Yves Achdou
email: achdou@math.jussieu.fr

Thibaut Deheuvels
email: thibaut.deheuvels@univ-rennes1.fr

Nicoletta Tchou
email: nicoletta.tchou@univ-rennes1.fr

JLip versus Sobolev Spaces on a Class of Self-Similar Fractal Foliages

For a class of self-similar sets Γ ∞ in R 2 , supplied with a probability measure µ called the self-similar measure, we investigate if the B q,q s (Γ ∞ ) regularity of a function can be characterized using the coefficients of its expansion in the Haar wavelet basis. Using the the Lipschitz spaces with jumps recently introduced by Jonsson, the question can be rephrased: when does B q,q s (Γ ∞ ) coincide with JLip(s, q, q; 0; Γ ∞ )? When Γ ∞ is totally disconnected, this question has been positively answered by Jonsson for all s, q, 0 < s < 1 and 1 ≤ q < ∞ (in fact, Jonsson has answered the broader question of characterizing B p,q s (Γ ∞ ), s > 0, 1 ≤ p, q < ∞, using possibly higher degree Haar wavelets coefficients). Here, we fully answer the question in the case when 0 < s < 1 and Γ ∞ is connected.

Introduction

There is a growing interest in analysis on self-similar fractal sets, see for instance Kigami [START_REF] Kigami | Analysis on fractals[END_REF], Strichartz [START_REF] Strichartz | Function spaces on fractals[END_REF][START_REF] Strichartz | Analysis on fractals[END_REF], Mosco [START_REF] Mosco | Energy functionals on certain fractal structures[END_REF][START_REF] Mosco | Dirichlet forms and self-similarity[END_REF] and references therein. These works aim at intrinsically defining function spaces using Dirichlet forms and a different metric from the Euclidean one. The results in this direction are often subject to the important assumption that the set is post-critically finite (or p.c.f.), see [START_REF] Kigami | Analysis on fractals[END_REF], page 23 for the definition. In a different direction, Jonsson has studied Lipschitz functions spaces on a self-similar fractal set S under a technical condition which yields a Markov inequality at any order, see the pioneering works [START_REF] Jonsson | Wavelets on fractals and Besov spaces[END_REF][START_REF] Jonsson | Haar wavelets of higher order on fractals and regularity of functions[END_REF]. More precisely, in [START_REF] Jonsson | Haar wavelets of higher order on fractals and regularity of functions[END_REF], Haar wavelets of arbitrary order on S were introduced and used for constructing a family of Lipschitz function spaces allowing jumps at some special points in S. These function spaces are named JLip(t, p, q; m; S), where t is a positive real number, p, q are two real numbers not smaller than 1 and m is an integer (m is the order of the Haar wavelets used for constructing the space). Here J stands for jumps, since the considered functions may jump at some points of S. The theory in [START_REF] Jonsson | Haar wavelets of higher order on fractals and regularity of functions[END_REF], which does not need the assumption that S be p.c.f., plays an important role in the present paper. It will be partially reviewed in § 4.1 (we will focus on the case when m = 0, p = q and 0 < t < 1). In the present work, for a class of self-similar sets contained in R 2 , we aim at studying the relationships between some JLip spaces and the more classical Besov spaces introduced and studied by Jonsson and Wallin [START_REF] Jonsson | Function spaces on subsets of R n[END_REF] for closed sets: consider a closed subset F of R n supplied with a Borel measure µ such that there exists a positive real number d and two positive constants c 1 and c 2 with c 1 r d ≤ µ(B(x, r)) ≤ c 2 r d , for all x ∈ F and r < 1 (here B(x, r) is the ball in F with center x and radius r, with respect to the Euclidean distance in R n ); the set F is said to be a d-set. In [START_REF] Jonsson | Function spaces on subsets of R n[END_REF], Sobolev and Besov spaces are defined on d-sets. For example, for 0 < s < 1, the Sobolev space B p,p s (F ) is defined as B p,p s (F ) = f ∈ L p µ (F );

x,y∈F,|x-y|<1

|f (x)f (y)| p |x -y| d+sp dµ(x)dµ(y) < ∞ ,

see the definition in [START_REF] Jonsson | Function spaces on subsets of R n[END_REF] page 103. In the same book, results on the extension (in R n ) of functions belonging to Besov and Sobolev spaces on F and trace results are proved using as a main ingredient Whitney extension theory. In particular, there exists a continuous trace operator from W 1,p (R n ) onto B p,p

1-n-d p (F ), if max(1, n -d) < p < ∞.
A more general trace theorem is available, see Theorem 1, page 141 in [START_REF] Jonsson | Function spaces on subsets of R n[END_REF].

The approach of Triebel [START_REF] Triebel | Fractals and spectra[END_REF] is somewhat different. In [START_REF] Triebel | Fractals and spectra[END_REF] chapter IV, paragraph 18, it is proved that the space of the traces on F of functions in B p,q n-d p (R n ) is L p µ (F ) for 0 < d < n, d/n < p < ∞ and 0 < q ≤ min(1, p); Besov spaces on F are then defined as spaces of the traces of Besov spaces on R n and embeddings properties are studied. In [START_REF] Jonsson | Haar wavelets of higher order on fractals and regularity of functions[END_REF], Jonsson has proved that if the self-similar set S is totally disconnected, then the JLip spaces coincide with Lipschitz or Besov spaces, more precisely that the spaces JLip(t, p, q; m; S) coincide with the Lipschitz spaces Lip(t, p, q; m; S) also introduced in [START_REF] Jonsson | Haar wavelets of higher order on fractals and regularity of functions[END_REF]. The latter are a generalization of the more classical spaces Lip(t, p, q; S) introduced in [START_REF] Jonsson | Function spaces on subsets of R n[END_REF] since Lip(t, p, q; [t]; S) = Lip(t, p, q; S). Note that Lip(t, p, q; [t]; S) = B p,q t (S), see [START_REF] Jonsson | The dual of Besov spaces on fractals[END_REF]. When the fractal set is not totally disconnected, the JLip space may not coincide with Lip or Besov spaces.

In the present work, we focus on a class of self-similar sets noted Γ ∞ below, see for example Figure 1. The set Γ ∞ is the unique compact subset of R 2 such that

Γ ∞ = f 1 (Γ ∞ ) ∪ f 2 (Γ ∞ ),
where f 1 and f 2 are two similitudes with rotation angles ±θ and contraction factor a, 0 < a ≤ a * (θ). As we shall see, Γ ∞ can be seen as a part of the boundary of a ramified domain Ω in R 2 , see Figure 1, and the restriction a ≤ a * (θ) allows for the construction of Ω as a union of nonoverlapping sub-domains, see [START_REF] Moran | Additive functions of intervals and Hausdorff measure[END_REF]. In § 2.2.3, we will recall the notion of self-similar measure µ defined in the triplet (Γ ∞ , f 1 , f 2 ), see [START_REF] Kigami | Analysis on fractals[END_REF]. With the Borel regular probability measure µ, Γ ∞ is a d-set where d ≡log 2/ log a is the Hausdorff dimension of Γ ∞ . The notion of traces on Γ ∞ for functions in W 1,q (Ω) has been defined in the earlier work [START_REF] Achdou | Neumann conditions on fractal boundaries[END_REF]. In [START_REF] Achdou | Trace theorems for a class of ramified domains with self-similar fractal boundaries[END_REF], some of the authors of the present paper have characterized the space of the traces on Γ ∞ of functions in W 1,q (Ω) as JLip(1 -2-d q , q, q; 0; Γ ∞ ), for 1 < q < ∞ (with d =log 2/ log a). Note that JLip(1 -2-d q , q, q; 0; Γ ∞ ) always contains Lip(1 -2-d q , q, q; 0; Γ ∞ ) = B q,q 1-2-d q (Γ ∞ ), and that

JLip(1 -2-d q , q, q; 0; Γ ∞ ) = Lip(1 -2-d q , q, q; 0; Γ ∞ ) if a < a * (θ). Therefore, the question considered here is to know for a = a * (θ), in which case the identity JLip(t, q, q; 0; Γ ∞ ) = Lip(t, q, q; 0; Γ ∞ ) = B q,q t (Γ ∞ )

holds, and if not, to find the parameters s such that JLip(t, q, q; 0; Γ ∞ ) ⊂ B q,q s (Γ ∞ ). The first part of the question covers the following one: when do the spaces containing the traces on Γ ∞ of the functions in W 1,q (Ω) and W 1,q (R 2 ) coincide? This is also linked to the possibility of constructing an extension operator from W 1,q (Ω) to W 1,q (R 2 ). Note that a partial answer was given in [START_REF] Achdou | Trace theorems on domains with self-similar fractal boundaries[END_REF] (before the characterization of the trace space as a JLip space was found) in the special case when q = 2 and for a special geometry (θ = π/4). We will see that two different situations occur:

• if there does not exist an integer p such that pθ = π/2 then f 1 (Γ ∞ ) ∩ f 2 (Γ ∞ ) is a singleton, and we will see that (2) holds if qt < d.

• Otherwise, the Hausdorff dimension of

f 1 (Γ ∞ ) ∩ f 2 (Γ ∞ ) is d/2 and (2) holds only for qt < d/2.
Finally, note that the question of extensions or traces naturally arises in boundary value or transmission problems in domains with fractal boundaries. Results in this direction have been given in [START_REF] Mosco | Variational problems with fractal layers[END_REF][START_REF] Lancia | Second order transmission problems across a fractal surface[END_REF][START_REF] Lancia | A transmission problem with a fractal interface[END_REF] for the Koch flake. There also, the assumption that the fractal set is p.c.f. is generally made. Boundary value problems posed in the domains Ω displayed in Figure 1 were studied in [START_REF] Achdou | Neumann conditions on fractal boundaries[END_REF].

The paper is organized as follows: the geometry is presented in Section 2. In Section 3, we recall some of the results of [START_REF] Achdou | Neumann conditions on fractal boundaries[END_REF] on the space W 1,q (Ω) and the construction of the trace operator. The theory proposed in [START_REF] Jonsson | Haar wavelets of higher order on fractals and regularity of functions[END_REF] is reviewed in Section 4, where we also recall the characterization of the trace space proved in [START_REF] Achdou | Trace theorems for a class of ramified domains with self-similar fractal boundaries[END_REF]. The main results of the paper are Theorems 6 and 7 which are stated in §5 and respectively proved in §5.2 and §5.3. For the ease of the reader, the geometrical lemmas, which are crucial but technical, are proved in the Appendix at the end of the paper.

The Geometry

2.1 The similitudes f 1 and f 2 and the self-similar set Γ ∞

Definitions

Consider four real numbers a, α, β, θ such that 0 < a < 1/ √ 2, α > 0, β > 0 and 0 < θ < π/2. Let f i , i = 1, 2 be the two similitudes in R 2 given by

f 1 x 1 x 2 = -α β + a x 1 cos θ -x 2 sin θ x 1 sin θ + x 2 cos θ , f 2 x 1 x 2 = α β + a x 1 cos θ + x 2 sin θ -x 1 sin θ + x 2 cos θ . (3)
The two similitudes have the same dilation ratio a and opposite angles ±θ. One can obtain f 2 by composing f 1 with the symmetry with respect to the axis {x 1 = 0}. We denote by Γ ∞ the self-similar set associated to the similitudes f 1 and f 2 , i.e. the unique compact subset of R 2 such that

Γ ∞ = f 1 (Γ ∞ ) ∪ f 2 (Γ ∞ ).
For n ≥ 1, we denote by

• A n the set containing all the 2 n mappings from {1, . . . , n} to {1, 2}

• A the set defined by

A = ∪ n≥1 A n • A ∞ = {1, 2} N\{0} the set of the sequences σ = (σ(i) ) i=1,...,∞ with values σ(i) ∈ {1, 2}.
Consider 1 ≤ m < n ≤ ∞ and σ ∈ A n : We say that σ m ∈ A m defined by σ m (i) = σ(i), i = 1, . . . , m is a prefix of σ. We also define for η ∈ A n and σ ∈ A k the sequence η + σ ∈ A n+k by η + σ = (η(1), . . . , η(n), σ(1), . . . , σ(k)).

