Elliptic equations with vertical asymptotes in the

Elliptic equations with vertical asymptotes in the nonlinear term Louis Dupaigne, Augusto C. Ponce and Alessio Porretta June 22, 2005 Abstract. We stu...

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Elliptic equations with vertical asymptotes in the nonlinear term Louis Dupaigne, Augusto Ponce, Alessio Porretta

To cite this version: Louis Dupaigne, Augusto Ponce, Alessio Porretta. Elliptic equations with vertical asymptotes in the nonlinear term. Journal d’analyse mathématique, Springer, 2005, ?.

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Elliptic equations with vertical asymptotes in the nonlinear term Louis Dupaigne, Augusto C. Ponce and Alessio Porretta June 22, 2005 Abstract. We study the existence of solutions of the nonlinear problem ( −∆u + g(u) = µ in Ω, u=0

(0.1)

on ∂Ω,

where µ is a bounded measure and g is a continuous nondecreasing function such that g(0) = 0. In this paper, we assume that the nonlinearity g satisfies lim g(t) = +∞. t↑1

(0.2)

Problem (0.1) need not have a solution for every measure µ. We prove that, given µ, there exists a “closest” measure µ∗ for which (0.1) can be solved. We also explain how assumption (0.2) makes problem (0.1) different compared to the case where g(t) is defined for every t ∈ R. (Paper to appear in J. Anal. Math.) Mathematics Subject Classification (2000). 28A78, 35B05, 35B45, 35B65, 35J60 Key words. Elliptic equations involving measures, nonlinearities with vertical asymptotes, reduced measures

1

Introduction

Let Ω ⊂ RN , N ≥ 2, be a smooth bounded domain. In this paper, we are interested in the existence of solutions of the following problem ( −∆u + g(u) = µ in Ω, (1.1) u = 0 on ∂Ω, where µ is a bounded measure in Ω and g : (−∞, 1) → R is a continuous nondecreasing function such that g(0) = 0 and lim g(t) = +∞. t↑1

1

(1.2)

By a solution u of (1.1) we mean that u ∈ L1 (Ω), u ≤ 1 a.e., g(u) ∈ L1 (Ω) and Z Z Z − u∆ζ + g(u)ζ = ζ dµ ∀ζ ∈ C 2 (Ω), ζ = 0 on ∂Ω. Ω





In particular, g(u) ∈ L1 (Ω) implies that u < 1 a.e. We observe that u, whenever it exists, is unique (see e.g. [4]). It has been proved by Boccardo [2] (in the spirit of Brezis-Strauss [7]) that, for every µ ∈ L1 (Ω), problem (1.1) has a solution. Moreover, Boccardo also shows that (1.1) has no solution if µ is a Dirac mass δa , with a ∈ Ω. Consequently, we say that µ is a good measure (relative to g) if (1.1) has a solution u. We shall denote by G(g) the set of good measures associated to g. Our goal in this paper is to investigate under what conditions on g and µ problem (1.1) admits a solution. We also point out to what extent assumption (1.2) makes this problem different compared to the case where g is a continuous function defined for every t ∈ R, which was recently studied by Brezis-MarcusPonce [4]. We shall assume henceforth that, in addition to (1.2), g satisfies ∀t ≤ 0.

g(t) = 0

(1.3)

In particular, this implies that nonpositive measures are good for any g. We denote by M(Ω) the space of bounded Radon measures in Ω, equipped with its standard norm k kM . Given ν ∈ M(Ω), we say that ν is diffuse if ν(A) = 0 for every Borel set A ⊂ Ω of zero H 1 -capacity (= Newtonian capacity). As we shall see, this capacity — which will be denoted throughout this paper by “cap” — plays an important role in the study of problem (1.1). The first consequence of (1.2) is that if (1.1) has a solution, then µ+ is diffuse (see Corollary 2 in Section 2 below). The converse is not true; more precisely, Theorem 1 Given any g, there exists a diffuse measure µ ≥ 0 such that µ 6∈ G(g). However, we shall see later on that every diffuse measure is good for some g (see Theorem 15). For a fixed nonlinearity g, a natural question is to characterize the set of good measures associated to g. The next result gives a sufficient condition for a measure to be good:

2

Theorem 2 Assume n o 2−β lim sup (1 − t) β g(t) > 0

(1.4)

t↑1

for some 0 < β < 2. If µ+  HN −2+β , then µ ∈ G(g). Here, Hs denotes the s-dimensional Hausdorff measure of a set. By µ+  Hs , we mean that µ+ (A) = 0 for every Borel set A ⊂ Ω such that Hs (A) = 0. We point out that the dimension s = N −2+β in the statement of the theorem cannot be improved. In fact, given β ∈ (0, 2), let g(t) =

1 (1 − t)

2−β β

− 1 ∀t ∈ [0, 1).

For any α < N −2+β, one can find a compact set Kα ⊂ Ω, with Hα (Kα ) ∈ (0, ∞), such that if θ > 0 is sufficiently large, then µ = θ Hα bKα is not good for g. This is easy to see if α ≤ N − 2 since in this case any compact set K ⊂ Ω such that Hα (K) < ∞ satisfies cap (K) = 0 (see e.g. [8]); thus, by Corollary 2 in Section 2, µ is not good. In the remaining case, namely N −2 < α < N −2+β, the construction of Kα is rather delicate and will be presented in Section 8 (see Theorem 18). Even though the existence of solutions of problem (1.1) may fail for some diffuse measures (by Theorem 1), L1 (Ω) is not the largest set where (1.1) has a solution for any g. For instance, let µ ∈ M(Ω) be such that v ≤ 1 a.e., where v is the unique solution of ( −∆v = µ in Ω, (1.5) v = 0 on ∂Ω. Then, µ is good for every g (see Proposition 7 in Section 7). The converse is also true if µ+ is singular with respect to the Lebesgue measure in RN . In fact, we have the following Theorem 3 Let µ ∈ M(Ω) be such that µ+ is singular. Then, µ ∈ G(g) for every g if and only if v ≤ 1 a.e., where v is given by (1.5). The characterization of the set of all measures in M(Ω) which are good for every g will be given in Section 7. Our method in the study of problem (1.1) starts with a standard procedure which consists in approximating g with bounded continuous functions defined on

3

the whole R. More precisely, let (gn ) be a sequence of bounded functions gn : R → R which are continuous, nondecreasing and satisfy the following conditions: 0 ≤ g1 (t) ≤ g2 (t) ≤ . . . ∀t ∈ R,

(1.6)

gn (t) → g(t)

∀t < 1

(1.7)

gn (t) → +∞

∀t ≥ 1.

(1.8)

and

Since each gn is bounded, there exists a unique solution un of ( −∆un + gn (un ) = µ in Ω, un = 0

(1.9)

on ∂Ω.

Passing to the limit as n tends to infinity we get the following result: Proposition 1 Given any µ ∈ M(Ω), then un ↓ u∗ in Ω as n ↑ +∞, where u∗ is the largest subsolution of (1.1). Moreover, we have kg(u∗ )kL1 ≤ kµkM and

Z ∗ ≤ 2kµkM kζkL∞ u ∆ζ

∀ζ ∈ C02 (Ω).

(1.10)

(1.11)



Here, we denote by n o C02 (Ω) = ζ ∈ C 2 (Ω) : ζ = 0 on ∂Ω . In the spirit of [4], we then define the reduced measure µ∗ as µ∗ = −∆u∗ + g(u∗ ), and we study the properties of µ∗ . First of all, since u∗ is the largest subsolution of (1.1), µ∗ is well-defined, independently of the sequence (gn ). Note that µ∗ ≤ µ; moreover, µ is a good measure if and only if µ = µ∗ . We have the following Theorem 4 For every µ ∈ M(Ω), there exist Borel sets Σ1 , Σ2 ⊂ Ω such that Σ1 ⊂ [u∗ = 1],

cap (Σ2 ) = 0,

and

4

 (µ − µ∗ ) Ω \ (Σ1 ∪ Σ2 ) = 0.

(1.12)

Note that, in the previous statement, the set [u∗ = 1] is well-defined up to sets of zero H 1 -capacity. Indeed, any function v ∈ L1 (Ω) such that ∆v ∈ M(Ω) admits a unique cap-quasicontinuous representative v˜ (see e.g. [1]); henceforth, we shall always identify v and v˜. We recall that v˜ is cap-quasicontinuous if, for every ε > 0, there exists an open set ωε ⊂ Ω such that cap (ωε ) < ε and v˜|Ω\ωε is continuous. We remark that, in Theorem 4, both sets Σ1 and Σ2 have zero Lebesgue measure, so that we deduce the following Corollary 1 For any measure µ, we have (µ∗ )a = µa , where “a” denotes the absolutely continuous part with respect to the Lebesgue measure.  In view of Theorem 4, if µ is diffuse and µ [u∗ = 1] = 0, then it follows that µ∗ = µ, hence µ is good. We use this idea in order to prove Theorem 2; in this case the main effort is thus to estimate the (N − 2 + β)-Hausdorff measure of the set [u∗ = 1]. This kind of estimate, which has an interest in its own, is given by Theorem 12 in Section 4. In Section 5, we present another approach based on energy estimates; in this case, the “smallness” of [u∗ = 1] is given in terms of (Sobolev) capacities. The next result says that µ∗ is the “best approximation” of µ in the class of good measures relative to g. More precisely, Theorem 5 Given µ ∈ M(Ω), we have kµ − µ∗ kM = min kµ − νkM . ν∈G

(1.13)

In addition, µ∗ is the unique good measure for which the minimum in (1.13) is attained. We recall that when the function g is defined for every t ∈ R, it has been shown in [4] that µ∗ is the largest good measure ≤ µ. In that case, the characterization of µ∗ given in Theorem 5 is then a straightforward consequence. We stress the following important difference in our case, namely there exist measures µ for which  the set λ ∈ G(g) : λ ≤ µ has no largest element (see Proposition 9 in Section 9). Thus, the fact that µ∗ is the unique measure which achieves the minimum in (1.13) needs a direct proof, which is more delicate. 5

Finally, two further differences with the case studied in [4] are worth being mentioned. When g(t) is defined for every t ∈ R, the set G of good measures is convex, and the mapping µ 7→ µ∗ is a contraction. As we shall see in Section 9 below, these properties are no longer true when g satisfies (1.2). In fact, for any such g we have (a) G is not convex; (b) the mapping µ 7→ µ∗ is not a contraction. We would like to emphasize that throughout this paper we assume that Ω is a domain of RN , with N ≥ 2. The case of dimension N = 1 is different and has been studied by V´ azquez [20]. We recall that in this case every measure is diffuse — since cap ({x}) > 0 for every x — and the solutions of (1.1) are Lipschitz continuous. In [20], V´ azquez proves that Z 1 (a0 ) if g = +∞, then every µ ∈ M(Ω) is good; 0 0

Z

1 +

g < +∞ and µ ∈ M(Ω) satisfies kµ kM

(b ) if

√ ≤2 2

Z

1/2

1

g

, then µ is

0

0

good. These two results have no counterpart when N ≥ 2. According to Theorem 1 above, for any g there exists a diffuse measure µ ≥ 0 such that µ is not good. As we shall see in Section 8, such µ can be chosen so that εµ is not good for any ε > 0. The plan of this paper is the following: 1. Introduction; 2. Proofs of Proposition 1 and Theorem 4; 3. The reduced measure is the closest good measure; 4. Proof of Theorem 2; 5. Capacitary estimates related to problem (1.1); 6. Every diffuse measure is good for some g; 7. Measures which are good for every g; 6

8. How to construct diffuse measures which are not good; 9. Further properties of µ∗ and G; References.

2

Proofs of Proposition 1 and Theorem 4

We start by recalling that every measure µ can be uniquely decomposed as (see e.g. [16]) µ = µd + µc , where µd is diffuse and µc is concentrated on a set of zero capacity. In particular, µ is diffuse if and only if µc = 0. A useful characterization of measures which are diffuse is given by the following Theorem 6 ([3, 17]) Let µ ∈ M(Ω). Then, µ is diffuse if and only if µ ∈ L1 (Ω) + H −1 (Ω). The next two results will be often used in this paper: Theorem 7 ([6]) Let v ∈ L1 (Ω) be such that ∆v ∈ M(Ω). Then, ∆v + ∈ Mloc (Ω) and (∆v + )d ≥ χ[v≥0] (∆v)d +

(−∆v )c =

(−∆v)+ c

in Ω,

(2.1)

in Ω.

