Directory UMM :Data Elmu:jurnal:J-a:Journal of Computational And Applied Mathematics:Vol105.Issue1-2.1999:

Journal of Computational and Applied Mathematics 105 (1999) 403–415

Continued fractions and Brjuno functions
Pierre Moussaa; ∗ , Andrea Cassab , Stefano Marmib
a

b

CEA=Saclay, Service de Physique Theorique, F-91191 Gif-sur-Yvette Cedex, France
Dipartimento 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 modied continued fraction expansion of the real number x dened by x = a0 +
0 x0 ; a0 ∈ Z, and, x−1
n−1 = an + n x n ; an ∈ N for n¿0, where  − 16n x n ¡; n = ±1, for n¿0, with x n ¿0. The usual
(Gaussian) case is  = 1, whereas  = 21 is the continued fraction to the nearest integer. The Brjuno function B() (x) is
P∞
then dened by B() (x) = −ln(x0 ) − n=1 x0 x1 · · · x n−1 ln(x n ). These functions were introduced by Yoccoz in the  = 1
and 12 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 to LpIoc (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 fractions pn =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
c 1999 Elsevier Science B.V. All rights reserved.
density, generalising the classical result of Gauss for the usual case.
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 particular 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 continued fractions, which uses for a real number a modied denition of integer and fractional parts.
∗

Corresponding author.

c 1999 Elsevier Science B.V. All rights reserved.

0377-0427/99/$ - see front matter
PII: S 0 3 7 7 - 0 4 2 7 ( 9 9 ) 0 0 0 2 9 - 1

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P. Moussa et al. / Journal of Computational and Applied Mathematics 105 (1999) 403–415

In Section 4, we describe in detail the connection between these continued fractions and the usual
Gaussian case. In Section 5, we dene 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. Let x ∈ R, and for n¿0 integer,
let
x−1
n = an+1 + n+1 x n+1 ;

starting with x = a0 + 0 x0 ;

(2.1)

where we assume a0 ∈ Z, and for n¿0; an ¿0 ∈ N. Moreover, we assume for n¿0; x n ¿0 and
n = ±1. The recursion can be followed with arbitrary coecients an and arbitrary signs n , as long
as we get x n ¿0. If we get x n = 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:
0
:
(2.2)
x = [(a0 ; 0 ); (a1 ; 1 ); : : : ; (an ; n ); x n ] = a0 +
1
a1 +
n−1
.
a2 + . . +
an +  n x n
Therefore x can be expressed as a fractional linear function of n x n , and by recursion over n, we
get from (2.1)
pn + pn−1 n x n
;

(2.3)
x=
qn + qn−1 n x n
where for n¿0, the coecients pn and qn are obtained using the following recursion relations:
pn+1 = an+1 pn + n pn−1 ;

qn+1 = an+1 qn + n qn−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+1 pn − pn+1 qn = (−1)n+1 0 1 · · · n ;

(2.5)

which shows that one cannot have simultaneously pn = qn = 0. Furthermore, if q 6= 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

pn − xqn
x n = (−n )
:
(2.6)
pn−1 − xqn−1

Therefore the numbers n = x0 x1 · · · x n satisfy for n¿0
n = x0 x1 · · · x n = (−1)n 0 1 · · · n (qn x − pn );

(2.7)

n−1 = an+1 n + n+1 n+1 :

(2.8)

and the recursion relations for n¿0 (with the convention −1 = 1)

P. Moussa et al. / Journal of Computational and Applied Mathematics 105 (1999) 403–415

405

We also easily get the following relations:
n (qn+1 + qn n+1 x n+1 ) = (−1)n 0 1 · · · n (qn x − pn )(qn+1 + qn n+1 x n+1 ) = 1;
qn n−1 + n qn−1 n = 1:

(2.9)
(2.10)

Finally we give here some notations for some algebraic numbers including the golden mean
√
√
√
√
−1 + 5
1+ 5
−1
g=
; G=g =
;
= −1 + 2;

=
−1 = 1 + 2:
(2.11)
2
2
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
dened by the condition
x = a0 + 0 x0 ;

and

∀n ¿0; x−1
n = an+1 + n+1 x n+1 ;

with ∀n¿0;  − 16n x n ¡:

