METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
arXiv:2306.08254v1 [math.NT] 14 Jun 2023
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
Abstract. We develop the geometry of Hurwitz continued fractions – a major tool in
understanding the approximation properties of complex numbers by ratios of Gaussian
integers. We obtain a detailed description of the shift space associated with Hurwitz
continued fractions and, as a consequence, we contribute significantly in establishing the
metrical theory of Hurwitz continued fractions, analogous to the well-established theory
of regular continued fractions for real numbers.
Let Φ ∶ N → R>0 be any function and an (z) denote the nth partial quotient in the
Hurwitz continued fraction of a complex number z. The main result of the paper is the
Hausdorff dimension analysis of the set
E(Φ)∶ = {z ∈ C ∶ ∣an (z)∣ ≥ Φ(n) for infinitely many n ∈ N} .
This study is the complex analogue of a well-known result of Wang and Wu [Adv. Math.
218 (2008), no. 5, 1319–1339].
Contents
1. Introduction
2
2. Frequently used notation
6
3. Basics on Hurwitz continued fractions
7
4. Cylinders
9
4.1. Symmetries of regular cylinders
12
4.2. Size of regular cylinders
14
5. Geometric constructions
16
5.1. First geometric construction
17
5.2. Second geometric construction
27
6. Approaching the irregular by the regular
36
7. Proof of Theorem 1.2
39
7.1. The functions gn
39
1
2
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
7.2. Properties of sB
43
7.3. Upper bound of Hausdorff dimension
50
7.4. Lower bound of Hausdorff dimension
51
8. A complex Luczak Theorem
66
8.1. Upper bound of Hausdorff dimension
66
8.2. Lower bound of Hausdorff dimension
72
9. Proof of Theorem 1.1
77
References
79
1. Introduction
Regular continued fractions is an important tool in understanding the complexities of
irrational numbers. A key property of continued fractions is that they solve the problem of
finding the best approximations of a given real number [6], but their uses and applications
go far beyond this problem. They enable us to draw connections with various branches of
mathematics, including dynamical systems, ergodic theory, and fractal geometry, allowing
us to unveil shared characteristics among different classes of irrational numbers [12, 28, 32].
The Borel-Bernstein Theorem [4, Theorem 1.11] and similar results give us families of null
sets determined by their continued fractions expansions and, thus, by their approximation
properties. A classical example is the set of badly approximable numbers. Hausdorff and
packing dimensions have been used to determine the size of such sets. The work in this
vein goes back as far as to V. Jarnı́k [22], and includes I. J. Good’s celebrated paper
[16] and several refinements and extensions [29, 33]. Other sets related to Diophantine
approximation, such as the Markov and Lagrange spectrum, have also been studied with
the aid of regular continued fractions [30].
Regular continued fractions link analytic and algebraic properties of irrational numbers.
In this sense, we have classical results such as Lagrange’s Theorem on quadratic surds
[24, Theorem 28] or J. Liouville’s construction of transcendental numbers by forcing fast
rational approximation [24, Section 9] (see [24] for the classical theory on regular continued
fractions and [4] for modern account). With the aid of Schmidt’s Subspace Theorem, B.
Adamczewski and Y. Bugeaud built families of badly approximable transcendental numbers
[1, 5].
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
3
It is, hence, highly desirable to have continued fractions with most of the aforementioned
properties in other contexts. In the complex plane, several continued fraction algorithms
have been suggested depending on the preferred properties [7, 20, 21, 27]. In particular, in [21], Adolf Hurwitz suggested a continued fraction algorithm which is an almost
straightforward analogue of an elementary definition of regular continued fractions.
Hurwitz continued fractions (defined in Section 3) gives a continued fraction expansion
for complex numbers that shares several desirable properties with regular continued fractions. For instance, the Hurwitz continued fraction of a complex number is infinite if and
only if this number is in C ∖ Q(i). The denominators of the convergents, also known as
continuants, have an exponential growth in complex absolute value [9, Corollary 5.2], and
the Hurwitz convergents give good approximations [25, Theorem 2]. As an application
of the approximation properties of Hurwitz continued fractions, R. Lakein [26] obtained
a new proof of L. Ford’s Theorem on approximation to complex numbers by elements of
Q(i).
Although it is far from being as well understood as their real countarpart yet, the
theory of Hurwitz continued fractions has recently being used in complex Diophantine
approximation. For example, the second author [14] constructed a family of transcendental
complex numbers giving an analogue of [5]. In [18], Y. He and Y. Xiong computed, for a
given function ψ with some regularity conditions, the Hausdorff dimension of set of complex
numbers z admitting rational approximation of order ψ whose Hurwitz continued fraction
differ from those of z. They also obtained a complex analogue of Jarnı́k’s Theorem on ψwell approximable numbers. In [17], the same authors calculated the Hausdorff dimension
of sets of complex numbers that can be approximated by quotients of Gaussian integers at
a given order ψ but no better. This is a complex version of a problem solved by Y. Bugeaud
[3]. S. G. Dani and A. Nogueira extended their work [8] on the image of Z2 on real binary
quadratic forms to complex numbers [9, Section 7]; and D. Hensley built complex badly
approximable numbers of degree 4 over Q(i), the field of quotients of Gaussian integers [19].
In [31], D. Simmons studied Diophantine approximation properties of Hurwitz continued
fractions restricted to real numbers comparing them with regular continued fractions.
In this paper, we compute the Hausdorff dimension of sets of complex numbers whose
Hurwitz continued fraction has a subsequence of partial quotients of fast growth. As a
consequence, we obtain a complex analogue of the Borel-Bernstein refinement by B. W.
Wang and J. Wu [33].
A significant part of this paper is dedicated to the study of the shift space associated
to Hurwitz continued fractions and to the geometry of the process. These two aspects are
intimately related. As we shall see, the space of sequences that are the Hurwitz continued
fraction of a complex number is not a closed subset of the sequences of Gaussian integers
Z[i]N (see the proof of Proposition 5.5). However, we will be able to replace this set by
4
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
the closure of a subset R of Z[i]N . The set R will be defined in simple topological terms in
Section 3. We describe the various possible forms of cylinders and study the intersection
of a disc and a finite union of cylinders. These auxiliary results, which are used in the
proofs of our main theorems, are of independent interest. All this shows that the shift
space associated to Hurwitz continued fractions is more complicated than the β-shift.
Let us define some notation in order to state our main results. For any complex number
z, we write ∥z∥∶ = max{∣R(z)∣, ∣I(z)∣}. For any t ∈ R, we let ⌊t⌋ be the integer part of t,
that is
⌊t⌋∶ = max{m ∈ Z ∶ m ≤ t}.
In other words, [z] is the Gaussian integer nearest to z. For each z ∈ C, define
1
1
[z]∶ = ⌊R(z) + ⌋ + i ⌊I(z) + ⌋ .
2
2
Define the set
F∶ = {z ∈ C ∶ [z] = 0}.
The set F is the unit square centered at the origin whose sides are parallel to the coordinate
axes including the left and the bottom sides and excluding the remaining ones. We restrict
to F, whose Gaussian integral translates form a partition of the complex plane. If z ∈ F,
we denote its Hurwitz continued fraction by [0; a1 (z), a2 (z), . . .], or simply [0; a1 , a2 , . . .].
For any B > 1, we write
ρ
⎧
⎫
⎪
⎪
1
⎪
⎪
sB = lim inf ⎨ρ ≥ 0 ∶
(
)
<
1
⎬,
∑
n
2
n→∞
⎪
⎪
B
∣q
(a
,
.
.
.
,
a
)∣
n
1
n
⎪
⎪
(a
,...,a
)∈R(n)
n
1
⎩
⎭
where R(n) is a special subset of D n (see section 4).
We let dimH (A) denote the Hausdorff dimension of a given set A ⊆ C.
Theorem 1.1. For a given a function Φ ∶ N → R>0 , define the set
E∞ (Φ)∶ = {z ∈ F ∶ ∥an (z)∥ ≥ Φ(n) for infinitely many n ∈ N}
and a number B ≥ 1 by
log B = lim inf
n→∞
log(Φ(n))
.
n
Then,
1. If B = 1, then dimH (E∞ (Φ)) = 2.
2. If 1 < B < ∞, then dimH (E∞ (Φ)) = sB .
3. When B = ∞, let b be given by log b = lim inf
i. If b = 1, then dimH (E∞ (Φ)) = 1.
ii. If 1 < b < ∞, then dimH (E∞ (Φ)) =
iii. If b = ∞, then dimH (E∞ (Φ)) = 0.
n→∞
2
1+b .
log log Φ(n)
.
n
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
5
Remark. The bulk of the proof is taken up by the case 1 < B < ∞. Under these circumstances, the function which determines dimH (E∞ (Φ)), B ↦ sB , is continuous and satisfies
lim sB = 2
B→1
and
lim sB = 1.
B→∞
For the proof of this claim, see Lemma 7.18.
Theorem 1.1 still holds when ∥ ⋅ ∥ is replaced with the complex absolute value ∣ ⋅ ∣. For
any given Φ ∶ N → R>0 , every positive real number c > 0 satisfies
log(Φ(n))
log(c Φ(n))
= lim inf
,
n→∞
n→∞
n
n
log log Φ(n)
log log (c Φ(n))
lim inf
= lim inf
;
n→∞
n→∞
n
n
and Theorem 1.1 yields dimH (E∞ (Φ)) = dimH (E∞ (cΦ)). Therefore, putting
lim inf
E(Φ)∶ = {z ∈ C ∶ ∣an (z)∣ ≥ Φ(n) for infinitely many n ∈ N}
we have
E∞ (2− 2 Φ) ⊆ E(Φ) ⊆ E∞ (Φ),
1
and thus dimH (E(Φ)) = dimH (E∞ (Φ)).
An important step towards Theorem 1.1 is the computation of the Hausdorff dimension
of certain sets. For each B > 1 define
F (B)∶ = {z = [0; a1 , a2 , . . .] ∈ F ∶ ∥an (z)∥ ≥ B n for infinitely many n ∈ N}.
Theorem 1.2. For each B > 1, we have dimH (F (B)) = sB .
In [29], T. Luczak calculated the Hausdorff measure of certain sets of real numbers
determined by their regular continued fraction. Given x ∈ (0, 1)∖Q, let [0; A1 (x), A2 (x), . . .]
denote its regular continued fraction. For any b, c > 1, define the sets
̃R (b, c)∶ = {ξ ∈ (0, 1) ∶ cbn ≤ An (ξ) for all n ∈ N},
Ξ
n
ΞR (b, c)∶ = {ξ ∈ (0, 1) ∶ cb ≤ An (ξ) for infinitely many n ∈ N}.
Theorem 1.3 (T. Luczak, [29]). For every pair of real numbers b, c > 1, we have
̃R (b, c) = dimH Ξ(b, c)R =
dimH Ξ
1
.
1+b
Our proof of Theorem 1.1 relies on a complex analogue of T. Luczak’s theorem. For any
b, c > 1 define
̃′ (b, c) = {z ∈ F ∶ cbn ≤ ∣an (z)∣ for all n ∈ N} ,
Ξ
Ξ′ (b, c) = {z ∈ F ∶ cb ≤ ∣an (z)∣ for infinitely many n ∈ N} .
n
Theorem 1.4. If b, c > 1, then
̃′ (b, c) = dimH Ξ′ (b, c) =
dimH Ξ
2
.
1+b
6
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
Observe that for any pair of real numbers b > 1 and c > 1, we have
n
{z ∈ F ∶ cb ≤ ∥an (z)∥ for all n ∈ N} ⊆ {z ∈ F ∶ lim ∣an (z)∣ = ∞} .
n→∞
Therefore, by Theorem 1.4
2
≤ dimH ({z ∈ F ∶ lim ∣an (z)∣ = ∞})
n→∞
1+b
and letting b → 1 we conclude
1 ≤ dimH ({z ∈ F ∶ lim ∣an (z)∣ = ∞}) .
n→∞
This is a considerably shorter proof of the lower bound in Good’s theorem for Hurwitz
continued fractions [13, Theorem 1.3].
Theorem 1.5. We have
dimH ({z ∈ F ∶ lim ∣an (z)∣ = ∞}) = 1.
n→∞
The paper is organized as follows. Section 2 gathers a list of frequently used notation. In
Section 3, we recall some basic facts about Hurwitz continued fractions. Section 4 studies
the symmetries and the size of regular cylinders (defined on Section 3). Section 5 contains
the proof of two estimates that will allow us to apply the mass distribution principle in the
proof of Theorem 1.2. In section 6, we show how can we approximate irregular numbers
by regular numbers (defined in section 3). Sections 7 and 8 are dedicated to the proofs of
Theorems 1.2 and 8, respectively. Lastly, Section 9 contains the proof of Theorem 1.1.
Acknowledgements. Yann Bugeaud is partially supported by project ANR-18-CE400018 funded by the French National Research Agency. The research of Mumtaz Hussain
and Gerardo González Robert is supported by the Australian Research Council Discovery Project (200100994). Gerardo González Robert was partially funded by the program
Estancias posdoctorales por México of the CONACyT, Mexico.
2. Frequently used notation
In this section, we collect some of the notation used frequently.
1. If X is a non-empty finite set, #X denotes the number of elements contained in X.
2. If X, Y are non-empty sets, A ⊆ X, and f ∶ X → Y is any function, the image of A
under f is f [A]; that is f [A]∶ = {f (a) ∶ a ∈ A}.
3. For any z ∈ C we write ∥z∥∶ = max{∣R(z)∣, ∣I(z)∣}.
4. For any z ∈ C, we write Pm(z)∶ = min{∣R(z)∣, ∣I(z)∣}.
√
5. D∶ = {a ∈ Z[i] ∶ ∣a∣ ≥ 2}.
√
6. E ∶ = {a ∈ D ∶ ∣a∣ ≥ 8}.
7. For M > 0, we define D(M )∶ = {a ∈ D ∶ ∥a∥ ≤ M }.
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
7
8. For any z ∈ C ∖ {0}, define ι(z)∶ = z −1 . For any a ∈ D, we write τa (z) = z + a for all
z ∈ C. The composition of functions τa and ι is denoted by juxtaposition; for example,
τa ι(z) = z −1 + a and ιτa (z) = (a + z)−1 for z ∈ C.
9. For n ∈ N and a ∈ D n , define the functions Qn ( ⋅ , a), Tn ( ⋅ ; a) by
Tn ( ⋅ ; a)∶ = τ−an ι⋯τ−a1 ι,
Qn ( ⋅ ; a)∶ = τa1 ι⋯τan ι.
10. The functions rot, mir1 , mir2 ∶ C → C are given by
rot(z) = iz,
mir1 (z) = z,
and
mir2 (z) = −z
for all z ∈ C.
11. For every A ⊆ C, we denote the closure of A by Cl(A), the interior of A by Int(A), and
the boundary of A by Fr(A).
12. Given two non-empty sets A, B ⊆ C, the distance between them is
dist(A, B)∶ = inf{∣a − b∣ ∶ a ∈ A, b ∈ B}.
13. For any z ∈ C and any ρ > 0, we write
D(z; ρ)∶ = {w ∈ C ∶ ∣z − w∣ < ρ} and D(z; ρ)∶ = {w ∈ C ∶ ∣z − w∣ ≤ ρ},
and also
C(z, ρ)∶ = {w ∈ C ∶ ∣z − w∣ = ρ}.
Following [25], we write
D(z)∶ = D(z; 1),
D(z)∶ = D(z; 1),
E(z)∶ = C ∖ D(z),
E(z)∶ = C ∖ D(z).
14. Given k ∈ N and a = (an )n≥1 ∈ D N , the k-th prefix of a
pref(a; k)∶ = (a1 , . . . , ak ).
3. Basics on Hurwitz continued fractions
For any real number t, let ⌊t⌋ be the largest integer less than or equal to t. For each
z ∈ C, we define its nearest Gaussian integer [z] by
1
1
[z] = ⌊R(z) + ⌋ + i ⌊I(z) + ⌋ .
2
2
The complex Gauss map, T ∶ F → F, associates to each z ∈ F the complex number T (z)
given by
⎧
⎪
⎪
⎪z −1 − [z −1 ], if z ≠ 0,
T (z) = ⎨
⎪
⎪
0, if z = 0.
⎪
⎩
As usual, T 0 represents the identity map on F and T n ∶ = T n−1 ○ T for all n ∈ N. For any
z ∈ F ∖ {0}, we define a1 (z)∶ = [z −1 ] and, if T n−1 (z) ≠ 0, we put an (z)∶ = a1 (T n−1 (z)). The
8
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
expression
1
[0; a1 (z), a2 (z), . . .]∶ =
1
a1 (z) +
a2 (z) +
1
⋱
is the Hurwitz continued fraction (hcf) of z and the terms of the sequence (an )n≥1 are
its partial quotients or elements. We define the sequences of Gaussian integers (pn )n≥0 ,
(qn )n≥0 from (an )n≥1 in the usual way (see [24, Section 2]):
⎛pn ⎞ ⎛pn−1 pn−2 ⎞ ⎛an ⎞
⎛p−1 p−2 ⎞ ⎛1 0⎞
for all n ∈ N0 .
=
and
=
⎝qn ⎠ ⎝qn−1 qn−2 ⎠ ⎝ 1 ⎠
⎝q−1 q−2 ⎠ ⎝0 1⎠
Following [9], we refer to ((pn )n≥0 , (qn )n≥0 ) as the Q-pair of (an )n≥1 .
Proposition 3.1. Take z ∈ F ∖ {0}, let (an )n≥1 be the sequence of its hcf elements and
((pn )n≥0 , (qn )n≥0 ) its Q-pair. The next assertions hold:
i. an ∈ D for every n.
ii. For every n ∈ N,
1
pn
= [0; a1 , a2 , . . . , an ] ∶=
qn
.
1
a1 +
1
a2 +
1
⋱+
an
iii. The sequence (an )n≥1 is infinite if and only if z ∈ F ∖ Q(i). In this case, we have that
lim [0; a1 , a2 , . . . , an ] = z.
n→∞
Thus, Hurwitz continued fractions provide an injection from F ∖ Q(i) into D N .
iv. The sequence
(∣qn ∣)n≥0 is strictly increasing.
√
√
v. If ψ∶ = 1+2 5 , then ∣qn ∣ ≥ ψ n−1 for all n ∈ N1 .
vi. If z = [0; a1 , a2 , . . .] ∈ F ∖ Q(i), for all n ∈ N we have
z=
(an+1 + [0; an+2 , an+3 , . . .])pn + pn−1
.
(an+1 + [0; an+2 , an+3 , . . .])qn + qn−1
vii. If z = [0; a1 , a2 , . . .] ∈ F ∖ Q(i), then every n ∈ N verifies
∣z −
pn
1
∣<
.
qn
∣qn ∣2
Proof. i. See [19, p. 73].
ii. This follows from the general theory of continued fractions.
iii. See [9, Theorem 6.1].
iv. See [21, p. 195].
v. This follows from [9, Corollary 5.3].
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
9
vi. See [9, Proposition 3.3]
vii. See [25, Theorem 1].
□
4. Cylinders
There are several differences between regular and Hurwitz continued fractions preventing
the proofs in the real case from being extended verbatim into the complex plane. The first
difference we point out is of a geometric nature.
Definition 4.1. Given n ∈ N and a = (a1 , . . . , an ) ∈ D n , the cylinder of level n based on
a is
Cn (a)∶ = {z ∈ F ∶ a1 (z) = a1 , . . . , an (z) = an }.
The prototype set Fn (a) is Fn (a)∶ = T n [Cn (a)].
We borrow the term prototype set from [17]. The cylinders of level 1 are depicted on
Figure 1.
1
0.5
−1 − i
−1 − 2i
−2 − 2i
−2 − i
−0.5
−2
−2 + i
−1 + i
−2i
−1 − 3i
1−i
1 − 2i
−3i
2 − 2i
2−i
3 − 2i
−3 − i
3−i
−3
2
3
0.5
0
−3 + i
3+i
−2 + 2i
−1 + 2i
−1 + 3i
3i
1 + 3i
2 + 2i
1 + 2i
2i
2+i
1+i
−0.5
Figure 1: Partition of F.
Figure 1. Cylinders of level 1.
Definition 4.2. Take n ∈ N. We say that a = (a1 , . . . , an ) ∈ D n is
i.
ii.
iii.
iv.
v.
vi.
valid if Cn (a) ≠ ∅;
regular if Cn (a) has non-empty interior;
irregular if it is valid and has empty interior;
extremely irregular if it contains exactly one element;
full if T n [Cn (a)] = F;
almost full if for every b ∈ D the segment (a1 , . . . , an , b) is regular.
10
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
We denote by Ω(n), R(n), Ir(n), EI(n), F(n), and AF(n) the set of valid, regular, irregular,
extremely irregular, full, almost full, respectively, words of length n.
We say that a = (an )n≥1 ∈ D N is
i.
ii.
iii.
iv.
v.
vi.
valid if it is the hcf of some z ∈ F;
regular if it is valid and (a1 , . . . , an ) ∈ R(n) for all n ∈ N;
irregular if it is valid and (a1 , . . . , an ) ∈ Ir(n) for some n ∈ N;
extremely irregular if it is valid and (a1 , . . . , an ) ∈ EI(n) for some n ∈ N.
full if it is valid and (a1 , . . . , an ) ∈ F(n) for all n ∈ N;
almost full if it is valid and (a1 , . . . , an ) ∈ AF(n) for all n ∈ N;
We denote by Ω, R, EI, F, and AF the set of valid, regular, extremely irregular, full, and
almost full words, respectively.
We extend in an obvious way the notions of being full, almost full, regular, irregular,
and extremely irregular to prototype sets. For example, given n ∈ N and a ∈ R(n), we say
that Fn (a) is regular if a ∈ Ω(n). Unlike cylinders, there are only finitely many prototype
sets. In fact, as noted in [10], when we disregard the boundary, there are only thirteen
regular prototype sets.
We denote the complex inversion by ι(z); that is, ι(z) = z −1 for all non-zero z ∈ C. Figure
2 depicts the set ι[F], the image of F under ι,
ι[F] = E(1) ∩ E(i) ∩ E(−1) ∩ E(−i).
This picture also shows translations of prototype sets of the form F1 (a).
3
2
1
−3
−2
−1
1
2
3
−1
−2
−3
Figure 2. The set ι[F].
We say that a number z = [0; a1 , a2 , . . .] ∈ F is full (resp. almost full, regular, irregular,
extremely irregular) if (an )n≥1 is full (resp. almost full, regular, irregular, extremely
irregular). From a measure theoretic perspective, almost every z ∈ F is regular (see [10]).
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
11
However, irregular numbers and extremely irregular numbers do exist. An example of
irregular cylinder is C2 (−2, 1 + 3i) and an example of an extremely irregular cylinder is
C3 (−2, 1 + 2i, −2 + i). Indeed, for each a ∈ Z[i], let τa ∶ C → C be the function given by
τa (z) = z + a for all z ∈ C. After some straightforward computations (or with the help
of a picture), we may see that ι[F1 (−2)] ∩ τ1+3i [F] is the line segment doing from 21 + 52 i
(included) to 12 + 72 i (excluded). Hence, the set
ι [T [C2 (−2, 1 + 3i)]] = ι [F1 (−2) ∩ C1 (1 + 3i)]
= ι [F1 (−2) ∩ F ∩ C1 (1 + 3i)]
= ι [F1 (−2) ∩ ιτ1+3i [F]]
and, thus, C2 (−2, 1 + 3i) has empty interior. Similarly, we may verify that
√
2− 3 i
F3 (−2, 1 + 2i, −2 + i) = {
− }.
2
2
In what follows, for all n ∈ N and a ∈ Ω(n), we write
C n (a)∶ = Cl(Cn (a)), Cn○ (a)∶ = Int(Cn (a)),
Fn (a)∶ = Cl(Fn (a)), F○n (a)∶ = Int(Fn (a)),
and F = Cl(F) and F○ = Int(F). For each n ∈ N and a ∈ R(n), let Tn ( ⋅ ; a), Qn ( ⋅ ; a) ∶ C → C
be given by
Tn (z; a)∶ = τ−an ι⋯τ−a1 ι(z)
and
Qn (z; a)∶ = ιτa1 . . . ιτan (z).
Clearly, Qn (z; a) and Tn (z; a) are mutual inverses and Tn ( ⋅ ; a) is a continuous extension
of T n restricted to Cn (a). The elementary theory of continued fractions tells us that
Qn (z; a) =
(−1)n z
wpn−1 + pn pn
=
+
,
wqn−1 + qn qn q 2 (1 + qn−1 z)
n
qn
and, for all z ∈ Fn (a), we have
Qn (z; a) = [0; a1 , a2 , . . . , an−1 , an + z].
We can avoid facing irregular and extremely irregular cylinders when we consider sequences
with sufficiently large partial quotients. This result is not new, but a complete proof has
not been given as far as we know.
Proposition 4.3. If a = (an )n≥1 ∈ D N satisfies ∣an ∣ ≥
E N ⊆ R.
√
8 for all n ∈ N, then a ∈ R; that is
Proof. Let a = (an )n≥1 be any element of E N and call (pn )n≥0 , (qn )n≥0 its Q-pair. Then, for
every n, k ∈ N, we have
∣
pn pn+k
1
1
−
∣=
≤
.
√
qn qn+k
2 (1 − 2 )
∣qn+k ∣2 (1 + qn−k−1
[0;
a
,
.
.
.
,
a
])
∣q
∣
n+1
n+k
n+k
qn−k
2
12
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
Then, because (∣qj ∣)j≥1 is strictly increasing, (pn /qn )n≥1 is a Cauchy sequence. Consider
pn
z∶ = lim .
n→∞ qn
√
From ∣a1 ∣ ≥ 8 and Figure 1, we obtain
Then, since ∣a2 ∣ ≥
√
8,
C 1 (a1 ) ⊆ F○ and F1 (a1 ) = F.
C 1 (a2 ) ⊆ F○ = F○1 (a1 ),
so
C 2 (a1 , a2 ) = Q1 (C 1 (a2 ); a1 ) ⊆ Q1 (F○ ; a1 ) = C ○ (a1 ).
