11institutetext:  DAMO Academy, Alibaba Group
11email<EMAIL_ADDRESS>
DMAC Group, DCD Lab, Zhejiang University
11email<EMAIL_ADDRESS>
University of Chinese Academy of Sciences
11email<EMAIL_ADDRESS>
# DAMO-NLP at NLPCC-2022 Task 2: Knowledge Enhanced Robust NER for Speech
Entity Linking

Shen Huang 11 Yuchen Zhai 22 Xinwei Long 33 Yong Jiang 11 Xiaobin Wang 11 Yin
Zhang 22 Pengjun Xie Corresponding author11112233

###### Abstract

Speech Entity Linking aims to recognize and disambiguate named entities in
spoken languages. Conventional methods suffer gravely from the unfettered
speech styles and the noisy transcripts generated by ASR systems. In this
paper, we propose a novel approach called Knowledge Enhanced Named Entity
Recognition (KENER), which focuses on improving robustness through painlessly
incorporating proper knowledge in the entity recognition stage and thus
improving the overall performance of entity linking. KENER first retrieves
candidate entities for a sentence without mentions, and then utilizes the
entity descriptions as extra information to help recognize mentions. The
candidate entities retrieved by a dense retrieval module are especially useful
when the input is short or noisy. Moreover, we investigate various data
sampling strategies and design effective loss functions, in order to improve
the quality of retrieved entities in both recognition and disambiguation
stages. Lastly, a linking with filtering module is applied as the final
safeguard, making it possible to filter out wrongly-recognized mentions. Our
system achieves 1st place in Track 1 and 2nd place in Track 2 of NLPCC-2022
Shared Task 2.

###### Keywords:

Entity Linking Robust NER

##  1 Introduction

Speech Entity Linking (SEL), which aims to recognize and disambiguate named
entities in spoken languages to a certain knowledge base (KB), is widely used
in search engines, voice assistants and other speech analysis services. It’s
quite annoying but very common, for example, when you want news of Yao Ming
who is the famous basketball player but the voice assistant returns you news
of another Yao Ming who is a composer. The spoken language, compared to the
written language, is much more informal for its unfettered styles and high-
frequency grammatical errors. All these bring great difficulties to SEL, not
to mention the noisy transcripts generated by Automatic Speech Recognition
(ASR) systems.

The NLPCC 2022 Shared Task 2 provides researchers an opportunity to study
Entity Linking (EL) problems with spoken languages and noisy texts. The task
includes two tracks: (1) Track 1 - Entity Recognition and Disambiguation; (2)
Track 2 - Entity Disambiguation-Only. The difference is that in Track 1 one
needs to process the transcripts to extract named entities and disambiguate
the extracted entities to the correct entry in a given knowledge base while in
Track 2 gold standard named entities are given as input and all one needs to
do is to disambiguate them.

Previous work on end2end EL problem show that is extraction is harder than
disambiguation. Following these lines of work, we perform some preliminary
experiments and verify it in this competition. To tackle this problem, we
propose a three-stage approach called Knowledge Enhanced Named Entity
Recognition (KENER). Firstly it retrieves candidate entities from a KB for an
input sentence without mentions via a dense retrieval module. Secondly the
retrieved entity descriptions and the sentence contexts are utilized as extra
information to help recognize mentions in the Named Entity Recognition (NER)
module. Thirdly a linking with filtering module is applied to link the
mentions to a KB while filtering out wrongly-recognized mentions. Finally,
with smart sampling strategy and ensemble methods, our system achieves 1st
place in track 1 and 2nd place in track 2 of NLPCC-2022 Shared Task 2.

Besides the system description, we make the following observations based on
our experiments.
1\. Our entity retrieval module can retrieve highly-related entities given a
certain sentence without mentions. (Section 3.1)
2\. Our NER module can effectively utilize the candidate entities to improve
the overall performance. (Section 3.2)
3\. Our linking module can tackle noisy mentions with the filtering mechanism
. (Section 3.3)

##  2 Related Work

Named Entity Recognition (NER): Most of the work takes NER as a sequence
labeling problems and applies the linear-chain CRF [1]. Recently, the
improvement of accuracy mainly benefits from stronger token representations
such as pretrained contextual embeddings such as BERT [2], Flair [3] and LUKE
[4]. Yu et al. [5] utilizes the strength of pretrained contextual embeddings
over long-range dependency and encodes the document-level contexts for token
representations. Wang et al. [6] proposes to mine external contexts of a
sentence by retrieving to improve NER performance.

Entity Disambiguation (ED): Most of the work focuses on contextual
representation learning or features based on the entity mention [7]. DPR [8]
and BLINK [9] calculate the match between the mention and entity through a bi-
encoder during retrieval and a cross-encoder is used for re-ranking [9].
Various context information is exploited, such as latent entity type
information in the immediate context of the mention [10].

End to End EL: There are some models that jointly processes NER and ED as
well. Kolitsas et al. [11] first proposes a neural end-to-end EL system that
jointly discovers and links entities in a text document. Martins et al. [12]
introduces a Stack-LSTM approach composed of SHIFT-REDUCE-OUT actions for
multi-task learning. Cao et al. [13] creatively formulate entity linking as a
sequence-to-sequence task, where an encoder-decoder architecture is used to
autoregressively generate terms in a prefix tree of entity title vocabulary.
Zhang et al. [14] proposes the EntQA model that combines progress in entity
linking with that in open-domain question answering and capitalizes on dense
entity retrieval and reading comprehension.

##  3 Methods

Figure 1: Our proposed KENER system architecture for the shared task.

Fig. 1 describes the architecture of our KENER system, which consists of three
modules sequentially: Candidate Entity Retriever, Knowledge Enhanced NER and
Linking with Filtering. We utilize the same linking module for Track 1 and
Track 2.

###  3.1 Candidate Entity Retriever

The Candidate Entity Retriever module retrieves top-K candidate entities that
might be mentioned in the input sentence. It adopts a variant of the bi-
encoder structure, where the representation of an input sentence and a
candidate entity from a KB are computed separately. The scores of candidate
entities can then be calculated by the representations, so as to find the
top-K candidate entities. We formulate the procedure as follows.

Given a sentence transcript $x\in\mathcal{V}^{L}$ and an entity
$e\in\mathcal{E}$, where $\mathcal{V}$ denotes the vocabulary and
$\mathcal{E}$ denotes the set of entities in a KB associated with text titles,
aliases, and descriptions. $L$ is the length of a sentence.

The candidate entity retrieval score $\mathrm{score}_{r1}(x,e)$ can be
computed by

$\displaystyle S_{1}$
$\displaystyle=\textbf{enc}_{S}([\mathrm{CLS}]~{}x~{}[\mathrm{SEP}]~{}\mathrm{ctx(x)}~{}[\mathrm{SEP}])$
$\displaystyle E_{1}$
$\displaystyle=\textbf{enc}_{E}([\mathrm{CLS}]~{}\mathrm{title(e)}~{}[\mathrm{SEP}]~{}\mathrm{alias(e)}~{}\mathrm{desc(e)}~{}[\mathrm{SEP}])$
$\displaystyle\mathrm{score}_{r1}(x,e)$ $\displaystyle=S_{1}^{T}E_{1}$

where the $\mathrm{ctx}(x)$ denotes the context sentences of $x$ in a speech
and $\mathrm{title(e)}$, $\mathrm{alias(e)}$, $\mathrm{desc(e)}$ denote the
text title, alias and description of the entity $e$.

During inference time, we use Faiss [15] with precomputed $E$ for each
$e\in\mathcal{E}$ for fast retrieval.

Multi-label NCE Loss. For the reason that one sentence may contain multiple
entities, we train the retriever using a multi-label variant of noise
contrastive estimation (NCE) following [14]. The optimization objective is

$\mathrm{max}\sum_{e\in\mathcal{E}(x)}\mathrm{log}(\frac{\mathrm{exp}(\mathrm{score}_{r1}(x,e))}{\mathrm{exp}(\mathrm{score}_{r1}(x,e))+\sum_{e^{\prime}\in\mathrm{N}(\mathcal{E},x)}\mathrm{exp}(\mathrm{score}_{r1}(x,e^{\prime}))})$

where $\mathrm{N}(\mathcal{E},x)\subset\mathcal{E}\setminus\mathcal{E}(x)$ is
a set of negative examples that excludes all gold entities $\mathcal{E}(x)$.

Hard Negative Sampling. To retrieve entities of high quality, the hard
negative sampling strategy is adopted. Instead of randomly sampling negative
examples, we use the model from the last iteration to generate negative
examples. It has been proved that the high-score negative examples can
significantly help polish up the model. For this shared task, the negative
examples are sampled randomly for the first iteration, while hard negative
examples are mined for the next two iterations.

###  3.2 Knowledge Enhanced NER

The Knowledge Enhanced NER module extracts entity mentions for a sentence. In
addition to the sentence itself, the contexts of the sentence in a speech and
the retrieved candidate entities are also used as input. We also tried to use
the audio but no significant improvement was revealed. We formulate NER as a
sequence labeling task and the classical model BERT-CRF is adopted in our
method.

Figure 2: BERT-CRF model with extra context and candidate entities as input.

