Computational Methods and Function Theory Volume 4 (2004), No. 2, 419–434

Quadrilaterals and John Disks Ole Jacob Broch (Communicated by Olli Martio) Abstract. Relations involving ordered quadruples of points on the boundary of a Jordan domain are investigated, and it is shown that these relations hold if and only if the domain is a John disk. In particular connections between absolute cross ratios and moduli of corresponding quadrilaterals are examined. A characterization of John disks based on a relation between diameters of subarcs of the boundary and bounds for moduli of associated quadrilaterals is also given. The results are one sided analogues to characterizations of quasidisks. Keywords. John disk, quadrilateral, reversed triangle inequality, cross ratio, quasim¨ obius map, modulus, harmonic doubling condition. 2000 MSC. Primary 30C20; Secondary 30C99.

1. Introduction Ptolemy’s identity states that if {z1 , z2 , z3 , z4 } is an ordered quadruple of points on a circle in R2 , then (1.1)

|z1 − z2 ||z3 − z4 | + |z2 − z3 ||z1 − z4 | = |z1 − z3 ||z2 − z4 |.

In other words the product of the two diagonals of the inscribed quadrilateral equals the sum of the two products of opposite sides. Equation (1.1) holds also in the case when the points lie on a circle in R2 . This can be seen by employing a M¨obius transformation, if necessary, mapping the circle onto a circle in R2 . Indeed, (1.1) states that the sum of two cross ratios equals 1, and cross ratios are invariant under M¨obius transformations. Furthermore, if (1.1) holds for every ordered quadruple of points on the boundary ∂D of a Jordan domain D, then D is a circle or a half plane; see [7] for a proof. If the equality in (1.1) is replaced by an inequality, we get a condition which is satisfied exactly by quasidisks. To be more precise: a Jordan domain D is a Received September 13, 2004, in revised form December 16, 2004. c 2004 Heldermann Verlag ISSN 1617-9447/$ 2.50 

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quasidisk if and only if there is a constant c such that for every ordered quadruple of points {z1 , z2 , z3 , z4 } ∈ ∂D we have (1.2)

|z1 − z2 ||z3 − z4 | + |z2 − z3 ||z1 − z4 | ≤ c|z1 − z3 ||z2 − z4 |.

See [18, 4.4], as well as [6] for a proof. The inequality (1.2) is called a reversed triangle inequality. In the present note we will replace Euclidean distances in (1.2) by internal distances (see (1.3) below). By doing this, we get a characterization for John disks; see Theorem 2.3. We then proceed, in Section 3, to exploit this relation to give a characterization of John disks in terms of quasim¨obius maps. In Section 4 we prove a result relating cross ratios of quadruples of points on the boundary to the moduli of the associated quadrilaterals in a Jordan John domain. The results on John disks are analogous to results for quasidisks. The characterizations of quasidisks are probably known already, but are not stated explicitly anywhere (to my knowledge). The formulations here have been included to emphasize the analogies between quasidisks and John disks. The results in Sections 2 to 4 in this paper state that John disks are exactly those domains which are “round” when measured with the internal metric, a fact which is of course well appreciated already. In Section 5 yet another characterization of John disks is given: a condition that involves diameters of boundary arcs and a fixed bound for the modulus of an associated quadrilateral. It is closely linked to the harmonic doubling condition, which characterizes John disks; [9], [10] and [8]. Throughout this paper we will consider only Jordan domains. The reason for this is that we will use moduli of quadrilaterals and harmonic measures of boundary arcs. Such quantities do not always behave well in non-Jordan domains. Notation and preliminaries Recall that a Jordan domain D ⊂ R2 is called a K-quasidisk if it is the image of a disk or a half plane under a K-quasiconformal self map of R2 . We will say simply quasidisk if we mean a K-quasidisk for some K. See [3], [4] and [6] for information on quasidisks. A simply connected proper subdomain D ⊂ R2 is said to be a c-John disk if it satisfies the following: for every pair z1 , z2 ∈ D there is a rectifiable arc α ⊂ D joining them, such that for every z ∈ α, min (α[zj , z]) ≤ c dist(z, ∂D),

j=1,2

where α[zj , z] denotes the subarc from zj to z and (β) denotes the length of an arc; see [14, 2.26]. By a John disk we will mean a c-John disk for some c. All quasidisks are John disks. Because John disks may have inward (but not outward) cusps, many properties of quasidisks become properties of John disks if we replace Euclidean distances

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by internal distances. If D is a Jordan domain, and if z1 , z2 ∈ D, the internal distance between z1 and z2 is the quantity (1.3)

λD (z1 , z2 ) = inf (β), β

where the infimum is taken over all rectifiable arcs joining z1 and z2 in D. Note that λD (z1 , z2 ) is defined also when one or both of z1 and z2 are rectifiably accessible boundary points of D. A point z ∈ ∂D is rectifiably accessible if there is a half open, rectifiable path in D ending at z. We write ∂r D for the subset of ∂D consisting of the rectifiably accessible points, and we let Dr = D ∪ ∂r D. It is easy to see that ∂r D is always dense in ∂D. Now λD is a metric in Dr . If D is a John disk, then ∂r D = ∂D; see e.g. [14] and [2, Section 2] for more details. In a convex domain D we have λD (z1 , z2 ) = |z1 − z2 |. The unit disk and its boundary are denoted by D and T, respectively, while H is the upper half plane. Euclidean diameters will be denoted by diam(A).