√
Therefore, z ∈ C ○ (a1 ) and a1 (z) = a1 . From ∣a2 ∣ ≥ 8 and F1 (a1 ) = F, we also obtain
F2 (a1 , a2 ) = T 2 [C2 (a1 , a2 )] = T [C1 (a2 ) ∩ F1 (a1 )] = F,
which means (a1 , a2 ) ∈ F(2). Relying on the identity
Cn+1 (a1 , . . . , an+1 ) = Cn (a1 , . . . , an ) ∩ Qn (C1 (an+1 ); (a1 , . . . , an )) for all n ∈ N
and using inductively the previous argument, we may conclude
(1)
(a1 , . . . , an ) ∈ R(n) for all n ∈ N.
and
(2)
z ∈ Cn○ (a1 , . . . , an ) for all n ∈ N.
From (2), we conclude z = [0; a1 , a2 , a3 , . . .]. Hence, a ∈ Ω and, by (1), we conclude a ∈
F(n).
□
In the previous proof, it is quite tempting to say that a ∈ Ω because (a1 , . . . , an ) ∈ R(n)
for all n ∈ N. However, there are sequences a = (an )n≥1 ∈ D N such that (a1 , . . . , an ) ∈ R(n)
for all n ∈ N but a ∈/ Ω (see the proof of Proposition 5.5). In symbolic terms, this means
that Ω is not a closed subspace of D N when endowed with the product topology.
This taxonomy is irrelevant for regular continued fractions, because every sequence of
natural numbers would be full.
4.1. Symmetries of regular cylinders. The symmetries of the Hurwitz continued fraction process lie at the heart of the geometric properties of Hurwitz continued fractions that
we develop. More precisely, consider the maps rot, mir1 ∶ C → C given by
rot(z)∶ = iz
and
mir1 (z)∶ = z
for all z ∈ C
and call Di8 the group of isometries generated by rot and mir1 along with the composition of
functions. Note that Di8 is isomorphic to the dihedral group of order 8. The juxtaposition
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
13
of elements of Di8 , ι, and the maps τa , a ∈ D, stands for their composition; for example,
rot mir1 (z) = iz for z ∈ C. Clearly, if f ∈ Di8 , then f [D] = D and ∥f (a)∥ = ∥a∥ for all a ∈ D.
Proposition 4.4. 1. rot ι = ι rot−1 , rot−1 ι = ι rot.
2. rot τa = τrot(a) rot, rot−1 τa = τrot−1 (a) rot−1 .
Proof. For z ∈ C, z ≠ 0, we have
1
1
rot ι(z) = i =
= ιrot−1 (z),
z −iz
so rotι = ιrot−1 . This equality implies rot−1 ι = ι rot. The second point is proven similarly.
□
Proposition 4.5. For all n ∈ N, a ∈ R(n), and f ∈ Di8 , there are some b ∈ R(n) and
f1 , . . . , fj ∈ Di8 such that
f [Cn○ (a)] = Cn○ (b)
and
bj = fj (aj ) for all j ∈ {1, . . . , n}.
We can replace Cn○ with C n in the previous equality.
Proof. It is enough to consider f ∈ {rot, mir1 }. Let us demonstrate that, if n ∈ N and
a = (a1 , . . . , an ) ∈ R(n), then
(3)
rot [Cn○ (a)] = C ○ (a) (rot(a1 ), rot−1 (a2 ), rot(a3 ), . . . , rot(−1) (an )) .
n
The proof is by induction on n. Given a1 ∈ D, we have z ∈ C ○ (a1 ) if and ony if there is
some w ∈ F○ (a1 ) such that
1
.
z=
a1 + w
Applying rot, we get
1
rot(z) =
,
rot−1 (a1 ) + rot−1 (w)
and, since rot(z) ∈ rot[F○ ] = F○ and rot−1 (a) ∈ D, we obtain
rot−1 (w) = ιrot(z) − rot−1 (a1 ) ∈ F○ .
Hence, rot(z) ∈ C1○ (rot−1 (a1 )) and rot[C1○ (a1 )] ⊆ C1○ (rot−1 (a1 )). If z ∈ C1○ (rot−1 (a1 )), there
is some w ∈ F○ such that
1
z=
−1
rot (a1 ) + w
−1
so, applying rot ,
1
rot−1 (z) =
.
a1 + rot(w)
As before, we conclude rot−1 (z) ∈ C1○ (a1 ) and z ∈ rot[C1○ (a1 )]. That is, (3) is true for n = 1.
Assume (3) for n = N ∈ N and let us show
(4)
rot [CN○ +1 (a)] = CN○ +1 (rot(a1 ), rot−1 (a2 ), rot(a3 ), . . . , rot(−1)
N +1
(aN +1 )) .
14
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
From
we obtain
(5)
CN○ +1 (a1 , . . . , aN +1 ) = CN○ (a1 , . . . , aN +1 ) ∩ QN (C1○ (aN +1 ); (a1 , . . . , aN )),
rot [CN○ +1 (a1 , . . . , aN +1 )] = rot [CN○ (a1 , . . . , aN )] ∩ rot ι τa1 ⋯ι τaN [C1○ (aN +1 )] .
The induction hypothesis tells us that
(6)
rot [CN○ (a1 , . . . , aN )] = CN○ (rot−1 (a1 ), . . . , rot(−1) (aN )) .
N
For the second term in (5), notice that every c ∈ D satisfies
rot ι τc = ι τrot−1 (c) rot−1
and
rot−1 ι τc = ι τrot(c) rot,
in view of Proposition 4.4. Then,
rot ι τa1 ⋯ι τaN [C1 (aN +1 )] =
= QN (C1○ (rot(−1)
N +1
(aN +1 )) ; (rot−1 (a1 ), rot(a2 ), . . . , rot(−1) (aN ))) .
N
We plug the previous equation and (6) into (5) and conclude (4).
A similar yet simpler procedure shows
(7)
mir1 [Cn○ (a)] = Cn○ (mir1 (a1 ), mir1 (a2 ), . . . , mir1 (an ))
for all n ∈ N and all a = (a1 , . . . , an ) ∈ R(n).
The assertion for closed cylinders is true because both mir1 and rot are homeomorphisms
and C n (a) = Cl (C ○ (a)) for any n ∈ N and a ∈ R(n).
□
For certain elements in Di8 , we can easily determine the functions f1 , . . . , fn given in
Proposition 4.5. For instance, consider mir2 ∶ = rot2 mir1 , whose correspondence rule is
mir2 (z) = −z for z ∈ C.
Corollary 4.6. If n ∈ N and a ∈ R(n), then
(8)
mir2 (Cn○ (a)) = Cn○ (mir2 (a1 ), mir2 (a2 ), . . . , mir2 (an )) .
4.2. Size of regular cylinders. We can estimate both the measure and the diameter of
regular cylinders with the continuants.
Proposition 4.7. There is a constant κ1 ∈ (0, 1) such that every n ∈ N and every a ∈ R(n)
satisfy
κ1
2
≤
∣C
(a)∣
≤
n
∣qn (a)∣2
∣qn (a)∣2
Proof. Lemma 2.7 in [17].
and
πκ1
π
≤
m(C
(a))
≤
.
n
∣qn (a)∣4
∣qn (a)∣4
□
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
15
For any m, n ∈ N, a = (a1 , . . . , an ) ∈ D n , and b = (b1 , . . . , bm ) ∈ D m , we write
ab∶ = (a1 , . . . , an , b1 , . . . , bm ) ∈ D n+m .
Proposition 4.8 (Lemma 2.2.(g), [17]). For all m, n ∈ N and any a ∈ Ω(n), b ∈ Ω(m)
such that ab ∈ Ω(n + m), we have
1
∣qn (a)∣∣qm (b)∣ ≤ ∣qn+m (ab)∣ ≤ 3∣qn (a)∣∣qm (b)∣.
5
Proposition 4.9. For any m, n ∈ N and a ∈ Ω(n), b ∈ Ω(m) such that ab ∈ Ω(m + n), we
have
1
√ ∥qn (a)∥∥qm (b)∥ ≤ ∥qn+m (ab)∥ ≤ 6∥qn (a)∥∥qm (b)∥.
5 2
Proof. Since ∥z∥ ≤ ∣z∣ ≤
√
2∥z∥ for any complex number z, Proposition 4.8 implies
∥qn+m (ab)∥ ≤ ∣qn+m (ab)∣
≤ 3∣qn (a)∣∣qm (b)∣
√
√
≤ 3( 2∥qn (a)∥)( 2∥qm (b)∥)
= 6∥qn (a)∥∥qm (b)∥
and
1
∥qn+m (ab)∥ ≥ √ ∣qn+m (ab)∣
2
1
≥ √ ∣qn (a)∣∣qm (b)∣
5 2
1
≥ √ ∥qn (a)∥∥qm (b)∥.
5 2
□
The combination of propositions 4.7 and 4.8 yield the following statement.
Proposition 4.10. There exists a constant κ2 > 1 such that for any r, s ∈ N and any
c ∈ R(r), d ∈ R(s) with cd ∈ R(r + s) we have
1
∣Cr (c)∣∣Cr (d)∣ ≤ ∣Cr+s (cd)∣ ≤ κ2 ∣Cr (c)∣∣Cr (d)∣.
κ2
For all z ∈ C, write Pm(z)∶ = min{∣R(z)∣, ∣I(z)∣}.
Proposition 4.11. Let n be a natural number and a ∈ R(n).
i. If Pm(an ) ≥ 3, then a ∈ F(n).
ii. If ∥an ∥ ≥ 3, then a ∈ AF(n).
iii. There exists d ∈ D such that Pm(d) = ∥d∥ = 3 and ad ∈ F(n).
16
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
Proof. i. We have to check case by case all the forms of regular prototype sets (see [10,
Section 2]).
ii. Same as above.
iii. Lemma 2.6(b) in [17] shows this claim for Pm(d) = ∥d∥ = 2 rather Pm(d) = ∥d∥ = 3, but
their argument also holds for 3.
□
5. Geometric constructions
In this section, we consider study two geometric constructions that will allow us to apply
the Mass Distribution Principle when showing Theorem 1.2. In both of them, we count
the number of sets pertaining to a particular family of subsets of F intersecting a given
disc.
Define the set
Esq∶ = {
√
1 + i −1 + i −1 − i 1 − i
,
,
,
}
2
2
2
2
and, writing ζ∶ = 12 + 2−2 3 i, consider
Cuad(ζ)∶ = {f (ζ) ∶ f ∈ Di8 } ,
√
1
3
Cuad∶ = Cuad(ζ) ∪ {f (− − i
) ∶ f ∈ Di8 }
2
6
Then,
4
Cuad(ζ) = Fr(F) ∩ ⋃ C(ij (1 + i), 1)
j=1
and Cuad is the set of points in Fr(F) where the closure of two cylinders coincide.
Proposition 5.1. For each n ∈ N and each a ∈ R(n), we have
⋃
C n+1 (ad) = C n (a) ∖ {[0; a1 , . . . , an ]}.
d∈D
ad∈R(n+1)
Proof. First, we show that
(9)
⋃ C 1 (d) = F ∖ {0}.
d∈D
For every d ∈ D and each z ∈ C1 (d), there is some w ∈ F such that z = (d + w)−1 , so
∣z∣ ≥
1
1
√ ,
≥
∣d∣ + ∣w∣ ∣d∣ + 2
2
and 0 ∈/ C 1 (d). Now pick any z ∈ F ∖ {0}. If z ∈ F, then z ∈ C1 (a1 (z)) ⊆ C 1 (a1 (z)).
Otherwise, z belongs to the upper side of F or to its right side; in either case, there is some
mir ∈ {mir1 , mir2 } such that mir(z) ∈ F, hence
mir(z) ∈ C 1 (a1 (mir(z)),
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
17
and, by either (7) or (8),
z ∈ mir(C 1 (a1 (mir(z)) = C 1 (mir(a1 (mir(z))).
Let n ∈ N and a ∈ R(n) be arbitrary. From (9) we get
Fn (a) ∩ ⋃ C 1 (d) = Fn (a) ∖ {0}.
d∈D
Since Qn ( ⋅ ; a) provides a homeomorphisn between Fn (a) and C n (a), the proposition follows.
□
Proposition 5.2. For each n ∈ N≥2 , we have
Esq ∩ ⋃ C n (b) = ∅.
a∈R(n)
Proof. The proof is by induction on n. Observe that
ι[Esq] = {τa (0) ∶ a ∈ {1 + i, −1 + i, −1 − i, 1 − i}} .
Then, if I∶ = {1 + i, 1 − i, −1 + i, −1 − i}, we have
⎡
⎤
⎢
⎥
⎢
ι ⎢Esq ∪ ⋃ C 2 (a)⎥⎥ =
ι[Esq] ∪ τa1 [F1 (a1 ) ∩ C 1 (a2 )]
⋃
⎢
⎥ (a1 ,a2 )∈R(2)
a∈R(2)
⎣
⎦
= ⋃ ι[Esq] ∪ τa1 [{0} ∩ F1 (a1 ) ∩ ⋃ C 1 (a2 )] .
a1 ∈I
a2 ∈D
Proposition 5.1 tells us that the last term is a union of empty sets, which gives
Esq ∪ ⋃ C 2 (a) = ∅.
a∈R(2)
□
5.1. First geometric construction. Theorem 5.3 is the main result of this section. It
says that a disc centered at a point belonging to the closure of a regular cylinder of level n,
say Cn (a), intersects at most six regular cylinders of level n as long as its radius is small
enough in terms of ∣Cn (a)∣.
Given n ∈ N and a ∈ R(n), the set of neighbors of a is
Vecn (a)∶ = {b ∈ R(n) ∶ b ≠ a and C n (a) ∩ C n (b) ≠ ∅} .
Furthermore, for any ρ > 0 and z ∈ F, define
Gn (z, a, ρ)∶ = {b ∈ R(n) ∶ D(z; ρ∣Cn (a)∣) ∩ C n (b) ≠ ∅}.
Theorem 5.3. There exists a constant ρ̃ > 0 such that for every k ∈ N, a ∈ R(n) and
z ∈ C n (a) we have
#Gn (z, a, ρ̃) ≤ 6,
Gn (z, a, ρ̃) ⊆ {a} ∪ Vecn (a)
18
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
and
D (z; ρ̃∣Cn (a)∣) ∩ F ⊆
⋃
b∈Gn (z,a,ρ̃)
C n (b).
Recall that for all z ∈ C, ρ > 0, and w ∈ D(z; ρ), we have
(10)
D (w; inf ∣w − (z + ρeiθ )∣) ⊆ D(z; ρ).
0≤θ<2π
Proposition 5.4. Let n ∈ N and a ∈ R(n) be arbitrary. If w ∈ F and 0 < r < 1 are such that
D(w; r) ⊆ D(0; 1), then Qn (D(w; r); a) is a disc in the complex plane and
⎛
⎞
r
D ⎜Qn (w; a);
∣Cn (a)∣⎟ ⊆ Qn (D(w; r); a).
√
√
2 (1 + 22 ) (2 + 22 )
⎝
⎠
Proof. Since Qn ( ⋅ ; a) is a Möbius transformation, the set Qn (D(w; r); a) is a disc on the
Riemann sphere. The strict growth of the continuants (Prop. 3.1, iv) implies that D(w; r)
does not contain the pole of Qn ( ⋅ ; a) and, hence, Qn (D(w; r); a) is in fact a disc in the
complex plane.
Write qn = qn (a) and qn−1 = qn−1 (a). For all θ ∈ [0; 2π], Proposition 4.7 yields
r
∣Qn (w + reiθ ; a) − Qn (w; a)∣ =
qn−1
iθ
∣qn ∣2 ∣1 + qn−1
qn w∣ ∣1 + qn (w + re )∣
r
≥
∣Cn (a)∣,
√
√
2
2 (1 + 2 ) (2 + 22 )
since r ≤ 1. The proposition now follows from (10).
□
The proof of Theorem 5.3 is long and involves cumbersome computations. One of its
many upshots, however, is that it provides a detailed description of neighboring regular
cylinders. We divide the argument into five parts.
Proposition 5.5. Consider
ρ̃1 ∶ =
1
20 (1 +
.
√
√
2
2
)
(2
+
)
2
2
Take any ξ ∈ Cuad(ζ). If n ∈ N and a ∈ R(n) are such that ξ ∈ C n (a), then
Gn (ξ, a, ρ̃1 ) ⊆ Vecn (a) ∪ {a},
#Gn (ξ, a, ρ̃1 ) ≤ 3,
and
D(ξ; ρ̃1 ∣Cn (a)∣) ∩ F ⊆
⋃
b∈Gn (ξ,a,ρ̃1 )
C n (b).
Proposition 5.6. There exists a constant ρ̃2 > 0 such that for all n ∈ N≥2 and all a ∈ R(n)
is such that C n (a) ∩ Fr(F) ≠ ∅, if z ∈ Fr(F) has the following properties:
1. There exists c ∈ R(n) such that a ≠ c and z ∈ C n (a) ∩ C n (c);
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
19
2. If z ∈ C 1 (a), then a = a1 ;
then,
Gn (z, a, ρ̃2 ) ⊆ Vecn (a) ∪ {a},
#Gn (z, a, ρ̃2 ) ≤ 3,
and
D(z; ρ̃2 ∣Cn (a)∣) ∩ F ⊆
⋃
b∈Gn (z,a,ρ̃2 )
C n (b).
Proposition 5.7. There exists a constant ρ̃3 > 0 such that for any n ∈ N, any a ∈ R(n)
with Fr(F) ∩ C n (a) ≠ ∅, and any z ∈ Fr(F) ∩ C n (a), we have
Gn (z, a, ρ̃3 ) ⊆ Vecn (a) ∪ {a},
#Gn (z, a, ρ̃3 ) ≤ 3,
and
D(z; ρ̃3 ∣Cn (a)∣) ∩ F ⊆
⋃
b∈Gn (z,a,ρ̃3 )
C n (b).
Proposition 5.8. There exists constant ρ̃5 > 0 such that if n ∈ N, a ∈ R(n), and z ∈
F○ ∩ Fr(Cn (a)), then
Gn (z, a, ρ̃5 ) ⊆ Vecn (a) ∪ {a},
#Gn (z, a, ρ̃5 ) ≤ 6,
and
D(z; ρ̃5 ∣Cn (a)∣) ∩ F ⊆
⋃
b∈Gn (z,a,ρ̃5 )
C n (b).
Proposition 5.9. There exists a constant ρ̃6 > 0 such that if n ∈ N, a ∈ R(n), and z ∈
Cn○ (a), then
Gn (z, a, ρ̃6 ) ⊆ Vecn (a) ∪ {a}, #Gn (z, a, ρ̃6 ) ≤ 6,
and
D(z; ρ̃6 ∣Cn (a)∣) ∩ F ⊆
⋃
b∈Gn (z,a,ρ̃6 )
C n (b).
Proof of Theorem 5.3. Define ρ̃∶ = min{ρ̃6 , ρ̃5 }. Pick any n ∈ N, a ∈ R(n), and z ∈ C n (a). If
z ∈ Cn○ (a), the theorem follows from Proposition 5.9. If z ∈ Fr(Cn (a)), the theorem follows
from Proposition 5.8.
□
We highlight two important consequences of the proofs of Propositions 5.5 and 5.6. First,
Corollary 5.10 tells us that two neighboring cylinders have a similar diameter. Second,
Corollary 5.11 says that the partial quotients determining two neighboring cylinders are of
similar size. We omit the proofs of both corollaries.
Corollary 5.10. There is a constant κ3 > 1 such that for any n ∈ N
1 ∣Cn (a)∣
≤
≤ κ3 for all a ∈ R(n) and b ∈ Vecn (a).
κ3 ∣Cn (b)∣
Corollary 5.11. Given n ∈ N and a ∈ R(n), every b ∈ Vecn (a) verifies
∥aj ∥ − 1 ≤ ∥b∥ ≤ ∥aj ∥ + 1 for all j ∈ {1, . . . , n}.
20
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
5.1.1. Proof of Proposition 5.5.
√
Proof. Let us further assume that ξ = ζ = 21 + 2−2 3 i. When n = 1, by Figure 3, we have
D (ι(ζ);
We apply τ2 to get
D (T1 (ζ; −2);
1
) ∩ ι [F] ⊆ (τ−2 [F] ∩ ι [F]) ∪ (τ−2+i [F] ∩ ι [F]) .
10
1
1
) ∩ T1 (F; −2) = D (τ2 ι(ζ); ) ∩ τ2 ι [F]
10
10
⊆ (F ∩ τ2 ι [F]) ∪ (τi [F] ∩ τ2 ι [F])
= (F ∩ T1 (F; −2)) ∪ (τi [F] ∩ T1 (F; −2)) .
Now, we apply Q1 (⋅; −2) = ιτ−2 and obtain
Q2 (D (T1 (ζ; −2);
1
) , −2) ∩ F ⊆ (Q1 (F; −2) ∩ F) ∪ (Q1 (F; −2 + i) ∩ F)
10
= C 1 (−2) ∪ C 1 (−2 + i).
We appeal to Proposition 5.4 to conclude that
D (ζ; ρ̃1 ∣C1 (−2)∣) ∩ F ⊆ C 1 (−2) ∪ C 1 (−2 + i).
This argument also allows us to conclude
D (ζ; ρ̃1 ∣C1 (−2 + i)∣) ∩ F ⊆ C 1 (−2) ∪ C 1 (−2 + i).
Assume that n = 2. Figure 4 depicts the set ι[F1 (−2)]. The red point is ι(T1 (ζ; −2)), the
1
).
Figure 3. The set ι[F] ∩ (τ−2 [F] ∪ τ−2+i [F] and the disc D (ι(ζ), 10
blue lines and the green arches are, respectively, the images of Fr(F) and C 1 (−2)∩C 1 (−2+i)
under the map ι τ2 ι. Hence, we have
D (ι (T1 (ζ; −2);
and, applying τ2i ,
D (T2 (ζ; −2, −2i);
1
)) ∩ ι τ2 ι [C 1 (−2)] ⊆ τ−2i [F] ∩ ι τ2 ι[C 1 (−2)]
10
1
) ∩ T2 (C1 (−2); (−2, −2i)) ⊆ F ∩ T2 (C 1 (−2); (−2, −2i)),
10
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
so, applying Q2 ( ⋅ , (−2, −2i)),
Q2 (D (T2 (ζ; (−2, −2i));
1
) , (−2, −2i)) ∩ C 1 (−2) ⊆ Q2 (F; (−2, −2i)) ∩ C 1 (−2)
10
= C 2 (−2, −2i)
and, in view of Proposition 5.4,
D(ζ; ρ̃1 ∣C2 (−2, −2i)∣) ∩ C 1 (−2)) ⊆ C 2 (−2, −2i).
In a similar fashion (see Figure 5), we may show that
D(ζ; ρ̃1 ∣C2 (−2 + i, −2 + i)∣) ∩ C 1 (−2 + i) ⊆ C 1 (−2 + i, 1 + 2i) ∪ C 1 (−2 + i, 2i).
For n = 3, we have
2
1
−2
−1
1
−1
−2
1
).
Figure 4. The set ι[F1 (−2)] and the disc D (ι(T1 (ζ, −2)); 10
Figure 5.
The set ι[F1 (−2 + i)] ∩ (τ2i [F] ∪ τ1+2i [F]) and the disc
1
).
D (ι(T1 (ζ; −2 + i)); 10
D (ιT2 (ζ; (−2, −2i);
1
) ∩ ι[F2 (2, −2i)] ⊆ ι[F2 (2, −2i)] ∩ τ2 [F]
10
and the previous argument conduces us to
D(ζ; ρ̃1 ∣C3 (−2, −2i, 2)∣) ∩ C2 (−2, −2i) ⊆ C 3 (−2, −2i, 2).
21
22
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
In an analogous way, we may conclude that
D(ζ; ρ̃1 ∣C3 (−2 + i, 2i, 2)∣) ∩ C2 (−2 + i, 2i) ⊆ C 3 (−2 + i, 2i, 2),
D(ζ; ρ̃1 ∣C3 (−2 + i, 1 + 2i, −2)∣) ∩ C2 (−2 + i, 1 + 2i) ⊆ C 3 (−2 + i, 1 + 2i, −2).
For n = 4, we have
D (ιT3 (ζ; (−2, −2i, 2);
1
) ∩ ι[F(−2, −2i, 2)] ⊆ τ2i [F] ∩ ι[F(−2, 2i, 2)]
10
and, arguing as before, we arrive at
D (ζ; ρ̃1 ∣C4 (−2, −2i, 2, 2i)∣) ∩ C3 (−2, −2i, 2) ⊆ C 4 (−2, −2i, 2, 2i).
Similarly,
D (ζ; ρ̃1 ∣C4 (−2 + i, 2i, 2, −2i)∣) ∩ C3 (−2 + i, 2i, 2) ⊆ C 4 (−2 + i, 2i, 2, −2i),
D (ζ; ρ̃1 ∣C4 (−2 + i, 1 + 2i, −2, −2i)∣) ∩ C3 (−2 + i, 1 + 2i, −2) ⊆ C4 (−2 + i, 1 + 2i, −2, −2i).
For n = 5, we use the same procedure to obtain
D (ζ; ρ̃1 ∣C4 (−2, −2i, 2, 2i, −2)∣) ∩ C4 (−2, −2i, 2, 2i) ⊆ C5 (−2, −2i, 2, 2i, −2).
Furthermore, both F1 (−2) and F5 (−2, −2i, 2, 2i, −2) are equal as sets and the distinguished
point, line segment, and arch coincide. As a consequence, if
d1 ∶ = (d1n )n≥1 ∶ = (−2, −2i, 2, 2i, −2, −2i, 2, 2i, . . .),
every n ∈ N0 verifies
(11)
D (ζ; ρ̃1 ∣Cn+1 (d11 , d12 , . . . , d1n+1 )∣) ∩ Cn (d11 , d12 , . . . , d1n ) ⊆ Cn+1 (d11 , d12 , . . . , d1n+1 ).