(2.2)

Moreover, if v ≥ 0 a.e., then (∆v)d ≥ 0

in [v = 0].

(2.3)

Theorem 8 ([14]) Let v ∈ L1 (Ω) be such that ∆v ∈ M(Ω). If v ≥ 0 a.e., then (∆v)c ≤ 0

in Ω.

(2.4)

As a result, we get a necessary condition in order that (1.1) admit a solution. Corollary 2 If µ is good, then µ+ is diffuse.

7

Proof. Applying Theorem 8 to v = 1 − u we get µc = (−∆u)c = (∆v)c ≤ 0. Thus, µ+ = (µd )+ = (µ+ )d and so µ+ is diffuse. Let us also recall that, given ν ∈ M(Ω), there exists a unique function v ∈ L (Ω) which satisfies Z Z − v∆ζ = ζ dν ∀ζ ∈ C02 (Ω). 1





This function is called Stampacchia’s solution of the problem (see [19]) ( −∆v = ν in Ω, v=0

(2.5)

on ∂Ω,

and it coincides with the notion of “renormalized solution” introduced in [9]. In particular, Theorems 2.33 and 10.1 of [9] provide the following useful Theorem 9 ([9]) Let v be the unique solution of (2.5). Let Φ ∈ W 2,∞ (R) be such that supp Φ00 is compact. Then, we have 0 + ∆Φ(v) = Φ0 (v)(∆v)d + Φ00 (v)|∇v|2 − Φ0 (+∞)(∆v)− c + Φ (−∞)(∆v)c

in Ω.

Here, we denote by Φ0 (±∞) the limit of Φ0 as |x| → ±∞. The proof of Proposition 1 follows along the same lines as in [4]. Below, we present the proof for the convenience of the reader. Proof of Proposition 1. Let un be the solution of (1.9). Since gn ≤ gn+1 , by a comparison principle (see e.g. [4, Appendix B]) we have un ≥ un+1 . Hence, we define u∗ such that un ↓ u∗ a.e. in Ω. Standard estimates imply that kgn (un )kL1 ≤ kµkM ;

(2.6)

thus, Z Z Z u ∆ζ = ζ dµ − g (u )ζ n n n ≤ 2kµkM kζkL∞ Ω





∀ζ ∈ C02 (Ω).



Clearly, u ∈ L (Ω) and (un ) converges strongly to u∗ in L1 (Ω). Moreover, it follows from (2.6) that u∗ ≤ 1 a.e. Then, by using Fatou’s lemma, we deduce from the previous estimates that 1

8

(i) g(u∗ ) ∈ L1 (Ω) and (1.10) holds; (ii) ∆u∗ ∈ M(Ω) and k∆u∗ kM ≤ 2kµkM . Finally, let v be any subsolution of (1.1), i.e. v ∈ L1 (Ω), v ≤ 1 a.e., g(v) ∈ L1 (Ω) and Z Z Z − v∆ζ + g(v)ζ ≤ ζ dµ ∀ζ ∈ C02 (Ω), ζ ≥ 0 in Ω. Ω





Since gn ≤ g, we have −∆v + gn (v) ≤ −∆v + g(v) ≤ µ = −∆un + gn (un )

 ∗ in C02 (Ω) ,

which yields v ≤ un a.e. Passing to the limit, we deduce that v ≤ u∗ . This proves that u∗ is the largest subsolution of (1.1). Let µ∗ = −∆u∗ + g(u∗ )

in D0 (Ω).

(2.7)

In view of Proposition 1, µ∗ ∈ M(Ω). The reduced measure µ∗ is uniquely determined by the weak∗ limit of gn (un ) in M(Ω). Indeed, comparing (2.7) with (1.9), and using that un → u∗ in L1 (Ω), we obtain the following Lemma 1 Let (un ) be the sequence defined in Proposition 1. Then, ∗

gn (un ) * g(u∗ ) + (µ − µ∗ )

weak∗ in M(Ω).

(2.8)

Note that, since ∆u∗ ∈ M(Ω), the function u∗ admits a unique cap-quasicontinuous representative, which we are going to use henceforth as our standard choice. In particular, we remark that the set [u∗ = 1] is uniquely defined up to sets of zero capacity. The main ingredient in the proof of Theorem 4 is the next Proposition 2 Let u∗ be given by Proposition 1 and let µ∗ be the reduced measure defined in (2.7). Then, we have 0 ≤ µ − µ∗ ≤ (µd )b[u∗ =1] + µ+ c

in Ω.

(2.9)

In particular,   (µ∗ )d = µd      (µ∗ )d ≥ 0  (µ∗ )c = −(µc )−      (µ∗ )− = µ−

9

in [u∗ < 1], in [u∗ = 1], in Ω, in Ω.

(2.10)

Proof. Step 1. Proof of (2.9).  Given δ > 0, let us define the function θδ (s) = min 1, 1δ (s − 1 + 2δ)+ . ApRs plying Theorem 9 with v = un and Φδ (s) = 0 θδ (ξ) dξ we get, for any ζ ∈ C02 (Ω), ζ ≥ 0 in Ω, Z Z Z Z gn (un ) θδ (un ) ζ dx ≤ θδ (un ) ζ dµd + ζ dµ+ + Φδ (un )∆ζ dx, (2.11) c Ω



which yields Z



Z gn (un ) ζ dx ≤



Z

ζ dµ+ c +

θδ (un ) ζ dµd + Ω



Z Φδ (un )∆ζ dx. Ω

[un >1−δ] ∞ Since (u+ n ) is bounded in L (Ω) and (∆un ) is bounded in M(Ω), the sequence (θδ (un )) is bounded in H01 (Ω), converges weakly to θδ (u∗ ) in H01 (Ω) and weak∗ in L∞ (Ω). Moreover, since µd ∈ L1 (Ω) + H −1 (Ω) (by Theorem 6), we have Z Z lim θδ (un ) ζ dµd = θδ (u∗ ) ζ dµd . n→+∞





Thus, Z

Z

θδ (u∗ ) ζ dµd +

gn (un ) ζ dx ≤

lim sup n→+∞



Z

ζ dµ+ c +



Z

Φδ (u∗ )∆ζ dx.



[un >1−δ]

Clearly, by dominated convergence we have, for a.e. δ > 0, Z Z lim gn (un ) ζ dx = g(u∗ ) ζ dx. n→+∞ [un ≤1−δ]

[u∗ ≤1−δ]

Therefore, Z lim sup gn (un ) ζ dx ≤ n→+∞ Ω Z Z Z Z ∗ ∗ + ≤ g(u ) ζ dx + θδ (u ) ζ dµd + ζ dµc + Φδ (u∗ )∆ζ dx, [u∗ ≤1−δ]







for a.e. δ > 0. Since u∗ < 1 a.e., we have Φδ (u∗ ) → 0 a.e. By dominated convergence, as δ → 0 we obtain Z Z Z Z lim sup gn (un ) ζ dx ≤ g(u∗ ) ζ dx + ζ dµd + ζ dµ+ c . n→+∞





[u∗ =1]

10



Comparing with (2.8) we get µ − µ∗ ≤ (µd )b[u∗ =1] + µ+ c . Clearly, by Fatou’s lemma, µ∗ ≤ µ. We thus obtain (2.9). Step 2. Proof of (2.10). From (2.9) we immediately deduce that (µ∗ )d = µd

in [u∗ < 1],

(µ∗ )d ≥ 0

in [u∗ = 1].

(2.12)

Since µd ≥ (µ∗ )d , (2.12) yields (µd )− = (µ∗ )− d

in Ω.

(2.13)

On the other hand, by (2.9), µc − (µ∗ )c ≤ µ+ c

in Ω;

that is (µ∗ )c ≥ −µ− c

in Ω.

Note that µ∗ ≤ µ and (µ∗ )c ≤ 0 (by Corollary 2); thus, (µ∗ )c ≤ −µ− c

in Ω.

(µ∗ )c = −µ− c

in Ω.

We deduce that (2.14)

Assertion (2.10) then follows from (2.12)–(2.14). As a consequence of the previous result we have the Proof of Theorem 4. Let Σ1 = [u∗ = 1] and let Σ2 ⊂ Ω be such that cap (Σ2 ) = 0 and µ+ c (Ω\Σ2 ) = 0. With this choice, (1.12) follows immediately from (2.9).

As a corollary of (2.10) we also have the Corollary 3 Let µ ∈ M(Ω). If µ ≥ 0, then µ∗ ≥ 0.

11

We give an alternative characterization of u∗ in the next Proposition 3 For every µ ∈ M(Ω), u∗ is the unique solution of the following problem:   v ∈ W01,1 (Ω), v ≤ 1 a.e., ∆v ∈ M(Ω), g(v) ∈ L1 (Ω),      (−∆v) + g(v) = µ in [v < 1], d d (2.15)   (−∆v)d ≤ µd in [v = 1],     (−∆v)c = −µ− in Ω. c Proof. From (2.7) and (2.10), it follows that u∗ is indeed a solution of (2.15). We now prove that the solution of (2.15) is unique. Assume that v1 , v2 both satisfy (2.15) and consider the function w = (v1 − v2 )+ . We first observe that ∆w is a measure and, by (2.3), (∆w)d ≥ 0 in [v1 ≤ v2 ]. (2.16) By (2.15), (∆v1 )d ≥ g(v1 ) − µd

in Ω.

(2.17)

On the other hand, since [v1 > v2 ] ⊂ [v2 < 1], we have (∆v2 )d = g(v2 ) − µd

in [v1 > v2 ].

(2.18)

Thus, by (2.17)–(2.18),   (∆w)d ≥ ∆(v1 − v2 ) d ≥ g(v1 ) − g(v2 ) ≥ 0

in [v1 > v2 ].

(2.19)

We deduce from (2.16) and (2.19) that (∆w)d ≥ 0

in Ω.

(2.20)

Since, by (2.2),  + (−∆w)c = − ∆(v1 − v2 ) c

in Ω,

we get  + (−∆w)c = (−∆v1 )c + (∆v2 )c = 0

in Ω.

(2.21)

From (2.20)–(2.21) we obtain that ∆w ≥ 0

in Ω.

Since w vanishes on ∂Ω, we have v1 ≤ v2 a.e. in Ω (see e.g. [4, Proposition B.1]). Reversing the roles of the two functions we finally obtain that v1 = v2 . 12

Until now, we have studied problem (1.1) by approximating the nonlinearity g using a sequence (gn ), with µ fixed. Another possible approach is to fix g and to approximate µ by ρn ∗ µ, where (ρn ) is a sequence of mollifiers. More precisely, ρn ∗ µ is given by Z (ρn ∗ µ)(x) = ρn (x − y) dµ(y) ∀x ∈ Ω. Ω

It turns out that the sequences of solutions in both cases converge to the same limit. More precisely, Theorem 10 Let µ ∈ M(Ω). For each n ≥ 1, let vn be the solution of ( −∆vn + g(vn ) = ρn ∗ µ in Ω, vn = 0

(2.22)

on ∂Ω.

