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arXiv:1710.10305v1 [math.CV] 27 Oct 2017

SQUEEZING FUNCTIONS AND CANTOR SETS.

L. AROSIO†, J. E. FORNÆSS††, N. SHCHERBINA, AND E. F. WOLD†††

Abstract. We construct “large” Cantor sets whose complements resemble the unit disk arbi-

trarily well from the point of view of the squeezing function, and we construct “large” Cantor

sets whose complements do not resemble the unit disk from the point of view of the squeezing

function. Finally we show that complements of Cantor sets arising as Julia sets of quadratic

polynomials have degenerate squeezing functions, despite of having Hausdorﬀ dimension arbi-

trarily close to two.

1. Introduction

Recently there have been many studies of the boundary behaviour of the squeezing function

(see Section 2 for the deﬁnition and references) in one and several complex variables. In one

complex variable there are two extremes:

(1) if γ⊂bΩ is an isolated boundary component of a domain Ω which is not a point, then

lim

Ω∋z→γSΩ(z) = 1,(1.1)

(2) if γ⊂bΩ is an isolated boundary component of a domain Ω which is a point, then

lim

Ω∋z→γSΩ(z) = 0.(1.2)

This suggests studying the boundary behaviour of SΩ(z) where Ω = P1\K, and Kis a Cantor

set. In [1] Ahlfors and Beurling showed that there exist Cantor sets in P1whose complement

admits a bounded injective holomorphic function. In particular, such complements admit a

non-degenerate squeezing function, and so this class of domains is nontrivial from the point of

view of the squeezing function.

Our ﬁrst result is the following.

Theorem 1.1. For any ǫ > 0there exists a Cantor set Q⊂I2with 2-dimensional Lebesgue

measure greater than 1−ǫ, such that

lim

Ω∋z→QSΩ(z) = 1,(1.3)

and, moreover, SΩ(z)≥1−ǫfor all z∈Ω, where Ω = P1\Q.

Date: October 31, 2017.

2010 Mathematics Subject Classiﬁcation. 32E20.

†† and ††† Supported by NRC grant no. 240569 .

†Supported by the SIR grant “NEWHOLITE - New methods in holomorphic iteration” no. RBSI14CFME.

This work was done during the international research program ”Several Complex Variables and Complex

Dynamics” at the Center for Advanced Study at the Academy of Science and Letters in Oslo during the academic

year 2016/2017.

1

2 L. AROSIO, J. E. FORNÆSS††, N. SHCHERBINA, AND E. F. WOLD

We also show that there exist Cantor sets with completely diﬀerent behaviour.

Theorem 1.2. There exists a Cantor set Q⊂P1such that the following hold

(1) for any point x∈Qand any neighbourhood Uof xwe have that U∩Qhas positive

2-dimensional Lebesgue measure, and

(2) SΩachieves any value between zero and one on U∩Ω, where Ω = P1\Q.

Finally we show that certain Julia sets arising in one dimensional complex dynamics are

Cantor sets which are degenerate from the point of view of the squeezing function, although

they can have Hausdorﬀ dimension arbitrarily close to two and thus their complements admit

bounded holomorphic functions. Recall that a compact set of Hausdorﬀ dimension strictly

larger than one has strictly positive analytic capacity, hence its complement admits bounded

holomorphic functions (see e.g. [15], part (B) of Theorem 64, page 74).

Theorem 1.3. Let fc(z) = z2+cwith c /∈ M. Then P1\ Jcdoes not admit a bounded injective

holomorphic function.

Here, Jcdenotes the Julia set for the function fc, and Mdenotes the Mandelbrot set, so that

Jcis a Cantor set if and only if c /∈ M.

Other Cantor sets of this type were constructed by Ahlfors and Beurling [1].