We may argue in an analogous manner that the sequences
d2 ∶ = (−2 + i, 2i, 2, −2i, −2, 2i, 2, −2i, −2, . . .),
d3 ∶ = (−2 + i, 1 + 2i, −2, −2i, 2, 2i, −2, −2i, 2, 2i, . . .)
also satisfy (10) and the proposition follows.
□
Remark. For each d = (dj )j≥1 ∈ D N and n ∈ N, define pref(d; n)∶ = (d1 , . . . , dn ). The proof
of Proposition 5.5 tells us that there is an absolute constant λ > 1 such that
1 ∣Cn (pref(dr ; n)∣
<
< λ for every n ∈ N and r, s ∈ {1, 2, 3}.
λ ∣Cn (pref(ds ; n)∣
Also, for such r, s we have ∥drn ∥ − 1 ≤ ∥dsn ∥ ≤ ∥drn ∥ + 1 for every n ∈ N. These observations
are the starting point for showing corollaries 5.10 and 5.11. We omit their proofs, for they
require arguments resembling the ones we have just presented.
5.1.2. Proof of Proposition 5.6.
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
23
Lemma 5.12. 1. Let n ∈ N and a ∈ R(n) be arbitrary. If b, d ∈ D are such that ab, ad ∈
R(n + 1), then
w ∈ Fr(Fn (a)) ∩ C 1 (b) ∩ C 1 (d) implies τb ι(w) ∈ Esq ∪ Cuad(ζ).
2. Given ξ ∈ Cuad, if a ∈ D is such that ι(ξ) ∈ τa [F], then T1 (ξ; a) ∈ Cuad(ζ).
Proof. The proof of the first claim amounts to checking all the possible forms of Fn (a).
The second claim can be readily seen on Figure 2.
□
Proof of Proposition 5.6. Let k ∈ {1, . . . , n − 1} be the minimal integer for which there is
some b ∈ Vecn (a) satisfying
pref(a, k) = pref(b, k)
and
ak+1 ≠ bk+1 .
Such a k exists, because we are assuming that z belongs to the closure of exactly one
cylinder of level 1. By Lemma 5.12 tells us that we have two options:
ι[Tk (z; (a1 , . . . , ak ))] ∈ τak+1 [Esq]
or
ι[Tk (z; (a1 , . . . , ak ))] ∈ τak+1 [Cuad(ζ)].
In the first case, k = n − 1 and there is only one b ∈ D for which (a1 , . . . , ak , b) ∈ R(n) and
D (Tk (z; (a1 , . . . , ak );
1
) ∩ ι[Fk (a1 , . . . , ak )] ⊆ τak+1 [F] ∪ τb [F].
10
Then, applying Qk ( ⋅ ; (a1 , . . . , ak )), Proposition 5.4, and ∣Cn (a)∣ ≤ ∣Ck (a1 , . . . , ak )∣, we conclude
D(z; ρ̃1 ∣Cn (a)∣) ∩ F ⊆ C n (a) ∪ C n (a1 , . . . , ak , b).
Assume that ξ∶ = ι[Tk (z; (a1 , . . . , ak )] ∈ τak+1 [Cuad(ζ)]. Figure 6 illustrates all the possibilities (modulo Di8 ) for the location of ξ (the red point) in ι[Fk (a)].
Let us further assume that n ≥ k + 2, a similar argument may be used when n = k + 1. In
case (a), by the proof of Proposition 5.5, we have
D(ξ; ρ̃1 ∣Cn−k−2 (ak+2 , . . . , an )∣) ∩ τak+1 [Fk (a1 , . . . , ak )] ⊆ τak+1 [Cn−k−2 (ak+2 , . . . , an )].
By symmetry, there is exactly one (bk+1 , . . . , bn ) ∈ R(n − k − 1) such that ∣ak+1 − bk+1 ∣ = 1,
there exists mir ∈ {mir1 , mir2 } such that bj = mir(aj ) for j ∈ {k + 2, . . . , n}, and the set
D (ξ; ρ̃1 ∣Cn−k−2 (ak+2 , . . . , an )∣) ∩ ι [Fk (a1 , . . . , ak )]
is contained in
ι [Fk (a1 , . . . , ak )] ∩ (τak+1 [C n−k−2 (ak+2 , . . . , an )] ∪ τbk+1 [C n−k−2 (bk+2 , . . . , bn )]) .
We now argue as in Proposition 5.5 to conclude the existence of a constant ρ̃2 > 0 such
that
D(ζ; ρ̃2 ∣Cn (a)∣) ∩ F ⊆ C n (a) ∪ C n (a1 , . . . , ak , bk+1 , . . . , bn ).
24
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
Cases (b) and (c) are handled in a similar way. However, in them we obtain conclude that
D(ζ; ρ̃2 ∣Cn (a)∣) ∩ F is contained in three cylinders of level n rather than two.
□
Case (a)
Case (b)
Case (c)
Figure 6.
Possible configurations for the position of ξ
ι[Fk−1 (a1 , . . . , ak−1 )].
in
5.1.3. Proof of Proposition 5.7.
Proof of Proposition 5.7. Let n, a, and z be as in the statement and define
√ 2 ⎫
⎧
⎪
2 ρ̃2 ⎪
⎪1
⎪
δ∶ = min ⎨ , κ1 (1 −
)
⎬.
⎪
10
2
2⎪
⎪
⎪
⎩
⎭
Note that z ∈ Fr(Cn (a)), because z ∈ Fr(F) ∩ C n (a). Then,
ι(Tn−1 (z; (a1 , . . . , an−1 )) ∈ Fr (ι(Fn−1 (a1 , . . . , an−1 ))
and
(12)
{d ∈ D ∶ τd [F] ∩ ι(Fn−1 (a1 , . . . , an−1 )) ∩ D(ι(Tn−1 (z; (a1 , . . . , an−1 ); δ) ≠ ∅}
contains at most two elements, one of them being an . This is shown by checking case by case
all the possible forms of ι(Fn−1 (a1 , . . . , an−1 )). For instance, Figure 7 depicts ι[F1 (1 + i)].
If the set in (12) contains only an , we argue as in the proof of Proposition 5.5 to conclude
⎛
⎞
δ
⎟ ∩ F ⊆ C n (a).
D ⎜z;
∣C
(a)∣
√
√
n
⎝ 2 (1 + 22 ) (2 + 22 )
⎠
Assume that the set in (12) has two elements, say an and b. Then, there is a unique
ξ ∈ Cuad ∪ Esq such that
D (ι(Tn−1 (z; (a1 , . . . , an−1 ); δ))) ∩ Fr(Fn−1 (a1 , . . . , an−1 )) ∩ τan [ F ] ∩ τb [ F ] = {τan (ξ)}.
Thus, ∣ξ − Tn (z; a)∣ < δ and, since Qn (ξ; a) ∈ F,
∣Qn (ξ; a) − z∣ = ∣Qn (ξ; a) − Qn (Tn (z; a); a)∣
=
∣ξ − Tn (z; a)∣
qn−1
∣qn (a)∣2 ∣1 − qn−1
qn Qn (ξ; a)∣ ∣1 − qn Qn (Tn (z; a); a)∣
δ
<
κ1 (1 −
<
√ 2 ∣Cn (a)∣
2
2 )
ρ̃2
∣Cn (a)∣.
2
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
25
Hence, for all positive r satisfying 2r < ρ̃2 , we have
D(z; r∣Cn (a)∣) ⊆ D(Qn (ξ; a); ρ̃2 ∣Cn (a)∣)
and the desired conclusion will hold for such r by Proposition 5.6. The proposition is done
if we choose
δ ρ̃2
ρ̃2
ρ̃3 ∶ =
< .
√
√
2 (1 + 22 ) (2 + 22 ) 2
□
Remark. Consider n ∈ N, a ∈ R(n), and ξ ∈ Cuad(ζ) ∪ Esq such that Qn (ξ; a) ∈ Fr(Cn (a)).
The choice of ρ̃3 yields
{Qn (ξ; a)} = Qn ((Cuad(ζ) ∪ Esq; a) ∩ D(Qn (ξ; a); ρ̃3 ) .
5.1.4. Proof of Proposition 5.8.
Proof. Let k ∈ {0, 1 . . . , n − 1} be such that
z ∈ Ck○ (a1 , . . . , ak ) ∩ Fr(Ck+1 (a1 , . . . , ak , ak+1 )).
Let κ1 and κ2 be as in propositions 4.7 and 4.10, respectively and define
√ 2
κ1
2
δ∶ =
(1 −
) .
κ2
2
The choice of k implies
(13)
ι(Tn−1 (z; (a1 , . . . , an−1 )) ∈ Fr (ι(Fn−1 (a1 , . . . , an−1 ))
Figure 7 depicts the situation for when Fk (a1 , . . . , ak ) = F1 (1 + i). In this picture, (13)
tells us that ιTk (z; (a1 , . . . , ak )) belongs in one of the dashed lines. We consider two cases
according to whether
(14)
D (ι(Tk (z; (a1 , . . . , ak ));
δ ρ̃3
∣Cn−k−2 (ak+2 , . . . , an )∣) ⊆ ι(F○k (a1 , . . . , ak )
2
holds or not. Assume (14). Let us further assume that n ≥ k + 2 and ιTk (z; (a1 , . . . , ak )) ∈/
τak+1 [Esq]. By Proposition 5.7, the set
D (ι(Tk (z; (a1 , . . . , ak ));
δ ρ̃3
∣Cn−k−2 (ak+2 , . . . , an )∣) ∩ τak+1 [F]
2
is contained in the image under τak+1 of at most three closed cylinders of order n − k − 2.
Moreover, the remark after Proposition 5.7 implies that the disc in (14) only intersects
one side of the square τak+1 [F]. Therefore, by symmetry, this disc is contained in the
Gaussian integral translations of at most six cylinders of level n − k − 2. As a consequence,
an application of Qk+1 (⋅; (a1 , . . . , ak+1 )) gives
⎛
⎞
1
D ⎜z,
∣Ck+1 (a1 , . . . , ak+1 )∣∣Cn−k−2 (ak+2 , . . . , an )∣⎟ .
√
√
⎝ 2 (1 + 22 ) (2 + 22 )
⎠
26
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
is contained in at most six closed cylinders C n (b) such that b ∈ Vecn (a) ∪ {a}. Writing
ρ̃4,1 ∶ =
1
2κ2 (1 +
,
√
√
2
2
)
(2
+
)
2
2
and invoking Proposition 4.10, we get
Gn (z, a, ρ̃4,1 ) ⊆ Vecn (a) ∪ {a},
#Gn (z, a, ρ̃4,1 ) ≤ 6,
and
D(z; ρ̃4,1 ∣Cn (a)∣) ∩ F ⊆
⋃
b∈Gn (z,a,ρ̃4,1 )
C n (b).
When we have n = k +1 or ιTk (z; (a1 , . . . , ak )) ∈ τak+1 [Esq] we may argue in a similar fashion
to obtain the result. However, in these cases, the constant 6 in the conclusion can replaced
with 4.
If (14) fails, there is some ξ ∈ Cuad(ζ) ∪ Esq such that
τak (ξ) ∈ Fr (ι(Fk (a1 , . . . , ak )))
and
δ ρ̃3
∣Cn−k−2 (ak+2 , . . . , an )∣.
2
Using Qk+1 (ξ; (a1 , . . . , ak+1 )) ∈ Fr(Ck+1 (a1 , . . . , ak+1 )) and arguing as in Proposition 5.7, we
obtain
∣ι(Tk (z; (a1 , . . . , ak ))) − τak (ξ)∣ <
∣z − Qk+1 (ξ; (a1 , . . . , ak+1 ))∣ =
= ∣Qk+1 (Tk+1 (z; (a1 , . . . , ak+1 ); (a1 , . . . , ak+1 )) − Qk+1 (ξ; (a1 , . . . , ak+1 ))∣
<
∣Tk+1 (z; (a1 , . . . , ak+1 ) − ξ∣
∣qk+1 (a1 , . . . , ak+1 )∣2 (1 −
<
∣Ck+1 (a1 , . . . , ak+1 )∣
κ1 (1 −
≤
δ ρ̃3
2κ1 (1 −
√
√
2
2 )
2
2 )
2
2
√ 2
2
2 )
∣Tk+1 (z; (a1 , . . . , ak+1 ) − ξ∣
∣Ck+1 (a1 , . . . , ak+1 )∣ ∣Cn−k−2 (ak+2 , . . . , an )∣
ρ̃3
∣Ck+1 (a1 , . . . , ak+1 )∣ ∣Cn−k−2 (ak+2 , . . . , an )∣
2κ2
ρ̃3
< ∣Cn (a)∣.
2
This implies
ρ̃3
D (z; ∣Cn (a)∣) ⊆ D (Qk+1 (ξ; (a1 , . . . , ak+1 ); ρ̃3 ∣Cn (a)∣)
2
and, on account of Proposition 5.7, the desired conclusions hold on D (z; ρ23 ∣Cn (a)∣). We
choose
ρ̃3
ρ̃4 ∶ = min {ρ̃4,1 , }
2
and the proof is done.
□
=
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
1
2
27
3
−1
−2
−3
Figure 7. The set ι[F1 (1 + i)].
5.1.5. Proof of Proposition 5.9.
Proof. Define
ρ̃5 ∶ =
√ 2
2
2 )
√
√
2
2
)
(2
+
2
2 )
κ1 (1 −
2 (1 +
ρ̃4
.
2
Let n, a, and z be as in the statement. If
√ 2
⎛
ρ̃4
2 ⎞
(15)
D Tn (z; a); κ1 (1 −
) ⊆ F○n (a1 , . . . , an ) = Tn (Cn○ (a); a) ,
2
2 ⎠
⎝
then
D (z; ρ̃5 ∣Cn (a)∣) ⊆ Cn○ (a) ⊆ C n (a).
If (15) is false, there is some ξ ∈ F such that
ξ ∈ Fr(Fn (a))
and
√ 2
2
ρ̃4
) .
∣ξ − Tn (z; a)∣ < κ1 (1 −
2
2
Calculating as in the proof of Proposition 5.8, we get
∣z − Qn (ξ; a)∣ <
ρ̃4
∣Cn (a)∣,
2
so
ρ̃4
∣Cn (a)∣) ⊆ D (Qn (ξ; a); ρ̃4 ∣Cn (a)∣) .
2
The desired conclusions follows from Proposition 5.8 and 0 < ρ̃5 = ρ̃24 .
D (z;
□
5.2. Second geometric construction. We dedicate this section to the proof of Theorem
5.14. First, we interpret it in geometric terms. Second, we show a simpler version of it
on ι[F]. And third, we conclude the proof using an adequate function Qk ( ⋅ ; c) and basic
properties of Möbius transformations.
Definition 5.13. For each n ∈ N, a ∈ R(n) and 0 < l < u, define
Sn (a; l, u)∶ =
⋃
l≤∥d∥≤u
ad∈R(n+1)
C n+1 (ad).
28
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
Theorem 5.14. There exists a constant ρ > 0 with the following property: for any two
l
28
positive real numbers l and u such that 11 < l < u and 22
50 < u < 50 , any a ∈ R(n), and
any z ∈ Sn (a; l, u), the disc D(z; ρ∣Sn (a; l, u)∣) intersects at most two sets of the form
Sn (c; l, u).
Definition 5.15. For every positive real numbers L, U with L < U , define
Treb(L, U )∶ = Cl ({z ∈ F ∖ {0} ∶
1
1
≤ ∥ι(z)∥ ≤ }) .
U
L
For each R > 0, define
LT(R)∶ = {z ∈ C ∶ ∥ι(z)∥ =
1
}.
2R
Observe that for any n ∈ N, d ∈ R(n), and l, u ∈ N such that l < u, the next equality
holds:
Sj (d; l, u) = Qn (Treb ((u + 21 )−1 , (l − 21 )−1 )) , d).
Proposition 5.16. For every R > 0, we have
√
dist (LT(R), LT(2R)) = ( 5 − 1)R.
Proof. We may assume that R = 1. The proposition for an arbitrary R > 0 follows from
applying the linear transformation z ↦ Rz. Let w ∈ LT(2) be such that I(w) ≥ 2. We first
show that
√
dist(w, LT(1)) ≥ 5 − 1.
Afterwards, we give a particular w0 for which the equality holds. For each j ∈ {1, 2, 3, 4},
put Cj ∶ = C(ij ; 1). An illustration of LT(1) and LT(2) is given in Figure 8.
4
3
2
C1
1
C2
−4
−3
−2
C4
−1
1
2
3
4
−1
−2
C3
−3
−4
Figure 8. The sets LT(1) and LT(2).
Since w ∈ LT(2) and I(w) ≥ 2, we can write w = x + i(2 +
√
4 − x2 ) for some x ∈ [−2, 2].
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
29
i. By elementary geometric means, we have that
dist (w, C1 ∩ LT(1)) = ∣w − i∣ − 1.
Therefore,
2
2
(dist (w, C1 ∩ LT(1)) + 1) = ∣w − i∣
= x2 + (1 +
√
√
2
4 − x2 ) = 5 + 2 4 − x2 ≥ 5,
and
dist (w, C1 ∩ LT(1)) ≥
√
5 − 1.
The equality holds if and only if x = 2 or x = −2, which means w = 2 + 2i or w = −2 + 2i.
ii. Because of dist(w, C2 ) = ∣w + 1∣ − 1,
2
(dist(w, C2 ) + 1) = ∣w + 1∣2
√
2
4 − x2 )
√
= x2 + 2x + 1 + 4 + 4 4 − x2 + 4 − x2
√
= 9 + 2x + 4 4 − x2 .
√
Let g ∶ [−2, −1] → R be given by g(t) = 9 + 2x + 4 4 − x2 . The function g is strictly
increasing so its minimum is attained at t = −2 and g(−2) = 5. Hence, −2 ≤ x ≤ −1
implies
√
dist(w, C2 ) ≥ 5 − 1.
√
√
If −1 ≤ x ≤ 0, then 2 + 3 ≤ 2 + 4 − x2 ≤ 4, and every z ∈ C2 ∈ LT(1) verifies
√
√
√
∣w − z∣ ≥ I(w) − I(z) ≥ 2 + 3 − 1 = 1 + 3 ≥ 5 − 1.
√
Now, assume that 0 ≤ x ≤ 2, then ∣z − w∣ ≥ 8 for all z ∈ C2 ∩ LT(1). This can be
shown, for instance, noting that
= (x + 1)2 + (2 +
C2 ∩ LT(1) ⊆ {z ∈ C ∶ R(z) ≤ −1 and I(z) ≤ 1},
that the considered part of LT(2) is contained in the upper closed semi-plane deter√
mined by the line containing 4i amd 2 + 2i, and that ∣ − 1 + i − (1 + 3i)∣ = 8.
iii. For each z ∈ C3 ∩ LT(1), we have ∣w − z∣ ≥ I(w) − I(z) ≥ 3.
iv. This case follows from the argument used for C2 .
□
Proposition 5.17. Let n ∈ N, c ∈ R(n) be arbitrary and let l, u ∈ N be such that 4 < l < u.
If z ∈ Sn (c; l, u), then
# {a ∈ R(n) ∶ z ∈ Sn (a; l, u)} ≤ 2.
Proof. Assume that n = 1. When a, b ∈ Z[i] are different and 0 < L < U < 41 , we have
τa [Treb(L, U )] ∩ τb [Treb(L, U )] = ∅.
30
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
In particular, for L0 = (u + 12 )−1 and U0 = (l − 21 )−1 , if a1 , b1 ∈ D satisfy
ι(z) ∈ τa1 [Treb(L0 , U0 )] ∩ τb1 [Treb(L0 , U0 )],
then a1 = b1 .
Suppose that n = 2. Pick a = (a1 , a2 ) and b = (b1 , b2 ) in R(2) such that z ∈ S2 (a; l, u) ∩
S2 (b; l, u). If a1 ≠ b1 , then
ι(z) ∈ (a1 + F) ∩ (b1 + F)
√
√
and ∣a1 − b1 ∣ ≤ 2. If we had ∣a1 − b1 ∣ = 2, we would have τ−a1 ι(z) ∈ Esq and there would
be some (a2 , a3 ) ∈ R(2) such that
(a1 , a2 , a3 ) ∈ R(3) and τ−a1 ι(z) ∈ C 2 (a2 , a3 ) ∩ Esq,
in contradiction with Proposition 5.2. Hence, either ∣a1 − b1 ∣ = 0 or ∣a1 − b1 ∣ ≤ 1. Assume
that ∣a1 − b1 ∣ = 1. Then, there is no c = (c1 , c2 ) ∈ R(2) different from both a and b such that
c1 = a1 and z ∈ S2 (c; l, u). If there was one, we would have
τ−a1 ι(z) ∈ Fr(F) ∩ C 1 (a2 ) ∩ C 1 (c2 ),
so τ−a2 ιτ−a1 ι(z) ∈ Cuad(ζ) ⊆ Fr(F) (see Lemma 5.12) and there would not be any d ∈ D
such that l ≤ ∥d∥ ≤ u and z ∈ C 3 (a1 , a2 , d), contradicting z ∈ S2 (a; l, u).
If there were two different Gaussian integers a2 , a′2 such that (a1 , a2 ), (a1 , a′2 ) ∈ R(2) and
z ∈ C 2 (a1 , a2 ) ∩ C 2 (a1 , a′2 ), we would have
τa1 ι(z) ∈ Fr(F1 (a1 )) ∩ C 1 (a2 ) ∩ C 1 (a′2 )
and, for all positive L, U with 0 < U < 41 , we would have
τ−a2 ιτ−a1 ι(z) ∈ Fr(F) ⊆ F ∖ Treb(L, U ),
which contradicts z ∈ S2 ((a1 , a2 ); l, u). Therefore, there is only one a2 ∈ D such that
z ∈ S2 ((a1 , a2 ); l, u). The same argument also shows the uniqueness of b2 ∈ D satisfying
z ∈ S2 ((b1 , b2 ); l, u). When ∣a1 − b1 ∣ = 0, we apply the case n = 1 on T1 (z; a1 ). The
proposition is now proven for n = 2.
Suppose that n = 3. Assume that for two different a, b ∈ R(3) we have z ∈ S3 (a; l, u) ∩
S3 (b; l, u). As discussed in the case n = 2, we must have ∣a1 −b1 ∣ ≤ 1. If ∣a1 −b1 ∣ = 1, we may
show as before that there there is no c ∈ R(3) such that c1 = a1 , c ≠ a, and z ∈ S3 (c; l, u).
Indeed, first, note that
τ−a1 ι(z) ∈ Fr(F1 (a1 )) ∩ S2 ((a2 , a3 ); l, u).
If we had two different a2 , a′2 ∈ D such that τ−a1 ι(z) ∈ Fr(F1 (a1 )) ∩ C 1 (a2 ) ∩ C 1 (a′2 ), we
would have
τ−a2 ιτ−a1 ι(z) ∈ Esq ∪ Cuad.
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
31
If τ−a2 ιτ−a1 ι(z) ∈ Esq, we would not be able to find (d1 , d2 ) ∈ R(2) such that z ∈ C 4 (a1 , a2 , d1 , d2 )
contradicting z ∈ S3 (a; l, u). If τ−a2 ιτ−a1 ι(z) ∈ Cuad, we would have
τ−a3 ιτ−a2 ιτ−a1 ι(z) ∈ Fr(F) ∩ Treb(L0 , U0 ) = ∅.
Therefore, a2 is unique.
If we had two different a3 , a′3 ∈ D such that
z ∈ S3 ((a1 , a2 , a3 ); l, u) ∩ S3 ((a1 , a2 , a′3 ); l, u) ,
we would have again
τ−a3 ιτ−a2 ιτ−a1 ι(z) ∈ Fr(F) ∩ Treb(L0 , U0 ) = ∅,
so, a3 is unique. The same argument provides the uniqueness of b. When ∣a1 − b1 ∣ = 0, we
apply the case n = 2 on T1 (z; a1 ). This shows the proposition holds for n = 3.
Suppose that we have shown the proposition for N ∈ N≥3 and consider n = N +1. Assume
that there is some c ∈ R(N + 1) such that z ∈ SN +1 (c; l, u). Pick a, b ∈ R(N + 1) with
z ∈ SN +1 (a; l, u) ∩ SN +1 (b; l, u).
As before, we must have ∣a1 −b1 ∣ ≤ 1 and ∣a1 −b1 ∣ = 1 forbids the existence of some c ∈ R(N +1)
with the properties: c1 = a1 , c ≠ a, and z ∈ SN +1 (c; l, u). Moreover, when ∣a1 − b1 ∣ = 1, we
may argue as in the case n = 3 to conclude that the uniqueness of the partial quotients
a2 , . . . , aN +1 and, similarly, the uniqueness of b2 , . . . , bN +1 .
If we had a1 = b1 , then
T1 (z; a1 ) = τ−a1 ι(z) ∈ SN ((a2 , . . . , aN +1 ; l, u)) ∩ SN ((b2 , . . . , bN +1 ; l, u))
and we conclude the proposition using the induction hypothesis.
□
Proposition 5.18. Let n ∈ N and a ∈ R(n). For each R ∈ (0, 41 ), the following estimates
hold
√
( 5 − 1)
R
≤ dist (Qn (LT(R) ∩ Fn (a); a), Qn (LT(2R) ∩ Fn (a); a))
(1 + 2R)(1 + 4R) ∣qn (a)∣2
√
( 5 − 1)
R
≤
,
(1 − 2R)(1 − 4R) ∣qn (a)∣2
and
∣LT(R) ∩ Fn (a)∣
∣LT(R) ∩ Fn (a)∣
≤ ∣Qn (LT(R) ∩ Fn (a); a)∣ ≤
2
(1 + 2R)(1 + 4R)∣qn (a)∣
(1 − 2R)(1 − 4R)∣qn (a)∣2
32
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
Proof. Both sets of inequalities are shown similarly, we only prove the first one. Write
qn = qn (a) and qn−1 = qn−1 (a1 , . . . , an−1 ). For α ∈ LT(R) and β ∈ LT(2R), we have
RRR
RRR
RRR
RRR
β
α
R
∣Qn (α; a) − Qn (β; a)∣ = RRR
−
RRR q 2 (1 − qn−1 α ) q 2 (1 − qn−1 β ) RRRRR
n
qn
qn
RR n
RR
∣α − β∣
1
=
∣qn ∣2 ∣(1 − qn−1 α ) (1 − qn−1 β )∣
qn
qn
and, since ∣α∣ ≤ 2R, ∣β∣ ≤ 4R,
1
1
∣α − β∣
∣α − β∣
≤ ∣Qn (α; a) − Qn (β; a)∣ ≤
.
