REDUCED MEASURES FOR OBSTACLE ... - Semantic Scholar
Advances in Differential Equations
Volume 10, Number 11 (2005), 1201–1234
REDUCED MEASURES FOR OBSTACLE PROBLEMS Ha¨ım Brezis Laboratoire Jacques-Louis Lions, Universit´e Pierre et Marie Curie Boˆıte courier 187, 75252 Paris Cedex 05, France and Rutgers University, Dept. of Mathematics, Hill Center, Busch Campus 110 Frelinghuysen Rd., Piscataway, NJ 08854 Augusto C. Ponce Institute for Advanced Study, Princeton, NJ 08540 (Submitted by: Reza Aftabizadeh)
1. Introduction Let Ω ⊂ RN , N ≥ 1, be a smooth bounded domain. In this paper, we study the problem −Δu + β(u) μ in Ω, (1.1) u = 0 on ∂Ω, where μ is a finite measure in Ω and β is a maximal monotone graph (m.m.g.) such that (0, 0) ∈ graph β. Throughout most of the Introduction, we assume that and
dom β = (−∞, a] β(t) = {0}
for some 0 ≤ a < ∞
∀t ≤ 0.
(1.2) (1.3)
(However, the case where dom β = [−b, a], b ≥ 0, is also of interest and will be discussed at the end of the Introduction. ) A typical example of a m.m.g. β satisfying (1.2)–(1.3) is the following ⎧ ⎪ ⎨ g(t) if t < a, β(t) = g(a), ∞ if t = a, ⎪ ⎩ if t > a, ∅ where g : (−∞, a] → [0, ∞) is any continuous nondecreasing function such that g(t) = 0, for all t ≤ 0. Accepted for publication: June 2005. AMS Subject Classifications: 35J85, 35R05. 1201
1202
Ha¨ım Brezis and Augusto C. Ponce
Let M(Ω) be the space of finite measures μ on Ω such that μ(∂Ω) = 0; M(Ω) can be also identified with C0 (Ω)]∗ . Given μ ∈ M(Ω), we say that u is a solution of the obstacle problem (1.1) if the following holds: (i) u ∈ L1 (Ω), u ≤ a almost everywhere and Δu ∈ M(Ω); (ii) there exists ν ∈ M(Ω) such that ν ∈ β(u) (in a sense to be explained below) and − uΔζ + ζ dν = ζ dμ ∀ζ ∈ C02 (Ω). (1.4) Ω
Here,
Ω
Ω
C02 (Ω) = ζ ∈ C 2 (Ω) ; ζ = 0 on ∂Ω .
Recall that if u ∈ L1 (Ω) is such that Δu ∈ M(Ω), then u admits a representative which is quasicontinuous with respect to the Newtonian (H 1 ) capacity, denoted cap (see, e.g., [1]). We may thus assume that u is well defined q.e. (= quasi-everywhere = outside a set of zero capacity). In particular, u is ν-measurable for any finite measure ν which does not charge sets of zero capacity. Given any such measure ν, we then say that ν ∈ β(u)
(1.5)
if the following holds: • νa ∈ β(u) almost everywhere; • νs is concentrated on [u = a] and νs ≥ 0. Here, we denote by νa and νs the absolutely continuous and the singular parts of ν with respect to the Lebesgue measure in RN . In Section 2 below, we discuss other equivalent definitions of (1.5). If the measure μ is an L1 function, it is known (see Brezis-Strauss [11]) that (1.1) has a unique solution (and in this case ν ∈ L1 (Ω)). As we shall see below, problem (1.1) still admits a solution for every measure μ in L1 + H −1 (see Theorem 1 below). In this case, the measure ν need not belong to L1 (Ω) (it is easy to construct examples in dimension N = 1). We say that μ ∈ M(Ω) is diffuse if μ(E) = 0 for every Borel set E ⊂ Ω such that cap (E) = 0. Equivalently, a measure μ is diffuse if and only if μ ∈ L1 + H −1 (see Grun-Rehomme [19] and Boccardo-Gallou¨et-Orsina [5]). In dimension N = 1, every measure is diffuse (since cap (E) = 0 if and only if E = ∅). However, when N ≥ 2, there are measures which are not diffuse, for instance Dirac masses. In this paper, we follow the same program as in Brezis-Marcus-Ponce [8]. The main difference is that here dom β = R and dom β is closed (the case
Reduced measures for obstacle problems
1203
where dom β = R and dom β is an open set has been studied by DupaignePonce-Porretta [16]). Our main concern in [8] was twofold: (a) Identify the good measures for problem (1.1), i.e., those for which (1.1) has a solution. (b) If μ is not a good measure, define some kind of “generalized” solution, i.e., a common limit of all natural approximation schemes. Our first result gives the complete answer to question (a): Theorem 1. Let μ ∈ M(Ω). Then, (1.1) has a solution if and only if μ+ is diffuse. Moreover, the solution is unique. In other words, μ is a good measure for (1.1) if and only if μ+ is diffuse (equivalently, μ+ ∈ L1 + H −1 ). Recall that any measure μ can be uniquely decomposed as (see, e.g., [17]) μ = μd + μc , where μd is a diffuse measure and μc is a measure concentrated on some set of zero H 1 -capacity. Note in particular that μ is diffuse if and only if μc = 0. Given μ ∈ M(Ω), set μ∗ = μ − (μc )+ .
(1.6)
An easy consequence of Theorem 1 is the Corollary 1. For every μ ∈ M(Ω), μ∗ is the largest good measure ≤ μ. Indeed, let λ be a good measure ≤ μ. By Theorem 1, we know that λc ≤ 0. On the other hand, from λ ≤ μ we deduce that λd ≤ μd and −(λc )− ≤ −(μc )− . Thus, λ = λd + (λc )+ − (λc )− ≤ μd − (μc )− = μ∗ . Notation. Given μ, we denote by u∗ the unique solution of −Δu∗ + β(u∗ ) μ∗ in Ω, u∗ = 0
on ∂Ω.
(1.7)
We say that v is a subsolution of (1.1) if v ∈ L1 (Ω), v ≤ a almost everywhere, Δv ∈ M(Ω) and there exists diffuse ν ∈ M(Ω) such that ν ∈ β(v) and −
ζ dν ≤
uΔζ + Ω
Ω
Ω
ζ dμ ∀ζ ∈ C02 (Ω), ζ ≥ 0 in Ω.
(1.8)
1204
Ha¨ım Brezis and Augusto C. Ponce
Since ν ≥ 0, v is a subsolution of (1.1) if and only if one can find f ∈ L1 (Ω) such that f ∈ β(v) almost everywhere and fζ ≤ ζ dμ ∀ζ ∈ C02 (Ω), ζ ≥ 0 in Ω. (1.9) − vΔζ + Ω
Ω
Ω
A “companion” to Corollary 1 is the following Proposition 1. For any μ ∈ M(Ω), u∗ is the largest subsolution of (1.1). Several authors have studied the obstacle problem (1.1) associated with measures. Given a ≥ 0, let β = βa be the m.m.g. defined as ⎧ ⎪ if t < a, ⎨{0}
βa (t) = (1.10) 0, ∞ if t = a, ⎪ ⎩ ∅ if t > a. When a = 0 and Ω = RN , N ≥ 3, Theorem 1 has been known for a long time to experts from potential theory (see Baxter [3] and the references therein). In [3], the following statement appears: Let λ1 , λ2 ∈ M(Ω) be two nonnegative measures and assume that Pot λ1 ≡ (−Δ)−1 λ1 < +∞ λ1 -a.e.;
(1.11)
then, (1.1) has a solution for μ = λ1 − λ2 . Note that (1.11) is equivalent to the condition λ1 is diffuse (1.12) + (see, e.g., [13, Lemma 3.1]). Hence, (1.11) implies that μ (≤ λ1 ) is diffuse, so that μ satisfies the assumption of Theorem 1. When a > 0 and β is given by (1.10), Theorem 1 and Proposition 1 are due to Dall’Aglio-Leone [14] and Dall’Aglio-Dal Maso [13]. More precisely, given any measure μ, Dall’AglioLeone [14] have proved that there exists a largest element u ˆ in the class of functions w such that ⎧ ⎪ ⎨ −Δw ≤ μ in Ω, w ≤ a in Ω, ⎪ ⎩ w = 0 on ∂Ω. Subsequently, Dall’Aglio-Dal Maso [13] have proved that u ˆ satisfies (1.1) ∗ relative to the measure μ . In other words, they have shown that u ˆ = u∗ . We now return to a general m.m.g. β satisfying (1.2)–(1.3) and we investigate part (b) of the program. Let (βn ) be a sequence of continuous, nondecreasing functions βn : R → R, βn (0) = 0, such that βn → β
in the sense of graphs;
(1.13)
Reduced measures for obstacle problems
1205
more precisely, for every (t, s) ∈ G ≡ graph β, there exists a sequence tn → t such that βn (tn ) → s. If N ≥ 2, then we also assume that each βn has subcritical growth; i.e.,
βn (t) ≤ C |t|p + 1 ∀t ∈ R, (1.14) for some constant C > 0 and some p < n). We suppose in addition that
N N −2
(both possibly depending on
βn (t) = 0 ∀t ≤ 0.
(1.15)
Given any μ ∈ M(Ω), there exists a unique function un such that (see B´enilan-Brezis [4]) −Δun + βn (un ) = μ in Ω, (1.16) un = 0 on ∂Ω. The limiting behavior of the sequence (un ) is given by the following Theorem 2. For every μ ∈ M(Ω), we have un → u∗
in L1 (Ω),
where un is the solution of (1.16) and u∗ is the solution of (1.7). We emphasize that—as in [8]—the limit of (un ) is independent of the approximating sequence (βn ). Another approximation scheme consists of keeping β fixed, and approximating μ by convolution. More precisely, given a sequence of mollifiers (ρn ), let vn denote the solution of −Δvn + β(vn ) ρn ∗ μ in Ω, (1.17) on ∂Ω. vn = 0 Then, the sequence (vn ) converges to the same limit u∗ . More precisely, Theorem 3. For every μ ∈ M(Ω), we have vn → u∗
in L1 (Ω),
where vn is the solution of (1.17) and u∗ is the solution of (1.7). In Section 7, we consider similar questions for the case of two obstacles, i.e., dom β = [−b, a] for some b ≥ 0. We define the notion of solution of −Δu + β(u) μ in Ω, (1.18) u = 0 on ∂Ω, by analogy with the case studied above. Our main results assert that
1206
Ha¨ım Brezis and Augusto C. Ponce
• Problem (1.18) has a solution if and only if μ is diffuse and, in this case, this solution coincides with the unique minimizer of the variational problem 1 min |∇v|2 + j(v) − v dμ , 2 Ω v∈H01 (Ω) Ω Ω −b≤v≤a a.e.
where ∂j = β and j(0) = 0. • Given any measure μ, the solutions of the “natural” approximation problems (1.16) and (1.17) converge to the solution of (1.18) with data μd (= diffuse part of μ). This last result is related to a theorem of Orsina-Prignet [20]. This paper is organized as follows. In Section 2, we examine the concept of ν ∈ β(u) for any m.m.g. β and we discuss other equivalent definitions. In Section 3, we study some properties related to equidiffuse sequences (see Definition 1 below). In Section 4, we prove that the sequences given by (1.16) and (1.17) both converge q.e. The proofs of Theorems 1–3 and Proposition 1 are then presented in Sections 5 and 6. In Section 7, we discuss problem (1.1) in the case of two obstacles; the proofs of these results are given in Section 8. 2. Various definitions of ν ∈ β(u) Let β be a m.m.g. with closed domain, let u be a quasicontinuous function in Ω such that u ∈ dom β q.e., and let ν ∈ M(Ω). Set a = sup dom β and b = − inf dom β . (2.1) We say that ν ∈ β(u) if the following conditions are satisfied: (1) ν is diffuse;
(2.1a)
(2) νa ∈ β(u) almost everywhere;
(2.1b)
(3) νs is concentrated on the set [u = a] ∪ [u = −b];
(2.1c)
(4) νs ≥ 0 on [u = a] and νs ≤ 0 on [u = −b].
(2.1d)
We first observe that this definition agrees with the cases we have already considered in the Introduction, namely dom β = (−∞, a] and dom β = R.
Reduced measures for obstacle problems
1207
In fact, recall that for every quasicontinuous function u, the set [u = ±∞] has zero capacity. Thus, for every diffuse measure ν, we have νs = 0
on [u = ±∞].
In particular, if dom β = R, then ν ∈ β(u)
if and only if ν ∈ L1 (Ω)
and ν ∈ β(u) a.e.
We also observe that, since dom β is closed, then u ∈ dom β q.e.
if and only if u ∈ dom β a.e.
We present some equivalent forms of the assertion “ν ∈ β(u)” in the next Proposition 2. Let u be a quasicontinuous function in Ω such that u ∈ dom β almost everywhere. Assume that ν ∈ M(Ω) is diffuse and u ∈ L1 (Ω; ν). Then, the following assertions are equivalent: (a) ν ∈ β(u); ∗ (b) j(u) ∈ L1 (Ω) and j(t) − j(u) ≥ ν (t − u) in C0 (Ω) , ∀t ∈ R; ∗ ≥ u (s − ν) in C0 (Ω) , ∀s ∈ R; (c) j ∗ (ν) ∈ M(Ω) and j∗ (s) − j ∗ (ν) (d) j(u) ∈ L1 (Ω) and Ω j(v) − Ω j(u) ≥ Ω (v − u) dν for every v ∈ H01 (Ω) ∩ L∞ ; (e) j ∗ (ν) ∈ M(Ω) and Ω j ∗ (σ) − Ω j ∗ (ν) ≥ Ω u d(σ − ν) for every such that u ∈ L1 (Ω; σ); σ ∈ M(Ω) diffuse (f) Ω j(u) + Ω j ∗ (ν) = Ω u dν. Here, j : R → [0, +∞] is a convex function such that ∂j = β and j(0) = 0; j ∗ denotes the convex conjugate of j. For any ν ∈ M(Ω), the measure j ∗ (ν) is a nonnegative Borel measure defined as j ∗ (ν) = j ∗ (νa ) + a(νs )+ + b(νs )− ,
(2.2)
where a and b are given by (2.1) (see, e.g., [6, 18, 21, 22]). Proof. (a) ⇒ (b), (c), (f). By assumption, j(t) − j(u) ≥ νa (t − u)
a.e. ∀t ∈ R.
(2.3)
Recall that νa ∈ β(u) almost everywhere is equivalent to u ∈ β −1 (νa ) = ∂j ∗ (νa )
a.e.
Thus, j ∗ (s) − j ∗ (νa ) ≥ u (s − νa ) ∗
j(u) + j (νa ) = uνa
a.e.
a.e. ∀s ∈ R,
(2.4) (2.5)
1208
Ha¨ım Brezis and Augusto C. Ponce
On the other hand, νs is ≥ 0 on [u = a] and ≤ 0 on [u = −b]; thus, uνs = aνs = a(νs )+
in [u = a],
−
uνs = −bνs = b(νs )
in [u = −b].
Here we use the convention that ∞ · 0 = 0. Since νs is concentrated on the set [u = a] ∪ [u = −b], we deduce that uνs = a(νs )+ + b(νs )− ≥ 0
in Ω.
(2.6)
Thus, in view of (2.2), j ∗ (ν) = j ∗ (νa ) + uνs
in Ω.
(2.7)
By (2.5)–(2.6), we have j(u) ∈ L1 (Ω) and j ∗ (ν) ∈ M(Ω). Therefore, (b) follows from (2.3) and (2.6); (c) follows from (2.4) and (2.7); (f) follows from (2.5) and (2.7). 1 ∞ (a) ⇒ (d). Let v ∈ H0 (Ω) ∩ L . If v > a or v < −b on a set of positive measure, then Ω j(v) = +∞ and (d) holds. We now assume that −b ≤ v ≤ a almost everywhere; hence, this also holds q.e. By (2.3), we have j(v) − j(u) ≥ νa (v − u) Thus,
j(v) − Ω
a.e.
j(u) ≥
Ω
νa (v − u).
(2.8)
Ω
Since, by assumption, −b ≤ v ≤ a q.e., it follows from (2.6) that (v − u) νs = (v − a) (νs )+ − (v + b) (νs )− ≤ 0 Thus,
in Ω.
(v − u) dνs ≤ 0.
(2.9)
Ω
Combining (2.8) and (2.9), we obtain (d). (a) ⇒ (e). Let diffuse σ ∈ M(Ω) be such that u ∈ L1 (Ω; σ). By (2.4), we have j ∗ (σa ) − j ∗ (νa ) ≥ u(σa − νa ) a.e. (2.10) On the other hand, −b ≤ u ≤ a q.e. implies j ∗ (σ) = j ∗ (σa ) + a(σs )+ + b(σs )− ≥ j ∗ (σa ) + uσs .
(2.11)
Combining (2.7), (2.10) and (2.11), we deduce (e). (b) ⇒ (a). By assumption, j(t) − j(u) ≥ ν (t − u)
in Ω,
∀t ∈ R.
