A FREE BOUNDARY PROBLEM, PROJECTIONS OF RANDOM CANTOR SETS, AND THE
    GEOMETRY OF CURVES WITH SMALL INTERSECTION WITH MANY LINES
                                       By
                               Dimitrios Vardakis
                               A DISSERTATION
                                   Submitted to
                           Michigan State University
                   in partial fulfillment of the requirements
                                for the degree of
                     Mathematics—Doctor of Philosophy
                                      2023


                                             ABSTRACT
Finding the geometric properties of a set is a very old problem. The present text consists of three
chapters where we study such properties with techniques involving Complex and Harmonic Analy-
sis, Probability, and Geometric Measure Theory. We specifically deal with a few considerations of
free boundary problems, we calculate the decay rate of the projections of a certain random Cantor
set, and we describe the shape of planar graphs which avoid having too many intersections with a
positive cone of lines.
     To begin with, we introduce Schwarz functions; holomorphic functions on open domains Ω
satisfying 𝑆(𝜁) = 𝜁 on Γ, part of Ω’s boundary. Sakai in 1991 gave a complete characterization
of the boundary of a domain admitting a Schwarz function. In fact, if Ω is simply connected
and Γ = 𝜕Ω ∩ 𝐷 (𝜁, 𝑟), then Γ has to be regular real analytic. Here, we attempt to describe
Γ when the boundary condition is slightly relaxed. In particular, three different scenarios over
a simply connected domain Ω are treated: when 𝑓1 (𝜁) = 𝜁 𝑓2 (𝜁) on Γ with 𝑓1 , 𝑓2 holomorphic
and continuous up to the boundary, when U/V equals certain real analytic function on Γ with
U, V positive and harmonic on Ω and vanishing on Γ, and when 𝑆(𝜁) = Φ(𝜁, 𝜁) on Γ with Φ a
holomorphic function of two variables. It turns out that the boundary piece Γ can be, respectively,
anything from real analytic to merely 𝐶 1 , regular except finitely many points, or regular except for
a measure zero set.
     For the second chapter, we consider a model of randomness for self-similar Cantor sets of finite
and positive 1-Hausdorff measure. We find the sharp rate of decay of the probability that a Buffon
needle lands 𝛿-close to a Cantor set of this particular randomness. Two quite different models of
randomness for Cantor sets, by Peres and Solomyak, and by Shiwen Zhang, appear to have the same
                                                       𝑐
order of decay for the Buffon needle probability:    log 1 . Here, we prove the same rate of decay for
                                                         𝛿
a third model of randomness, which asserts a vague feeling that any “reasonable” random Cantor
                                                                      𝑐
set of positive and finite length will have Favard length of order  log 1 for its 𝛿-neighbourhood. The
                                                                        𝛿
estimate from below was obtained long ago by Mattila.
     In the last chapter, we show the local Lipschitz property for a graph avoiding multiple-point


intersection with lines directed in a given cone. The assumption is much stronger than those of the
well-known Marstrand’s theorem, but the conclusion is much stronger too. Additionally, we find
that a continuous curve with a similar property is 𝜎-finite with respect to Hausdorff length, and we
give an estimate on the Hausdorff measure of each “piece”.


                                     ACKNOWLEDGEMENTS
Before anything else, I need to thank my parents for if it weren’t for their support, nothing from
what I have achieved in my life would have been possible.
    I also want to thank the professors that have taught me and shaped me as a mathematician. I want
to express my deepest gratitude and respect for my teacher Manolis G. Maragakis who taught me
that a person needs to have ethos before anything else. Also, I am grateful to my professors Nikos
Frantzikinakis and Themis Mitsis, who taught me how to think and showed me what Mathematics
are. Last, I am thankful to my advisor Alexander Volberg who introduced me to the great vastness
of the mathematical world of today.
    I want to thank all the friends I made here, who gave life a true flavour amidst the numbness of
East Lansing and of this foreign country. I am greatly grateful to my friend Stratos Tsoukanis who
bore with me throughout all of our conversations and our endless video calls. And I am especially
glad I have met here Estefanía who has given me a fresh view of the world.
    Of course, I want to thank my committee who allowed me to continue with my plans for the
future, as well as José Conde Alonso, Irina Holmes, Ilya Kachkovskiy, Ignacio Uriarte-Tuero, and
Alexander Volberg who wrote my reference letters and helped me make these plans a reality.
    Finally, I am grateful to the staff of the Michigan State University that have helped me all these
years to guide through the different tasks and obstacles those being administrative or related to
teaching. I am especially happy to have met Tsveta Sendova who really tried to take care of us all
graduate students and struggled alongside us in our first teaching steps.
                                                    iv


                               TABLE OF CONTENTS
CHAPTER 1      FREE BOUNDARY PROBLEMS VIA SAKAI’S THEOREM                         . . . . . .  1
     1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  1
     1.2 Polynomials &analytic functions . . . . . . . . . . . . . . . . . . .    . . . . . .  4
     1.3 Nevanlinna domains and inner functions . . . . . . . . . . . . . . .     . . . . . .  9
     1.4 Boundary behaviour of conformal maps in 𝐾𝜃 . . . . . . . . . . . .       . . . . . . 12
     1.5 Holomorphic functions in C2 . . . . . . . . . . . . . . . . . . . . .    . . . . . . 15
     1.6 The U-V problem . . . . . . . . . . . . . . . . . . . . . . . . . .      . . . . . . 19
     1.7 Some open “free boundary” problems in the spirit of Sakai . . . . .      . . . . . . 26
     BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .     . . . . . . 28
CHAPTER 2      THE BUFFON’S NEEDLE PROBLEM FOR RANDOM                         PLANAR
               DISK-LIKE CANTOR SETS . . . . . . . . . . . . . . .            . . . . . . . . 30
     2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
     2.2 Cantor Disks . . . . . . . . . . . . . . . . . . . . . . . . . . .   . . . . . . . . 32
     2.3 Favard Length . . . . . . . . . . . . . . . . . . . . . . . . . . .  . . . . . . . . 36
     2.4 Statement and use of the main lemma . . . . . . . . . . . . . .      . . . . . . . . 37
     2.5 Proving the main lemma . . . . . . . . . . . . . . . . . . . . .     . . . . . . . . 38
     2.6 Comparison with the other random models . . . . . . . . . . . .      . . . . . . . . 44
     BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . . . .     . . . . . . . . 46
CHAPTER 3      GEOMETRY OF PLANAR CURVES INTERSECTING MANY LINES
               IN A FEW POINTS . . . . . . . . . . . . . . . . . . . . . . . . . . .        . 48
     3.1 The statement of the problem . . . . . . . . . . . . . . . . . . . . . . . . . .   . 48
     3.2 An example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .   . 54
     3.3 Hausdorff measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .    . 57
     3.4 Higher dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .    . 66
     3.5 Relationships with perturbation theory . . . . . . . . . . . . . . . . . . . . .   . 67
     BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .     . 70
                                             v


                                                CHAPTER 1
                   FREE BOUNDARY PROBLEMS VIA SAKAI’S THEOREM
1.1   Introduction
    Let 𝐷 (𝜁0 , 𝑟) ⊂ C denote the open disk centred at 𝜁0 ∈ C and of radius 𝑟 > 0. Let Ω be an open
subset of 𝐷 (𝜁0 , 𝑟) where 𝜁0 ∈ Γ = 𝜕Ω ∩ 𝐷 (𝜁0 , 𝑟) is a non-isolated boundary point.
    A Schwarz function of Ω ∪ Γ is a function 𝑆 : Ω ∪ Γ → C holomorphic on Ω and continuous
on Ω ∪ Γ that satisfies
                                            𝑆(𝜁) = 𝜁        on Γ.                                (1.1.1)
    In his Acta Mathematica paper [13], Sakai proved that Schwarz functions completely charac-
terize the shape of Γ. One of the technical tools used was the Phragmén–Lindelöf principle in the
form below, but it is far from being the key to his proof; his paper is full of very subtle tricks.
Theorem 1.1.1. Let Ω be an open set in C and let 𝜁0 be a non-isolated boundary point of Ω. Let 𝑓
be a holomorphic function on Ω and 𝐷 (𝜁0 , 𝛿) a ball satisfying the following:
  (i) lim sup | 𝑓 (𝑧)| ≤ 1 while Ω ∋ 𝑧 → 𝜁 for every 𝜁 ∈ 𝜕Ω ∩ 𝐷 (𝜁0 , 𝛿) \ {𝜁0 } and
 (ii) | 𝑓 (𝑧)| ≤ 𝛼|𝑧 − 𝜁0 | −𝛽 in Ω ∩ 𝐷 (𝜁0 , 𝛿) for some positive constants 𝛼 and 𝛽.
Then,
                                             lim sup | 𝑓 (𝑧)| ≤ 1
while Ω ∋ 𝑧 → 𝜁0 .
    In particular, Sakai proved the following, see [13, Theorem 5.2].
Theorem 1.1.2. Let Ω ⊂ 𝐷 (𝜁0 , 𝑟) be a bounded open set in C and 𝜁0 an non-isolated point of its
boundary, Γ = 𝜕Ω ∩ 𝐷 (𝜁0 , 𝑟). Suppose 𝑆 is a Schwarz function on Ω ∪ Γ, that is,
  (i) 𝑆 is holomorphic on Ω,
 (ii) continuous on Ω ∪ Γ, and
                                                       1


 (iii) 𝑆(𝜁) = 𝜁 on Γ.
Then, for some small 0 < 𝛿 ≤ 𝑟 one of the following must occur (where we set 𝐷 = 𝐷 (𝜁0 , 𝛿)):
  (1) Ω ∩ 𝐷 is simply connected and Γ ∩ 𝐷 is a regular real analytic simple arc through 𝜁0 ;
(2a) Γ ∩ 𝐷 determines uniquely a regular real analytic arc through 𝜁0 ; Γ ∩ 𝐷 is either an infinite
       proper subset of this arc with 𝜁0 as an accumulation point or equal to it; also, Ω ∩ 𝐷 = 𝐷 \ Γ;
(2b) Ω ∩ 𝐷 = Ω1 ∪ Ω2 where Ω1 and Ω2 are (open) simply connected and 𝜕Ω1 ∩ 𝐷 and 𝜕Ω2 ∩ 𝐷
       are regular real analytic simple arcs through 𝜁0 and tangent at 𝜁0 ;
(2c) Ω ∩ 𝐷 is simply connected and Γ ∩ 𝐷 is a regular real analytic simple arc except for a cusp
       at 𝜁0 ; the cusp points into Ω.
Recall that a regular arc means a differentiable arc whose derivative never vanishes and simple
means that it is parametrized by an injective continuous function.
Remarks 1.1.3. Here is an example of a cusp of (2c) at 𝜁0 = 0 with Schwarz function. There
exist analytic functions 𝑇 on {|𝑧| ≤ 𝜂}, for some 𝜂 > 0, that have a zero of order 2 at 0, are
                                                
univalent on closed upper half-disk 𝐾𝜂 ≡ |𝑧| ≤ 𝜂 : Im(𝑧) ≥ 0 , and satisfy Γ ∩ 𝐷 ⊂ 𝑇 (−𝜂, 𝜂)
and 𝑇 (𝐾𝜂 ) ⊂ Ω ∪ Γ. In fact, it is easy to construct such functions. Every such 𝑇 leads to a Schwarz
function on the domain Ω = 𝑇 ({|𝑧| < 𝜂, Im 𝑧 > 0}), which has two analytic arcs forming a cusp Γ
at 0. In order to have 𝑆(𝜁) = 𝜁¯ on Γ, it suffices to have a function analytic in {|𝑧| < 𝜂, Im 𝑧 > 0} and
continuous up to (−𝜂, 𝜂) such that 𝐴(𝑥) = 𝑇 (𝑥), 𝑥 ∈ (−𝜂, 𝜂). Having such an 𝐴 we set 𝑆 = 𝐴 ◦ 𝑇 −1
on Ω. On the other hand, using that 𝑇 is analytic in the whole ball {|𝑧| ≤ 𝜂}, we can choose 𝐴 as
follows: 𝐴(𝑧) = 𝑇 ( 𝑧¯). Moreover, Sakai [13] showed that every Schwarz function on a cusp domain
appears because of an analytic function 𝑇 as above.
     The converse of this theorem also holds, in the sense that if any of the conditions (1), (2a), (2b),
or (2c) is satisfied, then Ω admits a Schwarz function.
     In order to distinguish between the cases, Sakai also showed an auxiliary result [13, Proposition
5.1], which we will also use here.
                                                      2


Theorem 1.1.4. Set 𝐷 ′ = 𝐷 (0, 𝑟). Let Ω′ ⊂ 𝐷 ′ be an open set and 0 an accumulation point of its
boundary, Γ′ = 𝜕Ω′ ∩ 𝐷 ′. Then, for some 𝑟 ′ ≤ 𝑟, either
(1) there exists a Schwarz function, 𝑆𝑡 , of (Ω′ ∪ Γ′) ∩ 𝐷 (0, 𝑟 ′) at 0 if and only if there exists a
     function Φ1 defined on (Ω′ ∪ Γ′) ∩ 𝐷 (0, 𝛿) for some 𝛿 > 0 such that
        (i) Φ1 is holomorphic and univalent in Ω′ ∩ 𝐷 (0, 𝛿),
       (ii) Φ1 is continuous on (Ω′ ∪ Γ′) ∩ 𝐷 (0, 𝛿),
      (iii) Φ1 (𝜁) = |𝜁 | 2 on Γ′ ∩ 𝐷 (0, 𝛿)
     or
(2) there exists a Schwarz function, 𝑆𝑡 , of (Ω′ ∪ Γ′) ∩ 𝐷 (0, 𝑟 ′) at 0 if and only if there exists a
     function Φ2 defined on (Ω′ ∪ Γ′) ∩ 𝐷 (0, 𝛿) for some 𝛿 > 0 such that
        (i’) Φ2 is holomorphic and univalent in Ω′ ∩ 𝐷 (0, 𝛿),
       (ii’) Φ22 is continuous on (Ω′ ∪ Γ′) ∩ 𝐷 (0, 𝛿),
      (iii’) Φ22 (𝜁) = |𝜁 | 2 on Γ′ ∩ 𝐷 (0, 𝛿),
      (iv’) Φ2 (Ω′ ∩ 𝐷 (0, 𝛿)) ∪ (−𝜖, 𝜖) contains a neighbourhood of 0 for 𝜖 > 0.
                                                                                        √︁
In particular, the functions Φ1 , Φ2 are related to 𝑆𝑡 by Φ1 (𝑧) = 𝑧𝑆𝑡 (𝑧) and Φ2 (𝑧) =    𝑧𝑆𝑡 (𝑧).
    Unfortunately, Theorem 1.1.4 is only valid around 0 in this form. Nevertheless, we can
“translate” the setup of Theorem 1.1.2 by setting Ω′ = Ω − 𝜁0 , Γ′ = Γ − 𝜁0 and 𝑆𝑡 (𝑧) = 𝑆(𝑧 + 𝜁0 ) − 𝜁 0
for 𝑧 ∈ Ω′. Then, 𝑆𝑡 is a Schwarz function on Ω′ ∪ Γ′ at 0. Cases (1) of the two theorems correspond
with one another as do (2a), (2b), and (2c) with (2).
    Sakai gave two applications of his results: the first one describes the local structure of the
boundary of quadrature domains, while the second one deals with a free boundary problem of
classical type, namely, what is the boundary of the set of positivity of a smooth non-negative
function in the disk such that Δ𝑢 = 1 on the set {𝑢 > 0}.
                                                   3


    It is natural to wonder how one can derive similar results for other forms of (1.1.1). In this
text, we examine three different scenarios for a simply connected domain Ω. In Sections 1.2 to 1.4
equation (1.1.1) is replaced by
                                   𝑓1 (𝜁) = 𝜁 𝑓2 (𝜁)   for all 𝜁 ∈ 𝜕Ω                              (1.1.2)
where 𝑓1 , 𝑓2 are holomorphic functions continuous up to the boundary. This is closely related to the
model subspaces 𝐾𝜃 and Nevanlinna domains, which will be important here. It is shown that there
are domains so that (1.1.2) holds for which 𝜕Ω is 𝐶 ∞ but not real analytic. Further, in Section 1.5
we replace the quantity 𝜁 𝑓2 (𝜁) with Φ(𝜁, 𝜁), where Φ is a holomorphic function of two variables, to
find that the boundary is locally composed of real analytic arcs. Finally, in Section 1.6 we consider
two positive harmonic functions U and V that are zero on a Jordan arc, Γ, of the boundary. If their
ratio on Γ is equal to a real analytic function of the form | 𝐴| 2 , where 𝐴 is holomorphic, then Γ is
real analytic itself with the possible exception of some cusps.
    Our interests to the problems considered below also was spurred by an application, which
originates from complex dynamics. A certain complex dynamics question naturally brought the
second author to another free boundary problem described in Section 1.6. After that it was very
natural to ask related questions, where the Sakai setup was generalized in yet two other ways. To
our surprise the answers were quite different and required different techniques: from the use of
Nevanlinna domains and pseudo-continuation to multivalued analytic functions.
1.2    Polynomials & analytic functions
    Let Ω be an open domain, 𝜁0 a non-isolated boundary point of Ω, and let Γ = 𝜕Ω ∩ 𝐷 (𝜁0 , 𝑟)
for some 𝑟 > 0. Suppose 𝑆 is a holomorphic function on Ω continuous on Ω ∪ Γ. We start with a
simple yet important case. Instead of (1.1.1), we consider
                                         𝑆(𝜁) = 𝜁 𝑝(𝜁)     on Γ,                                   (1.2.1)
                                                               𝑆(𝑧)
where 𝑝 is a polynomial. We will shortly show that 𝑓 (𝑧) =     𝑝(𝑧)  is, in fact, a Schwarz function on Γ.
                                                     4


Lemma 1.2.1. Assume that 𝑆 : Ω → C is holomorphic on Ω ⊂ 𝐷 (𝜁0 , 𝑟), continuous on Ω ∪ Γ,
and that it satisfies
                                        𝑆(𝜁) = 𝜁 (𝜁 − 𝜁0 ) 𝑛      on Γ.
Then, the function 𝑆𝑡 (𝑧) = 𝑆(𝑧 + 𝜁0 ) − 𝜁0 𝑧 𝑛 is holomorphic on Ω − 𝜁0 ⊂ 𝐷 (0, 𝑟), continuous on
(Ω − 𝜁0 ) ∪ (Γ − 𝜁0 ) and it satisfies
                                         𝑆𝑡 (𝜁) = 𝜁 𝜁 𝑛      on Γ − 𝜁0 .
Proposition 1.2.2. Assume 0 ∈ Γ is a non-isolated boundary point of Ω ⊂ 𝐷 (0, 𝑟) and suppose 𝑆
is a holomorphic function on Ω continuous on Ω ∪ Γ and satisfying
                                             𝑆(𝜁) = 𝜁 𝜁 𝑛      on Γ.
                                                  𝑆(𝑧)
Then, for any positive 𝛿 < 𝑟 the function          𝑧𝑛  is holomorphic on Ω ∩ 𝐷 (0, 𝛿) and continuous on
(Ω ∪ Γ) ∩ 𝐷 (0, 𝛿) \ {0}. Moreover, the following holds while 𝑧 ∈ Ω ∪ Γ \ {0}:
                                                       𝑆(𝑧)
                                                 lim        = 0.
                                                 𝑧→0 𝑧 𝑛
                        𝑆(𝑧)
Proof. The function      𝑧𝑛  is clearly holomorphic on Ω ∩ 𝐷 (0, 𝛿) and continuous on (Ω ∪ Γ) ∩
𝐷 (0, 𝛿) \ {0} for any 𝛿 ∈ (0, 𝑟). It remains to see what happens at 0.
    Fix 𝛿 ∈ (0, 𝑟). Since 𝑆 is bounded on Ω ∩ 𝐷 (0, 𝑟), say by 𝑚, we get
                                   𝑆(𝑧)
                                       𝑛
                                            ≤ 𝑚|𝑧| −𝑛       on Ω ∩ 𝐷 (0, 𝛿)
                                     𝑧
and additionally for any 𝜁 ∈ Γ ∩ 𝐷 (0, 𝛿) \ {0} we have
                                    𝑆(𝑧)
                               lim           = |𝜁 | ≤ 𝛿      while Ω ∋ 𝑧 → 𝜁 .
                                     𝑧𝑛
Hence, by the Phragmén-Lindelöf principle 1.1.1 we obtain
                                           𝑆(𝑧)
                                lim sup            ≤𝛿       while Ω ∋ 𝑧 → 0.
                                             𝑧𝑛
This last inequality is true for any positive 𝛿 < 𝑟 and therefore lim 𝑆(𝑧)  𝑧 𝑛 = 0 as 𝑧 → 0.         □
                                                         5


Corollary 1.2.3. Let 𝑝 be a complex polynomial. Assume that 𝜁0 ∈ Γ is a non-isolated boundary
point of Ω other than zero and suppose 𝑆 is a holomorphic function of Ω ⊂ 𝐷 (𝜁0 , 𝑟) continuous on
Ω ∪ Γ and satisfying
                                           𝑆(𝜁) = 𝜁 𝑝(𝜁)          on Γ.
Set 𝑓 (𝑧) = 𝑆(𝑧)/𝑝(𝑧) on Ω ∪ Γ \ {𝜁0 } and 𝑓 (𝜁0 ) = 𝜁0 . Then, 𝑓 is a Schwarz function of Ω ∪ Γ on
𝐷 (𝜁0 , 𝑟) for sufficiently small 𝑟 > 0.
Proof. Take 𝑟 so small that 𝑝 has no zeros on 𝐷 (𝜁0 , 𝑟) \ {𝜁0 }. If 𝑝(𝜁0 ) ≠ 0, the result is immediate.
    If 𝑝(𝜁0 ) = 0, we only need to show that 𝑓 is continuous on (Ω ∪ Γ) ∩ 𝐷 (𝜁0 , 𝑟). Denote by 𝑛
the order of 𝜁0 as a zero of 𝑝 and consider the function
                                                            (𝑧 − 𝜁0 ) 𝑛
                                            𝑆𝑛 (𝑧) = 𝑆(𝑧)               .
                                                               𝑝(𝑧)
𝑆𝑛 is holomorphic on Ω, is continuous on Ω ∪ Γ, and satisfies
                                        𝑆𝑛 (𝜁) = 𝜁 (𝜁 − 𝜁0 ) 𝑛       on Γ.
From Lemma 1.2.1 we get
                                        (𝑆𝑛 )𝑡 (𝜁) = 𝜁 𝜁 𝑛      on Γ − 𝜁0
and from Proposition 1.2.2 we deduce that while
                             (𝑆𝑛 )𝑡 (𝑧)                   𝑆𝑛 (𝑧 + 𝜁0 ) − 𝜁0 𝑧 𝑛
                         lim             = 0   =⇒   lim                         =0
                         𝑧→0    𝑧𝑛                  𝑧→0            𝑧𝑛
                                                           𝑆𝑛 (𝑧) − 𝜁0 (𝑧 − 𝜁0 ) 𝑛
                                               =⇒ lim                              =0
                                                    𝑧→𝜁0         (𝑧 − 𝜁0 ) 𝑛
                                                                            𝑆𝑛 (𝑧)
                                               =⇒ lim 𝑓 (𝑧) = lim                   = 𝜁0
                                                    𝑧→𝜁0           𝑧→𝜁0 (𝑧 − 𝜁0 ) 𝑛
𝑧 ∈ Ω, and the conclusion follows.                                                                     □
    Notice that the same proof works with 𝑝 replaced by any function 𝐹 that is analytic in a
neighbourhood of 𝜁0 . This along with Lemma 1.2.1 give us the following corollary.
                                                         6