(3.1)

We dene the modied integer part [x] and the modied 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
0 x0 = {x}

a0 = [x] ;

and

−1
∀n¿0; an+1 = [x−1

n ] ; n+1 x n+1 = {x n } :

(3.3)

Therefore the x n are generated by iterating the function A (x) = |{x } |, that is
−1

∀n¿0;

−1
−1
x n+1 = A (x n ) = |{x−1
n } | = |x n − [x n ] |:

(3.4)

A more detailed description states that the map A is made of the following branches:
branch k + :

A (x) =

1
−k
x

for

1
1
¡x6 ;
k +
k

(3.5)

1
x

for

1
1
¡x6
:
k
k +−1

(3.6)

branch k − : A (x) = k −

When 12 ¡61, the function A maps the interval [0; ) to itself, whereas when 0¡6 21 , 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 innitely 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 dened. For  =1, we get
the classical Gauss continued fraction expansion for which all signs n = +1, and for  = 21 , we have
the continued fractions to the nearest integer. When we compare the domain of denition of the map
with the possible domains of the various branches given in (3.4) and (3.5), √
we nd that the branch
1− never occurs for 0661, and that the branch 1+ occurs only if ¿g = ( 5 − 1)=2. In particular,
we always have an ¿2 whenever n = −1. Moreover, if ¿g, and j = −1, then aj+1 ¿3. Finally,
for 0¡61, the continued fraction which represent any rational number stops at nite order.

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P. Moussa et al. / Journal of Computational and Applied Mathematics 105 (1999) 403–415

The case  = 0 brings some dierent features: A0 maps (0; 1] to itself. It is still innitely dierentiable 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 x n = 0, so that the
expansion never stops. The rational numbers are represented by innite continued fraction expansion
with constant tail, more precisely, there exist N ¿0 such that, for all n¿N; x n = 1; an+1 = 2. In this
case, for n¿N , we also have x = (pn − pn−1 )=(qn − qn−1 ); pn = pN + (n − N )(pN − pN −1 ); qn =
qN + (n − N )(qN − qN −1 ). We quote now two lemmas.
Lemma 1. For 0¡61; then q1 ¿q0 ¿0; and for n¿1; qn ¿qn−1 ¿0; as long as the fraction does
not stop (that is xk 6= 0 for 06k¡n): For  = 0; we have for n¿0; qn ¿n + 1¿0; and qn+1 − qn ¿1:
Proof. We use recursively qn −qn−1 =(an −1)qn−1 +n−1 qn−2 , which gives the result when n−1 =+1.
Otherwise, we use an ¿2 if 6g, and an ¿3 if ¿g.
Lemma 2. Let 0¡61; and n¿1. If xk 6= 0 for k¡n; we have
1
1
1
6qn n−1 =
6 :
+1
1 + (qn−1 =qn )n x n 

(3.7)

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
have




1
1
1
1
 x − pn  6
6
6
:
(3.8)
6

2
(1 + )qn+1 (1 + )qn qn+1
qn  qn qn+1 qn2

If the fraction stops at some order n, we have x = pn =qn , otherwise qn → ∞, when n → ∞.
Therefore, when 0¡61, the sequence pn =qn converges to x. In the  = 0 case, from Lemma 2
we get |x − pn =qn |¡(qn (qn+1 − qn ))−1 , and the sequences pn =qn and (pn − pn−1 )=(qn − qn−1 ) both
converge to x. When ¿0; the rate of convergence is at least geometric, as shown by the estimate
of the following lemma.
Lemma 3. Let  6= 0; and L() = sup(; 1 − ); so that 12 6L()61: Then; (1) for g¡61; let  =
√
√
p
g = ( 5 − 1)=2; and (2) for 0¡6g; let  = sup(
; |1 − 2|); where
= 2 − 1: Then ∀n¿0; we
have the bounds