2. A reversed triangle inequality Definition 2.1. We will say that a Jordan domain D ⊂ R2 satisfies an internal reversed triangle inequality if there exists a constant b ≥ 1 such that (2.1)

λD (z1 , z2 )λD (z3 , z4 ) + λD (z2 , z3 )λD (z1 , z4 ) ≤ bλD (z1 , z3 )λD (z2 , z4 )

whenever {z1 , z2 , z3 , z4 } is an ordered quadruple of points in ∂r D (the order being inherited from ∂D). The two reversed triangle inequalities (1.2) and (2.1) are actually bounds for the sums of two absolute cross ratios; see Section 3 for the definition of absolute cross ratios. Example 2.2. Internal distances are, in general, difficult to compute. Even so, easy cases may provide good illustrations. For example, consider the domain given by: D = {x + iy : x < 0 or |y| > 1}. Then D is the outside of a half strip. Unfold the boundary of the half strip onto the real line, without stretching or compressing distances, using the map  sgn(y)(x + 1) if x = 0, f (x + iy) = y if x = 0. Thus |f (z1 ) − f (z2 )| = λH (f (z1 ), f (z2 )) = λD (z1 , z2 ) for every pair z1 , z2 ∈ ∂D, and f is an isometry with respect to internal distances. Now Ptolemy’s identity (1.1) states that |w1 − w2 ||w3 − w4 | + |w1 − w4 ||w2 − w3 | = |w1 − w3 ||w2 − w4 |

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for all ordered quadruples {w1 , w2 , w3 , w4 } ⊂ R = ∂H. This means that when D is the exterior of the half strip, (2.1) is satisfied with b = 1 (equality holds in this case). The complement of D, on the other hand, D∗ = {x + iy : x > 0, |y| < 1}, is convex, so that λD∗ (z1 , z2 ) = |z1 − z2 | there. Then D∗ cannot satisfy (2.1), for otherwise it would have to be a quasidisk (by [6, Section 2.3]), which it is not. The preceding example gives a clue about what holds for John disks in general; see also Remark 2.6. We have the following characterization for Jordan John disks. Theorem 2.3. A Jordan domain D ⊂ R2 is a c-John disk if and only if it satisfies the internal reversed triangle inequality (2.1) with constant b, where b and c depend only on each other. For the proof we need a version of the three point property for John disks, a proof of which is given in [2]. Lemma 2.4 ([2], Theorem 3.8). Let D be a Jordan domain. If D is a c-John disk, then there is a constant a such that (2.2)

λD (z1 , z2 ) ≤ aλD (z1 , z3 )

for any z1 , z3 ∈ ∂D, with z2 ∈ ∂D in the component of ∂D\{z1 , z3 } with the least Euclidean diameter. Conversely, suppose that there is a constant a such that whenever z1 , z3 ∈ ∂r D, inequality (2.2) holds for every z2 ∈ ∂r D in a component of ∂D\{z1 , z3 }. Then D is a c-John disk. The constants a and c depend only on each other. When ∂D is unbounded we have the following corollary of Lemma 2.4, which may serve as motivation for the more general result in Theorem 2.3. Corollary 2.5. A Jordan domain D with ∂D unbounded is a c-John disk if and only if there exists b ≥ 1 such that λD (z1 , z2 ) + λD (z2 , z3 ) ≤ bλD (z1 , z3 ) for any ordered triple {z1 , z2 , z3 } of points in ∂r D, i.e. z2 lies in the bounded component of ∂D \ {z1 , z3 }. The constants b and c depend only on each other. Proof. This follows immediately from Lemma 2.4 because the arc of smallest diameter is simply the bounded one. Proof of Theorem 2.3. We follow the proof of [6, Lemma 2.3.2].

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Necessity: Suppose that the Jordan domain D is a c-John disk. Choose any ordered quadruple of points {z1 , z2 , z3 , z4 } ⊂ ∂r D = ∂D. We may assume that (2.3)

λD (z1 , z3 ) ≤ λD (z2 , z4 ),

otherwise relabel the points. Let β2 be that component of ∂D\{z1 , z3 } which contains z2 and let β4 be the other component, which contains z4 . By relabeling once more (if necessary) we may assume further that diam β2 ≤ diam β4 . Since D is c-John we see from Lemma 2.4 that (2.4)

λD (z1 , z2 ) ≤ aλD (z1 , z3 ),

where a = a(c), and likewise that (2.5)

λD (z2 , z3 ) ≤ aλD (z1 , z3 ).

