(1)REMARKS ON THE STRONG MAXIMUM PRINCIPLE Ha¨ım Brezis and Augusto C

REMARKS ON THE STRONG MAXIMUM PRINCIPLE

Ha¨ım Brezis and Augusto C. Ponce

Laboratoire Jacques-Louis Lions, Universit´e Pierre et Marie Curie, B.C. 187 4 Pl. Jussieu, 75252 Paris Cedex 05, France

and

Rutgers University, Dept. of Math., Hill Center, Busch Campus 110 Frelinghuysen Rd, Piscataway, NJ 08854

1. Introduction

The strong maximum principle asserts that if u is smooth, u ≥ 0 and

−∆u ≥ 0 in a connected domain Ω ⊂ R^N, then either u ≡ 0 or u > 0 in Ω. The same conclusion holds when −∆ is replaced by −∆ +a(x) with a∈L^p(Ω),p > ^N₂ (this is a consequence of Harnack’s inequality; see e.g. [6], and also [7, Corollary 5.3]. Another formulation of the same fact says that ifu(x0) = 0 for some point x0 ∈Ω, then u ≡0 in Ω. A similar conclusion fails, however, when a 6∈ L^p(Ω), for any p > ^N₂. For instance, u(x) = |x|² satisfies−∆u+a(x)u= 0 inB₁ with a= _|x|^2N₂ 6∈L^N/2(Ω).

If u vanishes on a larger set, one may still hope to conclude, under some weaker condition on a, that u ≡ 0 in Ω. Such a result was obtained by B´enilan-Brezis [2, Appendix D] (with a contribution by R. Jensen) in the case wherea∈L¹(Ω) and suppuis a compact subset of Ω. Their maximum principle has been further extended by Ancona [1], who proved Theorem 1 below.

We recall that a function v : Ω → R is quasicontinuous if there exists a sequence of open subsets (ω_n) of Ω such that v|_Ω\ωn is continuous ∀n ≥ 1 and capωn→0 as n→ ∞, where capωn denotes theH¹-capacity ofωn. Theorem 1 ([1]). AssumeΩ⊂R^N is an open bounded set. Letu∈L¹(Ω), u ≥ 0 a.e. in Ω, be such that ∆u is a Radon measure on Ω. Then there exists u˜: Ω→R quasicontinuous such that u= ˜u a.e. inΩ.

Leta∈L¹(Ω), a≥0 a.e. in Ω. If

−∆u+au≥0 in Ω,

Accepted for publication: June 2002.

AMS Subject Classifications: 35B05, 35B50.

1

in the following sense Z

E

∆u≤ Z

E

au for every Borel set E⊂Ω, (1.1) and if u˜= 0on a set of positiveH¹-capacity in Ω, then u= 0a.e. in Ω.

The proof given by Ancona is purely based on Potential Theory, while ours is more direct in the spirit of PDE’s. We also discuss carefully the meaning of the condition−∆u+au≥0 in Ω.

The next two corollaries follow immediately from the theorem above:

Corollary 2. Letuandabe as in Theorem1,and suppose(1.1)is satisfied.

If u = 0 on a subset of Ω with positive measure, then u = 0 a.e. in Ω. If u is continuous in Ω and u= 0on a subset of Ωwith positive H¹-capacity, thenu≡0 in Ω.

Corollary 3. Let u and a be as in Theorem 1. Suppose that ∆u ∈L¹(Ω).

If

−∆u+au≥0 a.e. in Ω

and u= 0on a subset of Ωwith positive measure, then u= 0 a.e. inΩ.

The next corollary follows from Theorem 1 and Remark 3:

Corollary 4. Letu and a be as in Theorem1. Suppose that au∈L¹_loc(Ω).

If

−∆u+au≥0 in D⁰(Ω), i.e.,

Z

u∆ϕ≤ Z

auϕ ∀ϕ∈C₀^∞(Ω), ϕ≥0 in Ω,

and u= 0on a subset of Ωwith positive measure, then u= 0 a.e. inΩ.

