Characterizations of the Existence and Removable Singularities of Divergence-measure Vector Fields

Transcription

1 Characterizations of the Existence and Removable Singularities of Divergence-measure Vector Fields NGYEN CONG PHC & MONICA TORRES ABSTRACT. We study the solvability and removable singularities of the equation div F = µ, with measure data µ, in the class of continuous or L p vector fields F, where 1 p. In particular, we show that, for a signed measure µ, the equation div F = µ has a solution F L (R n ) if and only if µ() CH n 1 ( ) for any open set with smooth boundary. For non-negative measures µ, we obtain explicit characterizations of the solvability of div F = µ in terms of potential energies of µ for p, and in terms of densities of µ for continuous vector fields. These existence results allow us to characterize the removable singularities of the corresponding equation div F = µ with signed measures µ. 1. INTRODCTION The main goal of this paper is to characterize the solvability and removable singularities of the equation (1.1) div F = µ with measure data µ and continuous or L p vector fields F.Wededucesharpnecessary and sufficient conditions on the size of removable sets from explicit criteria for the solvability of equation (1.1) and fine properties of divergence-measure fields. Divergence-measure fields arise naturally in some areas of partial differential equations such as the field of non-linear conservation laws. It is known that entropy solutions u(t, x) to the system u t + div x f(u) = 0, x R d, 1573 Indiana niversity Mathematics Journal cfl, Vol. 57, No. 4 (2008)

2 1574 NGYEN CONG PHC & MONICA TORRES where u : R + R d R m and f = (f 1,...,f m )with f i : R m R d, i = 1,...,m, belong to some L p (R d+1 + ) space, 1 p, or they could even become measures. Moreover, u satisfies the entropy inequality η(u) t + div x q(u) 0 in the distributional sense, for any convex entropy-entropy flux pair (η, q) (see [12]). The entropy inequality and the Riesz representation theorem imply that there exists a measure µ η,q in R d+1 + such that div t,x (η(u), q(u)) = µ η,q and therefore F u := (η(u), q(u)) is a divergence-measure field. Divergence-measure fields have been investigated by several authors and we refer the readers to the papers Chen-Frid [7, 8], Chen-Torres [9], Chen-Torres- Ziemer [10], Ambrosio-Crippa-Maniglia [3], Šilhavý [26], and the references therein. For a more detailed explanation on the connection and applications of divergence-measure fields to conservation laws we refer the readers to Chen- Torres-Ziemer [11] and the references therein. We now discuss the solvability results in this paper for the L p case, 1 p. For p =, we show in Theorem 3.5 that the solvability of (1.2) div F = µ in R n,µsigned Radon measure, is strongly connected to the existence of normal traces, over boundaries of sets of finite perimeter, for divergence-measure fields. The trace theorem (see Theorem 2.6) obtained recently in Chen-Torres [9] and Chen-Torres-Ziemer [10] enables us to deal with signed measures in equation (1.2). In particular, we show that (1.2) has a global solution F L (R n, R n ) if and only if (1.3) µ() CH n 1 ( ), for any bounded open (or closed) set with smooth boundary. In fact, the new results obtained in Theorem 3.5 also characterize all signed measures µ BV(R n ). On the other hand, in Meyers-Ziemer ([22, Theorem 4.7]), the authors showed that property (1.3), with replaced by balls, characterizes all non-negative measures µ in BV(R n ). In Theorem 3.3, we prove that this condition also characterizes the solvability of equation (1.2) for nonnegative measures. That is, we show that for nonnegative measures µ, equation (1.2) has a global solution F L (R n, R n ) if and only if for any ball B r. µ(b r ) Cr n 1

3 Existence and Removable Singularities of Divergence-measure Fields 1575 It is still an open problem to us to characterize the solvability of (1.2) inthe class of L p vector fields F, 1 p<. We refer to [21] for a related and difficult problem involving signed measures (or even complex distributions) that has been solved by Maz ya and Verbitsky. In this paper, for p, we study the solvability of equation (1.2) for non-negative measures µ. sing the Gauss-Green formula in Theorem 2.10 and the boundedness of Riesz transform, we show that the equation (1.4) div F = µ in R n,µnon-negative measure, has a global solution in F L p (R n, R n ),1 p n/(n 1),ifandonlyifµ 0. Moreover, for n/(n 1) < p <, the equation (1.4) 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 n x y n 1 dµ(y), x Rn. The situation is more subtle if one looks for continuous vector fields F, i.e., if one considers the equation (1.5) div F = µ in, F C(). In the case dµ = f dx where f L n loc(), Brezis and Bourgain [5] showed that the existence of a solution F to (1.5) follows from the closed-range theorem. In a recent paper, motivated by [5], De Pauw and Pfeffer [14] proved that equation (1.5) has a solution if and only if µ is a strong charge, i.e., given ε>0anda compact set K, there is θ>0 such that ϕ dµ ε ϕ L 1 + θ ϕ L 1, for any smooth function ϕ compactly supported on K. Equivalently, equation (1.5) has a solution if and only if for each compactly supported sequence {E i } of BV sets in, (χ Ei ) dµ E i 0 whenever E i 0, where (χ Ei ) denotes the precise representation of χ Ei and E i, E i stand for the perimeter and the Lebesgue measure of E i respectively (see [14]). In this paper, when µ is a non-negative measure, we show that one can replace the BV sets E i with balls. That is, the equation (1.5), with non-negative measure µ, has a solution if and only if for each compact set K (1.6) lim r 0 µ(b r (x)) r n 1 = 0, x K

