KATO S INEQUALITY UP TO THE BOUNDARY. Contents 1. Introduction 1 2. Properties of functions in X 4 3. Proof of Theorem

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1 KATO S INEQUALITY UP TO THE BOUNDARY HAÏM BREZIS(1),(2) AND AUGUSTO C. PONCE (3) Abstract. We show that if u is a finite measure in then, under suitable assumptions on u near, u + is also a finite measure in. We also study properties of the normal derivatives + and on. Contents 1. Introduction 1 2. Properties of functions in X 4 3. Proof of Theorem Properties of 8 5. Proof of Theorem Kato s inequality up to the boundary Computing + for W 2,1 -functions 16 Appendix A. The measure u + need not be finite 18 Appendix B. Approximation by smooth functions in 19 Appendix C. Proof of Lemma References Introduction Let R N be a smooth bounded domain. Given u L 1 () with u L 1 (), Kato s inequality (see [9]; see also [4]) asserts that (1.1) u + χ u in D (). In particular, (1.1) implies that u + is a locally finite measure in. Our goal in this paper is to address the question whether u + is a finite measure up to the boundary of, i.e., whether u + <. In general, the answer is negative: one can even construct harmonic functions u C() H 1 () such that u + is not a finite measure in ; see Proposition A.1 below. With further assumptions on u (for instance if u W 2,1 () or if u vanishes on the boundary) we will see that the answer is positive. Date: November 26,

2 2 HAÏM BREZIS(1),(2) AND AUGUSTO C. PONCE (3) The following class of functions will play a central role. We say that u X if u W 1,1 () and if there exists a constant C > 0 such that (1.2) u ψ C ψ L ψ C1 (), in which case we set [u] X = sup u ψ. ψ C 1 () ψ L 1 Note that if u X, then there exists a unique T [ C() ] = M() such that T, ψ = u ψ ψ C 1 (). On the other hand, by the Riesz Representation Theorem any T M() admits a unique decomposition T, ψ = ψ dν + ψ dµ ψ C(), where µ M() and ν M( ). As usual, M() and M( ) denote the spaces of finite measures in and, respectively, equipped with the norm M ; measures in M() are identified with measures in which do not charge. When u X, we will denote µ = u and ν =. Throughout the paper, whenever u X we use the notation u and in the above sense. If u X, then u ψ = ψ ψ u ψ C 1 (), and consequently, u ψ u ψ = Also, note that if u X, then [u] X = ψ ψ u u +. ψ C 2 (). In particular, [ ] X defines a seminorm in X and [u] X = 0 if, and only if, u is constant in. In order to verify this last assertion, one may use the fact that for every h C () with h = 0, there exists ψ C () such that ψ = h in with ψ = 0 on. Clearly, any function u W 2,1 () belongs to X and our notation is consistent with the usual meaning of u and. Recall that, for any function u L1 (), u is well-defined as a distribution. When u X, the distribution u belongs to M(), but the converse is not true; see, e.g., Proposition A.1 below. We now present our main results.

3 KATO S INEQUALITY UP TO THE BOUNDARY 3 Theorem 1.1. If u X, then u + X and (1.3) [u + ] X [u] X. In other words, (1.4) u u +. Our next result gives additional properties when u vanishes on the boundary: Theorem 1.2. If u W 1,1 0 () and u M() (in the sense of distributions), then u X (hence u + X). Moreover, (1.5) u + u. In addition, L1 ( ) with (1.6) u. Note that assertions (1.5) (1.6) fail if u does not vanish on ; simply take = B 1, the unit ball in R N, and u(x) = x 1. We now state our extension of Kato s inequality up to the boundary: Theorem 1.3. Let u X be such that u L 1 () and L1 ( ). Then, (1.7) u + ψ Hψ Gψ ψ C 1 (), ψ 0 in, where G L 1 () and H L 1 ( ) are given by { u on [u > 0], on [u > 0], (1.8) G = and H = 0 on [u < 0], 0 on [u 0], min {, 0} on [u = 0]. Thus, u + G in, (1.9) + H on. We conclude this introduction with the following problems: Open Problem 1. Let u X. Is it true that (1.10) + on? This problem is open even under the additional assumption that u W 1,1 0 (). Open Problem 2. Assume that u X and L1 ( ). Is it true that + L 1 ( )? More precisely, does one have + (1.11) = H, where H is the function given by (1.8)?

