Continued fractions and Brjuno functions
Pierre Moussaa;∗, Andrea Cassab, Stefano Marmib a
CEA=Saclay, Service de Physique Theorique, F-91191 Gif-sur-Yvette Cedex, France
bDipartimento di Matematica “U. Dini”, Universita di Firenze, Viale Morgagni 67=A, I-50134 Florence, Italy
Received 2 October 1997; received in revised form 14 May 1998 Dedicated to Professor Haakon Waadeland on the occasion of his 70th birthday
Abstract
For 0661 given, we consider the modied continued fraction expansion of the real number x dened byx=a0+
0x0; a0∈Z, and, x
−1
n−1=an+nxn; an∈Nfor n¿0, where−16nxn¡; n=±1, forn¿0, with xn¿0. The usual (Gaussian) case is = 1, whereas = 1
2 is the continued fraction to the nearest integer. The Brjuno functionB ()
(x) is then dened by B()(x) =−ln(x0)−P
∞
n=1x0x1· · ·xn−1ln(xn). These functions were introduced by Yoccoz in the = 1
and 1
2 cases, in his work on the holomorphic conjugacy to a rotation, of an analytic map with an indierent xed point.
We will review some properties of these functions, namely, all these functions are 1-periodic, and belong toLpIoc(R), for
16p¡∞, and also to the space BMO(T). In this communication, we will mainly report on some of the technical tools
related to the continued fraction expansions required by the above mentioned results. These results deal with the growth of the denominator of the reduced fractionspn=qn of the above continued fraction expansion, which gives the maximal error
rate of approximation, the relation between these continued fraction and the usual Gaussian case, and nally the invariant density, generalising the classical result of Gauss for the usual case. c1999 Elsevier Science B.V. All rights reserved. Keywords:Continued fraction; Approximation of real numbers; Invariant measure; Brjuno function
1. Introduction
We consider a one-parameter family of continued fraction expansions, which includes as a par-ticular case the classical Gaussian case and the continued fractions to the nearest integer. We will introduce the corresponding Brjuno functions and discuss some of their properties. For that purpose, we rst describe several features of these continued fractions expansions, some of which we have been unable to nd in the literature. In Section 2, we recall some elementary algebraic relations concerning singular continued fractions. In Section 3, we describe a generalisation of the usual con-tinued fractions, which uses for a real number a modied denition of integer and fractional parts.
∗
Corresponding author.
In Section 4, we describe in detail the connection between these continued fractions and the usual Gaussian case. In Section 5, we dene the Brjuno functions, and we analyse their properties in Section 6. In Section 7, we discuss measures which are invariant with respect to the transformation which generates the generalised continued fractions considered here.
2. Continued fractions: algebraic properties
In this section, we recall some algebraic properties of some non-regular continued fractions. These continued fractions are generated by the following recursion relations. Letx∈R, and forn¿0 integer, let
x−1
n =an+1+n+1xn+1; starting withx=a0+0x0; (2.1) where we assume a0∈Z, and for n¿0; an¿0∈N. Moreover, we assume for n¿0; xn¿0 and
n=±1. The recursion can be followed with arbitrary coecients an and arbitrary signs n, as long
as we get xn¿0. If we get xn= 0, we say that the recursion stops at order n. If the recursion does
not stop before order n, we get the continued fraction expanded up to the nite order n:
x= [(a0; 0);(a1; 1); : : : ;(an; n); xn] =a0+
0
a1+
1
a2+. .. +
n−1
an+nxn
: (2.2)
Therefore x can be expressed as a fractional linear function of nxn, and by recursion over n, we
get from (2.1)
x=pn+pn−1nxn
qn+qn−1nxn
; (2.3)
where for n¿0, the coecients pn and qn are obtained using the following recursion relations:
pn+1=an+1pn+npn−1; qn+1=an+1qn+nqn−1; p0=a0; p−1= 1; q0= 1; q−1= 0: (2.4)
From (2.4), one easily deduces the following equality:
∀n¿0; qn+1pn−pn+1qn= (−1)n+101· · ·n; (2.5)
which shows that one cannot have simultaneously pn=qn= 0. Furthermore, if q6= 0, the fractions
pn=qn is irreducible. In that case, the reduced fraction pn=qn can be expressed as the nite length
continued fraction pn=qn = [(a0; 0);(a1; 1); : : : ;(an; n); 0]. In other words, if the fraction stops at
