = Uf + h (2 − n)σn n . where Uf is the Newtonian potential of f and h is harmonic on R+

10.3 Mixed Boundary Conditions for Laplacian This section will be limited to regions Ω that are spherical chips with Σ = int(∂Ω ∩ R0n ) or balls in Rn with Σ = ∅. Such regions are convex along *δ and are stratiﬁable. An integral representation of a solution to with the Ω Poisson’s equation Δu = f on a spherical chip using a kernel

10.3 Mixed Boundary Conditions for Laplacian

K0 (x, y) =

1 2σn (n − 2)

377

1 1 − n−2 |x − y| |x − y r |n−2

,

x, y ∈ Rn+

n will be derived next. If x0 ∈ Ω and ρ > 0, let Bt = Bx0 ,tρ , Bt+ = Bt ∩ R+ , + 0 n 0 and Bt = Bt ∩ R0 . Consider only those ρ > 0 for which ∂B2 ∼ B2 ⊂ Ω. The integral representation requires the use of a cutoﬀ function ψρ , depending upon ρ, which can be constructed as follows. Let ψ be a function in C ∞ (0, ∞) satisfying (i) 0 ≤ ψ ≤ 1, (ii) ψ = 1 on (0, 5/4], and (iii) ψ = 0 on (7/4, ∞); for example, take ψ = J1/4 χ where χ is the indicator function of the interval (0, 3/2) and J1/4 is the operator deﬁned in Section 2.5. For ρ > 0, as restricted above, let ψρ (y) = ψ( rρ ), y ∈ B2+ , where r = |x0 − y| for y ∈ Rn . It is easily seen that ψρ 0,B + ≤ 1, Dψρ 0,B + ≤ C/ρ, D2 ψρ 0,B + ≤ C/ρ2 , 2 2 2 and [ψρ ]2+α,B + ≤ C/ρ2+α . The following lemma is taken from [25]. The proof 2 requires the use of Green’s representation theorem, Theorem 1.4.2, with the ball B replaced by B2+ . Green’s theorem is valid for regions with a Lipschitz boundary (c.f.[48], p.121).

Lemma 10.3.1 If Ω is a spherical chip with Σ = int(∂Ω ∩ R0n ) and u ∈ C 1 (Ω ∪ Σ) ∩ C 2 (Ω), then u(x) = K0 (x, y)Δ(uψρ ) dy, x ∈ B1+ . B2+

Proof: For x, y ∈ B2+ , let r1 (x, y) = |x − y|−n+2 /σn (n − 2) and r2 (x, y) = |x − y r |−n+2 /σn (n − 2) so that K0 = (r1 − r2 )/2. By Green’s representation theorem, Theorem 1.4.2, u(x)ψρ (x) =

∂B2+

(r2 Dn (uψρ )−uψρ Dn r2 ) dσ(y)−

B2+

r2 Δ(uψρ ) dy, x ∈ B2+ .

Since ψρ = 0 and Dn (uψρ ) = 0 on ∂B2+ ∼ B20 and Dn = Dyn on B20 , u(x)ψρ (x) =

B20

(r2 Dyn (uψρ ) − uψρ Dyn r2 ) dσ(y)

− Similarly, if x, y r ∈ B2+ u(x)ψρ (x) =

(10.5)

B2+

r2 Δ(uψρ ) dy,

x∈

B2+ .

B20

(r1 D(yr )n (uψρ ) − uψρ D(yr )n r1 ) dσ(y r )

−

B2+

r1 Δ(uψρ ) dy r ,

x ∈ B2+ .

378

10 Apriori Bounds

Since D(yr )n = −Dyn , r1 = r2 , and y r = y on B20 , u(x)ψρ (x) = − (r2 Dyn (uψρ ) − uψρ Dyn r2 ) dσ(y) +

B20

B2+

(10.6) r1 Δ(uψρ ) dy,

x ∈ B2+ .

Adding Equations (10.5) and (10.6), for x ∈ B1+ 1 u(x) = u(x)ψρ (x) = (r1 − r2 )Δ(uψρ ) dy. 2 B2+ Lemma 10.3.2 If Ω is a spherical chip with Σ = int(∂Ω ∩ R0n ) and u ∈ C 1 (Ω ∪Σ)∩C 2 (Ω) satisﬁes Poisson’s equation Δu = f on Ω for f ∈ Hα (Ω), then |u; ρ|2+α,B + ≤ C(n, α)) u 0,B + + ρ2 |f ; ρ|α,B + . 1

2

2

Proof: By the preceding lemma, u= K0 (·, y)(uΔψρ + ψρ f + 2∇u · ∇ψρ ) dy B2+

on B1+ .

Using the fact that |x − y r | ≥ |x − y| for x, y ∈ B2+ , it is easily seen that K0 satisﬁes Inequalities (8.1) through (8.4) and is therefore a subnewtonian kernel. After writing the above integral as a sum of three integrals, denote + the three terms by u1 , u2 , and u3 , respectively. Since ψρ = 1 on B5/4 , u1 =

+ B2+ ∼B5/4

on B1+ .

K0 (·, y)uΔψρ dy

+ For x ∈ B1+ and y ∈ B2+ ∼ B5/4 , |x − y| ≥ ρ/4, making it easy to obtain bounds on the norms of the derivatives of u1 using the bounds on the

Dβ ψρ 0,B + with the result 2

|u1 ; ρ|2+α,B + ≤ C(n) u 0,B + . 1

(10.7)

2

By Lemma 8.2.10, [u2 ]2+0,B + + ρα [u2 ]2+α,B + ≤ C(n, α) ψρ f 0,B + + ρα [ψρ f ]α,B + . 1

1

2

2

Since [ψρ f ]α,B + ≤ ρ−α f 0,B + + [f ]α,B + , 2

2

2

[u2 ]2+0,B + + ρα [u2 ]2+α,B + ≤ C(n, α) f 0,B + + ρα [f ]α,B + . 1

1

2

2

10.3 Mixed Boundary Conditions for Laplacian

379

It follows from the deﬁnition of u2 that

u2 0,B + ≤ C(n)ρ2 f 0,B + 1

2

and [u2 ]1+0,B + = Du2 0,B + ≤ C(n)ρ f 0B + . 1

1

2

Therefore, |u2 ; ρ|2+α,B + ≤ C(n, α))ρ2 f 0,B + + ρα [f ]α,B + 1

2

2

2

= C(n, α))ρ |f ; ρ|α,B + .

(10.8)

2

Consider now u3 = 2

B2+

K0 (·, y)∇u · ∇ψρ dy =

n 2 i=1

B2+

K0 (·, y)

∂u ∂ψρ dy. ∂yi ∂yi

Letting vi denote the ith term of the sum, it suﬃces to estimate each |vi ; ρ|2+α,B + . For 1 ≤ i ≤ n, let τi denote the projection map 1

τi (y1 , . . . , yn ) = (y1 , . . . , yi−1 , 0, yi+1 , . . . , yn ). Letting dy i denote integration with respect to y1 , . . . , yi−1 , yi+1 , . . . , yn , + ζi ∂u ∂ψρ ∂u ∂ψρ K0 (·, y) dy = K0 (·, y) dyi dy i ∂yi ∂yi ∂yi ∂yi B2+ τi B2+ ζi− where ζi− and ζi+ are the ith components of point of intersection of the τi ysection of B2+ . Integrating by parts,

ζi+

ζi−

ζ + ζ +

i ∂u ∂ψρ ∂ψρ i ∂ ∂ψρ K0 (·, y) dy= K0 (·, y)u − u (·, y) K dyi . 0 ∂yi ∂yi ∂yi ζ − ∂yi ∂yi ζi− i

For 1 ≤ i ≤ n−1, (y1 , . . . , yi−1 , ζi± , yi+1 , . . . , yn ) is a point of ∂B2+ ∼ B20 where ∂ψρ /∂yi is equal to 0. When i = n, the same is true of (y1 , . . . , yn−1 , ζn+ ); at the point (y1 , . . . , yn−1 , ζn− ), K0 (·, y) = 0. The ﬁrst term on the right is thus equal to 0. Thus,

∂u ∂ψρ ∂ ∂ψρ K0 (·, y) dy = − u K0 (·, y) dy. ∂yi ∂yi ∂yi B2+ B2+ ∂yi + Since ∂ψρ /∂yi = 0 on B5/4 ,

∂ ∂ψρ ∂ 2 ψρ u K0 (·, y) + K (·, y) dy. 0 + ∂yi2 ∂yi ∂yi B2+ ∼B5/4

vi = −2

380

10 Apriori Bounds

+ As was the case for u1 , for x ∈ B1+ and y ∈ B2+ ∼ B5/4 , |x − y| ≥ ρ/4 making it easy to show that |vi ; ρ|2+α,B + ≤ C(n) u 0,B + . It follows that 1 2 |u3 ; ρ|2+α,B + ≤ C(n) u 0,B + . Combining this result with Inequalities (10.7) 1 2 and (10.8) completes the proof. Before stating the next result, it is necessary to deﬁne a weighted norm of a boundary function. Let Ω be a bounded open subset of Rn and let Σ be a nonempty, relatively open subset of ∂Ω. If d > 0 is a constant and g : Σ → R, let |g; d|1+α,Σ = inf {|u; d|1+α,Ω ; u ∈ H1+α (Ω; d), u|Σ = g}.

In using this deﬁnition in a proof, g should be replaced by a u ∈ H1+α (Ω; d) and then take the inﬁmum of the |u; d|1+α,Ω . For all practical purposes, this amounts to considering g as a function in H1+α (Ω; d). The following lemma applies to both spherical chips with Σ = int(∂Ω ∩ R0n ) and balls with Σ = ∅. The latter case is dealt with in Theorem 8.4.2 + 0 and is a special case of the following lemma with √ Bi = Bi and B6 = ∅. Although the condition on Ω, namely, ρ0 > ρ/ 2, in Lemma 8.5.6 is not required for the following lemma, the choice of the balls Bt ensures that the √ argument of g in the deﬁnition of ψ below is a point in Σ. The ρ0 > ρ 2 condition was imposed to achieve the same result. It follows that the estimates for ψ 0,B + , Dψ 0,B + , D2 ψ 0,B + , and [D2 ψ]α,B + are valid in the present 2 2 2 2 context. Lemma 10.3.3 If B6+ = Bx+0 ,6ρ ⊂ Ω, u ∈ C 1 (Ω ∪ Σ) ∩ C 2 (Ω), Δu = f on Ω for f ∈ Hα (Ω), α ∈ (0, 1), and M0 u = g on Σ for g ∈ H1+α (Σ), then |u; ρ|2+α,B + ≤ C( u 0,B + + ρ|g; ρ|1+α,B60 + ρ2 |f ; ρ|α,B + ) 1

2

2

where C = C(α, βn ). Proof: By Theorem 7.2.6, it can be assumed that g ∈ H1+α (Ω) and that n , the inequality follows from Theorem 8.4.2 |g|1+α,Σ = |g|1+α,Ω . If B2 ⊂ R+ without the ρ|g; ρ|1+α,B60 term. It can be assumed therefore that B2 ∩ R0n = ∅ so that x0n ≤ 2ρ where x0 = (x0 , x0n ). Consider the function ψ, modiﬁed by a constant factor 1/βn , deﬁned by Equation (8.30) using the extension g˜ of g to R0n and the function η ∈ C 2 (R0n ) having compact support in B0,1 deﬁned in the proof of Lemma 8.5.6. For (x , xn ) ∈ B2+ and y ∈ B0,1 ⊂ R0n , |x0 − (x − xn y , 0)| ≤ |x0 − (x0 , 0)| + |(x0 , 0) − (x , 0)| + |(x , 0) − (x − xn y , 0)| < 6ρ showing that (x − xn y , 0) ∈ Bx0 ,6ρ and therefore that x − xn y ∈ B60 ; that is, for x ∈ B2+ , the function ψ is deﬁned by xn ψ(x) = ψ(x , xn ) = g(x − xn y , 0)η(y ) dy . βn Rn0

10.3 Mixed Boundary Conditions for Laplacian

381

By Equations (8.31) to (8.35) and Inequality (8.36), ψ 0,B + ≤Cρ[g]0,B60 , 2

Dψ 0,B + ≤C[g]0,B60 , D2 ψ 0,B + ≤C[g]1+0,B60 , and [D2 ψ]α,B + ≤ C[g]1+α,B60 . 2 2 2 It follows that ψ ∈ H2+α (B2+ ) and that Δψ ∈ Hα (B2+ ). Combining these inequalities (10.9) |ψ; ρ|2+α,B + ≤ Cρ|g; ρ|1+α,B60 2

where C = C(n, βn ). Note that Dxi ψ = 0 for 1 ≤ i ≤ n − 1 when xn = 0 and Dxn ψ = g when xn = 0 so that M0 ψ = g on Σ. Letting v = u − ψ on B2+ , Δv = f − Δψ ∈ Hα (B2+ ) and M0 v = M0 u − M0 ψ = 0 on B20 . By the preceding lemma, Lemma 10.3.2, |v; ρ|2+α,B + ≤ C( v 0,B + + ρ2 |f − Δψ; ρ|α,B + ) 1

2

2

≤ C( u 0,B + + ψ 0,B + + ρ2 |f ; ρ|α,B + + ρ2 |Δψ; ρ|α,B + ) 2

2

2

2

where C = C(n, α). Since ψ 0,B + ≤C(n, βn )ρ[g]0,B60 ≤ C(n, βn )ρ|g; ρ|1+α,B60 2 and ρ2 |Δψ; ρ|α,B + ≤ nρ2 |D2 ψ; ρ|α,B + 2

2

≤ Cρ2 ( D2 ψ 0,B + + ρα [D2 ψ]α,B + ) 2

2

≤ Cρ2 ([ψ]2+0,B + + ρα [ψ]2+α,B + ) 2

2

≤ Cρ2 ([g]1+0,B60 + ρα [g]1+α,B60 ) ≤ Cρ|g; ρ|1+α,B60 . |v; ρ|2+α,B + ≤ C( u 0,B + + ρ|g; ρ|1+α,B60 + ρ2 |f ; ρ|α,B + ). 1

2

2

By Inequality (10.9), |u; ρ|2+α,B + ≤ |v; ρ|2+α,B + + |ψ; ρ|2+α,B + 1

1

1

≤ |v; ρ|2+α,B + + Cρ|g; ρ|1+α,B60 1

≤ C( u 0,B + + ρ|g; ρ|1+α,B60 + ρ2 |f ; ρ|α,B + ) 2

2

where C = C(n, α, βn ). Under the conditions of the preceding lemma, if b > −1, x0 ∈ Ω, and ρ = d*x0 /8, then (1+b) (10.10) ρ1+b |g; ρ|1+α,B60 ≤ 3|g|1+α,Σ and

(2+b)

ρ2+b |f ; ρ|α,B + ≤ 2|f |α,Ω . 2

(10.11)

To prove the ﬁrst inequality, prove it ﬁrst with Σ replaced by Ω and then (1+b) (1+b) use the fact that |g|1+α,Ω = |g|1+α,Σ . The second follows in a similar way.

382

10 Apriori Bounds

Theorem 10.3.4 If Ω is a spherical chip with Σ = int(∂Ω ∩ R0n ) or a ball in Rn with Σ = ∅, α ∈ (0, 1), b > −1, u ∈ C 1 (Ω ∪Σ)∩C 2 (Ω) satisﬁes Δu = f (2+b) (1+b) on Ω for f ∈ Hα (Ω), and M0 u = g on Σ for g ∈ H1+α (Σ), then there is a constant C0 = C0 (b, α, βn , d(Ω)) such that (b) (1+b) (2+b) |u|2+α,Ω ≤ C0 d* bu 0,Ω + |g|1+α,Σ + |f |α,Ω . (b)

(b)

(10.12)

(b)

Proof: Since |u|2+α,Ω ≤ |u|2+0,Ω + [u]2+α,Ω , Theorem 7.6.1 can be applied to the ﬁrst term to show that there is a constant C0 = C0 (b, α, d(Ω)) such that (b) (b) . ≤ C0 d* bu 0,Ω + [u] |u| 2+0,Ω

2+α,Ω

(b)

It therefore suﬃces to prove the inequality for [u]2+α,Ω . Consider any x ∈ Ω. *7ρ and B + = Letting ρ = d*x /8 and Bt = Bx,tρ , d*x = 8ρ > 7ρ so that x ∈ Ω 6 + + *ρ . If y ∈ B and b ≥ 0, then d*y ≥ d*x − |y − x| > d*x − 2ρ = (3/4)d*x Bx,6ρ ⊂ Ω 2 so that d*bx ≤ (4/3)b d*by ; if y ∈ B2+ and b < 0, then d*y ≤ d*x + |y − x| < 2d*x so that d*b < d*b /2b . In either case, there is a constant C = C(b) such that x

y

d*xb ≤ C d*yb ,

y ∈ B2+ .

(10.13)

(1+b)

Noting that g ∈ H1+α (Σ) implies g ∈ H1+α (B60 ) by Inequality (10.10) and applying Lemma 10.3.3, d*x2+b |D2 u(x)| ≤ d*bx (8ρ)2 D2 u 0,B + 1 b ≤ C d*x u 0,B + + ρ|g; ρ|1+α,B60 + ρ2 |f ; ρ|α,B + . 2

2

By Inequalities (10.10), (10.11), and (10.13) and the facts that B60 ⊂ Σρ , *ρ , for any x ∈ Ω B2+ ⊂ Ω d*x2+b |D2 u(x)| ≤ C d*xb u 0,B + + ρ1+b |g; ρ|1+α,B60 + ρ2+b |f ; ρ|α,B + 2 2 (1+b) (2+b) ≤ C d*b u 0,B + + |g|1+α,Σ + |f |α,Ω 2 (1+b) (2+b) b * (10.14) ≤ C d u 0,Ω + |g|1+α,Σ + |f |α,Ω . Consider now any x, y ∈ Ω, labeled so that d*x ≤ d*y . Deﬁne ρ and Bi relative to x as above. Applying Inequality (10.14) to d*x2+b |D2 u(x)| and d*y2+b |D2 u(y)| and applying Lemma 10.3.3 to [D2 u]α,B + , 1

10.3 Mixed Boundary Conditions for Laplacian

383

2 2 2 2 2+b+α |D u(x) − D u(y)| *2+b+α |D u(x) − D u(y)| χ|x−y|<ρ ≤ d d*xy x |x − y|α |x − y|α 2 |D u(x)| + |D2 u(y)| + d*x2+b+α χ|x−y|≥ρ ρα ≤ 82+b+α ρ2+b+α [D2 u]α,B + 1 α *2+b 2 2 * + 8 dx |D u(x)| + d2+b y |D u(y)| ≤ C ρb ( u 0,B + + ρ|g; ρ|1+α,B60 + ρ2 |f ; ρ|α,B + ) 2 2 (1+b) (2+b) + d*b u 0,Ω + |g|1+α,Σ + |f |α,Ω (1+b) (2+b) ≤ C d*b u 0,Ω + |g|1+α,Σ + |f |α,Ω .

*η , 0 < η < 1 and using the fact that Taking the supremum over x, y ∈ Ω *η , d*xy > η for x, y ∈ Ω (1+b) (2+b) η 2+b+α [u]2+α,Ω* η ≤ C d*b u 0,Ω + |g|1+α,Σ + |f |α,Ω . Taking the supremum over η, (b) (1+b) (2+b) [u]2+α,Ω ≤ C d*b u 0,Ω + |g|1+α,Σ + |f |α,Ω where c = C(α, b, βn ). The preceding results apply only to Poisson’s equation u = f . The ﬁrst step in passing to elliptic operators on general domains is to extend these results to elliptic operators L0 with constant coeﬃcients. Let T : Rn → Rn be the nonsingular transformation deﬁned by the matrix of coeﬃcients {aij } as in Section 9.3. If Ω ⊂ Rn and Σ ⊂ ∂Ω, let Ω ∗ = T (Ω) and Σ ∗ = T (Σ). Since T is one-to-one and onto, T preserves the set operations so that T (∂Ω ∼ Σ) = T (∂Ω) ∼ T (Σ) = ∂Ω ∗ ∼ Σ ∗ , (∂Ω)∗ = ∂(Ω ∗ ), Σ ∗ ⊂ ∂Ω ∗ , and * T (∂Ω) = ∂Ω ∗ . Recalling that d(x) = d(x, ∂Ω ∼ Σ), the d* notation will be augmented by adding the subscript Ω in order to deal with the two sets Ω and Ω ∗ simultaneously. Using the fact that T is one-to-one and onto, it is easily checked that there is a constant c = c(aij ; 1 ≤ i, j ≤ n) such that 1* dΩ (x) ≤ d*Ω ∗ (y) ≤ cd*Ω (x), c

y = T x, x ∈ Ω.