For a positive integer n and σ ∈ A n , we define the similitude f σ by

f σ = f σ(1) • . . . • f σ(n) . (5)
Similarly

, if σ ∈ A ∞ , f σ = lim n→∞ f σ(1) • . . . • f σ(n) = lim n→∞ f σn (6)
Let the subset Γ ∞,σ of Γ ∞ be defined by

Γ ∞,σ = f σ (Γ ∞ ). ( 7
)
The definition of Γ ∞ implies that for all n > 0, Γ ∞ = σ∈An Γ ∞,σ . We also define the set Ξ ∞ :

Ξ ∞ = f 1 (Γ ∞ ) ∩ f 2 (Γ ∞ ). ( 8
)
The following theorem was stated by Mandelbrot et al, [START_REF] Mandelbrot | The canopy and shortest path in a self-contacting fractal tree[END_REF] (a complete proof is given in [7]):

Theorem 1 For any θ, 0 < θ < π/2, there exists a unique positive number a * (θ) < 1/ √ 2, (which does not depend of (α, β) see [START_REF] Achdou | Trace theorems for a class of ramified domains with self-similar fractal boundaries[END_REF]) such that

0 < a < a * (θ) ⇒ Ξ ∞ = ∅ ⇒ Γ ∞ is totally disconnected, a = a * (θ) ⇒ Ξ ∞ = ∅ ⇒ Γ ∞ is connected, (from Th. 1.6.2 in [16]). (9)
The critical parameter a * (θ) is the unique positive root of the polynomial equation:

p-1 i=0 X i+2 cos iθ = 1 2 , ( 10
)
where p is the smallest integer such that pθ ≥ π/2.

Remark 1 From (10), it can be seen that θ → a * (θ) is a continuous and increasing function from (0, π/2) onto (1/2, 1/ √ 2) and that lim θ→0 a * (θ) = 1/2.

Hereafter, for a given θ, 0 < θ < π/2, we will write for brevity a * instead of a * (θ) and we will only consider a such that 0 < a ≤ a * .

Characterization of Ξ ∞

We aim at characterizing Ξ ∞ defined in [START_REF] Falconer | Techniques in fractal geometry[END_REF]. We already know that Ξ ∞ = ∅ if and only if a = a * . Let us denote by Λ the vertical axis: Λ = {x : x 1 = 0} and by O the origin O = (0, 0). Since

f 1 (Γ ∞ ) = Γ ∞ ∩ {x 1 ≤ 0} and f 2 (Γ ∞ ) = Γ ∞ ∩ {x 1 ≥ 0}, we immediately see that Ξ ∞ = Γ ∞ ∩ Λ.
It can be observed (see [7] for the proof) that the sequences σ ∈ A ∞ such that f σ (O) ∈ Λ and that σ(1) = 1 are characterized by the following property: for all n ≤ 1, the truncated sequence σ n achieves the maximum of the abscissa of f η (O) over all η ∈ A n such that η(1) = 1.

Let us make out two cases, according to the value of p defined in [START_REF] Jones | Quasiconformal mappings and extendability of functions in Sobolev spaces[END_REF]:

The case when pθ > π/2

Proposition 1 If pθ > π/2 and a = a * , then Ξ ∞ contains the single point

ξ = lim n→∞ f 1 • f 2 p+1 • (f 1 • f 2 ) n (O) = lim n→∞ f 2 • f 1 p+1 • (f 2 • f 1 ) n (O). ( 12
)
Proof. For brevity, we skip the proof, which is available in [START_REF] Mandelbrot | The canopy and shortest path in a self-contacting fractal tree[END_REF]7].

The case when pθ = π/2 We need some specific notation:

• for i = 1, 2, we define ī = 2 si i = 1 1 si i = 2 ,
• we introduce

F 1 = f 1 • f 2 and F 2 = f 2 • f 1 , (13)
• for η ∈ A k , we define η (1) , η (2) ∈ A 2(k+1)+p by:

           η (1) (1) = 1 and η (1) ( ) = 2 for all ∈ [2, p + 2],
η (2) (1) = 2 and η (2)

( ) = 1 for all ∈ [2, p + 2], ∀j, 1 ≤ j ≤ k, η (1)
(p + 2j + 1) = η(j) η (1) (p + 2(j + 1)) = η(j) and η (2)

(p + 2j + 1) = η(j) η (2) (p + 2(j + 1)) = η(j) (14)
In an equivalent manner,

         η (1) = (1, 2, . . . , 2 p+1 , η(1), η(1)
, . . . , η(k), η(k)),

η (2) = (2, 1, . . . , 1 p+1 , η(1), η(1), . . . , η(k), η(k)), (15)
which yields

       f η (1) = f 1 • f 2 • . . . • f 2 p+1 • F η(1) • . . . • F η(k) , f η (2) = f 2 • f 1 • . . . • f 1 p+1 • F η(1) • . . . • F η(k) . (16)
Proposition 2 If pθ = π/2 and a = a * , then

Ξ ∞ = lim n→∞ f σ (1) n (O) = lim n→∞ f σ (2) n (O); σ ∈ A ∞ . (17)
Moreover, for x ∈ Ξ ∞ , there exists a unique σ ∈ A ∞ such that

x = lim n→∞ f σ (1) n (O) = lim n→∞ f σ (2) n (O). ( 18
)
The set Ξ ∞ is not countable.

Proof. For brevity, we skip the proof, which is available in [7].

Ramified domains

The construction

Call P 1 = (-1, 0) and P 2 = (1, 0) and Γ 0 the line segment Γ 0 = [P 1 , P 2 ]. We impose that f 2 (P 1 ), and f 2 (P 2 ) have positive coordinates, i.e. that a cos θ < α and a sin θ < β.

We also impose that the open domain Y 0 inside the closed polygonal line joining the points P 1 , P 2 , f 2 (P 2 ), f 2 (P 1 ), f 1 (P 2 ), f 1 (P 1 ), P 1 in this order is convex. With [START_REF] Mandelbrot | The canopy and shortest path in a self-contacting fractal tree[END_REF], this is true if and only if (α -1) sin θ + β cos θ ≥ 0.

Under assumptions [START_REF] Mandelbrot | The canopy and shortest path in a self-contacting fractal tree[END_REF] and [START_REF] Maz'ja | Springer Series in Soviet Mathematics[END_REF], the domain Y 0 is either hexagonal or trapezoidal in degenerate cases, contained in the half-plane x 2 > 0 and symmetric w.r.t. the vertical axis x 1 = 0. We introduce K 0 = Y 0 . It is possible to glue together K 0 , f 1 (K 0 ) and f 2 (K 0 ) and obtain a new polygonal domain, also symmetric with respect to the axis {x 1 = 0}. The assumptions [START_REF] Mandelbrot | The canopy and shortest path in a self-contacting fractal tree[END_REF] and [START_REF] Maz'ja | Springer Series in Soviet Mathematics[END_REF]

imply that Y 0 ∩ f 1 (Y 0 ) = ∅ and Y 0 ∩ f 2 (Y 0 ) = ∅.
We also define the ramified open domain Ω, see Figure 1:

Ω = Interior K 0 ∪ ∪ σ∈A f σ (K 0 ) . ( 21
)
Note that Ω is symmetric with respect to the axis x 1 = 0, and that for a < 1/ √ 2, the measure of Ω is finite.

For a given θ, with a * defined as above, we shall make the following assumption on (α, β): Assumption 1 For 0 < θ < π/2, the parameters α and β satisfy [START_REF] Maz'ja | Springer Series in Soviet Mathematics[END_REF] and [START_REF] Mandelbrot | The canopy and shortest path in a self-contacting fractal tree[END_REF] for a = a * , and are such that

   i) for all a, 0 < a ≤ a * , the sets Y 0 , f σ (Y 0 ), σ ∈ A n , n > 0, are disjoint ii) for all a, 0 < a < a * , f 1 (Ω) ∩ f 2 (Ω) = ∅ iii) for a = a * , f 1 (Ω) ∩ f 2 (Ω) = ∅.
Remark 2 Assumption 1 implies that if a = a * , then f 1 (Ω)∩f 2 (Ω) = ∅; to prove this, we define the open set

T = Interior K 0 ∪ f 1 (K 0 ) ∪ f 2 (K 0 ) . It is easy to check that Ω = T ∪ ∪ σ∈A f σ (T ).
If f 1 (Ω) ∩ f 2 (Ω) = ∅, there exist x ∈ Ω, a positive number ρ, two positive integers n and n , and σ ∈ A n and σ ∈ A n with σ(1) = 1 and σ (1) = 2 such that B(x, ρ) ⊂ f σ (T ) ∩ f σ (T ). It is then easy to prove that this contradicts point i) in Assumption 1.

The following theorem proved in [START_REF] Achdou | Trace theorems for a class of ramified domains with self-similar fractal boundaries[END_REF] asserts that ∀θ, 0 < θ < π/2, there exists (α, β) satisfying Assumption 1.

Theorem 2 If θ ∈ (0, π/2), then for all α > a * cos θ, there exists β > 0 such that β > a * sin θ and (α -1) sin θ + β cos θ ≥ 0 and for all β ≥ β, (α, β) satisfies Assumption 1.

It has been proved in [START_REF] Achdou | Trace theorems on domains with self-similar fractal boundaries[END_REF] that if a < a * , then there exists > 0 and δ > 0 such that Ω is aδ domain as defined by Jones [START_REF] Jones | Quasiconformal mappings and extendability of functions in Sobolev spaces[END_REF], see also [START_REF] Jonsson | Function spaces on subsets of R n[END_REF] or in an equivalent manner a quasi-disk, see [START_REF] Maz'ja | Springer Series in Soviet Mathematics[END_REF]. On the contrary, if a = a * , then Ω is not aδ domain because from Propositions 1 and 2, it is possible to construct two sequences (x

(1) n ) n and (x (2) n ) n , x (1) n ∈ f 1 (Ω) and x (2) n ∈ f 2 (Ω) such that lim n→∞ |x (1) n -x (2)
n | = 0; then, any arc contained in Ω and joining x

n to x (2)
n has a length bounded from below by a positive constant.

The Moran condition

The Moran condition, (or open set condition), see [START_REF] Moran | Additive functions of intervals and Hausdorff measure[END_REF][START_REF] Kigami | Analysis on fractals[END_REF], is that there exists a nonempty bounded open subset ω of R 2 such that f 1 (ω) ∩ f 2 (ω) = ∅ and f 1 (ω) ∪ f 2 (ω) ⊂ ω. For a given θ ∈ (0, π/2), let (α, β) satisfy Assumption 1; for 0 < a ≤ a * , the Moran condition is satisfied with ω = Ω because

• f 1 (Ω) ∩ f 2 (Ω) = ∅, which stems from point ii) in Assumption 1 if a < a * ,

and from

Remark 2 if a = a * ;

• by construction of Ω, we also have

f 1 (Ω) ∪ f 2 (Ω) ⊂ Ω. The Moran condition implies that the Hausdorff dimension of Γ ∞ is dim H (Γ ∞ ) = d ≡ -log 2/ log a,
see [START_REF] Moran | Additive functions of intervals and Hausdorff measure[END_REF][START_REF] Kigami | Analysis on fractals[END_REF]. If 0 < θ < π/2, we have 0 < a ≤ a * < 1/ √ 2 and thus d < 2. It can also be seen that if pθ = π/2 and a = a * , then the Hausdorff dimension of Ξ ∞ is d/2.

The self-similar measure µ

To define traces on Γ ∞ , we recall the classical result on self-similar measures, see [START_REF] Falconer | Techniques in fractal geometry[END_REF][START_REF] Hutchinson | Fractals and self-similarity[END_REF] and [START_REF] Kigami | Analysis on fractals[END_REF] page 26:

Theorem 3 There exists a unique Borel regular probability measure µ on Γ ∞ such that for any

Borel set A ⊂ Γ ∞ , µ(A) = 1 2 µ f -1 1 (A) + 1 2 µ f -1 2 (A) . (22)
The measure µ is called the self-similar measure defined in the self-similar triplet (Γ ∞ , f 1 , f 2 ).

Proposition 3

The measure µ is a d-measure on Γ ∞ , with d =log 2/ log a, according to the definition in [START_REF] Jonsson | Function spaces on subsets of R n[END_REF], page 28: there exists two positive constants c 1 and c 2 such that

c 1 r d ≤ µ(B(x, r)) ≤ c 2 r d ,
for any r 0 < r < 1 and x ∈ Γ ∞ , where B(x, r) is the Euclidean ball in Γ ∞ centered at x and with radius r. In other words the closed set Γ ∞ is a d-set, see [START_REF] Jonsson | Function spaces on subsets of R n[END_REF], page 28.

Proof. The proof stems from the Moran condition. It is due to Moran [START_REF] Moran | Additive functions of intervals and Hausdorff measure[END_REF] and has been extended by Kigami, see [START_REF] Kigami | Analysis on fractals[END_REF], §1.5, especially Proposition 1.5.8 and Theorem 1.5.7. We define L p µ , p ∈ [1, +∞) as the space of the measurable functions v on Γ ∞ such that

Γ ∞ |v| p dµ < ∞, endowed with the norm v L p µ = Γ ∞ |v| p dµ 1/p .
We also introduce L ∞ µ , the space of essentially bounded functions with respect to the measure µ. A Hilbertian basis of L 2 µ can be constructed with e.g. Haar wavelets.