order n, we have x=pn=qn∈Q, that is x is rational. For the generalised expansions of the next
section, the converse is true whenever 6= 0. We also get
xn= (−n)
pn−xqn
pn−1−xqn−1
: (2.6)
Therefore the numbers n=x0x1· · ·xn satisfy for n¿0
n=x0x1· · ·xn= (−1)n01· · ·n(qnx−pn); (2.7)
and the recursion relations for n¿0 (with the convention −1= 1)
We also easily get the following relations:
n(qn+1+qnn+1xn+1) = (−1)n01· · ·n(qnx−pn)(qn+1+qnn+1xn+1) = 1; (2.9)
qnn−1+nqn−1n= 1: (2.10)
Finally we give here some notations for some algebraic numbers including the golden mean
g=−1 +
√
5
2 ; G=g
−1=1 +
√
5
2 ; =−1 +
√
2; =−1= 1 +√2: (2.11)
3. A generalisation of the classical continued fractions
The above elementary considerations apply to the following particular cases, which have been sometimes called “japanese continued fractions” [6 –8]. Let be a xed real number such that 0661. Then, given the starting number x, the coecients an and n are recursively uniquely
dened by the condition
x=a0+0x0; and ∀n¿0; x
−1
n =an+1+n+1xn+1; with ∀n¿0; −16nxn¡: (3.1)
We dene the modied integer part [x] and the modied fractional part {x} as follows:
[x]= [x−+ 1]1 and {x}={x−+ 1}1+−1; (3.2)
where [x]1 and{x}1 are the usual integer and fractional parts of x (so that 06{x}1¡1). With these notations, we can rewrite Eq. (3.1) as
a0= [x]; 0x0={x} and ∀n¿0; an+1= [x
−1
n ]; n+1xn+1={x
−1
n }: (3.3)
Therefore the xn are generated by iterating the function A(x) =|{x−1}|, that is
∀n¿0; xn+1=A(xn) =|{x
−1
n }|=|x
−1
n −[x
−1
n ]|: (3.4)
A more detailed description states that the map A is made of the following branches:
branchk+: A
(x) =
1
x −k for
1
k+¡x6
1
k; (3.5)
branchk−
: A(x) =k−
1
x for
1
k¡x6
1
k+−1: (3.6)
When 12¡61, the functionA maps the interval [0; ) to itself, whereas when 0¡612, it maps the
interval [0;1−] to itself. In both cases, it is convenient to set A(0) = 0, and we get a map which
is innitely dierentiable by pieces, with discontinuities accumulating to 0. The map is expanding, which means that its derivative is strictly greater than one everywhere it is dened. For = 1, we get the classical Gauss continued fraction expansion for which all signs n= +1, and for =12, we have
the continued fractions to the nearest integer. When we compare the domain of denition of the map with the possible domains of the various branches given in (3.4) and (3.5), we nd that the branch 1−
The case = 0 brings some dierent features: A0 maps (0;1] to itself. It is still innitely dieren-tiable by pieces, with discontinuities accumulating to 0. But the map is no longer expanding, since the derivative at one is equal to one. All coecients n=−1; and we never have xn= 0, so that the
expansion never stops. The rational numbers are represented by innite continued fraction expansion with constant tail, more precisely, there exist N ¿0 such that, for all n¿N; xn= 1; an+1= 2. In this
Proof. Use Eq. (2.9) and Lemma 1.
Using (2.7) and Lemma 2, we get the sign of x−pn=qn, and bounds on its absolute value. We
branch 3−, that is ¡x
j6(2 +)−1, which happens only if ¡−1−2 =. A simple exercise then
shows that if 0¡¡, then (2 +)−1¡√1−2, and therefore x
j¡ √
1−2. In all cases, we get either xj6= max(;
p
|1−2|), or xjxj+162, which proves part 2 of the lemma.
For ¿g, the value of =g in Lemma 3 is optimal since g is the largest xed point of the map
A. For ¡6g, the largest xed point is . The above lemma gives = for 661−. The
following theorem lls the apparently little gap between = 1−= 0:586: : : and =g= 0:618: : : .
Theorem 4. Let 661 and L() = sup(;1−). For g¡61; let =g= (√5−1)=2; and for 0¡6g; let ==√2−1. Then (3:9) holds.
For a proof see [5]. This proof is amazingly complicated when compared to the proof of Lemma 3. We do not know the optimal value of for6. The largest xed point in this case isg2, so that the optimalopt must satisfy g26opt6√1−2¡1−. We further know that for6(√3−1)=2 = 0:366::, there is a two cycle with values (3−√3)=3 and (3−√3)=2, and therefore the optimal value of is bigger or equal than the geometric mean of these two values, namely (2−√3)(1=2), but this number is nothing else than (1−2)(1=2) for the particular value = (√3−1)=2. The estimation of given by Lemma 3 is therefore optimal, and strictly greater than g2, for this value of .
