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p-Harmonic measure is not additive on null sets

JOSE´ G. LLORENTE, JUANJ. MANFREDI ANDJANG-MEIWU

Dedicated to the memory of Tom Wolff.

Without his work this note would not have been possible.

Abstract. When 1 < p < and p = 2 the p-harmonic measure on the boundary of the half planeR2+ is not additive on null sets. In fact, there are finitely many sets E1, E2,...,Eκ in R, of p-harmonic measure zero, such that E1E2...Eκ=R.

Mathematics Subject Classification (2000):31A15 (primary); 35J70, 60G46 (secondary).

1.

Introduction

We consider the p-harmonic measure associated to the operator Lp(u)=div

|∇u|p2u ,

the p-Laplacian of a function u, for 1 < p < ∞. A p-harmonic function in a domain

R

n(n≥2)is a weak solution ofLpu=0; that is,uWloc1,p()and

|∇u|p2u,∇ϕ d x =0

whenever ϕC0(). Weak solutions of Lp(u) = 0 are indeed in the class Cloc1, where α depends only on p andn ([DB], [L1].) A lower semicontinuous v :

R

∪ {∞}is p-superharmonic provided thatv ≡ ∞, and for each open DD and eachu continuous on D and p-harmonic in D, the inequality vuon∂Dimpliesvuin D.

Let E be a subset of . Consider the class

C

(E, ) of nonnegative p- superharmonic functionsvinsuch that

lim inf

X∈,X→ζv(X)χE(ζ )

Pervenuto alla Redazione il 20 gennaio 2005 e in forma definitiva il 18 maggio 2005.

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for allζ. The p-harmonic measureof the set Erelative to the domainis the functionωp(.,E, )whose value at any Xis given by

ωp(X,E, )=inf{v(X):v

C

(E, )} .

We often omit the variableXand the domainand writeωp(E, )or justωp(E). The functionωp(E, )is p-harmonic in, satisfies

0≤ωp(E, )≤1,

andωp(E, )has boundary values 1 at all regular points interior toEand boundary values 0 at all regular points interior to\E. For these and additional potential- theoretic properties of the p-Laplacian see [GLM] and the book [HKM].

When p=2 harmonic functions have the mean value property. Supposeis a Dirichlet regular domain, thenω2(X,·, )is a probability measure onand the

integral

f(ζ )dω2(X, ζ, )

gives the solution to the Dirichlet problem for a given boundary data function f. When p=2, due to the nonlinearity of thep-Laplacian,p-harmonic functions need not satisfy the mean value property and the sum of two p-harmonic functions need not be p-harmonic. Consequentlyωp(X,·, )is not additive on, hence not a measure.

Very little is known about measure-theoretic properties ofp-harmonic measure when p =2. Assume thatis Dirichlet regular. Then for all compact subsets E of the boundarywe have

ωp(E, )+ωp(∂\E, )=1; (1.1) and if EandFare both compact, disjoint, andωp(E, )=ωp(F, )=0 then

ωp(EF, )=0. (1.2)

These results can be found in [GLM] and also in [HKM].

Some conditions on the smallness of a compact set F in terms of Hausdorff dimension or capacity that implyωp(EF, )=ωp(E, )can be found in [AM], [K] and [BBS].

Martio asked in [M1] whether p-harmonic measure defines an outer measure on the zero level; i.e., whether (1.2) remains true when E and F are allowed to intersect and to be noncompact.

In this note we answer Martio’s question negatively by showing thatωpis not additive on null sets when p =2. We construct an example when=

R

2+is the upper half-space and=

R

. We may consider the point at infinity as a part of the boundary but it is not difficult to see thatωp({∞},

R

2+)= 0. Points in

R

2+ will be denoted by(x,y)orXinterchangeably.

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Theorem 1.1. Let 1 < p <and p = 2. Then there exist finitely many sets E1,E2, . . . ,Eκon

R

such that

ωp(Ek,

R

2+)=0, ωp

R

\Ek,

R

2+=1, and κ k=1

Ek =

R

.

Furthermore, the sets Ek verify|

R

\Ek| =0.

Here|.|stands for Lebesgue measure on the real line.

Corollary 1.2. There exist A and B

R

such that

ωp(A,

R

2+)=ωp(B,

R

2+)=0 and ωp(AB,

R

2+) >0. Thusωp(·,

R

2+)is not additive on null sets.