4 1576 NGYEN CONG PHC & MONICA TORRES and the limit is uniform on K. With these solvability results at hand, we show that singularities of the equation div F = µ in, F L p loc(), µ signed Radon measure can be removed if and only if they form a set of zero (1,p )-capacity, where p = p/(p 1) for n/(n 1) <p<, and of zero Hausdorff measure H n 1 in the case p = ; see Theorem 5.1 below. In particular, if E is a compact set with H n 1 (E) > 0, we find a bounded vector field F on such that div F = 0in \Ebut not in. This existence result was also obtained in [15] byadifferent method. It was also shown recently in [23] that there exists a vector field F L () C ( \ E) such that div F = 0in\Ebut not in. Finally, the new characterization with balls in (1.6) for the existence of continuous vectorfields allowsus to show that thehausdorff dimension of removable sets of the equation div F = µ in, F C(). cannot exceed n 1; see Theorem PRELIMINARIES In this section, we introduce our notation, as well as some definitions and earlier results needed for later development. If is an open set, M() is the set of all locally finite signed Radon measures in and M + () M() consists only of non-negative measures. If µ M(), then µ denotes the total variation of µ. The open ball of radius r centered at x R n will be denoted as B r (x). Given 1 p, we denote by p := p/(p 1) the conjugate of p. The capacity associated to the Sobolev space W 1,p (R n ) is defined by { } (2.1) cap 1,p (K) = inf ϕ p dx ϕ C 0 (R n ), χ K ϕ 1, R n for each compact set K R n. In the case 1 p <n, i.e., n/(n 1) <p this capacity is known to be locally equivalent to the capacity { } Cap 1,p (K) = inf ϕ p dx + ϕ p dx ϕ C 0 (R n ), χ K ϕ 1. However, cap 1,p ( ) = 0inthecasep n, whereas Cap 1,p is nondegenerate, i.e., Cap 1,p (K) > 0 for any nonempty compact set K R n when p >n(see [2, Proposition 2.6.1]). Remark 2.1. For p = 1 (i.e., p = ), cap 1,p (E) = 0ifandonlyif H n 1 (E) = 0(see[28, Theorem 3.5.5] and [2, Proposition 5.1.5]).

5 Existence and Removable Singularities of Divergence-measure Fields 1577 Definition 2.2. Let 1 <p<. We say that µ M + (R n )has finite (1,p)- energy if R n [I 1 µ(x)] p dx<, where I 1 is the Riesz potential of order 1 defined by 1 I 1 µ(x) = dµ(y). R n x y n 1 Remark 2.3. Since for any R>0, 1 (2.2) I 1 µ(x) x y n 1 dµ(y) cµ(b R(O)) ( x +R) n 1, B R (O) where O is the origin of R n,weseeinthecase1<p n/(n 1) that µ 0is the only measure in M + (R n ) that has finite (1,p)-energy. We recall that the space BV(R n ) consists of all functions u L 1 (R n ) such that the distributional gradient u of u is a signed Radon measure with finite total variation in R n. In whatfollows, we consider the space BV(R n ) with the norm u BV = u (R n )= u. The space BV R n c (R n ) consists of all bounded functions in BV(R n ) with compact support. Definition 2.4. Let E R n be an L n -measurable subset. We say that E is a set of finite perimeter if χ E BV(R n ).Thus, χ E is a measure with total variation χ E. The measure-theoretic interior of a set of finite perimeter E is denoted as E 1 anddefinedas { } (2.3) E 1 := y R n E B lim r (y) = 1, r 0 B r (y) Definition 2.5. Let E R n be a set of finite perimeter. The reduced boundary of E, denoted as E, is the set of all points y R n such that (i) χ E (B r (y)) > 0forallr>0; (ii) The limit ν E (y) := lim r 0 χ E (B r (y))/( χ E (B r (y))) exists and ν E (y) =1. The unit vector, ν E (y), is called the measure-theoretic interior unit normal to E at y. Wehave χ E =H n 1 E(see [4, Theorem 3.59]). The following Gauss-Green formula for bounded divergence-measure vector fields was proven in Chen-Torres [9] and Chen-Torres-Ziemer [10] (see also Šilhavý [26]).