(2.12)
Reduced measures for obstacle problems
1209
Taking the absolutely continuous parts of (2.12) with respect to Lebesgue measure, we obtain (2.3). Thus, νa ∈ β(u) almost everywhere. Given t ∈ (−b, a), it follows from the singular part of (2.12) that νs (t − u) ≤ 0
in Ω.
In other words, νs ≤ 0
in [u < t],
νs ≥ 0
in [u > t].
Since t ∈ (−b, a) was arbitrary, we deduce that νs = 0
in [−b < u < a],
νs ≤ 0
in [u = −b],
νs ≥ 0
in [u = a].
This establishes (a). (c) ⇒ (a). Comparing the absolutely continuous parts of both sides of (c), we obtain (2.4). Thus, (2.3) holds and so νa ∈ β(u) almost everywhere. From the singular parts of (c), we get a(νs )+ + b(νs )− ≤ uνs
in Ω.
Since −b ≤ u ≤ a q.e., we conclude that a(νs )+ + b(νs )− = uνs
in Ω.
(2.13)
Thus, νs is concentrated on [u = a] ∪ [u = −b], νs ≥ 0 in [u = a], and νs ≤ 0 in [u = −b]. We conclude that (a) holds. (f) ⇒ (a). We first observe that, by (f), we have j(u) ∈ L1 (Ω) and ∗ j(u) + j ∗ (ν) = uν in C0 (Ω) . (2.14) In fact, note that we always have j(u) + j ∗ (νa ) ≥ uνa
a.e.
and
a(νs )+ + b(νs )− ≥ uνs . Combining these two inequalities, we get ∗ j(u) + j ∗ (ν) ≥ uν in C0 (Ω) .
Using (f), (2.14) follows. From the absolutely continuous part of (2.14), we deduce that (2.5) holds, and so νa ∈ β(u) almost everywhere. Moreover, from the singular part of (2.14), we obtain (2.13). Therefore, (a) is satisfied.
1210
Ha¨ım Brezis and Augusto C. Ponce
(d) ⇒ (b). Since the right-hand side of (d) is finite and j ≥ 0, we have j(u) ∈ L1 (Ω). Let t ∈ (−b, a). We claim that j(t) − j(u) ≥ (t − u) dν ∀A ⊂ Ω Borel. (2.15) A
A
A
By outer regularity of Radon measures, it suffices to establish (2.15) when A is open and A ⊂⊂ Ω. Given an open subset ω ⊂⊂ A and k ≥ |t|, let v ∈ H01 (Ω), |v| ≤ k in Ω, be such that v = t in ω and v = Tk (u) in Ω\A, where Tk denotes the truncation at ±k. Note that j(v) ≤ j(u) in Ω\A. By (d), we then have j(v) − j(u) ≥ (v − u) dν. A
Ω
A
As ω ↑ A, we get j(t) − j(u) ≥ (t − u) dν + A
A
Tk (u) − u dν.
(2.16)
Ω\A
A
Since u ∈ L1 (Ω; ν), by monotone convergence we have
Tk (u) − u dν → 0 as k → ∞. Ω\A
Thus, as k → ∞ in (2.16), we get (2.15). Assertion (b) immediately follows from (2.15). (e) ⇒ (c). Given s ∈ R and a Borel set A ⊂ Ω, let σ = sLN A + νΩ\A , where LN denotes the Lebesgue measure in RN . Clearly, σ is a diffuse measure in Ω and u ∈ L1 (Ω; σ). By (e), we have ∗ ∗ j (s) − j (ν) ≥ u (s dLN − dν). A
A
A
Since A ⊂ Ω is arbitrary, we conclude that (c) holds. The proof of the proposition is complete. Remark 1. Assume ν ∈ β(u). The assumption that u ∈ L1 (Ω; ν) automatically holds in the following cases: • dom β = (−∞, a] and β(t) = {0}, ∀t ≤ 0; • dom β = [−b, a]. Note, however, that if ν ∈ β(u) and dom β = R, then it need not be true that u ∈ L1 (Ω; ν) (even if β is a continuous function); see however Lemma 2 below.
Reduced measures for obstacle problems
1211
As we mentioned in the Introduction, we are interested in the approximation of problem (1.1) with a sequence (βn ). Given a sequence of m.m.g. (βn ), then βn → β in the sense of graphs if, for every (t, s) ∈ graph β, there exists (tn , sn ) ∈ graph βn such that tn → t and sn → s. Since the notion of ν ∈ β(u) can be stated in terms of j, where ∂j = β, it is useful to have a characterization of the convergence of (βn ) in terms of the convergence of the primitives (jn ). This is given by the next Proposition 3. Let (βn ) be a sequence of m.m.g. Let jn : R → [0, +∞] be such that jn (0) = 0 and ∂jn = βn . Then, the following statements are equivalent: (a) βn → β in the sense of graphs; (b) for every t ∈ R, if tn → t, then j(t) ≤ lim inf jn (tn ); for every t ∈ R, n→∞
there exists tn → t such that j(t) = lim jn (tn ). n→∞
We refer the reader to Attouch [2, Theorem 3.66] for the proof of Proposition 3. In the literature, assertion (b) is called the Mosco-convergence of jn to j. We conclude this section with the standard Proposition 4. Let (βn ) be a sequence of m.m.g. such that βn → β in the sense of graphs. Let (tn , sn ) ∈ graph βn . If tnk → t and snk → s, then (t, s) ∈ graph β. Proof. Given (x, y) ∈ graph β, let (xn , yn ) ∈ graph βn be such that xn → x and yn → y. From the monotonicity of βnk we have (snk − ynk )(tnk − xnk ) ≥ 0. Thus, as k → ∞, we get (s − y)(t − x) ≥ 0
∀(x, y) ∈ graph β.
From the maximality of β, we conclude that (t, s) ∈ graph β. 3. Equidiffuse sequences of measures We begin this section with the following Definition 1. Let (λn ) ⊂ M(Ω). Given ε > 0, we say that (λn ) is εequidiffuse if (i) (λn ) is bounded in M(Ω);
1212
Ha¨ım Brezis and Augusto C. Ponce
(ii) there exists δ > 0 such that cap (A) < δ
=⇒
|λn |(A) < ε
∀n ≥ 1,
for every Borel set A ⊂ Ω. The sequence (λn ) is equidiffuse if it is ε-equidiffuse for every ε > 0. Clearly, if (λn ) is equidiffuse, then each measure λn is diffuse. We also observe that any finite set of diffuse measures is equidiffuse. Here is another example. Let μ ∈ M(Ω) and let (ρn ) be a sequence of mollifiers. If μc M < ε, then (ρn ∗μ) is ε-equidiffuse. It is then easy to see that (ρn ∗μ) is equidiffuse if and only if μ is diffuse. The main result of this section is the Theorem 4. Let (wn ) be a sequence of quasicontinuous functions in Ω. Let (βn ) be a sequence of m.m.g. Let (νn ) be a sequence of diffuse measures with νn ∈ βn (wn ) in the sense of (2.1a)–(2.1d). Assume that (i) wn → w q.e.; (ii) βn → β in the sense of graphs; ∗ (iii) νn ν weak∗ in M(Ω); (iv) for every ε > 0, there exists a compact set Kε ⊂ Ω such that cap (Kε ) < ε and (νn ) is ε-equidiffuse in Ω \ Kε . Then, w ∈ dom β a.e. and νd ∈ β(w). (3.1) Remark 2. Assumption (iv) is clearly weaker then the assumption “(νn ) is equidiffuse”; moreover, the measure ν need not be diffuse. For example, for any measure μ (not necessarily diffuse), the sequence (ρn ∗ μ) always satisfies (iv). Before turning to the proof of Theorem 4, we present two lemmas. The first one is reminiscent of part of the Dunford-Pettis theorem: Lemma 1. Let (λn ) ⊂ M(Ω) be a sequence of diffuse measures such that ∗
λn λ
weak∗ in M(Ω).
Let (wn ) be a bounded sequence of quasicontinuous functions in Ω such that wn → w q.e. as n → ∞. If (λn ) is ε-equidiffuse, then lim sup wn ζ dλn − wζ dλd ≤ CεζL∞ ∀ζ ∈ C0 (Ω), (3.2) n→∞
Ω
Ω
for some constant C > 0 depending on sup λn M and sup wn L∞ .
Reduced measures for obstacle problems
1213
Notice that since (wn ) is quasicontinuous in Ω and wn → w q.e., then w is also quasicontinuous. Proof. Let δ > 0 be such that cap (A) < δ implies |λn |(A) < ε, for all n ≥ 1, and |λ|(A) ≤ ε. In particular, λc M ≤ ε.
(3.3)
Let ζ ∈ C0 (Ω). Since wζ is quasicontinuous on Ω and ζ = 0 on ∂Ω, there exists ψ ∈ C0 (Ω) such that
cap |ψ − wζ| > ε < δ. (3.4) Thus,
Ω
|ψ − wζ| |λd | ≤ ελM +
|ψ − wζ| |λd | ≤ Cε.
(3.5)
[|ψ−wζ|>ε]
Moreover, since wn → w q.e., by (3.4) we have
< δ. lim sup cap |ψ − wn ζ| > ε n→∞
Proceeding as above, we get
|ψ − wn ζ| |λn | ≤ Cε.
lim sup n→∞
(3.6)
Ω ∗
Finally, since ψ ∈ C0 (Ω) and λn λ in M(Ω), ψ dλn = ψ dλ. lim n→∞ Ω
(3.7)
Ω
Combining (3.3), (3.5)–(3.7), we get wζ dλd ≤ 2CεζL∞ . lim sup wn ζ dλn − n→∞ Ω
Ω
This establishes Lemma 1. Our next lemma is a variant of Proposition 2, where the condition u ∈ L1 (Ω; ν) is not assumed: Lemma 2. Let u be a quasicontinuous function in Ω such that u ∈ dom β almost everywhere, and assume that ν ∈ M(Ω) is diffuse. Then, ν ∈ β(u) if and only if ∗ j(t) − j(u) − ν (t − u) Φk (u) ≥ 0 in C0 (Ω) , ∀t ∈ R, ∀k > 0, (3.8) where Φk ∈ Cc∞ (R) is such that Φk ≥ 0 in R and Φk (t) = 1 if |t| ≤ k.
1214
Ha¨ım Brezis and Augusto C. Ponce
Proof. The implication “⇒” is established as in the proof of Proposition 2. We now prove the converse statement. For simplicity, we only consider the case dom β = R. By (3.8), we have j(t) − j(u) ≥ ν (t − u) in |u| ≤ k , ∀t ∈ R, ∀k > 0. Thus, j(t) − j(u) ≥ νa (t − u)
and νs (t − u) ≤ 0 ∀t ∈ R.
(3.9)
We conclude that νa ∈ β(u) almost everywhere and νs = 0; in particular, ν = νa ∈ L1 (Ω). Therefore, ν ∈ β(u). We now present the Proof of Theorem 4. We split the proof into two steps: Step 1. w ∈ dom β almost everywhere. It suffices to consider the case where dom β = R. Assume for simplicity that dom β = (−∞, a] (the other cases are similar). Given t0 > a, by Proposition 4 there exists a sequence rn → +∞ such that if (t0 , s) ∈ graph βn , then s ≥ rn . Recall that (νn ) is a bounded sequence in M(Ω). Thus, in view of the assumption (νn )a ∈ βn (wn ) almost everywhere, we have 1 C [wn ≥ t0 ] ≤ 1 (νn )a ≤ |νn | ≤ . rn Ω rn Ω rn We conclude that [w > t0 ] ≤ lim inf [wn ≥ t0 ] = 0 ∀t0 > a. n→∞
This implies
[w > a] = 0.
In other words, w ≤ a almost everywhere and so w ∈ dom β almost everywhere. Step 2. νd ∈ β(w). Given ε > 0, let Kε be the compact set given by (iv). Let Φk be as in the statement of Lemma 2. Since wn Φk (wn ) → wΦk (w)
q.e.,
we can apply Lemma 1 to get wΦk (w)ζ dνd ≤ Ck ε ζL∞ lim sup wn Φk (wn )ζ dνn − n→∞
Ω
Ω
(3.10)
Reduced measures for obstacle problems
1215
for every ζ ∈ C0 (Ω\Kε ). Let t ∈ R be such that j(t) < ∞. Since βn → β in the sense of graphs, by Proposition 3 there exists a sequence tn → t such that jn (tn ) → j(t) as n → ∞. (3.11) Moreover, by Fatou, j(w)ζ ≤ lim inf jn (wn )ζ ∀ζ ∈ C0 (Ω), ζ ≥ 0 in Ω. (3.12) Ω
n→∞
Ω
On the other hand, by Lemma 2 we have jn (tn ) − jn (wn ) − νn (tn − wn ) Φk (wn ) ≥ 0 Combining (3.10)–(3.13), we conclude that j(t)−j(w)−νd (t−w) Φk (w)ζ ≥ −Ck ε ζL∞
∗ in C0 (Ω) .
(3.13)
∀ζ ∈ C0 (Ω\Kε ), ζ ≥ 0.
Ω
As ε → 0, we obtain j(t) − j(w) − νd (t − w) Φk (w) ≥ 0
∗ in C0 (Ω) .
By Lemma 2, this implies νd ∈ β(w). 4. A basic tool concerning q.e.-convergence Given μn ∈ M(Ω), let un be such that −Δun = μn in Ω, on ∂Ω. un = 0
(4.1)
In this section, we study the pointwise convergence of (some subsequence of) (un ) as n → ∞. Assume that Δun M ≤ C.
(4.2)
Using standard linear regularity theory, one can show that there exists a subsequence (uni ) such that uni → u strongly in W 1,p , for p < 2, and thus uni → u capW 1,p -q.e.
∀p < 2.
(4.3)
On the other hand, given a measure ν, diffuse with respect to the H 1 capacity, we can also extract a subsequence (unj ) (depending on ν) such that unj → u ν-a.e. (4.4) A natural question is whether unk → u q.e. (i.e., capW 1,2 -q.e.), for some subsequence (unk ).
(4.5)
1216
Ha¨ım Brezis and Augusto C. Ponce
It turns out that the answer is negative, even if the sequence (μn ) is equidiffuse (see Appendix below). Note however that (4.5) is true for some special sequences (μn ), e.g., μn = ρn ∗ μ for some fixed measure μ. A basic tool used in the proof of our main results is the following Theorem 5. Given μ ∈ M(Ω), let un ∈ L1 (Ω) be the solution of −Δun + βn (un ) ρn ∗ μ in Ω, on ∂Ω, un = 0 where (i) (βn ) is a sequence of m.m.g. such that (0, 0) ∈ graph βn ; (ii) (ρn ) is a sequence of mollifiers. If un → u in L1 (Ω), then u is quasicontinuous and cap |un − u| > δ → 0 as n → ∞, ∀δ > 0.
(4.6)
(4.7)
In particular, there exists a subsequence (unk ) such that unk → u q.e.
(4.8)
Warning. The reader might think that the conclusion (4.8) in Theorem 5 holds if (ρn ∗ μ) is replaced by an equidiffuse sequence (μn ) in (4.6). As we have already mentioned, the answer is negative even in the simple case βn ≡ 0 (see Appendix). In order to prove Theorem 5 we need the following Lemma 3. Assume un ∈ L1 (Ω) satisfies −Δun + βn (un ) μn un = 0
in Ω, on ∂Ω,
(4.9)
where (i) (βn ) is a sequence of m.m.g. such that (0, 0) ∈ graph βn ; (ii) (μn ) is a bounded sequence in M(Ω). Suppose (μn ) is ε-equidiffuse in Ω\F for some compact set F ⊂ Ω and some ε > 0. Then, for every open set ω ⊃ F , there exists n0 ≥ 1 such that
μn + Δun n≥n0 is 2ε-equidiffuse in Ω \ ω. (4.10) Proof. Let νn = μn + Δun . By assumption, νn is diffuse and νn ∈ βn (un ). Thus,
sign un νn = |νn |.
Reduced measures for obstacle problems
1217
By Kato’s inequality (see [9]), we have
Δ|un | d ≥ sign un (Δun )d = sign un νn − (μn )d ≥ |νn | − |μn |d and
Δ|un | c = −|Δun |c = −|μn |c .