Example

We make the choice

θ = π/4, α = 1 -a/ √ 2, β = 1 + a/ √ 2. Hence p = 2. The critical parameter a * (π/4) is the unique positive solution of X 3 + √ 2X 2 - √ 2/2 = 0, i.e.
a ≤ a * (π/4) 0.593465. The construction described in § 2.2.1 with the critical value a = a * (π/4) leads to the domain Ω shown in the left part of Figure 1. If a > 1/2, the Hausdorff dimension of Γ ∞ is larger than one. For instance, if a = a * (π/4), then dim H (Γ ∞ ) 1.3284371. In the right part of Figure 1, we show a similar construction with θ = π/5 (for which p = 3) and a = a * (π/5) 0.56658. Note the difference between the two cases: in the former case p(θ) • θ = π/2 and the set Ξ ∞ defined in ( 8) is not countable whereas in the latter case, p(θ) • θ > π/2 and the set Ξ ∞ is a singleton.

= π/4, a = a * (π/4), α = 1 -a * / √ 2, β = 1 + a * / √ 2.
Right: a similar construction for θ = π/5 and a = a * (π/5).

Additional notations

We define the sets

Γ σ = f σ (Γ 0 ) and Γ N = ∪ σ∈A N Γ σ . The one-dimensional Lebesgue measure of Γ σ for σ ∈ A N and of Γ N are |Γ σ | = a N |Γ 0 | and |Γ N | = (2a) N |Γ 0 |.
We will sometimes use the notation or to indicate that there may arise constants in the estimates, which are independent of the index n in Γ n , or of the index σ in Γ σ or Γ ∞,σ . We may also write A B if A B and B A.

3 The space W 1,q (Ω)

Hereafter, we take θ in (0, π/2) and suppose that the parameters (α, β) satisfy Assumption 1.

Basic facts For a real number q ≥ 1, let W 1,q (Ω) be the space of functions in L q (Ω) with first order partial derivatives in L q (Ω). The space W 1,q (Ω) is a Banach space with the norm

u q L q (Ω) + ∂u ∂x 1 q L q (Ω) + ∂u ∂x 2 q L q (Ω)
1 q , see for example [START_REF] Adams | Sobolev spaces[END_REF], p 60. Elementary calculus shows

that u W 1,q (Ω) ≡ u q L q (Ω) + ∇u q L q (Ω) 1 q is an equivalent norm, with ∇u q L q (Ω) ≡ Ω |∇u| q and |∇u| = | ∂u ∂x 1 | 2 + | ∂u ∂x 2 | 2 .
The spaces W 1,q (Ω) as well as elliptic boundary value problems in Ω have been studied in [START_REF] Achdou | Neumann conditions on fractal boundaries[END_REF], with, in particular Poincaré inequalities and a Rellich compactness theorem. The same results in a similar but different geometry were proved by Berger [START_REF] Berger | Eigenvalue distribution of elliptic operators of second order with Neumann boundary conditions in a snowflake domain[END_REF] with other methods.

Traces We first discuss very briefly the less interesting case when a < 1/2. If a < 1/2, then d < 1 and Γ ∞ is totally disconnected, see [START_REF] Falconer | The geometry of fractal sets[END_REF], Lemma 4.1 page 54. This implies that [START_REF] Kigami | Analysis on fractals[END_REF], theorem 1.6.2 page 33. The results of Jones [START_REF] Jones | Quasiconformal mappings and extendability of functions in Sobolev spaces[END_REF] and of Jonsson and Wallin [START_REF] Jonsson | Function spaces on subsets of R n[END_REF] can be combined to prove that if q > max(1, 2d), then the space of the traces on Γ ∞ of the functions v ∈ W 1,q (Ω) is B q,q 1-2-d q (Γ ∞ ) (see the introduction for the definition). We will see in Theorem 4 below that in this case, B q,q

f 1 (Γ ∞ ) ∩ f 2 (Γ ∞ ) = ∅, see
1-2-d q (Γ ∞ ) = JLip(1-2-d q , q, q; 0; Γ ∞ ).
Since the case a < 1/2 is understood, in the remaining part of the paper, we will take a such that 1/2 ≤ a ≤ a * , so the Hausdorff dimension d of Γ ∞ is not smaller than 1. We recall the construction of the trace operator made in [START_REF] Achdou | Neumann conditions on fractal boundaries[END_REF] by taking advantage of the self-simililarity; this trace operator, called ∞ below, is well defined even if a = a * . We first construct a sequence ( n ) n of approximations of the trace operator: consider the sequence of linear operators n :

W 1,q (Ω) → L q µ , n (v) = σ∈An 1 |Γ σ | Γ σ v dx 1 fσ(Γ ∞ ) , (23)
where

|Γ σ | is the one-dimensional Lebesgue measure of Γ σ .
Proposition 4 The sequence ( n ) n converges in L(W 1,q (Ω), L q µ ) to an operator that we call ∞ .

Proof. See [START_REF] Achdou | Neumann conditions on fractal boundaries[END_REF].

Remark 3 For a given θ, 0 < θ < π/2, let (α, β) satisfy Assumption 1 and Ω be constructed as in § 2.2.1, with 1/2 ≤ a ≤ a * ; in a work in progress [START_REF] Achdou | Extensions and traces for a class of domains with self-similar fractal boundaries[END_REF], we prove that Ω is a 2-set as defined in e.g. [START_REF] Jonsson | Function spaces on subsets of R n[END_REF] page 205, i.e. there exist three positive constants r 0 , c 1 and c 2 such that for any closed ball

B(P, r), P ∈ Ω, 0 < r ≤ r 0 , c 1 r 2 ≤ m 2 (B(P, r) ∩ Ω) ≤ c 2 r 2 , where m 2 is the Lebesgue measure in R 2 .
Since Ω is a 2-set, there is a classical definition of a trace operator on ∂Ω, see for instance [START_REF] Jonsson | Function spaces on subsets of R n[END_REF] page 206.

Although it has no bearing on the present paper, it is interesting to compare the operator ∞ , whose construction is based on the self-similarity properties, with the latter classical trace operator. In [START_REF] Achdou | Extensions and traces for a class of domains with self-similar fractal boundaries[END_REF], one of the goals is to prove that if q > 1, the two definitions of the trace of a function u ∈ W 1,q (Ω) coincide µ-almost everywhere. At the present stage of our investigations, we have proved the result for all θ ∈ (0, π/2)\{π/2k, 0 < k ∈ N} and 0 < a ≤ a * , (and also for all θ = π/2k, 1 < k ∈ N and 0 < a < a * ).

4 The spaces JLip(t, q, q; 0; Γ ∞ ) for 0 < t < 1 and the trace theorem

In [START_REF] Jonsson | Haar wavelets of higher order on fractals and regularity of functions[END_REF], A. Jonsson has introduced Haar wavelets of arbitrary order on self-similar fractal sets and has used these wavelets for constructing a family of Lipschitz spaces. These function spaces are named JLip(t, p, q; m; S), where S is the fractal set, t is a nonnegative real number, p, q are two real numbers not smaller than 1 and m is an integer (m is the order of the Haar wavelets used for constructing the space). Here J stands for jumps, since the considered functions may jump at some points of S. If the fractal set S is totally disconnected, then these spaces coincide with the Lipschitz spaces Lip(t, p, q; m; S) also introduced in [START_REF] Jonsson | Haar wavelets of higher order on fractals and regularity of functions[END_REF]. The latter are a generalization of the more classical spaces Lip(t, p, q; S) introduced in [START_REF] Jonsson | Function spaces on subsets of R n[END_REF] since Lip(t, p, q; [t]; S) = Lip(t, p, q; S). Note that Lip(t, p, q; [t]; S) = B p,q t (S), see [START_REF] Jonsson | The dual of Besov spaces on fractals[END_REF]. We will focus on the case when S = Γ ∞ , m = 0 and p = q, since this is sufficient for what follows.

4.1 Definition of JLip(t, q, q; 0; Γ ∞ ) for 0 < t < 1.

The definition of JLip(t, q, q; 0; Γ ∞ ) presented below is adapted to the class of fractal sets Γ ∞ considered in the present paper. It was proved in [START_REF] Achdou | Trace theorems for a class of ramified domains with self-similar fractal boundaries[END_REF] that this definition coincides with the original and more general one that was proposed in [START_REF] Jonsson | Haar wavelets of higher order on fractals and regularity of functions[END_REF]. Consider a real number t, 0 < t < 1. Following [START_REF] Jonsson | Haar wavelets of higher order on fractals and regularity of functions[END_REF], it is possible to characterize JLip(t, q, q; 0; Γ ∞ ) by using expansions in the standard Haar wavelet basis on Γ ∞ . Consider the Haar mother wavelet g 0 on Γ ∞ ,

g 0 = 1 f 1 (Γ ∞ ) -1 f 2 (Γ ∞ ) , (24)
and for n ∈ N, n > 0, σ ∈ A n , let g σ be given by

g σ | Γ ∞,σ = 2 n/2 g 0 • f -1 σ , and g σ | Γ ∞ \Γ ∞,σ = 0. ( 25
)
It is proved in [START_REF] Jonsson | Wavelets on fractals and Besov spaces[END_REF] §5 that a function f ∈ L q µ can be expanded on the Haar basis as follows:

f = P 0 f + β 0 g 0 + n≥1 σ∈An β σ g σ ,
where P 0 f = Γ ∞ f dµ. Let b 0 be a real number and (b σ ) σ∈A be a sequence of real numbers; we define (b 0 , (b σ )) b q,q t :

(b 0 , (b σ )) b q,q t = |b 0 | q + ∞ n=1 2 qtn/d 2 n(1/2-1/q)q σ∈An |b σ | q 1 q . = |b 0 | q + ∞ n=1 a -qtn 2 n(1/2-1/q)q σ∈An |b σ | q 1 q . ( 26
)
Definition 1 A function f ∈ L q µ belongs to JLip(t, q, q; 0; Γ ∞ ) if and only if the norm

f JLip(t,q,q;0;Γ ∞ ) = |P 0 f | + |f | JLip(t,q,q;0;Γ ∞ ) (27)
is finite, where |f | JLip(t,q,q;0;Γ ∞ ) = (β 0 , (β σ )) b q,q t . (28)

Remark 4 An equivalent definition of JLip(t, q, q; 0; Γ ∞ ) can be given using projection of f on constants on Γ ∞,σ , see [START_REF] Jonsson | Haar wavelets of higher order on fractals and regularity of functions[END_REF][START_REF] Achdou | Trace theorems for a class of ramified domains with self-similar fractal boundaries[END_REF].

If the fractal set Γ ∞ is totally disconnected, then JLip(t, q, q; 0; Γ ∞ ) coincides with a more classical function space:

Theorem 4 (Jonsson) If a < a * , then f 1 (Γ ∞ ) ∩ f 2 (Γ ∞
) is empty and

JLip(t, q, q; 0; Γ ∞ ) = Lip(t, q, q; 0; Γ ∞ ) = B q,q t (Γ ∞ ),
where the Lipschitz space Lip(t, q, q; 0; Γ ∞ ) and the Sobolev space B q,q t (Γ ∞ ) are defined in [START_REF] Jonsson | Function spaces on subsets of R n[END_REF].

Proof. This is a particular case of Theorem 2 in [START_REF] Jonsson | Haar wavelets of higher order on fractals and regularity of functions[END_REF], see also [START_REF] Jonsson | Wavelets on fractals and Besov spaces[END_REF] for a partial proof.

4.2 Characterization of the traces on Γ ∞ of the function in W 1,q (Ω)

The following theorem was proved in [START_REF] Achdou | Trace theorems for a class of ramified domains with self-similar fractal boundaries[END_REF].

Theorem 5 For a given θ, 0 < θ < π/2, let (α, β) satisfy Assumption 1 and Ω be constructed as in § 2.2.1, with 1/2 ≤ a ≤ a * ; then for all q, 1 < q < ∞,

∞ W 1,q (Ω) = JLip(1 - 2 -d q , q, q; 0; Γ ∞ ). (29)
A first consequence of Theorem 5 is that if 1/2 ≤ a < a * , then d ≥ 1 and from Theorem 4,

∞ W 1,q (Ω) = Lip(1 - 2 -d q , q, q; 0; Γ ∞ ) = B q,q 1-2-d q (Γ ∞ ), ∀q ∈ (1, +∞). ( 30
)
Remark 5 Note that (30) has been proved in [START_REF] Achdou | Trace theorems on domains with self-similar fractal boundaries[END_REF], without relying on the JLip spaces: indeed Ω is aδ domain and Γ ∞ is a d-set; in this case, the extension result of Jones [START_REF] Jones | Quasiconformal mappings and extendability of functions in Sobolev spaces[END_REF] (from W 1,q (Ω) to W 1,q (R 2 )) and the trace result of Jonsson and Wallin [START_REF] Jonsson | Function spaces on subsets of R n[END_REF] (from

W 1,q (R 2 ) onto B q,q 1-2-d q (Γ ∞ )) can be combined to obtain (30).
In what follows, we will see that when a = a * , then (30) does not hold for every q ∈ (1, +∞).

