Uniformly perfect sets, Hausdorff dimension, and conformal capacity

Using the definition of uniformly perfect sets in terms of convergent sequences, we apply lower bounds for the Hausdorff content of a uniformly perfect subset $E$ of $\mathbb{R}^n$ to prove new explicit lower bounds for the Hausdorff dimension of $E.$ These results also yield lower bounds for capacity test functions, which we introduce, and enable us to characterize domains of $\mathbb{R}^n$ with uniformly perfect boundaries. Moreover, we show that an alternative method to define capacity test functions can be based on the Whitney decomposition of the domain considered.


Introduction
Conformal invariants, in particular the modulus of a curve family and the conformal capacity of a condenser, are fundamental tools of geometric function theory and quasiconformal mappings [9,10,11,13,32].For applications, many upper and lower bounds for conformal invariants have been derived in terms of various geometric functionals.All this research shows that the metric structure of the boundary has a strong influence on the intrinsic geometry of the domain of the mappings studied.Indeed, many results originally proven for functions defined in the unit ball B n of R n , n ≥ 2, can be extended to the case of subdomains G ⊂ R n if the boundary ∂G is "thick enough" in the sense of capacity, or, more precisely, if the boundary is uniformly perfect.The thickness of the boundary has a strong influence on the intrinsic geometry of the domain and thus it also gives a restriction on the oscillation of a function defined in G.We give several new characterizations of uniformly perfect sets.
A condenser is a pair (G, E) where G ⊂ R n , n ≥ 2, is a domain and E ⊂ G is a compact set [9,11,13].A compact set E ⊂ R n is of conformal capacity zero if, for some closed ball B ⊂ R n \ E , the condenser (R n \ B, E) has capacity zero, written as cap (E) = 0 with notations of Definition 4.1.Sets of capacity zero are very thin, their Hausdorff dimensions are zero [27,p.120,Cor. 2], [13,Lemma 9.11], and they often have the role of negligible exceptional sets in potential theory or geometric function theory.Note that, due to the Möbius invariance of the conformal capacity, the notions of zero and positive capacity immediately extend to compact subsets of the Möbius space R n = R n ∪ {∞}.
Here our goal is to study those subsets of R n that have a positive capacity instead.However, the structure of sets of positive capacity can sometimes be highly dichotomic, for instance, in the case of E 1 ∪ E 2 where E 1 ⊂ R 2 is a segment and E 2 is a point not contained in E 1 .This kind of a dichotomy makes working with these sets difficult, but a subclass of sets with positive capacity, uniformly perfect sets, has certain natural properties useful for our purposes.During the past two decades, uniformly perfect sets have become ubiquitous for instance in geometric function theory [3,10], analysis on metric spaces [14,19], hyperbolic geometry [4,17] and in the study of complex dynamics and Kleinian groups [7,28].
We begin by giving a variant of the definition of uniformly perfect sets in terms of convergent sequences as follows.For 0 < c < 1, let UP n (c) denote the collection of compact sets E in R n with card (E) ≥ 2 satisfying the condition {x ∈ E : cr < |x − a| < r} ∅ for all a ∈ E and 0 < r < d(E)/2 .We say that a set is uniformly perfect if it is in the class UP n (c) for some c ∈ (0, 1).
The first lemma has an important role in the sequel.Moreover, the Hausdorff dimension dim H (E) of E is at least (log 2)/ log(3/c).
The explicit bounds we obtain in this paper depend on the above result of Sugawa, however, we prove it here with the above refined form of the constant β.Moreover, we also apply ideas from the work of Reshetnyak [26,27] and Martio [20], see also Remark 5.4, but now the constants are explicit which is crucial for what follows.In the study of uniformly perfect sets, similar methods were also applied by Järvi and Vuorinen [15, Thm 4.1, p. 522].Our results here yield explicit constants for several characterizations of uniform perfectness such as the following main result.Theorem 1.2.Let E ∈ UP n (c) for some 0 < c < 1.Then for every a ∈ E, a ∞, and all r ∈ (0, d(E)) the following lower bound for the conformal capacity cap(a, E, r) of the condenser (B n (a, 2r), E ∩ B n (a, r)) holds cap(a, E, r) ≥ where M 1 (n, β), given in (5.10), is an explicit constant depending only on n and β.
There are many equivalent definitions of uniformly perfect sets.The definition given in [15] says that a setE is α-uniformly perfect, if the moduli of the ring domains separating the set have the upper bound α.We show in Section 6 that this definition is quantitatively equivalent with the definition of UP n (c).Moreover, we prove this equivalence with explicite constants, a fact which would enable one to give explicit constants for instance in some earlier results such as in [15,Thm 4.1].