Thus, −Δ|un | + |νn | ≤ |μn | in Ω. Since |un | ∈ have
W01,1 (Ω),
then for every superharmonic function ζ ∈ C02 (Ω) we ζ |νn | ≤ ζ |μn |. (4.11) Ω
Ω
Given K ⊂ Ω, let ζK denote the capacitary potential of K. By density, (4.11) also holds with ζ = ζK and then we have ζK |νn | ≤ ζK |μn |. (4.12) |νn |(K) ≤ Ω
Ω
Assume by contradiction that (νn ) is not 2ε-equidiffuse in Ω \ ω for n sufficiently large. By inner regularity, there exists (νnj ) and a sequence of compact sets (Kj ) in Ω \ ω such that 3ε (4.13) and cap (Kj ) → 0. 2 Thus, ζKj → 0 in H01 (Ω) and uniformly on F . Passing to a subsequence, we may assume that ζKj → 0 q.e. By Lemma 1, we then have lim sup ζKn |μn | ≤ ε. (4.14) |νnj |(Kj ) ≥
n→∞
Ω
Combining (4.12)–(4.14), we get a contradiction. Therefore, there exists n0 ≥ 1 such that (4.10) holds. The proof of the lemma is complete. Proof of Theorem 5. By standard estimates, we have Δun M ≤ 2ρn ∗ μM ≤ 2μM . Thus, Δu ∈ M(Ω); hence u is quasicontinuous. Moreover, note that (4.8) follows from (4.7) by using a standard diagonalization argument. We now split the proof of (4.7) into two steps: Step 1. Proof of (4.7) when μc is concentrated on a compact set K ⊂ Ω of zero capacity. Let fn = (ρn ∗ μ) + Δun , so that fn ∈ L1 (Ω) and fn ∈ βn (un ) almost everywhere, for all n ≥ 1. We first establish the following
1218
Ha¨ım Brezis and Augusto C. Ponce
Claim 1. For every δ > 0 and for every open set ω ⊃ K, we have |fn | → 0 as n → ∞,
(4.15)
(An ∪Bn )\ω
where
An = 0 ≤ un ≤ u − δ and Bn = u + δ ≤ un ≤ 0 .
(4.16)
We show (4.15) for |fn | integrated over An \ ω; the term coming from Bn \ ω can be estimated in a similar way. We consider two cases, depending on whether u+ is bounded or not: Case 1. u+ ∈ L∞ (Ω). Let M = u+ L∞ . We claim that there exists Cδ > 0 such that if (tn , sn ) ∈ graph βn and tn ≤ M − δ, then sn ≤ Cδ . Assume by contradiction that there exists a sequence (tnk , snk ) ∈ graph βnk such that tnk ≤ M − δ
and snk → ∞.
Since fn L1 ≤ μM and fn ∈ β(un ) almost everywhere, we have [un > M − δ] sn ≤ fnk ≤ C. k k [unk ≥tnk ]
Thus,
[u > M − δ] ≤ lim inf [un > M − δ] = 0. k k→∞
We then deduce that u ≤ M − δ almost everywhere, a contradiction. Therefore, we have 0 ≤ fn ≤ Cδ on An \ ω. Since χAn → 0 in L1 (Ω), we conclude that 0≤ |fn | ≤ Cδ |An | → 0. An \ω
Case 2. u+ ∈ L∞ (Ω). Recall the standard estimate ∇Tk (u)2 ≤ kΔuM
∀k > 0.
Ω
Thus,
1 cap [u > k] ≤ 2 k
Ω
∇Tk (u)2 → 0
as k → ∞.
Reduced measures for obstacle problems
Given η > 0, let k0 ≥ 1 be such that
cap [u > k] < η
1219
∀k ≥ k0 .
Clearly, (ρn ∗ μ) is equidiffuse in Ω \ ω 0 for every open set ω0 ⊃ K. Thus, by Lemma 3, the sequence (fn ) is equidiffuse in Ω \ ω. Therefore, given ε > 0, one can take k0 ≥ 1 large enough such that |fn | < ε ∀k ≥ k0 . (4.17) [u>k]\ω
On the other hand, as in the proof of Case 1 above, there exists Ck0 > 0 such that if (tn , sn ) ∈ graph βn and tn ≤ k0 , then sn ≤ Ck0 . From this and (4.17), we then have |fn | ≤ |fn | + |fn | ≤ Ck0 |An | + ε. An \ω
An ∩[u≤k0 ]
[u>k0 ]\ω
Thus,
lim sup n→∞
|fn | ≤ ε.
An \ω
Since ε > 0 was arbitrary, we conclude that |fn | = 0. lim n→∞ An \ω
The same argument shows that the conclusion holds when integrating |fn | on Bn \ ω. The proof of (4.15) is complete. Claim 2. For every δ > 0 and for every open set ω ⊃ K, we have ∇(un − u)2 → 0 as n → ∞, (4.18) Zn \ω
where Zn = δ < |un − u| < 2δ . Let hδ : R → R be the function given by ⎧ ⎪ 0 if − δ ≤ t ≤ δ, ⎪ ⎪ ⎪ ⎪ ⎪ if t ≥ 2δ, ⎨1 hδ (t) = −1 if t ≤ −2δ, ⎪ ⎪ 1 ⎪ ⎪ δ (t − δ) if δ < t < 2δ, ⎪ ⎪ ⎩ 1 (t + δ) if − 2δ < t < −δ. δ
1220
Ha¨ım Brezis and Augusto C. Ponce
Note that given ψ ∈ C ∞ (Ω), ψ ≥ 0 in Ω, and v ∈ W01,1 (Ω) such that Δv ∈ M(Ω), we have ∇hδ (v)2 ψ ≤ − (Δv)d hδ (v)ψ+ |Δv|c ψ− ∇v·∇ψ hδ (v). (4.19) δ Ω
Ω
Ω
Ω
(This estimate clearly holds if v is smooth; the case of a general v ∈ W01,1 then follows by approximation.) Given open sets ω1 ⊃⊃ ω0 ⊃ K, we take ψ ∈ C ∞ (Ω) such that ψ = 1 on Ω \ ω1 and ψ = 0 on ω 0 . It easily follows from Lemma 3 that (Δu)c is supported in K. Thus, Δ(un − u) ψ = |Δu|c ψ = 0. (4.20) c
We now apply (4.19) to v = un − u. By (4.20), we get ∇hδ (v)2 ψ δ Ω ∇(un − u), (4.21) λn hδ (un − u) + C ≤ − fn hδ (un − u) ψ + Ω Ω Ω where λn = ρn ∗ μ − (Δu)d ψ. The last integral converges to 0 as n → ∞ because un → u in W01,1 (Ω). Since hδ (un − u) 0
weakly in H01 (Ω)
as n → ∞,
and λn = 0 on ω0 , it is not difficult to see that λn hδ (un − u) = 0. lim
(4.22)
n→∞ Ω
(To prove (4.22) one can proceed as in the proof of Proposition 2.1 in [9].) We now show that fn hδ (un − u) ψ ≥ 0. (4.23) lim inf n→∞
Ω
In fact, note that the integrand is ≥ 0 on Ω\(An ∪ Bn ), where An , Bn are given by (4.16). Thus, fn hδ (un − u) ψ ≥ fn hδ (un − u) ψ ≥ − |fn |. Ω
An ∪Bn
(An ∪Bn )\ω 0
By Claim 1 and this estimate, we conclude that (4.23) holds. Combining (4.21)–(4.23), we obtain (4.18) with ω = ω1 . We now conclude the proof of (4.7). Given ε > 0, we fix an open set ω0 ⊃ K satisfying cap (ω 0 ) < ε. Let ψ ∈ C ∞ (Ω) be such that ψ = 1 in
Reduced measures for obstacle problems
1221
Ω \ ω0 and ψ = 0 in some neighborhood of K. Recall that, by standard estimates (see, e.g., [23]), we have in W01,p (Ω) ∀p
2δ . Since hδ (un − u)ψ ≥ 1
(4.24)
(4.25)
on Wn \ ω 0 ,
2 cap (Wn \ ω 0 ) ≤ ∇ hδ (un − u) ψ .
we have
Ω
On the other hand, by subadditivity of cap, cap (Wn ) ≤ cap (Wn \ ω 0 ) + cap (ω 0 ) ≤ cap (Wn \ ω 0 ) + ε. Therefore,
lim sup cap (Wn ) ≤ ε. n→∞
Since ε > 0 was arbitrary, (4.7) follows (with δ replaced by 2δ). Step 2. Proof of the theorem completed. Clearly, it suffices to establish (4.7) for a subsequence (unj ). Let (μk ) be a sequence in M(Ω) so that μk → μ strongly in M(Ω) and so that each measure (μk )c is supported in some compact set of zero capacity. Let vn,k ∈ L1 (Ω) be the solution of −Δvn,k + βn (vn,k ) ρn ∗ μk in Ω, on ∂Ω. vn,k = 0 Clearly, we have Δvn,k M ≤ C. Thus, there exist an increasing sequence of integers (nj ) and a sequence of functions (vk ) in L1 (Ω) such that vnj ,k → vk
in L1 (Ω) as j → ∞,
∀k ≥ 1.
By standard estimates, we have Δ(vn ,k − un ) ≤ 2 ρn ∗ |μk − μ| ≤ 2μk − μM . j j j Ω
Thus,
Ω
Ω
∇Tδ (vn
j ,k
2 − unj ) ≤ 2δμk − μM .
(4.26)
1222
Ha¨ım Brezis and Augusto C. Ponce
As j → ∞, we also have ∇Tδ (vk − u)2 ≤ 2δμk − μM .
(4.27)
Ω
Combining (4.26) and (4.27), we get 4 cap |vnj ,k − unj | > δ + cap |vk − u| > δ ≤ μk − μM . δ Thus, 4 cap |unj − u| > 3δ ≤ cap |vnj ,k − vk | > δ + μk − μM . δ For k ≥ 1 fixed, we apply the previous step to (vnj ,k ). We deduce that 4 ≤ μk − μM . lim sup cap |unj − u| > 3δ δ j→∞ As k → ∞, we obtain (4.7) for the subsequence (unj ). The proof of the theorem is complete. Remark 3. An easy inspection of the proof of Theorem 5 shows that (4.7)– (4.8) are still valid if the sequence (ρn ∗ μ) is replaced be any sequence of measures (μn ) of the form μn = fn + Tn + ρn ∗ ν, where fn ∈
L1 (Ω),
Tn ∈
H −1 (Ω)
satisfy
fn → f in L1 (Ω) and Tn → T in H −1 (Ω), and ν is a measure concentrated on a set of zero H 1 -capacity. We will also need the following variant of Theorem 5: Theorem 6. Given μ ∈ M(Ω), let un ∈ L1 (Ω) be the solution of −Δun + βn (un ) = μ in Ω, un = 0 on ∂Ω,
(4.28)
where βn : R → R is a continuous, nondecreasing function with subcritical growth and such that βn (0) = 0. If un → u in L1 (Ω), then u is quasicontinuous and (4.29) cap |un − u| > δ → 0 as n → ∞, ∀δ > 0. In particular, there exists a subsequence (unk ) such that unk → u q.e.
(4.30)
Reduced measures for obstacle problems
1223
Proof. Let (ρk ) be a sequence of mollifiers. Let (un,k ) be the solution of −Δun,k + βn (un,k ) = ρk ∗ μ in Ω, on ∂Ω. un,k = 0 Since βn has subcritical growth, un,k → un in L1 (Ω) as k → ∞. By the previous theorem, we have cap |un,k − un | > δ → 0 as k → ∞. Thus, one can find an increasing sequence of integers (kn ) such that 1 cap |un,kn − un | > δ ≤ n and un,kn → u in L1 (Ω). 2 Therefore, 1 cap |un − u| > 2δ ≤ cap |un,kn − u| > δ + n . 2 Applying Theorem 5 to the sequence (un,kn ), the result follows. 5. Proof of Theorem 1 We first establish the following Lemma 4. Given any μ ∈ M(Ω), problem (1.1) has at most one solution. Proof. Assume u1 , u2 both satisfy (1.1). We claim that Δ(u1 − u2 )+ ≥ 0
in D (Ω).
(5.1)
In fact, let νi = μ + Δui , for i = 1, 2. Recall that each νi is diffuse. Thus, Δ(u1 − u2 ) is also diffuse. Applying Kato’s inequality (see [9, Theorem 1.1 and Remark 1]), we deduce that Δ(u1 − u2 )+ is a diffuse measure and Δ(u1 − u2 )+ ≥ χ[u1 >u2 ] Δ(u1 − u2 ) = χ[u1 >u2 ] (ν1 − ν2 ) in Ω.
(5.2)
We now observe that χ[u1 >u2 ] (ν1 − ν2 ) ≥ 0
in Ω.
(5.3)
Indeed, recall that (νi )a ∈ β(ui )
a.e.
Thus, χ[u1 >u2 ] (ν1 − ν2 )a ≥ 0
a.e.
(5.4)
1224
Ha¨ım Brezis and Augusto C. Ponce
On the other hand, since (ν2 )s is concentrated on the set [u2 = a] and [u1 > u2 ] ∩ [u2 = a] = ∅, we have (ν2 )s = 0
on [u1 > u2 ].
Thus, (5.5) χ[u1 >u2 ] (ν1 − ν2 )s = χ[u1 >u2 ] (ν1 )s ≥ 0 in Ω. Combining (5.4)–(5.5), we obtain (5.3). It then follows from (5.2)–(5.3) that (5.1) holds. Since u1 , u2 ∈ W01,1 (Ω), we get (see, e.g., [8, Proposition B.1]) (u1 − u2 )+ = 0
a.e.
In other words, u1 ≤ u2 almost everywhere. Reversing the roles of u1 and u2 , we conclude that u1 = u2 almost everywhere. Therefore, problem (1.1) has at most one solution. Lemma 5. Let μ ∈ M(Ω) be such that μ+ is diffuse. Given a sequence (βn ) satisfying (1.13)–(1.15), let un be the solution of (1.16). Then, un → u
in L1 (Ω),
(5.6)
where u satisfies (1.1). In particular, ∗
weak∗ in M(Ω).
βn (un ) μ + Δu
Proof. By standard estimates (see, e.g., [8]), we have βn (un ) ≤ μM and Δun M ≤ 2μM .
(5.7)
(5.8)
Ω
Passing to a subsequence if necessary, one can find u ∈ L1 (Ω) and ν ∈ M(Ω) such that un → u In particular,
∗
in L1 (Ω)
−
and βn (un ) ν
uΔζ + Ω
ζ dν =
Ω
weak∗ in M(Ω).
Ω
ζ dμ ∀ζ ∈ C02 (Ω).
(5.9)
Clearly, the sequence (un ) satisfies the assumptions of Theorem 6. Thus, passing to a further subsequence if necessary, we may also assume that un → u q.e. Our goal is to show that u satisfies (1.1). By Lemma 4, this will imply that the limit u is actually independent of the subsequence, so that the entire sequence converges to the same limit u. We first establish the following
Reduced measures for obstacle problems
1225
Claim. The sequence βn (un ) is equidiffuse. Let vn be the solution of −Δvn + βn (vn ) = μ+ in Ω, on ∂Ω. vn = 0
Since μ+ is diffuse, it follows from Lemma 3 that βn (vn ) is equidiffuse (note that in this case K = ∅). On the other hand, since un ≤ vn almost everywhere and βn (t) = 0, for all t ≤ 0, we have 0 ≤ βn (un ) ≤ βn (vn ) a.e.
We conclude that βn (un ) is also equidiffuse. We can now apply Theorem 4 to conclude that u ≤ a almost everywhere and ν ∈ β(u). In other words, u is the (unique) solution of (1.1). The proof of the lemma is complete. Proof of Theorem 1. (⇐) This follows from Lemma 5 above. By Lemma 4, the solution of (1.1) is unique. (⇒) Assume (1.1) has a solution u. In particular, there exists a diffuse measure ν such that −Δu = μ − ν in D (Ω). Since u ≤ a almost everywhere, it follows from the “Inverse” maximum principle that (−Δu)c ≤ 0 (see [15]). Thus,
μc = (μ − ν)c ≤ 0. In other words, μ+ is diffuse. Corollary 2. Let μi ∈ M(Ω), i = 1, 2, be such that μ+ i is diffuse. Let (ui , νi ) be the solution of (1.1) associated to μi . Then, (ν1 − ν2 )+ ≤ (μ1 − μ2 )+ . (5.10) Ω
Ω
Proof. Let ui,n be the solution of −Δui,n + βn (ui,n ) = μi ui,n = 0
in Ω, on ∂Ω,
where (βn ) satisfies (1.13)–(1.15). Then, by Lemma 5, we have ∗
βn (ui,n ) νi so that
∗
weak∗ in M(Ω),
βn (u1,n ) − βn (u2,n ) ν1 − ν2
weak∗ in M(Ω).
(5.11)
1226
Ha¨ım Brezis and Augusto C. Ponce
On the other hand, by standard estimates (see, e.g., [8, Corollary B.1]) we have + βn (u1,n ) − βn (u2,n ) ≤ (μ1 − μ2 )+ . Ω
Ω
As n → ∞, we conclude that (5.10) holds. 6. Proofs of Proposition 1 and Theorems 2, 3 Proof of Proposition 1. Let v be a subsolution of (1.1). In particular, there exists f ∈ L1 (Ω) such that f ∈ β(u) almost everywhere and −Δv + f ≤ μ in (C02 )∗ . Since v ≤ a almost everywhere, it follows from the “Inverse” maximum principle that (−Δv)c ≤ 0. Thus, −Δv + f ≤ μ∗
in (C02 )∗ .
Let (βn ) be a sequence satisfying (1.13)–(1.15) and such that βn ≤ β, for all n ≥ 1. In particular, −Δv + βn (v) ≤ −Δv + f ≤ μ∗ In other words, v is a subsolution of −Δwn + βn (wn ) = μ∗ wn = 0
in (C02 )∗ .
in Ω, on ∂Ω.