5 Embedding of the JLip spaces in Sobolev spaces for a = a *

Main results

Since a = a * , it is not possible to apply Theorem 4. Similarly, Ω is not anδ domain, so Jones extension result (from W 1,q (Ω) to W 1,q (R 2 )) does not hold for all q ∈ [1, +∞). Note that a = a * > 1/2 implies that d > 1. We are going to make out two cases: with p defined in [START_REF] Jones | Quasiconformal mappings and extendability of functions in Sobolev spaces[END_REF], the simpler case is when pθ > π/2, so Ξ ∞ is made of a single point; the case when pθ = π/2 will turn out to be more difficult because Ξ ∞ is not countable.

Theorem 6 Assume that a = a * and pθ > π/2.

1. For all t ∈ (0, 1) and s > d q , JLip(t, q, q; 0; Γ ∞ ) ⊂ B q,q s (Γ ∞ ).

2. If 0 < t < min(d/q, 1), then JLip(t, q, q; 0; Γ ∞ ) = Lip(t, q, q; 0; Γ ∞ ) = B q,q t (Γ ∞ ).

The following corollary stems from Theorem 6:

Corollary 1 Assume that a = a * and pθ > π/2. For all q > d and t ∈ [d/q, 1), JLip(t, q, q; 0; Γ ∞ ) ⊂ B q,q s (Γ ∞ ) with a continuous injection, for all s, 0 < s < d q .

As an easy consequence of Theorem 6, it is possible to find some relationships between JLip(1 -

2-d

q , q, q; 0; Γ ∞ ), the trace space of W 1,q (Ω), see (29), and some Sobolev spaces:

Corollary 2 Assume that a = a * and pθ > π/2.

1. If q ≥ 2, then (a) JLip(1 -2-d q , q, q; 0; Γ ∞ ) ⊂ B q,q s (Γ ∞ ), for all s, 0 < s < d q . (b) JLip(1 -2-d
q , q, q; 0; Γ ∞ ) ⊂ B q,q s (Γ ∞ ), for all s > d q .

2. If 1 ≤ q < 2, then JLip(1 -2-d q , q, q; 0; Γ

∞ ) = Lip(1 -2-d q , q, q; 0; Γ ∞ ) = B q,q 1-2-d q (Γ ∞ ).
Theorem 7 Assume that a = a * and pθ = π/2.

1. For all t ∈ (0, 1) and s > d 2q , JLip(t, q, q; 0; Γ ∞ ) ⊂ B q,q s (Γ ∞ ).

2. If 0 < t < min(d/(2q), 1), then JLip(t, q, q; 0; Γ ∞ ) = Lip(t, q, q; 0; Γ ∞ ) = B q,q t (Γ ∞ ).

Corollary 3 Assume that a = a * and pθ = π/2. For all q > d/2 and t ∈ [d/(2q), 1), JLip(t, q, q; 0; Γ ∞ ) ⊂ B q,q s (Γ ∞ ) with a continuous injection, for all s, 0 < s < d 2q .

Here again, it is possible to find some relationships between the trace space JLip(1 -

2-d

q , q, q; 0; Γ ∞ ) and some Sobolev spaces:

Corollary 4 Assume that a = a * and pθ = π/2.

1. If q ≥ 2 -d/2, then (a) JLip(1 -2-d q , q, q; 0; Γ ∞ ) ⊂ B q,q s (Γ ∞ ), for all s, 0 < s < d 2q . (b) JLip(1 -2-d q , q, q; 0; Γ ∞ ) ⊂ B q,q s (Γ ∞ ), for all s > d 2q . 2. If 1 ≤ q < 2-d/2, then JLip(1-2-d q , q, q; 0; Γ ∞ ) = Lip(1-2-d q , q, q; 0; Γ ∞ ) = B q,q 1-2-d q (Γ ∞ ).
Hereafter, when dealing with a = a * , we will always write a.

5.2 Proof of Theorem 6

Geometrical lemmas

The proofs of the lemmas below are given in appendix.

For two subsets X and Y of R 2 , we define d(X, Y ) = inf x∈X, y∈Y |y -x|. We will need to estimate

d(Γ ∞,σ , Γ ∞,τ ) for σ, τ ∈ A n , n ≥ 1.
We start by estimating the distance between Γ σ and the horizontal line H tangent to the upper part of Γ ∞ , i.e. H = {x : x 2 = h}, where

h = sup x∈Ω x 2 = max x∈Γ ∞ x 2 = β + a(α sin θ + β cos θ) 1 -a 2 . ( 31
)
Lemma 1 sup

x∈f 1 •f 1 (Ω) x 2 = sup x∈f 2 •f 2 (Ω)
x 2 < h.

Lemma 2 Take n ≥ 1 and σ ∈ A n . Let k be the largest integer such that 2k ≤ n and for all j ∈ {1, . . . , k}, σ(2j -1) = σ(2j). We have

d(Γ ∞,σ , H) + a n a 2k . ( 32
)
Remark 6 Note that Lemmas 1 and 2 hold if pθ = π/2.

Definition 2 Let us define the mapping Π : A → N as follows : for σ ∈ A n , n ≥ 1,

• if n < p + 4 or σ p+2 ∈ {(1, 2, . . . , 2
), (2, 1, . . . , 1)}, then Π(σ) = 0,

• else,

Π(σ) = max k ≥ 0, ∀j ∈ {1, . . . , k}, σ(p + 2j + 1) = 1 and σ(p + 2 + 2j) = 2 if σ(1) = 1, Π(σ) = max k ≥ 0, ∀j ∈ {1, . . . , k}, σ(p + 2j + 1) = 2 and σ(p + 2 + 2j) = 1 if σ(1) = 2.
Here, in other words, Π(σ) is the largest integer k ≥ 0 such that p + 2 + 2k ≤ n and

f σ p+2+2k = f 1 • f 2 p+1 • (f 1 • f 2 ) k or f σ p+2+2k = f 2 • f 1 p+1 • (f 2 • f 1 ) k .
Therefore, if n < p + 4 then Π(σ) = 0 and if n ≥ p + 4, then Π(σ) takes its values in {0, . . . , [(np -2)/2]}.

Definition 3 For σ ∈ A n , n ≥ 1, we say that Π(σ) is maximal if n < p + 4 (in this case Π(σ) = 0) or if n ≥ p + 4 and Π(σ) = [(n -p -2)/2].
The following lemma shows that the distance of Γ ∞,σ to the vertical axis Λ = {x : x 1 = 0} can be estimated in terms of Π(σ):

Lemma 3 Take n ≥ 1 and σ ∈ A n ; for d σ defined by

d σ = d(Γ ∞,σ , Λ), if Π(σ) is not maximal, d σ = d(Γ ∞,σ , Λ) + a n , if Π(σ) is maximal, (33)
we have

d σ a 2Π(σ) . ( 34
)
Lemma 4 Take n ≥ 1 and σ, τ ∈ A n such that σ(1) = τ (1); we have

d(Γ ∞,σ , Γ ∞,τ ) + a n a 2 min(Π(σ),Π(τ )) . (35)
Remark 7 From Lemma 3, we also have that for all σ, τ ∈ A with σ(1) = τ (1),

d(Γ ∞,σ , Γ ∞,τ ) a 2 min(Π(σ),Π(τ )) .
Definition 4 • Lemma 3 implies that there exists a positive constant c 1 , such that, for all n ≥ 1, σ ∈ A n and x ∈ Γ ∞,σ ,

c 1 a 2Π(σ) < d(x, Λ) if Π(σ) is not maximal, ( 36
)
and for all η ∈ A such that Π(η) = 0 and f η is a similitude with rotation angle 0,

d(f η (Λ), Λ) > c 1 . (37)
We must have

c 1 < d(Λ, f 1 • f 2 (Λ)), because Π((1, 2)) = 0.
• Let us define the positive number c 2 > 0 by

c 2 = d(Λ, f 1 • f 2 (Λ)) a 2 . ( 38
)
Note that

max x∈Γ ∞ x 1 = max x∈Γ ∞ d(x, Λ) = d(Λ, f 1 • f 2 (Λ)) a 2 = c 2 . ( 39
)
• Finally, from (34), we know that there exists a constant c 3 , such that for all

x ∈ Γ ∞,σ , σ ∈ A n , d(x, Λ) ≤ c 3 a 2Π(σ) . ( 40
)
We must have c 3 ≥ c 2 .

From (36) and (40), we deduce that for all n ≥ 1, σ ∈ A n and x ∈ Γ ∞,σ ,

d(x, Λ) ≤ c 3 a 2Π(σ) , if Π(σ) is maximal, c 1 a 2Π(σ) < d(x, Λ) ≤ c 3 a 2Π(σ) , if Π(σ) is not maximal. ( 41
)
Lemma 5 For any η ∈ A such that f η is a similitude with rotation angle 0,

d(Λ, f η (Λ)) > c 1 a 2Π(η) , (42)
where c 1 satisfies (36) and (37).

For what follows, we will need to partition f 1 (Γ ∞ ) into a sequence of subsets (X i ) i∈N . The measure of the set X i ⊂ f 1 (Γ ∞ ) and its distance to the axis Λ will be decreasing as i grows. By similarity, (f η (X i )) i∈N will be a partition of

f η • f 1 (Γ ∞ ):
Definition 5

• Let us define the subsets of Γ ∞ :

X i = {x ∈ f 1 (Γ ∞ ), c 1 a 2i ≤ d(x, Λ) < c 1 a 2(i-1) }, ∀i ≥ 1, X 0 = {x ∈ f 1 (Γ ∞ ), c 1 ≤ d(x, Λ) ≤ c 2 }, (43)
where c 1 satisfies (36) and (37) and c 2 is given by (38), see Figure 2. The

Λ X 1 X 0 Figure 2: θ = π/3: the domain f 1 • f p+1 2 (Ω), the fractal set f 1 • f p+1 2 (Γ ∞ )
and parts of X 0 and X 1 : X 0 (resp. X 1 ) is the intersection of Γ ∞ with the dark grey half-plane (resp. light grey strip).

• For ≥ 0, we define the class Z of subsets of Γ ∞ :

Z = {f η (X i ), η ∈ A n , n + 2i = }. (44)
• Let φ ∈ [0, 2π) be such that there exists a similitude f σ , σ ∈ A, with rotation angle φ.

For n ≥ 0, we define A n,φ = {η ∈ A n , f η is a similitude of angle φ}, and the class Z ,φ of subsets of Γ ∞ :

Z ,φ = {f η (X i ), η ∈ A n,φ , n + 2i = }. ( 45
)
Lemma 6 For all i ≥ 1,

µ(X i ) 2 -2i . ( 46
)
Remark 8 A direct consequence of Lemma 6 is that for all Y ∈ Z , µ(Y ) 2 -.

Lemma 7 For all nonnegative integers n, m, i, j such that n + 2i = m + 2j and η ∈ A n,φ , ν ∈ A m,φ , the sets Y = f η (X i ) and Z = f ν (X j ) are disjoint if (n, i, η) = (m, j, ν).

Remark 9 We will see that when pθ = π/2, the definition of Π differs, but once Π is defined, the definitions of c 1 and c 2 are the same. In that case, Lemma 5 and Lemma 7 are still true; by contrast, Lemma 6 does not hold, see Lemma 13.

Lemma 8 If pθ > π/2, then for any ≥ 0, any x ∈ Γ ∞ , there are at most a finite number of (i, η), 0 ≤ 2i ≤ , η ∈ A -2i such that x ∈ f η (X i ), and this number is independent of .

Remark 10

Although it seems clear that for a given Z ∈ Z , there is a unique

(i Z , η Z ) such that 0 ≤ 2i Z ≤ , η Z ∈ A -2i Z and Z = f η Z (X i Z ),
we have not found a short proof of this assertion. For what follows, it will be enough to use the following weaker result which stems from Lemma 8: there is at most a finite number of pairs (i, η) with 0 ≤ 2i ≤ , η ∈ A -2i and Z = f η (X i ), and this number is independent of Z and .