6 1578 NGYEN CONG PHC & MONICA TORRES Theorem 2.6. Let F L (, R n ) with div F = µ for some signed Radon measure µ M(). Then, for every bounded set of finite perimeter E, there exist functions F i ν L ( E) and F e ν L ( E) such that and div F = E 1 E 1 E div F = E E (F i ν)(y) dh n 1 (y) (F e ν)(y) dh n 1 (y). Moreover, F i ν L ( E) F and F e ν L ( E) F. A proof of the following result can be found in Chen-Frid [7, Proposition 3.1]. Theorem 2.7. Let F L (, R n ) with div F = µ for some signed Radon measure µ M(). Then, µ H n 1 in ; thatis,ifb is a Borel set satisfying H n 1 (B) = 0,then µ (B) = 0. The L p analogue of the above theorem is stated as follows. Theorem 2.8. Let F L p (, R n ), 1 <p<,withdiv F = µ forsomesigned Radon measure µ M(). Then, µ Cap 1,p in ; thatis,ifb is a Borel set satisfying Cap 1,p (B) = 0,then µ (B) = 0. Proof. Let K be a compact set such that Cap 1,p (K) = 0. It is enough to show that µ(k) = 0. By Corollary 2.39 in [18], we see that cap 1,p (K, O) = 0for any open set O containing K, where { } cap 1,p (K, O) = inf ϕ p dx ϕ C 0 (O), χ K ϕ 1. R n Let {O j } be a sequence of open sets containing K such that O 1 O 2 and j O j = K. Since cap 1,p (K, O j ) = 0 for each j, wecanfindϕ j C 0 (O j), 0 ϕ j 1, ϕ j 1 in a neighborhood of K such that (2.4) ϕ j L p 0 as j. Since div F = µ,wehaveµ(k)+ ϕ j dµ = F ϕ j F L p () ϕ j L p. Thus \K µ(k) µ (O j \ K) + F L p () ϕ j L p 0 as j by (2.4). We refer to Theorem 3.96 in [4] for a proof of the following chain rule.

7 Existence and Removable Singularities of Divergence-measure Fields 1579 Theorem 2.9. Let u BV() and f : R R be a Lipschitz function satisfying f(0) =0if =.Then,v=f ubelongs to BV() and v M u where M = sup f(z). z We also need the following Gauss-Green formula for L 1 vector fields (see loc [13, Theorem 5.4]): Theorem Let F L 1 loc(, R n ) such that div F = µ, for some signed Radon measure µ M(R n ). Then for each x R n and for almost every r>0, B r (x) div F = F(y) ν(y)dh n 1 (y). B r (x) In this Gauss-Green formula, F(y) ν(y) denotes the standard inner product of F (whichisdefinedon B r (x) for almost every r>0) and the outer unit normal ν(y). The next result, originally due to Gustin [19], is known as the boxing inequality (see [29, Chapter 5, Lemma 5.9.3]). Theorem Let n>1and 0 <τ< 1 2. Suppose E is a set of finite perimeter such that lim sup r 0 E B r (x) / B r (x) >τwhenever x E. Then there exists a constant C = C(r,n) and a sequence of closed balls Bri (x i ) with x i E such that and i=1 E r n 1 i B ri (x i ) i=1 CH n 1 ( E). Remark If E is an open set, the proof of Theorem 2.11 actually shows that we can take τ = 1 2. Moreover, the covering {B r i } canbechoseninsuchaway that B ri /5 E = 1 B ri /5 2. This fact will be used in the proof of Theorem 4.4 below. 3. THE L p CASE In this section, we study the solvability of the equation div F = µ, where µ M + (R n ) and F L p (R n, R n ) with 1 p. Our first result shows that n/(n 1) is a critical exponent of the problem since for 1 p n/(n 1) the equation admits no solution unless µ 0.

8 1580 NGYEN CONG PHC & MONICA TORRES Theorem 3.1. Assume that 1 p n/(n 1). If F L p (R n,r n )satisfies div F = µ, for some µ M + (R n ),thenµ 0. Proof. The case p = 1 will be treated slightly different from the case 1 <p n/(n 1). We first note that Fubini s theorem implies, for 1 <p< : µ(b r (x)) µ(b I 1 µ(x) = (n 1) 0 r n dr = (n 1) lim r (x)) ε 0 + ε r n dr. Thus by Theorem 2.10 and polar coordinates for L 1 functions we have I 1 µ(x) = (n 1) lim ε 0 + ε = (1 n) lim ε 0 + = (1 n) lim ε 0 + ε 1 F ν dh n 1 (y) dr B r (x) (x y) F(y) B r (x) x y n+1 dhn 1 (y) dr r n F(y) x y >ε (x y) dy. x y n+1 The last limit is known to exist for almost every x R n and is equal to c(n) n R j f j (x), j=1 where F = (f 1,f 2,...,f n )and R j f j denotes the j th -Riesz transform of the function f j (see [27, formula (5) on page 57]). Moreover, since for 1 <p<,and (see [27, Chapter II]), we conclude that R j f L p C f L p R j f L 1, C f L 1 n (3.1) I 1 µ L p = c(n) R j f j (x) C F L p L p < + j=1 for 1 <p<,and n (3.2) I 1 µ L 1, = c(n) R j f j (x) C F L 1, L 1 < +. j=1 Here L 1, is the weak L 1 space defined as L 1, ={f f measurable on R n, f L 1, < },

