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The Divergence Theorem and Sets of Finite Perimeter
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The Divergence Theorem and Sets of Finite Perimeter
Washek F. Pfeffer University of California, Davis University of Arizona, Tucson USA
© 2012 by Taylor & Francis Group, LLC K14657_FM.indd 5
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CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2012 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Version Date: 20120312 International Standard Book Number-13: 978-1-4665-0721-0 (eBook - PDF) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com
© 2012 by Taylor & Francis Group, LLC
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To Lida for her love and a lifetime of companionship
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“Book˙2011” — 2012/2/26 — 9:58 — page ix — #3
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Contents
Preface Part 1.
xiii Dyadic figures
1
1.
Preliminaries 1.1. The setting 1.2. Topology 1.3. Measures 1.4. Hausdorff measures 1.5. Differentiable and Lipschitz maps
3 3 7 9 13 16
2.
Divergence theorem for dyadic figures 2.1. Differentiable vector fields 2.2. Dyadic partitions 2.3. Admissible maps 2.4. Convergence of dyadic figures
21 21 24 27 31
3.
Removable singularities 3.1. Distributions 3.2. Differential equations 3.3. Holomorphic functions 3.4. Harmonic functions 3.5. The minimal surface equation 3.6. Injective limits
35 35 37 38 39 39 41
Part 2. 4.
Sets of finite perimeter
47
Perimeter 4.1. Measure-theoretic concepts 4.2. Essential boundary 4.3. Vitali’s covering theorem 4.4. Density 4.5. Definition of perimeter
49 49 51 53 54 56 ix
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x
5.
BV 5.1. 5.2. 5.3. 5.4. 5.5. 5.6. 5.7. 5.8. 5.9. 5.10.
6.
Line sections Lipeomorphisms functions Variation Mollification Vector valued measures Weak convergence Properties of BV functions Approximation theorem Coarea theorem Bounded convex domains Inequalities Lipschitz maps
Locally BV sets 6.1. Dimension one 6.2. Besicovitch’s covering theorem 6.3. The reduced boundary 6.4. Blow-up 6.5. Perimeter and variation 6.6. Properties of BV sets 6.7. Approximating by figures
Part 3.
The divergence theorem
58 68 73 73 76 79 86 92 98 101 106 110 117 121 121 123 126 131 137 142 146 149
7.
Bounded vector fields 7.1. Approximating from inside 7.2. Relative derivatives 7.3. The critical interior 7.4. The divergence theorem 7.5. Lipschitz domains 7.6. BV vector fields
151 151 155 158 160 166 179
8.
Unbounded vector fields 8.1. Minkowski contents 8.2. Controlled vector fields 8.3. Integration by parts
181 181 185 190
9.
Mean divergence 9.1. The derivative 9.2. The critical variation
193 193 197
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Contents 4.6. 4.7.
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Contents
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xi
10. Charges 10.1. Continuous vector fields 10.2. Localized topology 10.3. Locally convex spaces 10.4. Duality 10.5. The space BVc (Ω) 10.6. Streams
205 205 207 209 212 213 216
11. The divergence equation 11.1. Background 11.2. Solutions in Lp (Ω; Rn ) 11.3. Continuous solutions
219 219 221 224
Bibliography
231
List of symbols
235
Index
237
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Washek F. Pfeffer is Professor Emeritus of Mathematics at the University of California in Davis. He was born in Prague, Czech Republic, where he studied mathematics at Charles University (1955–60). He immigrated to the United States in 1965, and in 1966 received his Ph.D. from the University of Maryland in College Park. Dr. Pfeffer has worked at the Czechoslovak Academy of Sciences in Prague, and has taught at the Royal Institute of Technology in Stockholm, George Washington University, University of California in Berkeley, University of Ghana in Accra, and King Fahd University in Dhahran, Saudi Arabia. In 1994–95 he was a Fulbright Lecturer at Charles University. His primary research areas are analysis and topology. Dr. Pfeffer is a member of the American and Swedish Mathematical Societies, and an honorary member of the Academic Board of the Center for Theoretical Study at Charles University. Presently, he is a Research Associate in the Mathematics Department of the University of Arizona. He has written the books Integrals and Measures (Marcel Dekker, 1977), The Riemann Approach to Integration, and Derivation and Integration (Cambridge University Press, 1993 and 2001).
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“Book˙2011” — 2012/2/26 — 9:58 — page xiii — #7
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Preface
The divergence theorem and the resulting integration by parts formula belong to the most frequently used tools of mathematical analysis. In its elementary form, that is for smooth vector fields defined in a neighborhood of some simple geometric object such as rectangle, cylinder, ball, etc., the divergence theorem is presented in many calculus books. Its proof is obtained by a simple application of the one-dimensional fundamental theorem of calculus and iterated Riemann integration. Appreciable difficulties arise when we consider a more general situation. Employing the Lebesgue integral is essential, but it is only the first step in a long struggle. We divide the problem into three parts. (1) Extending the family of vector fields for which the divergence theorem holds on simple sets. (2) Extending the family of sets for which the divergence theorem holds for Lipschitz vector fields. (3) Proving the divergence theorem when the vector fields and sets are extended simultaneously. Of these problems, part (2) is unquestionably the most complicated. While many mathematicians contributed to it, the Italian school represented by Caccioppoli, De Giorgi, and others obtained a complete solution by defining the sets of bounded variation (BV sets). A major contribution to part (3) is due to Federer, who proved the divergence theorem for BV sets and Lipschitz vector fields. While parts (1)–(3) can be combined, treating them separately illuminates the exposition. We begin with sets that are locally simple — finite unions of dyadic cubes, called dyadic figures. Combining ideas of Henstock and McShane with a combinatorial argument of Jurkat, we establish the divergence theorem for very general vector fields defined on dyadic figures. The proof involves only basic properties of the Lebesgue integral and Hausdorff measures. An easy corollary of the divergence theorem is a powerful integration by parts formula. It yields results on removable sets for the Cauchy-Riemann, Laplace, and minimal surface equations. The next goal is to move from dyadic figures to BV sets. To enhance the intuition, our starting point is the geometric definition of perimeter. The perimeter of a set is the codimension one Hausdorff measure of its essential xiii
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xiv
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Preface
boundary. Several properties of sets with finite perimeter are derived directly from the definition. Deeper results rest on the equivalent analytic definition. Following the standard presentation, we say that an integrable function has bounded variation, or is a BV function, if its distributional gradient is a vector-valued measure of finite variation. The variation of a BV function is defined as the variation of its distributional gradient. A set whose indicator is a BV function is called a BV set. The variation of a BV set is the variation of its indicator. Although we are mainly interested in BV sets, it is neither possible nor desirable to separate them from BV functions. It is often easier to prove a result about BV functions first, and state the corresponding result about BV sets as a corollary. The link between BV sets and BV functions is the coarea formula, which connects the variation of a function with that of its level sets. Our objective is to show the equivalence of the geometric and analytic definitions by equating the perimeter of a measurable set with its variation. A variety of useful results concerning BV sets follows from the interplay between the two definitions. Throughout, we derive properties of BV functions directly from the definition, without referring to corresponding properties of Sobolev spaces. Sobolev spaces are not discussed in this text, and no a priori knowledge about them is required. Once the BV sets are defined and their main properties established, it is relatively easy to apply the divergence theorem we proved for dyadic figures to BV sets. The main tool, due to Giacomelli and Tamanini, consists of approximating arbitrary BV sets by their BV subsets with special properties. At the end, we extend the divergence theorem to a family of unbounded vector fields with controlled growth. We pay particular attention to continuous vector fields and their weak divergence. Elaborating on ideas of Bourgain and Brezis, we characterize the distributions F for which the divergence equation div v = F has a continuous weak solution — a recent joint work of T. De Pauw and the author. All of our results and proofs rely entirely on the Lebesgue integral. No exotic integrals, akin to the generalized Riemann integral of Henstock and Kurzweil, are involved. Notwithstanding, some techniques we use are inspired by investigations of these integrals. We strove to give complete and detailed proofs of all our claims. Only a few standard facts are quoted without proofs, in which case we always provide precise references. The book has three parts, roughly corresponding to parts (1)–(3) listed above. We trust that the titles of the chapters and sections are sufficiently descriptive. Results and comments we consider marginal are presented in small print. However, marginal does not mean unimportant; a useful enhancement of the main text can be found in the small print.
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Preface
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xv
The first two chapters, which deal with dyadic figures, are quite elementary. Except for very basic properties of Hausdorff measures, they should be accessible to the beginning graduate students. The rest of the book presupposes the knowledge equivalent to the first year graduate course in analysis. In addition, some familiarity with Hausdorff measures and distributions is expected. Rudimentary results from functional analysis are employed in the last two chapters. Our presentation owes much to the excellent textbooks [29, 75] and monographs [1, 33], which can serve as useful references. During the preparation of this text I largely benefited from discussions with L. Ambrosio, P. Bouafia, G.D. Chakerian, T. De Pauw, D.B. Fuchs, ˇ N. Fusco, R.J. Gardner, G. Gruenhage, Z. Nashed, M. Silhav´ y, S. Solecki, V. Sverak, B.S. Thomson, and M. Torres. I am obliged to W.G. McCallum who offered me a position of Research Associate in the Mathematics Department of the University of Arizona; it gave me access to university facilities, in particular to the university library. Editorial help provided by the publisher was invaluable. In this regard my thanks belong to K. Craig, M. Dimont, S. Kumar, S. Morkert, and R. Stern. W.F.P. Tucson, Arizona February 2012
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Part 1
Dyadic figures
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Chapter 1 Preliminaries
We establish the notation and terminology, and present some basic facts that will be used throughout the book. Several well-known theorems are stated without proofs; however, those results for which we found no convenient references are proved in detail. In general, the reader is expected to have some prior knowledge of the concepts introduced in this chapter.
1.1. The setting The sets of all integers and of all positive integers are denoted by Z and N, respectively. Symbols Q, R, and C denote, respectively, the sets of all rational, real, and complex numbers. The sets of all positive real numbers and of all positive rationals numbers are denoted by R+ and Q+ , respectively. Unless specified otherwise, by a number we always mean a real number. Elements of R := R ∪ {±∞} are called the extended real numbers. In R we consider the usual order and topology, and define the following algebraic operations: a + ∞ : = +∞ + a := +∞ for a > −∞,
a − ∞ : = −∞ + a := −∞ for a < +∞, ±∞ if a > 0, a · (±∞) : = ∓∞ if a < 0, 0 if a = 0.
At various places we write P := Q instead of P = Q to stress the fact that P is defined as equal to Q. Throughout, the symbol ∞ stands for +∞. Unless specified otherwise, ε → 0 means ε → 0+. Finite and countably infinite sets are called countable. We say that a � family E of sets covers a set E, or is a cover of E, if E ⊂ E. For any pair of sets A and B, the set A � B := (A − B) ∪ (B − A) = A ∪ B − A ∩ B is called the symmetric difference of A and B. 3
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4
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1. Preliminaries
By a function we always mean an extended real-valued function. A finite function is real-valued. If a function f equals identically to c ∈ R, we write f ≡ c. When no confusion can arise, the same symbol denotes a function f defined on a set A and its restriction f � B to a set B ⊂ A. For a function f defined on a set E and t ∈ R, we let � � {f > t} := x ∈ E : f (x) > t , and define the sets {f ≥ t}, {f < t}, etc. similarly. The set {f = 0} is called the null set of f . Further, we let f + := max{f, 0} and
f − := max{−f, 0},
and note that f = f + − f − and |f | = f + + f − . The value of f at x ∈ E is denoted interchangeably by f (x), f [x], and �f, x�. For m ∈ N, and x := (ξ1 , . . . , ξm ) and y := (η1 , . . . , ηm ) in Rm , x · y :=
m �
ξi ηi
and
i=1
|x| :=
√
x · x.
In Rm we use exclusively the Euclidean metric induced by the norm |x|. The diameter, closure, interior, and boundary of a set E ⊂ Rm are denoted by d(E), cl E, int E, and ∂E, respectively. The distance between sets A, B ⊂ Rm is denoted by dist(A, B), or dist(x, B) if A = {x} is a singleton. Given E ⊂ Rm and r ∈ R+ , we let � � U (E, r) : = x ∈ Rm : dist(x, E) < r , � � B(E, r) : = x ∈ Rm : dist(x, E) ≤ r . If E = {x} is a singleton, the sets � � U (x, r) := U {x}, r
and
� � B(x, r) := B {x}, r
are, respectively, the open and closed ball in Rm of radius r centered at x. For a pair of sets A, B ⊂ Rm , the symbol A � B indicates that cl A is a compact subset of int B. Given E ⊂ Rm and s ∈ N, we denote by C(E; Rs ) the linear space of all continuous maps φ : E → Rs . We let C(E) := C(E; R), and note that according to this definition, all continuous function are real-valued. The Urysohn function associated with a pair A, B of closed disjoint subsets of Rm is a function uA,B ∈ C(Rm ) defined by the formula uA,B (x) :=
dist (x, B) , dist (x, A) + dist (x, B)
x ∈ Rm .
(1.1.1)
Theorem 1.1.1 (Tietze). Let C ⊂ Rm be a closed set. Each continuous map φ : C → Rs has a continuous extension ψ : Rm → Rs .
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1.1. The setting
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5
Proof. As φ has a continuous extension if and only if each of its coordinates does, it suffices to show that every continuous function f : C → R has a continuous extension g : Rm → R. Now R is homeomorphic to the open interval (−1, 1), e.g., via the strictly increasing continuous function ϕ : x �→ 2 −1 x : R → (−1, 1). Hence we may assume that f : C → (−1, 1). Let π tan uC1− ,C1+ be the Urysohn function associated with � � C1± := x ∈ C : ±f (x) ≥ 1/3 .
If g1 := 1/3 − (2/3)uC1− ,C1+ , then � � �g1 (x)� ≤ 3−1 for all x ∈ Rm , � � �f (x) − g1 (x)� ≤ 3−1 for all x ∈ C.
Next let f1 = f −g1 , and let uC2− ,C2+ be the Urysohn function associated with � � C2± := x ∈ C : ±f1 (x) ≥ 1/32 . If g2 := 1/32 − (2/32 )uC2− ,C2+ , then � � �g2 (x)� ≤ 3−2 for all x ∈ Rm , � � �f (x) − g1 (x) − g2 (x)� ≤ 3−2 for all x ∈ C.
m k Proceeding by recursion, k| ≤ 3 � � we kdefine functions gk ∈ C(R ) such that |g � � k ∞ and �f − j=1 (gj � C)� ≤ 3 for k = 1, 2, . . . . It is clear that g := k=1 gk belongs to C(Rm ) and extends f .
Corollary 1.1.2. Let Ω ⊂ Rm be an open set, and let C ⊂ Ω be a closed set. Each continuous map φ : C → Rs has a continuous extension θ : Rm → Rs such that cl {θ �= 0} ⊂ Ω.
Proof. By Titze’s theorem φ has a continuous extension ψ : C → Rs . Let f = uC,Rm −Ω be the Urysohn function associated with C and Rm − Ω, and let D = {f ≤ 1/2}. Note C ⊂ Rm − D ⊂ cl (Rm − D) ⊂ Ω. If uC,D is the Urysohn function associated with C and D, then θ = uC,D ψ is the desired extension. IfΩ ⊂ Rm is an open set and k ∈ N, then C k (Ω; Rs ) denotes the linear space of all maps φ = (f1 , . . . , fs ) from Ω to Rs such that each fi : Ω → R has continuous partial derivatives of orders less than or equal to k. We let ∞
s
C (Ω; R ) =
∞ �
C k (Ω; Rs ),
k=1
and refer to elements of C (Ω; R ) and C ∞ (Ω; Rs ), respectively, as C k and C ∞ maps from Ω to Rs . Instead of C(E; R), C k (Ω; R), and C ∞ (Ω; R), we write C(E), C k (Ω), and C ∞ (Ω), respectively. The elements of C k (Ω) and C ∞ (Ω) are called, respectively, the C k and C ∞ functions defined on Ω. k
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1. Preliminaries
Let E ⊂ Rm and φ : E → Rs . The indicator χE of E and the zero extension φ of φ are defined by the formulae � � 1 if x ∈ E, φ(x) if x ∈ E, χE (x) := φ(x) := m 0 if x ∈ R − E, 0 if x ∈ Rm − E, respectively. The support of φ is the set spt φ := cl {φ �= 0}. LetΩ ⊂ Rm be an open set. The linear space of all φ ∈ C(Ω; Rs ) with spt φ � Ω is denoted by Cc (Ω; Rs ); the spaces Cc (Ω), Cck (Ω; Rs ), Cc∞ (Ω; Rs ), etc., are defined similarly. We always identify φ ∈ Cc (Ω; Rs ) with its zero extension φ ∈ Cc (Rm ; Rs ). This simple convention, which will cause no confusion, legitimizes the inclusions Cc (U ; Rs ) ⊂ Cc (Ω; Rs ) ⊂ Cc (Rm ; Rs ),
Cck (U ; Rs ) ⊂ Cck (Ω; Rs ) ⊂ Cck (Rm ; Rs )
(1.1.2)
where U ⊂ Ω is an open set and k = 1, 2, . . . , ∞. Throughout this book, the ambient space is Rn where n ≥ 1 is a fixed integer. By {e1 , . . . , en } we denote the standard base in Rn , i.e., ei := (0, . . . ,
1
, . . . , 0),
i-th place
i = 1, . . . , n.
The projection in the direction of ei is the linear map � (x · ej )ej : Rn → Rn . πi : x �→ j�=i
As the setΠ i := πi (R ) is a linear subspace of Rn with bases n
e1 , . . . , ei−1 , ei+1 , . . . , en , it is isometric to Rn−1 . For each x ∈ Πi , the set πi−1 (x) is isometric to R. Thus whenever convenient, we tacitly identify the spaceΠ i with Rn−1 , and the set πi−1 (x) with R. Note that if n = 1, then π1 (x) = 0 for each x ∈ R1 , and hence π1 (R1 ) = {0}. �n A cell in Rn is the set A := j=1 [aj , bj ] where aj < bj are real numbers. If b1 − a1 = · · · = bn − an , the cell A is called a cube. A figure is a finite, possibly empty, union of cells. A k-cube is a cube n � �
j=1
ij 2−k , (ij + 1)2−k
�
where k and i1 , . . . , in are integers. The family of all k-cubes is denoted by � Dk , and the elements of the union DC := k∈Z Dk are called dyadic cubes. A dyadic figure is a finite, possibly empty, union of dyadic cubes. The family of all dyadic figures in Rn is denoted by DF. At places we employ unspecified positive constants depending on certain parameters, such as the dimension n. If κ is a constant depending only on
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1.2. Topology
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7
parameters p1 , . . . , pk , we write κ = κ(p1 , . . . , pk ). With a few exceptions, we use no universal constants. Symbols denoting constants are tied to the context: distinct constants appearing in different contexts are often denoted by the same symbol.
1.2. Topology All topologies considered in this book are assumed to be Hausdorff. If T and S are topologies in a set X and T ⊂ S, we say that S is larger than T, or equivalently that T is smaller than S. In a topological space (X, T), the closure of E ⊂ X is denoted by cl T E. Unless specified otherwise, each Y ⊂ X is given the subspace topology. Let X be a topological space. A set E ⊂ X is a Gδ set if it is the intersection of countably many open subsets of X. Borel sets in X are elements of the smallest σ-algebra in X containing all open subsets of X. A map φ from X to a topological space Y is called Borel measurable, or merely Borel , if � � φ−1 (B) := x ∈ X : φ(x) ∈ B is a Borel subset of X for every Borel set B ⊂ Y . A subset E of a topological space X is called sequentially closed if each sequence {xk } in E that converges in X converges to x ∈ E. Each closed subset of X is sequentially closed but not vice versa; see Example 1.2.1 below. If the converse is true, i.e., if every sequentially closed set E ⊂ X is closed, the space X is called sequential . All closed and all open subsets of a sequential space are sequential. A map φ from a sequential space X to any topological space Y is continuous whenever lim φ(xk ) = φ(lim xk ) for every convergent sequence {xk } in X. Each first countable space is sequential, but the converse is false; see Example 10.3.4. Example 1.2.1. Let ω1 be the first uncountable ordinal, and let X be the space of all ordinals smaller than or equal to ω1 equipped with the order topology. The set E = X − {ω1 } is sequentially closed but not closed. Unless specified otherwise, a linear space is a linear space over R. Let X be a linear space. The zero element of X is denoted by 0, and A + B := {x + y : x ∈ A and y ∈ B}
for A, B ⊂ X and t ∈ R. As usual −A := (−1)A
and
tA := {tx : x ∈ A}
and x + A := {x} + A
for x ∈ X. A set C ⊂ X is called
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8
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1. Preliminaries � • absorbing if X = {tC : t ∈ R}, • symmetric if −C = C, • convex if tC + (1 − t)C ⊂ C for each 0 ≤ t ≤ 1.
The linear hull , or convex hull , of a set E ⊂ X is, respectively, the intersection of all linear subspaces of X containing E, or the intersection of all convex subsets of X containing E. A topology in X for which the maps (x, y) �→ x + y : X × X → X
and
(t, x) �→ tx : R × X → X
are continuous is called linear . Since each linear topology is induced by a uniformity [28, Example 8.1.17], all topological linear spaces are completely regular [28, Theorem 8.1.21]. A locally convex topology is a linear topology that has a neighborhood base at zero consisting of convex sets. A linear space equipped with a linear, or locally convex, topology is called a topological linear space, or a locally convex space, respectively. In this book we encounter only locally convex spaces. A seminorm in X is a functional p : X → R such that p(x + y) ≤ p(x) + p(y)
and
p(tx) = |t|p(x)
for all x, y ∈ X and each t ∈ R. Observe that p(0) = 0 ≤ p(x) for each x ∈ X. A norm in X is a seminorm p such that p(x) = 0 implies x = 0. A family P of seminorms is called separating if p(x) = 0 for all p ∈ P implies x = 0. A separating family P defines a locally convex topology in X; the neighborhood base at zero is given by convex symmetric sets � � � � Up1 ,...,pk ;ε := x ∈ X : max p1 (x), . . . , pk (x) < ε
where p1 , . . . , pk are in P and ε > 0. Conversely, each locally convex topology in X is induced by a separating family P of seminorms [64, Remark 1.38, (b)]. The separating property of P guarantees that the topology defined by P is Hausdorff. A locally convex topology induced by a countable separating family {pk : k ∈ N} of seminorms is metrizable; for instance, by the metric ρ(x, y) :=
∞ �
k=1
2−k
pk (x − y) . 1 + pk (x − y)
A Fr´echet space is a completely metrizable locally convex space. Even if a topology in X is defined by an uncountable family of seminorms, there may exist another family of seminorms in X that is countable and defines the same topology. The next example illustrates the situation. Example 1.2.2. LetΩ ⊂ Rm be an open set. The topology T of locally uniform convergence in C(Ω; Rs ) is defined by the seminorms � � pK (φ) := sup �φ(x)� x∈K
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9
where K ⊂ Ω is a compact set and φ ∈ C(Ω; Rs ). Each x ∈ Ω ∩ Qm has an open neighborhood Ux � Ω. Organize {Ux : x ∈ Ω ∩ Qm } into a sequence �j U1 , U2 , . . . , and let Vj = i=1 Ui . Since Vj � Ω, and since each compact set K ⊂ Ω is contained in some Vj , it is clear that T is defined by the seminorms � � qj (φ) := sup �φ(x)�, j = 1, 2 . . . . x∈Vj
It follows that T is the metrizable � topology of� uniform convergence on the sets Vj . Thus T is complete, and C(Ω; Rm ), T is a Fr´echet space.
Let (X, T) be a locally convex space. A set E ⊂ X is called bounded if for each convex neighborhood U of zero there is t > 0 such that E ⊂ tU . Every compact set E ⊂ X is bounded. In general, E ⊂ X is bounded if and only if lim tk xk = 0 whenever {xk } is a sequence in E and {tk } is a sequence in R converging to zero [64, Theorem 1.30]. If the topology T is induced by a family P of seminorms, then E ⊂ X is bounded if and only if for each p ∈ P, � � sup p(x) : x ∈ E < ∞.
The dual space of X, abreviated as the dual of X, is the linear space X ∗ of all continuous linear functionals x∗ : X → R. 1 To begin with, X ∗ is just a linear space with no topology. However, two locally convex topologies in X ∗ are easy to introduce: • the weak* topology W∗ defined by seminorms � � x∗ �→ ��x∗ , x�� : X ∗ → R
where x ∈ X; • the strong topology S∗ defined by seminorms �� � � �x∗ �B := sup ��x∗ , x�� : x ∈ B where x∗ ∈ X ∗ and B ⊂ X is a bounded set.
Since each singleton {x} ⊂ X is a bounded set, the weak* topology is smaller than the strong topology.
1.3. Measures A measure 2 in an arbitrary set X is a function µ defined on all subsets of X that satisfies the following conditions: 1 A notable exception to the notation X ∗ is the space D� of distributions defined in Section 3.1 below. 2 Our concept of measure is often called “outer measure”, and the term “measure” is reserved for the restriction of “outer measure” to the family of all measurable sets. For our purposes, such distinction is superfluous.
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1. Preliminaries
(i) µ(∅) = 0; (ii) µ(B) ≤ µ(A) whenever B ⊂ A ⊂ X; �∞ �∞ (iii) µ( k=1 Ak ) ≤ k=1 µ(Ak ) whenever Ak ⊂ X for k = 1, 2, . . . .
Throughout this section, µ is a measure in a set X ⊂ Rm . The reduction of µ to a set Y ⊂ X is a measure µ Y in X defined by (µ
Y )(A) := µ(A ∩ Y )
for each A ⊂ X. If µ = µ Y , we say that µ lives in Y . A set E ⊂ X is called µ measurable whenever µ=µ
E+µ
(X − E).
The support of µ is the set �� � spt µ := X − U ⊂ X : U is open in X and µ(U ) = 0 .
Since each subset of Rm has the Lindel¨of property [28, Section 3.8 and Corollary 4.1.16], we have µ(X − spt µ) = 0. In accordance with the standard terminology, the measure µ is called �∞ • σ-finite if X = k=1 Ek and µ(Ek ) < ∞ for k = 1, 2, . . . , • Borel if each relatively Borel subset of X is µ measurable, • Borel regular if µ is a Borel measure and each E ⊂ X is contained in a relatively Borel subset B of X such that µ(B) = µ(E), • Radon if µ is a Borel regular measure and µ(K) < ∞ for each compact set K ⊂ X, • metric if µ(A ∪ B) ≥ µ(A) + µ(B) for each pair A, B ⊂ X such that dist (A, B) > 0.
A set E ⊂ X is called µ σ-finite if the reduced measure µ E is σ-finite. The next two theorems are proved in [29, Sections 1.1 and 1.9]. Theorem 1.3.1. Let Ω ⊂ Rm be an open set. Each metric measure in Ω is a Borel measure. If µ is a Borel regular measure in Ω, then µ E is a Radon measure for each µ measurable set E ⊂ Ω with µ(E) < ∞. If µ is a Radon measure in Ω, then the following conditions hold: (1) For each set A ⊂ Ω, � � µ(A) = inf µ(U ) : U ⊂ Ω is open and A ⊂ U . (2) For each µ measurable set A ⊂ Ω, � � µ(A) = sup µ(K) : K ⊂ A is compact .
A set E ⊂ X with µ(E) = 0 is called µ negligible. Sets A, B ⊂ X are µ equivalent if µ(A � B) = 0; they are µ overlapping if µ(A ∩ B) > 0. Maps φ and ψ from a set E ⊂ X to a set Y are µ equivalent if the set � � {φ �= ψ} := x ∈ E : φ(x) �= ψ(x)
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11
is µ negligible. When the measure µ is clearly understood from the context, we indicate the equivalence by symbols A ∼ B and φ ∼ ψ. Let E ⊂ X be a µ measurable set. A map φ : E → Rs is called µ measurable if the set φ−1 (B) is µ measurable for every Borel set B ⊂ Rs . The linear space of all µ measurable maps φ : E → Rs is denoted by L0 (E, µ; Rs ).
The essential support of φ ∈ L0 (E, µ; Rs ) is the set � � ess spt φ := spt µ {φ �= 0} .
Unlike the support of φ, the essential support of φ depends only on the µ equivalence class of φ. If φ ∈ L0 (E, µ; Rs ), we also define � � � � � � ess sup�φ(x)� := inf sup �ψ(x)� : ψ ∈ L0 (E, µ; Rs ) and ψ ∼ φ . x∈E
x∈E
For each φ ∈ L (E, µ; R ), there is ψ ∼ φ such that � � � � ess spt φ = spt ψ and ess sup�φ(x)� = sup �ψ(x)�. 0
s
x∈E
x∈E
Convention 1.3.2. As is customary, we do not explicitly distinguish between an individual set E ⊂ X, or an individual map φ, and the µ equivalence class determined by E, or by φ, respectively. However, the reader should be aware of the following custom: we think of the space L0 (E, µ; Rs ) as consisting of equivalence classes, but when we write φ ∈ L0 (E, µ; Rs ), we view φ as a specific representative of its equivalence class. In particular, when writing φ ∈ L0 (E, µ; Rs ), we always assume that � � � � ess spt φ = spt φ and ess sup�φ(x)� = sup �φ(x)�. x∈E
x∈E
The next two theorems are standard tools of measure theory. Their proofs can be found in [29, Section 1.2]. Theorem 1.3.3 (Egoroff). Let µ be a finite measure in X ⊂ Rm , and let {φk } be a sequence in L0 (X, µ; Rs ) that converges pointwise. Given ε > 0, there is a µ measurable set E ⊂ X such that µ(X − E) < ε and the sequence {φk � E} converges uniformly. Theorem 1.3.4 (Luzin). Let µ be a finite Borel regular measure in X ⊂ Rm , and let φ ∈ L0 (X, µ; Rs ). Given ε > 0, there is a compact set K ⊂ X such that µ(X − K) < ε and the restriction φ � K is continuous. Given a µ measurable set E ⊂ X, we let �� �1/p �φ�Lp (E,µ;Rs ) : = |φ|p dµ if 1 ≤ p < ∞, E
� � �φ�L∞ (E,µ;Rs ) : = ess sup�φ(x)� x∈E
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for each φ ∈ L0 (E, µ; Rs ), and for 1 ≤ p ≤ ∞, define
� � Lp (E, µ; Rs ) : = f ∈ L0 (E, µ; Rs ) : �f �Lp (E,µ;Rs ) < ∞ .
Let 1 ≤ p ≤ ∞ and 1 ≤ q ≤ ∞ be such that 1/p + 1/q = 1 where we define 1/∞ := 0. The H¨ older inequality �f g�L1 (E,µ;Rs ) ≤ �f �Lp (E,µ;Rs ) �g�Lq (E,µ;Rs )
(1.3.1)
holds for each f, g ∈ L0 (E, µ; Rs ); see [63, Theorem 3.5]. Let µ be a Borel measure in an open setΩ ⊂ Rm . For 1 ≤ p ≤ ∞, we denote by Lploc (Ω, µ; Rs ) the linear space of all maps φ ∈ L0 (Ω, µ; Rs ) such that φ � U belongs to Lp (U, µ; Rs ) for each open set U � Ω. Unless stated otherwise, throughout we assume that Lploc (Ω, µ; Rs ) has been equipped with the Fr´echet topology defined by seminorms φ �→ �φ � U �Lp (U,µ;Rs ) : Lploc (Ω, µ; Rs ) → R where U � Ω is an open set; cf. Example 1.2.2. We write Lp (E, µ) and Lploc (Ω, µ) instead of Lp (E, µ; R) and Lploc (Ω, µ; R), respectively. The following theorem is essential for establishing Theorem 2.3.7 and Proposition 7.4.3 below; in addition, it simplifies proofs of some differentiation results (Theorems 4.3.4 and 6.2.3 below). We call it the Henstock lemma, but the name Saks-Henstock lemma is also used — cf. [44] and [37]. Theorem 1.3.5. Let µ be a Radon measure in X ⊂ Rm , let E ⊂ X be a µ measurable set with µ(E) < ∞, and let f ∈ L1 (E, µ) be real-valued. Given ε > 0, there is δ : E → R+ satisfying the following condition: for every collection {E1 , . . . , Ep } of µ measurable µ nonoverlapping subsets of E, and for every set of points {x1 , . . . , xp } ⊂ E, the inequality � p � � � �f (xi )µ(Ei ) − � i=1
Ei
� � f dµ�� < ε
� � holds whenever d Ei ∪ {xi } < δ(xi ) for i = 1, . . . , p.
Proof. Choose ε > 0, and using the Vitali-Carath´eodory theorem [63, Theorem 2.25], find functions g and h defined on E that are, respectively, upper and lower semicontinuous, and satisfy � (h − g) dµ < ε. g ≤ f ≤ h and E
There is δ : E → R+ such that g(y) < f (x) + ε and h(y) > f (x) − ε for all x, y ∈ E with |x − y| < δ(x). If {E1 , . . . , Ep } and {x1 , . . . , xp } satisfy the
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“Book˙2011” — 2012/2/26 — 9:58 — page 13 — #23
1.4. Hausdorff measures
13
conditions of the proposition, then � � g dµ ≤ �
Ei
Ei
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Ei
f dµ ≤
�
h dµ,
Ei
g dµ − εµ(Ei ) ≤ f (xi )µ(Ei ) ≤
�
h dµ + εµ(Ei )
Ei
for i = 1, . . . , p. Consequently � � � � p � p �� � � � �f (xi )µ(Ei ) − � f dµ� ≤ (h − g) dµ + εµ(Ei ) � i=1
Ei
i=1
≤
�
E
�
Ei
(h − g) dµ + εµ(E)
� < ε 1 + µ(E) .
Lebesgue measure in Rm is denoted by Lm . For each subset E of the ambient space Rn , we let |E| := Ln (E).
Sets A, B ⊂ Rn are called overlapping if they are Ln overlapping, that is to say if |A ∩ B| > 0. Unless specified otherwise, all concepts connected with measures, such as “measurable”, “negligible”, etc., as well as the expressions “almost all” and “almost everywhere”, refer to the measure Ln in Rn . For a measurable set E ⊂ Rn , we let Lp (E; Rs ) := Lp (E, Ln ; Rs )
and
Lp (E) := Lp (E, Ln ).
p p s IfΩ ⊂ Rn is an open set, the meanings of � Lloc (Ω; R ) �and Lloc (Ω) are� obvious. When no confusion is possible, we write E f (x) dx or E f instead of E f dLn .
1.4. Hausdorff measures We define Hausdorff measures in Rn , and state some of their elementary properties. Select a fixed s ≥ 0, and let � �s � ∞ Γ 12 s−1 −t �. Γ (s) := t e dt and α(s) := � s Γ 2 +1 0
Recall that Γ : t �→ Γ (t) is the classical Euler’s gamma function [62, Definition 8.17]. Using Fubini’s theorem and induction, we obtain �� �� α(n) = Ln x ∈ Rn : |x| ≤ 1 ;
a more advanced calculation can be found in [56, Chapter 1, Equation 1.1.7]. The function α : t �→ α(t) maps [0, ∞) to [1, 5), has only one local maximum
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1. Preliminaries
and one local minimum, attained at 5 < tmax < 6 and tmin = 0, respectively; in addition α(t) → 0 as t → ∞. For E ⊂ Rn and δ > 0, let �s � ∞ � d(Ck ) α(s) (1.4.1) Hδs (E) := inf 2 k=1
where the infimum is taken over all sequences {Ck } of subsets of Rn such that �∞ E ⊂ k=1 Ck and d(Ck ) < δ for k = 1, 2, . . . ; here we define 00 := 1 and d(∅)s := 0. Letting Hs (E) := sup Hδs (E) = lim Hδs (E), δ>0
δ→0
the function Hs : E �→ Hs (E), defined for each E ⊂ Rn , is a measure in Rn , called the s-dimensional Hausdorff measure. Since Hs (A ∪ B) = Hs (A) + Hs (B) for every pair of sets A, B ⊂ Rn with dist(A, B) > 0, it follows from Theorem 1.3.1 that Hs is a Borel measure in Rn . In addition, the measure Hs is Borel regular by Proposition 1.4.2 below. However, Hs is not a Radon measure in Rn when s < n. It is easy to verify that H0 is the counting measure in Rn . The constant α(s)/2s in the definition of Hδs (E) implies Hn = Ln . This equality follows (nontrivially) from the isodiametric inequality �n � d(E) n L (E) ≤ α(n) (1.4.2) 2
which holds for every E ⊂ Rn ; see [29, Section 2.2]. As the diameters of sets are invariant with respect to isometric transformations, so are the Hausdorff measures. Moreover, for each E ⊂ Rn and every t > 0, Hs (tE) = ts Hs (E).
Remark 1.4.1. The value of Hδs (E), and a fortiori that of Hs (E), does not change when the sequence {Ck } in the defining equality (1.4.1) is assumed to have one of the following additional properties: (1) Each Ck is convex ; since the diameters of Ck and its convex hull are the same. (2) Each Ck is closed ; since the diameters of Ck and its closure cl Ck are the same. find rk > 0 such that (3) Each Ck is �open; since� given ε > � �s 0, we can s d U (Ck , rk ) < δ and d U (Ck , rk ) < d(Ck ) + ε2−k for k = 1, 2, . . . . (4) Each Ck is contained in E; since E is covered by the family {Ck ∩ E} and d(Ck ∩ E) ≤ d(Ck ) for k = 1, 2, . . . .
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By (4), the value Hs (E) depends on Rn only to the extent to which Rn defines the metric in E. In particular, if 1 ≤ m < n is an integer, then Hs restricted to the subsets of Rm is the s-dimensional Hausdorff measure in Rm . Proposition 1.4.2. Given E ⊂ Rn , there is a Gδ set B ⊂ Rn such that E ⊂ B and Hs (E) = Hs (B). In particular, Hs is a Borel regular measure. Proof. Assume Hs (E) < ∞, since otherwise it suffices to let B := Rn . Fix k ∈ N. By Remark 1.4.1, (3), there are open sets Uk,j ⊂ Rn such that �∞ d(Uk,j ) < 1/k for j = 1, 2, . . . , Uk = j=1 Uk,j contains E, and �s � ∞ � d(Uk,j ) 1 s α(s) < Hs (E) + . H1/k (Uk ) ≤ 2 k j=1
s The first inequality follows directly from the definition of H1/k . The intersec�∞ tion B = k=1 Uk is a Gδ set containing E, and s s H1/k (B) ≤ H1/k (Uk ) < Hs (E) +
1 . k
Letting k → ∞ yields Hs (B) ≤ Hs (E). Proposition 1.4.3. Let E ⊂ Rn , and let 0 ≤ s < t. If the measure Hs is σ-finite, then Ht (E) = 0. Moreover, Hs ≡ 0 for each s > n.
E
Proof. In proving the first claim, we may assume that Hs (E) < ∞. Given δ > 0, there is a sequence {Ck } of subsets of Rn of diameters smaller than δ �∞ such that E ⊂ k=1 Ck and �s � ∞ � d(Ck ) α(s) < Hs (E) + 1. 2 k=1
Consequently
Hδt (E)
�t � �t−s �s � ∞ α(t) � d(Ck ) δ d(Ck ) ≤ α(t) ≤ α(s) 2 2 α(s) 2 k=1 k=1 � �t−s � α(t) � s δ H (E) + 1 , ≤ 2 α(s) ∞ �
�
and it suffices to let δ → 0. If s > n, it suffices to show that Hs (Q) = 0 for a 0-cube Q. For k ∈ N, √ each k-cube has diameter δk := 2−k n, and Q is the union of 2kn such cubes. Thus � √ �s � �s δk n = α(s) 2k(n−s) , Hδsk (Q) ≤ α(s)2kn 2 2 and letting k → ∞ yields the desired result. Next we relate Hausdorff measures in Rn to covers consisting of dyadic cubes. Recall that for k ∈ Z, the family of all k-cubes is denoted by Dk .
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Proposition 1.4.4. Let s ≥ 0 and E ⊂ Rn . Given ε > 0 and p�∈ Z,�there is � � a family Q ⊂ k≥p Dk such that E meets each Q ∈ Q, E ⊂ int Q , and � � � d(Q)s ≤ β Hs (E) + ε Q∈Q
where β = β(n) > 0. Proof. If s > n, the proposition holds with β = 1; the proof is analogous to that of the second part of Proposition 1.4.3. Hence assume s ≤ n, and choose ε > 0 and p ∈ Z. There is a cover {Cj } of E such that the diameter of each Cj is smaller than δ = 2−p , and �s � ∞ � d(Cj ) α(s) ≤ Hδs (E) + ε ≤ Hs (E) + ε. 2 j=1 Find an integer pj ≥ p with 2−pj −1 ≤ d(Cj ) < 2−pj , and note √ d(Cj ) < d(Q)/ n = 2−pj ≤ 2d(Cj ) for every pj -cube Q. Select a pj -cube Q with Q ∩ Cj �= ∅, and denote by Q1,j , . . . , Q3n ,j all pj -cubes which meet Q, including Q itself. It follows that ��3n � ��∞ �3n � Cj ⊂ int i=1 Qi,j , and hence E ⊂ int j=1 i=1 Qi,j . Moreover, n
∞ � √ �s � � (4n)s � s H (E) + ε . d(Qi,j )s ≤ 3n · 2 n d(Cj )s ≤ 3n α(s) j=1 i=1 j=1
∞ � 3 �
� � Since 0 ≤ s ≤ n implies α(s) ≥�min 1, α(n) , the desired inequality holds � with β := (12n)n max 1, 1/α(n) . Finally, replacing Q by a smaller family {Q ∈ Q : Q ∩ E �= ∅} completes the proof. Additional properties of Hausdorff measures in Rn can be found in [30] and [46]. Hausdorff measures defined in general metric spaces are investigated in [59].
1.5. Differentiable and Lipschitz maps LetΩ ⊂ Rn be an open set. A map φ : Ω → Rm is differentiable at x ∈ Ω if there is a linear map L : Rn → Rm such that � � �φ(y) − φ(x) − L(y − x)� lim = 0. y→x |y − x|
If such a map L exists, it is unique. We call it the derivative of φ at x, denoted by Dφ(x). If a map φ = (f1 , . . . , fn ) from Ω to Rn is differentiable at x ∈ Ω,
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1.5. Differentiable and Lipschitz maps
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17
then the divergence of φ at x is the real number div φ(x) :=
n �
Di fi (x)
i=1
where Di := ∂/∂ξi is the usual partial derivative operator . Let E ⊂ Rn be any set, and let φ : E → Rm . The Lipschitz constant of φ is the extended real number �� � � �φ(x) − φ(y)� Lip φ := sup : x, y ∈ E and x �= y . |x − y|
When Lip φ < ∞, the map φ is called Lipschitz . If Ω ⊂ Rn is an open set, we call a map φ : Ω → Rm locally Lipschitz whenever the restriction φ � U is Lipschitz for each open set U � Ω. The linear space of all Lipschitz maps φ : E → Rm is denoted by Lip(E; Rm ). The symbols Lip(E), Lipc (Ω; Rm ), Liploc (Ω; Rm ), etc., have the obvious meaning. For a Lipschitz map φ : E → Rm and s ≥ 0, we obtain � � Hs φ(E) ≤ (Lip φ)s Hs (E). (1.5.1) A bijective Lipschitz map whose inverse is also Lipschitz is called a lipeomorphism.
Observation 1.5.1. Let Ω ⊂ Rn be an open set. If φ ∈ Liploc (Ω; Rm ), then φ � K ∈ Lip(C; Rm ) for each compact set K ⊂ Ω. In particular, Liploc (Ω; Rm ) ∩ Cc (Ω; Rm ) = Lipc (Ω; Rm ). Proof. Suppose there is a compact set K ⊂ Ω such that φ is not Lipschitz in K. There are sequences {xk } and {yk } in K such that � � �φ(xk ) − φ(yk )� > k|xk − yk | > 0, k = 1, 2, . . . .
Passing to subsequences, still denoted by {xk } and {yk }, we obtain the limit points x = lim xk and y = lim yk in K. The continuity of φ implies � � � � ∞ > �φ(x) − φ(y)� = lim�φ(xk ) − φ(yk )� ≥ lim sup k|xk − yk |,
and consequently x = y. Since φ is Lipschitz in a neighborhood of x, there is 0 < c < ∞ such that for all sufficiently large k, � � c|xk − yk | ≥ �φ(xk ) − φ(yk )� > k|xk − yk | > 0. A contradiction follows.
Proposition 1.5.2. Let E ⊂ Rn and φ ∈ Lip(E; Rm ). There is a map ψ ∈ Lip(Rn ; Rm ) such that ψ(x) = φ(x) for each x ∈ E, √ Lip ψ ≤ m Lip φ, and �ψ�L∞ (Rn ;Rm ) ≤ �φ�L∞ (E;Rm ) .
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“Book˙2011” — 2012/2/26 — 9:58 — page 18 — #28
18
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1. Preliminaries
Proof. By [29, Section 3.1.1, Theorem 1], there is θ ∈ Lip(Rn ; Rm ) such that √ θ(x) = φ(x) for each x ∈ E and Lip θ ≤ m Lip φ. As there is nothing to prove otherwise, assume c := �φ�L∞ (E;Rm ) belongs to R+ . Define γ : Rm → Rm by � y if |y| > c, c |y| γ(y) := y if |y| ≤ c. Since Lip γ = 1 and |γ|L∞ (Rm ;Rm ) ≤ c, the composition ψ := γ ◦ θ is the desired extension of φ. With a considerable effort, one can improve on Proposition 1.5.2 by showing that Lip ψ = Lip φ. This stronger result is called Kirschbraun’s theorem [33, Theorem 2.10.43]. The next well-known theorem has several proofs of various levels of sophistication, e.g., [33, Theorem 3.1.6], [1, Theorem 2.14], or [29, Section 6.2, Theorem 2]. For a proof with minimal prerequisites we refer to [29, Section 3.1.2]. Theorem 1.5.3 (Rademacher). Each φ ∈ Lip(Rn ; Rm ) is differentiable at almost all x ∈ Rn . older constant at x ∈ E of a map Let E ⊂ Rn and 0 ≤ s ≤ 1. The s-H¨ φ : E → Rm is the extended real number � � �φ(y) − φ(x)� Hs φ(x) := lim sup . y→x |y − x|s y∈E
Clearly, H0 φ(x) < ∞ if and only if φ is bounded in a neighborhood of x, and H0 φ(x) = 0 if and only if φ is continuous at x. If Hs φ(x) < ∞ and 0 ≤ t < s, then Ht φ(x) = 0. We call Lip φ(x) := H1 φ(x) the Lipschitz constant of φ at x, and say that φ is Lipschitz at x whenever Lip φ(x) < ∞. We say that φ is pointwise Lipschitz in a set C ⊂ E if it is Lipschitz at each x ∈ C. A pointwise Lipschitz map in C need not be Lipschitz in C, even if C is compact [51, Section 1.6]. Theorem 1.5.4 (Stepanoff). Let Ω ⊂ Rn be an open set, and assume that φ : Ω → Rm is pointwise Lipschitz in a set E ⊂ Ω. Then φ is differentiable at almost all x ∈ E. For a proof of this generalization of Rademacher’s theorem we refer to [33, Theorem 3.1.9], or to Section 6.2 below where a slightly more general theorem is proved in detail; see Remark 7.2.4.
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“Book˙2011” — 2012/2/26 — 9:58 — page 19 — #29
1.5. Differentiable and Lipschitz maps
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19
Theorem 1.5.5 (Whitney). Let K ⊂ Rn be a compact set, and let f ∈ C(K) and v ∈ C(K; Rn ) satisfy the following condition: given ε > 0, we can find δ > 0 so that � � �f (y) − f (x) − v(x) · (y − x)� ≤ ε|y − x| for all x, y ∈ K with |y − x| < δ. There is g ∈ C 1 (Rn ) such that g(x) = f (x)
and
Dg(x) = v(x)
for each x ∈ K. Theorem 1.5.5 is a special case of Whitney’s extension theorem. Proofs of the general Whitney’s result, which implies the special case, can be found in [29, Section 6.5] or in [70, Chapter 6, Section 2]. Let φ = (f1 , . . . , fn ) be a Lipschitz map from Rn to Rn . Then Df1 ... Dfn
is an n × n matrix, whose determinant is denoted by det Dφ. The Jacobian of φ is the function Jφ = | det Dφ| defined almost everywhere in Rn by Rademacher’s theorem. In view of (1.5.1), the inequality �Jφ �L∞ (Rn ) ≤ (Lip φ)n
(1.5.2)
is a consequence of [29, Section 3.3, Lemma 1]. The next result is called interchangeably the area theorem or change of variables theorem. It follows from [29, Section 3.3, Theorem 2]. Theorem 1.5.6. Let φ : Rn → Rn be a Lipschitz map. If g ∈ L0 (Rn ) and �� g ≥ 0, then y �→ g(x) : x ∈ φ−1 (y)} is a measurable function on Rn and � � � g(x)Jφ (x) dx = g(x) dy. (1.5.3) Rn
Rn x∈φ−1 (y)
Employing Hausdorff measures and more elaborete Jacobians, formulas similar to (1.5.3) hold for Lipschitz maps φ : Rn → Rm where m �= n. On a few occasions when such formulas are used, we refer the reader to the appropriate sections of [29, Chapter 3].
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© 2012 by Taylor & Francis Group, LLC
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“Book˙2011” — 2012/2/26 — 9:58 — page 21 — #31
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Chapter 2 Divergence theorem for dyadic figures
Using the idea of W.B. Jurkat, we give an elementary proof of a fairly general divergence theorem for dyadic figures. While this is only a preliminary version of the divergence theorem we intend to establish, it is already a useful tool for studying removable singularities of some classical partial differential equations (Chapter 3 below).
2.1. Differentiable vector fields If A is a figure, then for Hn−1 almost all x ∈ ∂A there is a unique unit exterior normal of A at x, denoted by νA (x). The map νA : x �→ νA (x) : ∂A → Rn
is defined Hn−1 almost everywhere, has only finitely many values, and it is Hn−1 measurable. Let E ⊂ Rn , and assume that v : E → Rn belongs to L1 (∂A, Hn−1 ; Rn ) for each figure A ⊂ E. A real-valued function � v · νA dHn−1 (2.1.1) F : A �→ ∂A
defined on all figures A ⊂ E is called the flux of v. In this context it is customary to call v a vector field . The term “flux” is derived from a physical example: if v is the vector field of velocities of a fluid moving in the set E, then F (A) is the amount of fluid that flows out of the figure A ⊂ E in the unit of time. The next observation says that the flux of a vector field is an additive function with respect to nonoverlapping figures. Its simple verification is left to the reader. Observation 2.1.1. Let E ⊂ Rn , and assume that v : E → Rn belongs to L∞ (∂A, Hn−1 ; Rn ) for each figure A ⊂ E. Then � � � v · νA∪B dHn−1 = v · νA dHn−1 + v · νB dHn−1 ∂(A∪B)
∂A
∂B
for each pair A, B ⊂ E of nonoverlapping figures. 21
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© 2012 by Taylor & Francis Group, LLC
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“Book˙2011” — 2012/2/26 — 9:58 — page 22 — #32
22
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2. Divergence theorem for dyadic figures
Proposition 2.1.2. Let A be a cell, and let v ∈ C(A; Rn ) be differentiable at each x ∈ int A. If div v belongs to L1 (A), then �
�
div v(x) dx = A
∂A
Proof. Let v = (v1 , . . . , vn ) and A =
v · νA dHn−1 .
�n
− + i=1 [ai , ai ].
If
± A± i := {ai } × πi (int A),
then νA (x) = ±ei whenever x ∈ A± i , i = 1, . . . , n. The boundary ∂A differs �n − + n−1 from i=1 (Ai ∪ Ai ) by an H negligible set. Fix i and for x ∈ int A, write x = (u, t) where u = πi (x) and t = x · ei . By Fubini’s theorem and the fundamental theorem of calculus, � d vi (u, t) dt du Di vi (x) dx = dt A πi (int A) a− i � � � − = vi (u, a+ i ) − vi (u, ai ) du ��
�
�
=
�
a+ i
πi (int A)
A+ i
v · νA dHn−1 +
�
A− i
v · νA dHn−1 .
Summing up these equalities over i = 1, . . . , n completes the proof. Corollary 2.1.3. Let E ⊂ Rn , x ∈ E, and let {Ck } be a sequence of cubes such that lim d(Ck ) = 0. Assume that Ck ⊂ E and x ∈ Ck for k = 1, 2, . . . , and that v : E → Rm belongs to L1 (∂C, Hn−1 ; Rn ) for each cube C ⊂ E. (1) If 0 ≤ s ≤ 1, then 1 lim sup d(Ck )n−1+s
�
∂Ck
v · νCk dHn−1 ≤ 2nHs v(x).
(2) If x ∈ int E and v is differentiable at x, then 1 lim |Ck |
�
∂Ck
v · νCk dHn−1 = div v(x).
Proof. Choose ε > 0 — you can never go wrong by doing so. We may assume Hs v(x) < ∞, and find δ > 0 so that � � � � �v(y) − v(x)� ≤ Hs v(x) + ε · |y − x|s ✐
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“Book˙2011” — 2012/2/26 — 9:58 — page 23 — #33
2.1. Differentiable vector fields
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23
for each y ∈ E ∩ U (x,δ ). Denote by F the flux of v, and use Proposition 2.1.2 to show that for all sufficiently large k, �� � � � � � � � n−1 �F (Ck )� = � v(y) − v(x) · νCk (y) dH (y)�� � ∂Ck � � � ≤ Hs v(x) + ε |y − x|s dHn−1 (y) �
�
∂Ck
≤ Hs v(x) + ε d(Ck )s Hn−1 (∂Ck ) � � ≤ 2n Hs v(x) + ε d(Ck )n−1+s .
If x ∈ int E and v is differentiable at x, let � � w : y �→ v(x) + Dv(x) (y − x) : Rn → Rn
and observe that div w(y) = div v(x) for each y ∈ Rn . There is η > 0 such that U (x,η ) ⊂ E and � � �v(y) − w(y)� ≤ ε|y − x| for every y ∈ U (x,η ). As w ∈ C ∞ (Rn ; Rn ), Proposition 2.1.2 yields � � � ��� � � � � � � n−1 v(y) − w(y) · νCk (y) dH (y)�� �F (Ck ) − div v(x)|Ck |� = � ∂C � k ≤ε |y − x| dHn−1 (y) ∂Ck
≤ εd(Ck )Hn−1 (∂Ck ) = 2n3/2 ε|Ck |
for all sufficiently large k. Letting k → ∞, the corollary follows from the arbitrariness of ε. We prove the divergence theorem for closed balls. While this is not essential for the logical development of our exposition, it will facilitate an early presentation of examples. If B := B(x, r), then νB (y) := (y − x)/r is the unit exterior normal of B at y ∈ ∂B. Since the induced map νB : ∂B → Rn is continuous, a finite integral � v · νB dHn−1 ∂B
exists for each v ∈ L1 (∂B, Hn−1 ; Rn ).
Proposition 2.1.4. Let B ⊂ Rn be a closed ball, and let v ∈ C(B; Rn ) be differentiable in each x ∈ int B. If div v belongs to L1 (B), then � � div v(x) dx = v · νB dHn−1 . B
∂B
Proof. The proof is similar to that of Proposition 2.1.2. In view � of translation invariance, we may assume B = B(0, r). Let U := πn (int B) and g(u) = r2 − |u|2 for each u ∈ U . If (∂B)± := {x ∈ ∂B : ±x · en > 0}, then the bijections � � φ± : u �→ u, ±g(u) : U → (∂B)± are continuously differentiable and have the same Jacobian � � �2 r J = 1 + �Dg � = ; g
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“Book˙2011” — 2012/2/26 — 9:58 — page 24 — #34
24
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2. Divergence theorem for dyadic figures
see [29, Section 3.3.4, B]. Let vn := v · en and νn := νB · en . Observe νn ◦ φ± = (νB ◦ φ± ) · en =
φ± · e n g =± . r r
� � � If Uk = πn U (0, 1 − 2−k ) , then the maps φ± � Uk are Lipschitz and U = ∞ k=1 Uk . Thus applying [29, Section 3.3.4, B] to each Uk , we obtain � � � � � � � (vn νn ) ◦ φ± J dLn = ± vn νn dHn−1 = vn u, ±g(u) du. (∂B)±
U
U
Now Fubini’s theorem and the fundamental theorem of calculus imply � � � �� g(u) d Dn vn (x) dx = vn (u, t) dt du −g(u) dt B U � � � � � � vn u, g(u) du − vn u, −g(u) du = U �U � = vn νn dHn−1 + vn νn dHn−1 (∂B)+
=
�
(∂B)−
vn νn dHn−1 ,
∂B
since the boundary ∂B differs from (∂B)+ ∪ (∂B)− by an Hn−1 negligible set. The proposition follows from symmetry.
2.2. Dyadic partitions A partition is a finite (possibly empty) collection � � P := (E1 , x1 ), . . . , (Ep , xp )
where {E1 , . . . , Ep } is a collection of nonoverlapping subsets of Rn such that �p xi ∈ Ei for i = 1, . . . , p. The body of P is the union [P ] := i=1 Ei , and P is called a partition in a set A ⊂ Rn if [P ] ⊂ A. Given a set E ⊂ Rn and δ : E → R+ , we say that P is δ-fine if xi ∈ E and d(Ei ) < δ(xi ) for i = 1, . . . , p. When each set Ei is a dyadic cube, then P is called a dyadic partition. If dyadic cubes A and B overlap, then either A ⊂ B or B ⊂ A. Consequently, every family C of dyadic cubes has a nonoverlapping subfamily Q such � � that Q = C. Dyadic cubes A and B are called adjacent if d(A) = d(B) and A ∩ B �= ∅. Every dyadic cube is adjacent to 3n dyadic cubes, including itself. Recall that for an integer k the family of all k-cubes is denoted by Dk . Given a family E of subsets of Rn and x ∈ Rn , we let St(x, E) := {E ∈ E : x ∈ E}. For each x ∈ Rn and each k ∈ Z, the collection St(x, Dk ) consists of at most � St(x, Dk ). A star cover of 2n k-cubes, and x belongs to the interior of
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“Book˙2011” — 2012/2/26 — 9:58 — page 25 — #35
2.2. Dyadic partitions
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25
E ⊂ Rn is a family Q of dyadic cubes such that for each x ∈ E there is kx ∈ Z with St(x, Dkx ) ⊂ Q; in this case �� � �� � � E⊂ int St(x, Dkx ) ⊂ int Q . x∈E
It follows that a star cover Q of a compact set K ⊂ Rn has a finite nonoverlapping subcover. Lemma 2.2.1. Let δ be a positive function defined on a set E ⊂ Rn , and let 0 ≤ t ≤ n. Given ε > 0, the set E has a star cover C which satisfies the following conditions: (1) for each C ∈ C there is xC ∈ C ∩ E such that d(C) < δ(xC ); � t � � t (2) C∈C d(C) ≤ κ H (E) + ε where κ := κ(n) > 0.
Proof. To avoid trivialities, assume E �= ∅. Denote by B the family of all dyadic cubes C satisfying condition (1). For k ∈ N and x ∈ Rn , let � � � D≥k := {Di : i ≥ k} and Bk := x ∈ Rn : St(x, D≥k ) ⊂ B . �∞ Clearly {Bk } is an increasing sequence. Moreover E ⊂ k=1 Bk , since � √ � x ∈ E : δ(x) > 2−k n ⊂ Bk . Claim. Rn − Bk = is a Borel set.
� (D≥k − B) for every k ∈ N. In particular, each Bk
Proof . If x �∈ Bk , some Cx ∈ St(x, D≥k ) does not belong to B. Hence � x ∈ Cx and Cx ∈ D≥k − B. It follows that x ∈ (D≥k − B). Conversely, if � x ∈ (D≥k −B) then x ∈ Dx for some Dx ∈ D≥k −B. Thus St(x, D≥k ) �⊂ B, which means x �∈ Bk . If E1 := E ∩ B1 and Ek := E ∩ (Bk − Bk−1 ) for k = 2, 3, . . . , then E=
∞ �
Ek
t
and H (E) =
k=1
∞ �
Ht (Ek ).
k=1
Select Ek �= ∅. By Proposition 1.4.4, there is a cover Qk ⊂ D≥k of Ek such that Ek ∩ Q �= ∅ for each Q ∈ Qk , and � � � d(Q)t ≤ β Ht (Ek ) + ε2−k Q∈Qk
where β = β(n) > 0. If Ck consists of all dyadic cubes that meet Ek and are adjacent to some Q ∈ Qk , then Ck is a star cover of Ek , and � � � � d(C)t ≤ 3n d(Q)t ≤ 3n β Ht (Ek ) + ε2−k . C∈Ck
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Q∈Qk
© 2012 by Taylor & Francis Group, LLC
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2. Divergence theorem for dyadic figures
Now Ck ⊂ D≥k and Ek ⊂ Bk . Since each C ∈ Ck meets Ek , the definition of �∞ Bk implies C ∈ B. Thus Ck ⊂ B, and we see that the family C := k=1 Ck is a star cover of E that satisfies condition (1). Letting κ := 3n β, we obtain �
C∈C
d(C)t ≤
∞ � �
d(C)t
k=1 C∈Ck ∞ � � t
≤κ
k=1
� � � H (Ek ) + ε2−k = κ Ht (E) + ε .
Proposition 2.2.2. Let E be a family of disjoint subsets of a dyadic figure A, and for each E ∈ E select real numbers 0 ≤ tE ≤ n and εE > 0. Given δ : A → R+ , there is a δ-fine dyadic partition � � P := (C1 , x1 ), . . . , (Cp , xp ) such that [P ] = A, and with a fixed κ = κ(n) > 0, the inequality � � � d(Ci )tE ≤ κ HtE (E) + εE xi ∈E
holds for each E ∈ E. Proof. There is k ∈ N such that A is the union of k-cubes. Enlarging E � √ and making δ smaller, we may assume E = A and δ(x) < 2−k n for each x ∈ A. Let CE be a star cover of E ∈ E associated with δE := δ � E, tE , and εE according to Lemma 2.2.1. For every C ∈ CE , select xC ∈ E ∩ C with � d(C) < δE (xC ). Since C := E∈E CE is a star cover of the compact set A, �p there are nonoverlapping cubes C1 , . . . , Cp in C such that A ⊂ i=1 Ci . It follows that � � P := (Ci , xCi ) : |Ci ∩ A| > 0
is a δ-fine dyadic partition with A ⊂ [P ]. As our assumption about δ implies Ci ⊂ A whenever |Ci ∩ A| > 0, we obtain [P ] = A. Since E is a disjoint family, {Ci : xCi ∈ E} ⊂ CE for each E ∈ E. Hence with the same κ as in Lemma 2.2.1, the inequality � � � � d(Ci )tE ≤ d(C)tE ≤ κ HtE (E) + εE xCi ∈E
C∈CE
holds for every E ∈ E. Remark 2.2.3. Lemma 2.2.1 and Proposition 2.2.2 are due to W.B. Jurkat [42, Section 4]. The classical Cousin’s lemma [17] or [51, Lemma 2.6.1], as well as its generalization obtained by E.J. Howard [41, Lemma 5], are immediate consequences of Proposition 2.2.2.
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2.3. Admissible maps
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2.3. Admissible maps Definition 2.3.1. Let E ⊂ Rn be any set. A map φ : E → Rm is called admissible if there are numbers 0 ≤ sk < 1, and disjoint, possibly empty, sets �∞ Ek ⊂ E such that φ is pointwise Lipschitz in E − k=1 Ek and for k = 1, 2, . . . , the following conditions hold: (i) Ek is Hn−1+sk σ-finite, and Hsk φ(x) < ∞ for each x ∈ Ek ; (ii) Hn−1+sk (Ek ) > 0 implies Hsk φ(x) = 0 for each x ∈ Ek .
The family of all admissible maps from the set E to Rm is denoted by Adm(E; Rm ), and we write Adm(E) instead of Adm(E; R). Note that in Definition 2.3.1 no topological restrictions are placed on the exceptional sets Ek . � Remark 2.3.2. If Hn−1+sk (Ek ) = ∞, then Ek = j∈N Ek,j where Ek,j are disjoint sets with Hn−1+sk (Ek,j ) < ∞. Thus replacing each � � pair (Ek , sk ) with Hn−1+sk (Ek ) = ∞ by the collection (Ek,j , sk ) : j ∈ N , condition (i) of Definition 2.3.1 can be replaced by the condition: (i*) Hn−1+sk (Ek ) < ∞, and Hsk φ(x) < ∞ for each x ∈ Ek .
This observation will simplify future arguments.
Remark 2.3.3. Let E ⊂ Rn and φ ∈ Adm(E; Rm ). Then H0 φ(x) < ∞ for all x ∈ E, and H0 φ(x) = 0 for all x ∈ E − T where T ⊂ E is Hn−1 negligible. Thus φ is locally bounded in E and continuous in E − T . It follows that if E is Hn−1 measurable then so is φ, and if E is compact then φ is bounded. Since each set Ek is negligible, Stepanoff’s theorem implies that φ is differentiable at almost all x ∈ int E. The restriction φ � B belongs to Adm(B; Rm ) for each B ⊂ A. Remark 2.3.4. A commonly encountered map φ : E → Rm is locally bounded in E, continuous outside an Hn−1 negligible set T ⊂ E, and pointwise Lipschitz outside an Hn−1 σ-finite set S ⊂ E [53, Theorem 2.9]. Letting sk := 0 for k = 1, 2, . . . , E1 := T,
E2 := S − T,
and
Ek = ∅
for k = 3, 4, . . . , we see that φ is admissible. Since locally bounded and pointwise Lipschitz are extreme points of the scale represented by H¨older constants, considering admissible maps is natural. Proposition 2.3.5. Let E ⊂ Rn . With respect to pointwise addition and multiplication, Adm(E) is a commutative ring, and Adm(A; Rm ) is a module over Adm(E).
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“Book˙2011” — 2012/2/26 — 9:58 — page 28 — #38
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2. Divergence theorem for dyadic figures
Proof. Choosing φ,ψ ∈ Adm(E; Rm ) and g ∈ Adm(E), it suffices to show that φ + ψ and gφ belong to Adm(E; Rm ). As the other proof is similar, we show only that θ = gφ belongs to Adm(E; Rm ). Let {rk } and {sk } be sequences in [0, 1), and {Ak } and {Bk } be sequences of disjoint subsets of E, associated with φ and g, respectively, according to Definition 2.3.1. Further � � let A0 := E − k∈N Ak , B0 := E − k∈N Bk , and r0 = s0 = 1. Define ti,j := min{ri , sj }
and
Ei,j := Ai ∩ Bj
for i, j = 0, 1, . . . , and observe that 0 ≤ ti,j < 1 whenever (i, j) �= (0, 0), and that E is the union of a disjoint collection {Ei,j : i, j = 0, 1, . . . }. By Remark 2.3.3, both φ and g are locally bounded in E. A direct calculation shows that there are functions a, b : E → R+ such that Hti,j θ(x) ≤ a(x)Hri φ(x) + b(x)Hsj g(x) for each x ∈ Ei,j and i, j = 0, 1, . . . . Thus θ is pointwise Lipschitz in �� � E0,0 = E − Ei,j : i, j = 0, 1, . . . and (i, j) �= (0, 0) ,
and for each pair (i, j) �= (0, 0), the following conditions hold:
(i) Ei,j is Hn−1+ti,j σ-finite, and Hti,j θ(x) < ∞ for each x ∈ Ei,j ; (ii) Hn−1+ti,j (Ei,j ) > 0 implies Hti,j θ(x) = 0 for each x ∈ Ei,j .
This verifies that θ is an admissible map.
Lemma 2.3.6. Let A be a dyadic figure, let v ∈ Adm(A; Rn ), and define f : A → R by the formula � div v(x) if x ∈ int A and v is differentiable at x, f (x) := 0 otherwise. For each δ : A → R+ , there is a δ-fine dyadic partition � ε > 0 and each � P := (C1 , x1 ), . . . , (Cp , xp ) such that [P ] = A and � �� � � � p n−1 � � f (xi )|Ci | − v · νA dH � 0 implies Hsk v(x) = 0 for each x ∈ Ek .
