# Global and local minima of $\alpha$-Brjuno functions

A. Bakhtawar The first author acknowledge the support of the Centro di Ricerca
Matematica Ennio de Giorgi. The first and the third authors acknowledge the
support of UniCredit Bank R&D group for financial support through the
‘Dynamics and Information Theory Institute’ at the Scuola Normale Superiore.
C. Carminati  The second author is partially supported by the PRIN Grant
2022NTKXC of the Ministry of University and Research (MUR), Italy. The second
author acknowledges the support of the MIUR Excellence Department Project
awarded to the Department of Mathematics, University of Pisa, CUP
I57G22000700001. S. Marmi

###### Abstract

The aim of this article is to analyze some peculiar features of the global
(and local) minima of $\alpha$-Brjuno functions $B_{\alpha}$ where
$\alpha\in[\frac{1}{2},1].$ Our starting point is the result by
Balazard–Martin (2020), who showed that the minimum of $B_{1}$ is attained at
$g:=\frac{\sqrt{5}-1}{2}$; analyzing the scaling properties of $B_{1}$ near
$g$ we shall deduce that all preimages of $g$ under the Gauss map are also
local minima for $B_{1}$. Next we consider the problem of characterizing
global and local minima of $B_{\alpha}$ for other values of $\alpha$: we show
that for $\alpha\in(g,1)$ the global minimum is again attained at $g$, while
for $\alpha=1/2$ the function $B_{1/2}$ attains its minimum at
$\gamma:=\sqrt{2}-1$.

##  1 Introduction

Let $x\in R\setminus\mathbb{Q}$ and let
$\left\\{\frac{p_{n}}{q_{n}}\right\\}_{n\geq 0}$ be the sequence of its
convergents of its continued fraction expansion. A Brjuno number is an
irrational number $x$ such that $\sum^{\infty}_{n=0}\frac{\log
q_{n+1}}{q_{n}}<\infty.$ Almost all real numbers are Brjuno numbers, since for
all Diophantine numbers one has $q_{n+1}=\mathbb{O}(q^{\tau+1}_{n})$ for some
$\tau\geq 0.$ But some Liouville numbers also verify the Brjuno condition,
e.g. $\sum^{\infty}_{n=0}10^{-n!}.$ The importance of Brjuno numbers comes
from the study of one-dimensional analytic small divisor problems in dimension
one. In the case of germs of holomorphic diffeomorphisms of one complex
variable with an indifferent fixed point, extending a previous result of
Siegel [12], Brjuno proved [3] that all germs with linear part
$\lambda=e^{2\pi ix}$ are linearizable if $x$ is a Brjuno number.

The most important results are due to Yoccoz [13], who proved that the Brjuno
condition is optimal for the problem of linearization of germs of analytic
diffeomorphisms with a fixed point (and also for linearizing analytic
diffeomorphisms of the circle provided that they are sufficiently close to a
rotation [14]).

The set of Brjuno numbers is invariant under the action of the modular group
$PGL(2,\mathbb{Z})$ and it can be characterized as the set where the Brjuno
function $B:R\setminus\mathbb{Q}\to\mathbb{R}\cup\\{+\infty\\}$ is finite.
This arithmetical function is $\mathbb{Z}$\- periodic and satisfies a
remarkable functional equation which allows B to be interpreted as a cocycle
under the action of the modular group. The Brjuno function gives the size
(modulus $L^{\infty}$ functions) of the domain of stability around an
indifferent fixed point [13] and it conjecturally plays the same role in many
other small divisor problems [6, 9, 10].

###  1.1 $\alpha$-continued fractions

Let $\alpha\in[{1}/{2},1]$ and let $A_{\alpha}:[0,\alpha]\to[0,\alpha]$ be the
transformation of $\alpha$-continued fraction defined by $A_{\alpha}(0)=0$ and

$A_{\alpha}(x)=\left|\frac{1}{x}-\left[\frac{1}{x}-\alpha+1\right]\right|,\;\;\mbox{for
}x\neq 0$ (1.1)

where $[x]$ is the integer part of $x$.

Given any $\alpha\in[1/2,1],$ each $x\in(0,\alpha]\setminus\mathbb{Q}$ has an
infinite $\alpha$-continued fraction obtained by iterating the transformation
$A_{\alpha}$ as follows. For $n\geq 0$ let

$x_{0}=|x-[x-\alpha+1]|,a_{0}=[x-\alpha+1],x_{n+1}=A_{\alpha}(x_{n})=A^{n+1}_{\alpha}(x),a_{n+1}=\left[\frac{1}{x_{n}}-\alpha+1\right]\geq
1.$

Then $x_{n}^{-1}=a_{n+1}+\epsilon_{n+1}x_{n+1}$ and

$x=a_{0}+\epsilon_{0}x_{0}=a_{0}+\frac{\epsilon_{0}}{a_{1}+\epsilon_{1}x_{1}}=\cdots=a_{0}+\frac{\epsilon_{0}}{a_{1}+\frac{\epsilon_{1}}{a_{2}+\cdots+\frac{\epsilon_{n-1}}{a_{n}+\epsilon_{n}x_{n}}}}.$

We will denote the $\alpha$-expansion of $x$ as
$x=[(a_{0},\epsilon_{0});(a_{1},\epsilon_{1}),\cdots,(a_{n},\epsilon_{n}),\cdots].$
Since throughout this article we will often assume $x\in[0,\alpha]$ then
$(a_{0},\epsilon_{0})=(0,+1)$ and therefore we write
$x=[(a_{1},\epsilon_{1}),\cdots,(a_{n},\epsilon_{n}),\cdots]$. Note that when
$\alpha=1$ we recover the standard continued fraction expansion defined by the
iteration of Gauss map and in that case $\epsilon_{n}=1$ for all $n.$ When
$\alpha=1/2,$ we obtain the so called nearest integer continued fraction, and
in this case $a_{n}\geq 2$ for all $n\geq 1.$

The $nth$-convergent is defined by

$\frac{p_{n}}{q_{n}}=[(a_{0},\epsilon_{0});(a_{1},\epsilon_{1}),\cdots,(a_{n},\epsilon_{n})].$

The sequences $p_{n}$ and $q_{n}$ are recursively determined by the following
recursion relation

$p_{n}=a_{n}p_{n-1}+\epsilon_{n-1}p_{n-2},\quad
q_{n}=a_{n}q_{n-1}+\epsilon_{n-1}q_{n-2},\quad p_{-1}={q_{-2}}=1,\quad
p_{-2}=q_{-1}=0.$

For all $n,$ we have
$q_{n}p_{n-1}-p_{n}q_{n-1}=(-1)^{n}\epsilon_{0}\cdots\epsilon_{n-1}$ and

$\displaystyle
x=\frac{p_{n}+p_{n-1}\epsilon_{n}x_{n}}{q_{n}+q_{n-1}\epsilon_{n}x_{n}}\text{
and }x_{n}=-\epsilon_{n}\frac{q_{n}x-p_{n}}{q_{n-1}x-\epsilon_{n}p_{n-1}}.$

Let $\beta_{n}:=\prod^{n}_{i=0}x_{i}$ be the product of the iterates along the
$A_{\alpha}$ orbit satisfying

$\displaystyle\beta_{n}=\prod^{n}_{i=0}A^{i}_{\alpha}(x)=\prod^{n}_{i=0}x_{i}=(-1)^{n}(\epsilon_{0}\cdots\epsilon_{n-1})(q_{n}x-p_{n})\text{
for }n\geq 0,\text{ with }\beta_{-1}=1.$ (1.2)

Then

$x_{n}=\frac{\beta_{n}}{\beta_{n-1}}.$ (1.3)

Following Yoccoz [13], one can introduce the following (generalised) Brjuno
function (see [5, 7, 8, 11]) defined as $B_{\alpha}(x)=+\infty$ for
$x\in\mathbb{Q}$ and for irrational values as

$B_{\alpha}(x)=\sum^{\infty}_{j=0}\beta_{j-1}(x)\log
x^{-1}_{j}=\sum^{\infty}_{j=0}\beta_{j-1}(x)\log(1/A_{\alpha}^{j}(x)),$ (1.4)

where the $x_{n}$ follow $x_{0}=x$ by repeated iterations of the map
$A_{\alpha}$ as defined in (1.1) and the $\beta_{n}$’s are given by (1.2) with
$\beta_{-1}=1.$ The Brjuno function satisfies the functional equation

$B_{\alpha}(x)=-\log(x)+xB_{\alpha}(A_{\alpha}(x))\text{ for all
}x\in(0,\alpha)$ (1.5)

and more generally,

$B_{\alpha}(x)=B_{\alpha}^{(K)}(x)+\beta_{K}(x)B_{\alpha}(A_{\alpha}^{K+1}(x))\quad(K\in\mathbb{N},x\in(0,1)\setminus\mathbb{Q}),$
(1.6)

where $B_{K}$ denotes the partial sum w.r.t $\alpha$-continued fraction

$B_{\alpha}^{(K)}(x)=\sum_{j=0}^{K}\beta_{j-1}{(x)}\log(1/A_{\alpha}^{j}(x)).$
(1.7)

The set of the Brjuno numbers is a subset of the irrational numbers. If a real
number $x$ is rational, or if the denominators $q_{n}$ of the reduced
fractions increase extremely fast then $x$ is not a Brjuno number. If $x$
satisfies a diophantine condition, which typically says that we have a lower
bound for $|x-\frac{p_{n}}{q_{n}}|$ which is proportional to $q_{n},$ with
$\mu>1,$ then from (1.2) one deduces that $q_{n+1}q^{1-\mu}_{n}$ is bounded
and therefore the series $\sum^{\infty}_{n=0}\frac{\log(q_{n+1})}{q_{n}}$
converges. Therefore diophantine irrationals, and in particular algebraic
numbers, are Brjuno numbers.