Moreover, there are also constants c = c(aij ; 1 ≤ i, j ≤ n) and k = k(aij ; 1 ≤ i, j ≤ n) such that ∗) *δ ) ⊂ (Ω T (Ω (10.15) cδ and

∗ ) ⊂ T (Ω *kδ ). (Ω δ

384

10 Apriori Bounds

If u is any function on Ω, let u∗ (y) = u(x) for y = T (x), x ∈ Ω. Clearly,

u 0,Ω = u∗ 0,Ω ∗ . In particular, u 0,Ω* δ = u∗ 0,T (Ω* δ ) . By Relation (10.15), there is a constant c = c(aij ; 1 ≤ i, j ≤ n) such that .

u 0,Ω* δ = u∗ 0,T (Ω* δ ) ≤ u∗ 0,(Ω ∗) cδ

As was shown in Section 9.3, ∂u∗ 1 ∂u √ oij = ∂xj ∂yi λi i=1 n

and

n 1 √ 1 √ oij ≤ √ |oij | ≤ c n λ λi i i=1

Therefore, *δ }

Dxj u 0,Ω* δ = sup {|Dxj u(x)|; x ∈ Ω ≤ C sup { = C sup {

n i=1 n

*δ } |Dyi u∗ (y)|; T −1 y ∈ Ω *δ )} |Dyi u∗ (y)|; y ∈ T (Ω

i=1

≤ C sup {

n

∗) } |Dyi u∗ (y)|; y ∈ (Ω cδ

i=1

≤ C Du∗ 0,(Ω ∗)

cδ

Similar results hold for D2 u∗ 0,(Ω and [D2 u∗ ]α,(Ω ∗) ∗ ) . Thus, there are δ δ constants C and c, not depending on Ω, such that 1 |u| ≤ |u∗ |2+α,(Ω ˜ ˜ cδ . ∗ ) ≤ C|u|2+α,Ω δ C 2+α,Ωδ/c Multiplying both sides by δ 2+α+b , adjusting constants, and taking the supremum over δ, 1 (b) (b) (b) |u| ≤ |u∗ |2+α,Ω ∗ ≤ C|u|2+α,Ω (10.16) C 2+α,Ω where C = C(aij , α, b). If u ∈ C 2 (Ω) or u ∈ Hα (Ω), then u∗ ∈ C 2 (Ω ∗ ) or u∗ ∈ Hα (Ω ∗ ), respec(1+b) (1+b) tively; moreover, if g ∈ H1+α (Σ), then g ∗ ∈ H1+α (Σ ∗ ). If u ∈ C 2 (Ω), f ∈ (2+b)

(1+b)

Hα (Ω), and g ∈ H1+α (Σ), as in Section 9.3 the oblique derivative boundary value problem L0 u = f on Ω, M0 u = g on Σ

10.4 Nonconstant Coeﬃcients

385

is transformed by the transformation T into the problem Δu∗ = f ∗ on Ω ∗ , M1 u∗ = g ∗ on Σ ∗ where M1 u∗ (y) =

n

βj∗ Dyj u∗ (y) + γu∗ (y)

j=1

with βj∗ =

n

βi oji

λj ,

j = 1, . . . , n.

i=1

The following lemma is an immediate consequence of Theorem 10.3.4 and Inequality (10.16). Lemma 10.3.5 If Ω is a spherical chip with Σ = int(∂Ω ∩ R0n ) or a ball in Rn with Σ = ∅, α ∈ (0, 1), b > −1, and u ∈ C 1 (Ω ∪ Σ) ∩ C 2 (Ω) satisﬁes (2+b) (1+b) L0 u = f on Ω for f ∈ Hα (Ω), and M0 u = g on Σ for g ∈ H1+α (Σ), then there is a constant C0 = C0 (aij , b, α, βi , γ, d(Ω)) such that (b) (1+b) (2+b) |u|2+α,Ω ≤ C0 d*b u 0,Ω + |g|1+α,Σ + |f |α,Ω .

10.4 Nonconstant Coeﬃcients As in the preceding section, Ω will denote a spherical chip with Σ = int(∂Ω ∩ R0n ) or a ball in Rn with Σ = ∅. In this section, the general elliptic operator L as deﬁned in Section 9.1 will be assumed to satisfy Inequalities (9.2). Consider (2+b) the oblique derivative problem Lu = f on Ω for f ∈ Hα (Ω) subject to the (1+b) boundary condition Mu = g on Σ for g ∈ H1+α (Σ). If x0 is a ﬁxed point of Ω, deﬁne operators L0 and M0 with constant coeﬃcients by the following equations: n ∂2u L0 u(x) = aij (x0 ) (x), x∈Ω (10.17) ∂xi ∂xj i,j=1 and M0 u(x , 0) =

n i=1

βi (x0 )

∂u (x , 0) + γ(x0 )u(x , 0), ∂xi

x ∈ Σ.

(10.18)

These operators will be used as crude approximations to L and M. The following inequalities, with appropriate restrictions on the parameters α, b, b , along with repeated applications of Inequalities (7.25) and (7.26) will be used to justify these approximations.

386

10 Apriori Bounds (b+b )

|f g|α,Ω

(b)

(b )

≤ C|f |α,Ω |g|α,Ω ,

(b+b )

(b)

(10.19)

(b )

|f g|1+α,Ω ≤ C|f |1+α,Ω |g|1+α,Ω .

(10.20)

The proof of Inequality (10.19) is similar, but easier, than the proof of Inequality (10.20) and only the latter will be proved. Since |f g|1+α,Ω* δ = [f g]0,Ω* δ + [f g]1+0,Ω* δ + [f g]1+α,Ω* δ ≤ [f ]0,Ω* δ [g]0,Ω* δ + [f ]0,Ω* δ [g]1+0,Ω* δ + [f ]1+0,Ω* δ [g]0,Ω* δ + sup

*δ x,y∈Ω

|f (x)Dg(x) + g(x)Df (x) − f (y)Dg(y) − g(y)Df (y)| |x − y|α

≤ [f ]0,Ω* δ [g]0,Ω* δ + [f ]0,Ω* δ [g]1+0,Ω* δ + [f ]1+0,Ω* δ [g]0,Ω* δ + [f ]0,Ω* δ [g]1+α,Ω* δ + [f ]α,Ω* δ [g]1+0,Ω* δ + [f ]1+α,Ω* δ [g]0,Ω* δ +

[f ]1+0,Ω* δ [g]α,Ω* δ .

Multiplying both sides by δ b+b +1+α , using the fact that only δ < 1 need be considered, and using Inequality (7.33),

δ b+b +1+α |f g|1+α,Ω* δ

≤ δ b [f ]0,Ω* δ δ b [g]0,Ω* δ + δ b [f ]0,Ω* δ δ 1+b [g]1+0,Ω* δ

+ δ 1+b [f ]1+0,Ω* δ δ b [g]0,Ω* δ + δ b [f ]0,Ω* δ δ 1+b +α [g]1+α,Ω* δ

+ δ b+α [f ]α,Ω* δ δ 1+b [g]1+0,Ω* δ + δ 1+α+b [f ]1+α,Ω* δ δ b [g]0,Ω* δ

+ δ 1+b [f ]1+0,Ω* δ δ b +α [g]α,Ω* δ (b)

(b )

(b)

(b )

(b)

(b )

≤ +[f ]0,Ω [g]0,Ω + [f ]0,Ω [g]1+0,Ω + [f ]1+0,Ω [g]0,Ω (b)

(b )

(b)

(b )

(b)

(b )

+ [f ]0,Ω [g]1+α,Ω + [f ]α,Ω [g]1+0,Ω + [f ]1+α,Ω [g]0,Ω (b)

(b )

+ [f ]1+0,Ω [g]α,Ω (b)

(b )

≤ C|f |1+α,Ω |g|1+α,Ω . Inequality (10.20) follows by taking the supremum over δ > 0. It will be assumed throughout this section that the coeﬃcients of L and M satisfy the inequality (0) (1) (2) A = sup

ai,j 0,Ω + |aij |α,Ω + |bi |α,Ω + |c|α,Ω 1≤i,j≤n

(0) (1) + βi 0,Σ + |βi |1+α,Σ + γ 0,Σ + |γ|1+α,Σ < +∞.

10.4 Nonconstant Coeﬃcients

387

Theorem 10.4.1 If Ω is any spherical chip with Σ = int(∂Ω ∩R0n ) or a ball (b) B ⊂ Rn with Σ = ∅, α ∈ (0, 1), b ∈ (−1, 0), u ∈ H2+α (Ω) satisﬁes Lu = f (2+b) (1+b) (Ω) and Mu = g for g ∈ H1+α (Σ), then there is a constant for f ∈ Hα C = C(aij , bi , α, b, βi , γ, d(Ω)) such that (b) (1+b) (2+b) |u|2+α,Ω ≤ C d*b u 0,Ω + |g|1+α,Σ + |f |α,Ω . Proof: Consider any ρ > 0, let μ < 1/2 be a constant to be speciﬁed later, ˜2ρ , and let B = Bx0 ,2μρ . Then B ∩ Ω ⊂ Ω ˜ρ since |y − x0 | < 2μρ let x0 , y0 ∈ Ω implies that d*y ≥ d*x0 − 2μρ > 2ρ − 2μρ > ρ. For y0 ∈ Bx0 ,μρ ∩ Ω, (2ρ)2+b+α

2+b+α 2 (μρ)2+b+α |u|2+α,(Ω∩B) μ μρ

2+b+α 2 (b) ≤ |u|2+α,Ω∩B . μ (10.21)

|D2 u(x0 ) − D2 u(y0 )| ≤ |x0 − y0 |α

Since |y0 − x0 | ≥ μρ for y0 ∈ Bx0 ,μρ ∩ Ω, (2ρ)2+b+α

2 2 |D2 u(x0 ) − D2 u(y0 )| 2+b+α (|D u(x0 )| + |D u(y0 )|) ≤ (2ρ) |x0 − y0 |α μα ρα α+1 2 (b) ≤ α |u|2+0,Ω . μ

By Theorem 7.6.1, there is a constant C0 = C0 (b, α, μ, d(Ω)) such that (b) (b) |u|2+0,Ω ≤ C0 d*b u 0,Ω + μ2α |u|2+α,Ω .

(10.22)

Therefore, for y0 ∈ Bx0 ,μρ ∩ Ω |D2 u(x0 ) − D2 u(y0 )| 2α+1 (b) ≤ C0 d*b u 0,Ω + 2α+1 μα |u|2+α,Ω . |x0 − y0 |α μα (10.23) Combining Inequalities (10.21) and (10.23) and using the fact that x0 , y0 ∈ ˜2ρ , Ω (2ρ)2+b+α

(b) |u|2+α,Ω

2+b+α 2 2α+1 (b) (b) ≤ |u|2+α,Ω∩B + α C0 d*b u 0,Ω + 2α+1 μα |u|2+α,Ω . μ μ

Further restricting μ so that 2α+1 μα < 1/2, |u|2+α,Ω ≤ C0 d*b u 0,Ω + 2 (b)

2+b+α 2 (b) |u|2+α,Ω∩B μ

(10.24)

388

10 Apriori Bounds (b)

where C0 = C0 (b, α, μ). Consider the term |u|2+α,Ω∩B . Deﬁning operators L0 and M0 as in Equations (10.17) and (10.18), the equations Lu = f and Mu = g can be written L0 u = F = (L0 u − Lu) + f and M0 u = G = (M0 u − Mu) + g respectively. Applying Theorem 10.3.5 to L0 and M0 , (b) (1+b) (2+b) |u|2+α,Ω∩B ≤ C0 d*b u 0,Ω∩B + |G|1+α,Σ∩B + |F |α,Ω∩B where C0 = C0 (n, aij , b, α, βi , γ). Thus, (b) (1+b) (2+b) |u|2+α,Ω∩B ≤ C0 d*b u 0,Ω + |g|1+α,Σ + |f |α,Ω

(1+b) (2+b) + |M0 u − Mu|1+α,Σ∩B + |L0 u − Lu|α,Ω∩B .

(10.25)

Note that L0 u − Lu =

n

(aij (x0 ) − aij )Dij u −

i,j=1

and M0 − Mu =

n

bi Di u − cu

i=1

n

(βi (x0 ) − βi )Di u + (γ(x0 ) − γ)u.

i=1 (2+b)

Consider |(aij (x0 )−aij )Dij u|α,Ω∩B . By Inequality (10.19), there is a constant C = C(b, α) such that (2+b)

(0)

(2+b)

|(aij (x0 ) − aij )Dij u|α,Ω∩B ≤ C|aij (x0 ) − aij |α,Ω∩B |Dij u|α,Ω∩B (0)

(b)

≤ C|aij (x0 ) − aij |α,Ω∩B |u|2+α,Ω∩B . Since

(0)

|aij (x0 ) − aij |α,Ω∩B = sup α |aij (x0 ) − aij |α,(Ω∩B) ,

>0

*ρ , (Ω ∩ B) = ∅ for > 2μρ, and (Ω ∩ B) ⊂ Ω ∩ B ⊂ Ω (0)

(0)

|aij (x0 ) − aij |α,Ω∩B ≤ 2α μα ρα |aij (x0 ) − aij |α,Ω˜ ρ ≤ 2α μα |aij (x0 ) − aij |α,Ω . Note also that (0)

(0)

(0)

(0)

|aij (x0 ) − aij |α,Ω ≤ |aij (x0 )|α,Ω + |aij |α,Ω ≤ aij 0,Ω + |aij |α,Ω .

10.4 Nonconstant Coeﬃcients

389

Thus, (2+b) (b) (b) |(aij (x0 ) − aij )Dij u|α,Ω∩B ≤ Cμα aij 0,Ω + |aij |α,Ω |u|2+α,Ω∩B where C = C(b, α). Using Inequalities (7.25) and (7.26) and arguments similar to the above, (2+b)

(1)

(b)

(b)

|bi Di u|α,Ω∩B ≤ Cμα |bi |α,Ω |u|2+α,Ω∩B ≤ CAμα |u|2+α,Ω∩B (2+b)

(2)

(b)

(b)

|cu|α,Ω∩B ≤ Cμα |c|α,Ω |u|2+α,Ω∩B ≤ CAμα |u|2+α,Ω∩B . (b)

Since |u|2+α,Ω = sup0<δ<1 δ 2+b+α |u|2+α,Ω* δ < +∞, u and its derivatives of *δ without order two or less have continuous extensions to the closures of each Ω (b) (b) increasing norms and therefore |u|2+α,Σ∩B ≤ |u|2+α,Ω∩B . Using arguments similar to the above, (1+b) (0) (b) |(βi (x0 ) − βi )Di u|1+α,Σ∩B ≤ Cμα βi 0,Σ + |βi |1+α,Σ |u|2+α,Ω∩B (b)

(1+b)

|(γ(x0 ) − γ)u|1+α,Σ∩B

≤ CAμα |u|2+α,Ω∩B (1) (b) ≤ Cμα γ 0,Σ + |γ|1+α,Σ |u|2+α,Ω∩B (b)

≤ CAμα |u|2+α,Ω∩B . It follows that (2+b)

(1+b)

((b)

|L0 u − Lu|α,Ω∩B + |M0 u − M|1+α,Σ∩B ≤ CAμα |u|2+α,Ω∩B . Returning to Inequality (10.25), (b) (1+b) (2+b) (b) |u|2+α,Ω∩B ≤ C0 d*b u 0,Ω + |g|1+α,Σ + |f |α,Ω + Aμα |u|2+α,Ω∩B . Further restricting μ so that C0 d(Ω)2+α Aμα < 1/2, (b) (1+b) (2+b) |u|2+α,Ω∩B ≤ C0 d*b u 0,Ω + |g|1+α,Σ + |f |α,Ω . α Choosing μ = μ0 so that μ0 < 1/2, 2α+1μα 0 < 1/2, C0 Aμ0 < 1/2 and returning to Inequality (10.24), (b) |u|2+α,Ω ≤ C0 d*b u 0,Ω

2+b+α 2 (1+b) (2+b)

d*b u 0,Ω + |g|1+α,Σ + |f |α,Ω + C0 μ0 (1+b) (2+b) b ≤ C d* u 0,Ω + |g|1+α,Σ + |f |α,Ω .

Chapter 11

Oblique Derivative Problem

11.1 Introduction The Dirichlet problem for an elliptic operator was solved in Chapter 9 by morphing a solution of the Dirichlet problem for the Laplacian on a ball into a solution of the Dirichlet problem for an elliptic operator on a ball. In applying the method of continuity, it was necessary to show that an elliptic operator with a suitably restricted domain and range is a bounded operator. The ﬁrst step in mimicking this procedure is to show that the solution of a mixed boundary value problem for the Laplacian on a spherical chip, as in Theorem 8.5.7, can be morphed into a solution of the corresponding problem for an elliptic operator with an oblique derivative condition rather than a normal derivative condition. This results in the existence of local solutions of the oblique derivative problem. By adapting the Perron-Wiener-Brelot procedure, it will be shown that the oblique derivative problem has a solution for regions satisfying appropriate geometric conditions.

11.2 Boundary Maximum Principle Let L denote an elliptic operator on Ω ⊂ Rn deﬁned by the equation Lu(x) =

n i,j=1

∂2u ∂u (x) + bi (x) (x) + c(x)u(x), ∂xi ∂xj ∂x i i=1 n

aij (x)

x ∈ Ω,

with c ≤ 0 and m|x|2 ≤

n

aij (x)xi xj ≤ M |x|2 ,

x ∈ Ω,

(11.1)

i,j=1

L.L. Helms, Potential Theory, Universitext, c Springer-Verlag London Limited 2009

391

392

11 Oblique Derivative Problem

for positive constants m, M . Also let M be a boundary operator deﬁned for functions on the relatively open subset Σ of ∂Ω by the equation Mu(x) =

n

βi (x)

i=1

∂u (x) + γ(x)u(x), ∂xi

x ∈ Σ,

where γ ≤ 0 and β is a vector with β · ν > 0 at points of Σ and ν = −n is the inner unit normal at points of Σ. It will be assumed throughout this chapter that the coeﬃcients of L and M satisfy the condition (0) (1) (2)

aij 0,Ω + |aij |α,Ω + |bi |0,Ω + |bi |α,Ω + |c|0,Ω + |c|α,Ω A = sup 1≤i,j≤n (11.2) (0) (1) + βi 0,Σ + |βi |1+α,Σ + γ 0,Σ + |γ|1+α,Σ < +∞. It is customary to assume that the vector of coeﬃcients β(x)=(β1 (x), . . ., βn (x)) in the equation Mu(x , 0) = g(x ) is normalized by requiring that |β(x)| = 1 for each x ∈ Σ. If this is not the case, the equation Mu(x , 0) = g(x ) can be normalized by multiplying both sides of the equation by 1/|β(x)|. Justiﬁcation of this procedure requires showing that the coeﬃcients βi /|β| and γ/|β| satisfy the above inequality and that the forcing function g/|β| satisﬁes conditions required of the forcing function; namely, that g/|β| ∈ (1+b) (1+b) H1+α (Σ) if g ∈ H1+α (Σ). This can be done under the above additional conditions on the βi and |β|. It will be assumed in this chapter that |β(x)| = 1 for all x ∈ Σ. In discussions of the oblique derivative problem Lu = f on Ω and Mu = g on Σ when Σ = ∂Ω and the problem Lu = f on Ω,

Mu = g on Σ, and u = h on ∂Ω ∼ Σ,

when ∅ = Σ = ∂Ω, the symbols f, g, and h will be used solely for functions (2+b) (1+b) in Hα (Ω), H1+α (Σ), and C 0 (∂Ω ∼ Σ), respectively. The norm symbols will be simpliﬁed by omitting the domain of the functions, if clear from the (0) (0) (1) context; for example, |aij |α,Ω will be denoted simply by |aij |α and |γ|1+α,Σ (1)

will be denoted by |γ|1+α . It also will be assumed that b ∈ (−1, 0), α ∈ (0, 1), c ≤ 0, and γ ≤ 0 without further mention in this chapter. Some sections of this chapter may require additional restrictions on Ω and Σ which will be prescribed at the beginning of each such section. In order to proceed further, it is necessary to estimate the term d*b u 0,Ω in Inequality (10.12). The ﬁrst lemma of this section, known as the boundary point lemma, is valid if Ω is any bounded, convex, open subset of Rn and Σ is a nonempty, relatively open subset of ∂Ω having a unit normal vector

11.2 Boundary Maximum Principle

393

at each point of Σ. After the proof of this lemma and the following corollary, it will be assumed that Ω is a spherical chip with Σ = int(∂Ω ∩ R0n ) or a ball in Rn with Σ = ∅. Lemma 11.2.1 (Boundary Point Lemma, Hopf [33]) Let u ∈ C 0 (Ω ∪ Σ) ∩ C 2 (Ω) satisfy Lu ≥ 0 on Ω. If for x0 ∈ Σ, (i) u(x0 ) > u(x) for all x ∈ Ω and (ii) the interior sphere condition is satisﬁed at x0 , then Mu(x0 ) < 0, if deﬁned, whenever u(x0 ) ≥ 0. Proof: Choose a local coordinate system (z1 , . . . , zn ) with origin at x0 so that (0, . . . , 0, 1) is the inner unit normal to ∂Ω at x0 . Let By,ρ0 be the ball with center at y = (0, . . . , 0, ρ0 ) of radius ρ0 specifying the interior sphere condition at x0 ∈ ∂By,ρ0 . For z ∈ Ω, let r = |z − y|. Fix 0 < ρ1 < ρ0 and − ∼ By,ρ1 by putting deﬁne w on the closed annulus Γ = By,ρ 0 2

2

w(z) = e−kr − e−kρ0 where k is a constant to be chosen as follows. As in the proof of Lemma 9.5.5 and using the bound given by (11.2), √ 2 Lw(z) ≥ e−kr (4k 2 mρ21 − 2knA − 2k nAρ0 − A) and k > 0 can be chosen so that Lw ≥ 0 on Γ . As in the same proof, there is an > 0 such that u − u(x0 ) + w ≤ 0 on Γ . Since γ(x0 )u(x0 ) ≤ 0 whenever u(x0 ) ≥ 0 and w(x0 ) = 0, for x0 + tβ ∈ Γ u(x0 + tβ) − u(x0 ) u(x0 + tβ) − u(x0 ) + γ(x0 )u(x0 ) ≤ t t w(x0 + tβ) ≤ − t w(x0 + tβ) − w(x0 ) = − t and it follows that Mu(x0 ) ≤ − Dβ w(x0 ). If it can be shown that Dβ w(x0 ) > 0, it would follow that Mu(x0 ) < 0 whenever u(x0 ) ≥ 0. Letting β = (β1 , . . . , βn ) relative to the local coordinate system, Dβ w(x0 ) = Dβ w|z=0 = β · ∇w|z=0 n 2 = βi (−2k(zi − yi )e−kr )|z=0 . i=1

Since yi = 0, i = 1, . . . , n − 1 and yn = ρ0 , when z = 0 2

Dβ w(x0 ) = Dβ w|z=0 = 2kρ0 e−kρ0 β · ν > 0.

