Quasilinear Lane-Emden equations with absorption and measure data

0.31 MB·English

by  Marie-Françoise Bidaut-Véron

Quasilinear Lane-Emden equations with
absorption and measure data
3
1
0
2 Marie-Fran¸coise Bidaut-V´eron∗

n Nguyen Quoc Hung†
a
J Laurent V´eron‡

5
1 Laboratoire de Math´ematiques et Physique Th´eorique,
Universit´e Franc¸ois Rabelais, Tours, FRANCE
]
P
A
.
h
t Contents
a
m
[ 1 Introduction 2

2
2 Lorentz spaces and capacities 4
v
4 2.1 Lorentz spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1
2.2 Wolff potentials, fractional and η-fractional maximal operators . . . 5
3
6 2.3 Estimates on potentials . . . . . . . . . . . . . . . . . . . . . . . . . 5
. 2.4 Approximation of measures . . . . . . . . . . . . . . . . . . . . . . . 16
2
1
2 3 Renormalized solutions 19
1
3.1 Classical results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
:
v 3.2 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
i
X
4 Equations with absorption terms 23
r
a
4.1 The general case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
4.2 Proofs of Theorem 1.1 and Theorem 1.2 . . . . . . . . . . . . . . . . 26
2010 Mathematics Subject Classification. 35J92,35R06,46E30.
Key words: quasilinear elliptic equations, Wolff potential, maximal functions, Borel measures, Lorentz
spaces,Lorentz-Besselcapacities.
∗E-mail address: [email protected]
†E-mail address: [email protected]
‡E-mail address: [email protected]
1
Abstract We study the existence of solutions to the equation −∆ u+g(x,u) = µ when
p
g(x,.) is a nondecreasing function and µ a measure. We characterize the good measures,
i.e. the ones for which the problem has a renormalized solution. We study particularly the
caseswhereg(x,u)=|x|−β|u|q−1uandg(x,u)=sgn(u)(eτ|u|λ−1). Theresultsstatethata
measure is good if it is absolutely continuous with respect to an appropriate Lorentz-Bessel
capacities.
1 Introduction
Let Ω ⊂ RN be a bounded domain containing 0 and g : Ω×R → R be a Caratheodory
function. We assume that for almost all x ∈ Ω, r 7→ g(x,r) is nondecreasing and odd. In
this article we consider the following problem
−∆ u+g(x,u)=µ in Ω
p (1.1)
u=0 in ∂Ω
where∆ u=div |∇u|p−2∇u ,(1<p<N),isthe p-Laplacianandµaboundedmeasure.
p
A measureforwh(cid:16)ichthe probl(cid:17)emadmits asolution,in anappropriateclass,is calleda good
measure. When p = 2 and g(x,u) = g(u) the problem has been considered by Benilan and
Brezis [4] in the subcritical case that is when any bounded measure is good. They prove
that such is the case if N ≥3 and g satisfies
∞
g(s)s−NN−−12ds<∞. (1.2)
Z1
Thesupercriticalcase,alwayswithp=2,hasbeenconsideredbyBarasandPierre[3]when
g(u) = |u|q−1u and q > 1. They prove that the corresponding problem to (1.1 ) admits a
solution (always unique in that case) if and only if the measure µ is absolutely continuous
with respect to the Bessel capacity C (q′ = q/(q−1)). In the case p 6= 2 it is shown
2,q′
by Bidaut-V´eron [6] that if problem (1.1 ) with β = 0 and g(s) = |s|q−1s (q > p−1 > 0)
admits a solution, then µ is absolutely continuous with respect to any capacity Cp, q +ǫ
q+1−p
for any ǫ>0.