4 4 HAÏM BREZIS(1),(2) AND AUGUSTO C. PONCE (3) The answer to both Open Problems 1 and 2 is positive if u W 2,1 (); see Theorem 7.1 below. Addendum. Recently, A. Ancona informed us that he gave a positive answer to Open Problems 1 and 2 in full generality. His argument strongly relies on tools from Potential Theory; see [2]. 2. Properties of functions in X In this section, we investigate properties satisfied by elements in X. We first show that condition (1.2) required for a function to belong to X can be replaced by (2.1) u ζ C ζ L ζ C2 N(), where (2.2) C 2 N() = { ζ C 2 (); } ζ = 0 on. Proposition 2.1. Let u L 1 (). Then, u X if, and only if, (2.3) sup u ζ <. ζ C 2 N () ζ L 1 Moreover, (i) the quantity in (2.3) equals [u] X ; (ii) u W 1,p () for every 1 p < N N 1 ; moreover, u L p () C[u] X. In the proof of Proposition 2.1, we need the following variant of the classical De Giorgi-Stampacchia estimate (see [7, 8]) for the Neumann problem: Lemma 2.1. Given F C0 (; R N ), let w be the unique solution of w = div F in, (2.4) w = 0 on, such that w = 0. Then, for every q > N we have (2.5) w L C F L q. We present a sketch of the proof of Lemma 2.1 in Appendix C. Proof of Proposition 2.1. Note that if u X, then (2.6) u ζ = u ζ [u] X ζ L ζ C 2 N(). This gives the implication. We now assume that (2.3) holds. We split the proof of the converse into two steps: Step 1. u W 1,p () for every 1 p < where K denotes the quantity in (2.3). N N 1 and u L p () CK,

5 KATO S INEQUALITY UP TO THE BOUNDARY 5 Clearly, we may assume that 1 < p < N 1. Given F C 0 (; R N ), let w be the unique solution of (2.4) such that w = 0. By (2.3) and (2.5), we have u div F = u w K w L KC F L F p C 0 (; R N ). The conclusion follows by duality. Step 2. u X and [u] X = K. It suffices to show that (2.7) u ψ K ψ L N ψ C1 (). Indeed, this implies u X and [u] X K. Since by (2.6), K [u] X, equality must hold. We now turn ourselves to the proof of (2.7). Given ψ C 2 (), we first show that there exists a sequence (ζ k ) such that (2.8) ζ k C 2 N(), ζ k L C, ζ k ψ uniformly in and (2.9) ζ k ψ a.e. in. Indeed, let Φ C 0 (R) and η C 2 () with η = 0 on be such that Take Φ(t) = t t [ 1, 1] and η = ψ on. ζ k = ψ 1 Φ(kη) in. k Clearly, (2.8) holds. On the other hand, [ ] 1 k Φ(kη) = Φ (kη) η χ [η=0] η in. Since η = 0 a.e. on the set [η = 0], (2.9) follows. For every k 1, we thus have u ζ k = u ζ k K ζ k L. As k, we obtain (2.7) with test functions ψ C 2 (). argument, one then gets (2.7). The proof is complete. Using a density Remark 2.1. Using Proposition 2.1, one deduces that given measures µ M() and ν M( ), the Neumann problem u = µ in, (2.10) = ν on, has a solution u X if, and only if, (2.11) µ() + ν( ) = 0. The solution is unique up to an additive constant and belongs to u W 1,p () for every 1 p < N N 1. In particular, if u = 0, then u W 1,p () C[u] X. The following result complements Proposition 2.1:

6 6 HAÏM BREZIS(1),(2) AND AUGUSTO C. PONCE (3) Proposition 2.2. Let u L 1 () be such that (2.12) u ζ ζ dν + ζ dµ ζ CN(), 2 ζ 0 in for some µ M() and ν M( ). Then, u X, ) (2.13) [u] X 2 ( µ + M() + ν + M( ) and (2.14) Proof. By (2.12), we have (2.15) u ζ u µ in, ν on. ζ dν + + ζ dµ + ζ CN(), 2 ζ 0 in. For every ζ CN 2 (), we apply (2.15) with test functions ζ L ± ζ to get ( (2.16) u ζ 2 µ + M() + ν + M( ) ) ζ L. By Proposition 2.1, it follows that u X and (2.13) holds. Proceeding as in Step 2 of the proof of Proposition 2.1 (more precisely, using (2.8) (2.9)), one deduces from (2.12) that Therefore, This gives (2.14). u ψ ψ dν + ψ dµ ψ C 2 (), ψ 0 in. ψ ψ u ψ dν + ψ dµ ψ C 2 (), ψ 0 in. 3. Proof of Theorem 1.1 We begin by establishing the following lemma: Lemma 3.1. If u C 2 (), then (3.1) u + ψ ψ ψ u ψ C 1 (), ψ 0 in. Proof. We first prove the Claim. If u C 2 () and Φ C 2 (R) is convex, then (3.2) Φ(u) ψ ψ Φ (u) ψ Φ (u) u ψ C 1 (), ψ 0 in. Note that and, by the convexity of Φ, Φ(u) = Φ (u) on Φ(u) Φ (u) u in.