394

11 Oblique Derivative Problem

Corollary 11.2.2 (Strong Maximum Principle) If u ∈ C 0 (Ω − )∩C 2 (Ω) satisﬁes Lu ≥ 0 on Ω, Mu ≥ 0 on Σ, the interior sphere condition is satisﬁed at each point of Σ, and either (i) u ≤ 0 on ∂Ω ∼ Σ when nonempty or (ii) c ≡ 0 on Ω or γ ≡ 0 on Σ when Σ = ∂Ω, then either u = 0 on Ω or u < 0 on Ω and, in particular, u ≤ 0 on Ω. Proof: If u is a nonpositive constant, the conclusion is trivial in both the ∂Ω ∩ Σ = ∅ and Σ = ∂Ω cases; likewise, u cannot be a positive constant if ∂Ω ∼ Σ = ∅ and if Σ = ∂Ω, for in the latter case Lu = cu < 0 at some point of Ω or Mu = γu < 0 at some point of Σ, contrary to the hypotheses. It therefore can be assumed that u is not a constant function. The two cases Σ = ∂Ω and Σ = ∂Ω will be considered separately. Consider the latter case ﬁrst. If u attains a negative maximum value at a point x0 ∈ Ω, then u ≤ u(x0 ) < 0; on the other hand, since u is nonconstant, it cannot attain a nonnegative maximum value at a point of Ω by the strong maximum principle, Lemma 9.5.6, and must do so at a point x0 ∈ ∂Ω, which implies that Mu(x0 ) < 0 by the boundary point lemma, Lemma 11.2.1, a contradiction. Therefore, u < 0 in the nonconstant case. Suppose now that Σ = ∂Ω. As above, if u attains a negative maximum at a point x0 ∈ Ω, then u < 0 on Ω. Since u is not a constant, it cannot attain a nonnegative maximum value at a point of Ω and must do so at a point x0 ∈ ∂Ω. The point x0 ∈ Σ, for otherwise Mu(x0 ) < 0, a contradiction. The nonnegative maximum must occur at x0 ∈ ∂Ω ∼ Σ and so u < u(x0 ) ≤ 0 on Ω. It will be assumed throughout this section that Ω is a spherical chip with Σ = int(∂Ω ∩ R0n ) or a ball B in Rn with Σ = ∅. A function w will be constructed that will be used to estimate d*b u 0,Ω for Ω a ball or a particular n kind of spherical chip. Let Ω = By,ρ with Σ = ∅ or Ω = By,ρ ∩ R+ with Σ = By,ρ ∩ R0n , yn < 0, let K0 be a positive constant satisfying n

(|bi |0 + |βi |0 ) ≤ K0 ,

i=1

let t ∈ (0, 1), let r = |x − y| for x ∈ Ω, and let w = (ρ2 − r2 )t . Then for x ∈ Ω, Lw(x) = 4t(t − 1)(ρ2 − r2 )t−2

n

aij (xi − yi )(xj − yj )

i,j=1

−2t(ρ2 − r2 )t−1

n

(aii + bi (xi − yi )) + c(ρ2 − r2 )t .

i=1

Since c ≤ 0, the right side will be increased if the last term is omitted. This is also the case if the term involving the aii is omitted since the aii are nonnegative. By Inequality (11.1), n i,j=1

aij (xi − yi )(xj − yj ) ≥ m|x − y|2 = mr2 .

11.2 Boundary Maximum Principle

395

Since |xi − yi | ≤ |x − y|, n n bi (xi − yi ) ≤ |bi |0 |xi − yi | ≤ K0 r i=1

i=1

from which it follows that

n

i=1 bi (xi

− yi ) ≥ −K0 r. Thus, for x ∈ Ω

Lw(x) ≤ 4t(t − 1)(ρ2 − r2 )t−2 mr2 + 2t(ρ2 − r2 )t−1 K0 r ≤ −t(ρ2 − r2 )t−2 Q(r) where Q(r) = r 4(1 − t)mr − 2K0 ρ2 + 2K0r2 . Since Q(r) = 0 on (0, ρ) when r = r0 =

16(1 − t)2 m2 + 16K02 ρ2 4K0

−4(1 − t)m +

and Q(ρ) = 4(1 − t)mρ2 > 0, Q(r) ≥ q0 > 0 for r ∈ [(ρ + r0 )/2, ρ] where q0 = Q((ρ + r0 )/2). For x ∈ Ω with r ∈ [(ρ + r0 )/2, ρ], Lw(x) ≤ −q0 t(ρ2 − r2 )t−2 ≤ −q0 (2ρ)t−2 t(ρ − r)t−2 ≤ −C0 t(ρ − r)t−2 (11.3) −2

where C0 = C0 (m, bi , βi , ρ) = q0 (2ρ) min (1, 2ρ). At this point, it is convenient to prove the next lemma in the case that *δ = Ωδ . Suppose u ∈ Ω is a ball with Σ = ∅. In this case, d* = d, and Ω (2+b) 0 − 2 C (Ω ) ∩ C (Ω) satisﬁes Lu = f on Ω for f ∈ Hα (Ω) and u = 0 on ∂Ω. Consider any x ∈ Ω and r ∈ ((ρ + r0 )/2, ρ) for which x ∈ Ωρ−r . Letting v =u−

(2−t)

|f |0 w, C0 t (2−t)

Lv(x) ≥ f (x) + |f |0

(ρ − r)t−2 .

(2−t)

Since |f |0 ≥ (ρ − r)2−t |f |0,Ωρ−r , Lv(x) ≥ f (x) + |f |0,Ωρ−r ≥ 0. Since x can be any point of Ω, Lv(x) ≥ 0 on Ω; since u = 0 on ∂Ω by hypothesis, v ≤ 0 on Ω by the Strong Maximum Principle, Theorem 9.5.6; that is, u(x) ≤

(2−t)

(2−t)

(2ρ)t |f |0 |f |0 (ρ2 − r2 )t ≤ (ρ − r)t . C0 t C0 t

Since −u satisﬁes the same hypotheses as u, (2−t)

|u(x)| ≤

|f |0

(2ρ)t (ρ − r)t . C0 t

* −t ≤ (ρ − r)−t and using Inequality (7.26), Noting that d(x) |d*−t (x)u(x)| ≤ C1 |f |0

(2−t)

≤ C1 |f |(2−t) . α

396

11 Oblique Derivative Problem

(2+b) Letting b = −t and taking the supremum over x ∈ Ω, d*b u 0,Ω ≤ C1 |f |α where C1 = C1 (m, bi , βi , b, d(Ω)). This is the conclusion of the next lemma when Ω is a ball. Returning to the case of a spherical chip, let B be a ball of radius ρ with center at y = (0, −ρ0 ), ρ0 ∈ (0, ρ), let Ω = B + = ∅, and let Σ = B ∩ R0n . The constant ρ0 will be speciﬁed later. As above Lw ≤ −C1 t(ρ − r)t−2 for x ∈ Ω, r ∈ [(ρ + r0 )/2, ρ]. As before, the quantity on the right side of

Mw(x , 0) =

n

βi (−2t)(ρ2 − r2 )t−1 (xi − yi ) + γ(ρ2 − r2 )t ,

x ∈ Σ,

i=1

is increased by omitting the last term since γ ≤ 0. Since y = (0, . . . , 0, −ρ0 ), Mw(x , 0) ≤ −2t(ρ2 − r2 )t−1

n−1

βi xi + βn ρ0 .

i=1

Since |xi | ≤ |x | ≤ ρ2 − ρ20 , 1 ≤ i ≤ n − 1, n−1

βi xi ≥ −K0

+ ρ2 − ρ20 .

i=1

+ Mw(x , 0) ≤ −2t(ρ2 − r2 )t−1 − K0 ρ2 − ρ20 + βn ρ0 . Since K0 ρ/ βn2 + K02 < ρ, ρ0 ∈ (0, ρ) can be chosen so that

ρ K0 ρ ρ > ρ0 ≥ max √ , . (11.4) 2 βn2 + K02 For this choice of ρ0 , −K0 ρ2 − ρ20 + βn ρ0 > 0 and Mw(x , 0) ≤ −C2 t(ρ2 − r2 )t−1 for x ∈ Σ. Since r < ρ for x = (x , 0) ∈ Σ, with this choice of ρ0 and

Lw(x) ≤ −Ct(ρ − r)t−2

Mw(x , 0) ≤ −Ct(ρ − r)

t−1

x ∈ Ω, r ≥ ρ0

x ∈ Σ, r ≥ ρ0

(11.5) (11.6)

where C = C(m, bi , βi , d(Ω)) Following Lieberman [39], spherical chips fulﬁlling the above conditions will play a signiﬁcant role in the following. n Deﬁnition 11.2.3 The spherical chip Ω is admissible if Ω = By,ρ ∩ R+ where y = (0, . . . , −ρ0 ), ρ0 ∈ (0, ρ), and ρ0 satisﬁes Inequality (11.4).

In the course of the preceding proof, it was shown that there is an admissible √ spherical chip corresponding to each ρ > 0. The requirement that ρ0 > ρ/ 2 in Inequality (11.4) validates the use of Theorem 8.5.7 for admissible chips.

11.2 Boundary Maximum Principle

397

Lemma 11.2.4 If Ω is an admissible spherical chip with Σ = int(∂Ω ∩ R0n ) or a ball in Rn with Σ = ∅, b ∈ (−1, 0), α ∈ (0, 1), u ∈ C 0 (Ω − ) ∩ C 2 (Ω) satis(2+b) (1+b) ﬁes Lu = f on Ω for f ∈ Hα (Ω), Mu = g on Σ for g ∈ H1+α (Σ), and u = 0 on ∂Ω ∼ Σ, then there is a constant C = C(n, m, bi , b, α, βi , γ, d(Ω)) such that (1+b) (2+b) + |g|1+α .

d*b u 0,Ω ≤ C |f |α Proof: Since the inequality was proven above for a ball B ⊂ Rn with Σ = ∅, only the spherical chip case need be considered. Deﬁne t ∈ (0, 1), ρ0 and w as (1+b) above. Consider any x ∈ Ω and any ϕ ∈ H1+α (Ω) satisfying the condition ϕ = g on Σ. If v =u−

(2−t)

|f |0

(1−t)

+ |ϕ|0 Ct

w, (2−t)

then by Inequality (11.5) and the fact that |f |0 (2−t)

Lv(x) ≥ f (x) + |f |0

≥ (ρ − r)2−t |f |0,Ω* ρ−r ,

(ρ − r)t−2 ≥ f (x) + |f |0,Ω* ρ−r ≥ f (x) − f (z)

*ρ−r . Noting that f is continuous at x and x is a limit point of for all z ∈ Ω * Ωρ−r , Lv(x) ≥ f (x) − f (x), and therefore Lv ≥ 0 on Ω. Consider now any (x , 0) ∈ Σ. By Inequality (11.6)and a similar continuity argument, (1−t)

Mv(x , 0) ≥ ϕ(x , 0) + |ϕ|0

(ρ − r)t−1

≥ ϕ(x , 0) + (ρ − r)1−t |ϕ|0,Ω* ρ−r (ρ − r)t−1 ≥ ϕ(x , 0) + |ϕ|0,Ω* ρ−r ≥ 0. Since v = 0 on ∂Ω ∼ Σ, v ≤ 0 on Ω by the Strong Maximum Principle, Corollary 11.2.2; that is, u(x) ≤

(2−t)

|f |0

(1−t)

+ |ϕ|0 Ct

(ρ2 − r2 )t .

Since −u also satisﬁes the hypotheses of the lemma with f and ϕ replaced by −f and −ϕ, respectively, |u(x)| ≤

(2−t)

|f |0

(1−t)

+ |ϕ|0 Ct

(ρ2 − r2 )t ≤

(2−t)

|f |0

(1−t)

+ |ϕ|0 Ct

* = ρ − r, Using Lemma 7.6.7 and the fact that d(x) * −t |u(x)| ≤ C |f |(2−t) + |ϕ|(1−t) d(x) 0 0 (1−t) + |ϕ|1+α . ≤ C |f |(2−t) α

(2ρ)t (ρ − r)t .

398

11 Oblique Derivative Problem

Letting b = −t and taking the supremum over Ω, (1+b) (2+b)

d*b u 0,Ω ≤ C |f |α + |ϕ|1+α . (1+b)

Taking the inﬁmum over ϕ ∈ H1+α (Ω), the conclusion is established. The requirement in the hypotheses of the following theorem that the function f satisﬁes a weighted H¨ older norm on a superset is not an essential requirement in so far as future developments are concerned. In applying the theorem, starting with an f satisfying a H¨ older requirement on a spherical chip, an admissible spherical chip satisfying the requirement can be constructed. Only the k = 0 case of the following theorem will be proved in detail for spherical chips to illustrate the techniques. The theorem is stated only for spherical chips but is also applicable for balls with Σ = ∅. + where y = (0, . . . , 0, −ρ0 ), 0 < ρ0 < ρ, is Theorem 11.2.5 If Ω = By,ρ a spherical chip with Σ = int(∂Ω ∩ R0n ) α ∈ (0, 1), b ∈ (−1, 0), and f ∈ (2−k+b) Hk+α (Ω ) where Ω = By+ ,ρ with y = (0, . . . , 0, −ρ0 ), 0 < ρ0 < ρ0 < ρ, k = 0, 1, then there is a sequence {fj } in Hk+α (Ω ∪ Σ) that converges * − , 0 < δ < 1, and |fj |(2−k+b) ≤ C|f |(2−k+b) , j ≥ 1. uniformly to f on each Ω δ k+α,Ω k+α,Ω + Proof: Let Ω = By,ρ where y = (0, . . . , o, −ρ0 ), 0 < ρ0 < 1, let m be the (2+b)

(Ω). Since δ 2+b+α |f |α,Ω* δ < +∞ * − , denoted by the for each 0 < δ < 1, f has a continuous extension to Ω δ − * same symbol, and therefore to 0<δ<1 Ωδ = Ω ∪ Σ. Consider any x ∈ Ω and the point x = x − (xn /(xn − yn ))(x − y) at which the line joining y to x intersects R0n . Letting r1 = |x − x| = (xn /(xn − yn ))|x − y|, if * *δ = B + , then B + x ∈ Ω x ,r1 ⊂ Ωδ . Note also that if B is any ball in y,ρ−δ + * Bx ,r1 , then B ⊂ Ωδ . For each x ∈ Ω, let B(x, rj ) be a ball with center at ξ = ξ(x) = x − (xn /j(xn − yn ))(x − y) of radius rj = rj (x) = |ξ − x| = j −1 (xn /(xn −yn ))|x−y|. If j > 2ρ/ρ0 > 2, then ξn = xn (1−j −1 ) > xn /2 and the nth coordinate of the lowest point on the boundary of the ball B(x, rj ) is ξn − j −1 (xn /(xn − yn ))|x − y| > 0, showing that the ball B(x, rj ) ⊂ Bx+ ,r1 *δ if x ∈ Ω *δ . For j ≥ 1, let ϕj be the whenever j ≥ 2ρ/ρ0 . Thus, B(x, rj ) ⊂ Ω truncation function ϕj (x) = −j for x ≤ −j, ϕj (x) = x for −j < x < j, and ϕj (x) = j for x ≥ j and let ψj = J1/j ϕj be the smoothened version of ϕj . The latter function has the following properties: molliﬁer of Section 2.5, and let f ∈ Hα

(i) ψj ∈ C ∞ (R), (ii) ψj (x) = x if −j + 1j < x < j − 1j , ψj (0) = 0, |ψj | ≤ j, |ψj | ≤ 1, and (iii) |ψj (x) − ψj (y)| ≤ |x − y| for all x, y ∈ R. For j > 2ρ/ρo and x ∈ Ω ∪ Σ, let ⎧ ⎪ ⎪ ψj (f )(ξ + rj z)m(z) dz ⎪ ⎨ fj (x) = 1 1 ⎪ ⎪ ⎪ ψ (f )(z)m (z − ξ) dz. j ⎩ n

rj

rj

(11.7)

11.2 Boundary Maximum Principle

399

*δ and j ≥ 2ρ/ρ0 , then fj (x) is the weighted average of Note that if x ∈ Ω *δ . When xn = 0, fj (x) = ψj (f )(x). It will be shown ψj (f ) over B(x, rj ) ⊂ Ω * − . Since f now that the sequence {fj } converges uniformly to f on each Ω δ * − , given > 0 there is an η > 0 such that is uniformly continuous on Ω δ * − . Choose j0 ≥ 2ρ/ρ0 such |f (z) − f (z )| < whenever |z − z | < η, z, z ∈ Ω δ 1 that |f |0,Ω* δ < j0 − j0 and j > 2ρ(ρ − ρ0 )/ρ0 η for all j ≥ j0 . If z ∈ B(x, rj ), then |z − x| < 2rj < η and it follows from the second form for fj (x) in Equation (11.7) that

1 1 (z − ξ) dz |ψj (f )(z) − f (x)|m |fj (x) − f (x)| ≤ n rj rj

1 1 ≤ n (z − ξ) dz < |f (z) − f (x)|m rj rj *δ and j ≥ j0 ; that is, the sequence {fj } converges simultaneously for all x ∈ Ω − * . It will be shown next that each fj ∈ Hα (Ω). Using the uniformly to f on Ω δ ﬁrst form for fj (x) in Equation (11.7) and the fact that |ψj | ≤ j, [fj ]0,Ω < +∞. Consider any x, x ∈ Ω. Using the ﬁrst form for fj (x) and fj (x ) and the fact that |ψ(z) − ψ(z )| ≤ |z − z |, |fj (x) − fj (x )| ≤ |(ψj (f )(ξ(x) + rj (x)z) − ψj (f )(ξ(x ) + rj (x )z)|m(z) dz ≤ |f (ξ(x) + rj (x)z) − f (ξ(x ) + rj (x )z)|m(z) dz. Using the easily checked fact that |ξ(x) + rj (x)z) − ξ(x ) − rj (x )z)| ≤ |x − x |, |fj (x) − fj (x )| |x − x |α ≤ [f ]α,(Ω , )

ρ−ρ 0

≤ C[f ]α,(Ω , )

ρ−ρ0

|ξ(x) + rj (x)z) − ξ(x ) − rj (x )z)|α m(z) dz |x − x |α < +∞.