InthisarticleweintroduceanewclassofBesselcapacitieswhicharemodelledonLorentz
spaces Ls,q instead of Lq spaces. If G is the Bessel kernel of order α > 0, we denote by
α
Lα,s,q(RN)theBesovspacewhichisthespaceoffunctionsφ=G ∗f forsomef ∈Ls,q(RN)
α
and we set kφk = kfk (a norm which is defined by using rearrangements). Then we
α,s,q s,q
set
C (E)=inf{kfk : f ≥0, G ∗f ≥1 on E} (1.3)
α,s,q s,q α
for any Borel set E. We say that a measure µ in Ω is absolutely continuous with respect to
the capacity C if ,
α,s,q
∀E ⊂Ω, E Borel , C (E)=0=⇒|µ|(E)=0. (1.4)
α,s,q
We also introduce the Wolff potential of a positive measure µ∈M (RN) by
+
W [µ](x)= ∞ µ(Bt(x)) s−11 dt (1.5)
α,s tN−αs t
Z0 (cid:18) (cid:19)
2
if α > 0, 1 < s < α−1N. When we are dealing with bounded domains Ω ⊂ B and
R
µ∈M (Ω), it is useful to introduce truncated Wolff potentials.
+
WR [µ](x)= R µ(Bt(x)) s−11 dt (1.6)
α,s tN−αs t
Z0 (cid:18) (cid:19)
We prove the following existence results concerning
−∆ u+|x|−βg(u)=µ in Ω
p (1.7)
u=0 in ∂Ω
Theorem 1.1 Assume 1 < p < N, q > p−1 and 0 ≤ β < N and µ is a bounded Radon
measure in Ω.
1- If g(s)=|s|q−1s, then (1.7 ) admits a renormalized solution if µ is absolutely continuous
with respect to the capacity C .
p, Nq , q
Nq−(p−1)(N−β)) q+1−p
2- If g satisfies
∞
g(s)s−q−1ds<∞ (1.8)
Z1
then (1.7 ) admits a renormalized solution if µ is absolutely continuous with respect to the
capacity C .
p, Nq ,1
Nq−(p−1)(N−β))
Furthermore, in both case there holds
−cW2diam(Ω)[µ−](x)≤u(x)≤cW2diam(Ω)[µ+](x) for almost all x∈Ω. (1.9)
1,p 1,p
where c is a positive constant depending on p and N.
Inordertodealwithexponentialnonlinearitiesweintroducefor0<α<N thefractional
maximal operator (resp. the truncated fractional maximal operator), defined for a positive
measure µ by
µ(B (x)) µ(B (x))
M [µ](x)=sup t , resp M [µ](x)= sup t , (1.10)
α tN−α α,R tN−α
t>0 (cid:18) 0<t<R (cid:19)
and the η-fractional maximal operator (resp. the truncated η-fractional maximal operator)
µ(B (x)) µ(B (x))
Mη[µ](x)=sup t , resp Mη [µ](x)= sup t , (1.11)
α tN−αh (t) α,R tN−αh (t)
t>0 η (cid:18) 0<t<R η (cid:19)
where η ≥0 and
(−lnt)−η if 0<t< 1
h (t)= 2 (1.12)
η ( (ln2)−η if t≥ 21
Theorem 1.2 Assume 1 < p < N, τ > 0 and λ ≥ 1. Then there exists M > 0 depending
on N,p,τ and λ such that if a measure in Ω, µ=µ+−µ− can be decomposed as follows
µ+ =f +ν and µ− =f +ν , (1.13)
1 1 2 2
3
where f ∈L1(Ω) and ν ∈Mb (Ω) (j =1,2), and if
j + j +
(p−1)(λ−1)
Mp,2diaλm(Ω)[νj] <M, (1.14)
(cid:13) (cid:13)L∞(Ω)
(cid:13) (cid:13)
(cid:13) (cid:13)
there exists a renormalized solu(cid:13)tion to (cid:13)
−∆ u+sign(u) eτ|u|λ −1 =µ in Ω
p
(1.15)
(cid:16) u(cid:17)=0 in ∂Ω.
and satisfies (1.9 ).
OurstudyisbasedupondelicateestimatesonWolffpotentialsandη-fractionalmaximal
operators which are developed in the first part of this paper.