(6.1)
By comparison (see [8, Corollary B.2]), we then have v ≤ wn a.e. ∀n ≥ 1, where wn is the solution of (6.1). Since, by Lemma 5, wn → u∗
in L1 (Ω),
we conclude that v ≤ u∗ almost everywhere. Therefore, u∗ is the largest subsolution of (1.1). Proof of Theorem 2. Since Δun M ≤ 2μM , one can find a subsequence (unk ) such that unk → v
in L1 (Ω),
for some v ∈ L1 (Ω). By Theorem 6, we also have unk → v q.e.
Reduced measures for obstacle problems
1227
∗
Let λ ∈ M(Ω) be such that βnk (unk ) λ weak∗ in M(Ω). We now show that the sequence βnk (unk ) satisfies assumption (iv) of Theorem 4. In fact, given ε > 0, let F ⊂ Ω be a compact set such that cap (F ) = 0 and |μc |(Ω \ F ) < ε. Let ω ⊃ F be an
open set such that cap (ω) < ε. By Lemma 3, the sequence βnk (unk ) is 2ε-equidiffuse in Ω \ ω, as claimed. Applying Theorem 4, we deduce that v ≤ a almost everywhere and λd ∈ β(v). We now show that λc = (μ+ )c . In fact, note that λ ≥ 0.
(6.2)
Let w denote the solution of −Δw = −μ− w=0
in Ω, on ∂Ω.
By comparison, we have w ≤ unk , for all k ≥ 1. Thus, w ≤ v almost everywhere. By the “Inverse” maximum principle, we then have −(μ− )c = (−Δw)c ≤ (−Δv)c = μc − λc .
(6.3)
Similarly, v ≤ a almost everywhere implies μc − λc = (−Δv)c ≤ 0.
(6.4)
Combining (6.2)–(6.4), we conclude that λc = (μ+ )c . In other words, v satisfies
− vΔζ + ζ dν = ζ μ − (μ+ )c = ζ dμ∗ Ω
Ω
Ω
where ν = λd ∈ β(v). Therefore, v = the subsequence (unk ), we must have un → u∗
Ω
u∗ .
∀ζ ∈ C02 (Ω),
Since the limit v is independent of
in L1 (Ω).
This establishes Theorem 2. The proof of Theorem 3 is similar and will be omitted. Note that in this case one should apply Theorem 5 instead of Theorem 6.
1228
Ha¨ım Brezis and Augusto C. Ponce
7. The case of two obstacles We shall assume throughout this section that β is any m.m.g. such that dom β = [−b, a]
and (0, 0) ∈ graph β.
(7.1)
For simplicity, we restrict ourselves to the case where a = b = 1. Given μ ∈ M(Ω), we say that u is a solution of (1.18) if u ∈ L1 (Ω), |u| ≤ 1 almost everywhere, Δu ∈ M(Ω) and there exists a diffuse measure ν ∈ β(u) such that − uΔζ + ζ dν = ζ dμ ∀ζ ∈ C02 (Ω). (7.2) Ω
Ω
Ω
We refer the reader to Section 2 for the definition of ν ∈ β(u) in the case of two obstacles. By Proposition 2, ν ∈ β(u) if and only if j(v) − j(u) ≥ (v − u) dν ∀v ∈ H01 (Ω) ∩ L∞ , (7.3) Ω
Ω
Ω
where ∂j = β and j(0) = 0. The counterpart of Theorem 1 is given by the following Theorem 7. Let μ ∈ M(Ω). Then, (1.18) has a solution if and only if μ is diffuse. Moreover, this solution is unique. Solutions of (1.18) can be also obtained via minimization. More precisely, we have Theorem 8. Given a diffuse measure μ, let u be the solution of (1.18). Then, u ∈ H01 (Ω) and u coincides with the solution of the minimization problem 1 2 min |∇v| + j(v) − v dμ . (7.4) 2 Ω v∈H01 (Ω) Ω Ω |v|≤1 a.e.
Given any measure μ in Ω, let μ∗ = μd denote its diffuse part and let u∗ be the unique solution of −Δu∗ + β(u∗ ) μ∗ in Ω, (7.5) u∗ = 0 on ∂Ω. The analogs of Theorems 2 and 3 are given by the Theorem 9. Let (βn ) be a sequence satisfying (1.13)–(1.14). Given μ ∈ M(Ω), let wn denote the solution of (1.16) or (1.17). Then, wn → u ∗
in L1 (Ω),
Reduced measures for obstacle problems
1229
where u∗ is the solution of (7.5). Finally, we make a connection with another concept of generalized solution which was proposed in Brezis-Serfaty [10]. This is based on a duality principle mentioned in [7]. Given f ∈ L2 (Ω), consider the two minimization problems: 1 2 I1 (f ) = min |∇v| + j(v) − vf , (P) v∈H01 (Ω) 2 Ω Ω Ω |v|≤1 a.e.
I2 (f ) =
min
w∈H01 (Ω) Δw∈M(Ω)
1 2 ∗ |∇w| + j (f + Δw) . 2 Ω Ω
(P∗ )
Here, j ∗ denotes the convex conjugate of j. It is not difficult to prove (see [7]) that (P) and (P∗ ) admit the same minimizers. Moreover, I1 (f ) + I2 (f ) = 0 ∀f ∈ L2 (Ω). These two assertions remain valid when f = μ is a diffuse measure. Note that for general measures, problem (P) is not well posed; however, problem (P∗ ) is well defined and admits a unique minimizer U (μ). The function U (μ) was thus regarded in [10] as a kind of generalized solution for the original problem (P) (or (1.18)). The following result establishes the connection between the concept introduced in [10] and the notion of generalized solution in this paper: Theorem 10. For every μ ∈ M(Ω), we have I2 (μ) = −I1 (μd ) + μc M .
(7.6)
Moreover, the minimizers of I1 (μd ) and I2 (μ) coincide; i.e., U (μ) = u∗ . 8. Proofs of Theorems 7–10 Proofs of Theorems 7 and 9. Step 1. Equation (1.18) has, at most, one solution. The argument is exactly the same as in the proof of Lemma 4. We leave the details to the reader. Step 2. If (1.18) has a solution, then μ is diffuse. Since u is bounded and Δu ∈ M(Ω), we have u ∈ H01 (Ω). This easily implies that Δu is diffuse. Since ν is also diffuse, we deduce that μ = −Δu+ν is diffuse as well. Step 3. Proof of Theorem 9 completed.
1230
Ha¨ım Brezis and Augusto C. Ponce
Given μ ∈ M(Ω), let wn denote the solution of (1.17) (respectively, (1.16)). Thus, for some subsequence (nk ) we have unk → v
in L1 (Ω)
∗
and νnk λ
weak∗ in M(Ω).
By Theorem 5 (respectively, Theorem 6), we then have unk → v q.e. By Lemma 3, the sequence (νnk ) satisfies the assumptions of Theorem 4. We conclude that λd ∈ β(v). We claim that λc = μc . In fact, note that v ∈ H01 (Ω). Therefore, Δv is diffuse. Since −Δv +λ = μ, the claim follows by comparing the concentrated parts of both sides with respect to cap. We conclude that v satisfies (7.5), with ν = λd . By Step 1, the solution of (7.5) is unique. Hence, un → u∗
in L1 (Ω).
Step 4. Proof of Theorem 7 completed. By Step 2, if (1.18) has a solution, then μ is diffuse. Conversely, if μ is diffuse, then the existence of a solution of (1.18) follows from Step 3. By Step 1, the solution is unique. Proof of Theorem 8. Since both problems (1.18) and (7.4) have unique solutions, it suffices to show that if u satisfies (1.18), then u minimizes (7.4). By Proposition 2, we have j(v) − j(u) ≥ (v − u) dν ∀v ∈ H01 (Ω), |v| ≤ 1, (8.1) Ω
Ω
Ω
where ν = μ + Δu. On the other hand, we have 1 1 2 (v − u) Δu = − ∇(v − u) · ∇u ≥ |∇u| − |∇v|2 . (8.2) 2 Ω 2 Ω Ω Ω Combining (8.1) and (8.2), the result follows. Proof of Theorem 10. Note that for every w ∈ H01 (Ω) such that Δw ∈ M(Ω), it follows that Δw is a diffuse measure. Thus, j ∗ (μ + Δw) = j ∗ (μd + Δw) + |μc |. This immediately implies that I2 (μ) = I2 (μd ) + μc M . In particular, the minimizers of I2 (μ) and I2 (μd ) are equal; i.e., U (μ) = U (μd ). Therefore, it suffices to establish the theorem for μ diffuse. We now split the proof into two steps: Step 1. The minimizers of I1 (μ) and I2 (μ) coincide.
Reduced measures for obstacle problems
1231
In view of Theorem 8, it suffices to show that if u is the solution of (1.18), then u minimizes I2 (μ). By the equivalence (a) ⇔ (e) in Proposition 2, we have ∗ ∗ j (σ) − j (μ + Δu) ≥ u (σ − μ − Δu) ∀σ ∈ M(Ω) diffuse. (8.3) Ω
Ω 1 H0 (Ω)
Ω
Let w ∈ with Δw ∈ M(Ω). In particular, Δw is diffuse. Applying (8.3) to σ = μ + Δw, we get 1 |∇w|2 + j ∗ (μ + Δw) 2 Ω Ω 1 2 |∇w| + u(Δw − Δu) + j ∗ (μ + Δu) ≥ 2 Ω Ω Ω 1 2 |∇u| + j ∗ (μ + Δu). ≥ 2 Ω Ω Thus, u is the minimizer of I2 (μ). Step 2. Proof of the theorem completed. It remains to establish (7.6). Since we are assuming that μ is diffuse, and the solution u of (1.18) minimizes both I1 (μ) and I2 (μ), then we only need to show that 2 ∗ |∇u| + u dμ = 0, j(u) + j (μ + Δu) − Ω
Ω
Ω
which is equivalent to assertion (f) in Proposition 2. The proof of Theorem 10 is complete. 9. Appendix Our goal in this Appendix is to show that there are sequences such that (un ) is uniformly bounded and (Δun ) is equidiffuse, but (un ) need not have any subsequence converging q.e. For simplicity, we will assume throughout this Appendix that N ≥ 3. We first establish the following Proposition 5. Let (μn ) ⊂ M(RN ) be a sequence of nonnegative measures. Let dμn (y) 1 vn (x) = ∀x ∈ RN . (9.1) (N − 2)|B1 | RN |x − y|N −2 If (vn ) is uniformly bounded, then (μn ) is equidiffuse. Proof. Clearly, it suffices to show that for every Borel set A ⊂ RN we have μn (A) ≤ C cap (A) ∀n ≥ 1.
(9.2)
1232
Ha¨ım Brezis and Augusto C. Ponce
We only need to establish (9.2) when A = K is a compact set. Let ζK denote the capacitary potential of K. Since vn is bounded, μn is a diffuse measure. Thus, μn (K) ≤ ζK dμn RN =− ζK Δvn = − vn ΔζK ≤ vn L∞ |ΔζK | ≤ C cap (K). RN
RN
RN
This establishes (9.2) when A is compact; the general case easily follows by approximation. In conjunction with Proposition 5, we recall that there is an example of functions (wn ) constructed by Cioranescu-Murat [12, Example 2.1] such that (i) (ii) (iii) (iv)
0 ≤ wn ≤ 1 a.e. in Ω; (Δwn ) is equidiffuse; wn 1 weakly in H 1 (Ω); for every ω ⊂⊂ Ω, we have
lim inf cap [wn = 0] ∩ ω > 0. n→∞
Properties (i), (iii) and (iv) are clear from their construction. We only need to check that (ii) holds. By construction, (Δwn ) is bounded in M(Ω). Write −Δwn = μn − νn , where μn , νn ≥ 0 and μn , νn are mutually singular. We then extend μn , νn to be identically zero outside Ω. Let vn be defined as in (9.1). An easy inspection from [12] shows that (vn ) is uniformly bounded in RN . By Proposition 5, (μn ) is equidiffuse. Similarly, (νn ) is also equidiffuse. We deduce that (ii) holds. It now follows from (iii) and (iv) that (wn ϕ) has no subsequence con
verging q.e., for any ϕ ∈ Cc∞ (Ω), ϕ ≡ 0. Note however that Δ(wn ϕ) is equidiffuse in view of (ii). Acknowledgments. The first author (H.B.) is partially sponsored by an EC Grant through the RTN Program “Front-Singularities”, HPRN-CT2002-00274. H.B. is also a member of the Institut Universitaire de France. The second author (A.C.P.) is supported by the NSF grant DMS-0111298 and Sergio Serapioni, Honorary President of Societ` a Trentina Lieviti—Trento (Italy).
Reduced measures for obstacle problems
1233
References [1] A. Ancona, Une propri´et´e d’invariance des ensembles absorbants par perturbation d’un op´erateur elliptique, Comm. Partial Differential Equations, 4 (1979), 321–337. [2] H. Attouch, “Variational Convergence for Functions and Operators,” Applicable Mathematics Series, Pitman, Boston, MA, 1984. [3] J. R. Baxter, Inequalities for potentials of particle systems, Illinois J. Math., 24 (1980), 645–652. [4] Ph. B´enilan and H. Brezis, Nonlinear problems related to the Thomas-Fermi equation, J. Evol. Equ., 3 (2004), 673–770. Dedicated to Ph. B´enilan. [5] L. Boccardo, T. Gallou¨et, and L. Orsina, Existence and uniqueness of entropy solutions for nonlinear elliptic equations with measure data, Ann. Inst. H. Poincar´e Anal. Non Lin´eaire, 13 (1996), 539–551. [6] H. Brezis, Int´egrales convexes dans les espaces de Sobolev, Israel J. Math., 13 (1972), 9–23 (1973). Proceedings of the International Symposium on Partial Differential Equations and the Geometry of Normed Linear Spaces (Jerusalem, 1972). [7] H. Brezis, Probl`emes unilat´eraux, J. Math. Pures Appl., 51 (1972), 1–168. [8] H. Brezis, M. Marcus, and A. C. Ponce, Nonlinear elliptic equations with measures revisited, to appear in Annals of Math. Studies, Princeton University Press. Part of the results were announced in a note by the same authors: A new concept of reduced measure for nonlinear elliptic equations, C. R. Acad. Sci. Paris, Ser. I, 339 (2004), 169–174. [9] H. Brezis and A. C. Ponce, Kato’s inequality when Δu is a measure, C. R. Math. Acad. Sci. Paris, Ser. I, 338 (2004), 599–604. [10] H. Brezis and S. Serfaty, A variational formulation for the two-sided obstacle problem with measure data, Commun. Contemp. Math., 4 (2002), 357–374. [11] H. Brezis and W. A. Strauss, Semilinear second-order elliptic equations in L1 , J. Math. Soc. Japan 25 (1973), 565–590. [12] D. Cioranescu and F. Murat, A strange term coming from nowhere, In: Topics in the mathematical modelling of composite materials. Progr. Nonlinear Differential Equations Appl., vol. 31, Birkh¨ auser, Boston, 1997, pp. 45–93. [13] P. Dall’Aglio and G. Dal Maso, Some properties of the solutions of obstacle problems with measure data, Ricerche Mat., 48 (1999), 99–116. Papers in memory of Ennio De Giorgi. [14] P. Dall’Aglio and C. Leone, Obstacles problems with measure data and linear operators, Potential Anal. 17 (2002), 45–64. [15] L. Dupaigne and A. C. Ponce, Singularities of positive supersolutions in elliptic PDEs, Selecta Math. (N.S.), 10 (2004), 341–358. [16] L. Dupaigne, A. C. Ponce, and A. Porretta, Elliptic equations with vertical asymptotes in the nonlinear term, to appear in J. Anal. Math. [17] M. Fukushima, K. Sato, and S. Taniguchi, On the closable parts of pre-Dirichlet forms and the fine supports of underlying measures, Osaka J. Math., 28 (1991), 517–535. [18] C. Goffman and J. Serrin, Sublinear functions of measures and variational integrals, Duke Math. J., 31 (1964), 159–178. [19] M. Grun-Rehomme, Caract´erisation du sous-diff´ erential d’int´egrandes convexes dans les espaces de Sobolev, J. Math. Pures Appl., 56 (1977), 149–156.
1234
Ha¨ım Brezis and Augusto C. Ponce
[20] L. Orsina and A. Prignet, Non-existence of solutions for some nonlinear elliptic equations involving measures, Proc. Roy. Soc. Edinburgh Sect. A, 130 (2000), 167–187. [21] R. T. Rockafellar, Integrals which are convex functionals, Pacific J. Math., 24 (1968), 525–539. [22] R. T. Rockafellar, Integrals which are convex functionals, II, Pacific J. Math., 39 (1971), 439–469. ´ [23] G. Stampacchia, “Equations elliptiques du second ordre ` a coefficients discontinus, S´eminaire de Math´ematiques Sup´erieures,” no. 16 (´et´e, 1965), Les Presses de l’Universit´e de Montr´eal, Montr´eal, Qu´ebec, 1966.
