NUMERICAL VALUES OF THE HAUSDORFF AND PACKING MEASURES

FOR LIMIT SETS OF ITERATED FUNCTION SYSTEMS

James Edward Reid

Dissertation Prepared for the Degree of

DOCTOR OF PHILOSOPHY

UNIVERSITY OF NORTH TEXAS

August 2017

APPROVED:

Mariusz Urbański, Major Professor Lior Fishman, Committee Member Stephen Jackson, Committee Member Charles Conley, Chair of the Department of Mathematics David Holdeman, Dean of the College of Arts and Science Victor R. Prybutok, Dean of the Toulouse Graduate School Reid, James Edward. Numerical Values of the Hausdorff and Packing Measures

for Limit Sets of Iterated Function Systems. Doctor of Philosophy (Mathematics), August

2017, 107 pp., 56 numbered references.

In the context of fractal geometry, the natural extension of volume in Euclidean

space is given by Hausdorff and packing measures. These measures arise naturally in the

context of iterated function systems (IFS). For example, if the IFS is finite and conformal,

then the Hausdorff and packing dimensions of the limit sets agree and the corresponding

Hausdorff and packing measures are positive and finite. Moreover, the map which takes

the IFS to its dimension is continuous. Developing on previous work, we show that the map which takes a finite conformal IFS to the numerical value of its packing measure is continuous. In the context of self-similar sets, we introduce the super separation condition.

We then combine this condition with known density theorems to get a better handle on finding balls of maximum density. This allows us to extend the work of others and give exact formulas for the numerical value of packing measure for classes of Cantor sets,

Sierpinski N-gons, and Sierpinski simplexes.

Copyright 2017 by James Edward Reid

ii ACKNOWLEDGMENTS

A special thanks to my advisor Professor Urba´nski.Thank you! And of course, thank you to my family for their unconditional support: Mom, Dad, Ryan, Geemaw, Pa, Granny, Poppa, Matt, Richard, Sherri, Victoria, Katherine, Mark, Teresa, and Cooper. Thank you!

iii TABLE OF CONTENTS

Page

ACKNOWLEDGMENTS iii

CHAPTER 1 INTRODUCTION1 1.1. Hausdorff and Packing Measures1 1.2. Iterated Function Systems4 1.3. Conformal Mappings5 1.4. Conformal Iterated Function Systems9

CHAPTER 2 CONTINUITY OF INVARIANT SETS, INVARIANT MEASURES, AND HAUSDORFF/PACKING DIMENSION 13 2.1. The Hausdorff Metric and Continuity of Limit Sets 13 2.2. The Space of Measures and Metrics in the Weak*-Topology 19 2.3. The Transfer Operator and Continuity of Invariant Measures 22 2.4. The Pressure Function and Continuity of Hausdorff/Packing Dimension 36

CHAPTER 3 CONTINUITY OF THE NUMERICAL VALUE OF HAUSDORFF AND PACKING MEASURES FOR FINITE ITERATED FUNCTION SYSTEMS 42 3.1. General Density Theorems 43 3.2. Density Theorems for Self-Similar Sets 45 3.3. Density Theorems for Self-Conformal Sets 48 3.4. Continuity of the Numerical Value for Conformal IFS 67

CHAPTER 4 THE NUMERICAL VALUE OF PACKING MEASURE FOR SUPER SEPARATED ITERATED FUNCTION SYSTEMS 80 4.1. The Super Separation Condition 81

iv 4.2. A Density Theorem for Super Separated IFS 83 4.3. Packing Measure of Super Separated Cantor Sets 85 4.4. Packing Measure of Super Separated Sierpi´nskiN-gons 91 4.5. Packing Measure of Super Separated Sierpi´nskiSimplexes 102

BIBLIOGRAPHY 104

v CHAPTER 1

INTRODUCTION

1.1. Hausdorff and Packing Measures

We first discuss Hausdorff and packing measures, which are fundamental to geometric measure theory and fractal geometry (see for example [17, 18, 31, 35]). In this section we will assume that (X, ρ) is a metric space.

Definition 1.1.1. The diameter of a nonempty set A ⊆ X is defined to be

|A| = sup{ρ(x, y): x, y ∈ A},

and we define |∅| = 0.

Definition 1.1.2. Suppose 0 < δ ≤ ∞.A δ-covering of a set A ⊆ X is a countable [ collection U of subsets of X with diameters at most δ such that A ⊆ U. U∈U Definition 1.1.3. Let s ≥ 0 be some real number. For every 0 < δ ≤ ∞, the s-dimensional δ-Hausdorff premeasure of a set A ⊆ X is given by ( ) s X s Hδ (A) = inf |U| : U is a δ-covering of A . U∈U The s-dimensional Hausdorff measure of a set A ⊆ X is given by

s s H (A) = sup Hδ (A) . δ>0

The Hausdorff dimension of A is defined to be

HD (A) = inf {s ≥ 0 : Hs (A) = 0} = sup {s ≥ 0 : Hs (A) = ∞} ∪ {0}.

The integer-dimensional Hausdorff measures are important cases when X = Rd be- cause they correspond with our usual notion of length, area, and volume (see [18, 31] for more precise statements):

(i) H0 is the counting measure,

1 (ii) H1 (Γ) is the classical notion of length for a rectifiable curve Γ, (iii) if E is a k-dimensional C1-surface with 1 ≤ k ≤ d − 1, Hk (E) is a constant multiple of the classical k-dimensional area of E, (iv) Hd is a constant multiple of the d-dimensional Lebesgue measure.

In addition, for any s ≥ 0, if 0 < Hs (A) < ∞, then s = HD (A). Therefore, nice sets such as k-dimensional C1-surfaces retain their usual dimension. However, Hausdorff dimension allows for non-integer dimensions, and many interesting fractal sets have non- integer Hausdorff dimension and can even have positive and finite Hausdorff measure. For these reasons, we say that Hausdorff measure extends the notion of volume in Rd, and Hausdorff dimension extends the usual notion of dimension. In fractal geometry, Hausdorff dimension can be considered an invariant because it is preserved under bi-Lipschitz maps.

Another extension of volume in Rd called packing measure was introduced by Sullivan [47] and Taylor and Tricot [49, 50, 51]. Packing measures exhibit many of the same properties as Hausdorff measures. However, Sullivan demonstrated that for geometric limit sets arising from Kleinian groups, sometimes Hausdorff measure is the natural measure to consider, and sometimes packing measure is the natural measure to consider. More specifically, he showed that sometimes the Hausdorff measure of the limit set is 0, but the packing measure is positive and finite. Taylor and Tricot were interested in Brownian trajectories and showed that the packing measure with a certain gauge function is positive and finite.

Packing measure is a sort of dual to Hausdorff measure in the sense that Hausdorff measure is defined in terms of efficient coverings, while packing measure is defined in terms of bountiful packings.

Definition 1.1.4. Suppose 0 < δ < ∞.A δ-packing of a set A ⊆ X is a countable collection

∞ of closed balls {B(xi, ri)}i=1 with centers in A, radii at most δ, and ρ(xi, xj) > ri + rj for all i 6= j.

Definition 1.1.5. Let s ≥ 0 be some real number. For every 0 < δ ≤ ∞, the s-dimensional

2 δ-packing premeasure of a set A ⊆ X is given by

( ) s X s Pδ (A) = sup |B| : B is a δ-packing of A . B∈B

The s-dimensional packing premeasure of a set A ⊆ X is given by

s s P0 (A) = inf Pδ (A) . δ>0

The s-dimensional packing measure of a set A ⊆ X is given by

( ∞ ∞ ) s X s [ P (A) = inf P0 (Ai): A ⊆ Ai . i=1 i=1

The packing dimension of A is defined to be

PD (A) = inf {s ≥ 0 : Ps (A) = 0} = sup {s ≥ 0 : Ps (A) = ∞} ∪ {0}.

For any set A ⊆ Rd and s ≥ 0, Hs (A) ≤ Ps (A) (Theorem 5.12 [35]). However, it is actually rare for equality to hold. It turns out that 0 < Hs (A) = Ps (A) < ∞ if and only if s is an integer and A is sufficiently rectifiable (see Theorem 17.11 [35]). In particular, we see that packing measure is an extension of volume just like Hausdorff measure. However, the numerical value of their measures may behave quite differently for fractal sets.

We will be especially interested in the exact numerical value of packing measure. However, the definition is given in three stages, and this can make it awkward to work with at times. Fortunately, we have the following result by Feng, Hua, and Wen [20], which will be of great use in this thesis.

Theorem 1.1.1. Let K be a compact subset of Rd, and let s ≥ 0.

s s s (1) If P0 (K) < ∞, then P (K) = P0 (K). s (2) If P0 (K) = ∞, then for any ε > 0 there exists a compact subset F of K such that s s s s P (F ) = P0 (F ) and P (F ) ≥ P (K) − ε.

3 1.2. Iterated Function Systems

Let (X, ρ) be a nonempty compact metric space. Fix a countable (finite or infinite) set I with at least two elements. We will refer to the set I as the alphabet. An iterated

function system (IFS) is a collection Φ = {ϕi}i∈I , where ϕi : X → X is a contraction for

all i ∈ I and supi∈I Lip (ϕi) < 1. We will almost always use a capital letter to denote an IFS and the corresponding lowercase letter for the mappings in the IFS. The collection of all iterated function systems on the space X with alphabet I is denoted IFS (X,I). A finite word of length n ≥ 1 in the alphabet I is a finite sequence of symbols (a

tuple) denoted by ω = ω1 . . . ωn. The length of the word ω is denoted by |ω| = n. For each n ≥ 1, we define In to be the set of all finite words in the alphabet I of length n. Define

∗ S∞ n ∗ I = n=1 I be the collection of all finite words. For each ω ∈ I , we define

ϕω = ϕω1 ◦ ϕω2 ◦ · · · ◦ ϕω|ω| .

∗ Notice that ϕω is continuous for every ω ∈ I , and therefore, ϕω(X) is compact. In addition,

  |ω| ρ(ϕω(x), ϕω(y)) ≤ Lip (ϕω1 ) ··· Lip ϕω|ω| ρ(x, y) ≤ (sup Lip (ϕi)) ρ(x, y) i∈I

|ω| ∗ for all x, y ∈ X. Thus, |ϕω(X)| ≤ (supi∈I Lip (ϕi)) |X| for every ω ∈ I . An infinite word in the alphabet I is a sequence of symbols in I indexed by the

∞ positive integers denoted by ω = ω1ω2ω3 .... Define I to be the collection of all infinite

∞ words in the alphabet I. For every ω ∈ I , we define ω|n = ω1 . . . ωn for any n ≥ 1.

∞ ∞ Observe now that given ω ∈ I , {ϕω|n (X)}n=1 forms a descending sequence of com- pact sets whose diameters converge to zero since

n |ϕω|n (X)| ≤ (sup Lip (ϕi)) |X|. i∈I T∞ This implies that n=1 ϕω|n (X) contains exactly one element, which we denote by π(ω). Therefore, this formula defines a map π : I∞ → X which we will call the codingmap. We equip I∞ with the topology generated by the cylinders

∞ Cτ = {ω ∈ I : ω||τ| = τ}

4 ∗ for all τ ∈ I . Note that Cτ is the collection all infinite extensions of the word τ. With this

∞ topology, the map π is continuous. If I is finite, then I is compact, and consequently, JΦ is compact. When I is infinite, the limit set JΦ is not necessarily closed.

The main object of interest is the limit set of Φ, denoted JΦ, defined by

∞ ∞ [ \ JΦ = π(I ) = ϕω|n (X). ω∈I∞ n=1

∞ ∞ ∞ Define the left shift map σ : I → I by σ(ω) = ω2ω3 ... for all ω ∈ I . Notice that

π(ω) = ϕω1 (π(σ(ω)))

for every ω ∈ I∞. Then,

∞ [ ∞ [ ∞ [ JΦ = π(I ) = ϕi(π(σ(I ))) = ϕi(π(I )) = ϕi(JΦ). i∈I i∈I i∈I Iterating this process, we see that

[ JΦ = ϕω(JΦ) ω∈In for all n ≥ 1. The IFS Φ is said to satisfy the Open Set Condition (OSC) if there exists a nonempty

open set O ⊆ X such that ϕi ⊆ O for every i ∈ I and ϕi(O) ∩ ϕj(O) = ∅ for every i, j ∈ I, i 6= j.

1.3. Conformal Mappings

In order to talk about conformal iterated function systems, we first need to discuss

conformal mappings. Suppose U is a nonempty open connected subset of Rd and f : U → Rd is differentiable. Recall that for every x ∈ U, the derivative of f at x is a linear map

f 0(x): Rd → Rd such that |f(y) − f(x) − f 0(x)(y − x)| lim . y→x |y − x|

Moreover, the map f 0(x) is given by the Jacobian matrix of partial derivatives of f at x. We

0 denote the determinant of the Jacobian matrix by Jf (x), and we denote the norm of f (x)

5 by |f 0(x)(v)|  |f 0(x)| = sup : v 6= 0 . |v|

Definition 1.3.1. Suppose U and V are nonempty open connected subsets of Rd. A dif- feomorphism f : U → V is said to be conformal if the derivative of f at each point of U is a similarity map. That is, for every x ∈ U there is some constant c(x) > 0 with

|f 0(x)(v)| = c(x)|v| for every v ∈ Rd. It follows then that c(x) = |f 0(x)|.

In the case d = 1, the definition of conformal only requires that f 0(x) 6= 0 for all x ∈ U. Therefore the conformal maps for d = 1 are precisely the strictly monotone diffeomorphisms of open intervals. For d = 2, conformal means holomorphic or antiholomorphic. When d ≥ 3, it turns out that conformal maps are M¨obiustransformations. This fact is known as Liouville’s theorem. We will now give a more precise overview of the cases d = 2 and d ≥ 3. Recall that the orthogonal group is defined by

T O(d) = {A ∈ GL(d, R): A A = I}, where GL(d, R) is the group of invertible d×d matrices (called the general linear group) and AT is the transpose of A. A more geometric description of O(d) is

d O(d) = {A ∈ GL(d, R): hAv, Awi = hv, wi for all v, w ∈ R }, which means O(d) is the group of all isometries of Rd that fix the origin. Thus we immediately conclude the following proposition.

Proposition 1.3.1. A function f : Rd → Rd is a similarity if and only if f(x) = λAx + b for some λ > 0, A ∈ O(d), and b ∈ Rd. It is already evident that similarities are conformal, but we can easily see it from the above proposition. If f(x) = λAx + b is a similarity, then Df(x) = λA. Since A is an isometry, it is immediate that f is conformal and |f 0(x)| = λ.

6 Next we will describe (hyper)sphere inversion. Consider a sphere S centered at a ∈ Rd of radius r > 0. That is, S = {x ∈ Rd : |x − a| = r}. Consider any point x 6= a and the ray ax−→ starting at a and passing through x. Then there is exactly one pointx ˆ on ax−→ with |x−a||xˆ−a| = r2. We callx ˆ the inversion (or reflection) of the point x through S. Apparently,

x − a xˆ = r2 + a. |x − a|2

As x approaches S,x ˆ approaches S. Clearly if x ∈ S thenx ˆ = x. As x approaches a,x ˆ approaches infinity. For this reason it is convenient to say that the inversion of a is ∞ and vice-versa.

Define Rbd = Rd ∪ {∞}.

d d d Definition 1.3.2. Let a ∈ R and r > 0. Define ia,r : Rb → Rb by  a if x = ∞   ia,r(x) = ∞ if x = a   r2 x−a + a otherwise. |x−a|2

We say ia,r is the inversion of a sphere centered at a of radius r.

Proposition 1.3.2. Let a ∈ Rd and r > 0. Then for all x ∈ Rd \{a} and v ∈ Rd, r2 i0 (x)(v) = R (v), a,r |x − a|2 x−a

where hv, x − ai R (v) = v − 2 (x − a) x−a |x − a|2 is the reflection of v through the hyperplane orthogonal to the vector x − a. In particular,

sphere inversions are conformal in Rd \{a} with r2 |i0 (x)| = . a,r |x − a|2

Definition 1.3.3. A function f : Rbd → Rbd is called a M¨obiustransformation if f = λA◦i+b, where λ > 0, A ∈ O(d), i is either the identity map or a sphere inversion, and b ∈ Rd.

7 In other words, a M¨obiustransformation is a composition of similarities and sphere inversions. Since similarities and sphere inversions are conformal, M¨obiustransformations are conformal. When d ≥ 3, every conformal map is a M¨obius transformation. This re- markable fact is known as Liouville’s theorem. For a more thorough treatment of M¨obius transformations, see [9].

Assume momentarily that d = 2. The conjugation map z 7→ z¯ is not holomorphic since it does not satisfy the Cauchy-Riemann equations at any point. The conjugation map is conformal as it is a reflection across the real axis, but it does not preserve orientation.

For any complex function f, denote f¯(z) = f(z). A function f : U → C is said to be antiholomorphic on U if f¯ is holomorphic on U.

Theorem 1.3.3. Let f : U → V be a diffeomorphism between nonempty open connected sets in C. Then f is conformal if and only if Jf never vanishes and either f is holomorphic or antiholomorphic.

In the case d ≥ 3, the only conformal maps are M¨obiustransformations. The case of d = 3 was proven in 1850 by Liouville [28]. The case of d ≥ 3 and C1 conformal maps was shown in 1958 by Hartman [23]. A simple proof of Liouville’s theorem when f is a C4-diffeomorphism is given by Nevanlinna [41]. Full proofs of Liouville’s theorem can be found in [10, 11, 12].

Theorem 1.3.4 (Liouville). Let d ≥ 3 and let f : U → V be a conformal diffeomorphism between nonempty open connected sets in Rd. Then f can be extended to a M¨obiustransfor- mation on Rbd. That is, f = λA ◦ i + b, where λ > 0, A ∈ O(d), i is either the identity map or a sphere inversion, and b ∈ Rd.

We finish this section with a quick geometric/measure-theoretic observation. Let µ be d-dimensional Lebesgue measure. Recall that for any measurable set A ⊆ Rd and x ∈ U,

0 µ(f (x)(A)) = |Jf (x)|µ(A).

8 If f is conformal, then µ(f 0(x)(A)) = |f 0(x)|dµ(A)

0 0 d for all x ∈ U since f (x) is a similarity map. Therefore |f (x)| = |Jf (x)| for all x ∈ U. We conclude by the change of variable formula that for all conformal maps f : U → V and measurable sets A ⊆ Rd Z µ(f(A)) = |f 0(x)|ddµ(x). A This motivates the concept of conformal measure for a conformal IFS that is presented in the next section.

1.4. Conformal Iterated Function Systems

In this section, X will always be a nonempty compact subset of Rd for some d ≥ 1. The theory of conformal iterated function systems was developed by Mauldin and Urba´nski [36, 37, 38]. In this section, we will state the definitions and results most relevant to this thesis.

Definition 1.4.1. Suppose X is a nonempty compact subset of Rd for some d ≥ 1 , I is a countable alphabet, and Φ ∈ IFS (X,I). Then, Φ is called conformal (CIFS) if the following conditions are satisfied:

(1) X is compact, connected, and X = ClRd (IntRd (X)) (closure of the interior of X in Rd).

(2) ϕi : X → X is injective for all i ∈ I.

d (3) There exists an open connected set V ⊆ R such that X ⊂ V and ϕi extends to a C1 conformal diffeomorphism of V into V .

(4) Open Set Condition: Φ satisfies the OSC with O = IntRd (X). That is, for all i, j ∈ I, i 6= j,

ϕi(IntRd (X)) ∩ ϕj(IntRd (X)) = ∅.

(5) Cone Condition: There exists α, ` > 0 such that for every x ∈ ∂X ⊆ Rd there

exists an open cone Con(x, ux, α, `) ⊆ IntRd (X) with vertex x, direction vector ux, central angle of Lebesgue measure α, and altitude `.

9 (6) Bounded Distortion Property (BDP): There exists K ≥ 1 such that

0 0 |ϕω(y)| ≤ K|ϕω(x)|

for every ω ∈ I∗ and x, y ∈ V .

The collection of all conformal iterated function systems will be denoted CIFS (X,I).

The most important results we present in this thesis require that I be finite. In this

1+ε case, the BDP is automatically satisfied when ϕi : V → V is C for all i ∈ I. For additional sufficient conditions for the BDP to hold, consult [36].

Proposition 1.4.1. Let X ⊆ Rd be a nonempty compact set, I a finite alphabet, and Φ ∈

CIFS (X,I). Suppose for all i ∈ I, ϕi : V → V satisfies

0 0 ε ||ϕi(x)| − |ϕi(y)|| ≤ C|x − y| for some C > 0 and ε > 0. Then, there is a constant c > 0 such that for all ω ∈ I∗ and x, y ∈ V ,

0 c|x−y|ε 0 |ϕω(y)| ≤ e |ϕω(x)|.

In particular, assuming V is bounded,

K = ec|V |ε

∗ 0 Proof. Let ω ∈ I and x, y ∈ V . Let n = |ω|, m = mini∈I infz∈V |ϕi(z)|, and L = maxi∈I Lip (ϕi). Notice, m > 0 and log is Lipschitz on [m, ∞) with Lipschitz constant 1/m. Notice,

0 n |ϕω(y)| X 0 0 log ≤ log |ϕ (ϕ j (y))| − log |ϕ (ϕ j (x))| |ϕ0 (x)| ωj σ (ω) ωj σ (ω) ω j=1 n 1 X 0 0 ≤ ||ϕ (ϕ j (y))| − |ϕ (ϕ j (x))|| m ωj σ (ω) ωj σ (ω) j=1 n C X ε ≤ |ϕ j (y) − ϕ j (x)| m σ (ω) σ (ω) j=1

10 n C ε X

ε ≤ |x − y| Lip ϕ j m σ (ω) j=1 n C X ≤ |x − y|ε (Lε)n−j m j=1 C 1 ≤ |x − y|ε . m 1 − Lε

Therefore the result follows by letting

C c = . m(1 − Lε)



In R2, we can use the Koebe Distortion Theorem to guarantee BDP holds.

Theorem 1.4.2 (Koebe Distortion). If g : B(0, 1) → C is a univalent holomorphic (or antiholomorphic) function, then for all z ∈ B(0,R)

1 − |z| |g0(z)| 1 + |z| ≤ ≤ (1 + |z|)3 |g0(0)| (1 − |z|)3

In Rd for d ≥ 3, we can use the following estimate found in [48] as Theorem 3.1.

Theorem 1.4.3. Suppose V is a nonempty open connected subset of Rd, where d ≥ 3 and F ⊂ V is a bounded set such that F ⊂ V . If ϕ : V → Rd is a conformal map (in particular, ϕ−1(∞) ∈/ V ), then for all x, y ∈ F ,

|ϕ0(y)|  |F | 2 0 ≤ 1 + c , |ϕ (x)| dinf (F,V )

c c where dinf (F,V ) = inf{|x − y| : x ∈ F, y ∈ V }.

Definition 1.4.2. Let X ⊆ Rd be a nonempty compact set, I a countable alphabet, and Φ ∈ CIFS (X,I). Given t ≥ 0, a Borel probability measure m is said to be t-conformal if m(JΦ) = 1, m(ϕi(X) ∩ ϕj(X)) = 0 for all i, j ∈ I with i 6= j, and Z 0 t m(ϕi(A)) = |ϕi| dm A for every Borel set A ⊆ X. We will call Φ regular if a t-conformal measure exists for Φ.

11 The topological pressure of a conformal IFS Φ is given by

1 X 0 t P (t) = lim log kϕωk∞. n→∞ n ω∈In P 0 t Let θ = inf{t ≥ 0 : i∈I kϕik∞ < ∞}. Then, the topology pressure function is non- increasing on [0, ∞), strictly decreasing on [θ, ∞), and convex and continuous on (θ, ∞). Mauldin and Urba´nski[36] showed that a t-conformal measure exists if and only if P (t) = 0.

Moreover, t = HD (JΦ). Therefore, if Φ is regular, then Φ admits an HD (JΦ)-conformal measure. In this case, we will denote this measure by mΦ and refer to it as the conformal measure. We will mostly be interested in the case that I is finite. In this case, the situation is quite nice as demonstrated by the next theorem (see Lemma 3.14 of [36]).

