Regularity estimates for nonlinear elliptic and parabolic problems : Cetraro, Italy 2009

Lecture Notes in Mathematics Editors: J.-M. Morel, Cachan B. Teissier, Paris For further volumes: http://www.springer...

Author: Lewis J. | Lindqvist P. | Manfredi J. | et al.

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Lecture Notes in Mathematics Editors: J.-M. Morel, Cachan B. Teissier, Paris

For further volumes: http://www.springer.com/series/304

2045

Fondazione C.I.M.E., Firenze C.I.M.E. stands for Centro Internazionale Matematico Estivo, that is, International Mathematical Summer Centre. Conceived in the early fifties, it was born in 1954 in Florence, Italy, and welcomed by the world mathematical community: it continues successfully, year for year, to this day. Many mathematicians from all over the world have been involved in a way or another in C.I.M.E.’s activities over the years. The main purpose and mode of functioning of the Centre may be summarised as follows: every year, during the summer, sessions on different themes from pure and applied mathematics are offered by application to mathematicians from all countries. A Session is generally based on three or four main courses given by specialists of international renown, plus a certain number of seminars, and is held in an attractive rural location in Italy. The aim of a C.I.M.E. session is to bring to the attention of younger researchers the origins, development, and perspectives of some very active branch of mathematical research. The topics of the courses are generally of international resonance. The full immersion atmosphere of the courses and the daily exchange among participants are thus an initiation to international collaboration in mathematical research. C.I.M.E. Director Pietro ZECCA Dipartimento di Energetica “S. Stecco” Universit`a di Firenze Via S. Marta, 3 50139 Florence Italy e-mail: [email protected]

C.I.M.E. Secretary Elvira MASCOLO Dipartimento di Matematica “U. Dini” Universit`a di Firenze viale G.B. Morgagni 67/A 50134 Florence Italy e-mail: [email protected]

For more information see CIME’s homepage: http://www.cime.unifi.it CIME activity is carried out with the collaboration and financial support of: - INdAM (Istituto Nazionale di Alta Matematica) - MIUR (Ministero dell’Universita’ e della Ricerca)

John Lewis • Peter Lindqvist Juan J. Manfredi • Sandro Salsa

Regularity Estimates for Nonlinear Elliptic and Parabolic Problems Cetraro, Italy 2009 Editors: Ugo Gianazza John Lewis

123

John Lewis University of Kentucky Lexington, KY USA

Peter Lindqvist Norwegian University of Science and Technology Trondheim, Norway

Juan J. Manfredi University of Pittsburgh PA, USA

Sandro Salsa Politecnico di Milano Italy

ISBN 978-3-642-27144-1 e-ISBN 978-3-642-27145-8 DOI 10.1007/978-3-642-27145-8 Springer Heidelberg Dordrecht London New York Lecture Notes in Mathematics ISSN print edition: 0075-8434 ISSN electronic edition: 1617-9692 Library of Congress Control Number: 2012933109 Mathematics Subject Classification (2010): 5J70, 35J75, 35J92, 35K65, 35K67, 35K86, 35K92, 35Q91, 35R11, 35R35, 49K20, 49N60 c Springer-Verlag Berlin Heidelberg 2012 This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer. Violations are liable to prosecution under the German Copyright Law. The use of general descriptive names, 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 protective laws and regulations and therefore free for general use. Printed on acid-free paper Springer is part of Springer Science+Business Media (www.springer.com)

Preface

This volume collects the notes of the CIME course Regularity Estimates for Nonlinear Elliptic and Parabolic Problems held in Cetraro (Italy) on June 22–27, 2009. The school consisted in ﬁve series of lectures, delivered by Emmanuele DiBenedetto (Vanderbilt University, Nashville, USA) John Lewis (University of Kentucky, Lexington, USA) Peter Lindqvist (Norwegian University of Science and Technology, Trondheim, Norway) Juan J. Manfredi (University of Pittsburgh, Pittsburgh, USA) Sandro Salsa (Politecnico di Milano, Milano, Italy).

The issue of regularity has obviously played a central role in the theory of Partial Diﬀerential Equations, almost since its inception, and despite the tremendous development, it still remains a very fruitful research ﬁeld. In particular regularity estimates for degenerate and singular elliptic and parabolic equations have developed considerably in the last years, in many unexpected and challenging directions. Because of all these recent results, it seemed timely to trace an overview that would highlight emerging trends and issues of this fascinating research topic in a proper and eﬀective way. The course aimed at showing the deep connections among all these topics and at opening new research directions, through the contribution of leading experts in all these ﬁelds. Emmanuele DiBenedetto gave a course on Introduction to Regularity Theory for Degenerate Parabolic Equations in Divergence Form discussing some techniques recently introduced to investigate the local and global behavior of solutions to degenerate parabolic equations when their principal part fails to be coercive. The equations have to be regarded in their own intrinsic geometry, and the solutions have a limited degree of regularity. DiBenedetto showed how identifying regularity classes as functions v

vi

Preface

of the degenerate and/or singular structure of the equation is part of an emerging theory which promises to yield an understanding of degeneracy and/or singularity in Partial Diﬀerential Equations. Unfortunately there are no notes of this course. The course of John Lewis on Applications of Boundary Harnack Inequalities for p-Harmonic Functions and Related Topics discussed applications of recent work and techniques concerning the boundary behavior of positive p-harmonic functions vanishing on a portion of the boundary of Lipschitz, chord arc, and Reifenberg ﬂat domains. At ﬁrst fundamental properties of p-harmonic functions and elliptic measure were presented. Then the dimension of p-harmonic measure was dealt with. The ﬁnal part of the course ﬁrst considered boundary Harnack inequalities and the Martin boundary problem in Reifenberg ﬂat and Lipschitz domains, and at the end uniqueness and regularity both in free boundary and inverse type problems. Peter Lindqvist presented in his course Regularity of Supersolutions a general theory for supersolutions of the p-Laplace Equation. Indeed the regularity theory for solutions to the parabolic p-laplacian is a well-developed topic, but when it comes to (semicontinuous) supersolutions and subsolutions a lot remains to be done. Supersolutions are often auxiliary tools as in the celebrated Perron method, for example, but they are also interesting in their own right. Therefore, the lectures were entirely focused on this important issue. Juan J. Manfredi delivered a series of lectures on Introduction to random Tug-of-War games and PDEs providing an introduction to the connection between the theory of stochastic tug-of-war games and nonlinear equations of p-Laplacian type in the Euclidean and discrete cases. The fundamental contributions of Kolmogorov, Ito, Kakutani, Doob, Hunt, L´evy, and many others have shown the profound and powerful connection between classical linear potential theory and probability theory. The idea behind the classical interplay is that harmonic functions and martingales share a common cancellation property that can be expressed by using mean value properties. In his lectures, Manfredi showed how this approach turns out to be very useful in the nonlinear theory as well. Sandro Salsa taught a course on The Problems of the Obstacle in Lower Dimension and for the Fractional Laplacian giving a somewhat self-contained presentation of the results concerning the analysis of the solution and the free boundary of the thin obstacle problem and more generally of the obstacle for the fractional Laplacian. He started from the thin obstacle problem, considering the case of zero obstacle. In

Preface

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this case, the main ideas and tools were clearly seen and developed without too many technicalities and in a somewhat self-contained fashion. Later, he extended the results on the optimal regularity and the analysis of the regular s part of the free boundary to the general case for (−Δ) . This series of lectures attracted approximately 50 participants, largely PhD students or post-docs, and also senior researchers; we are sure that this CIME course was rich of useful suggestions and ideas for inspiring new developments, and opening new research prospects in the near future. We wish to thank all the lecturers for their active participation and their valuable contribution, and the CIME foundation, in particular the director Prof. Pietro Zecca and the secretary Prof. Elvira Mascolo, for their helpful support and for the organization of such a remarkable event in Cetraro. Pavia, Italy Lexington, KY

Ugo Gianazza John Lewis

•

Contents

Applications of Boundary Harnack Inequalities for p Harmonic Functions and Related Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . J. Lewis 1 Outline of the Course . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Ode to the p Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 My Introduction to p Harmonic Functions . . . . . . . . . . . . . . . . . . . . . 2 Basic Estimates for the p Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 p Harmonic Functions in NTA Domains . . . . . . . . . . . . . . . . . . . . . . . . 2.2 The p Laplacian and Elliptic PDE .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Degenerate Elliptic Equations .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 p Harmonic Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 p Harmonic Measure in Simply Connected Domains . . . . . . . . . . 3.2 Preliminary Reductions for Theorem 2.6 . . . . . . . . . . . . . . . . . . . . . . . 3.3 Proof of Theorem 2.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 The Final Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 p Harmonic Measure in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Open Problems for p Harmonic Measure .. . . . . . . . . . . . . . . . . . . . . . 4 Boundary Harnack Inequalities and the Martin Boundary Problem for p Harmonic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 History of Theorem 3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Proof of Step 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Proof of Step 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Proof of Step 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Proof of Step 4 and Theorem 3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 More on Boundary Harnack Inequalities . . . . . . . . . . . . . . . . . . . . . . . 4.7 The Martin Boundary Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8 Proof of Theorem 3.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.9 Further Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 1 2 2 4 6 7 9 15 15 16 19 21 22 23 24 26 27 30 33 37 38 42 46

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Contents

5 Uniqueness and Regularity in Free Boundary: Inverse Type Problems .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 History of Theorem 4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Proof of Theorem 4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Further Uniqueness Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Boundary Regularity of p Harmonic Functions . . . . . . . . . . . . . . . . 5.5 Proof of Theorem 4.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Proof of Theorem 4.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7 Proof of Theorem 4.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8 Regularity in a Lipschitz Free Boundary Problem . . . . . . . . . . . . 5.9 History of Theorem 4.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.10 Proof of Theorem 4.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.11 Enlargement of the Cone of Monotonicity in the Interior .. . . . 5.12 Enlargement of the Cone of Monotonicity at the Free Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.13 An Application of Theorem 4.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.14 Proof of (161) .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.15 Closing Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Regularity of Supersolutions.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Peter Lindqvist 1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 The Stationary Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 The Evolutionary Equation .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Deﬁnitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Bounded Supersolutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Unbounded Supersolutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Reduction to Zero Boundary Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Weak Supersolutions are Semicontinuous.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 The Equation with Measure Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Pointwise Behaviour .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 The Stationary Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 The Evolutionary Equation.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Introduction to Random Tug-of-War Games and PDEs . . . . . . . . Juan J. Manfredi 1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Probability Background .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 The p-Laplacian Gambling House . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 p-harmonious Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Directed Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Epilogue .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

46 46 49 50 51 52 55 57 59 60 60 61 61 63 65 68 69 73 73 78 91 92 94 102 109 111 122 123 123 125 130 133 133 133 141 144 147 150 151

Contents

The Problems of the Obstacle in Lower Dimension and for the Fractional Laplacian . . . . . . . . . . . . . . . . . . . . . . . Sandro Salsa 1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 The Zero Obstacle Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Setting of the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Lipschitz Continuity and Semiconvexity .. . . . . . . . . . . . . . . . . . . . . . . 2.3 Local C1,α Estimate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Optimal Regularity for Tangentially Convex Global Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Almgren’s Frequency Formula .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Asymptotic Proﬁles and Optimal Regularity.. . . . . . . . . . . . . . . . . . 2.7 Lipschitz Continuity of the Free Boundary at Stable Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8 Boundary Harnack Principles and C 1,α Regularity of the Free Boundary at Stable Points.. . . . . . . . . . . . . . . . . . . . . . . . . 2.9 Structure of the Singular Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Obstacle Problem for the Fractional Laplacian . . . . . . . . . . . . . . . . . . . . . . . 3.1 Construction of the Solution and Basic Properties . . . . . . . . . . . . 3.2 Lipschitz Continuity, Semiconvexity and C 1,α Estimates . . . . . 3.3 Thin Obstacle for the Operator La : Local C 1,α Estimates. . . . 3.4 Minimizers of the Weighted Rayleigh Quotient and a Monotonicity Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Optimal Regularity for Tangentially Convex Global Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Frequency Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 Blow-up Sequences and Optimal Regularity . . . . . . . . . . . . . . . . . . . 3.8 Nondegenerate Case: Lipschitz Continuity of the Free Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.9 Boundary Harnack Principles and C 1,α Regularity of the Free Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Appendix A: The Fractional Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Deﬁnition and Basic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Supersolutions and comparison .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Appendix B: The Operator La . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Deﬁnition and Preliminary Facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Harnack inequality, Liouville theorem and mean value property . . . . Poincar´e inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . s Appendix C: Relation between (−Δ) and La . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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153 153 160 160 162 166 170 173 177 180 183 188 201 202 203 204 206 207 211 217 225 227 231 231 232 234 234 236 240 240 243

List of Participants. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245

Applications of Boundary Harnack Inequalities for p Harmonic Functions and Related Topics J. Lewis

1

Outline of the Course

This course will be concerned with applications of recent work—techniques concerning the boundary behavior of positive p harmonic functions vanishing on a portion of the boundary of Lipschitz, chord arc, and Reifenberg ﬂat domains. An optimistic outline follows: 1. Fundamental properties of p harmonic functions and elliptic measure. 2. The dimension of p harmonic measure. 3. Boundary Harnack inequalities and the Martin boundary problem in Reifenberg ﬂat and Lipschitz domains. 4. Uniqueness and regularity in free boundary—inverse type problems. The lectures concerning 2 will be drawn from [6, 8, 48, 63]. Lectures involving 3 will be based on [49, 50, 52, 62]. Lectures on 4 will be concerned with [55–60] and [49–54].

1.1

Ode to the p Laplacian

I used to be in love with the Laplacian so worked hard to please her with beautiful theorems. However she often scorned me for the likes of Albert Baernstein, Bj¨orn Dahlberg, Carlos Kenig, and Thomas Wolﬀ. Gradually I became interested in her sister the p Laplacian, 1 < p < ∞, p = 2. I did not ﬁnd her as pretty as the Laplacian and she was often diﬃcult to handle J. Lewis () Mathematics Department, University of Kentucky, Lexington, KY 40506, USA e-mail: [email protected]

J. Lewis et al., Regularity Estimates for Nonlinear Elliptic and Parabolic Problems, Lecture Notes in Mathematics 2045, DOI 10.1007/978-3-642-27145-8 1, © Springer-Verlag Berlin Heidelberg 2012

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2

J. Lewis

because of her nonlinearity. However over many years I took a shine to her and eventually developed an understanding of her disposition. Today she is my girl and the Laplacian pales in comparison to her.

1.2

My Introduction to p Harmonic Functions

I was trained in function theory—subharmonic functions by my advisor, Maurice Heins and postdoctoral advisor Matts Ess´en. My ﬁrst paper on elliptic PDE and p harmonic functions (see [47]) was entitled ‘Capacitary Functions in Convex Rings.’ The catalyst for this paper was a problem in [31] which read as follows: ‘If D is a convex domain in space of 3 or more dimensions can we assert any inequalities for the Green’s function g(P, Q) which generalize the results for two dimensions, that follow from the classical inequalities for schlicht functions. Gabriel [29] has shown that the level surfaces of g(P, Q) = λ are convex but his proof is long. It would be interesting to ﬁnd a simpler proof.’ I tried to ﬁnd a simpler proof of Gabriel’s result but failed so eventually read his paper. In contrast to the author of the problem, I found Gabriel’s proof easy to follow and quite ingenious. Thus instead of ﬁnding a diﬀerent proof I found a diﬀerent PDE, the p Laplacian, to use Gabriel’s technique on. Moreover in writing the above paper I was forced to learn some classical PDE techniques (M¨ oser iteration, Schauder techniques) in order to deal with this degenerate nonlinear divergence form elliptic PDE.

2

Basic Estimates for the p Laplacian

We shall be working in Euclidean n space Rn . Points will be denoted by ¯ ∂E, be x = (x1 , . . . , xn ) or (x , xn ) where x = (x1 , . . . , xn−1 ) ∈ Rn−1 . Let E, the usual closure, boundary, of the set E and d(y, E) = the distance from y to E. ·, · denotes the standard inner product on Rn and |x| = x, x1/2 is the Euclidean norm of x. B(x, r) = {y ∈ Rn : |x − y| < r} and dx denotes Lebesgue n measure on Rn . Let ei be the i unit coordinate vector. If O ⊂ Rn is open and 1 ≤ q ≤ ∞, let W 1,q (O), denote the usual Sobolev space of equivalence classes of functions f with distributional gradient ∇f = (fx1 , . . . , fxn ), both of which are qth power integrable on O with Sobolev norm, f 1,q = f q + |∇f | q . Next let C0∞ (O) be inﬁnitely diﬀerentiable functions with compact support in O and let W01,q (O) be the closure of C0∞ (O) in the norm of W 1,q (O). Given G a bounded domain (i.e., a connected open set) and 1 < p < ∞, we say that u is p harmonic in G provided u ∈ W 1,p (G) and

Applications of Boundary Harnack Inequalities for p Harmonic Functions

3

|∇u|p−2 ∇u, ∇θ dx = 0

(1)

whenever θ ∈ W01,p (G) . Observe that if u is smooth and ∇u = 0 in G, then ∇ · (|∇u|p−2 ∇u) ≡ 0 in G

(2)

so u is a classical solution in G to the p Laplace partial diﬀerential equation. Equation (1) arises in the study of the following Dirichlet problem: Given g ∈ W 1,p (Rn ) let F = {h : h − g ∈ W01,p (G)}. Find |∇h|p dx.

inf

h∈F

(3)

G

it is well known that the inﬁmum in (3) occurs for a unique function u with u − g ∈ W01,p (G). Moreover u satisﬁes (1) as follows from the fact that u is a minimum and the usual calculus of variations type argument. v is said to be subpharmonic (superpharmonic) in G if v ∈ W 1,p (G) and whenever θ ≥ 0 ∈ W01,p (G), |∇v|p−2 ∇v, ∇θ dx ≤ 0 (≥ 0)

(4)

Lemma 1.1. (Boundary Maximum Principle) If v is subpharmonic in G, while w is superpharmonic in G with min{v −w, 0}∈W01,p (G), then v −w ≤ 0 a.e in G. Lemma 1.2. (Interior Estimates for u) Given p, 1 < p < ∞, let u be a positive p harmonic function in B(w, 2r). Then (i) Caccioppoli Inequality: |∇u|p dx ≤ c ( max u)p ,

rp−n

B(w,r)

B(w,r/2)

(ii) Harnack’s Inequality: max u ≤ c min u.

B(w,r)

B(w,r)

Furthermore, there exists α = α(p, n) ∈ (0, 1) such that if x, y ∈ B(w, r/2) then (iii) H¨ older Continuity: |u(x) − u(y)| ≤ c

|x−y| r

α max u. B(w,r)

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J. Lewis

Lemma 1.3. (Interior Estimates for ∇u) Let 1 < p < ∞ and suppose u is p harmonic in B(w, 2r). Then u has a representative in W 1,p (B(w, 2r)) with H¨ older continuous partial derivatives in B(w, 2r). In particular there exists σ ∈ (0, 1], depending only on p, n, such that if x, y ∈ B(w, r/2), then for some c = c(p, n), c−1 |∇u(x) − ∇u(y)| ≤ (|x − y|/r)σ max |∇u| B(w,r)

≤ c r−1 (|x − y|/r)σ max |u|. B(w,2r)

Also if ∇u(x) = 0, then u is C ∞ near x. For a proof of Lemmas 1.1 and 1.2, see [69]. Numerous proofs have been given of H¨ older continuity of ∇u in Lemma 1.3. Perhaps the ﬁrst was due to Ural’tseva for p > 2 while DiBenedetto, myself, and Tolksdorﬀ all gave proofs independently and nearly at the same time (1983, 1984) for 1 2 it is known that u (as above) need not be C 2 locally. For 1 < p < 2 this question is still open in Rn , n ≥ 3. In two dimensions, Iwaniec and Manfredi [36] showed that solutions are C k where k = k(p) ≥ 2 when 1 < p < 2 and k→∞ as p→1.

2.1

p Harmonic Functions in NTA Domains

Definition A. A domain Ω is called non tangentially accessible (NTA), if there exist M ≥ 2 and 0 < r0 ≤ ∞ such that the following are fulﬁlled, (i) Corkscrew condition: For any w ∈ ∂Ω, 0 < r < r0 , there exists ar (w) ∈ Ω satisfying M −1 r < |ar (w) − w| < r, d(ar (w), ∂Ω) > M −1 r, ¯ satisﬁes the corkscrew condition, (ii) Rn \ Ω (iii) Uniform condition: If w ∈ ∂Ω, and w1 , w2 ∈ B(w, r0 ) ∩ Ω, then there is a rectiﬁable curve γ : [0, 1]→Ω with γ(0) = w1 , γ(1) = w2 , and (a) H 1 (γ) ≤ M |w1 − w2 |, (b) min{H 1 (γ([0, t])), H 1 (γ([t, 1])) } ≤ M d(γ(t), ∂Ω). In Deﬁnition A, H 1 denotes length or Hausdorﬀ one measure. Often in our applications Ω will at least be an NTA domain with constants M, r0 while p is ﬁxed, 1 0 is p harmonic in Ω ∩ B(w, 2r) with u = 0 on ∂Ω ∩ B(w, 2r) in the usual Sobolev sense. We extend u to B(w, 2r) by putting u = 0 on B(w, 2r) \ Ω. Under this scenario we state

Applications of Boundary Harnack Inequalities for p Harmonic Functions

5

Lemma 1.4. Let u, p, w, Ω be as above. Then u ∈ W 1,p (B(w, 2r)) and |∇u|p dx ≤ c ( max u)p .

rp−n

B(w,r)

B(w,r/2)

Moreover there exists β = β(p, n, M ) ∈ (0, 1) such that u has a H¨ older continuous representative in B(w, 2r) with |u(x) − u(y)| ≤ c (|x − y|/r)β max u B(w,r)

whenever x, y ∈ B(w, r). Lemma 1.5. Let u, p, w, Ω, r, be as in Lemma 1.4. There exists c such that if rˆ = r/c, then max u ≤ cu(arˆ(w)). B(w,ˆ r)

2.1.1

Outline of Proofs

The ﬁrst display in Lemma 1.4 is a standard subsolution type inequality (use u times a cutoﬀ as a test function in (1)). As for the last display in Lemma 1.4 if p > n, this display is a consequence of Morrey’s Theorem and the ﬁrst display. If 1 < p ≤ n, then from the interior estimates in Lemma 1.2, we deduce that it suﬃces to consider only the case when y ∈ ∂Ω ∩ B(w, r). One then shows for some θ = θ(p, n, M ), 0 < θ < 1, that max u ≤ θ max u

B(z,ρ/2)

(5)

B(z,ρ)

whenever 0 < ρ < r/4 and z ∈ ∂Ω ∩ B(w, r). Equation (5) can then be iterated to get H¨older continuity in Lemma 1.4 for x, y as above. To prove (5) one uses the fact that (Rn \ Ω) ∩ B(z, ρ/2) and B(z, ρ/2) have comparable p capacities, as well as estimates for subsolutions to elliptic partial diﬀerential equations of p Laplacian type. These estimates are due to [68] for the p Laplacian (see also [30]). Lemma 1.5 for harmonic functions is often called Carleson’s lemma although apparently it could be due to Domar. This lemma for uniformly elliptic PDE in divergence form is usually attributed to [14]. All proofs use only Harnack’s inequality and H´older continuity near the boundary. Thus Lemma 1.5 is also valid for solutions to many PDE’s including the p Laplacian. In our study of p harmonic measure we shall outline a proof of a similar inequality when the geometry is considerably more complicated. That is when Ω ⊂ R2 is only a bounded simply connected domain.

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J. Lewis

2.2

The p Laplacian and Elliptic PDE

Let u be a solution to the p Laplace equation in (2) and suppose ∇u is nonzero n as well as suﬃciently smooth in a neighborhood of x ∈ Ω. Let η ∈ R with p−2 |η| = 1 and put ζ = ∇u, η . Then diﬀerentiating ∇ · |∇u| ∇u = 0 with respect to η one gets that ζ is a strong solution at x to Lζ = ∇ · [(p − 2)|∇u|p−4 ∇u, ∇ζ ∇u + |∇u|p−2 ∇ζ] = 0.

(6)

Clearly, (7) Lu = (p − 1) ∇ · |∇u|p−2 ∇u = 0 at x.

(7)

Equation (6) can be rewritten in the form Lζ =

n i,j=1

∂ [ bij (x)ζxj (x) ] = 0, ∂xi

where bij (x) = |∇u|p−4 [(p − 2)uxi uxj + δij |∇u|2 ](x),

(8)

(9)

for 1 ≤ i, j ≤ n, and δij is the Kronecker δ. In many of our applications it is of fundamental importance that u, derivatives of u, both satisfy the same divergence form PDE in (6), (7). For example, in several of our papers we integrate by parts functions of u, ∇u and the bad terms drop out because both functions satisfy the same PDE. Thus we study (8), (9). We note that if ξ ∈ Rn , then min{p − 1, 1}|ξ|2 |∇u(x)|p−2 ≤

n

bik ξi ξk

i,k=1

(10)

≤ max{1, p − 1}|∇u(x)|p−2 |ξ|2 . Observe from (10) that L can be degenerate elliptic if ∇u = 0. Thus in many of our papers we also prove the fundamental inequality: c−1 u(x)/d(x, ∂Ω) ≤ |∇u(x)| ≤ cu(x)/d(x, ∂Ω),

(11)

for some constant c and x near ∂Ω. Note that (10), (11), and Harnack’s inequality for u imply that (bik (x)) are locally uniformly elliptic in Ω. Behaviour near the boundary, such as boundary Harnack inequalities, are more involved. The easiest case for our methods to work is when ∂Ω is suﬃciently ﬂat in the sense of Reifenberg (to be deﬁned later). In this case we will be able to show that |∇u|p−2 is an A2 weight (also to be deﬁned). Thus

Applications of Boundary Harnack Inequalities for p Harmonic Functions

7

we list some theorems on degenerate elliptic equations whose degeneracy is given in terms of an A2 weight.

2.3

Degenerate Elliptic Equations

Let w ∈ Rn , 0 < r and let λ be a real valued Lebesgue measurable function deﬁned almost everywhere on B(w, 2r). λ is said to belong to the class A2 (B(w, r)) if there exists a constant γ such that −2n

λ dx ·

r˜

B(w,˜ ˜ r)

λ−1 dx ≤ γ

B(w,˜ ˜ r)

whenever w ˜ ∈ B(w, r) and 0 < r˜ ≤ r. If λ(x) belongs to the class A2 (B(w, r)) then λ is referred to as an A2 (B(w, r))-weight. The smallest γ such that the above display holds is referred to as the constant of the weight. Once again let Ω ⊂ Rn be a NTA domain with NTA-constants M, r0 . Let w ∈ ∂Ω, 0 < r < r0 , and consider the operator ˆ= L

n ∂ ˆ ∂ bij (x) ∂xi ∂xj i,j=1

(12)

in Ω ∩ B(w, 2r). We assume that the coeﬃcients {ˆbij (x)} are bounded, Lebesgue measurable functions deﬁned almost everywhere on B(w, 2r). Moreover, ˆbij = ˆbji for all i, j ∈ {1, .., n}, and n

c−1 λ(x)|ξ|2 ≤

ˆbij (x)ξi ξj ≤ c|ξ|2 λ(x)

(13)

i,j=1

for almost every x ∈ B(w, r), where λ ∈ A2 (B(w, r)). If O ⊂ B(w, 2r) is open ˜ 1,2 (O) be the weighted Sobolev space of equivalence classes of functions let W v with distributional gradient ∇v and norm

v ˜

21,2 = v 2 λdx + |∇v|2 λdx < ∞. O

O

˜ 1,2 (O) be the closure of C ∞ (O) in the norm W ˜ 1,2 (O). We say that v Let W 0 0 1,2 ˆ = 0 in O provided v ∈ W ˜ (O) and is a weak solution to Lv O

whenever φ ∈ C0∞ (O).

i,j

ˆbij vx φx dx = 0 i j

(14)

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J. Lewis

The following three lemmas, Lemmas 1.6–1.8, are tailored to our situation and based on the results in [21–23]. Lemma 1.6. Let Ω ⊂ Rn be a NTA-domain with constant M , w ∈ ∂Ω, 0 < r < r0 , and let λ be an A2 (B(w, r))-weight with constant γ. Suppose that ˆ = 0 in Ω ∩ B(w, 2r). Then there exists a v is a positive weak solution to Lv constant c, 1 ≤ c < ∞, depending only on n, M and γ, such that if w ˜∈Ω and B(w, ˜ 2˜ r) ⊂ Ω ∩ B(w, r), then (i) c

−1 2

2

|v|2 λdx,

|∇v| λdx ≤ c

r˜

B(w,˜ ˜ r)

B(w,2˜ ˜ r)

(ii) max v ≤ c min v.

B(w,˜ ˜ r)

B(w,˜ ˜ r)

Furthermore, there exists α = α(n, M, γ) ∈ (0, 1) such that if x, y ∈ B(w, ˜ r˜) then (iii)

|v(x) − v(y)| ≤ c

|x−y| r˜

α max v.

B(w,2˜ ˜ r)

Lemma 1.7. Let Ω ⊂ Rn be a NTA-domain with constant M , w ∈ ∂Ω, 0 < r < r0 , and let λ be an A2 (B(w, r))-weight with constant γ. Suppose ˆ = 0 in Ω ∩ B(w, 2r) and that v = 0 that v is a positive weak solution to Lv on ∂Ω ∩ B(w, 2r) in the weighted Sobolev sense. Extend v to B(w, 2r) by ¯ Then v ∈ W ˜ 1,2 (B(w, 2r)) and there exists setting v ≡ 0 in B(w, 2r) \ Ω. c˜ = c˜(n, M, γ), 1 ≤ c˜ < ∞, such that the following holds with r˜ = r/˜ c. (i) r2

|∇v|2 λdx ≤ c˜

Ω∩B(w,r/2)

|v|2 λdx,

Ω∩B(w,r)

(ii) max

Ω∩B(w,˜ r)

v ≤ c˜v(ar˜(w)).

Furthermore, there exists α = α(n, M, γ) ∈ (0, 1) such that if x, y ∈ Ω ∩ B(w, r˜), then (iii)

|v(x) − v(y)| ≤ c

|x−y| r

α max

Ω∩B(w,2˜ r)

v.

Lemma 1.8. Let Ω ⊂ Rn be a NTA-domain with constant M , w ∈ ∂Ω, 0 < r < r0 , and let λ be an A2 (B(w, r))-weight with constant γ. Suppose ˆ = 0 in Ω ∩ B(w, 2r) and that v1 and v2 are two positive weak solutions to Lv v1 = 0 = v2 on ∂Ω ∩ B(w, 2r) in the weighted Sobolev sense. Then there exist

Applications of Boundary Harnack Inequalities for p Harmonic Functions

9

c = c(n, M, γ), 1 ≤ c < ∞, and α = α(n, M, γ) ∈ (0, 1) such that if r˜ = r/c, and y1 , y2 ∈ Ω ∩ B(w, r/c), then

α

v1 (y1 ) v1 (y2 )

v1 (y1 ) |y1 − y2 |

.

v2 (y1 ) − v2 (y2 ) ≤ c v2 (y1 ) r Note: The last display implies v1 /v2 is H¨ older continuous, as well as bounded above and below by its value at any one point in Ω ∩ B(w, r/c). We refer to the last display as a boundary Harnack inequality.

3

p Harmonic Measure

If γ > 0 is a positive function on (0, r0 ) with lim γ(r) = 0 deﬁne H γ r →0 Hausdorﬀ measure on Rn as follows:For ﬁxed 0 < δ < r0 and E ⊆ R2 , let L(δ) = {B(zi , ri )} be such that E ⊆ B(zi , ri ) and 0 < ri < δ, i = 1, 2, ... Set φγδ (E) = inf γ(ri ). L(δ)

Then H γ (E) = lim φγδ (E). δ→0

k

k

γ

In case γ(r) = r we write H for H . Next let Ω ⊂ Rn , n ≥ 2, be a bounded domain, p ﬁxed, 1 < p < ∞, and N an open neighborhood of ∂Ω. Let v be p harmonic in Ω ∩ N and suppose that v is positive on Ω ∩ N with boundary value zero on ∂Ω, in the W 1,p Sobolev sense. Extend v to a function in W 1,p (N ) by setting v ≡ 0 on N \ Ω. Then there exists (see [33]) a unique positive Borel measure ν on Rn with support ⊂ ∂Ω, for which p−2 |∇v| ∇v, ∇φ dx = − φ dν (15) whenever φ ∈ C0∞ (N ). In fact if ∂Ω, |∇v|, are smooth dν = |∇v|p−1 dH n−1 on ∂Ω. Existence of ν follows if one can show for φ ≥ 0 as above, |∇v|p−2 ∇v, ∇φ dx ≤ 0.

(16)

Indeed, assuming (15) one can deﬁne a positive operator on the space of continuous functions and using basic Caccioppoli inequalities—the Riesz

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J. Lewis

representation theorem, get the existence of ν. If v has continuous boundary value zero one can get (16) as follows. Let θ = [(η + max[v − , 0]) − η ] φ. Then one can show that θ may be used as a test function in (1). Doing this we deduce {v≥ }∩N [(η + max[v − , 0]) − η ] × |∇v|p−2 ∇v, ∇φ dx ≤ 0. Using dominated convergence, letting η and then →0, we get (16). If p = 2 and v is the Green’s function with pole at x0 ∈ Ω, then ν = ω(·, x0 ) is harmonic measure with respect to x0 ∈ Ω. Green’s functions can be deﬁned for the p Laplacian when 1 < p < ∞, but are not very useful due to the nonlinearity of the p Laplacian when p = 2. Instead we often study the ¯ 0 , r), measure, μ, associated with a p capacitary function, say u, in Ω \ B(x ¯ where B(x0 , 4r) ⊂ Ω. That is, u is p harmonic in Ω \ B(x0 , r) with continuous boundary values, u ≡ 1 on ∂B(x0 , r) and u ≡ 0 on ∂Ω. Remark 1. μ is diﬀerent from the so called p harmonic measure introduced by Martio, which in fact is not a measure (see [65]). Deﬁne the Hausdorﬀ dimension of μ by H-dim μ = inf{k : there exists E Borel ⊂ ∂Ω with H k (E) = 0 and μ(E) = μ(∂Ω)}. Remark 2. We discuss for a ﬁxed p, 1 < p < ∞, what is known about H-dim μ when μ corresponds to a positive p harmonic function u in Ω ∩ N with boundary value 0 in the W 1,p Sobolev sense. It turns out that H-dim μ is independent of u as above. Thus we often refer to H-dim μ as the dimension of p harmonic measure in Ω. For p = 2, n = 2, and harmonic measure, Carleson [15] used ideas from ergodic theory and boundary Harnack inequalities for harmonic functions to deduce H-dim ω = 1 when Ω ⊂ R2 is a ‘snowﬂake’ type domain and H-dim ω ≤ 1 when Ω ⊂ R2 is the complement of a self similar Cantor set. He was also the ﬁrst to recognize the importance of |∇gn | log |∇gn | dH 1 ∂Ωn

(gn is Green’s function for Ωn with pole at zero and (Ωn ) is an increasing sequence of domains whose union is Ω). Wolﬀ [72] used Carleson’s ideas and brilliant ideas of his own to study the dimension of harmonic measure, ω, with respect to a point in domains bounded by ‘Wolﬀ snowﬂakes’ ⊂ R3 . He constructed snowﬂakes for which H-dim ω > 2 and snowﬂakes for which H-dim ω < 2. In [61] we constructed Wolﬀ Snowﬂakes, for which the harmonic measures on both sides of the snowﬂake were of H-dim < n − 1 and also a snowﬂake for which the harmonic measures on both sides were of H-dim > n − 1. Soon

Applications of Boundary Harnack Inequalities for p Harmonic Functions

11

after we ﬁnished this paper, Bj¨ orn Bennewitz became my Ph.D. student. We began studying the dimension of the measure μ as in (15) for ﬁxed p, 1 < p < ∞, p = 2. We tried to imitate the Carleson–Wolﬀ construction in order to produce examples of snowﬂakes where we could estimate H-dim μ, when Ω ⊂ R2 is a snowﬂake. To indicate the diﬃculties involved we note that Wolﬀ showed Carleson’s integral over ∂Ωn can be estimated at the nth step in the construction of certain snowﬂakes ⊂ R3 . His calculations make key use of a boundary Harnack inequality for positive harmonic functions vanishing on a portion of the boundary of a NTA domain. Thus we proved u1 /u2 ≤ c on B(z, r/2) ∩ Ω whenever z ∈ ∂Ω and 0 < r ≤ r0 . Here u1 , u2 > 0 are p harmonic in B(z, r) ∩ Ω and vanish continuously on B(z, r) ∩ ∂Ω. Using our boundary Harnack inequality, we were able to deduce that μ had a certain weak mixing property and consequently, arguing as in Carleson–Wolﬀ, we obtained an ergodic measure ≈ μ on ∂Ω. Applying the ergodic theorem of Birkhoﬀ and entropy theorem of Shannon–McMillan–Breiman it followed that lim r →0

log μ[B(x, r)] = H-dim μ for μ almost every x ∈ ∂Ω. log r

(17)

Wolﬀ uses H¨older continuity of the ratio and other arguments in order to make eﬀective use of (17) in his estimates of H-dim μ. We ﬁrst tried to avoid many of these estimates by a ﬁnesse type argument. However, later this argument fell through because of a calculus mistake. Finally we decided that instead of Wolﬀ’s argument we should use the divergence theorem and try to ﬁnd a partial diﬀerential equation for which u is a solution and log |∇u| is a subsolution (supersolution) when p > 2 (1 < p < 2). We succeeded, in fact the PDE is given in (8), (9): Lζ(x) =

n i,j=1

∂ (bij ζxj )(x) ∂xi

bij (x) = |∇u|p−4 [(p − 2)uxi uxj + δij |∇u|2 ](x), 1 ≤ i, j ≤ n. Moreover for domains ⊂ R2 whose boundary is a quasi circle. we were able to prove the fundamental inequality in (11): c−1 u(x)/d(x, ∂Ω) ≤ |∇u(x)| ≤ cu(x)/d(x, ∂Ω), for some constant c and x near ∂Ω. Thus interior estimates for uniformly elliptic non divergence form PDE could be applied to solutions of L. Armed with this knowledge we eventually proved in [6]: Theorem 2.1. Fix p, 1 0 be p harmonic in Ω ∩N ⊂ R2 with u = 0 continuously on ∂Ω. If ∂Ω is a snowﬂake and 1 1 while if 2 < p < ∞, then H-dim μ < 1.

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J. Lewis

Theorem 2.2. Let p, u, μ be as in Theorem 2.1. If ∂Ω ⊂ R2 is a self similar Cantor set and 2 < p < ∞, then H-dim μ < 1. Theorem 2.3. Let p, u, μ be as in Theorem 2.2. If ∂Ω ⊂ R2 is a k quasicircle, then H-dim μ ≤ 1 for 2 < p < ∞, while H-dim μ ≥ 1 for 1 < p < 2. To outline the proof of Theorem 2.1 we note that the boundary Harnack inequality mentioned earlier implies that all measures associated with functions in Theorems 2.1–2.3, have the same Hausdorﬀ dimension. Also, since the p Laplacian is translation, dilation, and rotation invariant we may ¯ 1) where d(0, ∂Ω) = 4. assume that u is the p capacitary function for Ω \ B(0, ¯ That is u is p harmonic in Ω \ B(0, 1) with continuous boundary values, u ≡ 1 on ∂B(0, 1) and u ≡ 0 on ∂Ω. Let μ be the measure associated with u as in (15). Let Ωn ⊂ Ω be a sequence of approximating domains constructed in the usual way and for large n let un be the p capacitary function for ¯ 1). Then one ﬁrst proves: Ωn \ B(0, Lemma 2.4. For ﬁxed p, 1 0 then H-dim μ < 1 while if η < 0, then H-dim μ > 1. To prove Lemma 2.4 we followed Carleson–Wolﬀ but our argument was necessarily more complicated, due to the non-linearity of the p Laplacian. To prove Theorem 2.1 let un , Ωn be as in Lemma 2.4. We note that one ¯ 1) and if v = log |∇un |, p = 2, 1 < p < ∞, can show ∇un = 0 in Ωn \ B(0, that 2 Lv ≥ min(1, p − 1) |∇u|p−4 (uxi xj )2 . (18) p−2 i,j=1 where L is as in (8), (9). ¯ 1) to Next we apply the divergence theorem for large n in On = Ωn \ B(0, the vector ﬁeld whose ith component, i = 1, 2, is un

2

bik vxk − v

k=1

2

bik (un )xk .

k=1

We get

un Lv dx =

On

2

bik ξi [ un vxk − v (un )xk ] dH 1 x + O(1)

(19)

∂Ωn i,k=1

where ξ denotes the outer unit normal to ∂Ωn . Using the fact that ξ = −∇un /|∇un | and the deﬁnition of (bik ), we ﬁnd that

Applications of Boundary Harnack Inequalities for p Harmonic Functions

2

13

bik ξi [un vxk − v (un )xk ] dH 1 x

∂Ωn i,k=1

(20)

|∇un |p−1 log |∇un | dH 1 x + O(1).

= (p − 1) ∂Ωn

From (18)–(20), and Lemma 2.4 we conclude that in order to prove Theorem 2.1 it suﬃces to show ⎞ ⎛ un |∇un |p−4 (un )2xi xj dx dx⎠ > 0. (21) lim inf ⎝ n−1 n→∞ On i,j=1

To prove (21) we showed the existence of λ ∈ (0, 1) such that if x ∈ Ωn \ B(0, 2) and d(x, ∂Ωn ) ≥ 3−n , then c On ∩B(x,λd(x,∂Ωn ))

un |∇un |p−4 (un )2yi yj dy ≥ μn (B(x, 2d(x, ∂Ωn ))

(22)

where c depends on p and the k quasi-conformality of Ω. Covering {3−m−1 ≤ d(x, ∂Ωn ) ≤ 3−m } by balls and summing over 1 ≤ m ≤ n − 1 we obtain ﬁrst (21) and then Theorem 2.1. To prove Theorem 2.3 let w(x) = max(v − c, 0) when 1 2. Here c is chosen so large that |v| ≤ c on B(0, 2). Following Makarov [66] we prove: Lemma 2.5. Let m be a nonnegative integer. There exists c+ = c+ (k, p) ≥ 1 such that for 0 < t < 1, |∇u|p−1 w2m dH 1 x ≤ cm+1 m! [log(2/t)]m . + {x:u(x)=t}

To outline the proof of Lemma 2.5 let Ω(t) = Ω \ {x : u(x) ≤ t}, whenever 0 < t < 1. Using the fact that Lw ≤ 0 for 1 0, one computes, L(w2m )(x) ≤ 2m(2m − 1)p |∇u|p−2 (x) w2m−2 (x) |∇w|2 (x).

(23)

Next we use (23) and apply the divergence theorem in Ω(t) to the vector ﬁeld whose ith component for i = 1, 2 is (u − t)

2 j=1

bij (w2m )xj − w2m

2 j=1

bij uxj .

14

J. Lewis

We get

(p − 1)

{x:u(x)=t}

|∇u|p−1 w2m dH 1 x (24)

≤ 2m(2m − 1)p

u |∇u|

p−2

w

2m−2

2

|∇w| dx.

Ω(t)

Using interior estimates for solutions to the p Laplace equation from Section 2, the coarea formula and once again and our fundamental inequality, |∇u(·)| ≥ u(·)/d(·, ∂Ω), we deduce from (24) that Im (t) =

{x:u(x)=t}

|∇u|p−1 w2m dH 1 x ≤ 2m(2m − 1)c

1

Im−1 (τ ) τ −1 dτ

t

where c = c(p, k). Lemma 2.5 then follows from an inductive type argument, using I0 ≡ constant on (0,1). Dividing the display in Lemma 2.5 by (2c+ )m m! logm (2/t) and summing we see for 0 < t < 1 that w2 dH 1 x ≤ 2 c+ . |∇u|p−1 exp (25) 2c+ log(2/t) {x:u(x)=t} Using (25) and weak type estimates it follows that if λ(t) =

4 c+ log(2/t) log(− log t) for 0 < t < e−2 , (26)

F (t) = {x : u(x) = t and w(x) ≥ λ(t)} then

F (t)

|∇u|p−1 dH 1 x ≤

2c+ log2 (1/t)

(27)

One can use (27) to show that if Hausdorﬀ measure (denoted H γ ) is deﬁned with respect to aλ(r) re when 1 2. and a is suﬃciently large then μ is absolutely continuous with respect to H γ when 1 2.

(30)

Clearly (29), (30) imply Theorem 2.3.

Applications of Boundary Harnack Inequalities for p Harmonic Functions

3.1

15

p Harmonic Measure in Simply Connected Domains

Recently in [63] we have managed to prove the following theorem. Theorem 2.6. Fix p, 1 0 be p harmonic in Ω ∩ N, where Ω is simply connected, ∂Ω is compact, and N is a neighborhood of ∂Ω. Suppose u has continuous boundary value 0 on ∂Ω and let μ be the measure associated with u as in (1). If λ, γ, are as in (26), (28), then (29), (30) are valid for a = a(p) suﬃciently large. Hence Theorem 2.3 remains valid in simply connected domains. We note that Makarov in [66] proved for harmonic measure (i.e., p = 2) the stronger theorem: Theorem 2.7. Let ω be harmonic measure with respect to a point in the simply connected domain Ω. Then (a) ω is concentrated on a set of σ ﬁnite H 1 measure (b) ω is absolutely continuous with respect to H γˆ measure deﬁned relative to γˆ(r) = r exp[A log 1/r log log log 1/r] for A suﬃciently large. √ 24−3 The best known value of A in the deﬁnition of γˆ appears to be A = 6 5 given in [32].

3.2

Preliminary Reductions for Theorem 2.6

To outline the proof of Theorem 2.6 we ﬁrst claim, as in the proof of Theorem 2.5, that all measures associated with functions satisfying the hypotheses of Theorem 2.6, will have the same Hausdorﬀ dimension. Indeed let u1 , u2 > 0 be p harmonic functions in Ω ∩ N and let μ1 , μ2 , be the corresponding measures as in (15). Observe from the maximum principle for p harmonic functions and continuity of u1 , u2 , that there is a neighborhood N1 ⊂ N of ∂Ω with M −1 u1 ≤ u2 ≤ M u1 in N1 ∩ Ω.

(31)

One can also show there exists r0 > 0, c = c(p) < ∞, such that whenever w ∈ ∂Ω, 0 < r ≤ r0 , and i = 1, 2, c−1 rp−2 μi [B(w, r/2)] ≤ max up−1 ≤ c rp−2 μi [B(w, 2r)]. i B(w,r)

(32)

Using (31), (32), and a covering argument it follows that μ1 , μ2 are mutually absolutely continuous. Mutual absolute continuity is easily seen to imply H-dim μ1 = H-dim μ2 . Thus we may assume, as in the proof of Theorem 2.3,

16

J. Lewis

¯ 1) and d(0, ∂Ω) = 4. that u is the p capacitary function for D = Ω \ B(0, The major obstacle to proving Theorem 2.6 in our earlier paper was that we could not prove the fundamental inequality in (11). That is, in our new paper, we prove Theorem 2.8. If u is the p capacitary function for D, then there exists c = c(p) ≥ 1, such that c|∇u|(z) ≥

u(z) whenever z ∈ D. d(z, ∂Ω)

Given Theorem 2.8 we can copy the argument leading to (27) in the proof of Theorem 2.3. However one has to work harder in order to deduce (29), (30) from (27) as previously we used the doubling property of μ in (32) and this property is not available in the simply connected case. Still we omit the additional measure theoretic argument and shall regard the proof of Theorem 2.6 as complete once we sketch the proof of Theorem 2.8.

3.3

Proof of Theorem 2.8

To prove Theorem 2.8 we assume, as we may, that ∂Ω is a Jordan curve, since otherwise we can approximate Ω in the Hausdorﬀ distance sense by Jordan domains and use the fact that the constant in Theorem 2.8 depends only on p to eventually get this theorem for Ω. We continue under √ this assumption and shall use complex notation. Let z = x + iy, where i = −1 and for a, b ∈ D, let ρ(a, b) denote the hyperbolic distance from a, b ∈ D to ∂Ω. Fact A. u is real analytic in D, ∇u = 0 in D, and uz = (1/2)(ux − iuy ), is k = k(p) quasi-regular in D. Consequently, log |∇u| is a weak solution to a divergence form PDE for which a Harnack inequality holds. That is, if h ≥ 0 is a weak solution to this PDE in B(ζ, r) ⊂ D, then max h ≤ c˜ min h, B(ζ,r/2)

B(ζ,r/2)

where c˜ = c˜(p). From Fact A and Lemma 1.3 one can show that |∇u(z)| ≤ cu(z)/d(z, ∂Ω) in D

(33)

¯ 1). Next we use Fact A and and that Theorem 2.8 is valid in B(0, 2) \ B(0, (33) to show that Theorem 2.8 for x ∈ D \ B(0, 2) is a consequence of the following lemma. Lemma 2.9. There is a constant c = c(p) ≥ 1 such that for any point z1 ∈ D \ B(0, 2), there exists z ∈ D \ B(0, 2) with u(z ) = u(z1 )/2 and ρ(z1 , z ) ≤ c.

Applications of Boundary Harnack Inequalities for p Harmonic Functions

17

Assuming Lemma 2.9 one gets Theorem 2.8 from the following argument. Let Γ be the hyperbolic geodesic connecting z1 to z ∗ and suppose that Γ ⊂ D. From properties of ρ one sees for some c = c(p) that H 1 (Γ ) ≤ cd(z1 , ∂Ω) and d(Γ, ∂Ω) ≥ c−1 d(z1 , ∂Ω). Thus

(34)

1 u(z1 ) ≤ u(z1 ) − u(z ) ≤ |∇u(z)||dz| 2 Γ ≤ cH 1 (Γ ) max |∇u| ≤ cd(z1 , ∂Ω) max |∇u|. Γ

∗

Γ

∗

So for some ζ ∈ Γ and c = c (p) ≥ 1, c |∇u(ζ)| ≥

u(z1 ) . d(z1 , ∂Ω)

(35)

Also from (35), we deduce the existence of Whitney balls {B(wj , rj }, with wj ∈ Γ, rj ≈ d(z1 , ∂Ω), connecting ζ to z1 and |∇u(z)| ≤ cu(z1 )/d(z1 , ∂Ω) when z ∈ B(wj , rj ) . (36) j

From (35), (36), we see that if c = c(p) is large enough and h(z) = log

c u(z1 ) d(z1 , ∂Ω) |∇u(z)|

for z ∈

B(wi , ri )

j

then h > 0 in ∪i B(wi , ri ) and h(ζ) ≤ c. From Fact A we see that Harnack’s inequality can be applied to h in successive balls of the form B(wi , ri /2). Doing this we obtain h(z1 ) ≤ c where c = c (p). Clearly, this inequality implies Theorem 2.8. We note that if ∂Ω is a quasicircle one can choose z ∗ to be a point on the line segment connecting z1 to w ∈ ∂Ω where |w − z1 | = d(z1 , ∂Ω) The proof uses H¨older continuity of u near ∂Ω and the fact that for some c = c(p, k), cu(z1 ) ≥ max u. (see Lemmas 1.4 and 1.5). This inequality B(z1 ,2d(z1 ,∂Ω))

need not hold in a Jordan domain and so we have to give a more complicated argument to get Lemma 2.9. To this end, we construct a Jordan arc σ : (−1, 1)→D with σ(0) = z1 , σ(±1) = lim σ(t) ∈ ∂Ω, and σ(1) = σ(−1). t→±1 Moreover, for some c = c(p), (α) H 1 (σ) ≤ cd(z1 , ∂Ω) (37) (β) u ≤ cu(z1 ) on σ.

18

J. Lewis

Let Ω1 be the component of Ω \ σ not containing B(0, 1). Then we also require that there is a point w0 on ∂Ω ∩ ∂Ω1 with |wo − z1 | ≤ cd(z1 , ∂Ω) and d(w0 , σ) ≥ c−1 d(z1 , ∂Ω).

(38)

Finally we shall show the existence of a Lipschitz curve τ : (0, 1)→Ω1 with τ (0) = z1 , τ (1) = w0 , satisfying the cigar condition: min{H 1 (τ [0, t]), H 1 (τ [t, 1])} ≤ cˆd(τ (t), ∂Ω),

(39)

for 0 < t < 1 and some absolute constant cˆ. To get Lemma 2.9 from (37)–(39) let u1 = u in Ω1 and u1 ≡ 0 outside of Ω1 . From PDE estimates, (37) (β), and (38) one ﬁnds θ > 0, c < ∞ such that max u1 ≤ cu(z1 )

B(w0 ,t)

t d(z1 , ∂Ω)

θ for 0 < t < d(w0 , σ).

(40)

From (40), (39) we conclude the existence of z ∗ with ρ(z1 , z ∗ ) ≤ c and u(z ∗ ) = 1/2, which is Lemma 2.9. To construct σ, τ let f be the Riemann mapping function from the upper half plane, H, onto Ω with f (i) = 0 and f (a) = z1 , where a = is for some s, 0 < s < 1. Note that f has a continuous extension ¯ since ∂Ω is a Jordan curve. Let I(b) = [ Re b − Im b, Re b + Im b] to H, whenever b ∈ H. We need the following lemmas. Lemma 2.10. There is a set E(b) ⊂ I(b) such that for x ∈ E(b) Im b 0

|f (x + iy)|dy ≤ c d(f (b), ∂Ω)

for some absolute constant c , and also H 1 (E(b)) ≥ (1 − 10−100 )H 1 (I(b)). Lemma 2.11. Given 0 < δ < 10−1000 , there is an absolute constant cˆ such that if δ = e−ˆc/δ then, whenever x ∈ E(b) there is an interval J = J(x) centered at x with 2δ Im b ≤ H 1 (J) ≤ cδ 1/2 Im b ≤

Im b 10000

(for some absolute constant c) and a subset F = F (x) ⊂ J with H 1 (F ) ≥ (1 − 10−100 )H 1 (J) so that δ Im b 0

|f (t + iy)|dy ≤ δd(f (b), ∂Ω) for every t ∈ F.

Applications of Boundary Harnack Inequalities for p Harmonic Functions

Lemma 2.12. Let Fˆ = Im b , then H 1 (L) ≥ 100

x∈E(b)

19

F (x). If L ⊂ I(b) is an interval with

Im b . H 1 (E(b) ∩ Fˆ ∩ L) ≥ 1000 Moreover, if {τ1 , τ2 , . . . , τm } is a subset of I(b), then there exists τm+1 in E(b) ∩ Fˆ ∩ L with |f (τm+1 ) − f (τj )| ≥

d(f (b), ∂Ω) whenever 1 ≤ j ≤ m. 1010 m2

To construct σ, τ from Lemma 2.12 we put b = a = is, and deduce for given δ, 0 < δ < 10−1000 , the existence of x1 , x2 , x3 ∈ E(a) with −s < x1 < −s/2, − 81 s < x3 < 18 s, and 12 s < x2 < s. Moreover, 0

δ∗ s

|f (xj + iy| dy ≤ δd(z1 , ∂Ω) for 1 ≤ j ≤ 3,

min{|f (x1 ) − f (x3 )|, |f (x2 ) − f (x3 )|} ≥ 10−11 d(z1 , ∂Ω)

(41) (42)

˜ Let Q(a) be the rectangle whose boundary in H, ξ, consists of the horizontal line segment from x1 + is to x2 + is, and the vertical line segments from xj to xj + is, for j = 1, 2. Put σ = f (ξ) and note from (41) that (37) (α) is valid. To construct τ we put t0 = 0, s0 = s, a0 = t0 + is0 . Let s1 = δ∗ s0 , t1 = x3 , and a1 = t1 + is1 . By induction, suppose am = sm + itm has been deﬁned for 1 ≤ m ≤ k − 1. We then choose tk ∈ E(ak−1 ) so that the last display in Lemma 2.11 holds with t = tk . Set sk = δ∗ sk−1 and ak = sk + itk . Let λk be the curve consisting of the horizontal segment from ak−1 to tk + isk−1 and the vertical line segment from ak−1 to ak . Put λ = λk and τ = f (λ). From our construction we deduce that τ satisﬁes the cigar condition in (39) for δ > 0 small. Also x0 = limt→1 λ(t) exists, |x0 | < 1/4, and (38) holds for w0 = f (x0 ), thanks to (42) and our construction.

3.4

The Final Proof

It remains to prove u ≤ cu(z1 ) on σ which is (37) (β). The proof is by contradiction. Suppose u > Au(z1 ) on σ. We shall obtain a contradiction if A = A(p) is suitably large. Our argument is based on the recurrence type scheme mentioned after Lemma 1.5 (often attributed to Carleson–Domar in

20

J. Lewis

the complex world and Caﬀarelli et al. in the PDE world). Given the rectangle ˜ Q(a) we let bj,1 = xj + iδ∗ Im a, j = 1, 2, and note that bj,1 , j = 1, 2, are ˜ points on the vertical sides of Q(a). These points will spawn two new boxes ˜ j,1 ), j = 1, 2, which in turn will each spawn two more new boxes, and so Q(b ˜ 1,1 ). on. Without loss of generality, we focus on Q(b ˜ This box is constructed in the same way as Q(a) and we also construct, using Lemma 2.12 once again, a polygonal path λ1,1 from b1,1 to some point x0,1 ∈ I(b1,1 ). λ11 is deﬁned relative to b1,1 in the same way that λ was deﬁned relative to a. Also in view of Lemma 2.12 we can require that λ1,1 ⊂ { Re z < Re b1,1 }. λ2,1 with endpoints, b2,1 , x0,2 is constructed similarly to lie in { Re z > Re b2,1 }. Next let Λ be a Harnack constant such that max{u(f (b1,1 )), u(f (b2,1 ))} ≤ Λu(z1 ).

(43)

From Harnack’s inequality for u and Lemma 2.12 with δ ﬁxed, it is clear that Λ in (43) can be chosen to depend only on p, so can also be used in further iterations. Let U = u ◦ f. By the maximum principle, since A > Λ and λ1,1 , λ2,1 ˜ lie outside of Q(a), there will be a point z ∈ λ1,1 ∪ λ2,1 such that U (z) > AU (a) = Au(z1 ). Suppose z ∈ λ1,1 . The larger the constant A, the closer z will be to R. More precisely, if A > Λk then Im z ≤ δ k Im a, as follows from Harnack’s inequality for u, and the construction of λ1,1 . In fact we can show that |f (z) − f (x0,1 )| ≤ Cδ k−1 d(f (b1,1 ), ∂Ω). The argument now is similar to the argument showing the existence of z ∗ ˜ 1,1 ) which is in H and let σ1,1 = given σ, τ. Let ξ1,1 be the boundary of Q(b f (ξ1,1 ). Set w0,1 = f (x0,1 ). Then ˜ 1,1 )). B(w0,1 , d(w0,1 , σ1,1 )) ⊂ f (Q(b and since d(w0,1 , σ1,1 ) ≈ d(f (b1,1 ), ∂Ω) it follows from H¨ older continuity of u that U (z) ≤ Cδ θk max U. ˜ 1,1 ) Q(b

Choose k, depending only on p, to be the least positive integer such that Cδ θk < Λ−1 . This choice of k determines A (say A = 2Λk ) which therefore also depends only on p. With this choice of A we have max U > ΛU (z) > ΛAU (a). ξ1,1

(44)

Applications of Boundary Harnack Inequalities for p Harmonic Functions

21

Since U (b1,1 ) ≤ ΛU (a) we see from (44) that we can now repeat the above ˜ ˜ 1,1 ) playing the role of Q(a). That is, we ﬁnd b1,2 on argument with Q(b ˜ 1,2 ) with ˜ 1,1 ) with Im b1,2 = δ 2 Im a and a box Q(b the vertical sides of Q(b

boundary ξ1,2 such that max U > Λ2 AU (a) ≥ AU (b1,2 ). ξ1,2

Continuing by induction we get a contradiction because U = 0 continuously on ∂Ω. The proof of (37) (β), Theorem 2.8, and Theorem 2.6 is now complete.

3.5

p Harmonic Measure in Space

In [48] we proved Theorem 2.13. Let p, u, μ, be as in Theorem 2.3. There exists k0 (p) > 0 such that if ∂Ω is a k quasicircle, 0 < k < k0 (p), then (a) μ is concentrated on a set of σ ﬁnite H 1 measure when p ¿ 2. (b) There exists A = A(p), 0 < A(p) < ∞, such that if 1 < p < 2, ˆ ˆ then μ is absolutely continuous with respect to H λ where λ(r) = r exp[A log 1/r log log log 1/r]. In [8] we prove an analogue of Theorem 2.13 when p ≥ n. To be more speciﬁc we need a deﬁnition. Definition B. Let Ω ⊂ Rn be a (δ, r0 ) NTA domain and 0 < r ≤ r0 . Then Ω and ∂Ω are said to be (δ, r0 ), Reifenberg ﬂat provided that whenever w ∈ ∂Ω, there exists a hyperplane, P = P (w, r), containing w such that (a) Ψ (∂Ω ∩ B(w, r), P ∩ B(w, r)) ≤ δr (b) {x ∈ Ω ∩ B(w, r) : d(x, ∂Ω) ≥ 2δr} ⊂ one component of Rn \ P . In Deﬁnition B, Ψ (E, F ) denotes the Hausdorﬀ distance between the sets E and F deﬁned by Ψ (E, F ) = max(sup{d(y, E) : y ∈ F }, sup{d(y, F ) : y ∈ E}) Theorem 2.14. Let Ω ⊂ Rn , n ≥ 3, be a (δ, r0 ) Reifenberg ﬂat domain, w ∈ ∂Ω, and p ﬁxed, n ≤ p < ∞. Let u > 0 be p harmonic in Ω with u = 0 continuously on ∂Ω. Let μ be the measure associated with u as in (15). There ˆ n) > 0, such that if 0 < δ ≤ δ, ˆ then μ is concentrated on a set exists, δˆ = δ(p, of σ ﬁnite H n−1 measure. To outline the proof of Theorem 2.14, we shall need the following result from [50]: Theorem 2.15. Let Ω be (δ, r0 ) Reifenberg ﬂat, 1 0, a p harmonic function in Ω with u ≡ 0 on ∂Ω. Then there exists, δ0 > 0, c1 ≥ 1,

22

J. Lewis

depending only on p, n, such that if 0 < δ ≤ δ0 and x ∈ Ω, then u ∈ C ∞ (Ω) and (a) c−1 1 |∇u(x)| ≤ u(x)/d(x, ∂Ω) ≤ c1 |∇u(x)|, x ∈ Ω, (b) |∇u|p−2 extends to an A2 weight on Rn with constant ≤ c1 . An outline of the proof of Theorem 2.15 will be given in the next lecture. From Theorem 2.15 we see that (bik (x)) in (8), (9) are locally uniformly elliptic in Ω with ellipticity constants given in terms of an A2 weight on Rn . Thus Lemmas 1.6–1.8 can be used. To prove Theorem 2.14 we need a key lemma: Lemma 2.16. Let u, Ω, be as in Theorem 2.14 and p ≥ n. Then Lv ≥ 0 where L is as in (8), (9), and v = log |∇u|. Using Lemma 2.16 and Theorem 2.15 we can essentially repeat the proof in [48] which in turn was based on the a proof in [66]. The main diﬃculty involves showing that if Θ = {y ∈ ∂Ω : v(x)→ − ∞ as x→y nontangentially} then μ(Θ) = 0. To accomplish this we use some results on elliptic PDE whose degeneracy is given in terms of an A2 weight.

3.6

Open Problems for p Harmonic Measure

Note. In problems (1)–(8) the surrounding space is R2 . 1. Can Theorem 2.6 for simply connected domains be generalized to: (a) μ is concentrated on a set of σ ﬁnite H 1 measure whenever p > 2. (b) If a = a(p) > 1 is large enough and 1 2 and Ω is any planar domain. For harmonic measure this result is in [39, 71]. 3. What is the exact value of H-dim μ for a given p when ∂Ω is the Van Koch snowﬂake and p = 2? 4. For a given p, what is the supremum (p < 2) or inﬁmum (p > 2) of H-dim μ taken over the class of quasi-circles and/or simply connected domains?. 5. Is H-dim μ continuous and/or decreasing as a function of p when ∂Ω is the Van Koch snowﬂake? Regarding this question, the proof of Theorem 2.1 gives that H-dim μ = 1 + O(|p − 2|) as p→2 for a snowﬂake domain. 6. Are the p harmonic measures deﬁned on each side of a snowﬂake mutually singular? The answer is yes when p = 2 as shown in [9].