2. A “large” Cantor Set whose complement resembles the unit disk - Proof of

Theorem 1.1

We give some deﬁnitions. Let △ ⊂ Cdenote the unit disc, and let Br(p)⊂Cdenote the disk

of radius rcentered at p. Let Ω ⊂P1be a domain and let x∈Ω. If ϕ: Ω → △ is an injective

holomorphic function such that ϕ(x) = 0, we set

SΩ,ϕ(x) := sup{r > 0 : Br(0) ⊂ϕ(Ω)},(2.1)

and we set

SΩ(x) := sup

ϕ

{SΩ,ϕ(x)},(2.2)

where the supremum is taken over all injective holomorphic functions ϕ: Ω → △ such that

ϕ(x) = 0. The function SΩis called the squeezing function. If the domain Ω does not admit any

bounded injective holomorphic functions, then the squeezing function is called degenerate. The

concept of squeezing function goes back to work by Liu-Sun-Yau, see [12] (2004), [13] (2005)

and S.-K- Yeung [17] (2009). More recently, Deng-Guan-Zhang, see [2] (2012) initiated a basic

study of the squeezing function. After that the squeezing function has been investigated by

several authors, among them, Fornæss-Wold [7] (2015), Nikolov-Trybula-Andreev [14] (2016),

Deng-Guan-Zhang [3](2016), Joo-Kim [10] (2016), Kim-Zhang[11] (2016), Zimmer [18] (2017),

Fornæss-Rong [5] (2017), Fornæss-Shcherbina [6] (2017), Diederich-Fornæss [4] and Fornæss-

Wold [8]. We will introduce an auxiliary function Rthat will enable us to bound the squeezing

function from below on the limit of a certain increasing sequence of domains. Let Ω ⊂P1be a

domain which admits an injective holomorphic map ψ: Ω ֒→ △. Then for any point x∈Ω it is

known (see, for example, Theorem 1 in [16]) that Ω also admits a circular slit map, that is an

injective holomorphic map ϕ: Ω ֒→ △ onto a circular slit domain S, such that ϕ(x) = 0. By

deﬁnition, Sis a circular slit domain if △ \ Sconsists of arcs lying on concentric circles centred

3

at the origin (the arcs may degenerate to points). If x∈Ω, we let Slitx(Ω) denote the set of all

circular slit maps that sends xto the origin. For a domain Ω ⊂P1we deﬁne

RΩ(x) := sup

ϕ∈Slitx(Ω)

{SΩ,ϕ(x)}.(2.3)

Notice that by deﬁnition RΩ≤SΩ.

Deﬁnition 2.1. Let {Ωj}j∈Nbe a sequence of domains in P1, and set Kj:= P1\Ωj. We say

that Ωjconverges strongly to a domain Ω ⊂P1with K:= P1\Ω, if the compact sets Kjconverge

to Kin the Hausdorﬀ distance, and we write Ωjs

→Ω. If xj∈Ωjand if xj→x∈Ω we write

(Ωj, xj)s

→(Ω, x).

Proposition 2.2. Let Ω⊂P1be a ﬁnitely connected domain such that no boundary component

of Ωis a point. Let N∈N, and let {Ωj}be a sequence of domains, where each Ωjis mj-connected

with mj≤N. Assume that (Ωj, xj)s

→(Ω, x). Then RΩj(xj)→RΩ(x).

Proof. Let K1, ..., Kmdenote the complementary components of Ω. Then for each 1 ≤k≤m

there is a unique slit map ϕk: Ω → △ such that ϕkidentiﬁes Kkwith b△,ϕk(x) = 0 and

ϕ′

k(x)>0 (see, for example, Theorem 7 in [16]). In particular, RΩ(x) is realised by one (or

more) of these maps.

Similarly, each Ωjhas complementary components Kj

kfor 1 ≤k≤mj, and for the Kj

k’s

that are not points, there are unique slit maps ϕj

kidentifying Kj

kwith b△,ϕj

k(xj) = 0 and

(ϕj

k)′(xj)>0.