Sobolev regularity of the Haar wavelet g 0

The following proposition will imply regularity results for the Haar wavelet g 0 :

Proposition 5 We have f 1 (Γ ∞ ) f 2 (Γ ∞ ) 1 |x -y| γ dµ(y) dµ(x) < +∞, if 0 ≤ γ < 2d, (47)
f 1 (Γ ∞ ) f 2 (Γ ∞ ) 1 |x -y| γ dµ(y) dµ(x) = +∞, if γ > 2d. ( 48
)
Proof. Take n ≥ p + 2 and let κ be the largest integer such that n ≥ p + 2 + 2κ. We have

f 1 (Γ ∞ ) f 2 (Γ ∞ ) dµ(y) dµ(x) (|x -y| + a n ) γ = σ∈An σ(1)=1 τ ∈An τ (1)=2 Γ ∞,σ Γ ∞,τ dµ(y) dµ(x) (|x -y| + a n ) γ 2 -2n σ∈An σ(1)=1 τ ∈An τ (1)=2 1 (d(Γ ∞,σ , Γ ∞,τ ) + a n ) γ , because if x ∈ Γ ∞,σ and y ∈ Γ ∞,τ , then |x -y| + a n d(Γ ∞,σ , Γ ∞,τ ) + a n . Thus, from Lemma 4, we have f 1 (Γ ∞ ) f 2 (Γ ∞ ) dµ(y) dµ(x) (|x -y| + a n ) γ S 1 + S 2 , with S 1 = 2 -2n σ,τ ∈An σ(1)=1, τ (1)=2 Π(σ)≥Π(τ ) 1 a 2γΠ(τ ) and S 2 = 2 -2n σ,τ ∈An σ(1)=1, τ (1)=2 Π(σ)<Π(τ ) 1 a 2γΠ(σ) .
We can write S 1 as follows:

S 1 = 2 -2n 0≤ ≤k≤κ σ∈An σ(1)=1 Π(σ)=k τ ∈An τ (1)=2 Π(τ )= 1 a 2γ . (49)
On the other hand, the number of σ ∈ A n such that Π(σ) = k is of the order of 2 n-2k . Therefore, (49) leads to

S 1 2 -2n 0≤ ≤k≤κ 2 n-2k 2 n-2 1 a 2γ = κ k=0 2 -2k k =0 2 2 (-1+ γ d ) κ k=0 2 -2k max(k, 2 2k(-1+ γ d ) ),
and

S 1 κ k=0 2 2k(-2+ γ d ) if γ > 2d.
The same is true for S 2 . Therefore, if γ > 2d, then

f 1 (Γ ∞ ) f 2 (Γ ∞ ) 1 (|x -y| + a n ) γ dµ(y) dµ(x) 2 n(-2+ γ d ) , -→ n→∞ + ∞,
which yields (48).

On the other hand, if γ < 2d, then

f 1 (Γ ∞ ) f 2 (Γ ∞ ) 1 (|x -y| + a n ) γ dµ(y) dµ(x) ∞ =0 2 -2 max( , 2 2 (-1+ γ d ) ) < ∞,
which yields (47) from the monotone convergence theorem.

Corollary 5 For any q, 1 ≤ q < ∞,

g 0 ∈ B q,q s (Γ ∞ ) if 0 ≤ s < d q and g 0 / ∈ B q,q s (Γ ∞ ) if s > d q .
Proof. The result follows from the identity

|g 0 /2| q B q,q s (Γ ∞ ) = f 1 (Γ ∞ ) f 2 (Γ ∞ ) 1
|x-y| d+qs dµ(y) dµ(x) and from Proposition 5.

Two lemmas

Lemma 9 (discrete Hardy inequalities, [START_REF] Jonsson | Function spaces on subsets of R n[END_REF], page 121, Lemma 3) For any γ ∈ R, any p ≥ 1 there exists a constant C such that, for any sequence of positive real numbers

(c k ) k∈N , n∈N 2 γn   k≤n c k   p ≤ C n∈N 2 γn c n p if γ < 0, ( 50
) n∈N 2 γn   k≥n c k   p ≤ C n∈N 2 γn c n p if γ > 0. ( 51
)
Lemma 10 For any γ > d, we have

f 1 (Γ ∞ ) f 2 (Γ ∞ ) |v(x)| q |x -y| γ dµ(y) dµ(x) f 1 (Γ ∞ ) |v(x)| q d(x, Λ) γ-d dµ(x), ∀v ∈ L q (Γ ∞ ).
Proof. For any n ≥ p + 2, let κ be the largest integer such that n ≥ p + 2 + 2κ.

f 1 (Γ ∞ ) f 2 (Γ ∞ ) |v(x)| q (|x -y| + a n ) γ dµ(y) dµ(x) ≤ σ,τ ∈An σ(1)=1, τ (1)=2 Γ ∞,σ Γ ∞,τ |v(x)| q (d(Γ ∞,σ , Γ ∞,τ ) + a n ) γ dµ(y) dµ(x) = 2 -n σ,τ ∈An σ(1)=1, τ (1)=2 1 (d(Γ ∞,σ , Γ ∞,τ ) + a n ) γ Γ ∞,σ |v(x)| q dµ(x) S 1 + S 2
where, from Lemma 4,

S 1 = 2 -n 0≤ ≤k≤κ σ∈An σ(1)=1 Π(σ)=k τ ∈An τ (1)=2 Π(τ )= 1 a 2γ Γ ∞,σ |v(x)| q dµ(x), S 2 = 2 -n 0≤k≤ ≤κ σ∈An σ(1)=1 Π(σ)=k τ ∈An τ (1)=2 Π(τ )= 1 a 2γk Γ ∞,σ |v(x)| q dµ(x).
Since the number of τ ∈ A n such that τ (1) = 2 et Π(τ ) = is of the order of 2 n-2 , we have

S 1 0≤ ≤k≤κ 2 -2 a 2γ σ∈An σ(1)=1 Π(σ)=k Γ ∞,σ |v(x)| q dµ(x) = κ k=0 k =0 1 a 2 (γ-d) σ∈An σ(1)=1 Π(σ)=k Γ ∞,σ |v(x)| q dµ(x) κ k=0 σ∈An σ(1)=1 Π(σ)=k Γ ∞,σ |v(x)| q a 2k(γ-d) dµ(x) f 1 (Γ ∞ ) |v(x)| q (d(x, Λ) + a n ) γ-d dµ(x), from Lemma 3. Similarly, S 2 f 1 (Γ ∞ ) |v(x)| q (d(x, Λ) + a n ) γ-d dµ(x).
Finally, we obtain the desired estimate by having n tend to ∞ and using the monotone convergence theorem.

Proof of Theorem 6

Proof of 1) Point 1) in Theorem 6 stems from Corollary 5 and from the fact that the wavelet g 0 belongs to JLip(t, q, q; 0; Γ ∞ ) for all q,t, 1 ≤ q < ∞, 0 < t < 1.

Proof of 2) Consider t, 0 < t < min(d/q, 1).

Γ ∞ Γ ∞ |v(x) -v(y)| q |x -y| d+qt dµ(x) dµ(y) - 2 i=1 f i (Γ ∞ )×f i (Γ ∞ ) |v(x) -v(y)| q
|x -y| d+qt dµ(x) dµ(y)

I 1 + I 2 + I 3
where

I 1 = 2 x∈f 1 (Γ ∞ ) y∈f 2 (Γ ∞ ) |v(x) -v f 1 (Γ ∞ ) | q |x -y| d+qt dµ(y) dµ(x), (52)
I 2 = 2| v f 2 (Γ ∞ ) -v f 1 (Γ ∞ ) | q f 1 (Γ ∞ ) f 2 (Γ ∞ ) 1 |x -y| d+qt dµ(y) dµ(x), (53)
I 3 = 2 x∈f 1 (Γ ∞ ) y∈f 2 (Γ ∞ ) | v f 2 (Γ ∞ ) -v(y)| q |x -y| d+qt dµ(x) dµ(y). ( 54
)
By iterating this argument and using Fatou lemma, we obtain that

Γ ∞ Γ ∞ |v(x) -v(y)| q |x -y| d+qt dµ(x) dµ(y) I 1 + I 2 + I 3 + n≥1 η∈An (I 1,η + I 2,η + I 3,η )
where

I 1,η = 2 x∈fη•f 1 (Γ ∞ ) y∈fη •f 2 (Γ ∞ ) |v(x) -v fη •f 1 (Γ ∞ ) | q |x -y| d+qt dµ(y) dµ(x), (55)
I 2,η = 2| v fη •f 2 (Γ ∞ ) -v fη •f 1 (Γ ∞ ) | q fη•f 1 (Γ ∞ ) fη•f 2 (Γ ∞ )
1 |x -y| d+qt dµ(y) dµ(x), (56)

I 3,η = 2 x∈fη•f 1 (Γ ∞ ) y∈fη •f 2 (Γ ∞ ) | v fη •f 2 (Γ ∞ ) -v(y)| q |x -y| d+qt dµ(x) dµ(y). ( 57
)
Let us estimate I 1 + n≥1 η∈An I 1,η : the change of variables x = f η (x ) and y = f η (y ) yields

I 1,η = 2a n(d-qt) x ∈f 1 (Γ ∞ ) y ∈f 2 (Γ ∞ ) |v • f η (x ) -v • f η f 1 (Γ ∞ ) | q |x -y | d+qt dµ(y ) dµ(x ).
From Lemma 10, I 1,η a n(d-qt)

x ∈f 1 (Γ ∞ ) |v • f η (x ) -v • f η f 1 (Γ ∞ ) | q d(x , Λ) qt dµ(x ). Let β 0 , (β σ ) σ∈A
be the coefficients in the Haar basis of v: v = P 0 v + β 0 g 0 + k≥1 σ∈A k β σ g σ . Note that for any

η ∈ A n , v • f η -v • f η Γ ∞ = 2 n 2 β η g 0 + k≥1 σ∈A k
β η+σ g σ , where η + σ ∈ A n+k is the sequence (η(1), . . . , η(n), σ(1), . . . , σ(k)). Thus,

I 1,η a n(d-qt) 2 nq 2 x∈f 1 (Γ ∞ ) k 0 σ∈A k β η+σ g σ (x) q d(x, Λ) qt dµ(x) i≥0 2 n( q 2 + qt d -1) X i a -2iqt k≥0 σ∈A k β η+σ g σ (x) q dµ(x),
where X i is defined in (43). We are led to estimate

I 1 + n≥1 η∈An I 1,η n≥0 η∈An i≥0 2 n( q 2 + qt d -1) a -2iqt X i k≥0 σ∈A k β η+σ g σ (x) q dµ(x) S 1 + S 2 , (58)
where

S 1 = n≥0 η∈An i≥0 2 n( q 2 + qt d -1) a -2iqt X i k 2i σ∈A k β η+σ g σ (x) q dµ(x), (59)
S 2 = n≥0 η∈An i≥0 2 n( q 2 + qt d -1) a -2iqt X i k>2i σ∈A k β η+σ g σ (x) q dµ(x), (60)
with the convention that if n = 0, then A n = {0}, f 0 = Id and 0 + σ = σ. It is convenient to rewrite S 2 as follows:

S 2 = n≥0 η∈An i≥0 a -(2i+n)qt fη(X i ) k>2i σ∈A k ,σ(1)=1 β η+σ g η+σ (x) q dµ(x). ( 61
)
We have

S 2 = ≥0 a -qt [ /2] i=0 η∈A -2i fη(X i ) k>2i σ∈A k ,σ(1)=1 β η+σ g η+σ (x) q dµ(x).
From the definition of Z in (44),

S 2 ≥0 2 qt d Y ∈Z Y k≥ ν∈A k β ν g ν (x) q dµ(x) (62)
Remark 11 Note that in (62), the sign has been used instead of =, because we did not prove that there exists a unique pair (i, η) such that Y ∈ Z coincide with f η (X i ), but only that the number of such pairs is bounded, see Remark 10.

Then from a triangle inequality,

S 2 ≥0 2 qt d    k≥   Y ∈Z Y ν∈A k β ν g ν (x) q dµ(x)   1 q    q = = ≥0 2 qt d    k≥   ν∈A k |β ν | q Y ∈Z Y |g ν (x)| q dµ(x)   1 q    q ≥0 2 qt d    k≥   ν∈A k |β ν | q 2 -k 2 q k 2   1 q    q .
The latter inequality comes from Lemma 8, because any point x ∈ Γ ∞ belongs to at most a finite number of sets Y ∈ Z (this number is independent of ).