As depicted in Fig. 2, we simply concatenate a sentence, its context and
retrieved candidate entities (using its title, aliases and descriptions) as
input to the BERT encoder:

$\hat{x}=[\mathrm{CLS}]~{}x~{}[\mathrm{SEP}]~{}\mathrm{ctx}(x)~{}[\mathrm{SEP}]~{}\mathcal{\hat{E}}_{1}(x)~{}\mathcal{\hat{E}}_{2}(x)~{}...~{}\mathcal{\hat{E}}_{M}(x)~{}[\mathrm{SEP}]$

where
$\mathcal{\hat{E}}_{i}(x)=\mathrm{title}(\mathcal{E}_{i}(x))~{}\mathrm{alias}(\mathcal{E}_{i}(x))~{}\mathrm{desc}(\mathcal{E}_{i}(x))$
denotes the text representation of the $i$th retrieved candidate entity of
$x$.

Experimental results prove that simple method works as well. The extra
information brought by contexts and candidate entities can help generate
better sentence representation. Then we drop other embeddings from contexts or
candidate entities, and only keep the raw sentence embeddings for decoding. A
CRF layer is used as the decoder to model the transition of tags.

###  3.3 Linking with Filtering

A common linking module focuses on linking and disambiguating the given
mentions to entities from a KB. In addition to that, our proposed Linking with
Filtering module also tries to solve the problem when the given mentions are
noisy and might contain errors.

Following the mainstream framework, a bi-encoder is used for retrieval and a
cross-encoder is used for disambiguation or ranking. The bi-encoder
independently embeds the mention context and the entity descriptions so
candidate entities can be retrieved in a dense space. Each candidate is then
examined more carefully with a cross-encoder, that concatenates the mention
and entity text for encoding their correlation.

Two additional entities are added to the provided KB: $\mathrm{NIL}$ and
$\mathrm{ERROR}$. $\mathrm{NIL}$ is used for those mentions that are actually
named entities but cannot be found in the given KB. $\mathrm{ERROR}$ is used
for some other mentions that are not named entities. This is designed to
filter out the possible prediction errors generated by NER modules. As gold
mentions are given, $\mathrm{ERROR}$ is not applied in Track 2.

Hard Negative Sampling. We also used the same hard negative sampling strategy
as we do in Section 3.1 for training the bi-encoder, despite of their inputs
for the encoder differ.

List-wise Loss. Instead of point-wise or pair-wise losses, we adopt the list-
wise loss for training the cross-encoder, which computes the Kullback-Leibler
divergence (KL divergence) to explicitly measure the difference between the
predicted and target distributions.

Dynamic Sampling. Aiming to reduce the size of candidates for disambiguation
while keeping the candidate entities hard to disambiguate, we propose a
dynamic sampling strategy that samples candidates with different
probabilities. The higher retrieval score one candidate gets, the higher
probability it is sampled.

###  3.4 Ensemble Methods

Ensemble methods are used twice in the end2end entity linking procedure. The
first time it is used to gather NER models to generate better results and the
second time it works for the disambiguation models to re-rank the candidate
entities.

We ensemble the NER models in two strategies using voting. One is for better
F1 score and the other is for better recall. We count the predicted tags for
each token from all models and determine the final tag by the distribution.
The F1 strategy takes “O” as a normal tag when voting wile the recall strategy
sets a threshold for “O” so that it’s less probable to predict a “O” tag. With
our proposed Linking with Filtering module, there’s no need to worry about the
noise.

The linking models are ensembled in a hybrid way. Not only the last ranking
scores are used, but also the retrieval scores are taken into consideration.
We re-rank the candidates with the weighted sum of retrieval and ranking
scores and obtain the final linking results.

##  4 Experiments

###  4.1 Data Analysis and Experimental Settings

The task provides annotated training data, unlabeled test data and the
corresponding audios. All corpus provided are obtained from TED
talks111https://www.ted.com/talks. It also provides a knowledge base from
Wikidata222http://www.wikidata.org/.

Track 1 and Track 2 share a training set consisting of 1936 talks. The
statistics of the training and test data is shown in Table 1.

Table 1: Dataset statistics. Data | sentence number | avg. sentence length | avg. mention number
---|---|---|---
train | 223348 | 32.16 | 1.40
track1 test | 2905 | 37.99 | -
track2 test | 2923 | 38.64 | 1.49

No validation set is provided. So we manually split the full training set into
4:1 for training and validation. The tuning and ablation studies are all
conducted on our validation set.

The provided knowledge base has a high recall of 95%. We check the rest 5%
entities which are linked to NIL and find their corresponding wiki pages don’t
exists veritably. So the official knowledge base is the only KB used in our
system.

The main pretrained models we use for NER and ED are RoBERTa-large
333https://huggingface.co/roberta-large and co-condenser-wiki
444https://huggingface.co/Luyu/co-condenser-wiki respectively.

In our NER model, the top-16 candidate entities are utilized as extra context
for it achieves a relatively high recall of 93% and the transformer structure
cannot handle too long inputs.

###  4.2 Main Results

The evaluation results on test sets of Track 1 and Track 2 are released by the
organizer. As shown as Table 2 and Table 3, our system achieves 1st place in
Track 1 and 2nd place in Track 2 of NLPCC-2022 Shared Task 2.

As we can see that, the performance of our system in Track 2 is comparable
(-0.16pt) with the 1st place system. However in Track 1 of Entity Recognition
and Disambiguation, we beat the 2nd and 3rd place system by 2.9pt and 10.7pt,
which proves the advantage of our proposed KENER method. Due to the task
difference of Track 1 and Track 2, we can conclude that this advantage stems
from the knowledge introduced in the NER stage.

Table 2: Track1 submission results. System Name | F1
---|---
Ours | 0.7460
HITsz_TMG | 0.7172
xmu-nlp-team | 0.6390
Table 3: Track2 submission results. System Name | F1
---|---
HITsz_TMG | 0.8884
Ours | 0.8868

###  4.3 Ablation Study

We conduct three ablation studies to validate the effectiveness of individual
components of our framework. Concretely, we first investigate the influence of
retrieval methods. The results are shown in Table 4, where stage 1 and stage 3
denote retrieval without mentions and with mentions respectively. We can see
that: (1) The hard negative mining is critical for both the candidate entity
retriever and the entity linking. The iterative hard negatives can lead to
further improvements. (2) It is important to excluding gold entities in the
normalization term of NCE, which leads to a massive increase in Recall@1
(38.44).

To further understand how the retrieval module affects our framework, we
investigate the NER performances for different retrieval-augmented strategies.
We train and evaluate the model with various contexts. The results are shown
in Table 5. We can observe that all these contexts leads to significantly
improvements, especially for the external TED context. This indicates that the
document-level information is more important for NER.

Table 6 shows an ablation study for the entity linking module. We report the
Accuracy on the validation set of the Track 2 dataset. We find that the Hybrid
Ensemble method helps boost the results by a large margin, and the List-wise
Loss and the Dynamic Sampling Strategy brings additional gains. This
experiment result shows that the retriever is critical for the final re-
ranking performance.

Table 4: Retrieval experiments. Stage | Methods | Rec@1 | Rec@16 | Rec@32 | Rec@64 | Rec@128
---|---|---|---|---|---|---
1 | Baseline | 25.34 | 53.91 | 90.16 | 95.64 | 96.82
1 | w/ Multi-label NCE Loss | 38.44 | 62.19 | 92.22 | 95.89 | 97.04
1 | w/ Hard Negative (HN) | 59.60 | 93.24 | 95.11 | 96.32 | 97.66
3 | Baseline | 56.62 | 91.73 | 97.42 | 98.92 | 99.12
3 | w/ First Stage HN | 82.19 | 95.92 | 98.14 | 99.16 | 99.35
3 | w/ Second Stage HN | 83.23 | 97.65 | 98.54 | 99.31 | 99.47
Table 5: NER experiments. Methods | F1
---|---
Ours | 70.31
w/o Context | 69.63
w/o Hard Negative | 68.19
w/o Candidate Entities | 68.32
w/o Context & Candidate Entities | 67.21
w/ Ensemble | 71.09
Table 6: Disambiguation experiments. Methods | F1
---|---
Ours | 85.71
w/o List-wise Loss | 85.33
w/o Dynamic Sampling | 85.15
w/ Hybrid Ensemble | 86.98

##  5 Conclusions

In this paper, we propose a novel knowledge enhanced method called KENER,
which aims to leverage knowledge in the NER phase to improve the overall
performance of entity linking. It consists of three-stage modules, each
focusing on one question we raise in Section 1. We attribute KENER’s
outperformance to the following reasons: (1) The hard-negative sampling
strategy and multi-label NCE loss can contribute to better retrieval results.
(2) The NER model benefits greatly from the retrieved candidate entities. (3)
The filtering mechanism is good when the input contains noisy mentions. There
are also scopes for improvement in our approach. Currently the three stages
are trained separately, which ignores the internal correlation among them. We
are looking forward to a unified solution that formulates the NER and ED in
one framework.

## References

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# The Littlewood conjecture for a restricted class of pairs of real numbers

Youssef Lazar<EMAIL_ADDRESS>
###### Abstract.

We prove that the Littlewood conjecture is satisfied for a restricted class of
pairs $(\alpha,\beta)$ of badly approximable numbers. We use the localization
of the roots of a cubic equation with coefficients depending on the
diophantine properties for the considered pair $(\alpha,\beta)$. The estimates
of the roots rely on the properties of the denominators of the convergents of
the continued fraction expansion of $\alpha$ and $\beta$.

