Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 30, 2005, 261–302

HYPERBOLIC AND UNIFORM DOMAINS IN BANACH SPACES

Jussi Väisälä

Helsingin yliopisto, Matematiikan ja tilastotieteen laitos PL 68, FI-00014 Helsinki, Finland; jvaisala@cc.helsinki.fi

Abstract. A domain G in a Banach space is said to be δ -hyperbolic if it is a Gromov δ -hyperbolic space in the quasihyperbolic metric. Then G has the Gromov boundary ∂^{∗}G and the norm boundary ∂G. We show that the following properties are quantitatively equivalent: (1) G is C -uniform. (2) G is δ -hyperbolic and there is a natural bijective map G ∪ ∂^{∗}G → G ∪ ∂G, which is η -quasimöbius rel ∂^{∗}G. (3) G is δ -hyperbolic and there is a natural η -quasimöbius homeomorphism ∂^{∗}G → ∂G. In a euclidean space, this improves a result of Bonk–Heinonen– Koskela, whose estimates depend on dimension and on a base point.

1. Introduction

Let E be a real Banach space and let G ⊊ E be a domain (open connected nonempty set). We say that G is δ -hyperbolic, δ ≥ 0, if G is a δ -hyperbolic metric space in the sense of M. Gromov when equipped with the quasihyperbolic metric k_{G}, defined by the element of length |dx|/d(x, ∂G).

In a recent paper [BHK], M. Bonk, J. Heinonen and P. Koskela study, among other things, relations between hyperbolic and uniform domains in R^{n}. They show that the following properties are equivalent for a bounded domain G ⊂ R^{n}:

(1) G is C -uniform,

(2) G is δ -hyperbolic and there is a natural η -quasisymmetric homeomorphism between the Gromov boundary ∂^{∗}G and the euclidean boundary ∂G of G,

(3) G is δ -hyperbolic and c-linearly locally connected.

The terminology is recalled in Section 2. The metric d_{p,ε} of ∂^{∗}G in (2) depends on a base point p ∈ G and on a real parameter ε, 0 < ε ≤ 1 ∧ (1/5δ).

The purpose of this paper is to study whether this result can be extended from R^{n} to arbitrary Banach spaces. We show that in this general case, (1) and (2) are still equivalent and that (1) implies (3). By an example we show that (3) does not imply (1).

We also improve the result of [BHK] in two other directions:

(a) Instead of bounded domains we consider arbitrary domains. ([BHK] considers arbitrary domains in the spherical metric.) The quasisymmetry must then be replaced by quasimöbius. This has the further advantage that the result is independent of the base point.

2000 Mathematics Subject Classification: Primary 30C65, 53C23.

(b) In (2), we get a natural bijective map between the Gromov closure G^{∗} = G ∪ ∂G and the norm closure \overline{G}, and this map is η -quasimöbius rel ∂^{∗}G.

The main results are given in 2.12, 2.29 and 3.27. A summary with a precise formulation of the quantitativeness of the results is given in 3.1. In particular, we obtain a dimension-free version of the result (1) ⇔ (2) of [BHK].

The proof for (1) ⇒ (2) is a variation of the proof of the corresponding result in [BHK]. On the other hand, the proof for (2) ⇒ (1) is entirely different. In fact, the proof for this in [BHK] is definitely n-dimensional, as it makes use of the modulus of a path family.

The metric space (G, k_{G}) need not be geodesic, and it is not locally compact if dim E = ∞. Therefore we cannot directly make use of the theory of Gromov hyperbolic spaces given in standard textbooks like [GdH] and [BH]. An exposition of the basic theory of hyperbolic spaces in the more general setting is given in the author’s article [Vä8].

It turns out that the lack of geodesics and local compactness is somewhat inconvenient but not a real difficulty. The hard problem in general Banach spaces compared with the euclidean space is the lack of Lebesgue measure and integration. Balls have no volumes. If the proof of an n-dimensional result makes use of volume or space integration, the proof of its Banach version (if true) requires a new method.

On the other hand, line integrals are available, and the quasihyperbolic metric of a domain can be defined as in R^{n}.

We prove that (1) ⇒ (2) in Section 2 and the converse in Section 3. In Section 4 we give a counterexample proving that (3) ⇏ (1). In Section 5 we give a lower bound for the hyperbolicity constant of a domain.

2. Uniform domains are hyperbolic with a boundary condition

In this section we show that if G is a uniform domain, then G is hyperbolic and there is a natural bijection G^{∗} → \overline{G}, which is quasimöbius rel ∂^{∗}G. The result is given in 2.12 and 2.29.

2.1. Notation and terminology. Throughout the paper we assume that E is a real Banach space with dim E ≥ 2. The norm of a point x ∈ E is written as |x|.

The basic notation is fairly standard and the same as in [Vä8]. We let R, Z, N denote the sets of real numbers, integers and positive integers, respectively. Balls and spheres in a metric space X are written as

\begin{aligned}B(a, r) & = \{x : |x − a| < r\}, \overline{B}(a, r) = \{x : |x − a| ≤ r\}, \\ S(a, r) & = \{x : |x − a| = r\}. \end{aligned}

In a vector space, we may drop the center a if a = 0. More generally, if ∅ ≠ A ⊂ X, we set

\overline{B}(A, r) = \{x ∈ X : d(x, A) ≤ r\}.

The distance between nonempty sets A, B is d(A, B), and the diameter of a set A is d(A). If there is a symbol denoting the metric, say k, we use the notation k(A, B) and k(A) for the distance and the diameter, and we may use the notation \overline{B}_{k}(a, r) etc. for balls and r-neighborhoods. In a vector space, [a, b] is the closed line segment between points a and b, and [a, b) = [a, b] \ {b}.

We write α: x ↷ y if α is an arc joining x and y. If needed, this notation also gives an orientation for α from x to y. For an arc α, we let α[u, v] denote the closed subarc of α between points u, v ∈ α, and for half open subarcs we write α[u, v). Occasionally, we consider a singleton {x} as an arc α: x ↷ x. For real numbers s, t we set s ∧ t = min{s, t}, s ∨ t = max{s, t}. By an expression like ab/cde we mean (ab)/(cde). To simplify expressions we often omit parentheses writing fx = f(x) etc.

In hyperbolic spaces we use the terminology and notation given in [Vä8]. However, the general distance |x − y| will usually be replaced by the quasihyperbolic distance k(x, y); see 2.7.

2.2. Quasihyperbolic metric. Let G ⊂ E be a domain. Without further notice, we always assume that G ≠ E. For x ∈ G we write d(x) = d(x, ∂G). The quasihyperbolic length of a rectifiable arc α ⊂ G is defined by

l_{k}(α) = ∫_{α} \frac{|dx|}{d(x)},

and the quasihyperbolic distance between points x, y ∈ G is the number

(2.3)  k(x, y) = k_{G}(x, y) = inf{l_{k}(α) | α: x ↷ y, α ⊂ G}.

Then l_{k}(α) is the length of α in the metric k; see [BHK, A.7]. Hence the space (G, k) is intrinsic, but it need not be geodesic; see [Vä6, 3.5]. We shall also consider the norm distance |x − y| and the distance

j(x, y) = j_{G}(x, y) = log(1 + \frac{|x − y|}{d(x) ∧ d(y)}).

We recall that always

log \frac{d(x)}{d(y)} ≤ j(x, y) ≤ k(x, y);

see [Vä6, 3.7(1)]. These inequalities will be frequently used without special reference.

2.6. Uniform domains. We recall two approaches to uniform domains: through quasihyperbolic metric and through uniform arcs.

Let C ≥ 1. A domain G is quasihyperbolically C -uniform or briefly QH C -uniform if k(x, y) ≤ Cj(x, y) for all x, y ∈ G.

An arc α: x ↷ y in G is C -uniform in G if

(1) l(α[x, z]) ∧ l(α[z, y]) ≤ Cd(z) for all z ∈ α (cigar condition),

(2) l(α) ≤ C|x − y| (turning condition). The domain G is C -uniform if each pair of points in G can be joined by a C - uniform arc in G.

The properties C -uniform and QH C -uniform are quantitatively equivalent; see [Vä6, 10.17].

2.5. Quasiconvexity and quasigeodesics. A metric space is c-quasiconvex if each pair of points x, y can be joined by an arc with length at most c|x − y|. An arc α in a domain G is a c-quasigeodesic in G if it is c-quasiconvex in the quasihyperbolic metric, that is, l_{k}(α[x, y]) ≤ ck(x, y) for all x, y ∈ α. Quasigeodesics have been called neargeodesics in my earlier papers.

The following results are from [Vä6, 9.4 and 10.9].

2.6. Lemma. (1) Let c > 1 and let x, y be points in a domain G. Then there is a c-quasigeodesic α: x ↷ y.

(2) A c-quasigeodesic in a C -uniform domain is a C_{1} -uniform arc with C_{1} = C_{1}(C, c).

2.7. Hyperbolic domains. Let δ ≥ 0. We say that a domain G ⊂ E is δ -hyperbolic if (G, k) is a δ -hyperbolic metric space. This means that

(2.8)  (x | z)_{p} ≥ (x | y)_{p} ∧ (y | z)_{p} − δ

for all x, y, z, p ∈ G, where (x | y)_{p} is the Gromov product, defined by

2(x | y)_{p} = k(p, x) + k(p, y) − k(x, y).

The constant δ cannot be arbitrarily close to 0; see 5.3.

We shall apply the theory of hyperbolic spaces given in [Vä8] to the space (G, k). In doing so, one must replace the distance |x − y| and the length l(α) by k(x, y) and l_{k}(α); in the present paper |x − y| and l(α) denote distance and length in the norm metric. An arc α: x ↷ y in G is h-short, h ≥ 0, if

l_{k}(α) ≤ k(x, y) + h.

Every subarc of an h-short arc is also h-short.

I recall the standard estimate [Vä8, 2.33] for an h-short arc α: x ↷ y in a δ -hyperbolic domain G:

(2.9)  k(p, α) − 2δ − h ≤ (x | y)_{p} ≤ k(p, α) + h/2

for all p ∈ G.

If an arc α ⊂ G is an h-short 2-quasigeodesic, we briefly say that α is an h-arc.

2.10. Lemma. Let h > 0 and let x and y be distinct points in a domain G. Then there is an h-arc α: x ↷ y.

Proof. Choose a c-quasigeodesic α: x ↷ y with c = 2∧(1+h/k(x, y)). Then l_{k}(α) ≤ ck(x, y) ≤ k(x, y) + h.

2.11. Examples. We show in 2.12 that every uniform domain is hyperbolic. Since hyperbolicity is preserved by quasi-isometries [Vä8, 3.18], it follows that every quasiconformal image of a uniform domain G ⊂ R^{n} is hyperbolic. In particular, every simply connected proper subdomain of the plane is δ -hyperbolic with a universal constant δ.

Some examples of nonhyperbolic domains are:

(1) G = R^{2} \ {ne_{1} : n ∈ Z}.

(2) G = {x ∈ R^{3} : 0 < x_{3} < 1}.

(3) More generally, G = E_{1} × B_{2} ⊂ E = E_{1} × E_{2} , where E_{1} is a Banach space of dimension at least two and B_{2} is a ball in E_{2} .

The following result is a free version of the first part of Theorem 3.6 of [BHK]. The proof is essentially the same.

2.12. Theorem. A C -uniform domain G ⊂ E is δ -hyperbolic with δ = δ(C).

Proof. Let 𝒜 be the family of all 2-quasigeodesics in G. By 2.10 this family satisfies the conditions in [Vä8, 2.26].

Let h > 0. By [Vä8, 2.34] it suffices to show that (G, k) is (δ, h, 𝒜)-Rips with δ = δ(C). Let Δ be an h-short triangle in G with sides α, β, γ in 𝒜 and with opposite vertices a, b, c. Let x ∈ γ. It suffices to show that

(2.13)  k(x, α ∪ β) ≤ δ(C).

Set s = l(γ[a, x]). We may assume that s ≤ l(γ[x, b]). By Lemma 2.6(2), the arcs α, β, γ are C_{1}(C)-uniform in G. Hence

(2.14)  2s ≤ l(γ) ≤ C_{1}|a − b| ≤ C_{1}(|a − c| + |c − b|) ≤ C_{1}(l(β) + l(α)).

We next show that there is a point y ∈ α ∪ β such that

(2.15)  |x − y| ≤ 3s, d(y) ≥ s/2C_{1}^{2}.

Case 1. l(β) ≤ s/C_{1} . Now l(α) ≥ s/C_{1} by (2.14). Hence there is y ∈ α with l(α[c, y]) = s/2C_{1} ≤ l(α[y, b]). Then

|x − y| ≤ s + l(β) + l(α[c, y]) ≤ s + s/C_{1} + s/2C_{1} ≤ 3s.

Since α is C_{1} -uniform, we have d(y) ≥ l(α[c, y])/C_{1} = s/2C_{1}^{2} .

Case 2. l(β) ≥ s/C_{1} . Now we can choose y ∈ β with l(β[a, y]) = s/2C_{1} ≤ l(β[y, c]). Then

|x − y| ≤ s + s/2C_{1} ≤ 2s.

The second inequality of (2.15) follows as in Case 1.

Since γ is C_{1} -uniform, we have d(x) ≥ s/C_{1} . As G is QH C_{2} -uniform with C_{2} = C_{2}(C), these estimates yield

k(x, y) ≤ C_{2} log(1 + \frac{|x − y|}{d(x) ∧ d(y)}) ≤ C_{2} log(1 + \frac{3s}{s/2C_{1}^{2}} ) = C_{2} log(1 + 6C_{1}^{2}),

which implies (2.13).

2.16. Remark. A modification of the proof above shows that, more generally, all inner uniform domains are hyperbolic. A domain G is inner C -uniform if each pair x, y ∈ G can be joined by an arc α satisfying the cigar condition 2.4(1) and the inner turning condition

(2i) l(α) ≤ Cl(x, y) = C inf{l(γ) | γ: x ↷ y, γ ⊂ G}.

A domain G is a C -John domain if each pair x, y ∈ G can be joined by an arc α satisfying the cigar condition 2.4(1). Obviously

C -uniform ⇒ inner C -uniform ⇒ C -John.

The planar domain R^{2} \{(x, 0) : x ≥ 0} is inner uniform but not uniform, and R^{2} \ {(n, 0) : n ∈ N} is a John domain but not inner uniform and not hyperbolic.

In the rest of this section, we show that if G is a uniform domain, we can extend the identity map of G to a bijection between the Gromov closure and the norm closure of G and that this map is quasimöbius rel boundary.

2.17. The one-point extension. Let X be a metric space and let ∞ be an element not in X. The one-point extension of X is the set

Ẋ = X ∪ {∞}.

The topology of Ẋ consists of all open sets in X and of all sets U containing ∞ such that Ẋ \ U = X \ U is closed and bounded in X. Then Ẋ is a Hausdorff space, and the subspace X ⊂ Ẋ has its original topology. In fact, Ẋ is metrizable but we do not need this fact. The space Ẋ is compact if and only if X is proper.

The one-point extension of a metric space was already considered by F. Hausdorff [Ha, p. 285] in 1914, but it is fairly seldom mentioned in the literature, except for the case where Ẋ is compact, in which case it agrees with the familiar one-point compactification.

2.18. Terminology and notation. We recall some theory from [Vä8, Section 5]. Let G ⊂ E be a δ -hyperbolic domain. We fix a base point p ∈ G and write

(x | y) = (x | y)_{p}

for x, y ∈ G. A sequence x̄ = (x_{i}) in G is a Gromov sequence if (x_{i} | x_{j}) → ∞ as i, j → ∞. Two Gromov sequences x̄ and ȳ are equivalent, x̄ ∼ ȳ, if (x_{i} | y_{i}) → ∞. We let x̂ denote the equivalence class containing x̄. The set of equivalence classes is the Gromov boundary ∂^{∗}G of G, and G^{∗} = G ∪ ∂^{∗}G is the Gromov closure of G. The product (x | y) is extended to G^{∗} as in [Vä8, (5.8)]. Then the basic inequality (2.8) holds for all x, y, z ∈ G^{∗} and p ∈ G. Moreover, (x | y) = ∞ if and only if x = y ∈ ∂G.

Let

(2.19)  0 < ε ≤ 1 ∧ (1/5δ)

and write ϱ_{p,ε}(x, y) = ϱ_{ε}(x, y) = e^{−ε(x | y)} for x, y ∈ G^{∗} with the convention e^{−∞} = 0. To the function ϱ_{p,ε} we associate a function d_{p,ε} = d_{ε} with

(2.20)  d_{ε} ≤ ϱ_{ε} ≤ 2d_{ε}

as in [Vä8, (5.15)]. The function d_{ε} is a metametric of G^{∗}, that is, it satisfies the axioms of a metric except that d_{ε}(x, x) may be positive. In fact, d_{ε}(x, y) = 0 if and only if x = y ∈ ∂^{∗}G. Hence d_{ε} defines a metric in ∂^{∗}G.

The metametric d_{ε} defines a topology 𝒯 ^{∗} in G^{∗}; see [Vä8, 4.2]. In this topology, the points of G are isolated, and a sequence x̄ in G converges to a point a ∈ ∂^{∗}G if and only if x̄ is a Gromov sequence and x̄ ∈ a.