Hardy inequality (51) in Lemma 9 can be used because qt d > 0: this yields

S 2 ≥0 2 ( qt d + q 2 -1) ν∈A |β ν | q |v| q JLip(t,q,q;0;Γ ∞ ) . ( 63
)
Let us turn to S 1 defined in (59): we have, using a triangle inequality,

S 1 ≤ i≥0 2 2i qt d    k≤2i   n≥0 η∈An 2 n( q 2 + qt d -1) X i σ∈A k β η+σ g σ (x) q dµ(x)   1 q    q = i≥0 2 2i qt d    k≤2i   n≥0 η∈An 2 n( q 2 + qt d -1) σ∈A k |β η+σ | q X i |g σ (x)| q dµ(x)   1 q    q ,
because the supports of g σ , σ ∈ A k are disjoint (up to a negligeable set). This implies that

S 1 i≥0 2 2i qt d    k≤2i   n≥0 η∈An 2 n( q 2 + qt d -1) 2 k q 2 σ∈A k |β η+σ | q µ(supp g σ ∩ X i )   1 q    q .
From Definition 3, Lemma 3, if Π(σ) is not maximal, then i > 1 and d(Γ ∞,σ , Λ) > c 1 a 2Π(σ) > c 1 a 2(i-1) thanks to Definition 5, and µ(supp g σ ∩ X i ) = 0. Hence, if

P k = {σ ∈ A k , σ(1) = 1, Π(σ) maximal}, S 1 n≥0 η∈An 2 n( q 2 + qt d -1) |β η | q + i≥1 2 2i( qt d -1)    k≤2i   n≥0 η∈An 2 n( q 2 + qt d -1) 2 k q 2 σ∈P k |β η+σ | q   1 q    q ≤ n≥0 η∈An 2 n( q 2 + qt d -1) |β η | q + j≥1 2 j( qt d -1)    k≤j   n≥0 η∈An 2 n( q 2 + qt d -1) 2 k q 2 σ∈P k |β η+σ | q   1 q    q n≥0 η∈An 2 n( q 2 + qt d -1) |β η | q + j≥1 n≥0 2 (n+j)( q 2 + qt d -1)
η∈An σ∈P j

|β η+σ | q ,
by Hardy's inequality (50) in Lemma 9, because qt < d. For all ν ∈ A, there exist at most N = p + 4 pairs (η, σ), η, σ ∈ A such that ν = η + σ and Π(σ) is maximal. Therefore, for all ν ∈ A, β ν appears in the latter sum at most N times. Hence,

S 1 m≥0 2 m( q 2 + qt d -1) ν∈Am |β ν | q |v| JLip(t,q,q;0;Γ ∞ ) . (64)
From the bounds (63) and (64), we immediately deduce that

I 1 + n≥1 η∈An I 1,η |v| q JLip(t,q,q;0;Γ ∞ ) , (65)
and the same argument shows that

I 3 + n≥1 η∈An I 3,η |v| q JLip(t,q,q;0;Γ ∞ ) . (66)
We are left with estimating I 2 + n≥1 η∈An I 2,η . From (47) in Proposition 6 and easy scaling arguments,

fη•f 1 (Γ ∞ ) fη•f 2 (Γ ∞ ) 1 |x -y| d+qt dµ(y) dµ(x) 2 n( qt d -1) , ∀η ∈ A n . On the other hand, | v fη •f 2 (Γ ∞ ) -v fη •f 1 (Γ ∞ ) | = 2 n 2 +1 |β η |.
Combining these two observations, we have that for all t < min(d/q, 1),

I 2 + n≥1 η∈An I 2,η n≥0 2 n( qt d -1) 2 qn 2 η∈An |β η | q |v| q JLip(t,q,q;0;Γ ∞ ) . (67)
From ( 65), (66), (67), we obtain the desired result.

Lemma 13 For all integers i ≥ 1, the sets X i defined in (43) are such that

µ (X i ) 2 -i . (70)
Remark 12 It can be seen that the set X i is made of O(2 i ) disjoint connected components whose measure is of the order of 2 -2i .

Sobolev regularity of the Haar wavelet g 0

The following proposition, which should be compared to Proposition 5, will imply regularity results for the Haar wavelet g 0 :

Proposition 6 We have f 1 (Γ ∞ ) f 2 (Γ ∞ ) 1 |x -y| γ dµ(y) dµ(x) < ∞, if γ < 3d 2 , (71)
f 1 (Γ ∞ ) f 2 (Γ ∞ ) 1 |x -y| γ dµ(y) dµ(x) = ∞, if γ > 3d 2 . (72)
Proof. For any n ≥ p + 2, let κ be the largest integer such that such that p + 2 + 2κ ≤ n. We have

f 1 (Γ ∞ ) f 2 (Γ ∞ ) dµ(y) dµ(x) (|x -y| + a n ) γ = σ∈An σ(1)=1 τ ∈An τ (1)=2 Γ ∞,σ Γ ∞,τ 1 (|x -y| + a n ) γ dµ(y) dµ(x) σ∈An σ(1)=1 τ ∈An τ (1)=2 1 2 2n . 1 (d(Γ ∞,σ , Γ ∞,τ ) + a n ) γ κ k=0 (σ,τ )∈P k n 1 2 2n . 1 a 2kγ , from Lemma 12. It is easy to see that P k n has 2 k • 2 n-2k • 2 n-2k = 2 2n-3k elements. Therefore, f 1 (Γ ∞ ) f 2 (Γ ∞ ) 1 (|x -y| + a n ) γ dµ(y) dµ(x) κ k=0 2 k( 2γ d -3) . Thus f 1 (Γ ∞ ) f 2 (Γ ∞ ) 1 (|x -y| + a n ) γ dµ(y) dµ(x) ∞ k=0 2 k( 2γ d -3) < ∞, if γ < 3d 2 , f 1 (Γ ∞ ) f 2 (Γ ∞ ) 1 (|x -y| + a n ) γ dµ(y) dµ(x) 2 n 2 ( 2γ d -3) -→ n→∞ ∞, if γ > 3d 2 ,
and the result follows by the monotone convergence theorem.

The following should be compared to Corollary 5:

Corollary 6 For any q, 1 ≤ q < ∞, g 0 ∈ B q,q s (Γ ∞ ) if 0 ≤ s < d 2q and g 0 / ∈ B q,q s (Γ ∞ ) if s > d 2q .

Two Lemmas

Lemma 14 For all γ > d, we have

f 1 (Γ ∞ ) f 2 (Γ ∞ ) |v(x)| q |x -y| γ dµ(y) dµ(x) f 1 (Γ ∞ ) |v(x)| q d(x, Λ) γ-d dµ(x), ∀v ∈ L q µ .
Proof. Consider n ≥ p + 2. For any σ ∈ A n , define

P k σ = τ ∈ A n , (σ, τ ) ∈ P k n (see Definition 7). Lemma 12 implies that f 1 (Γ ∞ ) f 2 (Γ ∞ ) |v(x)| q (|x -y| + a n ) γ dµ(y) dµ(x) ≤ σ∈An σ(1)=1 τ ∈An τ (1)=2 Γ ∞,σ Γ ∞,τ |v(x)| q (d(Γ ∞,σ , Γ ∞,τ ) + a n ) γ dµ(y) dµ(x) 2 -n σ∈An σ(1)=1 Π(σ) k=0 τ ∈P k σ 1 a 2γk Γ ∞,σ |v(x)| q dµ(x).
It is easy to see that since σ ∈ A n , P k σ has 2 n-2k elements. Therefore,

f 1 (Γ ∞ ) f 2 (Γ ∞ ) |v(x)| q (|x -y| + a n ) γ dµ(y) dµ(x) σ∈An σ(1)=1   Π(σ) k=0 2 -2k a 2γk   Γ ∞,σ |v(x)| q dµ(x) σ∈An σ(1)=1 1 a 2Π(σ)(γ-d) Γ ∞,σ |v(x)| q dµ(x) σ∈An σ(1)=1 Γ ∞,σ |v(x)| q (d(x, Λ) + a n ) γ-d dµ(x) = f 1 (Γ ∞ ) |v(x)| q (d(x, Λ) + a n ) γ-d dµ(x).
The desired result is obtained by letting n tend to ∞, by monotone convergence.

Remark 13 Although the statements of Lemma 10 and 14 are similar, the proofs differ.

We define X i by (43) where c 1 is the constant appearing in (41).

Lemma 15 For any

i ≥ 1, k ≥ 1 and σ ∈ A k , X i |g σ | q dµ 2 k 2 (q-1) 2 -i , if k < 2i, (73)
X i |g σ | q dµ 2 k( q 2 -1) , if k ≥ 2i, (74)
Proof. Assume that k < 2i. It is easy to see that for σ ∈ A 1 , X i |g σ | q dµ ≤ 2 q 2 µ(X i ); this is exactly (73) for k = 1. If k ≥ 2, let σ ∈ A k
and τ ∈ A k be such that σ = τ , Π(σ) and Π(τ ) are maximal; then Γ ∞,σ and Γ ∞,τ can be obtained from each other by a translation with a vertical vector (parallel to Λ). Hence, Π(σ) and Π(τ ) are maximal

⇒ µ(X i ∩ Γ ∞,σ ) = µ(X i ∩ Γ ∞,τ ). ( 75
) Moreover, since µ(Γ ∞,τ ∩ Γ ∞,σ ) = 0 for σ, τ ∈ A k , we get µ(X i ) ≥ τ ∈A k Π(τ ) maximal µ(X i ∩ Γ ∞,τ ). (76)
On the other hand, if τ ∈ A k and Π(τ ) is not maximal, then Π(τ ) ≤ k/2 -1 and, from Lemma 11, 1) .

d(Γ ∞,τ , Λ) ≥ c 1 a 2Π(τ ) ≥ c 1 a (k-2) > c 1 a 2(i-
Therefore, if τ ∈ A k and Π(τ ) is not maximal, then X i ∩ Γ ∞,τ = ∅. The latter observation, (75) and (76) imply that for any

σ ∈ A k , µ(X i ∩ Γ ∞,σ ) 2 -k/2 µ(X i ) 2 -i-k/2
, from Lemma 13. Therefore,

X i |g σ | q dµ = 2 kq 2 µ(X i ∩ Supp(g σ )) = 2 kq 2 µ(X i ∩ Γ ∞,σ ) 2 k 2 (q-1)-i ,
and we have proved (73). On the other hand,

X i |g σ | q dµ ≤ 2 kq 2 µ(Supp(g σ )) = 2 k( q 2 -1) ,
and we have proved (74).

Proof of Theorem 7

Proof of 1.) The result stems from the fact that g 0 ∈ JLip(t, q, q; 0; Γ ∞ ) and from Corollary 6.

Proof of 2.) Exactly as in the proof of Theorem 6,

Γ ∞ Γ ∞ |v(x) -v(y)| q |x -y| d+qt dµ(x) dµ(y) I 1 + I 2 + I 3 + n≥1 η∈An (I 1,η + I 2,η + I 3,η )
where I 1 , I 2 , I 3 , I 1,η , I 2,η and I 3,η are respectively given by (52) (53) (54) (55) (56) (57). As above, we get that I 1 + n≥1 η∈An I 1,η S 1 + S 2 , where S 1 and S 2 are given by ( 59) and (60). Let us first find a bound on S 1 : exactly as in the proof of Theorem 6, we see that

S 1 n≥0 η∈An 2 n( q 2 + qt d -1) |β η | q + i≥1 2 2i qt d    k≤2i   n≥0 η∈An 2 n( q 2 + qt d -1) σ∈P k |β η+σ | q X i |g σ | q dµ   1 q    q ,
where P k = {σ ∈ A k , σ(1) = 1, Π(σ) maximal}. Thus, from (73),

S 1 n≥0 η∈An 2 n( q 2 + qt d -1) |β η | q + i≥1 2 2i qt d    k≤2i   n≥0 η∈An 2 n( q 2 + qt d -1) 2 k 2 (q-1) 2 -i σ∈P k |β η+σ | q   1 q    q ≤ n≥0 η∈An 2 n( q 2 + qt d -1) |β η | q + j≥1 2 j( qt d -1/2)    k≤j   n≥0 η∈An 2 n( q 2 + qt d -1) 2 k 2 (q-1) σ∈P k |β η+σ | q   1 q    q n≥0 η∈An 2 n( q 2 + qt d -1) |β η | q + j≥1 n≥0
2 (n+j)( q by Hardy's inequality (50) in Lemma 9. For all ν ∈ A, there exist at most N = p + 4 pairs (η, σ), η, σ ∈ A such that ν = η + σ and Π(σ) is maximal. Therefore, for all ν ∈ A, β ν appears in the latter sum at most N times. Hence,

S 1 m≥0 2 m( q 2 + qt d -1)
ν∈Am |β ν | q |v| q JLip(t,q,q;0;Γ ∞ ) .

We now consider S 2 . Since π/θ = 2p is an integer, the rotation angles of the similitudes f η can take only a finite number of values in [0, 2π). Call Θ the finite set of all possible angles: Θ = {iθ, 0 ≤ i < 4p}. It is convenient to split S 2 as follows

S 2 = φ∈Θ S 2,φ , with S 2,φ = n≥0 η∈A n,φ i≥0 a -(2i+n)qt fη(X i ) k>2i σ∈A k ,σ(1)=1
β η+σ g η+σ (x) q dµ(x).