Corollary 1.2.4. Assume 𝜁0 ∈ Γ is a non-isolated boundary point of Ω ⊂ 𝐷 (𝜁0 , 𝑟). Suppose 𝐹
is a function analytic around 𝜁0 and 𝑆 is a holomorphic function on Ω continuous on Ω ∪ Γ and
satisfying
                                           𝑆(𝜁) = 𝜁 𝐹 (𝜁)        on Γ.
Set 𝑓 (𝑧) = 𝑆(𝑧)/𝐹 (𝑧) on (Ω∪ Γ) ∩ 𝐷 (𝜁0 , 𝛿) \ {𝜁0 } for some sufficiently small 𝛿 > 0 and 𝑓 (𝜁0 ) = 𝜁0 .
Then, 𝑓 is a Schwarz function of Ω ∪ Γ on 𝐷 (𝜁0 , 𝛿).
    The converse of this corollary also holds true in the sense that if Γ has certain shape, in
particular, if it satisfies (1), (2a), (2b), or (2c) of 1.1.2, then there is a Schwarz function 𝑓 of Ω ∪ Γ
at 𝜁0 such that 𝑆(𝜁) = 𝜁 𝐹 (𝜁) on Γ where 𝑆 = 𝐹 𝑓 .
    In fact, we can slightly modify the same proof to get a little more, again through the Phragmén-
Lindelöf principle 1.1.1.
Corollary 1.2.5. Let 𝑝 be a polynomial, 𝐹 a function analytic in a neighbourhood of Ω̄, and 𝑆 a
function holomorphic on the (bounded) set Ω and continuous on Ω ∪ Γ. Suppose that for all 𝜁 ∈ Γ
we have
                                              𝑆(𝜁) = 𝑝(𝜁)𝐹 (𝜁).
Then, for every non-isolated point 𝜁0 of the boundary Γ for which 𝑝′ (𝜁0 ) ≠ 0, there is some 𝛿 > 0
such that the function 𝑝 −1 (𝑆/𝐹) is a Schwarz function of Ω ∪ Γ on 𝐷 (𝜁0 , 𝛿).
    We wish to examine what happens in the more general case where 𝑝 in (1.2.1) is replaced
with any analytic function of Ω continuous on its boundary, but not necessarily analytic on that
boundary. More specifically, suppose that 𝑓1 and 𝑓2 are functions analytic on Ω, continuous on
Ω ∪ Γ, and satisfying
                                          𝑓1 (𝜁) = 𝜁 𝑓2 (𝜁)      on Γ.                              (1.2.2)
As above, if 𝑓2 (𝜁0 ) ≠ 0, the function 𝑓 = 𝑓1 / 𝑓2 is a Schwarz function around 𝜁0 ∈ Γ and no issues
arise. However, if 𝑓2 (𝜁0 ) = 0, the situation is very complicated in general.
    We start with a lemma analogous to Lemma 1.2.1:
                                                        7


Lemma 1.2.6. Assume that 𝑓1 , 𝑓2 : Ω → C are holomorphic on Ω ⊂ 𝐷 (𝜁0 , 𝑟), continuous on
Ω ∪ Γ, and that they satisfy
                                               𝑓1 (𝜁) = 𝜁 𝑓2 (𝜁)     on Γ.
Then, there exist functions ( 𝑓1 )𝑡 and ( 𝑓2 )𝑡 holomorphic on Ω − 𝜁0 , continuous on (Ω − 𝜁0 ) ∪ (Γ − 𝜁0 )
and such that
                                      ( 𝑓1 )𝑡 (𝜁) = 𝜁 ( 𝑓2 )𝑡 (𝜁)     on Γ − 𝜁0 .
If additionally 𝑓2 (𝜁0 ) = 0, then ( 𝑓2 )𝑡 (0) = 0.
Proof. Define ( 𝑓1 )𝑡 by
                                      ( 𝑓1 )𝑡 (𝑧) = 𝑓1 (𝑧 + 𝜁0 ) − 𝜁0 𝑓2 (𝑧 + 𝜁0 ).
Then for 𝜁 ∈ Γ − 𝜁0 we have
                                 ( 𝑓1 )𝑡 (𝜁) = 𝑓1 (𝜁 + 𝜁0 ) − 𝜁0 𝑓2 (𝜁 + 𝜁0 )
                                               = 𝜁 + 𝜁0 𝑓2 (𝜁 + 𝜁0 ) − 𝜁0 𝑓2 (𝜁 + 𝜁0 )
                                               = 𝜁 𝑓2 (𝜁 + 𝜁0 )
Setting ( 𝑓2 )𝑡 (𝑧) = 𝑓2 (𝑧 + 𝜁0 ), we have the desired identity.
     Clearly, ( 𝑓1 )𝑡 (0) = 0 and also if 𝑓2 (𝜁0 ) = 0, ( 𝑓2 )𝑡 (0) = 0.                                □
Abusing the notation, we denote these new functions again by 𝑓1 and 𝑓2 .
     It remains to show a result analogous to Corollary 1.2.3 with 𝑝 replaced by 𝑓2 . In particular,
we would like to show that the function 𝑓 = 𝑓1 / 𝑓2 is holomorphic on Ω, continuous on Γ, and that
it satisfies
                                                   𝑓1 (𝜁)
                                       𝑓 (𝜁) =            =𝜁      for all 𝜁 ∈ Γ.
                                                   𝑓2 (𝜁)
However, the limit of 𝑓1 (𝑧)/ 𝑓2 (𝑧) as Ω ∋ 𝑧 → 0 may even fail to exist when 𝑓2 (0) = 0, and
we cannot apply the Phragmén-Lindelöf principle here. We will need to see this problem from a
different scope.
                                                              8


1.3     Nevanlinna domains and inner functions
      We recall that a bounded simply connected domain Ω is called a Nevanlinna domain if there
exist bounded holomorphic functions 𝑓1 , 𝑓2 in Ω such that
                                                        𝑓1 (𝜑(𝑧))
                                              𝜑(𝑧) =
                                                        𝑓2 (𝜑(𝑧))
for almost every 𝑧 ∈ T = {𝑧 : |𝑧| = 1}, where 𝜑 is a conformal mapping of the unit disk onto Ω.
Note that this definition does not imply any additional regularity (for instance, continuity) of the
functions 𝑓1 , 𝑓2 on 𝜕Ω.
      We will restrict the above situation, and suppose there are holomorphic functions 𝑓1 , 𝑓2 : Ω → C
continuous up to the boundary that satisfy
                                       𝑓1 (𝜁) = 𝜁 𝑓2 (𝜁)      for 𝜁 ∈ Γ.                           (1.3.1)
      In order to better understand the situation, we rewrite (1.3.1) as
                                                  𝑓1 (𝜁)
                                                          = 𝜁,                                    (1.3.1’)
                                                  𝑓2 (𝜁)
which is now fulfilled almost everywhere on Γ except for the closed set Γ ∩ 𝑓2−1 {0}, which has zero
measure. Then, Ω is what we call a strong Nevanlinna domain and if such 𝑓1 and 𝑓2 exist, the ratio
 𝑓1 / 𝑓2 is unique thanks to the Lusin-Privalov uniqueness theorem.
      Let 𝜙 : D → Ω be a conformal map and consider the functions 𝐹1 = 𝑓1 ◦ 𝜙 and 𝐹2 = 𝑓2 ◦ 𝜙.
Formulas (1.3.1) and (1.3.1’) transform respectively to
                                             𝐹1 (𝜁) = 𝜙(𝜁)𝐹2 (𝜁)                                   (1.3.2)
and
                                                𝐹1 (𝜁)
                                                        = 𝜙(𝜁)                                    (1.3.2’)
                                                𝐹2 (𝜁)
both of which hold true in the sense of angular boundary values almost everywhere on T, because 𝜙
may fail to extend “nicely” to D̄. By the factorization theorem, we can write 𝐹1 and 𝐹2 in D as
                                       𝐹1 = 𝜃 1 F1    and    𝐹2 = 𝜃 2 F2                           (1.3.3)
                                                       9


where the F𝑖 are the outer factors of 𝐹𝑖 and the 𝜃 𝑖 are their inner factors. Since 𝐹1 , 𝐹2 ∈ 𝐻 ∞ , also
F1 , F2 ∈ 𝐻 ∞ , and from (1.3.2’) we get
                                              𝜃 1 (𝜁) F1 (𝜁)
                                                              = 𝜙(𝜁),                                (1.3.4)
                                              𝜃 2 (𝜁) F2 (𝜁)
almost everywhere on T in the sense of angular boundary values. We distinguish between two
cases: either 𝜃 2 divides 𝜃 1 , that is, 𝜃 1 /𝜃 2 ∈ 𝐻 ∞ , or it does not.
1.3.1    𝜃2 | 𝜃1.
     Let ℎ = 𝜃 1 /𝜃 2 ∈ 𝐻 ∞ . Then, the function (ℎF1 )/F2 belongs to the class 𝑁 + , defined as
                                                                                 
                               +       𝑓               ∞
                              𝑁 =         : 𝑓 , 𝑔 ∈ 𝐻 , 𝑔 is an outer function ,
                                      𝑔
and its (angular) boundary values are equal almost everywhere on T to the (angular) boundary
values of 𝜙. However, since Ω is bounded, we see that 𝜙 ∈ 𝐿 ∞ (T, 𝑚) where 𝑚 is the normalized
Lebesgue measure on T. Smirnov’s Theorem tells us that in fact (ℎF1 )/F2 ∈ 𝐻 ∞ . Therefore, we
have a bounded holomorphic function on the disk that is equal to 𝜙 almost everywhere on T. This
is impossible whenever 𝜙 is a bounded holomorphic function on D.
     We are necessarily left with the other case.
1.3.2    𝜃2 ∤ 𝜃1.
     We begin with some notation and definition which will be important for the rest of this text.
     Let D𝑒 = Cb \ D̄. For any function ℎ : D → C we define e          ℎ as
                                                  ℎ(𝑧) = ℎ(1/𝑧).
                                                  e
The notation 𝐻   e will stand for a function 𝐻     e : D𝑒 → C and we will write 𝐻 instead of 𝐻     e for the
                                                                                                   e
function 𝐻 e(1/𝑧). Observe that ℎ ∈ 𝐻 ∞ if and only if e          ℎ ∈ 𝐻 ∞ (D𝑒 ), and ℎ(0) = 0 if and only if
ℎ(∞) = 0.
e
     We will also consider the backward shift operator, B : 𝐻 𝑝 → 𝐻 𝑝 , for 𝑝 ∈ [1, ∞), that is
                                                         𝑓 (𝑧) − 𝑓 (0)
                                             B : 𝑓 ↦→                  .
                                                                𝑧
                                                         10


Definition 1.3.1. Let 𝑓 be a meromorphic function on D. We say that 𝑓 admits pseudo-continuation
(across T) if there exists another meromorphic function 𝑔 on D𝑒 such that 𝑓 = 𝑔 almost everywhere
(on T) in the sense of non-tangential limits.
    The pseudo-continuation of 𝑓 is called of bounded type or a Nevanlinna-type pseudo-continua-
tion if 𝑔 is of the form 𝑔 = ℎ1 /ℎ2 for some ℎ1 , ℎ2 ∈ 𝐻 ∞ (D𝑒 ).
Definition 1.3.2. A function 𝑓 ∈ 𝐻 𝑝 is called a cyclic vector for B, or simply cyclic for B if the set
{B 𝑛 𝑓 }∞
        𝑛=0 spans the space 𝐻 .
                                𝑝
    The following important result is due to Douglas, Shapiro, and Shields.
Theorem 1.3.3. Consider 1 ≤ 𝑝 < ∞. A function 𝑓 ∈ 𝐻 𝑝 is not cyclic for B if and only if 𝑓 has a
pseudo-continuation of bounded type.
    In the case when 𝑝 = 2, it is known that any non-cyclic function of B belongs to a proper B-
invariant subspace. As a consequence of Beurling’s theorem, these spaces are of the form (𝜃𝐻 2 ) ⊥
and are known as model spaces and denoted by 𝐾𝜃 . Here we will need the fact that
                                  𝐾𝜃 = (𝜃𝐻 2 ) ⊥ = 𝐻 2 (T) ∩ 𝜃𝐻02 (T),
where in the last identity we mean the boundary values of the corresponding functions and where
𝐻02 = { 𝑓 ∈ 𝐻 2 : 𝑓 (0) = 0}.
    Now, we can proceed with the case when 𝜃 2 ∤ 𝜃 1 :
    After dividing both 𝜃 1 and 𝜃 2 by their greatest common divisor, we may assume that 𝜃 1 and 𝜃 2
have no common zeros and that the Borel supports of their singular measures are disjoint. Much
as above, we see that the function 𝐹 = (𝜃 1 F1 )/F2 = 𝐹1 /F2 belongs the class 𝑁 + and thus 𝐹 ∈ 𝐻 ∞ ,
because 𝜃 2 𝜙 ∈ 𝐿 ∞ (T, 𝑚). Then the following is true in the sense of angular boundary values for
                                                   11


almost every 𝜁 ∈ T:
                                                     𝜃 1 (𝜁)F1 (𝜁)       𝐹 (𝜁)
                                             𝜙(𝜁) =                 =
                                                     𝜃 2 (𝜁)F2 (𝜁) 𝜃 2 (𝜁)
                                     ⇐⇒ 𝜙(𝜁) = 𝜃 2 (𝜁)𝐹 (𝜁)                                      (1.3.5)
                                                      e(𝜁)
                                                      𝐹
                                     ⇐⇒ 𝜙(𝜁) =               .                                   (1.3.6)
                                                     𝜃 2 (𝜁)
                                                     e
Since 𝐹,
       ee  𝜃 2 ∈ 𝐻 ∞ (D𝑒 ), we see that 𝜙 ∈ 𝐻 ∞ ⊂ 𝐻 2 admits pseudo-continuation across T of bounded
type, and Theorem 1.3.3 shows that 𝜙 is not cyclic for B. So, it has to belong to some model space
𝐾𝜃 . See [7, Theorem 1] for more details. In fact, from (1.3.5) and because we “need” to have
𝐹 (0) = 0, it follows that either
                            𝜙 ∈ 𝐾𝜃 2 if 𝜃 1 (0) = 0,    or     𝜙 ∈ 𝐾 𝑧𝜃 2 if 𝜃 1 (0) ≠ 0.
1.4   Boundary behaviour of conformal maps in 𝐾𝜃
    In this section we show that Theorem 1.1.2 fails when condition (iii) is replaced by (1.3.1). To
this end, we will find a simply connected domain Ω and a conformal map 𝜙 : D → Ω continuous up
the boundary that has a pseudo-continuation of bounded type and is smooth but not real analytic on
T. The functions participating in this pseudo-continuation will also be continuous on the boundary.
First, we go one step back and work with Nevanlinna domains. Thanks to [7, Theorem 1] by
Fedorovskiy, this is equivalent to studying the model subspaces, 𝐾𝜃 , for different inner functions 𝜃.
    If 𝜃 (𝑧0 ) = 0 for some 𝑧0 ∈ D, the function
                                                    1
                                       𝜙(𝑧) =             ∈ 𝐾𝜃 ∩ 𝐶 ∞ (T)
                                                 1 − 𝑧0 𝑧
has bounded type pseudo-continuation across T and thus 𝜙(D) is a Nevanlinna domain. In fact, 𝜙
can be analytically extended on the whole closed disk, D̄, and 𝜙(T) is real analytic. On the other
hand, in a series of papers, [10, 4, 7, 11, 12, 3], it has been shown that the boundary of a Nevanlinna
domain can be “arbitrarily bad”. In particular, it can be nowhere analytic [10], of class 𝐶 1 but not
in any 𝐶 1,𝛼 for no 𝛼 > 0 [7], or even non-rectifiable [11]. We refer also to the Belov-Fedorovskiy
paper [2], where the description is given of model spaces that contain bounded univalent functions.
                                                        12


We mention that the Hausdorff dimension of the accessible boundary of a Nevanlinna domain can
be any number between 1 and 2 as shown in [3], another construction can be found in [12].
    However, in all the above work the inner function 𝜃 is a Blaschke product or has a Blaschke
part. Moreover, in order to compare with Sakai’s theorem, we have to consider the case where the
functions 𝐹      e2 ∈ 𝐻 ∞ (D𝑒 ) for which 𝜙 = 𝐹
            e1 , 𝐹                              e1 /𝐹
                                                    e2 on T are continuous up to T. This is not always
possible when 𝜃 is not purely singular (see [4, Example 5.8]).
    Therefore, in this section 𝜃 will be a singular inner function of the form
                                                    ∫                
                                                        𝜁 +𝑧
                                     𝜃 (𝑧) = exp −            𝑑𝜇𝜃 (𝜁)
                                                        𝜁 −𝑧
                                                     T
with 𝜇𝜃 supported on a Carleson set, 𝐸 ⊂ T. We will show that there is a conformal map 𝜙 ∈ 𝐾𝜃
continuous on D̄ which is in 𝐶 ∞ (T) but not real analytic on T.
    In view of [6, Theorem 2.1], since supp(𝜇𝜃 ) is Carleson, the space 𝐾𝜃 then contains a non-trivial
function from some smoothness class, for example a function 𝑔 ∈ 𝐻 ∞ ∩ 𝐶 ∞ (T) (or in a Bergman
space, i.e., 𝑔 ∈ 𝐴 𝑝,1 for some 𝑝 > 1). Since 𝑔 ∈ 𝐾𝜃 , it admits a bounded type pseudo-continuation
of the form
                                  𝑔 = 𝐺/
                                       e e𝜃     almost everywhere on T,
where 𝐺 e ∈ 𝐻 ∞ (D𝑒 ) vanishes at infinity (see [5, Theorem 5.1.4]). Additionally, 𝑔 has an analytic
continuation, say G, to C   b \ supp(𝜇). Of course, G = 𝐺/   e e𝜃 on D𝑒 and observe that G cannot be
bounded in D𝑒 ; otherwise 𝑔 would be constant, as G|D = 𝑔 and G|D𝑒 coincide almost everywhere
on T.
    Now, consider 𝛼 ∈ D𝑒 with 𝜃 (1/𝛼) ≠ 0 and the following aggregate:
                                                   G(𝑧) − G(𝛼)
                                           𝜙(𝑧) =               .
                                                       𝑧−𝛼
We will show that 𝜙 ∈ 𝐾𝜃 ∩ 𝐶 ∞ (T) and 𝜙 is conformal in D̄.
    Clearly, 𝜙 is inside 𝐻 2 (D) and also
                                     𝜃 (𝜁)𝑔(𝜁) − 𝜃 (𝜁)G(𝛼) 𝐺    e(𝜁) − e 𝜃 (𝜁)G(𝛼)
                       𝜃 (𝜁)𝜙(𝜁) =                           =                     .
                                              𝜁 −𝛼                     𝜁 −𝛼
                                                     13


For 𝑧 ∈ D𝑒 the function
                                                                           
                             e(𝑧) − e
                             𝐺       𝜃 (𝑧)G(𝛼)       1           𝜃
                                                                 e (𝑧)
                                                =         e(𝑧) −
                                                          𝐺            e(𝛼)
                                                                       𝐺
                                   𝑧−𝛼             𝑧−𝛼           𝜃 (𝛼)
                                                                 e
is analytic around 𝛼 and vanishes at infinity. Hence, 𝜙 ∈ 𝐾𝜃 .
    Furthermore, 𝜙 is univalent in D̄. Indeed, suppose it is not. Then, there exist 𝑧, 𝑤 ∈ D̄ with
𝑧 ≠ 𝑤 and 𝜙(𝑧) = 𝜙(𝑤) or equivalently
                          𝑔(𝑧) − G(𝛼) 𝑔(𝑤) − G(𝛼)
                                          =
                               𝑧−𝛼              𝑤−𝛼
                           𝑔(𝑧)     𝑔(𝑤)      G(𝛼) G(𝛼)                    𝑧−𝑤
                    ⇐⇒           −          =       −        = G(𝛼)
                          𝑧−𝛼 𝑤−𝛼 𝑧−𝛼 𝑤−𝛼                             (𝑧 − 𝛼)(𝑤 − 𝛼)
                               𝑔(𝑧) − 𝑔(𝑤)       𝑔(𝑧) − 𝑔(𝑤)
                    ⇐⇒ − 𝛼                    +𝑤              − 𝑔(𝑤) = G(𝛼).
                                   𝑧−𝑤               𝑧−𝑤
The left-hand side is bounded, because 𝑔 ∈ 𝐶 ∞ ( D̄), whereas we can pick 1 < 𝛼 < 2 so that |G(𝛼)|
is arbitrarily large (recall G|D𝑒 is not bounded), a contradiction, and therefore 𝜙 is univalent in D̄.
    Consequently, if 𝑔 ∈ 𝐾𝜃 ∩ 𝐶 ∞ (T) and 𝜃 is a singular inner function, then 𝜙 is univalent in D̄ and
𝜙 ∈ 𝐾𝜃 ∩ 𝐶 ∞ (T). Also see [1, Section 4] for more details. At the same time, note that G cannot be
analytically extended to the whole D̄, because it is unbounded near the unit circle, and thus neither
can 𝜙; this fails exactly on the Carleson set 𝐸.
    Now, since 𝜙 ∈ 𝐾𝜃 , we can write
                                           𝜙 = 𝜃𝐹 ⇐⇒ 𝜃𝜙 = 𝐹                                     (1.4.1)
almost everywhere on T for some function 𝐹 ∈ 𝐻 2 with 𝐹 (0) = 0. In fact, 𝐹 ∈ 𝐻 ∞ because
𝜙 ∈ 𝐶 ∞ ( D̄).
    It is known that there exists some analytic function, H , with H|𝐸 = 0 such that both H and
H 𝜃 are Lipschitz on D̄. In fact, we can further consider H to be an outer function in 𝐶 ∞ (T).
Multiplying by H in (1.4.1), we get
                                               (H 𝜃)𝜙 = H 𝐹                                     (1.4.2)
almost everywhere on T. In particular, the left-hand side is now smooth on the whole T and the
same therefore holds true for the right-hand side. In a sense, H “annihilates” the singularities of 𝜃
as (1.4.1) fails exactly on the support, 𝐸, of 𝜇𝜃 .
                                                    14


     At this point, set 𝐹1 = H 𝐹, 𝐹2 = H 𝜃, and 𝑓 𝑗 = 𝐹 𝑗 ◦ 𝜙−1 for 𝑗 = 1, 2. Then, (1.4.2) becomes
                                                 𝐹1 = 𝜙𝐹2 ,
which now is fulfilled on the whole boundary T, and in turn
                                    𝑓1 (𝜁) = 𝜁 𝑓2 (𝜁)   for all    𝜁 ∈ Γ.
This is exactly the setup we were looking for, albeit it contrasts with Sakai’s result: Even though
Γ = 𝜙(T) is 𝐶 ∞ -smooth, 𝜙 cannot be analytic on the Carleson set 𝐸 and thus neither can Γ.
     It is worth mentioning that there are examples of Nevanlinna domains that come from singular
inner functions with particularly irregular boundaries. Namely, in [3] one can find examples
of univalent functions in a Paley-Wiener space such that they map the upper half-plane onto a
Nevanlinna domain whose boundary can have any dimension between 1 and 2.
1.5     Holomorphic functions in C2
     In this section we attempt to replace the function 𝜁 𝑓0 (𝜁) in (1.3.1) with a more general formula.
     For some positive 𝑟 > 0, let Ω ⊂ 𝐷 (𝜁0 , 𝑟) be a simply connected open set, let Γ = 𝜕Ω∩𝐷 (𝜁0 , 𝑟),
and let 𝜁0 ∈ Γ. Here, we will also need the extra assumption that Γ is a Jordan arc (or possibly a
union of Jordan arcs).
     Let Φ be a holomorphic function of two variables, that is, a function of the form
                                                      +∞
                                                      ∑︁
                                        Φ(𝑧, 𝑤) =         𝑏 𝑛𝑚 𝑧 𝑛 𝑤 𝑚
                                                    𝑛,𝑚=0
where each of the functions Φ(𝑧, · ) and Φ( · , 𝑤) is itself holomorphic. Suppose there exists a
function 𝑅 which is
   (i) holomorphic on Ω,
  (ii) continuous on Ω̄, and
 (iii) satisfies 𝑅(𝜁) = Φ(𝜁, 𝜁) on Γ.
                                                     15