Thus [fj ]α,Ω < +∞. Combining this inequality with the inequality [fj ]0,Ω < +∞, |fj |α,Ω < +∞ and fj ∈ Hα (Ω). It remains in the k = 0 case to show (2+b) (2+b) *δ . Since B(x , r) ⊂ Ω *δ , it that |fj |α,Ω ≤ C|f |α,Ω . Consider any x ∈ Ω follows from the ﬁrst form for fj (x) that [fj ]0,Ω* δ ≤ [f ]0,Ω* δ . Consider any *δ . Using the fact that |ψj (z) − ψj (z )| ≤ |z − z | for all z, z ∈ R, x, x ∈ Ω |fj (x) − fj (x )| ≤ ≤

|ψj (f )(ξ(x) + rj (x)z) − ψj (f )(ξ(x ) + rj (x )z)|m(z) dz |f (ξ(x) + rj (x)z) − f (ξ(x ) + rj (x )z)|m(z) dz

400

11 Oblique Derivative Problem

≤ [f ]α,Ω* δ

|ξ(x) + rj (x)z − ξ(x ) − rj (x )z|α m(z) dz

≤ C[f ]α,Ω* δ (|ξ(x) − ξ(x )| + |rj (x) − rj (x )|)

α

≤ C[f ]α,Ω* δ |x − x |α and it follows that [fj ]α,Ω* δ ≤ C[f ]α,Ω* δ . Combining this inequality with the inequality [fj ]0,Ω* δ ≤ [f ]0,Ω* δ , |fj |α,Ω* δ ≤ C|f |α,Ω* δ . Multiplying by δ 2+b+α (2+b)

and taking the supremum over 0 < δ < 1, |fj |α,Ω

(2+b)

≤ C|f |α,Ω . When (1+b) f ∈ H1+α (Ω), the fj (x) are deﬁned by putting fj (x) = ψj (ψj (f )(ξj (x) + rj (x)z)m(z) dz, x ∈ Ω. As above, it can be seen that [fj ]0,Ω ≤ j, [fj ]1+0,Ω ≤ 1, and [fj ]1+α,Ω ≤ 2C|m|0,B0,1 d(Ω)1−α so that |fj |1+α,Ω < +∞ and fj ∈ H1+α (Ω). Also, [fj ]0,Ω* δ ≤ [f ]1+α,Ω* δ , and [fj ]1+α,Ω* δ ≤ C[ψj (ψj )]2+0,R [f ]αΩ* δ ≤ C[f ]1+α,Ω* δ . Adding corresponding members, |fj |1+α,Ω* δ ≤ C[f ]1+α,Ω* δ . Multiplying both sides by δ 2+b+α and taking the supremum over 0 < δ < 1, (1+b) (1+b) |fj |1+α,Ω ≤ C|f |1+α,Ω . As above, the sequence {fj } converges uniformly to * − , 0 < δ < 1. f on each Ω δ

It is assumed in the following lemma that f and g satisfy the hypotheses of the preceding theorem. Lemma 11.2.6 If Ω is an admissible spherical chip with Σ = int(∂Ω ∩ R0n ) (2+b) or a ball in Rn with Σ = ∅, α ∈ (0, 1), b ∈ (−1, 0), f ∈ Hα (Ω), and (1+b) 0 − 1 g ∈ H1+α (Σ), then there is a unique u ∈ C (Ω ) ∩ C (Ω ∪ Σ) ∩ C 2 (Ω) satisfying u = f on Ω, Dn u = g on Σ, and u = 0 on ∂Ω ∼ Σ with (b) (2+b) (1+b) |u|2+α,Ω ≤ C(|f |α,Ω + |g|1+α,Σ ). Proof: Assume ﬁrst that Ω is an admissible spherical chip with Σ = ∅. Let {fj } be the sequence in Hα (Ω) of the preceding theorem that converges to (2+b) (2+b) f uniformly on compact subsets of Ω with |fj |α,Ω ≤ C|f |α,Ω . Similarly, there is a sequence {gj } of functions on Σ such that each gj ∈ H1+α (Σ), the sequence {gj } converges uniformly to g on each compact subset of Σ, and (1+b) (1+b) |gj |1+α,Σ ≤ C|g|1+α,Σ . By Theorem 8.5.7, for each j ≥ 1, there is a unique uj ∈ C 0 (Ω − ) ∩ C 1 (Ω ∪ Σ) ∩ C 2 (Ω) such that uj = fj on Ω, Dn uj = gj on Σ, and uj = 0 on ∂Ω ∼ Σ. By Theorem 10.3.4 and Lemma 11.2.4, (b) (2+b) (1+b) (2+b) (1+b) |uj |2+α,Ω ≤ C |fj |α,Ω + |gj |1+α,Σ ≤ C |f |α,Ω + |g|1+α,Σ . By the subsequence selection principle of Section 7.2, there is a subsequence {ujk } that converges uniformly on compact subsets of Ω to a unique function (b) u ∈ H2+α (Ω) such that u = f on Ω, Dn u = g on Σ, and u = 0 on ∂Ω ∼ Σ. In the case of a ball with Σ = ∅, the proof is the same except for using Theorem 8.3.1 to assert the existence of a unique uj satisfying the equation uj = fj subject to the condition uj = 0 on ∂Ω.

11.2 Boundary Maximum Principle

401

Lemma 11.2.7 Let Ω be a bounded open subset of Rn , let Ω be an open set containing Ω − , and let u ∈ C 0 (Rn ) vanish outside Ω . If η > 0, α ∈ (b) (2+b) (0, 1) and b > −1, then |Jη u|2+α,Ω ≤ C|u|0,Ω , |LJη u|α,Ω ≤ C|u|0,Ω , and (1+b)

|MJη u|1+α,Σ ≤ C|u|0,Ω where C = C(aij , bi , c, α, b, βi , γ, d(Ω)). Proof: Using the function m ∈ C0∞ (Rn ) deﬁned by Equation (2.4), if 0 < *δ , then δ < 1, j ≥ 0 and x, x ∈ Ω j j − D(x) m y−x D(x) m y−x η η 1 |Dj Jη u(x) − Dj Jη u(x )| ≤ n dy. |u(y)| α α |x − x | η |x − x | Letting z = (y − x)/η, z = (y − x )/η, for some point z = (y − x )/η on the line segment joining z to z j j y−x j j m − D D(x) m y−x η η (x) 1 |D(z) m(z) − D(z) m(z )| ≤ |x − x | η j+1 |z − z | 1 j m(z )| = j+1 |∇(z) D(z) η

y − x j m = ∇(x) D(x) η so that |Dj Jη u(x) − Dj Jη u(x )| |x − x |α ≤ |u|0,Ω d(Ω)1−α (Ω)

1 ηn

∇(x) Dj m y − x dy (x) η

≤ |u|0,Ω d(Ω)1−α νn [m]j+1+0,Rn ≤ C|u|0,Ω *δ , [Jη u] where C=C(j, α, d(Ω)). Taking the supremum over x, x ∈Ω *δ = 2+α,Ω j [D Jη u]α,Ω* δ ≤ C|u|0,Ω . Using interpolation as in Theorem 7.3.5, |Jη u|2+α,Ω* δ ≤ C|u|0,Ω . Multiplying by δ 2+b+α and taking the supremum over 0 < δ < (b) 1, |Jη u|2+α,Ω ≤ C|u|0,Ω . Similarly, |LJη u|α,Ω* δ = aij Dxi ,xj Jη u + bi Dxi Jη u + cJη uα,Ω*

δ

i,j

i

≤ A(|D2 Jη u|α,Ω* δ + |D1 Jη u|α,Ω* δ + |Jη u|α,Ω* δ ) ≤ C|u|0,Ω . (2+b)

Multiplying by δ 2+b+α and taking the supremum over 0 < δ < 1, |LJη u|α,Ω (1+b)

≤ C|u|0,Ω . A similar argument shows that |MJη u|1+α,Σ ≤ C|u|0,Ω .

402

11 Oblique Derivative Problem

It is assumed in the following theorem that the f and g satisfy the hypotheses of Theorem 11.2.5. Theorem 11.2.8 If Ω is an admissible spherical chip with Σ = int(∂Ω∩R0n ) (2+b) or a ball in Rn with Σ = ∅, b ∈ (−1, 0), α ∈ (0, 1), f ∈ Hα (Ω), g ∈ (1+b) H1+α (Σ), and h ∈ C 0 (∂Ω ∼ Σ), then there is a unique u ∈ C 0 (Ω − ) ∩ C 2 (Ω ∪ Σ) satisfying Lu = f on Ω, Mu = g on Σ, and u = h on ∂Ω ∼ Σ. Moreover, there is a constant C = C(aij , bi , c, α, b, βi , γ, d(Ω)) such that (b) (2+b) (1+b) (11.8) |u|2+α,Ω ≤ C |f |α,Ω + |g|1+α,Σ + |h|0,∂Ω∼Σ . Proof: The existence and uniqueness in the homogeneous case h = 0 will be proved ﬁrst. Consider the Banach space (b)

B = {u ∈ H2+α (Ω) : u = 0 on ∂Ω ∼ Σ}, (b)

with norm |u|B = |u|2+α,Ω , the normed linear space (1+b)

X = Hα(2+b) (Ω) × H1+α (Σ) (2+b)

(1+b)

with norm |(f, g)|X = |f |α,Ω + |g|1+α,Σ , and the family of operators from B to X deﬁned by L0 u = (u, Dn u) L1 u = (Lu, Mu) Lt u = (1 − t)L0 u + tL1 u,

0 ≤ t ≤ 1.

As in the proof of Theorem 9.6.5, each of the operators (1− t)Δ+ tL and (1 − t)Dn + tM satisﬁes the general conditions previously imposed on coeﬃcients. Solvability of the oblique derivative problem Lu = f on Ω, Mu = g on Σ, and u = 0 on ∂Ω ∼ Σ is equivalent to showing that the map L1 : B → X is invertible. If u ∈ B and Lt u = (f, g) ∈ X, then (1 − t)u + tLu = f on Ω (1 − t)Dn u + tMu = g on Σ u = 0 on ∂Ω ∼ Σ. By Theorem 10.4.1, for such u, (b) (2+b) (1+b) |u|B = |u|2+α,Ω ≤ C d*b u 0,Ω + |f |α,Ω + |g|1+α,Σ . (b) *δ ) for 0 < δ < 1 which Note that u ∈ H2+α (Ω) implies that u ∈ H2+α (Ω * − so in turn implies that u, Du, and D2 u have continuous extensions to Ω δ 2 that u, Du, and D u have continuous extensions to Ω ∪ Σ. In addition, it

11.2 Boundary Maximum Principle

403

*δ , d*b (x)|u(x)| ≤ δ b [u] * ≤ is also true that u ∈ C 0 (Ω − ) since for x ∈ Ω 0,Ωδ (b) (b) (b) |u| ≤ |u| < +∞; therefore, d*b (x)|u(x)| ≤ |u| for all x ∈ Ω and 0,Ω

2+α,Ω

2+α,Ω

limx→x0 ∈∂Ω∼Σ u(x) = 0 since limx→x0 ∈∂Ω∼Σ d*b (x) = +∞. This shows that u ∈ C 0 (Ω − ) ∩ C 2 (Ω ∪ Σ) so that Lemma 11.2.4 is applicable and (1+b) (2+b) + |g|1+α = |(f, g)|X = C|Lt u|X , |u|B ≤ C |f |α

where C is independent of t. By Theorem 9.2.2, L1 maps B onto X if and only if L0 = (u, Dn u) maps B onto X. According to Lemma 11.2.6, there is a unique u ∈ B satisfying u = f on Ω, Dn u = g on Σ, and u = 0 on ∂Ω ∼ Σ. Thus, L0 maps B onto X and so L1 maps B onto X; that is, there (b) is a u ∈ H2+α (Ω) such that Lu = f on Ω,

Mu = g on Σ,

u = 0 on ∂Ω ∼ Σ. (b)

It follows from Corollary 11.2.2 that u is unique. As above, u ∈ H2+α (Ω) implies that u ∈ C 0 (Ω − ) ∩ C 2 (Ω ∪ Σ). It remains only to show that the condition u = 0 on ∂Ω ∼ Σ can be replaced by u = h on ∂Ω ∼ Σ. By the Tietze Extension Theorem, there is a continuous function on Rn that agrees with h on ∂Ω ∼ Σ, denoted by the same symbol, with sup norm no greater than that of h on ∂Ω ∼ Σ and vanishing outside the ball By,2ρ , n or Ω = By,ρ . For j ≥ 1, let hj = J1/j h ∈ C0∞ (Rn ). where Ω = By,ρ ∩ R+ The sequence {hj } converges uniformly to h on Rn . For each j ≥ 1, consider the problem Lvj = f − Lhj on Ω,

Mvj = g − Mhj on Σ,

vj = 0 on ∂Ω ∼ Σ. (11.9) By the ﬁrst part of the proof, this problem has a unique solution vj ∈ (b) H2+α (Ω) which implies that vj ∈ C 0 (Ω − ) ∩ C 2 (Ω ∪ Σ) as above. It will (b) be shown now that the sequence {|vj |2+α,Ω } is bounded. By Lemma 11.2.4 and Lemma 11.2.7, (2+b) (1+b)

d*b vj 0,Ω ≤ C |f − Lhj |α,Ω + |g − Mhj |1+α,Σ (2+b) (1+b) ≤ C |f |α,Ω + |g|1+α,Σ + |h|0,∂Ω∼Σ . It follows from Lemma 10.4.1 that (b) (2+b) (1+b) |vj |2+α,Ω ≤ C |f |α,Ω + |g|1+α,Σ + |h|0,∂Ω∼Σ ≤ K0 < +∞, j ≥ 1. (11.10) (b) Letting uj = vj + hj , uj ∈ H2+α (Ω) since this is true of hj and Luj = f on Ω,

Muj = g on Σ,

uj = hj on ∂Ω ∼ Σ.

(11.11)

404

11 Oblique Derivative Problem

Note that uj ∈ C 0 (Ω − ) ∩ C 2 (Ω ∪ Σ) since this is true of vj and hj . For the diﬀerences uj − uk , L(uj −uk ) = 0 on Ω, M(uj −uk ) = 0 on Σ, uj −uk = hj −hk on ∂Ω ∼ Σ. It can be assumed that uj − uk attains a nonnegative, absolute maximum value at a point x0 of Ω − , for if not, replace it by its negative. If x0 ∈ Ω, then uj − uk is a constant function by Lemma 9.5.6, and since uj − uk = hj − hk on ∂Ω ∼ Σ, uj − uk 0,Ω = hj − hk 0,∂Ω∼Σ . If x0 is not a point of Ω, then x0 ∈ ∂Ω ∼ Σ, for if x0 ∈ Σ, it would follow that M(uj − uk )(x0 ) < 0 by the boundary point lemma, Lemma 9.5.5, a contradiction. In either case,

uj − uk 0,Ω ≤ hj − hk 0,∂Ω∼Σ . As the sequence {hj } converges uniformly to h, it is Cauchy in the · 0,∂Ω∼Σ norm and so it follows that the sequence {uj } is Cauchy in the · 0,Ω norm. Thus there is a u ∈ C 0 (Ω − ) which is the uniform limit of the {uj } sequence. (b) Using Inequality (11.10) and the fact that |hj |2+α,Ω ≤ C|h|0,∂Ω∼Σ for all (b)

j ≥ 1 according to Lemma 11.2.7, the sequence {|uj |2+α,Ω } is bounded by a *δ = ∅. Then constant K. Consider any 0 < δ < 1 for which Ω δ 2+b+α uj 0,Ω* δ + [Duj ]0,Ω* δ + [D2 uj ]0,Ω* δ + [D2 uj ]α,Ω* δ ≤ K. * − . MoreAs above, the uj , Duj , and D2 uj have continuous extensions to Ω δ 2 over, the sequences {uj }, {Duj }, and {D uj } are bounded and equicontin*δ ∪ Σδ and can be assumed to converge to u, Du, and *− ⊃ Ω uous on Ω δ *δ ⊃ Ω *δ ∪ Σδ by passing to a subsequence if necesD2 u, respectively, on Ω 2 * sary. Thus, u ∈ C (Ωδ ∪ Σδ ) for all δ as described above, and it follows that u ∈ C 0 (Ω − ) ∩ C 2 (Ω ∪ Σ). Letting j → ∞ in Equations (11.11), u satisﬁes the equations Lu = f on Ω, Mu = g on Σ, and u = h on ∂Ω ∼ Σ. Uniqueness follows from Corollary 11.2.2. By Inequality (11.10), (b)

(b)

(b)

|uj |2+α,Ω ≤ |vj |2+α,Ω + |hj |2+α,Ω (2+b) (1+b) ≤ C |f |α,Ω + |g|1+α,Σ + |h|0,∂Ω∼Σ . Thus for 0 < δ < 1, (2+b) (1+b) δ 2+b+α |uj |2+α,Ω* δ ≤ C |f |α,Ω + |g|1+α,Σ + |h|0,∂Ω∼Σ . By expanding the left side, letting j → ∞, and then taking the supremum over 0 < δ < 1, Inequality (11.8) is established.

11.3 Curved Boundaries

405

11.3 Curved Boundaries The procedure for dealing with curved boundaries involves the use of transformations to “straighten the boundary” as follows. Let Ω be a bounded open subset of Rn and let Σ be a nonempty, relatively open subset of ∂Ω, possibly equal to ∂Ω. For each x ∈ Σ, let Ux be a neighborhood of x and let τx be a map of Ux into Rn with continuous partials on Ux such that τx (x) = 0 ∈ Rn . If the map τx has an inverse, then facts known for spherical chips can be carried back to Ux by means of τx−1 . A suﬃcient condition for the existence of an inverse for τx is that the Jacobian of τx not vanish at x; that is, if τx = (τ1 , . . . , τn ), then ∂τ1 (x) · · · ∂τ1 (x) ∂x1 ∂xn .. .. = 0. J(τx )(x) = ... . . ∂τn (x) n (x) ∂x · · · ∂τ∂x 1 n If this condition is satisﬁed, then there is a neighborhood O of 0 ∈ Rn and a neighborhood Vx of x such that τx−1 is deﬁned on O, Vx = τx−1 (O), τx is one-to-one on Vx , τx−1 (τx ) = x, and τx−1 has continuous ﬁrst partials on O (c.f. [1]). Deﬁnition 11.3.1 If τ is a mapping from Ω into Rn , let τ (x) = (τ1 (x), . . . , τn (x))

if x = (x1 , . . . , xn ) ∈ Ω.

The map τ is said to be diﬀerentiable of class Ca on Ω if each component τj ∈ Ha (Ω). If U and V are open subsets of Rn and τ : U → V is a homeomorphism such that τ and τ −1 are of class Ca , then τ is said to be a Ca diﬀeomorphism of U onto V . Let Ω be a bounded, convex, open subset of Rn , let Λ be a neighborhood of Ω − , and let τ be a Ck+α diﬀeomorphism, k ≥ 1, 0 < α ≤ 1, which maps Λ onto a bounded, convex, open subset Λ of Rn . Using the mean value theorem, it is easy to see that there is a constant K > 0 such that 1 |x − y| ≤ |τ (x) − τ (y)| ≤ K|x − y|, K

x, y ∈ Ω

(11.12)

where K depends only upon τ and Ω. Deﬁnition 11.3.2 A relatively open subset Σ of the boundary of the open set Ω ⊂ Rn is class Ca , where a = k + α, k ≥ 1, 0 < α ≤ 1, if for each x ∈ Σ there is a neighborhood U of x and a Ca diﬀeomorphism τx of U onto a ball B = By,ρ ⊂ Rn of radius ρ with center at y = (0, . . . , 0, −ρ0 ) such that

406

11 Oblique Derivative Problem

(i) U ∩ ∂Ω ⊂ Σ, (ii) ∂(U ∩ Ω) ⊂ Ω ∪ Σ, (iii) τx (U ∩ Σ) = B 0 , and (iv) τx (U ∩ Ω) = B + . The relation Σ ∈ Ca signiﬁes that the Ha (Ω) norms of the τx and τx−1 , x ∈ Σ, are uniformly bounded. Lemma 11.3.3 Let τ be a diﬀeomorphism of class Ck+α , k ≥ 1, 0 < α ≤ 1 of Ω onto Ω ∗ . If u ∈ Hj+β (Ω), u˜(y) = u(τ −1 (y)), y ∈ Ω ∗ , and j +β ≤ k +α, 1 ≤ j ≤ k, 0 < β ≤ 1, then there is a constant C > 0 such that 1 |u|j+β,Ω ≤ |˜ u|j+β,Ω ∗ ≤ C|u|j+β,Ω C where C depends only upon β, τ, Ω. Proof: The proof will be carried out only in the j = 1 case. It follows from the deﬁnition of u ˜ that ˜ u 0,Ω ∗ = u 0,Ω ≤ |u|1+β,Ω . By the chain rule for diﬀerentiation, Dyi u˜(y) = Dyi u(τ −1 (y)) =

n

−1 Dm u(τ −1 (y))Dyi (τm (y)|.

m=1

Thus, [˜ u]1+0,Ω ∗ ≤ C[u]1+0,Ω where C is a constant depending only upon τ . By Inequality (11.12), |Dyi u ˜(y) − Dyi u ˜(x)| β |x − y| n n −1 −1 | m=1 Dm u(τ −1 (y))Dyi (τm (y)) − m=1 Dm u(τ −1 (x))Dyi (τm (x))| = β |x − y| n −1 −1 |(Dm u(τ (y)) − Dm u(τ −1 (x))Dyi (τm (y))| ≤C −1 −1 β |τ (y) − τ (x)| m=1 n −1 −1 |(Dm u(τ −1 (x))(Dyi (τm (y) − Dyi (τm (x))| β |x − y| m=1 ≤ C [u]1+β,Ω [τ −1 ]1+0,Ω ∗ + [u]1+0,Ω [τ −1 ]1+β,Ω ∗

+

≤ C[u]1+β,Ω [τ −1 |1+β,Ω ∗ ≤ C[u]1+β,Ω . Therefore, |˜ u|1+β,Ω ∗ ≤ C|u|1+β,Ω . Interchanging the roles of u, τ and u˜, τ −1 , u|1+β,Ω ∗ . The assertion follows there is a constant K such that |u|1+β,Ω ≤ K|˜ by replacing C and K by the maximum of the two. Consider a relatively open subset Σ of ∂Ω of class C2+α and the diﬀeomorphisms τx , x ∈ Σ, satisfying (i) through (iv) of Deﬁnition 11.3.2. For each

11.3 Curved Boundaries

407

x ∈ Σ, let N (x) be the set of neighborhoods U of x associated with x in the deﬁnition. For x ∈ Ω, N (x) will denote the set of balls Bx,ρ with closure in Ω; the τx associated with Bx,ρ will be the identity map. Consider any z ∈ Ω ∪ Σ, U ∈ N (z), and let τ be the diﬀeomorphism associated with z satisfying the above conditions mapping U onto the ball ˜(y) = u(x). Under the map B ⊂ Rn . Let y = τ (x) = (τ1 (x), . . . , τn (x)) and u * u = f˜ on B + where τ , the equation Lu = f on Ω ∩ U becomes the equation L˜ ˜ u(y) = L˜

n

a ˜pq (y)Dpq u ˜(y) +

p,q=1

n

˜bp Dp u ˜(y) + c˜(y)˜ u(y)

p=1

and a ˜pq (y) = ˜bp (y) =

n ∂τp ∂τq aij (x) ∂x i ∂xj i,j=1

p, q = 1, . . . , n

n

n ∂ 2 τp ∂τp aij (x) + bi (x) ∂xi ∂xj ∂xi i,j=1 i=1

c˜(y) = c(x),

p = 1, . . . , n

f˜(y) = f (x).