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“Book˙2011” — 2012/2/26 — 9:58 — page 29 — #39
2.3. Admissible maps
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29
�∞ By Stepanoff’s theorem, A − i=1 Ek is the union of disjoint sets E0 and D ⊂ int A such that Hn (E0 ) = 0 and v is differentiable at each x ∈ D. Thus A is the union of disjoint � � sets D, E0 , E1 , . . . , and we let s0 = 1. The family (Ek , sk ) : k = 0, 1, . . . is the disjoint union of subfamilies � � � � and (Ek , sk ) : Hn−1+sk (Ek ) = 0 , (Ek , sk ) : Hn−1+sk (Ek ) > 0 � � 0 0 � � which we enumerate as (Ei+ , s+ i ) : i ≥ 1 and (Ei , si ) : i ≥ 1 , respectively. For i, j ∈ N, let � � 0 := x ∈ Ei0 : j − 1 ≤ Hs0i v(x) < j Ei,j
+ 0 0 and define t+ i := n − 1 + si and ti := n − 1 + si . Now A − D is the union + 0 of disjoint sets Ei+ and Ei,j . Select ci > Hti (Ei+ ) and choose ε > 0. By Corollary 2.1.3, there is γ : A → R+ such that for each cube C ⊂ A, the following conditions are satisfied: � � (1) �f (x)|C| − F (C)� ≤ ε|C| if d(C) < γ(x) for some x ∈ D ∩ C, � � + t+ i if d(C) < γ(x) for some x ∈ E (2) �F (C)� ≤ ε2−i c−1 i d(C) i ∩ C, � � 0 (3) �F (C)� ≤ 2nj d(C)ti if d(C) < γ(x) for some x ∈ E 0 ∩ C. i,j
Next choose δ : A → R+ . With no loss of generality, we may assume that δ ≤ γ. 2.2.2, there is a δ-fine dyadic partition � According to Proposition � P := (C1 , x1 ), . . . , (Cp , xp ) such that [P ] = A and for κ = κ(n) > 0, � � + 0 d(Ck )ti ≤ κci and d(Ck )ti ≤ εj −1 2−i−j . 0 xk ∈Ei,j
xk ∈Ei+
Since f (x) = 0 for each x ∈ A − D, these inequalities, conditions (1)–(3), and Observation 2.1.1 imply the lemma: � �� � � � p n−1 � � f (xk )|Ck | − v · νA dH � � k=1
∂A
� � �� � ≤ �f (xk )|Ck | − F (Ck )� + xk ∈D
≤ε
�
xk ∈D
xk ∈A−D
� ∞ �� � � � � � � � � � � � |Ck | + F (Ck ) + F (Ck )
≤ ε|A| + ε
i≥1
��
∞ � i=1
0 j=1 xk ∈Ei,j
xk ∈Ei+
2−i c−1 i
i≥1
≤ ε|A| + εκ
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� � � �F (Ck )�
�
xk ∈Ei+
2−i + 2nε
+
d(Ck )ti + 2n
∞ �
i,j=1
∞ � j=1
j
�
0
d(Ck )ti
0 xk ∈Ei,j
�
� � 2−i−j = ε |A| + κ + 2n .
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2. Divergence theorem for dyadic figures
Theorem 2.3.7. Let A be a dyadic figure. If v ∈ Adm(A; Rn ) is such that div v belongs to L1 (A), then � � div v dLn = v · νA dHn−1 . A
∂A
Proof. Defining f as in Lemma 2.3.6, we have f ∈ L1 (A) and � � f dLn = div v dLn . A
A
Choose ε > 0, and select a function δ : A → R+ associated with ε and f according to Henstock’s lemma. By Lemma 2.3.6, there is a δ-fine partition � � P := (C1 , x1 ), . . . , (Cp , xp ) such that [P ] = A and �� � � �� � p � � � � � n−1 � n � div v dLn − � v · νA dH f (xi )|Ci |�� � � ≤ � f dL − A
∂A
A
�� � � p f (x)|Ci | − + �� i=1
∂A
i=1
� � v · νA dHn−1 �� < 2ε.
Remark 2.3.8. Some comments are in order.
(1) The assumptions of Theorem 2.3.7 are met if v ∈ Lip(A; Rn ), since �div v�L∞ (A) ≤ nLip v. (2) Let v(0) := 0, and v(x) := x cos |x|−n−1 for x ∈ Rn − {0}. Then v ∈ Adm(Rn ; Rn ), but div v does not belong to L1 (A) if A is a figure containing 0. Still, the flux of v can be calculated from div v by an averaging process which extends the Lebesgue integral. For a deeper analysis of this phenomenon, we refer the interested reader to [49, 51]; also see Chapter 9 below.
(3) Assume n = 1, and let v : R → R be differentiable almost everywhere and such that � b v � dL1 = v(b) − v(a) a
for each dyadic cell [a, b] ⊂ R. Since [29, Section 2.4.3] implies �� �� Hs x ∈ R : Hs v(x) > 0 = 0
for each 0 ≤ s < 1, condition (ii) of Definition 2.3.1 cannot be omitted. The Cantor-Vitali function (Example 9.2.5 below) and its multidimensional analogue [54] provide another rationale for the definition of admissible vector fields. (4) It is clear that using essentially the same arguments, the divergence theorem can be established for arbitrary figures. We employed dyadic figures merely for convenience. Theorem 2.3.7 is only a preliminary
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“Book˙2011” — 2012/2/26 — 9:58 — page 31 — #41
2.4. Convergence of dyadic figures
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31
result, sufficient for proving a satisfactory integration by parts theorem stated below. Theorem 2.3.9 (Integration by parts). Let Ω ⊂ Rn be an open set, and let both v : Ω → Rn and g : Ω → R be locally bounded and pointwise Lipschitz almost everywhere in Ω. Assume div v ∈ L1loc (U ) and Dg ∈ L1loc (Ω; Rn ). If gv ∈ Adm(Ω; Rn ) and spt (gv) � Ω, then � � g(x) div v(x) dx = − Dg(x) · v(x) dx. Ω
Ω
Proof. Let w = gv, and let A ⊂ Ω be a dyadic figure with spt w ⊂ int A. By the assumptions, v and g are measurable and bounded in A. Thus div w = g div v + Dg · v
belongs to L (A). Since w � ∂A = 0, Theorem 2.3.7 yields � � � div w(x) dx = g(x) div v(x) dx + Dg(x) · v(x) dx. 0= 1
A
A
(∗)
A
Let x ∈ Ω − A. Then div w(x) = 0, since w vanishes in the open setΩ − A. Thus g(x) div v(x) = −Dg(x) · v(x) and either g(x) = 0 or v(x) = 0. This shows that spt (g div v) and spt (Dg · v) are subsets of A, and the theorem follows from (∗). Remark 2.3.10. The integration by parts theorem is usually applied when both g and v are admissible, and either g or v has compact support contained in Ω; see Proposition 2.3.5.
2.4. Convergence of dyadic figures As figures are too specialized for applications, it is desirable to extend the divergence theorem to a larger family of sets. With the sole purpose of enhancing intuition, we describe the first step of the most obvious approach to this problem. Our main results will be obtained from less obvious but more efficient ideas of R. Caccioppoli [14] and E. De Giorgi [18, 19]. Lemma 2.4.1. Let {Ai } be a sequence of measurable set, and let E ⊂ Rn be any set. If lim |E � Ai | = 0, then E is measurable and for each f ∈ L1 (Rn ), � � f (x) dx = f (x) dx. lim Ai
E
Proof. Since lim |E − Ai | = lim |Ai − E| = 0, passing to a subsequence if necessary, we may assume that |E − Ai | ≤ 2−i and |Ai − E| ≤ 2−i for i = 1, 2, . . . . Letting I := lim inf Ai =
∞ ∞ � �
j=1 i=j
Aj
and
S := lim sup Ai =
∞ ∞ � �
Aj ,
j=1 i=j
we infer |E − I| = |S − E| = 0. As I ⊂ S implies |E � I| = |E � S| = 0, the set E is measurable and lim �χE − χAi �L1 (Rn ) = 0. Any subsequence {Bi } of {Ai } has a
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“Book˙2011” — 2012/2/26 — 9:58 — page 32 — #42
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2. Divergence theorem for dyadic figures
subsequence {Ci } such that lim χCi = χE almost everywhere. Thus for f ∈ L1 (Rn ), the dominated convergence theorem yields � �� � � � � � � � � �χE (x) − χC (x)� · �f (x)� dx = 0. f (x) dx − f (x) dx�� ≤ lim lim�� i E
Rn
Ci
The lemma follows from the arbitrariness of {Bi }.
Recall that the family of all dyadic figures in Rn is denoted by DF. We say that a sequence {Ai } in DF converges to a set E ⊂ Rn if the following conditions are satisfied: (i) Each Ai is contained in a fixed compact set K ⊂ Rn . (ii) lim |Ai � E| = 0 and sup Hn−1 (∂Ai ) < ∞.
By Lemma 2.4.1, the set E is measurable. Given dyadic figures A and B, we define nonoverlapping dyadic figures � A � B := cl (A − B) and A ⊙ B = cl int (A ∩ B)], and observe that A = (A � B) ∪ (A ⊙ B).
Proposition 2.4.2. Let F be the flux of v ∈ C(Rn ; Rn ), and let {Ai } be a sequence in DF converging to a set E ⊂ Rn . There exists a finite limit F�(E) := lim F (Ai ),
which does not depend on the choice of the sequence {Ai }. Proof. Let K ⊂ Rn be a compact set containing all figures Ai , and let c = sup Hn−1 (∂Ai ). Choose ε > 0, and use the Stone-Weierstrass theorem [62, Theorem 7.32] to find a vector field w ∈ C 1 (Rn ; Rn ) such that �v − w�L∞ (K;Rn ) ≤ ε. According to Theorem 2.3.7, �� � � � � � n−1 � �F (Ai � Aj )� ≤ � (v − w) · ν dH Ai �Aj � � ∂(Ai �Aj ) � �� � � w · νAi �Aj dHn−1 �� + �� ∂(Ai �Aj ) �� � � � � � div w(x) dx�� ≤ εHn−1 ∂(Ai � Aj ) + �� Ai �Aj
≤ 2cε + �div w�L∞ (K) |Ai � Aj |;
since ∂(Ai � Aj ) ⊂ ∂Ai ∪ ∂Aj . By Observation 2.1.1, � �� � � � �� �F (Ai ) − F (Aj )� = �� F (Ai � Aj ) + F (Ai ⊙ Aj) − F (Aj � Ai ) + F (Aj ⊙ Ai ) �� � � � � ≤ �F (Ai � Aj )� + �F (Aj � Ai )� � � ≤ �div w�L∞ (K) |Ai � Aj | + |Aj � Ai | + 4cε
= �div w�L∞ (K) |Ai � Aj | + 4cε. � It follows that F (Ai ) is a Cauchy sequence. The value �
F�(E) := lim F (Ai )
does not depend on the sequence {Ai } of dyadic figures converging to E. Indeed, if {Bi } is another sequence of dyadic figures converging to E, then so does the interlaced sequence {A1 , B1 , A2 , B2 , . . . }, and consequently lim F (Ai ) = lim F (Bi ). Denote by DF the family of all sets E ⊂ Rn , necessarily measurable, for which there is a sequence {Ai } in DF converging to E. Clearly DF ⊂ DF, and since
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2.4. Convergence of dyadic figures
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33
∂(A ∪ B) ∪ ∂(A � B) ∪ ∂(A ⊙ B) ⊂ ∂A ∪ ∂B, it is easy to verify that the family DF is closed with respect to unions, intersections, and set differences. It follows from Proposition 2.4.2 that the flux F of v ∈ C(Rn ; Rn ) defined on figures in Section 2.1 has a unique extension F� : DF → R, still called the flux of v. Proposition 2.4.3. If F� : DF → R is the flux of v ∈ C(Rn ; Rn ), then F�(A ∪ B) = F�(A) + F�(B)
for each pair of nonoverlapping sets A, B ∈ DF.
Proof. Let {Ak } and {Bk } be sequences in DF that converge to A and B, respectively. From (Ak − Bk ) − A ⊂ Ak − A and A − (Ak − Bk ) = (A − Ak ) ∪ (A ∩ Bk ) ⊂ (A − Ak ) ∪ (A ∩ B) ∪ (Bk − B), � � we infer lim�A � (Ak � Bk )� = 0. As ∂(Ak � Bk ) ⊂ ∂Ak ∪ ∂Bk , the sequences {Ak � Bk } � � and (Ak � Bk ) ∪ Bk = {Ak ∪ Bk } converge to A and A ∪ B, respectively. Thus � � F�(A ∪ B) = lim F (Ak � Bk ) ∪ Bk = lim F (Ak � Bk ) + lim F (Bk ) = F�(A) + F�(B).
Proposition 2.4.4. Let F� : DF → R be the flux of v ∈ C(Rn ; Rn ). If v is admissible and div v belongs to L1loc (Rn ), then for each E ∈ DF, � div v(x) dx. F�(E) = E
Proof. If {Ai } is a sequence of dyadic figures converging to E, then Proposition 2.4.2, Theorem 2.3.7, and Lemma 2.4.1 imply � � F�(E) = lim F (Ai ) = lim div v(x) dx = div v(x) dx. Ai
E
Proposition 2.4.4 establishes the divergence theorem for sets in DF, which are the desired generalization of dyadic figures (cf. Corollary 6.7.4 below). However, the flux F� : DF → R does not share the geometric content of the flux F : DF → R defined by formula (2.1.1). This will be remedied in Chapters 4–6 below, albeit with a substantial effort. We show that each set E ∈ DF has an “essential boundary” ∂∗ E ⊂ ∂E and a “unit exterior normal” νE , defined Hn−1 almost everywhere on ∂∗ E, such that the flux F� of a vector field v ∈ C(Rn ; Rn ) is calculated by the formula � v · νE dHn−1 F�(E) = ∂∗ E
analogous to (2.1.1); see formula (6.5.1) below.
Remark 2.4.5. Using the convergence of dyadic figures, it is possible to define a sequential topology T in DF that is induced by a uniformity, and show that DF is the sequential completion of the space (DF, T); see Chapter 10, in particular Section 10.6. Since the flux F : (DF, T) → R of v ∈ C(Rn ; Rn ) is uniformly continuous by additivity, it has a unique continuous extension F� : DF → R — a fact we proved directly in Proposition 2.4.2.
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Chapter 3 Removable singularities
We will study removable singularities for the Cauchy-Riemann, Laplace, and minimal surface equations. As these equations are in the divergence form � � div φ(Du) = 0, the integration by parts theorem established in the previous chapter is a natural tool. We define removable sets by means of Hausdorff measures, mostly without any topological restrictions. The results are established by short and simple arguments, which rely on the relationship between weak and strong solutions of partial differential equations. A few basic facts about distributions and weak solutions are stated without proofs. We made no attempt to survey the long history concerning removable singularities.
3.1. Distributions A multi-index is an n-tuple α := (α1 , . . . , αn ) where αi are nonnegative inte�n gers. Let |α| := i=1 αi and �α1 �αn � � ∂ ∂ α1 α αn D := D1 · · · Dn = ··· . ∂ξ1 ∂ξn Note that if |α| = 0, then Dα f = f for any f : Rn → C. LetΩ ⊂ Rn be an open set. Employing convention (1.1.2), we say that a sequence {ϕi } in Cc∞ (Ω; C) converges to zero in the sense of test functions if the following conditions hold: (i) {ϕi } is a sequence in Cc∞ (U ; C) for an open set U � Ω; (ii) lim �Dα ϕi �L∞ (Ω;C) = 0 for each multi-index α.
The complex linear space Cc∞ (Ω; C) equipped with this convergence is denoted by D(Ω; C), and the elements of D(Ω; C) are called test functions. The real linear subspace of D(Ω; C) consisting of all real-valued test functions is denoted by D(Ω). A distribution is a linear functional L : D(Ω; C) → C such that lim L(ϕi ) = 0 35
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3. Removable singularities
for each sequence {ϕi } in D(Ω; C) that converges to zero in the sense of test functions. The complex linear space of all distributions is denoted by D� (Ω; C). The real linear space D� (Ω) is defined analogously. Remark 3.1.1. In Example 3.6.5 below we define a locally convex topology S in the spaces D(Ω) of real-valued test functions so that the space D� (Ω) of distributions is the dual space � of D(Ω), S). The reader familiar with complex locally convex spaces will recognize instantly that a similar topology can be defined in the space D(Ω; C) of complex-valued test functions [64, Section 6.2].
Example 3.1.2. Let f ∈ L1loc (Ω; C), let µ be a Radon measure in Ω, and let v ∈ L1loc (Ω; Rn ). The linear functionals � f ϕ dx : D(Ω; C) → C, (1) Lf : ϕ �→ �Ω Lµ : ϕ �→ ϕ dµ : D(Ω; C) → C, (2) Ω � Fv : ϕ �→ − v · Dϕ dx : D(Ω) → R (3) Ω
are examples of distributions. Distribution Fv is called the distributional divergence of v, since if v ∈ C 1 (Ω; Rn ) integration by parts yields � ϕ div v dx = Ldiv v (ϕ) Fv (ϕ) = Ω
for each test function ϕ ∈ D(Ω).
Let α be a multi-index. If f ∈ C |α| (Ω; C), then repeated applications of the integration by parts theorem yield � α �LD f , ϕ� = ϕ(x)Dα f (x) dx Ω � = (−1)|α| f (x)Dα ϕ(x) dx = (−1)|α| �Lf , Dα ϕ� Ω
for each ϕ ∈ D(Ω; C). Since for any distribution L, the linear functional ϕ �→ (−1)|α| �L, Dα ϕ� : D(Ω; C) → C
is a distribution, the previous identity suggests to define a distribution Dα L by the formula �Dα L,ϕ � := (−1)|α| �L, Dα ϕ� for each test function ϕ ∈ D(Ω; C). Observe that D α L f = LD α f
whenever f ∈ C |α| (Ω; C). Additional information about test functions and distributions can be found in many standard textbooks, for instance in [64, Chapter 6] or [27, Chapter 5].
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3.2. Differential equations
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37
3.2. Differential equations A linear differential operator with constant coefficients is the expression � Λ := cα D α |α|≤k
where k ∈ N and cα ∈ C for each multi-index α with |α| ≤ k. In an open set Ω ⊂ Rn we consider two types of solutions of the partial differential equation Λu = 0. k • A � strong � solution is a complex-valued function u ∈ C (Ω; C) such that Λu, x = 0 for each x ∈ Ω. • A weak solution is a complex-valued function u ∈ L1 (Ω; C) such that ΛLu = 0 where Lu is defined in Example 3.1.2, (1).
Thus u ∈ L1 (Ω; C) is a weak solution of Λu = 0 if the equality � � � � |α| (−1) cα u(x)Dα ϕ(x) dx = ΛLu , ϕ = 0 Ω
|α|≤k
holds for each test function ϕ ∈ D(Ω; C). For a multi-index α = (α1 , . . . , αn ) and x = (ξ1 , . . . , ξn ) in Rn , let xα := ξ1α1 · · · ξnαn where ξ αi = 1 when αi = 0. A complex-valued function � pΛ : x �→ c α xα : R n → C |α|=k
is called the characteristic polynomial of Λ. If pΛ (x) �= 0 for each x in Rn −{0}, the operator Λ is called elliptic. Note that the ellipticity of Λ is determined by the leading coefficients, i.e., by cα with |α| = k. Example 3.2.1. The following linear differential operators are elliptic. (1) The Laplace operator � := D12 + · · · + Dn2 , since p� (x) = |x|2 for each x ∈ Rn . √ (2) For n = 2 and i := −1, the holomorphic operator ∂¯ := D1 + iD2 , since p∂¯(x) = ξ1 + iξ2 for each x = (ξ1 , ξ2 ) in R2 .
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3. Removable singularities
Clearly, each strong solution of the equation Λu = 0 is also a weak solution. While the converse is generally false, the next theorem, proved in [64, Corollary of Theorem 8.12], asserts that strong and weak solutions are essentially the same when Λ is an elliptic operator. Theorem 3.2.2. Let Ω ⊂ Rn be an open set, let Λ be a linear differential operator with constant coefficients, and let uw ∈ L1 (Ω; C) be a weak solution of the equation Λu = 0. If Λ is elliptic, there is a strong solution us ∈ C ∞ (Ω; C) such that us (x) = uw (x) for almost all x ∈ Ω.
3.3. Holomorphic functions Throughout this section, we assume that Ω is an open subset of the complex plane C. A complex-valued function f ∈ C 1 (Ω; C) is called holomorphic if the Cauchy-Riemann equation ¯ (z) = 0 ∂f is satisfied for all z ∈ Ω, or equivalently, if the complex derivative f � (z) exists at each z ∈ Ω. Making the obvious identification between C and R2 , we can talk about admissibility of complex-valued functions in the sense of Definition 2.3.1. Explicitly, f : Ω → C is admissible if there are numbers 0 ≤ sk < 1 and disjoint, �∞ possibly empty, sets Ek such that f is pointwise Lipschitz inΩ − k=1 Ek and for each k = 1, 2, . . . , the following conditions hold: (i) Ek is H1+sk σ-finite, and Hsk f (z) < ∞ for each z ∈ Ek ; (ii) H1+sk (Ek ) > 0 implies Hsk f (z) = 0 for each z ∈ Ek .
The meaning of the symbol Adm(Ω; C) is clear. The following theorem generalizes the well-known classical result of A.S. Besicovitch [2]. Theorem 3.3.1. Let f ∈ Adm(Ω; C) have the complex derivative f � (z) at almost all z ∈ Ω. Then f can be redefined at the points of discontinuity so that it becomes holomorphic. Proof. Since f � (z) exists at almost all z ∈ Ω, the Cauchy-Riemann condition ¯ (z) = 0 holds for almost all z ∈ Ω. If z = x + iy and f = u + iv, then ∂f � � ∂y ∂v ∂u ∂v ¯ − +i + = div(u, −v) + i div(v, u) ∂f = ∂x ∂y ∂x ∂y where the vector fields (u, −v) and (v, u) belong to Adm(Ω; R2 ). After a simple calculation, the integration by parts theorem yields � � ¯ =− ¯ =0 f ∂ϕ ϕ ∂f (∗) Ω
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“Book˙2011” — 2012/2/26 — 9:58 — page 39 — #49
3.5. The minimal surface equation
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39
for each test function ϕ ∈ D(Ω; C). Indeed, (∗) follows from Remark 2.3.10, ¯ = 0. since ϕ ∈ Adm(Ω; C) and spt ϕ � Ω. Thus f is a weak solution of ∂u ¯ As the holomorphic operator ∂ is elliptic, Theorem 3.2.2 implies that there is a holomorphic function g : Ω → C which is equal to f almost everywhere. In particular, g(z) = f (z) for each z ∈ Ω at which f is continuous.
3.4. Harmonic functions LetΩ ⊂ Rn be an open set. A function u ∈ C 2 (Ω) is called harmonic if the Laplace equation �u(x) = 0
is satisfied for all x ∈ Ω. As the coefficients of the Laplace operator � are real numbers, in testing for real-valued weak solutions of the Laplace equation, it suffices to use only real-valued test functions. Theorem 3.4.1. Let C ⊂ Ω be an Hn−1 negligible set which is relatively closed in Ω, and let u ∈ Liploc (Ω) be differentiable at each point of Ω − C. If Du belongs to Adm(Ω − C; Rn ) and �u(x) = 0 for almost all x ∈ Ω, then u is harmonic. Proof. Select a test function ϕ ∈ D(Ω), and observe u,ϕ ∈ Adm(Ω) and Dϕ ∈ Adm(Ω; Rn ). As u is locally Lipschitz, the assumptions about C imply that the map φ : Ω → Rn defined by the formula � Du(x) if x ∈ Ω − C, φ(x) := 0 if x ∈ C
belongs to Adm(Ω; Rn ). Observe that for almost all x ∈ Ω, Du(x) = φ(x)
and
div φ(x) = �u(x) = 0.
Since �ϕ = div Dϕ, integrating by parts twice, we obtain � � � u�ϕ=− φ · Dϕ = ϕ � u = 0. Ω
Ω
Ω
Thus u is a weak solution of the equation �u = 0. As � is elliptic, Theorem 3.2.2 implies that there is a harmonic function h defined on Ω that equals u almost everywhere. Since u is continuous, u ≡ h.
3.5. The minimal surface equation LetΩ ⊂ Rn be an open set and u ∈ Liploc (Ω). The set � � G(u) := (x, y) ∈ Rn+1 : y = u(x)
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40
3. Removable singularities � � is called the graph of u, and the extended real number S(u) := Hn G(u) is called the surface area of G(u). It follows from [29, Section 3.3.4, B] that � � � �2 S(u) = 1 + �Du(x)� dx. Ω
Seeking a local minimum of the functional
S : u �→ S(u) : Liploc (Ω) → R, we assume that there is u ∈ C 2 (Ω) such that S(u) < ∞ and � � d S(u + tϕ) =0 dt t=0 for each ϕ ∈ D(Ω). Differentiating under the integral sign and integrating by parts, we obtain that for every ϕ ∈ D(Ω), � � � � Du Du · Dϕ d � 0= S(u + tϕ) = =− ϕ div � . 2 dt 1 + |Du| 1 + |Du|2 Ω Ω t=0 Consequently for all x ∈ Ω,
Du(x) div � � �2 = 0. 1 + �Du(x)�
(∗)
Theorem 3.5.1. Let Ω ⊂ Rn be a bounded open set whose boundary ∂Ω is a C 2 manifold, and let g ∈ C(∂Ω). There is a unique u ∈ C(cl Ω) such that u is C 2 in Ω, u � ∂Ω = g, and (∗) holds for each x ∈ Ω. If v ∈ C(cl Ω) is C 2 in Ω and v � ∂Ω = g, then S(u) ≤ S(v). Theorem 3.5.1 is proved in [36, Sections 13]. Together with the previous calculations, it explains why the equation div �
Du 1 + |Du|2
=0
(MSE)
is called the minimal surface equation. Unless stated otherwise, by a solution of (MSE) we mean a solution in Ω. Strong solutions of (MSE) are defined as in Section 3.2; weak solutions are defined differently, since (MSE) is a nonlinear equation. • A strong solution of (MSE) is a function u ∈ C 2 (Ω) such that equality (∗) holds for every x ∈ Ω. • A weak solution of (MSE) is an almost everywhere differentiable function u : Ω → R such that Du ∈ L1loc (Ω) and � Du · Dϕ � =0 1 + |Du|2 Ω for each ϕ ∈ Cc1 (Ω).
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41
It is clear that each strong solution of (MSE) is a weak solution. A partial converse stated below is due to E. De Giorgi [20]. Theorem 3.5.2 (De Giorgi). If u ∈ Liploc (Ω) is a weak solution of (MSE), then it is a strong solution. We use Theorem 3.5.2 in lieu of Theorem 3.2.2, and employ the following result of L. Simon; see [68] or [36, Theorem 16.9]. Theorem 3.5.3 (Simon). Let C ⊂ Ω be an Hn−1 negligible set which is relatively closed in Ω. Every strong solution of (MSE) in Ω − C can be extended to a strong solution in Ω. Theorem 3.5.4. Let C ⊂ Ω be an Hn−1 negligible set which is relatively closed in Ω, and let u : Ω − C → R be differentiable at each point of Ω − C. If Du belongs to Adm(Ω − C; Rn ) and u satisfies (∗) for almost all x ∈ Ω − C, then u can be extended to Ω so that it is a strong solution of (MSE). Proof. Select ϕ ∈ Cc1 (Ω − C), and observe that both ϕ and x �→ �
Du(x) � �2 : Ω − C → R 1 + �Du(x)�
belong to Adm(Ω − C). Integrating by parts yields �
Ω−C
Du · Dϕ � =− 1 + |Du|2
�
Ω−C
ϕ div �
Du 1 + |Du|2
= 0,
and u is a weak solution of (MSE) inΩ − C. As Du is admissible inΩ − C by our assumption, Du is locally bounded by Remark 2.3.3. Thus u is locally Lipschitz, and De Giorgi’s theorem implies that it is a strong solution of (MSE) inΩ − C. An application of Simon’s theorem completes the proof. A nontrivial strong solution of (MSE) in Rn is called a Bernstein function. The following theorem shows that in higher dimensions, solutions of (MSE) may not correspond to our intuition about minimizing the surface area [36, Section 17]. Theorem 3.5.5. In dimension n ≤ 7, each Bernstein function is linear, i.e., its graph is a hyperplane in Rn+1 . If n > 7, there exist nonlinear Bernstein functions.
3.6. Injective limits We describe a general procedure for obtaining a new locally convex topology in a locally convex space X by restricting the original topology to a suitable family of subspaces. This construction, called the internal injective limit, has many applications. In Example 3.6.5 below, we use it to define a topology S in the space D(Ω) of test functions mentioned in Remark 3.1.1. Another application is given in Section 5.3 below.
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3. Removable singularities
A family A of sets is called directed upward by inclusion, or simply directed, if for each pair A, B ∈ A there is C ∈ A containing both A and B. For any family A of sets and any set E, we define the family A
E := {A ∩ E : A ∈ A}.
In this notation, if (X, T) is a topological space and Y ⊂ X, then T subspace topology of Y inherited from (X, T).
Y is the usual
Theorem 3.6.1. Let (X, T) be a locally convex space, and let A = {Xα : α ∈ A} be � a directed family of subspaces of X such that X = α∈A Xα . There is a unique locally convex topology TA in X that satisfies the following conditions: (i) TA Xα ⊂ T Xα for every α ∈ A, (ii) given a locally convex space Y , a linear map φ : (X, TA ) → Y is continuous whenever each restriction φ � (Xα , T Xα ) → Y is continuous. The topology TA has the following properties: (1) T ⊂ TA and T Xα = TA Xα for every α ∈ A. (2) TA is the largest topology among locally convex topologies S in X for which all inclusion maps iα : x �→ x : (Xα , T Xα ) → (X, S) are continuous. (3) TA has a neighborhood base B ⊂ TA at zero consisting of all absorbing symmetric convex sets U ⊂ X such that U ∩ Xα ∈ T Xα for every α ∈ A. Proof. If S is a locally convex topology in X that satisfies conditions (i) and (ii), then a straightforward verification reveals that the identity map id : x �→ x : (X, TA ) → (X, S) is a homeomorphism. Hence S = TA , which establishes the uniqueness of TA . (1) By condition (ii), the map id : (X, TA ) → (X, T) is continuous and so T ⊂ TA . Now fix α ∈ A, and observe that the diagram (Xα , TA
Xα )
��� ��⊂ ��� jα ��� � � (X, T) (X, TA ) iα ⊂
id
commutes. Since the maps iα and id are continuous, so is the inclusion map jα . Given U ∈ T Xα , there is V ∈ T with V ∩ Xα = U . Consequently U = j−1 α (V ) belongs to TA Xα . Now condition (i) implies T Xα = TA Xα . (2) It follows from (1) that all inclusions iα : (Xα , T Xα ) → (X, TA ) are continuous. Let S be a locally convex topology in X such that all inclusions iα : (Xα , T Xα ) → (X, S) are continuous. Then the identity map id : (X, TA ) → (X, S) is continuous by condition (ii), and we conclude S ⊂ TA . (3) Let B be the family of all absorbing, symmetric, and convex sets U ⊂ X such that for every α ∈ A the following condition holds: i−1 α (U ) is an open neighborhood of zero in the topology T Xα , or equivalently, U ∩ Xα ∈ T Xα . If U ∈ B and t ∈ R − {0}, then tU ∩ Xα = t(U ∩ Xα ) belongs to T Xα for each α ∈ A. Thus tU ∈ B, and as U is convex, 1 U + 12 U ⊂ U . Since the family B is closed with respect to finite intersections, it is a 2 neighborhood base at zero for a locally convex topology U in X, provided we show that U is Hausdorff. To this end, recall that in accordance with the general assumption made in Section 1.2, the topology T is Hausdorff. Since each symmetric convex T neighborhood of
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43
zero belongs to B, the topology U is also Hausdorff. From the definition of B it is clear that U is the largest among locally convex topologies S in X for which all inclusions Xα ) → (X, S)
iα : (Xα , T
are continuous. Thus U = TA by property (2). The topology TA defined in Theorem 3.6.1 is called the internal injective limit of the topologies T Xα ; see [27, Section 6.3]. The customary notation is TA = inj lim(T α∈A
Xα )
and
(X, TA ) = inj lim(Xα , T α∈A
Xα ).
Families A and B of sets are called interlacing if each A ∈ A is contained in some B ∈ B, and each B ∈ B is contained in some A ∈ A. Proposition 3.6.2. Let (X, T) be a locally convex space, and let A = {Xα : α ∈ A} and B = {Xβ : β ∈ B} be interlacing families of subspaces of X. If A is directed and � � X = α∈A Xα , then B is directed, X = β∈B Xβ , and inj lim(T α∈A
Xα ) = inj lim(T β∈B
Xβ ).
Proof. Let TA = inj limα∈A (T Xα ) and T B = inj limβ∈B (T find β ∈ B with Xα ⊂ Xβ . The diagram i
α −→ (Xα , T Xα ) −−−− ⊂ ⊂�
(Xβ , T
(X, T B ) � ⊂
Xβ ) −−−−−→ (Xβ , T B id
Xβ ). Choose α ∈ A and
Xβ )
commutes, and the vertical maps are clearly continuous. By Theorem 2.6.1, condition (i), the map id is continuous as well. Thus the map iα is continuous, and we conclude from Theorem 2.6.1, condition (ii), that id : (X, TA ) → (X, T B ) is continuous. The proposition follows by symmetry. A sequence {xk } in a topological linear space X is called Cauchy if for each neighborhood U of zero, xk ∈ U for all but finitely many k. If each Cauchy sequence in X converges, the space X is called sequentially complete. A closed subspace of a sequentially complete space is sequentially complete. Theorem 3.6.3. Let (X, T) be a locally convex space, and let J = {Xj : j ∈ N} be an � increasing sequence of closed subspaces of X such that X = j∈N Xj . Then (X, S) = inj lim(Xj , T j→∞
Xj )
has the following properties: (1) A sequence {xk } in X converges in (X, S) if and only if {xk } is a sequence in some Xj that converges in (X, T). (2) E ⊂ X is S bounded if and only if E is a T bounded subset of some Xj . (3) If each (Xj , T Xj ) is sequential, then a linear map φ from (X, S) to a locally convex space Y is continuous whenever lim φ(xk ) = 0 for each sequence {xk } that converges to zero in (X, S). (4) If each (Xj , T Xj ) is sequentially complete, then so is (X, S).
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44
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3. Removable singularities
Proof. (1) If {xk } is a sequence converging to zero in (X, S), then {xk } converges to zero in (X, T) by Theorem 3.6.1, (1). Seeking a contradiction, assume {xk } is not contained in any Xj , and construct recursively a subsequence {yj } of {xk } so that yj ∈ / Xj for j = 1, 2, . . . . Since Xj is closed, there is a convex symmetric neighborhood Uj ∈ T of zero such that yj � does not belong to Wj = Xj + Uj ; see [64, Theorem 1.10]. The intersection W = ∞ j=1 Wj is convex and symmetric. Since Xk ⊂ Xj ⊂ Wj whenever j ≥ k, � Xk ∩ W = Xk ∩ {Wj : j = 1, . . . , k} (∗) �k
j=1 Wj belongs to T, and hence Xk ∩W belongs to � T Xk . Given x ∈ X, find k ∈ N and t ∈ R so that x ∈ Xk and tx ∈ kj=1 Wj . As tx ∈ Xk , equality (∗) implies tx ∈ W . Thus W is an absorbing set, and hence an S neighborhood of zero by Theorem 3.6.1, (3). Since no yj belongs to W , this is a contradiction. The converse follows from Theorem 3.6.1, (1).
for k = 1, 2, . . . . The finite intersection
(2) If E ⊂ X is S bounded, it is T bounded by Theorem 3.6.1, (1). Assume that E is not contained in any Xj , and construct recursively a sequence {xj } in E so that xj ∈ / Xj � � for j = 1, 2, . . . . Since 1j xj converges to zero in (X, S), part (1) of the theorem shows that all xj belong to some Xk , a contradiction. (3) Assume that lim φ(xk ) = 0 for each {xk } in X that converges to zero in (X, S). If {xk } is a sequence in Xj that converges to zero in (Xj , T Xj ), then lim φ(xk ) = 0 by part (1) of the theorem. By our assumption, φ � Xj : (Xj , T Xj ) → Y is continuous, and Theorem 3.6.1, (ii) implies that φ : (X, S) → Y is continuous. (4) If {xk } is a Cauchy sequence in (X, S), then the set {xk : k ∈ N} is S bounded. Part (2) of the theorem and Theorem 3.6.1, (1) imply that {xk } is a Cauchy sequence in some (Xj , T Xj ). By our assumption, {xk } converges in (Xj , T Xj ), and hence in (X, S) by part (1) of the theorem.
Example 3.6.4. LetΩ ⊂ Rn be an open set, and let O be the family of all open sets U � Ω. Denote by T the topology in X = Cc (Ω; Rm ) defined by the L∞ norm, and let � � XU := v ∈ X : spt v ⊂ U
for each U ∈ O. There are compact sets Kj ⊂ Ω such that Kj � Kj+1 for each j ∈ N, and � j∈N Kj = Ω. Define closed subspaces � � Xj := v ∈ X : spt v ⊂ Kj of (X, T), and note the families {XU : U ∈ O} and {Xj : j ∈ N} are interlaced. Thus S = inj lim(T j→∞
Xj ) = inj lim(T U ∈O
XU )
by Proposition 3.6.2. In particular, S does not depend on the choice of {Kj }. Although the metrizable space (X, T) is not complete, the space (X, S) is sequentially complete according to Theorem 3.6.3, (4). On the other hand, the space (X, S) is not metrizable. To see this, select disjoint balls B(xk , rk ) ⊂ Ω, k = 0, 1, . . . so that the closure of {x0 , x1 , . . . } meets ∂Ω. If ϕk ∈ X are such that ϕk (xk ) = 1 and spt ϕk ⊂ B(xk , rk ), then in (X, S) the following is true: � � lim k1 ϕ0 = 0 and lim 1j ϕk + k1 ϕ0 = k1 ϕ0 , k = 1, 2, . . . . k→∞
j→∞
� � However, there is no diagonal sequence j1 ϕk + k1 ϕ0 that converges to zero in (X, S). k This shows that S does not have a countable neighborhood base at zero.
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45
According to Theorem 3.6.1, (ii), the dual space (X, S)∗ consists of all linear functionals F : X → R such that F � XU is T continuous for each open set U � Ω. Alternatively, a linear functional F : X → R belongs to (X, S)∗ if and only if � � �F �U = sup F (v) : v ∈ XU and �v�L∞ (Ω;Rm ) ≤ 1 < ∞ for each open set U � Ω. Denote by S the collection of all sequences s = {sj } in R+ that converge to zero. Let K0 = ∅, and for s ∈ S and v ∈ X define � � � � �v(x)� : x ∈ Ω − Kj−1 . ps (v) := sup sup s−1 j j∈N
It not difficult to verify that every ps is a seminorm in X.
Claim. The topology S in X is defined by the family of seminorms {ps : s ∈ S}.
Proof. Denote by U the topology in X defined by the seminorms ps . Given s ∈ S, � the set V = v ∈ X : ps (v) < 1} is absorbing, symmetric, and convex. If j ∈ N and t = min{s1 , . . . , sj }, then � � V ∩ Xj = Xj ∩ v ∈ X : �v�L∞ (Ω;Rm ) < t .
Thus V ∈ S. Conversely, choose an S neighborhood W of zero and find 0 < s1 < 1 so that every v ∈ X with spt v ⊂ K1 and �v�L∞ (Ω;Rm ) < s1 is contained in W . Next find 0 < s2 < s1 /2 so that every v ∈ X with spt v ⊂ K2 and �v�L∞ (Ω;Rm ) < s2 is contained in W . Proceeding recursively, construct a decreasing sequence s = {sj } in R+ such that lim sj = 0, and every v ∈ X with spt v ⊂ Kj and �v�L∞ (Ω;Rm ) < sj is contained in W . It � follows that v ∈ X : ps (v) < 1} ⊂ W , and we conclude S = U. If Y is a subspace of X, let YU = XU ∩ Y for each U ∈ O. In view of the claim, it is clear that the topology SY = inj limU ∈O (T YU ) in Y is defined by the family of seminorms {ps � Y : s ∈ S}. Thus SY = S Y . Example 3.6.5. Let Ωand O be as in Example 3.6.4. For k ∈ N, let pk (ϕ) := max �D α ϕ�L∞ (Ω) |α|≤k
where α is a multi-index. Each pk is a norm in D(Ω), and we denote by T the topology in � D(Ω) defined by the family pk : k ∈ N}; see Section 1.2. Letting � � S = inj lim T D(U ) , U ∈O
it follows from Theorem 3.6.1 that a linear functional L : D(Ω) → R is a distribution if and only if it is S continuous. Arguing as in Example 3.6.4, we can show that the space � � D(Ω), S is sequentially complete but not metrizable. The seminorms which define the topology S in D(Ω) are similar in spirit to those described in Example 3.6.4, but technically more complicated [27, Section 6.3].
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“Book˙2011” — 2012/2/26 — 9:58 — page 47 — #57
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Part 2
Sets of finite perimeter
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“Book˙2011” — 2012/2/26 — 9:58 — page 49 — #59
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Chapter 4 Perimeter
Our goal is to extend the divergence theorem from dyadic figures to more general sets, called the sets of finite perimeter. These sets have two equivalent definitions: geometric, which is intuitive but difficult to work with, and analytic, which is effective but nonintuitive. Both definitions are essential, as they complement each other. In this chapter, we introduce the geometric definition of perimeter, and derive elementary properties of sets whose perimeter is finite. Some of these properties will motivate the analytic concept of variation presented in the next chapter.
4.1. Measure-theoretic concepts We wish to define the flux of a vector field v from a set E ⊂ Rn so that it resembles the flux of v from a figure, and satisfies the divergence theorem when v and E are sufficiently regular. To this end, we need a “boundary” B and a “unit exterior normal” ν of a measurable set E such that the equality � � div v dLn = v · ν dHn−1 E
B
holds under suitable assumptions. As the left side of this equality depends only on the equivalence class of E, so must the right side. Thus ignoring temporarily the issue of “exterior normal”, we will seek a definition of the “boundary” B of E which depends only on the equivalence class of E. Such a task cannot be accomplished by topological means only — in particular, we cannot let B = ∂E. Fortunately, there is a suitable measure-theoretic analogue of the topological boundary. For any E ⊂ Rn , the sets � � � � �E ∩ B(x, r)� n � =0 , ext∗ E := x ∈ R : lim �� r→0 B(x, r)� int∗ E := ext∗ (Rn − E),
cl∗ E := Rn − ext∗ E
∂∗ E := ∂∗ E − int∗ E, 49
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50
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4. Perimeter
depend only on the equivalence class of E. We call them, respectively, the essential exterior , essential interior , essential closure, and essential boundary of E. A direct verification shows int∗ E ⊂ cl∗ E,∂
∗E
= cl∗ E ∩ cl∗ (Rn − E) = ∂∗ (Rn − E),
and from ext∗ (A ∪ B) = ext∗ A ∩ ext∗ B, we obtain int∗ (A ∩ B) = int∗ A ∩ int∗ B
and
cl∗ (A ∪ B) = cl∗ A ∪ cl∗ B.
The relationship to the corresponding topological concepts is given by int E ⊂ int∗ E,
cl∗ E ⊂ cl E,
and
∂∗ E ⊂ ∂E.
Moreover, ∂∗ E = ∂E if and only if cl∗ E = cl E and int∗ E = int E. As the converse is obvious, assume ∂∗ E = ∂E and observe that cl E = int E ∪ ∂E ⊂ int∗ E ∪ ∂∗ E = cl∗ E ⊂ cl E implies cl∗ E = cl E, and hence int∗ E = int E. The obvious inclusion � � � � �E ∩ B(x, r)� n � =1 int∗ E ⊂ x ∈ R : lim �� r→0 B(x, r)� becomes equality when E is a measurable set.
For x ∈ Rn , let x� := πn (x) and xn := x · en . Given x ∈ Rn and positive numbers r and h, we define an open cylinder � � C(x; r, h) := πn U (x, r) × (xn − h, xn + h).
Definition 4.1.1. An open setΩ ⊂ Rn is called a Lipschitz domain if for each point x ∈ ∂Ω there are a cylinder C(x; r, h), a function g ∈ Lip(Rn−1 ), and a rotation φ of Rn about x such that � � φ(Ω) ∩ C(x; r, h) = y ∈ C(x; r, h) : g(y � ) < yn .
IfΩ ⊂ Rn is a Lipschitz domain, then so is Rn − cl Ω. The interiors of figures are Lipschitz domains, and it follows from [66, Theorem 1.5.1] or [29, Section 6.3, Theorem 1] that so are open convex subsets of Rn . Proposition 4.1.2. If Ω is a Lipschitz domain, then ∂∗ Ω = ∂Ω. Proof. Choose x ∈ ∂Ω, and let yn := y · en and y � := πn (y) for each y ∈ Rn . As both ∂Ω and ∂∗ Ω are invariant with respect to translations and rotations, we may assume that x = 0, and that there are a cylinder C = C(0; r, h) and a Lipschitz function g : Rn−1 → R such that � � Ω ∩ C = y ∈ C : g(y � ) < yn .
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4.2. Essential boundary
51 � � Observe g(0) = 0, and define cones C± := y ∈ Rn : (Lip g)|y � | ≤ ±yn . Clearly C+ ∩ C ⊂ Ω ∩ C and C− ∩ C ∩ Ω = ∅. Since for each s > 0, � � �C± ∩ B(0, s)� � � =c �B(0, s)� where 0 < c ≤ 1/2, we see that x ∈ ∂∗ Ω.
4.2. Essential boundary We establish some elementary facts about the essential boundaries of arbitrary subsets of Rn . Observation 4.2.1. If A, B ⊂ Rn , then ∂∗ (A ∪ B) ∪ ∂∗ (A ∩ B) ∪ ∂∗ (A − B) ⊂ ∂∗ A ∪ ∂∗ B. Proof. By a direct calculation, ∂∗ (A ∪ B) = cl∗ (A ∪ B) − int∗ (A ∪ B) ⊂ cl∗ A ∪ cl∗ B − int∗ A ∪ int∗ B ⊂ (cl∗ A − int∗ A) ∪ (cl∗ B − int∗ B) = ∂∗ A ∪ ∂∗ B,
∂∗ (A ∩ B) = cl∗ (A ∩ B) − int∗ (A ∩ B) ⊂ cl∗ A ∩ cl∗ B − int∗ A ∩ int∗ B
⊂ (cl∗ A − int∗ A) ∪ (cl∗ B − int∗ B) = ∂∗ A ∪ ∂∗ B. � � Since ∂∗ (A − B) = ∂∗ A ∩ (Rn − B) and ∂∗ (Rn − B) = ∂∗ B, the observation follows. Proposition 4.2.2. If Ω ⊂ Rn is an open set, then for each E ⊂ Rn , int∗ E ∩ Ω = int∗ (E ∩ Ω) ∩ Ω,
cl∗ E ∩ Ω = cl∗ (E ∩ Ω) ∩ Ω,
∂∗ E ∩ Ω = ∂∗ (E ∩ Ω) ∩ Ω. Proof. Since a direct verification shows int∗ E ∩ Ω ⊂ int∗ E ∩ int∗ Ω = int∗ (E ∩ Ω) ⊂ int∗ E, cl∗ E ∩ Ω ⊂ cl∗ (E ∩ Ω) ⊂ cl∗ E,
we obtain the first two equalities by intersecting with Ω. The equality for essential boundaries follows. The shape of a bounded set C ⊂ Rn is the number � |C| if d(C) > 0, n s(C) := d(C) 0 otherwise. From the isodiametric inequality (1.4.2) we obtain s(C) ≤ α(n)/2n . The bound is attained when C is a ball.
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4. Perimeter
n n Lemma 4.2.3.� Let {Ck } be �a sequence � � of subsets of R and x n∈ R . If lim d Ck ∪ {x} = 0 and inf s Ck ∪ {x} > 0, then for each E ⊂ R , � 0 if x ∈ ext∗ E, |E ∩ Ck | = lim |Ck | 1 if x ∈ int∗ E.
� � Proof. Let �dk = d �Ck ∪ {x} � and Bk = B(x, dk ). Observe Ck ⊂ Bk and � inf |Ck |/|Bk | ≥ inf s Ck ∪ {x} /α(n) > 0. If x ∈ ext∗ E, then � � |E ∩ Bk | |E ∩ Ck | |Ck | ≥ lim sup · 0 = lim |Bk | |Ck | |Bk | |Ck | |E ∩ Ck | inf ≥0 ≥ lim sup |Ck | |Bk | � � and hence lim |E ∩ Ck |/|Ck | = 0. If x belongs to int∗ E = ext∗ (Rn − E), then by the first part of the proof, � m � �(R − E) ∩ Ck � |Ck | − |E ∩ Ck | ≥ lim sup 0 = lim |Ck | |Ck | |E ∩ Ck | ≥ 0. = 1 − lim inf |Ck | Proposition 4.2.4. If A, B ⊂ Rn , then cl∗ A ∩ int∗ B ⊂ cl∗ (A ∩ B).
Proof. Choose x ∈ cl∗ A ∩ int∗ B. There is a sequence {rk } in R+ such that lim rk = 0 and � � �A ∩ B(x, rk )� � = a > 0. lim �� B(x, rk )� Let Bk := B(x, rk ), and observe that
� � |A ∩ Bk | α(n) a α(n) |A ∩ Bk | > n · s (A ∩ Bk ) ∪ {x} ≥ = n · (2rk )n 2 |Bk | 2 2
for all sufficiently large k. Applying Lemma 4.2.3, we obtain � � � � �(A ∩ B) ∩ Bk � �B ∩ (A ∩ Bk )� |A ∩ Bk | lim = lim · lim = a. |Bk | |A ∩ Bk | |Bk | Consequently x ∈ cl∗ (A ∩ B).
Corollary 4.2.5. If A and B are measurable sets, then ∂∗ (A ∩ B) ⊃ (∂∗ A ∩ int∗ B) ∪ (int∗ A ∩ ∂∗ B)
and
∂∗ (A ∩ B) ⊂ (∂∗ A ∩ int∗ B) ∪ (int∗ A ∩ ∂∗ B) ∪ (∂∗ A ∩ ∂∗ B).
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4.3. Vitali’s covering theorem
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53
Proof. Proposition 4.2.4 implies ∂∗ A ∩ int∗ B = (cl∗ A − int∗ A) ∩ int∗ B
= cl∗ A ∩ int∗ B − int∗ (A ∩ B)
⊂ cl∗ (A ∩ B) − int∗ (A ∩ B) = ∂∗ (A ∩ B), and by symmetry, ∂∗ B ∩ int∗ A ⊂ ∂∗ (A ∩ B). On the other hand, ∂∗ (A ∩ B) ⊂ cl∗ (A ∩ B) ⊂ cl∗ A ∩ cl∗ B
= (int∗ A ∪ ∂∗ A) ∩ (int∗ B ∪ ∂∗ B)
= (∂∗ A ∩ int∗ B) ∪ (int∗ A ∩ ∂∗ B) ∪ (∂∗ A ∩ ∂∗ B), since int∗ (A ∩ B) and ∂∗ (A ∩ B) are disjoint sets.
4.3. Vitali’s covering theorem We prove an important combinatorial result which has many applications. In particular, it will allow us to proof some deeper properties of essential boundary. Throughout the remainder of this book, we adhere to the following convention. Convention 4.3.1. When no attributes are added, a ball is always a closed ball B(x, r) where x ∈ Rn and r ∈ R+ . Given a ball B = B(x, r), we let B • := B(x, 5r). Theorem 4.3.2 (Vitali). Let B be a family of balls. If d(B) ≤ d < ∞ for each B ∈ B, then there is a disjoint family C ⊂ B such that � � B ⊂ {C • : C ∈ C}. Proof. The family B is the union of disjoint subfamilies � � Bi := B ∈ B : 2−i d < d(B) ≤ 2−i+1 d , i = 1, 2, . . . .
By Zorn’s lemma [43, Chapter 0, Theorem 25], the family B1 contains a maximal disjoint subfamily C1 . We use recursion to construct families Ci , i = 2, 3, . . . , so that Ci is a maximal disjoint subfamily of � � i−1 � �� � n Ci = ∅ . B ∈ Bi : B ⊂ R − �∞
j=1
Observe that C := i=1 Ci is a disjoint subfamily of B, and choose a ball B ∈ B. As B ∈ Bk for an integer k ≥ 1, the maximality of Ck implies that B �k meets a ball C ∈ i=1 Ci . The inclusion B ⊂ C • follows from the inequality d(B) ≤ 2−k+1 d < 2d(C).
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4. Perimeter
� � Remark 4.3.3. As the family B = B(0, k) : k ∈ N contains no�nonempty disjoint subfamily, it is essential to assume that {d(B) : B ∈ B is a bounded set. In the proof of Vitali’s theorem, we used Zorn’s lemma merely for convenience. A slightly longer constructive proof is available. Theorem 4.3.4. Let Ω ⊂ Rn be an open set. If f ∈ L1loc (Ω), then � 1 � f (y) dy = f (x) lim �� r→0 B(x, r)� B(x,r) for almost all x ∈ Ω.
Proof. With no loss of generality, we �may assume that Ω is bounded and f is a real-valued function. Let F (B) := B f for every ball B ⊂ Ω, and denote by N the set of all x ∈ Ω at which the limit � � F B(x, r) � lim � r→0 �B(x, r)� either does not exist, or differs from f (x). Given x ∈ N , find γx > 0 so that for each η > 0, there is a positive r < η such that � � � � � ��� � � (∗) B(x, r) ⊂ Ω and �F B(x, r) − f (x)�B(x, r)�� ≥ γx �B(x, r)�. � � It suffices to show that each Nk := x ∈ N : γx > 1/k is a negligible set. Fix k ∈ N and choose ε > 0. There is δ : Ω → R+ such that p � � � ε � � (∗∗) �f (xi )|Bi | − F (Bi )� < n 5 k i=1 whenever Bi ⊂ Ω are disjoint balls, xi ∈ Bi , and d(Bi ) < δ(xi ) for i = 1, . . . , p; see Henstock’s lemma. For every x ∈ Nk select a positive rx < δ(x) so that (∗) holds for B(x, rx ). By Vitali’s theorem there are disjoint balls Bi = B(xi , rxi ) � satisfying Nk ⊂ i Bi• . Combining (∗) and (∗∗), we obtain � � � ��� � |Nk | ≤ |Bi• | = 5n |Bi | ≤ 5n k �f (xi )|Bi | − F (Bi )� < ε , i
i
i
and the theorem follows from the arbitrariness of ε.
4.4. Density Let E ⊂ Rn and x ∈ Rn . If the limit
� � �E ∩ B(x, r)� � Θ(E, x) := lim �� r→0 B(x, r)�
exists, it is called the density of E at x. Clearly, � � ext∗ E = x ∈ Rn : Θ (E, x) = 0 .
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4.4. Density
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Observation 4.4.1. Let E ⊂ Rn and θ ∈ R. The set � � E(θ) := x ∈ Rn : Θ (E, x) = θ
is Borel, and so are the sets ext∗ E, int∗ E, cl∗ E, and ∂∗ E. Proof. Enumerate all positive rationals as r1 , r2 , . . . , and note that � � �E ∩ B(x, r)� � : Rn → R fr : x �→ �� B(x, r)�
is a continuous function for every r > 0. Since the functions � � f ∗ : = lim sup fr = lim sup fri : ri < 1/k , k→∞ r→0 � � f∗ : = lim inf fr = lim inf fri : ri < 1/k r→0
k→∞
are Borel, so is the set E(θ) = {f ∗ ≤ θ} ∩ {f∗ ≥ θ}.
Theorem 4.4.2. If E ⊂ Rn is a measurable set, then Θ(E, x) = χE (x) for almost all x ∈ R , and the sets E, int∗ E, and cl∗ E are equivalent. n
Proof. Since χE ∈ L1loc (Rn ) and
1 � Θ(E, x) = lim �� r→0 B(x, r)�
�
χE (y) dy, B(x,r)
Θ(E, x) = χE (x) for almost all x ∈ Rn by Theorem 4.3.4. This implies ext∗ E ∼ Rn − E, and consequently int∗ E ∼ E and cl∗ E ∼ E. Lemma 4.4.3. Let µ be a Radon measure in an open set Ω ⊂ Rn , and let E ⊂ Ω. There is a Borel set B ⊂ Ω such that E ⊂ B and µ(E ∩ C) = µ(B ∩ C) for each µ measurable set C ⊂ Ω. Proof. There is an increasing sequence {Ki } of compact sets whose union equals Ω. Let Ei := E ∩ Ki , and find a Borel set Ai ⊂ Ω so that Ei ⊂ Ai ⊂ Ki �∞ and µ(Ai ) = µ(Ei ). Letting Bi := j=i Aj , we obtain Ei ⊂ Bi ⊂ Ki . If C ⊂ Ω is a µ measurable set, then µ(Ei ) = µ(Ei ∩ C) + µ(Ei − C) ≤ µ(Bi ∩ C) + µ(Bi − C) = µ(Bi ) ≤ µ(Ai ) = µ(Ei ) < ∞.
�∞ This implies µ(Ei ∩ C) = µ(Bi ∩ C). The union B := i=1 Bi is a Borel set containing E. Moreover µ(E ∩ C) = µ(B ∩ C), since {Ei } and {Bi } are increasing sequences. Corollary 4.4.4. If E ⊂ Rn , then Θ(E, x) = 1 for almost all x ∈ E.
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4. Perimeter
Proof. Let B be as in Lemma 4.4.3 applied to Ω= Rn and µ = Ln . Then E ⊂ B, and Theorem 4.4.2 implies that Θ(E, x) = Θ (B, x) = 1 for almost all x ∈ B. Theorem 4.4.5. A set E ⊂ Rn is measurable if and only if |∂∗ E| = 0. Proof. Corollary 4.4.4 implies |E − cl∗ E| = 0 and � � |int∗ E − E| = �(Rn − E) − cl∗ (Rn − E)� = 0.
If |∂E| = 0 then cl∗ E ∼ int∗ E. Thus E ∼ cl∗ E is measurable by Observation 4.4.1. The converse follows from Theorem 4.4.2.
4.5. Definition of perimeter Definition 4.5.1. The perimeter of E ⊂ Rn in an open set Ω ⊂ Rn is the extended real number P(E, Ω) := Hn−1 (∂∗ E ∩ Ω).
The perimeter P(E, Rn ) of E in Rn is called the perimeter of E, denoted by P(E). LetΩ ⊂ Rn be an open set. A set E ⊂ Ω is called
(i) a set of finite perimeter in Ωif |E| + P(E, Ω) < ∞, (ii) a set of locally finite perimeter in Ωif P(E, U ) < ∞ for each open set U � Ω.
The family of all sets E ⊂ Ω of finite, or locally finite, perimeter in Ω is denoted by P(Ω), or Ploc (Ω), respectively. Clearly P(Ω) ⊂ Ploc (Ω),
and if E ⊂ Rn belongs to Ploc (Ω), then so does Rn − E. Note that in the above terminology a set E ⊂ Ω such that |E| = ∞ and P(E, Ω) < ∞ is called a set of locally finite, rather than finite, perimeter in Ω. IfΩ ⊂ Rn is an open set and E ⊂ Rn , Proposition 4.2.2 implies P(E, Ω) = P(E ∩ Ω, Ω) ≤ P(E ∩ Ω).
(4.5.1)
It follows that when studying perimeters in an open setΩ ⊂ Rn , we may restrict our attention to subsets of Ω. The next proposition is a direct consequence of Observation 4.2.1. Proposition 4.5.2. If Ω ⊂ Rn is open and A, B ⊂ Ω, then � � max P(A ∪ B, Ω), P(A ∩ B, Ω), P(A − B, Ω) ≤ P(A, Ω) + P(B, Ω).
Proposition 4.5.3. Let Ω ⊂ Rn be an open set. Each set E ⊂ Ω that belongs to Ploc (Ω) is measurable.
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4.5. Definition of perimeter
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Proof. Suppose that E ⊂ Ω is not measurable. As Ω is the union of countably many open balls U � Ω, there is an open ball V � Ω for which E ∩ V is not a measurable set. Proposition 4.2.2 implies ∂∗ E ∩ V = ∂∗ (E ∩ V ) ∩ V = ∂∗ (E ∩ V ) − ∂V, since ∂∗ (E ∩ V ) ⊂ cl V . According to Theorem 4.4.5, � � |∂∗ E ∩ V | = �∂∗ (E ∩ V )� > 0 and P(E, V ) = ∞ by Proposition 1.4.3.
In terms of the measure Hn−1 , the essential boundary ∂∗ E of a nonmeasurable set E ⊂ Rn is enormous: it follows from [30, Thorem 5.6] and [59, Theorems 59] that ∂∗ E contains uncountably many disjoint subsets whose Hn−1 measure is infinite.
Proposition 4.5.4. Let Ω ⊂ Rn be an open set. Then E ⊂ Ω belongs to Ploc (Ω) if and only if P(E ∩ A) < ∞ for each A � Ω with P(A) < ∞. Proof. Assume E ∈ Ploc (Ω), and choose A � Ω with P(A) < ∞. There is a figure whose interior U satisfies A � U � Ω. By Proposition 4.5.2 and Corollary 4.2.5, P(E ∩ A) = P(E ∩ U ∩ A) ≤ P(E ∩ U ) + P(A) ≤ P(E, U ) + P(U ) + P(A) < ∞.
Conversely, choose an open set U � Ω and find a figure whose interior V satisfies U � V � Ω. As inequality (4.5.1) and our assumption imply P(E, U ) ≤ P(E, V ) ≤ P(E ∩ V ) < ∞, we see that E ∈ Ploc (Ω). Lemma 4.5.5. If µ is a Radon measure in Rn , then each family E of µ nonoverlapping µ measurable sets of positive measure is countable. � � Proof. As µ is Radon and B(0, r) is compact, µ B(0, r) < ∞ for every r ∈ R+ . It follows that for j, k = 1, 2, . . . , the family � � � � Ej,k = E ∈ E : µ E ∩ B(0, j) > 1/k is finite. Thus E =
�∞
j,k=1
Ej,k is a countable family.
Proposition 4.5.6. Let E ⊂ Rn . If n ≥ 2, then for all but countably many balls B ⊂ Rn , each pair � � ∂∗ (E ∩ B), (∂∗ E ∩ int B) ∪ (int∗ E ∩ ∂B) , � � ∂∗ (E − B), (∂∗ E − cl B) ∪ (int∗ E ∩ ∂B)
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consists of Hn−1 equivalent sets, and P(E ∩ B) = P(E, int B) + Hn−1 (int∗ E ∩ ∂B)
P(E − B) = P(E, Rn − cl B) + Hn−1 (int∗ E ∩ ∂B). Proof. Let B be a ball with Hn−1 (∂B ∩ ∂∗ E) = 0. Corollary 4.2.5 shows that the first pair consists of Hn−1 equivalent sets. Replacing B by Rn − B, we see that the same is true for the second pair. As n ≥ 2, the boundaries of two distinct balls have Hn−1 negligible intersection. The proposition follows from Lemma 4.5.5. Remark 4.5.7. Proposition 4.5.6 holds in any dimension for concentric balls, as well as for concentric cubes. Proposition 4.5.8. A Lipschitz domain in Rn belongs to Ploc (Rn ). It belongs to P(Rn ) whenever it is bounded. Proof. If Ω is a Lipschitz domain, then ∂∗ Ω = ∂Ω by Proposition 4.1.2. According to the definition, ∂Ω is locally the graph G of a Lipschitz function g defined on a bounded open set V ⊂ Rn−1 . Thus � � φ : u �→ u, g(u) : V → G
is a Lipschitz bijection, and inequality (1.5.1) implies Hn−1 (G) < ∞. Since G is a relatively open subset of ∂Ω, we have Hn−1 (K) < ∞ for any compact set K ⊂ ∂Ω. In particular, P(Ω, U ) < ∞ for any U � Rn . If Ω is bounded, there is U � Rn with clΩ ⊂ U .
4.6. Line sections Useful information about the perimeters of measurable sets is obtained by studying their intersections with lines. A line is a one-dimensional affine subspace of Rn . Via translation and rotation, a line L can be identified with R. This identification defines Lebesgue measure L1 in L, which is equal to the Hausdorff measure H1 restricted to subsets of L. For distinct points x and y in Rn , the sets � � (xy) := tx + (1 − t)y : 0 < t < 1 and [xy] := (xy) ∪ {x, y}
are called, respectively, the open and closed segments determined by the twopoint set {x, y}. Each segment is an uncountable set. Lemma 4.6.1. Let L be a line, and let X, Y, Z be disjoint sets whose union is equal to L. Suppose that Z is contained in cl X ∩ cl Y and contains no segment. If z ∈ Z and δ > 0, then there are x ∈ X and y ∈ Y such that |x − y| < δ and z ∈ (xy).
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Proof. We may assume L = R. Choose z ∈ Z and δ > 0. The open interval (z − δ/2, z + δ/2) contains points x ∈ X and y ∈ Y , and no generality is lost by assuming x < y. If y < z, there is u ∈ X ∪ Y such that z < u < z + δ/2, since (z, z + δ/2) �⊂ Z. Either (yu) or (xu) is the desired segment. The case z < x is argued similarly. Lemma 4.6.2. Let L be a line, and let X, Y, Z be disjoint sets whose union is equal to L. Suppose the following conditions hold: (a) there are increasing sequences {Xk } and {Yk } of closed subsets of L such that Xk ∩ Yk = ∅ for each k ∈ N, and � � X ⊂ {Xk : k ∈ N} and Y ⊂ {Yk : k ∈ N};
(b) given x ∈ X and y ∈ Y , the intersections (x˜ x) ∩ (X ∪ Z)
and
(y y˜) ∩ (Y ∪ Z)
are nonempty for all x ˜, y˜ ∈ L such that x ˜ �= x and y˜ �= y.
If x ∈ X and y ∈ Y , then (xy) ∩ Z �= ∅.
Proof. As we may assume L = R, open segments are nonempty open intervals, and condition (b) states that each x ∈ X is a two-sided cluster point of X ∪ Z, and each y ∈ Y is a two-sided cluster point of Y ∪ Z. Choose x1 ∈ X, y1 ∈ Y and, with no loss of generality, suppose x1 < y1 . Working toward a contradiction, assume (x1 , y1 ) ∩ Z = ∅. Select k1 ∈ N so that x1 ∈ Xk1 and y1 ∈ Yk1 , and define a1 := sup{x ∈ Xk1 : x < y1 }
and
b1 := inf{y ∈ Yk1 : y > a1 }.
Then a1 ∈ Xk1 , b1 ∈ Yk1 , x1 ≤ a1 < b1 ≤ y1 , and the intersection (a1 , b1 ) ∩ (Xk1 ∪ Yk1 ) is empty. As a1 and b1 are two-sided cluster points of X ∪ Z and Y ∪ Z, respectively, and (a1 , b1 ) ∩ Z = ∅, there are x2 ∈ X ∩ (a1 , b1 ) and y2 ∈ Y ∩ (x2 , b1 ). Thus we have x 1 < x 2 < y2 < y1
and
[x2 , y2 ] ∩ (Xk1 ∪ Yk1 ) = ∅.
By recursion, we construct a strictly increasing sequence {ki } of positive integers, and points xi ∈ Xki , yi ∈ Yki such that for i = 1, 2, . . . , xi < xi+1 < yi+1 < yi and [xi+1 , yi+1 ] ∩ (Xki ∪ Yki ) = ∅. �∞ The nonempty intersection i=1 [xi , yi ] is disjoint from X ∪ Y as well as from Z — a contradiction. A hyperplane is an (n − 1)-dimensional affine subspace of Rn . To a hyperplane Π corresponds the orthogonal projection π : Rn → Π, and we let Lu := π −1 (u) for each u ∈ Π. Via translation and rotation, Π can be identified with Rn−1 . This identification defines Lebesgue measure Ln−1 inΠ, which is equal to the Hausdorff measure Hn−1 restricted to subsets ofΠ.