7 KATO S INEQUALITY UP TO THE BOUNDARY 7 Thus, for every ψ C 1 (), ψ 0 in, Φ(u) ψ = ψ Φ(u) ψ Φ(u) This establishes the claim. ψ Φ (u) ψ Φ (u) u. We now apply (3.2) with Φ = Φ k, where (Φ k ) is a sequence of smooth convex functions such that Φ k (0) = 0, Φ k L 1 and satisfying { Φ 1 if t 0, k(t) 0 if t < 0. As k, we obtain (3.1). We now prove a special case of Theorem 1.1 for functions in C 2 (): Lemma 3.2. Let u C 2 (). Then, u + X and (3.3) [u + ] X [u] X. Proof. Note that u + W 1,1 (). In order to establish the lemma, it thus suffices to show that (3.4) u + ψ [u] X ψ L ψ C 1 (). For this purpose, given ψ C 1 () we apply (3.1) with ψ = ψ L + ψ. We then get ( ) (3.5) u + ψ u ψ L + ψ ψ u. Since estimate (3.5) becomes ( u + ψ ( [u<0] u = [u<0] [u<0] + [u<0] ) u ψ L + u, ψ ) + u ψ L = [u] X ψ L. ψ u This relation holds for every ψ C 1 (). Replacing ψ by ψ, we obtain (3.4). This establishes the lemma. Proof of Theorem 1.1. Since u X, u ψ = ψ ψ u Taking ψ = 1 as a test function, we get (3.6) = u. ψ C 1 ().

8 8 HAÏM BREZIS(1),(2) AND AUGUSTO C. PONCE (3) Let (µ k ) C () and (ν k ) C ( ) be two sequences such that µ k u weak in M() and µ k L 1 () u M(), ν k weak in M( ) and ν k L 1 ( ). M( ) In view of (3.6) we may also assume that ν k = µ k k 1. For each k 1, let u k C 2 () be the unique function such that u k = µ k in, k = ν k on, and u k = u. Then, by Remark 2.1 applied to u k u, the sequence (u k) is bounded in W 1,p () for every 1 p < N N 1. Since u k u a.e., one deduces that u + k u+ weakly in L 1 (). On the other hand, applying Lemma 3.2 to u k, we get u + k ψ [u + k ] X ψ L [u k ] X ψ L ψ C 1 (). As k, we obtain from which the conclusion follows. u + ψ [u] X ψ L 4. Properties of ψ C 1 (), We start with a result which seems intuitively true, but still requires a proof: Proposition 4.1. Let u W 1, (). Then, u X if, and only if, u M() (in the sense of distributions). In this case, L ( ) and (4.1) u L (). L ( ) If u C 1 () X, then coincides with the standard normal derivative on. Proof. We first assume that u W 1, () and u M(). Given a sequence of mollifiers (ρ k ) such that supp ρ k B 1/k, let u k (x) = ρ k (x y)u(y) dy x. Note that if d(x, ) > 1/k, then u k (x) = ρ k (x y) u(y) dy and u k (x) = ρ k (x y) u(y) dy.

9 KATO S INEQUALITY UP TO THE BOUNDARY 9 Denote (4.2) δ = { x ; d(x, ) > δ } ; for δ 0 > 0 small enough, δ is smooth for every δ (0, δ 0 ). For every k 1 and δ (0, δ 0 ) such that 1/k < δ we then have (4.3) k u k L ( δ ) u L (). L ( δ ) Thus, for every ψ C 1 (), (4.4) ψ u k + ψ u k u L () ψ L 1 ( δ ). δ δ Note that for a.e. δ (0, δ 0 ) (4.5) δ u = 0; hence, for any such δ > 0, ψ u k δ ψ u δ as k. Indeed, this is a general fact (see, e.g., [5, Theorem 1, p.54]): if µ M() and µ ( δ ) = 0, then ψ (ρ k µ) ψ dµ ψ C 0 ( δ ). δ δ For any δ (0, δ 0 ) verifying (4.5), as k in (4.4) we get (4.6) ψ u + ψ u u L () ψ L 1 ( δ ) δ δ ψ C 1 (). From this estimate, one deduces that for every ψ C 1 (), ψ u u M() ψ L ( δ ) + u L () ψ L 1 ( δ ) δ As δ 0, we conclude that u X. ( u M() + u L () δ ) ψ L (). In order to prove that L ( ), we return to estimate (4.6). Given φ C 1 ( ), we fix an extension ψ C 1 () of φ; note that ψ L 1 ( δ ) φ L 1 ( ) + Cδ δ (0, δ 0 ), for some constant C > 0. Insert this test function ψ in (4.6). As δ 0 we obtain, by dominated convergence, ψ u + ψ u u L () φ L 1 ( ). Hence, φ u L () φ L1 ( ) Therefore, by duality L ( ) and (4.1) holds. φ C 1 ( ).