Remark 1. In view of Corollary 4 above, it would seem natural to replace condition (1.1) in Theorem 1 by

Z

u∆ϕ≤ Z

auϕ ∀ϕ∈C₀^∞(Ω), ϕ≥0 in Ω, (1.2) which makes sense even if au 6∈ L¹_loc(Ω) (note that auϕ ≥ 0 a.e., so that the right-hand side is always well-defined, possibly taking the value +∞).

However, the strong maximum principle is no longer true in general. See Remark 4.

There are several interesting questions related to Theorem 1:

Open problem 1. In the statement of Theorem 1, suppose in addition that suppu ⊂ Ω is a compact set. Can one replace the assumption a∈ L¹_loc by a weaker condition, for example a^1/2 ∈L¹_loc (ora^1/2∈L^p_loc for some p >1), and still conclude thatu= 0 a.e. in Ω?

Note that one cannot hope to go belowL^1/2. For instance theC²-function u given by

u(x) = (

1− |x|²4

for|x| ≤1 0 for|x|>1,

satisfies−∆u+au≥0 for some functiona(x) such thata(x)∼ ¹

1−|x|² for

|x|.1. Here,a^α ∈L¹ ∀α <1/2, but a^1/26∈L¹. Here is another question:

Open problem 2. Assume u ∈C⁰, u ≥ 0, and a ∈ L^q_loc for some q ≥1, a≥0 a.e., satisfy (1.1). Suppose thatu= 0 on a setE with cap_1,2q(E)>0, where cap_1,2qrefers to the capacity associated with the Sobolev spaceW^1,2q. Can one conclude thatu≡0?

Theorem 1 above shows that the answer is positive whenq = 1. It is also true when q > ^N₂ by the strong maximum principle mentioned above (note that ifq > ^N₂ and x₀ is any point, then cap_1,2q {x₀}

>0).

2. Some comments about condition (1.1)

Since in the statement of Theorem 1 it may happen that au 6∈ L¹_loc(Ω), and soau is not necessarily a distribution, one should be careful in order to give a precise meaning to the inequality

∆u≤au in Ω.

More generally, letµbe a Radon measure on Ω andf a measurable function, f ≥0 a.e. in Ω. Here are two possible definitions for the inequalityµ≤f in Ω:

Definition 1. We shall writeµ≤₁ f in Ω if Z

E

dµ≤ Z

E

f for every Borel set E⊂Ω.

Definition 2. We shall writeµ≤₂ f in Ω if Z

ϕ dµ≤ Z

f ϕ ∀ϕ∈C₀^∞(Ω), ϕ≥0 in Ω.

In the first definition, we view f as the nonnegative measure f dx, while in the second onef is treated as if it were a distribution.

Remark 2. Ifµ≤₁ f in Ω, then µ≤₂f in Ω. However, the converse is not true in general. See Remark 4 below.

Remark 3. If we assume in addition thatf ∈L¹_loc(Ω), thenµ≤₁ f in Ω if, and only if,µ≤₂ f in Ω.

Remark 4. Theorem 1 above is no longer true in general (even for the case where ∆u∈L¹(Ω)) if we replace (1.1) by

−∆u+au≥₂0 in Ω, i.e., if

Z

u∆ϕ≤ Z

auϕ ∀ϕ∈C₀^∞(Ω), ϕ≥0 in Ω.

In fact, let N ≥2. Takev ∈L¹(R^N), v ≥0 a.e. in R^N, such that suppv⊂ B₁, ∆v ∈ L¹(R^N), but v is unbounded (this is possible since N ≥ 2). In particular, there existsb∈L¹(R^N),b≥0 a.e. inR^N, such thatbv6∈L¹(R^N).

Let (xj)⊂B1 be a dense sequence inB1 and, for eachj ≥1, let γj := min

n1

j,1− |x_j| 2

o .

We define u(x) :=

X∞ j=1

1 2^jγ_j^N−2 v

x−xj

γ_j

, a(x) :=

X∞ j=1

1 2^jγ_j^N b

x−xj

γ_j

.