1
pn+1  (L())2 (1 + ) 2(n+1)
1
n


n 6L() ; qn+1 ¿
6
; x −
:
(3.9)
L()(1 + ) n
qn+1 


Proof. Due to (3.7) and (2.7), we only need to prove n 6L()n . Assume rst g¡61; and observe
that either xj 6g; or xj ¿g¿(1 + )−1 , therefore xj+1 = xj−1 − 1¡g; xj xj+1 = 1 − xj ¡1 − g = g2 . From
this, one easily deduces that when x n 6g; n ¡gn+1 ; and if x n ¿g; n 6gn , which proves part 1 of
the lemma. Assume now
66g. Then either xj 6
, or xj ¿
and there are two possibilities. First
xj is in the domain of the branches 2+ or 2− , in which case we have xj xj+1 = |1 − 2xj |, and therefore,
since
¡xj 6max(; 1 − ), we get xj xj+1 6max(
2 ; |1 − 2|). Second, xj is in the domain of the

P. Moussa et al. / Journal of Computational and Applied Mathematics 105 (1999) 403–415

407

−1
branch 3− , that is
¡xj 6(2 + )−1 , which
2 =
. A simple exercise then
√ happens only if ¡
−√
−1
shows that if 0¡¡
,
then
(2
+
)
¡
1
−
2,
and
therefore
x
¡
1 − 2. In all cases, we get
j
p
either xj 6 = max(
; |1 − 2|), or xj xj+1 62 , 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.
2
We do not know the optimal value √
of  for 6
. The largest xed point in this case
√ is g , so that the
opt
2
opt
optimal  must satisfy g 6 6 1√− 2¡1−. We√further know that for 6( 3−1)=2 = 0:366::,
the optimal value of  is
there is a two cycle with values (3 − 3)=3 and (3 − 3)=2, and therefore √
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 };

(4.1)

i¿0;

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 sequence Sn 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 ]
= a0 −

k − (k − 1)xk
(k + 1) − kxk

with xk = (ak+1 + k+1 xk+1 )−1 :

(4.2)
(4.3)

In the left-hand side of (4.2), the number 2 occurs k times. The proof is straightforward by recursion
over k. The case k = 0 is just the elementary identity
A−

1
1
= (A − 1) + 1 +
B+X
(B − 1) + X


−1

:

(4.4)

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P. Moussa et al. / Journal of Computational and Applied Mathematics 105 (1999) 403–415

Denition. 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 satised: (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 from Si to S i , which consists in (i) suppressing the k +2 elements Sj ; Sj+1 ; : : : ; Sj+k+1 ,
and (ii) replacing them by the three elements S j ; S j+1 ; Sj+2 , where
S j = {+1; aj − 1; 1 − xj ; (1 − xj )j−1 ; pj − pj−1 ; qj − qj−1 };
(

)

xj+k
S j+1 = +1; k + 1;
; j+k ; pj+k ; qj+k ;
1 − xj+k

(4.5)

S j+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; S i = Si . The terms after Sj+k+1 , remain
unchanged up to a shift, namely ∀p¿0; S j+2+p = Sj+k+1+p .
The conditions in the above denition just mean that the minus sign j is followed by exactly k
successive iterations of the branch 2− , which give xj+k starting from xj . Using Lemma 1, one easily
gets
(k − l) − (k − l − 1)xj+k
∀l; 06l6k; xj+l =
:
(4.6)
(k − l + 1) − (k − l)xj+k

However xj+k does not belong to the branch 2− , therefore 0¡xj+k 6 12 . Using (4.6), one gets for
06l6k; 1 − (k − l + 1)−1 ¡xj+l 61 − (k − l + 2)−1 . Since j+l = −1, we have 06xj+l 61 − , 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 ¿ 21 . We have k61 for ¿ 13 .

Unbounded desingularisation lengths can occur only for  = 0. The length is innite in case of the
innite 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 positions j(n), with corresponding length k(n), in such a way that j(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
Pn−1
+ ∞. For 16n6N , let j(n)=j(n)+ j=1
(1−k(j)). It is easy to see that the sequence j(n) is strictly
increasing with n. Now we dene the sequence Sin by successive application of the desingularisation
n
procedure. We start with Si0 = Si , and set Sin+1 = S i , which means that in order to get Sin+1 we
apply the desingularisation to the sequence Sin , 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 Sjn corresponding to ¡1 and x xed; coincides with the Gaussian
sequence Gn for the same value of x; up to order j(n + 1) − 1 included. In particular; we have
Pj(n) = pj(n)+k(n) − pj(n)+k(n)−1 ;

