Contemporary Mathematics
Non-uniform Porosity for a Subset of Some Julia Sets.
Kimberly A. Roth
Abstract. Using similar techniques to McMullen’s paper [Mc1], it is shown
(z−3)
that the boundary of the basin of infinity of fθ (z) = e2πiτ (θ) z 2 1−3z , called
Jθ , is non-uniformly porous for all irrational θ ∈ (0, 1).
1. Introduction
We consider Herman’s Blaschke product fθ (z) = e2πiτ (θ) z 2 (z−3)
1−3z for all irrational numbers θ, where the number τ (θ) is chosen to give the map fθ the rotation
number θ when we restrict fθ to C(0, 1), the unit circle centered at the origin. The
restriction of fθ to C(0, 1) is an orientation-preserving homeomorphism which is
conjugate to an irrational rotation.
We consider Jθ the boundary of the immediate basin of infinity. This is a subset
of the Julia set of fθ . Formally, we can construct Jθ by removing the preimages
−n
of the interior of the unit disk. In other words, we define Jθ = Jfθ \ ∪∞
n=0 fθ (D),
where D is the unit disk centered at the origin. See Figure 1.
2000 Mathematics Subject Classification. Primary 37F50; Secondary 54F45.
√
Figure 1. Jθ FOR θ = 21 ( 5 − 1) FROM [Pet].
c
0000
(copyright holder)
1
2
KIMBERLY A. ROTH
One technique for bounding the Hausdorff dimension of the Julia set is to show
that the set is porous. A compact set Λ in the complex plane is called porous (or
shallow ) if there exists a positive integer N such that, for any z ∈ Λ and 0 < < 1,
when given a square of side length centered at z and subdivided into N 2 squares
of side length /N , at least one of these squares is disjoint from Λ. Equivalently,
a set Λ is porous if there exists a 0 < K < 1 such that for all z ∈ Λ and every
0 < r < 1, there is a ball B of radius s, Kr < s, disjoint from Λ and contained in
the ball of radius r about z.
The polynomial P (z) = e2πiθ z + z 2 is related to our map Fθ when θ is an
irrational number of a particular type. We can relate this polynomial to Herman’s Blaschke product by means of the Ghys construction, which uses quasiconformal surgery, thus relating Jθ to the quadratic polynomials P (z) for irrationals
of bounded type. This construction can be extended to a larger class of irrationals.
Each P is conjugate to an irrational rotation near the origin and has an irrationally
neutral fixed point at the origin. Therefore JP has a Siegel disk. For these particular maps, it can be shown that the critical point c0 lies on the boundary of the
Siegel disk. McMullen [Mc1] showed for irrational numbers of bounded type that
JP is porous and thus has Hausdorff dimension less than two. Peterson and Zakeri
[PZ] extended the class on which the conjugacy exists and showed that P is locally
connected and Lebesgue measure zero in the class.
We have not yet been able to show porosity for Jfθ , but we were able to show
a weaker condition called non-uniform porosity. A compact set Λ is called nonuniformly porous if for any z ∈ Λ there is a a sequence of radii {rn } with lim rn = 0
t→n
and a constant 0 < K < 1 such that there is a ball B of radius s, Krn < s, disjoint
from Λ and contained in the ball of radius rn about z.
Using techniques based on McMullen’s [Mc1] proofs, we will show:
Theorem 1.1. * The boundary of the immediate basin of infinity, Jθ , is nonuniformly porous for all irrational numbers θ ∈ (0, 1].
Non-uniform porosity is a stronger condition than having Lebesgue measure
zero. For example, the set of rational numbers has measure zero, but fails to
be non-uniformly porous. The fact that non-uniform porosity implies Lebesgue
measure zero can be seen by showing that non-uniform porosity at a point implies
the point is not a Lebesgue density point. Thus the theorem implies Jθ has measure
zero. The measure of Jθ was previously shown to be zero for irrationals of bounded
type by C. L. Peterson [Pet] in 1996 and by Yampolsky [Ya] for all irrationals in
1999, using the technique of complex bounds.
Not much is known about the full Julia set Jfθ . Peterson [Pet] proved that it
was locally connected in 1996. We hope that our theorems about Jθ will eventually
yield more information about Jfθ .
The author would like to thank her advisor Greg Świa̧tek for his guidance and
Luke Roth for his help with the graphics. The picture of Jfθ was drawn using
software of McMullen [Mc2]. The picture of Jθ is from Peterson [Pet].
2. Preliminaries
For a map f on C(0, 1) with an irrational rotation number θ, we can relate the
continued fraction expansion θ to the closest return on the circle of the image of
the critical point, which for our fθ is at 1.
NON-UNIFORM POROSITY FOR A SUBSET OF SOME JULIA SETS.
3
Figure 2. THE PREIMAGE f0−1 (C(0, 1)).