Moreover, the boundary condition Mu = g on U ∩ ∂Ω becomes the condition , u = g˜ on B ∩ Rn where M˜ 0 , u(y ) = M˜

n

˜(y) + γ˜ (y )˜ u(y , 0) β˜p (y )Dp u

p=1

and β˜p (y ) =

n

βi (x)

i=1

∂τp , γ˜ (y) = γ(x). ∂xi

* is strongly elliptic on B + , a few facts related to In order to show that L quadratic forms will be reviewed. Consider the quadratic function Q(y) =

n

cpq yp yq

p,q=1

where the n × n matrix [cpq ] is a real symmetric matrix. The quadratic form Q is said to be positive deﬁnite if Q(y) ≥ 0 for all y ∈ Rn and Q(y) = 0 only when y = 0. The form Q is positive deﬁnite if and only if there is a constant m such that Q(y) ≥ m|y|2 for all y ∈ Rn (cf. [43], p. 210). Consider the quadratic form

408

11 Oblique Derivative Problem

Q(y) =

n

a ˜pq yp yq =

n n ∂τp ∂τq aij (x) yp yq ∂xi ∂xj p,q=1 i,j=1

=

n n ∂τp ∂τq yp yq aij (x) ∂xi ∂xj i,j=1 p,q=1

p,q=1

=

n n ∂τp i,j=1

p=1

∂xi

yp

n ∂τq yq aij (x). ∂xj q=1

Since the matrix [aij ] is positive deﬁnite, Q(y) ≥ 0 and Q(y) = 0 implies that n ∂τp p=1

∂xi

yp = 0,

i = 1, . . . , n;

but since the Jacobian of the map τ does not vanish on U ∩ Ω by the multiplication theorem for Jacobians (c.f. [1]), this system of linear equations can * is only have the trivial solution y = 0. Thus, Q(y) is positive deﬁnite and L strongly elliptic on B + . Suppose N (x) contains a convex U with associated τx mapping U onto a ball B = By,ρ ⊂ Rn . Letting K denote the uniform bound on the H2+α norms of the τx and τx−1 and using the convexity of U , it is easily seen that 1 |y − z| ≤ |τx (y) − τx (z)| ≤ K|y − z|, K

y, z ∈ U.

(11.13)

Each U ∈ N (x) contains a V ∈ N (x), not necessarily convex, for which V ∩ Ω is the inverse image under τx of an admissible spherical chip and satisﬁes Inequalities (11.13). This can be seen as follows. Suppose τx (U ) = By,ρ , let Bx, be a ball with Bx, ⊂ U , let W = τx (Bx, ), and let By ,ρ be a ball for which By ,ρ ⊂ W and By ,ρ ∩ Rn+ is an admissible spherical chip. Let V = τx−1 (By ρ ) ⊂ Bx, ⊂ U . The convexity of Bx, can be used to show that V satisﬁes Inequalities (11.13) even though V need not be convex. Deﬁnition 11.3.4 If x ∈ Σ, U ∈ N (x) and τx is the associated diﬀeomorphism, U is an admissible neighborhood of x if it is the inverse image under τx of a ball deﬁning an admissible spherical chip and satisﬁes Inequality (11.13). The nonempty collection of admissible neighborhoods of x ∈ Σ will be denoted by A(x) with A(x) = N (x) for x ∈ Ω. It can be assumed that each admissible neighborhood is the inverse image of a spherical chip for which the conclusion of Theorem 11.2.8 is valid as the functions f and g will satisfy global weighted H¨ older norms. Consider any x ∈ Σ and U ∈ A(x) with associated diﬀeomorphism τx mapping U onto B. Letting d*U∩Ω (y) = d(y, ∂(U ∩ Ω) ∼ (U ∩ Σ)) and d*B + (z) = d(z, ∂B + ∼ B 0 ), if z = τx (y), then 1 * dU∩Ω (y) ≤ d*B + (z) ≤ K d*U∩Ω (y). K

11.3 Curved Boundaries

409

(b)

Moreover, if u ∈ H2+α (U ∩ Ω) and u˜ = u ◦ τx−1 , then as in Section 10.3 there is a constant C > 0 such that 1 (b) (b) (b) |u| ≤ |* u|2+α,B + ≤ C|u|2+α,U∩Ω . C 2+α,U∩Ω Since Theorem 7.6.7 and Inequality (7.33) apply to spherical chips Ω with Σ = int(Ω ∩ R0n ), if 0 ≤ a ≤ a, b ≤ b < 0, a + b ≥ 0, and b, b are not negative integers or 0, then (b)

(b )

|u|a ,U∩Ω ≤ C|u|a,U∩Ω

(11.14)

where C = C(a, a , b, b , K, A, d(Ω)). Although no mention is made of an interior sphere condition in the following versions of the Boundary Point Lemma and the Strong Maximum Principle, a version of the condition is implicit in the requirement that Σ be of class C2+α . Lemma 11.3.5 (Boundary Point Lemma) If x ∈ Σ ∈ C2+α , U ∈ N (x), u ∈ C 0 (cl(U ∩ Ω)) ∩ C 2 (U ∩ Ω), Lu ≥ 0 on U ∩ Ω, y0 ∈ U ∩ Σ, u(y0 ) > u(y) for all y ∈ U ∩ Ω, then Mu(y0 ) < 0, if deﬁned, whenever u(y0 ) ≥ 0. Proof: Using the above notation, the function u ˜ on B + induced by the *u ≥ 0 diﬀeomorphism τx associated with x and U will satisfy the conditions L˜ on B + , u˜(z0 ) > u ˜(z) for all z ∈ B + where z0 = τx (y0 ) ∈ B 0 and z = τx (y), y ∈ U ∩Ω. Since the interior sphere condition is satisﬁed at z0 , it follows , u(z0 ) < 0 whenever u ˜(z0 ) ≥ 0, and consequently that Mu(y0 ) < 0 that M˜ whenever u(y0 ) ≥ 0. The following corollary is proved in the same way Corollary 11.2.2 is proved, using the above boundary point lemma. Corollary 11.3.6 (Strong Maximum Principle) If Σ ∈ C2+α , u ∈ C 0 (Ω − ) ∩C 2 (Ω) satisﬁes Lu ≥ 0 on Ω, Mu ≥ 0 on Σ, and either (i) u ≤ 0 on ∂Ω ∼ Σ when nonempty or (ii) c ≡ 0 on Ω or γ ≡ 0 on Σ when Σ = ∂Ω, then either u = 0 on Ω or u < 0 on Ω and, in particular, u ≤ 0 on Ω. It should be noted that the f and g of the following theorem satisfy global weighted H¨ older norms so that the conditions of Theorems 11.2.5 and 11.2.8 are satisﬁed. Theorem 11.3.7 If b ∈ (−1, 0), α ∈ (0, 1), Ω is a bounded open subset of (2+b) (Ω), and Rn , Σ is a relatively open subset of ∂Ω of class C2+α , f ∈ Hα (1+b) g ∈ H1+α (Σ), then for each x ∈ Ω ∪ Σ, U ∈ A(x), and h ∈ C 0 (∂(U ∩ Ω) ∼ (U ∩ Σ)) there is a unique function u ∈ C 0 (cl(U ∩ Ω)) ∩ C 2 (U ∩ (Ω ∪ Σ)) satisfying Lu = f on U ∩ Ω, Mu = g on U ∩ Σ, and u = h on ∂(U ∩ Ω) ∼ (U ∩ Σ).

410

11 Oblique Derivative Problem

Moreover, there is a constant C = C(aij , bi , c, α, b, βi , γ, d(Ω)) such that (b) (2+b) (1+b) |u|2+α,U∩Ω ≤ C |f |α,U∩Ω + |g|1+α,U∩Σ + |h|0,∂(U∩Ω)∼(U∩Σ) . (11.15) Proof: If x ∈ Ω and U is a ball, the result is covered by Theorem 11.2.8. Consider any x ∈ Σ, U ∈ A(x), τx the diﬀeomorphism associated with x and U , and the ball B = τx (U ). Under the map τx , the ﬁrst two of the above equations become , u = g˜ on B 0 . * u = f˜ on B + and M˜ L˜ Since the components of τx and τx−1 are in H2+α (U ) and H2+α (B), respectively, the maps τx and τx−1 have continuous extensions, denoted by the same symbols, to U − and B − , respectively. It is easily seen that the extensions are inverses to each other and are one-to-one. It is also easily seen that the regions ∂(U ∩ Ω) ∼ (U ∩ Σ) and ∂B + ∼ B 0 are mapped onto each other. For z ∈ ∂B + ∼ B 0 , let ˜ h(z) = h(τx−1 (z)). Under the map τx , the equation u = h ˜ on ∂B + ∼ B 0 . By Theorem 11.2.8, the on ∂(U ∩Ω) ∼ (U ∩Σ) becomes u ˜=h transformed equations have a unique solution u ˜ ∈ C 0 (cl(B + ) ∩ C 2 (B + ∪ B 0 ). The function u(y) = u ˜(τx (y)), y ∈ U ∩ Ω, satisﬁes the above equations. The inequality follows from Inequality (11.8).

11.4 Superfunctions for Elliptic Operators Just as superharmonic functions and superfunctions played a role in proving the existence of a solution to the Dirichlet problems for the Laplacian and elliptic operators, respectively, the concept of superfunction can be used to prove the existence of a solution to the oblique derivative boundary problem. As usual Ω will denote a bounded open subset of Rn , α and b will be parameters with α ∈ (0, 1), b ∈ (−1, 0), Σ will denote a nonempty, relatively open subset of ∂Ω of class C2+α . No further conditions will be imposed on Ω and Σ in this section. Throughout this section f, g, and h will denote functions (2+b) (1+b) in Hα (Ω), H1+α (Σ), and C 0 (∂Ω ∼ Σ), respectively. By Theorem 11.3.7, for each x ∈ Σ and U ∈ A(x) there is a unique function w ∈ C 0 (cl(U ∩ Ω)) ∩ C 2 (U ∩ (Ω ∪ Σ)) satisfying the equations Lw = f on U ∩ Ω, Mw = g on U ∩ Σ, and w = φ on ∂(U ∩ Ω) ∼ (U ∩ Σ) for φ ∈ C 0 (∂(U ∩ Ω) ∼ (U ∩ Σ)). The function w will be called the local solution associated with x, U , and φ and will be denoted by LS(x, U, φ). It will be understood that if the domain of the function φ contains ∂(U ∩ Ω) ∼ (U ∩ Σ), then the third argument of LS(x, U, φ) is the restriction of φ to the latter set. It will be seen that LS(x, U, φ) plays the same role as the Poisson integral and might justiﬁably be denoted by PI(x, U, φ).

11.4 Superfunctions for Elliptic Operators

411

Lemma 11.4.1 (i) If x ∈ Ω ∪ Σ, U ∈ A(x), u ∈ C 0 (cl(U ∩ Ω)) ∩ C 2 (U ∩ Ω), Lu = 0 on U ∩ Ω, Mu = 0 on U ∩ Σ, and u ≥ 0 on ∂(U ∩ Ω) ∼ (U ∩ Σ), then u ≥ 0 on U ∩ Ω. (ii) If ui ∈ C 0 (cl(U ∩ Ω)) ∩ C 2 (U ∩ Ω), Lui = f on U ∩ Ω, Mui = g on U ∩ Σ, and ui = hi on ∂(U ∩ Ω) ∼ (U ∩ Σ), i = 1, 2, with h1 ≥ h2 , then u1 ≥ u2 on U ∩ Ω. Proof: The second assertion is an immediate consequence of the ﬁrst. To prove (i), let u satisfy the above conditions and assume that u attains a strictly negative minimum at a point x0 in cl(U ∩Ω). If x0 is an interior point of U ∩ Ω, then it must be a constant by Theorem 9.5.7; that is, u = k < 0 on U ∩ Ω and must be equal to k on ∂(U ∩ Ω) ∼ (U ∩ Σ), contradicting the fact that u ≥ 0 on ∂(U ∩ Ω) ∼ (U ∩ Σ). By the Boundary Point Lemma, x0 can not be a point of U ∩ Σ. Therefore, x0 ∈ ∂(U ∩ Ω) ∼ (U ∩ Σ) with u(x0 ) < 0, contradicting the fact that u ≥ 0 on ∂(U ∩ Ω) ∼ (U ∩ Σ). Therefore, u ≥ 0 on U ∩ Ω. Deﬁnition 11.4.2 A function u ∈ Cb0 (Ω) is a superfunction on Ω if for each x ∈ Ω ∪ Σ there is a Vx,u ∈ A(x) such that for all U ∈ A(x), U ⊂ Vx,u , v ∈ C 0 (cl(U ∩ Ω)) ∩ C 2 (U ∩ Ω) Lv = f on U ∩ Ω, Mv = g on U ∩ Σ, and u ≥ v on ∂(U ∩ Ω) ∼ (U ∩ Σ) (11.16) implies that u ≥ v on U ∩Ω. The set of all such superfunctions will be denoted by U(f, g). There is a corresponding deﬁnition of subfunction and L(f, g). Remark 11.4.3 Using the above notation, u ∈ Cb0 (Ω) is a superfunction on Ω if and only if there is a Vx,u ∈ A(x) such that u ≥ LS(x, U, u) on U ∩ Ω for all U ⊂ Vx,u , U ∈ A(x).

(11.17)

This can be seen as follows. Assume u is a superfunction on Ω, x ∈ Ω ∪ Σ, U ⊂ Vx,u , U ∈ A(x), and let v = LS(x, U, u), U ⊂ Vx,u . Then Lv = f on U ∩ Ω, Mv = g on U ∩ Σ, and v = u on ∂(U ∩ Ω) ∼ (U ∩ Σ). Thus, u ≥ v = LS(x, U, u) on U ∩ Ω for all U ⊂ Vx,u , U ∈ A(x). Suppose now that (11.17) holds. Consider any v satisfying (11.16). By (ii) of Lemma 11.4.1, u ≥ LS(x, U, u) ≥ LS(x, U, v) = v on U ∩Ω so that u is a superfunction on Ω. Lemma 11.4.4 If u ∈ C 0 (Ω − ) ∩ C 2 (Ω ∪ Σ) satisﬁes Lu ≤ f on Ω and Mu ≤ g on Σ and k > 0, then u and u + k are superfunctions on Ω. Proof: Suppose Lu ≤ f on Ω and Mu ≤ g on Σ. Consider any y ∈ Ω∪Σ, U ∈ A(y), and a function v such that Lv = f on U ∩Ω and Mv = g on U ∩Σ with u ≥ v on ∂(Ω ∩ U ) ∼ (U ∩ Σ). Then L(u − v) ≤ 0 on U ∩ Ω, M(u − v) ≤ 0 on U ∩ Σ, and u − v ≥ 0 on ∂(U ∩ Ω) ∼ (U ∩ Σ). It follows from Corollary 11.3.6

412

11 Oblique Derivative Problem

that u ≥ v on U ∩ Ω, proving that u is a superfunction. If k > 0, then L(u + k) ≤ Lu ≤ f and M(u + k) ≤ Mu ≤ g, and it follows from the ﬁrst part of the proof that u + k is a superfunction. Lemma 11.4.5 (Lieberman [39]) (i) If u and v are superfunctions on Ω, then min (u, v) is a superfunction on Ω. (ii) If u is a superfunction on Ω, y ∈ Ω ∪ Σ, and U ⊂ Vy,u , U ∈ A(y), the function u on Ω ∼ U uU = LS(y, U, u) on U ∩ Ω is a superfunction on Ω. If u, v are superfunctions on Ω and u ≥ v on Ω, then uU ≥ vU on Ω. (iii)If Σ = ∂Ω, either c < 0 on Ω or γ < 0 on Σ, u ∈ L(f, g), and v ∈ U(f, g), then u ≤ v on Ω. Proof: The ﬁrst assertion (i) follows from the deﬁnition of a superfunction. Suppose u and uU are as described in (ii). Since u ≥ LS(y, U, u) = uU on U ∩ Ω and uU = u on Ω ∼ U, u ≥ uU on Ω. Consider any z ∈ Ω ∪ Σ and W ⊂ Vz,u , W ∈ A(z) and the functions v=

u

on (∂(U ∩ Ω) ∼ (U ∩ Σ)) ∩ W

uU on (∂(W ∩ Ω) ∼ (W ∩ Σ)) ∩ U −

⎧ ⎨ LS(z, W, uU ) on (∂(U ∩ Ω) ∼ (U ∩ Σ)) ∩ W w=

⎩

uU

on (∂(W ∩ Ω) ∼ (W ∩ Σ)) ∩ U − .

Since u ≥ LS(z, W, u) ≥ LS(z, W, uU ) on W ∩ Ω, v ≥ w on (∂(U ∩ Ω) ∼ (U ∩ Σ)) ∩ W ). Since LS(y, U, u) = u = v on (∂(U ∩ Ω) ∼ (u ∩ Σ)) ∩ W and LS(y, U, u) = uU = v on (∂(W ∩ Ω) ∼ (W ∩ Σ)) ∩ U − , LS(y, U, u) is the unique local solution on U ∩ W ∩ Ω corresponding to the boundary function v on ∂(U ∩ W ∩ Ω) ∼ (U ∩ W ∩ Σ). Since LS(z, W, uU ) = w on (∂(U ∩ Ω) ∼ (U ∩ Σ)) ∩ W and LS(z, W, uU ) = uU = w on (∂(W ∩ Ω) ∼ (W ∩ Σ)) ∩ U − , LS(z, W, uU ) is the unique local solution on U ∩ W ∩ Ω corresponding to the boundary function w on ∂(U ∩ W ∩ Ω) ∼ (U ∩ W ∩ Σ). Note that v = u ≥ LS(z, W, u) ≥ LS(z, W, uU ) = w on (∂(U ∩ Ω) ∼ (U ∩ Σ)) ∩ W and v = uU = w on (∂(W ∩ Ω) ∼ (W ∩ Σ)) ∩ U − so that v ≥ w on ∂(U ∩ W ∩ Ω) ∼ (U ∩ W ∩ Σ). Thus, uU = LS(y, U, u) ≥ LS(z, W, uU ) on U ∩ W ∩ Ω. On (W ∼ U ) ∩ Ω, uU = u ≥ LS(z, W, u) ≥ LS(z, W, uU ).

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413

Therefore, uU ≥ LS(z, W, uU ) on W ∩ Ω for all W ∈ A(z), proving that uU is a superfunction. To prove (iii), consider any u ∈ L(f, g) and any v ∈ U(f, g). Letting uˆ(x) = lim sup u(y) and vˆ(x) = lim inf v(y), y→x,y∈Ω

y→x,y∈Ω

for x ∈ Ω − , u ˆ and vˆ are u.s.c. and l.s.c. on Ω − , respectively. If x ∈ Ω ∪ Σ, let Vx,u and Vx,v be the neighborhoods associated with the deﬁnitions of u and v, respectively. Let μ = sup {ˆ u(x) − vˆ(x); x ∈ Ω − }, assume μ > 0, and let Γ = {y ∈ Ω − ; uˆ(y) − vˆ(y) = μ}. Since both Vx,u and Vx,v contain arbitrarily small neighborhoods of x in A(x), for each x ∈ Ω ∪ Σ there is a Wx ⊂ Vx,u ∩ Vx,v , Wx ∈ A(x). It will be shown ﬁrst that Wx ∩ ∂Ω = ∅ for some x ∈ Γ . Suppose Wx ∩ ∂Ω = ∅ for all x ∈ Γ . In this case, Γ is a compact subset of Ω and there is a point y ∈ Γ nearest to ∂Ω. Note that Wy− ⊂ Ω by choice of Vy,u and Vy,v when y ∈ Ω. Let wy = LS(y, Wy , u) − LS(y, Wy , v) on Wy . Then Lwy = 0 on Wy , wy = u − v on ∂Wy , and wy (y) = LS(y, Wy , u)(y) − LS(y, Wy , v)(y) ≥ u(y) − v(y) = μ, the latter inequality holding since u is a subfunction and v is a superfunction. Since wy = u − v ≤ μ on ∂Wy and wy (y) ≥ μ, wy attains a nonnegative maximum value at an interior point of Wy and must be equal to a constant on Wy , namely μ, by the Strong Maximum Principle, Theorem 9.5.6. Thus, there are points of Γ that are closer to ∂Ω than y, a contradiction. Therefore, there is a point x0 ∈ Γ such that Wx0 ∩Ω = ∅. Letting w0 = LS(x0 , Wx0 , u)− LS(x0 , Wx0 , v), as above, Lw0 = 0 on Wx0 ∩ Ω, Mw0 = 0 on Wx0 ∩ Σ, w0 = u−v on ∂(Wx0 ∩Ω) ∼ (Wx0 ∩Σ), and w0 (x0 ) ≥ u(x0 )−v0 (x0 ) = μ. If w0 attains its nonnegative maximum value at an interior point of Wx0 ∩Ω, then it must be a constant, namely μ, on Wx0 ∩ Ω. If c < 0, then 0 = Lw0 = cμ < 0 on Wx0 ∩ Ω, a contradiction. Therefore, w0 cannot attain its nonnegative maximum value on Wx0 ∩ Ω and must do so at x0 . By the Boundary Point Lemma, Lemma 11.2.1, Mw0 (x0 ) < 0, a contradiction. The assumption that μ > 0 is untenable and so u ≤ v on Ω. The function uU of (ii) in the preceding lemma will be called the lower or lowering of u on U . The corresponding v U for v a subfunction on Ω will be called the lift or lifting of v on U .