The norm boundary ∂G and the norm closure \overline{G} = G ∪ ∂G of G are taken in the extended space Ė. Thus ∞ ∈ ∂G if and only if G is unbounded. We say that a sequence x̄ in E converges in norm to a point b ∈ Ė if it converges to b in the topology of Ė, that is, |x_{i} − b| → 0 if b ≠ ∞ and |x_{i}| → ∞ if b = ∞.

2.21. Natural maps. Suppose that G ⊂ E is a hyperbolic domain. Since 𝒯 ^{∗} | G is discrete, the identity map id: G → G is continuous from the topology 𝒯 ^{∗} to the norm topology. If it has a continuous extension φ: G^{∗} → \overline{G}, we say that φ is a natural map.

2.22. Lemma. The natural map φ: G^{∗} → \overline{G} exists if and only if every Gromov sequence x̄ in G has a limit b(x̄) in norm. Moreover, φ(a) = b(x̄) ∈ ∂G for each a ∈ ∂^{∗}G and x̄ ∈ a.

Proof. If x̄ is a Gromov sequence in G, then x̄ converges to the equivalence class x̂ in the topology 𝒯 ^{∗} by [Vä8, 5.21]. If φ exists, then x_{i} = φ(x_{i}) → φ(x̂) in Ė.

Conversely, assume that each Gromov sequence x̄ has a limit b(x̄) in Ė. Then k(x_{i}, p) = (x_{i} | x_{i}) → ∞, whence b(x̄) ∈ ∂G. If x̄ and ȳ are equivalent Gromov sequences, then z̄ = (x_{1}, y_{1}, x_{2}, y_{2}, …) is a Gromov sequence, and thus b(x̄) = b(z̄) = b(ȳ). Setting φ(x̂) = b(x̄) we therefore obtain a well-defined continuous extension φ: G^{∗} → \overline{G} of id.

2.23. Remark. If φ exists, it defines a continuous map ψ from the metric space ∂^{∗}G into the set ∂G ⊂ Ė. Also ψ is called a natural map.

2.24. Some constants. In the study of a C -uniform domain, we let C_{1}, C_{2}, … denote constants C_{j} ≥ C depending only on C. By the results mentioned in 2.4 and 2.6, we can fix constants C_{1} and C_{2} such that if G is C -uniform, then

(1) every 2-quasigeodesic in G is a C_{1} -uniform arc,

(2) G is QH C_{2} -uniform, that is,

k(x, y) ≤ C_{2}j(x, y) = C_{2} log(1 + \frac{|x − y|}{d(x) ∧ d(y)})

for all x, y ∈ G.

One can choose C_{2} = 4C^{2} by [BHK, (2.16)].

Furthermore, we fix a constant δ_{1} = δ_{1}(C) such that every C -uniform domain is δ_{1} -hyperbolic; see 2.12.

The notation C_{1} , C_{2} , δ_{1} will be fixed for the rest of this section.

2.25. Lemma. Suppose that G is a C -uniform domain and that a ∈ ∂G. Then (x | y) → ∞ as x, y → a in G in the topology of Ė.

Proof. Assume first that a ≠ ∞. Fix h = 1, let r > 0, and let x, y ∈ B(a, r) ∩ G. Choose an h-arc α: x ↷ y and let z ∈ α. By (2.9) it suffices to find an estimate k(p, z) ≥ M(r) → ∞ as r → 0. We have

d(z) ≤ |z − a| ≤ |z − x| + |x − a| ≤ C_{1}|x − y| + r ≤ (2C_{1} + 1)r,

which implies that

k(p, z) ≥ log \frac{d(p)}{d(z)} ≥ log \frac{d(p)}{(2C_{1} + 1)r} = M(r).

Next assume that a = ∞. By translation we may assume that p = 0. Let R > 0 and let x, y ∈ G \ B(R). Choose again an h-arc α: x ↷ y and let z ∈ α. Now we need an estimate k(p, z) ≥ M(R) → ∞ as R → ∞.

Fix a point b ∈ ∂G \ {∞} and let R > C_{1}|b|. We have

R − |z| ≤ |z − x| ∧ |z − y| ≤ C_{1}d(z) ≤ C_{1}|z − b| ≤ C_{1}|z| + C_{1}|b|,

whence |z| ≥ (R − C_{1}|b|)/(1 + C_{1}). Since k(p, z) ≥ log(1 + |z|/d(p)), this gives the desired estimate.

2.26. Proposition. If G is a C -uniform domain, then the natural map φ: G^{∗} → \overline{G} exists and is bijective. Moreover, a sequence x̄ in G converges to a ∈ ∂G in norm if and only if x̄ is a Gromov sequence with φ(x̂) = a.

Proof. By translation we may assume that p = 0. Let x̄ be a Gromov sequence in G. We show that x̄ converges in norm. By 2.22, this will imply that φ exists.

Case 1. x̄ is bounded. We show that x̄ is Cauchy in the norm metric. Assume that r_{ij} = |x_{i} − x_{j}| > 0, let h > 0, and choose an h-arc α: x_{i} ↷ x_{j}. Let z ∈ α be the point with l(α[x_{i}, z]) = r_{ij}/2. Then d(z) ≥ r_{ij}/2C_{1} . Choose M > 0 such that |x_{i}| ≤ M for all i. Since |z − x_{i}| ≤ r_{ij}/2 ≤ M, we have

|z| ≤ |z − x_{i}| + |x_{i}| ≤ 2M.

By (2.9) we obtain

(x_{i} | x_{j})−h/2 ≤ k(p, z) ≤ C_{2} log(1+\frac{|z|}{d(p) ∧ d(z)}) ≤ C_{2} log(1+\frac{2M}{d(p) ∧ (r_{ij}/2C_{1})}).

Since (x_{i} | x_{j}) → ∞ as i, j → ∞, this implies that r_{ij} → 0, whence x̄ is Cauchy.

Case 2. x̄ is unbounded. We show that |x_{i}| → ∞. If this is not true, then there is R > 0 such that |x_{i}| ≤ R for infinitely many i. Let α: x_{i} ↷ x_{j} be an h-arc as in Case 1, where i < j, |x_{i}| ≤ R and |x_{j}| ≥ 3R. Pick a point z ∈ α with |z| = 2R. Then d(z) ≥ R/C_{1} by uniformity, whence

k(p, α) ≤ k(p, z) ≤ C_{2} log(1 + \frac{2R}{d(p) ∧ (R/C_{1})}) = K.

By (2.9) this yields (x_{i} | x_{j}) ≤ K + h/2. As x̄ is a Gromov sequence, this gives a contradiction.

We have proved that the natural map φ: G^{∗} → \overline{G} exists.

To show that φ is injective, let x̄ and ȳ be Gromov sequences with φ(x̂) = φ(ŷ) = a. Then (x_{i} | y_{i}) → ∞ by 2.25, and hence x̂ = ŷ.

Finally, let a ∈ ∂G and choose a sequence x̄ in G converging to a in norm. By 2.25 we have (x_{i} | x_{j}) → ∞ as i, j → ∞. Thus x̄ is a Gromov sequence and φ(x̂) = a, whence φ is surjective.

The last statement of the lemma follows from the proof.

2.27. Notation. In view of 2.26, we simplify the notation by writing φx = x for x ∈ ∂^{∗}G if G is a uniform domain.

2.28. Quasimöbius maps in the extended space. Let X be a metric space and let Q = (x, y, z, w) be a quadruple in Ẋ = X ∪ {∞} with x ≠ z, y ≠ w. For finite points, the cross ratio of Q is defined by

cr Q = \frac{|x − y| |z − w|}{|x − z| |y − w|}

as in [Vä8, 4.5]. If one of the points is ∞, the cross ratio cr Q is defined by deleting the distances containing ∞. For example, cr(x, y, z, ∞) = |x − y|/|x − z|.

The definition of quasimöbius maps and its relative version (see [Vä8, 4.8]) can now be extended in an obvious manner to the case where the spaces may contain the point ∞. One could consider extended metametric spaces but they are not needed, because the metametric space (G^{∗}, d_{p,ε}) is bounded.

We are ready to formulate the second main result of this section; cf. [BHK, 3.6].

2.29. Theorem. Suppose that G is a C -uniform domain. Then the natural bijective map φ: G^{∗} → \overline{G}, which exists by 2.26, is η -quasimöbius rel ∂^{∗}G with respect to the metametric d_{p,ε} of G^{∗} and the norm metric of \overline{G} where 0 < ε ≤ ε_{0}(C). The function η depends only on C and ε but not on p.

We first give a version of [BHK, 3.14].

2.30. Lemma. Suppose that G is a C -uniform domain. Let a, b, z ∈ G with |a − b| = s > 0 and |z − a| ≥ 2s. Let h > 0 and let α: z ↷ a and β: a ↷ b be h-arcs. Define a point y ∈ α by l(α[y, a]) = s. Then

(a) k(y, β) ≤ C_{3} ,

(b) k(z, β) − C_{3} ≤ k(z, y) ≤ k(z, β) + C_{4} + h.

The constants C_{3} and C_{4} depend only on C.

Proof. The first inequality of (b) follows from (a). It remains to prove (a) and the second inequality of (b).

Let x ∈ β be the point bisecting the length of β. Recall the constants C_{1} , C_{2} from 2.24. Since α and β are C_{1} -uniform, we have d(x)∧d(y) ≥ s/2C_{1} . Since |x − y| ≤ |x − a| + |a − y| ≤ C_{1}s + s, we get

k(x, y) ≤ C_{2} log(1 + 2C_{1}(C_{1} + 1)) = C_{3},

which proves (a).

To prove (b) assume that u ∈ β. We must find an estimate

(2.31)  k(z, y) ≤ k(z, u) + C_{4} + h.

Case 1. l(β[a, u]) ∧ l(β[u, b]) ≥ s/2. Now d(u) ≥ s/2C_{1} , |u − y| ≤ C_{1}s + s, and we obtain an estimate k(u, y) ≤ C_{3} as above. This implies (2.31).

Case 2. l(β[a, u]) ∧ l(β[u, b]) ≤ s/2. Choose an h-arc γ: z ↷ u. Then
l(γ) ≥ |u − z| ≥ |a − z| − s − |b − u| ∧ |a − u| ≥ 2s − s − s/2 = s/2.

Choose a point u′ ∈ γ with l(γ[u, u′]) = s/4. Then d(u′) ≥ s/4C_{1} . Since

|u′ − y| ≤ |u′ − u| + |u − a| + |a − y| ≤ s/4 + C_{1}s + s ≤ 3C_{1}s,

we get

k(u′, y) ≤ C_{2} log(1 + 12C_{1}^{2}) = C_{4}.

As γ is h-short, this implies that

k(z, u) ≥ k(z, u′) − h ≥ k(z, y) − k(u′, y) − h ≥ k(z, y) − C_{4} − h,

which is (2.31).

2.32. Definition. A homeomorphism f: G → G′ between domains G and G′ is M -quasihyperbolic if f is M -bilipschitz in the quasihyperbolic metrics of G and G′.

2.33. Inversion. For a Banach space E, the inversion of E (in the unit sphere) is the map u: Ė → Ė, defined by u(x) = x/|x|^{2} for 0 ≠ x ≠ ∞, u(0) = ∞, u(∞) = 0. Then u is a homeomorphism with u^{−1} = u. We recall the following properties of u:

(1) u is η -quasimöbius with η(t) = 81t, [Vä6, 6.22].

(2) For each domain G ⊂ E \ {0}, the map u_{1}: G → uG defined by u is M -quasihyperbolic with a universal constant M. This was proved with M = 36 in [Vä6, 5.14] and with M = 12 in [Vä4, 2.9].

(3) u maps each C -uniform domain G ⊂ E \ {0} onto a C′ -uniform domain with C′ = C′(C) ≥ C. This follows from (1) and from [Vä6, 10.22].

It follows from (3) and 2.12 that u maps each C -uniform domain G ⊂ E \{0} onto a δ_{2} -hyperbolic domain, δ_{2} = δ_{2}(C) = δ_{1}(C′) ≥ δ_{1}(C). By (2) and by [Vä8, 5.38], u extends to a homeomorphism ū_{1}: G^{∗} → (uG)^{∗}, which is θ-quasimöbius in the metametrics d_{ε} with arbitrary base points for all ε ≤ 1∧(1/5δ_{2}) with θ = θ_{C}.

2.34. Proof of Theorem 2.29. Recall from 2.26 that the natural map φ: G^{∗} → \overline{G} is bijective and that φ∂^{∗}G = ∂G. It suffices to show that the map φ^{−1}: \overline{G} → G^{∗} is θ-quasimöbius rel ∂G with θ = θ_{C,ε}. As before, we write φx = x for x ∈ G^{∗}, and then G^{∗} = \overline{G} as a set. Let Q = (a, b, c, w) be a quadruple of distinct points in \overline{G} with a, w ∈ ∂G. We want to find an estimate

(2.35)  cr(Q, d_{p,ε}) ≤ θ(cr(Q, norm)),

where θ(s) → 0 as s → 0.

By 2.33 we may use translations and the inversion to normalize the situation so that G is unbounded and w = ∞. From 2.33(3) it follows that we can replace C by a larger constant, still denoted by C, so that G is C -uniform after the normalization. The domain G is δ_{1} -hyperbolic and we set ε_{0}(C) = 1 ∧ (1/5δ_{1}). We show that (2.35) holds for 0 < ε ≤ ε_{0} .

Choose sequences (a_{i}), (b_{i}), (c_{i}), (w_{i}) in G converging in norm to a, b, c, w, respectively. If b ∈ G [or c ∈ G], we choose b_{i} = b [or c_{i} = c] for all i. By 2.26, these sequences converge also in the metametric d_{p,ε} (to the same limits). We may assume that for each i, the points a_{i}, b_{i}, c_{i}, w_{i} are distinct and that |a_{i} − b_{i}| ∨ |a_{i} − c_{i}| ≤ |w_{i} − a_{i}|/2.

Writing

Q_{i} = (a_{i}, b_{i}, c_{i}, w_{i}), d_{i} = d_{w_{i},ε}, ϱ_{i} = ϱ_{w_{i},ε}

we have d_{i}(w_{i}, x) = 1 for all x ∈ \overline{G}, and cr(Q_{i}, d_{p,ε}) → cr(Q, d_{p,ε}). Since cr(Q_{i}, d_{p,ε}) ≤ 16 cr(Q_{i}, d_{i}) by [Vä8, 5.28], we get by (2.20)

(2.36)  cr(Q_{i}, d_{p,ε}) ≤ 16\frac{d_{i}(a_{i}, b_{i})}{d_{i}(a_{i}, c_{i})} ≤ 32\frac{ϱ_{i}(a_{i}, b_{i})}{ϱ_{i}(a_{i}, c_{i})} = 32e^{−εμ_{i}} ≤ 32e^{|μ_{i}|},

where μ_{i} = (a_{i} | b_{i})_{w_{i}} − (a_{i} | c_{i})_{w_{i}}. Setting t_{i} = |a_{i} − b_{i}|/|a_{i} − c_{i}|, t = |a − b|/|a − c| we have t = cr(Q, norm), and t_{i} → t as i → ∞. We want to get a lower bound for μ_{i}.

Fix h = 1 and choose h-arcs α_{i}: w_{i} ↷ a_{i}, β_{i}: a_{i} ↷ b_{i}, γ_{i}: a_{i} ↷ c_{i}. Let y(b_{i}) ∈ α_{i} be the point chosen as in 2.30 where now a = a_{i}, b = b_{i}, z = w_{i}, α = α_{i}, β = β_{i}, and let y(c_{i}) ∈ α_{i} be the corresponding point for c_{i}. Since α_{i} is C_{1} -uniform, we have

(2.37)  d(y(b_{i})) ≥ |a_{i} − b_{i}|/C_{1}, d(y(c_{i})) ≥ |a_{i} − c_{i}|/C_{1}.

We consider two cases.

Case 1. t ≥ 1. We may assume that t_{i} ≥ 1/2 for all i. By (2.37) we have

d(y(b_{i})) ∧ d(y(c_{i})) ≥ |a_{i} − c_{i}|/2C_{1}

for all i. Since

|y(b_{i}) − y(c_{i})| ≤ |y(b_{i}) − a_{i}| + |a_{i} − y(c_{i})| ≤ |a_{i} − b_{i}| + |a_{i} − c_{i}| ≤ 3|a_{i} − b_{i}|,

we get

k(y(b_{i}), y(c_{i})) ≤ C_{2} log(1 + 6C_{1}t_{i}).

By (2.9) and 2.30 we have

\begin{aligned}|μ_{i}| & ≤ |k(w_{i}, β_{i}) − k(w_{i}, γ_{i})| + 2δ_{1} + 2h \\ & ≤ |k(w_{i}, y(b_{i})) − k(w_{i}, y(c_{i}))| + C_{3} + C_{4} + 2δ_{1} + 3h. \end{aligned}

Since α_{i} is h-short and since h = 1, it follows that

|μ_{i}| ≤ k(y(b_{i}), y(c_{i})) + C_{5} ≤ C_{2} log(1 + 6C_{1}t_{i}) + C_{5}.

As i → ∞, this and (2.36) give (2.35) with θ(t) = 32 exp[C_{2} log(1 + 6C_{1}t) + C_{5}].