From (2.5) and (2.3) we now get: λD (z3 , z4 ) ≤ ≤ ≤ =

λD (z2 , z3 ) + λD (z2 , z4 ) aλD (z1 , z3 ) + λD (z2 , z4 ) aλD (z2 , z4 ) + λD (z2 , z4 ) (a + 1)λD (z2 , z4 ).

By (2.4) and (2.3) we obtain similarly λD (z4 , z1 ) ≤ (a + 1)λD (z2 , z4 ). Then by (2.4) λD (z1 , z2 )λD (z3 , z4 ) ≤ λD (z1 , z2 )(a + 1)λD (z2 , z4 ) ≤ aλD (z1 , z3 )(a + 1)λD (z2 , z4 ) = a(a + 1)λD (z1 , z3 )λD (z2 , z4 ), and, likewise, by (2.5) we have λD (z2 , z3 )λD (z4 , z1 ) ≤ a(a + 1)λD (z1 , z3 )λD (z2 , z4 ). Adding the last two inequalities gives λD (z1 , z2 )λD (z3 , z4 ) + λD (z2 , z3 )λD (z4 , z1 ) ≤ 2a(a + 1)λD (z1 , z3 )λD (z2 , z4 ). Thus (2.1) holds with b = 2a(a + 1). Sufficiency: Now suppose that the internal reversed triangle inequality (2.1) holds with constant b, but that D is not a John disk, so that the three point property (Lemma 2.4, (2.2)) does not hold with this constant. Then there exist

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z1 , z3 ∈ ∂r D such that we can find z2 ∈ β2 ∩ ∂r D and z4 ∈ β4 ∩ ∂r D, where β2 and β4 are components of ∂D\{z1 , z3 }, with λD (z1 , z2 ) > bλD (z1 , z3 ) λD (z1 , z4 ) > bλD (z1 , z3 ). Then we have bλD (z1 , z3 )λD (z2 , z4 ) ≤ bλD (z1 , z3 )[λD (z2 , z3 ) + λD (z3 , z4 )] = bλD (z1 , z3 )λD (z3 , z4 ) + bλD (z1 , z3 )λD (z2 , z3 ) < λD (z1 , z2 )λD (z3 , z4 ) + λD (z1 , z4 )λD (z2 , z3 ). This is a contradiction. Thus we must have either λD (z1 , z2 ) ≤ bλD (z1 , z3 ) for all z2 ∈ β2 ∩ ∂r D or λD (z1 , z4 ) ≤ bλD (z1 , z3 ) for all z4 ∈ β4 ∩ ∂r D or both. In any case Lemma 2.4 shows that D must be a c-John disk, with c = c(b). Remark 2.6. One of the interesting aspects of the reversed triangle inequality (1.2) is that it is satisfied with c = 1 if and only if the domain is a disk or a half plane. The three point property does not characterize disks or half planes in this way, i.e. the three point property is not numerically sharp; see [7] and [12, Example 4.11]. If we have b = 1 in (2.1) we do not necessarily have a disk or a half plane, as shown by Example 2.2. This is because we have a wider class of isometries when working with internal distances.

3. Quasim¨ obius maps We give the definitions of absolute cross ratios and quasim¨obius maps. If (X, d) is a metric space, we will use the notation |z1 , z2 , z3 , z4 |d =

d(z1 , z3 ) d(z2 , z4 ) d(z1 , z4 ) d(z2 , z3 )

for absolute cross ratios. (Other authors may use a slightly different notation.) For instance, (2.1) now reads |z1 , z4 , z2 , z3 |λ + |z1 , z2 , z4 , z3 |λ ≤ b. For convenience we omit D in the notation |z1 , z2 , z3 , z4 |λ , assuming that it will not cause any confusion. When d is the standard metric in Euclidean space we will write only |z1 , z2 , z3 , z4 |. We use the convention that |z1 , z2 , z3 , ∞| =

|z1 − z3 | . |z2 − z3 |

Following V¨ais¨al¨a, [19, p. 219], we say that a homeomorphism f : (X, d1 ) → (Y, d2 )

