Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual of

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1 Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual of BV Monica Torres Purdue University Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 1/49

2 The equation divf = T The solvability of the equation divf = T is connected to the problem of characterizing BV, the dual of the space of functions of bounded variation, BV. Phuc-Torres, Annali de la Scuola Normale Superiore di Pisa, When the right hand side T is a measure µ, then F is called a divergence-measure field. Divergence-measure fields naturally appear in the field of hyperbolic conservation laws. Characterizations of the solvability of divf = µ in the spaces L p (R N ) or C(U): Phuc-T., IUMJ If F L p (R N ) and divf = µ, we can study properties of F. In particular, the existence of normal traces and Gauss-Green formulas for divergence-measure fields: Anzellotti, Anal. Mat. Pure Appl., 1983; Chen-Frid, ARMA 1999, Comm. Math. Phys., 2003; Chen-T., ARMA 2005; Chen-T.-Ziemer, CPAM, 2009; Šilhavý, Rend. Sem. Mat. Padova, 2005; Ambrosio-Crippa-Maniglia, Ann. Fac. Sci. Toulouse, 2005; Frid, 2014; Comi-Chen-Torres, 2017 Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 2/49

3 The space BV 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 1 (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 u BV (Ω) = u L 1 (Ω) + Du (Ω). For the case when Ω = R N we will equip BV(R N ) with the norm given by u BV (R N ) = Du (RN ). (1) Another BV -like space is BV N (R N ), defined as the space of all functions in L N (R N ) such that Du is a finite vector-valued measure. The space BV N (R N ) is a Banach space when equipped with the norm u BV N (R N ) = Du (RN ). We remark that BV(R N ) is not a Banach space with the norm (1). Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 3/49

4 What is known about BV All positive measures in BV(R N ) were characterized by Meyers- Ziemer ( Amer. J. Math., 1977): µ BV(R N ) if and only if µ(b(x,r)) Mr for every r > 0 and x R N. All signed measures in BV(R N ) where characterized by Phuc-T. (IUMJ, 2008): A closed subspace of BV N (R N ), denoted as CH 0, was characterized in De Pauw-T. (Proceedings of the Royal Society of Edinburgh, 2011): Thierry De Pauw (IUMJ, 1998) studied the dual of SBV, the space of special functions of bounded variation. In Phuc-Torres (Annali de la Scuola Normale Superiore di Pisa, 2017) and motivated by image processing, we continued the analysis of BV (including bounded domains). In Fusco-Spector, submitted, 2017, the authors give an integral representation of elements of BV, but under the assumption of the continuum hypothesis. Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 4/49

5 Image processing and BV The space BV is relevant in the field of image processing since the dual norm in BV is needed to estimate the noise of an image (see "Oscillating Patterns in Image Proccessing and Nonlinear Evolution Equations" by Yves Meyer ). Indeed, if f represents an image and the functions u, v are the cartoon and noise respectively, then the function f u v is the texture. The following expression needs to be computed f u v L 2 +λ u BV +µ v BV. Proper characterizations of BV would allow to precisely compute the noise v BV. Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 5/49

6 Solvability of divf = f and the limiting cases. If u solves u = f L p (U) then N < p < u C 0,1 N p (U). McMullen and Preiss have shown that for some f s in L (even continuous f) the equation div F = f has no Lipschitz solution. For some f s in L 1, equation divf = f has no solution in BV and not even in L N/ (Bourgain-Brezis, JAMS 2003). The equation u = f L N admits a solution u W 2,N and therefore divf = f L N has a solution F W 1,N. However, since W 1,N is not contained in L (limiting case in the Sobolev imbbeding Theorem) one can not conclude that F L. Nirenberg gave the example u(x) = ϕ(x)x 1 log x α, 0 < α < N 1 N where ϕ is smooth supported near 0. One checks that u L N (R N ) and u L loc (RN ) Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 6/49

7 Solvability of the equation divf = T Solvability of divf = f L N # ([0,1]N ) in the class L ([0,1] N ;R N ) (Bourgain-Brezis, JAMS 2003). Solvability of divf = f L N # ([0,1]N ) in the class L ([0,1] N ;R N ) W 1,N ([0,1] N ;R N ) (Bourgain-Brezis, JAMS 2003). Solvability of divf = T in the space of continuous vector fields C(U;R N ) (De Pauw-Pfeffer, CPAM 2008). They showed that: The equation has a continuous solution if and only if T belongs to a class of distributions, the space of strong charges, denoted as CH N (U). Moreover, L N (U) CH N (U) Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 7/49