Volume 10, Number 11 (2005), 1201–1234
REDUCED MEASURES FOR OBSTACLE PROBLEMS Ha¨ım Brezis Laboratoire Jacques-Louis Lions, Universit´e Pierre et Marie Curie Boˆıte courier 187, 75252 Paris Cedex 05, France and Rutgers University, Dept. of Mathematics, Hill Center, Busch Campus 110 Frelinghuysen Rd., Piscataway, NJ 08854 Augusto C. Ponce Institute for Advanced Study, Princeton, NJ 08540 (Submitted by: Reza Aftabizadeh)
1. Introduction Let Ω ⊂ RN , N ≥ 1, be a smooth bounded domain. In this paper, we study the problem −Δu + β(u) μ in Ω, (1.1) u = 0 on ∂Ω, where μ is a finite measure in Ω and β is a maximal monotone graph (m.m.g.) such that (0, 0) ∈ graph β. Throughout most of the Introduction, we assume that and
dom β = (−∞, a] β(t) = {0}
for some 0 ≤ a < ∞
∀t ≤ 0.
(1.2) (1.3)
(However, the case where dom β = [−b, a], b ≥ 0, is also of interest and will be discussed at the end of the Introduction. ) A typical example of a m.m.g. β satisfying (1.2)–(1.3) is the following ⎧ ⎪ ⎨ g(t) if t < a, β(t) = g(a), ∞ if t = a, ⎪ ⎩ if t > a, ∅ where g : (−∞, a] → [0, ∞) is any continuous nondecreasing function such that g(t) = 0, for all t ≤ 0. Accepted for publication: June 2005. AMS Subject Classifications: 35J85, 35R05. 1201
1202
Ha¨ım Brezis and Augusto C. Ponce
Let M(Ω) be the space of finite measures μ on Ω such that μ(∂Ω) = 0; M(Ω) can be also identified with C0 (Ω)]∗ . Given μ ∈ M(Ω), we say that u is a solution of the obstacle problem (1.1) if the following holds: (i) u ∈ L1 (Ω), u ≤ a almost everywhere and Δu ∈ M(Ω); (ii) there exists ν ∈ M(Ω) such that ν ∈ β(u) (in a sense to be explained below) and − uΔζ + ζ dν = ζ dμ ∀ζ ∈ C02 (Ω). (1.4) Ω
Here,
Ω
Ω
C02 (Ω) = ζ ∈ C 2 (Ω) ; ζ = 0 on ∂Ω .
Recall that if u ∈ L1 (Ω) is such that Δu ∈ M(Ω), then u admits a representative which is quasicontinuous with respect to the Newtonian (H 1 ) capacity, denoted cap (see, e.g., [1]). We may thus assume that u is well defined q.e. (= quasi-everywhere = outside a set of zero capacity). In particular, u is ν-measurable for any finite measure ν which does not charge sets of zero capacity. Given any such measure ν, we then say that ν ∈ β(u)
(1.5)
if the following holds: • νa ∈ β(u) almost everywhere; • νs is concentrated on [u = a] and νs ≥ 0. Here, we denote by νa and νs the absolutely continuous and the singular parts of ν with respect to the Lebesgue measure in RN . In Section 2 below, we discuss other equivalent definitions of (1.5). If the measure μ is an L1 function, it is known (see Brezis-Strauss [11]) that (1.1) has a unique solution (and in this case ν ∈ L1 (Ω)). As we shall see below, problem (1.1) still admits a solution for every measure μ in L1 + H −1 (see Theorem 1 below). In this case, the measure ν need not belong to L1 (Ω) (it is easy to construct examples in dimension N = 1). We say that μ ∈ M(Ω) is diffuse if μ(E) = 0 for every Borel set E ⊂ Ω such that cap (E) = 0. Equivalently, a measure μ is diffuse if and only if μ ∈ L1 + H −1 (see Grun-Rehomme [19] and Boccardo-Gallou¨et-Orsina [5]). In dimension N = 1, every measure is diffuse (since cap (E) = 0 if and only if E = ∅). However, when N ≥ 2, there are measures which are not diffuse, for instance Dirac masses. In this paper, we follow the same program as in Brezis-Marcus-Ponce [8]. The main difference is that here dom β = R and dom β is closed (the case
Reduced measures for obstacle problems
1203
where dom β = R and dom β is an open set has been studied by DupaignePonce-Porretta [16]). Our main concern in [8] was twofold: (a) Identify the good measures for problem (1.1), i.e., those for which (1.1) has a solution. (b) If μ is not a good measure, define some kind of “generalized” solution, i.e., a common limit of all natural approximation schemes. Our first result gives the complete answer to question (a): Theorem 1. Let μ ∈ M(Ω). Then, (1.1) has a solution if and only if μ+ is diffuse. Moreover, the solution is unique. In other words, μ is a good measure for (1.1) if and only if μ+ is diffuse (equivalently, μ+ ∈ L1 + H −1 ). Recall that any measure μ can be uniquely decomposed as (see, e.g., [17]) μ = μd + μc , where μd is a diffuse measure and μc is a measure concentrated on some set of zero H 1 -capacity. Note in particular that μ is diffuse if and only if μc = 0. Given μ ∈ M(Ω), set μ∗ = μ − (μc )+ .
(1.6)
An easy consequence of Theorem 1 is the Corollary 1. For every μ ∈ M(Ω), μ∗ is the largest good measure ≤ μ. Indeed, let λ be a good measure ≤ μ. By Theorem 1, we know that λc ≤ 0. On the other hand, from λ ≤ μ we deduce that λd ≤ μd and −(λc )− ≤ −(μc )− . Thus, λ = λd + (λc )+ − (λc )− ≤ μd − (μc )− = μ∗ . Notation. Given μ, we denote by u∗ the unique solution of −Δu∗ + β(u∗ ) μ∗ in Ω, u∗ = 0
on ∂Ω.
(1.7)
We say that v is a subsolution of (1.1) if v ∈ L1 (Ω), v ≤ a almost everywhere, Δv ∈ M(Ω) and there exists diffuse ν ∈ M(Ω) such that ν ∈ β(v) and −
ζ dν ≤
uΔζ + Ω
Ω
Ω
ζ dμ ∀ζ ∈ C02 (Ω), ζ ≥ 0 in Ω.
(1.8)
1204
Ha¨ım Brezis and Augusto C. Ponce
Since ν ≥ 0, v is a subsolution of (1.1) if and only if one can find f ∈ L1 (Ω) such that f ∈ β(v) almost everywhere and fζ ≤ ζ dμ ∀ζ ∈ C02 (Ω), ζ ≥ 0 in Ω. (1.9) − vΔζ + Ω
Ω
Ω
A “companion” to Corollary 1 is the following Proposition 1. For any μ ∈ M(Ω), u∗ is the largest subsolution of (1.1). Several authors have studied the obstacle problem (1.1) associated with measures. Given a ≥ 0, let β = βa be the m.m.g. defined as ⎧ ⎪ if t < a, ⎨{0}
βa (t) = (1.10) 0, ∞ if t = a, ⎪ ⎩ ∅ if t > a. When a = 0 and Ω = RN , N ≥ 3, Theorem 1 has been known for a long time to experts from potential theory (see Baxter [3] and the references therein). In [3], the following statement appears: Let λ1 , λ2 ∈ M(Ω) be two nonnegative measures and assume that Pot λ1 ≡ (−Δ)−1 λ1 < +∞ λ1 -a.e.;
(1.11)
then, (1.1) has a solution for μ = λ1 − λ2 . Note that (1.11) is equivalent to the condition λ1 is diffuse (1.12) + (see, e.g., [13, Lemma 3.1]). Hence, (1.11) implies that μ (≤ λ1 ) is diffuse, so that μ satisfies the assumption of Theorem 1. When a > 0 and β is given by (1.10), Theorem 1 and Proposition 1 are due to Dall’Aglio-Leone [14] and Dall’Aglio-Dal Maso [13]. More precisely, given any measure μ, Dall’AglioLeone [14] have proved that there exists a largest element u ˆ in the class of functions w such that ⎧ ⎪ ⎨ −Δw ≤ μ in Ω, w ≤ a in Ω, ⎪ ⎩ w = 0 on ∂Ω. Subsequently, Dall’Aglio-Dal Maso [13] have proved that u ˆ satisfies (1.1) ∗ relative to the measure μ . In other words, they have shown that u ˆ = u∗ . We now return to a general m.m.g. β satisfying (1.2)–(1.3) and we investigate part (b) of the program. Let (βn ) be a sequence of continuous, nondecreasing functions βn : R → R, βn (0) = 0, such that βn → β
in the sense of graphs;
(1.13)
Reduced measures for obstacle problems
1205
more precisely, for every (t, s) ∈ G ≡ graph β, there exists a sequence tn → t such that βn (tn ) → s. If N ≥ 2, then we also assume that each βn has subcritical growth; i.e.,
βn (t) ≤ C |t|p + 1 ∀t ∈ R, (1.14) for some constant C > 0 and some p < n). We suppose in addition that
N N −2
(both possibly depending on
βn (t) = 0 ∀t ≤ 0.
(1.15)
Given any μ ∈ M(Ω), there exists a unique function un such that (see B´enilan-Brezis [4]) −Δun + βn (un ) = μ in Ω, (1.16) un = 0 on ∂Ω. The limiting behavior of the sequence (un ) is given by the following Theorem 2. For every μ ∈ M(Ω), we have un → u∗
in L1 (Ω),
where un is the solution of (1.16) and u∗ is the solution of (1.7). We emphasize that—as in [8]—the limit of (un ) is independent of the approximating sequence (βn ). Another approximation scheme consists of keeping β fixed, and approximating μ by convolution. More precisely, given a sequence of mollifiers (ρn ), let vn denote the solution of −Δvn + β(vn ) ρn ∗ μ in Ω, (1.17) on ∂Ω. vn = 0 Then, the sequence (vn ) converges to the same limit u∗ . More precisely, Theorem 3. For every μ ∈ M(Ω), we have vn → u∗
in L1 (Ω),
where vn is the solution of (1.17) and u∗ is the solution of (1.7). In Section 7, we consider similar questions for the case of two obstacles, i.e., dom β = [−b, a] for some b ≥ 0. We define the notion of solution of −Δu + β(u) μ in Ω, (1.18) u = 0 on ∂Ω, by analogy with the case studied above. Our main results assert that
1206
Ha¨ım Brezis and Augusto C. Ponce
• Problem (1.18) has a solution if and only if μ is diffuse and, in this case, this solution coincides with the unique minimizer of the variational problem 1 min |∇v|2 + j(v) − v dμ , 2 Ω v∈H01 (Ω) Ω Ω −b≤v≤a a.e.
where ∂j = β and j(0) = 0. • Given any measure μ, the solutions of the “natural” approximation problems (1.16) and (1.17) converge to the solution of (1.18) with data μd (= diffuse part of μ). This last result is related to a theorem of Orsina-Prignet [20]. This paper is organized as follows. In Section 2, we examine the concept of ν ∈ β(u) for any m.m.g. β and we discuss other equivalent definitions. In Section 3, we study some properties related to equidiffuse sequences (see Definition 1 below). In Section 4, we prove that the sequences given by (1.16) and (1.17) both converge q.e. The proofs of Theorems 1–3 and Proposition 1 are then presented in Sections 5 and 6. In Section 7, we discuss problem (1.1) in the case of two obstacles; the proofs of these results are given in Section 8. 2. Various definitions of ν ∈ β(u) Let β be a m.m.g. with closed domain, let u be a quasicontinuous function in Ω such that u ∈ dom β q.e., and let ν ∈ M(Ω). Set a = sup dom β and b = − inf dom β . (2.1) We say that ν ∈ β(u) if the following conditions are satisfied: (1) ν is diffuse;
(2.1a)
(2) νa ∈ β(u) almost everywhere;
(2.1b)
(3) νs is concentrated on the set [u = a] ∪ [u = −b];
(2.1c)
(4) νs ≥ 0 on [u = a] and νs ≤ 0 on [u = −b].
(2.1d)
We first observe that this definition agrees with the cases we have already considered in the Introduction, namely dom β = (−∞, a] and dom β = R.
Reduced measures for obstacle problems
1207
In fact, recall that for every quasicontinuous function u, the set [u = ±∞] has zero capacity. Thus, for every diffuse measure ν, we have νs = 0
on [u = ±∞].
In particular, if dom β = R, then ν ∈ β(u)
if and only if ν ∈ L1 (Ω)
and ν ∈ β(u) a.e.
We also observe that, since dom β is closed, then u ∈ dom β q.e.
if and only if u ∈ dom β a.e.
We present some equivalent forms of the assertion “ν ∈ β(u)” in the next Proposition 2. Let u be a quasicontinuous function in Ω such that u ∈ dom β almost everywhere. Assume that ν ∈ M(Ω) is diffuse and u ∈ L1 (Ω; ν). Then, the following assertions are equivalent: (a) ν ∈ β(u); ∗ (b) j(u) ∈ L1 (Ω) and j(t) − j(u) ≥ ν (t − u) in C0 (Ω) , ∀t ∈ R; ∗ ≥ u (s − ν) in C0 (Ω) , ∀s ∈ R; (c) j ∗ (ν) ∈ M(Ω) and j∗ (s) − j ∗ (ν) (d) j(u) ∈ L1 (Ω) and Ω j(v) − Ω j(u) ≥ Ω (v − u) dν for every v ∈ H01 (Ω) ∩ L∞ ; (e) j ∗ (ν) ∈ M(Ω) and Ω j ∗ (σ) − Ω j ∗ (ν) ≥ Ω u d(σ − ν) for every such that u ∈ L1 (Ω; σ); σ ∈ M(Ω) diffuse (f) Ω j(u) + Ω j ∗ (ν) = Ω u dν. Here, j : R → [0, +∞] is a convex function such that ∂j = β and j(0) = 0; j ∗ denotes the convex conjugate of j. For any ν ∈ M(Ω), the measure j ∗ (ν) is a nonnegative Borel measure defined as j ∗ (ν) = j ∗ (νa ) + a(νs )+ + b(νs )− ,
(2.2)
where a and b are given by (2.1) (see, e.g., [6, 18, 21, 22]). Proof. (a) ⇒ (b), (c), (f). By assumption, j(t) − j(u) ≥ νa (t − u)
a.e. ∀t ∈ R.
(2.3)
Recall that νa ∈ β(u) almost everywhere is equivalent to u ∈ β −1 (νa ) = ∂j ∗ (νa )
a.e.
Thus, j ∗ (s) − j ∗ (νa ) ≥ u (s − νa ) ∗
j(u) + j (νa ) = uνa
a.e.
a.e. ∀s ∈ R,
(2.4) (2.5)
1208
Ha¨ım Brezis and Augusto C. Ponce
On the other hand, νs is ≥ 0 on [u = a] and ≤ 0 on [u = −b]; thus, uνs = aνs = a(νs )+
in [u = a],
−
uνs = −bνs = b(νs )
in [u = −b].
Here we use the convention that ∞ · 0 = 0. Since νs is concentrated on the set [u = a] ∪ [u = −b], we deduce that uνs = a(νs )+ + b(νs )− ≥ 0
in Ω.
(2.6)
Thus, in view of (2.2), j ∗ (ν) = j ∗ (νa ) + uνs
in Ω.
(2.7)
By (2.5)–(2.6), we have j(u) ∈ L1 (Ω) and j ∗ (ν) ∈ M(Ω). Therefore, (b) follows from (2.3) and (2.6); (c) follows from (2.4) and (2.7); (f) follows from (2.5) and (2.7). 1 ∞ (a) ⇒ (d). Let v ∈ H0 (Ω) ∩ L . If v > a or v < −b on a set of positive measure, then Ω j(v) = +∞ and (d) holds. We now assume that −b ≤ v ≤ a almost everywhere; hence, this also holds q.e. By (2.3), we have j(v) − j(u) ≥ νa (v − u) Thus,
j(v) − Ω
a.e.
j(u) ≥
Ω
νa (v − u).
(2.8)
Ω
Since, by assumption, −b ≤ v ≤ a q.e., it follows from (2.6) that (v − u) νs = (v − a) (νs )+ − (v + b) (νs )− ≤ 0 Thus,
in Ω.
(v − u) dνs ≤ 0.
(2.9)
Ω
Combining (2.8) and (2.9), we obtain (d). (a) ⇒ (e). Let diffuse σ ∈ M(Ω) be such that u ∈ L1 (Ω; σ). By (2.4), we have j ∗ (σa ) − j ∗ (νa ) ≥ u(σa − νa ) a.e. (2.10) On the other hand, −b ≤ u ≤ a q.e. implies j ∗ (σ) = j ∗ (σa ) + a(σs )+ + b(σs )− ≥ j ∗ (σa ) + uσs .
(2.11)
Combining (2.7), (2.10) and (2.11), we deduce (e). (b) ⇒ (a). By assumption, j(t) − j(u) ≥ ν (t − u)
in Ω,
∀t ∈ R.