9 Existence and Removable Singularities of Divergence-measure Fields 1581 where f L 1, = sup t {x R n : f(x) >t}. t>0 Thus, in view of (2.2), (3.1)and(3.2), µ must be identically zero. We next consider the case where 1 <p <n, i.e., n/(n 1) <p<. Theorem 3.2. Assume that n/(n 1) <p<.iff L p (R n,r n )satisfies div F = µ, for some µ M + (R n ),thenµhas finite (1,p)-energy. Conversely, if µ M + (R n )has finite (1,p)-energy, then there is a vector field F L p (R n, R n ) such that div F = µ. Proof. The necessity part follows from (3.1) in the proof of the Theorem 3.1. To prove the sufficiency part, we consider the space w 1,p defined as the completion of D(R n ) under the Dirichlet norm u L p, where p = p/(p 1). Inthe case 1 <p <n, one has w 1,p = h 1,p, where h 1,p is the completion of D(R n ) under the norm 1/2 u L p (see [20, Section 7.1.2, Theorem 2]). Moreover, by Sobolev s imbedding theorem, in this case w 1,p can also be realized as w 1,p ={u L np /(n p ) (R n ) u L p (R n )}. Let X = w 1,p and Y = L p (R n, R n ). We define an operator A : X Y by A(u) = u, u X. Obviously, Au Y = u X and in particular A is a bounded and injective linear operator. Thus its adjoint A is surjective, where A : Y X, with Y = L p (R n, R n ) and X = (w 1,p ). We now note that for any u D(R n ), A F,u X,X = F,Au Y,Y = F u, R n which implies that A F = div F in D (R n ). In our situation, since µ M + (R n )has finite (1,p)-energy, using the pointwise estimate u(x) CI 1 ( u )(x), u D(R n ),

10 1582 NGYEN CONG PHC & MONICA TORRES (see [27, formula (18) on page 125]), we deduce that u dµ C I 1 ( u ) dµ R n R n = C u I 1 (µ) dx R ( n C u p dx R n ) 1/p by Hölder s inequality. Thus, by density µ X and this yields the desired result since A is surjective. TheanalogueofTheorem3.2 for the case F L (R n, R n ) is the following theorem. Theorem 3.3. Let µ M + (R n )with the property (3.3) µ(b r (x)) Mr n 1, for all r>0,x R n, for some constant M independent of x and r. Then there exists F L (R n, R n ) satisfying div F = µ. Conversely, if F L (R n, R n ) is such that div F = µ for some µ M + (R n ),thenµhas the property (3.3). Proof. For the necessity part, if F L (R n, R n ) satisfies div F = µ then, for any ball B r (x), we have by Gauss-Green formula (Theorem 2.6), µ(b r (x)) = div F = (F i ν)(y) dh n 1. B r (x) B r (x) Now since F i ν L ( B r (x)) F L we obtain µ(b r (x)) C(n) F L r n 1, as desired. To prove the sufficiency part, as before we consider the space w 1,1 defined as the completion of D(R n ) under the norm u L 1. Let X = w 1,1, Y = L 1 (R n, R n ) and let A : w 1,1 L 1 (R n, R n ) be an operator defined by A(u) = u. Since Au Y = u X,weseethatAis bounded and injective. Thus its adjoint A is surjective, where A : L (R n, R n ) (w 1,1 ) is given by A F = div F in D (R n ).

11 Existence and Removable Singularities of Divergence-measure Fields 1583 Therefore, to conclude the proof of the theorem it is enough to show that if µ satisfies property (3.3), then µ (w 1,1 ), i.e., u dµ C u dx, for all u D(R n ). This fact can be proved using the boxing inequality (Theorem 2.11) and the coarea formula as it can be found in [22, Theorem 4.7]. Thus the proof of the theorem is completed. Lemma 3.4. BVc (R n ) is dense in BV(R n ). Proof. We will prove that BVc (Rn ) is dense in BV(R n ) by showing that BV c is dense in BV and BVc is dense in BV c.weletu BV(R n ) and ϕ k C0 (Rn ) a sequence of smooth functions satisfying: (3.4) 0 ϕ k 1, ϕ k 1onB k (0), ϕ k 0onB 2k (0), and ϕ k C k. The product rule for BV functions gives that ϕ k u BV(R n ) and (ϕ k u) = ϕ k u + u ϕ (as measures). Thus (3.5) (ϕ k u u) = ϕ k u u+u ϕ k R n R n ϕ k 1 u + u ϕ k. We let k in (3.5) anduse(3.4) and the dominated convergence theorem to conclude lim (ϕ k u u) =0, k R n which shows that BV c (R n ) is dense in BV(R n ).Foru BV + c u k := u k, k = 1, 2,.... We will show that u k u in BV(R n ). The coarea formula yields we define (u u k ) dx = H n 1 ( {u u k >t})dt R n 0 = H n 1 ( {u k>t})dt 0 = 0 Hn 1 ( {u>k+t})dt = H n 1 ( {u>s})ds. k