Qj(n) = qj(n)+k(n) − qj(n)+k(n)−1 ;

(4.7)

P. Moussa et al. / Journal of Computational and Applied Mathematics 105 (1999) 403–415

Pj(n)+1 = pj(n)+k(n) ;
Xj(n) =

Qj(n)+1 = qj(n)+k(n) ;

1 − xj(n)+k(n)
= 1 − xj ;
k + 1 − kxj(n)+k(n)

Bj(n) = (1 − xj(n) )j(n)−1 ;

409

(4.8)

Xj(n)+1 =

xj(n)+k(n)
;
1 − xj(n)+k(n)

(4.9)

Bj(n)+1 = j(n)+k(n) :

(4.10)

If j(n + 1)¿j(n) + k(n) + 2; we also have Gj(n)+‘ = Sj(n)+‘ ; for all k(n) + 26‘6j(n + 1) − j(n) − 1.
In fact, this theorem summarises the results of the successive desingularisations. Note that when
¿ 12 , all desingularisations have zero length, and therefore the sequence (i ; pi ; qi ); i¿0, is a subsequence of the analogous Gaussian 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:
B=

∞
X

n−1 ln((x−1
n ):

(5.1)

n=0

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 x n = 0) being innite. Given ; 0661,
we apply this denition to the expansion (3.1), and the above series dene the Brjuno functions
B (x):
B (x) =

∞
X

n−1 ln(x−1
n ):

(5.2)

n=0

We give now two results on the Brjuno functions:
Theorem 8. For 0¡61; there exist a constant C1 ()¿0


∞

X
ln(qn+1 ) 

 ¡C1 ():
B (x) −

qn 

(5.3)

n=0

Proof. Using (2.10), we have
B (x) −

∞
X
ln(qn+1 )
n=0

qn

∞
X

n
=−
n−1 ln
n−1
n=0
=−

∞
X
n=0





−

n−1 ln(n qn+1 ) +

∞ 
X
n=0

∞
X
n=0

qn−1
n ln(qn+1 )
n−1 + n
qn

n−1 ln(n−1 ) −



∞
X
i=0

n

qn−1
n ln(qn+1 ):
qn

(5.4)

In the last equation, the modulus of the rst sum is bounded using (3.7) and the geometric
bound
P∞
(3.9) on n . Using (3.7), the second sum is bounded in modulus if and only if n=0 n ln(qn+1 ),
is bounded, in which case the third sum is also bounded in modulus, since qn−1 6qn . Finally, still

410

P. Moussa et al. / Journal of Computational and Applied Mathematics 105 (1999) 403–415

using (3.7), ∞
n=0 n ln(qn+1 ) is bounded in modulus if and only if
But this is a straightforward consequence of the bound (3.9) on qn .
P

P∞

n=0

−1
qn+1
ln(qn+1 ), is bounded.

Theorem 9. For 0¡61; we have |B (x) − B1 (x)|¡C2 () for some C2 ()¿0.

where the series B
Proof. We rst estimate the change in the Brjuno series
j B(x) = B(x) − B(x)

and B are associated to Si and S i 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

j B(x) = j−1

xj+k
ln(1 − xj ) + (1 − xj ) ln
1 − xj+k

!

−

∞
X
j+l−1
l=0

j−1

!

ln(xj+l ) :

(5.5)

Using (4.6), we can express xj+l and 1 − xj in terms of xj+k . We nd
1 − xj =

1 − xj+k
k + 1 − kxj+k

and

j+l−1 k − l + 1 − (k − l)xj+k
=
j−1
k + 1 − kxj+k

for 06l6k;

(5.6)

and we get after some calculations, where the unbounded terms proportional to ln(xj+k ) cancel
k−1
X

j B(x)
xj+k ln(xj+k )
j+l−1
=−
− ln(k + 1 − kxj+k ) + xj ln(1 − xj+k ) −
ln(xj+l ):
j−1
k + 1 − kxj+k
j−1
l=0

However, we have 0¡xj+k 6 21 , and therefore |xj+k ln(xj+k )|6e−1 6 21 . Furthermore,
06j6k − 1. Therefore we get
|
j B(x)|6j−1