##  1\. Introduction

### The Littlewood conjecture

Around 1930, J.E. Littlewood asked the question whether given any two distinct
real numbers $\alpha$ and $\beta$, the sequence
$(n|\sin(n\alpha)\sin(n\beta)|)_{n\geq 0}$ can take arbitrary small values
([10] Problem 5, p.19). The values of such sequences strongly depend on the
diophantine properties of the pair $(\alpha,\beta)$. If we denote by $\|.\|$
the function distance to the nearest integer, $\|x\|=d(x,\mathbb{Z})$. The
problem of Littlewood amounts to solving the following conjecture which is
called the Litllewood conjecture in the literature.

######  Conjecture 1.1 (Littlewood).

For any $(\alpha,\beta)\in\mathbb{R}^{2}$, one has

$~{}~{}~{}~{}~{}~{}\liminf_{n\rightarrow\infty}~{}~{}n\|n\alpha\|\|n\beta\|=0.$

The conjecture is obviously true if $\alpha$ or $\beta$ have unbounded partial
quotients in their continued fraction expansion or if $1,\alpha$ and $\beta$
are linearly dependent over the rationals. Thus, it remains to prove the
conjecture when $\alpha,\beta\in\mathbb{B}$ where $\mathbb{B}$ is the set of
real numbers which have uniformly bounded partial quotients and when
$1,\alpha$ and $\beta$ are linearly independent over the rationals.

This problem remains unsolved but it has been proved for some specific classes
of pairs $(\alpha,\beta)$.

  * •

For the pairs $(\alpha,\beta)$, where $\alpha$ and $\beta$ belongs to the same
cubic field. (Cassels-Swninerton-Dyer, 1955, [5]). It is not known whether
this result is relevant or not, indeed it is believed that cubic irrationals
are not badly approximable numbers.

  * •

For pairs $(\alpha,\beta)$ of badly approximable numbers, such that $\beta\in
B(\alpha)$ where $B(\alpha)$ is a subset of $\mathbb{B}$ with
$\dim_{H}B(\alpha)=1$. (Pollington-Velani, 2000, [16]). This is the first
result proving the existence of badly approximable solutions. An effective
version of this result has been proved by Bugeaud [3] (2014).

  * •

The first construction of explicit pairs of badly approximable numbers have
been discovered by De Mathan [6] (2003) and Adamczewski-Bugeaud [1] (2006).
Both results gives examples in terms of their continued fraction expansion.

  * •

The set of exceptional pairs which do not satisfy the conjecture has been
proved to be of Hausdorff dimension zero (Einsiedler, Katok, Lindenstrauss,
2006, [7]). This is the strongest evidence towards the conjecture. The full
conjecture is implied by a deep conjecture of Margulis on the distribution of
orbits under diagonal flows acting on
$\mathrm{SL}_{3}(\mathbb{R})/\mathrm{SL}_{3}(\mathbb{Z})$.

  * •

Lindenstrauss ([12]) suggested to exploit the result of (Einsiedler, Katok,
Lindenstrauss, 2006) in order to find a criterion on a real number $\alpha$
such that any pair $(\alpha,\beta)$ satisfies the conjecture. This leads to
the notion of combinatorial entropy of a real number. Lindenstrauss gives the
following criterion, if the entropy of a real number $\alpha$ is positive then
any pair $(\alpha,\beta)$ satisfies the Litllewood conjecture. Similarily to
the dynamical case, the set of reals with null entropy has Hausdorff dimension
zero.

For more precise information about the conjecture one can look at the
references ([17], [19]) and ([20]) for the dynamical point of view.

###  1.1. The main result

We give sufficent conditions which implies the validity of the conjecture for
a pair of badly approximable numbers. The conditions are given in terms of the
denominators of the convergents of the continued fraction expansion of
$\alpha$ and $\beta$.

######  Theorem 1.2.

Let $(\alpha,\beta)\in\mathbb{B}^{2}$. If one can find a sequence
$(\eta_{n})_{n\geq 1}$ of real numbers with $0\leq\eta_{n}<1/3$ for every $n$,
such that the two following conditions hold for $n$ large enough

  1. (1)

$q_{2n}(\beta)^{11/12+\eta_{n}/4}\leq q_{2n}(\alpha)\leq q_{2n}(\beta)$.

  2. (2)

$\mathrm{lcm}(q_{2n}(\alpha),q_{2n}(\beta))\leq q_{2n}(\beta)^{1+\eta_{n}}$.

Then the pair $(\alpha,\beta)$ satisfies the Littlewood conjecture.

Remarks. (1) The proof relies on the properties of an auxiliary cubic form
$F_{n}(t)$ whose coefficients depend on the diophantine properties of $\alpha$
and $\beta$. The conditions (1) and (2) of the theorem have been set in order
to ensure, given an $\varepsilon>0$ , the existence of an integer $t_{n}$
which is a multiple of $\mathrm{lcm}(q_{2n}(\alpha),q_{2n}(\beta))$ and such
that $|F_{n}(t_{n})|\leq\varepsilon$.

(2) It is quite easy to find a pair of badly approximable numbers which
satisfies the first condition. However the second condition is less flexible
due to the lack of information on the divisibility properties of the
denominators of the convergents. We discuss the special case of quadratic
irrationals in section §5. An interesting question is to prove whether or not
the Hausdorff dimension of the set of pairs satisfying the theorem is positive
and, in the latter case, to estimate its value.

(3) We expect that this result can be improved by refining the method for
finding a multiple of $\mathrm{lcm}(q_{2n}(\alpha),q_{2n}(\beta))$. Indeed, we
have used the fact the length of the interval $I_{j}$ is larger than
$\mathrm{lcm}(q_{2n}(\alpha),q_{2n}(\beta))$ which is asking too much. A
better estimation of the roots of $F_{n}(t)=\varepsilon$ may lead to the
discarding of the constrains. This can be done by using the algebraic
expression of the roots of a cubic instead of the trigonometric one.

(4) A similar problem concerns the values of indefinite quadratic forms at
integers instead of the product of linear forms. The analog of the Littlewood
conjecture is the Oppenheim conjecture which is now a theorem due to G.A.
Margulis [14],[15]. Recently using the same method we gave a proof of
Margulis’ Theorem for indefinite irrational diagonal forms in dimension 3
([12]).

##  2\. Rational lines of approximations and cubic equations

###  2.1. Continued fractions expansion of a real number

Let us recall some of the main properties of the theory of continued fractions
we are going to use, see for instance Chapter 3 [8] and [19].

The error in the approximation of $\beta$ by $c_{n}(\beta)$ is denoted by
$e_{n}(\beta)$, i.e. $e_{n}(\beta)=\beta-c_{n}(\beta)$, it is exactly given by

$e_{n}(\beta)=\dfrac{(-1)^{n}}{\beta_{n+1}q_{n}(\beta)+q_{n+1}(\beta)}$

where $\beta_{n+1}$ is such that
$\beta=[b_{0};b_{1},\ldots,b_{n},\beta_{n+1}]$ (see Lemma 3 E, [19]). In
particular, $e_{2n}(\beta)$ are all positive and one has ( [8] ex. 3.1.5 p. 76
)

$\dfrac{1}{2q_{2n+1}(\beta)^{2}}\leq
e_{2n}(\beta)\leq\dfrac{1}{q_{2n}(\beta)^{2}}.$ (1)

#### Badly approximable numbers

The elements of $\mathbb{B}$ are the real numbers with bounded partial
quotient in their continued fractions representation. It is not difficult to
show that

$\mathbb{B}:=\\{\beta\in\mathbb{R}~{}|~{}\inf_{q\geqslant
1}q\|q\beta\|>0~{}\mathrm{for}~{}\mathrm{every}~{}q\in\mathbb{N}\\}.$

Dirichlet’s theorem applied to a badly approximable number $\beta$ is
illustrated by the following estimate

$\left||q\beta\right||\asymp\dfrac{1}{q}.$

One has for any $\beta\in\mathbb{B}$ that $q_{2n+1}(\beta)\ll q_{2n}(\beta)$
for every $n$ thus using (1) we get the important estimate where the constant
involved depends on the maximum value $M$ among the $b_{i}$ for all $i\geq 0$,

$e_{2n}(\beta)\asymp\dfrac{1}{q_{2n}(\beta)^{2}}.$ (2)

###  2.2. Small values of the Littlewood cubic form

Let us fix a pair $(\alpha,\beta)$ of positive real numbers and let us
introduce the ternary cubic form deduced from the conjecture

$f(x,y,z):=x(\alpha x-y)(\beta x-z).$ (3)

This cubic is degenerate in the sense that it is the product of three linear
forms. The set of zeros of $f$ is the union of the three planes of respective
equations $x=0$, $y=\alpha x$ and $z=\beta x$. If we assume that $\alpha$ and
$\beta$ are two irrational numbers, then no nonzero integral vector could lie
in the set $f=0$. The Littlewood conjecture amounts to proving that

$m(f)=\inf_{v\in\mathbb{Z}^{3},v_{1}\neq 0}|f(v)|=0.$ (4)

This is equivalent to the assertion that for any real $\varepsilon>0$, there
exists $(x,y,z)\in\mathbb{Z}^{3}$, $x\neq 0$ such that

$|f(x,y,z)|\leq\varepsilon.$

Let us fix an arbitrary small real number $\varepsilon>0$ and set

$\mathcal{D}(\varepsilon):=\left\\{(x,y,z)\in\mathbb{R}^{3}:|f(x,y,z)|\leq\varepsilon\right\\}.$

If $x$ is outside the planes $x=0$, $y=\alpha x$ and $z=\beta x$, then
$\mathcal{D}(\varepsilon)$ is bounded by the two parametric surfaces

$\alpha x-\dfrac{\varepsilon}{x(\beta x-z)}\leq y\leq\alpha
x+\dfrac{\varepsilon}{x(\beta x-z)}$

and

$\beta x-\dfrac{\varepsilon}{x(\alpha x-y)}\leq z\leq\beta
x+\dfrac{\varepsilon}{x(\alpha x-y)}.$

One can check that $\mathcal{D}(\varepsilon)$ contains the slant asymptote
$\mathbb{R}(1,\alpha,\beta)$ intersection of the two planes $y=\alpha x$ and
$z=\beta x$. The set $\mathcal{D}(\varepsilon)$ is quite complicated since it
consists in $2^{3}$ connected components; each component is contained in one
of the intersection of the half-spaces bounded by the three planes $x=0$,
$y=\alpha x$ and $z=\beta$. Finding one lattice point in any of these 8
connected component will suffices to ensure the existence of a lattice point
in all other components by performing a change of sign on each cooordinate.
Usual methods of the geometry of numbers fails to apply in this situation due
to the irregular shape of the domain $\mathcal{D}(\varepsilon)$. We are going
to look for lattice points lying in rational line segments in
$\mathcal{D}(\varepsilon)$.