Case 2. t < 1. We may assume that t_{i} < 1 for all i. Applying (2.9) and 2.30 as in Case 1 we get

μ_{i} ≥ k(w_{i}, y(b_{i})) − k(w_{i}, y(c_{i})) − C_{3} − C_{4} − 2δ_{1} − 3h.

Since t_{i} < 1, we have l(α_{i}[y(b_{i}), a_{i}]) < l(α_{i}[y(c_{i}), a_{i}]) by the definition of y in 2.30. Hence the point y(c_{i}) lies between w_{i} and y(b_{i}) on α_{i}. As α_{i} is 1-short, it follows that μ_{i} ≥ k(y(b_{i}), y(c_{i})) − C_{6} . Writing r_{i} = |a_{i} − a| we have

d(y(b_{i})) ≤ |y(b_{i}) − a| ≤ |y(b_{i}) − a_{i}| + r_{i} ≤ |a_{i} − b_{i}| + r_{i}.

As in Case 1 we have d(y(c_{i})) ≥ |a_{i} − c_{i}|/C_{1} . Hence

k(y(c_{i}), y(b_{i})) ≥ log \frac{d(y(c_{i}))}{d(y(b_{i}))} ≥ − log(C_{1}t_{i} + C_{1}r_{i}/|a_{i} − c_{i}|).

Here r_{i}/|a_{i} − c_{i}| → 0 as i → ∞. Combining the estimates with (2.36) and letting i → ∞ yields (2.35) with θ(t) = 32e^{C_{6}}(C_{1}t)^{ε} → 0 as t → 0.

2.38. Bounded domains. A θ-quasimöbius map between bounded spaces is η -quasisymmetric, and a relative version of this is also true; see [Vä6, 6.29]. However, we cannot choose η depending only on θ. In order to get a quantitative result we must normalize the map in a suitable way; see [Vä6, 6.29 and 6.31]. These ideas and 2.29 give Theorem 2.39 below. An alternative proof is obtained by modifying the proof of 2.29 as in [BHK, 3.6]. I omit the details.

2.39. Theorem. Suppose that G is a bounded C -uniform domain with a base point p such that d(x) ≤ cd(p) for all x ∈ G. Then the natural bijective map φ: G^{∗} → \overline{G}, which exists by 2.26, is η -quasisymmetric rel ∂^{∗}G with respect to the metametric d_{p,ε} of G^{∗} and the norm metric of \overline{G}, where 0 < ε ≤ ε_{0}(C). The function η depends only on (C, c, ε).

2.40. An application to the free quasiworld. In the free quasiworld (see [Vä6]) we study (with different terminology) domains G ⊂ E and G′ ⊂ E′ in Banach spaces and homeomorphisms f: G → G′ that are (λ, μ)-quasi-isometries in the quasihyperbolic metric, as well as some subclasses of these maps. If G and G′ are C -uniform, then f extends to a homeomorphism f̄: \overline{G} → \overline{G}′, and f̄ is η -quasimöbius rel ∂G with η depending only on (C, λ, μ); see [Vä6, 11.8].

We remark that a new proof for this result is obtained by combining Theorems [Vä8, 5.38] and 2.29. In fact, the extension f̄ is obtained, somewhat more generally, for maps f that need not be homeomorphisms, only weakly surjective in the sense of [Vä8, 5.32]. The induced map ∂G → ∂G′ will nevertheless be a quasimöbius homeomorphism.

In [Vä5, Section 5] we considered domains of the type G = E_{1} × B_{2} in a product space E_{1} × E_{2} , where B_{2} ⊂ E_{2} is a ball. If dim E_{1} ≥ 2, these domains are not hyperbolic, but each homeomorphic (λ, μ)-quasi-isometry between such domains still extends to a homeomorphism f̄ between the closures, and f̄ is η - quasisymmetric rel ∂G with η depending on (λ, μ). I do not know whether this result can be obtained with the aid of Gromov hyperbolicity.

Furthermore, since hyperbolicity is preserved by quasi-isometries [Vä8, 3.18], it follows from 2.12 that there is no quasihyperbolic quasi-isometry from a uniform domain onto a nonhyperbolic domain, for example, from a ball onto a domain between two parallel hyperplanes in a space of dimension at least three. In particular, these domains are not (freely) quasiconformally equivalent. Again, [Vä5, Section 5] contains further results on domains G = E_{1} × B_{2} that do not directly follow from the results of this paper.

3. Hyperbolic domains with a boundary condition are uniform

3.1. Introduction to Section 3. We proved in Section 2 that a C -uniform domain G ⊂ E has the following properties:

(a) G is δ -hyperbolic.

(b) The natural map φ: G^{∗} → \overline{G} exists and φ is bijective and η -quasimöbius rel ∂^{∗}G from the metametric d_{p,ε} into the norm metric.

In this section we show that also the converse is true in the stronger form where (b) is replaced by

(b′) The natural map φ: G^{∗} → \overline{G} exists and its restriction ψ: ∂^{∗}G → ∂G is bijective and η -quasimöbius.

These results are quantitative in the following sense:

If G is C -uniform, then (a) holds with δ = δ(C) and (b) holds for all p ∈ G, for all 0 < ε ≤ ε_{0}(C) and for η depending only on C and ε.

If (b′) holds, then G is C -uniform with C = C(δ, η, ε). Observe that the result is independent of the base point p. The case where E = R^{n} and G is bounded (or more generally, G is a domain

in S^{n}) was proved in [BHK, 1.11 and 7.11] with C depending on δ, η, ε, n. However, in [BHK] the quasimöbius condition was replaced by quasisymmetry, and the base point p had to be chosen so that d(p) is maximal. The proof in [BHK] made use of the modulus of a path family, and it cannot be extended to general Banach spaces. Our proof makes use of h-short arcs and (μ, h)-biroads in G as well as of the rough starlikeness of G. Several times we shall make use of the idea of [BHK] to divide the points of a domain into annulus points and arc points.

As mentioned before, we assume throughout the paper that dim E ≥ 2. The case dim E = 1 is trivial: A domain is an open interval (bounded or unbounded), and each domain G ≠ E is 1-uniform and 0-hyperbolic.

The main result is given in 3.27. We start with various auxiliary results.

3.2. Lemma. If α: x ↷ y is a rectifiable arc in a domain G, then l(α) ≤ (e^{l_{k}(α)} − 1)d(x).

Proof. Set L = l_{k}(α) and let φ: [0, L] → α be the parametrization of α by quasihyperbolic length with φ(0) = x. Then d(φ(t)) ≤ e^{t}d(x), whence

l(α) = ∫_{0}^{L} d(φ(t)) dt ≤ (e^{L} − 1)d(x).

We recall that a length map between arcs is a map preserving the length of each subarc. For arcs in the space (G, k), such a map is called a quasihyperbolic length map.

We next give an estimate for the change of ordinary length in a quasihyperbolic length map.

3.3. Lemma. Let α and β be arcs in a domain G with l_{k}(α) ≤ l_{k}(β), and let f: α → β be a quasihyperbolic length map. If k(fx, x) ≤ c for all x ∈ α, then e^{−c}l(α) ≤ l(fα) ≤ e^{c}l(α).

Proof. We may assume that fα = β. By symmetry, it suffices to prove the second inequality. Since

log \frac{d(fx)}{d(x)} ≤ k(fx, x) ≤ c,

we have d(fx) ≤ e^{c}d(x) for all x ∈ α. Set L = l_{k}(α) = l_{k}(β), and let φ: [0, L] → α be the parametrization of α by quasihyperbolic length. Then

l(β) = ∫_{0}^{L} d(fφ(t)) dt ≤ e^{c} ∫_{0}^{L} d(φ(t)) dt = e^{c}l(α).

The following basic result on the geometry of normed spaces is from [Sc, Theorem 4J, p. 18]:

3.4. Theorem. Let E be a normed space with dim E ≥ 2. Then every sphere in E is 2-quasiconvex.

We next study the behavior of an h-short arc in a domain containing a large ring.

3.5. Lemma. Suppose that α: x_{1} ↷ x_{2} is an h-short arc with h ≤ 1/10 in a domain G and that a ∈ ∂G.

(1) If B(a, 16t)\\overline{B}(a, t/16) ⊂ G and if |x_{i} −a| ≥ 8t, i = 1, 2, then |z −a| > t for all z ∈ α.

(2) If B(a, 7s) \ \overline{B}(a, s) ⊂ G and if |x_{i} − a| ≤ 4s, i = 1, 2, then |z − a| < 64s for all z ∈ α.

Proof. (1) Assume that there is a point z ∈ α with |z −a| ≤ t. Choose points u_{1}, u_{2} ∈ α ∩ S(a, 8t) such that z ∈ α[u_{1}, u_{2}]. By 3.4, there is an arc β: u_{1} ↷ u_{2} in S(a, 8t) with l(β) ≤ 32t. For all x ∈ S(a, 8t) we have d(x) ≥ (8 − 1/16)t. Hence

k(u_{1}, u_{2}) ≤ l_{k}(β) ≤ 32/(8 − 1/16) = 4.031 ….

On the other hand, |u_{i}−z| ≥ 7t, d(z) ≤ |z−a| ≤ t, and we obtain the contradiction

k(u_{1}, u_{2}) ≥ k(u_{1}, z) + k(z, u_{2}) − h ≥ 2 log(1 + 7) − 1/10 = 4.058 ….

(2) Assume that z ∈ α and that |z − a| ≥ 64s. Choose u_{1}, u_{2} ∈ S(a, 4s) with z ∈ α[u_{1}, u_{2}]. Now 3.4 gives k(u_{1}, u_{2}) ≤ 16/3. Since |u_{i} − z| ≥ 60s and d(u_{i}) ≤ |u_{i} − a| = 4s, we get the contradiction k(u_{1}, u_{2}) ≥ 2 log(1 + 15) − 1/10 = 5.44 ….

3.6. Endcuts and crosscuts. Let G ⊂ E be a domain. An arc γ: a ↷ b in \overline{G} is an endcut of G if γ∩∂G = {b}, and γ is a crosscut of G if γ∩∂G = {a, b}. Such arcs are called c-quasigeodesics if l_{k}(γ[u, v]) ≤ ck(u, v) for all u, v ∈ γ ∩ G. The definition of a C -uniform arc (see 2.4) makes also sense for endcuts and crosscuts with finite endpoints.

If dim E < ∞, then each point x_{0} ∈ G can be joined to a nearest point b ∈ ∂G by the segmental geodesic endcut [x_{0}, b]. In the general case, a nearest boundary point need not exist. However, there always exists a quasigeodesic endcut consisting of a sequence of line segments. This was proved in [Vä3, 3.10]. I give a more precise formulation of this result in 3.9. In the proof, it is possible to make use of the following result of [BHK]:

3.7. Lemma. Suppose that G is a domain and that α: a ↷ b is a rectifiable arc in \overline{G} such that

(1) α \ {a, b} ⊂ G,

(2) l(α[a, x]) ∧ l(α[x, b]) ≤ Cd(x) for all x ∈ α \ {a, b},

(3) α is c-quasiconvex in norm.

Then α is a c_{1} -quasigeodesic with c_{1} = 4cC.

Proof. Let u, v ∈ α ∩ G and set α_{1} = α[u, v]. Then also α_{1} satisfies the conditions of the lemma. By [BHK, (2.15) and (2.12)] we get

l_{k}(α_{1}) ≤ 4C log(1 + \frac{l(γ)}{d(u) ∧ d(v)}) ≤ 4C log(1 + c\frac{|u − v|}{d(u) ∧ d(v)}) ≤ 4Cck(u, v).

3.8. Remark. If the cigar condition (2) is replaced by the carrot condition (2′) l(α[x, b]) ≤ Cd(x),

the proof of [BHK, (2.15)] shows that 3.7 holds with c_{1} = 2cC.

3.9. Proposition. Suppose that x_{0} is a point of a domain G, that 0 < s ≤ 1/4, and that a ∈ ∂G with |a − x_{0}| < (1 + s/2)d(x_{0}). Let x_{1} be the unique point in [x_{0}, a] ∩ S(x_{0}, d(x_{0})). Then there is an endcut γ: x_{0} ↷ b ∈ ∂G such that:

(1) l(γ) < (1 + s)d(x_{0}).

(2) γ ⊂ B(x_{0}, (1 + s)d(x_{0})).

(3) |a − b| < \frac{31}{30}sd(x_{0}) < 2sd(x_{0}).

(4) [x_{0}, x_{1}] ⊂ γ and γ \ [x_{0}, x_{1}] ⊂ B(x_{1}, sd(x_{0})).

(5) l(γ[x, b]) ≤ C_{0}d(x) for all x ∈ γ \ {b} with a universal C_{0} ≤ 47/15 < 4.

(6) γ is c_{0} -quasiconvex in norm with a universal c_{0} ≤ 4.

(7) γ is a λ_{0} -quasigeodesic with a universal λ_{0} ≤ 26.

Proof. I recall the construction of γ in [Vä3, 3.10]. This arc is the union of a finite or infinite sequence of line segments γ_{i} = [x_{i−1}, x_{i}) and b. The points x_{1}, x_{2}, … are found inductively as follows. Set s_{i} = 2^{−i}s, y_{1} = a. Assume that the points x_{0}, …, x_{i} have been chosen. If x_{i} ∈ ∂G, the process stops. If x_{i} ∈ G, we choose a point y_{i+1} ∈ ∂G with |y_{i+1} − x_{i}| < (1 + s_{i+1})d(x_{i}) and let x_{i+1} be the unique point in [x_{i}, y_{i+1}] ∩ S(x_{i}, d(x_{i})).

Setting d_{i} = d(x_{i}) we have

(3.10)  l(γ_{i+1}) = |x_{i} − x_{i+1}| = d_{i} ≤ |x_{i} − y_{i}| < s_{i}d_{i−1}

for i ≥ 1. Hence

d_{i+k} ≤ s_{i+1}s_{i+2} ⋯ s_{i+k}d_{i} ≤ s_{i+1}^{k}d_{i}

for i ≥ 0, k ≥ 1. From (3.10) it also follows that the sequence (x_{i}) converges to a point b ∈ ∂G.

We proved in [Vä3, 3.10] that γ is an arc, but the conditions (1)–(7) make sense if we interpret γ as a path in an obvious way, and then (6) implies that γ is an arc. Writing r_{i} = l(γ[x_{i}, b]) we get

(3.11)  r_{i} = ∑_{j≥i} d_{j} ≤ d_{i}/(1 − s_{i+1})

for i ≥ 0. In particular, l(γ) = r_{0} ≤ d_{0}/(1 − s_{1}) < (1 + s)d_{0} , which implies (1) and (2). Furthermore,

\begin{aligned}r_{1} ≤ d_{1}/(1 − s_{2}) & ≤ s_{1}d_{0}/(1 − 1/16) = 8sd_{0}/15 < sd_{0}, \\ |a − b| ≤ |a − x_{1}| + r_{1} & < s_{1}d_{0} + 8sd_{0}/15 = 31sd_{0}/30, \end{aligned}

and we obtain (4) and (3).

(5): Assume that x ∈ γ_{i}. Then d(x) ≥ |x − x_{i}| and

l(γ[x, b]) = |x − x_{i}| + r_{i} ≤ |x − x_{i}| + d_{i}/(1 − s_{i+1}) ≤ |x − x_{i}| + 16d_{i}/15,

because s_{i+1} ≤ s_{2} ≤ 1/16. If |x − x_{i}| ≥ d_{i}/2, then the right-hand side is at most (1 + 32/15)d(x), and (5) follows. If |x − x_{i}| ≤ d_{i}/2, then d(x) ≥ d_{i}/2, and we again obtain (5).

(6): Assume that x ∈ γ_{i}, y ∈ γ_{j} with i < j, and set ϱ = l(γ[x, y])/|x − y|. We must show that ϱ ≤ 4. If j = i + 1, we make use of some results in the appendix. Applying Lemma A.3(3) with (x, y, z) ↦ (x_{i−1}, x_{i}, y_{i+1}) we see that the deviation (see A.2) between the rays from x_{i} through x_{i−1} and x_{i+1} is at least 1, and hence ϱ ≤ 3 by A.4.

If j ≥ i + 2, then

l(γ[x, y]) ≤ l(γ[x, b]) = |x − x_{i}| + d_{i} + r_{i+1}.

By (3.10) we have d_{i+1} ≤ s_{i+1}d_{i} ≤ d_{i}/16. As s_{i+2} ≤ s_{3} ≤ 1/32, (3.11) implies that

r_{i+1} ≤ d_{i+1}/(1 − s_{i+2}) ≤ 2d_{i}/31 < d_{i}/15.

Since |y − x_{i+1}| ≤ r_{i+1} and since γ_{i} ∪ γ_{i+1} is 3-quasiconvex, we obtain

|x − y| ≥ |x − x_{i+1}| − r_{i+1} ≥ (|x − x_{i}| + d_{i})/3 − d_{i}/15.

Consequently,

ϱ ≤ 3\frac{|x − x_{i}| + 16d_{i}/15}{|x − x_{i}| + 4d_{i}/5}.

This expression is maximal for x = x_{i}, whence ϱ ≤ 4.

(7): This follows from (5), (6) and 3.8 with λ_{0} = 2c_{0}C_{0} < 26. In fact, (7) holds with λ_{0} = 17; see [Vä3, 3.10].

