Interpolation sets for Hardy-Sobolev spaces on the boundary of the unit ball of ${bf C}^n$

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Interpolation sets for Hardy-Sobolev spaces on the boundary of the unit ball of C n .Jaume Gudayol ?April 23,1998Abstract We study the interpolation sets for the Hardy-Sobolev spaces de?ned on the unit ball of C n .We begin by giving a natural extension to C n of the condition that is known to be necessary and su?tient for interpolation sets lying on the boundary of the unit disk.We show that under this condition the restriction of a function in the Hardy-Sobolev space to the set always exists,and lies in a Besov space.We then show that under the assumption that there is an holomorphic distance function for the set,there is an extension operator from these Besov spaces to the Hardy-Sobolev ones.1Introduction In this work we study the boundary interpolation sets for Hardy-Sobolev spaces de?ned on the unit ball of C n .The study of interpolation sets for di?erent spaces is one of the classical subjects of S.C.V.analysis.But in the previous works there are serious restrictions:one considers either sets contained in varieties or sets that have dimension less than one.In this work we study sets not having such restrictions.Even though there are other kinds of restrictions,we believe that one can ?nd here a (perhaps small)step towards the general case.The study of interpolation sets was begun by Carleson and Rudin (See

[Rud,80],chapter 10,for references).They showed (independently)that,for n =1,interpolation sets for the ball algebra were precisely those of zero Lebesgue 71b1b63083c4bb4cf7ecd126ter,and also for n =1,interpolation sets for A ∞(D )were described by Alexander,Taylor and Williams in [ATW,71].In this case the interpolation sets are those satisfying that for any arc I ?T ,

1d (e it ,E )dt ≤C log

1

?Partially supported by MEC grant PB95-0956-c02-01and CIRIT grant GRQ94-2014.

1

Interpolation sets for the spaces Aα(D)were caracterized by Dynkin in[Dyn,80]

and Bruna in[Bru,81].We will say that a closed set E?X satis?es the Uniform

Hole Condition(UHC-sets,for short)with respect to X if there exists0

sup{d(y,E),y∈B(x,r)}≥Cr.(1) The UHC as related to interpolation properties was introduced by Kotochigov, but other equivalent de?nitions have been introduced by other authors in di?er-

ent contexts.The de?nition says that a set has holes of a?xed size when looked at at any scale.Dynkin and Bruna proved that,for the spaces Aα(D),E is

an interpolation set i?E is a UHC-set.This characterization was obtained by

Dynkin forα∈N and by Bruna for all0<α<+∞.Later Dynkin([Dyn,84]) proved that a set is an interpolation set for the Hardy-Sobolev spaces i?it is a

UHC-set.

For n>1no caracterization of boundary interpolation sets is known,not

even for the ball algebra.This does not mean that there is no information about

interpolation sets.For the ball algebra,Rudin in[Rud,80]devotes all of chapter 10to these sets,that in this case are the same as peak sets and zero sets.There some examples are given,and one can?nd some background on the problem.

Also for the ball algebra,Nagel in[Nag,76]proved that any subset of a complex-tangential manifold is an interpolation set.On the other hand,Davie and?ksendal(see[Rud,80],section10.5)proved that any set that has,in a sense,dimension less than1is an interpolation set.Both results point out to the fact that an interpolation set can be as large as one wants in the complex-tangential directions,but has to be small in the other ones.

The study of zero sets and interpolation sets for spaces other than the ball

algebra has been done by several authors.In the case of Aα(B n)and Hardy-

Sobolev spaces results concerning sets contained in varieties were given by Bruna and Ortega in[B-O,86],[B-O,91],and[B-O,93].These works have provided us with our main inspiration.Chaumat and Chollet obtained several results for the space A∞(B n)in,for example,[C-C,86],and for the Gevrey classes in[C-C,88].

Our goal was to study interpolation sets for Hardy-Sobolev spaces.But in

this case,the?rst problem was to know,given an f in a Hardy-sobolev space, to which function space de?ned on the set would the restriction belong.This question,which is in most cases trivial,in this case is not so.However,the results in[B-O,86]showed clearly that the space of the restrictions should be some Besov space.But even in the real case,no general result on restrictions of functions to Besov spaces de?ned on arbitrary sets is known(however,Jonsson and Wallin in[J-W,84]and Jonsson in[Jon,94]give some partial results).In this article we give a restriction theorem for a general set E?S.For the restriction to exist,we impose that the Uniform Hole Condition(1)holds.We show that this condition in equivalent to other conditions that will be useful later,and in

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particular,that is equivalent to the fact that the set has,in a sense,dimension less than the dimension of S.