2 Lorentz spaces and capacities
2.1 Lorentz spaces
Let (X,Σ,α) be a measured space. If f : X →R is a measurable function, we set S (t) :=
f
{x∈X :|f|(x)>t} andλ (t)=α(S (t)). The decreasingrearrangementf∗ off is defined
f f
by
f∗(t)=inf{s>0:λ (s)≤t}.
f
It is well known that (Φ(f))∗ = Φ(f∗) for any continuous and nondecreasing function Φ :
R →R . We set
+ +
1 t
f∗∗(t)= f∗(τ)dτ ∀t>0.
t
Z0
and, for 1≤s<∞ and 1<q ≤∞,
∞tqs(f∗∗(t))qdt 1q if q <∞
t
kfkLs,q = (cid:18)Z0 (cid:19) (2.1)
 supessts1f∗∗(t) if q =∞
t>0
It is known that Ls,q(X,α) isa Banach space when endowed with the norm k.k . Fur-
Ls,q
thermore there holds (see e.g. [12])
s
t1sf∗ Lq(R+,dt) ≤kfkLs,q ≤ s−1 t1sf∗ Lq(R+,dt), (2.2)
(cid:13) (cid:13) t (cid:13) (cid:13) t
(cid:13) (cid:13) (cid:13) (cid:13)
the left-hand side i(cid:13)nequa(cid:13)lity being valid only if s >(cid:13)1. Fi(cid:13)nally, if f ∈ Ls,q(RN) (with
1 ≤ q,s < ∞ and α being the Lebesgue measure) and if {ρ } ⊂ C∞(RN) is a sequence of
n c
mollifiers, f∗ρ →f and(fχ )∗ρ →f inLs,q(RN), whereχ is the indicatorfunction
oftheballB cnenteredattheoBrniginonfradiusn. InparticularC∞B(nRN)isdenseinLs,q(RN).
n c
4
2.2 Wolff potentials, fractional and η-fractional maximal operators
If D is either a bounded domain or whole RN, we denote by M(D) (resp Mb(D)) the set
of Radon measure (resp. bounded Radon measures) in D. Their positive cones are M (D)
+
and Mb (D) respectively. If 0<R≤ ∞ and µ ∈M (D) and R ≥diam(D), we define, for
+ +
α>0 and 1<s<α−1N, the R-truncated Wolff-potential by
WR [µ](x)= R µ(Bt(x)) s−11 dt for a.e. x∈RN. (2.3)
α,s tN−αs t
Z0 (cid:18) (cid:19)
If h (t) = min{(−lnt)−η,(ln2)−η} and 0 < α < N, the truncated η-fractional maximal
η
operator is
µ(B (x))
Mη [µ](x)= sup t for a.e. x∈RN. (2.4)
α,R tN−αh (t)
0<t<R η
If R=∞, we drop it in expressions (2.3 ) and (2.4 ). In particular
µ(B (x))≤tN−αh (t)Mη [µ](x). (2.5)
t η α,R
We also define G the Bessel potential of a measure µ by
α
G [µ](x)= G (x−y)dµ(y) ∀x∈RN, (2.6)
α α
ZRN
where G is the Bessel kernel of order α in RN.
α
Definition 2.1 We denote by Lα,s,q(RN)the Besov space the space of functions φ=G ∗f
α
for some f ∈Ls,q(RN) and we set kφk =kfk . If we set
α,s,q s,q
C (E)=inf{kfk : f ≥0, G ∗f ≥1 on E}, (2.7)
α,s,q s,q α
then C is a capacity, see [1].
α,s,q
2.3 Estimates on potentials
In the sequel, we denote by |A| the N-dimensional Lebesgue measure of a measurable set
A and, if F,G are functions defined in RN, we set {F >a} := {x ∈ RN : F(x) > a},
{G≤b}:={x∈RN :G(x)≤b} and {F >a,G≤b}:={F >a}∩{G≤b}. The following
result is an extension of [14, Th 1.1]
Proposition 2.2 Let 0≤η <p−1, 0<αp<N and r >0. There exist c >0 depending
0
on N,α,p,η and ǫ > 0 depending on N,α,p,η,r such that, for all µ ∈ M (RN) with
0 +
diam(supp(µ))≤r and R∈(0,∞], ǫ∈(0,ǫ ], λ> µ(RN) p−11 l(r,R) there holds,
0
WαR,p[µ]>3λ,(Mηαp,R[µ])p−11 ≤ǫλ (cid:0) (cid:1)
(cid:12)n p−1o(cid:12) (2.8)
(cid:12)(cid:12) ≤c0exp − p4−(p1−−1η) p−1−η(cid:12)(cid:12)αpln2ǫ−p−p−1−1η {WαR,p[µ]>λ} .
