Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 27, 2002, 291–306

MAXIMAL INEQUALITY IN (s, m)-UNIFORM DOMAINS

Petteri Harjulehto

University of Helsinki, Department of Mathematics P.O. Box 4 (Yliopistonkatu 5), FIN-00014 Helsinki, Finland; petteri.harjulehto@helsinki.fi

Abstract. We define a class of bounded domains Ω ⊂ R^{n} which we call (s, m)-uniform, s ≥ 1 and 0 < m ≤ 1. In this class we show that every Sobolev function u ∈ W^{1,p}(Ω), 1 ≤ p ≤ ∞, satisfies

|u(x) − u(y)| ≤ C|x − y|^{α}(ℳ ∇u(x) + ℳ ∇u(y))

for almost every x, y ∈ Ω with

α = \frac{m}{s} (n − s(n − 1)).

Our result extends the previous result for Sobolev extension domains by P. Hajłasz. Classical bounded uniform domains or equivalently bounded (ε, ∞) domains form a proper subclass of the (s, m)-uniform domains, when s > 1 or 0 < m < 1, but our class of domains allows more irregular behavior for the boundary than in the classical case.

1. Introduction

P. Hajłasz showed that if Ω ⊂ R^{n} is a Sobolev extension domain or Ω = R^{n}, then every u ∈ W^{1,p}(Ω), 1 ≤ p ≤ ∞, satisfies

(1.1)  |u(x) − u(y)| ≤ C|x − y|^{α}(ℳ ∇u(x) + ℳ ∇u(y))

for almost every x, y ∈ Ω with α = 1, [H2]. Here ℳ ∇u is the Hardy–Littlewood maximal operator of a weak gradient of a function u. Hajłasz and O. Martio proved that under a weak geometric condition the inequality (1.1) with α = 1 implies that the domain Ω is a Sobolev extension domain for 1 < p ≤ ∞, [HM]. A variant of the inequality (1.1) in the domain whose boundary is locally a graph of a Lipschitz continuous function, and also the case Ω = R^{n}, has been studied in [DS], [H1] and [HM].

We define a new class of bounded domains which we call (s, m)-uniform, s ≥ 1 and 0 < m ≤ 1. The special case s = m = 1 is the class of bounded uniform domains defined by Martio and J. Sarvas, [MS] or equivalently the class of bounded (ε, ∞) domains defined by P.W. Jones, [J]. An example of (s, 1)-uniform domains in the plane is an s-cusp, {(x, y) ∈ R^{2} : 0 < x < 1, 0 < y < x^{s}}, with s ≥ 1. The class of (s, m)-uniform domains is a proper subclass of the class of s-John domains. We prove that if Ω is bounded and its boundary is locally a graph of a λ-Hölder continuous function, 0 < λ ≤ 1, then Ω is (1/λ, λ)-uniform. In the case λ = 1 this result seems to be well known, although we have not been able to find a reference. The converse does not hold. There exists a bounded domain which is even a (1, 1)-uniform domain, but whose boundary fails to be a graph of a continuous function.

2000 Mathematics Subject Classification: Primary 46E35.

Our main theorem shows that if Ω ⊂ R^{n} is a bounded (s, m)-uniform domain, 1 ≤ s < n/(n − 1) and 0 < m ≤ 1, then every u ∈ W^{1,p}(Ω), 1 ≤ p ≤ ∞, satisfies the inequality (1.1) for almost every x, y ∈ Ω with α = m(n−s(n−1))/s. Hajłasz and Martio proved the case s = 1, [HM, Lemma 14, p. 243]. Our proof is based on their proof. We calculate an upper bound for the exponent α of the inequality (1.1) in the class of (s, m)-uniform domains: if 1 < s < n/(n − 1) then

0 < α ≤ \frac{s(n − 1) + 1}{n}(n − s(n − 1)) < 1

and if s ≥ n/(n − 1) then the inequality does not hold with any α > 0 for every 1 < p < ∞.

Acknowledgements. I wish to thank my teacher R. Hurri-Syrjänen for her helpful guidance and kind advice.

2. Notation

Throughout this paper C will denote a constant which may change even in a single string of an estimate. We write C(M) to denote that the constant C depends on M. We let Ω and D be bounded domains in the Euclidean n-space R^{n}, n ≥ 2. We denote the boundary of a domain Ω by ∂Ω. By an open ball centered at x and with a radius r > 0 we mean the set B^{n}(x, r) = {y ∈ R^{n} : |y − x| < r}. We write kB for the ball with the same center as B and dilated by a factor k > 0. We let Ā denote the closure of a set A in R^{n}. The Lebesgue n-measure of a set A ⊂ R^{n} is denoted by |A|.

Following J. Väisälä [V] we say that γ is a curve if it is either a path or an arc. A path is a continuous mapping from a closed interval to Ω ⊂ R^{n}. A set in Ω is an arc if it is homeomorphic to a closed interval. We assume that every curve is rectifiable. A length of a curve γ is denoted by |γ|. If γ_{1} is a curve from a point x to a point z and γ_{2} is a curve from a point z to y then by γ_{1} ∪ γ_{2} we denote a curve from x to y via γ_{1} and γ_{2} .

The set of p-integrable functions in D is denoted by L^{p}(D), 1 ≤ p ≤ ∞. We denote by W^{1,p}(D), 1 ≤ p ≤ ∞, the class of all functions in L^{p}(D) whose first weak derivatives are in L^{p}(D). We equip the Sobolev space W^{1,p}(D) with the norm ∥u∥_{W^{1,p}(D)} = ∥u∥_{L^{p}(D)} + ∥∇u∥_{L^{p}(D)} , where ∇u is the weak gradient.

