REGULAR DOMAINS IN HOMOGENEOUS GROUPS

ROBERTO MONTI AND DANIELE MORBIDELLI

Abstract. We study John, uniform and non–tangentially accessible domains in homogeneous groups of step 2 and 3. We show that C^{1,1} domains in groups of step 2 are non–tangentially accessible and we give an explicit condition which ensures the John property in groups of step 3.

1. Introduction

This paper deals with the study of John, uniform and non-tangentially accessible domains in homogeneous groups (Carnot groups) of step 2 and 3 endowed with the Carnot–Carathéodory distance induced by a system of generators of their Lie algebra. Such regular domains are strongly related to the global embedding properties of Sobolev spaces and to the study of the boundary behavior of harmonic functions with respect to sub-elliptic Laplacians.

John domains have been introduced by John [Joh]. It is known from the classical theory of Sobolev spaces that if Ω ⊂ ℝ^{n} is a John domain and ∇u ∈ L^{p}(Ω), then u ∈ L^{pn/(n−p)}(Ω) (see the references [Be] and [Bo]). In the setting of homogeneous groups it is known that if Ω is a John domain with respect to the Carnot–Carathéodory distance the following Sobolev-Poincaré inequality holds

( ∫_{Ω} |u − u_{Ω}|^{p^{∗}} dx)^{1/p^{∗}} ≤ C( ∫_{Ω} |Xu|^{p} dx)^{1/p}  (1.1)

for all u ∈ C^{1}(ℝ^{n}), where u_{Ω} is the average of u over Ω, Xu = (X_{1}u, ..., X_{m}u) is the sub-elliptic gradient of u with respect to the vector fields X_{1}, ..., X_{m} which generate the Lie algebra of the group, the integer Q is the homogeneous dimension of the group, 1 ≤ p < Q and p^{∗} = pQ/(Q − p). Different forms of inequality (1.1) have been proved in [FLW], [GN1] and [HK] and they are a consequence of the Poincaré inequality proved by Jerison in [J] and of a chaining argument (relations between John condition and chaining properties are studied in [BKL], [GN1] and [HK]).

Uniform domains (also known as (ε, δ) domains) are a sub-class of John domains. The definition of uniform domain is due to Martio and Sarvas [MS] and to Jones [Jon]. In the latter paper an extension theorem for Sobolev functions in uniform domains is proved, theorem generalized in [VG] and [GN2] to the setting of Carnot–Carathéodory spaces.

Key words and phrases. John domains, non-tangentially accessible domains, homogeneous groups.

First author was supported by University of Trento.

2 ROBERTO MONTI AND DANIELE MORBIDELLI

A sub-class of uniform domains are non-tangentially accessible domains (briefly NTA domains) which, in the Euclidean case, were introduced by Jerison and Kenig [JK] in connection with the study of the boundary behavior of harmonic functions. The notion of NTA domain can be formulated in terms of the Carnot-Carathéodory metric associated with a sub-elliptic Laplacian L = ∑ X_{j}^{2} and used to prove a Fatou type theorem for non negative weak solutions u of the equation Lu = 0 (see [CG] and see also [W]). In connection with harmonic measures for sub-elliptic Laplacians a class of “cuspidal Harnack domains” has been recently studied in [FF1] and [FF2].

It is known that inequality (1.1) may fail even for smooth domains in Carnot– Carathéodory spaces. See [J, Section 6], where a counterexample is given. The natural problem that arises is to find explicit examples of domains where inequality (1.1) holds. This problem becomes very intriguing for homogeneous groups of step greater than 2 (in the step-two case more refined properties will be established, see the discussion below). In this paper we give a first answer to this problem.

In groups of step 3 we have found a differential condition on the boundary near characteristic points that seems to control in a sharp way the flatness behavior of the surface in order to have regularity. Our result gives a sufficient condition for the John condition and thus for (1.1). Precisely, consider two vector fields X_{1} and X_{2} in ℝ^{4} generating the Lie algebra a homogeneous group of step 3 with non–trivial commutators [X_{1}, X_{2}] = X_{3} and [X_{1}, X_{3}] = [X_{2}, X_{3}] = X_{4}. Here X_{j} = ∂/∂x_{j}, j = 1, …, 4, at the origin (a complete description of this structure is given in Section 4). In this setting we prove the following result:

Theorem 1.1. If Ω ⊂ ℝ^{4} is an admissible domain, then it is a John domain (with respect to the Carnot–Carathéodory distance generated by X_{1} and X_{2}).

The definition of admissible domain will be given in a detailed way in Section 4. We observe here that the key property that an admissible domain Ω should enjoy is the following “flatness” condition. If Ω = {x ∈ ℝ^{4} : Φ(x_{1}, x_{2}, x_{3}, x_{4}) > 0} we require that for all points in ∂Ω the following estimate holds

|X_{1}^{2}Φ| + |X_{2}^{2}Φ| + |(X_{1}X_{2} + X_{2}X_{1})Φ| ≤ k(|X_{1}Φ|^{1/2} + |X_{2}Φ|^{1/2} + |X_{3}Φ|) (1.2)

for a suitable positive constant k (see Lemma 4.3 and Theorem 4.4). This condition implies that at characteristic points of second type, that is characteristic points where also X_{3}Φ vanishes (see section 4), the second derivatives of Φ along X_{1} and X_{2} must also vanish. In particular, if (1.2) holds and Ω agrees near the origin with the set {x ∈ ℝ^{4} : x_{4} > φ(x_{1}, x_{2}, x_{3})}, φ(0) = 0, ∇φ(0) = 0, then (see Lemma 4.3) it should be |φ(x_{1}, x_{2}, x_{3})| ≤ k(|x_{1}| + |x_{2}| + |x_{3}|^{1/2})^{3}. This flatness condition is not ensured by Euclidean regularity. In Example 5.2 we also explicitly construct admissible domains.

On the other side we have the following necessary condition, whose proof is contained in Proposition 5.6:

REGULAR DOMAINS IN HOMOGENEOUS GROUPS 3

Theorem 1.2. If a domain Ω locally agrees near the origin with the set {x ∈ ℝ^{4} : x_{4} > φ(x_{1}, x_{2}, x_{3})} and |φ(x_{1}, x_{2}, x_{3})| ≤ k(|x_{1}| + |x_{2}| + |x_{3}|^{1/2})^{γ} for some γ strictly greater than 3, then the Sobolev-Poincaré inequality (1.1) does not hold in Ω.

This shows two surprising facts: (i) if a domain Ω locally agrees near the origin with the plane x_{4} > 0, then (1.1) can not hold; (ii) the Sobolev–Poincaré inequality is false in the the homogeneous ball (x_{1}^{2} +x_{2}^{2})^{6} +x_{3}^{6} +x_{4}^{4} < 1. The latter fact disproves a conjecture stated in [CG, p. 429].

In a forthcoming paper we shall show how conditions similar to (1.2) can be used to identify regular domains in the different context of diagonal vector fields of the form X_{j} = λ_{j}(x)\frac{∂}{∂x_{j}}, j = 1, …, n, where the λ_{j}’s are suitable functions.

In the setting of step 2 groups the situation is different. It has been proved that Carnot-Carathéodory balls in the Heisenberg group are uniform domains (see [VG]) but not NTA domains (see [CG]); examples of uniform domains in the Heisenberg group can also be found by means of quasiconformal mappings (see [CT]); half spaces and cubes centered at the origin in the Heisenberg group are uniform domains (see [G]). It was conjectured in [GN1], [CG] and [CGN1] that in any group of step 2 a connected, bounded open set with boundary of class C^{1,1} is a NTA domain. A partial positive answer to this conjecture has been given in [CG] for sets with cylindrical symmetry near each characteristic point. An improvement has been announced to the authors by Luca Capogna and Nicola Garofalo for sets with “strongly isolated” characteristic set in in the Heisenberg group. In the present paper we prove the full statement:

Theorem 1.3. Any C^{1,1} domain in a step 2 homogeneous group is NTA.

Theorem 1.3 is sharp in the following sense: in groups of step 2 for any α ∈ (0, 1) there are open sets of class C^{1,α} for which the Sobolev–Poincaré inequality (1.1) does not hold (see Example 5.1).

Our Theorem 1.3 has been recently used in cooperation with the results of [DGN] to give a complete characterization (in C^{2} domains) of the trace space for Sobolev functions in step 2 homogeneous groups (see [DGN, Thm. 13.5]).

Since the proofs in the paper are technical we would like to sketch here the key ideas in the simple situation of the Heisenberg group. We shall briefly discuss only the John property, requiring the uniform condition a deeper analysis. Consider the vector fields X_{1} = ∂_{x_{1}} − x_{2}∂_{x_{3}} and X_{2} = ∂_{x_{2}} + x_{1}∂_{x_{3}} in ℝ^{3}. Here ℝ^{3} is equipped with the group law x · y = (x_{1} + y_{1}, x_{2} + y_{2}, x_{3} + y_{3} + x_{1}y_{2} − x_{2}y_{1}), with the homogeneous norm ∥x∥ = ((x_{1}^{2}+x_{2}^{2})^{2}+x_{3}^{2})^{1/4} and with the left invariant distance d(x, y) = ∥x^{−1}·y∥. We consider an open set Ω = {x_{3} > φ(x_{1}, x_{2})}, where φ(0, 0) = 0 and φ is a smooth function. Since rotations around the x_{3}-axis are isometries of the group, there is no loss of generality in requiring ∂_{x_{1}}φ(0, 0) = −ν with ν ≥ 0, and ∂_{x_{2}}φ(0, 0) = 0. We shall explain how to construct a John curve, i.e. a curve satisfying condition (1.3) below, starting from the origin. By left translation this will also produce John curves starting from any point of the boundary of Ω. Moreover, since the map (x_{1}, x_{2}, x_{3}) ↦

4 ROBERTO MONTI AND DANIELE MORBIDELLI

(x_{1}, x_{2}, x_{3} + α), α ∈ ℝ, is a left translation the natural John curve starting from any point (x_{1}, x_{2}, φ(x_{1}, x_{2}) + α), α ≥ 0, can be consequently obtained.

Our curve γ will be constructed by two pieces:

γ(t) = \begin{cases}(t, 0, 0) & if 0 < t ≤ σν := t_{1}, \\ (t_{1}, 0, t − t_{1}) & if t ≥ t_{1}. \end{cases}

Here σ > 0 is a parameter depending on the given function φ (it must not depend on ν) and will be fixed during the argument. We note that if ν = 0, i.e. ∇φ(0, 0) = 0 (the origin is a characteristic point), then t_{1} = 0, the first piece of the path disappears and we simply have γ(t) = (0, 0, t): this curve is not rectifiable.

We show that γ satisfies the following “John property”

dist(γ(t); ∂Ω) ≥ λd(γ(t), 0) for all t > 0  (1.3)

for some small constant λ > 0 depending on φ, but not on ν (in Section 2 (1.3) is compared with similar conditions involving rectifiable curves). Subtracting from φ its linear part we can write φ(x) = −νx_{1} + ψ(x_{1}, x_{2}) where, in view of the smoothness of φ, the remainder ψ satisfies for some uniform constant C_{0} > 0 the quadratic growth estimate

|ψ(x_{1}, x_{2})| ≤ C_{0}(x_{1}^{2} + x_{2}^{2})

(the constant C_{0} is essentially the Lipschitz constant of ∇φ). Points of the ball B(γ(t), λd(γ(t), 0)) = γ(t)·B(0, λ∥γ(t)∥) are of the form γ(t)·h, with ∥h∥ < λ∥γ(t)∥. Thus (1.3) is equivalent to γ(t) · h ∈ Ω, that is

(γ(t) · h)_{3} ≥ φ(γ(t) · h)  (1.4)

for all h ∈ ℝ^{3}, ∥h∥ < λ∥γ(t)∥. By the quadratic estimate for ψ, (1.4) is ensured by

(γ(t) · h)_{3} + ν(γ(t) · h)_{1} ≥ C_{0} [(γ(t) · h)_{1}^{2} + (γ(t) · h)_{2}^{2}] .  (1.5)

Inequality (1.5) will be first checked for t ≤ t_{1} = σν. Note that

γ(t) · h = (t, 0, 0) · (h_{1}, h_{2}, h_{3}) = (t + h_{1}, h_{2}, h_{3} + th_{2}).

Then (1.5) becomes h_{3} + th_{2} + ν(t + h_{1}) ≥ C_{0}[(t + h_{1})^{2} + h_{2}^{2}], that is

ν(t + h_{1}) ≥ −h_{3} − th_{2} + C_{0}[(t + h_{1})^{2} + h_{2}^{2}].

Since ∥h∥ ≤ λt the terms in the right hand side can be estimated as follows: −h_{3} ≤ |h_{3}| ≤ (λt)^{2}, −th_{2} ≤ |th_{2}| ≤ λt^{2}, and the terms in square bracket can be estimated by Ct^{2}, with C absolute constant. Moreover in the left hand side we have

ν(t + h_{1}) ≥ ν(t − ∥h∥) ≥ \frac{1}{2}νt,

as soon as λ ≤ 1/2. We ultimately get the inequality

νt ≥ C_{0}t^{2},  (1.6)

REGULAR DOMAINS IN HOMOGENEOUS GROUPS 5

where C_{0} is an absolute constant possibly greater than the ones written above. This inequality holds as soon as t ≤ σν, where σ is any positive fixed constant satisfying σ ≤ 1/C_{0}.

We now check (1.5) for t ≥ t_{1} = σν. Note that

γ(t) · h = (t_{1}, 0, t − t_{1}) · h = (t_{1} + h_{1}, h_{2}, t − t_{1} + h_{3} + t_{1}h_{2}).

Then (1.5) becomes t − t_{1} + h_{3} + t_{1}h_{2} + ν(t_{1} + h_{1}) ≥ C_{0}[(t_{1} + h_{1})^{2} + h_{2}^{2}], that is

t − t_{1} + ν(t_{1} + h_{1}) ≥ −h_{3} − t_{1}h_{2} + C_{0}[(t_{1} + h_{1})^{2} + h_{2}^{2}].  (1.7)

This inequality has to be checked for ∥h∥ ≤ λ∥γ(t)∥ = λ(t_{1}^{4} + (t − t_{1})^{2})^{1/4}, or, letting a = (t − t_{1})^{1/2}, for ∥h∥ ≤ λ(t_{1}^{4} + a^{4})^{1/4}. The terms in the right hand side of (1.7) can be estimated as follows: −h_{3} ≤ |h_{3}| ≤ Cλ(t_{1}^{2} + a^{2}) and −t_{1}h_{2} ≤ |t_{1}h_{2}| ≤ t_{1}∥h∥ ≤ Cλ(t_{1}^{2} + t_{1}a). Moreover

(t_{1} + h_{1})^{2} ≤ (t_{1} + λ(t_{1}^{4} + a^{4})^{1/4})^{2} ≤ (Ct_{1} + Cλa)^{2},

while h_{2}^{2} ≤ ∥h∥^{2} ≤ λ^{2}(t_{1}^{4} + a^{4})^{1/2} ≤ C(t_{1}^{2} + λ^{2}a^{2}). The left hand side can be estimated as follows

ν(t_{1} + h_{1}) ≥ νt_{1} − ν∥h∥ ≥ νt_{1} − νλC(t_{1} + a) ≥ \frac{1}{2}νt_{1} − λCνa,

as soon as λ ≤ \frac{1}{2C}.

Then (1.7) will be proved if the following inequality holds:

a^{2} + \frac{1}{2}νt_{1} − λCνa ≥ Cλt_{1}^{2} + Cλt_{1}a + +Cλa^{2} + Ct_{1}^{2} + Cλ^{2}a^{2}.

Choosing λ small enough the terms Cλa^{2} and Cλ^{2}a^{2} in the right hand side can be absorbed in the left hand side. Note also that Cλt_{1}a ≤ \frac{C}{2} λt_{1}^{2} + \frac{C}{2} λa^{2}, so that \frac{C}{2} λa^{2} can be absorbed in the left hand side, too. Then we find the stronger inequality a^{2} + νt_{1} ≥ C(t_{1}^{2} + λνa + λt_{1}a) which is ensured by a^{2} + νt_{1} ≥ C(t_{1}^{2} + λνa), and recalling now that t_{1} = σν we finally get

a^{2} + σν^{2} ≥ C_{1}(σ^{2}ν^{2} + λνa).  (1.8)

We slightly modify the choice of σ requiring C_{1}σ^{2} ≤ σ/2. Then, in order to prove (1.8) it is enough to verify that

a^{2} + \frac{1}{2}σν^{2} ≥ C_{1}λ(\frac{1}{2}ν^{2} + \frac{1}{2}a^{2}).

This inequality is satisfied for any a ≥ 0 provided that C_{1}λ/2 ≤ 1 and C_{1}λ ≤ σ. Indeed, such a λ can be found and it does not depend on ν, as required.

The described argument proves that smooth domains in the Heisenberg group enjoy the John property. They also enjoy the uniform property: this can be proved taking suitable cones having as core the John curves constructed above and showing that these cones meet appropriately (see Theorem 3.2).

Before closing this introduction we briefly explain our notations. In the paper we denote by C (or k) absolute positive constants (they may depend on the surface we are considering). By a ≲ b we mean a ≤ Cb, and by a ≃ b we mean C^{−1}a ≤ b ≤ Ca.

6 ROBERTO MONTI AND DANIELE MORBIDELLI

By ε_{0} and C_{0} we denote, as well, absolute positive constants which are respectively smaller and greater than 1. The small parameter λ is used to denote the “aperture” of cones and in sections 3 and 4 we shall several times write λ instead of o(1), as λ → 0. Finally, if x, y ∈ ℝ^{n} then ⟨x, y⟩ = ∑_{i=1}^{n} x_{i}y_{i} denotes the usual Euclidean inner product.

Acknowledgments. We would like to thank Nicola Garofalo who has drawn our attention on the problems studied in Section 3, Olli Martio who has discussed with us some general properties of John and uniform domains and Nicoletta Ripamonti for her help with the computer. Finally, we wish to thank Bruno Franchi and Luca Capogna for having pointed out to us some references.

2. John, uniform and NTA domains. Some general facts

In this section we recall the basic definitions and some general known results concerning John, uniform and NTA domains (see, for example, [MS], [V], [CT], [CG], [HK]) We consider a metric space (M, d). If γ : [0, 1] → M is a curve, we denote by length(γ) the length of γ and by γ_{[a,b]} : [0, b − a] → M the curve γ_{[a,b]}(t) = γ(t + a), 0 ≤ a ≤ b ≤ 1. A curve is rectifiable if length(γ) < +∞.