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11 Oblique Derivative Problem

Deﬁnition 11.4.6 A function u ∈ Cb0 (Ω) is a supersolution for the problem Lv = f on Ω, Mv = g on Σ, and v = h on ∂Ω ∼ Σ if u is a superfunction for f and g and lim inf y→x,y∈Ω u(y) ≥ h(x) for x ∈ ∂Ω ∼ Σ. The set of all such supersolutions will be denoted by U(f, g, h). There is a corresponding deﬁnition of subsolution and L(f, g, h). When Σ = ∂Ω, the concept of supersolution is the same as that of superfunction. The following theorem eliminates the requirement that γ ≤ −γo < 0 on Σ in [40]. Lemma 11.4.7 (Σ = ∂Ω) (i)If u, v∈U(f, g, h), then min (u, v) ∈ U(f, g, h). (ii) If u ∈ U(f, g, h), y ∈ Ω ∪ Σ, and U ∈ A(y), then uU ∈ U(f, g, h). (iii) If u ∈ L(f, g, h) and v ∈ U(f, g, h), then u ≤ v on Ω. Proof: To prove (i), it need only be noted that lim inf y→x,y∈Ω min (u, v)(y) ≥ h(x) on ∂Ω ∼ Σ. To prove (ii), it suﬃces to show that lim inf y→x,y∈Ω uU (y) ≥ h(x) on ∂Ω ∼ Σ since uU is a superfunction according to Lemma 11.4.5. Since ∂(U ∩ Ω) ⊂ Ω ∪ Σ by (ii) of Deﬁnition 11.3.2, (U ∩ Ω)− ⊂ Ω ∪ Σ. This implies that for x ∈ ∂Ω ∼ Σ, there is a neightborhood V of x such that V ∩ (U ∩ Ω)− = ∅, which in turn implies that uU = u on V and therefore lim inf y→x,y∈Ω uU (y) = lim inf y→x,y∈Ω u(y) ≥ h(x), completing the proof of (ii). Consider any u ∈ L(f, g, h) and v ∈ U(f, g, h). Letting u ˆ(x) = ˆ and vˆ are lim supy→x,y∈Ω u(y) and vˆ(x) = lim inf y→x,y∈Ω v(y) for x ∈ Ω − , u u.s.c. and l.s.c. on Ω − , respectively, u ˆ(x) ≤ h(x) ≤ vˆ(x) for x ∈ ∂Ω ∼ Σ, and uˆ − vˆ ≤ 0 on ∂Ω ∼ Σ. To prove (iii), it suﬃces to show that u ≤ v on each component of Ω. It therefore can be assumed that Ω is connected. Let μ = sup {ˆ u(x) − vˆ(x); x ∈ Ω − }, assume μ > 0, and let Γ = {x ∈ Ω − ; uˆ(x) − vˆ(x) = μ} = ∅. If x0 ∈ Ω∩Γ and Wx0 ∈ A(x0 ), then Wx−0 ⊂ Ω. Letting w0 = LS(x0 , Wx0 , u)− LS(x0 , Wx0 , v) on Wx0 ∩ Ω, Lw0 = 0 on Wx0 ∩ Ω, w0 = u − v ≤ μ on ∂(Wx0 ∩ Ω), and w0 (x0 ) = LS(x0 , Wx0 , u)(x0 ) − LS(x0 , Wx0 , v)(x0 ) ≥ u(x0 ) − v(x0 ) = μ. Thus, w0 attains its nonnegative maximum value at an interior point of Wx0 . By the Strong Maximum Principle, Theorem 9.5.6, w0 is constant on Wx0 ∩Ω and since w0 ≤ u − v ≤ μ on ∂(Wx0 ∩ Ω), w0 = μ on Wx0 ∩ Ω. Therefore, Wx0 = Wx0 ∩ Ω ⊂ Γ . This shows that Ω ∩ Γ is an open subset of Ω. Since Ω ∼ Γ is an open subset of Ω, either Ω ∩ Γ = ∅ or Ω ∼ Γ = ∅ by the connectedness of Ω. Suppose Ω ∼ Γ = ∅. Then Γ ⊃ Ω so that u − v = μ on Ω or u = v + μ on Ω. If x is a point in ∂Ω ∼ Σ, then h(x) ≥ lim sup u(y) = lim sup(v(y) + μ) ≥ lim inf v(y) + μ ≥ h(x) + μ, y→x

y→x

y→x

11.4 Superfunctions for Elliptic Operators

415

a contradiction. Thus, under the assumption that μ > 0, Ω ∼ Γ = ∅. Suppose now that Ω ∩ Γ = ∅ and consider any y0 ∈ Γ ⊂ ∂Ω, Wy0 ∈ A(y0 ), and let w1 = LS(y0 , Wy0 , u) − LS(y0 , Wy0 , v). Then Lw1 = 0 on Wy0 ∩ Ω, Mw1 = 0 on Wy0 ∩ Σ, w1 = u − v on ∂(Wy0 ∩ Ω) ∼ (Wy0 ∩ Σ), and w1 (y0 ) ≥ u(y0 ) − v(y0 ) = μ as above. If w1 attains its nonnegative maximum value at an interior point of Wy0 , then it must be a constant and equal to μ on Wy0 ∩ Ω. Since w1 = u − v on ∂(Wy0 ∩ Ω) ∼ (Wy0 ∩ Σ), there would be points of Ω where u − v = μ, a contradiction. Thus, w1 < μ = w1 (y0 ) on Wy0 ∩ Ω. By the Boundary Point Lemma, Lemma 9.5.5, Mw1 (y0 ) < 0, contradicting the fact that Mw1 (y0 ) = 0. Thus, under the assumption μ > 0, Ω ∩ Γ = ∅ and, as noted above, Ω ∼ Γ = ∅; but this contradicts the fact that Ω is connected. Thus, the assumption that μ > 0 is untenable and u ≤ v on Ω. Lemma 11.4.8 If x ∈ Ω ∪Σ, U ∈ A(x), and {vk } is a sequence in C 0 (cl(U ∩ Ω)) ∩ C 2 (U ∩ (Ω ∪ Σ)) with supk vk 0,∂(U∩Ω)∼(U∩Σ) < +∞ satisfying the equations Lvk = f on U ∩ Ω,

Mvk = g on U ∩ Σ,

(2+b)

k ≥ 1,

(11.18)

(1+b)

(Ω) and g ∈ H1+α (Σ), then there is a subsequence {vk } that for f ∈ Hα converges uniformly on U ∩ (Ω ∪ Σ) to v ∈ C 0 (cl(U ∩ Ω)) ∩ C 2 (U ∩ (Ω ∪ Σ)) satisfying Lv = f on U ∩ Ω, Mv = g on U ∩ Σ. (11.19) Proof: Letting K0 = supk vk 0,∂(U∩Ω)∼(U∩Σ) and using Theorem 11.3.7, (b) (2+b) (1+b) (11.20) |vk |2+α,U∩Ω ≤ C K0 + |f |α,U∩Ω + |g|1+α,U∩Σ . For 0 < < d(U ∩ Ω), (2+b) (1+b) 2+b+α |vk |2+α,(U∩Ω) ≤ C K0 + |f |α,Ω + |g|1+α,Σ .

(11.21)

It follows from this inequality that the sequences {vk }, {Di vk }, and {Dij vk } are uniformly bounded on (U ∩ Ω) . Since the term [vk ]2+α,(U∩Ω) is included in the expression on the left side, the {Dij vk } sequences are uniformly (b) equicontinuous on (U ∩ Ω) and therefore on cl(U ∩ Ω) . Since |vk | ≤ (b)

1+α,U∩Ω

C|vk |2+α,U∩Ω by Inequality (11.14), the same argument is applicable to the sequences {Di vk }; that is, the sequences {Di vk } are uniformly equicontinu(b) ous on cl(U ∩ Ω) . Since |vk |α,U∩Ω need not be deﬁned, the same argument cannot be applied to the {vk } sequence. By considering the images v*k of the vk under the map τx , the mean value theorem can be applied to show that the diﬀerence quotients |vk (y) − vk (z)|/|y − z| are uniformly bounded on (U ∩ Ω) so that the sequence {vk } is also uniformly equicontinuous on cl(U ∩ Ω) . Therefore, the sequences {vk }, {Di vk }, and {Dij vk } are uniformly

416

11 Oblique Derivative Problem

bounded and equicontinuous on the compact set cl(U ∩ Ω) . Consider a sequence { m } strictly decreasing to zero. It follows from the Arzela-Ascoli (1) Theorem, Theorem 0.2.3, that there is a subsequence {vk } of the {vk } se(1) (1) (1) quence such that the sequences {vk }, {Di vk }, and {Dij vk } converge uniformly to continuous functions on cl (U ∩ Ω) . Applying the same argument 1

(1)

(2)

(1)

to the {vk } sequence, there is a subsequence {vk } of the {vk } sequence (2) (2) (2) such that the sequences {vk }, {Di vk }, and {Dij vk } converge uniformly to continuous functions on cl (U ∩ Ω) , etc. Using the Cantor diagonaliza 2

()

tion method, the sequence {vk } with vk = v has the property that the sequences {vk }, {Di vk }, and {Dij vk } converge to continuous functions on ∞ %

cl (U ∩ Ω) m = U ∩ (Ω ∪ Σ).

m=1

Letting v = lim→∞ vk , replacing vk in Inequality (11.21) by vk , letting → ∞ and then taking the supremum over 0 < < d(U ∩ Ω), v satisﬁes (b) (b) Inequality (11.20). Since |v|−b,U ∩Ω = |v|−b,U ∩Ω ≤ |v|2+α,U∩Ω ≤ C(K0 + (2+b)

(1+b)

|f |α,U∩Ω + |g|1+α,U∩Σ ), |v(x) − v(y)| ≤ K|x − y|−b from which it follows that v has a continuous extension to cl(U ∩ Ω) and that v ∈ C 0 (cl(U ∩ Ω)). Since the Dij vkl converge to Dij v on each cl (U ∩ Ω) m , v ∈ C 2 (U ∩ (Ω ∪ Σ)). Equation (11.19) follows easily by taking limits in Equation (11.18). Lemma 11.4.9 u ∈ U(f, g, h) if and only if −u ∈ L(−f, −g, −h). Proof: For y ∈ Ω ∪ Σ, let LS+ (y, U, u) and LS− (y, U, u) denote local so⊕ lutions relative to (f, g, h) and (−f, −g, −h), respectively. Also, let Vy,u and Vy,u be the elements of A(y) relative to (f, g, h) and (−f, −g, −h) described ⊕ in Deﬁnition 11.4.2. Consider any Wy ⊂ Vy,u ∩ Vy,−u , Wy ∈ A(y). It will be shown ﬁrst that LS+ (y, Wy , u) = −LS− (y, Wy , −u).

(11.22)

Since L(LS+ (y, U, u)+LS− (y, U, −u)) = f −f = 0 on Wy ∩Ω, M(LS+ (y, U, u) +LS− (y, U, −u)) = g −g = 0 on Wy ∩Σ, and LS+ (y, U, u)+LS− (y, U, −u) = u − u = 0 on ∂Wy ∩ Ω, LS+ (y, U, u) + LS− (y, U, −u) = 0 on Wy ∩ Ω by Corollary 11.2.2, establishing Equation 11.22. If u ∈ U(f, g, h), then u ≥ LS+ (y, Wy , u) on Wy ∩ Ω. Thus, −u ≤ −LS+ (y, Wy , u) = LS− (y, Wy , −u) on Wy ∩ Ω.

11.4 Superfunctions for Elliptic Operators

417

Since lim sup −u(y) = − lim inf u(y) ≤ −h(x), y→x,y∈Ω

y→x,y∈Ω

−u ∈ L(−f, −g, −h) whenever u ∈ U(f, g, h). Recall the standing assumption that |β(x)| = 1 for all x ∈ Σ. Part of the hypotheses of the following two lemmas is an exception to the standard requirement that f and g be ﬁxed functions. The following theorem was proved for ﬁxed f and g by Lieberman [39] assuming that γ ≤ −γ0 < 0. Theorem 11.4.10 If Ω is a bounded open subset of Rn , Σ is a relatively open subset of ∂Ω, ν has a continuous extension to Σ − , β · ν ≥ > 0 on Σ, c ≤ 0 on Ω, γ ≤ 0 on Σ, F ⊂ Cb0 (Ω) with supf ∈F f 0,Ω ≤ M < +∞, and G ⊂ Cb0 (Σ) with supg∈G g 0,Σ ≤ N < +∞, then '

U(f, g) ∩ C 0 (Ω − ) ∩ C 2 (Ω ∪ Σ) = ∅

f ∈F ,g∈G

and

'

L(f, g) ∩ C 0 (Ω − ) ∩ C 2 (Ω ∪ Σ) = ∅;

f ∈F ,g∈G

moreover, if Σ = ∂Ω and h ∈ C 0 (∂Ω ∼ Σ), then the same conclusions are valid for U(f, g, h) and L(f, g, h). Proof: For each x ∈ Σ − , let V (x) be a neighborhood of x such that OscV (x)∩Σ ν ≤ (1 − ρ) /2 where 0 < ρ < 1, Since Σ − is compact, there are points x1 , . . . , xp ∈ Σ − such that Σ − ⊂ ∪pi=1 V (xi ). Let {ψi }pi=1 be a p partition of unity; that is, ψi ∈ C0∞ (V (xi )), 0 ≤ ψi ≤ 1, and i=1 ψi (x) = 1 for all x ∈ Σ − . For i = 1, . . . , p, let Mi u(x) = ψi (x)Mu(x), x ∈ Σ, noting that Mi u(x) = 0 if x ∈ supp ψi , the support of ψi . Note also that p (i) Mu(x) = i=1 Mi u(x), x ∈ Σ. For each i = 1, . . . , p, let x0 be a ﬁxed (i) point V (xi ) ∩ Σ − and let a(i) = ν(x0 ). Note that |a(i) | = 1 since ν is a unit vector. For x ∈ supp ψi ∩ Σ, (i) (i) < ≤ β(x) · ν(x0 ) + β(x) · (ν(x) − ν(x0 ) ≤ β(x) · a(i) + (1 − ρ) /2 2 from which it follows that β(x) · a(i) ≥ ρ /2. Suppose now that supp ψi ∩ supp ψj ∩ Σ = ∅ for j = i. For any x ∈ supp ψi ∩ Σ and zij ∈ supp ψi ∩ supp ψj ∩ Σ (j) ≤ β(x) · ν(x0 ) + β(x) · ν(x) − ν(zij ) + ν(zij ) − ν(xj0 ) ≤ β(x) · a(j) + (1 − ρ) from which it follows that β(x) · a(j) ≥ ρ . For i = 1, . . . , p and λ > 0, let

418

11 Oblique Derivative Problem

vi (x) =

(i) M + N λ2 e − eλa ·x , p

x ∈ Ω,

p and let v(x) = i=1 vi (x), x ∈ Ω. Let λ > 0 be any real number satisfying the conditions |x| ≤ λ for all x ∈ Ω, −(λρ /2) + |γ|0 < −1, and −mλ2 + λ |b| 0 + |c|0 < −1, Then Lvi (x) =

⎛ M +N ⎝ 2 −λ p

n

(i) (i)

i = 1, . . . , p.

ajk (x)aj ak − λ

n

⎞ (i)

bj (x)aj − c(x)⎠ eλa

(i)

·x

j=1

j,k=1

M + N λ2 e p (i) M +N 2 −λ m + λ |b| 0 + |c|0 eλa ·x ≤ p M + N −λ2 ≤− . e p 2 ≤ −(M + N ) on Ω. Simultaneously for all f ∈ F, Therefore, L veλ 2 λ L ve ≤ −M ≤ − f 0 ≤ f on Ω. For x ∈ supp ψi , by choice of λ + c(x)

M +N Mi vi (x) = −λ p

n

(i) ψi (x)βk (x)ak

eλa

(i)

·x

+ ψi (x)γ(x)vi (x)

k=1

(i) −λρ M +N ≤ ψi (x) + |γ|0 eλa ·x p 2 2 M +N ψi (x)e−λ . ≤− p

For j = i and x ∈ supp ψi , M +N Mi vj (x) = −λ p

n

(j) ψi (x)βk (x)ak

eλa

(j)

·x

+ ψi (x)γ(x)vj (x)

k=1

(j) M +N ψi (x)(−λρ + |γ|0 )eλa ·x p 2 M +N ψi (x)e−λ . ≤− p

≤

2

Summing over j = 1, . . . , p, Mi v(x) ≤ −(M + N )ψi (x)e−λ and by summing 2 2 over i = 1, . . . , p, Mv(x) ≤ −(M + N )e−λ . Thus, M(veλ ) ≤ −(M + N ) ≤ −N ≤ − g 0 ≤ g simultaneously for all g ∈ G. Since |x| ≤ λ, each vi ≥ 0

11.4 Superfunctions for Elliptic Operators

419

2

on Ω and so w = veλ ≥ 0 on Ω. Thus w is a superfunction simultaneously for all f ∈ F, g ∈ G. If Σ = ∂Ω and h ∈ Cb0 (∂Ω ∼ Σ), then w + h 0 is a superfunction simultaneously for all f ∈ F, g ∈ G and w + h 0 ≥ h on ∂Ω ∼ Σ. Lemma 11.4.11 Under the conditions of the preceding theorem, if u ∈ C 0 (Ω − ) ∩ C 2 (Ω ∪ Σ) satisﬁes the equations Lu = f on Ω for f ∈ Cb0 (Ω) ∩ (2+b) (1+b) Hα (Ω), Mu = g for g ∈ Cb0 (Σ) ∩ H1+α (Σ) and u = h on ∂Ω ∼ Σ, when 0 nonempty, for h ∈ C (∂Ω ∼ Σ), then

u 0,Ω ≤

C( f 0 + g 0 ) if Σ = ∂Ω C( f 0 + g 0 + h 0 ) if Σ = ∂Ω

where C = C(βi , d(Ω). Proof: Consider the case ∂Ω ∼ Σ = ∅ and the function v = w + h 0 ≥ 0 of the preceding proof for which Lv ≤ f on Ω, Mv ≤ g on Σ, and v ≥ h on ∂Ω ∼ Σ. Letting u∗ = u − v, Lu∗ ≥ 0 on Ω, Mu∗ ≥ 0 on Σ, and u∗ ≤ 0 on ∂Ω ∼ Σ. By the Strong Maximum Principle, Corollary 11.2.2, u∗ ≤ 0 on Ω; that is, u ≤ v = w + h 0 . Since w ≤ C( f 0 + g 0 ) on Ω, u ≤ C( f 0 + g 0 + h 0 ). As the same result applies to −u, −f, −g and −h, the inequality holds for −u and for |u|. Taking the supremum over Ω,

u 0 ≤ C( f 0 + g 0 + h 0 ). The following theorem was proved by Lieberman [39] assuming that γ ≤ −γ0 < 0 for some constant γ0 . Theorem 11.4.12 If Σ = ∂Ω, β · ν ≥ > 0 for some > 0, and either c < 0 on Ω or γ < 0 on Σ, then u = inf {v; v ∈ U(f, g)} ∈ Cb0 (Ω)∩C 2 (Ω ∪Σ), Lu = f on Ω, and Mu = g on Σ. Proof: Consider any v + ∈ U(f, g) and v − ∈ L(f, g), the two sets being nonempty by Theorem 11.4.10. Since v + ∈ C 0 (Ω − ) and u ≤ v + , u is bounded above. Since v − ≤ v for all v ∈ U(f, g) by (iii) of Lemma 11.4.5, u is bounded below and therefore bounded on Ω. Consider any y ∈ Ω ∪ Σ and any U ∈ A(y). By Choquet’s lemma, Lemma 2.2.8, there is a sequence {vj } in U(f, g) with the property that if w is a l.s.c. function on U ∩ Ω satisfying w ≤ inf {vj ; j ≥ 1} on U ∩ Ω, then w ≤ u on U ∩ Ω. Since the minimum of two functions in U(f, g) is again in U(f, g) by Lemma 11.4.5, it can be assumed that the sequence {vj } is decreasing by replacing vj by min (v1 , . . . , vj ), if necessary. For each j ≥ 1, let wj = (vj )U∩Ω , the lowering of vj over U ∩ Ω. The sequence {wj } is also decreasing by Lemma 11.4.5. Note that wj = LS(y, U, vj ) on U ∩ Ω implies that Lwj = f on U ∩ Ω, Mwj = g on U ∩ Σ, and that wj ∈ C 0 (cl(U ∩ Ω) ∩ C 2 (U ∩ (Ω ∪ Σ)) according to Theorem 11.3.7. If v − is any element of L(f, g), then v1 ≥ wj ≥ v − on Ω for all j ≥ 1 and therefore wj 0,∂(U∩Ω)∼(U∩Σ) ≤ max ( v1 0,Ω , v − 0,Ω ) < +∞ for all j ≥ 1. It follows from Lemma 11.4.8 that there is a subsequence {wji } of

420

11 Oblique Derivative Problem

the {wj } sequence that converges uniformly on U ∩ (Ω ∪ Σ) to a function w ∈ C 0 (cl(U ∩ Ω)) ∩ C 2 (U ∩ (Ω ∪ Σ)) with Lw = f on U ∩ Ω and Mw = g on U ∩ Σ. Since w ≤ inf {vj ; j ≥ 1} on U ∩ Ω, w ≤ u on U ∩ Ω and therefore w = u on U ∩Ω. Thus, u ∈ C 0 (cl(U ∩Ω))∩C 2 (U ∩(Ω ∪Σ)), Lu = f on U ∩Ω, and Mu = g on U ∩ Σ. Since y and U are arbitrary, u ∈ C 2 (Ω ∪ Σ), Lu = f on Ω, and Mu = g on Σ. The following theorem is proved in the same way by using Lemma 11.4.7 in place of Lemma 11.4.5. Theorem 11.4.13 If Σ = ∂Ω, β · ν ≥ > 0 for some > 0, c ≤ 0 on Ω, γ ≤ 0 on Σ, and ν has a continuous extension to Σ − , then u = inf {v; v ∈ U(f, g, h)} ∈ Cb0 (Ω)∩C 2 (Ω ∪Σ), Lu = f on Ω, and Mu = g on Σ. As in Section 9.7, the function u = sup {w; w ∈ L(f, g, h)} will be called the Perron subsolution and the corresponding v = inf {w : w ∈ U(f, g, h)} − the Perron supersolution. These two functions will be denoted by Hf,g,h + + − and Hf,g,h , respectively. If Hf,g,h = Hf,g,h , then h is said to be (L, M)resolutive and the common value is denoted by Hf,g,h . If it can be shown ± (y) = h(x) for all x ∈ ∂Ω ∼ Σ, then it would follow that limy→x,y∈Ω Hf,g,h from Theorems 11.4.12, 11.4.13, and the strong maximum principle, Corol+ − lary 11.3.6, that Hf,g,h = Hf,g,h on Ω and that h is (L, M)-resolutive. As in the classical case, it is a question of ﬁnding conditions under which points of ∂Ω ∼ Σ are regular boundary points. As was the case in Section 9.7, approximate barriers will be used to answer this question. As before, a region Ω ⊂ Rn will be called (L, M)-regular if limy→x,y∈Ω Hf,g,h (y) = h(x) for all x ∈ ∂Ω ∼ Σ and all h ∈ C 0 (∂Ω ∼ Σ).