The class of λ-Hölder continuous functions, 0 < λ ≤ 1, in a domain D is denoted by C^{0,λ}(D): u ∈ C^{0,λ}(D) if there exists a constant C > 0 such that

|u(x) − u(y)| ≤ C|x − y|^{λ}

Maximal inequality in (s, m)-uniform domains 293

for every x, y ∈ D. If λ = 1 we say that the function u is a Lipschitz-continuous function.

For a measurable function defined in a set A, |A| > 0, we write

\overline{∫}_{A} u(x) dx = \frac{1}{|A|} ∫_{A} u(x) dx.

Let v ∈ L^{1}(D) and x ∈ D. We put v = 0 in the complement of the domain D. For every 0 < R ≤ ∞ we define

ℳ_{R}v(x) = sup_{0<r<R} \overline{∫}_{B^{n}(x,r)} |v(z)| dz.

We let ℳ u denote ℳ_{∞}u. The operator ℳ is the classical Hardy–Littlewood maximal operator. Recall that for 1 < p ≤ ∞ we have ∥ℳ u∥_{L^{p}(D)} ≤ A∥u∥_{L^{p}(D)} , where the constant A depends only on the dimension n and p, [St, Theorem 1, p. 6].

3. (s, m)-uniform domains

We define a new class of domains. The definition was suggested to the author by P. Hajłasz.

3.1. Definition. Let s ≥ 1 and 0 < m ≤ 1. A bounded domain Ω ⊂ R^{n} is an (s, m)-uniform domain if there exists a constant M ≥ 1 such that each pair x, y of points in Ω can be joined by a rectifiable curve γ: [0, l] → Ω parametrized by arclength, such that γ(0) = x, γ(l) = y,

(3.2)  l ≤ M|x − y|^{m}

and

(3.3)  min(t, l − t)^{s} ≤ M dist(γ(t), ∂Ω).

The idea of (s, m)-uniform domains is that every two points in Ω can be joined by a twisted double cusp inside the domain Ω. The exponent s describes which kind of outer peaks are allowed and the exponent m which kind of inner peaks. The special case s = m = 1 is the class of bounded uniform domains defined by Martio and J. Sarvas, [MS]. The class of bounded uniform domains, and thus the class of (1, 1)-uniform domains, coincides with the class of bounded (ε, ∞) domains defined by P.W. Jones, [J]. It is easy to see that the class of (s, m)- uniform domains is a proper subset of the class of (s′, m′)-uniform domains if s < s′ and m′ ≤ m or if s ≤ s′ and m′ < m. The standard examples in the plane are an s-cusp, {(x, y) ∈ R^{2} : 0 < x < 1, 0 < y < x^{s}}, with s ≥ 1 which is (s, 1)-uniform, and the interior of its complement with respect to the ball B^{2}(0, 1), which is (1, 1/s)-uniform.

We say that ∂Ω is λ-Hölder, 0 < λ ≤ 1, if for every point x ∈ ∂Ω there exists r(x) = (r_{1}(x), …, r_{n}(x)), r_{i}(x) > 0 for every i, and a λ-Hölder continuous function ϕ: R^{n−1} → R such that, upon rotating and relabeling the coordinate axes such that x is at the origin, we have

Ω ∩ U(x, r(x)) = {y ∈ R^{n} : ϕ(y_{1}, …, y_{n−1}) > y_{n}} ∩ U(x, r(x))

and

\frac{1}{2}r_{n}(x) > ϕ > −\frac{1}{2}r_{n}(x)

where U(x, r(x)) = {y ∈ R^{n} : |y_{i} −x_{i}| < r_{i}(x), i = 1, …, n} is an open rectangle. If λ = 1 we say that ∂Ω is Lipschitz.

In the case λ = 1 the following lemma seems to be well known, although we have not been able to find a reference.

3.4. Lemma. Let 0 < λ ≤ 1 and let Ω ⊂ R^{n} be a bounded domain. If ∂Ω is λ-Hölder then the domain Ω is (1/λ, λ)-uniform.

The converse does not hold. There exists even a (1, 1)-uniform domain, whose boundary is not locally a graph of a continuous function at any point. An example is the Koch snowflake domain. In Example 5.2 we construct for every s ≥ 1 an (s, 1)-uniform domain whose boundary fails to be a graph of a continuous function.

Proof. Since ∂Ω is bounded we may choose a finite covering of open rectangles {U(z_{i}, r(z_{i}))}_{i=1}^{k}. Let ϕ_{i} be a λ-Hölder continuous function with a constant L_{i} related to U(z_{i}, r(z_{i})). We write L = max_{1≤i≤k}{L_{i}}. For technical reasons we assume that diam(Ω) = 1.

First we prove that every pair of points inside each U(z_{i}, r(z_{i})) ∩ Ω can be joined by a curve satisfying the conditions (3.2) and (3.3). Let x = (x_{1}, …, x_{n}) and y = (y_{1}, …, y_{n}) be in U(z_{i}, r(z_{i})) ∩ Ω. We fix a two-coordinate axis in R^{n} so that x is the point (0, x_{n}) and y is the point (l, y_{n}),

l = \sqrt{(x_{1} − y_{1})^{2} + … + (x_{n−1} − y_{n−1})^{2}} .

We may assume that x_{n} ≥ y_{n}. Let I_{1} be a curve

{(ξ_{1}, ξ_{2}) : 0 ≤ ξ_{1} ≤ l, ξ_{2} = −Lξ_{1}^{λ} + x_{n}}

and I_{2} a curve

{(ξ_{1}, ξ_{2}) : 0 ≤ ξ_{1} ≤ l, ξ_{2} = −L|ξ_{1} − l|^{λ} + y_{n}},

Figure 1. The curves I_{1} and I_{2} .