Theorem 1.4.4. Let X ⊆ Rd be a nonempty compact set, I a finite alphabet, and Φ ∈ CIFS (X,I). Then, Φ is regular and there exists C ≥ 1 such that m (B(x, r)) C−1 ≤ Φ ≤ C rhΦ

hΦ hΦ for all x ∈ JΦ and 0 < 2r < |X|. In particular, 0 < H (JΦ) ≤ P (JΦ) < ∞. When Φ consists only of similarities, it is a well-known result due to Hutchinson in his seminal paper [24] that the above theorem holds and

X 0 hΦ (1) kϕik∞ = 1. i∈I Moran [40] had already shown this formula for a class of self-similar sets prior to Hutchinson, so we refer to formula (1) as the Hutchinson-Moran formula.

hΦ hΦ If Φ is regular and 0 < H (JΦ) ≤ P (JΦ) < ∞, then for all Borel sets A ⊆ X

hΦ hΦ H (A ∩ JΦ) P (A ∩ JΦ) mΦ(A) = h = h . H Φ (JΦ) P Φ (JΦ)

12 CHAPTER 2

CONTINUITY OF INVARIANT SETS, INVARIANT MEASURES, AND HAUSDORFF/PACKING DIMENSION

Definition 2.0.3. Suppose (X, ρ) is a metric space. Then, we make the following defini- tions:

(1) dinf (x, A) = inf{ρ(x, y): y ∈ A} for any x ∈ X and nonempty A ⊆ X

(2) dinf (A, B) = inf{ρ(x, y): x ∈ A, y ∈ B} for any nonempty A, B ⊆ X

(3) dsup(x, A) = sup{ρ(x, y): y ∈ A} for any x ∈ X and nonempty A ⊆ X

(4) dsup(A, B) = sup{ρ(x, y): x ∈ A, y ∈ B} for any nonempty A, B ⊆ X

(5)( A)ε = {x ∈ X : dinf (x, A) < } for any ε > 0 and nonempty A ⊆ X

(6)[ A]ε = {x ∈ X : dinf (x, A) ≤ } for any ε ≥ 0 and nonempty A ⊆ X

(7) If (X, ρ) is a bounded metric space, we equip C(X,X) with the metric ρ∞(f, g) =

supx∈X .ρ(f(x), g(x))

Definition 2.0.4. Let I be a countable alphabet, and let (X, ρ) be a bounded complete metric space. We equip IFS (X,I) with sup-metric D∞ defined by

D∞(Φ, Ψ) = sup ρ∞(ϕi, ψi). i∈I

For each 0 < c < 1, we define

IFSc (X,I) = {Φ ∈ IFS (X,I) : sup Lip (ϕi) ≤ c}. i∈I

2.1. The Hausdorff Metric and Continuity of Limit Sets

The goal of this section is to show that the map Φ 7→ JΦ which takes an IFS to its limit set is continuous. In fact, we will show that it is locally Lipschitz. Many of the results encountered here are either considered folklore or have been shown in less general settings (see for example [14] or [8]). The results of this section are well-established when the alphabet I is finite and (X, ρ) is a compact metric space. We generalize by allowing I to be countable.

13 Suppose (X, ρ) is a metric space. Let S(X) be the collection of all nonempty subsets of X, B(X) be the collection of all nonempty closed and bounded subsets of X, and C(X) be the collection of all nonempty compact subsets of X. It is well known that

H(A, B) = inf{ε ≥ 0 : A ⊆ [B]ε,B ⊆ [A]ε}

defines a psuedometric on S(X) and a metric on both B(X) and C(X). We call H the Haus- dorff (psuedo)metric. If (X, ρ) is a complete metric space, it is well known that (B(X),H) and (C(X),H) are complete metric spaces. In fact, C(X) is a closed subspace of B(X). If X is compact, then B(X) = C(X) and (C(X),H) is a compact metric space.

Proposition 2.1.1. Let A and B be nonempty subsets of X. Then,

(1) [A]0 = A (2) H(A, B) = 0 if and only if A = B (3) H(A, B) = H(A, B)

Proof. To prove (1), notice that x ∈ [A]0 if and only if dinf (x, A) = 0 if and only if there is

∞ a sequence {yn}n=1 ⊆ A such that limn→∞ ρ(x, yn) = 0 if and only if x ∈ A.

Now, H(A, B) = 0 if and only if A ⊆ [B]0 and B ⊆ [A]0. By (1), [A]0 = A and

[B]0 = B. The result follows since A ⊆ B and B ⊆ A if and only if A ⊆ B and B ⊆ A. By the triangle inequality and two applications of (2),

H(A, B) ≤ H(A, A) + H(A, B) + H(B,B) = H(A, B).

Using the same trick as above,

H(A, B) ≤ H(A, A) + H(A, B) + H(B, B) = H(A, B).

Therefore, H(A, B) = H(A, B). 

Definition 2.1.1. For any IFS Φ, we define Φ:b B(X) → S(X) by [ Φ(b A) = ϕi(A) i∈I for all nonempty closed and bounded sets A.

14 Theorem 2.1.2. Let (X, ρ) be a metric space. If the alphabet I is finite or X is bounded, then   Φb is a contraction mapping on B(X) with Lip Φb ≤ supi∈I Lip (ϕi). Moreover, if I is finite   or X is compact, then Φb|C(X) is contraction mapping on C(X) with Lip Φb ≤ supi∈I Lip (ϕi).

Proof. If the alphabet I is finite or X is bounded, then clearly Φ:b B(X) → B(X). Fix A, B ∈ B(X). By Proposition 2.1.1 (3), ! ! [ [ [ [ H(Φ(b A), Φ(b B)) = H ϕi(A), ϕi(B) = H ϕi(A), ϕi(B) . i∈I i∈I i∈I i∈I S If ε ≥ 0, ϕi(A) ⊆ [ϕi(B)]ε, and ϕi(B) ⊆ [ϕi(A)]ε for all i ∈ I, then i∈I ϕi(A) ⊆ S S S [ i∈I ϕi(B)]ε and i∈I ϕi(B) ⊆ [ i∈I ϕi(A)]ε. Thus, ! [ [ H ϕi(A), ϕi(B) ≤ sup H(ϕi(A), ϕi(B)) ≤ sup Lip (ϕi) H(A, B). i∈I i∈I i∈I i∈I S If I is finite and A is compact, then i∈I ϕi(A) is compact since each ϕi(A) is compact. If

X is compact, then B(X) = C(X). Therefore, if I is finite or X is compact, then Φb|C(X) is a contraction mapping on C(X). 

We get the next corollary by combining the Banach contraction principle with The- orem 2.1.2.

Corollary 2.1.3. Let (X, ρ) be a complete metric space. If the alphabet I is finite or X

n ∞ is bounded, then Φb has a unique fixed point and {Φb (A)}n=1 converges to the fixed point for every A ∈ B(X) . Moreover, if I is finite or X is compact, then the unique fixed point of Φb is compact.

S Proposition 2.1.4. If A ∈ S(X) and A = i∈I ϕi(A), then A is a fixed point of Φb.

Proof. Notice, [ [ A = ϕi(A) ⊆ ϕi(A) = Φ(b A). i∈I i∈I

By continuity of each ϕi, we obtain

[ [ Φ(b A) = ϕi(A) ⊆ ϕi(A). i∈I i∈I

15 Since [ [ ϕi(A) ⊆ ϕi(A) i∈I i∈I we have that [ [ ϕi(A) ⊆ ϕi(A) = A. i∈I i∈I Therefore, we have shown

A ⊆ Φ(b A) ⊆ A,

and we conclude that A is a fixed point of Φ.b 

In Chapter 1, we defined JΦ via the coding map. Hutchinson [24] was working in the finite alphabet case and worked in a complete metric space instead of a compact metric S space. Combining Corollary 2.1.3 with the fact that JΦ = i∈I ϕi(JΦ) we obtain the following

corollary. This allows us to refer to the unique invariant set of Φb as JΦ in a complete metric space.

Corollary 2.1.5. Let (X, ρ) be a complete metric space. If X is compact, then the unique

fixed point of Φb is JΦ. Moreover, if I is finite, then JΦ is the unique nonempty compact set S such that JΦ = i∈I ϕi(JΦ) . We also obtain a generalization of Barnsley’s Collage Theorem [8] (Chapter X).

Corollary 2.1.6 (Collage Theorem). Let (X, ρ) be a complete metric space. If the alphabet I is finite or X is compact, then for every A ∈ S(X) S
H A, i∈I ϕi(A) H(A, JΦ) ≤ . 1 − supi∈I Lip (ϕi) Proof. By Property 2.1.1 (3), the triangle inequality, Theorem 2.1.2, and Corollary 2.1.5

H(A, JΦ) = H(A, JΦ)     ≤ H A, Φ(b A) + H Φ(b A), JΦ     = H A, Φ(b A) + H Φ(b A), Φ(b JΦ) ! [
≤ H A, ϕi(A) + (sup Lip (ϕi))H A, JΦ i∈I i∈I

16 ! [ = H A, ϕi(A) + (sup Lip (ϕi))H (A, JΦ) . i∈I i∈I Rearranging, we obtain ! [ H(A, JΦ)(1 − sup Lip (ϕi)) ≤ H A, ϕi(A) . i∈I i∈I



Lemma 2.1.7. Suppose (X, ρ) is a bounded complete metric space. The map (IFS (X,I) ,D∞) →

(C(B(X), B(X)),H∞) given by Φ 7→ Φb is Lipschitz continuous with Lipschitz constant 1.

Proof. Assume Φ, Ψ ∈ IFS (X,I) and fix A ∈ B(X). Using the same argument as in Theorem 2.1.2, ! [ [ H(Φ(b A), Ψ(b A)) = H ϕi(A), ψi(A) i∈I i∈I ! [ [ = H ϕi(A), ψi(A) i∈I i∈I

≤ sup H(ϕi(A), ψi(A)). i∈I

For each i ∈ I, let εi = supx∈A ρ(ϕi(x), ψi(x)). It is clear that ψi(A) ⊆ [ϕi(A)]εi and

ϕi(A) ⊆ [ψi(A)]εi for each i ∈ I. Then, for each i ∈ I, we have

H(ϕi(A), ψi(A)) ≤ sup ρ(ϕi(x), ψi(x)) ≤ ρ∞(ϕi, ψi). x∈A

Thus,

H(Φ(b A), Ψ(b A)) ≤ sup ρ∞(ϕi, ψi) = D∞(Φ, Ψ). i∈I Therefore,

H∞(Φb, Ψ)b = sup H(Φ(b A), Ψ(b A)) ≤ D∞(Φ, Ψ). A∈B(X) 

We now present the main result of this section, which says that the map Φ 7→ JΦ is locally Lipschitz.

17 Theorem 2.1.8. Suppose (X, ρ) is a compact metric space. Fix Φ ∈ IFS (X,I). Then, for every Ψ ∈ IFS (X,I),

D∞(Φ, Ψ) H(JΦ,JΨ) = H(JΦ, JΨ) ≤ . 1 − supi∈I Lip (ϕi)

In particular, if 0 < c < 1 and Φ, Ψ ∈ IFSc (X,I), then

D (Φ, Ψ) H(J ,J ) = H(J , J ) ≤ ∞ . Φ Ψ Φ Ψ 1 − c

Proof. By Property 2.1.1 (3), H(JΦ,JΨ) = H(JΦ, JΨ). Combining Property 2.1.1 (3) and Corollary 2.1.5, we obtain

! [   H JΨ, ϕi(JΨ)) = H JΨ, Φ(b JΨ) i∈I   = H Ψ(b JΨ), Φ(b JΨ)

≤ H∞(Φb, Ψ)b .

Then by Lemma 2.1.7, ! [ H JΨ, ϕi(JΨ)) ≤ D∞(Φ, Ψ). i∈I

Using the Collage Theorem (Corollary 2.1.6), we conclude

S
H JΨ, i∈I ϕi(JΨ) D∞(Φ, Ψ) H(JΦ, JΨ) ≤ ≤ . 1 − supi∈I Lip (ϕi) 1 − supi∈I Lip (ϕi)

The second part of the theorem holds by simply noting that

1 1 ≤ . 1 − supi∈I Lip (ϕi) 1 − c



Corollary 2.1.9. Suppose (X, ρ) is a compact metric space. Then, the map (IFS (X,I) ,D∞) →

(C(X),H) given by Φ 7→ JΦ is locally Lipschitz.

18 2.2. The Space of Measures and Metrics in the Weak*-Topology

Suppose (X, ρ) is a metric space. Let M(X) be the collection of all finite Radon

measures on X, and let M1(X) be the collection of all Radon probability measures on X. If (X, ρ) is a complete separable metric space, every Borel measure on X is automatically a Radon measure (see [13] or [15]), which simply means that for every Borel set B

µ(B) = sup{µ(K): K is compact and K ⊆ B}.

It is a basic topological fact that if (X, ρ) is a compact metric space, then it is automatically a complete separable metric space. Since we will be assuming (X, ρ) is a compact metric

space, the spaces M(X) and M1(X) are the collections of all Borel measures on X and Borel probability measures on X, respectively.

We now collect some basic facts about M1(X) that will be useful in this section. Most importantly, we need a notion of convergence for our space of measures. The weak*- topology on M(X) is the smallest topology making each of the maps µ 7→ R fdµ continuous, where f : X → R is some bounded continuous function. A sequence of Borel measures µn converges to µ in the weak*-topology if and only if

lim µn(f) = µ(f) n→∞ for every bounded continuous function f : X → R. In the case that (X, ρ) is compact, we can drop the boundedness condition since f(X) is automatically compact and hence bounded in

R by the Heine-Borel theorem. We will simply say that µn converges weakly to µ instead of saying “converges weak-star-ly”. Some authors use other terminology such as “converges narrowly” [15] or “converges vaguely”. We now state the well-known Alexandroff’s Theorem (see [1,2,3] or Theorem 8.2.3 [13]), which is often referred to as the “portmanteau theorem”. A very general treatment of weak*-convergence can be found in Bogochev’s Measure Theory book (Volume II, Chapter 8) [13], and on page 453 of he writes “The English word ‘portmanteu’ (originally a French word, meaning coat-hanger) has the archaic meaning of a large traveling bag and may also denote multi-purpose or multi-function objects or concepts. I do not know who invented

19 such a nonsensical name for Alexandroff’s Theorem. It seems there is no need to attach a meaningless label without any mnemonic content to a result with obvious and generally recognized authorship, rather than just calling it by the inventor’s name”.

∞ Theorem 2.2.1. Suppose (X, ρ) is a metric space, {µn}n=1 is a sequence of Borel proba- bility measures on X, and µ is a Borel probability measure on X. Then, the following are equivalent:

∞ (1) {µn}n=1 converges weakly to µ,

(2) lim supn→∞ µn(F ) ≤ µ(F ) for every closed set F ⊆ X,

(3) lim infn→∞ µn(U) ≥ µ(U) for every open set U ⊆ X,

(4) limn→∞ µn(A) = µ(A) for every Borel set A with µ(∂A) = 0,

(5) lim supn→∞ µn(f) ≤ µ(f) for every upper semicontinuous function f bounded above,

(6) lim infn→∞ µn(f) ≥ µ(f) for every lower semicontinuous function f bounded below.

Notice that M1(X) is convex since for all p ∈ [0, 1] and µ, ν ∈ M1(X), (pµ + (1 − p)ν)(X) = pµ(X) + (1 − p)ν(X) = 1. Then, combining the next well-known theorem

(Corollary 2.5.29 of [15]) with the convexity of M1(X) allows to use tools like the Schauder- Tychonoff fixed point theorem.

Theorem 2.2.2. If (X, ρ) is a compact metric space, then M1(X) is compact in the weak*- topology.

We now define some useful metrics on M1(X) which coincide with the weak*-topology.

Definition 2.2.1. Suppose (X, ρ) is a metric space. For each γ > 0, define for all µ, ν ∈

M1(X),

Lγ(µ, ν) = sup{µ(f) − ν(f): kfk∞ ≤ γ and f ∈ Lip1 (X, R)},

and

P (µ, ν) = inf{ε > 0 : ν(B) ≤ µ([B]ε) + ε, µ(B) ≤ ν([B]ε) + ε for all Borel B}.

We refer to Lγ as the Hutchinson γ-metric because of its use in [24], and we will refer to P as the L´evy-Prokhorov metric (see for example [13]). The Hutchinson metric is sometimes called

20 the Monge-Kantorovich metric by Hutchinson, himself [25] or sometimes the Kantorovich- Rubinshtein metric (see for example [13]).

For proofs that Lγ and P define metrics on M1(X) that generate the weak*-topology, consult Theorem 8.3.2 in [13] and Chapter 2.5 of [15] respectively. Note that our notation differs from that of Bogachev and Edgar. Bogachev uses the norm k · k0 instead of L1 and

+ Mσ (X) to denote the set of all nonnegative Baire measures. In the case of metric spaces, there is no difference between Baire measures and Borel measures by Bogachev’s definition

(see page 12 of [13]). Edgar uses the notation ργ in place of Lγ.

Proposition 2.2.3. Suppose that (X, ρ) is a bounded metric space. Then, for all γ ≥ |X| and µ, ν ∈ M1(X),

Lγ(µ, ν) = sup{µ(f) − ν(f) : Lip (f) ≤ 1}.

Proof. Suppose (X, ρ) is bounded, fix c ∈ X, and let γ ≥ |X|. It is obvious from the definition that for any µ, ν ∈ M1(X),

Lγ(µ, ν) ≤ sup{µ(f) − ν(f) : Lip (f) ≤ 1}.

Now let f be Lipschitz with Lip (f) ≤ 1, and define g(x) = f(x) − f(c). Then, Lip (g) ≤ 1 and g(c) = 0. Now,

|g(x) − g(y)| ≤ ρ(x, y) ≤ |X| ≤ γ, and hence |g(X)| ≤ γ. Since g(c) = 0, we conclude that kgk∞ ≤ γ. For any µ, ν ∈ M1(X),

µ(g) − ν(g) = (µ(f) − µ(f(c))) − (ν(f) − ν(f(c)))

= (µ(f) − f(c)) − (ν(f) − f(c))

= µ(f) − ν(f).

Therefore we conclude that

sup{µ(f) − ν(f) : Lip (f) ≤ 1} ≤ Lγ(µ, ν).



21 In the case that (X, ρ) is a bounded metric space, Proposition 2.2.3 allows us to drop

the γ from the definition of Lγ and instead work with the metric

L(µ, ν) = sup{µ(f) − ν(f) : Lip (f) ≤ 1}.

We will refer to L as the Hutchinson metric. We also see from the proof of Proposition2 .2.3 that we can always assume that there is some c ∈ X such that for all f in the definition of L, f(c) = 0. The next theorem follows immediately from Theorem 8.10.43 [13].

Theorem 2.2.4. Suppose (X, ρ) is a metric space. Then,

2P (µ, ν)2 ≤ L (µ, ν) ≤ 3P (µ, ν). 2 + P (µ, ν) 1

If (X, ρ) is complete, then (M1(X),L1) and (M1(X),P ) are complete metric spaces which generate the weak*-topology.

2.3. The Transfer Operator and Continuity of Invariant Measures

Assume (X, ρ) is a compact metric space. Recall that (C(X, R), k · k∞) is a Banach 1 L space, and so the ` -sum i∈I C(X, R) is a Banach space with the norm k·k1. More precisely, P suppose g = {gi}i∈I , where each gi : X → R is continuous and satisfies i∈I kgik∞ < ∞. Then, we can define X kgk1 = kgik∞, i∈I L and k · k1 is a norm on i∈I C(X, R). We make the next definition in an analog to the work of Walters [54].

Definition 2.3.1. Suppose (X, ρ) is a compact metric space and I is some countable al-

phabet. Then, we define the set W (X,I) to be the collection of g = {gi}i∈I satisfying the following properties:

(1) for all i ∈ I, gi : X → R is continuous and nonnegative, X (2) kgk1 = kgik∞ < ∞, i∈I (3) infx∈X gi(x) > 0 for some i ∈ I.

22 We will call g a vector if gi is constant for all i ∈ I, and we will call g a probability vector

if it is a vector and kgk1 = 1.

Proposition 2.3.1. Suppose (X, ρ) is a compact metric space, I is some countable alphabet, and g ∈ W (X,I). Then, for all ε > 0, there is a δ > 0 such that for all x, y ∈ X, if X ρ(x, y) < δ, then |gi(x) − gi(y)| < ε. i∈I

Proof. Let ε > 0. Since kgk1 < ∞, there is some nonempty finite set F ⊆ I so that

X ε kg k < . i ∞ 4 i∈I−F Therefore, for all x, y ∈ X we have that

X X X |gi(x) − gi(y)| ≤ |gi(x)| + |gi(y)| i∈I−F i∈I−F i∈I−F X ≤ 2 kgik∞ i∈I−F ε < . 2

Since X is compact, gi is uniformly continuous for all i ∈ I by the Heine-Cantor theorem. Choose δ > 0 so that for all x, y ∈ X with ρ(x, y) < δ and i ∈ F ,

ε |g (x) − g (y)| < . i i 2#F Thus,

X X X |gi(x) − gi(y)| = |gi(x) − gi(y)| + |gi(x) − gi(y)| i∈I i∈F i∈I−F ε ε < #F + 2#F 2

= ε



We now make a definition of the transfer operator (or Perron-Frobenius operator) that makes sense for our setting. The definition is motivated by the approaches presented in [36, 54].

23 Definition 2.3.2. Suppose (X, ρ) is a compact metric space, I is a countable alphabet, and

Φ ∈ IFS (X,I). For each g ∈ W (X,I) and f ∈ C(X, R), we define the transfer operator by

X LΦ (g)(f)(x) = gi(x)f(ϕi(x)) i∈I for all x ∈ X.

Theorem 2.3.2. Suppose (X, ρ) is a compact metric space, I is a countable alphabet, and Φ ∈ IFS (X,I). Fix g ∈ W (X,I). Then,

(1) LΦ (g): C(X, R) → C(X, R) is a continuous linear map,

(2) if g is a vector, then LΦ (g) : Lip (X, R) → Lip (X, R), and the Lipschitz constant

of LΦ (g)(f) is at most kgk1Lip (f) supi∈I Lip (ϕi) for every f ∈ Lip (X, R),

(3) if g is a probability vector, then LΦ (g) : Lip1 (X, R) → Lip1 (X, R).

Proof. We will show (1) first. Let ε > 0, f ∈ C(X, R), and x ∈ X. Since X is compact, f must be uniformly continuous by the Heine-Cantor theorem. Then, there is some δ1 > 0, such that for all a, b ∈ X with ρ(a, b) < δ1,

ε |f(a) − f(b)| < . 2kgk1

Choose δ2 > 0 as a witness to Proposition 2.3.1 replacing ε with ε/(2kfk∞).

Let δ = min{δ1, δ2}. Suppose y ∈ X and ρ(x, y) < δ. Now, for all i ∈ I, we have that

ρ(ϕi(x), ϕi(y)) ≤ Lip (ϕi) ρ(x, y) ≤ ρ(x, y) < δ1.

Thus, for all i ∈ I, we have that

ε |f(ϕi(x)) − f(ϕi(y))| < . 2kgk1

For convenience, denote fL = LΦ (g)(f). Notice,

X X |fL(x) − fL(y)| ≤ fL(x) − gi(y)f(ϕi(x)) + gi(y)f(ϕi(x)) − fL(y) i∈I i∈I

X X = (gi(x) − gi(y))f(ϕi(x)) + gi(y)(f(ϕi(x)) − f(ϕi(y))) i∈I i∈I

24 X X ≤ |gi(x) − gi(y)| |f(ϕi(x))| + |gi(y)| |f(ϕi(x)) − f(ϕi(y))| i∈I i∈I X ε X < kfk∞ |gi(x) − gi(y)| + kgik∞ 2kgk1 i∈I i∈I ε ε ≤ kfk∞ + kgk1 2kfk∞ 2kgk1 = ε.

Therefore, we have shown that LΦ (g)(f) is continuous, and hence LΦ (g): C(X, R) →

C(X, R). Now, the fact that LΦ (g) is linear is straightforward, and

X kLΦ (g)(f) k∞ ≤ kgik∞kf ◦ ϕik∞ i∈I X ≤ kgik∞kfk∞ i∈I

= kgk1kfk∞.