4. Recovering the Gaussian case from the general case
Let 06¡1 and x∈R, and let us denote in a sequence of six-dimensional arrays
Si={i; ai; xi; i; pi; qi}; i¿0; (4.1)
the various quantities associated with the continued fraction expansion of x described in the previous section. We shall describe a procedure which applied to the sequenceSn leads to the similar sequence for the Gaussian case = 1. The procedure consists in changing successively every minus signs
n=−1 in a nite number of plus signs using what is usually called a desingularisation [1]. In the
following, we shall denote in the Gaussian case = 1;Gi={i= +1; Ai; Xi;Bi; Pi; Qi} the respective quantities {i; ai; xi; i; pi; qi} of Si. We rst need the following lemma.
Lemma 5. Using notations of (2:2) above; we have
[(a0;−1);(2;−1);(2;−1); : : : ;(2;−1);(ak+1; k+1); xk+1]
= [(a0−1;+1);(k+ 1;+1);(ak+1−1; k+1); xk+1] (4.2)
=a0−
k−(k−1)xk
(k+ 1)−kxk
with xk= (ak+1+k+1xk+1)
−1: (4.3)
In the left-hand side of (4.2), the number 2 occursk times. The proof is straightforward by recursion over k. The case k = 0 is just the elementary identity
A− 1
B+X = (A−1) +
1 + 1 (B−1) +X
−1
Denition. In the sequence Si, we say that we have a desingularisation position at order j¿0 with length k¿0, if the following conditions are satised: (i) n =−1 for j6n6j +k, (ii) an = 2
for j + 16n6j +k (condition void if k = 0), (iii) if j 6= 0, and j−1 =−1, then aj 6= 2, and
(iv) if j+k+1=−1, we have aj+k+1 6= 2. Then the desingularisation at order j (with length k), is a transformation fromSi to Si, which consists in (i) suppressing thek+2 elements Sj;Sj+1; : : : ;Sj+k+1, and (ii) replacing them by the three elements Sj;Sj+1;Sj+2, where
Sj={+1; aj−1; 1−xj; (1−xj)j−1; pj−pj−1; qj−qj−1};
Sj+1=
(
+1; k+ 1; xj+k
1−xj+k
; j+k; pj+k; qj+k
)
; (4.5)
Sj+2={j+k+1; aj+k+1−1; xj+k+1; j+k+1; pj+k+1; qj+k+1}:
The terms before Sj remain unchanged, that is for i¡j; Si =Si. The terms after Sj+k+1, remain unchanged up to a shift, namely ∀p¿0; Sj+2+p=Sj+k+1+p.
The conditions in the above denition just mean that the minus sign j is followed by exactly k
successive iterations of the branch 2−, which give x
j+k starting from xj. Using Lemma 1, one easily
gets
∀l; 06l6k; xj+l=
(k−l)−(k−l−1)xj+k
(k−l+ 1)−(k−l)xj+k
: (4.6)
However xj+k does not belong to the branch 2−, therefore 0¡xj+k612. Using (4.6), one gets for
06l6k;1−(k−l+ 1)−1¡x
j+l61−(k−l+ 2)−1. Since j+l=−1, we have 06xj+l61−, and
therefore 1−(k−l+ 1)−1¡1−, and for l= 0; (k+ 1)−1¿ which proves the following lemma.
Lemma 6. The desingularisation length k is smaller than −1
−1. In particular; we always have
k= 0 for ¿1
2. We have k61 for ¿ 1 3.
Unbounded desingularisation lengths can occur only for = 0. The length is innite in case of the innite tail which occur for rational number x, as already mentioned at the end of Section 3 above. In the sequence Si corresponding to values of and x xed, let us label by n∈N; n¿0 the desingularisation positionsj(n), with corresponding length k(n), in such a way thatj(n) is increasing with n. The value taken by n are from 1 to N, where we understand that N can be either nite or +∞. For 16n6N, letj(n)=j(n)+Pn−1
j=1(1−k(j)). It is easy to see that the sequencej(n) is strictly increasing with n. Now we dene the sequence Sn
i by successive application of the desingularisation
procedure. We start with S0
i =Si, and set Sin+1=S n
i, which means that in order to get Sin+1 we
apply the desingularisation to the sequence Sn
i, at the rst available desingularisation position, that
is j(n), still with length k(n). We immediately observe that the coecients j of the sequence Sin
are all equal to +1, up to order j(n+ 1)−1 included. Therefore we have the following theroem.