Definition 2.1. Let (M, d) be a metric space. A bounded open set Ω ⊂ M is a John domain if there exist x_{0} ∈ Ω and λ > 0 such that for every x ∈ Ω there exists a continuous rectifiable curve γ : [0, 1] → Ω, such that γ(0) = x, γ(1) = x_{0} and

dist(γ(t); ∂Ω) ≥ λ length(γ_{[0,t]})

for all t ∈ [0, 1].

Definition 2.2. Let (M, d) be a metric space. An open set Ω ⊂ M is a uniform domain if there exists λ > 0 such that for every x, y ∈ Ω there exists a continuous rectifiable curve γ : [0, 1] → Ω connecting them and such that

length(γ) ≤ \frac{1}{λ}d(x, y),  (2.1)

and

dist(γ(t); ∂Ω) ≥ λ min{length(γ_{[0,t]}), length(γ_{[t,1]})}  (2.2)

for all t ∈ [0, 1].

Let Ω be an open set in the metric space M and let k ≥ 1. A ball B(x, r) ⊂ Ω is k−non tangential if k^{−1}r ≤ dist(B; ∂Ω) ≤ kr. Given x, y ∈ Ω a Harnack chain joining x to y is a family B_{1}, ..., B_{n} of k–non tangential balls in Ω (for some k ≥ 1) such that x ∈ B_{1}, y ∈ B_{n} and B_{j} ∩ B_{j+1} ≠ ∅. The integer n is the length of the chain.

Definition 2.3. Let (M, d) be a metric space. A bounded domain Ω ⊂ M is non– tangentially accessible if there exist r_{0} > 0 and k ≥ 1 such that

REGULAR DOMAINS IN HOMOGENEOUS GROUPS 7

(i) Corkskrew condition. For all x ∈ ∂Ω and r ∈ (0, r_{0}) there exists y ∈ Ω such that k^{−1}r ≤ d(x, y) ≤ r and dist(y; ∂Ω) ≥ k^{−1}r.

(ii) The set M \ Ω̄ satisfies the corkscrew condition. (iii) Harnack chain condition. If ε > 0 and x, y ∈ Ω with dist(x; ∂Ω) > ε, dist(y; ∂Ω) >

ε and d(x, y) < Cε, then there exists a Harnack chain of k–non–tangential balls joining x to y with length depending on C but not on ε.

This paper deals with homogeneous groups. Endowed with their Carnot-Carathéo-dory distance homogeneous groups are metric spaces with geodesics, and moreover, if μ denotes their Haar measure (which is Lebesgue measure), then there are positive constants C and Q ∈ ℕ such that μ(B(x, r)) = Cr^{Q} for all x belonging to the group and for all r > 0 (see [FS]). More generally, a metric space (M, d) endowed with a Borel measure μ such that 0 < μ(B(x, 2r)) ≤ δμ(B(x, r)) < +∞ for all x ∈ M and r > 0 is called doubling metric space, and δ > 0 is its doubling constant.

We establish some propositions that will be needed in the next sections.

Proposition 2.4. Let (M, d) be a doubling metric space with geodesics. Let Ω ⊂ M be a bounded open set and for any r > 0 define Ω_{r} = {y ∈ Ω : dist(y; ∂Ω) > r}. Assume that there exist r > 0 and λ > 0 such that Ω_{r} is arcwise connected and such that for any x ∈ Ω there exists a continuous curve γ_{x} : [0, 1] → Ω such that γ_{x}(0) = x, γ_{x}(1) ∈ Ω_{r} and

dist(γ_{x}(t); ∂Ω) ≥ λd(γ_{x}(t), x)  (2.3)

for all t ∈ [0, 1]. Then Ω is a John domain.

Proof. We show that there exist x_{0} ∈ Ω and C > 0 such that for all x ∈ Ω there exists a continuous curve γ : [0, 1] → Ω such that γ(0) = x, γ(1) = x_{0} and (2.3) holds. In the terminology of [HK] this means that Ω is a weak John domain. Since (M, d) is a doubling metric space with geodesics from [HK, Proposition 9.6] it follows that Ω is a John domain.

Fix x_{0} ∈ Ω_{r} and for any y ∈ Ω_{r} denote by γ_{y,x_{0}} : [0, 1] → Ω_{r} a continuous path such that γ_{y,x_{0}}(0) = y and γ_{y,x_{0}}(1) = x_{0}. Then dist(γ_{y,x_{0}}(t); ∂Ω) ≥ r and d(γ_{y,x_{0}}(t), x) ≤ diam(Ω) for all t ∈ [0, 1]. If x ∈ Ω let γ_{x} be as in the statement of the Proposition, write ȳ = γ_{x}(1) and consider γ_{ȳ,x_{0}}. The path γ sum of γ_{x} and γ_{ȳ,x_{0}} enjoys property (2.3) (possibly with a new λ).

Proposition 2.5. Let (M, d) be a doubling metric space with geodesics. Let Ω ⊂ M be a bounded open set. If there exist λ > 0 and r > 0 such that for any z ∈ ∂Ω and for all x, y ∈ B(z, r) ∩ Ω there exists a continuous curve γ : [0, 1] → Ω such that γ(0) = x, γ(1) = y,

diam(γ) ≤ \frac{1}{λ}d(x, y)  (2.4)

and

dist(γ(t); ∂Ω) ≥ λ min{diam(γ_{[0,t]}), diam(γ_{[t,1]})}  (2.5)

for all t ∈ [0, 1], then Ω is an uniform domain.

Proof. The proof relies upon two facts. First, using the argument in [V, Theorem 4.1], which holds in any metric space, it can be proved that for all x, y ∈ Ω there is a continuous curve γ such that (2.4) and (2.5) hold (the constant λ may be different).

Second, curves can be taken rectifiable and satisfying (2.1) and (2.2). We sketch the proof. Let x, y ∈ Ω and fix a continuous curve γ such that γ(0) = x and γ(1) = y, and such that (2.4) and (2.5) hold. Let t̄∈ (0, 1) be such that diam(γ_{[0,t̄]}) = diam(γ_{[t̄,1]}). Consider the path γ̃_{x} = γ_{[0,t̄]}. By the choice of t̄ and by (2.5) γ̃_{x} satisfies dist(γ̃_{x}(t); ∂Ω) ≥ λdiam((γ̃_{x})_{[0,t]}). By Lemma 2.7 in [MS], which can be proved in any doubling metric space with geodesics, there exists a rectifiable path γ_{x} such that γ_{x}(0) = x, γ_{x}(1) = γ(t̄) and dist(γ_{x}(t); ∂Ω) ≥ λ′length(γ_{[0,t]}) (here λ′ depends on λ and on the doubling constant). A rectifiable path γ_{y} can be analogously constructed. The sum of the paths γ_{x} and γ_{y} gives the required path and proves that Ω is actually a uniform domain in the sense of Definition 2.2.

A continuous curve γ : [0, 1] → Ω such that γ(0) = x and dist(γ(t); ∂Ω) ≥ λd(γ(t), x) will be called a John curve starting from x and with constant λ > 0.

All the John curves constructed in Section 3 and 4 are such that the function t ↦ d(γ(t), γ(0)) is (equivalent to) a monotonic increasing function and satisfies

d(γ(0), γ(t)) ≃ diam(γ_{[0,t]}).  (2.6)

In section 3 we shall refer to the following proposition.

Proposition 2.6. Let (M, d) be a doubling metric space with geodesics and let Ω ⊂ M be a bounded open set. Assume that there exists λ > 0 and r > 0 such that for any z ∈ ∂Ω and for all x, y ∈ B(z, r) ∩ Ω there exist two John curves γ_{x}, γ_{y} : [0, 1] → Ω starting respectively from x and y, with John constant λ, such that γ_{x}(1) = γ_{y}(1) and

max\{diam(γ_{x}), diam(γ_{y})\} ≤ \frac{1}{λ}d(x, y).

Assume also that γ_{x} and γ_{y} verify (2.6). Then Ω is a uniform domain.

Proof. Let γ be the curve sum of γ_{x} and γ_{y} parameterized over [0, 2]. First of all

diam(γ) ≤ diam(γ_{x}) + diam(γ_{y}) ≤ \frac{2}{λ}d(x, y).

Consider now a point γ(t) and assume that γ(t) = γ_{x}(t). Then

dist(γ(t); ∂Ω) = dist(γ_{x}(t); ∂Ω) ≥ λd(γ_{x}(t), x)
≃ λdiam((γ_{x})_{[0,t]}) ≥ λ min {diam(γ_{[0,t]}), diam(γ_{[t,2]})}.

If γ(t) is in γ_{y} the estimate is the same. The claim follows from Proposition 2.5.

It is easy to see that if both Ω and M \ Ω satisfy the hypotheses of Proposition 2.6, then the interior and the exterior corkskrew conditions are satisfied. The following result contains another useful connection between uniform and NTA domains.

Proposition 2.7. Let (M, d) be a metric space. If Ω ⊂ M is a uniform domain then Ω satisfies the Harnack chain condition (iii) in Definition 2.3.

Proof. See [CT, Proposition 4.2]

3. Uniform and NTA domains in groups of step 2

In this section we study uniform and NTA domains in homogeneous groups of step 2. We shall work in ℝ^{n} endowed a left invariant metric induced by a system of vector fields X = (X_{1}, ..., X_{m}) which generates a stratified Lie algebra of step 2. In ℝ^{n} = ℝ^{m} × ℝ^{q} we denote x = (x′, x′′) ∈ ℝ^{m} × ℝ^{q} and by abuse of notation we shall write x′ = (x′, 0) and x′′ = (0, x′′). We say that x′ are the variables of the first slice and that x′′ are the variables of the second slice.

The vector fields can be assumed to be of the form

X_{j} = ∂_{j} + ∑_{k=m+1}^{n} q_{jk}∂_{k}, j = 1, ..., m,

where q_{jk} = q_{jk}(x′) are homogeneous polynomials of degree 1 in the variables x′ (see [FS]). Introduce the group law

\begin{aligned}x · y & = x + y + Q(x, y), \\ & = (x_{1} + y_{1}, …, x_{m} + y_{m}, x_{m+1} + y_{m+1} + Q_{m+1}(x, y), …, x_{n} + y_{n} + Q_{n}(x, y)), \end{aligned} (3.1)

where Q = (Q_{1}, ..., Q_{n}) with Q_{1} = ... = Q_{m} = 0, and Q_{j} = Q_{j}(x′, y′), j = m + 1, ..., n, are homogeneous polynomials of degree 2 (the fact that the Q_{j}’s do not depend on the variables of the second slice will be used several times). Moreover, they can be assumed to satisfy

|Q_{j}(x′, y′)| ≤ C|x′||y′|.  (3.2)

The vector fields X_{1}, …, X_{m} are by assumption left invariant with respect to the introduced law.

We denote by d the Carnot-Carathéodory distance induced on ℝ^{n} by X_{1}, ..., X_{m} and by B(x, r) the open ball centered at x ∈ ℝ^{n} with radius r ≥ 0. We also introduce in ℝ^{n} the following continuous homogeneous norm

∥x∥ = |x′| + |x′′|^{1/2}.  (3.3)

By a standard argument it can be proved that d(x, y) ≃ ∥y^{−1} · x∥. Thus, letting Box(x, r) = {x · y ∈ ℝ^{n} : ∥y∥ ≤ r} there exists c > 1 such that for all x ∈ ℝ^{n} and r ≥ 0

Box(x, c^{−1}r) ⊂ B(x, r) ⊂ Box(x, cr).

Definition 3.1. Let S ⊂ ℝ^{n} be a hypersurface of class C^{1} given in a neighborhood U of x_{0} ∈ S by the local equation Φ = 0 where Φ ∈ C^{1}(U). The point x_{0} is characteristic if X_{1}Φ(x_{0}) = ... = X_{m}Φ(x_{0}) = 0.

10 ROBERTO MONTI AND DANIELE MORBIDELLI

We denote by e_{j} the j−th coordinate versor and if x = ∑_{i=1}^{n} x_{i}e_{i} ∈ ℝ^{n} and j ∈ {1, …, n} we write

x̂_{j} = ∑_{1≤i≤n,i≠j} x_{i}e_{i}.

Let Ω ⊂ ℝ^{n} be an open set, x_{0} ∈ ∂Ω and let U ⊂ ℝ^{n} be a neighborhood of x_{0} such that ∂Ω ∩ U = {x ∈ U : Φ(x) = 0} for some defining function Φ ∈ C^{1}(U). If x ∈ ∂Ω ∩ U then, possibly shrinking U, the translated surface x^{−1} · (∂Ω ∩ U) can be expressed in parametric form by an equation of the type y_{j} = φ(ŷ_{j}) for ŷ_{j} belonging to a neighborhood of the origin in ℝ^{n−1} and φ of class C^{1}. If x_{0} is a characteristic point we must choose j ∈ {m + 1, ..., n}, otherwise we can choose j ∈ {1, ..., m}. In the next theorem we shall assume Φ ∈ C^{1,1} and so φ ∈ C^{1,1}. Actually, in the proof we shall need the Lipschitz continuity only of the derivatives ∂_{1}φ, ..., ∂_{m}φ.

Theorem 3.2. Any connected, bounded open set Ω ⊂ ℝ^{n} of class C^{1,1} is a NTA domain in the metric space (ℝ^{n}, d).

Proof. The proof will be split in several numbered small steps.

1. We claim that for all x_{0} ∈ ∂Ω there exists a neighborhood U of x_{0} such that for all x, y ∈ U ∩ Ω there exist continuous curves γ_{x} and γ_{y} : [0, 1] → Ω satisfying the hypotheses of Proposition 2.6. The proof will show that conditions (i) and (ii) in Definition 2.3 are verified, and by Proposition 2.7 Ω will be a NTA domain.

2. Let U be a neighborhood of x_{0} and let Φ ∈ C^{1}(U) be a defining function such that ∂Ω ∩ U = {x ∈ U : Φ(x) = 0}. We shall distinguish two cases: (C1) |X_{1}Φ(x_{0})| = ... = |X_{m}Φ(x_{0})| = 0 (x_{0} is a characteristic point of ∂Ω); (C2) |X_{1}Φ(x_{0})| + ... + |X_{m}Φ(x_{0})| > 0 (x_{0} is a non characteristic point of ∂Ω). Let x_{j} = φ(x̂_{j}) be a local parameterization of ∂Ω ∩ U around x_{0}. We assume without loss of generality that x_{0} = 0.

3. Case 1. We consider an open set {y ∈ ℝ^{n} : y_{j} > φ(ŷ_{j})} where j > m and φ ∈ C^{1}(ℝ^{n−1}) is a function such that φ(0) = 0, |∇φ| ≤ k and ∂_{1}φ, ..., ∂_{m}φ are k−Lipschitz continuous functions, where k > 0 is a fixed given constant. Define

ν_{i} = −∂_{i}φ(0), for i = 1, ..., m, and ν = (ν_{1}, ..., ν_{m}, 0, ..., 0).

Write also

φ(ŷ_{j}) = − ∑_{i=1}^{m} ν_{i}y_{i} + ψ(ŷ_{j})

where ψ can be written by Taylor formula in the following form

ψ(ŷ_{j}) = φ(ŷ_{j}) − ∑_{i=1}^{m} ∂_{i}φ(0)y_{j} = ∑_{i>m,i≠j} ∂_{i}φ(0)y_{i} + O(|ŷ_{j}|^{2}),

and satisfies the growth estimate

|ψ(ŷ_{j})| ≲ ∥ŷ_{j}∥^{2}.  (3.4)

REGULAR DOMAINS IN HOMOGENEOUS GROUPS 11

Here we used the homogeneous norm introduced in (3.3) and the Lipschitz continuity of ∂_{1}φ, ..., ∂_{m}φ.

Our construction will take place in two steps. In the first step we define “canonical” John curves starting from points near the boundary. In the second step we join points near the boundary by curves satisfying the hypotheses of Proposition 2.6.

4. First step. Define

N_{1} = \frac{ν_{1}}{|ν|}, ..., N_{m} = \frac{ν_{m}}{|ν|}, and N = (N_{1}, ..., N_{m}, 0, ..., 0),

and if ν = 0 simply set N = 0. For σ > 0 let t_{1} = σ|ν|. Fix x = x_{j}e_{j} with x_{j} ≥ 0 and define the continuous curve γ : [0, 1] → ℝ^{n}

γ(t) = \{ \begin{matrix}x · tN = tN + x_{j}e_{j}, & if 0 ≤ t ≤ t_{1}, \\ x · ((t_{1}N) + (t − t_{1})e_{j}) = t_{1}N + (t − t_{1} + x_{j})e_{j}, & if t_{1} ≤ t ≤ 1. \end{matrix}  (3.5)

5. We claim that there exist σ, λ ∈ (0, 1) such that for all t ∈ [0, 1]

dist(γ(t); ∂Ω) ≥ λd(γ(t), x).  (3.6)

If 0 ≤ t ≤ t_{1} then d(γ(t), x) ≃ ∥x^{−1} · γ(t)∥ = ∥tN∥ = t, and (3.6) is equivalent to

Box(γ(t), λt) ∩ {y_{j} = φ(ŷ_{j})} = ∅,  (3.7)

which is implied by

⟨ν, y⟩ + y_{j} ≥ |ψ(ŷ_{j})|, for all y ∈ Box(γ(t), λt).  (3.8)

Points in Box(γ(t), λt) are of the form

γ(t) · h = (tN + x_{j}e_{j}) · h = tN + x_{j}e_{j} + h + Q (tN, h)  (3.9)

with ∥h∥ ≤ λt (Q does not depend on the variables on the second slice) and thus we have to check that

⟨ν, tN + h⟩ + x_{j} + h_{j} + Q_{j}(tN, h) ≥ |ψ({ \atop (γ(t) · h)_{j}})|

which is guaranteed by

t|ν| + ⟨ν, h⟩ + x_{j} ≥ |h_{j}| + |Q_{j}(tN, h)| + |ψ({ \atop (γ(t) · h)_{j}})|.

Now, since |⟨ν, h⟩| ≤ λ|ν|t then t|ν| + ⟨ν, h⟩ ≳ t|ν| as soon as λ < 1/2. Moreover |h_{j}| ≤ t^{2} and by (3.4)

\begin{aligned}|ψ({ \atop (γ(t) · h)_{j}})| & ≲ ∥{ \atop (γ(t) · h)_{j}}∥^{2} = ∥tN + ĥ_{j} + Q̂_{j}(tN, h)∥^{2} \\ & ≲ t^{2} + ∥h∥^{2} + ∥Q(tN, h)∥^{2} ≲ t^{2}. \end{aligned}

Moreover |Q_{j}(tN, h)| ≲ λt^{2} ≲ t^{2}. Thus (3.7) is implied by

ε_{0}(t|ν| + x_{j}) ≥ t^{2},  (3.10)

where ε_{0} is a small but absolute constant. Since x_{j} ≥ 0, (3.10) holds for all t ≤ σ|ν| as soon as σ ≤ ε_{0}. Our claim is proved if 0 ≤ t ≤ t_{1}.