After re-grouping to simplify notation, we may assume that there is a sequence sj≤mjsuch

that the compact set Kj

1∪ · · · ∪ Kj

sjconverges in the Hausdorﬀ distance to a complementary

component of Ω, say K1, and groups of the other Kj

i’s converge to the other complementary

components of Ω. Since the diameter of K1is strictly positive, we may assume that there is a

lower bound for the diameters of the sets {Kj

1}.

We ﬁrst claim that there exists a constant c > 0 such that (ϕj

1)′(xj)> c for all j. Notice

that, in view of Koebe’s 1

4-theorem, our claim implies that all slits are bounded away from zero.

Assume by contradiction that there is such a sequence (ϕj

1)′(xj)→0. For any convergent sub-

sequence of the sequence (ϕj

1), the limit map is constantly equal to 0. Choose such a convergent

subsequence and denote it still by (ϕj

1). After possibly having to pass to a subsequence, we may

now choose a nontrivial loop ˜γ:= bBr(0), where 0 < r < 1, which is contained in ϕj

1(Ωj) for all

j, such that the Kobayashi length of ˜γin ϕj

1(Ωj) is uniformly bounded from above.

Let Ube any (small) open neighbourhood of K1. Then for jsuﬃciently large, we have that

ϕj

1(Ωj\U) is contained in the disk bounded by ˜γ. Set ˜γj:= (ϕj

1)−1(˜γ). Then since ϕj

1identiﬁes

Kj

1with b△, we have that Ωj\Uis on one side of ˜γjand Kj

1is on the other. Then the

spherical lengths of the ˜γj’s are bounded uniformly from below, since the diameters of the Kj

1’s

are bounded uniformly from below. But then the Kobayashi length of ˜γjin Ωjgoes to inﬁnity,

a contradiction. So we may extract a subsequence from ϕj

1converging uniformly on compact

subsets of Ω to an injective map ˜ϕ: Ω → △.

4 L. AROSIO, J. E. FORNÆSS††, N. SHCHERBINA, AND E. F. WOLD

We claim that ˜ϕmaps Ω onto a slit domain. First we show that the slits cannot close up

to a circle of radius strictly less than one. Indeed, ﬁx a compact set Lin Ω.Then we can

assume that the sequence converges uniformly on L. This implies that if there is a slit Sof

minimal radius r < 1 which closes up to a circle in the limit, then eventually all the images of L

must be contained in the disc of radius r. Then arguing as above we can pick a circle of radius

r < s < 1 so that on the preimages of the circle, the Kobayashi length is arbitrarily large. This

is impossible.

We can assume that the slits converge. The complement of the limiting slits is connected.

Pick any compact subset Fof the complement of the limiting slits. Consider the inverse maps

of the ϕj

k. This is a normal family. Indeed, we can remove a small disc around a point where

the sequence ϕj

kis uniformly convergent. After this removal the family of inverses is a normal

family. The limit map of the inverses is then the inverse of a slit map on Ω, which proves that

the limit map ˜ϕis a slit map.

So ϕis, up to rotation, the unique slit map which identiﬁes K1with b△, and by choosing

other complementary components than K1in the above construction, all the possible slit maps

ϕk: Ω → △ may be obtained as such limits. So we would arrive at a contradiction if we did not

have RΩj(xj)→RΩ(x).

Lemma 2.3. Let Qj= [aj, bj]×[cj, dj]⊂Cbe pairwise disjoint cubes for j= 1, ..., m, and set

Ω := P1\(Q1∪ · · · ∪ Qm).(2.4)

For each j, set

Γj:= {aj+ (1/2)(bj−aj)} × [cj, dj],(2.5)

and for k∈Ndenote by Γj(1/k)the open 1

k-neighborhood of Γj, and by Ql

j,k and Qr

j,k the left

and right connected components of Qj\Γj(1/k)respectively. Set

Ωk:= P1\(Ql

1,k ∪Qr

1,k ∪ · · · ∪ Ql

m,k ∪Qr

m,k).(2.6)

Then for any ǫ > 0there exist δ > 0and N∈Nsuch that, for all k≥N,

RΩk(z)≥1−ǫif z∈Ωk∩Qj(δ)for some j, (2.7)

and

|RΩk(z)−RΩ(z)|< ǫ if z /∈Qj(δ)for all j, (2.8)

where Qj(δ)denotes the δ-neighborhood of Qj.