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between metric spaces (X, d1 ) and (Y, d2 ) is ϑ-quasim¨ obius (ϑ-QM) if there is an increasing homeomorphism ϑ : [0, ∞) → [0, ∞) such that |f (x1 ), f (x2 ), f (x3 ), f (x4 )|d2 ≤ ϑ (|x1 , x2 , x3 , x4 |d1 ) for all x1 , x2 , x3 , x4 ∈ X. Quasim¨obius maps are akin to the well known quasisymmetric maps. Every quasisymmetric map is quasim¨obius ([19, Theorem 5.2], [17, Proposition 5.14]), but unlike quasisymmetric maps, a quasim¨obius map may send bounded sets to unbounded sets. Also notice that quasim¨obius maps remain invariant under M¨obius maps of Rn , while quasisymmetric maps are invariant under similarity maps of Rn . For the record we recall that f : (X, d1 ) → (Y, d2 ) is ηquasisymmetric (η-QS), for some increasing homeomorphism η : [0, ∞) → [0, ∞), if   d1 (x1 , x2 ) d2 (f (x1 ), f (x2 )) ≤η for all x1 , x2 , x3 ∈ X, x2 = x3 . d2 (f (x2 ), f (x3 )) d1 (x2 , x3 ) Observe that compositions and inverses of QM maps are QM maps, [19, p. 219]. Quasicircles are images of T under quasiconformal self maps of R2 . In fact, a Jordan curve is a quasicircle if and only if it is the image of T under a quasim¨obius self map of R2 ; see e.g. [17, Propositions 5.10, 5.11 and 5.14]. The following lemma is an analogous one sided statement for John disks. Lemma 3.1. Let D be a Jordan domain in R2 . Then D is a c-John disk if and only if every conformal map f : D → D has an extension f : D → D such that f : f −1 (∂r D) → (∂r D, λD ) is ϑ-QM. Moreover, c and ϑ depend only on each other. By f : f −1 (∂r D) → (∂r D, λD ) we mean that we use the Euclidean metric in f −1 (∂r D) ⊂ T and the internal metric in (∂r D, λD ). Proof of Lemma 3.1. Suppose first that D is c-John. If ∂D is bounded, then the statement is easily proved, using Lemma 4.4 in [2] and the fact that QS maps are QM, by Theorem 3.2 in [19]. If ∂D is unbounded, there is, by Lemma 4.4 in [2], a conformal map g : H → D such that the boundary extension g : R → (∂D, λD ) is QS. By composing this with an honest M¨obius map from D to H, we obtain the desired map, noting that M¨obius transformations preserve all cross ratios. All QS or QM maps involved here depend only on c. On the other hand, if D is the image of D under a map ϕ which is ϑ-QM on the boundary, then pick an ordered quadruple {z1 , z2 , z3 , z4 } ⊂ ∂r D, with preimages ζj in T. Then |z1 , z4 , z2 , z3 |λ + |z1 , z2 , z4 , z3 |λ ≤ ϑ (|ζ1 , ζ4 , ζ2 , ζ3 |) + ϑ (|ζ1 , ζ2 , ζ4 , ζ3 |) ≤ 2ϑ(1).

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The last inequality holds because in a disk or half plane we have the identity (1.1), so that |ζ1 , ζ4 , ζ2 , ζ3 | < 1 and |ζ1 , ζ2 , ζ4 , ζ3 | < 1. Then D is a c-John disk, c = c(ϑ), by Theorem 2.4. Remark 3.2. (i) We did not need a conformal map for the sufficiency in Lemma 3.1, as shown by the proof, but in Theorem 4.3 we do need the quasim¨obius map onto D to be conformal. (ii) In [19, Section 4] V¨ais¨al¨a proved that the image of a uniform domain under a quasim¨obius map is uniform. (iii) The proof of the lemma above can easily be adapted to show that every internal QM image of a John disk is a John disk.

4. Cross ratios and moduli of quadrilaterals We now propose a way to characterize quasidisks and John disks via a relation between cross ratios and moduli of quadrilaterals. Let us first recall a few definitions. A quadrilateral Q is a Jordan domain D ⊂ R2 together with an ordered quadruple of points {z1 , z2 , z3 , z4 } ⊂ ∂D; these points are the vertices of Q, and they divide ∂D into four arcs, the sides of Q. If D is a Jordan domain, we write D(z1 , z2 , z3 , z4 ) for the quadrilateral determined by D and an ordered quadruple {z1 , z2 , z3 , z4 } of points in ∂D; see Section 3.10 of [6]. Given a family Γ of curves in R2 , the modulus of Γ, mod Γ, is defined as  mod Γ = inf (z)2 dm, 

R2

where the infimum is taken over all non-negative Borel measurable functions in R2 with  (z) |dz| ≥ 1 γ

for all γ ∈ Γ. If Γ is the family of all curves in a quadrilateral Q = D(z1 , z2 , z3 , z4 ) joining the two sides from z1 to z2 and from z3 to z4 , we write mod Q = mod Γ and say that this is the modulus of the quadrilateral Q. See for example [13], [1] or [6] for more about modulus. In the unit disk there is an interesting and important connection between cross ratios and the modulus of the associated quadrilateral. More precisely, we have the following classical result ([13, II.2], [5, Lemma 1.2], [6]).