Theorem 7. The sequence Sn
j corresponding to ¡1 and x xed; coincides with the Gaussian
sequence G
n for the same value of x; up to order j(n+ 1)−1 included. In particular; we have
Pj(n)+1=pj(n)+k(n); Qj(n)+1=qj(n)+k(n); (4.8)
In fact, this theorem summarises the results of the successive desingularisations. Note that when
¿1
2, all desingularisations have zero length, and therefore the sequence (i; pi; qi); i¿0, is a sub-sequence of the analogous Gaussian sub-sequence (Bi; Pi; Qi).
5. The Brjuno functions and the Brjuno condition
We associate to the general continued fraction expansion (2.1) the following Brjuno series with positive terms:
We set B= +∞, either if the series diverges, or if the fraction stops, in which case the series is replaced by a nite sum, the last term (corresponding to xn= 0) being innite. Given ; 0661,
we apply this denition to the expansion (3.1), and the above series dene the Brjuno functions
B(x):
We give now two results on the Brjuno functions:
Theorem 8. For 0¡61; there exist a constant C1()¿0
In the last equation, the modulus of the rst sum is bounded using (3.7) and the geometric bound (3.9) on n. Using (3.7), the second sum is bounded in modulus if and only if P
∞
using (3.7), P∞ But this is a straightforward consequence of the bound (3.9) on qn.
Theorem 9. For 0¡61; we have |B(x)−B1(x)|¡C2() for some C2()¿0.
Proof. We rst estimate the change in the Brjuno series jB(x) =B(x)−B(x) where the series B
and B are associated to Si and Si respectively, this change resulting from a single desingularisation at position j with length k, as described above in Section 4. Using notations of (4.5), we get
jB(x) =j−1 ln(1−xj) + (1−xj) ln
and we get after some calculations, where the unbounded terms proportional to ln(xj+k) cancel
jB(x)
by Lemma 3, and the theorem is proved.
Denition. A real number x is a Brjuno number if the series P∞
n=0Q
−1
n ln(Qn+1) converges [2]
(remember that the Qn are the denominators of the Gaussian reduced continued fraction). We now
have the following obvious corollary. is bounded; for any arbitrary 0¡61.
The Brjuno function has been introduced by Yoccoz [9] in the = 1
2 case. The Brjuno numbers appear in the dynamical systems theory, as a characterisation of the rotation number for which an analytic map which behaves as a rotation in the vicinity of a xed point, can always be analytically conjugated to a rotation. The set of the Brjuno numbers is a subset of the irrational numbers. The real numberx is not a Brjuno number, either if it is rational, or if the denominators qn of the reduced
we have a lower bound for |x−pn=qn| which is proportional to q−n, with ¿1, then from (2:7)
one deduces that qn+1q1n− is bounded, and therefore the series
P
nq
−1
n ln(qn+1) converges. Therefore diophantine irrationals, and in particular algebraic numbers, are Brjuno numbers.
6. Properties of the Brjuno functions
Let 0¡61, and consider the Brjuno function B(x) dened above in (5:2). Let
L() = sup(;1−); l() = inf (;1−) = 1−L(): (6.1)
The following proposition is an immediate consequence of the denitions.
Proposition 11. The function B(x)has the following properties:
(1) B(x) =B(x+ 1);
(2) for |x|¡l(); B(x) =B(−x) =B(1−x);
(3) B(x) = ln(x
−1
0 ) +x0B(1x1); so that B(x) = ln(|x|−1) +|x|B(|x|−1); whenever −16x¡;
(4)
for0¡x¡; B(x) =xB(x
−1) + ln(x−1);
for0¡x61−; B(−x) =xB(x
−1) + ln(x−1):
It is convenient to introduce the Brjuno operators T acting on the space X made of functions
on R, and taking values on (0;∞]. We will discuss later various possible norms on these spaces.
(1) For ¿1
2, consider the space X of 1-periodic functions f(x), such that f(x) =f(−x) for
|x|6l() = 1−, and dene the function ln(x)∈X, which coincides with ln(x) on (0; L() =).