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Theorem 4.6.3. Let A be a measurable set, and let Π be a hyperplane. For Ln−1 almost all u ∈ Π the following conditions hold: (i) if z ∈ Lu ∩ ∂∗ A and δ > 0, then there are x ∈ Lu ∩ ext∗ A and y ∈ Lu ∩ int∗ A such that |x − y| < δ and z ∈ (xy); (ii) if x ∈ Lu ∩ ext∗ A and y ∈ Lu ∩ int∗ A, then (xy) ∩ ∂∗ A �= ∅.
Proof. With no loss of generality we may assume that Π =Π n . Let π = πn , and x� = x·en for each x ∈ Rn . Since Rn is the disjoint union of ext∗ A, int∗ A, and ∂∗ A, the line Lu is the disjoint union of Xu := Lu ∩ ext∗ A,
Yu := Lu ∩ int∗ A,
and
Zu := Lu ∩ ∂∗ A
for each u ∈ Π. Our proof follows those presented in [33, Theorem 4.5.11] and [29, Section 5.11, Theorem 1]: we denote byΠ 0 the set of all u ∈ Π such that Xu , Yu , and Zu satisfy the assumptions of Lemmas 4.6.1 and 4.6.2, and show that Ln−1 (Π − Π0 ) = 0. We split the argument into four separate claims. For u ∈ Π and r > 0, let Bu,r = Π ∩ B(u, r). Denote byΠ E the set of all u ∈ Π for which there exist δu > 0 and zu ∈ Lu ∩ ∂∗ E such that � � x ∈ Lu : 0 < |x − zu | < δu ⊂ ext∗ A. Replacing ext∗ A by int∗ A, a set ΠI is defined similarly. Let � � Π∗ := u ∈ Π : H1 (Lu ∩ ∂∗ E) > 0
andΠ 1 :=Π E ∪ ΠI ∪ Π∗ . As H1 negligible sets contain no segments, the sets Xu , Yu , and Zu satisfy the assumptions of Lemma 4.6.1 for all u ∈ Π − Π1 . Claim 1. Ln−1 (Π1 ) = 0.
Proof. Fubini’s theorem implies Ln−1 (Π∗ ) = 0, since |∂∗ A| = 0 by Theorem 4.4.5. Seeking a contradiction, we assume Ln−1 (ΠE ) > 0. For each (x, j) ∈ Qn × N, denote byΠ x,j the set of all u ∈ ΠE such that zu ∈ U (x, 1/j)
and
B(x, 1/j) ⊂ U (zu , δu ).
The set Nx,j = {zu : u ∈ Πx,j } is negligible by Fubini’s theorem, and B(x, 1/j) ∩ π −1 (Πx,j ) − Nx,j ⊂ ext∗ A.
(∗)
We fix a pair (x, j) ∈ Q × N with Ln−1 (Πx,j ) > 0; such a pair exists, because �� � ΠE = Πx,j : (x, j) ∈ Qn × N .
Applying Corollary 4.4.4 to Π= Rn−1 , find a point w ∈ Πx,j with Ln−1 (Bw,r ∩ Πx,j ) = 1. r→0 Ln−1 (Bw,r ) lim
Let r > 0 be so small that the closed cylinder � � � |≤r Cr := y ∈ π −1 (Bw,r ) : |y � − zw
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is contained in B(x, 1/j). Inclusion (∗) shows that Cr ∩π −1 (Nx,j ) is contained in ext∗ A, up to the negligible set Nx,j . Hence � � 2rLn−1 (Bw,r ∩ Πx,j ) = �Cr ∩ π −1 (Πx,j )� ≤ |Cr ∩ ext∗ A|. As |Cr | = 2rLn−1 (Bw,r ) for each r > 0, we obtain
|Cr ∩ ext∗ A| Ln−1 (Bw,r ∩ Πx,j ) ≥ lim = 1. r→0 r→0 |Cr | Ln−1 (Bw,r ) lim
The previous inequality and inclusion B(zw , r) ⊂ Cr yield
|Cr − ext∗ A| |Cr ∩ ext∗ A| = lim |Cr | |Cr | � � � � �B(zw , r) − ext∗ A� �B(zw , r) − ext∗ A� α(n) � � = lim . ≥ lim �B(zw , r)� r→0 |Cr | 2α(n − 1) r→0
0 ≥ 1 − lim
We conclude that contrary to our choice, zw belongs to
ext∗ (Rn − ext∗ A) = ext∗ (cl∗ A) = ext∗ A ⊂ Rn − ∂∗ A.
This and a similar argument show that Ln−1 (ΠE ∪ ΠI ) = 0.
For each k ∈ N, define closed sets � � � � �int∗ A ∩ B(x, r)� α(n − 1) 3 n Ek : = x ∈ R : , ≤ for all 0 < r < rn 3n+1 k � � � � �ext∗ A ∩ B(x, r)� α(n − 1) 3 , Ik : = x ∈ R n : ≤ for all 0 < r < rn 3n+1 k
and observe that the sequences {Ek } and {Ik } are increasing and � ext∗ A = ext∗ (int∗ A) ⊂ {Ek : k ∈ N}, � int∗ A = ext∗ (ext∗ A) ⊂ {Ik : k ∈ N}.
(∗∗)
Claim 2 . Ek ∩ Ik = ∅ for each k ∈ N.
Proof . A direct calculation shows that assuming Ek ∩ Ik �= ∅ leads to an inequality α(n) ≤ 2 · 3−n−1 α(n − 1). This is impossible, since � � � � α(n) = �B(0, 1)� = 2 1 − |u|2 du >2
�
B(0,1/3)∩Π
B(0,1)∩Π
�
1 − |u|2 du > 12 · 3−n−1 α(n − 1).
For u ∈ Π and k ∈ N, let Xu,k := Lu ∩ Ek and Yu,k := Lu ∩ Ik . It follows from (∗∗) and Claim 2 that for each u ∈ Π, the sequences {Xu,k } and {Yu,k } fulfill condition (a) of Lemma 4.6.2 with respect to the sets Xu and Yu , respectively. Denote by S± (x, j) the open segment determined by points x and x ± (3/j)en . For k, j ∈ N, define sets
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and let E0±
� � ± : = x ∈ Ek : S± (x, j) ⊂ int∗ A , Ek,j � � ± : = x ∈ Ik : S± (x, j) ⊂ ext∗ A , Ik,j � � ± ± := k,j∈N Ek,j , I0± := k,j∈N Ik,j , and Π2 = π(E0+ ∪ E0− ∪ I0+ ∪ I0− ).
Claim 3 . For each u ∈ Π − Π2 , the sets Xu , Yu , and Zu satisfy condition (b) of Lemma 4.6.2. Proof . Select u ∈ Π − Π2 . If x ∈ Xu , there is k ∈ N such that x belongs � + − to Ek − j∈N (Ek,j ∪ Ek,j ). Thus no segment (x˜ x) with x ˜ ∈ Lu and x ˜ �= x is x) ∩ (Xu ∪ Zu ) �= ∅ for all x ˜ ∈ Lu − {x}. contained in int∗ A. In other words, (x˜ Similarly, if y ∈ Yu then (y y˜) ∩ (Yu ∪ Zu ) �= ∅ for all y˜ ∈ Lu − {y}. Claim 4 . Ln−1 (Π2 ) = 0.
Proof. Fix k, j ∈ N, and for p ∈ Z, let � � p p+1 + Np := y ∈ Ek,j . : ≤ y� < j j If Np �= ∅, choose u ∈ π(Np ) and 0 < r < min{1/k, 1/j}. Find a point z ∈ Np ∩ π −1 (Bu,r ) so that � � r z � > sup y � : y ∈ Np ∩ π −1 (Bu,r ) − . 2 It is easy to verify that the ball B(z, 3r) contains the cylinder � � � � C := x ∈ π −1 π(Np ) ∩ Bu,r : z � + r/2 < x� < z � + r . Given x ∈ C, there is y ∈ Np ∩ π −1 (x). Then x� > z � + r/2 > y � by the choice of z. Since x� < z � + r and |y � − z � | < 1/j, we obtain 0 < x� − y � < 2/j. Hence + , we conclude that x ∈ S+ (y, j). As S+ (y, j) ⊂ int∗ A by the definition of Ek,j C ⊂ int∗ A ∩ B(z, 3r). Consequently � α(n − 1) � � r n−1 � L (3r)n π(Np ) ∩ Bu,r = |C| ≤ �int∗ E ∩ B(z, 3r)� ≤ 2 3n+1 where the last inequality holds because z ∈ Ek and 0 < 3r < 3/k. Thus � � Ln−1 π(Np ) ∩ Bu,r 2 ≤ lim sup n−1 L (Bu,r ) 3 r→0+ � � n−1 n−1 and Corollary 4.4.4 applied to Π= R yields L π(Np ) = 0. Since � + + Ek,j = p∈Z Np , the sets π(Ek,j ) are Ln−1 negligible for all k, j ∈ N, and so + is π(E0 ). Similarly we show that π(E0− ) and π(I0± ) are Ln−1 negligible sets.
The theorem follows by lettingΠ mas 4.6.1 and 4.6.2.
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0
= Π − Π1 ∪ Π2 and applying Lem-
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Corollary 4.6.4. Let n = 1. Then A ∈ Ploc (R) if and only if ∂∗ A is a locally finite set. In this case, there is a countable collection J of open intervals with disjoint closures such that � � int∗ A = J and ∂∗ A = {∂J : J ∈ J}. Proof. Since H0 is the counting measure in R, the first assertion is obvious. If ∂∗ A is locally finite, we can organize it into a bidirectional sequence · · · < a−1 < a0 < a1 < . . . of real numbers, which can be finite or infinite in either direction but has no cluster points in R. Theorem 4.6.3 shows that any interval Jk = (ak , ak+1 ) is contained either in ext∗ A or in int∗ A. Moreover, if Jk is a subset of ext∗ A, or int∗ A, then Jk−1 ∪ Jk+1 is a subset of int∗ A, or ext∗ A, respectively. Theorem 4.6.5. Let A ⊂ Rn be an Hn−1 measurable set, and let Π be a hyperplane. If Hn−1 A is a Radon measure, then u �→ H0 (Lu ∩ A) : Π → R is an Ln−1 measurable function and � H0 (Lu ∩ A) dLn−1 (u) ≤ Hn−1 (A). Π
Proof. Let π : R → Π be the orthogonal projection. Define n
fE : u �→ H0 (Lu ∩ E) : Π → [0, ∞]
for each E ⊂ Rn . Since A is Hn−1 measurable and Hn−1 A is a Radon measure, there are disjoint Hn−1 measurable sets Ai such that Hn−1 (Ai ) < ∞ � for i = 1, 2, . . . , and A = i∈N Ai . According to Theorem 1.3.1, there are � compact sets Ki,j ⊂ Ai such that each set Ai − j∈N Ki,j is Hn−1 negligible. � If Kp = i,j≤p Ki,j for p = 1, 2, . . . , then {Kp } is an increasing sequence of compact subsets of A, and the set � � �� � A− Kp ⊂ Ki,j Ai − p∈N
i∈N
j∈N
� � � is Hn−1 negligible. As Lip π = 1, the projection N = π A − p∈N Kp is Ln−1 negligible by (1.5.1), and for each u ∈ Π − N , fA (u) = lim fKp (u).
(∗)
Let K ⊂ Rn be a compact set, and choose j ∈ N and t > 0. If Uj is the set of all u ∈ Π such that Lu ∩ K can be covered by fewer than t open sets, each of diameter smaller than 1/j, then � � (∗∗) u ∈ Π : fK (u) < t} = {Uj : j ∈ N}.
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Indeed, because H0 takes only integer values, equality (∗∗) follows from Remark 1.4.1, (3). Fix a point u ∈ Uj , and find open sets V1 , . . . , Vp so that �p p < t, d(Vi ) < 1/j for i = 1, . . . , p, and V = i=1 Vi contains Lu ∩ K. Let Bk = Π ∩ B(u, 1/k) and observe that Ck := π −1 (Bk ) ∩ (K − V ) is a compact �∞ set. Since k=1 Ck = ∅, there is an integer p ≥ 1 such that Cp = ∅. Consequently Bp ⊂ Uj , and we see that Uj is a relatively open subset ofΠ . This and (∗∗) show that fK is a Borel function, and the Ln−1 measurability of fA follows from (∗). For every k ∈ N, there is a cover {Bk,j : j ∈ N} of A such that each Bk,j has diameter smaller than 1/k, and �n−1 � ∞ � d(Bk,j ) 1 n−1 α(n − 1) < H1/k (A) + . 2 k j=1 In view of Remark 1.4.1, (3), we may assume that Bk,j are open sets. It follows that each π(Bk,j ) is a relatively open subset ofΠ , and we define a Borel function ∞ � gk : u �→ χπ(Bk,j ) (u) : Π → [0, ∞]. j=1
Choose u ∈ Π, and observe that gk (u) is the number of elements in the � set Nk = {j : Lu ∩ Bk,j �= ∅}. Since Lu ∩ A ⊂ j∈Nk Bk,j , we see that 0 H1/k (Lu ∩ A) ≤ gk (u) for k = 1, 2, . . . . Consequently 0 H0 (Lu ∩ A) = lim H1/k (Lu ∩ A) ≤ lim inf gk (u).
Fatou’s lemma and the isodiametric inequality (1.4.2) imply � � H0 (Lu ∩ A) dLn−1 (u) ≤ lim inf gk dLn−1 Π
Π
= lim inf k→∞
∞ � i=1
� � Ln−1 π(Bi,k )
�n−1 d(Bi,k ) k→∞ 2 i=1 � � 1 n−1 = Hn−1 (A). ≤ lim H1/k (A) + k→∞ k ≤ lim inf
∞ �
α(n − 1)
�
Let E be a subset of a line L. Identifying L with R, the symbols extL ∗ E,
intL ∗ E,
clL ∗ E,
and
∂∗L E
denote, respectively, the essential exterior, essential interior, essential closure, and essential boundary of E relative to L.
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Theorem 4.6.6. Let E ∈ Ploc (Rn ). Given a hyperplane Π, the function u �→ H0 (Lu ∩ ∂∗ E) : Π → R is Ln−1 measurable and � H0 (Lu ∩ ∂∗ E) dLn−1 ≤ P(E). (4.6.1) Π
n−1
negligible set N ⊂ Π such that for each u ∈ Π − N , the set There is an L Lu ∩ ∂∗ E is locally countable, and u Lu ∩ int∗ E = intL ∗ (Lu ∩ E)
and
Lu ∩ ∂∗ E = ∂∗Lu (Lu ∩ E).
Proof. Note that ∂∗ E is a Borel set by Observation 4.4.1. Since by definition, Hn−1 (∂∗ E ∩ U ) = P(E, U ) < ∞ for each open set U � Rn , it follows from Proposition 1.4.2 that Hn−1 ∂∗ E is a Radon measure. Now the first assertion is a direct consequence of Theorems 4.4.5. Next let Uk = U (0, k), and apply Theorem 4.4.2 to the set ∂∗ E ∩ Uk . There is an Ln−1 negligible set Nk ⊂ Π such that Lu ∩ (∂∗ E ∩ Uk ) is a finite �n set for each u ∈ Π − Nk . The set N = k=1 Nk is still negligible, and the set Lu ∩ ∂∗ E is locally finite for each u ∈ Π − N . Enlarge N so that for each u ∈ Π − N , the assertions of Theorem 4.6.3 hold for the set E. The theorem follows by the same argument we used in the proof of Corollary 4.6.4. Proposition 4.6.7. Let E ∈ Ploc (Rn ), and let ϕ ∈ Lipc (Rn ) be such that �ϕ�L∞ (Rn ) ≤ 1. For i = 1, . . . , n, � � Di ϕ(x) dx ≤ H0 (Lu ∩ ∂∗ E) dLn−1 (u). E
Πi
Proof. Since E and int∗ E are equivalent sets by Theorem 4.4.2, Fubini’s theorem shows that Lu ∩ E and Lu ∩ int∗ E are L1 equivalent for Ln−1 almost all u ∈ Πn . Thus with no loss of generality, we may assume that E = int∗ E. Applying Theorem 4.4.2 and Corollary 4.6.4, we find an Ln−1 negligible set N ⊂ Πn such that for each point u ∈ Πn − N there is a strictly increasing finite or infinite bidirectional sequence of extended real numbers · · · < au,j < bu,j < au,j+1 < bu,j+1 < . . . , j ∈ Ju and Ju ⊂ Z, with no cluster points in R, which satisfies �� � (au,j bu,j ) : j ∈ Ju , Lu ∩ E = {u} × � � Lu ∩ ∂∗ E = (u, au,j ), (u, bu,j ) : j ∈ Ju ) ∩ Rn .
Select ϕ ∈ Lipc (Rn ) with �ϕ�L∞ (Rn ) ≤ 1, and for each u ∈ Πn , let � ϕ(u, t) if t ∈ R, fu (t) := 0 if t = ±∞.
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Since f ∈ Lipc (R), � �� Dn ϕ(u, t) dt = Lu ∩E
j∈Ju
=
��
j∈Ju
≤
bu,j au,j
fu� (t) dt
fu (bu,j ) − fu (au,j )
�
� �� � � �� �fu (bu,j )� + �fu (au,j )� ≤ H0 (Lu ∩ ∂∗ E),
j∈Ju
for each u ∈ Πn − N , and Fubini’s theorem implies � � � �� Dn ϕ(x) dx = Dn ϕ(u, t) dt dLn−1 (u) E
Lu ∩E
Πn
≤
�
Πn
H0 (Lu ∩ ∂∗ E) dLn−1 (u).
The proposition follows from symmetry. Remark 4.6.8. By Proposition 4.6.7 and Theorem 4.6.6, �� � n sup div v(x) dx : v ∈ Lipc (R ) and �v�L∞ (Rn ;Rn ) ≤ 1 ≤ nP(E) E
for each measurable set E. Indeed, this is true if E ∈ Ploc (Rn ). Otherwise P(E) = ∞ and there is nothing to prove. While the above inequality is useful at this stage of development, we show later that �� � sup div v(x) dx : v ∈ Lipc (Rn ) and �v�L∞ (Rn ;Rn ) ≤ 1 = P(E) E
for each measurable set E; see Theorem 6.5.5 below. The (n − 1)-dimensional sphere in Rn is the set � � S n−1 := e ∈ Rn : |e| = 1 .
For each pair (e, t) in S n−1 × R we define Πe,t := {x ∈ Rn : x · e = t}
and
� � He,t := x ∈ Rn : x · e > t .
The open half-spaces He,t and H−e,−t are the connected components of Rn − Πe,t , and the hyperplaneΠ e,t is their common boundary. Lemma 4.6.9. Let E ∈ P(Rn ). For each pair (e, t) in S n−1 × R, Ln−1 (int∗ E ∩ Πe,t ) ≤ Hn−1 (∂∗ E ∩ H±e,±t ).
Proof. In view of invariance with respect to rotations and translations, it suffices to prove the lemma for Π =Π en ,0 and H = Hen ,0 . Since |E| < ∞, Fubini’s theorem shows that Lu ∩ H meets ext∗ E for Ln−1 almost all u ∈ Π. By Theorem 4.6.3, (ii), the intersection (Lu ∩ H) ∩ ∂∗ E is not empty for Ln−1 almost all u ∈ int∗ E ∩ Π. If π : H → Π is the orthogonal projection,
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then int∗ E ∩ Π is contained in π(∂∗ E ∩ H) up to an Ln−1 negligible set. As Lip π = 1, we obtain � � Ln−1 (int∗ E ∩ Π) ≤ Ln−1 π(∂∗ E ∩ H) ≤ Hn−1 (∂∗ E ∩ H). The lemma follows from symmetry.
A set C ⊂ Rn is called a polytop if it is the convex hull of a finite set S ⊂ Rn and int C �= ∅. Each polytop is a compact set. Proposition 4.6.10. Let E ∈ P(Rn ). If C is a polytop, then P(E ∩ C) ≤ P(E).
�k Proof. There are (vj , tj ) in S n−1 ×R such that C = j=1 Hvj ,tj . To simplify the notation, let Hj± := H±vj ,±tj . By Corollary 4.2.5 and Lemma 4.6.9, P(E ∩ H1+ ) ≤ Hn−1 (∂∗ E ∩ H1+ ) + Hn−1 (int∗ E ∩ ∂H1+ ) + Hn−1 (∂∗ E ∩ ∂H1+ )
≤ Hn−1 (∂∗ E ∩ H1+ ) + Hn−1 (∂∗ E ∩ H1− ) + Hn−1 (∂∗ E ∩ ∂H1+ ) = P(E).
The proof is completed by induction, since � � P (E ∩ H1+ ) ∩ H2+ ≤ P(E ∩ H1+ ) ≤ P(E).
Proposition 4.6.10 still holds when polytops are replaced by arbitrary convex sets (Proposition 6.6.4 below). Proposition 4.6.11. Let E ∈ P(Rn ). If {Ck } is a sequence of cubes such that lim d(Ck ) = ∞, then lim P(E − Ck ) = 0.
Proof. With no loss of generality, we may assume that Ck ⊂ int Ck+1 for k = 1, 2, . . . . Indeed, each subsequence of {Ck } contains a subsequence which satisfies this assumption. By Corollary 4.2.5, ∂∗ (E − Ck ) ⊂ (∂∗ E − Ck ) ∪ (int∗ E ∩ ∂Ck ) ∪ (∂∗ E ∩ ∂Ck ). n−1
(∂∗ E) < ∞ and {∂∗ E − Ck } is a decreasing sequence with empty Since H intersection, lim Hn−1 (∂∗ E − Ck ) = 0. As ∂Ck are disjoint sets, ∞ �
k=1
and lim H
n−1
Hn−1 (∂∗ E ∩ ∂Ck ) ≤ Hn−1 (∂∗ E) < ∞
(∂∗ E ∩ ∂Ck ) = 0. If Ck =
int∗ E ∩ ∂Ck ⊂
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n � �
i=1
�n
i=1 [ak,i , bk,i ],
then
(int∗ E ∩ Πei ,ak,i ) ∪ (int∗ E ∩ Πei ,bk,i )
�
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and lim bk,i = − lim ak,i = ∞ for i = 1, . . . , n. As {Hei ,bk,i } is a decreasing sequence with empty intersection, Lemma 4.6.9 implies lim Ln−1 (int∗ E ∩ Πei ,bk,i ) ≤ lim Hn−1 (∂∗ E ∩ Hei ,bk,i ) = 0 for i = 1, . . . , n. The proposition follows by symmetry. The set E = Rn and cubes C = [−k, k]n , k = 1, 2, . . . , show that in Proposition 4.6.11, the assumption E ∈ P(Rn ) cannot be replaced by a weaker assumption P(E) < ∞.
4.7. Lipeomorphisms We show that a lipeomorphic image of a set of finite perimeter is again a set of finite perimeter. The proof is nontrivial and relies on topological arguments. Proposition 4.7.1. Let B ⊂ Rn be a closed ball, and let ψ : B → Rn be a homeomorphism. Then int ψ(B) = ψ(int B). Proof. By the Jordan-Brouwer separation theorem [69, Chapter 4, Section 7, Theorem 15], the open set Rn − ψ(∂B) has two connected components I and E such that ∂I = ∂E = ψ(∂B). By Brouwer’s invariance of domain theorem [Theorem 16, ibid.], the connected set O = ψ(int B) is open. Being connected and disjoint from ψ(∂B), the set O is contained in one of the components of Rn − ψ(∂B), say O ⊂ I. Assume there is z ∈ I ∩ ∂O and find a sequence {xi } in int B such that lim ψ(xi ) = z. Passing to a subsequence, we may assume that {xi } converges to x ∈ B. Then z = ψ(x), which is impossible because z∈ / O ∪ ψ(∂B). Thus I ∩ ∂O = ∅, and O is simultaneously a relatively open and relatively closed subset of I. As O �= ∅, we infer O = I. The proposition follows, since ψ(B) = O ∪ ψ(∂B) and ∂O = ψ(∂B). Throughout this section, for r > 0 we employ the notation Br = B(0, r)
and
Ur = U (0, r).
Proposition 4.7.2. Let B ⊂ Rn be a closed ball, and let ψ : B → Rn be a homeomorphism. If K ⊂ ψ(int B) is a compact set and φ : B → Rn is a continuous map satisfying � � � � sup �ψ(x) − φ(x)� < dist K,ψ (∂B) , x∈B
then K ⊂ φ(int B).
� � Proof. � If K and� φ are as in the proposition, choose 0 < δ < K,ψ (∂B) so that �ψ(x) − φ(x)� ≤ δ for each x ∈ B. Given y ∈ K, we show y = φ(x) for
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some x ∈ int B. Using translations, �we may assume y = 0. Proposition 4.7.1 � implies Bδ ⊂ ψ(int B). Let γ := φ ◦ ψ −1 � Bδ , and observe � � � � � � �� �x − γ(x)� = ��ψ ψ −1 (x) − φ ψ −1 (x) �� ≤ δ for each x ∈ Bδ . Thus θ : x �→ x − γ(x) maps Bδ into itself. By Brouwer’s fixed point theorem [69, Chapter 4, Section� 7, Theorem 5], there is z ∈ Bδ � such that θ(z) = z. Hence y = 0 = γ(z) = φ ψ −1 (z) and ψ −1 (z) ∈ int B.
The following result was obtained by Z. Buczolich [11]. However, our proof follows that presented in [35, Section A.2.4]. If A ⊂ Rn , then a lipeomorphism φ : A → Rn has a unique continuous extension to cl A. This extension is again a lipeomorphism, still denoted by φ.
Theorem 4.7.3. Let C ⊂ Rn , x �∈ cl C, and� let φ : C → Rn be a lipeomorphism. If Θ(C, �x) = 0 then� Θ φ(C), φ(x) = 0. If C is measurable and Θ(C, x) = 1, then Θ φ(C), φ(x) = 1; in particular, ∂∗ φ(C) = φ(∂∗ C).
Proof. If D = φ(C), then cl D = φ(cl C) and there are a, b ∈ R+ such that � � a|u − v| ≤ �φ(u) − φ(v)� ≤ b|u − v| for all u, v ∈ cl C. Choose x ∈ cl C and let y = φ(x). Observe that � � D ∩ B(y, r) ⊂ φ C ∩ B(x, r/a)
for each r > 0. Thus Θ(C, x) = 0 implies Θ (D, y) = 0, since � � � � �D ∩ B(y, r)� � b �n �C ∩ B(x, r/a)� � ≤ � = 0. lim sup �� lim sup �� a B(y, r)� B(x, r/a)� r→0 r→0
The rest of the proof is harder. Using translations, assume x = y = 0 and Θ(C, 0) = 1. In view of Proposition 1.5.2, the lipeomorphism φ : cl C → cl D extension φ� : Rn → Rn . Making b larger, the inequality �has a Lipschitz � � � � ≤ b|u − v| still holds for all u, v ∈ Rn . For r > 0, define �φ(u) − φ(v) � ψr : x �→ r−1 φ(rx) : Rn → Rn ,
Cr = r−1 C, and Dr = r−1 D. Clearly ψr (0) = 0, ψr (Cr ) = Dr , and � � a|u − v| ≤ �ψr (u) − ψr (v)� for all u, v ∈ cl Cr , � � �ψr (u) − ψr (v)�≤ b|u − v| for all u, v ∈ Rn .
(∗)
Claim. If 0 < c < a, then each sequence {rk } in R+ converging to zero has a subsequence {sk } such that Bc ⊂ ψsk (B1 ) for k = 1, 2, . . . . Proof. From Θ(C, 0) = 1 and |C ∩ Br |/|Br | = |Cr ∩ B1 |/|B1 |, we obtain lim
r→0
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|Cr ∩ B1 | = 1. |B1 |
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4. Perimeter
A sequence {rk } in R+ converging to zero has a subsequence {sk } such that |Csk ∩ B1 | > 1 − 2−k |B1 |
(∗∗)
for k = 1, 2, . . . . In view of (∗), Ascoli’s theorem [61, Chapter 9, Section 8] shows that {ψsk } has a subsequence, still denoted by {ψsk }, converging uniformly on B1 to a map ψ : B1 → Rn that satisfies � � ψ(0) = 0 and �ψ(u) − ψ(v)�≤ b|u − v| (†) �∞ �∞ for all u, v ∈ B1 . Letting Ej := k=j (Csk ∩ B1 ) and E := j=1 Ej , we deduce � � from (∗) that a|u − v| ≤ �ψsk (u) − ψsk (v)� whenever u, v ∈ Ej and k ≥ j. Since {Ej } is an increasing sequence, � � a|u − v| ≤ �ψ(u) − ψ(v)� (††) for all u, v ∈ E. Observe
B1 − E =
∞ �
(B1 − Ej ) =
j=1
∞ � ∞ �
j=1 k=j
(B1 − Csk ).
The measurability of C and (∗∗) imply |B1 − Csk | ≤ 2−k |B1 |, and so |B1 − E| ≤ |B1 − Ej | ≤ 2−j+1 |B1 | for j = 1, 2, . . . . Consequently |B1 − E| = 0; in particular E is a dense subset of B1 . Inequalities (†) and (††) together with the continuity of ψ show that ψ : B1 → Rn is a lipeomorphism. By Proposition 4.7.1, the� set ψ(U1� ) is open and ∂ψ(U1 ) = ψ(∂U1 ). As 0 ∈ ψ(U1 ) and a = a|z − 0| ≤ �ψ(z) − 0� for each z ∈ ∂B1 , we see that Bc ⊂ Ua ⊂ ψ(U1 ). Since lim ψsk = ψ uniformly in B1 , passing to a subsequence, still denoted by {ψsk }, we obtain � � sup �ψsk (z) − ψ(z)� < dist (Bc , ψ(∂B1 ) z∈B1
for k = 1, 2, . . . . The claim follows from Proposition 4.7.2.
Returning to the main proof, select 0 < c < a and a sequence {rk } in R+ converging to zero. Find a subsequence {sk } of {rk } according to the claim. � As Bc is contained in each ψsk (B1 ) = s−1 k φ(sk B1 ), there are inclusions � s ) ⊂ φ(B � s − C) ∪ φ(C) � � s − C) ∪ D. Bcsk ⊂ φ(B = φ(B k k k
� s − C) ∪ (D ∩ Bcs ), and consequently Hence Bcsk ⊂ φ(B k k � � � � � s − C)� �D ∩ Bcs � �φ(B k k + 1≤ |Bcsk | |Bcsk |
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4.7. Lipeomorphisms
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71
for k = 1, 2, . . . . This inequality and � � � s − C)� � b �n �φ(B |Bsk − C| k lim ≤ lim |Bcsk | c |Bsk | � �n � � |C ∩ Bsk | b = =0 lim 1 − c |Bsk | � � imply lim |D ∩ Bcsk |/|Bcsk | = 1. As the sequence {rk } has been arbitrary, the equality Θ(D, 0) = 1 follows. Corollary 4.7.4. Let A ⊂ Rn , and let φ : A → Rm be a lipeomorphism. If A belongs to P(Rn ), then so does φ(A) and � � P φ(A) ≤ (Lip φ)n−1 P(A).
If A belongs to Ploc (Rn ) then so does φ(A).
Proof. The claim for A ∈ P(Rn ) follows from Theorem 4.7.3 and inequality (1.5.1). If A ∈ Ploc (Rn ), choose x ∈ cl A and let y = φ(x). For r > 0 and a = 1/Lip (φ−1 ), Theorem 4.7.3 implies � � ∂∗ φ(A) ∩ U (y, r) = φ(∂∗ A) ∩ U (y, r) ⊂ φ ∂∗ A ∩ U (x, r/a) . By inequality (1.5.1), � � � � Hn−1 ∂∗ φ(A) ∩ U (y, r) ≤ (Lip φ)n−1 Hn−1 ∂∗ A ∩ U (x, r/a) < ∞.
As r is arbitrary, φ(A) belongs to Ploc (Rn ).
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Chapter 5 BV functions
In this chapter we present the analytic approach to the perimeter of measurable sets. We define BV functions and BV sets and establish their main properties. The emphasis is on BV functions — a connection with BV sets is provided by the coarea theorem. Sobolev’s and Poincar´e’s inequalities are proved directly within the framework of BV functions; their validity for C ∞ functions is not presupposed. Throughout this chapter Ω denotes an open subset of Rn .
5.1. Variation Definition 5.1.1. Let E ⊂ Rn be such that E ∩ Ω is measurable. The variation of E in Ω is the extended real number � �� n χE div v(x) dx : v ∈ Lipc (Ω; R ) and �v�L∞ (Ω;Rn ) ≤ 1 , sup Ω
denoted by V(E, Ω). If E is measurable, the variation V(E, Rn ) of E in Rn is called the variation of E, denoted by V(E).
A measurable set E ⊂ Ω is called
(i) a set of bounded variation in Ω, or a BV set in Ω, if |E| + V(E, Ω) < ∞,
(ii) a set of locally bounded variation in Ω, or a locally BV set in Ω, if V(E, U ) < ∞ for each open set U � Ω.
The family of all BV sets in Ω, or all locally BV sets in Ω, is denoted by BV(Ω), or BVloc (Ω), respectively. Clearly BV(Ω) ⊂ BVloc (Ω). Proposition 5.1.2. If E ⊂ Ω is a measurable set, then V(E, Ω) = V(Ω − E, Ω). Proof. Choose v ∈ Lipc (Ω; Rn ), and find a dyadic figure A ⊂ Ω containing spt v. In view of Remark 2.3.8, (1), the equality � � � div v(x) dx + div v(x) dx = div v(x) dx = 0 E
Ω−E
Ω
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5. BV functions
follows from Theorem 2.3.7. If �v�L∞ (Ω;Rn ) ≤ 1, then � � � � div −v(x) dx = div v(x) dx ≤ V(E, Ω). Ω−E
E
From the arbitrariness of v, we obtain V(Ω − E, Ω) ≤ V(E, Ω). The reverse inequality follows by symmetry. If E ⊂ Rn is such that E ∩ Ω is measurable, then V(E, Ω) = V(E ∩ Ω, Ω) ≤ V(E ∩ Ω).
(5.1.1)
In view of the equality V(E, Ω) = V(E ∩ Ω, Ω), it suffices to consider V(E, Ω) only when E is a measurable subset of Ω. Proposition 5.1.3. P(Rn ) ⊂ BV(Rn ) and Ploc (Rn ) ⊂ BVloc (Rn ). Proof. Remark 4.6.8 implies the first inclusion. Let E ∈ Ploc (Rn ), and choose an open set U � Rn . If V is the interior of a figure containing U , then E ∩ V belongs to P(Rn ) by Proposition 4.5.4. Thus V(E, U ) ≤ V(E, V ) ≤ V(E ∩ V ) < ∞, and E ∈ BVloc (Rn ). Although the variation V(E, Ω) has no obvious geometric meaning at this point, our goal is to show that V(E, Ω) = P(E, Ω). Since proving this equality in full generality requires a substantial effort, it is instructive to prove it first when Ω= Rn and E is a dyadic figure. Example 5.1.4. Let A be a dyadic figure. If v ∈ Lipc (Rn ; Rn ), then � � div v(x) dx = v · νA dH−1 ≤ �v�L∞ (Rn ;Rn ) Hn−1 (∂A) A
∂A
by Theorem 2.3.7, and we infer V(A) ≤ P(A). To obtain the reverse inequality, choose ε > 0, and in each (n − 1)-dimensional face Ai of A select an (n − 1)-dimensional cell Ci such that
(i) Ci meets no (n − 2)-dimensional face of A, � � � (ii) Hn−1 ∂A − i Ci ≤ ε. � The normal νA restricted to C = i Ci is a Lipschitz vector field, which can be extended to w ∈ Lip(Rn ; Rn ) so that �w�L∞ (Rn ;Rn ) ≤ 1; see Theorem 1.5.4. Select an open ball U containing A, and construct a function ϕ ∈ Lipc (Rn ) so that 0 ≤ ϕ ≤ 1 and ϕ(x) = 1 for each x ∈ U . Then v := ϕw belongs to Lipc (Rn ; Rn ), �v�L∞ (Rn ;Rn ) ≤ 1, and v(x) = νA (x) for each x ∈ C. Applying
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5.1. Variation
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75
Theorem 2.3.7 again, we obtain � � div v(x) dx = v · νA dHn−1 V(A) ≥ A ∂A � � = |νA |2 dHn−1 + v · νA dHn−1 C
∂A−C
≥ Hn−1 (C) − Hn−1 (∂A − C)
≥ Hn−1 (∂A) − 2ε = P(A) − 2ε. The inequality V(A) ≥ P(A) follows from the arbitrariness of ε. Example 5.1.5. Assume n = 1. Let E ⊂ R be a measurable set such that |E| > 0 and |R − E| > 0. By Theorem 4.4.2, there are x ∈ int∗ E and y ∈ ext∗ E. We may assume that x < y. Given 0 < η < 1, find 0 < ε < (y − x)/2 so that � � � � �E ∩ B(y,ε )� �E ∩ B(x,ε )� � � ≥ 1 − η and � � ≤ η. �B(x,ε )� �B(y,ε )� A function ϕ : R → [0, 1] defined by the formula 1 (t − x + ε) if x − ε ≤ t ≤ x + ε, 2ε 1 (−t + y + ε) if y − ε ≤ t ≤ y + ε, ϕ(t) := 2ε if x + ε < t < y − ε, 1 0 otherwise
is Lipschitz and has compact support. Since � � 1 �� 1 �� ϕ� (t) dt = E ∩ B(x,ε )| − E ∩ B(y,ε )� ≥ 1 − 2η, V(E) ≥ 2ε 2ε E
the arbitrariness of η implies V(E) ≥ 1. This lower estimate cannot be improved. Indeed if E = [0, ∞), then V(E) ≤ P(E) = 1 by Remark 4.6.8.
We now extend the concept of variation from measurable sets to locally integrable functions. Definition 5.1.6. The variation of a function f ∈ L1loc (Ω) in Ω is the extended real number �� � sup f (x) div v(x) dx : v ∈ Lipc (Ω; Rn ) and �v�L∞ (Ω;Rn ) ≤ 1 , Ω
denoted by V(f, Ω). If Ω= Rn , the variation V(f, Rn ) of f in Rn is called the variation of f , denoted by V(f ).
Let f be a function defined in Ω, and let U ⊂ Ω be an open set. If f � U belongs to L1loc (U ), we define V(f, U ) := V(f � U, U ). If E ⊂ R is such thatΩ ∩ E is measurable, then n
V(E, Ω) = V(χE , Ω).
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5. BV functions
A function f ∈ L1loc (Ω) is called
(i) a function of bounded variation in Ω, or a BV function in Ω, if �f �L1 (Ω) + V(f, Ω) < ∞,
(ii) a function of locally bounded variation in Ω, or a locally BV function in Ω, if V(f, U ) < ∞ for each open set U � Ω.
The family of all BV functions in Ω, or all locally BV functions in Ω, is denoted by BV (Ω), or BVloc (Ω), respectively. Note BV (Ω) ⊂ BVloc (Ω), � � BV(Ω) = E ⊂ Ω : χE ∈ BV (Ω) , � � BVloc (Ω) = E ⊂ Ω : χE ∈ BVloc (Ω) .
For f, g ∈ L1loc (Ω) and c ∈ R, a direct verification shows V(f + g, Ω) ≤ V(f, Ω) + V(g, Ω)
and
V(cf, Ω) = |c|V(f, Ω).
It follows that BV (Ω) and BVloc (Ω) are linear spaces, and that the functional f �→ V(f, Ω) is a seminorm in BV (Ω). For f ∈ L1 (Ω), let �f �BV (Ω) := �f �L1 (Ω) + V(f, Ω)
(5.1.2)
and note that f �→ �f �BV (Ω) is a norm in BV (Ω). Proposition 5.1.7. Let {fk } be a sequence in L1loc (Ω) that converges in L1loc (Ω) to a function f ∈ L1loc (Ω). Then V(f, Ω) ≤ lim inf V(fk , Ω). Proof. If v ∈ Cc1 (Ω; Rn ) and �v�L∞ (Ω;Rn ) ≤ 1, then � � f (x) div v(x) dx = lim fk (x) div v(x) dx ≤ lim inf V(fk , Ω), Ω
Ω
and the proposition follows from the arbitrariness of v.
5.2. Mollification A mollifier is a nonnegative function η ∈ Cc∞ (Rn ) such that � η(x) dx = 1, spt η ⊂ B(0, 1), and η(−x) = η(x) Rn
for each x ∈ Rn . For 0 < ε ≤ 1, if η is a mollifier then so is ηε : x �→ ε−n η(ε−1 x) : Rn → R.
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5.2. Mollification
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77
Example 5.2.1. Define a function η ∈ Cc∞ (Rn ) by the formula � � �−1 γ exp |x|2 − 1 if |x| < 1, η(x) := 0 if |x| ≥ 1, � where γ > 0 is such that Rn η(x) dx = 1. It is immediate that η is a mollifier, often called the standard mollifier ; cf. [29, Section 4.2.1]. For Ω and ε > 0, we define an open set � � Ωε := x ∈ Ω : B(x,ε ) ⊂ Ω .
Given f ∈ L1loc (Ω) and a mollifier η, the convolutions � ηε ∗ f : x �→ ηε (x − y)f (y) dy : Ωε → R, (Dηε ) ∗ f : x �→
�
Ω
Ω
(Dηε )(x − y)f (y) dy : Ωε → Rn
are C ∞ maps, and ηε ∗ f = f ∗ ηε . If f ∈ L1 (Ω), then its zero extension f belongs to L1 (Rn ) and we define ηε ∗ f (x) := ηε ∗ f (x) for all x ∈ Rn . For a map φ = (f1 , . . . , fm ) in L1loc (Ω; Rm ), we let ηε ∗ φ := (ηε ∗ f1 , . . . , ηε ∗ fm ). The next lemma is proved in [29, Sections 4.2.1 and 4.2.3]. The equality ηε ∗ (Df ) = (Dηε ) ∗ f of assertion (4) follows from Theorem 2.3.9 and Remark 2.3.10. Lemma 5.2.2. Let f ∈ L1loc (Ω) and let η be a mollifier. If ε > 0, then (1) (2) (3) (4)
�ηε ∗ f �L∞ (Ωε ) ≤ �f �L∞ (Ω) , �ηε ∗ f �L1 (U ) ≤ �f �L1 [B(U,ε)] for each open set U � Ωε , � � (η ∗ f )g = Ω f (ηε ∗ g) for each g ∈ L1 (Ω) with spt g ⊂ Ωε , Ω ε D(ηε ∗ f ) = ηε ∗ (Df ) = (Dηε ) ∗ f when f ∈ Liploc (Ω).
Let U � Ω. If {εk } is a sequence in R+ such that lim εk = 0 and U � Ωεk for k = 1, 2, . . . , then (5) lim �f − ηεk ∗ f �L1 (U ) = 0, (6) lim �f − ηεk ∗ f �L∞ (U ) = 0 whenever f ∈ C(Ω). The variation V(f, Ω) is usually defined by means of vector fields v which belong to Cc1 (Ω; Rn ), or to Cc∞ (Ω; Rn ), rather than to Lipc (Ω; Rn ). Utilizing mollifiers, we show that it makes no difference which space is used. In our exposition we employ all Lipc (Ω; Rn ), Cc1 (Ω; Rn ), and Cc∞ (Ω; Rn ), depending on the task at hand.
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5. BV functions
Proposition 5.2.3. If f ∈ L1loc (Ω), then � V(f, Ω) = sup f (x) div v(x) dx v
where v ∈
Cc∞ (Ω; Rn )
Ω
and �v�L∞ (Ω;Rn ) ≤ 1.
Proof. Denote by I the right side of the desired equality, and assume that I < V(f, Ω). There is v ∈ Lipc (Ω; Rn ) with �v�L∞ (Ω;Rn ) ≤ 1 and � f (x) div v(x) dx. I< Ω
Choose a mollifier η and ε > 0 so that B(spt v,ε ) ⊂ Ω. If vk := ηε/k ∗ v for k ∈ N, then vk ∈ Cc∞ (Ω; Rn ), �vk �L∞ (Ω;Rn ) ≤ 1, div vk = ηε/k ∗ div v, and �div vk �L∞ (Ω) ≤ �div v�L∞ (Ω) ≤ n Lip v, lim �div vk − div v�L1 (Ω) = 0.
Thus {vk } has a subsequence {vkj } such that lim div vkj (x) = div v(x) for almost all x ∈ Ω. The dominated convergence theorem yields a contradiction: � � f (x) div v(x) dx = lim f (x) div vkj (x) dx ≤ I. Ω
Ω
The inequality I ≤ V(f, Ω) follows from Cc∞ (Ω; Rn ) ⊂ Lipc (Ω; Rn ).
Remark 5.2.4. As Cc∞ (Ω; Rn ) ⊂ Cc1 (Ω; Rn ) ⊂ Lipc (Ω; Rn ), Proposition 5.2.3 holds when Cc∞ (Ω; Rn ) is replaced by Cc1 (Ω; Rn ). Example 5.2.5. Using an argument similar to that employed in Example 5.1.4, we show that V(B) = P(B) for each ball B. It suffices to consider the ball B := B(0, r). By Proposition 2.1.4, for each v ∈ Cc∞ (Rn ; Rn ), � � div v(x) dx = v · νB dHn−1 ≤ �v�L∞ (Rn ;Rn ) Hn−1 (∂B). B
∂B
Thus V(B) ≤ P(B). For the reverse inequality, let A := B(r + ε, 0) − B(0, r − ε) and ϕ := ηε ∗ χA where η is a mollifier and 0 < ε < r/2. The formula � ϕ(x)|x|−1 x if x ∈ Rn − {0}, w(x) := 0 if x = 0 defines w ∈ Cc∞ (Rn ; Rn ) such that �w�L∞ (Rn ;Rn ) ≤ 1 and w � ∂B = νB . Thus � � V(B) ≥ div w(x) dx = w · νB dHn−1 = P(B) B
∂B
by another application of Propositions 2.1.4.
Example 5.2.6 (Caviar and Swiss cheese). Assume n ≥ 2, and choose a countable dense set D in a closed ball B ⊂ Rn . Select recursively xi ∈ D and ri > 0 so that the balls � Bi := B(xi , ri ) are disjoint, C := ∞ i=1 Bi is a dense subset of B, and |C| < |B|. According to Example 5.2.5 and Proposition 5.1.7, � � �� k ∞ V(C) ≤ lim inf V Bi ≤ V(Bi ) < ∞. k→∞
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i=1
i=1
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Thus C is a BV set, and so is S := B − E. Moreover |S| > 0. Colloquially, the sets C and S are called “caviar” and “Swiss cheese”, respectively. More complicated BV sets can be obtained by placing smaller caviar balls into each hole of a Swiss cheese, then turning each new caviar ball into a Swiss cheese, and so on. Denoting the k-th Swiss cheese by Sk , the � union ∞ k=1 Sk is still a BV set, provided the diameters of the cheese holes and caviar balls are sufficiently small.
5.3. Vector valued measures Further study of BV functions relies on measures whose values are elements of Rn . In this section we define such measures, and establish their basic properties. Throughout this section, we fix Ω and denote by B the family of all Borel sets B � Ω. A division of B ∈ B is a countable disjoint collection C ⊂ B whose union equals B. An Rm -valued measure (in Ω, if emphasizing is desirable) is a map ν : B → Rm such that �� ν(B) = ν(C) : C ∈ C} for each B ∈ B and each division C of B. An R-valued measure is called a signed measure. A map ν = (ν1 , . . . , νm ) from B to Rm is an Rm -valued measure if and only if each component νi is a signed measure.
Example 5.3.1. Let µ be a Radon measure. If h = (h1 , . . . , hm ) belongs to L1loc (Ω, µ; Rm ), then the map �� � � µ h : B �→ h1 dµ, . . . , hm dµ : B → Rm B
B
is an Rm -valued measure. Proposition 5.3.9 below shows that each Rm -valued measure is of this form.
Lemma 5.3.2. A nonnegative signed measure ν has a unique extension to a Radon measure µ. Proof. Since all compact subsets of Ω belong to B, Theorem 1.3.1 shows that there is at most one Radon measure µ with µ � B = ν. It remains to prove the existence. For every E ⊂ Ω, define �� � µ(E) := inf ν(A) : A ∈ A A
where A ⊂ B is a countable cover of E. We show first that µ : E �→ µ(E) is a measure. As ν is real-valued, ν(∅) = ν(∅ ∪ ∅) = ν(∅) + ν(∅)
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�∞ yields ν(∅) = 0, and hence µ(∅) = 0. Let E ⊂ i=1 Ei ⊂ Ω, and assume µ(Ei ) < ∞ for every i ∈ N. Choose ε > 0 and select Bi,j ∈ B so that �∞ Ei ⊂ j=1 Bi,j and ∞ � ν(Bi,j ) < µ(Ei ) + ε2−i j=1
for i = 1, 2, . . . . From E ⊂ µ(E) ≤
∞ �
�∞
Bi,j , we obtain
i,j=1
∞ � ∞ �
ν(Bi,j ) =
i,j=1
ν(Bi,j )
0, find disjoint open sets U, V ⊂ Ω so that A ⊂ U and B ⊂ V . Given ε > 0, there is a sequence {Ci } in B such that �∞ A ∪ B ⊂ i=1 Ci and µ(A ∪ B) + ε > =
∞ � i=1
∞ � i=1
∞ � � � ν(Ci ) ≥ ν Ci ∩ (U ∪ V ) i=1
ν(Ci ∩ U ) +
≥ µ(A) + µ(B).
∞ � i=1
ν(Ci ∩ V )
It follows that µ is a metric, and hence Borel, measure. Given E ⊂ Ω with µ(E) < ∞, there are sequences {Bj,k } in B such that �∞ �∞ �∞ �∞ E ⊂ j=1 Bj,k and j=1 ν(Bj,k ) < µ(E) + 1/k. The set B = k=1 j=1 Bj,k �∞ is Borel, E ⊂ B ⊂ j=1 Bj,k , and µ(E) ≤ µ(B) ≤
∞ �
ν(Bj,k ) < µ(E) + 1/k
j=1
for k = 1, 2, . . . . This implies that µ is Borel regular. Since µ � B = ν, we conclude that µ is a Radon measure. Convention 5.3.3. In view of Lemma 5.3.2, we tacitly identify a nonnegative signed measure in Ω with its unique extension to a Radon measure in Ω, and denote both by the same symbol. Corollary 5.3.4. If µ is a Radon measure in Ω, then so is µ measurable set A ⊂ Ω, and µ A = µ χA .
A for each µ
Proof. There is an increasing sequence {Ak } of µ measurable sets such that �∞ A = k=1 Ak and µ(Ak ) < ∞ for all k ∈ N. Each µ Ak is a Radon measure by Theorem 1.3.1. Given E ⊂ Ω, there are Borel sets Bk ⊂ Ω such that
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81
E ⊂ Bk and (µ Ak )(E) = (µ a Borel set containing E, and (µ
A)(E) = lim(µ
✐
Ak )(Bk ). The intersection B =
Ak )(E) = lim(µ
Ak )(B) = lim(µ
�
k∈N
Bk is
A)(B).
As (µ A)(K) < ∞ for every compact set K ⊂ Ω, the previous equality implies that µ A is a Radon measure. The corollary follows from Lemma 5.3.2, since for each C ∈ B, � χA dµ = (µ χA )(C). (µ A)(C) = C
Lemma 5.3.5. Let {x1 , . . . , xp } ⊂ Rm . There are a set C ⊂ Rm and constant β = β(m) > 0 such that �� � p � � � � � ≥ β x |xj |. j� � xj ∈C
j=1
Proof. For y ∈ S m−1 and By = B(y, 1/2), denote by Cy the intersection of all closed cones C in Rm such that the vertex of C is at the origin and By ⊂ C. The axis of Cy is the ray emanating from the origin and passing through y. Denote by � π the � orthogonal projection from Cy onto its axis, and observe that |z| ≤ 2�π(z)� for each z ∈ Cy . Thus k � i=1
|zi | ≤ 2
� k � �� � � �� �� k k � � � �� � � � k � � �π(zi )� = 2� � � � � ≤ 2 π(z ) = 2 π z zi �� i � i � � � � i=1
i=1
i=1
i=1
for each collection {z1 , . . . , zk } ⊂ Cy . Since S m−1 is compact, there are points �q �q y1 , . . . , yq in S m−1 such that S m−1 ⊂ i=1 Byi , and thus Rm = i=1 Cyi . There is C = Cyi such that � � p �� � 1 � 1 1� � � · ≥ x |x | ≥ |xj |. j� j � 2 2 q j=1 xj ∈C
xj ∈C
Let ν be an Rm -valued measure. For B ∈ B, define � ��� � �ν(C)� : C ∈ C �ν�(B) := sup
(5.3.1)
C
where C is a division of B.
Proposition 5.3.6. If ν is an Rm -valued measure, then �ν� : B �→ �ν�(B) : B → R is a nonnegative signed measure, called the variation of ν.
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82
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5. BV functions
Proof. Let B ∈ B be the union of a sequence {Bi } in B consisting of disjoint sets. If C is a division of B, then {Bi ∩ C : C ∈ C} is a division of Bi and {Bi ∩ C : i = 1, 2, . . . } is a division of C ∈ C. Thus �∞ � ∞ �� � �� � � � �� � � � �ν(C)� = �ν(Bi ∩ C)� ν(B ∩ C) ≤ i � � C∈C
=
C∈C i=1 ∞ � � i=1 C∈C
C∈C i=1
� � �ν(Bi ∩ C)� ≤
∞ � i=1
�ν�(Bi ),
�∞ and so �ν�(B) ≤ i=1 �ν�(Bi ). Next choose a division Ci of Bi , and observe �∞ that C = i=1 Ci is a division of B. Hence ∞ � � � � �ν(C)� ≤ �ν�(B). i=1 C∈Ci
�∞ From this inequality it is easy to infer i=1 �ν�(Bi ) ≤ �ν�(B). It remains to show that �ν�(B) < ∞ for each B ∈ B. Seeking a contradiction, assume there is B ∈ B with �ν�(B) = ∞. Let {Cj } be a division of B such that p � � � � �� � �ν(Cj )� > 1 1 + �ν(B)� β j=1
where β is the constant from Lemma 5.3.2. After a suitable reordering, � � q � �� �� p � � � � � � � � � � q �ν � �ν(Cj )� > 1 + �ν(B)� � � C ν(C ) ≥ β = j � j � � � j=1
j=1
j=1
�q for a positive integer q ≤ p; see Lemma 5.3.2. Letting C = j=1 Cj , we obtain � � C ⊂ B, �ν(C)� > 1, and � � � � � � � � �ν(B − C)� = �ν(B) − ν(C)� ≥ �ν(C)� − �ν(B)� > 1.
As �ν�(B) = ∞, either �ν�(C) = ∞ or �ν�(B − C) = ∞. It follows there are disjoint Borel subsets B1 and C1 of B such that � � �ν�(B1 ) = ∞ and �ν(C1 )� > 1.
Replacing B by B1 , find disjoint Borel subsets B2 and C2 of B1 with � � �ν�(B2 ) = ∞ and �ν(C2 )� > 1.
of disjoint Proceeding recursively, we � obtain � a sequence {Cj } in B consisting �∞ subsets of B such that �ν(Cj )� > 1 for j = 1, 2, . . . . Thus j=1 ν(Cj ) does �∞ not converge. Since i=1 Cj belongs to B, this is a contradiction. If ν is a signed measure in Ω, then � � and ν+ := 12 �ν� + ν
✐
ν− =
1 2
�
�ν� − ν
�
(5.3.2)
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83
are nonnegative signed measures. They are called, respectively, the positive and negative parts of ν, and the identity ν = ν+ − ν− is referred to as the Jordan decomposition of ν. Proposition 5.3.7. Let µ be a Radon measure, and let ν = µ h ∈ L1loc (Ω, µ; Rm ). Then �ν� = µ |h|. Proof. If {Ai } is a division of A ∈ B, then � � � ∞ �� ∞ � � � � � � �ν�(A) ≤ h dµ� ≤ |h| dµ = |h| dµ = (µ � i=1
Ai
i=1
Ai
A
h for an
|h|)(A).
For the reverse inequality, we fix B ∈ B and consider two cases.
Case 1. Assume h(Ω) = {y1 , y2 , . . . }, and observe that the sets � � Bi = x ∈ B : h(x) = yi , i = 1, 2, . . . ,
form a division of B. Hence
∞ ∞ �� � � � � � � � � �ν�(B) ≥ ν(Bi ) = �
=
i=1 ∞ � i=1
i=1
|yi |µ(Bi ) =
�
B
Bi
� � h dµ��
|h| dµ.
Case 2. For an arbitrary h ∈ L1loc (Ω, µ; Rm ), choose ε > 0� and construct � a map f ∈ L1loc (Ω, µ; Rm ) with countably many values so that �h(x) − f (x)� < ε for all x ∈� Ω. Note that a grid in Rm obtained by translating the half√ �m facilitates the construction of f . Define an Rm -valued open cube 0, ε m measure λ := � µ f , and find a division {Ci } of B satisfying the inequality �∞ �� � > �λ�(B) − ε. Since λ(C ) i i=1 �∞ � ∞ ∞ ��� � � � � �� � � � �ν(Ci )� − �λ(Ci )�� ≤ �ν(Ci ) − λ(Ci )� � � i=1
i=1
i=1
≤
∞ � � i=1
Ci
|h − f | dµ �λ(Ci )� − εµ(B) > �λ�(B) − ε 1 + µ(B) i=1
�
B
i=1
�
� |f | dµ − ε 1 + µ(B) >
�
B
� � |h| dµ − ε 1 + 2µ(B) .
The desired inequality follows from the arbitrariness of ε.
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84
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5. BV functions
Proposition 5.3.8. Let µ be a Radon measure, and h ∈ L1loc (Ω, µ; Rm ). If C ⊂ Rm is a closed set such that the average � 1 (h)B,µ := h dµ µ(B) B
belongs to C for each Borel set B � Ω with µ(B) > 0, then h(x) belongs to C for µ almost all x ∈ Ω.
Proof. There are countable sequences {Bi } in B and {B(yj , rj )} of closed �∞ �∞ balls in Rm such that Ω= i=1 Bi and Rm − C = j=1 B(yj , rj ). Proceeding toward a contradiction, assume that the set ∞ � � � � � x ∈ Ω : h(x) ∈ Rm − C = x ∈ Bi : h(x) ∈ B(yj , rj ) i,j=1
is not µ negligible. It follows that there are B ∈ B with µ(B) > 0 and B(y, r) ⊂ Rm − C such that h(x) ∈ B(y, r) for each x ∈ B. Since �� � � � � � � � �(h)B,µ − y � = 1 � h(x) − y dµ(x)�� � µ(B) B � � � 1 �h(x) − y � dµ(x) ≤ r, ≤ µ(B) B we see that (h)B,µ belongs to B(y, r) ⊂ Rm − C, a contradiction.
Proposition 5.3.9. Let ν be an Rm -valued measure. Up to a �ν� equivalence, �there� is a unique s ∈ L1loc (Ω, �ν�; Rn ) such that ν = �ν� s. In addition �s(x)� = 1 for �ν� almost all x ∈ Ω.
Proof. If ν = (ν1 , . . . , νm ) then (νi )± ≤ �νi � ≤ �ν�; see (5.3.2). By the Radon-Nikodym theorem [29, Section 1.6.2, Theorem 2], there are functions si,± ∈ L1loc (Ω, �ν�) such that for each B ∈ B, � � si,+ d�ν� − si,− d�ν�; νi (B) = (νi )+ (B) − (νi )− (B) = B
B
see Convention 5.3.3. If si = si,+ − si,− , then s = (si , . . . , sm ) belongs to L1loc (Ω, �ν�; Rm ) and ν = �ν� s. To establish the uniqueness, assume ν = �ν� h where h = (h1 , . . . , hm ) � is a map in L1loc (Ω, �ν�; Rm ). Then B (si − hi ) d�ν� = 0 for i = 1, . . . , m and to each B ∈ B. It follows that� hi is �ν�� equivalent � � si for i = 1, . . . , m. Let B ∈ B and Nk = x ∈ B : �s(x)� ≤ k−1 for all k ∈ N. If {Ci } is a k division of Nk , then � � ∞ ∞ �� � � � � � � � �ν(Ci )� = s d�ν��� ≤ |s| d�ν� � i=1
≤
✐
i=1 ∞ � i=1
Ci
Ci
k−1 k−1 �ν�(Ci ) = �ν�(Nk ) k k
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85
k−1 and consequently � k ) ≤� k ��ν�(Nk ). Thus Nk is �ν� negligible, and so � �ν�(N �∞ � < 1 . As Ω is the union of countably many sets is k=1 Nk = x ∈ B� : �s(x) � � � B ∈ B, we see that s(x) ≥ 1 for �ν� almost all x ∈ Ω. On the other hand, if B ∈ B and �ν�(B) > 0, then � � �� � �ν(B)� � 1 �� � ≤ 1. s d�ν�� = (s)B,�ν� = �ν�(B) � B �ν�(B) � � Hence �s(x)� ≤ 1 for �ν� almost all x ∈ Ω by Proposition 5.3.8.
Let ν be an Rm -valued measure. The identity ν = �ν�
s,
(5.3.3)
whose existence was established in Proposition 5.3.9, is called the polar decomposition of ν. Since �ν� is a Radon measure in Ω, we may assume that s is a Borel map. Using the polar decomposition, we define � � v · dν := v · s dµ (5.3.4) B
B
for v ∈ L∞ (Ω, µ; Rm ) and B ∈ B.
Remark 5.3.10. � Let �ν = �ν� s be the polar decomposition of a signed measure ν. Since �s(x)� = 1 for �ν� almost all x ∈ Ω by Proposition 5.3.9, � � |s| ± s = �ν� s± . ν± = 12 �ν� (1 ± s) = 12 �ν� � Proposition 5.3.11. Let ν be an Rn -valued measure. If Ω v · dν = 0 for each v ∈ Cc∞ (Ω; Rn ), then ν ≡ 0. Proof. Let ν = µ s be the polar decomposition of ν. If v belongs to Cc (Ω; Rn ), construct by mollification a sequence {vj } in Cc∞ (Ω; Rn ) such that lim �v − vj �L∞ (Ω;Rn ) = 0 and spt vj ⊂ C for a compact set C ⊂ Ω and j = 1, 2, . . . . By our assumption, �� � �� � � � � � � v · s dµ� = � (v − vj ) · s dµ� ≤ �v − vj �L∞ (Ω;Rn ) µ(C) � � � � Ω
Ω
� and hence Ω v · s dµ = 0. Next choose an open set U � Ω and ε > 0. By Luzin’s theorem there is a compact set K ⊂ U such that s � K is continuous and µ(U − K) < ε. In view of Corollary 1.1.2, the map s � K extends to v ∈ Cc (Ω; Rn ) such n that �v�L∞ (Ω;R � ) ≤ 1 and� spt v ⊂ U . Since the first part of the proof yields � v · s dµ = v · s dµ = Ω v · dν = 0, we obtain U Ω � � � µ(K) = v · s dµ = v · s dµ − v · s dµ < ε . K
U
U −K
Hence µ(U ) ≤ µ(K) + µ(U − K) < 2ε, and the arbitrariness of ε implies µ(U ) = 0. As Ω is covered by countably many open sets U � Ω, we conclude µ ≡ 0 and consequently ν ≡ 0.
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86
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5. BV functions
The reduction of an Rm -valued measure to a Borel set A ⊂ Ω is an Rm valued measure ν A defined by the formula A)(B) := ν(A ∩ B)
(ν for each B ∈ B. If ν = ν
A we say that ν lives in A.
Observation 5.3.12. Let ν be an Rm -valued measure, and let A ⊂ Ω be a Borel set. If ν = �ν� s is the polar decomposition of ν, then � � s ν A = �ν� A A. In particular �ν
is the polar decomposition of ν Proof. Since (ν
A)(B) =
s d�ν� = A∩B
�
B
�
�
� s d �ν�
�
A.
�
A
A = �ν� A s. As � A� = �ν� A |s| = �ν� A
for each B ∈ B, we see that ν �ν
�
A� = �ν�
by Propositions 5.3.7 and 5.3.9, the observation follows from the uniqueness part of Propositions 5.3.9
5.4. Weak convergence Throughout this section we again fix an open setΩ � Rn . Definition 5.4.1. A sequence {νk } of Rn -valued measures converges weakly to an Rn -valued measure ν if � � v · dνk = v · dν lim Ω
Ω
for every v ∈ Cc (Ω; Rm ).
If a sequence {νk } of Rn -valued measures converges weakly to ν, we write w-lim νk = ν.
Observation 5.4.2. If {µk } is a weakly convergent sequence of Radon measures, then µ = w-lim µk is a Radon measure. Proof. Let �µ = �µ� s be the polar decomposition of µ, and assume that � �µ� {s < 0} > 0. Find a compact set K ⊂ {s < 0} with �µ�(K) > 0, and ϕj ∈ � Cc (Ω) so that � 0 ≤ ϕj ≤ 1 for j = 1, 2, . . . , and lim ϕj = χK . Since lim Ω ϕj s d�µ� = K s d�µ� < 0, there is ϕ = ϕj such that � � � 0> ϕs d�µ� = ϕ dµ = lim ϕ dµk ≥ 0. Ω
Ω
Ω
This contradiction implies µ = �µ�.
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87
Theorem 5.4.3. If µ and µj , j = 1, 2, . . . , are Radon measures, then the following conditions are equivalent: (1) w-lim µj = µ; (2) µ(U ) ≤ lim inf µj (U ) and lim sup µj (K) ≤ µ(K) for each open set U ⊂ Ω and each compact set K ⊂ Ω; (3) lim µj (B) = µ(B) for each Borel set B � Ω such that µ(∂B) = 0. Proof. (1) ⇒ (2) Choose an open set U ⊂ Ω and a compact set K ⊂ U , and find ϕ ∈ Cc (Ω) with χK ≤ ϕ ≤ χU . Then � � ϕ dµ = lim ϕ dµj ≤ lim inf µj (U ), µ(K) ≤ Ω � �Ω lim sup µj (K) ≤ lim ϕ dµj = ϕ dµ ≤ µ(U ). Ω
Ω
Since U and K are arbitrary and µ is a Radon measure, condition (2) follows from Theorem 1.3.1. (2) ⇒ (3) If B � Ω is a Borel set and µ(∂E) = 0, then µ(E) = µ(int E) ≤ lim inf µj (int E)
≤ lim sup µj (cl E) ≤ µ(cl E) = µ(E).
(3) ⇒ (1) From Lemma 4.5.5 we deduce that only countably many hyperplanes perpendicular to the vectors e1 , . . . , ek have positive µ measure. It follows that given ρ > 0, there is a positive r < ρ such that µ(∂C) = 0 for each half-open cube C=
n � �
ji r, (ji + 1)r
i=1
�
(∗)
where� j1 , . . . , jn are � integers. Given ϕ ∈ Cc (Ω) and ε > 0, find δ > 0 so that �ϕ(x) − ϕ(y)� ≤ ε for all x, y ∈ Ω with |x − y| < δ. There is a positive √ r < δ / n and disjoint half-open cubes C1 , . . . , Cp , defined in (∗), such that �p spt ϕ ⊂ j=1 Cj ⊂ Ω and µ(∂Cj ) = 0 for j = 1, . . . , p. If sj = inf ϕ(x)
and
x∈Cj
then 0 ≤ Sj − sj ≤ ε and sj µk (Cj ) ≤ sj µ(Cj ) ≤
✐
Sj = sup ϕ(x), x∈Cj
�
Cj
�
Cj
ϕ dµk ≤ Sj µk (Cj ), ϕ dµ ≤ Sj µ(Cj ).
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88
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5. BV functions
� �p With D = j=1 Cj and β = max |sj | + |Sj | : j = 1, . . . , p}, we obtain �� � � � � � p �� � � � � � ϕ dµk − � � ϕ dµ� ≤ ϕ dµk − ϕ dµ�� � � Ω
Ω
≤
j=1
Cj
Cj
p � � j=1
�� � �� [Sj − sj ]µk (Cj ) + |sj | + |Sj | �µk (Cj ) − µ(Cj )�
� � ≤ εµk (D) + β �µk (D) − µ(D)�.