10 10 HAÏM BREZIS(1),(2) AND AUGUSTO C. PONCE (3) We now assume that u C 1 () X and we denote by h the normal derivative of u in the standard sense. By Lemma B.1 and Remark B.1, there exists a sequence (u k ) C () satisfying (B.2) (B.3) and such that In particular, Thus, (4.7) u k u k h Hence, the normal derivative in C 1 (). uniformly on. u ψ + ψ u = h ψ ψ C 1 (). in the sense of the space X coincides with h. When u X the measure need not be an L1 -function. Surprisingly, this is always true if u vanishes on : Proposition 4.2. Let u W 1,1 0 (). Then, u X if, and only if, u M() in the sense of distributions. Moreover, L1 ( ) and (4.8) u M(). L 1 ( ) Proof. We split the proof into two steps: Step 1. Proof of (4.8) if u is smooth in a neighborhood of. Under this assumption, is a smooth function on. Denote by v 1 and v 2 the solutions of { v1 = µ + in, { v2 = µ in, v 1 = 0 where µ = u. In particular, on, u = v 1 v 2 in. v 2 = 0 on, Since µ is smooth in a neighborhood of, µ + and µ are Lipschitz continuous near. Hence, v 1 and v 2 are of class C 2 near. Moreover, v 1 0 in and v 1 = 0 on ; thus, It follows that Similarly, Therefore, v 1 v 1 0 = v 1 + v 2 = Step 2. Proof of the proposition completed. on. v 1 = µ +. µ. v 2 = (µ + + µ ) = u.

11 KATO S INEQUALITY UP TO THE BOUNDARY 11 Let (ϕ k ) C 0 () be a sequence of test functions such that 0 ϕ k 1 in and ϕ k (x) = 1 if d(x, ) 1 k. Take µ k = ϕ k u, k 1. Then, (µ k ) M() is a sequence of measures such that supp µ k and, by dominated convergence, (4.9) µ k u strongly in M(). For each k 1, let u k be the unique solution of { uk = µ k in, u k = 0 on. Note that u k is harmonic in a neighborhood of. We claim that (4.10) φ k φ φ C1 ( ). Indeed, since u k u in L 1 () and ( u k ) is bounded in W 1,p 0 () for every 1 p < N N 1, (see [10]) we have (4.11) ψ u k ψ u ψ C 1 (). Assertion (4.10) then follows from (4.9) and (4.11). Applying Step 1 to the function u i u j, we have i j µ i µ j M() i, j 1. L1 ( ) In view of the strong convergence of (µ k ) in M(), ( k ) is a Cauchy sequence in L 1 ( ). Hence, this sequence converges in L 1 ( ) to some function h. By (4.10), h = ; hence, k in L1 ( ). Moreover, since (4.8) holds for every u k, it also holds for u. The proof is complete. We now show that if u W 1,1 () and u BV () then the normal derivative in the sense of the space X coincides with the function n u on defined in the sense of traces: Proposition 4.3. Assume that u W 1,1 () and u BV (); hence, u = div ( u) M(). Then, u X and coincides with n u on, where u is understood in the sense of traces. In particular, L1 ( ) and (4.12) C u BV (). L 1 ( ) In the proof of Proposition 4.3 we use the notion of strict convergence in BV (A), where A R N is a Lipschitz domain. We recall that a sequence (f n ) BV (A) converges strictly to f BV (A) if f n f strongly in L 1 (A) and Df n Df. A A

12 12 HAÏM BREZIS(1),(2) AND AUGUSTO C. PONCE (3) By [1, Theorem 3.88], the trace operator f BV (A) f A L 1 ( A) is continuous from BV (A) (under strict convergence) into L 1 ( A) (under strong convergence). Proof of Proposition 4.3. By Lemma B.1 and Remark B.1, there exists a sequence (u k ) C () satisfying (B.1) (B.3) and (B.12). Since ( u k ) converges strictly to u in BV ( ), we have (4.13) u k u in L 1 ( ). Hence, ( ) u ψ + ψ u = n u ψ ψ C 1 ( ). This implies that L1 ( ) and equals n u. By the BV -trace theory, (4.12) holds. 5. Proof of Theorem 1.2 We first establish Theorem 1.2 for functions in CD 2 (), where { } (5.1) CD() 2 = ζ C 2 (); ζ = 0 on. Lemma 5.1. Let u C 2 D (). Then, u+ M() and (5.2) u + M u L 1. Proof. Apply (3.3) with u + a, where a > 0. We deduce that [ (5.3) (u + a) + ] [u + a] X X = [u] X. Since (u + a) + = u + a in a neighborhood of, (5.4) Note that (u + a)+ = on. [ (u + a) + ] = X (u + a) + M() + (u + a)+ [u] X = u L1 () +. L1 ( ) By (5.3) (5.4) we then have (u + a) + M u L 1 a > 0. L 1 ( ) The result follows from the lower semicontinuity of the norm M with respect to the weak convergence as a 0. Proof of Theorem 1.2. Since u X, u M(). Take a sequence (µ k ) C () such that µ k u weak in M() and µ k L 1 µ M. For each k 1, let u k CD 2 () be the solution of u k = µ k in.,