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Lemma 4.1. Let {ζ1 , ζ2 , ζ3 , ζ4 } be an ordered quadruple of points in T. Then 1 < |ζ1 , ζ2 , ζ3 , ζ4 | < ∞ and  2  mod D(ζ1 , ζ2 , ζ3 , ζ4 ) = μ |ζ1 , ζ2 , ζ3 , ζ4 |−1/2 π where √ 4 π K( 1 − r2 ) log μ(r) = 2 K(r) r as r → 0 and where K(r) is the complete elliptic integral of the first kind  π/2 dt  . K(r) = 0 1 − r2 sin2 t Observe that the map μ is decreasing with t, and hence increasing with 1/t. Could it be that a relation similar to the one in Lemma 4.1 holds in every Jordan domain? For example, could there be a bound for the modulus of a quadrilateral depending of the cross ratio of the vertices? Unfortunately the answer is no. An infinite half strip provides an easy counterexample. The half strip is not a John disk (hence not a quasidisk), a fact which is easy to prove directly from the definition. In quasidisks, however, we do have a bound for the modulus in the spirit of Lemma 4.1. The following characterization of quasidisks is probably known, but I have found no reference for the result in the present formulation. Theorem 4.2. A Jordan domain D ⊂ R2 is a K-quasidisk if and only if there exists an increasing homeomorphism ν : [0, ∞) → [0, ∞) such that mod D(z1 , z2 , z3 , z4 ) ≤ ν(|z1 , z2 , z3 , z4 |) for every ordered quadruple {z1 , z2 , z3 , z4 } ⊂ ∂D. Here K and ν depend only on each other. We will prove the following analogue for Jordan John disks. (Theorem 4.2 can be proved in a similar way.) Theorem 4.3. A Jordan domain D ⊂ R2 is a c-John disk if and only if there exists an increasing homeomorphism ν : [0, ∞) → [0, ∞) such that mod D(z1 , z2 , z3 , z4 ) ≤ ν(|z1 , z2 , z3 , z4 |λ ) for every ordered quadruple {z1 , z2 , z3 , z4 } ⊂ ∂r D. Here c and ν depend only on each other. Proof. First, let D be a Jordan c-John disk and suppose that {z1 , z2 , z3 , z4 } is an ordered quadruple in ∂r D. Let ϕ : D → D be a conformal map. Then ϕ extends as a homeomorphism of the boundaries, so that ϕ : (∂D, λD ) → T is quasim¨obius by Lemma 3.1. Thus there is an increasing homeomorphism of the positive real

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line ϑ depending only on c such that [ζ1 , ζ2 , ζ3 , ζ4 ] ≤ ϑ(|z1 , z2 , z3 , z4 |λ ), where ζj = ϕ(zj ). Lemma 4.1 then implies

2 −1/2 . mod D(ζ1 , ζ2 , ζ3 , ζ4 ) = μ |ζ1 , ζ2 , ζ3 , ζ4 |λ π But then mod D(z1 , z2 , z3 , z4 ) = mod D(ζ1 , ζ2 , ζ3 , ζ4 )  2  ≤ μ ϑ(|z1 , z2 , z3 , z4 |λ )−1/2 π = ν(|z1 , z2 , z3 , z4 |λ ), because the modulus is a conformal invariant. We define ν by the last inequality. Conversely, suppose that D is a Jordan domain and that there exists an increasing homeomorphism ν : [0, ∞) → [0, ∞) such that mod D(z1 , z2 , z3 , z4 ) ≤ ν(|z1 , z2 , z3 , z4 |λ ) whenever {z1 , z2 , z3 , z4 } is an ordered quadruple of points in ∂r D. By using a conformal map ϕ : D → D, which extends as a homeomorphism to the boundary, and writing ζj = ϕ(zj ), we see that mod D(ζ1 , ζ2 , ζ3 , ζ4 ) = mod D(z1 , z2 , z3 , z4 ) ≤ ν(|z1 , z2 , z3 , z4 |λ ). As before we then have  2  μ |ζ1 , ζ2 , ζ3 , ζ4 |−1/2 = mod D(ζ1 , ζ2 , ζ3 , ζ4 ) ≤ ν(|z1 , z2 , z3 , z4 |λ ). π Because μ is decreasing we see that



−1/2 −1 π −1 π mod D(ζ1 , ζ2 , ζ3 , ζ4 ) ≥ μ ν(|z1 , z2 , z3 , z4 |λ ) . =μ |ζ1 , ζ2 , ζ3 , ζ4 | 2 2 Then π

−2  ν(|z1 , z2 , z3 , z4 |λ ) = ϑ(|z1 , z2 , z3 , z4 |λ ). |ζ1 , ζ2 , ζ3 , ζ4 | ≤ μ−1 2 Defining ϑ by the last equality, we have |ϕ(z1 ), ϕ(z2 ), ϕ(z3 ), ϕ(z4 )| ≤ ϑ(|z1 , z2 , z3 , z4 |λ ), so that ϕ : (∂r D, λD ) → T is quasim¨obius. Because inverses of QM maps are QM, D is a c-John disk by Lemma 3.1, and we see that c depends only on ν.