12 ROBERTO MONTI AND DANIELE MORBIDELLI

6. We study the case t ≥ t_{1}. Notice that in this case

d(γ(t), x) ≃ ∥x^{−1} · γ(t)∥ ≃ t_{1} + (t − t_{1})^{1/2} =: δ(t).  (3.11)

Let a = (t−t_{1})^{1/2} so that δ(t) = t_{1}+a. We shall sometimes write δ instead of δ(t). We claim that there exists λ ∈ (0, 1) such that the John property Box(γ(t), λδ(t))∩{y_{j} = φ(ŷ_{j})} = ∅ holds for all t ≥ t_{1}.

Points in Box(γ(t), λδ) are of the form

\begin{aligned}γ(t) · h & = (t_{1}N + (t − t_{1} + x_{j})e_{j}) · h \\ & = t_{1}N + (t − t_{1} + x_{j})e_{j} + h + Q(t_{1}N, h), \end{aligned}  (3.12)

with ∥h∥ ≤ λδ. Thus, the John property is ensured by

⟨ν, t_{1}N + h⟩ + (t − t_{1}) + x_{j} + Q_{j}(t_{1}N, h) ≥ |ψ(t_{1}N + ĥ_{j} + Q̂_{j}(t_{1}N, h))|,

which (write t − t_{1} = a^{2}) is a consequence of the following stronger inequality

t_{1}|ν| + a^{2} + x_{j} ≥ |ν|∥h∥ + |Q_{j}(t_{1}N, h)| + |ψ(z)|,  (3.13)

where z denotes the argument of ψ in the previous inequality.

Now, |ν|∥h∥ ≲ λ|ν|t_{1} + λ|ν|a and λ|ν|t_{1} can be absorbed in the left hand side of (3.13) as soon as λ ≤ \frac{1}{2}. We also note that |Q_{j}(t_{1}N, h)| ≤ t_{1}∥h∥ ≤ t_{1}λδ ≤ t_{1}^{2} + λδ^{2}. Moreover

∥z∥ ≲ t_{1} + ∥h∥ + ∥Q̂_{j}(t_{1}N, h)∥ ≲ t_{1} + λδ + (t_{1}λδ)^{1/2} ≃ t_{1} + λδ,

and by (3.4)

|ψ(z)| ≲ ∥z∥^{2} ≲ t_{1}^{2} + λδ^{2} + λt_{1}δ ≃ t_{1}^{2} + λδ^{2} ≃ t_{1}^{2} + λa^{2}.

Since the term λa^{2} can be absorbed in the left hand side and x_{j} ≥ 0, then (3.13) will follow if we prove that for all a ≥ 0 it is ε_{0} (t_{1}|ν| + a^{2}) ≥ t_{1}^{2} + λ|ν|a where ε_{0} > 0 is a small but absolute constant. Replacing t_{1} = σ|ν| we get

ε_{0} (σ|ν|^{2} + a^{2}) ≥ σ^{2}|ν|^{2} + λ|ν|a.  (3.14)

Now, since σ^{2}|ν|^{2} + λ|ν|a ≤ (σ^{2} + λ/2)|ν|^{2} + λa^{2}/2 (3.14) holds for all a ≥ 0 provided σ^{2} + λ/2 < ε_{0}σ and λ/2 ≤ ε_{0}.

7. Second step. We prove that, given x and y in the open set {z_{j} > φ(ẑ_{j})} there exists a continuous curve connecting them and satisfying (2.4) and (2.5). Without loss of generality we can assume that x = x_{j}e_{j} with x_{j} ≥ 0 and y = y_{j}e_{j} + ŷ_{j} with y_{j} > φ(ŷ_{j}). In the first step (see (3.5)) the “canonical” John curve starting from x has been defined. The parameters ν, N and t_{1} = σ|ν| are defined as in the first step and are relative to x. The constant σ does not depend on x.

8. Our next task is to write the curve starting from y. We first explain the “intrinsic” argument which provides the parameters ν_{j}’s relative to x. Letting Φ(ξ) = ξ_{j} − φ_{(}ξ̂_{j}), we have for i = 1, ..., m

X_{i}Φ(ξ) = −∂_{i}φ(ξ̂_{j}) + ∑_{k>m} q_{ik}(ξ)∂_{k}Φ(ξ),

REGULAR DOMAINS IN HOMOGENEOUS GROUPS 13

and hence

ν_{i} = −∂_{i}φ(0) = X_{i}Φ(0).  (3.15)

Let now w = ŷ_{j} + φ(ŷ_{j})e_{j}. We look for the parameters ν_{i}, i = 1, …, m of the curve starting from w^{−1} · y = (y_{j} − φ(ŷ_{j}))e_{j} relatively to the translated boundary w^{−1} · {z_{j} = φ(ẑ_{j})}. Denote these parameters by ν̄_{1}, ..., ν̄_{m}. Then we find by left invariance

\begin{aligned}ν̄_{i} & = (X_{i}Φ)(ŷ_{j} + φ(ŷ_{j})e_{j}) = −∂_{i}φ(ŷ_{j}) + ∑_{k>m} q_{ik}(y)\frac{∂}{∂ξ_{k}} (ξ_{j} − φ(ξ̂_{j}))|_{ξ̂_{j}=ŷ_{j},ξ_{j}=φ(ŷ_{j})} \\ & = −∂_{i}φ(ŷ_{j}) + q_{ij}(y′) − ∑_{k>m,k≠j} q_{ik}(y)∂_{k}φ(ŷ_{j}). \end{aligned} (3.16)

Define

N̄ = (\frac{ν̄_{1}}{|ν̄|}, ..., \frac{ν̄_{m}}{|ν̄|}, 0, ..., 0), and t̄_{1} = σ|ν̄|.

The “canonical” John curve γ_{y} starting from y can be defined (by left translation of (3.5)) in the following way. If 0 ≤ t ≤ t̄_{1} let

\begin{aligned}γ_{y}(t) & = (ŷ_{j} + φ(ŷ_{j})e_{j}) · (tN̄ + (y_{j} − φ(ŷ_{j}))e_{j}) \\ & = ŷ_{j} + tN̄ + y_{j}e_{j} + Q(y, tN̄), \end{aligned}  (3.17)

and if t ≥ t̄_{1} let

\begin{aligned}γ_{y}(t) & = (ŷ_{j} + φ(ŷ_{j})e_{j}) · (t̄_{1}N̄ + (t − t̄_{1} + (y_{j} − φ(ŷ_{j}))e_{j}) \\ & = ŷ_{j} + t̄_{1}N̄ + (t − t̄_{1} + y_{j})e_{j} + Q(y, t̄_{1}N̄). \end{aligned}  (3.18)

9. Denote by γ_{x} and γ_{y} the curves starting from x and y. The curves γ_{x} and γ_{y} can not be expected to meet as Proposition 2.6 requires. Thus we enlarge the curve γ_{x} by constructing a curvilinear cone around it. Define

δ(t) = \{ \begin{matrix}t & if 0 ≤ t ≤ t_{1}, \\ t_{1} + (t − t_{1})^{1/2} & if t ≥ t_{1}, \end{matrix}

and recall that δ(t) ≃ d(γ_{x}(t), x). For λ > 0 let U(λ) = {h ∈ ℝ^{n} : ∥h∥ ≤ λ}, and if h = (h′, h′′) ∈ U(λ) define h_{t} = (δ(t)h′, δ(t)^{2}h′′). As h ∈ U(λ) the family of curves

γ_{x}^{h}(t) = γ_{x}(t) · h_{t} = \begin{cases}tN + x_{j}e_{j} + h_{t} + Q(tN, h_{t}) & if 0 ≤ t ≤ t_{1}, \\ t_{1}N + (t − t_{1} + x_{j})e_{j} + h_{t} + Q(t_{1}N, h_{t}) & if t ≥ t_{1} \end{cases}

forms a curvilinear cone with core γ_{x}. By the triangle inequality, if λ is small enough then for any h ∈ U(λ), the curve t ↦ γ_{x}^{h}(t) is a John curve starting from x. ¿From now on we assume that λ has been fixed small enough in order to ensure this property.

10. Two cases must be distinguished: (A) d(x, y) ≤ η|ν|; (B) d(x, y) > η|ν|.

14 ROBERTO MONTI AND DANIELE MORBIDELLI

The parameter η ∈ (0, 1) will be fixed later. Note that if 0 is a characteristic point, then Case A is empty. In Case A the curves γ_{x}^{h} and γ_{y} will meet in their first (rectifiable) piece. In Case B they will meet in their second (non rectifiable) piece.

11. Study of Case A. We claim that there exist η > 0 and M > 1 such that for all x and y there exists h ∈ U(λ) such that γ_{y}(Md(x, y)) = γ_{x}^{h}(Md(x, y)). A correct choice of η ∈ (0, 1) and M > 1 will show that the two curves meet in their first tract (see condition (3.27)).

Without loss of generality we can assume |ν| ≤ |ν̄| (otherwise the roles of x and y should be interchanged). If t ≤ t_{1} = σ|ν| then t ≤ t̄_{1} = σ|ν̄| and γ_{y}(t) = γ_{x}^{h}(t) reads

ŷ_{j} + tN̄ + y_{j}e_{j} + Q(y, tN̄) = tN + x_{j}e_{j} + h_{t} + Q(tN, h_{t}).  (3.19)

We have to show that the solution h = (h′, h′′) of this equation belongs to U(λ) if t = Md(x, y) and M is great enough.

As t ≤ t_{1} then δ(t) = t and h_{t} = (th′, t^{2}h′′). Projecting (3.19) along the first m components we get the equation y′ + tN̄ = tN + h′_{t} that is

th′ = y′ + t(N̄ − N).  (3.20)

Replacing t = Md(x, y) we find that the solution h′ satisfies

|h′| ≤ \frac{|y′|}{Md(x, y)} + |N − N̄|.  (3.21)

First of all notice that d(x, y) ≃ ∥(−x_{j}e_{j}) · (y_{j}e_{j} + ŷ_{j})∥ ≥ |y′|, which gives \frac{|y′|}{Md(x,y)} ≲ \frac{1}{M}. Moreover, using the inequality |\frac{v}{|v|} − \frac{w}{|w|} | ≤ 2\frac{|v−w|}{|v|} if v, w ∈ ℝ^{n} \ \{0\}, and the explicit form (3.15) and (3.16) of ν and ν̄ we get

\begin{aligned}|N − N̄| & ≤ 2\frac{|ν − ν̄|}{|ν|} \\ & ≤ \frac{2}{|ν|} ∑_{i=1}^{m} |∂_{i}φ(0) − ∂_{i}φ(ŷ_{j}) + q_{ij}(y) − ∑_{k>m, k≠j} q_{ik}(y)∂_{k}φ(ŷ_{j})| \\ & ≲ \frac{1}{|ν|}(|ŷ_{j}| + |y′|) ≲ \frac{d(x, y)}{|ν|}. \end{aligned}  (3.22)

The last string of estimates follows from the boundedness of ∂_{i}φ, i = m + 1, ..., n and i ≠ j, from the Lipschitz continuity of ∂_{i}φ, i = 1, ..., m, and from the inequality

d(x, y) ≃ ∥(−x_{j}e_{j}) · (ŷ_{j} + y_{j}e_{j})∥ = ∥ŷ_{j} + (y_{j} − x_{j})e_{j}∥ ≥ ∥ŷ_{j}∥ ≳ |ŷ_{j}|,  (3.23)

which holds because y lies in a bounded set.

Putting (3.22) into (3.21) and using Case A we get

|h′| ≲ \frac{1}{M} + \frac{d(x, y)}{|ν|} ≤ \frac{1}{M} + η.  (3.24)

This shows that |h′| ≤ λ as soon as M is great enough and η is small enough. We project now (3.19) along the components of the second slice obtaining

ŷ′′_{j} + y_{j}e_{j} + tQ(y, N̄) = x_{j}e_{j} + h′′_{t} + tQ(N, h′_{t}).

Here h′′_{t} = t^{2}h′′ and h′_{t} = th′ where h′ is the vector determined in (3.20) and satisfies the estimate (3.24). The last equation has a unique solution h′′ which satisfies

|h′′| ≤ \frac{|ŷ′′_{j}| + |y_{j} − x_{j}|}{t^{2}} + \frac{1}{t}|Q(y, N̄)| + |Q(N, h′)|.

Here we have to replace t = Md(x, y) but first we notice that, as in (3.23),

d(x, y) ≃ ∥ŷ_{j} + (y_{j} − x_{j})e_{j}∥ ≥ |ŷ′′_{j}|^{1/2} + |y_{j} − x_{j}|^{1/2}.  (3.25)

Moreover |Q(y, N̄)| ≲ |y′| ≲ d(x, y) and, by (3.24), |Q(N, h′)| ≲ |h′| ≲ \frac{1}{M} +η. Putting all these estimates together we find

|h′′| ≲ \frac{1}{M^{2}} + \frac{1}{M} + η.  (3.26)

Thus |h′′| ≤ λ as soon as M is great enough and η is small enough.

Our claim will be proved if we show that the choice of M and η is compatible with the condition Md(x, y) ≤ t_{1} = σ|ν|. As we are in Case A then d(x, y) ≤ η|ν| and we find the stronger condition

Mη ≤ σ,  (3.27)

which can be satisfied, taking if necessary a smaller η.

12. In view of Proposition 2.6 we have to estimate the diameter of the curves γ_{x}^{h} and γ_{y}. First of all by (3.5) we have diam(γ_{x}) ≃ Md(x, y). Moreover, if 0 ≤ s, t ≤ Md(x, y) and ∥h∥ ≤ 1

\begin{aligned}d(γ_{x}^{h}(s), γ_{x}^{h}(t)) & ≤ d(γ_{x}^{h}(s), γ_{x}(s)) + d(γ_{x}(s), γ_{x}(t)) + d(γ_{x}(t), γ_{x}^{h}(t)) \\ & ≲ ∥h_{s}∥ + diam(γ_{x}) + ∥h_{t}∥ ≲ d(x, y), \end{aligned}

and thus diam(γ_{x}^{h}) ≲ d(x, y).

13. Study of Case B. In this case the points x and y satisfy d(x, y) ≥ η|ν| where η > 0 is from now on a fixed constant. Recall that t_{1} = σ|ν| and t̄_{1} = σ|ν̄|, and for R > 0 let

t_{x} = t_{1} + R^{2}d(x, y)^{2} and t_{y} = t̄_{1} + R^{2}d(x, y)^{2}.

As above let U(λ) = {h ∈ ℝ^{n} : ∥h∥ ≤ λ} and write h_{t} = (δ(t)h′, δ(t)^{2}h′′) where now δ(t) = t_{1} + (t − t_{1})^{1/2} ≃ d(γ(t), x) for t ≥ t_{1}.

14. We claim that there exists R > 0 such that for all x, y there exists h ∈ U(λ) (λ is the parameter fixed at the end of 9) such that γ_{y}(t_{y}) = γ_{x}^{h}(t_{x}) (the times t_{x} and t_{y} depend on R).

This equation gives

ŷ_{j} + t̄_{1}N̄ + (t_{y} − t̄_{1} + y_{j})e_{j} + Q(y, t̄_{1}N̄) = t_{1}N + (t_{x} − t_{1} + x_{j})e_{j} + h_{t_{x}} + Q(t_{1}N, h_{t_{x}}).

Replacing t̄_{1} = σ|ν̄|, t_{1} = σ|ν|, t_{y} − t̄_{1} = R^{2}d(x, y)^{2} and t_{x} − t_{1} = R^{2}d(x, y)^{2} we find

\begin{aligned}ŷ_{j} & + σν̄ + (R^{2}d(x, y)^{2} + y_{j})e_{j} + σ|ν̄|Q(y, N̄) \\ & = σν + (R^{2}d(x, y)^{2} + x_{j})e_{j} + h_{t_{x}} + σ|ν|Q(N, h_{t_{x}}). \end{aligned}  (3.28)

16 ROBERTO MONTI AND DANIELE MORBIDELLI

Projecting this equation along the coordinates of the first slice we get

y′ + σν̄ = σν + h′_{t_{x}},  (3.29)

and the solution h′_{t_{x}} satisfies

|h′_{t_{x}}| ≤ |y′| + σ|ν| + σ|ν̄|.

We use |y′| ≤ d(x, y) and σ|ν| ≤ σd(x, y)/η (this is Case B). By (3.16)

\begin{aligned}|ν̄_{i}| & ≤ |∂_{i}φ(ŷ_{j})| + |q_{ij}(y)| + ∑_{k>m,k≠j} |q_{ik}(y)∂_{k}φ(ŷ_{j})| \\ & ≲ |∂_{i}φ(0)| + |∂_{i}φ(0) − ∂_{i}φ(ŷ_{j})| + |y′| \\ & ≲ |ν| + d(x, y) ≲ \frac{d(x, y)}{η}, \end{aligned}  (3.30)

because ∂_{k}φ, k > m and k ≠ j, are bounded functions, ∂_{i}φ, i = 1, ..., m are Lipschitz continuous, and |∂_{i}φ(ŷ_{j}) − ∂_{i}φ(0)| ≲ |ŷ_{j}| ≲ d(x, y) by (3.23). Ultimately, we obtain for some great but absolute constant C_{0}

|h′_{t_{x}}| ≤ C_{0} \frac{d(x, y)}{η} = C_{0}d(x, y)  (3.31)

(the parameter η has been fixed in 11 and can be considered from now on an absolute constant).

Projecting (3.28) along the coordinates of the second slice we have

ŷ′′_{j} + y_{j}e_{j} + σ|ν̄|Q(y, N̄) = x_{j}e_{j} + h′′_{t_{x}} + σ|ν|Q(N, h_{t_{x}}).

Thus

h′′_{t_{x}} = ŷ′′_{j} + (y_{j} − x_{j})e_{j} + σ|ν̄|Q(y, N̄) − σ|ν|Q(N, h′_{t_{x}}),

where h′_{t_{x}} satisfies (3.31). Notice that by (3.25) |ŷ′′_{j}|+|y_{j}−x_{j}| ≲ d(x, y)^{2} and moreover, taking into account (3.31) and Case B

σ|ν||Q(N, h′_{t_{x}})| ≲ |ν||h′_{t_{x}}| ≲ d(x, y)^{2}.