Corollary 1.3. Let1 < p <and p =2. Thenωp(X,·,

R

2+)is not a Choquet capacity for each X. In fact there exists an increasing sequence of sets B1

B2⊆ · · · ⊆Bj ⊆ · · · ⊆

R

so that

jlim→∞ωp(Bj) < ωp

j=1

Bj

.

To prove Corollary 1.2, choosek0 = min{k: ωp(E1E2. . .Ek) > 0}and let A= E1E2. . .Ek01,B=Ek0.

Corollary 1.3 follows from Theorem 1.1 as in the tree case given in [KLW].

The definition of Choquet capacity can be found in [HKM].

Both the Theorem and its corollaries can be extended to

R

n+(n≥3)by adding n−2 dummy variables.

Until recently, there has been no ground for conjecturing the answer to Mar- tio’s and some other questions about p-harmonic measures. A sequence of papers [CFPR], [KW], [ARY] and [KLW], is devoted to studying p-harmonic measure and Fatou theorem for bounded p-harmonic functions in an overly simplified model – forward directed regularκ-branching trees. On such trees, Theorem 1 is proved and for each fixed pthe exact value of the minimum of Hausdorff dimension of Fatou sets over all bounded p-harmonic functions is given in [KW] and [KLW].

In [KLW] the construction of the sets in Theorem 1 for trees starts with a basic p-harmonic functionuthat does not satisfy the mean value property, follows with a Riesz product and then a stopping time argument. It is really quite simple. In

R

2+ we follow the same procedures. The basic p-harmonic function is infinitely more complicated and is provided by remarkable examples of Wolff for 2 < p < ∞, and of Lewis for 1 < p< 2 ([Wo1], [Wo2] and [L2]). On a tree there is a perfect independence among branches and the Riesz product includes all generations; in

R

2+we obtain an approximate independence by introducing large gaps in the Riesz product. Finally, instead of a stopping time argument, we use an ingenious lemma of Wolff [Wo1] on gap series of p-harmonic functions, to estimate the p-harmonic function whose boundary values are given by an infinite product.

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ACKNOWLEDGEMENTS. First author partially supported by grants DGICYT BFM 2002-04072-102-02 1 and SGR 149275, 149255. Second author supported in part by NSF award DMS-0500983. Third author supported in part by NSF award DMS- 0400810.

This research was initiated when the first author visited the University of Illinois and the University of Pittsburgh in February 2004. He wishes to thank both institutions for their hospitality.

2.

Preliminaries

In this section we recall several properties of p-harmonic functions which are needed in the proofs.

If u(X)is p-harmonic andc

R

, thenc+u(X), cu(X)andu(c X)are p- harmonic. Ifuis a nonnegative p-harmonic function inandBis a ball such that 2B ⊆ , then supBuCinfBu for someC = C(n,p) > 0 (Harnack inequal- ity). A nonconstant p-harmonic function in a domain cannot attain its supremum or infimum (Strong Maximum Principle). If a sequence of p-harmonic functions converges uniformly then the limit is also p-harmonic.

We list now some basic properties of p-harmonic measure.

1. Ifωp(X,E, )=0 at someXthenωp(Y,E, )=0 for any otherY by Harnack inequality.

2. If E1E2thenωp(E1, )ωp(E2, )(monotonicity).

3. If12and E12 thenωp(E, 1)ωp(E, 2)(Carleman’s principle).

4. If E1E2⊇, . . . ,⊇Ej. . . are closed sets on, then ωp

j=1

Ej, = lim

j→∞ωp(Ej) (upper semicontinuity on closed sets).

See chapter 11 in [HKM] for these properties.

We follow [Wo1] and letWpbe the class of all functions f :

R

2+

R

which areλ-periodic in thex variable(f(x+λ,y)= f(x,y))and satisfy

fpp=

[0,λ)×(0,∞)|∇f(x,y)|p d x d y<,

where the gradient is taken in the sense of distributions. If fWp then the function f has a well-defined trace on

R

; and among the functions g such that gfWphas trace 0 on

R

, there is a uniqueg, denoted by fˆ, which minimizes gp. The function fˆis the unique p-harmonic function in

R

2+ with boundary values f on

R

. Moreover, there existsξ

R

so that

| ˆf(x,y)ξ| ≤2e−γy λ f

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for someγ = γ (p) > 0, [Wo1]. Extend then fˆto

R

by its boundary values. The comparison principle holds in this setting: let f,gWp satisfy fg in the Sobolev sense on

R

, then fˆ≤ ˆgin

R

2+([Ma], [Wo1]).