5. Diameters and moduli In this last section we exhibit another characterization of Jordan John disks, relating diameters of boundary arcs to an absolute bound for the modulus of a quadrilateral. We will not consider internal distances as such, but rather study one sided conditions in the spirit of Pommerenke [16].

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We first introduce some notation. Let D be a Jordan domain. Given any ordered quadruple {z1 , z2 , z3 , z4 } ⊂ ∂D, let a1 denote the arc from z1 to z2 , b1 the arc from z2 to z3 , a2 the arc from z3 to z4 and finally b2 the arc from z4 to z1 . We will use this notation throughout the section. The letters a and b are meant to correspond to the “a-sides” and “b-sides” of quadrilaterals, [1, p. 6]. If γ ⊂ ∂D is an arc, then the harmonic measure of γ with respect to the point z0 ∈ D will be denoted by ω(z0 , γ; D). According to Jerison and Kenig [8] (see also Kim and Langmeyer [10, Section 2] and Kim [9]) a Jordan domain is said to satisfy the harmonic doubling condition if for some z0 ∈ D and some constant c0 > 0, ω(z0 , a; D) ≤ c0 ω(z0 , b; D) for each pair of consecutive arcs a and b in ∂D with diam(a) ≤ 2 diam(b); diam(A) denotes the Euclidean diameter of A. Then we have the following lemma. Lemma 5.1 ([10], Theorem 2.3). A bounded Jordan domain D is a c-John disk with center z0 if and only if it satisfies the harmonic doubling condition with constant c0 , where c and c0 depend only on each other. Remark 5.2. (i) Jerison and Kenig, [8, Theorem 2.7], showed that a bounded Jordan domain D is a quasidisk if and only if D and D∗ = R2 \ D both satisfy the harmonic doubling condition. (ii) The assumption that ∂D be bounded is essential. The reason is that boundary arcs with large Euclidean diameters in unbounded domains need not carry much harmonic measure. For example, let D be the right half plane, and consider boundary arcs ay = [0, iy] and by = [iy, i2y]. We have diam(ay ) = diam(by ), but ω(1, ay ; D) arctan y = lim = ∞, y→∞ ω(1, by ; D) y→∞ arctan 2y − arctan y lim

i.e. there is no harmonic doubling condition even in the half plane. For convenience we introduce the following concept. Definition 5.3. A Jordan domain D has the diameter-modulus property if there is a constant c1 such that for each ordered quadruple {z1 , z2 , z3 , z4 } in ∂D we have mod D(z1 , z2 , z3 , z4 ) ≤ c1 , if diam(a1 ) ≤ 2 minj=1,2 diam(bj ). Lemma 5.4. Suppose that D is a bounded Jordan domain that satisfies the harmonic doubling condition with constant c0 relative to z0 ∈ D. Then D has the diameter-modulus property with c1 = c1 (c0 ).

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Proof. Assume that D satisfies the harmonic doubling condition. Take an ordered quadruple {z1 , z2 , z3 , z4 } ⊂ ∂D and assume that diam(a1 ) ≤ 2 min diam(bj ). j=1,2

Let f : D → D be a conformal map with f (z0 ) = 0. We write ζj = f (zj ) ∈ T. Clearly f extends as a homeomorphism of the boundaries. We will show that there is a constant b = b(c0 ) such that |ζ1 , ζ2 , ζ3 , ζ4 | ≤ b.

(5.1) By Lemma 4.1 we have then

 2  μ |ζ1 , ζ2 , ζ3 , ζ4 |−1/2 . π Hence the properties of μ and by the conformal invariance of the modulus, (5.1) will imply that  2  mod D(z1 , z2 , z3 , z4 ) ≤ μ b−1/2 .  −1/2  π 2 . It remains to prove (5.1). Thus the lemma holds with c1 = π μ b mod D(ζ1 , ζ2 , ζ3 , ζ4 ) =

We see that since diam(a1 ) ≤ 2 diam(bj ), j = 1, 2, we have ω(z0 , a1 ; D) ≤ c0 ω(z0 , bj ; D),

j = 1, 2,

by the harmonic doubling condition on D. Harmonic measure is a conformal invariant, so if we let αi = f (ai ), βj = f (bj ) we get ω(0, α1 ; D) ≤ c0 ,

ω(0, βj ; D),

j = 1, 2.

Recall also that the harmonic measure of an arc in the unit circle T with respect to the origin is equal to the normalized length of that arc. Thus for any arc γ ⊂ T with end points w1 , w2 we have ω(0, γ; D) ≤ |w1 − w2 | ≤ 2π ω(0, γ; D)

(5.2)

whenever γ lies in a half circle. Note that the first inequality is not true in general. Assume first that none of the arcs β1 or β2 are longer than a half circle. Then we have (5.3)

|ζ1 − ζ2 | ≤ 2π ω(0, α1 ; D) ≤ 2πc0 ω(0, β1 ; D) ≤ 2πc0 |ζ2 − ζ3 |,

and (5.4)

|ζ1 − ζ2 | ≤ 2π ω(0, α1 ; D) ≤ 2πc0

ω(0, β2 ; D) ≤ 2πc0 |ζ4 − ζ1 |.