By (3.30) σ|ν̄||Q(y′, N̄)| ≲ |ν̄||y′| ≲ d(x, y)^{2}, and hence |h′′_{t_{x}}| ≲ d(x, y)^{2}. Finally

∥h∥ = \frac{∥h_{t_{x}}∥}{δ(t_{x})} = \frac{|h′_{t_{x}}| + |h′′_{t_{x}}|^{1/2}}{δ(t_{x})} ≲ \frac{d(x, y)}{δ(t_{x})} = \frac{d(x, y)}{t_{1} + (t_{x} − t_{1})^{1/2}} ≤ \frac{1}{R},

and ∥h∥ ≤ λ as soon as R ≥ C_{0}/λ where C_{0} is a great but absolute constant.

15. The estimates for diam(γ_{x}^{h}) and diam(γ_{y}) can be obtained as in 12. Our claims are proved and the proof of the Theorem in the characteristic case is ended.

16. Case 2. We now study the non characteristic case. Assume without loss of generality that Ω = {y ∈ ℝ^{n} : y_{j} > φ(ŷ_{j})} where j ∈ {1, ..., m} and φ ∈ C^{1}(ℝ^{n−1}) is a function such that φ(0) = 0, |∇φ| ≤ k and ∂_{i}φ, i = 1, ..., m and i ≠ j, are k−Lipschitz continuous functions. Let ν_{i} = −∂_{i}φ(0) if i = 1, ..., m with i ≠ j, and ν_{j} = 1. Finally write ν = (ν_{1}, ..., ν_{m}, 0, ..., 0).

17. First step. We construct John curves starting from near the boundary. The function ψ defined by

ψ(ŷ_{j}) = φ(ŷ_{j}) + ∑_{i=1,...,m,i≠j} ν_{i}y_{i}  (3.32)

satisfies

\begin{aligned}|ψ(ŷ_{j})| & = |φ(ŷ_{j}) − ∑_{i=1,...,m,i≠j} ∂_{i}φ(0)y_{i}| \\ & = | ∑_{i>m} ∂_{i}φ(0)y_{i} + O(|ŷ_{j}|^{2})| ≲ ∥ŷ_{j}∥^{2}, \end{aligned}  (3.33)

because ∂_{i}φ, i > m, are bounded and ∂_{i}φ, i = 1, ..., m and i ≠ j, are Lipschitz continuous.

Fix a point x ∈ Ω of the form x = x_{j}e_{j} with x_{j} > 0. For t ≥ 0 define the curve starting from x

γ_{x}(t) = x · tν = x · (te_{j} + t ∑_{i=1,...m,i≠j} ν_{i}e_{i}).  (3.34)

Note first that d(γ(t), x) ≃ ∥tν∥ = t|ν| ≃ t.

18. We claim that there exist t_{0} > 0 and λ ∈ (0, 1) such that for all t ≤ t_{0}

dist(γ(t); ∂Ω) ≥ λt.  (3.35)

The John condition (3.35) is equivalent to Box(γ(t), λt) ∩ ∂Ω = ∅.

Points in Box(γ(t), λt) are of the form

\begin{aligned}x · tν · h & = x_{j}e_{j} · (tν + h + tQ(ν, h′)) \\ & = x_{j}e_{j} + tν + h + tQ(ν, h′) + Q(x_{j}e_{j}, tν + h′) ≡ z, \end{aligned}

where h ∈ ℝ^{n} and ∥h∥ ≤ λt. We have to check that (z is defined in the last equation)

x_{j} + t + h_{j} > φ(ẑ_{j}) = − ∑_{i≤m,i≠j} ν_{i}(tν_{i} + h_{i}) + ψ(ẑ_{j}),

by (3.32). Since |h_{k}| < λt, k = 1, …, m, if λ > 0 is small enough, the last inequality is ensured by

x_{j} + (1 − λ)t + t ∑_{i≤m,i≠j}(ν_{i}^{2} − λ|ν_{i}|) ≥ |ψ(ẑ_{j})|,

which is implied by

ε_{0}(x_{j} + t) ≥ |ψ(ẑ_{j})|.  (3.36)

The right hand side of (3.36) can be estimated by (3.33)

|ψ(ẑ_{j})| ≲ ∥ẑ_{j}∥^{2} = ∥tν̂_{j} + ĥ_{j} + tQ(ν, h′) + Q(x_{j}e_{j}, tν + h′)∥^{2}
≲ t^{2} + λt^{2} + ∥tQ(ν, h′)∥^{2} + ∥Q(x_{j}e_{j}, tν + h′)∥^{2}
≲ t^{2} + (t|ν|λt) + (x_{j}|tν + h′|) ≲ t^{2} + x_{j}t,

18 ROBERTO MONTI AND DANIELE MORBIDELLI

where we used |ν| ≲ 1. Then (3.36) is ensured by

ε_{0}(x_{j} + t) ≥ t^{2} + x_{j}t,

where ε_{0} > 0 is a small but absolute constant. This inequality is trivially satisfied as soon as t ≤ ε_{0}.

19. Second step. We prove the uniform condition. Given two points x, y ∈ Ω we have to connect them by curves γ_{x} and γ_{y} satisfying the hypotheses of Proposition 2.6. Assume that x = x_{j}e_{j} with x_{j} > 0 and write y = ŷ_{j} + y_{j}e_{j} with y_{j} > φ(ŷ_{j}).

We first notice that if d(x, y) < dist(x; ∂Ω) then x and y can be connected simply by a geodesic. Therefore, without loss of generality we can assume that

d(x, y) ≥ dist(x; ∂Ω).  (3.37)

20. We claim that there exists a constant C_{0} > 0 such that

x_{j} ≤ C_{0}d(x, y)  (3.38)

for all x = x_{j}e_{j}, y ∈ Ω satisfying (3.37) and lying in a bounded set (say the unit Euclidean ball centered at the origin). Indeed, if ξ = ξ̂_{j} + φ(ξ̂_{j})e_{j} ∈ ∂Ω then

d(x, ξ) ≃ ∥(−x_{j}e_{j}) · (ξ̂_{j} + φ(ξ̂_{j})e_{j})∥
≃ |φ(ξ̂_{j}) − x_{j}| + |ξ̂′_{j}| + |ξ′′ + Q( − x_{j}e_{j}, ξ̂_{j} + φ(ξ̂_{j})e_{j})|^{1/2}
= |φ(ξ̂_{j}) − x_{j}| + |ξ̂′_{j}| + |ξ′′ + Q( − x_{j}e_{j}, ξ̂′_{j})|^{1/2}.

We used here the bilinearity of Q and the property 0 = (−e_{j}) · e_{j} = −Q(e_{j}, e_{j}). In order to prove (3.38) it will be enough to show that

x_{j} ≤ C_{0}(|φ(ξ̂_{j}) − x_{j}| + |ξ̂′_{j}| + |ξ′′ + Q( − x_{j}e_{j}, ξ̂′_{j})|^{1/2}).  (3.39)

By the Lipschitz continuity of φ we find

x_{j} ≤ |x_{j} − φ(ξ̂_{j})| + |φ(ξ̂_{j})| ≲ |x_{j} − φ(ξ̂_{j})| + |ξ̂_{j}|
≃ |x_{j} − φ(ξ̂_{j})| + |ξ̂′_{j}| + |ξ′′|
≲ |x_{j} − φ(ξ̂_{j})| + |ξ̂′_{j}| + |ξ′′ + Q(−x_{j}e_{j}, ξ̂′_{j})| + |Q(−x_{j}e_{j}, ξ̂′_{j})|
≲ |x_{j} − φ(ξ̂_{j})| + |ξ̂′_{j}| + |ξ′′ + Q(−x_{j}e_{j}, ξ̂′_{j})|^{1/2} + x_{j}|ξ̂′_{j}|
≲ |x_{j} − φ(ξ̂_{j})| + |ξ̂′_{j}| + |ξ′′ + Q(−x_{j}e_{j}, ξ̂′_{j})|^{1/2}.

We used here the fact that all the involved vectors lie in a bounded set. Our claim (3.38) is proved.

21. Our next step is to compute the “canonical” John curve starting from a generic point y ∈ Ω. The point y and the boundary of Ω will be translated by a suitable vector η ∈ ℝ^{n} in such a way that η · y lies in the half axis {αe_{j} : α > 0}. Using the equation of the translated surface the correct vector of parameters ν̄ can be computed and the curve starting from y will be defined as γ_{y}(t) = η^{−1} · (η · y) · (tν̄) = y · (tν̄) for t ≥ 0.

22. We claim that there exist ϱ > 0 and C_{0} > 1 such that for all y ∈ Ω ∩ {|y| ≤ ϱ} there exists η ∈ ℝ^{n} such that:

REGULAR DOMAINS IN HOMOGENEOUS GROUPS 19

(i) η · ∂Ω contains the origin; (ii) η · y belongs to {αe_{j} : α > 0}; (iii) |η| ≤ C_{0}|ŷ_{j}|.

We look for η = (η′, η′′). If η′′ is given, and we define η′ by the equation

η′ = −ŷ′_{j} − φ(ŷ′_{j} − η′′)e_{j},  (3.40)

then (i) is satisfied. Indeed, z := ŷ′_{j} − η′′ + φ(ŷ′_{j} − η′′)e_{j} ∈ ∂Ω and

η · z = η′ + η′′ + ŷ′_{j} − η′′ + φ(ŷ′_{j} − η′′)e_{j} + Q(η′, ŷ′_{j} + φ(ŷ′_{j} − η′′)e_{j}) = 0,

by (3.40).

We shall soon prove that the implicit equation

η′′ + y′′ + Q(−ŷ′_{j} − φ(ŷ′_{j} − η′′)e_{j}, ŷ′_{j} + y_{j}e_{j}) = 0,  (3.41)

has a solution η′′. Then, the choice of η′′ solution of (3.41) and of η′ as in (3.40) ensures that the vector η = (η′, η′′) satisfies (ii). Indeed

\begin{aligned}η · y & = ( − ŷ′_{j} + η′′ − φ(ŷ′_{j} − η′′)e_{j}) · (ŷ′_{j} + y′′ + y_{j}e_{j}) \\ & = y′′ + η′′ + (y_{j} − φ(ŷ′_{j} − η′′))e_{j} + Q( − ŷ′_{j} − φ(ŷ′_{j} − η′′)e_{j}, ŷ′_{j} + y_{j}e_{j}), \end{aligned}

which belongs to the j−th axis if and only if (3.41) holds.

We prove the existence of a solution η′′. First notice that by the bilinearity of Q

Q( − ŷ′_{j} − φ(ŷ′_{j} − η′′)e_{j}, y′) = Q( − ŷ′_{j} − y_{j}e_{j} + (y_{j} − φ(ŷ′_{j} − η′′))e_{j}, y′)
= Q((y_{j} − φ(ŷ′_{j} − η′′))e_{j}, y′)
= (y_{j} − φ(ŷ′_{j} − η′′))Q(e_{j}, y′).

The map y′ ↦ Q(e_{j}, y′) is linear and does not depend on y_{j}. Thus (3.41) is equivalent to the following

η′′ + y′′ + (y_{j} − φ(ŷ′_{j} − η′′))Q(e_{j}, ŷ′_{j}) = 0.  (3.42)

We show that there exists ϱ > 0 such that if y ∈ Ω and |y| ≤ ϱ then (3.42) has a solution η′′ satisfying

|η′′| ≤ 2|ŷ_{j}|.  (3.43)

We use a fixed point argument. Letting F (η′′) = −y′′ − (y_{j} − φ(ŷ′_{j} − η′′))Q(e_{j}, ŷ′_{j}) equation (3.42) becomes F (η′′) = η′′. Let D = {η′′ : |η′′| ≤ 2|ŷ_{j}|}. If we show that F (D) ⊂ D, then the continuous map F has a fixed point by Brouwer theorem. Indeed

|F (η′′)| ≤ |y′′| + |Q(e_{j}, ŷ′_{j})||(y_{j} − φ(ŷ′_{j} − η′′))|
≤ |ŷ_{j}| + C|ŷ′_{j}|(|y_{j}| + |ŷ′_{j}| + |η′′|)
≤ |ŷ_{j}|(1 + 4C|y|) ≤ 2|ŷ_{j}|,

as soon as |y| ≤ ϱ = 1/(4C) (here the constant C depends only on the surface). Moreover by (3.40) and by (3.43)

|η′| = |ŷ′_{j}| + |φ(ŷ′_{j} − η′′)| ≲ |ŷ′_{j}| + |η′′| ≲ |ŷ_{j}|.

This proves claim (iii).

20 ROBERTO MONTI AND DANIELE MORBIDELLI

23. We compute ν̄ by a left translation argument. Let Φ(y) = y_{j} − φ(ŷ_{j}). The parameters ν at the point y = 0 are given by ν_{i} = X_{i}Φ(0), i = 1, ..., m. Then for any point ξ = ξ̂_{j} + φ(ξ̂_{j})e_{j} belonging to the surface {Φ = 0} the parameters ν_{i} = ν_{i}(ξ) are given by

\begin{aligned}ν_{i}(ξ) & = (X_{i}Φ)(ξ) \\ & = \begin{cases}(∂_{j} + ∑_{k>m} q_{jk}(ξ′)∂_{k})Φ(ξ) = 1 − ∑_{k>m} q_{jk}(ξ′)∂_{k}φ(ξ̂_{j}) & if i = j, \\ (∂_{i} + ∑_{k>m} q_{ik}(ξ′)∂_{k})Φ(ξ) = −∂_{i}φ(ξ̂) − ∑_{k>m} q_{ik}(ξ′)∂_{k}φ(ξ̂_{j}) & if i ≤ m, i ≠ j. \end{cases} \end{aligned}

Let η ∈ ℝ^{n} be a vector relative to y as in claims (i), (ii) and (iii) of 22. The correct value of the parameters is given by the evaluation of the previous equation at the point −η (this is because the point −η is taken to the origin by the left translation τ_{η}). Define ν̄_{i} = ν_{i}(−η). Set ν̄ = (ν̄_{1}, ..., ν̄_{m}, 0, ..., 0). We claim that

|ν − ν̄| ≲ |ŷ_{j}|  (3.44)

If i ≠ j, by the Lipschitz continuity of φ and by claim (iii)

|ν̄_{i} − ν_{i}| = | − ∂_{i}φ(−η̂_{j}) − ∑_{k=m+1}^{n} q_{ik}(−η′)∂_{k}φ(−η̂_{j}) + ∂_{i}φ(0)|

≲ |∂_{i}φ(0) − ∂_{i}φ(−η̂_{j})| + ∑_{k=m+1}^{n} |q_{ik}(−η′)||∂_{k}φ(−η̂_{j})|

≲ |η̂_{j}| + |η′| ≃ |η| ≲ |ŷ_{j}|.

The estimate of the j−th component of ν − ν̄ is easier and we do not prove it.

24. Let γ_{x} be the curve starting from x = x_{j}e_{j} defined in (3.34) and let γ_{y} be the curve starting from y ∈ Ω defined for t ≥ 0 by

γ_{y}(t) = y · (tν̄) = ŷ′_{j} + tν̄ + y_{j}e_{j} + y′′ + Q(y′, tν̄),

where ν̄ is the vector of parameters discussed above. We now construct a cone with core γ_{x}. For λ > 0 let U(λ) = {h ∈ ℝ^{n} : ∥h∥ ≤ λ} and for t ≥ 0 define h_{t} = (th′, t^{2}h′′). Note that ∥h_{t}∥ = t∥h∥ ≃ d(γ_{x}(t), γ_{x}(0))∥h∥. Finally let

\begin{aligned}γ_{x}^{h}(t) & = x_{j}e_{j} · (tν) · h_{t} = x_{j}e_{j} · (tν + h_{t} + Q(tν, h′_{t})) \\ & = x_{j}e_{j} + tν + h_{t} + tQ(ν, h′_{t}) + Q(x_{j}e_{j}, tν + h′_{t}). \end{aligned}

25. We claim that there exist M > 0 and ϱ > 0 such that for all x = x_{j}e_{j} ∈ Ω and for all y ∈ Ω such that |ŷ_{j}| ≤ ϱ there exists h ∈ U(λ) such that γ_{x}^{h}(Md(x, y)) = γ_{y}(Md(x, y)). Here λ is a parameter small enough to ensure that for all h ∈ U(λ) γ_{x}^{h} is a John curve with constant λ.

Equality γ_{y}(t) = γ_{x}^{h}(t) reads

ŷ′_{j}+tν̄ + y_{j}e_{j} + y′′ + tQ(y′, ν̄) = x_{j}e_{j} + tν + h_{t} + tQ(ν, h′_{t}) + x_{j}Q(e_{j}, tν + h′_{t}). (3.45)

Projecting this equation along the coordinates of the first slice we get

ŷ′_{j} + tν̄ + y_{j}e_{j} = x_{j}e_{j} + tν + th′  (3.46)

and the solution h′ satisfies |h′| ≤ \frac{1}{t}\{|ŷ′_{j}|+|y_{j} −x_{j}|+t|ν −ν̄|\}. Replacing t = Md(x, y) we find

|h′| ≤ \frac{|ŷ′_{j}| + |y_{j} − x_{j}|}{Md(x, y)} + |ν − ν̄|.

By the equivalence

d(x, y) ≃ ∥(−x) · y∥ ≃ |y_{j} − x_{j}| + |ŷ′_{j}| + |y′′ + Q(−x_{j}e_{j}, ŷ′_{j})|^{1/2},  (3.47)

and by (3.44) we obtain for some absolute constant C_{0}

|h′| ≤ C_{0}(\frac{1}{M} + ϱ)  (3.48)

as soon as |ŷ_{j}| ≤ ϱ.

We project now (3.45) along the coordinates of the second slice obtaining

y′′ + tQ(y′, ν̄) = h′′_{t} + tQ(ν, h′_{t}) + x_{j}Q(e_{j}, tν + h′_{t}),

where h′_{t} = th′ and h′ satisfies (3.48). We deduce that

|h′′_{t}| ≤ |y′′| + t|Q(y′, ν̄)| + t|Q(ν, h′_{t})| + x_{j}|Q(e_{j}, tν + h′_{t})|.

We estimate separately each term in the right hand side. By (3.47) and (3.38)

|y′′| ≤ |y′′ + Q(−x_{j}e_{j}, ŷ′_{j})| + |Q(−x_{j}e_{j}, ŷ′_{j})| ≲ d(x, y)^{2} + x_{j}|ŷ′_{j}| ≲ d(x, y)^{2},

Moreover |Q(y′, ν̄)| ≲ |y′| ≲ d(x, y) and by (3.48) |Q(ν, h′_{t})| ≲ |h′_{t}| ≲ t(\frac{1}{M} + ϱ). The vectors ν and ν̄ are bounded. Finally, again by (3.38), x_{j}|Q(e_{j}, tν + h′_{t})| ≲ td(x, y). Then

|h′′_{t}| ≲ d(x, y)^{2} + td(x, y) + t^{2}(\frac{1}{M} + ϱ),

and replacing t = Md(x, y) we finally get

|h′′| ≲ \frac{1}{M^{2}} + \frac{1}{M} + ϱ,

which shows that ∥h∥ ≤ λ if M is great and ϱ is small enough.