Suppose now that G ⊂ R n is a domain, its boundary ∂G is of positive capacity, and define u α : G → (0, ∞) by for z ∈ G where d(z, ∂G) is the distance from z to ∂G .We call u α (z) the capacity test function of G at the point z ∈ G .The numerical value of the capacity test function depends clearly on G, z, and α , but we omit G from the notation because it is usually understood from the context.Clearly, the capacity test function is invariant under similarity transformations.We will also show that it is continuous as a function of both z ∈ G and α ∈ (0, 1) .For the purpose of this paper, it is enough to choose e.g.
Analysing the capacity test function u α (z) further, we show that, for a fixed α, it satisfies the Harnack inequality as a function of z, a property which has a number of consequences.First, for every z 0 ∈ G, we see that u α (z 0 ) > 0 , because the boundary was assumed to be of positive capacity.Second, by fixing z 0 ∈ G, we see by a standard chaining argument [13, p. 96 Lemma 6.23 and p. 84] that u α (z)/u α (z 0 ) has a positive explicit minorant for a large class of domains, so called ϕ-uniform domains.This minorant, depending on ϕ, d(z, ∂G) and Harnack parameters, shows that u α (z) cannot approach 0 arbirarily fast when z moves far away from z 0 or when z → ∂G .Under the stronger requirement that ∂G be uniformly perfect, it follows that u α (z) is bounded from below by a constant c > 0. These observations lead to the following new characterization of uniformly perfect sets.
Theorem 1.3.The boundary of a domain G ⊂ R n is uniformly perfect if and only if there exists a constant s > 0 such that u α (x) ≥ s for all x ∈ G.
Many characterizations are known for plane domains with uniformly perfect boundaries and often these characterizations are given in terms of hyperbolic geometry [10, pp. 342-344], [17], [4].Because the hyperbolic geometry cannot be used in dimensions n ≥ 3, we use here another tool, the Whitney decomposition of a domain G ⊂ R n , which has numerous applications to geometric function theory and harmonic analysis [6,10,29].The Whitney decomposition represents G as a countable union of non-overlapping cubes with edge lengths equal to 2 −k , k ∈ Z, where the edge length is proportial to the distance from a cube to the boundary of the domain [29].Martio and Vuorinen [21] applied this decomposition to establish upper bounds for the metric size of the boundary ∂G in terms of N k , the number of cubes with edge length equal to 2 −k .Their method was based on imposing growth bounds for N k when k → ∞ and depending on the growth rate, the conclusion was either an upper bound for the Minkowski dimension of the boundary or a sufficient condition for the boundary to be of capacity zero.In our next main result, Theorem 1.4, we use Whitney decomposition "in the opposite direction".Indeed, we employ Whitney cubes as test sets for the capacity structure of the boundary and obtain the following characterization of uniform perfectness in all dimensions n ≥ 2. Whitney cubes also have applications to the study of surface area estimation of the level sets of the distance function [16].
Theorem 1.4.The boundary of a domain G ⊂ R n is uniformly perfect if and only if there exists a constant s > 0 such that, for every Whitney cube By definition, see the property 8.1(3) below, every Whitney cube where s > 0 is a constant depending only on the dimension n and the uniform perfectness parameters of E and ∂G.Acknowledgements.The authors are grateful to Prof. Akihiro Munemasa for information about the kissing numbers of combinatorial geometry.We are indebted to Prof. Don Marshall for a permission to use his software for plotting Whitney cubes.The second author would like to express his thanks to the Department of Mathematics and Statistics, University of Turku, for its hospitality and support during the visit to Turku, Finland in 2022.

Preliminary results
In this preliminary section we recall some basic facts about metrics and quasiconformal homeomorphisms.Moreover, we prove a few propositions which are results of technical character, essential for the proofs of the main theorems in subsequent sections.
The following notations will be used: The Euclidean diameter of the non-empty set J is d(J) = sup{|x − y| | x, y ∈ J}.The Euclidean distance between two non-empty sets y ∈ K} and the distance from a point x to the set J is d(x, J) = d({x}, J).Thus, for all points x in a domain G ⊊ R n , the Euclidean distance from x to the boundary ∂G is denoted by d(x, ∂G), the Euclidean open ball with a center x and a radius r by B n (x, r) = {y ∈ R n : |x − y| < r}, the corresponding closed ball by B n (x, r) and their boundary by S n−1 (x, r) = ∂B n (x, r).If the center x or the radius r are not otherwise specified, assume that x = 0 and r = 1.The unit ball is denoted by B n .We denote by Ω n the volume of the unit ball B n and by ω n−1 the area of the unit sphere S n−1 .As well known, ω n−1 = nΩ n = 2π n/2 /Γ (n/2).For x ∈ R n , 0 < a < b, we use the following notation for an annulus centered at x The first proposition shows that an annulus and its translation are both subsets of a larger annulus.This larger annulus is a superannulus for both of the two smaller annuli, i.e. the smaller annuli are separating the two boundary components of the larger annulus (Fig. 1).