We prove in 3.13 that quasigeodesic endcuts produce Gromov sequences. The proof makes use of the following corollary of the stability theorem [Vä8, 3.11]:

3.12. Stability lemma. Let G be a δ -hyperbolic domain and let γ and γ′ be arcs in G with common endpoints. If γ is a c-quasigeodesic and if γ′ is h-short, then the quasihyperbolic Hausdorff distance k_{H}(γ, γ′) is bounded by a constant M(δ, c, h).

3.13. Lemma. Suppose that γ: x_{0} ↷ b ∈ ∂G is a quasigeodesic endcut of a hyperbolic domain G and that x̄ is a sequence on γ ∩ G converging to b in norm. Then x̄ is a Gromov sequence. If ȳ is another such sequence, then x̄ ∼ ȳ.

Proof. Assume that G is δ -hyperbolic and that γ is c-quasigeodesic. It suffices to show that (x_{i} | y_{j}) → ∞ as i, j → ∞. Fix h > 0 and choose h-short arcs α_{ij}: x_{i} ↷ y_{j}. Then k_{H}(α_{ij}, γ[x_{i}, y_{j}]) ≤ M(δ, c, h) by 3.12. By (2.9) we obtain

(x_{i} | y_{j}) ≥ k(p, α_{ij}) − 2δ − h ≥ k(p, γ[x_{i}, y_{j}]) − M − 2δ − h.

Since γ is an endcut, this implies that (x_{i} | y_{j}) → ∞ as i, j → ∞.

3.14. Remarks. 1. Lemma 3.13 is, in fact, a special case of [Vä8, 6.31].

2. It follows from 3.13 that a quasigeodesic endcut γ: x_{0} ↷ b determines an element of ∂^{∗}G, which will also be written as b.

3. Similarly, a quasigeodesic crosscut γ: a ↷ b determines two elements of ∂^{∗}G, which will be written as a and b.

3.15. Lemma. Suppose that x̄ is a sequence in a domain G and that x̄ converges in norm to an isolated point b_{0} ∈ ∂G. Then x̄ is a Gromov sequence in the quasihyperbolic metric.

Proof. We prove the case b_{0} = ∞. The case of a finite b_{0} is rather similar, and it is not needed in this paper. Choose a number R > 0 such that ∂G \ {∞} ⊂ B(R). Let a, b ∈ G with 2R ≤ |a| ≤ |b| and set z = |b|a/|a|. By 3.4 there is an arc β: z ↷ b in S(|b|) with l(β) ≤ 4|b|. Then α = [a, z] ∪ β is an arc joining a and b. Since |d(x) − |x|| ≤ R for x ∈ α, we obtain

k(a, b) ≤ l_{k}(α) ≤ log \frac{|b| − R}{|a| − R} + \frac{4|b|}{|b| − R} ≤ log \frac{|b|}{|a|/2} + 8 ≤ log \frac{|b|}{|a|} + 9,
k(a, b) ≥ log \frac{d(b)}{d(a)} ≥ log \frac{|b|/2}{2|a|} ≥ log \frac{|b|}{|a|} − 2.

Choose the base point p with |p| = 2R. By the definition of the Gromov product we have

2(a | b) ≥ log \frac{|a|}{2R} + log \frac{|b|}{2R} − log \frac{|b|}{|a|} − 13 = 2 log \frac{|a|}{2R} − 13,

and the lemma follows.

3.16. Lemma. Let G be a hyperbolic domain and let U be an open set in Ė meeting ∂G. Then there is a Gromov sequence x̄ in G such that x̄ converges in norm to a point b ∈ U ∩ ∂G.

Proof. If U ∩ ∂G = {∞}, the lemma follows from 3.15. If U ∩ ∂G ≠ {∞}, there is a quasigeodesic endcut γ: x_{0} ↷ b ∈ U ∩ ∂G by 3.9, and the lemma follows from 3.13.

3.17. Annulus and arc points, anchors. We adopt the terminology of [BHK, p. 65] with a slight change, because a nearest boundary point need not exist.

Let 0 < λ ≤ 1/2. A point x in a domain G is said to be a λ-annulus point of G if there is a point a ∈ ∂G such that for t = |x − a|, the annulus B(a, t/λ) \ \overline{B}(a, λt) is contained in G. Observe that (1 − λ)t ≤ d(x) ≤ t.

If x is not a λ-annulus point of G, it is a λ-arc point of G.

Let x_{0} ∈ G and let c ≥ 1. A crosscut τ: a ↷ b of G is a c-anchor of x_{0} in G if x_{0} ∈ τ and if

(1) τ is a c-quasigeodesic,

(2) τ is c-quasiconvex in norm,

(3) l(τ[a, x]) ≤ cd(x) for all x ∈ τ[x_{0}, a) and l(τ[b, x]) ≤ cd(x) for all x ∈ τ[x_{0}, b).

Observe that these conditions imply that a ≠ ∞ ≠ b. Moreover, a c-anchor τ: a ↷ b has the following properties:

(4) τ is c-uniform in G,

(5) l(τ) ≤ 2cd(x_{0}),

(6) d(x_{0}) ≤ l(τ)/2 ≤ c|a − b|/2.

By an important result of [BHK, 7.2], every λ-arc point of a domain G ⊂ R^{n} has a c-anchor in G with c = c(λ). We next extend this result to all Banach spaces, and it will be used in several proofs of this article.

3.18. Anchor lemma. If x_{0} is a λ-arc point of a domain G, then x_{0} has a c-anchor in G with c = c(λ).

Proof. I follow the idea of [BHK, 7.2], but the proof is more complicated, because segmental endcuts are replaced by arcs given by 3.9 and, moreover, the geometry of normed spaces is not so simple as the euclidean geometry.

We choose a small positive number s. To be safe, we take s = λ/100 ≤ 1/200. Set d = d(x_{0}) and choose a point a ∈ ∂G with

|a − x_{0}| = t < (1 + s/2)d.

We may assume that a = 0. Let γ: x_{0} ↷ b be an endcut of G given by 3.9 for these a and s. Since x_{0} is a λ-arc point, there is a point y ∈ ∂G with λt < |y| < t/λ. Let L be the ray from 0 through y, and let z be the unique point in L ∩ S(t). Let T be a 2-dimensional plane containing 0, x_{0}, z (unique unless z = ±x_{0} ), and let α be the shorter arc between x_{0} and z in S(t) ∩ T. If z = x_{0} , the arc α degenerates to {x_{0}}.

Let β: x_{0} ↷ y be the arc β = α ∪ [z, y] and set r = λt/3 ≤ t/6. Since y ∈ ∂G and since r < d, there is a first point x′ of β, x′ ≠ x_{0} , such that d(x′) = r. Choose a point a′ ∈ ∂G with |a′ − x′| < (1 + s/2)r and then a point x′_{0} ∈ β[x_{0}, x′] such that |a′ − x′_{0}| = d(a′, β[x_{0}, x′]). Setting d′ = d(x′_{0}) we have

r ≤ d′ ≤ |a′ − x′_{0}| ≤ |a′ − x′| < (1 + s/2)r < (1 + s/2)^{2}d/6 < d.

Let γ′: x′_{0} ↷ b′ be an endcut of G given by 3.9 with the substitution (x_{0}, a, s) ↦ (x′_{0}, a′, s). Setting β_{0} = β[x_{0}, x′_{0}] we show that

τ = γ ∪ β_{0} ∪ γ′: b ↷ b′

is the desired anchor. It is not yet quite clear that τ is an arc, but the conditions (1)–(3) of 3.17 make sense in an obvious way. For example, for points u ∈ β_{0} and v ∈ γ′, condition (2) means that l(β_{0}[u, x′_{0}]) + l(γ′[x′_{0}, v]) ≤ c|u − v|. We show that τ has properties (1)–(3), and then (2) implies that τ is an arc.

We first verify (3). The case x ∈ γ ∪ γ′ follows from 3.9(5). If x ∈ β_{0} , then d(x) ≥ r ≥ λd/3. Furthermore, |z − y| ≤ t/λ − t and l(α) ≤ 4t by 3.4. Thus l(β_{0}) ≤ 3t+t/λ. Since l(γ′) ≤ (1+s)d′ ≤ (1+s)d by 3.9(1) and since t < (1+s)d, we have l(β_{0} ∪ γ′) ≤ (1 + s)(4 + 1/λ)d, and (3) follows.

Property (2) will be proved in A.6 in the appendix, and (1) follows from (2) and (3) by 3.7.

3.19. Roads and biroads. Recall from [Vä8, Section 6] that for μ ≥ 0, h ≥ 0, a (μ, h)-road in a δ -hyperbolic domain G is a sequence ᾱ of h-short arcs α_{i}: y_{i} ↷ u_{i} such that l_{k}(α_{i}) ↗ ∞ and such that for i ≤ j, the quasihyperbolic length map g_{ij}: α_{i} → α_{j} with g_{ij}y_{i} = y_{j} satisfies k(g_{ij}x, x) ≤ μ for all x ∈ α_{i}. The sequence (u_{i}) is then a Gromov sequence defining an element û ∈ ∂^{∗}G, and we write ᾱ: ȳ ↷ û. Each pair y ∈ G, b ∈ ∂^{∗}G can be joined by a (μ, h)-road with μ = 4δ + 2h.

Furthermore, each pair a, b ∈ ∂^{∗}G can be joined by a (μ, h)-biroad ᾱ: a ↷ b for each h > 0 and μ = 12δ + 10h. This means that ᾱ is a sequence of arcs α_{i}: u_{i} ↷ v_{i} in G and that for i ≤ j there are quasihyperbolic length maps g_{ij}: α_{i} → α_{j} with the following properties:

(1) Each α_{i} is h-short.
(2) u_{i} → a, v_{i} → b.
(3) g_{ii} = id, g_{ik} = g_{ij} ∘ g_{jk} for i ≤ j ≤ k.
(4) k(g_{ij}x, x) ≤ μ for all i ≤ j and x ∈ α_{i}.

The locus |ᾱ| of a road or a biroad ᾱ is the union of all arcs α_{i}.

We recall the extended standard estimate [Vä8, 6.20] for a (μ, h)-biroad ᾱ: a ↷ b in a δ -hyperbolic domain:

(3.20)  k(p, |ᾱ|) − 4δ − h ≤ (a | b)_{p} ≤ k(p, |ᾱ|) + μ + h/2.

We next show that hyperbolic domains are roughly starlike. The proof is a variation of [BHK, 7.8]. Recall from [Vä8, 6.33] that a hyperbolic domain G is said to be (K, μ)-roughly starlike with respect to a point y ∈ G if for each x ∈ G and for each h > 0 there is a (μ, h)-road ᾱ: y ↷ b ∈ ∂^{∗}G with k(x, |ᾱ|) ≤ K.

Furthermore, G is (K, μ)-roughly starlike with respect to a boundary point a ∈ ∂^{∗}G if for each x ∈ G and for each h > 0 there is a (μ, h)-biroad ᾱ: a ↷ b ∈ ∂^{∗}G with k(x, |ᾱ|) ≤ K.

3.12. Lemma. Let 0 < r < s, let G be a hyperbolic domain containing the annulus B(s) \ B̄(r), and let x̄ and ȳ be equivalent Gromov sequences in G. If |x_{i}| ≤ r for all i, then |y_{i}| < s for large i. If |y_{i}| ≥ s for all i, then |x_{i}| > r for large i.

Proof. Assume that |x_{i}| ≤ r, |y_{i}| ≥ s. Let h = 1 and choose h-short arcs γ_{i}: x_{i} ↷ y_{i}. Let p be a base point of G with |p| = t = (r + s)/2. Since γ_{i} meets S(t) and since k(S(t)) ≤ 4t/(r − s) by 3.4, the standard estimate 2.9 gives

(x_{i} | y_{i}) ≤ k(p, γ_{i}) + h/2 ≤ 4t/(r − s) + 1/2.

As x̄ ∼ ȳ, we have (x_{i} | y_{i}) → ∞, and the lemma follows.

3.22. Theorem. A δ -hyperbolic domain G is (K, μ_{1})-roughly starlike with respect to each point of G and (K, μ_{2})-roughly starlike with respect to each point of ∂^{∗}G, where K = K(δ), μ_{1} = 4δ + 1, μ_{2} = 12δ + 1.

Proof. We prove the second part, which is needed later. The proof for the first part is rather similar.

Let a ∈ ∂^{∗}G, x_{0} ∈ G, 0 < h ≤ 1/10. It suffices to find a (μ_{2}, h)-biroad σ̄ from a with k(x_{0}, |σ̄|) ≤ K(δ).

Case 1. x_{0} is a \frac{1}{3} -arc point. Let τ: b ↷ c be a c-anchor of x_{0} given by 3.18; now c = c(1/3) is a universal constant. By Remark 3.14.3, the points b and c can be considered as elements of the Gromov boundary ∂^{∗}G. Let ᾱ: b ↷ c be a (μ_{2}, h)-biroad given by [Vä8, 6.13]. By the extended stability theorem [Vä8, 6.32] there is a point x_{1} ∈ |ᾱ| with k(x_{0}, x_{1}) ≤ K_{1}(δ). If a ∈ {b, c}, we may put σ̄ = ᾱ. If a ∉ {b, c}, we choose (μ_{2}, h)-biroads β̄: a ↷ c and γ̄: a ↷ b. By the extended Rips condition [Vä8, 6.24] there is x_{2} ∈ |β̄| ∪ |γ̄| with k(x_{1}, x_{2}) ≤ K_{2}(δ). Then k(x_{0}, x_{2}) ≤ K_{1} + K_{2} , whence either β̄ or γ̄ is the desired biroad.

Case 2. x_{0} is a \frac{1}{3} -annulus point. There is a point b ∈ ∂G such that B(b, 3t)\ \overline{B}(b, t/3) ⊂ G for t = |x_{0}−b|. Choose a Gromov sequence ū ∈ a. Since k(x_{0}, u_{i}) → ∞, we may assume that either |u_{i} − b| < t/2 for all i or |u_{i} − b| > 2t for all i. Assume that |u_{i} − b| < t/2 for all i. By 3.16 we find a Gromov sequence v̄ such that |v_{i} − b| > 2t for all i. Choose a (μ_{2}, h)-biroad σ̄: a ↷ v̂. From 3.21 it follows that σ_{i} meets S(b, t) for large i. Since k(S(b, t)) ≤ 8 by 3.4, we have k(x_{0}, |σ̄|) ≤ 8.

The case |u_{i} − b| > 2t is treated similarly, choosing a Gromov sequence v̄ with |v_{i} − b| < t/2 for all i.

In the next three lemmas we estimate the Gromov product (a | b) = (a | b)_{p} for an anchor τ: a ↷ b of a point in a hyperbolic domain with a base point p. Since τ is a quasigeodesic, the points a, b can be considered as elements of ∂^{∗}G; see Remark 3.14.3. Then (a | b) is defined.

3.23. Lemma. Suppose that τ: a ↷ b is a c-anchor of a point x_{0} in a δ -hyperbolic domain G. Let α: p ↷ u be an h-short arc containing a point y_{0} with k(x_{0}, y_{0}) ≤ μ. Then (a | b) ≤ (a | u) + K(δ, c, μ, h).

Proof. Choose sequences (x_{i}) on τ[x_{0}, a) and (y_{i}) on τ[x_{0}, b) such that x_{i} → a, y_{i} → b. Fix i and set t = (x_{i} | y_{0}). As t ≤ k(p, y_{0}), we can choose a point z ∈ α[p, y_{0}] with k(p, z) = t. By [Vä8, 2.8(6)] we have (z | y_{0}) ∧ (z | u) ≥ t − h/2. Hence

(3.24)  (x_{i} | u) ≥ (x_{i} | y_{0}) ∧ (y_{0} | z) ∧ (z | u) − 2δ ≥ t − h/2 − 2δ.

Choose h-short arcs β: x_{i} ↷ y_{0} and γ: x_{i} ↷ y_{i}. By [Vä8, (2.23) and 2.24] we find a point z_{1} ∈ β with k(z_{1}, z) ≤ 4δ+4h. Since τ is a c-quasigeodesic, it follows from the stability lemma 3.12 that k(x_{0}, γ) ≤ M(δ, c, h), whence k(y_{0}, γ) ≤ M +μ. By the second ribbon lemma [Vä8, 2.18] this yields k(z_{1}, γ) ≤ K_{1}(δ, c, μ, h), and we obtain k(z, γ) ≤ 4δ + 4h + K_{1} . Since

(x_{i} | y_{i}) ≤ k(p, γ) + h/2 ≤ t + k(z, γ) + h/2

by 2.9, this and (3.24) imply that (x_{i} | y_{i}) ≤ (x_{i} | u) + K_{1} + 6δ + 5h. As i → ∞, this gives the lemma.

3.25. Lemma. Let τ: a ↷ b be a c-anchor of a point x_{0} of a δ -hyperbolic domain G. Then (a | b) ≤ k(p, x_{0}) + K(δ, c).

Proof. Choose sequences (x_{i}) and (y_{i}) on τ as in the proof of 3.23. Set h = 1 and let γ_{i}: x_{i} ↷ y_{i} be an h-short arc. It follows from (2.9) and stability 3.12 that

(x_{i} | y_{i}) ≤ k(p, γ_{i}) + h/2 ≤ k(p, x_{0}) + K_{1}(δ, c).