For $\alpha=1$ it is not difficult to prove that the Brjuno function is lower
semicontinuous (and in fact the same also holds for all rational values of
$\alpha$, see Proposition 3.2). Other local properties of $B_{1}$ were studied
in [1] where the authors showed that the Lebesgue points of the Brjuno
function $B_{1}$ are exactly the Brjuno numbers and the multifractal analysis
for $B_{1}$ is carried out in [4].

Figure 1: The graph of $B_{1}$ associated to Gauss map.

The aim of this paper is to characterize the local and global minima of
$B_{1}$ by means of a fine analysis of its scaling properties near $x=g$,
which is the point where $B_{1}$ attains its minimum. and try understand what
can be said of global and local minima of $B_{\alpha}$ for other values of
$\alpha\in[1/2,1].$ After giving a general discussion of the setup of the
problem in Section 2 and 3, in Section 4 we will consider the case when
$\alpha=1.$ In Section 5 we will study the global minima of $B_{\alpha}$ for
$\alpha\in(g,1).$ In Section 6, we deal with the case $\alpha=1/2.$

##  2 Notations and preliminary results

Let us define some notions for some algebraic numbers including the golden
ratio and the silver ratio as follows

$g=\frac{\sqrt{5}-1}{2}\quad
G=g^{-1}=\frac{\sqrt{5}+1}{2},\quad\gamma=\sqrt{2}-1\quad\Gamma=\gamma^{-1}=\sqrt{2}+1.$

Note that $1-2\gamma=\gamma^{2}$ and $\gamma^{-1}=2+\gamma.$

######  Proposition 2.1 ([7]).

Given $\alpha\in[1/2,1],$ for all $x\in\mathbb{R}\setminus\mathbb{Q}$ and for
all $n\geq 1$ one has

  * (i)

$q_{n+1}>q_{n}>0;$

  * (ii)

$p_{n}>0$ when $x>0$ and $p_{n}<0$ when $x<0;$

  * (iii)

$|q_{n}x-p_{n}|=\frac{1}{q_{n+1}+\epsilon_{n+1}q_{n}{x_{n+1}}}$ so that
$\frac{1}{1+\alpha}<\beta_{n}q_{n+1}<\frac{1}{\alpha};$

  * (iv)

if $\alpha>g,$ $\beta_{n}\leq\alpha g^{n};$

  * (v)

if $\alpha\leq g,$ $\beta_{n}\leq\alpha\gamma^{n};$

Recall that a sequence
$((a_{0},\epsilon_{0});(a_{1},\epsilon_{1}),\cdots,(a_{n},\epsilon_{n}))$ is
an admissible sequence if
$[(a_{0},\epsilon_{0});(a_{1},\epsilon_{1}),\cdots,(a_{n},\epsilon_{n}),\cdots,]$
is the $\alpha$-continued fraction expansion of
$x\in(0,\alpha)\setminus\mathbb{Q}.$ For any $n\geq 1,$ the admissible block
$S=:((a_{0},\epsilon_{0});(a_{1},\epsilon_{1}),\cdots,(a_{n},\epsilon_{n}))$
is obtained via finite truncation. For any $n\geq 1,$ let

$\mathcal{L}_{n}=\\{[(a_{0},\epsilon_{0});(a_{1},\epsilon_{1}),\cdots,(a_{n},\epsilon_{n})]:S=((a_{0},\epsilon_{0});(a_{1},\epsilon_{1}),\cdots,(a_{n},\epsilon_{n}))\text{
is an admissible block, }x\in(0,\alpha)\setminus\mathbb{Q}\\}$

and

$\mathcal{L}=\bigcup_{n=1}^{\infty}\mathcal{L}_{n}.$

Given any $n\geq 1$ and $S\in\mathcal{L},$ we define $I_{S}$ the $n$-th
cylinder generated by $S$ i.e. the set of all real numbers in
$x\in(0,\alpha)\setminus\mathbb{Q}$ whose $\alpha$-continued fraction
expansion begins with the string $S.$ Let us also set

$S\cdot x=\begin{pmatrix}\epsilon_{0}\quad a_{0}\\\ 0\quad
1\end{pmatrix}\begin{pmatrix}{0}\quad{1}\\\ \epsilon_{1}\quad
a_{1}\end{pmatrix}\cdots\begin{pmatrix}{0}\quad{1}\\\ \epsilon_{n}\quad
a_{n}\end{pmatrix}\cdot x$ (2.1)

Note that in the above definition the action of the fractional transformation
corresponding to the digit of index zero is different from all the others, and
this is the reason why we use a semicolumn (rather than a comma) to separate
the digit of index zero from all the others. Let $\mathcal{L}^{*}$ denote the
set of admissible blocks without digit of index zero; for $\in\mathcal{L}^{*}$
we shall use the same notation as in (2.1) (just omitting the first matrix);
there hardly is any risk of confusion since in this case the digit of index
zero would correspond to the identity matrix.

If $r\in\mathbb{Q}$ then the $\alpha$-expansion of $r$ is finite: indeed to
determine the first partial quotient of $r\in\mathbb{Q}$ we write
$r=a_{0}+\epsilon_{0}r_{0}$ with $r_{0}\in[0,\alpha)$, we can then compute
iteratively the other partial quotients applying repeatedly the map
$A_{\alpha}$ to $r_{0}:$ we get a (finite) sequence of rational values
$A_{\alpha}^{k}(r_{0})=\frac{|p_{k}|}{q_{k}}$ with

$|p_{k}|<q_{k},\ \ \ \ \ \alpha-1\leq\frac{q_{k-1}}{p_{k-1}}-a_{k}<\alpha,\ \
\ \ 1\leq q_{k}=|p_{k-1}|<q_{k-1}.$

The last property shows that this algorithm will end in a finite number of
steps: indeed by the pidgeon’s hole principle we will eventually get
$q_{k}=1,$ $p_{k}=0$, that in $A_{\alpha}^{k}(r_{0})=0$.

Actually, as in the case $\alpha=1$, for all $\alpha\in(0,1)$ every
$r\in\mathbb{Q}$ admits exactly two $\alpha$ continued fraction expansions
i.e.

######  Lemma 2.2.

Given $r\in\mathbb{Q}$ there exists $S,S^{\prime}\in\mathcal{L}$ such that
$S\cdot 0=r=S^{\prime}\cdot 0.$ Moreover, considering the following maps

$\varphi_{S}:y\mapsto S\cdot y\;\;\mbox{ and
}\;\;\varphi_{S^{\prime}}:y\mapsto S^{\prime}\cdot y,$

  1. (i)

one is orientation preserving and the other is orientation reversing on a
right neighbourhood of zero,

  2. (ii)

if $y>0$ is sufficiently small, all values of the form $S\cdot y$ (resp.
$S^{\prime}\cdot y$) have an $\alpha$-expansion starting with $S$ (resp.
$S^{\prime},$)

  3. (ii)

$S$ and $S^{\prime}$ can be used to parametrize the corresponding cylinders,
i.e. there exist $\delta,\delta^{\prime}>0$ such that

$I_{S}=\\{S\cdot
y,\,\,\,0<y<\delta\\}\,\,\,\,\,\,\,I_{S^{\prime}}=\\{S^{\prime}\cdot
y,\,\,\,0<y<\delta^{\prime}\\}.$

Hence $r$ is the separation point between the cylinders $I_{S}$ and
$I_{S^{\prime}}$.