(2.12)
Reduced measures for obstacle problems
1209
Taking the absolutely continuous parts of (2.12) with respect to Lebesgue measure, we obtain (2.3). Thus, νa ∈ β(u) almost everywhere. Given t ∈ (−b, a), it follows from the singular part of (2.12) that νs (t − u) ≤ 0
in Ω.
In other words, νs ≤ 0
in [u < t],
νs ≥ 0
in [u > t].
Since t ∈ (−b, a) was arbitrary, we deduce that νs = 0
in [−b < u < a],
νs ≤ 0
in [u = −b],
νs ≥ 0
in [u = a].
This establishes (a). (c) ⇒ (a). Comparing the absolutely continuous parts of both sides of (c), we obtain (2.4). Thus, (2.3) holds and so νa ∈ β(u) almost everywhere. From the singular parts of (c), we get a(νs )+ + b(νs )− ≤ uνs
in Ω.
Since −b ≤ u ≤ a q.e., we conclude that a(νs )+ + b(νs )− = uνs
in Ω.
(2.13)
Thus, νs is concentrated on [u = a] ∪ [u = −b], νs ≥ 0 in [u = a], and νs ≤ 0 in [u = −b]. We conclude that (a) holds. (f) ⇒ (a). We first observe that, by (f), we have j(u) ∈ L1 (Ω) and ∗ j(u) + j ∗ (ν) = uν in C0 (Ω) . (2.14) In fact, note that we always have j(u) + j ∗ (νa ) ≥ uνa
a.e.
and
a(νs )+ + b(νs )− ≥ uνs . Combining these two inequalities, we get ∗ j(u) + j ∗ (ν) ≥ uν in C0 (Ω) .
Using (f), (2.14) follows. From the absolutely continuous part of (2.14), we deduce that (2.5) holds, and so νa ∈ β(u) almost everywhere. Moreover, from the singular part of (2.14), we obtain (2.13). Therefore, (a) is satisfied.
1210
Ha¨ım Brezis and Augusto C. Ponce
(d) ⇒ (b). Since the right-hand side of (d) is finite and j ≥ 0, we have j(u) ∈ L1 (Ω). Let t ∈ (−b, a). We claim that j(t) − j(u) ≥ (t − u) dν ∀A ⊂ Ω Borel. (2.15) A
A
A
By outer regularity of Radon measures, it suffices to establish (2.15) when A is open and A ⊂⊂ Ω. Given an open subset ω ⊂⊂ A and k ≥ |t|, let v ∈ H01 (Ω), |v| ≤ k in Ω, be such that v = t in ω and v = Tk (u) in Ω\A, where Tk denotes the truncation at ±k. Note that j(v) ≤ j(u) in Ω\A. By (d), we then have j(v) − j(u) ≥ (v − u) dν. A
Ω
A
As ω ↑ A, we get j(t) − j(u) ≥ (t − u) dν + A
A
Tk (u) − u dν.
(2.16)
Ω\A
A
Since u ∈ L1 (Ω; ν), by monotone convergence we have
Tk (u) − u dν → 0 as k → ∞. Ω\A
Thus, as k → ∞ in (2.16), we get (2.15). Assertion (b) immediately follows from (2.15). (e) ⇒ (c). Given s ∈ R and a Borel set A ⊂ Ω, let σ = sLN A + νΩ\A , where LN denotes the Lebesgue measure in RN . Clearly, σ is a diffuse measure in Ω and u ∈ L1 (Ω; σ). By (e), we have ∗ ∗ j (s) − j (ν) ≥ u (s dLN − dν). A
A
A
Since A ⊂ Ω is arbitrary, we conclude that (c) holds. The proof of the proposition is complete. Remark 1. Assume ν ∈ β(u). The assumption that u ∈ L1 (Ω; ν) automatically holds in the following cases: • dom β = (−∞, a] and β(t) = {0}, ∀t ≤ 0; • dom β = [−b, a]. Note, however, that if ν ∈ β(u) and dom β = R, then it need not be true that u ∈ L1 (Ω; ν) (even if β is a continuous function); see however Lemma 2 below.
Reduced measures for obstacle problems
1211
As we mentioned in the Introduction, we are interested in the approximation of problem (1.1) with a sequence (βn ). Given a sequence of m.m.g. (βn ), then βn → β in the sense of graphs if, for every (t, s) ∈ graph β, there exists (tn , sn ) ∈ graph βn such that tn → t and sn → s. Since the notion of ν ∈ β(u) can be stated in terms of j, where ∂j = β, it is useful to have a characterization of the convergence of (βn ) in terms of the convergence of the primitives (jn ). This is given by the next Proposition 3. Let (βn ) be a sequence of m.m.g. Let jn : R → [0, +∞] be such that jn (0) = 0 and ∂jn = βn . Then, the following statements are equivalent: (a) βn → β in the sense of graphs; (b) for every t ∈ R, if tn → t, then j(t) ≤ lim inf jn (tn ); for every t ∈ R, n→∞
there exists tn → t such that j(t) = lim jn (tn ). n→∞
We refer the reader to Attouch [2, Theorem 3.66] for the proof of Proposition 3. In the literature, assertion (b) is called the Mosco-convergence of jn to j. We conclude this section with the standard Proposition 4. Let (βn ) be a sequence of m.m.g. such that βn → β in the sense of graphs. Let (tn , sn ) ∈ graph βn . If tnk → t and snk → s, then (t, s) ∈ graph β. Proof. Given (x, y) ∈ graph β, let (xn , yn ) ∈ graph βn be such that xn → x and yn → y. From the monotonicity of βnk we have (snk − ynk )(tnk − xnk ) ≥ 0. Thus, as k → ∞, we get (s − y)(t − x) ≥ 0
∀(x, y) ∈ graph β.
From the maximality of β, we conclude that (t, s) ∈ graph β. 3. Equidiffuse sequences of measures We begin this section with the following Definition 1. Let (λn ) ⊂ M(Ω). Given ε > 0, we say that (λn ) is εequidiffuse if (i) (λn ) is bounded in M(Ω);
1212
Ha¨ım Brezis and Augusto C. Ponce
(ii) there exists δ > 0 such that cap (A) < δ
=⇒
|λn |(A) < ε
∀n ≥ 1,
for every Borel set A ⊂ Ω. The sequence (λn ) is equidiffuse if it is ε-equidiffuse for every ε > 0. Clearly, if (λn ) is equidiffuse, then each measure λn is diffuse. We also observe that any finite set of diffuse measures is equidiffuse. Here is another example. Let μ ∈ M(Ω) and let (ρn ) be a sequence of mollifiers. If μc M < ε, then (ρn ∗μ) is ε-equidiffuse. It is then easy to see that (ρn ∗μ) is equidiffuse if and only if μ is diffuse. The main result of this section is the Theorem 4. Let (wn ) be a sequence of quasicontinuous functions in Ω. Let (βn ) be a sequence of m.m.g. Let (νn ) be a sequence of diffuse measures with νn ∈ βn (wn ) in the sense of (2.1a)–(2.1d). Assume that (i) wn → w q.e.; (ii) βn → β in the sense of graphs; ∗ (iii) νn ν weak∗ in M(Ω); (iv) for every ε > 0, there exists a compact set Kε ⊂ Ω such that cap (Kε ) < ε and (νn ) is ε-equidiffuse in Ω \ Kε . Then, w ∈ dom β a.e. and νd ∈ β(w). (3.1) Remark 2. Assumption (iv) is clearly weaker then the assumption “(νn ) is equidiffuse”; moreover, the measure ν need not be diffuse. For example, for any measure μ (not necessarily diffuse), the sequence (ρn ∗ μ) always satisfies (iv). Before turning to the proof of Theorem 4, we present two lemmas. The first one is reminiscent of part of the Dunford-Pettis theorem: Lemma 1. Let (λn ) ⊂ M(Ω) be a sequence of diffuse measures such that ∗
λn λ
weak∗ in M(Ω).
Let (wn ) be a bounded sequence of quasicontinuous functions in Ω such that wn → w q.e. as n → ∞. If (λn ) is ε-equidiffuse, then lim sup wn ζ dλn − wζ dλd ≤ CεζL∞ ∀ζ ∈ C0 (Ω), (3.2) n→∞
Ω
Ω
for some constant C > 0 depending on sup λn M and sup wn L∞ .
Reduced measures for obstacle problems
1213
Notice that since (wn ) is quasicontinuous in Ω and wn → w q.e., then w is also quasicontinuous. Proof. Let δ > 0 be such that cap (A) < δ implies |λn |(A) < ε, for all n ≥ 1, and |λ|(A) ≤ ε. In particular, λc M ≤ ε.
(3.3)
Let ζ ∈ C0 (Ω). Since wζ is quasicontinuous on Ω and ζ = 0 on ∂Ω, there exists ψ ∈ C0 (Ω) such that
cap |ψ − wζ| > ε < δ. (3.4) Thus,
Ω
|ψ − wζ| |λd | ≤ ελM +
|ψ − wζ| |λd | ≤ Cε.
(3.5)
[|ψ−wζ|>ε]
Moreover, since wn → w q.e., by (3.4) we have
< δ. lim sup cap |ψ − wn ζ| > ε n→∞
Proceeding as above, we get
|ψ − wn ζ| |λn | ≤ Cε.
lim sup n→∞
(3.6)
Ω ∗
Finally, since ψ ∈ C0 (Ω) and λn λ in M(Ω), ψ dλn = ψ dλ. lim n→∞ Ω
(3.7)
Ω
Combining (3.3), (3.5)–(3.7), we get wζ dλd ≤ 2CεζL∞ . lim sup wn ζ dλn − n→∞ Ω
Ω
This establishes Lemma 1. Our next lemma is a variant of Proposition 2, where the condition u ∈ L1 (Ω; ν) is not assumed: Lemma 2. Let u be a quasicontinuous function in Ω such that u ∈ dom β almost everywhere, and assume that ν ∈ M(Ω) is diffuse. Then, ν ∈ β(u) if and only if ∗ j(t) − j(u) − ν (t − u) Φk (u) ≥ 0 in C0 (Ω) , ∀t ∈ R, ∀k > 0, (3.8) where Φk ∈ Cc∞ (R) is such that Φk ≥ 0 in R and Φk (t) = 1 if |t| ≤ k.
1214
Ha¨ım Brezis and Augusto C. Ponce
Proof. The implication “⇒” is established as in the proof of Proposition 2. We now prove the converse statement. For simplicity, we only consider the case dom β = R. By (3.8), we have j(t) − j(u) ≥ ν (t − u) in |u| ≤ k , ∀t ∈ R, ∀k > 0. Thus, j(t) − j(u) ≥ νa (t − u)
and νs (t − u) ≤ 0 ∀t ∈ R.
(3.9)
We conclude that νa ∈ β(u) almost everywhere and νs = 0; in particular, ν = νa ∈ L1 (Ω). Therefore, ν ∈ β(u). We now present the Proof of Theorem 4. We split the proof into two steps: Step 1. w ∈ dom β almost everywhere. It suffices to consider the case where dom β = R. Assume for simplicity that dom β = (−∞, a] (the other cases are similar). Given t0 > a, by Proposition 4 there exists a sequence rn → +∞ such that if (t0 , s) ∈ graph βn , then s ≥ rn . Recall that (νn ) is a bounded sequence in M(Ω). Thus, in view of the assumption (νn )a ∈ βn (wn ) almost everywhere, we have 1 C [wn ≥ t0 ] ≤ 1 (νn )a ≤ |νn | ≤ . rn Ω rn Ω rn We conclude that [w > t0 ] ≤ lim inf [wn ≥ t0 ] = 0 ∀t0 > a. n→∞
This implies
[w > a] = 0.
In other words, w ≤ a almost everywhere and so w ∈ dom β almost everywhere. Step 2. νd ∈ β(w). Given ε > 0, let Kε be the compact set given by (iv). Let Φk be as in the statement of Lemma 2. Since wn Φk (wn ) → wΦk (w)
q.e.,
we can apply Lemma 1 to get wΦk (w)ζ dνd ≤ Ck ε ζL∞ lim sup wn Φk (wn )ζ dνn − n→∞
Ω
Ω
(3.10)
Reduced measures for obstacle problems
1215
for every ζ ∈ C0 (Ω\Kε ). Let t ∈ R be such that j(t) < ∞. Since βn → β in the sense of graphs, by Proposition 3 there exists a sequence tn → t such that jn (tn ) → j(t) as n → ∞. (3.11) Moreover, by Fatou, j(w)ζ ≤ lim inf jn (wn )ζ ∀ζ ∈ C0 (Ω), ζ ≥ 0 in Ω. (3.12) Ω
n→∞
Ω
On the other hand, by Lemma 2 we have jn (tn ) − jn (wn ) − νn (tn − wn ) Φk (wn ) ≥ 0 Combining (3.10)–(3.13), we conclude that j(t)−j(w)−νd (t−w) Φk (w)ζ ≥ −Ck ε ζL∞
∗ in C0 (Ω) .
(3.13)
∀ζ ∈ C0 (Ω\Kε ), ζ ≥ 0.
Ω
As ε → 0, we obtain j(t) − j(w) − νd (t − w) Φk (w) ≥ 0
∗ in C0 (Ω) .
By Lemma 2, this implies νd ∈ β(w). 4. A basic tool concerning q.e.-convergence Given μn ∈ M(Ω), let un be such that −Δun = μn in Ω, on ∂Ω. un = 0
(4.1)
In this section, we study the pointwise convergence of (some subsequence of) (un ) as n → ∞. Assume that Δun M ≤ C.
(4.2)
Using standard linear regularity theory, one can show that there exists a subsequence (uni ) such that uni → u strongly in W 1,p , for p < 2, and thus uni → u capW 1,p -q.e.
∀p < 2.
(4.3)
On the other hand, given a measure ν, diffuse with respect to the H 1 capacity, we can also extract a subsequence (unj ) (depending on ν) such that unj → u ν-a.e. (4.4) A natural question is whether unk → u q.e. (i.e., capW 1,2 -q.e.), for some subsequence (unk ).
(4.5)
1216
Ha¨ım Brezis and Augusto C. Ponce
It turns out that the answer is negative, even if the sequence (μn ) is equidiffuse (see Appendix below). Note however that (4.5) is true for some special sequences (μn ), e.g., μn = ρn ∗ μ for some fixed measure μ. A basic tool used in the proof of our main results is the following Theorem 5. Given μ ∈ M(Ω), let un ∈ L1 (Ω) be the solution of −Δun + βn (un ) ρn ∗ μ in Ω, on ∂Ω, un = 0 where (i) (βn ) is a sequence of m.m.g. such that (0, 0) ∈ graph βn ; (ii) (ρn ) is a sequence of mollifiers. If un → u in L1 (Ω), then u is quasicontinuous and cap |un − u| > δ → 0 as n → ∞, ∀δ > 0.
(4.6)
(4.7)
In particular, there exists a subsequence (unk ) such that unk → u q.e.
(4.8)
Warning. The reader might think that the conclusion (4.8) in Theorem 5 holds if (ρn ∗ μ) is replaced by an equidiffuse sequence (μn ) in (4.6). As we have already mentioned, the answer is negative even in the simple case βn ≡ 0 (see Appendix). In order to prove Theorem 5 we need the following Lemma 3. Assume un ∈ L1 (Ω) satisfies −Δun + βn (un ) μn un = 0
in Ω, on ∂Ω,
(4.9)
where (i) (βn ) is a sequence of m.m.g. such that (0, 0) ∈ graph βn ; (ii) (μn ) is a bounded sequence in M(Ω). Suppose (μn ) is ε-equidiffuse in Ω\F for some compact set F ⊂ Ω and some ε > 0. Then, for every open set ω ⊃ F , there exists n0 ≥ 1 such that
μn + Δun n≥n0 is 2ε-equidiffuse in Ω \ ω. (4.10) Proof. Let νn = μn + Δun . By assumption, νn is diffuse and νn ∈ βn (un ). Thus,
sign un νn = |νn |.
Reduced measures for obstacle problems
1217
By Kato’s inequality (see [9]), we have
Δ|un | d ≥ sign un (Δun )d = sign un νn − (μn )d ≥ |νn | − |μn |d and
Δ|un | c = −|Δun |c = −|μn |c .
