TRACE THEOREMS FOR VECTOR FIELDS

ROBERTO MONTI AND DANIELE MORBIDELLI

Abstract. In the setting of Carnot-Carathéodory spaces we prove some trace theorems for Sobolev functions. We consider the trace on a non characteristic surface for Hörmander vector fields of step r ≥ 1 and the trace on the boundary of a class of domains in the Grushin plane.

1. Introduction

In the last years the theory of the functional spaces related to vector fields has been deeply developed in several directions. Sobolev-Poincaré-type inequalities have been widely studied and applied to the analysis of solutions of second order Partial Differential Equations. An important tool in the study of boundary value problems is the estimate of the trace on the boundary ∂Ω of a Sobolev function u defined in an open set Ω. Only few results concerning this problem are known in the degenerate setting of vector fields. In this paper we give a contribution to the research in this direction.

In order to introduce our discussion let us recall the following classical result: if 1 < p < +∞ and Ω ⊂ ℝ^{n} is a bounded open set with regular boundary ∂Ω, then there exists a constant C > 0 such that for any u ∈ W^{1,p}(Ω)

∫_{∂Ω×∂Ω} \frac{|u(x) − u(y)|^{p}}{|x − y|^{n−1+ps}} dH^{n−1}(x)dH^{n−1}(y) ≤ C ∫_{Ω} |∇u(x)|^{p} dx,  (1.1)

where s = 1 − \frac{1}{p} is the fractional order of differentiability of the trace u = u_{|∂Ω}. We are interested in generalizations of (1.1) of the form

[u]_{s,p}(∂Ω) ≤ C ∑_{j=1}^{m} ( ∫_{Ω} |X_{j}u(x)|^{p}dx)^{1/p},  (1.2)

for a family of vector fields X_{j} = ∑_{k=1}^{n} a_{jk}(x)∂_{k}, j = 1, …, m, x ∈ ℝ^{n}. The left hand side of (1.2) will be some kind of fractional semi-norm, whose form will be discussed later.

Let us begin with some considerations about the history of the problem. In the paper [F] Franchi studies the trace problem for anisotropic Sobolev spaces related to diagonal vector fields X_{j} = λ_{j}(x)∂_{j}, j = 1, ..., n, and proves optimal trace estimates on corners of cubes using a semi-norm constructed by a sum of one dimensional fractional derivatives with suitable weights. The proof relies on representation formulas modeled on the geometry of the vector fields and does not seem to work in non diagonal situations.

The authors were supported by GNAFA, Italy and by MURST, Italy.

Berhanu and Pesenson [BP] prove a trace and lifting theorem for a family of Hörmander vector fields of step two. Actually, their result holds only for the trace on a non characteristic surface with one transversal vector field and the “projections” of the other ones satisfying the Hörmander condition of step two relatively to the surface. The approach of this paper is not completely satisfactory since the definition of the semi-norm, which involves increments along the integral curves of the projected vector fields, does not apply, for instance, to simple situations when the projected vector fields vanish on the surface.

Bahouri, Chemin and Xu [BCX] using the Weyl Hörmander calculus prove a lifting theorem and a trace theorem on non characteristic surfaces for Sobolev spaces associated with a system of vector fields of step 2. The case of isolated characteristic points in the framework of the Heisenberg group is also studied.

In the recent remarkable paper [DGN2], which continues the project started in [DGN1], Danielli, Garofalo and Nhieu prove the following trace theorem for Hörmander vector fields: if Ω is a (ε, δ)−domain and μ is a Borel measure supported in ∂Ω such that μ(B(x, r)) ≤ C|B(x, r)|/r for any Carnot-Carathéodory ball centered at x ∈ ∂Ω with radius 0 < r < r_{0}, then for any p > 1 the space W_{X}^{1,p}(Ω) is continuously embedded in B^{1− \frac{1}{p} ,p}(∂Ω, dμ), where the last Besov space is defined by the semi–norm (1.4) letting s = 1 − 1/p. Conversely, if Ω is a bounded open set, |∂Ω| = 0 and μ is a Borel measure supported in ∂Ω such that μ(B(x, r)) ≥ C|B(x, r)|/r for all x ∈ ∂Ω and 0 < r < r_{0}, then, given a function u ∈ B^{1−1/p,p}(∂Ω), there exists a function Eu ∈ W_{X}^{1,p}(Ω) which extends u and such that the operator E is continuous between the expected spaces. The trace result is proved by extending a function u ∈ W_{X}^{1,p}(Ω) to a Sobolev function on the the whole space and then by a restriction technique. The fact that Ω is an extension domain is guaranteed by the subelliptic version of Jones’ extension theorem for (ε, δ) domains [GN]. The notion of (ε, δ) or uniform domain in ℝ^{n} has been introduced by Martio and Sarvas [MS] and Jones [J]. Several properties of (ε, δ) domains are studied by Väisälä [V].

Finally, although not strictly related to our work, we would like to mention the paper by Hajłasz and Martio [HM] where the trace problem on general subsets of ℝ^{n} is treated in the Euclidean setting from a “metric” point of view.

Before stating our results we introduce the basic definitions. Given a family X = (X_{1}, ..., X_{m}) of vector fields with X_{j}(x) = ∑_{i=1}^{n} a_{ij}(x)∂_{i} (j = 1, ..., m) and a_{ij} ∈ Lip(ℝ^{n}) (j = 1, ..., m, i = 1, ..., n), we call subunit a Lipschitz continuous curve γ : [0, T] ⟶ ℝ^{n} such that

γ̇(t) = ∑_{j=1}^{m} h_{j}(t)X_{j}(γ(t)), and ∑_{j=1}^{m} h_{j}^{2}(t) ≤ 1 for a.e. t ∈ [0, T],

with h_{1}, ..., h_{m} measurable coefficients. Define the Carnot-Carathéodory (briefly C-C) distance between the points x, y ∈ ℝ^{n}

d(x, y) = inf{T ≥ 0 : there exists a subunit path γ : [0, T] → ℝ^{n}
such that γ(0) = x and γ(T) = y}.

We define B(x, r) = {y ∈ ℝ^{n} : d(x, y) < r}, for x ∈ ℝ^{n} and r > 0. The function d is finite and is a distance in the following two cases, that are the object of our study:

(1) The family X_{1}, …, X_{m} ∈ C^{∞}(ℝ^{n}, ℝ^{n}) satisfies the Hörmander condition

rank L(X_{1}, ..., X_{m})(x) = n for every x ∈ ℝ^{n},

where L(X_{1}, ..., X_{m}) is the Lie algebra generated by the vector fields.

(2) The vector fields are of the form

X_{1} = ∂_{x}, and X_{2} = |x|^{α}∂_{y} in ℝ^{2},  (1.3)

with α > 0. Let Ω ⊂ ℝ^{n} be an open set with boundary ∂Ω of class C^{1}. Denote ν(x) the

Euclidean unit normal to ∂Ω at x ∈ ∂Ω and define

|Xν(x)| = ( ∑_{j=1}^{m}⟨X_{j}(x), ν(x)⟩^{2})^{1/2}.

A point x ∈ ∂Ω is said to be characteristic if |Xν(x)| = 0. The natural surface measure that takes into account characteristic points in the boundary is

μ := |Xν|H^{n−1} ∂Ω,

that is the (n − 1)−dimensional Hausdorff measure restricted to the boundary with the weight |Xν|. The measure μ is the variational surface measure associated with the vector fields X_{1}, ..., X_{m} and μ(∂Ω) equals the Minkowski content of ∂Ω in the metric space (ℝ^{n}, d) (see [MSC]). Define the fractional semi-norm

[u]_{s,p}(∂Ω) = ( ∫_{∂Ω×∂Ω} \frac{|u(x) − u(y)|^{p}}{d(x, y)^{ps}μ(B(x, d(x, y)))} dμ(x)dμ(y))^{1/p}.  (1.4)

The fact that the semi–norm (1.4) gives the correct left hand side in the trace inequality has been discovered in [DGN2]. A “solid” version of (1.4) has been studied in [M].

In this note we study the trace inequality (1.2) using the semi-norm (1.4). We shall first consider the case when the boundary is non characteristic with respect to a family of Hörmander vector fields. We develop a technique inspired by the original paper of Gagliardo [G] which relies upon the possibility of connecting points on the boundary ∂Ω by means of sub-unit curves lying in Ω. The construction of such paths is a byproduct of a structure theorem for the restriction of C-C balls to non characteristic surfaces, theorem that seems to be of independent interest (see Section 2). When the boundary contains characteristic points the analysis is more difficult. Nonetheless, our technique still works in some situations and we focused our attention on the simple but significant case of the Grushin plane. Our main results can be summarized in the following way. Let 1 < p < ∞ and s = 1 − \frac{1}{p}. If

(1) X_{1}, …, X_{m} are Hörmander vector fields in ℝ^{n} and Ω ⊂ ℝ^{n} is a bounded open set whose boundary is of class C^{∞} and does not contain characteristic points, or alternatively

(2) X_{1}, X_{2} are of the form (1.3) and Ω ⊂ ℝ^{2} is a C^{1} bounded open set which is α-admissible (see Definition 4.3), then there exist constants C, δ_{0} > 0 such that

∫_{∂Ω×∂Ω∩\{d(x,y)<δ_{0}\}} \frac{|u(x) − u(y)|^{p} dμ(x)dμ(y)}{d(x, y)^{ps}μ(B(x, d(x, y)))} ≤ C ∫_{Ω} |Xu(x)|^{p} dx  (1.5)

for all u ∈ C^{1}(Ω) ∩ C(\overline{Ω}).

These trace estimates are optimal. This follows from the extension Theorem 4.1 in [DGN2] whose hypotheses are verified in Corollary 2.7 and Lemma 4.4 below. In case (1) if Ω is any smooth bounded set, inequality (1.5) continues to hold provided the integration in the left hand side takes place on a fixed compact set K ⊂ ∂Ω which does not contain characteristic points.

It seems reasonable that both non characteristic domains for Hörmander vector fields and our admissible domains in the Grushin plane enjoy the (ε, δ)−property. If this were the case our trace theorems could be obtained using the results by Danielli, Garofalo and Nhieu [DGN2]. However, the study of this property probably involves difficulties comparable to those one has to face attempting a direct proof of the trace theorem.

In Section 2 we prove the mentioned structure theorem for C-C balls. Section 3 deals with Case (1). Our results are the natural generalization to vector fields of arbitrary step r ≥ 2 of the trace results obtained in [BP] and [BCX] for vector fields of step 2. Section 4 is devoted to Case (2). The condition of α-admissibility identifies a large class of domains for which the trace theorem holds in relation with the “flatness” of their boundary at characteristic points. At the end of Section 4 this condition will be shown to be necessary.

In the paper we will denote by C a generic constant which may change even in a single string of estimates. We write u ≃ v to state that there exist two positive constants c_{1} and c_{2} such that c_{1}u ≤ v ≤ c_{2}u. A vector field X = ∑_{k=1}^{n} a_{k}(x)∂_{k} will be identified with the vector function (a_{1}(x), …, a_{n}(x)).