Since LΦ (g) is a bounded linear operator, it must be continuous. We will now show (2) and (3). Assume g is a vector and f is Lipschitz. Then,

X X |fL(x) − fL(y)| = kgik∞f(ϕi(x)) − kgik∞f(ϕi(y)) i∈I i∈I X ≤ kgik∞|f(ϕi(x)) − f(ϕi(y))| i∈I X ≤ kgik∞Lip (f) ρ(ϕi(x), ϕi(y)) i∈I X ≤ kgik∞Lip (f) Lip (ϕi) ρ(x, y) i∈I

≤ kgk1Lip (f) sup Lip (ϕi) ρ(x, y). i∈I

This finishes the proof of (2). In the case that Lip (f) ≤ 1 and kgk1 = 1, it follows immedi- ately that LΦ (g)(f) belongs to Lip1 (X, R) since we always have that supi∈I Lip (ϕi) < 1. 

By the Riesz Representation theorem, we identify the dual space C(X, R)∗ with ∗ ∗ ∗ M(X). Thus, we think of the dual operator LΦ (g): C(X, R) → C(X, R) as acting

25 on the space of finite Borel measures M(X).

Definition 2.3.3. Suppose (X, ρ) is a compact metric space and Φ ∈ IFS (X,I). For each

g ∈ W (X,I) and µ ∈ M1(X), define the normalized dual transfer operator by L∗ (g)(µ)(f) L∗ (g)(µ)(f) = Φ bΦ ∗ 1 LΦ (g)(µ)( ) for all f ∈ C(X, R). Note that the constant 1 function 1 : X → R is defined by 1(x) = 1 for all x ∈ X.

Because of assumption (3) in Definition 2.3.1, we have that for any µ ∈ M1(X), Z ∗ X 1 LΦ (g)(µ)(X) = gi · ( ◦ ϕi)dµ i∈I X ≥ µ(X) inf gi(x) x∈X i∈I

> 0,

1 ∗ where is the constant 1 function. Therefore, LbΦ (g): M1(X) → M1(X) is well-defined. ∗ We will be interested in probability measures invariant under LΦ (g).

Theorem 2.3.3. Suppose (X, ρ) is a compact metric space and I is a countable alphabet. Then,

(1) for all g ∈ W (X,I) and f ∈ Lip (X, R), the map (IFS (X,I) ,D∞) → (C(X, R), k ·

k∞) given by Φ 7→ LΦ (g)(f) is Lipschitz continuous with Lipschitz constant kgk1Lip (f),

(2) for all g ∈ W (X,I) and f ∈ C(X, R), the map (IFS (X,I) ,D∞) → (C(X, R), k·k∞)

given by Φ 7→ LΦ (g)(f) is uniformly continuous,

(3) for all Φ ∈ IFS (X,I) and f ∈ C(X, R), the map (W (X,I), k·k1) → (C(X, R), k·k∞)

given by g 7→ LΦ (g)(f) is Lipschitz continuous with Lipschitz constant kfk∞.

Proof. First we prove (1). Let Φ, Ψ ∈ IFS (X,I), g ∈ W (X,I), and f ∈ Lip (X, R). Notice,

X X kL (g)(f) − L (g)(f)k = g · (f ◦ ϕ ) − g · (f ◦ ψ ) Φ Ψ ∞ i i i i i∈I i∈I ∞

26

X = gi · (f ◦ ϕi − f ◦ ψi) i∈I ∞ X ≤ kgik∞ k(f ◦ ϕi − f ◦ ψi)k∞ i∈I X ≤ kgik∞ Lip (f) ρ∞(ϕi, ψi) i∈I X ≤ kgik∞ Lip (f) D∞(Φ, Ψ) i∈I

= kgk1 Lip (f) D∞(Φ, Ψ).

We now prove (2). Let g ∈ W (X,I), and f ∈ C(X, R). Let ε > 0. Since X is compact, f must be uniformly continuous by the Heine-Cantor theorem. Then, there is some δ > 0 such

that for all a, b ∈ X, if ρ(a, b) < δ, then |f(a) − f(b)| < ε/kgk1. Suppose Φ, Ψ ∈ IFS (X,I)

with D∞(Φ, Ψ) < δ. Then, for all x ∈ X and i ∈ I we have

ρ(ϕi(x), ψi(x)) < δ,

and hence, ε |f(ϕi(x)) − f(ψi(x))| < . kgk1 Therefore, for all i ∈ I ε kf ◦ ϕi − f ◦ ψik∞ ≤ . kgk1 Notice,

X X kL (g)(f) − L (g)(f)k = g · (f ◦ ϕ ) − g · (f ◦ ψ ) Φ Ψ ∞ i i i i i∈I i∈I ∞

X = gi · (f ◦ ϕi − f ◦ ψi) i∈I ∞ X ≤ kgik∞ k(f ◦ ϕi − f ◦ ψi)k∞ i∈I X ε ≤ kgik∞ kgk1 i∈I ε = kgk1 kgk1

27 = ε.

Finally, we show (3). Let Φ ∈ IFS (X,I) and f ∈ C(X, R). Now, for all g, h ∈ W (X,I),

X X kLΦ (g)(f) − LΦ (h)(f) k∞ = gi · (f ◦ ϕi) − hi · (f ◦ ϕi) i∈I i∈I ∞

X = (gi − hi)(f ◦ ϕi) i∈I ∞ X ≤ kgi − hik∞kf ◦ ϕik∞ i∈I X ≤ kgi − hik∞kfk∞ i∈I

= kfk∞kg − hk1.



Corollary 2.3.4. Suppose (X, ρ) is a compact metric space, I is a countable alphabet, and f ∈ C(X, R). Then, the map (IFS (X,I) ,D∞) × (W (X,I), k · k1) → (C(X, R), k · k∞) given

by (Φ, g) 7→ LΦ (g)(f) is continuous.

Proof. Let g ∈ W (X,I) and ε > 0. By Theorem 2.3.3 (2) and (3), for all Φ, Ψ ∈ IFS (X,I),

there is some δ1 > 0 so that if D∞(Φ, Ψ) < δ, then ε kL (g)(f) − L (g)(f) k < . Φ Ψ ∞ 2

If kfk∞ = 0, let δ = δ1. Otherwise, let δ = min{δ1, ε/(2kfk∞)}. Then, for all Φ, Ψ ∈

IFS (X,I) and h ∈ W (X,I) with D∞(Φ, Ψ) < δ and kg − hk1 < δ, we have

kLΦ (g)(f) − LΨ (h)(f) k∞ ≤ kLΦ (g)(f) − LΨ (g)(f) k∞ + kLΨ (g)(f) − LΨ (h)(f) k∞

≤ kLΦ (g)(f) − LΨ (g)(f) k∞ + kfk∞kg − hk1 ε ε < + 2 2 = ε.



28 Theorem 2.3.5. Suppose (X, ρ) is a compact metric space, I is a countable alphabet, Φ ∈ IFS (X,I), and g ∈ W (X,I). Then,

∗ (1) the function LbΦ (g): M1(X) → M1(X) is continuous in the weak-topology, ∗ (2) if g is a vector, then LbΦ (g):(M1(X),L) → (M1(X),L) is a contraction with

Lipschitz constant supi∈I Lip (ϕi).

∞ Proof. (1) Suppose {µn}n=1 is a sequence of Borel probability measures on X converging

weakly to µ. Let f ∈ C(X, R). By Theorem 2.3.2 (1), LΦ (g)(f) ∈ C(X, R). Then, by definition of weak convergence,

∗ lim LΦ (g)(µn)(f) = lim µn (LΦ (g)(f)) n→∞ n→∞

= µ (LΦ (g)(f))

∗ = LΦ (g)(µ)(f) .

In particular when we consider the constant 1 function 1, we have that

∗ 1 ∗ 1 lim LΦ (g)(µn)( ) = LΦ (g)(µ)( ) . n→∞

Therefore,

∗ ∗ LΦ (g)(µn)(f) lim LbΦ (g)(µn)(f) = lim n→∞ n→∞ ∗ 1 LΦ (g)(µn)( ) L∗ (g)(µ)(f) = Φ ∗ 1 LΦ (g)(µ)( )

∗ = LbΦ (g)(µ)(f) .

∗ ∞ ∗ Since f was arbitrary, we conclude that {LbΦ (g)(µn)}n=1 converges weakly to LbΦ (g)(µ). This finishes the proof of (1).

(2) Now suppose that g is a vector. Let µ, ν ∈ M1(X). First, notice that Z ∗ 1 X 1 LΦ (g)(µ)( ) = gi( ◦ ϕi)dµ = kgk1µ(X) = kgk1. i∈I Similarly, ∗ 1 LΦ (g)(ν)( ) = kgk1.

29 Let f : X → R be Lipschitz with Lip (f) ≤ 1. Notice that for any i ∈ I,

−1
−1 Lip (Lip (ϕi)) f ◦ ϕi ≤ (Lip (ϕi)) Lip (f) Lip (ϕi) ≤ 1.

Now,

L∗ (g)(µ)(f) L∗ (g)(ν)(f) L∗ (g)(µ)(f) − L∗ (g)(ν)(f) = Φ − Φ bΦ bΦ ∗ 1 ∗ 1 LΦ (g)(µ)( ) LΦ (g)(ν)( )

−1 = kgk1 (µ (LΦ (g)(f)) − ν (LΦ (g)(f))) Z Z ! −1 X X = kgk1 gi · (f ◦ ϕi)dµ − gi · (f ◦ ϕi)dν i∈I i∈I Z Z  −1 X = kgk1 kgik∞ (f ◦ ϕi)dµ − (f ◦ ϕi)dν i∈I

−1 X −1 −1
≤ kgk1 kgik∞Lip (ϕi) µ((Lip (ϕi)) f ◦ ϕi) − ν((Lip (ϕi)) f ◦ ϕi) i∈I

−1 X ≤ kgk1 kgik∞Lip (ϕi) L(µ, ν) i∈I

−1 = kgk1 kgk1 sup Lip (ϕi) L(µ, ν) i∈I

= sup Lip (ϕi) L(µ, ν). i∈I

Therefore, by definition of the Hutchinson metric,

∗ ∗ L(LbΦ (g)(µ) , LbΦ (g)(ν)) ≤ sup Lip (ϕi) L(µ, ν). i∈I



Hutchinson [24] (Section 4.4) showed the following corollary in the case that I is finite, and the case for infinite I follows from the work of Mauldin and Urba’nski [36]. When

∞ g is a vector, we can define the Bernoulli measure τg on I by defining for all finite words ω

kgω1 ··· gω|ω| k∞ τg(Cω) = , kgk1 where Cω is the set of all words with initial segment ω.

30 Corollary 2.3.6. Suppose (X, ρ) is a compact metric space, I is a countable alphabet, Φ ∈ IFS (X,I), and g ∈ W (X,I) is a vector. Then, there is a unique Borel probability measure m such that

∗ (1) LbΦ (g)(m) = m, ∗ n ∞ (2) {(LbΦ (g)) (µ)}n=1 converges weakly to m for all µ ∈ M1(X), −1 ∞ (3) m = τg ◦ πΦ , where τg is the Bernoulli measure on I ,

(4) m(JΦ) = 1.

Proof. Since (M1(X),L) is a compact metric space, it is also complete. Therefore (1) and (2) follow immediately from Theorem 2.3.5 (2) and the Banach contraction principle. For

∞ ∞ each i ∈ I, define σi : I → I by

σi(ω) = iω for all ω ∈ I∞. Notice that

πΦ ◦ σi = ϕi ◦ πΦ

for all i ∈ I.

Recall that for each i ∈ I the cylinder Ci is defined by

∞ Ci = {iω : ω ∈ I }.

∞ S Moreover, Ci ∩ Cj = ∅ for all i 6= j and I = i∈I Ci. Notice,

−1 ∞ −1 τg(σi (Ci)) = τg(I ) = kgik∞ τg(Ci)

for all i ∈ I.

Suppose f ∈ C(X, R). Then, Z ∗ −1
X −1 LbΦ (g) τg ◦ πΦ (f) = kgik∞f ◦ ϕidτg ◦ πΦ X i∈I Z X = kgik∞f ◦ ϕi ◦ πΦdτg ∞ I i∈I Z X = kgik∞f ◦ πΦ ◦ σidτg ∞ I i∈I

31 X Z = kgik∞f ◦ πΦ ◦ σidτg ∞ i∈I I Z X −1 = kgik∞f ◦ πΦdτg ◦ σi i∈I Ci X Z = f ◦ πΦdτg i∈I Ci Z = f ◦ πΦdτg I∞ Z −1 = fdτg ◦ πΦ X

−1 = (τg ◦ πΦ )(f).

Therefore,

∗ −1
−1 LbΦ (g) τg ◦ πΦ = τg ◦ πΦ .

−1 By uniqueness of (1), we conclude that m = τg ◦ πΦ . Finally,

−1 ∞ m(JΦ) = τg(πΦ (JΦ)) = τg(I ) = 1.



Lemma 2.3.7. Suppose (X, ρ) is a compact metric space and I is a countable alphabet. Let Φ ∈ IFS (X,I), g ∈ W (X,I), and µ ∈ M(X). Then,

(1) for all Ψ ∈ IFS (X,I) and f ∈ Lip (X, R),

∗ ∗ |LΦ (g)(µ)(f) − LΨ (g)(µ)(f)| ≤ µ(X)kgk1Lip (f) D∞(Φ, Ψ),

(2) for all h ∈ W (X,I) and f ∈ C(X, R),

∗ ∗ |LΦ (g)(µ)(f) − LΦ (h)(µ)(f)| ≤ µ(X)kfk∞kg − hk1.

Proof. (1) Let Ψ ∈ IFS (X,I) and f ∈ Lip (X, R). Using Theorem 2.3.3 (1), we obtain

∗ ∗ |LΦ (g)(µ)(f) − LΨ (g)(µ)(f)| = |µ (LΦ (g)(f)) − µ (LΨ (g)(f))| Z

= LΦ (g)(f) − LΨ (g)(f) dµ

32 Z ≤ kLΦ (g)(f) − LΨ (g)(f) k∞dµ

= µ(X)kLΦ (g)(f) − LΨ (g)(f) k∞

≤ µ(X)kgk1Lip (f) D∞(Φ, Ψ).

(2) Let h ∈ W (X,I) and f ∈ C(X, R). Using Theorem 2.3.3 (3), we obtain

∗ ∗ |LΦ (g)(µ)(f) − LΦ (h)(µ)(f)| = |µ (LΦ (g)(f)) − µ (LΦ (h)(f))| Z

= LΦ (g)(f) − LΦ (h)(f) dµ

Z ≤ kLΦ (g)(f) − LΦ (h)(f) k∞dµ

= µ(X)kLΦ (g)(f) − LΦ (h)(f) k∞

≤ µ(X)kfk∞kg − hk1.



Theorem 2.3.8. Suppose (X, ρ) is a compact metric space and I is a countable alphabet.

Then, the map (IFS (X,I) ,D∞) × (W (X,I), k · k1) × (M1(X),L) → (M1(X),L) given by

∗ (Φ, g, µ) 7→ LbΦ (g)(µ) is continuous.

∞ Proof. Let {Φn}n=1 be a sequence in IFS (X,I) converging to Φ ∈ IFS (X,I) in the D∞ ∞ metric, {gn}n=1 be a sequence in W (X,I) converging to g ∈ W (X,I) in the k · k1 norm, and ∞ {µn}n=1 be a sequence in M1(X) converging weakly to µ ∈ M1(X).

Let f ∈ Lip (X, R). By definition of weak convergence,

lim µn(f) = µ(f) n→∞

Notice, Z ! ∗ 1 X 1 X LΦ (g)(µ)( ) = gi · ( ◦ ϕi)dµ = µ gi i∈I i∈I and for all n ≥ 1 Z ! ∗ 1 X 1 X LΦ (g)(µn)( ) = gi · ( ◦ ϕi)dµn = µn gi . i∈I i∈I

33 P It follows from Proposition 2.3.1 that i∈I gi ∈ C(X, R). So, by definition of weak conver- gence, ! ! X X lim µn gi = µ gi . n→∞ i∈I i∈I By Theorem 2.3.5 (1),

∗ ∗ lim LbΦ (g)(µn)(f) = LbΦ (g)(µ)(f) . n→∞ Therefore, ! ∗ X ∗ lim LΦ (g)(µn)(f) = lim µn gi LbΦ (g)(µn)(f) n→∞ n→∞ i∈I ! X ∗ = µ gi LbΦ (g)(µ)(f) i∈I

∗ = LΦ (g)(µ)(f) .

By Lemma 2.3.7, for all n ≥ 1 we have

∗ ∗ LΦn (gn)(µn)(f) − LΦ (gn)(µn)(f) ≤ kgnk1Lip (f) D∞(Φn, Φ) and

∗ ∗ |LΦ (gn)(µn)(f) − LΦ (g)(µn)(f)| ≤ kfk∞kgn − gk1.

Therefore, by the triangle inequality

∗ ∗ LΦn (gn)(µn)(f) − LΦ (g)(µn)(f) ≤ kgnk1Lip (f) D∞(Φn, Φ) + kfk∞kgn − gk1.

Hence,

∗ ∗ lim LΦ (gn)(µn)(f) − LΦ (g)(µn)(f) = 0. n→∞ n Using the above observations, we conclude that

∗ ∗ ∗ ∗ lim LΦ (gn)(µn)(f) − LΦ (g)(µn)(f) + |LΦ (g)(µn)(f) − LΦ (g)(µ)(f)| = 0, n→∞ n and so by the triangle inequality

∗ ∗ lim LΦ (gn)(µn)(f) − LΦ (g)(µ)(f) = 0. n→∞ n

34 In particular, ∗ 1 ∗ 1 lim LΦ (gn)(µn)( ) − LΦ (g)(µ)( ) = 0. n→∞ n Therefore,

∗ ∗ LΦn (gn)(µn)(f) lim LbΦ (gn)(µn)(f) = lim n→∞ n n→∞ L∗ (g )(µ )(1) Φn n n L∗ (g)(µ)(f) = Φ ∗ 1 LΦ (g)(µ)( )

∗ = LbΦ (g)(µ)(f) .

By definition of the Hutchinson metric,

 ∗ ∗  lim L LbΦ (gn)(µn) , LbΦ (g)(µ) = 0. n→∞ n



Corollary 2.3.9. Suppose (X, ρ) is a compact metric space and I is a countable alphabet.

∞ Let {Φn}n=1 be a sequence in IFS (X,I) converging to Φ ∈ IFS (X,I) in the D∞ metric, and ∞ let {gn}n=1 be a sequence in W (X,I) converging to g ∈ W (X,I) in the k · k1 norm. Suppose ∞ {mn}n=1 is a sequence in M1(X) converging weakly to m ∈ M1(X) such that

∗ LΦn (gn)(mn) = mn for all n ≥ 1. Then,

∗ LΦ (g)(m) = m.

∗ Proof. By Theorem 2.3.8 and the assumption that LΦn (gn)(mn) = mn for all n ≥ 1, mn ∗ converges weakly to LΦ (g)(m). Since mn also converges weakly to m, we conclude that

∗ LΦ (g)(m) = m.



Corollary 2.3.10. Suppose (X, ρ) is a compact metric space and I is a countable alphabet.

∞ Let {Φn}n=1 be a sequence in IFS (X,I) converging to Φ ∈ IFS (X,I) in the D∞ metric, ∞ and let {gn}n=1 be a sequence of probability vectors in W (X,I) converging to the probability

35 ∞ vector g ∈ W (X,I) in the k · k1 norm. Suppose {mn}n=1 is the sequence of unique invariant ∞ probability measures corresponding to {(Φn, gn)}n=1 given by Corollary 2.3.6. Then, there is ∞ a subsequence {mnk }k=1 that converges weakly to the unique invariant measure corresponding to (Φ, g).

Proof. For all n ≥ 1, we have that Z ∗ 1 X 1 LΦn (gn)(mn)( ) = gn,i( ◦ ϕn,i)dmn = kgnk1mn(X) = 1, i∈I and hence

∗ LΦn (gn)(mn) = mn.

∞ Since M1(X) is compact, there is a subsequence {mnk }k=1 which converges to some m ∈

M1(X). By Corollary 2.3.9,

∗ LΦ (g)(m) = m.

By Corollary 2.3.6, m is the unique invariant measure corresponding to (Φ, g). 

2.4. The Pressure Function and Continuity of Hausdorff/Packing Dimension

We now turn our attention to conformal iterated function systems. The goal of the section is to establish continuity of the Hausdorff and packing dimensions for finite conformal iterated function systems. These results have already been shown by Roy and Urba´nski[45].

Definition 2.4.1. Let I be a finite alphabet, and let X ⊆ Rd be a nonempty compact set. We equip CIFS (X,I) with the metric D defined by

0 0 D(Φ, Ψ) = max{max kϕi − ψik∞, max kϕi − ψik∞}. i∈I i∈I for all Φ, Ψ ∈ CIFS (X,I).

Note that in the notation of the previous section,

D∞(Φ, Ψ) = max kϕi − ψik∞. i∈I

∞ Recall that πΦ : I → X denotes the coding map associated to the IFS Φ. When we

∞ topologize I with the product topology, the coding map is continuous. Therefore, πΦ

36 belongs to the space of continuous functions from I∞ to X, which we endow with the

supremum norm. The next lemma establishes continuity of the map Φ 7→ πΦ.

Lemma 2.4.1. Let I be a finite alphabet, and let X ⊆ Rd be a nonempty compact set. Then, ∞ the map (CIFS (X,I) ,D) → (C(I ,X), k · k∞) given by Φ 7→ πΦ is continuous. Moreover, for all Φ, Ψ ∈ CIFS (X,I),

 1 1  kπΦ − πΨk∞ ≤ min 0 , 0 D∞(Φ, Ψ). 1 − supi∈I kϕik∞ 1 − supi∈I kψik∞

Proof. Without loss of generality we will assume that I = {1,...,N}. We will show by induction that for every x ∈ X, n ≥ 1, ω ∈ In, and Φ, Ψ ∈ CIFS (X,I)

n−1 X 0 k (2) |ψω(x) − ϕω(x)| ≤ D∞(Φ, Ψ) (sup kϕik∞) . i∈I k=0 The base case n = 1 is obvious. Now suppose that2 holds for some n ≥ 1. Let x ∈ X and ω ∈ In+1. By our inductive hypothesis,

|ψω(x) − ϕω(x)| ≤ |ψω1 (ψσ(ω)(x)) − ϕω1 (ψσ(ω)(x))| + |ϕω1 (ψσ(ω)(x)) − ϕω1 (ϕσ(ω)(x))|

0 ≤ D∞(Φ, Ψ) + kϕω1 k∞|ψσ(ω)(x) − ϕσ(ω)(x)|

0 ≤ D∞(Φ, Ψ) + (sup kϕik∞)|ψσ(ω)(x) − ϕσ(ω)(x)| i∈I n−1 0 X 0 k ≤ D∞(Φ, Ψ) + (sup kϕik∞)D∞(Φ, Ψ) (sup kϕik∞) i∈I i∈I k=0 n X 0 k = D∞(Φ, Ψ) (sup kϕik∞) . i∈I k=0

∞ Since for every τ ∈ I , πΨ(τ) = limn→∞ ψτ|n (x) and πΦ(τ) = limn→∞ ϕτ|n (x), letting n → ∞ we obtain

∞ X 0 k |πΦ(τ) − πΨ(τ)| ≤ D∞(Φ, Ψ) (sup kϕik∞) i∈I k=0 1 = 0 D∞(Φ, Ψ) 1 − supi∈I kϕik∞ 1 ≤ 0 D(Φ, Ψ). 1 − supi∈I kϕik∞

37 The same arguments hold with Φ and Ψ interchanged. 

Recall that the pressure function PΦ : [0, ∞) → R is given by

1 X 0 t PΦ(t) = lim log kϕωk∞ n→∞ n ω∈In

∞ for all t ≥ 0. Define the function ζΦ : I → R by

0 ζΦ(ω) = log |ϕω(πΦ(σ(ω)))| for all ω ∈ I∞. Then then classical topological pressure of the function tζ (in the sense of [38]) is given by n−1 ! 1 X X P (tζ) = lim log exp sup tζ(τ) , n→∞ n τ∈C ω∈In ω j=0

where Cω is the cylinder of all infinite words with initial block ω. Note that P (tζ) = PΦ(t) since

n−1 ! 1 X X P (tζΦ) = lim log exp sup tζΦ(τ) n→∞ n τ∈C ω∈In ω j=0

n−1 ! 1 X X 0 = lim log exp sup t log |ϕω(πΦ(σ(τ)))| n→∞ n τ∈C ω∈In ω j=0

1 X 0 t = lim log sup |ϕω(πΦ(σ(τ)))| n→∞ n τ∈C ω∈In ω

1 X 0 t = lim log kϕωk∞ n→∞ n ω∈In

= PΦ(t).