12 1584 NGYEN CONG PHC & MONICA TORRES Since H n 1 ( {u>s})ds <, the Lebesgue dominated convergence theorem implies that 0 (3.6) (u u k ) 0 as k. R n If u BV c,wewriteu=u + u.wedefinef k =u + kand g k = u k. Thus f k g k BV c and (u (f k g k )) = u + u f k + g k R n R n (u + f k ) + (u g k ) 0 as k, due to (3.6). That completes the proof of the lemma. We next proceed to prove an analogue of Theorem 3.3 for signed measures. Theorem 3.5. Let µ M(R n ). The following are equivalent: (i) There exists a vector field F L (R n, R n ) such that div F = µ. (ii) For any bounded set of finite perimeter E, there exist functions F i ν and F e ν in L ( E) such that µ(e 1 ) = (F i ν)(y) dh n 1 (y) and µ(e 1 E) = E E (F e ν)(y) dh n 1 (y). Moreover, F i ν L ( E) F and F e ν L ( E) F. (iii) There is a constant C such that max{ µ(e 1 ), µ( E) } CH n 1 ( E) for any bounded set of finite perimeter E. (iv) There is a constant C such that µ() CH n 1 ( ) for any smooth bounded open (or closed) set with H n 1 ( ) < +. (v) H n 1 (A) = 0 implies µ(a) = 0 for all Borel sets A, and there is a constant C such that, for all u BVc (Rn ), µ, u := u dµ C u dx,

13 Existence and Removable Singularities of Divergence-measure Fields 1585 where u is the representative in the class of u that is defined H n 1 -almost everywhere. (vi) µ BV(R n ). The action of µ on any u BV(R n ) is defined (uniquely) as µ, u := lim µ, u k, k where u k BV c (R n ) converges to u in BV(R n ). Moreover, if µ is a nonnegative measure then, for all u BV(R n ), µ, u = u dµ. R n (vii) There is a constant C such that u dµ C u dx for all u C0 (Rn ). Proof. Suppose (i) holds. If F L (R n, R n ) satisfies div F = µ then, for any bounded set of finite perimeter E, the Gauss-Green formula (Theorem 2.6) yields, and µ(e 1 E) = E 1 E µ(e 1 ) = div F = E 1 div F = E which are the equalities in (ii). The estimates E (F e ν)(y) dh n 1 (y) (F i ν)(y) dh n 1 (y), F e ν L ( E) F L and F i ν L ( E) F L give and Therefore, µ(e 1 E) = µ(e 1 ) + µ( E) F L H N 1 ( E) µ(e 1 ) F L H N 1 ( E). µ( E) F L H N 1 ( E) + µ(e 1 ) 2 F L H N 1 ( E), which gives (iii). We note that (iii) (iv) since for any bounded open (resp. closed) set with smooth boundary we have = 1 (resp. = 1 ). We proceed now to show that (iv) (v). Since (iv) gives µ(b r (x)) Cr n 1 for every ball B r (x), a standard convering argument (see e.g., Theorem 2.56 in

14 1586 NGYEN CONG PHC & MONICA TORRES [4]) shows that µ H n 1 (see also Theorem 2.7). We let u BV c (R n ) and we consider its positive and negative parts u + and u. We note that and u + = f max (u), u = f min (u), where f max (x) = max(0,x) where f min (x) = min(0,x). Since f max and f min are Lipschitz functions, Theorem 2.9 implies that both u + and u belong to BV c (R n ). Let ρ ε be a stardard sequence of mollifiers. We consider the convolutions ρ ε u + and define A ε t := {ρ ε u + > t}. Since ρ ε u + C 0 (Rn ), it follows that A ε t is smooth for a.e. t and thus, for each ε and almost every t: (3.7) A ε t = (Aε t )1. Since ρ ε u + 0 we can now compute: ρ ε u + dµ = µ(a ε R n t ) dt = µ((a ε (3.8) t )1 ) dt 0 0 CH n 1 ( A ε t ) dt 0 = C (ρ ε u + ) dx R n C u + dx C u dx. In the same way we obtain (3.9) ρ ε u dµ C u dx. We let u denote the precise representative of u. Wehavethat(see[4, Chapter 3, Corollary 3.80]): (3.10) ρ ε u u H n 1 -almost everywhere. From (3.8) and(3.9)weobtain (3.11) ρ ε u dµ = (ρ ε u + ρ ε u ) dµ 2C u. R n We now let ε 0in(3.11). Since u is bounded, (3.10), Theorem 2.7 and the dominated convergence theorem yield (3.12) u dµ C u dx,