Then

u∈L¹(R^N), u≥0 a.e. in R^N,

∆u∈L¹(R^N),

a∈L¹(R^N), a≥0 a.e. in R^N, and

Z

u∆ϕ≤ Z

auϕ ∀ϕ∈C₀^∞(Ω), ϕ≥0 in Ω,

(note that the integral in the right-hand side is either 0 or +∞), but suppu⊂ B₁andu6≡0 inR^N. On the other hand, in view of Theorem 1, the inequality

∆u≤₁ auis not satisfied.

From now on, we shall always consider the inequality ∆u≤auin the sense of Definition 1. In particular we shall omit the subscript 1 in the symbol≤₁.

3. Proof of the quasicontinuity statement of Theorem 1 Before proving the first part of Theorem 1 (see Lemma 1 below), we make the following remark:

Remark 5. If v ∈ H_loc¹ (Ω), then there exists ˜v : Ω → R quasicontinuous such thatv= ˜va.e. in Ω (see e.g. [5]). In addition, ˜vis well-defined modulo polar subsets of Ω, i.e., if ˜v1 and ˜v2 are two quasicontinuous functions such that ˜v₁ = v = ˜v₂ a.e. in Ω, then there exists a polar set P ⊂Ω such that

˜

v1(x) = ˜v2(x) ∀x∈Ω\P (see [3]).

Notation. Givenk >0, we denote by T_k :R→Rthe truncation function Tk(s) :=







k if s≥k, s if −k < s < k,

−k if s≤ −k.

The existence of a quasicontinuous function ˜u : Ω→ R such that u = ˜u a.e. in Ω as in the statement of Theorem 1 is a consequence of Lemma 1 below (see [1]):

Lemma 1. LetΩ⊂R^N be an open set. Assumeu∈L¹(Ω)is such that ∆u is a Radon measure on Ω. Then

T_k(u)∈H_loc¹ (Ω) ∀k >0, (3.1) and, for each open subset A⊂⊂ Ω, there existsCA >0 so that

Z

A

∇T_k(u)²≤k Z

Ω

|∆u|+C_A Z

Ω

|u|

∀k >0. (3.2) Moreover, there existsu˜: Ω→R quasicontinuous such that u= ˜ua.e. inΩ.

Proof. We shall split the proof of Lemma 1 into two steps:

Step 1. Proof of (3.1) and (3.2). We first extenduto the wholeR^N so that u≡0 outside Ω. Let ρ∈C₀^∞(B₁) be a radial, nonnegative, mollifier. Set

u_ε(x) :=ρ_ε∗u(x) = Z

Ω

ρ_ε(x−y)u(y)dy ∀x∈Ω.

For k >0 fixed, we have T_k(u_ε)∈H¹(Ω) and

∇T_k(u_ε) =∇u_εχ_[|uε|<k], (3.3) whereχ_[|uε|<k] denotes the characteristic function of the set [|u_ε|< k].

Given an open set A⊂⊂ Ω, letϕ∈C₀^∞(Ω) be such that 0≤ϕ≤1 in Ω and ϕ≡1 onA. On the one hand, using (3.3) and integrating by parts, we have Z ∇T_k(uε)²ϕ=

Z

∇Tk(uε)·(∇uε)ϕ

=− Z

T_k(u_ε)(∆u_ε)ϕ− Z

T_k(u_ε)∇u_ε· ∇ϕ.

(3.4)

On the other hand, Z

Tk(uε)∇uε· ∇ϕ=− Z

uε∇Tk(uε)· ∇ϕ− Z

uεTk(uε)∆ϕ

=− Z

Tk(uε)∇Tk(uε)· ∇ϕ− Z

uεTk(uε)∆ϕ

=−1 2

Z

∇

Tk(uε)2

· ∇ϕ− Z

uεTk(uε)∆ϕ

= 1 2

Z T_k(u_ε)2

∆ϕ− Z

u_εT_k(u_ε)∆ϕ

=− Z

Tk(uε) uε−1

2Tk(uε)

∆ϕ≥ −k Z

|uε||∆ϕ|.