In view of Lemma 1.2.6, we may assume that 𝜁0 = 0 and 𝑏 00 = 0 so that 𝑅(0) = Φ(0, 0) = 0. Notice
that 𝑅(𝑧) and Φ(𝑧, 𝑧) are bounded on Ω̄ and thanks to the Phragmén-Lindelöf Principle 1.1.1, we
may assume without loss of generality that there exists some non-negative integer 𝑘 for which
                           𝜕                     𝜕 𝑘−1                          𝜕𝑘
              Φ(0, 0) =      Φ(0, 0) = · · · =           Φ(0, 0) = 0     and         Φ(0, 0) ≠ 0    (1.5.1)
                          𝜕𝑤                    𝜕𝑤 𝑘−1                         𝜕𝑤 𝑘
otherwise Φ would be identically zero.
      We would like to use the Weierstraß approximation theorem for the function Φ(𝑧, 𝑤) − 𝑅(𝑧)
around 0, but 𝑅 is not holomorphic on the boundary. But since it is continuous by (ii) and Γ is
Jordan, we can use Mergelyan’s theorem to get a sequence of polynomials 𝑝 𝑛 that converge to 𝑅
uniformly on Ω̄. And we can pick this sequence so that 𝑝 𝑛 (0) = 0 for every 𝑛 = 0, 1, . . . .
      Next, we define the functions
                        Ψ(𝑧, 𝑤) = Φ(𝑧, 𝑤) − 𝑅(𝑧)         and   Ψ𝑛 (𝑧, 𝑤) = Φ(𝑧, 𝑤) − 𝑝 𝑛 (𝑧).
The Ψ𝑛 are holomorphic on C2 and converge uniformly to Ψ on Ω̄ × C. Observe that for all 𝑛 we
have Ψ𝑛 (0, 0) = Φ(0, 0) − 𝑝 𝑛 (0) = 0 and also
                                     𝜕𝜅         𝜕𝜅
                                         Ψ𝑛 =       Φ        for all integers 𝜅 ≥ 1
                                    𝜕𝑤 𝜅       𝜕𝑤 𝜅
and all points (𝑧, 𝑤). Then, from (1.5.1) and from the Weierstraß approximation theorem, there
exist unique holomorphic functions 𝑎 0;𝑛 , . . . , 𝑎 𝑘−1;𝑛 : C → C and 𝑐 𝑛 : C2 → C with 𝑎 𝑗;𝑛 (0) = 0
and 𝑐 𝑛 (0, 0) ≠ 0 such that
                                                                                          
                          Ψ𝑛 (𝑧, 𝑤) = 𝑐 𝑛 (𝑧, 𝑤) 𝑤 𝑘 + 𝑎 𝑘−1;𝑛 (𝑧)𝑤 𝑘−1 + · · · + 𝑎 0;𝑛 (𝑧) .
      Following the proof of the Weierstraß theorem and since the convergence Ψ𝑛 → Ψ is uniform
on Ω̄ × C, we can find sufficiently small 𝛿 and 𝜌 with 𝜌 ≥ 𝛿 > 0 so that 𝑎 0;𝑛 , . . . , 𝑎 𝑘−1;𝑛 and the 𝑐 𝑛
                                                                  

converge uniformly on Ω̄ ∩ 𝐷 (0, 𝛿) and Ω̄ ∩ 𝐷 (0, 𝛿) × 𝐷 (0, 𝜌), respectively, to some functions
𝑎 0 , . . . , 𝑎 𝑘−1 and 𝑐 with 𝑎 𝑗 (0) = 0 and 𝑐(0, 0) ≠ 0. Note that the functions 𝑎 𝑗 are holomorphic on
Ω ∩ 𝐷 (0, 𝛿) and continuous on Ω̄ ∩ 𝐷 (0, 𝛿). Subsequently, we get
                                                                                            
                         Φ(𝑧, 𝑤) − 𝑅(𝑧) = 𝑐(𝑧, 𝑤) 𝑤 𝑘 + 𝑎 𝑘−1 (𝑧)𝑤 𝑘−1 + · · · + 𝑎 0 (𝑧) .          (1.5.2)
                                                           16


    Let us write
                                   𝑃(𝑧, 𝑤) = 𝑤 𝑘 + 𝑎 𝑘−1 (𝑧)𝑤 𝑘−1 + · · · + 𝑎 0 (𝑧)
for the polynomial factor. From (iii), (1.5.2) and since 𝑐(0, 0) ≠ 0, we have
                       𝑘                  𝑘−1
         𝑃(𝜁, 𝜁) = 𝜁 + 𝑎 𝑘−1 (𝜁)𝜁             + · · · + 𝑎 0 (𝜁) = 0     for all 𝜁 ∈ Γ ∩ 𝐷 (0, 𝛿).       (1.5.3)
Remark. Functions of the form
                                    𝑃(𝑧, 𝑧) = 𝑧 𝑘 + 𝑎 𝑘−1 (𝑧)𝑧 𝑘−1 + · · · + 𝑎 0 (𝑧),
where 𝑎 𝑗 are polynomials, are called polyanalytic polynomials. One can find more details on these
in [8, 11] or [14].
    We are interested in the roots of the polynomial 𝑃(𝑧, · ) when 𝑧 ∈ Ω̄ ∩ 𝐷 (0, 𝛿). In other words,
we will study the equation (in 𝑤)
                         𝑃(𝑧, 𝑤) = 0 ⇐⇒ 𝑤 𝑘 + 𝑎 𝑘−1 (𝑧)𝑤 𝑘−1 + · · · + 𝑎 0 (𝑧) = 0
                                                                                                        (1.5.4)
                                                                  when 𝑧 ∈ Ω̄ ∩ 𝐷 (0, 𝛿).
    Let D (𝑧) be the discriminant of 𝑃(𝑧, · ) (for any fixed 𝑧). Then, D (𝑧) is a polynomial of
                                                                                                𝜕
the coefficients 𝑎 0 (𝑧), . . . , 𝑎 𝑘−1 (𝑧) and is equal to 0 if, and only if, 𝑃(𝑧, 𝑤) and     𝜕𝑤 𝑃(𝑧, 𝑤) share
a common factor. The roots of 𝑃(𝑧, · ) are given by a multivalued holomorphic function, W,
depending on 𝑎 1 , . . . , 𝑎 𝑘−1 , and the points where W changes a branch inside Ω ∩ 𝐷 (0, 𝛿) are
exactly the zeros of D (in Ω ∩ 𝐷 (0, 𝛿)).
    We distinguish between two cases: when D is identically 0 and when it is not.
    Before moving on, let us note that the set M (Ω, Γ, 𝛿) of all meromorphic functions on Ω ∩
𝐷 (0, 𝛿) continuous up to (Ω ∪ Γ) ∩ 𝐷 (0, 𝛿) except possibly a (closed) measure zero subset of Γ is
a field with the usual operations of addition and multiplication.
1.5.1   D≠0
    Here 𝑃(𝑧, 𝑤) is irreducible over M (Ω, Γ, 𝛿). Since D is continuous on (Ω ∪ Γ) ∩ 𝐷 (0, 𝛿),
the set D −1 {0} ∩ Γ ∩ 𝐷 (0, 𝛿) is closed and of zero harmonic measure. Now, we decompose
                        

 Γ \ D −1 {0} ∩ 𝐷 (0, 𝛿) into countably many open connected arcs.
              

                                                             17


      Let 𝛾 be one of these arcs. Then, there exists a simply connected set 𝐷 ⊂ Ω ∩ 𝐷 (0, 𝛿)
such that 𝜕𝐷 ∩ 𝜕Ω = 𝛾. Since D has no zeros on 𝐷 ∪ 𝛾, by the monodromy theorem the
multivalued function W “splits” into 𝑘 distinct holomorphic functions, 𝑊 𝑗 ( 𝑗 = 1, . . . , 𝑘), and let
𝐶 𝑗 = {𝜁 ∈ 𝛾 : 𝑊 𝑗 (𝜁) = 𝜁 }. Notice that the 𝐶 𝑗 ’s are closed (in 𝛾), they cover 𝛾, and any two of
them intersect at a (closed) set of zero harmonic measure.
      Unfortunately, 𝐶 𝑗 need not be connected, but we can further decompose each 𝐶˚ 𝑗 (whenever it
is non-empty) into countably many open arcs as in 𝐶˚ 𝑗 = ∪𝑖 𝛾 𝑖𝑗 , for 𝑗 = 1, . . . , 𝑘. Again, around
each 𝛾 𝑖𝑗 we consider a neighbourhood 𝐷 𝑖𝑗 ⊂ 𝐷 with 𝜕𝐷 𝑖𝑗 ∩ 𝜕𝐷 = 𝛾 𝑖𝑗 (these can, but need not be
simply connected) and let 𝑊 𝑖𝑗 = 𝑊 𝑗 |𝐷 𝑖 ∪𝛾 𝑖 .
                                             𝑗  𝑗
      Then, for each 𝑗 = 1, . . . , 𝑘 and 𝑖 = 1, 2 . . . the functions 𝑊 𝑖𝑗 are holomorphic on 𝐷 𝑖𝑗 , continuous
on 𝐷 𝑖𝑗 ∪ 𝛾 𝑖𝑗 and satisfy 𝑊 𝑖𝑗 (𝜁) = 𝜁 for all 𝜁 ∈ 𝛾 𝑖𝑗 ; in other words, they are Schwarz functions on
𝐷 𝑖𝑗 ∪ 𝛾 𝑖𝑗 . Since Γ is Jordan, all 𝛾 𝑖𝑗 are also Jordan and from Theorem 1.1.2 we conclude that each
𝛾 𝑖𝑗 is, in fact, a regular real analytic simple arc except possibly some cusps.
1.5.2      D=0
      In this case, 𝑃(𝑧, 𝑤) has to be reducible over M (Ω, Γ, 𝛿). In particular, we can write 𝑃(𝑧, 𝑤) =
𝑃1 (𝑧, 𝑤) · · · 𝑃e𝑘 (𝑧, 𝑤) for some e 𝑘 ≤ 𝑘 where each 𝑃𝜅 (𝑧, 𝑤) has now coefficients in M (Ω, Γ, 𝛿) and
is irreducible, i.e., D𝜅 ≠ 0 where D𝜅 is the discriminant of 𝑃𝜅 (𝑧, · ). Since 𝑃(𝜁, 𝜁) = 0 for all
𝜁 ∈ Γ, we can split (Γ \ 𝐸) ∩ 𝐷 (0, 𝛿), where 𝐸 is some closed zero-(harmonic)-measure set, into
open sets 𝑂 𝜅 for 𝜅 = 1, . . . , e𝑘 so that 𝑃𝜅 (𝜁, 𝜁) = 0 for all 𝜁 ∈ 𝑂 𝜅 . Notice that 𝑂 𝜅 ∩ 𝑂 𝜅 ′ = ∅ when
𝑃𝜅 and 𝑃𝜅 ′ are different.
      Observe that, since 𝑃(𝑧, 𝑤) factors into the polynomials 𝑃𝜅 (𝑧, 𝑤) (over M (Ω, Γ, 𝛿)) and the
roots of 𝑃(𝑧, · ) are given by the multivalued holomorphic function W, the roots of each 𝑃𝜅 (𝑧, · ) are
also given by a multivalued holomorphic function W𝜅 whose branches are comprised of branches
of W.
      Working as above for each 𝜅 = 1, . . . , e   𝑘, we separate 𝑂 𝜅 \ D𝜅−1 {0} into countably many open
arcs and for each such arc, 𝛾, we find some simply connected neighbourhood, 𝐷 ⊂ Ω, with
𝜕𝐷 ∩ 𝜕Ω = 𝛾 so that W𝜅 “splits” into its different branches. Again following the above arguments,
                                                          18


we can decompose 𝛾 — minus a zero-measure set — into countably many open arcs over which
𝑊 𝑗 (𝜁) = 𝜁 for some branch 𝑊 𝑗 of W𝜅 . Constructing appropriate neighbourhoods, we conclude
that except a zero-measure set, 𝛾 is a countable union of regular real analytic simple arcs except
possibly some cusps.
     In either case, the cusps (if they exist) point into Ω and may only accumulate on the endpoints
of each open arc.
     Now we formulate the above results into a theorem.
Theorem 1.5.1. Let Ω be a bounded simply connected domain such that Γ = 𝜕Ω ∩ 𝐷 (𝜁0 , 𝑟) is
a (union of) Jordan arc(s). Also, let Φ be a (non-trivial) holomorphic function of two variables
defined in 𝐷 (𝜁0 , 𝑟) × 𝐷 ( 𝜁¯0 , 𝑟), and suppose there exists a function 𝑅
   (i) holomorphic on Ω,
  (ii) continuous on Ω̄, and such that
 (iii) 𝑅(𝜁) = Φ(𝜁, 𝜁) for all 𝜁 ∈ Γ.
Then, there exists a closed set, 𝐸 ⊂ Γ, of zero harmonic measure so that Γ \ 𝐸 is a countable union
of regular real analytic simple arcs except possibly for some cusps. The cusps (if they exist) point
into Ω and may only accumulate on 𝐸.
1.6    The U-V problem
     In this section, we are interested in the following setup.
     Let Ω be a simply connected open set in C and let 𝜁0 ∈ 𝜕Ω be a boundary point of Ω. Assume
that for some 𝜌 > 0 the connected component, Γ, of 𝜕Ω ∩ 𝐷 (𝜁0 , 𝜌) containing 𝜁0 is a Jordan curve.
Note that 𝜌 ≥ dist(𝜁0 , 𝜕Ω\ Γ) > 0. For convenience we will write simply Ω to denote Ω∩ 𝐷 (𝜁0 , 𝜌).
     Let 𝐴 be an analytic function in a neighbourhood, 𝐷 (𝜁0 , 𝜖), of 𝜁0 and suppose we have two
functions U and V defined on Ω that are not proportional and have the following properties:
    I) U and V are positive and harmonic on Ω,
                                                     19


   II) they are continuous on Ω ∪ Γ,
  III) U = V = 0 on Γ, and
        U (𝜁)
 IV)    V (𝜁) = | 𝐴(𝜁)| 2 ≠ const for 𝜁 ∈ Γ.
Notice that since U ≠ 𝑐V, the function | 𝐴| needs to be non-constant. Otherwise, we could have
U = 𝑐V and all our conditions work trivially for any Γ. Also, we may assume that 𝜌 < 𝜖 without
loss of generality (so that 𝐴 is defined over the whole Ω) to avoid unnecessary technical difficulties.
     Formula (IV) is to be understood in the sense of limits, i.e., the limit of U (𝑧)/V (𝑧) as
Ω ∋ 𝑧 → 𝜁 ∈ Γ exists and is equal to | 𝐴(𝜁)| 2 . In fact, this limit always exists when Ω is simply
connected and Γ is Jordan (see Remark 1.6.1), so the only assumption here is the values it takes.
     Consider a conformal map from the Poincaré plane to Ω, 𝜙 : H → Ω. Since Γ is connected
and Jordan, Carathéodory’s theorem implies that 𝜙 extends conformally to a function (abusing the
notation) 𝜙 : H ∪ 𝛾 → Ω ∪ Γ which we can pick so that 𝛾 ⊂ R is some bounded open interval with
𝜙(𝛾) = Γ and 𝜙(0) = 𝜁0 . Utilizing this 𝜙, we can “transfer” the information about U and V over
Ω to information over H. Define
                               𝑢 ≡ U ◦ 𝜙,    𝑣 ≡V◦𝜙      and    𝑎 ≡ 𝐴◦𝜙
and note that 𝑎 is analytic on H and continuous on H ∪ 𝛾. As above, we have
   i) 𝑢 and 𝑣 are positive and harmonic on H,
  ii) they are continuous on H ∪ 𝛾,
 iii) 𝑢 = 𝑣 = 0 on 𝛾, and
 iv)   𝑢
       𝑣 = |𝑎| 2 on 𝛾.
Again, (iv) is to be understood in the sense of limits.
                                                   20


      Now, harmonically extend 𝑢 and 𝑣 to H ∪ 𝛾 ∪ H− by
                                  
                                                                               
                                                                                
                                                  ∈                                        𝑧∈H
                                                                               
                                  
                                  
                                   𝑢(𝑧),       𝑧   H                           
                                                                                
                                                                                 𝑣(𝑧),
                                  
                                                                               
                                                                                
                                  
                                                                               
                                                                                
                       𝑢 ∗ (𝑧) = 0,             𝑧 ∈ 𝛾          and 𝑣 ∗ (𝑧) = 0,            𝑧∈𝛾
                                  
                                                                               
                                                                                
                                  
                                                                               
                                                                                
                                                      −
                                                                                 −𝑣(𝑧), 𝑧 ∈ H−
                                                                               
                                  
                                  
                                   −𝑢(𝑧),      𝑧 ∈ H                           
                                                                                
                                                                               
and let ℎ be the function
                                                    
                                                       𝑢 ∗ (𝑧)
                                                     𝑣 ∗ (𝑧) , 𝑧 ∈ H ∪ H− ,
                                                    
                                                    
                                                    
                                                    
                                             ℎ(𝑧) =
                                                      𝑢 ∗𝑦 (𝑧)
                                                     𝑣 ∗𝑦 (𝑧) , 𝑧 ∈ 𝛾.
                                                    
                                                    
                                                    
      We claim that ℎ is well defined and, in fact, real analytic on H∪ 𝛾 ∪H− . Indeed, using Harnack’s
inequality, for any (𝑥, 0) ∈ 𝛾 there exists a constant 𝑐 > 0 (dependent on 𝑣 ∗ ) such that
                                  𝑦                        2−𝑦
                              𝑐        ≤ 𝑣 ∗ (𝑥, 𝑦) ≤ 𝑐             for every 0 < 𝑦 < 1, or
                                2−𝑦                            𝑦
                                1          ∗
                                          𝑣 (𝑥, 𝑦)       2−𝑦
                            𝑐        ≤              ≤𝑐 2 .                                       (1.6.1)
                              2−𝑦             𝑦              𝑦
Recall that 𝑣 ∗ (𝑥, 0) = 0 and take limits as 𝑦 → 0+ . Since 𝑣 ∗ is harmonic on H ∪ 𝛾 ∪ H− , (1.6.1)
guarantees that 𝑣 ∗𝑦 > 0 on 𝛾 (the same holds true for 𝑢 ∗ ) and therefore the limit
                                                                       ∗
                                                    𝑢 ∗ (𝑥, 𝑦) 𝑢 𝑦 (𝑥, 0)
                                                lim               =
                                                𝑦→0 𝑣 ∗ (𝑥, 𝑦)      𝑣 ∗𝑦 (𝑥, 0)
exists and is finite. Hence, ℎ is a well-defined continuous function on H ∪ 𝛾 ∪ H− . In fact, since
𝑢 ∗𝑦 and 𝑣 ∗𝑦 are real analytic and non-zero around 𝛾, ℎ is also real analytic on H ∪ 𝛾 ∪ H− . What is
more is that
                                    𝑢 ∗𝑦 (𝜉)            𝑢(𝑧)
                           ℎ(𝜉) =             = lim              = |𝑎(𝜉)| 2       for any 𝜉 ∈ 𝛾  (1.6.2)
                                    𝑣 ∗𝑦 (𝜉)    H∋𝑧→𝜉   𝑣(𝑧)
because of (iv) and therefore |𝑎| 2 is also real analytic on 𝛾.
                                                                                                U
Remark 1.6.1. The above is the reason why relation (IV) is meaningful. When we write            V on Γ,
it really means the limit of ℎ ◦ 𝜙−1 as we approach Γ from the inside of Ω. This limit always exist
on a Jordan arc Γ when Ω is simply connected thanks to Harnack’s inequality.
      It is worth mentioning the work of Jerison and Kenig who showed [9, Theorems 5.1 and 7.9]
that equation (IV) makes sense whenever Ω is assumed to be a non-tangentially accessible (NTA)
domain.
                                                               21


     Next, consider ℎ |𝛾 . Its power series around 0 ∈ 𝛾 is given by
                                                         ∞
                                                        ∑︁
                                             ℎ |𝛾 (𝑥) =     𝑏𝑛𝑥𝑛
                                                        𝑛=0
for some real numbers 𝑏 0 , 𝑏 1 , . . . This readily extends to a complex analytic function, say 𝑟, on
some open neighbourhood, 𝐷 (0, 𝜖 ′):
                                                        ∞
                                                       ∑︁
                                              𝑟 (𝑧) =      𝑏𝑛 𝑧𝑛 ,
                                                       𝑛=0
where we can choose 𝜙 and 𝜖 ′ so that 𝛾 ⊂ 𝐷 (0, 𝜖 ′). Of course, by construction and from (1.6.2)
we get 𝑟 |𝛾 = ℎ |𝛾 = |𝑎| 2 .
     At this point, we want to “shift” everything back at Ω. We set
                                         𝑉 ≡ 𝜙(H ∩ 𝐷 (0, 𝜖 ′)) ⊂ Ω
and observe that 𝜕𝑉 is a closed Jordan arc such that Γ ⊊ 𝜕𝑉 ∩ 𝐷 (𝜁0 , 𝜌). Define a new function
                                             𝑅 ≡ 𝑟 ◦ (𝜙−1 ) |𝑉∪Γ ,                             (1.6.3)
which is holomorphic on 𝑉, continuous on 𝑉 ∪ Γ, and on Γ it satisfies 𝑅(𝜁) = | 𝐴(𝜁)| 2 .
     Now, consider the function Φ(𝑧, 𝑤) = 𝐴(𝑧) 𝐴(𝑤). Φ is holomorphic on 𝐷 (𝜁0 , 𝜖) × 𝐷 (𝜁 0 , 𝜖) and
it satisfies Φ(𝜁, 𝜁) = 𝐴(𝜁) 𝐴(𝜁) = | 𝐴(𝜁)| 2 when 𝑧 = 𝑤 = 𝜁 ∈ Γ. As a corollary to Theorem 1.5.1,
the next theorem follows.
Theorem 1.6.2. Let Ω be a bounded simply connected domain in C and let Γ be an open Jordan arc
of its boundary with 𝜁0 ∈ Γ. Suppose there are two positive non-proportional harmonic functions
U and V on Ω continuous on Ω ∪ Γ and such that
                                                    U (𝜁)
                    U (𝜁) = V (𝜁) = 0       and            = | 𝐴(𝜁)| 2 for all 𝜁 ∈ Γ,
                                                    V (𝜁)
where 𝐴 is a non-trivial analytic function on a neighbourhood of Ω.
     Then, there exists some neighbourhood 𝐷 of 𝜁0 and a closed set 𝐸 ⊂ Γ of zero harmonic
measure so that (Γ \ 𝐸) ∩ 𝐷 is a countable union of regular real analytic simple arcs except
possibly for some cusps. The cusps (if they exist) point into Ω and may only accumulate on 𝐸 ∩ 𝐷.
                                                      22