###  2.1 Behaviour of $B_{\alpha}$ near rational points

######  Lemma 2.3.

Let $\frac{p}{q}\in\mathbb{Q},$ then $B_{\alpha}(x)\to\infty\text{ as
}x\to\frac{p}{q}$ for all $\alpha.$

###### Proof.

Let $S,S^{\prime}\in\mathcal{L}$ be the two expansions of $\frac{p}{q}$ as in
Lemma 2.2 such that $\frac{p}{q}=S\cdot 0=S^{\prime}\cdot 0$ and
$I_{S}\cup\frac{p}{q}\cup I_{S^{\prime}}$ is a (punctured) neighbourhood of
$\frac{p}{q}.$ Moreover, suppose $x\in I_{S},$ we can write $x=S\cdot y$ where
$y>0.$ Then

$B_{\alpha}(x)\geq\beta_{n-1}(x)\log x_{n}^{-1}=\beta_{n-1}(S\cdot y)\log
y^{-1}.$ (2.2)

Applying Lagrange mean value theorem we can find $\xi\in(0,y)$ such that

$\displaystyle\left|x-\frac{p}{q}\right|=\left|S\cdot y-S\cdot
0\right|=\left|\varphi^{{}^{\prime}}_{S}(\xi)\right|y\geq\min_{\xi\in(0,y)}\frac{1}{(q_{n}+q_{n-1}\epsilon_{n}\xi)^{2}}y\geq\frac{1}{4q_{n}^{2}}y.$

Therefore

$y^{-1}\geq\frac{|x-\frac{p}{q}|^{-1}}{4q^{2}}$ (2.3)

since $q_{n}=q.$ By using (3.5), (2.3) and the fact that $\beta_{n-1}(S\cdot
y)>\frac{1}{(\alpha+1)q}$ we obtain

$B_{\alpha}(x)\geq\frac{1}{(\alpha+1)q}(\log|x-\frac{p}{q}|^{-1}-\log(4q^{2}))\to\infty,\text{
as }x\to\frac{p}{q}.$

The same argument applies on $I_{S^{\prime}}$ gaining the same result on the
other half neighbourhood of $p/q$. ∎

###  2.2 Mean value property for $B_{\alpha}$ when $\alpha\in\mathbb{Q}$

Even though the Bruno function $B_{\alpha}$ is never continuous, nonetheless
it satisfies the mean value property (at least when $\alpha\in Q$).

######  Proposition 2.4.

[Mean Value Theorem for $B_{\alpha}$] Let $\alpha\in\mathbb{Q}.$ If
$B_{\alpha}(a)=\lambda,$ $B_{\alpha}(b)=\nu,$ then for all $\gamma$ between
$\lambda$ and $\nu$ there exists $\xi\in(a,b)$ such that
$B_{\alpha}(\xi)=\gamma.$

The property is interesting in itself, but before giving a proof of the above
proposition, let us state (and prove) the following somewhat surprising
consequence:

######  Corollary 2.5.

Let $\alpha\in\mathbb{Q}\cap(0,1).$ Then

$\overline{\left\\{(x,y)\in\mathbb{R}^{2}:y=B_{\alpha}(x)\right\\}}=\left\\{(x,y)\in\mathbb{R}^{2}:y\geq
B_{\alpha}(x)\right\\}.$

###### Proof.

Let $B_{\alpha}(\xi)<+\infty,$ $\gamma>B(\xi).$ For any $n\in\mathbb{N}$ we
can pick a rational value $r$ such that $\xi<r<\xi+\frac{1}{n},$
$B_{\alpha}=+\infty.$ This implies there exists
$\xi_{n}\in(\xi,\xi+\frac{1}{n}),$ and $B_{\alpha}(\xi_{n})=\gamma$, so that
$\left(\xi_{n},B(\xi_{n})\right)\to(\xi,\gamma)$ as $n\to\infty$. ∎

In order to prove Proposition 2.4 we shall use the following auxiliary lemma

######  Lemma 2.6.

Let $\tilde{B}(x)=\sum^{\infty}_{k=0}\frac{\log a_{k+1}}{q_{k}},$ where
$x=[a_{0};a_{1},a_{2},\cdots,a_{n},a_{n+1},\cdots]$ is the regular continued
fraction expansion of $x$ (i.e. the one associated to the Gauss map). Let
$\xi$ be such that $\tilde{B}_{\alpha}(\xi)<+\infty.$ Then for all
$\epsilon>0$ there exists $\xi^{+}\in(\xi,\xi+\epsilon)$ and
$\xi^{-}\in(\xi-\epsilon,\xi)$ such that

$\left|\tilde{B}_{\alpha}(\xi^{\underset{-}{+}})-\tilde{B}_{\alpha}(\xi)\right|<\epsilon.$

###### Proof.

Let us consider the case $\xi^{+}.$ Let us fix $\epsilon>0$ and let us pick
$N$ such that

$\sum^{\infty}_{k=N}\frac{\log a_{k+1}}{q_{k}}<\frac{\epsilon}{2}.$

Then we will pick an even $n\geq N$ such that $\frac{\log
2}{q_{n-1}}<\frac{\epsilon}{2}$ and set
$\xi^{+}=[a_{0};a_{1},a_{2},\cdots,a_{n},a_{n+1},1,1,\cdots].$ Then

$\displaystyle|\tilde{B}(\xi)-\tilde{B}(\xi^{+})|$
$\displaystyle=\frac{\log(a_{n}+1)-\log(a_{n})}{q_{n-1}}+\sum^{\infty}_{k=n}\frac{\log
a_{k+1}}{q_{k}}$ $\displaystyle\leq\frac{\log
2}{q_{n-1}}+\frac{\epsilon}{2}<\epsilon.$

Note that in above proof we consider $\xi^{+}.$ The similar argument holds for
$\xi^{-}$ with $n$ odd.

∎

The main tool to prove Theorem 4.5 is the following

######  Lemma 2.7.

Let $\xi$ such that $B_{\alpha}(\xi)<+\infty.$ Then for all $\epsilon>0$ there
exists $\xi^{+}\in(\xi,\xi+\epsilon)$ and $\xi^{-}\in(\xi-\epsilon,\xi)$ such
that

$\left|B_{\alpha}(\xi^{\pm})-B_{\alpha}(\xi)\right|<\epsilon.$

###### Proof.

Again, we prove the statement just for $|\xi^{+}$; let
$\xi=[a_{0};a_{1},a_{2},\cdots,a_{n},a_{n+1},\cdots]$ such that such that
$B_{\alpha}(\xi)<+\infty$, and let us fix $\epsilon>0$ and choose $N$ such
that

$\beta_{N}B_{\alpha}(A_{\alpha}^{N+1}(\xi))<\frac{\epsilon}{2}.$

Since $B_{\alpha}(\xi)<+\infty$ the point $\xi$ is irrational, and since
$\alpha\in\mathbb{Q}$ there exists a neighbourhood $U$ of the point $\xi$ such
that the truncated Bruno function $B^{N}_{\alpha}$ (defined in (1.7)),
$\beta_{n}$, and $A_{\alpha}^{N+1}$ are all continuous on $U$.

We shall set $\xi^{+}=[a_{0};a_{1},a_{2},\cdots,a_{n},a_{n+1}+1,1,1,\cdots]$
for a suitable $n\geq N,$ and we only consider even $n$ in order to have
$\xi\leq\xi^{+}$.

$\displaystyle\left|B_{\alpha}(\xi)-B_{\alpha}(\xi^{+})\right|\leq\left|B^{N}_{\alpha}(\xi)-B^{N}_{\alpha}(\xi^{+})\right|+\left|\beta_{N}(\xi)B_{\alpha}(A_{\alpha}^{N+1}\xi)-\beta_{N}(\xi^{+})B_{\alpha}(A_{\alpha}^{N+1}\xi^{+})\right|$

Since $B^{N}_{\alpha}$ is continuous at irrational points, the term
$\left|B^{N}_{\alpha}(\xi)-B^{N}_{\alpha}(\xi^{+})\right|$ is arbitrarily
small if $n$ is large. The same is true for the other term in the sum since

$\displaystyle\left|\beta_{N}(\xi)B_{\alpha}(A_{\alpha}^{N+1}\xi)-\beta_{N}(\xi^{+})B_{\alpha}(A_{\alpha}^{N+1}\xi^{+})\right|\leq\beta_{N}(\xi^{+})\left|B_{\alpha}(A_{\alpha}^{N+1}\xi)-B_{\alpha}(A_{\alpha}^{N+1}\xi^{+})\right|+|\beta_{N}(\xi)-\beta_{N}(\xi^{+})|B_{\alpha}(A_{\alpha}^{N+1}\xi)$