As µ(∂D) = 0, letting k → ∞ yields � � � � � � � ≤ εµ(D). �lim ϕ dµ − ϕ dµ k � � Ω
Ω
The proposition follows from the arbitrariness of ε.
We show that Rn -valued measures possess a compactness property with respect to weak convergence. The argument relies on the Riesz representation theorem, stated below without proof. The reader interested in proving the Riesz theorem is referred to [29, Section 1.8]. Recall from (1.1.2) that for an open set U ⊂ Ω there is a legitimate inclusion Cc (U ; Rm ) ⊂ Cc (Ω; Rm ). Theorem 5.4.4 (Riesz). Let F : Cc (Ω; Rm ) → R be a linear functional such that for each open set U � Ω, � � sup F (v) : v ∈ Cc (U ; Rm ) and �v�L∞ (U ;Rm ) ≤ 1 < ∞. There is a unique Rm -valued measure ν in Ω satisfying � F (v) = v · dν Ω
for all v ∈ Cc (Ω; Rm ). In addition, for every open set U � Ω, � � �ν�(U ) = sup F (v) : v ∈ Cc (U ; Rm ) and �v�L∞ (Ω;Rm ) ≤ 1 .
Remark 5.4.5. Let S be the injective limit topology in Cc (Ω; Rm ) defined in Example 3.6.4. Denote by M (Ω; Rm ) the linear space of all Rm -valued measures inΩ , and observe that for each ν ∈ M (Ω; Rm ), the linear functional � � � v · dν : Cc (Ω, Rm ), S → R Fν : v �→ Ω
is continuous. With this notation, the Riesz theorem can be reformulated as follows.
Theorem. The map ν �→ Fν is a linear isomorphism from M (Ω; Rm ) onto the dual space � �∗ Cc (Ω; Rm ), S , and �Fν �U = �ν�(U ) for each open set U � Ω.
Corollary 5.4.6. If ν is an Rm valued measure, then �� � m �ν�(U ) = sup v · dν : v ∈ Cc (U ; R ) and �v�L∞ (Ω;Rm ) ≤ 1 Ω
for every open set U ⊂ Ω.
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89
Proof. Since it is clear that the linear functional � F : v �→ v · dν : Cc (Ω; Rm ) → R Ω
satisfies the conditions of the Riesz theorem, the desired equality holds for each open set V � Ω. If U is any open subset of Ω, then � � �ν�(U ) = sup �ν�(V ) : V is open and V � U �� � m = sup v · dν : v ∈ Cc (V ; R ), �v�L∞ (Ω;Rm ) ≤ 1, V � U Ω �� � = sup v · dν : v ∈ Cc (U ; Rm ) and �v�L∞ (Ω;Rm ) ≤ 1 . Ω
Proposition 5.4.7. Let {νk } be a sequence of Rm -valued measures that converges weakly to an Rm -valued measure ν. Then �ν�(U ) ≤ lim inf �νk �(U ) for each open set U ⊂ Ω. Proof. Let U ⊂ Ω be an open set, and let v ∈ Cc (U ; Rm ) be such that �v�L∞ (Ω;Rm ) ≤ 1. Using our assumption and the polar decompositions of νk , we obtain � � v · dν = lim v · dνk ≤ lim inf �νk �(U ). Ω
Ω
As v is arbitrary, the proposition follows from Corollary 5.4.6.
Theorem 5.4.8. Let {νk } be a sequence of Rm -valued measures. If lim sup �νk �(U ) < ∞
for each open set U � Ω, then {νk } has a subsequence which converges weakly to an Rm -valued measure. �∞ Proof. There is a sequence {Ki } of compact sets such that Ω= i=1 Ki and Ki � Ki+1 for i = 1, 2, . . . . Construct recursively subsequences si = {νi,j } of {νk } so that supj �νi,j �(Ki ) < ∞ and si+1 is a subsequence of si . Then {νj,j } is a subsequence of {νk } and supj �νj,j �(Ki ) < ∞ for i = 1, 2, . . . . Since every Borel set B � Ω is contained in some Kj , we obtain that sup �νj,j �(B) < ∞ for each Borel set B � Ω. To simplify the notation, νk,k for k = 1, 2,�. . . . � let µk := m each C := v ∈ C (Ω; R ) : spt v ⊂ Ki by Topologize Cc (Ω; Rm ) and i c � � the L∞ norm. Choose D = vp ∈ Cc (Ω; Rm ) : p ∈ N so that D ∩ Ci is dense in Ci for i = 1, 2, . . . . By the previous paragraph, �� � � � � � � sup � vp · dµk � ≤ sup |vp | d�µk � k∈N k∈N Ω
Ω
≤ �vp �L∞ (Ω;Rm ) sup �µk �(spt vp ) < ∞ k∈N
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“Book˙2011” — 2012/2/26 — 9:58 — page 90 — #100
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5. BV functions
�∞ �� for every p ∈ N. Thus each sequence v · dµk k=1 has a convergent Ω p subsequence. Using a diagonal argument similar to that employed in the previous paragraph, construct a subsequence {µkj } of {µk } so that for every p ∈ N there exists a finite limit � G(vp ) := lim vp · dµkj . j→∞
Ω
Denote by Hi the linear hull of D ∩ Ci , and note that G � (D ∩ Ci ) has a unique linear extension Gi : Hi → R. Observe that � � ti := sup �µk �(Ki ) : k ∈ N < ∞
and Gi (v) ≤ ti �v�L∞ (Ω;Rm ) for every v in Hi . Since Hi is a dense subspace of Ci , the linear functional Gi has a unique continuous extension Fi : Ci → R, necessarily linear and satisfying Fi (v) ≤ ti �v�L∞ (Ω;Rm ) for each v ∈ Ci . As the uniqueness implies Fi+1 � Ci = Fi , there is a linear functional F defined �∞ on Cc (Ω; Rm ) = i=1 Ci whose restriction to Ci equals Fi for i = 1, 2, . . . . If U � Ω, then U is contained in some Ki . Thus F � U = Fi , and consequently � � sup F (v) : v ∈ Cc (U ; Rm ) and �v�L∞ (U ;Rm ) ≤ 1 ≤ ti < ∞. � By the Riesz theorem, there is an Rm -valued measure µ with F (ϕ) = Ω v · dµ for every v ∈ Cc (Ω; Rm ). Choose v ∈ Cc (Ω; Rm ) and ε > 0. As v ∈ Ci for some i ∈ N, there is vp ∈ D ∩ Ci with �v − vp �L∞ (Ω;Rm ) < ε. Note �� � �� � � � � � � � vp · dµk − � � vp · dµ� = � vp · dµkj − G(vp )�� < ε j � Ω
Ω
Ω
for all sufficiently large integers j. Since �� � � � � � � v · dµk − � v · dµ |v − vp | d�µkj � ≤ εti , j = 1, 2, . . . , p kj � ≤ j � Ω Ω Ki �� � � � � � � vp · dµ − �≤ v · dµ |vp − v| d�µ� ≤ ε�µ�(Ki ), � � Ω
Ω
Ki
we obtain that for all sufficiently large integers j, �� � � � � � � � v · dµk − � ≤ ε ti + 1 + �µ�(Ki ) . v · dµ j � � Ω Ω � � Hence Ω v · dµ = lim Ω v · dµkj for each v ∈ Cc (Ω; Rm ).
Proposition 5.4.9. Let ν and νk be Rm -valued measures, and let µ be a Radon measure in Ω. Assume that w-lim νk = ν
and
w-lim �νk � = µ.
Then �ν� ≤ µ, and for each Borel set B � Ω with µ(∂B) = 0, lim νk (B) = ν(B).
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© 2012 by Taylor & Francis Group, LLC
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“Book˙2011” — 2012/2/26 — 9:58 — page 91 — #101
5.4. Weak convergence
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91
Proof. If U ⊂ Ω is an open set, find open sets U1 � U2 � · · · � U so that �∞ U = i=1 Ui . There are ϕi ∈ Cc (Ω) such that χUi ≤ ϕi ≤ χU for i = 1, 2, . . . . By Proposition 5.4.7, � � ϕi d�νk � = lim inf ϕi dµ ≤ µ(U ) �ν�(Ui ) ≤ lim inf �νk �(Ui ) ≤ lim inf k→∞
k→∞
k→∞
Ω
Ω
and hence �ν�(U ) = lim �ν�(Ui ) ≤ µ(U ). As �ν� and µ are Radon measures, the inequality �ν� ≤ µ follows. Choose a Borel set B � Ω with µ(∂B) = 0. If νk = (νk1 , . . . , νkm )
and ν = (ν 1 , . . . , ν m ),
1 1 then w-lim νki = ν i for i = 1, . . . , m. Let νk1 = νk+ − νk− be the Jordan 1 1 1 decomposition of νk ; see (5.3.3). As νk± ≤ �νk � ≤ �νk �, Theorem 5.4.8 shows that there is a subsequence {νs1 (k) } of {νk } such that the sequences {νs11 (k)± } converge weakly to Radon measures µ± . Clearly
µ+ − µ− = w-lim(νs11 (k)+ − νs11 (k)− ) = w-lim νs11 (k) = ν 1 , although this need not be the Jordan decomposition of ν 1 . In view of Observation 5.4.2, the inequalities νs11 (k)± ≤ �νs1 (k) �, k = 1, 2, . . . , yield µ± ≤ µ. In particular µ± (∂B) = 0, and so µ± (B) = lim νs11 (k)± (B) by Theorem 5.4.3. Hence � � ν 1 (B) = µ+ (B) − µ− (B) = lim νs11 (k)+ (B) − νs11 (k)− (B) = lim νs11 (k) (B).
Next find a subsequence {νs2 (k) } of {νs1 (k) } with ν 2 (B) = lim νs22 (k) (B). Proceeding recursively, it is obvious that {νsm (k) } is a subsequence of {νk } such that ν(B) = lim νsm (k) (B). However, more is true: each subsequence {νs(k) } of {νk } contains a subsequence {νs(kj ) } such that ν(B) = lim νs(kj ) (B). This implies ν(B) = lim νk (B).
Example 5.4.10. Let hk (t) := sin 2k t for t ∈ R and k ∈ N, and define νk := L1 hk . If f ∈ Cc (R), choose ε > 0 and find an open set U � R containing spt f . There is ϕ ∈ Cc1 (U ) with �f − ϕ�L∞ (R) < ε. Thus � � �� � �� � � � � � � �f (t) − ϕ(t)� dt + 2−k � ϕ� (t) cos 2k t dt� � f d νk � ≤ � � � � R
U
R
≤ ε|U | + 2−k �ϕ� �L1 (R) ,
and hence w-lim νk = 0. As a direct verification shows �νk � = (2/π)L1 for k = 1, 2, . . . , we see that � w-lim νk � < w-lim �νk �.
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© 2012 by Taylor & Francis Group, LLC
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5. BV functions
5.5. Properties of BV functions Theorem 5.5.1. Let f ∈ BVloc (Ω). There is a unique Rn -valued measure Df in Ω such that for each v ∈ Lipc (Ω; Rn ), � � f (x) div v(x) dx = − v · d(Df ). (5.5.1) Ω
Ω
If U ⊂ Ω is an open set, then �Df �(U ) = V(f, U ).
Proof. For every open set U ⊂ Ω, equip the linear spaces CU := Cc (U ; Rn )
and
LipU := Lipc (U ; Rn )
with the L∞ norm. Select open setsΩ 1 � Ω2 � · · · whose union is Ω. The linear functional � f (x) div v(x) dx : LipΩ → R (∗) G : v �→ − Ω
satisfies G(v) ≤ V(f, Ωi )�v�L∞ (Ωi ;Rn ) for every v ∈ LipΩi and i = 1, 2, . . . . As LipΩi is a dense subspace of CΩi , the restriction G � LipΩi has a unique continuous extension Gi : CΩi → R, necessarily linear and satisfying Gi (v) ≤ V(f, Ωi )�v�L∞ (Ωi ;Rn )
for each v ∈ CΩi . By the uniqueness Gi+1 � CΩi = Gi . It follows that G �∞ extends to a linear functional F defined on CΩ = i=1 CΩi . As each open set U � Ω is contained in someΩ i , the functional F satisfies the assumptions of the Riesz theorem. Thus there is an Rn -valued measure Df with � F (v) = v · d(Df ) Ω
for each v ∈ CΩ , and (∗) implies (5.5.1). For an open set U ⊂ Ω, � � �Df �(U ) = sup F (v) : v ∈ CU and �v�L∞ (U ;Rn ) ≤ 1
by the Riesz theorem and Corollary 5.4.6. Since LipU is dense in CU , � � �Df �(U ) = sup F (v) : v ∈ LipU and �v�L∞ (U ;Rn ) ≤ 1 = V(f, U ).
Finally, if ν is an Rn -valued measure satisfying � � f (x) div v(x) dx = − v · dν Ω
Ω
� for each v ∈ LipΩ , then arguing as above, we verify that F (v) = Ω v · dν for every v ∈ CΩ . The equality ν = Df follows from the uniqueness assertion of the Riesz theorem. Remark 5.5.2. In the proofs of Theorems 5.4.8 and 5.5.1, extending the linear functionals can be simplified by employing the injective limit topology defined in Example 3.6.4; see Remark 5.4.5.
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“Book˙2011” — 2012/2/26 — 9:58 — page 93 — #103
5.5. Properties of BV functions
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93
The Rn -valued measure Df of Theorem 5.5.1 is called the variational measure of f ∈ BVloc (Ω). The components of Df are denoted by D1 f, . . . , Dn f . Theorem 5.5.1 shows that Df is the distributional gradient of f , and that D1 f, . . . , Dn f are the distributional partial derivatives of f ; see Section 3.1, in particular Example 3.1.2. Note that while the variation V(f, Ω) is defined for each f ∈ L1loc (Ω), the symbol �Df �(Ω) has meaning only when f ∈ BVloc (Ω). Proposition 5.5.3. Let f ∈ L1loc (Ω), and let ν be an Rn -valued measure in Ω such that � � f (x) div v(x) dx = − v · dν Ω
Ω
for each v ∈ Cc1 (Ω; Rn ). Then f ∈ BVloc (Ω) and ν = Df .
Proof. Proposition 5.2.3 implies that V(f, U ) ≤ �ν�(U ) < ∞ for each open set U � Ω. Hence f ∈ BVloc (Ω). Since Cc1 (U ; Rn ) is a dense subspace of Lipc (Ω; Rn ) equipped with the L∞ norm, an argument similar to that employed in the proof of Theorem 5.5.1 shows that the equality � � f (x) div v(x) dx = − v · dν Ω
Ω
is valid for every v ∈ Lipc (Ω; Rn ). The proposition follows from the uniqueness assertion of Theorem 5.5.1. Let f : Ω → R be pointwise Lipschitz and Df ∈ L1loc (Ω; Rn ). Given v ∈ Lipc (Ω; Rn ), integration by parts (Theorem 2.3.9) yields � � f (x) div v(x) dx = − Df (x) · v(x) dx. Ω
Ω
Thus f ∈ BVloc (Ω), and Propositions 5.5.3 and 5.3.7 imply Df = Ln
Df
and �Df � = Ln
|Df |.
(5.5.2)
This standard notation is ambiguous: the same symbol Df denotes both the classical gradient of f as well as the distributional gradient of f . The symbols D1 f, . . . , Dn f have similar double meaning. Fortunately, the correct interpretation will be always clear from the context. Proposition 5.5.4. Let {fk } be a sequence in BVloc (Ω) converging in L1loc (Ω) to f ∈ BVloc (Ω). If lim sup �Dfk �(U ) < ∞ for each U � Ω then w-lim Dfk = Df. If in addition lim �Dfk �(Ω) = �Df �(Ω) < ∞, then lim sup �Dfk �(C) ≤ �Df �(C) for each relatively closed set C ⊂ Ω, and w-lim �Dfk � = �Df �.
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“Book˙2011” — 2012/2/26 — 9:58 — page 94 — #104
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5. BV functions
Proof. For each v ∈ Lipc (Ω; Rn ), Theorem 5.5.1 implies � � � � lim v · d(Dfk ) = − lim fk div v = − f div v = v · d(Df ). Ω
Ω
Ω
Ω
Since Lipc (Ω, R ) is a dense subspace of Cc (Ω; R ) equipped with the L∞ norm, it is easy to infer that w-lim Dfk = Df . Proposition 5.1.7 shows that �Df �(U ) ≤ lim inf �Dfk �(U ) for each open set U ⊂ Ω. Thus under the additional assumption lim �Dfk �(Ω) = �Df �(Ω) < ∞, we obtain n
n
lim sup �Dfk �(C) = lim �Dfk �(Ω) − lim inf �Dfk �(Ω − C)
≤ �Df �(Ω) − �Df �(Ω − C) = �Df �(C)
for every relatively closed set C ⊂ Ω. An application of Theorem 5.4.3 completes the proof. Example 5.5.5. Let Ω = (0, π) and gk (t) = 2−k cos 2k t for t ∈ Ω and k ∈ N. Each gk belongs to BV (Ω) and lim �gk �L∞ (Ω) = 0. Moreover, �Dgk �(Ω) = 2 for k = 1, 2, . . . by Example 5.4.10. Proposition 5.5.4 implies w-lim Dgk = 0. Lemma 5.5.6. Let f ∈ L1loc (Ω) and U � Ω. If η is a mollifier, then � � �D(ηε ∗ f )�(U ) ≤ V(f, Ω) for each ε > 0 with B(U,ε ) ⊂ Ω. Moreover, � � �D(ηε ∗ f )�(Rn ) ≤ V(f )
for each ε > 0 when Ω = Rn .
Proof. Choose ε > 0 with B(U,ε ) ⊂ Ω, and select v ∈ Cc1 (U ; Rn ) so that �v�L∞ (U ;Rn ) ≤ 1. Integration by parts yields � � � D(ηε ∗ f ) · v = (ηε ∗ f ) div v = (ηε ∗ f ) div v − U Ω �U = f div (ηε ∗ v) ≤ V(f, Ω). Ω
� � As v is arbitrary, �D(ηε ∗ f )�(U ) ≤ V(f, Ω). If Ω= Rn , then �� � � � � �D(ηε ∗ f )�(Rn ) = sup �D(ηε ∗ f )�(V ) : V � Rn is open ≤ V(f )
for every ε > 0 by the first part of the proof. Proposition 5.5.7. If f, g ∈ L1loc (Ω), then
V(f g, Ω) ≤ �f �L∞ (Ω) V(g, Ω) + �g�L∞ (Ω) V(f, Ω). In particular, BV (Ω) ∩ L∞ (Ω) and BVloc (Ω) ∩ L∞ loc (Ω) are algebras.
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5.5. Properties of BV functions
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Proof. Choose an open set V � Ω and a mollifier η. For a small ε > 0, fk := η1/k ∗ f and gk := η1/k ∗ g � belong to L U (V,ε ) for each k ∈ N, and lim �fk gk − f g�L1 (V ) = 0. Thus by Proposition 5.1.7 and Lemma 5.5.6, � � � � � V(f g, V ) ≤ lim inf D(fk gk ) (V ) ≤ lim inf |fk Dgk + gk Dfk | V � � ≤ lim inf �fk �L∞ (V ) �Dgk �(V ) + �gk �L∞ (V ) �Dfk �(V ) � 1
≤ �f �L∞ (Ω) V(g, Ω) + �g�L∞ (Ω) V(f, Ω).
The proposition follows, since V is arbitrary and � � V(f g, Ω) = sup V(f g, V ) : V � Ω is open . Example 5.5.8. Assume n = 2, and for tk = k −3/2 , define � � Uk := x ∈ R2 : t2k+1 < |x| < t2k .
Following Example 5.2.5, observe V(Uk ) = 2π(t2k+1 + t2k ). This and Propo�∞ sition 5.1.7 imply that U = k=1 Uk belongs to BV(R2 ). Let if |x| ≥ 1, 0 −1/2 g(x) := |x| − 1 if 0 < |x| < 1, ∞ if x = 0,
� � � and calculate γ = R2 �Dg(x)� dx < ∞. The functions gi = min{g, i} are Lipschitz for i = 1, 2, . . . , and by (5.5.2), � � � �Dgi (x)� dx ≤ γ. V(gi ) = R2
As lim �gi − g�L1 (R2 S) = 0, Proposition� 5.1.7 shows g ∈ BV (R2 ), and a � 1 fortiori g � U ∈ BV (U ). Find ϕ ∈ C 1] so that ϕ(t2k ) = 1, and � c �R; [0, −1 ϕ(t) = 0 for t ≤ t2k+1 . Since v : x �→ ϕ |x| |x| x belongs to Cc1 (R2 ; R2 ) and �v�L∞ (R2 ;R2 ) ≤ 1, Proposition 2.1.4 yields � � � V(gχUk ) ≥ χUk g div v = div (gv) − v · Dg R2 Uk Uk � � = gv · νUk dH1 − v · Dg ∂Uk Uk � � = g(t2k ) dH1 − v · Dg ≥ 2π(2k)−3/4 − γ. �∞
∂U (0,t2k )
Uk
Thus k=1 V(gχUk ) = ∞. As the sets Uk have disjoint closures, it is easy to infer that V(gχU ) = ∞.
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© 2012 by Taylor & Francis Group, LLC
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“Book˙2011” — 2012/2/26 — 9:58 — page 96 — #106
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5. BV functions
Proposition 5.5.9. Let f ∈ BVloc (Ω) be such that �Df �(Ω) = 0. If Ω is connected, then f is constant almost everywhere in Ω. Proof. Observe that Ω is the union of a countable family � � U = U (x, r) ⊂ Ω : x ∈ Ω ∩ Qn and r ∈ Q .
Given U ∈ U, find ε > 0 with B(U,ε ) ⊂ Ω. If η is a mollifier, then limk→∞ �f − ηε/k ∗ f �L1 (U ) = 0. By Lemma 5.5.6, loc � � � � � �D(ηε/k ∗ f )� = �D(ηε/k ∗ f )�(U ) ≤ �Df �(Ω) = 0 U
for k = 1, 2, . . . . As ηε/k ∗ f are functions in C ∞ (U ), they are constant. It follows that there are cU ∈ R and a negligible set NU ⊂ U such that f (x) = cU for all x ∈ U − NU . If sets U, V ∈ U meet, then cU = cV since |U ∩ V | > 0. Thus g : x �→ cU for all x ∈ U and all U ∈ U defines a locally constant function in Ω. As Ω is connected, g is constant. Hence f is constant outside � the negligible set U ∈U NU . Lemma 5.5.10. If f ∈ L1loc (Rn ) and y ∈ Rn , then � � � �f (x − y) − f (x)� dx ≤ |y|V(f ). Rn
Proof. If f ∈ C 1 (Rn ), then � 1 � 1 � d� f (x − y) − f (x) = f (x − ty) dt = − y · Df (x − ty) dt 0 dt 0 for each x ∈ Rn . By Fubini’s theorem and (5.5.2), � � � 1 �� � � � � �f (x − y) − f (x)� dx ≤ |y| �Df (x − ty)�dx dt Rn
Rn
0
= |y| · �Df �L1 (Rn ;Rn ) = |y| · �Df �(Rn ).
If f ∈ L1loc (Rn ), select a mollifier η and let fk := η1/k ∗ f . Then � � � � � � �f (x − y) − f (x)� dx ≤ �f (x − y) − fk (x − y)� dx B(0,r)
B(0,r)
+
+
�
�R
n
� � �fk (x − y) − fk (x)� dx
B(0,r)
� � �fk (x) − f (x)� dx
≤ 2�f − fk �L1 [B(0,r)] + |y| · �Dfk �(Rn )
≤ 2�f − fk �L1 [B(0,r)] + |y|V(f )
by the first part of the proof and Lemma 5.5.6. The lemma follows by letting k → ∞, and then r → ∞.
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© 2012 by Taylor & Francis Group, LLC
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“Book˙2011” — 2012/2/26 — 9:58 — page 97 — #107
5.5. Properties of BV functions
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97
Corollary 5.5.11. If f ∈ BV (Rn ) has compact support, then �f �L1 (Rn ) ≤ 12 d(spt f )�Df �(Rn ).
Proof. Choose y ∈ Rn so that |y| > d(spt f ), and observe that � � � �f (x − y) − f (x)� dx ≤ |y| · �Df � 2�f �L1 (Rn ) = Rn
by Lemma 5.5.10. Letting |y| → d(spt f ) completes the proof.
Theorem 5.5.12. Let K ⊂ Rn be a compact set and γ ∈ R+ . Suppose {fk } is a sequence in BV (Rn ) such that �Dfk �(Rn ) ≤ γ and spt fk ⊂ K for k = 1, 2, . . . . There are f ∈ BV (Rn ) and subsequence {fkj } satisfying lim �fkj − f �L1 (Rn ) = 0. � Proof. Let β = γ[d(K) + 1 . Choose a mollifier η and 0 < ε < 1. Define fk,ε := ηε ∗ fk for k = 1, 2, . . . . By Lemma 5.5.10, � � � �� � � � � �fk,ε − fk �L1 (Rn ) = (y)f (x − y) − η (y)f (x) dy �� dx η ε k ε k � Rn Rn �� � � � � �fk (x − y) − fk (x)� dx dy ≤ ηε (y) n Rn �R ≤ �Dfk �(Rn ) ηε (y)|y| dy ≤ εβ. B(0,ε)
Claim. For each δ > 0, there is a subsequence {fki } such that �fki − fkj �L1 (Rn ) ≤ δ,
i, j = 1, 2, . . . .
Proof. Letting ε := δ/(3β), the previous inequality yields sup �fk,ε − fk �L1 (Rn ) ≤ for k = 1, 2, . . . . By Lemma � � � �fk,ε (x)� ≤ ε−n −n
≤ε
δ 3
(∗)
5.5.6 and Corollary 5.5.11, � � � � x−y � � �η ε fk (y) dy �
Rn
�η�L∞ (Rn ) �fk �L1 (Rn ) ≤ ε−n �η�L∞ (Rn ) β,
�� � � � � � � � � � �� x−y � �Dfk,ε (x)� = �(fk ∗ Dηε )(x)� ≤ ε−n � f D η (y) dy k ε � n � R � � � � � x−y � � ≤ ε−n−1 �Dη ε fk (y)� dy ≤ε
−n−1
Rn
�Dη�L∞ (Rn ;Rn ) �fk �L1 (Rn )
≤ ε−n−1 �Dη�L∞ (Rn ;Rn ) β for k = 1, 2, . . . . The last two inequalities show that {fk,ε } is uniformly bounded and equicontinuous. Ascoli’s theorem [64, Appendix A5] implies
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© 2012 by Taylor & Francis Group, LLC
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“Book˙2011” — 2012/2/26 — 9:58 — page 98 — #108
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5. BV functions
that {fk,ε } has a uniformly Cauchy subsequence. In particular, {fk,ε } has a subsequence {fki ,ε } such that for i, j = 1, 2, . . . , � � �fki ,ε − fkj ,ε �L1 (Rn ) ≤ �fki ,ε − fkj ,ε �L∞ (Rn ) �B(K, 1)� < 3δ .
Combining this with (∗) establishes the claim.
Using the claim construct recursively subsequences si = {fi,k } of {fk } so that si+1 is a subsequence of si and �fi,j − fi,k �L1 (Rn ) ≤ 1/i,
i, j, k = 1, 2, . . . .
Clearly {fk,k } is a subsequence of {fk } that is Cauchy in L1 (Rn ). As L1 (Rn ) is complete, {fk,k } converges in L1 (Rn ) to f ∈ L1 (Rn ). Since V(f ) ≤ γ by Proposition 5.1.7, we see that f ∈ BV (Rn ).
5.6. Approximation theorem Let E be a family of subsets of Ω and A ⊂ Ω. We define St(A, E) := {E ∈ E : A ∩ E �= ∅}. � � Recall from Section 2.2 that St {x}, E = St(x, E) for every x ∈ Ω. The family E is called point-finite if St(x, E) is finite for each x ∈ Ω, and locally finite if every x ∈ Ω has a neighborhood U such that St(U, E) is finite. When there is p ∈ N such that St(x, E) contains at most p sets for every x ∈ Ω, the family E is called point-p-finite. Observation 5.6.1. Let µ be a measure in Ω, and let E be a point-p-finite � family of µ measurable subsets of Ω. If A = E, then � µ(E) ≤ pµ(A). E∈E
� Proof. If A ⊂ E is a finite family, then E∈A χE ≤ pχA . Thus � �� � µ(E) = χE dµ ≤ p χA dµ = pµ(A), E∈A
E∈A
A
A
and the observation follows from the arbitrariness of A. If C ⊂ Rn is a cube with center x and ε > 0, let C ε := (1 + ε)[int C − x] + x. Note that C ε is an open cube of diameter (1 + ε)d(C) concentric with C, and hence C � C ε . If C is a family of cubes, we let Cε := {C ε : C ∈ C}.
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5.6. Approximation theorem
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99
Lemma 5.6.2 (Whitney). Assume ∂Ω �= ∅. There is a family C of nonoverlapping dyadic cubes such that � (i) Ω = C; (ii) d(C) < dist (C,∂ Ω) ≤ 4d(C) for each C ∈ C.
Any family C of nonoverlapping dyadic cubes which satisfies conditions (i) and (ii) has the following properties: (1) If C, D ∈ C and C ∩ D �= ∅, then 1 4 d(D)
≤ d(C) ≤ 4d(D).
(2) For each C ∈ C, the number of cubes belonging to St(C, C) does not exceed N = (12)n . In particular, the family C is point-N -finite. (3) If 0 < ε ≤ 1/4 and C ∈ C, then C ε � Ω and the number of cubes belonging to St(C ε , Cε ) does not exceed N 2 . In particular, the family Cε is locally finite. Proof. Recall that Dk denotes the family of all k-cubes, and let � √ √ � Ck := C ∈ Dk : C ⊂ Ω and 2−k n < dist (C,∂ Ω) ≤ 2−k+2 n � for k ∈ Z. There is a nonoverlapping family C ⊂ k∈Z Ck whose union is � the same as that of k∈Z Ck . By the definition of Ck , each C ∈ C satisfies condition (ii). Choose x ∈ Ω. There are a unique k ∈ Z with √ √ 2−k+1 n < dist (x,∂ Ω) ≤ 2−k+2 n, and a k-cube C containing x. The previous inequality implies √ √ 2−k n < dist (C,∂ Ω) ≤ 2−k+2 n, and hence C ∩ ∂Ω = ∅. Since C is connected and x ∈ C ∩ Ω, we see that C ⊂ Ω. This establishes condition (i). (1) Let C, D ∈ C and C ∩ D �= ∅. By condition (ii),
d(D) < dist (D,∂ Ω) ≤ dist (C,∂ Ω) + d(C) ≤ 5d(C).
As D and C are dyadic cubes, d(D)/d(C) = 2k for k ∈ Z. From 2k < 5, we infer k ≤ 2. Thus d(D) ≤ 4d(C) and by symmetry, d(C) ≤ 4d(D).
(2) By property (1), the smallest cubes in C that meet C ∈ C are of diameter d(C)/4. Hence the number of cubes in St(C, C) does not exceed the number of dyadic cubes of diameter d(C)/4 contained in the union of all dyadic cubes adjacent to C. Since there are 3n dyadic cubes adjacent to C, and since each of them contains 4n dyadic cubes of diameter d(C)/4, property (2) follows. � (3) Choose C ∈ C, and observe that C ε ⊂ int St(C, C) � Ω according to properties (1) and (2). We infer that for each pair D, Q ∈ C, the open cube Dε meets Q if and only if D ∩ Q �= ∅. In particular, if D ∈ C meets
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100
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5. BV functions
no Q ∈ St(C, C), then Dε ∩ Q = ∅ for each Q ∈ St(C, C), and consequently Dε ∩ C ε = ∅. In other words, if Dε ∈ St(C ε , Cε ), then D ∈ St(Q, C) for some Q ∈ St(C, C). An application of property (2) completes the argument. A collection C of dyadic cubes which satisfies conditions (i) and (ii) of Lemma 5.6.2 is called a Whitney division of Ω. We apply it to approximating functions in BVloc (Ω) by C ∞ functions (Theorem 5.6.3 below). For other applications consult [70, Chapter 4]. The number N = (12)n in Lemma 5.6.2 is unnecessarily large. A better estimate can be obtained by counting more judiciously. However, what matters is that N is a finite number depending only on the dimension n. Theorem 5.6.3. Let f ∈ L1loc (Ω). There is {fi } in C ∞ (Ω) such that lim �fi − f �L1 (Ω) = 0
and
lim V(fi , Ω) = V(f, Ω).
Proof. Let N = (12)n , and select a nonempty compact set C0 ⊂ Ω that will be specified below. Let U0 := Ω, and let V0 � Ω be an open set containing C0 . Choose a Whitney division {Ck : k ∈ N} ofΩ − C0 , and define 1/4
Uk := Ck
1/8
and Vk = Ck ,
k = 1, 2, . . . .
Mollifying the indicators χVk of Vk , we obtain functions ψk ∈ Cc∞ (Rn ) such that χCk ≤ ψk ≤ 1 and spt ψk ⊂ Uk for k = 0, 1, . . . . By Lemma 5.6.2, (3), �∞ the function ψ = k=0 ψk belongs to C ∞ (Rn ) and 1 ≤ ψ(x) ≤ N 2 + 1 for all �∞ x ∈ Ω. Thus each function ϕk := ψk /ψ belongs to Cc∞ (Ω) and k=0 ϕk = 1. �∞ Since {Uk : k = 0, 1, . . . } is a locally finite family, k=0 Dϕk = 0. Now select a mollifier η, and let φθ := ηθ ∗ φ for each φ ∈ L1 (Ω; Rm ) and every θ > 0. Choose ε > 0, and find εk > 0 so that the following conditions are satisfied for k = 0, 1, . . . : (a) spt (f ϕk )εk ⊂ Uk ; � � (b) �(f ϕk )εk − f ϕk �L1 (Ω) ≤ ε2−k−1 ; � � (c) �(f D ϕk )εk − f D ϕk �L1 (Ω) ≤ ε2−k−1 .
By Lemma 5.6.2, (3), the family {Uk : k = 0, 1, . . . } is locally finite. Hence �∞ f ε := k=0 (f ϕk )εk belongs to C ∞ (Ω) by (a). Moreover, �f ε − f �L1 (Ω) ≤ ε
∞ � � � �(f ϕk )ε − f ϕk � 1 t}, Ω : R → [0, ∞]
is an L1 measurable function. � � Proof. The set E := (x, t) ∈ Ω × R : f (x) − t > 0 subset of Rn+1 . Since ∅ � � (x, t) ∈ Ω × R : χ{f >t} (x) > s = E Ω × R
is an Ln+1 measurable if s ≥ 1,
if 0 ≤ s < 1,
if s < 0.
(x, t) �→ χ{f >t} (x) : Ω × R → R is an Ln+1 measurable function. Given v ∈� Cc1 (Ω; Rn ) and k �∈ N, the function (x, t) �→ χ{f >t} (x) div v(x) belongs to L1 Ω × [−k, k], Ln+1 . By Fubini’s theorem, the function � F : t �→ χ{f >t} (x)(div v)± (x) dx, Ω
1
defined L almost everywhere in [−k, k], is L1 measurable. As k is arbitrary, F is an L1 measurable function defined for almost all t ∈ R, and the corollary follows from Lemma 5.7.1. Theorem 5.7.3 (Coarea theorem). If f ∈ L1loc (Ω), then � � � V(f, Ω) = V {f > t}, Ω dt. R
Proof. Case 1. Let f ∈ C 1 (Ω) and �Df �(Ω) < ∞. Then � � � �Df (x)� dx g : t �→ {f 0 define if s ≤ t, 0 ϕh (s) := 1 if s ≥ t + h, (s − t)/h if t < s < t + h. Choose v ∈ Cc1 (Ω; Rn ) with �v�L∞ (Ω;Rn ) ≤ 1. Since ϕh ◦ f ∈ Lip(Ω) and t ∈ R − N , integrating by parts gives � � � 1 (ϕh ◦ f ) div v = − (ϕ�h ◦ f )Df · v = − Df · v h {tt} (x) − χ{f >t} (x)�= 0. Hence � � � � � �χ{f >t} (x) − χ{f >t} (x)� dt = �fk (x) − f (x)�, k R
and for k = 1, 2, . . . , Fubini’s theorem yields � � �� � � � � � �fk (x) − f (x)� dx = �χ{f >t} (x) − χ{f >t} (x)� dt dx k Ω Ω R � � �� � � � � χ{fk >t} (x) − χ{f >t} (x) dx dt. = R
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Ω
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104
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5. BV functions
By (∗∗), there is a subsequence of {fk }, still denoted by {fk }, such that lim �χ{fk >t} − χ{f >t} �L1 (Ω) = 0 for L1 almost all t ∈ R. In view of Proposition 5.1.7, � � � � V {f > t}, Ω ≤ lim inf V {fk > t}, Ω
for L1 almost all t ∈ R. Fatou’s lemma and Case 1 yield � � � � � � V {f > t}, Ω dt ≤ lim inf V {fk > t}, Ω dt R
R
≤ lim �Dfk �(Ω) = V(f, Ω).
(†)
To complete the argument, observe that for each x ∈ Ω, � ∞ � ∞ f + (x) = χ[0,f + (x)) (t) dt = χ{f + >t} (x) dt. 0
Given v ∈ �
0
Cc1 (Ω; Rn )
with �v�L∞ (Ω;Rn ) ≤ 1, Fubini’s theorem implies � � �� ∞ f + (x) div v(x) dx = χ{f + >t} (x) dt div v(x) dx Ω Ω 0 � � ∞ �� = χ{f + >t} (x) div v(x) dx dt Ω 0 � ∞ � + � ≤ V {f > t}, Ω dt. 0
� � As v is arbitrary, V(f , Ω) ≤ 0 V {f + > t}, Ω dt. Similarly � ∞ � � − V(f , Ω) ≤ V {f − > t}, Ω dt �∞
+
=
�
0
0
−∞
�
�
V {f < t}, Ω dt =
�
0
−∞
� � V {f > t}, Ω dt,
� � since {f < t} ∪ {f = t} = Ω − {f > t}, and �{f = t}� = 0 for all but countably many t ∈ R; see Proposition 5.1.2 and Lemma 4.5.5. In view of (†), the theorem follows: � � � V {f > t}, Ω dt ≤ V(f, Ω) ≤ V(f + , Ω) + V(f − , Ω) R � � � = V {f > t}, Ω dt. R
Remark 5.7.4. The last inequality of the previous proof shows that � � V |f |, Ω ≤ V(f + , Ω) + V(f − , Ω) = V(f, Ω)
holds for each f ∈ L1loc (Ω). In particular, if f belongs to BV (Ω), then so do f ± and |f |. A similar assertion holds for BVloc (Ω).
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5.7. Coarea theorem
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105
Proposition 5.7.5. Let f ∈ BVloc (Ω). The set Et = {f > t} belongs to BVloc (Ω) for L1 almost all t ∈ R, and � �Df �(B) = �DχEt �(B) dt (5.7.1) R
for each �Df � measurable set B ⊂ Ω. Proof. For an open set U ⊂ Ω, the coarea theorem and (5.1.1) imply � V(Et , U ) dt = �D(f � U )�(U ) = �Df �(U ), R
(∗)
since Et ∩ U = {f � U > t}. Find open setsΩ 1 � Ω2 � · · · � Ω so that �∞ Ω = k=1 Ωk . By the previous equality � V(Et , Ωk ) dt = �Df �(Ωk ) < ∞ R
for k = 1, 2, . . . . There is an L1 negligible set N ⊂ R such that V(Et , Ωk ) < ∞ for each t ∈ R − N and k = 1, 2, . . . . If U � Ω is an open set, then U is a subset of someΩ k , and we see that V(Et , U ) ≤ V(Et , Ωk ) < ∞
(∗∗)
for every t ∈ R − N . In other words, Et ∈ BVloc (Ω) for all t ∈ R − N , and (∗) shows that equality (5.7.1) holds for each open set U ⊂ Ω. If K ⊂ Ω is a compact set, there is a decreasing sequence of open sets �∞ Uj � Ω such that K = j=1 Uj . Then �DχEt �(Uj ) is a decreasing sequence converging to �DχEt �(K) for each t ∈ R − N , and by (∗∗), � � �DχEt �(K) dt = lim �DχEt �(Uj ) dt R
R
= lim �Df �(Uj ) = �Df �(K).
If S ⊂ Ω is a �Df � negligible set, use Theorem 1.3.1, (1) to find a decreas�∞ ing sequence {Vk } of open subsets of Ω so that S is contained in V = k=1 Vk and �Df �(Vk ) < 1/k for k = 1, 2, . . . . Since � �DχEt �(V1 ) = �Df �(V1 ) < ∞, R
�DχEt �(V1 ) < ∞ for L1 almost all t ∈ R, and hence � � �DχEt �(V ) dt = lim �DχEt �(Vk ) = lim �Df �(Vk ) = 0. R
R
Thus �DχEt �(S) = 0 for L almost all t ∈ R. Let B � Ω be �Df � measurable. According to Theorem 1.3.1, (2), there is an increasing sequence {Kj } of compact subsets of B such that the set 1
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106 S=B−
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5. BV functions �∞
j=1
Kj is �Df � is negligible. By the previous paragraphs, � � �DχEt �(B) dt = lim �DχEt �(Kj ) R
R
= lim �Df �(Kj ) = �Df �(B).
Finally, if B ⊂ Ω is an arbitrary �Df � measurable set, we apply the previous paragraph to the sets B ∩ Ωk , k = 1, 2, . . . .
5.8. Bounded convex domains If Ω is bounded and convex, the space BV (Ω) exhibits additional useful properties. Such properties remain valid for larger families of open sets [1, 75], e.g., for Lipschitz domains with compact boundary (Section 7.5 below). However, specializing to open sets that are bounded and convex is sufficient for our present tasks and simplifies the proofs. Lemma 5.8.1. Let Ω be bounded and convex, and let g ∈ BV (Ω). There is a sequence {gk } in C ∞ (Ω) ∩ Lip(Ω) such that lim �gk − g�L1 (Ω) = 0
lim �Dgk �(Ω) = �Dg�(Ω).
and
Proof. Translating Ω, we may assume 0 ∈ Ω. It follows from Theorem 5.6.3 that there is a sequence {fk } in C ∞ (Ω) such that lim �fk − g�L1 (Ω) = 0
lim �Dfk �(Ω) = �Dg�(Ω).
and
For k = 1, 2, . . . and x ∈ R , let φk (x) := kx/(k +1) andΩ k := Ω is convex and bounded,Ω � Ωk and the function � � hk : x �→ fk φk (x) : Ωk → R n
(∗)
φ−1 k (Ω).
Since
belongs to C ∞ (Ωk ). Thus gk := hk � Ω belongs to C ∞ (Ω) ∩ Lip(Ω). In particular gk ∈ L1 (Ω), since Ω is bounded. For k = 1, 2, . . . , � � � k ��� � �fk − gk �L1 (Ω) = �fk (x) − fk k+1 x � dx Ω
� � �� 1 � � � k+t �� � d = �� fk k+1 x dt�� dx Ω 0 dt � � 1 �� � � k+t ��� � 1 ≤ k+1 |x| · �Dfk k+1 x � dx dt 0
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≤
1 k+1
≤
1 k+1
�� 1 � 0
·
Ω
k+1 k+t
�n+1
� k+1 �n+1 k
dt
���
Ω
� � |y| · �Dfk (y)� dy
�
d(Ω) �Dfk �(Ω),
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5.8. Bounded convex domains � � � � � � kx ��� � dx �Dgk �(Ω) = ��D fk k+1 � Ω � � � � �n−1 �Dfk (x)� dx = k+1 k ≤
� k+1 �n−1 k
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107
φk (Ω)
�Dfk �(Ω)
by the Fubini and change of variables theorems. These inequalities, (∗), and Proposition 5.1.7 imply lim �gk − g�L1 (Ω) = 0 and lim sup �Dgk �(Ω) ≤ lim �Dfk �(Ω) = �Dg�(Ω) ≤ lim inf �Dgk �(Ω). Recall from (5.1.2) that a norm in BV (Ω) is defined by �f �BV (Ω) = �f �L1 (Ω) + �Df �(Ω) for each f ∈ BV (Ω). Lemma 5.8.2. Let Ω be convex, and let U ⊂ Rn be an open set for which Ω � U . There is a constant κ > 0 such that each g ∈ Lip(Ω) has an extension h ∈ Lip(Rn ) satisfying the following conditions: (i) spt h ⊂ U , (ii) �h�L∞ (Rn ) ≤ �g�L∞ (Ω) , (iii) �h�BV (Rn ) ≤ κ�g�BV (Ω) .
Proof. Select z ∈ Ω, and let φ : cl Ω− {z} → Rn − Ω be the radial reflection from z across ∂Ω. More precisely: for x ∈ Ω − {z}, the ray � � �x := z + t(x − z) : t ∈ R+
emanating from z and passing through x meets ∂Ω at a unique point x� , and we let φ(x) := 2x� − x. There is a compact set K ⊂ Ω such that z ∈ int K � and the open set V := clΩ ∪ φ(Ω − K satisfiesΩ � V ⊂ U . The restriction ψ = φ � (Ω − K) is a lipeomorphism fromΩ − K onto V − clΩ . Since g has a Lipschitz extension to cl Ω, still denoted by g, � g(x) if x ∈ clΩ, g1 (x) := � −1 � g ψ (x) if x ∈ V − clΩ defines a Lipschitz function g1 on V which extends g and satisfies �g1 �L∞ (V ) ≤ �g�L∞ (Ω) ,
�g1 �L1 (V ) ≤ β�g�L1 (Ω) , �Dg1 �L1 (V ) ≤ β�Dg�L1 (Ω)
(∗)
where β > 0 is a constant depending only on Lip (ψ). Select open sets V1 , V2 so thatΩ � V1 � V2 � V , and find ϕ ∈ C 1 (Rn ) with χV1 ≤ ϕ ≤ χV2 . We see at once that h : Rn → R defined by the formula � g1 (x)ϕ(x) if x ∈ V , h(x) := 0 if x ∈ Rn − V
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108
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5. BV functions
is a Lipschitz extension of f satisfying (i) and (ii). By inequalities (∗), �h�BV (Rn ) ≤ �g1 �L1 (V ) + �Dϕ�L∞ (V ) �g1 �L1 (V ) + �Dg1 �L1 (V ) � � ≤ β 1 + �Dϕ�L∞ (V ) �g�L1 (Ω) + β�Dg�L1 (Ω) ≤ κ�g�BV (Ω) � � where κ := β 1 + �Dϕ�L∞ (Rn ) .
Theorem 5.8.3. Let U ⊂ Rn be an open set, and let Ω � U be convex. There is a constant κ > 0 such that each g ∈ BV (Ω) has an extension h ∈ BV (Rn ) that satisfies spt h ⊂ U and �h�BV (Rn ) ≤ κ�g�BV (Ω) . Proof. Select an open set V so thatΩ � V � U . Combining Lemmas 5.8.1 and 5.8.2, find a sequence {hk } in Lip(Rn ) such that (i) lim �hk − g�L1 (Ω) = 0 and lim �Dhk �(Ω) = �Dg�(Ω), (ii) spt hk ⊂ V and �hk �BV (Rn ) ≤ κ�hk �BV (Ω) for k = 1, 2, . . . .
It follows that sup �Dhk �(Rn ) < ∞. By Theorem 5.5.12, the sequence {hk } has a subsequence, still denoted by {hk }, which converges in L1 (Rn ) to a function h ∈ BV (Rn ). This and (i) imply that a subsequence of {hk } converges to h almost everywhere in Rn , and to g almost everywhere in Ω. By redefining h on a negligible set, we obtain spt h ⊂ U and h � Ω = g. Moreover, �h�BV (Rn ) = �h�L1 (Rn ) + �Dh�(Rn )
≤ lim �hk �L1 (Rn ) + lim inf �Dhk �(Rn )
≤ lim inf �hk �BV (Rn )
≤ κ lim �hk �BV (Ω) = κ�g�BV (Ω)
by Proposition 5.1.7. Corollary 5.8.4. Let Ω be bounded and convex, and let {fk } be a sequence in BV (Ω) such that sup �fk �BV (Ω) < ∞. There is a subsequence of {fk } converging in L1 (Ω) to a function f ∈ BV (Ω). Proof. Choose an open set U withΩ ⊂ U � Rn . Using Theorem 5.8.3, find hk in BV (Rn ) so that hk � Ω = fk , spt hk ⊂ U , and sup �Dhk �(Rn ) < ∞. By Theorem 5.5.12, there is a subsequence {hkj } and h ∈ BV (Rn ) such that lim �h − hkj �L1 (Rn ) = 0. Now it is clear that {fkj } and f := h � Ω satisfy the corollary’s assertion. Corollary 5.8.5. Let Ω be bounded. If f ∈ BV (Ω) has an extension to a BV function in Rn , then it has an extension to a BV function in Rn with compact support. Proof. Assume g ∈ BV (Rn ) extends f , and select U (0, r) containing cl Ω. By Theorem 5.8.3, there is h ∈ BV (Rn ) such that h � U = g � U and spt h ⊂ U (0, r + 1). Clearly h is the desired extension of f .
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5.8. Bounded convex domains
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109
The following examples show that various assumptions of Corollary 5.8.4 cannot be omitted. Example 5.8.6. Given a nonempty Ω that is bounded and convex, the functions fk := kχΩ � Ω belong to BV (Ω) and �Dfk �(Ω) = 0 for k = 1, 2, . . . . However, no subsequence of {fk } converges in L1 (Ω) to a function f ∈ BV (Ω), since lim �fk �L1 (Ω) = ∞. Example 5.8.7. Let n = 1. The function fk (x) = sin kx defined on an interval Ω = (0, π) belongs to BV (Ω), and a direct calculation shows that �b �fk �L1 (Ω) = 2 and �Dfk �(Ω) = 2k for k = 1, 2, . . . . In addition, lim a fk = 0 for each interval (a, b) ⊂ Ω. If a subsequence of {fk } converges in L1 (Ω) to �b f ∈ L1 (Ω), then �f �L1 (Ω) = 2 and a f = 0 for each (a, b) ⊂ Ω, which is a contradiction. Example 5.8.8. Assume n ≥ 2, and choose a closed ball B ⊂ Rn . LetΩ be the union of a family {Uk ⊂ B : k ∈ N} of open balls whose closures are disjoint (cf. Example 5.2.6). Functions fk := |Uk |−1 χUk � Ω belong to BV (Ω) and �fk �BV (Ω) = 1 for k = 1, 2, . . . . If some {fkj } converges in L1 (Ω) to f ∈ L1 (Ω), then �f �L1 (Ω) = 1 and a subsequence of {fkj } converges to f almost everywhere in Ω. Since lim fk (x) = 0 for each x ∈ Ω, we have a contradiction. Proposition 5.8.9. Let Ω be arbitrary, and let {fk } be a sequence in L1loc (Ω) such that lim sup �fk �BV (U ) < ∞ for each open ball U � Ω. There is a subsequence of {fk } that converges in L1loc (Ω), as well as almost everywhere, to a function f ∈ BVloc (Ω). �∞ Proof. There are open balls Ui � Ω such that Ω= i=1 Ui . Using Corollary 5.8.4, construct recursively subsequences si = {gi,k } of {fk } so that si+1 is a subsequence of {si } and {gi,k � Ui } converges in L1 (Ui ), as well as almost everywhere, to a function gi ∈ BV (Ui ). Then {gk,k } is a subsequence of {fk } which converges in L1 (Ui ), as well as almost everywhere, to gi for i = 1, 2, . . . . There is a negligible set N ⊂ Ω and a function f defined onΩ such that f (x) = gi (x) for each x ∈ Ui − N and i = 1, 2, . . . . In particular, lim gk,k (x) = f (x) for all x ∈ Ω − N . As every U � Ω is covered by finitely many balls Ui , we see that f ∈ BVloc (Ω) and lim gk,k = f in L1loc (Ω). Example 5.8.10. Let {Bk } be a sequence of disjoint closed balls, each of radius 1. Then {χBk } is a sequence in BV (Rn ) which converges to 0 in L1loc (Rn ), but does not converge in L1 (Rn ), since i �= j implies �χBi − χBj �L1 (Rn ) = |Bi | + |Bj | = 2α(n) > 0. Thus L1loc convergence is substantially weaker than L1 convergence.
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110
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5. BV functions
5.9. Inequalities Definition 5.9.1. Let E be a measurable set such that 0 < |E| < ∞. The mean value of f ∈ L1 (E) is the number � 1 f (x) dx. (f )E := |E| E Throughout this book, when using the symbol (f )E , we tacitly assume that E is a measurable set with 0 < |E| < ∞ and f ∈ L1 (E). Lemma 5.9.2. Let Ω be bounded and convex. There is β > 0 such that � � �f − (f )Ω � 1 ≤ β�Df �(Ω) L (Ω)
for each f ∈ BV (Ω).
Proof. Seeking a contradiction, suppose there are fk ∈ BV (Ω) with � � �fk − (fk )Ω � 1 > k�Dfk �(Ω) L (Ω)
(∗)
for k = 1, 2, . . . . Define gk ∈ BV (Ω) by letting
fk (x) − (fk )Ω � gk (x) := � �fk − (fk )Ω � 1
L (Ω)
for each x ∈ Ω, and observe that �gk �L1 (Ω) = 1,
�Dfk �(Ω) � �Dgk �(Ω) = � �fk − (fk )Ω � 1
�
g Ω k
L (Ω)
0, we let Cr := z + r(C − z). Corollary 5.9.3. Let Ω be bounded, convex, and centrally symmetric. There is β > 0 such that for each r > 0 and each f ∈ BV (Ωr ), � � �f − (f )Ω � 1 r L (Ω ) ≤ βr�Df �(Ωr ). r
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111
Proof. As this is our first scaling argument, we present it in detail. Via translation, we may assume that 0 is the center of Ω. Given r > 0 and f ∈ BV (Ωr ), let φ(x) := rx and g := f ◦ φ. Clearly g ∈ L1 (Ω), and we show that g ∈ BV (Ω). To this end, choose v ∈ Cc1 (Ω; Rn ) with �v�L∞ (Ω;Rn ) ≤ 1, and observe that w := v ◦ φ−1 belongs to Cc1 (Ωr ; Rn ) and �w�L∞ (Ωr ;Rn ) ≤ 1. Using the area theorem, � � � � −n g div v = r (g ◦ φ−1 ) · (div v) ◦ φ−1 Ω Ω � r = r−n f · [r div w] ≤ r−n+1 �Df �(Ωr ). Ωr
The arbitrariness of v implies V(g, Ω) ≤ r−n+1 �Df �(Ωr ) < ∞, which means g ∈ BV (Ω). By the area theorem and Lemma 5.9.2, �f − (f )Ωr �L1 (Ωr ) = �f − (g)Ω �L1 (Ωr )
= rn �g − (g)Ω �L1 (Ω)
≤ rn β�Dg�(Ω) ≤ rβ�Df �(Ωr ) where β > 0 is a constant independent of f . Lemma 5.9.4. Let E ⊂ Rn . If U and U ∩ E are measurable sets, then � � � �1 min |U ∩ E|, |U − E| p ≤ 2�χE − (χE )U �Lp (U ) for each real number p ≥ 1. �p � � Proof. Let I := U �χE (x) − (χE )U � dx and � � � �p � |U ∩E| � I= �χE (x) − |U | � dx + U ∩E
calculate � � �χE (x) −
U −E �p � |U ∩E| |U |U |
�p � ∩E| |U ∩ E| + − E| = 1 − |U|U | � �p �p � −E| ∩E| = |U|U |U ∩ E| + |U|U |U − E|. | |
�p
|U ∩E| � |U | �
dx
If |U − E| ≥ |U ∩ E| then 2|U − E| ≥ |U ∩ E| + |U − E| = |U |, and hence �p � � �p � �p � � −E| |U ∩ E| ≥ 12 |U ∩ E| = 12 min |U ∩ E|, |U − E| . I ≥ |U|U |
Similarly, if |U ∩ E| ≥ |U − E| then 2|U ∩ E| ≥ |U |, and hence �p � � �p � � ∩E| |U − E| ≥ 12 min |U ∩ E|, |U − E| . I ≥ |U|U |
Lemma 5.9.5. Let Ω be a bounded centrally symmetric convex set. There is β > 0 such that � � 2r V(E, Ωr ) min (χE )Ωr , 1 − (χE )Ωr ≤ β |Ωr | for each r > 0 and every E ⊂ Rn for which Ωr ∩ E is measurable.
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5. BV functions
Proof. Choose r > 0 and E ⊂ Rn so thatΩ r ∩ E is measurable. AsΩ r is bounded, χE ∈ BV (Ωr ). By Lemma 5.9.4 with p = 1 and Corollary 5.9.3, there is β > 0 such that � � � � 1 min |Ωr ∩ E|, |Ωr − E| min (χE )Ωr , 1 − (χE )Ωr = |Ωr | � 2 � �χE − (χE )Ωr � 1 ≤ L (Ωr ) |Ωr | 2r 2r �DχE �(Ωr ) = β V(E, Ωr ). ≤β |Ωr | |Ωr | Theorem 5.9.6 (Isoperimetric inequality). Let n ≥ 2. There is a constant γ = γ(n) > 0 such that for each measurable set E, � n−1 � min |E|, |Rn − E| n ≤ γV(E). If V(E) < ∞, then either E or Rn − E is a BV set.
Proof. As there is nothing to prove otherwise, assume V(E) < ∞. Hence E ∈ BVloc (Rn ), and we can use the variational measure �DχE �. For x = (x1 , . . . , xn ) in Rn and r > 0, a cube Qx,r :=
n �
i=1
(xi − r, xi + r)
is an open centrally symmetric convex set, and Qx,r = x+(Q0,1 )r . A constant β > 0 associated with Q0,1 according to Lemma 5.9.5 depends only on the dimension n. By translation we obtain � � β �DχE �(Qx,r ) (∗) min (χE )Qx,r , 1 − (χE )Qx,r ≤ (2r)n−1 � � 1 for each open cube Qx,r . Let Qx := Qx,R where R := 12 3β V(E) n−1 . For this choice of R, inequality (∗) yields � � 1 1 min (χE )Qx , 1 − (χE )Qx ≤ �DχE �(Qx ) ≤ . 3V(E) 3 Thus for each x ∈ Rn , we have two mutually exclusive cases (χE )Qx < 1/2
or
(χE )Qx > 1/2,
and the case that holds for one x ∈ Rn , holds for all x ∈ Rn , since x �→ (χE )Qx : Rn → [0, 1/2) ∪ (1/2, 1]
is a continuous function. If (χE )Qx < 1/2, inequality (∗) implies |Qx ∩ E| β = (χE )Qx ≤ �DχE �(Qx ), n (2R) (2R)n−1
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5.9. Inequalities
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113
or equivalently, |Qx ∩ E| ≤ 2Rβ�DχE �(Qx ). There is a sequence {xi } such � � that Qxi is a disjoint family which covers almost all of Rn . Thus |E| =
∞ � i=1
|Qxi ∩ E| ≤ 2Rβ
∞ � i=1
�DχE �(Qxi ) ≤ 2Rβ V(E), n
1
n
and the choice of R yields |E| ≤ γV(E) n−1 where γ := 3 n−1 β n−1 . When (χE )Qx > 1/2, inequality (∗) implies β |Qx − E| = 1 − (χE )Qx ≤ �DχE �(Qx ), (2R)n (2R)n−1 or equivalently, |Qx − E| ≤ 2Rβ�DχE �(Qx ). Proceeding as in the previous n case, we obtain |Rn − E| ≤ γV(E) n−1 . Corollary 5.9.7. Let n ≥ 2. If E ⊂ Rn and P (E) < ∞, then either E or Rn − E has finite measure. The corollary follows from Proposition 4.5.3 and Remark 4.6.8. The set R+ ⊂ R1 shows that it is false in dimension one. Theorem 5.9.8 (Sobolev’s inequality). Let n ≥ 2, and let γ be the constant from the isoperimetric inequality. If f ∈ BV (Rn ), then n ≤ γ�Df �(Rn ). �f �L n−1 (Rn )
Proof. Suppose first that f ≥ 0. For t ∈ R+ , the function ft := min{f, t} n belongs to L n−1 (Rn ). Indeed, � � � � � � � n � n � n ft (x) n−1 dx = ft (x) n−1 dx + ft (x) n−1 dx Rn
{f t} �L n−1 (Rn ) (Rn ) � � n−1 = h�{f > t}� n .
This and the isoperimetric inequality yield � � n−1 � � 0 ≤ g � (t) ≤ �{f > t}� n ≤ γV {f > t} ✐
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5. BV functions
for almost all t ≥ 0. By the coarea theorem (Theorem 5.7.3), � ∞ � ∞ � � � n = g (t) dt ≤ γ V {f > t} dt = γV(f ). �f �L n−1 (Rn ) 0
0
If f is arbitrary, then Remark 5.7.4 implies
n n n ≤ �f + �L n−1 + �f − �L n−1 �f �L n−1 (Rn ) (Rn ) (Rn )
=γV(f + ) + γV(f − ) = γV(f ) = �Df �(Rn ). The following corollary improves on Corollary 5.5.11. Corollary 5.9.9. If f ∈ BV (Rn ) has compact support, then 1
�f �L1 (Rn ) ≤ γ|spt f | n �Df �(Rn )
where γ = γ(n) > 0. Proof. Choose an open set U ⊂ Rn containing spt f . If n ≥ 2, then the H¨ older and Sobolev inequalities yield n �f �L1 (Rn ) = �f �L1 (U ) ≤ �1�Ln (U ) �f �L n−1 (U ) 1
1
n = |U | n �f �L n−1 ≤ γ(n)|U | n �Df �(Rn ) (Rn )
where γ(n) is the constant from the isoperimetric � � � inequality. Let n = 1. If �t g ∈ Cc1 (R), then g(t) = −∞ g � . Hence �g(t)� ≤ R |g � |, and � |g � | = |U | · �Dg�(U ) �g�L1 (U ) ≤ |U | U
whenever spt g ⊂ U . In view of this, �f �L1 (U ) ≤ |U | · �Df �(U ) by Theorem 5.6.3, since f � U belongs to BV (U ). Thus with γ(1) = 1, 1
�f �L1 (U ) ≤ γ(n)|U | n �Df �(Rn ) holds in each dimension n ≥ 1. The corollary follows from the arbitrariness of U and Theorem 1.3.1, (1). Example 5.9.10. Assume n ≥ 2, and let Ω be a bounded set that is the union of disjoint open balls Uk = U (xi , k −2 ), k = 1, 2, . . . . Define a function f ∈ BV (Ω) by letting f (x) = k 2n−2 for each x ∈ Uk . Since � n f n−1 (x) dx = α(n) > 0, Uk
n = ∞. Suppose f has an extension g ∈ BV (Rn ). In we see �f �L n−1 (U ) view of Corollary 5.8.5, we may assume that g has compact support, and a contradiction follows from Sobolev’s inequality: n n ≤ �g�L n−1 ≤ γ�Dg�(Rn ) < ∞. ∞ = �f �L n−1 (U ) (Rn )
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5.9. Inequalities
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115
Theorem 5.9.11 (Poincar´e’s inequality). Let n ≥ 2, and let Ω be convex, bounded, and centrally symmetric. There is θ > 0 such that � � �f − (f )U � n ≤ θ �Df �(U ) L n−1 (U ) for each U = x + Ωr and each f ∈ BV (U ).
Proof. If g is a BV function in Ω, then so is g − (g)Ω . By Theorem 5.8.3, g − (g)Ω has an extension h ∈ BV (Rn ) with compact support such that � � �h�BV (Rn ) ≤ κ�g − (g)Ω �BV (Ω)
where κ > 0 is a constant independent of g. From Sobolev’s inequality and Lemma 5.9.2, we obtain � � � � �h� n �g − (g)Ω � n = ≤ γ�Dh�(Rn ) n−1 L (Ω) L n−1 (Rn ) � � ≤ γ�h�BV (Rn ) ≤ γκ�g − (g)Ω �BV (Ω) �� � � = γκ �g − (g)Ω �L1 (Ω) + �Dg�(Ω) ≤ γκ(β + 1)�Dg�(Ω).
Here γ = γ(n) > 0 and β > 0 are constants independent of g, and so is the constant θ := γκ(β + 1). Let U = x+Ωr and f ∈ BV (U ). In view of translation invariance, we may assume that x = 0. For a map φ : x �→ rx : Ω → Ωr , the scaling argument, similar to that used in the proof of Corollary 5.9.3, shows that g = f ◦ φ belongs to BV (Ω) and �Dg�(Ω) ≤ r−n+1 �Df �(Ωr ). Thus � � �g − (g)Ω � n ≤ θ�Dg�(Ω) ≤ r−n+1 �Df �(Ωr ) L n−1 (Ω) by the previous paragraph. Since the area theorem yields � � � � � � �g − (g)Ω � n n n = �g − (f )Ωr �L n−1 = r−n+1 �f − (f )Ωr �L n−1 , L n−1 (Ω) (Ω) (Ω ) r
Poincar´e’s inequality follows.
Theorem 5.9.12 (Relative isoperimetric inequality). Let n ≥ 2, and let Ω be bounded, convex, and centrally symmetric. There is η > 0 such that � � n−1 min |U ∩ E|, |U − E| n ≤ η�DχE∩U �(U )
for each U = x + Ωr and each set E ⊂ Rn with E ∩ U ∈ BV(U ).
n , and Poincar´e’s inequality, Proof. By Lemma 5.9.4 with p = n−1 n−1 � � � � n ≤ 2θ�DχE∩U �(U ) min |U ∩ E|, |U − E| n ≤ 2�χE − (χE )U �L n−1 (U )
where θ > 0 is the constant occurring in Poincar´e’s inequality.
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5. BV functions
Corollary 5.9.13. Let n ≥ 2, and let U ⊂ Rn be either an open ball or the interior of a cube. There is κ = κ(n) > 0 such that � � �f − (f )U � n ≤ κ�Df �(U ), L n−1 (U ) � � n−1 min |U ∩ E|, |U − E| n ≤ κ�DχE∩U �(U )
for each f ∈ BV (U ) and each E ⊂ Rn with E ∩ U ∈ BV(U ). Proof. Let κ be the largest of the constants θU , θQ and ηU , ηQ associated with the ball U = U (0, 1) and cube Q = (0, 1)n in the Poincar´e and relative isoperimetric inequalities, respectively. Since each open ball in Rn has the form x + Ur and each open cube in Rn has the form x + Qr , the corollary follows from Theorems 5.9.11 and 5.9.12. Considering Us := (−s, s) and Es,t := (0, s + t) for s, t ∈ R+ shows that the relative isoperimetric inequality provides no immediate information when n = 1. For a more detailed treatment of isoperimetric inequalities, we refer to [75, Section 5.11] and [1, Section 3.4]. Lemma 5.9.14. If E ∈ BVloc (Ω), then for each x ∈ ∂∗ E ∩ Ω, � � �DχE � B(x, r) lim sup > 0. rn−1 r→0 Proof. Choose x ∈ ∂∗ E ∩ Ω, and note that the numbers � � � � �B(x, r) ∩ E � �B(x, r) − E � � � � � and s := lim sup � t := lim sup � B(x, r)� B(x, r)� r→0 r→0
are positive. By Corollary 5.9.13, � � � � n−1 � � � �� n−1 B(x, r) ∩ E � �B(x, r) − E � n �DχE � U (x, r) α(n) n � � � � min ≥ �B(x, r)� , �B(x, r)� rn−1 κ whenever U (x, r) ⊂ Ω. Hence � � � � �DχE � B(x, r) �DχE � U (x, r) lim sup ≥ lim sup rn−1 rn−1 r→0 r→0 α(n) = κ
n−1 n
min{t, s}
n−1 n
> 0.
Definition 5.9.15. If µ and ν are measures in an open setΩ ⊂ Rn , we say that ν is absolutely continuous with respect to µ, and write ν � µ, if µ(E) = 0 implies ν(E) = 0 for each E ⊂ Ω. If ν is an Rm -valued measure in Ω, we say that ν is absolutely continuous with respect to µ, and write again ν � µ, whenever �ν� � µ. A measure, or an Rm -valued measure, in Ω that is absolutely continuous with respect to Lebesgue measure Ln in Ω is called absolutely continuous.