(cid:18) (cid:16) (cid:17) (cid:19)(cid:12) (cid:12)
where l(r,R) = N−αp min{r,R}−Np−−α1p −R−Np−−α1p if R < ∞(cid:12), l(r,R) = N−(cid:12)αpr−Np−−α1p if
p−1 p−1
R=∞. Furthermore,(cid:16)if η =0, ǫ is independent of(cid:17)r and (2.8 ) holds for all µ∈M (RN)
0 +
with compact support in RN and R∈(0,∞], ǫ∈(0,ǫ ], λ>0.
0
5
Proof. Case R=∞. Let λ>0; since W [µ] is lower semicontinuous, the set
α,p
D :={W [µ]>λ}
λ α,p
is open. By Whitney covering lemma, there exists a countable set of closed cubes {Q }
i i
o o
such that D =∪ Q , Q ∩Q =∅ for i6=j and
λ i i i j
diam(Q)≤dist(Q,Dc)≤4diam(Q).
i i λ i
Let ǫ > 0 and Fǫ,λ = Wα,p[µ]>3λ,(Mηαp[µ])p−11 ≤ǫλ . We claim that there exist c0 =
c0(N,α,p,η) > 0 and nǫ0 = ǫ0(N,α,p,η,r) > 0 such thatofor any Q ∈ {Qi}i, ǫ ∈ (0,ǫ0] and
λ> µ(RN) p−11 l(r,∞) there holds
(cid:0) (cid:1) p−1
|Fǫ,λ∩Q|≤c0exp − p−1−η p−1−η ǫ−p−p−1−1ηαpln2 |Q|. (2.9)
4(p−1)
(cid:18) (cid:19) !
The first we show that there exists c > 0 depending on N,α,p and η such that for any
1
Q∈{Q } there holds
i i
F ∩Q⊂E ∀ǫ∈(0,c ],λ>0 (2.10)
ǫ,λ ǫ,λ 1
where
Eǫ,λ = x∈Q:Wα5,dpiam(Q)[µ](x)>λ,(Mαηp[µ](x))p−11 ≤ǫλ . (2.11)
Infact, take Q ∈ {Q }n such that Q∩F 6= ∅ and let x ∈ Dc such thoat dist(x ,Q) ≤
i i ǫ,λ Q λ Q
4diam(Q) and W [µ](x )≤λ. For k ∈N, r =5diam(Q) and x∈F ∩Q, we have
α,p Q 0 ǫ,λ
2k+1r0 µ(Bt(x)) p−11 dt
=A+B
tN−αp t
Z2kr0 (cid:18) (cid:19)
where
2k1+1+2k2+k1r0 µ(Bt(x)) p−11 dt 2k+1r0 µ(Bt(x)) p−11 dt
A= and B = .
Z2kr0 (cid:18) tN−αp (cid:19) t Z2k1+1+2k2+k1r0(cid:18) tN−αp (cid:19) t
Since
µ(B (x))≤tN−αph (t)Mη [µ](x)≤tN−αph (t)(ǫλ)p−1. (2.12)
t η αp η
Then
2k+1r0 tN−αphη(t)(ǫλ)p−1 p−11 dt 2k+1r0 1 dt
B ≤Z2k1+1+2k2+k1r0(cid:18) tN−αp (cid:19) t =ǫλZ2k1+1+2k2+k1r0(hη(t))p−1 t
Replacing h (t) by its value we obtain B ≤ c ǫλ2−k after a lengthy computation where c
η 2 2
depends onlyonpandη. Since δ :=(2k2+k1)Np−−α1p,then1−δ ≤c32−k where c3 depends only
on N−αp, thus
p−1
(1−δ)A≤c 2−k 2k+1r0 µ(Bt(x)) p−11 dt
3 tN−αp t
Z2kr0 (cid:18) (cid:19)
≤c32−kǫλ 2k+1r0(hη(t))p−11 dt
t
Z2kr0
≤c 2−kǫλ,
4
6
where c =c (N,α,p,η)>0.