Acknowledgments. It is now a pleasure to acknowledge with gratitude Ermanno Lanconelli for his encouragement, Nicola Garofalo for some helpful conversations concerning the paper [DGN2], and Bruno Franchi for having drawn our attention on some references.

2. Structure of balls restricted to non characteristic surfaces

Following the basic ideas contained in the classic paper [NSW] and the generalization in [M], we shall represent (in Theorem 2.5) C-C balls restricted to non characteristic surfaces by means of suitable exponential maps which are “small perturbations” of the exponential of the commutators of the vector fields (see Lemma 2.3). These maps enjoy a “factorization property” (see Lemma 2.4) which is crucial in the proof of the trace theorem.

First we recall that a non characteristic surface can be made flat by a diffeomorphism. A resulting transversal vector field can be orthogonalized and the other ones can be made lie on the surface.

Lemma 2.1. Let U ⊂ ℝ^{n} be a neighborhood of 0 ∈ ℝ^{n} and let X ∈ C^{∞}(U; ℝ^{n}) be a vector field such that ⟨X(0), e_{n}⟩ ≠ 0. Let x_{n} = g(x_{1}, …, x_{n−1}) = g(x′) be a function of class C^{∞} such that g(0) = 0 and ∇g(0) = 0. Possibly shrinking U, there exists a diffeomorphism Φ ∈ C^{∞}(U; ℝ^{n}) such that dΦ(x)X(x) = e_{n} for all x ∈ U and Φ(x′, g(x′)) = (x′, 0) for all (x′, g(x′)) ∈ U.

The proof of Lemma 2.1 can be essentially found in [FW, p. 83] where even less regularity is required.

Remark 2.2. Let X̃_{1}, ..., X̃_{m} ∈ C^{∞}(ℝ^{n}; ℝ^{n}) satisfy the Hörmander condition and induce the C-C metric d̃. Write (x, t) ∈ ℝ^{n−1} × ℝ and assume the vector fields are of the form

X̃_{j} = b_{j}(x, t)∂_{t} + ∑_{i=1}^{n−1} a_{ij}(x, t)∂_{i}, j = 1, ..., m − 1, X̃_{m} = ∂_{t}.

The new family of vector fields

X_{j} = ∑_{i=1}^{n−1} a_{ij}(x, t)∂_{i}, j = 1, ..., m − 1, X_{m} = ∂_{t}.  (2.6)

still satisfies the Hörmander condition. Moreover, if d is the corresponding C-C metric and K ⊂ ℝ^{n} is a compact set,

there exist c_{1} and c_{2} such that

c_{1}d̃≤ d ≤ c_{2}d̃ and c_{1}|X̃u| ≤ |Xu| ≤ c_{2}|X̃u|

for all u ∈ C^{1}. A proof of the equivalence between d and d̃ can be found in [FW, p.87]. Actually, it can be proved that each of the previous equivalences implies the other one (see [HK, Theorem 11.11]).

Consider m vector fields X_{1}, ..., X_{m} ∈ C^{∞}(ℝ^{n}; ℝ^{n}) of the form (2.6) and satisfying the Hörmander condition. We shall write X_{m} = T. For any multi-index I = (i_{1}, …, i_{k}), 1 ≤ i_{j} ≤ m and k ∈ ℕ, let

X_{[I]} = [X_{i_{1}}, [X_{i_{2}}, ⋯ [X_{i_{k−1}}, X_{i_{k}}] ⋯]],

where [X, Y] denotes the commutator of the vector fields X and Y. If I = (i_{1}, …, i_{k}) we set |I| = k and we say that the commutator X_{[I]} has length or degree d(X_{[I]}) = k.

For any commutator Y ≠ T and for small s ∈ ℝ we shall define a map exp_{T} (sY) : ℝ^{n−1} → ℝ^{n−1}. We proceed by induction on d(Y). If d(Y) = 1 and Y = X_{j} with j ∈ {1, ..., m − 1} define for x ∈ ℝ^{n−1}

exp_{T} (sY)(x) = \begin{cases}exp(−sT) exp(s(X_{j} + T))(x) & if s ≥ 0, \\ exp(s(X_{j} + T)) exp(−sT)(x) \\ = exp_{T} (|s|Y)^{−1}(x) & if s < 0. \end{cases}  (2.7)

The map is well defined provided x belongs to a compact set and s is small. We also set exp_{T} (sT) = exp(sT). Suppose now d(Y) = k, Y = X_{[J]} with |J| = k, and J = (j_{1}, ..., j_{k}). Set J′ = (j_{2}, ..., j_{k}) and define

exp_{T} (sY)(x) = \begin{cases}exp_{T} (s^{\frac{k−1}{k}} X_{[J′]})^{−1} exp_{T} (s^{\frac{1}{k}} X_{j_{1}})^{−1} \\ · exp_{T} (s^{\frac{k−1}{k}} X_{[J′]}) exp_{T} (s^{\frac{1}{k}} X_{j_{1}})(x) & if s ≥ 0, \\ exp_{T} (|s|Y)^{−1}(x) & if s < 0. \end{cases}  (2.8)

Some useful features of the maps exp_{T} are described in the following two lemmas. In Lemma 2.3, which is a generalization of [NSW, Lemma 2.21], we shall use the Campbell-Hausdorff formula

exp(u) exp(v) = exp (u + v − \frac{1}{2}[u, v] + S(u, v)),

where u and v are non commuting indeterminates and S is a formal series of commutators of u and v of length at least 3. We refer to the Appendix of [NSW] for a discussion and for the related references.

Lemma 2.3. For any commutator X_{[J]}, J = (j_{1}, …, j_{k}), of length k ≥ 1

exp_{T} (sX_{[J]}) = exp (sX_{[J]} + sgn(s) ∑_{|I|>k} c_{J,I}|s|^{|I|/k}X_{[I]}),  (2.9)

where the c_{J,I} are suitable constants.

The formal equality (2.9) means that, if x belongs to a compact set K and p > k is an integer, then

| exp_{T} (sX_{[J]})(x) − exp (sX_{[J]} + sgn(s) ∑_{k<|I|≤p} c_{J,I}s^{|I|/k}X_{[I]})(x)|

≤ Cs^{(p+1)/k}.

Proof. We proceed by induction. Consider first a commutator of length 1, i.e. a vector field X_{j}, j = 1, …, m. Applying the Campbell-Hausdorff formula to (2.7) we get for s > 0

exp_{T} (sX_{j}) = exp(−sT) exp(s(X_{j} + T))

= exp ( − sT + s(X_{j} + T) + \frac{1}{2}s^{2}[T, X_{j} + T] + ⋯)

= exp (sX_{j} + ∑_{|I|>1} c_{(j),I}s^{|I|}X_{[I]}).

For s < 0 note that

exp_{T} (sX_{j}) = exp_{T} (|s|X_{j})^{−1} = exp ( − |s|X_{j} − ∑_{|I|>1} c_{(j),I}|s|^{|I|}X_{[I]}).

We prove now the inductive step. Recall first that an application of the Campbell-Hausdorff formula asserts that, if u and v are non commuting indeterminates, then

exp(v)^{−1} exp(u)^{−1} exp(v) exp(u) = exp([u, v] + R),

where R = R(u, v) denotes a formal series containing commutators (of u and v) of length at least 3. Let k > 1, J = (j_{1}, ..., j_{k}), J′ = (j_{2}, ..., j_{k}) and s ≥ 0. Let also

\begin{aligned}u & = s^{1/k}X_{j_{1}} + ∑_{|I|>1} c_{(j_{1}),I}s^{|I|/k}X_{[I]} and \\ v & = s^{(k−1)/k}X_{[J′]} + ∑_{|I|>k−1} C_{J′,I}s^{|I|/k}X_{[I]}. \end{aligned}

Note that [u, v] = sX_{[J]} + R̃, where R̃ is a series containing commutators of order at least k + 1 of the original fields. Thus, by the inductive hypothesis

exp_{T} (sX_{[J]}) = exp_{T} (s^{\frac{k−1}{k}} X_{[J′]})^{−1} exp_{T} (s^{\frac{1}{k}} X_{j_{1}})^{−1}
· exp_{T} (s^{\frac{k−1}{k}} X_{[J′]}) exp_{T} (s^{\frac{1}{k}} X_{j_{1}})

= exp(v)^{−1} exp(u)^{−1} exp(v) exp(u)

= exp([u, v] + R))

= exp(sX_{[J]} + R̃ + R))

= exp (sX_{[J]} + ∑_{|I|>k} c_{J,I}s^{|I|/k}X_{[I]}),

for suitable constants c_{J,I}. We used the fact that the series R is actually a series of commutators of length at least k + 1 of the original fields. If s < 0, formula (2.9) follows analogously. □

Define for λ > 0 and for any vector field X

S_{1}(λ, X) = exp(λ(X − T)) exp(λT),

S_{2}(λ, X) = exp(−λT) exp(λ(X + T)). (2.10)

Theorem 2.4 (Factorization). Let Y = X_{[J]}, J = (j_{1}, …, j_{k}). The map exp_{T} (sY), s ∈ ℝ, can be factorized as the composition of a finite number of factors of the form S_{1}(h|s|^{\frac{1}{k}} , τX_{j}) and S_{2}(h|s|^{\frac{1}{k}} , τX_{j}), where τ ∈ \{−1, 1\}, j = 1, ..., m and 1 ≤ h ≤ k. Moreover, the number of factors depends only on k.

Proof. Since S_{2}(h|s|^{\frac{1}{k}} , τX_{j}) = S_{1}(h|s|^{\frac{1}{k}} , −τX_{j})^{−1}, if we prove the claim for s > 0 it will also follows for s < 0. Without loss of generality we can suppose s = 1. First notice that

S_{1}(h, τX) exp(T) = exp(h(τX − T)) exp(hT) exp(T)
= exp(T) exp(T)^{−1} exp(−τX + T)S_{1}(h + 1, τX)
= exp(T)S_{2}(1, −τX)S_{1}(h + 1, τX)

(2.11)

and

S_{2}(h, τX) exp(T) = exp(hT)^{−1} exp(h(τX + T)) exp(T)
= exp(T)S_{2}(h + 1, τX) exp(−τX − T) exp(T)
= exp(T)S_{2}(h + 1, τX)S_{1}(1, −τX).

(2.12)

The proof is by induction on k = d(Y). If k = 1 the claim follows directly from definition (2.7) with h = 1. Let k = d(Y) > 1 and let Y = X_{[J]} with J = (j_{1}, ..., j_{k}). If j_{1} ≠ m the claim follows directly from (2.8) and the inductive hypothesis on X_{[J′]}, J′ = (j_{2}, ..., j_{k}). Suppose j_{1} = m and by the inductive hypothesis write

exp_{T} (X_{[J′]}) = ∏_{i=1}^{p} S_{σ_{i}}(h_{i}, τ_{i}X_{j_{i}})

with σ_{i} ∈ {1, 2}, τ_{i} ∈ {−1, 1}, p ∈ ℕ less than a constant depending on k, and 1 ≤ h_{i} ≤ k − 1. Write

exp_{T} (X_{[J]}) = exp_{T} (X_{[J′]})^{−1} exp(T)^{−1} exp_{T} (X_{[J′]}) exp(T)

= exp_{T} (X_{[J′]})^{−1} exp(T)^{−1} ∏_{i=1}^{p} S_{σ_{i}}(h_{i}, τ_{i}X_{j_{i}}) exp(T).