We can write any irrational number θ ∈ (0, 1] uniquely in the form
θ=
1
a1 +
1
a2 + a
= [a1 , a2 , a3 . . . ]
1
3 +...
where each ai is chosen so ai ≥ 1. Let pqnn = [a1 , a2 , a3 . . . an ] be the nth rational
convergent of θ. An irrational number θ is of bounded type or constant type if
sup ai < ∞.
In the nth rational convergent of the continued fraction expansion of θ, qn and
pn are related to an by qn = an qn−1 + qn−2 and pn = an pn−1 + pn−2 . General
information about continued fractions can be found in [HW].
Let the preimages of the critical point be denoted xk = f −k (1), and let the
images of the critical point, the critical values, be denoted ck = f k (1). It is easiest
to consider these on the lifting to the interval (− 12 , 12 ] instead of on the circle, on
which 0 represents the critical point. We define the closest return to the critical
point to be f −n (0), such that for all 0 < k < n, |f −k (0) − 0| > |f −n (0) − 0|.
We can order the closest returns in the following manner:
xqn−1 < 0 < xqn < cqn −1 < x(an −1)qn−1 +qn−2 < · · · < xqn−1 +qn−2 < xqn−2 .
We will see that in our case the xn will play the role of the intersection points
of C(0, 1) with the other portions of Jθ .
To accurately describe portions of the Julia set, we must establish a few definitions. Again, the orbit of the critical point is of importance to us. As before
we let the preimages of the critical point 1 be called xn = f −n (1). In order to
describe some other portions related to the xn on Jfθ and Jθ , we will consider first
the properties of Jfθ .
We begin with the degree three Blaschke product, f0 (z) = z 2 (z−3)
1−3z . The map
f0 (z) leaves C(0, 1) invariant, and is also symmetric in this circle. It has three
critical points: 0, 1, and ∞. If we examine (f0 )−1 (C(0, 1)), we note there are two
preimages of the circle. The first is the circle itself, and the second is a curve passing
through the critical point 1. See figure 2.
Similarly, we can consider fθ (z) = e2πiτ (θ) f0 (z), which has the above properties
as well. Also, for θ irrational, C(0, 1) belongs to the Julia set of fθ . This follows
from the fact that the point 1 is in the Julia set, the forward orbit of 1 is dense in
C(0, 1), and the Julia set is forward invariant. Since C(0, 1) is part of the Julia set,
the closure of the preimages of C(0, 1) must be the whole Julia set.
−n
The boundary of the immediate basin of infinity, Jθ = Jfθ \ ∪∞
n=0 fθ (D), has
the property that its points fall in three classes: points on C(0, 1), points on finite
preimages of C(0, 1), and the points which are limits of preimages of C(0, 1), which
we will refer to as limit points. This property follows from the properties of Jfθ .
4
KIMBERLY A. ROTH
Figure 3. THE JULIA SET Jfθ .
Examine fθ−1 (C(0, 1)) ∩ Jθ . It now consists of two parts, the first being C(0, 1)
and the second being a teardrop shaped curve, a bulb, intersecting C(0, 1) at the
critical point 1, which we will call the primary preimage of C(0, 1), C 0 . In fact all
preimages of C(0, 1) contained in Jθ will be made up of such bulbs. The bulbs which
intersect C(0, 1) will do so at the preimage of the critical point f −n (1), denoted
xn , for some n. These bulbs that intersect the circle will be called main bulbs with
basepoint xn .
We consider the point xn . If we remove the point xn from Jθ , we get two
connected components, one of which contains points in C(0, 1), the other component
unioned with xn will be called the limb Xn , with basepoint nn . We call the limb
with the basepoint x0 = 1, the primary limb, X0 . The primary preimage of C(0, 1)
is called C 0 or the main bulb of X0 . The main bulb of a limb Xn will be the primary
nth preimage of the circle, namely the preimage in Xn which contains the point
xn . Note that f n (Xn ) = X0 . This property allows us to examine behavior on all
limbs by examining behavior on the primary limb, X0 . The basepoints xqn are the
closest returns to the critical point at 1 and by the properties of preimages of the
critical point discussed above, we have the following ordering seen in Figure 4.
Frequently, it will be useful to consider the hyperbolic metric on the complex
plane minus the unit disk centered at the origin. Because f maps its domain into
a subset of itself, f is contracting in the hyperbolic metric. This implies that f −1
is expanding. The hyperbolic metric is comparable to the Euclidean metric in our
case. Thus, knowing that the hyperbolic metric is expanding will be helpful.
NON-UNIFORM POROSITY FOR A SUBSET OF SOME JULIA SETS.
5
Figure 4. THE ORDERING OF THE LIMBS NEAR 1.