Once we have done that,we show that under some restrictions,there is an extension operator,thus proving that the given set is interpolating.The restriction we impose is that we assume that there is a holomorphic function behaving like the distance to the set.We give some examples of such functions. 2De?nitions and statement of results

The upper dimension of a set

Let(X,ρ)be a compact pseudo-metric space,with diam(X)<+∞(this means thatρsatis?es the triangle inequality with a constant).For x∈X,R>0 and k≥1,let N(x,R,k)be the maximum number of points lying in B(x,kR) separated by a distance greater or equal than R.

De?nition1We will say that(X,d)∈Υγif there exists C(γ)=C(X,d,γ) such that,for all x∈X and all0

N(x,R,k)≤C(γ)kγ.(Υγ)

We de?ne the upper dimensionΥ(X)as

Υ(X)=inf{γ,(X,d)∈Υγ}.

This dimension was?rst introduced by Larman under the name of uniform metric dimension.

We will say that a probability measureμlies in Uγ=Uγ(X,ρ)if there exists C(γ)so that for all x∈X and all0

μ(B(x,kR))≤Ckγμ(B(x,R)).(Uγ)

Note that,by taking k=1/R,(Uγ)implies the weaker condition that,for all x∈X and all0

μ(B(x,R))≥CRγ.(U′γ) Notice that if for someγ,μ∈Uγ,then suppμ=X.Moreover,in this caseμis a doubling measure,that is,there exists C>0for which:

μ(B(x,2r))≤Cμ(B(x,r)).

Let U=∪γUγ.It is easily seen(see[V-K,88])that U is precisely the set of all doubling measures with support on X.

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The lower dimension of a set

De?nition2We will say that(X,d)∈Λγif there exists C(γ)=C(X,d,γ) such that,for x∈X and0

N(x,R,k)≥C(γ)kγ.(Λγ) Then we de?ne the lower dimensionΛ(X)as:

Λ(X)=sup{γ,(X,d)∈Λγ}.

This dimension was?rst de?ned by Larman under the name of minimal dimen-sion.

We will say that a doubling measureμbelongs to Lγ=Lγ(X,ρ)if there exists C(γ)so that for all x∈X and all0

μ(B(x,kR))≥Ckγμ(B(x,R)).(Lγ) As before,by taking k=1/R,condition(Λγ)implies

μ(B(x,R))≤CRγ.(L′γ)

Note that L0poses no restriction onμ∈U.

The following improvement of Volberg and Konyagin’s theorem1in[V-K,88] can be found in[B-G,98]:

Theorem3Let(X,ρ)be a pseudo-metric space.Let(X,ρ)∈Υυ∩Λλ,for some0<λ≤υ<+∞.Then for anyυ′>υandλ′<λ(orλ′=0if Λ(E)=0)there existsμ∈Uυ′∩Lλ′.

The uniform hole condition

We are now going to restrict ourselves to closed subsets E?S.On S we will use the pseudo-metric given by d(x,y)=|1?x

B n×

2(d(x,y)+d(y,z))(for this and the following, see[Rud,80],chapter5).

If we considerρ(x,y)=d(x,y)1

For n=1,Bruna([Bru,81]),using also results from Dynkin([Dyn,80]) proved that E satis?ed the UHC i?there was a s<1so that E satis?edΣs, and that both conditions were equivalent to the boundedness of certain integrals. We are going to extend Dynkin and Bruna’s results to C n,n≥71b1b63083c4bb4cf7ecd126ly,we are going to prove the following:

Theorem5Let E?S be a closed set.Then the following statements are equivalent:

(a)E satis?es the Uniform Hole Condition;

(b)There is a s

(c)Υ(E)

(d)There are an a>0and C>0so that for any x∈S and R>0,

B(x,R)d(z,E)?a dσ(z)≤CR n?a;

(e)There is a C>0so that for any x∈S and R>0,

B(x,R)log d(z,E)?1 dσ(z)≤σ(B(x,R))log1

δ)δn?a

dδ

We will work with functions f ∈H p β(B n ).Recall that if f ∈H ol (B n ),we can de?ne its radial derivative as Nf (z )= z j ?