By (2.11) and (2.12) exp(T) can be shifted p times from right to left cancelling exp(T)^{−1} and the claim follows. □

From now on fix a bounded open set Ω_{0} ⊂ ℝ^{n} and let {Y_{1}, ..., Y_{q}} be a fixed enumeration of the commutators of length ≤ k, where k is large enough to ensure that span{X_{[I]}(x, t) : |I| ≤ k} has dimension n at each point (x, t) ∈ Ω_{0}. Assume also

that Y_{q} = T.

Introduce the family of multi-indices I = {I = (i_{1}, …, i_{n−1}) : 1 ≤ i_{j} ≤ q − 1}. Given a multi-index I ∈ I, set d(I) = d(Y_{i_{1}}) + ⋯ + d(Y_{i_{n−1}}) and for h̃ = (h, h_{n}) ∈ ℝ^{n−1} × ℝ “small enough” define

Φ_{I,x}(h) = exp_{T} (h_{n−1}Y_{i_{n−1}}) ⋯ exp_{T} (h_{1}Y_{i_{1}})(x, 0),
Φ̃_{I,x}(h̃) = exp(h_{n}T) exp_{T} (h_{n−1}Y_{i_{n−1}}) ⋯ exp_{T} (h_{1}Y_{i_{1}})(x, 0)
= (Φ_{I,x}(h), h_{n}).

(2.13)

The form of the fields (2.6) guarantees that Φ_{I,x}(h) ∈ {(x, t) ∈ ℝ^{n} : t = 0} for h ∈ ℝ^{n−1}. Let also

||h||_{I} = max_{l=1,...,n−1} |h_{l}|^{1/d(Y_{i_{l}})} and
λ_{I}(x) = det(Y_{i_{1}}(x, 0), …, Y_{i_{n−1}}(x, 0)),

where the vectors Y_{i_{l}} are thought of as vectors in ℝ^{n−1}.

If I ∈ I define Ĩ = (I, q) and set d(Ĩ) = d(I) + 1. If h̃ = (h, h_{n}) and (x, t) ∈ Ω_{0} define

||h̃||Ĩ = max{||h||_{I}, |h_{n}|} and
λ̃Ĩ(x, t) = det(Y_{i_{1}}(x, t), …, Y_{i_{n−1}}(x, t), Y_{n}(x, t))

where Y_{n} = T and the vectors are thought of as vectors in ℝ^{n}.

Let d be the C-C metric induced by the vector fields (2.6) on ℝ^{n} and consider the balls B((x, 0), r) = \{(y, t) ∈ ℝ^{n} : d((x, 0), (y, t)) < r\} and \overline{B}(x, r) = \{y ∈ ℝ^{n−1} : d((x, 0), (y, 0)) < r}. We now state and prove the structure theorem for the restricted balls \overline{B}.

Theorem 2.5. Let Ω_{0} ⊂ ℝ^{n} be a bounded open set. There exist r_{0} > 0 and 0 < a < b < 1 such that for any (x, 0) ∈ Ω_{0}, I ∈ I and 0 < r < r_{0} such that the inequality

|λ_{I}(x)|r^{d(I)} ≥ \frac{1}{2} max_{J∈I} |λ_{J}(x)|r^{d(J)}  (2.14)

is satisfied, we have

(i) \frac{1}{4}|λ_{I}(x)| ≤ |J_{h}Φ_{I,x}(h)| = |Jh̃Φ̃_{I,x}(h̃)| ≤ 4|λ_{I}(x)| for every ||h̃||Ĩ < br, where

J_{h}Φ_{I,x}(h) = det \frac{∂}{∂h}Φ_{I,x}(h).

(ii) B((x, 0), ar) ⊂ Φ̃_{I,x}({||h̃||Ĩ < br}) ⊂ B((x, 0), r).

(iii) \overline{B}(x, ar) ⊂ Φ_{I,x}(\{||h||_{I} < br\}) ⊂ \overline{B}(x, r).

(iv) The map Φ̃_{I,x} is one to one on {||h̃||Ĩ < br}.

Remark 2.6. Inclusions (iii) for the restricted balls are immediate consequence of (ii) and of the structure (2.13) of the map Φ̃. Indeed, starting from (ii) we get

\overline{B}(x, ar) ⊂ Φ̃_{I,x}(\{||h̃||Ĩ < br\}) ∩ \{t = 0\} = Φ_{I,x}(\{||h||_{I} < br\}).

The opposite inclusion is analogous.

Proof of Theorem 2.5. Since λ_{I}(x) = λ̃Ĩ(x, 0), if (2.14) is verified for some (n − 1)−tuple I ∈ I then the n−tuple Ĩ = (I, q) satisfies

|λ̃Ĩ(x, 0)|r^{d(Ĩ)} ≥ \frac{1}{2} max_{J∈I} |λ̃J̃(x, 0)|r^{d(J̃)}.  (2.15)

In [NSW, Theorem 7] it is proved that if Y_{j_{1}}, …, Y_{j_{n}} are commutators of degrees d_{1}, …, d_{n} which satisfy (2.15), then the map Φ̃_{I,x}^{∗} defined by Φ̃_{I,x}^{∗}(h̃) = exp(h_{1}Y_{j_{1}} + ⋯ + h_{n}Y_{j_{n}})(x, 0) satisfies (i), (ii) and (iv). Moreover in [M, Lemmas 3.2-3.6] the following is proved. Assume that the exponential of any commutator Y_{j} can be approximated by a map E(sY_{j}) in the sense that

E(sY_{j}) = exp (sY_{j} + sgn(s) ∑_{|I|>d(Y_{j})} k_{(j),I}|s|^{|I|/d(Y_{j})}X_{[I]}),

where the k_{(j),I} are constants and assume also that for a n-tuple of commutators Y_{j_{1}}, …, Y_{j_{n}} (2.15) holds at a point (x, 0) and for a radius r. Then the map

Φ̃_{I,x}(h̃) = E(h_{n}Y_{j_{n}}) ⋯ E(h_{1}Y_{j_{1}})(x, 0)

satisfies (i), (ii) and (iv). In view of Lemma 2.3 this assertion can be applied to the map E = exp_{T} and the Theorem is proved. We also note that the estimate

μ(B((x, 0), r)) ≃ ∑_{I∈I} |λ_{I}(x)|r^{d(I)}  (2.16)

holds. □

Corollary 2.7. Let Ω ⊂ ℝ^{n} be an open set with C^{∞} boundary. Let K ⊂ ∂Ω be a compact set of non characteristic points with respect to the vector fields X_{1}, ..., X_{m} ∈ C^{∞}(ℝ^{n}, ℝ^{n}) satisfying the Hörmander condition. If μ = H^{n−1} ∂Ω, then there exist r_{0} > 0, 0 < m_{1} < m_{2} such that

m_{1} \frac{|B(x, r)|}{r} ≤ μ(B(x, r)) ≤ m_{2} \frac{|B(x, r)|}{r}  (2.17)

for all x ∈ K and for all 0 < r < r_{0}.

Proof. In view of Lemma 2.1 and Remark 2.2 X_{1}, ..., X_{m} can be assumed to be of the form (2.6) and K ⊂ ∂Ω ⊂ {(x, t) ∈ ℝ^{n−1} × ℝ : t = 0}. The Lemma follows from (ii) and (iii) in Theorem 2.5. □

3. Trace for Hörmander vector fields

Let X_{1}, ..., X_{m} ∈ C^{∞}(ℝ^{n}; ℝ^{n}) be a family of vector fields satisfying the Hörmander condition and let Ω ⊂ ℝ^{n} be a bounded open set with ∂Ω of class C^{∞} and non characteristic. In this section we prove that Ω supports a trace theorem.

Recall first the Hardy inequality. Let 0 < r ≤ +∞. If 1 < p < ∞ and if f ∈ L^{p}(0, r) then

∫_{0}^{r} (\frac{1}{t} ∫_{0}^{t} |f(x)| dx)^{p} dt ≤ (\frac{p}{p − 1})^{p} ∫_{0}^{r} |f(x)|^{p} dx.  (3.18)

Next, we shall need the following formula for integration of “radial functions”.

Lemma 3.1. Let d : ℝ^{n} → [0, ∞) be a Lipschitz function such that |{x ∈ ℝ^{n} : d(x) < λ}| = σλ^{Q} for some Q > 0, σ > 0, for all λ > 0, and |∇d(x)| ≠ 0 for a.e. x ∈ ℝ^{n}. Then

∫_{{d(x)<r}} φ(d(x)) dx = σQ ∫_{0}^{r} φ(λ)λ^{Q−1} dλ  (3.19)

for all measurable functions φ ≥ 0, r > 0.

Proof. For ε > 0 let g_{ε}(x) = χ_{{|∇d|>ε}}(x) and by the coarea formula write

∫_{\{d(x)<λ\}} g_{ε}(x)φ(d(x)) dx = ∫_{0}^{λ} φ(r) ∫_{\{d(x)=r\}} \frac{g_{ε}(x)}{|∇d(x)|}dH^{n−1}(x) dr.

Since H^{n−1}({d(x) = r} ∩ {∇d(x) = 0}) = 0 for a.e. r > 0, by monotone convergence we get

∫_{\{d(x)<λ\}} φ(d(x)) dx = ∫_{0}^{λ} φ(r) ∫_{\{d(x)=r\}} \frac{1}{|∇d(x)|}dH^{n−1}(x) dr,

for all λ > 0. Choosing φ = 1 we find

σλ^{Q} = |\{x ∈ ℝ^{n} : d(x) < λ\}| = ∫_{0}^{λ} ∫_{\{d(x)=r\}} \frac{1}{|∇d(x)|}dH^{n−1}(x) dr,

and taking the derivative we obtain for a.e. λ > 0

σQλ^{Q−1} = ∫_{\{d(x)=λ\}} \frac{1}{|∇d(x)|}dH^{n−1}(x),

which gives the proof. □

We now prove the basic trace theorem. Let X_{1}, ..., X_{m} be a family of smooth vector fields on ℝ^{n} satisfying the Hörmander condition, assume they are of the form (2.6) and let d be the C-C metric induced by them. We shall write (x, t) ∈ ℝ^{n−1} × ℝ and for the sake of simplicity we contract the notation by writing x = (x, 0). Let μ = H^{n−1} {t = 0} be the Lebesgue measure on ℝ^{n−1}.