###  2.3. Dirichlet lattice points near $\mathcal{D}(\varepsilon)$

For any arbitrary $\varepsilon>0$, the line $\mathbb{R}(1,\alpha,\beta)$ lies
in the domain $\mathcal{D}(\varepsilon)$, and it may happen that one could
find a lattice point on it and solving the problem. But a non-trivial pair
must satisfy that $1,\alpha$ and $\beta$ are algebraically independent over
$\mathbb{Q}$, this breaks the hope of finding a lattice point on
$\mathbb{R}(1,\alpha,\beta)$. Nevertheless, Dirichlet’s approximation theorem
tells us that it is always possible to find a lattice point lying arbitrarily
near the line in question with a certain level of precision. The crucial issue
is that the speed of convergence is not strong enough in order to ensure that
such integral vectors are in $\mathcal{D}(\varepsilon)$. More precisely, if we
consider a sequence of integers given by $(N_{n})_{1\leqslant i\leqslant n}$.
The two-dimensional version of Dirichlet’s approximation Theorem tells us that
for every $n\geq 1$ there exists an integral vector
$M_{n}=(x_{n},y_{n},z_{n})\in\mathbb{Z}^{3}$ with $1\leq x_{n}\leq N_{n}$ such
that

$\left\\{\begin{array}[]{c}\displaystyle|\alpha x_{n}-y_{n}|\leqslant
N_{n}^{-1/2}\\\ \displaystyle|\beta x_{n}-z_{n}|\leqslant
N_{n}^{-1/2}.\end{array}\right.$ (5)

Geometrically, this says that the lattice point $M_{n})$ approaches the line
$\mathbb{R}(1,\alpha,\beta)$ as $n$ increases with a rate of convergence given
by $1/\sqrt{N_{n}}$. Applying this to the cubic form $f$ we get that

$|f(M_{n})|=|x_{n}||\alpha x_{n}-y_{n}||\beta
x_{n}-z_{n}|\leq\dfrac{x_{n}}{N_{n}}\leq 1.$

Not surprisingly, Dirichlet’s Theorem is not enough for proving the Littlewood
conjecture. It can happen that $|f(M_{n})|\leq\varepsilon$ for some positive
integer $n$, in this case the theorem is proved. Thus from now on, we assume
that it is not the case, meaning that for every positive integer $n$,

$|f(M_{n})|\geq\varepsilon.$

Using (5), we obtain

$\varepsilon\leq|f(M_{n})|=|x_{n}||\alpha x_{n}-y_{n}||\beta
x_{n}-z_{n}|\leq\frac{x_{n}}{N_{n}}.$

This yields the crucial bound on $x_{n}$,

$\varepsilon N_{n}\leq x_{n}\leq N_{n}.$ (6)

Recall that we are assuming that $\alpha$ and $\beta$ are badly approximable
numbers, which amounts to saying that

$C:=\min\\{\inf_{q\geq 1}q\|q\alpha\|,\inf_{q\geq 1}q\|q\beta\|\\}>0.$

In particular, this information together with (5) give the following
inequalities

$\left\\{\begin{array}[]{c}\displaystyle\frac{C}{x_{n}}<\|\alpha
x_{n}\|\leq|\alpha x_{n}-y_{n}|\leq 1/\sqrt{N_{n}}\\\ \\\
\displaystyle\frac{C}{x_{n}}<\|\beta x_{n}\|\leq|\beta x_{n}-z_{n}|\leq
1/\sqrt{N_{n}}.\end{array}\right.$ (7)

To sum up, given an arbitrary positive real number $\varepsilon$ and positive
integers $N_{1},N_{2},\ldots$ we have found a sequence of lattice points
$M_{n}=(x_{n},y_{n},z_{n})\in\mathbb{Z}^{3}$ with $\varepsilon N_{n}<x_{n}\leq
N_{n}$ with $M_{n}\notin\mathcal{D}(\varepsilon)$ but lying within a distance
$1/\sqrt{N_{n}}$ to the line $\mathbb{R}(1,\alpha,\beta)$.

###  2.4. Line approximation

Using the lattice points $(M_{n})_{n\geqslant 1}$ arising from Dirichlet’s
theorem, we construct a family of rational lines which are nearly parallel to
the line $\mathbb{R}(1,\alpha,\beta)$ in $\mathbb{R}^{3}$ and which converges
towards it. To do so, we introduce the line passing through $P_{n}$ and
directed by the vector given by $(1,c_{2n}(\alpha),c_{2n}(\beta))$, that is,

$L^{n}_{\alpha,\beta}=M_{n}+\mathbb{R}(1,c_{2n}(\alpha),c_{2n}(\beta))$

where $c_{2n}(\alpha)$ (resp. $c_{2n}(\beta)$) is the $2n^{\mathrm{th}}$
convergent of $\alpha$ (resp. $\beta$).

A parametrization of the line segment given by
$v_{n}:\mathbb{R}\rightarrow\mathbb{R}^{3}$ is defined as

$v_{n}(t):=M_{n}-t(1,c_{2n}(\alpha),c_{2n}(\beta))=(x_{n}(t),y_{n}(t),z_{n}(t))~{}~{}$

where

$(L^{n}_{\alpha,\beta}):\left\\{\begin{array}[]{ccc}x_{n}(t)&=&x_{n}-t\\\
y_{n}(t)&=&y_{n}-tc_{2n}(\alpha)\\\
z_{n}(t)&=&z_{n}-tc_{2n}(\beta).\end{array}\right.$

The strategy for proving the main theorem consists in showing the existence of
an integer $t_{n}$ satisfying the three conditions

  1. (1)

(Non vanishing condition) $x_{n}(t_{n})\neq 0$.

  2. (2)

(Geometric condition) For every $\varepsilon>0$,
$v_{n}(t_{n})\in\mathcal{D}(\varepsilon)$ i.e.
$0<|f(v_{n}(t_{n}))|\leq\varepsilon$.

  3. (3)

(Arithmetic conditon) $t_{n}$ is an integral multiple of
$l_{n}:=\mathrm{lcm}(q_{2n}(\alpha),q_{2n}(\beta))$.

We have the following obvious bound on $l_{n}$

$\max\\{q_{2n}(\alpha),q_{2n}(\beta)\\}\leq l_{n}\leq
q_{2n}(\alpha)q_{2n}(\beta).$ (8)

Since the denominators for the convergents of a badly approximable number have
a geometrical growth, we can assume that there exists an integer $n_{0}$ large
enough such that $q_{2n}(\alpha)\leq q_{2n}(\beta)$ for every $n\geq n_{0}$.
In other words, for $n\geq n_{0}$ (8) is just

$q_{2n}(\beta)\leq l_{n}\leq q_{2n}(\alpha)q_{2n}(\beta).$ (9)

Let us introduce the logarithmic ratio between $q_{2n}(\beta)$ and
$q_{2n}(\alpha)$ as

$\gamma_{n}:=\dfrac{\ln q_{2n}(\alpha)}{\ln q_{2n}(\beta)}.$

Thus one has $0<\gamma_{n}\leqslant 1$ for every $n$ greater than $n_{0}$, and
we can now write

$q_{2n}(\alpha)=q_{2n}(\beta)^{\gamma_{n}}.$ (10)

For $n$ large enough, the bounds (9) can be rewritten

$q_{2n}(\beta)\leq l_{n}\leq q_{2n}(\beta)^{1+\gamma_{n}}.$ (11)

The assumptions of Theorem 1.2 tell us that we have the bounds

$\dfrac{11}{12}+\dfrac{\eta_{n}}{4}<\gamma_{n}\leq 1$ (12)

under the condition $0\leq\eta_{n}<1/3$ when $n\geq n_{0}$.