8 The space BV N (R N ) We let BV N (R N ) denote the linear subspace of L(R N N ) consisting of those functions u whose distributional gradient Du is a finite vector valued measure; i.e., Du M(R N ;R N ). { } Du M = Du (R N ) = sup udivf : F D(R N ;R N ) and F 1 R N <. BV N (R N ) is a Banach space with the norm u BV N (R N ) = u L N (R N )+ Du (R N ). We will work with the equivalent norm u BV N (R N ) := Du M. Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 8/49

9 The space of charges vanishing at infinity CH 0 (R N ) Definition. Given a sequence{u j } inbv N (R N ) we writeu j 0 whenever sup j Du j M < ; u j 0 weakly inl N (R N ). A charge vanishing at infinity is a linear functionalf : BV N (R N ) R such that u j,f 0 wheneveru j 0. The collection of these is denotedch 0 (R N ). The compactness in BV N (R N ) implies F CH0 := sup{ u,f : u BV N (R N ) and Du M 1} <. and hence CH0 is a norm on CH 0 (R N ). CH 0 (R N ) BV N (R N ) CH 0 (R N ) with the norm CH0 is a Banach space. Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 9/49

10 The divergence operator Define: div : C 0 (R N ;R N ) CH 0 (R N ) div(f) : BV N (R N ) R, F C 0 (R N ;R N ) div(f)(u) = F,d(Du), u BV N (R N ). R N Proposition. GivenF C 0 (R N ;R N ) one hasdiv(f) CH 0 (R N ) and div : C 0 (R N ;R N ) CH 0 (R N ) is a bounded linear operator: div(f) CH0 F. Question: Is the divergence operator on-to? Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 10/49

11 Characterization of the solvability of divf = T in C 0 (R N,R N ) Theorem[DePauw-T.] There exists F C 0 (R N ;R N ) such that divf = T if and only if T CH 0 (R N ). Since CH 0 (R N ) BV N (R N ), we have then characterized a closed subspace of BV N (R N ). Since L N (R N ) CH 0 (R N ) the theorem implies that to each f L N (R N ) there corresponds a continuous vector field, vanishing at infinity, F C 0 (R N ;R N ) such that divf = f weakly. Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 11/49

12 Proof of the theorem CH 0 (R N ) ev div C 0 (R N ;R N ) BV N (R N ) D M(R N ;R N ) Let {α j } be a sequence in CH 0 (R N ) such that div (α j ) µ. Let {u j } in BV N (R N ) such that α j = ev(u j ). div (α j ) M = (div ev)(u j ) M = Du j M. Since {Φ (α j )} is bounded then sup j Du j M <. Then, there exists a subsequence {u jk } and u BV N (R N ) such that u u jk 0. F,d(Du) = udivf = lim R N R N k u jk divf = lim R N k F,d(Du jk ). R N Then R N F,d(Du) = R N F,dµ since Du j µ. From here we conclude Du = µ. Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 12/49

13 BVc (R N ) is dense in BV N (R N ) Lemma (Phuc-T). Let u BV N (R N ), u 0, and φ k C c (RN ) be a sequence of smooth functions satisfying: 0 φ k 1, φ k 1 onb k (0), φ k 0 onr N \B 2k (0) and Dφ k c/k. Then and for each fixed k > 0 we have lim k (φ ku) u BV N (R N ) = 0, lim j (φ ku) j φ k u BV N (R N ) = 0. In particular, BV c (R N ) is dense in BV N (R N ). Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 13/49

14 Proof of the density The product rule for BV functions gives that φ k u BV N (R N ) and D(φ k u) = φ k Du+uDφ k (as measures). Thus R N D(uφ k u) = φ k Du Du+uDφ k R N φ k 1 Du + R N R N φ k 1 Du + c k R N φ k 1 Du + c k R N φ k 1 Du +c R N supp(dφ k ) u B 2k \B k ( ( B 2k \B k u B 2k \B k u u Dφ k ) N N ) N N. B 2k \B k 1 N Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 14/49

