Characterizations of signed measures in the dual of BV and related isometric isomorphisms
aa r X i v : . [ m a t h . A P ] M a r CHARACTERIZATIONS OF SIGNED MEASURES IN THE DUAL OF BV AND RELATED ISOMETRIC ISOMORPHISMS
NGUYEN CONG PHUC AND MONICA TORRES
Abstract.
We characterize all (signed) measures in BV nn − ( R n ) ∗ , where BV nn − ( R n ) is definedas the space of all functions u in L nn − ( R n ) such that Du is a finite vector-valued measure. Wealso show that BV nn − ( R n ) ∗ and BV ( R n ) ∗ are isometrically isomorphic, where BV ( R n ) is definedas the space of all functions u in L ( R n ) such that Du is a finite vector-valued measure. Asa consequence of our characterizations, an old issue raised in Meyers-Ziemer [16] is resolved byconstructing a locally integrable function f such that f belongs to BV ( R n ) ∗ but | f | does not.Moreover, we show that the measures in BV nn − ( R n ) ∗ coincide with the measures in ˙ W , ( R n ) ∗ ,the dual of the homogeneous Sobolev space ˙ W , ( R n ) , in the sense of isometric isomorphism. Fora bounded open set Ω with Lipschitz boundary, we characterize the measures in the dual space BV (Ω) ∗ . One of the goals of this paper is to make precise the definition of BV (Ω) , which is thespace of functions of bounded variation with zero trace on the boundary of Ω . We show that themeasures in BV (Ω) ∗ coincide with the measures in W , (Ω) ∗ . Finally, the class of finite measuresin BV (Ω) ∗ is also characterized. Introduction
It is a challenging problem in geometric measure theory to give a full characterization of the dualof BV , the space of functions of bounded variation. Meyers and Ziemer characterized in [16] thepositive measures in R n that belong to the dual of BV ( R n ) . They defined BV ( R n ) as the spaceof all functions in L ( R n ) whose distributional gradient is a finite vector-measure in R n with normgiven by k u k BV ( R n ) = k Du k ( R n ) . They showed that the positive measure µ belongs to BV ( R n ) ∗ if and only if µ satisfies the condition µ ( B ( x, r )) ≤ Cr n − for every open ball B ( x, r ) ⊂ R n and C = C ( n ) . Besides the classical paper by Meyers and Ziemer,we refer the interested reader to the paper by De Pauw [9], where the author analyzes SBV ∗ , thedual of the space of special functions of bounded variation.In Phuc-Torres [17] we showed that there is a connection between the problem of characterizing BV ∗ and the study of the solvability of the equation div F = T . Indeed, we showed that the (signed)measure µ belongs to BV ( R n ) ∗ if and only if there exists a bounded vector field F ∈ L ∞ ( R n , R n ) such that div F = µ . Also, we showed that µ belongs to BV ( R n ) ∗ if and only if(1.1) | µ ( U ) | ≤ C H n − ( ∂U ) for any open (or closed) set U ⊂ R n with smooth boundary. The solvability of the equation div F = T , in various spaces of functions, has been studied in Bourgain-Brezis [5], De Pauw-Pfeffer [10], DePauw-Torres [11] and Phuc-Torres [17] (see also Tadmor [19]).In De Pauw-Torres [11], another BV -type space was considered, the space BV nn − ( R n ) , definedas the space of all functions u ∈ L nn − ( R n ) such that Du , the distributional gradient of u , is a finitevector-measure in R n . A closed subspace of BV nn − ( R n ) ∗ , which is a Banach space denoted as CH , Key words and phrases.
BV space, dual of BV, measures. was characterized in [11] and it was proven that T ∈ CH if and only if T = div F , for a continuousvector field F ∈ C ( R n , R n ) vanishing at infinity.In this paper we continue the analysis of BV ( R n ) ∗ and BV nn − ( R n ) ∗ . We show that BV ( R n ) ∗ and BV nn − ( R n ) ∗ are isometrically isomorphic (see Corollary 3.3). We also show that the measures in BV nn − ( R n ) ∗ coincide with the measures in ˙ W , ( R n ) ∗ , the dual of the homogeneous Sobolev space ˙ W , ( R n ) (see Theorem 4.7), in the sense of isometric isomorphism. We remark that the space ˙ W , ( R n ) ∗ is denoted as the G space in image processing (see Meyer [15]), and that it plays a keyrole in modeling the structured component of an image.It is obvious that if µ is a locally finite signed Radon measure then k µ k ∈ BV ( R n ) ∗ implies that µ ∈ BV ( R n ) ∗ . The converse was unknown to Meyers and Ziemer as they raised this issue in theirclassical paper [16, page 1356]. In Section 5, we show that the converse does not hold true in generalby constructing a locally integrable function f such that f ∈ BV ( R n ) ∗ but | f | 6∈ BV ( R n ) ∗ .In this paper we also study these characterizations in bounded domains. Given a bounded openset Ω with Lipschitz boundary, we consider the space BV (Ω) defined as the space of functions ofbounded variation with zero trace on ∂ Ω . One of the goals of this paper is to make precise thedefinition of this space (see Theorem 6.10) . We then characterize all (signed) measures in Ω thatbelong to BV (Ω) ∗ . We show that a locally finite signed measure µ belongs to BV (Ω) ∗ if and onlyif (1.1) holds for any smooth open (or closed) set U ⊂⊂ Ω , and if and only if µ = div F for a vectorfield F ∈ L ∞ (Ω , R n ) (see Theorem 7.4). Moreover, we show that the measures in BV (Ω) ∗ coincidewith the measures in W , (Ω) ∗ (see Theorem 7.6), in the sense of isometric isomorphism.In the case of BV (Ω) , the space of functions of bounded variation in a bounded open set Ω withLipschitz boundary (but without the condition of having zero trace on ∂ Ω ), we shall restrict ourattention only to measures in BV (Ω) ∗ with bounded total variation in Ω , i.e., finite measures. Thisis in a sense natural since any positive measure that belongs to BV (Ω) ∗ must be finite due to thefact that the function ∈ BV (Ω) . We show that a finite measure µ belongs to BV (Ω) ∗ if and only if(1.1) holds for every smooth open set U ⊂⊂ R n , where µ is extended by zero to R n \ Ω (see Theorem8.2). 2. Functions of bounded variation
In this section we define all the spaces that will be relevant in this paper.2.1.
Definition.
Let Ω be any open set. The space M (Ω) consists of all finite (signed) Radonmeasures µ in Ω ; that is, the total variation of µ , denoted as k µ k , satisfies k µ k (Ω) < ∞ . The space M loc (Ω) consists of all locally finite Radon measures µ in Ω ; that is, k µ k ( K ) < ∞ for every compactset K ⊂ Ω . Note here that M loc (Ω) is identified with the dual of the locally convex space C c (Ω) (the spaceof continuous real-valued functions with compact support in Ω ) (see [7]), and thus it is a real vectorspace. For µ ∈ M loc (Ω) , it is not required that either the positive part or the negative part of µ hasfinite total variation in Ω .2.2. Definition.
Let Ω be any open set. The space of functions of bounded variation, denoted as BV (Ω) , is defined as the space of all functions u ∈ L (Ω) such that the distributional gradient Du is a finite vector-valued measure in Ω . The space BV (Ω) is a Banach space with the norm (2.1) k u k BV (Ω) = k u k L (Ω) + k Du k (Ω) , where k Du k (Ω) denotes the total variation of the vector-valued measure Du over Ω . For the casewhen Ω = R n we will equip BV ( R n ) with the homogeneous norm given by (2.2) k u k BV ( R n ) = k Du k ( R n ) . Another BV -like space is BV nn − ( R n ) , defined as the space of all functions in L nn − ( R n ) such that Du is a finite vector-valued measure. The space BV nn − ( R n ) is a Banach space when equipped with EASURES IN THE DUAL OF BV the norm k u k BV nn − ( R n ) = k Du k ( R n ) . Remark. BV ( R n ) is not a Banach space under the norm (2.2) . Also, we have k Du k (Ω) = sup (cid:26) ˆ Ω u div ϕdx : ϕ ∈ C c (Ω) and | ϕ ( x ) | ≤ ∀ x ∈ Ω (cid:27) , where ϕ = ( ϕ , ϕ , ..., ϕ n ) and | ϕ ( x ) | = ( ϕ ( x ) + ϕ ( x ) + · · · + ϕ n ( x ) ) / . In what follows, weshall also write ´ Ω | Du | instead of k Du k (Ω) . We will use the following Sobolev’s inequality for functions in BV ( R n ) whose proof can be foundin [3, Theorem 3.47]:2.4. Theorem.
Let u ∈ BV ( R n ) . Then (2.3) k u k L nn − ( R n ) ≤ C ( n ) k Du k ( R n ) . Inequality (2.3) immediately implies the following continuous embedding(2.4) BV ( R n ) ֒ → BV nn − ( R n ) . We recall that the standard Sobolev space W , (Ω) is defined as the space of all functions u ∈ L (Ω) such that Du ∈ L (Ω) . The Sobolev space W , (Ω) is a Banach space with the norm(2.5) k u k W , (Ω) = k u k L (Ω) + k Du k L (Ω) = ˆ Ω h | u | + ( | D u | + | D u | + · · · + | D n u | ) i dx. However, we will often refer to the following homogeneous Sobolev space. Hereafter, we let C ∞ c (Ω) denote the space of smooth functions with compact support in a general open set Ω .2.5. Definition.