Lemma 2.4.2. Let I be a finite alphabet, and let X ⊆ Rd be a nonempty compact set. Then,

for every t ≥ 0, the map (CIFS (X,I) ,D) → R given by Φ 7→ P (tζΦ) is continuous.

0 0 Proof. Let Φ ∈ CIFS (X,I), γ ∈ (0, infi∈I kϕik∞), and δ > 0 such that infi∈I kϕik∞ ≥ γ for all Ψ ∈ B(Φ, δ). Then, for every i ∈ I and x ∈ X, we have that

0 0 −1 0 0 −1 |log |ϕi(x)| − log |ψi(x)|| ≤ γ |ϕi(x) − ψi(x)| ≤ γ D(Φ, Ψ).

38 Thus, for all ω ∈ I∞, we get that

0 0 |t log |ψω1 (πΨ(σ(ω)))| − t log |ϕω1 (πΦ(σ(ω)))|

0 0 = |t|| log |ψω1 (πΨ(σ(ω)))| − log |ϕω1 (πΦ(σ(ω)))|

0 0 ≤ |t|(| log |ψω1 (πΨ(σ(ω)))| − log |ϕω1 (πΨ(σ(ω)))|

0 0 + | log |ϕω1 (πΨ(σ(ω)))| − log |ϕω1 (πΦ(σ(ω)))|)

−1 −1 ≤ |t|(γ D(Φ, Ψ) + γ |πΨ(σ(ω)) − πΦ(σ(ω))|)

−1 0 −1 ≤ γ |t|(1 + (1 − (sup kϕik∞) )D(Φ, Ψ). i∈I

Therefore,

−1 0 −1 ktζΦ − tζΨk∞ ≤ γ |t|(1 + (1 − (sup kϕik∞) )D(Φ, Ψ). i∈I

Since the pressure function is continuous, the proof is complete. 

In the context of finite alphabet CIFS, it was shown by Mauldin and Urba´nski(Corol-

lary 4.18 [36]) that HD (JΦ) = PD (JΦ). Therefore, the next theorem establishes continuity

of the Hausdorff and packing dimension function. We will write hΦ = HD (JΦ) = PD (JΦ) to simplify notation.

Theorem 2.4.3. Let I be a finite alphabet, and let X ⊆ Rd be a nonempty compact set. The Hausdorff (and packing) dimension function h : (CIFS (X,I) ,D) → (0, d] given by

h(Φ) = hΦ for any Φ ∈ CIFS (X,I) is continuous.

Proof. Let Φ ∈ CIFS (X,I) and ε > 0. By Lemma 2.4.2, there is some δ > 0 such that if Ψ ∈ B(Φ, δ), then

1 |P ((h − ε)ζ ) − P ((h − ε)ζ )| ≤ min {P ((h − ε)ζ ), −P ((h + ε)ζ )} Φ Ψ Φ Φ 2 Φ Φ Φ Φ and

1 |P ((h + ε)ζ ) − P ((h + ε)ζ )| ≤ min {P ((h − ε)ζ ), −P ((h + ε)ζ )} . Φ Ψ Φ Φ 2 Φ Φ Φ Φ

39 Therefore,

1 P ((h − ε)ζ ) ≥ P ((h − ε)ζ ) − P ((h − ε)ζ ) Φ Ψ Φ Φ 2 Φ Φ 1 = P ((h − ε)ζ ) 2 Φ Φ > 0.

Hence, hΦ − ε < hΨ. Similarly,

1 P ((h + ε)ζ ) ≤ P ((h + ε)ζ ) − P ((h + ε)ζ ) Φ Ψ Φ Φ 2 Φ Φ 1 = P ((h + ε)ζ ) 2 Φ Φ < 0.

Hence, hΨ < hΦ + ε. Therefore, |hΨ − hΦ| < ε. 

It was established by Mauldin and Urba´nski[36] that for every Φ ∈ CIFS (X,I) where

I is a finite alphabet, there exists a unique conformal measure mΦ such that

∗ 0 hΦ
LΦ {|ϕi| }i∈I (mΦ) = mΦ.

hΦ hΦ Moreover, 0 < H (JΦ) < ∞, 0 < P (JΦ) < ∞, and

hΦ hΦ H (A ∩ JΦ) P (A ∩ JΦ) mΦ(A) = h = h H Φ (JΦ) P Φ (JΦ)

for all Borel sets A ⊆ Rd. Combining Theorem 2.4.3 and Corollary 2.3.9, we obtain the following.

Corollary 2.4.4. Let I be a finite alphabet, and let X ⊆ Rd be a nonempty compact set. ∞ Let {Φn}n=1 be a sequence in CIFS (X,I) converging to Φ ∈ CIFS (X,I) in the D metric. ∞ ∞ Suppose {mΦn }n=1 is the sequence of unique conformal measures corresponding to {Φn}n=1. Then, there is a subsequence {m }∞ that converges weakly to the unique conformal mea- Φnk k=1

sure mΦ corresponding to Φ.

40 Proof. Since M (X) is compact, there is a subsequence {m }∞ that converges weakly 1 Φnk k=1

to some m ∈ M1(X). By Corollary 2.3.9,

∗ 0 hΦ
LΦ {|ϕi| }i∈I (m) = m.

By uniqueness of mΦ, mΦ = m. 

41 CHAPTER 3

CONTINUITY OF THE NUMERICAL VALUE OF HAUSDORFF AND PACKING MEASURES FOR FINITE ITERATED FUNCTION SYSTEMS

Suppose X ⊆ Rd is nonempty and compact and I is a finite alphabet. In Chapter 2, we saw that if Φ ∈ CIFS (X,I), then the following maps are continuous:

(1) (CIFS (X,I) ,D) → (C(X),H) given by Φ 7→ JΦ

(2) (CIFS (X,I) ,D) → [0, d] given by Φ 7→ hΦ

(3) (CIFS (X,I) ,D) → (M1(X),L) given by Φ 7→ mΦ.

It is therefore natural to ask whether or not the maps (CIFS (X,I) ,D) → (0, ∞)

hΦ hΦ given by Φ 7→ H (JΦ) and Φ 7→ P (JΦ) are also continuous.

Even if we restrict to IFS consisting of similarities, Ayer and Strichartz [4] showed by

hΦ way of counterexample when X = [0, 1] that Φ 7→ H (JΦ) is not continuous if the domain is all IFSs satisfying the OSC . Similarly, Feng’s formulas [19] provide a counterexample when

hΦ X = [0, 1] (discussed in Chapter 4) that Φ 7→ P (JΦ) is not continuous if the domain is all IFSs satisfying the OSC.

In the case of finite IFSs consisting of similarities, Olsen [43] showed that if we restrict the domain to IFSs satisfying the strong separation condition (SSC), then the map

hΦ Φ 7→ H (JΦ) is continuous. Qiu [44] later showed the analogous result for packing measure. We collect these two results in the following theorem.

Theorem 3.0.5. Let X ⊆ Rd be a nonempty compact set, and let I be a finite alphabet. Let

SIFSSSC(X,I) be the subspace of (CIFS (X,I) ,D) where each IFS consists of similarities and satisfies the strong separation condition. Then, the maps (SIFSSSC(X,I),D) → (0, ∞)

hΦ hΦ given by Φ 7→ H (JΦ) and Φ 7→ P (JΦ) are continuous.

Definition 3.0.2. Let X ⊆ Rd be a nonempty compact set, and let I be a countable alphabet. Define CIFSSSC (X,I) to be the subset of CIFS (X,I) where each IFS satisfies the

42 strong separation condition. That is, for every Φ ∈ CIFSSSC (X,I),

ϕi(JΦ) ∩ ϕj(JΦ) = ∅ for all i, j ∈ I and i 6= j.

Assuming the strong separation condition, Szarek, Urba´nski,and Zdunik [48] estab-

hΦ lished continuity of the map Φ 7→ H (JΦ) for finite conformal IFS. Urba´nski and Zdunik [53] established H¨oldercontinuity for parametrized families of conformal iterated function systems in Rd, d ≥ 3, linear iterated function systems in Rd, d ≥ 1, and analytic families of conformal expanding repellers in the complex plane C. The goal of this chapter is to establish continuity of the numerical value of packing measure for conformal iterated function systems satisfying the strong separation condition. We will do this by way of density theorems analogous to those in Olsen [42, 43] and Qiu [44]. Along the way, we will also establish continuity in the case of Hausdorff measure.

3.1. General Density Theorems

The main tool for calculating Hausdorff and packing measures is through the concept of density. These next results can be viewed as extensions of the Lebesgue density theorem, which says that for every Lebesgue measurable set A ⊆ Rd µ(A ∩ B(x, r)) lim = 1 r→0 µ(B(x, r)) for µ-a.e. x ∈ A, where µ is Lebesgue measure in Rd. The first theorem can be found as Theorem 2.3 in [16].

Theorem 3.1.1. Let s ≥ 0. If A ⊆ Rd and Hs (A) < ∞, then  s  H (A ∩ U) d lim sup : x ∈ U, 0 < |U| ≤ β, U ⊆ R = 1 β→0 |U|s for Hs-a.e. x ∈ A. The sup in Theorem 3.1.1 can be taken over all sets U which are closed and convex since |U| = |U| and every set is contained in a convex set of equal diameter. Then next theorem can be found as Theorem 6.10 in [35].

43 Theorem 3.1.2. Let s ≥ 0. If A ⊆ Rd and Ps (A) < ∞, then

Ps A ∩ B(x, r)
lim inf = 1 r→0 (2r)s for Ps-a.e. x ∈ A.

Suppose X ⊆ Rd is a nonempty compact set and I is a countable alphabet. If

Φ ∈ CIFS (X,I) is regular, then there exists a unique conformal measure mΦ associated to Φ. As discussed at the end of Chapter 2, if I is finite, then hΦ = HD (JΦ) = PD (JΦ)

hΦ hΦ and 0 < H (JΦ) ≤ P (JΦ) < ∞. When I is infinite, it is possible (see [36]) that

HD(JΦ) PD(JΦ) H (JΦ) = 0 and P (JΦ) = ∞. Because we are interested in the numerical values of these measures, we will always assume

hΦ hΦ 0 < H (JΦ) ≤ P (JΦ) < ∞,

and consequently

hΦ hΦ H (A ∩ JΦ) P (A ∩ JΦ) mΦ(A) = h = h H Φ (JΦ) P Φ (JΦ)

for every Borel set A ⊆ Rd. We will mostly be interested in the case when I is finite, but we can obtain a few precise density results even when I is infinite. We define the density of a

d set A ⊆ R with mΦ(A) > 0 to be

|A|hΦ denΦ (A) = . mΦ(A)

If mΦ(A) = 0, then we will define denΦ (A) = ∞. The following corollaries are immediate from Theorems 3.1.1 and 3.1.2.

Corollary 3.1.3. Let X ⊆ Rd be a nonempty compact set, and let I be a countable alphabet.

HD(JΦ) Suppose Φ ∈ CIFS (X,I) is regular with H (JΦ) < ∞. Then,

hΦ  d H (JΦ) = lim inf denΦ (U): x ∈ U, 0 < |U| ≤ β, U ⊆ R β→0

for mΦ-a.e. x ∈ JΦ.

44 Corollary 3.1.4. Let X ⊆ Rd be a nonempty compact set, and let I be a countable alphabet.

PD(JΦ) Suppose Φ ∈ CIFS (X,I) is regular with P (JΦ) < ∞. Then,

hΦ
P (JΦ) = lim sup denΦ B(x, r) r→0 for mΦ-a.e. x ∈ JΦ.

3.2. Density Theorems for Self-Similar Sets

The main density result for iterated function systems found in the literature is con- tained in the next theorem. It can be seen for example in [4, 33, 39, 42]. It appears as Corollary 7 and Theorem 10 in [39].

Theorem 3.2.1. Let X ⊆ Rd be a nonempty compact set, and let I be a finite alphabet. Suppose Φ ∈ CIFS (X,I) consists of similarities. Then,

hΦ (1) H (JΦ) = inf {denΦ (F ): F ⊆ X is closed},

hΦ 
(2) P (JΦ) = sup denΦ B(x, r) : x ∈ JΦ, r > 0, B(x, r) ⊆ O .

A proof that Theorem 3.2.1 can be extended for Hausdorff measure to an infinite alphabet was shown in [52] as Theorem 5.1. We modify their proof here to show the analogous result for packing measure.

Theorem 3.2.2. Let X ⊆ Rd be a nonempty compact set, and let I be a countable alphabet.

hΦ Suppose Φ ∈ CIFS (X,I) is regular with 0 < H (JΦ) < ∞, and Φ consists of similarities. Then,

hΦ H (JΦ) = inf {denΦ (F ): F ⊆ X is closed} .

Theorem 3.2.3 (♣). Let X ⊆ Rd be a nonempty compact set, and let I be a countable

hΦ alphabet. Suppose Φ ∈ CIFS (X,I) is regular with 0 < P (JΦ) < ∞, and Φ consists of similarities. Then,

hΦ 
P (JΦ) = sup denΦ B(x, r) : x ∈ JΦ, r > 0, B(x, r) ⊆ O .

45 Proof. Define


D = sup denΦ B(x, r) : x ∈ JΦ, r > 0, B(x, r) ⊆ O .

Fix n ≥ 1. Let x0 ∈ J and r0 > 0 be such that B(x0, r0) ⊆ O and 1 den B(x , r )
≥ D − . Φ 0 0 n

n0 There is some n0 ≥ 1 and τ ∈ I such that 1 den B(y, r )
≥ D − Φ 0 n

for all y ∈ ϕτ (X). Let Fn = ϕτ (X). Define

j −1 Zn(ω) = {j ≥ 0 : σ (ω) ∈ πΦ (Fn)}.

∞ −1 Let me be the unique Borel probability measure on I given by Corollary 2.3.6 with me ◦πΦ = m. By Theorem 3.8 [36], there is a unique ergodic σ-invariant probability measure m∗

∗ ∞ equivalent to me . By Birkhoff’s Ergodic Theorem, for m -a.e. ω ∈ I 1 n Z X 1 j 1 ∗ ∗ −1 lim π−1(F ) ◦ σ (ω) = π−1(F )dm = m (πΦ (Fn)) > 0. n→∞ n Φ n Φ n j=1 Thus, for m∗-a.e. ω ∈ I∞, there are infinitely many j ≥ 1 with

j 1 −1 ◦ σ (ω) = 1. πΦ (Fn)

∗ Therefore m (Γ1/n) = 1 where

∞ Γ1/n = {ω ∈ I : Zn(ω) is infinite}.

If ω ∈ Γ1/n and j ∈ Zn(ω), then

j πΦ(ω) = ϕω|j (πΦ(σ (ω))) ∈ ϕω|j (Fn),

and hence,

0 hΦ (2|ϕ |r0)  0  ω|j denΦ B(πΦ(ω), |ϕ |r0) = ω|j m (B(π (ω), |ϕ0 |r )) Φ Φ ω|j 0 (2|ϕ0 |r )hΦ ω|j 0 = −1 |ϕ0 |hΦ m (B(ϕ (π (ω)), r )) ω|j Φ ω|j Φ 0

46 (2r )hΦ = 0 m (B(ϕ−1 (π (ω)), r )) Φ ω|j Φ 0   = den B(ϕ−1 (π (ω)), r )) Φ ω|j Φ 0 1 ≥ D − . n

Thus, for all ω ∈ Γ1/n,
1 lim sup denΦ B(πΦ(ω), r) ≥ D − . r→0 n T∞ ∗ Now, Γ = n=1 Γ1/n satisfies m (Γ) = 1 and for all ω ∈ Γ we have

lim sup denΦ B(πΦ(ω), r) ≥ D. r→0

Therefore, for all ω ∈ Γ,
lim sup denΦ B(πΦ(ω), r) = D. r→0

Now, m(πΦ(Γ)) ≥ me (Γ) = 1. By Corollary 3.1.4, there is some x ∈ πΦ(Γ) such that

hΦ
P (JΦ) = lim sup denΦ B(x, r) = D. r→0



Recall that the strong separation condition (SSC) on Φ means that for all i, j ∈ I,i 6= j,

ϕi(JΦ) ∩ ϕj(JΦ) = ∅.

Llorente and Mor´an(Theorem 3.3 [29]) showed that for finite alphabet IFSs consisting of similarities and satisfying the SSC, the infimum (resp. supremum) in Theorem 3.2.2 (resp. 3.2.3) is obtained. This result is stated below.

Theorem 3.2.4. Let X ⊆ Rd be a nonempty compact set, and let I be a finite alphabet. Suppose Φ ∈ CIFS (X,I) consists of similarities and satisfies the strong separation condition.

0 Let `min = mini6=j inf {|x − y| : x ∈ ϕi(JΦ), y ∈ ϕj(JΦ)} and cmin = mini∈I |ϕi|. Then,

hΦ (1) H (JΦ) = min {denΦ (F ): F ⊆ X is closed with `min ≤ |F | ≤ |JΦ|},

hΦ 
(2) P (JΦ) = max denΦ B(x, r) : x ∈ JΦ, `mincmin/2 ≤ r ≤ |JΦ| .

47 3.3. Density Theorems for Self-Conformal Sets

In this section, when there is only a single IFS Φ and the context is clear, we will

write J = JΦ, h = hΦ, m = mΦ, and den = denΦ. When Φ is finite and satisfies the SSC, we define

`Φ,min = min inf {|x − y| : x ∈ ϕi(JΦ), y ∈ ϕj(JΦ)} , i6=j

and when the context is clear, we write `min = `Φ,min.

Definition 3.3.1. Suppose U ⊆ Rd is a nonempty open set. If f : U → Rd is differentiable,

then for any nonempty set A ⊆ U the distortion of f|A is defined by |f 0(y)| κ(f|A) = sup 0 . x,y∈A |f (x)|

For shorthand, we write κ(f) = κ(f|U ).

For each β > 0 and Φ ∈ CIFS (X,I), we will use the notation KΦ,β to mean

∗ KΦ,β = sup{κ(ϕω|B(x,r)): ω ∈ I , x ∈ J, 0 < r ≤ β, B(x, r) ⊆ O}.

We will also define

∗ KΦ,β = sup{κ(ϕω|F ): ω ∈ I , 0 < |F | ≤ 2β, F ⊆ X is closed}.

When the context is clear, we will suppress the Φ and simply write Kβ or Kβ. Notice,

1 ≤ Kβ ≤ Kβ ≤ K, where K is the distortion constant from the Bounded Distortion Property for Φ. When Φ consists of similarities,

Kβ = Kβ = K = 1.

When Φ is finite and d ≥ 2, it follows from Theorems 1.4.2 and 1.4.3 that

(3) lim Kβ = lim Kβ = 1. β→0 β→0

When d = 1, we must assume (3). Notice that Proposition 1.4.1 provides nice sufficient conditions for (3) to hold.

48 Proposition 3.3.1. Let X ⊆ Rd be a nonempty compact set, I be a countable alphabet, and Φ ∈ CIFS (X,I) be regular. For each finite word ω ∈ I∗ and Borel set F ⊆ V ,

0 h 0 h inf |ϕω(y)| m(F ) ≤ m(ϕω(F )) ≤ sup |ϕω(y)| m(F ). y∈F y∈F

Proof. Recall, since m is a conformal measure, Z 0 h m(ϕω(F )) = |ϕω| dm. F Then, Z 0 h 0 h m(ϕω(F )) ≥ inf |ϕω(y)| dm = inf |ϕω(y)| m(F ) y∈F F y∈F and Z 0 h 0 h m(ϕω(F )) ≤ sup |ϕω(y)| dm = sup |ϕω(y)| m(F ). y∈F F y∈F 

Proposition 3.3.2. Let X ⊆ Rd be a nonempty compact set, I be a countable alphabet, and Φ ∈ CIFS (X,I) be regular. For each finite word ω ∈ I∗ and convex set F ⊆ V ,

−h h κ (ϕω|F )den (F ) ≤ den (ϕω(F )) ≤ κ (ϕω|F )den (F ) .

Proof. By Proposition 3.3.1 and the Mean Value Inequality, we obtain

h |ϕω(F )| den(ϕω(F )) = m(ϕω(F )) 0 h (infy∈F |ϕω(y)||F |) ≥ 0 h supy∈F |ϕω(y)| m(F ) |F |h = κ−h(ϕ | ) ω F m(F )

−h = κ (ϕω|F )den (F ) .

Similarly,

h |ϕω(F )| den(ϕω(F )) = m(ϕω(F ))

49 0 h (supy∈F |ϕω(y)||F |) ≤ 0 h infy∈F |ϕω(y)| m(F ) |F |h = κh(ϕ | ) ω F m(F )

h = κ (ϕω|F )den (F ) .



Proposition 3.3.3. Let X ⊆ Rd be a nonempty compact set, I be a countable alphabet, and Φ ∈ CIFS (X,I) be regular. Fix k ≥ 1. Then, for all Borel sets A ⊆ O, ! [ −h m ϕω(A) ≥ K m(A). ω∈Ik Proof. Since Φ satisfies the OSC,

ϕω(A) ∩ ϕτ (A) = ∅ for all ω, τ ∈ Ik. Now, for all ω ∈ Ik, Z 0 m(ϕω(A)) = |ϕω|dm A Z ≥ inf |ϕ0(y)|h dm y∈A A ≥ inf |ϕ0(y)|hm(A) y∈X

−h 0 h = κ (ϕω) sup |ϕ (y)| m(A) y∈X Z −h 0 ≥ K m(A) |ϕω|dm X

−h = K m(A)m(ϕω(X))

Notice, ! [ X m ϕω(A) = m(ϕω(A)) ω∈Ik ω∈Ik

50 X −h ≥ K m(A)m(ϕω(X)) ω∈Ik ! −h [ ≥ K m(A)m ϕω(X) ω∈Ik

= K−hm(A)m(J)

= K−hm(A).



The next theorem is a generalization of Olsen’s Corollary 1.2 [42].

Theorem 3.3.4 (♣). Let X ⊆ Rd be a nonempty compact set, I be a countable alphabet, and Φ ∈ CIFS (X,I) be regular with 0 < Hh (J) < ∞. Let r > 0. Then, for any closed convex set F ⊆ V with 0 < |F | ≤ 2r,

h H (J ∩ F ) h ≤ K . |F |h r

Proof. Let F ⊆ X be a closed convex set such that 0 < |F | ≤ 2r. If m(F ) = 0, then the result is trivial, so we may assume m(F ) > 0. Suppose to the contrary

h h h H (J ∩ F ) > |F | Kr .

Then, there is some ε > 0 such that

h h h H (J ∩ F ) (1 − ε) > |F | Kr .

Recall that for any ω, τ ∈ Ik, k ≥ 1,

m(ϕω(F ) ∩ ϕτ (F )) = 0.