Proof. Since all cases are similar, to avoid notation, we prove (2.7) for j= 1. We may also

assume that Q1= [−1,1] ∪[−1,1].

For each k∈Nthere is a unique conformal map φk:P1\Ql

1,k ∪Qr

1,k →P1such that the

image is the complement of two closed disks B1

kand B2

k, normalized by the condition

φk(z) = z+

∞

X

j=1

ak

j(1/z)j(2.9)

near inﬁnity (see e.g. [9], Theorem 2, page 237). Then by uniqueness, φk(z) = −φk(−z), so

the two disks have the same size. Moreover, since each φkis normalized to have derivative

one at inﬁnity, the radii of the disks have to be bounded from above and from below: we can

5

assume that the centers and the radii (in the spherical metric) converge. Indeed, by the Koebe

1/4-theorem the discs must all be in a bounded region of C.Hence the radii are bounded above.

Next we assume that the radii converge to 0.Let p, q denote the limits of the centers. Then

the inverses are a normal family in the complement of the two points, hence the limit must be

constant. This is only possible if p, q =∞contradicting the uniform boundedness of the discs.

So by scaling and rotation, we may then assume that

B1

k=B1(−1−δk) and B2

k=B1(1 + δk),(2.10)

for some δk>0, where in general Br(p) denotes the disk of radius rcentered at p(however, we

have now possibly destroyed the normalization condition).

We now show that δk→0. Otherwise, consider the circles |z−(1 + δk)|= 1 + δk.These have

uniformly bounded Kobayashi length. However their preimage goes around one of the rectangles

and passes between the rectangles, where the Kobayashi metric is arbitrarily large. Hence their

Kobayashi length is unbounded, contradiction.

Since we may assume that the sequence {φk}converges to a conformal map, we get (2.7)

from Lemma 2.4 below. And since all cases are similar, we conclude that (2.7) holds for any

j= 1, ..., m. Finally (2.8) follows from Proposition 2.2.

Lemma 2.4. Set Ω := P1\(B1(−1) ∪B1(1) ∪K)be a domain, with Ka compact set with

ﬁnitely many connected components, disjoint from B1(−1) ∪B1(1). Let δjց0, and suppose

that

Ωj:= P1\(B1(−1−δj)∪B1(1 + δj)∪Kj) (2.11)

is a sequence of domains such that Kj→Kwith respect to Hausdorﬀ distance, and such that

the number of connected components of Kjis uniformly bounded. Then for any ǫ > 0there

exists η > 1such that RΩj(z)≥1−ǫfor all jlarge enough such that B1(−1−δj)⊂Bη(−1)

and B1(1 + δj)⊂Bη(1), and for all z∈(Bη(−1) ∪Bη(1)) ∩Ωj.

Proof. Assume to get a contradiction that there exist ǫ > 0 and sequences ηkց1, jk→ ∞,

such that

B1(−1−δjk)⊂Bηk(−1), B1(1 + δjk)⊂Bηk(1)

and a sequence zk∈(Bηk(−1) ∪Bηk(1)) ∩Ωjksuch that RΩjk(zk)<1−ǫ. We may assume that

Re(zk)≥0 for all k.

Set fk(z) := z−(1 + δjk), Ω′

jk:= fk(Ωjk), and z′

k:= fk(zk). Note that 1 <|z′

k| ≤ 2ηk−1 and

that fk(−2−δjk) = −3(1 + (2/3)δjk). Next set gk(z) := 1/z, Ω′′

jk:= gk(Ω′

jk), and z′′

k:= gk(z′

k).