11.5 Regularity of Boundary Points It was shown in Theorem 2.6.29 that the existence of a cone C ⊂∼ Ω with vertex at x ∈ ∂Ω implies that x is a regular boundary point for the Dirichlet problem u = 0 on Ω and u = g on ∂Ω. More generally, it was shown in Theorem 9.8.8 that the existence of such a cone implies that x is a regular boundary point for the elliptic boundary value problem Lu = f on Ω and u = g on ∂Ω. The boundary behavior of a solution to the oblique derivative problem Lu = f on Ω, Mu = g on Σ, and u = h on ∂Ω ∼ Σ is more complicated. The behavior of u at points x ∈ Σ is determined by the existence theorems, Theorems 11.4.12 and 11.4.13, which assert that u ∈ C 2 (Ω ∪ Σ) and Mu = g on Σ. For points x ∈ ∂Ω ∼ Σ, it is necessary to distinguish two cases: (i) x ∈ ∂Ω ∼ Σ, d(x, Σ) > 0 (ii) x ∈ ∂Ω ∼ Σ, d(x, Σ) = 0.

11.5 Regularity of Boundary Points

421

The existence of a cone C as described above suﬃces to show in the ﬁrst case that x is a regular boundary point. Only a minor modiﬁcation of the proof of Lemma 9.8.8 is needed to prove this assertion, namely, by replacing M therein by M + N . To show that x ∈ ∂Ω ∼ Σ is a regular boundary point when d(x, Σ) = 0 requires the assumption of a wedge in ∼ Ω rather than a cone and the construction of an approximate barrier as deﬁned in Deﬁnition 9.8.1. The imposition of a wedge condition at such boundary points was introduced by Lieberman in [40]. In addition to the assumption of the preceding section that |β(x)| = 1 for all x ∈ Σ, it will be assumed that ν and the βi , i = 1, . . . , n, have continuous extensions to Σ − . Let (y1 , . . . , yn ) = (y1 , y2 , y ) be a local coordinate system at x0 chosen so that ν(x0 ) is in the direction of the y2 -axis. By means of a translation followed by a rotation of axes, the x and y coordinate systems are related by an equation y = (x − x0 )O where O is an orthogonal matrix. ˜ Letting ν˜ (y) = ν(x) and β(y) = β(x), the condition ν(x) · β(x) ≥ δ > 0 ˜ on Σ carries over to the condition ν˜(y) · β(y) ≥ δ > 0 since an orthogonal transformation preserves inner products. In particular, ˜ ˜(0) · β(0) = β˜2 0 < δ ≤ ν(x0 ) · β(x0 ) = ν

(11.23)

since ν˜ (0) = (0, 1, 0, . . . , 0) in the y-coordinate system. Under the transformation from the x-coordinate system to the y-coordinate system, the operator L is transformed as in Equation (9.33). It will be assumed that the operator L is deﬁned on a neighborhood of Ω − in the remainder of this chapter. Let (r, θ) denote polar coordinates in the y1 y2 -plane so that y1 = r cos θ,

y2 = r sin θ.

Also let W (θ1 , θ2 ) denote a wedge in Rn that is cylindrical in the y -direction and deﬁned by the equation W (θ1 , θ2 ) = {y : θ1 < θ < θ2 }; if θ1 > 0, W (−θ1 , θ1 ) will be denoted by W (θ1 ). The function f (θ) in the following lemma was taken from [40]. Lemma 11.5.1 If (i) x0 ∈ ∂Ω ∼ Σ with d(x0 , Σ) = 0, (ii) there is a neighborhood V of x0 and a wedge W (θ1 , θ2 ) ⊂∼ Ω, with θ2 − θ1 < 2π, (iii) there is a δ > 0 such that ν(x) · β(x) ≥ δ > 0 for all x ∈ Σ, and (iv) there is a wedge W (ψ0 ) such that Σ ∩ V ⊂ W (ψ0 ) where 0 < ψ0 < arccos (1 − δ 2 /2), then there are constants λ0 > 0, r0 > 0, and a function f ∈ C 2 ([0, 2π]) such that L(rλ f (θ)) ≤ −1, rλ−2

0 < |λ| < λ0 , r < r0 , θ1 < θ < θ2

(11.24)

422

and

11 Oblique Derivative Problem

M(rλ f (θ)) ≤ −1, rλ−1

0 < |λ| < λ0 , r < r0 , |θ| < ψ0 .

(11.25)

Proof: It can be assumed that r ≤ 1 and that λ0 < 1. Consider the function f (θ) = 2e4πq − peqθ , where

p = e−qθ1 ,

0 ≤ θ ≤ 2π,

q = 4|˜ a12 |0 + 1,

and a ˜12 is the coeﬃcient in Equation (9.33). Note that f (θ) > e4πq for 0 < θ < 2π. Assuming for the time being that m = 1 and using the Miller representation, Theorem 9.8.5, with n = 2 ˜11 λ(λ − 1)f + 2˜ L(rλ f (θ)) ≤ rλ−2 a a12 (λ − 1)fθ + a ˜22 (fθθ + λf ) + rλ−1 ˜b1 λf + ˜b2 fθ + c˜rf . Since a ˜22 ≥ m = 1, lim sup λ→0

1 rλ−2

a12 fθ + fθθ L(rλ f (θ)) − rλ−1 (˜b2 fθ + c˜rf ) ≤ −2˜ ≤ q(−2|˜ a12 |0 − 1)

uniformly for r < 1 and θ1 < θ < θ2 . Thus, there are constants c0 > 0 and λ0 > 0 such that L(rλ f (θ)) ≤ −c0 rλ−2 + rλ−1 (˜b2 fθ + c˜rf ) for |λ| < λ0 uniformly for r < 1, θ1 < θ < θ2 . Letting c1 = |˜b2 |0 |fθ |0 + |˜ c|0 |f |0 , L(rλ f (θ)) ≤ −c0 rλ−2 + c1 rλ−1 ,

|λ| < λ0

uniformly for r < 1, θ1 < θ < θ2 . There is therefore a constant r0 < 1 such that L(rλ f (θ)) ≤ −c0 , |λ| < λ0 , r < r0 , θ1 < θ < θ2 , (11.26) rλ−2 where c0 = c0 /2. Now consider M(rλ f (θ)). By Equations (9.29) and (9.30), M(rλ f (θ)) = rλ−1 (β˜1 (y) cos θ + β˜2 (y) sin θ)λf (11.27) +(−β˜1 (y) sin θ + β˜2 (y) cos θ)fθ + γ˜ rf . Let ω(θ) = (cos θ, sin θ, 0, . . . , 0) be a point of W (ψ0 ). Since (cos (θ + π/2), sin (θ + π/2)) = (− sin θ, cos θ),

11.5 Regularity of Boundary Points

423

• !(µ) y1

Ã0 W(Ã0)

Σ

˜ ν(0)

• ! (µ)

y2

Fig. 11.1 y1 y2 -section of Σ

the point ω (θ) = (− sin θ, cos θ, 0, . . . , 0) is in the wedge obtained by rotating W (ψ0 ) about its edge through π/2 radians as depicted in Figure 11.1. The distance from ν˜(0) to ω (θ) is a maximum when θ = ψ0 so that |˜ ν (0) − ω (θ)|2 ≤ 2(1 − cos ψ0 ). Choose 0 < < δ such that

(δ − )2 δ2 ψ0 < arccos 1 − < arccos 1 − . 2 2 Using the fact that ν˜ is continuous at 0, there is a constant 0 < r0 < 1 such that |˜ ν (y) − ν˜(0)| < , r < r0 . Since ˜ · ν˜(y) = β(y) ˜ · (˜ ˜ · (˜ ˜ · ω (θ), 0 < δ ≤ β(y) ν (y) − ν˜(0)) + β(y) ν (0) − ω (θ)) + β(y) √ ˜ · ω (θ) ≥ (δ − ) − 2(1 − cos ψ0 )1/2 . β(y) Returning to Equation (11.27), if |θ| < ψ0 and r < r0 , then √ M(rλ f (θ)) ≤ rλ−1 λf + ((δ − ) − 2(1 − cos ψ0 )1/2 )fθ + γ˜ rf . √ Since (δ − ) − 2(1 − cos ψ0 )1/2 > 0 and fθ ≤ −pq exp (−qψ0 ) for |θ| < ψ0 , there are positive numbers c2 , c3 , and c4 such that

424

11 Oblique Derivative Problem

M(rλ f (θ)) ≤ rλ−1 (c2 λ − c3 + c4 r). By further reducing r0 , there is a λ0 > 0 and a positive constant c1 such that M(rλ f (θ)) ≤ −rλ−1 c1 ,

r < r0 , λ < λ0 .

(11.28)

The Inequalities (11.24) and (11.25) follow from the Inequalities (11.26) and (11.28) by replacing f / min (c0 , c1 ) by f , assuming that m = 1. Apply this result to the operators (1/m)L and (1/m)M and then replace f /m by f to get the general result. Deﬁnition 11.5.2 If Ω and Σ satisfy the hypotheses of Theorem 11.5.1 at x0 ∈ ∂Ω ∼ Σ = ∅, then Ω and Σ are said to satisfy wedge conditions at x0 . Theorem 11.5.3 If (i) Ω is a bounded open subset of Rn , (ii) Σ is a relatively open subset of ∂Ω of class C2+α with Σ = ∂Ω, (iii) β · ν ≥ > 0 on Σ for some 0 < < 1, (iv) there is an exterior cone at each point x ∈ ∂Ω ∼ Σ with d(x, Σ) > 0, (v) wedge conditions are satisﬁed at each (2+b) (Ω) with point x ∈ ∂Ω ∼ Σ with d(x, Σ) = 0, (vi) F ⊂ Cb0 (Ω) ∩ Hα (1+b) supf ∈F f 0,Ω ≤ M < +∞, g ∈ Cb0 (Σ) ∩ H1+α (Σ), and h ∈ C 0 (∂Ω ∼ Σ), then for each x0 ∈ ∂Ω ∼ Σ lim

y→x0 ,y∈Ω

± Hf,g,h (y) = h(x0 )

uniformly for f ∈ F.

Proof: The proof assuming (iv) is the same as the proof of Lemma 9.8.8. Consider any x0 ∈ ∂Ω ∼ Σ satisfying (v). As in Lemma 9.8.2, it suﬃces to prove there is an approximate barrier {wj± } at x0 . Consider any φ∈

' A(f, g, h) ∩ C 0 (Ω − ) ∩ C 2 (Ω ∪ Σ) , f ∈F

which is nonempty by Theorem 11.4.10. Note that φ is a supersolution simultaneously for all f ∈ F. It can be assumed that h(x0 ) + φ is a supersolution, that h(x0 ) + φ ≥ h on ∂Ω ∼ Σ, and that φ(x0 ) ≥ 0, for if not, replace φ by 2|h|0,∂Ω∼Σ +|φ|0,Ω +φ since h(x0 )+2|h|0,∂Ω∼Σ +|φ|0,Ω +φ is a superfunction by Lemma 11.4.4. For each j ≥ 1, let Uj be a neighborhood of x0 such that (i) h(x) < h(x0 ) + 1j , x ∈ Uj ∩ (∂Ω ∼ Σ), (ii) L(rλ f (θ)) < −1 on Uj ∩ Ω, (iii) M(rλ f (θ)) < −1 on Uj ∩ Σ where w = rλ f (θ) is the function of the preceding lemma for some ﬁxed λ satisfying |λ| < λ0 < 1. For each j ≥ 1, choose k(j) ≥ 1 so that

11.5 Regularity of Boundary Points

425

& 1 1 (iv) k(j) > max M + |c|0 h(x0 ) + j , g − M(h(x0 ) + j ) 0,Σ & 1 1 ≥ max f − L(h(x0 ) + j ) g − M(h(x0 ) + j ) for all f ∈ F. (v) 1j + k(j)w > φ on ∂Uj ∩ Ω.

0,Ω

0,Σ

The latter condition is possible, since f (θ) ≥ e4πq on [0, 2π], as noted in the proof of the preceding lemma. On Uj ∩ Ω,

1 1 L h(x0 ) + + k(j)w ≤ L h(x0 ) + − k(j) j j

1 1 ≤ L h(x0 ) + − f − L h(x0 ) + j j

1 1 ≤ L h(x0 ) + + f − L h(x0 ) + j j = f. Similarly, on Uj ∩ Σ,

1 M h(x0 ) + + k(j)w ≤ g. j It follows that the function h(x0 ) + 1/j + k(j)w is a superfunction for f and g on Uj ∩ Ω by Lemma 11.4.4. Now let min {h(x0 ) + 1j + k(j)w, h(x0 ) + φ} on Uj ∩ Ω + wj = on Ω ∼ Uj . h(x0 ) + φ By (i) of Lemma 11.4.5, wj+ is a superfunction on Uj ∩ Ω and on Ω ∼ cl Uj since h(x0 )+φ is a superfunction on Ω. Consider any point z ∈ ∂Uj ∩Ω. Since h(x0 )+1/j +k(j)w(z) > h(x0 )+φ(z), there is a neighborhood V ∈ A(z) such that h(x0 ) + 1/j + k(j)w > h(x0 ) + φ on cl V . Consider any U ⊂ V, U ∈ A(z). Then wj+ (z) = h(x0 ) + φ(z) ≥ L(z, U, h(x0 ) + φ) = L(z, U, wj+ ) which shows that wj+ is a superfunction on Ω. On (∂Ω ∼ Σ) ∩ Uj , h(x0 ) + 1/j + k(j)w ≥ h(x0 ) + 1/j > h and h(x0 ) + φ ≥ h so that wj+ ≥ h on (∂Ω ∼ Σ) ∩ Uj . On (∂Ω ∼ Σ) ∼ Uj , wj+ = h(x0 ) + φ ≥ h. Then, wj+ ≥ h on ∂Ω ∼ Σ and wj+ is a supersolution on Ω. At x0 , 1 wj+ (x0 ) = min {h(x0 ) + , h(x0 ) + φ(x0 )} j

426

and

11 Oblique Derivative Problem

lim wj+ (x0 ) = min {h(x0 ), h(x0 ) + φ(x0 )} = h(x0 ).

j→∞

The convergence is uniform for f ∈ F since wj+ depends only on M . This completes the proof that {wj+ } is an upper approximate barrier at x0 . Only minor modiﬁcations of the above arguments are needed to construct a lower approximate barrier {wj− } at x0 . Theorem 11.5.4 Under the conditions of the preceding theorem, the oblique derivative problem Lu = f on Ω,

Mu = g on Σ, and u = h on ∂Ω ∼ Σ

has a unique solution u ∈ C 0 (Ω − ) ∩ C 2 (Ω ∪ Σ). ± ∈ C 2 (Ω ∪ Σ) and Proof: It follows from Theorem 11.4.13 that Hf,g,h ± that limy→x0 ,y∈Ω Hf,g,h = h(x0 ) for all x0 ∈ ∂Ω ∼ Σ as in Lemma 9.8.2. + − = Hf,g,h so that h is By the remarks at the end of Section 11.4, Hf,g,h (L, M)-resolutive. Letting u = Hf,g,h , Lu = f on Ω, Mu = g on Σ, and u = h on ∂Ω ∼ Σ. Uniqueness follows from the strong maximum principle, Corollary 11.3.6.

Example 11.5.5 If Ω is a convex polytope in Rn and Σ is a union of noncontiguous faces and β is nontangential, then it is easily seen that Ω satisﬁes the wedge conditions at all points x of ∂Ω ∼ Σ with d(x, Σ) = 0 and that h is (L, M)-resolutive for each f, g as described above. In the Σ = ∂Ω case, the theorem corresponding to this theorem is Theorem 11.4.12 which asserts the existence of a solution to the equation Lu = f subject to the boundary condition Mu = g assuming c < 0 or γ < 0 with uniqueness following in this case from Corollary 11.3.6. In the c = 0 and γ = 0 case, some additional criteria is required for uniqueness. Both Theorems 11.4.11 and 11.4.12 provide for the existence of a solution to the oblique derivative boundary value problem in Cb0 (Ω) ∪ C 2 (Ω ∪ Σ). In order to show that the solution has additional properties, a global version of Inequality (11.4) is needed. Moreover, the hypotheses that f ∈ (2+b) (1+b) (Ω) and g ∈ Cb0 (Σ) ∩ H1+α (Σ) call for modiﬁcations of Cb0 (Ω) ∩ Hα the corresponding norms. (b)

Deﬁnition 11.5.6 If f ∈ Cb0 (Ω) ∩ Hk+α (Ω), k + b + α ≥ 0, let ∗(b)

(b)

|f |k+α,Ω = max ( f 0,Ω , |f |k+α,Ω ) (b)

and for g ∈ Cb0 (Σ) ∩ Hk+α (Σ), let ∗(b)

(b)

|g|k+α,Σ = max ( g 0,Σ , |g|k+α,Σ ).