13 KATO S INEQUALITY UP TO THE BOUNDARY 13 Then, by standard elliptic estimates, u k u in L 1 (). On the other hand, it follows from Lemma 5.1 that u + k Thus, As k we obtain u + k M u k L 1. M() and u + k ζ u k L 1 ζ L = µ k L 1 ζ L ζ C 2 D(). u + ζ u M ζ L ζ C 2 D(). This gives (1.5). From Proposition 4.2, we know that holds., + 6. Kato s inequality up to the boundary L1 ( ) and (1.6) Before proving Theorem 1.3, we first present some variants of Kato s inequality when u and are not necessarily L1 -functions but only finite measures. We prove for instance the following companion to [3, Proposition 4.B.5]: Proposition 6.1. Let u L 1 () be such that (6.1) u ζ hζ + gζ ζ CN(), 2 ζ 0 in for some g L 1 () and h L 1 ( ). Then, u W 1,1 () and (6.2) u + ψ hψ + gψ ψ C 1 (), ψ 0 in. Proof. By Proposition 2.2, u X. Moreover, u g in, (6.3) h on. We now split the proof into two steps: Step 1. Let Φ C 2 (R) be a nondecreasing convex function such that Φ L (R). Then, (6.4) Φ(u) ψ ψ Φ (u)h + ψ Φ (u)g for every ψ C 1 () such that ψ 0 in. Let (g k ) C () and (h k ) C ( ) be such that g k g in L 1 () and a.e. and h k h in L 1 ( ) and a.e. Next, take (µ k ) C () and (ν k ) C ( ) such that µ k u weak in M() and ν k weak in M( ).

14 14 HAÏM BREZIS(1),(2) AND AUGUSTO C. PONCE (3) In view of (6.3) and we may assume that = u, µ k g k in, ν k h k on and ν k = µ k k 1. Let u k C () be the unique solution of u k = µ k in, k = ν k on, such that u k = u. By Remark 2.1, the sequence (u k) is bounded in W 1,p () for every 1 p <. Passing to a subsequence if necessary, we have N N 1 Φ(u k ) Φ(u) weakly in L 1 (). Let ψ C 1 (), ψ 0 in. As in Lemma 3.1, for every k 1 we have Φ(u k ) ψ ψ Φ (u k ) k ψ Φ (u k ) u k ψ Φ (u k )h k + ψ Φ (u k )g k. By dominated convergence we obtain (6.4) as k. Step 2. Proof of the proposition completed. Apply (6.4) with Φ = Φ k, where (Φ k ) is a sequence of smooth convex functions such that Φ k (0) = 0, 0 Φ k 1 and { Φ 1 if t 0, k(t) 0 if t < 0. The result follows as we let k. The following variant of Proposition 6.1 will be needed below: Proposition 6.2. Let u L 1 () be such that (6.5) u ζ hζ + ζ dµ ζ CN(), 2 ζ 0 in for some µ M(), µ 0, and h L 1 ( ). Then, u W 1,1 () and (6.6) u + ψ hψ + ψ dµ ψ C 1 (), ψ 0 in. Proof. One can proceed as in the proof of Proposition 6.1. In Step 1, one should replace (6.4) by (6.4 ) Φ(u) ψ ψ Φ (u)h + Φ L ψ dµ. Inequality (6.4 ) is easily obtained by approximation, where the sequence (g k ) C () is chosen so that g k µ weak in M().