The following lemma of Wolff ([Wo1]) is a substitute for a “local comparison principle” (unknown for p = 2)for p-harmonic functions. It is not difficult to prove (2.1) below for y < 1 and (2.3) below for y > 1. However, a much deeper analysis is needed to obtain (2.1) and (2.3) below on the two sides of the liney = −αfor some 0< α <1. We shall need the full force of Wolff’s lemma.

Wolff’s Lemma[Wo1]. Let 1 < p <. Define α = 1−2/p if p ≥ 2 and α = 1− p/2if p < 2. Let > 0 and0 < M <. Then there are small A= A(p, ,M) > 0and largeν0 = ν0(p, ,M) <so that the following are true:

Ifν > ν0is an integer, f , g, qLi p1(

R

)are periodic with periods1,1,ν1 respectively, and

max(f,g,q,f||Li p1,gLi p1, ν1qLi p1)M, then for(x,y)

R

2+we have

|(q f

+g)(x,y)(qˆ(x,y)f(x)+g(x))|< if y< −α. (2.1) If, in addition to the above,qˆ(x,y)→0as y→ ∞, then

|(q f

+g)(x,Aν−α)g(x)|< (2.2) and

|(q f

+g)(x,y)− ˆg(x,y)|< if y > −α. (2.3)

The key to [Wo1] and [L2] is the existence of a basic functionwhich shows the failure of the mean value property for periodic p-harmonic functions in the class Wp(

R

2+)when p=2. The mean of(x,0)on [0,1] equals the limit ofat∞ when p=2.

Theorem 2.1. (Wolff and Lewis [Wo1], [L2]) For1 < p <and p = 2there exists a Lipschitz function:

R

2+

R

such that Lp=0,has period1in the x variable(x+1,y)=(x,y),

[0,1)×(0,∞)|∇|pd xd y<+∞, 1

0

(x,0)d x >0, but (x,y)→0 as y→ ∞.

Note that when p = 2, the p-harmonic function|X|p−np−1 if p = n , or log|X| if p=n, fails to have the mean value property on any sphere or half plane in

R

n\ {0} (n≥2.) But these functions are not periodic.

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3.

Proofs

Proof of Theorem 1.1. Fix p =2, 1< p < ∞. Letbe the basic function of Wolff and Lewis. Note that(x,0)must take both positive and negative values by the comparison principle. Replacingbyc (c>0 small constant), if necessary, we may assume

< 1

2 (3.1)

and 1

0

log(1+(x,0))d x >0. Fix a positive integerκsuch that

κ k=1

ak >0 and κ k=1

(1+ak) >1, where

ak =min

(x,0):xk−1

κ ,k κ

(3.2) Let

L= Li p1, and fix >1 and an integern0>5 so that

1< <κ

k=1

(1+ak)1κ (3.3)

and

3n0<min

1+max{ak} −,L κ

. (3.4)

For convenience we write f(x)for f(x,0)andωp(E)forωp(E,

R

2+)from now on.

We shall chooseinductivelyan increasing sequence of positive powers of the integerκ

1< ν1< ν2< . . .

and shall define for eachk∈[1, κ] two sequences of functions on

R

q1k(x)=

x+ k−1 κ

, f1k(x)=1+q1k(x) (3.5) and

qkj(x)=

νjx+ k−1 κ

, fjk(x)= fkj1(x)(1+qkj(x)). (3.6)

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After these are defined, we observe from (3.2), (3.3) and the periodicity of (x)that

κ k=1

fjk(x)= j i=1

κ k=1

1+

νix+k−1 κ

> κj for all x. (3.7)

Next, it follows from (3.1) that for j ≥1 qkj< 1

2, (3.8)

2j < fjk <

3 2

j

, (3.9)

qkjLi p1j, (3.10)

and

fjkLi p1j2j. (3.11) We then define for eachk∈[1, κ] a set

Ek = {x

R

: fjk(x) > j for infinitely many js}. Observe that (3.7) implies

κ k=1

Ek =

R

.

To finish the proof we need to establish ωp(Ek)=0, ωp

R

\Ek,

R

+2=1, and |

R

\Ek| =0

for eachk.