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5.10. Lipschitz maps
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117
Proposition 5.9.16. For every E ∈ BVloc (Ω), Hn−1
(∂∗ E ∩ Ω) � �DχE �
(∂∗ E ∩ Ω).
Proof. Select A ⊂ ∂∗ E ∩ Ω with �DχE �(A) = 0. Letting � � � � �DχE � B(x, r) 1 , > Ak := x ∈ A : lim sup rn−1 k r→0 �∞ Lemma 5.9.14 implies A = k=1 Ak . Fix k ∈ N, and choose ε > 0. By Theorem 1.3.1, (1), there is an open set U ⊂ Ω such that Ak ⊂ U and �DχE �(U ) < ε. Select δ > 0, and denote by B the family of all closed balls B(x, r) ⊂ U satisfying � x ∈ Ak , 0 < r < δ /10, and �DχE � B(x, r)] ≥ rn−1 /k. By Vitali’s theorem (Theorem 4.3.2), there are disjoint balls B(xi , ri ) in B � such that Ak ⊂ i B(xi , 5ri ). Thus � α(n − 1)(5ri )n−1 Hδn−1 (Ak ) ≤ i
n−1
≤5
kα(n − 1)
� i
� � �DχE � B(xi , ri )
≤ 5n−1 kα(n − 1)�DχE �(U ) < 5n−1 kα(n − 1)ε. As δ and ε are arbitrary, Hn−1 (Ak ) = 0 and the proposition follows.
5.10. Lipschitz maps Throughout this section, φ = (f1 , . . . , fn ) is a Lipschitz map from Rn to Rn . Let g ∈ L1 (Rn ). Since g sign det Dφ belongs to L1 (Rn ), it follows from the area theorem (Theorem 1.5.6) that the function � �� � � φ# g : y �→ g(x) sign det Dφ(x) : x ∈ φ−1 (y) (5.10.1) belongs to L1 (Rn ). Indeed, inequality (1.5.2) implies � � � � �g(x)� dy �φ# g�L1 (Rn ) ≤ =
�
Rn x∈φ−1 (y)
Rn
� � �g(x)�Jφ (x) dx ≤ (Lip φ)n �g�L1 (Rn ) .
(5.10.2)
We show that if g ∈ BV (Rn ) has compact support, then φ# g ∈ BV (Rn ). To this end, denote by adj Dφ the adjoint matrix of Dφ, and recall that (adj Dφ) · Dφ = Dφ · (adj Dφ) = (det Dφ)I
(5.10.3)
where I is the identity n × n matrix [40, Section 5.4].
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118
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5. BV functions
Lemma 5.10.1. Let u ∈ C 1 (Rn ; Rn ) and v = (adj Dφ)(u ◦ φ). Then � � � � � � v · d(Dg) = g(x) det Dφ(x) (div u) φ(x) dx. − Rn
Rn
n
for each g ∈ BV (R ) with compact support.
Proof. Assume first that φ ∈ C 2 (Rn ; Rn ). For i, j = 1, . . . , n, denote by ai,j the elements of adj Dφ so that the vector fields Ri = (ai,1 , . . . , ai,n )
and
Cj = (a1,j . . . , an,j )
are the i-th row and j-th column of adj Dφ, respectively. A straightforward but tedious calculation establishes that div Cj = 0 for j = 1, . . . , n. Letting u = (u1 , . . . , un ) and v = (v1 , . . . , vn ), we calculate �� � div v = Di Ri · (u ◦ φ) + Ri · Di (u ◦ φ) i
=
�� i
=
�
j
uj div Cj +
j
=
� i
=
Ri ·
�� i
= = =
Di ai,j uj +
�� k
ai,j
j
� �� j
k
j
k
� �� �� j
�
� i
�
i
Ri · Di (u ◦ φ)
Ri · Di (u ◦ φ)
� (Dk u) ◦ φ Di fk
�� k
�
� (Dk uj ) ◦ φ Di fk
(Dk uj ) ◦ φ
��
ai,j Di fk
i
� (Dk vj ) ◦ φ Cj · Dfk
� � � (Dj uj ) ◦ φ (det Dφ) = (det Dφ) (div u) ◦ φ ;
the equality before last follows from (5.10.3). Since g has compact support, Theorem 5.5.1 implies � � − v · d(Dg) = g(x) div v(x) dx n Rn �R � � � � = g(x) det Dφ(x) (div u) φ(x) dx. Rn
If φ is merely Lipschitz, select a mollifier η and define φk := η1/k ∗ φ and wk := (adj Dφk )(u ◦ φk ), k = 1, 2, . . . . By the first part of the proof, � � � � � � − wk · d(Dg) = g(x) det Dφk (x) (div u) φk (x) dx Rn
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Rn
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5.10. Lipschitz maps
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119
for k = 1, 2, . . . . As the convergence lim φk = φ and lim Dφk = Dφ is uniform on a compact neighborhood of spt g, the desired equality follows. Theorem 5.10.2. If g ∈ BV (Rn ) has compact support, then V(φ# g) ≤ (Lip φ)n−1 �Dg�(Rn );
in particular φ# g ∈ BV (Rn ).
Proof. Choose u ∈ C 1 (Rn ; Rn ) with �u�L∞ (Rn ;Rn ) ≤ 1. If v = (adj Dφ)(u ◦ φ),
then �v�L∞ (Rn ;Rn ) ≤ (Lip φ)n−1 by the polar decomposition theorem for linear maps [29, Section 3.2, Theorem 2]. Since � �� � � � � g(x) sign det Dφ(x) (div u) φ(x) : x ∈ φ−1 (y) φ# g(y) div u(y) =
for each y ∈ Rn , the area theorem and Lemma 5.10.1 yield � � � � � � φ# g(y) div u(y) dy = g(x) sign det Dφ(x) (div u) φ(x) Jφ (x) dx Rn
Rn
=
�
Rn
=−
�
� g(x) det Dφ(x)(div u) φ(x)] dx
Rn
v · d(Dg) ≤ (Lip φ)n−1 �Dg�(Rn ).
Thus V(φ# g) ≤ (Lip φ)n−1 �Dg�(Rn ) by the arbitrariness of u.
Corollary 5.10.3. Let E ∈ BV(Rn ) be bounded. If φ � E is a lipeomorphism, then φ(E) ∈ BV(Rn ). Proof. As χE belongs to BV (Rn ) and has compact support, Theorem 5.10.2 shows φ# χE ∈ BV (Rn ). Thus |φ# χE | belongs to BV (Rn ) by Remark 5.7.4. Since φ � E is a lipeomorphism, det Dφ(x) �= 0 for almost all x ∈ E. It follows that |φ# χE | = χφ(E) almost everywhere. Remark 5.10.4. Observe that sign det Dφ(x) equals 1 or −1 according to whether the vectors Dφ(x)e1 , . . . , D φ(x)en form a base in Rn whose orientation is, respectively, the same as or opposite to the orientation of the standard base e1 , . . . , en . We abbreviate this by saying that the point x has, respectively, a positive or negative multiplicity with respect to φ . Thus given E ⊂ Rn and y ∈ Rn , the function φ# χE sums the multiplicities of points x ∈ E that are mapped by φ onto y. If E is measurable and bounded, then inequality (5.10.2) shows that φ# χE (y) is an integer for almost all y ∈ Rn . In Section 7.5 we show that Theorem 5.10.2 and Corollary 5.10.3 hold, respectively, for any function g ∈ BV (Rn ) and any set E ∈ BV(Rn ).
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Chapter 6 Locally BV sets
This chapter is devoted to deeper results about locally BV sets obtained by E. De Giorgi [18, 19]. We prove that the families of all locally BV sets and all sets with locally finite perimeter coincide. In addition, we show that the variational measure DχE of a locally BV set E lives in ∂∗ E and has the polar decomposition (Hn−1 ∂∗ E) (−νE ) where νE � ∂∗ E is the measuretheoretic exterior normal of E. Throughout this chapter Ω denotes an open subset of the ambient space Rn .
6.1. Dimension one Although the approach developed in Sections 6.2–6.5 below is valid for all dimensions, it is beneficial to present separately a substantially simpler proof of the one-dimensional case. Proposition 6.1.1. Let E ∈ BVloc (Ω). There is a locally finite family J of open intervals J ⊂ Ω with disjoint closures such that int∗ E =
�
J
and
∂∗ E ∩ Ω =
�
{∂∗ J ∩ Ω : J ∈ J}.
Let νE (x) = 0 if x ∈ Ω−∂∗ E, and let νE (x) equal 1 or −1 according to whether x ∈ ∂∗ E ∩ Ω is the right or left endpoint of an interval J ∈ J, respectively. Then DχE = H0 (−νE ) and �DχE � = H0 ∂∗ E. Proof. As it suffices to prove the proposition for each connected component of Ω, we may assume from the onset that Ω = (a, b) where −∞ ≤ a < b ≤ ∞. In view of translation invariance, we may also assume that a < 0 < b. Let DχE = µ1 − µ2 be the Jordan decomposition of the signed measure DχE ; see (5.3.2). For i = 1, 2, the formulae � � � µi [0, x) fi (x) := � � −µi [x, 0)
if 0 ≤ x < b,
if a < x < 0
121
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122
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6. Locally BV sets
define increasing left-continuous functions fi : Ω → R. For v ∈ Cc1 (Ω), Fubini’s theorem yields �� � � b � b fi (x)v � (x) dx = v � (x) dµ(y) dx a
= and similarly,
�
0 a
�
a
[0,b)
��
[0,x)
b
y
�
�
v (x) dx dµi (y) = −
fi (x)v � (x) dx = −
�
�
v(y) dµi (y), [0,b)
v(y) dµi (y). (a,0)
Summing the previous equalities and letting f := f1 − f2 , we obtain � � f (x)v � (x) dx = − v d(DχE ). Ω
Ω
Proposition 5.5.3 shows that f ∈ BVloc (Ω) and DχE = Df . By Proposition 5.5.9, there are a real number c and a negligible set N ⊂ Ω such that χE (x) = f (x) + c for every x ∈ Ω − N . Each x ∈ Ω is a two-sided cluster point ofΩ − N , and f (x) + c equals 0 or 1 for every x ∈ Ω − N . Since the function f + c is left-continuous, f (x) + c equals 0 or 1 for every x ∈ Ω. In other words, f + c = χA where A ⊂ Ω is equivalent to E. From the left-continuity of χA it is easy to deduce that A is the union of nonempty intervals (x, y] ∩ (a, b) whose closures are disjoint. The interiors of these intervals form a countable family J = {Jk : k ∈ N} whose union is equivalent to E. Let Jk = (ak , bk ), and choose an open interval U = (α,β ) so that U � Ω. Replacing α by α� < α, we may assume inf{ak : ak ∈ U } > α. Since χA is left-continuous at ak ∈ U , there is Bk = B(ak , rk ) such that Bk ⊂ U and Bk ∩ A ⊂ Jk . Define � |x−ak | − 1 if x ∈ Bk , rk vk (x) := 0 if x ∈ U − Bk , � and let v := ak ∈U vk . As {Bk : ak ∈ U } is a disjoint family, v ∈ Lip(U ) and �v�L∞ (U ) = 1. The choice of α shows that spt v ⊂ U . Thus � � � ak +rk � � �DχE �(U ) ≥ v (x) dx = vk� (x) dx = − vk (ak ). E
ak ∈U
ak
ak ∈U
As �DχE �(U ) < ∞ and vk (ak ) = −1 for each ak ∈ U , we see that U meets � only finitely many intervals Jk . This implies that int∗ E = k∈N (ak , bk ) and ∂∗ E ∩ Ω = {ak , bk : k ∈ N} ∩ Ω. Define νE as in the proposition, and select ϕ ∈ Cc1 (Ω). There is p ∈ N such that Jk ∩ spt ϕ = ∅ for all integers k > p, and an open interval U � Ω
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6.2. Besicovitch’s covering theorem
123
containing spt ϕ. By the previous paragraph, � p � � � � ϕ(bj ) − ϕ(aj ) χE (x)ϕ (x) dx = Ω
j=1
=
�
✐
ϕ(x)νE (x) =
x∈∂∗ E∩Ω
�
Ω
� ϕ d H0
νE ).
Thus DχE = H0 (−νE ) by Proposition 5.5.3, and �DχE � = H0 follows from the definition of νE .
∂∗ E
Remark 6.1.2. An immediate consequence of Proposition 6.1.1 is that each E ∈ BV(R) is equivalent to a one-dimensional figure. In this case, νE defined in Proposition 6.1.1 is the zero extension of the exterior normal νE of E defined in Section 2.1.
6.2. Besicovitch’s covering theorem An extension of Section 6.1 to higher dimensions is not easy. To begin with, we need a covering theorem, similar to Vitali’s theorem (Theorem 4.3.2), that holds for any Borel measure in Rn . This result, obtained by A.S. Besicovitch [3, 4], is based on a combinatorial lemma, quoted below without proof. The lemma is proved in many standard texts, e.g., in [29, Section 1.5.2, Theorem 2] or [75, Theorem 1.3.5]. A very detailed and transparent presentation is given in [46, Chapter 2]. The most general version can be found in [33, Theorem 2.8.14]. Lemma 6.2.1. Let E ⊂� Rn , and let B�be a family of closed balls whose centers cover E. If sup d(B) : B ∈ B < ∞, then there are a positive integer N = N (n) and subfamilies E1 , . . . , EN of B such that each Ei is a �N disjoint family and i=1 Ei covers E. � � To see that the assumption sup d(B) : B ∈ B < ∞ is essential, consider � the set N ⊂ R covered by the family B(k, k) : k ∈ N}.
Theorem 6.2.2 (Besicovitch). Let µ be a Borel measure in Ω, and let E ⊂ Ω. Suppose B is a family of closed balls contained in Ω such that for each x ∈ E and each η > 0 there is B(x, r) ∈ B with r < η. If µ(E) < ∞, there is a disjoint family C ⊂ B satisfying � � � µ E− C = 0.
Proof. As there is nothing to prove otherwise, assume µ(E) > 0. Eliminating larger balls, we may assume that d(B) ≤ 1 for every B ∈ B. By
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124
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6. Locally BV sets
Lemma 6.2.1, there are subfamilies E1 , . . . , EN of B, each consisting of disjoint sets, whose union covers E. Thus µ(E) ≤
N � � � � µ E∩ Ei , i=1
� � � and there is an integer k with 1 ≤ k ≤ N and µ(E)/N ≤ µ E ∩ Ek . The family Ek contains a finite subfamily C1 such that � � � 1 µ(E) < µ E ∩ C1 . N +1 � Since C1 := C1 is a closed set, and since µ is a Borel measure, 1 µ(E) + µ(E − C1 ). N +1 Hence µ(E − C1 ) < cµ(E) where c = N/(N + 1). µ(E) = µ(E ∩ C1 ) + µ(E − C1 ) >
If µ(E − C1 ) = 0, then letting C := C1 completes the argument. If µ(E − C1 ) > 0, let B1 := {B ∈ B : B ∩ C1 = ∅}. As C1 is a closed set, B1 satisfies the hypothesis of the theorem with respect to E1 = E −C1 . Applying the previous result, we obtain a finite disjoint family C2 ⊂ B1 such that for � C2 = C 2 , � � µ E − (C1 ∪ C2 ) = µ(E1 − C2 ) < cµ(E1 ) < c2 µ(E).
The set C1 ∪ C2 is closed, and C1 ∪ C2 is a disjoint subfamily of B. Proceeding � recursively, we construct finite subfamilies C1 , C2 , . . . of B so that C := i Ci is a disjoint family, and � p �� � �� � � � µ E− C ≤µ E− Ci < cp µ(E) i=1
� � � for p = 1, 2, . . . . As 0 < c < 1, we conclude µ E − C = 0.
With some extra work, Besicovitch’s theorem for a Radon measure µ can be proved without the assumption µ(E) < ∞. Theorem 6.2.3. Let µ be a Radon measure in Ω. If f ∈ L1loc (Ω, µ), then � 1 � f dµ = f (x) (6.2.1) lim � r→0 µ B(x, r) B(x,r)
for µ almost all x ∈ Ω.
Proof. The proof is similar to that of Theorem 4.3.4. We may assume that � f is a real-valued function. Let F (B) := B f dµ for each ball B ⊂ Ω, and denote by E the set of all x ∈ Ω at which the limit � � F B(x, r) � � lim r→0 µ B(x, r)
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6.2. Besicovitch’s covering theorem
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125
either does not exist, or differs from f (x). Given x ∈ E, find γx > 0 so that for each η > 0, there is B = B(x, r) with r < η and � � �F (B) − f (x)µ(B)� ≥ γx µ(B). (∗) �∞ Select open setsΩ k � Ω so that Ω= k=1 Ωk . It suffices to show that each Ek := {x ∈ E ∩ Ωk : γx > 1/k} is a µ negligible set. To this end, fix k ∈ N and choose ε > 0. By Henstock’s lemma there is δ : Ωk → R+ such that p � � � �f (xi )µ(Bi ) − F (Bi )� < ε
(∗∗)
i=1
whenever Bi ⊂ Ωk are disjoint balls with xi ∈ Bi and d(Bi ) < δ(xi ) for i = 1, . . . , p. Let B be the family of all closed balls B(x, r) ⊂ Ωk such that x ∈ Ek , r < δ(x), and inequality (∗) holds for B = B(x, r). As B and Ek satisfy the hypotheses of Besicovitch’s theorem, µ almost all of Ek is covered by disjoint balls B1 , B2 , . . . from B. In view of (∗) and (∗∗), � �� � �f (xi )µ(Bi ) − F (Bi )� < kε , µ(Ek ) ≤ µ(Bi ) ≤ k i
i
and µ(Ek ) = 0 by the arbitrariness of ε.
Corollary 6.2.4. Let µ be a Radon measure in Ω, and let 1 ≤ p < ∞. If f ∈ Lploc (Ω, µ), then for µ almost all x ∈ Ω, � � � 1 �f (y) − f (x)�p dµ(y) = 0. � (6.2.2) lim � r→0 µ B(x, r) B(x,r)
Proof. For all nonnegative real numbers a and b, � �p (a + b)p ≤ 2 max{a, b} = 2p max{ap , bp } ≤ 2p (ap + bp ).
(∗)
Thus the function |f − t|p belongs to L1loc (Ω, µ) for each t ∈ R. Since Q is a countable set, it follows from Theorem 6.2.3 that there is a µ negligible set E ⊂ Ω such that for each t ∈ Q and each x ∈ Ω − E, � � �p 1 � |f − t|p dµ = �f (x) − t� . (∗∗) lim � r→0 µ B(x, r) B(x,r)
�Fix x ∈ Ω�p− E, and for r > 0, let Br := B(x, r). Given ε > 0, find t ∈ Q with �f (x) − t� < ε. By inequality (∗), � � � �f (y) − f (x)�p dµ(y) Br � � � �p � � p p � � �t − f (x)�p dµ(y) ≤2 f (y) − t dµ(y) + 2 B Br � r � � p �f (y) − t� dµ(y) + 2p εµ(Br ) ≤ 2p Br
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6. Locally BV sets
for each r > 0. From this and inequality (∗∗), we obtain � � � � � 1 �f (y) − f (x)�p dµ(y) ≤ 2p �f (x) − t�p + 2p ε ≤ 2p+1 ε lim sup r→0 µ(Br ) Br and the corollary follows from the arbitrariness of ε.
Each point x ∈ Ω at which equality (6.2.2) holds is called the Lebesgue point of f ∈ Lp (Ω, µ). Remark 6.2.5. Since � � � � � 1 � µ B(x, r)�
B(x,r)
� � f dµ − f (x)�� ≤
1 � µ B(x, r) �
�
B(x,r)
� � �f (y) − f (x)� dµ(y),
it is clear that equality (6.2.1) holds at every Lebesgue point of f ∈ L1loc (Ω, µ). However, the converse is false: it suffices to define f (0) = 0 and f (x) = x/|x| for x ∈ R − {0}, and consider the point x = 0 with respect to L1 .
Proposition 6.2.6. Let σ and µ be Radon measures in Ω. If E ⊂ spt µ is a Borel set and � � σ B(x, r) � =1 lim � r→0 µ B(x, r) for each x ∈ E, then σ
E=µ
E.
Proof. By Theorem 1.3.1, it suffices to show σ(K) = µ(K) for each compact set K ⊂ E. Select a compact set K ⊂ E, choose ε > 0, and find an open set U so that K ⊂ U � Ω. Denote by B the family of all B(x, r) ⊂ U such that � � � � σ B(x, r) ≤ (1 + ε)µ B(x, r)
for each x ∈ K. Besicovitch’s � shows that there are disjoint balls � theorem � B1 , B2 , . . . in B such that σ K − i Bi = 0. Consequently, � � σ(K) ≤ σ(Bi ) ≤ (1 + ε) µ(Bi ) ≤ (1 + ε)µ(U ) i
i
and σ(K) ≤ µ(K) by the arbitrariness of U and ε. As our assumptions imply E ⊂ spt σ, the reverse inequality follows by symmetry.
6.3. The reduced boundary While the results of Sections 6.3–6.7 are correct in any dimension, the proofs we present assume tacitly that the dimension is greater than one. The proofs in dimension n = 1 are left to the reader: they are either trivial or follow from Proposition 6.1.1. Throughout this section we fix a set E ∈ BVloc (Ω) of positive measure. We let µE := �DχE �, and define a µE almost everywhere unique Borel map
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6.3. The reduced boundary
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127
νE : Rn → Rn so that DχE = µE (−νE ) is the polar decomposition of DχE ; see Proposition 5.3.9. Theorem 5.5.1 shows that for each v ∈ Lipc (Ω; Rn ), � � div v(x) dx = v · νE dµE . (6.3.1) E
Ω
Aside from simplifying notation, the introduction of symbols µE and νE enhances the geometric essence of the measure DχE , whose complete description is given in Theorem 6.5.2 below. Verifying equalities µΩ−E = µE
and
νΩ−E = −νE
(6.3.2)
is easy: the first � is a consequence � of Proposition 5.1.2; the second follows from (6.3.1), since Ω−E div v = − E div v for each v ∈ Lipc (Ω) by Theorem 2.3.7.
Remark 6.3.1. In Section 2.1, we denoted by νA and νB , respectively, the unit exterior normal of a figure A and a ball B in Ω= Rn . This is consistent with the above notation, provided νA and νB are extended arbitrarily to Rn . Indeed, let C be a figure or a ball, and let νC be the unit exterior normal of C whose extension to Rn is denoted by ν˜C . By Propositions 2.1.2 and 2.1.4, � � � div v(x) dx = v · νC dHn−1 = v · ν˜C d(Hn−1 ∂C) C
Rn
∂C
Cc1 (Rn ; Rn ).
n−1
Hence µC = H ∂C and DχC = µC for each v ∈ the polar decomposition of DχC ; see Propositions 5.5.3 and 5.3.7.
(−˜ νC ) is
∗ Definition 6.3.2 (De Giorgi). � � The reduced boundary of E is the set ∂ E of all x ∈ spt µE such that �νE (x)� = 1 and � 1 � � lim νE (y) dµE (y) = νE (x). (6.3.3) r→0 µE B(x, r) B(x,r)
Remark 6.3.3. Note that |νE (x)| = 1 for µE almost all x ∈ Ω by Proposition 5.3.9, and that equality (6.3.3) holds for µE almost all x ∈ Ω by Theorem 6.2.3. Consequently (6.3.4) µE (Ω − ∂ ∗ E) = 0. In particular, ∂ ∗ E is µE measurable, and so is νE � ∂ ∗ E. In addition, ∂ ∗ (Ω − E) = ∂ ∗ E
(6.3.5)
by the identities (6.3.2). In the line of motivation, we describe the geometric meaning of νE , established in Corollary 6.4.4 below. For each x ∈ ∂ ∗ E, denote by � � H± (E, x) := y ∈ Rn : ±νE (x) · (y − x) ≥ 0 (6.3.6) the closed half-spaces determined by the hyperplane � � H(E, x) := y ∈ Rn : νE (x) · (y − x) = 0
✐
(6.3.7)
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6. Locally BV sets
perpendicular to νE (x) and passing through x. We show that νE (x) is the unit exterior normal of E at the point x ∈ ∂ ∗ E by proving that H(x) is tangent to E in the following measure-theoretic sense: � � � � Θ H+ (E, x) ∩ E, x = Θ H− (E, x) − E, x = 0.
Recall that the density Θ(A, x) was defined for any set A ⊂ Rn in Section 4.4. Observation 6.3.4. If A � Ω is a measurable set, then V(A, Ω) = V(A).
Proof. Find an open set U so that A � U � Ω, and find ϕ ∈ C 1 (Rn ) so that χA ≤ ϕ ≤ χU . If v ∈ Cc1 (Rn ; Rn ) and �v�L∞ (Rn ;Rn ) ≤ 1, then � � div v(x) dx = div (ϕv)(x) dx ≤ V(A, Ω). A
A
The arbitrariness of v implies V(A) ≤ V(A, Ω), and the reverse inequality is obvious. Lemma 6.3.5. If x ∈ Ω, then � � � � � V E ∩ B(x, r) ≤ µE [B(x, r) + Hn−1 ∂B(x, r)
for all r > 0 with B(x, r) ⊂ Ω, and � � � � � V E ∩ B(x, r) ≤ µE [B(x, r) + Hn−1 E ∩ ∂B(x, r) for L1 almost all r > 0 with B(x, r) ⊂ Ω. If x ∈ ∂ ∗ E, then � � � µE [B(x, r) ≤ 2Hn−1 E ∩ ∂B(x, r)
(1)
(2)
(3)
for L1 almost all sufficiently small r > 0.
Proof. If ϕ ∈ Lipc (Ω) and v ∈ Cc1 (Rn ; Rn ), then by (6.3.1), � � (ϕv) · νE dµE = div (ϕv) dLn Ω E � � n = ϕ div v dL + v · Dϕd Ln . E
(∗)
E
In view of translation invariance, we may assume that x = 0. Given a ball Br := B(0, r) contained in Ω, choose ε > 0 so that Br+ε ⊂ Ω. Define if |y| ≤ r, 1 � � ε 1 ϕ (y) := ε r + ε − |y| if r < |y| < r + ε, 0 if |y| ≥ r + ε,
and observe that ϕε ∈ Lipc (Ω), and that for each y ∈ Ω, � y if r < |y| < r + ε, − 1ε |y| ε Dϕ (y) = 0 if |y| < r or |y| > r + ε.
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6.3. The reduced boundary
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129
Consider (∗) with ϕ = ϕε and �v�L∞ (Rn ;Rn ) ≤ 1. Employing the spherical coordinates [29, Section 3.4.4, A], calculate � (ϕε v) · νE dµE Ω � � 1 y ε n dy = ϕ div v dL − v(y) · ε |y| E E∩(Br+ε −Br ) �� � � � 1 r+ε ϕε div v dLn − v · νBs dHn−1 ds = (∗∗) ε r E E∩∂Bs �� � � � 1 r+ε ε n n−1 ≥ ϕ div v dL − dH ds ε r E ∂Bs � 1 = ϕε div v dLn − |Br+ε − Br |. ε E Note that �ϕε �L∞ (Rn ) ≤ 1, limε→0 ϕε (y) = χBr (y) for all y ∈ Rn , and d 1 lim |Br+ε − Br | = |Br | = Hn−1 (∂Br ). ε dr
ε→0
Thus letting ε → 0 in inequality (∗∗), we obtain � � v · νE dµE ≥ div v dLn − Hn−1 (∂Br ), �
Br
Br
v · νE dµE =
�
E∩Br
E∩Br
div v dLn −
�
E∩∂Br
v · νBr dHn−1
(†)
where the inequality holds for all r > 0 with Br ⊂ Ω, and the equality (†) holds for L1 almost all r > 0 with Br ⊂ Ω. The qualifier “almost all” appears because we applied Theorem 4.3.4 to the second term in the line opposite to the marker (∗∗). Hence � div v dLn ≤ µE (Br ) + Hn−1 (∂Br ), E∩Br � div v dLn ≤ µE (Br ) + Hn−1 (E ∩ ∂Br ), E∩Br
and inequalities (1) and (2) follow from the arbitrariness of v. Now assume x = 0 lies in ∂ ∗ E. According to identity (6.3.3), � � �2 1 νE dµE = �νE (0)� = 1. νE (0) · lim r→0 µE (Br ) B r Select δ > 0 so that 0 < r < δ implies Br ⊂ Ω and � 1 νE (0) · νE dµE ≥ µE (Br ). 2 Br
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130
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6. Locally BV sets
There is w ∈ Cc1 (Ω; Rn ) such that �w�L∞ (Ω;Rn ) ≤ 1 and w(y) = νE (0) for each y ∈ Bδ . Equality (†) with v = w implies � � 1 µE (Br ) ≤ νE (0) · νE dµE = w · νE 2 Br Br � =− w · νBr dHn−1 ≤ Hn−1 (E ∩ ∂Br ) E∩∂Br
1
for L almost all positive r < δ. This establishes inequality (3). Lemma 6.3.6. There are positive constants βi = βi (n), i = 1, 2, 3, such that given x ∈ ∂ ∗ E, the inequalities � � �E ∩ B(x, r)� � � β1 ≤ � ≤ 1 − β1 (1) B(x, r)� � � µE B(x, r) ≤ β3 (2) β2 ≤ rn−1 � � V E ∩ B(x, r) ≤ β3 (3) rn−1 hold for all sufficiently small r > 0. Proof. Choose x ∈ ∂ ∗ E, and let Br := B(x, r). By Lemma 6.3.5, (3), there is δ > 0 such that Bδ ⊂ Ω and µE (Br ) ≤ 2Hn−1 (E ∩ ∂Br ) ≤ 2Hn−1 (∂Br ) = 2nα(n)rn−1 for L1 almost all positive r < δ. As r �→ µE (Br ) is a right-continuous function defined in (0, δ), the inequality µE (Br ) ≤ 2nα(n)rn−1 holds for all sufficiently small r > 0. In view of Lemma 6.3.5, (1), so does � � max µE (Br ), V(E ∩ Br ) ≤ 3nα(n)rn−1 .
With β3 = 3nα(n), this establishes inequality (3) and the second inequality in (2). Inequalities (2) and (3) of Lemma 6.3.5 yield V(E ∩ Br ) ≤ 3Hn−1 (E ∩ ∂Br )
(∗)
for L1 almost all sufficiently small r > 0. Since � r Hn−1 (E ∩ ∂Bs ) ds, |E ∩ Br | = 0
inequality (∗) and the isoperimetric inequality (Theorem 5.9.6) imply � 1 1 d� |E ∩ Br | n = |E ∩ Br | n −1 Hn−1 (E ∩ ∂Br ) dr n−1 1 1 ≥ |E ∩ Br |− n V(E ∩ Br ) ≥ 3 3γ
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6.4. Blow-up
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131
for L1 almost all sufficiently small r > 0; here γ = γ(n) > 0 is the constant appearing in the isoperimetric inequality. By integration, 1 1 |E ∩ Br | ≥ |Br | rn = (∗∗) (3γ)n (3γ)n α(n) � �−1 and applying for all sufficiently small r > 0. Letting β1 = (3γ)n α(n) inequality (∗∗) to Ω − E yield � � (†) β1 |Br | ≤ �(Ω − E) ∩ Br � = |Br | − |E ∩ Br | for all sufficiently small r > 0. Inequality (1) follows from (∗∗) and (†). Combining (∗∗) and (†) with the relative isoperimetric inequality of Corollary 5.9.13, we obtain that for all sufficiently small r > 0, � � � � n−1 � � n µE U (x, r) 1 1 µE (Br ) ≥ ≥ min |B ∩ E|, |B − E| r r rn−1 rn−1 κ rn � � �� n−1 n |Br ∩ E| |Br ∩ E| 1 ,1 − = min α(n) κ |Br | |Br | � n−1 1� 1 α(n)β1 n = (3γ)1−n = β2 ; ≥ κ κ here κ = κ(n) > 0 is the constant from Corollary 5.9.13. This establishes the first inequality in (2).
Proposition 6.3.7. ∂ ∗ E ⊂ ∂∗ E ∩ Ω and Hn−1 (∂∗ E ∩ Ω − ∂ ∗ E) = 0. Proof. The inclusion ∂ ∗ E ⊂ ∂∗ E ∩ Ω follows from Lemma 6.3.6, (1). Since µE = µE ∂ ∗ E by (6.3.4), we have µE (∂∗ E ∩ Ω − ∂ ∗ E) = 0. An application of Proposition 5.9.16 completes the argument.
6.4. Blow-up In this section, we again fix a set E ∈ BVloc (Ω). For z ∈ Ω and r > 0, we define a blow-up of E about z by r as the set Ez,r := z + r−1 (E − z). We say blow-up because we are interested only in small r > 0. Clearly Ez,r ⊂ Ωz,r , and it follows by translation � � and scaling that Ez,r belongs to BVloc (Ωz,r ). Observe B(z, s) = B(z, rs) z,r for each s > 0. Lemma 6.4.1. Let z ∈ ∂ ∗ E. Then equalities � � � � �B(z, s) ∩ Ez,r � = 1 �B(z, rs) ∩ E �, n r � � � � 1 µEz,r B(z, s) = n−1 µE B(z, rs) , r
✐
(1) (2)
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132
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6. Locally BV sets �
νEz,r dµEz,r = B(z,s)
1 rn−1
�
νE dµE
(3)
B(z,rs)
hold for each pair r, s ∈ R+ for which B(z, rs) ⊂ Ω. Proof. We may assume z = 0. To simplify the notation, let Ur = U (0, r),
Br = B(0, r),
Ωr = Ω0,r ,
Er = E0,r .
Observe that the linear map φ : x �→ r−1 x maps Urs onto Ur , Brs onto Br , Ω ontoΩ r , and E onto Er . Equality (1) is a direct consequence of the area theorem, and holds for all r, s ∈ R+ . For the proof of remaining equalities, fix r, s ∈ R+ so that Brs ⊂ Ω and consequently Bs ⊂ Ωr . Choose v ∈ Cc1 (Us ; Rn ) with �v�L∞ (Us ;Rn ) ≤ 1. Since χE = χEr ◦ φ, � � 1 χEr div v = n (χEr div v) ◦ φ r Urs Us � (6.4.1) 1 = n−1 χE div (v ◦ φ) ≤ µE (Urs ). r Urs Consequently µEr (Us ) ≤ µE (Urs ), and µEr (Us ) = µE (Urs ) by symmetry. �∞ �∞ Since Bs = k=1 Us(k+1)/k and Brs = k=1 Urs(k+1)/k , and since both µEr and µE are Radon measures inΩ r and Ω, respectively, identity (2) follows. Denote by ν1 �and νr,1 �the first component of νE and νEr , respectively. Find {ϕk } in Lipc Ωr�; [0, 1] that � converges pointwise to χBs . Then {ϕk ◦ φ} is a sequence in Lipc Ω; [0, 1] that converges pointwise to χBrs . Since the vector field wk := ϕk e1 belongs to Lipc (Ωr , Rn ), the dominated convergence theorem and the equality in (6.3.1) imply � � νr,1 dµEr = lim ϕk νr,1 dµEr Bs Ωr � � = lim wk · νEr dµEr = lim div wk Ωr Er � � 1 1 = n lim (div wk ) ◦ φ = n−1 lim div (wk ◦ φ) r r E E � 1 = n−1 lim (wk ◦ φ) · νE dµE r � �Ω 1 1 = n−1 lim (ϕk ◦ φ)ν1 dµE = n−1 ν1 dµE . r r Ω Brs Identity (3) follows by symmetry. Lemma 6.4.2. The half-space � � H− = (ξ1 , . . . , ξn ) ∈ Rn : ξn ≤ 0
belongs to BVloc (Rn ) and µH− = Hn−1
✐
∂H− .
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6.4. Blow-up
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133
Proof. The first assertion follows from Propositions 4.5.8 and 5.1.3. Select an open set U � Rn , v ∈ Lipc (U ; Rn ) with �v�L∞ (U ;Rn ) ≤ 1, and a dyadic figure B ⊂ U such that spt v ⊂ int B. Since B ∩ H− is still a dyadic figure, Theorem 2.3.7 implies � � χH− div v(x) dx = div v(x) dx U
B∩H−
= =
�
�
∂(B∩H− )
U ∩∂H−
v · νB∩H− dHn−1
v · en dHn−1 ≤ Hn−1 (U ∩ ∂H− ).
As v is arbitrary, µH− (U ) ≤ Hn−1 (∂H− ∩ U ). For the reverse inequality, choose ε > 0 and find w ∈ Lipc (U ; Rn ) so that �w�L∞ (U ;Rn ) ≤ 1 and �� �� < ε; Hn−1 x ∈ ∂H− ∩ U : w(x) �= en
cf. Example 5.1.4. If B ⊂ U is a dyadic figure containing spt w, then calculating as above but backward, we obtain H
n−1
(U ∩ ∂H− ) ≤ =
�
�
U ∩∂H− U
w · en dHn−1 + 2ε
χH− div w(x) dx + 2ε ≤ µH− (U ) + 2ε.
The lemma follows from the arbitrariness of ε and Theorem 1.3.1. Theorem 6.4.3. Let z ∈ ∂ ∗ E. Then lim χEz,r = χH− (E,z)
r→0
in L1loc (Rn ), and for each B(z, s) ⊂ Ωz,r , � � lim µEz,r B(z, s) = α(n − 1)sn−1 . r→0
Proof. In view of the translation and rotation invariance, we may assume that z = 0 and νE (0) = en . We simplify the notation by letting Ur = U (0, r),
Br = B(0, r),
H− = H− (E, 0).
Recall from (6.3.6) that H− is the same as that defined in Lemma 6.4.2. Select R > 0, and let φr (x) := r−1 x for x ∈ Rn and r > 0. Given v ∈ Cc1 (UR ; Rn ) with �v�L∞ (UR ;Rn ) ≤ 1, the equality in (6.4.1) and Lemma 6.3.6, (3) yield � � 1 χE0,r div v = n−1 χE div (v ◦ φr ) r UR BrR ≤
✐
V(E ∩ BrR ) ≤ β3 Rn−1 rn−1
© 2012 by Taylor & Francis Group, LLC
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134
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6. Locally BV sets
for all sufficiently small r > 0; here β3 = β3 (n) > 0. As v is arbitrary, lim sup �χE0,r �BV (UR ) ≤ α(n)Rn + β3 Rn−1 < ∞.
(∗)
r→0
Choose a sequence {rk } in R+ converging to zero. Since R is arbitrary, Proposition 5.8.9 shows that {rk } has a subsequence, still denoted by {rk }, such that {χE0,rk } converges in L1loc (Rn ), as well as almost everywhere, to a function f ∈ BVloc (Rn ). The convergence almost everywhere shows that f is equivalent to χA for a set A in BVloc (Rn ). As we have started with an arbitrary sequence {rk }, the first assertion of the theorem will be established by showing that A is equivalent to H− . We split the proof into three claims. To simplify the notation further, we letΩ k = Ω0,rk and Ek = E0,rk for k = 1, 2, . . . . Claim 1. lim µEk (Bs ) = µA (Bs ) for all but countably many s > 0, and νA (x) = en for µA almost all x ∈ Rn .
Proof. Choose R > 0. Eliminating finitely many members of {rk }, we may assume that UR � Ωk for k = 1, 2, . . . . As each Ek belongs to BVloc (Ωk ), our assumption implies that each χEk belongs to BV (UR ). Now we have lim �χEk − χA �L1 (UR ) = 0, and lim sup µEk (UR ) ≤ lim sup �χEk �BV (UR ) < ∞ according to inequality (∗). It follows from Proposition 5.5.4 that w-lim(µEk
ν Ek ) = µ A
νA .
Theorem 5.4.8 shows that {Ek } has a subsequence, still denoted by {Ek }, such that {µEk } converges weakly in UR to a Radon measure µ. Select 0 < s < R so that µ(∂Bs ) = 0. Then lim µEk (Bs ) = µ(Bs )
(∗∗)
by Theorem 5.4.3. Proposition 5.4.9 yields µA ≤ µ and lim(µEk
νEk )(Bs ) = (µA
νA )(Bs ).
Writing out the n-th component of the previous equality gives � � lim en · νEk dµEk = en · νA dµA . Bs
(†)
Bs
As each Bsrk is contained in Ω, we can apply Lemma 6.4.1: dividing the n-th component of (3) by (2), we obtain � � 1 1 en · νEk dµEk = en · νE dµE . µEk (Bs ) Bs µE (Bsrk ) Bsr k
Since 0 ∈ ∂ ∗ E and νE (0) = en , Definition 6.3.2 implies � � 1 1 en · νEk dµEk = en · lim νE dµE lim µEk (Bs ) Bs µE (Bsrk ) Bsr k
= en · νE (0) = 1.
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6.4. Blow-up
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135
According to (∗∗), the previous equality, and (†), � en · νEk dµEk µA (Bs ) ≤ µ(Bs ) = lim µEk (Bs ) = lim Bs � = en · νA dµA ≤ µA (Bs ). Bs
Summarizing, we have established that lim µEk (Bs ) = µA (Bs )
and
�
Bs
en · νA dµA = µA (Bs )
(††)
for each positive s < R with µ(Bs ) = 0. However, µ(∂Bs ) = 0 for all but countably many positive s < R; see Lemma 4.5.5. Thus (††) holds for all but countably many positive s < R. Since the second equality in (††) implies νA (x) = en for µA almost all x ∈ Bs , we see that νA (x) = en for almost all x ∈ UR . Claim 1 follows from the arbitrariness of R. Claim 2. µA (Us ) ≤ lim µEk (Bs ) ≤ µA (Bs ) for each s > 0.
Proof. Choose s > 0 and ε > 0. Denote by D the set of all s ∈ R+ such that µA (Bs ) = lim µEk (Bs ). Since R+ − D is a countable set by Claim 1, there are sequences {tj } and {uj } in D such that {tj } is strictly increasing, {uj } is decreasing, and lim tj = lim uj = s. As � � Us = {Btj : j ∈ N} and Bs = {Buj : j ∈ N}, there is p ∈ N satisfying
µA (Us ) < µA (Btp ) + ε = lim µEk (Btp ) + ε ≤ lim µEk (Bs ) + ε, k→∞
k→∞
µA (Bs ) > µA (Bup ) − ε = lim µEk (Bup ) − ε ≥ lim µEk (Bs ) − ε. k→∞
k→∞
The claim follows from the arbitrariness of ε. Claim 3. There is c ∈ R such that A is equivalent to the set � � (ξ1 , . . . , ξn ) ∈ Rn : ξn < c .
Proof. Choose a mollifier η and let fk := η1/k ∗ χA for k = 1, 2, . . . . Then {fk } is a sequence in C 1 (Rn ) which converges in L1loc (Rn ) to χA . The standard diagonal construction (see the proof of Proposition 5.8.9) produces a subsequence of {fk }, still denoted by {fk }, which converges to χA almost everywhere in Rn . If ϕ ∈ Cc1 (Rn ), then � � � ϕDi fk = (ϕei ) · Dfk = − fk div (ϕei ) Rn Rn Rn � � � � � � =− χA div (η1/k ∗ ϕ)ei = (η1/k ∗ ϕ)ei · d(DχA ) n Rn �R =− (η1/k ∗ ϕ)ei · νA dµA Rn
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© 2012 by Taylor & Francis Group, LLC
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136
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6. Locally BV sets
and consequently, � � 0 ϕDi fk = � Rn − Rn η1/k ∗ ϕ dµA
if i = 1, . . . , n − 1, if i = n.
As Di fk is continuous and ϕ is arbitrary, we infer Dn fk ≤ 0 and Di fk = 0 for i = 0, . . . , n − 1. Consequently fk is constant with respect to the first n − 1 variables, and decreasing with respect to the n-th variable. By Fubini’s theorem, there is an Ln−1 negligible set M ⊂ Rn−1 so that for each point x ∈ Rn−1 − M , the function t �→ χA (x, t) is decreasing L1 almost everywhere in R. Thus for every x ∈ Rn−1 − M , there is a tx ∈ R such that χA (x, t) = 1 for L1 almost all t < tx and χA (x, t) = 0 for L1 almost all t > tx . Using Fubini’s theorem again, we find an L1 negligible set N ⊂ R so that for each t ∈ R − N , the function x �→ χA (x, t) is constant Ln−1 almost everywhere in Rn−1 . This implies that the function x �→ tx is constant Ln−1 almost everywhere in Rn−1 − M , and Claim 2 follows. If c �= 0, select a positive s < |c| and observe that either |Bs ∩ A| = 0 or |Bs ∩ A| = |Bs |. Observe that lim χEk = χA in L1loc implies lim |Bs ∩ Ek | = |Bs ∩ A|. Using Lemma 6.4.1, (1), we obtain lim
|Brk ∩ E| |Brk s ∩ E| 1 |Brk s ∩ E| = lim = lim |Brk | |Brk s | |Bs | rn 1 |Bs ∩ A| lim |Bs ∩ Ek | = = 0 or 1. = |Bs | |Bs |
As 0 ∈ ∂ ∗ E, a contradiction follows from Lemma 6.3.6, (1). Thus c = 0 and A is equivalent to H− , which is the first assertion of the theorem. An immediate consequence of Lemma 6.4.2 is that µH− (Bs ) = µH− (Us ) = α(n − 1)sn−1 for every s > 0. An application of Claim 2 completes the proof. Corollary 6.4.4. If z ∈ ∂ ∗ E, then � � � � Θ H+ (E, z) ∩ E, z = Θ H− (E, z) − E, z = 0, Θ(E, z) = 21 , � � µE B(z, r) = 1. lim r→0 α(n − 1)r n−1
(1) (2) (3)
Proof. If B± (z, r) := B(z, r) ∩ H± (E, z), then Lemma 6.4.1, (1) with s = 1 and Theorem 6.4.3 imply � � �B+ (z, r) ∩ E � � � Θ H+ (E, z) ∩ E, z = lim r→0 α(n)rn
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6.5. Perimeter and variation
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137
� � 1 lim �B+ (z, 1) ∩ Ez,r � α(n) r→0 � 1 �� B+ (z, 1) ∩ H− (E, z)� = 0, = α(n) � � and similarlyΘ H− (z, E) − E, z = 0. Now (2) follows from (1) and the obvious equality: � � � � � � � � �B(z, r) ∩ E � = �B− (z, r)� − �B− (z, r) − E �+�B+ (z, r) ∩ E �. =
Finally, by Lemma 6.4.1, (2) and Theorem 6.4.3, � � � � 1 lim n−1 µE B(z, r) = lim µEx,r B(z, 1) = α(n − 1). r→0 r r→0
6.5. Perimeter and variation As in the previous section, a set E ∈ BVloc (Ω) is fixed. We employ the notation introduced in Section 6.3. In particular, we use the polar decomposition DχE = µE νE , and the symbols defined in (6.3.6) and (6.3.7). Lemma 6.5.1. Let n ≥ 2. For k = 1, 2, . . . , the functions � � �B(z, r) ∩ H+ (E, z) ∩ E � � � : ∂ ∗ E → R, fk,+ : z �→ sup �B(z, r)� 0 0, � � � 1 �� B(z, r)� = �B(z, r) ∩ H− (E, z)� 2 � � � � = �B(z, r) ∩ H− (E, z) ∩ E � + �B(z, r) ∩ H+ (E, z) − E �.
Thus if 0 < r < 1/k, inequalities (1) and (2) imply � � n � � � � �B(z, r) ∩ H− (E, z) ∩ E � > 1 − ε �B(z, r)�, n+2 2 2 � � n � � � � �B(z, r)�. �B(z, r) ∩ H+ (E, z) − E � > 1 − ε 2 2n+2
(3) (4)
Claim 1. For each pair x, y ∈ K with |y − x| < 1/(2k), � � �νE (x) · (y − x)� ≤ ε|y − x|.
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© 2012 by Taylor & Francis Group, LLC
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6.5. Perimeter and variation
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139
Proof. Seeking a contradiction, assume there are x, y ∈ K such that � � 0 < |y − x| < 1/(2k) and �νE (x) · (y − x)� > ε|y − x|. � � Suppose first νE (x) · (y − x) > ε|y − x|. If u ∈ B y,ε |x − y| , then νE (x) · (u − x) = νE (x) · (u − y) + νE (x) · (y − x) > −|u − y| + ε|y − x| ≥ 0
and consequently u ∈ H+ (E, x). As ε < 1, � � � � B y,ε |x − y| ⊂ B x, 2|x − y| ∩ H+ (E, x)
and intersecting this inclusion with E, we obtain � � � � B y,ε |x − y| ∩ E ⊂ B x, 2|x − y| ∩ H+ (E, x) ∩ E.
(∗)
For z = x and r = 2|x − y|, inequality (1) yields n � � � � � � �B x, 2|x − y| ∩ H+ (E, x) ∩ E � < ε α(n) 2|x − y| n n+2 2 � �n = 14 α(n) ε|x − y| ,
and for z = y and r = ε|x − y| inequality (3) yields � � � � � � � � �B y,ε |x − y| ∩E � ≥ �B y,ε |x − y| ∩H− (E, y) ∩ E � � � � �n 1 εn − > α(n) ε|x − y| 2 2n+2 � �n > 14 α(n) ε|x − y| ,
since ε < 1. The previous two inequalities contradict inclusion (∗). If νE (x) · (y − x) < −ε|y − x|, then � � � � B y,ε |x − y| ⊂ B x, 2|x − y| ∩ H− (E, x) by the same calculation as above. Subtracting E gives � � � � B y,ε |x − y| − E ⊂ B x, 2|x − y| ∩ H− (E, x) − E,
and a contradiction follows by letting z = x and r = 2|x − y| in inequality (2), and z = y and r = ε|x − y| in inequality (4). Claim 1 is thus established.
In view of Claim 1, Whitney’s extension theorem (Theorem 1.5.5) applies to the function f ≡ 0 on K and v = νE � K. There is g ∈ C 1 (Rn ) such that g(x) = 0 and Dg(x) = νE (x) for each x ∈ K. By the implicit function theorem [62, Theorem 9.28], the set � � S := x ∈ Rn : g(x) = 0 and Dg(x) > 1/2 is an (n − 1)-dimensional C 1 submanifold of Rn without boundary (in the sense of differential topology). Note that K ⊂ S, and that Dg(x) is the normal vector of S at each x ∈ S. Claim 2. µE (B) = Hn−1 (B) for each Borel set B ⊂ K.
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Proof. Fix x ∈ K. Via translation and rotation, we may assume that x = 0 and Dg(x) = en . Let Π =Π � n and π = �πn be as in Section 1.1. There 1 are R > 0 and a function ϕ ∈ C U (0, R) ∩ Π such that the map � � � � φ : u �→ u,ϕ (u) : U (0, R) ∩ Π → S ∩ π −1 U (0, R) ∩ Π is a bijective C 1 diffeomorphism. Letting Br := B(0, r), the inequality � � � � �� Hn−1 π(S ∩ Br ) ≤ Hn−1 (S ∩ Br ) ≤ Hn−1 φ π(Br )
(∗)
holds for 0 < r < R. By [29, Section 3.3.4, B], � � � � � �2 �� n−1 H φ π(Br ) = 1 + �Dϕ(u)� du. �
�
Br ∩Π
Since Dg(0) = en and g u,ϕ (u) = 0 for each u in U (0, R) ∩ Π, we calculate Dϕ(0) = 0. The continuity of Dϕ implies � � �� Hn−1 φ π(Br ) lim = 1. (∗∗) r→0 α(n − 1)rn−1 � � In addition, �ϕ(y � )� ≤ ε(r)|y � | where limr→0 ε(r) = 0. It follows that for � s(r) = r/ 1 + ε(r)2 , we have Bs(r) ∩ Π ⊂ π(S ∩ Br ) for 0 < r < R. Thus � � � � Hn−1 π(Bs(r) ) Hn−1 π(S ∩ Br ) lim ≥ lim = 1. (†) r→0 r→0 α(n − 1)r n−1 α(n − 1)rn−1 Summarizing our results, � � (Hn−1 S) B(x, r) =1 lim r→0 α(n − 1)rn−1
and
� � µE B(x, r) lim = 1. r→0 α(n − 1)r n−1
Indeed, the first equality follows from (∗), (∗∗), and (†), and the second is implied by Corollary 6.4.4, (3). Consequently � � (Hn−1 S) B(x, r) � � lim = 1. r→0 µE B(x, r) As x ∈ K is arbitrary, Claim 2 follows from Proposition 6.2.6.
By condition (i), each Borel set B ⊂ ∂∗ E ∩ Ω is the union of disjoint Borel sets Bi ⊂ Ki , i = 1, 2, . . . , and the set � � B ∩ N ∪ (∂∗ E ∩ Ω − ∂ ∗ E) ,
which is both µE and Hn−1 negligible by (6.3.4) and Propositions 5.9.16 and 6.3.7. In view of identity (6.3.4) and Proposition 6.3.7, the theorem follows from Claim 2.
Remark 6.5.3. In proving Theorem 6.5.2, we have shown that up to an Hn−1 negligible set, ∂∗ E ∩ Ω is the union of countably many pieces of (n − 1)dimensional C 1 manifolds (the compact sets K1 , K2 , . . . ). Sets of this type
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6.5. Perimeter and variation
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141
are called rectifiable. For the precise definition and terminology we refer to [33, Section 3.2.14]. An instant benefit afforded by Theorems 6.5.2 is that in ∂∗ E ∩ Ω we can switch freely between the measures µE and Hn−1 . We will do it at will throughout the remainder of this book, often tacitly without any references. If v ∈ L∞ (Ω, Hn−1 ; Rn ) has compact support, we call a function � F :B→ v · νB dHn−1 : BVloc (Ω) → R (6.5.1) ∂∗ B∩Ω
the flux of v. This extends the concept of flux defined in Section 2.1 for figures A ⊂ Ω. The next divergence theorem is an immediate consequence of Theorems 5.5.1 and 6.5.2: it follows at once from identity (6.3.1) by substituting Hn−1 (∂∗ E ∩ Ω) for the measure µE . Theorem 6.5.4. If E ∈ BVloc (Ω) and v ∈ Lipc (Ω; Rn ), then � � div v(x) dx = v · νE dHn−1 . E
∂∗ E∩Ω
Theorem 6.5.4 is due to H. Federer [31, 32]. For Lipschitz vector fields, it achieves the maximal generality with respect to the integration domains. More general integration domains are available for continuously differentiable vector fields [39, 38], but we will not pursue this direction. Theorem 6.5.5. BV(Ω) = P(Ω) and BVloc (Ω) = Ploc (Ω). Moreover, V(B, Ω) = P(B, Ω) for each B ⊂ Rn such that B ∩ Ω is measurable. Proof. The inclusion BV(Ω) ⊂ P(Ω) is a direct consequence of Theorem 6.5.5. Given B ∈ P(Ω), choose a vector field v ∈ Cc1 (Ω; Rn ) with �v�L∞ (Ω;Rn ) ≤ 1, and find a figure A ⊂ Ω so that spt v ⊂ int A. As ∂∗ (B ∩ A) ⊂ (∂∗ B ∩ Ω) ∪ ∂A, we see that B ∩ A belongs to P(Rn ) and by Proposition 5.1.3, also to BV(Rn ). By Theorem 6.5.4, �
div v(x) dx = B
≤ =
�
� �
div v(x) dx =
B∩A
∂∗ (B∩A)
∂∗ (B∩A)
∂∗ B∩Ω
�
|v| dH
n−1
≤
�
v · νB∩A dHn−1
(∂∗ B∩Ω)∪∂A
|v| dHn−1
|v| dHn−1 ≤ Hn−1 (∂∗ B ∩ Ω).
Since v is arbitrary, V(B, Ω) ≤ P(B, Ω) < ∞ and hence B ∈ BV(Ω). Thus BV(Ω) = P(Ω) and consequently BVloc (Ω) = Ploc (Ω). Now select B ⊂ Rn
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so that B ∩ Ω is measurable, and observe that V(B ∩ Ω, Ω) = V(B, Ω)
and P(B ∩ Ω, Ω) = P(B, Ω)
according to (5.1.1) and (4.5.1). Thus if V(B, Ω) < ∞ or P(B, Ω) < ∞, then B ∩ Ω ∈ BV(Ω). In this case, the equality V(B, Ω) = P(B, Ω) follows from Theorem 6.5.2. Proposition 6.5.6. Let E ∈ BV(Ω) and ε > 0. There is an open set U ⊂ Ω such that the following conditions are satisfied: (1) (2) (3) (4)
∂∗ E ∩ U = ∂∗ E ∩ Ω; � � P(U ) ≤ β P(E, Ω) + ε where β = β(n) > 0; the sets U and E ∩ U belong to BV(Rn ); the measures �DχE � and �DχE∩U � coincide on subsets of ∂∗ E ∩ Ω.
Proof. For each x ∈ ∂∗ E ∩ Ω, select 0 < δx ≤ 1 with B(x,δ x ) ⊂ Ω. By Lemma 2.2.1, there is a family C of dyadic cubes such that (a) C is a star cover of ∂∗ E ∩ Ω; (b) for each C ∈ C there is xC ∈ C ∩ ∂∗ E ∩ Ω with d(C) < δxC ; � � � n−1 (c) ≤ κ Hn−1 (∂∗ E ∩ Ω) + ε where κ = κ(n) > 0. C∈C d(C) �� � Let U := int C . Conditions (a) and (b) imply, respectively, the inclusions ∂∗ E ∩ Ω ⊂ U and U ⊂ Ω. Thus ∂∗ E ∩ Ω = ∂∗ E ∩ U . Since � � {int C : C ∈ C} ⊂ U ⊂ {C : C ∈ C}, � the sets U and C are equivalent. By Theorem 6.5.5 and Propositions 5.1.7, �� � � � � P(U ) = V C ≤ V(C) = P(C) = 2n d(C)n−1 C∈C
C∈C
C∈C
and (c) shows that (2) is satisfied with β := 2nκ. Thus U ∈ BV(Rn ), since (c) implies |U | < ∞. From the inclusion ∂∗ (E ∩ U ) ⊂ (∂∗ E ∩ Ω) ∪ ∂∗ U we obtain E ∩ U ∈ BV(Rn ). As ∂∗ E ∩ Ω ⊂ ∂∗ (E ∩ U ), condition (4) follows. Proposition 6.5.6 allows us to derive many properties of sets in BV(Ω) from those in BV(Rn ). Accordingly, in the remainder of this book we concentrate on the family BV(Rn ).
6.6. Properties of BV sets The additivity of flux, which is trivial for figures (Observation 2.1.1), holds also for BV sets, albeit nontrivially.
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143
Theorem 6.6.1. If A, B ∈ BVloc (Rn ) and |A ∩ B| = 0, then � � � v · νA∪B dHn−1 = v · νA dHn−1 + v · νB dHn−1 ∂∗ (A∪B)
∂∗ A
� 1
for each v ∈ L ∂∗ A ∪ ∂∗ B, Hn−1 ; R
∂∗ B
� n
.
Proof. From |A ∩ B| = 0, we obtain χA∪B = χA + χB almost everywhere, and hence DχA∪B = DχA + DχB . As Theorem 6.5.2 implies � � DχA∪B = Hn−1 ∂∗ (A ∪ B) νA∪B , DχA = (Hn−1
∂∗ A)
νA , D Bχ = (Hn−1
∂∗ B)
νB ,
the theorem follows by integrating over Rn the zero extension v of v. Remark 6.6.2. Let A, B ∈ BVloc (Rn ) and |A ∩ B| = 0. Since Θ(A ∪ B, x) = Θ (A, x) + Θ (B, x) for each x ∈ R , using Corollary 6.4.4, (2), it is easy to show that the sets ∂∗ (A ∪ B) and ∂∗ A � ∂∗ B are Hn−1 equivalent. By the uniqueness of polar decomposition, the equalities n
νA∪B � (∂∗ A − ∂∗ B) = νA � (∂∗ A − ∂∗ B),
νA∪B � (∂∗ B − ∂∗ A) = νB � (∂∗ B − ∂∗ A),
νA � (∂∗ A ∩ ∂∗ B) = −νB � (∂∗ A ∩ ∂∗ B)
hold Hn−1 almost everywhere. From this we see that the geometric reasons for the additivity of flux for BV sets are the same as those for figures. The next proposition provides a useful and intuitive identity. Proposition 6.6.3. If A, B ∈ BV(Rn ) and B ⊂ A, then P(A − B) = P(A) + P(B) − 2Hn−1 (∂∗ A ∩ ∂∗ B). Proof. If E ∈ BVloc (Rn ), then Proposition 6.3.7 and Corollary 6.4.4, (2) imply that for Hn−1 almost all x ∈ Rn , the density Θ (E , x) attains only the values 0, 1/2, and 1. Since Θ(A − B, x) = Θ (A, x) − Θ(B, x) for each x ∈ Rn , a direct calculation shows that the sets ∂∗ (A − B)
and
(∂∗ A − cl∗ B) ∪ (int∗ A ∩ ∂∗ B)
are Hn−1 equivalent. As ∂∗ A ∩ int∗ B = ∅ and ∂∗ B ⊂ cl∗ A, we obtain (∂∗ A − cl∗ B) ∪ (int∗ A ∩ ∂∗ B) = (∂∗ A − ∂∗ B) ∪ (∂∗ B − ∂∗ A) and the proposition follows.
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Proposition 6.6.4. Let E ∈ BV(Rn ). If C ⊂ Rn is a convex set, then P(E ∩ C) ≤ P(E). Proof. With no loss of generality, we may assume that C is open. Enumerate C ∩ Qn as {x1 , x2 , . . . }, and let p ≥ n + 1 be the least integer such that the convex hull of {x1 , . . . , xp } is a polytop (see the paragraph following Lemma 4.6.9). Denote by Ck the convex hull of {x1 , . . . , xp+k }, and observe that {Ck } is an increasing sequence of polytops whose union contains C ∩ Qn . As C is an open set, each x ∈ C has a neighborhood U (x, r) ⊂ C. Since U (x, r) ∩ Qn contains a set {xi0 , . . . , xij } whose convex hull contains x, it is �∞ clear that C = k=1 Ck . Thus P(E ∩ C) ≤ lim inf P(E ∩ Ck ) ≤ P(E)
by Propositions 4.6.10 and 5.1.7 and Theorem 6.5.2. A different proof of Proposition 6.6.4 is given in [51, Corollary 1.9.4]. It is based on calculating perimeters via the (n − 1)-dimensional integral-geometric measure; see [46, Section 5.14] or [47, Section 2.4]. Proposition 6.6.5. If E ∈ BVloc (Rn ), then n � � � � P(E) ≤ H0 ∂∗Lu (Lu ∩ E) dLn−1 (u), i=1
P(E) ≥
�
Πi
Πi
� � H0 ∂∗Lu (Lu ∩ E) dLn−1 (u),
i = 1, . . . , n.
Proof. Since Theorem 4.6.6 and Proposition 4.6.7 imply � n � � � � div v(x) dx ≤ H0 ∂∗Lu (Lu ∩ E) dLn−1 (u) E
i=1
Πi
for each v ∈ Lip(Rn ; Rn ) with �v�L∞ (Rn ;Rn ) ≤ 1, the first inequality follows directly from Definition 5.1.1 and Theorem 6.5.5. The second inequality is a mere reformulation of inequality (4.6.1). Proposition 6.6.6. Let E ∈ BVloc (Rn ), and assume P(E) < ∞. If {Ek } is a sequence in BV(Rn ) such that lim P(Ek ) = 0, then lim P(Ek ∩ E) = lim P(Ek − E) = 0. Proof. For i = 1, . . . , n and k = 1, 2, . . . , � � � � Ek+ : = u ∈ Πi : H0 ∂∗Lu (Lu ∩ Ek ) > 0 � � � � = u ∈ Πi : H0 ∂∗Lu (Lu ∩ Ek ) ≥ 1 .
By Theorem 4.6.6, for Ln−1 almost all u ∈ Πi , the intersection Lu ∩ Ek is L1 equivalent to the union of finitely many open segments. It follows that for
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6.6. Properties of BV sets
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145
Ln−1 almost all u ∈ Πi − Ek+ , the set Lu ∩ Ek is L1 negligible, and so is the set Lu ∩ Ek ∩ E. Using this and Proposition 6.6.5, � � � H0 ∂∗Lu (Lu ∩ Ek ∩ E) dLn−1 (u) Πi � � � = H0 ∂∗Lu (Lu ∩ Ek ∩ E) dLn−1 (u) Ek+
≤
�
Ek+
� � H0 ∂∗Lu (Lu ∩ Ek ) ∪ ∂∗Lu (Lu ∩ E) dLn−1 (u)
≤ P (Ek ) +
�
Ek+
(∗)
� � H0 ∂∗Lu (Lu ∩ E) dLn−1 (u).
Applying Proposition 6.6.5 again, � � � + n−1 (Ek ) ≤ H0 ∂∗Lu (Lu ∩ Ek ) dLn−1 (u) ≤ P(Ek ). L Πi
According to our assumption, lim Ln−1 (Ek+ ) = 0. As P(E) < ∞, � � � lim H0 ∂∗Lu (Lu ∩ E) dLn−1 (u) = 0. Ek+
From this and (∗) we obtain n � � � � H0 ∂∗Lu (Lu ∩ (Ek ∩ E) dLn−1 (u) = 0, lim Πi
i=1
and Proposition 6.6.5 implies lim P(Ek ∩ E) = 0. The remaining equality lim P(Ek − E) = 0 follows from the identity Ek − E = Ek − (Ek ∩ E) by Observation 4.2.1.
Considering sets E = B(0, 1) and Ek = Rn − B(0, 1/k), we see that Proposition 6.6.6 is false for Ek ∈ Ploc (Rn ). The following example shows that the assumption P(E) < ∞ cannot be omitted either. Example 6.6.7. For j, k ∈ N, let xk = (k, 0, . . . , 0) and � � 1 1 Bk,j := x ∈ Rn : < |x − xk |n−1 < . (2j + 1)k 2jk �∞ Since each sum j=1 P(Bk,j ) diverges, there are integers pk ≥ 1 such that �pk � ∞ �p k that B := k=1 j=1 Bk,j is a locally BV set j=1 P(Bk,j ) ≥ k. It follows � � with P(B) = ∞. Now lim P U (xk , 1/k) = 0 and �
�
lim P U (xk , 1/k) ∩ B = lim
k→∞
✐
k→∞
pk � j=1
P(Bk,j ) = ∞.
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6.7. Approximating by figures In Section 2.4 we extended the divergence theorem from dyadic figures to the family DF, which was the completion of DF with respect to the convergence of dyadic figures introduced in Section 2.4. We denote by BVc (Rn ) the family of all bounded BV sets in Rn , and show that DF = BVc (Rn ). The convergence of dyadic figures defined in Section 2.4 extends verbatim to a convergence of bounded BV sets. We say that a sequence {Ek } in BVc (Rn ) converges to a set E ⊂ Rn if the following conditions are satisfied: (i) each Ek is contained in a fixed compact set K ⊂ Rn ; (ii) lim |Ek � E| = 0 and sup P(Ek ) < ∞.
Proposition 6.7.1. Let {Ek } be a sequence in BVc that converges to a set E ⊂ Rn . Then E ∈ BVc (Rn ), and for each v ∈ C(Rn ), � � v · νEk dHn−1 = v · νE dHn−1 . lim ∂∗ Ek
∂∗ E
Proof. Since each Ek is contained in a fixed compact set K ⊂ Rn , so is E. By Lemma 2.4.1, the set E is measurable and lim χEk = χE in L1 (Rn ). Since there is w ∈ Cc (Rn ; Rn ) such that w � K = v � K, the proposition follows from Proposition 5.5.4 and Theorem 6.5.2. Proposition 6.7.1 shows that DF ⊂ BVc (Rn ), and that on DF the flux F� defined in Section 2.4 coincides with that defined by formula (6.5.1). Lemma 6.7.2. Let Q be a cube, and let E ⊂ Q be a BV set with |E| ≤ |Q|/2. Then Hn−1 (∂Q ∩ ∂∗ E) ≤ β Hn−1 (int Q ∩ ∂∗ E) where β = β(n) > 0. Proof. As the lemma follows from Proposition 6.1.1 when n = 1, suppose n ≥ 2. By translation invariance, we may assume Q = [0, h]n . The intersection S := Q ∩ Πn is an (n − 1)-dimensional face of Q, and we estimate Hn−1 (S ∩ ∂∗ E). Letting � � −1 (u) ∩ E ∩ Q] = h , A : = u ∈ S : H1 [πn � � −1 (u) ∩ E ∩ Q] < h , B : = u ∈ S : 0 < H1 [πn it is clear that S ∩ ∂∗ E ⊂ A ∪ B and B = πn (int Q ∩ ∂∗ E). Since |E| ≤ |Q − E|, the relative isoperimetric inequality (Corollary 5.9.13) implies 1
hHn−1 (A) ≤ |E| n |E|
n−1 n
1
≤ κ|Q| n Hn−1 (int Q ∩ ∂∗ E) = κhHn−1 (int Q ∩ ∂∗ E).