4. John domains in a group of step 3

In this section we study John domains in groups of step 3. In order to make explicit computations we shall study the simplest Carnot group of step 3 whose Lie algebra has the lowest dimension, which is 4.

Consider in ℝ^{4} the vector fields

\begin{aligned}X_{1} & = ∂_{1} − \frac{1}{2}x_{2}∂_{3} − \{\frac{1}{12}(x_{1}x_{2} + αx_{2}^{2}) + \frac{1}{2}x_{3}\}∂_{4}, \\ X_{2} & = ∂_{2} + \frac{1}{2}x_{1}∂_{3} + \{\frac{1}{12}(x_{1}^{2} + αx_{1}x_{2}) − \frac{α}{2}x_{3}\}∂_{4}, \\ X_{3} & = ∂_{3} + \frac{1}{2}(x_{1} + αx_{2})∂_{4} \\ X_{4} & = ∂_{4}, \end{aligned}

where α ∈ ℝ is a real parameter. The commutation relations are

[X_{1}, X_{2}] = X_{3}, [X_{1}, X_{3}] = X_{4}, [X_{2}, X_{3}] = αX_{4},

and all other commutators vanish. Thus, for any α ∈ ℝ the vector fields X_{1}, X_{2} are generators of a Lie algebra of differential operators in ℝ^{4} of step 3. It can be checked that the following group law on ℝ^{4} makes X_{1}, X_{2}, X_{3} and X_{4} left invariant:

\begin{matrix}x · y = ( & x_{1} + y_{1}, x_{2} + y_{2}, x_{3} + y_{3} + \frac{1}{2}(x_{1}y_{2} − x_{2}y_{1}), \\ x_{4} + y_{4} + \frac{1}{12}\{(y_{1} + αy_{2})(x_{2}y_{1} − x_{1}y_{2}) + (x_{1} + αx_{2})(x_{1}y_{2} − x_{2}y_{1})\} \\ + \frac{1}{2}\{(x_{1}y_{3} − x_{3}y_{1}) + α(x_{2}y_{3} − y_{2}x_{3})\}). \end{matrix} (4.1)

Notice that x^{−1} = −x. Introduce the abbreviations

\begin{aligned}q_{1}(x_{1}, x_{2}, x_{3}) & = −\{\frac{1}{12}(x_{1}x_{2} + αx_{2}^{2}) + \frac{1}{2}x_{3}\} \\ q_{2}(x_{1}, x_{2}, x_{3}) & = \{\frac{1}{12}(x_{1}^{2} + αx_{1}x_{2}) − \frac{α}{2}x_{3}\} \\ q_{3}(x_{1}, x_{2}) & = \frac{1}{2}(x_{1} + αx_{2}), \end{aligned}  (4.2)

and

\begin{aligned}Q_{3}(x_{1}, x_{2}, y_{1}, y_{2}) & = \frac{1}{2}(x_{1}y_{2} − x_{2}y_{1}) \\ Q_{4}(x_{1}, x_{2}, x_{3}, y_{1}, y_{2}, y_{3}) & = \frac{1}{12}\{(y_{1} + αy_{2})(x_{2}y_{1} − x_{1}y_{2}) + (x_{1} + αx_{2})(x_{1}y_{2} − x_{2}y_{1})\} \\ & + \frac{1}{2}\{(x_{1}y_{3} − x_{3}y_{1}) + α(x_{2}y_{3} − y_{2}x_{3})\}, \end{aligned} (4.3)

in such a way that

\begin{matrix}x · y = (x_{1} + y_{1}, x_{2} + y_{2}, x_{3} + y_{3}+ & Q_{3}(x_{1}, x_{2}, y_{1}, y_{2}), \\ x_{4} + y_{4} + Q_{4}(x_{1}, x_{2}, x_{3}, y_{1}, y_{2}, y_{3})). \end{matrix}

We denote by d the Carnot-Carathéodory distance induced on ℝ^{4} by X_{1} and X_{2} and by B(x, r) the open ball centered at x ∈ ℝ^{4} with radius r ≥ 0. Define also the following homogeneous norm in ℝ^{4}

∥x∥ = |x_{1}| + |x_{2}| + |x_{3}|^{1/2} + |x_{4}|^{1/3}.

REGULAR DOMAINS IN HOMOGENEOUS GROUPS 23

By a standard argument it can be proved that d(x, y) ≃ ∥y^{−1} · x∥. Define the Box

Box(x, r) = {x · y ∈ ℝ^{4} : ∥y∥ ≤ r}.  (4.4)

Then it follows that there exists c > 1 such that for all x ∈ ℝ^{n} and r ≥ 0

Box(x, c^{−1}r) ⊂ B(x, r) ⊂ Box(x, cr).

Let S ⊂ ℝ^{4} be a 3−dimensional surface of class C^{1}. If x_{0} ∈ S there exists a neighborhood U of x_{0} in ℝ^{4} and there exists Φ ∈ C^{1}(U; ℝ) such that S ∩ U = {x ∈ U : Φ(x) = 0} and ∇Φ ≠ 0 on S ∩ U. A point x ∈ S ∩ U is said to be characteristic if and only if X_{1}Φ(x) = X_{2}Φ(x) = 0. From a geometric point of view this means that X_{1} and X_{2} belong to the tangent spaces to S at x.

Definition 4.1. A characteristic point x ∈ S ∩ U is of first type if X_{3}Φ(x) ≠ 0. If X_{3}Φ(x) = 0 then it is of second type.

If x ∈ S ∩ U is a characteristic point of second type then X_{4}Φ(x) = ∂_{4}Φ(x) can not be 0. Otherwise it would be X_{1}Φ = ⋯ = X_{4}Φ = 0 at x and this is impossible because ∇Φ ≠ 0 and X_{1}, …, X_{4} are independent at each point.

We are interested in expressing S as a graph in a neighborhood of a point x̄ ∈ S after a translation that takes x̄ to the origin. Notice that x̄ is a characteristic point (of first, second type) of S if and only if 0 is a characteristic point (of first, second type) of the translated surface x̄^{−1} · S. This is an easy consequence of the left invariance of the X_{j}’s.

Now, let S = ∂Ω be the boundary of an open set of class C^{1} and let x̄ ∈ ∂Ω. Assume that x̄ is a characteristic point of first type. Then, for any point x̃ in a neighborhood of x̄, the surface (x̃^{−1} ·∂Ω)∩V, where V ⊂ ℝ^{4} is a suitable neighborhood of the origin, can be parameterized by a function x_{3} = φ(x_{1}, x_{2}, x_{4}), where φ ∈ C^{1}(D) and D is a neighborhood of the origin in ℝ^{3}.

If x̄ is a characteristic point of second type then the variable x_{4} must be given in terms of the variables x_{1}, x_{2}, x_{3}. Then, for any x̃ near x̄, the surface (x̃^{−1} · ∂Ω) ∩ V, can be parameterized by a function x_{4} = φ(x_{1}, x_{2}, x_{3}), where φ ∈ C^{1}(D) and D is a neighborhood of the origin in ℝ^{3}.

Definition 4.2. Let Ω be a C^{1} bounded connected open set. Denote by Σ_{2} ⊂ ∂Ω the characteristic set of second type. We say that Ω is admissible if

(i) there are a neighborhood A in ∂Ω of Σ_{2}, a neighborhood D of the origin in ℝ^{3} and constants ε_{0}, k > 0 such that for any z ∈ A

z^{−1} · ∂Ω ∩ (D ×]−ε_{0}, ε_{0}[) = {x ∈ ℝ^{4} : x_{4} = φ_{z}(x′), x′ = (x_{1}, x_{2}, x_{3}) ∈ D},

where the function φ_{z} satisfies

\begin{aligned}|φ_{z}(x′) & − ⟨∇φ_{z}(0), x′⟩| \\ & ≤ k(∥x′∥^{3} + (|∂_{1}φ_{z}(0)|^{1/2} + |∂_{2}φ_{z}(0)|^{1/2} + |∂_{3}φ_{z}(0)|)(x_{1}^{2} + x_{2}^{2})) \end{aligned}  (4.5)

(here ∥x′∥ = |x_{1}| + |x_{2}| + |x_{3}|^{1/2} and ∇ denotes the Euclidean gradient); (ii) ∂Ω is of class C^{1,1} away from Σ_{2}.

The requirement (4.5) near Σ_{2} is the key point. The natural question now is how to check it for a given surface. An answer is contained in the following Lemma 4.3.

Write x = (x_{1}, x_{2}, x_{3}, x_{4}) = (x′, x_{4}) ∈ ℝ^{3} × ℝ. Thus we can split the group law (4.1) as follows

x · y = ((x′ ∘ y′), x_{4} + y_{4} + Q_{4}(x′, y′)),

where ∘ denotes the composition law in the Heisenberg group. Denote by ϱ the control distance in the Heisenberg group.

Lemma 4.3. Let D_{0} ⊂ ℝ^{3} be a neighborhood of the origin. Consider a C^{1} function φ : D_{0} → ℝ and its graph Γ = {x_{4} = φ(x_{1}, x_{2}, x_{3}) = φ(x′)}. Assume that φ(0, 0, 0) = 0 and that there exists k > 0 such that

(i) the second derivatives X_{i}X_{j}φ(x′), i, j = 1, 2, exist at any point x′ ∈ D and the functions X_{i}X_{j}φ, i, j = 1, 2, satisfy

|X_{i}X_{j}φ(x′) − X_{i}X_{j}φ(y′)| ≤ kϱ(x′, y′), x′, y′ ∈ D_{0}, i, j = 1, 2;  (4.6)

(ii) Written Φ(x) = Φ(x_{1}, x_{2}, x_{3}, x_{4}) = φ(x_{1}, x_{2}, x_{3}) − x_{4}, for all x = (x′, φ(x′)) ∈ Γ the following estimate holds

\begin{matrix}|X_{1}^{2}Φ(x)| + |X_{2}^{2}Φ(x)| + |(X_{1}X_{2} + X_{2}X_{1})Φ(x)| \\ ≤ k(|X_{1}Φ(x)|^{1/2} + |X_{2}Φ(x)|^{1/2} + |X_{3}Φ(x)|). \end{matrix}  (4.7)

Then there is D ⊂ D_{0} such that, fixed any point z′ ∈ D, the surface (z′, φ(z′))^{−1} · Γ is the graph of a function φ_{z} defined on (z′)^{−1} · D_{0} which satisfies (4.5) for all x′ ∈ D.

In Example 5.2 we will give an example of admissible set by means of this lemma. Proof. Fix a neighborhood of the origin D ⊂ D_{0} such that D ⊂ (z′)^{−1} ∘ D_{0} for all z′ ∈ D (i.e. D ∘ D ⊂ D_{0}). Take a point z′ ∈ D. It is easy to check that the set (z′, φ(z′))^{−1} · Γ is defined by

\begin{aligned}y_{4} & = φ(z′ ∘ y′) − φ(z′) + Q_{4}(−z′, z′ ∘ y′) \\ & = φ(z′ ∘ y′) − φ(z′) − Q_{4}(z′, y′) := φ_{z}(y), y ∈ (z′)^{−1} ∘ D_{0}. \end{aligned}

The equality Q_{4}(−z′, z′ ∘ y′) = −Q_{4}(z′, y′) is an immediate consequence of the associative property (y′, y_{4}) = ((z′, z_{4})^{−1} · (z′, z_{4})) · (y′, y_{4}) = (z′, z_{4})^{−1} · ((z′, z_{4}) · (y′, y_{4})), y, z ∈ ℝ^{4}.

By (4.2) and (4.3) the function Q_{4}(z, y) can be written as

\begin{aligned}Q_{4}(z′, y′) & = q_{1}(z′)y_{1} + q_{2}(z′)y_{2} + q_{3}(z′)y_{3} \\ & + X_{1}q_{1}(z′)\frac{y_{1}^{2}}{2} + X_{2}q_{2}(z′)\frac{y_{2}^{2}}{2} + (X_{1}q_{2}(z′) + X_{2}q_{1}(z′))\frac{y_{1}y_{2}}{2} . \end{aligned}  (4.8)

We now claim that the function y′ ↦ φ(z′ ∘ y′) − φ(z′) admits the following “homogeneous Taylor expansion”:

\begin{aligned}φ(z′ ∘ y′) & − φ(z′) = X_{1}φ(z′)y_{1} + X_{2}φ(z′)y_{2} + X_{3}φ(z′)y_{3} + \frac{1}{2}X_{1}^{2}φ(z′)y_{1}^{2} \\ & + \frac{1}{2}X_{2}^{2}φ(z′)y_{2}^{2} + \frac{1}{2}(X_{1}X_{2} + X_{2}X_{1})φ(z′)y_{1}y_{2} + O(∥y′∥^{3}), \end{aligned}  (4.9)

where O(∥y′∥^{3}) ≤ C∥y′∥^{3} for all y ∈ (z′)^{−1} ∘ D_{0} (and thus for all y′ ∈ D). The constant C does not depend on z′ ∈ D. Write now

φ_{z}(y′) = φ_{z}(y_{1}, y_{2}, y_{3}) − φ_{z}(y_{1}, y_{2}, 0) + φ_{z}(y_{1}, y_{2}, 0) − φ_{z}(0, 0, 0).

We examine first the first term. By the mean value theorem

φ_{z}(y_{1}, y_{2}, y_{3}) − φ_{z}(y_{1}, y_{2}, 0) = φ(z′ ∘ (y_{1}, y_{2}, y_{3})) − φ(z′ ∘ (y_{1}, y_{2}, 0))

= (∂_{3}φ)(z′ ∘ (y_{1}, y_{2}, ϑy_{3}))y_{3} = (X_{3}φ)(z′ ∘ (y_{1}, y_{2}, ϑy_{3}))y_{3}

= X_{3}φ(z′)y_{3} + {X_{3}φ(z′ ∘ (y_{1}, y_{2}, ϑy_{3})) − X_{3}φ(z′)}y_{3},

where ϑ ∈ ]0, 1[. By the Lipschitz continuity of X_{3}φ (recall that X_{3} = X_{1}X_{2} − X_{2}X_{1}) we get |{X_{3}φ(z′ ∘ (y_{1}, y_{2}, ϑy_{3})) − X_{3}φ(z′)}| ≤ k∥y′∥, which multiplied by y_{3} can be estimated by ∥y′∥^{3}.

We now look at the second piece. Write g(t) = φ(z′∘(ty_{1}, ty_{2}, 0)) = φ(exp(t(y_{1}X_{1}+ y_{2}X_{2}))(z)), t ∈ [0, 1]. Condition (4.6) ensures that g is of class C^{2} and that

g′(t) = ∑_{i=1}^{2} y_{i}X_{i}φ(z′ ∘ (ty_{1}, ty_{2}, 0)) and g′′(t) = ∑_{i,j=1}^{2} y_{i}y_{j}X_{i}X_{j}φ(z′ ∘ (ty_{1}, ty_{2}, 0)).

Thus, by the Taylor formula

φ(z′ ∘ (ty_{1}, ty_{2}, 0)) − φ(z′) = g(1) − g(0) = g′(0) + \frac{1}{2}g′′(ϑ)

= ∑_{i=1}^{2} X_{i}φ(z′)y_{i} + \frac{1}{2} ∑_{i,j=1}^{2} X_{i}X_{j}φ(z′)y_{i}y_{j}

+ \frac{1}{2} ∑_{i,j=1}^{2}\{X_{i}X_{j}φ(z′ ∘ (ϑy_{1}, ϑy_{2}, 0)) − X_{i}X_{j}φ(z′)\}y_{i}y_{j}.

In view of the Lipschiz continuity of X_{i}X_{j}φ the last line can be easily estimated with ∥y′∥. Thus we have proved that (4.9) holds.

Subtracting (4.8) from (4.9) we find the Taylor expansion of φ_{z}:

φ_{z}(y) = ∑_{j=1}^{3}\{X_{j}φ(z′) − q_{j}(z′)\}y_{j} + \frac{1}{2} ∑_{i,j=1}^{2}\{X_{i}X_{j}φ(z′) − X_{i}q_{j}(z′)\}y_{i}y_{j} + O(∥y′∥^{3}).

Then

∂_{i}φ_{z}(0) = X_{i}φ(z′) − q_{i}(z′) = X_{i}Φ(z′, φ(z′)), i = 1, 2, 3,
∂_{ij}^{2}φ_{z}(0) = \frac{1}{2}(X_{i}X_{j}φ(z′) − X_{i}q_{j}(z′) + X_{j}X_{i}φ(z′) − X_{j}q_{i}(z′))

= \frac{1}{2}(X_{i}X_{j} + X_{j}X_{i})Φ(z′, φ(z′)), i, j = 1, 2.

26 ROBERTO MONTI AND DANIELE MORBIDELLI

Thus, assuming (4.7) we immediately see that (4.5) holds. This ends the proof of the lemma.

We are now ready to prove our main result.

Theorem 4.4. If Ω ⊂ ℝ^{4} is an admissible domain then it is a John domain in (ℝ^{4}, d).

Proof. We prove the theorem using Proposition 2.4. We shall construct “canonical” John curves starting from points near the boundary ∂Ω. The proof will be split in several numbered small steps.