Applications of Boundary Harnack Inequalities for p Harmonic Functions

23

7. Is it always true for 1 < p < ∞ that H-dim μ < Hausdorﬀ dimension of ∂Ω when ∂Ω is a snowﬂake or a self similar Cantor set? The answer is yes when p = 2 for the snowﬂake as shown in [40]. The answer is also yes for self similar Cantor sets when p = 2. This question and continuity questions for H-dim ω on certain four cornered Cantor sets are answered by Batakis in [3–5]. 8. We noted in Remark 2 that H-dim μ was independent of the choice of u vanishing on ∂Ω. However in more general scenarios we do not know whether H-dim μ is independent of u. For example, suppose x0 ∈ ∂Ω and u > 0 is p harmonic in Ω∩B(x0 , r) with u = 0 on ∂Ω∩B(x0 , r) in the W 1,p sense. If ∂Ω ∩B(x0 , r) has positive p capacity, then there exists a measure μ satisfying (15) with φ ∈ C0∞ (N ) replaced by φ ∈ C0∞ (B(x0 , r)). Is H-dim μ|B(x0 ,r/2) independent of u? If Ω is simply connected and p = 2, then I believe the answer to this question is yes. In general this problem appears to be linked with boundary Harnack inequalities. 9. Is it true for p ≥ n that H-dim μ ≤ 1 whenever Ω ⊂ Rn ? If not is there a more general class of domains than Reifenberg ﬂat domains (see Theorem 2.14) for which this inequality holds? Compare with Problem 2. 10. What can be said about the dimension of Wolﬀ snowﬂakes? Regarding this question it appears that we can perturb oﬀ the p = 2 case (see [61]) in order to construct Wolﬀ snowﬂakes for 0 < |p − 2| < , > 0 small, for which the H-dim of the corresponding p harmonic measures on both sides of the snowﬂake are < n − 1 and also examples for which the H-dim of these measures are > n − 1. 11. What can be said for the dimension of p harmonic measure, p > 3 − log 4/ log 3, or even just harmonic measure in Ω = R3 \ J where J is the Van Koch snowﬂake? 12. The existence of a measure μ, corresponding to a weak solution u with vanishing boundary values, as in (2), exists for a large class of divergence form partial diﬀerential equations. What can be said about analogues of Theorems 2.1–2.3 or Theorems 2.6, 2.8, 2.14, 2.15 for the measures corresponding to these solutions? What can be said about analogues of problems (1)–(11)?

4

Boundary Harnack Inequalities and the Martin Boundary Problem for p Harmonic Functions

Recall from Section 3 the deﬁnition of nontangentially accessible and Reifenberg ﬂat domains. We note that a Lipschitz domain (i.e., a domain which is locally the graph of a Lipschitz function) is NTA. Also a Reifenberg ﬂat domain need not have a rectiﬁable boundary or tangent planes in the geometric measure sense anywhere (e.g., the Van Koch snowﬂake, in two dimensions). In [50] we prove

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J. Lewis

Theorem 3.1. Let Ω ⊂ Rn be a (δ, r0 )-Reifenberg ﬂat domain. Suppose that u, v are positive p-harmonic functions in Ω ∩ B(w, 4r), that u, v are ¯ ∩ B(w, 4r) and u = 0 = v on ∂Ω ∩ B(w, 4r). There exists continuous in Ω ˜ δ, σ > 0 and c1 ≥ 1, all depending only on p, n, such that

σ

log u(y1 ) − log u(y2 ) ≤ c1 |y1 − y2 |

v(y1 ) v(y2 )

r ˜ whenever y1 , y2 ∈ Ω ∩ B(w, r/c1 ). Here w ∈ ∂Ω, 0 < r < r0 , and 0 < δ < δ. Observe that the last display in Theorem 3.1 is equivalent to:

σ

u(y1 ) u(y1 ) |y1 − y2 | u(y2 )

v(y1 ) − v(y2 ) ≤ c1 v(y1 ) r whenever y1 , y2 ∈ Ω ∩ B(w, r/c1 ) so Theorem 3.1 is a boundary Harnack inequality for positive p harmonic functions vanishing on a portion of a suﬃciently ﬂat Reifenberg domain.

4.1

History of Theorem 3.1

The term boundary Harnack inequality for harmonic functions was ﬁrst introduced by Kemper in [41]. He attempted to show the ratio of two positive harmonic functions vanishing on a portion of a Lipschitz domain was bounded. Unfortunately Kemper’s proof was not correct, as Brelot later pointed out. This inequality for harmonic functions in Lipschitz domains was later proved independently and at about the same time in [2, 17, 73]. Jerison and Kenig in [37] proved Theorem 3.1 for NTA domains. Moreover boundary Harnack inequalities for solutions to linear divergence form uniformly elliptic PDE are proved in [14] while these inequalities for degenerate linear divergence form elliptic PDE whose degeneracy is speciﬁed in terms of an A2 weight were proved by [21–23], as mentioned earlier. Theorem 3.1 will be proved in the following steps: Step 1:

We prove Theorem 3.1 for Q = {x : |xi | < 1, 1 ≤ i ≤ n − 1, 0 < xn < 2}.

Step 2: (The ‘fundamental inequality’ for |∇u|) In this step, for u as in ¯ = λ(p, ¯ n), such that if Theorem 3.1, we show there exist cˆ = cˆ(p, n) and λ 0 < δ ≤ δ1 , then

Applications of Boundary Harnack Inequalities for p Harmonic Functions

¯−1 λ

u(y) ¯ u(y) ≤ |∇u(y)| ≤ λ d(y, ∂Ω) d(y, ∂Ω)

25

(45)

whenever y ∈ Ω ∩ B(w, r/ˆ c). Step 3: In this step we show that |∇u|p−2 extends to an A2 -weight locally with constants depending only on p, n (provided δ is small enough). Step 4: (Deformation of p harmonic functions). Let u, v be as in Theorem 3.1, r∗ = r/c , c large and for 0 ≤ τ ≤ 1, let u ˜(·, τ ) be the p harmonic function in Ω ∩ B(w, 4r∗ ) with continuous boundary values, u ˜(y, τ ) = τ v(y) + (1 − τ )u(y) whenever y ∈ ∂(Ω ∩ B(w, 4r∗ )) and τ ∈ [0, 1]. To simplify matters assume that ¯ ∩ B(w, 4r∗ ), 0 ≤ u ≤ v/2 and v ≤ c in Ω (46) where c, as in the rest of this lecture, may depend only on p, n. Then from the maximum principle for p harmonic functions (Lemma 1.1) we have 0≤

u ˜(·, τ2 ) − u˜(·, τ1 ) ≤c τ2 − τ1

(47)

˜(·, τ ) has whenever 0 ≤ τ1 , τ2 ≤ 1. Proceeding operationally we note that if u partial derivatives with respect to τ on B(w, 4r∗ ), then diﬀerentiating the p Laplace equation: ∇ · (|∇˜ u(x)|p−2 ∇˜ u(x, τ )) = 0 with respect to τ one ﬁnds that u ˜τ (x, τ ) is a solution for x ∈ Ω ∩ B(w, 4r∗ ) to the PDE ˜ = Lζ

n i,j=1

∂ ˜ ( bij (x, τ )ζxj (x, τ ) ) = 0 ∂xi

(48)

˜(·, τ ). Thus, u ˜(·, τ ) and where (˜bij )(·, τ ) are deﬁned as in (9) relative to u u ˜τ (·, τ ), both satisfy the same PDE. Now log

u ˜(x, 1) v(x) = log = u(x) u ˜(x, 0)

0

1

u ˜τ (x, τ ) dτ u ˜(x, τ )

(49)

when x ∈ Ω ∩ B(w, 4r∗ ). Observe also from (47) that u ˜τ ≥ 0 with continuous boundary value zero on ∂Ω ∩ B(w, 4r∗ ). From this fact, (49), we see that to prove Theorem 3.1 it suﬃces to prove boundary Harnack inequalities for the PDE in (48) with constants independent of τ ∈ [0, 1]. Moreover, from Steps 1–2, we see that u(·, τ ) is a solution to a uniformly elliptic PDE whose degeneracy is given in terms of an A2 weight with A2 constant independent of τ. Thus Lemma 1.8 can be applied to u ˜τ , u ˜ in order to conclude Theorem 3.1.

26

4.2

J. Lewis

Proof of Step 1

Step 1 is stated formally as Lemma 3.2. Let Q = {x : |xi | < 1, 1 ≤ i ≤ n − 1, 0 < xn < 2}. ¯ with Given p, 1 0 be p harmonic in Q, continuous in Q u ≡ v ≡ 0 on ∂Q ∩ {x : xn = 0}. Then for some c = c(p, n) ≥ 1,

log u(z) − log u(y) ≤ c|z − y|σ

v(z) v(y)

whenever z, y ∈ Q ∩ B(0, 1/16), where σ is the exponent in Lemma 1.3. Proof. To begin the proof of Lemma 3.2, observe that xn is p harmonic and vanishes when xn = 0. Thus from the triangle inequality, it suﬃces to prove Lemma 3.2 when v = xn . To prove that u/v is bounded in Q ∩ B(0, 1/2) we use barrier estimates. Given x ∈ B(0, 1/2) with xn ≤ 1/100 let x ˆ = (x , 1/8) ¯ x, 1/100) with continuous and let f be p harmonic in D = B(ˆ x, 1/8) \ B(ˆ boundary values, f = u(en /4) on ∂B(ˆ x, 1/100) while f ≡ 0 on ∂B(ˆ x, 1/8). From Lemma 1.1 we see that for some a, b, f (x) =

a|x − x ˆ|(p−n)/(p−1) + b, p = n, a ln |x − x ˆ| + b, p = n,

(50)

Also f ≤ cu in D thanks to Harnack’s inequality in Lemma 1.2. Using these facts and (50) it follows from direct calculation that cu(en /4) ≤

u(x) xn

(51)

for some c = c(p, n) when xn ∈ Q∩B(0, 1/2) and xn ≤ 1/100. From Harnack’s inequality we see that this inequality holds on Q ∩ B(0, 1/2). Next we extend u to Q = {x : |xi | < 1, 1 ≤ i ≤ n − 1, |xn | < 2} by putting u(x , xn ) = −u(x , −xn ) when xn < 0 (Schwarz Reﬂection). It is easily shown that u is p harmonic in Q’. We can now use Lemmas 1.3 and 1.5 for u to deduce for x ∈ Q ∩ B(0, 1/8), that |∇u(x)| ≤ c max u. ≤ c2 u(en /4).

(52)

B(0,1/4)

From (52) and the mean value theorem we get cu(en /4) ≥

u(x) xn

(53)

Applications of Boundary Harnack Inequalities for p Harmonic Functions

27

when x ∈ Q ∩ B(0, 1/8). Combining (53), (51), we obtain c−1 u(en /4) ≤ u(x)/xn ≤ cu(en /4).

(54)

H¨older continuity of the above ratio follows from (54) and Lemma 1.3. We omit the details.

4.3

Proof of Step 2

In the proof of (45) we shall need the following comparison lemma. Lemma 3.3. Let O be an open set, w ∈ ∂O, r > 0, and suppose that u ˆ, vˆ are positive p harmonic functions in O ∩ B(w, 4r). Let a ≥ 1, x ∈ O, and suppose that vˆ(x) vˆ(x) a−1 ≤ |∇ˆ v (x)| ≤ a . d(x, ∂Ω) d(x, ∂Ω) If ˜ −1 = (ca)(1+σ)/σ , where σ is as in Lemma 1.3, then for c = c(p, n) suitably large, the following statement is true. If ˆ ˆ ≤ vˆ ≤ (1 + ˜)L (1 − ˜)L u ˆ ˆ 0
max

|∇ˆ u(·)|

B(x,td(x,∂O))

(55) 2 σ

≤ c t u ˆ(x)/d(x, ∂O). Here c depends only on p, n. From (55) we conclude that we only have to prove bounds from below for the gradient of u ˆ at x. To do this, suppose for some small ζ > 0 (to be chosen) that, |∇ˆ u(x)| ≤ ζ u ˆ(x)/d(x, ∂O).

(56)

28

J. Lewis

From (55) with z = z1 , x = z2 and (56) we deduce |∇ˆ u(z)| ≤ [ζ + c2 tσ ] u ˆ(x)/d(x, ∂O)

(57)

for z ∈ B(x, td(x, ∂O)). Integrating, it follows that if y ∈ ∂B(x, td(x, ∂O)), with |x − y| = td(x, ∂O), t = ζ 1/σ , then |ˆ u(y) − u ˆ(x)| ≤ c ζ 1+1/σ u ˆ(x).

(58)

Constants in (57), (58) depend only on p, n. On the other hand (55) also ∇ˆ v (x) holds with u ˆ replaced by vˆ. Let λ = |∇ˆ ˆ and the v (x)| . Then from (55) for v non-degeneracy assumption on |∇ˆ v | in Lemma 3.3, we ﬁnd ¯ ζ 1/σ d(x, ∂O)), ∇ˆ v (z), λ ≥ (1 − c aζ)|∇ˆ v (x)| in B(x, where c = c(p, n). If ζ ≤ (2ca)−1 , where c is the constant in the above display, then we can integrate, to get for y = x + ζ 1/σ d(x, ∂O)λ, that v (y) − vˆ(x)) ≥ a−1 ζ 1/σ vˆ(x) c∗ (ˆ

(59)

with a constant c∗ depending only on p, n. From (59), (58), we see that if ˜ is as in Lemma 3.3, then u ˆ(y) 1 + c ζ 1+1/σ u ˆ(x) ≤ vˆ(y) 1 + ζ 1/σ /(ac∗ ) vˆ(x) 1 + c ζ 1+1/σ ˆ < (1 − ˜)L ˆ L ≤ (1 + ˜) 1 + ζ 1/σ /(ac∗ )

ˆ≤ (1 − ˜ )L

(60)

provided 1/(a˜ c)1/σ ≥ ζ 1/σ ≥ a˜ c˜ for some large c˜ = c˜(p, n). This inequality and (59) are satisﬁed if ˜−1 = (˜ ca)(1+σ)/σ and ζ −1 = c˜a. Moreover, if the hypotheses of Lemma 3.3 hold for this ˜, then in order to avoid the contradiction in (60) it must be true that (56) is false for this choice of ζ. Hence Lemma 3.3 is true. As an application of Lemmas 3.2 and 3.3 we note that if u ˆ is p harmonic in D = B(ζ, ρ) ∩ {y : yn > ζn } and u ˆ has continuous boundary value zero on ∂D ∩ {y : yn = ζn }, then there exists c = c(p, n) ≥ 1 such that c−1

u ˆ(x) u ˆ(x) ≤ |∇ˆ u(x)| ≤ c d(x, ∂D) d(x, ∂D)

(61)

˜ = uˆ(z)/ˆ in D ∩ B(ζ, ρ/c). Indeed, let vˆ = xn − ζn and put L v(z) where z is a ﬁxed point in D ∩ B(ζ, ρ/c+ ). Using Lemma 3.2 we see that

ˆ(y)

L ˜−u ˜ ≤ c c−σ + L, when y ∈ D ∩ B(ζ, ρ/c+ ).

vˆ(y)

(62)

Applications of Boundary Harnack Inequalities for p Harmonic Functions

29

Choosing c+ large enough we deduce that Lemma 3.2 applies, so (61) is true. Equation (61) could also be proved more or less directly using barrier arguments, Schwarz reﬂection, and Lemma 1.3. Next we use Lemma 3.3 to get the nondegeneracy property in (45). We restate this property as, Lemma 3.4. Let Ω be (δ, r0 ) Reifenberg ﬂat. Let u > 0 be p harmonic in Ω ∩B(w, 2r), continuous in B(w, 2r) with u ≡ 0 in B(w, 2r)\ Ω. There exists δ ∗ = δ ∗ (p, n), c1 = c1 (p, n), such that if 0 < δ ≤ δ ∗ and y ∈ Ω ∩ B(w, r/c1 ), then u(y) u(y) ≤ |∇u(y)| ≤ c1 . c−1 1 d(y, ∂Ω) d(y, ∂Ω) Proof. Let c∗ be the constant in (61) and choose c ≥ 1000c∗ so that if x ∈ Ω ∩ B(w, r/c ), s = 4c∗ d(x, ∂Ω), and z ∈ ∂Ω with !x − z| = d(x, ∂Ω), then max u ≤ cu(x) (63) B(z,4s)

for some c = c(p, n), which is possible thanks to Lemma 1.5. From Deﬁnition B with w, r replaced by z, 4s, there exists a plane P = P (z, 4s) with Ψ (∂Ω ∩ B(z, 4s), P ∩ B(z, 4s)) ≤ 4δs. Since the p Laplacian is invariant under rotations, and translations, we may assume that z = 0 and P = {y : yn = 0}, Also, if G = {y ∈ B(0, 2s) : yn > 8δs} then we may assume G ⊂ Ω ∩ B(0, 2s). Let v be the p harmonic function in G with boundary values in the Sobolev sense as follows: v ≤ u on ∂G and v(y) = u(y) when y ∈ ∂G and yn > 32δs, yn − 1) u(y) when y ∈ ∂G and 16δs < yn ≤ 32δs . v(y) = ( 16δs v(y) = 0 when y ∈ ∂G and yn < 16δs . From Lemma 1.1 (i.e., the maximum principle for p harmonic functions) we have v ≤ u in G. Also, since each point of ∂G where u(x) = v(x) lies within 100δs of a point where u is zero, it follows from (63) and Lemma 1.4 that u ≤ v + cδ β u(x) on ∂G. Using Lemma 1.1 we conclude that v ≤ u ≤ v + cδ α u(x) in G.

30

J. Lewis

Thus, 1≤

u(y) ≤ (1 − cδ α )−1 , y ∈ B(x, d(x, ∂G)/4). v(y)

(64)

From (64) and (61) with u ˆ replaced by v, we conclude that the hypotheses of Lemma 3.3 are satisﬁed with O = G. Applying Lemma 3.3 and using d(x, ∂G) ≈ d(x, ∂Ω), we obtain Lemma 3.4.

4.4

Proof of Step 3

In the proof of Step 3 we shall need the following lemma. Lemma 3.5. Let Ω ⊂ Rn be (δ, r0 ) Reifenberg ﬂat. Let w ∈ ∂Ω, 0 < r < r0 , and suppose u > 0 is p harmonic in Ω ∩ B(w, 2r), continuous in B(w, 2r), ˆ n, ) > 0 and with u ≡ 0 on B(w, 2r) \ Ω. Given > 0, there exist δˆ = δ(p, c = c(p, n, ), 1 ≤ c < ∞, such that c−1

1− 1+ rˆ rˆ u(arˆ(w)) ≤c ≤ r u(ar (w)) r

whenever 0 < δ ≤ δˆ and 0 < rˆ < r/4. Proof. Lemma 3.5 can be proved by a barrier type argument, using barriers which vanish on the boundary of certain cones or by an iterative type argument using Reifenberg ﬂatness of ∂Ω. We omit the details. Lemma 3.6. Let Ω be (δ, r0 ) Reifenberg ﬂat. Let w ∈ ∂Ω, 0 < r < r0 , and let u > 0 be p harmonic in Ω ∩ B(w, 2r), continuous in B(w, 2r), with u ≡ 0 on B(w, 2r) \ Ω. There exists δ = δ (p, n), c = c(p, n, M ) ≥ 1 such that if 0 < δ < δ , and rˆ = r/c, then |∇u|p−2 extends to an A2 (B(w, rˆ)) weight with constant depending only on p, n. Proof. We use Lemma 3.5 to prove Lemma 3.6. Let {Qj (xj , rj )} be a Whitney ¯ into open cubes with center at xj and sidelength rj . decomposition of Rn \ Ω Then ¯ ¯ j , rj ) = Rn \ Ω ∪j Q(x Q(xj , rj ) ∩ Q(xi , ri ) = ∅ i = j 10−4n d(Qj , ∂Ω) ≤ rj ≤ 10−2n d(Qj , ∂Ω). Let rˆ = r/c2 , where c is so large that c−1

u(x) u(x) ≤ |∇u(x)| ≤ c d(x, ∂Ω) d(x, ∂Ω)

(65)

Applications of Boundary Harnack Inequalities for p Harmonic Functions

31

whenever x ∈ Ω ∩ B(w, cˆ r ). Existence of c follows from Lemma 3.4. We also ¯ j ⊂ B(w, cˆ assume c is so large that if Qj ∩ B(w, 4ˆ r ) = ∅, then Q r ) and there exists wj ∈ Ω ∩ B(w, cˆ r ) with |xj − wj | ≈ d(xj , ∂Ω) ≈ d(wj , ∂Ω).

(66)

Existence follows from the fact that Ω is an NTA domain. Let λ(x) = |∇u(x)|p−2 , x ∈ Ω ∩ B(w, 4ˆ r) r ). λ(x) = |∇u(wj )|p−2 , x ∈ Qj ∩ B(w, 4ˆ From (66) we see that λ(x) = λ(wj ) ≈ λ(z)

(67)

whenever x ∈ Qj and z ∈ B(wj , d(wj , ∂Ω)/2). To complete the proof of Lemma 3.6 we prove that λ satisﬁes the A2 condition given in Sect. 2.2. Let w ˜ ∈ B(w, rˆ) and 0 < r˜ < rˆ. We consider several cases. If r˜ < d(w, ˜ ∂Ω)/2, then the A2 condition follows from (65) and Harnack’s inequality. On the other hand, if r˜ ≥ d(w, ˜ ∂Ω)/2 then we choose z ∈ ∂Ω with d(w, ˜ ∂Ω) = |w ˜ − z| and thus B(w, ˜ r˜) ⊂ B(z, 3˜ r) ⊂ B(w, ˜ 8˜ r). First suppose p > 2. From H¨older’s inequality, Lemmas 1.2, 1.3, and (67) we see that λdx ≤ λdx ≤ c |∇u|p−2 dx B(w,˜ ˜ r)

Ω∩B(z,c∗ r˜)

B(z,3˜ r)

⎛ ⎜ ≤ c⎝

⎞(1−2/p)

⎟ |∇u|p dx⎠

r˜2n/p

(68)

Ω∩B(z,c∗ r˜)

≤ cu(ar˜(z))p−2 r˜n+2−p . Let η = min{1, |p − 2|−1 }/20. To estimate the integral in involving λ−1 observe from Lemma 3.5 and once again Harnack’s inequality, that if y ∈ Ω ∩ B(z, c∗ r˜), and δ is small, then cu(y) ≥ u(ar˜(z))

d(y, ∂Ω) r˜

1+η .

(69)

32

J. Lewis

Therefore, using (67) and (69) we obtain

λ−1 dx ≤ c r˜(1+η)(p−2) u(ar˜(z))2−p

B(w,˜ ˜ r)

(70)

−η(p−2)

×

d(y, ∂Ω)

dy.

Ω∩B(z,c∗ r˜)

To estimate the integral involving the distance function in (70) set I(z, s) =

d(y, ∂Ω)−η(p−2) dy

Ω∩B(z,s)

whenever z ∈ ∂Ω ∩ B(w, r), 0 < s < r. Let Ek = Ω ∩ B(z, s) ∩ {y : d(y, ∂Ω) ≤ δ k s} for k = 1, 2, . . . We claim that dy ≤ ck+ δ k sn for k = 1, 2, . . .

(71)

Ek

where c+ = c+ (p, n). Indeed, from δ Reifenberg ﬂatness it is easily seen that this statement holds for E1 . Moreover, E1 can be covered by at most c/δ n−1 balls of radius 100δs with centers in ∂Ω ∩ B(z, s). We can then repeat the argument in each ball to get that (71) holds for E2 . Continuing in this way we get (71) for all positive integers k. Using (71) and writing I(z, s) as a sum over Ek \ Ek+1 , k = 1, 2, . . . we get I(z, s) ≤ csn−η(p−2) + δ η(p−2)−n

∞ k=1

(ck+ δ k s)n−η(p−2) < c˜sn−η(p−2) ,

where c˜ = c˜(p, n), provided δ is small enough. Using this estimate with s = r˜, we can continue our calculation in (70) and conclude that

λ−1 dx ≤ c r˜n+p−2 u(ar˜(z))(2−p)

(72)

B(w,˜ ˜ r)

To complete the proof of Lemma 3.6 in the case p > 2, we simply combine (68) and (72). Note that the case p = 2 is trivial and in case p < 2 the argument above can be repeated with p − 2 replaced by 2 − p < p. Hence Lemma 3.6 and Step 3 are complete.

Applications of Boundary Harnack Inequalities for p Harmonic Functions

4.5

33

Proof of Step 4 and Theorem 3.1

To justify the claims in Step 4, ﬁrst choose c = c (p, n) so large that if r∗ = r/c , then max ∗ h ≤ ch(ar∗ (w)) (73) Ω∩B(w,4r )

for some c = c(p, n) whenever h > 0 is p harmonic in Ω ∩ B(w, 2r). This choice is possible as we see from Lemma 1.5 and Harnack’s inequality in Lemma 1.2. We also suppose c > 1000c1 , where c1 is the constant in Lemma 3.4. Hence from Lemma 3.4 we have c−1

k(y) k(y) ≤ |∇k(y)| ≤ c . d(y, ∂Ω) d(y, ∂Ω)

(74)

whenever y ∈ Ω ∩ B(w, r∗ /c ) and k > 0 is p harmonic in Ω ∩ B(w, 2r∗ ), continuous in B(w, 2r∗ ), with k ≡ 0 on B(w, 2r∗ )\Ω. We temporarily assume that 0 < u ≤ v/2 ≤ c on Ω ∩ B(w, 4r∗ ). (75) and also that

c−1 ≤ u(ar∗ (w)) .

(76) ∗

Next if t ∈ [0, 1], let u ˜(·, t) be the p harmonic function in Ω ∩ B(w, 2r ) with continuous boundary values, u ˜(·, t) = (1 − t)u + tv

(77)

on ∂[Ω ∩ B(w, 2r∗ )]. Extend u ˜(·, t), t ∈ (0, 1), to be H¨older continuous in B(w, 2r∗ ) by putting u˜(·, t) ≡ 0 on B(w, 2r∗ ) \ Ω. Let r = r∗ /c and observe that (74) holds whenever k = u(·, t), t ∈ [0, 1], on Ω ∩ B(w, r ). Thus from Lemma 1.3, we see that u˜(·, t) is inﬁnitely diﬀerentiable in B(w, r ) and so u(·, t) = 0 ∇ ·|∇˜ u(·, t)|p−2 ∇˜

(78)

in Ω ∩ B(w, r ). Set U (x) = U (x, t, τ ) =

u ˜(x, t) − u ˜(x, τ ) . t−τ

and note from (75), (76), for ﬁxed t, τ ∈ [0, 1], t = τ, that 0 ≤ v/2 ≤ U (x) = v − u ≤ v ≤ c(p, n)

(79)

on ∂(Ω ∩ B(w, 2r∗ )), so by Lemma 1.1 we have 0 ≤ U ≤ c in Ω ∩ B(w, r ), U ≡ 0 on ∂Ω ∩ B(w, r ).

(80)

34

J. Lewis

From (80) we see for ﬁxed x ∈ Ω ∩ B(w, 2r∗ ) that t→˜ u(x, t), is Lipschitz with norm ≤ c. Thus u ˜τ (x, ·) exists almost everywhere in [0,1]. Let (xν ) be a dense sequence of Ω ∩B(w, 2r∗ ) and let W be the set of all t ∈ [0, 1] for which u ˜t (xm , ·) exists, in the sense of diﬀerence quotients, whenever xm ∈ (xν ). We note that H 1 ([0, 1]\ W ) = 0 where H 1 is one-dimensional Hausdorﬀ measure. Next we note from (79) that for t ∈ (0, 1] u ˜(·, t)/2 ≤ U (·, t, τ ) ≤ t−1 u ˜(·, t)

(81)

on ∂Ω ∩ B(w, 2r∗ ), so by Lemma 1.1, this inequality also holds in Ω ∩ B(w, 2r∗ ). To ﬁnd a divergence form PDE that U satisﬁes let ξ = (ξ1 , . . . , ξn ), w = (w1 , . . . , wn ) ∈ Rn \ {0}, and 1 ≤ i ≤ n. Then |ξ|p−2 ξi − |w|p−2 wi

1

= 0

=

n

d dλ {|λξ

+ (1 − λ)w|p−2 [λξi + (1 − λ)wi ]}dλ

(ξ − w)j

j=1

1

0

aij [λξ + (1 − λ)w]dλ ,

where for 1 ≤ i, j ≤ n, η ∈ Rn \ {0}, aij (η) = |η|p−4 [(p − 2)ηi ηj + δij |η|2 ].

(82)

In this display δij , once again, denotes the Kronecker delta. Using (78), (80)– (82) we ﬁnd for ﬁxed t, τ that if Aij (x) = Aij (x, t, τ )

1

= 0

aij [λ∇˜ u(x, t) + (1 − λ)∇˜ u(x, τ )] dλ,

then, for x ∈ Ω ∩ B(w, r ), t, τ ∈ [0, 1], ˜ (x) = LU

n ∂ [Aij (x) Uxj ] = 0. ∂xi i,j=1

(83)

Moreover, if x ∈ Ω ∩ B(w, r ), then c−1 |ξ|2 ||∇˜ u(x, t)| + |∇˜ u(x, τ )||p−2 n ≤ Aij (x)ξi ξj whenever ξ ∈ Rn \ {0}. i,j=1

(84)

Applications of Boundary Harnack Inequalities for p Harmonic Functions

Also,

n

|Aij (x)| ≤ c||∇˜ u(x, t)| + |∇˜ u(x, τ )||p−2 ,

35

(85)

i,j=1

where c depends only on p, n. From Lemma 1.3, (82)–(85), and (74) for u ˜(·, t), u ˜(·, τ ), we see that U is a solution on Ω∩B(w, r ) to a locally uniformly elliptic divergence form PDE with C ∞ coeﬃcients. Since u ˜(x, t) − u ˜(x, τ ) = (t−τ )(v(x)−u(x)) on ∂(Ω ∩B(w, 2r)) it follows from the maximum principle for p harmonic functions that u˜(x, τ )→˜ u(x, t) uniformly in the closure of Ω∩B(x, 2r∗ ) as τ →t. Also from Lemma 1.2 we deduce that ∇˜ u(·, τ )→∇˜ u(·, t) on compact subsets of Ω ∩ B(w, 2r∗ ). Using these facts and (74) we see for 1 ≤ i, j ≤ n, Ai,j (x, t, τ )→˜bij (x) (86) as τ →t uniformly on compact subsets of Ω ∩ B(w, r ), where ˜bij are deﬁned as in (9) relative to u ˜(·, t). Finally we note from linear elliptic PDE theory that U (·, t, τ ) is locally in W 1,2 and locally H¨ older continuous on Ω ∩B(w, r ) with norms independent of τ. From (81) we see that U (·, t, τ ), t ∈ (0, 1], has a H¨ older continuous extension to B(w, r ) obtained by putting U ≡ 0 in B(w, r ) \ Ω. Also H¨older constants can be chosen independent of τ for τ near t. Using these facts we see for ﬁxed t ∈ (0, 1], that there is a sequence U (·, t, τk )→f (·, t) on Ω ∩ B(w, r ) as τk →t. Put f ≡ 0 on B(w, r ) \ Ω. Then from (80)–(86), Lemma 1.2, and Schauder estimates, we conclude that f has the following properties: (a) (b) (c) (d) (e)

˜ = 0 in Ω ∩ B(w, r ) where L ˜ is as in (48), Lf f is continuous in B(w, r ) with f ≡ 0 on B(w, r ) \ Ω, ˜t (xm , t) when xm ∈ (xν ), t ∈ W, f (xm , t) = u u/2 ≤ u ˜(·, t)/2 ≤ f (·, t) ≤ c in Ω ∩ B(w, r ), f ∈ C ∞ [Ω ∩ B(w, r )].

(87)

From (87) (c) we get ln

v(xm ) u(xm )

= ln

u ˜(xm , 1) u ˜(xm , 0)

= 0

1

f (xm , t) dt u ˜(xm , t)

(88)

whenever xm ∈ (xν ). Since this sequence is dense in Ω ∩ B(w, r ), we conclude from (87), (88) that Claims (48), (89) are true. From (88) and Lemmas 3.4, 3.6 we see that u ˜τ (·, τ ), u ˜(·, τ ) are solutions to a degenerate divergence form elliptic PDE whose degeneracy is given in terms of an A2 weight. Thus Lemma 1.8 can be used with r replaced by r˜ = min(r , rˆ), v1 = f (·, t), v2 = u˜(·, t), where f is as in (87), (88). Let r = r˜/c where c is the constant in Lemma 1.8. From (88)(d), (76), and Harnack’s inequality we get

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J. Lewis

c−1 ≤

f (ar (w), t) ≤c u ˜(ar (w), t)

(89)

where c depends only on p, n. Using this fact and Lemma 1.8, we ﬁnd that f (y, t) ≤ c. u ˜(y, t) whenever y ∈ Ω ∩ B(w, r ) and thereupon

α

f (z, t) f (y, t)

|z − y|

− ≤c

u ˜(z, t) u ˜(y, t)

r

(90)

whenever y, z ∈ Ω ∩ B(w, r ). Hence,

log v(y) − log v(z)

u(y) u(z)

≤

0

1

α

f (z, t) f (y, t)

dt ≤ c |z − y| .

−

u ˜(z, t) u ˜(y, t)

r

(91)

From (91) we conclude that Theorem 3.1 is valid under assumptions (75), (76). To remove (75), (76), suppose u, v are as in Theorem 1.3. We assume as we may that 1 = u(ar∗ (w)) = v(ar∗ (w)) (92) since otherwise we divide u, v by their values at this point and use invariance of the p Laplacian under scaling. Let u˜, v˜ be p harmonic functions in Ω ∩ ¯ B(w, 2r∗ ), continuous in B(w, 2r∗ ) with u ˜ = min(u, v), v˜ = 2 max(u, v) on ∂[B(w, 2r∗ ) \ Ω]. Let r∗∗ = r∗ /c. From Lemmas 1.1, 1.5, and (92) we see that u ˜ ≤ min(u, v) ≤ max(u, v) ≤ v˜/2 ≤ c in Ω ∩ B(w, 2r∗∗ ). Also, using (92), Harnack’s inequality, and Lemmas 1.4, 1.5 we see that (76) is valid with r∗ replaced by r∗∗ . Hence (75), (76), are valid with r∗ , replaced by r∗∗ so Theorem 3.1 can be applied to get max[v/u, u/v] ≤ v˜/˜ u≤c

(93)

in Ω ∩B(w, r ). It follows for suitably chosen c, that u, cv satisfy (75), (76) in Ω ∩ B(w, r ). Consequently Theorem 3.1 is valid without assumptions (75), (76).

Applications of Boundary Harnack Inequalities for p Harmonic Functions

4.6

37

More on Boundary Harnack Inequalities

In [62] we generalize Theorem 3.1 to weak solutions u of ∇ · A(x, ∇u(x)) = 0, where A = (A1 , ..., An ) : Rn × Rn → Rn . Also A = A(x, η) is continuous in Rn × (Rn \ {0}) and A(x, η), for ﬁxed x ∈ Rn , is continuously diﬀerentiable in ηk , for every k ∈ {1, ..., n}, whenever η ∈ Rn \ {0}. Moreover, (i) −1

α (ii)

|η|

p−2

n ∂Ai |ξ| ≤ (x, η)ξi ξj , ∂ηj i,j=1 2

∂Ai

≤ α|η|p−2 , 1 ≤ i, j ≤ n, (x, η)

∂ηj

(iii) |A(x, η) − A(y, η)| ≤ β|x − y|γ |η|p−1 , (iv) A(x, η) = |η|p−1 A(x, η/|η|). Under these assumptions, we prove Theorem 3.7. Let Ω ⊂ Rn be a (δ, r0 )-Reifenberg ﬂat domain. Suppose that u, v are positive A-harmonic functions in Ω ∩ B(w, 4r), that u, v are continuous in B(w, 4r) and u = 0 = v on B(w, 4r) \ Ω. For ﬁxed p, α, β, γ, ˜ σ > 0 and c1 ≥ 1, all depending only on p, n, α, β, γ such that there exist δ,

σ

log u(y1 ) − log u(y2 ) ≤ c1 |y1 − y2 |

v(y1 ) v(y2 )

r ˜ whenever y1 , y2 ∈ Ω ∩ B(w, r/c1 ). Here w ∈ ∂Ω, 0 < r < r0 , and 0 < δ < δ. For completeness we now give the deﬁnition of a Lipschitz domain. Definition C. Ω ⊂ Rn is said to be a bounded Lipschitz domain provided there exists a ﬁnite set of balls {B(xi , ri )}, with xi ∈ ∂Ω and ri > 0, such that {B(xi , ri )} constitutes a covering of an open neighbourhood of ∂Ω and such that, for each i, Ω ∩ B(xi , 4ri ) = {y = (y , yn ) ∈ Rn : yn > φi (y )} ∩ B(xi , 4ri ), ∂Ω ∩ B(xi , 4ri ) = {y = (y , yn ) ∈ Rn : yn = φi (y )} ∩ B(xi , 4ri ), in an appropriate coordinate system and for a Lipschitz function φi . The Lipschitz constants of Ω are deﬁned to be M = maxi |∇φi | ∞ and r0 = min ri .

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J. Lewis

In [52] we prove Theorem 3.8. Let Ω ⊂ Rn be a bounded Lipschitz domain with constant M . Given p, 1 < p < ∞, w ∈ ∂Ω, 0 < r < r0 , suppose that u and v are positive p harmonic functions in Ω ∩B(w, 2r). Assume also that u and v are continuous in B(w, 2r) and u = 0 = v on B(w, 2r) \ Ω. Under these assumptions there exist c1 , 1 ≤ c1 < ∞, and α, α ∈ (0, 1), both depending only on p, n, and M , such that if y1 , y2 ∈ Ω ∩ B(w, r/c2 ) then

α

log u(y1 ) − log u(y2 ) ≤ c1 |y1 − y2 | .

v(y1 ) v(y2 )

r Remark. To prove Theorem 3.7 we argue as in Steps 1–4. The techniques and arguments are similar to the proof of Theorem 3.1. In the proof of Theorem 3.8 we follow a somewhat similar game plan although in this case the techniques are necessarily more sophisticated. For example there is no readily available comparison function (xn in Theorems 3.1 or 3.7) from which we can extrapolate the fundamental inequality in Step 2. Thus we study p harmonic capacitary functions in starlike Lipschitz ring domains and show these functions satisfy the fundamental inequality near boundary points where they vanish. Also, in the Lipschitz case simple examples show |∇u|p−2 need not extend to an A2 weight locally, so that Lemma 1.8 cannot be applied. Instead for given w, r ∈ ∂Ω we show that |∇u|p−2 satisﬁes a Carleson measure condition in a certain starlike Lipschitz ˜ ⊂ Ω. This fact and a theorem in [42] (see also [35]) can then subdomain Ω be used to conclude that a boundary Harnack inequality (as in Lemma 1.8) holds for solutions to (8), (9) in the Lipschitz case. Finally we note that Theorems 3.7 and 3.8 are weaker than the corresponding results for p = 2. Thus for example does Theorem 3.8 hold in an NTA domain? Can the assumptions on A in Theorem 3.7 be weakened?

4.7

The Martin Boundary Problem

The Martin boundary for harmonic functions was ﬁrst introduced by Martin [67]. Over the years, it has been of considerable interest to researchers in potential theory. Unfortunately Martin did not receive many accolades for his contribution, as according to math. sci. net, the above is the only paper he ever wrote. In order to deﬁne the p Martin boundary of a NTA domain, we need to ﬁrst deﬁne a minimal positive p harmonic function. Definition D. Fix p, 1 0 is p harmonic in Ω and u ˆ has continuous boundary value 0 on ∂Ω \ {w}. u ˆ is said to be unique up to constant multiples if vˆ = λˆ u, for some constant λ,

Applications of Boundary Harnack Inequalities for p Harmonic Functions

39

whenever vˆ is a minimal positive p harmonic function relative to w ∈ ∂Ω. The p Martin boundary of Ω is equivalence classes of all minimal positive p harmonic functions deﬁned relative to boundary points of Ω. Two minimal positive p harmonic functions are in the same equivalence class if and only if they are constant multiples of each other. We say that the p Martin boundary of Ω can be identiﬁed with ∂Ω provided each w ∈ ∂Ω corresponds to a unique (up to constant multiples) minimal positive p harmonic function. Theorem 3.9. Let Ω, δ, p, r0 , be as in Theorem 3.1. There exists δ+ = δ+ (p, n) such that if 0 < δ < δ+ , then the p Martin boundary of Ω can be identiﬁed with ∂Ω. Remark. Theorem 3.9 for p = 2, i.e., harmonic functions, in an NTA domain G is an easy consequence of the boundary Harnack inequality for harmonic functions in NTA domains. Indeed, if w ∈ ∂G and if u, v are minimal harmonic functions corresponding to w, one ﬁrst uses the boundary Harnack inequality for harmonic functions to show that γ = inf G u/v > 0. Next one applies this result to u − γv, v in order to conclude that u = γv. Note however that this argument depends heavily on linearity of the Laplacian and thus the argument fails for the p Laplacian when p = 2. We also note that if r0 = ∞ in Theorem 3.1, i.e., Ω is an unbounded Reifenberg ﬂat NTA domain, and u, v are minimal positive p harmonic functions relative to ∞, then we can let r→∞ in the conclusion of Theorem 3.1, to get u = λv, λ = constant. To make this idea work when w ∈ ∂Ω \ {∞}, we need to prove an analogue of Theorem 3.1 for positive p harmonic functions in Ω \ B(w, r ), (r small) vanishing on ∂Ω \ B(w, r ). We could get this analogue, by arguing as in the proof of Theorem 3.1, except we do not know a priori, that our functions have the fundamental nondegeneracy property (45) in an appropriate domain. We overcome this deﬁciency, using arguments from the proof of Theorem 3.1, as well as an induction—bootstrap type argument. We start by showing that if one such p harmonic function has the fundamental nondegeneracy property then all such functions have this property. Lemma 3.10. Let Ω be (δ, r0 ) Reifenberg ﬂat. Let u ˆ, vˆ > 0 be p harmonic in Ω \ B(w, r ), continuous in Rn \ B(w, r ), with u ˆ ≡ vˆ ≡ 0 on Rn \ [Ω ∪ B(w, r )]. Suppose for some r1 , r < r1 < r0 , and A ≥ 1, that A−1

u ˆ(x) u ˆ(x) ≤ |∇ˆ u(x)| ≤ A . d(x, ∂Ω) d(x, ∂Ω)

whenever x ∈ Ω ∩[B(w, r1 )\B(w, r )]. There exists α > 0, λ, c ≥ 1, depending on p, n, A, such that if 0 < δ < δˆ (δˆ as in Theorem 3.1), λ−1

vˆ(x) vˆ(x) ≤ |∇ˆ v (x)| ≤ λ . d(x, ∂Ω) d(x, ∂Ω)

for x ∈ Ω ∩ [B(w, r1 /c) \ B(w, cr )]. Moreover,

40

J. Lewis

α

ˆ(z) u ˆ(y)

r

log u − log ≤c

vˆ(z) vˆ(y)

min(r1 , |z − w|, |z − y|) whenever z, y ∈ Ω \ B(w, cr ). Proof. We assume that r /r1 << 1, since otherwise there is nothing to prove. Let r˜ = cˆr . If cˆ = cˆ(p, n) is large enough, we may assume uˆ ≤ vˆ/2 ≤ cˆu in Ω \ B(w, r˜),

(94)

as we see from Theorem 3.1, Harnack’s inequality, and the maximum principle. As in Step 4 of Sect. 1, let u(·, t), t ∈ [0, 1], be p harmonic in Ω \ B(w, r˜), with continuous boundary values, u(·, t) = (1 − t)ˆ u + tˆ v on ∂[Ω \ B(w, r˜)].

(95)

Extend u(·, t), t ∈ [0, 1], to be continuous on Rn \ [Ω ∪ B(w, r˜)] by setting u(·, t) ≡ 0 on this set. Next we note from Lemma 3.3 that there exists 0 = 0 (p, n, M, A) such that if r˜ ≤ s1 < ρ1 /4 ≤ r1 /16, τ ∈ (0, 1], and ˜ ≤ u(·, τ )/ˆ ˜ (1 − 0 )L u ≤ (1 + 0 )L,

(96)

˜ then in Ω ∩ [B(w, 2ρ1 ) \ B(w, s1 )], for some L, ˆ u(x, τ ) ˆ −1 u(x, τ ) ≤ |∇u(x, τ )| ≤ λ λ d(x, ∂Ω) d(x, ∂Ω)

(97)

ˆ = λ(p, ˆ n, A). Observe from (94), (95), in Ω ∩ [B(w, ρ1 ) \ B(w, 2s1 )] where λ that if τ1 , τ2 ∈ [0, 1], u(·, τ2 ) − u(·, τ1 ) τ2 − τ1 = v − u ≤ c u(·, τ1 )

c−1 u(·, τ1 ) ≤ U (·, τ1 , τ2 ) =

(98)

on ∂[Ω \ B(w, r˜)], so from the maximum principle this inequality also holds in Ω \ B(w, r˜). Thus for 0 as in (96), there exists 0 , 0 < 0 ≤ 0 , with the same dependence as 0 , such that if |τ2 − τ1 | ≤ 0 , then 1 − 0 /2 ≤

u(·, τ2 ) ≤ 1 + 0 /2 in Ω \ B(w, r˜). u(·, τ1 )

(99)

Divide [0,1] into closed intervals, disjoint except for endpoints, of length 0 /2 except possibly for the interval containing 1 which is of length ≤ 0 /2. Let ξ1 = 0 < ξ2 < ... < ξm = 1 be the endpoints of these intervals. Thus [0,1] is divided into {[ξk , ξk+1 ]}m 1 . Next suppose for some l, 1 ≤ l ≤ m − 1, that (97)

Applications of Boundary Harnack Inequalities for p Harmonic Functions

41

is valid whenever τ ∈ [ξl , ξl+1 ] and x ∈ Ω ∩ [B(w, ρ1 ) \ B(w, 2s1 )]. Under this assumption we claim for some cˆ1 , cˆ2 , α, depending only on p, n, A that

α

s1

log u(z, ξl+1 ) − log u(y, ξl+1 ) ≤ cˆ1

u(z, ξl ) u(y, ξl )

min{|z − w|, |w − y|}

(100)

whenever z, y ∈ Ω ∩ [B(w, ρ1 /ˆ c2 ) \ B(w, cˆ2 s1 )]. Indeed we can retrace the argument in Theorem 3.1 to get for z, y ∈ Ω ∩ [B(w, ρ1 /c) \ B(w, cs1 )], that there exists f as in (87) with

log u(z, ξl+1 ) − log u(y, ξl+1 )

u(z, ξl ) u(y, ξl )

≤

ξl+1

ξl

α (101)

f (z, τ ) f (y, τ ) s 1

u(z, τ ) − u(y, τ ) dτ ≤ c min{|z − w|, |y − w|} .

The last inequality in (101) follows from a slightly more general version of Lemma 1.8. We now proceed by induction. Observe from (99) as well as u(·, ξ1 ) = u ˆ that (96) holds whenever τ ∈ [ξ1 , ξ2 ]. Thus (97) and consequently (100) are true for l = 1 with s1 = r˜, ρ1 = r1 /4. Let s2 = cˆ2 s1 , ρ2 = ρ1 /ˆ c2 . By induction, suppose for some 2 ≤ k < m,

α

sk

log u(z, ξk ) − log u(y, ξk ) ≤ (k − 1)ˆ c (102) 1

u ˆ(z) u ˆ(y)

min{|z − w|, |y − w|} whenever z, y ∈ Ω ∩ [B(w, ρk ) \ B(w, sk )], where α, cˆ1 are the constants in (100). Choose sk ≥ 2sk , so that

u(z, ξk ) u(y, ξk )

≤ η u(z, ξk )

−

u ˆ(z) u ˆ(y)

u ˆ(z) whenever z, y ∈ Ω ∩ [B(w, ρk ) \ B(w, sk )]. Fix z as above and choose η > 0 so small that (1 − 0 )

u(y, τ ) u(z, ξk ) u(z, ξk ) ≤ ≤ (1 + 0 ) u ˆ(z) u ˆ(y) u ˆ(z)

(103)

whenever y ∈ Ω ∩ [B(w, ρk ) \ B(w, sk )] and τ ∈ [ξk , ξk+1 ]. To see the size of η observe for τ ∈ [ξk , ξk+1 ] that u(y, τ ) u(y, ξk ) u(y, τ ) = · u ˆ(y) u(y, ξk ) u ˆ(y) u(z, ξk ) . ≤ (1 + 0 /2)(1 + η) u ˆ(z)

42

J. Lewis

Thus if η = 0 /4 (0 small), then the right hand inequality in (103) is valid. A similar argument gives the left hand inequality in (103) when η = 0 /4. Also since k ≤ 2/0 , and 0 , α depend only on p, n, M, A, we deduce from (102) that one can take sk = cˆ3 sk for cˆ3 = cˆ(p, n, M, A) large enough. From (103) ˜ = u(z,ξk ) in Ω ∩ [B(w, ρk ) \ B(w, s )]. From we ﬁnd that (96) holds with L k u ˆ(z) ρk (97) we now get that (100) is valid for l = k in Ω ∩ [B(w, 2ˆ ) \ B(w, cˆ2 sk )]. c2 ρk Let sk+1 = cˆ3 cˆ2 sk and ρk+1 = 2ˆc2 . Using (100) and the induction hypothesis we have

log u(z, ξk+1 ) − log u(y, ξk+1 ) ≤ log u(z, ξk+1 ) − log u(y, ξk+1 )

u ˆ(z) u ˆ(y)

u(z, ξk ) u(y, ξk )

α (104)

) ) u(z, ξ u(y, ξ s k k

k+1 +

log − log ≤ kˆ c1 u ˆ(z) u ˆ(y)

min{|z − w|, |w − y| whenever z, y ∈ Ω ∩ [B(w, ρk+1 ) \ B(w, sk+1 )]. Thus by induction we get (102) with k = m. Since u(·, ξm ) = vˆ and sm ≤ cr , ρm ≥ r1 /c, for some large c = c(p, n, A), we can now argue as in (103) to ﬁrst get (96) with u(·, τ ) replaced by vˆ and then (97) for vˆ. We conclude that Lemma 3.10 is valid for z, y ∈ Ω ∩ [B(w, r1 /c) \ B(w, cr )] provided c is large enough. Using the maximum principle it follows that the last display in Lemma 3.49 is also valid for z, y ∈ Ω \ B(w, r1 /c).

4.8

Proof of Theorem 3.9

Note that if u ˆ is a minimal p harmonic function satisfying (45) in Ω∩B(w, r1 ), then one can let r →0 in Lemma 3.10 to get Theorem 3.9. Thus to complete the proof of Theorem 3.9 it suﬃces to show the existence of a minimal positive p harmonic function u ˆ relative to w ∈ ∂Ω and 0 < r1 < r0 for which the fundamental nondegeneracy property in (45) holds in Ω ∩ B(w, r1 ). To this end we introduce, ˜ ⊂ Ω a non tangential approach region at w ∈ ∂Ω Definition E. We call Ω ˜ if for some η˜ > 0, d(x, ∂Ω) ≥ η˜|x − w| for all x ∈ Ω. ˜ If we wish to emphasize w, η˜ in Deﬁnition E we write Ω(w, η˜). Now let u ˆ be a minimal positive p harmonic function in Ω relative to w ∈ ∂Ω and 0 < δ < δ ∗ . Then we can apply Lemma 3.3 to conclude for each xˆ ∈ ∂Ω \ {w} that for some c˜ = c˜(p, n) ≥ 1, c˜−1

u ˆ(x) u ˆ(x) ≤ |∇ˆ u(x)| ≤ c˜ d(x, ∂Ω) d(x, ∂Ω)

(105)

Applications of Boundary Harnack Inequalities for p Harmonic Functions

43

whenever x ∈ ∂Ω ∩ B(ˆ x, |ˆ x − w|/˜ c) ∩ B(w, r0 ). Using this fact we see that if 0 < δ+ < δ ∗ in Theorem 3.9 then there exists η˜ depending only on p, n, such that ˜ u ˆ satisﬁes (105) in [Ω \ Ω(w, η˜)] ∩ B(w, r0 ). (106) From (106) we see that if (105) holds in Ω(w, η˜) ∩ B(w, r1 ), then uˆ can be used in Lemma 3.10 for each small r > 0 so Theorem 3.9 is true. To prove the above statement, we ﬁrst extend u ˆ continuously to Rn \ {w} by letting u ˆ=0 in Rn \ Ω. Let 0 < r < r0 /n and σ = 100nδ. Using translation, rotation invariance of the p Laplacian and Reifenberg ﬂatness in Deﬁnition B, we assume as we may that w = 0 and for given σ > 0 (suﬃciently small) that B(0, nr) ∩ {y : yn ≥ σr} ⊂ Ω (107) B(0, nr) ∩ {y : yn ≤ −σr} ⊂ Rn \ Ω Extend u ˆ to a continuous function in Rn \ {0} by putting u ˆ ≡ 0 on Rn \ (Ω ∪ {0}). Let √ Q = {y : |yi | < r, 1 ≤ i ≤ n − 1} ∩ {y : σr < yn < r} \ B(0, σ r ) and let v1 be the p harmonic function in Q with the following continuous boundary values, v1 (y) = uˆ(y) if y ∈ ∂Q ∩ {y : 2σr ≤ yn }, v1 (y) =

(yn − σr) u ˆ(y) if y ∈ ∂Q ∩ {y : σr ≤ yn < 2σr} . σr

Comparing boundary values and using the maximum principle for p harmonic functions, we deduce v1 ≤ u ˆ in Q. (108) Let σ() = exp(−1/). To complete the proof of Theorem 3.9 we will make use of the following lemmas. Lemma 3.11. Let 0 < ≤ ˆ, let σ = σ() be as above and let η˜ be as in ˆ n), 0 < θˆ ≤ 1/4, such (106). If ˆ is small enough, then there exists θˆ = θ(p, 1/2−θˆ that if ρˆ = σ r, then 1 ≤ u ˆ(y)/v1 (y) ≤ 1 + ˜ η˜/16) ∩ [B(0, ρˆ) \ B(0, 4√σr)]. whenever y ∈ Ω(0, ˆ r, σ, be as in Lemma 3.11 and let η˜ be as in Lemma 3.12. Let v1 , , ˆ, θ, ˆ (106). If ˆ is small enough, then there exist θ = θ(p, n), 0 < θ ≤ θ/10, and 1/2−4θ −θ c = c(p, n) > 1 such that if ρ = σ r, b = σ , then

44

J. Lewis

c−1

v1 (x) v1 (x) ≤ |∇v1 (x)| ≤ c d(x, ∂Ω) d(x, ∂Ω)

˜ η˜/4) ∩ [B(0, bρ) \ B(0, ρ/b)] and 0 < ≤ ˆ. whenever x ∈ Ω(0, Before proving Lemmas 3.11 and 3.12 we indicate how the proof of Theorem 3.9 follows from these lemmas. Indeed, using Lemmas 3.3, 3.11, 3.12 we see for ˆ suﬃciently small and ﬁxed, 0 < ≤ ˆ, that there exists c˜ > 1, depending only on p, n, such that c˜−1

u ˆ(x) u ˆ(x) ≤ |∇ˆ u(x)| ≤ c˜ d(x, ∂Ω) d(x, ∂Ω)

(109)

˜ η˜/2)∩[B(0, b1/2 ρ)\B(0, ρ/b1/2 )]. With > 0 now ﬁxed it follows from in Ω(0, (109), (106), and arbitrariness of ρ < r0 /c that u ˆ can be used in Lemma 3.10. As mentioned earlier, Theorem 3.9 follows from Lemma 3.10. Proof of Lemma 3.11. From (108) we observe that it suﬃces to prove the righthand inequality in Lemma 3.11. We note that if y ∈ ∂Q and u ˆ(y) = v1 (y), then y lies within 4σr of a point in ∂Ω. Also, max∂B(0,t) u is nonincreasing as a function of t > 0 as we see from the maximum principle for p harmonic functions. Using these notes and Lemmas 1.2–1.5 we see that √ u ˆ ≤ v1 + cσ α/2 u( σen )

(110)

on ∂Q. By Lemma 1.1 this inequality also holds in Q. Using Lemmas 1.2–1.5 we also ﬁnd that there exist β = β(p, n) ≥ 1 and c = c(p, n) > 1 such that β

max{ψ(z), ψ(y)} ≤ c (d(z, ∂Q)/d(y, ∂Q)) min{ψ(z), ψ(y)}

(111)

whenever z ∈ Q, y ∈ Q ∩ B(z, 4d(z, ∂Q)) and ψ = uˆ or v1 . Also from Lemmas 1.2–1.5 applied to v1 we deduce √ √ ˆ( σ ren ). v1 (2 σ ren ) ≥ c−1 u

(112)

˜ η˜/16)∩ Let ρˆ, θˆ be as in Lemma 3.11. From (110)–(112) we see that if y ∈ Ω(0, √ [B(0, ρˆ) \ B(0, 4 σ r)], then √ ˆ uˆ(y) ≤ v1 (y) + cσ α/2 u( σen ) ≤ (1 + c2 σ α/2−θβ )v1 (y) ≤ (1 + )v1 (y) (113) ˆ = α/4. Thus Lemma 3.11 is true. provided ˆ is small enough and θβ Proof of Lemma 3.12. Using Lemmas 1.2–1.5 we note that there exist γ = γ(p, n) > 0, 0 < γ ≤ 1/2, and c = c(p, n) > 1 such that u ˆ(x) ≤ c(s/t)γ u ˆ(sen )

(114)

Applications of Boundary Harnack Inequalities for p Harmonic Functions

45

−1 provided x ∈ Rn \ B(0, √ t), t ≥ s, and sen ∈ Ω with d(sen , ∂Ω) ≥ c s. Using (114) with t = r, s = σr, we ﬁnd that

√ √ ¯ v1 ≤ c σ γ/2 u ˆ( σren ) on ∂Q \ B(0, σ r ),

(115)

where c depends only on p, n. Let v˜ be the p harmonic function in Q with ¯ √σ r) and v˜ = v1 on ∂Q ∩ continuous boundary values v ˜ = 0 on ∂Q \ B(0, √ ∂B(0, σr). From Lemma 1.1 and (115) it follows that √ 0 ≤ v˜ ≤ v1 ≤ v˜ + cσ γ/2 u( σren ) in Q.

(116)

From Lemmas 1.2–1.5 we observe that √ √ √ v˜(2 σren ) ≥ c−1 v1 ( σren ) = c−1 u( σren ).

(117)

Using (116), (117), and (111) applied to ψ = v˜ we obtain for ρ = σ 1/2−4θ r, θ small, b = σ −θ , and ˆb = 8b2 , that v ≤ (1 + )˜ v v˜ ≤ v1 ≤ (1 + cσ γ/2−6θβ )˜

(118)

˜ η˜/8) ∩ [B(0, ˆbρ) \ B(0, ρ/ˆb)], provided ˆ is small enough and θ = on Ω(0, ˆ min{γ/(24β), θ/10}. Next let v be the p harmonic function in √ √ ¯ Q = {y : |yi | < r, 1 ≤ i ≤ n − 1} ∩ {y : σr < yn < r} \ B(2 σ ren , σ r ) ¯ √σen , √σ r) while v = 1 with continuous boundary values v = 0 on ∂Q \ B(2 √ √ on ∂B(2 σ ren , σ r ). One can show that √ v(x) ≤ c2 σ ren − x, ∇v(x)

(119)

when x ∈ Q where c = c(p, n). Clearly this inequality implies that there exists c = c(p, n, η) ≥ 1, for given η, 0 < η ≤ 1/2, such that c−1

v(x) v(x) ≤ |∇v(x)| ≤ c d(x, ∂Q ) d(x, ∂Q )

(120)

˜ (0, η)\B(0, 10√σ r ) where Q ˜ (0, η) is the non-tangential approach region in Q deﬁned relative to 0, η, Q . Using Theorem 1.3 and (120) for√ suitable η = η(p, n) we conclude that (120) actually holds in Q \ B(0, 10 √σr). We now use Lemma 3.10 applied to v, v˜ with Ω, r , replaced by Q , 10 σr, in order to get, for some a = a(p, n) > 0 and c = c(p, n) > 1, that c−1

v˜(x) v˜(x) ≤ |∇˜ v (x)| ≤ c d(x, ∂Ω) d(x, ∂Ω)

(121)

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J. Lewis

in [B(0, r/c) \ B(0, c) r)]. Finally, note √ that if 0 ≤ ≤ ˆ and if ˆ is suﬃciently small, then r/c > b2 ρ > ρ/b2 > c σ r. Hence, if ˆ is small enough then we can use (121), (118), and Lemma 3.3 to conclude that Lemma 3.12 is valid. The proof of Theorem 3.9 is now complete.

4.9

Further Remarks

We note that Theorem 3.9 has been generalized in [62] to weak solutions of ∇ · A(x, ∇u) = 0 where A is as in Theorem 3.7. Also in [52] we show that the conclusion of Theorem 3.9 holds when Ω is convex or the complement of a convex domain. Finally the conclusion of Theorem 3.9 is valid when Ω ⊂ R2 is a Lipschitz domain. This problem for Lipschitz domains remains open when n ≥ 3. However the same argument as in Theorem 3.9 yields that a p Martin function in a Lipschitz domain is unique (up to constant multiples) at each boundary point where a tangent plane exists.

5

Uniqueness and Regularity in Free Boundary: Inverse Type Problems

We begin this section by outlining the proof of Theorem 4.1. Let E be a compact convex set, a > 0, and p ﬁxed, 1 0, then there is a unique solution to the following free boundary problem: Find a bounded domain D with E ⊂ D and u, p harmonic in D \ E, satisfying (a) (b) (c) (d)

u has continuous boundary values 1 on E and 0 on ∂D, u is p harmonic in D \ E, μ = ap−1 H n−1 |∂D where μ is the measure associated with u as in (15), For some positive c, r0 , and all x ∈ ∂D μ(B(x, r)) ≤ crn−1 , 0 < r ≤ r0 .

5.1

History of Theorem 4.1

My interest in free boundary problems of the above type started in ≈1989 when Andrew Vogel (my former Ph.D. student) was a graduate student at the University of Kentucky. He went to a conference where the following problem was proposed:

Applications of Boundary Harnack Inequalities for p Harmonic Functions

5.1.1

47

The Ball Problem

Let g be the Green’s function for a domain D with smooth boundary and pole at 0 ∈ D. If |∇g| = a = constant on ∂D, show that D is a ball with center at 0. We came up with the following proof: From the above assumption and properties of g one has ω(∂D) = 1 = aH n−1 (∂D) where ω is harmonic measure on ∂D relative to 0. Thus H n−1 (∂D) = 1/a.

(122)

Choose B(0, R) ⊂ D so that y ∈ ∂B(0, R) ∩ ∂D. Let G be the Green’s function for B(0, R) with pole at 0. If b = |∇G|(y), then since |∇G| = b on ∂B(0, R) we have, as in (122), H n−1 (B(0, R)) = 1/b.

