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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