1. Fix a point x̄ ∈ ∂Ω, let U ⊂ ℝ^{4} be a neighborhood of x̄ and let Φ ∈ C^{2}(U; ℝ) be a local equation for ∂Ω ∩ U. We shall distinguish three cases:

(C1) X_{1}Φ(x̄) = X_{2}Φ(x̄) = 0, and X_{3}Φ(x̄) ≠ 0 (x̄ is a characteristic point of first

type); (C2) X_{1}Φ(x̄) = X_{2}Φ(x̄) = X_{3}Φ(x̄) = 0 (x̄ is a characteristic point of second type); (C3) |X_{1}Φ(x̄)| + |X_{2}Φ(x̄)| > 0 (x̄ is a non characteristic point of ∂Ω);

2. Case 1. After a translation 0 ∈ ∂Ω can be assumed to be near x̄. Thus, in a neighborhood of 0, ∂Ω can be written in the form y_{3} = φ(y_{1}, y_{2}, y_{4}), where the function φ = φ(y_{1}, y_{2}, y_{4}) of class C^{1,1} and φ(0) = 0. Define

ν_{1} = −∂_{1}φ(0), ν_{2} = −∂_{2}φ(0), ν = (ν_{1}, ν_{2}), N_{1} = \frac{ν_{1}}{|ν|}, N_{2} = \frac{ν_{2}}{|ν|},

and if ν = 0 simply set N_{1} = N_{2} = 0. Moreover let ψ(y) = φ(y) + ν_{1}y_{1} + ν_{2}y_{2}. Since φ is C^{1,1}, we have

|ψ(y)| = |φ(y) + ν_{1}y_{1} + ν_{2}y_{2}| ≲ y_{1}^{2} + y_{2}^{2} + |y_{4}|.  (4.10)

Consider now a point x = (0, 0, x_{3}, 0) ∈ Ω with 0 < x_{3} ≤ 1. We shall define a continuous path γ : [0, 1] → Ω such that γ(0) = x and the John property dist(γ(t); ∂Ω) ≥ λd(γ(t), x) holds for all t ∈ [0, 1] and for some λ > 0 depending only on Ω. The path will be made by two pieces.

3. First piece. Let σ > 0 and define t_{1} = σ|ν|. For t ∈ [0, t_{1}] we define

γ(t) = (0, 0, x_{3}, 0) · (tN_{1}, tN_{2}, 0, 0).

Notice that d(γ(t), x) ≃ t and that the first piece degenerates if ν = 0.

4. We claim that there exist σ, λ ∈ (0, 1) absolute constants such that for all t ≤ t_{1}

Box(γ(t), λt) ⊂ Ω.  (4.11)

REGULAR DOMAINS IN HOMOGENEOUS GROUPS 27

Condition (4.11) is equivalent to the John property for γ in this first piece. Points in Box(γ(t), λt) are of the form

γ(t) · h = (0, 0, x_{3}, 0) · (tN_{1}, tN_{2}, 0, 0) · (h_{1}, h_{2}, h_{3}, h_{4})
= (0, 0, x_{3}, 0) · (tN_{1} + h_{1}, tN_{2} + h_{2}, h_{3} + Q_{3}(tN_{1}, tN_{2}, h_{1}, h_{2}),
, h_{4} + Q_{4}(tN_{1}, tN_{2}, 0, h_{1}, h_{2}, h_{3}))
= (tN_{1} + h_{1}, tN_{2} + h_{2}, x_{3} + h_{3} + Q_{3}(tN_{1}, tN_{2}, h_{1}, h_{2}),
, h_{4} + Q_{4}(tN_{1}, tN_{2}, 0, h_{1}, h_{2}, h_{3})
+ Q_{4}(0, 0, x_{3}, tN_{1} + h_{1}, tN_{2} + h_{2}, h_{3} + Q_{3}(tN_{1}, tN_{2}, h_{1}, h_{2})))

with h = (h_{1}, h_{2}, h_{3}, h_{4}) and ∥h∥ ≤ λt.

Now, γ(t) · h ∈ Ω provided that (recall that φ(z) = −ν_{1}z_{1} − ν_{2}z_{2} + ψ(z))

\begin{matrix}x_{3}+ & h_{3} + Q_{3}(tN_{1}, tN_{2}, h_{1}, h_{2}) ≥ −ν_{1}(tN_{1} + h_{1}) − ν_{2}(tN_{2} + h_{2}) + \\ + ψ(tN_{1} + h_{1}, tN_{2} + h_{2}, h_{4} + Q_{4}(tN_{1}, tN_{2}, 0, h_{1}, h_{2}, h_{3}) \\ + Q_{4}(0, 0, x_{3}, tN_{1} + h_{1}, tN_{2} + h_{2}, h_{3} + Q_{3}(tN_{1}, tN_{2}, h_{1}, h_{2}))). \end{matrix} (4.12)

Since ν_{1}N_{1} + ν_{2}N_{2} = |ν| this inequality is guaranteed by

x_{3} + |ν|t ≥ |h_{1}||ν_{1}| + |h_{2}||ν_{2}| + |h_{3}| + |Q_{3}(tN_{1}, tN_{2}, h_{1}, h_{2})| + |ψ(z)|,

where z = (z_{1}, z_{2}, z_{4}) denotes the argument of ψ in (4.12). Note that |h_{1}||ν_{1}| + |h_{2}||ν_{2}| ≤ λ|ν|t and this term can be absorbed in the left hand side if λ is small. Moreover |h_{3}| ≤ λt^{2} and |Q_{3}(tN_{1}, tN_{2}, h_{1}, h_{2})| ≲ λt^{2}. Then, in order to prove inclusion (4.11) it will be enough to show that

ε_{0}(x_{3} + |ν|t) ≥ λt^{2} + |ψ(z)|

for some ε_{0} > 0 small but absolute. We estimate z_{1}, z_{2} and z_{4}. Clearly, |z_{1}| = |tN_{1} + h_{1}| ≲ t and |z_{2}| = |tN_{2} + h_{2}| ≲ t. Moreover,

|z_{4}| = |h_{4} + Q_{4}(tN_{1}, tN_{2}, 0, h_{1}, h_{2}, h_{3})
+ Q_{4}(0, 0, x_{3}, tN_{1} + h_{1}, tN_{2} + h_{2}, h_{3} + Q_{3}(tN_{1}, tN_{2}, h_{1}, h_{2}))|
≲ λt^{3} + x_{3}t

because Q_{4}(0, 0, x_{3}, ξ_{1}, ξ_{2}, ξ_{3}) = 1/2{(−x_{3}ξ_{1}) + α(−ξ_{2}x_{3})}.

Thus by (4.10)

|ψ(z)| ≲ z_{1}^{2} + z_{2}^{2} + |z_{4}| ≲ t^{2} + λt^{3} + x_{3}t ≃ t^{2} + x_{3}t,

because λt^{3} ≲ t^{2} (we assume t ≤ 1).

We finally have to prove the inequality

ε_{0}(x_{3} + |ν|t) ≥ t^{2} + x_{3}t,  (4.13)

which holds if t ≤ σ|ν| with σ > 0 small depending only on Ω.

28 ROBERTO MONTI AND DANIELE MORBIDELLI

5. Second piece. From now up to the end of Case 1 t_{1} = σ|ν| will be fixed. For t ≥ t_{1} define

γ(t) = (0, 0, x_{3}, 0) · (t_{1}N_{1}, t_{1}N_{2}, t − t_{1}, 0),

and note that d(γ(t), x) ≃ t_{1} + (t − t_{1})^{1/2}. Write b = (t − t_{1})^{1/2} and δ(t) = t_{1} + b. 6. We claim that there exists a positive λ < 1 such that for all t_{1} ≤ t ≤ 1

Box(γ(t), λδ(t)) ⊂ Ω.  (4.14)

Condition (4.14) is equivalent to the John property for γ in its second piece. Points in Box(γ(t), λδ(t)) have the form

γ(t) · h = (0, 0, x_{3}, 0) · (t_{1}N_{1}, t_{1}N_{2}, b^{2}, 0) · (h_{1}, h_{1}, h_{3}, h_{4})
= (t_{1}N_{1}, t_{1}N_{2}, x_{3} + b^{2}, Q_{4}(0, 0, x_{3}, t_{1}N_{1}, t_{1}N_{2}, b^{2})) · (h_{1}, h_{2}, h_{3}, h_{4})
= (t_{1}N_{1} + h_{1}, t_{1}N_{2} + h_{2}, x_{3} + b^{2} + h_{3} + Q_{3}(t_{1}N_{1}, t_{1}N_{2}, h_{1}, h_{2}),

, Q_{4}(0, 0, x_{3}, t_{1}N_{1}, t_{1}N_{2}, b^{2}) + h_{4} + Q_{4}(t_{1}N_{1}, t_{1}N_{2}, x_{3} + b^{2}, h_{1}, h_{2}, h_{3})),

with h = (h_{1}, h_{2}, h_{3}, h_{4}) and ∥h∥ ≤ λδ(t). Now, γ(t) · h ∈ Ω provided that

\begin{matrix}x_{3}+ & b^{2} + h_{3} + Q_{3}(t_{1}N_{1}, t_{1}N_{2}, h_{1}, h_{2}) ≥ −ν_{1}(t_{1}N_{1} + h_{1}) − ν_{2}(t_{1}N_{2} + h_{2}) \\ + ψ(t_{1}N_{1} + h_{1}, t_{1}N_{2} + h_{2}, Q_{4}(0, 0, x_{3}, t_{1}N_{1}, t_{1}N_{2}, b^{2}) + h_{4} \\ + Q_{4}(t_{1}N_{1}, t_{1}N_{2}, x_{3} + b^{2}, h_{1}, h_{2}, h_{3})), \end{matrix} (4.15)

which is implied by

t_{1}|ν| + x_{3} + b^{2} ≥ |ν_{1}||h_{1}| + |ν_{2}||h_{2}| + |h_{3}| + |Q_{3}(t_{1}N_{1}, t_{1}N_{2}, h_{1}, h_{2})| + |ψ(z)|, (4.16)

where z = (z_{1}, z_{2}, z_{4}) is the argument of ψ in (4.15). In order to prove (4.16) note that |ν_{1}||h_{1}| + |ν_{2}||h_{2}| ≲ λ|ν|δ(t) ≃ λ|ν|t_{1} + λ|ν|b. The term λ|ν|t_{1} can be put in the left hand side. Moreover |h_{3}| ≤ λδ^{2}(t) ≲ λt_{1}^{2} + λb^{2} and |Q_{3}(t_{1}N_{1}, t_{1}N_{2}, h_{1}, h_{2})| ≤ λt_{1}δ(t) ≲ λt_{1}^{2} + λb^{2}. The term λb^{2} can also be absorbed in the left hand side.

Claim (4.14) will be proved if we show that for t_{1} ≤ t ≤ 1

ε_{0}(t_{1}|ν| + x_{3} + b^{2}) ≥ λ|ν|b + λt_{1}^{2} + |ψ(z)|.  (4.17)

We estimate z_{1}, z_{2} and z_{4}. First of all |z_{1}| = |t_{1}N_{1} + h_{1}| ≲ t_{1} + λδ(t) ≲ t_{1} + λb and the same estimate holds for |z_{2}|. Moreover, writing δ instead of δ(t) and recalling (4.3)

|z_{4}| = |Q_{4}(0, 0, x_{3}, t_{1}N_{1}, t_{1}N_{2}, b^{2}) + h_{4} + Q_{4}(t_{1}N_{1}, t_{1}N_{2}, x_{3} + b^{2}, h_{1}, h_{2}, h_{3})|
≲ x_{3}t_{1} + λδ^{3} + λt_{1}^{2}δ + λt_{1}δ^{2} + λ(x_{3} + b^{2})δ
≃ x_{3}t_{1} + λ(t_{1} + b)^{3} + λt_{1}^{2}(t_{1} + b) + λt_{1}(t_{1} + b)^{2} + λ(x_{3} + b^{2})(t_{1} + b)
≃ x_{3}t_{1} + λt_{1}^{3} + λb^{3} + λx_{3}t_{1} + λx_{3}b ≃ x_{3}t_{1} + λt_{1}^{3} + λb^{3} + λx_{3}b.

Then by (4.10)

|ψ(z)| ≲ |z_{1}|^{2} + |z_{2}|^{2} + |z_{4}| ≤ t_{1}^{2} + λb^{2} + x_{3}t_{1} + λt_{1}^{3} + λb^{3} + λx_{3}b.

Thus (4.17) is implied by

\begin{aligned}ε_{0}(t_{1}|ν| + x_{3} + b^{2}) & ≥ λ|ν|b + t_{1}^{2} + λb^{2} + x_{3}t_{1} + λt_{1}^{3} + λb^{3} + λx_{3}b \\ & ≃ λ|ν|b + t_{1}^{2} + λb^{2} + x_{3}t_{1} + λb^{3} + λx_{3}b. \end{aligned}  (4.18)

Inequality (4.18) holds for b = 0. This has been proved in (4.13) with t = t_{1}.

Taking a smaller constant in the left hand side of (4.13) we can assert that (4.18) is guaranteed by

ε_{0}(t_{1}|ν| + x_{3} + b^{2}) ≥ λ|ν|b + λb^{2} + λb^{3} + λx_{3}b.  (4.19)

We can estimate the right hand side using |ν| ≤ 1 and b ≤ 1 getting

λ(|ν|b + b^{2} + b^{3} + x_{3}b) ≤ λ(|ν|^{2} + b^{2} + x_{3}).

Recalling now that t_{1} = σ|ν| inequality (4.19) is proved for all b ∈ (0, 1) if λ is small enough.

7. Case 2. Let x̄ ∈ ∂Ω be a characteristic point of second type. Take a point near x̄. Translate it to the origin and write locally ∂Ω in the form y_{4} = φ(y_{1}, y_{2}, y − 3). Since the domain is admissible (4.5) holds. Write

\begin{aligned}ν_{1} & = −∂_{1}φ(0), ν_{2} = −∂_{2}φ(0), ν_{3} = −∂_{3}φ(0), ν = (ν_{1}, ν_{2}), \\ N_{1} & = \frac{ν_{1}}{|ν|}, N_{2} = \frac{ν_{2}}{|ν|}, N_{3} = \frac{ν_{3}}{|ν_{3}|} = sgn(ν_{3}). \end{aligned}

If ν = 0 simply set N_{1} = N_{2} = 0. If ν_{3} = 0 set N_{3} = 0. Moreover let ψ(y) = φ(y) + ν_{1}y_{1} + ν_{2}y_{2} + ν_{3}y_{3}. By (4.5) ψ satisfies the following growth condition

|ψ(y)| ≲ ∥y∥^{3} + (|ν|^{1/2} + |ν_{3}|)(y_{1}^{2} + y_{2}^{2}).  (4.20)

We shall now construct the John curve starting from x = x_{4}e_{4}, x_{4} ≥ 0. Without loss of generality (the map z ↦ z + μe_{4}, μ ∈ ℝ, is a left translation) assume that x = 0 ∈ ∂Ω. We have to define a continuous path γ : [0, 1] → Ω such that γ(0) = 0 and dist(γ(t); ∂Ω) ≥ λd(γ(t), 0) for all t ∈ [0, 1] and for some λ > 0 depending only on Ω. We split the path in three pieces.

8. First piece. For σ > 0 let

t_{1} = \begin{cases}σ min \{|ν|^{1/2}, |ν|/|ν_{3}|\} & if ν_{3} ≠ 0, \\ σ|ν|^{1/2} & if ν_{3} = 0, \end{cases}  (4.21)

and if t ∈ [0, t_{1}] define

γ(t) = (N_{1}t, N_{2}t, 0, 0).

Note that d(γ(t), 0) = t.

9. We claim that there exist positive constants σ, λ ∈ (0, 1) such that for all t ∈ [0, t_{1}] the following John property holds

Box(γ(t), λt) ⊂ Ω.  (4.22)

30 ROBERTO MONTI AND DANIELE MORBIDELLI

Points in Box(γ(t), λt) are of the form

γ(t) · h = (N_{1}t, N_{2}t, 0, 0) · (h_{1}, h_{2}, h_{3}, h_{4})
= (N_{1}t + h_{1}, N_{2}t + h_{2}, h_{3} + Q_{3}(N_{1}t, N_{2}t, h_{1}, h_{2}),
, h_{4} + Q_{4}(N_{1}t, N_{2}t, 0, h_{1}, h_{2}, h_{3})),

with h = (h_{1}, h_{2}, h_{3}, h_{4}) and ∥h∥ ≤ λt. Now, γ(t) · h ∈ Ω if

h_{4} + Q_{4}(N_{1}t, N_{2}t, 0, h_{1}, h_{2}, h_{3}) > − ν_{1}(N_{1}t + h_{1}) − ν_{2}(N_{2}t + h_{2})
− ν_{3}(h_{3} + Q_{3}(N_{1}t, N_{2}t, h_{1}, h_{2}))
+ ψ(N_{1}t + h_{1}, N_{2}t + h_{2}, h_{3} + Q_{3}(N_{1}t, N_{2}t, h_{1}, h_{2})),

which is implied by

\begin{aligned}|ν|t ≥ |ν_{1}||h_{1}| & + |ν_{2}||h_{2}| + |ν_{3}||h_{3}| + |ν_{3}||Q_{3}(N_{1}t, N_{2}t, h_{1}, h_{2})| \\ & + |ψ(N_{1}t + h_{1}, N_{2}t + h_{2}, h_{3} + Q_{3}(N_{1}t, N_{2}t, h_{1}, h_{2}))| \\ & + |h_{4}| + |Q_{4}(N_{1}t, N_{2}t, 0, h_{1}, h_{2}, h_{3})|. \end{aligned}  (4.23)

Recall that |ν_{1}||h_{1}| + |ν_{2}||h_{2}| ≤ λ|ν|t, |h_{3}| ≤ λt^{2}, |Q_{3}(N_{1}t, N_{2}t, h_{1}, h_{2})| ≤ λt^{2}, |h_{4}| ≤ λt^{3} and |Q_{4}(N_{1}t, N_{2}t, 0, h_{1}, h_{2}, h_{3})| ≤ λt^{3}. If z = (z_{1}, z_{2}, z_{3}) is the argument of ψ in (4.23) then we get

∥z∥ = ∥(N_{1}t + h_{1}, N_{2}t + h_{2}, h_{3} + Q_{3}(N_{1}t, N_{2}t, h_{1}, h_{2}))∥ ≲ t + λt ≃ t,

and by (4.20)

|ψ(z)| ≲ ∥z∥^{3} + (|ν|^{1/2} + |ν_{3}|)(z_{1}^{2} + z_{2}^{2}) ≤ t^{3} + (|ν|^{1/2} + |ν_{3}|)t^{2}.

We finally get the following inequality which is stronger than (4.23)

ε_{0}|ν|t ≥ λ|ν|t + λ|ν_{3}|t^{2} + t^{3} + (|ν|^{1/2} + |ν_{3}|)t^{2},

where ε_{0} < 1 is an absolute constant. Dividing by t we have to show that

ε_{0}|ν| ≥ t^{2} + (|ν|^{1/2} + |ν_{3}|)t.  (4.24)

(λ|ν| has been absorbed in the left hand side). It will be enough to determine all t that solve the following two inequalities

t^{2} < ε_{0}|ν| and t(|ν|^{1/2} + |ν_{3}|) < ε_{0}|ν|.

The first one gives t ≤ ε_{0}|ν|^{1/2} and the second one is consequently solved by t|ν_{3}| ≤ ε_{0}|ν|. Claim (4.22) is proved if t_{1} is as in (4.21) for a small absolute constant σ > 0.