15 KATO S INEQUALITY UP TO THE BOUNDARY 15 The rest of the argument remains unchanged. We now prove the Proposition 6.3. Let u X. If L1 ( ), then + on [u > 0], (6.7) 0 on [u < 0], min {, 0} on [u = 0]. Proof. Denoting by µ = ( u) + and h =, we have u ζ hζ + ζ dµ ζ CN(), 2 ζ 0 in. Therefore, by Proposition 6.2, u + satisfies (6.8) u + ψ hψ + ψ dµ ψ C 1 (), ψ 0 in. By Theorem 1.1, we know that u + X. It thus follows that + (6.9) χ h = χ on. Given a > 0, we now apply (6.8) with u replaced by u a. As a 0, we obtain (6.10) u + ψ u + ψ hψ + ψ dµ ψ C 1 (), ψ 0 in. Hence, In particular, [u>0] + χ [u>0]h = χ [u>0] + (6.11) 0 on [u = 0]. Assertion (6.7) follows by combining (6.9) and (6.11). on. We state the following consequence of Proposition 6.3: Corollary 6.1. Let u X W 1,1 0 (). If u 0 in, then 0 on. Proof. Since u = u + in and u = 0 on, applying Proposition 6.3 above we get { } = + min, 0 0 on. We now present the Proof of Theorem 1.3. By Theorem 1.1, u + X. Applying Kato s inequality to u a, we have (6.12) (u a) + χ [u a] u in for every a R. As a 0 in (6.12) we get u + χ [u>0] u = G in.

16 16 HAÏM BREZIS(1),(2) AND AUGUSTO C. PONCE (3) By this estimate and (6.7), for every ψ C 1 () with ψ 0 in, u + ψ = ψ + ψ u + Hψ The proof is complete. Gψ. 7. Computing + for W 2,1 -functions Our goal in this section is to give a positive answer to Open Problems 1 and 2 under the additional assumption that u W 2,1 (): Theorem 7.1. If u W 2,1 (), then u + BV () (so that, u + X by Proposition 4.3) and + on [u > 0], (7.1) = 0 on [u < 0], min {, 0} on [u = 0]. We first prove the Lemma 7.1. If v W 1,1 () and v BV (), then v v(x) v ( x tn(x) ) (7.2) (x) = lim H N 1 -a.e. on. t 0 t In (7.2), we identify v with its precise representative, which is well-defined outside a set of zero H N 1 -Hausdorff measure; see [5, Section 4.8, Theorem 1 and Section 5.6, Theorem 3]. Proof. Since v W 1,1 (), for H N 1 -a.e. x the function t (0, δ) v ( x tn(x) ) is well-defined for some δ > 0 and belongs to W 1,1 (0, δ). Thus, (7.3) v ( x tn(x) ) v(x) t = n(x) 1 0 v ( x stn(x) ) ds. Moreover, since v BV (), for H N 1 -a.e. x the function r (0, δ) v ( x rn(x) ) belongs to BV (0, δ) L (0, δ) and (see [1, Theorem 3.108]) (7.4) lim r 0 v ( x rn(x) ) = v (x). We deduce from (7.3) (7.4) that ( ) v x tn(x) v(x) lim = n(x) v (x). t 0 t By Proposition 4.3 above, v = n v and the conclusion follows. We also need the following elementary lemma whose proof is left to the reader:

17 KATO S INEQUALITY UP TO THE BOUNDARY 17 Lemma 7.2. Let v : [0, δ] R be such that v(0) v(t) (7.5) lim = α R. t 0 t Then, v + (0) v + (t) (7.6) lim = t 0 t We now present the α if v(0) > 0, 0 if v(0) < 0, min {α, 0} if v(0) = 0. Proof of Theorem 7.1. We split the proof into three steps: Step 1. Proof of the assertion: u + BV (). (7.7) Extending u to R N, we may assume that u W 2,1 (R N ). We claim that 2 u + e 2 χ 2 u e 2 in D (R N ) for every e R N \ {0}. Indeed, let (Φ k ) be a sequence of smooth convex functions such that Φ k (0) = 0, Φ k L 1 and { (7.8) Φ 1 if t 0, k(t) 0 if t < 0. Then, 2 Φ k (u) e 2 ( ) 2 = Φ k(u) 2 u e 2 + Φ k(u) Φ e k(u) 2 u e 2 in R N. As k, we obtain (7.7). It follows from (7.7) that 2 u + e 2 M() for every e R N \ {0}. Applying the conclusion with e = e i, e j, e i + e j for every i, j {1,..., N} we deduce that D 2 u + is a finite measure in. Thus, u + BV (). Step 2. Proof of (7.1). By Lemma 7.1, for H N 1 -a.e. x, u satisfies ( ) u(x) u x tn(x) (7.9) lim = t 0 t (x). Hence, by (7.2) applied to u + and by (7.6) applied to v(t) = u(x tn(x)), + u + (x) u +( x tn(x) ) (x) if u(x) > 0, (x) = lim = 0 if u(x) < 0, t 0 t min { (x), 0} if u(x) = 0, for every x for which (7.9) holds. Since this is true H N 1 -a.e. on, (7.1) follows. The proof of Theorem 7.1 is complete.