Theorem 3.2. Let 1 < p < ∞, s = 1 − \frac{1}{p} and let U ⊂ ℝ^{n−1} be a bounded open set. If λ > 0 and t_{0} > 0 there exist C > 0 and δ_{0} > 0, such that

∫_{U×U∩\{d(x,y)<δ_{0}\}} \frac{|u(x, 0) − u(y, 0)|^{p} dxdy}{d(x, y)^{ps}μ(B(x, d(x, y)))} ≤ C ∫_{U_{λ}×(0,t_{0})}|Xu(x, t)|^{p} dxdt  (3.20)

for all u ∈ C^{1}(U_{λ} × (0, t_{0})) ∩ C(U_{λ} × [0, t_{0})), where U_{λ} = {y ∈ ℝ^{n−1} : dist(y, U) < λ}.

Proof. Let U ⊂ Ω_{0} for some bounded open set Ω_{0} ⊂ ℝ^{n} and let k ∈ ℕ be the minimal length of the commutators which ensures the Hörmander condition on Ω_{0}. Fix r_{0} > 0 and 0 < a < b by Theorem 2.5. Define

N(p, δ_{0}; U) = ∫_{U×U∩\{d(x,y)<δ_{0}\}} \frac{|u(x, 0) − u(y, 0)|^{p}}{d(x, y)^{ps}μ(B(x, d(x, y)))} dxdy.

Let I be the set of the multi-indices I defined in Section 2 and write

\begin{aligned}N(p, δ_{0}; U) & = & ∫_{U} dx  ∫_{U∩\{d(x,y)<δ_{0}\}} \frac{|u(x, 0) − u(y, 0)|^{p}}{d(x, y)^{ps}μ(B(x, d(x, y)))} dy \\ & ≤ & ∑_{I∈I} ∫_{U} dx  ∫_{U∩A_{I}(x)∩\{d(x,y)<δ_{0}\}} \frac{|u(x, 0) − u(y, 0)|^{p} dy}{d(x, y)^{ps}μ(B(x, d(x, y)))} \\ & = & ∑_{I∈I} ∫_{U} f_{I}(x) dx, \end{aligned}  (3.21)

where f_{I} is defined by the last equality and we introduced the annulus

\begin{aligned}A_{I}(x) & := \{y ∈ ℝ^{n−1} : |λ_{I}(x)|(2d(x, y)/a)^{d(I)} \\ & ≥ \frac{1}{2} max_{J∈I} |λ_{J}(x)|(2d(x, y)/a)^{d(J)}\}. \end{aligned}

Fix δ_{0} ≤ ar_{0}/2. By Theorem 2.5 the map y = Φ_{I,x}(h) is one-to-one on the set {h ∈ ℝ^{n−1} : ||h||_{I} < (2b/a)d(x, ȳ)} where ȳ ∈ A_{I}(x) is such that d(x, ȳ) = min{δ_{0}, max_{y∈A_{I}(x)} d(x, y)} (the condition d(x, ȳ) ≤ δ_{0} amounts to 2d(x, ȳ)/a < r_{0} and ensures that Theorem 2.5 can be applied), and moreover Φ_{I,x}({h ∈ ℝ^{n−1} : ||h||_{I} < (2b/a)d(x, ȳ)}) ⊃ \overline{B}(x, 2d(x, ȳ)) ⊃ A_{I}(x). By the same theorem statement (iii)

\begin{aligned}\overline{B}(x, 2d(x, y)) & ⊂ Φ_{I,x}(\{h ∈ ℝ^{n−1} : ||h||_{I} < (2b/a)d(x, y)\}) \\ & ⊂ \overline{B}(x, 2d(x, y)/a) \end{aligned}

for all y ∈ A_{I}(x) and d(x, y) < δ_{0}, i.e. 2d(x, y)/a < r_{0}. Thus

||h||_{I} < \frac{2b}{a}d(x, Φ_{I,x}(h)) ≤ \frac{2b}{a}δ_{0} ≤ br_{0}.  (3.22)

Set H_{I,δ_{0}}(x) = Φ_{I,x}^{−1}(U ∩A_{I}(x)∩{d(x, y) < δ_{0}}). Thus, by the first inequality of (3.22),

f_{I}(x) ≤ C ∫_{H_{I,δ_{0}}(x)} \frac{|u(x, 0) − u(Φ_{I,x}(h), 0)|^{p} |J_{h}Φ_{I,x}(h)|}{||h||_{I}^{ps}μ(B(x, C||h||_{I}))} dh.  (3.23)

Note that (2.16) furnishes the estimate μ(B(x, C||h||_{I})) ≥ C|λ_{I}(x)|||h||^{d(I)}. Letting η = 2bδ_{0}/a and recalling that |J_{h}Φ_{I,x}(h)| ≃ |λ_{I}(x)| from (3.21) and (3.23) we get

\begin{aligned}N(p; δ_{0}; U) & ≤ C ∑_{I∈I} ∫_{U} dx ∫_{\{||h||_{I}<η\}} \frac{|u(x, 0) − u(Φ_{I,x}(h), 0)|^{p}}{||h||_{I}^{ps+d(I)}} dh \\ & = C ∑_{I∈I} ∫_{\{||h||_{I}<η\}} \frac{dh}{||h||_{I}^{ps+d(I)}} ∫_{U} |u(x, 0) − u(Φ_{I,x}(h), 0)|^{p} dx. \end{aligned}  (3.24)

If I = (i_{1}, ..., i_{n−1}) and ||h||_{I} < η set z_{0}(x) = x and define z_{l}(x) = ∏_{j=1}^{κ} exp_{T} (h_{j}Y_{i_{j}})(x) for κ = 1, ..., n − 1, in such a way that z_{n−1}(x) = Φ_{I,x}(h). Thus, fixed a constant λ′, 0 < λ′ < λ

∫U|u(x, 0) − u(Φ_{I,x}(h), 0)|^{p} dx

≤ C ∑_{κ=1}^{n−1} ∫_{U} |u(z_{κ−1}(x), 0) − u(z_{κ}(x), 0)|^{p} dx

≤ C ∑_{κ=1}^{n−1} ∫_{U} |u( ∏_{j=1}^{l−1} exp_{T} (h_{j}Y_{i_{j}})(x), 0)

− u( exp_{T} (h_{l}Y_{i_{l}}) ∏_{j=1}^{l−1} exp_{T} (h_{j}Y_{i_{j}})(x), 0)|^{p} dx

≤ C ∑_{κ=1}^{n−1} ∫_{U_{λ′}} |u(ξ, 0) − u(exp_{T} (h_{κ}Y_{i_{κ}})(ξ), 0)|^{p} dξ,

(3.25)

where in each integral we performed the change of variable ξ = z_{κ−1}(x) which has Jacobian greater than a positive constant. Moreover, ξ ∈ U_{λ′} if δ_{0} is small enough. Then, we have to estimate a finite number of integrals of the form

∫_{U_{λ′}} |u(x, 0) − u(exp_{T} (t(h)Y)(x), 0)|^{p} dx,

with d(Y) ≤ k and |t(h)|^{1/d(Y)} ≤ ||h||_{I}. By Lemma 2.4 we can write exp_{T} (tY) = ∏_{i=1}^{p} S_{σ_{i}}(q_{i}|t|^{1/d(Y)}, τ_{i}X_{j_{i}}) with σ_{i} ∈ {1, 2}, τ_{i} ∈ {−1, 1}, 1 ≤ q_{i} ≤ k, p less than an absolute constant and S_{1}, S_{2} as in (2.10). With triangle inequalities and changes of variable quite similar to the ones in (3.25) we are led to the estimate of integrals of one of the two types

∫_{U_{λ′}} |u(exp(q|t(h)|^{1/d(Y)}T)(x), 0) − u(x, 0)|^{p}dx  or
∫_{U_{λ′}} |u(exp(q|t(h)|^{1/d(Y)}(τX_{j} + T))(x), 0) − u(x, 0)|^{p}dx,

(3.26)

with j = 1, ..., m − 1, 1 ≤ q ≤ k and |t(h)|^{1/d(Y)} ≤ ||h||_{I}. If we consider, for instance, an integral of the second type with τ = 1 the computation in (3.24) can be concluded in the following way (recall that ps + d(I) = p − 1 + d(I)):

∫_{\{||h||_{I}<η\}} \frac{dh}{||h||_{I}^{ps+d(I)}} ∫_{U_{λ′}} |u(exp(q|t(h)|^{1/d(Y)}(X_{j} + T)(x, 0))) − u(x, 0)|^{p} dx

≤ C ∫_{\{||h||_{I}<η\}} \frac{dh}{||h||_{I}^{ps+d(I)}} ∫_{U_{λ′}} ( ∫_{0}^{k||h||_{I}}|Xu(exp(t(X_{j} + T))(x, 0)| dt)^{p} dx

≤ C ∫_{\{||h||_{I}<η\}} \frac{dh}{||h||_{I}^{ps+d(I)}} ( ∫_{0}^{k||h||_{I}}( ∫_{U_{λ′}} |Xu(exp(t(X_{j} + T))(x, 0)|^{p}dx)^{\frac{1}{p}} dt)^{p}

= C∫_{0}^{kη} \frac{dr}{r^{p}}( ∫_{0}^{r} ( ∫_{U_{λ′}} |Xu(exp(t(X_{j} + T))(x, 0)|^{p} dx)^{\frac{1}{p}} dt)^{p}

≤ C ∫_{0}^{kη}∫_{U_{λ′}} |Xu(exp(t(X_{j} + T))(x, 0))|^{p} dx dt.

We used the Minkowski inequality, formula (3.19) and the Hardy inequality (3.18).

Finally, write exp(t(X_{j} + T))(x, 0) = Θ(x, t) and perform the change of variable (ξ, τ) = Θ(x, t). Since Θ(x, 0) = (x, 0) then

\frac{∂Θ(x, t)}{∂x∂t}|_{t=0} = (\begin{matrix}I_{n−1} & X_{j}(x, 0) \\ 0 & 1 \end{matrix})

and thus Θ is a change of variable on the rectangle U_{λ′} × (0, ϱ_{0}), where ϱ_{0} is suitably small. Choosing δ_{0} small we obtain kη ≤ ϱ_{0} and Θ(x, t) ∈ U_{λ} × (0, t_{0}) for all (x, t) ∈ U_{λ′} × (0, kη). Then

∫_{U_{λ′}×(0,kη)} |Xu(Θ(x, t))|^{p} dxdt ≤ C ∫_{U_{λ}×(0,t_{0})} |Xu(ξ, τ)|^{p} dξdτ.

Integrals of the first type in (3.26) can be treated in the same way and the proof of the Theorem is concluded. □

Corollary 3.3. Let X_{1}, ..., X_{m} ∈ C^{∞}(ℝ^{n}; ℝ^{n}) satisfy the Hörmander condition and let Ω ⊂ ℝ^{n} be a bounded open set with ∂Ω of class C^{∞} and non characteristic. Let 1 < p < ∞ and s = 1 − \frac{1}{p}. There exist constants C, δ_{0} > 0 such that

∫_{∂Ω×∂Ω∩\{d(x,y)<δ_{0}\}}\frac{|u(x) − u(y)|^{p} dμ(x)dμ(y)}{d(x, y)^{ps}μ(B(x, d(x, y)))} ≤ C ∫_{Ω} |Xu(x)|^{p} dx

for all u ∈ C^{1}(Ω) ∩ C(\overline{Ω}), where μ = H^{n−1} ∂Ω.