We start by discussing the choice of {νj}and two other sequences{rj} and {tj}; we always assume{νj}are positive powers ofκ, and{rj}and{tj}are negative powers ofκ.

Setr0 = t0 = 1 and ν1 = 1. After {ν1, ν2, . . . , νj},{r0,r1, . . . ,rj1} and {t0,t1, . . . ,tj1}are chosen, the functions

{q1k,q2k, . . . ,qkj} and

{f1k, f2k, . . . , fkj}

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are then defined by (3.5) and (3.6) for eachk ∈[1, κ]. We then chooserj > 0 so that

rj <min{tj1, (Lνj6j+1)1} (3.12) and that

|fjk(x,y)fjk(x)|<3j1 if 0≤yrj (3.13) for allk∈[1, κ].

Let f = g= fjk,q =qkj+1, M = j2j and =3j1in Wolff’s lemma;

thenνj+1andtj can be chosen from (2.1) and (2.3) so that

νj+11<tj <rj (3.14)

|fjk+1(x,y)fkj(x)(1+qkj+1(x,y))|<3j1 if 0<ytj (3.15) and

|fjk+1(x,y)fkj(x,y)|<3j1 if ytj (3.16) for all k ∈ [1, κ]. The fact that 0 < α < 1 in Wolff’s lemma is needed here to ensure that we can always find a tj such that νj+11 < tj < rj.We also need the fact that qkj+1(x,y) → 0 as y → ∞ to obtain (3.16). This ends the induction procedure.

For each k ∈ [1, κ] the sequence {fjk}converges to a p-harmonic function fk on

R

2+ uniformly on compact subsets. Since{tj}is decreasing, it follows from (3.16) that

|fNk(x,y)fkj(x,y)|<3j if ytj (3.17) for allNjandk∈[1, κ]; and from (3.15) and (3.17) that

fNk(x,y) > 1

2fkj(x)−3j if tj+1ytj (3.18) for all Nj +1 andk ∈[1, κ]. To see (3.18), observe that, sinceytj+1, we get by (3.17),

|fNk(x,y)fjk+1(x,y)|<3j1. On the other hand, sinceytj, by (3.15) and (3.1) we have

fjk+1(x,y) > 1

2fjk(x)−3j1.

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We are now ready to proveωp(Ek)=0 andωp(

R

\Ek)=1 for allk∈[1, κ].

In view of the Harnack inequality and the strong maximum principle, it is enough to prove ωp(X0,Ek,

R

2+) = 0 and ωp

X0,

R

\Ek,

R

2+ = 1 for a fixed point X0

R

2+. We take X0 = (0,1). We fixk and from now on, we omitk in the subscripts and superscripts ofEk, qkj and fkj. LetGj = {x : fj(x) > j}, so that we have

E =

n=1

j=n

Gj.

By monotonicity we getωp(E)ωp

j=n

Gj

. Therefore to showωp(E)=0 it suffices to prove that for someC >0,

ωp

X0,

j=n

Gj

Cn for all n>n0. (3.19)

In fact it is enough to show that for someC >0, ωp

X0,

N j=n

Gj

<Cn for all N >n>n0 (3.20)

Let us see how (3.20) implies (3.19). Observe that

R

\N

j=nGj, Nn is a decreasing sequence of closed sets on

R

. Since the characteristic function of an open set is bounded and lower semicontinous, it is resolutive. Thus, we have

ωp

N

j=n

Gj =1−ωp

R

\

N j=n

Gj

and

ωp

j=n

Gj =1−ωp

R

\

j=n

Gj

(See (9.31) and (11.4) of [HKM].) By the upper semicontinuity of p-harmonic measure on closed sets, we can letN go to∞to get

lim

N→∞ωp

N

j=n

Gj =1−ωp

R

\

j=n

Gj .

Therefore we conclude

Nlim→∞ωp

N

j=n

Gj =ωp

j=n

Gj .

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By monotonicity we have ωp(

R

\ E)ωp

R

\

j=nGj

; the equality ωp(

R

\E)=1 follows again from (3.20).

We need to establish (3.20). Define for each j >n0a set Hj = I:κ-adic closed interval of lengthtj, max

xI fj(x)j −3j1

and let

Tj = Hj×[0,tj]. Observe that from the definition of Hj we have

fj(x) < j −3j1 on Hj\ Hoj (3.21) where Hoj is the relative interior ofHj. Hence, it follows that

GjGjHojHj.