10. Second piece. From now on t_{1} is fixed as in (4.21). For η > 0 let

t_{2} = \begin{cases}η max\{|ν|, |ν_{3}|^{2}\} & if ν_{3} ≠ 0, \\ 0 & if ν_{3} = 0. \end{cases}  (4.25)

and if t ∈ [t_{1}, t_{1} + t_{2}] define

γ(t) = (t_{1}N_{1}, t_{1}N_{2}, (t − t_{1})N_{3}, 0).

Notice that

δ(t) := t_{1} + (t − t_{1})^{1/2} ≃ d(γ(t), 0).

REGULAR DOMAINS IN HOMOGENEOUS GROUPS 31

In the sequel we shall sometimes write δ instead of δ(t). Moreover, let b = (t − t_{1})^{1/2}.

11. We claim that there exist positive constants η, λ < 1 such that for all t ∈ [t_{1}, t_{1} + t_{2}] the following John property for γ holds

Box(γ(t), λδ(t)) ⊂ Ω.  (4.26)

Points in Box(γ(t), λδ) are of the form

γ(t) · h = (t_{1}N_{1}, t_{1}N_{2}, (t − t_{1})N_{3}, 0) · (h_{1}, h_{2}, h_{3}, h_{4})
= (t_{1}N_{1} + h_{1}, t_{1}N_{2} + h_{2}, (t − t_{1})N_{3} + h_{3} + Q_{3}(t_{1}N_{1}, t_{1}N_{2}, h_{1}, h_{2}),
, h_{4} + Q_{4}(t_{1}N_{1}, t_{1}N_{2}, (t − t_{1})N_{3}, h_{1}, h_{2}, h_{3}),

with h = (h_{1}, h_{2}, h_{3}, h_{4}) and ∥h∥ ≤ λδ. Now, γ(t) · h ∈ Ω if

h_{4} + Q_{4}(t_{1}N_{1}, t_{1}N_{2}, b^{2}N_{3},h_{1}, h_{2}, h_{3}) > −ν_{1}(t_{1}N_{1} + h_{1}) − ν_{2}(t_{1}N_{2} + h_{2})
− ν_{3}(b^{2}N_{3} + h_{3} + Q_{3}(t_{1}N_{1}, t_{1}N_{2}, h_{1}, h_{2}))
+ ψ(t_{1}N_{1} + h_{1}, t_{1}N_{2} + h_{2}, b^{2}N_{3} + h_{3} + Q_{3}(t_{1}N_{1}, t_{1}N_{2}, h_{1}, h_{2})),

which is implied by

\begin{aligned}|ν|t_{1} + |ν_{3}|b^{2} & > |h_{4}| + |Q_{4}(t_{1}N_{1}, t_{1}N_{2}, b^{2}N_{3}, h_{1}, h_{2}, h_{3})| \\ & + |ν_{1}||h_{1}| + |ν_{2}||h_{2}| + |ν_{3}||h_{3}| + |ν_{3}||Q_{3}(t_{1}N_{1}, t_{1}N_{2}, h_{1}, h_{2})| \\ & + |ψ(t_{1}N_{1} + h_{1}, t_{1}N_{2} + h_{2}, b^{2}N_{3} + h_{3} + Q_{3}(t_{1}N_{1}, t_{1}N_{2}, h_{1}, h_{2}))|. \end{aligned} (4.27)

We estimate now the right hand side: |h_{4}| ≤ λδ^{3}, |Q_{4}(t_{1}N_{1}, t_{1}N_{2}, b^{2}, h_{1}, h_{2}, h_{3})| ≤ λδ^{3} + b^{2}λδ ≃ λδ^{3}, |ν_{1}||h_{1}| + |ν_{2}||h_{2}| ≤ λ|ν|δ, |h_{3}| ≤ λδ^{2} and |Q_{3}(t_{1}N_{1}, t_{1}N_{2}, h_{1}, h_{2})| ≤ λt_{1}δ ≤ λδ^{2}.

Let z = (z_{1}, z_{2}, z_{3}) be the argument of ψ in (4.27). Then |z_{1}| = |t_{1}N_{1} +h_{1}| ≲ t_{1} +λδ and analogously |z_{2}| ≲ t_{1} + λδ. Moreover, as b ≤ δ and t_{1} ≤ δ

\begin{aligned}∥(z_{1}, z_{2}, z_{3})∥ & = ∥(t_{1}N_{1} + h_{1}, t_{1}N_{2} + h_{2}, b^{2}N_{3} + h_{3} + Q_{3}(t_{1}N_{1}, t_{1}N_{2}, h_{1}, h_{2})∥ \\ & ≃ t_{1} + λδ + b + λδ + (t_{1}λδ)^{1/2} ≲ δ. \end{aligned}

By (4.20) this furnishes

\begin{aligned}|ψ(z)| & ≲ δ^{3} + (|ν|^{1/2} + |ν_{3}|)(t_{1} + λδ)^{2} \\ & ≲ δ^{3} + |ν|^{1/2}t_{1}^{2} + λ|ν|^{1/2}δ^{2} + |ν_{3}|t_{1}^{2} + λ|ν_{3}|δ^{2}, \end{aligned}

and (4.27) is guaranteed by

|ν|t_{1} + |ν_{3}|b^{2} ≥ λδ^{3} + λ|ν|δ + λ|ν_{3}|δ^{2} + δ^{3} + |ν|^{1/2}t_{1}^{2} + λ|ν|^{1/2}δ^{2} + |ν_{3}|t_{1}^{2} + λ|ν_{3}|δ^{2}. (4.28)

Replacing δ = t_{1} + b we get

\begin{aligned}ε_{0}(|ν|t_{1} + |ν_{3}|b^{2}) ≥ (t_{1} + b)^{3} & + λ|ν|(t_{1} + b) + λ|ν_{3}|(t_{1} + b)^{2} + |ν|^{1/2}t_{1}^{2} \\ & + λ|ν|^{1/2}(t_{1} + b)^{2} + |ν_{3}|t_{1}^{2}, \end{aligned}

where ε_{0} is a small but absolute constant. Possibly changing ε_{0} it will be enough to show that

\begin{aligned}ε_{0}(|ν|t_{1} + |ν_{3}|b^{2}) ≥ t_{1}^{3} + b^{3} & + λ|ν|t_{1} + λ|ν|b + λ|ν_{3}|t_{1}^{2} + λ|ν_{3}|b^{2} \\ & + |ν|^{1/2}t_{1}^{2} + λ|ν|^{1/2}t_{1}^{2} + λ|ν|^{1/2}b^{2} + |ν_{3}|t_{1}^{2}. \end{aligned}

32 ROBERTO MONTI AND DANIELE MORBIDELLI

Now, λ|ν|t_{1} and λ|ν_{3}|b^{2} can be absorbed in the left hand side, and λ|ν_{3}|t_{1}^{2} + |ν_{3}|t_{1}^{2} ≃ |ν_{3}|t_{1}^{2}. Then

ε_{0}(|ν|t_{1} + |ν_{3}|b^{2}) ≥ t_{1}^{3} + b^{3} + λ|ν|b + |ν_{3}|t_{1}^{2} + |ν|^{1/2}t_{1}^{2} + λ|ν|^{1/2}b^{2}.  (4.29)

Inequality (4.29) holds with b = 0 by (4.24) with t = t_{1}. It will be enough to show that

ε_{0}(|ν|t_{1} + |ν_{3}|b^{2}) ≥ b^{3} + λ|ν|b + λ|ν|^{1/2}b^{2}.  (4.30)

12. In order to prove (4.30) the following two cases must be distinguished: (2A) |ν_{3}| ≤ |ν|^{1/2}; (2B) |ν_{3}| > |ν|^{1/2}.

13. Case 2A. In this case t_{1} = σ|ν|^{1/2} and (4.30) becomes (with a smaller ε_{0})

ε_{0}(|ν|^{3/2} + |ν_{3}|b^{2}) ≥ b^{3} + λ|ν|b + λ|ν|^{1/2}b^{2}.

By the trivial estimate |ν_{3}|b^{2} ≥ 0 and letting λ = 1 in the right hand side we get the stronger inequality ε_{0}|ν|^{3/2} ≥ b^{3} + |ν|b + |ν|^{1/2}b^{2}. Setting b = |ν|^{1/2}a (this can be done because in Case 2A it should be ν ≠ 0) we find ε_{0} ≥ a^{3} + a^{2} + a which holds for all 0 ≤ a < a_{0}. Then (4.30) holds for all 0 ≤ b ≤ a_{0}|ν|^{1/2} and consequently our claim (4.26) holds for all t ≤ t_{1} + a_{0}^{2}|ν|.

14. Case 2B. Here t_{1} = σ|ν|/|ν_{3}|. The term λ|ν|^{1/2}b^{2} in the right hand side of (4.30) is less than ε_{0}|ν_{3}|b^{2} and can be absorbed in the left hand side. Then we get the inequality (with a possibly smaller ε_{0})

ε_{0}(\frac{|ν|^{2}}{|ν_{3}|} + |ν_{3}|b^{2}) ≥ b^{3} + λ|ν|b

that is

ε_{0}(|ν|^{2} + |ν_{3}|^{2}b^{2}) ≥ b^{3}|ν_{3}| + λ|ν||ν_{3}|b.

Now, λ|ν||ν_{3}|b ≤ \frac{λ}{2}|ν|^{2} + \frac{λ}{2}|ν_{3}|^{2}b^{2} and both these terms can be absorbed in the left hand side if λ is suitable. Thus it suffices to solve

ε_{0}(|ν|^{2} + |ν_{3}|^{2}b^{2}) ≥ b^{3}|ν_{3}|.

Setting |ν| = 0 we find b ≤ ε_{0}|ν_{3}| which gives the correct choice t_{2} = ε_{0}^{2}|ν_{3}|^{2}, as declared in (4.25). Claim (4.26) is proved in Case 2B too.

15. Third piece. From now on t_{2} is fixed as in (4.25). If t ≥ t_{1} + t_{2} define

γ(t) = (t_{1}N_{1}, t_{1}N_{2}, t_{2}N_{3}, t − (t_{1} + t_{2})),

and notice that

δ(t) := t_{1} + t_{2}^{1/2} + (t − (t_{1} + t_{2}))^{1/3} ≃ d(γ(t), 0).

As before we shall sometimes write δ instead of δ(t). Moreover, let a = (t−(t_{1}+t_{2}))^{1/3}.

16. We claim that there exists λ < 1 such that the following John property for γ holds for all t_{1} + t_{2} ≤ t ≤ 1

Box(γ(t), λδ(t)) ⊂ Ω.  (4.31)

Points in Box(γ(t), λδ(t)) are of the form

γ(t) · h = (t_{1}N_{1}, t_{1}N_{2}, t_{2}N_{3}, a^{3}) · (h_{1}, h_{2}, h_{3}, h_{4})
= (t_{1}N_{1} + h_{1}, t_{1}N_{2} + h_{2}, t_{2}N_{3} + h_{3} + Q_{3}(t_{1}N_{1}, t_{1}N_{2}, h_{1}, h_{2}),
a^{3} + h_{4} + Q_{4}(t_{1}N_{1}, t_{1}N_{2}, t_{2}N_{3}, h_{1}, h_{2}, h_{3})),

where h = (h_{1}, h_{2}, h_{3}, h_{4}) and ∥h∥ ≤ λδ. Now, γ(t) · h ∈ Ω if

a^{3} + h_{4} + Q_{4}(t_{1}N_{1}, t_{1}N_{2}, t_{2}N_{3}, h_{1}, h_{2}, h_{3}) ≥ −ν_{1}(t_{1}N_{1} + h_{1}) − ν_{2}(t_{1}N_{2} + h_{2})
− ν_{3}(t_{2}N_{3} + h_{3} + Q_{3}(t_{1}N_{1}, t_{1}N_{2}, h_{1}, h_{2}))
+ |ψ(t_{1}N_{1} + h_{1}, t_{1}N_{2} + h_{2}, t_{2}N_{3} + h_{3} + Q_{3}(t_{1}N_{1}, t_{1}N_{2}, h_{1}, h_{2}))|.

As usual, we find the stronger inequality

|ν|t_{1} + |ν_{3}|t_{2} + a^{3} ≥ |h_{4}| + |Q_{4}(t_{1}N_{1}, t_{1}N_{2}, t_{2}N_{3}, h_{1}, h_{2}, h_{3})|
+ |ν_{1}||h_{1}| + |ν_{2}||h_{2}| + |ν_{3}||h_{3}| + |ν_{3}||Q_{3}(t_{1}N_{1}, t_{1}N_{2}, h_{1}, h_{2}))|

+ |ψ(t_{1}N_{1} + h_{1}, t_{1}N_{2} + h_{2}, t_{2}N_{3} + h_{3} + Q_{3}(t_{1}N_{1}, t_{1}N_{2}, h_{1}, h_{2}))|.

In the right hand side we can estimate |h_{4}|, |Q_{4}| ≲ λδ^{3}, |ν_{1}||h_{1}| + |ν_{2}||h_{2}| ≤ λ|ν|δ, |h_{3}| ≤ λδ^{2} and |Q_{3}(t_{1}N_{1}, t_{1}N_{2}, h_{1}, h_{2})| ≲ λt_{1}δ ≲ λδ^{2}.

Let z = (z_{1}, z_{2}, z_{3}) be the argument of ψ. Then |z_{1}| = |t_{1}N_{1} + h_{1}| ≤ t_{1} + λδ and |z_{2}| ≤ t_{1} + λδ. Moreover

\begin{aligned}∥z∥ & = ∥(t_{1}N_{1} + h_{1}, t_{2}N_{2} + h_{2}, t_{2}N_{3} + h_{3} + Q_{3}(t_{1}N_{1}, t_{1}N_{2}, h_{1}, h_{2}))∥ \\ & ≲ t_{1} + λδ + t_{2}^{1/2} + λδ + (t_{1}λδ)^{1/2} ≃ t_{1} + λδ + t_{2}^{1/2}. \end{aligned}

By (4.20)

|ψ(z)| ≤ ∥z∥^{3} + (|ν|^{1/2} + |ν_{3}|)(z_{1}^{2} + z_{2}^{2})
≤ (t_{1} + λδ + t_{2}^{1/2})^{3} + (|ν|^{1/2} + |ν_{3}|)(t_{1} + λδ)^{2}
≃ t_{1}^{3} + t_{2}^{3/2} + λδ^{3} + t_{1}^{2}|ν|^{1/2} + t_{1}^{2}|ν_{3}| + λ|ν|^{1/2}δ^{2} + λ|ν_{3}|δ^{2}.

Ultimately, we have to show that

\begin{aligned}ε_{0}(t_{1}|ν| + t_{2}|ν_{3}| + a^{3}) ≥ λδ^{3} & + λ|ν|δ + λ|ν_{3}|δ^{2} + t_{1}^{3} + t_{2}^{3/2} \\ & + t_{1}^{2}|ν|^{1/2} + t_{1}^{2}|ν_{3}| + λ|ν|^{1/2}δ^{2}. \end{aligned}

Notice that |ν|^{1/2}δ^{2} ≤ \frac{1}{2}(|ν|δ + δ^{3}) and thus the term λ|ν|^{1/2}δ^{2} in the right hand side can be deleted. Now, writing δ = t_{1} + t_{2}^{1/2} + a we get

\begin{aligned}ε_{0}(t_{1}|ν| + t_{2}|ν_{3}| + a^{3}) ≥ λt_{1}^{3} & + λt_{2}^{3/2} + λa^{3} + λ|ν|t_{1} + λ|ν|t_{2}^{1/2} + λ|ν|a + λ|ν_{3}|t_{1}^{2} \\ & + λ|ν_{3}|t_{2} + λ|ν_{3}|a^{2} + t_{1}^{3} + t_{2}^{3/2} + t_{1}^{2}|ν|^{1/2} + t_{1}^{2}|ν_{3}|, \end{aligned},

and letting absorbe λ|ν|t_{1}, λ|ν_{3}|t_{2} and λa^{3} by the left hand side we find the stronger inequality

\begin{aligned}ε_{0}(t_{1}|ν| + t_{2}|ν_{3}| + a^{3}) ≥ t_{1}^{3} & + t_{2}^{3/2} + λ|ν|t_{2}^{1/2} + λ|ν|a \\ & + λ|ν_{3}|a^{2} + t_{1}^{2}|ν|^{1/2} + t_{1}^{2}|ν_{3}|. \end{aligned}

Such inequality holds if a = 0 (let b^{2} = t_{2} in (4.29)). Thus, it will be enough to prove that for a small but absolute constant ε_{0}

ε_{0}(t_{1}|ν| + t_{2}|ν_{3}| + a^{3}) ≥ λ|ν|a + λ|ν_{3}|a^{2},  (4.32)

for all a ≥ 0.

We distinguish Case 2A and Case 2B.

17. Case 2A. In this case |ν_{3}| ≤ |ν|^{1/2}, t_{1} = σ|ν|^{1/2} and t_{2} = η|ν|. Using t_{2}|ν_{3}| ≥ 0 in the left hand side of (4.32), replacing t_{1} = σ|ν|^{1/2} and using also |ν_{3}| ≤ |ν|^{1/2} in the right hand side we get the stronger inequality

ε_{0}(|ν|^{3/2} + a^{3}) ≥ λ(|ν|a + |ν|^{1/2}a^{2})

which holds for all a ≥ 0 if λ is small enough (3 and 3/2 are Hölder conjugate exponents).

18. Case 2B. Here |ν_{3}| > |ν|^{1/2}, t_{1} = σ|ν|/|ν_{3}| and t_{2} = η|ν_{3}|^{2}. In the left hand side of (4.32) we use t_{1}|ν| ≥ 0 and put t_{2} = η|ν_{3}|^{2}. In the right hand side we estimate |ν| ≤ |ν_{3}|^{2}. Thus we find the stronger inequality

ε_{0}(|ν_{3}|^{3} + a^{3}) ≥ λ|ν_{3}|^{2}a + λ|ν_{3}|a^{2},

which holds for all a ≥ 0 if λ is small enough.

19. Case 3. This is the non characteristic case and can be analyzed as in Theorem 3.2.

5. Examples

Example 5.1. In the setting of the Heisenberg group we give an example of domain of class C^{1,α} with α ∈ (0, 1) which is not a John domain. To this aim we construct a counterexample to the Sobolev-Poincaré inequality (1.1).

We write (x, y, t) = (z, t) ∈ ℝ^{2}×ℝ = ℍ^{1} and consider the vector fields X_{1} = ∂_{x}+y∂_{t} and X_{2} = ∂_{y} − x∂_{t}, (x, y, t) = (z, t) ∈ ℝ^{3}. We shall write X = (X_{1}, X_{2}). It is well known that these vector fields are associated with a homogeneous group with dimension Q = 4.