Notice, for any k ≥ 1, ! h [ X h H J ∩ ϕω(F ) = H (J ∩ ϕω(F )) ω∈Ik ω∈Ik

X 0 h h ≥ inf |ϕω(y)| H (J ∩ F ) y∈F ω∈Ik

51 X 0 h −1 h h > inf |ϕω(y)| (1 − ε) |F | Kr y∈F ω∈Ik

X −h 0 h −1 h h = κ (ϕω|F ) sup |ϕω(y)| (1 − ε) |F | Kr y∈F ω∈Ik

X −h 0 h −1 h h ≥ Kr sup |ϕω(y)| (1 − ε) |F | Kr y∈F ω∈Ik

−1 X 0 h h = (1 − ε) sup |ϕω(y)| |F | y∈F ω∈Ik

−1 X h ≥ (1 − ε) |ϕω(F )| . ω∈Ik S For every δ > 0, there is some kδ ≥ 1 such that {ϕω(F )}ω∈Ikδ is a δ-cover of J ∩ ω∈Ik ϕω(F ). Define   1 [ η = εHh J ∩ ϕ (F ) . δ 2  ω  ω∈Ikδ S For each δ > 0, let Uδ be a δ-cover of J − ω∈Ik ϕω(F ) such that ! X h h [ |U| ≤ Hδ J − ϕω(F ) + ηδ. U∈U ω∈Ik S Then, {ϕω(F )}ω∈Ikδ Uδ forms a δ-cover of J. Now,

h X h X h Hδ (J) ≤ |ϕω(F )| + |U| i∈Ikδ U∈U     h [ h [ ≤ (1 − ε)H J ∩ ϕω(F ) + Hδ J − ϕω(F ) + ηδ ω∈Ikδ ω∈Ikδ   1 [ = Hh (J) − εHh J ∩ ϕ (F ) . 2  ω  ω∈Ikδ By Proposition 3.3.3, we conclude that

ε Hh (J) ≤ Hh (J) − m(F ). δ 2Kh

Letting δ → 0, we have ε Hh (J) ≤ Hh (J) − m(F ), 2Kh which is a contradiction since m(F ) > 0. 

52 Corollary 3.3.5 (♣). Let X ⊆ Rd be a nonempty compact set, I be a countable alphabet, and Φ ∈ CIFS (X,I). If β > 0, then

−h h h Kβ H (J) ≤ inf den (F ) ≤ H (J) F ⊆X closed, convex F ∩J6=∅,|F |≤2β

Proof. The first inequality is immediate from the previous theorem. The second inequality

follows from Corollary 3.1.3. 

For each Φ ∈ CIFSSSC (X,I) with I finite, define

Vmin = min inf {|x − v| : x ∈ ϕi(X), v ∈ ϕi(∂V )} . i∈I

For each i ∈ I, define [ Ui = B(x, 2βΦ),

x∈ϕi(J) S where βΦ = min{`min/4,Vmin/4}. Let U = i∈I Ui. Notice, for all i 6= j, Ui ∩Uj = ∅. Clearly,

for each i ∈ I, ϕi(J) ⊂ Ui ⊆ ϕi(V ) and Φ satisfies the Open Set Condition with open set U.

Suppose F ⊆ X is a nonempty Borel set with F ∩J 6= ∅ and |F | < 2βΦ. Then, F ⊆ Uj

∗ for exactly one j ∈ I. Let ω ∈ I be the longest word (possibly empty) such that F ⊆ ϕω(U).

−1 −1 Then, ϕjω (F ) is not contained in Ui for any i ∈ I. In particular, |ϕjω (F )| ≥ 2βΦ.

Proposition 3.3.6. Let X ⊆ Rd be a nonempty compact set, I be a finite alphabet, and

Φ ∈ CIFSSSC (X,I). Fix M ≥ 1 with

1 0 ≤ min inf |ϕi(y)|. M i∈I y∈X

Let 0 < β < βΦ. Suppose F ⊆ X is a closed convex set with 0 < |F | < 2β/M and m(F ) > 0.

∗ −1 If ω ∈ I is the shortest word with |ϕω (F )| ≥ 2β/M, then

2β ≤ |ϕ−1(F )| ≤ 2β, M ω

and

−1
h den ϕω (F ) ≤ Kβden (F ) .

53 ∗ −1 Proof. Let ω ∈ I be the shortest word with |ϕω (F )| ≥ 2β/M. Let n = |ω|. Suppose to −1 the contrary |ϕω (F )| > 2β. Then,

−1 2β < |ϕω (F )|

= |ϕ−1(ϕ−1 (F ))| ωn ω|n−1

≤ sup |(ϕ−1)0(y)||ϕ−1 (F )| ωn ω|n−1 y∈ϕωn (X) 1 ≤ |ϕ−1 (F )| 0 ω|n−1 infy∈X |ϕωn (y)| 1 2β < 0 infy∈X |ϕωn (y)| M ≤ 2β, which is a contradiction. Now,

−1 h −1
|ϕω (F )| den ϕω (F ) = −1 m(ϕω (F )) 0 h sup −1 |ϕ (y)| h y∈ϕω (F ) ω |F | ≤ 0 h inf −1 |ϕ (y)| m(F ) y∈ϕω (F ) ω

h = κ (ϕ | −1 )den (F ) ω ϕω (F )

h ≤ Kβden (F ) .



Lemma 3.3.7 (♣). Let X ⊆ Rd be a nonempty compact set, I be a finite alphabet, and

Φ ∈ CIFSSSC (X,I). Fix M as in Proposition 3.3.6. Then for all 0 < β < βΦ,

−h h h h Kβ H (J) ≤ inf den (F ) ≤ KβH (J) F ⊆X closed 2β F ∩J6=∅, M ≤|F |≤2β

Proof. Let 0 < β < βΦ. We first show the second inequality. Let ε > 0. By Corollary 3.3.5, there is some closed convex F ⊂ X with m(F ) > 0 and 0 < |F | ≤ 2β such that

den(F ) ≤ Hh (J) + ε.

54 If |F | ≥ 2β/M, then we are done. Assume |F | < 2β/M. Let ω ∈ I∗ be the shortest word

−1 with |ϕω (F )| ≥ 2β/M. By Proposition 3.3.6,

2β ≤ |ϕ−1(F )| ≤ 2β M ω

and

−1
h h h den ϕω (F ) ≤ Kβden (F ) ≤ Kβ(H (J) + ε).

Therefore, the second inequality holds.

Now assume 2β/M ≤ |F | ≤ 2β is closed and m(F ) > 0. By the definition of βΦ, there is some i ∈ I such that F ⊆ Ui. In fact, F ⊆ B(x, 2βΦ) ⊆ Ui for any x ∈ J ∩ F . Let

C be the smallest convex set containing F so that |F | = |C| and C ⊆ B(x, 2βΦ) ⊆ V . By Theorem 3.3.4, h h H (J ∩ F ) H (J ∩ C) h ≤ ≤ K . |F |h |C|h β Thus,

−h h Kβ H (J) ≤ den (F ) .



Theorem 3.3.8 (♣). Let X ⊆ Rd be a nonempty compact set, I be a finite alphabet, and

Φ ∈ CIFSSSC (X,I). Fix M as in Proposition 3.3.6. Then,

Hh (J) = lim inf den (F ) . β→0 F ⊆X closed 2β F ∩J6=∅, M ≤|F |≤2β

Proof. Follows immediately from Lemma 3.3.7 since limβ→0 Kβ = 1. 

For each finite word ω ∈ I∗ and B(x, r) ⊆ V , we define

Rω(x, r) = inf{s > 0 : ϕω(B(x, r)) ⊂ B(ϕω(x), s)}

and

Rω(x, r) = sup{s > 0 : B(ϕω(x), s) ⊂ ϕω(B(x, r))}.

55 It follows from the Mean Value Inequality that

0 Rω(x, r) ≤ sup |ϕω(y)|r y∈B(x,r) and

0 Rω(x, r) ≥ inf |ϕω(y)|r y∈B(x,r) for every finite word ω ∈ I∗ and B(x, r) ⊆ V . We will discuss this argument here briefly. If B(x, r) ⊆ V , then by the Mean Value Inequality, for all z ∈ B(x, r),

0 0 |ϕω(x) − ϕω(z)| ≤ sup |ϕω(y)||x − z| ≤ sup |ϕω(y)|r. y∈B(x,r) y∈B(x,r)

Notice, ∂B(ϕω(x),Rω(x, r)) ∩ ∂ϕω(B(x, r)) 6= ∅. Now,

−1 −1 0 ϕω (B(ϕω(x),Rω(x, r))) ⊆ B(ϕω(x), sup |(ϕω ) (y)|Rω(x, r)) y∈B(ϕω(x),Rω(x,r))) Notice, 1 sup |(ϕ−1)0(y)| ≤ sup |(ϕ−1)0(y)| = . ω ω inf |ϕ0 (y)| y∈B(ϕω(x),Rω(x,r)) y∈ϕω(B(x,r) y∈B(x,r) ω Then,

Rω(x, r) 0 ≥ r, infy∈B(x,r) |ϕω(y)| and we are done.

Proposition 3.3.9. Let X ⊆ Rd be a nonempty compact set, I be a countable alphabet, and Φ ∈ CIFS (X,I). For each finite word ω ∈ I∗, x ∈ X, and r > 0 with B(x, r) ⊆ V ,

Rω(x, r) ≤ κ(ϕω|B(x,r))Rω(x, r).

Proof. Notice,

0 Rω(x, r) ≤ sup |ϕω(y)|r y∈B(x,r)

0 = κ(ϕω|B(x,r)) inf |ϕω(y)|r y∈B(x,r)

≤ κ(ϕω|B(x,r))Rω(x, r).



56 Proposition 3.3.10. Let X ⊆ Rd be a nonempty compact set, I be a countable alphabet, and Φ ∈ CIFS (X,I) be regular. For each finite word ω ∈ I∗, x ∈ J, and r > 0 with B(x, r) ⊆ O,

h (i) den(B(ϕω(x), Rω(x, r))) ≤ κ (ϕω|B(x,r))den(B(ϕω(x),Rω(x, r)))

h (ii) den(ϕω(B(x, r))) ≤ κ (ϕω|B(x,r))den(B(ϕω(x),Rω(x, r)))

2h (iii) den(B(x, r)) ≤ κ (ϕω|B(x,r))den(B(ϕω(x),Rω(x, r)))

h (iv) den(B(ϕω(x), Rω(x, r))) ≤ κ (ϕω|B(x,r))den(ϕω(B(x, r)))

2h (v) den(B(ϕω(x), Rω(x, r))) ≤ κ (ϕω|B(x,r))den(B(x, r))

Proof. To show (i), apply Proposition 3.3.9 to obtain

h (2Rω(x, r)) den(B(ϕω(x), Rω(x, r))) = m(B(ϕω(x), Rω(x, r))) h (2κ(ϕω| )R (x, r)) ≤ B(x,r) ω m(B(ϕω(x), Rω(x, r))) h h (2Rω(x, r)) ≤ κ (ϕω|B(x,r)) m(B(ϕω(x),Rω(x, r)))

h = κ (ϕω|B(x,r))den(B(ϕω(x),Rω(x, r))).

To show (ii), apply Proposition 3.3.9 and Proposition 3.3.1 to obtain

h |ϕω(B(x, r))| den(ϕω(B(x, r))) = m(ϕω(B(x, r))) (2R (x, r))h ≤ ω m(ϕω(B(x, r))) h (2κ(ϕω| )R (x, r)) ≤ B(x,r) ω m(ϕω(B(x, r))) h h (2Rω(x, r)) ≤ κ (ϕω|B(x,r)) m(ϕω(B(x, r))) h h (2Rω(x, r)) ≤ κ (ϕω|B(x,r)) m(B(ϕω(x),Rω(x, r)))

57 h = κ (ϕω|B(x,r))den(B(ϕω(x),Rω(x, r)))

To show (iii), combine Proposition 3.3.2 and part (ii) to obtain

h 2h den(B(x, r)) ≤ κ (ϕω|B(x,r))den(ϕω(B(x, r))) ≤ κ (ϕω|B(x,r))den(B(ϕω(x),Rω(x, r))).

To show (iv), apply Proposition 3.3.9 and Proposition 3.3.1 to obtain

h (2Rω(x, r)) den(B(ϕω(x), Rω(x, r))) = m(B(ϕω(x), Rω(x, r))) h (2κ(ϕω| )R (x, r)) ≤ B(x,r) ω m(B(ϕω(x), Rω(x, r))) h h |ϕω(B(x, r))| ≤ κ (ϕω|B(x,r)) m(B(ϕω(x), Rω(x, r))) h h |ϕω(B(x, r))| ≤ κ (ϕω|B(x,r)) m(ϕω(B(x, r)))

h = κ (ϕω|B(x,r))den(ϕω(B(x, r))).

To show (v), combine Proposition 3.3.2 and part (iv) to obtain

h 2h den(B(ϕω(x), Rω(x, r))) ≤ κ (ϕω|B(x,r))den(ϕω(B(x, r))) ≤ κ (ϕω|B(x,r))den(B(x, r)).



Corollary 3.3.11. Let X ⊆ Rd be a nonempty compact set, I be a countable alphabet, and Φ ∈ CIFS (X,I) be regular. For each finite word ω ∈ I∗, x ∈ J, and r > 0 with B(x, r) ⊆ O,

−2h (i) κ (ϕω|B(x,r))den(B(ϕω(x), Rω(x, r))) ≤ den(B(x, r)),

2h (ii) den(B(x, r)) ≤ κ (ϕω|B(x,r))den(B(ϕω(x),Rω(x, r))),

−3h (iii) κ (ϕω|B(x,r))den(B(ϕω(x), Rω(x, r))) ≤ den(ϕω(B(x, r))),

3h (iv) den(ϕω(B(x, r))) ≤ κ (ϕω|B(x,r))den(B(ϕω(x),Rω(x, r))).

58 Proof. Combine Proposition 3.3.2 with Proposition 3.3.10 (iii) and (v). 

∗ Now suppose B(x, r) ⊆ ϕω(V ) for some ω ∈ I . Then, we define

−1 −1 −1 Rω (x, r) = inf{s > 0 : ϕω (B(x, r)) ⊂ B(ϕω (x), s)} and

−1 −1 −1 Rω (x, r) = sup{s > 0 : B(ϕω (x), s) ⊂ ϕω (B(x, r))}.

It follows from the Mean Value Inequality that

−1 −1   R (x, r) ≤ inf |ϕ0 (y)| r ω −1 ω y∈ϕω (B(x,r)) and !−1 −1 0 Rω (x, r) ≥ sup |ϕω(y)| r −1 y∈ϕω (B(x,r))

Proposition 3.3.12. Let X ⊆ Rd be a nonempty compact set, I be a countable alphabet, ∗ and Φ ∈ CIFS (X,I). For each finite word ω ∈ I , x ∈ X, and r > 0 with B(x, r) ⊆ ϕω(V ),

−1 −1 R (x, r) ≤ κ(ϕ | −1 )R (x, r). ω ω ϕω (B(x,r)) ω

Proof. Notice,

−1 −1   R (x, r) ≤ inf |ϕ0 (y)| r ω −1 ω y∈ϕω (B(x,r)) !−1 0 = κ(ϕ | −1 sup |ϕ (y)| r ω ϕω (B(x,r)) ω y∈B(x,r)

−1 ≤ κ(ϕ | −1 )R (x, r). ω ϕω (B(x,r)) ω



Proposition 3.3.13. Let X ⊆ Rd be a nonempty compact set, I be a countable alphabet, and Φ ∈ CIFS (X,I) be regular. For each finite word ω ∈ I∗, x ∈ J, and r > 0 with

B(x, r) ⊆ ϕω(V ),

−1 −1 h −1 (i) den(B(ϕ (x), R (x, r))) ≤ κ (ϕ | −1 )den(B(ϕ (x),R (x, r))) ω ω ω ϕω (B(x,r)) ω ω

59 −1 h −1 −1 (ii) den(ϕ (B(x, r))) ≤ κ (ϕ | −1 )den(B(ϕ (x),R (x, r))) ω ω ϕω (B(x,r)) ω ω

2h −1 −1 (iii) den(B(x, r)) ≤ κ (ϕ | −1 )den(B(ϕ (x),R (x, r))) ω ϕω (B(x,r)) ω ω

−1 −1 h −1 (iv) den(B(ϕ (x), R (x, r))) ≤ κ (ϕ | −1 )den(ϕ (B(x, r))) ω ω ω ϕω (B(x,r)) ω

−1 −1 2h (v) den(B(ϕ (x), R (x, r))) ≤ κ (ϕ | −1 )den(B(x, r)) ω ω ω ϕω (B(x,r))

Proof. To show (i), apply Proposition 3.3.12 to obtain

−1 h −1 −1 (2Rω (x, r)) den(B(ϕω (x), Rω (x, r))) = −1 −1 m(B(ϕω (x), Rω (x, r))) −1 h (2κ(ϕω| −1 )R (x, r)) ≤ ϕω (B(x,r))) ω −1 −1 m(B(ϕω (x), Rω (x, r))) −1 h h (2Rω (x, r)) ≤ κ (ϕω| −1 ) ϕω (B(x,r)) −1 −1 m(B(ϕω (x),Rω (x, r)))

h −1 −1 = κ (ϕ | −1 )den(B(ϕ (x),R (x, r))). ω ϕω (B(x,r)) ω ω

To show (ii), apply Proposition 3.3.12 and Proposition 3.3.1 to obtain

|ϕ−1(B(x, r))|h den(ϕ−1(B(x, r))) = ω ω −1 m(ϕω (B(x, r))) −1 (2R (x, r))h ≤ ω −1 m(ϕω (B(x, r)) −1 h (2κ(ϕω| −1 )R (x, r)) ≤ ϕω (B(x,r)) ω −1 m(ϕω (B(x, r))) −1 h h (2Rω (x, r)) ≤ κ (ϕω| −1 ) ϕω (B(x,r)) −1 m(ϕω (B(x, r))) −1 h h (2Rω (x, r)) ≤ κ (ϕω| −1 ) ϕω (B(x,r)) −1 −1 m(B(ϕω (x),Rω (x, r)))

h −1 −1 = κ (ϕ | −1 )den(B(ϕ (x),R (x, r))) ω ϕω (B(x,r)) ω ω

60 To show (iii), combine Proposition 3.3.2 and part (ii) to obtain

h −1 den(B(x, r)) ≤ κ (ϕ | −1 )den(ϕ (B(x, r))) ω ϕω (B(x,r)) ω

2h −1 −1 ≤ κ (ϕ | −1 )den(B(ϕ (x),R (x, r))). ω ϕω (B(x,r)) ω ω

To show (iv), apply Proposition 3.3.12 and Proposition 3.3.1 to obtain

−1 h −1 −1 (2Rω (x, r)) den(B(ϕω (x), Rω (x, r))) = −1 −1 m(B(ϕω (x), Rω (x, r))) −1 h (2κ(ϕω| −1 )R (x, r)) ≤ ϕω (B(x,r)) ω −1 −1 m(B(ϕω (x), Rω (x, r))) −1 h h |ϕω (B(x, r))| ≤ κ (ϕω|ϕ−1(B(x,r))) ω −1 −1 m(B(ϕω (x), Rω (x, r))) h h |ϕω(B(x, r))| ≤ κ (ϕω|B(x,r)) m(ϕω(B(x, r)))

h −1 = κ (ϕ | −1 )den(ϕ (B(x, r))). ω ϕω (B(x,r)) ω

To show (v), combine Proposition 3.3.2 and part (iv) to obtain

−1 −1 h −1 den(B(ϕ (x), R (x, r))) ≤ κ (ϕ | −1 )den(ϕ (B(x, r))) ω ω ω ϕω (B(x,r)) ω

2h ≤ κ (ϕ | −1 )den(B(x, r)). ω ϕω (B(x,r))



Corollary 3.3.14. Let X ⊆ Rd be a nonempty compact set, I be a countable alphabet, and Φ ∈ CIFS (X,I) be regular. For each finite word ω ∈ I∗, x ∈ J, and r > 0 with

B(x, r) ⊆ ϕω(V ),

−2h −1 −1 (i) κ (ϕ | −1 )den(B(ϕ (x), R (x, r))) ≤ den(B(x, r)), ω ϕω (B(x,r)) ω ω

2h −1 −1 (ii) den(B(x, r)) ≤ κ (ϕ | −1 )den(B(ϕ (x),R (x, r))), ω ϕω (B(x,r)) ω ω

−3h −1 −1 −1 (iii) κ (ϕ | −1 )den(B(ϕ (x), R (x, r))) ≤ den(ϕ (B(x, r))), ω ϕω (B(x,r)) ω ω ω

61 −1 3h −1 −1 (iv) den(ϕ (B(x, r))) ≤ κ (ϕ | −1 )den(B(ϕ (x),R (x, r))). ω ω ϕω (B(x,r)) ω ω

Proof. Combine Proposition 3.3.2 with Proposition 3.3.13 (iii) and (v). 

We now prove the analogous result for packing measure, which is a generalization of Qiu’s Theorem 1.2 [44].

Theorem 3.3.15 (♣). Let X ⊆ Rd be a nonempty compact set, I be a finite alphabet, and Φ ∈ CIFS (X,I). Fix x ∈ J. If r > 0 with B(x, r) ⊆ O, then

Ph J ∩ B(x, r)
≥ K−h. (2r)h r

Proof. Suppose to the contrary there is some x ∈ J and r > 0 with B(x, r) ⊆ O such that

h
h −h P J ∩ B(x, r) < (2r) Kr .

Then, there is some ε > 0 such that

h
h −h (1 + ε)P J ∩ B(x, r) < (2r) Kr .

Notice, for any k ≥ 1, ! h [ X h
P J ∩ ϕω(B(x, r)) = P J ∩ ϕω(B(x, r)) ω∈Ik |ω|=k

X 0 h h
≤ sup |ϕω(y)| P J ∩ B(x, r) |ω|=k y∈B(x,r)

X 0 h −1 h −h < sup |ϕω(y)| (1 + ε) (2r) Kr |ω|=k y∈B(x,r)

X 0 h h −1 h −h = inf |ϕω(y)| κ (ϕω|B(x,r))(1 + ε) (2r) Kr y∈B(x,r) |ω|=k

X 0 h h −1 h −h ≤ inf |ϕω(y)| Kr (1 + ε) (2r) Kr y∈B(x,r) |ω|=k

−1 X 0 h h = (1 + ε) inf |ϕω(y)| (2r) y∈B(x,r) |ω|=k

−1 X h ≤ (1 + ε) (2Rω(x, r)) |ω|=k

62 −1 X h = (1 + ε) |B(ϕω(x),Rω(x, r))| . |ω|=k

For each δ > 0 there is some kδ large enough so that {B(ϕω(x),Rω(x, r))}|ω|=kδ is a δ-packing of J. Define [ F0 = J − ϕω(B(x, r)). ω∈Ikδ

Case1 : F0 = ∅ In this case, J ∩ S ϕ (B(x, r)) = J. Notice that for each δ > 0, |ω|=kδ ω

h X h Pδ (J) ≥ |B(ϕω(x),Rω(x, r))| ω∈Ik ! h [ ≥ (1 + ε)P J ∩ ϕω(B(x, r)) ω∈Ik

= (1 + ε)Ph (J) .

Case2 : F0 6= ∅ For each n ≥ 1, define

[ Fn = J − (ϕω(B(x, r)))2/n

|ω|=kδ

h h Since each Fn is compact, P0 (Fn) = P (Fn) by Theorem 1.1.1. Since F1 ⊆ F2 ⊆ S∞ F3 ... and n=1 Fn = F0, we have that

h h h lim P0 (Fn) = lim P (Fn) = P (F0) . n→∞ n→∞

Define   1 [ η = εPh J ∩ ϕ (B(x, r)) . 2  ω  ω∈Ikδ

Choose n large enough so that 1/n < δ, Fn 6= ∅, and

η Ph (F ) ≥ Ph (F ) − n 0 2

63 Choose a 1/n-packing B of Fn such that

X η |B|h ≥ Ph (F ) − . n 2 B∈B

Notice, {B(ϕω(x),Rω(x, r))}|ω|=kδ ∪ B forms a δ-packing of J. Now,

h X h X h Pδ (J) ≥ |B(ϕω(x),Rω(x, r))| + |B| ω∈Ikδ B∈B   [ η ≥ (1 + ε)Ph J ∩ ϕ (B(x, r)) + Ph (F ) −  ω  n 2 ω∈Ikδ   h [ h ≥ (1 + ε)P J ∩ ϕω(B(x, r)) + P (F0) − η ω∈Ikδ   h [ h = P J ∩ ϕω(B(x, r)) + P (F0) ω∈Ikδ   h [ + εP J ∩ ϕω(B(x, r)) − η ω∈Ikδ

= Ph (J) + 2η − η   1 [ = Ph (J) + εPh J ∩ ϕ (B(x, r)) . 2  ω  ω∈Ikδ

By Property 3.3.3, we conclude that

ε Ph (J) ≥ Ph (J) + Ph J ∩ B(x, r)
. δ 2Kh

Combining Cases 1 and 2, there is some constant C > 0 such that

h h Pδ (J) ≥ P (J) + C.