11.5 Regularity of Boundary Points

427

Although there is a claim in the literature (c.f. [40]), without proof, that a global version of Inequality (11.15) holds under the usual assumptions regarding the operators L and M, the straightening of Σ, and the f, g, and h, it seems unlikely that a proof can be found without some kind of uniformity *δ regions. In the following deﬁnition, δ0 will be a ﬁxed condition on the Ω *δ0 = ∅. number in (0, 1) for which Ω Deﬁnition 11.5.7 Ω has the homogeneity property if there is a σ > 0 such that for each 0 < δ < δ0 there is a collection of admissible neighborhoods * − } such that Bx,σδ ⊂ U (x). {U (x); x ∈ Ω δ The essential point of the deﬁnition is that σ does not depend upon δ for 0 < δ < δ0 , but may depend upon δ0 . Example 11.5.8 If Ω is a convex polytope and Σ consists of noncontiguous faces of Ω, then Ω has the homogeneity property. This can be seen as follows. Without loss of generality, it can be assumed that Σ is just one face of Ω. As the points of Σ are the troublesome points, only such points will be considered. For each x ∈ Σ, choose a local coordinate system centered at x with the positive xn -axis orthogonal to Σ and intersecting the interior of * *δ0 ∩ Σ. Then d(x) = d(x, ∂Ω ∼ Σ) ≥ δ0 Ω. Fix δ0 ∈ (0, 1). Suppose x ∈ Ω + * and Bx,δ0 /2 ⊂ Ωδ0 /2 . Letting ρ˜ = δ0 /2 and choosing 0 < ρ˜0 < ρ˜ so that Inequality (11.4) is satisﬁed, U (x) = By˜,ρ˜ ∈ A(x) where y˜ = (0, . . . , 0, −ρ˜0 ) ˜ is replaced δ ρ˜, and and Bx,ρ− ˜ ρ˜0 ⊂ U (x). Note that if δ0 is replaced by δδ0 , ρ ρ˜0 is replaced by δ ρ˜0 , then Bx,ρ− is replaced by B ˜ ρ˜0 x,δ(ρ− ˜ ρ˜0 ) ⊂ By˜,δ ρ˜ where y˜ = (0, . . . , 0, −δ ρ˜0 ). Taking σ = ρ˜ − ρ˜0 , and U (x) = By˜,δρ˜, Bx,σδ ⊂ U (x). Thus, Ω has the homogeneity property. The homogeneity concept applies to both the Σ = ∂Ω and the Σ = ∂Ω cases as does the following theorem, without the third term in the former case. Theorem 11.5.9 If the conditions of Theorem 11.5.3 are satisﬁed, Ω has the homogeneity property, and u ∈ C 0 (Ω − ) ∩ C 2 (Ω ∪ Σ) satisﬁes the equations (2+b) (Ω), Mu = g on Σ for g ∈ Cb0 (Σ) ∩ Lu = f on Ω for f ∈ Cb0 (Ω) ∩ Hα (1+b) H1+α (Σ), and u = h on ∂Ω ∼ Σ for h ∈ C 0 (∂Ω ∼ Σ), then ∗(2+b−k) ∗(2+b) ∗(1+b) k = 0, 1, 2. |u|k+α,Ω ≤ C |f |α,Ω + |g|1+α,Σ + |h|0,∂Ω∼Σ Proof: Only the k = 2 case will be proved, the k = 0, 1 cases are proved in *δ0 = ∅ and let the same using Inequality (11.14). Fix δ0 ∈ (0, 1) for which Ω *−} σ > 0 be the associated scaling factor. For 0 < δ < δ0 , let {U (z); z ∈ Ω δ be the family of admissible neighborhoods of z such that Bz,σδ ⊂ U (z). Con* − , then *δ . If x ∈ Bz,σδ/2 ⊂ Bz,δ ⊂ U (z) = U for some z ∈ Ω sider any x ∈ Ω δ d*U∩Ω (x) > σδ/2 and x ∈ (U ∩ Ω)σδ/2 . By Inequalities (11.14) and (11.15) and Lemma 11.4.11,

428

11 Oblique Derivative Problem

δ 2+b+α (|u(x)| + |Du(x)| + |D2 u(x)|) 22+b (b) 22+b (b) ≤ 2+b |u|2+0,U∩Ω ≤ 2+b |u|2+α,U∩Ω ≤ δ 2+b |u|2+0,(U∩Ω) σ σ σδ/2 (2+b) (1+b) ≤ C |f |α,U∩Ω + |g|1+α,U∩Σ + |u|0,∂(U∩Σ)∼(U∩Σ) (2+b) (1+b) ≤ C |f |0,Ω + |f |α,Ω + |g|0,Σ + |g|1+α,Σ + |h|0,∂Ω∼Σ ∗(2+b) ∗(1+b) ≤ C |f |α,Ω + |g|1+α,Σ + |h|0,∂Ω∼Σ .

(11.29)

*δ and |y − x| < σδ/4, then y ∈ Bz,3σδ/4 ⊂ U and If y ∈ Ω δ 2+b+α

2 2 |D2 u(x) − D2 u(y)| 2+b+α |D u(x) − D u(y)| ≤ δ χ|x−y|<σδ/4 |x − y|α |x − y|α (11.30) 2 2 2+b+α |D u(x)| + |D u(y)| +δ . (σδ/4)α

Note that y ∈ Bx,3σδ/4 implies that d*U∩Ω (y) > σδ/4, which in turn means that y ∈ (U ∩ Ω)σδ/4 . Using Inequalities (11.15) and Lemma 11.4.11, the ﬁrst term on the right is dominated by

2+b+α 4 (σδ/4)2+b+α |u|2+α,(U∩Ω) σ σδ/4

2+b+α 4 (b) ≤ |u|2+α,U∩Ω σ (2+b) (1+b) ≤ C |f |α,U∩Ω + |g|1+α,U∩Σ + |u|0,∂(U∩Ω)∼(U∩Σ) ∗(2+b) ∗(1+b) ≤ C |f |α,U∩Ω + |g|1+α,U∩Σ + |h|0,∂Ω∼Σ .

(11.31)

By Inequality (11.29), the second term on the right in Inequality (11.30) is dominated by

α 4 ∗(2+b) ∗(1+b) δ 2+b (|D2 u(x)| + |D2 u(y)|) ≤ C |f |α,Ω + |g|1+α,Σ + |h|0,∂Ω∼Σ . σ (11.32) Therefore, δ 2+b+α

|D2 u(x) − D2 u(y)| ∗(2+b) ∗(1+b) ≤ C |f |α,Ω + |g|1+α,Σ + |h|0,∂Ω∼Σ . (11.33) α |x − y|

*δ in Inequality (11.29), Taking the supremum over x ∈ Ω ∗(2+b) ∗(1+b) δ 2+b+α |u|2+0,Ω* δ ≤ C |f |α,Ω + |g|1+α,Σ + |h|0,∂Ω∼Σ ;

(11.34)

11.5 Regularity of Boundary Points

429

*δ in Inequality (11.33), taking the supremum over x, y ∈ Ω ∗(2+b) ∗(1+b) δ 2+b+α [u]2+α,Ω* δ ≤ C |f |α,Ω + |g|1+α,Σ + |h|0,∂Ω∼Σ .

(11.35)

The assertion is established by combining the last two inequalities and taking the supremum over 0 < δ < 1. (b) The condition that f ∈ H2+α (Ω) can be relaxed by imposing the global condition that f ∈ C 0 (Ω − ). Theorem 11.5.10 If conditions (i) through (v) of Theorem 11.5.3 are sat(1+b) isﬁed, Ω has the homogeneity property, f ∈ C 0 (Ω − ), g ∈ Cb0 (Σ) ∩ H1+α (Ω), 0 0 − 2 and h ∈ C (∂Ω ∼ Σ), then there is a unique u ∈ C (Ω ) ∩ C (Ω ∪ Σ) such that Lu = f on Ω, Mu = g on Σ, and u = h on ∂Ω ∼ Σ. Proof: By the Tietze extension theorem, f can be extended to all of Rn without increasing its norm so that it vanishes outside a bounded open set Ω ⊃ Ω. By Lemma 11.2.7, if fn = J1/n f , then |fn |α,Ω ≤ C f 0,Ω for all n ≥ 1 and the sequence {fn } converges uniformly to f on Ω − . By Theo(b) rem 11.5.4, for each n ≥ 1 there is a unique un ∈ H2+α (Ω) for which Lun =fn on Ω, Mun = g on Σ, and un = h on ∂Ω ∼ Σ. By Theorem 11.5.9, ∗(b) ∗(2+b) ∗(1+b) |un |2+α,Ω ≤ C |fn |α,Ω + |g|1+α,Σ + |h|0,∂Ω∼Σ ∗(1+b) ≤ C f 0,Ω + |g|1+α,Σ + |h|0,∂Ω∼Σ . (b)

Since the sequence {un } is bounded in H2+α (Ω), by the use of the subsequence selection principle as in the proof of Theorem 7.5.3, there is a subsequence of the {un } sequence, which can be assumed to be the sequence itself, and a function u on Ω ∪ Σ such that un → u, Dun → Du, and * − , and it follows that un → u, Dun → Du, D2 un → D2 u uniformly on each Ω δ 2 2 * − . Therefore, Lu = f and D un → D pointwise on Ω ∪ Σ = ∪0<δ<1 Ω δ on Ω and Mu = g on Σ. Since un = Hfn ,g,h on Ω by Theorem 11.5.4 and limy→x un (y) = limy→x Hfn ,g,h = h(x) uniformly in n for each x ∈ ∂Ω ∼ Σ, limy→x u(y) = h(x), x ∈ ∂Ω ∼ Σ. Thus, u has a continuous extension to ∂Ω ∼ Σ and u = h on ∂Ω ∼ Σ.

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Index

a.e., 66 admissible, 396 admissible neighborhood, 408 alternating method, 213 analytic, 175 approximate barrier, 358 approximate lower barrier, 358 approximate upper barrier, 358 apriori inequality, 371 Arzel´ a-Ascoli Theorem, 4, 5 averaging principle, 13 Bˆ ocher’s theorem, 109 balayage, 115 Banach space, 6 barrier, 82, 211 barrier, approximate, 358 barrier, approximate lower, 358 barrier, approximate upper, 358 Borel measure, 3, 5 Borel sets, 5 Bouligand, 82, 155, 215 boundary point lemma, 345, 392, 409 bounded linear operator, 334 bounded map, 334 bounded variation, 5 Brelot, 103, 155, 201, 232, 234, 239 capacitable, 166 capacitary distribution, 164 capacitary potential, 164 capacity, 164, 166 capacity, inner, 165 capacity, outer, 165 Cartan, 135, 139, 167, 238, 256 Choquet, 178 Choquet’s lemma, 57

concave, 73 continuity method, 334 continuum, 86 contraction map, 227, 334 convex, 64, 257 countably strongly subadditive, 170 d-system, 100 derivate, 92 diﬀeomorphism, 405 diﬀerentiable of class Ca , 405 Dini’s formula, 41 directed set, 5 Dirichlet problem, 24, 75 Dirichlet region, 81 Dirichlet solution, 78 divergence, 8 divergence theorem, 8 domination principle, 117, 166 Dynkin system, 100 elliptic, 338 elliptic operator, 333 elliptic, strictly, 338 energizable, 244 energy, 241, 250 energy principle, 248 equicontinuity, 4 Evans and Vasilesco, 144 exterior Dirichlet problem, 197 exterior sphere condition, 85 Fatou, 99 ﬁne closure, 224 ﬁne interior, 224 ﬁne limit, 224 ﬁne limsup, 224

435

436 ﬁne neighborhood, 224 ﬁne topology, 223 ﬁne topology on Rn ∞ , 237 Frostman, 246 fundamental harmonic function, 10 Gauss, 246 Gauss’ averaging principle, 23 Gauss’ Integral Theorem, 10 Gauss’s integral, 248 Gauss-Frostman, 246 gradient, 3 greatest harmonic minorant, 116 Green function, 16, 110 Green function, half space, 203 Green function, Neumann problem, 46 Green function, second kind, 46 Green potential, 122 Green’s Identity, 9 Green’s representation theorem, 10 Greenian set, 111 H¨ older continuous, 309 H¨ older spaces, 268 H¨ ormander, 276 H¨ ormander’s convexity theorem, 278 harmonic, 9, 198 harmonic function, 9 harmonic measure, 217 harmonic minorant, 115 Harnack’s Inequality, 54 Herglotz, 24 Herglotz theorem, 31 Herv´ e, 360 homogeneity, 427 Hopf, 345–347 hyperharmonic, hypoharmonic, 76, 206 indicator function, 101 inner capacity, 165 inner polar set, 150 inner q.e., 150 inner quasi everwhere, 150 interior estimates, 338 interior sphere condition, 345 interpolate, 276 interpolation inequalities, 280 interpolation, global, 280 interpolation, local, 288 inverse, 14 inversion, 14, 15, 32 irregular boundary point, 81, 128 Jensen’s Inequality, 64

Index Kelvin operator, 33 Kelvin transformation, 32, 207 kernel, 303 L-regular boundary point, 359 L-resolutive, 357 l.s.c., 3 Laplacian, 8 Lebesgue, 86 Lebesgue measure, 5 Lebesgue points, 97 Lebesgue spine, 128 left-directed, 56 lift, 413 lifting, 70, 355, 413 linear map, 334 linear operator, 6 Lipschitz continuous, 309 local barrier, 82, 211 local interpolation, 288 local property, 154 local solution, 410 locally H¨ older continuous, 309 locally integrable, 5 locally super-mean-valued, 60 logarithmic potential, 127 Lord Kelvin, 14 lower, 413 lower class, 76 lower regularization, 4 lower semicontinuous, 3 lowered, 205 lowering, 70, 355, 413 Maria-Frostman, 166 maximum principle, 20 maximum principle, strong, 346, 409 maximum principle, weak, 344 mean value property, 12 mean value theorem, 4 mean valued, 13, 60 measurable map, 227 method of continuity, 334, 352, 402 Miller, 365 minimum principle, 20 Mokobodzki, 186 molliﬁer, 71 multi-index, 2 mutual energy, 241, 250 Myrberg’s theorem, 182 net, 5 Neumann problem, 39 Neumann problem for a ball, 42

Index Newtonian potential, 127, 316 nonnegative deﬁnite, 343 nontangential limit, 98 norm, 5, 6 normed linear space, 6 nucleus, 175 oblique derivative problem, 333 outer capacity, 165 outer unit normal, 8 partition of unity, 132 paved space, 175 paving, 175 peaks, 229 Perron lower solution, 77 Perron subsolution, 357, 420 Perron supersolution, 357, 420 Perron upper solution, 77 Perron-Wiener method, 75 Picard, 20 Poincar´ e, 85, 115, 360 point at inﬁnity, 197 Poisson Integral, 25, 34, 201 Poisson Integral for half-space, 33 Poisson Integral Formula, 17 Poisson’s equation, 316 polar cap, 89 polar representation, 365 polar set, 88, 150, 206 pole, 10 positive deﬁnite, 343, 407 potential, 122 projection, 258 Proof of Theorem 2.6.28, 232 Property B, 166 PWB method, unbounded regions, 204 q.e., 150 quasi everywhere, 150 reciprocity law, 135 reduction, 116, 159 reﬂection, 38 reﬂection principle, 326 regular, 420 regular Borel measure, 5 regular boundary point, 81, 211 regular region, 81 regular,(L,M)-, 420 regularized reduction, 159 resolutive, 207 resolutive boundary function, 78 resolutive,(L,M)-, 420

437 Riesz Decomposition Theorem, 142 Riesz measure, 142 Riesz potential, 304 Riesz Representation Theorem, 142 right-directed, 56 saturated, 75, 205 Schauder interior estimates, 338 Schwarz, 26, 155, 213 Schwarz’s Reﬂection Principle, 326 seminorm, 6 Sibony, 186 Sierpinski, 177 signed measure, 5 spherical chip, 288 spherical chip, admissible, 396 spherical coordinates, 1 spherical harmonic, 52 spine, Lebesgue, 128 Stolz domain, 98 straighten boundary, 405 stratiﬁable, 296 strictly elliptic, 338 strong barrier, 82 strong convergence, 253 strong maximum principle, 346, 409 strongly subadditive, 170 strongly subadditive, countably, 170 subfunction, 354, 411 subharmonic, 60, 198 subnewtonian kernel, 304 subsequence selection principle, 269 subsolution, 356 super-mean-valued, 60 superfunction, 354, 411 superharmonic, 60, 62, 63, 198 superharmonic, alternative deﬁnition, 62 superharmonic, yet another deﬁnition, 63 supersolution, 356, 414 Suslin scheme, 175 sweeping, 189 sweeping out, 115 swept measure, 189 symmetric derivate, 89 thin, 225, 237 Tonelli, 23 total, 135 total variation, 5 u.s.c., 3 uniform equicontinuity, 4 uniformly H¨ older continuous, 309

438 upper class, 76 upper inner normal derivate, 345 upper semicontinuous, 3 vague convergence, 253 Vasilesco, 155 vertex, 259 Vitali covering, 92 w*-topology, 5

Index weak convergence, 253 weak maximum principle, 344 wedge conditions, 424 Wiener, 81, 164 Wiener’s test, 261 Zaremba, 86 Zaremba’s cone condition, 86, 265, 367 zero-one law, 194

Notation

A(f : x, δ), 60 A(u : y, ρ), 14 A0 , 313 Af , 224 A(x), 408 A(F ), 175 Bx,ρ , 1 (b) H2+α (Ω; φ), 272 Bx , 112 B(X), 5 C, 294 C(n, a, b), 312 C(Γ ), 164 C 0 (Ω), 3 C 0 (Ω − ), 3 C k (Ω), 3 C k (Ω − ), 3 C0k (Ω), 3 C k (Ω ∪ Σ), 305 Cz,α , 88 C∗ , 165 C ∗ , 165 C, 166 Ca , 405 clf A, 224 Di , 2 Dij , 2 Dxi , 3 Dxi xj , 3 β D(x) ,3

Dβ , 3 D j u, 280 Dν+ u, 345 D(A), 6 Ds μ, 89 Dμ, 92

d, 272 d(Ω), 272 d(x, y), 271 d(F ), 92 d(F0 ), 100 dx , 272 * d(x), 284 d*x , 284 d*xy , 284 * y), 284 d(x, div v, 8 E, 250 E + , 242 F −, 1 ∂F , 1 cl F , 1 int F , 1 ∼ F, 1 F (Γ ), 259 f ∗ , 33 f-lim, 224 f-limsup, 224 GΛ Ω , 189 GΩ f , 129 G1/2 , 376 GΩ μ, 122 GΛ Ω (x, ·), 189 G, 111 GΩ , 110 GB , 16 G1/2 , 372 ghmΩ h, 116 Hf , 78 H f , 77 H f , 77 Hf,g,h, 420

439

440 − Hf,g,h , 420

+ , 420 Hf,g,h Hk+α (Ω), 268 (b) Hk+α (Ω), 285 Hk+α,loc (Ω), 268 Hf,g , 357 H+ f,g , 357

H− f,g , 357 ∗(b)

Hk+α (Ω), 427 h, 372 intf A, 224 Jh , 71 K, 305 K + , 135 KB (x, z), 46 Kz,α , 89 K0+ , 135 K + (Ω), 135 K(Ω), 164 k(x, y), 303 k(| x − y |), 312 L, 333 L0 , 335 L(f : x, δ), 60 LS(x, B, g), 354 LS(x, U, φ), 410 L(u : y, ρ), 14 Lx , 100 L(Ω), 60 Lf , 76 LΩ f , 77 L(f, g), 356, 411 L(f, g, h), 414 M, 333 M0 , 372 M , 338 m, 338 m, M , 336 N ∞ , 175 N (x), 407 N (F, ), 92 n, 175 n(x), 8 O(Ω), 164 PI(c, μ, Ω), 35 PI(f : B), 25 PI(c, f, Ω), 35 PI(f, ∼ By,ρ ), 201 PI(μ : B), 25 PI(μ : y, ρ), 25 Pm,n , 51 P (d(Ω)), 277

Notation R , 164 RΓ , 164 ˆ Γ , 164 R ˆ , 164 R ˆ u , 159 R Λ Ru Λ , 159 R(A), 6 Rn , 1 Rn ∞ , 197 Rn 0 , 313 Rn ± , 313 ˆ Λ (u, ·), 189 R (r, θ), 1 S(Ω), 60 S + (Ω), 60 Sx,ρ , 1 Sz,θ , 98 Uf , 316 Uμ , 23, 127 Uf , 76 UΩ f , 77 U(f, g), 356, 411 U(f, g, h), 414 u ˆ, 4 u∞ , 115 u∞,Λ , 115 uB , uB , 70, 355 uU , 413 u∼B , 206 u(| x − y |), uy (x), 10 uy (x), 10 Vx,u , 411 v U , 413 wj+ , 358 wj− , 358 wj± , 358 x, 1 x · y, 1 (x , xn ), 204 x , 204 x∗ , 14, 32 Yn,m , 52 y r , 38 βt , 373 β!, 2 βn , 373 γj , 309 χC , 101 δΩ (x, ·), Λ (·, u), δΩ Λ δΩ (μ, ·), Λ δΩ (x, ·),

189 193 193 189

Notation ε, 1 j , 273 Λr , 38 λ0 , 392 μ, μ+ , μ− , | μ |, 5 μn , 5 μF , 257 μΓ , 164 μF (Γ ) , 259 μΩ x , 217 μx , 100 νn , 2 νn (ρ), 2 ν, 392 Ω, 3 Ωδ , 272, 309 * δ , 284 Ω Ω0 , 326 Ω± , 326 ∂∞ Ω, 198 Φu Λ , 158 ψ, 330 σn , 2 σn (ρ), 2 ∗(b)

| f |k+α,Ω , 427 ∗(b)

| g |k+α,Σ , 427 f 0,Ω , 3 f k , 268 | g |1+α,Σ , 271 | g; d |1+α,Σ , 380 (1+b) | g |1+α,Σ , 287

441 |u|k+0,Ω , 268 [u]k+α,Ω , 268 |u|k+α,Ω , 268 [u]k,Ω , 268 |u|k,Ω , 268 |u; φ|k+α,Ω , 272 (b)

[u; φ]k+0,Ω , 271 (b)

|u; φ|k+0,Ω , 271 (b)

[u; φ]k+α,Ω , 271 (b)

[u; φ]k,Ω , 271 (b)

|u; φ|k,Ω , 271 (b)

|u; φ|k+α,Ω , 271 (b)

[u]k+α,Ω , 285 (b)

|u|k+α,Ω , 285 |x|, 1 |β|, 2 |λ|2e , 250 | μ |e , 250 |μ|2e , 241 · X , 6 Δu, 8 Δ(x) , 8 ∧, 284 ∂∞ Ω, 198 ∞, 197 [λ1 , λ2 ]e , 250 [μ, ν]e , 241 ∇, 3 θ − limx→z u(x), 98