So, the fact that this term can be made arbitrarily small (choosing $n$ big)
is a consequence of the continuity of $\beta_{N}$ and $A_{\alpha}^{N+1}$ un
$U$, by the fact that $\beta_{N}$ is small, and the fact that the term
$\left|B_{\alpha}(A_{\alpha}^{N+1}\xi)-B_{\alpha}(A_{\alpha}^{N+1}\xi^{+})\right|$
is bounded since $B_{\alpha}-\tilde{B}$ is bounded and

$\left|B_{\alpha}(A_{\alpha}^{N+1}\xi)-B_{\alpha}(A_{\alpha}^{N+1}\xi^{+})\right|\leq\left|B_{\alpha}(A_{\alpha}^{N+1}\xi)-\tilde{B}(A_{\alpha}^{N+1}\xi)\right|+\left|\tilde{B}(A_{\alpha}^{N+1}\xi)-\tilde{B}(A_{\alpha}^{N+1}\xi^{+})\right|+\left|\tilde{B}(A_{\alpha}^{N+1}\xi^{+})-B_{\alpha}(A_{\alpha}^{N+1}\xi^{+})\right|.$

∎

##  3 Lower Semicontinuity of generalized Brjuno functions

In this section we will discuss the lower semicontinuity of generalized Brjuno
function as defined above for different choices of $\alpha.$ First we prove
that if $\alpha$ is rational than $B_{\alpha}$ is lower semi-continuous. Next
we show that the hypothesis $\alpha\in\mathbb{Q}$ cannot be dropped since
there exists irrational values $\alpha\in(1/2,1)$ such that $B_{\alpha}$ is
not lower semi-continuous.

###  3.1 Lower semi-continuity of $B_{\alpha}$ for
$\alpha\in[1/2,1]\cap\mathbb{Q}.$

######  Lemma 3.1.

For all $\alpha\in[1/2,1]\cap\mathbb{Q}$ the partial sum $B^{K}_{\alpha}(x)$
defined in (1.7) is smooth on every $K$-cylinder .

###### Proof.

Let $S\in\mathcal{L}_{K},$ since
$\varphi_{S}:(0,\delta)\xrightarrow{\sim}I_{S}$ it is enough to prove that
$B^{K}_{\alpha}(S\cdot y)$ is smooth in $y.$ For all $0\leq j\leq K,$
$A_{\alpha}^{j}(x)=(\sigma^{j}S)\cdot y$ is smooth in $y$ hence so is
$\beta_{j}(x).$ Thus $\sum^{K}_{j=0}\beta_{j-1}(S\cdot
y)\log(1/(\sigma^{j}S)\cdot y)$ is smooth in $y\in(0,\delta).$ ∎

######  Proposition 3.2.

Let $\alpha\in\mathbb{Q}\cap[1/2,1].$ Then $B_{\alpha}(x)$ is lower semi
continuous for all $x\in(0,1).$

###### Proof.

Let $D_{c}=\\{x\in(0,1):B_{\alpha}(x)\leq c\\}$ where $c>0$. To show that
$B_{\alpha}(x)$ is lower semicontinuous we need to show $D_{c}$ is closed for
all $c\in\mathbb{R}.$ Note that
$B_{\alpha}(x)=\sup_{K\to\infty}B^{K}_{\alpha}(x)$ where $B^{(K)}_{\alpha}(x)$
is defined in (1.7).

Therefore we can rewrite $D_{c}=\bigcap_{K\in\mathbb{N}}D_{K,c}$ where
$D_{K,c}=\\{x\in(0,1):B_{\alpha}^{(K)}(x)\leq c\\}.$ Thus it is enough to show
$D_{K,c}$ is closed for all $c$ and for all $K\in\mathbb{N}.$ Indeed if
$(x_{n})_{n\in\mathbb{N}}\in D_{K,c}$ such that $x_{n}\to\bar{x}$ then by
Lemma 2.3 $\bar{x}\notin\mathbb{Q}.$ That implies $A^{j}_{\alpha}$ is
continuous at $\bar{x}$ for all $1\leq j\leq K.$ Hence $B^{K}_{\alpha}$ is
continuous at $\bar{x}$ and $\bar{x}\in I_{S}$ for some $K$-cylinder. Since
$I_{S}$ is open and $x_{n}\to\bar{x}$ there exist $n_{0}$ such that $x_{n}\in
I_{S}$ for every $n\geq n_{0}.$ By the fact that $x,\bar{x}\in I_{S}$ we can
write $\bar{x}=S\cdot\bar{y},$ $x_{n}=S\cdot y_{n},$ with $y_{n}\to\bar{y}$
for all $n\geq n_{0}.$ By the continuity of $B^{K}_{\alpha}$ at $\bar{x}$ it
follows that $B^{K}_{\alpha}(S\cdot y_{n})\to B^{K}_{\alpha}(S\cdot\bar{y})$
as $n\to\infty.$ Hence $B^{K}_{\alpha}(x_{n})\to B^{K}_{\alpha}(\bar{x})$ and
$B^{K}_{\alpha}(\bar{x})\leq c$ which implies $\bar{x}\in D_{K,c}.$ Thus
$D_{K,c}$ is closed and consequently $D_{c}$ is closed. ∎

Since $B_{\alpha}$ is lower semicontinuous and 1-periodic, it can be
considered as a function on the circle (which is compact), hence admits an
absolute minimum. Without loss of generality, we can think that this minimum
belongs to the period $[\alpha-1,\alpha)$, and by the symmetry on
$[\alpha-1,1-\alpha]$ one can find a global minimum on $(0,\alpha)$.

######  Corollary 3.3.

The Brjuno function $B_{\alpha}$ has a global minimum on $[0,\alpha].$

###  3.2 Remark on the lower semi-continuity of Brjuno function when $\alpha$
is irrational

It is worth pointing out that the hypothesis $\alpha\in\mathbb{Q}$ in
Proposition 3.2 is not a technical assumption; indeed there exist irrational
$\alpha$ for which the Brjuno function $B_{\alpha}$ is not lower semi-
continuous, as the following example shows (see also Figure 2)

######  Example 3.4.

Suppose $\hat{\alpha}:=\frac{1}{1+\frac{1}{a+g}}$ where $a\geq 2$ is a
positive integer. Clearly $\hat{\alpha}\in[0,1]\setminus\mathbb{Q}.$ Then

  1. (i)

$B_{\hat{\alpha}}(\hat{\alpha})=B_{\hat{\alpha}}(1-\hat{\alpha})=\log(a+1+g)+\frac{1}{a+1+g}B_{\hat{\alpha}}(g).$

  2. (ii)

$\lim\inf_{x\to\hat{\alpha}^{+}}B_{\hat{\alpha}}(x)=B_{\hat{\alpha}}(\hat{\alpha}).$

  3. (iii)

$\lim\inf_{x\to\hat{\alpha}^{-}}B_{\hat{\alpha}}(x)<B_{\hat{\alpha}}(\hat{\alpha}).$

Figure 2: The graph showing lower semicontinuity fails when we take $\alpha$
to be irrationa. Here a=2 and $\hat{\alpha}=\frac{2+g}{3+g}.$

###### Proof.

  1. (i)

Clearly, follows from the definition (1.5) and the fact that
$A_{\hat{\alpha}}(1-\hat{\alpha})=g.$

  2. (ii)

Suppose $x\geq\hat{\alpha}.$ Then we can write
$x=\frac{1}{1+\frac{1}{a+g+\epsilon}}$ and $1-x=\frac{1}{a+1+g+\epsilon}$
where $\epsilon>0.$ Therefore

$\displaystyle\liminf_{x\to\hat{\alpha}^{+}}B_{\hat{\alpha}}(x)$
$\displaystyle=\lim\inf_{\epsilon\to
0^{+}}\left(\log(a+1+g+\epsilon)+\frac{1}{a+1+g+\epsilon}\log
B_{\hat{\alpha}}(g+\epsilon)\right)$
$\displaystyle=\log(a+1+g)+\frac{1}{a+1+g}\lim\inf_{\epsilon\to 0^{+}}\log
B_{\hat{\alpha}}(g+\epsilon),$

and the required results follows since $\lim\inf_{\epsilon\to 0^{+}}\log
B_{\hat{\alpha}}(g+\epsilon)=B_{\alpha}(g).$

  3. (iii)

Suppose $x\leq\hat{\alpha}$ and write $x=\frac{1}{1+\frac{1}{a+g-\epsilon}}$
where $\epsilon>0.$ Therefore

$\displaystyle\liminf_{x\to\hat{\alpha}^{-}}B_{\hat{\alpha}}(x)$
$\displaystyle=\lim\inf_{\epsilon\to
0^{-}}\left(\log(a+1+g-\epsilon)-\log(a+g-\epsilon)+xB_{\hat{\alpha}}(\frac{1}{a+g-\epsilon})\right)$
$\displaystyle=\lim\inf_{\epsilon\to
0^{-}}\left(\log(a+1+g-\epsilon)-\log(a+g-\epsilon)+x\left(\log(a+g-\epsilon)+\frac{1}{a+g-\epsilon}B_{\hat{\alpha}}(g)\right)\right)$
$\displaystyle=B_{\hat{\alpha}}+\log(a+g)[-1+\hat{\alpha}]<B_{\hat{\alpha}}(\hat{\alpha}),$

the last inequality follows by the fact that $\log(a+g)[-1+\hat{\alpha}]<0.$

∎

###  3.3 On the lower semi-continuity of other more general versions of Brjuno
function

Unlike the results of Section 4 and the following sections, the argument to
prove lower semi-continuity is quite general and with minor modifications it
applies also to other variants of Brjuno function.