1
+ ln(k + 1) + ln(2) + k ln(2) 6K j−1 ;
k +2


(5.7)

1
¡xj+l ¡1
2

for

(5.8)

where K = 1 + ln(−1 ) + −1 ln(2), since after Lemma 4, k + 1¡−1 . Now we can bound the
dierence BP
 (x) − B1 (x) by summing the contribution of each desingularisation, and we get |B (x) −
P
P
m
−1
B1 (x)|6K Nn=1 j(n)−1 6L()K Nn=1 j(n)−1 6L()K ∞
m=0  = (1 − ) L()K , where  is given
by Lemma 3, and the theorem is proved.
∞
−1
Denition. A real number x is a Brjuno number if the series
n=0 Qn 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.

P

Corollary 10. The real number x is a Brjuno number if and only if B (x) is bounded;
for any
P
−1
arbitrary but xed 0¡61. The real number x is a Brjuno number if and only if ∞
q
ln(q
n+1 )
n=0 n
is bounded; for any arbitrary 0¡61.
The Brjuno function has been introduced by Yoccoz [9] in the  = 21 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 number x is not a Brjuno number, either if it is rational, or if the denominators qn of the reduced
fractions increase suciently fast. If x satises a diophantine condition, which typically says that

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411

we have a lower bound for |x − pn =qn | which is proportional P
to qn− , with ¿1, then from (2:7)
1−
one deduces that qn+1 qn is bounded, and therefore the series n qn−1 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) dened above in (5:2). Let
L() = sup(; 1 − ); l() = inf (; 1 − ) = 1 − L():

(6.1)

The following proposition is an immediate consequence of the denitions.
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(x0−1 ) + x0 B (1 x1 ); so that B (x) = ln(|x|−1 ) + |x|B (|x|−1 ); whenever  − 16x¡;
(4)
for 0¡x¡;

B (x) = xB (x−1 ) + ln(x−1 );

for 0¡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 ¿ 12 , consider the space X of 1-periodic functions f(x), such that f(x) = f(−x) for
|x|6l() = 1 − , and dene the function ln (x) ∈ X , which coincides with ln(x) on (0; L() = ).
Then, for 0¡x¡L() = , we set
(T f)(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 T f ∈ X . Thus we have
(1 − T )B (x) = −ln (x):

(6.3)

(2) For 0¡¡ 21 , consider the space X of 1-periodic functions f(x), such that f(x) = f(−x) for
|x|¡l() = , and dene 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
(T f)(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 denition of X , so that T f ∈ X . Let now B (x) = B (−x) (note that B ∈ X implies B (x) ∈ X ). Thus
we have
(1 − T )B (x) = −ln (x):
Now, we need to inverse the operator 1 − T . We have the following theorem.

(6.5)

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P. Moussa et al. / Journal of Computational and Applied Mathematics 105 (1999) 403–415

Theorem 12. Let 0¡61. Assume that there exists a measure m on (0; L()); which is invariant
under the transformation A . Then X becomes a Banach space X;p ; taking for f ∈ X the norm
kfkp equal to the Lp -norm of f on (0; L()); with respect to the measure m . Then the spectral
radius of T in this Banach space is bounded by the constant ¡1 of Lemma 3 above; and therefore
(1−T ) is invertible in this space. If this measure is equivalent to the Lebesgue measure on (0; L());
the function B belongs to the space X;p .
Proof. Note rst that due to the parity properties, the norm is equivalent to the norm Lp (T) with respect 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 ¿ 12 , we get from (6:2) for the nth iterate T(n) of T ; (T(n) f)(x) =
n−1 f(A(n) (x)), and since the measure is invariant, kT(n) fk6kfkmaxx∈(0;L()) [n−1 (x)]6n kfk, 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 function B belongs to Lp (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  = 21 , the map A1=2 is continuous everywhere in (0; 12 ); 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 )−1 f 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¡x6

1
2

(6.6)

by adding to the right-hand side a C 1 function, the solution (even and 1-periodic) is modied by a
C 1=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. Let B1+ and B1− be the even and odd part of B1 . B1+ and B1− are periodic; so that they
are determined by their values in [0; 12 ]. We have
for x ∈ [0; 12 ]; B1− (x) =
B1+ (x)