(123)

On the other hand from G ≤ g one gets b ≤ a or 1/a ≤ 1/b. Combining this inequality with (122), (123), we deduce H n−1 (∂D) ≤ H n−1 (∂B(0, R)) which in view of the isoperimetric inequality or the fact that H n−1 decreases under a projection implies that B(0, R) = D. Later we found out that numerous other authors, including H. Shahgholian, had also obtained that D is a ball, under various smoothness assumptions on g, ∂D. In fact Henrot and Shahgholian in [34], generalized this problem, by proving Theorem 4.2. Let E be a compact convex set, a > 0, and p ﬁxed, 1 0, then there is a unique solution to the following free boundary problem: Find a bounded domain D with E ⊂ D and u, p harmonic in D \E, satisfying (α) u has continuous boundary values 1 on E and 0 on ∂D, (β) u is p harmonic in D \ E, (γ) |∇u|(x)→a as x→∂D. Moreover D is convex and ∂D is C ∞ . To prove uniqueness in Theorem 4.2, given existence, one can argue as in the ball problem using the nearest point projection onto a convex set. In fact

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J. Lewis

if u , D are solutions to the above problem for a given a, p, one can show there exists a Henrot–Shahgholian domain D ⊂ D with ∂D ∩ ∂D = ∅. Using the nearest point projection onto a convex set as in the ball problem, it follows that D = D . We note that if E is a ball in Theorem 4.2, then necessarily D is a ball, since radial solutions exist. Also, Henrot–Shahgholian domains will play the same role in the proof of Theorem 4.1, as balls did in the ball problem. Andy and I considered other generalizations of the ball problem. We ﬁrst rephrased the problem as in (c) of Theorem 4.1 by ω = aH n−1 |∂D

(124)

where ω is harmonic measure with respect to 0, ∂D. However some examples of Keldysh and Larrentiev in two dimensions showed that this assumption was not enough to guarantee that D was a ball, so we also assumed that |∇g| ≤ M < ∞ near ∂D. Using the Riesz representation formula for subharmonic functions one can show that boundedness of |∇g| near ∂D, is equivalent to ω(B(x, r)) ≤ crn−1 for all x ∈ ∂D and 0 < r ≤ r0

(125)

(i.e., condition (d) in Theorem 4.1). Given (124), (125), and the added assumption that H n−1 almost every point of ∂D lies in the reduced boundary of D in the sense of geometric measure theory,

(126)

we were able to show that lim sup |∇g| ≤ a x→∂D

(127)

which then allowed us to repeat the argument in the smooth case and get that D is a ball. Our paper appeared in [56]. In this paper we listed a number of symmetry problems including whether hypothesis (126) was needed for the ball theorem as well as analogues for the p Laplacian. During this period we also wrote [55]. This paper used a technique in [72] to construct a bounded domain D in Rn , n ≥ 3 for which ∂D is homeomorphic to a sphere and (124), (126) hold, but D = ball (so necessarily (125) is false). We improved this result in [58] where we showed that there exists quasi-spheres = ball for which (124), (126), held. Finally in [57] we were able to show that our generalization of the ball theorem remained valid without (126). Thus we obtained an endpoint result

Applications of Boundary Harnack Inequalities for p Harmonic Functions

49

for harmonic functions. The key new idea in removing (126) was the following square function—Carleson measure estimate: g D∩B(x,r)

n

(gyi yj )2 dy ≤ c rn−1 , 0 < r ≤ r1 ,

(128)

i,j=1

for some positive c, r1 , and all x ∈ ∂D. Let d(E, F ) denote the distance between the sets E, F. Then (128) allowed us to conclude for given > 0 (in a qualitative H n−1 sense) that there were ‘lots’ of tangent balls B(x, d(x, ∂D)) where ∇g had oscillation ≤ in B(x, (1−)d(x, ∂D)). Moreover using this fact and subharmonicity of |∇g| we could also conclude that if Λ = lim sup |∇g|, y →D then in a certain percentage of these balls, Λ − ≤ |∇g|(x) ≤ Λ +

(129)

Using (124), (129), and an asymptotic argument, we then concluded that Λ ≤ a, which as earlier implies D = B(0, R).

5.2

Proof of Theorem 4.1

To prove Theorem 4.1 for ﬁxed p, 1 < p < ∞, we ﬁrst showed that (d) in Theorem 4.1 implies, |∇u| ≤ M < ∞ near ∂D.

(130)

To prove (130) for x near ∂D we used the inequality c d(x,∂D)

u(x) ≤ c

0

μ[B(w, t)] tn−p

1/(p−1)

dt , t

(131)

(d) of Theorem 4.1, and the interior estimate, |∇u(x)| ≤ cˆu(x)/d(x). Equation (131) is proved in [46]. Armed with (130) we then proved a square function estimate similar to (128) with g replaced by u. As in the p = 2 case, the square function estimate enabled us to conclude that ∇u had small oscillation on ‘lots’ of tangent balls in the H n−1 measure sense. In retrospect the subharmonicity of |∇g| in the ball problem allowed us to use the Poisson integral formula on certain level sets to get a boundary type integral for which estimates could be made in terms of g, ω. To make this method work in the

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J. Lewis

proof of Theorem 4.1, it was ﬁrst necessary to ﬁnd a suitable divergence form partial diﬀerential equation for which u is a solution and |∇u|2 is a subsolution. For a long time we did not think there was any such PDE and that the lack of such a PDE resulted in some rather deep questions involving absolute continuity of μ with respect to H n−1 measure. Finally we discovered the PDE in (8), (9). It is easily checked that |∇u|2 is a subsolution to (8), (9). Using this discovery, we were able to follow the general outline of the proof in the harmonic case and after some delicate asymptotics eventually get ﬁrst (127) with g replaced by u and then Theorem 4.1. 2

5.3

Further Uniqueness Results

We note that the derivation of (8), (9) depends heavily on the fact that the p Laplacian is homogeneous in ∇u so does not work for general PDE of p Laplace type. Our earlier investigations before (8), (9) led to [60] where we consider symmetry—uniqueness problems similar to those in Theorem 4.1 for non homogeneous PDE of p Laplace type. In this case u, |∇u|2 , are not a solution, subsolution, respectively of the same divergence form PDE. Thus we were forced to tackle some rather diﬃcult questions involving absolute continuity of elliptic measure with respect to H n−1 measure on ∂Ω. To outline our eﬀorts, for these PDE’s we could still prove a square function estimate for u similar to the one in (128). This estimate together with the stronger assumption that c−1 rn−1 ≤ μ(B(x, r)) ≤ crn−1 , x ∈ ∂Ω, 0 < r ≤ r0

(132)

(where μ is the measure related to a solution u by way of an integral identity similar to (15)), enabled us to conclude that ∂Ω is uniformly rectiﬁable in the sense of [18]. At one time we hoped that uniform rectiﬁability of ∂Ω would imply absolute continuity of a certain elliptic measure with respect to H n−1 measure on ∂Ω. Eventually however we found an illuminating example in [10]. The example showed that harmonic measure for Laplace’s equation need not be absolutely continuous with respect to H 1 measure in a uniformly rectiﬁable domain. This example appeared to provide a negative end for our eﬀorts. Later we observed that in order to obtain the desired analogue of (129) it suﬃces to make absolute continuity type estimates for the above elliptic measure on the boundary of a certain subdomain Ω1 ⊂ Ω, with ∂Ω1 uniformly rectiﬁable. Here Ω1 is obtained by adding to Ω certain balls on which |∇u| is ‘small’. With this intuition we ﬁnally were able to use a rather involved stopping time argument in order to ﬁrst establish the absolute continuity of our elliptic measure with respect to H n−1 |∂Ω1 and second get an analogue of (127).

Applications of Boundary Harnack Inequalities for p Harmonic Functions

51

From 2004–2006, Bj¨orn Bennewitz was my Ph.D. student. In his thesis he generalized the nonuniqueness results of Andy and I to p harmonic functions when Ω ⊂ R2 . More speciﬁcally for ﬁxed p, 1 < p < ∞, he constructed an Ω = a Henrot–Shahgholian domain with ∂Ω a quasi circle, E ⊂ Ω, and for which the corresponding p harmonic u satisﬁed all the conditions of Theorem 4.1 except (d). His construction makes important use (as in Sect. 2) of the fact that in R2 , v = log |∇u| is a subsolution to (8), (9), when p ≥ 2 and a supersolution to (8), (9), when 1 < p ≤ 2. Reference [7] is based on his thesis.

5.4

Boundary Regularity of p Harmonic Functions

Next we discuss boundary regularity and corresponding inverse problems for positive p harmonic functions vanishing on a portion of a Lipschitz domain. We ﬁrst introduce some more or less standard notation for Lipschitz domains. Let Ω ⊂ Rn be a bounded Lipschitz domain with w ∈ ∂Ω. If 0 < ρ < r0 let Δ(w, ρ) = ∂Ω ∩ B(w, ρ) and for given b, 0 b|x − y|} ∩ B(w, 4r)}. Given a measurable function k on ∪x∈Δ(w,2r) Γ (x) we deﬁne the non tangential maximal function N (k) : Δ(w, 2r)→R for k as N (k)(x) = sup |k|(y) whenever x ∈ Δ(w, 2r). y∈Γ (x)

Let σ be H n−1 measure on ∂Ω and let Lq (Δ(w, 2r)), 1 ≤ q ≤ ∞, be the space of functions which are qth power integrable, with respect to σ, on Δ(w, 2r). Furthermore, given f : Δ(w, 2r)→R, we say that f is of bounded mean oscillation on Δ(w, r), f ∈ BM O(Δ(w, r)), if there exists A, 0 < A < ∞, such that |f − fΔ |2 dσ ≤ A2 σ(Δ(x, s)) (133) Δ(x,s)

whenever x ∈ Δ(w, r) and 0 < s ≤ r. Here fΔ denotes the average of f on Δ = Δ(x, s) with respect to σ. The least A for which (133) holds is denoted by f BMO(Δ(w,r)) . Finally we say that f is of vanishing mean oscillation on Δ(w, r), f ∈ V M O(Δ(w, r)), provided for each > 0 there is a δ > 0 such that (133) holds with A replaced by whenever 0 < s < min(δ, r) and x ∈ Δ(w, r). Theorem 4.3. Let Ω ⊂ Rn be a bounded Lipschitz domain with Lipschitz constant M . Given p, 1 < p < ∞, w ∈ ∂Ω, 0 < r < r0 , suppose that u is a positive p harmonic function in Ω ∩ B(w, 4r) and u is continuous in B(w, 4r) with u ≡ 0 on B(w, 4r) \ Ω. Then

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J. Lewis

lim

∇u(y) = ∇u(x)

y∈Γ (x),y→x

For σ almost every x ∈ Δ(w, 4r). Furthermore there exist q = q(p, n, M ) > p and a constant c = c(p, n, M ) ≥ 1, such that (i) N (|∇u|) ∈ Lq (Δ(w, 2r)) p−1−q |∇u|q dσ ≤ cr(n−1)( p−1 ) (ii) Δ(w,2r)

q/(p−1) |∇u|p−1 dσ

Δ(w,2r)

(iii) log |∇u| ∈ BM O(Δ(w, r)), log |∇u| BMO(Δ(w,r)) ≤ c. Next we state Theorem 4.4. Let Ω, M, p, w, r and u be as in the statement of Theorem 4.3. If, in addition, ∂Ω is C 1 regular then log |∇u| ∈ V M O(Δ(w, r)).

Theorem 4.5. Let Ω, M, p, w, r and u be as in the statement of Theorem 4.3. If log |∇u| ∈ V M O(Δ(w, r)), then the outer unit normal to Δ(w, r) is in V M O(Δ(w, r/2)). To put these results into historical perspective, we note that for harmonic functions, i.e., p = 2, Theorem 4.3 was proved in [17]. Theorem 4.4 for harmonic functions was proved by [37]. This theorem for harmonic functions was generalized to vanishing chord arc domains in [43]. Also a version of Theorem 4.5 in vanishing chord arc domains was proved in [44] An improved version of this theorem in chord arc domains with small constants was proved by these authors in [45]. Currently we are in the process of generalizing Theorems 4.4, 4.5 to the more general setting of chord arc domains.

5.5

Proof of Theorem 4.3

To prove Theorem 4.3 we shall need several lemmas. Lemma 4.6. Let Ω ⊂ Rn be a bounded Lipschitz domain with constant M. Given p, 1 0 is p harmonic in Ω ∩ B(w, 4r) and continuous in B(w, 4r) with u = 0 on B(w, 4r) \ Ω. Let μ be the measure corresponding to u as in (15). There exists c = c(p, n, M ) such that r¯p−n μ(Δ(w, r¯)) ≈ u(ar¯(w))p−1 whenever 0 < r¯ ≤ r/c.

Applications of Boundary Harnack Inequalities for p Harmonic Functions

53

Proof. Lemma 4.6 was essentially proved in [20]. Step (ii) in Theorem 4.3 for smooth domains (i.e., the reverse H¨ older inequality) follows from Lemma 4.6 and a Rellich inequality for the p Laplacian (as in the case p = 2). For example suppose Ω ∩ B(w, 4r) = {(x , xn ) : xn > ψ(x )} ∩ B(w, 4r) where ψ ∈ C0∞ (Rn−1 ). (134) Then from results in [64] and Schauder type theory it follows that u is smooth near ∂Ω so we can apply the divergence theorem to |∇u|p φ en . Here φ ∈ C0∞ (B(w, 2r)) with φ ≡ 1 on B(w, r) and |∇φ| ≤ cr−1 . Using p harmonicity of u - Lipschitzness of ∂Ω, we get r1−n |∇u|p dσ ≤ cr−n |∇u|p dx. (135) ∂Ω∩B(w,r)

Ω∩B(w,2r)

Also, using Lemmas 1.4 and 4.6, and the fact that dμ = |∇u|p−1 dσ we ﬁnd that −n r |∇u|p dσ ≤ cr−p u(ar (w))p Ω∩B(w,2r)

≤ c2

r1−n

p/(p−1)

(136)

|∇u|p−1 dσ ∂Ω∩B(w,r)

where constants depend only on p, n, M. Combining (135), (136) we get (ii) in Theorem 4.3 with q replaced by p. However this reverse H¨older inequality has a self improving property so actually implies the higher integrability result in (ii), as follows from a theorem originally due to Gehring. Approximating u by certain p harmonic functions in smooth domains, applying (ii), and taking weak limits it follows that dμ = kdσ where μ is the measure corresponding to u as in (15). Moreover for some q > p/(p − 1) Δ(w,¯ r)

⎛

k q dσ ≤ c¯ r(n−1)(

p−1−q p−1

)

⎜ ⎝

⎞1/q ⎟ kdσ ⎠

, 0 < r¯ ≤ r/c.

(137)

Δ(w,¯ r)

Note that we still have to prove k = |∇u|p−1 on ∂Ω (σ a. e.). Lemma 4.7. Let Ω, M, p, w, r and u be as in Theorem 4.3. Then there exists ˜ ⊂ Ω ∩ B(w, 2r), with center at a point w a starlike Lipschitz domain Ω ˜ ∈ Ω ∩ B(w, r), d(w, ˜ ∂Ω) ≥ c−1 r, such that ˜ ∩ Δ(w, r)) ≥ rn−1 . (a) c σ(∂ Ω −1 −1 ˜ (b) c r u(w) ˜ ≤ |∇u(x)| ≤ cr−1 u(w), ˜ for x ∈ Ω.

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J. Lewis

˜ as in Lemma Proof. Using Lemma 4.6 and (137) one can show there exists Ω 4.7 for which (a), (b) hold with |∇u(x)| replaced by u(x)/d(x, ∂Ω). To ﬁnish oﬀ the proof of Lemma 4.7 we need to prove the fundamental inequality for u. In fact, assuming (134), the following stronger version is available in Ω : There exists c = c(p, n, M ) ≥ 1, and z ∈ B(w, r/c) such that |∇u(z)| ≈ uxn (z), ≈ u(z)/d(z, ∂Ω).

(138)

(138) is a consequence of a boundary Harnack inequality that was stated in Theorem 3.8. An improved version of Lemma 4.7 will be used in the proof of Theorem 4.5. Next from (138), we see as in (6)–(9) that u, uxi , 1 ≤ i ≤ n, both satisfy (α) Lζ =

n ∂ ∂ζ bij ∂xi ∂xj i,j=1

(139)

(β) bij (x) = |∇u|p−4 [(p − 2)uxi uxj + δij |∇u|2 ](x) and for ξ ∈ Rn , c−1 |∇u|p−2 |ξ|2 ≤

n

bij (x)ξi ξj ≤ c|∇u|p−2 |ξ|2 .

(140)

i,j=1

From (138)–(140) we see once again that u, uxi , 1 ≤ i ≤ n, are locally solutions to a uniformly elliptic PDE in divergence form. ˜ M, p, w, r, u, be as in Lemma 4.7. Deﬁne, for y ∈ Ω, ˜ Lemma 4.8. Let Ω, Ω, the measure ⎧ ⎫ n ⎨ ⎬ ˜ d˜ γ (y) = d(y, ∂ Ω) max |∇bij (x)|2 dy. ˜ ⎩ ⎭ B(y, 12 d(y,∂ Ω)) i,j=1

˜ and 0 < s < r, then If z ∈ ∂ Ω 2p−4 ˜ ∩ B(z, s)) ≤ csn−1 (u(w)/r) ˜ . γ˜ (Ω

Proof. We get Lemma 4.8 from Lemma 4.7 and integration by parts. ˜ To continue the γ˜ in Lemma 4.8 is said to be a Carleson measure on Ω. proof of Theorem 4.3, let ω ˜ (w, ˜ ·) be elliptic measure deﬁned with respect to ˜ and w L, Ω, ˜ as above. ˜ w, Lemma 4.9. Let u, Ω, ˜ be as in Lemma 4.7 and L as in (139), (140). Then there exist c ≥ 1 and θ, 0 < θ ≤ 1, such that

Applications of Boundary Harnack Inequalities for p Harmonic Functions

ω ˜ (w, ˜ E) ≤c ˜ ∩ B(z, s)) ω ˜ (w, ˜ ∂Ω

σ(w, ˜ E) ˜ ∩ B(z, s) σ(w, ˜ ∂Ω

55

θ

˜ 0 < s < r, and E ⊂ ∂ Ω ˜ ∩ B(z, s) a Borel set. We say that ω for z ∈ ∂ Ω, ˜ is ˜ Lemma 4.9 is a direct consequence an A∞ weight with respect to σ, on ∂ Ω. of Lemma 4.8 and a theorem in [42]. 2 Next we prove by a contradiction argument that ∇u has non tangential limits for σ almost every y ∈ Δ(w, 4r). To begin suppose there exists a set F ⊂ Δ(w, 4r), σ(F ) > 0, such that if y ∈ F then the limit of ∇u(z), as z → y with z ∈ Γ (y), does not exist. Let y ∈ F be a point of density for F with respect to σ. Then t1−n σ(Δ(y, t) \ F )→0 as t→0, so if t > 0 is small enough, ˜ ∩ Δ(y, t) ∩ F ) ≥ tn−1 where Ω ˜ ⊂ Ω is the starlike Lipschitz then cσ(∂ Ω domain deﬁned in Lemma 4.7 with w, w, ˜ r replaced by y, y˜, t. From (139), ˜ (140), we see that uxi , 1 ≤ i ≤ n, is a weak solution to Lζ = 0 in Ω. We now apply a theorem in [14] to deduce that uxk , 1 ≤ k ≤ n, has ˜ almost everywhere with respect to ω nontangential limits in Ω, ˜ (·, w). ˜ Recall from Lemma 4.9 that ω ˜ and σ are mutually absolutely continuous. Thus these limits also exist almost everywhere with respect to σ. Since nontangential ˜ agree with those in Ω, for σ almost every point in F, we have limits in Ω reached a contradiction. Thus ∇u has nontangential limits almost everywhere in Ω. Step (i) follows from (137), Lemmas 4.6 and 1.3. Finally we use nontangential limits of ∇u, the fact that for small t, {u = t} is Lipschitz, the implicit function theorem, as well as (i) of Theorem 4.3, to take limits as t→0 in order to conclude that k = |∇u| in (137). The proof of Theorem 4.3 is now complete. 2

5.6

Proof of Theorem 4.4

To prove Theorem 4.4 it suﬃces by way of an argument of Sarason (see [38]) to show that there exist 0 < 0 and r˜ = r˜(), for ∈ (0, 0 ), such that whenever y ∈ Δ(w, r) and 0 < s < r˜() we have p/(p−1) p p−1 − |∇u| dσ ≤ (1 + ) − |∇u| dσ . Δ(y,s)

(141)

Δ(y,s)

Here − f dσ denotes the average of f on E with respect to σ. The E

proof of (141) is by contradiction. Otherwise there exist two sequences ∞ {ym }∞ 1 , {sm }1 satisfying ym ∈ Δ(w, r) and sm →0 as m→∞ such that (141) is false with y, s replaced by ym , sm for m = 1, 2, . . . . Let A = e1/ and put ym = ym + Asm n(ym ), where n(ym ) is the inner unit normal to

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Ω at ym . Since ∂Ω is C 1 we see for > 0 small and m = m() large that if Ω(ym ) is constructed by drawing all line segments from points in B(ym , Asm /4) to points in Δ(ym , Asm ), then Ω(ym ) is starlike Lipschitz ¯ with respect to ym . Let Dm = Ω(ym ) \ B(ym , Asm /8). and let um be the p harmonic function in Dm that is continuous in Rn with um ≡ 1 on B(ym , Asm /8) and u ≡ 0 on Rn \ Ω(ym ). From the boundary Harnack inequality in Theorem 3.8 with w, r, u, v replaced by ym , Asm /100, u, um we deduce that if w1 , w2 ∈ Ω ∩ B(ym , 2sm ) then

log um (w1 ) − log um (w2 ) ≤ cA−α .

u(w1 ) u(w2 )

(142)

Letting w1 , w2 →z1 , z2 ∈ Δ(ym , 2sm ) in (142) and using Theorem 4.3, we get, σ almost everywhere, that

log |∇um (z1 )| − log |∇um (z2 )| ≤ cA−α

|∇u(z1 )| |∇u(z2 )|

or equivalently that (1 − c˜A−α )

|∇u(z1 )| |∇um (z1 )| |∇um (z1 )| ≤ ≤ (1 + c˜A−α ) . |∇um (z2 )| |∇u(z2 )| |∇um (z2 )|

(143)

Using (143) and the fact that (141) is false we obtain −

|∇um |p dσ

Δ(ym ,sm )

−

p/(p−1)

|∇um |p−1 dσ

Δ(ym ,sm )

−

(144) |∇u|p dσ

Δ(ym ,sm )

≥ (1 − cA−α )

−

−α p/(p−1) ≥ (1 − cA )(1 + ) .

|∇u|p−1 dσ

Δ(ym ,sm )

Let Tm be a conformal aﬃne mapping of Rn which maps the origin and en onto ym and ym respectively and which maps W = {x ∈ Rn : xn = 0} onto the tangent plane to ∂Ω at ym . Let Dm , um be such that Tm (Dm ) = Dm and um (Tm x) = um (x) whenever x ∈ Dm . Since the p Laplace equation is invariant under translations, rotations, and dilations, we see that um is p harmonic in Dm . Letting m→∞ one can show that um converges uniformly

Applications of Boundary Harnack Inequalities for p Harmonic Functions

57

on Rn to u where u is continuous on Rn and p harmonic in D = Ω \ B(en , 1/8) with u ≡ 1 on B(en , 1/8) and u ≡ 0 on Rn \Ω . Here Ω is obtained by drawing all line segments connecting points in B(0, 1) ∩ W to points in B(en , 1/4). Changing variables in (144) and using Rellich type inequalities one gets − |∇um |p dσ ∂D ∩B(0,1/A)

(1 − cA−α )(1 + ) ≤ lim sup m m→∞ −

p/(p−1)

|∇um |p−1 dσ

∩B(0,1/A) ∂Dm

−

(145) |∇u |p dσ

W ∩B(0,1/A)

≤ −

p/(p−1)

|∇u |p−1 dσ

W ∩B(0,1/A)

Finally from interior estimates for p harmonic functions and Schwarz reﬂection one ﬁnds for z ∈ B(0, 1/A) that (1 − cA−θ )|∇u (0)| ≤ |∇u (z)| ≤ (1 + cA−θ )|∇u (0)| which in view of (145) yields −

|∇u |p dx

W ∩B(0,1/A)

(1 + cA−θ ) ≥ −

−α p/(p−1) ≥ (1 − cA )(1 + ).

|∇u |p−1 dx

W ∩B(0,1/A)

Clearly this inequality cannot hold for small since A = e1/ . The proof of Theorem 4.4 is now complete.

5.7

Proof of Theorem 4.5

In this section we prove Theorem 4.5. We shall need the following reﬁned version of Lemma 4.7. Lemma 4.10. Given Ω, w, p, n, M, u as in Theorem 4.3. If log |∇u| ∈ V M O(Δ(w, r)), then for each > 0 there exists, 0 < r˜ = r˜() < r and c = c(p, n, M ), 1 ≤ c < ∞, such that if 0 < r ≤ r˜ then the following :

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˜ ⊂ Ω ∩ B(w, r ), with center at a point There is a starlike Lipschitz domain Ω w ˆ ∈ Ω ∩ B(w, r ), d(w, ˆ ∂Ω) ≥ r /c, and Lipschitz constant ≤ c, satisfying (a) ˜ ∩ Δ(w, r )) σ(∂ Ω ≥ 1 − . σ(Δ(w, r )) (b) (1 − )bp−1 ≤

μ(Δ(y, s)) ≤ (1 + )bp−1 σ(Δ(y, s))

˜ ∩ Δ(w, r ), and log b is the average of log |∇u| on where 0 < s ≤ r , y ∈ ∂ Ω Δ(w, 4r ). Moreover, (c) u(w) ˆ u(w) ˆ ˜ ≤ |∇u(x)| ≤ c for all x ∈ Ω. r r Proof. Lemma 4.10 is proved in [51] as Lemma 4.1. To begin the proof of Theorem 4.5 let n denote the outer unit normal to ∂Ω and put c−1

η = lim

sup

r˜→0 w∈Δ(w,r/2) ˜

n BMO(Δ(w,˜ ˜ r)) .

(146)

To prove Theorem 4.5 it is enough to prove that η = 0. To do this we argue by contradiction and assume that (146) holds for some η > 0. This assumption implies that there exist a sequence of points {wj }, wj ∈ Δ(w, r/2), and a sequence of scales {rj }, rj → 0, such that n BMO(Δ(wj ,rj )) → η as j → ∞. To get a contradiction we use a blow-up argument. In particular, let u be as in the statement of Theorem 4.5 and extend u to B(w, 4r) by putting u = 0 in B(w, 4r) \ Ω. For {wj }, {rj } as above we deﬁne Ωj = {rj−1 (x − wj ) : x ∈ Ω} and uj (z) = λj u(wj + rj z) whenever z ∈ Ωj . Using properties of Lipschitz domains, one can show that subsequences of {Ωj }, {∂Ωj } converge to Ω∞ , ∂Ω∞ , in the Hausdorﬀ distance sense, where Ω∞ is an unbounded Lipschitz domain with Lipschitz constant bounded by M . Moreover, from (15), Lemmas 1.2–1.5, and 4.10 we deduce for an appropriate choice of (λj ), that a subsequence of (uj ) converges uniformly on compact subsets of Rn to u∞ , a positive p harmonic function in Ω∞ vanishing continuously on ∂Ω∞ . If dμj = |∇uj |p−1 dσ|∂Ωj , it also follows that a subsequence of (μj ) converges weakly as Radon measures to μ∞ where

|∇u∞ |p−2 ∇u∞ , ∇φdx = − Rn

∂Ω∞

φdμ∞

(147)

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59

whenever φ ∈ C0∞ (Rn−1 ) Moreover, using Lemma 4.10, one can show that μ∞ and u∞ , satisfy, (a) μ∞ = σ on ∂Ω∞ , (148) (b) c−1 ≤ |∇u∞ (z)| ≤ 1 whenever z ∈ Ω∞ . Finally one shows that (147), (148) imply Ω∞ is a halfspace

(149)

which in turn implies that η = 0, a contradiction to (146). If M is suﬃciently small, then (149) follows directly from a theorem in [1]. For large M we needed to use our generalization in [53] of the work in [11]. We discuss this work further in the next section.

5.8

Regularity in a Lipschitz Free Boundary Problem

We begin our discussion of two phase free boundary problems for p harmonic functions with some notation. Let D ⊂ Rn be a bounded domain and suppose that u is continuous on D. Put D+ (u) = {x ∈ D : u(x) > 0} F (u) = ∂D + (u) ∩ D D− (u) = D \ [D+ (u) ∪ F (u)]. F (u) is called the free boundary of u in D. Let G > 0 be an increasing function on [0, ∞) and suppose for some N > 0 that s−N G(s) is decreasing when s ∈ (0, ∞). Let u+ = max(u, 0) and u− = − min(u, 0). Definition F. We say that u satisﬁes weakly the two sided boundary condition |∇u+ | = G(|∇u− |) on F (u) provided the following holds. Assume that w ∈ F (u) and there is a ball B(w, ˆ ρˆ) = {y : |y − w| ˆ < ρˆ} ⊂ D+ (u) ∪ D− (u) with w ∈ ∂B(w, ˆ ρˆ). If ν = (w ˆ − w)/|w ˆ − w| and B(w, ˆ ρˆ) ⊂ D+ (u), then (∗) u(x) = αx − w, ν+ − βx − w, ν− + o(|x − w|), as x → w non-tangentially while if B(w, ˆ ρˆ) ⊂ D− (u), then (∗) holds with x − w replaced by w − x, where, α, β ∈ [0, ∞] and α = G(β). We note that if If F (u), u are suﬃciently smooth in D, then at w, α = |∇u+ | = G(β) = G(|∇u− |).

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In [51] we prove, Theorem 4.11. Let u be continuous in D, p harmonic in D \ F (u), 1 < p < ∞, p = 2, and a weak solution to |∇u+ | = G(|∇u− |) on F (u). Suppose B(0, 2) ⊂ D, 0 ∈ F (u), and F (u) coincides in B (0, 2) with the graph of a Lipschitz function. Then F (u) ∩ B(0, 1) is C 1,σ where σ depends only on p, n, N and the Lipschitz constant for the graph function.

5.9

History of Theorem 4.11

For p = 2, i.e., harmonic functions, Caﬀarelli developed a theory for general two-phase free boundary problems in [11–13]. In [11] Lipschitz free boundaries were shown to be C 1,σ -smooth for some σ ∈ (0, 1) and in [13] it was shown that free boundaries which are well approximated by Lipschitz graphs are in fact Lipschitz. Finally, in [12] the existence part of the theory was developed. We also note that the work in [11] was generalized in [70] to solutions of fully nonlinear PDEs of the form F (∇2 u) = 0, where F is homogeneous. Further analogues of [11] were obtained for a class of nonisotropic operators and for fully nonlinear PDE’s of the form F (∇2 u, ∇u) = 0, where F is homogeneous in both arguments, in [24, 25]. Extension of the results in [11] were made to non-divergence form linear PDE with variable coeﬃcients in [16], and generalized in [26] to fully nonlinear PDE’s of the form F (∇2 u, x) = 0. Finally generalizations of the work in [11] (also [13]) to linear divergence form PDE’s with variable coeﬃcients were obtained in [27, 28].

5.10

Proof of Theorem 4.11

To outline the proof of Theorem 4.11 we need another deﬁnition. Definition G. We say that a real valued function h is monotone on an open set O in the direction of τ ∈ Rn , provided h(x − τ ) ≤ h(x) whenever x ∈ O. If x, y ∈ Rn , let θ(x, y) be the angle between x and y. Given θ0 , 0<θ0 <π, 0 >o, and ν with |ν| = 1, put Γ (ν, θ0 , 0 ) = {y ∈ Rn : |θ(y, ν)| < θ0 , 0 < |y| < 0 }. We note from elementary geometry that if h is monotone on O with respect to the directions in Γ (ν, θ0 , 0 ), then (2)

sup B(x−τ,|τ | sin(θ0 /2))

h ≤ h(x)

Applications of Boundary Harnack Inequalities for p Harmonic Functions

61

whenever x ∈ O and τ ∈ Γ (ν, θ0 /2, 0 /2). To establish Theorem 4.11 we ﬁrst show the existence of a cone of monotonicity. To this end, we assume as we may, that Ω ∩ B(0, 2) = {(x , xn ) : xn > ψ(x )} ∩ B(w, 4r), ψ Lipschitz on Rn−1 . (150) If M is the Lipschitz norm of ψ, then as in (138) we see that Theorem 3.8 and (150) imply there exists c = c(p, n, M ) ≥ 1, such that whenever z ∈ B(0, r1 ), r1 = 1/c, |∇u(z)| ≈ uxn (z), ≈ u(z)/d(z, ∂Ω).

(151)

Clearly (151) implies the existence of θ0 ∈ (0, π/2], 0 > 0, and c > 1, depending only on p, n, M, such that u is monotone in B(0, r1 ) with respect to the directions in the cone Γ (en , θ0 , 0 ).

5.11

(152)

Enlargement of the Cone of Monotonicity in the Interior

Let τ ∈ Γ (en , θ0 /2, 0) for θ0 , 0 , as in (152), put = |τ | sin(θ0 /2) and set v (x) = v ,τ (x) =

sup

u(y − τ )

y∈B(x, )

We note from (152) that v (x) ≤ u(x), when x ∈ B(0, r1 ). Next we show that if ρ, μ > 0 are small enough, depending only on p, n, and the Lipschitz constant for ∂Ω ∩ B(0, 1), and ν = ∇u( r18en ))/|∇u( r18en )|, then v(1+μλ) (x) ≤ (1 − μλ)u(x) whenever x ∈ B( r18en , ρr1 ).

(153)

where λ = cos(θ0 /2 + θ(ν, τ )), and 0 < |τ | ≤ 0 ρr1 . The proof of (153) is essentially the same as in [11], thanks to basic interior estimates for p harmonic functions, Theorem 3.8, and (152).

5.12

Enlargement of the Cone of Monotonicity at the Free Boundary

In this part of the proof we show there exists μ ¯ > 0, depending only on p, n, M , such that if τ, are as deﬁned above (153), then

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J. Lewis

v(1+¯μλ) (x) ≤ u(x) whenever x ∈ B(0, r1 /100),

(154)

It is shown in [11] that (154) implies the existence of ω, |ω| = 1, θ¯ ∈ (0, π/2], c− , c+ > 1, such that ¯ 0 /c+ ) u is monotone in Γ (ω, θ,

(155)

¯ where π/2 − θ¯ = c−1 − (π/2 − θ0 ), Γ (en , θ0 , 0 ) ⊂ Γ (ω, θ, 0 /c+ ), and all constants depend only on p, n, M. Using (155) as well as invariance of the p Laplace equation under scalings and translations, we can replace u(x) by u(x0 + ηx)/η and then repeat our argument in (153), (154), in order to eventually conclude the C 1,σ -smoothness of F (u) ∩ B(0, 1/2). Hence to prove Theorem 4.11 we have to prove (154). To do this, given an n × n symmetric matrix M let P (M ) = inf

n

A∈Ap

aij Mij .

i,j=1

where Ap denotes the set of all symmetric n × n matrices A = {aij } which satisfy n min{p − 1, 1} |ξ|2 ≤ aij ξi ξj ≤ max{p − 1, 1} |ξ|2 i,j=1

whenever ξ ∈ R \ {0}. Next we state Lemma 4.12. Let φ > 0 be in C 2 (D), ∇φ L∞ (D) ≤ 1/2, p ﬁxed, 1 < p < ∞, and suppose that n

φ(x)P (∇2 φ(x)) ≥ 50pn |∇φ(x)|2 whenever x ∈ D. Let u be continuous in an open set O containing the closure of x∈D B(x, φ(x)) and deﬁne v(x) =

max

¯ B(x,φ(x))

u

whenever x ∈ D. If u is p-harmonic in O \ {u = 0}, then v is continuous and ¯ ⊂ D. a p-subsolution in {v = 0} ∩ G whenever G is an open set with G n Theorem 4.13. Let Ω ⊂ R be a Lipschitz graph domain with Lipschitz constant M . Given p, 1 0, suppose that u ˆ and vˆ are non-negative p-harmonic functions in Ω ∩ B(w, 2r) with vˆ ≤ u ˆ. Assume also that u ˆ, vˆ, are continuous in B(w, 2r) with u ˆ ≡ 0 ≡ vˆ on B(w, 2r) \ Ω. There exists c ≥ 1, depending only on p, n, M, such that if y, z ∈ Ω ∩ B(w, r/c), then u ˆ(y) − vˆ(y) u ˆ(z) − vˆ(z) ≤c . vˆ(y) vˆ(z)

Applications of Boundary Harnack Inequalities for p Harmonic Functions

63

Remark. Lemma 4.12 is not much more diﬃcult than the corresponding lemma in [11], thanks to translation, dilation and rotational invariance of the p Laplacian. Theorem 4.13 uses the full toolbox developed in [49–52] and [62]. For p = 2 Theorem 4.13 is equivalent to a boundary Harnack inequality for harmonic functions while for p = 2, it is stronger than the boundary Harnack inequality in Theorem 3.8. That is, Theorem 4.13 implies not only boundedness of u ˆ/ˆ v , but also H¨older continuity of the ratio in Ω ∩B(w, r/c∗ ), ∗ ∗ for some c − c (p, n, M ) ≥ 1. Theorem 4.13 is our main contribution to the proof of Theorem 4.11. To prove (154), using Lemma 4.12 and Theorem 4.13, let v˜t (x) := sup u(y − τ ) for t ∈ [0, 1] y∈B(x, φμλt (x))

and x ∈ B(0, r1 ) \ B(r1 en /8, ρr1 ). Here {φt }, is a family of C 2 functions, each satisfying the hypotheses of Lemma 4.12 in B(0, r1 ) \ B(r1 en /8, ρr1 ). Moreover, (a) (b) (c) (d)

φt ≡ 1 on B(0, r1 ) \ B(0, r1 /2) 1 ≤ φt ≤ 1 + tγ in B(0, r1 ) \ B(r1 en /8, ρr1 ) φt ≥ 1 + htγ in B(0, r1 /100) |∇φt | ≤ γt.

In (c), h > 0 depends on ρ, p, n while γ > 0 is a parameter to be chosen suﬃciently small. From (c) one sees that (154) holds if v˜t ≤ u in B(0, r1 ) \ B(r1 en /8, ρr1 ) whenever t ∈ [0, 1].

(156)

From Step 1 and (b) above, this inequality holds when t = 0. One can now use a method of continuity type argument to show that if (156) is false then there exist t ∈ [0, 1] for which v˜t ≤ u, w ∈ F (u) ∩ {˜ vt = 0}, and w, ˆ ρˆ > 0 with B(w, ˆ ρˆ) ⊂ {˜ vt > 0} ⊂ D+ (u) and w ∈ ∂B(w, ˆ ρˆ). One then uses in this tangent ball, the asymptotic free boundary condition for u, similar asymptotics for v˜t , (153), Theorem 4.13, and a Hopf maximum principle type argument to get a contradiction. Thus (156) and so (154) are valid. This completes our outline of the proof of Theorem 4.11.

5.13

An Application of Theorem 4.11

Recall that the proof of Theorem 4.5 was by contradiction. Indeed assuming that η = lim sup

ν BMO(Δ(w,˜ (157) ˜ r)) = 0 r˜→0 w∈Δ(w,r/2) ˜

we obtained u∞ , a positive p-harmonic function in a Lipschitz graph domain, Ω∞ , which is H¨ older continuous in Rn with u∞ ≡ 0 on Rn \ Ω∞ . We also

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J. Lewis

had

|∇u∞ |p−2 ∇u∞ , ∇ψdx = −

Rn

whenever ψ ∈

ψdσ∞

(158)

∂Ω∞

C0∞ (Rn )

and

c−1 ≤ |∇u∞ (z)| ≤ 1 whenever z ∈ Ω∞ .

(159)

where σ∞ is surface area on ∂Ω∞ . To get a contradiction to (157) we needed to show that (158), (159) imply Ω∞ is a halfspace. This conclusion follows easily from applying the argument in Theorem 4.11 to u∞ (Rx)/R and letting R→∞, once it is shown that u∞ is a weak solution to the free boundary problem in Theorem 4.11 with G(s) = 1 + s for s ∈ (0, ∞). That is, u is a weak solution to the ‘one phase free boundary problem’ |∇u+ | ≡ 1 and |∇u− | ≡ 0 on F (u).

(160)

ˆ ρˆ), To prove (160) assume w ∈ F (u∞ ) and that there exists a ball B(w, w ˆ ∈ Rn \ ∂Ω∞ and ρˆ > 0, such that w ∈ ∂B(w, ˆ ρˆ). Let P be the plane through w with normal ν = (w ˆ − w)/|w ˆ − w|. We claim that P is a tangent plane to Ω∞ at w in the usual sense. That is given > 0 there exists rˆ() > 0 such that Ψ (P ∩ B(w, r), ∂Ω∞ ∩ B(w, r)) ≤ r (161) whenever 0 < r ≤ rˆ(). Once (161) is proved we can show that (i)

If B(w, ˆ ρˆ) ⊂ Ω∞ then u+ ∞ (x) = x − w, ν + o(|x − w|) in Ω∞

If B(w, ˆ ρˆ) ⊂ Rn \ Ω∞ then u+ ∞ (x) = w − x, ν + o(|x − w|) in Ω∞ . (162) To prove (162) (given (161)) we assume that w = 0, ν = en , and ρˆ = 1. This assumption is permissible since linear functions and the p-Laplacian are invariant under rotations, translations, and dilations. Then w ˆ = en and either ¯ ∞ . We assume that B(en , 1) ⊂ Ω∞ , since B(en , 1) ⊂ Ω∞ or B(en , 1) ⊂ Rn \ Ω ¯∞ , is handled similarly. Let {rj } be the other possibility, B(en , 1) ⊂ Rn \ Ω a sequence of positive numbers tending to 0 and let u ˆj (z) = u∞ (rj z)/rj n ˆ whenever z ∈ R . Let Ωj = {z : rj z ∈ Ω∞ } be the corresponding blowˆj and H¨ up regions. Then u ˆj is p-harmonic in Ω older continuous in Rn with ˆj . Moreover, (159) is valid for each j with u∞ replaced u ˆj ≡ 0 on Rn \ Ω by u ˆj . Using these facts, assumption (161), and Lemmas 1.2–1.5 we see that a subsequence of {ˆ uj }, denoted {uj }, converges uniformly on compact subsets of Rn , as j→∞, to a H¨older continuous function u∞ . Moreover, u∞ is a nonnegative p-harmonic function in H = {x : xn > 0} with u∞ ≡ 0 on ˆ j } corresponding to {u }. From Rn \ H. Let {Ωj } be the subsequence of {Ω j (161) we see that Ωj ∩ B(0, R) converges to H ∩ B(0, R) whenever R > 0, in the sense of Hausdorﬀ distance as j→∞. Finally we note that ∇uj →∇u∞ (ii)

Applications of Boundary Harnack Inequalities for p Harmonic Functions

65

uniformly on compact subsets of H and hence c−1 ≤ |∇u∞ | ≤ 1

(163)

where c is the constant in (159). Next we apply the boundary Harnack inequality in Theorem 4.1 with Ω = H, u ˆ(x) = u∞ (x), and vˆ(x) = xn . Letting r→∞ in Theorem 4.1, it follows that u∞ (x) = lxn

(164)

for some nonnegative l. From (163) and the above discussion we conclude that c−1 ≤ l ≤ 1. (165) Next using (158) we see that if σj is surface area on Ωj , σ surface area on H, and φ ≥ 0 ∈ C0∞ (Rn ), then ∂{uj >0}

φdσj = −

Rn

→−

Rn

|∇uj |p−2 ∇uj , ∇φdx |∇u∞ |p−2 ∇u∞ , ∇φdx

=l

p−1

(166) φdσ

{xn =0}

as j → ∞. Moreover, using the divergence theorem we ﬁnd that ∂{uj >0}

φdσj

≥−

∇ · (φen )dx → −

{uj >0}

{u∞ >0}

∇ · (φen )dx =

φdσ {xn =0}

(167) as j→∞. Combining (166), (167) we obtain ﬁrst that l ≥ 1 and thereupon from (165) that l = 1. Thus any blowup sequence of u∞ , relative to zero, n tends to x+ n uniformly on compact subsets of R , and the corresponding gradients tend uniformly to en on compact subsets of H. This conclusion is easily seen to imply (162). Hence (161) implies (162).

5.14

Proof of (161)

The proof of (161) is again by contradiction. We continue under the assumption that w = 0, ν = w ˆ = en , and ρˆ = 1. First suppose that B(en , 1) ⊂ Ω∞ .

(168)

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J. Lewis

If (161) is false, then there exists a sequence {sm } of positive numbers and δ > 0 with lim sm = 0 and the property that m→∞ Ω∞ ∩ ∂B(0, sm ) ∩ {x : xn ≤ −δsm } = ∅

(169)

for each m. To get a contradiction we show that (169) leads to lim sup t−1 u∞ (ten ) = ∞ t→0

(170)

which in view of the mean value theorem from elementary calculus, contradicts (159). For this purpose let f be the p-harmonic function in B(en , 1) \ ¯ n , 1/2) with continuous boundary values, B(e f ≡ 0 on ∂B(en , 1) and f ≡

min

¯ n ,1/2) B(e

u∞ on ∂B(en , 1/2).

Recall that n f can be written explicitly in the form, f (x) =

A|x − en |(p−n)/(p−1) + B when p = n, −A log |x − en | + B when p = n,

where A, B are constants. Doing this we see that lim t−1 f (ten ) > 0.

t→0

(171)

From the maximum principle for p-harmonic functions we also have ¯ n , 1/2). u∞ ≥ f in B(en , 1) \ B(e

(172)

¯ s) ∩ B(en , 1), for Next we show that if 0 < s < 1/4, and u∞ ≥ kf in B(0, some k ≥ 1, then there exists ξ = ξ(p, n, M, δ) > 0 and s , 0 < s < s/2, such that ¯ s ) ∩ B(en , 1). u∞ ≥ (1 + ξ)kf in B(0, (173) Clearly (171)–(173) and an iterative argument yield (170). To prove (173) we observe from a direct calculation that ¯ n , 1/2), |∇f (x)| ≈ f (x)/(1 − |x − en |) when x ∈ B(en , 1) \ B(e

(174)

where proportionality constants depend only on p, n. Also, we observe from (169) and Lipschitzness of ∂Ω∞ that if m0 is large enough, then there exists a sequence of points {tl }∞ m0 in Ω∞ ∩ {x : xn = 0} and η = η(p, n, M, δ) > 0 such that for l ≥ m0 ,

Applications of Boundary Harnack Inequalities for p Harmonic Functions

ηsl ≤ |tl | ≤ η −1 sl and d(tl , ∂Ω∞ ) ≥ η|tl |.

67

(175)

−1 |tm | ≤ s/100. If ρ = d(tm , ∂Ω∞ ), then from Choose tm ∈ {tl }∞ m0 such that η (175), and Lemmas 1.2–1.5 for u∞ we deduce for some C = C(p, n, M, δ) ≥ 1 that Cu∞ (tm ) ≥ max u∞ . (176) ¯ B(0,4|t m |)

From (176), the assumption that kf ≤ u∞ , Lemmas 1.2–1.5 for kf, and the fact that tm lies in the tangent plane to B(en , 1) through 0, we see there exists λ = λ(p, n, M, δ), 0 < λ ≤ 10−2 , and m1 ≥ m0 such that if m ≥ m1 and tm = tm + 3λρen , then ¯ , ρλ). B(tm , 2ρλ) ⊂ B(en , 1) and (1 + λ)kf ≤ u∞ on B(t m

(177)

¯ m , ρλ) Let f˜ be the p-harmonic function in G = B(0, 4|tm |) ∩ B(en , 1) \ B(t with continuous boundary values f˜ = kf on ∂[B(en , 1) ∩ B(0, 4|tm |)] while f˜ = (1 + λ)kf on ∂B(tm , ρλ). Using (174), Theorem 1.13, and Harnack’s inequality for f˜ − kf, kf as in the proof of (154), we deduce the existence of τ > 0, c¯ ≥ 1, with (1 + τ λ)kf ≤ f˜ (178) ¯ |tm |/¯ in B(en , 1) ∩ B(0, c) where τ = τ (p, n, M, δ), 0 < τ < 1, and c¯ = c¯(p, n, M ) ≥ 1. Moreover, using the maximum principle for p-harmonic functions we see from (177) that f˜ ≤ u∞ in G.

(179)

Combining (178), (179), we get (173) with ξ = τ λ and s = |tm |/¯ c. As mentioned earlier, (173) leads to a contradiction. Hence (161) is true when ¯∞ we proceed similarly. That is, if (161) is (168) holds. If B(en , 1) ⊂ Rn \ Ω false, then there exists a sequence {sm } of positive numbers and δ > 0 with lim sm = 0 and the property that m→∞ ¯∞ ∩ ∂B(0, sm ) ∩ {x : xn ≤ −δsm } = ∅ Rn \ Ω

(180)

for each m. To get a contradiction one shows that (180) leads to lim inf t−1 max u∞ = 0 t→0 B(0,t)

(181)

which in view of Lipschitzness of Ω∞ and the mean value theorem from elementary calculus, again contradicts (159). We omit the details.

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J. Lewis

Closing Remarks

To state our likely results in [54] we need a deﬁnition. Definition H. Given > 0 we say that u is -monotone in O ⊂ Rn , with respect to the directions in the cone Γ˜ (ν, θ0 ) = {y ∈ Rn : |y| = 1 and θ(ν, y) < θ0 }, if sup

B(x, sin θ0 )

u(y − ν) ≤ u(x)

whenever ≥ and x ∈ O with B(x− ν, sin θo ) ⊂ O. Moreover, u is said to be monotone or fully monotone in O ⊂ Rn , with respect to the directions in the cone Γ (ν, θ0 ), provided the above inequality holds whenever > 0. Plausible Theorem 4.14. Let D, u, D+ (u), D− (u), F (u), G be as in Theo¯ p, n, N ) such that if u is rem 4.11. If θ¯ ∈ (π/4, π/2), then there is a ¯ = ¯(θ, ¯ for monotone on B(0, 2) with respect to the directions in the cone Γ (en , θ), some ∈ (0, ¯ ), then u is monotone in B(0, 1/2) with respect to the directions in the cone Γ (en , θ¯1 ), where θ¯1 has the same dependence as ¯. Equivalently, F (u) ∩ B(0, 1/2) is the graph of a Lipschitz function with Lipschitz norm ¯ p, n, N. depending on θ, We note that we can use Theorems 4.14 and 4.11 to conclude that monotonicity of u implies F (u) is C 1,σ provided is small enough. Plausible Theorem 4.15. Let D, F (u), D+ (u), D− (u), G be as in Plausible Theorem 4.14 except for the following changes: ¯ (a) Assume only that u+ is monotone in Γ (en , θ), −1 + (b) Assume 0 < δ ≤ |∇u| ≤ δ on D (u) ∩ B(0, 2), (c) G is also Lipschitz continuous. There exists ˆ > 0 and θˆ ∈ (π/4, π/2), both depending on p, n, δ, N, such that if θˆ < θ¯ ≤ π/2, and 0 < ≤ ˆ , then u+ is monotone in B(0, 2) with respect to the directions in the cone Γ (en , θ1 ) for some θ1 > 0, depending on ˆ p, n, δ, N, ˆ, θ. As a corollary to Plausible Theorem 4.15 we also plan to show that Plausible Corollary 4.16. Replace the monotonicity assumption in Plausible Theorem 4.15 by Ψ (F (u) ∩ B(0, 2), Λ ∩ B(0, 2)) ≤ , where Λ is the graph of a Lipschitz function with Lipschitz norm ≤ tan ˆ Then the same conclusion holds as in Plausible Theorem 4.15. (π/2 − θ). One can also ask if Theorem 4.11 generalizes to equations of p Laplace type, as in the p = 2 case. In this case, the analogue of Lemma 4.12 may be diﬃcult since the proof of this lemma made important use of the invariance of the

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p Laplace equation under rotations and translations. Also, the analogue of Theorem 4.13 could be diﬃcult. Acknowledgements Work partially supported by NSF DMS-0900291.

References 1. H.W. Alt, L.A. Caﬀarelli, A. Friedman, A free boundary problem for quasilinear elliptic equations. Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4) 11(1), 1–44 (1984) 2. A. Ancona, Principe de Harnack ` a la Fronti` ere et Th´eor` eme de Fatou pour un Op´eet Elliptique Dans un Domain Lipschitzien. Ann. Inst. Fourier (Grenoble) 28(4), 169–213 (1978) 3. A. Batakis, Harmonic measure of some cantor type sets. Ann. Acad. Sci. Fenn. 21(2), 255–270 (1996) 4. A. Batakis, A continuity property of the dimension of harmonic measure under perturbations. Ann. Inst. H. Poincar´e Probab. Stat. 36(1), 87–107 (2000) 5. A. Batakis, Continuity of the dimension of the harmonic measure of some cantor sets under perturbations. Annales de l’ Institut Fourier 56(6), 1617–1631 (2006) 6. B. Bennewitz, J. Lewis, On the dimension of p harmonic measure. Ann. Acad. Sci. Fenn. 30, 459–505 (2005) 7. B. Bennewitz, Nonuniqueness in a free boundary problem. Rev. Mat. Iberoam. 24(2), 567–595 (2008) 8. J. Lewis, K. Nystr¨ om, A. Vogel, p harmonic measure in space (submitted) 9. C. Bishop, L. Carleson, J. Garnett, P. Jones, Harmonic measures supported on curves. Pac. J. Math. 138, 233–236 (1989) 10. C. Bishop, P. Jones, Harmonic measure and arclength. Ann. Math. (2) 132(3), 511–547 (1990) 11. L. Caﬀarelli, A Harnack inequality approach to the regularity of free boundaries. Part I, Lipschitz free boundaries are C 1,α . Revista Mathematica Iberoamericana 3, 139–162 (1987) 12. L. Caﬀarelli, A Harnack inequality approach to the regularity of free boundaries. Part III. Existence theory, compactness, and dependence on X. Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4) 15(4), 583–602 (1988) 13. L. Caﬀarelli, Harnack inequality approach to the regularity of free boundaries. Part II. Flat free boundaries are Lipschitz. Comm. Pure Appl. Math. 42(1), 55–78 (1989) 14. L. Caﬀarelli, E. Fabes, Mortola, S. Salsa, Boundary behavior of nonnegative solutions of elliptic operators in divergence form. Indiana J. Math. 30(4), 621–640 (1981) 15. L. Carleson, On the support of harmonic measure for sets of cantor type. Ann. Acad. Sci. Fenn. 10, 113–123 (1985) 16. M.C. Cerutti, F. Ferrari, S. Salsa, Two-phase problems for linear elliptic operators with variable coeﬃcients: Lipschitz free boundaries are C 1,γ . Arch. Ration. Mech. Anal. 171, 329–448 (2004) 17. B. Dahlberg, On estimates of harmonic measure. Arch. Ration. Mech. Anal. 65, 275– 288 (1977) 18. G. David, S. Semmes, Analysis of and on uniformly rectiﬁable sets. Am. Math. Soc. Surv. Mono. 38 (1993) 19. E. DiBenedetto, Degenerate Parabolic Equations, Universitext (Springer, New York, 1993) 20. A. Eremenko, J. Lewis, Uniform limits of certain A harmonic functions with applications to quasiregular mappings. Ann. Acad. Sci. Fenn. AI Math. 16, 361–375 (1991)

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47. J. Lewis, Capacitary functions in convex rings. Arch. Ration. Mech. Anal. 66, 201–224 (1977) 48. J. Lewis, Note on p harmonic measure. Comput. Meth. Funct. Theor. 6(1), 109–144 (2006) 49. J. Lewis, K. Nystr¨ om, Boundary behavior for p harmonic functions in Lipschitz and ´ starlike Lipschitz ring domains. Ann. Sci. Ecole Norm. Sup. (4) 40(4), 765–813 (2007) 50. J. Lewis, K. Nystr¨ om, Boundary behavior of p harmonic functions in domains beyond Lipschitz domains. Adv. Calc. Var. 1, 1–38 (2008) 51. J. Lewis, K. Nystr¨ om, Regularity and free boundary regularity for the p Laplacian in Lipschitz and C 1 domains. Ann. Acad. Sci. Fenn. 33, 1–26 (2008) 52. J. Lewis, K. Nystr¨ om, Boundary Behaviour and the Martin Boundary Problem for p Harmonic Functions in Lipschitz Domains (with Kaj Nystr¨ om), Annals of Mathematics 172, 1907–1948 (2010) 53. J. Lewis, K. Nystr¨ om, Regularity of Lipschitz Free Boundaries in Two Phase Problems for the p Laplace Operator, Advances in Mathematics 225, 2565–2597 (2010) 54. J. Lewis, K. Nystr¨ om, Regularity of Flat Free Boundaries in Two Phase Problems for the p Laplace Operator, to appear Annales de l’Institut Henri Poincare, Analyse non lineaire. 55. J. Lewis, A. Vogel, On pseudospheres. Revista Mathematica Iberoamericana 7, 25–54 (1991) 56. J. Lewis, A. Vogel, On Some Almost Everywhere Symmetry Theorems. Nonlinear Diffusion Equations and Their Equilibrium, vol. 3 (Birkh¨ auser, Basel, 1992), pp. 347–374 57. J. Lewis, A. Vogel, A symmetry theorem revisited. Proc. Am. Math. Soc. 130(2), 443–451 (2001) 58. J. Lewis, A. Vogel, On pseudospheres that are quasispheres. Revista Mathematica Iberoamericana 17, 221–255 (2001) 59. J. Lewis, A. Vogel, Uniqueness in a free boundary problem. Comput. Meth. Funct. Theor. 31, 1591–1614 (2006) 60. J. Lewis, A. Vogel, Symmetry theorems and uniform rectiﬁability. Bound. Value Probl. 2007, 1–59 (2007) 61. J. Lewis, G. Verchota, A. Vogel, On Wolﬀ snowﬂakes. Pac. J. Math. 218(1), 139–166 (2005) 62. J. Lewis, N. Lunstr¨ om, K. Nystr¨ om, Boundary Harnack inequalities for operators of p Laplace type in Reifenberg ﬂat domains. Proc. Symp. Pure Math. 79, 229–266 (2008) 63. J. Lewis, K. Nystr¨ om, P. Poggi Corradini, p Harmonic Measure in Simply Connected Domains, Ann. Inst. Fourier Grenoble 61, 2, 689–715 (2011) 64. G. Lieberman, Boundary regularity for solutions of degenerate elliptic equations. Nonlinear Anal. 12(11), 1203–1219 (1988) 65. J. Llorente, J. Manfredi, J.M. Wu, p harmonic measure is not Additive on null sets. Ann. Scuola Norm. Sup. Pisa Cl. Sci. (5) 4(2), 357–373 (2005) 66. N. Makarov, Distortion of boundary sets under conformal mapping. Proc. Lond. Math. Soc. 51, 369–384 (1985) 67. R.S. Martin, Minimal positive harmonic functions. Trans. Am. Math. Soc. 49, 137–172 (1941) 68. V.G. Maz’ya, The continuity at a boundary point of the solutions of quasilinear elliptic equations (Russian). Vestnik Leningrad. Univ. 25(13), 42–55 (1970) 69. J. Serrin, Local behavior of solutions of quasilinear elliptic equations. Acta Math. 111, 247–302 (1964) 70. P. Wang, Regularity of free boundaries of two-phase problems for fully non-linear elliptic equations of second order. Part 1: Lipschitz free boundaries are C 1,α . Comm. Pure Appl. Math. 53, 799–810 (2000) 71. T. Wolﬀ, Plane harmonic measures live on sets of σ ﬁnite length. Ark. Mat. 31(1), 137–172 (1993)

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72. T. Wolﬀ, in Counterexamples with harmonic gradients in R3 . Essays in honor of Elias M. Stein. Princeton Mathematical Series, vol. 42 (Princeton University Press, Princeton, 1995), pp. 321–384 73. J.M. Wu, Comparisons of kernel functions, boundary Harnack principle and relative Fatou theorem on Lipschitz domains. Ann. Inst. Fourier (Grenoble) 28(4), 147–167 (1978)

Regularity of Supersolutions Peter Lindqvist

1

Introduction

The regularity theory for solutions of certain parabolic diﬀerential equations of the type ∂u = div A(x, t, u,∇u) (1) ∂t is a well developed topic, but when it comes to (semicontinuous) supersolutions and subsolutions a lot remains to be done. Supersolutions are often auxiliary tools as in the celebrated Perron method, for example, but they are also interesting in their own right. They appear as solutions to obstacle problems and variational inequalities. As a mnemonic rule vt ≥ div A(x, t, v,∇v) for smooth supersolutions. Our supersolutions are required to be lower semicontinuous but are not assumed to be diﬀerentiable in any sense: part of the theory is to prove that they have Sobolev derivatives. If one instead studies weak supersolutions that by deﬁnition belong to a Sobolev space, then one has the task to prove that they are semicontinuous. Unfortunately, the weak supersolutions do not form a good closed class under monotone convergence. For bounded functions the deﬁnitions yield the same class of supersolutions. The modern theory of viscosity solutions, created by Lions, Crandall, Evans, Ishii, Jensen, and others, relies on the appropriately deﬁned viscosity supersolutions, which are merely lower semicontinuous functions by their deﬁnition. For second order equations, these are often the same functions as those supersolutions that are encountered in potential theory. The P. Lindqvist () Department of Mathematical Sciences, Norwegian University of Science and Technology, 7491 Trondheim, Norway e-mail: [email protected]

J. Lewis et al., Regularity Estimates for Nonlinear Elliptic and Parabolic Problems, Lecture Notes in Mathematics 2045, DOI 10.1007/978-3-642-27145-8 2, © Springer-Verlag Berlin Heidelberg 2012

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link enables one to study the regularity properties also of the viscosity supersolutions. This is the case for the so-called p-Laplace equation. As an example of what we have in mind, consider the Laplace equation Δu = 0 and recall that a superharmonic function is a lower semicontinuous function satisfying a comparison principle with respect to the harmonic functions. An analogous deﬁnition comes from the super meanvalue property. General superharmonic functions are not diﬀerentiable in the classical sense. Nonetheless, the following holds. Proposition 1. Suppose that v is a superharmonic function defined in Rn . Then the Sobolev derivative ∇v exists and |∇v|q dx < ∞ BR

whenever 0 < q <

n n−1 .

Moreover, Rn

∇v, ∇η dx ≥ 0

for η ≥ 0, η ∈ C0∞ (Rn ). The fundamental solution v(x) = |x|2−n (= − log(|x|), when n = 2) is a superharmonic function showing that the summability exponent q is sharp. We seize the opportunity to mention that the superharmonic functions are exactly the same as the viscosity supersolutions of the Laplace equation. In other words, a viscosity supersolution has a gradient in Sobolev’s sense. As an example, the Newtonian potential v(x) =

cj , |x − qj |n−2

where the rational points qj are numbered and the cj ’s are convergence factors, is a superharmonic function, illustrating that functions in the Sobolev space can be inﬁnite in a dense set. The proof of the proposition follows from Riesz’s representation theorem, a classical result according to which we have a harmonic function plus a Newtonian potential. This was about the Laplace equation. A similar theorem holds for the viscosity supersolutions (= the p-superharmonic functions) of the so-called p-Laplace equation ∇· |∇v|p−2 ∇v = 0 but now 0 < q < n(p−1) n−1 . (Strictly speaking, we obtain a proper Sobolev space 1 only for p > 2 − n , because q < 1 otherwise.) The principle of superposition is not valid and, in particular, Riesz’s representation theorem is no longer

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available. The original proof in [20] was based on the obstacle problem in the calculus of variations and on the so-called weak Harnack inequality. At present, the simplest proof seems to rely upon an approximation with socalled inﬁmal convolutions |x − y|2 vε (x) = inf v(y) + , ε > 0. y 2ε At each point vε (x) v(x). They are viscosity supersolutions, if the original v is. Moreover, they are (locally) Lipschitz continuous and hence diﬀerentiable a.e. Therefore the approximants satisfy expedient a priori estimates, which, to some extent, can be passed over to the original function v itself. Another kind of results is related to the pointwise behaviour. The viscosity supersolutions are pointwise deﬁned. At each point we have v(x) = ess liminf v(y) y→x

where essential limes inferior means that sets of measure zero are neglected in the calculation of the lower limit. In the linear case p = 2 the result seems to be due to Brelot, cf. [2]. So much about the p-Laplace equation for now. The theory extends to a wider class of elliptic equations of the type div A(x, u,∇u) = 0. For parabolic equations like

p−2 ∂u ∂ ∂u ∂u 2 ∂u = , ak,m ai,j ∂t ∂xi ∂xk ∂xm ∂xj i,j k,m

where the matrix (ai,j ) satisﬁes the ellipticity condition

ai,j ξi ξj ≥ γ|ξ|2 ,

the situation is rather similar, although technically much more demanding. Now the use of inﬁmal convolutions as approximants oﬀers considerable simpliﬁcation, at least in comparison with the original proofs in [15]. We will study the Evolutionary p-Laplace Equation ∂u = div(|∇u|p−2 ∇u) ∂t

(2)

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where u = u(x, t), restricting ourselves to the slow diﬀusion case p > 2. The celebrated Barenblatt solution1 ⎧ p−1 p p−1 p−2 ⎪ ⎨ −n/λ |x| p−2 1/(1−p) C− p λ t if t > 0 1/λ t Bp (x, t) = (3) + ⎪ ⎩ 0 if t ≤ 0 where λ = n(p − 2) + p is the leading example of a viscosity supersolution (= p-supercaloric function). It has a compact support in the x-variable for each ﬁxed instance t. Disturbances propagate with ﬁnite speed and an interface (moving boundary) appears. Notice that T 0

|x|<1

|∇Bp (x, t)|p dx dt = ∞

due to the singularity at the origin. Thus Bp fails to be a weak supersolution in a domain containing the origin.2 Our main theorem is: Theorem 2. Let p > 2 and suppose that v = v(x, t) is a viscosity supersolution in the domain ΩT in Rn × R. Then v ∈ Lqloc (ΩT ) whenever q < p − 1 +

p n

and the Sobolev derivative ∇v =

∂v ∂v ,··· , ∂x1 ∂xn

exists and ∇v ∈ Lqloc (ΩT ) whenever q < p − 1 +

1 . n+1

The summability exponents are sharp. Moreover, −vηt + |∇v|p−2 ∇v,∇η dx dt ≥ 0 ΩT

for all η ≥ 0, η ∈ C0∞ (ΩT ). 1 “Einen wahren wissenschaftlichen Werth erkenne ich—auf dem Felde der Mathematik — nur in concreten mathematischen Wahrheiten, oder sch¨ arfer ausgedr¨ uckt, ‘nur in mathematischen Formeln’. Diese allein sind, wie die Geschichte der Mathematik zeigt, das Unverg¨ angliche. Die verschiedenen Theorien f¨ ur die Grundlagen der Mathematik (so die von Lagrange) sind von der Zeit weggeweht, aber die Lagrangesche Resolvente ist geblieben!” Kronecker 1884. 2 In my opinion, a deﬁnition of “supersolutions” that excludes the fundamental solution cannot be regarded as entirely satisfactory.