15 Proof of the density On the other hand, the coarea formula for BV functions yields R N D(φ k u (φ k u) j) = = = = j H ( {φ k u (φ k u) j > t})dt H ( {φ k u j > t})dt H ( {φ k u > j +t})dt H ( {φ k u > s})ds. Here F stands for the reduced boundary of a set F. Since 0 H ( {φ k u > s})ds <, the Lebesgue dominated convergence theorem yields that the limit, as j, is 0 (for each fixed k > 0). By the triangle inequality, each nonnegativeu BV N (R N ) can be approximated by a function in BV c (R N ). By considering the positive and negative parts of a function u BV N (R N ), the density of BV c (RN ) in BV N (R N ) follows. Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 15/49

16 BV(R N ) and BV N (R N ) are isometrically isomorphic BV(R N ) BV N (R N ) since u N L (R N ) C(N) Du (R N ), u BV(R N ) We define the map as S : BV N (R N ) BV(R N ) S(T) = T LBV(R N ) S is an isometry as a consequence of our Lemma: BV c (R N ) is dense inbv N (R N ) Notice also that BV c (RN ) is dense inbv(r N ). Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 16/49

17 The Homogeneous Sobolev space Ẇ1,1 (R N ) Definition: Let Ẇ 1,1 (R N ) denote the space of all functions u L N (R N ) such that Du L 1 (R N ). Equivalently, the space Ẇ1,1 (R N ) can also be defined as the closure of C c (R N ) in BV N (R N ) (i.e. in the norm Du L 1 (R N ) ). Thus, u Ẇ1,1 (R N ) if and only if there exists a sequence u k C c (R N ) such that R N D(u k u) 0. We remark that Ẇ 1,1 (R N ) is denoted as G by Y. Meyer in his book Oscillating Patterns in Image Processing and Nonlinear Evolution Equations. The space G plays a key role in modeling the oscillatory nature of noise and texture in images. We have Ẇ 1,1 (R N ) BV N (R N ). It is noted in Meyer s book that, since the space G is simpler than BV N (R N ), the space G = Ẇ1,1 (R N ) is used instead of BV N (R N ). Our main theorem shows that the measures in G coincide with the measures in BV N (R N ). Moreover, all the (signed) measures in BV N (R N ) can be characterized. Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 17/49

18 Measures in Ẇ1,1 (R N ) and BV N (R N ) Definition: We let M loc Ẇ 1,1 (R N ) := {T Ẇ 1,1 (R N ) : T(ϕ) = R N ϕdµ, µ M loc (R N ), ϕ C c (RN )}. Therefore, if µ M loc (R N ) Ẇ1,1 (R N ), the action < µ,u > can be uniquely defined for all u Ẇ1,1 (R N ) (because of the density of C c (RN ) in Ẇ1,1 (R N )). Definition: We let M loc BV N (R N ) := {T BV N (R N ) : T(ϕ) = R N ϕ dµ, µ M loc, ϕ BV c }, where ϕ is the precise representative in BV c (R N ). Thus, if µ M loc BV N (R N ), the action < µ,u > can be uniquely defined for all u BV N (R N ) Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 18/49

19 Characterization of G = Ẇ1,1 (R N ) Theorem: The distribution T belongs to Ẇ1,1 (R N ) if and only if T = divf for some vector field F L (R N,R N ). Moreover, T Ẇ1,1 (R N ) = min{ F L (R N,R N ) }, where the minimum is taken over all F L (R N,R N ) such that divf = T. Here we use the norm F L (R N,R N ) := (F 2 1 +F F2 N )1/2 L (R N ) for F = (F 1,...,F N ). Proof: It is easy to see that if T = divf where F L (R N,R N ) then T Ẇ1,1 (R N ) with T Ẇ1,1 (R N ) F L (R N,R N ). Conversely, let T Ẇ1,1 (R N ). Define A : Ẇ 1,1 (R N ) L 1 (R N,R N ), A(u) = Du, Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 19/49

20 Let T Ẇ1,1 (R N ). We have A : Ẇ 1,1 (R N ) L 1 (R N,R N ), A(u) = Du. The range of A, denoted as R(A), is a closed subspace of L 1 (R N,R N ) since Ẇ 1,1 (R N ) is complete. Define T 1 : R(A) R, T 1 (Du) = T(u), for each Du R(A). Then we have T 1 R(A) = T Ẇ1,1 (R N ). By Hahn-Banach Theorem there exists a norm-preserving extension T 2 of T 1 to all L 1 (R N,R N ). On the other hand, by the Riesz Representation Theorem for vector valued functions there exists a vector field F L (R N,R N ) such that T 2 (v) = R N F v, for every v L 1 (R N,R N ), and F L (R N,R N ) = T 2 L 1 (R N,R N ) = T 1 R(A) = T Ẇ1,1 (R N ). Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 20/49