We have

S 2,φ = ≥0 a -qt [ /2] i=0 η∈A -2i,φ fη(X i ) k>2i σ∈A k ,σ(1)=1 β η+σ g η+σ (x) q dµ(x).
We can rewrite S 2,φ as follows:

S 2,φ = ≥0 2 qt d Y ∈Z ,φ Y k≥ ν∈A k β ν g ν (x) q dµ(x). (78)
Thus, by the triangle inequality and the fact that the supports of g ν , ν ∈ A k are disjoint,

S 2,φ ≤ ≥0 2 qt d    k≥   ν∈A k |β ν | q Y ∈Z ,φ Y |g ν (x)| q dµ(x)   1 q    q .
From Remark 9, for all ≥ 0 and φ ∈ Θ, the sets Y ∈ Z ,φ are disjoint. Therefore,

S 2,φ ≥0 2 qt d    k≥   ν∈A k |β ν | q 2 -k 2 q k 2   1 q    q ,
because any x ∈ Γ ∞ belongs to at most one set Y ∈ Z ,φ . Hardy inequality (51) in Lemma 9 can be used because qt d > 0: this yields

S 2,φ ≥0 2 ( qt d + q 2 -1) ν∈A |β ν | q |v| q JLip(t,q,q;0;Γ ∞ ) .
Since this is true for all φ ∈ Θ and since Θ is a finite set, we get S 2 |v| q JLip(t,q,q;0;Γ ∞ ) . From this and (77), we immediately deduce (65) and the same argument yields (66). The conclusion of the proof is identical as that of Theorem 6. Remark 14 For s > 1 -2-d q , q < 2, it is interesting to construct a function u ∈ W 1,q (Ω) whose trace ∞ (u) does not belong to B q,q s (Γ ∞ ). One can take the following example: let χ ∈ W 1,q (Y 0 ) be such that χ |Γ 0 = 0, χ |f 1 (Γ 0 ) = 1 and χ |f 2 (Γ 0 ) = 0. For ρ > 0, we build u by the following iterative process:

• u |Y 0 = χ;
• let the polygonal open domain Y n be obtained by stopping the construction at step n + 1:

Y n = Interior K 0 ∪ n p=1 σ∈Ap f σ (K 0 ) . Let us also introduce Y σ = f σ (Y 0 ).
If u is already defined in Y n-1 , we define u |Y σ , σ ∈ A n as follows:

u |Y σ = 1 + ρu |Y n-1 • f -1 σ(n) if Π(σ) is maximal, u |Yσ = γ σ otherwise,
where γ σ = u |Γ σ for σ ∈ A (note that the function u is constant on the lines Γ σ ).

It is possible to prove that if ρ = 2

2-q

dq -1 2q -, for > 0, then u ∈ W 1,q (Ω) and that for any s > 1 -2-d q , one may choose small enough such that ∞ (u) / ∈ B q,q s (Γ ∞ ).

A Proofs of the geometrical lemmas in the case pθ > π/2

Proof of Lemma 1 We have sup

x∈f 1 •f 1 (Ω) x 2 = sup x∈f 2 •f 2 (Ω)
x 2 = sup

x∈f 1 •f 1 •f 2 •f 2 (Y 0 )
x 2 + a 4 h.

On the other hand, with F 1 defined in [START_REF] Jonsson | Haar wavelets of higher order on fractals and regularity of functions[END_REF],

h = sup x∈F 1 •F 1 (Y 0 )
x 2 + a 4 h.

Easy algebra shows that

f 2 1 •f 2 2 x 1 x 2 = -α + a(-α cos θ -β sin θ) + a 2 (α cos 2θ -β sin 2θ) + a 3 (α cos θ -β sin θ) + a 4 x 1 β + a(-α sin θ + β cos θ) + a 2 (α sin 2θ + β cos 2θ) + a 3 (α sin θ + β cos θ) + a 4 x 2 ,
and that

F 1 • F 1 x 1 x 2 =
-α + a(α cos θβ sin θ)a 2 α + a 3 (α cos θβ sin θ) + a 4 x 1 β + a(α sin θ + β cos θ) + a 2 β + a 3 (α sin θ + β cos θ) + a 4 x 2 .

Thus, the desired result will be a consequence of the inequality -α sin θ+β cos θ+a(α sin 2θ+β cos 2θ)+a 2 (α sin θ+β cos θ) < α sin θ+β cos θ+aβ+a 2 (α sin θ+β cos θ)

which is true, since α(-2 sin θ + a sin 2θ) + aβ(cos 2θ -1) < 0.

Proof of Lemma 2 From the definition of k in the statement of Lemma 2,

• if k = [n/2] (where we denote by [z] the integer part of z), then at least one of the two points lim

m→∞ f σ 2k • F m 1 (O), lim m→∞ f σ 2k • F m 2 (O) belongs to Γ ∞,σ , which implies that d(Γ ∞,σ , H) = 0. This implies that d(Γ ∞,σ , H) + a n a 2k .
• otherwise, let us define i = σ(2k + 1) = σ(2k + 2), and σ = σ 2k+2 . We have by selfsimilarity that

d(f σ (Ω), H) = a 2k d(f i • f i (Ω), H) > 0, f 1 • f p 2 (H) Λ f 1 • f p+1 2 (H) pθ -π 2 π 2 -(p -1)θ Λ d Γ ∞,σ lies here pθ -π 2 h > d h sin " π 2 -(p -1)θ " f 1 • f p 2 (H) f 1 • f p+1 2 (H) π 2 -(p -1)θ d = d " Γ ∞,σ , f 1 • f p+1 2 (H) " d Figure 3: Left: the lines f 1 • f 2 p (H) and f 1 • f 2 p+1 (H) (for θ = π/5). Right: d(Γ ∞,σ , Λ) ≥ sin π 2 -(p -1)θ d(Γ ∞,σ , f 1 • f 2 p+1 (H)). • If (i, j) = (2, 1), then Π(σ) = max{m ≥ 0 : ∀ , 1 ≤ ≤ m, τ (2 -1) = τ (2 )}. From Lemma 2 and since Π(σ) < κ, we have that d(Γ ∞,τ , H) a 2Π(σ) . Thus d(Γ ∞,σ , f 1 • f 2 p (H)) a p+1+2Π(σ) . (82)
Combining the second inequality in (80) and (82) yields that d(Γ ∞,σ , Λ) a 2Π(σ) .

In both cases, we have proved that d(Γ ∞,σ , Λ) a 2Π(σ) . For the opposite inequality, recall that Ξ ∞ = {ξ} where the point ξ is defined by [START_REF] Jonsson | Wavelets on fractals and Besov spaces[END_REF]. There exists

ζ ∈ Γ ∞ such that ξ = f 1 • f p+1 2 • F Π(σ) 1 (ζ). Let η ∈ A n-p-2-2Π(σ) be such that f σ = f 1 • f p+1 2 • F Π(σ) 1 • f η . We have d(Γ ∞,σ , Λ) = d(Γ ∞,σ , ξ) = a p+2+2Π(σ) d(Γ ∞,η , ζ) ≤ a p+2+2Π(σ) Diam(Ω), which yields that d(Γ ∞,σ , Λ) a 2Π(σ) .
Proof of Lemma 4 We may suppose that n ≥ p + 4. Let κ be the largest integer such that p + 2 + 2κ ≤ n. If Π(σ) = Π(τ ) = κ, then d(Γ ∞,σ , Γ ∞,τ ) = 0 which yields (35). Otherwise, min(Π(σ), Π(τ )) < κ: since Γ ∞,σ and Γ ∞,τ are separated by Λ, we have

d(Γ ∞,σ , Γ ∞,τ ) ≥ max(d(Γ ∞,σ , Λ), d(Γ ∞,τ , Λ)) a 2 min(Π(σ),Π(τ )) , (83)
from Lemma 3. On the other hand, defining k = p+2+2 min(Π(σ), Π(τ )), we see that for any ≤ k, σ( ) = τ ( ). Thus, Γ ∞,σ k and Γ ∞,τ k are symmetric with respect to Λ. Hence,

d(Γ ∞,σ k , Γ ∞,τ k ) = 2(Γ ∞,σ k , Λ) a 2 min(Π(σ),Π(τ )) , which implies d(Γ ∞,σ , Γ ∞,τ ) ≤ d(Γ ∞,σ k , Γ ∞,τ k ) + 2a k Diam(Ω). (84)
From ( 83) and (84), we obtain (35).

Proof of Lemma 5

• If Π(η) = 0, then we use (37).

• If Π(η) > 0, then Π(η) cannot be maximal: indeed, if Π(η) was maximal, then η would be of the form η = σ (1) or η = σ (1) + 1 or η = σ (1) + 2, where σ ∈ A, and σ (1) is defined as in ( 14); hence, the angle of f η would be pθ, (p -1)θ or (p + 1)θ, so it would not be an integer multiple of of 2π. Since Π(η) is not maximal, the result stems from (41) and the fact that Λ and f η (Λ) are parallel.

Proof of Lemma 6 Take

i ≥ 1. Since Γ ∞ is symmetric w.r.t. Λ, we can estimate µ({x ∈ Γ ∞ , c 1 a 2i < d(x, Λ) ≤ c 1 a 2(i-1) }) instead of µ(X i ).
Consider n, n > 2I + 3 + p where 1) . We know that Γ ∞ = ∪ σ∈An Γ ∞,σ , so there exists σ ∈ A n such that x ∈ Γ ∞,σ . We have d(Γ ∞,σ , Λ) > c 1 a 2i . The upper bounds in (41) imply that we must have Π(σ) ≤ I, which implies that Π(σ) is not maximal. Then, the lower bound in (41) implies that Π(σ) ≥ i. Hence,

I ≡ i + log " c 1 c 3 " 2 log a . Let us first estimate µ{x ∈ Γ ∞ , c 1 a 2i + a n Diam(Γ ∞ ) < d(x, Λ) ≤ c 1 a 2(i-1) }. Take x ∈ Γ ∞ and assume that c 1 a 2i + a n Diam(Γ ∞ ) < d(x, Λ) ≤ c 1 a 2(i-
µ({x ∈ Γ ∞ , c 1 a 2i + a n Diam(Γ ∞ ) < d(x, Λ) ≤ c 1 a 2(i-1) }) ≤ µ   I Π(σ)=i Γ ∞,σ   ≤ 2 -n I =i #{σ ∈ A n , Π(σ) = } 2 -n I =i 2 n-2 2 -2i ,
which yields (46) by letting n tend to infinity (monotone convergence).

Proof of Lemma 7 Let κ be the maximal integer k such that ν k = η k for all k ≤ κ. We must have either κ < min(n, m) or (κ = min(n, m) and n = m) otherwise (m, j, ν) = (n, i, η).

• Assume that κ < min(n, m): this implies that there exist σ

∈ A κ , ν ∈ A m-κ , η ∈ A n-κ such that ν = σ + ν , η = σ + η and ν (1) = η (1)
, with the notation defined in (4): f η (Γ ∞ ) and f ν (Γ ∞ ) lie on two different sides of Λ. We may assume that f ν (Γ ∞ ) lies on the right side of Λ and that f η (Γ ∞ ) lies on the left side of Λ.

-If f ν (Γ ∞ ) lies strictly on the right side of Λ, we get the desired result. This happens in particular if Π(ν ) is not maximal.

-If f η (Γ ∞ ) lies strictly on the left side of Λ, we get the desired result. This happens in particular if Π(η ) is not maximal.

-Assume that f ν (Γ ∞ ) ∩ Λ = ∅ and f η (Γ ∞ ) ∩ Λ = ∅. * if Π(ν )
is positive then it is maximal, and f ν is a similitude whose angle can be -(p -1)θ, -pθ or -(p + 1)θ. If the angle is -pθ or -(p + 1)θ, then f ν (X j ) does not intersect Λ (because f ν (X j ) is on the left of f ν (Λ)), which yields the desired result. If the angle is -(p -1)θ, then the similitude f η has the same angle and f η (X i ) does not intersect Λ (because f η (X i ) is on the left of f η (Λ)), which yields the desired result. * Similarly, if Π(η ) is positive then it is maximal, and f η is a similitude whose angle can be (p-1)θ, pθ or (p+1)θ. If the angle is (p-1)θ, then f η (X i ) does not intersect Λ which yields the desired result. If the angle is pθ or (p + 1)θ, then the similitude f ν has the same angle and f ν (X j ) does not intersect Λ which yields the desired result. * We are left with the case where Π(η ) = 0 and Π(ν ) = 0: it can be shown that there are only three pairs (ν , η ) such that the related similitudes have the same angle, Π(η ) = 0, Π(ν ) = 0, f ν (Γ ∞ ) ∩ Λ = ∅ and f η (Γ ∞ ) ∩ Λ = ∅: 1) η = (1) and ν = (2, 1, 1) 2) η = (1, 2) and ν = (2, 1), 3) η = (1, 2, 2) and ν = (2). In these three cases, the desired result follows easily.