Let $u:(0,1)\to\mathbb{R}^{+}$ be a positive $C^{1}$ function such that
$\lim_{x\to 0^{+}}u(x)=\infty$ and $\nu\in\mathbb{N}$ be fixed. Then one can
define the following class of generalized Brjuno functions which includes
those studied in [5]

$B_{\alpha,\nu,u}(x)=\sum^{\infty}_{j=0}\beta^{\nu}_{j-1}(x)u(x_{j}).$ (3.1)

Definition (3.1) is more general as for different choices of $\nu$ and $u$ it
implies various classical Brjuno functions. For example if we take $\nu=1,$
$u(x)=-\log(x)$ it implies classical $\alpha$-Brjuno function as defined in
(1.4). For other choices of the singular behaviour of $u$ at zero the
condition $B_{\alpha,\nu,u}<\infty$ leads to different diophantine condition.
For instance choose $u(x)=x^{-1/\sigma}$ where $\sigma>2.$ Then it is not
difficult to check that if
$B_{1,1,\sigma}(x)=\sum^{\infty}_{j=0}\beta_{j-1}(x)x_{j}^{-1/\sigma}<\infty$
then $x$ is Diophantine number i.e. $x\in$
$CD(\sigma):=\\{x\in\mathbb{R}\setminus\mathbb{Q}:q_{n+1}=O(q^{1+\sigma}_{n})\\}.$
For more details regarding the function $B_{1,1,\sigma}$ we refer the reader
to [5].

######  Remark 3.5.

Let $B_{\alpha,u,\nu}(x)$ be as defined in (3.1). Then
$B_{\alpha,u,\nu}(x)\to\infty$ as $x\to\frac{p}{q}.$

The proof is on the similar lines as of Lemma 2.3. Indeed we will have
$B_{\alpha,u,\nu}(x)=\sum^{\infty}_{j=0}\beta^{\nu}_{j-1}(x)u(x_{j})\geq\beta^{\nu}_{n-1}(x)u(x_{n})=\beta^{\nu}_{n-1}(S\cdot
y)u(y).$ By using the value of $y$ from (2.3), $y\to 0$ as $x\to\frac{p}{q},$
and therefore by the definition of $u,$$u(y)\to\infty.$ Consequently
$B_{\alpha,u,\nu}(x)\to\infty.$ Since $\nu$ is fixed positive integer the term
$\beta^{\nu}_{n-1}\geq\frac{1}{(\alpha+1)^{\nu}q^{\nu}_{n}}$ will become
smaller and smaller but remain non-zero.

By Remark 3.5 and the fact that the partial sum $B^{K}_{\alpha,u,\nu}(x)$ is
smooth on every $K$-cylinder we have the following.

######  Remark 3.6.

The function $B_{\alpha,u,\nu}$ is lower semi-continuous for all
$\alpha\in\mathbb{Q}\cap[1/2,1].$

##  4 Scaling properties and local minima of $B_{1}$

Throughout this section we will focus on the classical case $\alpha=1.$ In [2]
Balazard-Martin studied Brjuno function for $\alpha=1$ and showed that the
global minimum of $B_{1}$ is attained at the golden number $`g$.’ In fact they
prove the following:

######  Theorem 4.1 ([2]).

$\min_{x\in[0,1]}B_{1}(x)=B_{1}(g).$

It is natural to ask about the minima on other intervals, for instance using
Theorem 4.1, it is not difficult to prove that,

######  Corollary 4.2.

$\min_{x\in(0,1/2)}B_{1}(x)\geq B_{1}(1-g)=B_{1}(g^{2}).$

The proof of Corollary 4.2 is a routine computation, for completeness we add
its proof in Subsection 4.2.

Note that both $g$ and $g^{2}$ are “noble” numbers in the sense of the
following definition.

######  Definition 4.3.

Let $A_{1}$ be the Gauss map, the set of noble numbers $\mathcal{N}$ consists
of all inverse images of $g$ under the Gauss map:

$\mathcal{N}=\cup_{k}A_{1}^{-k}(g)$

In other words, a number $\nu$ is noble if its regular continued fraction
expansion is of the form

$\nu=[0;a_{1},a_{2},\cdots,a_{n},\bar{1}].$ (4.1)

Equivalently, using the notation introduced by equation (2.1) in Section 2,
$\nu$ is noble if there is $S=(a_{1},...,a_{n})$ such that $\nu=S\cdot g$.

Theorem 4.1 together with Corollary 4.2 and some numerical evidence naturally
lead to the following conjecture.

######  Conjecture 4.4.

Let $\mathcal{M}$ be the set of local minima of $B_{1}$. Then

$\mathcal{M}=\mathcal{N}.$

In the following we shall give a partial answer to this conjecture, indeed we
will show the inclusion $\mathcal{N}\subset\mathcal{M}$. The first step in
order to get this result is to analize the behaviour of $B_{1}$ near its
absolute minimum $g$: we aim to show that the Brjuno function $B_{1}$ has a
cusp-like minimum at $g$, namely

######  Theorem 4.5.

There exists $c>0$ such that

$B_{1}(x)-B_{1}(g)\geq c|x-g|^{1/2}\text{ for all }x\in(0,1).$ (4.2)

Before going into the proof of this statement, let us point out that the
property of being a cusp-like local minimum propagates from $g$ to all other
noble numbers.

######  Corollary 4.6.

Let $\nu$ be a noble number. Then $\nu$ is a local minimum of $B_{1}.$

###### Proof.

Let $\nu=S\cdot g$ with $|S|=K$, and let us use the map $x\mapsto S\cdot x$ to
parametrize a nbd of $\nu$

$\displaystyle B_{1}(S\cdot x)-B_{1}(S\cdot g)$
$\displaystyle=B^{(K)}_{1}(S\cdot x)-B^{(K)}_{1}(S\cdot g)+(S\cdot
x)B_{1}(x)+(S\cdot g)B_{1}(g)$ $\displaystyle\geq B^{(K)}_{1}(S\cdot
x)-B^{(K)}_{1}(S\cdot g)+(S\cdot x-S\cdot g)B_{1}(g)+c(S\cdot x)|x-g|^{1/2}.$

Let us denote the right hand side of last inequality by $\phi(x)$; it is easy
to see that there is an open neighbourhood $U$ of $g$ such that $\phi$ is
continuous on $U$ and differentiable on $U\setminus\\{g\\}$, and $\lim_{x\to
g^{\pm}}\phi^{\prime}(x)=\pm\infty$. Therefore $\phi$ has a minimum at $g$,
hence $B_{1}$ has a local minimum at $\nu=S\cdot g$. ∎

###  4.1 Proof of Theorem 4.5

The proof of Theorem 4.5 is relies on some scaling properties of $B_{1}$ near
its absolute minimum $g$; Lemma 4.7 and Lemma 4.8 will give a precise
description of this scaling property, which is actually very clearly visible
from numerical data (see also Figure 3).