(3.5)

It follows from (3.4) and (3.5) that Z

A

∇T_k(u_ε)²≤Z ∇T_k(u_ε)²ϕ

≤k Z

suppϕ

|∆uε|+k∆ϕkL^∞

Z

suppϕ

|uε|

.

In particular, for every 0< ε <dist (suppϕ, ∂Ω), Z

A

∇T_k(u_ε)²≤k Z

Ω

|∆u|+k∆ϕk_L^∞ Z

Ω

|u|

.

Letting ε↓ 0, we conclude that T_k(u) ∈H¹(A) and (3.2) holds with C_A = k∆ϕkL^∞.

Step 2. We prove that, under the assumptions of the lemma, there exists a function ˜u: Ω→Rquasicontinuous such that u= ˜u a.e. in Ω.

By (3.1) and Remark 5, for each k > 0 there exists ^Tk(u) : Ω → R quasicontinuous such thatTk(u) =T^k(u) a.e. in Ω.

Letvk := ¹_kTk(u), so that

v_k →0 inL^q(Ω) ∀q∈[1,∞)

and, by (3.2),

Z

A

|∇v_k|²→0 ∀A⊂⊂Ω.

In particular,v_k →0 inH_loc¹ (Ω), which implies there exists a polar setP ⊂Ω such that

˜

vk(x) = 1 k

^Tk(u)(x)→0 ∀x∈Ω\P.

We conclude that

caphT^k(u) > k

2 i

= cap h

|˜vk|> 1 2 i

→0. (3.6)

Set

w(x) :=



 sup

k∈N

nT^_k(u)(x)o

if sup

k∈N

^T_k(u)(x) <∞,

0 otherwise,

(3.7) so that w =u a.e. in Ω. By (3.6) and the quasicontinuity of the functions T^_k(u), it is easy to see that w is quasicontinuous in Ω. This concludes the proof of the lemma.

4. A variant of Kato’s inequality when ∆u is a Radon measure We start with the following (see [1])

Lemma 2. LetΩ⊂R^N be an open set. Assumeu ∈L¹(Ω),u ≥0 a.e. in Ω, is such that ∆u is a Radon measure on Ω. Then

∆Tk(u) is a Radon measure ∀k >0.

Moreover, for any a∈L^∞(Ω), a≥0 a.e. in Ω, we have

∆T_k(u)−aT_k(u)≤ ∆u−au+

in D⁰(Ω). (4.1) Proof. We shall use the same notation as in the proof of Lemma 1. By the standard L¹-version of Kato’s inequality (see [4]) we have (note thatTk|_R+ is concave)

∆T_k(u_ε)≤t_k(u_ε)∆u_ε in Ω ∀ε >0, (4.2) where the function t_k :R₊→Ris given by

tk(s) :=

(1 if 0≤s≤k, 0 ifs > k.

SinceTk(s)≥tk(s)s ∀s≥0 anda≥0 a.e. in Ω, it follows from (4.2) that

∆Tk(uε)−aTk(uε)≤tk(uε) ∆uε−auε

≤(∆uε−auε

+

inD⁰(Ω).

In other words, we have Z

T_k(u_ε)∆ϕ−aT_k(u_ε)ϕ≤ Z

∆u_ε−au_ε+

ϕ ∀ϕ∈C₀^∞(Ω), ϕ≥0 in Ω.

(4.3) For λ >0, let Ωλ:=

x∈Ω : dist (x, ∂Ω)> λ . Thus, if 0< ε < λ, we get

∆uε−auε= ∆u−au

ε+ (au)ε−auε

≤ρε∗ ∆u−au+

+(au)_ε−au+au−auε

in Ωλ. Therefore, for anyϕ ∈C₀^∞(Ω), ϕ≥0 in Ω, and 0 < ε < dist (suppϕ, ∂Ω), we may write

Z

∆u_ε−au_ε+

ϕ≤ Z

ρ_ε∗ ∆u−au+

ϕ

+kϕk_L^∞n(au)_ε−au

L¹+kak_L^∞u−u_ε

L¹

o

= Z

(ρε∗ϕ) ∆u−au+

+o(1).