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The Barenblatt solution shows that the exponents are sharp. Notice that the time derivative is not included in the statement. Actually, the time derivative need not be a function, as the example v(x, t) = 0, when t ≤ 0, and v(x, t) = 1, when t > 0 shows. Dirac’s delta appears! It is worth our while to emphasize that the gradient ∇v is not present in the deﬁnition of the viscosity supersolutions (= the p-supercaloric functions). An important feature is that the viscosity supersolutions are deﬁned at each point, not just almost everywhere in their domain. When it comes to the pointwise behaviour, one may even exclude all future times so that only the instances τ < t are used for the calculation of v(x, t), as in the next theorem. (It is also, of course, valid without restriction to the past times.) Theorem 3. Let p ≥ 2. A viscosity supersolution of the Evolutionary p-Laplace Equation satisfies v(x, t) = ess liminf v(y, τ ) (y,τ )→(x,t) τ
at each interior point (x,t). In the calculation of essential limes inferior sets of (n + 1)-dimensional Lebesgue measure zero are neglected. We mention an immediate consequence, which does not seem to be easily obtained by other methods. Corollary 4. Two viscosity supersolutions that coincide almost everywhere are equal at each point. A general comment about the method employed in these notes is appropriate. We do not know about proofs for viscosity supersolutions that would totally stay within that framework. It must be emphasized that the proofs are carried out for those supersolutions that are deﬁned as one does in Potential Theory, namely through comparison principles, and then the results are valid even for the viscosity supersolutions, just because, incidentally, they are the same functions. The identiﬁcation3 of these two classes of “supersolutions” is not a quite obvious fact. This limits the applicability of the method. In passing, we also treat the measure data equation ∂v − ∇ · (|∇v|p−2 ∇v) = μ ∂t where the right-hand side is a Radon measure. It follows quite easily from Theorem 2 that each viscosity supersolution induces a measure and is a 3 Unfortunately,

the proof of this fundamental identification is not included in these notes,

only the reference [10] is given. There is an as yet unpublished much simpler proof by P. Juutinen.

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solution to the measure data equation. (The reversed problem, which starts with a given measure μ instead of a given function v, is a much investigated topic, cf. [1].) Some other equations that are susceptible of this kind of analysis are the Porous Medium Equation4 ∂u = Δ(|u|m−1 u) ∂t and

∂(|u|p−2 u) = ∇· |∇u|p−2 ∇u , ∂t but it does not seem to be known which equations of the form ∂u = F (x, t, u, ∇u, D2 u) ∂t enjoy the property of having their viscosity supersolutions in some local Sobolev x-space. I hope that this could be a fruitful research topic for the younger readers. I thank T. Kuusi and M. Parviainen for a careful reading of the manuscript.

2

The Stationary Equation

For reasons of exposition,5 we begin with the stationary equation Δp u ≡ div(|∇u|p−2 ∇u) = 0,

(4)

which oﬀers some simpliﬁcations not present in the time dependent situation. In principle, here we keep p ≥ 2, although the theory often allows that 1 n, p = n, and p < n often require separate proofs. We sometimes skip the borderline case p = n. Some general references are [8], [9], and [21]. 1,p Definition 5. We say that u ∈ Wloc (Ω) is a weak solution in Ω, if

4 The Porous Medium Equation is not well suited for the viscosity theory (it is not “proper”), but the comparison principle works well. It is not ∇v but ∇(|v|m−1 v) that is guaranteed to exist. 5 Chapter “Introduction to Random Tug-of-War Games and PDEs” is rather independent of the present chapter.

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|∇u|p−2 ∇u, ∇η dx = 0 Ω

for all η ∈ C0∞ (Ω). If, in addition, u is continuous, it is called a p-harmonic function. The weak solutions can, in accordance with the elliptic regularity theory, be made continuous after a redeﬁnition in a set of Lebesgue measure zero. The H¨older continuity estimate |u(x) − u(y)| ≤ L |x − y|α

(5)

holds when x, y ∈ B(x0 , r), B(x0 , 2r) ⊂⊂ Ω; here α depends on n and p but L also depends on the norm u p,B(x0 ,2r) . We omit the proof. The continuous weak solutions are called p-harmonic functions.6 In fact, even the gradient 1,α is continuous. One has u ∈ Cloc (Ω), where α = α(n, p). This deep result of N. Ural’tseva will not be needed here. According to [27] positive solutions obey the Harnack inequality. Lemma 6 (Harnack’s Inequality). If the p-harmonic function u is nonnegative in the ball B2r = B(x0 , 2r), then max u ≤ Cn,p min u . Br

Br

The p-Laplace equation is the Euler-Lagrange equation of a variational integral. Let us recall the Dirichlet problem in a bounded domain Ω. Let f ∈ C(Ω) ∩ W 1,p (Ω) represent the boundary values. Then there exists a unique function u in C(Ω) ∩ W 1,p (Ω) such that u − f ∈ W01,p (Ω) and

|∇u|p dx ≤ Ω

|∇(u + η)|p dx Ω

for all η ∈ C0∞ (Ω). The minimizer is p-harmonic. If the boundary ∂Ω is regular enough, the boundary values are attained in the classical sense: lim u(x) = f (ξ), ξ ∈ Ω.

x→ξ

When it comes to the super- and subsolutions, several deﬁnitions are currently being used. We need the following ones: (1) Weak supersolutions (test functions under the integral sign) 6 Thus the 2-harmonic functions are the familiar harmonic functions encountered in Potential Theory.

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(2) p-superharmonic functions (deﬁned via a comparison principle) (3) Viscosity supersolutions (test functions evaluated at points of contact) The p-superharmonic functions and the viscosity supersolutions are exactly the same functions, see [10]. They are not assumed to have any derivatives. In contrast, the weak supersolutions are by their deﬁnition 1,p required to belong to the Sobolev space Wloc (Ω) and therefore their Caccioppoli estimates are at our disposal. As we will see, locally bounded p-superparabolic functions (= viscosity supersolutions) are, indeed, weak supersolutions, having Sobolev derivatives as they should. To this one may add that the weak supersolutions are p-superharmonic functions, provided that the issue of semicontinuity be properly handled. Definition 7. We say that a function v : Ω → (−∞, ∞] is p-superharmonic in Ω, if (i) v is finite in a dense subset (ii) v is lower semicontinuous (iii) In each subdomain D ⊂⊂ Ω v obeys the comparison principle: if h ∈ C(D) is p-harmonic in D, then the implication v|∂D ≥ h|∂D

⇒

v≥h

is valid. Remarks. For p = 2 this is the classical deﬁnition of superharmonic functions due to F. Riesz. It is suﬃcient7 to assume that v ≡ ∞ instead of (i). The fundamental solution |x|(p−n)/(p−1) , is not a weak supersolution in Rn , merely because it fails to belong to the right Sobolev space, but it is p-superharmonic. The next deﬁnition is from the theory of viscosity solutions. See [5] and [17] for an overview. One deﬁnes them as being both viscosity super- and subsolutions, since it is not practical to do it in one stroke. Definition 8. Let p ≥ 2. A function v : Ω → (−∞, ∞] is called a viscosity supersolution, if (i) v is finite in a dense subset (ii) v is lower semicontinuous (iii) Whenever x0 ∈ Ω and φ ∈ C 2 (Ω) are such that v(x0 ) = φ(x0 ) and v(x) > φ(x) when x = x0 , we have div |∇φ(x0 )|p−2 ∇φ(x0 ) ≤ 0. 7 This

is not quite that simple in the parabolic case.

Regularity of Supersolutions

81

Remarks. The diﬀerential operator is evaluated only at the point of contact. The singular case 1 < p < 2 requires a modiﬁcation,8 if it so happens that ∇φ(x0 ) = 0. Notice that each point has its own family of test functions. If there is no test function touching from below at x0 , then there is no requirement: the point passes for free. Please, notice that nothing is said about the gradient ∇v, it is ∇φ(x0 ) that appears. As we mentioned, both deﬁnitions lead to the same class of “supersolutions”. Some examples are the following functions. − n−p p−1

(n − p)|x| V (x) =

V (x) =

(n = p),

V (x) =

cj |x − qj |(n−p)/(p−1)

(y) dy |x − y|(n−p)/(p−1)

(y) dy (p = 2, n ≥ 3), |x − y|n−2

(2 < p < n), (2 < p < n), v(x) = min{v1 , v2 , · · · , vm }.

The ﬁrst example is the fundamental solution, which fails to belong to the 1,p “natural” Sobolev space Wloc (Rn ).9 The second is the Newtonian potential. In the third example the cj ’s are positive convergence factors and the qj ’s are the rational points; the superposition of fundamental solutions is credited to Crandall and Zhang, cf. [3]. The last example says that one may take the pointwise minimum of a ﬁnite number of p-superharmonic functions, which is an essential ingredient in the celebrated Perron method. The functions in the next lemma, the continuous weak supersolutions, form a more tractable subclass, when it comes to a priori estimates, since they are diﬀerentiable. Lemma 9. Let v ∈ C(Ω) ∩ W 1,p (Ω). Then the following conditions are equivalent: (i) Ω |∇v|p dx ≤ Ω |∇(v + η)|p dx when η ≥ 0, η ∈ C0∞ (Ω), (ii) Ω |∇v|p−2 ∇v, ∇η dx ≥ 0 when η ≥ 0, η ∈ C0∞ (Ω), (iii) v is p-superharmonic. Proof. The equivalence of (i) and (ii) is plain. So is the necessity of (iii), stating that the comparison principle must hold. The crucial part is the suﬃciency of (iii), which will be established by the help of an obstacle problem in the calculus of variations. The function v itself will act asan obstacle for the admissible functions in the minimization of the p-energy D |∇v|p dx and it also induces the boundary values in the subdomain D. If D is a regular 8 There

is no requirement when ∇φ is 0, see [10].

9 Therefore

it is not a weak supersolution, but it is a viscosity supersolution and a p-superharmonic function.

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subdomain of Ω, then there exists a unique minimizer, say wv , in the class Fv = {w ∈ C(D) ∩ W 1,p (D)| w ≥ v, w = v on ∂D}. The crucial part is the continuity of wv cf. [24]. The solution of the obstacle problem automatically has the property (i), and hence also (ii). We claim that wv = v in D, from which the desired conclusion thus follows. The minimizer is a p-harmonic function in the open set {wv > v} where the obstacle does not hinder. On the boundary of this set wv = v. Hence the comparison principle, which v is assumed to obey, can be applied. It follows that wv ≤ v in the same set. To avoid a contradiction it must be the empty set. The conclusion is that wv = v in D, as desired. One can now deduce that (iii) is suﬃcient. 2 A function, whether continuous or not, belonging to W 1,p (Ω) and satisfying (ii) in the previous lemma is called a weak supersolution. For completeness we record below that weak supersolutions are semicontinuous “by nature”. Proposition 10. A weak supersolution v ∈ W 1,p (Ω) is lower semicontinuous (after redefinition in a set of measure zero). We can define v(x) = ess liminf v(y) y→x

pointwise. Proof. The case p > n is clear, since then the Sobolev space contains only continuous functions (Morrey’s inequality). In the range p < n we claim that v(x) = ess liminf v(y) y→x

at a.e. x ∈ Ω. The proof follows from this, because the right-hand side is always lower semicontinuous. We omit two demanding steps. First, it is required to establish that v is locally bounded. This is standard regularity theory. Second, for non-negative functions we use “the weak Harnack estimate”

q1 1 q v dx ≤ C ess inf v, Br |B2r | B2r when q < n(p − 1)/(n − p), C = C(n, p, q). This comes from the celebrated Moser iteration, cf. [27].10 Taking q = 1 and using the non-negative function 10 Harnack’s

inequality can be replaced by the more elementary estimate ess sup (v(x0 ) − v(x))+ ≤ Br

C |B2r |

B(x0 ,2r)

(v(x0 ) − v(x))+ dx

as a starting point for the proof. It follows immediately that also

Regularity of Supersolutions

83

v(x) − m(2r), where m(r) = ess inf v, Br

we have 1 0≤ |B2r | =

1 |B2r |

v dx − m(2r) B2r

(v(x) − m(2r)) dx ≤ C(m(r) − m(2r)). B2r

Since m(r) is monotone, m(r) − m(2r) → 0 as r → 0. It follows that 1 ess liminf v(y) = lim m(2r) = lim v(x) dx y→x0 r→0 r→0 |B2r | B(x ,2r) 0 at each point x0 . Lebesgue’s diﬀerentiation theorem states that the limit of the average on the right-hand side coincides with v(x0 ) at almost every point x0 . 2 Lemma 11 (Caccioppoli). Let v ∈ C(Ω) ∩ W 1,p (Ω) be a p-superharmonic function. Then p p p p ζ |∇v| dx ≤ p (osc v) |∇ζ|p dx ζ=0

Ω

Ω

holds for non-negative ζ ∈ C0∞ (Ω). If v ≥ 0, then p p p −1−α p ζ v |∇v| dx ≤ v p−1−α |∇ζ|p dx α Ω Ω when α > 0. Proof. To prove the ﬁrst estimate, ﬁx ζ and let L = sup v taken over the set where ζ = 0. Use the test function η = (L − v(x))ζ(x)p in Lemma 9(ii) and arrange the terms.

ess sup (v(x0 ) − v(x)) ≤ Br

C |B2r |

B(x0 ,2r)

|v(x0 ) − v(x)| dx.

If x0 is a Lebesgue point, the integral approaches zero as r → 0 and it follows that ess liminf v(x) ≥ v(x0 ). x→x0

The opposite inequality holds for “arbitrary” functions at their Lebesgue points. (See the end of chapter “The Problems of the Obstacle in Lower Dimension and for the Fractional Laplacian.”)

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To prove the second estimate, ﬁrst replace v(x) by v(x) + ε, if needed, and use η = v −α ζ p . 2

The rest is clear.

The special case α = p − 1 is appealing, since the right-hand member of the inequality p p ζ p |∇ log v|p dx ≤ |∇ζ|p dx (6) p−1 Ω Ω is independent of the non-negative function v itself. We aim at approximating v with functions for which Lemma 11 is valid. To this end, let v be lower semicontinuous and bounded in Ω: 0 ≤ v(x) ≤ L.

Deﬁne vε (x) = inf

y∈Ω

v(y) +

|x − y|2 2ε

,

ε > 0.

(7)

Then • • • •

vε (x) v(x) as ε → 0+ vε (x) − |x|2 /2ε is locally concave in Ω vε is locally Lipschitz continuous in Ω The Sobolev gradient ∇vε exists and belongs to L∞ loc (Ω)

The last assertion follows from Rademacher’s theorem about Lipschitz functions, cf. [7]. Thus these “inﬁmal convolutions” are rather regular. A most interesting property for a bounded viscosity supersolution is the following: Proposition 12. The approximant vε is a viscosity supersolution in the open subset of Ω where √ dist(x, ∂Ω) > 2Lε. Proof. First, notice that for x as required above, the inﬁmum is attained at some point y = x comprised in Ω. The possibility that x escapes to the boundary of Ω is prohibited by the inequalities |x − x |2 |x − x |2 ≤ + v(x ) = vε (x) ≤ v(x) ≤ L, 2ε 2ε √ |x − x | ≤ 2Lε < dist(x, ∂Ω). This explains why the domain shrinks a little. Now we give two proofs. Viscosity proof: Fix a point x0 so that also x 0 ∈ Ω. Assume that the test function ϕ touches vε from below at x0 . Using

Regularity of Supersolutions

85

ϕ(x0 ) = vε (x0 ) = ϕ(x) ≤ vε (x) ≤

|x0 − x 0 |2 + v(x 0 ) 2ε |x − y|2 + v(y) 2ε

we can verify that the function ψ(x) = ϕ(x + x0 − x 0 ) −

|x0 − x 0 |2 2ε

touches the original function v from below at the point x 0 . Since x 0 is an interior point, the inequality div |∇ψ(x 0 )|p−2 ∇ψ(x 0 ) ≤ 0 holds by assumption. Because ∇ψ(x 0 ) = ∇ϕ(x0 ), D2 ψ(x 0 ) = D2 ϕ(x0 ), we also have that

div |∇ϕ(x0 )|p−2 ∇ϕ(x0 ) ≤ 0

at the original point x0 , where ϕ was touching vε . Thus vε fulﬁlls the requirement in the deﬁnition. Proof by Comparison: Denote √ Ωε = x ∈ Ω| dist(x, ∂Ω) > 2Lε . We have to verify the comparison principle for vε . To this end, let D ⊂⊂ Ωε be a subdomain and suppose that h ∈ C(D) is a p-harmonic function so that vε (x) ≥ h(x) on the boundary ∂D or, in other words, |x − y|2 + v(y) ≥ h(x) when x ∈ ∂D, y ∈ Ω. 2ε Thus, writing y = x + z, we have w(x) ≡ v(x + z) +

|z|2 ≥ h(x), x ∈ ∂D 2ε

whenever z is a small ﬁxed vector. But also w = w(x) is a p-superharmonic function in Ωε . By the comparison principle w(x) ≥ h(x) in D. Given any point x0 in D, we may choose z = x 0 − x0 . This yields vε (x0 ) ≥ h(x0 ). Since x0 was arbitrary, we have veriﬁed that vε (x) ≥ h(x), when x ∈ D. This concludes the proof. We record the following result.

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Corollary 13. The approximant vε is a weak supersolution in Ωε , i.e., |∇vε |p−2 ∇vε , ∇η dx ≥ 0 (8) Ωε

when η ≥ 0, η ∈ C0∞ (Ωε ). Proof. This is a combination of the Proposition and Lemma 9 provided that vε obeys the comparison principle. This is clear, if v is p-superharmonic. However, when the original assumption is that v is a viscosity supersolution, then one needs the piece of knowledge that v is also p-superharmonic. This is proved in [10]. 2 The Caccioppoli estimate for vε reads ζ p |∇vε |p dx ≤ (pL)p |∇ζ|p dx, Ω

Ω

when ε is so small that the support of ζ is in Ωε . By a compactness argument (a subsequence of) ∇vε is locally weakly convergent in Lp (Ω). We conclude that ∇v exists in Sobolev’s sense and that ∇vε ∇v weakly in Lploc (Ω). By the weak lower semicontinuity of the integral also ζ p |∇v|p dx ≤ (pL)p |∇ζ|p dx. Ω

Ω

We have proved the ﬁrst part of the next theorem. Theorem 14. Suppose that v is a bounded viscosity supersolution in Ω. 1,p Then the Sobolev gradient ∇v exists and v ∈ Wloc (Ω). Moreover, |∇v|p−2 ∇v, ∇η dx ≥ 0 (9) Ω

for all η ≥ 0, η ∈ C0∞ (Ω). Proof. To conclude the proof, we show that the convergence ∇vε → ∇v is strong in Lploc (Ω), so that we may pass to the limit under the integral sign in (8). To this end, ﬁx a function θ ∈ C0∞ (Ω), 0 ≤ θ ≤ 1 and use the test function η = (v − vε )θ in the equation for vε . Then |∇v|p−2 ∇v − |∇vε |p−2 ∇vε , ∇((v − vε )θ) dx Ω

≤

|∇v|p−2 ∇v, ∇((v − vε )θ) dx −→ 0, Ω

Regularity of Supersolutions

87

where the last integral approaches zero because of the weak convergence. The ﬁrst integral splits into the sum θ|∇v|p−2 ∇v − |∇vε |p−2 ∇vε , ∇(v − vε ) dx Ω

(v − vε )|∇v|p−2 ∇v − |∇vε |p−2 ∇vε , ∇θ dx.

+ Ω

The last integral approaches zero because its absolute value is majorized by

1/p (p−1)/p (p−1)/p p p ∇θ , (v − vε ) dx |∇v| dx + |∇vε | dx p

D

D

D

where D contains the support of θ and v − vε p approaches zero. Thus we have established that θ|∇v|p−2 ∇v − |∇vε |p−2 ∇vε , ∇(v − vε ) dx Ω

approaches zero. Now the strong convergence of the gradients follows from the vector inequality 22−p |b − a|p ≤ |b|p−2 b − |a|p−2 a, b − a

(10) 2

valid for p > 2.

It also follows that the Caccioppoli estimates in Lemma 11 are valid for locally bounded p-superharmonic functions. The case when v is unbounded can be reached via the truncations vk = min{v(x), k},

k = 1, 2, 3, . . . ,

because Theorem 14 holds for these locally bounded functions. Aiming at a local result, we may just by adding a constant assume that v ≥ 0 in Ω. The situation with v = 0 on the boundary ∂Ω oﬀers expedient simpliﬁcations. We shall describe an iteration procedure, under this extra assumption. See [13]. Lemma 15. Assume that v ≥ 0 and that vk ∈ W01,p (Ω) when k = 1, 2, . . . Then p |∇vk | dx ≤ k |∇v1 |p dx Ω

Ω

and, in the case 1 < p < n Ω

n

n−p v α dx ≤ Cα 1 + |∇v1 |p dx

Ω

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P. Lindqvist

whenever α <

n(p−1) n−p .

Proof. Let j be a large index and use the test functions ηk = (vk − vk−1 ) − (vk+1 − vk ),

k = 1, 2, · · · , j − 1

in the equation for vj , i.e., |∇vj |p−2 ∇vj , ∇ηk dx ≥ 0. Ω

Indeed, ηk ≥ 0. We obtain |∇vj |p−2 ∇vj , ∇vk+1 − ∇vk dx Ak+1 =

Ω

≤

|∇vj |p−2 ∇vj , ∇vk − ∇vk−1 dx = Ak . Ω

Thus Ak+1 ≤ A1 =

|∇v1 |p dx Ω

and hence A1 + A2 + · · · + Aj ≤ jA1 . The “telescoping” sum becomes |∇vj |p dx ≤ j |∇v1 |p dx. Ω

Ω

This was the ﬁrst claim. If 1 < p < n, it follows from Tshebyshev’s and Sobolev’s inequalities that j|j ≤ v ≤ 2j|

1 p∗

≤ Ω

p∗ v2j

1

p∗

p1 1 1 p dx ≤S |∇v2j | dx ≤ S(2j) p A1p ,

Ω

where p∗ = np/(n − p). We arrive at the estimate |j ≤ v ≤ 2j| ≤ Cj −

n(p−1) n−p

n

A1n−p

for the measure of the level sets. To conclude the proof we write

v α dx =

Ω

Since

v α dx + v≤1

∞ j=1

2j−1
v α dx.

Regularity of Supersolutions

2j−1
89

v α dx ≤ C 2jα 2−(j−1)

n(p−1) n−p

n

A1n−p , 2

the series converges when α is as prescribed.

It remains to abandon the restriction about zero boundary values and to estimate |∇v1 |p dx. Ω

The reduction to zero boundary values is done locally in a ball B2r ⊂⊂ Ω. Suppose ﬁrst that v ∈ C(B2r ) ∩ W 1,p (B2r ), v ≥ 0, and deﬁne w=

v in Br h in B2r \Br

where h is the p-harmonic function in the annulus having outer boundary values zero and inner boundary values v. Now h ≤ v. The so deﬁned w is p-superharmonic in B2r , which follows by comparison. It is quite essential that the original v was deﬁned in a domain larger than Br ! We also have |∇w|p dx ≤ Crn−p (max w)p B2r

B2r

after some estimation.11 Finally, if v ∈ W 1,p (B2r ) is semicontinuous and bounded (but not necessarily continuous), then we ﬁrst modify the approximants vε deﬁned as in (10) and obtain p-superharmonic functions wε . Since 0 ≤ wε ≤ vε ≤ v, the previous estimate becomes |∇wε |p dx ≤ Crn−p (max v)p B2r

B2r

and, by the weak lower semicontinuity of the integral, we can pass to the limit as ε approaches zero. We end up with a p-superharmonic function w ∈ W01,p (B2r ) such that w = v in Br and, in particular,

|∇v| dx ≤

|∇w|p dx ≤ Crn−p (max v)p .

p

Br

11 It

B2r

B2r

is important to include the whole B2r . Of course, the Caccioppoli estimate (Lemma 11) will do over any smaller ball B , < 2r. To get the missing estimate, say over the boundary annulus B2r \ B3r/2 , the test function η = ζh works in Deﬁnition 5, where ζ = 1 in the annulus and = 0 on ∂Br . The zero boundary values of the weak solution h were essential.

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P. Lindqvist

This is the desired modiﬁed function. Now, repeat the procedure with every function min{v(x), k} in sight. We obtain

|∇v1 |p dx ≤

Br

|∇w1 |p dx ≤ Crn−p 1p B2r

for the modiﬁcation of v1 = min{v(x), 1}. We have achieved that the bounds in the previous lemma hold for the modiﬁed function over the domain B2r and a fortiori for the original v, estimated only over the smaller ball Br . Such a local estimate is all that is needed in the proof of the theorem below. Theorem 16. Suppose that v is a viscosity supersolution (= p-superharmonic function) in Ω. Then v ∈ Lqloc (Ω), whenever q <

n(p − 1) n−p

in the case 1 n. Moreover, ∇v exists in Sobolev’s sense12 and ∇v ∈ Lqloc (Ω),

whenever q <

n(p − 1) n−1

in the case 1 n we have ∇v ∈ Lploc (Ω). Finally, |∇v|p−2 ∇v, ∇η dx ≥ 0

(11)

Ω

when η ≥ 0, η ∈ C0∞ (Ω). Proof. In view of the local nature of the theorem we may assume that v > 0. According to the previous construction we can further reduce the proof to the case vk ∈ W01,p (B2r ) for each truncation at height k. The ﬁrst part of the theorem is included in Lemma 15 when 1 n is related to the fact that then all functions in the Sobolev space W 1,p are continuous. We proceed to the estimation of the gradient. First we keep 1 2 −

B

1 n

so that q ≥ 1. This can be circumvented.

Regularity of Supersolutions

91

Take q < n(p − 1)/(n − 1) and ﬁx α so that n(p − 1) (1 + α)q < . p−q n−p Continuing, the Caccioppoli estimate yields the majorant ≤ B

(1+α)q p−q

vk

1− pq dx C 2B

vkp−1−α

pq dx .

(12)

We can take v ≥ 1. Then let k −→ ∞. Clearly, the resulting majorant is ﬁnite (Lemma 15). This concludes the case 1 n we obtain that |∇ log vk |p dx ≤ Crn−p Br

from (6), where C is independent of k. Hence log vk is continuous. So is v itself. Now p |∇vk | dx = vkp |∇ log vk |p dx ≤ C v p∞ rn−p Br

Br

implies the desired p-summability of the gradient.

2

It stands to reason that the lower semicontinuous solutions of (9) are psuperharmonic functions. However, this is not known under the summability assumption ∇v ∈ Lqloc (Ω) accompanying the diﬀerential equation, if q < p and p < n. In fact, an example of Serrin indicates that even for solutions to linear equations strange phenomena occur, cf. [26]. False solutions appear, when the a priori summability of the gradient is too poor. About this topic there is nowadays a theory credited to Iwaniec, cf. [19].13

3

The Evolutionary Equation

This chapter is rather independent of the previous one. See [6], [11], [12], [28], [30], and [31] for some background. After some deﬁnitions we ﬁrst treat bounded supersolutions and then the unbounded ones. As a mnemonic rule, vt ≥ Δp v for smooth supersolutions, ut ≤ Δp u for smooth subsolutions. We need the following classes of supersolutions: (1) Weak supersolutions (test functions under the integral sign) (2) p-supercaloric functions (deﬁned via a comparison principle) (3) Viscosity supersolutions (test functions evaluated at points of contact) 13 The

“pathological solutions” of Serrin are now called “very weak solutions”.

92

P. Lindqvist

The weak supersolutions do not form a good closed class under monotone convergence.

3.1

Definitions

We ﬁrst deﬁne the concept of solutions, p-supercaloric functions and viscosity supersolutions. Then we state the main theorem. The section ends with an outline of the procedure for the proof. Suppose that Ω is a bounded domain in Rn and consider the space-time cylinder ΩT = Ω × (0, T ). Its parabolic boundary consists of the portions Ω × {0} and ∂Ω × [0, T ]. Definition 17. In the case14 p ≥ 2 we say that u ∈ Lp (0, T ; W 1,p (Ω)) is a weak solution of the Evolutionary p-Laplace Equation, if T 0

(−uφt + |∇u|p−2 ∇u,∇φ) dx dt = 0

(13)

Ω

for all φ ∈ C01 (ΩT ). If the integral in (13) is ≥ 0 for all test functions φ ≥ 0, we say that u is a weak supersolution. In particular, one has the requirement T 0

(|u|p + |∇u|p ) dx dt < ∞. Ω

By the regularity theory one may regard a weak solution u = u(x, t) as continuous.15 For simplicity we call the continuous weak solutions for pcaloric functions.16 The interior H¨ older estimate 17 takes the following form for solutions according to [6]. In the subdomain D × (δ, T − δ) |u(x1 , t1 ) − u(x2 , t2 )| ≤ γ u L∞(ΩT ) |x1 − x2 |α + |t1 − t2 |α/p ,

(14)

where the positive exponent α depends only on n and p, while the constant γ depends, in addition, on the distance to the subdomain. 14 The singular case 1 < p < 2 requires an extra a priori assumption, for example, u ∈ L∞ (0, T ; L2 (Ω)) will do. 15 The weak supersolutions are lower semicontinuous according to [18], see chapter “The Problems of the Obstacle in Lower Dimension and for the Fractional Laplacian.” 16 One

may argue that this is more adequate than “p-parabolic functions”, which is in use.

17 This

is weaker than the estimate in [6]. See also [29] for intrinsic scaling.

Regularity of Supersolutions

93

Definition 18. We say that a function v : ΩT → (−∞, ∞] is p-supercaloric in ΩT , if (i) v is finite in a dense subset (ii) v is lower semicontinuous (iii) In each cylindrical subdomain D×(t1 , t2 ) ⊂⊂ ΩT v obeys the comparison principle: if h ∈ C(D × [t1 , t2 ]) is p-caloric in D × (t1 , t2 ), then v ≥ h on the parabolic boundary of D × (t1 , t2 ) implies that v ≥ h in the whole subdomain. The next deﬁnition is from the theory of viscosity solutions. One deﬁnes them as being both viscosity super- and subsolutions, since it is not practical to do it in one stroke. Definition 19. Let p ≥ 2. A function v : ΩT → (−∞, ∞] is called a viscosity supersolution, if (i) v is finite in a dense subset (ii) v is lower semicontinuous (iii) Whenever (x0 , t0 ) ∈ ΩT and φ ∈ C 2 (ΩT ) are such that v(x0 , t0 ) = φ(x0 , t0 ) and v(x, t) > φ(x, t) when (x, t) = (x0 , t0 ), we have φt (x0 , t0 ) ≥ ∇ · |∇φ(x0 , t0 )|p−2 ∇φ(x0 , t0 ) . The p-supercaloric functions are exactly the same as the viscosity supersolutions. For a proof of this fundamental equivalence we refer to [10]. The leading example is the Barenblatt solution, which is a viscosity supersolution in the whole Rn+1 . Another example is any function of the form v(x, t) = g(t), where g(t) is an arbitrary monotone increasing lower semicontinuous function. We also mention ε v(x, t) + , 0 < t < T, T −t v(x, t) = min{v1 (x, t), . . . , v2 (x, t)}. Finally, if v ≥ 0 is a viscosity supersolution, so is the function obtained by redeﬁning v(x, t) = 0 when t ≤ 0. –These were some examples. An outline of the procedure is the following. We aim at proving the summability results (Theorem 2) for a viscosity supersolution v. Step 1. Assume ﬁrst that v is bounded. Step 2. Approximate v locally with inﬁmal convolutions vε . These are diﬀerentiable.

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P. Lindqvist

Step 3. The inﬁmal convolutions are viscosity supersolutions and they are shown to be weak supersolutions of the equation (with test functions under the integral sign). Step 4. Estimates of the Caccioppoli type for vε are extracted from the equation. Step 5. The Caccioppoli estimates are passed over from vε to v. This concludes the proof for bounded functions. Step 6. The unbounded case is reached via the bounded viscosity supersolutions vk = min{v, k}, k = 1, 2, · · · , for which the results in Step 5 already are available. Step 7. An iteration with respect to the index k is designed so that the ﬁnal result does not blow up as k → ∞. This works well when the parabolic boundary values (in the subdomain studied) are zero. Step 8. An extra construction is performed to reduce the proof to the situation of zero parabolic boundary values (so that the iterated result in Step 7 is at our disposal).

3.2

Bounded Supersolutions

We aim at proving Theorem 2. The ﬁrst step is to consider bounded viscosity supersolutions. We want to prove that they are weak supersolutions. We ﬁrst approximate them with their inﬁmal convolutions. Then estimates mainly of the Caccioppoli type are proved for these approximants. Finally, the so obtained estimates are passed over to the original functions. Assume therefore that 0 ≤ v(x, t) ≤ L, (x, t) ∈ ΩT = Ω × (0, T ). The approximants vε (x, t) =

inf

(y,τ )∈ΩT

|x − y|2 + |t − τ |2 v(y, τ ) + 2ε

,

ε > 0,

have the properties • • • •

vε (x, t) v(x, t) as ε → 0+ 2 2 vε (x, t) − |x|2ε+t is locally concave in ΩT vε is locally Lipschitz continuous in ΩT ∞ ε The Sobolev derivatives ∂v ∂t and ∇vε exist and belong to Lloc (ΩT )

The last assertion follows from Rademacher’s theorem about Lipschitz functions. Thus these “inﬁmal convolutions” are rather regular. The existence of the time derivative is very useful. A most interesting property for a bounded viscosity supersolution is the following:

Regularity of Supersolutions

95

Proposition 20. The approximant vε is a viscosity supersolution in the open subset of ΩT where dist((x, t), ∂ΩT ) >

√ 2Lε.

Proof. First, notice that for (x, t) as required above, the inﬁmum is attained at some point (y, τ ) = (x , t ) comprised in ΩT . The possibility that (x , t ) escapes to the boundary of Ω is prohibited by the inequalities |x − x |2 + |t − t |2 |x − x |2 + |t − t |2 ≤ + v(x , t ) 2ε 2ε

= vε (x, t) ≤ v(x, t) ≤ L, √ |x − x |2 + |t − t |2 ≤ 2Lε < dist((x, t), ∂ΩT ).

Thus the domain shrinks a little. Again there are two proofs. Viscosity proof: Fix a point (x0 , t0 ) so that also (x 0 , t 0 ) ∈ ΩT . Assume that the test function ϕ touches vε from below at (x 0 , t 0 ). Using ϕ(x0 , t0 ) = vε (x0 , t0 ) = ϕ(x, t) ≤ vε (x, t)

≤

|x0 − x 0 |2 + |t0 − t 0 |2 + v(x 0 , t 0 ) 2ε |x − y|2 + |t − τ |2 + v(y, τ ) 2ε

we can verify that the function ψ(x, t) = ϕ(x + x0 − x 0 , t + t0 − t 0 ) −

|x0 − x 0 |2 + |t0 − t 0 |2 2ε

touches the original function v from below at the point (x 0 , t 0 ). Since (x 0 , t 0 ) is an interior point, the inequality div |∇ψ(x 0 , t 0 )|p−2 ∇ψ(x 0 , t 0 ) ≤ ψt (x 0 , t 0 ) holds by assumption. Because ψt (x 0 , t 0 ) = ϕt (x0 , t0 ), ∇ψ(x 0 , t 0 ) = ∇ϕ(x0 , t0 ), D2 ψ(x 0 , t 0 ) = D2 ϕ(x0 , t0 ) we also have that div |∇ϕ(x0 , t0 )|p−2 ∇ϕ(x0 , t0 ) ≤ ϕt (x0 , t0 ) at the original point (x0 , t0 ), where ϕ was touching vε . Thus vε fulﬁlls the requirement in the deﬁnition.

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P. Lindqvist

Proof by Comparison: We have to verify the comparison principle for vε √ in a subcylinder Dt1 ,t2 having at least the distance 2Lε to the boundary of ΩT . To this end, assume that h ∈ C(Dt1 ,t2 ) is a p-caloric function such that vε ≥ h on the parabolic boundary. It follows that the inequality |x − y|2 + |t − τ |2 + v(y, τ ) ≥ h(x, t) 2ε is available when (y, τ ) ∈ ΩT and (x, t) is on the parabolic boundary of Dt1 ,t2 . Fix an arbitrary point (x0 , t0 ) in Dt1 ,t2 . Then we can take y = x + x 0 − x0 and τ = t + t 0 − t0 in the inequality above so that w(x, t) ≡ v(x + x 0 − x0 , t + t 0 − t0 ) +

|x0 − x 0 |2 + |t0 − t 0 |2 ≥ h(x, t) 2ε

when (x, t) is on the parabolic boundary. But the translated function w is p-supercaloric in the subcylinder Dt1 ,t2 . By the comparison principle w ≥ h in the whole subcylinder. In particular, vε (x0 , t0 ) = w(x0 , t0 ) ≥ h(x0 , t0 ). This proves the comparison principle for vε .

2

The “viscosity proof” did not contain any comparison principle while the “proof by comparison” required the piece of knowledge that the original v obeys the principle. To proceed, this comparison is unevitable. It is needed for vε . As a viscosity supersolution vε obeys the comparison principle in its domain according to [10]. This parabolic comparison principle allows comparison in space-time cylinders. We will encounter domains of a more general shape, but the following elliptic version of the principle turns out to be enough for our purpose. Instead of the expected parabolic boundary, the whole boundary (the “Euclidean” boundary) appears. Proposition 21. Given a domain Υ ⊂⊂ Ωε and a p-caloric function h ∈ C(Υ ), then vε ≥ h on the whole boundary ∂Υ implies that vε ≥ h in Υ . Now Υ does not have to be a space-time cylinder and ∂Υ is the total boundary in Rn+1 . Proof. It is enough to realize that the proof is immediate when Υ is a ﬁnite union of space-time cylinders Dj × (aj , bj ). To verify this, just start with the earliest cylinder(s) and pay due attention to the passages of t over the aj ’s and bj ’s. Then the general case follows by exhausting Υ with such unions. Indeed, given α > 0 the compact set {h(x, t) ≥ vε (x, t)} is contained in an open ﬁnite union Dj × (aj , bj )

Regularity of Supersolutions

97

comprised in Υ so that h < vε + α on the (Euclidean) boundary of the union. It follows that h ≤ vε + α in the whole union. Since α was arbitrary, we conclude that vε ≥ h in Υ . 2 The above elliptic comparison principle does not acknowledge the presence of the parabolic boundary. The reasoning above can easily be changed so that the latest portion of the boundary is exempted. For this improvement, suppose that t < T for all (x, t) ∈ Υ ; in this case ∂Υ may have a plane portion with t = T . It is now suﬃcient to verify that vε ≥ h

when t < T

on ∂Υ

in order to conclude that vε ≥ h in Υ . To see this, just use vε +

σ −t

T

in the place of vε and then let σ → 0+. This variant of the comparison principle is convenient for the proof of the following conclusion. Lemma 22. The approximant vε is a weak supersolution in the shrunken domain, i.e., T 0

−vε

Ω

∂φ + |∇vε |p−2 ∇vε , ∇φ dx dt ≥ 0 ∂t

(15)

whenever φ ∈ C0∞ (Ωε × (ε, T − ε)), φ ≥ 0. Proof. We show that in a given subdomain Dt1 ,t2 = D × (t1 , t2 ) of the “shrunken domain” our vε coincides with the solution of an obstacle problem. The solutions of the obstacle problem are per se weak supersolutions. Hence, so is vε . Consider the class of all functions ⎧ p 1,p ⎪ ⎪ ⎨w ∈ C(Dt1 ,t2 ) ∩ L (t1 , t2 ; W (D)), w ≥ vε in Dt1 ,t2 , and ⎪ ⎪ ⎩w = v on the parabolic boundary of D . ε

t1 ,t2

The function vε itself acts as an obstacle and induces the boundary values. There exists a (unique) weak supersolution wε in this class satisfying the variational inequality

t2

t1

∂ψ + |∇wε |p−2 ∇wε , ∇(ψ − wε ) dx dt (ψ − wε ) ∂t D 1 (ψ(x, t2 ) − wε (x, t2 ))2 dx ≥ 2 D

98

P. Lindqvist

for all smooth ψ in the aforementioned class. Moreover, wε is p-caloric in the open set Aε = {wε > vε }, where the obstacle does not hinder. We refer to [4]. On the boundary ∂Aε we know that wε = vε except possibly when t = t2 . By the elliptic comparison principle we have vε ≥ wε in Aε . On the other hand wε ≥ vε . Hence wε = vε . To ﬁnish the proof, let ϕ ∈ C0∞ (Dt1 ,t2 ), ϕ ≥ 0, and choose ψ = vε + ϕ = wε + ϕ above. An easy manipulation yields (15). 2 Recall that 0 ≤ v ≤ L. Then also 0 ≤ vε ≤ L. An estimate for ∇vε is provided in the well-known lemma below. Lemma 23 (Caccioppoli). The inequality T

T ζ |∇vε | dx dt ≤ CL p

0

p

T p ∂ζ |∇ζ| dx dt + CL dx dt Ω Ω ∂t 0

p 0

Ω

2

p

holds whenever ζ ∈ C0∞ (Ωε × (ε, T − ε)), ζ ≥ 0. Proof. Use the test function φ(x, t) = (L − vε (x, t))ζ p (x, t).

2

The Caccioppoli estimate above leads to the conclusion that, keeping 0 ≤ v ≤ L, the Sobolev gradient ∇v ∈ Lploc exists and ∇vε ∇v weakly in Lploc , at least for a subsequence. For v the Caccioppoli estimate T 0

T ζ p |∇v|p dx dt ≤ CLp

Ω

0

|∇ζ|p dx dt + CL2

Ω

T p ∂ζ dx dt, 0 Ω ∂t

is immediate, because of the lower semicontinuity of the integral under weak convergence. However the corresponding passage to the limit under the integral sign of T 0

Ω

−vε

∂φ + |∇vε |p−2 ∇vε , ∇φ dx dt ≥ 0 ∂t

requires a justiﬁcation, as ε → 0. The elementary vector inequality p−2 |b| b − |a|p−2 a ≤ (p − 1)|b − a| (|b| + |a|)p−2 , p > 2, and H¨older’s inequality show that it is suﬃcient to establish that ∇vε −→ ∇v strongly in Lp−1 loc ,

(16)

Regularity of Supersolutions

99

to accomplish the passage. Notice the exponent p−1 in place of p. This strong convergence is given in the next theorem, where the sequence is renamed to vk . The proof is from [22]. Theorem 24. Suppose that v1 , v2 , v3 , . . . is a sequence of Lipschitz continuous weak supersolutions, such that 0 ≤ vk ≤ L in ΩT = Ω × (0, t), vk → v in Lp (ΩT ). Then ∇v1 , ∇v2 , ∇v3 , . . . is a Cauchy sequence in Lp−1 loc (ΩT ). Proof. The central idea is that the measure of the set where |vj − vk | > δ is small. Given δ > 0, we have, in fact, mes{|vj − vk | > δ} ≤ δ −p vj − vk pp

(17)

according to Tshebyshef’s inequality. Fix a test function θ ∈ C0∞ (ΩT ), 0 ≤ θ ≤ 1. From the Caccioppoli estimate we can extract a bound of the form |∇vk |p dx dt ≤ Ap , k = 1, 2, . . . , {θ=0}

since the support is a compact subset. Fix the indices k and j and use the test function ϕ = (δ − wjk )θ where wjk

⎧ if vj − vk > δ ⎨ δ, = vj − vk , if |vj − vk | ≤ δ ⎩ −δ, if vj − vk < −δ

in the equation T ∂φ + |∇vj |p−2 ∇vj , ∇φ dx dt ≥ 0. −vj ∂t Ω 0 Since |wjk | ≤ δ, we have ϕ ≥ 0. In the equation for vk use ϕ = (δ + wjk )θ. Add the two equations and arrange the terms:

θ|∇vj |p−2 ∇vj − |∇vk |p−2 ∇vk , ∇vj − ∇vk dx dt T ≤ δ 0 Ω |∇vj |p−2 ∇vj + |∇vk |p−2 ∇vk , ∇θ dx dt

|vj −vk |≤δ

100

P. Lindqvist

−

T

wjk |∇vj |p−2 ∇vj − |∇vk |p−2 ∇vk , ∇θ dx dt T T ∂ (θwjk ) dx dt − δ 0 Ω (vj + vk ) ∂θ + 0 Ω (vj − vk ) ∂t ∂t dx dt 0

Ω

= I − II + III − IV. The left-hand side is familiar from inequality (22). As we will see, the righthand side is of the magnitude O(δ). We begin with term III, which contains time derivatives that ought to be avoided. Integration by parts yields T

2 T wjk ∂θ θ ∂θ ∂t ( 2 ) dx dt + 0 Ω (vj − vk )wjk ∂t dx dt T T ∂θ 2 ∂θ = − 12 0 Ω wjk ∂t dx dt + 0 Ω (vj − vk )wjk ∂t dx dt.

III =

0

Ω

We obtain the estimate |III| ≤

1 2 δ θt 1 + 2Lδ θt 1 ≤ δC3 . 2

For the last term we immediately have |IV | ≤ 2δL θt 1 = δC4 . The two ﬁrst terms are easy, |I| ≤ δC1 ,

|II| ≤ δC2 .

Summing up, |I| + |II| + |III| + |IV | ≤ Cδ. Using the vector inequality (10) to estimate the left hand side, we arrive at

|vj −vk |≤δ

θ|∇vj − ∇vk |p dx dt ≤ 2p−2 δC, 1

|vj −vk |≤δ

θ|∇vj − ∇vk |p−1 dx dt = O(δ 1− p ).

We also have in virtue of (17) |vj −vk |>δ

θ|∇vj − ∇vk |p−1 dx dt

≤ δ −1 vj − vk p ( ∇vj p + ∇vk p )

p−1

≤ (2A)p−1 δ −1 vj − vk p . Finally, combining the estimates over the sets |vj − vk | ≤ δ and |vj − vk | > δ, we have an integral over the whole domain:

Regularity of Supersolutions

T 0

101

1

θ|∇vj − ∇vk |p−1 dx dt ≤ O(δ 1− p ) + C5 δ −1 vj − vk p .

Ω

Since the left-hand side is independent of δ, we can make it as small as we please, by ﬁrst ﬁxing δ small enough and then taking the indices large enough. 2 We have arrived at the following result for bounded supersolutions. Theorem 25. Let v be a bounded viscosity supersolution of vt = Δp v in ΩT , p ≥ 2. Then

∂v ∂v ∇v = ,..., ∂x1 ∂xn exists in Sobolev’s sense, ∇v ∈ Lploc (ΩT ), and T 0

−v

Ω

∂φ + |∇v|p−2 ∇v, ∇φ dx dt ≥ 0 ∂t

for all non-negative compactly supported test functions ϕ. Notice that so far the exponent is p, as it should for v bounded. Remark. It was established that the bounded viscosity supersolutions are also weak supersolutions. In the opposite direction, according to [18] every weak supersolution is lower semicontinuous upon a redeﬁnition in a set of (n+1)-dimensional measure 0. Moreover, the representative obtained as ess liminf v(y, τ )

(y,τ )→(x,t)

will do. A proof is given in chapter “The Problems of the Obstacle in Lower Dimension and for the Fractional Laplacian.” We need a few auxiliary results. Lemma 26 (Sobolev’s inequality). If u ∈ Lp (0, t; W01,p (Ω)), then T 0

2

|u|p(1+ n ) dx dt ≤ C

Ω

T 0

Ω

np |∇u|p dx dt ess sup |u(x, t)|2 dx . 0
Ω

Proof. See for example [6, Chap. 1, Proposition 3.1]. If the test function φ is zero on the lateral boundary ∂Ω × [t1 , t2 ], then the diﬀerential inequality for the weak supersolution takes the form t2

∂φ p−2 −v + |∇v| ∇v, ∇φ dx dt ∂t t1 Ω + Ω v(x, t2 )φ(x, t2 ) dx ≥ Ω v(x, t1 )φ(x, t1 ) dx.

102

P. Lindqvist

Thus, if v is zero on the lateral boundary, we may take φ = v above. We obtain t2 1 1 2 p v(x, t1 ) dx ≤ |∇v| dx dt + v(x, t2 )2 dx, (18) 2 Ω 2 Ω t1 Ω which estimates the past in terms of the future and an “energy term”.

3.3

Unbounded Supersolutions

We proceed to study an unbounded viscosity supersolution (= p-supercaloric function) v. Let us brieﬂy describe the method taken from [14]. The starting point is to apply Theorem 24 on the functions vk = min{v, k} so that estimates depending on k = 1, 2, · · · are obtained. Then an iterative procedure is used to gradually increase the summability exponent of v. First, we achieve that v α for some small exponent α < p− 2. That result is iterated, again using the vk ’s till we come close to the exponent α = p − 1 − 0. Then the passage over p − 1 requires a special, although simple, device. At the end we will reach the desired summability for the function v itself. From this it is not diﬃcult to obtain the corresponding result also for the gradient ∇v. Again the vk ’s are employed. The considerations are in a bounded subdomain, which we again call ΩT = Ω × (0, T ), for simplicity. The situation is easier when the function is zero on the parabolic boundary: v(x, 0) = 0 when x ∈ Ω,

v = 0 on ∂Ω × [0, T ].

We may assume that v ≥ 0. The functions vk (x, t) = min{v(x, t), k},

k = 0, 1, 2, . . .

cut oﬀ at the height k are bounded, whence the previous results apply for them. Fix a large index j. We may use the test functions φk = (vk − vk−1 ) − (vk+1 − vk ), k = 1, 2, . . . , j in the equation τ 0

τ |∇vj |p−2 ∇vj , ∇φj dx dt +

Ω

0

φk Ω

∂vj dx dt ≥ 0, ∂t

where 0 < τ ≤ T. Indeed, φk ≥ 0. The “forbidden” time derivative can be avoided through an appropriate regularization. In principle vj is ﬁrst replaced by its convolution with a molliﬁer and later the limit is to be taken. We

Regularity of Supersolutions

103

postpone this complication in order to keep the exposition more transparent. The insertion of the test function yields τ 0

Ω

≤

|∇vj |p−2 ∇vj , ∇(vk+1 − vk ) + (vk+1 − vk )

τ 0

∂vj dx dt ∂t

|∇vj |p−2 ∇vj , ∇(vk − vk−1 ) + (vk − vk−1 )

Ω

∂vj dx dt, ∂t

succinctly written as ak+1 (τ ) ≤ ak (τ ). It follows that

j

ak (τ ) ≤ ja1 (τ )

k=1

and, since the sum is “telescoping”, we have the result below. Lemma 27. If each vk ∈ Lp (0, T ; W01,p (Ω)) and vk (x, 0) = 0 when x ∈ Ω, then τ 1 |∇vj |p dx dt + v 2 (x, τ ) dx 2 Ω j 0 Ω τ p ≤j |∇v1 | dx dt + j vj (x, τ ) dx 0

Ω

Ω

holds for a.e. τ in the range 0 < τ ≤ T . Before continuing, we justify the use of the time derivative in the previous reasoning. Regularisation of the equation. We use the convolution (f ρε )(x, t) =

∞

−∞

f (x, t − s)ρε (s) ds,

(19)

where ρε is, for instance, Friedrich’s molliﬁer deﬁned as ρε (t) =

C −ε2 /(ε2 −t2 ) , εe

|t| < ε,

0,

|t| ≥ ε.

If the function vj is extended as 0 when t ≤ 0, x ∈ Ω, the new function is p-supercaloric in Ω × (−∞, T ), indeed. To see this, one has only to verify the comparison principle. We have, when τ ≤ T − ε

∂ (|∇vj |p−2 ∇vj ) ρε , ∇ϕ + ϕ (vj ρε ) dx dt ≥ 0 ∂t −∞ Ω

τ

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P. Lindqvist

for all test functions ϕ ≥ 0 vanishing on the lateral boundary. Replace vk in the previous proof by v˜k = min{vj ρε , k} and choose ϕk = (˜ vk − v˜k−1 ) − (˜ vk+1 − v˜k ). Since the convolution with respect to the time variable does not aﬀect the zero boundary values on the lateral boundary, we conclude that v˜k vanishes on the parabolic boundary of Ω × (−δ/2, T − δ/2), when ε < δ/2 and δ can be taken as small as we wish. (The functions vk ρε instead of the employed (v ρε )k do not work well in this proof.) The same calculations as before yield ˜k (τ ) a ˜k+1 (τ ) ≤ a

j

and

a ˜k (τ ) ≤ j a ˜1 (τ ),

k=1

where a ˜k (τ ) =

τ

−δ/2 Ω

τ

(|∇vj |p−2 ∇vj ) ρε , ∇(˜ vk − v˜k−1 ) dx dt

+ −δ/2 Ω

(˜ vk − v˜k−1 )

∂ (vj ρε ) dx dt. ∂t

Summing up, we obtain j k=1

a ˜k (τ ) =

τ

−δ/2 Ω

(|∇vj |p−2 ∇vj ) ρε , ∇˜ vj dx dt

τ

+ −δ/2 Ω

v˜j

∂ (vj ρε ) dx dt, ∂t

where the last integral can be written as 1 (vj ρε )2 (x, τ ) dx. 2 Ω Also for a ˜1 (τ ) we get an expression free of time derivatives. Therefore we can safely ﬁrst let ε → 0 and then δ → 0. This leads to the lemma. Let us return to the lemma. Provided that we already have a majorant for the term T

0

|∇v1 |p dx dt, Ω

Regularity of Supersolutions

we see that

105

T 0

|∇vj |p dx dt = O(j 2 ).

Ω

Yet, the right magnitude is O(j). Lemma 28. Suppose that vj ∈ Lp (0, T ; W01,p(Ω)) and T 0

|∇vj |p dx dt ≤ Kj 2 ,

j = 1, 2, 3, . . .

Ω

Then v ∈ Lqloc (ΩT ), whenever q 2.) Proof. (Recall that the desired bound is p − 1 + np and not the above p − 2.) The assumption and Sobolev’s inequality will give us a bound on the measure of the level sets Ej = {(x, t)| j ≤ v(x, t) ≤ 2j} so that the integral can be controlled. To this end, denote κ=1+

2 . n

We have T

j |Ej | ≤ κp

Ej

κp v2j

T ≤C

0

Ω

dx dt ≤

0

Ω

κp v2j dx dt

np p 2 |∇v2j |p dx dt · ess sup v2j dx ≤ CKj 2 4|Ω|j 2 n . 0
It follows that

Ω

|Ej | ≤ Constant j 2−p .

We use this to estimate the Lq -norm using a dyadic division of the domain. Thus T v dx dt ≤ T |Ω| + q

0

Ω

∞ j=1

≤ T |Ω| +

∞

v q dx dt

E2j−1

2jq |E2j−1 |

j=1

≤ T |Ω| + C

∞

2j(q+2−p) ,

j=1

which is a convergent majorant when q < p − 2.

2

106

P. Lindqvist

Remark. If the majorant Kj 2 in the assumption is replaced by a better Kj γ , then the procedure yields that |Ej | ≈ j γ−p resulting in q 2. To improve the exponent, we start from Lemma 27 and write the estimate in the form

t1

0

T |∇vj | dx dt ≤ j

Ω

|∇v1 | dx dt + j

p

p

0

vj (x, τ ) dx,

Ω

Ω

where 0 < t1 ≤ τ ≤ T . Integrate with respect to τ over the integral [t1 , T ]: (T − t1 )

0

t1

T

|∇vj | dx dt ≤ j(T − t1 )K + j p

Ω

≤ j(T − t1 )K + j

2−ε

vj (x, t) dx dt

t1

T

v ε (x, t) dx dt. t1

Thus we have reached the estimate t1 |∇vj |p dx dt ≤ j 2−ε K1 , 0

Ω

Ω

(t1 < T ).

(20)

Ω

This is an improvement from j 2 to j 2−ε , but we have to obey the restriction that ε ≤ 1, because the term j(T − t1 )K was absorbed. Estimating again the measures |Ej |, but starting with the bound j γ K, γ = 2 − ε in place of j 2 K, yields |Ej | ≈ j 2−p−ε 0 < ε ≤ 1. The result is that t1 v q dx dt < ∞ 0

when

0 < q p − 1. 18 It

does not work for the Heat Equation!

Regularity of Supersolutions

107

We have to stop, because the previous exponent (k − 1)(p − 2) + ε has to obey the rule not to become larger than 1. This way we can reach that v ∈ L1 (ΩT ), with a T < T , which will do to proceed. In fact, adjusting we can reach any exponent strictly below p − 1, but the passage over the exponent p − 1 requires a special device. Since only a ﬁnite number of steps were involved, we can take T as close to T as we wish. We use inequality (18) in the form 1 2

vj (x, t)2 dx ≤

Ω

τ |∇vj |p dx dt +

0

Ω

1 2

vj (x, τ )2 dx,

(21)

Ω

where t < τ . For t1 < τ < T it follows that

vj (x, t) dx ≤ 2

ess sup 0
τ

2

Ω

vj (x, τ )2 dx

|∇vj | dx dt + p

0

Ω

Ω

τ ≤ 2j

0

≤ 2j

|∇v1 |p dx dt + 2j Ω

vj (x, τ ) dx Ω

T

|∇v1 |p dx dt + 2j

0

vj (x, τ ) dx,

Ω

Ω

where the second step is from Lemma 27. We integrate the resulting inequality with respect to τ over the interval [t1 , T ], which aﬀects only the last integral. Upon division by T − t1 , the last term is replaced by 2j T − t1

T vj dx dt. t1

Ω

We can combine this and the earlier estimate 0

t1

T |∇vj |p dx dt ≤ j

Ω

0

|∇v1 |p dx dt + Ω

j 2−ε T − t1

T v ε dx dt, t1

Ω

taking ε = 1, so that we ﬁnally arrive at 0

t1 Ω

≤ 3j

|∇vj |p dx dt + ess sup T 0

0
|∇v1 |p dx dt + Ω

vj (x, t)2 dx

Ω

3j T − t1

T v dx dt. t1

Ω

(22)

108

P. Lindqvist

The majorant is now O(j), which is of the right order, as the following lemma shows with its sharp exponents. Lemma 29. If T

vj (x, t)2 dx ≤ jK

|∇vj |p dx dt + ess sup

0

0
Ω

Ω

when j = 1, 2, . . . then v ∈ Lq (ΩT )

whenever

∇v ∈ Lq (ΩT )

whenever

p , n 1 . 0
Proof. The ﬁrst part is a repetition of the proof of Lemma 28. Denote again 2 Ej = {(x, t)| j ≤ v(x, t) ≤ 2j}, κ = 1 + . n We have as before j κp |Ej | ≤ Ej

κp v2j dx dt ≤

T ≤C

T 0

Ω

κp v2j dx dt

np p p 2 |∇v2j | dx dt · ess sup v2j dx ≤ CK 1+ n (2j)1+ n . p

0

Ω

0
It follows that

Ω

p

|Ej | ≤ Constant × j 1−p− n .

We estimate the Lq -norm using the subdivision of the domain. Thus T v dx dt ≤ T |Ω| + q

0

Ω

∞ j=1

≤ T |Ω| +

∞

v q dx dt

E2j−1

2jq |E2j−1 |

j=1

≤ T |Ω| + C

∞

p

2j(q+1−p− n ) ,

j=1

which converges in the desired range for q. Thus the ﬁrst part is proved. For the summability of the gradient, we use the bound on the measure of the level sets Ej and also the growth assumed for the energy of the truncated functions. Fix a large index k and write, using that |∇vk | ≤ |∇v2j | on E2j−1 :

Regularity of Supersolutions

109

T 0

|∇vk |q dx dt Ω

∞ j=1

≤

|∇vk |q dx dt E2j−1

∞

≤

j=1

≤

∞

0

Ω

|∇vk |p dx dt

E2j−1

j=1

∞ T

pq

pq

q

|E2j−1 |1− p

q

p

|∇v2j | dx dt

2(j−1)(1− p )(1−p− n )

q

q

p

p

2(j−1)(1− p )(1−p− n ) (2j K) p ,

j=1

where the geometric series converges provided that q < p − 1 + 1/(n + 1). Strictly speaking, the “ﬁrst term” K q/p (T |Ω|)1−q/p ought to be added to the sum, since the integral over the set {0 < v < 1} was missing. Now we may let k go to inﬁnity. 2

3.4

Reduction to Zero Boundary Values

It remains to reduce the situation to a p-supercaloric function with zero boundary values and to estimate the p-energy of the ﬁrst truncated function so that the majorant in Lemma 27 is ﬁnite. The constructions have to be performed only on the truncated functions vk and so we are dealing only with bounded functions. A ﬁrst reduction comes from the fact that if v is a nonnegative p-supercaloric function, so is the function obtained by redeﬁning v(x, t) as 0 when t ≤ δ. Thus the initial values can be regarded as zero. Given v so that v ≥ 0 in, say, the closure of B2r × (0, T ), we ﬁrst redeﬁne v(x, t) as 0 when t ≤ δ, where δ is small. Then we want to deﬁne the modiﬁed function ⎧ ⎨ v(x, t), if (x, t) ∈ Br × (δ, T ) w = h(x, t), if (x, t) ∈ (B2r \ Br ) × (δ, T ) ⎩ 0, if t ≤ δ where h is the p-caloric function in the domain (B2r \ Br ) × (0, T ) having 0 (this is on ∂B2r × (0, T )), v (this is on ∂Br × (0, T )), Initial values 0 (this is when t = 0). Outer boundary values Inner boundary values

By the maximum principle, h(x, t) = 0 when t < δ. Thus also w(x, t) = 0 when t < δ.

110

P. Lindqvist

1,p This construction is easily done for v ∈ C(ΩT ) ∩ Lp (0, T ; Wloc (Ω) It is suﬃcient to do this modiﬁcation for each of the truncated functions appearing in Sect. 3.3. Thus we may regard v as bounded and hence as belonging to the required Sobolev space (Theorem 25). Finally, to reach the case of a discontinuous v, we ﬁrst perform the above construction on the inﬁmal convolution vε and we may arrange it in advance19 so that this is a viscosity supersolution in the whole domain B2r × (0, T ). Now 0 ≤ vε ≤ v so that vε (x, t) = 0 when t ≤ δ. The corresponding hε takes the given initial and boundary values and it is easily seen that wε obeys the comparison principle. For the veriﬁcation it is decisive to use the fact that vε is deﬁned also outside the inner cylinder. By Theorem 25 wε is a weak supersolution. We need a uniform bound as ε → 0. To this end, choose a test function ζ = ζ(x) so that ζ = 0 on ∂Br and ζ = 1 on B2r \ B3r/2 . Taking into account that hε has outer boundary values zero, we get a Caccioppoli estimate of the form

T 0

T |∇hε |p dx dt ≤ C

B2r\B3r/2

0

B2r\Br

hpε |∇ζ|p dx dt +

B2r\Br

h2ε (x, T ) dx.

Indeed, to see this use the test function ζ p hε and take into account that we deal with a solution. The Caccioppoli estimate for wε can be written as 0

T

|∇wε |p dx dt ≤ C max(wε )p rn−p + max(wε )2 rn .