?z j

+?B n ),be a vector ?eld.We de?ne its weight ω(X )as 1/2if X is complex-tangential,i.e. a j V (B (z,δ)∩B n |X f (ζ)|dV (ζ),

where z ∈S .

In this context,we want to know under which conditions there is a reasonable way of de?ning the restriction

X f |E and in which space of functions it lies.For doing so,we use the results of [B-O,93],where the following is proved:

Lemma 6Let μbe a measure on S satisfying L ′d ,for some d

n ?d V (B (ζ,δ))

B (ζ,δ)

X f (ζ)dV (ζ),and they are equal.For a function f ∈H p β(B n )we can de?ne,as in [B-O,86]and [B-O,93],the non-isotropic Taylor polynomial at a point ζ∈E .This Taylor polynomial T αζf

is twice as long in the complex-tangential directions.The non-isotropic Taylor polynomial can be de?ned in an intrinsic way,using the covariant di?erentials of f ,as in [B-O,86],or in an explicit way,using local coordinates,as in [B-O,93].

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Let us express Tαζf(z)in coordinates:for a pointζ∈S,let w n(z,ζ)=1?z

2(γ1+

...+γn?1)<α,

Dγf(ζ):=lim

r→1?|γ|

V(B(ζ,δ)) B(ζ,δ)?|γ|

γ!

Dγf(ζ)w(z,ζ)γ,

where z∈B.

In view of that,we de?ne the holomorphic jets of class B pα(μ)as those collections F=(Fγ)ω(γ)<αof L p(μ)functions such that

ω(γ)<α Fγ p L p(dμ)+ ω(γ)<α E×E|Fγ(x)?Dγ(Tαy F)(x)|pμ[x,y]2(2) is?nite,whereμ[x,y]=μ(B(x,d(x,y))).

Note that if E is Alhfors regular the corresponding Besov space is given by the norm:

F B pα(μ)= ω(γ)<α Fγ L p(dμ)+ E×E|Fγ(x)?Dγ(Tαy F)(x)|p

2. Hence,onΓ1

d(x,y)αp?dμ(B(x,d(x,y)))2≈|x?y|αp?1

2

(αp?1

so that these spaces are of the Besov kind,but they have di?erent regularity in Γ1and Γ2.

Another useful fact,that we will use later without further comment,is that μ[x,y ]≈μ[y,x ].This is so because

μ(B (x,d (x,y )))≤μ(B (y,4d (x,y )))≤Cμ(B (y,d (x,y )))

because of U s ,and if we exchange x for y we get the reverse inequality.Hence our de?nition is symetric with respect to x and y .

We have seen that for f ∈H p

β(B n )and α=β?n ?d

p and β?n ?Υ(E )

2.Take n >s ≥Υ(E )

and d ≤Λ(E )close enough so that the same is true for β?n ?d p .Take any μ∈U s ∩L d .Then for any f ∈H p

β(B n )its restriction to E lies in

B p α(μ),where α=β?n ?d

ζ)?N in a suitable way,plus the bounding of certain integrals.The restriction on β?n ?

d p is mor

e or less natural.Notice that our spaces are de?ned using only ?rst di?erences.In the case o

f an Ahlfors-regular set,the restriction is that 2α∈N ,which is the natural one in this case.Thus our restriction is related to the use of ?rst di?erences.

The extension theorem

Consider H p

β(B n )and B p α(μ),with α=β?n ?d

f∈H p

β(B n)so that,forω(γ)<α,(Dγf)|E=Fγin L p(dμ).In this chapter we

will introduce a condition under which it holds:

De?nition8Let E?S be a closed set,and let h∈H ol(B)∩C∞(

B n\E;

2.for any di?erential operator X there is a constant C(X)so that

|X h(z)|≤Cd(z,E)1?ω(X)

for all z∈

q

dσ(z)<+∞.

It is well known that HF p,2

β(B n)=H p

β

(B n),and also that HF p,q1

β?HF p,q2βif

q2≥q1.