The curves I_{1} and I_{2} are presented in Figure 1 with L = 1, λ = 0.5, x_{n} = 1, y_{n} = 0.75 and l = 0.5.

If the curve I_{1} intersects the curve I_{2} , as in Figure 1, we let J be a curve connecting x and y via I_{1} and I_{2} . Let ξ be a point in I_{1} with dist(I_{1}, y) = dist(ξ, y). Otherwise we let J be a curve connecting x to y via I_{1} and a line segment from ξ to y. It is easy to see that l(J) ≤ C|x − y|^{λ} here C is a constant, depending on s, L and diam(Ω), and l(J) is the length of the curve J. Let J^{∗} be a curve from x to y via the curves J_{1}^{∗} = {(ξ_{1}, ξ_{2}) : ξ_{1} = 0, ξ_{2} ≤ x_{n}}, J′ and J_{2}^{∗} = {(ξ_{1}, ξ_{2}) : ξ_{1} = |x − y|, ξ_{2} ≤ y_{n}}. Here J′ is defined as follows: if (ξ_{1}, ξ_{2}) ∈ J then (ξ_{1}, ξ_{2} − \frac{1}{10}|x − y|) ∈ J′. If necessary we replace a part of J^{∗} by a line segment in the hyperplane

\{(ξ_{1}, ξ_{2}) ∈ U(z_{i}, r(z_{i})) ∩ Ω : ξ_{2} = −\frac{3}{4}r_{n}(z_{i})\}.

This yields

dist(ξ, ∂Ω) ≥ C(L)|x_{n} − ξ|^{1/λ}

for every ξ ∈ J_{1}^{∗},

dist(ξ, ∂Ω) ≥ C(L)|y_{n} − ξ|^{1/λ}

for every ξ ∈ J_{2}^{∗} and

dist(ξ, ∂Ω) ≥ min\{\frac{1}{10}|x − y|, \frac{1}{4}r_{n}(z_{i})\}

for every ξ ∈ J′. It is easy to see that J^{∗} satisfies the conditions (3.2) and (3.3) with s = 1/λ, m = λ and a constant M depending on L, diam(U(z_{i}, r(z_{i}))) and r_{n}(z_{i}).

Let 𝒲_{0} be a Whitney composition of Ω, [St, Theorem 1, p. 167]. Let 𝒲 be a collection of cubes Q_{i} from 𝒲_{0} dilated by a factor \frac{9}{8} with Q_{i} ⊄ ⋃_{i=1}^{k} U(x_{i}, r(z_{i})).

There exists ε > 0 depending on the collection {U(z_{i}, r(z_{i}))}_{i=1}^{k} such that for every w ∈ Ω we have B^{n}(w, ε) ⊂ \frac{9}{8}Q_{j} for some \frac{9}{8}Q_{j} ∈ 𝒲 or B^{n}(w, ε) ⊂ U(z_{i}, r(z_{i})) for some i = 1, …, k. Since every cube is a (1/λ, λ)-uniform domain we see that each pair of points x, y ∈ Ω with |x − y| < ε can be joined by a curve satisfying the conditions (3.2) and (3.3) with the constant M.

To complete the proof we use the same method as in [HK1, Theorems 2.4 and 3.3, pp. 175 and 178].

Let x, y ∈ Ω with |x − y| ≥ ε. An elementary covering argument shows that there exists a positive integer N, depending on diam(Ω), ε and n, such that Ω can be covered by balls B_{i}, i = 1, …, N, with radius \frac{1}{4}ε. Now there exists a chain of balls B_{i}, i ∈ {1, …, K} and K ≤ N, such that x ∈ B_{1} , y ∈ B_{K} and B_{i} ∩ B_{i+1} ∩ Ω ≠ ∅ for each j = 1, …, K − 1. We set x = z_{1} , y = z_{K} and choose z_{i} ∈ B_{i} ∩ Ω. Since |z_{i} − z_{i+1}| < ε, there exists a curve γ_{i} joining z_{i} to z_{i+1} in Ω with l(γ_{i}) ≤ M|z_{i} − z_{i+1}|^{λ} < Mε^{λ}. Thus we obtain

l(γ) = l( ⋃_{i=1}^{K} γ_{i}) ≤ KMε^{λ} ≤ KM|x − y|^{λ}.

We choose points w_{1} = x, w_{2}, …, w_{l} = y on the curve γ satisfying

(\frac{ε}{2M} )^{1/λ} ≤ |w_{i} − w_{i+1}| < (\frac{ε}{M} )^{1/λ}

for i = 1, 2, …, l − 1. Let β_{i} be a curve joining w_{i} to w_{i+1} as in the definition of (s, m)-uniform domains, hence l(β_{i}) ≤ M|w_{i} − w_{i+1}|^{λ} < ε and

l( ⋃_{i=1}^{l−1} β_{i}) ≤ \frac{KM|x − y|^{λ}}{(\frac{ε}{2M} )^{1/λ}} ε ≤ 2^{1/λ}KM^{1+1/λ}ε^{1−1/λ}|x − y|^{λ}.

By the definition of (s, m)-uniform domains every curve β_{i} has arclength as its parameter. We choose b_{i} to be the arclength midpoint of β_{i}. Since |b_{i} − b_{i+1}| < ε there exists a curve α_{i} joining b_{i} to b_{i+1} as in the definition of (s, m)-uniform domains. We denote by β_{i}(ξ_{1}, ξ_{2}) that part of the curve β_{i} from the point ξ_{1} to the point ξ_{2} . We write

α = β_{1}(x, b_{1}) ∪ α_{1} ∪ … ∪ α_{l−2} ∪ β_{l−1}(b_{l}, y).