15 Existence and Removable Singularities of Divergence-measure Fields 1587 which proves (v). From (v) we obtain that the linear operator T(u):= µ, u = u dµ, u BV R n c (Rn ) is continuous and hence it can be uniquely extended, since BV c (Rn ) is dense in BV(R n ) (Lemma 3.4), to the space BV(R n ). Assume now that µ is non-negative. We take u BV(R n ) and consider the positive and negative parts (u ) + and (u ) of the representative u. With the notation of Lemma 3.4 we find that the sequence u k := ϕ k ((u ) + k) belongs to BVc (R n ) and converges to u in the BV norm. Therefore, the monotone convergence theorem yields T ((u ) + ) = lim u k dµ = (u ) + dµ. k We proceed in the same way for (u ) and thus we conclude T(u) = T ((u ) + ) T ((u ) ) = (u ) + (u ) dµ = u dµ. It is clear that (vi) (vii). We note now that if µ M(R n )satisfies (vii), then µ belongs to the space (w 1,1 ) defined in Theorem 3.3, and since A : L (R n, R n ) (w 1,1 ) is surjective, we obtain (i); that is, there is a vector field F L (R n, R n ) satisfying div F = µ. Remark 3.6. In Meyers-Ziemer ([22, Theorem 4.7]), the authors showed that property (3.3) characterizes all positive measures in BV(R n ). In Theorem 3.3, we have shown that property (3.3) also characterizes the solvability of the equation div F = µ. Moreover, the results obtained in Theorem 3.5 allowustocharacterize even signed measures µ BV(R n ) and in particular include the Meyers-Ziemer result for non-negative measures. 4. THE CONTINOS CASE We begin this section by quoting the following corollary of a result due to De Pauw and Pfeffer [14] (see also [25]). Theorem 4.1. Let µ be a signed Radon measure in a nonempty open set R n. Then the following properties are equivalent. (i) The equation div F = µ has a continuous solution F : R n. (ii) Given ε>0and a compact set K,thereisθ>0such that ϕ dµ ε ϕ dx + θ ϕ dx for all ϕ C 0 (Rn ) with supp ϕ K.

16 1588 NGYEN CONG PHC & MONICA TORRES (iii) For each compactly supported sequence {E i } of BV sets in, (χ Ei ) dµ E i 0 whenever E i 0. Here (χ Ei ) denotes the precise representation of χ Ei, whereas E i and E i stand for the perimeter and the Lebesgue measure of E i respectively. Our goal in this section is to consider the case where µ is a non-negative measure in the above theorem and simplify property (iii) there by replacing BV sets E i with balls. Thereby, we obtain new criteria for the solvability of div F = µ in the class of continuous vector fields F when µ is a non-negative measure. We remark that De Pauw-Pfeffer s result in [14] deals with distributions called strong charges and Radon measures are not necessarily strong charges. In order to obtain our main result, Theorem 4.5 below, we first show the following result. Theorem 4.2. Let R n be an open set and let µ be a signed Radon measure in. Suppose that for any compact set K, lim sup δ 0 + x 0 K { sup } u dµ : u C0 (B δ(x 0 )), u L 1 1 = 0. R n Then, given ε>0and a compact set K,thereisθ>0such that (4.1) ϕ dµ ε ϕ dx + θ ϕ dx R n for all ϕ C 0 (Rn ) with supp ϕ K. Proof. We let ε>0andϕ C 0 (Rn )with supp ϕ K. Let d(k) = dist(k, ), andset { K d(k)/2 = x 0 dist(x 0,K) d(k) 2 By hypothesis, there exists 0 <δ=δ(ε) < d(k)/2 such that for any x 0 K d(k)/2 }. (4.2) u dµ ε u dx, for all u C 0 (B δ(x 0 )). Let η C 0 (B 1(0)) be a cut-off function such that 0 η(x) 1, η(x) = 1 for x 1 2,and η(x) C(n). We next choose x i R n, i = 1, 2,...,

17 Existence and Removable Singularities of Divergence-measure Fields 1589 so that {x i } form a cubic lattice with grid distance δ/(3 n) and let η i (x) = η(x x i )/δ. Note that η i C 0 (B δ(x i )), η i (x) = 1for x x i <δ/2, and (4.3) η i C δ. Let (4.4) ϕ(x) = i η i (x), where the sum is taken over a finite number of indices i such that K d(k)/2 i B δ/2 (x i ). Since each point x K d(k)/2 is contained in at most N = N(n) balls B δ/2 (x i ), we have that 1 ϕ(x) N(n) on K d(k)/2,andby(4.3)and(4.4), (4.5) ϕ(x) i η i (x) C(n) δ on K d(k)/2. We now define (4.6) ξ i (x) = η i(x) ϕ(x) and note that i ξ i (x) = 1onK. Since each ξ i ϕ C0 (B δ(x i )), from(4.2) we obtain ϕ dµ = ξ i ϕ dµ ε (ξ i ϕ) dx R i i n ε ξ i ϕ dx+ε ϕ ξ i dx R i n R i n = ε ϕ dx + ε ϕ ξ i dx. R i n R n i On the other hand, using (4.5)and(4.6) we can estimate the last sum by ξ i (x) i i i C 1(n) δ η i (x) ϕ(x) η i (x) ϕ(x) + i + C 2(n) δ ϕ(x) η i (x) (ϕ(x)) 2 + ϕ(x) ϕ(x) = C(n) δ.