Figure 1.An annulus centered at x 1 and its translation centered at x 2 (marked with solid and dash-dot markers, resp.) are both subsets of a larger annulus (marked with thick marker) centered at the midpoint w = (x 1 + x 2 )/2.The larger annulus is a common superannulus of the two smaller annuli.
Proof.The claims follow from the triangle inequality and we prove here only the second one.Without loss of generality we may assume that j = 1.Fix u ∈ R(x 1 , τb, a/τ).
where the last inequality holds because τ ≥ 2. Similarly, . The last assertion also follows easily from the triangle inequality.□ Topology in R n or in its one point compactification R n = R n ∪ {∞} is the metric topology defined by the chordal metric.The chordal (spherical) metric is the function Thus, for instance, an unbounded domain G ⊂ R n has ∞ as one of its boundary points.
For the sake of convenient reference we record the following simple inequality.
Proof.The inequality clearly holds for all large values of t, larger than |x|, where x is a root of (x − a) 2 − λ 2p (x + 1) = 0.By the quadratic formula we see that both roots have absolute value less than λ 2p + λ p √ 1 + a + a. □ For two points in a domain, the next result gives an estimate for the number of annular domains separating these points.This result has an important role in the proof of one of our main results in Section 7.
log λ holds.To find a lower bound for p, we observe that the inequality By Proposition 2.3, this holds for all u > λ 2 + 2λ + 2. By (2.5), we see that this yields also the desired lower bound for p. □ 2.6.Quasiconformal maps and moduli of curve families.Quasiconformal homeomorphisms f : commonly defined in terms of moduli of curve families.For the basic properties of the modulus M(Γ ) of a curve family Γ , the reader is referred to [2,14,11], [32, 6.1, p. 16], [13].According to Väisälä's book [32], K-quasiconformal maps are characterized by the inequality The following monotonicity of the moduli of curve families is quite useful in various estimates.Let Γ 1 , Γ 2 be two curve families in R n .We say that Γ 2 is minorized by Γ 1 and write Γ 2 > Γ 1 for it, if every curve γ ∈ Γ 2 has a subcurve belonging to Γ 1 .For instance, (2) For a collection of curve families Moreover, equality holds if the curve families are separate.
Here the families Γ i are said to be separate if they are contained in pairwise disjoint Borel sets E i (see also [11, §4.2.2]).
Let G be a domain in R n and E, F ⊂ G.In what follows, ∆(E, F; G) will stand for the family of all the curves that are in G except for the endpoints, and that have one endpoint in the set E and another endpoint in F [32, pp. 11-25], [13].
where ω n−1 is the (n − 1)-dimensional surface area of the unit sphere S n−1 .
The function τ n often occurs as a lower bound for moduli of curve families like in the following lemma, based on the spherical symmetrization of condensers.This lemma has found many applications because it provides, for a pair of non-degenerate continua E and F, an explicit connection between the geometric quantity d(E, F)/ min{d(E), d(F)} and the modulus of the family of all curves joining the continua.Also a similar upper bound holds [33, 7.42], [13,Rmk 9.30], but the upper bound will not be needed here.
Lemma 2.11.Let E and F be continua in [13, 7.22] (or [33, 5.33]).If d(E; F) > 0, the proof of(1) follows from [13,Lemma 9.26] (or [33, 7.38]) and the proof of ( 2) from ( 1) and [13,Lemma 7.14] (or [33, 5.22]).□ 2.12.Quasiconformal self-homeomorphism of a domain.For a proper subdomain G of R n and for a fixed point x ∈ G, we define a homeomorphism f : and, as shown in [32, 16.2], the maximal dilatation of this map is By [32,Thm 35.1,p. 118], the dilatation of the mapping g is same as the one of h.Fix now x in G and let r = d(x, ∂G).Let f be a radial K-quasiconformal map defined by We summarize the above arguments as a lemma.
Lemma 2.14.For a proper subdomain G of R n and for a fixed point x ∈ G, there exists a quasiconformal homeomorphism f : where a is the number in (2.13).