18 18 HAÏM BREZIS(1),(2) AND AUGUSTO C. PONCE (3) Appendix A. The measure u + need not be finite In this appendix, we construct a harmonic function in dimension 2 such that u+ = : Proposition A.1. Let = { (x, y) R 2 ; x 2 + y 2 < 1 and x > 0 }. There exists a harmonic function u C() H 1 () with u W 1,1 ( ) such that (i) u X and u + X; (ii) u + 0 in the sense of distributions; (iii) u + is not a finite measure in. Proof. Let u be the function in given in polar coordinates by (A.1) u(r, θ) = r a k sin (a k θ) k=1 where (a k ) (0, 1) is a sequence such that ka k <. Since u(r, θ) k=1 sin (a k θ) π 2 k=1 a k, it follows that u C() and u is harmonic in (u is a series of harmonic functions). Note that u 2 = a j a k r aj+ak 2 cos ( (a j a k )θ ). Thus, u 2 π j,k=1 j,k=1 a j a k a j + a k 2π j,k=1 j k k=1 a j a k a j + a k 2π ka k < ; in other words, u H 1 (). Denoting by τ the tangential unit vector of u on, we have τ = 4 ( π ) sin a k 2π a k < ; 2 k=1 hence, u W 1,1 ( ). Since u is harmonic in, u + is subharmonic. Thus, u + 0 in. We show that u + is not a finite measure in. Note that u vanishes only on the x-axis. Denoting by dx (= dr) the 1-dimensional Lebesgue measure on the segment (0, 1) {0}, we then have u + = y (x, 0) dx = 1 r θ (r, 0) dr = a k r ak 1 dr. Therefore, u + = k=1 1 0 a k r a k 1 dr = k=1 k=1 k=1 1 =. k=1

19 KATO S INEQUALITY UP TO THE BOUNDARY 19 Hence, u + X and, by Theorem 1.1, this means that u X. Remark A.1. This example also shows that given ϕ W 1,1 ( ), it is in general not possible to construct a function v W 2,1 () such that v = ϕ. This is in contrast with the well-known result of Gagliardo [6] which asserts that the map w W 1,1 () w L 1 ( ) is surjective. Indeed, take ϕ = u, where u is given by (A.1). Suppose by contradiction that there exists some v W 2,1 () such that v = ϕ. Applying Proposition 4.2 to u v W 1,1 0 (), we would deduce that (u v) L1 ( ). But v W 2,1 () implies v L1 ( ) and therefore = v (u v) + L1 ( ), a contradiction. Appendix B. Approximation by smooth functions in In this appendix, we establish the following Lemma B.1. Given u X, there exists a sequence (u k ) C () such that (B.1) (B.2) and (B.3) u k u in W 1,1( ), ψ u k ψ u ψ C 1 () ψ k ψ ψ C1 (). Proof. We split the proof into two steps: Step 1. Given x 0, there exist δ > 0 and a sequence (v k ) C () such that (B.4) (B.5) v k u in W 1,1( B δ (x 0 ) ), ψ v k ψ u ψ C 1 () with supp ψ B δ (x 0 ). Since is smooth, there exist δ 1 > 0 and an open cone T R N (with vertex at 0 R N ) such that (B.6) (x + T ) B δ1 (x) x B δ1 (x 0 ). Let δ = δ 1 /2 and ρ C 0 (B δ ), ρ 0, be such that B δ ρ = 1 and (B.7) Set supp ρ T. ρ k (x) = k N ρ(kx) x R N. We show that the sequence (v k ) C () given by (B.8) v k (x) = ρ k (x y)u(y) dy x satisfies (B.4) (B.5). Note that given any x B δ (x 0 ), by (B.7) v k (x) depends only on the values of

20 20 HAÏM BREZIS(1),(2) AND AUGUSTO C. PONCE (3) u on a compact subset of (x + T ) B δ1 (x). In fact, from (B.6) (B.7) and a change of variable, we can rewrite (B.8) as (B.9) v k (x) = ρ k ( z)u(x + z) dz x B δ (x 0 ). Therefore, T B δ1 (0) (B.10) v k = ρ k ( u) and v k = ρ k ( u) in B δ (x 0 ). In particular, (B.4) (B.5) hold. Step 2. Proof of the proposition completed. By compactness of, we can cover this set with finitely many balls B δ (x 1 ),..., B δ (x t ) such that (B.4) (B.5) hold on each ball B δ (x i ) for some sequence (v i k ) C (). We now take (v 0 k ) C () and ω such that \ t i=1 B δ(x i ) ω, v 0 k u in W 1,1 (ω) and v 0 k u weak in M(ω) (such sequence can be obtained via convolution of u). Let (ϕ i ) be a partition of unity subordinated to the covering ω, B δ (x 1 ),..., B δ (x t ) of. One verifies that (B.1) (B.2) hold for the sequence (u k ) given by u k = t ϕ i vk. i i=0 Assertion (B.3) immediately follows from (B.1) (B.2). Remark B.1. An inspection of the proof of Lemma B.1 shows that (i) if u C 1 (), then (B.11) u k u in C 1 (); (ii) if u BV (), then (B.12) D 2 u k L 1 () D 2 u M(). Appendix C. Proof of Lemma 2.1 The proof of Lemma 2.1 we present below follows the lines of [8, Lemma 7.3] (see also [7, Theorem 8.15]) with some minor modifications. We first need the following variant of the Gagliardo-Nirenberg inequality: Proposition C.1. Let (C.1) A = Then, { v W 1,1 (); } [v = 0]. 3 (C.2) v L N N 1 C v L 1 v A. We denote by E the Lebesgue measure of a set E R N.