Then, vn → u∗ in L1 (Ω), where u∗ is the function given by Proposition 1. Proof. By standard estimates we have kg(vn )kL1 (Ω) ≤ kρn ∗ µkM(Ω) ≤ kµkM(Ω) . Thus, ∆vn is bounded in L1 (Ω) and there exist v ∈ L1 (Ω) and ν ∈ M(Ω) such that, for a subsequence (still denoted (vn )), we have vn → v

strongly in L1 (Ω) and a.e.



g(vn ) * g(v) + ν

weak∗ in M(Ω).

By Fatou’s lemma, we have ν ≥ 0. Moreover, it follows that v satisfies −∆v + g(v) = µ − ν

in Ω.

(2.23)

We now follow the outline of the proof of Proposition 2. Take θδ (vn )ζ as a test function in (2.22), where ζ ∈ C02 (Ω). We get the analog of (2.11), namely Z Z Z Z + g(vn ) θδ (vn ) ζ dx ≤ θδ (vn ) (ρn ∗µd ) ζ dx+ (ρn ∗µc ) ζ dx+ Φδ (vn )∆ζ dx. Ω







As in Step 1 of Proposition 2 we obtain, as n tends to infinity, that ν ≤ (µd )b[v=1] + µ+ c

in Ω.

(2.24)

Thanks to (2.23)–(2.24), we obtain that v is a solution of (2.15) (see Step 2 of Proposition 2). From the uniqueness result of Proposition 3, we conclude that v = u∗ . In particular, the whole sequence (vn ) converges to u∗ .

13

3

The reduced measure is the closest good measure

We start with the following simple result: Proposition 4 Let µ ∈ M(Ω). If µ ∈ G(g) and ν ≤ µ, then ν ∈ G(g). Proof. Let (un ) be the sequence of functions satisfying (1.9). By standard estimates (see e.g. [4]), we have Z Z gn (un ) − gn (u) ≤ g(u) − gn (u) . Ω



Thus, Z

gn (un ) − g(u) ≤ 2



Z

g(u) − gn (u) → 0

as n → +∞.



We conclude that gn (un ) → g(u) in L1 (Ω). Let (vn ) be the sequence associated to ν. By comparison, ν ≤ µ implies vn ≤ un a.e.; thus, gn (vn ) ≤ gn (un ) a.e. Applying the Dominated convergence theorem, we conclude that gn (vn ) → g(v ∗ ) in L1 (Ω). We then deduce that v ∗ is the solution of (1.1) with data ν, and so ν is a good measure. We also have the following Lemma 2 For every µ, ν ∈ M(Ω), we have Z

∗ g(u ) − g(v ∗ ) + (µ − µ∗ ) − (ν − ν ∗ )

M



≤ kµ − νkM ,

(3.1)

where u∗ , v ∗ are the solutions of (1.1) with respect to µ∗ , ν ∗ , resp. Proof. Let vn denote the solution of ( −∆vn + gn (vn ) = ν vn = 0

in Ω, on ∂Ω.

By Lemma 1, we get ∗

gn (un ) − gn (vn ) * g(u∗ ) − g(v ∗ ) + (µ − µ∗ ) − (ν − ν ∗ )

weak∗ in M(Ω).

Since µ − µ∗ and ν − ν ∗ are both singular with respect to the Lebesgue measure (see Corollary 1), we have Z





g(u )−g(v ∗ )+(µ−µ∗ )−(ν −ν ∗ ) = g(u )−g(v ∗ ) + (µ−µ∗ )−(ν −ν ∗ ) . M M Ω

14

On the other hand,

gn (un ) − gn (vn ) 1 ≤ kµ − νkM L Thus, Z



g(u ) − g(v ∗ ) + (µ − µ∗ ) − (ν − ν ∗ )

M



∀n ≥ 1.

≤ lim inf gn (un ) − gn (vn ) L1 n→+∞

≤ kµ − νkM , which gives (3.1). Our next result is the following Theorem 11 For every µ, ν ∈ M(Ω), we have kµ∗ − ν ∗ kM ≤ 2kµ − νkM .

(3.2)

Proof. Let µ, ν ∈ M(Ω). By Lemma 2, we have

(µ − µ∗ ) − (ν − ν ∗ ) ≤ kµ − νkM . M Applying the triangle inequality, we obtain (3.2). We now present the Proof of Theorem 5. We shall split the proof into two steps. Step 1. Proof of (1.13). Given ν ∈ G, we have ν = ν ∗ . It then follows from Lemma 2 that Z ∗ g(u ) − g(v) + kµ − µ∗ kM ≤ kµ − νkM ,

(3.3)



where v is the solution of (1.1) with measure ν. In particular, kµ − µ∗ kM ≤ kµ − νkM , which gives (1.13). Step 2. µ∗ is the unique good measure which achieves the minimum in (1.13). We now assume that ν ∈ G satisfies kµ − νkM = kµ − µ∗ kM . 15

(3.4)

By (3.3), we have Z

∗ g(u ) − g(v) = 0.



Thus, g(u∗ ) = g(v)

a.e.

(3.5)

We next observe that ν ≤ µ. In fact, note that inf {µ, ν} = µ − (µ − ν)+ .

(3.6)

Moreover, due to Proposition 4, inf {µ, ν} ≤ ν implies that inf {µ, ν} is also a good measure. It then follows from (3.6) and the minimality of ν that



kµ − νkM ≤ µ − inf {µ, ν} M = (µ − ν)+ M . Therefore, (µ − ν)− = 0; in other words, ν ≤ µ. In particular, v is a subsolution of (1.1), so that v ≤ u∗ a.e. by Proposition 1. We now split the proof into two cases:  Case 1. cap [u∗ = 1] = 0. By Theorem 4, this implies (µ − µ∗ )d = 0. Thus, νd ≤ µd = (µ∗ )d . On the other hand, since v ≤ u∗ a.e., it follows from Theorem 8 that νc = (−∆v)c ≤ (−∆u∗ )c = (µ∗ )c . We conclude that ν ≤ µ∗ ≤ µ. By (3.4), we must have ν = µ∗ .  Case 2. cap [u∗ = 1] > 0. We first show that u∗ = v on a set of positive Lebesgue measure. By contradiction, suppose that v < u∗ a.e. Let α0 , β0 ∈ [0, 1] be such that α0 < β0 and g   is increasing on [α0 , β0 ]. Since (3.5) holds and v < u∗ a.e., the set α0 < u∗ < β0 has zero Lebesgue measure. Let  w = min β0 , max {α0 , u∗ } − α0 . Thus, w ∈ H01 (Ω) and w achieves only the values 0 and β0 − α0 . We conclude  that w = 0 a.e. In other words, u∗ ≤ α0 a.e. Since cap [u∗ = 1] > 0, we get a contradiction. 16

We now proceed with the proof of Case 2. Given ε > 0, let α, β ∈ (1 − ε, 1), α < β, be such that g is increasing in [α, β]. Let Φε : R → R be a smooth function such that Φε (t) = t if t ≤ α, Φε (t) = 1 if t ≥ β and Φ0 (t) ≥ 0, ∀t ∈ R. We now establish the following Claim. For every ε > 0, we have   −∆ Φε (u∗ ) − Φε (v) ≥ 0

in D0 (Ω).

(3.7)

In fact, by Theorem 9,   ∆Φε (u∗ ) d = Φ0ε (u∗ )(∆u∗ )d + Φ00ε (u∗ )|∇u∗ |2   = Φ0ε (u∗ ) g(u∗ ) − (µ∗ )d + Φ00ε (u∗ )|∇u∗ |2

(3.8)

and, similarly,     ∆Φε (v) d = Φ0ε (v) g(v) − νd + Φ00ε (v)|∇v|2 .

(3.9)

By construction of Φε , we have Φ0ε (u∗ ) = Φ0ε (v)

a.e.

(3.10)

This is clear if v ≤ u∗ ≤ α or β ≤ v ≤ u∗ . Finally, if α < u∗ and v < β, then u∗ = v a.e. since g is increasing in [α, β] and g(u∗ ) = g(v) a.e. We conclude that (3.10) holds. By (3.5) and (3.10) we then have Φ0ε (u∗ )g(u∗ ) − Φ0ε (v)g(v) = 0

a.e.

(3.11)

Note that Φ00ε (u∗ ) = Φ00ε (v)

a.e.

In addition, on the set where Φ00ε (u∗ ) 6= 0, we have u∗ = v a.e., so that ∇u∗ = ∇v

  a.e. in Φ00ε (u∗ ) 6= 0 .

Thus, Φ00ε (u∗ )|∇u∗ |2 − Φ00ε (v)|∇v|2 = 0

a.e.

(3.12)

Finally, since Φ0ε (1) = 0 and (µ∗ )d = µd on the set [u∗ < 1] (by Theorem 4), we have Φ0ε (u∗ )(µ∗ )d = Φ0ε (u∗ )µd in Ω. 17

Moreover, Φ0ε ≥ 0 and ν ≤ µ imply Φ0ε (v)νd ≤ Φ0ε (v)µd

in Ω.

Therefore,   Φ0ε (u∗ )(µ∗ )d − Φ0ε (v)νd ≥ Φ0ε (u∗ ) − Φ0ε (v) µd = 0 in Ω.

(3.13)

Subtracting (3.9) from (3.8), and then applying (3.11)–(3.13), we conclude that    ≥ 0 in Ω. (3.14) − ∆ Φε (u∗ ) − Φε (v) d

On the other hand, since u∗ ≥ v a.e., we have Φε (u∗ ) − Φε (v) ≥ 0 a.e. It then follows from Theorem 8 that    ≥ 0 in Ω. (3.15) − ∆ Φε (u∗ ) − Φε (v) c

Combining (3.14) and (3.15), we obtain (3.7). This concludes the proof of the claim. According to the previous claim, the function Φε (u∗ )−Φε (v) is superharmonic. Moreover, since it is nonnegative and Φε (u∗ ) = Φε (v) a.e. on a set of positive (Lebesgue) measure, we deduce from the strong maximum principle (see [1]; see also [5]) that Φε (u∗ ) = Φε (v) a.e. in Ω. Since this holds true for every ε > 0, as we let ε ↓ 0 we conclude that u∗ = v a.e. Thus, µ∗ = ν. The proof of Theorem 5 is complete.

4

Proof of Theorem 2

In order to establish Theorem 2, we shall assume the next result which will be proved afterwards: Theorem 12 Let v ∈ L1 (Ω), v ≤ 1 a.e., be such that ∆v ∈ M(Ω). Assume g satisfies (1.4) for some 0 < β < 2. If g(v) ∈ L1 (Ω), then  HN −2+β [v = 1] = 0.

18

(4.1)

Proof of Theorem 2. Clearly, it suffices to establish the theorem for µ ≥ 0. Let u∗ be the function given by Proposition 1. Since ∆u∗ ∈ M(Ω) and g(u∗ ) ∈ L1 (Ω), it follows from Theorem 12 that  HN −2+β [u∗ = 1] = 0. By assumption, we have µ  HN −2+β . Thus, µ is diffuse and  µ [u∗ = 1] = 0. We deduce from Theorem 4 that µ∗ = µ. In other words, µ ∈ G. This concludes the proof of Theorem 2. We shall split the proof of Theorem 12 into two cases, whether 0 < β < 1 or 1 ≤ β < 2. We first consider the case where 0 < β < 1. An important ingredient is the following Lemma 3 Let ν ∈ M(Ω) and let v be the solution of ( −∆v = ν in Ω, v=0

(4.2)

on ∂Ω.