###  2.5. The cubic polynomial $F_{n}(t)$ associated to the Dirichlet lattice
points $(M_{n})_{n\geqslant 1}$

In order to study the existence of lattice points in
$\mathcal{D}(\varepsilon)$, we will restrict to study their eventual presence
on the lines $(L_{\alpha,\beta}^{n})=\\{v_{n}(t)~{}:~{}t\in\mathbb{R}\\}$.
First, we evaluate the cubic $f$ on $(L_{\alpha,\beta}^{n})$. One has, for any
real $t$,

$f(v_{n}(t))=(x_{n}-t)\left\\{\alpha(x_{n}-t)-(y_{n}-tc_{2n}(\alpha)\right\\}\left\\{\beta(x_{n}-t)-(z_{n}-tc_{2n}(\beta))\right\\}$
$=(x_{n}-t)\left\\{(\alpha x_{n}-y_{n})-te_{2n}(\alpha)\right\\}\left\\{(\beta
x_{n}-z_{n})-te_{2n}(\beta)\right\\}$

Let us denote $\delta_{n}(\alpha):=\alpha x_{n}-y_{n}$ and
$\delta_{n}(\beta):=\beta x_{n}-z_{n}$, without loss of generality and
changing signs of $y_{n}$ and $z_{n}$ or of $f$ if necessary, we can assume
that $\delta_{n}(\alpha)$ and $\delta_{n}(\beta)$ are both positive. Thus, we
get

$f(v_{n}(t))=(x_{n}-t)\left\\{\delta_{n}(\alpha)-te_{2n}(\alpha)\right\\}\left\\{\delta_{n}(\beta)-te_{2n}(\beta)\right\\}$

Also we set $t_{n}(\alpha)=\delta_{n}(\alpha)/e_{2n}(\alpha)$,
$t_{n}(\beta)=\delta_{n}(\beta)/e_{2n}(\beta)$ and
$A_{n}=e_{2n}(\alpha)e_{2n}(\beta)$. Hence,

$f(v_{n}(t))=-A_{n}(t-x_{n})(t-t_{n}(\alpha))(t-t_{n}(\beta)).$

The expression of $f(v_{n}(t))$ defines a cubic polynomial in $t$ which we
denote $F_{n}(t)$. It assumes three real (positive)111The positivity of roots
is not important but it allows us to work without taking care of the sign in
the estimates and inequalities. At the end we are going to solve
$|f(v)|<\varepsilon$ with the asbolute value. roots, namely $x_{n}$,
$t_{n}(\alpha)$ and $t_{n}(\beta)$.

$x$$y$$0$$N_{n}$$x_{n}$$x(t_{n}(\varepsilon))$$\displaystyle y=\alpha
x-\dfrac{\varepsilon}{\beta x^{2}}$$\displaystyle y=\alpha
x+\dfrac{\varepsilon}{\beta x^{2}}$$\displaystyle y=\alpha
x$$(L^{n}_{\alpha,\beta})$$\bullet(x_{n},y_{n},0)$$\bullet(N_{n},\alpha
N_{n},0)$ Figure 1. A view of the picture in the plane $z=0$. The red line
$(L^{n}_{\alpha,\beta})$ passes through the Dirichlet lattice vector and cuts
$\mathcal{D}(\varepsilon)$ at $x(t_{n}(\varepsilon))$. The boundary of
$\mathcal{D}(\varepsilon)$ is colored in blue.

###  2.6. Estimates of the coefficients of $F_{n}$

Let us introduce the elementary symmetric polynomials of the roots $x_{n}$,
$t_{n}(\alpha)$ and $t_{n}(\beta)$ of $F_{n}$.

$\left\\{\begin{array}[]{ccc}\Sigma_{1}&=&x_{n}+t_{n}(\alpha)+t_{n}(\beta)\\\
\Sigma_{2}&=&x_{n}t_{n}(\alpha)+x_{n}t_{n}(\beta)+t_{n}(\alpha)t_{n}(\beta)\\\
\Sigma_{3}&=&x_{n}t_{n}(\alpha)t_{n}(\beta).\\\ \end{array}\right.$ (13)

We define a sequence of real numbers $(\delta_{n})_{n\geqslant 1}$ such that

$N_{n}=q_{2n}(\beta)^{\delta_{n}}.$ (14)

In order words $\delta_{n}$ is just the quotient $\log N_{n}/\log
q_{2n}(\beta)$. The following proposition gives asymptotic bounds depending on
$\varepsilon$ for each $\Sigma_{i}$ ($i=1,2,3$).

######  Proposition 2.1.

For $n$ large enough, one has

  1. (1)

$q_{2n}(\beta)^{\delta_{n}}+q_{2n}(\beta)^{2-\delta_{n}}\ll_{\varepsilon}\Sigma_{1}\ll_{\varepsilon}q_{2n}(\beta)^{\delta_{n}}+q_{2n}(\beta)^{2-\delta_{n}/2}.$

  2. (2)

$q_{2n}(\beta)^{2}+q_{2n}(\beta)^{2(1+\gamma_{n})-\delta_{n}}\ll_{\varepsilon}\Sigma_{2}\ll_{\varepsilon}q_{2n}(\beta)^{2(1+\gamma_{n})-\delta_{n}}+q_{2n}(\beta)^{2+\delta_{n}/2}$.

  3. (3)

$\Sigma_{3}\asymp_{\varepsilon}q_{2n}(\beta)^{2(1+\gamma_{n})}.$

Proof of the proposition. (1) From 1, 6 and 7 we get

$\varepsilon N_{n}<x_{n}\leq N_{n},$ (15)

$\dfrac{q_{2n}(\beta)^{2}}{x_{n}}\ll
t_{n}(\beta)\ll\dfrac{q_{2n}(\beta)^{2}}{\sqrt{N_{n}}},$ (16)

$\dfrac{q_{2n}(\alpha)^{2}}{x_{n}}\ll
t_{n}(\alpha)\ll\dfrac{q_{2n}(\alpha)^{2}}{\sqrt{N_{n}}}.$ (17)

An estimate of $\Sigma_{1}$ is thus given by,

$N_{n}+\dfrac{1}{N_{n}}\left(q_{2n}(\alpha)^{2}+q_{2n}(\beta)^{2}\right)\ll_{\varepsilon}\Sigma_{1}\ll_{\varepsilon}N_{n}+\dfrac{1}{\sqrt{N_{n}}}\left(q_{2n}(\alpha)^{2}+q_{2n}(\beta)^{2}\right).$
(18)

Using the notation 14,

$q_{2n}(\beta)^{\delta_{n}}+q_{2n}(\beta)^{2\gamma_{n}-\delta_{n}}+q_{2n}(\beta)^{2-\delta_{n}}\ll_{\varepsilon}\Sigma_{1}\ll_{\varepsilon}q_{2n}(\beta)^{\delta_{n}}+q_{2n}(\beta)^{2\gamma_{n}-\delta_{n}/2}.+q_{2n}(\beta)^{2-\delta_{n}/2}.$
(19)

Since $0<\gamma_{n}\leq 1$ then, we are reduced to

$q_{2n}(\beta)^{\delta_{n}}+q_{2n}(\beta)^{2-\delta_{n}}\ll_{\varepsilon}\Sigma_{1}\ll_{\varepsilon}q_{2n}(\beta)^{\delta_{n}}+q_{2n}(\beta)^{2-\delta_{n}/2}.$
(20)

(2) Let us consider $\Sigma_{2}$. The asymptotic estimates (16), (17) allow us
to obtain

$q_{2n}(\alpha)^{2}\ll x_{n}t_{n}(\alpha)\ll\sqrt{N_{n}}q_{2n}(\alpha)^{2},$
(21)

$q_{2n}(\beta)^{2}\ll x_{n}t_{n}(\beta)\ll\sqrt{N_{n}}q_{2n}(\beta)^{2}.$ (22)

For the product $t_{n}(\alpha)t_{n}(\beta)$, the estimates (16) and (17) do
not provide optimal bounds. Instead we use the assumption that
$|f(x_{n},y_{n},z_{n})|>\varepsilon$, which we assume to hold otherwise the
conjecture is proved. This condition is equivalent to

$\varepsilon<x_{n}\delta_{n}(\alpha)\delta_{n}(\beta)$ (23)

or also,

$\dfrac{\varepsilon}{A_{n}}<x_{n}t_{n}(\alpha)t_{n}(\beta).$ (24)

Using 2 again, we arrive to

$\dfrac{\varepsilon q_{2n}(\alpha)^{2}q_{2n}(\beta)^{2}}{x_{n}}\ll
t_{n}(\alpha)t_{n}(\beta).$ (25)

Thus, we obtain the following refined estimate

$\dfrac{q_{2n}(\alpha)^{2}q_{2n}(\beta)^{2}}{x_{n}}\ll_{\varepsilon}t_{n}(\alpha)t_{n}(\beta)\ll_{\varepsilon}\dfrac{q_{2n}(\alpha)^{2}q_{2n}(\beta)^{2}}{N_{n}}.$
(26)

From (21) (22), (26),

$\dfrac{q_{2n}(\alpha)^{2}q_{2n}(\beta)^{2}}{N_{n}}+\left(q_{2n}(\alpha)^{2}+q_{2n}(\beta)^{2}\right)\ll_{\varepsilon}\Sigma_{2}\ll_{\varepsilon}\dfrac{1}{N_{n}}q_{2n}(\alpha)^{2}q_{2n}(\beta)^{2}+\sqrt{N_{n}}\left(q_{2n}(\alpha)^{2}+q_{2n}(\beta)^{2}\right).$
(27)

As $N_{n}=q_{2n}(\beta)^{\delta_{n}}$, we arrive to

$q_{2n}(\beta)^{2(1+\gamma_{n})-\delta_{n}}+q_{2n}(\beta)^{2\gamma_{n}}+q_{2n}(\beta)^{2}\ll_{\varepsilon}\Sigma_{2}\ll_{\varepsilon}q_{2n}(\beta)^{2(1+\gamma_{n})-\delta_{n}}+q_{2n}(\beta)^{2\gamma_{n}+\delta_{n}/2}+q_{2n}(\beta)^{2+\delta_{n}/2}.$
(28)

Thus,

$q_{2n}(\beta)^{2}+q_{2n}(\beta)^{2(1+\gamma_{n})-\delta_{n}}\ll_{\varepsilon}\Sigma_{2}\ll_{\varepsilon}q_{2n}(\beta)^{2(1+\gamma_{n})-\delta_{n}}+q_{2n}(\beta)^{2+\delta_{n}/2}.$
(29)

and this shows (2).