Harnack inequality and capacity test functions
In the dimensions n ≥ 3, quasihyperbolic distances defined below are widely used as substitutes of hyperbolic distances.
3.1.Quasihyperbolic metric.For a domain G ⊊ R n , the quasihyperbolic metric k G is defined by [11, p. 39], [13, p.68] where Γ is the family of all rectifiable curves in G joining x and y.This infimum is attained when γ is the quasihyperbolic geodesic segment joining x and y .The hyperbolic metric of B n can be also defined in terms of a similar length minimizing property, with the weight function 2/(1 − |x| 2 ) .In many ways the quasihyperbolic metric is similar to the hyperbolic metric, see [13,Chapter 5], but unfortunately its values are known only in a few special cases.Fortunately, some lower bounds can be given in terms of the j G metric and upper bounds can be given for a large class of domains as we will now show.
The distance ratio metric is defined in a domain G ⊊ R n as the function ).
The lower bound j G (x, y) ≤ k G (x, y) holds for an arbitrary domain G ⊊ R n and all x, y ∈ G [13, Cor.5.6, p.69].
For the upper bound we introduce a class of domains for which we have a simple upper bound of the quasihyperbolic distance.This upper bound combined with the above lower bound provide handy estimates for many applications.

ϕuniform domains. We say that a domain
The special case ϕ(t) = c log(1 + t), c > 1 , yields the so called uniform domains which are ubiquitous in geometric function theory [13, p.84], [14].For instance, balls and halfspaces and their images under quasiconformal mappings of R n belong to this class of domains.It is easy to check that all convex domains are ϕ-uniform with ϕ(t) ≡ t .The strip domain {z ∈ C : 0 < Imz < 1} is ϕ-uniform but not uniform.[13, p. 96].Let G ⊂ R n be a domain and let u : G → (0, ∞) be a continuous function.We say that u is a Harnack function with parameters (s, C),

Harnack functions
It follows easily from the definition of the quasihyperbolic metric, see [13,  (1) Then We next start our study of the capacity test function and examine its dependence on α and z when the other argument is fixed.It turns out that the dependence of the capacity test function on α is controlled by standard ring domain capacity estimates from 2.12 whereas, as a function of z, it is continuous and satisfies a Harnack condition. where In other words, Proof.The first inequality follows from Lemma 2.7 and, by using the quasiconformal map f of 2.12, we see that the second inequality holds.□ The above result shows that for a fixed x ∈ G, u α (x) is continuous with respect to the parameter α because a → 1 when β → α .The next result shows, among other things, that for a fixed α ∈ (0, 1), u α (x) is continuous as a function of x .This continuity follows from the domain monotonicity of the capacity (4.4) and Lemma 3.6.
Similarly, by the inequalities (3.8) and (3.10), Therefore, we have where β = (α + s)/(1 − s) .In particular, with the help of Lemma 3.6, we obtain where D = [log β/ log α] n−1 .The first inequality is nothing but the required Harnack inequality.Since C < D, we obtain Now the continuity of u α follows because D → 1 as s → 0. □ Let G ⊂ R n be a domain with cap (∂G) > 0, z 1 ∈ G and fix α = 1/4 .Theorem 3.7 and Lemma 3.4 show that, perhaps surprisingly, the speed of decrease of the function u α (z)/u α (z 1 ) to 0 when k G (z 1 , z) → ∞ or z → ∂G is controlled from below by the Harnack parameters given by Theorem 3.7 and by k G (z 1 , z) .

Capacity test function
Various capacities are widely applied in geometric function theory to investigate the metric size of sets [10,9].We use here the conformal capacity of condensers and prove several lemmas involving this capacity.We begin by pointing out the connection between the condenser capacity and the modulus of a curve family.These lemmas, together with the superannulus Proposition 2.2, are applied to prove Lemma 4.8, which will be a key tool for the proof of a main result in Section 7.
A domain D ⊂ R n is called a ring if its complement R n \D has exactly two components C 0 and C 1 .Sometimes, we write D = R(C 0 , C 1 ).We say that a ring As in [13, 7.16, p. 120], the (conformal) modulus of a ring D = R(C 0 , C 1 ) is defined by and its capacity is cap (D) = M(∆(C 0 , C 1 )).For the definition of ACL and ACL n mappings, see [32, Def.26.2, p. 88; Def.26.5 p. 89], [11, 6.4].It is useful to recall the close connection between the modulus of a curve family and capacity, because many properties of curve families yield similar properties for the capacity.E = (A, C) can also be expressed in terms of a modulus of a curve family as follows: Proof.The proof follows immediately from Remark 4.2 and Lemma 2.14.□ Let E = (A, C) be a condenser and A 1 a domain with A ⊂ A 1 .It follows readily from the definition of the capacity (and also from Lemma 2.7) that the following domain monotonicity property holds For a compact set E ⊂ R n , x ∈ R n , and r > 0, we introduce the notation The condition cap(x, E, r) > 0 gives information about the size of the set E in a neighborhood of the point x.The next lemma shows that there is a substantial part of the set E, in the sense of capacity, in an annulus centered at x. Lemma 4.6.Let E ⊂ R n be a compact set, x ∈ R n , and r > 0 and suppose that .