21 In particular, for each ϕ C c (RN ) we have T(ϕ) = T 1 (Dϕ) = T 2 (Dϕ) = R N F Dϕ, which yields with T = div( F), F L (R N,R N ) = T Ẇ 1,1 (R N ). Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 21/49

22 Characterization of measures in BV N (R N ). Theorem 1 (Phuc-T.) 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 divf = µ in the sense of distributions. (ii) There is a constant C such that µ(u) CH ( U) for any smooth bounded open (or closed) set U with H ( U) < +. (iii) H (A) = 0 implies µ (A) = 0 for all Borel sets A and there is a constant C such that, < µ,u > := u dµ C Du, u BV R N R N c (R N ), where u is the representative in the class of u that is defined H -almost everywhere. (iv) µ BV N (R N ). The action of µ on any u BV N (R N ) is defined (uniquely) as < µ,u >:= lim < µ,u k >= u k R N k dµ, whereu k BVc (R N ),u k u inbv N (R N ) Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 22/49

23 In particular, if u BV c (RN ) then < µ,u >= R N u dµ, and moreover, if µ is a non-negative measure then, for all u BV N (R N ), < µ,u >= R N u dµ. Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 23/49

24 The measures in BV N (R N ) coincide with the measures in Ẇ1,1 (R N ) Theorem 1.2 (Phuc,T.): Let E := M loc (R N ) BV N (R N ) and F := M loc (R N ) Ẇ1,1 (R N ). Then E and F are isometrically isomorphic. Proof: We define S : E F, S(T) = T LẆ1,1. Clearly, T is a linear map. Assume that S(T) = 0 F for some T E. Then T(u) = 0, for all u Ẇ 1,1 (R N ). (2) Thus, if µ is the measure associated to T E, then R N ϕdµ = T(ϕ) = 0 for all ϕ C c (R N ), which implies thatµ = 0. (3) Now, by definition of E, we have T(u) = R N u dµ = 0, for all u BV c (RN ), and by density: T 0 on BV N (R N ). Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 24/49

25 S is on-to Take H F. Thus, there exists µ M loc (R N ) such that ϕdµ = H(ϕ), for all ϕ C R N c (RN ), Since H Ẇ1,1 (R N ), there exists a bounded vector field F L (R N,R N ) such that divf = µ in the sense of distributions and H Ẇ1,1 (R N ) = µ Ẇ1,1 (R N ) = F L (R N,R N ) From our Theorem1: µ << H and µ(u) F H ( U), for all open sets U R N, (4) and u dµ F R N L (R N,R N ) u BVc (RN ), for all u BV c (R N ), (5) Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 25/49

26 S is an isometry Hence, µ BV c (R N ) = F L (R N,R N ). From Theorem1, it follows that µ can be extended to a continuous linear functional ˆµ BV N (R N ) and clearly, S(ˆµ) = µ, which implies that T is surjective. This extension preserves the operator norm and thus S(ˆµ) Ẇ1,1 (R N ) = ˆµ BV N (R N ) = F L (R N,R N ), which shows that E and F are isometrically isomorphic. Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 26/49

27 A formula to compute µ BV N (R N ) The space Ẇ1,1 (R N ) is denoted as the G space in image processing, and it plays a key role in modeling the noise of an image. It is more convenient to work with G instead of BV N (R N ). Indeed, the full characterization of BV N (R N ) is unknown. However, G can be easily characterized. Our previous results show that, when restricted to measures, both spaces coincide. Moreover, the norm of any (signed) measure µ G can be computed as µ G = sup U R N µ(u) H ( U), (6) where the sup is taken over all open sets U R N with smooth boundary and H ( U) < +. Hence, our results give an alternative to the more abstract computation of µ G : µ G = min{ F L (R N,R N ) }, where the minimum is taken over all F L (R N,R N ) such that divf = T. We refer the reader to Kindermann-Osher-Xu for an algorithm based on the level set method to compute (6) for the case when µ is a function f L 2 (R 2 ) with zero mean. Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 27/49