18 1590 NGYEN CONG PHC & MONICA TORRES Hence ϕ dµ ε ϕ dx + εc(n) δ R n ϕ dx, which gives inequality (4.1) with θ=εc(n)/δ. This completes the proof of the theorem. Remark 4.3. Theorem 4.2 holds also for general distributions µ with the same proof. We next prove the following result. Theorem 4.4. Let R n be an open set and let µ be a non-negative measure in. Suppose that for any compact set K, lim sup δ 0 + x 0 K Then, for any compact set K, lim sup δ 0 + x 0 K { sup { } µ(br (x sup 0 )) r n 1, 0 <r <δ =0. } u dµ : u C0 (B δ(x 0 )), u L 1 1 = 0. Proof. Let K be a compact set. Let ε>0andd(k) = dist(k, ). We define K d(k)/2 ={x 0 dist(x 0,K) d(k)/2}. By hypothesis, there exists 0 <δ 1 < d(k)/2 such that (4.7) µ(b 2r (x 0 )) εr n 1, for all x 0 K d(k)/2 and 0 <r <δ 1.Nowletu C 0 (B δ(x 0 )) with u L 1 1, δ<δ 1 /10 and x 0 K. We consider u + and u, the positive and negative parts of u, which are continuous functions. By applying boxing inequality (Theorem 2.11) to the open set {u + >t}we can find a covering { Bri,t (x i,t )} of {u + >t} such that (4.8) i r n 1 i,t C(n)H n 1 ( {u + >t}), where C(n) is independent of t. Note that since u + is compactly supported in B δ (x 0 ), the set {u + >t} B δ (x 0 ). Also, from the proof of the boxing inequality (see Remark 2.12) wehave that the covering { Bri,t (x i,t )} is chosen in such a way that 2 B ri,t /5(x i,t ) {u + >t} = B ri,t /5(x i,t ).

19 Existence and Removable Singularities of Divergence-measure Fields 1591 But since {u + >t} B δ (x 0 ), we conclude that each radius r i,t is less than 10δ. That is, we have x i,t K d(k)/2 and r i,t < 10δ <δ 1, which implies, in view of (4.7), that (4.9) µ(b 2ri,t (x i,t )) εr n 1 i,t, for all i. Therefore, from (4.8), (4.9) and the coarea formula we obtain (4.10) u + dµ = µ({u + >t})dt R n 0 µ(b 2ri,t (x i,t )) dt 0 i ε r n 1 i,t dt 0 i C(n)ε H n 1 ( {u + >t})dt 0 =C(n)ε u + dx C(n)ε u dx C(n)ε. In the same way we obtain (4.11) u dµ C(n)ε u dx C(n)ε u dx C(n)ε. R n From (4.10) and(4.11) we conclude (4.12) u dµ = (u + u ) dµ 2Cε. which yields the theorem. We can now put together the previous results to obtain the following equivalences. Theorem 4.5. Let µ be a non-negative measure on a nonempty open set R n. Then the following properties are equivalent. (i) The equation div F = µ has a continuous solution F : R n. (ii) Given ε>0and a compact set K,thereisθ>0such that ϕ dµ ε ϕ dx + θ ϕ dx, R n for all ϕ C 0 (Rn ) with supp ϕ K.

20 1592 NGYEN CONG PHC & MONICA TORRES (iii) For any compact set K, lim sup δ 0 + x 0 K (iv) For any compact set K, lim sup δ 0 + x 0 K { sup { } µ(br (x sup 0 )) r n 1, 0 <r <δ =0. } u dµ : u C0 (B δ(x 0 )), u L1 1 = 0. R n Proof. By Theorem 4.1 we have (i) (ii). Thus it is enough to show that (ii) (iii) since by Theorem 4.4 and Theorem 4.2 we have (iii) (vi) and (vi) (ii). To this end, let ε>0andk. As before, we define d(k) = dist(k, ) and set { K d(k)/2 = x 0 dist(x 0,K) d(k) 2 By property (ii), there exists θ(ε) > 0 such that (4.13) ϕ dµ ε ϕ dx + θ(ε) ϕ dx R n for all ϕ C0 (Rn ) with supp ϕ K d(k)/2. Let x 0 K and let 0 <r <δ where δ = min{d(k)/4,ε/(2θ(ε))}. We next choose a cut-off function ϕ C0 (B 2r (x 0 )) with 0 ϕ 1, ϕ = 1onB r (x 0 ),and ϕ(x) C(n)/r. Since supp ϕ K d(k)/2, we can use it to test (4.13) to obtain µ(b r (x 0 )) ϕ dµ B 2r (x 0 ) ε ϕ dx + θ(ε) B 2r (x 0 ) εc(n)r n 1 + θ(ε)c(n)r n εc(n)r n 1, }. ϕ dx B 2r (x 0 ) by our choice of δ. Thus, we get (ii) and the theorem is completely proved. 5. REMOVABLE SINGLARITIES We give in this section an application of the previous results to the removability of singularities for the equation div F = µ, for both F L p loc, n/(n 1) <p, andfcontinuous. As it turns out, removable sets of such equations can be characterized by the capacity associated to the Sobolev space W 1,p (R n ) defined in (2.1). To emphasize our next result for p =, we recall from Remark 2.1 that cap 1,1 (E) = 0ifandonlyifH n 1 (E) = 0.