     Of course, Theorems 1.5.1 and 1.6.2 are somewhat far from Sakai’s result. Nevertheless,
because of the special form of the function Φ(𝑧, 𝑤) = 𝐴(𝑧) 𝐴(𝑤), we can actually say more in this
case.
Proposition 1.6.3. Let Ω be a bounded simply connected domain in C and let Γ be an open Jordan
arc of its boundary with 𝜁0 ∈ Γ. Suppose there are two positive non-proportional harmonic
functions U and V on Ω continuous on Ω ∪ Γ and such that
                                                U (𝜁)
                    U (𝜁) = V (𝜁) = 0    and            = | 𝐴(𝜁)| 2  for all 𝜁 ∈ Γ
                                                V (𝜁)
where 𝐴 is a non-trivial analytic function on a neighbourhood of Γ.
     Then, there exists a neighbourhood 𝐷 of 𝜁0 and a function 𝑅 satisfying the following:
   (i) 𝑅 is holomorphic on Ω ∩ 𝐷,
  (ii) 𝑅 is continuous on (Ω ∪ Γ) ∩ 𝐷 and
 (iii) 𝑅(𝜁) = | 𝐴(𝜁)| 2 for 𝜁 ∈ Γ ∩ 𝐷.
Additionally, for any 𝜁0 ∈ Γ with 𝐴′ (𝜁0 ) ≠ 0 either
(1) there exist a function Ψ1 holomorphic and univalent on Ω ∩ 𝐷 such that Ψ1 is continuous on
      (Ω ∪ Γ) ∩ 𝐷, and Ψ1 (𝜁) = | 𝐴(𝜁) − 𝐴(𝜁0 )| 2 for 𝜁 ∈ Γ ∩ 𝐷, or
(2) there exist a function Ψ2 holomorphic and univalent on Ω ∩ 𝐷 such that Ψ22 is continuous on
      (Ω ∪ Γ) ∩ 𝐷, and Ψ22 (𝜁) = | 𝐴(𝜁) − 𝐴(𝜁0 )| 2 for 𝜁 ∈ Γ ∩ 𝐷.
Proof. We have already established the existence of such a function 𝑅 in (1.6.3).
     For the rest, 𝐴′ (𝜁0 ) ≠ 0 and we may assume without loss of generality that 𝐴 is conformal on
a neighbourhood of Ω̄. Recall that 𝑉 from the definition of 𝑅 in (1.6.3) is such that 𝜕𝑉 is Jordan
and Γ ⊊ 𝜕𝑉 ∩ 𝐷 (𝜁0 , 𝜌) when Ω ⊂ 𝐷 (𝜁0 , 𝜌). Since 𝐴 is continuous and injective on 𝑉, ¯ there exists
some small 𝛿, 0 < 𝛿 ≤ 𝜌, such that 𝜕 ( 𝐴(𝑉)) ∩ 𝐷 (𝜁0′ , 𝛿) ⊂ 𝐴(Γ).
                                                  23


      Now, let 𝐴(𝜁0 ) = 𝜁0′ , Ω′ = 𝐴(𝑉) ∩ 𝐷 (𝜁0′ , 𝛿), and Γ′ = 𝜕Ω′ ∩ 𝐷 (𝜁0′ , 𝛿). The function
                                                     1
                                            𝑆(𝑧) ≡     𝑅 ◦ 𝐴−1 (𝑧)                               (1.6.4)
                                                     𝑧
is a Schwarz function of Ω′ ∪ Γ′ in 𝐷 (𝜁0′ , 𝛿):
   (i) 𝑆 is holomorphic on Ω′,
  (ii) it is continuous on Ω′ ∪ Γ′, and
 (iii) 𝑆(𝜁) = 1𝜁 𝑅( 𝐴−1 (𝜁)) = 𝜁 on Γ′.
Notice that from (1.6.1) the functions 𝑎 = 𝐴 ◦ 𝜙 and 𝐴 are always non-zero and thus 𝑆 is a
well-defined holomorphic function, because 0 cannot be a point of Ω̄′.
      Finally, consider the function 𝑆𝑡 (𝑧) = 𝑆(𝑧 + 𝜁0′ ) − 𝜁0′ , which is a Schwarz function on (Ω′ −
𝜁0′ ) ∪ (Γ′ − 𝜁0′ ) at 0. From Theorem 1.1.4, we know that one of the functions Φ1 (𝑧) = 𝑧𝑆𝑡 (𝑧) and
             √︁
Φ2 (𝑧) = 𝑧𝑆𝑡 (𝑧) is univalent on (Ω′ − 𝜁0′ ) ∩ 𝐷 (0, 𝛿′) for some 𝛿′ ≤ 𝛿. Changing variables to
get back to our initial domain Ω, we find that one of the following functions, Ψ1 or Ψ2 , has to be
univalent on Ω ∩ 𝐷 ′:
                                                                              
                                                               𝑅(𝑧)
                                Ψ1 (𝑧) = ( 𝐴(𝑧) − 𝐴(𝜁0 ))             − 𝐴(𝜁0 )
                                                               𝐴(𝑧)
and
                                         √︄                                     
                                                                  𝑅(𝑧)
                                Ψ2 (𝑧) =    ( 𝐴(𝑧) − 𝐴(𝜁0 ))            − 𝐴(𝜁0 )
                                                                  𝐴(𝑧)
for 𝑧 ∈ Ω ∩ 𝐷 ′, where 𝐷 ′ = 𝐴−1 (𝐷 (𝜁0′ , 𝛿′)). The rest of the desired properties are obvious.      □
      In the above proof, Γ′ is the image of a Jordan arc under the (conformal) map 𝐴. Therefore, the
existence of a Schwarz function, 𝑆, along with Theorem 1.1.2 imply that Γ′, and in turn Γ, satisfy
(1) or (2c) of Theorem 1.1.2. Case (1) corresponds to (1) of Proposition 1.6.3 and (2c) to (2), that
is, Γ′ (respectively, Γ) has a cusp if, and only if, the function
                                             √︃
                                                𝑧 𝑆(𝑧 + 𝜁0′ ) − 𝜁0′
                                                                    

                                                       24


is univalent on (Ω′ − 𝜁0′ ) ∩ 𝐷 (0, 𝛿′) (respectively, Ψ2 on Ω ∩ 𝐷).
     As a consequence, we have the following theorem, which is the main result of this section.
Theorem 1.6.4. Let Ω be a bounded simply connected domain in C and let Γ be an open Jordan arc
of its boundary with 𝜁0 ∈ Γ. Suppose there are two positive non-proportional harmonic functions
U and V on Ω continuous on Ω ∪ Γ and satisfying
                                                  U (𝜁)
                    U (𝜁) = V (𝜁) = 0      and          = | 𝐴(𝜁)| 2    for all 𝜁 ∈ Γ,
                                                  V (𝜁)
where 𝐴 is a non-trivial analytic function on a neighbourhood of Γ.
     Then, for all but possibly finitely many points 𝜁0 ∈ Γ there exists some small neighbourhood 𝐷
of 𝜁0 such that the following holds:
     Γ ∩ 𝐷 is a regular real analytic simple arc through 𝜁0 except possibly a cusp at 𝜁0 .    (1.6.5)
The finitely many points around which (1.6.5) might fail are the points 𝜁 ∈ Γ where 𝐴′ (𝜁) = 0, i.e.,
where 𝐴 might not be invertible.
     There is a cusp at 𝜁0 if and only if (2) of Proposition 1.6.3 holds true.
     Of course, one can ask at this point whether it is possible to actually have a cusp. The answer
is yes as the next example shows.
Example 1.6.5. Let Ω be open and Γ = 𝜕Ω ∩ 𝐷 (0, 𝜌) (with 𝜌 ≤ 1 sufficiently small) be such
that Γ has a cusp at 0 (i.e., 𝜁0 = 0). Then, from Remarks 1.1.3, for some 𝜂 > 0, there is a
holomorphic function 𝑇 defined on {|𝑧| ≤ 𝜂} that maps conformally the closed upper half-disk
𝐾𝜂 = {|𝑧| ≤ 𝜂 : Im(𝑧) ≥ 0} into Ω ∪ Γ and Γ ∩ 𝐷 ⊂ 𝑇 (−𝜂, 𝜂) for some small neighbourhood 𝐷 of
0. Also, 𝑇 (0) = 0 with order 2. By dilating appropriately, we may assume that everything happens
in the unit disk, that is, 𝜂 = 1, 𝑇 is defined on D̄ and is univalent on 𝐾1 = {|𝑧| ≤ 1 : Im(𝑧) ≥ 0},
𝑇 (𝐾1 ) ⊂ Ω ∪ Γ, and Γ ∩ 𝐷 (0, 𝜌) ⊂ 𝑇 (−1, 1).
     Next, consider two positive harmonic functions, 𝑢 and 𝑣, on the upper half-disk D ∪ H that are
zero on (−1, 1). As we saw in the beginning of this section, 𝑢 and 𝑣 can be extended on the whole
                                                    25


disk and the ratio 𝑢/𝑣 is a positive analytic function on (−1, 1). Therefore, on (−1, 1) we can write
𝑢/𝑣 = |𝑎| 2 for some function 𝑎 holomorphic on D.
    Finally, construct the functions
                         U = 𝑢 ◦ 𝑇 −1 ,   V = 𝑣 ◦ 𝑇 −1 ,   and   𝐴 = 𝑎 ◦ 𝑇 −1 .
Then, 𝐴 is holomorphic around the cusp at 0, and U, V are positive harmonic functions on
Ω ∩ 𝐷 (0, 𝜌) and zero on the boundary Γ. Moreover, U and V satisfy U/V = | 𝐴| 2 on Γ.
1.7   Some open “free boundary” problems in the spirit of Sakai
    All problems treated above are examples of the so-called free boundary problems (non-
variational free boundary problems).
    We would like to call the attention of the reader to one open question: what can one say for
the boundary of a domain Ω that is not simply connected but admits positive harmonic functions
vanishing on its boundary and whose ratio is “nice” on that boundary? Finitely connected situations
present no difficulties, but what if, for example, Γ is a Cantor set and Ω = D \ Γ? Suppose we
know that the ratio of two positive harmonic (non-proportional) functions U, V in Ω vanishing on
the Cantor set Γ has a well-defined ratio on Γ (this happens for a wide class of Cantor sets Γ’s, for
example for all regular Cantor sets of positive Hausdorff dimension). Suppose this ratio is equal to
| 𝐴(𝜁)| 2 ≠ const for 𝜁 ∈ Γ, where 𝐴 is a holomorphic function on D. What we can say about the
Cantor set Γ? The “desired” answer is that this is impossible to happen on any Cantor set.
    This type of problems (we may call them “one-phase free boundary problems”) appear naturally
in certain problems of complex dynamics, see, e.g., [15]. If we would know the aforementioned
answer (we conjecture that no Cantor set would allow such a triple (U, V, 𝐴)), then a long-standing
problem about the dimension of harmonic measure on Cantor repellers would be solved.
    Another similar one-phase boundary problem concerns functions in R𝑛 for 𝑛 > 2. Let Ω be a
bounded domain in R𝑛 , 𝑛 > 2, and let Γ = 𝜕Ω ∩ 𝐷 (𝑥, 𝑟), where 𝑥 ∈ 𝜕Ω. Again, let U, V be two
positive (non-proportional) harmonic functions in Ω vanishing continuously on Γ. If Ω is assumed
                                                  26


to be a Lipschitz domain, then [9] claims that U/V makes sense on Γ and is additionally a Hölder
function on Γ (boundary Harnack principle).
    Here is a question. Let 𝑅 be a real analytic function on 𝐷 (𝑥, 𝑟), 𝑥 ∈ Γ, and let U/V = 𝑅 on
Γ ∩ 𝐷 (𝑥, 𝑟). Is it true that Γ ∩ 𝐷 (𝑥, 𝑟) is real analytic, maybe with the exception of some lower
dimensional singular set?
                                                   27


                                      BIBLIOGRAPHY
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     1740. DOI: 10.1070/sm2011v202n12abeh004205.
 [2] Y. S. Belov and K. Y. Fedorovskiy. “Model spaces containing univalent functions”. In:
     Russian Mathematical Surveys 73.1 (Feb. 2018), pp. 172–174. DOI: 10.1070/rm9807.
 [3] Y. Belov, A. Borichev, and K. Fedorovskiy. “Nevanlinna domains with large boundaries”.
     In: Journal of Functional Analysis 277.8 (Oct. 2019), pp. 2617–2643. DOI: 10.1016/j.jfa.
     2018.12.015.
 [4] J. J. Carmona, P. V. Paramonov, and K. Y. Fedorovskiy. “On uniform approximation by
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     10.1090/conm/393/07371.
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     10.1134/s0081543806020155.
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     fiable contours in C𝑛 ”. In: Mathematical Notes 59 (4 Apr. 1, 1996), pp. 435–439. DOI:
     10.1007/BF02308692.
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     10.1016/0001-8708(82)90055-x.
[10] M. Y. Mazalov. “An example of a nonconstant bianalytic function vanishing everywhere
     on a nowhere analytic boundary”. In: Mathematical Notes 62.4 (Oct. 1997), pp. 524–526.
     DOI: 10.1007/bf02358988.
[11] M. Y. Mazalov. “Example of a non-rectifiable Nevanlinna contour”. In: St. Petersburg
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[13] M. Sakai. “Regularity of a boundary having a Schwarz function”. In: Acta Mathematica
     166.0 (1991), pp. 263–297. DOI: 10.1007/bf02398888.
[14] C. L. Siegel. Topics in complex function theory - Elliptic Functions and Uniformization
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                                            29


                                                  CHAPTER 2
 THE BUFFON’S NEEDLE PROBLEM FOR RANDOM PLANAR DISK-LIKE CANTOR
                                                      SETS
2.1     Introduction
    Let 𝐸 be a subset of the unit disk, D. The Buffon needle problem wants to determine the
probability with which a random needle or line intersects 𝐸 provided that it already intersects the
unit disk. At the same time, let 𝑙 𝜃 be the line passing through the origin and forming angle 𝜃 with
the horizontal axis. The Favard length of 𝐸 is the average length of the projection of 𝐸 onto 𝑙 𝜃
when averaging over all angles 𝜃. It turns out these two quantities are proportional.
    Now, consider the following picture: let us have 𝐿 many (𝐿 ≥ 3) disjoint closed disks
(𝐷 1 , . . . , 𝐷 𝐿 ) of diameter 1/𝐿 and strictly inside D. These are disks of the first generation.
Consider also a piecewise affine map 𝑓 = ( 𝑓1 , . . . , 𝑓 𝐿 ) from those disks onto D. Then, 𝑓 −1 (D) =
𝐷 1 ∪ · · · ∪ 𝐷 𝐿 . Furthermore, 𝑓 −1 (𝐷 1 ∪ · · · ∪ 𝐷 𝐿 ) is consists of 𝐿 2 disks (groups of 𝐿 many disks in
each 𝐷 𝑖 ); we call those disks of the second generation. We can iterate this procedure: denoting by
𝑈𝑛 the union of disks of the 𝑛-th generation, where 𝑈1 := 𝐷 1 ∪ · · · ∪ 𝐷 𝐿 , we form the self-similar
Cantor set K = ∞
                      Ñ
                        𝑛=1 𝑈𝑛 . This has positive and finite 1-dimensional Hausdorff measure; thus it is
completely unrectifiable in the sense of Besicovitch [8]; and thus its Favard length is zero [8].
    Of course, the disks can be replaced by other shapes. For example, 𝑈1 can consist of 𝐿 disjoint
squares with side-length 1/𝐿 inside the unit square [0, 1] 2 (where the word “strictly” can be omitted
but “disjoint” cannot). One of such Cantor sets is a rather “famous”, namely the 1/4-corner Cantor
set, K1/4 (see [7]).
    The 𝐿 −𝑛 -neighbourhood of such sets is roughly 𝑈𝑛 , and therefore its Favard length
                                          Fav(𝑈𝑛 ) → 0,        as 𝑛 → ∞.
But what is Fav(𝑈𝑛 ), or what is the speed with which Fav(𝑈𝑛 ) decreases? Nobody knows exactly,
but there has been considerable interest in recent years. It is now clear that the answer may depend
on several factors; the magnitude of 𝐿; the geometry of 𝑈1 ; the subtle algebraic and number
                                                        30


theoretic properties of a certain trigonometric sum built by the centres of the disks of the first
generation. See [2, 3, 4, 6, 10] and the survey paper [5].
    For the 1/4-corner Cantor set K1/4 in particular, the best known estimate from above for its
4−𝑛 -neighbourhood is
                                                       𝐶𝜖
                                 Fav(𝑁4−𝑛 (K1/4 )) ≤    1
                                                             ,     ∀𝜖 > 0,
                                                      𝑛 6 −𝜖
for all large 𝑛. We suspect that this estimate can be improved to
                                                       𝐶𝜖
                                 Fav(𝑁4−𝑛 (K1/4 )) ≤         ,     ∀𝜖 > 0,
                                                      𝑛1+𝜖
but at this moment this is only a conjecture.
    On the other hand, there is a universal estimate from below obtained in [9] for every self-similar
Cantor set constructed as above:
                                                               𝑐
                                          Fav(𝑁4−𝑛 (K)) ≥        .                             (2.1.1)
                                                               𝑛
For any concrete set, this bound from below could be improved. In fact, it is proven in [1] that for
the same 1/4-corner Cantor set K1/4
                                                            𝑐 log 𝑛
                                      Fav(𝑁4−𝑛 (K1/4 )) ≥            .
                                                                 𝑛
    For random Cantor sets the situation should be simpler. With large probability, Mattila’s lower
estimate (2.1.1) is met by the same estimate from above (with a different constant). The problem is
that in general there can be many different models of randomness.
    In this note, we are interested in an analogue of the random Cantor set appearing in [11] and in
[14]. In our case, this will come from the random Cantor disks constructed below at Section 2.2.
The model of randomness presented here is somewhat different from the ones in the above two
papers, but it amazingly exhibits the same behaviour, as we’ll see below in our main Theorem 2.3.1,
which we contrast with [11, Theorem 2.2] and [14, Theorem 1].
    In particular, we prove an analogue of [14, Theorem 1]. Unfortunately, the randomness of the
disk model we study here is not equivalent to that of the random (square) Cantor set R = ∞
                                                                                             Ñ
                                                                                                𝑛=0 R 𝑛
                                                  31


from [11], but it is nonetheless closer compared to the one constructed in [14]. The essential
                                                                                             𝑛−1
difference between [14] and our consideration are the angles 𝜔1𝑛 , 𝜔2𝑛 , . . . , 𝜔4𝑛             , which are here
allowed to be distinct and independent whereas in [14] are all equal. So, our model is a little “more
random” than the random Cantor sets of Zhang in [14].
    We introduce our notations —some borrowed from [14]— in the next Section 2.2. The problem
of interest, namely the Favard length of a random planar disk-like Cantor set, is explained in
Section 2.3. Our results and their proofs are postponed to Sections 2.4 and 2.5. In Section 2.6, we
compare the differences and difficulties between our work and that of Peres and Solomyak’s and
Zhang’s.
2.2   Cantor Disks
    Our work will be heavy on notation; without any ado let us introduce our basic “vocabulary”.
    The letter 𝑛 will stand for a (large) positive integer.
    The letter 𝜔 will be used to denote angles with values inside the interval [0, 𝜋2 ]. Now, let us
consider a word of length 𝑛 made of the alphabet of angles in [0, 𝜋2 ], i.e. a word of the form
𝜔1 𝜔2 · · · 𝜔𝑛 . The subscript in 𝜔 𝑘 denotes the position of the angle 𝜔 𝑘 within such a word of length
𝑛. We refer to the position of an angle within a word as the depth of that angle.
    Our operators, which we will introduce below, are such that every choice of an angle, say,
𝜔1 necessitates four different independent choices for the angle 𝜔2 ; every choice of the angle
𝜔2 necessitates four different independent choices for the angle 𝜔3 ; and so on up until depth 𝑛
where we will have 4𝑛−1 different angles 𝜔𝑛 . In order to differentiate between all those, for each
                                           𝑗
𝑗 𝑘 = 1, 2, . . . , 4 𝑘−1 we write 𝜔 𝑘𝑘 for the 𝑗 𝑘 -th choice of an angle 𝜔 𝑘 at depth 𝑘. Notice there are
4 𝑘−1 such choices. Therefore, a typical word from our alphabet of angles looks as follows, where
                    𝑗
we note that 𝜔 𝑘𝑘 ∈ [0, 𝜋2 ]:
                                                         𝑗1 = 1 ,        𝑗 𝑘 = 1, 2, . . . , 4 𝑘−1 ,
                     𝑗   𝑗        𝑗          𝑗
                  𝜔11 𝜔22 · · · 𝜔 𝑘𝑘 · · · 𝜔𝑛𝑛 where 𝑗2 = 1, 2, 3, 4 ,       ···
                                                            ···          𝑗 𝑛 = 1, 2, . . . , 4𝑛−1 .
                                                         32


     At certain instances, we need to consider cumulatively all angles of a certain depth; given a
collection of words of length 𝑛, for each 𝑘 = 1, 2, . . . , 𝑛 let 𝜔′𝑘 be the collection of all 4 𝑘−1 many
angles at depth 𝑘, that is 𝜔′𝑘 = (𝜔1𝑘 , . . . , 𝜔4𝑘
                                                             𝑘−1
                                                                 ). With this notation, we may use the symbols 𝜔1 ,
𝜔11 , 𝜔11 , and 𝜔′1 interchangeably as these all refer to the same single angle.
   𝑗
     All the above give to our angles the structure of a rooted tree of height 𝑛 with root 𝜔1 and
such that each parent has four children. The vertexes have values in [0, 𝜋2 ], and are independent
from each other and from their predecessors and ancestors. This tree we denote by 𝜔′1 · · · 𝜔′𝑛 ; the
trimmed tree with root 𝜔1 and height 𝑘 we denote as 𝜔′1 · · · 𝜔′𝑘 (for any 𝑘 = 1, 2, . . . , 𝑛). For the
                                                       𝑗
subtree of height 𝑛 − 𝑘 + 1 with root 𝜔 𝑘𝑘 , which reaches up to the leaves (that is, from depth 𝑘 till
                                        𝑗                     𝑗
depth 𝑛 with starting vertex 𝜔 𝑘𝑘 ) we write 𝜔¯ 𝑘𝑘 . Later on, we will be working with rooted subtrees
                 𝑗                                𝑗                                    𝑗
of the form 𝜔¯ 𝑛−𝑘+1
                   𝑛−𝑘+1
                         . To reiterate, 𝜔¯ 𝑛−𝑘+1   𝑛−𝑘+1
                                                           consists of the angle 𝜔𝑛−𝑘+1  𝑛−𝑘+1
                                                                                               (as its root) along with all
                                                                               𝑗
the angles from depth 𝑛 − 𝑘 + 1 till depth 𝑛 (which have 𝜔𝑛−𝑘+1                  𝑛−𝑘+1
                                                                                       as an ancestor). This has height
                         𝑗
𝑘. Alternatively, 𝜔¯ 𝑛−𝑘+1 𝑛−𝑘+1
                                 is the collection of all the words (from our alphabet of angles) which have
                                         𝑗
depth 𝑘 and the first letter is 𝜔𝑛−𝑘+1     𝑛−𝑘+1
                                                 . There are 4𝑛−𝑘 such words.
     Next, we will need to introduce certain operators and sets. The main objects of interest will be
the operators D 𝑘 (𝑘 = 0, 1, . . . , 𝑛) which will act on trees of angles of depth 𝑘. To understand these
we need some auxiliary constructions first.
     For any angle 𝜔 and for 𝛼 = 0, 1, 2, 3 consider the transformations
                                                                1      3    𝜋
                                                   𝑇𝛼𝜔 (𝑧) =      𝑧 + 𝑒 (𝛼 2 −𝜔)𝑖
                                                                4      4
where 𝑧 is any number on the complex plane C. Observe that if D is the unit disk, 𝑇00 (D), 𝑇10 (D),
𝑇20 (D), and 𝑇30 (D) are disks of radius 1/4 centred respectively at (3/4, 0), (0, 3/4), (−3/4, 0), and
(0, −3/4). Introducing an angle 𝜔 in 𝑇𝛼𝜔 (D), rotates (about (0, 0)) the aforementioned disks by
angle 𝜔 in the clockwise direction.
                                           𝑗                             𝑗
     Moreover, given an angle 𝜔 𝑘𝑘 from depth 𝑘 let Ω 𝑘𝑘 be the set
                                                              3
                                                      𝑗
                                                            Ø       1 𝜔 𝑘𝑗𝑘
                                                    Ω 𝑘𝑘  =            𝑇 (D).
                                                                    𝑘−1 𝛼
                                                            𝛼=0
                                                                  4
                                                                  33