Let ˙ W , ( R n ) denote the space of all functions u ∈ L nn − ( R n ) such that Du ∈ L ( R n ) . Equivalently, the space ˙ W , ( R n ) can also be defined as the closure of C ∞ c ( R n ) in BV nn − ( R n ) (i.e., in the norm k Du k L ( R n ) ) . Thus, u ∈ ˙ W , ( R n ) if and only if there exists a sequence u k ∈ C ∞ c ( R n ) such that ´ R n | D ( u k − u ) | dx = 0 , and moreover, ˙ W , ( R n ) ֒ → BV nn − ( R n ) . Definition.
Given a bounded open set Ω , we say that the boundary ∂ Ω is Lipschitz if for each x ∈ ∂ Ω , there exist r > and a Lipschitz mapping h : R n − → R such that, upon rotating andrelabeling the coordinate axes if necessary, we have Ω ∩ B ( x, r ) = { y = ( y , . . . , y n − , y n ) : h ( y , . . . , y n − ) < y n } ∩ B ( x, r ) . Remark.
Let Ω be a bounded open set with Lipschitz boundary. We denote by W , (Ω) theSobolev space consisting of all functions in W , (Ω) with zero trace on ∂ Ω . Then it is well-knownthat C ∞ c (Ω) is dense in W , (Ω) . One of the goals of this paper is to make precise the definitionof BV (Ω) , the space of all functions in BV (Ω) with zero trace on ∂ Ω (see Theorem 6.10). In thispaper we equip the two spaces, BV (Ω) and W , (Ω) , with the equivalent norms (see Theorem 6.11)to (2.1) and (2.5) , respectively, given by k u k BV (Ω) = k Du k (Ω) , and k u k W , (Ω) = ˆ Ω | Du | dx. Definition.
For any open set Ω , we let BV c (Ω) denote the space of functions in BV (Ω) withcompact support in Ω . Also, BV ∞ (Ω) and BV ∞ (Ω) denote the space of bounded functions in BV (Ω) and BV (Ω) , respectively. Finally, BV ∞ c (Ω) is the space of all bounded functions in BV (Ω) withcompact support in Ω . We will use the following result (see [13, Proposition 1.13]). We include the proof here for thesake of completeness.
NGUYEN CONG PHUC AND MONICA TORRES
Lemma.
Suppose { u k } is a sequence in BV (Ω) such that u k → u in L loc (Ω) and (2.6) lim k →∞ ˆ Ω | Du k | = ˆ Ω | Du | . Then for every open set A ⊂ Ω , ˆ A ∩ Ω | Du | ≥ lim sup k →∞ ˆ A ∩ Ω | Du k | . In particular, if ´ ∂A ∩ Ω | Du | = 0 , then (2.7) ˆ A | Du | = lim k →∞ ˆ A | Du k | . Proof.
Consider the open set B = Ω \ A . Since u k → u in L loc (Ω) , by the lower semicontinuityproperty we have(2.8) ˆ A | Du | ≤ lim inf k →∞ ˆ A | Du k | , and ˆ B | Du | ≤ lim inf k →∞ ˆ B | Du k | . On the other hand, ˆ A ∩ Ω | Du | + ˆ B | Du | = ˆ Ω | Du | = lim k →∞ ˆ Ω | Du k | = lim k →∞ (cid:18) ˆ A ∩ Ω | Du k | + ˆ B | Du k | (cid:19) , by (2.6) = lim sup k →∞ (cid:18) ˆ A ∩ Ω | Du k | + ˆ B | Du k | (cid:19) ≥ lim sup k →∞ ˆ A ∩ Ω | Du k | + lim inf k →∞ ˆ B | Du k |≥ lim sup k →∞ ˆ A ∩ Ω | Du k | + ˆ B | Du | , by (2.8) , and hence ˆ A ∩ Ω | Du | ≥ lim sup k →∞ ˆ A ∩ Ω | Du k | . In particular, if ´ ∂A ∩ Ω | Du | = 0 then we obtain from the last inequality ˆ A | Du | = ˆ A ∩ Ω | Du | ≥ lim sup k →∞ ˆ A ∩ Ω | Du k | ≥ lim sup k →∞ ˆ A | Du k | ≥ lim inf k →∞ ˆ A | Du k | , and since, by (2.8), ˆ A | Du | ≤ lim inf k →∞ ˆ A | Du k | ≤ lim sup k →∞ ˆ A | Du k | , clearly (2.7) follows. (cid:3) The following theorem from functional analysis (see [18, Theorem 1.7] ) will be fundamental inthis paper:2.10.
Theorem.
Let X be a normed linear space and Y be a Banach space. Suppose T : D → Y isa bounded linear transformation, where D ⊂ X is a dense linear subspace. Then T can be uniquelyextended to a bounded linear transformation ˆ T from X to Y . In addition, the operator norm of T is c if and only if the norm of ˆ T is c . The following formula will be important in this paper.
EASURES IN THE DUAL OF BV Lemma.
Let µ ∈ M loc ( R n ) and f be a function such that ´ R n | f | d k µ k < + ∞ . Then ˆ R n f dµ = ˆ ∞ µ ( { f ≥ t } ) dt − ˆ −∞ µ ( { f ≤ t } ) dt. The same equality also holds if we replace the sets { f ≥ t } and { f ≤ t } by { f > t } and { f < t } ,respectively.Proof. We write f = f + − f − , where f + ≥ and f − ≥ are the positive and negative parts of f .Then ˆ R n f dµ = ˆ R n ( f + − f − ) dµ = ˆ ∞ µ ( { f + ≥ t } ) dt − ˆ ∞ µ ( { f − ≥ t } ) dt = ˆ ∞ µ ( { f ≥ t } ) dt − ˆ ∞ µ ( {− f ≥ t } ) dt = ˆ ∞ µ ( { f ≥ t } ) dt − ˆ ∞ µ ( { f ≤ − t } ) dt = ˆ ∞ µ ( { f ≥ t } ) dt − ˆ −∞ µ ( { f ≤ s } ) ds, by making the change of variables t = − s, which is the desired result. (cid:3) BV ∞ c ( R n ) is dense in BV nn − ( R n ) Theorem.
Let u ∈ BV nn − ( R n ) , u ≥ , and φ k ∈ C ∞ c ( R n ) be a nondecreasing sequence ofsmooth functions satisfying: (3.1) ≤ φ k ≤ , φ k ≡ on B k (0) , φ k ≡ on R n \ B k (0) and | Dφ k | ≤ c/k. Then (3.2) lim k →∞ k ( φ k u ) − u k BV nn − ( R n ) = 0 , and for each fixed k > we have (3.3) lim j →∞ k ( φ k u ) ∧ j − φ k u k BV nn − ( R n ) = 0 . In particular, BV ∞ c ( R n ) is dense in BV nn − ( R n ) .Proof. As BV nn − ( R n ) ⊂ BV loc ( R n ) , the product rule for BV loc functions gives that D ( φ k u ) = φ k Du + uDφ k (as measures) (see [3, Proposition 3.1]) and hence φ k u ∈ BV ( R n ) ⊂ BV nn − ( R n ) .Thus ˆ R n | D ( uφ k − u ) | = ˆ R n | φ k Du − Du + uDφ k |≤ ˆ R n | φ k − || Du | + ˆ R n ∩ supp ( Dφ k ) | u || Dφ k |≤ ˆ R n | φ k − || Du | + ck ˆ B k \ B k | u |≤ ˆ R n | φ k − || Du | + ck ˆ B k \ B k | u | nn − ! n − n | B k \ B k | n ≤ ˆ R n | φ k − || Du | + c ˆ B k \ B k | u | nn − ! n − n . (3.4) NGUYEN CONG PHUC AND MONICA TORRES
We let k → ∞ in (3.4) and use (3.1) and the dominated convergence theorem together with the factthat u ∈ L nn − to obtain (3.2).On the other hand, the coarea formula for BV functions yields ˆ R n | D ( φ k u − ( φ k u ) ∧ j ) | = ˆ ∞ H n − ( ∂ ∗ { φ k u − ( φ k u ) ∧ j > t } ) dt = ˆ ∞ H n − ( ∂ ∗ { φ k u − j > t } ) dt = ˆ ∞ H n − ( ∂ ∗ { φ k u > j + t } ) dt = ˆ ∞ j H n − ( ∂ ∗ { φ k u > s } ) ds. Here ∂ ∗ E stands for the reduced boundary of a set E . Since ´ ∞ H n − ( ∂ ∗ { φ k u > s } ) ds < ∞ , theLebesgue dominated convergence theorem yields the limit (3.3) for each fixed k > .By the triangle inequality and (3.2)-(3.3), each nonnegative u ∈ BV nn − ( R n ) can be approximatedby a function in BV ∞ c ( R n ) . For a general u ∈ BV nn − ( R n ) , let u + be the positive part of u . Fromthe proof of [3, Theorem 3.96], we have u + ∈ BV loc ( R n ) and k Du + k ( A ) ≤ k Du k ( A ) for any openset A ⋐ R n . Thus k Du + k ( R n ) ≤ k Du k ( R n ) < + ∞ and u + belongs to BV nn − ( R n ) . Likewise, wehave u − ∈ BV nn − ( R n ) . Now by considering separately the positive and negative parts of a function u ∈ BV nn − ( R n ) , it is then easy to see the density of BV ∞ c ( R n ) in BV nn − ( R n ) . (cid:3) We have the following corollaries of Theorem 3.1:3.2.
Corollary. BV ∞ c ( R n ) is dense in BV ( R n ) .Proof. This follow immediately from (2.4) and Theorem 3.1. (cid:3)
Corollary.