This yields

l(α) ≤ C|x − y|^{λ},

where the constant C depends on M, ε, λ, L, diam(U(z_{i}, r(z_{i}))) and r_{n}(z_{i}) for each i = 1, …, k. Since |β_{i}| ≥ \frac{1}{2}ε and since the point b_{i} is the arclength midpoint of β_{i} we obtain

dist(b_{i}, ∂Ω) ≥ \frac{1}{M} (\frac{1}{4}ε)^{1/λ}.

Hence it is easy to see that the curve α satisfies the conditions (3.2) and (3.3). This completes the proof of Lemma 3.4.

Let s ≥ 1. A domain Ω ⊂ R^{n} is an s-John domain if there exists a distinguished point x_{0} ∈ Ω and a constant C ≥ 1 such that each point x ∈ Ω can be joined to x_{0} by a rectifiable curve γ: [0, l] → Ω parametrized by arclength, such that γ(0) = x, γ(l) = x_{0} ,

l ≤ C

and

t^{s} ≤ C dist(γ(t), ∂Ω).

The definition implies that every s-John domain is bounded. When s = 1 these domains coincide with the class of John domains defined by Martio and Sarvas [MS]. The s-John domains for s > 1 are much wider than John domains. If a domain Ω ⊂ R^{n} is an s-John domain with a distinguished point x_{0} ∈ Ω then it is an s-John also with any other point x ∈ Ω. This means that the distinguished point can be changed. Note that the constant C depends on the distance between the distinguished point and the boundary of Ω. For more information about s-John domains we refer to [SS], [HK2] and [KM].

3.5. Lemma. Let s ≥ 1 and 0 < m ≤ 1. A bounded (s, m)-uniform domain is an s-John domain.

The case s = 1 of Lemma 3.5 is proved by F.W. Gehring and Martio, [GM, Lemma 2.18, p. 209]. The case s > 1 is similar.

4. Main theorem

First we prove a chain condition for (s, m)-uniform domains. This is a modification of the standard chaining argument for uniform domains and John domains, see [HM] and [HK2].

4.1. Lemma. Let Ω ⊂ R^{n} be a bounded (s, m)-uniform domain. Let x, y ∈ Ω. Then there exists a sequence of balls {B_{i}}_{i=−∞}^{∞}, where B_{i} = B^{n}(x_{i}, r_{i}), and constants C, d ≥ 1 with the following properties:

(1) |B_{i} ∪ B_{i+1}| ≤ C|B_{i} ∩ B_{i+1}|,

(2) dist(x, B_{i}) ≤ dr_{i}^{1/s}, B_{i} ⊂ B^{n}(x, C|x−y|^{m/s}) if i ≤ 0 and r_{i} → 0 as i → −∞,

(3) dist(y, B_{i}) ≤ dr_{i}^{1/s}, B_{i} ⊂ B^{n}(y, C|x − y|^{m/s}) if i ≥ 0 and r_{i} → 0 as i → ∞,

(4) no point of the domain Ω belongs to more than C balls B_{i}. The constants depend only on s, m, the dimension n and the uniform constant M of the domain Ω.

Proof. We may assume that diam(Ω) ≤ 1. Fix x, y ∈ Ω and let γ be a curve joining x and y as in the definition of (s, m)-uniform domains, γ(0) = x and γ(l) = y. Fix x_{0} = γ(\frac{1}{2}l). Let B′_{0} = B^{n}(x_{0}, \frac{1}{4} dist(x_{0}, ∂Ω ∪ \{x\})). We let γ′ be the subcurve of γ from x to x_{0} . We cover γ′ \ {x} with balls as follows. Consider the collection of balls \overline{B^{n}}(γ(t), \frac{1}{4} dist(γ(t), ∂Ω ∪ \{x\})), t ∈ (0, \frac{1}{2}l), and \overline{B′_{0}} . By Besicovitch covering theorem [M, Theorem 2.7, p. 30] we find a sequence of closed balls \overline{B′_{0}}, \overline{B′_{1}}, \overline{B′_{2}}, … that cover γ′ \ \{x\} and have uniformly bounded overlap depending only on n.

We define open balls B_{i} = 2B′_{i}, i = 0, 1, 2, …. Here 2B′_{i} is the ball with same center as B′_{i} but twice the radius of the ball B′_{i}. We write x_{i} = γ(t_{i}) and r_{i} = \frac{1}{2} dist(x_{i}, ∂Ω ∪ \{x\}).

If r_{i} = \frac{1}{2}|x_{i} − x| then dist(x, B_{i}) = 2r_{i} ≤ 2r_{i}^{1/s}. If r_{i} = \frac{1}{2} dist(x_{i}, ∂Ω) then the definition of an (s, m)-uniform domain yields

dist(x, B_{i}) ≤ dist(x, x_{i}) ≤ t_{i} ≤ M^{1/s} dist(x_{i}, ∂Ω)^{1/s} ≤ 2M^{1/s}r_{i}^{1/s}.

We choose d = max{2, 2M^{1/s}}. Since r_{i} ≤ t_{i} properties of (s, m)-uniform domains imply

\begin{aligned}dist(x, B_{i}) + 2r_{i} & ≤ dr_{i}^{1/s} + 2r_{i} ≤ (d + 2)r_{i}^{1/s} \\ & ≤ \frac{1}{2}(d + 2)t_{i}^{1/s} ≤ \frac{1}{2}M^{1/s}(d + 2)|x − y|^{m/s}. \end{aligned}

Hence, we obtain B_{i} ⊂ B^{n}(x, C|x − y|^{m/s}) for every i, i = 0, 1, …, where C = \frac{1}{2}M^{1/s}(d + 2).