Let Ω be the space C \ D(0, 1) endowed with the hyperbolic metric dΩ . If we
consider any portion of Ω minus some disk centered at infinity, Ω0 , we can estimate
the element of the hyperbolic metric ρΩ (z)|dz| as follows:
Theorem 2.1. *[McMullen [Mc2] 2.3] For some constant K > 1,
1
1
≤ ρΩ (z) ≤ K
.
Kd(z, ∂Ω)
d(z, ∂Ω)
We will use the notation ρΩ for the element of the hyperbolic metric on Ω. We
will use dΩ (., .) for the hyperbolic distance on Ω, Dz for the derivative with respect
to the hyperbolic metric, and ||.||Ω when we require the norm with respect to the
hyperbolic metric on Ω.
3. Nearby visits to C(0, 1)
In order to show Jθ is non-uniformly porous, for θ irrational, we show that all
but a countable number of points map near to the circle C(0, 1). Since points in
Jθ are either on C(0, 1), are on a finite preimage of C(0, 1), or are limit points,
we can deal with each in turn. Points on C(0, 1) and finite preimages if C(0, 1)
are certainly not Lebesgue density points, so the only class of points that must
be considered are the limit points. We will deal with the limit points by finding
univalent mappings with bounded distortion which map a ball near each limit point
to a region that contains a point on C(0, 1). We will be able to find these mappings
for all but a countable number of limit points, which is enough.
So the main result of this section will be to show the following:
Theorem. For all but a countable number of limit points x ∈ Jθ , there is
a sequence of positive integers nk , nk → ∞ such that, for each nk there is an
embedded hyperbolic ball Bnk centered at a point ynk such that for some ball B 0 and
p ∈ C(0, 1)
f nk +1 : (Bnk , ynk ) → (B 0 , p).
For the constant ρ from the cone K and a constant κ depending on ρ κ1 ||Dz (f −nk )(x)||Ω ≤
radiusΩ (Bnk )
ρ
nk
mapping f
≤ κ||Dz (f −nk )(x)||Ω , and dΩ (x, ynk ) < 2ρκ||Dz (f −nk )(x)||Ω . The
is univalent and has bounded distortion on Bnk .
The fact that Jθ has measure zero for all irrational θ will follow shortly from
Theorem 3.7, as will the non-uniform porosity proved in the next section.
6
KIMBERLY A. ROTH
Figure 5. THE RECTANGLE Rn AND THE CONE K.
We will define a cone K = {z ∈ Ω|dΩ (z, C 0 ) ≤ ρ}, where C 0 is the primary
preimage of C(0, 1) and ρ is a constant that will be chosen later. Once K is
defined, we will define some related regions, Rn . This cone will be the place where
we are guaranteed to have the mappings for the above theorem. Since each point in
C 0 maps to C(0, 1) and a ball of finite hyperbolic radius centered at a point in C 0
maps conformally onto C(0, 1), if we can map all other points in Jθ into the cone,
we will have similar mappings.
Since inside the cone K we have our mappings, we want all points to eventually
enter the cone. What we show is that all but a countable number of limit points
enter the cone infinitely many times on their orbits.
Let Rn be the open quadrilateral between Xqn and Xqn−2 .The four sides of the
quadrilateral will be defined as follows. The first side is the segment on C(0, 1)
between xqn and xqn−2 , which does not contain 1. Let ln be the line segment
between xqn and ∞. The second side of Rn is the piece of ln between xqn and the
point yqn which is the first intersection of ln and the boundary of cone K. Define
the third side in the same manner as the second, except using xqn−2 . The last side
is the segment of the boundary of cone between yqn and yqn−2 . See figure 5.
Proposition 3.1. For some constant ρ, the cone K = {z ∈ Ω|dΩ (z, C 0 ) ≤ ρ}
has the following properties: for each n, Rn contains an open region Gn , such that
Gn contains all points of X0 in Rn and f qn−1 : Gn → (Gn ∪ K).
Proof. Let n be given.
In order to increase clarity we will break the proof into several lemmas. We
first show we can pick a region, which we will call B(n,k) , where the height is
always a fixed proportion of the length of the dynamical interval. Second, we
show this region, B(n,k) , contains no points of X0 for all but a finite number of
n. Third, we construct the subregion with the invariant boundary. Once we have
these three lemmas we will be able to choose the cone and the region Gn and verify
its properties.
Define a closed box B(n,k) , 0 ≤ k ≤ an − 1; its boundary consists of four sides.
The first side is the segment on C(0, 1) between xkqn−1 +qn−2 and x(k+1)qn−1 +qn−2 ,
which does not contain 1. Let ln be the line segment between xkqn−1 +qn−2 and ∞.
The second side of B(n,k) is the piece of ln between xkqn−1 +qn−2 and the intersection
of ln with circle of radius 1 + (n,k) centered at the origin, where (n,k) > 0 will
be chosen later. Define the third side in the same manner as the second, except
NON-UNIFORM POROSITY FOR A SUBSET OF SOME JULIA SETS.