Since J is compact, we obtain by Theorem 1.1.1,

h h h h P (J) = P0 (J) = inf Pδ (J) ≥ P (J) + C, δ>0 which is a contradiction. 

64 If Φ ∈ CIFS (X,I) consists of similarities, then Kβ = 1 for all β > 0. Thus, as a corollary, we obtain the following generalization of Theorem 3.2.1 (2).

Corollary 3.3.16 (♣). Let X ⊆ Rd be a nonempty compact set, I be a finite alphabet, and Φ ∈ CIFS (X,I). If β > 0, then

h
h h P (J) ≤ sup den B(x, r) ≤ Kβ P (J) x∈J,0

Proof. The first inequality follows from Corollary 3.1.4. The second inequality is immediate

from the previous theorem. 

Suppose x ∈ J and 0 < r < βΦ with B(x, r) ⊆ V . Then, B(x, r) ⊆ Uj for exactly one

∗ j ∈ I. Let ω ∈ I be the longest word (possibly empty) such that B(x, r) ⊆ ϕω(U). Then,

−1 −1 ϕjω (B(x, r)) is not contained in Ui for any i ∈ I. In particular, |ϕjω (B(x, r))| ≥ 2βΦ. Thus,

−1 −1 −1 βΦ ≤ R (x, r) ≤ κ(ϕjω| −1 )R (x, r) ≤ KR (x, r). jω ϕjω (B(x,r)) jω jω

This guarantees that the ω in the next proposition exists.

Proposition 3.3.17. Let X ⊆ Rd be a nonempty compact set, I be a finite alphabet, and

Φ ∈ CIFSSSC (X,I). Fix M ≥ 1 with

1 −1 0 ≤ K min inf |ϕi(y)|. M i∈I y∈V

∗ Let 0 < β < βΦ. Suppose x ∈ J and 0 < r < β. If ω ∈ I is the shortest word with

−1 Rω (x, r) ≥ β/M, then β ≤ R−1(x, r) ≤ β, M ω and

−1
−2h
den B(ϕω(x),Rω (x, r) ≥ Kβ den B(x, r) .

∗ −1 −1 Proof. Let ω ∈ I be the shortest word with Rω (x, r) ≥ β/M. Notice, ϕω (x) ∈ J and

−1 −1 −1 B(ϕω (x),Rω (x, r)) ⊆ ϕω (B(x, r)) ⊆ U.

65 By Corollary 3.3.11,

−1 −1 −2h −2h den(B(ϕ (x),R (x, r))) ≥ κ (ϕ | −1 )den(B(x, r)) ≥ K den(B(x, r)). ω ω ω ϕω (B(x,r)) β

−1 Suppose to the contrary Rω (x, r) > β. Then,

−1 2β < 2Rω (x, r)

−1 ≤ |ϕω (B(x, r))|

= |ϕ−1(ϕ−1 (B(x, r)))| ωn ω|n−1

−1 0 ≤ sup |(ϕωn ) (y)||ϕω|n−1 (B(x, r))| y∈ϕ−1 (B(x,r)) ω|n−1

−1 0 −1 ≤ sup |(ϕωn ) (y)|2Rω|n−1 (x, r) y∈ϕωn (U)  −1 0 −1 ≤ inf |ϕω (y)| 2KR (x, r) y∈U n ω|n−1  −1 0 −1 ≤ min inf |ϕi(y)| 2KR (x, r) i∈I y∈V ω|n−1

≤ MK−12KR−1 (x, r) ω|n−1 β < M2 M = 2β,

which is a contradiction. 

Lemma 3.3.18 (♣). Let X ⊆ Rd be a nonempty compact set, I be a finite alphabet, and

Φ ∈ CIFSSSC (X,I). Fix M as in Proposition 3.3.17. Then for all 0 < β < βΦ,

−2h h
h h Kβ P (J) ≤ sup den B(x, r) ≤ Kβ P (J) β x∈J, M ≤r≤β

Proof. Let 0 < β < βΦ. The second inequality follows immediately from Corollary 3.3.16. Let ε > 0. By Corollary 3.3.16, there is some x ∈ J and 0 < r ≤ β such that

den(B(x, r)) ≥ Ph (J) − ε.

66 If r ≥ β/M, then we are done. Assume r < β/M. Let ω be the shortest word with −1 Rω (x, r) ≥ β/M. By Proposition 3.3.17, β ≤ R−1(x, r) ≤ β. M ω Thus, by Proposition 3.3.17

−1 −1 −2h −2h h den(B(ϕω (x),Rω (x, r))) ≥ Kβ den(B(x, r)) ≥ Kβ (P (J) − ε).

Since ε > 0 was arbitrary, the first inequality follows. 

Theorem 3.3.19 (♣). Let X ⊆ Rd be a nonempty compact set, d ≥ 2, I be a finite alphabet, and Φ ∈ CIFSSSC (X,I). Fix M as in Proposition 3.3.17. Then,

Ph (J) = lim sup den(B(x, r)). β→0 β x∈J, M ≤r≤β

Proof. Follows immediately from Lemma 3.3.18 since limβ→0 Kβ = 1. 

3.4. Continuity of the Numerical Value for Conformal IFS

hΦ hΦ The goal of this section is to show that the maps Φ 7→ H (JΦ) and Φ 7→ P (JΦ) are continuous. We will assume that the open connected set V in the definition of conformal IFS is the same for each IFS Φ. Since we will establish continuity at each point Φ, we really only need to be able to choose V uniformly for a sufficiently small ball around Φ (in the D-metric). In view of the density theorems from the previous section, we will need to assume that the distortion constants are well behaved. In other words, for every ε > 0, conformal

IFS Φ, and 0 < β < βΦ, there is some δ > 0 and M ≥ 1 such that if D(Φ, Ψ) < δ, then

(1) |KΨ,β − KΦ,β| < ε,

(2) |KΨ,β − KΦ,β| < ε,

1 −1 0 (3) ≤ KΨ min inf |ψi(y)|. M i∈I y∈X These assumptions are quite natural and easily satisfied. In R2, we can appeal to the Koebe Distortion Theorem (Theorem 1.4.2), and in Rd for d ≥ 3 we can appeal to Theorem 1.4.3. The real issue is with R, where the distortions can behave badly. In this case, we need to

67 assume the above conditions. If, for example, we restrict ourselves to only C1+ε maps (see Proposition 1.4.1), then the above conditions are satisfied.

∗ In order to emphasize the assumptions discussed above, we will let CIFSSSC (X,I) ⊆

CIFSSSC (X,I) be the subspace that satisfies our assumptions on the distortion constants and V .

We first prove Lemma 4.4 of [43].

Lemma 3.4.1. Let a, A, κ, ε > 0 with κ < aε and a < A. Then, there exists 0 < γ < a such that x + κ x ≤ + ε y − γ y

for all x, y ∈ [a, A].

Proof. We may choose A0 ≥ A such that (εa − κ)/(A0(1 + ε)) < 1. Let

a(εa − κ) γ = < a. A0(1 + ε)

Then, for all x, y ∈ [a, A] y(εy − κ) γ ≤ . x + yε

This inequality is equivalent to

y(x − κ) ≤ (x + yε)(y − γ).

Since y ≥ a > γ, we conclude that y − γ > 0. Thus,

x + κ x + yε x ≤ = + ε. y − γ y y



Lemma 3.4.2. Let X ⊆ Rd be a nonempty compact set, I be a finite alphabet, and let Φ ∈ ∗ ∗ CIFSSSC (X,I). Then, there is some δ > 0 and C > 0 such that for all Ψ ∈ CIFSSSC (X,I),

hΨ if D(Φ, Ψ) < δ, then H (JΨ) < C.

68 Proof. Let M be as in Proposition 3.3.6. Fix 0 < β < min{1/2, βΦ}. There is constant c > 0 such that for every y ∈ JΦ

mΦ(B(y, β/(4M))) > c.

By continuity of the limit sets, there is some δ1 > 0 such that if D(Φ, Ψ) < δ1, then β H(J ,J ) < . Φ Ψ 2M Let 0 < ε < min{c/2, β/(4M)}. By the continuity of conformal measures for finite conformal

IFS and the L´evy-Prokhorov metric (see Theorem 2.2.4), there is some δ2 > 0 such that for

all y ∈ JΦ

mΦ(B(y, β/(4M)))) ≤ mΨ(B(y, β/(4M) + ε)) + ε ≤ mΨ(B(y, β/(2M))) + ε

∗ for all Ψ ∈ CIFSSSC (X,I) with D(Φ, Ψ) < δ2.

Choose δ3 > 0 so that if D(Φ, Ψ) < δ3, then

hΦ hΨ |KΦ,β − KΨ,β| < ε.

∗ Let δ = min{δ1, δ2, δ3}. Fix Ψ ∈ CIFSSSC (X,I) and x ∈ JΨ. Then, there is some

y ∈ JΦ ∩ B(x, β/(2M)). Notice,

B(y, β/(2M)) ⊂ B(x, β/M).

Then,

mΦ(B(y, β/(4M))) ≤ mΨ(B(y, β/(2M))) + ε ≤ mΨ(B(x, β/M))) + ε.

Hence,

mΨ(B(x, β/M)) > c − ε.

By Corollary 3.3.5,

h hΨ Ψ
H (JΨ) ≤ KΨ,βdenΨ B(x, β/M)

hΨ hΦ (2β/M) ≤ (KΦ,β + ε) mΨ(B(x, β/M))

h 1 ≤ (K Φ + ε) . Φ,β c − ε

69 Therefore, the lemma holds by taking

h 2K Φ C = Φ,β + 1. c



We now establish continuity for the numerical value of Hausdorff measure which was first shown by Szarek, Urba´nskiand Zdunik [48] in a more general setting. It should be noted that Urba´nskiand Zdunik [53] established were able to establish H¨oldercontinuity in certain cases. It seems possible that their results extend to the packing measure setting as well, but this would require additional machinery not developed here. We instead adapt the argument given by Olsen [43] for the self-similar case to the conformal case.

Theorem 3.4.3. Let X ⊆ Rd be a nonempty compact set and I be a finite alphabet. Then

∗ hΦ the map (CIFSSSC (X,I) ,D) → (0, ∞) given by Φ 7→ H (JΦ) is continuous.

∗ Proof. Fix Φ0 = {ϕ0,i}i∈I ∈ CIFSSSC (X,I). Let 0 < ε < 1. By Lemma 3.4.2 there is some ∗ δ0 > 0 and constant C0 > 0 so that for all Φ ∈ CIFSSSC (X,I) with D(Φ0, Φ) < δ0 we have

hΦ H (JΦ) < C0.

−1 Let 0 < ε1 < min{C0ε, C0 ε}. Fix 0 < ∆ < βΦ0 such that

h ε K Φ0 ≤ 1 + 1 . Φ0,∆ 16

∗ Choose δ1 > 0 and M ≥ 1 such that for all Φ ∈ CIFSSSC (X,I) with D(Φ0, Φ) < δ1 we have the following:

(1) |hΦ − hΦ0 | < ε1,

(2) βΦ > ∆, h h ε (3) |K Φ − K Φ0 | < 1 , Φ,∆ Φ0,∆ 16 1 0 (4) ≤ min inf |ϕi(y)|. M i∈I y∈X

Fix 0 < β < ∆. Then there are constants 0 < c < C0 ≤ C such that

c ≤ |B|hΦ ≤ C

70 for all D(Φ0, Φ) < δ1 and Borel sets B ⊆ X such that 2β/M ≤ |B| ≤ 2β. Let a = c/(2+2C),

A = 1 + C, and κ = aε1/4.

Choose 0 < ρ < ∆ − β such that if D(Φ0, Φ) < ρ and F,G ⊆ X with |F |, |G| ∈ [2β/M, 2∆] and H(F,G) < ρ, then

h h κ (4) ||F | Φ0 − |G| Φ | < . 2 By Lemma 3.4.1, there is some 0 < γ < min{a, ρ} such that x + κ x ε C (5) ≤ + 1 0 y − γ y 16 for all x, y ∈ [a, A].

By the L´evy-Prokhorov metric (see Theorem 2.2.4), there is δ2 > 0 such that if

D(Φ0, Φ) < δ2 and D(Φ0, Ψ) < δ2, then

(6) mΦ(B) ≤ mΨ([B]ρ) + γ for all Borel sets B ⊆ X. Finally, choose δ3 > 0 so that if D(Φ0, Φ) < δ3, then

(7) H(JΦ0 ,JΦ) < ρ/2.

∗ Let δ = min{ρ, δ0, δ1, δ2, δ3}. It suffices to show that for all Φ, Ψ ∈ CIFSSSC (X,I), if

D(Φ0, Φ) < δ and D(Φ0, Ψ) < δ, then

hΦ hΨ H (JΦ) ≤ H (JΨ) + ε.

∗ Fix Φ, Ψ ∈ CIFSSSC (X,I) such that D(Φ0, Φ) < δ and D(Φ0, Ψ) < δ. By Lemma 3.3.7,

there is some closed F ⊆ X such that 2β/M ≤ |F | ≤ 2β, mΦ(F ) > 0, and

hΦ ε1C0 den (F ) ≤ K HhΦ (J ) + Φ Φ,β Φ 8

hΦ ε1C0 ≤ K HhΦ (J ) + Φ,∆ Φ 8 h ε ε C Φ0 1 hΦ 1 0 ≤ (K + )H (JΦ) + Φ0,∆ 16 8

ε1 ε1C0 ≤ (1 + )HhΦ (J ) + 8 Φ 8

ε1C0 ≤ HhΦ (J ) + . Φ 4

71 Thus,

ε1C0 (8) den (F ) ≤ HhΦ (J ) + . Φ Φ 4 Notice, by (7),

H(JΦ,JΨ) ≤ H(JΦ,JΦ0 ) + H(JΦ0 ,JΨ) < ρ.

Since F ∩ JΦ 6= ∅, we conclude that [F ]ρ ∩ JΨ 6= ∅. Moreover, |[F ]ρ| ≤ |F | + 2ρ ≤ 2∆. By Corollary 3.3.5,

h hΨ Ψ H (JΨ) ≤ KΨ,∆denΨ ([F ]ρ) .

hΦ Combining H (JΦ) ≤ C0 ≤ C and (8), we have that

|F |hΦ c mΦ(F ) ≥ ≥ = 2a. hΦ 3ε1 H (JΦ) + 4 C + 1 Now, by (6),

mΨ([F ]ρ) ≥ mΦ(F ) − γ ≥ 2a − a = a > 0.

Notice,

a < c ≤ |F |hΦ ≤ C < A and

a < 2a ≤ mΦ(F ) ≤ 1 < A.

Therefore, by (5)

hΦ hΦ |F | + κ |F | ε1C0 ε1C0 ≤ + = denΦ (F ) + . mΦ(F ) − γ mΦ(F ) 16 16

Since D(Φ0, Φ) < δ ≤ ρ, D(Φ0, Ψ) < δ ≤ ρ, and H(F, [F ]ρ) ≤ ρ, we conclude by two applications of (4) that

h h κ h |[F ] | Ψ ≤ |F | Φ0 + ≤ |F | Φ + κ. ρ 2 Finally,

h hΨ Ψ H (JΨ) ≤ KΨ,∆denΨ ([F ]ρ)

hΨ hΨ |[F ]ρ| ≤ KΨ,∆ mΨ([F ]ρ)

72 hΦ hΨ |F | + κ ≤ KΨ,∆ mΦ(F ) − γ

hΦ0 ε1 ε1C0 ≤ (K + )(denΦ (F ) + ) Φ0,∆ 16 16

ε1 ε1C0 ε1C0 ≤ (1 + )(HhΦ (J ) + + ) 8 Φ 4 16 2 5ε1C0 ε1C0 5ε C0 ≤ HhΦ (J ) + + + 1 Φ 16 8 128

hΦ ≤ H (JΦ) + C0ε1

hΦ ≤ H (JΦ) + ε.



We now work towards the analogous proof for packing measure by establishing some lemmas. The first lemma is Lemma 3.3 of [44].

Lemma 3.4.4. Let a, A, κ, ε > 0 with κ < aε and a < A. Then, there exists 0 < γ < a such that x − κ x ≥ − ε y + γ y for all x, y ∈ [a, A].

Proof. Let γ = a(aε − κ)/A. For simplicity, assume that x/y − ε > 0. Then,

y(εy − κ) γ ≤ . x − yε

Hence

(x − yε)γ ≤ y(εy − κ),

which implies

κy + xγ ≤ εy(y + γ).

Thus, dividing the above inequality by y(y + γ) gives

κy + xγ ≤ ε. y(y + γ)

73 Thus, x x − κ − ≤ ε. y y + γ 

Lemma 3.4.5. Let X ⊆ Rd be a nonempty compact set, I be a finite alphabet, and let Φ ∈ ∗ ∗ CIFSSSC (X,I). Then, there is some δ > 0 and C > 0 such that for all Ψ ∈ CIFSSSC (X,I),

hΨ if D(Φ, Ψ) < δ, then P (JΨ) > C.

Proof. Let 0 < ε < 1. Fix 0 < β < min{1/2, βΦ} and M as in Proposition 3.3.17. Using continuity of packing dimension, choose δ > 0 so that if D(Φ, Ψ) < δ, then

(1) |hΦ − hΨ| < ε,

(2) β < βΨ,

hΦ hΨ (3) |KΦ,β − KΨ,β| < ε.

∗ Fix Ψ ∈ CIFSSSC (X,I) so that D(Φ, Ψ) < δ. By Lemma 3.3.18, for any x ∈ JΨ,

hΨ −hΨ
P (JΨ) ≥ KΨ,β denΨ B(x, β)

hΨ −hΨ (2β) = KΨ,β mΨ(B(x, β))

hΦ −1 hΦ+ε > (KΦ,β + ε) (2β) .

The lemma follows by letting (2β)hΦ+1 C = . hΦ KΦ,β + 1 

Lemma 3.4.6. Let X ⊆ Rd be a nonempty compact set, I be a finite alphabet, and let Φ ∈ ∗ ∗ CIFSSSC (X,I). Then, there is some δ > 0 and C > 0 such that for all Ψ ∈ CIFSSSC (X,I),

hΨ if D(Φ, Ψ) < δ, then P (JΨ) < C.

Proof. Let M be as in Proposition 3.3.17. Fix 0 < ∆ < βΦ, and let 0 < β < min{1/2, ∆}.

There is constant c > 0 such that for every y ∈ JΦ

mΦ(B(y, β/(4M))) > c.

74 By continuity of the limit sets, there is some δ1 > 0 such that if D(Φ, Ψ) < δ1, then

β H(J ,J ) < . Φ Ψ 2M

Let 0 < ε < min{c/2, β/(4M)}. By the continuity of conformal measures for finite conformal

IFS and the L´evy-Prokhorov metric (see Theorem 2.2.4), there is some δ2 > 0 such that for

all y ∈ JΦ

mΦ(B(y, β/(4M)))) ≤ mΨ(B(y, β/(4M) + ε)) + ε ≤ mΨ(B(y, β/(2M))) + ε

∗ for all Ψ ∈ CIFSSSC (X,I) with D(Φ, Ψ) < δ2. ∗ Choose δ3 > 0 and M ≥ 1 such that for all Φ ∈ CIFSSSC (X,I) with D(Φ, Ψ) < δ3 we have the following:

(1) βΦ > ∆,

2hΦ 2hΦ (2) |KΨ,β − KΦ,β | < ε, 1 −1 0 (3) ≤ KΨ min inf |ψi(y)|. M i∈I y∈X ∗ Let δ = min{δ1, δ2, δ3}. Fix Ψ ∈ CIFSSSC (X,I) and x ∈ JΨ. Then, there is some

y ∈ JΦ ∩ B(x, β/(2M)). Notice,

B(y, β/(2M)) ⊂ B(x, β/M).

Then,

mΦ(B(y, β/(4M))) ≤ mΨ(B(y, β/(2M))) + ε ≤ mΨ(B(x, β/M))) + ε.

Hence, for all x ∈ JΨ and β/M ≤ r ≤ β,

mΨ(B(x, r)) > c − ε.

Then, for all x ∈ JΨ and β/M ≤ r ≤ β,

h
(2r) Ψ 1 denΨ B(x, r) = ≤ . mΨ(B(x, r)) c − ε By Lemma 3.3.18,

hΨ 2hΨ −1 2hΦ −1 P (JΨ) ≤ KΨ,β (c − ε) ≤ (KΦ,β + ε)(c − ε) .

75 The lemma holds by taking 2K2hΦ C = Φ,β + 1. c



We can now establish continuity for the numerical value of packing measure. The self- similar case has already been shown by Qiu [44], who adapted the proof given by Olsen [43] for Hausdorff measure. We adapt the argument given by Qiu to the more general conformal setting.

Theorem 3.4.7 (♣). Let X ⊆ Rd be a nonempty compact set and I be a finite alphabet.

∗ hΦ Then the map (CIFSSSC (X,I) ,D) → (0, ∞) given by Φ 7→ P (JΦ) is continuous.

∗ Proof. Fix Φ0 = {ϕ0,i}i∈I ∈ CIFSSSC (X,I). Let 0 < ε < 1. By Lemma 3.4.5, there is some δ0 > 0 and constant C0 > 0 such that if D(Φ0, Φ) < δ0, then

hΦ P (JΦ) < C0.

−1 Let 0 < ε1 < min{C0ε, C0 ε}. Fix 0 < ∆ < βΦ0 such that

−2h ε K Φ0 ≥ 1 − 1 . Φ0,∆ 16

∗ Choose δ1 > 0 and M ≥ 1 such that for all Φ ∈ CIFSSSC (X,I) with D(Φ0, Φ) < δ1 we have the following:

(1) |hΦ − hΦ0 | < ε1,

(2) βΦ > ∆, −2h ε (3) |K−2hΦ − K Φ0 | < 1 , Φ,∆ Φ0,∆ 16 1 −1 0 (4) ≤ KΦ min inf |ϕi(y)|. M i∈I y∈X

Then, there are constants c and C with 0 < c < C0 ≤ C such that

c ≤ (2s)hΦ ≤ C

∗ for all Φ ∈ CIFSSSC (X,I) such that D(Φ0, Φ) < δ1 and ∆/M ≤ s ≤ ∆. It follows from the

continuity of conformal measures (see the proof of Lemma 3.4.2) that there exist δ2 > 0 and

76 d > 0 such that if D(Φ0, Φ) < δ2, then

mΦ(B(x, s)) > d

for all x ∈ JΦ and ∆/M ≤ s ≤ ∆.

Let a = min{c, d}, A = a + C + 1, κ = aε1/4. Choose 0 < ρ < ∆/(2M) such that if

D(Φ0, Φ) < ρ and s, t ∈ [∆/(2M), ∆] with |s − t| ≤ ρ, then

h h κ (9) |s Φ0 − t Φ | < . 2

By Lemma 3.4.4, there is some γ > 0 such that

x − κ x ε C ≥ − 1 0 . y + γ y 16

By the L´evy-Prokhorov metric, there is δ3 > 0 such that if D(Φ0, Φ) < δ3 and D(Φ0, Ψ) < δ3, then

(10) mΨ(B) ≤ mΦ([B]ρ/2) + γ

for all Borel sets B ⊆ V . Finally, choose δ4 > 0 so that if D(Φ0, Φ) < δ4, then

(11) H(JΦ0 ,JΦ) < ρ/4.

∗ Let δ = min{ρ, δ0, δ1, δ2, δ3, δ4}. It suffices to show that for all Φ, Ψ ∈ CIFSSSC (X,I),

if D(Φ0, Φ) < δ and D(Φ0, Ψ) < δ, then

hΨ hΦ (12) P (JΨ) ≥ P (JΦ) − ε.

∗ Fix Φ, Ψ ∈ CIFSSSC (X,I) such that D(Φ0, Φ) < δ and D(Φ0, Ψ) < δ. By Lemma 3.3.18,

there is some x ∈ JΦ and ∆/M ≤ r ≤ ∆ such that

ε1C0 den B(x, r)
≥ K−2hΦ PhΦ (J ) − Φ Φ,∆ Φ 8 −2h ε ε C Φ0 1 hΦ 1 0 ≥ (K − )P (JΦ) − Φ0,∆ 16 8

ε1 ε1C0 ≥ (1 − )PhΦ (J ) − 8 Φ 8

ε1C0 ≥ PhΦ (J ) − . Φ 4

77 Thus,

ε1C0 (13) den B(x, r)
≥ PhΦ (J ) − . Φ Φ 4

Notice,

a ≤ c ≤ (2r)hΦ ≤ C < A and

a ≤ d < mΦ(B(x, r)) ≤ 1 < A.