We renumber the balls. Let B_{0} be as above. If we have chosen balls B_{i}, i = 0, 1, …, m, then we choose a ball B_{m+1} that is the ball for which x_{j} ∈ B_{m} and t_{j} < t_{m}. We recall that γ′(t_{j}) = x_{j} and γ′(t_{m}) = x_{m}. Hence r_{i} → 0 and x_{i} → x, as i → ∞.

Next we prove that every point in the domain Ω belongs to a finite number of balls B_{i} only. The point x does not belong to any ball. Let x′ be an arbitrary point in the domain Ω. Let r = |x′ − x|. The point x′ cannot belong to those balls B_{i} for which r_{i} ≤ \frac{1}{2}|x_{i} − x| < \frac{1}{2}r. If x′ ∈ B_{i} then dist(x, B_{i}) < r and furthermore |x − x_{i}| ≤ 2r. Thus we obtain that if x′ ∈ B_{i} then \frac{1}{2}r ≤ r_{i} ≤ r. The construction of the Besicovitch covering theorem [M, Theorem 2.7, p. 30] implies that balls with radius of \frac{1}{4} of original balls are disjoint. Thus x′ belongs to less than or equal to

C \frac{|B^{n}(x′, 2r)|}{|B^{n}(0, \frac{1}{8}r)|} = 16^{n}C

balls B_{i}. The constant C is from the Besicovitch covering theorem.

Finally we prove the property (1). Assume that r_{i} = \frac{1}{2} dist(x_{i}, ∂Ω) and r_{i+1} = \frac{1}{2} dist(x_{i+1}, ∂Ω). Since x_{i+1} ∈ B(x_{i}, r_{i}) we obtain dist(x_{i+1}, ∂Ω) ≥ r_{i}. This yields

\frac{|B_{i}|}{|B_{i+1}|} ≤ ( \frac{r_{i}}{\frac{1}{2}r_{i}} )^{n} = 2^{n}.

If r_{i} = \frac{1}{2}|x_{i} − x| and r_{i+1} = \frac{1}{2}|x_{i+1} − x| then

\frac{|B_{i}|}{|B_{i+1}|} ≤ ( \frac{r_{i}}{\frac{1}{2}r_{i}} )^{n} = 2^{n}.

If r_{i} = \frac{1}{2} dist(x_{i}, ∂Ω) and r_{i+1} = \frac{1}{2}|x_{i+1} − x| we obtain

\frac{|B_{i}|}{|B_{i+1}|} = (\frac{r_{i}}{r_{i+1}} )^{n} ≤ ( \frac{\frac{1}{2}|x_{i} − x|}{r_{i+1}})^{n} = 2^{n}.

Similarly if r_{i} = \frac{1}{2}|x_{i} − x| and r_{i+1} = \frac{1}{2} dist(x_{i+1}, ∂Ω) then

\frac{|B_{i}|}{|B_{i+1}|} ≤ 2^{n}.

We have proved that |B_{i}| ≤ 2^{n}|B_{i+1}|. Similar arguments imply that |B_{i}| ≥ 3^{−n}|B_{i+1}|. This yields |B_{i} ∪ B_{i+1}| ≤ C|B_{i} ∩ B_{i+1}|; here the constant C depends only on the dimension n.

Using again the same arguments for the point y imply Lemma 4.1.

Next we prove our main theorem. In the proof we need only the chain of balls constructed in Lemma 4.1, the Lebesgue differentiation theorem, the Poincaré inequality in a ball and properties of the Riesz potential.

4.2. Theorem. Let 1 ≤ s < n/(n − 1), 0 < m ≤ 1 and 1 ≤ p ≤ ∞. If Ω ⊂ R^{n} is a bounded (s, m)-uniform domain then there exists a constant C > 0 such that every u ∈ W^{1,p}(Ω) satisfies the inequality

(4.3)  |u(x) − u(y)| ≤ C|x − y|^{α}(ℳ ∇u(x) + ℳ ∇u(y)),

for almost every x, y ∈ Ω with α = m(n − s(n − 1))/s. Here ℳ ∇u is the Hardy– Littlewood maximal operator of the function ∇u. The constant C depends only on n, s, m and the uniform constant of Ω.

Hajłasz and Martio proved that if Ω ⊂ R^{n} is a bounded uniform domain then every u ∈ W^{1,p}(Ω) satisfies the inequality (4.3) for every 1 ≤ p ≤ ∞, with α = 1, [HM, Lemma 14, p. 243]. Our proof is a modification of the proof of Hajłasz and Martio.

Proof. We may assume that diam(Ω) ≤ 1. Let {B_{i}}_{i=−∞}^{∞} be a chain of balls from the point x ∈ Ω to the point y ∈ Ω as in Lemma 4.1. Then by the Lebesgue differentiation theorem [St, Chapter 1, Section 1.8] we have u_{B_{i}} → u(x), whenever i → −∞, and u_{B_{i}} → u(y), whenever i → ∞, for almost every x, y ∈ Ω. Thus we have

|u(x) − u(y)| ≤ ∑_{i=−∞}^{∞} |u_{B_{i}} − u_{B_{i+1}}|

≤ ∑_{i=−∞}^{∞}(|u_{B_{i}} − u_{B_{i}∩B_{i+1}}| + |u_{B_{i+1}} − u_{B_{i}∩B_{i+1}}|)

≤ ∑_{i=−∞}^{∞}(\overline{∫}_{B_{i}∩B_{i+1}} |u − u_{B_{i}}| + \overline{∫}_{B_{i}∩B_{i+1}} |u − u_{B_{i+1}}|)

and furthermore by Lemma 4.1

|u(x) − u(y)| ≤ ∑_{i=−∞}^{∞}(\frac{1}{|B_{i} ∩ B_{i+1}|} ∫_{B_{i}∩B_{i+1}} |u − u_{B_{i}}|