(4.4)

Sinceρ_ε∗ϕ→ϕuniformly in Ω and ∆u−au+

is a Radon measure in Ω, by letting ε↓0 in (4.3) and (4.4), we conclude that

Z

Tk(u)∆ϕ−aTk(u)ϕ≤ Z

∆u−au+

ϕ ∀ϕ∈C₀^∞(Ω), ϕ≥0 in Ω, so thatT_k(u) is a Radon measure (take for instancea≡0) and (4.1) holds.

Lemma 3. LetΩ⊂R^N be an open set. Assumeu ∈L¹(Ω),u ≥0 a.e. in Ω, is such that ∆u is a Radon measure on Ω. Leta∈L¹(Ω), a≥0 a.e. in Ω. If

−∆u+au≥0 in Ω, in the following sense

Z

E

∆u≤ Z

E

au for every Borel set E⊂Ω, (4.5) then

−∆Tk(u) +aTk(u)≥0 in D⁰(Ω) ∀k >0. (4.6) Proof. By the preceding lemma applied with ai := Ti(a), where i is a positive integer, we know that

∆Tk(u)−aiTk(u)≤ ∆u−aiu+

inD⁰(Ω). (4.7)

On the other hand, from (4.5) we get Z

E

(∆u−a_iu)≤ Z

E

(a−a_i)u for every Borel setE⊂Ω. (4.8) Since (a−ai)u≥0 a.e. in Ω, (4.8) implies that

0≤ Z

E

(∆u−aiu)⁺≤ Z

E

(a−ai)u for every Borel setE ⊂Ω. (4.9) Hence, (∆u−a_iu)⁺is a nonnegative measure which is absolutely continuous with respect to the Lebesgue measure. Therefore, we have

(∆u−a_iu)⁺∈L¹(Ω) ∀i= 1,2, . . . . (4.10) We now return to (4.9) to conclude that

0≤(∆u−aiu)⁺≤(a−ai)u a.e. in Ω.

In particular,

(∆u−aiu)⁺↓0 a.e. in Ω as i↑ ∞. (4.11) It follows from (4.10) and (4.11) that

(∆u−aiu)⁺→0 inL¹(Ω) asi→ ∞. (4.12) Finally, for any ϕ∈C₀^∞(Ω),ϕ≥0 in Ω, by (4.7) and (4.12) we have

Z

T_k(u)∆ϕ−aT_k(u)ϕ≤ Z

T_k(u)∆ϕ−a_iT_k(u)ϕ≤ Z

∆u−a_iu+

ϕ→0 asi→ ∞, so that (4.6) holds.

5. Proof of Theorem 1 completed

It follows from Lemma 1 in Section 2 that, under the hypotheses of the theorem, there exists ˜u: Ω →Rquasicontinuous such that u= ˜u a.e. in Ω.

Let us assume that ˜u = 0 on a set of positive capacity E ⊂ Ω. We shall prove thatu= 0 a.e. in Ω.

We split the proof into two steps:

Step 1. Under the hypotheses of the theorem, if we assume in addition that u∈L^∞(Ω), then u= 0 a.e. in Ω.

Sinceu∈L^∞(Ω), we haveau∈L¹(Ω). It follows from (1.1) and Remark 3 that

−∆u+au≥0 inD⁰(Ω).

Recall that for ε, λ > 0 we have defined Ωλ :=

x ∈ Ω : dist (x, ∂Ω) > λ and

u_ε(x) :=ρ_ε∗u(x) = Z

Ω

ρ_ε(x−y)u(y)dy ∀x∈Ω, whereρ∈C₀^∞(B₁),ρ≥0 inB₁, is a radial mollifier.