There are two reasons for working with the Triebel-Lizorkin norms.The ?rst one is that it is technically simpler to work with integer powers of R than to work with Rβ,forβ∈N.On the other hand,the results we get are more general.

Theorem9Let E?S be a closed set,withΥ(E)

β?n?Λ(E)

p

lies no integer multiple of1

p

and

β?n?s

p .Assume that there is a

holomorphic distance function h for E.Then for each jet(Fγ)ω(γ)<α∈B pα(μ)

there is an f∈HF p,1

β

so that,forω(γ)<α,we have(Dγf)|E=Fγin L p(dμ).

This theorem gives us directly that E is an interpolation set for the Hardy-Sobolev spaces.

To prove the theorem,we will?rst construct a function g satisfying the required growth and interpolation properties,and then we will correct it using a

On the other hand,it is easily checked that the condition that between α=β?n?d

p

lies no integer multiple of1

B n),had already been proved.For these spaces,Bruna and Ortega in[B-O,86]gave the following:

Theorem10(Bruna-Ortega)LetΓbe a transverse curve,and let E?ΓwithΥ(E)<1.Then E is an interpolation set for Aα,forα∈R+\N.

In reading the proof,it is easy to check that the fact that E is contained in a transverse curve is used at two points of it:when,in theorem4.3,it is proved that for such a set there is a holomorphic distance function;and in lemma5.7, where it is proved that for such a set,condition(d)in theorem5is satis?ed. But in the proof of the theorem what is used is thatΥ(E)

Consider now the Besov spaces A p q,α(B n).It is well kwown that if we have

two pairs(q1,α1)and(q2,α2)then A p q

1,α1=A p q

2,α2

wheneverα2?α1=(q2?

q1)/p.Another remarkable fact about these spaces is that,in a limit sense, A p0,α=H pα.

Let M={z∈B n+1,z n+1=0}≈B n.As M=B n,we have on

M the spaces A p q,α(M).We consider on B n+1the spaces H p

β(B n+1).Beat-

rous in[Bea,86]proves that there exists a bounded restriction operator R: H p

β

(B n+1)→A p q,α(B n)forβ=α?(q?1)/p,and that in this case there is

also an extension operator E:A p q,α(B n)→H p

β(B n+1)such that R?E=Id.

Therefore,using theorems7and9,we can obtain similar results for these Besov spaces.Moreover,in the process of passing from C n to C n+1we drop the conditionΥ(E)

Examples of holomorphic distance functions

Here we are going to give some examples of sets E for which there is a holomor-phic distance function.

The Chaumat-Chollet example:Our?rst example is the one given by Chaumat and Chollet in[C-C,88],where they construct a holomorphic distance function on E wheneverΥ(E)<1.

They proceed as follows:they begin with any set satisfying condition(f)in theorem5for some n?1

10

E by balls {B (ζj,k ,2?k ),j =1,...,N k }.Then,by de?ning

φ(z )=∞ k =1N k j =12?k (n ?a )ζj,k )

they get a holomorphic function φsuch that |φ(z )|≈d (z,E )a ?n and ?(φ(z ))>0.Hence h (z )=φ(z )

1(1?z

ζ)q ≥C q |1?z

F (z,?(x ))p

where,for

n 2+12?p .In particular,M is an interpolation set for H p β(B n ).If K ?M is a compact set,then it is an interpolation set for the ball algebra,and also for A α(B n ).For if we have a function f on K ,as M is totally real,we can extend it by any real method to the whole manifold,and then extend it from the manifold to the ball.On B p α(μ),though,there was no

known general result on the extension of functions from subsets of R n to R n .But in [Gud,98]one can ?nd the necessary results,so that we will be able to extend any function ?rst to M ,and then from M to H p β(B n ).Therefore,any compact subset of M is an interpolation set for H p β(B n ),with the usual restrictions on the indices.

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An interpolation set of Hausdor?dimension n?δ:For each0<δ<1 we can build an interpolation set with Hausdor?dimension n?δ.To do so,we consider the variety

Γ={z∈S,?(z1)=...=?(z n)=0}.