Then, for 0¡x¡L() =, we set
(Tf)(x) =xf(x−1) =xf(A(x)); (6.2)
and we extend the domain of Tf using the periodicity and parity properties embedded in the de-nition of X, so that Tf∈X. Thus we have
(1−T)B(x) =−ln(x): (6.3)
(2) For 0¡¡12, consider the space X of 1-periodic functions f(x), such that f(x) =f(−x) for |x|¡l() =, and dene the function ln(x) which coincides with ln(x) on (0; L() = 1−], and
such that ln(x)∈X. Then, for 0¡x61−=L(), we set
(Tf)(x) =xf(−x
−1) =xf(A
(x)); (6.4)
and we extend the domain of Tf using the periodicity and parity properties embedded in the deni-tion of X, so that Tf∈X. Let now B(x) =B(−x) (note that B∈X implies B(x)∈X). Thus
we have
(1−T)B(x) =−ln(x): (6.5)
Theorem 12. Let0¡61. Assume that there exists a measure m on (0; L());which is invariant
under the transformation A. ThenX becomes a Banach spaceX;p;taking forf∈X the norm kfkp equal to theLp-norm off on (0; L());with respect to the measurem
. Then the spectral
radius of T in this Banach space is bounded by the constant ¡1of Lemma3above;and therefore
(1−T) is invertible in this space. If this measure is equivalent to the Lebesgue measure on(0; L());
the functionB belongs to the space X;p.
Proof. Note rst that due to the parity properties, the norm is equivalent to the normLp(T) with
re-spect to the same measure, suitably completed by parity. If the measure is equivalent to the Lebesgue measure on either T or (0; L()), the function ln∈X;p. Therefore we just have to check the bound
on the spectral radius. When ¿1
2, we get from (6:2) for the nth iterate T (n)
of T; (T(n)f)(x) =
n−1f(A(n)(x)), and since the measure is invariant, kT(n)fk6kfkmaxx∈(0;L())[n−1(x)]6nkfk, due to (3:9). This shows that the spectral radius of T is smaller than .
In the next section, we shall display for ¿ invariant measures which are absolutely continuous with respect to Lebesgue measure. Indeed, due to Theorem 9, it is sucient to use the invariant measure in the Gaussian case = 1. Therefore we have the following corollary.
Corollary 13. For any 0¡61;the Brjuno functionB belongs toLp(T).
In [5], using still spectral radius evaluations, it is proved that the operator 1−T1=2 is invertible in the space of even periodic functions on R, with a norm naturally associated to the BMO (bounded
mean oscillation) seminorm. It follows that the function B1=2 belongs to BMO(R), and since for 0¡61, the dierence between B and B1=2 is bounded, B also belongs to BMO(R).
When =1
2, the map A1=2 is continuous everywhere in (0; 1
2);0 expected. In this case, we mention, following [5] that the operator 1−T1=2 can be inverted if we use now some Holder continuity norm on X1=2. More precisely, if f is continuous and Holder continuous with exponent greater than 12, then (1−T1=2)−1f is Holder continuous with exponent 12. In other words, if we perturb the functional
equation
B1=2(x)−xB1=2 1
x
=−ln(x); 0¡x61
2 (6.6)
by adding to the right-hand side a C1 function, the solution (even and 1-periodic) is modied by a
C1=2 additive contribution. This observation is useful for explaining the numerical results of Marmi [4] on the size of stability domains in holomorphic maps. We end this section by a curious result which states that the odd part of B1 is a simple explicit continuous function.
Theorem 14. LetB+
1 andB
−
1 be the even and odd part ofB1. B+1 andB
−
1are periodic; so that they are determined by their values in [0;1
2]. We have
forx∈[0;1 2]; B
−
1(x) =
x
2ln 1
−x x
; (6.7)
B+1(x) =xB+1 1
x
with; still forx∈[0;1
which leads to (6.8) since we already know B−
1.
Finally, we mention a stronger result from [5], namely that the dierence B+
1(x)−B1=2, can be extended as an even 1-periodic continuous function which is Holder continuous with exponent 1
2.
7. Invariant measures and maps associated to continued fractions
We now give some results on the invariant measures under the map A given in (3.4).
(iii)if 1
The theorem can be proven by direct calculations, which are very tedious for general values of . It is much easier to derive the rst three cases from similar results of Nakada [6,7], who considers the maps A=|x−1| −[|x−1|], instead of A=|A|. The fourth case has been obtained by Cassa [9].
It is easy to see from the above results, that the measure given there have everywhere a nonzero density, and therefore are equivalent to Lebesgue measure. In [3], the reader can nd the results in the case =2
5, and = 1−g=g
2, with the dierence that the density in each interval of denition
is expressed as an innite series of terms of the kind (Ak −x)−1, and not as a nite sum as in
Theorem 15. This drastic change around = is not understood. Is there a relation with the change in the behaviour of in (3.9)?
Acknowledgements
We thank Professor J.-C. Yoccoz for his help and warm encouragements. We thank Prof. Lorentzen, Njastad, and RHnning, for allowing the rst author to participate to the present meeting in honour
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