B3r/2

Adding the estimates we get an integral over the whole B2r × (0, T ), and hence T |∇wε |p dx dt ≤ C v p∞ rn−p + v 2∞ rn . 0

B2r

Now it follows that, as ε → 0, wε approaches a weak supersolution w. Indeed, w is semicontinuous and being the limit of an increasing sequence it satisﬁes the comparison principle. Since w is also bounded, Theorem 25 applies again. By the weak lower semicontinuity of the integral also 0

T B2r

|∇w|p dx dt ≤ C v p∞ rn−p + v 2∞ rn .

Moreover, w ∈ Lp (0, T ; W01,p(B2r )) and w(x, 0) = 0 when x ∈ B2r , 1,p provided, of course, that w ∈ Lp (0, T ; Wloc (Ω)). Performing the above reduction for the function20 v1 , which is v cut at level 1, we obtain 19 Start with a nonnegative v in a larger domain, say B 3r × (−δ, T ) and put v = 0 up to time t = δ. 20 To

clarify the notation, this is not vε for ε = 1.

Regularity of Supersolutions

0

111

T

|∇w1 |p dw dt ≤ C(rn−p + rn ) B2r

as required in Lemma 27. We know that w1 = v1 in the smaller set. The procedure is to perform the previous reduction with v replaced by each of the truncated functions vk . Repeating the whole iteration of the previous section, we obtain the results for the functions wk and thereby also for vk , although in the smaller set where the functions are unaﬀected by the reduction. This is enough for our local results.

4

Weak Supersolutions are Semicontinuous

Are the weak supersolutions p-supercaloric functions (=viscosity supersolutions)? To qualify they have to obey the comparison principle and to be semicontinuous. The comparison principle is rather immediate. The semicontinuity is a delicate issue. For a weak supersolution deﬁned in the classical way with test functions under the integral sign (Deﬁnition 16) the Sobolev derivative is assumed to exist, but the semicontinuity, which now is not assumed, has to be established. The proof requires parts of the classical regularity theory.21 We will use a variant of the Moser iteration, for practical reasons worked out for weak subsolutions bounded from below. Our proof of the theorem below is essentially the same as in [18], but we avoid the use of inﬁnitely stretched inﬁnitesimal space-time cylinders. Theorem 30. Suppose that v = v(x, t) is a weak supersolution of the Evolutionary p-Laplace equation. Then it is locally bounded from below and at almost every point (x0 , t0 ) it holds that v(x0 , t0 ) = ess

lim inf

(x,t)→(x0 ,t0 )

v(x, t).

In particular, v is lower semicontinuous after a redefinition in a set of measure zero. Functions like esslim inf v(x, t) are lower semicontinuous, if they are bounded from below. Thus the problem is the formula. The hardest part of the proof is to establish that the supremum norm of a non-negative weak subsolution is 10 bounded (Lemma 34) and 20 bounded in terms of quantities that can carry information from the Lebesgue points (Theorem 36). With such estimates the proof follows easily (at the end of this section). Before entering into the semicontinuity proof we address the comparison principle. 21 The

preface of Giuseppe Mingione’s work [25] is worth reading as an enlightenment.

112

P. Lindqvist

Proposition 31 (Comparison Principle). Suppose that v is a weak supersolution and u a weak subsolution, u, v ∈ Lp (0, T ; W 1,p (Ω)), satisfying lim inf v ≥ lim sup u on the parabolic boundary. Then v ≥ u in the whole domain ΩT . Proof. This is well-known and we only give a formal proof. For a nonnegative test function ϕ ∈ C0∞ (ΩT ) the equations T

(−vϕt + |∇v|p−2 ∇v, ∇ϕ) dx dt ≥ 0

0

Ω

0

Ω

T

(+uϕt − |∇u|p−2 ∇u, ∇ϕ) dx dt ≥ 0

can be added. Thus T (u − v)ϕt + |∇v|p−2 ∇v − |∇u|p−2 ∇u, ∇ϕ dx dt ≥ 0. 0

Ω

These equations remain true if v is replaced by v + ε, where ε is any constant. To complete the proof we choose (formally) the test function to be ϕ = (u − v − ε)+ η, where η = η(t) is a cut-oﬀ function; even η(t) = T − t will do here. We arrive at

T 0

η(|∇v|p−2 ∇v − |∇u|p−2 ∇u, ∇v − ∇u) dx dt u≥v+ε

T

≤ =

0

1 2

=−

Ω

T

0

1 2

(u − v − ε)2+ η dx dt +

0

Ω

1 2

T

η 0

Ω

∂ (u − v − ε)2+ dx dt ∂t

(u − v − ε)2+ η dx dt

T Ω

(u − v − ε)2+ dx dt ≤ 0.

Since the ﬁrst integral is non-negative by the structural inequality (22), the last integral is, in fact, zero. Hence the integrand (u − v − ε)2+ = 0 almost everywhere. But this means that u≤v+ε almost everywhere. Since ε > 0 we have the desired inequality v ≥ u a.e. 2

Regularity of Supersolutions

113

We need some estimates for the semicontinuity and begin with the well-known Caccioppoli estimates, which are extracted directly from the diﬀerential equation. Lemma 32 (Caccioppoli Estimates). For a non-negative weak subsolution u in Ω × (t1 , t2 ) we have the estimates

ζ p uβ+1 dx ≤

ess sup t1
Ω

+2p−1 β 2−p

t2 t1

t2

∂ uβ+1 ζ p dx dt ∂t Ω

t1

p

up−1+β |∇ζ| dx dt

Ω

and

t2

t1

p−1+β p p−2 p ) dx dt ≤ Cβ ∇(ζu

Ω

t1

t2

∂ uβ+1 ζ p dx dt ∂t Ω

p

up−1+β |∇ζ| dx dt,

+C t1

t2

Ω

where the exponent β ≥ 1, C = C(p), and ζ ∈ C ∞ (Ω × [t1 , t2 )), ζ(x, t1 ) = 0, ζ ≥ 0. Proof. Use the test function ϕ = uβ ζ p in the equation τ t1

−uϕt + |∇u|p−2 ∇u, ∇ϕ dx dt

Ω

u(x, τ )ϕ(x, τ ) dx ≤

+ Ω

u(x, t1 )ϕ(x, t1 ) dx = 0, Ω

where t1 < τ ≤ t2 . (The intermediate τ is needed to match the supremum in the ﬁrst estimate.) Strictly speaking, the “forbidden” time derivative ut is required at the intermediate steps. This can be handled through a regularization, which we omit. Proceeding, integration by parts leads to τ

−uϕt dx dt +

t1

Ω

1 = β+1

u(x, τ )ϕ(x, τ ) dx Ω p

β+1

ζ(x, τ ) u(x, τ ) Ω

1 dx − β+1

τ t1

valid for a.e. τ . To treat the “elliptic term”, we use ∇ϕ = βζ p uβ−1 ∇u + pζ p−1 uβ ∇ζ

∂ uβ+1 ζ p dx dt ∂t Ω

114

P. Lindqvist

and obtain τ 1 p β+1 ζ(x, τ ) u(x, τ ) dx + β ζ p uβ−1 |∇u|p dx dt β+1 Ω t1 Ω τ τ ∂ 1 uβ+1 ζ p dx dt + p ζ p−1 uβ |∇u|p−1 |∇ζ| dx dt. ≤ β + 1 t1 Ω ∂t t1 Ω As much as possible of the last integral must be absorbed by the double integral in the left-hand member. It is convenient to employ Young’s inequality aq bp ab ≤ + q p to achieve the splitting ζ p−1 uβ |∇u|p−1 |∇ζ| a

b

p−1 p−1 p p p−1 p−1+β β p ζ p−1 u(β−1) p |∇u|p−1 × u p |∇ζ| = p β

p−1 p−1 β 1 p p β−1 p ζ u ≤ |∇u| + up−1+β |∇ζ|p , p p p β which has to be multiplied by p and integrated. Absorbing one integral into the left-hand member, we arrive at the fundamental estimate β τ ζ p uβ−1 |∇u|p dx dt p Ω t1 Ω p−1 τ τ ∂ p 1 uβ+1 | ζ p | dx dt + up−1+β |∇ζ|p dx dt. ≤ β + 1 t1 Ω ∂t β t1 Ω 1 β+1

ζ(x, τ )p u(x, τ )β+1 dx +

Since the integrands are positive it follows that 1 ζ(x, τ )p u(x, τ )β+1 dx β+1 Ω t2 ∂ 1 uβ+1 ζ p dx dt ≤ β + 1 t1 Ω ∂t p−1 t2 p up−1+β |∇ζ|p dx dt, + β t1 Ω where the majorant now is free from τ . Taking the supremum over τ we obtain the ﬁrst Caccioppoli inequality.

Regularity of Supersolutions

115

To derive the second Caccioppoli inequality, we start from β p

t2 t1

≤

ζ p uβ−1 |∇u|p dx dt

Ω

t2 ∂ 1 uβ+1 ζ p dx dt β + 1 t1 Ω ∂t p−1 t2 p up−1+β |∇ζ|p dx dt + β t1 Ω

and notice that ζ p uβ−1 |∇u|p =

p p−1+β

p |ζ∇u

p−1+β p

|p .

Then the triangle inequality |∇(ζu

p−1+β p

)| ≤ |ζ∇u

p−1+β p

| + |u

p−1+β p

∇ζ| 2

and a simple calculation yield the desired result.

In the following version of Sobolev’s inequality the exponents are adjusted to our need. For a proof, see [6, Chap. 1]. Proposition 33 (Sobolev). For ζ ∈ C ∞ (ΩT ) vanishing on the lateral boundary ∂Ω × [0, T ] we have ≤S

T

T

ζ pγ |u|p−2+(β+1)γ dx dt

0

Ω

|∇(ζ|u|

0

p−1+β p

np )|p dx dt ess sup ζ p |u|β+1 dx , 0
Ω

Ω

where γ = 1 + np . Now we can control the right-hand member in the Sobolev inequality by the quantities in the Caccioppoli estimates. Thus

t2

pγ p−2+(β+1)γ

ζ u t1

Ω

≤ Cβ

(2−p)p n+p

β p−2

γ1 dx dt

t2

t1

u Ω

t2 ∂ p p p−1+β u |∇ζ| dx dt . ζ dx dt + ∂t t1 Ω

β+1

We select the test function ζ so that it is equal to 1 in the cylinder BR−ΔR × (T + ΔT, t2 ), ζ(x, T ) = 0, and so that ζ(x, t) = 0 when x is outside BR .

116

P. Lindqvist

Then we can write

γ1

t2

u T +ΔT

≤ Cβ

p−2+(β+1)γ

dx dt

BR−ΔR

(2−p)p n+p

β p−2 ΔT

t2

u T

β+1

BR

1 dx dt + p (ΔR)

t2

u T

p−1+β

dx dt ,

BR

where C is a new constant. Recall that γ > 1. This is a reverse H¨ older inequality, which is most transparent for p = 2. It will be important to keep ΔT = (ΔR)p . This is the basic inequality for the celebrated Moser iteration, which we will employ. The power of u increases to p − 2 + (β + 1)γ, but the integral is taken over a smaller cylinder. In order to iterate over a chain of shrinking cylinders Uk = B(x0 , Rk ) × (Tk , t2 ), starting with T U0 = B(x0 , 2R) × ( , t2 ) 2 and ending up with an estimate over the cylinder U∞ = B(x0 , R) × (T, t2 ), we introduce the quantities R , 2k T Tk = T − kp+1 , 2

R 2k+1 T Tk+1 − Tk = (k+1)p s, 2

Rk − Rk+1 =

Rk = R +

ω=

(ΔRk )p Rp = Ts ΔTk

s=

2p−1 − 1 . 2

We remark that ω is independent of the index k. Further, we write α = β + 1, so that α ≥ 2. Thus

γ1 u

p−2+αγ

dx dt

Uk+1

2(k+1)p β ≤C Rp

(2−p)p n+p

β

p−2

ω

u dx dt + Uk

α

u

p−2+α

dx dt .

(23)

Uk

It is inconvenient to deal with two diﬀerent integrals in the majorant. For simplicity we will perform two iteration procedures, depending on which integral is dominating. For the first procedure we assume that

Regularity of Supersolutions

117

ω ≤ up−2 . Then we have the simpler expression

γ1 u

p−2+αγ

(p−2)

2kp α γ ≤ C1 Rp

dx dt

up−2+α dx dt.

Uk+1

Uk

We start the iteration with α = 2 and k = 0. Thus

γ1 u

p−2+2γ

dx dt

(p−2)

20p 2 γ ≤ C1 Rp

up dx dt.

U1

U0

Then take α = 2γ and k = 1 so that

u

p−2+2γ 2

1 γ2

≤

dx dt

21p (2γ) C1 Rp

(p−2) γ

γ1

u

U2

p−2+2γ

dx dt

U1

≤

C1

21p (2γ) Rp

1 (p−2) γ

(p−2)

20p 2 γ ×C1 Rp

γ

up dx dt. U0

The result of the next step is

u

p−2+2γ 3

1 γ3

dx dt

U3

≤

p−2

C1 2 γ Rp

1+ γ1 +

1 γ2

2

p( γ1 + γ22 ) (p−2)( γ12 + γ23 )

up dx dt.

γ

U0

Continuing the chain and noticing that the geometric series 1+ and the series arrive at

1 1 n 1 + 2 + 3 + ··· = 1 + γ γ γ p

kγ −k appearing in the exponents converge, since γ > 1, we

118

P. Lindqvist

u

p−2+2γ k+1

1 γ k+1

≤ KR

dx dt

−p(1+ γ1 + γ12 +···+

1 γk

)

up dx dt.

Uk+1

U0

Here K is a numerical constant. As k → ∞, we obtain the ﬁnal estimate ess sup (u2 ) ≤

BR ×(T,t2 )

K Rn+p

t2 T 2

up dx dt =

B2R

K ωs T Rn

t2 T 2

up dx dt,

B2R

where the square came from the factor 2 in 2γ k+1 . The sum of the geometric series determined the power of R. Finally, if the assumption ω ≤ up−2 is relaxed to u ≥ 0, we can apply the previous estimate to the function 1

u(x, t) + ω p−2 = u(x, t) +

1 Rp p−2

Ts

.

A simple calculation gives us the bound in the next lemma. Lemma 34. Suppose that u ≥ 0 is a weak supersolution in the cylinder B2R × ( T2 , t2 ). Then ess sup {u2 } ≤ C

BR ×(T,t2 )

p 2 R p−2 T

+

T 1 Rp T Rn

t2 T 2

up dx dt

,

B2R

where C = C(n, p). We can extract the following piece of information. Corollary 35. A weak supersolution that is bounded from above, is locally bounded from below. Proof. Use u(x, t) = L − v(x, t).

2

The estimate in the lemma suﬀers from the defect that it is not sharp when u ≈ 0 because of the presence of the constant term. Our remedy is a second iteration procedure, this time under the assumption that 0 ≤ u ≤ j, where we take j so large that also j p−2 ≥ ω. Read j p−2 as max{ω, j p−2 }. The previous lemma shows that j is ﬁnite, but the point now is that u is not bounded away from zero. Then the ﬁrst integral

Regularity of Supersolutions

119

in the majorant of (23) is dominating and we can begin with the bound

γ1 u

p−2+αγ

≤Cj

dx dt

p−2 2

kp

(p−2)

α γ Rp

Uk+1

uα dx dt. Uk

We start the iteration with α = p and k = 0. Thus

γ1 u

p−2+pγ

(p−2)

≤ Cj p−2

dx dt

20p p γ Rp

U1

up dx dt. U0

Then take α = p − 2 + pγ, which is < nγ 2 , and k = 1 so that

2

u(p−2)(1+γ)+pγ dx dt

1 γ2

≤ Cj p−2

21p (nγ 2 ) Rp

U2

(p−2) γ

γ1

up−2+pγ dx dt U1

(p−2) 1 γ 21p (nγ 2 ) γ ≤ Cj p−2 p R (p−2) 0p γ p−2 2 (nγ) × Cj up dx dt. p R

U0

At the next step α = (p − 2)(1 + γ) + pγ 2 < nγ 3 and k = 2. The result is

u(p−2)(1+γ+γ

2

)+pγ

3

1 γ3

dx dt

U3

≤

Cj p−2 n Rp

p−2 γ

1+ γ1 +

1 γ2

2

p( γ1 + γ22 ) (p−2)( γ1 + γ22 + γ33 )

up dx dt.

γ

U0

Continuing like this we end up with an estimate integrated over Uk+1 with the power αk+1 = p − 2 + αk γ, where αk = (p − 2)(1 + γ + γ 2 + · · · + γ k−1 ) + pγ k n(p − 2) k 2n k (γ − 1) + pγ k ≈ n + p − γ = p p and αk < nγ k+1 . As k → ∞ we ﬁnd that

120

P. Lindqvist

ess sup {u

n+p− 2n p

BR ×(T,t2 )

}≤C

j

(p−2)(n+p) p

Rn+p

t2 T 2

up dx dt. B2R

We can summarize the result. Theorem 36. A weak supersolution u that is non-negative in the cylinder U = B(x0 , 2R) × (t0 − 3T /2, t0 + T ) has the bound ess sup BR ×(t0 −T,t0 +T )

{u

n+p− 2n p

Rp }≤K

T

+ u p−2 ∞ T Rn

1+ np

t0 +T t0 − 3T 2

up dx dt, B2R

(24)

where 0 ≤ u ≤ u ∞ in U . We need the fact that the positive part (u)+ of a weak subsolution is again a weak subsolution. Here the proof has to avoid the comparison principle, which is not yet available. It reduces to the following lemma. Lemma 37. If v is a weak supersolution, so is vL = min{v, L}. Proof. Formally, the test function22 ϕ = min{k(L − v)+ , 1}ζ = χk ζ inserted into

T

0

−vϕt + |∇v|p−2 ∇v, ∇ϕ dx dt ≥ 0

Ω

implies the desired inequality

T

0

−vL ζt + |∇vL |p−2 ∇vL , ∇ζ dx dt ≥ 0

Ω

at the limit k = ∞. As usual, ζ ∈ C0∞ (ΩT ), ζ ≥ 0. The explanation is that lim χk = the characteristic function of the set {v < L}. Under the assumption that the “forbidden” time derivative ut is available at the intermediate steps we have 0

T

χk −vζt + |∇v|p−2 ∇v, ∇ζ dx dt

Ω

≥k 1 L− k
22 This

T

ζ|∇v|p dx dt +

is from Lemma 2.109 on page 122 of [23].

vζ 0

Ω

∂ χk dx dt ∂t

Regularity of Supersolutions

121

T ∂ 1 ∂ 2 χk dx dt = − ζ (χk ) dx dt ∂t 2k 0 Ω ∂t Ω 0 T 1 2 (χk ) ζt dx dt −→ 0. =+ 2k 0 Ω

T

≥

vζ

The formula ∂χk /∂t = −vk or = 0 was used. The result follows. Finally, to handle the problem with the time derivative, one has ﬁrst to regularize the equation and then to use the test function ϕε = min{k(L − v ε )+ , 1}ζ = χk ζ, where v ε is the convolution in (19). The term

T

−v ε

0

Ω

∂ϕε dx dt ∂t

can be written so that the derivative ∂v ε /∂t disappears. Then one may safely let ε → 0. The result follows as before. 2 Proof of Theorem 30. Let (x0 , t0 ) be a Lebesgue point for the weak supersolution v. Then

T Rn →0

1 T Rn

lim n

1 T Rn

lim

t0 +T

t0 −2T

|v(x0 , t0 ) − v(x, t)|p dx dt = 0.

B2R

A fortiori

T R →0

t0 +T

t0 −2T

B2R

(v(x0 , t0 ) − v(x, t))p+ dx dt = 0.

(25)

We claim that v(x0 , t0 ) ≤ ess

lim inf

(x,t)→(x0 ,t0 )

v(x, t).

(26)

It is suﬃcient to establish that ess lim sup (v(x0 , t0 ) − v(x, t))+ = 0, (x,t)→(x0 ,t0 )

since those points where v(x, t) ≥ v(x0 , t0 ) can do no harm to inequality (26). To this end, notice that the function v(x0 , t0 )−v(x, t) is a weak subsolution and so is its positive part, the function u(x, t) = (v(x0 , t0 ) − v(x, t))+

122

P. Lindqvist

by Lemma 37. It is locally bounded according to Lemma 34. Thus the essliminf is > −∞ in (26). Use Theorem 36 and let T Rn → 0, keeping Rp /T ≤ Constant. In virtue of (24) it follows that ess lim sup

u(x, t)n+p−

2n p

=0

(x,t)→(x0 ,t0 )

and the exponent can be erased. This proves the claim (26) at the given Lebesgue point. Furthermore, the Lebesgue points have the property that v(x0 , t0 ) ≤ ess

lim inf

(x,t)→(x0 ,t0 )

≤

lim

T Rn →0

1 T Rn

v(x, t)

t0 +T

t0 −2T

v(x, t) dx dt = v(x0 , t0 ). B2R

Since almost every point is a Lebesgue point, we have established that v(x0 , t0 ) = ess

lim inf

(x,t)→(x0 ,t0 )

v(x, t)

almost everywhere. The right-hand member is a semicontinuous function. 2

5

The Equation with Measure Data

There is a close connexion between supersolutions and equations where the right-hand side is a Radon measure. The Barenblatt solution has the Dirac measure (multiplied by a suitable constant) as the right-hand side, and hence it is, indeed, a solution to an equation. The equation ∂v − ∇ · (|∇v|p−2 ∇v) = μ ∂t with a Radon measure μ has been much studied. For example, in [1] a summability result is given for the spatial gradient ∇v of the solution. There the starting point was the given measure and the above equation. However, we can do the opposite and produce the measure. Indeed, every viscosity supersolution (or p-supercaloric function) induces a Radon measure μ ≥ 0. This follows from our summability theorem, combined with the Riesz Representation Theorem for linear functionals. Theorem 38. Let v be a viscosity solution in Ω × (0, T ). Then there exists a non-negative Radon measure μ such that

T 0

∂ϕ p−2 + |∇v| ∇v, ∇ϕ dx dt = −v ϕ dμ ∂t Ω Ω×(0,T )

Regularity of Supersolutions

123

for all ϕ ∈ C0∞ (Ω × (0, T )). Proof. We already know that v, ∇v ∈ Lp−1 loc (Ω × (0, T )). In order to use Riesz’s Representation Theorem we deﬁne the linear functional Λv : C0∞ (Ω × (0, T )) −→ R, T ∂ϕ + |∇v|p−2 ∇v, ∇ϕ dx dt. −v Λv (ϕ) = ∂t 0 Ω Now Λv (ϕ) ≥ 0 for ϕ ≥ 0 according to Theorem 2. Thus the functional is positive and the existence of the Radon measure follows from Riesz’s theorem, cf. [7, Sect. 1.8]. 2 Some further results can be found in [16].

6

Pointwise Behaviour

The viscosity supersolutions are deﬁned at each point, not only almost everywhere. Actually, the results in this section imply that two viscosity supersolutions that coincide almost everywhere do so at each point.

6.1

The Stationary Equation

We begin with the stationary case. At each point a p-superharmonic function v satisﬁes v(x) ≤ lim inf v(y) ≤ ess lim inf v(y) y→x

y→x

by lower semicontinuity. Essential limes inferior means that sets of Lebesgue measure zero be neglected in the calculation of the lower limit. The reverse inequalities also hold. To see this, we start by a lemma, which requires a pedantic formulation. Lemma 39. Suppose that v is p-superharmonic in the domain Ω. If v(x) ≤ λ at each point x in Ω and if v(x) = λ at almost every point x in Ω, then v(x) = λ at each point x in Ω. Proof. The proof is trivial for continuous functions and the idea is that v is everywhere equal to a p-harmonic function, which, of course, must coincide with the constant λ. We approximate v by the inﬁmal convolutions vε . We can assume that the function v is bounded also from below in a given ball B2r , strictly interior in Ω. We may even take 0 ≤ v ≤ λ by adding a constant. We approximate v by the inﬁmal convolutions vε . Replace vε in Br by the

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P. Lindqvist

p-harmonic function hε having boundary values vε . Thus we have the function wε =

hε in Br vε in B2r \Br

As we have seen before, also wε is p-superharmonic. By comparison wε ≤ vε ≤ v pointwise in B2r . As ε approaches zero via a decreasing sequence, say 1, 1/2, 1/3, · · · , the hε ’s converge to a p-harmonic function h, which is automatically continuous because the family is uniformly equicontinuous so that Ascoli’s theorem applies. The equicontinuity is included in the H¨ older estimate (8), because 0 ≤ hε ≤ λ. Thus h≤v≤λ at each point in Br . Since λ − vε ≥ λ − v ≥ 0, the Caccioppoli estimate Br

|∇hε |p dx ≤

|∇vε |p dx ≤ pp B2r(λ − vε )p |∇ζ|p dx ≤ Cr−p B2r(λ − vε )p dx Br

is valid. The weak lower semicontinuity of the integral implies that p p −p |∇h| dx ≤ lim |∇hε | dx ≤ Cr (λ − v)p dx = 0. Br

ε→0

Br

B2r

The conclusion is that h is constant almost everywhere, and hence everywhere by continuity. The constant must be λ, because it has boundary values λ in Sobolev’s sense. We have proved that also v(x) = λ at each point in the ball Br . The result follows. 2 Lemma 40. If v is p-superharmonic in Ω and if v(x) > λ for a.e. x in Ω, then v(x) ≥ λ for every x in Ω. Proof. If λ = −∞, there is nothing to prove. Applying the previous lemma to the p-superharmonic function deﬁned by min{v(x), λ} we obtain the result in the case λ > −∞. Theorem 41. At each point a p-superharmonic function v satisfies v(x) = ess lim inf v(y). y→x

2

Regularity of Supersolutions

125

Proof. Fix an arbitrary point x ∈ Ω. We must show only that λ = ess lim inf v(y) ≤ v(x), y→x

since the opposite inequality was clear. Given any ε > 0, there is a δ such that v(y) > λ − ε for a.e. y ∈ B(x, δ). By the lemma v(y) ≥ λ − ε for each such y. In particular, v(x) ≥ λ − ε. Because ε was arbitrary, we have established that v(x) ≥ λ. 2

6.2

The Evolutionary Equation

We turn to the pointwise behaviour for the Evolutionary p-Laplacian Equation. At each point in its domain a lower semicontinuous function satisﬁes v(x, t) ≤ lim inf v(y, τ ) ≤ ess lim inf v(y, τ ) ≤ ess lim inf v(y, τ ). (y,τ )→(x,t)

(y,τ )→(x,t)

(y,τ )→(x,t) τ
We show that for a viscosity supersolution also the reverse inequalities hold, thus establishing Theorem 3 in the Introduction. In principle, the proof is similar to the stationary case, but now a delicate issue of regularization arises. We ﬁrst consider a non-positive viscosity supersolution v = v(x, t) which is equal to zero at almost each point and, again, we show that locally it coincides with the viscosity solution having the same boundary values, now in a spacetime cylinder. Then one has to conclude that v was identically zero. We seize the opportunity to describe a useful procedure of regularizing by taking the convolution23 1 u (x, t) = σ

t

e(s−t)/σ u(x, s) ds,

σ > 0.

0

The notation hides the dependence on the parameter σ. For continuous and for bounded semicontinuous functions u the averaged function u is deﬁned at each point. We will stay within this framework. Observe that σ

∂u

+ u = u. ∂t

Some of its properties are listed in the next lemma. 23 The

origin of this function is unknown to me. In connexion with the Laplace transform it would be the convolution of u and σ−1 e−t/σ .

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P. Lindqvist

Lemma 42.

(i) If u ∈ Lp (DT ), then u Lp (DT ) ≤ u Lp(DT )

and

∂u

u − u

= ∈ Lp (DT ). ∂t σ Moreover, u → u in Lp (DT ) as σ → 0. (ii) If, in addition, ∇u ∈ Lp (DT ), then ∇(u ) = (∇u) componentwise, ∇u Lp (DT ) ≤ ∇u Lp(DT ) , and ∇u → ∇u in Lp (DT ) as σ → 0. (iii) Furthermore, if uk → u in Lp (DT ) then also u k → u

and

∂u

∂u k → ∂t ∂t

in Lp (DT ). 1. (iv) If ∇uk → ∇u in Lp (DT ), then ∇u k → ∇u in Lp (DT ). 2. (v) Finally, if ϕ ∈ C(DT ), then ϕ (x, t) + e−t/σ ϕ(x, 0) → ϕ(x, t) uniformly in DT as σ → 0. Proof. We leave this as an exercise. (Some details are worked out on page 7 of [15].) 2 The averaged equation for a weak supersolution u in DT reads as follows: T p−2

∂ϕ dx dt + (|∇u| ∇u) , ∇ϕ − u u (x, T )ϕ(x, T ) dx ∂t 0 D D 1 T −s/σ u(x, 0) ϕ(x, s)e ds dx ≥ σ 0 D valid for all test functions ϕ ≥ 0 vanishing on the lateral boundary ∂D ×[0, T ] of the space-time cylinder. For solutions one has equality. Notice the typical diﬃculty with obtaining (|∇u|p−2 ∇u) and not |∇u |p−2 ∇u , except in the linear case. The averaged equation follows from the equation for the retarded supersolution u(x, t − s), where 0 ≤ s ≤ T :

Regularity of Supersolutions

T s

127

|∇u(x, t − s)|p−2 ∇u(x, t − s), ∇ϕ(x, t) − u(x, t − s)

D

u(x, T − s)ϕ(x, T ) dx ≥

+ D

∂ϕ (x, t) dx dt ∂t

u(x, 0)ϕ(x, s) dx. D

Notice that (x, t − s) ∈ DT when 0 ≤ s ≤ t ≤ T . Multiply by σ −1 e−s/σ , integrate over [0, T ] with respect to s, and, ﬁnally, interchange the order of integration between s and t. This yields the averaged equation above. The advantage of this procedure over more conventional convolutions is that no values outside the original space-time cylinder are evoked. We begin with a simple situation. Lemma 43. Suppose that v is a viscosity supersolution in a domain containing the closure of BT = B × (0, T ). If (i) v ≤ 0 at each point in BT and (ii) v = 0 at almost every point in BT , then v = 0 at each point in B × (0, T ]. Proof. We may assume that v is bounded. Construct the inﬁmal convolution vε with respect to a larger domain than BT . Fix a small time t > 0 and let hε be the p-caloric function with boundary values induced by vε on the parabolic boundary of the cylinder B × (t , T ) and deﬁne the function wε =

hε , vε

in B × (t , T ] otherwise.

To be on the safe side concerning the validity at the terminal time T we may solve the boundary value problem in a slightly larger domain with terminal time T > T . Also wε is a viscosity supersolution. By comparison wε ≤ vε ≤ 0 pointwise in

BT .

We let ε go to zero through a monotone sequence, say 1, 12 , 13 , · · · . Then the limit h = lim hε ε→0

exists pointwise and it follows from the uniform H¨older estimates (14) that this h is continuous without any correction made in a set of measure zero. It is important to preserve the information at each point. Thus h is a p-caloric function. The so obtained function h, in B × (t , T ) w= v otherwise

128

P. Lindqvist

is a viscosity supersolution. For the veriﬁcation of the semicontinuity and the comparison principle, which proves this, the fact that h ≤ v is essential. We know that w ≤ v ≤ 0 everywhere in a domain containing B × (0, T ). In particular, h ≤ v ≤ 0 everywhere in B × (0, T ). We claim that h = 0 at each point. The claim immediately implies that v = 0 at each point in B × (0, T ). Concerning the statement at the terminal time T , we notice that v ≥ h and v(x, T ) ≥ h(x, T ) = lim h(x, t) = 0, t→T −

since h is continuous. On the other hand v(x, T ) ≤ 0 by the lower semicontinuity. Thus also v(x, T ) = 0. Therefore it is suﬃcient to prove the claim. To conclude that h is identically zero we use the averaged equation for w and write T 0

(|∇w|p−2 ∇w) , ∇ϕ + ϕ

B

≥

w(x, 0) B

1 σ

T

∂w dx dt ∂t

ϕ(x, s)e−s/σ ds dx,

0

where the test function vanishes on the parabolic boundary (an integration by parts has been made with respect to time.) Select the test function ϕ = (vε − wε ) and let ε approach zero. Taking into account that ϕ = 0 when t < t , we arrive at T t

(|∇h|p−2 ∇h) , ∇v − ∇h + (v − h )

B

≥

v(x, 0) B

1 σ

T

t

∂h dx dt ∂t

(v (x, s) − h (x, s))e−s/σ ds dx.

The last integral (which could be negative) approaches zero as the regularization parameter σ goes to zero, because t > 0, so that the exponential decays. Integrating (v − h )

∂h

∂(v − h ) ∂v

= −(v − h ) + (v − h ) ∂t ∂t ∂t

Regularity of Supersolutions

129

we obtain T

− h ) ∂h ∂t dx dt 1 = − 2 B (v (x, T ) − h (x, T ))2 dx T + t D (v − h ) ∂v ∂t dx dt.

t B (v

because v (x, t ) − h (x, t ) = 0. The last integral is zero because v and ∂v ∂t are zero almost everywhere according to property (i) in Lemma 42. Erasing this integral and letting the regularization parameter σ go to zero (so that the ’s disappear) we ﬁnally obtain T

1 |∇h| dx dt + 2 B

h2 (x, T ) dx ≤ 0

p

t

i.e.

= 0.

B

In fact,24 the proof guarantees this only for almost all values of T in the range t < T < T . From this it is not diﬃcult to conclude that h is identically zero. Thus our claim has been proved. 2 Lemma 44. Suppose that v is a viscosity supersolution in a domain containing BT = B × (0, T ). If v(x, t) > λ for almost every (x, t) ∈ BT , then v(x, t) ≥ λ for every (x, t) ∈ B × (0, T ]. Proof. The auxiliary function u(x, t) = min{v(x, t), λ} − λ in place of v satisﬁes the assumptions in the previous lemma. Hence u = 0 everywhere in B × (0, T ]. This is equivalent to the assertion. 2 Proof of Theorem 3. Denote λ = ess

lim inf

(x,t)→(x0 ,t0 )

v(x, t).

t
According to the discussion in the beginning of this section, it is suﬃcient to prove that λ ≤ v(x0 , t0 ). Thus we can assume that λ > −∞. 24 It

is the validity of lim

σ→0

1 2

B

(v (x, T ) − h (x, T ))2 dx =

1 2

B

h2 (x, T ) dx

that requires some caution. We know that v is zero almost everywhere but with respect to the (n + 1)-dimensional measure.

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P. Lindqvist

First, we consider the case λ < ∞. Given ε > 0, we can ﬁnd a δ > 0 and a ball B with centre x0 such that the closure of B × (t0 − δ, t0 ) is comprised in the domain and v(x, t) > λ − ε for almost every (x, t) ∈ B × (t0 − δ, t0 ). According to the previous lemma v(x, t) ≥ λ − ε for every (x, t) ∈ B × (t0 − δ, t0 ]. In particular, we can take (x, t) = (x0 , t0 ). Hence v(x0 , t0 ) ≥ λ − ε. Since ε was arbitrary, we have proved that λ ≤ v(x0 , t0 ), as desired. Second, the case λ = ∞ is easily reached via the truncated functions vk = min{v(x, t), k}, k = 1, 2, · · · . Indeed, v(x0 , t0 ) ≥ vk (x0 , t0 ) ≥ min{∞, k} = k, in view of the previous case. This concludes the proof of Theorem 3.

2

References 1. L. Boccardo, A. Dall’Aglio, T. Gallou¨ et, L. Orsina, Nonlinear parabolic equations with measure data. J. Funct. Anal. 147, 237–258 (1997) ´ ements de la Th´eorie Classique du potential, 3e ´ 2. M. Brelot, El´ edition, Les cours de Sorbonne, 3e cycle, Centre de Documentation Universitaire, Paris, 1965 3. M. Crandall, J. Zhang, Another way to say harmonic. Trans. Am. Math. Soc. 355, 241–263 (2003) 4. H.-J. Choe, A regularity theory for a more general class of quasilinear parabolic partial diﬀerential equations and variational inequalities. Diﬀer. Integr. Equat. 5, 915–944 (1992) 5. M. Crandall, H. Ishii, P.-L. Lions, User’s guide to viscosity solutions of second order partial diﬀerential equations. Bull. Am. Math. Soc. 27, 1–67 (1992) 6. E. DiBenedetto, Degenerate Parabolic Equations (Springer, Berlin, 1993) 7. L. Evans, R. Gariepy, Measure Theory and Fine Propeties of Functions (CRC Press, Boca Raton, 1992) 8. E. Giusti, Metodi diretti nel calcolo delle variazioni (Unione Matematica Italiana, Bologna, 1994) 9. E. Giusti, Direct Methods in the Calculus of Variations (World Scientiﬁc, Singapore, 2003) 10. P. Juutinen, P. Lindqvist, J. Manfredi, On the equivalence of viscosity solutions and weak solutions for a quasi-linear equation. SIAM J. Math. Anal. 33, 699–717 (2001) 11. S. Kamin, J.L. Vasquez, Fundamental solutions and asymptotic behaviour for the p-Laplacian equation. Revista Mathematica Iberoamericana 4, 339–354 (1988) 12. T. Kilpel¨ ainen, P. Lindqvist, On the Dirichlet boundary value problem for a degenerate parabolic equation, SIAM J. Math. Anal. 27, 661–683 (1996) 13. T. Kilpel¨ ainen, J. Mal´ y, Degenerate elliptic equations with measure data and nonlinear potentials. Annali della Scuola Normale Superiore di Pisa Cl. Sci. (4) 19, 591–613 (1992)

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14. J. Kinnunen, P. Lindqvist, Summability of semicontinuous supersolutions to a quasilinear parabolic equation. Annali della Scuola Normale Superiore di Pisa Cl. Sci. (5) 4, 59–78 (2005) 15. J. Kinnunen, P. Lindqvist, Pointwise behaviour of semicontinuous supersolutions to a quasilinear parabolic equation. Annali di Matematica Pura ed Applicata (4) 185, 411–435 (2006) 16. J. Kinnunen, T. Lukkari, M. Parviainen, An existence result for superparabolic functions. J. Funct. Anal. 258, 713–728 (2010) 17. S. Koike, in A Beginner’s Guide to the Theory of Viscosity Solutions. MSJ Memoirs, vol. 13 (Mathematical Society of Japan, Tokyo, 2004) 18. T. Kuusi, Lower semicontinuity of weak supersolutions to nonlinear parabolic equations. Diﬀer. Integr. Equat. 22, 1211–1222 (2009) 19. J. Lewis, On very weak solutions of certain elliptic systems. Comm. Part. Diﬀer. Equat. 18, 1517–1537 (1993) 20. P. Lindqvist, On the deﬁnition and properties of p-superharmonic functions. Journal f¨ ur die Reine und Angewandte Mathematik (Crelles Journal) 365, 67–79 (1986) 21. P. Lindqvist, Notes on the p-Laplace equation. University of Jyv¨ askyl¨ a, Report 102 (2006) 22. P. Lindqvist, J. Manfredi, Viscosity supersolutions of the evolutionary p-Laplace equation. Diﬀer. Integr. Equat. 20, 1303–1319 (2007) 23. J. Maly, W. Ziemer, in Fine Regularity of Solutions of Elliptic Partial Diﬀerential Equations. Math. Surveys Monogr., vol. 51 (AMS, Providence, 1998) 24. J. Michael, W. Ziemer, Interior regularity for solutions to obstacle problems. Nonlinear Anal. 10, 1427–1448 (1986) 25. G. Mingione, Regularity of minima: An invitation to the dark side of the calculus of variations. Appl. Math. 51, 355–426 (2006) 26. J. Serrin, Pathological solutions of elliptic diﬀerential equations. Annali della Scuola Normale Superiore di Pisa Cl. Sci. (3) 18, 385–387 (1964) 27. N. Trudinger, On Harnack type inequalities and their application to quasilinear elliptic equations. Comm. Pure Appl. Math. 20, 721–747 (1967) 28. N. Trudinger, Pointwise estimates and quasilinear parabolic equations. Comm. Pure Appl. Math. 21, 205–226 (1968) 29. J. Urbano, The Method of Intrinsic Scaling: A Systematic Approach to Regularity for Degenerate and Singular PDEs. Lecture Notes in Mathematics, vol. 1930 (Springer, Berlin, 2008) 30. Z. Wu, J. Zhao, J. Yin, H. Li, Non-linear Diﬀusion Equations (World Scientiﬁc, Singapore, 2001) 31. C. Yazhe, H¨ older continuity of the gradient of the solutions of certain degenerate parabolic equations. Chin. Ann. Math. Ser. B 8(3), 343–356 (1987)

Introduction to Random Tug-of-War Games and PDEs Juan J. Manfredi

1

Introduction

The fundamental contributions of Kolmogorov, Ito, Kakutani, Doob, Hunt, L´evy, and many others have shown the profound and powerful connection between classical linear potential theory and probability theory. The idea behind the classical interplay is that harmonic functions and martingales share a common cancelation property that can be expressed by using mean value properties. In these lectures, we will see how this approach turns out to very useful in the nonlinear theory as well. The objective of this course is to provide an introduction to the connection between the theory of stochastic tug-of-war games and non-linear equations of p-Laplacian type in the Euclidean and discrete cases. These notes will provide the student with background to read [16] and [17]. Most of the material e based on the joint papers [11–13] with Mikko Parviainen and Julio Rossi, and on the 2010 doctoral thesis of Alexander Sviridov [18]. I am grateful to Alex for correcting many misprints and for suggesting changes that have improved the readability of the manuscript.

2

Probability Background

We present a quick introduction to Doob’s Optimal Sampling Theorem and to Kolmogorov’s construction of inﬁnite product of measures. We follow the presentation in the book [21]. We refer to [21] for the basic probability deﬁnitions and the proofs. However, we have chosen to emphasize certain J.J. Manfredi () Department of Mathematics, University of Pittsburgh, Pittsburgh, PA 15260 e-mail: [email protected]

J. Lewis et al., Regularity Estimates for Nonlinear Elliptic and Parabolic Problems, Lecture Notes in Mathematics 2045, DOI 10.1007/978-3-642-27145-8 3, © Springer-Verlag Berlin Heidelberg 2012

133

134

J.J. Manfredi

details in the proofs that are likely to help student of analysis going through this material for the ﬁrst time. We are given a set Ω endowed with a σ-algebra F and probability measure P on F . The triplet (Ω, F , P) is a probability space. Definition 2.1. If G ⊂ F is a sub-σ-algebra of F and f ∈ L1 (Ω, P) the conditional expectation of f given G is the only G-measurable function g = E[f |G] such that f dP = A

gdP A

for all sets A ∈ G. Example 2.2. Suppose that we have a ﬁnite partition Ω = A1 ∪ A2 ∪ · · · ∪ An where all the Ai ∈ F have positive probability P(Ai ) > 0. Let G the σ-algebra generated by this partition. For f ∈ L1 (Ω, P), the function g = E[f |G] has a constant value gi on each Ai given by f dP gi = Ai

The notion of (discrete) martingale will be key to our developments later on. Definition 2.3. Let Fi ⊂ Fi+1 ⊂ · · · ⊂ F be a ﬁltration of σ algebras of F and Xi : Ω → F be an Fi -measurable random variable (or Fi -measurable function.) 1. The sequence of random variables {Xi } is a martingale if Xi = E[Xi+1 |Fi ]. 2. The sequence of random variables {Xi } is a submartingale if Xi ≤ E[Xi+1 |Fi ]. 3. The sequence of random variables {Xi } is a supermartingale if Xi ≥ E[Xi+1 |Fi ]. These relations are supposed to hold a.e. with respect to P. When needed we will make explicit the σ-algebras by writing {Xi , Fi }. If {Xi } is a martingale, we have E[Xi+1 ] = E[Xi ] = · · · = E[X1 ] = c. The random variables Yi+1 = Xi+1 − Xi are the martingales diﬀerences. We clearly have E[Yi ] = 0. Example 2.4. If X ∈ L1 (Ω, P), the sequence Xi = [X|Fi ] is a martingale. It turns that this is the most general L1 -martingale (see [21].) Example 2.5. Let {Yi } be a collection of independent random variables with mean zero E[Yi ] = 0. Set Fi = σ(Y1 , Y2 , . . . , Yi ). Then, the sequence Xi = c +

i j=1

is a martingale.

Yj

Introduction to Random Tug-of-War Games and PDEs

135

The following lemma is a consequence of Jensen’s inequality for conditional expectations. Lemma 2.6. Let {Xi , Fi } be a martingale and 1 ≤ p < ∞. Then {|Xi |p , Fi } is a submartingale provided that E[|Xi |p ] < ∞. We present Doob’s weak type 1-1 inequality for ﬁnite martingales. The simple proof contains all the ingredients of the more general cases. Theorem 2.7. Let {Xi }ni=1 be a martingale and l > 0. Then we have 1 P ω : sup |Xi (ω)| ≥ l ≤ |Xn | dP l {sup |Xi |>l} 1≤i≤n 1 |Xn | dP. ≤ l Ω Proof. The maximum random variable S is S(ω) = sup{|Xi (ω) : 1 ≤ i ≤ n}. For each 1 ≤ i ≤ n deﬁne the set describing the ﬁrst time we go above l, Ei = {ω : |X1 (ω)| < l, |X1 (ω)| < l, . . . , |Xi−1 (ω)| < l, |Xi (ω)| ≥ l} and the set when the maximum is above l E = {ω : S(ω) ≥ l}. We have a disjoint union E=

n

Ei

1

and the basic estimate P(Ei ) ≤

1 l

|Xi | dP.

(1)

Ei

Since |Xi | is a submartingale we have E [ |Xn | | Fi ] ≥ |Xi | for a.e. ω. Since Ei ∈ Fi we then have E [ χEi |Xn | | Fi ] = χEi E [ |Xn | | Fi ] ≥ χEi |Xi |. Taking expectations we get

|Xn | dP ≥

Ei

|Xi | dP Ei

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J.J. Manfredi

and using (1) we get P(Ei ) ≤

1 l

|Xn | dP. Ei

A well-known argument using distribution functions (see [21]) gives the maximal theorem Corollary 2.8. (Doob’s maximal theorem) Let {Xi }ni=1 be a martingale and 1 < p < ∞. For S = sup{|Xi : 1 ≤ i ≤ n} we have E[S p ] ≤ (

p p ) E[|Xn |p ]. p−1

In addition to martingales, stopping times play a crucial role in stochastic games. Definition 2.9. The random variable τ : Ω → N ∪ {0, ∞} is a stopping time with respect to the ﬁltration {Fi } if {ω : τ (ω) ≤ n} ∈ Fn for all n ≥ 0. Example 2.10. The following are examples of stopping times. The reader would beneﬁt from checking in detail that they are indeed stopping times. 1. τ (ω) = k, where k is constant. 2. If τ is a stopping time and f is monotone such that f (t) ≥ t, then τ = f ◦τ is a stopping time. 3. If τ1 and τ2 are stopping times so are max {τ1 , τ2 } and min {τ1 , τ2 }. We see then that the truncated stopping time τn = min{τ, n} is also a stopping time, so that we have the following useful fact. Corollary 2.11. Every stopping time is the limit of an increasing sequence of bounded stopping times We next formalize the notion of information available up to time τ . Definition 2.12. Let τ be a stopping time respect to the ﬁltration {Fi }. Set Fτ = {A : A ∈ F and A ∩ {ω : τ (ω) ≤ n} ∈ Fn for all n} Lemma 2.13. (Basic properties of Fτ ) 1. 2. 3. 4. 5.

Fτ is a σ-field. If τ (ω) = k constant, then Fτ = Fk . τ1 ≤ τ2 =⇒ Fτ1 ⊂ Fτ2 . τ is Fτ measurable. {ω : τ (ω) < ∞} ∈ Fτ .

Introduction to Random Tug-of-War Games and PDEs

The last part follows from {ω : τ (ω) < ∞} = martingale by deﬁning Xτ as follows

137

n {ω :

τ (ω) ≤ n}. We stop a

Xτ (ω) = Xτ (ω) (ω). Lemma 2.14. The stopped martingale Xτ is Fτ measurable on the set where τ is finite {ω : τ (ω) < ∞}. Proof. We need to check that for λ ∈ R we have Eλ = {ω : τ (ω) < ∞, Xτω (ω) > λ} ∈ Fτ , which follows from the expression {Xn (ω) > λ} ∩ {τ (ω) = n} Eλ = n

The cornerstone of the applications of martingale to partial diﬀerential equations is Doob’s Optional Sampling Theorem. Theorem 2.15. Let {Xn } be a submartingale with respect to the filtration {Fn }. Let 0 ≤ τ1 ≤ τ2 be bounded stopping times. Then we have E [Xτ2 | Fτ1 ] ≥ Xτ1 . If {Xn } is supermartingale we get instead E [Xτ2 | Fτ1 ] ≤ Xτ1 , and If {Xn } is martingale we get the equality E [Xτ2 | Fτ1 ] = Xτ1 , By making τ1 equals to zero we get the following corollary. Corollary 2.16. Let {Xn } be a martingale. For any stopping time we have E[Xτ ] = E[X0 ]. Proof. Let us prove the theorem in the martingale case with τ1 = τ and τ2 = k. We need to establish that E[Xk | Fτ ] = Xτ , or equivalently that for all A ∈ Fτ we have Xk dP = A

Xτ dP.

(2)

A

Set Ei = {ω : τ (ω) = i} and decompose Ω = ∪k1 Ei as a disjoint union. Note that if A ∈ Fτ then A ∩ Ei ∈ Fi . We then have

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J.J. Manfredi

Xk dP =

A∩Ei

Xi dP =

A∩Ei

Xτ dP, A∩Ei

where the ﬁrst equality follows by the martingale property and the second by the deﬁnition of Ei . By adding in i we obtain the required (2). We now change gears and consider the product of probability spaces. Suppose that we have two probability spaces (Ω1 , B1 , P1 ) and (Ω2 , B2 , P2 ). A rectangle is a set A = A1 × A2 where A1 ∈ B1 and A2 ∈ B2 . We denote by F the ﬁeld of all ﬁnite disjoint unions of rectangles. The product σ-algebra is the σ algebra generated by F B = B1 × B2 = σ(F ). We deﬁne the product probability P = P1 × P2 on B as follows: (i) (ii) (iii) (iv)

P(A1 × A2 ) = P1 (A1 ) · P2 (A2 ) for rectangles A1 × A2 . Extend P to F as ﬁnitely additive measure. P is in fact countably additive on F . Extend P to B by using Caratheodory’s theorem.

A ﬁnite product of probability spaces is deﬁned similarly. We next consider inﬁnite products. We are given probability measures Pn on (Rn , B(Rn )), where B(Rn )) is the Borel σ-algebra in Rn . The projection πn : Rn+1 → Rn is given by πn (x1 , x2 , . . . , xn , xn+1 ) = (x1 , x2 , . . . , xn ). The family {Pn }n is consistent if for all n all rectangles A1 × . . . × An we have Pn+1 (A1 × . . . × An × R) = Pn (A1 × . . . × An ). We also write this equation as Pn+1 πn−1 = Pn and say that the marginal probability of Pn+1 on Rn is Pn . From now on Ω will be the inﬁnite cartesian product Ω = R∞ = {ω : ω = (xn )n∈N }. A cylinder with base A ∈ B(Rn ) is a set of the form C = {ω : (x1 , x2 , . . . , xn ) ∈ A}. The set of all cylinders form a ﬁeld F . They generate a σ-algebra Σ = σ(F ).

Introduction to Random Tug-of-War Games and PDEs

139

Theorem 2.17. (Kolmogorov) Let {Pn } be a consistent family of probabilities in (Rn , B(Rn )). Then, there exists a unique probability in (Ω, Σ) such that P πn−1 = Pn . That is, for all rectangles A1 × A2 × . . . × An we have P(A1 × A2 × . . . An × R . . . × R × . . .) = Pn (A1 × A2 × . . . An ). Proof. For a cylinder C ∈ F with base A ∈ B(Rn ) deﬁne P(C) = Pn (A). The ﬁrst observation is that P is well-deﬁned in F by the consistency hypothesis. The key point is to establish the P is countably additive on F . Then we can extend P to Σ by the Caratheodory’s procedure. We need to show that if Bn ∈ F , Bn+1 ⊂ Bn , and ∩n Bn = ∅, then lim P(Bn ) = 0.

n→∞

The proof is by contradiction. Suppose that for some δ > 0 we have that P(Bn ) > δ for all n ∈ N. Since Bn is a cylinder we write Bn = πn−1 (An ). Suppose for the moment that all basis An are compact and write Bn = An1 × An2 × . . . × Ann × R × R . . . × An+1 × . . . × An+1 × An+1 Bn+1 = An+1 n 1 2 n+1 × R × R . . . ⊂ We see that for each j we have a nested sequence of compact subsets An+1 j Anj . By the ﬁnite intersection property we get A∞ j =

∞

Anj = ∅.

n=1

But then we would get a contradiction since ∞

∞ Bn = A∞ 1 × A2 × . . . = ∅.

n=1

In the general case An is only a Borel set. For each n select a compact set Kn ⊂ An such that δ Pn (An \ Kn ) ≤ n+2 . 2

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Write Dn = πn−1 (Kn ) ⊂ Bn and observe that En = some compact set Fn . We then have

n j=1

Dj = πn−1 (Fn ) for

P(Bn \ En ) = P Bn \ ∩nj=1 Dj ⎞ ⎛ n = P ⎝ (Bn \ Dj )⎠ j=1

⎛ ≤ P⎝

n

⎞ (Bj \ Dj )⎠

j=1

≤

n

P (Bj \ Dj ))

j=1

≤

n

P (Aj \ Kj ))

j=1

≤

n j=1

≤

δ 2j+2

δ . 4

From which we deduce that P(En ) ≥ P(Bn ) − P(Bn \ En ) ≥δ− ≥

δ 4

δ , 2

reducing the problem to the compact case. Let see how Kolmogorov’s theorem can be used to give a quick construction of the Lebesgue measure. Consider the simplest case of ternary trees. We follow the formalism developed in [5]. A directed tree T with regular 3-branching consists of the empty set ∅ as the top vertex, 3 sequences of length 1 with terms chosen from the set X = {0, 1, 2}, 9 sequences of length 2 with terms chosen from the set X 2 = {0, 1, 2},. . . , 3r sequences of length r with terms chosen from the et X r = {0, 1, 2} and so on. A vertex br at level r is labeled by a sequence of digits d1 d2 . . . dr , where dj ∈ X for all 1 ≤ j ≤ r. A branch of T is an inﬁnite sequence of vertices, each followed by one of its immediate successors. We denote a branch b starting at the vertex b1 as follows b = (b1 , b2 , . . . , br , . . .).

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141

The collection of all branches forms the boundary of the tree T and is denoted by ∂T . A branch b determines a real number in the interval [0, 1] by means of the ternary expansion ∞ bk g(b) = . (3) 3k k=1

Note that the set of all branches that start at a given vertex b = d1 d2 . . . dr is the ternary interval Ib = [0.d1 d2 . . . dr , 0.d1 d2 . . . dr + 3−r ] of length 3−r , where the expansions are in base 3. Note also that the classical Cantor set C is the subset of ∂T formed by branches that don’t go through any vertex labeled 1. Let Pn be the uniform probability in X n . The family {Pn } is a consistent family of probabilities, so that by Theorem 2.17 there exists a unique probability in X ∞ such that P(A1 × A2 × . . . An × X . . . × X × . . .) = Pn (A1 × A2 × . . . An ). If we take A1 = {d1 }, A2 = {d1 },..., Ar = {dr } we get P({b} × X . . . × X . . .) = Pr ({b}) = 3−r = |Ib | Corollary 2.18. Consider the probability spaces (X ∞ , Σ, P) and ([0, 1], B, L), where L is the Lebesgue measure in the interval [0, 1]. Let g be the ternary expansion mapping (3). Then we have P g −1 = L.

3

The p-Laplacian Gambling House

Start with a set X endowed with a σ-algebra B. Decompose X = X ∪Y as a disjoint union of two non-empty sets X and Y . We shall call X the interior and Y the boundary. For each point x ∈ X we have a nonempty set S(x) ⊂ X of successors of x. For points y ∈ Y we require that S(y) = {y}. Moreover, the set S(x) comes equipped with a probability measure supported in S(x) denoted by μ(x). For points y ∈ Y on the boundary we have that μ(y) = δy . We are given non-negative numbers α and β so that α + β = 1 and a pay-oﬀ function F : Y → R. At every point x ∈ X we have a family of probability measures Γ (x) in (X, B) given by

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J.J. Manfredi

Γ (x) =

α 2

(δxI + δxII ) + β μ(x) : xI , xII ∈ S(x)

(4)

To play a Tug-of-War game with noise starting at a point x0 ∈ X, choose a probability γ0 [x0 ] ∈ Γ (x0 ). The next position x1 ∈ S(x0 ) is selected according to γ0 [x0 ]. Once x0 and x1 are chosen, we pick a probability γ1 [x0 , x1 ] ∈ Γ (x1 ) to determine the next game position x2 ∈ S(x1 ). In this manner we determine a particular history x = (x0 , x1 , x2 , . . . ) ∈ X × X × · · · × X × · · · = X∞ . The game ends when we reach the boundary Y since once xj ∈ Y we have xj+1 ∈ S(xj ) = {xj }. We write τ (x) = inf{k : xk ∈ Y } for the ﬁrst time we hit the boundary with the understanding that τ (x) = ∞ if the boundary is never reached. If the game ends at a point y ∈ Y the pay-oﬀ value is F (y). We now apply a variant of the Kolmogorov’s construction. Let us denote by B j the product σ-algebra in Xj and by B ∞ the σ-algebra in X∞ generated by the cylinder sets A0 × A1 × · · · × Aj × X × X × · · · , where Ak ∈ Bk for k = 0, 1, . . . , j. We deﬁne a sequence of probability measures Pxσ0 ,k on (Xk , B k ) uniquely determined by the following properties: (i) Pσx0 ,1 = γ0 [x0 ], (ii) Pxσ0 ,k+1 has marginal probability Pxσ0 ,k on (Xk , B k ), and (iii) Pxσ0 ,k+1 has conditional probabilities γk [x0 , x1 , . . . , xk−1 ] on the ﬁbers (x0 , x1 , . . . , xk−1 ) × X; that is, for every rectangle (A0 × A1 × . . . × Ak ) in (Xk+1 , B k+1 ) we have Pxσ0 ,k+1(A0 ×A1 ×. . .×Ak ) =

γk [x0 , x1 , . . . , xk−1 ](Ak ) dPxσ0 ,k

(A0 ×A1 ×...×Ak−1 )

Under these conditions an extension of Kolmogorov’s construction due to Tulcea [19, 20] shows that there exists a unique probability measure Pxσ0 = limk→∞ Pxσ0 ,k in (X∞ , B ∞ ) with transition probabilities Pxσ0 ({xj+1 ∈ A} | Bj+1 ) = γj [x0 , x1 , . . . , xj ] See Chap. 4 in [21] for more details.

(5)

Introduction to Random Tug-of-War Games and PDEs

143

We call the collection of probability measures σ = (γ0 [x0 ], γ1 [x0 , x1 ] . . . , γk [x0 , x1 . . . xk ], . . .) a strategy. This formalism, coming from [14], is equivalent to the presentation in [16]. In this paper a strategy S is a collection of mappings σj : Xj+1 → X indicating the next move xj+1 = σj (x0 , x1 , . . . , xj ) given the partial history (x0 , x1 , . . . , xj ). A pair of strategies SI and SII and a starting point determine a family of measures {PxS0I ,SII }x0 ∈X that describe the game played under this pair of strategies. That is, the players choose either xI or xII to move there in case they win the coin toss. Their choices determine the probability measures γ[x0 , x1 , . . . , xk ] given (x0 , x1 , . . . , xk−1 ) and vice versa. Player I will try to choose points xI to maximize the pay-oﬀ while player II will try to choose points xII to minimize the pay-oﬀ. Each pair of strategies (SI , SII ), SI for player I and SII for player II as in [16], determine a strategy in this sense and vice versa. We write σ = (SI , SII ) Having ﬁxed a strategy σ and assuming, as we do from now on, that the game ends a.s. Pxσ0 (τ (x) < ∞) = 1, (6) we average with respect to Pxσ0 to obtain the expected pay-oﬀ for the Tugof-War game starting at x0 uσ (x0 ) = Exσ0 [F (xτ )].

(7)

To write down the mean value property satisﬁed by uσ we condition on the ﬁrst move using (5) with j = 0. Lemma 3.1. ([14], Chap. 2) The value function uσ (x) satisfies the mean value property α

uσ[xI ] (xI ) + uσ[xII ] (xII ) + β uσ (x) = uσ[y] (y) dμ(y) (8) 2 S(x0 ) Here the conditional strategy σ[y0 ] is deﬁned as follows for y0 ∈ S(x0 ) σ[y0 ] = (γ1 [x0 , y0 ], γ2 [x0 , y0 , y1 ] . . . , γk [x0 , y0 , y1 . . . yk ], . . .) 0 so that Pyσ[y is the conditional distribution of (x2 , x3 , . . . ) given that x1 = y0 . 0]

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J.J. Manfredi

Let us stop and consider the particular case when α = 0 and β = 1. In this case –the linear case– the strategies are irrelevant since Γ (x) is always μ(x) so that there is only one family of measures {Px0 }x0 ∈X . We recover the classical mean value formula u(x) = u(y) dμ(y). S(x)

But the case of interest to us is when we have α = 0. In this case the value function for player I is uI (x) = sup inf Exσ [F (xτ )] SI SII

and for player II uII (x) = sup inf Exσ [F (xτ )]. SII SI

Player I lets Player II choose a strategy, presumably to decrease Exσ0 [F (xτ )], and then do as best a possible. Notice that we always have uI (x) ≤ uII (x)

for all

x ∈ X.

It turns out that in many cases the game has a value; that is uI (x) = uII (x)

for all

x ∈ X,

(9)

and that this function satisﬁes a version of the Mean Value Property (3.1) given by α u(x) = 2

sup u(y) + inf u(y) y∈S(x)

y∈S(x)

+β

u(y) dμ(y).

(10)

S(x)

Equation (10) is the Dynamic Programming Principle or DPP for short. Next, we will present two scenarios in which all the details above have been worked out.

4

p-harmonious Functions

Consider a bounded Lipschitz domain Ω ⊂ Rn and ﬁx ε > 0. To prescribe boundary values, let us denote the compact boundary strip of width ε by Γε = {x ∈ Rn \ Ω : dist(x, ∂Ω) ≤ ε}.

Introduction to Random Tug-of-War Games and PDEs

145

Let X = Ω with the Borel σ-algebra, X = Ω \ Γε and Y = Γε . The successors of x are S(x) = B ε (x) = {y ∈ Rn : |y − x| ≤ ε} and the measure μ(x) is just the Lebesgue measure restricted to S(x) and normalized so that μ(x)(S(x)) = 1. As it will be clear later on, we take α and β to be α=

p−2 , p+n

and β =

2+n . p+n

(11)

Notice that since α ≥ 0 we necessarily have p ≥ 2. We are given a bounded Borel pay-oﬀ function F : Γε → R and play the Tug-of-War game with parameters α and β and obtain value functions uεI and uεII , where we have chosen to emphasize the dependence on the step size ε. The following results are from [12]: Theorem 4.1. The value functions uεI and uεII are p-harmonious in Ω with boundary values F : Γε → R; that is, they both satisfy α uε (x) = 2

sup uε + inf uε B ε (x)

B ε (x)

+β Bε (x)

uε dy

for every

x ∈ Ω, (12)

and uε (x) = F (x),

for every

x ∈ Γε .

The existence of p-harmonious functions with given boundary values is obtained by playing the Tug-of-War games with noise. Uniqueness follows by using martingales, although the equation is not linear. This was ﬁrst proved to the best of my knowledge in [16] for p = ∞. For ﬁnite p whether the original Tug-of-War game with noise described in [17] has a value is an open problem. For our modiﬁed version of the p-game we do have a value. The key is to judiciously choose strategies so that we can bring martingales into play. Lemma 4.2. (Key Lemma) Let vε be p-harmonious such that F ≤ vε on Γε . Player I chooses an arbitrary strategy SI and player II chooses a 0 strategy SII that almost minimizes vε , vε (xk ) ≤

inf

y∈B ε (xk−1 )

vε (y) + η2−k .