+ \frac{1}{|B_{i} ∩ B_{i+1}|} ∫_{B_{i}∩B_{i+1}} |u − u_{B_{i+1}}|)

≤ ∑_{i=−∞}^{∞}(\frac{1}{|B_{i} ∩ B_{i+1}|} ∫_{B_{i}} |u − u_{B_{i}}|

+ \frac{1}{|B_{i} ∩ B_{i+1}|} ∫_{B_{i+1}} |u − u_{B_{i+1}}|)

≤ ∑_{i=−∞}^{∞}(\frac{C}{|B_{i}|} ∫_{B_{i}} |u − u_{B_{i}}| + \frac{C}{|B_{i+1}|} ∫_{B_{i+1}} |u − u_{B_{i+1}}|)

≤ 2 · C ∑_{i=−∞}^{∞} \overline{∫}_{B_{i}} |u − u_{B_{i}}|.

The Poincaré inequality in a ball with a radius r_{i}, [GT, 7.45, p. 157], yields

|u(x) − u(y)| ≤ C ∑_{i=−∞}^{∞} r_{i} \overline{∫}_{B_{i}} |∇u| ≤ C ∑_{i=−∞}^{∞} ∫_{B_{i}} \frac{|∇u|}{r_{i}^{n−1}} .

Lemma 4.1 implies that for each z ∈ B_{i}, |x − z| ≤ (d + 2)r_{i}^{1/s} and B_{i} ⊂ B^{n}(x, C|x − y|^{m/s}), when i ≤ 0 and |y − z| ≤ (d + 2)r_{i}^{1/s} and, when i ≥ 0, B_{i} ⊂ B^{n}(y, C|x − y|^{m/s}). We obtain

|u(x) − u(y)| ≤ C ∑_{i=−∞}^{0} ∫_{B_{i}} \frac{|∇u(z)|}{|x − z|^{s(n−1)}} dz + C ∑_{i=0}^{∞} ∫_{B_{i}} \frac{|∇u(z)|}{|y − z|^{s(n−1)}} dz

≤ C ∫_{B^{n}(x,C|x−y|^{m/s})} \frac{|∇u(z)|}{|x − z|^{s(n−1)}} dz

+ C ∫_{B^{n}(y,C|x−y|^{m/s})} \frac{|∇u(z)|}{|y − z|^{s(n−1)}} dz.

We put |∇u| = 0 in the complement of the domain Ω. Since s(n − 1) < n we obtain by [Z, Lemma 2.8.3, p. 85] that

|u(x) − u(y)| ≤ C(|x − y|^{m(n−s(n−1))/s}ℳ_{C|x−y|^{m/s}}∇u(x)
+ |x − y|^{m(n−s(n−1))/s}ℳ_{C|x−y|^{m/s}}∇u(y))
= C|x − y|^{m(n−s(n−1))/s}(ℳ_{C|x−y|^{m/s}}∇u(x) + ℳ_{C|x−y|^{m/s}}∇u(y)).

This completes the proof of Theorem 4.2.

5. Sharpness of Theorem 4.2

Assume that a bounded domain Ω ⊂ R^{n} satisfies the inequality (4.3) for all 1 < p < ∞ with some exponent α > 0. We obtain by the inequality (4.3) that

(5.1)

|u(x) − \overline{∫}_{Ω} u(y) dy| ≤ \overline{∫}_{Ω} |u(x) − u(y)| dy
≤ C diam(Ω)^{α}(M∇u(x) + \overline{∫}_{Ω} M∇u(y) dy)

≤ C diam(Ω)^{α}(M∇u(x) + (\overline{∫}_{Ω}(M∇u(y))^{p} dy)^{1/p})

and the boundedness of the Hardy–Littlewood maximal operator, [St, Theorem 1, p. 6], yields

∥u − u_{Ω}∥_{L^{p}(Ω)} ≤ C diam(Ω)^{α}∥M∇u∥_{L^{p}(Ω)} ≤ C diam(Ω)^{α}∥∇u∥_{L^{p}(Ω)}

as in [H2, Lemma 2, p. 407]. Thus Theorem 4.2 implies that a bounded (s, m)- uniform domain Ω ⊂ R^{n}, 1 ≤ s < n/(n − 1) and 0 < m ≤ 1, is a p-Poincaré domain for every 1 < p < ∞. W. Smith and D. Stegenga showed that an s-John domain is a p-Poincaré domain for every 1 < p < ∞, if 1 ≤ s ≤ n/(n − 1), [SS, Theorem 10, p. 86]. Hajłasz and Koskela proved with a “mushroom” example that the limit is sharp in the sense that s cannot be greater than n/(n − 1), [HK2, Corollary 6].

We show that if s > n/(n − 1) then an (s, 1)-uniform domain is not necessarily a p-Poincaré domain for every 1 < p < ∞. The following rooms and passages example is by R. Hurri [Hu, Chapter 5, p. 17].

5.2. Example. Let Ω = ⋃_{i=1}^{∞}(R_{2i−1} ∪ P_{2i}), where the sets R_{2i−1} and P_{2i} are defined as follows. Let a ≥ 1. Let h_{i} = 2^{−i}, δ_{2i} = 2·2^{−2ai} and d_{i} = ∑_{j=1}^{i} 2^{−j} for every i = 1, 2, …. We define

\begin{aligned}R_{2i−1} & = (d_{2i−1} − h_{2i−1}, d_{2i−1}) × (−\frac{1}{2}h_{2i−1}, \frac{1}{2}h_{2i−1})^{n−1}, \\ P_{2i} & = [d_{2i−1}, d_{2i−1} + h_{2i}] × (−\frac{1}{2}δ_{2i}, \frac{1}{2}δ_{2i})^{n−1}. \end{aligned}

By Hurri [Hu, Remark 5.9, p. 19] the domain Ω is a p-Poincaré domain if and only if p ≥ (n − 1)(a − 1).