4 4
Byachangeofvariablesandusingthatforanyx∈F ∩Qandt∈[r (1+2k),r (1+2k+1)],
ǫ,λ 0 0
B (x)⊂B (x ), we get
2kt t Q
1+2k
1
δA= r0(1+2k+1) µ(B12+k2tk)(x) p−1 dt ≤ r0(1+2k+1) µ(Bt(xQ)) p−11 dt.
 tN−αp  t tN−αp t
Zr0(1+2k) Zr0(1+2k) (cid:18) (cid:19)
 
Therefore
2k+1r0 µ(Bt(x)) p−11 dt ≤c 2−kǫλ+ r0(1+2k+1) µ(Bt(xQ)) p−11 dt,
tN−αp t 5 tN−αp t
Z2kr0 (cid:18) (cid:19) Zr0(1+2k) (cid:18) (cid:19)
with c =c (N,α,p,η)>0. This implies
5 5
∞ µ(Bt(x)) p−11 dt ∞ µ(Bt(xQ)) p−11 dt
≤2c ǫλ+ ≤(1+2c ǫ)λ, (2.13)
tN−αp t 5 tN−αp t 5
Zr0 (cid:18) (cid:19) Z2r0 (cid:18) (cid:19)
since W [µ](x )≤λ. If ǫ∈(0,c ] with c =(2c )−1 then
α,p Q 1 1 5
∞ µ(Bt(x) p−11 dt
≤2λ
tN−αp t
Zr0 (cid:18) (cid:19)
which implies (2.10 ).
Now,weletλ> µ(RN) p−11 l(r,∞). LetB beaballwithradiusrsuchthatsupp(µ)⊂B .
1 1
We denote B by the ball concentric to B with radius 2r. Since x∈/ B ,
2 1 2
(cid:0) (cid:1)
W [µ](x)= ∞ µ(Bt(x)) p−11 dt ≤ µ(RN) p−11 l(r,∞).
α,p tN−αp t
Zr (cid:18) (cid:19)
(cid:0) (cid:1)
Thus, we obtain D ⊂B . In particular, r =5diam(Q)≤20r.
λ 2 0
Next we set m = max(1,ln(40r)), so that 2−mr ≤2−1 if m≥m . Then for any x∈E
0 ln2 0 0 ǫ,λ
r0 µt(NB−t(αxp)) p−11 dtt ≤ǫλ r0 (hη(t))p−11dtt
Z2−mr0(cid:16) (cid:17) Z2−mr0
2−m0r0 −η dt r0 −η dt
≤ǫλ (−lnt)p−1 +ǫλ (ln2)p−1
t t
Z2−mr0 Z2−m0r0
≤m ǫλ+ (p−1)((m−m0)ln2)1−p−η1ǫλ.
0 p−1−η
Forthelastinequalitywehaveuseda1−p−η1−b1−p−η1 ≤(a−b)1−p−η1 validforanya≥b≥0.
Therefore,
r0 µ(Bt(x)) p−11 dt ≤ 2(p−1) m1−p−η1ǫλ ∀m∈N,m>mp−p−1−1η. (2.14)
tN−αp t p−1−η 0
Z2−mr0(cid:18) (cid:19)
7
Set
2−i+1r0 µ(Bt(x)) p−11 dt
g (x)= ,
i tN−αp t
Z2−ir0 (cid:18) (cid:19)
then
Wαr0,p[µ](x)≤ p2−(p1−−1)ηm1−p−η1ǫλ+Wα2−,pmr0[µ](x)
∞
≤ 2(p−1) m1−p−η1ǫλ+ gi(x)
p−1−η
i=m+1
X
p−1
for all m>mp−1−η. We deduce that, for β >0,
0
∞
|Eǫ,λ|≤ x∈Q: gi(x)> 1− 2(p−1) m1−p−η1ǫ λ
(cid:12)(cid:12)( i=Xm+1 (cid:18) p−1−η (cid:19) )(cid:12)(cid:12)
(cid:12) ∞ (cid:12)
≤ (cid:12)(cid:12)x∈Q: gi(x)>2−β(i−m−1)(1−2−β) 1− 2(p(cid:12)(cid:12)−1) m1−p−η1ǫ λ
(cid:12)(cid:12)( i=Xm+1 (cid:18) p−1−η (cid:19) )(cid:12)(cid:12)
(cid:12) ∞ (cid:12)
≤(cid:12)(cid:12) x∈Q:gi(x)>2−β(i−m−1)(1−2−β) 1− 2(p−1) m1−p−η1ǫ λ (cid:12)(cid:12).