• If κ = min(n, m), for example κ = n < m, then Y = f η (X i ) and Z = f η • f ν (X j ), ν ∈ A m . We have to prove that X i ∩ f ν (X j ) = ∅. The angle of the similitude f ν is 0 and 2i = 2j + m .

-If f ν (1) = 2, then f ν (X j ) lies on the right side of Λ and X i strictly lies on the left side of Λ, which yields the result. -If f ν (1) = 1, then d(f ν (Λ), Λ) > c 1 a 2Π(ν ) from Lemma 5. Therefore, from the definition of X i , d(f ν (Λ), X i ) > c 1 a 2Π(ν )c 1 a 2i-2 > 0.

Proof of Lemma 8 We can assume > 1.

• Suppose first that = 2i, i > 0, x ∈ X i . Since we are interested in finding j < i and η ∈ A -2j such that the set Z = f η (X j ) contains x, we can suppose that η(1) = 1. If Π(η) is not maximal, then X i ∩Z = ∅: indeed, from (41), Z ⊂ f η (Γ ∞ ) and d(f η (Γ ∞ ), Λ) > c 1 a 2Π(η) . Therefore d(Z, X i ) ≥ d(f η (Γ ∞ ), Λ)c 1 a 2(i-1) > c 1 (a 2Π(η)a 2(i-1) ). But 2Π(η) < 2(i -1), so d(Z, X i ) > 0 and x / ∈ Z. We now focus on the Z = f η (X j ) ∈ Z such that η(1) = 1 and Π(η) is maximal. Since there are a finite number, namely 2 p+2 , of η such that Π(η) = 0 and Π(η) is maximal, we can suppose that Π(η) > 0. We make out two cases:

-If p is even, then η ∈ A p+2+2Π(η) , and the angle of the similitude f η is pθ. Since the sets Z ∈ Z ,pθ are pairwise disjoint, only one of them can contain x. -If p is odd, then η ∈ A p+3+2Π(η) . Since η + 1 ∈ A p+4+2Π(η) , Π(η + 1) = Π(η) is not maximal. The facts that Z = f η (X j ) ⊂ f η+1 (Γ ∞ ) and d(f η+1 (Γ ∞ ), Λ) > c 1 a 2Π(η) > c 1 a 2(i-1) imply that X i ∩ Z = ∅, so x / ∈ Z.

• Suppose x ∈ Y = f ν (X i ), Y ∈ Z , with ν ∈ A n Y , n Y > 0 and x ∈ Z, Z ∈ Z , with n Z < n Y . On the one hand, the number of the sets f ν+ν (X j ) ∈ Z containing x coincides with the number of the sets f ν (X j ) containing f -1 ν (x) ∈ X i ; this number has been estimated above. On the other hand, if a set Z = f η (X j ) ∈ Z , with n Z ≥ n Y , is such that η = ν +ν , then calling κ the maximal integer such that ν k = η k for all k ≤ κ, we know that κ < n Y and that f η (X j ) and f ν (X i ) lie on different sides of f νκ (Λ), so their intersection is empty: Z does not contain x.

B Proofs of the geometrical lemmas in the case pθ = π/2

Proof of Lemma 11 It is enough to consider n ≥ p + 4.

• If σ p+2 / ∈ {(1, 2, . . . , 2), (2, 1, . . . , 1)}, then there exists a constant c independent of σ such that d(Γ ∞,σ , Λ) > c. Since d(Γ ∞,σ , Λ) ≤ Diam(Ω), we obtain the desired result in this case.

• If σ p+2 ∈ {(1, 2, . . . , 2), (2, 1, . . . , 1)}, for example σ p+2 = (1, 2, . . . , 2), then Γ ∞,σ = f 1 • f p+1 2 Γ ∞,τ , for some τ ∈ A n-p-2 . Since we also have Λ =

f 1 • f p+1 2 (H), d(Γ ∞,σ , Λ) = d(f 1 • f p+1 2 (Γ ∞,τ ), f 1 • f p+1 2 (H)) = a p+2 d(Γ ∞,τ , H). (85)
We also have that Π(σ) = max{k; ∀j ≤ k, τ (2j -1) = τ (2j)}. Thus from Lemma 2, d(Γ ∞,τ ), H) + a n-p-2 a 2Π(σ) . Combining this and (85) yields the desired result.

Proof of Lemma 12

From the definition of k, it is clear that Π(σ) ≥ k and that Π(τ ) ≥ k.

Let κ be the largest integer such that p + 2 + 2κ ≤ n. With the notation defined in [START_REF] Jonsson | The dual of Besov spaces on fractals[END_REF], let η ∈ A k be such that σ p+2+2k = η (1) and τ p+2+2k = η (2) .

• If k = κ, then Γ ∞,η (1) is the symmetric of Γ ∞,η (2) w.r.t. Λ, and d(Γ ∞,η (1) , Γ ∞,η (2) ) = 0. This implies (69).

• If k < κ, then d(Γ ∞,σ , Γ ∞,τ ) ≤ d(Γ ∞,η (1) , Γ ∞,η (2) ) + 2Diam(Γ ∞,η (1) ) a 2k , because d(Γ ∞,η (1) , Γ ∞,η (2) ) = 0. For the opposite inequality, since κ > k, we can define σ = σ p+4+2k and τ = τ p+4+2k .

-If min(Π(σ), Π(τ )) = k then d(Γ ∞,σ , Γ ∞,τ ) ≥ d(Γ ∞,σ , Γ ∞,τ ) ≥ max d(Γ ∞,σ , Λ), d(Γ ∞,τ , Λ) a 2k .

-Otherwise, min(Π(σ), Π(τ )) > k. This implies that σ(p + 3 + 2k) = τ (p + 3 + 2k) = σ(p + 4 + 2k) = τ (p + 4 + 2k).

Without restriction, we may assume that σ(p + 3 + 2k) = τ (p + 3 + 2k) = 1 and that σ(p + 4 + 2k) = τ (p + 4 + 2k) = 2, thus σ = η (1) 12 and τ = η (2) 12.

For what follows we define δ = d(Γ 1,∞ ∩ H, Γ 2,∞ ∩ H) > 0 and E as the convex subset of R 2 located under the straight lines H, f 1 (H) and f 2 (H). It is clear that Ω ⊂ E. Therefore, Ω σ ⊂ f σ (E) and Ω τ ⊂ f τ (E), see

d(f σ (Γ ∞ ) ∩ Λ, f τ (Γ ∞ ) ∩ Λ) = d(f η (1) (F 1 (Γ ∞ )) ∩ Λ, f η (2) (F 1 (Γ ∞ )) ∩ Λ) = d(f η (1) (F 1 (Γ ∞ )) ∩ Λ, f η (1) (F 2 (Γ ∞ )) ∩ Λ).
By self-similarity, β = a p+2+2k δ. Therefore d(Γ ∞,σ , Γ ∞,τ ) ≥ d(Γ ∞,σ , Γ ∞,τ ) ≥ d(f σ (E), f τ (E)) a 2k . Proof of Lemma 13 The argument is the same as the one used for Lemma 6. The only difference is that #{σ ∈ A n , Π(σ) = } 2 n-, instead of 2 n-2 in the former case. Hence, with I defined as in the proof of Lemma 6,

f τ (E) f σ (E) β Γ η (2) Γ η (1) Γ σ Γ τ
µ({x ∈ Γ ∞ , c 1 a 2i + a n Diam(Γ ∞ ) < d(x, Λ) ≤ c 1 a 2(i-1) }) ≤ µ   i+I Π(σ)=i Γ ∞,σ   ≤ 2 -n i+I =i #{σ ∈ A n , Π(σ) = } 2 -n i+I =i 2 n- 2 -i ,
and (70) is obtained by Fatou lemma.

Figure 1 :

 1 Figure 1: Left: the ramified domain Ω for θ = π/4, a = a * (π/4), α = 1a * / √ 2, β = 1 + a * / √ 2. Right: a similar construction for θ = π/5 and a = a * (π/5).

Figure 4 .

 4 Elementary geometrical arguments lead tod(f σ (E), f τ (E)) = β sin θ, where β = d(f σ (Γ ∞ ) ∩ Λ, f τ (Γ ∞ ) ∩ Λ), see Figure 4. But

Figure 4 :

 4 Figure 4: θ = π/4: the case when σ = η (1) 12 and when τ = η (2) 12

+ qt d -1) η∈An σ∈P j |β η+σ | q ,

Proof of Theorem 7

We now consider the case when pθ = π/2, with p defined in [START_REF] Jones | Quasiconformal mappings and extendability of functions in Sobolev spaces[END_REF]. The situation is more complex because Ξ ∞ is a non countable set whose Hausdorff dimension is d/2.

Geometrical lemmas

We state several useful geometrical lemmas whose proofs are given in appendix. Here, we define the mapping Π : A → N as follows: Definition 6

• if σ ∈ A n with n < p + 4 then Π(σ) = 0,

In other words, with F 1 and F 2 defined in [START_REF] Jonsson | Haar wavelets of higher order on fractals and regularity of functions[END_REF],

One can estimate the distance of Γ ∞,σ to Λ as a function of Π(σ):

Lemma 11 Take n ≥ 1 and σ ∈ A n ; with d σ defined in (33), we have

= 2 must be done more carefully than in the case when pθ > π/2: indeed, in the present case, the quantity max

), because Γ ∞,σ and Γ ∞,τ may touch Λ without facing each other. This is why we have to make the following definition:

Definition 7 For any n ≥ p + 2 and any k ≥ 0 such that p + 2 + 2k ≤ n, let P k n be the set containing all the pairs (σ, τ ) such that    σ ∈ A n and τ ∈ A n , σ(1) = 1 and τ (1) = 2, k is the largest integer such that ∃η ∈ A k with σ p+2+2k = η (1) and τ p+2+2k = η (2) , (68) where η (1) and η (2) are defined by ( 14) or [START_REF] Jonsson | The dual of Besov spaces on fractals[END_REF]. For example, take σ = (1, 2, . . . ,

We have (σ, τ ) ∈ P 2 p+10 , with η = (1, 2) in (68).

Lemma 12 For any n ≥ p + 2 and k ≥ 0 such that p + 2 + 2k ≤ n, for any (σ, τ

Finally, as in § 5.2.1, there exist two positive constants c 1 ≤ c 2 such that (41) holds for all n ≥ 1, σ ∈ A n and x ∈ Γ ∞,σ ; the following lemma should be compared to Lemma 6.

by Lemma 1. Hence,

On the other hand,

We have proved that d(Γ ∞,σ , H) a 2k , which implies (32).

Proof of Lemma 3 We may suppose that n ≥ p + 4. Let κ be the largest integer such that p + 2 + 2κ ≤ n. Take σ ∈ A n . If Π(σ) = κ, then d(Γ ∞,σ , Λ) = 0, which yields (34). Suppose that Π(σ) < κ. Suppose first that σ p+2 ∈ {(1, 2, . . . , 2), (2, 1, . . . , 1)}. Without restriction, we can also suppose that σ(1) = 1: there exists j, 0 ≤ j ≤ p such that

We have that d(Γ ∞,σ , Λ) ≥ c. On the other hand d(Γ ∞,σ , Λ) ≤ Diam(Ω). We have obtained (34). We are left with considering the case when Π(σ) < κ and σ p+2 ∈ {(1, 2, . . . , 2), (2, 1, . . . , 1)}. Without restriction, we can also suppose that σ(1) = 1: there exists τ ∈ A n-p-2 and τ

1 2 i j . . . . . . τ , with i = σ(p + 3 + 2π(σ)) and j = σ(p + 4 + 2π(σ)). The definition of Π(σ) implies that (i, j) = (1, 2). We have obviously that

(79) Three observations will prove useful, see Figure 3 : a) by self similarity, the set f 1 (Γ ∞ ) lies on one side of the straight lines f 1 • f 2 p (H) and

H) makes an angle of pθ -π 2 > 0 with Λ. An elementary geometrical argument combining points a), b) and c), leads to

The geometrical argument for the first inequality in (80) is summarized in the right part of Figure 3. We make out two cases:

• If i = j then Π(σ) = max{m ≥ 0 : ∀ , 1 ≤ ≤ m, τ (2 -1) = τ (2 )}. From Lemma 2 and since Π(σ) < κ, we have that d(Γ ∞,τ , H) a 2Π(σ) . Thus

Combining the first inequality in (80) and (81) yields that d(Γ ∞,σ , Λ) a 2Π(σ) .