Dividing by h > 0 and using Hn−1 (B) ≤ Hn−1 (int Q ∩ ∂∗ E), we obtain Hn−1 (S ∩ ∂∗ E) ≤ (1 + κ)Hn−1 (int Q ∩ ∂∗ E). With β = 2n(1 + κ), the lemma follows by symmetry. Proposition 6.7.3. Let E ∈ BVc (Rn ), and for an integer k, let A be the collection of all � k-cubes Q such that |Q ∩ E| ≥ |Q|/2. If A := A then for γ = γ(n) > 0, P(A) ≤ γP(E)
✐
and
|A � E| ≤ 2−k γP(E).
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6.7. Approximating by figures
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147
Proof. We assume n ≥ 2, since for n = 1 the proposition follows from Proposition 6.1.1. � Denote by B the collection of all k-cubes Q such that 0 < |Q∩E| < |Q|/2, and let B = B. Note A and B are finite collections, A ∩ B = ∅, and A ∪ B covers E almost entirely. The relative isoperimetric inequality yields � � n−1 1 |Q − E| = |Q − E| n |Q − E| n |A − E| = Q∈A
≤κ |E − A| =
�
Q∈A
�
Q∈B
≤κ
Q∈A
|Q|
1 n
H
n−1
|Q ∩ E| =
�
Q∈B
|Q|
1 n
H
(int Q ∩ ∂∗ E) ≤ 2−k κP(E, int A),
�
Q∈B
n−1
1
|Q ∩ E| n |Q ∩ E|
n−1 n
(int Q ∩ ∂∗ E) ≤ 2−k κP(E, int B).
As int A ∩ int B = ∅,
� � |A � E| ≤ 2−k κ P(E, int A) + P(E, int B) ≤ 2−k κP(E).
Now let S be an (n − 1)-dimensional face of a cube Q ∈ A such that S ⊂ ∂A, and let Q� be the other k-cube whose (n − 1)-dimensional face is S. Clearly Q� ∈ / A, and � � � � � � � � S = S ∩ ∂∗ (Q − E) ∪ S ∩ ∂∗ (Q ∩ E) ⊂ ∂Q ∩ ∂∗ (Q − E) ∪ S ∩ ∂∗ (Q ∩ E) .
/ B then S ∩ ∂∗ (Q ∩ E) = S ∩ ∂∗ E ⊂ ∂∗ E. On the other hand, if Q� ∈ B then If Q� ∈ �� � � �� S ∩ ∂∗ (Q ∩ E) = S ∩ ∂∗ (Q ∩ E) ∩ ∂∗ (Q� ∩ E) ∪ ∂∗ (Q ∩ E) − ∂∗ (Q� ∩ E) � � �� � � = S ∩ ∂∗ (Q ∩ E) ∩ ∂∗ (Q� ∩ E) ∪ S ∩ ∂∗ E � � ⊂ ∂Q� ∩ ∂∗ (Q� ∩ E) ∪ ∂∗ E.
It follows that
∂A ⊂ ∂∗ E ∪
� �
Q∈A
� � � � ∂Q ∩ ∂∗ (Q − E) ∪ ∂Q ∩ ∂∗ (Q ∩ E) .
(∗)
Q∈B
From Corollary 4.2.5 we infer int Q ∩ ∂∗ (Q − E) = int Q ∩ ∂∗ (Q ∩ E) = int Q ∩ ∂∗ E for each cube Q ⊂ Rn . Thus (∗) and Lemma 6.7.2 imply � � Hn−1 (int Q ∩ ∂∗ E) + β Hn−1 (int Q ∩ ∂∗ E) P(A) ≤ P(E) + β Q∈A
Q∈B
≤ (1 + β)P(E),
and the proposition holds with γ := max{κ, 1 + β}. Corollary 6.7.4. DF = BVc (Rn ). Remark 6.7.5. Given the dimension n and the grid of k-cubes, Proposition 6.7.3 asserts that both P(A) and |A � E| depend only on P(E). The readers familiar with currents will recognize that Proposition 6.7.3 is a version of the deformation theorem for the top dimensional integral currents; see [33, Theorem 4.2.9] or [47, Section 5.1]. Theorem 6.7.6. Let K ⊂ Rn be a compact set, and let c ∈ R+ . The set � � X := χE : E ⊂ K and P(E) ≤ c
is a compact subset of L1 (Rn ).
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6. Locally BV sets
Proof. Although the theorem follows immediately from Theorems 5.5.12 and 6.5.5, a different proof based on Proposition 6.6.2 is instructive. From Propositions 4.5.3 and 5.1.7 we infer that X is a closed subset of L1 (Rn ). Since L1 (Rn ) is complete, so is X. Thus it suffices to show that X is totally bounded. Choose ε > 0, and recall that |A � B| = �χA − χB �L1 (Rn )
for any pair of measurable sets A, B ⊂ Rn . If γ is the constant from Proposition 6.7.3, select k ∈ N so that 2−k cγ 0 and find a compact set K ⊂ ∂∗ E with Hn−1 (∂∗ E − K) < ε. If U = Rn − K and B ∈ E, then ∂∗ B − ∂∗ E ⊂ ∂∗ B ∩ U ⊂ (∂∗ B − ∂∗ E) ∪ (∂∗ E − K) for each B ∈ E, and consequently G(B) ≤ P(B, U ) ≤ G(B) + ε. In view of Theorem 6.5.5 and Proposition 5.1.7, G(A) ≤ P(A, U ) ≤ lim inf P(Ak , U ) ≤ lim inf G(Ak ) + ε and the lemma follows from the arbitrariness of ε. We say that a set A ∈ E is a minimizer of F , or minimizes F , if � � F (A) = inf F (B) : B ∈ E .
If A minimizes F , then so does any set B ∈ E that is equivalent to A. In particular, replacing A by int∗ A or cl∗ A, we may assume that a minimizer of F satisfies A = int∗ A or A = cl∗ A, respectively. Lemma 7.1.2. The functional F has a minimizer. � � Proof. Let c := inf Fk (B) : B ∈ E . As ∅ ∈ E and F (∅) = β|E|, we see that c ≤ β|E|. Thus there is a sequence {Ak } in E such that c ≤ F (Ak ) < c + 1/k,
k = 1, 2, . . . .
From F (Ak ) ≤ β|E|+1 we obtain P(Ak ) ≤ β|E|+1+αP(E). Theorem 5.5.12 implies that there are a set A ∈ E and a subsequence of {Ak }, still denoted by {Ak }, which L1 converges to A. By Lemma 7.1.1, c ≤ F (A) ≤ lim F (Ak ) = c. Lemma 7.1.3. Let A be a minimizer of F such that A = int∗ A, and let x ∈ Rn . If Hn−1 (∂∗ A ∩ ∂Br ) = 0 for Br = B(x, r), then Hn−1 (∂∗ A ∩ Br ) − αHn−1 (∂∗ E ∩ ∂∗ A ∩ Br )
≤ Hn−1 (A ∩ ∂Br ) + β|A ∩ Br |.
Proof. Fix r > 0 so that Hn−1 (∂∗ A ∩ ∂Br ) = 0, and define a measure µ := Hn−1 − αHn−1
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7.1. Approximating from inside
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Observe that F (B) = β|E − B| + µ(∂∗ B) for each B ∈ E, and that the inequality we wish to prove is transformed to µ(∂∗ A ∩ Br ) ≤ Hn−1 (A ∩ ∂Br ) + β|A ∩ Br |.
(∗)
By Lemma 4.5.5, there is an increasing sequence {sk } of positive numbers such that lim sk = r and Hn−1 (∂∗ A ∩ ∂Bsk ) = 0 for k = 1, 2, . . . . As A minimizes F , for Ck := A − Bsk we obtain β|E − A| + µ(∂∗ A) = F (A) ≤ F (Ck ) = β|E − Ck | + µ(∂∗ Ck ), and consequently µ(∂∗ A) ≤ µ(∂∗ Ck ) + β|A − Ck |
= µ(∂∗ Ck ) + β|A ∩ Bsk | ≤ µ(∂∗ Ck ) + β|A ∩ Br |.
In addition ∂∗ A − Br = ∂∗ Ck − Br , since Bsk ⊂ int Br . Hence µ(∂∗ A ∩ Br ) = µ(∂∗ A) − µ(∂∗ A − Br )
≤ µ(∂∗ Ck ) + β|A ∩ Br | − µ(∂∗ Ck − Br ) = µ(∂∗ Ck ∩ Br ) + β|A ∩ Br |
for k = 1, 2, . . . . Thus to prove (∗), it suffices to verify lim µ(∂∗ Ck ∩ Br ) = Hn−1 (A ∩ ∂Br ).
(∗∗)
lim Hn−1 (∂∗ A ∩ Dk ) = lim Hn−1 (∂∗ A ∩ int Dk ) = 0.
(†)
To this end, let Dk := Br − Bsk , and note that Hn−1 (∂∗ A ∩ ∂Dk ) = 0 by the choice of r and sk . In particular, n
As A = int∗ A and Ck = A ∩ (R − Bsk ), Corollary 4.2.5 shows that the pairs � � � ∂∗ A ∩ Dk ), (∂∗ A ∩ Dk ) ∪ (A ∩ ∂Dk ) , (††) � � ∂∗ Ck , (∂∗ A − Bsk ) ∪ (A ∩ ∂Bsk )
consist of Hn−1 equivalent sets. Intersecting the sets of� the second pair by � Br produces a pair ∂∗ Ck ∩ Br , (∂∗ A ∩ Dk ) ∪ (A ∩ ∂Bsk ) of Hn−1 equivalent sets. In addition, A = int∗ A and A ⊂ E imply A ∩ ∂∗ E = ∅. Thus µ(∂∗ Ck ∩ Br ) = µ(∂∗ A ∩ Dk ) + µ(A ∩ ∂Bsk )
= µ(∂∗ A ∩ Dk ) + Hn−1 (A ∩ ∂Bsk ).
In view of this equality and (†), we prove (∗∗) by establishing lim Hn−1 (A ∩ ∂Bsk ) = Hn−1 (A ∩ ∂Br ). Cc1 (Rn ; Rn )
Via mollification, define v ∈ so that v(y) = (y − x)|y − x| every y ∈ cl Dk , and observe that � νA∩Dk (y) if y ∈ A ∩ ∂Br , v(y) = −νA∩Dk (y) if y ∈ A ∩ ∂Bsk .
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(♦) −1
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7. Bounded vector fields
Theorem 6.5.4 and (††) imply � � div v = v · νA∩Dk dHn−1 A∩Dk
= =
�
�
∂∗ (A∩Dk )
∂∗ A∩Dk ∂∗ A∩Dk
v · νA∩Dk dHn−1 +
�
A∩∂Dk
v · νA∩Dk dHn−1
v · νA∩Dk dHn−1 +
Hn−1 (A ∩ ∂Br ) − Hn−1 (A ∩ ∂Bsk ) � for k = 1, 2, . . . . Since lim A∩Dk div v = 0, and since � �� � � v · νA∩Dk dHn−1 �� ≤ lim Hn−1 (∂∗ A ∩ Dk ) = 0 lim�� ∂∗ A∩Dk
by (†), the desired equality (♦) is established.
Lemma 7.1.4. If A = int∗ A minimizes F , then cl∗ A = cl A. Proof. Since Proposition 6.1.1 shows that cl∗ A = cl A for any one-dimensiox� ∈ cl A, nal BV set A, we assume n ≥ 2 and prove that cl A ⊂ cl∗ A. Choose � and let Bs := B(x, s) for each s > 0. Select R > r > 0 so that �A∩BR � ≤ 12 |A| and Hn−1 (∂∗ A ∩ ∂Br ) = 0. By Lemma 4.5.5, the latter condition is satisfied for all but countably many 0 < r < R. The relative isoperimetric inequality and Lemma 7.1.3 yield (1 − α)κ|A ∩ Br |
n−1 n
≤ (1 − α)Hn−1 (∂∗ A ∩ Br )
≤ Hn−1 (∂∗ A ∩ Br ) − αHn−1 (∂∗ A ∩ Br ∩ ∂∗ E)
≤ Hn−1 (A ∩ ∂Br ) + β|A ∩ Br | 1
≤ Hn−1 (A ∩ ∂Br ) + β|Br | n |A ∩ Br |
n−1 n
where κ = κ(n) > 0. Observe that A = int∗ A and A ∩ U (x, r) �= ∅ implies n−1 |A ∩ Br | > 0. Dividing by |A ∩ Br | n , we obtain 1
1
(1 − α)κ − β|Br | n ≤ |A ∩ Br | n −1 Hn−1 (A ∩ ∂Br ). 1
Let γ := 12 (1 − α)κ, and make R so small that β|BR | n ≤ γ. Then 1
γ ≤ |A ∩ Br | n −1 Hn−1 (A ∩ ∂Br ) for all but countably many 0 < r < R. Since � r |A ∩ Br | = Hn−1 (A ∩ ∂Bs ) ds 0
[29, Section 3.4.4, Proposition 1], the previous inequality transforms to � 1 1 d� d γ ≤ |A ∩ Br | n −1 |A ∩ Br | = n |A ∩ Br | n dr dr
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7.2. Relative derivatives
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155
which holds for L1 almost all 0 < r < R. Integrating over the interval 1 [0, r] ⊂ [0, R] gives γr ≤ n|A ∩ Br | n for all 0 < r < R. Thus � �n |A ∩ Br | 1 γ ≤ 0< α(n) n |Br | for all sufficiently small r > 0. It follows that x ∈ cl∗ A.
Combining Lemmas 7.1.2 and 7.1.4, we obtain the aforementioned result of Tamanini and Giacomelli [72]. Theorem 7.1.5. Let E ∈ BV(Rn ). There exists a sequence {Ek } in BV(Rn ) such that Ek ⊂ E and cl Ek = cl∗ Ek for k = 1, 2, . . . , and lim |E − Ek | = lim P(E − Ek ) = 0. Proof. Assume E = cl∗ E. For k = 1, 2, . . . and A ∈ E, define
Fk (A) := k|E − A| + P (A) − αHn−1 (∂∗ A ∩ ∂∗ E).
By Lemmas 7.1.2 and 7.1.4, each Fk has a minimizer Ek which satisfies cl∗ Ek = cl Ek . Since αHn−1 (∂∗ A ∩ ∂∗ E) ≤ P(A) for every A ∈ E, and since Ek minimizes Fk , we obtain k|E − Ek | ≤ Fk (Ek ) ≤ Fk (E) ≤ (1 − α)P(E).
(∗)
Hence lim |E − Ek | = 0, and consequently P(E) ≤ lim inf P(Ek ) by Proposition 5.1.7. On the other hand, inequality (∗) yields P(Ek ) − αP(E) ≤ P(Ek ) − αHn−1 (∂∗ Ek ∩ ∂∗ E) ≤ F (Ek ) ≤ (1 − α)P(E).
Now deduce first that lim P(Ek ) = P(E), and then lim Hn−1 (∂∗ Ek ∩ ∂∗ E) = P(E). This and Proposition 6.6.3 imply lim P(E − Ek ) = 0.
7.2. Relative derivatives Let E ⊂ Rn . A map φ : E → Rm is relatively differentiable at a point x ∈ E ∩ int∗ E if there is a linear map L : Rn → Rm such that � � �φ(y) − φ(x) − L(y − x)� lim = 0. y→x |y − x| y∈E The linear map L, which is unique by Proposition 7.2.1 below, is called the relative derivative of φ at x, denoted by DE φ(x). Note that the concepts of relative differentiability and differentiability coincide whenever x ∈ int E; in
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7. Bounded vector fields
particular DE φ(x) = Dφ(x) for each x ∈ int E. If A ⊂ E and φ is differentiable at x ∈ A∩int∗ A relative to E, then the restriction φ � A is differentiable at x relative to A and DA (φ � A)(x) = DE φ(x).
(7.2.1)
Proposition 7.2.1. If the relative derivative exists, it is unique. Proof. Let x ∈ E ∩ int∗ E, and assume that there exist distinct linear maps L1 and L2 satisfying the following condition: given ε > 0, there is δ > 0 such that for each y ∈ E ∩ U (x,δ ) and k = 1, 2, � � �φ(y) − φ(x) − Lk (y − x)� < ε|y − x|. (∗) As L1 �= L2 , the linear map L := L1 −L2 has a positive norm N . Let ε := N/4, and find δ > 0 so that (∗) holds for each y ∈ E ∩ U (x,δ ). Define open sets � � � � and T := x + S. S := y ∈ Rn : �L(y)� > 2ε|y|
If S = ∅ then N ≤ 2ε = N/2, a contradiction. Being open, the set S has positive measure, and as 0 ∈ cl S, so does the intersection S ∩ B(0, 1). As rS = S for each r > 0, we obtain � � � � �S ∩ B(0, r)� �S ∩ B(0, 1)� � = � � lim �� �B(0, 1)� > 0, r→0 B(0, r)�
which means 0 ∈ cl∗ S. Consequently x ∈ cl∗ T , and Proposition 4.2.4 yields x ∈ cl∗ (T ∩ E); in particular, T ∩ E ∩ U (x,δ ) �= ∅. However, � � � � 2ε|y − x| < �L(y − x)� = �L1 (y − x) − L2 (y − x)� � � � � ≤ �L1 (y − x) − φ(y) + φ(x)� + �φ(y) − φ(x) − L2 (y − x)� < 2ε|y − x|
for each y in T ∩ E ∩ U (x,δ ) — a contradiction.
Using relatively differentiable maps, we extend Stepanoff’s theorem (Theorem 1.3.3) to pointwise Lipschitz maps defined on arbitrary subsets of Rn . The proof is almost identical to that of the original Stepanoff’s theorem [33, Theorem 3.1.9]. We begin with a preparatory lemma. Lemma 7.2.2. Let E ⊂ Rn , C ⊂ E ∩ int∗ E, and let φ : E → Rm satisfy the following conditions: (i) there are positive numbers c and δ such that � � �φ(x) − φ(y)� ≤ c|x − y|
for each y ∈ C and each x in E ∩ U (y,δ ); (ii) there is a Lipschitz map ψ : Rn → Rm such that ψ � C = φ � C.
If ψ is differentiable at a point z ∈ C ∩int∗ C, then φ is relatively differentiable at z and DE φ(z) = Dψ(z).
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7.2. Relative derivatives
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Proof. Suppose ψ is differentiable at z ∈ C ∩ int∗ C, and choose a positive ε ≤ 1. Making δ smaller, we may assume that for all x ∈ U (z,δ ), � � �ψ(x) − ψ(z) − Dψ(z)(x − z)� ≤ ε|x − z|. � � If� x ∈ U (z,δ ) and r := |x − z| > 0, then d U (x,εr ) ∪ {z} ≤ 2rx and x x � n s U (x,εr x ) ∪ {z} = α(n)(ε/2) . By Lemma 4.2.3, � � �C ∩ U (x,εr x )� � = 1. lim �� x→z U (x,εr x )�
Thus making δ still smaller, we may assume � that C ∩ U� (x,εr x ) �= ∅ for each x ∈ U (z,δ ) with x �= z. Select x ∈ E ∩ U (z,δ ) − {z} , and choose a point y ∈ C ∩ U (x,εr x ). Observing that |x − y| < ε|x − z| < δ, and that ψ(z) = φ(z) and ψ(y) = φ(y), we obtain � � � � �φ(x) − φ(z) − Dψ(z)(x − z)� = �φ(x) − ψ(z) − Dψ(z)(x − z)� � � � � ≤ �φ(x) − ψ(x)� + �ψ(x) − ψ(z) − Dψ(z)(z − z)� � � � � ≤ �φ(x) − φ(y)� + �ψ(y) − ψ(x)� + ε|x − z| ≤ (c + Lip ψ)|x − y| + ε|x − z| ≤ ε(c + Lip ψ + 1)|x − z|.
Theorem 7.2.3. Let E ⊂ Rn . If φ : E → Rm is pointwise Lipschitz in C ⊂ E ∩ int∗ E, then φ is relatively differentiable at almost all x ∈ C. Proof. For j = 1, 2, . . . , denote by Cj the set of all x ∈ C such that � � �φ(x) − φ(y)� ≤ j|x − y| (∗) �∞ for each y ∈ E∩U (x, 1/j). Then C = j=1 Cj , and each Cj is the union of sets Cj.k , k = 1, 2, . . . , of diameters smaller than 1/j. Observe that the sets Cj satisfy condition (i) of Lemma 7.2.2, and hence so do the sets Cj,k . In addition, inequality (∗) implies that the restrictions φ � Ci,j are Lipschitz maps. Using Proposition 1.5.2, extend φ � Ci,j to a Lipschitz map ψi,j : Rn → Rm . By Rademacher’s theorem (Theorem 1.5.3), each ψi,j is differentiable at almost all x ∈ Rn . Lemma 7.2.2 implies that φ is relatively differentiable almost everywhere in every Ci,j ∩ int∗ Ci,j , and hence almost everywhere in C by Corollary 4.4.4. Remark 7.2.4. If the set E in Theorem 7.2.3 is open, we obtain Stepanoff’s theorem quoted without proof in Theorem 1.5.4. Let E ⊂ Rn , and let f : E → R be relatively differentiable at a point x ∈ E ∩ int∗ E. The real numbers � � DE,i f (x) := Di DE f (x) , i = 1, . . . , n,
are called the relative partial derivatives of f at x. In other words, the relative partial derivatives DE,i f (x) are defined as the usual partial derivatives of the
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7. Bounded vector fields
linear map DE f (x) : Rn → R. The usual partial derivatives Di f of f may not exist unless x ∈ int E, in which case Di f (x) = DE,i f (x). If v = (v1 , . . . , vn ) is a vector field defined on a set E ⊂ Rn that is relatively differentiable at x ∈ E ∩ int∗ E, then n � divE v(x) := DE,i vi (x) i=1
is called the relative divergence of v at x. Clearly, divE v(x) is the trace of the linear map DE v(x) : Rn → Rn , and divE v(x) = div v(x) whenever x ∈ int E.
7.3. The critical interior The critical interior of a set E ⊂ Rn is the set � � � � Hn−1 ∂∗ E ∩ B(x, r) intc E := x ∈ int∗ E : lim = 0 . r→0 rn−1
In general, intc E is a proper subset of int∗ E even when E is a compact BV set; see Example 7.3.3 below. However, for locally BV sets, the difference between the essential and critical interiors is small. Proposition 7.3.1. If E ∈ BVloc (Rn ) then Hn−1 (int∗ E − intc E) = 0. Proof. It suffices to choose t > 0 and show that the set � � � � Hn−1 ∂∗ E ∩ B(x, r) A := x ∈ int∗ E : lim sup > t rn−1 r→0+
is Hn−1 negligible. To this end, choose positive ε and δ, and recall that the reduced measure Hn−1 ∂∗ E is Radon (Theorems 6.5.2). As A ∩ ∂∗ E = ∅, there is an open set U with A ⊂ U and Hn−1 (∂∗ E ∩ U ) ≤ ε. For each x ∈ A select 0 < rx < δ /10 so that � � Hn−1 ∂∗ E ∩ B(x, rx ) >t rxn−1 and B(x, rx ) ⊂ U . By Vitali’s � 4.3.2), there is a count� theorem (Theorem able set C ⊂ A such that B(x, rx ) : x ∈ C is a disjoint family and � A ⊂ x∈C B(x, 5rx ). We calculate: � Hδn−1 (A) ≤ α(n − 1)(5rx )n−1 x∈C
≤ α(n − 1)5n−1 t−1
�
x∈C
� � Hn−1 ∂∗ E ∩ B(x, rx )
≤ α(n − 1)5n−1 t−1 Hn−1 (∂∗ E ∩ U ) ≤ α(n − 1)5n−1 t−1 ε.
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7.3. The critical interior
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The arbitrariness of ε implies Hδs (A) = 0. As δ is also arbitrary, the proposition follows. Throughout the remainder of this book, ρ(n) := 18 n−3/2 is a fixed number. A bounded set E ∈ BV(Rn ) is called regular if |E| > 0 and |E| > ρ(n). d(E)P(E)
A direct calculation shows that each cube C ⊂ Rn is a regular BV set. We note that the value of ρ(n) is to a large extent arbitrary. Our choice simplifies the proof of Lemma 7.3.2 below, but any positive value of ρ(n) which makes cubes regular can be used; cf. Example 7.3.3 below. Recall from Section 4.2 that the shape s(E) of a bounded set E ⊂ Rn of positive diameter is defined as the ratio |E|/d(E)n . Suppose that a bounded set E ∈ BV(Rn ) is regular. If n = 1 then s(E) > ρ(n), since P(E) ≥ 2. If n ≥ 2, then the isoperimetric inequality implies |E|n−1 ≤ γ n P(E)n where γ = γ(n) > 0. Thus � �n � � |E|n−1 |E| |E| ρ(n) n −n s(E) ≥ γ −n · = γ > , P(E)n d(E)n d(E)P(E) γ and we see that the shapes of regular bounded BV sets are bounded away from zero.
Lemma 7.3.2. Let A ∈ BVloc (Rn ) and x ∈ intc A. If a cube C containing x is sufficiently small, then A ∩ C is a regular BV set. Proof. If r := d(C), then Corollary 4.2.5 implies � P(A ∩ C) ≤ Hn−1 ∂∗ A ∩ B[x, r]) + P(C), and we calculate:
� � rn Hn−1 ∂∗ A ∩ B(x, r) d(C)P(C) rP(A ∩ C) ≤ · + n−1 |C| |C| r |C| � � n−1 ∂∗ A ∩ B(x, r) H 1 . = nn/2 + n−1 r 4ρ(n)
By the definition of intc E and Lemma 4.2.3, � � Hn−1 ∂∗ A ∩ B(x, r) lim = 0 and r→0 rn−1 Hence for each sufficiently small r,
lim
r→0
|A ∩ C| = 1. |C|
d(A ∩ C)P(A ∩ C) rP(A ∩ C) |C| 1 1 ≤ · < ·2= |A ∩ C| |C| |A ∩ C| 2ρ(n) ρ(n)
and the lemma follows.
Example 7.3.3. Assume that n = 2, and choose an integer k ≥ 8. For i = 1, . . . , k, let Ui be the set � � (x, y) ∈ R2 : 0 < y < x2 tan 2π k and x > 0
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�k rotated counterclockwise by 2π(i − 1)/k. Then Ak := B(0, 1) − i=1 Ui is a compact BV set, and a direct calculation reveals that the origin 0 of R2 is a point of int∗ Ak which does not belong to intc Ak . If Cr := [−r, r]2 , then it is easy to see that lim
r→0
|Ak ∩ Cr | 1 √ → 0 as k → ∞. = d(Ak ∩ Cr )P(Ak ∩ Cr ) (4 + k) 2
Thus if k is sufficiently large, Ak ∩ Cr is not a regular set for all sufficiently small r > 0. It follows that the assumption x ∈ intc A of Lemma 7.3.2 cannot be omitted no matter how small the constant ρ(n) > 0 we select.
7.4. The divergence theorem Lemma 7.4.1. Let E ⊂ Rn , x ∈ cl∗ E, and v ∈ L∞ (cl∗ E, Hn−1 ; Rn ). Assume {Bk } is a sequence in BV(Rn ) such that Bk ⊂ cl∗ E and x ∈ Bk for k = 1, 2, . . . , and lim d(Bk ) = 0. Then � 1 v · νBk dHn−1 ≤ H0 v(x). (1) lim sup P(Bk ) ∂∗ Bk Assume in addition that each Bk is regular. For 0 ≤ s ≤ 1, �� � 1 � v · νBk dHn−1 ≤ βHs v(x) lim sup d(Bk )n−1+s � ∂∗ Bk
(2)
where β = β(n) > 0. If x ∈ int∗ E and v is relatively differentiable at x, then � 1 v · νBk dHn−1 = divE v(x). (3) lim |Bk | ∂∗ Bk
Proof. The proof is similar to that of Corollary 2.1.3. We may assume Hs v(x) < ∞. Choose ε > 0, and for B ∈ BV(Rn ) with B ⊂ cl∗ E, let � v · νB dHn−1 . F (B) := ∂∗ B
� � � � There is δ > 0 such that �v(y) − v(x)� ≤ Hs v(x) + ε |y − x|s for every y ∈ E ∩ U (x,δ ). By Theorem 6.5.4, � � � F (Bk ) = v(y) − v(x) · νBk dHn−1 (y) ∂∗ B k � � � (∗) ≤ Hs v(x) + ε |y − x|s dHn−1 (y) �
�
∂ ∗ Bk
≤ Hs v(x) + ε d(Bk )s P(Bk )
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161
for all sufficiently large k. If s = 0, inequality (1) follows from the arbitrariness of ε. If Bk is regular, then (∗) and the isodiametric inequality (1.4.2) imply � � 1 F (Bk ) ≤ Hs v(x) + ε d(Bk )s−1 |Bk | ρ(n) � α(n) � d(Bk )n−1+s . ≤ Hs v(x) + ε n 2 ρ(n) � � As ε is arbitrary, inequality (2) holds with β = α(n)/ 2n ρ(n) . Finally, assume that in addition x ∈ int∗ E and v is relatively differentiable at x. Letting � � w : y �→ v(x) + DE v(x) (y − x) : Rn → Rn , we have div w(y) = divE v(x) for each y ∈ Rn , and there is η > 0 such that � � �v(y) − w(y)� < ε|y − x|
for every y ∈ E ∩ U (x,η ). By Theorem 6.5.4, � � � ��� � � � � � � n−1 v(y) − w(y) · νBk (y) dH (y)�� �F (Bk ) − divE v(x)|Bk |� = � ∂ B �∗ k ≤ε |y − x| dHn−1 (y) ∂ ∗ Bk
≤ εd(Bk )P(Bk ) ≤
ε |Bk | ρ(n)
for all sufficiently large k. Equality (3) follows. Lemma 7.4.2. Let A ∈ BV(Rn ) be a compact set such that A = cl∗ A, and let v ∈ Adm(A; Rn ). Define f : A → R by � divA v(x) if x ∈ int∗ A and v is relatively differentiable at x, f (x) := 0 otherwise. For each ε > 0 and each δ : A → R+ , there is a δ-fine partition � � P = (C1 , x1 ), . . . , (Cp , xp )
such that Ci ∈ BV(Rn ) for i = 1, . . . , p, [P ] = A, and � �� � � � p n−1 � � f (xi )|Ci | − v · νA dH � 0 implies Hsk v(x) = 0 for each x ∈ Bk .
We apply separate arguments to the sets intc A and A − intc A. While the proof for intc A is similar to that of Lemma 2.3.6, new techniques are needed to deal with A − intc A. The intersections Ek = Bk ∩ intc A, k = 1, 2, . . . , satisfy conditions (i*) �∞ and (ii). Theorem 7.2.3 shows that intc A − k=1 Ek is the union of disjoint sets E0 and D such that Hn (E0 ) = 0 and v is relatively differentiable at each x ∈ D. Thus intc A� is the union of disjoint � sets D, E0 , E1 , . . . , and we let s0 = 1. The family (Ek , sk ) : k = 0, 1, . . . is the disjoint union of � � � � (Ek , sk ) : Hn−1+sk (Ek ) > 0 and (Ek , sk ) : Hn−1+sk (Ek ) = 0 . � � � 0 0 � We enumerate these subfamilies as (Ei+ , s+ i ) : i ≥ 1 and (Ei , si ) : i ≥ 1 , respectively. For j ∈ N, let � � 0 Ei,j := x ∈ Ei0 : j − 1 ≤ Hs0i v(x) < j
+ 0 0 and define t+ i = n − 1 + si and ti = n − 1 + si . Now E = intc A − D is the + 0 union of disjoint sets Ei and Ei,j . Choose ε > 0, and select c > �v�L∞ (A;Rn ) +
and ci > Hti (Ei+ ), i = 1, 2, . . . . By Lemma 7.4.1, there are β = β(n) > 0 and γ : A → R+ such that for each regular set C ∈ C, the following is true: � � (1) �f (x)|C| − F (C)� ≤ ε|C| if d(C) < γ(x) for some x ∈ D ∩ C, � � + t+ i if d(C) < γ(x) for some x ∈ E (2) �F (C)� ≤ ε2−i c−1 i d(C) i ∩ C, � � 0 (3) �F (C)� ≤ βj d(C)ti if d(C) < γ(x) for some x ∈ E 0 ∩ C. i,j
Making γ smaller, we may assume that the intersection A ∩ Q is regular for every cube Q such that d(Q) < γ(x) for some x ∈ intc A ∩ Q; see Lemma 7.3.2. Turning our attention to the set A − intc A, let N := (A − intc A) ∩
∞ � �
k=1
� Bk : sk = 0 and Hn−1 (Bk ) = 0
and M = (A − intc A) − N . Clearly Hn−1 (N ) = 0, and condition (ii) implies that H0 v(x) = 0 for each x ∈ M . By Proposition 6.6.6, there is η > 0 such that P(A ∩ S) ≤ ε/c for each set S ∈ BV(Rn ) with P(S) ≤ η. By making c larger, we achieve Hn−1 (M ) < c. Moreover, in view of Lemma 7.4.1, making γ smaller, we may assume that � � �F (C)� ≤ εc−1 P(C) (∗) for each C ∈ C with d(C) < γ(x) for some x ∈ M ∩ C.
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7.4. The divergence theorem
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163
Select a dyadic figure T containing A, and define δ : T → R+ by letting δ(x) = γ(x) if x ∈ A, and δ(x) = dist (x, A) if x ∈ T − A. Proposition 2.2.2 shows that there is a δ-fine dyadic partition � � P = (Q1 , x1 ), . . . , (Qp , xp )
such that [P ] = T , and for a fixed κ = κ(n) > 0, the following is true: � n−1 (a) ≤ (2n)−1 η, xk ∈N d(Qk ) � n−1 (b) ≤ κc, x ∈M d(Qk ) � k + ti ≤ κci , (c) xk ∈Bi+ d(Qk ) � 0 t −1 −i−j (d) 2 . 0 d(Qk ) i < ε j xk ∈Bi,j � � If Ck := A ∩ Qk , then R := (Ck , xk ) : xk ∈ A is a δ-fine partition such that [R] = A and Ck is regular whenever xk ∈ intc A. We distinguish four cases. � � Case 1 . If C = xk ∈N Ck and Q = xk ∈N Qk , then C = A ∩ Q. By inequality (a), � � P(Q) ≤ P(Qk ) = 2n d(Qk )n−1 ≤ η xk ∈N
xk ∈N
�and our � choice of η implies P(C) < ε/c. Since �v�L∞ (A;Rn ) < c, we obtain �F (C)� ≤ cP(C) ≤ ε. Case 2 . By (∗), Corollary 4.2.5, and inequality (b), � � � �F (Ck )| ≤ εc−1 P(Ck )
xk ∈M
xk ∈M
−1
≤ εc
≤ εc
� �
xk ∈M
−1
Hn−1 (∂∗ A ∩ int Qk ) + P(Qk ) �
P(A) + 2nεc−1
xk ∈M
�
� � d(Qk )n−1 ≤ ε P (A) + 2nκ .
Case 3 . Condition (1) yields � � � �� � |Ck | ≤ ε|A|. �f (xk )|Ck | − F (Ck )� ≤ ε xk ∈D
xk ∈D
Case 4 . Conditions (2) and (3) and inequalities (c) and (d) imply � � ∞ � � �� � � � � � � � �F (Ck )� ≤ �F (Ck )� + �F (Ck )� xk ∈E
i≥1
≤ε
��
✐
2−i c−1 i
i≥1
≤ εκ
0 j=1 xk ∈Ei,j
xk ∈Ei+
∞ � i=1
�
xk ∈Ei+
2−i + εβ
∞ �
+
d(Ck )ti + β
∞ � j=1
j
�
0 xk ∈Ei,j
0
d(Ck )ti
�
2−i−j = ε(κ + β).
i,j=1
© 2012 by Taylor & Francis Group, LLC
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164
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7. Bounded vector fields
Since A is the disjoint union of the sets N, M, D, and E, and since f (x) = 0 for each x ∈ A − D, the lemma follows by summing up the inequalities established in Cases 1–4. Proposition 7.4.3. Let A ∈ BV(Rn ), and let v : cl∗ A → Rn be a bounded admissible vector field. If divcl∗ A v belongs to L1 (A), then � � divcl∗ A v(x) dx = v · νA dHn−1 . A
∂∗ A
Proof. As Theorem 4.4.2 implies ∂∗ (cl∗ A) = ∂∗ A, no generality is lost by assuming A = cl∗ A. Define f : A → R as in Lemma 7.4.2, and observe that by our assumptions, f ∈ L1 (A) and � � f (x) dx = divA v(x) dx. A
A
Case 1. Let A = cl∗ A be a compact BV set. Choose ε > 0, and find to the Henstock lemma δ : A → R+ associated � with f and ε according � (Lemma 1.2.3). If P := (B1 , x1 ), . . . , (Bp , xp ) is a δ-fine partition for which Lemma 7.4.2 holds, then �� � �� � � p � � � � � n−1 � � divA v(x) dx − � � ≤ v · ν dH f (x) dx − f (x )|B | A i i � � � � A
∂∗ A
A
� p � �� + �� f (xi )|Bi | − i=1
∂∗ A
i=1
� � n−1 � v · νA dH � < 2ε.
The arbitrariness of ε implies the desired equality.
Case 2. Let A = cl∗ A be a bounded BV set. Use Theorem 7.1.5 to find a sequence {Ak } of compact BV sets such that Ak ⊂ A and Ak = cl∗ Ak for k = 1, 2, . . . , and lim |A − Ak | = lim P(A − Ak ) = 0. As divAk (v � Ak )(x) = divA v(x) for almost all x ∈ int Ak , Case 1 yields � � divA v(x) dx = v · νAk dHn−1 Ak
∂∗ Ak
for k = 1, 2, . . . . From Theorem 6.6.1, we obtain � �� � � � n−1 n−1 � � v · νA dH − v · νAk dH lim� � ∂∗ A ∂ ∗ Ak �� � � � = lim�� v · νA−Ak dHn−1 �� ∂∗ (A−Ak )
≤ �v�L∞ (A;Rn ) lim P(A − Ak ) = 0.
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7.4. The divergence theorem
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165
The proposition follows from the equality � � lim divA (x) dx = divA (x) dx. Ak
A
Case 3. Let A = cl∗ A be an arbitrary BV set. If Ck := [−k, k]n for k = 1, 2, . . . , then �� � � � n−1 � � lim� v · νA−Ck dH � ≤ �v�L∞ (A;Rn ) lim P(A − Ck ) = 0 ∂∗ (A−Ck )
by Proposition 4.6.11. Since � � lim divA v(x) dx = divA v(x) dx, A∩Ck
A
an application of Case 2 and Theorem 6.6.1 completes the proof.
Proposition 7.4.4 (Partition of unity). Let O be a family of open subsets of Rn . There are nonnegative functions ϕk ∈ Cc∞ (Rn ) such that
(1) for each k ∈ N there is O ∈ O with spt ϕk ⊂ O, � � (2) O, k∈N ϕk (x) = 1 for each x ∈ (3) the family {spt ϕk : k ∈ N} is locally finite. � Proof. Let Ω= O. Denote by U the family of all open balls U (x, r) such that x ∈ Ω ∩ Qn , r ∈ Q+ , and U (x, r) ⊂ O for some O ∈ O. Enumerate U as {Uk : k ∈ N}. If Uk = U (x, r), let Vk := U (x, r/2) and Bk := B(x, 2r/3). Mollifying the indicator χBk of Bk , we obtain a function ψk ∈ Cc∞ (Rn ) such that χVk ≤ ψk ≤ 1 and spt ψk ⊂ Uk . Let ϕ1 := ψ1 , and for k = 1, 2, . . . , define ϕk+1 ∈ Cc∞ (Rn ) by the formula ϕk+1 := (1 − ψ1 ) · · · (1 − ψk )ψk+1 .
(∗)
ϕ1 + · · · + ϕk = 1 − (1 − ψ1 ) · · · (1 − ψk ).
(∗∗)
ϕ1 (x) + · · · + ϕj (x) = 1
(†)
Since spt ϕk ⊂ Uk assertion (1) is proved. Moreover, for each k ∈ N, Indeed (∗∗) is true for k = 1, and assuming it holds for k ∈ N, then adding (∗) and (∗∗) shows that (∗∗) holds for k + 1. If x ∈ Vk then ψk (x) = 1, and for each integer j ≥ k by (∗∗). In particular, ϕj (x) = 0 when x ∈ Vk and j > k; this fact is also implied by (∗). Since Vk is an open set, spt ϕj ∩ Vk = ∅ for each j > k.
(††)
Let x ∈ Ω. Then x ∈ O for some O ∈ O, and there is r ∈ Q+ with B(x, r) ⊂ O. Find y ∈ U (x, r/3)∩Qn , and note U (y, 2r/3) ⊂ O. Thus U (y, 2r/3) = Ukx for an integer kx ≥ 1. As Vkx = U (y, r/3), we see that x ∈ Vkx , and (†) implies � k∈N ϕk (x) = 1. Since Vkx is a neighborhood of x, assertion (3) follows from identity (††).
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© 2012 by Taylor & Francis Group, LLC
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166
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7. Bounded vector fields
A family {ϕk : k ∈ N} of nonnegative functions defined on Rn which satisfies conditions (1)–(3) of Proposition 7.4.4 is called a partition of unity subordinated to O. Our proof follows that of [64, Theorem 6.20]; for continuous partitions of unity, it applies to any separable metric space. Theorem 7.4.5. Let A ∈ BVloc (Rn ), and let v : cl∗ A → Rn be a bounded admissible vector field. Then � � divcl∗ A v(x) dx = v · νA dHn−1 (7.4.1) A
∂∗ A
whenever divcl∗ A v ∈ L1 (A) and v � ∂∗ A ∈ L1 (∂∗ A, Hn−1 ; Rn ).
Proof. If spt v is compact, choose an open ball U with spt v ⊂ U . Equality (7.4.1) holds trivially when A is replaced by A−U , and when A is replaced by A ∩ U , it holds according to Proposition 7.4.3. The theorem follows from Theorem 6.6.1. If v is arbitrary, cover cl∗ A by a family U of open balls, and choose a partition of unity {ϕk : k ∈ N} ⊂ Cc∞ (Rn ) subordinated to U. If k ∈ N and vk := (ϕk � cl∗ A)v, then � � divcl∗ A vk (x) dx = vk · νA dHn−1 (∗) A
∂∗ A
by Proposition 2.3.5 and the first part of the proof. Since the family {spt ϕk } � is locally finite and k∈N vk = v, we have � � �� vk (x) = divcl∗ A vk (x) divcl∗ A v(x) = divcl∗ A k∈N
k∈N
at each x ∈ int∗ A at which v is relatively differentiable. If divcl∗ A v and v � ∂∗ A belong, respectively, to L1 (A) and L1 (∂∗ A, Hn−1 ; Rn ), the theorem follows by summing up equalities (∗).
7.5. Lipschitz domains We show that BV functions in Lipschitz domains have boundary values. Using this result, we prove the divergence theorem for vector fields whose components are BV functions. Proposition 7.5.1. Let Ω ⊂ Rn be an open set, and let f, fk ∈ BV (Ω), k = 1, 2, . . . . If lim �fk − f �L1 (Ω) = 0 and lim �Dfk �(Ω) = �Df �(Ω), then � � lim v · Dfk = v · d(Df ) Ω n
Ω
for each bounded v ∈ C(Ω; R ).
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7.5. Lipschitz domains
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167
Proof. Choose ε > 0, and find an open set U � Ω so that �Df �(B) ≤ ε for B :=Ω − U . Then lim sup �Dfk �(B) ≤ �Df �(B) < ε by Proposition 5.5.4. Assume first that v ∈ C 1 (Ω; Rn ) and β = �v�L∞ (Ω;Rn ) < ∞. Choose a function ϕ ∈ Cc1 (Rn ) so that χU ≤ ϕ ≤ 1 and spt ϕ ⊂ Ω, and calculate: � � � v · Dfk = ϕv · Dfk + (1 − ϕ)v · Dfk Ω Ω Ω � � =− fk div (ϕv) + (1 − ϕ)v · Dfk B �Ω ≤− fk div (ϕv) + β�Dfk �(B), − lim
�
�
Ω
� fk div (ϕv) = − f div (ϕw) = ϕw · d(Df ) Ω Ω Ω � � = w · d(Df ) + (ϕ − 1)v · d(Df ) B �Ω ≤ v · d(Df ) + β�Df �(B). Ω
Combining these inequalities with (∗) and (∗∗) implies � � v · Dfk ≤ v · d(Df ) + 2βε. lim Ω
Ω
As the same inequality holds for −v, the arbitrariness of ε yields � � lim v · Dfk = v · d(Df ). Ω
Ω
If v ∈ C(Ω; Rn ) is bounded, use Corollary 1.1.2 to extend v � cl U to a vector field u ∈ Cc (Rn ; Rn ), and let � � β = max �v�L∞ (Ω;Rn ) , �u�L∞ (Rn ;Rn ) .
A mollification of u provides w ∈ C 1 (Rn ; Rn ) such that �w�L∞ (Rn ;Rn ) ≤ β and �v − w�L∞ (U ;Rn ) < ε. We obtain �� � � � � � v · d(Df ) − � w · d(Df ) � � Ω Ω � � � � � � ≤ |v − w| d �Df � + |v − w| d �Df � U
B
� � ≤ ε�Df �(Ω) + 2β�Df �(B) ≤ ε �Df �(Ω) + 2β ,
and by a similar calculation, � �� � � � � � � w · Dfk �� ≤ ε �Df �(Ω) + 2β . lim� v · Dfk − Ω
✐
Ω
© 2012 by Taylor & Francis Group, LLC
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168
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7. Bounded vector fields
The proposition follows from the arbitrariness of ε, since � � lim w · Dfk = w · d(Df ) Ω
Ω
by the first part of the proof applied to w � Ω.
Lemma 7.5.2. Let Ω ⊂ Rn be an open set. If f ∈ BV (Ω), then � � �Df � Ω ∩ B(x, r) =0 lim r→0 rn−1 for Hn−1 almost all x ∈ ∂Ω.
Proof. For β > 0, let Aβ be the set of all x ∈ ∂Ω such that � � �Df � Ω ∩ B(x, r) >β. lim sup rn−1 r→0 Given δ > 0 and x ∈ Aβ , find 0 < rx < δ with � � �Df � Ω ∩ B(x, rx ) > β rxn−1 .
By Vitali’s theorem there are xi ∈ Aβ such that the balls B(xi , rxi ) are disjoint � and Aβ ⊂ i B(xi , 5rxi ). Thus � n−1 (Aβ ) ≤ α(n − 1)(5rxi )n−1 H10δ i
n−1 −1
≤5
β
α(n − 1)
n−1 −1
�
where γ = 5 β α(n − 1) andΩ � Ω = ∅, we infer δ>0 δ
i
δ
� � �Df � Ω ∩ B(xi , rxi ) ≤ γ�Df �(Ωδ )
= Ω ∩ B(∂Ω, δ). As �Df �(Ω) < ∞ and
n−1 Hn−1 (Aβ ) = lim H10δ (Aβ ) ≤ γ lim �Df �(Ωδ ) = 0. δ→0
δ→0
The lemma follows, since � � � � ∞ � �Df � Ω ∩ B(x, r) x ∈ ∂Ω : lim sup > 0 = A1/k . rn−1 r→0 k=1
Theorem 7.5.3. Let Ω ⊂ Rn be a Lipschitz domain. If f ∈ BVloc (Ω), then � 1 � f (y) dy : ∂Ω → R (7.5.1) T f : x �→ lim �� r→0 Ω ∩ B(x, r)� Ω∩B(x,r)
is defined for Hn−1 almost all x ∈ ∂Ω and belongs to L1loc (∂Ω, Hn−1 ). If f ∈ BV (Ω) and ∂Ω is compact, then � � � (T f )v · νΩ dHn−1 = f (x) div v(x) dx + v · d(Df ) (7.5.2) ∂Ω
Ω
Ω
for each continuous v ∈ Adm(cl Ω; Rn ) with div v ∈ L∞ (Ω).
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7.5. Lipschitz domains
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169
Proof. For z ∈ Rn , let z � = πn (z) and zn = z · en . Fix x ∈ ∂Ω. After a suitable rotation about x, there are an open cylinder � � C(x; r, h) = πn U (x, r) × (xn − h, xn + h) and a Lipschitz function g : Rn−1 → R such that � � Ω ∩ C(x; r, h) = y ∈ C(x; r, h) : g(y � ) < yn .
Letting L := max{1, Lip g}, and making we may assume h = 2Lr. � r smaller, � If Cx,r := C(x; r, 2Lr) and Ux,r := πn U (x, r) , then � � � � � �g(y ) − xn � = �g(y � ) − g(x� )� ≤ (Lip g)|y � − x� | < Lr
for each y � ∈ Ux,r . It follows that for each 0 ≤ s < t ≤ r, the intersection Ω ∩ Cx,r contains the open set � � Cx,r (s, t) := y ∈ Cx,r : g(y � ) + s < yn < g(y � ) + t .
We employ the Lipschitz maps � � φ : y � �→ y � , g(y � ) : Ux,r → ∂Ω ∩ Cx,r ,
ψ : (y � , t) �→ (y � , g(y � ) + t) : Ux,r × (0, r) → Ω ∩ Cx,r (0, r),
σ : (y, t) �→ (y � , yn + t) : (∂Ω ∩ Cx,r ) × (0, r) → Ω ∩ Cx,r (0, r), and for 0 < t ≤ r define a translation � � σ t : y �→ y � , yn + t : ∂Ω ∩ Cx,r → Ω ∩ Cx,r (0, r).
Note that φ and ψ are �surjective lipeomorphisms whose Jacobians satisfy 1 ≤ Jφ ≤ β where β = 1 + (Lip g)2 , and Jψ = Jψ−1 = 1. The diagram Ux,r × (0, r) ��� ��� ψ ��� φ×id ��� �� � � Ω ∩ Cx,r (0, r) (∂Ω ∩ Cx,r ) × (0, r) σ
commutes. Given u : Ω → R and 0 < t < r, let ut := u ◦ σ t ; explicitly � � ut : y �→ u y � , g(y � ) + t : ∂Ω ∩ Cx,r → R. Part 1. f ∈ C 1 (Ω) ∩ BV (Ω).
Case 1a. Let 0 < s < t < r, and observe that � � � � s � � �� �f (y) − f t (y)� = ��f y � , g(y � ) + s − f y � , g(y � ) + t �� �� t � � � � ∂f � � � � =� y , g(y ) + τ dτ �� s ∂xn � t � t� � � ��� � � � �(Df )τ (y)� dτ ≤ �Df y , g(y � ) + τ � dτ = s
✐
s
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170
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7. Bounded vector fields
for each y ∈ ∂Ω ∩ Cx,r . In view of [29, Section 3.3.3, Theorem 2] and Fubini’s theorem, the preceding inequality implies �f s − f t �L1 (∂Ω∩Cx,r ,Hn−1 ) � � s � �f (y) − f t (y)� dHn−1 (y) = ∂Ω∩Cx,r
≤
=
�
�
∂Ω∩Cx,r
Ux,r
≤β =β
�
�
��
s
t�
� �(Df )τ (y)� dτ
�
dHn−1 (y)
�� t � � � �� � τ � � �(Df ) φ(y ) � dτ Jφ (y � ) dy � s
Ux,r ×(s,t) Cx,r (s,t)
(∗)
� � ��� � �Df ψ(y � , τ ) � dy � dτ
� � � � �Df (y)� d(y) = β�Df � Cx,r (s, t) .
� Since 0 0 so that Bρ� ⊂ Cx,r . Note ρ� is the least radius such that Cz,ρ ⊂ Bρ� , and observe Ω ∩ Bρ�� ⊂ Cz,ρ(0, ρ) ⊂ Ω ∩ Cz,ρ ⊂ Ω ∩ Bρ� , ∂Ω ∩ Cz,ρ ⊂ ∂Ω ∩ Bρ� .
(†)
If 0 < s < t < ρ, then calculating as in (∗), � � �f s − f t �L1 (∂Ω∩Cz,ρ ,Hn−1 ) ≤ β�Df � Cz,ρ(s, t) .
In view of (†), fixing t so that 0 < t < ρ, and letting s → 0, we obtain � �Tx f − f t �L1 (∂Ω∩Cz,ρ ,Hn−1 ) ≤ β�Df � Ω ∩ Bρ� ).
(††)
Combining (†), Fubini’s theorem, [29, Section 3.3.3, Theorem 2], and (††),
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7.5. Lipschitz domains �
Ω∩Bρ��
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171
� � �Tx f (z) − f (y)� dy ≤
=
= ≤ = ≤
�
�
�
Cz,ρ (0,ρ)
Uz,ρ
Uz,ρ
��
� � �Tx f (z) − f (y)� dy
g(y � )
� �� ρ � � � ��� � � �Tx f (z) − f y , g(y ) + t � dt dy �
� ρ ��
0
� � � ��� � �Tx f (z) − f t φ(y � ) �Jφ (y � ) dy dt
0
Uz,ρ
0
∂Ω∩Cz,ρ
� ρ ��
� ρ �� 0
�
+ � ≤ρ
∂Ω∩Cz,ρ
ρ
0
� � �Tx f (z) − f (y � , t)� dt dy �
g(y � )+ρ �
� � � �Tx f (z) − f t (y)� dHn−1 (y) dt
� � � �Tx f (z) − Tx f (y)� dHn−1 (y) dt
�Tx f − f t �L1 (∂Ω∩Cz,ρ ,Hn−1 ) dt
∂Ω∩Bρ�
� � �Tx f (z) − Tx f (y)� dHn−1 (y)
+ ρβ�Df �(Ω ∩ Bρ� ).
Case 1c. Select a continuous v ∈ Adm(cl Ω; Rn ) so that div v belongs to L (Ω) and spt v ⊂ Cx,r . Choose 0 < t < r and let ∞
C t = Ω ∩ Cx,r − Cx,r (0, t).
Now f v � C t is continuous, bounded, and admissible, and div(f v) belongs to L1 (Ω). By Proposition 7.4.3, � � � div(f v) = (f v) · νC t dHn−1 = (f v) · νC t dHn−1 Ct
=
�
∂C t
σ t (∂Ω∩Cx,r )
(f v)t νΩ dHn−1 . ∂Ω∩Cx,r
The last equality holds, since (f v)t = (f v) ◦ σ t where σ t is the translation by t in the direction en . Because v is continuous, limt→0 v t (y) = v(y) for all y ∈ ∂Ω ∩ Cx,r . Letting t → 0, equality (∗∗) and the generalized dominated convergence theorem [29, Section1.3, Theorem 4] yield � � div(f v) = (Tx f ) v · νΩ dHn−1 . Ω∩Cx,r
∂Ω∩Cx,r
As spt v ⊂ Cx,r , the previous equality expands to � � � f div v + v · Df = (Tx f ) v · νΩ dHn−1 . Ω
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7. Bounded vector fields
Part 2. f ∈ BVloc (Ω).
Case 2a. By Theorem 5.6.3 there is a sequence {fk } in C 1 (Ω) such that lim �fk − f �L1 (Ω) = 0 and lim �Dfk �(Ω) = �Df �(Ω). Fix 0 < ε < r, and for each y ∈ ∂Ω ∩ Cx,r , define � 1 ε Fk (y) := (fk )t (y) dt. ε 0 By Fubini’s theorem and inequality (∗ ∗ ∗), � � � �Tx fk (y) − Fk (y)� dHn−1 (y) ∂Ω∩Cx,r
� � ε � �1 � � � t � = Tx fk (y) − (fk ) (y) dt�� dHn−1 (y) � ∂Ω∩Cx,r ε 0 � 1 ε �Tx fk − (fk )t �L1 (∂Ω∩Cx,r ,Hn−1 ) dt ≤ ε 0 � � � � � β ε ≤ �Dfk � Cx,r (0, t) dt ≤ β�Dfk � Cx,r (0, ε) . ε 0 �
In addition, [29, Section 3.3.3, Theorem 2] and Fubini’s theorem yield � � � �Fk (y) − Fj (y)� dHn−1 (y) ∂Ω∩Cx,r
1 = ε =
1 ε
β ≤ ε =
β ε
� �
∂Ω∩Cx,r
Ux,r
� �
��
0
� � �(fk )t (y) − (fj )t (y)� dt dHn−1 (y)
ε�
�� ε � � � � � �� � t � t � � φ(y φ(y ) ) − (f ) ) dt Jφ (y � ) dy � �(fk � j 0
Ux,r ×(0,ε)
� � � � ��� � �fk ψ(y � , t) − fj ψ(y � , t) � dy � dt
� � �fk (y) − fj (y)� dy ≤ β �fk − fj �L1 (Ω) . ε Cx,r (0,ε)
Combining the previous two inequalities with
|Tx fk − T fj | ≤ |Tx fk − Fk | + |Fk − Fj | + |Fj − Tx fj |, we obtain �Tx fk − Tx fj �L1 (∂Ω∩Cx,r ,Hn−1 )
� � β � � ≤ β�Dfk � Cx,r (0, ε) + �fk − fj �L1 (Ω) + β�Dfj � Cx,r (0, ε) ε for k, j = 1, 2, . . . and 0 < ε < r. Now choose η > 0. Since �� � Ω ∩ cl Cx,r (0, ε) : 0 < ε < r = ∅,
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173
� � there is 0 < ε < r with �Df � Ω ∩ �cl Cx,r (0, ε) < �η. Applying Proposition 5.5.4, find p ∈ N so that �Dfk � Ω ∩ cl Cx,r (0, ε) < η for each integer k ≥ p. Making the integer p larger, we may assume �fk − fj �L1 (Ω) ≤ εη whenever k, j ≥ p. Thus for all integers k, j ≥ p, �Tx fk − Tx fj �L1 (∂Ω∩Cx,r ,Hn−1 ) ≤ 3βη. As η is arbitrary, there is Tx f ∈ L1 (∂Ω ∩ Cx,r , Hn−1 ) such that lim �Tx fk − Tx f �L1 (∂Ω∩Cx,r ,Hn−1 ) = 0.
(†††)
Case 2b. Choose z ∈ ∂Ω ∩ Cx,r so that the following holds Tx f (z) = lim Tx fk (z), 1 |Ω ∩ Bs | = α(n), n s 2 Hn−1 (∂Ω ∩ Bs ) lim = α(n − 1), s→0 sn−1 �Df �(Ω ∩ Bs ) lim = 0, s→0 sn−1 � � � 1 �Tx f (z) − T f (y)� dHn−1 (y) = 0. lim n−1 s→0 H (Bs ∩ ∂Ω) Bs ∩∂Ω lim
s→0
(1) (2) (3) (4) (5)
In view of (†††), Corollary 6.4.4, and Lemma 7.5.2, equalities (1)–(4) hold for Hn−1 almost all z ∈ ∂Ω ∩ Cx,r . The same is true for equality (5). Indeed, applying Corollary 6.2.4 to the Radon measure Hn−1 ∂Ω and to the zero extension Tx f of Tx f , we infer that Hn−1 almost all z ∈ ∂Ω∩Cx,r are Lebesgue points of Tx f . Define ρ,ρ � , and ρ�� so that conditions (†) are satisfied. By Case 1b, � � � � � � �Tx fk (z) − fk (y)� dy ≤ ρ �Tx fk (z) − Tx fk (y)� dHn−1 (y) Ω∩Bρ��
∂Ω∩Bρ�
+ ρβ�Dfk �(Ω ∩ Bρ� )
for k = 1, 2, . . . . If k → ∞, condition (1) and Proposition 5.5.4 yield � � � � � � �Tx f (z) − f (y)� dy ≤ ρ �Tx f (z) − Tx f (y)� dHn−1 (y) Ω∩Bρ��
∂Ω∩Bρ�
+ ρβ�Df �(Ω ∩ Bρ� ).
Using this inequality and equalities (2)–(5), we calculate � � � 1 �Tx f (z) − f (y)� dy = 0 lim �� ρ →0 |Ω ∩ Bρ�� | Ω∩B �� ρ
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174
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7. Bounded vector fields
for Hn−1 almost all z ∈ ∂Ω ∩ Cx,r . For all such points z, � 1 Tx f (z) = lim f (y) dy s→0 |Ω ∩ Bs | Ω∩B s by Remark 6.2.5. As the right side of the previous equality does not depend on x, neither does Tx f (z). Hence we write T f instead of Tx f , and the first part of the theorem is proved. Case 2c. Assume that f ∈ L1 (Ω) and ∂Ω is compact. Then T f belongs to L1 (∂Ω, Hn−1 ). Select v ∈ C(cl Ω; Rn ) ∩ Adm(cl Ω; Rn ) such that div v belongs to L∞ (Ω), and observe that f div v, v, and (T f ) v · νΩ belong to L1 (Ω), L1 (Ω, Df ; Rn ), and L1 (∂Ω, Hn−1 ), respectively. If some Cx,r contains spt v, then by Case 1c, � � � fk div v + v · Dfk = (T fk ) v · νΩ dHn−1 Ω
Ω
∂Ω
for k = 1, 2, . . . . In view of Theorem 2.3.9, this is still true if spt v � Ω. We infer from Case 2a and Proposition 7.5.1 that � � � f div v + v · d(Df ) = (T f ) v · νΩ dHn−1 Ω
Ω
∂Ω
whenever spt v � Ω or spt v ⊂ Cx,r for some x ∈ ∂Ω. Note that we must use Proposition 7.5.1; as spt v ⊂ Cx,r does not imply v ∈ Cc (Ω; Rn ), we cannot use Proposition 5.5.4. Let {ϕj : j ∈ N} ⊂ Cc∞ (Rn ) be a partition of unity subordinated to the family {Ω} ∪ {Cx,rx : x ∈ ∂Ω} of open sets; see Proposition 7.4.4. Each vj = v(ϕj � cl Ω) is admissible by Proposition 2.3.5, and either spt vj ⊂ Cx,rx for some x ∈ ∂Ω, or spt vj � Ω. Thus � � � f div vj + vj · d(Df ) = (T f ) vj · νΩ dHn−1 Ω
Ω
∂Ω
�
for j = 1, 2, . . . . Since v(x) = j∈N vj (x) for each x ∈ cl Ω, and since � div v(x) = j∈N div vj (x) for each x ∈ Ω at which v is differentiable, equality (7.5.2) follows: � � � � ��� f div v + v · d(Df ) = f div vj + vj · d(Df ) Ω
Ω
j∈N
=
��� j∈N
=
�
Ω
Ω
∂Ω
Ω
(T f ) vj · νΩ dHn−1
�
(T f ) v · νΩ dHn−1 .
The function T f defined in Theorem 7.5.3 is called the trace of f . We view T f as the boundary value of f .
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175
Lemma 7.5.4. Let Ω ⊂ Rn be an open set, f ∈ BV (Ω), and ϕ ∈ C 1 (Ω). If ϕ and Dϕ are bounded, then f ϕ ∈ BV (Ω) and � � �D(f ϕ)�(Ω) ≤ �ϕ�L∞ (Ω) �Df �(Ω) + �f �L1 (Ω) �Dϕ�L∞ (Ω) .
Proof. By Theorem 5.6.3, there is a sequence {fi } in C ∞ (Ω) ∩ BV (Ω) such that lim �fi − f �L1 (Ω) = 0 and lim �Dfi �(Ω) = �Df �(Ω). It follows that lim �fi ϕ − f ϕ�L1 (Ω) = 0, and Proposition 5.1.7 implies � � � V (f ϕ ,Ω) ≤ lim inf �D(fi ϕ)�(Ω) = lim inf |ϕDfi + fi Dϕ| Ω � � ≤ lim inf �ϕ�L∞ (Ω) �Dfi �(Ω) + �fi �L1 (Ω) �Dϕ�L∞ (Ω) = �ϕ�L∞ (Ω) �Df �(Ω) + �f �L1 (Ω) �Dϕ�L∞ (Ω) < ∞.
In particular, f ϕ ∈ BV (Ω). Theorem 7.5.5. Let Ω ⊂ Rn be a Lipschitz domain with compact boundary. There is a constant γ > 0 such that �T f �L1 (∂Ω,Hn−1 ) ≤ γ�f �BV (Ω) for each f ∈ BV (Ω). Proof. Since |T f | ≤ T |f |, we may assume f ≥ 0. Choose x ∈ ∂Ω, and find a cylinder Cx = C(x; r, h) and a Lipschitz function g : Rn−1 → R such that � � Ω ∩ Cx = y ∈ Cx : g(y � ) < yn � where yn = y · en and y � = πn (y). Let β := 1 + (Lip g)2 , and deduce from Corollary 6.4.4, (1) that −en ·νΩ (y) ≥ 1/β for Hn−1 almost all y ∈ ∂Ω∩Cx . If spt f ⊂ Cx , then spt T f ⊂ ∂Ω ∩ Cx by (7.5.1). Thus applying equality (7.5.2) to f and v = −en , we obtain � � n−1 T f dH ≤β (T f )(−en ) · νΩ dHn−1 ∂Ω ∂Ω � = β (−en ) · d(Df ) ≤ β�Df �(Ω). Ω
If f ∈ BV (Ω) is arbitrary, let {ϕk : k ∈ N} ⊂ Cc∞ (Rn ) be a partition of unity subordinated to the family {Cx : x ∈ ∂Ω}. For k = 1, 2, . . . , � � � T (ϕk f ) dHn−1 ≤ β �D(ϕk f )�(Ω) (∗) ∂Ω ≤ �Df �(Ω) + �f �L1 (Ω) �Dϕk �L∞ (Ω) by Lemma 7.5.4 and the first part of the proof. Since ∂Ω is compact, and since the family {spt ϕk : k ∈ N} is locally finite, there is p ∈ N and a neighborhood U of ∂Ω such that spt ϕk ∩ U = ∅ for all integers k > p. By equality (7.5.1), g �→ T g : BV (Ω) → L1 (∂Ω, Hn−1 )
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7. Bounded vector fields
is a linear map. Thus by inequality (∗), � p � p � � � � �D(ϕk f )�(Ω) T f dHn−1 = T (ϕk f ) dHn−1 ≤ β ∂Ω
k=1
∂Ω
i=1
≤ βp�Df �(Ω) + β�f �L1 (Ω) �
≤β p+
p � i=1
�
p � i=1
�Dϕi �L∞ (Ω)
�Dϕi �L∞ (Ω) �f �BV (Ω)
� � �p and γ = β p + i=1 �Dϕi �L∞ (Ω) is the desired constant.
The following example shows that in Theorem 7.5.5, the compactness of the boundary ∂Ω cannot be relaxed to P(Ω) < ∞.