Given 0 < β < 1 and k ≥ 1, there exists a Borel set Ak ⊂ Ω such that v(x) − v(y) ≤ Ck |x − y|β and N −2+β H∞ (Ak ) ≤

∀x, y ∈ Ω \ Ak

C kνkM , k

(4.3)

(4.4)

for some constant C > 0 independent of k. α Given α ≥ 0, the Hausdorff content H∞ of a Borel set A ⊂ RN is defined as nX o [ α H∞ (A) = inf riα : A ⊂ Bri (xi ) , i

i

where the infimum is taken over all coverings of A with balls Bri (xi ) of radii ri . Note that we make no restriction on the size of such balls. In particular, for every α bounded set A we have H∞ (A) < ∞. It is easy to see that α H∞ (A) = 0

if and only if Hα (A) = 0.

19

(4.5)

Proof of Lemma 3. By linearity, it suffices to establish the lemma for ν ≥ 0. Let n o  Ak = x ∈ Ω : ν Br (x) ≥ k rN −2+β for some r > 0 . (4.6) (Here, ν is viewed as a measure in RN such that ν(RN \ Ω) = 0). We claim that (4.3) and (4.4) hold for Ak . We begin by establishing (4.4). For each x ∈ Ak , let rx > 0 be such that  ν Brx (x) ≥ k rxN −2+β .  Clearly, B5rx (x) x∈A is a covering of Ak . Applying Vitali’s covering lemma, we k  may extract a subcovering B5ri (xi ) of Ak such that the balls Bri (xi ) are all disjoint. We then have X N −2+β H∞ (Ak ) ≤ (5ri )N −2+β i

=C

X

riN −2+β

i

 C  C [ CX ≤ Bri (xi ) ≤ kνkM . ν Bri (xi ) = ν k i k k i This is precisely (4.4). We now turn to the proof of (4.3). We shall closely follow the argument presented in [8]. For simplicity, we assume N ≥ 3; the case N = 2 follows along the same lines. Clearly, it suffices to prove (4.3) for the function w defined as Z dν(z) 1 w(x) = ∀x ∈ Ω, N (N − 2)ωN Ω |z − x|N −2 where ωN = |B1 | is the measure of the unit ball in RN . It is not difficult to see that w can be rewritten as (see e.g. [18, Lemma 2])  Z ∞ ν Bs (x) 1 w(x) = ds. N ωN 0 sN −1 Given x, y ∈ Ω \ Ak , let δ = |x − y|. We then write Z ∞h  i ds 1 ν Bs (x) − ν Bs (y) w(x) − w(y) = N ωN 0 sN −1  Z 2δ Z ∞  1 = + . N ωN 0 2δ

20

(4.7)

Since x, y 6∈ Ak ,   ν Bs (x) , ν Bs (y) ≤ k sN −2+β

∀s > 0.

We then have Z





Z ≤ 0

0

 ds ν Bs (x) N −1 ≤ k s



Z 0

ds = Ck δ β . s1−β

(4.8)

On the other hand, for s ≥ 2δ, we have Bs−δ (x) ⊂ Bs (y); thus, Z ∞ Z ∞h  i ds ≤ ν Bs (x) − ν Bs−δ (x) sN −1 2δ   Z2δ∞  1 1 − ds. ≤ ν Bs (x) sN −1 (s + δ)N −1 δ Since

1 sN −1



1 δ ≤C N N −1 (s + δ) s

∀s ≥ δ,

we then get Z



Z



≤ Cδ 2δ

δ

 ds ν Bs (x) N ≤ Ck δ s

Z δ



ds s2−β

≤ Ck δ β .

(4.9)

It follows from (4.7)–(4.9) that w(x) − w(y) ≤ Ck δ β = Ck |x − y|β . Switching the roles between x and y we conclude that w satisfies (4.3). Since v − w is a harmonic function, v also verifies (4.3). The proof of the lemma is complete.

Given a Borel set A ⊂ RN , let A ∩ Bt (x) , Θ (x, A) = lim sup Bt (x) t→0 ∗

where | · | denotes the Lebesgue measure in RN . This function gives the density of points of A which are close to x. Clearly, 0 ≤ Θ∗ (x, A) ≤ 1. Another ingredient in the proof of Theorem 12 is the next Lemma 4 Given a Borel set A ⊂ RN , let   1 N ∗ F = x ∈ R : Θ (x, A) ≥ . 4 21

(4.10)

Then, for every 0 ≤ α ≤ N we have α α H∞ (F ) ≤ C H∞ (A),

(4.11)

for some C > 0 depending on N and α. Proof. If α = 0, then the conclusion is clear. We now assume α > 0. Given ε > 0,  let Bri (xi ) be a covering of A such that X α riα ≤ H∞ (A) + ε. (4.12) i

Let F1 = F ∩

h[

i B2ri (xi )

and F2 = F \

h[

i

i B2ri (xi ) .

i

Clearly, α (F1 ) ≤ H∞

X

h i α (2ri )α ≤ 2α H∞ (A) + ε .

(4.13)

i α (F2 ). Since (4.10) holds, for each y ∈ F2 We now prove a similar estimate for H∞ one can find sy > 0 sufficiently small so that

1 (4.14) (y) ≥ Bsy /2 (y) . 8  Applying Vitali’s covering lemma to B5sy (y) y∈F2 , we may extract a subcovering  B5sj (yj ) of F2 such that the balls Bsj (yj ) are disjoint. For each j, we define A ∩ B s

y /2

 Ij = i : xi ∈ Bsj (yj ) . In particular, the sets Ij are disjoint. We claim that X sα riα ∀j ≥ 1. j ≤ CN,α

(4.15)

i∈Ij

In order to establish (4.15), we first observe that [ A ∩ Bsj /2 (yj ) ⊂ Bri (xi ).

(4.16)

i∈Ij

In fact, given z ∈ A ∩ Bsj /2 (yj ), let i be such that z ∈ Bri (xi ). We claim that i ∈ Ij . Assume by contradiction that i ∈ 6 Ij , i.e. suppose xi 6∈ Bsj (yj ). Since sj ≤ d(xi , yj ) ≤ d(xi , z) + d(z, y) < ri +

22

sj 2

s

we deduce that 2j < ri and then d(xi , yj ) < 2ri . In other words, yj ∈ B2ri (xi ), which contradicts the definition of F2 , since yj ∈ F2 . This establishes (4.16). Applying (4.14) and (4.16), we have  s N j

2

=

X 8 8 X 1 riN . Bsj /2 (yj ) ≤ A ∩ Bsj /2 (yj ) ≤ Bri (xi ) = 8 ωN ωN ωN i∈Ij

i∈Ij

Since 0 < α ≤ N , we conclude that (4.15) holds. It now follows from (4.15) that α H∞ (F2 ) ≤

∞ X

(5sj )α ≤ 5α CN,α

j=1

∞ X X

riα ≤ C

X

h i α riα ≤ C H∞ (A) + ε . (4.17)

i

j=1 i∈Ij

Combining (4.13) and (4.17), we obtain h i α α H∞ (F ) ≤ C H∞ (A) + ε . Since ε > 0 was arbitrary, (4.11) follows. We now present the Proof of Theorem 12 when 0 < β < 1. Without loss of generality, we may assume that v = 0 on ∂Ω; the general case follows by taking vϕ, where ϕ is any function such that ϕ ∈ Cc∞ (Ω) and 0 ≤ ϕ ≤ 1 in Ω. Fix k ≥ 1 and let Ak be the set given by Lemma 3. We have [v = 1] ⊂ Ak ∪ Ek , where Ek = [v = 1] \ Ak . We further decompose Ek as Ek = Ek,1 ∪ Ek,2 , where Ek,1

  1 ∗ = x ∈ Ek : Θ (x, Ak ) ≥ 4

and Ek,2

  1 ∗ = x ∈ Ek : Θ (x, Ak ) < . 4

By Lemma 4, we have N −2+β N −2+β H∞ (Ek,1 ) ≤ C H∞ (Ak ).

(4.18)

We now claim that lim sup t→0

1 tN −2+β

Z g(v) > 0 ∀x ∈ Ek,2 . Bt (x)

23

(4.19)

In fact, given x ∈ Ek,2 , let t0 > 0 be sufficiently small so that Ak ∩ Bt (x) ≤ 1 Bt (x) 4

∀t ∈ (0, t0 ).

(4.20)

Recall that x ∈ Ω \ Ak and v(x) = 1. It follows from (4.3) that v(y) ≥ 1 − Ck |x − y|β ≥ 1 − Ck tβ

∀y ∈ Bt (x) \ Ak .

Since (1.4) holds, there exist C˜k > 0 and a sequence tn ↓ 0 such that g(1 − Ck tβn ) ≥

C˜k tn2−β

∀n ≥ 1.

Thus, for every n ≥ 1 sufficiently large, we get  C˜k g v(y) ≥ 2−β tn

∀y ∈ Btn(x) \ Ak .

Since (4.20) holds, we obtain Z

Z g(v) ≥

Btn(x)

g(v) ≥ Btn(x)\Ak

C˜k 3 −2+β , Btn(x) 2−β = Ck tN n 4 tn

which gives (4.19). It now follows from (4.19) that HN −2+β (Ek,2 ) = 0 (see e.g. [15, p.77]); equivalently, we have N −2+β H∞ (Ek,2 ) = 0.

(4.21)

We now deduce from (4.18) and (4.21) that N −2+β N −2+β H∞ (Ek ) ≤ C H∞ (Ak ).

Therefore,  C N −2+β N −2+β H∞ [v = 1] ≤ C H∞ (Ak ) ≤ kνkM . k Since this estimate holds true for every k ≥ 1, as we let k → +∞ we obtain  N −2+β H∞ [v = 1] = 0. In view of (4.5), the result follows. The proof of Theorem 12 in the case 1 ≤ β < 2 follows the same strategy, although it is more technical. For this reason we shall indicate the main steps in the proof. The counterpart of Lemma 3 is given by the following 24

Lemma 5 Let ν ∈ M(Ω) and let v be the solution of (4.2). Given 1 ≤ β < 2 and k ≥ 1, there exists a Borel set Ak ⊂ Ω such that (4.22) ) − v(x) − v(y) 2 v( x+y ≤ Ck |x − y|β 2 for every x, y ∈ Ω\Ak such that

x+y 2

∈ Ω \ Ak ; moreover,

N −2+β H∞ (Ak ) ≤

C kνkM , k

(4.23)

for some constant C > 0 independent of k. Proof. It suffices to consider the case where ν ≥ 0. Let Ak be given by (4.6). Proceeding as in the proof of Lemma 3, we obtain (4.23). We assume N ≥ 3. We now show that w defined by Z dν(z) w(x) = aN ∀x ∈ Ω, N −2 Ω |z − x| 1 where aN = N (N −2)ω , satisfies property (4.22). Let x, y ∈ Ω\Ak be such that N x+y ∈ Ω \ A . Set δ = |x − y|. We have k 2 2 w( x+y 2 ) − w(x) − w(y) ≤ Z 1 2 1 dµ(z). − ≤ aN − N −2 N −2 x+y N −2 |z − x| |z − y| Ω z− 2

We split this integral into two parts: 1 ) − w(x) − w(y) ≤ 2 w( x+y 2 aN

Z

Z +

|

z− x+y 2

|<2δ

. |

z− x+y 2

|≥2δ

 x+y

Note that B2δ 2 ⊂ B 5δ (x) ∩ B 5δ (y). Thus, 2 2 Z Z Z Z dν(z) dν(z) dν(z) ≤2 + + x+y N −2 N −2 |z − x| |z − y|N −2 |z − 2 | x+y x+y (x) (y) B B 5δ 5δ <2δ B z− | 2δ ( 2 ) 2 | 2 2 Z 5δ2 h i    ds ≤C 2ν Bs ( x+y 2 ) + ν Bs (x) + ν Bs (y) sN −1 0 ≤ Ck δ β . On the other hand, we have 2 1 1 δ2 ≤C − − N |z − x|N −2 |z − y|N −2 N −2 |z − x+y z − x+y 2 | 2 25

if z −

x+y 2



≥ 2δ. Therefore, Z |z− x+y 2 |≥2δ

≤ Cδ 2

Z

dν(z) N |z − x+y 2 |

|z− x+y 2 |≥2δ Z ∞  ds ν Bs ( x+y ≤ CN δ 2 2 ) sN +1 2δ Z ∞ ds 2 ≤ Ck δ β . ≤ CN k δ 3−β s 2δ

As in the proof of Lemma 3, we conclude that (4.22) holds. We now present the Proof of Theorem 12 completed. Assume 1 ≤ β < 2. Let Ek,1 and Ek,2 be defined as in the case 0 < β < 1; in particular, [v = 1] ⊂ Ak ∪ Ek,1 ∪ Ek,2 . By Lemma 4 we have N −2+β N −2+β H∞ (Ek,1 ) ≤ C H∞ (Ak ).