That is, Ω 𝑘𝑘 is a collection of four disks of radius 4−𝑘 with centres (0, ±3/4 𝑘 ) and (±3/4 𝑘 , 0) rotated
               𝑗
                       𝑗
clockwise by 𝜔 𝑘𝑘 .
                                                                                                              𝑗
      We also give an enumeration to all the disks for all depths. We number the disks of Ω 𝑘𝑘 so
                  𝑗𝑘
         1     𝜔
that         𝑇 𝑘 (D) is the (4 𝑗 𝑘 − 3 + 𝛼)-th disk at depth 𝑘. We call this the 𝑘-depth enumeration (of
       4 𝑘−1 𝛼
                                                                                                      𝑘−1
                                                                1    𝜔1        1   𝜔1        1    𝜔4𝑘
the disks lying at depth 𝑘). Illustratively, we note               𝑇 𝑘 (D),
                                                              4 𝑘−1 0
                                                                                  𝑇 𝑘 (D),
                                                                             4 𝑘−1 1
                                                                                                𝑇
                                                                                           4 𝑘−1 0
                                                                                                          (D), and
            𝑘−1
  1     𝜔4𝑘
      𝑇
4 𝑘−1 3
                 (D) are respectively the 1st, 2nd, (4 𝑘 − 3)-th, and 4 𝑘 -th disks of depth 𝑘. We retain this
enumeration as we translate these disks at different positions on the plane. This will be useful to
track down each disk at each step so that our subsequent constructions make better sense.
      Now, we are ready to introduce our main protagonists. The operator D 𝑘 acts on the collection
of trees (of angles) of height 𝑘 and for each such tree outputs a certain collection of 4 𝑘 disks of
radius 4−𝑘 . We define these inductively below.
      To begin with, set D0 = D to be the unit disk.
      Next, we define D1 by
                                                             3
                                                                   𝜔1
                                                            Ø
                                          D1 (𝜔′1 ) = Ω11 =      𝑇𝛼 1 (D0 ),
                                                            𝛼=0
that is,     D1 (𝜔′1 )   consists of four disks of radius 1/4 centred at (0, ±3/4) and (±3/4, 0) rotated
clockwise by 𝜔1 . Recall these disks are enumerated as in Ω11 .
      For the operator D2 , consider a tree of height 2, 𝜔′1 𝜔′2 , which consists of the angles 𝜔11 , and
𝜔12 , 𝜔22 , 𝜔32 , 𝜔42 . Then, we define D2 (𝜔′1 𝜔′2 ) to be the collection of disks constructed as follows:
Replace the 1st, 2nd, 3rd and 4th disk of D1 (𝜔′1 ) respectively by Ω12 , Ω22 , Ω32 and Ω42 . By replacing
                                     𝑗
we mean the translation of Ω2 in such a way that (0, 0) is translated to the centre of the 𝑗-th disk of
D1 (𝜔′1 ).
      Consequently, D2 (𝜔′1 𝜔′2 ) consists of 42 disks of radius 4−2 translated appropriately so that each
Ω22 replaces one the disks from D1 (𝜔′1 ). The set, say, Ω12 is in fact a subset of the 1st disk of
   𝑗
D1 (𝜔′); actually D2 (𝜔′1 𝜔′2 ) ⊂ D1 (𝜔′1 ). Again, the disks comprising D2 (𝜔′1 𝜔′2 ) are enumerated
to match Ω12 , Ω22 , Ω32 and Ω42 as we described above. Also see the Figure 2.1 below.
                                                         34


                                     D0
                                                                            𝜔22
                                        𝜔32
                                                                 𝜔1
                                                                               D2 (𝜔′1 𝜔′2 )
                                                            𝜔12
                                                                    Ω12
                                                                                   𝜔12
                                      D1 (𝜔′1 )
                                                    𝜔42
                          Figure 2.1 The collections D0 , D1 (𝜔′1 ), D2 (𝜔′1 𝜔′2 ) and Ω12 .
    Continuing inductively, the operator D 𝑘 acts on the tree 𝜔′1 · · · 𝜔′𝑘 in this manner: Consider the
collection D 𝑘−1 (𝜔′1 · · · 𝜔′𝑘−1 ). These are 4 𝑘−1 many (enumerated) disks. Replace the 1st of them
by Ω1𝑘 , the 2nd of them by Ω2𝑘 , etc., until every disk of D 𝑘−1 (𝜔′1 · · · 𝜔′𝑘−1 ) has been replaced by four
                                                                                            𝑗
smaller ones. This replacement is done so that (0, 0), as the “centre” of Ω 𝑘 , is translated to the centre
of the 𝑗-th disk of D 𝑘−1 (𝜔′1 · · · 𝜔′𝑘−1 ). That is, we substitute the 𝑗-th disk (from depth 𝑘 − 1) with
the (4 𝑗 − 3)-, (4 𝑗 − 2)-, (4 𝑗 − 1)-, and 4 𝑗-th disks of depth 𝑘. The resulting collection, which has
4 𝑘 many disks of radius 4−𝑘 , is D 𝑘 (𝜔′1 . . . 𝜔′𝑘 ). It holds that D 𝑘 (𝜔′1 · · · 𝜔′𝑘 ) ⊂ D𝑘−1 (𝜔′1 · · · 𝜔′𝑘−1 ).
    In the present work, we will study the collection of disks D𝑛 (𝜔′1 · · · 𝜔′𝑛 ) where the angles 𝜔 𝑘𝑘
                                                                                                                   𝑗
(for 𝑗 𝑘 = 1, . . . , 4 𝑘−1 and all 𝑘 = 1, 2, . . . , 𝑛) of the tree 𝜔′1 · · · 𝜔′𝑛 are chosen randomly with uniform
and independent distributions on the interval [0, 𝜋2 ]. So, let us describe this picture once more
before moving on further.
    The set D𝑛 (𝜔′1 · · · 𝜔′𝑛 ) consists of 4𝑛 disks of radius 4−𝑛 . These can be separated into 4𝑛−1
groups of four, which are copies of
                                                         3
                                                 𝑗
                                                        Ø      1 𝜔𝑛𝑗𝑛
                                             Ω𝑛𝑛 =                𝑇 (D0 )
                                                               𝑛−1 𝛼
                                                        𝛼=0
                                                             4
                                                              35


(for 𝑗 𝑛 = 1, . . . , 4𝑛−1 ) translated appropriately within the unit disk.
2.3    Favard Length
     Recall the Favard length of a planar set 𝐸 ⊂ C is the integral
                                                               ∫    𝜋
                                                            1
                                                 Fav(𝐸) =              proj𝜃 𝐸 𝑑𝜃
                                                            𝜋    0
where proj𝜃 𝐸 is the projection of 𝐸 onto the line with slope tan 𝜃 passing through the origin, and
| 𝐴| is the (1-dimensional) Lebesgue measure of 𝐴.
     Now, consider an infinite tree of angles from [0, 𝜋2 ] with root 𝜔1 and four branches at each
vertex, and let D be the limit set
                                                        Ù∞
                                                   D=       D𝑛 (𝜔′1 · · · 𝜔′𝑛 ).
                                                        𝑛=0
Notice that by construction, D a purely unrectifiable planar set. As such, Fav(D) = 0 and
by dominated convergence Fav(D𝑛 (𝜔′1 · · · 𝜔′𝑛 )) → 0 while 𝑛 → ∞. In fact, if the angles are
randomly chosen uniformly and independently over [0, 𝜋2 ], by dominated convergence and Fubini
E[Fav(D)] = 0 and E[Fav(D𝑛 (𝜔′1 · · · 𝜔′𝑛 ))] → 0 as 𝑛 → ∞, where the expectation is taken over
all such angles.
     The question arises as to the rate with which E[Fav(D𝑛 (𝜔′1 · · · 𝜔′𝑛 ))] goes to 0. This we answer
in the following theorem:
Theorem 2.3.1. Let 𝑛 ∈ N and consider a tree of angles of height 𝑛 with each vertex having
                                                         𝑗
four branches. Suppose that the angles 𝜔 𝑘𝑘 (for all 𝑗 𝑘 = 1, 2, . . . , 4 𝑘−1 and all 𝑘 = 1, 2, . . . , 𝑛)
are chosen randomly with uniform and independent distributions on the interval [0, 𝜋2 ]. Also set
𝜔′𝑘 = (𝜔1𝑘 , 𝜔2𝑘 , . . . , 𝜔4𝑘
                               𝑘−1
                                   ) for each 𝑘 = 1, 2, . . . , 𝑛. Then, there exists a constant 𝑐 > 0 such that
for any 𝜃 ∈ [0, 𝜋2 ] it holds that
                                                                             𝑐
                                   E𝜔1′ ··· 𝜔′𝑛 proj𝜃 D𝑛 (𝜔′1 · · · 𝜔′𝑛 ) ≤       ∀𝑛 ∈ N.                (2.3.1)
                                                                             𝑛
     Consequently,
                                                                              𝑐
                                   E𝜔1′ ··· 𝜔′𝑛 [Fav(D𝑛 (𝜔′1 · · · 𝜔′𝑛 ))] ≤      ∀𝑛 ∈ N                 (2.3.2)
                                                                              𝑛
                                                               36


and also
                       lim inf 𝑛 Fav(D𝑛 (𝜔′1 · · · 𝜔′𝑛 )) < ∞               ∀𝑛 ∈ N almost surely.                (2.3.3)
                         𝑛→∞
Clearly, (2.3.3) follows from (2.3.2) by an immediate application of Fatou’s lemma, whereas (2.3.2)
follows from (2.3.1) through Fubini.
2.4    Statement and use of the main lemma
    The present and the following sections are dedicated to the proof of (2.3.1). Towards this goal,
we need to introduce Lemma 2.4.1 below, which describes the decay of the average projection when
transitioning from depth 𝑘 to depth 𝑘 + 1. The main difficulty will come from obtaining the square
factor appearing in (2.4.1), which emanates from the naturally occurring overlap of the projections.
    From now on, suppose we are given a tree of angles of height 𝑛 with four branches at each
vertex where the angles are uniformly and independently distributed random variables on the
                                                            𝑗                                                      𝑗
interval [0, 𝜋2 ]. Recall that given such a tree 𝜔¯ 𝑛−𝑘+1     𝑛−𝑘+1
                                                                    is the subtree of height 𝑘 with the vertex 𝜔𝑛−𝑘+1𝑛−𝑘+1
as its root. Observe that 𝜔¯ 11 = 𝜔′1 · · · 𝜔′𝑛 is the full tree whilst 𝜔¯ 𝑛𝑛 = 𝜔𝑛𝑛 ( 𝑗 𝑛 = 1, 2, . . . , 4𝑛−1 ) are
                                 𝑗                                                  𝑗        𝑗
the its leaves, i.e. trees of height 1.
    For any 𝜃 ∈ [0, 𝜋2 ] and all 𝑘 = 1, 2, . . . , 𝑛, define the following quantities
                         𝑗                          𝑗
                      𝐷 1𝑛 = E𝜔¯ 𝑗𝑛 proj𝜃 D1 ( 𝜔¯ 𝑛𝑛 ) ,                          𝑗 𝑛 = 1, 2, . . . , 4𝑛−1
                                 𝑛
                    𝑗                                     𝑗
                  𝐷 𝑘𝑛−𝑘+1 = E𝜔¯ 𝑗𝑛−𝑘+1 proj𝜃 D 𝑘 ( 𝜔¯ 𝑛−𝑘+1𝑛−𝑘+1
                                                                  ) ,             𝑗 𝑛−𝑘+1 = 1, 2, . . . , 4𝑛−𝑘
                                 𝑛−𝑘+1
             𝐷 𝑛1 = 𝐷 1𝑛 = E𝜔¯ 𝑗1 proj𝜃 D𝑛 ( 𝜔¯ 11 ) ,
                 𝑗                                  𝑗
                                                                                  𝑗 1 = 1.
                                 1
                                                                                                               𝑗
Notice that, because we are averaging over the independent and identically distributed 𝜔 𝑘𝑘 ,
                                                         𝑛−𝑘
                            𝐷 1𝑘 = 𝐷 2𝑘 = · · · = 𝐷 4𝑘               for any 𝑘 = 1, 2, . . . , 𝑛.
Therefore, it suffices to work with 𝐷 1𝑘 ; the rest should be identical. Also, note that
                                      𝐷 1𝑛 = E𝜔1′ ··· 𝜔′𝑛 proj𝜃 D𝑛 (𝜔′1 · · · 𝜔′𝑛 ) .
    Now, we are ready to state a simple but important lemma. Also, see [14, Lemma 2.1].
                                                                37


                                                      𝑗
Lemma 2.4.1. With notation as above, if 𝜔 𝑘𝑘 are uniformly and independently distributed random
variables on [0, 𝜋2 ], there exits a constant 𝑐 ≥ 4 such that for any 𝑛 ∈ N (and any 𝜃 ∈ [0, 𝜋2 ])
                          𝐷 1𝑘+1 ≤ 𝐷 1𝑘 − 𝑐−1 (𝐷 1𝑘 ) 2        for all 𝑘 = 1, . . . , 𝑛 − 1.          (2.4.1)
In fact, we will see below that 𝑐 = 8.
     Provided this holds true we can give a very compact proof of Theorem 2.3.1 using induction:
Proof of Theorem 2.3.1. Let 𝑐 be as in Lemma 2.4.1 and note that 𝐷 12 ≤ 𝐷 11 < 2 ≤                 𝑐
                                                                                                   2.  Also,
𝐷 11 < 𝑐.
     Next, assume 𝐷 1𝑘 <     𝑐
                             𝑘 for some 2 ≤ 𝑘 ≤ 𝑛 − 1. From Lemma 2.4.1, and by the monotonicity
of the function 𝑥 − 𝑥 2 /𝑐 in [0, 2𝑐 ], we see that
                                                             𝑐     𝑐      𝑘 −1             𝑐
                         𝐷 1𝑘+1 ≤ 𝐷 1𝑘 − 𝑐−1 (𝐷 1𝑘 ) 2 <       − 2 =𝑐 2 <                     .
                                                             𝑘 𝑘           𝑘             𝑘 +1
Therefore, 𝐷 1𝑘 <   𝑐
                    𝑘 holds for all for 1 ≤ 𝑘 ≤ 𝑛 − 1 and thus for 𝑘 = 𝑛 we get
                                                                                     𝑐
                                 E𝜔1′ ··· 𝜔′𝑛 proj𝜃 D𝑛 (𝜔′1 · · · 𝜔′𝑛 ) = 𝐷 1𝑛 <       .
                                                                                     𝑛
This is (2.3.1). Equation (2.3.2) follows after integrating with respect to 𝜃, and (2.3.3) after applying
Fatou’s Lemma.                                                                                             □
2.5    Proving the main lemma
     Whatever follows is dedicated to the proof of (2.4.1).
     First, we rewrite the length of the projection of a set in more convenient way. Let 𝑙 𝜃 and
𝑙 𝜃⊥ be two lines through the origin so that 𝑙 𝜃 forms an angle 𝜃 with the horizontal axis and 𝑙 𝜃⊥ is
perpendicular to 𝑙 𝜃 . Also, let n be the unit normal vector of 𝑙 𝜃⊥ . The length of the projection of a
planar set 𝐸 ⊂ C onto the line 𝑙 𝜃 can be written as
                                                                             ∫
                       proj𝜃 𝐸 = {𝑡 ∈ R :        (𝑙 𝜃⊥  + 𝑡n) ∩ 𝐸 ≠ ∅} =                       𝑑𝑡.    (2.5.1)
                                                                               (𝑙 ⊥𝜃 +𝑡n)∩𝐸≠∅
     For brevity, we denote the line 𝑙 𝜃⊥ + 𝑡n by 𝑙 𝜃⊥ (𝑡) where 𝑡 ∈ R. Additionally, because of the
symmetry of our considerations, we can assume without loss of generality that 𝜃 = 0 —as we will
average over all 𝜃 at the end. So, we can simply omit writing 𝜃 altogether from now on.
                                                           38


    The idea behind Lemma 2.4.1 is to look at the collection D𝑛 (𝜔′1 · · · 𝜔′𝑛 ) at depth 𝑛 but “zoomed
in” so that it looks like depth 1. Then, we go one level up and look at the disks of depth 𝑛 − 1 and 𝑛
zooming in enough so that they to look like depth 2; and so forth. If we rewrite the projections in
the form of (2.5.1), the average overlap at each level is of at least a square factor compared to the
total average projection of the level above.
    This last comparison is paramount to the proof. It will follow from the fact that the disks in
our constructions never get too close to one another. In fact, this observation is not true in the case
of the random square Cantor sets, which is the reason why we cannot directly apply the arguments
here to the setting of [11].
    Let us proceed with the proof of (2.4.1).
    Fix some 𝑘 = 1, 2, . . . , 𝑛 and recall that by construction
                                                                 3
                                                                     𝜔1
                                                               Ø
                                           D1 (𝜔1𝑛−𝑘+1 )     =      𝑇𝛼 𝑛−𝑘+1 (D0 ).
                                                               𝛼=0
This means that each disk from the collection D𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) lies inside one of the above four disks,
and therefore we can separate D 𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) into four groups of disks depending on their positioning
at depth 1.
    More precisely, for each 𝛼 = 0, 1, 2, 3 define T𝛼𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) as
                                                         𝜔1
                                                                       Ù
                                 T𝛼𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) = 𝑇𝛼 𝑛−𝑘+1 (D0 )           D𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ).
That is, the set T𝛼𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) consists of those disks of D 𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) which lie inside the 41 -radius
           𝜔1
disk 𝑇𝛼 𝑛−𝑘+1 (D0 ). We can think of T𝛼𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) as the East, North, West, and South parts of
D 𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ), respectively for 𝛼 = 0, 1, 2, 3. From this definition, it is also clear that
                                                                Ø3
                                           D 𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) =      T𝛼𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ).                   (2.5.2)
                                                                𝛼=0
    In fact, T𝛼𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) depends only on the angle 𝜔1𝑛−𝑘+1 and the subtree 𝜔¯ 4·1−3+𝛼             ¯ 1+𝛼
                                                                                                𝑛−𝑘+2 = 𝜔 𝑛−𝑘+2 .
(Recall our enumeration of the angles in Section 2.2.) Thus, we can write T𝛼𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) as
                                      T𝛼𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) = T𝛼𝑘 (𝜔1𝑛−𝑘+1 , 𝜔¯ 1+𝛼 𝑛−𝑘+2 ).
                                                               39


2.5.1    Key observations
    There are two key observations regarding the sets T𝛼𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ). First, note that each point of
                                                                                                     𝜔1
the interval (−1, 1) can be covered by at most two of the projections proj 𝑇𝛼 𝑛−𝑘+1 (D0 ) for different
                                        𝜔1
𝛼’s. Since T𝛼𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) ⊂ 𝑇𝛼 𝑛−𝑘+1 (D0 ), the same holds true for proj T𝛼𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ); the intersection
Ñ             𝑘 ¯1
  𝛼 proj T𝛼 ( 𝜔     𝑛−𝑘+1 ) is empty when the intersection is over more than two values of 𝛼.
    Second, we can compare the average projections of T𝛼𝑘 ( 𝜔¯ 𝑛−𝑘+1 ) = T𝛼𝑘 (𝜔1𝑛−𝑘+1 , 𝜔¯ 1+𝛼                        𝑛−𝑘+2 ) and
D 𝑘−1 ( 𝜔¯ 1+𝛼
           𝑛−𝑘+2 ). Notice that both these collections consist of 4
                                                                                       𝑘−1 many disks, which in fact have
the same 𝑛-depth enumerations. This means that they correspond to same disks of the collection
D𝑛 (𝜔′1 · · · 𝜔′𝑛 ). The difference is that the disks of the former are translated according to D1 (𝜔1𝑛−𝑘+1 )
and have radius 4−𝑘 , whereas the ones of the latter have radius 4−(𝑘−1) .
    Consequently, T𝛼𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) is a shifted copy of D 𝑘−1 ( 𝜔¯ 1+𝛼           𝑛−𝑘+2 ) dilated by a factor of 1/4. As
such, the (average of the) projections of T𝛼𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) and D 𝑘−1 ( 𝜔¯ 1+𝛼               𝑛−𝑘+2 ) should also differ by a
factor of 1/4. In other words, for any 𝛼 = 0, 1, 2, 3 we have
          E𝜔¯ 1       proj T𝛼𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) = E𝜔1          E𝜔¯ 1+𝛼    proj T𝛼𝑘 (𝜔1𝑛−𝑘+1 , 𝜔¯ 1𝑛−𝑘+2 )
              𝑛−𝑘+1                                   𝑛−𝑘+1     𝑛−𝑘+2
                                                                                                                           (2.5.3)
                                                                          1
                                                                       = E𝜔¯ 1+𝛼 proj D 𝑘−1 ( 𝜔¯ 1+𝛼  𝑛−𝑘+2 ) .
                                                                          4 𝑛−𝑘+2
2.5.2    The estimates
    Utilising the above, we can now estimate 𝐷 1𝑘 in terms of 𝐷 1𝑘−1 :
         𝐷 1𝑘 = E𝜔¯ 1         proj D 𝑘1 ( 𝜔¯ 1𝑛−𝑘+1 )
                       𝑛−𝑘+1
                              ∑︁3
                ≤ E𝜔¯ 1            proj T𝛼𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) − E𝜔¯ 1          proj T0𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) ∩ proj T1𝑘 ( 𝜔¯ 1𝑛−𝑘+1 )
                        𝑛−𝑘+1                                       𝑛−𝑘+1
                              𝛼=0
                              3
                (2.5.3)   1 ∑︁
              ========            E𝜔¯ 1+𝛼   proj D 𝑘−1 ( 𝜔¯ 1+𝛼
                                                             𝑛−𝑘+2 )
                          4 𝛼=0
                                      𝑛−𝑘+2
                                                          − E𝜔¯ 1         proj T0𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) ∩ proj T1𝑘 ( 𝜔¯ 1𝑛−𝑘+1 )
                                                                  𝑛−𝑘+1
                                                                  40


where the inequality follows from the first observation above (see 2.5.1) and (2.5.2). This in turn
gives
           1 1
   𝐷 1𝑘 ≤     (𝐷       + 𝐷 2𝑘−1 + 𝐷 3𝑘−1 + 𝐷 4𝑘−1 ) − E𝜔¯ 1             proj T0𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) ∩ proj T1𝑘 ( 𝜔¯ 1𝑛−𝑘+1 )
           4 𝑘−1                                                𝑛−𝑘+1
                                                                                                                        (2.5.4)
        =  𝐷 1𝑘−1  − E𝜔¯ 1         proj T0 ( 𝜔¯ 1𝑛−𝑘+1 )
                                          𝑘
                                                         ∩ proj T1 ( 𝜔¯ 1𝑛−𝑘+1 )
                                                                     𝑘
                                                                                     ,
                           𝑛−𝑘+1
since 𝐷 1+𝛼          1
          𝑘−1 = 𝐷 𝑘−1 for any 𝛼 = 0, 1, 2, 3.
    The next step, is to estimate the overlap term proj T0𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) ∩ proj T1𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) from below.
For this, recall T0𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) and T1𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) depend (aside from 𝜔1𝑛−𝑘+1 ) respectively on 𝜔¯ 1𝑛−𝑘+2
and 𝜔¯ 2𝑛−𝑘+2 .
    First, we average with respect to the subtrees 𝜔¯ 1𝑛−𝑘+2 and 𝜔¯ 2𝑛−𝑘+2 , and afterwards we integrate
over their common ancestor 𝜔1𝑛−𝑘+1 . To simplify the notation, let us write 𝜔¯ 1,2                          𝑛−𝑘+2 for both the
subtrees 𝜔¯ 1𝑛−𝑘+2 and 𝜔¯ 2𝑛−𝑘+2 , and also 𝜓 for the angle 𝜔1𝑛−𝑘+1 . Then, we have
            E𝜔¯ 1,2     proj T0𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) ∩ proj T1𝑘 ( 𝜔¯ 1𝑛−𝑘+1 )
                 𝑛−𝑘+2
              = E𝜔¯ 1,2 proj T0𝑘 (𝜔1𝑛−𝑘+1 , 𝜔¯ 1𝑛−𝑘+2 ) ∩ proj T1𝑘 (𝜔1𝑛−𝑘+1 , 𝜔¯ 2𝑛−𝑘+2 )
                     𝑛−𝑘+2
                       ∫               
              (2.5.1)
            ========         P𝜔¯ 1,2 𝑙 ⊥ (𝑡) ∩ T0𝑘 (𝜔1𝑛−𝑘+1 ,𝜔¯ 1𝑛−𝑘+2 ) ≠ ∅
                                 𝑛−𝑘+2
                                                                                                              