2
2
∣qn ∣ ∣(1 + 2R) (1 + 4R)∣
∣qn ∣ ∣(1 − 2R) (1 − 4R)∣
We arrive at the desired conclusion after taking the infima over α and β and using Lemma
5.16.
□
Proposition 5.18 yields the following property of Treb sets.
Proposition 5.19. For any pair of positive numbers 0 < L < U < 81 , any n ∈ N, and
b ∈ R(n), we have
U
16 U
8
≤ ∣Qn (Treb(L, U ) ∩ Fn (b); b)∣ ≤
2
15 ∣qn (b)∣
5 ∣qn (b)∣2
Proposition 5.20. Let L, U > 0 be such that
U<
1
10
and
1 L 3
< < .
4 U 4
Define
√
100( 5 − 1)
(16)
ρ1 ∶ =
.
43 ⋅ 23
For every j ∈ N, d ∈ R(j), and z ∈ Qn (Treb(L, U ) ∩ Fn (d); d)), we have
(17)
D (z; ρ1
U
L
) ⊆ Qn (Treb ( , 2U ) ∩ Fn (d); d) .
2
∣qn (d)∣
2
We only need UL to be contained in a small neighborhood of 21 and U to be small enough.
However, we refrain from using unspecified bounds in order to avoid introducing new
parameters.
Proof. The hypotheses on L and U guarantee
2 (1 − 34 U ) (1 − 23 U ) 37 ⋅ 17 3 L
>
> > ,
(1 + 2U )(1 + 4U )
48 ⋅ 14 4 U
and hence
√
√
5−1
L
5−1
U
<
.
2
(1 − L)(1 − 2L) 2∣qj (d)∣
(1 + 2U )(1 + 4U ) ∣qj (d)∣2
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
33
Therefore, Proposition 5.18 gives
L
dist (Qj (LT ( ) ∩ Fn (d); d) , Qj (LT(L) ∩ Fn (d); a)) <
2
< dist (Qj (LT(U ) ∩ Fn (d); d) , Qj (LT(2U ) ∩ Fn (d); d)) .
Once again, by Proposition 5.18,
√
L
( 5 − 1)
L
dist (Qn (LT ( ) ∩ Fn (d); d), Qn (LT(L) ∩ Fn (d); d)) ≥
2
2(1 + L)(1 + 2L) ∣qn (d)∣2
√
U
( 5 − 1)
≥
3 ∣q (d)∣2
3
8 (1 + 40 ) (1 + 20 ) n
= ρ1
U
.
∣qn (d)∣2
Therefore, (17) holds for all z ∈ Qn (Treb(L, U ) ∩ Fn (d); d)).
□
We now return to the sets Sn (a; l, u).
Proposition 5.21. Assume that l, u ∈ N satisfy
1 −1 1
(l − ) <
2
10
and
1 l − 12 3
< .
<
4 u + 21 4
Define
1
√ .
50(2 + 2)
Given k ∈ N0 , a ∈ R(k), j ∈ N, and d, e ∈ R(j) such that d1 ≠ e1 and ad, ae ∈ R(j + k), for
any
z ∈ Sj+k (ad; l, u) ∩ Sj+k (ae; l, u)
ρ2 ∶ =
we have
⎛
⎞
l−1
l−1
ρ2
D Tk (z; a);
⊆
S
(d;
⌊
⌋
,
2(u
+
1))
∪
S
(e;
⌊
⌋ , 2(u + 1)) .
j
j
2
2
⎝
(l − 12 ) ∣qj (d)∣2 ⎠
Proof. We claim that for some mir ∈ {mir1 , mir2 }, we have
(18)
mir [C j−1 (d2 , . . . , dj )] = C j−1 (e2 , . . . , ej ).
Since both ad and ae are regular, the sets
τd1 [Cj−1 (d2 , . . . , dj )] ∩ ι[Fk (a)]
and
τe1 [Cj−1 (e2 , . . . , ej )] ∩ ι[Fk (a)]
have non-empty interior and, as argued before, we must have ∣d1 − e1 ∣ = 1. Let b, b′ ∈ D be
such that
(19)
and
l ≤ ∥b∥ ≤ u,
l ≤ ∥b′ ∥ ≤ u
Tk (z; a) ∈ C j+1 (db) ∩ C j+1 (eb′ ).
34
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
Let us further assume that e1 − d1 = 1. If
(20)
τd1 [mir1 [C j−1 (d2 , . . . , dj )]] ⊆ ι [Fk (a)]
was not true, the sets τe1 [F] ∩ ι[Fk (a)] and τd1 [F] ∩ ι[Fk (a)] would look like either case (b)
or case (c) in Figure 6. In this situation, the red point represents ι Tk (z; a). However, as
we saw in the proof of Proposition 5.5, this implies
∥e2 ∥ = . . . = ∥ej ∥ = ∥b∥ = ∥d2 ∥ = . . . = ∥dj ∥ = ∥b′ ∥ = 2,
which contradicts (19) and l ≥ 10. Hence, (20) is true.
We claim that
(21)
ι Tk (z; a) ∈ τd1 [C j−1 (d2 , . . . , dj−1 )] ∩ τe1 [mir1 [C j−1 (d2 , . . . , dj−1 )]] .
Since ι Tk (z; a) ∈ τd1 [F] ∩ τe1 [F] and e1 = d1 + 1, we have
1
R (τ−d1 ι Tk (z; a)) = R (Tk+1 (z; ad1 )) = ,
2
so
mir1 Tk+1 (z; ad1 ) = Tk+1 (z; ad1 ) − 1 = τ−1 τ−d ι Tk (z; a) = Tk+1 (z; ae1 ).
○
As such, if (zn )n≥1 is a sequence in Cj−1
(d2 , . . . , dj ) converging to Tk+1 (z; ad1 ), then
lim mir1 (zn ) = τ−e1 ι Tk (z; a)
n→∞
and, thus, ι Tk (z; a) ∈ τe1 [C j−1 (d2 , . . . , dj )]. Therefore, Proposition 5.17 and (21) yield
(18).
⌋ and u′ ∶ = 2(u + 1), the disc
Proposition 5.20 tells us that, for l ′ ∶ = ⌊ l−1
2
is contained in
⎞
⎛
ρ1
D ι(w);
1
⎝
(l ′ − 2 ) ∣qj−1 (d2 , . . . , dj )∣2 ⎠
τd1 [Qj−1 (Treb(l ′ , u′ ); (d2 , . . . , dj−1 )] ∪ τe1 [Qj (Treb(l ′ , u′ ); (e2 , . . . , ej−1 )] .
Lastly, applying ι and proceeding as in the proof of Theorem 5.3, we conclude
⎞
⎛
ρ2
D Tk (z; a);
⊆ Sj (d; l ′ , u′ ) ∪ Sj (e; l ′ , u′ ) .
1
2
⎝
(l − 2 ) ∣qj (d)∣ ⎠
□
Proposition 5.22. Assume that l, u ∈ N satisfy
1 −1 1
(l − ) <
2
10
and
1 l − 12 3
<
< .
4 u + 21 4
Define
ρ3 ∶ =
3ρ2
16 ⋅ 25 (1 +
√
√
2
2
)
(2
+
2
2 )
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
35
and put l ′ ∶ = ⌊ l−1
2 ⌋ and u∶ = 2(u + 1). For every a, d, e as in Proposition 5.21, every
z ∈ Sj+k (ae; l, u) ∩ Sj+k (ad; l, u) verifies
D (z; ρ3 ∣Sk+j (ad; l, u)∣) ⊆ Sj+k (ae; l ′ , u′ ) ∩ Sj+k (ad; l ′ , u′ ).
Proof. This is a consequence of propositions 5.4, 5.19, and 5.21.
□
We now have all the ingredients needed for the proof of Theorem 5.14.
Proof of Theorem 5.14. Choose ρ = ρ23 . The assumptions on l and u imply those imposed
on l, u, L = L(l), and U = U (u) in propositions 17, 5.21, and 5.22. When Fn (a) = F or
Fn (a) = F1 (ij (1 + 2i)) for some j ∈ {1, 2, 3, 4}, we pick L = (u + 21 )−1 and U = (l − 12 )−1 , and
the theorem follows from Proposition 5.18.
Suppose that Fn (a) = F1 (2), Fn (a) = F1 (1 + i) or any of their images under elements of
Di8 . Assume that
z ∈ Sn (a; l, u) ∩ Sn (b; l, u)
for two different a, b ∈ R(n). Let k ∈ {0, . . . , n − 1} such that pref(a; k) = pref(b; k) and
⌋ and u′ = 2(u + 1),
ak+1 ≠ bk+1 . By Proposition 5.22, for l ′ = ⌊ l−1
2
(22)
D (z; 2ρ∣Sn (a; l, u)∣) ⊆ Sn (a; l ′ , u′ ) ∪ Sn (b; l ′ , u′ ).
Take any w ∈ D(z; 2ρ∣Sn (a; l, u)∣) and suppose there existed some c ∈ R(n) different from
a and b such that w ∈ Sn (c; l, u). Then, for some c ∈ D with l ≤ ∥c∥ ≤ u, we have
○
w ∈ C n+1 (cc). Since C n+1 (cc) = Cl(Cn+1
(cc)), we may pick some
○
w′ ∈ D(z; 2ρ∣Sn (a; l, u)∣) ∩ Cn+1
(cc) ⊆ D(z; 2ρ∣Sn (a; l, u)∣) ∩ Cn○ (c).
However, the assumptions c ≠ a and c ≠ b yield
Cn○ (c) ∩ (Sn (a; l ′ , u′ ) ∪ Sn (b; l ′ , u′ )) ⊆ Cn○ (c) ∩ (C n (a) ∪ C n (b)) = ∅,
thus w′ ∈/ Sn (a; l ′ , u′ ) ∪ Sn (b; l ′ , u′ ), contrary to (22). In short, this shows that
{c ∈ R(n) ∶ Sn (c; l, u) ∩ D(z; 2ρ∣Sn (a; l, u)∣)} ⊆ {a, b}.
To finalize the proof, assume that z ∈ Sn (a; l, u) for exactly only one a ∈ R(n). If there
is some b ∈ R(n) different from a such that
D (z; ρ∣Sn (a; l, u)∣) ∩ Sn (b; l, u) ≠ ∅,
then D (z; ρ∣Sn (a; l, u)∣) contains some w ∈ Sn (a; l, u) ∩ Sn (b; l, u). The previous argument along with
D (z; ρ∣Sn (a; l, u)∣) ⊆ D (w; 2ρ∣Sn (a; l, u)∣) ⊆ Sn (a; l ′ , u′ ) ∪ Sn (b; l ′ , u′ )
drive us to the desired conclusion.
□
36
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
6. Approaching the irregular by the regular
We have seen that there are uncountably many irregular numbers, yet we have only
discussed about regular cylinders so far. Furthermore, in Theorem 1.1 we made no distinction between regular or irregular numbers. In this section, we explain why we can focus
on regular cylinders.
Theorem 6.1. For every n ∈ N≥2 , a = (a1 , . . . , an ) ∈ Ir(n), and f ∈ Di8 , there is some
b = (b1 , . . . , bn ) ∈ R(n) such that
f [Cn (a)] ⊆ C n (b)
and
∥aj ∥ = ∥bj ∥ for all j ∈ {1, . . . , n}.
Propositions 6.2 and 6.3 below are shown by checking all the possible cases for the
corresponding prototype sets. We omit their straightforward but long proofs.
Proposition 6.2. Let a = (a1 , a2 ) ∈ Ω(2) be such that C2 (a) ∩ Fr(F) ≠ ∅. Then, a1 ∈
{−2, 2i, −1 + i} if and only if a ∈ Ir(2); in this case, C2 (a) ⊆ Fr(F).
Proposition 6.3. Let n ∈ N and a ∈ R(n) be arbitrary. If b ∈ D is such that
ι[Fn (a)] ∩ τb [F] ≠ ∅,
then either
ι[F○n (a)] ∩ τb [F○ ] ≠ ∅
or
ι[Fn (a)] ∩ τb [F] ⊆ Fr (ι[Fn (a)]) .
In the second case, there exists c ∈ ι[F○n (a)] such that ∣c − b∣ = 1, ∥c∥ = ∥b∥, c = m or c = im
for some m ∈ Z with ∣m∣ ≥ 2, and
ι[Fn (a)] ∩ τb [F] = ι[Fn (a)] ∩ τc [F] ∩ τb [F].
Moreover, if ∥c∥ ≥ 3, then τc [F○ ] ⊆ ι[Fn (a)].
Proof of Theorem 6.1. By Proposition 4.5, we can think that f is the identity map without
losing any generality. The proof is by induction on n. Assume that n = 2 and pick
a = (a1 , a2 ) ∈ Ir(2). If we had C 2 (a) ⊆ Fr(F), then
(23)
a1 ∈ {−2, −1 + i, 2i} .
Suppose that a1 = −1 + i, the other possibilities are treated similarly. Since a ∈ Ir(2), we
must have
a2 ∈ {1 − im ∶ m ∈ N≥2 } ∪ {−m + i ∶ m ∈ N≥2 }
(see Figure 9). Choose
⎧
⎪
⎪
⎪−im, if a2 = 1 − im,
b2 ∶ = ⎨
⎪
⎪
−m, if a2 = −m + i.
⎪
⎩
Then, we have
τa2 [F] ∩ ι [C 1 (−1 − i)] ⊆ τb2 [F] ∩ ι [C 1 (−1 − i)] ,
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
37
C2 (a) ⊆ C 2 (b) and ∥b1 ∥ = ∥a1 ∥, ∥b2 ∥ = ∥a2 ∥. Observe that if a = (a1 , a2 ) ∈ Ω(2) is such
√
that C2 (a) is not contained in Fr(F), then a ∈ R(2). Indeed, when ∣a1 ∣ ≥ 8, we have
F1 (a1 ) = F, so
C2○ (a) = Q1 (C1○ (a1 ) ∩ F1 (a1 ); a1 ) = Q1 (C1○ (a1 ); a1 ) ≠ ∅.
√
When ∣a1 ∣ ≤ 5, we must check all the possible options separately. For example, if a1 =
−1 − i, then for b ∈ D we have τb [F○ ] ∩ ι[F○1 (a1 )] ≠ ∅ when R(b) ≤ 0 and I(b) ≥ 0 and
τb [F○ ] ∩ ι[F○1 (a1 )] = ∅ otherwise.
−5
−4
−3
−2
−1
−1
−2
−3
−4
−5
Figure 9. The set ι[F1 (−1 − i)].
Assume that Theorem 6.1 holds for N ∈ N≥2 . Let a ∈ Ir(N + 1) be arbitrary. Pick
k ∈ {1, . . . , N } such that
(a1 , . . . , ak ) ∈ R(k)
and
(a1 , . . . , ak , ak+1 ) ∈ Ir(k + 1).
Then, the non-empty set
Tk (Ck+1 (a1 , . . . , ak , ak+1 ); (a1 , . . . , ak )) = Fk (a1 , . . . , ak ) ∩ C1 (ak+1 )
has empty interior; thus, Proposition 6.3 implies
ι[Fk (a1 , . . . , ak )] ∩ τak+1 [F] ⊆ Fr (ι[Fk (a1 , . . . , ak )])
and there is some c ∈ ι[F○k (a1 , . . . , ak )] such that ∣c − ak+1 ∣ = 1 and
ι[Fk (a1 , . . . , ak )] ∩ τak+1 [F] = ι[Fk (a1 , . . . , ak )] ∩ τc [F] ∩ τak+1 [F].
Let us further assume that N ≥ k + 2, the case N = k + 1 is similar to the base of induction.
Let mir ∈ {mir1 , mir2 } be mir1 if ak+1 − c ∈ R and mir2 if ak+1 − c ∈ C ∖ R. Suppose that
(ak+2 , . . . , aN +1 ) ∈ R(N − k − 1). If
(24)
τc [(CN○ −k−1 (mir(ak+2 ), . . . , mir(aN +1 ))] ⊆ ι[Fk (a1 , . . . , ak )]
then b = (a1 , . . . , ak , c, mir(ak+2 ), . . . , mir(aN +1 ))) verifies the conclusions, because
τc [mir(CN○ −k−1 (ak+2 , . . . , aN +1 )] = τc [(CN○ −k−1 (mir(ak+2 ), . . . , mir(aN +1 ))] .
38
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
Condition (24) holds when ∥c∥ ≥ 3. Suppose that (24) is false, so ∥c∥ = 2. The symmetries
of the whole construction allow us to only consider c = 2i and ak+1 = 1 + 2i. In Figure 10,
the grey region of the square to the left represents ι[Fk (a1 , . . . , ak )] ∩ τc [F] and the square
to the right is τak+1 [F]. By Figure 10, when (24) fails we must have ak+2 = −1 + 2i and we
take bk+2 = 2. In this case, we have
Ck+2 (a1 , . . . , ak , ak+1 , ak+2 ) = Ck+2 (a1 , . . . , ak , 1 + 2i, −1 + 2i)
⊆ C k+2 (a1 , . . . , ak , 2i, 2) ∈ R(k + 2).
From this point, we argue as in the proof of Proposition 5.3 to show that
(a1 , . . . , ak , ak+1 , ak+2 , ak+3 , . . . , aN +1 ) = (a1 , . . . , ak , 1 + 2i, −1 + 2i, 2i, 2, −2i, −2, . . . , aN +1 ),
where (2i, 2, −2i, 2) appears periodically and that the finite sequence
(b1 , . . . , bk , bk+1 , bk+2 , bk+3 , . . . , bN +1 ) = (a1 , . . . , ak , 2i, 2, −2i, 2, . . . , bN +1 ),
where the pattern (2i, 2, −2i, 2) is repeated, verifies the conclusions.
When (ak+2 , . . . , aN +1 ) ∈ Ir(N −k−1), we may use the induction hypothesis on it to obtain
(ck+2 , . . . , cN +1 ) ∈ R(N − k − 1) such that
and
CN −k−1 (ak+2 , . . . , aN +1 ) ⊆ C N −k−1 (ck+2 , . . . , cN +1 )
∥aj ∥ = ∥cj ∥ for all j ∈ {k + 2, . . . , N + 1}.
Now, considering an appropriate mir ∈ {mir1 , mir2 }, we deal with (ck+2 , . . . , cN +1 ) as we did
with (ak+2 , . . . , aN +1 ) when we assumed it to be regular.
□
Figure 10. Configuration when (24) fails.
Theorem 6.1 gives us an insight on the interaction between the symbolic space Ω and
F∖Q(i). Let Λ ∶ Ω → F be the map given by (an )n≥1 ↦ [0; a1 , a2 , a3 , . . .]. When Ω is endowed
with the topology inherited from D N , the strict growth of (∣qn ∣)n≥1 and Proposition 3.1.v
imply the continuity of Λ. Denote by Λ∣R the restriction of Λ to R.
Corollary 6.4. There exists a unique continuous extension Λ of Λ∣R to the closure of R.
The image of Λ contains F ∖ Q(i).
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
39
Remark. Another approach to deal with irregular numbers is to show that they form a set
of Hausdorff dimension 1. This will allow us to ignore them while proving Theorem 1.1
and Theorem 1.2. The downside of this strategy, however, is the loss of Corollary 6.4.
7. Proof of Theorem 1.2
We calculate separately the lower and the upper bound for dimH (F (B)) for any B > 1.
The upper bound will be obtained using a natural covering of F (B), while we obtain the
lower bound relying on the Mass Distribution Principle (Lemma 7.1 below). The proof of
the Mass Distribution Principle can be found, for example, in[11, Proposition 2.1].
Lemma 7.1 (Mass Distribution Principle). Let F ⊆ C be non-empty and let µ be a finite
measure such that µ(F ) > 0. If there are constants c > 0, r0 > 0 and t ≥ 0 such that
µ (D(z; r)) ≤ crt
for all z ∈ F and all r ∈ (0, r0 ), then dimH (F ) ≥ t.
7.1. The functions gn . In this section, we define the functions gn , which correspond to
the functions fn of [33, p. 1322]. In what follows, we will use the next technical lemma
without any reference.
Proposition 7.2. Given a = (an )n≥1 ∈ Ω, let (qn )n≥0 be the corresponding sequence of its
Q-pair. Every n ∈ N satisfies
qn−1
1
∣1 +
∣≥1− √ .
an+1 qn
2
Proof. The sequence (∣qj ∣)j≥1 is strictly increasing by Proposition 3.1.iv, and ∣an ∣ ≥
Proposition 3.1.i, then
∣1 +
√
2 by
qn−1
qn−1
1
1
∣≥1−∣
∣≥1−
≥1− √ .
an+1 qn
an+1 qn
∣an+1 ∣
2
□
√
In what follows, both N1 ∶ = 11 and ψ∶ =
√
1+ 5
2
will remain fixed.
Definition 7.3. For any n ∈ N≥N1 and any non-empty A ⊆ D, define
∆n (A )∶ = A n ∩ R(n).
The set A is n-useful if ∆n (A ) ≠ ∅. The set A is useful if it is n-useful for every
n ∈ N≥N1 .
The useful sets we will use are D and D(M )∶ = {a ∈ D ∶ ∥a∥ ≤ M } for M ≥ 3.
40
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
Definition 7.4. Given B > 1, n ∈ N, and an n-useful set A ⊆ D, define
√
ρ
5 2
gn (ρ; A , B)∶ = ∑ ( n
) for any ρ > 0.
2
a∈∆n (A ) B ∥qn (a)∥
When A and B are fixed, we simply write gn (ρ) rather than gn (ρ; A , B).
Proposition 7.5. Every B > 1, n ∈ N≥N1 , and a ∈ Ω(n) satisfy
√
5 2
(25)
< 1.
B n ∥qn (a)∥2
In particular, if A ⊆ D is n-useful, then ρ ↦ gn (ρ; A , B) is non-increasing.
Proof. Take n ∈ N≥N1 and a ∈ Ω(n). In view of ∣qn (a)∣ > ψ n−1 ≥ ψ 10 > 10, we have
√
5 2 < ∥qn (a)∥ and (25) holds for any B > 1. The second assertion follows from the first
one.
□
Proposition 7.6. Given B > 1, n ∈ N≥N1 , and two n-useful sets A , A ′ such that A ⊆
A ′ ⊆ D, we have
gn (ρ; A , B) ≤ gn (ρ; A ′ , B) for all ρ > 0.
Proof. This follows from ∆n (A ) ⊆ ∆n (A ′ ).
□
Definition 7.7. For any n ∈ N, any B > 1, and any n-useful set A ⊆ D define
sn,B (A )∶ = inf{ρ > 0 ∶ gn (ρ; A , B) ≤ 1}.
Proposition 7.8. Let B > 1, n ∈ N≥N1 and two n-useful sets A , A ′ be given. If A ⊆ A ′ ,
then sn,B (A ) ≤ sn,B (A ′ ).
Proof. Take ε > 0 and put ρ = sn,B (A ′ ) + ε. By Proposition 7.5, the functions ρ ↦
gn (ρ; A , B) and ρ ↦ gn (ρ; A ′ , B) are non-increasing, and, by Proposition 7.6,
gn (ρ; A , B) ≤ gn (ρ; A ′ , B) < 1.
This implies sn,B (A ) ≤ sn,B (A ′ ).
□
For any j ∈ N, B > 1, ρ > 0, and a ∈ Ω(j), put
h̃j (a; ρ, B)∶ = (B j ∥qj (a)∥2 )−ρ .
Whenever B is understood, we only write h̃j (a; ρ).
Proposition 7.9. Take B > 1. For any m, n ∈ N, any a ∈ Ω(n), b ∈ Ω(m) such that
ab ∈ Ω(m + n), and ρ > 0, we have
√ 2ρ
1
h̃
(a;
ρ,
B)
h̃
(b;
ρ,
B)
≤
h̃
(ab;
ρ,
B)
≤
(5
2) h̃n (a; ρ, B)h̃m (b; ρ, B).
n
m
n+m
62ρ
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
41
Proof. Let B, m, n, a, b as in the statement. Proposition 4.9 yields
√ 2
1
1
1
1
≤
≤
(5
2)
.
2
2
2
2
2
6 ∥qn (a)∥ ∥qm (b)∥
∥qn+m (ab)∥
∥qn (a)∥ ∥qm (b)∥2
We now multiply by B −(n+m) , take the ρ-th power, and conclude.
□
Proposition 7.10. Let A be a useful set and B > 1. For every m, n ∈ N and ρ > 0, we
have
gn+m (ρ) ≤ gn (ρ)gm (ρ).
Proof. Let ρ > 0 be arbitrary. Proposition 7.9 yields
√
gn+m (ρ) = ∑ (5 2)2ρ h̃n+m (c; ρ)
≤
=
c∈∆n (A )
∑
√
√
(5 2)2ρ h̃n (a; ρ)(5 2)2ρ h̃m (b; ρ)
a∈∆n (A )
b∈∆n (A )
√
√
⎞⎛
⎞
⎛
∑ (5 2)2ρ h̃m (b; ρ)
∑ (5 2)2ρ h̃n (a; ρ)
⎠ ⎝b∈∆n (A )
⎠
⎝a∈∆n (A )
= gn (ρ)gm (ρ).
We might be introducing new terms in the above inequality.
Define κ4 ∶ =
□
√
3√ 2+1
.
2−1
Proposition 7.11. Take any B > 1. Assume that A is either D, D(M ) for some M ≥ 4,
or E . Every m, n ∈ N and every ρ > 0 satisfy
1800ρ
1
gn (ρ)gm (ρ) ≤ gn+m (ρ).