(3) For $\Sigma_{3}$, we already have from (26) that

$\dfrac{\varepsilon}{A_{n}}<x_{n}t_{n}(\alpha)t_{n}(\beta)\leq
q_{2n}(\alpha)^{2}q_{2n}(\beta)^{2}.$ (30)

That is,

$x_{n}t_{n}(\alpha)t_{n}(\beta)\asymp_{\varepsilon}q_{2n}(\alpha)^{2}q_{2n}(\beta)^{2}.$
(31)

Hence,

$\Sigma_{3}\asymp_{\varepsilon}q_{2n}(\beta)^{2(1+\gamma_{n})}.$ (32)

$\square$

###  2.7. Asymptotic shape of the graph $y=F_{n}(t)$

The derivative of $F_{n}(t)$ with respect to the real variable is

$F^{\prime}_{n}(t)=-A_{n}\left\\{3t^{2}-2\Sigma_{1}t+\Sigma_{2}\right\\}.$

To find the local extrema of $F_{n}$ we are then reduced to compute the zeroes
of $3t^{2}-2\Sigma_{1}t+\Sigma_{2}$. The fact that we have three roots for
$F_{n}$, a graphical argument shows one has two extrema so that the
discriminant of $F_{n}^{\prime}$ is nonnegative for every integer $n\geq 1$,

$\Delta_{n}=4\Sigma_{1}^{2}-12\Sigma_{2}=4(\Sigma_{1}^{2}-3\Sigma_{2}).$

The roots of the derivative $F^{\prime}_{n}(t)$ are then given by

$\tau_{n}^{\pm}=\Sigma_{1}\pm\dfrac{1}{2}\sqrt{\Delta_{n}}.$ (33)

$t$$y$$y=\varepsilon$$y=-\varepsilon$$\tau_{n}^{-}$$\tau_{n}^{+}$$y=F_{n}(t)$$t_{n}(\alpha)$$x_{n}$$t_{n}(\beta)$
Figure 2. The shape of $y=F_{n}(t)$

The values of the local maximum and minimum are respectively given by

$F_{n}(\tau_{n}^{\pm})=-A_{n}(\tau_{n}^{\pm}-x_{n})(\tau_{n}^{\pm}-t_{n}(\alpha))(\tau_{n}^{\pm}-t_{n}(\beta)).$

In fact we have much more precise than this, it requires the use of a result
due to Laguerre-Cesáro (see e.g. Theorem 6.5.1 [18]) which says in our case
that $\tau^{\pm}_{n}$ is in the middle third of the interval bounded by two
the critical numbers bounding $\tau^{\pm}_{n}$. In concrete words, if the root
are ordered as $\tau_{n}^{(1)}<\tau_{n}^{(2)}<\tau_{n}^{(3)}$ so that we have
the following configuration

$\tau_{n}^{(1)}<\tau_{n}^{-}<\tau_{n}^{(2)}<\tau_{n}^{+}<\tau_{n}^{(3)}.$

$\tau_{n}^{-}\in\left[\tau_{n}^{(1)}+\dfrac{\tau_{n}^{(2)}-\tau_{n}^{(1)}}{3},\tau_{n}^{(2)}-\dfrac{\tau_{n}^{(2)}-\tau_{n}^{(1)}}{3}\right]$

and

$\tau_{n}^{+}\in\left[\tau_{n}^{(2)}+\dfrac{\tau_{n}^{(3)}-\tau_{n}^{(2)}}{3},\tau_{n}^{(3)}-\dfrac{\tau_{n}^{(3)}-\tau_{n}^{(2)}}{1}\right].$

Also one has the equivalent

$|F_{n}(\tau_{n}^{\pm})|\asymp\dfrac{1}{q_{2n}(\alpha)^{2}q_{2n}(\beta)^{2}}|x_{n}-\tau_{n}^{\pm}||\tau_{n}^{\pm}-t_{n}(\alpha)||\tau_{n}^{\pm}-t_{n}(\beta)|.$

Given $\varepsilon>0$ arbitrary, replacing $F_{n}$ by $-F_{n}$ if necessary we
are reduced to the three possible scenarios

  1. (1)

$\max\\{|F_{n}(\tau_{n}^{-})|,|F_{n}(\tau_{n}^{+})|\\}<\varepsilon$ then
$J_{n}$ is a single interval.

  2. (2)

$F_{n}(\tau_{n}^{-})<-\varepsilon$ and $F_{n}(\tau_{n}^{+})\leq\varepsilon$
$J_{n}$ is the union of two intervals.

  3. (3)

$-\varepsilon<F_{n}(\tau_{n}^{-})$ and $\varepsilon<F_{n}(\tau_{n}^{+})$
$J_{n}$ is the union of two intervals.

  4. (4)

$\min\\{|F_{n}(\tau_{n}^{-})|,|F_{n}(\tau_{n}^{+})|\\}\geq\varepsilon$ $J_{n}$
is the union of three intervals.

Unfortunately, we are not able to estimate the factors in
$|F_{n}(\tau_{n}^{\pm})|$ due to the lack of a 2-dimensional analog of the
notion of continued fraction for real numbers. We are going to treat the
problem of the intervals by solving directly the equation
$F_{n}(t)=\varepsilon$ in §3. We finish the study of the shape of the graph of
$F_{n}$ by giving an upper bound on the measure of $t$ such that
$|F_{n}(t)|\leq\varepsilon$. This will follow from the study of the set of the
small values of the cubic polynomial $F_{n}(t)$. A key result due to H. Cartan
[4] given in the cubic case shows that the measure of set of small values
cannot be too large.

######  Theorem 2.2 (Theorem 3.1, VIII,§3 [11]).

Let $f(t)=(t-t_{1})(t-t_{2})(t-t_{3})$ be a monic polynomial in
$\mathbb{C}[t]$ and $\varepsilon$ be a positive real. Then, the set of $t$
such that

$|f(t)|\leq\varepsilon$

is contained in the union of a most 3 intervals such that the sum of the
lengths is bounded by $6e\varepsilon^{1/3}$.

Applying this result to the monic polynomial $p_{n}(t)=-A_{n}^{-1}F_{n}(t)$
tells us that there exists a covering of $\mathcal{D}(\varepsilon)$ consisting
in at most three intervals $I_{1},I_{2},I_{3}$ respectively with length
$l_{1},l_{2},l_{3}$ so that their sums is less or equal to
$6e|A_{n}|\varepsilon^{1/3}$. In particular, we have

$\lambda(t\in\mathbb{R}~{}:~{}0<|F_{n}(t)|\leq\varepsilon)\leq\lambda(I_{1}^{n})+\lambda(I_{2}^{n})+\lambda(I_{3}^{n})\leq
6e(|A_{n}^{-1}|\varepsilon)^{1/3}$

$\lambda(t\in\mathbb{R}~{}:~{}0<|F_{n}(t)|\leq\varepsilon)\ll(q_{2n}(\alpha)^{2}q_{2n}(\beta)^{2}\varepsilon)^{1/3}.$

We have obtained the following asymptotic estimate of the total length of the
interval(s)

######  Proposition 2.3.

For every positive real $\varepsilon$, there exists an integer $n_{0}$ such
that for all $n\geq n_{0}$,

$\lambda(t\in\mathbb{R}~{}:~{}0<|F_{n}(t)|\leq\varepsilon)\ll_{\varepsilon}(q_{2n}(\alpha)q_{2n}(\beta))^{2/3}.$

##  3\. Intervals for the solutions of $F_{n}(t)=\pm\varepsilon$

### Real solutions of the cubic equation $F_{n}(t)=\pm\varepsilon$

Now we need to find all the possible intersection points which correspond to
the values of $t$ such that $F_{n}(t)=\pm\varepsilon$. We have the expanded
form

$F_{n}(t)=-A_{n}\left\\{t^{3}-\Sigma_{1}t^{2}+\Sigma_{2}t-\Sigma_{3}\right\\}.$

The equation $F_{n}(t)=\pm\varepsilon$ takes the following form

$-A_{n}\\{t^{3}-\Sigma_{1}t^{2}+\Sigma_{2}t-\Sigma_{3}\\}\pm\varepsilon=0.$
(34)

We rather consider the equivalent equation

$G_{n}(\pm\varepsilon)(t):=t^{3}-\Sigma_{1}t^{2}+\Sigma_{2}t-\left(\Sigma_{3}\pm\dfrac{\varepsilon}{A_{n}}\right)=0.$
(35)

If we specializes $\varepsilon$ to be zero, we recover the solution of the
cubic equation $F_{n}(t)=0$. In particular we see that $F_{n}(t)$ and
$G_{n}(\pm\varepsilon)(t)$ share the same symmetric functions of the roots
$\Sigma_{1}$ and $\Sigma_{2}$, only the third differs by a factor
$\pm\dfrac{\varepsilon}{A_{n}}$.
We choose the trigonometric parametrization of the solution of this cubic
equation $G_{n}(\pm\varepsilon)(t)=0$ following the presentation of [2]
(Appendix of Chapter 4). The first step consists to perform the Vieta
transform $t=y+\dfrac{\Sigma_{1}}{3}$ to get the following cubic in the
reduced form (the $t^{2}$ term has vanished)