As i → ∞, this implies the lemma.

In the next lemma we show that under certain additional conditions, the inequality of 3.25 can be reversed.

3.26. Lemma. Suppose that τ: a ↷ b is a c-anchor of a point x_{0} in a δ -hyperbolic domain G. Let α: p ↷ y_{0} be an h-short arc with k(x_{0}, y_{0}) ≤ μ such that d(y_{0}) ≤ c_{1}d(x) for all x ∈ α. Then (a | b) ≥ k(p, x_{0}) − K(δ, c, c_{1}, μ, h).

Proof. Let K_{0}, K_{1}, … denote positive constants depending on (δ, c, c_{1}, μ, h). Let (x_{i}) be a sequence on τ[x_{0}, a) converging to a. Fix i ∈ N and choose hshort arcs β: x_{i} ↷ y_{0} and γ: x_{i} ↷ x_{0} . As in the proof of 3.23, we choose a point z ∈ α with k(p, z) = (y_{0} | x_{i}), and then k(z, β) ≤ 4δ + 4h. By the second ribbon lemma [Vä8, 2.18] we have k_{H}(β, γ) ≤ 8δ + 5μ + 5h. From these estimates and from stability 3.12 it follows that there is a point z′ ∈ τ[x_{0}, x_{i}] with k(z, z′) ≤ K_{1}(δ, c, μ, h). We have

log \frac{d(y_{0})}{d(z′)} ≤ log \frac{c_{1}d(z)}{d(z′)} ≤ k(z, z′) + log c_{1} ≤ K_{1} + log c_{1},

whence d(x_{0}) ≤ e^{μ}d(y_{0}) ≤ K_{2}d(z′). For each x ∈ σ = τ[z′, x_{0}], this and 3.17(3) yield

d(x_{0})/K_{2} ≤ d(z′) ≤ |z′ − a| ≤ l(τ[x, a]) ≤ cd(x).

Moreover, we have l(σ) ≤ cd(x_{0}) by 3.17(3). Integration along σ gives k(z′, x_{0}) ≤ c^{2}K_{2} , and we obtain

k(p, x_{0}) ≤ k(p, z) + k(z, z′) + k(z′, x_{0}) ≤ (x_{i} | y_{0}) + K_{1} + c^{2}K_{2} ≤ (x_{i} | x_{0}) + K_{3}.

As i → ∞, this gives k(p, x_{0}) ≤ (a | x_{0}) + K_{4} . Similarly k(p, x_{0}) ≤ (b | x_{0}) + K_{4} . Hence (a | b) ≥ (a | x_{0}) ∧ (b | x_{0}) − δ ≥ k(p, x_{0}) − K_{4} − δ; see [Vä8, 5.12].

We next state the main result of this section. Recall from 2.21–2.23 that for a δ -hyperbolic domain G, the natural map φ: G^{∗} → \overline{G} exists if and only if each Gromov sequence x̄ in G converges in norm to a limit b(x̄); then φx̂ = b(x̄). The map φ is continuous in the metametric d_{p,ε} of G^{∗} and the norm metric of \overline{G}, where p ∈ G and 0 < ε ≤ 1 ∧ (1/5δ). Moreover φ defines a continuous map ψ: ∂^{∗}G → ∂G between metric spaces.

3.27. Theorem. Suppose that G ⊂ E is a δ -hyperbolic domain such that the natural map φ: G^{∗} → \overline{G} exists in the sense of 2.12 and defines an η -quasi-möbius bijection ψ: ∂^{∗}G → ∂G in the metric d_{p,ε} and the norm metric for some p ∈ G and for some ε ≤ 1 ∧ (1/5δ). Then G is C -uniform with C = C(δ, η, ε).

3.28. Notation. Let δ > 0, 0 < ε ≤ 1 ∧ (1/5δ) and let η: [0, ∞) → [0, ∞) be a homeomorphism. We let Q(δ, η, ε) denote the set of all domains G in some Banach space such that (1) G is unbounded,

(2) G is δ -hyperbolic,

(3) the natural map φ: G^{∗} → \overline{G} exists and defines an η -quasimöbius homeomorphism ψ: ∂^{∗}G → ∂G in the metric d_{p,ε} and in the norm metric for all p ∈ G.

For G ∈ Q(δ, η, ε), we simplify the notation by identifying ∂^{∗}G and ∂G by the natural bijection ψ: ∂^{∗}G → ∂G writing ψx = x for x ∈ ∂^{∗}G.

We shall reduce Theorem 3.27 to the following result:

3.29. Proposition. If G ∈ Q(δ, η, ε), then G is C -uniform with C = C(δ, η, ε).

3.30. Lemma. Proposition 3.29 implies Theorem 3.27.

Proof. Suppose that G ⊂ E is a domain satisfying the conditions of 3.27 with some δ, η, ε, p. We may assume (by a translation) that 0 ∈ ∂G. Let u be the inversion of E; see 2.33. We show that the domain G′ = uG lies in Q(υ′) with some υ′ = (δ′, η′, ε′) depending only on υ = (δ, η, ε). By Proposition 3.29, this will imply that G′ is C′(υ′)-uniform, whence G is C(υ)-uniform by 2.33(3).

Since 0 ∈ ∂G, the domain G′ is unbounded. The inversion u defines homeomorphisms u_{1}: G → G′ and v_{1} = u_{1}^{−1}: G′ → G, and these maps are 12-quasi-hyperbolic by 2.33(2). From [Vä8, 3.18] it follows that G′ is δ′ -hyperbolic with some δ′(δ) ≥ δ.

It remains to show that G satisfies condition 3.28(3). Set ε′ = ε ∧ (1/5δ′). Then ε′ ≤ ε ≤ cε′ with c = c(δ) = δ′/δ. Let p′ ∈ G′ be an arbitrary base point. As v_{1} is 12-quasihyperbolic, it follows from [Vä8, 5.38] that v_{1} induces a homeomorphism ψ_{1} = ∂v_{1}: ∂^{∗}G′ → ∂^{∗}G. Moreover, ψ_{1} is η_{1} -quasimöbius in the metrics d_{p′,ε′} and d_{p,ε′} with η_{1} depending only on δ. Consequently, the natural map ψ′: ∂^{∗}G′ → ∂G′ exists and ψ′a = uψψ_{1}a for a ∈ ∂^{∗}G′. By [Vä8, 5.28], the identity map (∂^{∗}G, d_{p,ε′}) → (∂^{∗}G, d_{p,ε}) is θ-quasimöbius with θ(t) = 4^{c+1}(t ∨ t^{c}). By 2.33(1), these facts imply that ψ′ is η′ -quasimöbius with η′(t) = 81ηθη_{1}(t). Hence G′ ∈ Q(δ′, η′, ε′).

3.31. Outline of the proof of Proposition 3.29. Suppose that G ∈ Q(δ, η, ε) and let x_{1}, x_{2} ∈ G. If |x_{1} − x_{2}| ≤ d(x_{1}) ∨ d(x_{2}), then the line segment [x_{1}, x_{2}] is a 1-uniform arc. Hence we may assume that |x_{1}−x_{2}| ≥ d(x_{1})∨d(x_{2}). We shall show that any h-short arc γ: x_{1} ↷ x_{2} with h = 1/10 is C -uniform, C = C(δ, η, ε). For this we approximate γ by a (μ, h)-biroad ᾱ: a_{1} ↷ a_{2} such that k(x_{i}, |ᾱ|) ≤ C_{1} . In 3.54 we show that the members of ᾱ satisfy conditions close to uniformity. As a crucial step we prove the case a_{2} = ∞ in the length carrot lemma 3.40, which is preceded by a related result 3.36 on distance carrots.

The proof makes substantial use of the division of a domain into annulus points and arc points and of the anchor lemma 3.18. It will be completed in 3.37.

The following basic estimates follow almost immediately from (2.20):

3.32. Lemma. Suppose that G is a δ -hyperbolic domain and that x, y, a, b ∈ G^{∗}, ε ≤ 1 ∧ (1/5δ).

(1) If d_{ε}(x, y) ≤ cd_{ε}(a, b), then (a | b) ≤ (x | y) + K with K = K(c, ε) = ε^{−1} log(2c).

(2) If (a | b) ≤ (x | y) + K, then d_{ε}(x, y) ≤ cd_{ε}(a, b) with c = c(K) = 2e^{K}.

3.33. Lemma. Let G be a δ -hyperbolic domain, let h ≤ 1/10, and let ᾱ be a (μ, h)-road, α_{i}: z_{i} ↷ u_{i}, such that u_{i} → b ∈ ∂G in norm. Let i ∈ N and let x ∈ α_{i} be a λ-annulus point of G with λ^{−1} ≥ 16e^{μ} and let d(z_{i}) ≥ 9e^{2μ}d(x). Then for r = d(x) we have

(1) B(b, (λ^{−1} − 1)r) \ \overline{B}(b, 3λr) ⊂ G,

(2) α_{i}[x, u_{i}] ⊂ B(b, 9r),

(3) r ≤ |x − b| ≤ (1 + 4λ)r.

Proof. There is a point a ∈ ∂G such that G contains the annulus B(a, t/λ) \ \overline{B}(a, λt) for t = |x − a|. Then

(3.34)  (1 − λ)t ≤ r ≤ t.

Let j ≥ i. There is a point x_{j} ∈ α_{j} with k(x_{j}, x) ≤ μ. We have |x_{j} − x| ≤ (e^{μ} − 1)r ≤ (e^{μ} − 1)t, whence |x_{j} − a| ≤ e^{μ}t. Furthermore, k(z_{i}, z_{j}) ≤ μ implies that d(z_{j}) ≥ e^{−μ}d(z_{i}) ≥ 9e^{μ}r, whence

|z_{j} − a| ≥ d(z_{j}) ≥ 9e^{μ}r ≥ 8e^{μ}t

by (3.34).

If |y − a| ≥ 8e^{μ}t for some y ∈ α_{j}[x_{j}, u_{j}], then we may apply Lemma 3.5(1) to the arc α_{j}[z_{j}, y] with the substitution t ↦ e^{μ}t to conclude that |x_{j} − a| > e^{μ}t, a contradiction. Hence α_{j}[x_{j}, u_{j}] ⊂ B(a, 8e^{μ}t) for all j ≥ i. For j = i, we choose x_{i} = x and get the better estimate

(3.35)  α_{i}[x, u_{i}] ⊂ B(a, 8t).

Since u_{j} → b, we obtain |b − a| ≤ 8e^{μ}t < t/λ. As b ∈ ∂G, this implies that |b − a| ≤ λt. The lemma follows by easy estimates from this, (3.34) and (3.35).

3.36. Distance carrot lemma. Let G ∈ Q(δ, η, ε), let h = 1/10, let ᾱ: b_{0} ↷ ∞, α_{i}: u_{i} ↷ v_{i} be a (μ, h)-biroad and let x ∈ |ᾱ|. Then

|x − b_{0}| ≤ C(δ, η, μ)d(x).
Proof. Choose i with x ∈ α_{i} and set λ^{−1} = 16e^{μ}.

Case 1. x is a λ-annulus point. There is a point a ∈ ∂G such that G contains the annulus B(a, t/λ) \ \overline{B}(a, λt) where t = |x − a|. Since v_{i} → ∞, there is an integer j_{0} ≥ i such that |v_{j} − a| ≥ 8e^{μ}t for j ≥ j_{0} . Let j ≥ j_{0} and set x_{j} = g_{ij}x; then k(x_{j}, x) ≤ μ.

If |u_{j} − a| ≥ 8e^{μ}t, we may apply Lemma 3.5(1) with the substitution t ↦ e^{μ}t and get |x_{j} − a| > e^{μ}t. On the other hand,

|x_{j} − x| ≤ (e^{k(x_{j},x)} − 1)d(x) ≤ (e^{μ} − 1)t,

which yields |x_{j} − a| ≤ e^{μ}t, a contradiction. Hence |u_{j} − a| ≤ 8e^{μ}t for all j ≥ j_{0} , whence |b_{0} − a| ≤ 8e^{μ}t, which implies that |b_{0} − a| ≤ λt. Since d(x) ≥ (1 − λ)t, we obtain

|x − b_{0}| ≤ |x − a| + |a − b_{0}| ≤ (1 + λ)d(x)/(1 − λ) < 2d(x).

Case 2. x is a λ-arc point. By 3.18 there is a c(μ)-anchor τ: a ↷ b of x. Set μ_{0} = 12δ+1 and choose a (μ_{0}, h)-biroad β̄: a ↷ b. By (3.20) and by the extended version [Vä8, 6.32(2)] of the stability lemma we get (a | b)_{x} ≤ k(x, |β̄|) + μ + h/2 ≤ M(δ, μ). As ε ≤ 1, this yields d_{x,ε}(a, b) ≥ e^{−M}/2.

Furthermore, for j ≥ i we have (u_{j} | v_{j})_{x} ≤ k(x, α_{j}) + h/2 ≤ μ + 1, which yields d_{x,ε}(u_{j}, v_{j}) ≥ e^{−μ−1}/2 and so d_{x,ε}(b_{0}, ∞) ≥ e^{−μ−1}/2. Set Q = (a, b_{0}, b, ∞). As d_{x,ε} is bounded by 1, these estimates give

cr(Q, d_{x,ε}) ≤ 4e^{M+μ+1} = K(δ, μ).

Since the map id: (∂G, d_{x,ε}) → (∂G, norm) is η -quasimöbius, we obtain

|a − b_{0}|/|a − b| = cr(Q, norm) ≤ η(K).

By 3.17(3) we have |x − a| ≤ cd(x) and |a − b| ≤ 2cd(x), and we get the desired estimate

|x − b_{0}| ≤ |x − a| + |a − b_{0}| ≤ (1 + 2η(K))cd(x).

3.37. Lemma. Let 1 < c_{1} ≤ c_{2} < c_{3} with c_{1} − 1 ≤ c_{3} − c_{2} . Suppose that G is a domain containing an annulus B(b, c_{3}r) \ \overline{B}(b, r) and that x and y are points with c_{1}r ≤ |x − b| ≤ |y − b| ≤ c_{2}r. Then k(x, y) ≤ 5c_{2}/(c_{1} − 1).

Proof. We may assume that b = 0. Set z = |y|x/|x|. By 3.4 there is an arc α: z ↷ y in S(|y|) with l(α) ≤ 2|z −y| ≤ 4c_{2}r. The arc β = [x, z]∪α joins x and y with l(β) ≤ 5c_{2}r. For every u ∈ β we have d(u) ≥ (|u|−r)∧(c_{3}r−|u|) ≥ c_{1}r−r. Hence k(x, y) ≤ l_{k}(β) ≤ 5c_{2}/(c_{1} − 1).

3.38. Strings. Let ᾱ: a ↷ b, α_{i}: u_{i} ↷ v_{i}, be a (μ, h)-biroad in a hyperbolic domain G. We recall from [Vä8, 6.21] that the string str ᾱ of ᾱ is obtained by identifying the elements (x, i) and (g_{ij}x, j) in the disjoint union of all α_{i}. There are natural injective maps

(3.39)  π_{i}: α_{i} → str ᾱ,

defined by (x, i) ∈ π_{i}x, and a metric l_{k} in str ᾱ, defined by l_{k}(ξ, ζ) = l_{k}(α_{i}[x, z]) where π_{i}x = ξ, π_{i}z = ζ. Furthermore, there is a bijective isometry ω: str ᾱ → R, unique up to an additive constant, and ω defines a linear order in str ᾱ. The locus |ξ| of an element ξ ∈ str ᾱ is the set of all x ∈ G such that (x, i) ∈ ξ for some i. Then k(|ξ|) ≤ μ and |ᾱ| = ⋃{|ξ| : ξ ∈ str ᾱ}.

3.40. Length carrot lemma. Let G ∈ Q(δ, η, ε) and let ᾱ: b_{0} ↷ ∞, α_{i}: u_{i} ↷ v_{i} be a (μ, h)-biroad in G with h = 1/10. Then

l(α_{i}[u_{i}, x]) ≤ C(δ, η, ε, μ)d(x)

for all i and for all x ∈ α_{i}.

Proof. Let K_{1}, K_{2}, … denote positive constants depending only on (δ, η, ε, μ). In particular, let K_{1} be the constant C(δ, η, μ) of 3.36. Set

d^{∗}(ξ) = d(|ξ|, ∂G)

for ξ ∈ str ᾱ. Since k(|ξ|) ≤ μ, we have d(x) ≤ e^{μ}d(y) for all x, y ∈ |ξ|, whence

(3.41)  d^{∗}(ξ) ≤ d(x) ≤ e^{μ}d^{∗}(ξ)

for all x ∈ |ξ|. Hence d^{∗}(ξ) > 0 for all ξ ∈ str ᾱ.

We say that an element ξ ∈ str ᾱ tends to −∞ or to ∞ if the number ω(ξ) ∈ R tends to −∞ or to ∞, respectively. Moreover, we use obvious notation like [ξ_{1}, ξ_{2}] and (−∞, ξ_{0}] for intervals in str ᾱ. For a set A ⊂ str ᾱ we write

|A| = ⋃{|ξ| : ξ ∈ A}.