+ 20(3600κ24 )ρ
Proof. Take ρ > 0. By definition of gn ,
√
gm (ρ)gn (ρ) = gm (ρ)(5 2)2ρ ∑
(26)
√
= gm (ρ)(5 2)2ρ
a∈∆n (A )
∑
a∈∆n (A )
∥an ∥≥3
h̃n (a; ρ)
√
h̃n (a; ρ) + gm (ρ)(5 2)2ρ
∑
a∈∆n (A )
∥an ∥≤2
h̃n (a; ρ).
We now bound separately each of the two addends in (26). First, by Proposition 4.11.ii,
⎛
⎞
⎛
⎞
√ 2ρ ⎜
√ 4ρ ⎜
⎞
⎟
⎟⎛
(5 2) ⎜ ∑ h̃n (a; ρ)⎟ gm (ρ) = (5 2) ⎜ ∑ h̃n (a; ρ)⎟
h̃ (b; ρ)
⎜a∈∆ (A )
⎟
⎜a∈∆ (A )
⎟ ⎝b∈∆∑(A ) m
⎠
m
⎝ ∥ann∥≥3
⎝ ∥ann∥≥3
⎠
⎠
√
h̃n+m (ab; ρ)
≤ (5 2)4ρ 62ρ
∑
√
≤ (5 2)2ρ 62ρ
a∈∆n (A )∩AF(n)
b∈∆m (A )
∑
c∈∆n+m (A )
√
(5 2)2ρ h̃n+m (c; ρ);
42
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
that is
(27)
⎞
⎛
√ 2ρ ⎜
⎟
(5 2) ⎜ ∑ h̃n (a; ρ)⎟ gm (ρ) ≤ 1800ρ gn+m (ρ).
⎜a∈∆ (A )
⎟
⎝ ∥ann∥≥3
⎠
A more intricate argument is required when bounding the second term. For each a ∈ ∆n (A )
such that ∥an ∥ ≤ 2, choose d(a) ∈ D for which
Pm(d(a)) = ∥d(a)∥ = 3 and x(a)∶ = (a1 , . . . , an−1 , d) ∈ ∆n (A ) ∩ F(n)
(see Proposition 4.11). Then, we have
√
qn−2 (a) R
R
qn (x(a))
d(a)qn−1 (a) + qn−2 (a) RRRR d(a) + qn−1 (a) RRRR ∣d(a)∣ + 1 3 2 + 1
R≤
∣
= κ4 ,
∣=∣
∣ = RRR
≤ √
RRR an + qn−2 (a) RRRRR
qn (a)
an qn−1 (a) + qn−2 (a)
∣an ∣ − 1
2
−
1
qn−1 (a) R
R
which implies
√
∥qn (x(a))∥ ≤ ∣qn (x(a))∣ ≤ κ4 ∣qn (a)∣ ≤ κ4 2∥qn (a)∥,
and hence
(2κ24 )ρ
1
≤
.
(B n ∥qn (a)∥2 )ρ (B n ∥qn (x(a))∥2 )ρ
(28)
The function a ↦ x(a) might not be injective, but it is at most twenty to one. Indeed,
for any b = (b1 , . . . , bn−1 , bn ) ∈ ∆n (A ) ∩ F(n) such that Pm(bn ) = ∥bn ∥ = 3, the set of words
a = (a1 , . . . , an ) ∈ ∆n (A ) satisfying ∥an ∥ ≤ 2 and x(a) = b is contained in the set
{a ∈ ∆n (A ) ∶ (a1 , . . . , an−1 ) = (b1 , . . . , bn−1 ), ∥an ∥ ≤ 2},
which has at most 20 elements, because #{a ∈ D ∶ ∥a∥ ≤ 2} = 20. This observation and (28)
give
∑
a∈∆n (A )
∥an ∥≤2
h̃n (a; ρ) ≤ (2κ24 )ρ
≤ 20(2κ24 )ρ
≤ 20(2κ24 )ρ
and, by Proposition 7.9,
∑
a∈∆n (A )
∥an ∥≤2
h̃n (x(a); ρ)
∑
h̃n (c; ρ)
∑
h̃n (c; ρ),
c∈∆n (A )
Pm(cn )=∥cn ∥=3
c∈∆n (A )∩F(n)
⎛
⎞
√ 2ρ ⎜
√
⎛
⎞⎛
⎞
⎟
h̃n (a; ρ)
h̃m (b; ρ)
(5 2) ⎜ ∑ h̃n (a; ρ)⎟ gm (ρ) ≤ (5 2)4ρ 20(2κ24 )ρ
∑
∑
⎜a∈∆ (A )
⎟
⎝a∈∆n (A )∩F(n)
⎠ ⎝b∈∆m (A )
⎠
⎝ ∥ann∥≤2
⎠
√
√
≤ (5 2)2ρ 20(2κ24 )ρ 62ρ
∑ (5 2)2ρ h̃n+m (c; ρ).
c∈∆n+m (A )
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
43
Again, we might be adding new terms in the last inequality. The previous computations
show that
⎛
⎞
√ 2ρ ⎜
⎟
(5 2) ⎜ ∑ h̃n (a; ρ)⎟ gm (ρ) ≤ 20(3600κ24 )ρ gn+m (ρ).
⎟
⎜a∈∆ (A )
⎝ ∥ann∥≤2
⎠
(29)
The result follows from combining (26), (27), and (29).
□
7.2. Properties of sB . The function sB shares some properties with the corresponding
map defined in [33]. Lemmas 7.13, 7.15, and 7.18 correspond to [33, Lemma 2.4], [33,
Lemma 2.5], and [33, Lemma 2.6], respectively.
For the rest of this section, B > 1 will remain fixed.
Proposition 7.12. If p, q ∈ N are coprime and p < q, then
N≥p(q−1) ⊆ {j ∈ N ∶ j = pk + tq for some k, t ∈ N0 }.
Proof. See [2, Section 3. Problem 14].
□
Lemma 7.13. If A = D or A = D(M ) for some M ≥ 4, then the following limit exists
sB (A )∶ = lim sn,B (A ).
n→∞
Proof. Let A be as in the statement.
i. We claim that if m, n ∈ N≥N1 , then
(30)
sm+n,B ≤ max{sm,B (A ), sn,B (A )}.
Take ε > 0 and ρ = max{sm,B (A ), sn,B (A )} + ε. Then, gm (ρ) < 1, gn (ρ) < 1 and, by
Proposition 7.10,
gn+m (ρ) ≤ gn (ρ)gm (ρ) < 1;
hence,
sm+n,B (A ) < max{sm,B (A ), sn,B (A )} + ε
(31)
and (30) follows. In particular, whenever n, t ∈ N≥N1 and r, k ∈ N we have
srn+tk,B (A ) ≤ max{srn,B (A ), stk,B (A )} ≤ max{sn,B (A ), st,B (A )}.
ii. Define
s∶ = lim sup sn,B (A ).
n→∞
We claim that
(32)
#{n ∈ N ∶ sn,B (A ) ≥ s} = ∞.
If (32) were false, there would be some N0 ∈ N≥N1 satisfying
sn,B (A ) < s for all n ∈ N≥N0 .
44
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
Define
r∶ = max{sN0 +r,B ∶ r ∈ {0, 1, . . . , N0 − 1}} < s.
Given n ∈ N≥2N0 , let q ∈ N≥2 and r ∈ {0, 1, . . . , N0 − 1} be such that n = qN0 + r, then
sn,B (A ) = s(q−1)N0 +(N0 +r),B (A )
≤ max{s(q−1)N0 ,B (A ), sN0 +r,B (A )}
≤ max{s(q−1)N0 ,B (A ), r}
≤ max{sN0 ,B (A ), r} ≤ r < s.
However, this yields the contradiction
s = lim sup sn,B (A ) ≤ r < s,
n→∞
so (32) must hold.
iii. We claim that
(33)
sn,B (A ) ≤ max{sN1 ,B (D), sN1 +1,B (D)} for all n ∈ N≥N1 (N1 +1) .
Take any n ∈ N≥N1 (N1 +1) . First, by (31) and Proposition 7.12,
sn,B (D) ≤ max{sN1 ,B (D), sN1 +1,B (D)}.
Second, Proposition 7.6 gives sn,B (A ) ≤ sn,B (D) and (33) follows.
iv. Write
ŝ∶ = max{sN1 ,B (D), sN1 +1,B (D)},
κ(ŝ)∶ = (1800ŝ + 20(3600κ24 )ŝ )−1 .
By Proposition 7.11,
(34)
κ(ŝ)gn (ρ)gm (ρ) ≤ gn+m (ρ) for every ρ ∈ (0, ŝ) and m, n ∈ N.
Hence, for every ρ ∈ (0, ŝ), j ∈ N, and n ∈ N, we obtain
gjn (ρ) ≥ κ(ŝ)j−1 gn (ρ)j .
Then, if j ≥ 2, we have
gn (ρ) ≤
1
κ(ŝ)
1
1− 1j
gjn (ρ) j ≤
1
κ(ŝ)
1
1
2
gjn (ρ) j
and, using the definition of snj,B (A ), we get
(35)
gn (sjn,B (A ) + δ) ≤
1
1
κ(ŝ) 2
for all δ > 0.
v. Let us show that
(36)
sr,B (A ) ≥ s for all r ∈ N≥N1 (N1 +1) .
Assume that r ∈ N≥N1 (N1 +1) satisfies
sr,B (A ) < s.
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
45
Consider ε∶ = s − sr,B (A ) > 0. By (32), we may choose some p ∈ N satisfying
ε
sp,B (A ) ≥ s,
Then, taking δ =
and
B pψp 2
1
( √ ) > κ(ŝ) 2 .
5 2
ε
2
in (35) and recalling part v. of Proposition 3.1,
√
(5 2)2srp,B (A )+ε
gp (srp,B (A ) + ε) = ∑
p
2 srp,B (A )+ε
a∈∆p (A ) (B ∥qp (a)∥ )
√ ε
√
ε
5 2 2
(5 2)2srp,B (A )+ 2
≤ ( p p)
∑
ε
p
2 srp,B (A )+ 2
B ψ
a∈∆p (A ) (B ∥qp (a)∥ )
√ ε
5 2 2
ε
= ( p p ) gp (srp,B (A ) + ) < 1,
B ψ
2
Thus,
(37)
sp,B (A ) ≤ srp,B (A ) +
ε
2
and, by (31),
(38)
sp,B (A ) ≥ s = sr,B (A ) + ε ≥ srp,B (A ) + ε.
From (37) and (38) we get
ε
+ srp,B (A ) ≥ sp,B (A ) ≥ srp,B (A ) + ε,
2
which implies the false inequality 0 > 2ε . Therefore, (36) is true.
vi. Lastly, because of (36),
lim inf sn,B (A ) ≥ s = lim sup sn,B (A )
n→∞
n→∞
and the proof is done.
□
For any p ∈ N≥N1 , write sp,B ∶ = sp,B (D) and sp,B (M )∶ = sp,B (D(M )) for all M ∈ N≥2 .
Lemma 7.14. If p ∈ N≥N1 , then
lim sp,B (M ) = sp,B .
M →∞
Proof. Take p ∈ N≥N1 . Proposition 7.8 tells us that (sp,B (M ))M ≥4 is non-decreasing and
that each term satisfies sp,B (M ) ≤ sp,B . Then, (sp,B (M ))M ≥4 converges and
tp,B ∶ = lim sp,B (M ) ≤ sp,B .
M →∞
Take any ε > 0. The definition of sp,B and p ≥ N1 imply that gp (sp,B − ε; D, B) > 1. The
definition of sp,B also says that
gp (ρ; D, B) = lim gp (ρ; D(M ), B) for all ρ > 0.
M →∞
46
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
Therefore, every sufficiently large M satisfies
gp (sn,B − ε, D(M ), B) > 1,
so sp,B − ε ≤ sp,B (M ) ≤ tp,B , hence sp,B ≤ tp,B .
□
Write sB ∶ = sB (D) and sB (M )∶ = sB (D(M )) for all M ∈ N≥2 .
Lemma 7.15. lim sB (M ) = sB .
M →∞
̃=N
̃ (B, ε) ∈ N≥N be such that
Proof. Take ε > 0. Let N
1
B Ñ ψ Ñ 2
κ(ŝ) < ( √ ) .
5 2
ε
1
2
Consider any n ∈ N≥Ñ . Following the proof of (37), we may conclude–for A = D or
A = D(M ) with M ≥ 4–that every m ∈ N satisfies
ε
sn,B (A ) ≤ smn,B (A ) + .
2
Moreover, we obtain from (31) that smn,B (A ) ≤ sn,B (A ), so
∣sn,B (A ) − smn,B (A )∣ < ε.
We let m → ∞ and use Lemma 7.13 to get
(39)
∣sn,B (A ) − sB (A )∣ < ε
Moreover, Lemma 7.14 guarantees that every sufficiently large M satisfies
(40)
∣sn,B (M ) − sn,B ∣ < ε.
Inequalities (39), and (40) lead us to the desired conclusion:
∣sB (M ) − sB ∣ ≤ ∣sB (M ) − sn,B (M )∣ + ∣sn,B (M ) − sn,B ∣ + ∣sn,B − sB ∣ < 3ε.
□
Proposition 7.16. Define
N2 ∶ = max {N1 ,
log(50) − log π − log κ1
+ 1} .
2 log B
We have
sup {sn,B (M ) ∶ n ∈ N≥N2 , M ∈ N} ≤ 2.
In particular, the sequence (sn,B (M ))n≥1 is bounded for each M ∈ N≥4 .
Proof. For each n ∈ N≥N1 , the sequence (sn,B (M ))M ≥4 is non-decreasing and converges to
sn,B , so
sup {sn,B (M ) ∶ M ∈ N≥N2 } ≤ sn,B .
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
47
If n ≥ N2 , then sn,B < 2, because
√
2
5 2
gn (2; D, B) = ∑ ( n
)
2
a∈R(n) B ∥qn (a)∥
=
50
1
∑
2n
B a∈R(n) ∥qn (a)∥4
50
≤
∑
B 2n κ1 π a∈R(n)
m(Cn (a)) ≤
50
B 2n κ1 π
< 1.
□
Proposition 7.17. If n ∈ N and a ∈ R(n), then
n
1 n
√ ∏(∣aj ∣ − 1) ≤ ∥qn (a)∥ ≤ ∏(∣aj ∣ + 1).
2 j=1
j=1
Proof. Using mathematical induction on n, we may show that
n
n
∏(∣aj ∣ − 1) ≤ ∣qn (a)∣ ≤ ∏(∣aj ∣ + 1).
j=1
j=1
√
These inequalities along with ∥z∥ ≤ ∣z∣ ≤ 2∥z∥ for all z ∈ C imply the proposition.
Lemma 7.18. The function from (1, ∞) → R, B ↦ sB has the following properties:
1. lim sB = 2,
B→1
2. lim sB = 1,
B→∞
3. it is continuous.
Proof. 1. Consider 0 < ε < 2 and let B be such that 1 < B < ψ ε . If n ∈ N satisfies
n ≥ max {N2 ,
2ε log(ψ) + log π
},
ε log ψ − 2 log B
then
√
2−ε
5 2
gn (2 − ε; D, B) = ∑ ( n
)
2
a∈R(n) B ∥qn (a)∥
√
= (5 2)2−ε ∑ (
a∈R(n)
2−ε
1
)
B n ∥qn (a)∥2
∥qn (a)∥2ε
2n
4
a∈R(n) B ∥qn (a)∥
≥ ∑
≥
1
ψε n
1
(
) ∑
ε
2
4
(2ψ) B
a∈R(n) ∥qn (a)∥
≥
1
ψε n
1
(
) ∑
ε
2
4
(2ψ) B
a∈R(n) ∣qn (a)∣
≥
1
1
ψε n
1
ψε n 1
(
)
m(C
(a))
≥
(
)
> 1.
∑
n
(2ψ)ε B 2 a∈R(n) π
(2ψ)ε B 2 π
□
48
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
and hence 2 − ε < sn,B . Since sn,B < 2 holds by Proposition 7.14, we conclude that
sn,B → 2 as B → 1.
2. Take any n ∈ N≥N1 . First, we show that sn,B ≥ 1. Recall that E n ⊆ R(n). Then,
√
5 2
gn (1, D, B) = ∑
n
2
a∈R(n) B ∥qn (a)∥
√
n
5 2
1
≥ n ∑ ∏
B a∈R(n) j=1 (∣aj ∣ + 1)2
√
n
1
5 2
≥ n ∑ ∏
B a∈E n j=1 (∣aj ∣ + 1)2
√
√
n
n
1
2n (5 2)
1
5 2
)
≥
(
)
≥ n (∑
∑ 2 = ∞,
2
B
Bn
a∈E (∣a∣ + 1)
a∈E ∣a∣
thus, sn,B ≥ 1.
Take ε ∈ (0, 1) and choose N ∈ N≥N1 so large that n ∈ N≥N implies
n
n
⎛
⎞ ⎛
⎞
1
1
200 ∑
< ∑
.
2+2ε
2+ε
⎝∣a∣≥√2 (∣aj ∣ − 1)
⎠ ⎝∣a∣≥√2 (∣aj ∣ − 1) ⎠
For such n, we have
√
n
√ 1+ε √ 2+2ε
(5 2)1+ε
1
≤
(5
2)
(
2)
∑
∑
∏
2+2ε
2+2ε
a∈R(n) ∥qn (a)∥
a∈R(n) j=1 (∣aj ∣ − 1)
n
⎞
⎛
1
< 200 ∑
⎝∣a∣≥√2 (∣a∣ − 1)2+2ε ⎠
n
Therefore, if
⎛
⎞
1
< ∑
.
2+ε
⎝∣a∣≥√2 (∣a∣ − 1) ⎠
1
⎞ 1+ε
⎛
1
,
Bε ∶ = ∑
⎝∣a∣≥√2 (∣a∣ − 1)2+ε ⎠
we have gn (1 + ε; D, Bε ) < 1 and sn,Bε ≤ 1 + ε. Taking n → ∞, we conclude sBε ≤ 1 + ε.
Finally, since B ↦ sn,B is decreasing for every n ∈ N, the function B → sB is also
decreasing and we conclude that sB → 1 when B → ∞.
3. Let B0 > 1 be given. We are going to show the next statement: for each ε > 0 there exist
δ = δ(B0 , ε) > 0 and N (B0 , δ) ∈ N≥N2 such tha, if n ∈ N≥N (B0 ,δ) , then for all B > 1
∣B − B0 ∣ < δ implies ∣sn,B − sn,B0 ∣ < ε.
The previous statement and Proposition 7.13 yield the continuity of B ↦ sB at B0 .
Let ε ∈ (0, 1) be arbitrary. Put
δ1 ∶ =
1
1
1+ ε
(B0 − B0 4 ) ,
2
δ2 ∶ =
1
1+ ε
(B0 12 − B0 ) ,
2
δ∶ = min{δ1 , δ2 }
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
49
and
√
√
⎧
⎫
⎪
⎪
log(5 2)
log(5 2)
⎪
⎪
N ∶ = max ⎨N2 ,
,
⎬.
ε
ε
⎪
⎪
((1
)
)
((1
)
2
+
log(B
−
δ)
−
log
B
2
+
log
B
−
log(B
+
δ))
⎪
⎪
0
0
0
0
4
4
⎩
⎭
We chose the parameters so the next estimates hold for all n ∈ N≥N :
2n
B0
(
ε )
(B0 − δ)1+ 4
1
< √ ,
5 2
2n
⎛ B0 + δ ⎞
ε
⎝ B 1+ 4 ⎠
0
1
< √ .
5 2
Given n ∈ N≥N (B0 ,δ) , using sn,B0 < 2, we get
√
sn,B0 + 2ε
5 2
ε
)
gn (sn,B0 + ; D, B0 − δ) = ∑ (
n
2
2
a∈R(n) (B0 − δ) ∥qn (a)∥
√ ε
√
sn,B0
(5 2) 2
5 2
B0 nsn,B0
≤
)
)
∑ ( n
nε (
2
(B0 − δ) 2 B0 − δ
a∈R(n) B0 ∥qn (a)∥
√ ε
(5 2) 2
B0 nsn,B0
=
)
gn (sn,B ; D, B0 )
nε (
(B0 − δ) 2 B0 − δ
√
≤ 5 2(
2n
B0
ε )
(B0 − δ)1+ 4
< 1.
Therefore, since B ↦ sn,B is non-increasing,
ε
sn,B0 + ≥ sn,B0 −δ ≥ sn,B0 .
2
Similarly, we may obtain
ε
sn,B0 − ≤ sn,B0 +δ ≤ sn,B0 ;
2
in fact,
√
sn,B0 +δ + 2ε
ε
5 2
gn (sn,B0 +δ + ; D, B0 ) = ∑ ( n
)
2
2
a∈R(n) B0 ∥qn (a)∥
√ ε
√
sn,B0 +δ
(5 2) 2 B0 + δ nsn,B0 +δ
5 2
≤
(
)
)
∑ (
nε
n
2
B0
a∈R(n) (B0 + δ) ∥qn (a)∥
B02
2n
√ ⎛ B0 + δ ⎞
<5 2
ε
⎝ B 1+ 4 ⎠
0
< 1.
□
√
In the definition of s(B), we can replace the constant 5 2 with 1 and ∥ ⋅ ∥ with ∣ ⋅ ∣.
Indeed, for each n ∈ N define Gn ∶ R>0 → R by
Gn (ρ; B)∶ = ∑
a∈R(n)
(B n ∣q
1
2 ρ
n (a)∣ )
for all ρ > 0
and consider
Sn,B ∶ = inf {ρ > 0 ∶ Gn (ρ; B) ≤ 1} .
50
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
Corollary 7.19. We have
lim Sn,B = s(B).
n→∞
Proof. On the one hand,
√
1
5 2
≤
for all n ∈ N and a ∈ R(n),
B n ∣qn (a)∣2 B n ∥qn (a)∥2
so Gn (ρ; B) ≤ gn (ρ; B) for all ρ > 0 and, thus, Sn,B ≤ sn,B . On the other hand, for any
δ > 0, every large n ∈ N satisfies
√
5 2
1
≤ n
for all a ∈ R(n).
n
2
(B + δ) ∥qn (a)∥
B ∣qn (a)∣2
Therefore, gn (ρ; B + δ) ≤ Gn (ρ; B) and, thus, sn,B+δ ≤ Sn,B . Letting n → ∞, we obtain
s(B + δ) ≤ lim inf Sn,B ≤ lim sup Sn,B ≤ s(B).
n→∞
n→∞
Letting δ → 0 and using the third item of Lemma 7.18, we conclude Sn,B → s(B) as
n → ∞.
□
7.3. Upper bound of Hausdorff dimension. For each n ∈ N and each a = (a1 , . . . , an ) ∈
R(n) define
Jn (a)∶ = Cl ({z ∈ F∶ ak (z) = ak for all k ∈ {1, . . . , n} and ∥an+1 (z)∥ ≥ B n+1 }) .
Write
−1
1
λ1 ∶ = (1 − √ ) .
2
Lemma 7.20. For each n ∈ N and a ∈ R(n),
∣Jn (a)∣ ≤
2λ1 (λ1 + 1)
.
B n+1 ∥qn (a)∥2
Proof. The proof is similar to that of Proposition 5.18.
□
Recall that
F (B)∶ = {z = [0; a1 , a2 , . . .] ∈ F ∶ ∥an ∥ ≥ B n for infinitely many n ∈ N}.
Proposition 7.21. We have
F (B) ⊆ ⋂ ⋃
⋃ Jn (a).
m∈N n≥m a∈R(n)
Proof. Take z = [0; a1 , a2 , . . .] ∈ F (B). The result is obvious when a = (an )n≥1 ∈ R, so
we assume that a ∈ Ir. Theorem 6.1 ensures that for each n ∈ N there is some bn =
(bn,1 , . . . , bn,n ) ∈ R(n) such that z ∈ C n (b) and ∥bn,j ∥ = ∥aj ∥ for each j ∈ {1, . . . , n}. Hence,
if ∥am ∥ ≥ B m for some m ∈ N, then ∥bn,m ∥ ≥ B m for all n ∈ N≥m .
□
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
51
Proof of Theorem 1.2, Upper Bound. Let ε > 0 be arbitrary. By Lemma 7.13, we may pick
N ∈ N≥N1 so large enough that any n ∈ N≥N verifies sn,B < sB + 2ε , which gives sn,B + 2ε < sB +ε
and
gn (sB + ε; D, B) < 1.
For each η ≥ 0, let H η denote the η-Hausdorff measure. By Lemma 7.20, putting λ2 ∶ =
2λ1 (λ1 + 1), we get
H sB +2ε (F (B)) ≤ lim inf ∑ ∑ ∣Jn (a)∣sB +2ε
N →∞
≤
=
≤
=
n≥N a∈R(n)
λs2B +2ε lim inf
N →∞
sB +2ε
1
)
∑ ∑ ( n+1
∥qn (a)∥2
n≥N a∈R(n) B
√
sB +ε
λs2B +2ε
5 2
1
√
lim inf ∑ ∑ ( n
)
2
n+1
(B ∥qn (a)∥2 )ε
(5 2B)sB +ε N →∞ n≥N a∈R(n) B ∥qn (a)∥
√
sB +ε
ε
λs2B +2ε
2ψ
5 2
√
lim inf ∑ (
) ∑ (
)
(5 2B)sB +ε N →∞ n≥N B n+1 ψ n a∈R(n) B n ∥qn (a)∥2
√
sB +ε
2ε ψ ε λs2B +2ε
1
5 2
√
lim inf ∑
)
∑ (
B ε (5 2B)sB +ε N →∞ n≥N (B ε ψ ε )n a∈R(n) B n ∥qn (a)∥2
2ε ψ ε λs2B +2ε
1
√
lim inf ∑
=
g (s + ε; D, B)
ε
ε n n B
ε
s
+ε
N
→∞
B (5 2B) B
n≥N (B φ )
2ε ψ ε λs2B +2ε
1
√
≤
lim inf ∑
= 0.
ε
ε n
ε
s
+ε
N →∞
B (5 2B) B
n≥N (B ψ )
The last equality holds because B ε ψ ε > 1. Therefore, dimH (F (B)) ≤ sB .