$y^{3}+P_{n}y=Q_{n}(\pm\varepsilon)$ (36)

where

$P_{n}=\dfrac{3\Sigma_{2}-\Sigma_{1}^{2}}{3}$ and
$Q_{n}(\pm\varepsilon)=\dfrac{-9\Sigma_{1}\Sigma_{2}+2\Sigma_{1}^{3}}{27}+\left(\Sigma_{3}\pm\dfrac{\varepsilon}{A_{n}}\right).$

The discriminant of $G_{n}(\pm\varepsilon)$ is given by

$D_{n}(\pm\varepsilon):=-4P_{n}^{3}-27Q_{n}(\pm\varepsilon)^{2}.$

Let us solve (36), to do so we set $y=h_{n}z$ where $h_{n}=2\sqrt{|P_{n}|/3}$
and replace in (36)

$h_{n}^{3}z^{3}+P_{n}h_{n}z=\left(\dfrac{4|P_{n}|}{3}\right)^{3/2}z^{3}+P_{n}\left(\dfrac{4|P_{n}|}{3}\right)z=Q_{n}(\pm\varepsilon).$

Then we divide by $\left(\dfrac{4|P_{n}|}{3}\right)^{3/2}$

$z^{3}+\dfrac{3P_{n}}{4|P_{n}|}z=\left(\dfrac{4|P_{n}|}{3}\right)^{-3/2}Q_{n}(\pm\varepsilon).$

We are therefore reduced to solve the following equation,

$4z^{3}+3~{}\mathrm{sgn}(P_{n})z=C_{n}(\pm\varepsilon)$ (37)

where

$C_{n}(\pm\varepsilon)=\dfrac{1}{2}\left(\dfrac{3}{|P_{n}|}\right)^{3/2}Q_{n}(\pm\varepsilon).$

$t$$y$$y=\varepsilon$$y=-\varepsilon$$y=F_{n}(t)$$t_{n}(\alpha)$$t_{n}(\beta)$$x_{n}$
Figure 3. The root $x_{n}$ dominates $t_{n}(\alpha)$ and $t_{n}(\beta)$ as
soon as $\delta_{n}>4/3$.

If we replace $\varepsilon$ by zero, we recover the solution of the cubic
equation $F_{n}(t)=0$. The latter equation has 3 real solutions $x_{n}$,
$t_{n}(\alpha)$ and $t_{n}(\beta)$, thus $P_{n}<0$ ( $P_{n}$ does not depend
on $\varepsilon$). Therefore we are left with two cases.
Case 1.$\ \ \mathrm{sgn}(P_{n})<0,|C_{n}(\pm\varepsilon)|\geq 1$ in this case,
one real root one has to solve

$4z^{3}+3z=C_{n}(\pm\varepsilon).$

Using the formula $\cosh 3\theta=4\cosh^{3}\theta-3\cosh^{2}\theta$

$z=\pm\mathrm{cosh}\left(\dfrac{1}{3}{\mathrm{arccosh}}(C_{n}(\pm\varepsilon))\right)$

where the sign + is involved when $C_{n}(\pm\varepsilon)\geq 1$ and the sign -
when $C_{n}(\pm\varepsilon)\leq-1$. If we assume that, say
$C_{n}(\pm\varepsilon)\geq 1$ then we have a unique real root for equation
(36)

$y=2\sqrt{-P_{n}/3}\cosh\left(\dfrac{1}{3}\Phi_{n}(\pm\varepsilon)\right)$

where $\Phi_{n}(\pm\varepsilon)=\mathrm{arccosh}(C_{n}(\pm\varepsilon))$.
Finally, we get the solutions of the equation $F_{n}(t)=\pm\varepsilon$

$t(\pm\varepsilon)=2\sqrt{-P_{n}/3}\cosh\left(\dfrac{1}{3}\Phi_{n}(\pm\varepsilon)\right)+\dfrac{\Sigma_{1}}{3}.$
(38)

Thus $\mathcal{D}(\varepsilon)=\\{t\in\mathbb{R},|F_{n}(t)|\leq\varepsilon\\}$
is exactly the interval $I_{n}=[t({\varepsilon}),t({-\varepsilon})]$.

Case 2. $\ \ \mathrm{sgn}(P_{n})<0,|C_{n}(\pm\varepsilon)|<1$.

One is reduced to solve

$4z^{3}-3z=C_{n}(\pm\varepsilon).$

Taking advantage of the relation $\cos
3\theta=4\cos^{3}\theta-3\cos^{2}\theta$ we deduce the solutions

$z_{k}=\cos\left(\dfrac{1}{3}\arccos(C_{n}(\pm\varepsilon))+\dfrac{2(k-1)\pi}{3}\right)~{}~{}~{}~{}~{}(k=1,2,3).$

Thus in the case when $|C_{n}(\pm\varepsilon)|\leq 1$, one has three real
distinct roots for equation (36)

$y_{k}=2\sqrt{-P_{n}/3}\cos\left(\dfrac{1}{3}\Phi_{n}(\pm\varepsilon)+\dfrac{2(k-1)\pi}{3}\right)~{}~{}~{}~{}~{}(k=1,2,3)$

where $\Phi_{n}(\pm\varepsilon)=\arccos(C_{n}(\pm\varepsilon))$ lies in the
open interval $(0,\pi)$. Finally, we get the solutions of the equation
$F_{n}(t)=\pm\varepsilon$

$t_{k}(\pm\varepsilon)=2\sqrt{-P_{n}/3}\cos\left(\dfrac{1}{3}\Phi_{n}(\pm\varepsilon)+\dfrac{2(k-1)\pi}{3}\right)+\dfrac{\Sigma_{1}}{3}~{}\
\ \ \ \ \ \ ~{}~{}~{}~{}(k=1,2,3).$ (39)

Thus $\mathcal{D}(\varepsilon)=\\{t\in\mathbb{R},|F_{n}(t)|\leq\varepsilon\\}$
is the union of three the interval
$I_{1}^{n}=[t_{1}({\varepsilon}),t_{1}({-\varepsilon})]$,
$I_{2}^{n}=[t_{2}({-\varepsilon}),t_{2}({\varepsilon})]$,
$I_{3}^{n}=[t_{3}({\varepsilon}),t_{3}({-\varepsilon})]$. Hence we are in the
worst case where $\mathcal{D}(\varepsilon)$ is the union of three intervals.

### Asymptotic estimate for $-P_{n}$

The next lemma is crucial it gives an asymptotic estimate for $-P_{n}$ with
constants depending on $\varepsilon$.

######  Lemma 3.1.

Assume $\delta_{n}>4/3$ for every $n$, then as $n$ gets large, one has

$-P_{n}\asymp_{\varepsilon}q_{2n}(\beta)^{2\delta}.$

Proof. We have from Proposition 2.1

$q_{2n}(\beta)^{\delta_{n}}+q_{2n}(\beta)^{2-\delta_{n}}\ll_{\varepsilon}\Sigma_{1}\ll_{\varepsilon}q_{2n}(\beta)^{\delta_{n}}+q_{2n}(\beta)^{2-\delta_{n}/2}.$

First, one has

$\delta_{n}-(2-\dfrac{\delta_{n}}{2})=\delta_{n}-2+\dfrac{\delta_{n}}{2}=\dfrac{3\delta_{n}}{2}-2>0.$

Also,

$\delta_{n}-(2-\delta_{n})=2\delta_{n}-2=2(\delta_{n}-1)>0.$

Thus,

$q_{2n}(\beta)^{\delta_{n}}\ll_{\varepsilon}\Sigma_{1}\ll_{\varepsilon}q_{2n}(\beta)^{\delta_{n}}.$

By definition $-P_{n}=\dfrac{1}{3}\Sigma_{1}^{2}-\Sigma_{2}$ and it is
positive because $F_{n}(t)$ has three real roots. Therefore $\Sigma_{1}^{2}$
is the dominant term of $-P_{n}$, then

$-P_{n}\asymp_{\varepsilon}\Sigma_{1}^{2}\asymp_{\varepsilon}q_{2n}(\beta)^{2\delta_{n}}.$

##  4\. Proof of Theorem 1.2

$y$$y=\varepsilon$$y=-\varepsilon$$(F_{n}(t))$

$y$$y=\varepsilon$$y=-\varepsilon$$(F_{n}(t))$

Figure 4. One the left the worst scenario $|C_{n}(\pm\varepsilon)|<1$ (three
intervals) and one the right the best scenario $|C_{n}(\pm\varepsilon)|\geq 1$
(one interval).