Fact 1. d^{∗}(ξ) → 0 as ξ → −∞ and d^{∗}(ξ) → ∞ as ξ → ∞.

Choose an arbitrary base point p ∈ G and let r_{0} > 0. As the natural map φ: G^{∗} → \overline{G} is continuous at b_{0} by Definition 2.21, there is M_{0} > 0 such that |x − b_{0}| < r_{0} for all x ∈ G with (x | b_{0}) > M_{0} . From [Vä8, 6.8] it follows that there is ξ_{0} ∈ str ᾱ such that (x | b_{0}) > M_{0} whenever x ∈ |ξ| for some ξ ≤ ξ_{0} . Hence d(|ξ|, b_{0}) → 0 as ξ → −∞, and the first part of Fact 1 follows.

A similar argument shows that d(|ξ|, b_{0}) → ∞ as ξ → ∞. Since d^{∗}(ξ) ≥ d(|ξ|, b_{0})/K_{1} by 3.36, we obtain the second part of Fact 1.

Fact 2. For each ξ_{0} ∈ str ᾱ, the set |(−∞, ξ_{0}]| is bounded.

By 3.36 and (3.41), it suffices to show that d^{∗}(ξ) is bounded over ξ ∈ (−∞, ξ_{0}]. By Fact 1 there is ξ_{1} ≤ ξ_{0} such that d^{∗}(ξ) ≤ 1 for ξ ≤ ξ_{1} . Choose i ∈ N such that the natural image π_{i}α_{i} of α_{i} covers [ξ_{1}, ξ_{0}]. Since d(x) is bounded over x ∈ α_{i} by compactness, Fact 2 follows.

From Fact 1 it follows that for each t > 0, the set {ξ ∈ str ᾱ : d^{∗}(ξ) ≤ t} is nonempty and bounded from above. Hence it has a supremum in str ᾱ. We set

ζ_{n} = sup{ξ ∈ str ᾱ : d^{∗}(ξ) ≤ 2^{n}}

for n ∈ Z. Then (ζ_{n}) is an increasing sequence and str ᾱ is the union of the intervals [ζ_{n−1}, ζ_{n}], n ∈ Z.

We next show that

(3.42)  e^{−μ}2^{n} ≤ d^{∗}(ζ_{n}) ≤ 2^{n}

for all n ∈ Z. Assume that d^{∗}(ζ_{n}) < e^{−μ}2^{n}. Then there is (z, i) ∈ ζ_{n} with d(z) < e^{−μ}2^{n}. Choose an integer j ≥ i such that g_{ij}z ≠ v_{j}. Since k(g_{ij}z, z) ≤ μ, we have d(g_{ij}z) ≤ e^{μ}d(z) < 2^{n}. Choose a point x ∈ α_{j}(g_{ij}z, v_{j}] with d(x) < 2^{n}. Then π_{j}x > ζ_{n} and therefore d^{∗}(π_{j}x) > 2^{n}, which is impossible because x ∈ |π_{j}x|. Hence the first inequality of (3.42) is true.

Assume that d^{∗}(ζ_{n}) > 2^{n} and let t > 0. Since π_{i}v_{i} → ∞, there is ξ_{0} < ζ_{n} such that for no i ∈ N we have ξ_{0} < π_{i}v_{i} < ζ_{n}. Choose an element ξ ∈ str ᾱ with ξ_{0} < ξ < ζ_{n} such that d^{∗}(ξ) ≤ 2^{n} and such that l_{k}(ξ, ζ_{n}) < t. If α_{i} meets |ξ|, then α_{i} also meets |ζ_{n}|. Consequently, for each x ∈ |ξ| we have k(x, |ζ_{n}|) ≤ l_{k}(ξ, ζ_{n}) < t. Since k(x, |ζ_{n}|) ≥ log(d^{∗}(ζ_{n})/d(x)), we obtain d^{∗}(ζ_{n}) ≤ e^{t}d(x) for all x ∈ |ξ|, whence d^{∗}(ζ_{n}) ≤ e^{t}d^{∗}(ξ) ≤ e^{t}2^{n}. As t → 0, this implies the second inequality of (3.42).

By (3.41) and (3.42) we have

(3.43)  e^{−μ}2^{n} ≤ d(x) ≤ e^{μ}2^{n}

for all n ∈ Z and x ∈ |ζ_{n}|.

We prove in Fact 5 below that l_{k}(ζ_{n−1}, ζ_{n}) ≤ K_{2} for all n ∈ Z. We show now how this implies the lemma. Let i ∈ N and let x ∈ α_{i}. Choose integers s and n with s ≤ n such that π_{i}u_{i} ∈ (ζ_{s−1}, ζ_{s}] and π_{i}x ∈ (ζ_{n−1}, ζ_{n}]. Next choose m ≥ i such that [ζ_{s−1}, ζ_{n}] ⊂ π_{m}α_{m}. Writing g = g_{im} we have l(α_{i}[u_{i}, x]) ≤ e^{μ}l(α_{m}[gu_{i}, gx]) by 3.3. For each j = s − 1, …, n there is a point z_{j} ∈ π_{m}^{−1}ζ_{j}, and we obtain

l(α_{m}[gu_{i}, gx]) ≤ l(α_{m}[z_{s−1}, z_{n}]) ≤ ∑_{j=s}^{n} l(α_{m}[z_{j−1}, z_{j}]).

Since l_{k}(α_{m}[z_{j−1}, z_{j}]) = l_{k}(ζ_{j−1}, ζ_{j}) ≤ K_{2} and since d(z_{j−1}) ≤ e^{μ}2^{j−1} by (3.43), this and Lemma 3.2 yield

l(α_{m}[gu_{i}, gx]) ≤ e^{K_{2}+μ} ∑_{j=s}^{n} 2^{j−1} ≤ 2^{n}e^{K_{2}+μ}.

As π_{i}x > ζ_{n−1} , we have d(x) ≥ d^{∗}(π_{i}x) > 2^{n−1} . Hence the lemma holds with C = 2e^{K_{2}+2μ}.

Set λ = (20e^{2μ})^{−1} and λ_{1} = e^{−μ}λ = (20e^{3μ})^{−1} . We say that an element ξ ∈ str ᾱ is a λ-annulus element if |ξ| contains a λ-annulus point of G. All other elements of str ᾱ are called λ-arc elements. Each point in the locus of a λ-arc element is a λ-arc point.

Fact 3. If ξ ∈ str ᾱ is a λ_{1} -annulus element, then all points of |ξ| are λannulus points of G.

There are a λ_{1} -annulus point x_{1} ∈ |ξ| and a point a ∈ ∂G such that G contains the annulus B(a, t_{1}/λ_{1}) \ \overline{B}(a, λ_{1}t_{1}) where t_{1} = |x_{1} − a|. Let x ∈ |ξ| and set t = |x − a|. As k(|ξ|) ≤ μ, we have

|x − x_{1}| ≤ (e^{μ} − 1)(d(x) ∧ d(x_{1})) ≤ (e^{μ} − 1)(t ∧ t_{1}),

which implies that e^{−μ}t_{1} ≤ t ≤ e^{μ}t_{1} . Hence B(a, t/λ) \ \overline{B}(a, λt) ⊂ G, and Fact 3 is proved.

Fact 4. Let ξ_{1} < ξ_{2} be λ_{1} -arc elements of str ᾱ such that d^{∗}(ξ_{2}) ≤ 2e^{μ}d^{∗}(ξ_{1}) ≤ c_{1}d^{∗}(ξ) for all ξ ∈ str ᾱ with ξ ≥ ξ_{1} . Then

l_{k}(ξ_{1}, ξ_{2}) ≤ c_{2}(δ, η, ε, μ, c_{1}).

Fix a member α_{m} of ᾱ such that [ξ_{1}, ξ_{2}] ⊂ π_{m}α_{m}. Let x_{i} ∈ α_{m} be the point with π_{m}x_{i} = ξ_{i}, i = 1, 2. As these points are λ_{1} -arc points, we can choose c-anchors τ_{i}: a_{i} ↷ b_{i} of x_{i}, i = 1, 2, c = c(μ). We may assume that |b_{i} − b_{0}| ≤ |a_{i} − b_{0}|. Then 0 < |a_{i} − b_{i}| ≤ 2|a_{i} − b_{0}|. Hence cr(Q, norm) ≤ 2 for the quadruple Q = (a_{2}, b_{2}, b_{0}, ∞). Since the map ψ^{−1}: ∂G → ∂^{∗}G is η′ - quasimöbius with η′(t) = η^{−1}(t^{−1})^{−1} , this yields cr(Q, d_{p,ε}) ≤ η′(2) for each p ∈ G. If p ∈ |ᾱ|, then (3.20) gives (b_{0} | ∞) ≤ k(p, |ᾱ|) + μ + h/2 ≤ μ + 1, which implies that d_{p,ε}(b_{0}, ∞) ≥ e^{−μ−1}/2 by (2.20). Since d_{p,ε} ≤ 1, we obtain d_{p,ε}(a_{2}, b_{2}) ≤ 2e^{μ+1}η′(2)d_{p,ε}(a_{2}, b_{0}), which gives

(3.44)  (a_{2} | b_{0}) ≤ (a_{2} | b_{2}) + K_{3}

by 3.32 for each base point p ∈ |ᾱ|.

We have |a_{2} − x_{2}| ≤ cd(x_{2}) by 3.17(3) and |x_{2} − b_{0}| ≤ K_{1}d(x_{2}) by 3.36. By 3.17(6) we get d^{∗}(ξ_{1}) ≤ d(x_{1}) ≤ c|a_{1} − b_{1}|/2 ≤ c|a_{1} − b_{0}|, whence d(x_{2}) ≤ e^{μ}d^{∗}(ξ_{2}) ≤ 2e^{2μ}d^{∗}(ξ_{1}) ≤ 2e^{2μ}c|a_{1} − b_{0}|. Consequently,

|a_{2} − b_{0}| ≤ |a_{2} − x_{2}| + |x_{2} − b_{0}| ≤ K_{4}|a_{1} − b_{0}|.

Setting Q′ = (b_{0}, a_{2}, a_{1}, ∞) we thus have cr(Q′, norm) ≤ K_{4} , whence

(3.45)  cr(Q′, d_{p,ε}) ≤ η′(K_{4}).

To estimate this cross ratio we need a lower bound for d_{p,ε}(a_{1}, ∞). We may assume that μ ≥ 12δ + 1, and hence we can choose a (μ, h)-biroad ᾱ_{1}: a_{1} ↷ ∞; see 3.19. By the closeness lemma [Vä8, 6.9], there is an element κ ∈ str ᾱ such that κ > ξ_{2} and k(|κ|, |ᾱ_{1}|) ≤ 7δ + μ + 1/2. We fix a base point p ∈ |κ|. Then (3.20) gives

(a_{1} | ∞) ≤ k(p, |ᾱ_{1}|) + μ + h/2 ≤ 7δ + 2μ + 1 = K_{5}.

Hence d_{p,ε}(a_{1}, ∞) ≥ e^{−K_{5}}/2 and cr(Q′, d_{p,ε}) ≥ e^{−K_{5}}d_{p,ε}(b_{0}, a_{2})/2d_{p,ε}(b_{0}, a_{1}). By (3.45) and 3.32 we get

(3.46)  (a_{1} | b_{0}) ≤ (a_{2} | b_{0}) + K_{6}.

There is n_{0} ∈ N such that [ξ_{1}, κ] ⊂ π_{n}α_{n} for n ≥ n_{0} . Let n ≥ n_{0} and let x′_{1}, x′_{2}, p′ ∈ α_{n} be the points with π_{n}x′_{i} = ξ_{i}, π_{n}p′ = κ. Applying Lemma 3.23 with the substitution (τ, x_{0}, y_{0}, p, α) ↦ (τ_{1}, x_{1}, x′_{1}, p′, α_{n}[u_{n}, p′]) we obtain (a_{1} | b_{1})_{p′} ≤ (a_{1} | u_{n})_{p′} + K_{7} . Since k(p, p′) ≤ μ, this gives (a_{1} | b_{1}) ≤ (a_{1} | b_{0}) + K_{8} as n → ∞. By (3.44) and (3.46) it follows that

(3.47)  (a_{1} | b_{1}) ≤ (a_{2} | b_{2}) + K_{9}.

We have (a_{2} | b_{2})_{p′} ≤ k(p′, x′_{2}) + K_{10} by 3.25. If x ∈ α_{n}[x′_{1}, p′], and ξ = π_{n}x, then

d(x′_{1}) ≤ e^{μ}d^{∗}(ξ_{1}) ≤ e^{μ}c_{1}d^{∗}(ξ) ≤ e^{μ}c_{1}d(x).

Hence we may apply Lemma 3.26 with (τ, x_{0}, y_{0}, p, α, c_{1}) ↦ (τ_{1}, x_{1}, x′_{1}, p′, α_{n}[x′_{1}, p′], e^{μ}c_{1}) and obtain the estimate k(p′, x_{1}) ≤ (a_{1} | b_{1})_{p′} + C(δ, η, ε, μ, c_{1}). Since k(p, p′) ≤ μ and since

l_{k}(ξ_{1}, ξ_{2}) = l_{k}(α_{n}[x′_{1}, x′_{2}]) ≤ k(x′_{1}, p′) − k(x′_{2}, p′) + h,

these estimates and (3.47) imply Fact 4.

Fact 5. l_{k}(ζ_{n−1}, ζ_{n}) ≤ K_{2} for all n ∈ Z.

We consider two cases.

Case 1. ζ_{n−1} and ζ_{n} are λ_{1} -arc elements. Since d^{∗}(ζ_{n}) ≤ 2^{n} ≤ 2e^{μ}d^{∗}(ζ_{n−1}) by (3.42) and since d^{∗}(ξ) ≥ d^{∗}(ζ_{n−1}) for all ξ > ζ_{n−1} by the choice of ζ_{n−1} , the desired estimate follows from Fact 4.

Case 2. ζ_{n−1} or ζ_{n} is a λ_{1} -annulus element. By Fact 2 there is M_{1} such that d(x) ≤ M_{1} whenever x ∈ |ξ| for some ξ ≤ ζ_{n}. Pick m ∈ N such that π_{m}α_{m} covers [ζ_{n−1}, ζ_{n}] and such that d(v_{m}) ≥ 9e^{2μ}M_{1} . Choose points x_{j} ∈ π_{m}^{−1}ζ_{j}, j = n − 1, n and set r_{j} = d(x_{j}). By (3.43) we have

(3.48)  e^{−μ}2^{j} ≤ r_{j} ≤ e^{μ}2^{j}, e^{−2μ}/2 ≤ r_{n−1}/r_{n} ≤ e^{2μ}/2.

If ζ_{j} is a λ_{1} -annulus element, then x_{j} is a λ-annulus point by Fact 3. Since d(g_{mi}v_{m}) ≥ e^{−μ}d(v_{m}) ≥ 9e^{μ}d(x_{j}), we may apply Lemma 3.33 to the (μ, h)-road (α_{i}[u_{i}, g_{mi}v_{m}])_{i≥m} with i = m, x = x_{j} and obtain

(1) B(b_{0}, (λ^{−1} − 1)r_{j}) \ \overline{B}(b_{0}, 3λr_{j}) ⊂ G,

(2) α_{m}[u_{m}, x_{j}] ⊂ B(b_{0}, 9r_{j}),
(3) r_{j} ≤ |x_{j} − b_{0}| ≤ (1 + 4λ)r_{j} < 2r_{j}.

Subcase 2a. ζ_{n} is a λ_{1} -annulus element. By (2) we have |x_{n−1} − b_{0}| ≤ 9r_{n}. Since λ^{−1} − 1 ≥ 19e^{2μ} and 3λ ≤ e^{−2μ}/6, it follows that d(x_{n−1}) ≥ |x_{n−1} − b_{0}| − r_{n}/6. By (3.48) this yields |x_{n−1} −b_{0}| ≤ e^{2μ}r_{n}. In the other direction (3.48) gives |x_{n−1} − b_{0}| ≥ r_{n−1} ≥ e^{−2μ}r_{n}/2. In view of (3) it follows that the points x_{n−1} and x_{n} lie in the closed annulus \overline{B}(b_{0}, 2e^{2μ}r_{n}) \ B(b_{0}, e^{−2μ}r_{n}/2). By 3.37 we see that k(x_{n−1}, x_{n}) ≤ 30e^{4μ}. Hence l_{k}(ζ_{n−1}, ζ_{n}) ≤ k(x_{n−1}, x_{n}) + h ≤ 30e^{4μ} + 1.

Subcase 2b. ζ_{n−1} is a λ_{1} -annulus element and ζ_{n} is a λ_{1} -arc element. Choose a point a_{1} ∈ ∂G with |a_{1} − x_{n}| ≤ 2r_{n}. If |a_{1} − b_{0}| ≤ r_{n−1}/6, then (3.48) gives

|x_{n} − b_{0}| ≤ |x_{n} − a_{1}| + |a_{1} − b_{0}| ≤ 2r_{n} + r_{n−1}/6 ≤ 5e^{2μ}r_{n−1}

and |x_{n} − b_{0}| ≥ r_{n} ≥ 2e^{−2μ}r_{n−1} . Applying again Lemma 3.37 we get an upper bound l_{k}(ζ_{n−1}, ζ_{n}) ≤ K_{11} .