Since there exists a constant C > 0 so that \frac{1}{2}δ_{2i} ≥ C(1 − d_{2i−1})^{a} for every i = 1, 2, …, the domain Ω is an (a, 1)-uniform domain. Let ε > 0 be arbitrary. If a = (n/(n − 1)) + ε, then the domain Ω is not a p-Poincaré domain for any 1 ≤ p < 1 + ε(n − 1).

5.3. Corollary. Let s > n/(n − 1) and 0 < m ≤ 1. There exists a bounded (s, m)-uniform domain where the inequality (4.3) does not hold for all 1 < p < (s − 1)(n − 1) with any α > 0.

Proof. Let ε > 0. Let Ω ⊂ R^{n} be the bounded (s, m)-uniform domain, s = (n/(n − 1)) + ε and m = 1, constructed in Example 5.2. Assume that there exist constants C, α > 0 such that for every u ∈ W^{1,p}(Ω), 1 < p < ∞, we have

(5.4)  |u(x) − u(y)| ≤ C|x − y|^{α}(ℳ ∇u(x) + ℳ ∇u(y)),

for almost every x, y ∈ Ω. As in (5.1) this implies that the domain Ω is a p-Poincaré domain for all 1 < p < ∞.

In Example 5.2 we showed that the domain Ω is not a p-Poincaré domain for any 1 < p < 1 + ε(n − 1). Thus the inequality (5.4) cannot hold for all 1 < p < 1 + ε(n − 1) with any α > 0 in the domain Ω.

Following Hajłasz, [H2], we say that a domain D is δ -regular, δ > 0, if there exists a constant b > 0 such that

(5.5)  |B^{n}(x, r) ∩ D| ≥ br^{δ}

for every x ∈ D and for every 0 < r ≤ diam(D). It is easy to see that every bounded (s, m)-uniform domain is (s(n − 1) + 1)-regular.

Using the method of Hajłasz, [H2, Theorem 6, p. 410], it is easy to prove the following Sobolev–Poincaré inequality. In the proof we need only the inequality (4.3) and the property (5.5).

5.6. Lemma. Assume that Ω ⊂ R^{n} is a bounded δ -regular domain, δ > 1, which satisfies the inequality (4.3) with an exponent 0 < α ≤ 1. If 1 < p < δ/α, then for every u ∈ W^{1,p}(Ω) we have

(5.7)  ∥u − u_{Ω}∥_{L^{p^{∗}} (Ω)} ≤ C∥∇u∥_{L^{p}(Ω)},

with p^{∗} = δp/(δ − αp).

Proof. We may assume that diam(Ω) ≤ 1. Let

E_{k} = {x ∈ Ω : M∇u(x) ≤ 2^{k}}, k ∈ Z.

There exists a constant C > 0 such that

(5.8) C^{−1} ∑_{i=−∞}^{∞} 2^{kp}|E_{k} \ E_{k−1}| ≤ ∫_{Ω} M|∇u|^{p} dx ≤ C ∑_{i=−∞}^{∞} 2^{kp}|E_{k} \ E_{k−1}|.

Let a_{k} = ess sup_{x∈E_{k}}|u(x)|. We will estimate a_{k} in terms of a_{k−1} . Let x ∈ E_{k}. Let B^{n}(x, r) be a ball with a radius r = 2b^{−1/δ}|Ω \ E_{k−1}|^{1/δ}. We obtain by the δ -regularity property (5.5)

|B^{n}(x, r) ∩ Ω| ≥ br^{δ} > |Ω \ E_{k−1}|.

Hence there exists y ∈ B^{n}(x, r) ∩ E_{k−1} . By the inequality (4.3) the function u|_{E_{k}} is α-Hölder continuous with a constant C2^{k+1} . We obtain

|u(x)| ≤ |u(x)−u(y)|+|u(y)| ≤ C|x−y|^{α}2^{k+1}+a_{k−1} ≤ C|Ω\E_{k−1}|^{α/δ}2^{k+1}+a_{k−1}.

The definition of E_{k} yields

(5.9)  |Ω \ E_{k−1}|2^{kp} ≤ C∥ℳ ∇u∥_{L^{p}(Ω)}^{p};

hence we obtain that

(5.10)  \begin{aligned}a_{k} & ≤ C2^{−kpα/δ}∥ℳ ∇u∥_{L^{p}(Ω)}^{pα/δ}2^{k+1} + a_{k−1} \\ & ≤ C2^{k(1−(pα/δ))}∥ℳ ∇u∥_{L^{p}(Ω)}^{pα/δ} + a_{k−1}. \end{aligned}

We may assume that ℳ ∇u(x) > 0 for every x ∈ Ω since otherwise |∇u| = 0 which implies that u is a constant function almost everywhere in Ω. Let b_{k} = ess inf_{x∈E_{k}}|u(x)|. It is clear that b_{k} ≤ ∥u∥_{L^{p}(Ω)}|E_{k}|^{−1/p}. Since ℳ ∇u > 0 everywhere then there exists k_{0} such that |E_{k_{0}−1}| < \frac{1}{2}|Ω| and |E_{k_{0}}| ≥ \frac{1}{2}|Ω|. We obtain by the inequality (5.9) that

2^{k_{0}} ≤ C∥ℳ ∇u∥_{L^{p}(Ω)}|Ω \ E_{k_{0}−1}|^{−1/p}.