Proof. The proof follows from Theorem 3.2 using a standard covering argument, Lemma 2.1 and Remark 2.2. □

Example 3.4 (Trace on subgroups of ℍ^{n}). Consider the Heisenberg group ℍ^{n} = ℂ^{n} ×ℝ, n ≥ 1, whose elements are (z, t) ∈ ℍ^{n} with z = x + iy ∈ ℂ^{n}, x, y ∈ ℝ^{n}, and t ∈ ℝ, and whose group law is (ζ, τ)·(z, t) = (ζ +z, τ +t+2 Im(ζz̄)). The Lie algebra of the group is generated by the vector fields

X_{j} = ∂_{x_{j}} + 2y_{j}∂_{t} and Y_{j} = ∂_{y_{j}} − 2x_{j}∂_{t}, j = 1, ..., n,

which satisfy the Hörmander condition. The homogeneous norm ||(z, t)|| := (|z|^{4} + t^{2})^{1/4} is equivalent to the C-C metric d, i.e. c_{1}d((z, t), (ζ, τ)) ≤ ||(ζ, τ)^{−1} · (z, t)|| ≤ c_{2}d((z, t), (ζ, τ)) for all (z, t), (ζ, τ) ∈ ℍ^{n} and for some 0 < c_{1} < c_{2}. The integer Q = 2n + 2 is the “dimension” of ℍ^{n} and |B((z, t), r)| = cr^{Q} for some c > 0 and for all (z, t) ∈ ℍ^{n} and r ≥ 0.

Consider the half space Ω = {(x, y, t) ∈ ℍ^{n} : x_{j} > 0} for some j = 1, ..., n with boundary ∂Ω = {(x, y, t) ∈ ℍ^{n} : x_{j} = 0}. Actually, the hyperplane ∂Ω is a subgroup of ℍ^{n} and all its points are non characteristic. If μ = H^{2n} ∂Ω then μ(B((z, t), r)) = mr^{Q−1} for some m > 0 and for all (z, t) ∈ ∂Ω. Using the technique developed in this section it can be proved that there exists a constant C > 0 such that (1 < p < ∞ and s = 1 − \frac{1}{p})

∫_{∂Ω×∂Ω} \frac{|u(z, t) − u(ζ, τ)|^{p} dμ(z, t)dμ(ζ, τ)}{||(ζ, τ)^{−1} · (z, t)||^{ps+Q−1}} ≤ C ∫_{Ω} |∇_{ℍ^{n}}u(z, t)|^{p} dzdt

for all u ∈ C^{1}(Ω) ∩ C(\overline{Ω}), where ∇_{ℍ^{n}} = (X_{1}, ..., X_{n}, Y_{1}, ..., Y_{n}).

4. Trace Theorem in the Grushin Plane

4.1. Trace theorem. In this section we begin the study of the trace theorem when the boundary contains characteristic points. We focus our attention on the Grushin plane where we prove that the trace estimate holds for domains which are sufficiently “flat” at characteristic points.

Let d be the C-C metric induced on ℝ^{2} by the vector fields

X_{1} = ∂_{x} and X_{2} = |x|^{α}∂_{y}, α > 0.

If (x, y) ∈ ℝ^{2} and r ≥ 0 let B((x, y), r) = {(ξ, η) ∈ ℝ^{2} : d((x, y), (ξ, η)) < r}. Moreover, define the “box”

Box((x, y), r) = [x − r, x + r] × [y − r(|x| + r)^{α}, y + r(|x| + r)^{α}].

Such boxes are equivalent to C-C balls and the metric d can be evaluated rather explicitly. This is stated in the following Lemmas, whose proof is a consequence of the results in [FL].

Lemma 4.1. There exist constants 0 < c_{1} < c_{2} such that for all (x, y) ∈ ℝ^{2} and r ≥ 0

Box((x, y), c_{1}r) ⊂ B((x, y), r) ⊂ Box((x, y), c_{2}r).  (4.27)

Lemma 4.2. Let λ > 0. For all (x, y), (ξ, η) ∈ ℝ^{2} with |x| ≥ |ξ|

d((x, y), (ξ, η)) ≃ |x − ξ| + \frac{|y − η|}{|x|^{α}} if |x|^{α+1} ≥ λ|y − η|,  (4.28)
d((x, y), (ξ, η)) ≃ |x − ξ| + |y − η|^{\frac{1}{α+1}} if |x|^{α+1} < λ|y − η|,  (4.29)

where the equivalence constants depend on λ.

Definition 4.3. Let Ω ⊂ ℝ^{2} be an open set with ∂Ω of class C^{1}. A point (0, y_{0}) ∈ ∂Ω is said to be α−admissible, α > 0, if one of the following two conditions holds:

(i) (Non characteristic case). There exist δ > 0 and ψ ∈ C^{1}(y_{0} − δ, y_{0} + δ) such

that ψ(y_{0}) = 0 and
∂Ω ∩ (−δ, δ) × (y_{0} − δ, y_{0} + δ) = {(ψ(y), y) : |y − y_{0}|, |ψ(y)| < δ}.

(ii) (Characteristic case). There exist δ > 0 and c > 0 such that

∂Ω ∩ (−δ, δ) × (y_{0} − δ, y_{0} + δ) = {(x, φ(x)) ∈ ℝ^{2} : |x| < δ},

where φ ∈ C^{1}(−δ, δ) and |φ′(x)| ≤ c|x|^{α} for all x ∈ (−δ, δ). Finally, Ω is said to be α−admissible if all the points of ∂Ω ∩ {x = 0} are α−admissible.

Let Ω ⊂ ℝ^{2} be an open set of class C^{1} and let ν(x, y) be the unit normal to ∂Ω at (x, y) ∈ ∂Ω. Consider the modulus of the “projected” normal

\begin{aligned}|Xν(x, y)| & = (⟨X_{1}(x, y), ν(x, y)⟩^{2} + ⟨X_{2}(x, y), ν(x, y)⟩^{2})^{\frac{1}{2}} \\ & = (ν_{1}(x, y)^{2} + |x|^{2α}ν_{2}(x, y)^{2})^{\frac{1}{2}} , \end{aligned}

and define the measure μ = |Xν|H^{1} ∂Ω. The measure μ is the one that appears in the left hand side of the trace estimates.

In the sequel we shall use the equivalence

∫_{I} |ξ|^{α}dξ ≃ |I| max_{ξ∈I} |ξ|^{α}  (4.30)

for any interval I ⊂ ℝ, where the equivalence constants depend only on α > 0.

Lemma 4.4. Let Ω ⊂ ℝ^{2} be a bounded open set with ∂Ω of class C^{1} and suppose it is α−admissible. Then there exist 0 < m_{1} < m_{2} and r_{0} > 0 such that

m_{1} \frac{|B((x, y), r)|}{r} ≤ μ(B(x, y), r) ≤ m_{2} \frac{|B((x, y), r)|}{r}  (4.31)

for all (x, y) ∈ ∂Ω and for all 0 < r < r_{0},

Proof. Since away from the set {x = 0} we are essentially in a Euclidean situation it suffices to prove (4.31) for (x, y) ∈ ∂Ω belonging to a neighborhood of an α−admissible point.

Suppose first that (0, 0) ∈ ∂Ω is an α−admissible point of type (i) (non characteristic). In a neighborhood of the origin ∂Ω is the graph of a function ψ ∈ C^{1}(−δ, δ) in the variable y. If δ > 0 and r > 0 are small, then the graph of ψ meets ∂Box((ψ(y), y), r) on its horizontal edges. This is ensured by |ψ(y) − ψ(y − r(|ψ(y)| + r)^{α})| < r, which holds true provided y and r are small enough. Now

\begin{aligned}μ(Box((ψ(y), y), r)) & ≃ ∫_{y−r(|ψ(y)|+r)^{α}}^{y+r(|ψ(y)|+r)^{α}} dη \\ & = 2r(|ψ(y)| + r)^{α} = \frac{|Box((ψ(y), y), r)|}{2r}, \end{aligned}

and (4.1) gives the proof of the required estimate.

Suppose now that (0, 0) ∈ ∂Ω is an α−admissible point of type (ii). Let φ ∈ C^{1}(−δ, δ) be the function whose graph represents ∂Ω and such that |φ′(x)| ≤ c|x|^{α} for all |x| < δ and for some c ≥ 0. Then, if y = φ(x) and |x| ≤ δ/2

ν(x, y) = \frac{(φ′(x), −1)}{\sqrt{1 + φ′(x)^{2}}} , and |Xν(x, y)| = \begin{matrix}√ & \overline{|x|^{2α} + φ′(x)^{2}} \\ √ & \overline{1 + φ′(x)^{2}} \end{matrix} ≃ |x|^{α}.

By Lemma 4.1 μ(Box((x, y), c_{1}r)) ≤ μ(B((x, y), r)) ≤ μ(Box((x, y), c_{2}r)), and, supposing for instance 0 ≤ x ≤ δ/2 and 0 < r < δ/(2c_{2})

μ(Box((x, y), c_{2}r)) = ∫_{Box((x,y),c_{2}r)∩∂Ω} |Xν|dH^{1} ≤ C ∫_{x−c_{2}r}^{x+c_{2}r} |ξ|^{α} dξ

≤ 2Cc_{2}r(x + c_{2}r)^{α} ≃ \frac{|Box((x, y), c_{2}r)|}{r}.

The estimate from above in (4.31) follows by Lemma 4.1. In order to prove the opposite inequality assume without loss of generality that the constant c relative to φ is greater than 1 and that x ≥ 0. Introduce the new box

\overline{Box}((x, y), c_{1}r) := [x − \frac{c_{1}}{c} r, x + \frac{c_{1}}{c} r]
× [y − c_{1}r(x + c_{1}r)^{α}, y + c_{1}r(x + c_{1}r)^{α}]
⊂ Box((x, y), c_{1}r).

Since |φ(x + \frac{c_{1}}{c} r) − φ(x)| ≤ c_{1}r(x + c_{1}r)^{α}, the graph of φ meets ∂\overline{Box}((x, y), c_{1}r) on its left and right vertical edges. Thus

μ(B((x, y)), r)) ≥ μ(\overline{Box}((x, y), c_{1}r) = ∫_{\overline{Box}((x,y),c_{1}r)∩∂Ω} |Xν|dH^{1}

≥ C ∫_{x− \frac{c_{1}}{c} r}^{x+ \frac{c_{1}}{c} r} |ξ|^{α} dξ ≃ C \frac{c_{1}}{c} r(x + \frac{c_{1}}{c} r)^{α}

≃ \frac{|Box((x, y), c_{1}r)|}{r},

which is the required estimate. We also used (4.30). □

Theorem 4.5. Let X_{1} = ∂_{x} and X_{2} = |x|^{α}∂_{y}, α > 0. Let 1 < p < ∞ and s = 1 − \frac{1}{p}. If Ω ⊂ ℝ^{2} is a bounded open set of class C^{1} which is α−admissible, then there exist C > 0 and δ_{0} > 0 such that

∫_{∂Ω×∂Ω∩\{d(z,ζ)<δ_{0}\}} \frac{|u(z) − u(ζ)|^{p} dμ(z)dμ(ζ)}{d(z, ζ)^{ps}μ(B(z, d(z, ζ)))} ≤ C ∫_{Ω} |Xu(x, y)|^{p} dxdy

for all u ∈ C^{1}(Ω) ∩ C(\overline{Ω}).