Note from (3.8), (3.9), (3.10), (3.11), (3.12), and (3.14) that we have

|fj(x)fj(x)| ≤Lνj2jtj <3j61 if |xx| ≤tj. (3.22) Therefore the inequality

minHj fjj −3j21 (3.23) holds. Finally, from (3.13) and (3.14) we deduce

fj(x,y) > j−3j on Tj (3.24) We pause for a remark. If the statement

fN(x,y) >Cj on ∂Tj\Hoj for all Nj >n0 (3.25) were true, then it would follow from the comparison principle applied on the domain

R

2+\ ∪Nj=1Tj and the convergence of{fj}that

ωp

X0,N

j=n

Gjωp

X0,N

j=n

∂Tj\HojC1nfN(X0) <C(X0)n. This would give (3.20) and thus ωp(E) = 0. Since (3.25) need not be true on vertical edges in∂Tj, we need to modify the setsTj.

The connected components ofTj are mutually disjoint rectanglesQ of height tj and of widths integer multiples oftj. This class of rectangles is mapped to itself by the family of mappings(x,y)(mνj1+x,y),m

Z

.

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SupposeQ =[a,b]×[0,tj] is such a component. Then

fj(a), fj(b) < j−3j1 (3.26) by (3.21). There are two possibilities.

Case I: max

[a,b] fjj.

In this case defineQto be the empty set∅, and note from (3.26) and the definition ofGj that

Gj∩[a,b]= ∅. (3.27)

Case II: max

[a,b] fj > j.

In this case letIjQ=[a,a+tj] andJjQ =[b−tj,b], and note from (3.22), (3.23), and (3.26) that

j −3j < fj(x) < j −3j2 on IQjJjQ, so that we have

Gj(IjQJjQ)= ∅. (3.28) To modify Qin Case II, we need the following fact.

Fact. If I is a κ-adic closed interval of length t ( > n0)on which f(x)− 3, then I contains a κ-adic closed subinterval of length t+1 on which

f+1(x) > +1.

To see this, we write f+1=(1+q+1)fand note that I containstν+1pe- riods ofq+1. So from (3.2), the intervalI has at leasttν+1κ-adic subintervals of lengthκ1ν+11on whichq+1≥max{ak}. Let Ibe any one of such subintervals and let Ibe anyκ-adic subinterval ofIof lengtht+1. Then

f+1(−3)(1+max{ak}) > +1 on I by (3.4).

Therefore, we may choose two sequences ofκ-adic closed intervals:

IjQIQj+1IQj+2. . . and

JjQJjQ+1JjQ+2. . . such that|IQ| = |JQ| =tand

f(x) > −3 on IQJQ (3.29)

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for allj. Let

a=

=j

IQ and b =

=j

JQ. (3.30)

Clearly we have the inclusioin [a+tj,btj]⊆[a,b]⊆[a,b]. Finally replace Q by

Q=[a,b]×[0,tj] in Case II.

Set

Tj =

{Q: Q a component of Tj}, and

Hj =Tj∩ {y =0}.

Then it follows from (3.27) and (3.28) that

GjGjHojHjTjTj.

Claim. fN(x,y) > j/3 on∂Tj\ Hoj for allNj.

To establish the claim, note first that∂Tj\HojTj, so that (3.24) implies fj(x,y) > j −3j > j

3 on ∂Tj\Hoj .

Next assume Nj +1. On Tj∩ {tj+1ytj}, it follows from (3.18) and (3.23) that

fN(x,y) > 1

2fj(x)−3j > 1

2(j−3j21)−3j > j 3 .

The portion V = (∂Tj\ Hoj)∩ {0≤ ytj+1}consists of vertical line segments only. Suppose (x,y)V, then x = a orb, associated with some component [a,b]×[0,tj] ofTj, as defined in (3.30). If(x,y)V∩ {t+1yt}for some ∈[j+1,N −1], then

fN(x,y) > 1

2f(x)−3> 1

2(−3)−3> j 3 by (3.18) and (3.29). Finally, if(x,y)V∩ {0≤ ytN}, then

fN(x,y) > fN(x)−3N1> N −3N−3N1 > j 3 by (3.13), (3.14) and (3.29). This proves the claim.