Then the Hausdor?dimension ofΓis n?1,andΥ(Γ)=Λ(Γ)=n?1

2

,1

2(1+z21+...+z2n).Then clearly f is a peak function forΓ,so

that f t(z)=f(e?it z)is a peak function forΓt.Now theorem6.2in[B-O,86] says that if M is a complex-tangential variety of dimension n?1and f∈A∞is a peak function on M,the function h(z)=1?f(z)satis?es|h(z)|≈d(z,M). Then it is easily checked that h is a holomorphic distance function for M.Hence for each t∈C the function1?f t(z)satis?es|1?f t(z)|≈d(z,Γt).

Let E=∪t∈C

δΓt.We want to construct a function h so that|h(z)|≈d(z,E).

To do so,letμbe the Hausdor?measure on Cδ,and letδ

h q(z)= Cδ1

δ?q we have built a holomorphic distance function for E,so E is an interpolation set for H p

β

(B n).

We only have to check3,as the other inequality is proved in the same way. To begin with,we will check the upper inequality.

If I is an interval centered at some t∈Cδ,thenμ(I)≈|I|δ.Now?x z∈B, let t0∈E be so that d(z,E)=d(z,Γt0)and for each t∈Cδlet z t∈Γt be so that d(z,z t)=d(z,Γt).Let B k be the set de?ned by

B k={t∈Cδ,d(z,Γt)≤2k d(z,Γt

)}

for k≥0,and B?1=?.Then if s∈B k,the distance from s to t0is comparable to d(Γs,Γt

).Thus,and because of the triangle inequality,

|s?t0|≤Cd(z s,z t0)≤C(d(z,Γs)+d(z,Γt0))≤C2k d(z,E),

12

so that μ(B k )≤C 2kδd (z,E )δ.Now if we decompose the integral over E into the integrals over the coronae B k +1\B k ,and use the previous inequality,we obtain

|h q (z )|≤C ∞ k =02?qk d (z,E )?q μ(B k \B k ?1)≤Cd (z,E )?q +δ

as we wanted to see.

For the other bound in 3,we use that d (z,Γt )≤d (z,e i (t ?t 0)z t 0)≤C (d (z,E )+|t ?t 0|),

whence {t,|t ?t 0|≤d (z,E )}?{t,d (z,Γt )≤Cd (z,E )}.Now,we use that ?(1?f t )>0and q <1,so that we can use the bound ?(1?f t )q ≥C q |1?f t |q .Thus we can,modulo a constant,enter the modulus inside the integral.Then we can bound the integral by the integral over a smaller set where we can compare d (z,E )with d (z,Γt ),and obtain the result.?

2.1proof of theorem 5

We begin by proving that (a )implies (b ),which is the hardest.To prove it,we will use the following lemma,due to Sawyer and Wheeden ([S-W,92]):

Lemma 12Let (X,d )be a separable quasi-metric space,that is,d satis?es the triangle inequality with constant A 0.Then for λ=8A 50and for any m ∈Z ,there are points {x k j ,k ≥m,j =1,...,n j }and Borel sets {X k j ,k ≥m,j =1,...,n j }(where n j ∈N ∪{∞})such that i)B (x k j ,λk )?X k j ?B (x k j ,λ

k +1);ii)for any k ≥m ,∪n j j =1

X k j =X ;iii)given i ,j ,k ,and l ,with m ≤k ≤l ,either X k i ?X l j or X k i ∩X l j =?.

We will work in (S,ρ).Here ρis a metric,so we can apply the previous lemma with any A 0≥1.Note also that there is a constant C 0,depending only on n ,so that for any ball Q (x,R )and any y ∈Q (x,R ),if r

We take λ=max {8,4/(K 0C 0)}.Fix x ∈S ,R >0and 0<ελm ?1,and apply lemma 12with λand m ?4to the pseudo-metric space (Q (x,R ),ρ).Let {X k j ,k ≥m ?4,j =1,...,n j }be the dyadic decomposition of Q (x,R )given by the lemma.Let k 0∈Z be the only integer so that λk 0?1<2R ≤λk 0.This implies that X k 01=Q (x,R )whereas n k 0?2>1.Write U k 0?2=Q (x,R ).Then U k 0?2satis?es

13

1.U k =

X k j ?U k X k j ;