Then |z′′

k| ≥ 1

2ηk−1and |gk(fk(−2−δjk))|<1/3.

To sum up: Ω′′

jkis a domain obtained by removing a disk Dkand the compact set gk(fk(Kjk))

from the unit disk, the point qkon the boundary of Dkclosest to the origin is of modulus less

than one third, and there is a point z′′

k∈Ω′′

jkwith |z′′

k| ≥ 1

2ηk−1for which RΩ′′

jk

(z′′

k)<1−ǫ.

Clearly, the Poincar´e distances between z′′

kand qk, and z′′

kand gk(fk(Kjk)), goes to inﬁnity

as k→ ∞, so if we set ψk(z) := z−z′′

k

1−z′′

kz, after possibly having to pass to a subsequence and in

view of the following below sublemma, the domains ψk(Ω′′

jk) converge to a simply connected

domain with respect to strong convergence. Applying Proposition 2.2 and using one more time

the following sublemma, this implies that Rψk(Ω′′

jk)(0) →1 as k→ ∞ - a contradiction.

6 L. AROSIO, J. E. FORNÆSS††, N. SHCHERBINA, AND E. F. WOLD

Sublemma 2.5. We have that liminfk→∞ dP(z′′

k, Dk)>0, where dPdenotes the Poincar´e

distance.

Proof. Note that Re(z′

k)≥ −1−δjk, so that if we set γk:= {z∈C: Re(z) = −1−δjk}, then

any curve connecting z′′

kand Dkwill have to pass through ˜γk=gk(γk). So it is enough to ﬁnd

a lower bound for the Poincar´e distance between Dkand ˜γkfor large k. Now the real points on

˜γkare 0 and 1

−1−δjk

, and the real points on bDkare 1

−1−2δjk

and 1

−3(1+(2/3)δjk), and using the

fact that Poincar´e disks are Euclidean disks, it suﬃces to control the distance between 1

−1−δjk

and 1

−1−2δjk

, and between 0 and 1

−3(1+(2/3)δjk). The last distance is clearly bounded away from

zero, so we compute the ﬁrst. We have that

lim

k→∞ log

1 + 1

1+δjk

1−1

1+δjk

−log

1 + 1

1+2δjk

1−1

1+2δjk

= lim

k→∞ log

1−1

1+2δjk

1−1

1+δjk

= log 2.

Lemma 2.6. Let Ω⊂P1, x ∈Ω, and suppose that P1\Ωcontains at least three points. Suppose

that Ωjs

→Ω. Then

r:= lim sup

j→∞

SΩj(x)≤SΩ(x).

Proof. If r= 0 this is clear, so we assume that r > 0. Then, after possibly having to pass to a

subsequence, there exists a sequence ϕj: Ωj→ △ of embeddings, ϕj(x) = 0, Brj(0) ⊂ϕj(Ωj),

rj→r. Let aj∈P1be distinct points such that ai/∈Ω for i= 1,2,3. For any δ > 0 the ball

Bδ(ai) is not contained in Ωjfor all jlarge enough. So we may ﬁx 0 < δ << 1, and assume

that there exist points aj

i∈Bδ(ai) such that aj

i/∈Ωjfor all jand for i= 1,2,3. Since there

is a compact family of M¨obius transformations mapping the triples {aj

1, aj

2, aj

3}to the triple

{a1, a2, a2}, and since the complement of three points is Kobayashi hyperbolic, we may assume

that for all 0 < r′< r the sequence ϕ−1

j|Br′(0) is convergent. Hence the derivatives of ϕ′

j(x)

are uniformly bounded below and above. Therefore we can assume that the ϕjconverge to an

injective holomorphic map from Ω to △. Moreover the image contains the disc of radius r.