The spaces BV ( R n ) ∗ and BV nn − ( R n ) ∗ are isometrically isomorphic.Proof. We define the map S : BV nn − ( R n ) ∗ → BV ( R n ) ∗ as S ( T ) = T BV ( R n ) . First, we note the S is injective since S ( T ) = 0 implies that T BV ( R n ) ≡ . In particular, T BV ∞ c ( R n ) ≡ . Since BV ∞ c ( R n ) is dense in BV nn − ( R n ) and T is continuous on BV nn − ( R n ) ,it is easy to see that T BV nn − ( R n ) ≡ . We now proceed to show that S is surjective. Let T ∈ BV ( R n ) ∗ . Then T BV ∞ c ( R n ) is a continuous linear functional. Using again that BV ∞ c ( R n ) is dense in BV nn − ( R n ) , T BV ∞ c ( R n ) has a unique continuous extension ˆ T ∈ BV nn − ( R n ) ∗ andclearly S ( ˆ T ) = T . Moreover, for any T ∈ BV ( R n ) ∗ , the unique extension ˆ T to BV nn − ( R n ) has thesame norm (see Theorem 2.10), that is, k T k BV ( R n ) ∗ = (cid:13)(cid:13)(cid:13) ˆ T (cid:13)(cid:13)(cid:13) BV nn − ( R n ) ∗ , and hence (cid:13)(cid:13)(cid:13) S ( ˆ T ) (cid:13)(cid:13)(cid:13) BV ( R n ) ∗ = (cid:13)(cid:13)(cid:13) ˆ T (cid:13)(cid:13)(cid:13) BV nn − ( R n ) ∗ , which implies that S is an isometry. (cid:3) We now proceed to make precise our definitions of measures in ˙ W , ( R n ) ∗ and BV nn − ( R n ) ∗ . EASURES IN THE DUAL OF BV Definition.
We let M loc ∩ ˙ W , ( R n ) ∗ := { T ∈ ˙ W , ( R n ) ∗ : T ( ϕ ) = ˆ R n ϕdµ for some µ ∈ M loc ( R n ) , ∀ ϕ ∈ C ∞ c ( R n ) } . Therefore, if µ ∈ M loc ( R n ) ∩ ˙ W , ( R n ) ∗ , then the action < µ, u > can be uniquely defined for all u ∈ ˙ W , ( R n ) (because of the density of C ∞ c ( R n ) in ˙ W , ( R n )) . Definition.
We let M loc ∩ BV nn − ( R n ) ∗ := { T ∈ BV nn − ( R n ) ∗ : T ( ϕ ) = ˆ R n ϕ ∗ dµ for some µ ∈ M loc , ∀ ϕ ∈ BV ∞ c ( R n ) } , where ϕ ∗ is the precise representative of ϕ in BV ∞ c ( R n ) . Thus, if µ ∈ M loc ∩ BV nn − ( R n ) ∗ , then theaction < µ, u > can be uniquely defined for all u ∈ BV nn − ( R n ) (because of the density of BV ∞ c ( R n ) in BV nn − ( R n )) . We will study the normed linear spaces M loc ∩ ˙ W , ( R n ) ∗ and M loc ∩ BV nn − ( R n ) ∗ in the nextsection. In particular, we will show in Theorem 4.7 below that these spaces are isometrically isomor-phic. In Definition 3.5, if we use C ∞ c ( R n ) instead of BV ∞ c ( R n ) , then by the Hahn-Banach Theoremthere exist a non-zero T ∈ BV nn − ( R n ) ∗ that is represented by the zero measure, which would causea problem of injectivity in Theorem 4.7.4. Characterizations of measures in BV nn − ( R n ) ∗ The following lemma characterizes all the distributions in ˙ W , ( R n ) ∗ . We recall that ˙ W , ( R n ) is the homogeneous Sobolev space introduced in Definition 2.5.4.1. Lemma.
The distribution T belongs to ˙ W , ( R n ) ∗ if and only if T = div F for some vectorfield F ∈ L ∞ ( R n , R n ) . Moreover, k T k ˙ W , ( R n ) ∗ = min {k F k L ∞ ( R n , R n ) } , where the minimum is taken over all F ∈ L ∞ ( R n , R n ) such that div F = T . Here we use the norm k F k L ∞ ( R n , R n ) := (cid:13)(cid:13)(cid:13) ( F + F + · · · + F n ) / (cid:13)(cid:13)(cid:13) L ∞ ( R n ) for F = ( F , . . . , F n ) . Proof.
It is easy to see that if T = div F where F ∈ L ∞ ( R n , R n ) then T ∈ ˙ W , ( R n ) ∗ with k T k ˙ W , ( R n ) ∗ ≤ k F k L ∞ ( R n , R n ) . Conversely, let T ∈ ˙ W , ( R n ) ∗ . Define A : ˙ W , ( R n ) → L ( R n , R n ) , A ( u ) = Du, and note that the range of A is a closed subspace of L ( R n , R n ) since ˙ W , ( R n ) is complete. Wedenote the range of A by R ( A ) and we define T : R ( A ) → R as T ( Du ) = T ( u ) , for each Du ∈ R ( A ) . Then we have k T k R ( A ) ∗ = k T k ˙ W , ( R n ) ∗ . By Hahn-Banach Theorem there exists a norm-preserving extension T of T to all L ( R n , R n ) .On the other hand, by the Riesz Representation Theorem for vector valued functions (see [8, pp.98–100]) there exists a vector field F ∈ L ∞ ( R n , R n ) such that T ( v ) = ˆ R n F · v, for every v ∈ L ( R n , R n ) , NGUYEN CONG PHUC AND MONICA TORRES and k F k L ∞ ( R n , R n ) = k T k L ( R n , R n ) ∗ = k T k R ( A ) ∗ = k T k ˙ W , ( R n ) ∗ . In particular, for each ϕ ∈ C ∞ c ( R n ) we have T ( ϕ ) = T ( Dϕ ) = T ( Dϕ ) = ˆ R n F · Dϕ, which yields T = div ( − F ) , with k− F k L ∞ ( R n , R n ) = k T k ˙ W , ( R n ) ∗ . (cid:3) Theorem.
Let Ω ⊂ R n be any open set and suppose µ ∈ M loc (Ω) such that (4.1) | µ ( U ) | ≤ C H n − ( ∂U ) for any smooth open and bounded set U ⊂⊂ Ω . Let A be a compact set of Ω . If H n − ( A ) = 0 , then µ ( A ) = 0 .Proof. As H n − ( A ) = 0 , for any < ε < dist ( A, ∂ Ω) (or for any ε > , if Ω = R n ), we can find afinite number of balls B ( x i , r i ) , i ∈ I , with r i < ε such that A ⊂ S i ∈ I B ( x i , r i ) ⊂ Ω and(4.2) X i ∈ I r n − i < ε. Let W ε = S i ∈ I B ( x i , r i ) . Then A ⊂⊂ W ε ⊂ A ε : = { x ∈ R n : dist ( x, A ) < ε } . The first inclusion follows since A is compact and W ε is open; the second one follows since r i < ε and since we may assume that B ( x i , r i ) ∩ A = ∅ for any i ∈ I .We now claim that for each ε > there exists an open set W ′ ε such that W ′ ε has smooth boundaryand(4.3) ( A ⊂⊂ W ′ ε ⊂ A ε H n − ( ∂W ′ ε ) ≤ P ( W ε , Ω) , where P ( E, Ω) denotes the perimeter of a set E in Ω . Assume for now that (4.3) holds. Then, since A is compact, χ W ′ ε → χ A pointwise as ε → , and | µ ( W ′ ε ) | ≤ C H n − ( ∂W ′ ε ) , by our hypothesis (4.1) ≤ CP ( W ε , Ω) ≤ C X i ∈ I r n − i ≤ εC, by (4.2) . Thus, the Lebesgue dominated convergence theorem yields, after letting ε → , the desired result: | µ ( A ) | = 0 . We now proceed to prove (4.3). Let ρ be a standard symmetric mollifier: ρ ≥ , ρ ∈ C ∞ ( B (0 , , ˆ R n ρ ( x ) dx = 1 , and ρ ( x ) = ρ ( − x ) . Define ρ /k ( x ) = k n ρ ( kx ) and u k ( x ) = χ W ε ∗ ρ /k ( x ) = k n ˆ ρ ( k ( x − y )) χ W ε ( y ) dy EASURES IN THE DUAL OF BV for k = 1 , , . . . For k large enough, say for k ≥ k = k ( ǫ ) , it follows that u k ≡ on A, since A ⊂⊂ W ε , (4.4) u k ≡ on Ω \ A ε , since W ε ⊂ A ε . (4.5)We have P ( W ε , Ω) = | Dχ W ε | (Ω) ≥ | Du k | (Ω)= ˆ P ( F kt , Ω) dt, since ≤ u k ≤ , where F kt := { x ∈ Ω : u k ( x ) > t } . Note that for k ≥ k , and t ∈ (0 , we have, by (4.4) and (4.5), A ⊂⊂ F kt ⊂ A ε . For a.e. t ∈ (0 , the sets F kt have smooth boundaries. Thus we can choose t ∈ (0 , with thisproperty and such that P ( F kt , Ω) ≤ P ( W ε , Ω) , which is H n − ( ∂F kt ) ≤ P ( W ε , Ω) . Finally, we choose W ′ ε = F kt for any fixed k ≥ k . (cid:3) Corollary. If µ ∈ M loc (Ω) satisfies the hypothesis of Theorem 4.2, then k µ k << H n − in Ω ;that is, if A ⊂ Ω is any Borel measurable set such that H n − ( A ) = 0 then k µ k ( A ) = 0 .Proof. The domain Ω can be decomposed as Ω = Ω + ∪ Ω − , such that µ + = µ Ω + and µ − = µ Ω − ,where µ + and µ − are the positive and negative parts of µ , respectively. Let A ⊂ Ω be a Borel setsatisfying H n − ( A ) = 0 . By writing A = ( A ∩ Ω + ) ∪ ( A ∩ Ω − ) , we may assume that A ⊂ Ω + andhence k µ k ( A ) = µ + ( A ) . Moreover, since µ + is a Radon measure we can assume that A is compact.Hence, Theorem 4.2 yields k µ k ( A ) = µ + ( A ) = µ ( A ) = 0 . (cid:3) The following theorem characterizes all the signed measures in BV nn − ( R n ) ∗ . This result was firstproven in Phuc-Torres [17] for the space BV ( R n ) ∗ with no sharp control on the involving constants.In this paper we offer a new and direct proof of ( i ) ⇒ ( ii ) . We also clarify the first part of ( iii ) .Moreover, our proof of ( ii ) ⇒ ( iii ) yields a sharp constant that will be needed for the proof ofTheorem 4.7 below.4.4. Theorem.