□
According to Lemma 4.6 we can find, under the above assumptions, a substantial portion of the set E in the annulus B n (x, r) \ B n (x, r/λ) for x ∈ E. We need to use this type of annuli for two disjoint sets E and F, which are close enough to each other and then to find a lower bound for the modulus of the curve family joining the respective substantial portions of each set.These annuli are translated versions of each other and we can use Proposition 2.2 to find a common superannulus for both annuli and consider the joining curves in this superannulus.
(2) [33, p.63, 5.41 and 5.42] If F j ⊂ B n (z, s) ⊂ G, j = 1, 2, 3, 4, and and there exists t > 0 such that for all γ 13 ∈ Γ 13 and γ 24 ∈ Γ 24 Proof.The first claim (1) is proved in the cited reference.The second claim also follows easily from the cited reference but for clarity we include the details here.We apply the comparison principle of part (1) to get a lower bound for M(Γ 12 ).Because F j ⊂ B n (z, s) ⊂ G, j = 1, 2, 3, 4, it follows from Lemma 2.10 that the infimum in the lower bound of (1) is at least v 1 (n, t) ≡ 1 2 τ n (4m 2 + 4m), m = 2/t, and thus The next lemma will be a key tool in Section 7. It is based on three earlier results: (a) the superannulus Proposition 2.2, (b) the substantial subset selection Lemma 4.6, (c) the comparison principle for the modulus of a curve family, Lemma 4.7.
Lemma 4.8.Let n ≥ 2, c > 0 and λ = λ(n, c) ≥ 2 be as in Lemma 4. 6.Let E, F ⊂ R n be compact sets, x, y ∈ R n , and r > 0 and suppose that Proof.By Proposition 2.2, we see that and that R(w, λ 2 r, r/λ 2 ) is a common superannulus for the two smaller annuli.We apply the comparison principle for the modulus, Lemma 4.7(2), with Using Lemma 4.6 we obtain the numbers d = d 2 (n, c)/2 and τ = √ 2λ.□

Hausdorff content and lower estimate of capacity
In this section, we discuss lower bounds for the capacity in terms of the Hausdorff h-content.Our main references are O.Martio [20] and Yu.G.Reshetnyak [26], [27, pp.110-120].Reshetnyak also cites an earlier lemma of H. Cartan 1928 and gives its proof based on the work of L.V. Ahlfors [1] (cf.R. Nevanlinna [23, p.141]).We give here a short review of the earlier relevant results and, for the reader's benefit, outline sketchy proofs.
We start with the next covering lemma [18, p. 197].In the following, we denote by χ E the characteristic function of a set E ⊂ R n ; that is, χ E (x) = 1 if x ∈ E and χ E (x) = 0 otherwise.Lemma 5.1.Let n be an integer with n ≥ 2 and let A be a set in R n .Suppose that a radius r(x) > 0 is assigned for each point x ∈ A in such a way that sup x∈A r(x) < +∞.Then one can find a countable subset {x k } of A such that where and N n is a constant depending only on n.
The above inequalities mean that A is covered by the family of balls {B k } and the number of overlapping of the covering is at most N n .It is an interesting problem to find the best possible number N * n for the constant N n in the above lemma.For ϕ ∈ (0, π], we will say that a subset V of the unit sphere S n−1 = {x ∈ R n : |x| = 1} is ϕ-separated if the angle subtended by the two line segments P O and QO is at least ϕ for distinct points P , Q in V .We denote by ν(n, ϕ) the maximal cardinal number of ϕ-separated subsets V of S n−1 .For instance, ν(n, π) = 2.It is clear that ν(n, ϕ) ≤ ν(n, ϕ ′ ) for 0 < ϕ ′ < ϕ.The proof of the above lemma in p. 199 of [18] tells us that On the other hand, a standard compactness argument leads to the left continuity of ).Here we note that the number ν(n, π/3) is known as the kissing number κ(n) in dimension n [8].This number is closely related to other important issues such as sphere packing problems.For instance, it is known that κ(2) = 6, κ(3) = 12, κ(4) = 24.However, it is difficult to determine κ(n) in general.The true value of κ( 5) is not determined up to the present.By using the special nature of the lattices E 8 and Λ 24 , Viazovska determined κ( 8) and, later with her collaborators, κ (24) and won a Fields medal in 2022 [24].In summary, we can state the following.Let h(r) be a measure function, that is, a monotone increasing continuous function on 0 < r < +∞ with h(r) → 0 as r → 0 and h(r) → +∞ as r → +∞.The h-Hausdorff content of a set E ⊂ R n is defined as For a positive number β > 0, and for h(r) = r β , the h-Hausdorff content is called the β-dimensional Hausdorff content and denoted by Λ β (E).Recall that the Hausdorff dimension dim E of E is characterized as the infimum of β > 0 with Λ β (E) = 0.