Thus, −Δ|un | + |νn | ≤ |μn | in Ω. Since |un | ∈ have
W01,1 (Ω),
then for every superharmonic function ζ ∈ C02 (Ω) we ζ |νn | ≤ ζ |μn |. (4.11) Ω
Ω
Given K ⊂ Ω, let ζK denote the capacitary potential of K. By density, (4.11) also holds with ζ = ζK and then we have ζK |νn | ≤ ζK |μn |. (4.12) |νn |(K) ≤ Ω
Ω
Assume by contradiction that (νn ) is not 2ε-equidiffuse in Ω \ ω for n sufficiently large. By inner regularity, there exists (νnj ) and a sequence of compact sets (Kj ) in Ω \ ω such that 3ε (4.13) and cap (Kj ) → 0. 2 Thus, ζKj → 0 in H01 (Ω) and uniformly on F . Passing to a subsequence, we may assume that ζKj → 0 q.e. By Lemma 1, we then have lim sup ζKn |μn | ≤ ε. (4.14) |νnj |(Kj ) ≥
n→∞
Ω
Combining (4.12)–(4.14), we get a contradiction. Therefore, there exists n0 ≥ 1 such that (4.10) holds. The proof of the lemma is complete. Proof of Theorem 5. By standard estimates, we have Δun M ≤ 2ρn ∗ μM ≤ 2μM . Thus, Δu ∈ M(Ω); hence u is quasicontinuous. Moreover, note that (4.8) follows from (4.7) by using a standard diagonalization argument. We now split the proof of (4.7) into two steps: Step 1. Proof of (4.7) when μc is concentrated on a compact set K ⊂ Ω of zero capacity. Let fn = (ρn ∗ μ) + Δun , so that fn ∈ L1 (Ω) and fn ∈ βn (un ) almost everywhere, for all n ≥ 1. We first establish the following
1218
Ha¨ım Brezis and Augusto C. Ponce
Claim 1. For every δ > 0 and for every open set ω ⊃ K, we have |fn | → 0 as n → ∞,
(4.15)
(An ∪Bn )\ω
where
An = 0 ≤ un ≤ u − δ and Bn = u + δ ≤ un ≤ 0 .
(4.16)
We show (4.15) for |fn | integrated over An \ ω; the term coming from Bn \ ω can be estimated in a similar way. We consider two cases, depending on whether u+ is bounded or not: Case 1. u+ ∈ L∞ (Ω). Let M = u+ L∞ . We claim that there exists Cδ > 0 such that if (tn , sn ) ∈ graph βn and tn ≤ M − δ, then sn ≤ Cδ . Assume by contradiction that there exists a sequence (tnk , snk ) ∈ graph βnk such that tnk ≤ M − δ
and snk → ∞.
Since fn L1 ≤ μM and fn ∈ β(un ) almost everywhere, we have [un > M − δ] sn ≤ fnk ≤ C. k k [unk ≥tnk ]
Thus,
[u > M − δ] ≤ lim inf [un > M − δ] = 0. k k→∞
We then deduce that u ≤ M − δ almost everywhere, a contradiction. Therefore, we have 0 ≤ fn ≤ Cδ on An \ ω. Since χAn → 0 in L1 (Ω), we conclude that 0≤ |fn | ≤ Cδ |An | → 0. An \ω
Case 2. u+ ∈ L∞ (Ω). Recall the standard estimate ∇Tk (u)2 ≤ kΔuM
∀k > 0.
Ω
Thus,
1 cap [u > k] ≤ 2 k
Ω
∇Tk (u)2 → 0
as k → ∞.
Reduced measures for obstacle problems
Given η > 0, let k0 ≥ 1 be such that
cap [u > k] < η
1219
∀k ≥ k0 .
Clearly, (ρn ∗ μ) is equidiffuse in Ω \ ω 0 for every open set ω0 ⊃ K. Thus, by Lemma 3, the sequence (fn ) is equidiffuse in Ω \ ω. Therefore, given ε > 0, one can take k0 ≥ 1 large enough such that |fn | < ε ∀k ≥ k0 . (4.17) [u>k]\ω
On the other hand, as in the proof of Case 1 above, there exists Ck0 > 0 such that if (tn , sn ) ∈ graph βn and tn ≤ k0 , then sn ≤ Ck0 . From this and (4.17), we then have |fn | ≤ |fn | + |fn | ≤ Ck0 |An | + ε. An \ω
An ∩[u≤k0 ]
[u>k0 ]\ω
Thus,
lim sup n→∞
|fn | ≤ ε.
An \ω
Since ε > 0 was arbitrary, we conclude that |fn | = 0. lim n→∞ An \ω
The same argument shows that the conclusion holds when integrating |fn | on Bn \ ω. The proof of (4.15) is complete. Claim 2. For every δ > 0 and for every open set ω ⊃ K, we have ∇(un − u)2 → 0 as n → ∞, (4.18) Zn \ω
where Zn = δ < |un − u| < 2δ . Let hδ : R → R be the function given by ⎧ ⎪ 0 if − δ ≤ t ≤ δ, ⎪ ⎪ ⎪ ⎪ ⎪ if t ≥ 2δ, ⎨1 hδ (t) = −1 if t ≤ −2δ, ⎪ ⎪ 1 ⎪ ⎪ δ (t − δ) if δ < t < 2δ, ⎪ ⎪ ⎩ 1 (t + δ) if − 2δ < t < −δ. δ
1220
Ha¨ım Brezis and Augusto C. Ponce
Note that given ψ ∈ C ∞ (Ω), ψ ≥ 0 in Ω, and v ∈ W01,1 (Ω) such that Δv ∈ M(Ω), we have ∇hδ (v)2 ψ ≤ − (Δv)d hδ (v)ψ+ |Δv|c ψ− ∇v·∇ψ hδ (v). (4.19) δ Ω
Ω
Ω
Ω
(This estimate clearly holds if v is smooth; the case of a general v ∈ W01,1 then follows by approximation.) Given open sets ω1 ⊃⊃ ω0 ⊃ K, we take ψ ∈ C ∞ (Ω) such that ψ = 1 on Ω \ ω1 and ψ = 0 on ω 0 . It easily follows from Lemma 3 that (Δu)c is supported in K. Thus, Δ(un − u) ψ = |Δu|c ψ = 0. (4.20) c
We now apply (4.19) to v = un − u. By (4.20), we get ∇hδ (v)2 ψ δ Ω ∇(un − u), (4.21) λn hδ (un − u) + C ≤ − fn hδ (un − u) ψ + Ω Ω Ω where λn = ρn ∗ μ − (Δu)d ψ. The last integral converges to 0 as n → ∞ because un → u in W01,1 (Ω). Since hδ (un − u) 0
weakly in H01 (Ω)
as n → ∞,
and λn = 0 on ω0 , it is not difficult to see that λn hδ (un − u) = 0. lim
(4.22)
n→∞ Ω
(To prove (4.22) one can proceed as in the proof of Proposition 2.1 in [9].) We now show that fn hδ (un − u) ψ ≥ 0. (4.23) lim inf n→∞
Ω
In fact, note that the integrand is ≥ 0 on Ω\(An ∪ Bn ), where An , Bn are given by (4.16). Thus, fn hδ (un − u) ψ ≥ fn hδ (un − u) ψ ≥ − |fn |. Ω
An ∪Bn
(An ∪Bn )\ω 0
By Claim 1 and this estimate, we conclude that (4.23) holds. Combining (4.21)–(4.23), we obtain (4.18) with ω = ω1 . We now conclude the proof of (4.7). Given ε > 0, we fix an open set ω0 ⊃ K satisfying cap (ω 0 ) < ε. Let ψ ∈ C ∞ (Ω) be such that ψ = 1 in
Reduced measures for obstacle problems
1221
Ω \ ω0 and ψ = 0 in some neighborhood of K. Recall that, by standard estimates (see, e.g., [23]), we have in W01,p (Ω) ∀p
2δ . Since hδ (un − u)ψ ≥ 1
(4.24)
(4.25)
on Wn \ ω 0 ,
2 cap (Wn \ ω 0 ) ≤ ∇ hδ (un − u) ψ .
we have
Ω
On the other hand, by subadditivity of cap, cap (Wn ) ≤ cap (Wn \ ω 0 ) + cap (ω 0 ) ≤ cap (Wn \ ω 0 ) + ε. Therefore,
lim sup cap (Wn ) ≤ ε. n→∞
Since ε > 0 was arbitrary, (4.7) follows (with δ replaced by 2δ). Step 2. Proof of the theorem completed. Clearly, it suffices to establish (4.7) for a subsequence (unj ). Let (μk ) be a sequence in M(Ω) so that μk → μ strongly in M(Ω) and so that each measure (μk )c is supported in some compact set of zero capacity. Let vn,k ∈ L1 (Ω) be the solution of −Δvn,k + βn (vn,k ) ρn ∗ μk in Ω, on ∂Ω. vn,k = 0 Clearly, we have Δvn,k M ≤ C. Thus, there exist an increasing sequence of integers (nj ) and a sequence of functions (vk ) in L1 (Ω) such that vnj ,k → vk
in L1 (Ω) as j → ∞,
∀k ≥ 1.
By standard estimates, we have Δ(vn ,k − un ) ≤ 2 ρn ∗ |μk − μ| ≤ 2μk − μM . j j j Ω
Thus,
Ω
Ω
∇Tδ (vn
j ,k
2 − unj ) ≤ 2δμk − μM .
(4.26)
1222
Ha¨ım Brezis and Augusto C. Ponce
As j → ∞, we also have ∇Tδ (vk − u)2 ≤ 2δμk − μM .
(4.27)
Ω
Combining (4.26) and (4.27), we get 4 cap |vnj ,k − unj | > δ + cap |vk − u| > δ ≤ μk − μM . δ Thus, 4 cap |unj − u| > 3δ ≤ cap |vnj ,k − vk | > δ + μk − μM . δ For k ≥ 1 fixed, we apply the previous step to (vnj ,k ). We deduce that 4 ≤ μk − μM . lim sup cap |unj − u| > 3δ δ j→∞ As k → ∞, we obtain (4.7) for the subsequence (unj ). The proof of the theorem is complete. Remark 3. An easy inspection of the proof of Theorem 5 shows that (4.7)– (4.8) are still valid if the sequence (ρn ∗ μ) is replaced be any sequence of measures (μn ) of the form μn = fn + Tn + ρn ∗ ν, where fn ∈
L1 (Ω),
Tn ∈
H −1 (Ω)
satisfy
fn → f in L1 (Ω) and Tn → T in H −1 (Ω), and ν is a measure concentrated on a set of zero H 1 -capacity. We will also need the following variant of Theorem 5: Theorem 6. Given μ ∈ M(Ω), let un ∈ L1 (Ω) be the solution of −Δun + βn (un ) = μ in Ω, un = 0 on ∂Ω,
(4.28)
where βn : R → R is a continuous, nondecreasing function with subcritical growth and such that βn (0) = 0. If un → u in L1 (Ω), then u is quasicontinuous and (4.29) cap |un − u| > δ → 0 as n → ∞, ∀δ > 0. In particular, there exists a subsequence (unk ) such that unk → u q.e.
(4.30)
Reduced measures for obstacle problems
1223
Proof. Let (ρk ) be a sequence of mollifiers. Let (un,k ) be the solution of −Δun,k + βn (un,k ) = ρk ∗ μ in Ω, on ∂Ω. un,k = 0 Since βn has subcritical growth, un,k → un in L1 (Ω) as k → ∞. By the previous theorem, we have cap |un,k − un | > δ → 0 as k → ∞. Thus, one can find an increasing sequence of integers (kn ) such that 1 cap |un,kn − un | > δ ≤ n and un,kn → u in L1 (Ω). 2 Therefore, 1 cap |un − u| > 2δ ≤ cap |un,kn − u| > δ + n . 2 Applying Theorem 5 to the sequence (un,kn ), the result follows. 5. Proof of Theorem 1 We first establish the following Lemma 4. Given any μ ∈ M(Ω), problem (1.1) has at most one solution. Proof. Assume u1 , u2 both satisfy (1.1). We claim that Δ(u1 − u2 )+ ≥ 0
in D (Ω).
(5.1)
In fact, let νi = μ + Δui , for i = 1, 2. Recall that each νi is diffuse. Thus, Δ(u1 − u2 ) is also diffuse. Applying Kato’s inequality (see [9, Theorem 1.1 and Remark 1]), we deduce that Δ(u1 − u2 )+ is a diffuse measure and Δ(u1 − u2 )+ ≥ χ[u1 >u2 ] Δ(u1 − u2 ) = χ[u1 >u2 ] (ν1 − ν2 ) in Ω.
(5.2)
We now observe that χ[u1 >u2 ] (ν1 − ν2 ) ≥ 0
in Ω.
(5.3)
Indeed, recall that (νi )a ∈ β(ui )
a.e.
Thus, χ[u1 >u2 ] (ν1 − ν2 )a ≥ 0
a.e.
(5.4)
1224
Ha¨ım Brezis and Augusto C. Ponce
On the other hand, since (ν2 )s is concentrated on the set [u2 = a] and [u1 > u2 ] ∩ [u2 = a] = ∅, we have (ν2 )s = 0
on [u1 > u2 ].
Thus, (5.5) χ[u1 >u2 ] (ν1 − ν2 )s = χ[u1 >u2 ] (ν1 )s ≥ 0 in Ω. Combining (5.4)–(5.5), we obtain (5.3). It then follows from (5.2)–(5.3) that (5.1) holds. Since u1 , u2 ∈ W01,1 (Ω), we get (see, e.g., [8, Proposition B.1]) (u1 − u2 )+ = 0
a.e.
In other words, u1 ≤ u2 almost everywhere. Reversing the roles of u1 and u2 , we conclude that u1 = u2 almost everywhere. Therefore, problem (1.1) has at most one solution. Lemma 5. Let μ ∈ M(Ω) be such that μ+ is diffuse. Given a sequence (βn ) satisfying (1.13)–(1.15), let un be the solution of (1.16). Then, un → u
in L1 (Ω),
(5.6)
where u satisfies (1.1). In particular, ∗
weak∗ in M(Ω).
βn (un ) μ + Δu
Proof. By standard estimates (see, e.g., [8]), we have βn (un ) ≤ μM and Δun M ≤ 2μM .
(5.7)
(5.8)
Ω
Passing to a subsequence if necessary, one can find u ∈ L1 (Ω) and ν ∈ M(Ω) such that un → u In particular,
∗
in L1 (Ω)
−
and βn (un ) ν
uΔζ + Ω
ζ dν =
Ω
weak∗ in M(Ω).
Ω
ζ dμ ∀ζ ∈ C02 (Ω).
(5.9)
Clearly, the sequence (un ) satisfies the assumptions of Theorem 6. Thus, passing to a further subsequence if necessary, we may also assume that un → u q.e. Our goal is to show that u satisfies (1.1). By Lemma 4, this will imply that the limit u is actually independent of the subsequence, so that the entire sequence converges to the same limit u. We first establish the following
Reduced measures for obstacle problems
1225
Claim. The sequence βn (un ) is equidiffuse. Let vn be the solution of −Δvn + βn (vn ) = μ+ in Ω, on ∂Ω. vn = 0
Since μ+ is diffuse, it follows from Lemma 3 that βn (vn ) is equidiffuse (note that in this case K = ∅). On the other hand, since un ≤ vn almost everywhere and βn (t) = 0, for all t ≤ 0, we have 0 ≤ βn (un ) ≤ βn (vn ) a.e.
We conclude that βn (un ) is also equidiffuse. We can now apply Theorem 4 to conclude that u ≤ a almost everywhere and ν ∈ β(u). In other words, u is the (unique) solution of (1.1). The proof of the lemma is complete. Proof of Theorem 1. (⇐) This follows from Lemma 5 above. By Lemma 4, the solution of (1.1) is unique. (⇒) Assume (1.1) has a solution u. In particular, there exists a diffuse measure ν such that −Δu = μ − ν in D (Ω). Since u ≤ a almost everywhere, it follows from the “Inverse” maximum principle that (−Δu)c ≤ 0 (see [15]). Thus,
μc = (μ − ν)c ≤ 0. In other words, μ+ is diffuse. Corollary 2. Let μi ∈ M(Ω), i = 1, 2, be such that μ+ i is diffuse. Let (ui , νi ) be the solution of (1.1) associated to μi . Then, (ν1 − ν2 )+ ≤ (μ1 − μ2 )+ . (5.10) Ω
Ω
Proof. Let ui,n be the solution of −Δui,n + βn (ui,n ) = μi ui,n = 0
in Ω, on ∂Ω,
where (βn ) satisfies (1.13)–(1.15). Then, by Lemma 5, we have ∗
βn (ui,n ) νi so that
∗
weak∗ in M(Ω),
βn (u1,n ) − βn (u2,n ) ν1 − ν2
weak∗ in M(Ω).
(5.11)
1226
Ha¨ım Brezis and Augusto C. Ponce
On the other hand, by standard estimates (see, e.g., [8, Corollary B.1]) we have + βn (u1,n ) − βn (u2,n ) ≤ (μ1 − μ2 )+ . Ω
Ω
As n → ∞, we conclude that (5.10) holds. 6. Proofs of Proposition 1 and Theorems 2, 3 Proof of Proposition 1. Let v be a subsolution of (1.1). In particular, there exists f ∈ L1 (Ω) such that f ∈ β(u) almost everywhere and −Δv + f ≤ μ in (C02 )∗ . Since v ≤ a almost everywhere, it follows from the “Inverse” maximum principle that (−Δv)c ≤ 0. Thus, −Δv + f ≤ μ∗
in (C02 )∗ .
Let (βn ) be a sequence satisfying (1.13)–(1.15) and such that βn ≤ β, for all n ≥ 1. In particular, −Δv + βn (v) ≤ −Δv + f ≤ μ∗ In other words, v is a subsolution of −Δwn + βn (wn ) = μ∗ wn = 0
in (C02 )∗ .
in Ω, on ∂Ω.