7
Figure 6. THE TRAPEZOID T AND THE SHAVED LIMB U0 .
using x(k+1)qn−1 +qn−2 . The last side is the segment of the circle of radius 1 + (n,k)
centered at the origin between the points of intersection of the second and third
segments with this circle.
We also define a subregion of Xqn for any n. We will do this by choosing a
region of the primary limb X0 and taking its preimages.
Let the shaved limb U0 be a curve along X0 which connects 1 to the fixed
point on X0 . We may choose U0 so that a trapezoid T with an endpoint at 1, an
angle of 100 degrees at 1, and a small height h > 0 intersects U0 only at 1. Call
Un = f −n (U ).
Lemma 3.2. We can choose the (n,k) so that there exists an A > 0 independent
of n and k such that (n,k) = A|x(k+1)qn−1 +qn−2 − xkqn−1 +qn−2 |, such that Bn,k ∩
U(k+1)qn−1 +qn−2 = x(k+1)qn−1 +qn−2 . See Figure 7.
Proof. Consider the regions B(n,k) , 0 ≤ k ≤ an − 1, as above.
We wish to choose (n,an −1) so that B(n,an −1) ∩Uqn = xqn . Under f qn B(n,an −1)
maps to a region with one end point at 1. Now we must choose (n,an −1) to be small
enough, so that the image f qn (B(n,an −1) ) is contained in the trapezoid, T .
We can guarantee this because of the bounded geometry. This comes from the
Lemma 1.2 of Świa̧tek from [S] , which can be stated in the following manner.
Lemma 3.3 (Świa̧tek, [S]). Suppose that f is a real-analytic degree 1 circle
homeomorphism with an irrational rotation number. Let pqnn be a convergent of the
rotation number. Then there is a K ≥ 1 so that for all z ∈ C(0, 1),
K −1 d(f qn (z), z) ≤ d(f −qn (z), z) ≤ Kd(f qn (z), z).
Since our f is a real-analytic circle map with an irrational rotation, we can
apply the lemma.
So by Lemma 3.3, if (n,an −1) is small enough we can map B(n,an −1) so that it
is contained in the trapezoid T . This implies that f qn (B(n,an −1) ) ∩ U0 = 1, which
implies B(n,an −1) ∩ Uqn = xqn . Since the distortion under f qn of the box B(n,an −1)
8
KIMBERLY A. ROTH
Figure 7. THE RECTANGLE B(n,k) .
depends on the length |xqn − xqn −qn−1 | and the height of the trapezoid T is fixed,
then we can define (n,an −1) = A|xqn − xqn −qn−1 |, where A is a constant depending
on the height of the trapezoid T .
For all other 0 ≤ k < an − 1, we proceed in the same manner. We choose
the height of B(n,k) to be (n,k) so that f (an −(k+1))qn−1 (B(n,k) ) is contained in T ,
yielding Bn,k ∩ U(k+1)qn−1 +qn−2 = x(k+1)qn−1 +qn−2 . Then (n,k) has the property
that
(n,k) = A|x(k+1)qn−1 +qn−2 − xkqn−1 +qn−2 |
for the same reasons that (n,an −1) = A|xqn − xqn −qn−1 |.
So in general B(n,k) is a short quadrilateral which is on one side of the shaved
limb at X(k+1)qn−1 +qn−2 . See Figure 5.3.
Now we examine a B(n,k) for some k, 0 ≤ k ≤ an − 1. We want to show that
no points of X0 lie in the boxes.
Lemma 3.4. The boxes B(n,k) contain no points of X0 for all 0 ≤ k ≤ an − 1
and for all but a finite number of n.
Proof. In order to get a contradiction, we will assume there is a point z ∈
(X0 ∩ B(n,k) ). We will consider the case k = an − 1 first.
We will begin by constructing a region around some part of X0 for a z ∈
X0 ∩ B(n,an −1) and then will repeat a similar construction for B(n,k) . Then we will
show these lead to a contradiction.
Given z ∈ X0 ∩ B(n,an −1) , we will define a region relating to z. Take a path P
that connects z to some point z 0 on C(0, 1). Choose the path P , so that is contained
in B(n,an −1) \ Uqn . We can do this, since B(n,an −1) \ Uqn is connected and open
and so is path connected. Remember that Ω = C \ D(0, 1) and Ω contains Uqn .
Consider the set R = Ω \ (P ∪ X0 ), which has several components. Let W be the
NON-UNIFORM POROSITY FOR A SUBSET OF SOME JULIA SETS.
9
Figure 8. THE REGION W AND THE SHAVED LIMB Uqn .
closure of the connected component of R which contains U . Call ∂W the loop L.
See figure 8.