Since the function u|_{E_{k}} is α-Hölder continuous with a constant C2^{k+1} we obtain a_{k} ≤ b_{k} + 2^{k+1} diam(Ω)^{α}. This yields

(5.11) a_{k_{0}} ≤ ∥u∥_{L^{p}(Ω)}|E_{k_{0}}|^{−1/p} + C diam(Ω)^{α}∥ℳ ∇u∥_{L^{p}(Ω)}|Ω|^{−1/p}

≤ C|Ω|^{−1/p}(∥u∥_{L^{p}(Ω)} + diam(Ω)^{α}∥ℳ ∇u∥_{L^{p}(Ω)}).

Since p < δ/α, it follows, for k > k_{0} , by the inequality (5.10) and the monotonicity of a_{k} that

(5.12)  \begin{aligned}a_{k} & ≤ C∥ℳ ∇u∥_{L^{p}(Ω)}^{pα/δ}( ∑_{i=k_{0}}^{k} 2^{i(1−(pα/δ))}) + a_{k_{0}} \\ & ≤ C∥ℳ ∇u∥_{L^{p}(Ω)}^{pα/δ}( ∑_{i=−∞}^{k} 2^{i(1−(pα/δ))}) + a_{k_{0}} \\ & ≤ C∥ℳ ∇u∥_{L^{p}(Ω)}^{pα/δ}2^{k(1−(pα/δ))} + a_{k_{0}}. \end{aligned}

Since p^{∗} = pδ/(δ − αp) the inequalities (5.8), (5.11), (5.12) and the regularity property (5.5) yield that

(∫_{Ω} |u|^{p^{∗}})^{1/p^{∗}} ≤ ( ∑_{k=k_{0}+1}^{∞} a_{k}^{p^{∗}}|E_{k} \ E_{k−1}| + a_{k_{0}}^{p^{∗}}|E_{k_{0}}|)^{1/p^{∗}}

≤ C(∥ℳ ∇u∥_{L^{p}(Ω)}^{pαp^{∗}/δ} ∑_{k=−∞}^{∞}2^{k(1−(pα/δ))p^{∗}}|E_{k} \ E_{k−1}| + a_{k_{0}}^{p^{∗}}|Ω|)^{1/p^{∗}}

≤ C(∥ℳ ∇u∥_{L^{p}(Ω)}^{pαp^{∗}/δ}∥ℳ ∇u∥_{L^{p}(Ω)}^{p}

+ (C|Ω|^{−1/p}(∥u∥_{L^{p}(Ω)} + diam(Ω)^{α}∥ℳ ∇u∥_{L^{p}(Ω)})^{p^{∗}}|Ω|)^{1/p^{∗}}

≤ C(∥u∥_{L^{p}(Ω)} + ∥ℳ ∇u∥_{L^{p}(Ω)}).

Since u−u_{Ω} ∈ W^{1,p}(Ω), Ω is a p-Poincaré domain and the Hardy–Littlewood maximal operator is bounded, [St, Theorem 1, p. 5], we obtain

∥u − u_{Ω}∥_{L^{p^{∗}} (Ω)} ≤ C(∥u − u_{Ω}∥_{L^{p}(Ω)} + ∥ℳ ∇(u − u_{Ω})∥_{L^{p}(Ω)} ≤ C∥∇u∥_{L^{p}(Ω)}.

We write δ = s(n − 1) + 1. Hajłasz and P. Koskela have proved the inequality (5.7) for s-John domains with a better exponent. Let Ω ⊂ R^{n} be an s-John domain, s ≥ 1, then the inequality (5.7) holds with 1 ≤ p ≤ p^{∗} ≤ np/(δ − p), [HK2, Corollary 6, p. 20]. The limiting case p^{∗} = np/(δ − p) is by T. Kilpeläinen and J. Malý [KM]. The exponent is the best possible in the class of s-John domains, [HK2]. It is also the best possible in the class of (s, m)-uniform domains. Let s > 1. Using the (s, 1)-uniform domain constructed by Hurri, see Example 5.2, we obtain as in [Hu, Remark 5.8, p. 19], by replacing the exponent −n/p by the exponent −n/p^{∗}, that the exponent np/(δ − p) is the best possible.

5.13. Corollary. Let Ω ⊂ R^{n} be a bounded (s, m)-uniform domain, with 1 < s < n/(n − 1) and 0 < m ≤ 1. If there exists an α > 0 such that the inequality (4.3) holds for all 1 < p < ∞ then

α ≤ \frac{s(n − 1) + 1}{n}(n − s(n − 1)) < 1.

If s = n/(n − 1) then Ω does not satisfy the inequality (4.3) for all 1 < p < ∞ with any α > 0.

Proof. Let 1 ≤ s < n/(n − 1). Lemma 5.6 shows that the inequality (4.3) with an exponent α > 0 and the δ -regular property (5.5), δ = s(n−1)+1, implies the Sobolev–Poincaré inequality with p^{∗} = δp/(δ − αp).

The exponent δp/(δ − αp) has to be less than or equal to the best possible exponent np/(δ − p) for every 1 < p < ∞. This gives

α ≤ \frac{δ}{np}(n − δ + p)

for every 1 < p < ∞. As p → 1 we see that

α ≤ \frac{δ}{n}(n − δ + 1).

Let s = n/(n − 1). Assume that Ω is a bounded (s, m)-uniform domain which satisfies the inequality (4.3) with some α > 0 for every 1 < p < ∞. By Lemma 5.6 we obtain that Ω satisfies the Sobolev–Poincaré inequality with (n + 1)p/(n + 1 − αp). Thus we obtain

α ≤ (1 + \frac{1}{n})(1 − \frac{1}{p})

for every 1 < p < ∞. As p → 1 we see that α ≤ 0. Hence the domain Ω cannot satisfy the inequality (4.3) with any α > 0 for small p > 1. This completes the proof of Corollary 5.13.

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Received 6 June 2001