□
7.4. Lower bound of Hausdorff dimension.
7.4.1. Construction of Cantor sets. We now construct a family of Cantor set FM (B), M ∈
N≥4 , whose Hausdorff dimension will be lower than dimH (F (B)). Let M be a natural
number. Later, we will impose additional restrictions on M . Let (nk )k≥1 be a sequence of
natural numbers such that
(41)
(42)
B n1 −1 > 11,
nk+1
for all k ∈ N.
n1 + . . . + nk ≤
k+1
For each n ∈ N, let Dn be the set of words a = (a1 , . . . , an ) ∈ R(n) such that every
j ∈ {1, . . . , n} satisfies
i. If j = nk for some k ∈ N, then ⌊B nk ⌋ + 2 ≤ ∥ank ∥ ≤ 2⌊B nk ⌋ + 1;
ii. If j ∈ {1, . . . , n} ∖ {nk ∶ k ∈ N}, then ∥aj ∥ ≤ M .
52
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
We put D0 ∶ = ∅ and D∶ = ⋃n∈N0 Dn . For n ∈ N and a ∈ R(n), write
An (a)∶ =
⋃ C n+1 (ab).
b∈D
ab∈Dn+1
We refer to the sets An (a) as fundamental sets of level n. Define
FM (B)∶ = ⋂ ⋃ An (a).
n∈N a∈Dn
Note the each union in the definition of FM (B) is finite.
Lemma 7.22. If M ∈ N≥4 , then dimH (FM (B)) ≤ dimH (F (B)).
Proof. Fix any a ∈ D such that ∥a∥ ≤ M and C1 (a) is full; for example, a = 3 + 3i. Clearly,
dimH (FM (B) ∩ C1 (a)) ≤ dimH (FM (B)). We focus on showing
dimH (FM (B)) ≤ dimH (FM (B) ∩ C1 (a)).
The basic properties of Hausdorff dimension give
(43)
dimH (FM (B)) = max dimH (FM (B) ∩ C1 (b)) .
∥b∥≤M
√
For any c ∈ D such that ∣c∣ ≥ 8, the function ιτa τ−c ι provides a bi-Lipschitz map between
C 1 (c) ∩ FM (B) and C 1 (a) ∩ FM (B). Certainly, pick ξ ∈ FM (B) ∩ C 1 (c). Given n ∈ N,
choose (c2 , . . . , cn ) such that (c, c2 , . . . , cn ) ∈ Dn and ξ ∈ C n (c, c2 , . . . , cn ). Then, ιτb τ−c ι(ξ) ∈
C n (c, c2 , . . . , cn ); in other words, we have
ιτb τ−c ι [FM (B) ∩ C n (c, c2 , . . . , cn )] ⊆ ιτb τ−c ι [FM (B) ∩ C n (b, c2 , . . . , cn )] .
Therefore, ιτb τ−c ι [FM (B) ∩ C n (c, c2 , . . . , cn )] ⊆ ιτb τ−c ι [FM (B) ∩ C n (b, c2 , . . . , cn )].
rôles of b and c can be reversed, so
The
dimH (C 1 (a) ∩ FM (B)) = dimH (C 1 (c) ∩ FM (B)) .
√
On the contrary, when ∣c∣ ≤ 8 holds, ιτa τ−c ι establishes a bi-Lipschitz map between C 1 (c)∩
FM (B) and a strict subset of C 1 (a) ∩ FM (B), namely Q1 (F1 (c); a) ∩ FM (B), then
dimH (C 1 (c) ∩ FM (B)) ≤ dimH (C 1 (a) ∩ FM (B))
and we conclude
(44)
dimH (FM (B)) = dimH (FM (B) ∩ C1 (a)) .
To finish the proof, we show that
(45)
FM (B) ∩ C 1 (a) ⊆ F (B).
Take any ξ = [0; a, a2 , . . .] ∈ FM (B). Let (cn )n≥1 be a sequence such that for all n ∈ N we
have cn ∈ Dn and ξ ∈ C n (cn ). Pick k ∈ N.
i. If ξ ∈ Cnk (cnk ), then ∥ank ∥ ≥ ⌊B nk ⌋ + 2 > B nk .
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
53
ii. Assume that ξ ∈ C nk (cnk ) ∖ Cnk (cnk ) and write cnk = (cn1 k , . . . , cnnkk ). If ξ is regular,
then ξ ∈ C nk (cnk ) ∩ Cn (a1 , . . . , ank ) and
∥ank ∥ ≥ ∥cnnkk ∥ − 1 ≥ ⌊B nk ⌋ + 1 > B nk .
If ξ is irregular, there is some b = (b1 , . . . , bnk ) ∈ R(n) such that ξ ∈ C nk (b) and
∥bj ∥ = ∥aj ∥ for all j ∈ {1, . . . , nk }. Therefore, ξ ∈ C nk (cnk ) ∩ C nk (b) and
∥ank ∥ = ∥bnk ∥ ≥ ∥cnnkk ∥ − 1 ≥ ⌊B nk ⌋ + 1 > B nk .
Since k was arbitrary, ξ ∈ F (B) and we confirm (45).
□
7.4.2. A probability measure on FM (B). Let (mj )j≥1 be the sequence given by
m1 ∶ = n1 − 1
mj ∶ = nj − nj−1 − 1 for all j ∈ N≥2 .
and
We distribute a unit mass over the sets {An (a) ∶ n ∈ N, a ∈ Dn } to obtain a probability
measure on FM (B). First, take n ∈ {1, . . . , n1 }.
i. If n = n1 − 1, for each a ∈ Dn1 −1 , put
√
sm1 ,B (M )
5 2
)
.
µ (An1 −1 (a)) ∶ = ( m1
B ∥qm1 (a)∥2
ii. If n = n1 , for each a = (a1 , . . . , an1 ) ∈ Dn1 , put
µ (An1 (a)) ∶ =
µ (An1 −1 (a1 , . . . , an1 −1 ))
.
#{d ∈ D ∶ (a1 , . . . , an−1 , d) ∈ Dn }
iii. If n ∈ {1, . . . , n1 − 2}, for each a ∈ Dn , put
µ (An (a)) ∶ =
∑
b∈R(n1 −n−1)
ab∈Dn1 −1
µ (An1 −1 (ab)) .
Assume that, for a given k ∈ N, we have already defined µ(Aj (a)) for all j ∈ {1, . . . , nk − 1}
and a ∈ Dj . Consider n ∈ {nk , . . . , nk+1 − 1} and a = (a1 , . . . , an ) ∈ Dn .
i. If n = nk , put
µ (Ank (a)) ∶ =
µ (Ank −1 (a1 , . . . , ank −1 ))
.
#{d ∈ D ∶ ad ∈ Dnk }
ii. If n = nk+1 − 1, put
√
smk+1 ,B (M )
5 2
µ (Ank+1 −1 (a)) ∶ = µ(Ank (a1 , . . . , ank )) ( m
)
.
B k+1 ∥qmk+1 (ank +1 , . . . , ank+1 −1 )∥2
iii. If n ∈ {nk + 1, . . . , nk+1 − 2}, put
µ (An (a)) ∶ =
∑
b∈R(nk+1 −1−n)
ab∈Dnk+1 −1
µ (Ank+1 −1 (ab)) .
Finally, for each n ∈ N, for a, b ∈ Dn with a ≠ b we define µ(An (a) ∩ An (b)) = 0.
54
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
For all m ∈ N, the infimum defining sm,B (M ) is actually a minimum, because the involved
sum is finite. As a consequence, using the condition we imposed on µ when applied to
intersections, we have
µ
⎞
⎛
⋃ An1 −1 (a) = ∑ µ (An1 −1 (a)) = 1.
⎠ a∈Dn1 −1
⎝a∈Dn1 −1
Also, given a ∈ Dn1 −1 , we defined µ on the sets {An1 (ad) ∶ ad ∈ Dn1 } by uniformly distributing the mass µ(An1 −1 (a)) among them, so we have
µ(An1 −1 (a)) = ∑ µ(An1 (ad)).
d∈D
ad∈Dn1
The definition of µ ensures that the previous equation holds if we replace n1 with any
n ∈ {1, . . . , n1 − 1}. Moreover, it still holds for any n ∈ N. Therefore, by the DaniellKolmogorov consistency theorem (see, for example, [23, Corollary 8.22]), the function µ is
indeed a probability measure defined on FM (B).
7.4.3. Bounding µ(An (a)). In this section, we give an upper estimate of the µ measure
of the sets An (a) in terms of ∣An (a)∣ (Lemma 7.27). The proof relies on estimates of a
geometric (Propositions 7.23, 7.24) and arithmetic nature (Propositions 7.25, 7.26).
Proposition 7.23. If k ∈ N and a ∈ Dnk −1 , then
2⌊B nk ⌋2 ≤ #{d ∈ D ∶ ad ∈ Dnk } ≤ 14⌊B nk ⌋2 .
We make no attempt in getting optimal bounds. In fact, by taking n1 sufficiently large,
we could make the multiplicative constant in the upper (resp. lower) bound as close to 12
(resp. 3) and as we please.
Proof. Pick k ∈ N and, for notation’s sake, write B1 ∶ = ⌊B nk ⌋ + 2 and B2 ∶ = 2⌊B nk ⌋ + 1. The
proof is by cases. First, assume that a is almost full, then
#{d ∈ D ∶ ad ∈ Dnk } = {d ∈ D ∶ B1 ≤ ∥z∥ ≤ B2 }
and, since B1 ≥ 3,
(2B2 + 1)2 − (2B1 − 1)2 = 4(B2 − B1 )(B2 + B1 + 1)
= 4(⌊B nk ⌋ − 1)(3⌊B nk ⌋ + 3)
= 12⌊B nk ⌋2 (1 −
1
1
) (1 + n ) ≤ 14⌊B nk ⌋2 .
n
4⌊B k ⌋
⌊B k ⌋
Assume that, F○nk −1 (a) = F○1 (ir (1+i)) for some r ∈ {0, 1, 2, 3}. Then, the number of elements
in {d ∈ D ∶ ad ∈ Dnk } can be bounded as follows:
(B2 + 1)2 − (B12 + 1)2 = (B2 − B1 )(B2 + B1 + 2)
= (⌊B nk ⌋ − 1)(3⌊B nk ⌋ + 4)
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
= 3⌊B nk ⌋2 (1 −
4
1
) (1 +
)
n
⌊B k ⌋
3⌊B nk ⌋
≥ 3⌊B nk ⌋2 (1 −
1
) ≥ 2⌊B nk ⌋2
⌊B nk ⌋
55
When there is some r ∈ {0, 1, 2, 3} such that F○nk −1 (a) = F○1 (ir (1 + 2i)) or F○nk −1 (a) = F○1 (2ir ),
the previous bounds also hold and the result follows.
□
Define κ5 ∶ =
κ1
√
,
2
with κ1 as in Proposition 4.7, and κ6 ∶ =
20
3 .
Proposition 7.24. The following estimates hold for all n ∈ N:
i. If n = nk − 1 for some k ∈ N, then
κ5
κ6
≤ ∣Ank −1 (a)∣ ≤
for all a ∈ Dnk .
2
n
∥qnk −1 (a)∥ ⌊B k ⌋
∥qnk −1 (a)∥2 ⌊B nk ⌋
ii. If n = nk for some k ∈ N, then
κ6
κ5
≤
∣A
(a)∣
≤
for all a ∈ Dnk −1 .
n
k
∥qnk (a)∥2
∥qnk (a)∥2
iii. If nk−1 < n < nk for some k ∈ N, then
κ6
κ5
≤ ∣An (a)∣ ≤
for all a ∈ Dn .
2
∥qn (a)∥
∥qn (a)∥2
Proof. The first possibility follows from Proposition 5.18. Whenever n ∈ N falls in the last
two cases, we have ∣Cn (a)∣ = ∣An (a)∣ for all a ∈ Dn and we apply Proposition 4.7.
□
Proposition 7.25. For each n ∈ N, M ∈ N, and a ∈ Ω(n) ∩ D(M ), we have
∥qn (a)∥ ≤ (2M + 1)n .
Proof. This follows from the inequality ∥ξζ∥ ≤ 2∥ξ∥∥ζ∥ for every ζ, ξ ∈ C and induction on
n.
□
Proposition 7.26. Consider a = (an )n≥1 ∈ Ω, (tj )j≥1 a sequence in N and let (Tj )j≥1 be
the sequence given by Tj = t1 + . . . + tj . The following inequality holds for every n ∈ N:
6n−1 ∥qTn (a)∥ ≤ ∥qt1 (a1 , . . . , at1 )∥⋯∥qtn (aTn−1 +1 , . . . , aTn )∥.
Proof. This is an immediate consequence of Proposition 4.9.
Lemma 7.27. For each 0 < t < sB (M ), there are k0 ∈ N and c0 > 0 satisfying
µ(An (a)) ≤ c0 ∣An (a)∣t
for all k ∈ N≥k0 , n ∈ N≥nk , and a ∈ Dn .
□
56
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
Proof. Pick any 0 < t < sB (M ). Choose a positive number γ such that 1 < γ < ψ. Define ε
and ε̃ by
sB (M ) − t
log 2
ε∶ =
, ε̃∶ =
ε.
4
2 log(B) + 2 log(2M + 1)
Let N3 ∈ N≥N2 be such that
∣sn,B (M ) − sl,B (M )∣ < ε̃ for all n, l ∈ N≥N3 ,
t + 2ε < sn,B (M ) for all n ∈ N≥N3 .
Without any reference, we will use
t < t + ε < t + 2ε < sn,B (M ) ≤ 2
for all n ∈ N≥N3 .
Define κ7 ∶ = 81000 and let k0 ∈ N be such that every k ∈ N≥k0 +1 has the following properties:
(46)
mk ≥ N3 ,
(47)
nk
2ε
≤1+
,
mk
t + 2ε
(48)
k
log(B 2N3 (2M + 1)4N3 ) + log(68 ⋅ 350) − log κ25 − log κ(ŝ)
log κ7
1
+
≤
∑ nj
2kε log B
2kε log B
k − 1 j=k0 +1
Condition (48) is actually attained by every large k because (42) implies
1 k
∑ nj = ∞.
k→∞ k j=1
lim
For notational convenience, for any k ∈ N and any j ∈ {1, . . . , k}, we write qmj rather than
the cumbersome qmj (anj−1 +1 , . . . , anj −1 ). Lastly, for
c0 ∶ = (2M + 1)4nk0 B 4(n1 +...+nk0 ) ,
Proposition 7.25 guarantees
√
√
smj ,B (M )
t+ε
k0
1
5 2
5 2
)
≤ 1 ≤ c0 ∏ ( 2nj
) .
∏ nj ( mj
∥qmj ∥2
∥qmj ∥2
j=1 ⌊B ⌋ B
j=1 B
k0
(49)
Although all of these parameters depend on t, the order we used to define them is relevant.
We have the following dependencies: ε = ε(t), ε̃ = ε̃(ε), N3 = N3 (ε, ε̃), k0 = k0 (ε, N3 ), and
c0 = c0 (k0 ).
Take n ∈ N such that n ≥ nk0 +1 and a ∈ Dn . We now estimate µ(An (a)).
7.4.4. Case n = nk . For any a = (a1 , . . . , ank ) ∈ Dnk , the definition of µ and Proposition
7.23 yield
√
smj ,B (M )
1 k
1
5 2
)
µ(Ank (a)) ≤ k ∏ nj 2 ( mj
2 j=1 ⌊B ⌋ B ∥qmj ∥2
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
(50)
57
√
√
smj ,B (M )
smj ,B (M )
k
5 2
5 2
1 k0
1
1
(
= k ∏ nj 2 ( mj
)
)
∏
nj 2 B mj ∥q ∥2
2 j=1 ⌊B ⌋ B ∥qmj ∥2
mj
j=k0 +1 ⌊B ⌋
√
√
t+ε
t+2ε
k
1 k0
5 2
5 2
1
≤ k c0 ∏ ( 2nj
(
)
)
(by (46) and (49))
∏
nj 2 B mj ∥q ∥2
2
∥qmj ∥2
mj
j=1 B
j=k0 +1 ⌊B ⌋
√
√
t+2ε
t+ε
k
(5 2)2k0 k0
1
5 2
1
≤
c0 ∏ ( 2nj
(
)
)
∏
nj 2 B mj ∥q ∥2
2k
∥qmj ∥2
mj
j=1 B
j=k0 +1 ⌊B ⌋
1
50k0 k0
= k c0 ∏ 2nj
2
∥qmj ∥2 )t+ε
j=1 (B
√
t+2ε
5 2
(
)
∏
nj 2 B mj ∥q ∥2
mj
j=k0 +1 ⌊B ⌋
k
1
Let us focus on the last product:
√
√
t+2ε
t+2ε
k
k
5 2
1
5 2
22
(
)
≤ ∏
( mj
)
∏
nj 2 B mj ∥q ∥2
2nj
B ∥qmj ∥2
mj
j=k0 +1 ⌊B ⌋
j=k0 +1 B
√
t+2ε
k
k
1
5 2
)
= ∏ 4(
∏
2nj +mj (t+2ε)
∥qmj ∥2
j=k0 +1 B
j=k0 +1
√
t+2ε
k
k
1
5 2
)
≤ ∏ 4(
∏
(nj +mj )(t+2ε)
∥qmj ∥2
j=k0 +1 B
j=k0 +1
k
200
2(t+2ε)
j=k0 +1 ∥qmj ∥
≤ ∏
k
200
2(t+2ε)
j=k0 +1 ∥qmj ∥
≤ ∏
k
k
1
B (nj +mj )(t+2ε)
∏
j=k0 +1
k
1
∏
j=k0 +1
1
2(t+ε)
j=k0 +1 ∥qmj ∥
≤ 200k−k0 ∏
B 2nj (t+ε)
k
We substitute (51) in (50) to obtain
µ(Ank (a)) ≤
1
B 2nj (t+ε)
k
1
∏
B 2nj ε
j=k0 +1
.
k
k
50k0 k0
1
1
1
k−k0
c
200
∏
∏
∏
0
2n
2n
2n
k
2
t+ε
2
t+ε
j
j
2
∥qmj ∥ )
∥qmj ∥ ) j=k0 +1 B j ε
j=1 (B
j=k0 +1 (B
= 100k
≤ 100k
k
k
1
c0 k 0
1
1
∏ 2nj
∏
∏
2n
2n
k
2
t+ε
2
t+ε
j
0
4 j=1 (B ∥qmj ∥ ) j=k0 +1 (B ∥qmj ∥ ) j=k0 +1 B j ε
k
c0 k
1
1
∏
∏
2n
2n
k
2
t+ε
j
0
4 j=1 (B ∥qmj ∥ ) j=k0 +1 B j ε
k
ε
⎛ k
1
1 ⎞
≤ 100 c0 ∏ 2nj
∏
2
t+ε
∥qmj ∥ ) ⎝j=k0 +1 B 2nj ⎠
j=1 (B
k
(52)
j=k0 +1
1
2n
j
∥qmj ∥2 )t+ε
j=k0 +1 (B
= 200k−k0 ∏
(51)
k
∏
(by (47))
k
ε
⎛ k
1
1 ⎞
≤ 100 c0 ∏
∏
nj 2
2 t+ε
⎝j=k0 +1 B 2nj ⎠
j=1 (⌊B ⌋ ∥qmj ∥ )
k
58
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
Recall that for every j ∈ N we have ∥anj ∥ ≤ 2⌊B nj ⌋ + 1, so ∥anj ∥ ≤ 3⌊B nj ⌋ and
t+ε
k
k
1
32
=
)
(
∏
∏
nj 2
nj 2
2 t+ε
2
j=1 (⌊B ⌋ ∥qmj ∥ )
j=1 (3⌊B ⌋) ∥qmj ∥
t+ε
k
9
)
≤ ∏(
2
2
j=1 ∥anj ∥ ∥qmj ∥
2(t+ε)
k(t+ε)
≤9
1
)
( 2k−1
6
∥qnk (a)∥
(Proposition 7.26)
9t+ε k 2t
1
) 6
4t
6
∥qnk (a)∥2t+2ε
1
≤ 81k 64
.
∥qnk (a)∥2t+2ε
=(
Hence, plugging the previous inequality into (52), we have
ε
⎛ k
1 ⎞
1
µ(Ank (a)) ≤ c0 8100k 64 ∏
2n
j
⎝j=k0 +1 B ⎠ ∥qnk (a)∥2t+2ε
ε
⎛ k
1 ⎞ 1
∣Ank (a)∣t
≤ c0 8100 6
∏
⎝j=k0 +1 B 2nj ⎠ κt5
k 4
(Proposition 7.24)
ε
1 ⎛ k
1 ⎞
≤ c0 8100 6 t
∣Ank (a)∣t
∏
κ5 ⎝j=k0 +1 B 2nj ⎠
k 4
=
κk
c0 64 72
κ5
ε
⎛ k
1 ⎞
∣Ank (a)∣t
∏
2nj
B
⎝j=k0 +1
⎠
≤ c0 ∣Ank (a)∣t .
The last inequality follows from (48). This finishes the proof for the case n = nk . We state
the next corollary for reference.
Corollary 7.28. If r ∈ N≥k0 and a ∈ Dnr , then
√
smj ,B (M )
1
5 2
1 r
µ(Anr (a)) ≤ r ∏
(
)
2 j=k0 +1 ⌊B nj ⌋2 B mj ∥qmj ∥2
ε
≤
c0 64 κr7
⎛ r
1
1 ⎞
.
∏
2n
⎝j=k0 +1 B j ⎠ ∥qnr (a)∥2t+2ε
7.4.5. Case n = nk − 1. Take k ∈ Nk0 +1 , a ∈ Dnk −1 and let b ∈ D be such that ab ∈ Dnk .
Then, using Proposition 7.23 and Corollary 7.28,
µ(Ank −1 (a)) ≤ 14⌊B nk ⌋2 µ (Ank (ab))
√
√
smj ,B (M )
smk ,B (M )
⎞
14 ⎛ 1 k−1 1
5 2
5 2
≤ c0
(
)
(
)
∏
2 ⎝ 2k−1 j=1 ⌊B nj ⌋2 B mj ∥qmj ∥2
⎠ ∥qmk ∥2 B mk
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
59
ε
√
smk ,B (M )
⎛ k−1 1 ⎞
1
5 2
(
)
≤
∏
⎝j=k0 +1 B 2nj ⎠ ∥qnk−1 ∥2t+2ε ∥qmk ∥2 B mk
ε
√
t+2ε
⎛ k−1 1 ⎞
1
5 2
k−1 4
≤ 7c0 κ7 6
(
)
∏
⎝j=k0 +1 B 2nj ⎠ ∥qnk−1 ∥2t+2ε ∥qmk ∥2 B mk
4
7c0 κk−1
7 6
ε
⎛ k−1 1 ⎞
50
1
4
≤ 7c0 κk−1
∏
7 6
2n
2(t+ε)
2ε
j
∥qmk ∥ B mk (t+2ε)
⎝j=k0 +1 B ⎠ (∥qnk−1 ∥∥qmk ∥)
ε
⎛ k−1 1 ⎞
50
1
∏
2n
2(t+ε)
m
B k (t+2ε)
⎝j=k0 +1 B j ⎠ (∥qnk−1 ∥∥qmk ∥)
≤
4
7c0 κk−1
7 6
≤
4
7c0 κk−1
7 50 ⋅ 6
ε
2(t+ε)
⎛ k−1 1 ⎞
6
(
)
∏
2n
j
⎝j=k0 +1 B ⎠ ∥qnk −1 ∥
ε
1
B mk (t+2ε)
(Proposition 4.9)
⎛ k−1 1 ⎞
1
1
8
≤ c0 κk−1
350
⋅
6
∏
7
2n
2(t+ε)
j
B tnk
⎝j=k0 +1 B ⎠ ∥qnk −1 ∥
ε
≤
8
c0 κk−1
7 350 ⋅ 6
⎛ k−1 1 ⎞
1
∏
⎝j=k0 +1 B 2nj ⎠ (B nk ∥qnk −1 ∥2 )t
ε
⎛ k−1 1 ⎞
1
8
≤ c0 κk−1
∏
7 350 ⋅ 6
2n
n
j
k
⎝j=k0 +1 B ⎠ (⌊B ⌋∥qnk −1 ∥2 )t
ε
⎛ k−1 1 ⎞ 1
∣Ank −1 (a)∣t
∏
⎝j=k0 +1 B 2nj ⎠ κt5
≤
8
c0 κk−1
7 350 ⋅ 6
≤
8 −2
c0 κk−1
7 350 ⋅ 6 κ5
≤ c0 ∣Ank −1 (a)∣t
(Proposition 7.24)
ε
⎛ k−1 1 ⎞
∣Ank −1 (a)∣t
∏
2n
j
⎝j=k0 +1 B ⎠
7.4.6. Case nk−1 < n < nk . Put
l′ ∶ = nk − n − 1 and l∶ = n − nk−1 .