Proof. Since away from the set {x = 0} we are essentially in the Euclidean case, it suffices to prove the estimate in a neighborhood of an α−admissible point which may assumed to be the origin. Denote by U the intersection of ∂Ω with a small fixed neighborhood of (0, 0). Recalling that, by Lemma 4.4, d(z, ζ)^{ps}μ(B(z, d(z, ζ))) ≃ d(z, ζ)^{ps−1}|B(z, d(z, ζ))|, we have to prove that

\begin{aligned}N(p; U) & := ∫_{U×U} \frac{|u(z) − u(ζ)|^{p}}{d(z, ζ)^{ps−1}|B(z, d(z, ζ))|}dμ(z)dμ(ζ) \\ & ≤ C ∫_{Ω} |Xu(x, y)|^{p} dxdy. \end{aligned}

The α−admissible point can be of type (i) or of type (ii).

Type (i). We may assume that U = {(ψ(y), y) : |y| < δ} for some δ > 0 and ψ ∈ C^{1}(−δ, δ) with ψ(0) = 0, and that Ω lies in the region {x > ψ(y)}. Write z = (ψ(y), y) and ζ = (ψ(η), η), and notice that, by the doubling property of the Lebesgue measure, which follows from Lemma 4.1, |B(z, d(z, ζ))| ≃ |B(ζ, d(z, ζ))|. Thus the kernel is essentially symmetric and the integration can be performed without loss of generality on the set {|ψ(η)| < |ψ(y)|}

N(p; U) ≃ ∫_{\{|y|<δ, |η|<δ, |ψ(η)|<|ψ(y)|\}} \frac{|u(z) − u(ζ)|^{p}}{d(z, ζ)^{ps−1}|B(z, d(z, ζ))|}dydη

= ∫_{A} \frac{|u(z) − u(ζ)|^{p}dydη}{d(z, ζ)^{ps−1}|B(z, d(z, ζ))|} + ∫_{B} \frac{|u(z) − u(ζ)|^{p}dydη}{d(z, ζ)^{ps−1}|B(z, d(z, ζ))|}

:= I_{A} + I_{B},

where we let

\begin{aligned}A & = \{(y, η) : |y| < δ, |η| < δ, |ψ(η)| < |ψ(y)|, |ψ(y)|^{α+1} ≥ |y − η|\}, \\ B & = \{(y, η) : |y| < δ, |η| < δ, |ψ(η)| < |ψ(y)|, |ψ(y)|^{α+1} < |y − η|\}. \end{aligned}

We begin with the estimate of I_{A}. If (y, η) ∈ A then

\begin{aligned}d(z, ζ) & ≃ |ψ(y) − ψ(η)| + \frac{|y − η|}{|ψ(y)|^{α}} \\ & = \frac{|y − η|}{|ψ(y)|^{α}} (1 + |ψ(y)|^{α} \frac{|ψ(y) − ψ(η)|}{|y − η|}) ≃ \frac{|y − η|}{|ψ(y)|^{α}} \end{aligned}

and

|B(z, d(z, ζ))| ≃ d(z, ζ)^{2}(|ψ(y)| + d(z, ζ))^{α} ≃ d(z, ζ)^{2}|ψ(y)|^{α}.

Without loss of generality assume y > η. Let η = y − h and write (recall that 1 + ps = p)

\begin{aligned}I_{A} & ≃ ∫_{A} \frac{|u(ψ(y), y) − u(ψ(η), η)|^{p}}{|y − η|^{p}} |ψ(y)|^{pα−α}dydη \\ & ≤ ∫_{0}^{2δ} \frac{dh}{|h|^{p}} ∫_{\{|ψ(y)|^{α+1}>|h|, |y|<δ\}} |u(ψ(y), y) − u(ψ(y − h), y − h)|^{p}|ψ(y)|^{pα−α}dy. \end{aligned}

We shall connect the points (ψ(y), y) and (ψ(y − h), y − h) by the curves

\begin{aligned}γ_{1}(t) & := exp(t(X_{1} − bX_{2}))(ψ(y), y) \\ & = (ψ(y) + t, y − b ∫_{0}^{t} |ψ(y) + τ|^{α}dτ) := Ψ_{1}(t, y), \end{aligned}

where b = min{1, 1/L}, L := sup_{|y|<δ} |ψ′(y)|, and

\begin{aligned}γ_{2}(t) : & = exp(tX_{1})(ψ(y − h), y − h) \\ & = (ψ(y − h) + t, y − h) := Ψ_{2}(t, y − h). \end{aligned}

In order to reach the height y − h, the curve γ_{1} needs a time t_{1} such that

∫_{0}^{t_{1}} |ψ(y) + τ|^{α}dτ = \frac{|h|}{b}.  (4.32)

By (4.30) the left hand side is greater than Ct_{1}|ψ(y)|^{α} and then t_{1} ≤ C|h|/|ψ(y)|^{α}. The time t_{2} such that γ_{2}(t_{2}) = γ_{1}(t_{1}) can also be estimated by |h|/|ψ(y)|^{α}. Indeed

t_{2} = |ψ(y) + t_{1} − ψ(y − h)| ≤ L|h| + t_{1} ≤ C \frac{|h|}{|ψ(y)|^{α}} .

The choice of the parameter b guarantees that γ_{1}(t) ∈ Ω for all |y| < δ and 0 < t ≤ t_{1}. In fact this happens if and only if

ψ(y − b ∫_{0}^{t} |ψ(y) + τ|^{α}dτ) < ψ(y) + t.  (4.33)

This last inequality is a consequence of the following

|ψ(y − b ∫_{0}^{t} |ψ(y) + τ|^{α}dτ) − ψ(y)| ≤ Lb ∫_{0}^{t} |ψ(y) + τ|^{α}dτ < t.

Since Ψ_{1}(t_{1}, y) = Ψ_{2}(t_{2}, y − h) then |u(ψ(y), y) − u(ψ(y − h), y − h)| is less than

\begin{aligned}|u(ψ(y), y) & − u(Ψ_{1}(t_{1}, y))| + |u(ψ(y − h), y − h)) − u(Ψ_{2}(t_{2}, y − h))| \\ & ≤ C( ∫_{0}^{t_{1}} |Xu(Ψ_{1}(t, y))|dt + ∫_{0}^{t_{2}} |Xu(Ψ_{2}(t, y − h))|dt), \end{aligned}

and we find

I_{A} ≤ C[ ∫_{0}^{2δ} \frac{dh}{|h|^{p}}  ∫_{(−δ,δ)∩\{|ψ(y)|^{α+1}≥|h|\}} |ψ(y)|^{pα−α}( ∫_{0}^{t_{1}} |Xu(Ψ_{1}(t, y))|dt)^{p}dy

+ ∫_{0}^{2δ} \frac{dh}{|h|^{p}} ∫_{(−δ,δ)∩\{|ψ(y)|^{α+1}≥|h|\}} |ψ(y)|^{pα−α}( ∫_{0}^{t_{2}} |Xu(Ψ_{2}(t, y − h))|dt)^{p}dy]

:= C[I_{A}^{(1)} + I_{A}^{(2)}].

We shall estimate I_{A}^{(1)} and I_{A}^{(2)} by the same technique and we begin with I_{A}^{(1)}. Letting in the inner integral τ = |ψ(y)|^{α}t, recalling that t_{1} ≤ C|h|/ |ψ(y)|^{α} and using the Minkowski inequality we find

\begin{aligned}I_{A}^{(1)} & ≤ ∫_{0}^{2δ} \frac{dh}{|h|^{p}} ∫_{(−δ,δ)∩\{|ψ(y)|^{α+1}≥|h|\}} \frac{dy}{|ψ(y)|^{α}} ( ∫_{0}^{C|h|}|Xu(Ψ_{1}(τ/|ψ(y)|^{α}, y))|dτ)^{p} \\ & ≤ ∫_{0}^{2δ}(\frac{dh}{|h|} ∫_{0}^{C|h|}(  ∫_{(−δ,δ)∩\{|ψ(y)|^{α+1}≥|h|\}} \frac{|Xu(Ψ_{1}(τ/|ψ(y)|^{α}, y))|^{p}dy}{|ψ(y)|^{α}} )^{\frac{1}{p}} dτ)^{p}. \end{aligned}

Since {|ψ(y)|^{α+1} ≥ |h|} ⊂ {C|ψ(y)|^{α+1} ≥ τ} the last integral is estimated by an integral of the form ∫_{0}^{2δ} (\frac{1}{|h|} ∫_{0}^{C|h|} |f(τ)|dτ)^{p}dh with f not depending on h. So we can apply the Hardy inequality to get

\begin{aligned}I_{A}^{(1)} & ≤ C ∫_{0}^{2δ} ∫_{(−δ,δ)∩\{C|ψ(y)|^{α+1}≥τ\}} \frac{|Xu(Ψ_{1}(τ/|ψ(y)|^{α}, y))|^{p}}{|ψ(y)|^{α}} dy dτ \\ & ≤ C ∫_{−δ}^{δ} ∫_{0}^{Cδ} |Xu(Ψ_{1}(t, y))|^{p}dt dy. \end{aligned}

We let τ/|ψ(y)|^{α} = t and we used τ/|ψ(y)|^{α} ≤ C|ψ(y)| ≤ C|y| ≤ Cδ. The Jacobian matrix of Ψ_{1} is

\frac{∂Ψ_{1}(y, t)}{∂y∂t} = (\begin{matrix}1 & ψ′(y) \\ −b|ψ(y) + t|^{α} & 1 − b(|ψ(y) + t|^{α} − |ψ(y)|^{α})ψ′(y) \end{matrix}) .

By the same argument used in the proof of (4.33) we can see that if δ > 0 is small, then Ψ_{1}((0, Cδ) × (−δ, δ)) ⊂ Ω. Moreover |JΨ_{1}(t, y)| = |1 + bψ′(y)|ψ(y)|^{α}| ≃ 1. Then

I_{A}^{(1)} ≤ C ∫_{Ω} |Xu(x, y)|^{p}dxdy.