If |a_{1} − b_{0}| > r_{n−1}/6, then (1) yields |a_{1} − b_{0}| ≥ (λ^{−1} − 1)r_{n−1} ≥ 19e^{2μ}r_{n−1} . By connectedness we find points x′, z′ ∈ α_{m}[x_{n−1}, x_{n}] in the order x_{n−1}, x′, z′, x_{n} such that

(i) z′ is a λ_{1} -arc point,

(ii) all points of α_{m}[x_{n−1}, x′] are λ-annulus points,
(iii) l_{k}(α_{m}[x′, z′]) ≤ 1.

Set ξ′ = π_{m}x′, ζ′ = π_{m}z′, r′ = d(x′). Applying Lemma 3.33 to the points of α_{m}[x_{n−1}, x′] we see that G contains the annulus B(b_{0}, 19e^{2μ}r′)\\overline{B}(b_{0}, e^{−2μ}r_{n−1}/6), whence |a_{1} − b_{0}| ≥ 19e^{2μ}r′. By 3.36 and (3.48) we obtain

(3.49) 19e^{2μ}r′ ≤ |a_{1} − x_{n}| + |x_{n} − b_{0}| ≤ 2r_{n} + K_{1}r_{n} ≤ 2e^{2μ}(2 + K_{1})r_{n−1}.

Moreover, (3) gives r_{n−1} ≤ |x_{n−1} − b_{0}| ≤ 2r_{n−1} and similarly r′ ≤ |x′ − b_{0}| ≤ 2r′. Assuming ξ′ ≠ ζ_{n−1} we have r′ ≥ d^{∗}(ξ′) > 2^{n−1} ≥ e^{−μ}r_{n−1} . By Lemma 3.37 these estimates yield l_{k}(ζ_{n−1}, ξ′) ≤ k(x_{n−1}, x′) + h ≤ K_{12} .

For each ξ ≥ ζ′ we have d^{∗}(ξ) ≥ 2^{n−1} . In particular, d^{∗}(ζ_{n}) ≤ 2^{n} ≤ 2d^{∗}(ζ′). Moreover, by (iii), (3.48) and (3.49) we have

d^{∗}(ζ′) ≤ d(z′) ≤ er′ ≤ (2 + K_{1})e^{μ}2^{n−1}.

Since d^{∗}(ξ) ≥ 2^{n−1} for all ξ ≥ ζ′, we may apply Fact 4 with ξ_{1} ↦ ζ′, ξ_{2} ↦ ζ_{n}, c_{1} ↦ 2e^{2μ}(2 + K_{1}) and obtain l_{k}(ζ′, ζ_{n}) ≤ K_{13} . Hence l_{k}(ζ_{n−1}, ζ_{n}) ≤ K_{12} + 1 + K_{13} . Fact 5 and the lemma are proved.

We next show that a biroad with finite endpoints cannot go too far from the boundary.

3.50. Lemma. Let G ∈ Q(δ, η, ε) and let ᾱ: a_{1} ↷ a_{2} be a (μ, h)-biroad in G with μ = 12δ + 1, h = 1/10, a_{1} ≠ ∞ ≠ a_{2} . Then d(z) ≤ K(δ, η, ε)|a_{1} − a_{2}| for all z ∈ |ᾱ|.

Proof. We let K_{1}, K_{2}, … denote positive constants depending only on (δ, η, ε). Set λ^{−1} = 1 + 64e^{μ}.

Case 1. z is a λ-annulus point. There is a point a ∈ ∂G such that G contains the annulus A = B(a, t/λ)\\overline{B}(a, λt) where t = |z−a|. If a_{1} and a_{2} lie in different components of E \ A, then d(z) ≤ |z − a| = t, |a_{1} − a_{2}| ≥ (λ^{−1} − λ)t > t, and the lemma holds with K = 1.

Suppose that a_{1} and a_{2} lie in a component of E \A. We show that this leads to a contradiction.

If |a_{1} − a| ∨ |a_{2} − a| ≤ λt, we choose a member α_{m}: u_{m} ↷ v_{m} of ᾱ such that |u_{m} − a| ∨ |v_{m} − a| ≤ 4λt and such that α_{m} contains a point y with k(y, z) ≤ μ. By Lemma 3.5(2) we have d(y) ≤ |y − a| < 64λt. Since d(z) ≥ (1 − λ)t, we get the contradiction

μ ≥ k(y, z) ≥ log \frac{d(z)}{d(y)} > log \frac{λ^{−1} − 1}{64} = μ.

If |a_{1} − a| ∧ |a_{2} − a| ≥ t/λ, we choose α_{n} and a point y ∈ α_{n} such that |u_{n} − a| ∧ |v_{n} − a| ≥ t/2λ and k(y, z) ≤ μ. Now Lemma 3.5(1) with t ↦ t/16λ gives |y − a| > t/16λ. Since |y − z| ≤ (e^{μ} − 1)d(z) ≤ (e^{μ} − 1)t, we get t/16λ < |y − z| + |z − a| ≤ e^{μ}t, which contradicts the definition of λ.

Case 2. z is a λ-arc point. By the anchor lemma 3.18 there is an c-anchor τ: b_{1} ↷ b_{2} of z; now c = c(δ). By [Vä8, 6.13] we can choose (μ, h)-biroads ᾱ_{i}: a_{i} ↷ ∞ and β̄_{i}: b_{i} ↷ ∞, i = 1, 2. By the extended Rips condition [Vä8, 6.24] we may assume that k(z, |ᾱ_{1}|) ≤ K_{1}(δ) = 46δ + 11μ + 3. Hence there is an element ξ_{1} ∈ str ᾱ_{1} with d(z, |ξ_{1}|) ≤ K_{1} + μ. By the closeness lemma [Vä8, 6.9] we can find an element ζ_{1} > ξ_{1} in str ᾱ_{1} with k(|ζ_{1}|, |β̄_{i}|) ≤ 7δ + μ + 1/2, i = 1, 2. Pick a member α_{1m} of ᾱ_{1} whose natural image π_{1m}α_{1m} covers [ξ_{1}, ζ_{1}], and choose points x_{1}, p ∈ α_{1m} with π_{1m}x_{1} = ξ_{1} , π_{1m}p = ζ_{1} . We consider p as the base point of G.

By the extended standard estimate (3.20) we have

(b_{i} | ∞) ≤ k(p, |β̄_{i}|) + μ + h/2 ≤ K_{2}(δ) = 7δ + 2μ + 1.

As ε ≤ 1, this yields

(3.51)  d_{p,ε}(b_{i}, ∞) ≥ e^{−K_{2}}/2, i = 1, 2.

Furthermore,

(3.52)  d(z) ≤ |b_{i} − z| ≤ l(τ) ≤ c|b_{1} − b_{2}|, i = 1, 2.

Since |x_{1} − a_{1}| ≤ K_{3}d(x_{1}) by 3.36 and since k(x_{1}, z) ≤ K_{1} + 2μ, we obtain

\begin{aligned}|b_{i} − a_{1}| & ≤ |b_{i} − z| + |z − x_{1}| + |x_{1} − a_{1}| \\ & ≤ c|b_{1} − b_{2}| + e^{K_{1}+2μ}d(z) + K_{3}e^{K_{1}+2μ}d(z) ≤ K_{4}|b_{1} − b_{2}|. \end{aligned}

Let Q_{i} be the quadruple (b_{i}, a_{1}, b_{3−i}, ∞) in ∂G, i = 1, 2. Then cr(Q_{i}, norm) = |b_{i} − a_{1}|/|b_{1} − b_{2}| ≤ K_{4} , whence cr(Q_{i}, d_{p,ε}) ≤ η′(K_{4}). As d_{p,ε} ≤ 1, this and (3.51) yield d_{p,ε}(b_{i}, a_{1}) ≤ 2e^{K_{2}}η′(K_{4})d_{p,ε}(b_{1}, b_{2}), which gives

(3.53)  (b_{1} | b_{2}) ≤ (b_{i} | a_{1}) + K_{5}

by 3.32.

Choose an h-short arc σ: z ↷ p and let x ∈ σ. Since k(z, x_{1}) ≤ K_{1} + 2μ, we may apply the second ribbon lemma [Vä8, 2.18] to find a point x′ ∈ α_{1m}[x_{1}, p] with k(x, x′) ≤ K_{6}(δ) = 8δ + 5(K_{1} + 2μ) + 1. If |x_{1} − x′| ≤ d(x_{1})/2, then d(x′) ≥ d(x_{1})/2. If |x_{1} − x′| ≥ d(x_{1})/2, it follows from the length carrot lemma 3.40 that d(x_{1}) ≤ 2|x_{1} − x′| ≤ 2K_{7}d(x′). In both cases these estimates imply that d(z) ≤ K_{8}d(x) for all x ∈ σ with K_{8} = 2K_{7}e^{K_{1}+2μ+K_{6}} . By 3.26 and (3.53) this yields k(p, z) ≤ (b_{i} | a_{1}) + K_{9} , i = 1, 2. By (3.20) we get

(a_{1} | a_{2}) ≤ k(p, |ᾱ|) + μ + h/2 ≤ (b_{i} | a_{1}) + K_{10},

whence d_{p,ε}(b_{i}, a_{1}) ≤ K_{11}d_{p,ε}(a_{1}, a_{2}). Setting Q′_{i} = (a_{1}, b_{i}, a_{2}, ∞) we obtain by (3.51) cr(Q′_{i}, d_{p,ε}) ≤ 2K_{11}e^{K_{2}} = K_{12} , whence

|a_{1} − b_{i}|/|a_{1} − a_{2}| = cr(Q′_{i}, norm) ≤ η(K_{12}), i = 1, 2.

By (3.52) this gives the desired estimate d(z) ≤ 2cη(K_{12})|a_{1} − a_{2}|.

The next lemma shows that a (μ, h)-biroad in a domain G ∈ Q(δ, η, ε) has properties close to uniformity.

3.54. Lemma. Let G ∈ Q(δ, η, ε) and let ᾱ: a_{1} ↷ a_{2} be a (μ, h)-biroad with a_{1} ≠ ∞ ≠ a_{2} , μ = 12δ + 1, h = 1/10. Then:

(1) There is an element ξ_{α} ∈ str ᾱ such that if x_{1}, x_{2} ∈ α_{m} and if either π_{m}x_{1} ≤ π_{m}x_{2} ≤ ξ_{α} or π_{m}x_{1} ≥ π_{m}x_{2} ≥ ξ_{α}, then l(α_{m}[x_{1}, x_{2}]) ≤ C(δ, η, ε)d(x_{2}).

(2) If y_{1}, y_{2} ∈ α_{m} and if d(y_{1}) ∨ d(y_{2}) ≤ 2|y_{1} − y_{2}|, then l(α_{m}[y_{1}, y_{2}]) ≤ C′(δ, η, ε)|y_{1} − y_{2}|.

Proof. (1) We let C_{1}, C_{2}, … denote positive constants depending only on (δ, η, ε). Choose (μ, h)-biroads ᾱ_{i}: a_{i} ↷ ∞, i = 1, 2. By the extended tripod lemma [Vä8, 6.25] we can find elements ξ_{α} ∈ str ᾱ and ξ_{i} ∈ str ᾱ_{i}, i = 1, 2, such that the bijective length maps f_{1}: (−∞, ξ_{α}] → (−∞, ξ_{1}] and f_{2}: [ξ_{α}, ∞) → (−∞, ξ_{2}] satisfy k(|f_{i}ξ|, |ξ|) ≤ C_{1}(δ) for all possible ξ.

It suffices to consider the case π_{m}x_{1} ≤ π_{m}x_{2} ≤ ξ_{α}. Choose a member α_{1n} of α_{1} whose natural image π_{1n}α_{1n} covers the interval f_{1}π_{m}α_{m}. The length map π_{1n}^{−1} is defined in f_{1}π_{m}α_{m} and we obtain a quasihyperbolic length map g = π_{1n}^{−1}f_{1}π_{m}: α_{m} → α_{1n} satisfying k(gx, x) ≤ C_{1} + 2μ = C_{2}(δ). Then d(gx_{2}) ≤ e^{C_{2}}d(x_{2}) and l(α_{m}[x_{1}, x_{2}]) ≤ e^{C_{2}}l(α_{1n}[gx_{1}, gx_{2}]) by 3.3, and (1) follows from 3.40.

(2) We may assume that π_{m}y_{1} ≤ π_{m}y_{2} . If π_{m}y_{2} ≤ ξ_{α}, then (1) gives l(α_{m}[y_{1}, y_{2}]) ≤ Cd(y_{2}) ≤ 2C|y_{1} − y_{2}|, and the case π_{m}y_{1} ≥ ξ_{α} is treated similarly. Assume that π_{m}y_{1} < ξ_{α} < π_{m}y_{2} .

Let z ∈ α_{m} be the point in α_{m}[y_{1}, y_{2}] with π_{m}z = ξ_{α}. By (1) we have

(3.55)  l(α_{m}[y_{1}, y_{2}]) ≤ 2Cd(z).

Now the extended tripod lemma gives points z_{i} ∈ |ᾱ_{i}| with k(y_{i}, z_{i}) ≤ C_{2} . Then d(z_{i}) ≤ e^{C_{2}}d(y_{i}) and |y_{i} − z_{i}| ≤ e^{C_{2}}d(y_{i}). By the distance carrot lemma 3.36 we have |a_{i} − z_{i}| ≤ C_{3}d(z_{i}), whence

|a_{i} − y_{i}| ≤ |a_{i} − z_{i}| + |z_{i} − y_{i}| ≤ C_{4}d(y_{i})

where C_{4} = e^{C_{2}}(C_{3} + 1). By Lemma 3.50 we have

(3.56)  d(z) ≤ C_{5}|a_{1} − a_{2}|.

Consequently, if d(y_{1}) ∨ d(y_{2}) ≤ d(z)/4C_{4}C_{5} , then |a_{i} − y_{i}| ≤ |a_{1} − a_{2}|/4, whence |a_{1} − a_{2}| ≤ 2|y_{1} − y_{2}|, and (2) follows from (3.55) and (3.56). Finally, if d(y_{1}) ∨ d(y_{2}) ≥ d(z)/4C_{4}C_{5} , then d(z) ≤ 8C_{4}C_{5}|y_{1} − y_{2}|, and (2) follows from (3.55).

3.57. Proof of Theorem 3.27. Let G ∈ Q(δ, η, ε). By 3.30 it suffices to show that G is C(δ, η, ε)-uniform. We let C_{1}, C_{2}, … denote positive constants depending only on (δ, η, ε). Let x_{1}, x_{2} ∈ G. If |x_{1} − x_{2}| ≤ d(x_{1}) ∨ d(x_{2}), then the line segment [x_{1}, x_{2}] is a 1-uniform arc. Assume that |x_{1}−x_{2}| ≥ d(x_{1})∨d(x_{2}). Let h = 1/10 and let γ: x_{1} ↷ x_{2} be an h-short arc. We show that γ is C -uniform.

By 3.22 and by [Vä8, 6.35] there is a (μ, h)-biroad ᾱ: a_{1} ↷ a_{2} with μ = 12δ+1 such that k(x_{i}, ᾱ) ≤ C_{1}(δ), i = 1, 2. By the ribbon lemma [Vä8, 2.17] we find a member α_{m} of ᾱ and a quasihyperbolic length map f: γ → α_{m} such that k(fx, x) ≤ C_{2}(δ) for all x ∈ γ. We may assume that π_{m}fx_{1} ≤ π_{m}fx_{2} on str ᾱ.

We first verify the cigar condition. Let x ∈ γ. If a_{2} = ∞, we have l(α_{m}[fx_{1}, fx]) ≤ C_{3}d(fx) by the length carrot lemma 3.40. By 3.3 this yields l(γ[x_{1}, x]) ≤ C_{3}e^{2C_{2}}d(x), which implies the cigar condition. The case a_{2} = ∞ is similar.

Assume that a_{1} ≠ ∞ ≠ a_{2} . Let ξ_{α} ∈ str ᾱ be the element given by 3.54(1) and let z ∈ α_{m} be the point with π_{m}z = ξ_{α}. We have either π_{m}fx_{1} ≤ π_{m}fx ≤ ξ_{α} or ξ_{α} ≤ π_{m}fx ≤ π_{m}fx_{2} . In the first case, Lemma 3.54(1) gives l(α_{m}[fx_{1}, fx]) ≤ C_{4}d(fx), and in the second case, we similarly get l(α_{m}[fx, fx_{2}]) ≤ C_{4}d(fx). As above, these estimates yield

l(γ[x_{1}, x]) ∧ l(γ[x, x_{2}]) ≤ C_{4}e^{2C_{2}}d(x),

which is the cigar condition.

To prove the turning condition, we first assume that d(fx_{1})∨d(fx_{2}) ≤ 2|fx_{1}− fx_{2}|. If a_{1} ≠ ∞ ≠ a_{2} , then 3.54(2) gives l(α_{m}[fx_{1}, fx_{2}]) ≤ C_{5}|fx_{1} − fx_{2}|. But this is also true if a_{1} = ∞ or a_{2} = ∞, because then 3.40 implies that l(α_{m}[fx_{1}, fx_{2}]) ≤ C_{3}(d(fx_{1}) ∨ d(fx_{2})) ≤ 2C_{3}|fx_{1} − fx_{2}|. Since |fx_{i} − x_{i}| ≤ e^{C_{2}}d(x_{i}) ≤ e^{C_{2}}|x_{1} − x_{2}|, we obtain the turning condition

l(γ) ≤ e^{C_{2}}C_{5}|fx_{1} − fx_{2}| ≤ e^{C_{2}}C_{5}(1 + 2e^{C_{2}})|x_{1} − x_{2}|.