The loop L consists of three pieces. The first is a segment on the circle which
contains xqn in the interior. The second is the path P . The third is a path on X0
leading to z.
Apply f to W , and examine what happens to the pieces of the loop. The image
of the segment on the circle remains a segment on the circle which contains xqn −1
in its interior. The image of the path P remains a short segment connecting with
C(0, 1). The image of the path on X0 leading to z is a path on some other limb
Xm1 and some segment of the circle which is connected to the previous segment
on the circle. So overall, f (L) consists still of three pieces; a segment of the circle
containing xqn −1 in its interior which begins at xm1 and ends at f (z 0 ), the image
of the path P , and a path on some limb Xm .
By induction, if we apply f k to W and examine the boundary f k (L), we have
the following three pieces. The first is a segment of C(0, 1) containing xqn −k in the
interior and the endpoint of the segment are f k (z 0 ) and some basepoint xmk for
some other limb Xmk . The second is the image of the path f k (P ). The last is a
path along some limb Xmk .
10
KIMBERLY A. ROTH
Apply f qn to W . The image under f qn of the shaved limb Uqn ,which is contained in W , is mapped into U0 and it and f qn (z) are contained in f qn (W ). The
only point of f qn (U ) not contained in the interior of f qn (W ) is the basepoint 1.
Now examine f qn (L). By the previous argument, f qn (L) consists of a segment
of the circle containing 1 in its interior, the image of the path f qn (P ), and a path
along a limb Xm containing f qn (z). The segment of the circle has endpoints xmqn
and f qn (z 0 ).
Now we can construct similar regions for any point z ∈ B(n,k) . Using Uqn−1 +(k+1)qn−2
and using f qn−1 +(k+1)qn−2 to map the region W to the primary limb we construct
the regions. The property that the image of z 0 and the basepoint of the image of z
are on opposite sides of 1 hold, and the image of the loop still has the three types
of pieces. We will use this to achieve a contradiction.
We know that there is an external ray that lands on the point fixed under
iteration of f on X0 . The preimage of the external ray under f −qn must pass
through the loop L and so the external ray itself must pass through f qn (L). The
external ray cannot pass through the circle or any other limb, so the only possible
point of intersection is on P .
This is only possible for a finite number of n since the length of the image of
P must be less than a constant times (n,k) for all k and n and (n,k) goes to zero
as n goes to infinity. If for all n the external ray passes through the image of P ,
then the ray cannot land. This is a contradiction, therefore B(n,k) cannot contain
any points of X0 for all but a finite number of n and k.
Now that we have a region not containing any points of X0 , we wish to obtain
n −1
an invariant region contained in ∪ak=1
B(n,k) , so that we have an invariant region
containing no points of X0 .
n −1
We will now define a region contained in ∪ak=1
B(n,k) . Remember, the second
side of B(n,k) is the piece of ln between xqn−1 +kqn−2 and the intersection of ln and the
circle of radius 1 + (n,k) centered at the origin, where (n,k) > 0 and the third side
was determined in the same manner as the second, except using xqn−1 +(k+1)qn−2 .
n −1
Thus the region ∪ak=1
B(n,k) has one side with a segment of the ray between xqn
and ∞ and the other is a segment of the ray between xqn−2 and ∞.
Pick a point z belonging to the first side of B(n,an −1) and a height δ < (n,an −1) ,
which will be chosen later. Construct a curve γn as follows. Its first piece will be
the segment between z and f qn−1 (z). Subsequent pieces will be images of the first
piece under f qn−1 . End the last piece at the point of intersection with the second
side of B(n,0) .
Let the region Bn be defined by four pieces as follows. The first side will be the
segment of C(0, 1) between xqn and xqn−2 which does not contain 1. The second
side will be γn . The third side will be a segment of the ray between xqn and ∞,
going from xqn to z. The last side is the segment of the ray between xqn−2 and ∞,
going from xqn−2 to the intersection of γn with the ray.
Lemma 3.5. The region Bn contains no points of X0 and the top of Bn , γn ,
has the property
n −1
B(n,k) )) ⊂ γn .
(f qn−1 (γn ) ∩ (∪ak=1
Proof. Notice that Bn does not contain any points of X0 , as long as the B(n,k)
n −1
do not. So we must choose δ so that Bn is contained in ∪ak=1
B(n,k) .
NON-UNIFORM POROSITY FOR A SUBSET OF SOME JULIA SETS.
11
Figure 9. THE REGION Gn AND THE CURVE γn .
The Lemma 3.3 gives us that the distortion from mapping forward, as we did
to get the pieces of γn , is bounded. The bounded distortion implies that δ’s value
depends only on the heights , (n,k) , of the boxes, B(n,k) . Since (n,k) is a fixed
proportion of the length of the dynamical interval [x(k+1)qn−1 +qn−2 , xkqn−1 +qn−2 ],
we can choose δ > 0 dependent only on this proportion, so that Bn is contained in
n −1
∪ak=1
B(n,k) .