We estimate now I_{A}^{(2)}. Note first that if δ > 0 is small and (y, η) ∈ A, we have

|ψ(y)| ≤ 2|ψ(η)|.  (4.34)

Indeed L|ψ(y)|^{α+1} ≥ L|y − η| ≥ |ψ(y) − ψ(η)| ≥ |ψ(y)| − |ψ(η)|, and thus |ψ(η)| ≥ |ψ(y)|−L|ψ(y)|^{α+1} ≥ 1/2|ψ(y)| if δ > 0 is small. Taking (4.34) into account with η = y −h, recalling that t_{2} ≤ C|h|/|ψ(y)|^{α} ≤ C|h|/|ψ(y −h)|^{α} and letting τ = |ψ(y −h)|^{α}t in the inner integral we find that I_{A}^{(2)} is smaller than

∫_{0}^{2δ} \frac{dh}{|h|^{p}} ∫_{(−δ,δ)∩\{C|ψ(y−h)|^{α+1}≥|h|\}} \frac{dy}{|ψ(y − h)|^{α}} ( ∫_{0}^{C|h|} |Xu(Ψ_{2}(\frac{τ}{|ψ(y − h)|^{α}} , y − h))|dτ)^{p}

≤ ∫_{0}^{2δ}(\frac{dh}{|h|} ∫_{0}^{C|h|}( ∫_{(−δ,δ)∩\{C|ψ(y−h)|^{α+1}≥|h|\}} \frac{|Xu(Ψ_{2}(\frac{τ}{|ψ(y−h)|^{α}} , y − h))|^{p}}{|ψ(y − h)|^{α}} dy)^{1/p}dτ)^{p}

≤ ∫_{0}^{2δ}(\frac{dh}{|h|} ∫_{0}^{C|h|}( ∫_{(−3δ,δ)∩\{C|ψ(y)|^{α+1}≥|h|\}} \frac{|Xu(Ψ_{2}(τ/|ψ(y)|^{α}, y))|^{p}}{|ψ(y)|^{α}} dy)^{1/p}dτ)^{p}.

Since {C|ψ(y)|^{α+1} ≥ |h|} ⊂ {C|ψ(y)|^{α+1} ≥ τ}, we can apply the Hardy inequality to get

\begin{aligned}I_{A}^{(2)} & ≤ C ∫_{0}^{2δ} ∫_{(−3δ,δ)∩\{C|ψ(y)|^{α+1}≥τ\}} \frac{|Xu(Ψ_{2}(τ/|ψ(y)|^{α}, y))|^{p}}{|ψ(y)|^{α}} dy dτ \\ & ≤ C ∫_{−3δ}^{δ} ∫_{0}^{Cδ} |Xu(Ψ_{2}(t, y))|^{p}dt dy. \end{aligned}

Since |JΨ_{2}(t, y)| = 1 the estimate for I_{A}^{(2)} follows.

We now turn to the estimate of I_{B}. Writing again z = (ψ(y), y) and ζ = (ψ(η), η), if (y, η) ∈ B then

d(z, ζ) ≃ |ψ(y) − ψ(η)| + |y − η|^{1/(α+1)} ≃ |y − η|^{1/(α+1)}

because ψ ∈ C^{1} and |y − η| ≤ 2δ. Moreover starting from the inequality |ψ(y)| ≤ |y − η|^{1/(α+1)} which defines B, we find

|B(z, d(z, ζ))| ≃ d(z, ζ)^{2}(|ψ(y)| + d(z, ζ))^{α}
≃ |y − η|^{2/(α+1)}(|ψ(y)| + |y − η|^{1/(α+1)})^{α}
≃ |y − η|^{(α+2)/(α+1)}.

Assume η < y, let η = y − h and write

\begin{aligned}I_{B} & ≃ ∫_{B} \frac{|u(ψ(y), y) − u(ψ(η), η)|^{p}}{|y − η|^{1+ \frac{ps}{α+1}}} dydη \\ & ≤ C ∫_{0}^{2δ} \frac{dh}{|h|^{1+ \frac{ps}{α+1}}} ∫_{\{|ψ(y)|^{α+1}<|h|, |y|<δ\}} |u(ψ(y), y) − u(ψ(y − h), y − h)|^{p}dy. \end{aligned}

The points (ψ(y), y) and (ψ(y − h), y − h) can be connected by the curves γ_{1}(t) := exp(t(X_{1}−bX_{2}))(ψ(y), y) = Ψ_{1}(t, y) and γ_{2}(t) := exp(tX_{1})(ψ(y−h), y−h) = Ψ_{2}(t, y− h). In order to reach the height y − h, the curve γ_{1} needs a time t_{1} such that (4.32) holds. By (4.30)

∫_{0}^{t_{1}} |ψ(y) + τ|^{α}dτ ≃ t_{1} max_{τ∈[ψ(y),ψ(y)+t_{1}]} |τ|^{α} ≥ t_{1}(\frac{t_{1}}{2} )^{α}.

This yields t_{1} ≤ C|h|^{1/(α+1)}. The time t_{2} such that γ_{2}(t_{2}) = γ_{1}(t_{1}) can also be estimated by |h|^{1/(α+1)}. By the triangle inequality we get

I_{B} ≤ C[ ∫_{0}^{2δ} \frac{dh}{|h|^{1+ \frac{ps}{α+1}}} ∫_{−δ}^{δ} ( ∫_{0}^{t_{1}} |Xu(Ψ_{1}(t, y))|dt)^{p}dy

+ ∫_{0}^{2δ} \frac{dh}{|h|^{1+ \frac{ps}{α+1}}} ∫_{−δ}^{δ} ( ∫_{0}^{t_{2}} |Xu(Ψ_{2}(t, y + h))|dt)^{p}dy]

:= C[I_{B}^{(1)} + I_{B}^{(2)}].

Now, by the Minkowski inequality

I_{B}^{(1)} ≤ ∫_{0}^{2δ} \frac{dh}{|h|^{1+ \frac{ps}{α+1}}} ( ∫_{0}^{C|h|^{1/(α+1)}} ( ∫_{−δ}^{δ} |Xu(Ψ_{1}(t, y))|^{p}dy)^{1/p}dt)^{p}

≤ C ∫_{0}^{(2δ)^{1/(α+1)}} \frac{dr}{r^{p}}( ∫_{0}^{Cr} ( ∫_{−δ}^{δ} |Xu(Ψ_{1}(t, y))|^{p}dy)^{1/p}dt)^{p}

≤ ∫_{(0,(2δ)^{1/(α+1)})×(−δ,δ)} |Xu(Ψ_{1}(t, y))|^{p}dtdy.

We used s = 1 − 1/p, the change of variable r = h^{1/(α+1)} and the Hardy inequality.

The estimate of I_{B}^{(2)} is analogous to the one of I_{A}^{(2)}. This ends the trace estimates for α−admissible points of type (i).

Type (ii). Write U = {(x, φ(x)) ∈ ∂Ω : |x| < δ} for some φ ∈ C^{1}(−δ, δ) such that |φ′(x)| ≤ c|x|^{α} for some c ≥ 0 and for all x ∈ (−δ, δ). Write z = (x, φ(x)), ζ = (y, φ(y)), and observe that

N(p; U) ≃ ∫_{|x|<δ, |y|<δ} \frac{|u(z) − u(ζ)|^{p} |xy|^{α}}{d(z, ζ)^{ps−1}|B(z, d(z, ζ))|} dxdy.

Since the integrand is symmetric up to equivalence constants, the integration may take place on the set {|x| < |y| < δ}. Since |φ′(y)| ≤ c|y|^{α} we have |φ(y) − φ(x)| ≤ c|y − x| |y|^{α} ≤ 2c|y|^{α+1}. Then on the mentioned set the C-C metric behaves as

d(z, ζ) ≃ |y − x| + \frac{|φ(y) − φ(x)|}{|y|^{α}} ≃ |y − x|.

By Lemma 4.1

|B(z, d(z, ζ)| ≃ |y − x|^{2}(|x| + |y − x|)^{α} ≃ |y − x|^{2} |y|^{α},

and, since ps − 1 = p − 2, we get

N(p; U) ≃ ∫_{\{|x|<|y|<δ\}} \frac{|u(x, φ(x)) − u(y, φ(y))|^{p} |x|^{α}}{|y − x|^{p}} dxdy.

By symmetry it suffices to consider the integration on A_{1} := {0 < x < y < δ} and A_{2} := {x > 0, −δ < y < −x}. Set h = y − x and write

\begin{aligned}I_{A_{1}} & = ∫_{\{0<x<y<δ\}} \frac{|u(x, φ(x)) − u(y, φ(y))|^{p} |x|^{α}}{|y − x|^{p}} dxdy \\ & ≤ ∫_{0}^{δ} \frac{dh}{|h|^{p}} ∫_{0}^{δ} |u(x, φ(x)) − u(x + h, φ(x + h))|^{p} |x|^{α} dx. \end{aligned}

We shall connect the points (x, φ(x)) and (x + h, φ(x) + h) by the paths

\begin{aligned}γ_{1}(t) & := exp(t(bX_{1} + X_{2}))(x, φ(x)) \\ & = (x + bt, φ(x) + ∫_{0}^{t} |x + bτ|^{α}dτ) := Φ_{1}(x, t), \end{aligned}

for 0 ≤ t ≤ t_{1} := |h|/b (here b ∈ (0, 1) is a fixed number such that 2^{α+1}cb < 1), and

\begin{aligned}γ_{2}(t) : & = exp(t(X_{2}))(x + h, φ(x + h)) \\ & = (x + h, φ(x + h) + (x + h)^{α}t) := Φ_{2}(x + h, t). \end{aligned}

If t = t_{1}, γ_{1} reaches the height φ(x) + ∫_{0}^{|h|/b}(x + bτ)^{α}dτ. Thus the curve γ_{2} needs the time t_{2} = \frac{1}{(x+h)^{α}} |φ(x) − φ(x + h) + ∫_{0}^{|h|/b}(x + bτ)^{α}dτ| to reach the same height. The hypothesis on φ and (4.30) give the estimate t_{2} ≤ C|h|.

The choice of b ensures that γ_{1}(t) ∈ Ω for all t ∈ (0, t_{1}]. In fact this amounts to

φ(x + bt) < φ(x) + ∫_{0}^{t} |x + bτ|^{α}dτ.

In view of |φ(x + bt) − φ(x)| ≤ cbt(x + bt)^{α} and ∫_{0}^{t}(x + bτ)^{α}dτ ≥ ∫_{0}^{t/2}(x + bτ)^{α}dτ ≥ t/2(x + bt/2)^{α} the inequality is implied by cb(x + bt)^{α} < 1/2(x + bt/2)^{α} which holds true if 2^{α+1}cb < 1.

By the triangle inequality

I_{A_{1}} ≤ C[ ∫_{0}^{δ} \frac{dh}{|h|^{p}} ∫_{0}^{δ} ( ∫_{0}^{C|h|} |Xu(Φ_{1}(x, t))| dt)^{p}|x|^{α}dx+

+ ∫_{0}^{δ} \frac{dh}{|h|^{p}} ∫_{0}^{δ} ( ∫_{0}^{C|h|} |Xu(Φ_{2}(x + h, t))| dt)^{p}|x|^{α}dx]

:= C[I_{A_{1}}^{(1)} + I_{A_{1}}^{(2)}].