p−1−η
i=m+1(cid:12)(cid:26) (cid:18) (cid:19) (cid:27)(cid:12)
X (cid:12) (cid:12)
(cid:12) ((cid:12)2.15)
(cid:12) (cid:12)
Next we claim that
c (N,η)
|{x∈Q:g (x)>s}|≤ 6 2−iαp|Q|(ǫλ)p−1. (2.16)
i sp−1
To see that, we pick x ∈E and we use the Chebyshev’s inequality
0 ǫ,λ
1
|{x∈Q:g (x)>s}|≤ |g |p−1dx
i sp−1 i
ZQ
1 r02−i+1 µ(Bt(x)) p−11 dt p−1
= dx
sp−1 tN−αp t
ZQ Zr02−i (cid:18) (cid:19) !
1 µ(B (x))
≤ r02−i+1 :=A.
sp−1 (r 2−i)N−αp
ZQ 0
Thanks to Fubini’s theorem, the last term A of the above inequality can be rewritten as
1 1
A= χ (y)dµ(y)dx
sp−1(r02−i)N−αpZQZRN Br02−i+1(x)
1 1
= χ (x)dxdµ(y)
sp−1(r02−i)N−αpZQ+Br02−i+1(0)ZQ Br02−i+1(y)
1 1
≤ |B (y)|dµ(y)
sp−1(r 2−i)N−αp r02−i+1
0 ZQ+Br02−i+1(0)
1
≤c (N) 2−iαprαpµ(Q+B (0))
7 sp−1 0 r02−i+1
1
≤c (N) 2−iαprαpµ(B (x )),
7 sp−1 0 r0(1+2−i+1) 0
8
sinceQ+B (0)⊂B (x ). Usingthefactthatµ(B (x ))≤(ln2)−ηtN−αp(ǫλ)p−1
r02−i+1 r0(1+2−i+1) 0 t 0
for all t>0 and r =5diam(Q), we obtain
0
1 1
A≤c (N,η) 2−iαprαp(r (1+2−i+1))N−αp(ǫλ)p−1 ≤c (N,η) 2−iαp|Q|(ǫλ)p−1,
8 sp−1 0 0 9 sp−1
which is (2.16 ). Consequently, (2.15 ) can be rewritten as
∞
c (N,η)
|E |≤ 6 2−iαp(ǫλ)p−1|Q|
ǫ,λ i=m+1 2−β(i−m−1)(1−2−β) 1− 2(p−1)m1−p−η1ǫ λ p−1
X p−1−η
(cid:16) (cid:16) p−1 (cid:17) (cid:17) ∞
≤c (N,η)2−(m+1)αp ǫ |Q| 1−2−β −p+1 2(β(p−1)−αp)(i−m−1).
6 1− 2(p−1)m1−p−η1ǫ!
p−1−η (cid:0) (cid:1) i=Xm+1
(2.17)
If we choose β =β(α,p) so that β(p−1)−αp<0, we obtain
p−1
ǫ p−1
|E |≤c 2−mαp |Q| ∀m>mp−1−η (2.18)
ǫ,λ 10 1− 2(p−1)m1−p−η1ǫ! 0
p−1−η
where c = c (N,α,p,η) > 0. Put ǫ = min 1 ,c . For any ǫ ∈ (0,ǫ ] we
choose m10∈N s1u0ch that 0 (cid:26)p4(−p1−−1η)m0+1 1(cid:27) 0
p−1 p−1 p−1 p−1
p−1−η p−1−η 1 p−1−η p−1−η p−1−η 1 p−1−η
−1 −1<m≤ −1 .
2(p−1) ǫ 2(p−1) ǫ
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
Then
p−1
ǫ
≤1
1− 2(p−1)m1−p−η1ǫ!
p−1−η
and
p−1
2−mαp ≤2αp−αp(p2(−p1−−1η))p−p−1−1η(1ǫ−1)p−p−1−1η ≤2αpexp −αpln2 p−1−η p−1−η ǫ−p−p−1−1η .