In order to establish the theorem, we are left to prove (4.21). Given x ∈ E2,k , let Rx denote the reflexion with respect to x; namely, Rx (y) = 2x − y

∀y ∈ RN .

We claim that v(y) ≥ 1 − Ck tβ

∀y ∈ Bt (x) \ (Ak ∪ Rx Ak ).

(4.24)

In fact, for every y ∈ Bt (x)\(Ak ∪Rx Ak ), we have Rx (y) ∈ Ω\Ak . Since x ∈ Ω\Ak , v(x) = 1 and v ≤ 1, we get v(y) ≥ v(y) + v(Rx y) − 1 ≥ 1 − Ck |x − y|β ≥ 1 − Ck tβ , which is precisely (4.24). We now take t0 > 0 sufficiently small so that Ak ∩ Bt (x) ≤ 1 Bt (x) 4

∀t ∈ (0, t0 ).

Therefore, (Ak ∪ Rx Ak ) ∩ Bt (x) ≤ 1 Bt (x) 2 26

∀t ∈ (0, t0 ).

We can now proceed as in the case 0 < β < 1 to conclude that Z 1 lim sup N −2+β g(v) > 0 ∀x ∈ Ek,2 . t t→0 Bt (x) Thus, N −2+β H∞ (Ek,2 ) = 0.

As before, we deduce that (4.1) holds. The proof of Theorem 12 is complete.

5

Capacitary estimates related to problem (1.1)

In this section we prove some estimates on the capacity of the set [u∗ = 1]. They should be compared with the result of Theorem 12 concerning the Hausdorff measure of this set. We assume throughout this section that g satisfies a slightly stronger hypothesis than (1.4), namely o n 2−β lim inf (1 − t) β g(t) > 0. t↑1

(5.1)

Given p > 1 and a Borel set E ⊂ Ω, we shall denote by capp (E) the capacity of E associated to W01,p (Ω). Note that cap2 coincides with the H 1 -capacity, denoted by cap elsewhere in this paper. Our goal in this section is to establish the Theorem 13 Let v ∈ L1 (Ω), v ≤ 1 a.e., be such that ∆v ∈ M(Ω). Assume that g satisfies (5.1) for some β ∈ (0, 1]. If g(v) ∈ L1 (Ω) then  cap2−β [v = 1] = 0 (5.2) Note that β ∈ (0, 1] implies 2 − β ≥ 1, so that cap2−β is well-defined. Proof. Step 1. Proof of (5.2) if v ∈ W01,1 (Ω). Set η(s) = we have

s+ 1−s

and let Tk (s) = min {s, k}. Since v ∈ W01,1 (Ω), for every k ≥ 1 Z

 ∇v · ∇Tk η(v) dx ≤ k k∆vkM .

(5.3)



 Indeed, inequality (5.3) (which formally amounts to multiplying ∆v by Tk η(v) ) can be obtained by approximating v (e.g. through convolution) with smooth functions vn such that k∆vn kM ≤ k∆vkM . 27

We can rewrite (5.3) as |∇v + |2 dx ≤ k k∆vkM . (1 − v)2

Z

(5.4)

[η(v)
On the other hand, applying H¨ older’s inequality with exponents have Z |∇v + |2−β dx ≤ (1 − v)2(2−β)

2 2−β

and

2 β,

we

[η(v)
|∇v + |2 dx (1 − v)2

Z ≤

!1− β2

Z

[η(v)
[η(v)
! β2

1 (1 − v)

2(2−β) β

dx

. (5.5)

It then follows from (5.4)–(5.5) and the definition of η that Z

1− β2 |∇v + |2−β dx ≤ c k k∆vkM 2(2−β) (1 − v)

Z

η(v)

1+

[η(v)
[η(v)
! β2

2−β β

dx

2−β β

(1 − v)

. (5.6)

By assumption (5.1) there exists a constant c0 > 0 such that g(t)(1 − t)

2−β β

≥ c0

∀t ∈ ( 12 , 1).

From (5.6), we obtain Z |∇v + |2−β dx ≤ (1 − v)2(2−β) [η(v)
≤ c k k∆vkM

Z

1− β2

h

1 + g(v) η(v)

2−β β

i

! β2 dx

[η(v)
≤ ck 2−β k∆vkM 2

Z

(5.7)

1 + g(v) Tk (η(v))



k

2−β β

2−β β

! β2 dx

.

Since g(v) ∈ L1 (Ω) (which also implies that η(v) is finite a.e.) we have 2−β Z 1 + g(v) Tk (η(v)) β lim dx = 0. 2−β k→+∞ Ω k β We then deduce from (5.7) that Z lim k→+∞



∇Tk (η(v)) 2−β dx = 0. k 28

(5.8)

Note that

Tk (η(v)) ≥1 k

in [η(v) ≥ k].

Therefore, cap2−β

 [η(v) ≥ k] ≤

Z ∇Tk (η(v)) 2−β k→+∞ dx −−−−−→ 0. k Ω

Since [v = 1] = [η(v) = +∞] =

∞ \

[η(v) ≥ k],

k=1

we conclude that  cap2−β [v = 1] = 0 .

Step 2. Proof of Theorem 13 completed. We replace v with vϕ where ϕ ∈ Cc∞ (Ω) is a cut-off function, i.e. 0 ≤ ϕ ≤ 1 in Ω and ϕ = 1 on a compact set K ⊂ Ω. Since v and ∇v ∈ L1loc (Ω), it follows that ∆(vϕ) ∈ M(Ω). Moreover g(vϕ) ≤ g(v) a.e., hence g(vϕ) ∈ L1 (Ω). We can then  apply the previous step to vϕ to deduce that cap2−β [vϕ = 1] = 0. Thus,  cap2−β [v = 1] ∩ K = 0

for every compact K ⊂ Ω.

By subadditivity of cap2−β , we conclude that  cap2−β [v = 1] = 0.

It is well-known (see e.g. [15]) that cap1 (E) = 0 if and only if HN −1 (E) = 0. Thus, in the case β = 1, we recover Theorem 12 but with a totally different proof. On the other hand, for any p > 1, capp (E) = 0 implies Hs (E) = 0 for any s > N − p (but the converse is not true). Thus, for β ∈ (0, 1), Theorem 13 only  gives Hs [v = 1] = 0 for any s > N − 2 + β, which is not optimal in view of Theorem 12. However, it should be noticed that the proof of Theorem 13 only relies on energy estimates, which remain true for more general operators, for instance in  the inhomogeneous case. Namely, assume A(x) = ai,j (x) is an N × N -matrix with bounded measurable coefficients satisfying λ1 |ξ|2 ≤ A(x)ξ · ξ ≤ λ2 |ξ|2 29

∀ξ ∈ RN ,

for a.e. x ∈ Ω,

where 0 < λ1 ≤ λ2 . Proceeding as in the proof of Theorem 13, one deduces the following result: Theorem 14 Let v ∈ L1 (Ω) be such that div (A(x)∇v) is a bounded measure “in the sense of Stampacchia”, i.e. assume there exists µ ∈ M(Ω) such that Z Z ∗ − v div (A (x)∇ζ) dx = ζ dµ (5.9) Ω

Ω ∗

such that div (A (x)∇ζ) ∈ L∞ (Ω). Assume g satisfies for every ζ ∈ C0 (Ω) ∩ (5.1) for some β ∈ (0, 1]. If g(v) ∈ L1 (Ω), then we have  cap2−β [v = 1] = 0. H01

Proof. Let µn be a suitable smooth convolution of µ, and consider the solutions vn of  − div(A(x)∇v ) = µ in Ω, n n 1  vn ∈ H (Ω). 0  Multiplying this equation by Tk η(vn ) (see the definition of η(s) in Step 1 of Theorem 13), we get Z Z   |∇vn+ |2 λ1 dx ≤ A(x)∇vn · ∇Tk η(vn ) dx ≤ k kµn kM ≤ k kµkM . 2 (1 − vn ) Ω [η(vn )
Since the solutions in the sense of Stampacchia are unique and stable, vn converges to v in L1 (Ω). Therefore, as n → +∞, we obtain Z |∇v + |2 λ1 dx ≤ k kµkM . (1 − v)2 [η(v)
Henceforth, one can follow the proof of Step 1 of Theorem 13 in order to conclude. In particular, if v satisfies the assumptions of Theorem 14, then  Hs [v = 1] = 0 for any s > N − 2 + β, if β ∈ (0, 1).

(5.10)

Note that  HN −1 [v = 1] = 0

if β = 1.

(5.11)

It is an open problem whether (5.10) holds with s = N − 2 + β, where β ∈ (0, 2), β 6= 1. Note that, in the inhomogeneous case, it is not clear how to implement an approach based on H¨ older continuity, as used in the proof of Theorem 12. 30

Remark 1 In the same spirit, the proof of Theorem 13 extends to nonlinear operators, as e.g. the p-Laplacian, for functions v which satisfy − div(|∇v|p−2 ∇v) ∈ M(Ω) “in the renormalized sense” (see [9] for the precise definition). In this case, one can prove with the same method that if (5.1) holds true for some β ∈ (0, 1] and g(v) ∈ L1 (Ω), then  capq [v = 1] = 0 with q =

(2 − β)p . 2(1 − β) + βp

Note that if β = 1, then it still holds that   cap1 [v = 1] = 0 = HN −1 [v = 1] .

6

Every diffuse measure is good for some g

Our goal in this section is to establish the following Theorem 15 Let µ ∈ M(Ω) be such that µ+ is diffuse. Then, there exists some g such that µ ∈ G(g). We shall start with the Proposition 5 Let g1 , g2 be such that g1 ≤ g2 . Then, G(g1 ) ⊂ G(g2 ). Proof. Given µ ∈ G(g1 ), let u be the solution of ( −∆u + g1 (u) = µ in Ω, u=0

on ∂Ω.

Let µ∗ be the reduced measure relative to g2 and denote by u∗ the solution of ( −∆u∗ + g2 (u∗ ) = µ∗ in Ω, u∗ = 0

on ∂Ω.

Since µ∗ ≤ µ and g2 ≥ g1 , we have u∗ ≤ u (see [4, Corollary B.2]). In other words, u − u∗ ≥ 0 in Ω and u − u∗ = 0 on the set [u∗ = 1]. Thus, by (2.3) we have   (µ∗ − µ)d = ∆(u − u∗ ) d ≥ 0

in [u∗ = 1].

This implies (µ∗ )d = µd in [u∗ = 1]. On the other hand, by Theorem 4, (µ∗ )d = µd

in [u∗ < 1]. 31

We conclude that (µ∗ )d = µd .