Example 7.5.6. Assume n ≥ 2, and let s = 2/(2n − 3). Choose a sequence {xk } in Rn so that |xj −xk | ≥ 2 whenever j �= k, and let Uk = U (xk , rk ) where � rk = k −s . The open balls Uk are disjoint, and Ω= k∈N Uk is a Lipschitz domain such that |Ω| < ∞ and P(Ω) < ∞. Letting f=
∞ �
k s χ Uk ,
k=1
we calculate V(f, Ω) = 0 and �f �L1 (Ω) < ∞. Thus f ∈ BV (Ω). On the other hand, from (7.5.1) we see that for almost all x ∈ ∂Ω, � k s χ∂Uk (x) T f (x) = k=1
and consequently �T f �L1 (∂Ω,Hn−1 ) = ∞. Proposition 7.5.7. Let Ω ⊂ Rn be a Lipshitz domain with compact boundary. Assume f ∈ BV (Ω) and g ∈ BV (Rn − clΩ) , and let h : Rn → R be such that h � Ω = f and h � (Rn − clΩ ) = g. Then h ∈ BV (Rn ) and � |T f − T g| dHn−1 . �Dh�(Rn ) = �Df �(Ω) + �Dg�(Rn − clΩ ) + ∂Ω
Cc1 (Rn ; Rn ),
and apply equalProof. Note R − cl Ω = ext∗ Ω. Select v ∈ ity (7.5.2) to Ω and ext∗ Ω. As ∂(ext∗ Ω) = ∂Ω and νext∗ Ω = −νΩ , we obtain � � � h div v dx = − f div v dx − g div v dx − Rn Ω ext∗ Ω � � = v · d(Df ) − (T f ) v · νΩ dHn−1 (∗) Ω ∂Ω � � + v · d(Dg) + (T g) v · νΩ dHn−1 . n
ext∗ Ω
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7.5. Lipschitz domains
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177
If �v�L∞ (Rn ;Rn ) ≤ 1, then � � h div (−v) dx ≤ �Df �(Ω) + �Dg�(ext∗ Ω) + Rn
∂Ω
|T g − T f | dHn−1 .
n
The arbitrariness of v implies h ∈ BV (R ) and � �Dh�(Rn ) ≤ �Df �(Ω) + �Dg�(ext∗ Ω) +
∂Ω
|T g − T f | dHn−1 .
(∗∗)
� � Now h ∈ BV (Rn ) implies Rn v · d(Dh) = − Rn h div v dx. By equality (∗), � � � v · d(Dh) = v · d(Df ) + v · d(Dg) Rn Ω ext∗ Ω (†) � n−1 + (T g − T f ) v · νΩ dH ∂Ω
for each v ∈
Cc1 (Rn ; Rn ).
In particular, � � v · d(Dh) = v · d(Df ) Ω
Ω
when spt v ⊂ Ω. Thus viewing Df as a measure in Rn that lives in Ω, Proposition 5.3.11 shows that Dh Ω = Df . Similarly, viewing Dg as a measure in Rn that lives in ext∗ Ω, we show that Dh ext∗ Ω = Dg. Denoting the zero extension of T g − T f by T g − T f , equality (†) translates to � � � v · d(Dh) = v · d(Dh Ω) + v · d(Dh ext∗ Ω) Rn Rn Rn � � � + T g − T f v · νΩ dHn−1 Rn
for every v ∈ Cc1 (Rn ; Rn ). Another application of Proposition 5.3.11 yields � � T g − T f νΩ . Dh = Dh Ω + Dh ext∗ Ω + Hn−1 � � T g − T f νΩ , and consequently We infer Dh ∂Ω = Hn−1 � (T g − T f ) νΩ dHn−1 . Dh(∂Ω) = ∂Ω
According to Proposition 5.3.7, �Dh�(∂Ω) =
�
∂Ω
|T g − T f | dHn−1
and the proposition follows from Observation 5.3.12: � �Df �(Ω) + �Dg�(ext∗ Ω) + |T g − T f | dHn−1 ∂Ω
= �Dh�(Ω) + �Dh�(ext∗ Ω) + �Dh�(∂Ω) = �Dh�(Rn ).
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Corollary 7.5.8. Let Ω ⊂ Rn be a Lipschitz domain with compact boundary, and let f ∈ BV (Ω). The zero extension f of f belongs to BV (Rn ), and there is a constant κ > 0 independent of f such that � � �f � ≤ κ�f �BV (Ω) . BV (Rn ) Proof. Proposition 7.5.7 and Theorem 7.5.5 imply f ∈ BV (Rn ) and � � � � � � �f � = �f �L1 (Rn ) + �Df �(Rn ) BV (Rn ) = �f �L1 (Ω) + �Df �(Ω) + �T f �L1 (∂Ω,Hn−1 )
≤ �f �BV (Ω) + γ�f �BV (Ω) = (γ + 1)�f �BV (Ω) where γ > 0 is a constant independent of f . Remark 7.5.9. Corollary 7.5.8 generalizes Theorem 5.8.3. It follows that Corollary 5.8.4, and hence Lemma 5.9.2, holds when Ω is a bounded Lipschitz domain. The same is true about Poincar´e’s inequality (Theorem 5.9.11) and the relative isoperimetric inequality (Theorem 5.9.12), except thatΩ r may not be defined. Lemma 7.5.10. If g ∈ BV (Rn ) then lim �g − gχB(0,k) �BV (Rn ) = 0. Proof. If Uk = Rn − B(0, k) then g − gχBk = gχUk , and we may assume that g ≥ 0. Since for k = 1, 2, . . . , the unit exterior normal νUk : x �→ −x|x|−1 : ∂Uk → Rn
is Lipschitz with Lip νUk ≤ 2, it can be extended by Proposition 1.5.2 to a Lipschitz vector field vk : Rn → R so that √ �vk �L∞ (Rn ;Rn ) ≤ 1 and Lip vk ≤ 2 n; in particular �div vk �L∞ (Rn ) ≤ 2n3/2 . According to equality (7.5.2), � � � � n−1 T (g � Uk ) vk · νUk dHn−1 T (g � Uk ) dH = ∂Uk ∂U � k � = g(x) div vk (x) dx + vk · d(Dg) Uk Uk � ≤ 2n3/2 g(x) dx + �Dg�(Uk ). Since
�∞
k=1
Uk
Uk = ∅ and Proposition 7.5.7 implies � T (g � Uk ) dHn−1 , �D(gχUk )�(Rn ) = �Dg�(Uk ) + ∂Uk
the lemma follows.
The next proposition is the predicted generalization of Theorem 5.10.2, and by extrapolation, of Corollary 5.10.3.
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7.6. BV vector fields
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179
Proposition 7.5.11. Let φ : Rn → Rn be a Lipschitz map. If g belongs to BV (Rn ), then so does φ# g and �Dφ# g�(Rn ) ≤ (Lip φ)n−1 �Dg�(Rn ). Proof. For k = 1, 2, . . . , let Bk = B(0, k) and gk = gχBk . From (5.10.1) and (5.10.2) we see that lim φ# gk (y) = φ# g(y) for each y ∈ Rn , and sup �φ# gk �L1 (Rn ) ≤ (Lip φ)n �g�L1 (Rn ) .
Thus φ# g = lim φ# gk in L1 (Rn ), and the desired inequality V(φ# g) ≤ lim inf �Dφ# gk �(Rn )
≤ (Lip φ)n−1 lim �Dgk �(Rn ) = (Lip φ)n−1 �Dg�(Rn )
follows from Proposition 5.1.7, Theorem 5.10.2, and Lemma 7.5.10.
7.6. BV vector fields For an open setΩ ⊂ Rn define the linear space BV (Ω; Rn ) in the obvious way. Select v = (v1 , . . . , vn ) in BV (Ω; Rn ), and note that div v :=
n �
Di vi
i=1
is a signed measure in Ω; see Section 5.5. If µ+ and µ− are, respectively, the positive and negative parts of div v, let Ldiv v := Lµ+ − Lµ− where Lµ± are the distributions defined in Example 3.1.2, (2). Since for each test function ϕ ∈ D(Ω), Theorem 5.5.1 implies � n � � Ldiv v (ϕ) = ϕ d(div v) = ϕ d(Di vi ) Ω
=−
n � � i=1
i=1
Ω
vi Di ϕ = −
Ω
�
Ω
v · Dϕ
we see that Ldiv v is the distributional divergence Fv of v defined in Example 3.1.2, (3). If Ω is a Lipschitz domain, we define a vector field T v := (T v1 , . . . , T vn ), in L1 (∂Ω, Hn−1 ; Rn ), called the trace of v; see Theorem 7.5.3. Theorem 7.6.1. Let Ω ⊂ Rn be a Lipschitz domain with compact boundary. If w ∈ BV (Ω; Rn ), then � (T w) · νΩ dHn−1 . div w(Ω) = ∂Ω
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Proof. Substituting f := wi and v := ei into equality (7.5.2), we obtain � � ei · d(Dwi ) = (T wi )(ei · νΩ ) dHn−1 Di wi (Ω) = Ω
∂Ω
for i = 1, . . . , n. Summing up these equalities completes the proof.
Remark 7.6.2. Theorem 7.5.3 and its consequences hold for sets that are more general than Lipschitz domains [75, Definition 5.10.1]. The largest family of such sets is defined abstractly in [1, Definition 3.20]; its elements are called the extension domains. Remark 7.6.3. Let v ∈ L∞ (Rn ; Rn ) be a vector field whose distributional divergence is a signed measure µ; explicitly, let v be such that � � ϕ dµ = − v · Dϕ Rn
Rn
for each ϕ ∈ D(Rn ). It has been shown that for any bounded set A ∈ BV(Rn ), normalized by the condition A = cl∗ A, the vector field v has a “trace” T v which belongs to L∞ (∂∗ A, Hn−1 ; Rn ) and satisfies � (T v) · νA dHn−1 . µ(A) = ∂∗ A
This result is relatively recent. It is due to G.-Q. Chen, M. Torres, and W.P. ˇ Ziemer [15, 16], and independently to M. Silhav´ y [74], who proved it by a different method.
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Chapter 8 Unbounded vector fields
We show that the divergence theorem is still valid for admissible vector fields that are unbounded along compact sets whose upper Minkowski contents in a codimension larger than one are finite. The growth of the vector field must be controlled proportionately to the Minkowski content of the exceptional sets.
8.1. Minkowski contents Let E ⊂ Rn and t ≥ 0. The t-dimensional upper Minkowski content of E is the extended real number � � �B(E, r)� ∗t M (E) := lim sup .1 rn−t r→0 In general, the function M∗ t : E Example 8.1.3. Notwithstanding, information about subsets of Rn . M∗ t (E) = M∗ t (cl E) and M∗ t (E) compact set K ⊂ Rn , we have
� → M∗ t (E) is not a measure in Rn ; see upper Minkowski contents provide useful We apply it only to compact sets, since = ∞ whenever E is not bounded. For a
M∗ 0 (K) = α(n)H0 (K)
and
if t > n then M∗ t (K) = 0.
M∗ n (K) = |K|;
Remark 8.1.1. In order to achieve the equality Ht (E) = M∗ t (E) for some special sets, the definition of M∗ t (E) is often presented with a normalizing constant depending on n and t; see [46, Section 5.5] and[33, Section 3.2.37]. For our purposes this is not necessary, since we will be interested only in distinguishing the cases M∗ t = 0,
0 < M∗ t (E) < ∞,
M∗ t (E) = ∞.
Proposition 8.1.2. For each compact set K ⊂ Rn and each t ≥ 0, Ht (K) ≤ 2t
α(t) ∗ t M (K). α(n)
1 Replacing “lim sup” by “lim inf” defines the t-dimensional lower Minkowski content Mt∗ (E) of E — a concept not used in this book.
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Proof. It suffices to assume that K �= ∅ and 0 < t < n. For r > 0, the covering number N (K, r) is the least p ∈ N such that K can be covered by p closed balls of radius r, and the packing number P (K, r) is the largest q ∈ N such that there are q disjoint closed balls of radius r centered at points of K. � � Claim. N (K, 2r) ≤ P (K, r) and P (K, r)α(n)rn ≤ �B(K, r)�.
Proof. Let N := N (K, 2r) and P := P (K, r). If N > P , there are �P disjoint balls B(xi , r) with xi ∈ K, and x ∈ K − i=1 B(xi , 2r). Thus the balls B(x1 , r), . . . , B(xP , r), B(x, r) are disjoint, a contradiction. Since �P i=1 B(xi , r) ⊂ B(K, r), the second inequality follows. Choose δ > 0, and cover K by N (K, 2δ) closed balls of radius 2δ. Using the claim, we calculate α(t) t−n t H5δ δ (K) ≤ α(t)N (K, 2δ)(2δ)t ≤ 2t P (K,δ )α(n)δ n α(n) � � �B(K,δ )� t α(t) · ≤2 , α(n) δ n−t
and the proposition follows by letting δ → 0.
Example 8.1.3. Assume n = 2 and 0 < t < 2. Let x0 := (0, 0), and for j ∈ N, let xj := (1/j, 0) and rj := 12 |xj − xj+1 |. Consider compact sets K := {x0 , x1 , . . . } and Kj = [0, 1/j] × {0}. Given 0 < r ≤ 1/4, there is a unique j ∈ N with rj+1 < r ≤ rj . Since � � � � � � j �� j �� �B(K, r)� �B(Kj , rj )� � � B(xk , rj+1 )� B(xk , rj )� ≤ ≤ + , (rj )2−t r2−t (rj+1 )2−t (rj+1 )2−t k=1
k=1
∗ 1/2
(K) < ∞, and � � � �B(K, r)� 0 if t > 1/2, ∗t = M (K) = lim sup 2−t r r→0 ∞ if t < 1/2. � � Note that M∗ t {xj } = 0 for j = 0, 1, . . . , and that Ht (K) = 0.
we calculate 0 < M
Example 8.1.4. Let I ⊂ Rn be a closed segment of unit length. As the upper Minkowski volume is invariant with respect to rotations and translations, we may assume that � � I = (0, . . . , 0, t) ∈ Rn : 0 ≤ t ≤ 1 .
Let I0 and I1 be disjoint closed subsegments of I, each of length 1/3, such that I − (I0 ∪ I1 ) is an open segment, necessarily in the middle of I and of length 1/3. For k ∈ N, define recursively disjoint closed segments Ii1 ···ik , ij = 0, 1, each of length 3−k , so that Ii1 ···ik+1 ⊂ Ii1 ···ik and Ii1 ···ik − (Ii1 ···ik 0 ∪ Ii1 ···ik 1 )
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183
is an open segment, necessarily in the middle of Ii1 ···ik and of length 3−k−1 . The Cantor ternary discontinuum [62, Section 2.44] in I is the intersection �∞ D = k=1 Dk where � Dk := {Ii1 ···ik : ij = 0, 1 for j = 1, . . . , k}.
Furthermore, denote by Ii∗1 ···ik the closed segment of length 3 · 3−k that contains Ii1 ···ik in the middle, and let � Dk∗ := {Ii∗1 ···ik : ij = 0, 1 for j = 1, . . . , k}. � � For E ⊂ I and r > 0, let C(E, r) := πn B(0, r) × E. Given 0 < r < 1/3, find k ∈ N with 3−k−1 ≤ r < 3−k . The inclusions C(Dk+1 , 3−k−1 ) ⊂ B(D, r) ⊂ B(Dk , r) ⊂ C(Dk∗ , 3−k )
imply
� � � � �B(D, r)� �C(Dk+1 , 3−k−1 )� ≤ ≤ � �n−t rn−t 3−k
� � �C(D∗ , 3−k )� k � �n−t . 3−k−1
By a direct calculation, � � �C(Dk+1 , 3−k−1 )� = 2k+1 · 3−k−1 · (3−k−1 )n−1 α(n − 1), � � � � �C(Dk∗ , 3−k )� ≤ 2k · (3 · 3−k ) · 3−k n−1 α(n − 1).
Now let t := log 2/ log 3. Since 3t = 2, we obtain � � �B(D, r)� 3n+1 2 α(n − 1). α(n − 1) ≤ ≤ n n−t 3 r 2
As this inequality holds for all 0 < r < 1/3,
3n+1 2 ∗t α(n − 1). α(n − 1) ≤ M (D) ≤ 3n 2 We note that Ht (D) = α(t)/2t for t := log 2/ log 3. A proof of this equality involving only the definition of Ht is given in [30, Theorem 1.14]. Proposition 8.1.5. If K ⊂ Rn is a compact set, then B(K, r) belongs to BVc (Rn ) for L1 almost all r > 0, and � � P B(K, r) ≤ (n − t)M∗ t (K) lim inf r→0 rn−1−t for 0 ≤ t ≤ n − 1. Proof. Select r > 0, and observe that the function � � f : x �→ min r, dist(x, K) : Rn → [0, r]
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is Lipschitz with Lip (f ) ≤ 1. Clearly, {f ≤ s} = B(K, s) for 0 < s < r, and {f ≤ s} = Rn for s ≥ r. By the coarea theorem (Theorem 5.7.3), � � r � � � � � � �Df (x)� dx ≤ �B(K, r)�. P B(K, s) ds = P (r) := �
�
0
B(K,r)
Thus� P B(K,� s) < ∞ for L1 almost all 0 < s < r, and since r is arbitrary and �B(K, r)� < ∞, the first claim of the proposition is satisfied. As there is nothing more to prove otherwise, we assume M∗ t (K) < ∞. This implies limr→0 P (r)rt+1−n = 0. Given ε > 0, find δ > 0 so that � � �B(K, s)� ≤ M∗ t (K) + ε sn−t for all 0 < s ≤ δ. Integrating by parts, we obtain � � r � r � P B(K, s) t+1−n ds = P (r)r − (t + 1 − n) P (s)st−n ds n−1−t s 0 0 � � � � r �� �B(K, r)� B(K, s)� ≤r + (n − 1 − t) ds rn−t sn−t 0 � � ≤ r(n − t) M∗ t (K) + ε for each 0 < r ≤ δ. Hence � � � � � P B(K, r) 1 r P B(K, s) lim inf ≤ lim inf ds r→0 r→0 r 0 rn−1−t sn−1−t � � ≤ (n − t) M∗ t (K) + ε and the proposition follows from the arbitrariness of ε.
Let A ∈ BV(Rn ). For a compact set K ⊂ Rn and t ≥ 0, the t-dimensional upper Minkowski content of K relative to ∂∗ A is the extended real number � � Hn−1 B(K, r) ∩ ∂∗ A ∗t . M (K,∂ ∗ A) := lim sup rn−1−t r→0 Since Hn−1 (∂∗ A) < ∞, it is clear that for t ≥ n − 1,
M∗ t (K,∂ ∗ A) = Ht (K ∩ ∂∗ A).
Example 8.1.7 below shows that in general M∗ t (K,∂ ∗ A) differs from both Ht (K ∩ ∂∗ A) and M∗ t (K ∩ ∂∗ A), even when K ⊂ ∂∗ A. This difference cannot be eliminated by introducing a normalizing constant mentioned in Remark 8.1.1. Proposition 8.1.6. If A ∈ BV(Rn ), then the following is true � � (1) M∗ 0 {x}, ∂∗ A = 0 for Hn−1 almost all x ∈ Rn − ∂∗ A, � � (2) M∗ 0 {x}, ∂∗ A > 0 for each x ∈ ∂∗ A, � � (3) M∗ 0 {x}, ∂∗ A = α(n − 1) for each x ∈ ∂ ∗ A.
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Proof. Since ∂∗ A = ∂∗ (Rn − A) and
� � Hn−1 B(x, r) ∩ ∂∗ A , M {x}, ∂∗ A = lim sup rn−1 r→0 � � we see that M∗ 0 {x}, ∂∗ A = 0 if and only if x ∈ intc A ∪ intc (Rn − A). By Proposition 7.3.1, the sets intc A∪intc (Rn −A) and Rn −∂∗ A differ by an Hn−1 negligible set. This establishes claim (1). In view of Theorem 6.5.2, claims (2) and (3) follow, respectively, from Lemma 5.9.14 and Corollary 6.4.4, (3). ∗0
�
�
Example 8.1.7. Assume n = 2. For k = 1, 2, . . . , let � � Ak := x ∈ R2 : (k + εk )−2 ≤ |x| ≤ k −2 �∞ where 0 < εk < 1 and lim εk = ε. Then A = k=1 Ak is a BV set in R2 , and after some calculation we obtain Θ(A, 0) = ε. It follows that � � � � � 1 if 0 < ε < 1 and t = 0, ∗t ∗t M {0} ∩ ∂∗ A = H {0} ∩ ∂∗ A = 0 otherwise. On the other hand, for any value of ε we calculate ∞ if 0 ≤ t < 1/2, � � ∗t M {0}, ∂∗ A = 4π if t = 1/2, 0 if t > 1/2.
8.2. Controlled vector fields We consider unbounded vector fields v whose growth to infinity is controlled by an appropriate power of the distance from a compact set K ⊂ Rn . The controlling power is determined by M∗ t (K). In essence, we extrapolate from the fact that the function x �→| x|t is locally integrable in Rn when t > −n; cf. Example 8.2.9. Let A ⊂ E ⊂ Rn . Given f, g : E → R+ , we abbreviate lim
f (x) 0, then � � �φ(x)� = o(1)dist(x, Kj )tj +1−n as x → Kj .
Remark 8.2.2. In Definition 8.2.1, each Kj is called a singular set of φ, and tj is called the size of Kj . Note that the size of a singular set K is not determined by K alone. The upper Minkowski dimension of K � � dim K := inf s ≥ 0 : M∗ s (K) < ∞ ,
which is the optimal candidate for the size of K, can be used only when M∗ dim K (K) < ∞; see [46, Sections 4.8, 5.3 and 5.5].
Let φ : E → Rm be a controlled map. Since Hn−1 (E∞ ) = 0 by Proposition 8.1.2, it follows from Remark 2.3.3 that φ is Hn−1 measurable whenever E is Hn−1 measurable. Moreover, the restriction φ � (E − E∞ ) is Lipschitz at almost all x ∈ E − E∞ . As E∞ is a closed negligible set, the map φ is Lipschitz at almost all x ∈ E, and hence relatively differentiable at almost all x ∈ E by Theorem 7.2.3 and Corollary 4.4.4. Clearly, φ is admissible if and only if E ∩ E∞ = ∅, or equivalently, if and only if φ is locally bounded.
Proposition 8.2.3. Let E ⊂ Rn , and let φ : E → Rm be a controlled map. The family {Kj } of all singular sets of φ is locally finite. In particular, {Kj } is finite whenever E is bounded.
Proof. By Definition 8.2.1, (2) and (3), each singular set Kj has a neighborhood U such that φ is bounded at each x ∈ cl E ∩ (U − Kj ). Thus U ∩ Ki = ∅ whenever i �= j. Choose x ∈ cl E and B = B(x, r). For each Kj that meets B select xj ∈ Kj ∩ B. Since {xj : Kj ∩ B �= ∅} is a discrete subset of a compact set B, it is finite. Lemma 8.2.4. Let A ⊂ Rn be an Hn−1 measurable set, and let K be a singular set of a controlled map φ : A → Rm . There is a sequence {ri } in R+ such that lim ri = 0 and � |φ| dHn−1 = 0. lim ∂B(K,ri )∩A
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Proof. Let 0 ≤ t < n − 1 be the size of K. By Proposition 8.1.5, there is a decreasing sequence {ri } with � � P B(K, ri ) lim ri = 0 and lim ≤ nM∗ t (K). rin−1−t Choose ε > 0, and let δ(x) = dist (x, K) for x ∈ Rn . If M∗ t (K) = 0, there are a > 0 and ja ∈ N such that � � � � �φ(x)� ≤ aδ(x)t+1−n and P B(K, ri ) ≤ ε rn−1−t a i whenever x ∈ A, 0 < δ(x) ≤ rja , and i ≥ ja . Thus for each i ≥ ja , � � n−1 |φ| dH ≤a δ(x)t+1−n dHn−1 (x) ∂∗ B(K,ri )∩A
∂∗ B(K,ri )
=
�
arit+1−n P
� B(K, ri ) ≤ ε.
If M∗ t (K) > 0, select b ≥ nM∗ t (K) and find an integer jb ∈ N so that � � � � �φ(x)� ≤ ε δ(x)t+1−n and P B(K, ri ) ≤ brn−1−t i b whenever x ∈ A, 0 < δ(x) ≤ rjb , and i ≥ jb . Calculating as before, � |φ| dHn−1 ≤ ε ∂∗ B(K,ri )∩cl∗ A
for each i ≥ jb . The lemma follows from the arbitrariness of ε. A control function is a decreasing function β : R+ → R+ such that � 1 β(s) ds < ∞. 0
A control function need not�be bounded, but β(r) = o(1)r−1 as r → 0. Indeed r since β decreases, rβ(r) ≤ 0 β(s) ds → 0 as r → 0.
Definition 8.2.5. Let A ∈ BV(Rn ). A controlled map φ : cl∗ A → Rm with singular sets Kj of size tj is called fully controlled if the following additional conditions are satisfied for j = 1, 2, . . . : (i) M∗ tj (Kj , ∂∗ A) < ∞; (ii) there is a control function βj such that for ψ := φ � ∂∗ A, � � � � �ψ(x)� = O(1)βj dist (x, Kj ) dist (x, Kj )t+2−n as x → Kj .
Lemma 8.2.6. Let A ∈ BV(Rn ), and let K be a singular set of a fully controlled map φ : cl∗ A → Rm . There is a sequence {ri } in R+ such that lim ri = 0 and � lim
∂∗ [A∩B(K,ri )]
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|φ| dHn−1 = 0.
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Proof. Let 0 ≤ t < n − 1 be the size of K, and let δ(x) = dist (x, K) for x ∈ Rn . Choose c > M∗ t (K,∂ ∗ A) and 0 < r ≤ 1 so that � � � � � � Hn−1 B(K, s) ∩ ∂∗ A ≤ csn−1−t and �φ(x)� ≤ cβ δ(x) δ(x)t+2−n for each 0 < s ≤ r and each x ∈ B(K, r) ∩ (∂∗ A − K). The sets
Di = B(K, r2−i ) − B(K, r2−i−1 ) �∞ are disjoint and B(K, r) = K ∪ i=0 Di . Since Hn−1 (K) = 0, � ∞ � � � � n−1 |φ| dH ≤c β δ(x) δ(x)t+2−n dHn−1 (x) B(K,r)∩∂∗ A
≤c
∗ i=0 Di ∩∂ A ∞ � 2 −i
β(r2 )(r2−i−1 )t+2−n (r2−i )n−1−t
i=0
2 n−1−t
=c 2
∞ �
β(r2−i )r2−i−1
i=0
≤ c 2 2n
∞ � � i=0
r2−i
β(s) ds = c2 2n r2−i−1
�
r
β(s) ds. 0
From the definition of a control function, we obtain � lim |φ| Hn−1 = 0. r→0
(∗)
B(K,r)∩∂∗ A
According to Corollary 4.2.5, � � � � � � ∂∗ A ∩ B(K, r) ⊂ B(K, r) ∩ ∂∗ A ∪ ∂B(K, r) ∩ cl∗ A ,
and the lemma follows from (∗) and Lemma 8.2.4. �∞ �∞ Lemma 8.2.7. If j=1 P(Bj ) < ∞ and B = j=1 Bj , then � � ∞ ∞ � � P(B) ≤ P(Bj ) and Hn−1 ∂∗ B − ∂∗ Bj = 0. j=1
j=1
Proof. For k = 0, 1, . . . , Observation 4.2.1 yields � � � k ∞ � ∂∗ B − ∂ ∗ Bj ⊂ ∂∗ Bj . j=1
j=k+1
From Theorem 6.5.2 and Propositions 4.5.2 and 5.1.7, we obtain � � � �� � � � ∞ ∞ ∞ � Bj Bj ≤ P(Bj ). =P Hn−1 ∂∗ j=k+1
j=k+1
j=k+1
The lemma follows by letting k = 0 and k → ∞.
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Theorem 8.2.8. Let A ∈ BV(Rn ), and let v : cl∗ A → Rn be a fully controlled vector field. If divcl∗ A v belongs to L1 (A), then � � divcl∗ A v(x) dx = v · νA dHn−1 . A
∂∗ A
Proof. Let Kj be the singular sets of v. Choose ε > 0 and η > 0. As M∗ n−1 (Kj ) = 0, Propositions 8.1.5 and 6.6.6 show that there are BV sets Bj := A ∩ B(Kj , rj ) such that |Bj | ≤ η2−j
and
P(Bj ) ≤ η2−j .
(∗)
Making rj smaller, if necessary, Lemma 8.2.6 implies � |v| dHn−1 ≤ ε2−j . ∂∗ Bj
� By Lemma 8.2.7 and (∗), the union B := j Bj is a BV set, and � � �� |v| dHn−1 ≤ � |v| dHn−1 ≤ |v| dHn−1 ≤ ε. ∂∗ B
j
∂∗ Bj
j
∂ ∗ Bj
Since divcl∗ A v ∈ L1 (A), and since |B| ≤ η by (∗), we obtain � � � �divcl A v(x)� dx ≤ ε ∗ B
� � by making η sufficiently small. The restriction v � cl∗ (A − B) is admissible and bounded, and divcl∗ A v(x) = divcl∗ A−B v(x) for almost all x ∈ A − B in accordance with (7.2.1). Thus � � divcl∗ A v(x) dx = v · νA−B dHn−1 A−B
∂∗ (A−B)
by Proposition 7.4.3. In view of this equality and Theorem 6.6.1, � �� � � � n−1 � v · νA dH − divcl∗ A v(x) dx�� � ∂∗ A A �� � � � � n−1 � =� v · νB dH − divcl∗ A v(x) dx�� ∂ B B � ∗ � � � ≤ |v| dHn−1 + �divcl∗ A v(x)� dx < 2ε. ∂∗ B
B
The theorem follows from the arbitrariness of ε.
Example 8.2.9. Assume n = 2 and 0 ≤ s ≤ 1. If � x|x|−1−s if x ∈ R2 − {0}, v(x) := 0 if x = 0,
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8. Unbounded vector fields
then div v(x) = (1 − s)|x|−1−s for each �x ∈ �R2 − {0}. It is clear that {0} is the only singular set of v, and that M∗ 0 {0} = 1. For s < 1, the vector field v is fully controlled in B(0, 1), since � � �v(x)� = |x|−s = o(1)|x|−1 as x → 0.
Thus� Theorem 8.2.8 applies for s < 1. When s = 1, then div v ≡ 0 in R2 − {0} and ∂B v · νB dH1 = 2π. Now �let A ⊂ B(0, � 1) be the BV set defined in Example 8.1.7, and recall ∗t that M {0}, ∂∗ A < ∞ if and only if t ≥ 1/2. If s < 1/2, then w := v � cl∗ A is fully controlled and Theorem 8.2.8 applies. Indeed, for t = 1/2 and the 1 control function β : r �→ r− 2 −s , we have � � � � �v(x)� = β |x| |x|t for each x ∈ B(0, 1). On the other hand, if 1/2 ≤ s ≤ 1 then � � div w(x) dx < ∞ and w · νA dH1 = ∞. A
∂∗ A
8.3. Integration by parts Lemma 8.3.1. Let A ⊂ Rn be a measurable set, and let K be a singular set of a controlled map φ : A → Rm . Then � � � �φ(x)� dx = 0. lim r→0
A∩B(K,r)
Proof. The proof is similar to that of Lemma 8.2.6. Let 0 ≤ t < n − 1 be the size of K, and let δ(x) = dist (x, K) for each x ∈ Rn . There are c > M∗ t (K) and r > 0 such that � � � � �B(K, s)� ≤ csn−t and �φ(x)� ≤ cδ(x)t+1−n
for each 0 < s ≤ r and each x ∈ B(K, r) − K. The sets
Di := B(K, r2−i ) − B(K, r2−i−1 ) �∞ are disjoint, and B(K, r) = K ∪ i=1 Bi . Since |K| = 0, we obtain � ∞ � � � � �φ(x)� dx ≤ c δ(x)t+1−n dx A∩B(K,r)
≤c
≤c
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i=1 ∞ �
Di
� � (r2−i−1 )t+1−n �B(K, r2−i )�
i=1 ∞ � 2
(r2−i−1 )t+1−n (r2−i )n−t = c2 2n−t r.
i=1
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8.3. Integration by parts
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191
Corollary 8.3.2. Let Ω ⊂ Rn be an open set. If φ : Ω → Rm is a controlled map, then φ ∈ L1loc (Ω; Rm ). Proof. By Lemma 8.3.1, each x ∈ Ω ∩ Ω∞ has a neighborhood U ⊂ Ω with φ ∈ L1 (U ; Rm ). As φ � (Ω − Ω∞ ) is locally bounded, the same is true for every x ∈ Ω − Ω∞ . Theorem 8.3.3 (Integration by parts). Let Ω ⊂ Rn be an open set, let g : Ω → R be an admissible function, and let v : Ω → Rn be a controlled n vector field. If div v ∈ L1loc (Ω) and Dg ∈ L∞ loc (Ω; R ), then � � g(x) div v(x) dx = − Dg(x) · v(x) dx Ω
Ω
whenever spt (gv) � Ω.
Proof. Although the assumptions differ, the proof is similar to that of Theorem 2.3.9. Find a figure A ⊂ Ω whose interior contains the compact set C := spt (gv), and let w := gv � A. Since g � A is bounded by Remark 2.3.3, the vector field w is controlled, and if Kj are the singular sets of v, then C ∩Kj are the singular sets of w. Moreover, w is fully controlled, since C � A. By our assumptions and Corollary 8.3.2, div w = g div v + Dg · v
belongs to L1 (A). As w ≡ 0 on ∂A, Theorem 8.2.8 implies � � � div w(x) dx = g(x) div v(x) dx + Dg(x) · v(x) dx. 0= A
A
(∗)
A
Choose x ∈ Ω − A so that g and v are differentiable at x; this is true for almost all x ∈ Ω − A. Since gv ≡ 0 in the open setΩ − A, 0 = div (gv)(x) = g(x) div v(x) + Dg(x) · v(x). Hence g(x) div v(x) = −Dg(x) · v(x). As either g(x) = 0 or v(x) = 0, the theorem follows from (∗). Remark 8.3.4. In view of Theorem 8.3.3, the techniques introduced in Sections 3.3–3.4 can be applied to controlled vector fields. For illustration, consider an open setΩ ⊂ C containing the Cantor set D of Example 8.1.4, and a holomorphic function f : Ω − D → C such that � � �f (z)� ≤ dist(z, D)t−1 for log 2/ log 3 < t < 1 and each z ∈ U − D where U � Ω is an open set containing D. Following the proof of Theorem 3.3.1, we can show that f has a unique extension to a holomorphic function in Ω.
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Chapter 9 Mean divergence
The mean divergence of a vector field is the density of its flux. For controlled vector fields the mean and ordinary divergence coincide almost everywhere. In general, the mean divergence may exist even if the ordinary divergence does not. Viewing the flux of a vector field as an additive function of dyadic figures, we give a sufficient condition under which the mean divergence exists and determines the flux. This is accomplished by replacing the classical variation of an additive function by a suitable Borel measure — the idea originally introduced by B.S. Thomson [73] in the real line. Throughout this chapter, Ω is a fixed open subset of Rn .
9.1. The derivative By DC(Ω) and DF(Ω) we denote, respectively, the family of all dyadic cubes and dyadic figures contained in Ω. For F : DF(Ω) → R and x ∈ Ω, let DF (x) = sup inf η>0 C
F (C) |C|
and
DF (x) = inf sup η>0 C
F (C) |C|
where C ⊂ Ω is a dyadic cube with x ∈ C and d(C) < η. When DF (x) = DF (x) �= ±∞,
we call this common value the derivative of F at x, denoted by F � (x). n−1 Example 9.1.1. Let v ∈ L∞ ; Rn ), and let loc (Ω, H � F : C �→ v · νC dHn−1 : DF(Ω) → R. ∂C
If v is differentiable at x ∈ Ω, then Corollary 2.1.3 implies F � (x) = div v(x).
Observation 9.1.2. Let f ∈ L1loc (Ω). If � F : C �→ f (x) dx : DF(Ω) → R, C
then F � (x) = f (x) for almost all x ∈ Ω.
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9. Mean divergence
The proof of the previous observation is almost identical to that of Theorem 4.3.4. Properties of dyadic cubes are used in lieu of Vitali’s theorem. For illustration, we present it in the small print. Proof of Observation 9.1.2. Denote by N the set of all x ∈ Ω at which either F � (x) does not exist or F � (x) differs from f (x). Given x ∈ N � , find γx > 0 so� that for each η > 0, there is C ∈ DC(Ω) satisfying x ∈ C, d(C) < η, and �F (C) − f (x)|C|� ≥ γx |C|. Select open sets �∞ Ωk � Ω with Ω = k=1 Ωk , and let Nk := {x ∈ N ∩ Ωk : γx ≥ 1/k}. Choose ε > 0, and use Henstock’s lemma to find δ : Ωk → R+ so that q � � � ε � � �f (xi )|Qi | − F (Qi )� < k i=1 � � for each δ-fine dyadic partition (Q1 , x1 ), . . . , (Qq , xq ) inΩ k . The set Nk is covered by a family C of all C ∈ DC(Ωk ) such that d(C) < δ(xC ) for some xC ∈ C ∩ N . There are � � C. Since for p = 1, 2, . . . , the collections nonoverlapping cubes Ci ∈ C with i Ci = � � (C1 , xC1 ), . . . , (Cp , xCp ) are δ-fine dyadic partitions inΩ k , � � ��� � |Ci | ≤ k |Nk | ≤ �f (xi )|Ci | − F (Ci )� ≤ ε. i
i
As ε is arbitrary, Nk is a negligible set and so is N =
�∞
k=1
Nk .
Proposition 9.1.3. For each F : DF(Ω) → R, the functions DF : x �→ DF (x) : Ω → R
and
DF : x �→ DF (x) : Ω → R
are Borel measurable. � � Proof. Choose c ∈ R, and observe that E := x ∈ Ω : DF (x) < c consists of all x ∈ Ω which satisfy the following condition: there is j ∈ N such that for each k ∈ N we can find C ∈ DC(Ω) with x ∈ C,
d(C) < 1/k,
and
F (C) 1 0 for each x ∈ X. There are δ > 0 and a set E ⊂ X of positive measure such that G(Q) > 0 for each Q ∈ DC(Ω) satisfying d(Q) < δ
and
Q∗ ∩ E �= ∅.
Proof. For each x ∈ X there is a δx > 0 such that G(Q) > 0 for every Q ∈ DC(Ω) with x ∈ Q and d(Q) < δx . Since ∞� � � X= x ∈ X : δx > 1i i=1
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9.1. The derivative
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195
there is an integer j ≥ 1 such that Y := {x ∈ X : δx > 1/j} has positive measure, and the following holds: G(Q) > 0 for each Q ∈ DC(Ω) with Y ∩ Q �= ∅ and d(Q) < 1j .
(∗)
Let x ∈ int∗ Y . By Lemma 4.2.3 there is ηx > 0 such that |Y ∩ Q|/|Q| > 1/2 for each Q ∈ DC(Ω) with x ∈ Q and d(Q) < ηx . As int∗ Y =
∞ � �
j=1
x ∈ int∗ Y : ηx >
1 j
�
,
there is an integer k ≥ 1 such that the set E = {int∗ Y : ηx > 1/k} has positive measure, and the following is true: |Y ∩ Q|/|Q| >
1 2
for each Q ∈ DC(Ω) with E ∩ Q �= ∅ and d(Q) < k1 . (∗∗)
Now let δ := 12 min{1/j, 1/k}, and choose Q ∈ DC(Ω) with d(Q) < δ and Q∗ ∩ E �= ∅. As d(Q∗ ) = 2d(Q) < 2δ ≤ 1/k, the inequality |Y ∩ Q∗ | > 12 |Q∗ | = 2n−1 |Q|
(†)
follows from (∗∗). On the other hand, |Q∗ | = 2n |Q|. Thus if Y ∩ Q = ∅, then |Y ∩ Q∗ | ≤ 2n−1 |Q| contrary to (†). This shows that Y ∩ Q �= ∅, and since d(Q) < δ < 1/j, the lemma follows from (∗). A function F : DF(Ω) → R is called additive if F (A ∪ B) = F (A) + F (B) for each pair A, B ∈ DF(Ω) with |A∩B| = 0. For additive functions of dyadic figures, a version of the classical Ward theorem [65, Chapter 4, Theorem 11.15] is not difficult to prove. Theorem 9.1.5 (Ward). If F : DF(Ω) → R is an additive function, then F � (x) exists at almost all x ∈ Ω for which D|F |(x) < ∞. � � Proof. Let N = x ∈ Ω : D|F |(x) < ∞ , and observe that DF (x) and DF (x) are real numbers for every x ∈ N . Since ∞� � � � x ∈ N : DF (x) − DF (x) > 1i , x ∈ N : DF (x) < DF (x) =
�
i=1
we proceed to a contradiction by assuming that for some 0 < η < 1, the set � � Nη := x ∈ N : DF (x) − DF (x) > η
has positive measure. Choose 0 < ε < η /4 and for k ∈ Z, let � � Xk := x ∈ Ω1 : εk < DF (x) ≤ ε(k + 1) .
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9. Mean divergence
� As Nη = k∈Z Xk , there is an integer s such that the set X := Xs has positive measure. Letting G := F − εpLn , we calculate that for each x ∈ X, 0 < DG(x) ≤ ε
and
DG(x) > η .
(∗)
The remainder of the proof relies only on inequalities (∗). By Lemma 9.1.4, there are δ > 0 and a set E ⊂ X of positive measure such that G(Q) > 0 for each Q ∈ DC(Ω) with d(Q) < δ and Q∗ ∩ E �= ∅. Choose z ∈ E ∩ int∗ E, and using (∗) and Lemma 4.2.3, find Q ∈ DC(Ω) so that z ∈ Q, d(Q) < δ, and the following inequalities are satisfied: G(Q) < 2ε|Q|
and
|E ∩ Q| > (1 − ε)|Q| > 0.
(∗∗)
The second inequality of (∗) shows that for each x ∈ E ∩ int Q, there is a dyadic cube Cx ⊂ Q with x ∈ Cx and G(Cx ) > η|Cx |. Since � E ∩ int Q ⊂ {Cx : x ∈ E ∩ int Q},
� there are nonoverlapping Cxi such that E ∩ int Q ⊂ i Cxi ⊂ Q. Thus � |Ci | ≥ |E ∩ Q| > (1 − ε)|Q|, i
and there is an integer p ≥ 1 such that p �
G(Cxi ) > η
i=1
p � i=1
|Cxi | > η(1 − ε)|Q|.
(†)
� ��p �p If i=1 Cxi �= Q, let A = Q − int i=1 Cxi . Given y ∈ A, denote by Qy �p the largest dyadic cube such that y ∈ Qy and Qy does not overlap i=1 Cxi . As Qy � Q, we have Q∗y ⊂ Q. It follows from the maximality of Qy that Q∗y overlaps, and thus contains, some Cxi for 1 ≤ i ≤ p. In particular, � � xi ∈ Q∗y ∩ E, and if d = min d(Ci ) : i = 1, . . . , p , then d(Q∗x ) ≥ d. Moreover d(Q ) ≤ d(Q) < δ, and the choice of δ and E implies G(Qy ) > 0. Since � y y∈A Qy = A and d(Qy ) ≥ d/2 > 0, the set�A is the union of nonoverlapping p dyadic cubes Qy1 , . . . , Qyq where q ∈ N. If i=1 Cxi = Q, let q = 0. In either case, Q is the union of nonoverlapping dyadic cubes Cx1 , . . . , Cxp , Qy1 , . . . , Qyq where q ≥ 0 is an integer. Consequently, 2ε|Q| > G(Q) =
p � i=1
G(Cxi ) +
q � i=1
G(Qyi ) ≥
p � i=1
G(Cxi ) > η(1 − ε)|Q|
follows from (∗∗) and (†). We infer 2ε > η(1 − ε) > η /2, contrary to our choice of ε < η /4.
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9.2. The critical variation
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9.2. The critical variation Let F : DF(Ω) → R. Given E ⊂ Ω and δ : E → R+ , we let Vδ F (E) := sup P
p � � � �F (Ci )� i=1
� � where P = (C1� , x1 ), .�. . , (Cp , xp ) is a δ-fine dyadic partition in Ω; as usual, �p � we define i=1 F (Ci )� := 0 when P = ∅. We further let VF (E) := inf Vδ F (E) δ
where δ : E → R+ . The function VF : E �→ VF (E), defined for every E ⊂ Ω, is called the critical variation of F . Proposition 9.2.1. Let F : DF(Ω) → R be an additive function. Then � � �F (A)� ≤ VF (A) for each A ∈ DF(Ω), and VF (int A) ≤ F (A) if F is nonnegative.
Proof. Choose A ∈ DF(Ω) � and δ : A → R+ . By Proposition 2.2.2 there is � a δ-fine dyadic partition P = (A1 , x1 ), . . . , (Ap , xp ) with [P ] = A. Then � � �� p � � � � � p �≤ �F (Ci )� ≤ Vδ F (A), �F (A)| = � F (C ) i � � i=1
i=1
� and �F (A)| ≤ VF (A) by the arbitrariness of δ. Now assume F is nonnegative, and select σ : int A → R+ �so that σ(x) ≤ dist (x, ∂A) for each x ∈ int A. If � Q = (B1 , y1 ), . . �. , (B�s , ys ) is a σ-fine dyadic partition, then [Q] ⊂ A. Thus �s i=1 F (Bi ) = F [Q] ≤ F (A). As Q is arbitrary, VF (int A) ≤ Vσ F (int A) ≤ F (A).
Proposition 9.2.2. If F : DF(Ω) → R, then VF is a Borel measure in Ω. Proof. Clearly VF (∅) = 0. If A ⊂ B ⊂ Ω and δ : B → R+ , then VF (A) ≤ Vδ�A F (A) ≤ Vδ F (B) and VF (A) ≤ VF (B) by the arbitrariness of δ. Let E be the union of a sequence {Ek } of disjoint subsets of Ω, and let δk : Ek → R+ . Define a function δ : E → R+ by letting δ(x) � := δk (x) when x ∈�Ek . If P is a δ-fine dyadic partition in Ω, then Pk := (C, x) ∈ P : x ∈ Ek are δk -fine disjoint � dyadic partitions in Ω and P = k Pk . Thus � � � � � � � � �F (C)� = �F (C)� ≤ VF (E) ≤ Vδ F (E) ≤ Vδk F (Ek ), (C,x)∈P
✐
k (C,x)∈Pk
k
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9. Mean divergence
� and the arbitrariness of δk implies VF (E) ≤ k VF (Ek ). If Ek are any sets, �k−1 � then Ak := Ek − j=1 Ek are disjoint sets and E = k Ak . Hence � � VF (E) ≤ VF (Ak ) ≤ VF (Ek ). k
k
Finally, given sets A, B ⊂ Ω such that d := dist (A, B) > 0, choose a function δ : A ∪ B → (0, d/3) and observe that VF (A) + VF (B) ≤ Vδ�A F (A) + Vδ�B F (B) = Vδ F (A ∪ B). As δ is arbitrary, VF (A) + VF (B) ≤ VF (A ∪ B), and VF is a Borel measure by Theorem 1.3.1. Proposition 9.2.3. Let F : DF(Ω) → R. If the measure VF is absolutely continuous, it is Borel regular. � �� Proof. The set B := ∂Q : Q ∈ DC(Ω) is negligible and Borel. Thus to verify the Borel regularity of an absolutely continuous Borel measure, it suffices to consider a set E ⊂ Ω − B. As there is nothing to prove otherwise, assume that VF (E) < ∞. Fix an integer k ≥ 1, and find δk : E → R+ so that Vδk F (E) < VF (E) + 1/k. For j ∈ N define � � Ej = x ∈ E : δk (x) > 1/j . Let σ ≡ 1/j � be a constant function � on cl Ej − B, and choose a σ-fine dyadic partition (Q1 , y1 ), . . . , (Qp , yp ) in Ω. Since yj ∈ cl Ej ∩ int Qj , the intersecChoosing xj ∈ Ej ∩ int Qj , the dyadic partition tion Ej ∩ int Qj is not empty. � � (Q1 , x1 ), . . . , (Qp , xp ) is δk -fine. We infer VF (cl Ej − B) ≤ Vσ F (cl Ej − B) ≤ Vδk F (E) < VF (E) + 1/k. � Now Ck = j∈N (cl Ej − B) is a Borel set containing E, and �
VF (Ck ) = lim VF (cl Ej − B) ≤ VF (E) + 1/k.
Hence C = k∈N Ck is a Borel set containing E, and the proposition follows, since VF (C) ≤ VF (E). Lemma 9.2.4. Let F : DF(Ω) → R. If VF is absolutely continuous, then D|F |(x) < ∞ for almost all x ∈ Ω. � � Proof. By Proposition 9.1.3, the set E := x ∈ Ω : D|F |(x) = ∞ is Borel. Seeking a contradiction, suppose |E| > 0 and find a compact set K ⊂ E with |K| > 0. Replacing K by spt (Ln K) if necessary, we may assume that |K ∩V | > 0 for each open set V ⊂ Rn which meets K. Select a countable dense subset C of K, and find open sets Uk ⊂ Ω so that C ⊂ Uk and |Uk | < 1/k. �∞ If U = k=1 Uk , then D = K ∩ U is a negligible Gδ set, still dense in K by the Baire category theorem [61, Chapter 7, Section 7]. According to our assumptions VF (D) = 0.
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9.2. The critical variation
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Now choose σ : D → R+ , and using Baire’s theorem again, find t > 0 and an open set V ⊂ Ω such that D ∩ V �= ∅ and the set � � Dt := x ∈ D ∩ V : σ(x) > t
is dense in D ∩ V , and hence in K ∩ V . Denote by Q the family of all cubes Q ∈ DC(Ω) satisfying d(Q) < t
� � and �F (Q)� >
|Q| . |K ∩ V |
� Observe E ⊂ Q. Passing to a subfamily if necessary, we may assume that Q consists of nonoverlapping cubes. Since the boundary of each cube is a negligible set, the subfamily {Q ∈ Q : K ∩ V ∩ int Q �= ∅}, enumerated as Q1 , Q2 , . . . , covers K ∩ V almost entirely. Thus �� � � 1 �F (Qi )� > |Qi | ≥ 1, |K ∩ V | i i � �p �� � Qi contains a and consequently i=1 F (Qi ) >�1 for some p ∈ N. Each � point xi ∈ Dt , and the collection (Q1 , x1 ), . . . , (Qp , xp ) is a σ-fine dyadic partition in Ω. Hence Vσ F (D) > 1, and VF (D) ≥ 1 by the arbitrariness of σ — a contradiction. Example 9.2.5. Assume n = 1 and Ω= R. Let D be the Cantor ternary discontinuum in the interval I = [0, 1], and let U be the family of all connected components of I − D; see Example 8.1.4. A variant of the Cantor-Vitali function (“devil’s staircase”) is an increasing continuous function v : R → R such that if x < 0, 0 v(x) := (a + b)/2 if a < x < b and (a, b) ∈ U, 1 if x > 1; cf. [26, Section 4.2] and [54]. If F is the flux of v, then � � � F [a, b] = v · ν[a,b] dH0 = v(b) − v(a) ≥ 0 ∂([a,b])
� � for every interval [a, b] ⊂ R. If P = (A1 , x1 ), . . . , (Ap , xp ) is any dyadic �p partition, then i=1 F (Ai ) ≤ 1. Hence by Proposition 9.2.1, 1 = F (I) ≤ VF (I) ≤ VF (R) ≤ 1.
Now define σ : x �→ dist (x, D) : R − D → R+ , and observe that VF (R − D) ≤ Vσ F (R − D) = 0. Thus VF (D) = 1, and as |D| = 0, the measure VF is not absolutely continuous. Still F � (x) = v � (x) = 0 for each x in R − D.
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9. Mean divergence
Lemma 9.2.6. Let F : DF(Ω) → R be additive. If DF (x) ≥ 0 for VF almost all x ∈ Ω, then F is nonnegative. � � Proof. Let E = x ∈ Ω : DF (x) < 0 , and choose A ∈ DF(Ω) and ε > 0. There is σ : E → R+ such that Vσ F (E) < ε. Given x ∈ A − E, find ηx > 0 so that F (Q) > −ε|Q| for every cube Q ∈ DC(Ω) with x ∈ Q and d(Q) < ηx . Define a function δ : A → R+ by the formula � σ(x) if x ∈ E, δ(x) := it x ∈ A − E. ηx � � There is a δ-fine dyadic partition P = (C1 , x1 ), . . . , (Cp , xp ) with [P ] = A; see Proposition 2.2.2. By additivity, F (A) =
p � i=1
F (Ci ) ≥ −ε
�
xi ∈A−E
|Ci | −
�
xi ∈E
> −ε|A| − Vσ F (E) > −ε |A| + 1 and F (A) ≥ 0 by the arbitrariness of ε.
�� � �F (Ci )�
�
We denote by AC∗ (Ω) the linear space of all additive functions F : DF(Ω) → R whose critical variation VF is absolutely continuous. Theorem 9.2.7. Let F ∈ AC∗ (Ω). The derivative F � (x) exists at almost all x ∈ Ω, and the linear map F �→ F � : AC∗ (Ω) → L0 (Ω)
is injective. If F � ∈ L1loc (Ω) then for each A ∈ DF(Ω), � F (A) = F � (x) dx. A
Proof. Theorem 9.1.5 and Lemmas 9.2.4 and 9.2.6 imply the first two assertions. Suppose F � ∈ L1loc (Ω), and for each A ∈ DF(Ω), let � G(A) := F � (x) dx. A
Choose a negligible set N ⊂ Ω and ε > 0. Making N larger, we may assume that F � (x) exists for each x ∈ Ω − N . There are open setsΩ k � Ω such that � Ω = k∈N Ωk . Fix k ∈ N, let Nk = N ∩ Ωk , and define g ∈ L1 (Ωk ) by � F � (x) if x ∈ Ωk − Nk , g(x) := 0 if x ∈ Nk .
Find δ : Ωk → R+ that corresponds to g, Ωk , and ε according to Henstock’s lemma. We may assume that δ(x) < dist (x,∂ Ωk ) for each x ∈ Nk . Thus if
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9.2. The critical variation 201 � � P = (C1 , x1 ), . . . , (Cp , xp ) is a (δ � Nk )-fine dyadic partition, then [P ] ⊂ Ωk . By Henstock’s lemma, p p � � � � � � � � �G(Ci )�= �g(xi )|Ci | − G(Ci )� < ε i=1
i=1
and hence VG(Nk ) ≤ Vδ�Nk (Nk ) ≤ ε. The arbitrariness of ε shows that Nk � is VG negligible, and so is N = k∈N Nk . It follows that G ∈ AC∗ (Ω). Since G� (x) = F � (x) for almost all x ∈ Ω by Observation 9.1.2, the first part of the proof implies G = F . n−1 The flux of v ∈ L∞ ; Rn ) is the additive function loc (Ω, H � F : A �→ v · νA dHn−1 : DF(Ω) → R. ∂A
Let x ∈ Ω. If the derivative F � (x) exists, we call it the mean divergence of v at the point x, and denote it by div v(x). The following corollary is a mere reformulation of Theorem 9.2.7. n−1 ; Rn ) belong to AC∗ (Ω). Corollary 9.2.8. Let the flux F of v ∈ L∞ loc (Ω, H Then F is uniquely determined by the mean divergence div v, which is defined almost everywhere in Ω. If div v ∈ L1loc (Ω), then for each A ∈ DF(Ω), � � div v(x) dx = v · νA dHn−1 . (9.2.1) A
∂A
Remark 9.2.9. If v ∈ C(Ω; R ), then it follows from Sections 2.4 and 6.7 that under the assumptions of Corollary 9.2.8, equality (9.2.1) holds for each A � Ω such that A ∈ BV(Rn ). This fact will be made more explicit in the next chapter. n
Remark 9.2.10. Linear spaces akin to AC∗ (Ω) provide descriptive definitions of various forms of multidimensional conditionally convergent integrals [55, 5, 12, 6, 13, 7, 50]. It is easy to verify that in dimension one, the functions in AC∗ (R) coincide with Denjoy-Perron primitives, called ACG∗ functions in [65, Chapter 7, Section 8]; cf. [50]. However, the classical definition of ACG∗ functions depends in an essential way on the order structure of the real line, and its extension to higher dimensions is far from obvious [60]. Thus it is significant that the space AC∗ (Ω) is defined directly in any dimension. A typical element of AC∗ (Ω) is the flux of a pointwise Lipschitz vector field v : Ω → Rn whose divergence is not locally integrable; cf. Remark 2.3.8, (2). Although we do not use such vector fields in this book, we believe that they occur naturally and deserve attention. Under less restrictive assumptions, deeper properties of the space AC∗ (Ω) are established in [51]. The following two examples are due to Z. Buczolich.
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9. Mean divergence
Example 9.2.11. Let n = 2 and Ω= R2 . For r > 0 and x ∈ Rn , let � � � 2 if |x| < r, exp |x||x| 2 −ε2 ϕr (x) := 0 if |x| ≥ r, � � and ur (x) := ϕr (x), 0 . Clearly ur ∈ C ∞ (R2 ; R2 ), and we have � �ur �L∞ (R2 ;R2 ) = 1 and |div ur | ≤ 2πr. R2
Enumerating a dense countable subset of R , define recursively sequences {zk } in R2 and {rk } in R+ so that 2
(1) r1 ≤ 1/2 and rk+1 ≤ rk /2 for k = 1, 2, . . . , � � (2) the family B(zk , rk ) : k = 1, 2, . . . is disjoint, �∞ (3) the open set U = k=1 U (zk , rk ) is a dense in R2 .
Note that 0 < |U | < 2πr12 and P(U ) ≤ 4πr1 .
Claim. If Z = {zk : k = 1, 2, . . . } then R2 − U ⊂ cl Z.
Proof. Suppose there are x ∈ R2 − U and ε > 0 such that U (x,ε ) does not meet Z. As lim rk = 0, there exists an integer p ≥ 1 which satisfies U (x,ε/ 2) ∩ B(zk , rk ) = ∅ for k > p. Since x ∈ / U , and since the compact sets B(z1 , r1 ), . . . , B(zp , rp ) are disjoint, we see at once that the open set �p U (x,ε/ 2) − k=1 B(zk , εk ) is not empty and does not meet U . As U is dense in R2 , this is a contradiction. �∞ Let vk (x) := urk (x − zk ) for each x ∈ R2 , and v := k=1 vk . Clearly v is bounded, Borel measurable, and C ∞ in U . By the Claim, v is discontinuous at every x ∈ R2 − U . Let A ∈ BV(R2 ). For k = 1, 2, . . . , � � div vk (x) dx = vk · νA dH1 (∗) A
∂∗ A
by Theorem 6.5.4, and ∞ � ∞ � � � � |vk · νA | dH1 ≤ H1 ∂∗ A ∩ B(zk , rk ) ≤ P(A) k=1
∂∗ A
�∞
by (2). The function g =
k=1
∞ � �
k=1
R2
k=1
div v belongs to L1 (R2 ), since
|div vk | ≤ 2π
∞ �
k=1
rk ≤ 2π.
In view of the previous two inequalities and (∗), � � g(x) dx = v · νA dH1 . A
∂∗ A
It follows from Observation 9.1.2 that g(x) = div v(x) for almost all x ∈ R2 .
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9.2. The critical variation
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Example 9.2.12. Adhering to the notation of Example 9.2.11, we modify v : R2 → R2 so that it is continuous but not Lipschitz at any x ∈ R2 − U .
ε > 0 and p ∈ N, there is an Claim. Let A ⊂ R2 be a�bounded set. Given � integer q ≥ p such that dist x, {zp , . . . , zq } < ε for each point x ∈ A − U .
Proof. As A − U is bounded, it contains points x1 , . . . , xr such that �r A − U ⊂ i=1 U (xi , ε /2). By the claim of Example 9.2.11, each U (xi , ε /2) contains zki with ki ≥ p. Thus q := max{k1 , . . . , kr } is the desired integer. � � Let p(1) = 0, and define recursively a strictly increasing sequence p(i) of integers so that � � dist x, {zp(i)+1 , . . . , zp(i+1) < 2−i
for every x in U (0, i) − U, i = 1, 2, . . . . Given an integer k ≥ 1, there is a unique integer ik ≥ 1 with p(ik ) < k ≤ p(ik + 1), and we let ck = 1/ik . Observe that ik ≤ ik+1 for k = 1, 2, . . . and lim ik = ∞. �∞ If w := k=1 ck vk , then calculating as in Example 9.2.11, we obtain � div w(x) if x ∈ U , div v(x) = 0 if x ∈ R2 − U . Clearly w is C ∞ in U . Select x in R2 −U and ε > 0. There are an integer k ≥ 1 with ck < ε, and η > 0 such that U (x,η ) ∩ U (zj , rj ) = ∅ for j = 1, . . . , k − 1. Each y ∈ U (x,η ) is either in R2 − V , in which case w(y) = w(x) = 0, or in U (zj , rj ) where j ≥ k, in which case � � �w(y) − w(x)� = cj vj (y) ≤ cj ≤ ck < ε .
Consequently w is continuous at x. On the other hand, if x ∈ R2 − U then x ∈ U (0, ik ) − U for all sufficiently large k ∈ N. Since p(ik ) < k ≤ p(ik + 1), we obtain |zk − x| < 2−ik . Hence � � �w(zk ) − w(x)� 2ik ≥ 2ik ck vk (zk ) ≤ |zk − x| ik for all sufficiently large k ∈ N. It follows that w is not Lipschitz at x.
We note that V. Shapiro [67] studied the mean divergence of vector fields by a completely different method.
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Chapter 10 Charges
A charge is a distribution whose continuity properties resemble those of the distributional divergence of a continuous vector field. We define a locally convex topology T in the space D of all test functions so that charges are T continuous, and that the space BVc of all BV functions with compact support is the sequential completion of (D, T). In the sense of Mackey-Arens theorem, the space of all charges is in duality with BVc — a fact we employ in Chapter 11. Some properties of locally convex spaces are stated without proofs. In such cases, we provide references to standard texts.
10.1. Continuous vector fields Throughout this section,Ω ⊂ Rn is a fixed open set and � � D(Ω) := ϕ ∈ D(Rn ) : spt ϕ ⊂ Ω .
Thus when U is an open subset of Ω, we write D(U ) ⊂ D(Ω); see (1.1.2). The distributional divergence of v ∈ L1loc (Ω; Rn ) is the distribution � v(x) · Dϕ(x) dx : D(Ω) → R Fv : ϕ �→ − Ω
introduced in Example 3.2.1, (3). For v ∈ C(Ω; Rn ) we state the essential continuity property of Fv .
Observation 10.1.1. Let Fv be the distributional divergence of a vector field v ∈ C(Ω; Rn ). Given an open set U � Ω and ε > 0, there is θ > 0 such that Fv (ϕ) ≤ θ�ϕ�L1 (U ) + ε�Dϕ�L1 (U ;Rn ) for each test function ϕ ∈ D(U ). Proof. Choose an open set U � Ω and ε > 0. There is w ∈ C 1 (Ω; Rn ) such that �w − v�L∞ (U ;Rn ) < ε. Now select θ > �div w�L∞ (U ) and a test function ϕ ∈ D(U ). The following calculation, which involves integration by parts, completes the argument: 205
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10. Charges � � Fv (ϕ) = − w · Dϕ + (w − v) · Dϕ U � U ≤ ϕ div w + ε�Dϕ�L1 (U ;Rn ) U
≤ θ�ϕ�L1 (U ) + ε�Dϕ�L1 (U ;Rn ) .
Formulating the continuity property of Observation 10.1.1 without reference to a particular vector field leads to the following definition. Definition 10.1.2. A distribution F ∈ D� (Ω) is called a charge 1 in Ω if given an open set U � Ω and ε > 0, there is θ > 0 such that F (ϕ) ≤ θ�ϕ�L1 (U ) + ε�Dϕ�L1 (U ;Rn ) for each test function ϕ ∈ D(U ). The linear space of all charges in Ω is denoted by CH(Ω). We define a locally convex topology in CH(Ω) by seminorms � � (10.1.1) �F �U := sup F (ϕ) : ϕ ∈ D(U ) and �Dϕ�L1 (U ;Rn ) ≤ 1
where F ∈ CH(Ω) and U � Ω is an open set. Arguing as in Example 1.2.2, we see that CH(Ω) is a Fr´echet space. When the open setΩ ⊂ Rn is understood, we usually say only “charge” instead of “charge in Ω”. Observation 10.1.1 asserts that the distributional divergence Fv of a continuous vector field v : Ω → Rn is a charge. If U � Ω is an open set, then � Fv (ϕ) = − v · Dϕ ≤ �Dϕ�L1 (U ;Rn ) �v�L∞ (U ;Rn ) U
for each ϕ ∈ D(U ), and hence �Fv �U ≤ �v�L∞ (U ;Rn ) .
(10.1.2)
The following example and proposition show that charges may arise from other sources than continuous vector fields. However, see Theorem 11.3.8 below. Example 10.1.3. Assume n = 2. For x = (ξ,η ) in R2 , define � |x|−3/2 (−η,ξ ) if |x| > 0, v(x) := 0 if |x| = 0. 1 The
word “charge” has been used in the literature to describe various concepts. Some differ completely from our definition [58], while others are similar [51, 52, 22]. Our notion of “charge”, called “strong charge” in [23], was introduced without name in [51, Proposition 2.1.7]. It evolved from the concept of “continuous additive function” defined in [48].
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10.2. Localized topology
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Clearly v is discontinuous at x = 0, controlled, and div v(x) = 0 for each x �= 0; cf. Example 8.2.9. As Theorems 8.3.3 show that Fv (ϕ) = 0 for each ϕ ∈ D(R2 ), we see that Fv is a charge in R2 . Proposition 10.1.4. If f ∈ Lnloc (Ω), then � f ϕ : D(Ω) → R Lf : ϕ �→ Ω
is a charge. In addition, there is a constant γ = γ(n) > 0 such that
for each open set U � Ω.
�Lf �U ≤ γ�f �Ln (U )
Proof. Choose an� open� set U� � Ω and a test function ϕ ∈ D(Ω). For c > 0, � n let Uc := x ∈ U : �f (x)� ≥ c . By the H¨older and Sobolev inequalities, � n |ϕ| + �f �Ln (Uc ) �ϕ�L n−1 Lf (ϕ) ≤ c (Uc ) U −Uc (∗) ≤ c�ϕ�L1 (U ) + γ�f �Ln (Uc ) �Dϕ�L1 (U ;Rn )
where γ = γ(n) > 0. Observe limc→∞ |Uc | = 0, since χUc ≤ |f |n /c. Thus � n |f | ≤ (ε/γ)n for some θ > 0, and we obtain Uθ Lf (ϕ) ≤ θ�g�L1 (U ) + ε�Dϕ�L1 (U ;Rn ) .
This shows that Lf is a charge. Since Uc = U when c = 0, Lf (ϕ) ≤ γ�f �Ln (U ) �Dϕ�L1 (U ;Rn ) by inequality (∗). The proposition follows. Our next goal �is to define a topology T in D(Ω) so that CH(Ω) is the dual � space of D(Ω), T . This problem is best approached in the abstract setting of locally convex spaces. The definition of T resembles that of the internal injective limit topology discussed in Section 3.6. We follow ideas presented previously in [21, Section 6], [51, Section 1.2], and [24, Section 3].
10.2. Localized topology For a family A of sets and a set E, we defined in Section 3.6 the family A
E := {A ∩ E : A ∈ A}.
A topology U in a set X is called regular if each neighborhood of every x ∈ X contains a closed neighborhood of x. Equivalently, U is regular if each closed set C ⊂ X and every point x ∈ X − C have disjoint neighborhoods. Each locally convex topology in a linear space is regular [64, Theorem 1.10].
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10. Charges
Proposition 10.2.1. Let (X, U) be a topological space, and let X be the union of a family C of its subsets. There is a unique topology UC in X such that (i) UC C ⊂ U C for each C ∈ C, (ii) given a topological space Y , a map φ : (X, UC ) → Y is continuous whenever the restrictions φ � C : (C, U C) → Y are continuous for each C ∈ C.
The topology UC has the following properties:
(1) UC := {E ⊂ X : E ∩ C ∈ U C for each C ∈ C}; in particular U ⊂ UC and UC C = U C for every C ∈ C. (2) E ⊂ X is UC closed if and only if the intersection E ∩ C is U C closed for every C ∈ C. (3) If Y ⊂ X, then UC Y ⊂ (U Y )C Y and the equality occurs when Y is U open or U closed. Proof. If S and S� are topologies in X satisfying conditions (i) and (ii), then x �→ x : (X, S) → (X, S� )
is a homeomorphism. Thus S = S� , and the uniqueness of UC is proved. A direct verification validates property (1), which establishes the existence of the topology UC . Let E ⊂ X and C ∈ C. Clearly E ∩ C is U C closed if and only if (X − E) ∩ C = C − E ∩ C belongs to U C. It follows from property (1) that E ∩ C is closed for every C ∈ C if and only if X − E belongs to UC , or alternatively if and only if E is UC closed. Given Y ⊂ X, choose an arbitrary but fixed C ∈ C. Select A ∈ UC Y , and find U ∈ UC with U ∩ Y = A. By property (1), there is V ∈ U such that A ∩ (C ∩ Y ) = (U ∩ Y ) ∩ C = (U ∩ C) ∩ Y = (V ∩ C) ∩ Y. As V ∩ C ∈ U
C, the intersection A ∩ (C ∩ Y ) belongs to (U
C)
Y = (U
Y)
(C ∩ Y ).