Proof of Theorem 1.1:

Set Q1=I2. By alternating Lemma 2.3 and its horizontal analogue we obtain a decreasing

sequence Qjof disjoint unions of cubes, such that

(1) Q:= ∩j≥1Qjis a Cantor set with 2-dimensional Lebesgue measure arbitrarily close to

one,

(2) RP1\Qj≥1−ǫ, and

(3) For any sequence (zj) in P1\Qconverging to Qand any δ > 0 there exists an N∈N

such that RP1\Qi(zj)>1−δwhenever j≥Nand iis large enough (depending on j).

By (1) the two-dimensional Lebesgue measure of Qcan be arbitrarily close to one. By (2)

and by Lemma 2.6 it follows that SP1\Q(z)≥1−ǫ. By (3) and by Lemma 2.6 it follows that

SP1\Q(zj)≥1−δfor all j≥N, and hence limP1\Q∋z→QSP1\Q(z) = 1.

7

3. A “large” Cantor Set whose complement does not resemble the unit disk -

Proof of Theorem 1.2

We modify the construction in the previous section. For an inductive construction, assume

that we have constructed a family Qj:= {Qj

1,...,Qj

m(j)}of m(j) disjoint cubes. We may

choose m(j) closed loops Γj

i, each surrounding and being so close to one of the cubes, that

SΩj(z)≥1−1/j if z∈Γj

ifor some 1 ≤i≤m(j), where Ωj=P1\ ∪iQj

i. Further we may

choose a ﬁnite number of points pj

1, ..., pj

k(j)in Ωjsuch that we ﬁnd a point pj

ℓin any 1/j-

neighbourhood of any point in bΩj, and such that SΩ′

j(z)≥1−2/j if z∈ ∪1≤i≤m(j)Γj

i, where

we denote Ω′

j:= Ωj\ ∪1≤ℓ≤k(j){pj

ℓ}. The reason is that the removal of a set suﬃciently close to

the boundary of a domain, will essentially not disturb a lower bound for neither Snor R.

Then by Lemma 3.2 below and Proposition 2.2 we may choose an arbitrarily small δj>0

and arbitrarily small cubes ˜

Qj

ℓ⊂ {|z−pj

ℓ|< δj}such that

(i) SΩ′′

j(z)≤1/j if |z−pj

ℓ|=δjfor some 1 ≤ℓ≤k(j), and

(ii) SΩ′′

j(z)≥1−3/j if z∈Γj

ifor some 1 ≤i≤m(j),

where we denote Ω′′

j:= Ωj\ ∪ℓ˜

Qj

ℓ. By applying Lemma 2.3 twice we may divide each cube in

the collection

{Qj

1,...,Qj

m(j),˜

Qj

1,..., ˜

Qj

k(j)}

into four, creating a new collection of cubes Qj+1 such that

(i) SΩj+1 (z)≤2/j if |z−pj

ℓ|=δjfor some 1 ≤ℓ≤k(j), and

(ii) SΩj+1 (z)≥1−4/j for z∈Γj

ifor some 1 ≤i≤m(j),

where Ωj+1 denotes the complement of the cubes in Qj+1. The inductive step may be repeated

indeﬁnitely so as to ensure that for all k > 1 we still have that

(i’) SΩj+k(z)<3/j for |z−pj

ℓ|=δjfor some ℓ, and

(ii’) SΩj+k(z)>1−5/j for z∈Γj

ifor some i.

We now deﬁne

Q:= lim sup

j→∞ [

Qj

l∈Qj

Qj

l

If in each step of the construction the points pj

iwere chosen close enough to each of the previously

constructed cubes, it follows that any connected component of Qmust be a point (since the

diameters of the cubes go to zero), and no point will be isolated. Hence Qis a Cantor set. It

follows from Lemma 3.2 that we may arrange that statement corresponding to (i’) holds in the

limit. The statement corresponding to (ii’) holds in the limit by Lemma 2.6 since Ωjconverges

strongly to P1\Q.