The next lemma constitutes a key step in the proof of Martio's Theorem 5.7 below.In the proof below we give explicit estimates of the relevant constants.
Remark 5.4.The next lemma has a long history which goes back to H. Cartan and L.V. Ahlfors [1], [23, p. 141].Yu.G.Reshetnyak [26], [27,Lemma 3.7,p.115],extended their two dimensional work to the case of R n and applied the result to prove a lower bound for the capacity in terms of the Hausdorff content.O. Martio, in turn, made use of these results in his paper [20] which is one of our key references.Lemma 5.5 (Lemma 2.6 in [20]).Let σ be a positive finite measure on R n and h be a measure function.We denote by T the set of those points x ∈ R n for which the inequality σ (B n (x, r)) ≤ h(r) holds for all r > 0.
Proof.We choose r 0 > 0 so that h(r 0 ) = σ (R n ) and let A = R n \ T .For each x ∈ A, by definition, there is a positive r(x) such that σ (B n (x, r(x))) > h(r(x)).Note that r(x) ≤ r 0 .We now apply Lemma 5.1 to extract a countable set {x k } from A so that B k = B n (x k , r(x k )) satisfy (5.2).Then □ By making use of the preceding lemma, Martio proved the following result.For the proof, see Lemma 2.8 in [20].
Lemma 5.6.Let 1 < p ≤ n and α > 0. For a function u ≥ 0 in L p (R n ) with support in B n (0, r 1 ), the set F of points x ∈ R n satisfying the inequality where N n is the number in Lemma 5.1. Let The following theorem is a special case of Theorem 3.1 in [20] when p = n.Since an explicit form of the constant M 1 is not given in [20], we give an outline of the proof with a concrete form of M 1 .
Theorem 5.7.Suppose that a measure function h(r) satisfies the inequality for some constants A > 0 and r 0 > 0. Let E be a closed set in R n .Then where M 1 is the positive constant given by (5.9) Proof.We may assume x = 0 and write B r = B n (0, r) for short.Because E ∩ B r ⊂ B r ′ , r < r ′ , by the definition of the Hausdorff content it is clear that Λ h (E ∩ B r ) ≤ h(r).Since M 1 ≥ 1/K n , the required inequality holds trivially when cap (0, E, r) ≥ K n .Thus we may assume that cap (0, E, r) < K n .By the definition of capacity, for each r > 0 and a small enough ε > 0, we may choose a smooth function w ≥ 0 on R n with support in B 2r so that w > 1 on E ∩ B r and so that where dm denotes the Lebesgue measure.We apply Lemma 5.6 with r 1 = 2r and p = n to the function u = |∇w|/ω n−1 and by the above inequality we may choose α so that Since ∥∇w∥ n < K 1/n n , we have the inequality ∥∇w∥ n /ω n−1 < Ω 1/n−1 n /2, which enables us to estimate α as α = (n − 1)I(r) for 0 < r ≤ r 0 .By the representation formula (see [20, (2.2)]) we have the inequality Let F be as in Lemma 5.6.In particular, we have E ∩ B r ⊂ F and thus by Lemma 5.6 where Letting ε → 0, we obtain finally Now, in conjunction with Lemma 5.3, the required inequality follows.□ When h(r) = r β for some 0 < β ≤ n, we obtain 2r 0 Hence, in this case, the constant M 1 in (5.9) may be expressed by For instance, when n = 2, we have N * 2 = 7 and thus M 1 (2, β) = 28(2 β /β 2 ).