Let Ω = {(z, t) ∈ ℍ^{1} : |z|^{α+1} < t < 1} where α ≥ 0 is a real parameter. The domain ∂Ω is not smooth when |z| = t = 1 but this does not matter as we are interested in the characteristic point 0 ∈ ∂Ω. If α ≥ 1 then Ω is of class C^{2} in a neighborhood of 0 ∈ ∂Ω and it belongs to the regular class studied in Section 3.

We consider the case 0 ≤ α < 1. We let u(z, t) = t^{−γ} and look for an exponent γ > 0 such that

∫_{Ω} |Xu|^{p} dzdt < +∞ but ∫_{Ω} |u|^{q} dzdt = +∞,  (5.1)

where 1 ≤ p < Q = 4, for some q ≥ 1 which should be less than p^{∗} = 4p/(4 − p). We have

\begin{aligned}∫_{Ω} |Xu|^{p} dzdt & ≃ ∫_{0}^{1} t^{−p(γ+1)} ∫_{|z|<t^{1/(α+1)}} |z|^{p} dz dt \\ & ≃ ∫_{0}^{1} t^{−p(γ+1)+(2+p)/(α+1)} dt < +∞ ⇔ γ < \frac{3 + α − pα}{p(α + 1)}. \end{aligned}

On the other hand

∫_{Ω} |u|^{q} dzdt ≃ ∫_{0}^{1} t^{−qγ+4/(α+1)} dt = +∞ ⇔ γ ≥ \frac{3 + α}{q(α + 1)}.

An exponent γ ensuring (5.1) can be found if the following condition holds

\frac{3 + α}{q(α + 1)} < \frac{3 + α − pα}{p(α + 1)}  ⇔  q > \frac{p(3 + α)}{3 + α − pα}.  (5.2)

If α = 1 (5.2) becomes q > p^{∗} which is exactly what one should expect. If α < 1 we can find q < p^{∗} such that (5.2) holds (the function α ↦ (3 + α)p/(3 + α − pα) is increasing near α = 1). Thus if α < 1 the Sobolev–Poincaré inequality does not hold and Ω is not a John domain.

Example 5.2. Using Theorem 4.4 we construct an example of John domain Ω ⊂ ℝ^{4} with respect to the metric structure of the group of step 3 considered in section 4. Let g ∈ C^{2}(0, 1) ∩ C([0, 1]) be a function such that

g(t) = \begin{cases}1 − t^{1/4} & if 0 ≤ t ≤ 1/4, \\ (1 − t)^{1/4} & if 3/4 ≤ t ≤ 1. \end{cases}

Such a function can be chosen with the additional property g′(t) < 0 for all t ∈ (0, 1). Let

N(x_{1}, x_{2}, x_{3}) = (x_{1}^{2} + x_{2}^{2})^{6} + x_{3}^{6},  (5.3)

and define the open set

Ω = {x ∈ ℝ^{4} : |x_{4}| < g(N(x_{1}, x_{2}, x_{3}))}.  (5.4)

Notice that if N(x_{1}, x_{2}, x_{3}) ≥ 3/4 then ∂Ω has local equation (x_{1}^{2} +x_{2}^{2})^{6} +x_{3}^{6} +x_{4}^{4} = 1. If N(x_{1}, x_{2}, x_{3}) ≤ 1/4 then ∂Ω has equation |x_{4}| + [(x_{1}^{2} + x_{2}^{2})^{6} + x_{3}^{6}]^{1/4} = 1.

We first show that the points (0, 0, 0, ±1) ∈ ∂Ω are the unique characteristic points of second type of ∂Ω. Indeed, let Φ(x_{1}, x_{2}, x_{3}, x_{4}) = g(N(x)) − x_{4} and compute

\begin{aligned}X_{1}Φ(x) & = g′(N(x))X_{1}N(x) − q_{1}(x) \\ X_{2}Φ(x) & = g′(N(x))X_{2}N(x) − q_{2}(x) \\ X_{3}Φ(x) & = g′(N(x))X_{3}N(x) − q_{3}(x), \end{aligned}

where q_{1}, q_{2} and q_{3} are defined in (4.2). Note that x_{1}q_{1}(x) + x_{2}q_{2}(x) = −x_{3}q_{3}(x). Moreover

\begin{matrix}X_{1}N(x) = 12x_{1}(x_{1}^{2} + x_{2}^{2})^{5} − 3x_{2}x_{3}^{5} \\ X_{2}N(x) = 12x_{2}(x_{1}^{2} + x_{2}^{2})^{5} + 3x_{1}x_{3}^{5}, \end{matrix}

and thus x_{1}X_{1}N(x) + x_{2}X_{2}N(x) = 12(x_{1}^{2} + x_{2}^{2})^{6}. Then

x_{1}X_{1}Φ + x_{2}X_{2}Φ = g′(N(x))12(x_{1}^{2} + x_{2}^{2})^{6} + x_{3}q_{3} = 0,

if x is characteristic. Moreover, if x is of second type we have

x_{3}X_{3}Φ = g′(N(x))6x_{3}^{6} − x_{3}q_{3}(x) = 0.

Summing up the last two equations we finally get

g′(N(x))(12(x_{1}^{2} + x_{2}^{2})^{6} + 6x_{3}^{6}) = 0,

which implies x_{1} = x_{2} = x_{3} = 0, as g′(N(x)) ≠ 0.

We now check that hypotheses of Lemma 4.3 are satisfied. We start with (i). Consider the graph x_{4} = φ(x_{1}, x_{2}, x_{3}) = N(x)^{1/4}, x ∈ D_{0} = {(x_{1}, x_{2}, x_{3}) : N(x) < 1}. We need to show that the derivatives X_{i}X_{j}φ, i, j = 1, 2, are Lipschitz continuous on D_{0} (with respect to the distance ϱ in the Heisenberg group). This is equivalent to show that the derivatives X_{i}X_{j}X_{k}φ (i, j, k = 1, 2) are bounded. This is trivially true because these functions are smooth away from the origin and homogeneous of degree zero with respect to the dilations (x_{1}, x_{2}, x_{3}) ↦ (λx_{1}, λx_{2}, λ^{2}x_{3}) (we are taking derivatives of order 3 of the function φ = N^{1/4} which is homogeneous of degree 3).

In order to check hypothesis (ii) of Lemma 4.3 we have to prove that, letting Φ(x) = N(x)^{1/4} − x_{4}, there exists a constant k > 0 such that

|X_{1}^{2}Φ| + |X_{2}^{2}Φ| + |(X_{1}X_{2} + X_{2}X_{1})Φ| ≤ k(|X_{1}Φ|^{1/2} + |X_{2}Φ|^{1/2} + |X_{3}Φ|)

for all x ∈ ∂Ω such that 0 < N(x) ≤ 1/4. We note that away from the origin the function Φ(x) = N(x)^{1/4} − x_{4} is smooth and moreover it is homogeneous of degree 3 with respect to the dilations (x_{1}, x_{2}, x_{3}, x_{4}) → (λx_{1}, λx_{2}, λ^{2}x_{3}, λ^{3}x_{4}). Then the derivatives X_{1}Φ and X_{2}Φ are homogeneous of degree 2 and their square roots |X_{1}Φ|^{1/2} and |X_{2}Φ|^{1/2} are homogeneous of degree 1. Analogously, X_{1}^{2}Φ, X_{2}^{2}Φ, X_{3}Φ and (X_{1}X_{2} + X_{2}X_{1})Φ, being derivatives of degree 2, are homogeneous of degree 1. Then the function H = H(x_{1}, x_{2}, x_{3}) defined by

H = \frac{|X_{1}^{2}Φ| + |X_{2}^{2}Φ| + |(X_{1}X_{2} + X_{2}X_{1})Φ|}{|X_{1}Φ|^{1/2} + |X_{2}Φ|^{1/2} + |X_{3}Φ|}

is homogeneous of degree 0. We showed above that |X_{1}Φ(x)|^{1/2} + |X_{2}Φ(x)|^{1/2} + |X_{3}Φ(x)| > 0 for all N(x) > 0, and thus by 0−homogeneity

sup_{0<N(x)≤1/4} H(x) = max_{N(x)=1/4} H(x) = k < +∞.

Thus Lemma 4.3 can be applied and the set Ω is admissible.

REGULAR DOMAINS IN HOMOGENEOUS GROUPS 37

In the following examples we show that in groups of step 3 there are domains of class C^{∞} which are not John domains and for which the Sobolev-Poincaré inequality (1.1) does not hold.

We begin with a well known Lemma. Here X = (X_{1}, ..., X_{m}) is a system of generators of a homogeneous group on ℝ^{n} with homogeneous dimension Q. Let d be the left invariant metric induced by these vector fields.

Lemma 5.3. If Ω ⊂ ℝ^{n} is a bounded open set such that the Sobolev−Poincaré inequality (1.1) holds then there exist r_{0} > 0 and C > 0 such that

|Ω ∩ B(x, r)| ≥ Cr^{Q}

for all x ∈ \overline{Ω} and 0 ≤ r ≤ r_{0}.

Proof. If p = 1 (1.1) reads

( ∫_{Ω} |u − u_{Ω}|^{Q/(Q−1)} dx)^{(Q−1)/Q} ≤ C ∫_{Ω} |Xu| dx.  (5.5)

Fix x̄ ∈ \overline{Ω} and 0 < r < R. The function

u(x) = \begin{cases}1 & if d(x, x̄) < r, \\ 1 − \frac{d(x,x̄)−r}{R−r} & if r ≤ d(x, x̄) ≤ R, \\ 0 & if d(x, x̄) > R \end{cases}

can be used in (5.5). Moreover, taking R small enough to ensure |Ω \ B(x̄, R)| ≥ \frac{1}{2}|Ω| we can let u_{Ω} = 0 to find (for a possibly different constant)

( ∫_{Ω} |u|^{Q/(Q−1)} dx)^{(Q−1)/Q} ≤ C ∫_{Ω} |Xu| dx.

Define g(r) = |B(x̄, r) ∩ Ω|. First of all

( ∫_{Ω} |u|^{Q/(Q−1)} dx)^{(Q−1)/Q} ≥ g(r)^{(Q−1)/Q},

and on the other hand, since |Xd(x, x̄)| = 1 for almost every x

∫_{Ω} |Xu| dx = \frac{1}{R − r}|(B(x̄, R) \ B(x̄, r)) ∩ Ω| = \frac{g(R) − g(r)}{R − r}.

Thus we find

g(r)^{(Q−1)/Q} ≤ C \frac{g(R) − g(r)}{R − r},  0 < r < R ≤ r_{0}.  (5.6)

The function g is differentiable almost everywhere, being monotonic, and letting R → r in (5.6) we get g(r)^{(Q−1)/Q} ≤ Cg′(r) at every point of differentiability, or equivalently

\frac{d}{dr}g(r)^{1/Q} ≥ \frac{1}{C} .

Thus

g(r)^{1/Q} ≥ ∫_{0}^{r} \frac{d}{ds}g(s)^{1/Q} ds ≥ \frac{1}{C} r,

and the claim is proved.

Example 5.4. Consider now ℝ^{4} with the homogeneous group structure introduced in section 4 (we choose α = 0). The homogeneous dimension of the group is Q = 7. Let Ω ⊂ ℝ^{4} be a bounded open set of class C^{∞} such that for some open neighborhood U of the origin Ω ∩ U = {x ∈ ℝ^{4} : x_{4} > 0} ∩ U. We show that the Sobolev−Poincaré inequality (1.1) fails in Ω.

We localize our analysis in a neighborhood of 0. Take a point x̄ = (0, −b, 0, 0) ∈ ∂Ω, b ≠ 0, and translate it to the origin. The translated boundary is

\begin{aligned}\{x̄^{−1} · (y_{1}, y_{2}, y_{3}, 0)\} & = \{(y_{1}, y_{2} + b, y_{3} + Q_{3}(0, b, y_{1}, y_{2}), Q_{4}(0, b, 0, y_{1}, y_{2}, y_{3}))\} \\ & = \{(y_{1}, y_{2} + b, y_{3} − \frac{1}{2}by_{1}, \frac{b}{12}y_{1}^{2})\} = \{y_{4} = \frac{b}{12}y_{1}^{2}\}. \end{aligned} (5.7)

We write Ω_{0} = x̄^{−1} · Ω and assume that b = 12 (any b > 0 gives the same result). It will be enough to show that Ω_{0} does not support the Sobolev−Poincaré inequality. We can use the boxes (4.4) instead of balls to estimate for 0 < r < 1

\begin{aligned}|Box(0, r) ∩ Ω_{0}| & = ∫_{∥y∥<r, y_{4}>y_{1}^{2}} dy = ∫_{|y_{1}|<r} dy_{1} ∫_{|y_{2}|<r} dy_{2} ∫_{|y_{3}|<r^{2}} dy_{3} ∫_{|y_{4}|<r^{3}, y_{1}^{2}<y_{4}} dy_{4} \\ & = 4r^{3} ∫_{|y_{1}|<r, |y_{4}|<r^{3}, y_{1}^{2}<y_{4}} dy_{1}dy_{4}. \end{aligned}

On the integration domain we have |y_{1}| ≤ |y_{4}|^{1/2} ≤ r^{3/2}. Therefore

|Box(0, r) ∩ Ω_{0}| ≤ 4r^{3} ∫_{−r^{3/2}}^{r^{3/2}} dy_{1} ∫_{y_{1}^{2}}^{r^{3}} dy_{4} = 8r^{3} ∫_{0}^{r^{3/2}}(r^{3} − y_{1}^{2})dy_{1} = \frac{16}{3}r^{15/2}.

If the Sobolev–Poincaré inequality would hold the estimate |Box(0, r) ∩ Ω_{0}| ≥ Cr^{7} should be true, but this is not possible.

We note that 0 ∈ ∂Ω is a characteristic point of second type which is “flat”, but near 0 there are characteristic points of the same type x̄ = (0, −b, 0, 0), b ≠ 0, which are not “flat”: by (5.7) the parametric equation for the translated boundary x̄^{−1} · ∂Ω is y_{4} = \frac{b}{12}y_{1}^{2} and the flatness condition |y_{4}| ≤ C∥(y_{1}, y_{2}, y_{3})∥^{3} does not hold. This makes Ω a bad domain.

The following example is a refined version of the previous.

Example 5.5. In this example we disprove Conjecture 1 stated in [CG, p. 429]. We consider again the group of step 3 studied in section 4 (with α = 0). We prove that in the “gauge ball”

Ω = {x ∈ ℝ^{4} : (x_{1}^{2} + x_{2}^{2})^{6} + x_{3}^{6} + x_{4}^{4} < 1}

the Sobolev–Poincaré inequality does not hold. We actually show the following stronger fact.

Proposition 5.6. Assume that Ω ∩ U = {x_{4} > φ(x_{1}, x_{2}, x_{3})} ∩ U for some neighborhood U of the point (0, 0, 0, φ(0, 0, 0)). Assume also that for some γ > 3

|φ(x_{1}, x_{2}, x_{3}) − φ(0, 0, 0)| ≤ C∥x∥^{γ},  (5.8)

for all x near the origin. Then the Sobolev Poincaré inequality does not hold in Ω.

We note that in the particular case of the homogeneous ball Ω = {(x_{1}^{2} + x_{2}^{2})^{6} + x_{3}^{6} + x_{4}^{4} < 1} the function φ is φ(x_{1}, x_{2}, x_{3}) = −{1 − [(x_{1}^{2} + x_{2}^{2})^{6} + x_{3}^{6}]}^{1/4}. Thus (5.8) is satisfied.

Proof. As in Example 5.4 we show that the estimate of Lemma 5.3 is violated but here the choice of the center and of the radius of the balls which give the counterexample is more delicate.

Fix α ∈ ]1, \frac{γ}{3}[ (this open interval is non empty as soon as γ > 3). We claim that there exist ε ∈ (0, 1) and C > 0 such that

|Box((0, −b, 0, φ(0, −b, 0)), b^{α}) ∩ Ω| ≤ Cb^{7α+(α−1)/2} for all b ∈ (0, ε). (5.9)

Since α is strictly bigger than 1, this is not compatible as b → 0^{+} with the estimate

|Box((0, −b, 0, φ(0, −b, 0)), b^{α}) ∩ Ω| ≳ b^{7α}

which by Lemma 5.3 should be satisfied if the Sobolev–Poincaré inequality would hold.

If x̄ = (0, −b, 0, φ(0, −b, 0)), b > 0, let Ω_{b} = x̄^{−1} · Ω. It is easy to check that the surface ∂Ω_{b} := x̄^{−1} · ∂Ω contains the origin and can be locally parameterized by

x_{4} = φ(x_{1}, x_{2} − b, x_{3} + \frac{1}{2}bx_{1}) − φ(0, −b, 0) + \frac{1}{12}bx_{1}^{2} := φ_{b}(x_{1}, x_{2}, x_{3}).

If x belongs to Box(0, b^{α}), then |x_{1}|, |x_{2}| ≤ b^{α} and |x_{3}| < b^{2α}. Thus we have the estimate

|φ(x_{1}, x_{2} − b, x_{3} + \frac{1}{2}bx_{1})| ≲ |x_{1}|^{γ} + |x_{2} − b|^{γ} + |x_{3} + \frac{1}{2}bx_{1}|^{γ/2} ≲ b^{γ},

because α > 1. Moreover |φ(0, −b, 0)| ≤ b^{γ}.

This implies that if x ∈ Box(0, b^{α}) ∩ Ω_{b} then x_{4} > φ_{b}(x_{1}, x_{2}, x_{3}) and

\begin{aligned}\frac{1}{12}bx_{1}^{2} & ≤ |x_{4}| + |φ(…) − φ(…)| \\ & ≲ b^{3α} + b^{γ} ≃ b^{3α}, \end{aligned}

because 3α < γ. As a consequence |x_{1}| ≲ b^{(3α−1)/2} and

Box(0, b^{α}) ∩ Ω_{b} ⊂ {|x_{1}| < C_{0}b^{(3α−1)/2}, |x_{2}| < b^{α}, |x_{3}| < b^{2α}, |x_{4}| < b^{3α}}

for some C_{0} which stays bounded as b approaches 0. This proves (5.9).

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Roberto Monti: Mathematisches Institut, Universität Bern, Sidlerstrasse 5, 3012 Bern, Switzerland

  E-mail address: roberto.monti@math-stat.unibe.ch

Daniele Morbidelli: Dipartimento di Matematica, Università di Bologna, Piazza di Porta San Donato, 5, 40127 Bologna, Italy

  E-mail address: morbidel@dm.unibo.it