Let $\Phi(x)=\frac{1}{1+x}$ and for all $n\geq 1$ define the recursive
relation

$x_{n+1}=\Phi(x_{n})\quad\text{ where }x_{0}\in(0,1/2).$ (4.3)

Note that $B_{1}(g)=\frac{\log 1/g}{1-g}.$

We prove the scaling property of $\mathcal{E}_{n}:=B_{1}(x_{n})-B_{1}(g),$
where $\mathcal{E}_{0}\geq\left(3-\frac{1}{1-g}\right)\log\frac{1}{g}>0,$
namely

######  Lemma 4.7.

$\displaystyle\mathcal{E}_{2n+1}\geq\sigma\mathcal{E}_{1}g^{2n}\quad\text{
where
}\quad\sigma:=\exp(-\sum^{\infty}_{n=1}\log(g^{2}\frac{F_{2n+2}}{F_{2n}})).$
(4.4)

Figure 3:  Graph of $\left(\log|x-g|,\log(B_{1}(x)-B_{1}(g))\right)$ for
various values of $x$ including $x_{n}=\Phi^{n}(1-g).$

###### Proof.

Since $x_{1}=\frac{1}{1+x_{0}},x_{2}=\frac{1+x_{0}}{2+x_{0}},\cdots,$
continuing in this way (4.3) induces the recursive relation

$x_{n}=\frac{F_{n}+x_{0}F_{n-1}}{F_{n+1}+x_{0}F_{n}},\quad\forall{n\geq
1}\text{ and }x_{0}\in(0,1/2),$ (4.5)

where $F_{n}$ are Fibonacci numbers with $F_{-1}=1$ and $F_{0}=0.$ Using (1.5)
we write

$\displaystyle B_{1}(\Phi(x_{n}))$
$\displaystyle=-\log(\Phi(x_{n}))+\Phi(x_{n})B_{1}(x_{n})\text{ and}$
$\displaystyle B_{1}(\Phi(g))$ $\displaystyle=-\log(\Phi(g))+\Phi(g)B_{1}(g).$

Using (4.3) and the fact that $\Phi(g)=g,$ we have

$\displaystyle
B_{1}(x_{n+1})-B_{1}(g)=-\log\frac{x_{n+1}}{g}+x_{n+1}[B_{1}(x_{n})-B_{1}(g)]+B_{1}(g)[x_{n+1}-g].$
(4.6)

Setting $\mathcal{E}_{n}:=B_{1}(x_{n})-B_{1}(g)$ and $\delta_{n}:=x_{n}-g$ and
observing that $-\log t\geq 1-t$ we get

$\mathcal{E}_{n+1}\geq x_{n+1}\mathcal{E}_{n}-l\delta_{n+1}$ (4.7)

where $l:=\frac{1}{g}-B_{1}(g)$ and $l>0.$ Hence

$\displaystyle\mathcal{E}_{n+2}$ $\displaystyle\geq
x_{n+2}\mathcal{E}_{n+1}-l\delta_{n+2}\geq
x_{n+2}x_{n+1}\mathcal{E}_{n}-l(x_{n+2}\delta_{n+1}+\delta_{n+2}).$

Note that

$\displaystyle
x_{n+2}\delta_{n+1}+\delta_{n+2}=x_{n+2}(1+x_{n+1})-g(1+x_{n+2})=1-g(x_{n+2}),$

which is positive if and only if $x_{n+2}<g$ i.e when $n$ is even. Therefore
for all $n\geq 1,$

$\displaystyle\mathcal{E}_{2n+1}\geq\lambda_{n}\mathcal{E}_{2n-1},$ (4.8)

where $\lambda_{n}:=x_{2n+1}x_{2n}$ and by using (4.5), we can write

$\lambda_{n}=\frac{F_{2n+1}+x_{0}F_{2n-1}}{F_{2n+2}+x_{0}F_{2n+1}}.$

Further for all $n\geq 1$

$\frac{F_{2n}}{F_{2n+2}}\leq\lambda_{n}\leq\frac{F_{2n+2}}{F_{2n+4}},$

i.e. $\lambda_{n}\geq 1/3$ and $\lambda_{n}\to g^{2}$ as $n\to\infty$
(exponentially).

In order to obtain the optimal estimate first note that

$\displaystyle\mathcal{E}_{2n+1}$
$\displaystyle=\big{(}\prod^{n}_{k=1}\lambda_{k}\big{)}\mathcal{E}_{1}\text{
with}$ $\displaystyle\prod^{n}_{k=1}\lambda_{k}$
$\displaystyle=\exp(\sum^{n}_{k=1}\log\lambda_{k})=\exp(n\log
g^{2}+\sum^{n}_{k=1}\log\frac{\lambda_{k}}{g^{2}})$
$\displaystyle=g^{2n}\exp(\sum^{n}_{k=1}\log\frac{\lambda_{k}}{g^{2}}).$

The term $\sum^{\infty}_{k=1}\log\frac{\lambda_{k}}{g^{2}}$ in last equation
is an absolutely convergent series because $\lambda_{k}\to g^{2}$ act an
exponential rate, therefore we can write

$\displaystyle\mathcal{E}_{2n+1}\geq\sigma\mathcal{E}_{1}g^{2n}\quad\text{
where
}\quad\sigma:=\exp(-\sum^{\infty}_{n=1}\log(g^{2}\frac{F_{2n+2}}{F_{2n}})).$
(4.9)

∎

The estimate (4.9) implies that $B_{1}$ has a cusp like minima at $g,$ namely

$B_{1}(x)-B_{1}(g)\geq c(x-g)^{1/2}\quad\text{ with }$

(where the value of the exponent $1/2$ is obtained by the corresponding
estimate and comparing with the fact that $\delta_{n}\asymp g^{2n}$ for any
$n\geq 1$).

######  Lemma 4.8.

$B_{1}(x)-B_{1}(g)\geq\frac{\sigma\mathcal{E}_{1}}{2}{(x-g)^{1/2}}\text{ for
}x>g.$ (4.10)

###### Proof.

Define $t_{n}:=\Phi^{2n-1}(\frac{1}{2})=\frac{F_{2n+1}}{F_{2n+2}}$ such that
$t_{0}:=1,$ $t_{1}:=\frac{2}{3},$ $t_{2}:=\frac{5}{8}\cdots$ and
$I_{n}=\Phi^{2n-1}((0,\frac{1}{2}))=(t_{n},t_{n}-1).$ Observe that

$t_{n}-g=\frac{F_{2n+1}}{F_{2n+2}}-g=g^{4n+3}\frac{1+g^{2}}{1-g^{4n+4}}$
(4.11)

and

$1\leq\frac{1+g^{2}}{1-g^{4n+4}}\leq G.$ (4.12)

On the other hand, by (4.9) we get $x\in I_{n}\implies
B_{1}(x)-B_{1}(g)\geq\sigma\mathcal{E}_{1}g^{2n}.$ Therefore, if $t_{n}\leq
x\leq t_{n-1}$ then

$\displaystyle x-g\leq t_{n-1}-g\leq g^{4n-1}G=G^{2}g^{4n}\text{ and }$
$\displaystyle
B_{1}(x)-B_{1}(g)\geq\sigma\mathcal{E}_{1}g^{2n}=\frac{\sigma\mathcal{E}_{1}}{G}Gg^{2n}.$

Hence for all $x\in I_{n},$ and for all $n\geq 1$

$B_{1}(x)-B_{1}(g)\geq\frac{\sigma\mathcal{E}_{1}}{G}({x-g})^{1/2}.$

Since $\bigcup_{n\geq 1}I_{n}=(g,1),$ we obtain the required result, with a
constant which is slightly better. ∎

To complete the proof of Theorem 4.5 we will also include the case when
$1/2<x<g.$

######  Lemma 4.9.

$B_{1}(x)-B_{1}(g)\geq\frac{\sigma\mathcal{E}_{1}}{2}\sqrt{g-x}\text{ for
}1/2<x<g.$

###### Proof.

Let $1/2<x<g$ and write $x=\Phi(t)$ with $t\in(g,1).$ Then by repeating the
same argument used in the proof of last lemma we obtain

$\displaystyle B_{1}(\Phi(t))-B_{1}(\Phi(g))$
$\displaystyle\geq(\Phi(t)-\Phi(g))\left(B_{1}(g)-\frac{1}{g}\right)+\Phi(t)(B_{1}(t)-B_{1}(g))$
$\displaystyle\geq\Phi(t)\frac{\sigma\mathcal{E}_{1}}{G}\sqrt{t-g}.$

Observe that

$\displaystyle\Phi(g)-\Phi(t)$ $\displaystyle=\Phi^{\prime}(\xi)(g-t)\text{
for some }\xi\in(g,t)$ $\displaystyle=\frac{1}{(1+\xi)^{2}}(t-g)\text{ since
}\Phi^{\prime}(\xi)=-\frac{1}{(1+\xi)^{2}}$

hence $\sqrt{t-g}=(1+\xi)\sqrt{g-x}$ for some $\xi\in(g,t)$ and

$\displaystyle
B_{1}(x)-B_{1}(g)\geq\frac{1+\xi}{1+t}\frac{\sigma\mathcal{E}_{1}}{G}\sqrt{g-x}$

with $t\in(g,1),$ $\xi\in(g,t)$ so that

$\frac{1+\xi}{1+t}\geq\frac{1+g}{1+t}\geq\frac{1+g}{2}=\frac{G}{2}.$

This implies

$B_{1}(x)-B_{1}(g)\geq\frac{\sigma\mathcal{E}_{1}}{2}\sqrt{g-x}\text{ for
}1/2<x<g,$

hence proving the claim. ∎

###  4.2 Proof of Corollary 4.2

###### Proof.

$\displaystyle B_{1}(x)$ $\displaystyle=-\log x+xB_{1}(A_{1}(x))$
$\displaystyle\geq-\log x+xB(g)\text{ on }x\in[0,1).$