2.E ε∩Q (x,R )?U k ,

with k =k 0.Fix k ≤k 0?2and assume we have built U k satisfying (1)and

(2).Assume also that λk ≥2ε/(K 0C 0).We are going to see that we can build U k ?2also satisfying the previous properties and with mass less than a constant times the mass of U k .Take a j so that X k j ?U k .Then Q (x k j ,λk )?X k j .As x k j ∈Q (x,R )there is a ball Q (z,C 0λk )?Q (x k j ,λk )∩Q (x,R ).Inside Q (z,C 0λk )there must be a ball Q (w,K 0C 0λk )not intersecting E .If K 0C 0λk >2ε,then Q (w,K 0C 0Cλ3n σ(U k ).We begin with U k 0?2and we can go through the previous process while λk 0?2j ≥2ε/(C 0K 0).As m is the last integer satisfying this inequality,we can keep on doing it while k 0?2j ≥m ,that is,while 2j ≤k 0?m .Take j to be the last one ful?lling this inequality.For such a j ,we can bound σ(Q (x,R )∩E ε)by σ(U j ).Applying the previous bounds,we get that:

σ(U j )≤ 1?1

Cλ3n 1

On the other hand,λm?1<2ε/(K0C0)andλk0≥2R,so thatλk0?m+1≥2K0C0R/ε,that is,

(k0?m)logλ≥log K0C0ε.

If we write s=?1

Cλ3n )/logλ,then s>0and,because of the previous inequality,

1?12(k0?m)R2n=exp(?s(k0?m)logλ)R2n≤Cεs R2n?s so that(E,ρ)∈Σ2n?s,therefore(E,d)∈Σn?s

))≥CNR2n,

2

whence N≤Ck2s,hence(E,ρ)∈Υ2s and so(E,d)∈Υs.

The implication(c)?(g)is obtained in essentially the same way.

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Proof of (b)?(d):Fix x and R .If d (x,E )≥2√

2R ,then for any

y ∈B (x,R )we have that d (y,E )≤6R .Then if we decompose the integral we have to bound into a sum of integrals on coronae of decreasing radii,and apply the trivial bounds to each of these integrals,we obtain the result.

Proof of (d)?(a):Let S R (x )=sup {d (y,E ),y ∈B (x,R )}.Then for any z ∈B (x,R ),d (z,E )?a ≥S R (x )?a .Using it to get an inferior bound of the integral in (d)gives us the result.To see that (e)implies (a)we proceed in the same way.Proof of (b)?(e):Assume d (x,E )≤2

√2√d (z,E )a dV (z )≥CNεn +1?a

so that N ≤C (R/ε)n +1?a and then (E,d )∈Υn +1?a ,where a >1.With this statement we have ?nished the proof of the theorem.?3Technical lemmas

The following lemmas are going to be later.

Lemma 13Let z ∈

d (y,z )a

dμ(y )

B n \E ,and let a,b,c >0.Then E ×E

d (x,y )c

μ[x,y ]2≤Cd (z,E )c ?a ?b ,

whenever c >s ,c ?a ?b <0,c ?a

16

Proof:We split E×E into the sets,

(E×E)1={(x,y)∈E×E,d(y,z)≤d(x,z)}

and its complementary.We will bound ony the integral over(E×E)1,as the other is bounded exactly in the same way,changing the roles of x and y.Note that,in(E×E)1,d(x,y)≤2

√

2d(y,z)≤2

√

2d(y,z)≤d(x,y)≤2

√

2d(y,z)}.

Because of U s,

μ(B(y,d(y,z)))=μ(B(y,d(x,y)d(y,z)

d(x,y) s,

whereμ[y,x]=μ(B(y,d(y,x))).Using it,and that d(x,z)?b≤d(y,z)?b,the integral over(E×E)11can be bounded by:

E1μ(B(y,d(x,y)))dμ(x)dμ(y)

2d(y,z)≤d(y,x)≤2?j+1

√

2d(y,z)))

2d(y,z)))

dμ(y)

d(y,z)a+b?c

dμ(y)

2d(x,y),so d(x,z)≈d(x,y).Also2

√

d(y,z)

))≥Cμ(B(y,d(y,z))) d(x,y)

Lemma15Let a,b,c≥0.If c?a?b+n+1<0,c?a+n+1>0and c?b+n+1>0,then,for any z,w∈

|1?ζw|b dV(ζ)≤C|1?z

ζw|≤|1?