Now we have all the tools needed in order to prove the Proposition.
We will choose our cone K = {z ∈ Ω|dΩ (z, C 0 ) ≤ ρ}. Choose ρ so that two
properties are satisfied. First, three pieces of γn must be completely inside the
cone. These pieces consist of the piece between the starting point z and f qn−1 (z)
and the last two pieces of γn . If γn has three or less pieces, all of γn should be
inside the cone. Second, the cone K must contain all of the points of X0 in the
boxes which do not cause a contradiction in the Loop Lemma 3.4.
The first property can be assured from the way the height δ was chosen in
Lemma 3.5. The second is feasible since we can lower the cone any finite length
needed. Since we can achieve these two properties, we choose ρ so that properties
are satisfied and hence we have defined the cone K.
Note that f k(qn−1 ) (γn ∩ Rn ) ⊂ γn ∪ K for all k. Since the only regions of γn
that did not map into γn are in the cone K and the portion of γn not in the cone,
γn ∩ R, maps into itself and the cone by definition. Define Gn = Rn \ Bn . Now we
must verify that Gn maps into Gn ∪ K under f qn−1 .
Take a connected component of Gn . The boundary ∂Gn consists of some portion of the boundary of K and some portion of γn Since f qn−1 is univalent on Gn ,
the boundary maps to the boundary. In fact, from the definition of γn and the
choice of the cone K, f qn−1 (γn ∩ Rn ) ⊂ (γn ∪ K). Also since f qn−1 is orientation
preserving, the points of Gn near the curve γn remain in Gn under f qn−1 . Thus Gn
maps into Gn ∪ K under f qn−1 .
Now we have the cone K and the region Gn with the appropriate properties,
so we are done.
Note that the Proposition 3.1 guarantees that on Gn we have a well defined
inverse function f −(qn−1 ) from Gn into itself.
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KIMBERLY A. ROTH
Proposition 3.6. For all but a countable number of limit points z ∈ Jθ . The
iterates f n (z) are contained in K for infinitely many n.
Proof. Let x ∈ J be given. Let K be the cone from Proposition 3.1, K =
{z ∈ Ω|dΩ (z, C 0 ) ≤ ρ}.
Now assume x ∈ Xn for some n. Since x ∈ Xn then f n (x) ∈ X0 .
Now if f n (x) ∈ K, take f n+1 (x) which belongs to Xk for some k and begin the
process again.
If f n (x) ∈
/ K then we examine its position. It lies outside the cone in between
Xqk and Xqk−2 for some k. Since f n (x) ∈ X0 , then it must be in Gk . So we apply
f qk−1 repeatedly to f n (x) and examine its trajectory. There are two cases for the
trajectory. Either the point enters the cone or the point does not enter the cone.
If the point enters the cone K we apply f once more and begin the process again
as in the previous paragraph.
So all we have left is to show that only a countable number of points never
enter the cone under iteration. We will do this by showing for each k the set of
points that never leave Gk is at most one point.
Define S 0 to be the set of all points that remain forever in Gk under iteration
by f qk−1 . Let S be the set of all points that remain forever in Gk under iteration
by f qk−1 and the well-defined inverse f −(qk−1 ) .
First examine S. It is invariant under f qk−1 , meaning f qk−1 (S) = S. However
since f qk−1 is expanding in the hyperbolic metric, then f qk−1 must strictly expand
the diameter of the set S unless S has zero diameter. So S must have zero diameter,
which implies either S consists of one point or S is empty.
We would like to show S 0 = S. By definition S ⊂ S 0 . Thus, all we need to
show is that S 0 ⊂ S. In order to achieve a contradiction, assume there is an x ∈ S 0
such that x ∈
/ S.
Call f qk−1 = F and consider the sequence {F n (x)} Since Gk is compact, there
is a convergent subsequence {F nj (x)} converging to some x0 ∈ Gk . The point x0
must belong to S 0 , because
F m (x0 ) = lim F m (F nj (x)) = lim F m+nj (x),
j→∞
j→∞
0
which must belong S since x and all its images belong to S 0 . Furthermore, we can
show x0 ∈ S, since the above limits also hold if m is negative. Notice since x0 ∈ S
which is invariant and there is at most one point in S that x0 must be a fixed point
for F . Since {F nk (x)} converges to x0 , for all > 0 there is a n0 such that if
nk > n0 then dΩ (F nk (x), x0 ) < . However, since F is expanding in the hyperbolic
metric and x0 is fixed under F , dΩ (x, x0 ) ≤ dΩ (F nk (x), x0 ) < for all > 0 and
sufficiently large n. So x = x0 ∈ S, which is a contradiction. Therefore S = S 0 and
contains at most one point.