Whereas for $x\in(1/3,1/2),$ we have

$\displaystyle B_{1}(x)$ $\displaystyle=-\log
x+xB_{1}\left(\frac{1}{x}-2\right)$ $\displaystyle=-\log
x+x\left(\log\frac{x}{1-2x}+(1-2x)B(A_{1}^{2}(x))\right)$
$\displaystyle\geq-\log x+x\log\frac{x}{1-2x}+(1-2x)B(g).$

Setting,

$\displaystyle\varphi_{0}(x)$ $\displaystyle=-\log x+xB(g)\text{ and }$
$\displaystyle\varphi_{1}(x)$ $\displaystyle=-\log
x+x\log\frac{x}{1-2x}+(1-2x)B(g).$

We have $\varphi_{0}(g^{2})=\varphi_{1}(g^{2})=B_{1}(g^{2})=-3\log g.$ Next we
set

$\varphi(x)=\begin{cases}\varphi_{0}(x)&\text{ for }x<g^{2}\\\
\varphi_{1}(x)&\text{ for }x\geq g^{2}.\end{cases}$

Clearly $\varphi$ is continuous and for all $x\in(0,1/2)$ we have

$B_{1}(x)-B_{1}(g^{2})\geq\varphi(x)-B_{1}(g^{2})\geq d\quad\text{ where
}\quad d\approx 1.8.$

∎

##  5 Minima of $B_{\alpha}$ when $g<\alpha<1.$

In this section we show that for any real number $\alpha\in(g,1)$ the minimum
of $B_{\alpha}$ is attained at $g.$ It is worth mentioning that the method
introduced here works for every $\alpha$ including those irrational values for
which $B_{\alpha}$ is not even lower semi-continuous (which would be an
obstruction for the use of Balazard-Martin[2] arguments).

######  Theorem 5.1.

For all $\alpha\in(g,1],$ we have that $B_{\alpha}(g)=B_{1}(g)$ and

$\min_{x\in[0,\alpha]}B_{\alpha}(x)=B_{\alpha}(g).$

######  Corollary 5.2.

Let $g<\alpha<1.$ If $\nu=S\cdot g$ is a noble number such that

$A^{k}_{\alpha}(\nu)\neq\alpha\text{ for all }k\leq|S|.$

Then $\nu$ is a local cusp-like minimum for $B_{\alpha}$. For example
$\nu=g^{2}$ is a local minimum for $B_{\alpha}$ for all $g<\alpha<1.$

The proof of Corollary 5.2 is an immediate consequence of Theorem 5.1 and the
fact that $A^{k}_{\alpha}$ is smooth at $\nu$ for all $k\leq|S|$. In turn,
Theorem 5.1 is an immediate consequence of the following lemma:

######  Lemma 5.3.

For all $\alpha\in(g,1)$

$B_{\alpha}(x)\geq B_{1}(x).$

Indeed, if $\alpha\in(g,1)$ then $B_{\alpha}(g)=B_{1}(g),$ and using the fact
that $g$ is the minimum of $B_{1},$ inequality (5.3) implies

$B_{\alpha}(x)\geq B_{1}(x)\geq B_{1}(g)\text{ for all }x.$

Therefore to reach our goal we only have to prove the inequality (5.3):

###### Proof.

Recall that $\beta_{k}(x)=|xq_{k}-p_{k}|$ where $\frac{p_{k}}{q_{k}}$ is the
$k^{th}$ $\alpha$-convergent of $x.$ By using [7, Lemma 1.8.], we can write

$p_{k}=P_{n(k)}\quad q_{k}=Q_{n(k)},$

where $\frac{P_{n}}{Q_{n}}$ is the $n^{th}$ convergent of the regular
continued fraction of $x.$ Moreover one has that $n(k)-n(k-1)\in\\{1,2\\}$
i.e. either $n(k-1)=n(k)-1$ or $n(k-1)=n(k)-2.$ Therefore setting
$\tilde{\beta}_{n}=|Q_{n}x-P_{n}|$ we get $\beta_{k}=\tilde{\beta}_{n(k)}.$

$\displaystyle B_{\alpha}(x)$
$\displaystyle=\sum^{\infty}_{k=0}\beta_{k-1}(x)\log(1/A_{\alpha}^{k}(x))$
$\displaystyle=\sum^{\infty}_{k=0}\beta_{k-1}(x)[\log\beta_{k-1}(x)-\log\beta_{k}(x)]$
$\displaystyle=\sum^{\infty}_{k=0}\tilde{\beta}_{n(k-1)}(x)[\log\tilde{\beta}_{n(k-1)}(x)-\log\tilde{\beta}_{n(k)}(x)]$
(5.1)

Case1. If $n(k-1)=n(k)-1,$ the quantity in (5.1) coincides with
$\sum^{\infty}_{k=0}\tilde{\beta}_{n(k)-1}(x)[\log\tilde{\beta}_{n(k)-1}(x)-\log\tilde{\beta}_{n(k)}(x)],$
which is the sum producing $B_{1}(x).$

Case2. Else, it must be $n(k-1)=n(k)-2,$ and we can write (5.1) as

$\displaystyle B_{\alpha}(x)$
$\displaystyle=\sum^{\infty}_{k=0}\tilde{\beta}_{n(k)-2}(x)[\log\tilde{\beta}_{n(k)-2}(x)-\log\tilde{\beta}_{n(k)}(x)]$
$\displaystyle=\sum^{\infty}_{k=0}\tilde{\beta}_{n(k)-2}(x)[\log\tilde{\beta}_{n(k)-2}(x)-\log\tilde{\beta}_{n(k)-1}+\log\tilde{\beta}_{n(k)-1}-\log\tilde{\beta}_{n(k)}(x)]$
$\displaystyle\geq\sum^{\infty}_{k=0}\tilde{\beta}_{n(k)-2}(x)[\log\tilde{\beta}_{n(k)-1}(x)-\log\tilde{\beta}_{n(k)-1}]+\sum^{\infty}_{k=0}\tilde{\beta}_{n(k)-1}(x)[\log\tilde{\beta}_{n(k)-1}-\log\tilde{\beta}_{n(k)}(x)]$
$\displaystyle=B_{1}(x),$

the second last inequality follows by the fact that
$\tilde{\beta}_{n-2}\geq\tilde{\beta}_{n-1}.$

∎

##  6 Scaling properties and local minima for $B_{1/2}$

Now we will show that we can prove analogous results for other values of
$\alpha,$ for instance for $\alpha=1/2.$ In this case one cannot reduce to the
result for $B_{1},$ (in fact the absolute minimum will be
$\gamma=\sqrt{2}-1$). However we can use the same techniques as in Theorem 4.1
and Theorem 4.5 to get analogous statements.

Throughout this section we will assume $\alpha=1/2$ and corresponding to it we
will consider the Brjuno functions $B_{1/2}$ associated with the nearest
integer continued fractions. First we will study the global minima of Brjuno
function for $\alpha=1/2$ and then we will explore the scaling properties of
Brjuno functions for $\alpha=1/2.$

###  6.1 Minimum of the Brjuno functions $B_{1/2}$

In this section we will show that the minimum of $B_{1/2}$ is attained at the
silver number $\gamma.$ In fact we prove the following theorem.

######  Theorem 6.1.

Let $\gamma=\sqrt{2}-1.$ For $x\neq\gamma$ the
$\min_{x\in(0,1/2)}B_{1/2}(x)=B_{1/2}(\gamma).$

Figure 4: The graphs of $B_{1/2}$ with a zoom around the global minima at
$\gamma=\sqrt{2}-1.$

####  6.1.1 Proof of Theorem 6.1

In this subsection we will give the proof of Theorem 6.1 that is based on four
main propositions. We will adapt a similar method as in [2] to prove Theorem
6.1.

Define $C:=\inf_{x\in[0,1]}B_{1/2}(x).$ The following corollary is the
consequence of Lemma 3.2.