The arbitrariness of C and property (1) yield A ∈ (U Y )C Y . If Y ∈ U then U Y ⊂ U. Select B ∈ (U Y )C Y and using property (1), observe that B ∩ (C ∩ Y ) = B ∩ C belongs to (U Y ) (C ∩ Y ). In other words, there is W in U Y , and hence in U, such that B ∩ C = W ∩ (C ∩ Y ) = W ∩ C. Since C is arbitrary, B ∈ UC Y by property (1). In view of property (2), the case when Y is U closed is proved similarly.
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10.3. Locally convex spaces
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Throughout we adhere to the notation of Proposition 10.2.1, and call UC the localization of the original topology U by C, or simply the localized topology when the original topology U and localizing family C are understood from the context. Proposition 10.2.2. Let (X, U) be a regular space, and let X be the union of an increasing sequence C = {Ck } of closed subsets of (X, U). The localized topology UC has the following properties: (1) A sequence {xi } in X converges in (X, UC ) if and only if {xi } is a sequence in some Ck and {xi } converges in (X, U). (2) If U is sequential, then so is UC . Proof. (1) Let {xi } be a sequence in X converging to x ∈ X in (X, UC ). As U ⊂ UC , the sequence {xi } converges to x in (X, U). If {xi } is not a sequence in any Ck , construct recursively a subsequence {yk } of {xi } so that yk ∈ / Ck for k = 1, 2, . . . . By regularity, there are Uk ∈ U such that Ck ⊂ Uk and �∞ yk ∈ / Uk . If x ∈ Cp , let U := j=p Uj . Since {Ck } is increasing, � Ck if k ≤ p, Ck ∩ U = � Ck ∩ {Uj : p ≤ j < k} if k > p
for k = 1, 2, . . . . Thus U is a neighborhood of x in (X, UC ) that contains no yk with k ≥ p — a contradiction. Since each Ck is U closed, the converse follows from Proposition 10.2.1, (i). (2) By Proposition 10.2.1, (1), each Ck is UC closed. Suppose U is sequential, and E ⊂ X is sequentially closed in (X, UC ). Let {xi } be a sequence in E ∩ Ck converging in (X, U). Property (1) shows that {xi } converges to x ∈ Ck in (X, UC ). By our assumption, x ∈ E ∩ Ck . Thus E ∩ Ck is sequentially closed, and hence closed, in (X, U). As Ck is arbitrary, E is closed in (X, UC ) according to Proposition 10.2.1, (2). Remark 10.2.3. Sequences {Cj } and {Dk } of sets are called interlacing if each Cj is contained in some Dk , and each Dk is contained in some Cj . Let (X, U) be a topological space, and let C and D be interlacing increasing � � sequences of subsets of X. If X = C, then X = D and UC = UD by Proposition 10.2.1, (1); cf. Proposition 3.6.2.
10.3. Locally convex spaces The following lemma is a convenient tool for applying the results of Section 10.2 to locally convex spaces.
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10. Charges
Lemma 10.3.1. Let X be a linear space, and let U be any topology in X that has a neighborhood base at zero consisting of convex sets. If the maps z �→ x + z : X → X
and
z �→ tz : X → X
are continuous for every x ∈ X and every t ∈ R, then (X, U) is a locally convex space. Proof. According to our assumptions, for every t ∈ R − {0}, the map z �→ tz : (X, U) → (X, U) is a surjective homeomorphism. Thus if U is a convex neighborhood of zero, then so is tU for each t �= 0. In particular, U ∩ (−U ) is a neighborhood of zero, and U has a neighborhood base at zero consisting of convex symmetric sets. Our assumptions also show that for u, x ∈ X and t ∈ R, the maps (v, z) �→ (u + v, x + z)
and
(t, z) �→ (t, x + tz)
are surjective homeomorphisms of the spaces (X, U) × (X, U) and R × (X, U), respectively. Thus it suffices to show that the addition (x, y) �→ x + y is continuous at (0, 0), and the scalar multiplication (t, x) �→ tx is continuous at (t, 0) for each t ∈ R. Choose a convex symmetric neighborhood U ⊂ X of zero. Since 12 U + 12 U ⊂ U , the continuity of addition at (0, 0) follows. Next choose t ∈ R. If |t| < 1/2, then sx ∈ U for x ∈ U and |s − t| < 1/2, since 1 U and |s − t| < |t|, since |s| < 2|t|. |s| < 1. If |t| ≥ 1, then sx ∈ U for x ∈ 2|t| This proves the continuity of scalar multiplication at (t, 0). Definition 10.3.2. A localizing sequence 2 in a locally convex space X is an increasing sequence {Ck } of compact convex sets such that 0 ∈ C1 , and for each k ∈ N, t ∈ R, and x ∈ X there is k(t, x) ∈ N with �∞
x + tCk ⊂ Ck(t,x) .
Note that X = k=1 Ck , since x ∈ C1(0,x) for each x ∈ X, and that (X, U) is a regular space [64, Theorem 1.11]. According to Proposition 10.2.1, each localizing sequence C in a locally convex space (X, U) defines a localized topology UC . Theorem 10.3.3. If C = {Ck } is a localizing sequence in a locally convex space (X, U), then (1) UC is locally convex and sequentially complete, (2) E ⊂ X is UC bounded if and only if E is contained in some Ck . 2 The terms “localized topology” and “localizing sequence” were introduced in [24] by T. De Pauw.
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10.3. Locally convex spaces
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211
Proof. Let x ∈ X and t ∈ R. By Proposition 10.2.1, (ii), τx : z �→ x + z
and
µt : z �→ tz
are continuous maps of (X, UC ), since τx (Ck ) ⊂ Ck(1,x) and µt (Ck ) ⊂ Ck(t,0) for k = 1, 2, . . . . In view of Lemma 10.3.1, we prove that UC is a locally convex topology by showing that it has a neighborhood base at zero consisting of convex sets. To this end, choose U ∈ UC containing zero. Using Proposition 10.2.1, (1), find Uk ∈ U so that Ck ∩ U = Ck ∩ Uk for k = 1, 2, . . . , and observe that each Uk contains zero. There is a convex set V0 ∈ U such that 0 ∈ V0 and cl U V0 ⊂ U1 . As C1 ∩ cl U V0 is a compact convex set and C1 ∩ cl U V0 ⊂ C1 ∩ U1 = C1 ∩ U ⊂ C2 ∩ U = C2 ∩ U2 ⊂ U2 , there is a convex set V1 ∈ U such that C1 ∩ cl U V0 ⊂ V1 and cl U V1 ⊂ U2 ; see [64, Theorem 1.10]. Recursively, we define convex sets Vk ∈ U so that Ck ∩ cl U Vk−1 ⊂ Vk and cl U Vk−1 ⊂ Uk for each k ∈ N. In particular, we have Ck ∩ Vk−1 ⊂ Vk ⊂ Uk+1 and hence Ck ∩ Vk−1 ⊂ Ck+1 ∩ Vk ⊂ Ck+1 ∩ Uk+1 = Ck+1 ∩ U ⊂ U. �∞ As 0 ∈ C1 ∩ V0 , the union V = k=1 (Ck ∩ Vk−1 ) is a convex subset of U �∞ containing zero. In addition V ∈ UC , because Cj ∩ V = Cj ∩ k=j Vk for j = 1, 2, . . . . Let E ⊂ X be UC bounded. Proceeding to a contradiction, assume that E is contained in no Ck . It follows from Definition 10.3.2 that there is an increasing sequence {jk } in N such that kCk ⊂ Cjk for each k ∈ N. By our assumption, E contains a sequence {xk } such that xk ∈ / Cjk , or alternatively / Ck , for all k ∈ N. As E is UC bounded, the sequence {xk /k} converges xk /k ∈ to zero in (X, UC ), contrary to Proposition 10.2.2, (1). Since each Ck is compact, the converse is clear. As each Cauchy sequence in a topological linear space is bounded, it follows from the previous paragraph that a Cauchy sequence {xj } in (X, UC ) is a sequence in some Ck . By compactness, {xj } has a U convergent subsequence, and being Cauchy, it U converges itself. Thus {xj } converges in (X, UC ) by Proposition 10.2.2, (1). Example 10.3.4. Denote by R∞ the linear space of all sequences {xi } in R such that xi = 0 for all but finitely many i. Given x = {xi } and y = {yi } �∞ yi and give R∞ the locally convex topology U in R∞ , let x · y = i=1 xi√ induced by the norm |x| := x · x . Via the embedding (x1 , . . . , xk ) �→ (x1 , . . . , xk , 0, 0, . . . ) : Rk → R∞ ,
with a subspace of R∞ , still denoted by Rk . It is clear that we identify Rk � the sets Ck := x ∈ Rk : |x| ≤ k}, k = 1, 2, . . . , form a localizing sequence C := {Ck } in (R∞ , U). According to Proposition 10.2.2, (2), the space (X, UC )
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is sequential. We show next that it is not first countable. For k, j = 1, 2, . . . , let xk := (1/k, 0, 0, . . . ) and xk,j := (1/k, 1/j, . . . , 1/j , 0, 0, . . . ). � �� � k-times
∞
In (R , UC ), we have lim xk = 0 and limj→∞ xk,j = xk for k = 1, 2, . . . . On the other hand, it follows from Proposition 10.2.2, (1) that for no sequence {jk } of positive integers, the sequence {xk,jk } converges to zero in (R∞ , UC ). Thus (R∞ , UC ) is not first countable; in particular it is not metrizable. Using Section 3.6, observe that (R∞ , UC ) = inj lim (Rk , U k→∞
Rk ).
10.4. Duality Let X ∗ = (X, U)∗ be the dual space of a locally convex space (X, U). Each x ∈ X defines a linear functional fx : x∗ �→ �x∗ , x� : X ∗ → R.
It is clear that the weak* topology W∗ , defined at the end of Section 1.2, is the smallest locally convex topology in X ∗ such that the functional fx : (X ∗ , W∗ ) → R is continuous for each x ∈ X. It follows from [64, Theorem 3.10 and Section 3.14] that each continuous linear functional f : (X ∗ , W∗ ) → R has the form f = fx for some x ∈ X. Thus E : x �→ fx : X → (X ∗ , W∗ )∗
is a surjective linear map, called the evaluation map. The Hahn-Banach theorem shows that E is also injective [64, Theorem 3.4, (b)]. If V is a locally convex topology in X ∗ that is larger than W∗ , then (X ∗ , W∗ )∗ ⊂ (X ∗ , V)∗ . In this case, the evaluation map E : X → (X ∗ , V)∗
is still defined and injective, but it may not be surjective. The MackeyArens theorem [27, Theorem 8.3.2], stated below without proof, describes all topologies V in X ∗ such that (X ∗ , V)∗ = (X ∗ , W∗ )∗ , or equivalently, such that the evaluation map E : X → (X ∗ , V)∗ is bijective. The smallest locally convex topology in X for which all linear functionals in (X, U)∗ are still continuous is called the weak topology in X, denoted by W. In other words, W is the smallest locally convex topology in X such that (X, W)∗ = (X, U)∗ . A set E ⊂ X is called weakly compact if it is compact
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10.5. The space BVc (Ω)
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in (X, W). For emphasis, compact subsets of (X, U) are sometimes called originally compact. Theorem 10.4.1 (Mackey-Arens). Let (X, U) be a locally convex space, let W be the weak topology in X, and let V be a locally convex topology in the dual space X ∗ = (X, U)∗ . The evaluation map E : X → (X ∗ , V)∗ is bijective if and only if there is a family C consisting of weakly compact subsets of X such that V is defined by seminorms �� � � �x∗ �C := sup ��x∗ , x�� : x ∈ C
where x∗ ∈ X ∗ and C ∈ C.
For a locally convex space (X, U), the strong topology S∗ in (X, U)∗ is defined at the end of Section 1.2. It can be described as the topology of uniform convergence on each bounded subset of (X, U). Corollary 10.4.2. Let C = {Ck } be a localizing sequence in a locally convex space (X, U), and let UC be the localization of U by C. If X ∗ = (X, UC )∗ , then the evaluation map E : X → (X ∗ , S∗ )∗ is bijective. Proof. By Theorem 10.3.3, (2), each bounded subset of (X, UC ) is contained in some Ck . Thus the strong topology S∗ in X ∗ is defined by seminorms � �� � �x∗ �Ck = sup ��x∗ , x�� : x ∈ Ck
where x∗ ∈ X ∗ and k = 1, 2, . . . . As each Ck is originally compact, it is weakly compact. The corollary follows from Mackey-Arens theorem. In fact, the strong topology S∗ is the Arens topology in X ∗ ; see [27, Section 8.3.3].
In Bourbaki’s terminology, Corollary 10.4.2 states that the space (X, UC ) is semireflexive [9, page 87].
10.5. The space BVc (Ω) We assume thatΩ ⊂ Rn is a fixed open set, and let � � BVc (Ω) := f ∈ BV (Rn ) : spt f � Ω .
Remark 10.5.1. Since the elements of BV (Rn ) are equivalence classes, the definition of BVc (Ω) is meaningful only in light of Convention 1.3.2, according to which we assume that for each f ∈ BV (Ω), spt f = ess spt f.
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We define the original topology U in BVc (Ω) as the metric topology given �∞ by the L1 norm � · �L1 (Ω) . Choose open setsΩ k � Ωk+1 so that Ω= k=1 Ωk . Using Theorem 5.5.12, observe that � � BVk := g ∈ BVc (Ω) : spt g ⊂ clΩ k and �Dg�(Ω) ≤ k � � are compact subsets of a normed space BVc (Ω), � · �L1 (Ω) . It follows that {BVk } is a localizing sequence. Throughout the remainder of this book, we denote by T the localized topology U{BVk } . If X ⊂ BVc (Ω) and no confusion is possible, we write (X, T) instead of the cumbersome (X, T X). Letting Dk = BVk ∩ D(Ω), we have two topologies in D(Ω): � � T := T D(Ω) and T := U D(Ω) {D } . k
The inclusion T D(Ω) ⊂ T may be proper, and the topology T may not be locally convex. By Proposition 10.2.2 (2), the necessary condition for the equality T D(Ω) = T is that T D(Ω) is a sequential topology. It is easy to see that this condition is also sufficient. Whether it holds is unclear. Proposition 10.5.2. A linear functional F : D(Ω) → R is a charge if and only if it is T continuous. Proof. Assume F is a charge. Fix k ∈ N and choose ε > 0. By Definition 10.1.2, there is θ > 0 such that F (ϕ) ≤ θ�ϕ�L1 (Ω2k ) + 2εk for each ϕ ∈ D2k . If ϕ,ψ ∈ Dk and �ϕ − ψ�L1 (Ωk ) < ε, then ϕ − ψ ∈ D2k and F (ϕ) − F (ψ) = F (ϕ − ψ) ≤ (θ + 2k)ε.
Thus F � Dk is uniformly U continuous. By Lemma 5.5.6, the space (BVk , U) is the completion of (Dk , U). We infer that F has a unique U continuous extension to BVk . As this is true for each k ∈ N, the functional F has an extension to BVc (Ω), which is T continuous by Proposition 10.2.1, (ii). In particular, F is T continuous. Conversely, assume F is T continuous, and hence T continuous according to Proposition 10.2.1, (3). Choose ε > 0 and an open set U � Ω. Find k ∈ N with U ⊂ Ωk . Since F is U continuous in Dk , there is η > 0 such that F (ψ) ≤ ε for every ψ ∈ D(U ) satisfying �Dψ�L1 (U ;Rn ) ≤ 1 and �ψ�L1 (U ) ≤ η. Now let θ := ε/η, and select ϕ ∈ D(U ) so that �Dϕ�L1 (U ;Rn ) = 1. Letting c := �ϕ�L1 (U ) , we distinguish two cases. (i) If c ≤ η, then F (ϕ) ≤ ε = ε�Dϕ�L1 (U ;Rn ) . (ii) If c > η, let ψ := (η/c)ϕ. As �ψ�L1 (U ) = η and
�Dψ�L1 (U ;Rn ) = (η/c)�Dϕ�L1 (U ;Rn ) = η/c < 1, we obtain F (ψ) ≤ ε = ηθ, and consequently F (ϕ) = (c/η)F (ψ) ≤ cθ = θ�ϕ�L1 (U ) .
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In either case the inequality F (ϕ) ≤ θ�ϕ�L1 (U ) + ε�Dϕ�L1 (U ;Rn )
(∗)
holds, and it remains to remove the assumption �Dϕ�L1 (U ;Rn ) = 1. Assuming b = �Dϕ�L1 (U ;Rn ) > 0, inequality (∗) is valid for ϕ/b, and hence for ϕ. If �Dϕ�L1 (U ;Rn ) = 0, then ϕ = 0 and (∗) is satisfied, since spt ϕ � U implies V − spt ϕ �= ∅ for each connected component V of U . � � Proposition � 10.5.3. � The space BVc (Ω), T is the sequential completion of the space D(Ω), T . � � Proof. Lemma 5.5.6 shows that BVc (Ω), T is the sequential closure of � � D(Ω). Since BVc (Ω), T is a sequentially complete space by Theorem 10.3.3, the proposition is proved. Proposition 10.5.4. Each F ∈ CH(Ω) has a unique T continuous extension F� : BVc (Ω) → R,
which is linear. Given an open set U � Ω and ε > 0, there is θ > 0 such that F� (g) ≤ θ�g�L1 (U ) + ε�Dg�(U )
for each g ∈ BVc (U ). Moreover, for every open set U � Ω, � � �F �U = sup F� (g) : g ∈ BVc (U ) and �Dg�(U ) ≤ 1 . If Fv is the distributional divergence of v ∈ C(Ω; R ), then � �v (g) = − v · d(Dg) F
(10.5.1)
(10.5.2)
n
(10.5.3)
Ω
for each g ∈ BVc (Ω).
Proof. Let F ∈ CH(Ω). The first claim is an immediate consequence of Propositions 10.5.3 and 10.5.2. The linearity of F� follows from that of F and the continuity of F� . Next fix an open set U � Ω and g ∈ BVc (U ). By Lemma 5.5.6, there is {ϕj } in D(U ) such that lim �ϕj − g�L1 (U ) = 0
and
sup �Dϕj �L1 (U ;Rn ) ≤ �Dg�(U ).
Choose ε > 0 and find θ > 0 so that for j = 1, 2, . . . , F (ϕj ) ≤ θ�ϕj �L1 (U ) + ε�Dϕj �L1 (U ;Rn ) . � � Since {ϕj } converges to g in BVc (Ω), T , we obtain F� (g) = lim F (ϕk ) ≤ θ�g�L1 (U ) + ε�Dg�(U ),
F� (g) = lim F (ϕk ) ≤ �F �U �Dϕk �L1 (U ;Rn ) ≤ �F �U �Dg�(U ).
(∗) (∗∗)
Now (∗) is the inequality (10.5.1), and (∗∗) implies � � sup F� (g) : g ∈ BVc (U ) and �Dg�(U ) ≤ 1 ≤ �F �U . ✐
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Since the reverse inequality is obvious, equality (10.5.2) is proved. Finally, Proposition 5.5.4 yields w-lim Dϕk = Dg in Ω. As there is w ∈ Cc (Ω; Rn ) such that w � U = v � U , we conclude � � �v (g) = lim Fv (ϕk ) = − lim v · Dϕk = − v · d(Dg). F Ω
Ω
In view of Proposition 10.5.4, throughout we assume that all charges are � � continuous linear functionals defined on BVc (Ω), T . Accordingly we identify F with F� , and denote both by F .
Theorem 10.5.5. There are equalities � �∗ � �∗ CH(Ω) = D(Ω), T = BVc (Ω), T ,
and the evaluation map E : BVc (Ω) → CH(Ω)∗ , defined by � � E(g), F = �F, g�
(10.5.4)
for g ∈ BVc (Ω) and F ∈ CH(Ω), is linear and bijective.
Proof. The equalities follow from Propositions 10.5.2 and 10.5.4. Observe that for each F ∈ CH(Ω) and k = 2, 3, . . . , � � 1 1 k−1 sup F (g) : g ∈ BVk−1 } ≤ �F �Ωk ≤ k sup F (g) : g ∈ BVk }.
This and Theorem 10.3.3, (2) show that the topology in CH(Ω) defined by seminorms (10.1.1) is the strong topology S∗ ; see Section 1.2. The theorem follows from Corollary 10.4.2.
10.6. Streams Fix an open setΩ ⊂ Rn , and let
� � BVc (Ω) := E ⊂ Rn : χE ∈ BVc (Ω) .
Note that for Ω= Rn , the family BVc (Ω) coincides with the family BVc (Rn ) defined in Section 6.7. Via the injection E �→ χE : BVc (Ω) → BVc (Ω), we identify BVc (Ω) with a � � closed subset of BVc (Ω), T where T is the localized topology defined in Section 10.5. It � � � � follows from Proposition 6.7.3 that BVc (Ω), T is the sequential completion of DF(Ω), T . Let v ∈ C(Ω; Rn ). The flux of v is the additive function � v · νE dHn−1 : DF(Ω) → R φv : A �→ A
introduced in Section 2.1. If Fv is the distributional divergence of v, then φv (A) = Fv (χA )
for each A ∈ DF(Ω) according to Proposition 10.5.4. Thus given an open set U ⊂ Ω and ε > 0, there is θ > 0 such that for each A ∈ DF(Ω), � � �φv (A)� ≤ θ|A| + εP(A). This motivates a definition analogous to that of a charge (Definition 10.1.2).
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Definition 10.6.1. An additive function φ : DF(Ω) → R is called a stream in Ω if given an open set U � Ω and ε > 0, there is θ > 0 such that for each A ∈ DF(Ω), � � �φ(A)� ≤ θ|A| + εP(A).
Proposition 10.6.2. Each stream φ : DF(Ω) → R is T continuous and has a unique � � : BVc (Ω), T) → R. The extension is additive with respect to continuous extension φ nonoverlapping sets. Moreover, given an open set U � Ω and ε > 0, there is θ > 0 such that for each E ∈ BVc (U ), � � � �φ(E) � ≤ θ|E| + εP(E). (10.6.1)
Proof. Proving the T continuity of φ is similar to proving the T continuity of a charge; see � � � � the proof of Proposition 10.5.2. As BVc (Ω), T is a sequential completion of DF(Ω), T , � � � : BVc (Ω), T → R is obvious. the existence and uniqueness of the continuous extension φ � is the same as that of Proposition 2.4.3. Choose an open set The proof of additivity of φ U � Ω and ε > 0. Given E ∈ BVc (U ), Proposition 6.7.3 implies that there is a constant γ = γ(n) > 0 and a sequence {Ak } in DF(U ) such that sup P(Ak ) ≤ γP(E)
and
lim |E � Ak | = 0.
Find θ > 0 corresponding to U and ε/γ according to Definition 10.6.1. Then � � � � � �φ(E) � = lim�F (Ak )� ≤ θ lim |Ak | + ε lim sup P(Ak ) ≤ θ|E| + εP(E). γ
In view of Proposition 10.6.2, we assume that all streams are defined on BVc (Ω) and � It is clear that each charge F defines a stream φ = F � BVc (Ω). The identify φ with φ. converse is also true: in a slightly different context, the following theorem is proved in [51, Section 4.1]. Theorem 10.6.3. Let φ : BVc (Ω) → R be a stream. Then F : BVc (Ω) → R defined by � ∞ � ∞ � � � � φ {g > t} dt − φ {−g > t} dt F (g) := 0
0
for each g ∈ BVc (Ω) is a charge and F � BVc (Ω) = φ. Moreover, if G : BVc (Ω) → R is a charge such that G � BVc (Ω) = φ, then G = F . In accordance with our agreement, each stream is a T continuous additive function defined on BVc (Ω). The next example shows that the converse is false. Example 10.6.4. Let v(x) = x/|x| for x ∈ R2 − {0}, and v(0) = 0. Since v is admissible and div v ∈ L1loc (R2 ), Proposition 7.4.3 implies that � � div v(x) dx = v · νA dH1 A
for each A ∈ BVc
(R2 ).
∂∗ A
As the Lebesgue integral is absolutely continuous, � � � div v(x) dx : BVc (R2 ), T → R φ : A �→ A
is an additive T continuous function. On the other hand, for Bk = B(0, 1/k) we calculate � v · νBk dH1 = P(Bk ). φ(Bk ) = �
�
∂Bk
In particular lim φ(Bk )/P(Bk ) = 1. Proceeding toward a contradiction, assume that φ is a stream. Choose 0 < ε < 1, and let U = U (0, 2). There is θ > 0 such that � � �φ(B)� ≤ θ|B| + εP(B)
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for each B ∈ BVc (U ). Since {Bk } is a sequence in BVc (U ), we obtain a contradiction: 1 = lim
|Bk | φ(Bk ) ≤ θ lim + ε = ε. P(Bk ) P(Bk )
The following result, proved in [23, Section 6], describes the structure of T continuous additive functions defined on BVc (Ω). � � Theorem 10.6.5. Each continuous additive function F : BVc (Ω), T → R is the sum of a stream and an absolutely continuous signed measure.
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Chapter 11 The divergence equation
For bounded vector fields, and then for continuous vector fields, we characterize distributions which are their distributional divergences. These results, obtained jointly by T. De Pauw and the author [23], extend parts of an earlier work of J. Bourgain and H. Brezis [10]. As the proofs involve functional analysis in an essential way, some familiarity with basic properties of locally convex spaces is assumed. More specialized facts are presented with precise references, but often without proofs.
11.1. Background Throughout this chapter, n ≥ 2 andΩ ⊂ Rn is a fixed open set. Consistently we denote by p and q a conjugate pair of extended real numbers, i.e., we assume that 1 ≤ p ≤ ∞, 1 ≤ q ≤ ∞, and defining 1/∞ := 0, we also assume 1 1 + = 1. p q
The Sobolev conjugate of 1 ≤ p ≤ n is the extended real number � np if p < n, ∗ p := n−p ∞ if p = n. Let L ∈ D� (Ω) be any distribution. We want to find a vector field v in that is a weak solution of the divergence equation
L1loc (Ω; Rn )
div v = L. In other words, we wish to find v ∈ L1loc (Ω; Rn ) whose distributional divergence Fv equals L; see Example 3.1.2, (3). Depending on L we are interested in finding v that satisfies some regularity conditions, such as boundedness, or continuity, or both. An important special case is when L = Lf for a function f in Lploc (Ω); see Example 3.1.2, (1). As usual, instead of div v = Lf , we write div v = f.
(11.1.1)
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11. The divergence equation
The naive approach of solving equation (11.1.1) in Ω= Rn by letting � ξ1 f (t,ξ 2 , . . . , ξn ) dt v1 (ξ1 , . . . , ξn ) = 0
and v = (v1 , 0, . . . , 0) is not satisfactory. It provides regular solutions only to the extent to which f is regular in the variables ξ2 , . . . , ξn . Utilizing Poisson’s equation is more promising. Since � = div ◦ D, we find a weak solution u : Ω → R of Poisson’s equation �u = f
and let v := Du. Assuming that f ∈ Lp (Ω) has compact support, the regularity of Du depends on the relationship between p and the dimension n. Out of many successful applications, we quote two that emphasize the critical nature of the exponent p = n. ∗
(i) If 1 < p < n, then Du ∈ Lp (Ω; Rn ); see [71, Chapter 8, Section 4.2]. (ii) If p > n, then Du is continuous [45, Theorem 10.2]. If p = n, regular solutions of the divergence equation (11.1.1) cannot be obtained by solving Poisson’s equation. The following example of L. Nirenberg shows that Du need not be locally bounded when p = n. Example 11.1.1. For each x = (ξ1 , . . . , ξn ) in Rn , let � � �s ξ1 �log |x|� ϕ(x) if x �= 0, u(x) := 0 if x = 0
where 0 < s < (n − 1)/n and ϕ ∈ Cc∞ (Rn ) equals one in a neighborhood of 0. A direct calculation reveals that Du is not bounded about 0. Since �u has compact support, and since there is c > 0 such that � � � � � �u(x)� ≤ c �log |x|�s−1 |x| for each x ∈ Rn , we see that f := �u belongs to Ln (Rn ).
Notwithstanding Nirenberg’s example, Bourgain and Brezis have shown that a locally bounded, in fact continuous, solution of equation (11.1.1) exists [10, Proposition 1]. We elaborate on their ideas. In broad terms, our approach to solving the divergence equation can be described as follows. We consider a space X equal to Lp (Ω; Rn ), or to C(Ω; Rn ), and identify a linear space Y ⊂ D� (Ω) such that v �→ Fv : L1loc (Ω) → D� (Ω)
maps X into Y . Then, using functional analysis, we show that the map v �→ Fv : X → Y is surjective. With no reference to Poisson’s equation, we prove
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(i) if f ∈ Lp (Ω) and 1 < p ≤ n, then equation (11.1.1) has a weak solution ∗ in Lp (Ω; Rn ); see Theorem 11.2.3 below. (ii) if f ∈ Lnloc (Ω), then equation (11.1.1) has a continuous weak solution in L∞ (Ω; Rn ); see Theorem 11.3.9 below.
11.2. Solutions in Lp (Ω; Rn ) If v ∈ Lq (Ω; Rn ) then by H¨ older’s inequality, � v · Dϕ ≤ �v�Lq (Ω;Rn ) �Dϕ�Lp (Ω;Rn ) Fv (ϕ) = −
(11.2.1)
Ω
for each ϕ ∈ D(Ω). This suggests to define � � �L�p := sup L(ϕ) : ϕ ∈ D(Ω) and �Dϕ�Lp (Ω;Rn ) ≤ 1
for each L ∈ D� (Ω). Observe that � · �p is a norm in the linear space � � D�p (Ω) := L ∈ D� (Ω) : �L�p < ∞ .
Since inequality (11.2.1) implies
�Fv �p ≤ �v�Lq (Ω;Rn )
(11.2.2)
for every v ∈ L (Ω; R ), there is a continuous linear map � � � � v �→ Fv : Lq (Ω; Rn ), � · �Lq (Ω;Rn ) → D�p (Ω), � · �p . q
n
Proposition 11.2.1. Let p < ∞. The map
v �→ Fv : Lq (Ω; Rn ) → D�p (Ω)
is surjective. Moreover, given L ∈ D�p (Ω), there is v ∈ Lq (Ω; Rn ) such that Fv = L and �v�Lq (Ω;Rn ) = �L�p . � � Proof. Let X = Dϕ : ϕ ∈ D(Ω) . As the support of ϕ ∈ D(Ω) differs from each connected component of Ω, the map D : ϕ �→ Dϕ : D(Ω) → X
is a bijection. Choose L ∈ D�p (Ω), and define a linear functional G := L ◦ D−1 : X → R.
If w ∈ X, let ϕ = D−1 (w) and observe
G(w) = L(ϕ) ≤ �L�p �Dϕ�Lp (Ω;Rn ) = �L�p �w�Lp (Ω;Rn ) . By the Hahn-Banach theorem, the functional G extends to a linear functional � : Lp (Ω; Rn ) → R so that G � G(w) ≤ �L�p �w�Lp (Ω;Rn )
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(∗)
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“Book˙2011” — 2012/2/26 — 9:58 — page 222 — #232
222
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11. The divergence equation
for each w ∈ Lp (Ω; Rn ). The following diagram commutes: ⊂
� Lp (Ω; Rn ) X� �� �� ��G �� � bijection D G �� �� � �� D(Ω) R L Since p < ∞, the standard duality properties of Lp spaces [63, Theorem 6.16] show that there is v ∈ Lq (Ω; Rn ) such that � � v·w G(w) = − Ω
for each w ∈ L (Ω; R ), and that � � � �v�Lq (Ω;Rn ) = sup G(w) : w ∈ Lp (Ω; Rn ) and �w�Lp (Ω;Rn ) ≤ 1 ; p
n
in particular, �v�Lq (Ω;Rn ) ≤ �L�p by (∗). If ϕ ∈ D(Ω) then � � v · Dϕ = Fv (ϕ). L(ϕ) = G(Dϕ) = G(Dϕ) =− Ω
We conclude that L = Fv , and hence �v�Lq (Ω;Rn ) ≤ �Fv �p . The proposition follows from inequality (11.2.2). Corollary 11.2.2. Let q > 1. The equation div v = L has a weak solution v ∈ Lq (Ω; Rn ) if and only if L ∈ D�p (Ω). Moreover, the solution v can be selected so that �v�Lq (Ω;Rn ) = �L�p . In order to apply Corollary 11.2.2 to the equation div v = f for a function f ∈ Lp (Ω), we need the Gagliardo-Nirenberg-Sobolev inequality. Theorem 11.2.3. Let 1 ≤ r < n. There is κ = κ(n, r) > 0 such that �ϕ�Lr∗ (Ω) ≤ κ�Dϕ�Lr (Ω;Rn )
(GNS)
for each ϕ ∈ D(Ω). Proof. If r = 1, inequality (GNS) is the Sobolev’s inequality of Theorem 5.9.8. Thus assume 1 < r < n and find β = β(n, r) > 0 so that � r n � r =β − . r−1 r−1 n−1
Applying the Sobolev and H¨ older � n−1 �� � n � β n n−1 |ϕ| ) ≤γ Ω
Ω
≤ βγ
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inequalities to |ϕ|β , we obtain � � � �� � β � |ϕ|β−1 |Dϕ| �D |ϕ| � = βγ
��
Ω
Ω
�
|ϕ|
β−1
r � r−1
� r−1 �� r
Ω
|Dϕ|
r
� r1
.
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11.2. Solutions in Lp (Ω; Rn )
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223
A direct calculation reveals β = r(n − 1)/(n − r), and hence n r =β = r∗ . (β − 1) r−1 n−1
Thus letting κ = βγ, the previous inequality translates to �� �� � n−1 � r−1 �� � r1 n r r∗ r∗ r |ϕ| ≤κ |ϕ| |Dϕ| . Ω
Ω
Ω
The proposition follows, since 1 n−1 r−1 − = ∗. n r r Theorem 11.2.4. There is κ = κ(n, p) > 0 such that for each f ∈ Lp (Ω), ∗ 1 < p ≤ n, the equation div v = f has a weak solution v ∈ Lp (Ω; Rn ) with �v�Lp∗ (Ω;Rn ) ≤ κ�f �Lp (Ω) . Proof. If p = n then p∗ = ∞. By the H¨older and (GNS) inequalities, � n f ϕ ≤ �f �Ln (Ω) �ϕ�L n−1 ≤ κ�f �Ln (Ω) �Dϕ�L1 (Ω;Rn ) Lf (ϕ) = (Ω) Ω
for every ϕ ∈ D(Ω); here κ = κ(n, 1). It follows �Lf �1 ≤ κ�f �Ln (Ω) , and consequently Lf ∈ D�1 (Ω). According to Corollary 11.2.2, the equation div v = Lf has a weak solution v ∈ L∞ (Ω; Rn ) such that �v�L∞ (Ω;Rn ) = �Lf �1 ≤ κ�f �Ln (Ω) . n < p∗ < ∞ and there is 1 < r < n such that If 1 < p < n, then n−1 1 1 1 1 r + p∗ = 1. A direct calculation shows that r ∗ + p = 1. Given ϕ ∈ D(Ω), the H¨ older and (GNS) inequalities yield
Lf (ϕ) ≤ �f �Lp (Ω) �ϕ�Lr∗ (Ω) ≤ κ�f �Lp (Ω) �Dϕ�Lr (Ω,Rn ) where κ = κ(n, r). We infer �Lf �r ≤ κ�f �Lp (Ω) , and consequently Lf belongs to D�r (Ω). Corollary 11.2.2 implies that the equation div v = Lf has a weak ∗ solution v ∈ Lp (Ω; Rn ) such that �v�Lp∗ (Ω;Rn ) = �Lf �r ≤ κ�f �Lp (Ω) . Unlike in Corollary 11.2.2, the sufficient condition of Theorem 11.2.4 is by no means necessary. Example 11.2.5. Assume n = 2, and letΩ a := (0, a)2 where 0 < a ≤ 1. For (ξ,η ) ∈ Ωa , define f (ξ,η ) = ξ −η . If 1 ≤ p ≤ 1/a, we calculate � � 1 1 p f (ξ,η ) dξdη = t−1 at dt. p 1−ap Ωa
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224
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11. The divergence equation
Thus f ∈ Lp (Ωa ) if and only if 1 ≤ p < 1/a. On the other hand, � 1−η � ξ v : (ξ,η ) �→ , 0 : Ω1 → R2 1−η belongs to C ∞ (Ω1 ; R2 ) and div v(x) = f (x) for each x ∈ Ω1 . In particular, div v = f has a weak solution in L∞ (Ω1/2 , R2 ) for a function f that does not belong to L2 (Ω1/2 ) — a fact unobtainable from Theorem 11.2.4.
11.3. Continuous solutions n Throughout, we topologize C(Ω; Rn ) as a subspace of L∞ loc (Ω; R ). In view of n Section 1.5, with this topology C(Ω; R ) is a Fr´echet space. The distributional divergence of v ∈ C(Ω; Rn ) is denoted by Fv ; see Example 3.1.2, (3). It follows from (10.1.2) that there is a continuous linear map
Γ : v �→ Fv : C(Ω; Rn ) → CH(Ω)
(11.3.1)
and we show that Γ is�surjective.� To this end, we establish first that the range ofΓ , i.e., the spaceΓ C(Ω; Rn ) , is dense in CH(Ω). Then we prove that the range of the adjoint map Γ∗ : CH(Ω)∗ → C(Ω; Rn )∗ is closed in the strong topology S∗ of C(Ω; Rn )∗ , and infer the surjectivity of Γ form the close range theorem (Theorem 11.3.5 below). � � Lemma 11.3.1. The linear space Γ C(Ω; Rn ) is dense in CH(Ω).
Proof. Choose α ∈ CH(Ω)∗ so that α(Fv ) = 0 for each v ∈ C(Ω; Rn ). Using Theorem 10.5.5, find g ∈ BVc (Ω) with E(g) = α. By equality (10.5.4), � � � v · d(Dg) 0 = �α, Fv � = E(g), Fv = �Fv , g� = − Ω
for every v ∈ C(Ω; R ). It follows from Proposition 5.3.11 that Dg ≡ 0. Since |U −spt g| > 0 for every connected component U of Ω, Proposition 5.5.9 shows that g ≡ 0; see Convention 1.3.2. Thus α = E(g) = 0, and the lemma is a consequence of the Hahn-Banach theorem [64, Theorem 3.5]. n
Let U ⊂ Rn be an open set. An amiable subset of U is a compact set K ⊂ U such that each connected component V of U − K satisfies either d(V ) = ∞ or ∂V ∩ ∂U �= ∅. Observe that a compact set K is an amiable subset of Rn if and only if Rn − K is connected. Lemma 11.3.2. Let U ⊂ Rn be an open set. Each compact set C ⊂ U is contained in an amiable subset K of U .
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“Book˙2011” — 2012/2/26 — 9:58 — page 225 — #235
11.3. Continuous solutions
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225
Proof. Denote by V the collection of all bounded connected components V of U − C such that ∂V ⊂ C, and by W the collection of all other connected components of U − C. For V ∈ V, dist (V, ∂U ) = dist (∂V, ∂U ) ≥ dist (C, ∂U ) and since V is bounded, also d(V ) = d(∂V ) ≤ d(C). Consequently �� � �� � dist V, ∂ U ≥ dist (C, ∂U ) and d V ≤ 3d(C). � Thus V is a relatively closed subset of U whose closure is contained in U . � � It follows that V is a compact set, and so is K = C ∪ V. If W ∈ W is bounded, then ∂W is a subset of ∂(U − C) = ∂U ∪ ∂C, but not a subset of C. Thus each W ∈ W is either unbounded or ∂W ∩ ∂U �= ∅. Since W consists of all connected components of U − K, we conclude that K is an amiable subset of U . Lemma 11.3.3. Let g ∈ BV (Ω). If the support of Dg is contained in an amiable subset of Ω, then so is the support of g. Proof. Assume spt Dg is contained in an amiable subset K of Ω. Choose a connected component V ofΩ − K. If V ⊂ S then d(V ) < ∞, and the definition of K leads to a contradiction: ∅ �= ∂V ∩ ∂Ω ⊂ spt Dg ⊂ Ω. Thus V − spt Dg is a nonempty open set, and hence |V − spt Dg| > 0. We infer g(x) = 0 for almost all x ∈ V , since g is constant almost everywhere in V by Proposition 5.5.9. AsΩ − K has only countably many connected components, g(x) = 0 for almost all x ∈ Ω − K, and the lemma follows from Convention 1.3.2; cf. Remark 10.5.1. Lemma 11.3.4. Let {gi } be a sequence in BVc (Ω) such that � ��� � � � γ(B) := sup �� v · d(Dgi )�� : v ∈ B and i = 1, 2, . . . < ∞ Ω
for every bounded set B ⊂ C(Ω; Rn ). Then sup �Dgi �(Ω) < ∞, and there is a compact set K ⊂ Ω containing the support of each gi . � � Proof. As the set C = v ∈ C(Ω; Rn ) : �v�L∞ (Ω;Rn ) ≤ 1 is bounded in C(Ω; Rn ), Proposition 5.2.3 and Theorem 5.5.1 show that for i = 1, 2, . . . , �� � ∞ n �Dgi �(Ω) = sup gi div v : v ∈ C ∩ Cc (Ω; R ) Ω � � � ∞ n = sup − v · d(Dgi ) : v ∈ C ∩ Cc (Ω; R ) Ω � � ��� � � � � ≤ sup � v · d(Dgi )� : v ∈ C ≤ γ(C) < ∞. Ω
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226
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11. The divergence equation
�∞ Next find open setsΩ i such thatΩ i � Ωi+1 and i=1 Ωi = Ω. In view of Lemmas 11.3.2 and 11.3.3, it suffices to show that spt Dgi ⊂ Ωj for some j ∈ N and every i ∈ N. Suppose this is not true, and construct recursively subsequences of {gi } and {Ωi }, still denoted by {gi } and {Ωi }, so n that (Ωi+1 − clΩ i ) ∩ spt Dgi �= ∅. There are vi ∈ �� C(Ω; R )� such that �vi �L∞ (Ω;Rn ) ≤ 1, spt vi ⊂ Ωi+1 − clΩ i , and ai = � Ω vi · Dgi � > 0. Let � � −1 bi := max a−1 , and observe that 1 , . . . , ai � � B := v ∈ C(Ω; Rn ) : �v�L∞ (Ωi+1 ;Rn ) ≤ ibi for i = 1, 2, . . . is a bounded subset of C(Ω; Rn ) containing all wi = (ibi )vi ; see Section 1.2. On the other hand, �� �� � � � � � � � � � γ(B) ≥ � wi · d(Dgi )� = ibi � vi · d(Dgi )�� ≥ i Ω
Ω
for i = 1, 2, . . . , contrary to our assumption.
Let X ∗ and Y ∗ be the duals of locally convex spaces X and Y , respectively. If Φ: X → Y is a continuous linear map, then the formula � � � ∗ ∗ � (11.3.2) Φ (y ), x := y ∗ , Φ(x) ,
where x ∈ X and y ∗ ∈ Y ∗ , defines a linear mapΦ ∗ : Y ∗ → X ∗ , called the adjoint map of Φ. Recall that the weak* topology W∗ and the strong topology S∗ , both in X ∗ , are defined in Section 1.2, and that some of their properties were discussed in Seciton 10.4. The next important result is called the close range theorem. Theorem 11.3.5. Let X and Y be Fr´echet spaces. If Φ : X → Y is a continuous linear map, then the following conditions are equivalent: (1) (2) (3) (4)
Φ(X) is closed in Y ; Φ∗ (Y ∗ ) is closed in (X ∗ , W∗ ); Φ∗ (Y ∗ ) is closed in (X ∗ , S∗ ); Φ∗ (Y ∗ ) is sequentially closed in (X ∗ , S∗ ).
Proof. The equivalence of conditions (1), (2), and (3) is proved in [27, Theorem 8.6.13]. Analyzing the proof of implication (3) ⇒ (2) ibid., it is not difficult to show that (3) can be relaxed to (4). For the details we refer the reader to [24, Proposition 6.8]. � � Lemma 11.3.6. Γ C(Ω; Rn ) is a closed subset of CH(Ω).
Proof. Following Theorem 11.3.5, we consider the adjoint map Γ∗ : CH(Ω)∗ → C(Ω; Rn )∗ ,
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11.3. Continuous solutions 227 � � � � and show thatΓ ∗ CH(Ω)∗ is sequentially closed in C(Ω; Rn )∗ , S∗ . To this end, denote by Mc (Ω; Rn ) the linear space of all compactly supported Rn valued measures in Ω. For each µ ∈ Mc (Ω; Rn ), � v · dµ : C(Ω; Rn ) → R Lµ : v �→ − Ω
is a continuous linear functional, and there is a linear map Λ : µ �→ Lµ : Mc (Ω; Rn ) → C(U ; Rn )∗ . According to Theorem 10.5.5, the evaluation map E : BVc (Ω) → CH(Ω)∗ is a linear bijection. The mapsΓ ∗ , Λ, E, and D : g �→ Dg : BVc (Ω) → Mc (Ω; Rn ) are linked by the following commutative diagram: D
BVc (Ω) −−−−→ Mc (Ω; Rn ) bijection�E Λ�
(11.3.3)
Γ∗
CH(Ω)∗ −−−−→ C(Ω; Rn )∗
Indeed for g ∈ BVc (Ω) and v ∈ C(Ω; Rn ), the defining equalities (10.5.3), (10.5.4), and (11.3.2) yield � � � v · d(Dg) = �Fv , g� (Λ ◦ D)(g), v = �LDg , v� = − Ω � � � � � � = E(g), Fv = E(g), Γ(v) = (Γ∗ ◦ E)(g), v . For a bounded set B ⊂ C(Ω; Rn ) and S ∈ C(Ω; Rn )∗ , let �� � � S B := sup �S(v)� : v ∈ B
and note that · B are seminorms which define the strong topology S∗ in C(Ω; Rn )∗ ; see Section 1.2. � � Proceeding proof, select {αi } in CH(Ω)∗ such that Γ∗ (αi ) � to thenactual � converges in C(Ω; R )∗ , S∗ to some T ∈ C(Ω; Rn )∗ . Define gi := E−1 (αi ), and infer from diagram (11.3.3) that � � ∗ � Γ (αi ), v = − v · d(Dgi ) Ω
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228
11. The divergence equation � � for�each v ∈ C(Ω;�Rn ). Being convergent, the sequence Γ∗ (αi ) is bounded in C(Ω; Rn )∗ , S∗ . Thus for each bounded set B ⊂ C(Ω; Rn ), � ��� � � � � � sup � v · d(Dgi )� : v ∈ B and i = 1, 2, . . . Ω
��� � ��� � = sup � Γ∗ (αi ), v � : v ∈ B and i = 1, 2, . . . � � = sup Γ∗ (αi ) B : i = 1, 2, . . . < ∞.
Lemma 11.3.4 implies that sup �Dgi �(Ω) < ∞, and that there is a compact set K ⊂ Ω containing the support of each gi . By Theorem 5.5.12, the sequence {gi } has a subsequence, still denoted by {gi }, such that lim �g − gi �L1 (Ω) = 0 for some g ∈ BVc (Ω). Moreover Dg = w-lim Dgi by Proposition 5.5.4. Let α := E(g) and choose v ∈ C(Ω; Rn ). Since the supports of g and of all gi are contained in K, we can apply the weak convergence to v: � � � �T, v� = lim Γ∗ (αi ), v = − lim v · d(Dgi ) = Ω � � � =− v · d(Dg) = Γ∗ (α), v . Ω
From the arbitrariness of v, we conclude T = Γ∗ (α).
Remark 11.3.7. Although we did not need this for the proof of Lemma 11.3.6, using the Riesz theorem (Theorem 5.4.4), one can show that the map Λ : µ �→ Lµ : Mc (Ω; Rn ) → C(Ω; Rn )∗ is bijective [27, Theorem 4.10.1]. Loosely speaking, it means that D : BVc (Ω) → Mc (Ω; Rn )
and
Γ∗ : CH(Ω) → C(Ω; Rn )∗
are two representations of the same map. A geometric rationale supports our vague assertion. Identifying vectors in C(Ω; Rn ) with the corresponding (n−1)-dimensional differential forms, the weak divergence operator Γ becomes the weak exterior derivative. In this context, interpreting elements of BVc (Ω) as n-dimensional normal currents [33, Section 4.5.1], it is clear that the adjoint mapΓ ∗ is the boundary operator D. This reasoning suggests that diagram (11.3.3) may commute. The actual verification of it is routine.
The following theorem is an easy corollary of Lemmas 11.3.1 and 11.3.6. Theorem 11.3.8. Let L ∈ D� (Ω). The equation div v = L has a weak solution v ∈ C(Ω; Rn ) if and only if L is a charge in Ω. Let f ∈ Lnloc (Ω). Combining Theorem 11.3.8 and Proposition 10.1.4, we see at once that the equation div v = f has a weak solution v ∈ C(Ω; Rn ). In view of this and Theorem 11.2.4, the equation div v = f has bounded weak solutions, as well as continuous weak solutions. The next theorem shows that it has a solution v which is bounded and continuous simultaneously. The idea of the proof is due to T. De Pauw.
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11.3. Continuous solutions
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229
Theorem 11.3.9. Let f ∈ Ln (Ω). There is κ = κ(n) > 0 such that the equation div v = f has a weak solution v ∈ C(Ω; Rn ) with �v�L∞ (U ;Rn ) ≤ κ�f �Ln (Ω) . Proof. Avoiding a triviality assume �f �Ln (Ω) > 0, and let � � �Lf � := sup Lf (g) : g ∈ BVc (Ω) and �Dg�(Ω) ≤ 1 .
By H¨ older and Sobolev inequalities there is β = β(n) > 0 such that � n f g ≤ �f �Ln (Ω) �g�L n−1 ≤ β�f �Ln (Ω) �Dg�(Ω) Lf (g) = (Ω) Ω
n for each g ∈ BVc (Ω). Thus � �Lf � ≤ β�f �L (Ω) , and we claim �Lf � > 0. Indeed, �Lf � = 0 implies Ω f ϕ = 0 for each ϕ ∈ D(Ω), and consequently f (x) = 0 for almost all x ∈ Ω contrary to our assumption. Letting κ = 3β, it suffices to show that the nonempty convex sets � � A : = v ∈ C(Ω; Rn ) : �v�L∞ (Ω;Rn ) < 3�Lf � , � B : = w ∈ C(Ω; Rn ) :Γ( w) = Lf }
have nonempty intersection. Proceeding toward a contradiction, suppose the intersection A∩B is empty. Since A is an open set, the Hahn-Banach theorem implies that there are T ∈ C(Ω; Rn )∗ and γ ∈ R such that �T, v� < γ ≤ �T, w� for each v ∈ A and each w ∈ B; see [64, Theorem 3.4, (a)]. As 0 ∈ A, we see that γ > 0. Claim. Γ−1 (0) ⊂ T −1 (0).
Proof. Choose u ∈ Γ−1 (0) and w ∈ B. Observe that w + tu belongs to B for each t ∈ R. Thus T (w) + tT (u) = T (w + tu) ≥ γ for all t ∈ R. This is impossible unless T (u) = 0. By the claim, there is S ∈ CH(Ω)∗ such that the diagram Γ
� CH(Ω) C(Ω; Rn ) ��� ��� S ��� T ��� � R commutes. As the evaluation map E : BVc (Ω) → CH(Ω)∗ is bijective (Theorem 10.5.5), there is g ∈ BVc (Ω) with E(g) = S. By (10.5.3), � � � � � (∗) v · d(Dg) = �Fv , g� = E(g), Fv = S, Γ(v) = �T, v� − Ω
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230
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11. The divergence equation
for each v ∈ C(Ω; Rn ). If w ∈ B, then � � γ ≤ �T, w� = S, Γ(w) = �S, Lf � � � = E(g), Lf = �Lf , g� ≤ �Lf � · �Dg�(Ω).
(∗∗)
Now select u ∈ Cc1 (Ω; Rn ) with �u�L∞ (Ω;Rn ) ≤ 1, and observe that the vector field v := 2�Lf � u belongs to A ∩ Cc1 (Ω; Rn ). According to (∗), � � � g div u = g div v = − v · d(Dg) = �T, v� < γ . 2�Lf � Ω
Ω
Ω
� � As u is arbitrary, �Dg�(U ) ≤ γ/ 2�Lf � < γ /�Lf � contrary to (∗∗).
Remark 11.3.10. The following are some concluding comments.
(1) Recently P. Bouafia [8] proved that there exists no continuous map F �→ vF : CH(Ω) → C(Ω; Rn ) such that vF is a weak solution of div v = F for each F ∈ CH(Ω). (2) If p > n and f ∈ Lp (Ω) has compact support, then div v = f has solutions whose regularity is stronger than continuity. For instance, � � �v(x) − v(y)� ≤ |x − y|s
for all x, y ∈ Ω and 0 < s < 1 − (n/p); see [45, Theorem 11.2]. On the other hand, D. Preiss [57] constructed a function f ∈ Cc (Rn ) such that div v = f has no weak solution that is locally Lipschitz. (3) For more results related to Theorem 11.3.8 we refer to [10, 23, 24, 25].
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Bibliography
I not only use all the brains I have, but all that I can borrow. Woodrow Wilson
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“Book˙2011” — 2012/2/26 — 9:58 — page 235 — #245
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List of symbols
Euclidean spaces and related concepts R, R+ , R, Rm , 3
A � B, A ⊙ B, 32
N, Z, Q, Q+ , C, 3
C(x; r, h), 50
A � B, 3
B • , 53
x · y, |x|, 4, 205
(xy), [xy], 58
d(E), clE, int E, ∂E, 4
Lu , 59
dist (A, B), dist (x, B), 4
S n−1 , Πe,t , He,t , 66
U (E, r), U (x, r) 4
Ωε , 77
B(E, r), B(x, r) 4
C ε , 98
A�B 4
Ez,r , 131
e i , πi , Π i , 6
C ∗ 194
A + B, tA, −A, 7
R∞ , 211
Functions and maps f ≡ c, f � B, 4
Lf , Lµ , Fv , 36
{f = t}, {f > t}, {f �= t}, 4
V(f, Ω), V(f ), 75
f (x), f [x], �f, x�, 4
(f )E , 110
4
φ# g, 117
χE , φ, spt φ, 6
T f , 168
φ−1 (B), 7
T v, 178
Γ (s), α(s), 13
O(1), o(1), 185
Lip φ, 17
E, 212, 216
Lip φ(x), Hs φ(x), 18
φv , 216
f + , f − , |f |, 4
uA,B ,
ηε ∗ f , 77
Families of sets Dk , DC, DF, 6
B, 79
P, [P ], 24
St(A, E), Cε , 98
St(x, E), 24
BVc (Rn ), 146
DF, 32
N (K, r), P (K, r), 182
A
E
42
E∞ , 185
P(Ω), Ploc (Ω), 56
DC(Ω), DF(Ω) 193
BV(Ω), BVloc (Ω), 73
BVc (Ω), 216 235
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“Book˙2011” — 2012/2/26 — 9:58 — page 236 — #246
236
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List of symbols
Measures and related concepts µ, µ
�ν�, 81
Y, spt µ, 10
φ ∼ ψ, A ∼ B, 11
ν+ , ν− , 82
ess spt φ, ess sup φ, 11
(h)B,µ , 84
Ln , |E|, 13 � � � n E f (x) dx, E f, E f dL , 13
w-lim µk , 86
Hδs , Hs , 14
ext∗ E, int∗ E, cl∗ E,∂
∗ E,
49
µE , 126 ∂ ∗ E, 127 H± (E, x), H(E, x), 127
s(E), 51
intc E, 158
Θ(E, x), 54
ρ(n), 159
P(E, Ω), P(E), 56
M∗t (E), Mt∗ (E), 181
L L extL ∗ E, int∗ E, cl∗ E,∂
V(E, Ω), V(E), 73
µ
L ∗ E,
64
h, 79
M∗t (K,∂ ∗ A), 184 dim K, 186 Vδ F (E), V F (E),
197
Function spaces and norms C(E; Rs ), C(E), 4
Lipc (Ω), Liploc (Ω), 17
C k (Ω; Rs ), C ∞ (Ω; Rs ), 5
Adm(E; Rm ), Adm(E)
Cc (Ω; Rs ), 6
D(Ω; C), D(Ω), 35
L0 (E, µ; Rs ), 11
D� (Ω; C), D� (Ω), 36
� · �Lp (E,µ;Rs ) , 1 ≤ p ≤ ∞, 11
BV (Ω), � · �BV (Ω) , 76
Lp (E, µ; Rs ), 12
BVloc (Ω), 76
Lp (E, µ), � · �Lp (E,µ) , 12
BV (Ω; Rn ), 178
p
Lloc (Ω, µ; Rs ), 12
Lp (E; Rs ),
p
Lloc
(Ω; Rs ),
Lip(E; Rm ), Lip(E), 17
13
27
AC∗ (Ω), 200
CH(Ω), � · �U , 206
BVc (Ω), 213
Differentiation Dφ(x), 16
Di f , 93
Di f (x), div v(x), 17
adj Dφ, 117
det Dφ, Jφ , 19
DE φ(x), 155
Dα ,
35
DE,i f (x), 157
Λ, pΛ , 37 ¯ 37 �, ∂,
divE v(x), 158 DF (x), DF (x), F � (x), 193
Df, �Df �, 92
div v(x), 201
Topology cl T E, Gδ , 7 X∗,
∗
W , S , 9
�x∗ �B , 9
inj lim(Xα , T
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UC , 209
∗
W, 212 T, T, 214
Xα ), 43
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“Book˙2011” — 2012/2/26 — 9:58 — page 237 — #247
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Index
set in locally convex space, 9
ACG∗ function, 201 absolutely continuous measure, 116 absorbing set, 8 additive function, 21, 195 adjacent dyadic cubes, 24 adjoint map, 226 matrix, 117 admissible complex-valued function, 38 map, 27 ambient space, 6 amiable subset, 224 area theorem, 19 Arens topology, 213
Ck, C∞ function, 5 map, 5 Cantor discontinuum, 183 Cauchy-Riemann equation, 38 caviar, 79 cell, 6 change of variables theorem, 19 characteristic polynomial, 37 charge, 206 closed range theorem, 226 closure, 4, 7 essential, 50 relative to line, 64 coarea theorem, 120 compact set originally, 213 weakly, 212 complex derivative, 38 conjugate pair of numbers, 219 Sobolev, 219 constant H¨ older, 18 Lipschitz, 17 at point, 18 control function, 187 controlled map, 186 fully, 187 convergence from inside, 151 from outside, 151 of BV sets, 146 of figures, 32 of test functions, 35
BV function, 76 set, 73 ball closed, 4 open, 4 Bernstein function, 41 Besicovitch theorem, 38, 123 Borel set, 7 map, 7 Borel measurable map, 7 measure, 10 Borel regular measure, 10 boundary, 4 essential, 50 relative to line, 64 reduced, 127 operator, 228 bounded map at point, 185 237
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“Book˙2011” — 2012/2/26 — 9:58 — page 238 — #248
238 weak, 86 convex hull, 8 set, 8 center of, 110 centrally symmetric, 110 convolution, 77 countable set, 3 cover, 3 star, 24 covering number, 182 critical interior, 158 variation, 197 cube, 6 dyadic, 6 mother of, 194 current integral, 147 normal, 228 cylinder, 50 deformation theorem, 147 De Giorgi theorem, 41 Denjoy-Perron primitive, 201 density of set, 54 derivative, 16, 193 complex, 38 exterior, 228 partial, 17 relative, 155 diameter, 4 differentiable map, 16 relatively, 155 differential form, 228 directed family, 42 distance, 4 distribution, 35 distributional divergence, 36, 178 gradient, 93 partial derivative, 93 divergence, 17 distributional, 36, 178 equation, 219 mean, 201 relative, 158 divergence theorems, 30, 141, 166, 178, 189, 201 domain
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Index extension, 179 Lipschitz, 50 dual space, 9 dyadic cube, 6 figure, 6 partition, 24 Egoroff theorem, 11 equation Cauchy-Riemann, 38 divergence, 219 Laplace, 39 minimal surface, 40 Poisson, 220 equivalent maps, 10 sets, 10 essential boundary, 50 relative to line, 64 closure, 50 relative to line, 64 exterior, 50 relative to line, 64 interior, 50 relative to line, 64 support, 11 extension domain, 179 exterior derivative, 228 essential, 50 normal, 21, 23, 128 family of separating seminorms, 8 of sets directed, 42 locally finite, 98 point-finite, 98 point-p-finite, 98 figure, 6 dyadic, 6 flux, 21, 33, 141 Fr´ echet space, 8 function, 4 ACG∗ , 201 BV , 76 Ck, C∞, 5 additive, 21, 195
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“Book˙2011” — 2012/2/26 — 9:58 — page 239 — #249
Index critical variation of, 197 admissible complex-valued, 38 Bernstein, 41 control, 187 finite, 4 gamma, 13 graph of, 40 harmonic, 39 holomorphic, 38 mean value of, 110 of bounded variation, 76 real-valued, 4 test, 35 Urysohn, 4 variation of, 75 Gδ set, 7 gamma function, 13 (GNS) inequality, 222 graph of function, 40 surface area of, 40 half-space, 66 harmonic function, 39 Hausdorff measure, 14 topology, 7 Henstock lemma, 12 H¨ older constant, 18 inequality, 12 holomorphic function, 38 operator, 37 hull convex, 8 linear, 8 hyperplane, 59 indicator, 6 inequality Gagliardo-Nirenberg-Sobolev, 222 H¨ older, 12 isodiametric, 14 isoperimetric, 112 Poincar´ e, 115 relative isoperimetric, 115 Sobolev, 113 injective limit, 43 integer, 3
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239 integral current, 147 integral-geometric measure, 144 integration by parts theorem, 31, 191 interior, 4 critical, 158 essential, 50 relative to line, 64 interlacing families of sets, 43 sequences of sets, 209 isodiametric inequality, 14 isoperimetric inequality, 108 relative, 115 Jacobian, 19 Jordan decomposition, 83 Kirschbraun theorem, 18 Laplace equation, 39 operator, 37 Lebesgue integral, 13 measure, 13 point, 126 lemma Henstock, 12 Whitney, 99 line, 58 linear differential operator, 37 elliptic, 37 hull, 8 space, 7 topology, 8 lipeomorphism, 17 Lipschitz constant, 17 at point, 18 domain, 50 map, 17 at point, 18 localized topology, 209 localizing sequence, 210 locally BV function, 76 BV set, 73 convex space, 8 convex topology, 8
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“Book˙2011” — 2012/2/26 — 9:58 — page 240 — #250
240 finite family, 98 integrable function, 12 Lipschitz map, 17 Luzin theorem, 11 µ σ-finite set, 10 µ equivalent maps, 10 sets, 10 µ measurable map, 11 set, 10 µ negligible set, 10 µ overlapping sets, 10 map Ck, C∞, 5 µ measurable, 11 adjoint, 226 admissible, 27 bounded at point, 185 Borel, 7 Borel measurable, 7 controlled, 186 fully, 187 derivative of, 16 differentiable, 16 Lipschitz, 17 at point, 18 range of, 224 support of, 6 zero extension of, 6 mean divergence, 201 value of function, 110 measurable map, 11 set, 10 measure, 9 Rm -valued, 79 polar decomposition of, 85 variation of, 81 σ-finite, 10 absolutely continuous, 116 Borel, 10 Borel regular, 10 Hausdorff, 14 integral-geometric, 144 Lebesgue, 13 lives in, 10 metric, 10
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Index Radon, 10 reduction of, 10 signed, 79 Jordan decomposition of, 83 negative, positive parts of, 83 support of, 10 variational, 93 metric measure, 10 metrizable space, 8 minimal surface equation, 40 minimizer, 152 Minkowski content lower, 181 upper, 181 relative, 184 Minkowski dimension, 186 mollifier, 76 standard, 77 multi-index, 35 multiplicity of point, 119 positive, negative, 119 negligible set, 10 normal current, 228 null set, 4 number, 3 complex, 3 covering, 182 extended real, 3 finite, 3 packing, 182 rational, 3 real, 3 operator boundary, 228 linear differential, 37 elliptic, 37 holomorphic, 37 Laplace, 37 originally compact set, 213 overlapping sets, 13 packing number, 182 partial derivatives, 17 distributional, 93 relative, 157 δ-fine, 24 body of, 24 dyadic, 24
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“Book˙2011” — 2012/2/26 — 9:58 — page 241 — #251
Index in a set, 24 of unity, 165 partition of unity, 165 perimeter, 56 Poincar´ e inequality, 115 point-finite family, 98 point-p-finite family, 98 pointwise Lipschitz map, 18 Poisson equation, 220 polar decomposition, 85 polytop, 67 projection, 6 Rm -valued measure, 79 polar decomposition of, 85 variation of, 81 Rademacher theorem, 18 Radon measure, 10 rectifiable set, 141 reduction of measure, 10 regular set, 159 relative derivative, 155 divergence, 158 isoperimetric inequality, 115 partial derivatives, 157 upper Minkowski content, 184 Riesz representation theorem, 88 σ-finite measure, 10 set, 10 segment closed, 58 open, 58 seminorm, 8 semireflexive space, 213 sequential space, 7 topology, 7 sequentially complete space, 210 set BV, 73 Gδ , 7 µ σ-finite, 10 µ measurable, 10 µ negligible, 10 absorbing, 8 blow-up of, 131 Borel, 7
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241 bounded in locally convex space, 9 compact originally, 213 weakly, 212 convex, 8 center of, 110 centrally symmetric, 110 countable, 3 density of, 54 null, 4 of bounded variation, 73 of finite perimeter, 56 perimeter of, 56 rectifiable, 141 regular, 159 sequentially closed, 7 shape of, 51 singular, 186 size of, 186 symmetric, 8 variation of, 73 sets µ equivalent, 10 µ overlapping, 10 overlapping, 13 symmetric difference of, 3 signed measure, 79 Jordan decomposition of, 83 negative, positive parts of, 83 shape of set, 51 Simon theorem, 41 size of singular set, 186 Sobolev conjugate, 219 inequality, 113 solution strong, 37, 40 weak, 37, 40, 219 space ambient, 6 dual, 9 Fr´ echet, 8 linear, 7 locally convex, 8 metrizable, 8 semireflexive, 213 sequential, 7 sequentially complete, 210 topological, 7 sphere in Rn , 66
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“Book˙2011” — 2012/2/26 — 9:58 — page 242 — #252
242 standard base in Rn , 6 star cover, 24 Stepanoff theorem, 18, 157 stream, 217 strong topology, 9 subspace topology, 7 support of map, 6 essential, 11 of measure, 10 Swiss cheese, 79 symmetric difference of sets, 3 test function, 35 Tietze theorem, 4 theorem area, 19 Besicovitch, 38, 123 change of variables, 19 close range, 226 deformation, 147 De Giorgi, 41 divergence, 30, 141, 166, 178, 189, 201 Egoroff, 11 integration by parts, 31, 191 Kirschbraun, 18 Luzin, 11 Rademacher, 18 Riesz, 88 Simon, 41 Stepanoff, 18, 157 Tietze, 4 Vitali, 53 Ward, 195 Whitney, 19 topological space, 7 Frech´ et, 8 linear, 8 locally convex, 8
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Index metrizable, 8 sequential, 7 sequentially complete, 210 topology, 7 Arens, 213 Hausdorff, 7 localization of, 209 localized, 209 locally convex, 8 metrizable, 8 sequential, 7 strong, 9 weak, 212 weak*, 9 trace, 158, 174, 178 upper Minkowski content, 181 relative, 184 dimension, 186 Urysohn function, 4 variation critical, 197 of function, 75 of set, 73 vector field, 21 Vitali theorem, 53 Ward theorem, 195 weak convergence, 86 topology, 212 weak* topology, 9 Whitney division, 100 extension theorem, 19 zero extension of map, 6
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