We have shown that at most one point remains in Gk under iteration. This
implies that for each Gk all points, except perhaps one, enter the cone under iteration. Therefore, all but a countable number of points eventually map into the
cone.
Theorem 3.7. For all but a countable number of limit points x ∈ Jθ , there
is a sequence of positive integers nk , nk → ∞ such that, for each nk there is an
embedded hyperbolic ball Bnk centered at a point ynk such that for some ball B 0 and
NON-UNIFORM POROSITY FOR A SUBSET OF SOME JULIA SETS.
13
p ∈ C(0, 1)
f nk +1 : (Bnk , ynk ) → (B 0 , p).
For the constant ρ from the cone K and a constant κ depending on ρ, κ1 ||Dz (f −nk )(x)||Ω ≤
radiusΩ (Bnk )
ρ
nk
mapping f
≤ κ||Dz (f −nk )(x)||Ω , and dΩ (x, ynk ) < 2ρκ||Dz (f −nk )(x)||Ω . The
is univalent and has bounded distortion on Bnk .
Proof. Let x ∈ J, a limit point, be given. Let K be the cone from Proposition
3.1. We will show whenever f nk (x) ∈ K for some nk we get a mapping of hyperbolic
balls. From the Proposition we have an infinite number of these mappings with
nk → ∞ for all but a countable number of x, so this will yield our result.
Remember that C 0 is the primary preimage of C(0, 1). So for every y 0 ∈ C 0 ,
any hyperbolic ball of finite radius in Ω containing y 0 is disjoint from C(0, 1) and
f is univalent on the ball with y 0 mapping to some point p ∈ C(0, 1).
Now if f nk (x) ∈ K for some nk , we know there is a hyperbolic ball V 0 centered
at a point y 0 ∈ C 0 which contains f nk (x) and has radiusΩ (V 0 , y 0 ) = ρ and f :
(V 0 , y 0 ) → (B 0 , p) for some ball B 0 and p ∈ C(0, 1).
Now examine f −(nk ) on V 0 . The mapping f −(nk ) has bounded distortion on V 0 ,
because f −(nk ) is univalent on the ball of radius 2ρ centered at y 0 , which contains
V 0 . Call the distortion constant, κ, for f −(nk ) on V 0 . The distortion constant, κ,
depends only on ρ. From the bounded distortion, we have f −(nk ) (V 0 ) contains a
hyperbolic ball B centered at y with
radiusΩ (B)
1
||Dz (f −nk )(x)||Ω ≤
≤ κ||Dz (f −nk )(x)||Ω ,
κ
ρ
so f nk +1 : (B, yk ) → (B 0 , p) as desired. Also the original ball V 0 contains f nk (x),
so dΩ (x, yk ) ≤ 2κρ||Dz (f −nk )0 (x)||Ω .
Since from Proposition 3.6 we know that for all but a countable number of
points, the point enters the cone infinitely many times and we can repeat this
process whenever f nk (x) lands in K for some nk , we are done.
The Nearby Visits to C(0, 1) Theorem will be our main tool for showing nonuniform porosity, as we will see in the next section.
4. Non-uniform porosity
A compact set Λ is called porous (or shallow ) if there exists an N such that
for any z ∈ Λ and 0 < < 1, given a square of any side length centered at z and
subdivided into N 2 squares of side length /N , then at least one of these squares
is disjoint from Λ . Equivalently a set Λ is porous if there exists a 0 < K < 1 such
that for all z ∈ Λ and every 0 < r < 1, there is a ball B of radius s > Kr, disjoint
from Λ and contained in the ball of radius r about z.
It can be shown directly that if a set Λ is porous then it must have upper box
dimension less than 2 and hence have Hausdorff dimension less than 2.
This notion of porosity and the bounding of the upper box dimension using
porosity depends heavily on having a uniform constant for all points z ∈ Λ.
We will define a type of non-uniform porosity. A compact set Λ is called nonuniformly porous if for any z ∈ Λ there is a a sequence of radii rn with limn→∞ rn =
0 and a constant 0 < K < 1 such that there is a ball B of radius s > Krn , disjoint
from Λ and contained in the ball of radius rn about z.
14
KIMBERLY A. ROTH
√
Figure 10. Jθ FOR θ = 12 ( 5 − 1) FROM [Pet].
We can see from the definition that we have lost our uniformity and now cannot
currently say anything about the upper box dimension.
Theorem 4.1. Jθ is non-uniformly porous.
Proof. Let z ∈ Jθ , be given. We shall show Jθ is non-uniformly porous at z.
If z ∈ C(0, 1) then Jθ is non-uniformly porous at z, since C(0, 1) is in Jθ and
its interior is not in Jθ . Similarly if z is on a finite preimage of the circle, the same
holds.