Therefore, by Lemma 3.4.4 and (13)

(2r)hΦ − κ (2r)hΦ ε C ≥ − 1 0 mΦ(B(x, r)) + γ mΦ(B(x, r)) 16 ε C = den B(x, r)
− 1 0 Φ 16

5ε1C0 ≥ PhΦ (J ) − . Φ 16

By two applications of (9) we obtain

h h h (2r) Φ ≤ (2(r − ρ)) Φ0 + κ/2 ≤ (2(r − ρ)) Ψ + κ.

Thus,

(2(r − ρ))hΨ ≥ (2r)hΦ − κ.

Notice, by (11)

H(JΦ,JΨ) ≤ H(JΦ,JΦ0 ) + H(JΦ0 ,JΨ) < ρ/2.

Then, there is some y ∈ B(x, ρ/2) ∩ JΨ. Notice, r − ρ > 0 and

B(y, r − ρ/2) ⊆ B(x, r).

By (10), we obtain

(14) mΨ(B(y, r − ρ)) ≤ mΦ(B(y, r − ρ/2)) + γ ≤ mΦ(B(x, r)) + γ.

Therefore, by Lemma 3.3.18

hΨ −hΨ
P (JΨ) ≥ KΨ,∆ denΨ B(y, r − ρ)

78 hΨ −2hΨ (2(r − ρ)) ≥ KΨ,∆ mΨ(B(y, r − ρ))

hΦ −2hΨ (2r) − κ ≥ KΨ,∆ mΨ(B(y, r − ρ)) − γ

−2h ε 5ε C Φ0 1 hΦ 1 0 ≥ (K − )(P (JΦ) − ) Φ0,∆ 16 16

ε1 5ε1C0 ≥ (1 − )(PhΦ (J ) − ) 8 Φ 16

7ε1C0 ≥ PhΦ (J ) − Φ 16 7ε ≥ PhΦ (J ) − Φ 16

hΦ ≥ P (JΦ) − ε.



79 CHAPTER 4

THE NUMERICAL VALUE OF PACKING MEASURE FOR SUPER SEPARATED ITERATED FUNCTION SYSTEMS

The problem of calculating or estimating the numerical value of Hausdorff and packing measures for a limit set has been an active area of interest (see for example [4,5,6,7, 19, 21, 22, 26, 27, 29, 30, 32, 33, 46, 55, 56]) . In the context of finite iterated function systems in [0, 1] consisting of increasing similarities and satisfying the open set condition with O = (0, 1), Marion [32, 33], and independently, Ayer and Strichartz [4] created an algorithm for computing the Hausdorff measure of the limit set. In addition, they showed

hΦ that Φ 7→ H (JΦ) is not continuous if the domain is all IFSs satisfying the OSC. As shown in the previous chapter, continuity holds if we restrict to IFSs satisfying the strong separation condition (SSC).

Under the same conditions as Ayer and Strichartz, Feng [19] was able to find exact formulas for the packing measure of the limit set. Llorente and Moran [30] created an algorithm for estimating the packing measure of limit sets generated by IFSs in Rd under the strong separation condition, but they do not provide error bounds. These results, all using the same underlying density theorems, seem to indicate that obtaining formulas for the packing measure of fractals generated by IFSs may be a tractable problem despite the awkwardness of the definition of packing measure.

In this chapter, we consider the numerical value of the packing measure for fractals generated by IFSs under the super separation condition. We first introduce super separation and show that the packing measure for the limit set of an IFS satisfying super separation can be reduced to checking the densities of a finite number of balls around each point in the limit set. We then extend Feng’s result [19] by allowing for decreasing contraction maps along with super separation. Then, we give explicit formulas for the packing measures of limit sets based on regular polygons in the plane. Finally, we consider Sierpi´nskisimplexes as a higher dimensional analog to the fractals.

80 Throughout the chapter, we will always be talking about some nonempty compact

set X ⊆ Rd and a finite alphabet I = {1,...,N} for some N ≥ 2. The IFS Φ ∈ IFS (X,I)

will always consist of similarity maps, and we will denote ci = Lip (ϕi), J = JΦ, h = hΦ,

and m = mΦ. Since we will usually only care about the density of closed balls, we will write
den(x, r) = denΦ B(x, r) . We will also use the following notation:

(1) cmin = min{cj : 1 ≤ j ≤ N}

d d (2) dinf (x, A) = inf{|x − y| : y ∈ A} for any x ∈ R and A ⊆ R d (3) dinf (A, B) = inf{|x − y| : x ∈ A, y ∈ B} for any A, B ⊆ R d d (4) dsup(x, A) = sup{|x − y| : y ∈ A} for any x ∈ R and A ⊆ R d (5) dsup(A, B) = sup{|x − y| : x ∈ A, y ∈ B} for any A, B ⊆ R

(6) `i,j = dinf (ϕi(J), ϕj(J)) for all 1 ≤ i, j, ≤ N

(7) `j,min = min{`i,j : i 6= j} for all 1 ≤ j ≤ N

(8) `min = min{`j,min : 1 ≤ j ≤ N}.

4.1. The Super Separation Condition

Definition 4.1.1. An IFS Φ is said to be super separated if |ϕj(J)| ≤ `j,min for all 1 ≤ j ≤ N.

Intuitively, Φ is super separated if the gap between ϕj(J) and its nearest neighboring level set ϕi(J) is at least as big as the diameter of ϕj(J).

Proposition 4.1.1. If Φ is super separated, then Φ satisfies the SSC and cmin|J| ≤ `min.

Proof. Since |ϕj(J)|= 6 0 for all j = 1,...,N, it follows immediately from super separation that `j,min > 0 for all j = 1,...,N. Therefore ϕi(J) ∩ ϕj(J) = ∅ for all i 6= j, where i, j = 1,...,N, and hence Φ satisfies the SSC. Now, for all j = 1,...,N,

cmin|J| ≤ cj|J| ≤ `j,min.

Therefore, cmin|J| ≤ `min. 

The calculation of the exact packing measure of a self-similar set (under the OSC) can be boiled down to computing densities (see for example [29],[42]). We take advantage of

81 the following theorem which can be found in [30] as Theorem 1. Note that µ is used in place of m in [30] and is referred to as the natural probability measure. The natural probability measure µ is the unique probability measure (see [24]) satisfying

N X h −1 cj µ ◦ ϕj = µ. j=1

As shown in Chapter 2, the natural probability measure is equivalent to both the normalized Hausdorff measure and normalized packing measure.

Theorem 4.1.2. If Φ satisfies the SSC, then for any a ∈ (0, `min ] cmin

Ph(J) = max{den(x, r): x ∈ J, r ≤ a}

= max{den(x, r): x ∈ J, acmin ≤ r ≤ a}.

The following corollary is immediate from Proposition 4.1.1 and Theorem 4.1.2.

Corollary 4.1.3. If Φ is super separated, then

Ph(J) = max{den(x, r): x ∈ J, r ≤ |J|}

= max{den(x, r): x ∈ J, cmin|J| ≤ r ≤ |J|}.

The next lemma is an important scaling property of densities shown in [30] as Lemma 4.

Lemma 4.1.4. For every finite word ω and x ∈ J,

den(ϕω(x), cωr) = den(x, r).

If B(x, r) ∩ J ⊆ ϕω(J), then

−1 −1 den(ϕω (x), cω r) = den(x, r).

We now prove a slightly stronger form of Corollary 4.1.3 for super separated IFSs.

82 Proposition 4.1.5 (♣). If Φ is super separated, then

h P (J) = max{den(x, r): cj|J| ≤ r ≤ |J|, where x ∈ ϕj(J)

and either ∂B(x, r) ∩ J 6= ∅ or r = |J|}.

h Proof. By Corollary 4.1.3, there is some x0 ∈ J and cmin|J| ≤ r0 ≤ |J| with P (J) =

den(x0, r0). If ∂B(x0, r0) ∩ J = ∅ and r < |J|, then we may increase r0 until the boundary

intersects J or until r = |J|, which clearly increases den(x0, r0), contradicting maximality. Therefore we assume either ∂B(x, r) ∩ J 6= ∅ or r = |J|. Let ω be the longest word (possibly empty) such that

B(x0, r0) ∩ J ⊂ ϕω(J).

−1 −1 Then B(ϕω (x0), cω r0) ∩ J is not contained in ϕj(J) for any j. By Lemma 4.1.4,

−1 −1 den(x0, r0) = den(ϕω (x0), cω r0).

−1 −1 Thus, ϕω (x0) ∈ ϕj0 (J) for some j0 and cω r0 ≥ cj0 |J|. Moreover, since r0 ≤ cω|J|, we have −1 that cω r0 ≤ |J|. That is,

−1 cj0 |J| ≤ cω r0 ≤ |J|.



4.2. A Density Theorem for Super Separated IFS

d Definition 4.2.1. Let x ∈ R , r > 0, and k = 1,...,N. We will say that ϕk(J) straddles d B(x, r) if B(x, r)∩ϕk(J) 6= ∅ and (R −B(x, r))∩ϕk(J) 6= ∅. Then, ϕk(J) does not straddle d B(x, r) if and only if ϕk(J) ⊆ (R − B(x, r)) or ϕk(J) ⊆ B(x, r).

Definition 4.2.2. Suppose x ∈ ϕj(J) for some j = 1,...,N. Let R(x) be the collection of

r such that cj|J| ≤ r ≤ |J|, ∂B(x, r) ∩ (J − ϕj(J)) 6= ∅, and ϕk(J) does not straddle B(x, r) for all k = 1,...,N. In addition, if ∂B(x, |J|) ∩ J = ∅, then we also include |J| in R(x).

Suppose x ∈ ϕj(J) for some j = 1,...,N and r ∈ R(x). Recall that m(∂B(x, r)) = 0. If r = |J|, then m(B(x, r)) = 1. Assume r 6= |J|. Then, ∂B(x, r) ∩ J 6= ∅ and B(x, r) ∩ J =

83 S {ϕj(J)} ∪ i∈I ϕi(J), where the union is a disjoint union and I is some (possibly empty) subset of {1,...,N} depending only on x and r.

Now, m(B(x, r)) is easy to compute since

h X h m(B(x, r)) = cj + ci . i∈I

Moreover, R(x) contains at most N elements since any element is either |J| or of the form

dinf (x, ϕk(J)) for some k 6= j and k = 1,...,N. Therefore, the next theorem demonstrates that computing the packing measure of a super separated IFS with h < 1 is easier to handle.

Theorem 4.2.1 (♣). If Φ is super separated and h < 1, then

Ph(J) = max{den(x, r): x ∈ J and r ∈ R(x)}.

h Moreover, if P (J) = den(x0, r0) with cj|J| ≤ r0 ≤ |J| and x0 ∈ ϕj(J), then r0 ∈ R(x0).

h Proof. By Proposition 4.1.5, there is some x0 ∈ ϕk0 (J) and ck0 |J| ≤ r0 ≤ |J| with P (J) =

den(x0, r0) and either r0 = |J| or ∂B(x0, r0) ∩ J 6= ∅. If r0 = |J| and ∂B(x0, |J|) ∩ J = ∅, then we are done since r0 ∈ R(x0). Thus, we may assume ∂B(x0, r0) ∩ J 6= ∅.

Suppose to the contrary ∂B(x0, r0) ∩ J ⊂ ϕk0 (J). Then, r0 ≤ |ϕk0 (J)| since x0 ∈

ϕk0 (J). Since r0 ≥ ck0 |J| = |ϕk0 (J)| by assumption, we conclude that r0 = |ϕk0 (J)|. Since Φ

is super separated, r0 ≤ `k0,min, and hence m(B(x0, r0)) = m(ϕk0 (J)). Since ∂B(x0, r0)∩J ⊂

ϕk0 (J) and Φ satisfies the SSC, we may choose δ > 0 such that ∂B(x0, r0 + δ) ∩ J = ∅ and m(B(x0, r0)) = m(B(x0, r0 + δ)). In particular, den(x0, r0) < den(x0, r0 + δ), which

contradicts maximality. Therefore, we may assume ∂B(x0, r0) ∩ (J − ϕk0 (J)) 6= ∅.

Suppose to the contrary there is some k1 = 1,...,N such that k1 6= k0 and ϕk1 (J)

straddles B(x0, r0). Let d = dinf (x0, ϕk1 (J)). Since ϕk1 (J) straddles B(x0, r0), there is some

y ∈ ϕk1 (J) ∩ B(x0, r0) with d = |x0 − y|. In particular, d < r0.

Let ε = r0 − d. Notice, if x ∈ B(y, ε), then

|x0 − x| ≤ |x0 − y| + |y − x| ≤ d + ε = r0,

84 d which shows that B(y, ε) ⊆ B(x0, r0). Since ϕk1 (J) straddles B(x0, r0), we know (R −

B(x0, r0)) ∩ ϕk1 (J) 6= ∅, and thus, ε ≤ |ϕk1 (J)|. Then, by super separation, we have

ε ≤ |ϕk1 (J)| ≤ `k1,min,

and hence, B(y, ε) ∩ ϕj(J) = ∅ for all j 6= k1. Note that B(x0, d) ∩ ϕk1 (J) = ∅. Since the boundary of any ball hits J at a m-null set by [34],

m(B(x0, r0)) ≥ m(B(x0, d) ∪ B(y, ε)) = m(B(x0, d)) + m(B(y, ε)).

Using the assumption that h < 1,

h (2r0) den(x0, r0) = m(B(x0, r0)) (2(d + ε))h = m(B(x0, r0)) (2d)h + (2ε)h < m(B(x0, d)) + m(B(y, ε))  (2d)h (2ε)h  ≤ max , m(B(x0, d)) m(B(y, ε))

= max{den(x0, d), den(y, ε)}.

This contradicts the maximality of den(x0, r0) since y ∈ J and ε ≤ |J|. Therefore

ϕk1 (J) does not straddle B(x0, r0). Since our choice of k1 6= k0 was arbitrary, we have shown

that r0 ∈ R(x0). 

4.3. Packing Measure of Super Separated Cantor Sets

In this section we consider super separated IFSs in the interval [0, 1]. In particular, we have that h < 1 so that we may apply Theorem 4.2.1. Denote the intervals ϕj([0, 1]) = [aj, bj] and assume a1 < ··· < aN so that the intervals are in increasing order with respect to j. We also assume without loss of generality that 0 ∈ ϕ1(J) and 1 ∈ ϕN (J). Note that this implies a1 = 0 and bN = 1. The following theorem was shown by Feng [19] assuming the maps

ϕj were increasing and that the open intervals ϕj((0, 1)) were disjoint (a stronger version

85 of the OSC). Our result extends Feng’s because we do not assume that the similarities are increasing. However, super separation is much stronger than his separation condition.

Theorem 4.3.1 (♣). Suppose Φ is a super separated IFS on [0, 1] as described above. Then   max{R,S} if N = 2 Ph(J) =  max{R,S,T } if N ≥ 3, where

(2a )h R = max n 2≤n≤N Pn−1 h j=1 cj (2b )h S = max n 1≤n≤N−1 PN h j=n+1 cj a +b (a − b − 2d ( n2+1 n1 ,J))h T = max n2+1 n1 inf 2 . Pn2 h 1≤n1

h Proof. By Theorem 4.2.1 there is some x0 ∈ J and r0 ∈ R(x0) with P (J) = den(x0, r0).

h Suppose to the contrary that r0 = 1. Since m(B(x0, 1)) = 1, we have that den(x0, r0) = 2 . However, h h h (2a2) 2 (c1 + `1,2) h den(0, a2) = = h > 2 , m(B(0, a2)) c1

which contradicts maximality. Therefore, r0 < |J| = 1.

Suppose that x0 ∈ ϕ1(J). Consider the balls B(x, r0) with x ∈ ϕ1(J). As x ap-

proaches 0, it is clear that m(B(x, r0)) can not increase since 0 is the leftmost point in J.

In other words, den(x0, r0) ≤ den(0, r0). By maximality, we conclude that den(x0, r0) =

den(0, r0), and by Theorem 4.2.1, we know r0 ∈ R(0). Then r0 = an for some n ≥ 2. Since r0 < 1, we have (2r )h (2a )h Ph(J) = den(0, r ) = 0 = n , 0 Pn−1 h m(B(0, r0)) j=1 cj h which means that P (J) = R. By a symmetric argument, if x0 ∈ ϕN (J), then we can assume

h x0 = 1 and conclude that P (J) = S.

Now suppose x0 ∈ ϕk(J) for some 1 < k < N. Note that this immediately implies

that N ≥ 3. Let B(x0, r0) = [c, d]. Since r0 ∈ R(x0), c = bn or d = an for some n. We

86 assume without loss of generality that d = an for some 1 < n ≤ N. Suppose that c ∈ J.

Since r0 ∈ R(x0), it follows that c = bm for some 1 ≤ m < n − 1. Note that c 6= bn−1 since

an+bm an+bm that implies x0 ≤ c. Since x0 = 2 , it follows that dinf ( 2 ,J) = 0. Therefore,

(a − c)h (a − b − 2d ( an+bm ,J))h den(x , r ) = n = n m inf 2 , 0 0 Pn−1 h m(B(x0, r0)) j=m+1 cj which implies Ph(J) = T .

Now suppose that c∈ / J. First consider the case that c < 0. Then m([c, an]) = m([−an, an]), and since x0 ≥ 0, we know that −an ≤ c. Thus,

h h (an − c) (2an) den(x0, r0) = ≤ = den(0, an). m([c, an]) m([−an, an])

Notice, (2a )h den(0, a ) = n . n Pn−1 h j=1 cj

h By maximality, den(x0, r0) = den(0, an), and hence P (J) = R.

Since 0 ∈ J and c∈ / J, we can now assume c > 0. Since r0 ∈ R(x0), ϕj(J) must not straddle [c, an] for any j = 1,...,N. Therefore, we may assume that bm < c < am+1 for some 1 ≤ m < n − 1.

an+bm Let D = dinf ( 2 ,J), and let y ∈ J be such that |y − (an + bm)/2| = D. Suppose

(an + bm)/2 ≤ y ≤ x0. We will show that y = x0. Notice,

bm ≤ 2y − an ≤ c,

and 2y − an = c if and only if y = x0. Then,

h (an − c) den(x0, r0) = m([c, an]) (a − (2y − a ))h ≤ n n m([c, an]) 2h(a − y)h = n m([2y − an, an])

= den(y, an − y).

87 Now, equality holds if and only if y = x0. Since y ∈ J, we conclude that y = x0 by the

maximality of den(x0, r0). Therefore,

(a − c)h (a − b − 2d ( an+bm ,J))h den(x , r ) = n = n m inf 2 , 0 0 m([b , a ]) Pn−1 h m n j=m+1 cj and hence, Ph(J) = T .

Finally, suppose y < (an + bm)/2. Notice that

a + b c + a a + b c − b x − n m = n − n m = m . 0 2 2 2 2

Therefore, D ≤ (c − bm)/2. Then,

a + b c − b n m − y ≤ m , 2 2

or equivalently

an − c ≤ 2(y − bm).

Moreover, equality holds if and only if D = (c − bm)/2.

Now suppose towards a contradiction that D < (c − bm)/2. Then,

h (an − c) den(x0, r0) = m([c, an]) (2(y − b ))h < m m([bm, an]) 2h(y − b )h ≤ m m([bm, 2y − bm])

= den(y, y − bm).

Since y ∈ J, we have reached a contradiction since den(x0, r0) is maximal.

Suppose D = (c − bm)/2. Notice,

h (an − c) den(x0, r0) = m([c, an]) (2(y − b ))h = m m([bm, an]) 2h(y − b )h ≤ m m([bm, 2y − bm])

88 = den(y, y − bm).

Since y ∈ J, we conclude that den(x0, r0) = den(y, y − bm), and in particular, m([bm, an]) = m([bm, 2y − bm]). Therefore,

2h(y − b )h (a − b − 2d ( an+bm ,J))h den(y, y − b ) = m = n m inf 2 , m m([b , a ]) Pn−1 h m n j=m+1 cj

h and hence, P (J) = T . 

Corollary 4.3.2 (♣). Suppose Φ is a super separated IFS on [0, 1] consisting of increasing

linear maps such that c = cj for all j = 1,...,N and ` = `j,j+1 for all j = 1,...,N − 1. Then,

Ph(J) = N2h(c + `)h.

Proof. Suppose the maximum density is obtained by B(x0, r0) for some x0 ∈ J and r0 ∈

R(x0). We will show that den(x0, r0) = R as in Theorem 4.2.1. If den(x0, r0) = S, then

x0 = 1, den(0, r0) = den(1, r0) by symmetry, and thus, den(x0, r0) = R. Suppose that Ph(J) = T as in Theorem 4.2.1. Then the maximum density is obtained by an interval

an+bm an+bm of diameter 2r0 = an − bm − 2dinf ( 2 ,J) with center x0 ∈ J nearest to 2 for some

1 ≤ m < n − 1. Now, x0 ∈ ϕk0 (J) ⊂ [ak0 , bk0 ] for some m < k0 < n. Notice,

m([ak0 , x0 + r0]) = m([ak0 , an])

n−1 X h = cj j=k0 n − k = 0 . N

Since c = cj for all j = 1,...,N and ` = `j,j+1 for all j = 1,...,N − 1,

n−k0−1 X n − k0 m([0, a ]) = ch = . n−k0 j N j=0

Thus,

m([−an−k0 , an−k0 ]) = m([ak0 , x0 + r0]) ≤ m(B(x0, r0)).

89 Moreover, since c = cj for all j = 1,...,N and ` = `j,j+1 for all j = 1,...,N − 1,

2an−k0 = 2(an − ak0 ) ≥ 2r0.

We conclude that

den(x0, r0) ≤ den(0, an−k0 ).

h By maximality, den(x0, r0) ≤ den(0, an−k0 ). Therefore, we have shown that P (J) = R.

Now we can safely assume that x0 = 0. Thus, B(0, r0) = [−an, an] for some n ≥ 2. Therefore, (2a )h 2h((n − 1)c + (n − 1)`)h N2h(c + `)h den(0, r ) = n = = . 0 Pn−1 h n−1 (n − 1)1−h j=1 cj N

Since den(0, r0) is maximal and 1−h > 0, we must have n = 2, which proves the corollary. 

For the remainder of this section, we will assume the hypothesis of Corollary 4.3.2

1−Nc with a fixed N. Notice, we have the relation Nc + (N − 1)` = 1, and thus, ` = N−1 . Since 1 ` depends only on c, we can associate to each c ∈ (0, N ] the limit set Jc given by the IFS Φ with parameter c. Moreover, Φ is super separated exactly when c ≤ `, or equivalently,

1 log N c ≤ 2N−1 . By Hutchinson’s formula, we know the packing dimension of Jc is hc = − log c . Therefore, − log N  1 − c  log c Phc (J ) = N2hc (c + `)hc = N 2 . c N − 1 1 Fix N = 2. The limit set Jc in this case is the middle-(1 − 2c) Cantor set. If c ≤ 3 ,

hc hc we have that P (Jc) = 2(2(1 − c)) . Feng [19] showed that this formula actually holds for

1 1 hc 1 all 0 < c < 2 . It is worth noting that as c → 2 , hc → 1 so that P (Jc) → 2. When c = 2 ,

hc 1 1 P (Jc) = P ([0, 1]) = 1, so packing measure as a function of c is not continuous at c = 2 .

hc hc 0 As c → 0, hc → 0 so that P (Jc) → 2. When c = 0, P (Jc) = P ({0, 1}) = 2, so packing measure as a function of c is continuous at c = 0. Moreover,

  d h − log 2 log 2 log(2 − 2c) 2 log 2 P c (J ) = 2(2 − 2c) log c + dc c c(log c)2 (2 − 2c) log c

− log 2 log 2 = 2(2 − 2c) log c ((1 − c)(log 2 + log(1 − c)) + c log c) . c(1 − c)(log c)2

90 1 d hc In the interval (0, 2 ), dc P (Jc) = 0 exactly when

(c − 1) log 2 = c log c + (1 − c) log(1 − c). or equivalently when

21−ccc(1 − c)1−c = 1.

hc Thus, we see that P (Jc) obtains its maximum at about c ≈ 0.227.

hc 1 Fix N ≥ 3. We will show that P (Jc) is decreasing on (0, 2N−1 ], which demonstrates 1 1−a that N = 2 is a peculiar case. Suppose 0 < a ≤ b ≤ 2N−1 . Notice, 2 N−1 < 1 since N ≥ 3.

ha hb Then, P (Ja) ≥ P (Jb) is equivalent to

1−b
h log 2 a ≤ N−1 . h 1−a
b log 2 N−1

log 2 1−b 1−b 1−a ha ( N−1 ) It is clear that ha ≤ hb and log(2 ) ≤ log(2 ) < 0. Thus ≤ 1 and 1−a ≥ 1, N−1 N−1 hb log(2 N−1 ) and we are done.