Then Mk = vε (xk ) + η2−k is a supermartingale for any η > 0 and uεI ≤ vε . We can now see how the inequality at the boundary literally walks into the interior by using Doob’s optional stopping theorem for martingales

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J.J. Manfredi

uεI (x0 ) = sup inf ExSI ,SII [F (xτ )] SI SII

≤ sup ExS0I ,S 0 [vε (xτ ) + η2−τ ] II

SI

≤ sup ExS0I ,S 0 [Mτ ] II

SI

≤ sup M0 = v ε (x0 ) + η SI

An extension of the above technique gives the uniqueness of the value function. Theorem 4.3. [12] The game has a value. That is uεI = uεII . Most importantly for our purposes is the fact the p-harmonious functions satisfy the Strong Comparison Principle: Theorem 4.4. [12] Let Ω ⊂ Rn be a bounded domain and let uε and vε be p-harmonious with boundary data Fu ≥ Fv in Γε . Then if there exists a point x0 ∈ Ω such that uε (x0 ) = vε (x0 ), it follows that uε = vε in Ω, and, moreover, the boundary values satisfy Fu = Fv in Γε . To prove that p-harmonious functions converge to the unique solution of the Dirichlet problem for the p-Laplacian in Ω with ﬁxed continuous boundary values, we assume that Ω is bounded and satisﬁes the exterior cone condition. Theorem 4.5. [12] Consider the unique viscosity solution u to

div(|∇u|p−2 ∇u)(x) = 0, x ∈ Ω

(13)

x ∈ ∂Ω,

u(x) = F (x),

and let uε be the unique p-harmonious function with boundary values F . Then uε → u

uniformly in

Ω

as

ε → 0.

The above limit only depends on the values of F on ∂Ω, and therefore any continuous extension of F |∂Ω to Γε0 gives the same limit. The key to prove this theorem is to pass from the discrete setting of pharmonious functions to the continuous case of p-harmonic functions. This is done by means of a characterization of p-harmonic functions in terms of asymptotic mean value properties. Theorem 4.6. [11] Let u ∈ C(Ω) such that for all x ∈ Ω we have α 2

sup u + inf u B (x)

B (x)

u = u(x) + o(2 ),

+β B (x)

as → 0.

Then u is p-harmonic in Ω. Here α and β are chosen as in (11).

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147

The converse of this theorem holds if we weaken the asymptotic expansion to hold only in the viscosity sense. See [11] for details. Another approach to pass from the discrete to the continuous for fully-nonlinear equations has been given by Kohn and Serfaty [6] by using a deterministic control theory approach.

5

Directed Trees

Let T be a directed tree with regular 3-branching as in Sect. 1. Let u : T → R be a real valued function. The gradient of u at the vertex v is the vector in R3 ∇u(v) = (u(v0 ) − u(v), u(v1 ) − u(v), u(v2 ) − u(v)). The divergence of a vector X = (x, y, z) ∈ R3 is div(X) = x + y + z. A function u is harmonic if it satisﬁes the Laplace equation div(∇u) = 0.

(14)

Observe that a function u is harmonic if and only if it satisﬁes the mean value property 1 u(v) = (u(v0 ) + u(v1 ) + u(v2 )) 3 Set X = T ∪ ∂T , X = T and Y = ∂T . The measure μ(v) is the normalized counting measure in S(v) μ(v) =

1 (δv0 + δv1 + δv2 ) . 3

The pay-oﬀ function F : ∂T → R is deﬁned on the unit interval [0, 1]. We are ready to play games in T . Think of a random walk started at the top vertex ∅ and move downward by choosing successors at random with uniform probability. When you get at ∂T at the branch point b determined by the random walk, you get paid f (b) dollars. Every time we run the game we get a sequence of vertices v1 , v2 , . . . , vk , . . . that determine a point on b the boundary ∂T . The set of all boundary points that start at a given vertex vr at level r is a ternary interval of length 3−r that we denote by Ivr . Averaging out over all possible plays that start at vr we obtain the value function

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J.J. Manfredi

Evr [f (t)] = u(vr ) =

1 |Ivr |

f (b) db,

(15)

Ivr

which is indeed harmonic in T . Therefore we have the well-known Lemma 5.1. Dirichlet Problem in Trees (p = 2): Given a continuous (indeed in L1 ([0, 1])) function f : [0, 1] → R the unique harmonic function u : T → R such that lim u(br ) = f (b) r→∞

for every branch b = (br ) ∈ ∂T is given by (15). Let us now play a Tug-of-War game with noise. Choose α ≥ 0, β ≥ 0 such that α + β = 1. Start at ∅. With probability α the players play Tug-ofWar. With probability β move downward by choosing successors at random. When you get at ∂T at the point b player II pays f (b) dollars to player I. The value function u veriﬁes the dynamic programming principle or mean value property u(v )+ u(v )+ u(v ) α 0 1 2 u(v) = max{u(vi )}+ min{u(vi )} +β i 2 i 3

(16)

that we can interpret as a PDE on the tree by using the following formula for a generalized divergence depending on the parameters α and β. Definition 5.2. Let X = (x, y, z) be a vector in R3 . The (α, β)-divergence of X is given by divα,β (X) =

α (max{x, y, z} + min{x, y, z}) + β 2

x+y+z 3

.

Theorem 5.3. [18] We have the equivalence DPP ≈ MVP ≈ PDE in the sense that the function u satisfies (16) in the tree T if and only if divα,β (∇u) = 0

(17)

Some particular cases are:

(i) The Linear Case: α = 0, β = 1 that corresponds to the linear case p = 2 of harmonic functions (14).

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149

(ii) The Discrete ∞-Laplacian: α = 1, β = 0 that corresponds to the case p = ∞. In this case the divergence is 1 div∞ (X) = div1,0 (X) = (max{x, y, z} + min{x, y, z}) 2 and the equation is the discrete ∞-Laplacian div∞ (∇u) = 0.

(iii) The Discrete p-Laplacian: For α = 0 and β = 0 we can select p as in (11), but the role of n is not intrinsically deﬁned, to obtain the discrete p-Laplacian We remark that this is the non-divergence form of the p-Laplacian (17). A discrete version of the p-Laplacian in divergence form can be found in [5]. While the formula (15) for the solution to the Dirichlet problem for p = 2 is explicit, there are not such formulas to my knowledge for the case p = 2. However, the game theoretic interpretation allows us to ﬁnd explicit formulas in some special, but interesting cases. Suppose that f is monotonically increasing. In this case the best strategy SI for player I is always to move right and the best strategy SII for player II always to move left. Starting at the vertex v at level k v = 0.b1 b2 . . . bk ,

bj ∈ {0, 1, 2}

we always move either left (adding a 0) or right (adding a 1). In this case Iv is the Cantor-like set Iv = {0.b1 b2 . . . bk d1 d2 . . .}, dj ∈ {0, 2}. Theorem 5.4. [18] The (α, β)-harmonic function with boundary values f in the tree T is given by f (b)dPα,β u(v) = v db, Iv

is a probability in [0, 1]. where Pα,β v Moreover in the case α = 0, β = 1, which corresponds to p = 2 the measure P0,1 is just the Lebesgue measure, and in the case α = 1, β = 0, v which corresponds to the case p = ∞, the measure P1,0 v is a Cantor measure supported in Iv . To see why this theorem is true observe that [f (b)]. u(v) = sup inf EvSI ,SII [f (b)] = EvSI ,SII

SI SII

Since the strategies used are always the same, we are indeed in a linear situation. All we need to do is to compute the probability PvS ,S . I

II

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J.J. Manfredi

Epilogue

We take the opportunity to state two open problems for p-harmonic functions in Rn that have challenged the mathematical community for more than thirty years. Open Problem 6.1 (Strong Comparison Principle): Suppose that u and v are p-harmonic functions in BR (x0 ) such that

u(x)

≤ v(x) for x ∈ BR (x0 ), and

u(x0 )

= v(x0 ).

(18)

Does it follow that u ≡ v in BR (x0 )? Open Problem 6.2 (Unique Continuation Property): Suppose that u is p-harmonic in B2R (x0 ) and that u(x) = 0 for every x ∈ BR (x0 ). Does it follow that u ≡ 0 in B2R (x0 )? Of course, the answer to this problems is clearly yes in the linear case p = 2. The answer to both problems is also yes in the planar case n = 2 since complex methods are then available [1, 10]. It is natural to try to apply new techniques developed in Analysis to these problems in the hope of improving our understanding of them. When the viscosity theory was developed, it was ﬁrst proved that the notions of Sobolev weak solution and viscosity solution agree [3] allowing us to study the p-harmonic equation not only by variational methods for divergence form equations but also using viscosity methods for non-divergence form equations. Progress in various problems followed: The ∞-eigenvalue problem [4], unexpected superposition principles [2] and [7], and various proofs were extended and simpliﬁed by using sup-convolutions [8]. The non-linear potential theory on trees of Kauﬀman and Wu [5] proved essential to settle a long standing conjecture of Martio on the lack of subadditivity of p-harmonic measures even at the zero level [9]. In the opinion of this author, the connection between the p-harmonic equation and discrete stochastic games discovered in the case p = ∞ ﬁrst by Peres, Sheﬃeld, Schramm, and Wilson [16] and for ﬁnite p by Peres and Sheﬃeld [17] (see also [15]) that opens the door to the use of game theoretic and control theoretic methods to the study of the p-Laplace equation, will have substantial applications [15]. A distinguishing feature of the stochastic games approach is that, since it is based on discrete stochastic processes, it provides good discrete approximations to p-harmonic functions, making a direct connection with the analysis in trees of [5].

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References 1. G. Alessandrini, Critical points of solutions of elliptic equations in two variables, Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4) 14(2), 229–256 (1988) 2. M. Crandall and J. Zhang, Another way to say harmonic, Trans. Am. Math. Soc. 355, 241–263 (2002) 3. P. Juutinen, P. Lindqvist, J. Manfredi, On the equivalence of viscosity solutions and weak solutions for a quasi-linear elliptic equation, SIAM J. Math. Anal. 33, 699–717 (2001) 4. P. Juutinen, P. Lindqvist, J. Manfredi, The ∞-eigenvalue problem, Arch. Ration. Mech. Anal. 148 89–105 (1999) 5. R. Kaufman, J.G. Llorente, J.M. Wu, Nonlinear harmonic measures on trees, Ann. Acad. Sci. Fenn. Math. 28, 279–302 (2003) 6. R. Kohn, S. Serfaty, A deterministic-control-based approach to fully nonlinear parabolic and elliptic equations, Comm. Pure Appl. Math. 63(10), 1298–1350 (2010) 7. P. Lindqvist, J. Manfredi, Note on a remarkable superposition for a nonlinear equation, Proc. Am. Math. Soc. 136, 229–248 (2008) 8. P. Lindqvist, J. Manfredi, Viscosity solutions of the evolutionary p-Laplace equation, Diﬀer. Integr. Equat. 20, 1303–1319 (2007) 9. J.G. Llorente, J. Manfredi, J.M. Wu, p-harmonic measure is not subadditive, Annalli de della Scuola Normale Superiore di Pisa, Classe di Scienze IV(5), 357–373 (2005) 10. J. Manfredi, p-harmonic functions in the plane, Proc. Am. Math. Soc. 103(2), 473–479 (1988) 11. J.J. Manfredi, M. Parviainen, J.D. Rossi, An asymptotic mean value characterization for p-harmonic functions. Proc. Am. Math. Soc., 258, 713–728 (2010) 12. J.J. Manfredi, M. Parviainen, J.D. Rossi, On the deﬁnition and properties of p-harmonious functions. To appear in Ann. Sc. Norm. Super. Pisa Cl. Sci. 13. J.J. Manfredi, M. Parviainen, J.D. Rossi, Dynamic programming principle for tug-ofwar games with noise. ESAIM: Control, Optimisation and Calculus of Variations, 18, 81–90 (2012) 14. A.P. Maitra, W.D. Sudderth, Discrete gambling and stochastic games. Appl. Math. 32, Springer-Verlag, 1996 15. Y. Peres, G. Pete, S. Somersielle, Biased tug-of-war, the biased inﬁnity Laplacian and comparison with exponential cones, Calc. Var. Partial Diﬀerential Equations 38(3), 541–564 (2010) 16. Y. Peres, O. Schramm, S. Sheﬃeld, D. Wilson, Tug-of-war and the inﬁnity Laplacian, J. Am. Math. Soc. 22, 167–210 (2009) 17. Y. Peres, S. Sheﬃeld, Tug-of-war with noise: a game theoretic view of the p-Laplacian, Duke Math. J. 145(1), 91–120 (2008) 18. A. Sviridov, Elliptic Equations in Graphs via Stochastic Games, University of Pittsburgh 2010 doctoral dissertation 19. D.W. Stroock, S.R.S. Varadan, Multidimensional Diﬀusion Processes, Grundlehren der Mathematische Wissenschaften 233, Springer-Verlag, 1979 20. C.T. Ionescu Tulcea, Measures dans les espaces produits, Atti. Acad. Naz. Lincei. Rend. Cl. Sci. Fis. Mat. Nat. 7(8), 208–211 (1949) 21. S.R.S. Varadan, Probability Theory, Courant Lecture Notes in Mathematics 7, New York University, 2000

The Problems of the Obstacle in Lower Dimension and for the Fractional Laplacian Sandro Salsa

1

Introduction

The obstacle problem for a fractional power of the Laplace operator appears in many contexts, such as in the study of anomalous diﬀusion [5], in the so called quasi-geostrophic ﬂow problem [12], and in pricing of American options governed by assets evolving according to jump processes [13]. It can be stated in several ways. Given a smooth function ϕ : Rn → R, n > 1, with bounded support (or at least rapidly decaying at inﬁnity), we look for a continuous function u satisfying the following system: ⎧ u≥ϕ ⎪ ⎪ ⎨ s (−Δ) u ≥ 0 ⎪ (−Δ)s u = 0 ⎪ ⎩ u (x) → 0

in Rn in Rn when u > ϕ as |x| → +∞.

(1)

Here we consider only the case s ∈ (0, 1). The deﬁnition and some of the main s properties of (−Δ) are collected in Appendix A. The set Λ (u) = {u = ϕ} is called the contact or coincidence set. The boundary of the set Rn \Λ (u) is the free boundary, denoted by F (u). The main theoretical issues in a constrained minimization problem are optimal regularity of the solution and the analysis of the free boundary. If s = 1 and Rn is replaced by a bounded domain Ω our problem corresponds to the usual obstacle problem for the Laplace operator. The existence of a unique solution satisfying some given boundary condition u = g S. Salsa () Department of Mathematics, Politecnico di Milano, Piazza Leonardo Da Vinci 32, I-20133 Milano, Italy e-mail: [email protected]

J. Lewis et al., Regularity Estimates for Nonlinear Elliptic and Parabolic Problems, Lecture Notes in Mathematics 2045, DOI 10.1007/978-3-642-27145-8 4, © Springer-Verlag Berlin Heidelberg 2012

153

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S. Salsa

can be obtained by minimizing the Dirichlet integral in H 1 (Ω) under the constraint u ≥ ϕ. The solution is the least superharmonic function greater than equal to ϕ in Ω, with u ≥ g on ∂Ω, and inherits up to a certain level the 1,1 regularity of ϕ [15]. In fact, even if ϕ is smooth, u is only Cloc , which is the optimal regularity. A classical reference for the obstacle problem, including the regularity and the complete analysis of the free boundary is [7]. Analogously, the existence of a solution u for problem (1) can be obtained by variational methods as the unique minimizer of the functional

J (v) = Rn

2

|v (x) − v (y)|

Rn

n+2s

|x − y|

dxdy

over a suitable set of functions v ≥ ϕ. Also we can obtain u via a Perron type method, as the least supersolution of (−Δ)s such that u ≥ ϕ. By analogy (see also later the Signorini problem), when ϕ is smooth, we expect the optimal regularity for u to be C 1,s . This is indeed true as it is shown in [24] when the contact set {u = ϕ} is convex and in [11] in the general case. The case s = 1/2 is strongly related to the so called thin (or lower dimensional) obstacle problem for the Laplace operator. To keep a better connection with the obstacle problem for (−Δ)s , it is better to work in Rn+1 , writing X = (x, y) ∈ Rn × R. The thin obstacle problem concerns the case in which the obstacle is not anymore n + 1 dimensional, but supported instead on a smooth n-dimensional manifold M in Rn+1 . This problem and variation of it also arises in many applied contexts, such as ﬂow through semi-permeable membranes (Fig. 1), elasticity (known as Signorini problem), boundary control temperature or heat problems (see [14]). More precisely, let Ω be a domain in Rn+1 divided into two parts Ω + and − Ω by M. Let ϕ : M →R be the (thin) obstacle and g be a given function on ∂Ω satisfying g > ϕ on M∩∂Ω. ϕ(x) pressure free boundary ϕ=u @yu>0 ϕ
porous media u = u(x, y) pressure Δu = 0

Π

y Fig. 1 Stationary ﬂow through a semipermeable membrane of negligible thickness

The Problems of the Obstacle in Lower Dimension

155

The problem consists in the minimization of the Dirichlet integral 2 |∇v| J (v) = Ω

over the closed convex set K = v ∈ H 1 (Ω) : v = g on ∂Ω and v ≥ ϕ on M ∩ ∂Ω . Since we can perturb the solution u upwards and freely away from M, it is apparent that u is superharmonic in Ω and harmonic in Ω\M. One expects the continuity of the ﬁrst derivatives along the directions tangential to M, and the one sided continuity of normal derivatives [15]. In fact (see [8]), on M, u satisﬁes the following complementary conditions u ≥ ϕ, uν + + uν − ≤ 0, (u − ϕ) (uν + + uν − ) = 0 where ν ± are the interior unit normals to M from the Ω ± side. The free boundary here is given by the boundary of the set Ω\Λ (u) in the relative topology of M, and in general, we expect it is a (n−1)-dimensional manifold. A related problem is the boundary thin obstacle problem (or Signorini problem), in which the manifold M is part of ∂Ω and one has to minimize the Dirichlet integral over the closed convex set K = v ∈ H 1 (Ω) : v = g on ∂Ω\M and v ≥ ϕ on M . In this case, u is harmonic in Ω and on M satisﬁes the complementary conditions u ≥ ϕ, uν + ≤ 0, (u − ϕ) uν + = 0. If M is a hyperplane (say {y = 0}) and Ω is symmetric with respect to M, then the thin obstacle in Ω and the boundary obstacle problems in Ω + or Ω − are equivalent. It is this last case that is strongly related to the obstacle problem for (−Δ)1/2 . This is clearly explained through the following remarks. (a) Reduction to a global problem. Let Ω = B1 be the unit ball in Rn+1 and B1 = B1 ∩ {y = 0}. Let ϕ : Rn → R be a smooth obstacle, ϕ < 0 on ∂B1 and positive somewhere inside B1 . Consider the following Signorini problem in B1+ = B1 ∩ {y > 0}: ⎧ ⎪ −Δu = 0 ⎪ ⎪ ⎪ ⎪ ⎨u = 0 u (x, 0) ≥ ϕ (x) ⎪ ⎪ ⎪ ⎪ uy (x, 0) ≤ 0 ⎪ ⎩ u (x, 0) = 0 y

in B1+ on ∂B1 ∩ {y > 0} ¯ 1 in B ¯ 1 in B when u (x, 0) > ϕ (x) .

(2)

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We want to convert the above problem in B1 into a global one, that is in Rn × (0, +∞). To do this, let η be a radially symmetric cut-oﬀ function in B1 such that {ϕ > 0} {η = 1} and supp (η) ⊂ B1 . Extending ηu by zero outside B1 , we have η (x) u (x, 0) ≥ ϕ (x) and also (ηu)y (x, 0) ≤ 0 for every x ∈ Rn . Moreover, (ηu)y (x, 0) = 0 if η (x) u (x, 0) > ϕ (x). Let now v be the unique solution of the following Neumann problem in the upper half space, vanishing at inﬁnity:

Δv = Δ (ηu) in Rn × {y > 0} vy (x, 0) = 0 in Rn .

Then w = ηu − v is a solution of a global Signorini problem with ϕ − v as the obstacle. Thus, the regularity of u in the local setting can be inferred from the regularity for the global problem. The opposite statement is obvious. 1/2

(b) Realization of (−Δ) as a Dirichlet-Neumann map. Consider a smooth function u0 : Rn −→ R with rapid decay at inﬁnity. Let u : Rn × (0, +∞) −→ R be the unique solution of the Dirichlet problem

Δu = 0 in Rn × (0, +∞) u (x, 0) = u0 (x) in Rn

vanishing at inﬁnity. We call u the harmonic extension of u0 to the upper half space. Consider the Dirichlet-Neumann map T : u0 (x) −→ −uy (x, 0). We have: (T u0 , u0 ) =

Rn

−uy (x, 0) u (x, 0) dx

=

Rn ×(0,+∞)

=

Rn ×(0,+∞)

2 Δu (X) u (X) + |∇u (X)| dX |∇u (X)|2 dX ≥ 0

so that T is a positive operator. Moreover, since u0 is smooth and uy is harmonic, we can write: T ◦ T u0 = −∂y (−∂y )u (x, 0) = uyy (x, 0) = −Δu0 . We conclude that

T = (−Δ)1/2 .

The Problems of the Obstacle in Lower Dimension

157

As a consequence: 1. If u = u (X) is a solution of the Signorini problem in Rn × (0, +∞), that is Δu = 0 in Rn+1 , u (x, 0) ≥ ϕ, uy (x, 0) ≤ 0, and (u − ϕ) uy (x, 0) = 0 in 1/2 Rn , then u0 = u (·, 0) solves the obstacle problem for (−Δ) . 1/2 2. If u0 is a solution of the obstacle problem for (−Δ) , then its harmonic extension to Rn × (0, +∞) solves the corresponding Signorini problem. Therefore, the two problems are equivalent and any regularity result for one of them can be carried to the other one. More precisely, consider the 1/2 optimal regularity for the solution u0 of the obstacle problem for (−Δ) , 1,1/2 1,1/2 which is C . If we can prove a C regularity of the solution u of the Signorini problem up to y = 0, then the same is true for u0 . On the other hand, the C 1,α regularity of u0 extends to u, via boundary estimates for the Neumann problem. Similarly, the analysis of the free boundary in the Signorini problem carries to the obstacle problem for 1/2 (−Δ) as well and viceversa. Although the two problem are equivalent, there is a clear advantage in favor of the Signorini type formulation. This is due to the possibility of avoiding the direct use of the non local pseudodiﬀerential operator 1/2 (−Δ) , by localizing the problem and using local P.D.E. methods, such as monotonicity formulas and classiﬁcation of blow-up proﬁles. Concerning the Signorini problem, we will mainly refer to the papers [1, 3, 8, 18]. In [8] it is shown that u is C 1,α up to y = 0, for some α ≤ 1/2. For the zero-obstacle problem (ϕ = 0), the optimal regularity is shown in [1] while the analysis of the free boundary around a regular point is contained in [3]. A complete classiﬁcation of singular points, i.e. points of F (u) of vanishing density for Λ (u), is developed in [18] (also for non zero obstacles). At this point it is a natural question to ask whether there exists a P.D.E. s realization of (−Δ) for every s ∈ (0, 1), s = 12 . The answer is positive as it is shown in [10]. Indeed using the results in Sect. C3, in a weak sense we have that (−Δ)s u0 (x) = −κa lim y a uy (x, y) y→0+

for a suitable constant κa , where u is the solution of the problem

La u = div (y a ∇u) = 0 in Rn × (0, +∞) in Rn u (x, 0) = u0 (x)

vanishing at inﬁnity, where a = 2s− 1. Coherently, we call u the La -harmonic extension of u0 Thus problem (1) is equivalent to the following Signorini problem for the operator La = div(y a ∇),

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S. Salsa

u (x, 0) ≥ ϕ in Rn

(3)

La u = div (y a ∇u) = 0 in Rn × (0, +∞)

(4)

limy→0+ y a uy (x, y) = 0 when u (x, 0) > ϕ (x)

(5)

limy→0+ y a uy (x, y) ≤ 0 in Rn .

(6)

For y > 0, u is smooth so that (4) is understood in the classical sense. The equations at the boundary (5) and (6) should be understood in the weak sense (see Sect. A3). Since in [24] it is shown that u (x, 0) ∈ C 1,α for every α < s, for the range of values 2s − 1 < α < s, limy→0+ y a uy (x, y) can be understood in the classical sense too. The solution u of the above Signorini problem can be extended to the whole space by symmetrization, setting u (x, −y) = u (x, y). Then, by the results in [10] (see Sect. A3), condition (5) holds if and only if the extended u is a solution of La u = 0 across y = 0, where u (x, 0) > ϕ (x). On the other hand, condition (6) is equivalent to La u ≤ 0 in the sense of distributions. Thus, for the extended u, the Signorini problem translates into the following system: ⎧ in ⎪ ⎪ u (x, 0) ≥ ϕ (x) ⎨ u (x, −y) = u (x, y) in ⎪ La u = 0 in ⎪ ⎩ in La u ≤ 0

Rn Rn+1 Rn+1 \ {(x, 0) : u (x, 0) = ϕ (x)} Rn+1 , in the sense of distributions.

Again, we can exploit the advantages to analyze the obstacle problem for a nonlocal operator in P.D.E. form by considering a local version of it. Indeed, to study the optimal regularity properties of the solution we will focus on the following local version, where ϕ : B −→ R: ⎧ u (x, 0) ≥ ϕ (x) ⎪ ⎪ ⎨ u (x, −y) = u (x, y) ⎪ L u=0 ⎪ ⎩ a La u ≤ 0

in in in in

B1 B1 B1 \ {(x, 0) : u (x, 0) = ϕ (x)} B1 , in the sense of distributions.

The above problem can be thought as the minimization of the weighted Dirichlet integral a 2 Ja (v) = |y| |∇u (X)| dX B1

over the set a Ka = v ∈ W 1,2 (B1 , |y| ) : u (x, 0) ≥ ϕ (x) .

The Problems of the Obstacle in Lower Dimension

159

In a certain sense, this corresponds to an obstacle problem, where the obstacle is deﬁned in a set of codimension 1 + a, where a is not necessarily an integer. The operator La is degenerate elliptic, with a degeneracy given by the a n+1 weight |y| . This weight belongs to the Muckenhoupt class . NA

2 R A positive weight function w = w (X) belongs to A2 R if

1 |B|

w B

1 |B|

w

−1

≤C

B

for every ball B ⊂ RN . For the class of degenerate elliptic operators of the form Lu = div(A (X) ∇u), where 2

2

λw (X) |ξ| ≤ A (X) ξ · ξ ≤ Λw (X) |ξ| , there is a well established potential theory for solutions in the weighted Sobolev space W 1,2 (Ω, w) , Ω bounded domain in RN , deﬁned as the closure

∞ ¯ of C Ω in the norm

2

1/2

|∇v| w

v w+ Ω

2

Ω

(see [16]). Since w ∈ A2 the gradient of a function in W 1,2 (Ω, w) is well deﬁned in the sense of distributions1 and belongs to the weighted space L2 (Ω, w). The outline of the paper. We intend here to give a somewhat self-contained presentation of the results concerning the analysis of the solution and the free boundary of the thin obstacle problem and more generally of the obstacle for the fractional Laplacian, contained in the papers [1, 3, 8, 10, 11, 18, 24]. We will start from the thin obstacle problem, considering the case of zero obstacle. In this case the main ideas and tools are clearly seen and developed without too many technicalities and in a somewhat self-contained fashion. Afterwards, following the same strategy and using the results of [24] for the initial partial regularity of the solution, we extend the results on the optimal regularity and the analysis of the regular part of the free boundary s to the general case for (−Δ) . Since all the main diﬃculties in the analysis appear in the behavior of the solution near F (u), we will always assume that the origin belongs to the free boundary. 1

In fact, if {ϕk } ⊂ C0∞ (Ω) and ϕk L2 (Ω,w) → 0, ∇ϕk − vL2 (Ω,w) → 0, then v = 0.

160

S. Salsa

y

Δn+1 u = 0 @yu = 0

Λ(u) @y u ≤ 0

F (u)

xn

Fig. 2 Zero obstacle problem

2 2.1

The Zero Obstacle Problem Setting of the Problem

In this section we give a rather complete and self-contained analysis of the thin obstacle problem. To better emphasize the ideas, we consider in this section the zero obstacle problem: to minimize the Dirichlet integral (see Fig. 2) 2

|∇v| dX B1

over the set of function belonging to the closed convex set K = v ∈ H 1 (B1 ) : v = g on ∂B1 , v (x, 0) ≥ 0 in B1 where g is a smooth function on ∂B1 . It is easily seen that the minimizer u coincides with the least among all superharmonic functions v in B1 with v ≥ g on ∂B1 , and v (x, 0) ≥ 0 in B1 . Thus, {u > 0} is open, and the contact set Λ (u) = {(x, 0) : u (x, 0) = 0, x ∈ B1 } is closed. The free boundary is the boundary of the set {u (x, 0) > 0} ∩ B1 relative to the hyperplane y = 0 that is: F (u) = ∂ {x : u (x, 0) > 0} ∩ B1 .

The Problems of the Obstacle in Lower Dimension

161

In order to avoid trivial contact set and free boundary, we assume that g changes sign and that g (θ , 0) > 0 for θ ∈ ∂B1 . Moreover, we set Λ (u) = {x : u (x, 0) = 0} and Ω = B1 \Λ (u). Our analysis consists in proving the following steps: 1. Preliminary estimates: Lipschitz continuity and C 1,α local estimates for some 0 < α < 1. 2. Optimal regularity for tangentially convex global solutions. 3. Classiﬁcation of asymptotic blow-up proﬁles around a free boundary point. 4. Almgren’s frequency formula and optimal regularity (i.e. C 1,1/2 ) of the solution. 5. Analysis of the free boundary for non-degenerate proﬁles around stable points: Lipschitz continuity. 6. Boundary Harnack Principles and optimal regularity of the free boundary. 7. Structure of the subset of the non stable points of the free boundary at which Λ (u) has vanishing density (singular set ). It is worthwhile to start with an observation of Hans Levy in R2 that gives a clue of why C 1,1/2 is the optimal regularity. If u is a global solution of the zero obstacle problem, the complex function w = ux − iuy is analytic outside Λ (u), and therefore, there, ux uy =

1 Im(w2 ) 2

is harmonic and vanishes on y = 0. Thus, ux uy has an harmonic odd extension across y = 0 and w2 has an analytic extension. Then w ∈ C 1/2 which implies u ∈ C 1,1/2 . An important example of global solution is u = ρ3/2 cos 32 θ; this is a typical proﬁle that we call nondegenerate (Fig. 3). On the other hand, there are solutions like ρk+1/2 cos

2k + 1 θ 2

or ρ2k cos 2kθ, k ≥ 2,

vanishing of higher order at the origin. In these cases we cannot expect any regularity of the free boundary. In fact, for the free boundary, the regular (or nondegenerate or stable) points are those for which a suitable asymptotic proﬁle is nondegenerate in the above sense. Around this kind of points we expect regularity of F (u).

Fig. 3 Nondegenerate global proﬁle

162

S. Salsa

The non regular points of the free boundary can be divided in two classes: the set Σ (u) at which Λ (u) has a vanishing density (singular points), that is Σ (u) =

Hn (Λ (u) ∩ Br (x0 )) (x0 , 0) ∈ F (u) : lim = 0 , rn r→0+

and the set of non regular, non singular points. The following example (see [18]), given by the harmonic polynomial 1 p (x1 , x2 , y) = x21 x22 − (x21 + x22 )y 2 + y 4 3 shows that the entire free boundary could be composed by singular points. In fact F (p) = Λ (p) is given by the union of the lines x1 = y = 0 and x2 = y = 0. We shall see that, as in this example, the singular set is contained in the union of C 1 -manifold of suitable dimension [18].

2.2

Lipschitz Continuity and Semiconvexity

We start by proving some preliminary estimate and partial regularity. Notice that, without loss of generality, we can assume that u (x, −y) = u (x, y). Otherwise, we solve Δw = 0 in B1 w = g ∗ on B1 where g ∗ (x, y) = (g (x, y) − g (x, −y)) /2. Then, u − w is symmetric about y = 0 and solves the same kind of problem. Also, it is clear that u is harmonic outside Λ (u) and u (x, 0) > 0 in B1 \B1−d for some d0 depending on g. 0 Lemma 2.2.1. Let u be a solution of the zero thin obstacle problem in B1 . Then: (a) u+ and u− are subharmonic. (b) There exist constants C such that:

u H 1 (B1/2 ) ≤ C u L2 (B1 )

u L∞ (B1/2 ) ≤ C u L2 (B1 ). Proof. Since u is superharmonic, min {u, 0} = −u− is superharmonic. To show that u+ is subharmonic we perturb u as follows. Let η ∈ C0∞ (B1 ), η ≥ 0, and fδ (s) be a δ approximation of the Heaviside function, that is

The Problems of the Obstacle in Lower Dimension

163

⎧ s≤0 ⎨ 0 fδ (s) = s/δ 0 < s < δ ⎩ 1 s ≥ δ. For 0 < ε < δ, small, the function uε = u − εηfδ (u) belongs to K. Therefore we can write |∇uε |2 ≥ |∇u|2 B1

B1

which gives, letting ε → 0, after a division by ε: fδ (u) ∇u · ∇η ≤ − η∇u · ∇fδ (u) {0
B1

or, setting Fδ (u) =

u

0

fδ (s) ds,

∇Fδ (u)·∇η ≤ −

B1

{0
η∇u·∇fδ (u) = −

2

{0
η |∇u| fδ (u) ≤ 0.

Therefore, Fδ (u) is subharmonic. Now, if δ → 0, Fδ (u) u+ weakly in H 1 (B1 ) and we deduce that u+ is subharmonic too. (b) Since u± are subharmonic, Caccioppoli inequality gives

± 2 ∇u ≤ C

B1/2

(u± )2

B1

so that u H 1 (B1/2 ) ≤ C u L2 (B1 ) .

Moreover, since (u± )2 also are subharmonic, we can write

sup u B1/2

± 2

≤

(u± )2

B1

from which u L∞ (B1/2 ) ≤ C u L2 (B1 ) . Now we prove Lipschitz continuity and semi-convexity along tangential directions. By a tangential direction we mean any unit vector τ parallel to the hyperplane {y = 0}. Theorem 2.2.2. Let u be a solution of the zero thin obstacle problem in B1 . Then: (a) u is Lipschitz. Moreover

u Lip(B1/2 ) ≤ C u L2 (B1 ) . (b) inf uτ τ ≥ −C u L2 (B1 ) .

B1/2

164

S. Salsa 2

Proof. We enlarge the obstacle to hε (x, y) = − yε and let uε be the corresponding solution of the thick obstacle problem. Notice that u ≤ uε . Then: i) uε C 1,1 (B2/3 ) ≤ C (ε).

ii) Λε = {X : uε = hε } ⊂ {|y| < Cε} Indeed, i) is well known (see [15]). To prove ii) let m > 0 such that −m < min uε ∂B1

and take h ∈ C(B1+ ) such that ⎧ ⎨ Δh = 0 h=0 ⎩ h = −m

in B1+ in B1 on ∂B1+ ∩ {y > 0} .

Since h is Lipschitz up to B1−d , then h > hε if y > Cε and, since uε ≥ h, 0 ii) follows by symmetry. Let us prove (a). Since ∇uε is harmonic outside Λε and ∇uε = ∇hε on Λε , it follows from ii) and the maximum principle that, in B1 , |∇uε | ≤ C, with C = C (n, g) independent of ε. In particular, in B2/3 , we have |∇uε | ≤ c uε L2 (B1 ) . Letting ε → 0 and using b) in Lemma 2.2.1, we deduce (a).

For (b) we give two proofs. First proof. We use a penalization technique. Let βδ = βδ (s) a family of smooth increasing, concave functions such that βδ (s) = δ if s > 0 and as δ → 0, βδ converges to the graph β0 (s) = −∞ if s < 0, β0 (s) = 0 if s > 0. Let now solve the following penalized problem:

Δvε,δ = βδ (vε,δ − hε ) vε,δ = g

in B1 on ∂B1 .

From [6], we know that a smooth unique solution vε,δ exists and vε,δ H 2,p ≤ c ( g H 2−1/p,p ). Diﬀerentiating twice the equation along the direction τ , we get, setting w = ∂τ τ vε,δ : 2

Δw = βδ (vε,δ − hε ) [∂τ vε,δ ] + βδ (vε,δ − hε ) w ≤ βδ (vε,δ − hε ) w. Hence, since near ∂B1 we have vε,δ > hε and vε,δ is bounded, we have Δw = δw and w remains bounded. On the other hand, if w has a global minimum at a point x∗ ∈ B1 , then Δw (x∗ ) ≥ 0. Since βδ ≥ 0, we infer w ≥ 0. Using again the results in [6] and passing to the limit ﬁrst as δ → 0 and then for ε → 0, we conclude the ﬁrst proof.

The Problems of the Obstacle in Lower Dimension

165

Second proof. For global solutions (b) follows simply from the observation that u (X + hτ ) and u (X − hτ ) are admissible nonnegative superharmonic functions and therefore 1 [u (X + hτ ) + u (X − hτ )] ≥ u (X) . 2 In the local case, we modify the argument in the following way. Consider the ring S = {X ∈ B1 : 1 − d0 < |X| < 1} . Inside S, u is harmonic. Then, from interior estimates, we get

u C 2,α (S ∗ ) ≤ C u L2 (B1 ) 2 1 S = x ∈ B1 : 1 − d0 < |X| < 1 − d0 . 3 3

where

∗

Thus, wh (X) ≡

1 [u (X + hτ ) + u (X − hτ )] + Ch2 u L2 (B1 ) ≥ u (X) 2

for any X with |X| = 1 − d0 /2 and −d0 /6 < h < d0 /6. Observe now that wh is superharmonic in B1−d0 /2 and wh (x, 0) ≥ 0 in B1−d . On the other hand, u is the solution of the zero thin obstacle in 0 /2 B1−d0 /2 . Therefore, since wh is admissible for comparison with u in B1−d0 /2 we can write wh (X) ≥ u (X) in all B1−d0 /2 . This shows that inf B1/2 uτ τ ≥ −C u L2 (B1 ) .

Remark 2.2.1 As a consequence: uyy ≤ C in B1 \Λ (u) . In particular, the function uy − Cy is monotone and bounded. Thus we are allowed to deﬁne in B1 : σ (x) = lim uy (x, y) . y→0+

Since n 0 ≥ Δu = 2uy H|Λ(u)

in D (B1 ), we have σ (x) ≤ 0 in Λ (u) and by symmetry, σ (x) = 0 in B1 \Λ (u). We may summarize the property of the solution of a zero thin obstacle problem in complementary form as follows:

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⎧ ⎪ ⎪ Δu ≤ 0, uΔu = 0 ⎨ Δu = 0 ⎪ u (x, 0) ≥ 0, σ (x) ≤ 0, u (x, 0) σ (x) = 0 ⎪ ⎩ σ (x) = 0

in in in in

B1 B1 \Λ (u) B1 B1 \Λ (u)

Let us point out some consequences of the above estimates. Lemma 2.2.3. Let u be a solution of the zero thin obstacle problem in + : B1 , normalized by u L2 (B1 ) = 1. Then, for some universal constant, in B1/2 2

(a) y −→ u (x, y) − Cy 2 is concave and x −→ u (x, y) + C |x| is convex. In particular: uy (x, t) − uy (x, s) ≤ 2C (t − s) and

(t > s)

u (x, t) − u (x, 0) ≤ Ct2 .

(7)

(8)

(b) If u (x, t) ≥ h, then in the half ball (a ball if ∇x u (x, t) = 0) HBρ (x) = {z : |x − z| ≤ ρ, z − x, ∇x u (x, t) ≥ 0} we have u (z, t) ≥ h − Cρ2 . Proof. (a) follows directly from Theorem 2.2.2b. Then uy − 2Cy is decreasing and therefore uy (x, t) − 2Ct ≤ uy (x, s) − 2Cs which is (7). Integrating uy (x, s) ≤ Cs over 0, t gives (8). (b) We have, by convexity: 2

2

u (z, t) + C |z| ≥ u (x, t) + C |x| + z − x, ∇x u (x, t) + 2Cx . If u (x, t) ≥ h, in HBρ (x) we have 2

u (z, t) ≥ u (x, t) − C |x − z| ≥ h − Cρ2 .

2.3

Local C1,α Estimate

Let u be a solution of the zero thin obstacle problem in B1 , normalized by

u L2 (B1 ) = 1. We prove now a local C 1,α estimate. It is enough to show that σ ∈ C 0,α near the free boundary F (u). In fact, in the interior of Λ (u), u (x, 0) is smooth and so σ is. On the other hand, on Ω = B \Λ (u) , σ = 0. Thus if we show that σ is C 0,α in a

The Problems of the Obstacle in Lower Dimension

167

neighborhood of F (u), then u ∈ C 1,α from both sides of the free boundary by standard estimates for the Neumann problem. In particular it is enough to show uniform estimates around a free boundary point, say the origin. We distinguish two steps: Step 1: To show that near the free boundary we can locate large regions where −σ grows at most linearly (estimates in measure of the oscillation of −σ). Step 2: Using Poisson representation formula and concavity, we convert the estimate in average of the oscillation of −σ, done in step 1, into pointwise estimates, suitable for iteration. Step 1 : First a barrier. Lemma 2.3.1. Let x0 ∈ Ω = B1 \Λ (u). Deﬁne 2

hx0 (X) = |x − x0 | − ny 2 . Then, for any open set A such that (x0 , 0) ∈ A ⊂ B1 , sup (u − hx0 ) ≥ 0. ∂A

Proof. The function w = u − hx0 is harmonic in A\Λ (u) and w (x0 , 0) ≥ 0. Since x0 ∈ / Λ (u) we infer that sup (u − hx0 ) ≥ 0.

∂(A\Λ)

On the other hand, on Λ (u), hx0 > u, and therefore the lemma follows.

The point in the above lemma is that the supremum of w is not attained at a point on Λ (u) . Lemma 2.3.2. Let x0 ∈ Ω = B1 \Λ (u) and deﬁne Eγ = {x : σ (x) ≥ −γ}

(γ > 0) .

Then, for suitable positive constants C ∗ , C¯ and any suﬃciently small γ, there ∗ exists a ball BC ∗ γ (x ) such that ∗ BC ∗ γ (x ) ⊂ B ¯ (x0 ) ∩ Eγ . Cγ

In particular

n BCγ ¯ (x0 ) ∩ Eγ ≥ cγ .

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Proof. We apply Lemma 2.3.1, choosing A = BC (x0 ) × (−C2 γ, C2 γ) 1γ

with C1 C2 .

The maximum of u − hx0 can be achieved only on the lateral side of the cylinder or on one of the two bases. By symmetry, it is enough to consider the base BC (x0 ) × {y = C2 γ} . 1γ Case 1. sup∂A (u − hx0 ) is attained at a point (ξ, t) on the lateral side. In this case:

u (ξ, t) ≥ hx0 (ξ, t) ≥ C12 − nC22 γ 2 ≡ C3 γ 2 . Apply now Lemma 2.2.3 (b) with h = C3 γ 2 and ρ = C4 γ, with C4 to be chosen small enough. Then in the half ball HBρ (ξ) = {z : |ξ − z| ≤ ρ, z − ξ, ∇x u (ξ, t) ≥ 0} we have

u (z, t) ≥ C3 γ 2 − CC4 γ 2 ≡ C5 γ 2

if C4 is small enough. Notice that HBρ (ξ) ⊂ B(C (x0 ) , if C4 is small, 4 1 +C )γ ∗ and that HBρ (ξ) contains a ball Bρ/2 (x ). To prove the theorem it is enough to show that if z ∈ HBρ (ξ) then z ∈ Eγ . If not, σ (z) < −γ. Then u (z, 0) = 0 so that, since uyy ≤ C:

u (z, t) = u (z, t) − u (z, 0) ≤ tσ (z) + Ct2 ≤ −γt + Ct2 ≤ (−C2−1 + C)t2 < 0 if C2 is chosen small, since t ≤ C2 γ. This gives a contradiction and ends the proof in the ﬁrst case, choosing C¯ = C1 + C4 . Case 2. sup∂A (u − hx0 ) is attained at a point (ξ, C2 γ) on the base of the cylinder. Then 2

u (ξ, C2 γ) ≥ hx0 (ξ, C2 γ) ≥ −n (C2 γ) . As before, apply Lemma 2.2.3 (b) with h = −n (C2 γ)2 and ρ = C2 γ. Then in the half ball HBρ (ξ) = {z : |ξ − z| ≤ ρ, z − ξ, ∇x u (ξ, t) ≥ 0} we have

2

2

2

u (z, C2 γ) ≥ −n (C2 γ) − C (C2 γ) ≡ −C5 (C2 γ) . If σ (z) < −γ, we would have u (z, 0) = 0 and, using once more uyy ≤ C, we get

The Problems of the Obstacle in Lower Dimension 2

169 2

2

u (z, C2 γ)−u (z, 0) = C2 γσ (z)+C (C2 γ) ≤ −C2 γ 2 +C (C2 γ) < −C5 (C2 γ) if C2 is small. Contradiction. We complete the proof choosing C = C1 + C2 .

Step 2. We convert the above estimate into a pointwise estimate. We need the following remark, consequence of the Poisson formula. Remark 2.3.1. Let v be a positive harmonic function in B1 (x0 ) × (0, 1) continuous in B1 (x0 ) × [0, 1]. Assume that in Bδ (x∗ )

v (x, 0) ≥ 1

for some Bδ (x∗ ) ⊂ B1 (x0 ). Then, by Poisson formula, v (X) ≥ η (δ) > 0

(x0 ) × [1/4, 3/4] . in B1/2

Theorem 2.3.3. Let u be our normalized solution and x0 ∈ Ω . Then there exist α, 0 < α < 1, and a constant C, depending only on n and the number d0 (deﬁned at the beginning of the section) such that σ (x) ≥ −C |x − x0 |

for every x ∈ B2/3 .

α

) and u ∈ C 1,α (B1/2 ). In particular As a consequence, σ ∈ C 0,α (B2/3

u C 1,α B +

1/2

≤ C.

Proof. Recall that uy is bounded in B2/3 and uyy ≤ C in B1 \Λ. Moreover, σ ≤ 0. Thus, it is enough to prove that uy (X) ≥ −β k

in Bγ k (x0 ) × 0, γ k

(9)

for some 0 < γ < 1, 0 < β < 1, both to be chosen, and any k ≥ 0. The case k = 0 follows from the boundedness of uy in B2/3 . Now, assume that, inductively, we have proved (9) up to k : uy (X) ≥ −β k

in Bγ k (x0 ) × 0, γ k

for 0 < γ β < 1. Consider the function w (X) =

uy (X) + β k −C0 μγ k + β k

k . in Bμγ k (x0 ) × 0, μγ

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S. Salsa

If μ is ﬁxed, suﬃciently small, adjusting the constant2 C0 , by Lemma 2.3.2 there is a ball Bc ∗ μγ k (x∗ ) ⊂ Bμγ k (x0 ) ∩ EC0 μγ k . k Then, w is positive and harmonic in Bμγ k (x0 ) × (0, μγ ) and w (x, 0) ≥ 1 in ∗ Bc∗ μγ k (x ). Remark 2.3.1 gives

1 with η = η (c∗ ) . uy (X) ≥ −β k + η −C0 μγ k + β k ≥ −β k + ηβ k 2 for (x, y) ∈ B μ γ k (x0 ) × 14 μγ k , 34 μγ k , since γ β. Note that μ is 2 independent of γ and β. We have to ﬁll the gap from y = 0 to y = 14 μγ k . Using uyy < C, we obtain 1 C uy (x, y) − uy x, μγ k ≥ − μγ k 4 4 or

1 C uy (x, y) ≥ −β k + ηβ k − μγ k 2 4 3 k for (x, y) ∈ B μ2 γ k (x0 ) × 0, 4 μγ . We have to choose γ and β such that 1 C −β k + ηβ k − μγ k ≥ −β k+1 . 2 4 Choose γ < μ/2 and

β so that the left hand side is larger than γ − 1 − η4 β k . If β ≤ 1 − η4 then η k β ≥ −β k+1 uy (x, y) ≥ − 1 − 4 in Bγ k+1 (x0 ) × 0, γ k+1 and the inductive argument is complete.

2.4

Optimal Regularity for Tangentially Convex Global Solutions

In this section we consider global solution which are tangentially convex. Typically these solutions are obtained as a consequence of a blow up centered 2

¯ replaced by CC ¯ 0 μγ k and C ∗ γ We apply the Lemma 2.3.2 with γ replaced by C0 μγ k , Cγ ¯ −1 . replaced by C ∗ C0 μγ k . Take C0 = C

The Problems of the Obstacle in Lower Dimension

171

at a free boundary point. Notice that, in this case, the set {x : σ (x) < 0} is convex. We also assume it is non empty. We have already seen an explicit example, given by 3/2

u (X) = Re (xn + i |y|)

3 = ρ3/2 cos θ 2

where ρ = x2n + y 2 and tan θ = y/xn , which once more displays optimal growth at the origin. In this example the zero set for n = 1 is given by the half line xn ≤ 0 and the free boundary is given by the origin. Lemma 2.4.1. Let ∇θ denote the surface gradient on the unit sphere ∂B1 . Set ⎧ ⎫ 2 ⎨ ⎬

∂B1+ |∇θ w| dS + 1/2 − : w ∈ H λ0 = inf ∂B : w = 0 on (∂B ) 1 1 ⎩ ⎭ w2 dS ∂B + 1

−

where (∂B1 ) = {(x , xn ) ∈ Rn , xn < 0} . Then λ0 = Proof. Let ρ =

2n − 1 . 4

x2n + y 2 and tan ψ = y/xn . Deﬁne, for 0 ≤ ψ ≤ π, w0 (x, y) = ρ1/2 cos

ψ . 2

Then, w0 is the restriction to the surface of upper hemisphere of an harmonic function homogeneous of degree 1/2 of the form3 w (x, y) = r1/2 h (θ)

2

r2 = |x| + y 2 , θ ∈ ∂B1 .

Indeed, h is an eigenfunction of the Laplace-Beltrami operator −Δθ on ∂B1 . Computing −Δθ h we get, since the eigenvalue corresponding to the homogeneity α is α (α − 1) + nα: −Δθ h =

2n − 1 1 n h= h. − + 4 2 4

Since h is nonnegative and if we reﬂect it even, it is an eigenfunction in ∂B1 − with homogeneous Dirichlet data on (∂B1 ) , it follows that h is (a multiple) of the ﬁrst eigenfunction of the problem and therefore λ0 = 2n−1 4 . 3

For instance, if n + 1 = 3, θ = (ϕ, ψ), then: h (θ) = cos2 ψ cos2 ϕ + sin2 ψ

1/4

cos ψ/2.

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Now, a monotonicity formula. Lemma 2.4.2. Let w be continuous in Br+ , harmonic in Br+ , w (0, 0) = 0, w (x, 0) ≤ 0, w (x, 0) wy (x, 0) = 0 in Br . Assume that {x ∈ Br : w (x, 0) < 0} is a non empty and convex cone. Set β (r) =

1 r

|∇w (X)|

2

dX.

|X|n−1

Br+

Then, β (r) is bounded and increasing for r ∈ (0, 1/2]. 2

2

Proof. We have Δw2 = 2wΔw + 2 |∇w| = 2 |∇w| , so that β (r) =

1 r

|∇w (X)| |X|

Br+

2

n−1

dX =

1 2r

Δw2 |X|n−1

Br+

dX.

Now (letting ∂Br+ = (∂Br ∩ {y ≥ 0})): β (r) = −

1 2r2

Δw2

Br+

1 n−1 dX + rn |X|

∂Br+

|∇w|2 dS.

Since w (0) = 0 and wwy = 0 on y = 0, we can write 1 2r2

Δw2

Br+

n−1 dX

|X|

= =

≤

1

rn+1 1 rn+1

w2 dS

1 1 2 dX wwν dS − 2 ∇w · ∇ n−1 2r Br+ |X| ∂Br+ n−1 wwν dS + n+2 w2 dS + + 2r ∂Br ∂Br

1

1/2

2 rn

wν2 dS

1/2 +

2rn+2 ∂Br+ ∂Br+ 1 2n − 1 w2 dS + n w2 dS. ≤ n+2 4r r ∂Br+ ν ∂Br+ On the other hand, ∂Br+

2

|∇w| dS =

∂Br+

2

|∇θ w| dS +

n−1 2rn+2

∂Br+

wν2 dS

∂Br+

w2 dS

The Problems of the Obstacle in Lower Dimension

173

so that we obtain: β (r) ≥ −

2n − 1 4rn+2

∂Br+

w2 dS +

1 rn

2

∂Br+

|∇θ w| dS.

The convexity of {x ∈ Br : w (x, 0) < 0} implies that w vanishes at least on − (∂B1 ) so that, according to Lemma 2.4.1, the Rayleigh quotient ∂B1+

2

|∇θ w| dS w2 dS

∂B1+

must be greater than λ0 = particular, β (r) ≤ ϕ (1/2).

2n−1 4 .

Thus we conclude that β (r) ≥ 0 and in

Theorem 2.4.3 (Optimal regularity for global solutions). Let u be a global solution of a zero thin obstacle problem, which is convex along any tangential direction. Then u ∈ C 1,1/2 up to the coincidence set, from both sides. Proof. It is enough to prove that uy goes to zero in C 1/2 fashion when X approaches a free boundary point, which we take to be the origin. Set w = uy . Then, since Λ (u) is convex, w satisﬁes all the hypotheses of Lemma 2.4.2. Therefore, if the coincidence set is non empty, for r small, 1 rn

Br+

|∇w (X)|2 dX ≤

1 r

|∇w (X)|

Br+

|X|n−1

2

dX ≤ ϕ (1/2) .

On the other hand, since w vanishes on at least half of the ball Br , by Poincar´e inequality Br+

2

w2 ≤ Cr2

Br+

|∇w| ≤ C0 r.

Finally, since w2 is subharmonic across y = 0, we conclude that w2 + Br/2

2.5

≤

Br+

w2 ≤ cr.

Almgren’s Frequency Formula

Once the optimal regularity for global solutions has been established, a blow up analysis around a free boundary point allows to carry the same regularity to local solutions. A key tool is a variant of the so called frequency formula of Almgren. In the case of solutions to the zero thin obstacle problem, this kind of monotonicity formula takes the following form:

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S. Salsa

Theorem 2.5.1. Let u be a continuous function in B1 , harmonic in B1+ \Λ (u) and u ∈ C 1,α (B1+ ) ∩ C 1,α (B1− ). Assume that u (0, 0) = 0 and u (x, 0) uy (x, 0) = 0 in B1 . Deﬁne, for 0 < r<1: |∇u|2 V (r; u) Φ (r; u) = r Br 2 . ≡r H (r; u) u dS ∂Br Then, for 0 < r < 1, Φ (r; u) is nondecreasing. Moreover, let μ = lim Φ (r; u) . r→0+

Then Φ (r; u) ≡ μ in (0, 1) if and only if u is a homogeneous function of the form μ u (X) = |X| g (θ) θ ∈ ∂B1 . In particular, in this last case, if u is a solution of the zero thin obstacle problem (that is u (x, 0) ≥ 0), then μ≥

3 . 2

Remark 2.5.1. Thus, u gives the homogeneity degree of u at the origin. We know that, if u is the restriction of a global solution of the zero obstacle problem, then μ ≥ 3/2, from optimal regularity (Theorem 2.4.3). Finally, note the scaling property of D (r; u) : if ur (X) = u (rX) then: Φ (rR; u) = Φ (R, ur ) . Proof of Theorem 2.5.1. We have log Φ (r; u) = log r + log V (r; u) + log H (r; u) so that

d 1 V (r; u) H (r; u) log Φ (r; u) = + − . dr r V (r; u) H (r; u)

We have (ν is the exterior normal):

2

|∇u| dS

V (r; u) = ∂Br

n and H (r; u) = H (r; u) + 2 r

uuν dS (10) ∂Br

and therefore we want to prove that 1−n d log Φ (r; u) = + dr r

∂Br

|∇u|2 dS

Br

|∇u|

2

− 2 ∂Br

uuν dS

∂Br

u2 dS

≥ 0.

(11)

The Problems of the Obstacle in Lower Dimension

175

2 2 Now, since u (x, 0) uy (x, 0) = 0 and Δ u2 = 2uΔu + 2 |∇u| = 2 |∇u| in Br \Λ (u), we can write

1 |∇u| = 2 Br

To control

2

Δ u2 =

uuν dS.

Br

(12)

∂Br

2

∂Br

|∇u| dS, we use the divergence theorem in Br \Λ (u). Let 2 h (X) = div X |∇u| − 2 (X · ∇u) ∇u .

Under our hypotheses, in Br \Λ (u), 2

h (X) = (n − 1) |∇u| . On the other hand, since (X · ∇u) uy vanishes continuously on Λ, we can write (n − 1) |∇u|2 = h=r |∇u|2 dS − 2r (uν )2 dS Br

Br

or

2

|∇u| dS = ∂Br

∂Br

(n − 1) r

∂Br

2

(uν )2 dS.

|∇u| + 2 Br

(13)

∂Br

Inserting into (11), we get (uν )2 dS uuν dS d ∂Br log Φ (r; u) = 2 − 2 ∂Br 2 . dr uuν dS u dS ∂Br ∂Br By Schwarz inequality

2 uuν dS

∂Br

so that

d dr

2

≤

(uν ) dS ∂Br

u2 dS

∂Br

log Φ (r; u) ≥ 0.

The equality sign Φ (r; w) ≡ μ in (0, 1) holds if and only if w is proportional to wν on ∂Br for every 0 < r < 1, which implies that w must be of the form w (X) = f (|X|)g (θ)

θ ∈ ∂B1 . μ

From the radial formula for the Laplace operator, it must be f (|X|) = |X| . In fact, by unique continuation μ must be the same for all component of B1 \Λ (u) where g has constant sign.

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S. Salsa

If w is a solution of the zero obstacle problem, then, being homogeneous, can be extended to a global solution and therefore, from Theorem 2.4.3 we infer μ ≥ 3/2. Let us point out an important consequence of Theorem 2.5.1. First observe that if ϕ (r) = ϕ (r; u) = u2 . ∂Br+

then Φ (r; u) =

r d log ϕ (r; u) . 2 dr

(14)

Corollary 2.5.2. Let u be a solution of the zero obstacle problem and μ = lim Φ (r, u) . r→0+

Then: (a) The function r −→ r−2μ ϕ (r) is nondecreasing for 0 < r < 1 and in particular ϕ (r) ≤ r2μ ϕ (1) ≤ r2μ sup |u| . B1

(b)

d −2μ ϕ (r)] dr [r

≡ 0 in (0, 1) if and only if u (X) = |X|μ g (θ)

θ ∈ ∂B1

with μ ≥ 3/2. (c) Let 0 < r < R < 1. Given ε > 0, for every r, R ≤ r0 (ε), ϕ (R) ≤

R r

2(μ+ε) ϕ (r) .

Proof. (a) and (b) . We have: ϕ (r; u) =

d dr

∂Br

u2 dS = 2

uuν dS = ∂Br

2 |∂Br |

|∇u|

2

Br

so that 1 d −2μ 2 [r ϕ (r)] = −2μr−2μ−1 ϕ (r) + 2r−2μ |∇u| dr |∂Br | Br 2μr−2μ−1 2 r = |∇u| − μ u2 ≥ 0 |∂Br | Br ∂Br and (a) , (b) follow from Theorem 2.5.1.

The Problems of the Obstacle in Lower Dimension

177

(c) Let r0 = r0 (ε) such that Φ (r; u) ≤ μ + ε for r, R ≤ r0 , that is (recall (14)): r d Φ (r; u) = log ϕ (r) ≤ μ + ε. 2 dr An integration over (r, R) gives (c). Here is a ﬁrst step towards the optimal regularity of the solution. Corollary 2.5.3. Let u be a solution of the zero obstacle problem, with (0, 0) ∈ F (u). Then |u (X)| ≤ rμ sup |u| B1

∀X ∈ Br/2 .

(15)

Proof. Since (u+ )2 is subharmonic, we have, for any X ∈ Br/2

2 u+ (X) ≤

u+

∂Br (X)

2

2 dS ≤ r2μ sup u+ . B1

2

A similar inequality holds for (u− ) .

2.6

Asymptotic Profiles and Optimal Regularity

The last step to obtain optimal regularity requires the analysis of suitable blow-up sequences around a free boundary point. Also a careful classiﬁcation of this kind of limiting proﬁles is crucial for the study of the free boundary regularity. Given a solution u of our zero thin obstacle problem, we consider the blow-up family u (rX) vr (X) = (16) 1/2 [ϕ (r; u)] around (0, 0) ∈ F (u). Notice that the ”natural” rescaling u (rX) /rμ , where μ = limr→0+ Φ (r, u) , would not be appropriate, because on this kind of rescaling we have precise control of its behavior as r → 0 merely from oneside. Rescaling by an average over smaller and smaller balls provides the necessary adjustments for controlling the oscillations of u around the origin. Our purpose is to classify the limit of vr as r → 0. Observe that

vr L2 (∂B1 ) = 1.

(17)

Moreover,

2

|∇vr | dX = BR

Φ (R, vr ) R

∂BR

vr2 dS = |∂B1 | Rn−1 Φ (rR, u)

ϕ (rR; u) . ϕ (r; u)

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S. Salsa

Thus, for every R > 1 and every r such that rR ≤ r0 (ε), we can write

2

|∇vr | dX ≤ |∂B1 | (μ + ε) Rn−1+2(μ+ε) BR

and by Poincar´e inequality, vr2 dX ≤ |∂B1 | (μ + ε) Rn+1+2(μ+ε) . BR 1 and by the C 1,α estimates of Sect. 2.3, Therefore, vr is equibounded in Hloc 1,α also in Coc . Hence there exists a subsequence {vj }, vj = vrj , such that, in every compact subset of Rn+1 :

vj → v ∗ and ∇vj → ∇v ∗ , both uniformly. Because of (17), v ∗ is a nontrivial global solution of the zero thin obstacle problem. Moreover, since

Φ (rj ; u) = Φ 1; vrj → Φ (1; v ∗ ) = μ,

(18)

as rj → 0, it follows from Theorem 2.5.1 that v ∗ is homogeneous, that is: v ∗ (X) = |X| g (θ) μ

and from Theorem 2.4.3 that μ≥

θ ∈ ∂B1

3 . 2

From (18) and Corollary 2.5.3 we infer: Theorem 2.6.1. (Optimal regularity). Let u be a solution of the zero obstacle problem, with (0, 0) ∈ F (u). Then u is C 1,1/2 at the origin, in the sense that |u (X)| ≤ r3/2 sup |u| ∀X ∈ Br/2 . (19) B1 + − and B1/2 . Thus, u is C 1,1/2 in both B1/2 We are now ready for our classiﬁcation theorem. Theorem 2.6.2 (Blow-up limits). Let u be a solution of our zero thin obstacle problem and μ = lim Φ (r, u) . (20) r→0+

∗

We know that μ ≥ 3/2. Let v be the global solution deﬁned above. Up to a multiplicative constant, in a suitable system of coordinates, the following hold:

The Problems of the Obstacle in Lower Dimension

(a) Assume μ < 2. Then μ =

3 2

179

and

3 v ∗ (X) = ρ3/2 cos ψ 2 where ρ2 = x2n + y 2 and tan ψ = y/xn . (b) Assume μ = 2. Then v ∗ is a quadratic polynomial: v ∗ (X) =

!

ai x2i − cy 2 ,

ai ≥ 0.

i≤n

Proof. (a) From the tangential quasi-convexity of u, for every tangential direction τ , we have: rj2 ∂τ τ vj ≥ −c . (21) 1/2 [ϕ (rj )] Choose ε such that μ + ε < 2 in Corollary 2.5.2 (c). Then rj2 [ϕ (rj )]1/2

→ 0 as rj → 0

and we obtain vτ∗τ ≥ 0. Thus, v ∗ is tangentially convex and Λ (v ∗ ) is a thick convex cone (since v ∗ is nontrivial). Claim: w = vx∗n is the first eigenfunction for the spherical Laplacian, vanishing on ∂B1 ∩ Λ (v ∗ ). ∗ ≤ 0. Proof. Observe that on Λ (v ∗ ), v ∗ = 0, vy∗ ≤ 0 and for y = 0, vyy ∗ ∗ This implies that v (x, y) ≤ 0 if (x, 0) ∈ Λ (v ). Assume that the vector 0 −en belongs to (Λ (v ∗ )) . For any point X, consider the line LX = {X + ten }. For t large, negative, the function v ∗ (X + ten ) becomes nonpositive, from the previous remark. Since v ∗ is convex along LX it follows that the derivative w cannot become negative anywhere on LX , otherwise v ∗ (X + ten ) → +∞ as t → −∞. In particular, since X is arbitrary, w must be nonnegative in Rn+1 . On the other hand, w = 0 on Λ (v ∗ ) and wy = 0 on {y = 0} \Λ (v ∗ ), by symmetry. Thus, the restriction of w to the unit sphere must be the ﬁrst eigenfunction for the spherical Laplacian, with zero Dirichlet data on ∂B1 ∩ Λ (v ∗ ). This proves the claim.