For z a limit point apply Theorem 3.7 to z. Except for a countable number of
limit points z ∈ Jθ , we have a sequence of positive integers nk , nk → ∞ such that,
for each nk there is an embedded hyperbolic ball Bnk centered at a point ynk such
that for some ball B 0 and p ∈ C(0, 1),
f nk +1 : (Bnk , ynk ) → (B 0 , p).
For the constant ρ from the cone K and a constant κ depending on ρ
1
radiusΩ (Bnk )
||Dz (f −nk )(z)||Ω ≤
≤ κ||Dz (f −nk )(z)||Ω ,
κ
ρ
and dΩ (z, ynk ) < 2ρκ||Dz (f −nk )(z)||Ω . The mapping f nk is univalent and has
bounded distortion on Bnk .
So since from (McMullen,[Mc2], section 3.2) we know that ||Dz (f −nk (z))||Ω →
0, as nk → ∞. This means the hyperbolic radii of the balls and dΩ (z, yk ) go to zero
as k approaches infinity. So similarly the Euclidean radius and distance do as well,
due to the comparison between the hyperbolic and Euclidean metrics.
We will construct the sequence of radii in the following manner. We have a
sequence of hyperbolic balls Bnk . Begin with some radius r0 , such that the ball
Br0 (z) of Euclidean radius r0 , contains Bn0 . From the Theorem 3.7, we know
f n0 +1 (Bn0 ) maps to some (B 0 , p) such that p ∈ C(0, 1) . We take a ball U inside
NON-UNIFORM POROSITY FOR A SUBSET OF SOME JULIA SETS.
15
D(0, 1) ∩ B 0 and consider f −(n0 +1) (U ); it contains a ball of radius cd(x, yn0 ) for
some c and is disjoint from Jθ . We can then take a radius r1 < r0 such that Bn1 is
contained inside Br1 (z) and there is a ball of radius cd(x, yn1 disjoint from Jθ . We
repeat this process and we will have achieved a sequence of radii and a sequence of
disjoint balls, all taking the same fixed proportion of Brn (z).
If z is one of the countable number of limit points that are not covered by
Theorem 3.7, then we know z is a either a repelling periodic point or a preimage
of a repelling periodic point. If z is fixed point, we take a ball of some radius r
around z, Br (z), which is small enough that that it maps into itself under f −1 .
Since the Fatou set is non-empty Jθ must be nowhere dense. Thus there is a ball
U contained in Br (z) such that U ∩ Jθ = ∅. Under f −1 , (Br (z)) will contain a
ball of some smaller radius and the preimage of U will contain a ball that takes up
the same proportion of the ball. We can continue this process indefinitely, yielding
the sequence of radii as desired. If the point z is periodic with period k, the same
construction holds as well using f −k . If z is a preimage of a periodic point, we can
pull back the sequence of radii and balls disjoint from Jθ from the period point on
the forward orbit of z.
Non-uniform porosity implies Lebesgue measure zero. This fact follows easily
from the fact that a point at which there is non-uniform porosity cannot be a
Lebesgue density point. We can now see directly the following corollary.
Corollary 4.2. The Lebesgue measure of Jθ , for θ an irrational number is
zero.
5. Conclusions
Now that we have shown that Jθ is non-uniformly porous, many questions
remain. Is Jθ actually porous? While theoretically possible, it is unlikely that the
methods used in this paper could be adapted to show porosity. The only possible
way to achieve such an adaption would be to pick a cone so that all points of the
primary limb X0 lie inside it, which seems unlikely.
However, despite that porosity seems out of reach. it appears likely that Jθ
has Hausdorff dimension less than two. There are two possible ways of showing
this given our results. The first would be to show that non-uniform porosity itself
leads to a dimension bound. The second is to show that non-uniform porosity is
equivalent to mean porosity or weak mean porosity.
One other item of note about Jθ is about the immediate basin of infinity itself.
It appears not to be a John domain. A set Λ in Ĉ is a John domain if there is a
center z0 and an > 0 such that for all z1 ∈ Λ there exists an arc γ ⊂ Λ connecting
z0 and z1 and d(z, ∂Λ) > |z − z1 | for all z ∈ γ. If Λ is unbounded and its boundary
is compact then z0 = ∞. In order to prove that the immediate basin of infinity is
not a John domain, we would need to show that the set S for points that did not
enter the cone K under iteration was non-empty for an infinite number of n. This
would provide the fact that if we picked z1 close to the critical point 1, we would
not be able to find an appropriate path γ.
Another interesting question is what can be said about the full Julia set Jfθ .
Of the results proven here, the finding of the cone K, Proposition 3.1, is certainly
true for Jfθ . This comes from the fact that the filled in Julia set Kfθ has Jθ as its
16
KIMBERLY A. ROTH
boundary. The Corollary 3.7 is also true for Jfθ as well. These might be of some
use when attempting to find or estimate the measure of Jfθ which is unknown.
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