Lemma 3.1. Let K⊂P1be a compact set such that Ω = P1\Kadmits a bounded injective

holomorphic function. Let p1, ..., pm∈Ωbe distinct points, and set Ω′= Ω \ {p1, ..., pm}. Then

lim

Ω′∋zj→pj

SΩ′(z) = 0,(3.1)

8 L. AROSIO, J. E. FORNÆSS††, N. SHCHERBINA, AND E. F. WOLD

for j= 1, ..., m.

Proof. We consider p1. Assume to get a contradiction that there exists a sequence Ω′∋zj→p1

and injective holomorphic maps ϕj: Ω′→ △, ϕj(zj) = 0, and Br(0) ⊂ϕj(Ω′) for some r > 0. All

maps extend holomorphically across p1, ...., pm, and we may extract a subsequence converging

to a limit map ϕ, with ϕ(p1) = 0. Since |ϕ(p1)−ϕ(zj)| → 0 as j→ ∞, this leads to a

contradiction.

Lemma 3.2. Let K⊂P1be a compact set such that Ω = P1\Kadmits a bounded injective

holomorphic function. Let p1, ..., pm∈Ωbe distinct points, and let ǫ > 0. Then there exist

δ1>0(arbitrarily small) and 0< δ2<< δ1, such that for any domain Λ⊂P1with P1\Λ⊂

K(δ2)∪(∪m

j=1Bδ2(pj)) (with at least one complementary component in each Bδ2(pj)), we have that

SΛ(z)< ǫ for all z∈Λwith |z−pj|=δ1for some j. Here K(δ2)denotes the δ2-neighbourhood

of K.

Proof. Let 0 < µ << 1 (to be determined). Fix δ1>0 such that the Kobayashi length in

Ω′= Ω \ {p1, ..., pm}of each loop |z−pj|=δ1is strictly less than µ. Let fθ:△ → Ω′be

a continuous family of universal covering maps with fθ(0) = p1+δ1eiθ . Then the Kobayashi

metric gΩ′

K(p1+δ1eiθ) is equal to 1/|f′

θ(0)|. Fix any 0 < r < 1. Then for any domain Ω′′ ⊂P1

that covers the union ∪θfθ(△r), which is a compact subset of Ω′, we have that gΩ′′

K(p1+δ1eiθ)

is bounded from above by 1/|r·f′

θ(0)|. So for rsuﬃciently close to 1 the Kobayashi length of

the loop |z−p1|=δ1in Ω′′ is less than µfor any such domain. The same argument may be

applied to all points pj.

Now for any such domain Ω′′ we estimate the squeezing function with respect to µ. Write

SΩ′′ (pj+δ1eiθ) = s, let g: Ω′′ → △ be a map that realises the squeezing function at pj+δ1eiθ,

and let Γjdenote the loop |z−pj|=δ1. Then, since g(Γj) is a nontrivial loop in g(Ω′′) we have

that lK(Γj)≥log(1+s

1−s). Then s≤eµ−1

eµ+1 →0 as µ→0, and so the lemma follows.

4. Julia sets for quadratic polynomials - Proof of Theorem 1.3

Fix a quadratic polynomial fc(z) = z2+cand assume that c /∈ M, where Mdenotes the

Mandelbrot set. Then the critical point 0 is in the basin of attraction of inﬁnity Ω∞, and the

Julia set Jc=P1\Ω∞is a Cantor set. We let Gc(z) denote the negative Green’s function

associated to fc. It satisﬁes the following properties:

(1) Gcis continuous on Cand harmonic on C\ Jc,

(2) Gc(z) = −log |z|+O(1) near ∞, and

(3) Gc(fn(z)) = 2nGc(z) for all z∈C.