Let µ ∈ M loc ( R n ) be a locally finite signed measure. The following are equivalent: (i) There exists a vector field F ∈ L ∞ ( R n , R n ) such that div F = µ in the sense of distributions. (ii) There is a constant C such that | µ ( U ) | ≤ C H n − ( ∂U ) for any smooth bounded open (or closed) set U with H n − ( ∂U ) < + ∞ . (iii) H n − ( A ) = 0 implies k µ k ( A ) = 0 for all Borel sets A and there is a constant C such that,for all u ∈ BV ∞ c ( R n ) , | < µ, u > | := (cid:12)(cid:12)(cid:12)(cid:12) ˆ R n u ∗ dµ (cid:12)(cid:12)(cid:12)(cid:12) ≤ C ˆ R n | Du | , where u ∗ is the representative in the class of u that is defined H n − -almost everywhere. (iv) µ ∈ BV nn − ( R n ) ∗ . The action of µ on any u ∈ BV nn − ( R n ) is defined (uniquely) as < µ, u > := lim k →∞ < µ, u k > = lim k →∞ ˆ R n u ∗ k dµ, where u k ∈ BV ∞ c ( R n ) converges to u in BV nn − ( R n ) . In particular, if u ∈ BV ∞ c ( R n ) then < µ, u > = ˆ R n u ∗ dµ, and moreover, if µ is a non-negative measure then, for all u ∈ BV nn − ( R n ) , < µ, u > = ˆ R n u ∗ dµ. Proof.
Suppose (i) holds. Then for every ϕ ∈ C ∞ c ( R n ) we have(4.6) ˆ R n F · Dϕdx = − ˆ R n ϕdµ. Let U ⊂⊂ R n be any open set (or closed set) with smooth boundary satisfying H n − ( ∂U ) < ∞ .Consider the characteristic function χ U and a sequence of mollifications u k := χ U ∗ ρ /k , where { ρ /k } is as in the proof of Theorem 4.2. Then, since U has a smooth boundary, we have(4.7) u k ( x ) → χ ∗ U ( x ) pointwise everywhere , where χ ∗ U ( x ) is the precise representative of χ U (see [3, Corollary 3.80] given by χ ∗ U ( x ) = x ∈ Int ( U ) , x ∈ ∂U , x ∈ R n \ U .We note that χ ∗ U is the same for U open or closed, since both are the same set of finite perimeter (theydiffer only on ∂U , which is a set of Lebesgue measure zero). From (4.6), (4.7), and the dominatedconvergence theorem we obtain (cid:12)(cid:12)(cid:12)(cid:12) µ (Int( U )) + 12 µ ( ∂U ) (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) ˆ R n χ ∗ U dµ (cid:12)(cid:12)(cid:12)(cid:12) = lim k →∞ (cid:12)(cid:12)(cid:12)(cid:12) ˆ R n u k dµ (cid:12)(cid:12)(cid:12)(cid:12) (4.8) = lim k →∞ (cid:12)(cid:12)(cid:12)(cid:12) ˆ R n F · Du k dx (cid:12)(cid:12)(cid:12)(cid:12) ≤ lim k →∞ k F k ∞ ˆ R n | Du k | dx = k F k ∞ ˆ R n | Dχ U | = k F k ∞ H n − ( ∂U ) . We now let K := U .
For each h > we define the function F h ( x ) = 1 − min { d K ( x ) , h } h , x ∈ R n , where d K ( x ) denotes the distance from x to K , i.e., d K ( x ) = inf {| x − y | : y ∈ K } . Note that F h is aLipschitz function such that F h ( x ) ≤ , F h ( x ) = 1 if x ∈ K and F h ( x ) = 0 if d K ( x ) ≥ h . Moreover, F h is differentiable L n -almost everywhere and | DF h ( x ) | ≤ h for L n -a.e. x ∈ R n . By standard smoothing techniques, (4.6) holds for the Lipschitz function F h . Therefore, (cid:12)(cid:12)(cid:12)(cid:12) ˆ R n F h dµ (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) ˆ R n F · DF h dx (cid:12)(cid:12)(cid:12)(cid:12) . (4.9) EASURES IN THE DUAL OF BV Since F h → χ K pointwise, it follows from the dominated convergence theorem that(4.10) | µ ( K ) | = (cid:12)(cid:12)(cid:12)(cid:12) ˆ R n χ K dµ (cid:12)(cid:12)(cid:12)(cid:12) = lim h → (cid:12)(cid:12)(cid:12)(cid:12) ˆ R n F h dµ (cid:12)(cid:12)(cid:12)(cid:12) . On the other hand, using the coarea formula for Lipschitz maps, we have (cid:12)(cid:12)(cid:12)(cid:12) ˆ R n F · DF h dx (cid:12)(cid:12)(cid:12)(cid:12) ≤ k F k ∞ ˆ R n | DF h | dx (4.11) = k F k ∞ h ˆ {
Inequality (4.13) can also be obtained be means of the (one-sided) outer Minskowskicontent . Indeed, since | Dd K | = 1 a.e., we find (cid:12)(cid:12)(cid:12)(cid:12) ˆ R n F · DF h dx (cid:12)(cid:12)(cid:12)(cid:12) ≤ k F k ∞ ˆ R n | DF h | dx = k F k ∞ h |{ < d K < h }| . Now sending h → + and using (4.9) - (4.10) we have | µ ( K ) | ≤ k F k ∞ SM ( K ) = k F k ∞ H n − ( ∂K ) , where SM ( K ) is the outer Minskowski content of K (see [2, Definition 5] ), and the last equalityfollows from [2, Corollary 1] . This argument also holds in the case U only has a Lipschitz boundary.Note that in this case we can only say that the limit in (4.7) holds H n − -a.e., but this is enough for (4.8) since k µ k << H n − by (4.6) and [6, Lemma 2.25] . Remark. If F ∈ L ∞ ( R n , R n ) satisfies div F = µ then, for any bounded set of finite perimeter E , the Gauss-Green formula proved in Chen-Torres-Ziemer [6] yields, µ ( E ∪ ∂ ∗ E ) = ˆ E ∪ ∂ ∗ E div F = ˆ ∂ ∗ E ( F e · ν )( y ) d H n − ( y ) and µ ( E ) = ˆ E div F = ˆ ∂ ∗ E ( F i · ν )( y ) d H n − ( y ) . EASURES IN THE DUAL OF BV Here E is the measure-theoretic interior of E and ∂ ∗ E is the reduced boundary of E . The estimates k F e · ν k L ∞ ( ∂ ∗ E ) ≤ k F k L ∞ and k F i · ν k L ∞ ( ∂ ∗ E ) ≤ k F k L ∞ give | µ ( E ∪ ∂ ∗ E ) | = | µ ( E ) + µ ( ∂ ∗ E ) | ≤ k F k L ∞ H n − ( ∂ ∗ E ) and | µ ( E ) | ≤ k F k L ∞ H n − ( ∂ ∗ E ) . Therefore, | µ ( ∂ ∗ E ) | ≤ k F k L ∞ H n − ( ∂ ∗ E ) + | µ ( E ) | ≤ k F k L ∞ H n − ( ∂ ∗ E ) . We note that this provides another proof of (i) ⇒ (ii) (with C = k F k ∞ for both open and closedsmooth sets) since for any bounded open (resp. closed) set U with smooth boundary we have U = U (resp. U = U ∪ ∂ ∗ U ). We recall the spaces defined in Definitions 3.4 and 3.5. We now show the following new result.4.7.
Theorem.