4.4. Packing Measure of Super Separated Sierpi´nskiN-gons

We will now compute the packing measure for limit sets based on regular convex

2 1 polygons, so in this section we always work in R . Fix N ≥ 3 and c1, . . . , cN ∈ (0, N ]. Now let Φ be the IFS consisting of

ϕj(x) = cj · x + (1 − cj) · vj,

  2π(j−1)   2π(j−1)  where vj = cos N , sin N for j = 1,...,N. Note that the points vj form the

vertices of a regular N-gon and lie on the unit circle. The maps ϕj are contractions on X

where X is the regular N-gon with vertices vj. It is sometimes more useful to take X to be B(0, 1). In particular, if N = 3, then the limit set J is a Sierpi´nskitriangle. It then makes sense to refer to the limit set J as a Sierpi´nski N-gon in general. By Hutchinson’s formula, the packing dimension is the value h ≥ 0 that satisfies

N X h cj = 1. j=1

1 Note that h ≤ 1 since c1, . . . , cN ∈ (0, N ].

91 Lemma 4.4.1. Suppose Φ is a Sierpi´nski N-gon IFS as described above. Then Φ is super separated. Moreover, 4 |ϕ (0) − ϕ (0)| ≥ i i+1 N for all N ≥ 4 and 1 ≤ i < N.

Proof. Define `N,N+1 = `N,1 and `0,1 = `N,1. For any i 6= j, `i,j = |ϕi(vi0 ) − ϕj(vj0 )| for some i0, j0 = 1,...,N. Moreover, `j,min = min{`j,j+1, `j−1,j} for j = 1,...,N. Therefore, it is enough to show super separation for the case c = cj for all j = 1,...,N. Since v1, . . . , vN are evenly spaced around the unit circle, it is clear from the symmetry of J that `j,min = `1,2 for all j = 1,...,N. Let ` = `1,2. It then suffices to show c|J| ≤ `. First consider N = 3. The side length of the N-gon in our parametrization is √ π 1 2 sin( 3 ) = 3. Since |J| ≤ 2 and c ≤ 3 , we obtain √ √ 2 2 ` = 3 − 2c ≥ 3 − > ≥ c|J|. 3 3

Now, assume N ≥ 4. Notice that

` ≥ dinf (ϕ1(B(0, 1)), ϕ2(B(0, 1))) = |ϕ1(0) − ϕ2(0)| − 2c.

Now,

 2π 2  2π 2 |ϕ (0) − ϕ (0)|2 = (1 − c) − (1 − c) cos + (1 − c) sin 1 2 N N  2π  2π  2π  = (1 − c)2 1 − 2 cos + cos2 + sin2 N N N  2π  = 2(1 − c)2 1 − cos N  π  = 4(1 − c)2 sin2 . N

Thus,  π  |ϕ (0) − ϕ (0)| = 2(1 − c) sin . 1 2 N 1 Since c ≤ N , we have  1   π  2 ` ≥ 2 1 − sin − . N N N

92 2 Since c|J| ≤ N , it suffices to show for N ≥ 4,  π  (N − 1) sin ≥ 2. N The inequality clearly holds for N = 4. Now, d  π   π  π(N − 1) cos π
(N − 1) sin = sin − N . dN N N N 2 So, it is enough to know for N ≥ 4  π  π(N − 1) cos π
sin ≥ N , N N 2 or equivalently  π  π π tan ≥ − . N N N 2 π
π Since tan N ≥ N by the Taylor series at 0, we are done. 

Theorem 4.4.2 (♣). Suppose Φ is a Sierpi´nski 3-gon IFS (i.e. J is a Sierpi´nskitriangle).

1 Assume 0 < c1 ≤ c2 ≤ c3 ≤ 3 . Then √ √ √ ( h h h h h h ) h 2 ( 3(1 − c3)) 2 ( 3(1 − c3)) 2 ( 3(1 − c2)) P (J) = max h , h , h . c1 c2 c3

Proof. By Lemma 4.4.1 and Theorem 4.2.1, there is some x0 ∈ J and r0 ∈ R(x0) with

h P (J) = den(x0, r0). Moreover, x0 ∈ ϕk0 (J) for exactly one k0 = 1, 2, 3 and ck0 |J| ≤ r ≤ |J|.

Let y ∈ ϕj(J) for some j 6= k0. Since v1, v2, and v3 form the vertices of an equilateral triangle

and ϕk0 (B(0, 1)) ∩ ϕj(B(0, 1)) = ∅, we obtain by basic geometry that |x0 − y| < |vk0 − y|.

Thus, m(B(x0, r0)) ≥ m(B(vk0 , r0)), and hence den(x0, r0) ≤ den(vk0 , r0). By Theorem

4.2.1, we can assume that x0 = vk0 and r0 ∈ R(x0). Notice,

R(v1) ⊆ {dinf (v1, ϕ2(J)), dinf (v1, ϕ3(J)), |J|},

R(v2) ⊆ {dinf (v2, ϕ1(J)), dinf (v2, ϕ3(J)), |J|},

R(v3) ⊆ {dinf (v3, ϕ1(J)), dinf (v3, ϕ2(J)), |J|}.

Suppose r0 = |J|. Then

h (2r0) h h den(vk0 , r0) = = 2 |J| . m(B(vk0 , r0))

93 Let r1 = minj6=k0 dinf (vk0 , ϕj(J)). Notice r1 ∈ R(vk0 ), and since Φ is super separated by Lemma 4.4.1,

h (2r1) den(vk0 , r1) = m(B(vk0 , r1)) 2hrh = 1 ch k0 2h`h ≥ k0,min ch k0 2h(c |J|)h ≥ k0 ch k0 = 2h|J|h

= den(vk0 , r0).

By maximality of den(vk0 , r0), we are safe to assume that r0 6= |J|.

We will now show that den(vk0 , r0) must be equal to one of den(v1, dinf (v1, ϕ3(J))), √ den(v2, dinf (v2, ϕ3(J))), or den(v3, dinf (v3, ϕ2(J))). Since |vi − vj| = 2 sin(π/3) = 3 for all i 6= j, the result will follow by noticing that √ h h 2 ( 3(1 − c3)) den(v1, dinf (v1, ϕ3(J))) = h c1 √ h h 2 ( 3(1 − c3)) den(v2, dinf (v2, ϕ3(J))) = h c2 √ h h 2 ( 3(1 − c2)) den(v3, dinf (v3, ϕ2(J))) = h . c3

Notice that ϕ2(J) and ϕ3(J) do not straddle B(v1, dinf (v1, ϕ3(J))), ϕ1(J) and ϕ3(J) do

not straddle B(v2, dinf (v2, ϕ3(J))), and ϕ1(J) and ϕ2(J) do not straddle B(v3, dinf (v3, ϕ2(J))).

First consider B(v1, dinf (v1, ϕ2(J))). If c2 = c3, then

dinf (v1, ϕ2(J)) = dinf (v1, ϕ3(J)),

and we are done. If c2 < c3, then ϕ3(J) straddles B(v1, dinf (v1, ϕ2(J))), and by Theorem

4.2.1, B(v1, dinf (v1, ϕ2(J))) is not a ball of maximum density.

94 Now consider B(v2, dinf (v2, ϕ1(J))). If c1 = c3, then

dinf (v2, ϕ1(J)) = dinf (v2, ϕ3(J)),

and we are done. If c1 < c2 < c3, then ϕ3(J) straddles B(v2, dinf (v2, ϕ1(J))), and by Theorem

4.2.1, B(v2, dinf (v2, ϕ1(J))) is not a ball of maximum density.

Finally, consider B(v3, dinf (v3, ϕ1(J))). If c1 = c2, then

dinf (v3, ϕ1(J)) = dinf (v3, ϕ2(J)),

and we are done. If c1 < c2, then ϕ2(J) straddles B(v3, dinf (v3, ϕ1(J))), and by Theorem

4.2.1, B(v3, dinf (v3, ϕ1(J))) is not a ball of maximum density. This finishes the proof. 

It is more difficult to compute the packing measure for N ≥ 4. In the proof of Theorem 4.4.2, we assumed that the ball of maximum density was obtained at a vertex of

the triangle formed by v1, v2, and v3. This does not have to be the case for N ≥ 4. Suppose √ π 1 1 1 N = 4. Note that |v1 −v2| = 2 sin 4 = 2. Let c1 = c3 = 4 and let 0 < ε < 4 . Let c2 = 4 −ε, √ 3 and let r = dinf (ϕ1(v4), ϕ2(J)). Notice that r = |ϕ1(v4) − ϕ2(v4)|. Moreover, 4 2 < r and

h h m(B(ϕ1(v4), r)) ≤ m(ϕ1(J)) + m(ϕ4(J)) = c1 + c4 .

k Now, dinf (v1, ϕ2(J)) = |v1 − ϕ2(v1)| < r. Then for some large enough k ≥ 1, ϕ2 ◦ ϕ1(J) ⊆ k B(v1, r). Fix such a k and choose c4 so that 0 < c4 < c2c1. Then,

k m(B(v1, r)) ≥ m(ϕ1(J)) + m(ϕ2 ◦ ϕ1(J))

h k h = c1 + (c2c1)

h h > c1 + c4

≥ m(B(ϕ1(v4), r)).

Therefore, den(v1, r) < den(ϕ1(v4), r). This shows that we can not repeat the arguments used in Theorem 4.4.2 and assume that the ball of maximum density occurs at a vertex when

N ≥ 4 and the contraction ratios cj are not all the same.

95 When the contraction ratios are all the same, i.e. c = cj for j = 1,...,N, the symmetry of J will reflect the symmetry of the regular N-gon. We can exploit this symmetry to compute the packing measure of J provided c is small enough.

Let t ≥ 2. If N = 2t or N = 2t + 1, we define the following constants: 2π
1 − cos N (1) C1 = 2π
, 3 − cos N πt
 π(t−1)  sin N − sin N (2) C2 = , πt
 π(t−1)  2 + sin N − sin N π
sin N (3) C3 = π
. 3(sin N + 1)

Lemma 4.4.3. Suppose Φ is a Sierpi´nski N-gon IFS, 0 < c ≤ C1, and c = cj for all j = 1,...,N. Then,

 d ϕ2(B(0, 1)) ⊆ (x, y) ∈ R : −1 < x ≤ 1 − 2c, c < y ≤ 1 .

d In particular, if x0 ∈ ϕ1(J) ∩ {(x, y) ∈ R : y ≥ 0}, x0 6= v1, and y ∈ ϕ2(J), then

|x0 − y| < |v1 − y|.

Proof. By Lemma 4.4.1,

2 2 |ϕ (0) − ϕ (0)| − 2c > |ϕ (0) − ϕ (0)| − ≥ ≥ 2c. 2 N 2 1 N N

Thus, by the symmetry of J across the x-axis,

 d ϕ2(B(0, 1)) ⊆ (x, y) ∈ R : c < y ≤ 1 .

Now, to finish the proof we simply need to verify

ϕ2(0) + c ≤ 1 − 2c.

Then, by rearranging the inequality,

2π  (1 − c) cos + c ≤ 1 − 2c N

96 we obtain 2π
1 − cos N c ≤ 2π
= C1. 3 − cos N Therefore, by our assumptions, we are done.

d If x0 ∈ ϕ1(J) ∩ {(x, y) ∈ R : y ≥ 0}, x0 6= v1, and y ∈ ϕ2(J), then the angle ∠v1x0y is at least 90 degrees, and in particular,

|x0 − y| < |v1 − y|.



Lemma 4.4.4. Suppose Φ is a Sierpi´nski N-gon IFS, 0 < c ≤ C2, and c = cj for all

j = 1,...,N. Then, for all x0 ∈ ϕ1(J), r0 ∈ R(x0), y ∈ ϕ1(B(0, 1)), and i = 1,...,N, if

ϕi(B(0, 1)) ⊆ B(x0, r0), then ϕi(B(0, 1)) ⊆ B(y, r0).

Proof. Let x0 ∈ ϕ1(J), r0 ∈ R(x0), y ∈ ϕ1(B(0, 1)), and i = 1,...,N such that ϕi(B(0, 1)) ⊆

B(x0, r0). If r0 = |J|, then clearly ϕi(B(0, 1)) ⊆ B(y, r0). Assume r0 < |J|. Then

r0 = dinf (x0, ϕk0 ) for some k0 = 2,...,N. Hence, ϕi(J) is the right right of ϕk0 (J).

Since r0 ≥ |ϕ1(J)| by the definition of R(x0), the result holds for k0 = 2. We may then

assume k0 ≥ 3. Assume N = 2t or N = 2t + 1 for some t ≥ 2. Since J is symmetric about

the x-axis, we may assume that k0 ≤ t + 1 by reflecting x0 across the x-axis if necessary. Now,

r0 > |ϕ1(0) − ϕk0 (0)| − 2c.

In order to show ϕi(J) ⊆ B(y, r0), it suffices to show dsup(y, ϕi(J)) ≤ r0. Therefore it suffices to show

|ϕ1(0) − ϕi(0)| + 2c ≤ |ϕ1(0) − ϕk0 (0)| − 2c.

Since

|ϕ1(0) − ϕi(0)| + 2c ≤ |ϕ1(0) − ϕk0−1(0)| + 2c, it suffices to show

|ϕ1(0) − ϕk0−1(0)| + 2c ≤ |ϕ1(0) − ϕk0 (0)| − 2c.

97 In other words, we need to verify

2π(k − 2) 2π(k − 1) 2(1 − c) sin 0 + 2c ≤ 2(1 − c) sin 0 − 2c. N N

Rearranging, we obtain

 π(k0−1)   π(k0−2)  sin N − sin N c ≤ .  π(k0−1)   π(k0−2)  2 + sin N − sin N

By the mean value theorem, there is i − 1 ≤ ai ≤ i for all i = 3, . . . , t + 1 such that π2 π(a − 1) π(i − 1) π(i − 2) cos i = sin − sin . N 2 N N N

Since the left hand side is decreasing on [3, t+1], the minimum occurs at i = t+1. Therefore, it suffices to know πt
 π(t−1)  sin N − sin N c ≤ = C2, πt
 π(t−1)  2 + sin N − sin N which we have assumed. 

Lemma 4.4.5. Suppose Φ is a Sierpi´nski N-gon IFS, 0 < c ≤ C2, and c = cj for all

2 j = 1,...,N. Let N = 2t or N = 2t+1 for some t ≥ 2. If x0 ∈ ϕ1(J)∩{(x, y) ∈ R : y ≥ 0} and r0 ∈ R(x0), then either r0 = |J| or r0 = dinf (x0, ϕk0 (J)) for some k0 = 2, . . . , t + 1. Moreover, k −1 N [0 [ B(x0, r0) ∩ J = ϕ1(J) ∪ ϕi(J) ∪ ϕi(J).

i=2 i=N−k0+3 In particular, 2k − 3 m(B(x , r )) = 0 . 0 0 N

2 Proof. Suppose x0 ∈ ϕ1(J) ∩ {(x, y) ∈ R : y ≥ 0}, r0 ∈ R(x0), and r0 < |J|. Then, r0 =

dinf (x0, ϕk0 ) for some k0 = 2,...,N. If x0 is on the x-axis, then the result is immediate from

2 the symmetry of J and the definition of R(x0). Assume x0 ∈ ϕ1(J) ∩ {(x, y) ∈ R : y > 0}.

Let x0 be the reflection of x0 across the x-axis. Then x0 ∈ J, and by the symmetry of J, R(x0) = R(x0). For each r ∈ R(x0) define

Mr = {1 ≤ i ≤ N : ϕi(J) ⊆ B(x0, r)}

98 and

M r = {1 ≤ i ≤ N : ϕi(J) ⊆ B(x0, r)}.

By Lemma 4.4.4, Mr = M r for every r ∈ R(x0) since R(x0) = R(x0). By symmetry

across the x-axis, this implies that ϕi(J) ⊆ B(x0, r) if and only if ϕj(J) ⊆ B(x0, r), where

vi = vj. Then, we have k0 ≤ t + 1 and

k −1 N [0 [ B(x0, r0) ∩ J = ϕ1(J) ∪ ϕi(J) ∪ ϕi(J).

i=2 i=N−k0+3



Lemma 4.4.6. Suppose Φ is a Sierpi´nski N-gon IFS, 0 < c ≤ C3, and c = cj for all

j = 1,...,N. Then, for all x ∈ ϕ1(J),

2 d (x, ϕ (J)) ≥ |v − v |. inf 2 3 1 2

Proof. Let x ∈ ϕ1(J). Notice,

 π  d (x, ϕ (J)) ≥ |ϕ (0) − ϕ (0)| − 2c = 2(1 − c) sin − 2c, inf 2 1 2 N

and 2 2 2π  |v − v | = sin . 3 1 2 3 N Thus, we see that we must have

π
sin N c ≤ π
= C3. 3(sin N + 1)



We now make an observation about the geometry of Sierpi´nski N-gons. Suppose

1 N = 6 and c = 6 . Then,

dsup(v1, ϕj(J)) = |v1 − ϕ3(v4)| ≈ 1.74

and 5 d (v , ϕ (J)) = |v − ϕ (v )| = ≈ 1.67, inf 1 4 1 4 1 3

99 whence ϕ3(J) 6⊆ B(v1, dinf (v1, ϕ4(J))). In other words, ϕ3(J) straddles

B(v1, dinf (v1, ϕ4(J))), and by Theorem 4.2.1, B(v1, dinf (v1, ϕ4(J))) is not a ball of maxi- mum density. Thus, Theorem 4.2.1 allows us to ignore the difficult problem of computing m(B(v1, dinf (v1, ϕ4(J)))).

Theorem 4.4.7 (♣). Suppose Φ is a Sierpi´nski N-gon IFS, 0 < c ≤ min{C1,C2,C3}, and c = cj for all j = 1,...,N. Then,

h h h P (J) = N2 (dinf (v1, ϕ2(J))) .

Proof. By Lemma 4.4.1 and Theorem 4.2.1, there is some x0 ∈ J and r0 ∈ R(x0) with

h P (J) = den(x0, r0). Since c = cj for all j = 1,...,N and v1, . . . , vN are evenly spaced on the unit circle, we can assume without loss of generality that x0 ∈ ϕ1(J). Since J is

 d symmetric about the x-axis, we can assume that x0 ∈ ϕ1(J) ∩ (x, y) ∈ R : y ≥ 0 . The case N = 3 follows from Theorem 4.4.2. Let t ≥ 2 satisfying either N = 2t or N = 2t + 1. Let rj = dinf (x0, ϕj(J)) for all j = 2, . . . , t + 1. By Hutchinson’s formula,

log N h = − log c . Thus,

h (2r2) h h den(x0, r2) = = N2 (dinf (x0, ϕ2(J))) . m(B(x0, r2))

We will first show that den(x0, r2) ≥ den(x0, |J|) and den(x0, r2) ≥ den(x0, rj) for all rj ∈ R(x0),j = 2, . . . , t + 1. Suppose r0 = |J|. Note that

h (2|J|) h h den(x0, |J|) = = 2 |J| . m(B(x0, |J|))

Since Φ is super separated, we have that

|J|h (d (x , ϕ (J)))h ≥ `h ≥ ch|J|h = . inf 0 2 1,min N

Thus |J|h den(x , r ) ≥ N2h = 2h|J|h = den(x , |J|). 0 2 N 0

100 Fix j = 2, . . . , t+1 such that rj ∈ R(x0). By Lemma 4.4.5, we know that m(B(x0, rj)) =

2j−3 N . Thus, h h h (2rj) N2 rj den(x0, rj) = = . m(B(x0, rj)) 2j − 3 Therefore it suffices to show rh j ≤ rh. 2j − 3 2

Since h ≤ 1 and rj ≥ r2, it is enough to show

r j ≤ 2j − 3 r2

for all j = 2, . . . , t + 1. The case j = 2 clearly holds, so we may assume j ≥ 3. Notice,

rj ≤ |ϕ1(0) − ϕj(0)| π(j − 1) = 2(1 − c) sin N π(j − 1) < 2 sin N

= |v1 − vj|.

Thus, j−1 X rj < |v1 − vj| ≤ |vk − vk+1| = (j − 1)|v1 − v2|. k=1 Combining the above inequality with Lemma 4.4.6, we obtain

rj (j − 1)|v1 − v2| 3 ≤ 2 = (j − 1) ≤ 2j − 3, r2 3 |v1 − v2| 2

h and we are done. We conclude that P (J) = den(x0, r2).

Suppose to the contrary x0 6= v1. Let y ∈ ϕ2(J) such that

|v1 − y| = dinf (v1, ϕ2(J)).

Then, by Lemma 4.4.3,

r2 ≤ |x0 − y| < |v1 − y| = dinf (v1, ϕ2(J)).

101 Moreover, 1 m(B(x , r )) = = m(B(v , d (v , ϕ (J)))). 0 2 N 1 inf 1 2 Therefore,

den(x0, r2) < den(v1, dinf (v1, ϕ2(J))),

which contradicts the maximality of den(x0, r2). We conclude that x0 = v1 and

h h h P (J) = den(v1, r2) = N2 (dinf (v1, ϕ2(J))) .



4.5. Packing Measure of Super Separated Sierpi´nskiSimplexes

In this section, we take a quick look at some higher dimensional examples. We will

compute the packing measure for limit sets based on regular simplexes in Rd where d ≥ 2. d Let v1, . . . , vd+1 be the vertices of a regular simplex in R , d ≥ 2, centered at zero, so that 1 |vi − vj| = 1 for all i 6= j. Fix c ∈ (0, d+1 ] and let Φ be the IFS consisting of the maps

ϕj(x) = c · x + (1 − c) · vj

for j = 1, . . . , d+1. It makes sense to call the limit set J a Sierpi´nski d-simplex. For example, if d = 3, then J is a Sierpi´nskitetrahedron. By Hutchinson’s formula, the packing dimension

log (d+1) of J is h = − log c ≤ 1. Since |J| = 1 and `j,min = 1 − 2c for all j, we immediately have that 1 Φ is super separated since c ≤ 3 .

Theorem 4.5.1 (♣). The packing measure of the Sierpi´nski d-simplex defined above is given by

Ph(J) = (d + 1)2h(1 − c)h.

Proof. Once we know the formula for all 0 < h < 1, we can use continuity of packing

1 measure [44] to get the case h = 1. Assume c < 3 . By Theorem 4.2.1, there is some x0 ∈ J h and r0 ∈ R(x0) such that P (J) = den(x0, r0). By the symmetry of J, we may assume that x0 ∈ ϕ1(J). Let y ∈ ϕj(J) for some j = 2, . . . , d + 1. Since ϕ1(B(0, 1)) ∩ ϕj(B(0, 1)) = ∅,

|x0 − y| < |v1 − y|. Thus, m(B(x0, r0)) ≥ m(B(v1, r0)), and hence den(x0, r0) ≤ den(v1, r0).

102 By Theorem 4.2.1, we can assume x0 = v1 and r0 ∈ R(v1). Then, r0 = 1 − c or r0 = |J| = 1. Now, h (2|J|) h h h den(v1, |J|) = = 2 |J| = 2 , m(B(v1, |J|)) and h (2(1 − c)) h h den(v1, 1 − c) = = (d + 1)2 (1 − c) m(B(v1, |J|)) However,  1  (d + 1)2h(1 − c)h ≥ (d + 1)2h 1 − = d2h > 2h. d + 1 Therefore, Ph(J) = (d + 1)2h(1 − c)h.



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