Letting b run along the set {c ∈ R(l′ ) ∶ ac ∈ Dnk −1 }, applying Corollary 7.28, and writing
a[α, β] = (aα , aα+1 , . . . , aβ ) for natural numbers α < β, we obtain
µ(An (a)) = ∑ µ(Ank −1 (ab))
b
√
smk ,B (M )
5 2
= ∑ µ(Ank−1 (a1 , . . . , ank−1 )) ( m
)
B k ∥qmk (a[nk−1 + 1, n]b)∥2
b
ε
√
smk ,B (M )
k−1
⎞
⎛
1
5
2
1
k−1 4
≤ c0 κ7 6
)
∏
∑(
⎝j=k0 +1 B 2nj ⎠ ∥qnk−1 (a)∥2t+2ε b B mk ∥qmk (a[nk−1 + 1, n]b)∥2
ε
√
smk ,B (M )
⎛ k−1 1 ⎞
1
(5 2)2
k−1 4
≤ c0 κ7 6
)
∏
∑(
⎝j=k0 +1 B 2nj ⎠ ∥qnk−1 (a)∥2t+2ε b B mk ∥ql (a[nk−1 + 1, n])∥2 ∥ql′ (b)∥2
60
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
ε
√
√
smk ,B (M )
⎛ k−1 1 ⎞
5 2
(5 2)smk ,B(M )
≤
)
∑(
∏
⎝j=k0 +1 B 2nj ⎠ ∥qnk−1 (a)∥2t+2ε ∥ql (a[nk−1 + 1, n])∥2smk ,B (M ) b B mk ∥ql′ (b)∥2
ε
√
smk ,B (M )
2t+2ε
⎛ k−1 1 ⎞
1
5 2
k−1 4
≤ c0 50κ7 6
(
)
)
∏
∑ ( mk
B ∥ql′ (b)∥2
⎝j=k0 +1 B 2nj ⎠ ∥qnk−1 (a)∥ ∥ql (a[nk−1 + 1, n])∥
b
ε
√
smk ,B (M )
2t+2ε
⎛ k−1 1 ⎞
6
5 2
k−1 4
(
)
)
≤ c0 50κ7 6
∑ ( mk
∏
B ∥ql′ (b)∥2
⎝j=k0 +1 B 2nj ⎠ ∥qn (a)∥
b
ε
√
smk ,B (M )
⎛ k−1 1 ⎞
1
5 2
k−1 8
≤ c0 50κ7 6
)
∏
∑(
⎝j=k0 +1 B 2nj ⎠ ∥qn (a)∥2t+2ε b B mk ∥ql′ (b)∥2
4
c0 κk−1
7 6
(53)
ε
≤
8
c0 50κk−1
7 6
⎛ k−1 1 ⎞
1
gl′ (smk ,B (M ); D(M ), B).
∏
⎝j=k0 +1 B 2nj ⎠ ∥qn (a)∥2t+2ε
Then, we must give an upper estimate of gl (smk ,B (M ); D(M ), B) in order to bound (53).
Since l + l′ = mk , in view of (34), we have
κ(ŝ)gl′ (smk ,B (M ); D(M ), B)gl (smk ,B (M ); D(M ), B) ≤ gmk (smj ,B (M ); D(M ), B) = 1,
so
1
,
κ(ŝ)gl (smk ,B (M ); D(M ), B)
and we are after a lower bound of gl (smj ,B (M ); D(M ), B). We will get it taking two cases
into account: l ≤ N3 and l > N3 . If l ≤ N3 , then
√
smk ,B (M )
5 2
gl (smk ,B (M ); D(M ), B) = ∑ ( l
)
2
c∈∆l (M ) B ∥ql (c)∥
gl′ (smk ,B (M ); D(M ), B) ≤
1
≥
1
B N3 smk ,B (M ) c∈∆l (M ) ∥ql (c)∥2smk ,B (M )
∑
1
≥
B 2N3 (2M
+ 1)4N3
and, by (53),
ε
8 ⎛ k−1
c0 50κk−1
1 ⎞
1
7 6
B 2N3 (2M + 1)4N3
µ(An (a)) ≤
∏
2n
j
κ(ŝ)
⎝j=k0 +1 B ⎠ ∥qn (a)∥2t+2ε
ε
8 ⎛ k−1
t+ε
c0 50κk−1
1 ⎞ 2N3
7 6
4N3 ∣Cn (a)∣
≤
B
(2M
+
1)
∏
κ(ŝ)
κt+ε
⎝j=k0 +1 B 2nj ⎠
5
ε
8 ⎛ k−1
t+ε
c0 50κk−1
1 ⎞ 2N3
7 6
4N3 ∣An (a)∣
≤
B
(2M
+
1)
∏
κ(ŝ)
κt+ε
⎝j=k0 +1 B 2nj ⎠
5
ε
8 2
⎛ k−1 1 ⎞
c0 50κk−1
7 6 κ5 2N3
≤
B (2M + 1)4N3 ∏
∣An (a)∣t+ε
κ(ŝ)
⎝j=k0 +1 B 2nj ⎠
≤ c0 ∣An (a)∣t .
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
61
Now assume that l > N3 . The definitions of N3 and ε̃ > 0 yield
√
smk ,B (M )
5 2
)
gl (smj ,B (M ); D(M ), B) = ∑ ( l
2
c∈∆l (M ) B ∥ql (c)∥
√
sl,B (M )+ε̃
5 2
≥ ∑ ( l
)
2
c∈∆l (M ) B ∥ql (c)∥
√
sl,B (M )
1
5 2
≥ ∑
(
)
l
2l ε̃ B l ∥q (c)∥2
l
c∈∆l (M ) (B (2M + 1) )
1
=
(B l (2M
=
(B l (2M
+ 1)2l )ε̃
gl (sl,B (M ); D(M ), B)
1
+ 1)2l )ε̃
≥
1
1
≥ nε .
lε
γ
γ
The previous inequality and (53) imply
ε
8 ⎛ k−1
c0 50κk−1
1 ⎞
γ nε
7 6
µ(An (a)) ≤
∏
κ̂(s)
⎝j=k0 +1 B 2nj ⎠ ∥qn (a)∥2t+2ε
ε
8 ⎛ k−1
c0 50κk−1
γ nε
1 ⎞
1
7 6
≤
∏
2n
2ε
j
κ̂(s)
⎝j=k0 +1 B ⎠ ∥qn (a)∥ ∥qn (a)∥2t
ε
8 ε ⎛ k−1
c0 50κk−1
1 ⎞ γ nε
7 6 ψ
∣An (a)∣t
≤
∏
t
2nj
nε
κ(ŝ)κ5
B
ψ
⎝j=k0 +1
⎠
≤ c0 ∣An (a)∣t .
□
7.4.7. Estimate of µ(D(z, r)). In the context of regular continued fractions or Lüroth series, the measure of the balls is bounded by estimating the distance between two different
fundamental sets (see [33, Section 3.2]). The strategy must be modified for Hurwitz continued fractions. Let us illustrate why with a simple example. When M = 5 and n1 ≥ 3,
both a = 3 + 3i and 2 + 3i belong to D1 but A1 (a) ∩ A1 (b) ≠ ∅ (see Figure 11). We may
thus loose any hope of obtaining a non-trivial lower bound of the distance between any
two fundamental sets of the same level. This is when our discussion on the geometry of
Hurwitz continued fractions pays off.
Let ρ̃ and ρ be as in Theorems 5.3 and 5.14, respectively. Set ρ′ ∶ = min{ρ̃, ρ}. For each
a ∈ D N such that pref(a; n) ∈ Dn for all n ∈ N, define
Gn (a)∶ =
ρ′
∣An (a)∣,
κ3
where κ3 is as in Corollary 5.10. Let r0 be
r0 ∶ = min {Gn (a) ∶ n ∈ {1, . . . , nk0 }, a ∈ Dn } .
62
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
2 + 3i
3 + 3i
Figure 11. The sets ι[A1 (2 + 3i)] and ι[A1 (3 + 3i)].
Lemma 7.29. There is a constant ν = ν(M, B, t) > 0 such that µ(D(z; r)) ≤ νrt for every
z ∈ FM (B) and every positive number r < r0 .
Proof. Take a real number r with 0 < r < r0 , z ∈ FM (B), and a ∈ Dn such that z ∈
An (a1 , . . . , n) for all n ∈ N. Let n ∈ N be such that
Gn+1 (a1 , . . . , an , an+1 ) ≤ r < Gn (a1 , . . . , an ).
We examine three different cases for n.
7.4.8. Case n = nk − 1. First, let us further assume that either Fn (a) = F or F = F1 (1 + 2i)
or some of their images under any f ∈ Di8 . Then, the disc D(z; r) intersects exactly one
fundamental set of order n. We now consider two possibilites:
r ≤ κ3 ∣Ank (a1 , . . . , ank )∣ or κ3 ∣Ank (a1 , . . . , ank )∣ < r.
In the first option, D(z; r) intersects at most six cylinders of level nk , which are Cn (pref(a; n))
and five of its neighbors, so it intersects at most six fundamental sets of level nk . This
gives
κ3
µ(D(z; r)) ≤ 6κ5 c0 ∣Ank (a)∣t < 6κ5 c0 rt .
ρ̃
Now assume that κ3 ∣Ank (a1 , . . . , ank )∣ < r. We claim that every d ∈ D such that (a1 , . . . , ank −1 , d) ∈
Dnk satisfies
(54)
(55)
m (Cnk (a1 , . . . , ank −1 , d)) ≥
κ1
1
,
4
5
6 2 ∥qnk (a1 , . . . , ank −1 )∥4 ⌊B nk ⌋2
1 ∣Ank (a1 , . . . , ank −1 , ank )∣
≤
≤ ν1
ν1
∣Ank (a1 , . . . , ank −1 , d)∣
for some universal constant ν1 > 1. By Proposition (4.7) and the choice of n1 , we have
∣Ank (a1 , . . . , ank −1 , ank )∣ ∣Cnk (a1 , . . . , ank −1 , ank )∣
=
∣Ank (a1 , . . . , ank −1 , d)∣
∣Cnk (a1 , . . . , ank −1 , d)∣
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
63
κ1 ∣qnk (a1 , . . . , ank −1 , d)∣2
2 ∣qnk (a1 , . . . , ank −1 , ank )∣2
κ1 ⋅ 25 ⋅ 9 ∣qnk −1 (a1 , . . . , ank −1 )∣2 ∣d∣2
≤
2
∣qnk −1 (a1 , . . . , ank −1 )∣2 ∣ank ∣2
225κ1 ∣d∣2
=
2 ∣ank ∣2
∥d∥2
≤ 675κ1 .
≤ 225κ1
∥ank ∥2
≤
The last estimate holds because we have
1
3
≤
∥d∥
∥ank ∥
≤ 3 and ν1 ∶ = 675κ1 works.
Write ν2 ∶ = 1 + κν13 . We conclude from (55) that
Ank (a1 , . . . , ank −1 , d) ∩ D(z; r) ≠ ∅ implies Ank (a1 , . . . , ank −1 , d) ⊆ D(z; ν2 r).
Therefore, if
N∶ = # {d ∈ Dnk ∶ D(z; r) ∩ Ank (d) ≠ ∅} ,
we have
N ≤ 64 25 πν22 ∥qnk (a1 , . . . , ank −1 )∥4 ⌊B nk ⌋4 r2 ,
because, by Proposition 4.8,
κ1
4
nk (a1 , . . . , ank −1 , d)∥
κ1 6−4
≥ 2
2 ∥qnk −1 (a1 , . . . , ank −1 )∥4 ∥d∥4
κ1 6−4
≥
2
22 ∥qnk −1 (a1 , . . . , ank −1 )∥4 24 ⌊B nk ⌋2 (1 + 2⌊B1nk ⌋ )
m (Cnk (a1 , . . . , ank −1 , d)) ≥
=
We thus conclude
22 ∥q
1
κ1
.
4
7
6 2 ∥qnk −1 (a1 , . . . , ank −1 )∥4 ⌊B nk ⌋2
µ (Ank −1 (a1 , . . . , ank −1 ))
N
#{d ∈ D ∶ ad ∈ Dnk }
64 25 πν22
≤
µ (Ank −1 (a1 , . . . , ank −1 )) ∥qnk (a1 , . . . , ank −1 )∥4 ⌊B nk ⌋4 r2
2⌊B nk ⌋2
µ (D(z; r)) ≤
(56)
= 64 24 πν22 µ (Ank −1 (a1 , . . . , ank −1 )) ∥qnk (a1 , . . . , ank −1 )∥4 ⌊B nk ⌋2 r2 .
Moreover, since Ank −1 (a) is the only fundamental set of level nk − 1 that D(z; r) intersects,
we have
(57)
µ (D(z; r)) = µ (D(z; r) ∩ Ank −1 (a1 , . . . , ank −1 )) ≤ µ (Ank −1 (a1 , . . . , ank −1 )) .
Combining inequalities (56) and (57) and recalling that t < 2, we conclude that for some
constant ν3 we have
µ (D(z; r)) ≤ µ (Ank −1 (a1 , . . . , ank −1 )) min {1, 64 24 πν22 ∥qnk (a1 , . . . , ank −1 )∥4 ⌊B nk ⌋2 r2 }
64
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
t
t
≤ µ (Ank −1 (a1 , . . . , ank −1 )) 11− 2 (64 24 πν22 ∥qnk −1 (a1 , . . . , ank −1 )∥4 ⌊B nk ⌋2 r2 ) 2
≤ µ (Ank −1 (a1 , . . . , ank −1 )) ∥qnk −1 (a1 , . . . , ank −1 )∥2t ⌊B nk ⌋t rt
≤ c0 ν3 r t .
We now appeal to the second geometric construction. Assume that Fn (a) is equal to either
Fn (2), F1 (1 + i) or any of their images under elements of Di8 . Then, D(z; r) intersects at
most two fundamental sets of level nk −1 and they are subsets of neighboring nk−1 cylinders,
say Ank −1 (a1 , . . . , ank −1 ) and Ank −1 (b). Hence,
µ (D(z; r)) = µ (D(z; r) ∩ An (a1 , . . . , ank −1 )) + µ (D(z; r) ∩ An (b1 , . . . , bnk −1 )) .
Propping on the above argument we may then say that
µ (D(z; r)) ≤ ν3 c0 ∣Ank −1 (a1 , . . . , akn −1 )∣t + ν3 c0 ∣Ank −1 (b1 , . . . , bkn −1 )∣t
≤ c0 ν3 (1 + κ3 )rt .
7.4.9. Case n = nk − 2. Observe that for any b ∈ Dnk −2 and any c, d ∈ D such that ac, ad ∈
Dnk −1 , the definition of µ gives
smk ,B (M )
∥qmk (bd)∥2
µ (Ank −1 (bc))
=(
)
µ (Ank −1 (bd))
∥qmk (bc)∥2
≤ (1800
smk ,B (M )
∥d∥2
)
∥c∥2
smk ,B (M )
≤ (1800M 2 )
≤ 18002 M 4 .
(Proposition 7.15)
As a consequence, estimating #{x ∈ D ∶ bc ∈ Dnk −1 },
µ (Ank −2 (b)) ≤ 8κ5 18002 M 4 (2M + 1)2 c0 µ (Ank −1 (bc))
Finally, since D(z; r) intersects at most six fundamental sets of level nk − 2, one of them
being Ank −2 (a1 , . . . , ank −2 ) and the others are contained in the neighboring cylinders of
Cnk −2 (a1 , . . . , ank −2 ), we have
µ (D(z; r)) ≤ 48κ5 18002 M 4 (2M + 1)2 c0 rt .
7.4.10. Case nk ≤ n ≤ nk+1 − 3. For any b = (b1 , . . . , bn ) ∈ Dn and any c ∈ D, we may
express µ(An+1 (bc)) as follows (these sums run along the words c ∈ R(nk+1 − n − 1) such
that b c c ∈ Dnk+1 −1 ) ∶
µ (An+1 (bc)) = ∑ µ (Ank+1 −1 (bcc))
c
√
snk −n−2,B (M )
5 2
= ∑ µ (Ank+1 −1 (pref(b; nk−1 ))) ( m
)
B k+1 ∥qmk+1 (bnk , . . . , bn , c, c)∥2
c
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
65
√
µ (Ank+1 −1 (pref(b; nk−1 ))) (5 2)snk −n−2,B (M )
1
=
∑
2snk −n−2,B (M )
B mk+1 snk −n−2,B (M )
c ∥qmk+1 (bnk , . . . , bn , c, c)∥
On the one hand,
∑
c
1
≤
∥qmk+1 (bnk , . . . , bn , c, c)∥2snk −n−2,B (M )
2snk −n−2,B (M )
50
)
≤(
∥qn−nk +1 (bnk , . . . , bn )∥∥c∥
≤(
c
2snk −n−2,B (M )
50
)
κ5 ∥qn−nk +1 (bnk , . . . , bn )∥∥c∥
c
∑ ∣Cnk+1 −n−2 (c)∣snk −n−2,B (M ) .
c
On the other hand,
∑
1
∥qnk+1 −1−n (c)∥2snk −n−2,B (M )
∑
1
≥
∥qmk+1 (bnk , . . . , bn , c, c)∥2snk −n−2,B (M )
1
≥(
≥(
36∥qn−nk +1 (bnk , . . . , bn )∥∥c∥
2snk −n−2,B (M )
)
1
)
72∥qn−nk +1 (bnk , . . . bn )∥∥c∥
2snk −n−2,B (M )
∑
c
1
∥qnk+1 −1−n (c)∥2snk −n−2,B (M )
∑ ∣Cnk+1 −n−2 (c)∣snk −n−2,B (M ) .
c
Hence, if d ∈ D is such that bd also belongs to Dn+1 and we take ν5 ∶ = ν5 (M, B)∶ =
sup{sn,B (M ) ∶ n ∈ N} (see Proposition 7.15), then
∥d∥
µ (An+1 (bc))
≤ (3600
)
µ (An+1 (bd))
∥c∥
2ν
≤ 3600 M
2ν
2snk −n−2,B (M )
∑c ∣Cnk −n−2 (c)∣
∑d ∣Cnk −n+1 (d)∣
∑c ∣Cnk −n−2 (c)∣
snk −n−2,B (M )
∑d ∣Cnk −n+1 (d)∣
≤ 4 ⋅ 36002ν5 M 2ν5 .
snk −n−2,B (M )
snk −n−2,B (M )
snk −n−2,B (M )
The last inequality is due to the symmetries of cylinders and considering Fn+1 (bc) = F
in the numerator and Fn+1 (bc) = F1 (1 + i) in the denominator. We may now argue as in
the case n = nk − 2 to conclude the existence of a constant ν6 = ν6 (M, B, t) > 0 such that
µ (D(z; r)) ≤ ν6 rt .
□
Proof of Theorem 1.2. Lower bound. For any M ∈ N≥4 , the Mass Distribution Principle
applied to FM (B) tells us that sB (D(M )) ≤ dimH (FM (B)). Therefore, by Lemmas 7.22
and Lemma 7.15,
s = lim sB (M ) ≤ dimH (F (B)).
M →∞
□
The proof of Theorem 1.2 is flexible enough for us to assume that the terms of the
sequence (nj )j≥1 lie in a prescribed infinite subset of N. Let L ⊆ N be infinite and, for
66
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
each B > 1, define
E(B; L )∶ = {z ∈ F ∶ ∥an ∥ ≥ B n for infinitely many n ∈ L } .
Corollary 7.30. If L ⊆ N is infinite and B > 1, then dimH (E(B; L )) = sB .
8. A complex Luczak Theorem
This section is dedicated to the proof of Theorem 1.4. First, observe that we can replace
the complex absolute value ∣ ⋅ ∣ with ∥ ⋅ ∥. Indeed, let b, c > 1 be given. Define the sets
̃ c)∶ = {ξ ∈ F ∶ cbn ≤ ∥an (ξ)∥ for all n ∈ N},
Ξ(b,
n
Ξ(b, c)∶ = {ξ ∈ F ∶ cb ≤ ∥an (ξ)∥ for infinitely many n ∈ N}.
Since ∥z∥ ≤ ∣z∣ for all z ∈ C, we have
̃ c) ⊆ {ξ ∈ F ∶ cbn ≤ ∣an (ξ)∣ for all n ∈ N}.
Ξ(b,
Consider ε =
c−1
2
> 0. Then, c − ε > 1 and every large n ∈ N satisfies
1 n
n
(c − ε)b ≤ √ cb ,
2
therefore
n
{ξ ∈ F ∶ cb ≤ ∣an (ξ)∣ for infinitely many n ∈ N} ⊆ Ξ(b, c − ε).
Hence, Theorem 1.4 is a consequence of Theorem 8.1 below.
Theorem 8.1. If b, c > 1, then
̃ c)) =
dimH (Ξ(b, c)) = dimH (Ξ(b,
2
.
b+1
̃ c) is contained in Ξ(b, c). Then, in order to prove Theorem 8.1, it
It is clear that Ξ(b,
suffices to show
2
̃ c)) and dimH (Ξ(b, c)) ≤ 2 .
≤ dimH (Ξ(b,
b+1
b+1
8.1. Upper bound of Hausdorff dimension. The overall strategy resembles T. Luczak’s
proof. Nevertheless, as should be clear by now, Hurwitz continued fractions present certain
difficulties that prevent the arguments for real continued fractions to carry over into the
complex plane.
Proposition 8.2. Given ξ = [0; a1 , a2 , a3 , . . .] ∈ F and n ∈ N, we have
∣ξ −
pn
2
∣<
.
qn
∥qn ∥∥qn+1 ∥
Proof. This follows from
∣ξ −
pn+1
pn+1 pn
1
2
2
pn
1
∣ ≤ ∣ξ −
∣+∣
− ∣<
+
≤
≤
.
2
qn
qn+1
qn+1 qn
∣qn+1 ∣
∣qn+1 qn ∣ ∣qn ∣∣qn+1 ∣ ∥qn ∥∥qn+1 ∥
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
67
□
Let b, c be real numbers > 1. Let α and d be two real numbers such that
2
2
<
< α.
1+b 1+d
In what follows, given a complex number ξ = [0; a1 , a2 , a3 , . . .] ∈ F ∖ Q(i), we write qn (ξ)∶ =
qn (a1 , . . . , an ) for all n ∈ N. For each n ∈ N and each ξ ∈ F define
1<d<b
In,ξ ∶ = D (
and
pn (ξ)
2
;
),
qn (ξ) ∥qn (ξ)∥1+d
Kn,ξ ∶ = D (
2
pn (ξ)
;
),
qn (ξ) ∥qn (ξ)∥2(1+d)
and, for each q ∈ N,
dn
Iq ∶ = {In,ξ ∶ ξ ∈ F, q = ∥qn (ξ)∥ ≥ c 3 } ,
Kq ∶ = {Kn,ξ ∶ ξ ∈ F, ∥qn (ξ)∥ = q} .
Proposition 8.3. If ξ ∈ Ξ(b, c), then there are infinitely many n ∈ N such that
n
∥qn+1 (ξ)∥ ≥ max {∥qn (ξ)∥d , cd }
(58)
Proof. Given m ∈ N, we can pick k ∈ N>m such that
∥qm (ξ)∥ < cb
(59)
k dm−k
,
and
k
cb < ∥ak (ξ)∥.
Indeed, the first condition is attained for every large k because b > d > 1. The second condition is true for infinitely many natural numbers by the definition of Ξ(b, c). Furthermore,
the strict growth of the continuants allows us to further impose the condition
√
2
∣qk ∣
√ (1 −
) > 1.
2
2
Let fk ∶ N → R be the function given by
k dN −k
fk (N ) = cb
for all N ∈ N.
Thus, the first inequality in (59) is ∥qm (ξ)∥ < fk (m). Moreover, by Proposition 7.2,
√
∣qk ∣
2
∥qk (ξ)∥ ≥ ∣ak ∣ √ (1 −
) ≥ ∥ak ∥ > fk (k).
2
2
Let n ∈ {m, . . . , k − 1} be the largest integer less than k such that ∥qn (ξ)∥ < fk (n), then
d
∥qn+1 (ξ)∥ ≥ fk (n + 1) = (fk (n)) ≥ ∥qn (ξ)∥d .
n+1
Finally, b > d yields bk dn−k > dn , which implies fk (n)d > cd
d
and
n+1
∥qn+1 (ξ)∥ ≥ (fk (n)) ≥ max {∥qn (ξ)∥d , cd
}.
□
Lemma 8.4. We have
∞
∞
Ξ(b, c) ⊆ ⋂ ( ⋃ Iq ∪ ⋃ Kq ) .
m∈N
q=m
q=m
68
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
Proof. Pick any ξ ∈ Ξ(b, c) and m ∈ N. Take n ∈ N≥m satisfying (58) and q∶ = ∥qn (ξ)∥.
We may assume that q > n ≥ m, because (∥qj (ξ)∥)j≥0 grows exponentially. There are two
possibilities:
dn
dn
∥qn (ξ)∥ ≥ c 3
or ∥qn (ξ)∥ < c 3 .
If the first inequality holds, then, by Proposition 8.2,
∣ξ −
pn (ξ)
2
2
∣<
≤
qn (ξ)
∥qn (ξ)∥∥qn+1 (ξ)∥ ∥qn (ξ)∥1+d
and ξ ∈ In,ξ , thus ξ ∈ Iq . When the second inequality holds, the choice of n yields
∥qn (ξ)∥1+2d < c
dn
(1+2d)
3
< cd
n+1
≤ ∥qn+1 (ξ)∥.
Then, by Lemma 8.2, we have
∣ξ −
pn (ξ)
2
2
∣<
≤
,
qn (ξ)
∥qn (ξ)∥∥qn+1 (ξ)∥ ∥qn (ξ)∥2(1+d)
which means ξ ∈ Kn,ξ , thus ξ ∈ Kq .
□
√
Set β∶ = 5 2. For each m, k ∈ N, define
k
Cas(m, k)∶ = {a = (a1 , . . . , ak ) ∈ D k ∶ ∏ ∥aj ∥ ≤ mβ k−1 } .
j=1
Proposition 8.5. Given m, n ∈ N, let a ∈ D be such that ∥a∥ ≤ mβ k−1 . For any b ∈ D k−1 ,
we have ab ∈ Cas(m, k + 1) if and only if b ∈ Cas (⌊ mβ
∥a∥ ⌋ , k).
Proof. Write b = (b1 , . . . , bk ). Then, ab ∈ Cas(m, k + 1) if and only if
k
∏ ∥bj ∥ ≤
j=1
β k−1 βm
,
∥a∥
which is equivalent to b ∈ Cas (⌊ βm
∥a∥ ⌋ , k).
□
Proposition 8.6. If m, k ∈ N and m ≥ 2, then
Cas(m, k + 1) =
Furthermore,
⋃
∥a∥≤β k m
Cas (⌊
⌊β k m⌋
βm
⌋ , k) .
∥a∥
# Cas(m, k + 1) ≤ 8 ∑ j # Cas (⌊
j=1
βm
⌋ , k) .
j
Proof. The first assertion is obvious from the definition of Cas( ⋅ , ⋅ ) and Proposition 8.5.
The second assertion holds, because # {a ∈ D ∶ ∥a∥ = j} = 8j for all j ∈ N.