28 The dual of BV 0 (Ω) Let Ω be a bounded open set with Lipschitz continuous boundary Ω and let u BV(Ω). Then, there exists a function ϕ L 1 ( Ω) such that, for H -almost every x Ω, lim r 0 r N B(x,r) Ω u(y) ϕ(x) dy = 0. From the construction of the trace ϕ, we see that ϕ is uniquely determined. Therefore, we have a well defined operator γ 0 : BV(Ω) L 1 ( Ω). We now define the space BV 0 (Ω) as follows: Definition: Let BV 0 (Ω) = ker(γ 0 ). Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 28/49

29 Strict Convergence in BV(Ω) Definition: Let {u k } BV(Ω) and u BV(Ω). We say that u k converges to u in the sense of intermediate (or strict) convergence if u k u strongly inl 1 (Ω) and Ω Du k Ω Du. We also define another BV function space with a zero boundary condition. Definition: Let BV 0 (Ω) := C c (Ω), where the closure is taken with respect to the strict convergence of BV(Ω). The following well known results will be very useful: Theorem: The trace operator γ 0 is continuous from BV(Ω) equipped with the intermediate convergence onto L 1 ( Ω) equipped with the strong convergence. Theorem: The space C (Ω) BV(Ω) is dense in BV(Ω) equipped with the intermediate convergence. Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 29/49

30 The space BV 0 (Ω) Theorem 2 (Phuc-T.): BV 0 (Ω) = BV 0 (Ω). Proof: Let u BV 0 (Ω). Thus, there exists a sequence {u k } C c (Ω) such that u k u in L 1 (Ω) and Du k Ω Ω Du Since u k C c (Ω) then γ 0 (u k ) 0. The continuity of the trace operator implies γ 0 (u k ) γ(u) in L 1 ( Ω), and hence γ(u) = 0 and u BV 0 (Ω). On the other direction, let u BV 0 (Ω). The key point in this proof is to show that BV c (Ω) is dense in BV 0 (Ω) in the strong topology of BV(Ω). Thus, there exists a sequence u k BV c (Ω) such that u k u inl 1 (Ω), lim k Ω Du k Du = 0 Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 30/49

31 Given a sequence ε k 0, we consider the sequence of mollifications w k := u k ρ εk. We can choose ε k sufficiently small to have w k C c (Ω). Also, for each k, lim D(u k ρ ε ) = ε 0 Ω Ω Du k, lim u k ρ ε u k = 0. ε 0 Ω Thus we can choose ε k small enough so that, for each k, Ω D(u k ρ εk ) = Ω Du k +α(k), where α k 1 and 2k, Ω u k ρ εk u k 1 2 k Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 31/49

32 lim w k u lim w k u k + lim u k u = 0. k Ω k Ω k Ω Also, letting k we obtain lim Dw k = lim D(u k ρ εk ) = k Ω k Ω Ω Du. we conclude that w k u in the strict convergence which implies that u BV 0 (Ω). Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 32/49

33 BV c (Ω) is dense in BV 0 (Ω) Theorem: Let Ω be any bounded open set with Lipschitz boundary. Then BV c (Ω) is dense in BV 0 (Ω) in the strong topology of BV(Ω). Proof: We consider first the case u BV 0 (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, and supp(u) C R,T = supp(u) (B R {0}). A generic point in C R,T will be denoted by (x,t), with x B R and t (0,T). We can approximate u with a sequence of functions u k C (C R,T ) such that u k u inl 1 (C R,T ) and C R,T Du k C R,T Du. The condition supp(u) C R,T = supp(u) (B R {0}) implies that γ 0 (u k ) ( CR,T \(B R {0})) 0. By continuity of the trace operator, γ 0 (u k ) γ 0 (u) inl 1 ( C R,T ) and hence Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 33/49

34 and hence, γ 0 (u k ) (BR {0}) = u k (BR {0}) 0 inl1 (B R {0}). u k (x,x n ) u k (x,0) = xn u k (x,x n ) u k (x,0) + We integrate both sides to obtain: B R u k (x,x n ) dx 0 u k x n (x,t)dt, x B R, 0 x n T xn 0 B R u k (x,0) dx + u k,(x,t) x dt. n xn 0 B R Du k (x,t) dx dt. Letting k we obtain u(x,x n ) dx B R xn 0 B R Du = Du (B R (0,x n )) for a.e. 0 < x n < T. Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 34/49