Now, by Minkowski and Hardy

\begin{aligned}I_{A_{1}}^{(1)} & ≤ ∫_{0}^{δ} (\frac{1}{|h|} ∫_{0}^{C|h|} ( ∫_{0}^{δ} |Xu(Φ_{1}(x, t))|^{p}|x|^{α}dx)^{1/p}dt)^{p}dh \\ & ≤ C ∫_{(0,δ)×(0,δ)} |Xu(Φ_{1}(x, t))|^{p}|x|^{α}dxdt ≤ C ∫_{Ω} |Xu(x, y)|^{p} dxdy. \end{aligned}

The last inequality follows from the fact that if δ > 0 is small then Φ_{1} is one-to-one, Φ_{1}((0, δ) × (0, δ)) ⊂ Ω and

\frac{∂Φ_{1}(x, t)}{∂t∂x} = ( \begin{matrix}1 & b \\ φ′(x) + \frac{1}{b}[(x + bt)^{α} − x^{α}] & (x + bt)^{α} \end{matrix} ) .

Thus |JΦ_{1}(x, t)| = |x^{α} − bφ′(x)| ≥ |x|^{α} − b|φ′(x)| ≥ (1 − bc)|x|^{α} ≥ (1 − 2^{−(α+1)})|x|^{α}, and the estimate for I_{A_{1}}^{(1)} follows.

Analogously, recalling that t_{2} ≤ C|h| and |x| ≤ |x + h|

I_{A_{1}}^{(2)} ≤ ∫_{0}^{δ} (\frac{1}{h} ∫_{0}^{C|h|} ( ∫_{0}^{δ} |Xu(Φ_{2}(x + h, t))|^{p}|x + h|^{α}dx)^{1/p}dt)^{p}dh

≤ ∫_{0}^{δ} (\frac{1}{|h|} ∫_{0}^{C|h|} ( ∫_{0}^{2δ} |Xu(Φ_{2}(x, t))|^{p}|x|^{α}dx)^{1/p}dt)^{p}dh

≤ C ∫_{(0,2δ)×(0,δ)} |Xu(Φ_{2}(x, t))|^{p}|x|^{α}dxdt.

Since |JΦ_{2}(x, t)| = |x|^{α}, the change of variable (ξ, τ) = Φ_{2}(x, t) ends the estimate for I_{A_{1}}^{(2)}.

The integral on the set A_{2} = {0 < x < δ, −δ < y < −x} can be treated in the same way of I_{A_{1}}, letting y = x + h and using the curves

\begin{aligned}γ_{1}(t) & = exp(t(−bX_{1} + X_{2}))(x, φ(x)), \\ γ_{2}(t) & = exp(tX_{2})(x + h, φ(x + h)). \end{aligned}

□

4.2. Analysis of a counterexample. The hypothesis of α−admissibility for the domain Ω in Theorem 4.5 is necessary. More precisely, there exist domains of class C^{1} that are not α−admissible for which the trace estimate (4.5) fails.

Let α > 0, fix β ∈ (0, α + 1) and consider the domain

Ω = {(x, y) ∈ ℝ^{2} : |x|^{β} < y < 1}.

Except that at the points (±1, 1) the boundary ∂Ω is of class C^{1}. These points are not important, problems stem from the boundary point (0, 0) which is not α−admissible.

We shall consider the case p = 2. As usual write z = (x, y) and ζ = (ξ, η).

Proposition 4.6. Let α > 0 and β ∈ (0, α + 1). There exists γ > 0 such that the function u(x, y) = y^{−γ} satisfies

I := ∫_{Ω} |Xu|^{2} dxdy < +∞

and

N := ∫_{∂Ω×∂Ω} \frac{|u(z) − u(ζ)|^{2}}{d(z, ζ)μ(B(z, d(z, ζ)))}dμ(z)dμ(ζ) = +∞.

Proof. We compute first I. Indeed

I = γ^{2} ∫_{0}^{1} y^{−2γ−2}( ∫_{−y^{1/β}}^{y^{1/β}} |x|^{2α} dx)dy = \frac{2γ^{2}}{2α + 1} ∫_{0}^{1} y^{−2γ−2+(2α+1)/β} dy,

and

I < +∞ ⇔ −2γ − 2 + (2α + 1)/β > −1 ⇔ γ < \frac{2α + 1 − β}{2β} .  (4.35)

Now we shall estimate N but first some remarks on d(z, ζ) and μ(B(z, d(z, ζ))) are in order. Let z = (x, x^{β}) ∈ ∂Ω with 0 < x < 1 and let r > 0. Assume that

r ≥ x^{β/(α+1)}.  (4.36)

¿From (4.36) it follows that x^{β} ≤ r^{α+1} ≤ r(x + r)^{α} and thus x^{β} − r(x + r)^{α} ≤ 0. This means that

Box(z, r) ∩ {y ≤ 0} ≠ ∅,  (4.37)

i.e. the box Box(z, r) meets the lower half plane.

Analogously, since β < α + 1 we find x ≤ x^{β/(α+1)} ≤ r and thus x − r ≤ 0. This means that

Box(z, r) ∩ {x ≤ 0} ≠ ∅,  (4.38)

i.e. the box Box(z, r) meets the left half plane.

We now claim that, for r and x sufficiently small the right part {(t, t^{β}) : 0 < t < 1} of the boundary of Ω meets ∂Box(z, r) at its upper horizontal edge. This is equivalent to show that (x + r)^{β} ≥ x^{β} + r(x + r)^{α}, which holds because

(x + r)^{β} − x^{β} ≥ Cr(x + r)^{β−1} ≥ r(x + r)^{α}

for x, r ≤ σ_{0}, where σ_{0} is a suitable constant (we have used β < α + 1). We also note that the x-coordinate of the intersection point {(t, t^{β}) : 0 < t < 1} ∩ ∂Box(z, r) is (x^{β} + r(x + r)^{α})^{1/β}. Then from (4.37) and (4.38)

\begin{aligned}μ(Box(z, r)) & ≃ μ(Box(z, r) ∩ \{(ξ, η) : ξ ≥ 0\}) \\ & ≃ ∫_{0}^{(x^{β}+r(x+r)^{α})^{1/β}} |ξ|^{β−1} dξ ≃ x^{β} + r(x + r)^{α}. \end{aligned}

Since r ≤ x + r ≤ 2r then x + r ≃ r and μ(Box(z, r)) ≃ x^{β} + r^{α+1}. But r^{α+1} ≤ x^{β} + r^{α+1} ≤ 2r^{α+1} and this proves that if (4.36) holds then

μ(Box(z, r)) ≃ r^{α+1}.  (4.39)

We shall now briefly discuss d(z, ζ) where z = (x, x^{β}) and ζ = (ξ, ξ^{β}). Assume that 0 < x < ξ and that

ξ^{α+1} ≤ ξ^{β} − x^{β}.  (4.40)

¿From (4.29)

d(z, ζ) ≃ (ξ − x) + (ξ^{β} − x^{β})^{1/(α+1)},

and using the equivalence ξ^{β} − x^{β} ≃ (ξ − x)ξ^{β−1} we get

\begin{aligned}d(z, ζ) & ≃ (ξ − x)^{1/(α+1)}((ξ − x)^{α/(α+1)} + ξ^{(β−1)/(α+1)}) \\ & ≃ (ξ − x)^{1/(α+1)}ξ^{(β−1)/(α+1)}. \end{aligned}  (4.41)

In the last equivalence we used again β < α + 1.

Recalling (4.36) and (4.40) we define

\begin{aligned}D = \{(z, ζ) & ∈ ∂Ω × ∂Ω : 0 < x < ξ < σ_{0}, ξ^{α+1} ≤ ξ^{β} − x^{β}, \\ & σ_{0} ≥ d(z, ζ) ≥ x^{β/(α+1)}\}. \end{aligned}

Then, by (4.39) and Lemma 4.1

N ≥ ∫_{D} \frac{|u(z) − u(ζ)|^{2}}{d(z, ζ)μ(B(z, d(z, ζ)))}dμdμ ≃ ∫_{D} \frac{|u(z) − u(ζ)|^{2}}{d(z, ζ)^{α+2}} dμdμ := M.

By (4.41) there is a positive constant k > 0 such that

d(z, ζ) ≥ (\frac{(ξ − x)ξ^{β−1}}{k} )^{1/(α+1)}

and thus {(z, ζ) ∈ ∂Ω×∂Ω : 0 < x < ξ < σ_{0}, ξ^{α+1} ≤ ξ^{β}−x^{β}, (ξ−x)ξ^{β−1} ≥ kx^{β}} ⊂ D. Then, letting

E = {(x, ξ) : 0 < x < ξ < σ_{0}, ξ^{α+1} ≤ ξ^{β} − x^{β}, (ξ − x)ξ^{β−1} ≥ kx^{β}}.

we have

\begin{aligned}M & ≃ ∫_{E} \frac{|x^{−βγ} − ξ^{−βγ}|^{2}|xξ|^{β−1}}{((ξ − x)^{1/(α+1)}ξ^{(β−1)/(α+1)})^{α+2}} dxdξ \\ & ≃ ∫_{E} \frac{(ξ^{βγ} − x^{βγ})^{2}}{x^{2βγ−β+1}ξ^{φ(α,β,γ)}(ξ − x)^{(α+2)/(α+1)}} dxdξ, \end{aligned}

where φ(α, β, γ) = 2βγ − β + 1 + (α + 2)(β − 1)/(α + 1).

In order to separate the integration variables we perform in the last integral the change of variable x = ξt. The integration domain E changes in the following way. The relation 0 < x < ξ < σ_{0} gives 0 < t < 1, the relation (ξ − x)ξ^{β−1} ≥ kx^{β} gives (1 − t) ≥ kt^{β}, and finally the relation ξ^{α+1} ≤ ξ^{β} − x^{β} gives t^{β} ≤ 1 + ξ^{α−β+1} which is implied by the first one. This shows that in the new integral we may integrate on the square {(t, ξ) : 0 < t, ξ < δ} where δ > 0 is a small but positive constant. Thus we find

M ≥ ∫_{0}^{δ} \frac{dξ}{ξ^{φ(α,β,γ)−β+(α+2)/(α+1)}} ∫_{0}^{δ} \frac{(1 − t^{βγ})^{2}}{t^{2βγ−β+1}(1 − t)^{(α+2)/(α+1)}} dt.

If ψ(α, β, γ) := φ(α, β, γ) − β + (α + 2)/(α + 1) ≥ 1 then M = +∞, which implies N = +∞. Now, ψ(α, β, γ) = 2βγ−2β+β(α+2)/(α+1)+1, and hence ψ(α, β, γ) ≥ 1 if and only if γ ≥ α/(2α + 2). Finally

γ ≥ \frac{α}{2(α + 1)} ⇒ N = +∞.  (4.42)

Notice that if β ∈ (0, α + 1) then

\frac{α}{2(α + 1)} < \frac{2α + 1 − β}{2β} ,

and we can therefore choose

γ ∈ [\frac{α}{2(α + 1)}, \frac{2α + 1 − β}{2β} ).

The interval becomes empty when β = α+1, i.e. exactly when the domain Ω becomes α−admissible. With such a choice I < +∞ by (4.35) and N = +∞ by (4.42). □

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Roberto Monti: Dipartimento di Matematica, Università di Trento, Via Sommarive, 14. 38050 Povo (Trento), Italy

  E-mail address: rmonti@science.unitn.it

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

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