Using the above notation, for 0< ε < λ, we have in Ωλ that

∆u_ε ≤(au)_ε=au_ε+

(au)_ε−au_ε

≤au_ε+

(au)_ε−au_ε+

=:au_ε+f_ε. (5.1)

Since (au)_ε→au inL¹(Ω),u_ε→u a.e. in Ω andu is bounded,

f_ε →0 inL¹(Ω). (5.2)

Letδ >0 be a fixed number. Multiplying (5.1) by 1

uε+δ, we get

∆u_ε

u_ε+δ ≤a+f_ε

δ in Ω_λ ∀ε∈(0, λ). (5.3) We also remark that

∇uε

(uε+δ)² =−∇ 1 uε+δ

in Ω. (5.4)

Letϕ∈C₀^∞(Ω) and 0< ε <dist(suppϕ, ∂Ω). We now use (5.4), integration by parts, estimate (5.3) and Cauchy-Schwarz, to get

Z |∇uε|²

(uε+δ)²ϕ²=− Z

∇uε· ∇ 1 uε+δ

ϕ² =

Z ∆uε

uε+δϕ²+

Z 2ϕ∇ϕ· ∇uε

uε+δ

≤ Z

a+f_ε δ

ϕ²+1 2

Z |∇u_ε|²

(u_ε+δ)²ϕ²+ 2 Z

|∇ϕ|².

Therefore, 1 2

Z |∇uε|² (u_ε+δ)²ϕ²≤

Z

a+fε

δ

ϕ²+ 2 Z

|∇ϕ|².

Since

∇log u_ε δ + 1

= ∇u_ε u_ε+δ, the estimate above may be rewritten as

1 2

Z ∇log uε

δ + 1²ϕ²≤ Z

a+ fε

δ

ϕ²+ 2 Z

|∇ϕ|². (5.5)

We now letε↓0 in (5.5). It follows from (5.2) that (see also Lemma 1) logu

δ + 1

∈H_loc¹ (Ω) ∀δ >0 and

1 2

Z ∇logu δ + 1

2

ϕ² ≤ Z

aϕ²+ 2|∇ϕ|²

∀ϕ∈C₀^∞(Ω). (5.6) LetE ⊂Ω be a set of positive capacity such that ˜u= 0 onE. Without any loss of generality, we may assume that E ⊂ Ω_λ for someλ > 0 sufficiently small.

Assumeω ⊂⊂Ω is an open connected set containingE. Letϕ₀ ∈C₀^∞(Ω) be a fixed test function such thatϕ≡1 on ω.

By (5.6), we have Z

ω

∇logu δ + 1

2

≤2 Z

aϕ²₀+ 2|∇ϕ₀|²

. (5.7)

Since the quasicontinuous representative log^u

δ + 1

= log u˜ δ + 1 of log ^u_δ + 1

equals 0 on E ⊂Ω with capE > 0, it follows from a variant of Poincar´e’s inequality (easily proved by contradiction) that there exists C >0 (depending only onE and Ω) such that

Z

ω

log² u

δ + 1

≤C Z

ω

∇log

u

δ + 1² ∀δ >0. (5.8) (5.7) and (5.8) yield

Z

ω

log² u

δ + 1

≤2C Z

aϕ²₀+ 2|∇ϕ₀|²

∀δ >0. (5.9) In particular, the integral in the left-hand side remains bounded asδ↓0.

On the other hand, log²

u δ + 1

→+∞ a.e. inω\[u= 0] asδ ↓0. (5.10) By (5.9) and (5.10), we conclude thatu= 0 a.e. inω. Sinceωis an arbitrary connected neighborhood of E in Ωλ for all λ > 0 small, we conclude that u= 0 a.e. in Ω.

Step 2. Proof of Theorem 1 completed.

From Lemma 3, we know that ∆T1(u) is a Radon measure and

−∆T1(u) +aT1(u)≥0 inD⁰(Ω).

In addition,^T₁(u) =T₁(˜u) = 0 onE ⊂Ω with capE >0.

By Step 1, we haveT1(u) = 0 a.e. in Ω, and so u= 0 a.e. in Ω.

Acknowledgments. The authors thank A. Ancona for interesting discus- sions. The first author (H.B.) is partially supported by a European grant ERB FMRX CT98 0201. He is also a member of the Institut Universitaire de France. The second author (A.C.P.) is supported by CAPES, Brazil, under the grant BEX1187/99-6.

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