35 Consider a function ϕ C c (R) such that ϕ is decreasing in [0,+ ) and satisfies ϕ 1 on[0,1],ϕ 0 onr\[ 1,2],0 ϕ 1. We define ϕ k (t) = ϕ(kt), k = 1,2,... v k (x,t) = (1 ϕ k (t))u(x,t) 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 C R,T Dv k Du = = C R,T C R,T ( D x u ϕ k D x u, u t ϕ u ) ( k t kϕ (kt)u D x u, u ) t ( u ) ϕ k D x u, ϕ k t kϕ (kt)u. Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 35/49

36 Since ϕ k (t) = 0 for t > 2 k we have the following: ( ) Dv k Du C ϕ k Du + k ϕ (kx n ) u C R,T C R,T C R,T C C C = C = C 2/k 0 2/k 0 2/k 0 2/k 0 2/k 0 Du +Ck B R Du +Ck B R 2/k 0 2/k 0 B R u(x,t) dx dt Du (B R (0,t))dt, Du +Ck Du (B R (0,2/k)) B R Du +Ck 2 2/k B R k Du 0 B R Du. B R 2/k 0 dt Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 36/49

37 A Sobolev inequality for functions in BV 0 (Ω) As a corollary of our theorem we have: Theorem: Let u BV 0 (Ω). Then u L N (Ω) C Du (Ω) for a constant C = C(n). Remark 1: For the next theorems, we equip BV 0 (Ω) and W 1,1 0 (Ω) with the following equivalent norms: u BV0 (Ω) = Du (Ω), u W 1,1 0 (Ω) = Ω Du dx Remark 2: By truncating the function as before, we obtain that BVc (Ω) is dense in BV 0 (Ω). Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 37/49

38 Measures in W 1,1 0 (Ω) and BV 0 (Ω) Definition. For a bounded open set Ω with Lipschitz boundary, we let M loc (Ω) W 1,1 0 (Ω) := {T W 1,1 0 (Ω) : T(ϕ) = Ω ϕdµ,µ M loc (Ω), ϕ C c (Ω)}. Therefore, if µ M loc (Ω) W 1,1 0 (Ω), then the action < µ,u > can be uniquely defined for all u W 1,1 0 (Ω) (because of the density of Cc (Ω) in W 1,1 0 (Ω)). Definition For a bounded open set Ω with Lipschitz boundary, we let M loc (Ω) BV 0 (Ω) := {T BV 0 (Ω) : T(ϕ) = Ω ϕ dµ,µ M loc (Ω), ϕ BV c (Ω)}, where ϕ is the precise representative of ϕ. Thus, if µ M loc (Ω) BV 0 (Ω), then the action < µ,u > can be uniquely defined for all u BV 0 (Ω) (because of the density of BV c (Ω) in BV 0(Ω)). Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 38/49

39 Characterization of measures in BV 0 (Ω) Theorem 3 (Phuc,T.): 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 divf = µ. (ii) µ(u) C H n 1 ( U) for any smooth open (or closed) set U Ω with H n 1 ( U) < +. (iii) H n 1 (A) = 0 implies µ (A) = 0 for all Borel sets A Ω and there is a constant C such that, for all u BV c (Ω), < µ,u > := Ω u dµ C Ω Du, whereu is the representative in the class of u that is definedh n 1 -almost everywhere. (iv) µ BV 0 (Ω). The action of µ on any u BV 0 (Ω) is defined (uniquely) as < µ,u >:= lim < µ,u k >= lim u k dµ, k k Ω where u k BV c (Ω) converges to u in BV 0 (Ω). Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 39/49

40 In particular, if u BVc (Ω) then < µ,u >= Ω u dµ, and moreover, if µ is a non-negative measure then, for all u BV 0 (Ω), < µ,u >= u dµ. Ω Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 40/49

41 The measures in BV(Ω) coincide with the measures in W 1,1 0 (Ω) Theorem 3.1 (Phuc-T.): The normed spaces M loc (Ω) BV 0 (Ω) and M loc (Ω) W 1,1 0 (Ω) are isometrically isomorphic. The characterization of the measures in BV(Ω) is more challenging since, in this case, we need to consider the space BV(Ω) with the full norm: u BV (Ω) = u L ( Ω) + Du (Ω). However, we can prove the following: Theorem: Let µ be a finite signed measure in Ω. Extend µ by zero to R n \Ω by setting µ(r n \Ω) = 0. Then, µ BV(Ω) if and only if µ(u) C H n 1 ( U), for every smooth open set U R n and a constant C = C(Ω,µ). Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 41/49