We regard Gcas an exhaustion function of Ω∞. Let t0=Gc(0). The exhaustion may be

described as follows. For t < t0the level sets Γt={Gc=t}are smooth connected embeddings

of S1, shrinking around inﬁnity as tdecreases to −∞. Considering the picture in C, as tincreases

to t0, the family Γtis a decreasing family of embedded S1’s, decreasing to Γt0, which is a ﬁgure

eight, the origin being the ﬁgure eight crossing point. In general, the level sets Γ2−nt0consists of

2npairwise disjoint ﬁgure eights, and for 2−nt0< t < 2−n+1t0the level set Γtconsists of 2n+1

disjoint smoothly embedded copies of S1, one contained in each hole of a ﬁgure eight in Γ2−nt0.

9

We now assume to get a contradiction that there exists a bounded holomorphic injection

ϕ: Ω∞→ △, and we may assume that ϕ(∞) = 0. We will ﬁrst use the exhaustion just

described to get a description of ϕ(Ω∞) that will allow us to modify ϕin a useful way. Set

H=Gc◦ϕ−1, deﬁned on ϕ(Ω∞).

Start by choosing s0<< 0 and let D0be the disk bounded by γs0={H=s0}, a single closed

loop. Increasing sbetween s0and t0we get an increasing family of single loops γs, but when s

crosses the critical value t0it breaks into two loops, say γ1

s1, γ2

s1, for sclose to t0. One of these

loops is going to enclose the other, and we relabel it γs1. Next, increasing sbetween s1and

2t0we follow a path of loops starting from γs1, until scrosses 2t0, and it again breaks into two

loops, say γ1

s2and γ2

s2for s2close to 2t0. Again, single out the one enclosing the other, and

relabel it γs2. Continuing in this fashion, we obtain a family of loops γsjsuch that γsjencloses

γsj−1, and such that the disk Djbounded by γsjcontains the whole sublevel set {H < sj}. We

have that {Dj}is an increasing family of disk, we denote by Dits increasing union, and we let

ψ:D→ △ be the Riemann map satisfying ψ(0) = 0, ψ′(0) >0. Our modiﬁed map will be

˜ϕ:= ψ◦ϕ.

Next we will use the map fcto ﬁnd some other loops ˜γjin Ω∞, each one in the same free

homotopy class as ϕ−1(γsj). Start by deﬁning ˜γ0as the level set Gc=tfor some t < t0close to

t0. Then f−1

c(˜γ0) consists of two disjoint loops, one of them free homotopic to ϕ−1(γs1). Single

this out, and label it ˜γ1. Next f−1

c(˜γ1) consists of two disjoint loops, and one of them is free

homotopic to ϕ−1(γs2). Single it out, and denote it by ˜γ2. Continue in this fashion indeﬁnitely.

We are now ready to reach the contradiction. On the one hand, since the family ˜ϕ(˜γj) will

increase towards b△, it follows that the Kobayashi lengths of ˜γjin Ω∞will increase towards

inﬁnity. On the other hand, let C⊂Ω∞denote the forward and backward orbit of the critical

point 0. Then the Kobayashi length of each ˜γjin Ω∞\Cis longer than the Kobayashi length

in Ω∞. But fc: Ω∞\C→Ω∞\Cis a covering map, and so the Kobayashi lengths of all the

˜γj’s in Ω∞\Care the same. A contradiction.

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L. Arosio: Dipartimento Di Matematica, Universit`

a di Roma “Tor Vergata”, Via Della Ricerca

Scientifica 1, 00133, Roma, Italy

E-mail address:arosio@mat.uniroma2.it

J. E. Fornæss: Department of Mathematics, NTNU, Norway

E-mail address:john.fornass@ntnu.no

N. Shcherbina: Department of Mathematics, University of Wuppertal, 42097 Wuppertal, Ger-

many

E-mail address:shcherbina@math.uni-wuppertal.de

E. F. Wold: Department of Mathematics, University of Oslo, Postboks 1053 Blindern, NO-0316

Oslo, Norway

E-mail address:erlendfw@math.uio.no