Finally, assume that d(fx_{1}) ∨ d(fx_{2}) ≥ 2|fx_{1} − fx_{2}|. Now k(fx_{1}, fx_{2}) ≤ 1, whence k(x_{1}, x_{2}) ≤ 1 + 2C_{2} . As γ is h-short, this gives l_{k}(γ) ≤ 2 + 2C_{2} = C_{6} . By 3.2 we obtain l(γ) ≤ e^{C_{6}}d(x_{1}) ≤ e^{C_{6}}|x_{1} − x_{2}|.

4. A counterexample

In [BHK, 7.12] it was proved that a domain in S^{n} is uniform if and only if it is hyperbolic and linearly locally connected (LLC). Every C -uniform domain G ⊂ E is δ(C)-hyperbolic by 2.12 and 3C -LLC by Theorem 4.2 below. However, we give in 4.3 an example of a domain G ⊂ l_{2} that is hyperbolic and LLC but not uniform.

4.1. Definition. Let c ≥ 1. We recall that a domain G is c-linearly locally connected or briefly c-LLC if the following conditions hold for all x ∈ G and r > 0:

(LLC_{1} ) Each pair of points in G ∩ B(x, r) can be joined by an arc in G ∩ B(x, cr).

(LLC_{2} ) Each pair of points in G \ \overline{B}(x, r) can be joined by an arc in G \ \overline{B}(x, r/c).

It is well known that uniform domains are LLC; we recall the easy argument in 4.2. The planar domain G = R^{2} \ {(n, 0) : n ∈ Z} is 1-LLC but not uniform.

4.2. Theorem. Every C -John domain is c-LLC_{2} and every C -uniform domain is c-LLC with c = 3C.

Proof. Let x ∈ G and let a, b ∈ G \ \overline{B}(x, r). We may assume that x = 0. Suppose that G is a C -John domain. Choose a number s with r < s < |a| ∧ |b|. Let α: a ↷ b be an arc satisfying the C -cigar condition 2.4(1). If α∩S(s/3C) = ∅, there is nothing to prove. If there is a point z ∈ α ∩ S(s/3C), we have

d(z) ≥ (|z − a| ∧ |z − b|)/C > (s − s/3C)/C ≥ 2s/3C

by the cigar condition. Thus d(0) ≥ d(z)−|z| > s/3C, whence S(s/3C) ⊂ G. We can therefore join a and b by an arc γ ⊂ G \ B(s/3C) consisting of two subarcs of α and an arc β ⊂ S(s/3C).

Suppose that G is C -uniform. Let x ∈ G and let a, b ∈ G ∩ B(x, r). Choose a C -uniform arc α: a ↷ b in G. For each y ∈ α we have

|y − x| ≤ |y − a| + |a − x| < l(α) + r < 3Cr

by the turning condition. Hence α ⊂ B(x, 3Cr).

4.3. Example. Let E be an infinite-dimensional separable Hilbert space. Let W ⊂ E be a broken tube, considered first in [Vä2, 2.12]; a detailed treatment is given in [Vä7]. We recall the construction.

Choose an orthonormal basis (e_{j})_{j∈Z} of E, indexed by all integers, and set u_{j} = \sqrt{2} e_{j}. Let α be the line spanned by e_{0} , and set a_{j} = 2je_{0} for j ∈ Z. Let f: α → E be the map with fa_{j} = u_{j}, j ∈ Z, such that f is affine on each line segment α_{j} = [a_{j−1}, a_{j}]. Then f | α_{j} is an isometry onto β_{j} = [u_{j−1}, u_{j}]. Let U be the tubular neighborhood {x ∈ E : d(x, α) < 1/5} of α. In [Vä7] we give a detailed construction of a locally bilipschitz extension F: U → W of f onto a domain W satisfying the conditions

(1) W is hyperbolic,

(2) W is LLC,

(3) W is not a John domain and hence not uniform.

5. A lower bound to the hyperbolicity constant

5.1. A lower bound for δ. There are 0-hyperbolic metric spaces (metric trees). However, we show in 5.3 that the constant δ cannot be arbitrarily close to 0 for a δ -hyperbolic domain G ⊂ E. Remember that dim E ≥ 2.

5.2. Lemma. If E is a normed plane, then there are unit vectors x, y, z such that

|x − y| = |y − z|, |x − z| ≥ |x − y| + 1/2.

Proof. Choose a unit vector x and then a unit vector y such that |x − y| = |y + x| = s.

Case 1. s ≤ 3/2. Now x, y and z = −x are the desired points, because |x − y| = |y − z| = s and |x − z| = 2 ≥ s + 1/2.

Case 2. s ≥ 3/2. Choose unit vectors u and v such that the points x, u, y, v, −x are in this order in the unit circle and such that |x − u| = |u − y| = t_{1} , |y − v| = |v + x| = t_{2} . We may assume that t_{1} ≤ t_{2} . Since the length of a unit semicircle is at most 4 by Theorem 3.4, we have 2t_{1} + 2t_{2} ≤ 4, whence t_{1} ≤ 1. Since

|x − y| = s ≥ 3/2 ≥ t_{1} + 1/2,

the desired points are x, u, y.

5.3. Theorem. If G is a δ -hyperbolic domain, then δ ≥ 0.0027.

Proof. Fix a point p ∈ G. We normalize the situation by d(p) = 1. Let 0 < t ≤ 1/2. By 5.2 there are points x, y, z ∈ S(p, t) such that

|x − y| = |y − z|, |x − z| ≥ |x − y| + t/2.

For all a, b ∈ \overline{B}(p, t) we have

\frac{|a − b|}{1 + 2t} ≤ k(a, b) ≤ \frac{|a − b|}{1 − t}

by [Vä6, 3.7]. Hence

2(x | y)_{p} = k(x, p) + k(y, p) − k(x, y) ≥ \frac{2t}{1 + 2t} − \frac{|x − y|}{1 − t},

and a similar lower bound holds for 2(y | z)_{p}. Moreover,

2(x | z)_{p} = k(x, p) + k(z, p) − k(x, z) ≤ \frac{2t}{1 − t} − \frac{|x − z|}{1 + 2t}.

As |x − z| ≤ 2t, we obtain by combining the estimates that

δ ≥ (x | y)_{p} ∧ (y | z)_{p} − (x | z)_{p} ≥ \frac{t(1 − 22t)}{4(1 − t)(1 + 2t)}.

For t = 0.02 this gives δ ≥ 0.0027.

5.4. Remark. For Hilbert spaces one can get the bound δ ≥ 0.005. These bounds are presumably far from the best possible. In the other direction, we know that the half plane is δ -hyperbolic with δ = log 3 = 1.098 … [CDP, p. 12]. I conjecture that 5.3 holds with this δ.

Appendix. Quasiconvexity in normed spaces

In this appendix I give some results on quasiconvexity needed in Section 3. The main goal is Lemma A.6, which gives a rigorous and detailed proof for the quasiconvexity of the crosscut γ in Lemma 3.18. Moreover, Lemmas A.3 and A.4 were used in 3.9.

Throughout the appendix, we let E denote a normed space. We start with a useful elementary inequality.

A.1. Lemma. Let x, y, z, u ∈ E with |x − y| ≤ |x − z| and u ∈ [x, y]. Then

|y − z| ≤ 2|u − z|.

Proof. If |u − y| ≤ |y − z|/2, then |u − z| ≥ |y − z| − |u − y| ≥ |y − z|/2. If |u − y| ≥ |y − z|/2, then

|u − z| ≥ |x − z| − |x − u| ≥ |x − y| − |x − u| = |u − y| ≥ |y − z|/2.

A.2. Deviations. The deviation between nonzero vectors x, y ∈ E is defined by

dev(x, y) = |x/|x| − y/|y|| ∈ [0, 2].

If α and β are rays with common vertex v ∈ E, then dev(x − v, y − v) is independent of the points x ∈ α and y ∈ β, and we set

dev(α, β) = dev(x − v, y − v).

In an inner product space we have dev(α, β) = 2 sin(φ/2), where φ is the angle between the rays α and β.

A.3. Lemma. Let α and β be rays from the origin and let x ∈ α, y ∈ β.

(1) If |x| = |y| = r, then |x − y| = r dev(α, β).

(2) |x − y| ≥ (|x| ∨ |y|) dev(α, β)/2.

(3) Let x, y, z ∈ E with 0 < |y −z| ≤ |x−y| ≤ |x−z|. Then dev(x−y, z −y) ≥ 1.

Proof. Part (1) is trivial, and (2) follows from (1) and A.1. To prove (3), set r = |x − y| and let u be the point on the ray from y through z with |u − y| = r. Then |u − x| ≥ r by the convexity of the norm. Hence dev(x − y, z − y) = |u − x|/r ≥ 1.

A.4. Lemma. Let α ≠ β be rays with a common vertex. Then α ∪ β is c-quasiconvex with c = 4/ dev(α, β) − 1. The bound is sharp.

Proof. We may assume that the common vertex is the origin. Set r = dev(α, β) and let x ∈ α, y ∈ β. It suffices to show that |x|+|y| ≤ (4/r −1)|x−y|.

We may assume that 0 < |y| ≤ |x| = 1. Let b ∈ β be a unit vector and set t = |y − b| = 1 − |y|. We must show that

\frac{2 − t}{|x − y|} ≤ \frac{4}{r} − 1.

If t ≤ r/2, then |x − y| ≥ |x − b| − |y − b| = r − t, whence

\frac{2 − t}{|x − y|} ≤ \frac{2 − t}{r − t} ≤ \frac{2 − r/2}{r − r/2} = \frac{4}{r} − 1.

If t ≥ r/2, then |x − y| ≥ r/2 by A.3, and we obtain

\frac{2 − t}{|x − y|} ≤ \frac{2 − r/2}{r/2} = \frac{4}{r} − 1.

To prove the sharpness, let E be the plane with the norm ∥x∥ = |x_{1}| ∨ |x_{2}| and let α and β be the rays from the origin through the points (1, 0) and (1, 1), respectively. Now dev(α, β) = 1. For x = (2, 0) and y = (1, 1) we have ∥x∥ + ∥y∥ = 3 = 3∥x − y∥. Hence α ∪ β is not c-quasiconvex for any c < 3 = 4/ dev(α, β) − 1.

A.5. Lemma. Suppose that α: x ↷ y is a c-quasiconvex arc and that a ∈ E with |a − y| = d(a, α). Then β = α ∪ [y, a]: x ↷ a is a (2c + 1)-quasiconvex arc.

Proof. Let u ∈ α, v ∈ [y, a]. Then |v − y| = d(v, α) ≤ |u − v|, and we get

\begin{aligned}l(β[u, v]) & = l(α[u, y]) + |y − v| ≤ c|u − y| + |y − v| \\ & ≤ c|u − v| + c|v − y| + |y − v| ≤ (2c + 1)|u − v|. \end{aligned}

A.6. Lemma. The arc τ: b ↷ b′ constructed in the proof of Lemma 3.18 is c-quasiconvex in norm with c = c(λ).

Proof. We recall from 3.18 the basic inequalities

(A.7)  (1 − s/2)t ≤ d ≤ t ≤ (1 + s/2)d, r ≤ d′ ≤ (1 + s/2)r,

where d = d(x_{0}), d′ = d(x′_{0}), t = |x_{0}|, s = λ/100, r = λt/3, λ ≤ 1/2.

As in 3.9, we let x_{1} denote the unique point in [x_{0}, 0]∩S(x_{0}, d), and similarly x′_{1} ∈ [x′_{0}, a′] ∩ S(x′_{0}, d′). Then [x_{0}, x_{1}] ⊂ γ and γ_{0} = γ \ [x_{0}, x_{1}] ⊂ B(x_{1}, sd) ⊂ B(2sd) by 3.9(4). Similarly, [x′_{0}, x′_{1}] ⊂ γ′ and γ′_{0} = γ′ \ [x′_{0}, x′_{1}] ⊂ B(x′_{1}, sd′).

Let u, v ∈ τ be points with u ∈ τ[b, v]. We must get an estimate

(A.8)  l(τ[u, v]) ≤ c|u − v|

with c = c(λ). The arcs γ and γ′ are 4-quasiconvex by 3.9(6), and α is 2quasiconvex by 3.4. Applying Lemma A.5 we see that the arcs [x_{1}, x_{0}] ∪ α and β_{0} are 5-quasiconvex and that β_{0} ∪ [x′_{0}, x′_{1}] is 11-quasiconvex. Consequently, if {u, v} is contained in one of the arcs

γ, γ′, [x_{1}, x_{0}] ∪ α, β_{0} ∪ [x′_{0}, x′_{1}],

then (A.8) holds with c = 11.

There remain 4 cases and some subcases for u and v. We proved in 3.18 that τ satisfies condition 3.17(3), whence l(τ) ≤ c_{1}d ≤ c_{1}t with c_{1} = c_{1}(λ). Hence it suffices to get a lower bound

(A.9)  |u − v| ≥ qt

with q = q(λ) > 0.

Case 1. u ∈ γ_{0} , v ∉ γ. Now |u| ≤ 2sd ≤ 2st. Let w ∈ [x′_{0}, x′_{1}]. We show that

(A.10)  |w| ≥ λt.

We have |w| ≥ |x′_{0}| − d′, where d′ < (1 + s/2)r < λt and d′ ≤ |x_{0} − y′|. If x′_{0} ∈ α, then |w| ≥ t − λt ≥ λt. If x′_{0} ∈ [y, z], then |w| ≥ |x′_{0}| − |x′_{0} − y| = |y| ≥ λt, and (A.10) is proved. It follows that |v| ≥ λt − sd′ ≥ (λ − s)t, whence |u − v| ≥ (λ − 3s)t > λt/2, and (A.9) holds with q = λ/2.

Case 2. u ∈ [x_{1}, x_{0}], v ∈ [z, x′_{0}]. This case does not occur if x′_{0} ∈ α. We consider two subcases.

If |y| ≥ t we apply A.3 and obtain |u − v| ≥ |v| dev(u, v)/2 ≥ |x_{0} − z|/2. Since α is 2-quasiconvex, this implies that

l(τ[u, v]) ≤ |u − x_{0}| + 2|x_{0} − z| + |z − v| = 2|x_{0} − z| + |v| − |u| ≤ 5|u − v|.

Next assume that |y| < t, and let y_{1} ∈ [0, x_{0}] be the point with |y_{1}| = |y| ≥ λt. Since |y − x_{0}| ≥ d ≥ t(1 − s/2), we have

|y| + t(1 − s/2) ≤ |y| + |y − x_{0}| ≤ |y_{1}| + |y − y_{1}| + |y_{1} − x_{0}| = t + |y − y_{1}|,

whence |y −y_{1}| ≥ |y|−st/2. Consequently, dev(x_{0}, z) = |y −y_{1}|/|y| ≥ 1−st/2|y|. Here st/2|y| ≤ s/2λ = 1/200. By A.3 we get

|u − v| ≥ |v| dev(u, v)/2 ≥ |v|/4 ≥ |y|/4 ≥ λt/4.

302 Jussi Väisälä

Case 3. u ∈ [x_{1}, x_{0}], v ∈ [x′_{0}, x′_{1}]. We again consider two subcases.

First assume that |x′_{0}| ≥ t. By (A.7) we get

|x_{0} − x′_{0}| ≥ d − d′ ≥ (1 − s/2)t − (1 + s/2)r.

By A.1 we have |u − x′_{0}| ≥ |x_{0} − x′_{0}|/2. Since |v − x′_{0}| ≤ d′ ≤ (1 + s/2)r and since r = λt/3 ≤ t/6, these estimates yield (A.9) with q = 1/4 − 3s/8 > 1/5.

Next assume that |x′_{0}| < t. Now |y| < t and x′_{0} ∈ [y, z]. As in Case 2 we get dev(x_{0}, z) ≥ 1 − 1/200. Since

|x′_{0}| ≥ |y| + r ≥ λt + λt/3 = 4λt/3,

Lemma A.3 gives |x′_{0} − u| ≥ 2λt(1 − 1/200)/3. As |x′_{0} − v| ≤ d′ ≤ (1 + s/2)r ≤ (1 + 1/400)λt/3, this yields (A.9) with q = λ/4.

Case 4. u ∈ τ[x_{1}, x′_{0}] = [x_{1}, x_{0}] ∪ β_{0} , v ∈ γ′_{0} . If u ∈ [x_{1}, x_{0}], then |u − x′_{1}| ≥ λt/4 by Case 3. If u ∈ β_{0} , then |u − x′_{1}| ≥ d(x′_{1}, β_{0}) = |x′_{1} − x′_{0}| = d′ ≥ r = λt/3. Since

|v − x′_{1}| ≤ sd′ ≤ s(1 + s/2)t/6 < λt/500,

we obtain (A.9) with q = λt/5.

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Received 11 May 2004