42 f BV(R N ) but f / BV(R N ) Proposition: Let f(x) = ǫ x 1 ǫ sin( x ǫ )+(n 1) x 1 cos( x ǫ ), where 0 < ǫ < n 1 is fixed. Then f(x) = div[x x 1 cos( x ǫ )]. Moreover, there exists a sequence {r k } decreasing to zero such that B rk (0) f + (x)dx cr n 1 ǫ k for a constant c = c(n,ǫ) > 0 independent of k. Here f + is the positive part of f. Thus by Theorem 1 we see that f belongs to BV(R N ), whereas f does not. This result resolves a question raised by Meyers and Ziemer in their classical paper " Integral inequalities of Poincare and Wirtinger type for BV functions", Amer. J. Math., Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 42/49

43 A constructive approach of bounded solutions In the paper Hierarchical construction of bounded solutions in critical regularity spaces, CPAM), Tadmor presents another duality-based approach for the existence of bounded solutions to divf = f,f L N # ([0,1]N ). His approach is constructive, meaning that the solution F is constructed as a hierarchical sum, F = F j, where the {F j } s are computed recursively as appropiate minimizers, for j = 0,1,... F j+1 = arginf F F L +λ 1 2 j f div j k=1 F k divf N L N, Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 43/49

44 Solvability of divf = µ in L p (R N,R N ), p. Theorem (Phuc-T.) div F = µ µ non-negative measure, has a global solution in F L p (R N,R N ), 1 p N, if and only if µ 0. Moreover, for N < p <, the equation has a solution if and only if I 1 µ L p (R N ), where I 1 µ is the Riesz potential of order 1 of µ defined by 1 I 1 µ(x) = R x y dµ(y), x RN. N Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 44/49

45 Solvability for divf = µ in C(U;R N ) Theorem (Phuc-T.). Let µ be a non-negative measure on a nonempty open setu R N. Then the following are equivalent. The equationdivf = µ has a continuous solution F : U R N. Givenε > 0 and a compact setk U, there isθ > 0 such that ϕdµ ε ϕ dx+θ ϕ dx, ϕ C R R N N R N 0 (R N ), sptϕ K. For any compact setk U, lim r 0 µ(b r (x)) r = 0,x K, and this limit is uniform. Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 45/49

46 Removable singularities of divf = 0 Given the space of vector fields F and S R N a closed set, it is said that the set S is F-removable for the equation divf = 0 if whenever F F satisfies divf = µ in R N \S, then divf = µ in R N. Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 46/49

47 Application to the removable of singularities: L p case Theorem (Phuc-T): Let E be a compact set contained in an open set U R N. Let N µ M(U) such that µ(e) = 0, and let < p. If cap 1,p (E) = 0 then every solution F to divf = µ inu \E, F L p loc (U) (7) is a solution to divf = µ inu, F L p loc (U). (8) Conversely, assume there is at least one vector field F that solves (8) and suppose that every solution to (7) is also a solution to (8), then necessarily cap 1,p (E) = 0. We recall that cap 1,1 (E) = 0 if and only if H (E) = 0. Therefore, S is L -removable if and only if H (S) = 0. Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 47/49

48 Application to the removable of singularities: continuous case Theorem (Phuc-T): Let E be a compact set contained in an open set U R N. Let µ M(U) such that µ(e) = 0. If H (E) = 0 then every solution F to divf = µ inu \E, F C(U) (9) is a solution to divf = µ inu, F C(U). (10) Conversely, assume there is at least one vector field F that solves (10) and suppose that every solution to (9) is also a solution to (10), then H +ǫ (E) = 0 for any ǫ > 0. That is, the Hausdorff dimension of E cannot exceed N 1. Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 48/49

49 Related results in this direction Moonens, Real Anal. Exchange, Moonens-Valeriola, Removable sets for the flux of continuous vector fields, Proc. Amer. Math. Soc., 2010). Moonens-Russ-Tuominen, Removable singularities for div v = f in weighted Lebesgue spaces, Ponce, IUMJ, S is removable if and only if S is σ-finite with respect to H. Ponce-Spector, A boxing inequality for the fractional perimeter, Proved a fractional boxing inequality and used the existence theorems for divergence-measure fields to obtain a trace inequality in fractional Sobolev spaces. Outline of course, Part I: Divergence-measure fields, the solvability of divf = T and the dual ofbv p. 49/49