Now, since Λ (v ∗ ) is convex, it is strictly contained in a half plane. If it is not a half plane, from Lemma 2.4.1 we deduce that the homogeneity degree of w should be less than 1/2, since 1/2 corresponds to the case in which a half plane is removed. However this is a contradiction to μ ≥ 3/2. We conclude that Λ (v ∗ ) is a half plane, say {xn < 0, y = 0}, and w (X) = 1/2 ρ cos 12 ψ. This implies that, up to a multiplicative factor,

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3 v ∗ (X) = ρ3/2 cos ψ. 2 2

(b) Let now μ = 2. The limiting proﬁle v ∗ is of the form |X| g (θ) , θ ∈ ∂B1 , and Λ (v ∗ ) is a cone. Consider w = vy∗ . Then w is linearly homogeneous and w = 0 on {y = 0} \Λ (v ∗ ) . We reﬂect evenly w across the hyperplane y = 0, deﬁning w ˜ (X) =

w (x, y) y > 0 . w (x, −y) y < 0

Claim 1: w ˜ has constant sign. Suppose not. Then, since w ˜ is harmonic on his support and w (0, 0) = 0, we may apply the monotonicity formula of [4] to w+ and w− and, in particular, Corollary 12.4 in [9]. Accordingly, the homogeneity and the linear behavior of w ˜ forces w ˜ to be a two-planes solution with respect to a direction transversal to the plane y = 0, say en , due to the − even symmetry of w. ˜ Therefore w ˜ (X) = αx+ n − βxn . This is a contradiction, ∗ ∗ since w (x, 0) ≤ 0 on Λ (v ) and zero outside Λ (v ). The claim is proved. ˜ is the ﬁrst Claim 2: Λ (v ∗ ) has empty interior. Suppose not. Then w eigenfunction for the spherical Laplacian, with zero Dirichlet data on [{y = 0} \Λ (v ∗ )] ∩ ∂B1 . This forces a superlinear behavior of w ˜ at the origin, since linear behavior corresponds to a half sphere, giving again a contradiction with point (a). Thus, Λ (v ∗ ) has empty interior. Therefore, v ∗ is harmonic across Λ (v ∗ ) and must coincide with a quadratic polynomial. Remark 2.6.1. Let μ = 3/2. Observe that if τ = αen + βe, where e is tangential, e⊥en , and α2 + β 2 = 1, α > 0, then, outside a η-strip |y| < η, we have in B1 : vτ∗ ≥ C (α, η) > 0. (22)

2.7

Lipschitz Continuity of the Free Boundary at Stable Points

Form the knowledge of the optimal regularity of the solution and the classiﬁcation of blow-up proﬁles we can start the analysis of the free boundary. First of all we distinguish the points of the free boundary according to the following deﬁnition.

The Problems of the Obstacle in Lower Dimension

181

Definition 2.7.1. Let X0 ∈ F (u) and B (X ) μ (X0 ) = r r 0

2

|∇u| dX

∂Br (X0 )

u2 dS

.

We say that X0 is a regular or stable point if μ (X0 ) = 3/2. The examples in the introductory Sect. 2.1 reveal that any hope to achieve regularity of F (u) is restricted to a neighborhood of regular points. Therefore from now on we will focus on regular points. The ﬁrst step is to prove that near a regular point (e.g. the origin) the free boundary is a Lipschitz graph. Precisely: Theorem 2.7.1. Let u be a solution of the zero thin obstacle problem in B1 . Assume that (0, 0) ∈ F (u) and is a regular point. Then, with respect to a suitable coordinate system, there exists a neighborhood of the origin Bρ and a cone of tangential directions Γ (en , θ) with axis en and opening θ ≥ π/3 (say), such that for every τ ∈ Γ (en , θ) we have Dτ u ≥ 0. In particular, in that neighborhood, F (u) is the graph of a Lipschitz function xn = f (x1 , ..., xn−1 ). Proof. As in (16) let vrj (X) =

u (rj X) 1/2

.

[ϕ (rj ; u)]

From Theorem 2.6.2 we know that, for a suitable sequence {rj }, rj → 0, vrj → v ∗ and ∇vj → ∇v ∗ , both uniformly in compact sets, where 3 v ∗ (X) = ρ3/2 cos ψ 2 (ρ2 = x2n + y 2 and tan ψ = y/xn ). Let τ = αen + βe be a tangential direction (α2 + β 2 = 1). If we ﬁx

1/2 , then the set of all these directions α ≥ α0 > 0, so that β ≤ β0 ≡ 1 − α20 ﬁlls a cone Γ (en , θ) with θ = arctan β0 /α0 . Thus, if α0 is small then θ ≥ π/3. We want to show that Dτ vrj ≥ 0 along every τ ∈ Γ (en , θ) . For σ small and rj ≤ r0 (σ), we deduce from (22) that Dτ vrj enjoys the following properties in B5/6 : (i) Dτ vrj ≥ 0 outside the strip |y| ≤ σ, (ii) Dτ vrj ≥ c0 for |y| ≥ 12 , and from optimal regularity, (iii) Dτ vrj ≥ −cσ 1/2 inside the strip |y| ≤ σ.

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To get rid of the lack of positivity inside the strip |y| ≤ σ and to conclude the proof, we apply to h = Dτ vrj the following lemma. Lemma 2.7.2. Let u be as in Theorem 2.6.2. Assume h is a function with the following properties: (i) (ii) (iii) (iv)

Δh ≤ 0 in B1 \Λ (u) ; h ≥ 0 for |y| ≥ σ, h = 0 on Λ (u), with σ > 0, small; 1 h ≥ c0 for |y| ≥ 8n ; h ≥ −ω (σ), for |y| ≤ σ, where ω is the modulus of continuity of h.

There exists σ0 = σ0 (n, c0 , ω) such that, if σ ≤ σ0 then h ≥ 0 in B1/2 . Proof. Suppose there is Z = (z, t) ∈ B1/2 such that h (Z) < 0. Let 1 1 Q = (x, y) : |x − z| ≤ , |t − y| < 3 4n and

P (x, y) = |x − z|2 − ny 2 .

Deﬁne v (X) = h (X) + δP (X) where δ > 0, to be chosen later. We have: (a) v (Z) = h (Z) − δnt2 < 0; (b) Δv ≤ 0 in B1 \Λ (u) ; (c) v ≥ 0 on Λ (u), since h = 0, P ≥ 0 there. Thus v|Q must have anegativeminimum on ∂Q. 1 , we have However, on ∂Q ∩ |y| ≥ 8n v ≥ c0 − if δ = 16nc0 . On

1 1 |x − z| = , σ ≤ |y| ≤ 3 8n

we have h ≥ 0 so that

v≥δ

Finally, on

δ =0 16n

1 1 − ≥ 0. 9 64n

1 |x − z| = , |y| < σ 3

we have v ≥ −ω (σ) + δ

1 − nσ 2 ≥ 0 9

if σ is chosen small depending on n, c0 , ω.

The Problems of the Obstacle in Lower Dimension

183

Hence v ≥ 0 on ∂Q and we have reached a contradiction. Therefore h ≥ 0 in B1/2 .

2.8

Boundary Harnack Principles and C 1,α Regularity of the Free Boundary at Stable Points

We are now in position to show that if the origin is a regular point, then nearby, the free boundary is a C 1,α graph. Precisely, we are going to prove the following result. Theorem 2.8.1. Let u be a solution of the zero thin obstacle problem in B1 . Assume that (0, 0) ∈ F (u) and is a regular point. Then, with respect to a suitable coordinate system, there exists a neighborhood of the origin Bρ in which F (u) is the graph of a C 1,α function xn = f (x1 , ..., xn−1 ). There are two ways to prove the above theorem. One way, that we only sketch (see for the details [2]) is to use a bi-lipschitz transformation to map a neighborhood of the origin in B1 \Λ (u) onto a half ball, say, B + = {Z = (z, η) : |Z| < 1, η > 0} . The Laplace operator in transformed into a uniformly elliptic operator L in divergence form. Each tangential derivative Dτ u with τ belonging to the cone Γ (en , θ), deﬁned in Theorem 2.7.1, is mapped into a positive solution w (Z; τ ) of Lw = 0 in B + vanishing on {η = 0}. An application of Corollary 1 in [2] implies that the ratio w (Z; τ ) w (Z; en ) + . This in turn, implies that any is H¨ older continuous up to {η = 0} ∩ B2/3 quotient Dτ u/Den u, with τ ∈ Γ (en , θ), is H¨ older continuous up to F (u) in a neighborhood of the origin. This implies that the level sets of u are tangentially C 1,α surfaces and in particular that F (u) is a C 1,α graph.

The second way is a somewhat more direct proof, based on the next theorem, which is a generalization of by now well known boundary Harnack principles (see e.g. [14, 37]) and which could be interesting in itself. First, we deﬁne a class of domains, closely related to the Nontangentially Accessible Domains (brieﬂy NTA) in [37]. Let D be a subdomain of B1 and let ∂ ∗ D = ∂D ∩ B1 . Let Υ (X, Y ) be the set of regular curves contained in D with end points X, Y . We denote by dg (X, Y ) the geodesic distance in D of X and Y , that is dg (X, Y ) = inf {l (γ) : γ ∈ Υ (X, Y )} .

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Definition 2.8.1. We say that D is a Δ- N T A domain if the following properties hold: (1) For every X, Y ∈ D, dg (X, Y ) is ﬁnite. (2) Non tangential ball condition. Let Q be arbitrary on ∂ ∗ D. There exist positive numbers r0 = r0 (Q, D) and η = η (D) such that, for every r ≤ r0 there is a point Ar (Q) ∈ Br (Q) such that Bηr (Ar (Q)) ⊂ Br (Q) ∩ D. (3) Harnack chain condition. There is a constant M = M (D) such that, for all X, Y ∈ D, ε > 0 and k ∈ N satisfying d (X, ∂ ∗ D) > ε, d (Y, ∂ ∗ D) > ε, dg (X, Y ) < 2k ε, there exists a sequence of M k balls Br1 (P1 ) , ..., BrM k (PMk ) contained in D, such that X ∈ Br1 (P1 ) , Y ∈ BrM k (PMk ) and Brj (Pj ) ∩ Brj+1 (Pj+1 ) = ∅

(j = 1, ..., M k − 1)

and

1 rj < d Brj (Pj ) , ∂D∗ < 4rj 2

(j = 1, ..., M k − 1).

(4) Uniform capacity condition. Let Q be an arbitrary point on ∂ ∗ D. There exist positive numbers r0 = r0 (Q, D) and γ = γ (D) such that, for every r ≤ r0 ,

capΔ Br (Q) \Br/2 (Q) ∩ ∂ ∗ D ≥ γrn−1 where cap Δ (K) is the capacity of K in B1 with respect to the Laplace operator.4 Remark 2.8.1. Conditions (2) and (3) appears in the deﬁnition of NTA domains as given in [37]. Condition (4) replaces the exterior ball property in that deﬁnition. A simple example of Δ − N T A domains is an n-dimensional smooth manifold with Lipschitz boundary. Remark 2.8.2. Let L = div(A (X) ∇) be a uniformly elliptic operator with ellipticity constants λ, λ−1 , and bounded measurable coeﬃcients. Since for any K ⊂ B1 , capL (K) ∼cap· (K) with equivalence constant depending only on n, λ, the notion of Δ − N T A domain refers actually to the above 4

CapΔ (K) = inf ∇ϕL2 : ϕ ≥ 1 on K and ϕ ∈ C0∞ (B1 ) .

The Problems of the Obstacle in Lower Dimension

185

class of operators. From now on, it is understood that the constants c s and C s appearing in the statements and proofs are understood depend only on some of the relevant parameters n, η, γ, M and λ. According to Remark 2.8.2 we prove the following result. Theorem 2.8.2 Let D ⊂ B1 be a Δ − N T A domain. Suppose v and w are positive functions in D, continuously vanishing on ∂ ∗ D and satisfying Lv = Lw = 0 in D. Let X0 ∈ D ∩ B2/3 such that d (X0 , ∂ ∗ D) = d0 > 0 and v (X0 ) = w (X0 ) = 1. Then: (a) (Carleson estimate) For every Q ∈ ∂ ∗ D ∩ B1/2 : sup D∩B1/3 (Q)

v≤C

(b) (Boundary Harnack Principle) v ≤ C. w D∩B1/3 (Q) sup

(c) (H¨older continuity for quotients). The quotient in B1/2 ∩ D up to ∂ ∗ D.

v w

is H¨ older continuous

Proof. The proof follows by now standard lines (see for instance [9], Sect. 11.2, and [37]). We emphasize the main steps and diﬀerences. (a) Fix Q ∈ ∂ ∗ D ∩ B1/2 and let Y0 be such that v (Y0 ) =

sup B1/3 (Q)∩D

v ≡ N.

The interior ball condition and the Harnack chain condition plus the interior Harnack inequality imply that, if N is large then d (Y0 , ∂ ∗ D) ≡ |Y0 − Q0 | ≤ N −δ with Q0 ∈ ∂ ∗ D and δ = δ (n, λ, d0 , M ) > 0. Let r0 = |Y0 − Q0 |. Then, the uniform capacity condition implies that sup B2r0 (Q0 )∩D

v ≡ v (Y1 ) ≥ CN

with C > 1. Thus, for some Q1 ∈ ∂ ∗ D, d (Y1 , ∂ ∗ D) ≡ |Y1 − Q1 | ≤ (CN )−δ .

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Iterating the process, we construct a sequence of points {Ym } satisfying the following conditions: (i) v(Ym ) ≥ C m N (ii) d (Ym , ∂ ∗ D) ≤ (C m N )−δ (iii) |Ym − Ym+1 | ≤ 4(C m N )−δ . If N is large enough, we can make !

|Ym − Ym+1 | ≤

m≥0

1 16

and therefore {Ym } ⊂ D ∩ B2/3 . On the other hand, d (Ym , ∂ ∗ D) → 0 and v(Ym ) → +∞ which leads to a contradiction, since v (Ym ) → 0. This proves (a) . We prove (b). Let P ∈ ∂ ∗ D ∩ B1/3 and R0 = |X0 − P | + ≤ R0 ≤ 1. Deﬁne ΨR0 (P ) = BR0 (P ) ∩ D

3 2 d0

and

d0 2 .

Notice that

Σ0 = ∂BR0 (P ) ∩ BR0 (P ) , Σ1 = ∂ΨR0 (P ) \∂ ∗ D

Observe that Σ0 , Σ1 ⊂ D. X Let ωL be the L−harmonic measure in ΨR0 (P ) . Claim: for every X ∈ BR0 /8 (P ) ∩ D, X X (Σ1 ) ≤ CωL (Σ0 ) , ωL

(23)

which is a doubling property for the L−harmonic measure. To prove the claim, introduce the Green’s function G (X, X0 ) for the operator L in ΨR0 (P ) with pole at X0 . From H¨older continuity, on ∂Bd0 /3 (X0 ), we have X ωL (Σ0 ) ≥ c > 0 and by the estimates for the Green’s function (we restrict for simplicity to the case n + 1 > 2). G (X, X0 ) ≤ cd1−n . 0 X (Σ0 ) ≥ 0. On the other hand, on ∂ΨR0 (P ), we have G (X, X0 ) = 0 and ωL Therefore, by maximum principle, outside BR0 /3 (X0 ) we get X ωL (Σ0 ) . G (X, X0 ) ≤ cd1−n 0

(24)

X (Σ1 ). Let ϕ ∈ C0∞ (BR0 (P )) We need a bound from below of G in terms of ωL such that ϕ ≡ 0 in BR0 /4 (P ) , ϕ ≡ 1 outside BR0 /2 (P )

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187

and 0 ≤ ϕ ≤ 1 in C0 (P ) ≡ BR0 /2 (P ) \BR0 /4 (P ). Note that |∇ϕ| ≤ c/R0 . We have: X X ωL (Σ1 ) ≤ ϕdωL . ∂ΨR0 (P )

Fix X ∈ BR0 /8 (P ) ∩ D. Then we can write

0 = ϕ (X) = ∂ΨR0 (P )

X ϕdωL

−

C0 (P )∩D

A (X) ∇Y G (X, Y ) · ∇ϕ (Y ) dY.

Therefore, from Caccioppoli estimate and Carleson estimate, we have, in BR0 /8 (P ) ∩ D, X (Σ1 ) ≤ ωL

≤

(n−1)/2 CR0

(n−3)/2 CR0

C0 (P )∩D

2 ∇Y G (X, Y ) dY

1/2

1/2 G2 (X, Y ) dY

C0 (P )∩D

≤ CR0n−1 G (X, X0 ) . Comparing with (24) we conclude the proof of the claim. With the doubling condition at hand the proof of (a) is quickly ﬁnished. In fact, let Q ∈ ∂ ∗ D ∩ B1/2 . by the Carleson estimate and maximum principle, in D ∩ B1/3 (Q) we get X v (X) ≤ CωL (Σ1 ) . On the other hand, the maximum principle, interior Harnack inequality and the doubling property (23) give, in BR0 /8 (P ) ∩ D, X X (Σ1 ) ≤ CωL (Σ0 ) ≤ w (X) . v (X) ≤ CωL

Thus

v ≤ C. w D∩BR0 /8 (Q) sup

Using a covering argument and the interior Harnack inequality one can easily conclude the proof. At this point, the proof of (c) follows, for instance, as in [9, Sect. 11.2]. Proof of Theorem 2.8.1. We apply Theorem 2.8.2 with D = B1 \Λ (u), ∂ ∗ D = Λ (u) and v = Dτ u, w = Den u where τ ∈ Γ (en , θ). Since F (u) is Lipschitz,

capΔ Br (Q) \Br/2 (Q) ∩ Λ (u) ≥ γrn−1 . We obtain, in particular, that on {y = 0} \Λ (u) the quotient Dτ u/Den u, is H¨older continuous up to F (u) in a neighborhood of the origin. As already

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mentioned before, this implies that the level sets of u are tangentially C 1,α surfaces and in particular that F (u) is a C 1,α graph in B1/2 .

2.9 2.9.1

Structure of the Singular Set Main Statements and Strategy

As we have already seen, the non regular (or non stable) points of the free boundary can be divided in two classes: the set Σ (u) at which Λ (u) has a vanishing density (singular points), that is Σ (u) =

Hn (Λ (u) ∩ Br (x0 )) (x0 , 0) ∈ F (u) : lim+ = 0 , rn r→0

and the set of non regular, non singular points. In this section, we describe the results from [18] on the structure of the set of singular points of the free boundary. As we shall see, around these kind of points a precise analysis of the behavior of u and the structure of the free boundary can be carried out. The analysis of the free boundary around the other kind of points is still an open question. It is convenient to classify a point on F (u) according to the degree of homogeneity of u, given by the frequency formula centered at that point. In other words, set 2 Br (X0 ) |∇u| X0 Φ (r; u) = r 2 ∂Br (X0 ) u dS and deﬁne Fκ (u) = X0 ∈ F (u) : ΦX0 (0+; u) = κ Σκ (u) = Σ (u) ∩ Fκ (u) . According to these notations, X0 is a regular point if it belongs to F3/2 (u) . Since r −→ ΦX0 (r; u) is nondecreasing, it follows that the mapping X0 −→ ΦX0 (0+; u) is upper-semicontinuous. Moreover, since ΦX0 (0+; u) misses all the values in the interval (3/2, 2), it follows that F3/2 (u) is a relatively open subset of F (u). Before stating the structure theorems of Σ (u), it is necessary to examine the asymptotic proﬁles obtained at a singular point from the rescalings (16). Note that, in terms of these rescalings, saying that X0 = (x0 , y) ∈ Σ (u) is equivalent to

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189

lim Hn (Λ (vr ) ∩ B1 (x0 )) = 0.

r→0+

(25)

As we see immediately, this implies that any blow-up v ∗ at a singular point is harmonic in B1 . Moreover, it is possible to give a complete characterization of these blow-ups in terms of the value κ = ΦX0 (r; u) . In particular Σκ (u) = Fκ (u)

for κ = 2m, m ∈ N.

Theorem 2.9.1 (Blow-ups at singular points). Let (0, 0) ∈ Fκ (u). The following statements are equivalent: (i) (0, 0) ∈ Σκ (u) . (ii) Any blow-up of u at the origin is a non zero homogeneous polynomial pκ of degree κ satisfying Δpκ = 0,

pκ (x, 0) ≥ 0,

pκ (x, −y) = pκ (x, y) .

(iii) κ = 2m for some m ∈ N. Proof. (i) =⇒ (ii). Since u is harmonic in B1± , we have: n Δvr = 2(∂y vr )H|Λ(v r)

in D (B1 ) .

(26)

1 (B1 ) and (25) says that From Sect. 2.5, vr is equibounded in Hloc

Hn (Λ (vr ) ∩ B1 ) → 0 as r → 0. Thus (26) implies that Δvr → 0 in D (B1 ) and therefore any blow-up v ∗ must be harmonic in B1 . From Sect. 2.6 we know that v ∗ is homogeneous and non trivial and therefore can be extended to a harmonic function in all Rn+1 . Being homogeneous, v ∗ has at most a polynomial growth at inﬁnity, hence Liouville Theorem implies that v ∗ is a non trivial homogeneous harmonic polynomial pκ of integer degree κ. The properties of u imply that pκ (x, 0) ≥ 0, and pκ (x, −y) = pκ (x, y) in Rn+1 . (ii) =⇒ (iii). We must show that κ is an even integer. If κ is odd, the nonnegativity of pκ on y = 0, implies that pκ vanishes on the hyperplane y=0. On the other hand, from the even symmetry in y we infer that ∂y pκ (x, 0) ≡ 0 in Rn . Since pκ is harmonic, the Cauchy-Kowaleskaya Theorem implies that pκ ≡ 0 in Rn+1 . Thus κ = 2m, for some m ∈ N. (ii) =⇒ (i) . Suppose (0, 0) is not a singular point so that there exists a sequence rj → 0 such that Hn (Λ (vr ) ∩ B1 ) ≥ δ > 0. We may assume that vrj converges to a blow-up p∗ . We claim that

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Hn (Λ (p∗ ) ∩ B1 ) ≥ δ > 0. Indeed, otherwise, there exists an open set U ⊂ Rn with Hn (U ) < δ such that Λ (p∗ ) ∩ B 1 ⊂ U. Then, for j large, we must have Λ (vr ) ∩ B 1 ⊂ U which

is a contradiction, since Hn (Λ vrj ∩ B 1 ) ≥ δ > Hn (U ). This implies that p∗ (x, 0) ≡ 0 in Rn and consequently in Rn+1 , by the Cauchy-Kowaleskaya theorem. Contradiction to (ii). (iii) =⇒ (ii) . From Almgren’s formula, any blow-up is a κ-homogeneous n solution of the zero thin obstacle problem in Rn+1 . Then Δv = 2vy H|Λ(v) in n+1 R , with vy ≤ 0 on y = 0. Since κ = 2m, the following auxiliary lemma implies that Δv = 0 in Rn+1 and therefore v is a polynomial. n+1 1 n+1 satisfy Δv ≤ 0 in R and Δv = 0 in Lemma. Let v ∈ Hloc R Rn+1 \ {y = 0}. If v is homogeneous of degree κ = 2m then Δv = 0 in Rn+1 . Proof. By assumption, μ = Δv is a nonpositive measure, supported on {y = 0} . We have to show that μ = 0. Let q be a 2m-homogeneous harmonic polynomial, which is positive on {y = 0} \ (0, 0). For instance: q (X) =

n !

Re(xj + iy)2m .

j=1

Take ψ ∈ C0∞ (0, +∞) such that ψ ≥ 0 and let Ψ (X) = ψ (|X|). Then, we have: − μ, Ψ q = − Δv, Ψ q = = Rn+1

Rn+1

(−Ψ vΔq − v∇q · ∇Ψ + q∇v · ∇Ψ ) dX

=

Rn+1

(Ψ ∇v · ∇q + q∇v · ∇Ψ ) dX

[−Ψ vΔq − v

ψ (|X|) ψ (|X|) (X · ∇q) + q (X · ∇v)]dX |X| |X|

=0 since Δq = 0, X·∇q = 2mq, X ·∇v = 2mv. This implies that μ is supported at X = 0 that is μ = cδ(0,0) . On the other hand, δ(0,0) is homogeneous of degree − (n + 1) while μ is homogeneous of degree 2m − 2 and therefore μ = 0. Definition 2.9.1: We denote by Pκ the class of homogeneous harmonic polynomials of degree κ = 2m, defined in Theorem 2.9.1, that is: Pκ = {pκ : Δpκ = 0, ∇pκ · X = κpκ , pκ (x, 0) ≥ 0, pκ (x, −y) = pκ (x, y)} . (27) Via the Cauchy-Kovaleskaya Theorem, it is easily shown that the polynomials in Pκ can be uniquely determined from their restriction to the

The Problems of the Obstacle in Lower Dimension

191

hyperplane y = 0. Thus, if pκ ∈ Pκ is not trivial, then also its restriction to y = 0 must be non trivial. The next theorem gives an exact asymptotic behavior of u near a point X0 ∈ Σκ (u) . Theorem 2.9.2 (κ-diﬀerentiability at singular points). Let X0 ∈ Σκ (u), 0 with κ = 2m, m ∈ N. Then there exists a non trivial pX κ ∈ Pκ such that κ

0 u (X) = pX κ (X − X0 ) + o (|X − X0 | ) .

(28)

0 Moreover, the mapping X0 −→ pX κ is continuous on Σκ (u) . The proof is given in Sect. 2.9.3. Note that, since Pκ is a convex subset of the space of the homogeneous polynomial of degree κ, all the norms on Pκ are equivalent. Thus, the continuity in Theorem 2.9.2 can be understood, for instance, in the L2 (∂B1 ) norm. The structure of F (u) around a singular point X0 depends on the dimension of the singular set at that point, as deﬁned below in terms of 0 the polynomial pX κ :

Definition 2.9.3 (Dimension at a singular point). Let X0 ∈ Σκ (u) . The dimension of Σκ (u) at X0 is deﬁned as the integer n X0 n 0 . dX κ = dim ξ ∈ R : ξ · ∇x pκ (x, 0) = 0 for all x ∈ R n 0 Since pX κ (x, 0) is not identically zero on R , we have 0 0 ≤ dX κ ≤ n − 1.

For d = 0, 1, ..., n − 1 we deﬁne 0 Σκd (u) = X0 ∈ Σκ (u) : dX κ = d . Here is the structure theorem: Theorem 2.9.3 (Structure of the singular set). Every set Σκd (u), κ = 2m, m ∈ N, d = 0, 1, ..., n − 1, is contained in a countable union of d- dimensional C 1 manifolds. For the harmonic polynomial 1 p (x1 , x2 , y) = x21 x22 − (x21 + x22 )y 2 + y 4 3 considered in Sect. 2.1, it is easy to check that (0, 0) ∈ Σ40 (u) and the rest of the points on F (u) belongs to Σ21 (u). As in the classical obstacle problem, the main diﬃculty in the analysis consists in establishing the uniqueness of the Taylor expansion (28), which in turn is equivalent to establish the uniqueness of the limiting proﬁle obtained by the sequence of rescalings (16).

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A couple of monotonicity formulas, strictly related to Almgren’s formula and to formulas in [25] and [22], play a crucial role in circumventing these diﬃculties. 2.9.2

Monotonicity Formulas

We introduce here two main tools. We start with a one-parameter family of monotonicity formulas (see also [25]) based on the functional: WκX0

(r; u) =

1 rn−1+2κ

2

Br (X0 )

|∇u| dX −

κ rn+2κ

u2 dS.

∂Br (X0 )

where κ ≥ 0. If X0 = (0, 0) we simply write Wκ (r; u). The functionals WκX0 (r; u) and ΦX0 (r; u) are strictly related. Indeed, taking for brevity X0 = (0, 0), in terms of the notations in Sect. 2.5, we have: Wκ (r; u) =

1 rn−1+2κ

V (r) −

κ rn+2κ

H (r) =

H (r) [Φ (r; u) − κ] . rn+2κ

(29)

This formula shows that WκX0 (r; u) is particularly suited for the analysis of asymptotic proﬁles at points X0 ∈ Fκ (u). Moreover, for these points, since from Almgren’s frequency formula we have ΦX0 (r; u) ≥ ΦX0 (0+; u) = κ we deduce that WκX0 (r; u) ≥ 0. (30) The next theorem shows the main properties of Wκ (r; u) . Theorem 2.9.4. (W -type monotonicity formula). Let u be a solution of our zero obstacle problem in B1 . Then, for 0 < r < 1, 1 d Wκ (r; u) = n+2κ dr r

(X · ∇u − κu)2 dS.

∂Br

As a consequence, the function r −→ Wκ (r; u) is nondecreasing in (0, 1) . Moreover, Wκ (·; u) is constant if and only if u is homogeneous of degree κ. Proof. Using the identities (10), (12) and (13), we get: d Wκ (r; u) dr κ n − 1 + 2κ κ (n + 2κ) 1 V (r) − H (r) + H (r) = n−1+2κ V (r) − r r r r2 4κ 2κ2 1 2 2 uν dS − uuν dS + 2 u dS = n−1+2κ 2 r r ∂Br r ∂Br ∂Br 2 (X · ∇u − κu)2 dS. = n+1+2κ r ∂Br

The Problems of the Obstacle in Lower Dimension

193

The next one is a generalization of a formula in [22], based on the functional MκX0 (r; u, pκ ) =

1 rn+2κ

∂Br (X0 )

(u − pκ )2 dS.

(0,0)

(r; u, pκ ). Here κ = 2m and pκ is a As usual we set Mκ (r; u, pκ ) = Mκ polynomial in the class Pk deﬁned in (27). Since it measures the distance of u from an homogeneous polynomial of even degree, it is apparent that MκX0 (r; u, pκ ) is particularly suited for the analysis of blow-up proﬁles at points X0 ∈ Σκ (u). We have: Theorem 2.9.5. (M -type monotonicity formula). Let u be a solution of our zero obstacle problem in B1 . Assume (0, 0) ∈ Σκ (u), κ = 2m, m ∈ N. Then, for 0 < r < 1, the function r −→ Mκ (r; u, pκ ) is nondecreasing in (0, 1). Proof. We show that d 2 Mκ (r; u, pκ ) ≥ Wκ (r; u) ≥ 0. dr r Let w = u − pκ . We have: d 1 dr rn+2κ

w2 dS =

∂Br

= =

d 1 dr r2κ 1 r2κ+1

2

w (rY ) dS

∂B1

[w (rY ) [∇w (rY ) · rY − κw(rY )]dS ∂B1

2 rn+1+2κ

w(X · ∇w − κw)dS ∂Br

On the other hand, since Φ (r; pκ ) = κ, it follows that Wκ (r; pκ ) = 0 and we can write: Wκ (r; u) = Wκ (r; u) − Wκ (r; pκ ) 2 2 (|∇w| + 2∇w · ∇pκ )dX = n−1+2κ r Br κ (w2 + 2wpκ )dS − n+2κ r ∂Br κ 2 2 |∇w| dX − n+2κ w2 dS = n−1+2κ r r Br ∂Br

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S. Salsa

w(X · ∇pκ − κpκ )dS

+ ∂Br

=

rn−1+2κ

=− ≤

2 2

Br

rn−1+2κ 1

rn+2κ

κ

2

|∇w| dX −

rn+2κ

wΔwdX + Br

1

w2 dS

∂Br

rn+2κ

w(X · ∇w − κw)dS =

∂Br

w(X · ∇w − κw)dS ∂Br

r d Mκ (r; u, pκ ) 2 dr

since wΔw = (u − pκ ) (Δu − Δpκ ) = −pκ Δu ≥ 0. 2.9.3

Proofs of Theorems 2.9.2 and 2.9.3

With the above monotonicity formulas at hands we are ready to prove Theorems 2.9.2 and 2.9.3. Recall that, from the frequency formula, we have the estimate (Corollary 2.5.3): κ

|u (X)| ≤ c |X|

in B2/3

(31)

for any solution of our free boundary problem. At a singular point, we have also a control from below. Lemma 2.9.6. Let u be a solution of our zero obstacle problem in B1 . Assume (0, 0) ∈ Σκ (u). Then, sup |u| ≥ crκ

(0 < r < 2/3).

(32)

∂Br

Proof. Suppose that (32) does not hold. Then, for a sequence rj → 0 we have 1/2 2 hj = u dS = o rjκ . ∂Brj

We may also assume that (see Lemma 2.9.1) vj (X) =

u (rj X) → qκ (X) hj

uniformly on ∂B1 , for some qκ ∈ Pκ . Since ∂B1 qκ2 dS = 1, it follows that qκ is non trivial. Under our hypotheses, we have 1 1 2 2 Mκ (0+; u, qκ) = lim n+2κ (u − qκ ) dS = qκ dS = n+2κ qκ2 dS. j→∞ r rj ∂Brj ∂B1 ∂Brj j

The Problems of the Obstacle in Lower Dimension

Hence,

(u − qκ )2 dS ≥

∂Brj

qκ2 dS

∂Brj

or

195

(u2 − 2uqκ )dS ≥ 0. ∂Brj

Rescaling, we obtain hj rjκ

( ∂Brj

hj 2 v − 2vj qκ )dS ≥ 0. rjκ j

Dividing by hj rjκ and letting j → ∞ we get −

qκ2 dS ≥ 0

∂Brj

which gives a contradiction, since qκ is non trivial.

Given the estimates (31) and (32) around a point X0 ∈ Σκ (u) , it is natural to introduce the family of homogeneous rescalings given by u(κ) r (X) =

u (rX + X0 ) . rκ

From the estimate (31) we have that, along a sequence r = rj , uκr → u0 in 1,α (Rn ). We call u0 homogeneous blow-up. Lemma 2.9.6 assures that u0 is Cloc non trivial. The next results establishes the uniqueness of these asymptotic proﬁles and proves the ﬁrst part of Theorem 2.9.2. Theorem 2.9.7 (Uniqueness of homogeneous blow-up at singular points). Assume (0, 0) ∈ Σκ (u) . Then there exists a unique non trivial pκ ∈ Pκ such that u (rX) u(κ) → pκ (X) . r (X) = rκ As a consequence, (28) holds. Proof. Consider a homogeneous blow-up u0 . For any r > 0 we have: (κ)

Wκ (r; u0 ) = lim Wκ (r; uj ) = lim Wκ (rrj ; u) = lim Wκ (0+; u). rj →0

rj →0

rj →0

From Theorem 2.9.2 we infer that u0 is homogeneous of degree κ. The same arguments in the proof of Lemma 2.9.1 give that u0 must be a polynomial pκ ∈ Pκ . To prove the uniqueness of u0 , apply the M -monotonicity formula to u and u0 . We have:

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S. Salsa

Mκ (0+; u, u0) = cn lim

j→∞

(κ)

(uj

− u0 )2 dS = 0.

∂B1

In particular, by monotonicity, we obtain also that 2 (u(κ) cn r − u0 ) dS = Mκ (r; u, u0 ) → 0 ∂B1

as r → 0, and not just over a subsequence rj . Thus, if u0 is a homogeneous blow-up, obtained over another sequence rj → 0, we deduce that ∂B1

(u0 − u0 )2 dS = 0.

Since u0 and u0 are both homogeneous of degree κ, they must coincide in Rn+1 . The next lemma gives the second part of Theorem 2.9.2. Lemma 2.9.8 (Continuous dependence of the blow-ups). For X0 ∈ Σκ (u) 0 denote by pX κ the blow-up of u obtained in Theorem 2.9.7 so that: κ 0 u (X) = pX κ (X − X0 ) + o (|X − X0 | ) . 0 Then, the mapping X0 −→ pX κ from Σκ (u) to Pκ is continuous. Moreover, for any compact K ⊂ Σκ (u) ∩ B1 , there exists a modulus of continuity σK , σK (0+) = 0, such that

κ 0 u (X) − pX κ (X − X0 ) ≤ σK (|X − X0 |) |X − X0 | for every X0 ∈ K. Proof. As we have already observed, we endow Pκ with the L2 (∂B1 ) norm. The ﬁrst part of the lemma follows as in Theorem 2.9.7. Indeed, ﬁx ε > 0 and rε such that 1 0 0 2 MκX0 (rε ; u, pX ) = (u (X + X0 ) − pX κ κ ) dS < ε. rεn+2κ ∂Brε There exists δε such that if X0 ∈ Σκ (u) and |X0 − X0 | < δε , then X

0 Mκ 0 (rε ; u, pX κ )=

1 rεn+2κ

∂Brε

0 2 (u (X + X0 ) − pX κ ) dS < 2ε.

By monotonicity, we deduce that, for 0 < r < rε , 1 X 0 0 2 ) = (u (X + X0 ) − pX Mκ 0 (r; u, pX κ κ ) dS < 2ε. rn+2κ ∂Br

The Problems of the Obstacle in Lower Dimension

197

Letting r → 0 we obtain

X

0 Mκ 0 (0+; u, pX κ ) = cn

X

0 2 (pκ 0 − pX κ ) dS < 2ε

∂B1

and therefore the ﬁrst part of the lemma is proved. To show the second part, note that if |X0 − X0 | < δε and 0 < r < rε , we have: " " " " " " " X " X " " 0" ≤ "u (·+X0 )−pX + X0 −pκ 0 " "u (·+X0 )−pκ 0 " 2 κ L2 (∂Br ) "pκ 2 L (∂Br )

L (∂Br )

≤ 2 (2ε)1/2 rκ+(n−1)/2 . This is equivalent to " " X " " X0 "wr − pκ 0 "

L2 (∂B1 )

where

≤ 2 (2ε)1/2

(33)

u (rX + X0 ) . rκ Covering K with a ﬁnite number of balls Bδε (X ) (X0 ) for some X0 ∈ K, we X

wr 0 (X) =

0

obtain that (33) holds for all X0 ∈ K with r ≤ rεK . We claim that, if X0 ∈ K and 0 < r < rεK then " " " X0 X " "wr − pκ 0 "

L∞ (B1/2 )

≤ Cε

with Cε → 0 as ε → 0. X

(34)

X

To prove the claim, observe that the two functions wr 0 and pκ 0 are both ± solution of our zero thin obstacle problem, uniformly bounded in C 1,α (B 1 ). If (34) were not true, by compactness, we can construct a sequence of solutions converging to a non trivial zero trace solutions (from (33)). The uniqueness of the solution of the thin obstacle problem with Dirichlet data implies a contradiction. It is easy to check that the claim implies the second part of the lemma. We are now in position to prove Theorem 2.9.3. The proof uses the Whitney’s extension theorem (see [25] or [27]) and the implicit function theorem. We recall that the extension theorem prescribes the compatibility conditions under which there exists a C k function f in RN having prescribed derivatives up to the order k on a given closed set. Since our reference set is Σκ (u), we ﬁrst need to show that Σκ (u) is a countable union of closed sets (an Fσ set). This is done in the next Lemma. Lemma 2.9.9 (Topological structure of Σκ (u)). Σκ (u) is a Fσ set.

198

S. Salsa

Proof. Let Ej be the set of points X0 ∈ Σκ (u) ∩ B1−1/j such that 1 κ ρ ≤ sup |u (X)| < jρκ j |X−X0 |=ρ

(35)

for 0 < ρ < 1 − |X0 |. By non degeneracy and (31) we know that Σκ (u) ⊂ ∪j≥1 Ej . We want to show that Ej is a closed set. Indeed, if X0 ∈ E j then X0 satisﬁes (35) and we only need to show that X0 ∈ Σκ (u) i.e., from Theorem 2.9.1, that ΦX0 (0+; u) = κ. Since the function X → ΦX (0+; u) is upper-semicontinuous we deduce that ΦX0 (0+; u) = κ ≥ κ. If we had κ > κ, we would have |u (X)| ≤ |X − X0 |

κ

in B1−|X0 | (X0 )

which contradicts the estimate from below in (35). Thus κ = κ and X0 ∈ Σκ (u). We are now in position for the proof of Theorem 2.9.3. Proof of Theorem 2.9.3. We divide the proof into two steps. Recall that Σκ (u) = Fκ (u) if κ = 2m. Step 1. Whitney’s extension. For simplicity it is better to make a slight change of notations, letting y = xn+1 and X = (x1 , ..., xn , xn+1 ) . Let K = Ej be one of the compact subset of Σκ (u) constructed in Lemma 2.9.9. We can write ! aα (X0 ) 0 X α. pX κ (X) = α! |α|=κ

The coeﬃcients aα (X) are continuous on Σκ (u) by Theorem 2.9.2. Since u = 0 on Σκ (u) we have X pκ 0 (X − X0 ) ≤ σ (|X − X0 |) |X − X0 |κ

X ∈ K.

For every multi-index α, 0 ≤ |α| ≤ κ, deﬁne: fα (X) =

0 if 0 < |α| < κ aα (X) if |α| = κ

X ∈ Σκ (u) .

We want to construct a function f ∈ C κ Rn+1 , whose derivatives ∂ α f up to the order κ are prescribed and equal to fα on K. The Whitney extension theorem states that this is possible if, for all X, X0 ∈ K, the following coherence conditions hold for every multi-index α, 0 ≤ |α| ≤ κ :

The Problems of the Obstacle in Lower Dimension

!

fα (X) =

|β|≤κ−|α|

199

fα+β (X0 ) (X − X0 )β + Rα (X, X0 ) β!

(36)

with |Rα (X, X0 )| ≤ σαK (|X − X0 |) |X − X0 |

κ−|α|

(37)

where σαK is a modulus of continuity. Claim: Equations (36) and (37) hold in our case. Proof. Case |α| = κ. Then we have Rα (X, X0 ) = aα (X) − aα (X0 ) and therefore |Rα (X, X0 )| ≤ σα (|X − X0 |) by the continuity on K of the map X → pX κ . Case 0 ≤ |α| < κ. We have R0 (X, X0 ) = −

! γ>α,|γ|=κ

aγ (X0 ) 0 (X − X0 )γ−α = −∂ α pX κ (X − X0 ) . (γ − α)!

Now, suppose that there exists no modulus of continuity σα such that (37) holds for all X, X0 ∈ K. Then, there is δ > 0 and sequences X i , X0i ∈ K with i X − X0i = ρi → 0 and such that

α X ∂ p 0 (X − X0 ) ≥ δ X i − X i κ−|α|. 0 κ

(38)

Consider the rescalings

u X0i + ρi X w (X) = ρκi i

ξi =

X i − X0i . ρi

We may assume that X0i → X0 ∈ K and ξ i → ξ0 ∈ ∂B1 . From Theorem 2.9.8 we have that i Xi κ w (X) − pκ 0 (X) ≤ σ (ρi |X|) |X| Xi

and therefore wi (X) converges to pκ 0 (X) , uniformly in every compact subset of Rn+1 . Note that, since X i , X0i ∈ K, the inequalities (35) are satisﬁed there. Moreover, we also have that similar inequalities are satisﬁed for the rescaled function wi at 0 and ξ i .

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S. Salsa

Thus, passing to the limit, we deduce that 1 κ κ 0 ρ ≤ sup pX κ (X) < jρ j |X−X0 |=ρ

0 < ρ < +∞.

0 This implies that ξ0 is a point of frequency κ = 2m for the polynomial pX κ ξ0 so that, from Theorem 2.9.1, we infer that ξ0 ∈ Σκ (pκ ). In particular,

∂ α pξκ0 = 0

for |α| < κ. κ−|α|

However, dividing both sides of (38) by ρi obtain α ξ ∂ pκ0 ≥ δ,

and passing to the limit, we

a contradiction. This ends the proof of the claim. Step 2. Implicit function theorem. Applying Whitney’s Theorem we deduce

the existence of a function f ∈ C κ Rn+1 such that ∂ α f = fα

on Ej

for every |α| ≤ κ. Suppose now X0 ∈ Σκd (u). This means that 0 d = dim ξ ∈ Rn : ξ · ∇x pX κ (x, 0) ≡ 0 . Then there are n − d linearly independent unit vectors νi ∈ Rn , such that 0 νi · ∇x pX κ (x, 0) is not identically zero.

This implies that there exist multi-indices β i of order β i = κ − 1 such that i

0 ∂νi (∂ β pX κ (X0 )) = 0.

This can be written as i

∂νi (∂ β f (X0 )) = 0

i = 1, ..., n − d.

(39)

On the other hand, we have

i Σκd (u) ∩ Ej ⊂ ∪i=1,...,n−d ∂ β f = 0 . From (39) and the implicit function theorem, we deduce that Σκd (u) ∩ Ej is contained in a d-dimensional C 1 manifold in a neighborhood of X0 . Since Σκ (u) = ∪Ej we conclude the proof.

The Problems of the Obstacle in Lower Dimension

201

Comments and Further Reading Sections 2.2 and 2.3 follow closely [8]. The results of these subsections have been generalized in [23] to a Signorini problem for convex fully nonlinear uniformly elliptic operators . The optimal regularity for global solutions in Sect. 2.4 is taken from [1]. The optimal regularity of a local solution follows a diﬀerent strategy with respect to this paper, and makes use of the Almgren’s formula of Sect. 2.5. The optimal regularity for the Signorini problem for a non ﬂat smooth manifold M, a smooth non zero obstacle and a divergence form uniformly elliptic operators with smooth coeﬃcients has been proved in [19]. Sections 2.5–2.8 follow [3]. Lemma 2.7.2, crucial in the deduction of the monotonicity of tangential derivatives is a generalization of a corresponding lemma in [7]. Starting from a Lipschitz free boundary, the achievement of further regularity by means of Boundary Harnack Principles is by now a classical technique in both elliptic and parabolic free boundary problems (see once more [7] and [9]). Section 2.9 is taken from part I of [18]. In part II the case of non zero obstacle is considered. Also, some partial results on the structure of the free boundary around non regular, non singular points are conjectured. A quadratic Taylor expansion to classify singular points of the free boundary is used in [7] for the classical obstacle problem. The quadratic polynomial (which must be nonnegative!) is constructed through a sequence of approximating polynomials Qj . In order to control the oscillation of the Qj and assure the uniqueness of the expansion, the deep monotonicity formula in [4] plays a crucial role. In [25] a diﬀerent proof, based on a simpler monotonicity formula, based on the functional W2 (r; u), is given. More recently, another monotonicity formula based on the functional M2 (r; u, p2 ) has been proved in [22], which is well designed for studying singular points of the classical obstacle problem.

3

Obstacle Problem for the Fractional Laplacian

This section is devoted to the study of the fractional Laplacian obstacle problem that we recall below. Given a smooth function ϕ : Rn → R, with bounded support (or rapidly vanishing at inﬁnity), we look for a continuous function u satisfying the following conditions: • • • •

u ≥ ϕ in Rn s (−Δ) u ≥ 0 in Rn (−Δ)s u = 0 when u > ϕ u (x) → 0 as |x| → +∞.

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S. Salsa

The analysis of the problem follows the strategy used for the thin obstacle problem in Sect. 2, with the exception of the preliminary properties in Sects. 1–3, taken from [24]. In particular we borrow the C 1,α estimates without proof, for brevity. The main steps are the following ones: 1. 2. 3. 4. 5. 6. 7. 8.

Construction of the solution and basic properties. Lipschitz continuity, semiconvexity and C 1,α estimates. Reduction to the thin obstacle for the operator La . Optimal regularity for tangentially convex global solution. Classiﬁcation of asymptotic blow-up proﬁles around a free boundary point. Optimal regularity of the solution. Analysis of the free boundary at stable points: Lipschitz continuity. Boundary Harnack Principles and C 1,α regularity of the free boundary at stable points.

3.1

Construction of the Solution and Basic Properties

We start by proving the existence of a solution. Observe that the proof fails for s n = 1 and s > 1/2, because in this case it is impossible to have (−Δ) u ≥ 0 in R with u vanishing at inﬁnity. Let S be the set of rapidly decreasing C ∞ functions in Rn . We denote by s ˙ H the completion of S in the norm

f 2H˙ s

|f (x) − f (y)|

= Rn

Rn

n+2s

|x − y|

2

dxdy ∼

Rn

2 |ξ|2s fˆ (ξ) dξ.

With the inner product

f, gH˙ s =

Rn

(f (x) − f (y)) (g (x) − g (y))

Rn

n+2s

|x − y|

s

=2 Rn

f (x) (−Δ) g (x) dx ∼

Rn

dxdy

2s |ξ| fˆ (ξ) gˆ (ξ)dξ,

H˙ s is a Hilbert space. Since we are considering n ≥ 2 and s < n/2, it turns out that H˙ s coincides with the set of functions in L2n/(n−2s) , for which the H˙ s -norm is ﬁnite. The solution u0 of the obstacle problem is constructed as the unique minimizer of the strictly convex functional 2

J (v) = v H˙ s

over the closed, convex set Ks = v ∈ H˙ s : v ≥ ϕ .

The Problems of the Obstacle in Lower Dimension

203

In the following proposition we gather some basic properties of u. Proposition 3.1.1. Let u0 be the minimizer of the functional J over Ks . Then: (a) The function u0 is a supersolution, that is (−Δ)s u0 ≥ 0 in Rn in the sense of measures. Thus, is lower semicontinuous and in particular, the set {u0 > ϕ} is open. (b) u0 is actually continuous in Rn . (c) If u0 (x) > ϕ (x) in some ball B then (−Δ)s u0 = 0 in B. Proof. (a) Let h ≥ 0 be any smooth function with compact support. If t > 0, u0 + th ≥ ϕ so that u0 , u0 H˙ s ≤ u0 + th, u0 + thH˙ s or

0 ≤ 2t u0 , hH˙ s + t2 h, hH˙ s = 2t u0 , (−Δ) hL2 + t2 h, hH˙ s s

from which s

s

u0 , (−Δ) hL2 = (−Δ) u0 , hL2 ≥ 0. Therefore (−Δ)s u0 is a nonnegative measure and therefore is lower semicontinuous by Propositions A2. (b) The continuity follows Proposition A3. (c) For any test function h ≥ 0, supported in B the proof in (a) holds also s for t < 0. Therefore (−Δ) u0 = 0 in B. Corollary 3.1.2. The minimizer u0 of the functional J over Ks is a solution of the obstacle problem.

3.2

Lipschitz Continuity, Semiconvexity and C 1,α Estimates

Following our strategy, we ﬁrst show that, if ϕ is smooth enough then the solution of our obstacle problem is Lipschitz and semiconvex. We are mostly interested in the case ϕ ∈ C 1,1 . When ϕ has weaker regularity, u0 inherits corresponding weaker regularity (see [24]). We emphasize that the proof in this section depends only on maximum principle and translation invariance. s Lemma 3.2.1. The function u0 is the least supersolution of (−Δ) such s that u0 > ϕ and lim inf |x|→∞ u0 (x) ≥ 0. Proof. Let v such that (−Δ) v ≥ 0, v > ϕ and lim inf |x|→∞ v (x) ≥ 0. Let w = min {u0 , v}. Then w is lowersemicontinuous in Rn (by Proposition A.4) and is another supersolution above ϕ. We show that w ≥ u0 . Since ϕ ≤ w ≤ u0 , we have w (x) = u0 (x) for every x in the contact set s Λ (u0 ) = {u0 = ϕ} . In Ω = {u0 > ϕ}, u0 solves (−Δ) u0 = 0 and w is a

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S. Salsa

supersolution. By Proposition 3.1.1 (b) , u0 is continuous. Then w − u0 is lower-semicontinuous and w ≥ u0 from comparison. Corollary 3.2.2. The function u0 is bounded and sup u0 ≤ sup ϕ. If the obstacle ϕ has a modulus of continuity ω, then u0 has the same modulus of continuity. In particular, if ϕ is Lipschitz, then u0 is Lipschitz and Lip(u0 ) ≤Lip(ϕ). Proof. By hypothesis u0 ≥ 0. The constant function v (x) = sup ϕ is a supersolution that is above ϕ. By Lemma 3.2.1, u0 ≤ v in Rn . Moreover, since ω is a modulus of continuity for ϕ, for any h ∈ Rn , ϕ (x + h) + ω (|h|) ≥ ϕ (x) for all x ∈ Rn . Then, the function u0 (x + h)+ω (|h|) is a supersolution above ϕ (x). Thus u0 (x + h) + ω (|h|) ≥ u0 (x) for all x, h ∈ Rn . Therefore u0 has a modulus of continuity not larger than ω. Lemma 3.2.3. Let ϕ ∈ C 1,1 and assume that inf ∂τ τ ϕ ≥ −C, for any unit vector τ. Then ∂τ τ u0 ≥ −C too. Thus u0 is semiconvex. Proof. Since ∂τ τ ϕ ≥ −C, we have ϕ (x + hτ ) + ϕ (x − hτ ) + Ch2 ≥ ϕ (x) 2 for every x ∈ Rn and h > 0. Therefore: V (x) ≡

u0 (x + hτ ) + u0 (x − hτ ) + Ch2 ≥ ϕ (x) 2

and V is also a supersolution: (−Δ)s V ≥ 0. Thus, by Lemma 3.2.1, V ≥ u0 so that: u0 (x + hτ ) + u0 (x − hτ ) + Ch2 ≥ u0 (x) 2 for every x ∈ Rn and h > 0. This implies ∂τ τ u0 ≥ −C. From the results in [24] we can prove a partial regularity result, under the hypothesis that ϕ is smooth. The proof is long and non elementary, so we refer to the original paper [24]. Theorem 3.2.4. Let ϕ ∈ C 2 . Then u0 ∈ C 1,α for every α < s and s (−Δ) u0 ∈ C β for every β < 1 − s.

3.3

Thin Obstacle for the Operator La : Local C 1,α Estimates

To achieve optimal regularity we now switch to the equivalent thin obstacle problem for the operator La as mentioned in the introduction and that we restate here:

The Problems of the Obstacle in Lower Dimension

⎧ u (x, 0) ≥ ϕ (x) ⎪ ⎪ ⎨ u (x, −y) = u (x, y) ⎪ L u=0 ⎪ ⎩ a La u ≤ 0

in in in in

B1 B1 B1 \Λ (u) B1 , in the sense of distributions.

205

(40)

In the global setting (i.e. with B1 replaced by Rn ), u0 (x) = u (x, 0) is the s solution of the global obstacle problem for (−Δ) and, from Appendix C, s

(−Δ) u0 = κa lim y a uy (x, y) . y→0+

The estimates in Corollary 3.2.2 and Lemma 3.2.3 translate, after an appropriate localization argument and the use of boundary estimates for the operator La , into corresponding estimates for the solution of u. Namely: Lemma 3.3.1. Let ϕ ∈ C 2,1 (B1 ) and u be the solution of (40). Then

1. ∇x u (X) ∈ C α B 1/2 for every α < s; a 2. |y| uy (X) ∈ C α B1/2 for every α < 1 − s; 3. uτ τ (X) ≥ −C in B1/2 . Proof. From Corollary 3.2.2 and Lemma 3.2.3 we have that the above estimates holds on y = 0. Since ∂xj u and uτ τ also solve the equation La w = 0 in B1 \Λ (u), the estimates 1 and 3 extends to the interior. On the other hand a −a w (x, y) = |y| uy (X) solves the conjugate equation div(|y| ∇w (X)) and we obtain 2. Remark 3.3.1. Observe that u can only be C 1,α in both variables up to y = 0 only if a ≤ 0. If a > 0, since y a uy (X) has a non-zero limit for some x in the contact set, it follows that uy cannot be bounded. We close this section with a compactness result, useful in dealing with blow-up sequences. Lemma 3.3.2. Let {vj } be a bounded sequence of functions in a W 1,2 (B1 , |y| ). Assume that there exists a constant C such that, in B1 : |∇x vj (X)| ≤ C

and

|∂y vj (X)| ≤ C |y|−a

(41)

and that, for each small δ > 0, vj is uniformly C 1,α in B1−δ ∩ {|y| > δ}. exists a subsequence {vjk } strongly convergent in W 1,2 Then, athere

B1/2 , |y| . Proof. From the results in [20], there is a subsequence, that we still call a {vj }, that converges strongly in L2 B1/2 , |y| . Since for each δ > 0, vj is uniformly bounded in C 1,α in the set B1−δ ∩ {|y| > δ}, we can extract a subsequence so that ∇vj converges uniformly in B1−δ ∩ {|y| > δ} . Thus, ∇vj converges pointwise in B1 \ {y = 0}. −a Now, from (41) and the fact that C and |y| both belongs to L 2 B1/2 , |y|a ), the convergence of each partial derivative of vj in L2 B1/2 , |y|a follows from the dominated convergence theorem.

206

3.4

S. Salsa

Minimizers of the Weighted Rayleigh Quotient and a Monotonicity Formula

The next step towards optimal regularity is to consider tangentially convex global solutions. The following lemma is the analogue to Lemma 2.4.1. Lemma 3.4.1. Let ∇θ denote the surface gradient on the unit sphere ∂B1 . Set, for −1 < a < 1, ⎧ ⎨ λ0,a = inf

⎩

2

|∇θ w| y a dS

∂B1+

:w∈W

w2 y a dS

∂B1+

1/2

∂B1+ , y a dS : w = 0 on (∂B1 )

+

⎫ ⎬ ⎭

+

where (∂B1 ) = {(x , xn ) ∈ ∂B1 , xn > 0} . Then the ﬁrst eigenfunction, up to a multiplicative factor, is given by w (x, y) =

x2n + y 2 − xn

s

s = (1 − a) /2

and5

1−a (2n + a − 1) . 4 Lemma 3.4.2 Let w be continuous in B1 , w (0) = 0, w (x, 0) ≤ 0, w (x, 0) = 0 on Λ ⊂ {y = 0}, La w = 0 in B1 \Λ,. Assume that the set λ0,a =

{x ∈ Br : w (x, 0) < 0} is non empty and convex. Set β (r) = β (r; w) =

1 r1−a

Br+

y a |∇w (X)| n+a−1

|X|

Then, β (r) is bounded and increasing for r ∈ (0, 1/2]. 2

2

dX.

2

Proof. We have La w2 = 2wLa w + 2y a |∇w| = 2y a |∇w| , so that β (r) = Now:

1 r1−a

n+a−1

|X|

Br+

a−1 β (r) = 2−a 2r

y a |∇w (X)|

Br+

2

dX =

La w 2

2r1−a

1 n+a−1 dX + rn |X|

La w 2

1

Br+

2

∂Br+

n+a−1 dX.

|X|

y a |∇w|2 dS.

5 Formally, the ﬁrst eigenvalue can be obtained plugging α = s = (1 − a) /2 and n + a instead of n into the formula α (α − 1) + nα.

The Problems of the Obstacle in Lower Dimension

207

Since w (0, 0) = 0 and y a wy (x, y) w (x, y) → 0 as y → 0+ , we can write 1

La w 2

dX 2r2−a Br+ |X|n+a−1 1 1 1 a a 2 dX = n+1 y wwν dS − 2−a y ∇w · ∇ n+a−1 r 2r |X| ∂Br+ Br+ 1 (n + a − 1) = n+1 y a wwν dS + y a w2 dS r 2rn+2 ∂Br+ ∂Br+ 1/2 1/2 2 1−a a 2 a 2 y w dS y wν dS ≤ 2rn+2 ∂Br+ (1 − a) rn ∂Br+ (n + a − 1) + y a w2 dS 2rn+2 ∂Br+ 1 (2n + a − 1) a 2 y w dS + y a wν2 dS. ≤ n + + 2rn+2 (1 − a) r ∂Br ∂Br On the other hand, 2 y a |∇w| dS = ∂Br+

2

∂Br+

y a |∇θ w| dS +

∂Br+

y a wν2 dS

so that we obtain: β (r) ≥ − (1 − a)

(2n + a − 1) 4rn+2

∂Br+

y a w2 dS +

1 rn

∂Br+

y a |∇θ w|2 dS.

The convexity of {x ∈ Br : w (x, 0) < 0} implies that the Rayleigh quotient must be greater than λ0,a and therefore we conclude β (r) ≥ 0 and in particular, β (r) ≤ ϕ (1/2).

3.5

Optimal Regularity for Tangentially Convex Global Solutions

In this section we consider global solutions that represent possible asymptotic proﬁles, obtained by a suitable blow-up of the solution at a free boundary point. First of all we consider functions u : Rn → R, homogeneous of degree k, solutions of the following problem:

208

S. Salsa

⎧ ⎪ u (x, 0) ≥ 0 ⎪ ⎪ ⎪ ⎪ ⎨ u (x, −y) = u (x, y) La u = 0 ⎪ ⎪ ⎪ La u ≤ 0 ⎪ ⎪ ⎩ uτ τ ≥ 0

Rn Rn × R (Rn × R) \Λ the sense of distributions in Rn × R Rn × R, for every tangential unit vector τ

in in in in in

(42)

where Λ = Λ (u) = {(x, 0) : u (x, 0) = 0}. The following proposition gives a lower bound for the degree k, which implies the optimal regularity of the solution. Lemma 3.5.1. If there exists a solution u of problem (42), then k ≥ 1 + s = (3 − a) /2. Proof. Apply the monotonocity formula in Lemma 3.4.2 to w = uτ . Then, La w = 0 in (Rn × R) \Λ and, by symmetry, w (x, 0) wy (x, 0) = 0. Moreover, the contact set where w = 0 is convex, since uτ τ ≥ 0. Therefore w satisﬁes all the hypotheses of that lemma. Recall that we always assume that (0, 0) ∈ F (u) so that w (0, 0) = 0. Thus β (r; w) =

1 r1−a

y a |∇w (X)|

Br+

|X|

n+a−1

2

dX ≤ β (1, w) .

On the other hand, since w is homogeneous of degree k − 1, we have r2k−2 β (r; w) = 1−a r

B1+

y a |∇w (X)| n+a−1

|X|

2

dX =

r2k−2 β (1, w) . r1−a

This implies r2k−2 ≤ r1−a or k ≥ 1 + s. From Lemma 3.5.1 it would be possible to deduce the optimal regularity of the solution u. However, to study the free boundary regularity we need to classify precisely the solution to problem (42). Let n a Λ∗ = (x, 0) ∈ R : lim y uy (x, y) < 0 . y→0+

Notice that Λ∗ is the support of La u and since La u = 0 in (Rn × R) \Λ, we have Λ∗ ⊂ Λ (Λ is closed). Lemma 3.5.2. Let u be a solution of problem (42). If Hn (Λ∗ ) = 0 then u is a polynomial of degree k. a

Proof. We know from Lemma 3.3.1 that |y| uy (x, y) is locally bounded. If Hn (Λ∗ ) = 0, then lim |y|a uy (x, y) = 0 a.e. x ∈ Rn .

y→0

The Problems of the Obstacle in Lower Dimension

209

Thus limy→0 |y| uy (x, y) = 0 weak∗ in L∞ and from Lemma B1 we infer that u is a global solution of La u = 0 in Rn × R. Using Lemma B5 we conclude the proof. a

Lemma 3.5.3. Let u be a solution of problem (42). If Hn (Λ) = 0 then, either u ≡ 0 or k = 1 + s and Λ is a half n-dimensional space. Proof. First observe that if Hn (Λ∗ ) = 0, then u ≡ 0, otherwise, from Lemma 3.5.2, u (x, 0) would be a polynomial vanishing on a set of positive measure and therefore identically zero. Thus, the polynomial u must have the form u (x, y) = p1 (x) y 2 + ... + pj (x) y 2j . We have La u = Δp1 (x) y 2 + ... + Δpj (x) y 2j + 2p1 (x) + ... +(2j − 2)pj (x) y 2j−2 + ap1 (x) y + ... + pj (x) y 2j−1 =0 which implies p1 = 0. Iterating the computation of La we get p2 = ... = pj =0. Consider now the case Hn (Λ∗ ) = 0. Then Λ∗ is a thick convex cone. Assume that en is a direction inside Λ∗ such that a neighborhood of en is contained in Λ∗ . Therefore, for any x ∈ Rn , x + hen ∈ Λ∗ for h > 0, large enough. Let x/ |x| be close to en so that x ∈ Λ∗ and lim

y→0

u (x, y) − u (x, 0) |y|

−a

y

= lim |y|a uy (x, y) < 0. y→0

Then, u ≤ 0 near the direction en . On the other hand, if y > 0, then ∂y (y a uy (x, y)) = −Δx u ≤ 0, and therefore, for any X ∈ Rn+1 we have that u (X + hen ) ≤ 0 for h large enough. Since uτ τ ≥ 0, u is convex in the en direction. Thus w = uxn is increasing along the direction en and cannot be positive at any point X, because, otherwise, limh→+∞ u (X + hen ) = +∞, a contradiction. Moreover, w = 0 on Λ and La w (X) = 0 in (Rn × R) \Λ∗ ⊇ (Rn × R) \Λ. Thus w must coincide with eigenfunction of the weighted spherical the ﬁrst 2 a Laplacian, minimizer of S1 |∇θ v| |y| dS over all v vanishing on Λ and such a that S1 v 2 |y| dS = 1.