Uniform perfectness and capacity
In this section, we study the connection between potential theoretic thickness of sets, as expressed in terms of capacity, and uniform perfectness.The notion of uniform perfectness was first used by Beardon and Pommerenke [5] in two dimensions.Later, Pommerenke [25] found a characterization of uniform perfectness in terms of the logarithmic capacity.One of the main results of Järvi and Vuorinen [15] was a characterization of uniform perfectness for the general dimension in terms of the quantity cap (x, E, r) defined in Section 5.The novel feature in our work is to give an explicit form for cap (x, E, r) in terms of the dimension n and the parameter c of UP n (c) sets.Then we will prove Lemma 1.1 and Theorem 1.2 given in Introduction.Definition 6.1.For c ∈ (0, 1), let UP n (c) denote the collection of compact sets E in R n with card (E) ≥ 2 satisfying the condition A set is called uniformly perfect if it is of class UP n (c) for some c.Lemma 6.2 (Proposition 7.4 in [30]).Let I be the index set {1, 2, . . ., p} and 0 < c < 1 and r > 0. Suppose that a sequence of families of closed balls for k = 1, 2, 3, . . ., satisfies the following two conditions: Then the set where β = −(log p)/ log c, and its Hausdorff dimension is β.
The proof of (6.Proof.This result is contained in the proof of Theorem 3.3 in [12].□ According to [12], the constant A n admits the following majorant for n ≥ 2 1) .
The constant λ n is the so called Grötzsch constant [11,12,13].Observe that we take r in 0 < r < d(E)/2 in the above definition 6.1 of UP n (c), whereas in [12] it is required that 0 < r < d(E).
7. Proof of Theorems 1.5 and 1.3 In this section our goal is to prove one of the main results of this paper, Theorem 1.5, which gives a lower bound for cap(G, E) when G ⊂ R n is a domain and E ⊂ G is a compact set and both E and ∂G are uniformly perfect.The proof is based on the results given in earlier sections and it is divided into three cases: (a) d(E)/d(E, ∂G) is small (Lemma 7.3), (b) d(E)/d(E, ∂G) is large (Lemma 7.4), (c) neither (a) nor (b) holds.These three cases form the logical structure of the proof of Theorem 7.5 which immediately yields the proof of Theorem 1.5.For the case (b) we apply Proposition 2.4 and Lemma 4.8 to construct a sequence of separate annuli with the parameter λ adjusted so that each annulus contains a substantial portion of both E and ∂G.
Then using Theorem 1.5 we also prove Theorem 1.3.In Lemma 4.6 we proved that, for a compact set E ⊂ R n of positive capacity, the condition cap(x, E, r) = c > 0 implies the existence of λ = λ(n, c) such that the set is quite substantial.Now for a uniformly perfect set E and x ∈ E, we see by Theorem 6.5 that for 0 < r < d(E) the sets are substantial for all k = 1, 2, . . ., where r 1 = r, r k+1 = r k /λ.Observe that these sets are subsets of separate annuli centered at x.
Our first result in this section, Lemma 7.3, yields a lower bound for the modulus of the family of all curves joining for a pair of compact sets F 1 and F 2 , in terms of the respective capacities, the dimension n and a set separation parameter t ∈ (0, 1/2).This result is a counterpart of Lemma 2.11 which gives a similar lower bound for a pair of continua E and F. The parameter t now plays the role of d(E, F)/ min{d(E), d(F)}.Proposition 7.1.Let K > 1, n ≥ 2, and Then h(0) = 0, h(∞) = 0, and h has its only maximal value at u = n − 1, equal to h(n − 1) = ((n − 1)/e) n−1 .Setting u = log(K/x) and applying log(1 + t) ≤ t, t ≥ 0, yields The above upper bound for the function h shows that for all x ∈ (0, 1] which completes the proof.□ Lemma 7.3.Let t ∈ (0, 1/2] and let F j ⊂ B n ((j − 1)e 1 , t) be compact sets with δ j = cap((j − 1)e 1 , F j , t) > 0, j = 1, 2.
Then there exists a constant µ n > 0 depending only on n, such that 2 ), let F 3 = ∂G, and let ∆ j3 = ∆(F j , F 3 ; G), j = 1, 2 .Then G ⊂ B n (e 1 , 2), G ⊂ B n (0, 2) and Lemma 4.3 with the triple of radii {t, 2t, 2} yields By the choice of G for all γ j ∈ ∆ j3 we have and by Lemma 2.10 Finally, we estimate u(t) using (7.2) In conclusion, we can choose be a domain and let E ⊂ G be a compact set, and suppose that for all r ∈ (0, min{d(∂G), d(E)}) and for all z Observe first that F 1 ∪ F 2 ⊂ B n (x 0 , T ) and For the proof of the converse implication we may assume by Theorem 3.7 without loss of generality that α = 1/2 and suppose that u 1/2 (z) ≥ γ > 0, for all z ∈ G and write F = R n \ G for short.Take c ∈ (0, 1) so that .