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From the claim we deduce that the function u(x,y) = 3nfN(x,y) has valuesu(x,y) >1 on

N j=n

∂Tj∩ {y>0} = N j=n

(∂Tj\Hjo).

We can now pass to a subset to conclude u(x,y) >1 on

N

j=n

Tj

∩ {y>0}, for Nn>n0.

Repeat now the argument after (3.25).The statement (3.20) follows by apply- ing the comparison principle to the functionsuandωp

Nj=nGj

on the domain

R

2+\∪Nj=nTj.This completes the proof ofωp(Ek,

R

2+)=0 andωp(

R

\Ek,

R

2+)=1.

It remains to prove|

R

\Ek| =0 for allk∈[1, κ]. Defineon [0,1)so that (x)=log(1+a) on

−1 κ ,

κ

, 1≤κ,

and extendperiodically to

R

so that(x+1)=(x)for allx. Recall thata= min

(x):x ∈[κ1,κ]

. Define for eachk ∈[1, κ] a sequence of functionshk1, hk2,hk3, . . . so that

hkj(x)=

νjx+k−1 κ

m, wherem= 1κ κ

k=1

log(1+a).

Fixkin [1, κ]. Note thathkj is constant on each interval i1

κνj,κνij

,i an inte- ger, and has average zero with respect to the Lebesgue measureµon each interval

i −1 κνj1, i

κνj1

.

Here we have setν1 =κ1. Therefore the functionshk1,hk2,hk3, . . . are orthogonal inL2. Since the sequence is uniformly bounded, it has partial sums

hk1+hk2+ · · · +hkj =o(j3/4) µa.e. Since

log fjkj

=1

νx+k−1 κ

=m j+ j

1

hk(x)

and 1< <em, therefore forµ-almost everyx there exist an integer j(x) >0 so that

fjk(x) > j for all j > j(x).

This says that|

R

1\Ek| =0.

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4.

Questions and Comments

Many questions concerning p-harmonic measure andp-harmonic functions remain unanswered.

4.1.Are therecompactsetsA

R

andB

R

so that we have ωp(A,

R

2+)=ωp(B,

R

2+)=0, butωp(AB,

R

2+) >0?

4.2.Can the numberκof sets in Theorem 1.1 be as small as 2?

Based on a theorem of Baernstein [B], we conjecture that when pis close to 2 and p =2,κ =5 suffices. In the tree case,κmust be and can be any integer≥3 [KLW].

Theorem 4.1. (Baernstein [B])Let

D

be the unit disk in

R

2. For a set S

D

let S be the closed arc on∂

D

centered at1of length|S|. Suppose that E

D

is the union of two disjoint closed arcs of equal positive length, and that the two components of∂

D

\E have unequal length, then there exist p1and p2(depending on E) with1< p1<2< p2<such that

ωp(0,E,

D

) > ωp(0,E,

D

) for p1< p<2 (4.1) and

ωp(0,E,

D

) < ωp(0,E,

D

) for 2< p< p2. (4.2) If E

D

is the union of two disjoint closed arcs of unequal positive length for which the components of∂

D

\E do have equal length, then inequalities opposite to (4.1) and (4.2) are true.

According to Baernstein’s theorem, there exist 1 < p1 < 2 < p2 < ∞ so that for each p(p1,2)(2,p2), there is one set J among the four {eiθ : θ ∈ [0,45π]},{eiθ : θ ∈ [0,25π]∪[44π,65π]},{eiθ : θ ∈ [0,65π]}and{eiθ : θ ∈ [0,45π]∪[65π,85π]},which satisfies

ωp(0,J,

D

) <|J|/2π. (4.3) From this, a p-harmonic functionˆ on

D

having Lipschitz continuous boundary valuesmay be constructed so that(ˆ 0)=0 and

5 k=1

(ei(θ+k2π/5)) >c>0 for every θ ∈[0,2π]; (4.4)

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consequently,

1 2π

2π

0

(eiθ)dθ >c>0.

On the other hand, using p-capacity estimates we can show that if 1< p< 32 and J is an arc of the unit circle then (4.3) holds provided|J| < δ0(p). This implies that for 1 < p < 32, there existsˆ for which(ˆ 0) = 0 and (4.4) holds with 5 replaced by someκ=κ(p).