                                                            and 𝑙 ⊥ (𝑡) ∩ T1𝑘 (𝜔1𝑛−𝑘+1 , 𝜔¯ 2𝑛−𝑘+2 ) ≠ ∅ 𝑑𝑡
                ∫                                                            
            =        P𝜔¯ 1,2 𝑙 ⊥ (𝑡) ∩ T0𝑘 (𝜔1𝑛−𝑘+1 ,𝜔¯ 1𝑛−𝑘+2 ) ≠ ∅ ·
                         𝑛−𝑘+2
                                                                                                                 
                                                          ·P𝜔¯ 1,2 𝑙 ⊥ (𝑡) ∩ T1𝑘 (𝜔1𝑛−𝑘+1 , 𝜔¯ 2𝑛−𝑘+2 ) ≠ ∅ 𝑑𝑡
                                                               𝑛−𝑘+2
                ∫                                                            
            =        P𝜔¯ 1       𝑙 ⊥ (𝑡) ∩ T0𝑘 (𝜔1𝑛−𝑘+1 ,𝜔¯ 1𝑛−𝑘+2 ) ≠ ∅ ·
                         𝑛−𝑘+2
                                                                                                                 
                                                          ·P𝜔¯ 2        𝑙 ⊥ (𝑡) ∩ T1𝑘 (𝜔1𝑛−𝑘+1 , 𝜔¯ 2𝑛−𝑘+2 ) ≠ ∅ 𝑑𝑡
                                                               𝑛−𝑘+2
              =: E (𝜔1𝑛−𝑘+1 ).
The 3rd equality above holds because for a fixed angle 𝜔1𝑛−𝑘+1 the events
                                           {𝑙 ⊥ (𝑡) ∩ T𝛼𝑘 (𝜔1𝑛−𝑘+1 , 𝜔¯ 1+𝛼  𝑛−𝑘+2 ) ≠ ∅}
are independent.
                                                                 41


    It would be very nice if these two events would have the same probability. Then at the end, we
would use Hölder inequality to get
                                     ∫ h                                                           i 2
                    E (𝜔1𝑛−𝑘+1 )  =       P𝜔¯ 1         𝑙 ⊥ (𝑡) ∩ T0𝑘 (𝜔1𝑛−𝑘+1 , 𝜔¯ 1𝑛−𝑘+2 ) ≠ ∅         𝑑𝑡
                                              𝑛−𝑘+2
                                        ∫                                                             2
                                                              ⊥           𝑘     1        1
                                  ≥𝐶          P𝜔¯ 1          𝑙 (𝑡) ∩ T0 (𝜔𝑛−𝑘+1 , 𝜔¯ 𝑛−𝑘+2 ) ≠ ∅ 𝑑𝑡 .
                                                  𝑛−𝑘+2
However, this is not the case.
    For brevity, let us temporarily denote
                                                                               3
                                    𝜓 := 𝜔1𝑛−𝑘+1          and     𝑠(𝜓) :=        (1 − cos 𝜓).                     (2.5.5)
                                                                               4
Also, set
                                       𝐹 (𝑡) := {𝑙 ⊥ (𝑡) ∩ T0𝑘 (0, 𝜔¯ 1𝑛−𝑘+2 ) ≠ ∅}.                              (2.5.6)
For fixed 𝜓, the events
                     {𝑙 ⊥ (𝑡) ∩ T0𝑘 (𝜓, 𝜔¯ 1𝑛−𝑘+2 ) ≠ ∅}         and      {𝑙 ⊥ (𝑡) ∩ T1𝑘 (𝜓, 𝜔¯ 2𝑛−𝑘+2 ) ≠ ∅}
do not have the same probability; one should take into consideration that the probability of the non-
empty intersection with 𝑙 ⊥ (𝑡) for the first T has the same probability as the non-empty intersection
with 𝑙 ⊥ (𝑡 + 𝑠(𝜓))) with T0𝑘 (0, 𝜔¯ 1𝑛−𝑘+2 ). (Notice what happens with 𝜓!) And the probability of
the non-empty intersection with 𝑙 ⊥ (𝑡) for the second T has the same probability as the non-empty
intersection with 𝑙 ⊥ (𝑡 + 𝑠(𝜓 + 𝜋2 ))) for the event T0𝑘 (0, 𝜔¯ 1𝑛−𝑘+2 ). In fact, a simple geometric
consideration shows the following holds:
Lemma 2.5.1. With notation as above we have that
                                                                                                            
    P𝜔¯ 1       𝑙 ⊥ (𝑡) ∩ T0𝑘 (𝜓, 𝜔¯ 1𝑛−𝑘+2 ) ≠ ∅ = P𝜔¯ 1              𝑙 ⊥ (𝑡 + 𝑠(𝜓)) ∩ T0𝑘 (0, 𝜔¯ 1𝑛−𝑘+2 ) ≠ ∅ . (2.5.7)
        𝑛−𝑘+2                                                  𝑛−𝑘+2
                                                                 42


    In other words, Lemma 2.5.1 shows that
         E (𝜓) =
                 ∫                                                               
               =    P𝜔¯ 1        𝑙 ⊥ (𝑡 + 𝑠(𝜓)) ∩ T0𝑘 (0, 𝜔¯ 1𝑛−𝑘+2 ) ≠ ∅ ·
                        𝑛−𝑘+2
                                                                                 𝜋                             
                                                                 ⊥                          𝑘      1
                                                   ·P𝜔¯ 1       𝑙 (𝑡 + 𝑠(𝜓 + )) ∩ T0 (0, 𝜔¯ 𝑛−𝑘+2 ) ≠ ∅ 𝑑𝑡
                 ∫
                                                        𝑛−𝑘+2                     2
                                                             𝜋
               =     𝐹 (𝑡 + 𝑠(𝜓)) · 𝐹 (𝑡 + 𝑠(𝜓 + ))𝑑𝑡.
                                                             2
    Next, if we integrate over 𝜓 ∈ [0, 𝜋2 ], we get that the
                                             ∫                    ∫ ∫
                                                                                                           𝜋
      Expectation of the overlap =                E (𝜓)𝑑𝜓 =                 𝐹 (𝑡 + 𝑠(𝜓)) · 𝐹 (𝑡 + 𝑠(𝜓 +      ))𝑑𝜓𝑑𝑡.
                                                                                                           2
Let’s make this change of variables: 𝑢 = 𝑡 + 34 (1 − cos 𝜓) and 𝑣 = 𝑡 + 43 (1 − cos(𝜓 + 𝜋2 )). The
Jacobian of this change is at most 43 , and thus
                                                          ∫ ∫                                 ∫           2
                                                       4                                   4
             Expectation of the overlap ≥                          𝐹 (𝑢)𝐹 (𝑣)𝑑𝑢𝑑𝑣 =                𝐹 (𝑡)𝑑𝑡    .
                                                       3                                   3
    Since there is no dependence on 𝜔¯ 3𝑛−𝑘+1 or 𝜔¯ 4𝑛−𝑘+1 , we get
                         E𝜔¯ 1         proj T0𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) ∩ proj T1𝑘 ( 𝜔¯ 1𝑛−𝑘+1 )
                             𝑛−𝑘+1
                                      = Expectation of the overlap
                                           4                                     2
                                      ≥        E𝜔¯ 1       proj T0𝑘 ( 𝜔¯ 1𝑛−𝑘+1 )
                                           3       𝑛−𝑘+1
                                        (2.5.3) 4      1                                       2
                                      ======== ·            E𝜔¯ 1      proj D 𝑘−1 ( 𝜔¯ 1𝑛−𝑘+1 )
                                                 3 16            𝑛−𝑘+2
                                           1
                                      =       (𝐷 1𝑘−1 ) 2 .
                                          12
    Finally, combing the fact that
                                                                                          1
                    E𝜔¯ 1        proj T0𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) ∩ proj T1𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) ≥           (𝐷 1𝑘−1 ) 2
                         𝑛−𝑘+1                                                           12
with (2.5.4) and setting 𝑐 = 12 we get
                                               𝐷 1𝑘 ≤ 𝐷 1𝑘−1 − 𝑐−1 (𝐷 1𝑘−1 ) 2
and Lemma 2.4.1 is proved.
                                                               43


2.6    Comparison with the other random models
     The random Cantor set in [14] is a very close relative of the random Cantor set in this note,
the difference is that Zhang’s random construction of 𝑛 generations has 𝑛 independent rotations
involved, whereas our construction has 1 + · · · + 4𝑛−1 independent rotations. There the disks of
generation 𝑘 are rotated by the same angle 𝜔 𝑘 , while in this note we have 4 𝑘−1 independent rotations
of disks of generation 𝑘. Naturally, it is more difficult to work in a more chaotic model such as
ours, and the techniques here use independence in a more involved way than in [14]. It is just a
little harder to make sense of the combinatorics involved in our model.
     On the other hand, there are many “common places”: the use of overlap as the way to see the
rate of decays of successive approximations of the random Cantor set, the use of Lemma 2.4.1, as
well as the technical Lemma 2.5.1.
     Concerning [11], there are two main differences which create difficulties. The first is the fact
that at most two of the projections proj𝜃 T𝛼𝑘 ( 𝜔¯ 1𝑛−𝑘+1 ) can intersect at each point on the line 𝑙 𝜃 . This
is equivalent to line 𝑙 𝜃⊥ (𝑡) intersecting at most two of the disks for any 𝑡, and is key to the square
factor appearing in our calculations.
     However, this is simply not true in the case of squares. In fact, in the Peres and Solomyak case
the corresponding line 𝑙 𝜃⊥ (𝑡) can simultaneously intersect 3 squares of generation 𝑘 for any 𝑘 and
any 𝑡. Because of this, the inequalities appearing here cannot be translated directly in the square
setting.
     But even if this wasn’t an obstacle, the reader should pay attention to Lemma 2.5.1. Let’s
pretend that we can repeat everything before this lemma for the model of Peres and Solomyak. The
role of the angle 𝜔1𝑛−𝑘+1 will be played by the “Favard angle” 𝜃, the shift function 𝑠(𝜔1𝑛−𝑘+1 ) will
be replaced by
                                                           1
                                                   𝑆(𝜃) =    sin 𝜃,
                                                           2
and all seems to be following smoothly along the same lines. Also, the following equality
                   ∫                                      ∫
                      E1{𝑙 ⊥ (𝑡)∩T 𝑘 (𝜔¯ 1     )≠∅ } 𝑑𝜃 =    E1{𝑙 ⊥ (𝑡+𝑠(𝜃))∩T 𝑘 (𝜔¯ 1     )≠∅ } 𝑑𝜃,  (2.6.1)
                              𝜃    0     𝑛−𝑘+2                    𝜃           1      𝑛−𝑘+2
                                                          44


which would be the analogue of (2.5.7), makes sense in principle if we understand 𝜔’s as the random
variables in the Peres–Solomyak model, which assume the values 0, 1, 2, 3 (instead of values in the
interval [0, 𝜋2 ] as in our’s and Zhang’s models).
    But, there is a caveat. We reduced the function of two variables
                                                                                      
                             𝐺 (𝜓, 𝑡) := 𝑃𝜔¯ 1       𝑙 ⊥ (𝑡) ∩ T0𝑘 (𝜓, 𝜔¯ 1𝑛−𝑘+2 ) ≠ ∅
                                             𝑛−𝑘+2
to the composition with a function of one variable and the shift (see (2.5.6) for the definition of 𝐹):
                                 𝐺 (𝜓, 𝑡) = 𝐺 (0, 𝑡 + 𝑠(𝜓)) = 𝐹 (𝑡 + 𝑠(𝜓))                           (2.6.2)
thanks to (2.5.7). But looking at (2.6.1), we can notice that the function
                                      G(𝜃, 𝑡) := E1{𝑙 ⊥ (𝑡)∩T 𝑘 (𝜔¯ 1     )≠∅ }
                                                          𝜃    0    𝑛−𝑘+2
cannot be written as some F (𝑡 + 𝑆(𝜃)).
    As a result of this misfortune, we cannot write
                                                ∫                ∫ ∫
              Expectation of the overlap =          E (𝜃)𝑑𝜃 =               F (𝑡 + 𝑆(𝜃)) · F (𝑡)𝑑𝜃𝑑𝑡
                                                                                     ∫
as before. Working similarly, this would in turn bring about the term ( F 𝑑𝑡) 2 . Instead, we only
have that
                                           ∫                  ∫ ∫
           Expectation of the overlap =         E (𝜃)𝑑𝜃 =             G(𝜃, 𝑡) · G(𝜃, 𝑡 + 𝑆(𝜃)))𝑑𝜃𝑑𝑡,
and it is not clear (at least to us) how to estimate this integral from below as no change of variables
seems to be of help.
                                                          45


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     10.1112/plms/pdn037.
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                                                47


                                               CHAPTER 3
     GEOMETRY OF PLANAR CURVES INTERSECTING MANY LINES IN A FEW
                                                 POINTS
3.1    The statement of the problem
    The problem at hand is to better understand the structure of Borel sets in R2 that have a small
intersection with parallel shifts of lines from a whole cone. Here, we work only with sets that are
graphs and continuous curves. So we have strong assumptions. But the results claim some estimate
on the Hausdorff measure (not merely the Hausdorff dimension).
    Initially, we show that a function’s graph intersecting all parallel shifts of lines from a nonde-
generate cone in at most two points is locally Lipschitz and also present a counter-example showing
this fails if more intersection points are allowed.
    Next, we prove that any curve that has finitely many intersections with a cone of lines is 𝜎-finite
with respect to Hausdorff length and we find a bound on the Hausdorff measure of each “piece.”
    On the other hand, in [1] it was shown that, given countably many graphs of functions, there
is another function whose graph has only one intersection with all shifts of the given graphs but
whose graph has dimension 2.
    This result shows that there is a “thick” graph having only one intersection with all shifts of
countably many other graphs. In our turn, we show that the graph having finitely many intersection
with shifts of the whole cone of linear functions must be in fact very “thin”.
Proposition 3.1.1. Let 𝜆 > 0 be a fixed number and consider all the cones of lines with slopes
between 𝜆 and −𝜆 (containing the vertical line). If 𝑓 : (0, 1) → R is a continuous function such
that any line of these cones intersects its graph at at most two points, then 𝑓 is locally Lipschitz.
    Notice that our hypothesis implies that no three points of the graph of 𝑓 can lie on the same
line that is a parallel shift of a line from a given cone.
    For the proof we will need the following lemmas.
Lemma 3.1.2. Every convex (or concave) function on an open interval is locally Lipschitz.
                                                    48


           Figure 3.1 Each line from any cone intersects the graph at at most two points.
Lemma 3.1.3. If a function 𝑔 : (0, 1) → R is continuous and has a unique local extremum, 𝑥,             ˜
inside (0, 1), then it is strictly monotone in (0, 𝑥]  ˜ and [ 𝑥, ˜ 1) with opposite monotonicity on each
interval.
Proof of Lemma 3.1.3. Suppose 𝑥˜ is a local minimum for 𝑔. We will show that 𝑔 is strictly
monotone increasing in [ 𝑥, ˜ 1). Assume the contrary, i.e., consider two points 𝑥 1 < 𝑥2 ∈ [𝑥, ˜ 1) such
that 𝑔(𝑥1 ) ≥ 𝑔(𝑥 2 ). On the compact interval [𝑥 1 , 𝑥2 ], the function 𝑔 has to attain a minimum and
a maximum, which respectively are at 𝑥 2 and 𝑥1 otherwise the uniqueness of 𝑥˜ is contradicted. If
𝑥 1 = 𝑥,
      ˜ the point 𝑥˜ is not a local minimum and so 𝑥˜ < 𝑥 1 . Again, 𝑥˜ and 𝑥 1 must be the minimum
and maximum, respectively, of 𝑔 in [ 𝑥,  ˜ 𝑥 1 ], which in turn says 𝑥1 is a local maximum contradicting
the uniqueness of 𝑥. ˜ Therefore, 𝑔(𝑥 1 ) < 𝑔(𝑥 2 ) and 𝑔 is strictly monotone increasing on [ 𝑥,    ˜ 1).
Similarly, on (0, 𝑥]
                  ˜ 𝑔 is (strictly) monotone decreasing and the same arguments work for when 𝑥˜ is
local maximum.                                                                                          □
                                                          𝑓 (𝑥)− 𝑓 (𝑦)
Proof. Consider the slope function of 𝑓 , 𝑆(𝑥, 𝑦) =           𝑥−𝑦      , and note that
                                  𝑓 (𝑥) − 𝑓 (𝑦)
                      𝑆(𝑥, 𝑦) =                   = 𝜁 ⇐⇒ 𝑓 (𝑥) − 𝜁𝑥 = 𝑓 (𝑦) − 𝜁 𝑦.
                                      𝑥−𝑦
If for any two points 𝑥 < 𝑦 ∈ (0, 1) we have |𝑆(𝑥, 𝑦)| < 𝜆, then 𝑓 is Lipschitz (with Lipschitz
constant at most 𝜆).
                                                      49


Figure 3.2 𝑆(𝑥0 , 𝑦 0 ) = 𝜆′ ≥ 𝜆; The part of the graph of 𝑓 between 𝑥 0 and 𝑦 0 cannot lie on different
sides of 𝜖𝜆′ .
      Now suppose that there exist 𝑥0 , 𝑦 0 ∈ (0, 1) for which |𝑆(𝑥 0 , 𝑦 0 )| ≥ 𝜆 and consider the case
where 𝑆(𝑥 0 , 𝑦 0 ) = 𝜆′ ≥ 𝜆. Since 𝑆(𝑥, 𝑦) = 𝑆(𝑦, 𝑥), we may assume that 𝑥 0 < 𝑦 0 . We will denote
the line passing through (𝑥 0 , 𝑓 (𝑥 0 )) and (𝑦 0 , 𝑓 (𝑦 0 )) by 𝜖𝜆′ .
      If there are numbers 𝑥0 < 𝑎 < 𝑏 < 𝑦 0 such that
                                          (𝑆(𝑥 0 , 𝑎) − 𝜆′)(𝑆(𝑥 0 , 𝑏) − 𝜆′) ≤ 0,
                                                                                                            𝑓 (𝑥0 )− 𝑓 (𝑐)
then by the continuity of 𝑆(𝑥, · ) there has to exist a number 𝑐 ∈ [𝑎, 𝑏] such that                             𝑥 0 −𝑐     =
         𝑓 (𝑥 0 )− 𝑓 (𝑦 0 )
𝜆′ =         𝑥0 −𝑦 0        . But this means that (𝑥 0 , 𝑓 (𝑥 0 )), (𝑐, 𝑓 (𝑐)) and (𝑦 0 , 𝑓 (𝑦 0 )) are colinear, which
contradicts our hypothesis and therefore 𝑆(𝑥 0 , 𝑦) has to be constantly greater or constantly less than
𝜆′ for 𝑥0 < 𝑦 < 𝑦 0 (see Figure 3.2). For the same reasons 𝑆(𝑥 0 , 𝑦) has to be constantly greater or
constantly less than 𝜆′ also for 𝑦 > 𝑦 0 and the same holds for 𝑆(𝑥, 𝑦 0 ) for 𝑥 < 𝑥 0 .
      Graphically, this means that 𝜖𝜆′ separates 𝑓 in three parts that do not intersect 𝜖𝜆′ ; one before
𝑥 0 , one over (𝑥0 , 𝑦 0 ), and one after 𝑦 0 . We proceed to show that the part over (𝑥 0 , 𝑦 0 ) lies on a
different side of 𝜖𝜆′ from the other two.
      Let us consider the case when 𝑆(𝑥 0 , 𝑦) < 𝜆′ for 𝑥 0 < 𝑦 < 𝑦 0 . Then, the function 𝑓 (𝑥) − 𝜆′𝑥
defined on [𝑥 0 , 𝑦 0 ] attains a maximum at 𝑥0 and at 𝑦 0 (which also implies that 𝑆(𝑥, 𝑦 0 ) > 𝜆′ for
𝑥 0 < 𝑥 < 𝑦 0 ) and let 𝑦˜ ∈ (𝑥0 , 𝑦 0 ) be the point where 𝑓 (𝑥) − 𝜆′𝑥 attains a minimum (see Figure 3.4).
                                                              50