(6.1)

Finally, since µ is a good measure relative to g1 , we have µc ≤ 0. Thus, by (2.10), (µ∗ )c = −(µc )− = µc .

(6.2)

It follows from (6.1) and (6.2) that µ = µ∗ ∈ G(g2 ). This concludes the proof of the proposition. Related to the previous result, we point out the following Open Problem. Assume g1 ≤ g2 and G(g1 ) = G(g2 ). Is it true that g1 = g2 ? We now establish the Lemma 6 Let µ ∈ M(Ω) be a nonnegative diffuse measure. Given ε > 0, s0 ∈ (0, 1), and a continuous nondecreasing function g : (−∞, 1) → R satisfying (1.2)– (1.3), then there exists g˜ : (−∞, 1) → R with g˜ ≥ g

in (−∞, 1),

g˜ = g

in (−∞, s0 ],

(6.3)

and such that  µ [v = 1] < ε,

(6.4)

where v is the largest subsolution of the problem ( −∆u + g˜(u) = µ in Ω, u=0

on ∂Ω.

Proof. Fix t0 ∈ (s0 , 1). Let gk : (−∞, 1) → R be any increasing sequence of continuous, nondecreasing functions, such that    gk (t) = g(t) if t ≤ s0 , gk (t) ≥ k if t ≥ t0 ,    gk (t) ≥ g(t) ∀t ∈ (−∞, 1). For each k ≥ 1, let µk denote the reduced measure of µ relative to gk . We shall denote by vk the corresponding solution. In particular, by Proposition 1, Z |∆vk | ≤ 2kµk kM ≤ 2kµkM (6.5) Ω

32

and

Z gk (vk ) ≤ kµkM .

(6.6)



In view of (6.6), we have [vk ≥ t0 ] ≤ 1 k

Z gk (vk ) ≤ Ω

1 kµkM → 0 k

(6.7)

as k → +∞. On the other hand, the sequence (vk ) is non-increasing; thus, there exists v ∈ L1 (Ω) such that vk ↓ v in L1 (Ω). By (6.7), we have v ≤ t0 a.e. Moreover, since 0 ≤ vk ≤ 1 a.e., it follows from (6.5) that (vk ) is bounded in L∞ (Ω) ∩ H01 (Ω). We then conclude that vk → v µ-a.e. in Ω (see e.g. [6, Lemma 2.1]). Therefore,  µ [vk > t0 ] → 0 as k → +∞. The lemma then follows by taking g˜ = gk0 for some k0 ≥ 1 sufficiently large.

We now present the Proof of Theorem 15. We shall split the proof of the theorem into two steps. Step 1. Given µ ∈ M(Ω) diffuse and nonnegative, there exists g satisfying (1.2)– (1.3) such that µ ∈ G(g). t We begin by constructing a sequence (gk ) as follows. Let g0 (t) = 1−t , ∀t ∈ 1 1 [0, 1). Given gk , we apply Lemma 6 to g = gk , ε = 2k and s0 = 1 − 2k . Set gk+1 = g˜, where g˜ is the function given by Lemma 6. In particular, the sequence (gk ) is nondecreasing and  ∀k ≥ k0 . gk = gk0 in − ∞, 1 − 2k10

Set g(t) = lim gk (t) ∀t ∈ (−∞, 1). k→+∞

We claim that µ ∈ G(g). In fact, let µk denote the reduced measure of µ relative to gk . In particular, µk is also a diffuse measure. Since g ≥ gk , it follows from Proposition 5 that µk ∈ G(g) for every k ≥ 1. Let vk be the solution of ( −∆vk + gk (vk ) = µk in Ω, vk = 0

on ∂Ω.

By Theorem 4 and the choice of gk , we have  1 kµ − µk kM ≤ µ [vk = 1] ≤ k . 2 33

Thus, µk → µ strongly in M(Ω). Since G(g) is closed, we conclude that µ ∈ G(g) as claimed. Step 2. Proof of the theorem completed. Let µ ∈ M(Ω) be such that µc ≤ 0; in other words, µ+ is diffuse. We can then apply the previous step to µ+ to conclude that there exists g such that µ+ ∈ G(g). Since µ ≤ µ+ , by Proposition 4 we deduce that µ is also good for g.

7

Measures which are good for every g

In this section we characterize the set of measures which are always good. In order to do so, we first need to recall some notions about obstacle problems with measure data. Throughout this section, we denote by β any maximal monotone graph (m.m.g.) of the form   if t < 1,  b(t) (7.1) β(t) = [b(1), ∞) if t = 1,    ∅ if t > 1. where b : (−∞, 1] → R is a nondecreasing continuous function such that b(t) = 0 if t ≤ 0. Given a bounded measure µ in Ω, we say that w is a solution of ( −∆w + β(w) 3 µ in Ω, (7.2) w = 0 on ∂Ω, if w ∈ L1 (Ω), w ≤ 1 a.e., ∆w ∈ M(Ω), and there exists a nonnegative diffuse measure ν ∈ M(Ω) such that νa ∈ β(w) a.e., νs is concentrated on the set [w = 1], and Z Z Z − w∆ζ + ζ dν = ζ dµ ∀ζ ∈ C02 (Ω). (7.3) Ω





(Here, νa and νs denote the absolutely continuous and the singular parts of ν with respect to the Lebesgue measure in RN ). In particular, the measure µ + ∆w is diffuse and µ + ∆w = ν ≥ inf β(1)

in [w = 1].

(7.4)

Problem (7.2) has been studied by Dall’Aglio-Leone [11], Dall’Aglio-Dal Maso [10], Brezis-Ponce [6]; see also the references therein. It turns out that (7.2) has 34

a solution if and only if µ+ is diffuse; moreover, this solution is unique and is the largest solution of the problem   −∆v + b(v) ≤ µ in Ω,   (7.5) v ≤ 1 in Ω,    v = 0 on ∂Ω. Our goal in this section is to establish the following Theorem 16 Let µ ∈ M(Ω). Then, µ is good for every g if and only if µ+ is diffuse and µ + ∆w0 ∈ L1 (Ω), where w0 is the unique solution of the obstacle problem ( −∆w0 + β0 (w0 ) 3 µ in Ω, w0 = 0

(7.6)

on ∂Ω,

with β0 (s) = 0 if s < 1 and β0 (1) = [0, ∞). Remark 2 It is known from [2] that if µ ∈ L1 (Ω), then problem (1.1) has a solution for every g. This is consistent with Theorem 16. Indeed, let w0 be the solution of (7.6) with µ ∈ L1 (Ω). Then, in view of (2.3), we have µ + ∆w0 ≤ µ on [w0 = 1]. Since µ + ∆w0 is a nonnegative measure and it is concentrated on [w0 = 1], we conclude that 0 ≤ µ + ∆w0 ≤ µ in Ω. Hence, µ ∈ L1 (Ω) implies that µ + ∆w0 ∈ L1 (Ω). In the proof of Theorem 16 we shall need the next two lemmas: Lemma 7 Let β be a m.m.g. and let µ ∈ M(Ω) be such that µ+ is diffuse. If ∞ µ+ a ∈ L (Ω)

and

kµ+ a kL∞ < inf β(1),

(7.7)

then [w = 1] = 0, where w is the solution of (7.2). 35

(7.8)

Proof. By (2.3), we have (∆w)d ≤ 0 on the set [w = 1]. Thus, µd ≥ (µ + ∆w)d = µ + ∆w ≥ inf β(1)

in [w = 1].

Comparing the absolutely continuous part of both sides, we get µa = (µd )a ≥ inf β(1)

a.e. in [w = 1].

In view of (7.7), we deduce that (7.8) holds. Lemma 8 Let µ ∈ M(Ω) be such that µ+ is diffuse. Given two m.m.g. β1 , β2 , let wi be the solution of (7.2) associated to βi , i = 1, 2. If β1 ≥ β2 , then 0 ≤ µ + ∆w1 ≤ µ + ∆w2

in [w1 = 1].

(7.9)

Proof. By comparison, we have w2 − w1 ≥ 0 a.e. In particular, w2 − w1 = 0 in [w1 = 1]. Applying (2.3), we get   ∆(w2 − w1 ) d ≥ 0

in [w1 = 1].

Thus, on the set [w1 = 1], we have µ + ∆w1 = (µ + ∆w1 )d ≤ (µ + ∆w2 )d = µ + ∆w2 . Since w1 is the solution of (7.2) with β = β1 , we have ν1 = µ + ∆w1 ≥ 0 in Ω. We conclude that (7.9) holds. Proof of Theorem 16. Proof of (⇐). We shall establish a slightly more general result: Proposition 6 Let µ ∈ M(Ω) be such that µ+ is diffuse. Assume that µ + ∆w ∈ L1 (Ω),

(7.10)

where w is the unique solution of (7.2). Then, µ is good for every g such that g ≥ β. ∞ Proof. We first assume µ+ a ∈ L (Ω). Let β1 be a m.m.g. such that

β ≤ β1 ≤ g

and kµ+ a kL∞ < inf β1 (1).

36

Let w1 be the solution of (7.2) with obstacle β1 . We claim that µ + ∆w1 ∈ L1 (Ω).

(7.11)

In fact, since w1 is the solution of an obstacle problem, the measure (µ + ∆w1 )s is concentrated on the set [w1 = 1]. By Lemma 8 above, we have 0 ≤ (µ + ∆w1 )s ≤ (µ + ∆w)s = 0 in [w1 = 1]. This establishes (7.11). On the other hand, it follows from Lemma 7 that the set [w1 = 1] has zero Lebesgue measure. We conclude that µ + ∆w1 = b1 (w1 ) In other words, w1 verifies (

a.e.

−∆w1 + b1 (w1 ) = µ in Ω, w1 = 0

on ∂Ω.

Since [w1 = 1] = 0, by a variant of the De La Vall´ee-Poussin theorem (see [12, Remark 23] or [13, Theorem II.22]), one can find g1 satisfying (1.2) such that b1 ≤ g1 ≤ g

and g1 (w1 ) ∈ L1 (Ω).

Thus, in view of Proposition 4, µ is a good measure for g1 . By Proposition 5, we deduce that µ is also good for g. Since g ≥ β was arbitrary, the result follows ∞ when µ+ a ∈ L (Ω). In order to establish the proposition for any measure µ satisfying (7.10), we let  µn = min µa , n + µs

∀n ≥ 1.

Proceeding as in the proof of (7.11), for every n ≥ 1 we have µn + ∆wn ∈ L1 (Ω), ∞ where wn is the solution of (7.2) with data µn . Moreover, (µn )+ a ∈ L (Ω). Thus, µn is good for every g ≥ β. As n → +∞, we deduce that µ is also good for any such g.

Proof of (⇒). We shall need the following

37

Lemma 9 Let µ ∈ M(Ω). Assume that µ is good for every g, ∞ µ+ a ∈ L (Ω)

and

kµ+ a kL∞ ≤ inf β(1).

Let w be the solution of (7.2). Then, ( −∆w + b(w) = µ w=0

in Ω,

(7.12)

(7.13)

on ∂Ω.

Proof. By making a small perturbation of µ, it suffices to establish the result when kµ+ (7.14) a kL∞ < inf β(1). By Lemma 7 above, we know that [w = 1] = 0. Thus, one can find a continuous nondecreasing function H : (−∞, 1) → R, H ≥ b, satisfying (1.2) and such that H(w) ∈ L1 (Ω). We now take a sequence of functions (gn ) such that gn ≤ H

∀n ≥ 1

and gn ↓ b

as n ↑ +∞.

Since µ is good for every gn , there exists vn satisfying (1.1) with nonlinearity gn . Clearly, vn ↑ v, where v ∈ L1 (Ω) and v ≤ 1 a.e. By Fatou, v verifies (7.5); in particular, v ≤ w a.e. Thus, [v = 1] ≤ [w = 1] = 0.