2,by changing a and b for some a′and b′if

necessary.Recall that in B1,|1?wz|+(1?|ζ|)).Then if we write r=|ζ|,what we have to bound is

10(1?r2)c w|+1?r)a S1ζw|b dσ(ζ)dr.

But proposition1.4.10in[Rud,80]says that,for b>n,

S|1?r

wz|c+n+1?a?b ∞0t c?b+n wz|c+n+1?a?b,

the last integral being?nite if c?b+n>?1and c+n?a?bs?n?1,b0.Then the integral B d(z,E)a

Lemma 17Let 0|z |<1, 10 log 1|1?tz |a dt ≤C 1B n ,if we write T αy =T NI,αy ,we have:

T αy 1ζz )a (x )=? k =0 a +k ?1k [(x ?y )(1?

y ?1)(1?ζ?(x ζy k ?j .

Proof:In order to compute the non isotropic Taylor polynomial of weight αof a given F ,we begin by computing the isotropic Taylor polynomial of degree ?of F .We write it in terms of N = n i =1z i ??z j ?

2).Then we will keep only those terms with weight less

than α,and we will be done.Recall that if we make this development at a ?xed point y ,we must write the polynomial in terms of (N )y and (Y j )y (we are using here that the values of a tensor at a point depend only on the values of the coe?cients at that point).So we must write it in terms of N y = n

i =1y i

??z j |z =y ?

yN y .Thus A is complex tangential whereas B is not.Then a straightforward computation shows that:

(T I,?y F )(x )=? k =0

1

k !k j =0 k j

A k ?j

B j F.We compute now the non isotropic Taylor polynomial of weight α,for 2α/∈N .Let ?=[2α].We begin with the isotropic Taylor polynomial of degree ?,

19

and keep those terms with weight less thanα.Now the weight of A k?j B j is

k?j

2.So the terms we want are those with j≤k and j<2α?k,so

that j≤??k.For k≤?/2,the smaller of the two is k,so we kave to keep all the terms,whereas for k>?/2,we have to take away the terms from??k+1 up to k.Therefore,

T NI,αy F(x)=(T I,?

y

F)(x)?

?

k>?/21

ζz)?a.Note that:

N y F a=

n

i=1y i?(1?ζiζz)a+1 |z=y=ay y?1)y

?z i|z=y?ζi F a+1(y)?aζF a+1(y),

thus

AF a=a (x?y)y?1)

y?1)

ζ?(xζy j F a+j(y). By adding up these two things,we get:

A k?j

B j F a=(a+k?1)!

y?1)ζ?(xζy]k?j F a+k(y).

On the other hand,

(A+B)F a=a[(x?y)

ζ]j F a+j(y).

Using all of this in the formula4gives us the claim.?

20

Lemma 19Let f ∈H p β(B n )∩C (

B n ,and r ≥0,T αy f (x )=

C 1

0 B

(log 1(1?t B n ),for any z ∈

t )β?1R βf (tz )dt.

But theorem 7.1.4from

[Rud,80]says that for g ∈H p (B ),with p ≥1,and r ≥0,g (z )=C (n,r ) B

g (ζ)(1?|ζ|2)r

ζ)n +1+r dV (ζ).Then for α<β,we can di?erentiate under the integral and get the result.?Lemma 20Let β>0,α=β?n ?d

p and C <α.Then for r large enough

(r >(D ?α)p ?n ?1)),if we write:

I (x,y )= 1

0 log

1d (x,tζ)C +n +r +1d (y,tζ)D dV (ζ)dt we have the bound: E ×E I (x,y )p μ[x,y ]2≤C f p

H p (B ).

Proof:Take δ>0small enough (δ

10

(log 1d (x,tζ)(C +2δ)p +n +r +1d (y,tζ)(D ?4δ)p dV (ζ)dt times the integral

1

0(log 1

d (x,tζ)λp ′d (y,tζ)4δp ′dV (ζ)dt

p

p ′

.To bound the integral with respect

to t we obtain,we use that |1?t 2a |≈|1?ta |and then apply lemma 17.In this way we see that 5is bounded by d (z,E )?δp .21