######  Corollary 6.2.

$C=B_{1/2}(r)$ for some $r\in(0,1/2).$

######  Proposition 6.3.

Let $r\in(0,1/2).$ Then for all $K\in\mathbb{N},$ we have

$C=B_{1/2}(r)\geq\frac{B^{(K)}_{1/2}(r)}{1-\beta_{K}(r)}.$

###### Proof.

By using (1.6) for $\alpha=1/2$ we have

$\displaystyle
C=B_{1/2}(r)=B^{(K)}_{1/2}(r)+\beta_{K}(r)B_{1/2}(A_{1/2}^{K+1}(r))$
$\displaystyle\geq B_{1/2}^{(K)}(r)+C\beta_{K}(r)$
$\displaystyle=\frac{B^{(K)}_{1/2}(r)}{1-\beta_{K}(r)}.$

∎

######  Proposition 6.4.

For all $K\in\mathbb{N},$ we have

$B_{1/2}(\gamma)=\frac{B^{K}_{1/2}(\gamma)}{1-\beta^{K}_{1/2}(\gamma)}.$

###### Proof.

Again by using (1.6) now for $r=\gamma$ we obtain

$\displaystyle
B_{1/2}(\gamma)=B^{K}_{1/2}(\gamma)+\beta_{K}(\gamma)B_{1/2}(A_{1/2}^{K+1}(\gamma))$

since $A_{1/2}^{K+1}(\gamma)=A_{1/2}(\gamma)=\gamma$ for any $K\in\mathbb{N}$
therefore the required result follows. ∎

######  Proposition 6.5.

Let $r\in(0,1/2)$ such that $C=B_{1/2}(r).$ Then $r\geq\gamma.$

###### Proof.

From Proposition 6.3 and 6.4 with $K=0$ and the definition of $C,$ we have

$\frac{B^{(0)}_{1/2}(\gamma)}{1-\beta_{0}(\gamma)}=B_{1/2}(\gamma)\geq
C=B_{1/2}(r)\geq\frac{B^{(0)}_{1/2}(r)}{1-\beta_{0}(r)}.$

Next $0<x<1/2,$

$\displaystyle\frac{B^{(0)}_{1/2}(x)}{1-\beta_{0}(x)}=\frac{\ln\frac{1}{x}}{1-x}.$

Let $h(x)=\frac{\ln\frac{1}{x}}{1-x}.$ Then

$h^{\prime}(x)=\frac{-x^{-1}(1-x)+\ln(1/x)}{(1-x)^{2}}.$

$h^{\prime}(x)<0$ on $(0,1/2).$ Thus the function $h$ is strictly decreasing
on $(0,1/2).$ Therefore $\gamma\leq r.$

∎

######  Proposition 6.6.

Let $r\in(0,1/2)$ such that $C=B_{1/2}(r).$ Then $r=\gamma.$

###### Proof.

We need to show $\gamma>r.$ From Proposition 6.3 and Proposition 6.4 with
$K=1$ and by the definition of $C,$ we have

$\displaystyle\frac{B^{(1)}_{1/2}(\gamma)}{1-\beta_{1}(\gamma)}=B_{1/2}(\gamma)\geq
C=B_{1/2}(r)\geq\frac{B^{(1)}_{1/2}(r)}{1-\beta_{1}(r).}$

Note that for $2/5<x<1/2,$ $A_{1/2}=\frac{1}{x}-2.$ Therefore, let

$\displaystyle f(x)$
$\displaystyle=\frac{\ln(1/x)}{1-xA_{1/2}}+\frac{x\ln(\frac{1}{A_{1/2}(x)})}{1-xA_{1/2}(x)}$
$\displaystyle=\frac{\ln(1/x)}{2x}+\frac{1}{2}\ln\frac{x}{1-2x}.$

Then

$\displaystyle f^{\prime}(x)=\frac{(1-2x)\ln x+3x-1}{2x^{2}(1-2x)}.$

Since $(1-2x)>0$ for $x\leq 1/2$ therefore the sign of $f^{\prime}(x)$ depends
only on $g(x):=(1-2x)\ln x+3x-1.$ It is easy to see that $g$ is strictly
increasing for on the interval $(0,1/2]$ and consequently $f$ is increasing on
$(2/5,1/2]$ (because $g^{\prime}(x)$ and $f^{\prime}(x)$ are positive on
(2/5,1/2)). ∎

###  6.2 Scaling properties of $B_{1/2}$

######  Theorem 6.7.

$B_{1/2}(x)-B_{1/2}(\gamma)\geq c|x-g|^{1/2}.$ (6.1)

The proof of Theorem 6.7 follows almost exactly on the same line of
investigations as for the case $\alpha=1.$ There are some added arguments in
accordance with the settings of $\alpha=1/2$ which we will outline for
completeness.

###### Proof.

Let $\Psi(x)=\frac{1}{2+x}$ and define the recursive relation

$x_{n+1}=\Psi(x_{n})\quad\text{ where }x_{0}\in(0,2/5).$ (6.2)

Then from (1.5) we have

$B_{1/2}(\Psi(x_{n}))=-\log(\Psi(x_{n}))+\Psi(x_{n})B(x_{n}))$

and

$B_{1/2}(\Psi(\gamma))=-\log(\Psi(\gamma))+\Psi(\gamma)B((\gamma).$

Note that $\Psi(\gamma)=\gamma.$

Now

$\displaystyle
B_{1/2}(x_{n+1})-B_{1/2}(\gamma)=-\log\frac{x_{n+1}}{\gamma}+x_{n+1}[B_{1/2}(x_{n})-B_{1/2}(\gamma)]+B_{1/2}(\gamma)[x_{n+1}-\gamma].$
(6.3)

Setting $\mathcal{E}_{n}:=B_{1/2}(x_{n})-B_{1/2}(\gamma)$ and
$\delta_{n}:=x_{n}-\gamma$ and observing that $-\log t\geq 1-t$ we get

$\mathcal{E}_{n+1}\geq x_{n+1}\mathcal{E}_{n}-l\delta_{n+1}$ (6.4)

where $l:=\frac{1}{\gamma}-B_{1/2}(\gamma)$ and $l>0.$ Hence

$\displaystyle\mathcal{E}_{n+2}\geq
x_{n+2}x_{n+1}\mathcal{E}_{n}-l(x_{n+2}\delta_{n+1}+\delta_{n+2}).$

Note that

$\displaystyle x_{n+2}\delta_{n+1}+\delta_{n+2}$
$\displaystyle=x_{n+2}(1+x_{n+1})-\gamma(1+x_{n+2}),$
$\displaystyle=x_{n+2}(2+x_{n+1})-\gamma(2+x_{n+2})=1-\gamma(2+x_{n+2})+\gamma-
x_{n+2}$

Both these terms $1-\gamma(2+x_{n+2}),$ $\gamma-x_{n+2}$ are negative iff
$x_{n+2}>\gamma$ i.e. if $n$ is odd.

Therefore for all $n\geq 1,$

$\displaystyle\mathcal{E}_{2n+1}\geq x_{2n+1}x_{2n}\mathcal{E}_{2n-1}.$ (6.5)

Repeating the similar arguments as used for $B_{1},$ we have

$\mathcal{E}_{2n+1}\geq c_{3}\gamma^{2n}\text{ for some constant }c_{3}.$
(6.6)

This estimate shows that $B_{1/2}$ has a cusp like minima at $\gamma,$ namely

$B_{1/2}(x)-B_{1/2}(\gamma)\geq c|x-\gamma|^{\tau}\quad\text{ with }$

$\tau=\frac{1}{2}$ where the value of $\tau$ is obtained by using the fact
that $\delta_{2n+1}\asymp g^{4n}$ for any $n\geq 1$ and comparing it with the
estimates (6.6). ∎

Using the fact that lower semi-continuity holds when $\alpha$ is rational we
can repeat the same argument as in [2] to prove the following proposition.

######  Proposition 6.8.

For any rational parameter $\alpha\in(\gamma,g)$ we have that for any Brjuno
number $x$ with $x\neq\gamma$, $B_{\alpha}(x)>B_{\alpha}(\gamma).$

We will not give here the proof of this claim (since the proof is very much
the same as for $\alpha=1/2$); instead we give some numerical evidence of this
fact for $\alpha=3/5$ in Figure 5: one can see quite clearly that the minimum
of $B_{3/5}$ is attained at $\gamma.$

Figure 5: The graphs of $B_{3/5}$ where $\alpha<g$ with a zoom around the
global minima at $\gamma.$

Let us point out that the minimum of $B_{\alpha}$ changes abruptly when
$\alpha$ moves across the ’critical value’ $g$: indeed, when $\alpha>g$ the
minimum of $B_{\alpha}$ is attained at $g$ (in Figure 6 an instance of this
phenomenon in the case of $\alpha=2/3$), while for rational parameters
$\alpha\in(\gamma,g)$ the minimum is attained in $\gamma$.

Figure 6: The graphs of $B_{2/3}$ where $\alpha>g$ with a zoom around the
global minima at $g.$

As a final remark, let us mention that it is plausible that the minimum of
$B_{\alpha}$ is in $\gamma$ for every $\alpha\in(\gamma,g)$. However, in order
to give a rigorous proof of this claim, one has to find an alternative proof
(since lower semi-continuity might fail when $\alpha$ isn’t rational).

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