We now show that the annulus cr < |x − a| < r meets F for all a ∈ F and 0 < r < +∞.On the contrary, we suppose that there exist a ∈ F \ {∞} and r > 0 such that the annulus A = R(a, r, cr) separates F. It is easy to see that a is not an isolated point of F. Thus, by decreasing r if necessary, we may assume that there is a point ξ ∈ ∂G with |ξ − a| = cr.By assumption, cap (G, B) = u 1/2 (x 0 ) ≥ γ so that (− log c) n−1 < 2 n ω n−1 /γ, which is impossible by the choice of c.Hence, we have shown the required assertion and conclude that F ∈ UP n (c).□

Whitney Cubes and Uniform Perfectness
Next, we will study the condenser capacity by using Whitney cubes.
8.1.Whitney decomposition.If G ⊂ R 2 is a bounded domain, we can clearly represent it as a countable union of non-overlapping closed squares.By the Whitney decomposition theorem, we can choose these squares Q k j , k ∈ Z, 0 ≤ j ≤ N k , so that they have pairwise disjoint interiors and sides parallel to the coordinate axes and the following properties are fulfilled: (1) every cube ) .These kinds of squares Q k j are Whitney squares and this definition can be clearly extended to the general case G ⊂ R n , n ≥ 2, so that we will have n-dimensional closed hypercubes called Whitney cubes instead of just squares [29,Thm 1,p. 167].Note that the Whitney cubes of a domain G resemble in a way hyperbolic balls of B n with a constant radius (Fig. 2).and G = ∪Q k j be a Whitney decomposition.Then Q k j has a side length 2 −k and If m k j is the midpoint of Q k j , then clearly (8.4)

Definition 4 . 1 .
[13, Def.9.2, p. 150] A pair E = (A, C) where A ⊂ R n is open and nonempty, and C ⊂ A is compact and non-empty is called a condenser.The capacity of this condenser E is cap(E) = inf u R n |∇u| n dm, where the infimum is taken over the family of all non-negative ACL n functions u with compact support in A such that u(x) ≥ 1 for x ∈ C. A compact set E is of capacity zero, denoted by cap E = 0 , if cap (A, C) = 0 for some bounded domain A, C ⊂ A .Otherwise we denote cap E > 0 and say that E is of positive capacity.Note that the definition of capacity zero does not depend on the open bounded set A [13, pp.150-153].

Lemma 5 . 3 .
The minimal number N * n of the bound N n in Lemma 5.1 satisfies the inequality N * n ≤ κ(n) + 1, where κ(n) is the kissing number in dimension n.

Figure 2 .
Figure 2. Whitney decomposition of a polygon.The picture was generated with software by D. E. Marshall.
Thus we see that the next theorem generalizes Theorem 1.4.
Theorem 1.5.Let G ⊂ R n be a domain and E ⊂ G a compact set.If E and ∂G are uniformly perfect, then This theorem is well-known if both E and ∂G are continua, see [33, Lemma 7.38, Notes 7.60].
which is independent of the dimension n.Remark 6.4.Recall that the Cantor middle-third set K has Hausdorff dimension (log 2)/ log 3 ≈ 0.63093[22, p.60].One can check that K ∈ UP 2 (2/5) and the number 2/5 cannot be increased.The above corollary thus implies that dim H (K) ≥ (log 2)/ log(15/2) ≈ 0.34401.Theorem 6.5.Let E ∈ UP n (c) for some 0 < c < 1.Then for every a ∈ E, a ∞, [15]llows from Theorem 5.8 and (6.6).□ 6.8.Proofs of Lemma 1.1 and Theorem 1.2.Theorem 6.5 also proves Lemma 1.1 and Theorem 1.2.□Next we estimate the parameter c in terms of a cap (x, E, r) lower bound.Equivalent characterization of uniform perfectness.According to[15]a closed set E ⊂ R n with cardE ≥ 2 is α-uniformly perfect, α > 0, if there is no ring domain D separating E with modD > α.The following lemma shows that this notion is quantitatively equivalent to our notion of UP n (c), with explicit constants.Lemma 6.11.Let E be a closed set in R n containing at least two points.(1)IfE is α-uniformly perfect, then E ∈ UP n (e −α ).(2)If E ∈ UP n (c), then E is α-uniformly perfect for α < A n + log(3/c), where A n is a positive constant depending only on n.