Using (5.4) we obtain (5.5)

|ζ1 − ζ3 | ≤ |ζ1 − ζ2 | + |ζ2 − ζ3 | ≤ (2πc0 + 1)|ζ2 − ζ3 |,

while (5.5) gives (5.6)

|ζ2 − ζ4 | ≤ |ζ2 − ζ1 | + |ζ4 − ζ1 | ≤ (2πc0 + 1)|ζ4 − ζ1 |.

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Thus we have |ζ1 , ζ2 , ζ3 , ζ4 | =

|ζ1 − ζ3 | |ζ2 − ζ4 | ≤ (2πc0 + 1)2 . |ζ2 − ζ3 | |ζ4 − ζ1 |

If β1 is longer than a half circle, then |ζ1 − ζ3 | < |ζ2 − ζ3 |, and using (5.6) we have |ζ2 − ζ4 | ≤ |ζ2 − ζ1 | + |ζ1 − ζ4 | ≤ (2πc0 + 1)|ζ1 − ζ4 | as above. Thus we have |ζ1 , ζ2 , ζ3 , ζ4 | ≤ (2πc0 + 1) in this case. The proof in the case when β2 is longer than a half circle is similar. Finally note that by (5.3) and (5.4) it does not matter if αj is longer than a half circle. In any case we have the bound (5.1) with b = (2πc0 + 1)2 . The number 2 may be replaced by any number strictly greater than 1 in the definition of the harmonic doubling condition, and hence in the above lemmas. This can be seen from the proof of Theorem 2.3 in [10]. Of course we have to change the constants c0 and c1 accordingly. Likewise the point z0 in Lemma 5.1 may be moved around by changing the constant, by Remark 2.2 in [10]. Next we prove the converse of Lemma 5.4. We retain the notation ai , bj related to quadruples z1 , z2 , z3 , z4 from above. Lemma 5.5. Let D be a bounded Jordan domain, and suppose that D has the diameter-modulus property. Then D satisfies the harmonic doubling condition. Proof. We assume that D does not satisfy the harmonic doubling condition, and show that then it cannot have the diameter-modulus property either. Thus for some z0 ∈ D and all c0 > 0 there are adjacent arcs a, b ⊂ ∂D with diam(a) ≤ 2 diam(b) and ω(z0 , a; D) > c0 ω(z0 , b; D). Fix such a z0 ∈ D and a conformal map f : D → D with f (z0 ) = 0. Then for every natural number n there are adjacent arcs a1n , b1n ⊂ ∂D with diam(a1n ) ≤ 2 diam(b1n ) and ω(z0 , a1n ; D) > n ω(z0 , b1n ; D). Because harmonic measure is bounded by 1 this means that ω(z0 , b1n ; D) → 0 as n → ∞. It is not difficult to see that since z0 is fixed, diam(b1n ) → 0 as n → ∞. Then diam(a1n ) → 0 too, and finally we see that ω(z0 , a1n ; D) → 0. Let z1n and z2n be the end points of a1n , while z2n and z3n are the end points of b1n . As ∂D is compact, by considering subsequences if necessary, we see that (z1n ), (z2n ), (z3n ) all tend to the same point z (since |z1n − z2n |, |z2n − z3n | → 0). We may assume, possibly by considering subsequences once more and without any essential loss of generality, that all the b1n lie on the same side of a1n , and that a1n comes before b1n in the natural ordering of ∂D; see Figure 5. Let w ∈ ∂D be the point farthest away from z and let b2n be the subarc of ∂D from w to z1n . If n is large enough, diam(b2n ) > diam(a1n ). Then (5.7)

diam(a1n ) ≤ 2 min diam(bjn ). j=1,2

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z1n z2n

a1n

b1n z3n b2n

D

w Figure 1. Now let ζjn = f (zjn ) and α1n = f (a1n ), βjn = f (bjn ). By conformal invariance we have: ω(0, α1n ; D) = ω(z0 , a1n ; D),

ω(0, βjn ; D) = ω(z0 , bjn ; D).