(7.15)

On the other hand, note that gn (vn ) → b(v)

a.e. on [v < 1].

(7.16)

Thus, by (7.15)–(7.16), we have gn (vn ) → b(v)

a.e.

Since gn (vn ) ≤ H(w) a.e., it follows by dominated convergence that gn (vn ) → b(v)

in L1 (Ω).

We deduce that v satisfies (7.13). In particular, v is also a solution of (7.2). By uniqueness, we conclude that v = w. This establishes the lemma. We can now conclude the proof of Theorem 16. Let µ be a measure such that µ is good for every g. We assume in addition that + µa ∈ L∞ (Ω). Let bn : (−∞, 1] → R be a sequence of nondecreasing continuous 38

functions such that bn (t) = 0 if t ≤ 0, bn (1) = kµ+ a kL∞ and bn (t) ↓ 0 uniformly away from t = 1. By Lemma 9, equation (7.13) has a solution vn ≤ 1 associated to bn . Note that vn ↑ v, where v ∈ L1 (Ω), v ≤ 1 a.e. Moreover, passing to a subsequence if necessary, we have bn (vn ) * f

weakly in L∞ (Ω)

for some f ∈ L∞ (Ω) with kf kL∞ ≤ kµ+ a kL∞ . We claim that v = w0 a.e. In fact, note that v satisfies ( −∆v = µ − f in Ω, v=0

on ∂Ω.

Since 0 ≤ bn (vn ) ≤ bn (v)

a.e. ∀n ≥ 1,

as n → +∞ we obtain 0 ≤ f ≤ α χ[v=1]

a.e.,

where α = kµ+ a kL∞ . This implies that f is nonnegative and concentrated on the set [v = 1]. Therefore, v verifies problem (7.6) and so v = w0 as claimed. We conclude that µ + ∆w0 = f ∈ L∞ (Ω). ∞ This establishes the theorem under the additional assumption that µ+ a ∈ L (Ω). The general case easily follows by using an approximation argument.

Before proving Theorem 3, we start with the following Proposition 7 Given µ ∈ M(Ω), let v be the unique solution of ( −∆v = µ in Ω, v=0

(7.17)

on ∂Ω.

If v ≤ 1 a.e., then µ is good for every g. Proof. Let α < 1. Since v ≤ 1 a.e., αv is a supersolution of problem (1.1) with data αµ. Thus, by Proposition 4, αµ is good for every α < 1. Since G(g) is closed with respect to the strong topology in M(Ω), we deduce that µ ∈ G(g) for every g. We now present the 39

Proof of Theorem 3. The implication “⇐” already follows from Proposition 7 above. We now establish the reverse implication. Let µ be a singular measure which is good for every g. In particular, µ+ is diffuse. Let ν = µ + ∆w0 , where w0 is the solution of (7.6). In view of the definition of β0 , ν is a nonnegative diffuse measure concentrated on the set [w0 = 1]. By (2.3), on this set we have (∆w0 )d ≤ 0, so that 0 ≤ ν ≤ µ. On the other hand, by Theorem 16, we know that ν ∈ L1 (Ω). Since µ is singular, we conclude that ν = 0, hence w0 coincides with the unique solution v of (7.17). Since w0 ≤ 1 a.e., the result follows.

8

How to construct diffuse measures which are not good

Our goal in this section is to establish Theorem 1. The main ingredient is the following Lemma 10 Given g, there exists v ∈ C0 (Ω) such that  ∆v ∈ L1 (Ω), v ≤ 1 in Ω, cap [v = 1] > 0 and

g(v) ∈ L1 (Ω).

(8.1)

Proof. Let (`k ) be a decreasing sequence of positive numbers such that `k ≤ θ`k−1

∀k ≥ 2,

(8.2)

for some θ ∈ (0, 21 ). Let (kj ) be an increasing sequence of nonnegative integers. Both sequences (`k ) and (kj ) will be explicitly chosen later on. We now briefly recall the construction presented in [18] of the Cantor set F associated to the subsequence (`kj ). We shall assume for simplicity that Ω = Q1 , the unit cube centered at 0. We first define a decreasing sequence of sets (Fj )j≥0 as follows. Let F0 = Q1 , k0 = 0 and `0 = 1. We now proceed by induction. Assume Fj−1 , j ≥ 1, is the union of 2N kj−1 disjoint cubes of length `kj−1 . Let Qi be any component of Fj−1 , ˜ i ⊂ Qi be a smaller cube concentric to Qi (so that the ratio between their and let Q ˜ i , we select 2N (kj −kj−1 ) cubes Qi,s of length lengths is 12 + θ ∈ ( 12 , 1)). Inside Q ˜ i . Set `kj , uniformly distributed in Q [ Fj = Qi,s . i,s

40

Thus, Fj ⊂ Fj−1 and Fj is the union of 2N kj disjoint cubes of length `kj . The Cantor set associated to the subsequence (`kj ) is then defined as ∞ \

F =

Fj .

j=1

We now split the proof of the lemma into two cases, whether N ≥ 3 or N = 2: Case 1. N ≥ 3 We start with the following Claim 1. For every j ≥ 1, we have cap (Fj , Fj−1 ) ≤ Cθ 2N kj `kNj−2 ,

(8.3)

where cap (Fj , Fj−1 ) denotes the H 1 -capacity of the set Fj with respect to Fj−1 . Since Fj−1 has 2N kj−1 connected components, it suffices to show that cap (Fj ∩ Qi , Qi ) ≤ Cθ 2N (kj −kj−1 ) `kNj−2 ,

(8.4)

where Qi is any component of Fj−1 . Note that each component Qi,s of Fj ∩ Qi has length `kj and (see [18]) d(Qi,s , ∂Qi ) ≥

1 − 2θ diam Qi . 4

Hence, −2 cap (Qi,s , Qi ) ≤ Cθ `N . kj

(8.5)

Recall that Qi contains 2N (kj −kj−1 ) components Qi,s . By the subadditivity of the capacity, we conclude that (8.4) holds. This concludes the proof of the claim. By (8.3) and Theorem E.1 in [4], there exists vj ∈ Cc∞ (Fj−1 ) such that 0 ≤ vj ≤ 1 in Ω, vj = 1 on Fj , and Z |∆vj | ≤ C 2N kj `kNj−2 . (8.6) Ω

Our aim is to construct the function v of the form v=

∞ X

αj vj ,

(8.7)

j=1

where (αj ) is a sequence of positive numbers to be chosen later on such that ∞ X

αj = 1.

j=1

41

(8.8)

Clearly, v ∈ C0 (Ω) and v ≤ 1 in Ω. Moreover, v = 1 precisely on We claim that one can choose (`k ), (kj ) and (αj ) such that ∞ X

1

< ∞,

−2 2N kj `N kj j=1 ∞ X −2 αj 2N kj `N kj j=1

< ∞,

 j ∞ X X g αi 2N kj `N kj < ∞. j=1

T

j

Fj = F .

(8.9)

(8.10)

(8.11)

i=1

In fact, let αj = 3 · 2−2j

∀j ≥ 1,

so that (8.8) holds. Let (kj ) be any increasing sequence of positive numbers such that 1 g(1 − 2−2j ) ≤ j ∀j ≥ 1. N kj 2 N −2 2 Finally, we take (`k ) satisfying (8.2) (with, say, θ = 34 ) and −2 = 2j 2N kj `N kj

∀j ≥ 1.

It immediately follows that (8.9) and (8.10) hold. After some straightforward computation, the left-hand side of (8.11) can be estimated by  j  NN−2 ∞ X 2 g(1 − 2−2j ) g(1 − 2−2j ) 2kj ≤ , N 2 2 N −2 kj j=1 j=1

∞ X

which is finite in view of our choice of (kj ). We conclude that (8.9)–(8.11) hold. By construction, [v = 1] coincides with F . On the other hand, by [18], we know that cap (F ) > 0 if and only if ∞ X

1

−2 2N kj `N kj j=1

< ∞.

(8.12)

In view of (8.9), we deduce that cap (F ) > 0. Thus,  cap [v = 1] > 0.

(8.13)

By (8.6), (8.7) and (8.10), we have ∆v ∈ L1 (Ω). 42

(8.14)

Finally, it follows from (8.11) that g(v) ∈ L1 (Ω).

(8.15)

The proof of the lemma is complete when N ≥ 3. Case 2. N = 2. As in the previous case, we start with the Claim 2. For every j ≥ 1, we have cap (Fj , Fj−1 ) ≤ Cθ 4kj

 log

1 `kj

−1 .

(8.16)

The argument is similar to the proof of Claim 1. It suffices to observe that the analog of (8.5) is  −1 1 . (8.17) cap (Qi,s , Qi ) ≤ Cθ log `kj We now conclude the proof of the lemma. Let (αj ) be defined as before. Take an increasing sequence of positive integers (kj ) such that g(1 − 2−2j ) k 44 j



1 2j

∀j ≥ 1.

Finally, let (`k ) satisfying (8.2) and `kj = 4−4

kj

·2−j

∀j ≥ 1.

With such choices, one can easily check that ∞ X 1 1 log < ∞, k j 4 `kj j=1  −1 ∞ X 1 kj αj 4 log < ∞, `kj j=1

 j ∞ X X g αi 4kj `2kj < ∞. j=1

(8.18)

(8.19)

(8.20)

i=1

Let v be given by (8.7), where vj ∈ Cc∞ (Fj−1 ) is such that 0 ≤ vj ≤ 1 in Ω, vj = 1 on Fj , and −1  Z 1 . |∆vj | ≤ C 4kj log `kj Ω 43

(The existence of such vj follows from Claim 2 above.) In particular, [v = 1] = F . By (8.18), we have (see [18, Lemma 4])  cap [v = 1] > 0. Moreover, proceeding as before, we deduce from (8.19) and (8.20) that ∆v ∈ L1 (Ω)

and g(v) ∈ L1 (Ω).

This concludes the proof of the lemma. Using Lemma 10, we establish the following Proposition 8 For every g, there exist a nonnegative function h0 ∈ L1 (Ω) and a compact set K0 ⊂ Ω, with |K0 | = 0 and cap (K0 ) > 0, such that for any measure σ ≥ 0 supported in K0 we have (h0 + σ)∗ = h0 ,

(8.21)

where (h0 + σ)∗ is the reduced measure associated to h0 + σ. Proof. Take K0 = [v = 1] and f0 = −∆v + g(v), where v is the function constructed in Lemma 10. We begin with the following Claim. If λ is a good measure ≥ f0 , then λ(K0 ) = 0. We first observe that λ is a diffuse measure. In fact, since λ is good, we have λc ≤ 0 by Corollary 2. On the other hand, λ ≥ f0 implies λc ≥ 0. Thus, λc = 0, so that λ is diffuse. Let u be the solution of ( −∆u + g(u) = λ in Ω, u=0

on ∂Ω.

Clearly, u ≥ v a.e. Moreover, since v = 1 in K0 , we have u − v = 0 in K0 . Thus, by (2.3),   f0 − λ = ∆(u − v) = ∆(u − v) d ≥ 0 in K0 . In other words, λ ≤ f0 in K0 ; thus, λ = f0 in K0 . Since |K0 | = 0, we deduce that λ(K0 ) = 0. This establishes the claim. Let h0 = f0+ . We now show that h0 and K0 satisfy the desired properties. In fact, let σ ≥ 0 be a measure concentrated on K0 . Since h0 + σ ≥ 0, it follows from Corollary 3 that 0 ≤ (h0 + σ)∗ ≤ h0 + σ in Ω. 44

Moreover, by Corollary 1,   (h0 + σ)∗ a = h0

a.e. in Ω.