Let E := M loc ∩ BV nn − ( R n ) ∗ and F := M loc ∩ ˙ W , ( R n ) ∗ . Then E and F areisometrically isomorphic.Proof. We define a map S : E → F as S ( T ) = T ˙ W , . Clearly, S is a linear map. We need to show that S is 1-1 and on-to, and k S ( T ) k ˙ W , ( R n ) ∗ = k T k BV nn − ( R n ) ∗ for all T ∈ E . In order to show the injectivity we assume that S ( T ) = 0 ∈ F forsome T ∈ E . Then T ( u ) = 0 for all u ∈ ˙ W , ( R n ) . Thus, if µ is the measure associated to T ∈ E , then ˆ R n ϕdµ = T ( ϕ ) = 0 for all ϕ ∈ C ∞ c ( R n ) , which implies that µ = 0 . Now, by definition of E , we have T ( u ) = ˆ R n u ∗ dµ = 0 for all u ∈ BV ∞ c ( R n ) , which implies, by Theorem 2.10 and Theorem 3.1, that T ≡ on BV nn − ( R n ) . We now proceed to show the surjectivity and take H ∈ F . Thus, there exists µ ∈ M loc ( R n ) suchthat ˆ R n ϕdµ = H ( ϕ ) for all ϕ ∈ C ∞ c ( R n ) . From Lemma 4.1, since H ∈ ˙ W , ( R n ) ∗ , there exists a bounded vector field F ∈ L ∞ ( R n , R n ) suchthat(4.17) div F = µ in the distributional sense and k H k ˙ W , ( R n ) ∗ = k µ k ˙ W , ( R n ) ∗ = k F k L ∞ ( R n , R n ) . Now, from the proof of Theorem 4.4, ( i ) ⇒ ( ii ) ⇒ ( iii ) , it follows that k µ k << H n − , | µ ( U ) | ≤ k F k ∞ H n − ( ∂U ) for all closed and smooth sets U ⊂⊂ R n , and (cid:12)(cid:12)(cid:12)(cid:12) ˆ R n u ∗ dµ (cid:12)(cid:12)(cid:12)(cid:12) ≤ k F k L ∞ ( R n , R n ) k u k BV nn − ( R n ) for all u ∈ BV ∞ c ( R n ) . Hence, µ ∈ BV ∞ c ( R n ) ∗ and from (4.17) we obtain k µ k BV ∞ c ( R n ) ∗ = k F k L ∞ ( R n , R n ) = k µ k ˙ W , ( R n ) ∗ . From Theorem 2.10, it follows that µ can be uniquely extended to a continuous linear functional ˆ µ ∈ BV nn − ( R n ) ∗ and clearly, S (ˆ µ ) = µ, which implies that S is surjective. According to Theorem 2.10, this extension preserves the operatornorm and thus (cid:13)(cid:13) S − ( µ ) (cid:13)(cid:13) BV nn − ( R n ) ∗ = k ˆ µ k BV nn − ( R n ) ∗ = k µ k BV ∞ c ( R n ) ∗ = k µ k ˙ W , ( R n ) ∗ , which shows that E and F are isometrically isomorphic. (cid:3) On an issue raised by Meyers and Ziemer
In this section, using the result of Theorem 4.4, we construct a locally integrable function f suchthat f ∈ BV ( R n ) ∗ but | f | 6∈ BV ( R n ) ∗ . This example settles an issue raised by Meyers and Ziemerin [16, page 1356]. We mention that this kind of highly oscillatory function appeared in [14] in adifferent context.5.1. Proposition.
Let f ( x ) = ǫ | x | − − ǫ sin( | x | − ǫ ) + ( n − | x | − cos( | x | − ǫ ) , where < ǫ < n − isfixed. Then (5.1) f ( x ) = div [ x | x | − cos( | x | − ǫ )] . Moreover, there exists a sequence { r k } decreasing to zero such that (5.2) ˆ B rk (0) f + ( x ) dx ≥ c r n − − ǫk for a constant c = c ( n, ǫ ) > independent of k . Here f + is the positive part of f . Thus by Theorem4.4 we see that f belongs to BV ( R n ) ∗ , whereas | f | does not.Proof. The equality (5.1) follows by a straightforward computation. To show (5.2), we let r k =( π/ kπ ) − ǫ for k = 1 , , , . . . Then we have ˆ B rk (0) f + ( x ) dx = s ( n ) ˆ r k t n [ ǫ t − − ǫ sin( t − ǫ ) + ( n − t − cos( t − ǫ )] + dtt = s ( n ) ǫ ˆ ∞ r − ǫk x − nǫ [ ǫ x ǫ +1 ǫ sin( x ) + ( n − x ǫ cos( x )] + dxx ≥ s ( n )2 ∞ X i =0 ˆ π/ kπ +2 iππ/ kπ +2 iπ x − n +1 ǫ dx, where s ( n ) is the area of the unit sphere in R n . Thus using the elementary observation ˆ π/ kπ +2( i +1) ππ/ kπ +2 iπ x − n +1 ǫ dx ≤ ˆ π/ kπ +2 iππ/ kπ +2 iπ x − n +1 ǫ dx, EASURES IN THE DUAL OF BV we find that ˆ B rk (0) f + ( x ) dx ≥ s ( n )14 ∞ X i =0 ˆ π/ kπ +2 iππ/ kπ +2 iπ x − n +1 ǫ dx ≥ s ( n )14 ∞ X i =0 ˆ π/ kπ +2 iππ/ kπ +2 iπ x − n +1 ǫ dx + ˆ π/ kπ +2( i +1) ππ/ kπ +2 iπ x − n +1 ǫ dx ! ≥ s ( n )14 ∞ X i =0 ˆ π/ kπ +2( i +1) ππ/ kπ +2 iπ x − n +1 ǫ dx = s ( n )14 ˆ ∞ π/ kπ x − n +1 ǫ dx = s ( n ) ǫ n − − ǫ ) r n − − ǫk . This completes the proof of the proposition. (cid:3) The space BV (Ω) In this section we let Ω ⊂ R n be a bounded open set with Lipschitz boundary. We have thefollowing well known result concerning the existence of traces of functions in BV (Ω) (see for example[13, Theorem 2.10] and [4, Theorem 10.2.1]):6.1. Theorem.
Let Ω be a bounded open set with Lipschitz continuous boundary ∂ Ω and let u ∈ BV (Ω) . Then, there exists a function ϕ ∈ L ( ∂ Ω) such that, for H n − -almost every x ∈ ∂ Ω , lim r → r − n ˆ B ( x,r ) ∩ Ω | u ( y ) − ϕ ( x ) | dy = 0 . From the construction of the trace ϕ (see [13, Lemma 2.4], we see that ϕ is uniquely determined.Therefore, we have a well defined operator γ : BV (Ω) → L ( ∂ Ω) . We now define the space BV (Ω) as follows:6.2. Definition.
Let BV (Ω) = ker( γ ) . We also define another BV function space with a zero boundary condition.6.3. Definition.
Let BV (Ω) := C ∞ c (Ω) , where the closure is taken with respect to the intermediate convergence of BV (Ω) . By the intermediate convergence of BV (Ω) , we mean the following6.4. Definition.
Let { u k } ∈ BV (Ω) and u ∈ BV (Ω) . We say that u k converges to u in the sense ofintermediate (or strict) convergence if u k → u strongly in L (Ω) and ˆ Ω | Du k | → ˆ Ω | Du | . The following theorem can be found in [4, Theorem 10.2.2]:6.5.
Theorem.
The trace operator γ is continuous from BV (Ω) equipped with the intermediateconvergence onto L ( ∂ Ω) equipped with the strong convergence. The following theorem is well known and can be found in many standard references including[4, 12, 20, 13], but for completeness we will include the proof here.6.6.
Theorem.
The space C ∞ (Ω) ∩ BV (Ω) is dense in BV (Ω) equipped with the intermediate conver-gence. Moreover, if Ω is a Lipschitz domain then C ∞ (Ω) is also dense in BV (Ω) for the intermediateconvergence. Proof.
We note that C ∞ (Ω) ∩ BV (Ω) = C ∞ (Ω) ∩ W , (Ω) . For Lipschitz domains, it is proved,e.g., in [12, page 127] that C ∞ (Ω) is dense in W , (Ω) , equipped with the strong convergence. Thisactually holds even for domains that possess the so-called segment property (see [1, Theorem 3.18]).Thus, since the strong convergence implies the intermediate convergence it follows that C ∞ (Ω) isdense in C ∞ (Ω) ∩ BV (Ω) in the intermediate convergence. Therefore, if C ∞ (Ω) ∩ BV (Ω) is densein BV (Ω) for the intermediate convergence, the second statement of the theorem holds. Let ε > and u ∈ BV (Ω) . We decompose Ω as follows: Ω = ∞ [ i =0 Ω i , ˆ Ω \ Ω | Du | < ε and Ω i ⊂⊂ Ω i +1 . We consider the open cover { C i } defined as follows: C := Ω C i := Ω i +1 \ Ω i − , i ≥ . Let { ϕ i } be a partition of unity associated to { C i } ; that is, ϕ i ∈ C ∞ c ( C i ) , ≤ ϕ i ≤ , ∞ X i =1 ϕ i = 1 . Note that ϕ ≡ on Ω . Let ρ be a standard mollifier as in the proof of Theorem 4.2. For each i ,choose ε i > so that: spt ( ρ ε i ∗ ϕ i u ) ⊂ C i , ˆ Ω | ρ ε i ∗ ( uϕ i ) − uϕ i | < ε i , (6.1) ˆ Ω | ρ ε i ∗ ( uDϕ i ) − uDϕ i | < ε i , (6.2) (cid:12)(cid:12)(cid:12)(cid:12) ˆ Ω | ρ ε ∗ ( ϕ Du ) | dx − ˆ Ω | ϕ Du | (cid:12)(cid:12)(cid:12)(cid:12) < ε. (6.3)Define u ε := ∞ X i =1 ρ ε i ∗ ( uϕ i ) . Then ˆ Ω | u − u ε | dx ≤ ∞ X i =1 ˆ Ω | ρ ε i ∗ ( uϕ i ) − uϕ i | dx < ε, by (6.1) . We have Du ε = ∞ X i =1 ρ ε i ∗ ( ϕ i Du ) + ∞ X i =1 ρ ε i ∗ ( uDϕ i )= ∞ X i =1 ρ ε i ∗ ( ϕ i Du ) + ∞ X i =1 (cid:0) ρ ε i ∗ ( uDϕ i ) − uDϕ i (cid:1) . EASURES IN THE DUAL OF BV Then, on the one hand, (cid:12)(cid:12)(cid:12)(cid:12) ˆ Ω | ρ ε ∗ ( ϕ Du ) | dx − ˆ Ω | Du ε | (cid:12)(cid:12)(cid:12)(cid:12) ≤ ∞ X i =2 ˆ Ω | ρ ε i ∗ ( ϕ i Du ) | dx + ∞ X i =1 ˆ Ω | ρ ε i ∗ ( uDϕ i ) − uDϕ i | dx ≤ ∞ X i =2 ˆ Ω | ρ ε i ∗ ( ϕ i Du ) | + ε, by (6.2) ≤ ∞ X i =2 ˆ Ω | ϕ i Du | + ε, by a property of convolution ≤ ˆ Ω \ Ω | Du | + ε ≤ ε + ε = 2 ε. (6.4)On the other hand, (cid:12)(cid:12)(cid:12)(cid:12) ˆ Ω | ρ ε ∗ ( ϕ Du ) | dx − ˆ Ω | Du | (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) ˆ Ω | ρ ε ∗ ( ϕ Du ) | dx − ˆ Ω | ϕ Du | − ˆ Ω (1 − ϕ ) | Du | (cid:12)(cid:12)(cid:12)(cid:12) ≤ ε + ˆ Ω (1 − ϕ ) | Du | , by (6.3) ≤ ε + ˆ Ω \ Ω | Du | ≤ ε, since ϕ ≡ on Ω . (6.5)From (6.4) and (6.5): (cid:12)(cid:12)(cid:12)(cid:12) ˆ Ω | Du ε | − ˆ Ω | Du | (cid:12)(cid:12)(cid:12)(cid:12) < ε. (cid:3) Theorem.