By (5.2): ω(0, α1n ; D) |ζ1n − ζ2n | ≤ 2π , ω(0, β1n ; D) |ζ2n − ζ3n | because if n is large enough, then both α1n and β1n are contained in a half circle. Thus by assumption |ζ1n − ζ2n | = ∞. lim n→∞ |ζ2n − ζ3n | On the other hand f (w) is fixed and we have that |ζ2n − f (w)| = 1. n→∞ |ζ1n − f (w)| lim

In effect, by using that |ζ1n − ζ3n | > |ζ1n − ζ2n |, |ζ1n − ζ3n | |ζ2n − f (w)| = ∞. n→∞ |ζ2n − ζ3n | |ζ1n − f (w)| lim

By Lemma 4.1 and conformal invariance of the modulus we have lim mod D(z1n , z2n , z3n , w) = lim mod D(ζ1n , ζ2n , ζ3n , w) = ∞.

n→∞

n→∞

This means that D does not have the diameter-modulus property.

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Combining Lemmas 5.4 and 5.5 we obtain the following characterization of bounded Jordan John disks. Theorem 5.6. A bounded Jordan domain D ⊂ R2 is a John disk if and only if there is a constant c1 such that whenever {z1 , z2 , z3 , z4 } ⊂ ∂D is an ordered quadruple with diam(a1 ) ≤ 2 minj=1,2 diam(bj ) we have mod D(z1 , z2 , z3 , z4 ) ≤ c1 . Proof. Lemmas 5.4 and 5.5 say that D has the c1 -diameter-modulus property if and only if it satisfies the c0 -harmonic doubling condition, which happens if and only if D is a c-John disk by Lemma 5.1. Closing remarks According to Kim and Langmeyer [10, Section 1] the harmonic doubling condition (Lemma 5.1) may be regarded as a one sided version of the harmonic quasisymmetry property. Krzyˇz [11] proves this property to be characteristic for quasidisks. In their recent paper [5], Gehring and Hag relate the quasisymmetry property to Pfluger’s well known characterization of quasidisks [15] in terms of the conjugate quadrilateral inequality; see also [13], [5], [6]. They show that a Jordan domain D has the harmonic quasisymmetry property if and only if it satisfies the conjugate quadrilateral inequality. Lemmas 5.4 and 5.5 show that a Jordan domain D has the diameter-modulus property if and only if it has the harmonic doubling condition. Perhaps we may consider Theorem 5.6 a one sided version of the conjugate quadrilateral inequality? Acknowledgements. I want to thank my supervisor, Kari Hag, for all the encouragement with this paper. I am also indebted to Per Hag, whose many suggestions and remarks greatly improved both the content and exposition. Finally I am grateful to the referee for a thorough review of this paper.

References 1. L. V. Ahlfors, Lectures on Quasiconformal Mappings, Van Nostrand Company, 1966. 2. O. J. Broch, Extension of internally bilipschitz maps in John disks, submitted. 3. F. W. Gehring, Characteristic Properties of Quasidisks, Les presses de l’universit´e de Montr´eal, 1982. , Characterizations of quasidisks, Banach Center Publications 48 (1999), 11–41. 4. 5. F. W. Gehring and K. Hag, Sewing Homeomorphisms and Quasidisks, Comput. Methods Funct. Theory 3 (2003), 143–150. , The ubiquitous quasidisk, to appear. 6. 7. K. Hag, What is a disk?, Banach Center Publications 48 (1999), 43–53. 8. D. S. Jerison and C. E. Kenig, Boundary behaviour of harmonic functions in nontangentially accessible domains, Adv. Math. 46 (1982), 80–147. 9. K. Kim, Harmonic doubling condition and John disks, Comm. Korean Math. Soc. 10 (1995), 145–153. 10. K. Kim and N. Langmeyer, Harmonic measure and hyperbolic distance in John disks, Math. Scand. 83 (1998), 283–299.

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11. J. G. Krzyˇz, Quasicircles and harmonic measure, Ann. Acad. Sci. Fenn. (Ser. A. I. Math.) 12 (1987), 19–24. 12. N. Langmeyer, The Quasihyperbolic Metric, Growth, and John Domains, Ph. D. thesis, University of Michigan, 1996. 13. O. Lehto and K. I. Virtanen, Quasiconformal Mappings in the Plane, 2nd ed., Springer, 1973. 14. R. N¨ akki and J. V¨ ais¨al¨ a, John disks, Expo. Math. 9 (1991), 3–43. ¨ 15. A. Pfluger, Uber die Konstruktion Riemannscher Fl¨ achen durch Verheftung, J. Indian Math. Soc. 24 (1960), 401–412. 16. Ch. Pommerenke, One-sided smoothness conditions and conformal mapping, J. London Math. Soc. (2) 26 (1982), 77–88. , Boundary Behaviour of Conformal Maps, Springer, 1992. 17. 18. S. Rickman, Characterization of quasiconformal arcs, Ann. Acad. Sci. Fenn. (Ser. A I Math.) 395 (1966), 3–30. 19. J. V¨ais¨al¨ a, Quasi-M¨ obius maps, J. Analyse Math. 44 (1984/1985), 218–234. Ole Jacob Broch E-mail: [email protected] Address: Department of Mathematical Sciences, Norwegian University of Science and Technology, N-7491 Trondheim, Norway.