4(p−1)
(cid:18) (cid:19) !
Combining these inequalities with (2.18 ) and (2.10 ), we get (2.9 ).
In the case η =0 we still have for any m∈N, λ,ǫ>0 and x∈E
ǫ,λ
∞
Wr0 [µ](x)≤mǫλ+ g (x)
α,p i
i=m+1
X
Accordingly (2.18 ) reads as
p−1
ǫ
|E |≤c 2−mαp |Q| ∀m∈N,λ,ǫ>0 with mǫ<1.
ǫ,λ 10
1−mǫ
(cid:18) (cid:19)
9
Put ǫ = min{1,c }. For any ǫ ∈ (0,ǫ ] and m ∈ N satisfies ǫ−1−2 < m ≤ ǫ−1−1, we
0 2 1 0
finally get from (2.10 )
|F ∩Q|≤|E |≤c 22αpexp −αpǫ−1ln2 |Q|, (2.19)
ǫ,λ ǫ,λ 10
which ends the proof in the case R=∞. (cid:0) (cid:1)
Case R < ∞. For λ > 0, D = {WR > λ} is open. Using again Whitney covering
λ α,p
lemma, there exists a countable set of closed cubes Q := {Q } such that ∪ Q = D ,
i i i λ
o o
Q ∩Q =∅ fori6=j anddist(Q ,Dc)≤4diam(Q). IfQ∈Q:is suchthatdiam(Q)> R,
i j i λ i 8
there exists a finite number n of closed dyadic cubes {P }nQ such that ∪nQ P = Q,
Q j,Q j=1 j=1 j,Q
o o
P ∩P = ∅ if i 6= j and R < diam(P )≤ R. We set Q′ = Q∈Q:diam(Q)≤ R ,
i,Q j,Q 16 j,Q 8 8
Q′′ = P :1≤i≤n ,Q∈Q,diam(Q)> R and F =Q′∪Q′′.
i,Q Q 8 (cid:8) (cid:9)
For ǫ >(cid:8) 0 we denote again Fǫ,λ = WαR,p[µ]>(cid:9)3λ,(Mηαp,R[µ])p−11 ≤ǫλ . Let Q ∈ F such
that Fǫ,λ∩Q6=∅ and r0 =5diam(nQ). o
If dist(Dc,Q) ≤ 4diam(Q), that is if there exists x ∈ Dc such that dist(x ,Q) ≤
λ Q λ Q
4diam(Q) and WR [µ](x ) ≤ λ, we find, by the same argument as in the case R = ∞,
α,p Q
(2.13 ), that for any x∈F ∩Q there holds
ǫ,λ
R µ(Bt(x)) p−11 dt
≤(1+c ǫ)λ. (2.20)
tN−αp t 11
Zr0 (cid:18) (cid:19)
where c =c (N,α,p,η)>0.
11 11
If dist(Dc,Q) > 4diam(Q), we have R < diam(Q) ≤ R since Q ∈ Q′′. Then, for all
λ 16 8
x∈F ∩Q, there holds
ǫ,λ
R µ(Bt(x)) p−11 dt R tN−αp(ln2)−η(ǫλ)p−1 p−11 dt
≤
tN−αp t tN−αp t
Zr0 (cid:18) (cid:19) Z51R6 (cid:18) (cid:19)
=(ln2)−p−η1 ln16 ǫλ (2.21)
5
≤2ǫλ.
Thus, if we take ǫ∈(0,c ] with c =min{1,c−1}, we derive
12 12 11
F ∩Q⊂E , (2.22)
ǫ,λ ǫ,λ
where
1
E = Wr0 [µ]>λ, Mη [µ] p−1 ≤ǫλ .
ǫ,λ α,p αp,R
(cid:26) (cid:16) (cid:17) (cid:27)
Furthermore, since x∈/ B ,
2
WR [µ](x)= R µ(Bt(x)) p−11 dt ≤ µ(RN) p−11 l(r,R).
α,p tN−αp t
Zmin{r,R}(cid:18) (cid:19)
(cid:0) (cid:1)
10