Letn(eiθ)=(ei nθ)for integersn ≥1. It is not clear, and probably false, whethern(0) = 0. Therefore it is unclear how to adapt Wolff’s lemma to disks.

Unlike in the half plane, shortening the period of the boundary function on

D

complicates the p-harmonic solution in

D

.

4.3.Given any Lipschitz functionon

D

, letbe the p-harmonic function in

D

with boundary values, and letn(eiθ)=(ei nθ)shortening the period. Suppose ( 0)21π 2π

0 (eiθ)dθ. We ask whether (

0)n(0)≤ 1

2π 2π

0

(eiθ)dθ for n≥2;

and what the value lim

n→∞n(0)might be.

4.4.Not much is known about the structure of the sets havingp-harmonic measure zero. Sets E

R

nof absolute p-harmonic measure zero,ωp(E∂, )=0 for all bounded domains, are exactly those of p-capacity zero. There exist sets on

R

n+of Hausdorff dimensionn−1 that have zerop-harmonic measure with respect to

R

n+ whenp=2. There are also sufficient conditions on setsE

R

n+in terms of porosity, that implyωp(E,

R

n+)= 0. For these and more, see [HM], [M2] and [W].

Further questions and discussions on p-harmonic measures can be found in [B] and [HKM].

4.5.Given a functionuin

R

n+, denote by

F

(u)the Fatou set

x

R

n1: lim

y0u(x,y)exists and it is finite

.

Fatou’s Theorem states that

R

n1\

F

(u)has zero(n−1)-dimensional measure for any bounded 2-harmonic functionu in

R

n+. When 1 < p < ∞and p = 2, the Hausdorff dimension of the Fatou set of any bounded p-harmonic function in

R

n+is bounded below by a positive numberc(n,p)independent of the function [FGMS], [MW].

Deep and unexpected examples in [Wo1], [Wo2] and [L2] show that Fatou Theorem relative to the Lebesgue measure fails when p=2.

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Theorem 4.2. (Wolff and Lewis [Wo1], [L2])For1 < p <and p =2, there exists a bounded p-harmonic function u on

R

2+ such that the Fatou set

F

(u)has zero length, and there exists a bounded positive p-harmonic functionvon

R

2+such

that the set

x

R

: lim

y0supv(x,y) >0

has zero length.

Define the infimum of the dimensions of Fatou sets to be dimF(p)=inf

dim

F

(u):ubounded p-harmonic in

R

2+, and the dimension of the p-harmonic measure to be

dimωp =inf

dimE :E

R

1, ωp(E,

R

2+)=1 .

We ask what the values of dimF(p)and dimωpare, and conjecture that dimωp = dimF(p) <1 whenp=2.

The question and the conjecture are based on results in [KW]. In the case of forward directed regularκ-branching trees(κ >1)whose boundary is normalized to have dimension 1, the infimum of the dimensions of Fatou sets dimF(κ,p)is attained and is given by

dimF(κ,p)=min







 logκ

1

exj logκ :

κ 1

xj|xj|p2=0







;

furthermore 0 < dimF(κ,p) < 1 except when p = 2 or κ = 2, and in the exceptional case dimF(κ,p)=1.

References

[ARY] V. ALVAREZ, J. M. RODR´IGUEZand D. V. YAKUBOVICH,Estimates for nonlinear harmonic “measures” on trees, Michigan Math. J.48(2001), 47–64.

[AM] P. AVILES´ and J. J. MANFREDI,On null sets of p-harmonic measure, In: “Partial Differential Equations with minimal smoothness and applications”, Chicago, IL 1990, B. Dahlberg et al. (eds.), Springer Verlag, New York, 1992, 33–36.

[B] A. BAERNSTEIN,Comparison of p-harmonic measures of subsets of the unit circle, St. Petersburg Math. J.9(1998), 543–551.

[BBS] A. BJORN¨ , J. BJORN¨ and N. SHANMUGALINGAM,A problem of Baernstein on the equality of the p-harmonic measure of a set and its closure, Proc. AMS, to appear.

[DB] E. DIBENEDETTO, C1-local regularity of weak solutions of degenerate elliptic equations, Nonlinear Anal.,7(1983), 827–850.

[CFPR] A. CANTON´ , J. L. FERNANDEZ´ , D. PESTANAand J. M. RODR´IGUEZ,On harmonic functions on trees, Potential Anal.15(2001), 1999–244.

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