Let Ω be any bounded open set with Lipschitz boundary. Then BV c (Ω) is dense in BV (Ω) in the strong topology of BV (Ω) .Proof. We consider first the case u ∈ BV ( C R,T ) , where C R,T is the open cylinder C R,T = B R × (0 , T ) , B R is an open ball of radius R in R n − , and supp ( u ) ∩ ∂C R,T = supp ( u ) ∩ ( B R × { } ) . A genericpoint in C R,T will be denoted by ( x ′ , t ) , with x ′ ∈ B R and t ∈ (0 , T ) . From Theorem 6.6, we canapproximate u with a sequence of functions u k ∈ C ∞ ( C R,T ) such that u k → u in L ( C R,T ) and ˆ C R,T | Du k | → ˆ C R,T | Du | . (6.6)Notice that the condition supp ( u ) ∩ ∂C R,T = supp ( u ) ∩ ( B R × { } ) implies that γ ( u k ) ( ∂C R,T \ ( B R × { } )) ≡ . From Theorem 6.5, γ ( u k ) → γ ( u ) in L ( ∂C R,T ) and hence(6.7) γ ( u k ) ( B R × { } ) = u k (cid:12)(cid:12) ( B R ×{ } ) → in L ( B R × { } ) . For x ′ ∈ B R , ≤ x n ≤ T , we have u k ( x ′ , x n ) − u k ( x ′ ,
0) = ˆ x n ∂u k ∂x n ( x ′ , t ) dt, and hence,(6.8) | u k ( x ′ , x n ) | ≤ | u k ( x ′ , | + ˆ x n (cid:12)(cid:12)(cid:12)(cid:12) ∂u k ∂x n ( x ′ , t ) (cid:12)(cid:12)(cid:12)(cid:12) dt. We integrate both sides in (6.8) to obtain:(6.9) ˆ B R | u k ( x ′ , x n ) | dx ′ ≤ ˆ B R | u k ( x ′ , | dx ′ + ˆ x n ˆ B R | Du k ( x ′ , t ) | dx ′ dt. From (6.7) we have(6.10) lim k →∞ ˆ B R | u k ( x ′ , | dx ′ = 0 , and thus, letting k → ∞ in (6.9) and using (6.10), (6.6) and Lemma 2.9 (in particular, (2.7) with A := B R × (0 , x n ) for a.e. < x n < T ) we obtain(6.11) ˆ B R | u ( x ′ , x n ) | dx ′ ≤ ˆ x n ˆ B R | Du | = k Du k ( B R × (0 , x n )) for a.e. < x n < T. Consider a function ϕ ∈ C ∞ c ( R ) such that ϕ is decreasing in [0 , + ∞ ) and satisfies ϕ ≡ on [0 , , ϕ ≡ on R \ [ − , , ≤ ϕ ≤ . We define ϕ k ( t ) = ϕ ( kt ) , k = 1 , , . . .v k ( x ′ , t ) = (1 − ϕ k ( t )) u ( x ′ , t ) . (6.12)Clearly, v k → u in L ( C R,T ) . Also, if u ≥ then v k ↑ u since ϕ is decreasing in [0 , + ∞ ) . Moreover, ∂v k ∂t = (1 − ϕ k ) ∂u∂t − kϕ ′ ( kt ) u,D x ′ v k = (1 − ϕ k ) D x ′ u. Thus we have ˆ C R,T | Dv k − Du | = ˆ C R,T (cid:12)(cid:12)(cid:12)(cid:12)(cid:16) D x ′ u − ϕ k D x ′ u, ∂u∂t − ϕ k ∂u∂t − kϕ ′ ( kt ) u (cid:17) − (cid:16) D x ′ u, ∂u∂t (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) = ˆ C R,T (cid:12)(cid:12)(cid:12)(cid:12)(cid:16) − ϕ k D x ′ u, − ϕ k ∂u∂t − kϕ ′ ( kt ) u (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) . Since ϕ k ( t ) = 0 for t > k we have the following: ˆ C R,T | Dv k − Du | ≤ C (cid:18) ˆ C R,T ϕ k | Du | + ˆ C R,T k | ϕ ′ ( kx n ) || u | (cid:19) ≤ C ˆ /k ˆ B R | Du | + C k ˆ /k ˆ B R | u ( x ′ , t ) | dx ′ dt ≤ C ˆ /k ˆ B R | Du | + C k ˆ /k k Du k ( B R × (0 , t )) dt, by (6.11) ≤ C ˆ /k ˆ B R | Du | + C k · k Du k ( B R × (0 , /k )) · ˆ /k dt = C ˆ /k ˆ B R | Du | + C k · k · ˆ /k ˆ B R | Du | = C ˆ /k ˆ B R | Du | . (6.13)Since k Du k is a Radon measure and ∩ ∞ k =1 ( B R × (0 , k )) = ∅ we find ˆ /k ˆ B R | Du | → , as k → ∞ , EASURES IN THE DUAL OF BV which by (6.13) yields lim k →∞ ˆ C R,T | Dv k − Du | = 0 . Thus(6.14) v k → u in the strong topology of BV ( C R,T ) . We consider now the general case of a bounded open set Ω with Lipschitz boundary and let u ∈ BV (Ω) . For each point x ∈ ∂ Ω , there exists a neighborhood A and a bi-Lipschitz function g : B (0 , → A that maps B (0 , + onto A ∩ Ω and the flat part of ∂B (0 , + onto A ∩ ∂ Ω . A finitenumber of such sets A , A , . . . , A n cover ∂ Ω . By adding possibly an additional open set A ⊂⊂ Ω ,we get a finite covering of Ω . Let { α i } be a partition of unity relative to that covering, and let g i be the bi-Lipschitz map relative to the set A i for i = 1 , , . . . , N . For each i ∈ { , , . . . , N } thefunction U i = ( α i u ) ◦ g i belongs to BV ( B (0 , + ) , and has support non-intersecting the curved part of ∂B (0 , + . Thus,we can extend U i to the whole cylinder C , := B (0) × (0 , by setting U i equal to zero outside B (0 , + . By (6.14), for each ε > , we can find a function W i ∈ BV c ( C , ) such that(6.15) k W i − U i k BV ( C , ) ≤ ε, for i = 1 , , . . . , N . Letting now w i = W i ◦ g − i , i = 1 , , . . . , N, we have w i ∈ BV c ( A i ∩ Ω) and k D ( w i − α i u ) k ( A i ∩ Ω) = (cid:13)(cid:13) D ( W i ◦ g − i − (( α i u ) ◦ g i ) ◦ g − i ) (cid:13)(cid:13) ( A i ∩ Ω)= k D ( g i ( W i − ( α i u ) ◦ g i )) k ( A i ∩ Ω) ≤ C g i k D ( W i − ( α i u ) ◦ g i ) k ( A i ∩ Ω) , by [3, Theorem 3.16] = C ˆ g − i ( A i ∩ Ω) | D ( W i − U i ) | , by definition of g i acting on measures = C ˆ B (0 , + | D ( W i − U i ) |≤ C ε, by (6.15) . (6.16)Here C = max i { [ Lip ( g i )] n − } (see [3, Theorem 3.16]). Let w = α u . Then w ∈ BV c (Ω) . Define w = N X i =0 w i . We have w ∈ BV c (Ω) , and by (6.16) k D ( w − u ) k (Ω) ≤ N X i =0 k D ( w i − α i u ) k ( A i ∩ Ω)= N X i =1 k D ( w i − α i u ) k ( A i ∩ Ω) ≤ N C ε.