                             𝑆(𝑥 0 , 𝑦) > 𝜆′                                              𝑆(𝑥 0 , 𝑦) < 𝜆′
                                        Figure 3.3 The two cases when 𝑥 0 < 𝑦 < 𝑦 0 .
Figure 3.4 If 𝑆(𝑥 0 , 𝑦) < 𝜆′ for every 𝑦 ∉ (𝑥 0 , 1) \ {𝑦 0 }, by moving the line 𝜖𝜆′ slightly down, we get
three points of intersection.
Now, suppose additionally that 𝑆(𝑥0 , 𝑦) < 𝜆′ also for 𝑦 > 𝑦 0 .
      Pick a number 𝑘 with 𝑓 (𝑥0 ) − 𝜆′𝑥 0 > 𝑘 > max{ 𝑓 ( 𝑦˜ ) − 𝜆′ 𝑦˜ , 𝑓 (𝑦) − 𝜆′ 𝑦} for some 𝑦 > 𝑦 0 . Then,
we have simultaneously
                                                𝑓 ( 𝑦˜ ) − 𝜆′ 𝑦˜ < 𝑘 < 𝑓 (𝑥0 ) − 𝜆′𝑥 0 ,
                                               𝑓 ( 𝑦˜ ) − 𝜆′ 𝑦˜ < 𝑘 < 𝑓 (𝑦 0 ) − 𝜆′ 𝑦 0 ,
                                               𝑓 (𝑦) − 𝜆′ 𝑦 < 𝑘 < 𝑓 (𝑦 0 ) − 𝜆′ 𝑦 0 .
The continuity of 𝑓 and the above inequalities imply that there must exist numbers 𝑎, 𝑏, and 𝑐 in
(𝑥 0 , 𝑦˜ ), ( 𝑦˜ , 𝑦 0 ), and (𝑦 0 , 𝑦) respectively such that
                                         𝑓 (𝑎) − 𝜆′ 𝑎 = 𝑓 (𝑏) − 𝜆′ 𝑏 = 𝑓 (𝑐) − 𝜆′ 𝑐 = 𝑘
                                                                   51


Figure 3.5 If 𝑓 attains a local minimum at another point 𝑦˜ ′ > 𝑦˜ , we can find a line of slope greater
than 𝜆′ intersecting 𝑓 at three points.
which implies that (𝑎, 𝑓 (𝑎)), (𝑏, 𝑓 (𝑏)), and (𝑐, 𝑓 (𝑐)) are colinear, a contradiction, and therefore
𝑆(𝑥 0 , 𝑦) has to be greater than 𝜆′ for 𝑦 > 𝑦 0 . Working similarly, we see that 𝑆(𝑥, 𝑦 0 ) < 𝜆′ for
𝑥 < 𝑥0 .
      An identical argument gives us that 𝑦˜ is the only point in [𝑥 0 , 𝑦 0 ], and eventually in [𝑥0 , 1),
where 𝑓 (𝑥) − 𝜆′𝑥 attains a local minimum (see Figure 3.5) and from Lemma 3.1.3 we deduce that
 𝑓 (𝑥) − 𝜆′𝑥 has to be monotone increasing in [ 𝑦˜ , 1). Hence, for any 𝑥, 𝑦 ≥ 𝑦˜ we have:
                                                                    𝑥<𝑦
                          𝑥 < 𝑦 ⇐⇒ 𝑓 (𝑥) − 𝜆′𝑥 < 𝑓 (𝑦) − 𝜆′ 𝑦 ⇐=⇒ 𝑆(𝑥, 𝑦) > 𝜆′ .
      However, observe that for any 𝑥 and 𝑦 for which 𝑆(𝑥, 𝑦) > 𝜆′, the function 𝑆(𝑥, · ) has to be
1-1 otherwise our hypothesis fails in a similar way as above and, since it is continuous, it has to be
monotone in (𝑥, 1) for every 𝑥 ∈ [ 𝑦˜ , 1). Therefore, 𝑓 is either convex or concave in [ 𝑦˜ , 1) and thus
locally Lipschitz in ( 𝑦˜ , 1) thanks to Lemma 3.1.2.
      In particular, 𝑓 has to be convex in [ 𝑦˜ , 1). Indeed, assume 𝑓 is concave and let 𝑥 be any number
in ( 𝑦˜ , 𝑦 0 ), see Figure 3.6. By concavity, the point ( 𝑦˜ , 𝑓 ( 𝑦˜ )) has to lie below the line passing
through (𝑦 0 , 𝑓 (𝑦 0 )) with slope 𝜁 = 𝑆(𝑥, 𝑦 0 ) and, since 𝜁 = 𝑆(𝑥, 𝑦 0 ) > 𝑆(𝑥 0 , 𝑦 0 ) = 𝜆′ ≥ 𝜆, the point
(𝑥 0 , 𝑓 (𝑥 0 )) lies above. Hence, this line will intersect the graph of 𝑓 at some point (𝑐, 𝑓 (𝑐)) with
                                                        52


                   Figure 3.6 𝑆(𝑥 0 , 𝑦) has to be strictly monotone increasing in (𝑦 0 , 1).
𝑐 ∈ (𝑥 0 , 𝑦˜ ) and the points (𝑐, 𝑓 (𝑐)), (𝑥, 𝑓 (𝑥)), and (𝑦 0 , 𝑓 (𝑦 0 )) are colinear, a contradiction.
    If we instead assume that 𝑆(𝑥 0 , 𝑦) > 𝜆′ for 𝑥 0 < 𝑦 < 𝑦 0 , working similarly we conclude that
there must exist 𝑦˜ ∈ [𝑥 0 , 𝑦 0 ] such that 𝑓 is concave in (0, 𝑦˜ ].
    The case when there exist 𝑥0 , 𝑦 0 ∈ (0, 1) for which 𝑆(𝑥0 , 𝑦 0 ) = 𝜆′ ≤ −𝜆 is identical and gives
us the reverse implications.
    To sum up, we conclude that there are points 𝑥,          ˜ 𝑦˜ ∈ (0, 1) such that 𝑓 has some particular
convexity on (0, 𝑥]   ˜ and on [ 𝑦˜ , 1). These intervals cannot overlap, because otherwise 𝑓 would be a
line segment of slope at least 𝜆 (or at most −𝜆) on [ 𝑦˜ , 𝑥],   ˜ which contradicts our hypothesis and so
𝑥˜ ≤ 𝑦˜ . Let 𝑥˜ be the maximal point so that 𝑓 is, for instance, convex on (0, 𝑥],      ˜ and 𝑦˜ the minimal so
that 𝑓 is convex on [ 𝑦˜ , 1). When 𝑥˜ ≠ 𝑦˜ , for every points 𝑥, 𝑦 ∈ [ 𝑥,    ˜ 𝑦˜ ] we have |𝑆(𝑥, 𝑦)| ≤ 𝜆 and 𝑓
is Lipschitz in [𝑥,  ˜ 𝑦˜ ] with Lipschitz constant 𝜆.
    This concludes the proof.                                                                                  □
    Of course, any continuous function that satisfies the condition of the proposition and has
different convexity on (𝑎, 𝑥]    ˜ and on [ 𝑦˜ , 𝑏) has to additionally satisfy lim𝑥→𝑎+ ,𝑦→𝑏 − |𝑆(𝑥, 𝑦)| < 𝜆.
    Furthermore, notice that the fact that the cone is vertical (or at least that it contains the vertical
line) is essential to get the locally Lipschitz property. Indeed, if 𝐶 is a cone avoiding the vertical
                                        √
line, we can restrict the function 3 𝑥 to a sufficiently small interval around 0 so that it intersects all
                                                         √
the lines of the cone at at most two points. But 3 𝑥 is clearly not Lipschitz around 0. However, we
                                                         53


                     Figure 3.7 All the possible ways the graph of 𝑓 can look like.
do have the following corollary.
Corollary. Let 𝜆 1 > 0 > 𝜆 2 be some fixed numbers and consider all the cones of lines with
slopes between 𝜆 1 and 𝜆 2 (containing the vertical line). If 𝑓 : (0, 1) → R is a continuous function
satisfying the same condition as above, then it is locally Lipschitz.
Proof. The inequalities |𝑆(𝑥, 𝑦)| < 𝜆 and |𝑆(𝑥, 𝑦)| ≥ 𝜆 in this case correspond to 𝜆2 < 𝑆(𝑥, 𝑦) < 𝜆1
and 𝑆(𝑥, 𝑦) ≥ 𝜆 1 𝑜𝑟 𝑆(𝑥, 𝑦) ≤ 𝜆 2 , respectively. The proof is the same as before and on the regions
where 𝑓 is not convex or concave it is Lipschitz with Lipschitz constant the maximum of 𝜆 1 and
−𝜆 2 .                                                                                                □
Remark. All the above remains true for any interval (𝑎, 𝑏). It is not hard to see that the same
proof also works in the case where 𝑓 is defined on a closed interval, but Lemma 3.1.2 cannot be
used in this setting. However, if 𝑓 : [0, 1] → R, its restriction 𝑓 |(0,1) is locally Lipschitz.
3.2    An example
    It is natural then to ask whether our assumption still gives us the locally Lipschitz property when
we allow more points of intersection. It turns out this fails even for at most 3 points of intersection
in the sense that there can be infinitely many points around where the function cannot be locally
Lipschitz. Here, we construct such a function whose graph intersects a certain cone of lines at at
most three points.
                                                   54


                                           1
     Consider the sequence 𝑎 𝑘 =           2  − 21𝑘 for 𝑘 ≥ 1, and on the each of the intervals [𝑎 𝑘 , 𝑎 𝑘+1 ] define
a continuous function 𝑓 𝑘 with the following properties:
   i) 𝑓1 (0) = 0, 𝑓1 ( 41 ) = 𝑓2 ( 41 ) = 𝜆4 ;
  ii) 𝑓 𝑘+1 (𝑎 𝑘+1 ) = 𝑓 𝑘 (𝑎 𝑘+1 );
                     1
 iii) 𝑓 𝑘 (𝑎 𝑘+1 ) = 2 ( 𝑓 𝑘 (𝑎 𝑘 ) + 𝑓 𝑘−1 (𝑎 𝑘−1 ));
 iv) 𝑓2𝑘 is monotone decreasing and convex on [𝑎 2𝑘 , 𝑎 2𝑘+1 ] and 𝑓2𝑘−1 is monotone increasing and
      concave on [𝑎 2𝑘−1 , 𝑎 2𝑘 ];
  v) the tangent line to 𝑓 𝑘 at (𝑎 𝑘 , 𝑓 𝑘 (𝑎 𝑘 )) is vertical.
     Let 𝑓 : [0, 1] → R be the function given by
                                         
                                         
                                            𝑓 𝑘 (𝑥)                if 𝑥 ∈ [𝑎 𝑘 , 𝑎 𝑘+1 ),
                                         
                                         
                                         
                                         
                                         
                                         
                                         
                                         
                                𝑓 (𝑥) = 𝑓 𝑘 (1 − 𝑥)                if 𝑥 ∈ (1 − 𝑎 𝑘+1 , 1 − 𝑎 𝑘 ],
                                         
                                         
                                         
                                         
                                                                          1
                                          𝜆
                                         
                                         
                                         6                        if 𝑥 = 2
                                         
for all 𝑘 ≥ 1 (Figure 3.8), which is clearly continuous in (0, 1) \ { 12 } because of (ii). Observe that
                                                                                     𝑏 𝑘 +𝑏 𝑘−1
the sequence (𝑏 𝑘 ) = ( 𝑓 𝑘 (𝑎 𝑘 )) is recursively defined by 𝑏 𝑘+1 =                     2     (through property (iii)) and
                                                 𝑏 𝑘+1 −𝑏 𝑘
it converges. In particular, we have             𝑏 𝑘 −𝑏 𝑘−1 = − 12 and therefore
                          −1  𝑘−1                                                   −1  𝑘−1 
                                                                            1
          𝑏 𝑘+1 = 𝑏 𝑘 +               (𝑏 2 − 𝑏 1 ) =⇒ 𝑏 𝑘+1        = 𝑏2 − 1 −                       (𝑏 2 − 𝑏 1 ).    (3.2.1)
                             2                                              3             2
In our case, we have 𝑏 1 = 𝑓1 (0) = 0, 𝑏 2 = 𝑓2 ( 14 ) = 𝜆4 , and also
                                                                      −1  𝑘−1 
                                                             𝜆
                                                 𝑓 𝑘 (𝑎 𝑘 ) = 1 −                  ,
                                                             6          2
hence lim 𝑘→+∞ 𝑓 𝑘 (𝑎 𝑘 ) =         𝜆
                                    6.   But note that for every 𝑥 ∈ (0, 12 ) there is an 𝑛 ≥ 1 for which
𝑥 ∈ [𝑎 𝑛 , 𝑎 𝑛+1 ) and, since each 𝑓 𝑘 is monotone in [𝑎 𝑘 , 𝑎 𝑘+1 ) for every 𝑘, we get
                                                                             
                       min 𝑓𝑛 (𝑎 𝑛 ), 𝑓𝑛+1 (𝑎 𝑛+1 ) ≤ 𝑓 (𝑥) ≤ max 𝑓𝑛 (𝑎 𝑛 ), 𝑓𝑛+1 (𝑎 𝑛+1 ) .
                                                                 55


              Figure 3.8 At most 3 points of intersection with any line inside the cones.
Therefore, we have lim𝑥→ 1 − 𝑓 (𝑥) =          𝜆
                                              6 = 𝑓 ( 21 ), and similarly for 𝑥 ∈ ( 12 , 1), which means that 𝑓 is
                               2
also continuous at 12 .
    However, by construction 𝑓 is locally Lipschitz on (0, 1) \ { 12 } except at around 𝑎 𝑘 and 1 − 𝑎 𝑘 ,
                                                                    1                               1
𝑘 ≥ 1, and therefore it is not locally Lipschitz around             2  either, because 𝑎 𝑘 →        2 as 𝑘 → +∞.
    Now we proceed to show the graph of 𝑓 has at most 3 intersection points with any line inside a
vertical cone with slopes between 𝜆 and −𝜆.
    Each 𝑓 𝑘 is monotone and has certain concavity on [𝑎 𝑘 , 𝑎 𝑘+1 ], hence its graph is contained inside
the triangle 𝑇𝑘 with vertices (𝑎 𝑘 , 𝑓 (𝑎 𝑘 )), (𝑎 𝑘+1 , 𝑓 𝑘+1 (𝑎 𝑘+1 )), and (𝑎 𝑘 , 𝑓 (𝑎 𝑘+1 )) (see Figure 3.9)
and therefore any line intersecting the graph of 𝑓 (at at least two points) has to pass through some
of these triangles. Notice, however, that if a line passes through two nonconsecutive triangles, say
𝑇𝑘 and 𝑇𝑘+ 𝑗 ( 𝑗 > 1), then it falls outside the admissible cone of lines. In particular, (because of
properties (ii) through (iv)) each 𝑇𝑘+1 is half the size of 𝑇𝑘 and they are placed is such a way that
the maximum and minimum slope a line through them can have are respectively the maximum and
the minimum of the quantities
                        𝑓 𝑘+ 𝑗 (𝑎 𝑘+ 𝑗 ) − 𝑓 𝑘 (𝑎 𝑘 )           𝑓 𝑘+ 𝑗 (𝑎 𝑘+ 𝑗+1 ) − 𝑓 𝑘 (𝑎 𝑘+1 )
                                                        and                                       ,
                                 𝑎 𝑘+ 𝑗 − 𝑎 𝑘                            𝑎 𝑘+ 𝑗 − 𝑎 𝑘+1
when one of the numbers 𝑘 and 𝑘 + 𝑗 is even and the other is odd, and the maximum and minimum
                                                            56


of the quantities
                          𝑓 𝑘+ 𝑗 (𝑎 𝑘+ 𝑗+1 ) − 𝑓 𝑘 (𝑎 𝑘 )           𝑓 𝑘+ 𝑗 (𝑎 𝑘+ 𝑗 ) − 𝑓 𝑘 (𝑎 𝑘+1 )
                                                              and                                   ,
                                     𝑎 𝑘+ 𝑗 − 𝑎 𝑘                           𝑎 𝑘+ 𝑗 − 𝑎 𝑘+1
when 𝑘 and 𝑘 + 𝑗 are both even or both odd. Using (3.2.1) we can see that each of the above is
bounded in absolute value by 𝜆 whenever 𝑗 > 1.
     For the same reasons any admissible line passing through ( 12 , 𝜆6 ) intersects the graph only at
that point, because
                                                                𝜆
                                                  𝑓 𝑘 (𝑎 𝑘 ) −  6   𝜆
                                                             1
                                                                  =    < 𝜆.
                                                     𝑎𝑘 −    2
                                                                    3
     Therefore, the admissible lines intersecting the graph necessarily pass through two (or maybe
only one) consecutive triangles and each such line intersects the graph of 𝑓 𝑘 at at most two points
because of (iv). Furthermore, due to the difference in concavity of 𝑓 𝑘 and 𝑓 𝑘+1 , a line cannot
intersect both of their graphs at two points, because then it would need to have both negative and
positive slope, which is absurd.
     An example of a sequence ( 𝑓 𝑘 ) of functions with the above properties is the following:
                                                      −1  𝑘−1  (−1) 𝑘+1𝜆 √
                                             𝜆
                                 𝑓 𝑘 (𝑥) = 1 −                    +                  𝑥 − 𝑎𝑘 .
                                             6           2              2 2
                                                                           𝑘+1
3.3    Hausdorff measure
     Marstrand in [5, Theorem 6.5.III] proved that if a Borel set on the plane has the property that
  if the lines in a positive measure of directions intersect this Borel set at a set of Hausdorff
                                                                                                      (3.3.1)
  dimension zero, then the Hausdorff dimension of this Borel set is at most 1.
In particular, this happens if the intersections are at most countable. The Borel assumption is
essential.
     That said, Marstrand’s theorem does not in general guarantee the Hausdorff measure of the
Borel set is finite. Our next goal will be to deal with the Hausdorff measure of a continuous curve
and also generalise to arbitrarily many points of intersection with our cones (still finitely many,
though). It turns out that the curve has to always be 𝜎-finite with respect to the H 1 measure.
     In order to proceed we need set up things more rigorously:
                                                               57


                        Figure 3.9 The case when 𝑘 and 𝑘 + 𝑗 are both odd.
Notations. Let 𝐶 (𝜙, 0) = {(𝑥, 𝑦) ∈ R2 : |𝑦| ≥ tan(𝜙) |𝑥|} denote the vertical closed cone in
between the lines through the point (0, 0) with slopes tan(𝜙) and − tan(𝜙) (where 0 < 𝜙 < 𝜋2 ). By
𝐶+ (𝜙, 0) we will denote the upper half of the cone 𝐶 (𝜙, 0), that is 𝐶+ (𝜙, 0) = {(𝑥, 𝑦) ∈ R2 : |𝑦| ≥
tan(𝜙) |𝑥|, 𝑦 ≥ 0}, and by 𝐶− (𝜙, 0) its lower half. Let 𝐶 (𝜙, 𝜌) be the cone’s counter-clockwise
rotation by angle 𝜌, 𝐶 (𝜙, 0, ℎ) = 𝐵0 (ℎ) ∩ 𝐶 (𝜙, 0), where 𝐵𝑥 (𝑟) = 𝐵(𝑥, 𝑟) is the closed ball centred
at 𝑥 with radius 𝑟, and 𝐶𝑃 (𝜙, 0) the translation of 𝐶 (𝜙, 0) so that its vertex is the point 𝑃. Finally,
𝐶 ∗ will denote the dual cone of 𝐶, that is 𝐶 ∗ (𝜙, 0) = 𝐶 (𝜙, 0) 𝐶 . We will be combining different
notation in a natural way, for example 𝐶+ (𝜙, 𝜌, ℎ) is the upper half of the truncated and rotated
cone with vertex at 0.
    𝛾 : [0, 1] → R2 will be a continuous curve.
                                                   58


3.3.1   The main hypothesis
  Fix an integer 𝑘 ≥ 2. Fix an angle 𝜙 ∈ (0, 𝜋2 ) and a rotation 𝜌 ∈ [0, 2𝜋). A line contained
                                                                                               (3.3.2)
  inside the cone 𝐶𝑃 (𝜙, 𝜌) for any point 𝑃 ∈ R2 intersects the curve 𝛾 at at most 𝑘 points.
    Any such line will be called admissible. A cone consisting of only admissible lines will also be
called admissible.
3.3.2   𝛾 is 𝜎-finite
    For simplicity and without loss of generality we will assume the the curve 𝛾 : [0, 1] → R2 is
bounded inside the unit square and that (0, 0), (1, 1) ∈ 𝛾. We additionally assume that the cones of
our hypothesis are vertical, i.e., that 𝜌 = 0.
Theorem 3.3.1. 𝛾 can be split into countably many sets 𝛾𝑛 with finite H 1 measure. In particular,
𝛾 is 1-rectifiable.
    The following lemma plays a key role in the proof of this theorem, but we will postpone its
proof until later.
Lemma 3.3.2. For every point 𝑃 ∈ 𝛾 there exists an admissible cone 𝐶𝑃 (𝜃, 𝜌, ℎ) that avoids the
curve 𝛾 except at 𝑃, that is 𝐶𝑃 (𝜃, 𝜌, ℎ) ∩ 𝛾 = {𝑃}.
    In view of Lemma 3.3.2 — by slightly tilting 𝜌, enlarging 𝜃 and monotone decreasing ℎ — we
may assume the triplet (𝜃, 𝜌, ℎ) consists of rational numbers. If {(𝜃 𝑛 , 𝜌𝑛 , ℎ𝑛 )} is an enumeration
of all rational triples that still lie within our admissible set, then we can decomposed 𝛾 into the
countably many sets
                                      
                               𝛾𝑛 = 𝑃 ∈ 𝛾 : 𝐶𝑃 (𝜃 𝑛 , 𝜌𝑛 , ℎ𝑛 ) ∩ 𝛾 = {𝑃}
(see Figure 3.10). Note that 𝛾𝑛 are not necessarily disjoint for different values of 𝑛.
    We proceed to prove each one of them has finite H 1 measure. Note that this is not new
knowledge and it can be found, for example, in [2, Lemma 3.3.5] or [6, Lemma 15.13] in a more
general setup. Nevertheless, we present it here for completeness.
    For the rest of this section 𝑛 will be fixed.
                                                    59


                    Figure 3.10 The curve 𝛾 and its part 𝛾𝑛 for 𝜃 𝑛 , 𝜌𝑛 = 0, and ℎ𝑛 .
                                2𝑘
Lemma 3.3.3. H 1 (𝛾𝑛 ) <      cos(𝜃 𝑛 ) .
Proof. Without loss of generality we may assume the cone 𝐶𝑃 (𝜃 𝑛 , 𝜌𝑛 , ℎ𝑛 ) is vertical, i.e., that
𝜌𝑛 = 0. Let us now split the unit square into 𝑁 vertical strips, 𝑆 𝑗 ( 𝑗 = 1, 2, . . . , 𝑁), of base length
 1                                        1
𝑁  with 𝑁 sufficiently large so that      𝑁 < cos(𝜃 𝑛 ) ℎ𝑛 . Let 𝐽 be the set of indices 𝑗 for which
                                                 𝑆 𝑗 ∩ 𝛾𝑛 ≠ ∅
and for any point 𝑃 ∈ 𝛾 denote the connected component of 𝛾 inside 𝑆 𝑗 through 𝑃 ∈ 𝑆 𝑗 ∩ 𝛾 by
Γ𝑃∗ ( 𝑗).
                                                                 1
     Fix a 𝑗 ∈ 𝐽 and consider a point 𝑃 ∈ 𝑆 𝑗 ∩ 𝛾𝑛 . Since       𝑁 < cos(𝜃 𝑛 ) ℎ𝑛 , the sides of 𝑆 𝑗 necessarily
intersect both sides of the cone 𝐶𝑃 (𝜃 𝑛 , 0, ℎ𝑛 ) creating thus two triangles both contained inside
                  1
                          (see Figure 3.11). For any point 𝑃′ ∈ 𝑆 𝑗 ∩ 𝛾𝑛 other than 𝑃 there are two
                        

the ball 𝐵 𝑃 𝑁 cos(𝜃 𝑛)
                                                        60