□
Proposition 8.7. For m ∈ N≥2 and n ∈ N, we have
n−1
# Cas(m, n) ≤ m2 8n β 2(n−1) (1 + n log β + log m)
.
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
69
Proof. The proof is by induction on n. For all m ∈ N we have # Cas(m, 1) ≤ 8m2 . This is
obvious when m = 1. If m ≥ 2, then
1
) < 8m2 .
2m
# Cas(m, 1) = (2m + 1)2 − 5 ≤ 4m2 (1 +
Assume that the next inequality is true for any n = N ∈ N and any m ∈ N:
N −1
# Cas(m, N ) ≤ m2 8N β 2(N −1) (1 + N log β + log m)
.
Then, for n = N + 1,
⌊β N m⌋
# Cas(m, N + 1) ≤ 8 ∑ j # Cas (⌊
j=1
⌊β N m⌋
≤8 ∑ j
j=1
N +1 2N
=8
β
N +1 2N
<8
β
βm
⌋ , N)
j
β 2 m2 N 2(N −1)
βm N −1
8
β
(1
+
N
log
β
+
log
)
j2
j
2
m
⌊β N m⌋
∑
j=1
βm N −1
1
(1 + N log β + log
)
j
j
N −1
2
m (1 + (N + 1) log β + log m)
N −1
< 8N +1 β 2N m2 (1 + (N + 1) log β + log m)
N −1
< 8N +1 β 2N m2 (1 + (N + 1) log β + log m)
⌊β N m⌋
1
j
∑
j=1
(log(β N m) + 1)
(1 + (N + 1) log β + log m)
= 8N +1 β 2N m2 (1 + (N + 1) log β + log m) .
N
□
Proposition 8.8. If m, n ∈ N and m ≥ 2, then
n
n
# ⋃ Cas(m, k) ≤ 8m2 (8β 2 (1 + (n − 1) log β + log m)) .
k=1
Proof. By Proposition 8.7,
n
n
k=1
k=1
# ⋃ Cas(m, k) ≤ ∑ 8k β 2(k−1) m2 (1 + k log β + log m)k−1
n
k−1
= 8m2 ∑ (8β 2 (1 + k log β + log m))
k=1
n
k−1
≤ 8m2 ∑ (8β 2 (1 + n log β + log m))
k=1
Then, considering τ = 8β 2 (1 + n log β + log m) > 1,
n
n
k=1
k=1
# ⋃ Cas(m, k) ≤ 8m2 ∑ τ k−1
= 8m2 (
τn − 1
)
τ −1
≤ 8m2 τ n
.
70
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
n
= 8m2 (8β 2 (1 + n log β + log m)) .
□
Proposition 8.9. If m > c, then ⋃m
q=1 Iq contains at most
logd (3 logc m)+1
8m2 (8β 2 (1 + logd (3 logc m) log β + log m))
discs.
Proof. Consider n = ⌊logd (3 logc m)⌋. For every k ∈ {1, . . . , n} and a = (a1 , . . . , ak ) ∈ Ω(k),
by Proposition 4.9 we have
1 k
∏ ∥aj ∥ ≤ ∥qk (a)∥.
β k−1 j=1
Therefore, the set
dk
{a ∈ Ω(k) ∶ 1 ≤ k ≤ n, c 3 ≤ ∥qk (a)∥ ≤ m}
(60)
is contained in
k
n
j=1
k=1
{a ∈ D k ∶ 1 ≤ k ≤ n, ∏ ∥aj ∥ ≤ β k−1 m} = ⋃ Cas(m, k).
dk
By the choice of n, the inequality c 3 ≤ m implies dk log c ≤ 3 log m and, thus, k ≤ n. Then,
since each Ik,ξ ∈ ⋃m
q=1 Iq determines a single word a ∈ Ω(k) contained in the set in (60), we
get
m
n
# ⋃ Iq ≤ # ⋃ Cas(m, k)
q=1
k=1
n
≤ 8m2 (8β 2 (1 + n log β + log m))
logd (3 logc m)
≤ 8m2 (8β 2 (1 + n log β + log m))
logd (3 logc m)
≤ 8m2 (8β 2 (1 + logd (3 logc m) log β + log m))
.
□
Given a countable collection {Aq ∶ q ∈ N} of subsets of C, we write
Λα ({Aq ∶ q ∈ N})∶ = ∑ ∣Aq ∣α .
q≥1
Proposition 8.10. We have
∞
∞
lim Λα ( ⋃ Iq ∪ ⋃ Kq ) = 0.
m→∞
q=m
q=m
Proof. Let us show first that
(61)
∞
lim Λα ( ⋃ Iq ) = 0.
m→∞
q=m
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
Fix r ∈ N such that r ≥ max{2, c} and define c1 ∶ = logd logc r + log 3 and c2 =
arbitrary j ∈ N, then
logd (3 logc (rj )) = c1 + c2 log j.
71
1
log d .
Pick an
Observe that for all q ∈ {rj−1 , . . . , rj } and each Il,ζ ∈ Iq , we have
∣Il,ζ ∣ =
4
4
≤ (j−1)(1+d) .
1+d
∥ql (ζ)∥
r
As a consequence, using Proposition 8.9 with m = rj , we get
rj
⎞
⎛ rj
4α
Λα ⋃ Iq ≤ (j−1)(d+1)α # ⋃ Iq
⎝q=rj−1 ⎠ r
q=1
j
4α rα(d+1) 2j
2
j
j logd (3 logc r )
(8β
8r
(1
+
log
(3
log
r
)
log
β
+
log
r
))
d
c
rα(d+1)j
4α rα(d+1) 8
c2 log j+c1
= j(α(d+1)−2) (8β 2 (1 + logd (3 logc rj ) log β + j log r))
.
r
≤
Then, for
c3 ∶ = 8β 2 (1 + c1 ),
c5 ∶ = 8β 2 log r,
c4 ∶ = 8β 2 c2 ,
and
ε∶ = α(1 + d) − 2 > 0,
we have
Λα
⎛ rj
⎞ 4α rα(d+1) 8
c log j+c1
⋃ Iq ≤ j(α(d+1)−2) (c5 j + c4 log j + c3 ) 2
⎝q=rj−1 ⎠ r
= 4α rα(d+1) 8
(c5 j + c4 log j + c3 )
(rε )j
and, since,
c2 log j+c1 +1
(c5 j + c4 log j + c3 )
lim
j→∞
(rε )j
we conclude
∞
∑ Λα
j=1
c2 log j+c1
,
= 0,
⎛ rj
⎞
⋃ Iq < ∞
⎝q=rj−1 ⎠
and (61) follows.
We now show that
(62)
lim Λα ( ⋃ Kq ) = 0.
n→∞
q≥m
For each q ∈ N, the collection Kq has at most 8q 2 discs of diameter 4q −2(1+d) , so
Λα (Kq ) ≤
4α ⋅ 8q 2
.
q 2(1+d)α
Thus, for any m ∈ N and ε as above,
Λα ( ⋃ Kq ) ≤ 4α ⋅ 8 ∑
q≥m
which gives (62).
q≥m
1
q 2(1+d)α−2
= 4α ⋅ 8 ∑
q≥m
1
q 2+4ε
< ∞,
□
72
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
8.1.1. Proof of Theorem 8.1. Upper bound.
Proof of Theorem 1.2. Upper bound. By Proposition 8.4 and taking α arbitrarily close to
1/(1 + b), we arrive at
2
dimH (Ξ(b, ξ)) ≤
.
1+b
□
8.2. Lower bound of Hausdorff dimension. We obtain the desired bound applying
the Mass Distribution Principle. Take M0 ∈ N such that
√
√
⎛ 8 (1 + 22 ) (2 + 22 ) √2 ⎞
⎟.
M0 ≥ logb logc ⎜
+
√
2 ⎠
κ1 (1 − 22 )
⎝
Note that M0 ≥ logb logc (3). For every n ∈ N, define
̃ n ∶ = {d ∈ D ∶ cbM0 +n+1 ≤ ∥aj ∥ ≤ 2cbM0 +n+1 } ,
D
n
̃j .
Pre(n)∶ = ∏ D
j=1
Consider
J0 ∶ = ⋃ C 1 (a1 )
̃1
a1 ∈D
and, for n ∈ N and a ∈ Pre(n), put
M0 +n
Jn (a)∶ = Sn (a; cb
, 2cb
M0 +n
̃ n+1
b∈D
Define the set
E∶ = ⋂
) = ⋃ C n+1 (ab).
Jn (a).
⋃
n∈N a∈Pre(n)
̃ c), so it is enough to show that
By Proposition 4.3, E ⊆ Ξ(b,
(63)
dimH (E) ≥
2
.
1+b
̃ 1 , define
8.2.1. A measure on E. For every a1 ∈ D
µ̃(J1 (a1 ))∶ =
1
.
̃1
#D
Assume that we have already defined µ̃(J(a)) for all a ∈ Pre(n). For (a1 , . . . , an , an+1 ) ∈
Pre(n + 1), we put
µ̃ (Jn+1 (a1 , . . . , an , an+1 )) ∶ =
1
µ̃ (Jn (a1 , . . . , an )) .
̃
#Dn+1
We are distributing uniformly the mass on each step. Then, µ̃ actually defines a measure
by the Daniell-Kolmogorov Consistency Theorem.
8.2.2. Estimate of µ(Jn (a1 , . . . , an )). We estimate µ(Jn (a1 , . . . , an )) in terms of ∣Jn (a1 , . . . , an )∣.
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
73
Lemma 8.11. For k ∈ N, we have
M0 +k+1
̃ k ≥ 8c2b
#D
.
Proof. If k ∈ N, then
̃ k = (2 ⌊2c2b
#D
M0 +k+1
M0 +k+1
≥ (4cb
= 6c2b
M0 +k+1
= 12c2b
2
⌋ + 1) − (2 ⌈2c2b
2
− 1) − (2cb
(2c2b
M0 +k+1
M0 +k+1
M0 +k+1
+ 1)
− 1⌉ + 1)
2
2
− 2)
1
(1 −
M0 +k+1
M +k+1
cb 0
2 M +k+1
M +k+1
) > 12 c2b 0
= 8c2b 0 .
3
□
Write t∶ =
2
1+b .
Proposition 8.12. There exists some γ1 > 0 such that
γ1
for all n ∈ N and a ∈ Pre(n).
(64)
µ̃ (Jn (a)) ≤ tbM0 +n+1
c
∣qn (a)∣2t
Proof. Take any n ∈ N and a ∈ Pre(n). We obtain (64) by showing that the product
M0 +n+1
µ̃ (Jn (a)) ctb
∣qn (a)∣2t
is bounded above by some constant depending on b and c. Equivalently, if
M0 +n+1
log (µ̃ (Jn (a))) + log (ctb
∣qn (a)∣2t )
is bounded. We work each term of the sum separately. On the one hand,
n
1
)
̃j
#D
j=1
n
1
1
≤ ∑ log ( ) + log ( bM0 +j+1 )
8
c
j=1
log (µ̃ (Jn (a))) = ∑ log (
n
= −n log 8 − ∑ 2bM0 +j+1 log c
j=1
n
= −n log 8 − 2bM0 +2 log c ∑ bj−1
j=1
= −n log 8 − 2bM0 +2 (
bn
−1
) log c
b−1
1 − b−n
= −n log 8 − 2bM0 +2 bn−1 (
) log c
1 − b−1
1 − b−n
= −n log 8 − 2bM0 +n+1 (
) log c
1 − b−1
2b
1
= −n log 8 − bM0 +n+1
(1 − n ) log c.
b−1
b
74
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
On the other hand, writing a = (a1 , . . . , an ),
n
n √
n
n
1
n
n
∣qn (a)∣ ≤ ∏ ∣aj ∣ + 1 ≤ ∏ 2∥aj ∥ + 1 ≤ 2 2 ∏ ∥aj ∥ + √ ≤ 8 2 ∏ ∥aj ∥,
2
j=1
j=1
j=1
j=1
so
n
n
log ∣qn (a)∣ ≤ log 8 + ∑ log ∥aj ∥
2
j=1
and, hence,
M0 +n+1
log (ctb
∣qn (a)∣2t ) = tbM0 +n+1 log c + 2t log ∣qn (a)∣
n
n
≤ tbM0 +n+1 log c + 2t ( log 8 + ∑ log ∥aj ∥)
2
j=1
n
= tbM0 +n+1 log c + tn log 8 + 2t ∑ log ∥aj ∥
j=1
n
≤ tbM0 +n+1 log c + 3tn log 2 + 2t (∑ log 2 + bM0 +j+1 log c)
j=1
n
= tbM0 +n+1 log c + 3tn log 2 + 2tn log 2 + 2tbM0 +2 (∑ bj−1 ) log c
j=1
= tbM0 +n+1 log c + 5nt log 2 + 2tbM0 +2 (
bn
−1
) log c
b−1
1 − b−n
= tbM0 +n+1 log c + 5nt log 2 + 2tbM0 +n+1 (
) log c
1 − b−1
1 − b−n
= tbM0 +n+1 (1 + 2 (
)) log c + 5nt log 2
1 − b−1
2b
1
= tbM0 +n+1 (1 +
(1 − n )) log c + 5nt log 2.
b−1
b
As a consequence,
log (µ̃ (Jn (a))) + log (ctb
M0 +n+1
∣qn (a)∣2t ) <
1
2b
1
2b
(1 − n ) log c + tbM0 +n+1 (1 +
(1 − n )) log c + 5nt log 2
b−1
b
b−1
b
2tb
1
2b
1
= bM0 +n+1 log c (t +
(1 − n ) −
(1 − n )) + n(5t log 2 − log 8)
b−1
b
b−1
b
2
4b
1
2b
1
(65) = bM0 +n+1 log c (
+ 2
(1 − n ) −
(1 − n )) + n(5t log 2 − log 8).
b+1 b −1
b
b−1
b
< −n log 8 − bM0 +n+1
Since b > 1, we have
2
4b
2b
b−1
+ 2
−
= −2 (
)<0
b+1 b −1 b−1
b+1
so the coefficient of bM0 +n+1 in (65) is negative for large n, and (64) holds for some γ1 > 0. □
Corollary 8.13. There exists some γ2 > 0 such that
µ̃ (Jn (a)) ≤ γ2 ∣Jn (a)∣
for all large n ∈ N and all a ∈ Pre(n).
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
75
8.2.3. Estimates of µ̃(D(z; r)). Define
1
min{∣J1 (a)∣ ∶ a ∈ D1 }.
2
Proposition 8.14. There is a constant γ3 = γ3 (b, c) > 0 such that µ(D(z; r)) < γ3 rt whenever z ∈ E and 0 < r < r0 .
r0 ∶ =
Proof. Let z = [0; a1 , a2 , . . .] ∈ E be arbitrary and take any real number r such that 0 < r <
r0 . Let n ∈ N be the natural number determined by
∣Jn+1 (a1 , . . . , an , an+1 )∣ < r ≤ ∣Jn (a1 , . . . , an , an )∣.
Define
1
̃
γ∶ =
16 (1 +
.
√
√
2
2
)
(2
+
)
2
2
We consider two possible cases: either
(66)
∣Jn+1 (a1 , . . . , an , an+1 )∣ < r ≤ ̃
γ ∣Cn+1 (a1 , . . . , an , an+1 )∣
or
(67)
̃
γ ∣Cn+1 (a1 , . . . , an , an+1 )∣ < r ≤ ∣Jn (a1 , . . . , an , an )∣.
Suppose that (66) is true. We claim that D(z; r) intersects exactly one cylinder of level
n + 1 and, hence, only one set of the form Jn+1 (b1 , . . . , bn+1 ). The condition imposed on M0
implies
√
κ1 (1 − 22 )
∣T n+1 (z)∣ <
,
√
√
8 (1 + 22 ) (2 + 22 )
for
1
∣
T n+1 (z)
∣ ≥ ∥an+2 (z)∥ − ∣T n+2 (z)∣ ≥ cb
M0
√
√
√
2
2
8
(1
+
)
(2
+
2
2
2 )
−
>
.
√
2
κ1 (1 − 2 )
2
Therefore,
∣
pn+1 (z)
− z∣ = ∣Qn+1 (0; (a1 , . . . , an+1 ) − Qn+1 (T n+1 (z); (a1 , . . . , an , an+1 )∣
qn+1 (z)
∣T n+1 (z)∣
≤
qn
∣qn+1 (a1 , . . . , an+1 )∣2 ∣1 + qn+1
∣
≤
8 (1 +
√
1
∣Cn+1 (a1 , . . . , an )∣.
√
2
2
)
(2
+
)
2
2
As a consequence, for all w ∈ D(z; ̃
γ ∣Cn+1 (a1 , . . . , an+1 )∣) we have
∣
pn+1 (z)
pn+1 (z)
− w∣ ≤ ∣
− z∣ + ∣z − w∣
qn+1 (z)
qn+1 (z)
⎛
⎞
1
≤⎜
+̃
γ ⎟ ∣Cn+1 (a1 , . . . , an )∣
√
√
⎝ 8 (1 + 22 ) (2 + 22 )
⎠
76
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
<
4 (1 +
√
1
∣Cn+1 (a1 , . . . , an )∣.
√
2
2
)
(2
+
)
2
2
In other words, we obtain
⎛ pn+1 (z) ∣Cn+1 (a1 , . . . , an+1 )∣ ⎞
⎟.
;
D(z; ̃
γ ∣Cn+1 (a1 , . . . , an+1 )∣) ⊆ D ⎜
√
√
⎝ qn+1 (z) 4 (1 + 22 ) (2 + 22 ) ⎠
By Proposition 5.4 and
D(0; 2−1 ) ⊆ F○ = F○n+1 (a1 , . . . , an+1 ),
we have
⎛ pn+1 (z) ∣Cn+1 (a1 , . . . , an+1 )∣ ⎞
○
⎟ ⊆ Cn+1
(a1 , . . . , an+1 ).
D⎜
;
√
√
2
2
⎝ qn+1 (z) 4 (1 + 2 ) (2 + 2 ) ⎠
As a consequence, the next estimate holds
µ̃ (D(z; r)) = µ̃ (D(z; r) ∩ Jn+1 (a1 , . . . , an+1 ))
≤ µ̃ (Jn+1 (a1 , . . . , an+1 )) ≤ γ2 ∣Jn+1 (a1 , . . . , an+1 )∣t ≤ γ2 rt .
Let us now assume (67). Define
̃ n+1 ∶ Jn+1 (a1 , . . . , an , h) ∩ D(z; r) ≠ ∅} .
Ñ∶ = # {h ∈ D
In view of
⋃ C 1 (a) ⊆ D (0; 2−1 ) ⊆ F○ ,
̃ n+1
a∈D
̃ n+1 , we apply Qn+1 (⋅; (a1 , . . . , an , a)) and obtain
for each a ∈ D
○
(a1 , . . . , an , a).
Jn+1 (a1 , . . . , an , a) ⊆ Qn+1 (D (0; 2−1 ) ; (a1 , . . . , an , a)) ⊆ Cn+1
○
Therefore, since Cn+1
(a1 , . . . , an , a) contains only one set of the form Jn+1 (b1 , . . . , bn+1 ),
̃ n+1 ∶ Qn+1 (D (0; 2−1 ) ; (a1 , . . . , an , a)) ∩ D(z; r) ≠ ∅} .
Ñ ≤ # {a ∈ D
Moreover, if m is the Lebesgue measure on C, there is an absolute constant γ3 > 0 such
that
1
π
m (Qn+1 (D (0; 2−1 ) ; (a1 , . . . , an , a))) ≥ γ3
4 ∣qn+1 (a1 , . . . , an , a)∣4
π
1
≥ γ3
36 ∣qn (a1 , . . . , an )∣4 ∣a∣4
π
1
≥ γ3
,
36 ⋅ 4 ∣qn (a1 , . . . , an )∣4 c4bM0 +n+2
(see the proof of [15, Lemma 9]), so
Ñ
γ3 π
1
≤ πr2 ,
36 ⋅ 4 ∣qn (a1 , . . . , an )∣4 c4bM0 +n+2
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
77
which gives
36 ⋅ 4
M +n+2 2
∣qn (a1 , . . . , an )∣4 c4b 0
r .
γ3
< 12 , hence
Ñ ≤
From 0 < t < 1 we get 0 <
t
2
1
}
̃
#Dn+1
1
36 ⋅ 4 4bM0 +n+2
c
∣qn (a)∣4 r2 2bM0 +n+2 }
≤ µ̃(Jn (a1 , . . . , an ) min {1,
γ4
8c
µ̃ (D(z; r)) ≤ µ̃ (Jn (a1 , . . . , an )) min {1, Ñ
= µ̃ (Jn (a1 , . . . , an )) min {1, 18c2b
t
≤ µ̃ (Jn (a1 , . . . , an )) 18 2 ctb
< µ̃ (Jn (a1 , . . . , an )) 18cb
M0 +n+2
M0 +n+1
M0 +n+2
∣qn (a)∣4 r2 }
∣qn (a)∣2t rt
∣qn (a)∣2t rt
≤ 18γ2 rt .
□
8.2.4. Proof of Theorem 8.1. Lower bound.
Proof of Theorem 8.1. Lower bound. The Mass Distribution Principle and Proposition 8.14
̃ c),
imply 63. Hence, since E ⊆ Ξ(b,
2
̃ c)) .
≤ dimH (Ξ(b,
1+b
□
Remark. A slight modification of the proof of the lower bound in Theorem 8.1 tells us that
for every b, c > 1 and δ > 0 we have
̃ c) ∖ Ξ(b,
̃ c + δ)) = dimH (Ξ(b, c) ∖ Ξ(b, c + δ)) =
dimH (Ξ(b,
2
.
b+1
9. Proof of Theorem 1.1
Proof. 1. Assume that B = 1. Let ε > 0 be arbitrary and consider the infinite set
Lε ∶ = {n ∈ N ∶ Φ(n) ≤ (1 + ε)n } .
Then, we have
{z ∈ F ∶ ∥an ∥ ≥ (1 + ε)n for infinitely many n ∈ L } ⊆ E∞ (Φ),
so s1+ε ≤ dimH (E∞ (Φ)), by Corollary 7.30. In view of Proposition 7.18, we conclude
2 = lim s1+ε ≤ dimH (E∞ (Φ)) ≤ 2.
ε↘0
2. Assume that 1 < B < ∞. For each ε > 0, define the infinite set
Lε′ ∶ = {n ∈ N ∶ Φ(n) ≤ (B + ε)n } .
78
YANN BUGEAUD, GERARDO GONZÁLEZ ROBERT, AND MUMTAZ HUSSAIN
Hence,
and
{z ∈ F ∶ ∥an ∥ ≥ (B + ε)n for infinitely many n ∈ Lε′ } ⊆ E∞ (Φ)
E∞ (Φ) ⊆ {z ∈ F ∶ ∥an ∥ ≥ (B − ε)n for infinitely many n ∈ N} ;
therefore, sB−ε ≤ dimH (E∞ (Φ)) ≤ sB+ε and, since B ↦ sB is continuous (Proposition
7.18), dimH (E∞ (Φ)) = sB .
3. Assume that B = ∞. Pick ε > 0. The assumption b = 1 implies that the set
L ∶ = {n ∈ N ∶ Φ(n) ≤ e(1+ε) }
n
is infinite. Then, we have
{z ∈ F ∶ ∥an (z)∥ ≥ e(1+ε) for all n ∈ N} ⊆ {z ∈ F ∶ ∥an (z)∥ ≥ e(1+ε) for all n ∈ L }
n
n
⊆ E∞ (Φ)
and Theorem 8.1 implies
2
.
2+ε
Letting ε → 0, we conclude dimH (E∞ (Φ)) ≥ 1. Moreover, B = ∞ implies that for any
≥ log C holds for infinitely many n ∈ N, hence
C > B the inequality log Φ(n)
n
dimH (E∞ (Φ)) ≥
E∞ (Φ) ⊆ {z ∈ F ∶ C n ≤ ∥an (z)∥ for infinitely manyn ∈ N}
and dimH (E∞ (Φ)) ≤ sC . We let C → ∞ and recourse to Lemma 7.18 to conclude
dimH (E∞ (Φ)) ≤ 1, so
dimH (E∞ (Φ)) = 1.
Suppose that 1 < b < ∞. Then, for any ε > 0, the inequality Φ(n) ≤ e(b+ε) holds infinitely
n
often and Φ(n) ≥ e(b−ε) is true for every large n. As a consequence, since the set
n
{z ∈ F ∶ ∥an (z)∥ ≥ e(b+ε) for all n ∈ N}
n
is contained in
{z ∈ F ∶ ∥an (z)∥ ≥ e(b+ε) ≥ Φ(n) for infinitely many n ∈ N} ⊆ E∞ (Φ)
n
and
E∞ (Φ) ⊆ {z ∈ F ∶ ∥an (z)∥ ≥ Φ(n) ≥ e(b−ε) for infinitely many n ∈ N} ,
n
Theorem 8.1 tells us that
2
2
≤ dimH (E∞ (Φ)) ≤
.
b+1+ε
b+1−ε
2
Letting ε → 0, we arrive at dimH (E∞ (Φ)) = b+1
.
n
To finish the proof, assume that b = ∞. Then, Φ(n) ≥ ec for all c > 0 and every
sufficiently large n ∈ N; that is,
n
E∞ (Φ) ⊆ {z ∈ F ∶ ∥an (z)∥ ≥ ec for infinitely many n ∈ N}
METRICAL PROPERTIES OF HURWITZ CONTINUED FRACTIONS
An application of Theorem 8.1 leads us to dimH (E∞ (Φ)) ≤
conclude dimH (E∞ (Φ)) = 0.
2
1+c
79
and letting c → ∞, we
□
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I.R.M.A., UMR 7501, Université de Strasbourg et CNRS, 7 rue René Descartes, 67084
Strasbourg Cedex, France
Institut universitaire de France
Email address: bugeaud@math.unistra.fr
Department of Mathematical and Physical Sciences, La Trobe University, Bendigo
3552, Australia.
Email address: G.Robert@latrobe.edu.au
Department of Mathematical and Physical Sciences, La Trobe University, Bendigo
3552, Australia.
Email address: m.hussain@latrobe.edu.au