=

xB1+

1
x

 

x
1− x
ln
;
2
x


+ g(x) − ln x;



(6.7)
(6.8)

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413

with; still for x ∈ [0; 21 ];
1− x
x
g(x) = − ln
2
x




+

xB1−

1
:
x

 

Proof. For x ∈ [0; 12 ] ∩ [R \ Q], we have B1 (−x) = B1 (1 − x), and 1¡(1 − x)−1 ¡2, so that B1 (−x) =
−ln(1 − x) + (1 − x)B1 ((1 − x)−1 − 1). But
B1



1
x
− 1 = B1
1− x
1− x






= ln



1− x
x



+

x
1
B1
;
1− x
x
 

since 0¡x=(1 − x)¡1. Therefore B1 (−x) = −x ln(1 − x) − (1 − x) ln x + xB1 (x−1 ). Since we also
have B1 (x) = −ln x + xB1 (x−1 ), we easily get (6.7) by subtraction. By addition, we get
B1+ (x)

= xB1

1
x

 

x
1− x
− ln x − ln
2
x




= xB1

1
x

 

− ln x − B1− (x);

(6.9)

which leads to (6.8) since we already know B1− .
Finally, we mention a stronger result from [5], namely that the dierence B1+ (x) − B1=2 , can be
extended as an even 1-periodic continuous function which is Holder continuous with exponent 21 .

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).
Theorem 15. The dynamical system dened by the iteration of (3:4) preserves an absolutely continuous (with respect to Lebesgue) probability measure m with density c  (x); that is m (d x) =
c  (x) d x. The density is given by
(i) if g661; then 061 − 6A () = −1 − 16; and c−1 = ln(1 + ):
(−1 −1;) (x) (1−;−1 −1] (x)
+
 (x) =
+
1+x
1 + 2x



1
1
+
(0;1−] (x);
2+x 2− x


(7.1)

(ii) if 1 −
6¡g; then 06A () = 2 − −1 ¡1 −6A (1 − ) = (1 − )−1 − 2¡; and c−1 = ln(G):
 (x) =

((1−)−1 −2;) (x) (1−;(1−)−1 −2] (x)
+
G+x
2+x
1
1
+
+
(2−−1 ;1−] (x) +
x+2 G+1− x






1
1
+
(0;2−−1 ] (x);
x+2 2− x


(7.2)

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P. Moussa et al. / Journal of Computational and Applied Mathematics 105 (1999) 403–415

(iii) if 21 6¡1−
; then 06A ()=2−−1 6A (1−)=(1−)−1 −2¡1−6; and c−1 =ln(G):
(1−;) (x)
 (x) =
+
G+x
+





1
1
((1−)−1 −2;1−] (x)
+
G+x G+1− x


1
1
+
(2−−1 ;(1−)−1 −2] (x) +
x+2 G+1− x




1
1
+
(0;2−−1 ] (x);
x+2 2− x


(7.3)

(iv) if
6¡ 12 ; then 0¡A (1−)=(1−2)=(1−)6A ()=(1−2)=¡¡1−; and c−1 =ln(G):
 (x) =

1
(;1−] (x)
1
+ [(1−2)=;) (x)
+
G+1− x
G+x G+1− x




1
1
1
1
+
−
+ [(1−2)=(1−);(1−2)=) (x)
+
G+x G+1− x 2− x G+1+x


+ (0;(1−2)=(1−)) (x)





1
1
1
1
1
1
−
−
:
+
+
+
G+x G+1− x 2+x 2− x G+1+x G− x


(7.4)
where (a;b) (x) denotes the characteristic function of the interval (a; b).
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].
Note that if  = 1, one recovers Gauss’ result: c1 1 (x) = 1=(1 + x)ln 2, and if  = 12 , one nds
c1=2 1=2 (x) =

1
ln G



1
1
+
:
G+x G+1− x


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  = 52 , and  = 1 − g = g2 , with the dierence that the density in each interval of denition
is expressed as an innite 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
of Professor Waadeland.

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415

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University of Florence, Italy, 1995.
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