                                                                               1
                  Figure 3.11 Each cone intersects a strip of length           𝑁 < cos(𝜃 𝑛 ) ℎ𝑛 .
cases: either |𝑃 − 𝑃′ | ≤ ℎ𝑛 or |𝑃 − 𝑃′ | > ℎ𝑛 . In the first case, the sets Γ𝑃∗ ( 𝑗) and Γ𝑃∗ ′ ( 𝑗) are
both contained inside the two triangles 𝐶𝑃∗ (𝜃 𝑛 , 0) ∩ 𝑆 𝑗 . In the second, they are necessarily disjoint,
because 𝐶𝑃 (𝜃 𝑛 , 0, ℎ𝑛 ) is free from points of 𝛾 (other than 𝑃). These additionally imply that there
                             1
can be no more than     sin(𝜃 𝑛 ) ℎ 𝑛 such distinct paths inside 𝑆 𝑗 . In particular,
                                                                                                   
               𝑃 ∈ Γ𝑃∗ ( 𝑗) ⊂ 𝑆 𝑗 ∩ 𝛾 ∩ 𝐵 𝑃 (ℎ𝑛 ) ⊂ 𝐶𝑃∗ (𝜃 𝑛 , 0, ℎ𝑛 ) ∩ 𝑆 𝑗 ⊂ 𝐵 𝑃          1
                                                                                        𝑁 cos(𝜃 𝑛 )   .
    Now, let P 𝑗 be a maximal set of points in 𝑆 𝑗 ∩ 𝛾𝑛 such that the sets Γ𝑃∗ ( 𝑗) for 𝑃 ∈ P 𝑗 are
                                                                                  1
                                                                                        

all disjoint and observe that 𝑆 𝑗 ∩ 𝛾𝑛 is covered by the balls 𝐵 𝑃 𝑁 cos(𝜃          𝑛 )   with 𝑃 ∈ P 𝑗 . Indeed,
                                                              1
                                              Ð                      

if 𝑃0 ∈ 𝑆 𝑗 ∩ 𝛾𝑛 is not inside the set 𝑃∈P 𝑗 𝐵 𝑃 𝑁 cos(𝜃         𝑛 )   , then by construction it is also outside
                                                        61


         𝐵 𝑃 (ℎ𝑛 ) and therefore Γ𝑃∗ 0 ( 𝑗) and Γ𝑃∗ ( 𝑗) are disjoint for all 𝑃 ∈ P 𝑗 , which contradicts the
Ð
   𝑃∈P 𝑗
maximality of P 𝑗 . Moreover, due to the connectedness of 𝛾, the set {𝑃} has to be path-connected
with (0, 0) and (1, 1) and therefore each Γ𝑃∗ ( 𝑗) has to intersect at least one side of the strip 𝑆 𝑗 . Hence,
because of (3.3.2), there can be at most 2𝑘 of these paths, i.e., #(P 𝑗 ) ≤ min{2𝑘, sin(𝜃1𝑛 ) ℎ𝑛 } ≤ 2𝑘
for every 𝑗 ∈ 𝐽. Therefore,
                                Ø                                   Ø Ø                        
                                                  1                                       1
                     𝛾𝑛 ∩ 𝑆 𝑗 ⊂        𝐵𝑃    𝑁 cos(𝜃 𝑛 )    =⇒ 𝛾𝑛 ⊂              𝐵𝑃   𝑁 cos(𝜃 𝑛 )
                                𝑃∈P 𝑗                                 𝑗 ∈𝐽 𝑃∈P 𝑗
and the total sum of the radii of these balls is at most
                                                   1                 2𝑘
                                         2𝑘                #(𝐽) ≤           .
                                            𝑁 cos(𝜃 𝑛 )           cos(𝜃 𝑛 )
    Finally, if 𝛾˜ 𝑛 = {𝑃 ∈ 𝛾 : 𝐶𝑃 (𝜃 𝑛 , 0, ℎ𝑛 /2) ∩ 𝛾 = {𝑃}}, then 𝛾𝑛 ⊂ 𝛾˜ 𝑛 . Repeating the above
                      1               ℎ𝑛
construction with     𝑁  < cos(𝜃 𝑛 )  2 ,  we get a cover of 𝛾˜ 𝑛 — and thus of 𝛾𝑛 — consisting of balls
                                         2𝑘
with a total sum of radii at most     cos(𝜃 𝑛 ) . The result follows.                                          □
                                                                                                        Ð
Remark. In the above construction we are in fact able to cover the whole part of 𝛾 inside                 𝑗 ∈𝐽 𝑆𝑗
with the same balls, and not merely 𝛾𝑛 .
    Eventually, the curve 𝛾 has to be 𝜎-finite.
3.3.3    Cones free of 𝛾
    Here we prove Lemma 3.3.2.
    Fix 𝑃 ∈ 𝛾. Since 𝛾 is bounded, there must exist an ℎ˜ > 0 such that 𝐶𝑃 (𝜙, 0)∩𝛾 = 𝐶𝑃 (𝜙, 0, ℎ)∩𝛾.     ˜
If
                           𝐶𝑃 (𝜙′, 0) ∩ 𝛾 = {𝑃}            or 𝐶𝑃 (𝜙′, 0, ℎ) ∩ 𝛾 = {𝑃}
for some 𝜙′ ∈ [𝜙, 𝜋2 ) and some ℎ > 0, then we are done.
    Suppose this does not happen. Then, for all 𝜙′ ∈ [𝜙, 𝜋2 ) and for all sufficiently small ℎ > 0 we
have
                                          𝐶𝑃 (𝜙′, 0, ℎ) ∩ 𝛾 \ {𝑃} ≠ ∅.                                   (3.3.3)
Lemma 3.3.4. For any 𝑃 ∈ 𝛾 the set 𝐶𝑃 (𝜙, 0) ∩𝛾 has finitely many (closed) connected components.
                                                            62


Proof. Since 𝛾 is connected, every point of 𝐶𝑃 (𝜙, 0) ∩ 𝛾 has to be path-connected with the point 𝑃
through some part of the curve 𝛾. There are two possibilities: either that path is entirely contained
inside 𝐶𝑃 (𝜙, 0) or it has to pass through its sides. If a path does not intersect the sides, then it
necessarily has to pass through 𝑃 otherwise 𝛾 would not be connected. This yields precisely one
connected component — the one containing 𝑃 — and all the rest (if any) have to intersect the
sides of the cone. If these components are infinitely many, there have to exist also infinitely many
points of intersection on the sides of the cone; at least one for each connected component. But this
contradicts (3.3.2).                                                                                 □
Remark. The connected components of Lemma 3.3.4 total at most 2𝑘 and 𝑃 need not be a point
of the curve. This lemma is still valid regardless of the cone we are working with as soon as it is in
our admissible family of cones.
    Let Γ𝑃 (𝜙, 0) be the connected component of 𝐶𝑃 (𝜙, 0) ∩ 𝛾 that contains the point 𝑃, which
because of (3.3.3) cannot be precisely the point set {𝑃}. Because of Lemma 3.3.4, the set
𝐶𝑃 (𝜙, 0) ∩𝛾\Γ𝑃 (𝜙, 0) is compact and thus there exists ℎ0 > 0 such that 𝐶𝑃 (𝜙, 0, ℎ0 )∩𝛾 ⊂ Γ𝑃 (𝜙, 0).
Observe that 𝐶𝑃 (𝜙, 0) ∩ 𝛾 \ Γ𝑃 (𝜙, 0) could be empty in general in which case ℎ0 = ∞, however,
we can always assume that ℎ0 ≤ ℎ.    ˜
    Next, we bisect our cone into two new identical cones sharing one common side
                                 𝐶𝑃 (𝜙, 0) = 𝐶𝑃 (𝜙1 , 𝜌1 ) ∪ 𝐶𝑃 (𝜙1 , −𝜌1 ),
                 𝜙
where 𝜙1 =   𝜋
             4 + 2  and 𝜌1 =   𝜋
                               4 − 𝜙2 , and repeat the above arguments for each new cone: If
                         𝐶𝑃 (𝜙′, 𝜌1 ) ∩ 𝛾 = {𝑃}     or 𝐶𝑃 (𝜙′, 𝜌1 , ℎ) ∩ 𝛾 = {𝑃}
for some 𝜙′ ∈ [𝜙1 , 𝜋2 ) and some ℎ > 0, then we are done. Similarly for −𝜌1 in place of 𝜌1 .
    Suppose none of these happen. Then, for all 𝜙′ ∈ [𝜙1 , 𝜋2 ) and for all sufficiently small ℎ and ℎ′
we have
                𝐶𝑃 (𝜙′, 𝜌1 , ℎ) ∩ 𝛾 \ {𝑃} ≠ ∅     and 𝐶𝑃 (𝜙′, −𝜌1 , ℎ′) ∩ 𝛾 \ {𝑃} ≠ ∅.          (3.3.4)
                                                     63


Figure 3.12 Finding a cone free from points of 𝛾. The parameters 𝑟, 𝑑, and ℎ determine the radius.
We denote by Γ𝑃 (𝜙1 , 𝜌1 ) and Γ𝑃 (𝜙1 , −𝜌1 ) the connected component of
                                𝐶𝑃 (𝜙1 , 𝜌1 ) ∩ 𝛾     and 𝐶𝑃 (𝜙1 , −𝜌1 ) ∩ 𝛾
containing 𝑃, respectively. Then, the sets 𝐶𝑃 (𝜙1 , 𝜌1 ) ∩ 𝛾 \ Γ𝑃 (𝜙1 , 𝜌1 ) and 𝐶𝑃 (𝜙1 , −𝜌1 ) ∩ 𝛾 \
                                                                                          ˜ such that
Γ𝑃 (𝜙1 , −𝜌1 ) are compact (thanks to Lemma 3.3.4) and thus there exist ℎ1,0 , ℎ1,1 ∈ (0, ℎ]
𝐶𝑃 (𝜙1 , 𝜌1 , ℎ1,0 ) ∩ 𝛾 ⊂ Γ𝑃 (𝜙1 , 𝜌1 ) and 𝐶𝑃 (𝜙1 , −𝜌1 , ℎ1,1 ) ∩ 𝛾 ⊂ Γ𝑃 (𝜙1 , −𝜌1 ).
    We iterate this construction indefinitely (Figure 3.12). If at any step we get
                                           𝐶𝑃 (𝜙′, 𝜌, ℎ) ∩ 𝛾 = {𝑃}                             (3.3.5)
for some 𝜙′, 𝜌, and ℎ, then we have found our desired cone and we stop. Otherwise, we get an
                                                       64


infinite sequence of smaller and smaller cones satisfying the following:
                       {𝑃} ⊊ 𝐶𝑃 (𝜙𝑛 , 𝜌𝑛,𝑖 , ℎ𝑛,𝑖 ) ∩ 𝛾 ⊂ Γ𝑃 (𝜙𝑛 , 𝜌𝑛,𝑖 ) ⊂ 𝐶𝑃 (𝜙𝑛 , 𝜌𝑛,𝑖 )
                                                               for all 𝑖 = 0, 1, . . . , 2𝑛 − 1
for all 𝑛 ≥ 0 where
                              𝜋 𝜙                                                  𝜋 𝜙𝑛−1
       𝜙0 = 𝜙           𝜙1 =      +                                         𝜙𝑛 =       +
                              4 2                                                  4        2
                                         𝜋 𝜙                                                    2(𝜙𝑛 − 𝜙)
       𝜌0,0 = 0         𝜌1,0 = 𝜌1 =        −            𝜌1,1 = −𝜌1          𝜌𝑛,𝑖 = (𝜙𝑛 − 𝜙) − 𝑖
                                         4 2                                                     2𝑛 − 1
       ℎ0,0 = ℎ0                                                            0 < ℎ𝑛,𝑖 ≤ ℎ.  ˜
Note that at the 𝑛th iteration we have exactly 2𝑛 truncated closed cones separated by the lines
                                            
                              𝑙 𝑛,𝑖 = 𝑃 + (𝑥, 𝑦) : 𝑦 = tan(𝜋 − 𝜙𝑛 + 𝜌𝑛,𝑖 ) 𝑥
through 𝑃. The sets Γ𝑃 (𝜙𝑛 , 𝜌𝑛,𝑖 ) might intersect these lines, but this can happen at at most 𝑘 may
points due to (3.3.2). Let 𝑟 𝑛,𝑖 be the smallest distance between these points of intersection (if any)
and 𝑃, that is
                                                                                 

                                   𝑟 𝑛,𝑖 = dist 𝑃, 𝑙 𝑛,𝑖 ∩ Γ𝑃 (𝜙𝑛 , 𝜌𝑛,𝑖 ) \ {𝑃}
(again we can arbitrarily set some 0 < 𝑟 𝑛,𝑖 ≤ ℎ˜ if 𝑙 𝑛,𝑖 ∩ Γ𝑃 (𝜙𝑛 , 𝜌𝑛,𝑖 ) \ {𝑃} = ∅) and let
                    n                                                                                   o
        𝑑𝑛,𝑖 = min sup{𝑑 (𝑃, Γ𝑃+ (𝑡) \ 𝑃) : 𝑡 ∈ (0, 1]}, sup{𝑑 (𝑃, Γ𝑃− (𝑡) \ 𝑃) : 𝑡 ∈ (0, 1]}
where Γ𝑃+ (𝑡) and Γ𝑃− (𝑡) are parametrisations of the sets Γ𝑃 (𝜙𝑛 , 𝜌𝑛,𝑖 )∩𝐶𝑃+ (𝜙𝑛 , 𝜌𝑛,𝑖 ) and Γ𝑃 (𝜙𝑛 , 𝜌𝑛,𝑖 )∩
𝐶𝑃− (𝜙𝑛 , 𝜌𝑛,𝑖 ) respectively (which in general could be precisely the point set {𝑃}) with Γ𝑃+ (0) =
Γ𝑃− (0) = 𝑃. Finally, we set
                            ℎ𝑛 = min{𝑟 𝑛,𝑖 , 𝑑𝑛,𝑖 , ℎ𝑛,𝑖 : 𝑖 = 0, 1, . . . , 2𝑛 − 1}.
Since the above set is finite, ℎ𝑛 > 0. From this construction for every 𝑛 ≥ 0 we get a collection of
truncated cones 𝐶𝑃 (𝜙𝑛 , 𝜌𝑛,𝑖 , ℎ𝑛 ), for 𝑖 = 0, 1, . . . , 2𝑛 − 1, (see Figure 3.12) that have the following
                                                         65


property.
   There is a path (part of 𝛾) lying inside the cone that connects the point 𝑃 with at least one
   of the two arcs of length (𝜋 − 2𝜙𝑛 )ℎ𝑛 which bound the cone 𝐶𝑃 (𝜙𝑛 , 𝜌𝑛,𝑖 , ℎ𝑛 ). Moreover,
                                                                                                            (3.3.6)
   these paths avoid any other intersections with that cone’s boundary aside 𝑃 and the
   (closed) arc(s).
     Now, fix 𝑛 sufficiently large so that 2𝑛 ≥ 2𝑘 + 3. Then, we can find at least 𝑘 + 2 of the
cones 𝐶𝑃 (𝜙𝑛 , 𝜌𝑛,𝑖 , ℎ𝑛 ) that contain some path of those mentioned at (3.3.6) all lying on the same
half-cone, say on 𝐶𝑃+ (𝜙, 0, ℎ𝑛 ). Consider one of the sides of our initial cone 𝐶𝑃 (𝜙, 0), say
𝑙 = 𝑃 + {(𝑥, 𝑦) : 𝑦 = tan(𝜙) 𝑥}, fix 0 < 𝜖 < ℎ𝑛 sin(𝜋 − 2𝜙𝑛 ) and translate 𝑙 vertically by 𝜖:
𝑙 𝜖 = 𝑙 + (0, 𝜖). Then, 𝑙 𝜖 necessarily intersects all the 2𝑛 different sectors of the ball 𝐵 𝑃 (ℎ𝑛 ) inside
𝐶𝑃+ (𝜙, 0, ℎ𝑛 ), but only the right-most one, 𝐶𝑃+ (𝜙𝑛 , 𝜌𝑛,2𝑛 −1 , ℎ𝑛 ), at its arc-like part of the boundary.
In particular, 𝑙 𝜖 has to intersect the sides of at least 𝑘 + 1 sectors that contain the paths described in
(3.3.6) and therefore also intersects these paths. Hence, 𝑙 𝜖 is one of our admissible lines that has at
least 𝑘 + 1 intersections with 𝛾, a contradiction.
     Lemma 3.3.2 is proved.                                                                                      □
Remarks.          i) In the definition of ℎ𝑛 , three different parameters occur, 𝑟 𝑛,𝑖 , 𝑑𝑛,𝑖 , and ℎ𝑛,𝑖 . Without
       ℎ𝑛,𝑖 , (3.3.5) automatically fails; 𝑑𝑛,𝑖 is to ensure Γ𝑃 (𝜙𝑛 , 𝜌𝑛,𝑖 ) will always intersect the boundary
      of the corresponding cone and 𝑟 𝑛,𝑖 forces this intersection to avoid the sides.
  ii) In the above construction we bisected the initial cone into 2, 4, 8 etc. smaller cones every time.
      However, any possible way to cut the cones would still work as soon as it eventually yields an
      infinite sequence.
 iii) The same proof can be applied to any cone within our admissible set of directions.
3.4     Higher dimensions
     Mattila in [7, Lemma 6.4] generalised Marstrand’s results from [5] and showed the following.
Lemma 3.4.1 (Mattila). Let 𝐸 be an H 𝑠 measurable subset of R𝑛 with 0 < H 𝑠 (𝐸) < ∞. Then,
                                        dim(𝐸 ∩ (𝑉 + 𝑥)) ≥ 𝑠 + 𝑚 − 𝑛
                                                         66


for almost all (𝑥, 𝑉) ∈ 𝐸 × 𝐺 (𝑛, 𝑚).
    In particular, for a Borel set in, say, R2 we have:
     if any 2-dimensional plane in a positive measure of directions intersects this Borel set at
     a set of Hausdorff dimension at most 1, then the Hausdorff dimension of this Borel set is
     at most 2.
Furthermore, if every line in the direction of some 2-dimensional cone intersects a Borel set (not
merely the graph of some continuous function) at at most countably many points, then any 2-
dimensional plane in a positive measure of directions intersects this Borel set by a set of Hausdorff
dimension at most 1 (Marstrand) and then the Hausdorff dimension of this Borel set is at most 2
(Mattila).
    Of course, the same is also true in R𝑛 , that is, if a Borel set has countable intersection with a
certain cone of lines, then its dimension does not exceed 𝑛 − 1.
    Now, we restrict our attention to what happens with only 2 points of intersection in higher
dimensions and we would like to generalize Proposition 3.1.1 to R𝑛 .
    Suppose we have a continuous function 𝑧 = 𝑓 (𝑥, 𝑦), say, on a square in R2 , satisfying the
property that
  any line in the direction of a certain open cone with axis along a vector v ∈ R3 intersects
                                                                                                (3.4.1)
  the graph at at most two points.
Then, we would want 𝑓 to obey the same rule. Namely we ask the following:
Question. Is a continuous function on (−1, 1) 2 having property (3.4.1) locally Lipschitz?
3.5   Relationships with perturbation theory
    The problem we consider in this note grew from a question in perturbation theory of self-adjoint
operators (see [4]). The question was to better understand the structure of Borel sets in R𝑛 that
have a small intersection with a whole cone of lines. Marstrand’s and Mattila’s theorems in [5]
and [7], respectively, give a lot of information about the exceptional set of finite-rank perturbations
of a given self-adjoint operator. The exception happens when singular parts of unperturbed and
                                                   67


perturbed operators are not mutually singular. It is known that this is a rare event in the sense that
its measure is zero among all finite-rank perturbations. The paper [4] proves a stronger claim: the
dimension of a bad set of perturbations actually drops.
    Let us explain what was the thrust from [4] and why that paper naturally gives rise to the
questions considered above: what is the structure of Borel sets in R𝑛 that have a small intersection
with all the lines filling a whole cone and their parallel shifts?
    In [4], a family of finite rank (self-adjoint) perturbations, 𝐴𝛼 , of a self-adjoint (suppose bounded
for simplicity) operator 𝐴 in a Hilbert space H is considered:
                                             𝐴𝛼 := 𝐴 + 𝐵𝛼𝐵∗
parametrized by self-adjoint operators 𝛼 : C𝑑 → C𝑑 (i.e., Hermitian matrices). The operator
𝐵 : C𝑑 → H is a fixed injective and bounded operator. It is also assumed that range of 𝐵 is cyclic
with respect to 𝐴. In the case when 𝑑 = 1 (rank-one perturbations), the Aronszajn-Donoghue
theorem states that the singular parts of the spectral measures of 𝐴 and 𝐴𝛼 are always mutually
singular. However, it is known that for 𝑑 > 1 the singular parts of the spectral measures of
unperturbed and perturbed operators are not always mutually singular.
    Notice that the space of perturbations, that is the space 𝐻 (𝑑) of Hermitian (𝑑 × 𝑑) matrices, has
dimension 𝑑 2 . In [3], it was proved that, given a singular measure 𝜈, the scalar spectral measure
𝜇𝛼 of the perturbation 𝐴𝛼 is not singular with respect to 𝜈 for the set of 𝛼’s having zero Lebesgue
measure in 𝐻 (𝑑). Such 𝛼’s are called exceptional, and this result shows that even though the set of
exceptional 𝛼’s can be non-empty (for 𝑑 > 1), it is a thin set. But is it maybe thinner?
    In fact, the following result was proved in [3]. Fix 𝛼0 , 𝛼1 ∈ 𝐻 (𝑑) where 𝛼1 is in the cone of
positive Hermitian matrices and consider 𝛼(𝑡) = 𝛼0 + 𝑡𝛼1 . Then, for any such 𝛼0 , 𝛼1 there are
at most countably many 𝑡 ∈ R such that the 𝛼(𝑡) is exceptional. This extra information allowed
the authors in [4] to prove that the Hausdorff dimension of exceptional perturbations is actually at
most 𝑑 2 − 1.
    The reader might have noticed an underlying geometric measure theory fact: a Borel set in R𝑛
(here 𝑛 = 𝑑 2 ) that has an at most countable intersection with a whole cone of lines and their parallel
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shifts is, in fact, of dimension 𝑛 − 1.
    Thus the dimension drop detected in Marstrand’s and Mattila’s theorems was instrumental for
the drop in dimension for exceptional perturbations.
    It seems enticing to understand the structure of the sets that have even less than countable
intersection with all parallel shifts of all lines from a fixed cone. Suppose the Borel set under
investigation intersects only at at most two, or at most 𝑘 < ∞, points with these lines. What
additional knowledge one can obtain about this set? This question motivated the work presented in
the previous sections.
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                                    BIBLIOGRAPHY
[1] V. Eiderman and M. Larsen. “A “rare” plane set with Hausdorff dimension 2”. In: (Apr. 18,
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[2] H. Federer. Geometric measure theory. Springer, 1996, p. 676. ISBN: 3540606564.
[3] C. Liaw and S. Treil. “Matrix measures and finite rank perturbations of self-adjoint
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[7] P. Mattila. “Hausdorff dimension, orthogonal projections and intersections with planes”. In:
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    DOI: 10.5186/aasfm.1975.0110.
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