21 KATO S INEQUALITY UP TO THE BOUNDARY 21 Proof. By a variant of the Poincaré inequality (easily proved by contradiction), we have (C.3) v L 1 C v L 1 v A. On the other hand, by the standard Gagliardo-Nirenberg inequality and an extension argument, (C.4) v N L N 1 C ( ) v L 1 + v L 1 v W 1,1 (). Combining (C.3) (C.4), we obtain (C.2). Proof of Lemma 2.1. Replacing w by w a for some suitable constant a R if necessary, we may assume that (C.5) Given t > 0, let [w 0] 3 and [w 0] 3. (C.6) v t (x) = [ w(x) t ] + x. Using v t as a test function in (2.4), one shows that v t L 2 1 F L q [w > t] 2 1 q. On the other hand, by Hölder s inequality and Proposition C.1, 1 v t L 1 C v t L 2 [w > t] N. Thus, (C.7) v t L 1 C F L q [w > t] α t > 0, where α = N 1 q. Recall that (C.8) v t L 1 = 0 [vt > r] M dr = [w > s] ds, where M = w + L. Since α > 1, one deduces using (C.7) (C.8) that (C.9) w + L C F 1 α L q w α L 1. From (C.9) and w + L 1 w + L, we then have w + L C F L q. Replacing w by w, one obtains a similar estimate for w. Thus, t w L w + L + w L 2C F L q. Acknowledgment. The work of the first author (H. B.) is partially supported by NSF Grant DMS

22 22 HAÏM BREZIS(1),(2) AND AUGUSTO C. PONCE (3) References [1] L. Ambrosio, N. Fusco, and D. Pallara, Functions of bounded variation and free discontinuity problems. Oxford Mathematical Monographs, Oxford University Press, New York, [2] A. Ancona, Inégalité de Kato et inégalité de Kato jusqu au bord. C. R. Math. Acad. Sci. Paris 346 (2008), [3] H. Brezis, M. Marcus, and A. C. Ponce, Nonlinear elliptic equations with measures revisited. In: Mathematical Aspects of Nonlinear Dispersive Equations (J. Bourgain, C. Kenig, and S. Klainerman, eds.), Annals of Mathematics Studies, 163, Princeton University Press, Princeton, NJ, 2007, pp [4] H. Brezis and A. C. Ponce, Kato s inequality when u is a measure. C. R. Math. Acad. Sci. Paris, Ser. I 338 (2004), [5] L. C. Evans and R. F. Gariepy, Measure theory and fine properties of functions. Studies in Advanced Mathematics, CRC Press, Boca Raton, FL, [6] E. Gagliardo, Caratterizzazioni delle tracce sulla frontiera relative ad alcune classi di funzioni in n variabili. Rend. Sem. Mat. Univ. Padova 27 (1957), [7] D. Gilbarg and N. S. Trudinger, Elliptic partial differential equations of second order. Grundlehren der Mathematischen Wissenschaften, vol. 224, Springer-Verlag, Berlin, [8] P. Hartman and G. Stampacchia, On some non-linear elliptic differential-functional equations. Acta Math. 115 (1966), [9] T. Kato, Schrödinger operators with singular potentials. Israel J. Math. 13 (1972), (1973). Proceedings of the International Symposium on Partial Differential Equations and the Geometry of Normed Linear Spaces (Jerusalem, 1972). [10] G. Stampacchia, Équations elliptiques du second ordre à coefficients discontinus. Séminaire de Mathématiques Supérieures, no. 16 (été, 1965), Les Presses de l Université de Montréal, Montréal, Québec, (1) Rutgers University Department of Mathematics, Hill Center, Busch Campus 110 Frelinghuysen Rd. Piscataway, NJ 08854, USA (2) Technion Department of Mathematics Haifa, Israel (3) Université catholique de Louvain Département de mathématique Chemin du Cyclotron Louvain-la-Neuve, Belgium