Likewise, by (6.15) and a change of variables we have k w − u k L (Ω) ≤ N X i =0 k w i − α i u k L ( A i ∩ Ω) ≤ N X i =1 k w i − α i u k L ( A i ∩ Ω) ≤ N c ε.
Thus BV c (Ω) = BV (Ω) in the strong topology of BV (Ω) . (cid:3) Remark. By (6.12) and the construction of w in the proof of Theorem 6.7 above, we see thateach u ∈ BV (Ω) can be approximated by a sequence { u k } ⊂ BV c (Ω) such that u k = u in Ω \ N k fora set N k = { x ∈ Ω : d ( x, ∂ Ω) ≤ δ ( k ) } with δ ( k ) → as k → + ∞ . Moreover, if u ≥ then so is u k and u k ↑ u as k increases to + ∞ . We will also need the following density result.6.9.
Lemma. BV ∞ (Ω) is dense in BV (Ω) . Likewise, BV ∞ c (Ω) is dense in BV c (Ω) , and BV ∞ (Ω) is dense in BV (Ω) in the strong topology of BV (Ω) .Proof. We shall only prove the first statement as the others can be shown in a similar way. Let u ∈ BV +0 (Ω) and define u j := u ∧ j, j = 1 , , . . . Obviously, u j → u in L (Ω) . We will now show that k D ( u − u j ) k (Ω) → . The coarea formulayields ˆ Ω | D ( u − u j ) | = ˆ ∞ H n − (Ω ∩ ∂ ∗ { u − u j > t } ) dt = ˆ ∞ H n − (Ω ∩ ∂ ∗ { u − j > t } ) dt = ˆ ∞ H n − (Ω ∩ ∂ ∗ { u > j + t } ) dt = ˆ ∞ j H n − (Ω ∩ ∂ ∗ { u > s } ) ds. Since ´ ∞ H n − (Ω ∩ ∂ ∗ { u > s } ) ds < ∞ , the Lebesgue dominated convergence theorem implies that(6.17) ˆ Ω | D ( u − u j ) | → as j → ∞ . If u ∈ BV (Ω) , we write u = u + − u − and define f j = u + ∧ j and g j = u − ∧ j . Thus f j − g j ∈ BV (Ω) and ˆ Ω | D ( u − ( f j − g j )) | = ˆ Ω | Du + − Du − − Df j + Dg j |≤ ˆ Ω | D ( u + − f j ) | + ˆ Ω | D ( u − − g j ) |→ as j → ∞ , due to (6.17). That completes the proof of the lemma. (cid:3) We are now ready to prove the main theorem of this section that makes precise the definition ofthe space of functions of bounded variation in Ω with zero trace on the boundary of Ω .6.10. Theorem. BV (Ω) = BV (Ω) .Proof. Let Let u ∈ BV (Ω) . Then Definition 6.3 implies the existence of a sequence { u k } ∈ C ∞ c (Ω) such that u k → u in L (Ω) and ˆ Ω | Du k | → ˆ Ω | Du | . Since u k ∈ C ∞ c (Ω) , we have γ ( u k ) ≡ . Then Theorem 6.5 yields γ ( u k ) → γ ( u ) in L ( ∂ Ω) , and so γ ( u ) = 0 and u ∈ BV (Ω) . EASURES IN THE DUAL OF BV In the other direction, let u ∈ BV (Ω) . Then, from Theorem 6.7 there exists a sequence u k ∈ BV c (Ω) such that(6.18) lim k →∞ ˆ Ω | u k − u | = lim k →∞ ˆ Ω | Du k − Du | = 0 . Given a sequence ε k → , we consider the sequence of mollifications w k := u k ∗ ρ ε k . We can choose ε k sufficiently small to have w k ∈ C ∞ c (Ω) . Also, for each k , lim ε → ˆ Ω | D ( u k ∗ ρ ε ) | = ˆ Ω | Du k | , and lim ε → ˆ Ω | u k ∗ ρ ε − u k | = 0 . Thus we can choose ε k small enough so that, for each k ,(6.19) (cid:12)(cid:12)(cid:12)(cid:12) ˆ Ω | D ( u k ∗ ρ ε k ) | − ˆ Ω | Du k | (cid:12)(cid:12)(cid:12)(cid:12) ≤ k , and(6.20) ˆ Ω | u k ∗ ρ ε k − u k | ≤ k . Using (6.20) and (6.18) we obtain(6.21) lim k →∞ ˆ Ω | w k − u | ≤ lim k →∞ ˆ Ω | w k − u k | + lim k →∞ ˆ Ω | u k − u | = 0 . Also, letting k → ∞ in (6.19) and using (6.18), we obtain(6.22) lim k →∞ ˆ Ω | D ( u k ∗ ρ ε k ) | = ˆ Ω | Du | . From (6.21) and (6.22) we conclude that w k → u in the intermediate convergence which implies that u ∈ BV (Ω) . (cid:3) With Theorem 6.10 we can now prove the following Sobolev’s inequality for functions in BV (Ω) :6.11. Theorem.
Let u ∈ BV (Ω) , where Ω is a bounded open set with Lipschitz boundary. Then k u k L nn − (Ω) ≤ C k Du k (Ω) , for a constant C = C ( n ) .Proof. The Sobolev inequality for smooth functions states that(6.23) k u k L nn − ( R n ) ≤ C ˆ R n | Du | for each u ∈ C ∞ c ( R n ) . From Theorem 6.10 there exists a sequence u k ∈ C ∞ c (Ω) such that(6.24) u k → u in L (Ω) and ˆ Ω | Du k | → ˆ Ω | Du | . Since u k → u in L (Ω) then there exists a subsequence { u k j } of { u k } such that u k j ( x ) → u ( x ) for a.e. x ∈ Ω . Using Fatou’s Lemma and (6.23), we obtain(6.25) ˆ Ω | u | nn − ≤ lim inf j →∞ ˆ Ω | u k j | nn − ≤ lim inf j →∞ (cid:18) C ˆ Ω | Du k j | (cid:19) nn − . Finally, using (6.24) in (6.25) we conclude (cid:18) ˆ Ω | u | nn − (cid:19) n − n ≤ C ˆ Ω | Du | . (cid:3) By Theorem 6.11, we see that k u k BV (Ω) is equivalent to k Du k (Ω) whenever u ∈ BV (Ω) (or BV (Ω) ) and Ω is a bounded Lipschitz domain. Thus, for the rest of the paper we will equip BV (Ω) with the homogeneous norm: k u k BV (Ω) = k Du k (Ω) . From Theorem 6.7 and Lemma 6.9 we obtain6.12.
Corollary.
Let Ω be any bounded open set with Lipschitz boundary. Then BV ∞ c (Ω) is densein BV (Ω) . Characterizations of measures in BV (Ω) ∗ First, as in the case of R n , we make precise the definitions of measures in the spaces W , (Ω) ∗ and BV (Ω) ∗ .7.1. Definition.
For a bounded open set Ω with Lipschitz boundary, we let M loc (Ω) ∩ W , (Ω) ∗ := { T ∈ W , (Ω) ∗ : T ( ϕ ) = ˆ Ω ϕdµ for some µ ∈ M loc (Ω) , ∀ ϕ ∈ C ∞ c (Ω) } . Therefore, if µ ∈ M loc (Ω) ∩ W , (Ω) ∗ , then the action < µ, u > can be uniquely defined for all u ∈ W , (Ω) (because of the density of C ∞ c (Ω) in W , (Ω)) . Definition.
For a bounded open set Ω with Lipschitz boundary, we let M loc (Ω) ∩ BV (Ω) ∗ := { T ∈ BV (Ω) ∗ : T ( ϕ ) = ˆ Ω ϕ ∗ dµ for some µ ∈ M loc (Ω) , ∀ ϕ ∈ BV ∞ c (Ω) } , where ϕ ∗ is the precise representative of ϕ . Thus, if µ ∈ M loc (Ω) ∩ BV (Ω) ∗ , then the action < µ, u > can be uniquely defined for all u ∈ BV (Ω) (because of the density of BV ∞ c (Ω) in BV (Ω) by Corollary 6.12). We will use the following characterization of W , (Ω) ∗ whose proof is completely analogous tothat of Lemma 4.1.7.3. Lemma.
Let Ω be any bounded open set with Lipschitz boundary. The distribution T belongsto W , (Ω) ∗ if and only if T = div F for some vector field F ∈ L ∞ (Ω , R n ) . Moreover, k T k W , (Ω) ∗ = min {k F k L ∞ (Ω , R n ) } , where the minimum is taken over all F ∈ L ∞ (Ω , R n ) such that div F = T . Here we use the norm k F k L ∞ (Ω , R n ) := (cid:13)(cid:13)(cid:13) ( F + F + · · · + F n ) / (cid:13)(cid:13)(cid:13) L ∞ (Ω) for F = ( F , . . . , F n ) . We are now ready to state the main result of this section.
EASURES IN THE DUAL OF BV Theorem.