21 Existence and Removable Singularities of Divergence-measure Fields 1593 Theorem 5.1. Let E be a compact set contained in an open set R n. Let µ M() such that µ(e) = 0, and let n/(n 1) <p.ifcap 1,p (E) = 0; then every solution F to (5.1) div F = µ in \ E, F L p loc() is a solution to (5.2) div F = µ in, F L p loc(). Conversely, assume there is at least one vector field F that solves (5.2) and suppose that every solution to (5.1) is also a solution to (5.2), then necessarily cap 1.p (E) = 0. Proof. We prove first the sufficiency part and assume now that cap 1,p (E) = 0. Let F be a solution to (5.1). Thus, (5.3) F ϕdx= ϕdµ, ϕ C0 ( \ E). Since cap 1,p (E) = 0, we can find a sequence u k C0 () such that u k 1onE and u k L p 0. Moreover, u k can be chosen so that 0 u k 1, and u k 0 pointwise on, except possibly on a set N with cap 1,p (N ) = 0(see[24]). We need to show that (5.4) F ψdx= ψdµ, for all ψ C0 (). To this end, we approximate ψ by the sequence of functions (5.5) ψ k := ψ(1 u k ) C0 ( \ E). We have and hence (5.6) ψ k = ψ(1 u k ) ψ u k ψ k ψ L p = u k ψ ψ u k L p u k ψ L p + ψ u k L p 0. From (5.3) and(5.5)weget,forallk, (5.7) F ψ k = ψ k dµ.

22 1594 NGYEN CONG PHC & MONICA TORRES As k,hölder s inequality and (5.6)yield (5.8) F ψ k dx F ψdx. Moreover, since µ = div F on the open set \ E, Theorems 2.7 and 2.8 imply that µ H n 1 on \ E (for p = )andµ cap 1,p on \ E (for p< ). In any case, we have µ(n ) = 0sinceµ(E) = 0. Thus, ψ k ψµ-e.a. and the dominated convergence theorem then gives (5.9) ψ k dµ ψ dµ as k. Combining (5.7), (5.8)and(5.9)weobtain(5.4). We now proceed to prove the necessity part and we consider first the case p =.IfH n 1 (E) > 0, then Frostman s lemma (see [6, Theorem 1 on page 7]) gives the existence of a non-trivial positive measure σ supported on E such that for any ball B r, σ(b r ) Cr n 1. Thus by Theorem 3.3, there is F σ L (R n, R n ) such that div F σ = σ. Wenow let F = F + Fσ. Then div F = µ in D ( \ E) but div F = µ + σ µ in D (), whichgivesa contradiction. For the case p, we assume now that every solution to (5.1) is also a solution to (5.2) butcap 1,p (E) > 0. Since cap 1,p (E) > 0, there is a non-trivial non-negative measure σ supported on E such that σ has finite (1,p)- energy (see [2, Theorem 2.5.3]). By Theorem 3.2, there is F L p (R n, R n ) such that div F = σ. Ifwelet F= F+Fσ, then as before we obtain a contradiction since div F = µ in D ( \ E) but not in D (). This completes the proof of the theorem. Theorem 5.2. Let E be a compact set contained in an open set R n. Let µ M() such that µ(e) = 0. IfH n 1 (E) = 0, then every solution F to (5.10) div F = µ in \ E, F C() is a solution to (5.11) div F = µ in, F C(). Conversely, assume there is at least one vector field F that solves (5.11) and suppose that every solution to (5.10) is also a solution to (5.11);then H n 1+ε (E) = 0 for any ε>0. That is, the Hausdorff dimension of E cannot exceed n 1.

23 Existence and Removable Singularities of Divergence-measure Fields 1595 Proof. The proof of the sufficiency part is the same as that of Theorem 5.1 since F L loc(). To prove the necessity part, we let ε>0 and assume that H n 1+ε (E) > 0. Then by Frostman s lemma there exists a non-trivial positive measure σ supported on E such that for any ball B r, σ(b r ) Cr n 1+ε. Thus, σ(b lim r ) r 0 r n 1 = 0 which, in view of Theorem 4.5, implies that there is F σ C(R n,r n ) such that div F σ = σ. Wenowlet F= F+Fσ. Then div F = µ in D ( \ E) but div F = µ + σ µ in D (), whichgivesa contradiction. Remark 5.3. A similar result on removable singularities was obtained in [23] by a different method, where it is shown that if H n 1 (E) = 0 then every solution F to (5.12) div F = 0in\E, F L loc() C ( \ E) is a solution to (5.13) div F = 0in, F L loc() C ( \ E). Conversely, if H n 1 (E) > 0, then there exists a vector field F L loc() C ( \ E) that solves (5.12)butnot(5.13). This strengthens our result in Theorem 5.1 for p = in the necessity direction. Acknowledgments The authors would like to thank W. Ziemer and the anonymous referee for useful comments and suggestions. Monica Torres s research was supported in part by the NSF under grant DMS REFERENCES [1] D. R. ADAMS, A note on Choquet integrals with respect to Hausdorff capacity, Proc. Function Spaces and Applications, Lund, 1986, Lecture Notes in Math., vol. 1302, Springer, Berlin, 1988, pp MR (90c:26028) [2] D. R. ADAMS and L. I. HEDBERG, Function Spaces and Potential Theory, Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], vol. 314, Springer-Verlag, Berlin, 1996, ISBN MR (97j:46024) [3] L. AMBROSIO, G.CRIPPA,andS.MANIGLIA, Traces and fine properties of a BD class of vector fields and applications, Ann. Fac. Sci. Toulouse Math. (6) 14 (2005), MR (2007b:35040) (English, with English and French summaries)

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