Thus, f0 ≤ h0 ≤ (h0 + σ)∗ ≤ h0 + σ in Ω.   In particular, (h0 + σ)∗ s is concentrated on K0 . By our previous claim, we have   (h0 + σ)∗ s = 0 in K0 . Therefore,   (h0 + σ)∗ = (h0 + σ)∗ a = h0

in Ω.

Remark 3 A slight modification in the construction of v, given by Lemma 10, allows to obtain the following further property in the statement of Proposition 8: given any ε > 0 and any ball Br ⊂⊂ Ω, one can choose h0 and K0 such that kh0 kL1 < ε and K0 ⊂ Br .

Theorem 1 is now a consequence of Proposition 8: Proof of Theorem 1. Given g, let h0 , K0 be as in the statement of Proposition 8. Since cap (K0 ) > 0, there exists a diffuse measure σ ≥ 0 concentrated on K0 such that σ(K0 ) = 1 (see e.g. [8]). Let µ = h0 + σ. By Proposition 8, we have µ 6= µ∗ . Thus, µ 6∈ G(g). A slightly stronger version of Theorem 1 is the following Theorem 17 Given g, let h0 ∈ L1 (Ω) and K0 ⊂ Ω be given by Proposition 8. Let σ be a nonnegative diffuse measure supported in K0 . If σ is good, then kσkM < kh0 kL1 .

(8.22)

Proof. Assume σ is good. By Proposition 8, we have (h0 + σ)∗ = h0 . Recall that, by Theorem 5, (h0 + σ)∗ is the closest good measure to h0 + σ. Thus,



kσkM = (h0 + σ) − (h0 + σ)∗ M < (h0 + σ) − σ M = kh0 kL1 .

45

Corollary 4 Given g, there exists a diffuse measure µ ≥ 0 such that εµ is not good for any ε > 0. Proof. Using Remark 3, we can take sequences of disjoint compact sets (Kj ) in Ω and L1 -functions (hj ), such that each pair Kj , hj satisfies the assumptions of Proposition 8 and 1 khj kL1 ≤ j . 4 P 1 Let h = j hj ∈ L (Ω). For each j ≥ 1, we fix a diffuse measure σj ≥ 0 P concentrated on Kj such that kσj kM = 21j . Let µ = j σj ∈ M(Ω). Assume by contradiction that εµ is good for some ε > 0. Since εµ ≥ εσj

∀j ≥ 1,

then εσj is also good. By Theorem 17, this gives ε<

khj kL1 1 ≤ j kσj kM 2

∀j ≥ 1.

As j → +∞, we get a contradiction. Imposing some additional assumption on the nonlinearity g one can construct a measure µ of the form µ = θ Hα bK , for some α > N − 2, such that µ 6∈ G(g). To this purpose, one first needs a slight modification of Lemma 10: Lemma 11 Assume g is given by g(t) =

1 (1 − t)

2−β β

−1

∀t ∈ [0, 1),

(8.23)

where β ∈ (0, 2). Then, for any α ∈ (0, β), there exists v˜ ∈ C0 (Ω) such that ∆˜ v ∈ L1 (Ω),

v˜ ≤ 1 in Ω,

 HN −2+α [˜ v = 1] ∈ (0, ∞)

and

g(˜ v ) ∈ L1 (Ω). (8.24)

Proof. We just need to adapt the proof of Lemma 10. We shall consider both cases N ≥ 3 and N = 2 simultaneously. Let v˜ be given by (8.7). Using the same notation as before, we let αj = am 2−mN j , where

α (2 − α)β α
(8.25)

and the constant am is chosen so that (8.8) holds. Observe that the range of admissible m given by (8.25) is nonempty since 0 < α < β < 2. Next, we let kj = j and Nk `k = 2− N −2+α ∀k ≥ 1. (8.26) With (`k ) defined as above, one can show that (see e.g. [18]) HN −2+α (F ) ∈ (0, ∞), T where F = j Fj = [˜ v = 1]. We now prove (8.11) (or, equivalently, (8.20) if N = 2). Note that, with our choices of (αj ) and (`k ), the left-hand side of (8.11) reduces to ∞ ∞ X X 2−β 2−β N2j 2−α 2mN j β 2N j 2− N −2+α = 2N j (m β − N −2+α ) , j=1

j=1

which is finite, by (8.25). We now assume N ≥ 3. Note that (8.10) becomes ∞ X

2−mN j 2N j 2−

N (N −2)j N −2+α

=

j=1

∞ X

2−N j (m− N −2+α ) < ∞, α

j=1

which clearly holds in view of (8.25). Similarly, if N = 2, then one easily checks that (8.19) is also satisfied. Proceeding as in the proof of Lemma 10, we conclude that (8.24) holds. As a consequence, we have the following Theorem 18 Given β ∈ (0, 2), let g be given by (8.23). Then, for any α ∈ (0, β), there exist θ0 > 0 and K ⊂ Ω compact, HN −2+α (K) ∈ (0, ∞), such that θ HN −2+α bK ∈ G(g)

implies

θ < θ0 .

(8.27)

Proof. Let  + ˜ 0 = − ∆˜ h v + g(˜ v)

and K = [˜ v = 1],

where v˜ is given by Lemma 11 above. Proceeding as in the proof of Proposition 8, we have  ˜ 0 + θ HN −2+α bK ∗ = h ˜ 0 ∀θ > 0. h Therefore, if θ HN −2+α bK is good, then as in the proof of Theorem 17 we conclude that ˜ 0 kL1 . θ HN −2+α (K) < kh In other words, (8.27) holds with θ0 =

˜0k 1 kh L . HN −2+α (K)

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9

Further properties of µ∗ and G

In this section we prove some properties of the reduced measures, which should be compared with those in [4]. In particular, we start by showing that the reduced measure µ∗ need not be the largest good measure ≤ µ, contrarily to what happens when g is everywhere defined. In fact, we have Proposition 9 There exists µ ∈ M(Ω), µ ≥ 0, for which the set 

λ∈G:λ≤µ

(9.1)

has no largest element. Proof. Let K0 , h0 be given by Proposition 8. Let σ be the capacitary measure associated to K0 . In particular, σ is a nonnegative measure concentrated on K0 ; moreover, σ is good (see Proposition 7). Let µ = h0 + σ. By Proposition 8, µ 6∈ G(g). Assume by contradiction that the set given by (9.1) has a largest element, say ν ≤ µ. Clearly, ν ≥ h0 and ν ≥ σ. Thus, ν ≥ sup {h0 , σ} = h0 + σ = µ. We deduce that ν = µ, so that µ is a good measure. This is a contradiction.

Note that the same argument can be used to establish the next results (in what follows, σ is the capacitary measure associated to K0 , with K0 and h0 being given by Proposition 8). Proposition 10 There exist good measures µ, ν ≥ 0 such that sup {µ, ν} is not good. Proof. Take µ = h0 , ν = σ and use Proposition 8. Proposition 11 There exist diffuse measures µ, ν ≥ 0 such that ν ≤ µ but µ∗ −ν ∗ is not ≥ 0. Proof. Take µ = h0 + σ and ν = σ. Similarly, the mapping µ 7→ µ∗ is not a contraction. More precisely,

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Proposition 12 There exist diffuse measures µ, ν ≥ 0 such that kµ − νkM < kµ∗ − ν ∗ kM . Proof. Take µ = h0 + σ and ν = σ. We conclude with the following Proposition 13 The set G is not convex. Proof. By Theorem 1, there exists µ diffuse such that µ 6∈ G. Applying Theorem 3 in [4], we can decompose µ as µ = f + ∆v, where f ∈ L1 (Ω), v ∈ H01 (Ω) ∩ C(Ω) and kvkL∞ ≤ 13 . In particular, 2f ∈ G and ∆(2v) ∈ G; however, 2f + ∆(2v) = µ 6∈ G. 2

Acknowledgments The second author (A.C.P.) was supported by the NSF grant DMS-0111298 and Sergio Serapioni, Honorary President of Societ`a Trentina Lieviti—Trento (Italy). A.C.P. gratefully acknowledges the invitation and the warm hospitality of the Math Department at the University of Rome 1. The third author (A.P.) warmly thanks H. Brezis for the invitation at Rutgers University. Part of this work was carried out during these two occasions.

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] L. Boccardo, On the regularizing effect of strongly increasing lower order terms. J. Evol. Equ. 3 (2003), 225–236. [3] 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. 49

[4] 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. [5] H. Brezis and A. C. Ponce, Remarks on the strong maximum principle. Differential Integral Equations 16 (2003), 1–12. [6] H. Brezis and A. C. Ponce, Kato’s inequality when ∆u is a measure. C. R. Math. Acad. Sci. Paris, Ser. I 338 (2004), 599–604. [7] H. Brezis and W. A. Strauss, Semilinear second-order elliptic equations in L1 . J. Math. Soc. Japan 25 (1973), 565–590. [8] L. Carleson, Selected problems on exceptional sets. Van Nostrand Mathematical Studies, No. 13, Van Nostrand, Princeton, 1967. [9] G. Dal Maso, F. Murat, L. Orsina, and A. Prignet, Renormalized solutions of elliptic equations with general measure data. Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4) 28 (1999), 741–808. [10] 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. [11] P. Dall’Aglio and C. Leone, Obstacles problems with measure data and linear operators. Potential Anal. 17 (2002), 45–64. [12] C. De La Vall´ee Poussin, Sur l’int´egrale de Lebesgue. Trans. Amer. Math. Soc. 16 (1915), 435–501. [13] C. Dellacherie and P.-A. Meyer, Probabilit´es et potentiel. Chapitres I `a IV, Publications de l’Institut de Math´ematique de l’Universit´e de Strasbourg, No. XV, Actualit´es Scientifiques et Industrielles, No. 1372, Hermann, Paris, 1975. [14] L. Dupaigne and A. C. Ponce, Singularities of positive supersolutions in elliptic PDEs. Selecta Math. (N.S.) 10 (2004), 341–358. [15] L. C. Evans and R. F. Gariepy, Measure theory and fine properties of functions. Studies in Advanced Mathematics, CRC Press, Boca Raton, FL, 1992.

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[16] M. Fukushima, K. Sato, and S. Taniguchi, On the closable parts of preDirichlet forms and the fine supports of underlying measures. Osaka J. Math. 28 (1991), 517–535. [17] 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. [18] A. C. Ponce, How to construct good measures. In: Elliptic and Parabolic Equations (C. Bandle, H. Berestycki, B. Brighi, A. Brillard, M. Chipot, J.M. Coron, C. Sbordone, I. Shafrir, V. Valente, and G. Vergara-Caffarelli, eds.), Progress in Nonlinear Differential Equations and their Applications, 63, Birkh¨ auser, Basel, 2005, pp. 335–348. A special tribute to the work of Ha¨ım Brezis. [19] G. Stampacchia, Le probl`eme de Dirichlet pour les ´equations elliptiques du second ordre ` a coefficients discontinus. Ann. Inst. Fourier (Grenoble) 15 (1965), 189–258. [20] J. L. V´azquez, On a semilinear equation in R2 involving bounded measures. Proc. Roy. Soc. Edinburgh Sect. A 95 (1983), 181–202.

L. Dupaigne Laboratoire Ami´enois de Math´ematique Fondamentale et Appliqu´ee Universit´e Picardie Jules Verne Facult´e de Math´ematiques et d’Informatique 33, rue Saint-Leu 80039 Amiens Cedex 1 France e-mail: [email protected] A.C. Ponce Institute for Advanced Study Princeton, NJ 08540 USA e-mail: [email protected] A. Porretta Dipartimento di Matematica Universit`a di Roma “Tor Vergata”

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Via della Ricerca Scientifica 1 00133 Roma Italy e-mail: [email protected]

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