(6.1)
By comparison (see [8, Corollary B.2]), we then have v ≤ wn a.e. ∀n ≥ 1, where wn is the solution of (6.1). Since, by Lemma 5, wn → u∗
in L1 (Ω),
we conclude that v ≤ u∗ almost everywhere. Therefore, u∗ is the largest subsolution of (1.1). Proof of Theorem 2. Since Δun M ≤ 2μM , one can find a subsequence (unk ) such that unk → v
in L1 (Ω),
for some v ∈ L1 (Ω). By Theorem 6, we also have unk → v q.e.
Reduced measures for obstacle problems
1227
∗
Let λ ∈ M(Ω) be such that βnk (unk ) λ weak∗ in M(Ω). We now show that the sequence βnk (unk ) satisfies assumption (iv) of Theorem 4. In fact, given ε > 0, let F ⊂ Ω be a compact set such that cap (F ) = 0 and |μc |(Ω \ F ) < ε. Let ω ⊃ F be an
open set such that cap (ω) < ε. By Lemma 3, the sequence βnk (unk ) is 2ε-equidiffuse in Ω \ ω, as claimed. Applying Theorem 4, we deduce that v ≤ a almost everywhere and λd ∈ β(v). We now show that λc = (μ+ )c . In fact, note that λ ≥ 0.
(6.2)
Let w denote the solution of −Δw = −μ− w=0
in Ω, on ∂Ω.
By comparison, we have w ≤ unk , for all k ≥ 1. Thus, w ≤ v almost everywhere. By the “Inverse” maximum principle, we then have −(μ− )c = (−Δw)c ≤ (−Δv)c = μc − λc .
(6.3)
Similarly, v ≤ a almost everywhere implies μc − λc = (−Δv)c ≤ 0.
(6.4)
Combining (6.2)–(6.4), we conclude that λc = (μ+ )c . In other words, v satisfies
− vΔζ + ζ dν = ζ μ − (μ+ )c = ζ dμ∗ Ω
Ω
Ω
where ν = λd ∈ β(v). Therefore, v = the subsequence (unk ), we must have un → u∗
Ω
u∗ .
∀ζ ∈ C02 (Ω),
Since the limit v is independent of
in L1 (Ω).
This establishes Theorem 2. The proof of Theorem 3 is similar and will be omitted. Note that in this case one should apply Theorem 5 instead of Theorem 6.
1228
Ha¨ım Brezis and Augusto C. Ponce
7. The case of two obstacles We shall assume throughout this section that β is any m.m.g. such that dom β = [−b, a]
and (0, 0) ∈ graph β.
(7.1)
For simplicity, we restrict ourselves to the case where a = b = 1. Given μ ∈ M(Ω), we say that u is a solution of (1.18) if u ∈ L1 (Ω), |u| ≤ 1 almost everywhere, Δu ∈ M(Ω) and there exists a diffuse measure ν ∈ β(u) such that − uΔζ + ζ dν = ζ dμ ∀ζ ∈ C02 (Ω). (7.2) Ω
Ω
Ω
We refer the reader to Section 2 for the definition of ν ∈ β(u) in the case of two obstacles. By Proposition 2, ν ∈ β(u) if and only if j(v) − j(u) ≥ (v − u) dν ∀v ∈ H01 (Ω) ∩ L∞ , (7.3) Ω
Ω
Ω
where ∂j = β and j(0) = 0. The counterpart of Theorem 1 is given by the following Theorem 7. Let μ ∈ M(Ω). Then, (1.18) has a solution if and only if μ is diffuse. Moreover, this solution is unique. Solutions of (1.18) can be also obtained via minimization. More precisely, we have Theorem 8. Given a diffuse measure μ, let u be the solution of (1.18). Then, u ∈ H01 (Ω) and u coincides with the solution of the minimization problem 1 2 min |∇v| + j(v) − v dμ . (7.4) 2 Ω v∈H01 (Ω) Ω Ω |v|≤1 a.e.
Given any measure μ in Ω, let μ∗ = μd denote its diffuse part and let u∗ be the unique solution of −Δu∗ + β(u∗ ) μ∗ in Ω, (7.5) u∗ = 0 on ∂Ω. The analogs of Theorems 2 and 3 are given by the Theorem 9. Let (βn ) be a sequence satisfying (1.13)–(1.14). Given μ ∈ M(Ω), let wn denote the solution of (1.16) or (1.17). Then, wn → u ∗
in L1 (Ω),
Reduced measures for obstacle problems
1229
where u∗ is the solution of (7.5). Finally, we make a connection with another concept of generalized solution which was proposed in Brezis-Serfaty [10]. This is based on a duality principle mentioned in [7]. Given f ∈ L2 (Ω), consider the two minimization problems: 1 2 I1 (f ) = min |∇v| + j(v) − vf , (P) v∈H01 (Ω) 2 Ω Ω Ω |v|≤1 a.e.
I2 (f ) =
min
w∈H01 (Ω) Δw∈M(Ω)
1 2 ∗ |∇w| + j (f + Δw) . 2 Ω Ω
(P∗ )
Here, j ∗ denotes the convex conjugate of j. It is not difficult to prove (see [7]) that (P) and (P∗ ) admit the same minimizers. Moreover, I1 (f ) + I2 (f ) = 0 ∀f ∈ L2 (Ω). These two assertions remain valid when f = μ is a diffuse measure. Note that for general measures, problem (P) is not well posed; however, problem (P∗ ) is well defined and admits a unique minimizer U (μ). The function U (μ) was thus regarded in [10] as a kind of generalized solution for the original problem (P) (or (1.18)). The following result establishes the connection between the concept introduced in [10] and the notion of generalized solution in this paper: Theorem 10. For every μ ∈ M(Ω), we have I2 (μ) = −I1 (μd ) + μc M .
(7.6)
Moreover, the minimizers of I1 (μd ) and I2 (μ) coincide; i.e., U (μ) = u∗ . 8. Proofs of Theorems 7–10 Proofs of Theorems 7 and 9. Step 1. Equation (1.18) has, at most, one solution. The argument is exactly the same as in the proof of Lemma 4. We leave the details to the reader. Step 2. If (1.18) has a solution, then μ is diffuse. Since u is bounded and Δu ∈ M(Ω), we have u ∈ H01 (Ω). This easily implies that Δu is diffuse. Since ν is also diffuse, we deduce that μ = −Δu+ν is diffuse as well. Step 3. Proof of Theorem 9 completed.
1230
Ha¨ım Brezis and Augusto C. Ponce
Given μ ∈ M(Ω), let wn denote the solution of (1.17) (respectively, (1.16)). Thus, for some subsequence (nk ) we have unk → v
in L1 (Ω)
∗
and νnk λ
weak∗ in M(Ω).
By Theorem 5 (respectively, Theorem 6), we then have unk → v q.e. By Lemma 3, the sequence (νnk ) satisfies the assumptions of Theorem 4. We conclude that λd ∈ β(v). We claim that λc = μc . In fact, note that v ∈ H01 (Ω). Therefore, Δv is diffuse. Since −Δv +λ = μ, the claim follows by comparing the concentrated parts of both sides with respect to cap. We conclude that v satisfies (7.5), with ν = λd . By Step 1, the solution of (7.5) is unique. Hence, un → u∗
in L1 (Ω).
Step 4. Proof of Theorem 7 completed. By Step 2, if (1.18) has a solution, then μ is diffuse. Conversely, if μ is diffuse, then the existence of a solution of (1.18) follows from Step 3. By Step 1, the solution is unique. Proof of Theorem 8. Since both problems (1.18) and (7.4) have unique solutions, it suffices to show that if u satisfies (1.18), then u minimizes (7.4). By Proposition 2, we have j(v) − j(u) ≥ (v − u) dν ∀v ∈ H01 (Ω), |v| ≤ 1, (8.1) Ω
Ω
Ω
where ν = μ + Δu. On the other hand, we have 1 1 2 (v − u) Δu = − ∇(v − u) · ∇u ≥ |∇u| − |∇v|2 . (8.2) 2 Ω 2 Ω Ω Ω Combining (8.1) and (8.2), the result follows. Proof of Theorem 10. Note that for every w ∈ H01 (Ω) such that Δw ∈ M(Ω), it follows that Δw is a diffuse measure. Thus, j ∗ (μ + Δw) = j ∗ (μd + Δw) + |μc |. This immediately implies that I2 (μ) = I2 (μd ) + μc M . In particular, the minimizers of I2 (μ) and I2 (μd ) are equal; i.e., U (μ) = U (μd ). Therefore, it suffices to establish the theorem for μ diffuse. We now split the proof into two steps: Step 1. The minimizers of I1 (μ) and I2 (μ) coincide.
Reduced measures for obstacle problems
1231
In view of Theorem 8, it suffices to show that if u is the solution of (1.18), then u minimizes I2 (μ). By the equivalence (a) ⇔ (e) in Proposition 2, we have ∗ ∗ j (σ) − j (μ + Δu) ≥ u (σ − μ − Δu) ∀σ ∈ M(Ω) diffuse. (8.3) Ω
Ω 1 H0 (Ω)
Ω
Let w ∈ with Δw ∈ M(Ω). In particular, Δw is diffuse. Applying (8.3) to σ = μ + Δw, we get 1 |∇w|2 + j ∗ (μ + Δw) 2 Ω Ω 1 2 |∇w| + u(Δw − Δu) + j ∗ (μ + Δu) ≥ 2 Ω Ω Ω 1 2 |∇u| + j ∗ (μ + Δu). ≥ 2 Ω Ω Thus, u is the minimizer of I2 (μ). Step 2. Proof of the theorem completed. It remains to establish (7.6). Since we are assuming that μ is diffuse, and the solution u of (1.18) minimizes both I1 (μ) and I2 (μ), then we only need to show that 2 ∗ |∇u| + u dμ = 0, j(u) + j (μ + Δu) − Ω
Ω
Ω
which is equivalent to assertion (f) in Proposition 2. The proof of Theorem 10 is complete. 9. Appendix Our goal in this Appendix is to show that there are sequences such that (un ) is uniformly bounded and (Δun ) is equidiffuse, but (un ) need not have any subsequence converging q.e. For simplicity, we will assume throughout this Appendix that N ≥ 3. We first establish the following Proposition 5. Let (μn ) ⊂ M(RN ) be a sequence of nonnegative measures. Let dμn (y) 1 vn (x) = ∀x ∈ RN . (9.1) (N − 2)|B1 | RN |x − y|N −2 If (vn ) is uniformly bounded, then (μn ) is equidiffuse. Proof. Clearly, it suffices to show that for every Borel set A ⊂ RN we have μn (A) ≤ C cap (A) ∀n ≥ 1.
(9.2)
1232
Ha¨ım Brezis and Augusto C. Ponce
We only need to establish (9.2) when A = K is a compact set. Let ζK denote the capacitary potential of K. Since vn is bounded, μn is a diffuse measure. Thus, μn (K) ≤ ζK dμn RN =− ζK Δvn = − vn ΔζK ≤ vn L∞ |ΔζK | ≤ C cap (K). RN
RN
RN
This establishes (9.2) when A is compact; the general case easily follows by approximation. In conjunction with Proposition 5, we recall that there is an example of functions (wn ) constructed by Cioranescu-Murat [12, Example 2.1] such that (i) (ii) (iii) (iv)
0 ≤ wn ≤ 1 a.e. in Ω; (Δwn ) is equidiffuse; wn 1 weakly in H 1 (Ω); for every ω ⊂⊂ Ω, we have
lim inf cap [wn = 0] ∩ ω > 0. n→∞
Properties (i), (iii) and (iv) are clear from their construction. We only need to check that (ii) holds. By construction, (Δwn ) is bounded in M(Ω). Write −Δwn = μn − νn , where μn , νn ≥ 0 and μn , νn are mutually singular. We then extend μn , νn to be identically zero outside Ω. Let vn be defined as in (9.1). An easy inspection from [12] shows that (vn ) is uniformly bounded in RN . By Proposition 5, (μn ) is equidiffuse. Similarly, (νn ) is also equidiffuse. We deduce that (ii) holds. It now follows from (iii) and (iv) that (wn ϕ) has no subsequence con
verging q.e., for any ϕ ∈ Cc∞ (Ω), ϕ ≡ 0. Note however that Δ(wn ϕ) is equidiffuse in view of (ii). Acknowledgments. The first author (H.B.) is partially sponsored by an EC Grant through the RTN Program “Front-Singularities”, HPRN-CT2002-00274. H.B. is also a member of the Institut Universitaire de France. The second author (A.C.P.) is supported by the NSF grant DMS-0111298 and Sergio Serapioni, Honorary President of Societ` a Trentina Lieviti—Trento (Italy).
Reduced measures for obstacle problems
1233
References [1] A. Ancona, Une propri´et´e d’invariance des ensembles absorbants par perturbation d’un op´erateur elliptique, Comm. Partial Differential Equations, 4 (1979), 321–337. [2] H. Attouch, “Variational Convergence for Functions and Operators,” Applicable Mathematics Series, Pitman, Boston, MA, 1984. [3] J. R. Baxter, Inequalities for potentials of particle systems, Illinois J. Math., 24 (1980), 645–652. [4] Ph. B´enilan and H. Brezis, Nonlinear problems related to the Thomas-Fermi equation, J. Evol. Equ., 3 (2004), 673–770. Dedicated to Ph. B´enilan. [5] L. Boccardo, T. Gallou¨et, and L. Orsina, Existence and uniqueness of entropy solutions for nonlinear elliptic equations with measure data, Ann. Inst. H. Poincar´e Anal. Non Lin´eaire, 13 (1996), 539–551. [6] H. Brezis, Int´egrales convexes dans les espaces de Sobolev, Israel J. Math., 13 (1972), 9–23 (1973). Proceedings of the International Symposium on Partial Differential Equations and the Geometry of Normed Linear Spaces (Jerusalem, 1972). [7] H. Brezis, Probl`emes unilat´eraux, J. Math. Pures Appl., 51 (1972), 1–168. [8] H. Brezis, M. Marcus, and A. C. Ponce, Nonlinear elliptic equations with measures revisited, to appear in Annals of Math. Studies, Princeton University Press. Part of the results were announced in a note by the same authors: A new concept of reduced measure for nonlinear elliptic equations, C. R. Acad. Sci. Paris, Ser. I, 339 (2004), 169–174. [9] H. Brezis and A. C. Ponce, Kato’s inequality when Δu is a measure, C. R. Math. Acad. Sci. Paris, Ser. I, 338 (2004), 599–604. [10] H. Brezis and S. Serfaty, A variational formulation for the two-sided obstacle problem with measure data, Commun. Contemp. Math., 4 (2002), 357–374. [11] H. Brezis and W. A. Strauss, Semilinear second-order elliptic equations in L1 , J. Math. Soc. Japan 25 (1973), 565–590. [12] D. Cioranescu and F. Murat, A strange term coming from nowhere, In: Topics in the mathematical modelling of composite materials. Progr. Nonlinear Differential Equations Appl., vol. 31, Birkh¨ auser, Boston, 1997, pp. 45–93. [13] P. Dall’Aglio and G. Dal Maso, Some properties of the solutions of obstacle problems with measure data, Ricerche Mat., 48 (1999), 99–116. Papers in memory of Ennio De Giorgi. [14] P. Dall’Aglio and C. Leone, Obstacles problems with measure data and linear operators, Potential Anal. 17 (2002), 45–64. [15] L. Dupaigne and A. C. Ponce, Singularities of positive supersolutions in elliptic PDEs, Selecta Math. (N.S.), 10 (2004), 341–358. [16] L. Dupaigne, A. C. Ponce, and A. Porretta, Elliptic equations with vertical asymptotes in the nonlinear term, to appear in J. Anal. Math. [17] M. Fukushima, K. Sato, and S. Taniguchi, On the closable parts of pre-Dirichlet forms and the fine supports of underlying measures, Osaka J. Math., 28 (1991), 517–535. [18] C. Goffman and J. Serrin, Sublinear functions of measures and variational integrals, Duke Math. J., 31 (1964), 159–178. [19] M. Grun-Rehomme, Caract´erisation du sous-diff´ erential d’int´egrandes convexes dans les espaces de Sobolev, J. Math. Pures Appl., 56 (1977), 149–156.
1234
Ha¨ım Brezis and Augusto C. Ponce
[20] L. Orsina and A. Prignet, Non-existence of solutions for some nonlinear elliptic equations involving measures, Proc. Roy. Soc. Edinburgh Sect. A, 130 (2000), 167–187. [21] R. T. Rockafellar, Integrals which are convex functionals, Pacific J. Math., 24 (1968), 525–539. [22] R. T. Rockafellar, Integrals which are convex functionals, II, Pacific J. Math., 39 (1971), 439–469. ´ [23] G. Stampacchia, “Equations elliptiques du second ordre ` a coefficients discontinus, S´eminaire de Math´ematiques Sup´erieures,” no. 16 (´et´e, 1965), Les Presses de l’Universit´e de Montr´eal, Montr´eal, Qu´ebec, 1966.