Let Ω be any bounded open set with Lipschitz boundary and µ ∈ M loc (Ω) . Then,the following are equivalent: (i) There exists a vector field F ∈ L ∞ (Ω , R n ) such that div F = µ . (ii) | µ ( U ) | ≤ C H n − ( ∂U ) for any smooth open (or closed) set U ⊂⊂ Ω with H n − ( ∂U ) < + ∞ . (iii) H n − ( A ) = 0 implies k µ k ( A ) = 0 for all Borel sets A ⊂ Ω and there is a constant C suchthat, for all u ∈ BV ∞ c (Ω) , | < µ, u > | := (cid:12)(cid:12)(cid:12)(cid:12) ˆ Ω u ∗ dµ (cid:12)(cid:12)(cid:12)(cid:12) ≤ C ˆ Ω | Du | , where u ∗ is the representative in the class of u that is defined H n − -almost everywhere. (iv) µ ∈ BV (Ω) ∗ . The action of µ on any u ∈ BV (Ω) is defined (uniquely) as < µ, u > := lim k →∞ < µ, u k > = lim k →∞ ˆ Ω u ∗ k dµ, where u k ∈ BV ∞ c (Ω) converges to u in BV (Ω) . In particular, if u ∈ BV ∞ c (Ω) then < µ, u > = ˆ Ω u ∗ dµ, and moreover, if µ is a non-negative measure then, for all u ∈ BV (Ω) , < µ, u > = ˆ Ω u ∗ dµ. Proof.
Suppose (i) holds. Then for every ϕ ∈ C ∞ c (Ω) we have ˆ Ω F · Dϕdx = − ˆ Ω ϕdµ. Let U ⊂⊂ Ω be any open (or closed) set with smooth boundary satisfying H n − ( ∂U ) < ∞ . Weproceed as in Theorem 4.4 and consider the characteristic function χ U and the sequence u k := χ U ∗ ρ /k . Since U is strictly contained in Ω , for k large enough, the support of { u k } are containedin Ω . We can then proceed exactly as in Theorem 4.4 to conclude that | µ ( U ) | ≤ C H n − ( ∂U ) , where C = k F k L ∞ (Ω) for closed sets U and C = 3 k F k L ∞ (Ω) for open sets U .If µ satisfies (ii) with a constant C > , then Corollary 4.3 implies that k µ k << H n − . We let u ∈ BV ∞ c (Ω) and { ρ ε } be a standard sequence of mollifiers. Consider the convolution ρ ε ∗ u andnote that ρ ε ∗ u ∈ C ∞ c (Ω) , for ε small enough. Then as in the proof of Theorem 4.4 we have, for ε small enough, (cid:12)(cid:12)(cid:12)(cid:12) ˆ Ω ρ ε ∗ udµ (cid:12)(cid:12)(cid:12)(cid:12) ≤ C ˆ Ω | Du | . Sending ε to zero and using the dominated convergence theorem yield (cid:12)(cid:12)(cid:12)(cid:12) ˆ Ω u ∗ dµ (cid:12)(cid:12)(cid:12)(cid:12) ≤ C ˆ Ω | Du | , with the same constant C as in (ii) . This gives (ii) ⇒ (iii) .From (iii) we obtain that the linear operator(7.1) T ( u ) := < µ, u > = ˆ Ω u ∗ dµ, u ∈ BV ∞ c (Ω) is continuous and hence it can be uniquely extended, since BV ∞ c (Ω) is dense in BV (Ω) (Corollary6.12), to the space BV (Ω) .Assume now that µ is non-negative. We take u ∈ BV (Ω) and consider the positive and negativeparts ( u ∗ ) + and ( u ∗ ) − of the representative u ∗ . By Remark 6.8, there is an increasing sequence of nonnegative functions { v k } ⊂ BV c (Ω) that converges to ( u ∗ ) + pointwise and in the BV norm.Therefore, using (7.1) we have T ( v k ∧ j ) = ˆ Ω v k ∧ jdµ, j = 1 , , . . . We first send j to infinity and then k to infinity. Using the continuity of T , (6.17), and the monotoneconvergence theorem we get T (( u ∗ ) + ) = ˆ Ω ( u ∗ ) + dµ. We proceed in the same way for ( u ∗ ) − and thus by linearity we conclude T ( u ) = T (( u ∗ ) + ) − T (( u ∗ ) − ) = ˆ Ω ( u ∗ ) + − ( u ∗ ) − dµ = ˆ Ω u ∗ dµ. Finally, to prove that (iv) implies (i) we take µ ∈ BV (Ω) ∗ . Since W , (Ω) ⊂ BV (Ω) then ˜ µ := µ W , (Ω) ∈ W , (Ω) ∗ , and therefore Lemma 7.3 implies that there exists F ∈ L ∞ (Ω , R n ) such that div F = ˜ µ and thus,since C ∞ c ⊂ W , (Ω) , we conclude that div F = µ in the sense of distributions. (cid:3) Remark. If Ω is a bounded domain containing the origin then the function f given in Propo-sition 5.1 belongs to BV (Ω) ∗ but | f | does not. Theorem 7.4 and Lemma 7.3 immediately imply the following new result which states that theset of measures in BV (Ω) ∗ coincides with that of W , (Ω) ∗ .7.6. Theorem.
The normed spaces M loc (Ω) ∩ BV (Ω) ∗ and M loc (Ω) ∩ W , (Ω) ∗ are isometricallyisomorphic. The proof of Theorem 7.6 is similar to that of Theorem 4.7 but this time one uses Theorem 7.4and Corollary 6.12 in place of Theorem 4.4 and Theorem 3.1, respectively. Thus we shall omit itsproof. 8.
Finite measures in BV (Ω) ∗ In this section, we characterize all finite signed measures that belong to BV (Ω) ∗ . Note thatthe finiteness condition here is necessary at least for positive measures in BV (Ω) ∗ . By a measure µ ∈ BV (Ω) ∗ we mean that the inequality (cid:12)(cid:12)(cid:12)(cid:12) ˆ Ω u ∗ dµ (cid:12)(cid:12)(cid:12)(cid:12) ≤ C k u k BV (Ω) holds for all u ∈ BV ∞ (Ω) . By Lemma 6.9 we see that such a µ can be uniquely extended to be acontinuous linear functional in BV (Ω) .We will use the following result, whose proof can be found in [20, Lemma 5.10.14]:8.1. Lemma.
Let Ω be an open set with Lipschitz boundary and u ∈ BV (Ω) . Then, the extensionof u to R n defined by u ( x ) = ( u ( x ) , x ∈ Ω0 , x ∈ R n \ Ω satisfies that u ∈ BV ( R n ) and k u k BV ( R n ) ≤ C k u k BV (Ω) , where C = C (Ω) . EASURES IN THE DUAL OF BV Theorem.
Let Ω be an open set with Lipschitz boundary and let µ be a finite signed measurein Ω . Extend µ by zero to R n \ Ω by setting µ ( R n \ Ω) = 0 . Then, µ ∈ BV (Ω) ∗ if and only if (8.1) | µ ( U ) | ≤ C H n − ( ∂U ) for every smooth open set U ⊂ R n and a constant C = C (Ω , µ ) .Proof. Suppose that µ ∈ BV (Ω) ∗ . Let u ∈ BV ∞ c ( R n ) and assume that u is the representative thatis defined H n − -almost everywhere. Consider v := uχ Ω and note that v Ω ∈ BV ∞ (Ω) since Dv isa finite vector-measure in R n given by Dv = uDχ Ω + χ Ω Du, and therefore, ˆ Ω | Dv | = ˆ Ω | uDχ Ω + χ Ω Du | ≤ ˆ Ω | u || Dχ Ω | + ˆ Ω | Du | = ˆ Ω | Du | ≤ ˆ R n | Du | = k u k BV ( R n ) < + ∞ . (8.2)Since µ ∈ BV (Ω) ∗ there exists a constant C = C (Ω , µ ) such that(8.3) (cid:12)(cid:12)(cid:12)(cid:12) ˆ Ω vdµ (cid:12)(cid:12)(cid:12)(cid:12) ≤ C k v k BV (Ω) . Then, (cid:12)(cid:12)(cid:12)(cid:12) ˆ R n udµ (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) ˆ Ω udµ (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) ˆ Ω vdµ (cid:12)(cid:12)(cid:12)(cid:12) ≤ C k v k BV (Ω) , by (8.3) = C k v k L (Ω) + C ˆ Ω | Dv |≤ C k v k L (Ω) + C ˆ R n | Du | , by (8.2) ≤ C k v k L nn − (Ω) + C ˆ R n | Du | , since Ω is bounded = C k u k L nn − ( R n ) + C ˆ R n | Du |≤ C ˆ R n | Du | = k u k BV ( R n ) , by Theorem . , which implies that µ ∈ BV ( R n ) ∗ . Thus, Theorem 4.4 gives | µ ( U ) | ≤ C H n − ( ∂U ) for every open set U ⊂ R n with smooth boundary.Conversely, assume that µ satisfies condition (8.1). Then Theorem 4.4 yields that µ ∈ BV ( R n ) ∗ .Let u ∈ BV ∞ (Ω) and consider its extension u ∈ BV ( R n ) as in Lemma 8.1. Then, since u ∈ BV ∞ c ( R n ) , there exists a constant C such that(8.4) (cid:12)(cid:12)(cid:12)(cid:12) ˆ R n ( u ) ∗ dµ (cid:12)(cid:12)(cid:12)(cid:12) ≤ C k u k BV ( R n ) . Now, Lemma 8.1 yields k u k BV ( R n ) ≤ C k u k BV (Ω) and since u ≡ on R n \ Ω and u ≡ u on Ω ,we obtain from (8.4) the inequality(8.5) (cid:12)(cid:12)(cid:12)(cid:12) ˆ Ω u ∗ dµ (cid:12)(cid:12)(cid:12)(cid:12) ≤ C k u k BV (Ω) , which means that µ ∈ BV (Ω) ∗ . (cid:3) Remark.
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