210

S. Salsa

Since Λ is convex, Λ ∩ B1 is contained in half of the sphere B1 ∩ {y = 0}. If it were exactly half of the sphere then it would be given by the ﬁrst eigenfunction deﬁned in Lemma 3.4.1, up to a multiplicative constant, by he explicit expression w (x, y) =

x2n + y 2 − xn

s

s = (1 − a) /2.

On the other hand, the above function is not a solution across {y = 0, xn ≥ 0}. Therefore, if Λ ∩ B1 is strictly contained in half of the sphere B1 ∩ {y = 0} , there must be another eigenfunction corresponding to a smaller eigenvalue and consequently to a degree of homogeneity smaller than s. This would imply k < 1 + s, contradicting Lemma 3.5.1. The only possibility is therefore k = 1 + s, with Λ = {y = 0, xn ≥ 0}. The next theorem gives the classiﬁcation of asymptotic proﬁles. Theorem 3.5.4. Let u be a non trivial solution of problem (42). There are only two possibilities: (1) k = 1 + s, Λ is a half n-dimensional space and u depends only on two variables. Up to rotations and multiplicative constants u is unique and there is a unit vector τ such that Λ = {(x, 0) : x · τ ≥ 0} and # uτ (x, y) =

s

2

(x · τ ) + y 2 − (x · τ )

(2) k is an integer greater than equal to 2, u is a polynomial and Hn (Λ) = 0. Proof. If Hn (Λ) = 0, from Lemma 3.5.3 we deduce that, up to rotations and multiplicative constants, there is a unique solution of problem (42), homogeneous of degree k = 1+s. Moreover, for this solution the free boundary F (u) is ﬂat, that is there is a unit vector (say) en such that Λ = {(x, 0) : xn ≥ 0} and uxn (x, y) =

x2n + y 2 − xn

s .

Integrating uxn from F (u) along segments parallel to en we uniquely determine u (x, 0) = u (xn , 0). If we had another solution v, homogeneous of degree 1 + s, with v (x, 0) = u (x, 0), then necessarily (see the proof of the Liouville-type Lemma B5), for some constant c and y = 0, we have s

v (x, y) − u (x, y) = c |y| y. But the constant c must be zero, otherwise v − u cannot be solution across {y = 0} \Λ.

The Problems of the Obstacle in Lower Dimension

211

As a consequence, if u is a solution homogeneous of degree 1 + s, with en normal to F (u), then u = u (xn , y). Indeed, translating in any direction orthogonal to xn and y we get another global solution with the same free boundary. By uniqueness, u must be invariant in those directions. If Hn (Λ) = 0, then Hn (Λ∗ ) = 0 and from Lemma 3.5.2 we conclude that u is a polynomial and k ≥ 2.

3.6

Frequency Formula

As in the zero thin obstacle case, a crucial tool in order to achieve optimal regularity is given by a frequency formula of Almgren type. If the obstacle were zero, then the frequency formula states that the quantity a 2 |y| |∇u| dX Da (r; u) = r Br a 2 |y| u dS ∂Br is bounded and monotonically increasing. The proof mimics the case a = 0 and the conclusion is the following. Theorem 3.6.1. Let u be a solution of the zero thin obstacle for the operator La in B1 . Then Da (r; u) is monotone nondecreasing in r for r < 1. Moreover, Da (r; u) is constant if and only if u is homogeneous. When the obstacle ϕ is non zero we cannot reduce to that case. Instead, assuming that ϕ ∈ C 2,1 , we let u ˜ (x, y) = u (x, y) − ϕ (x) +

Δϕ (0) 2 y 2 (1 + a)

so that La u ˜ (0) = 0. Moreover Λ = Λ (u) = {˜ u = 0}. The function u ˜ is a solution of the following system: ⎧ u ˜ (x, 0) ≥ 0 in ⎪ ⎪ ⎨ u (x, −y) = u (x, y) in a ⎪ L u ˜ (x, y) = |y| g (x) in ⎪ ⎩ a a La (x, y) ≤ |y| g (x) in

B1 B1 B1 \Λ B1 , in the sense of distributions

where g (x) = Δϕ (x) − Δϕ (0) is Lipschitz. a Notice that |y| g (x) → 0 as x → 0 and a

˜ (x, y)| ≤ C |y| g (x) . |La u

(43)

212

S. Salsa

What we expect is a small variation of Almgren’s formula. Since u − u ˜ is a C 2,1 function, it is enough to prove any regularity result for u ˜ instead of u. In order to simplify the notation we will still write u for u ˜. Deﬁne a 2 a F (r) = F (r; u) = u2 |y| dσ = rn+a (u (rX)) |y| dS. ∂Br

∂B1

We have:

F (r) = (n+a) r

(u (rX))2 |y|a dS

n+a−1

∂B1

2u (rX) ∇u (rX) · X |y|a dS

+rn+a

∂B1

= (n + a) r

−1

2

a

a

(u (X)) |y| dS + ∂Br

2u (X) uν (X) |y| dS ∂Br

Thus have that log F (r) is diﬀerentiable for r > 0 and: a 2uuν |y| dS d F (r) n+a r log F (r) = = + ∂B . a dr F (r) r u2 |y| dS ∂Br Note that the monotonicity of Da (r; u) when ϕ = 0 amounts to say that the function d r −→ r log F (r) dr is increasing, since in this case a a 2 2uuν |y| dS = La (u2 ) = (|y| |∇u| + 2uLa u)dX ∂Br

Br

Br a

2

2 |y| |∇u| dX.

= Br

We will use the following modiﬁcation:

d log max F (r) , rn+a+4 . Φ (r) = Φ (r; u) = r + c0 r2 dr

(44)

Then: Theorem 3.6.2 (Monotonicity formula). Let u be a solution of problem (43). Then, there exists a small r0 and a large c0 , both depending only on a, n, ϕ C 2,1 , such that Φ (r; u) is monotone nondecreasing for r < r0 .

The Problems of the Obstacle in Lower Dimension

213

We ﬁrst need a Poincar´e type estimate. Recall that u (0, 0) = 0 since the origin belongs to the free boundary. To prove the theorem we need two lemmas. Lemma 3.6.3. Let u be a solution of problem (43), u (0, 0) = 0. Then

2

(u (X)) |y| dS ≤ Cr ∂Br

2

a

2

a

|∇u (X)| |y| dX + c (a, n) r6+a+n

a

Br

and

2

a

(u (X)) |y| dX ≤ Cr

2

Br

|∇u (X)| |y| dX + c (a, n) r7+a+n

Br

where c, C depend only on a, n and ϕ C 2,1 . Proof. By Lemma B7 we have 2 a a (u (X)) |y| dS ≤ Cr |∇u (X)| |y| dX + u ¯ ∂Br

Br

a

u (X) |y| dS

∂Br

so that we have to prove that a u (X) |y| dS ≤ c (a, n) r6+a+n . ∂Br

From Lemma B6 we have (here k = 1): 1 0 = u (0) ≥ ωn+a rn+a and therefore

+

u (X) |y| dS − Cr3 a

∂Br

u− (X) |y| dS + Cr3+a+n .

a

a

u (X) |y| dS ≤ ∂Br

(45)

∂Br

We estimate u− by integrating along the straight line from (x, 0) to (x, y) and applying Cauchy-Schwarz inequality. Since u (x, 0) ≥ 0, we can write u− (x, y) = u− (x, y) − u− (x, 0) ≤ ≤ ≤ Cy

y

0

|∇u (x, t)|2 ta dt

1−a 2

y 0

0

y

|∇u (x, t)| dt

1/2

y

t−a dt

0

2

|∇u (x, t)| ta dt

1/2 .

1/2

214

S. Salsa

Integrating on ∂Br and using Cauchy-Schwarz inequality again, we get, r observing that on ∂Br we have dS = |y| dx:

−

u− (X) |y|

a

u (X) |y| dS = ∂Br

∂Br

|y|

≤ Cr ∂Br

≤ Cr

y

0

0

1/2

2

|∇u (x, t)| ta dt

dx 1/2

2 a

|∇u (x, t)| t dtdx

1/2 |y|a−1 dx

× ∂Br

≤ Cr

rdx

√r2 −|x|2

Br

a−1 2

a−1

(n+a+1)/2

2

a

1/2

|∇u (X)| |y| dX

.

Br

Combining with (45) we obtain: a

u (X) |y| dσ ≤ Cr

(n+a+1)/2

∂Br

2

a

1/2

|∇u (X)| |y| dX

+ Cr3+a+n .

Br

Therefore, 2 1 a |u (X)| |y| dS ≤ Cr |∇u (X)| |y| dX + u (X) |y| dS ωn+a rn+a ∂Br Br ∂Br a ≤ Cr |∇u (X)| |y| dX + Cr6+a+n .

2

a

a

Br

This gives the ﬁrst inequality. The second one can be obtained by integrating in r.

Lemma 3.6.4. The following identity holds for any r ≤ 1. 2 a 2 a 2 |∇θ u| −uν |y| dS = [(n+a−1) |∇u (X)| −2 X, ∇u g (X)] |y| dX r

∂Br

Br

(46)

where ∇θ u denotes the tangential gradient. Proof. Consider the vector ﬁeld 1 2 F = y a |∇u| X − y a X, ∇u ∇u 2

(y > 0) .

The Problems of the Obstacle in Lower Dimension

215

We have:

1 2 divF = (n + a − 1)y a |∇u| − X, ∇u La u. 2 Since X, ∇u is a continuous function on Br that vanishes on Λ = {u = 0} we have that X, ∇u La u has no singular part and coincides with X, ∇u |y|a g (x). An application of the divergence theorem gives (46). Proof of Theorem 3.6.2. First we observe that by taking the maximum in (44) However, it may happen that we get a non diﬀerentiable functions. 1,1 max F (r) , rn+a+4 is absolutely continuous (it belongs to Wloc (0, 1)) and in any case, the jump in the derivative will be in the positive direction. When F (r) ≤ rn+a+4 we have

d log rn+a+4 Φ (r) = r + c0 r2 dr and Φ (r) = (n + a + 4)c0 > 0. Thus we can concentrate on the case F (r) > rn+a+4 where

d log F (r) . Φ (r) = r + c0 r2 dr We have: a

∂Br 2uuν |y| dS 2 Φ (r) = r + c0 r + (1 + c0 r) (n + a) a 2 ∂Br u |y| dS ≡ 2Ψ (r) + (1 + c0 r) (n + a). We show that the ﬁrst term is increasing, by computing its logarithmic derivative. We ﬁnd:

d d a a 2 2 log Ψ (r) = log r + c0 r +log uuν |y| dS −log u |y| dS dr dr ∂Br ∂Br a a d 2uuν |y| dS n+a c0 1 dr ∂Bruuν |y| dS ∂Br + . − − = + a a 2 r 1+c0r r ∂Bruuν |y| dS ∂Br u |y| dS We estimate

d dr

∂Br

a

uuν |y| dS from below. Since

a

uuν |y| dS = ∂Br

(|y|a |∇u|2 + uLa u)dX

Br a

we can write, recalling that |La u| ≤ c |y| |x| ,

216

d dr

S. Salsa

a

∂Br

2

a

uuν |y| dS =

(|y| |∇u| + uLa u)dS ∂Br

2

|y| |∇u| dS − cr(n+a+2)/2 a

≥

∂Br

u2 |y| dS a

1/2

∂Br

|y|a |∇u|2 dS − cr(n+a+2)/2 [F (r)]1/2 .

= ∂Br

We now use Lemma 3.6.4 to estimate

2

a

|y| |∇u| dσ = ∂Br

∂Br

|y|a |∇u|2 dσ from below.

2 a |uτ | + u2ν |y| dσ

∂Br

u2ν |y| dS + a

=2 ∂Br

1 r

[(n + a − 1) |∇u (X)|

2

Br a

−2 X, ∇u g (X)] |y| dX n+a−1 a a u2ν |y| dS + uuν |y| dS =2 r ∂Br ∂Br 1 a [(n + a − 1)u − 2 X, ∇u]g (X) |y| dX. − r Br Therefore n+a−1 d a a a 2 uuν |y| dS ≥ 2 uν |y| dS + uuν |y| dS dr ∂Br r ∂Br ∂Br G (r) + r H (r) + rF (r) −crn+a+1 where

u2 |y| dX a

G (r) =

and

2

Br

a

|∇u| |y| dX.

H (r) = Br

Collecting all the above estimates, we can write: d log Ψ (r) = P (r) + Q (r) dr with 2 P (r) =

∂Br

∂Br

a

a

r − ∂B

u2ν |y| dS

uuν |y| dS

a

2uuν |y| dS

∂Br

a

u2 |y| dS

≥0

The Problems of the Obstacle in Lower Dimension

217

and G (r) + rF (r) + r H (r) a uuν |y| dσ ∂Br G (r) + rF (r) + r H (r) c0 (n+a+1)/2 − cr ≥ . 1 + c0 r H (r) − r(n+a+2)/2 G (r)

c0 − cr(n+a+1)/2 Q (r) = 1 + c0 r

First we estimate F, G, H. Since F (r) > rn+4+a , from the Poincar´e Lemma 3.6.3 we have: rn+4+a < F (r) ≤ Cr |∇u (X)|2 |y|a dX + c (a, n) r6+a+n Br

= CrH (r) + c (a, n) r6+a+n . Integrating the above inequalities in r, we get: G (r) = 0

r

F (s) ds ≤ Cr2 H (r) + c (a, n) r7+a+n .

This means that, for small enough r0 and r < r0 : F (r) ≤ crH (r)

and

G (r) ≤ Cr2 H (r)

so that: r H (r) c0 (n+a+1)/2 √ . − cr Q (r) ≥ 1 + cr H (r) − r(n+a+4)/2 Hr Since rn+4+a < F (r) ≤ CrH (r), we also have H (r) ≥ crn+a+3 and for r0 small: Q (r) ≥ ≥

c0 r − cr(n+a+1)/2 1 + c0 r H (r) − r(n+a+4)/2 n+a+1 n+a+3 c0 c0 − cr 2 +1− 2 = −c 1 + c0 r 1 + cr

which is positive if c0 is large and r0 small.

3.7

Blow-up Sequences and Optimal Regularity

The optimal regularity of the solution can be obtained by a careful analysis of the possible values of Φ (0+). When Φ is constant and the obstacle is zero, Φ (0+) − n − a represents the degree of homogeneity at the origin.

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S. Salsa

Thus, by a suitable blow-up of the solution, we will be able to classify the possible asymptotic behaviors at the origin, using the results of Sect. 5. The ﬁrst result is the following (compare with Theorem 2.6.2). Theorem 3.7.1. Let u be a solution of problem (43), u (0, 0) = 0. Then either Φ (0+, u) + a + 2 (1 + s) or Φ (0+, u) ≥ n + a + 4. To prove the lemma, we introduce a rescaling similar to (16) in the case of the zero obstacle problem. We set: ur (X) =

u (rX) dr

(47)

where dr =

r−(n+a)

u2 |y| dσ a

1/2

1/2 = r−(n+a) F (r) .

∂Br

Two things can occur: dr r→0 r2

lim inf

= +∞ < +∞

ﬁrst case second case.

(48)

The next lemma takes care of the ﬁrst case. Lemma 3.7.2. Let u be as in Theorem 3.7.1. Assume that lim inf

r→0

dr = +∞. r2

Then, there is a sequence rk → 0 and a function u0 : Rn+1 → R, non identically zero, such that:

1. urk → u0 in W 1,2 B1/2 , |y a | 2. urk → u0 uniformly in B1/2 . 3. ∇x urk → ∇x u0 uniformly in B1/2 . a a 4. |y| ∂y urk → |y| ∂y u0 uniformly in B1/2 . Moreover, u0 is a solution of system (42) and its degree of homogeneity is [Φ (0+; u) − n − a]/2. Proof. First of all, observe that ur L2 (∂B1 ,|y|a ) = 1 for every r. Claim: ur is bounded in W 1,2 (B1 , |y a |) . To prove the claim we use the frequency formula. Since lim inf r→0 drr2 = +∞, then F (r) > rn+a+4 for r small enough. In particular we may assume r < r0 where r0 is deﬁned in Theorem 3.6.2.

The Problems of the Obstacle in Lower Dimension

219

Thus we can write (Φ (r) = Φ (r, u)):

d log max F (r) , rn+a+4 Φ (r0 ) ≥ Φ (r) ≥ r + c0 r2 dr 2 a uL dX u + |∇u| |y| a

B r ≥ r + c0 r 2 + (n + a) (1 + c0 r) a 2 u |y| dS ∂Br a

∂Br 2uuν |y| dS 2 + (n + a) (1 + c0 r) . ≥ 2 r + c0 r (49) a u2 |y| dS ∂Br a

Recalling that |La u| ≤ C |y| |x| in Rn+1 \Λ (u) and that u = 0 on Λ (u), we get: 1/2 a (n+a+3)/2 2 uL udX ≤ cr u |y| dX . a Br

Br

From Lemma 3.6.4: a u2 |y| dX ≤ Cr2 Br

2

|∇u (X)| |y| dX + cr7+a+n a

Br

so that

Br

uLa udX ≤ Cr(n+a+5)/2

2

a

1/2

|∇u (X)| |y| dX

+ cr5+a+n . (50)

Br

From the same lemma: 1 2 a a |∇u (X)| |y| dX ≥ u2 |y| dσ − cr5+a+n r Br ∂Br but, since dr /r2 → +∞, we deduce

u2 |y|a dσ ≥ C1 rn+a+4

∂Br

with a constant C1 as large as we wish as r 0. Therefore, for r small: C1 3+a+n 2 a r |∇u (X)| |y| dX ≥ C1 rn+a+3 − cr5+a+n ≥ (51) 2 Br and

Br

uLa udX ≤ Cr

Br

2

a

|∇u (X)| |y| dX.

(52)

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S. Salsa

Comparing (50) with (52) we see that, for r small:

2

uLa u + |∇u| |y|

a

dX ≥

Br

1 2

2

a

|∇u (X)| |y| dX. Br

Continuing from (49) we get:

Φ (r0 ) ≥ r + c0 r

2

2

Br

2

≥

a

|∇u (X)| |y| dX + (n + a) (1 + c0 r) a 2 ∂Br u |y| dσ a

|∇ur (X)| |y| dX. B1

a Combining the above inequality with ∂Br u2r |y| dS = 1 we infer that ur is 1,2 a bounded in W (B1 , |y |) . − Now, u+ r and ur are subsolutions of the equation a

La u ≥ −cr |y| |x|

and therefore (see [16]) ur is bounded in L∞ B3/4 . We also know that u is semiconvex in the x variable, that is uτ τ ≥ −C for every tangential direction τ . This implies a similar bound for ur , namely: ∂τ τ ur =

r2 r2 uτ τ (rX) ≥ −C . dr dr

(53)

The functions ur are solutions of a uniformly elliptic equation with smooth coeﬃcients in B1 ∩ {|y| > δ} for every δ > 0. Then the ur are uniformly bounded in C 1,α in B1−δ ∩ {|y| > δ} for any α, δ ≤ 1. Therefore, we can apply Lemma 3.3.2 to obtain a subsequence urk strongly convergent in

W 1,2 B1/2 , |y a | to some function u0 as rk 0. From Theorem 3.6.2, we know that Φ (r, u) is monotone and converges to Φ (0+; u) as r 0. We have: Φ (rs, u) ∼ rs

2

Brs

∂Brs

=r

a

u2 |y| dS 2

Br

a

|∇u (X)| |y| dX

+ (n + a)

a

|∇us (X)| |y| dX + (n + a) . a 2 ∂Br us |y| dS

We want to set s = rk 0 and pass to the limit in the above expression to obtain: 2 a |∇u0 (X)| |y| dX . (54) Φ (0+, u) − (n + a) = r Br a 2 ∂Br u0 |y| dS

The Problems of the Obstacle in Lower Dimension

221

This is possible since ∂Br u2s |y|a dS ≥ c > 0. In fact, if δ > 0 is small and r ≥ 1 − δ, from the Poincar´e inequality (64) we have

a u2s |y|

∂B(1−δ)r

u2s |y| dS a

dS ≥ c ∂Br

where c depends only on δ, n and Φ (1; u) . k Iterating this inequality k times until (1 − δ) < r, we obtain the desired uniform bound from below. From (53), we have that ∂τ τ urk ≥ −C

rk2 →0 drk

and therefore u0 is tangentially convex. On the other hand, each ur satisﬁes the following conditions: (a) ur (x, 0) ≥ 0 in B1 . (b) La ur =

r2−a r2 a La u (rX) = |y| g (rx) dr dr

(c) |La ur | ≤ Since

r2 a |y| |g (rx)| dr

in B1 \Λ (ur )

in B1 .

r2 a r2 a |y| |g (rx)| ≤ c |y| r |x| → 0 dr dr

as r → 0

it follows that u0 is a solution of the homogeneous problem ⎧ u0 (x, 0) ≥ 0 in ⎪ ⎪ ⎨ u (x, −y) = u (x, y) in ⎪ La u0 = 0 in ⎪ ⎩ in La (x, y) ≤ 0

⎫ B1 ⎪ ⎪ ⎬ B1 ⎪ B1 \Λ ⎪ ⎭ B1 , in the sense of distributions

For this problem, the frequency formula holds as in Theorem 3.7.1 without any error correction. Thus we conclude that u0 is homogeneous in B1/2 and its degree of homogeneity is exactly (Φ (0+; u0 ) − (n + a)) /2. Since it is homogeneous, then it can be extended to Rn+1 as a global solution of the homogeneous problem. Finally, from the a priori estimates in Lemma 3.3.1, it follows that we can a choose rk so that the sequences urk , ∇urk and |y| ∂urk converge uniformly in B1/2 .

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S. Salsa

Proof of Theorem 3.7.1. In the first case of (48), we use Lemma 3.7.2 and Theorem 3.5.4 to ﬁnd the blow-up proﬁle u0 and to obtain that the degree of homogeneity of u0 is 1 + s or at least 2. Therefore Φ (0+; u) = Φ (0+; u0 )+a+2 (1 + s)

or Φ (0+; u) = Φ (0+; u0 ) ≥ n+a+4.

Consider now the second case of (48). If F (rk ; u) < rkn+a+4 for some sequence rk → 0, then, for these values of rk , Φ (rk ; u) = (n + a + 4) (1 + c0 rk ) so that Φ (0+; u0 ) + a + 4. On the other hand, assume that F (r; u) ≥ rn+a+4 for r small. Since we are in the second case, for some sequence rj 0 we have drj /rj2 ≤ C so that rjn+a+4 ≤ F (rj ; u) ≤ Crjn+a+4 . Taking logs in the last inequality, we get (n + a + 4) log rj ≤ log F (rj ; u) ≤ C + (n + a + 4) log rj . We want to show that Φ (0+; u) ≥ n + a + 4 in this case. By contradiction, assume that for small rj we have Φ (rj ; u) ≤ n + a + 4 − ε0 . Take rm < rj 1 and write: (n + a + 4) (log rj − log rm ) − C ≤ log F (rj ; u) − log F (rm ; u)

rj

= rm

d log F (s; u) ds ≤ dr

rj

r + c0 r 2

rm

−1

Φ (s; u) ds ≤

rj

r−1 Φ (rj ; u) ds

rm

≤ (n + a + 4 − ε0 ) (log rj − log rm ) which gives a contradiction if we make (log rj − log rm ) → +∞.

From the classiﬁcation of the homogeneity of a global proﬁle, we may proceed, using Theorem 3.6.4, to identify u0 , modulus rotations. We have: Lemma 3.7.3 Let u be as in Lemma 3.7.1. Assume that Φ (0+; u) + a + 2 (1 + s) There is a family of rotations Ar , with respect to x, such that ur ◦ Ar converges to the unique proﬁle u0 of degree 1 + s, where u0 is deﬁned in Theorem 3.6.4. More precisely:

1. ur ◦ Ar → u0 in W 1,2 B1/2 , |y a | 2. ur ◦ Ar → u0 uniformly in B1/2 . 3. ∇x (ur ◦ Ar ) → ∇x u0 uniformly in B1/2 . a a 4. |y| ∂y (ur ◦ Ar ) → |y| ∂y u0 uniformly in B1/2 .

The Problems of the Obstacle in Lower Dimension

223

We now control the decay of u at (0, 0) in terms of the decay of F (r; u). Lemma 3.7.4. If F (r; u) ≤ crn+a+2(1+α)

(55)

for every r < 1, then u (0, 0) = 0, |∇u (0, 0)| = 0 and u is C in the sense that 1+α |u (X)| ≤ C1 |X|

1,α

at the origin

for |X| ≤ 1/2 and C1 = C1 (C, n, a). Proof. Consider u+ and u− , the positive and negative parts of u. We have La u+ ≥ |y| g (x) ≥ −C |y| |x| a

a

La u+ ≥ − |y| g (x) ≥ −C |y| |x| . a

a

For some r > 0, let U be the La -harmonic replacement of u+ in Br . Note that 2 |X| − r2 . 0 = La U ≤ La U + C 2 (n + a + 1) Hence, by comparison principle U ≥ u+ − C r 2 . From (55) we have:

U 2 |y| dσ =

a

∂Br

(u+ )2 |y| dσ ≤ crn+a+2(1+α) . a

∂Br

Since w (X) = |y|a is a A2 weight, from the local L∞ estimates (see [16]) we conclude sup |U (X)| ≤ C1 |r|1+α . B1/2

Then u+ ≤ U + r2 ≤ Cr1+α in B1/2 . A similar estimate holds for u− and the proof is complete. The frequency formula provides the precise control of F (r) from above. Lemma 3.7.5. If Φ (0+; u) = μ then F (r; u) ≤ crμ for any r < 1 and c = c (F (1; u) , c0 ) . Proof. Let f (r) = max F (r) , rn+a+4 ≥ F (r). Since Φ is nondecreasing:

d log f (r) μ = Φ (0+) ≤ Φ (r) = r + c0 r2 dr

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S. Salsa

and then

d μ log f (r) ≥ . dr (r + c0 r2 )

An integration gives:

1

μ ds 2 r (s + c0 s ) 1 μ 1 − ds ≥ −μ log r − μ s (s + c0 s2 ) r

log f (1) − log f (r) ≥

≥ −μ log r − C1 μ so that log f (r) ≤ log f (1) + μ log r + C1 μ. Taking the exponential of the two sides we infer F (r) ≤ f (r) ≤ Crμ

with C = f (1) eC1 μ .

From Lemmas 3.7.4 and 3.7.5 the optimal regularity of a solution u of (43) follows easily. Precisely: Theorem 3.7.6. Let u be a solution of (43), with u (0, 0) = 0. Then 1+s

|u (X)| ≤ C |X|

sup |u| B1

where C = C n, a, g Lip . Proof. From Lemma 3.7.1, μ = Φ (0+) ≥ n+a+2 (1 + s) and the conclusion comes from Lemmas 3.7.5 and 3.7.4. Corollary 3.7.7. Let ϕ ∈ C 2,1 . Then the solution u of the obstacle problem for the operator (−Δ)s belongs to C 1,s (Rn ). s

Proof. Using the equivalence between the obstacle problem for (−Δ) and the thin obstacle for La , Theorem 3.7.6 shows that u − ϕ has the right decay at free boundary points. This is enough to prove that u ∈ C 1,s (Rn ). Remark 3.7.1. Observe that it is not true that the solution of the thin obstacle problem for La is C 1,s in both variables x and y. It is quite interesting however, that the optimal decay takes place in both variables at a free boundary point. In any case, we have: Corollary 3.7.7. Let u be a solution of one of the systems (40) or (43). Then u (x, y0 ) ∈ C 1,s (B1/2 ) for every y0 ∈ (0, 1/2) .

The Problems of the Obstacle in Lower Dimension

3.8

225

Nondegenerate Case: Lipschitz Continuity of the Free Boundary

As we have already seen from the examples in Sect. 2.1, we cannot assure any minimal growth of u at a free boundary point. Moreover, the free boundary could be composed entirely by singular points. Indeed, in analogy with what happen in the zero obstacle problem, the regularity of the free boundary can be inferred for points around which u has an asymptotic proﬁle corresponding to the optimal homogeneity degree Φ (0+) + a + 2 (1 + s). Accordingly, we say that X0 ∈ F (u) is regular or stable if μ (X0 ) = Φ (0+) + a + 4. As always, we refer to the origin (X0 = (0, 0)). Theorem 3.8.1. Assume μ = Φ (0+) + a + 4. Then, there exists a neighborhood of the origin Bρ and a tangential cone Γ (θ, en ) ⊂ Rn × {0} such that, for every τ ∈ Γ (θ, en ), we have ∂τ u ≥ 0 in Bρ . In particular, the free boundary is the graph xn = f (x1 , ..., xn−1 ) of a Lipschitz function f. The theorem follows by applying the following lemma (compare with Lemma 2.7.2) to a tangential derivative h = ∂τ ur , where ur is the blow-up family (47) that deﬁnes the limiting proﬁle u0 , for r small. Lemma 3.8.2. Let Λ be a subset of Rn × {0} . Assume h is a continuous function with the following properties: 1. 2. 3. 4.

La h ≤ γ |y|a in B1 \Λ. h ≥ 0 for |y| ≥ σ√> 0, h = 0 on Λ. h ≥ c0 for |y| ≥ 1 + a/8n. h ≥ −ω (σ) for |y| < σ, where ω is the modulus of continuity of h.

There exist σ0 = σ0 (n, a, c0 , ω) and γ0 = γ0 (n, a, c0 , ω) such that, if σ < σ0 and γ < γ0 , then h ≥ 0 in B1/2 . Proof. Suppose X0 = (x0 , y0 ) ∈ B1/2 and h (X0 ) < 0. Let √ 1 1+a Q = (x, y) : |x − x0 | < , |y| ≥ 3 4n and

2

P (x, y) = |x − x0 | −

n 2 y . a+1

Observe that La P = 0. Deﬁne v (X) = h (X) + δP (X) −

γ y2. 2(a + 1)

226

S. Salsa

Then: γ y02 < 0 • v (X0 ) = h (X0 ) + δP (X0 ) − 2(a+1) • v (X) ≥ 0 on Λ a • La v = La h + δLa P − γ |y| ≤ 0 outside Λ.

Thus, v must have a negative minimum on ∂Q. However, on √ 1+a ∂Q∩ |y| ≥ 8n we have v ≥ c0 − On

γ − 32n > 0 if δ, γ are small depending on c0 and n. √ 1 1+a |x − x0 | = , σ ≤ |y| ≤ 3 8n

δ 16n

we have v≥δ

1 1 − 9 64n

−

γ >0 128n2

if γ is small compared to δ. Finally, on 1 |x − x0 | = , |y| ≤ σ 3

we have v ≥ −ω (σ) +

n2 2 1 − σ 9 1+a

−

γσ 2 >0 2+a

if σ is small, depending on δ, γ and ω. Hence v ≥ 0 on ∂Q and we have reached a contradiction. Therefore h ≥ 0 in B1/2 . Proof of Theorem 3.8.1. Since μ = Φ (0+) < n + a + 4, Lemma 3.7.1 gives μ + a + 2 (1 + s). Moreover, the blow-up sequence ur converges (modulus subsequences) to the global proﬁle u0 , whose homogeneity degree is 1 + s and whose free boundary is ﬂat. Let us assume that en is the normal to the free boundary of u0 . Then ∂n u0 (x, y) = c

s x2n + y 2 − xn .

For some θ0 > 0, let σ any vector orthogonal to y and xn such that |σ| < θ0 . From Theorem 3.6.4 we know that u0 is constant in the direction of σ and therefore, if τ = en + σ, ∂τ u0 = ∂n u0 . On the other hand, ∇x ur → ∇x u0 uniformly in every compact subset of Rn+1 . Thus, for every δ0 , there is an r for which |∂τ u0 − ∂τ ur | ≤ δ0

The Problems of the Obstacle in Lower Dimension

227

where τ = en + σ. If we diﬀerentiate the equation La ur (X) =

r2 a |y| g (rX) dr

we get La [∂τ ur (X)] =

r2 a a |y| r∂τ g (rX) ≤ Cr |y| dr

in B1 \Λ (ur )

(56)

and the right hand side tends to zero as r → 0. Thus, for r small enough, ∂τ ur satisﬁes all the hypotheses of Lemma 3.8.2 and therefore is nonnegative in B1/2 . This implies that near the origin, the free boundary is a Lipschitz graph.

3.9

3.9.1

Boundary Harnack Principles and C 1,α Regularity of the Free Boundary Growth Control for Tangential Derivatives

We continue to examine the regularity of F (u) at stable points. As we have seen, at these point we have an exact asymptotic picture and this fact allows us to get a minimal growth for any tangential derivative when ur is close to the blow-up limit u0 . This is needed in extending the Carleson estimate and the Boundary Harnack principle in our non-homogeneous setting. First, we need to reﬁne Lemma 3.8.2. −1/2s

Lemma 3.9.1. Let δ0 = (12n) following properties:

. Let v be a function satisfying the

1. La v (X) ≤ ε0 for X ∈ B1 × (0, δ0 ) 2. v (X) ≥ 0 for X ∈ B1 × (0, δ0 ) 1 3. v (x, δ0 ) ≥ 4n for x ∈ B1 . Then, there exists ε0 = ε0 (n, a) > 0 such that 2s

v (x, y) ≥ C |y|

in B1/2 × [0, δ0 ] .

× [0, δ0 ]. We compare v with Proof. Let X0 = (x0 , y0 ) ∈ B1/2 2 ε0 2 |x − x0 | y − + y 2s . w (x, y) = 1 + 2 n

Inside B1 × (0, δ0 ), we have La w (X) = ε0 ≥ La v.

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S. Salsa

For y = 0, w (x, 0) ≤ 0 ≤ v (x, 0) . For y = δ0 , ε0 2 1 δ0 + δ02s ≤ 3δ02s = w (x, δ0 ) ≤ 1 + ≤ v (x, δ0 ) . 2 4n Finally, if |x − x0 | = 1/2, we have w (x, y) ≤ 3δ02s −

1 ≤ 0 ≤ v (x, y) . 4n

Therefore, by comparison, we deduce that w ≤ v in B1 ×(0, δ0 ) . In particular, v (x0 , y0 ) ≥ w (x0 , y0 ) ≥ y 2s .

Corollary 3.9.2. Let u be a solution of (43) with |g| , |∇g| ≤ ε0 . Let u0 the usual asymptotic nondegenerate proﬁle and assume that |∇x u − ∇u0 | ≤ ε0 . Then, if ε0 is small enough, there exists c = c (n, a) such that 2s

uτ (X) ≥ cdist (X, Λ)

for every X ∈ B1/2 and every tangential direction τ such that |τ − en | < 1/2. Proof. From (56) and Theorem 3.8.1, we know that uτ is positive in B1/2 . Applying Lemma 3.9.1 we get uτ ≥ c |y|2s

in B1/4 .

Let now X = (x, y) ∈ B1/8 and d =dist(X, Λ) . Consider the ball Bd/2 (X). At the top point of this ball, say (xT , yT ) we have yT ≥ d/2. Therefore uτ (xT , yT ) ≥ cd2s . By Harnack inequality, uτ (x, y) ≥ cuτ (xT , yT ) ≥ cd2s .

3.9.2

Boundary Harnack Principles

Using the growth control from below provided by Lemma 3.9.2 it is easy to extend the Carleson estimate to our nonhomogeneous setting. Deﬁnition 2.8.1 of Δ − N T A domain can be generalized to La − N T A domains replacing Δ-capacity by La -capacity. Again a ﬂat n-dimensional manifold with Lipschitz free boundary is a simple example of La − N T A domain.

The Problems of the Obstacle in Lower Dimension

229

We keep the same notation of Sect. 2.8: • D ⊂ B1 is a La − N T A domain • ∂ ∗ D = ∂D ∩ B1 . • For Q ∈ ∂ ∗ D, Ar (Q) is a point such that Bηr (Ar (Q)) ⊂ Br (Q) ∩ D. In particular, |Ar (Q) − Q| ∼ r. Lemma 3.9.3 (Carleson estimate). Let D ⊂ B1 be a La − N T A domain. Suppose w is a positive function in D vanishing on ∂ ∗ D. Assume that: 1. |La w| ≤c |y|a 2. nondegeneracy: w (X) ≥ CdβX for some β ∈ (0, 2), where dX =dist(X, ∂ ∗ D). Then, for every Q ∈ ∂ ∗ D ∩ B1/2 and r small: sup Br (Q)∩D

w ≤ C (n, a, D) w (Ar (Q)) .

Proof. Let w∗ be the harmonic replacement of w in B2r (Q) ∩ D, r small. From Theorem 2.8.2 adapted to the operator La , with identical proof, we get w∗ (X) ≤ Cw∗ (Ar (Q)) in Br (Q) ∩ D. On the other hand, we claim that |w∗ − w| ≤ cr2

in B2r (Q) ∩ D.

2 In fact, compare w with the function w∗ (X) + C |X − Q| − r2 . On ∂ [B2r (Q) ∩ D] we have: w∗ (X) + C |X − Q|2 − r2 ≤ w (X) . Also:

2 a La w∗ (X) + C |X − Q| − r2 = C |y| ≥ La w.

Thus w − w∗ ≤ Cr2 . Similarly we obtain the other inequality. The claim follows. Collecting our estimate we can write w (X) ≤ C w (Ar (Q)) + cr2

in Br (Q) ∩ D.

From the nondegeneracy condition w (Ar (Q)) ≥ crβ and since β < 2, the theorem follows.

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S. Salsa

The following theorem expresses a boundary Harnack principle valid in our nonhomogeneous setting. Here D = B1 \Λ where Λ is a piece of the hyperplane y = 0. We require that at each point of Λ, the La capacity of Br (Q) ∩ Λ has a natural minimal growth of order rn+a−1 . This assures that D is a La − N T A domain. Theorem 3.9.3 (Boundary Harnack principle). Let D = B1 \Λ where Λ ⊂ {y = 0}. Assume that for every X ∈ Λ and r < 1/2, capLa (Br (X) ∩ Λ) ≥ κrn+a−1 . Let v, w positive functions in D satisfying the hypotheses 1 and 2 of Lemma 3.9.3 and symmetric in y. Then there is a constant c = c (n, a, κ) such that

v 0, 12 v (X) ≤ c 1 w (X) w 0, 2

in B1/2 .

Moreover, the ratio v/w is H¨ older continuous in B1/2 , uniformly up to Λ.

Proof. Let us normalize v, w setting v 0, 12 = w 0, 12 = 1. From Carleson estimate and Harnack inequality, for any δ > 0 we get: v (X) ≤ C

in B3/4

and w (X) ≥ c

in B3/4 ∩ {|y| > δ} .

This implies that, for a constant s small enough, v − sw fulﬁlls the conditions of Lemma 3.9.2. Therefore v − sw ≥ 0 in B1/2 or, in other words: v (X) ≤s w (X)

in B1/2 .

At this point, the rest of proof of (c) follows again, for instance, as in [9, Sect. 11.2].

3.9.3

C 1,α Regularity of the Free Boundary

As in the case of the thin-obstacle from the Laplace operator, the C 1,α regularity of the free boundary follows by applying Theorem 3.9.3 to the quotient of two positive tangential derivatives. Precisely we have:

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231

Theorem 3.9.4. Let u be a solution of (43). Assume ϕ ∈ C 2,1 and Φ (0) < n+a+4. Then the free boundary is a C 1,α (n − 1)-dimensional surface around the origin. Proof. See the proof of Theorem 2.8.1.

Comments and Further Reading Sections 3.1 and 3.2 follow [24], to which we refer for the missing proofs. We emphasizes that, in this paper, the optimal regularity of the solution is proved in the case that the contact set is convex. The rest of the section follows basically the paper [11] except for Sect. 3.4 and part of Sect. 3.5. In particular, the optimal regularity of global solutions follows a diﬀerent approach, similar to the corresponding proof for the zero obstacle problem. Lemma 3.8.2 parallels Lemma 2.6.2 and once more is crucial to achieve the monotonicity of tangential derivatives and, consequently, the Lipschitz continuity of the free boundary. Due to the non homogeneous right hand side in the equation, the boundary Harnack principle in Theorem 3.9.3 is somewhat weaker than the corresponding result in Theorem 2.8.2. Notice the less than quadratic decay to zero of the solution at the boundary, necessary to control the eﬀect of the right hand side.

Appendix A: The Fractional Laplacian Definition and Basic Properties In this section we brieﬂy recall the deﬁnition and some elementary properties of the fractional Laplace operator (see [24]). We denote by S the Schwartz space of rapidly decreasing C ∞ functions in Rn . Its dual S is the space of tempered distributions in Rn . s

Definition A1. For s ∈ (− n2 , 1] and f ∈ S, we deﬁne (−Δ) f through the formula s 2s (−Δ) f (ξ) = |ξ| f$(ξ) . / S for s < 1, because of Observe that (−Δ) f ∈ C ∞ (Rn ) but (−Δ) f ∈ 2s the singularity at ξ = 0 of its Fourier transform. If s ≤ − n2 , |ξ| is not a tempered distribution. 0 1 Clearly, (−Δ) = Id and (−Δ) = −Δ. We are mainly interested in the s case 0 < s < 1. In this case, we can also compute (−Δ) f using a singular integral: s

s

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S. Salsa

f (x) − f (y)

s

(−Δ) f (x) = cn,s P.V. For s = −σ, 0 < σ < Potential:

|x − y|n+2s

Rn

dy.

−σ

n 2,

the operator (−Δ) can be computed as a Riesz f (y) −σ (−Δ) f (x) = cn,−σ n−2σ dy. Rn |x − y|

In particular, the last formula indicates that the fundamental solution for (−Δ)s is the Riesz kernel ϕ (x) = cn,−σ

1

|x|

n−2s . s

to a large class of

We can extend by duality the deﬁnition of (−Δ) distributions.

Definition A2. Let

n+2s Ps = f ∈ C ∞ (Rn ) : (1 + |x| )Dk f is bounded, ∀k ≥ 0 In Ps introduce the topology induced by the seminorms n+2s )Dk f . [f ]k,s = sup (1 + |x| Denote by Ps the dual of Ps . s

s

If f ∈ S then (−Δ) f ∈ Ps . The symmetry of (−Δ) allows us to deﬁne s (−Δ) u for u ∈ Ps : s

s

(−Δ) u, f = u, (−Δ) f

for every f ∈ S.

Supersolutions and Comparison The deﬁnition of supersolution is deﬁned for u ∈ Ps , and the meaning is that it is a nonnegative measure: Definition A3. Let u ∈ Ps . We say that (−Δ) u ≥ 0 in an open set Ω s if for every nonnegative test function ϕ ∈ C0∞ (Ω) then (−Δ) u, ϕ ≥ 0. s

Every supersolution shares some properties of superharmonic functions. For instance, u is lower-semicontinuous. s

Proposition A1. Let (−Δ) u ≥ 0 in an open set Ω, then u is lowersemicontinuous in Ω.

The Problems of the Obstacle in Lower Dimension

233 s

Moreover, if the restriction of u on supp((−Δ) u) is continuous then, u is continuous everywhere. Precisely, we have: Proposition A2. Let v be a bounded function in Rn such that s s (−Δ) u ≥ 0. If E =supp((−Δ) u) and v|E is continuous, then v is n continuous in R . s

Due to the nonlocal nature of (−Δ) , a comparison theorem in a domain Ω must take into account what happens outside Ω. Indeed we have: Proposition A3. (Comparison). Let Ω Rn be an open set. Let (−Δ)s u ≥ 0 and (−Δ)s v ≤ 0 in Ω, such that u ≥ v in Rn \Ω and u − v is lower-semicontinuous in Ω. Then u ≥ v in Rn . Moreover, if x ∈ Ω and u (x) = v (x) then u = v in Rn . Also, the set of supersolutions is a directed set, as indicated by the following proposition. Proposition A4. Let Ω Rn be an open set. Let (−Δ)s u1 ≥ 0 and (−Δ)s u2 ≥ 0 in Ω, such that u ≥ v in Rn \Ω. Then u = min {u1 , u2 } is a supersolution in Ω. The proof of the above results are based on a characterization of supersolutions somewhat similar to the super-meanvalue formula for superharmonic functions. s 1 Let ϕ (x) = cn,−σ |x|n−2s be the fundamental solution of (−Δ) . Let P = P (|x|) a paraboloid, tangent from below to ϕ along ∂B1 . Deﬁne Φ (x) = P (|x|) inside B1 , and Φ = ϕ outside B1 . Then Φ ∈ C 1,1 (Rn ). For λ > 1, set Φλ (x) =

x . Φ λn−2s λ 1

s

Note that Φλ1 ≥ Φλ1 if λ1 ≥ λ2 . It turns out that γλ (x) = (−Δ) Φλ (x) is a continuous and positive function in L1 (Rn ) and that is an approximation of the identity, in the sense that (u ∗ γλ ) (x) = u (y) γλ (x − y) dy → u (x) as λ → 0. Rn

In analogy with what happens for the Laplace operator we have: Proposition A5. If (−Δ)s u is continuous at a point x ∈ Rn , then s

(−Δ) u (x) = lim

λ→0

C [u (x) − (u ∗ γλ ) (x)] λn s

where C = C (s, n). In particular, (−Δ) u ≥ 0 in an open set Ω if and only if u is lower-semicontinuous and (u ∗ γλ ) (x) ≤ u (x)

in Ω.

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S. Salsa

Appendix B: The Operator La Definition and Preliminary Facts For y > 0 and −1 < a < 1, we deﬁne the operator La u = div (y a ∇u) . The equation La u = 0 can be written in the form a Δx u + uy + uyy = 0 y

(57)

that induces some analogy with partially symmetric harmonic functions. Indeed, for a integer, let u: Rn × R1+a be radially symmetric with respect to the y variable, that is u (x, y) = U (x, |y|). Letting |y| = r, the Laplace operator in the variables x and r is given by a Δu = Δx U + Ur + Urr r which is identical to the left hand-side of (57). If we allow non integer values of a then we may think to the operator La as a sort of Laplace operator in dimension n + 1 + a. Although this is clearly only an analogy, as we shall see later, the solutions of (57) share many properties with harmonic functions. • Fundamental solution. The ﬁrst issue of this analogy is given by the fundamental solution of La , given by Γa (X) = Cn,a

1 |X|

n+a−1

if n + a − 1 > 1, for an appropriate constant Cn,a . In fact, a direct computation gives that Γa is a solution of (57) for y = 0. Note that, for an appropriate constant Cn+a−1 , Γa (x, 0) is the fundamental solution of (−Δ)s , s = (1 − a)/2, so that s

lim y a ∂y Γa (X) = (−Δ) Γa (x, 0) = −δ0 .

y→0+

(58)

• Reflection. If u is a solution of La u = 0 in the upper half space and its “conormal derivative” limy→0+ y a ∂y u (x, y) vanishes in a ball BR ⊂ Rn then, reﬂecting u evenly with respect to y, we obtain a solution in the n + 1 dimensional ball BR . We ﬁrst consider the case in which u is smooth.

The Problems of the Obstacle in Lower Dimension

235

Lemma B1. Let u : Rn × [0, +∞) be a smooth solution of La u = 0 in the upper half space, such that lim y a ∂y u (x, y) = 0

y→0+ . Then uniformly in BR

u ˜ (x, y) =

u (x, y) u (x, −y)

y≥0 y<0

a

is a weak solution of div(|y| ∇u) = 0 in BR . Proof. We show that a

|y| ∇˜ u · ∇h dX = 0

(59)

BR

for every h ∈ C0∞ (BR ) . Let ε > 0 and write:

a

|y| ∇˜ u · ∇h dX = BR

BR \{|y|<ε}

···+

BR ∩{|y|<ε}

···

div (|y|a h∇˜ u) dX

= BR \{|y|<ε}

a

+ =−

BR ∩{|y|<ε}

BR ∩{|y|=ε}

|y| ∇˜ u · ∇h dX εa h (x, ε) u ˜y (x, ε) dx

a

+ BR ∩{|y|<ε}

|y| ∇˜ u · ∇h dX. a

u∈ When ε → 0, the integral over BR ∩ {|y| < ε} goes to zero, since |y| ∇˜ L1loc Rn+1 . The integral over BR ∩ {|y| = ε} goes to zero since εa h (x, ε) u ˜y (x, ε) → 0. If u is not smooth then the meaning of limy→0+ y a ∂y u (x, y) = 0 is given by (59). Precisely: Definition B1. We say that limy→0+ y a ∂y u (x, y) = 0 (resp. ≤ 0) in BR, in a weak sense, if |y|a ∇˜ u · ∇h dX = 0 BR

for every h ∈ C0∞ (BR ), (resp. h ≥ 0).

(60)

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S. Salsa

From the proof of Lemma B1, it follows that if limy→0+ y a ∂y u (x, y) = 0 in a weak sense, then lim y a uy (x, y) ϕ (x) dx = 0 ∀ϕ ∈ C0∞ (BR ). (61) y→0+

BR

On the other hand, if y a ∂y u (x, y) ∈ L1loc (Rn+1 ) and (61) holds then lim y a ∂y u (x, y) = 0

y→0+

in BR, in a weak sense. • La -harmonic extension and conjugate equation. Let f ∈ S and u : Rn × [0, +∞) → R be the solution of the following problem:

La u = 0 u (x, 0) = f (x)

in Rn × (0, +∞) in Rn

(62)

We call u the La -harmonic extension of f in the upper half space. The function w (x, y) = y a uy (x, y) is a solution of the conjugate equation

div y −a ∇w = 0.

(63)

Indeed, we have:

div y −a ∇w = Δx uy + ∂y y −a ∂y (y a uy ) = Δx uy + uyyy + a∂y (y −1 uy ) a = ∂y Δx u + uyy + uy = 0. y

Harnack Inequality, Liouville Theorem and Mean Value Property As we have already noted, the operator La is a particular case of the class of degenerate elliptic operators considered in [16]. For the following result see Theorems 2.3.8 and 2.3.12 in that paper. Theorem B2. Assume La v = f (X) in Br . Then: (1) (Harnack inequality) sup v ≤ C Br/2

inf v + r2 sup |f | ;

Br/2

Br

The Problems of the Obstacle in Lower Dimension

237

(2) v ∈ C 0,α Br/2 for some α ≤ 1 and if f = 0

v C 0,α (Br/2 ) ≤

C oscv. rα Br

Using the translational invariance of the equation in the x variable, we obtain the following result. Lemma B3 (Shauder type estimates). Assume La v = 0 in Br . Then, for every k ≥ 1, integer: " k " C "Dx v " ∞ ≤ k oscv L (Br/2 ) r Br and6

k Dx v

C 0,α (Br/2 )

≤

C

oscv.

rk+α Br

Using the above Theorem and the equation in the form a Δx v = −vyy − vy y we get: Lemma B4. Assume La v = 0 in Br . Then, for every r ≤ 1: " " " " C "vyy + a vy " ≤ 2 oscv. " y "L∞ (Br/2 ) r Br The next is a Liouville type result. Lemma B5. Let v a solution of La v (X) = 0 in Rn+1 . Assume that v (x, y) = v (x, −y)

and

|v (X)| ≤ C (1 + |X|γ ) , γ ≥ 0.

Then v is a polynomial. Proof. We use induction and the following elementary fact: if ∇x v is a polynomial and v (0, y) is a polynomial in y, then v is a polynomial. Assume γ ≤ 1. Taking r → ∞ in Lemma B4 we get a vyy + vy = 0 y 6 |w| C 0,α (D)

in Rn+1 .

denotes the seminorm sup

x,y∈D

|w (x) − w (y)| . |x − y|α

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S. Salsa

For x ﬁxed, the solution of this o.d.e. is given by a

v (x, y) = b (x) y |y| + c (x) . Since v is symmetric in y we have b = 0 and therefore v (x, y) = c (x). On the other hand, taking r → ∞ in Lemma B3 with k = 2, we get Dx2 v = 0 in Rn+1 . Therefore, for each ﬁxed y, v is a ﬁrst order polynomial, so that, if k ≤ 1, v (x, y) = mx + q. Assume now that if γ ≤ k − j, j = 1, ..., k − 1, then v is a polynomial of degree k − j. Let k − 1 < γ ≤ k. From Lemma B3, k = 1 we have γ−1 |∇x v (X)| ≤ C 1 + |X| . Moreover, every x-partial derivative is a global solution of the same equation, symmetric in y. Thus, by the inductive hypothesis, we infer that ∇x v is a polynomial of degree k − 1. From Lemma B4, vyy + a vy ≤ C 1 + |X|γ−2 . y Since

a vyy + vy = −Δx v y

and since Δx commutes with La , we have that vyy + ay vy satisﬁes the same equation as v. Therefore vyy + ay vy is a polynomial of degree k − 2. In particular, for x = 0, a vyy (0, y) + vy (0, y) = p (y) y where p is an even polynomial of degree at most k − 2. We can write p (y) = a0 + a2 y 2 + ... + a2d y 2d . Then, v (0, y) must be a

v (0, y)= c+by |y| +

a0 a0 a2d y2 + y 4 + ... + y 2d+2 . 2 (1+a) 4 (3+a) (2d+2) (2d+1+a)

Since v is even in y, we must have b = 0 and v (0, y) is a polynomial. Since ∇x v is a polynomial and v (0, y) is a polynomial in y, then v is a polynomial of degree k. Remark B1. Notice that the symmetry hypothesis is necessary. Namely −a v (x, y) = y |y| is a global solution.

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239

We now state a mean value property for supersolution of a nonhomogeneous solution: Lemma B6 (Mean value property). Let v be a solution of k

La v (X) ≤ C |y a | |X|

in B1 .

Then, for every r ≤ 1, v (0) ≥

1 ωn+a rn+a

a

v (X) |y| dS − Crk+2 ∂Br

where

|y|a dS.

ωn+a = ∂B1

Proof. The proof mimic the case of harmonic functions. Consider ﬁrst C = 0. Deﬁne % & 1 1 Γ (X) = Cn,a max n+a+1 − rn+a+1 , 0 |X| −1

−1 . Note that Γ is supported in Br , Γ ≥ 0 and where cn,a = (n + a + 1) ωn+a a direct calculation yields

La Γ (X) = −δ0 + μ −1 |y| dS. Thus where μ is the measure supported on ∂Br given by r−n−a ωn+a a

0≤

−La v (X) Γ (X) dX

Br

Br

a

−∇v (X) · ∇Γ (X) |y| dX

=

−v (X) · La Γ (X) dX

= Br

= v (0) −

1 ωn+a rn+a

a

v (X) |y| dS ∂Br

which proves the lemma when C = 0. In the case C > 0, we apply the above computation to k+2

w (X) = v (X) − since La w (X) ≤ 0.

C |X| (k + 2) (k + n + a + 1)

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S. Salsa

Poincar´ e Inequalities Poincar´e type inequalities in the context of weighted Sobolev spaces can also be found in [16]. In our case, letting 1 a v¯ = v (X) |y| dσ, ωn+a rn+a ∂Br we have7 : a Lemma B7 (Poincar´e inequalities). Let v ∈ W 1,2 (B1 , |y| ) . Then, for r≤1 : a a 2 (v (X) − v¯) |y| dσ ≤ C (a, n) r |∇v (X)| |y| dX. ∂Br

and

Br

(v (X) − v (rX))2 |y| dσ ≤ C (a, n, r)

a

a

|∇v (X)| |y| dX.

∂B1

(64)

B1

The ﬁrst inequality is standard, The second one can be proved by integrating ∇v along the lines sX with s ∈ (r, 1) .

Appendix C: Relation Between (−Δ)s and La Let f : Rn → R and u : Rn × [0, +∞) → R be the La -harmonic extension of f in the upper half space that is the solution of the following problem:

in Rn × (0, +∞) in Rn

La u = 0 u (x, 0) = f (x)

vanishing at inﬁnity. Lemma C1. Let f ∈ S and u be its La -harmonic extension and 2s = 1 − a. Then, for a suitable constant κa , a

κa lim y uy (x, y) = (−Δ) f (x) = y→0+

f (y) − f (x)

s

Rn

|y − x|n+2s

dy

Proof. We use the Fourier transform in x. Equation (57) transforms into a 2 ˆy (ξ, y) + uˆyy (ξ, y) = 0 ˆ (ξ, y) + u − |ξ| u y function v ∈ W 1,2 (B1 , |y|a ) has a trace in L2 (∂B1 , |y|a ) and the trace operator is compact.

7A

The Problems of the Obstacle in Lower Dimension

241

with u ˆ (ξ, 0) = fˆ (ξ) . We claim that u ˆ (ξ, y) = fˆ (ξ) φ (|ξ| y) where φ : [0, +∞) → R is the minimizer of the functional

+∞

J (v) =

2 y a (v ) + v 2 dy

0

over the class of functions v ∈ H 1 (0, +∞), with v (0) = 1. Then φ is the solution of the following problem ⎧ a ⎪ y>0 ⎨ φ (y) + y φ (y) − φ (y) = 0 φ (0) = 1 ⎪ ⎩ φ (+∞) = 0 The scaling y −→ |ξ| y shows the claim. Then (X = (x, y))

2

y a |∇u| dX =

Rn ×[0,+∞)

Rn 0

=

Rn 0

2 2 2 y a |ξ| |ˆ u| +|ˆ uy | dydξ

+∞

2 2 2 2 y a |ξ| fˆ (ξ) |φ (|ξ| y)| +|φ (|ξ| y)| dydξ

+∞

ˆ 2 1−a (|ξ| y = z) = f (ξ) |ξ| Rn

2 2 z a |φ (z)| +|φ (z)| dzdξ

+∞ 0

ˆ 2 2s ˆ 2 2s = J (φ) f (ξ) |ξ| dξ ≡ κa f (ξ) |ξ| dξ. Rn

Rn

We conclude that the Euler equations of the two functionals ˆ 2 2s 2 y a |∇u| dX and f → κa u→ f (ξ) |ξ| dξ Rn ×[0,+∞)

Rn

coincide, that is, for every test function h : Rn × [0, +∞) → R with compact support: Rn

0

+∞

y a ∇u · ∇h dxdy = κa

Rn

2s ˆ (ξ, 0)dξ |ξ| fˆ (ξ) h

= κa

s

h (x, 0) (−Δ) u (x) dx. Rn

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S. Salsa

Since div(y a ∇u) = 0, Rn

0

+∞

y a ∇u · ∇h dxdy = −κa

Rn

h (x, 0) lim y a uy (x, y) dx y→0+

from which −κa limy→0+ y a uy (x, y) = (−Δ)s f (x) .

• Poisson formula. To ﬁnd an explicit formula for u, notice that, form the previous paragraphs, the function P (x, y) = −Cn,a y −a ∂y Γ−a (x, y) solves div(y a ∇P ) = 0 for y > 0 and, for an appropriate constant Cn,a , P (x, y) → δ0 as y → 0+. Thus y 1−a P (x, y) = Cn,a (n−a+1)/2 2 |x| + y 2 must be the Poisson kernel for La . Therefore, the solution of (62) is given by u (x, y) = P (x − z, y) f (z) dz. Rn

Remark C1. An alternative way to see that −κa limy→0+ y a uy (x, y) is 1−a y a realization of (−Δ)s is to make the change of variable z = 1−a representation which reduces La u = 0 to Δx u ˜ + z αu ˜zz = 0 and limy→0+ y a uy (x, y) to u˜z (x, 0) for α = −2a/ (1 − a). The Poisson kernel becomes z P˜ (x, z) = Cn,a (n−a+1)/2 2 2 |x| + (1 − a) z 2/(2−a)

and u ˜ (x, z) =

Rn

P˜ (x − z, y) f (z) dz.

Using directly this representation, we have 1 P˜ (x − z, y) [f (z) − f (x)] dz u ˜z (x, 0) = lim z→0+ z Rn f (y) − f (x) s = cn,a p.v. n+2s dy = − (−Δ) f (x) . n |y − x| R

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243

Comments and Further Reading Standard references for the properties of the fractional Laplacian are [21] and [47]. Appendix A is taken from [24], to which we refer for the proof of Propositions A1–A5. Appendix B1 follows [10], while Appendices B2 and B3 are taken from [11]. Appendix C is taken from [10]. In these paper a connection between the operator La u and the operator z α Δx u + uzz , where α = 2a/ (1 − a), is also shown. Operators of this type are known as Baouendi–Grushin operators. For α > 0, frequency formulas for these operators appeared before in [17] in the context of unique continuation.

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17. N. Garofalo, Unique continuation for a class of elliptic operators which degenerate on a manifold of arbitrary codimension. J. Diﬀ. Eq. 104(1), 117–146 (1993) 18. N. Garofalo, A. Petrosian, Some new monotonicity formulas and the singular set in the lower dimensional obstacle problem, Inv. Math. 177, 415–464 (2009) 19. N. Guillen, Optimal regularity for the Signorini problem, Calculus of Variations and Partial Diﬀerential Equations, 36(4), 533–546 (2009) 20. T. Kilpel¨ ainen, Smooth approximation in weighted Sobolev spaces. Commentat. Math. Univ. Carol., 38(1), 29–35 (1997) 21. N.S. Landkof, Foundation of Modern Potential Theory, Die Grundlehren der Mathematischen Wissenschaften, Band 180. Springer-Verlag, New York-Heidelberg, 1972 22. R. Monneau, On the number of singularities for the obstacle problem in two dimensions, J. Geom. Anal. 13(2), 359–389 (2003) 23. E. Milakis, L. Silvestre, Regularity for the non linear Signorini problem, Adv. Math. 217, 1301–1312 (2008) 24. L. Silvestre, Regularity of the obstacle problem for a fractional power of the Laplace operator, C.P.A.M. 60(1), 67–112 (2007) 25. G. Weiss A homogeneity improvement approach to the obstacle problem, Inv. Math 138(1), 23–50 (1999) 26. H. Whitney, Analytic extensions of diﬀerentiable functions deﬁned in closed sets, Trans. A.M.S. 36(1), 63–89 (1934) 27. W.P. Ziemer, Weakly diﬀerentiable functions, Graduate Text in Math, Springer-Verlag New York, 1989

List of Participants

1. Argiolas Roberto [email protected] 2. Avelin Benny [email protected] 3. Bianchini Chiara [email protected] 4. Bianchini Massimiliano [email protected] 5. Bonfanti Giovanni [email protected] 6. Bonforte Matteo [email protected] 7. Candito Pasquale [email protected] 8. Cellina Arrigo [email protected] 9. Charro Caballero Fernando [email protected] 10. Chiricotto Maria [email protected] 11. D’Ambrosio Roberta [email protected] 12. D’ Ambrosio Lorenzo [email protected] 13. DiBenedetto Emmanuele (lecturer) [email protected] 14. Di Castro Agnese [email protected] 15. Esposito Pierpaolo [email protected] J. Lewis et al., Regularity Estimates for Nonlinear Elliptic and Parabolic Problems, Lecture Notes in Mathematics 2045, DOI 10.1007/978-3-642-27145-8, © Springer-Verlag Berlin Heidelberg 2012

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16. Ferrari Fausto [email protected] 17. Fimiani Michele [email protected] 18. Floridia Giuseppe [email protected] 19. Fornaro Simona [email protected] 20. Fugazzola Andrea [email protected] 21. Gianazza Ugo (editor) [email protected] 22. Grillo Gabriele [email protected] 23. Iagar Razvan Gabriel [email protected] 24. Kokocki Piotr [email protected] 25. Laleoglu Rojbin Ozlem [email protected] 26. Leonori Tommaso [email protected] 27. Lewis John (editor) [email protected] 28. Lindqvist Peter (lecturer) [email protected] 29. Livrea Roberto [email protected] 30. Lundstrom Niklas [email protected] 31. Manfredi Juan (lecturer) [email protected] 32. Mannucci Paola [email protected] 33. Marcellini Paolo [email protected] 34. Mascolo Elvira [email protected] 35. Masson Mathias [email protected] 36. Mazzola Marco [email protected] 37. Pilarczyk Dominika [email protected]

List of Participants

List of Participants

38. Piro Grimaldi Anna [email protected] 39. Salsa Sandro (lecturer) [email protected] 40. Siljander Juhana [email protected] 41. Sirakov Boyan [email protected] 42. Stinga Pablo Ra` ul [email protected] 43. Surnachev Mikhail [email protected] 44. Terracini Susanna [email protected] 45. Urbano Jos Miguel [email protected] 46. Vazquez Juan Luis [email protected] 47. Verzini Gianmaria [email protected] 48. Vespri Vincenzo [email protected]

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