Let us fix $\varepsilon>0$, and $N_{n}=q_{2n}(\beta)^{\delta_{n}}$ where
$\delta_{n}>4/3$ is chosen such that

$\delta_{n}\in\left(\dfrac{4}{3},\dfrac{1}{5}(3+4\gamma_{n}-\eta_{n})\right]\neq\emptyset$
(40)

for $n$ large enough. The nonemptyness in (40) is ensured by condition (12),
indeed the length of
$\left(\dfrac{4}{3},\dfrac{1}{5}(3+4\gamma_{n}-\eta_{n})\right]$ is given by

$\dfrac{1}{5}(3+4\gamma_{n}-\eta_{n})-\dfrac{4}{3}=\dfrac{4}{5}\gamma_{n}-\dfrac{1}{5}\eta_{n}-\dfrac{11}{15}>\dfrac{4}{5}(\dfrac{11}{12}+\dfrac{\eta_{n}}{4})-\dfrac{1}{5}\eta_{n}-\dfrac{11}{15}=0.$

We assume we are in the worst case222The shape of $\mathcal{D}(\varepsilon)$
can convince the reader that the best case is when the line cuts he boundary
of $\mathcal{D}(\varepsilon)$ only twice., namely the case 2 when
$|C_{n}(\pm\varepsilon)|<1$. Thus $F_{n}(t)=\pm\varepsilon$ has three real
roots. In that case we have seen in 39 that the three (trigonometric)
solutions are given by

$t_{k}(\pm\varepsilon)=2\sqrt{-P_{n}/3}\cos\left(\dfrac{1}{3}\Phi_{n}(\pm\varepsilon)+\dfrac{2(k-1)\pi}{3}\right)+\dfrac{\Sigma_{1}}{3}~{}\
\ \ \ \ \ \ ~{}~{}~{}~{}(k=1,2,3).$

Thus, the set of times $t$ such that $|F_{n}(t)|\leqslant\varepsilon$ is the
union of three intervals $I_{1}^{n}$, $I_{2}^{n}$ and $I_{3}^{n}$ bounded
respectively by the roots $t_{1}(\pm\varepsilon),t_{2}(\pm\varepsilon)$ and
$t_{3}(\pm\varepsilon)$.

Since the lengths of each interval $I_{1}^{n}$,$I_{2}^{n}$ and $I_{3}^{n}$ are
all equal, we only need to focus on the length of $I_{1}^{n}$. One has

$t_{1}(\pm\varepsilon)=2\sqrt{-P_{n}/3}\cos\left(\dfrac{1}{3}\Phi_{n}(\pm\varepsilon)\right)+\dfrac{\Sigma_{1}}{3}.$

The length of $I_{1}^{n}$ is given by

$\lambda(I_{1}^{n})=\mathrm{length}(I_{1}^{n})=|t_{1}(\varepsilon)-t_{1}(-\varepsilon)|=2\sqrt{-P_{n}/3}\left|\cos\dfrac{\Phi_{n}(\varepsilon)}{3}-\cos\dfrac{\Phi_{n}(-\varepsilon)}{3}\right|.$

A lower bound for the length of $I_{1}^{n}$ is given by,

######  Proposition 4.1.

For $n$ large enough,

$\lambda(I_{1}^{n})\gg_{\varepsilon}\dfrac{q_{2n}(\beta)^{4(1+\gamma_{n})}}{(-P_{n})^{5/2}}.$

Proof of the Proposition. Using the identity
$\displaystyle\cos\alpha-\cos\beta=-2\sin(\frac{\alpha+\beta}{2})\sin(\frac{\alpha-\beta}{2})$,
we get

$\lambda(I_{1}^{n})=4\sqrt{-P_{n}/3}\left|\sin\left(\dfrac{\Phi_{n}(\varepsilon)+\Phi_{n}(-\varepsilon)}{6}\right)\sin\left(\dfrac{\Phi_{n}(\varepsilon)-\Phi_{n}(-\varepsilon)}{6}\right)\right|.$

Remark that, one has
$|(\Phi_{n}(-\varepsilon)\pm\Phi_{n}(\varepsilon))/6|\in[0,\pi/3]$, and thus
one can apply the inequality $\sin x\geqslant x/2$ which is valid for all
$x\in[0,\pi/2]$. We obtain

$\lambda(I_{1}^{n})\geq\dfrac{1}{36}\sqrt{-P_{n}/3}~{}\left|\Phi_{n}(\varepsilon)^{2}-\Phi_{n}(-\varepsilon)^{2}\right|.$

or more precisely,

$\lambda(I_{1}^{n})\geq\dfrac{1}{36}\sqrt{-P_{n}/3}~{}\left|\arccos^{2}(C_{n}(\varepsilon))-\arccos^{2}(C_{n}(-\varepsilon))\right|.$

######  Lemma 4.2.

For all $x,y\in[-1,1]$,

$|\arccos^{2}x-\arccos^{2}y|\geq|x-y|^{2}.$

Proof. One remarks that

$\left|\arccos^{2}x-\arccos^{2}y\right|=\left|\arccos x-\arccos
y\right||\arccos x+\arccos y|.$

Noting the fact that the function arcos assumes positive values on the open
interval $(-1,1)$, we can write that

$|\arccos x+\arccos y|=\arccos x+\arccos y\geq|\arccos x-\arccos y|.$

Thus,

$\left|\arccos^{2}x-\arccos^{2}y\right|\geq\left|\arccos x-\arccos
y\right|^{2}.$

Moreover, the Lipschitz property for the cosine yields

$|\cos u-\cos v|\leq|u-v|.$

The latter applied to $u=\arccos x$ and $v=\arccos y$ we get

$|x-y|\leq|\arccos x-\arccos x|.$ (41)

Hence,

$\left|\arccos^{2}x-\arccos^{2}y\right|\geq\left|x-y\right|^{2}.$

$\square$

Remembering that $\Phi_{n}(\pm\varepsilon)=\arccos(C_{n}(\pm\varepsilon))$ the
previous lemma gives us

$|C_{n}(\varepsilon)-C_{n}(-\varepsilon)|^{2}\leq|\Phi_{n}(\varepsilon)^{2}-\Phi_{n}(-\varepsilon)^{2}|.$

Therefore,

$\lambda(I_{1}^{n})\geq\dfrac{1}{36\sqrt{3}}\sqrt{-P_{n}}~{}\left|\Phi_{n}(\varepsilon)^{2}-\Phi_{n}(-\varepsilon)^{2}\right|\gg\sqrt{-P_{n}}\left|C_{n}(\varepsilon)-C_{n}(-\varepsilon)\right|^{2}.$

Thus, we obtain

$\lambda(I_{1}^{n})\gg\sqrt{-P_{n}}\left|C_{n}(\varepsilon)-C_{n}(-\varepsilon)\right|^{2}.$

Concerning the right hand side, one has

$C_{n}(\varepsilon)-C_{n}(-\varepsilon)=\dfrac{1}{2}\left(\dfrac{3}{-P_{n}}\right)^{3/2}\left(Q_{n}(\varepsilon)-Q_{n}(-\varepsilon)\right)=\dfrac{1}{2}\left(\dfrac{3}{-P_{n}}\right)^{3/2}\dfrac{2\varepsilon}{A_{n}}=\dfrac{3\sqrt{3}\varepsilon}{(-P_{n})^{3/2}A_{n}}.$

Thus, for $n$ large enough

$|C_{n}(\varepsilon)-C_{n}(-\varepsilon)|\asymp\dfrac{q_{2n}(\beta)^{2(1+\gamma_{n})}}{(-P_{n})^{3/2}}$

since $A_{n}=(e_{2n}(\alpha)e_{2n}(\alpha))^{-1}\sim
q_{2n}(\alpha)^{2}q_{2n}(\beta)^{2}=q_{2n}(\beta)^{2(1+\gamma_{n})}$ by (2).
Hence, we get the lower bound for the length

$\lambda(I_{1}^{n})\gg_{\varepsilon}\sqrt{-P_{n}}\left|C_{n}(\varepsilon)-C_{n}(-\varepsilon)\right|^{2}\gg_{\varepsilon}\dfrac{q_{2n}(\beta)^{4(1+\gamma_{n})}}{(-P_{n})^{5/2}}.$
(42)

$\square$

The previous proposition combined with Lemma 3.1 yields,

$\lambda(I_{1}^{n})\gg_{\varepsilon}q_{2n}(\beta)^{4(1+\gamma_{n})-5\delta_{n}}.$
(43)

We are ready to complete the proof of the theorem. One hypothesis of the
theorem tells us that the least common multiple $l_{n}$ of $q_{2n}(\alpha)$
and $q_{2n}(\beta)$ satisfies

$l_{n}\leq q_{2n}(\beta)^{1+\eta_{n}}$

with some $0\leq\eta_{n}\leq 1/3$. Using the latter with Lemma 4.1,

$\dfrac{\lambda(I_{1}^{n})}{l_{n}}\gg_{\varepsilon}\dfrac{q_{2n}(\beta)^{4(1+\gamma_{n})-5\delta_{n}}}{q_{2n}(\beta)^{1+\eta_{n}}}$
(44)

which reduces to

$\dfrac{\lambda(I_{1}^{n})}{l_{n}}\gg_{\varepsilon}q_{2n}(\beta)^{3+4\gamma_{n}-\eta_{n}-5\delta_{n}}.$
(45)

For each $n$, let us consider $j_{n}\in\\{1,2,3\\}$ to be the index of one of
two intervals $I_{j}^{n}$ which do not contain $x_{n}$. The initial choice of
the sequence $\delta_{n}$ in (40) implies that
$3+4\gamma_{n}-\eta_{n}-5\delta_{n}$ is positive for $n$ large enough. In
particular,

$\lim_{n}\dfrac{\lambda(I_{1}^{n})}{l_{n}}=\infty.$ (46)

The same holds for $\lambda(I_{j_{n}}^{n})=\lambda(I_{1}^{n})$ and therefore
$I_{j_{n}}^{n}$ always contains at least one multiple of $l_{n}$. For each $n$
large enough, let $t_{n}=k_{n}l_{n}\in I_{j_{n}}$ be a multiple of $l_{n}$,
and by definition of $j_{n}$, $x_{n}\neq t_{n}$. Moreover, if we set
$u_{n}=v_{n}(t_{n})$, we have

$u_{n}=(x_{n}-t_{n},y_{n}-k_{n}\lambda_{n}c_{2n}(\alpha),z_{n}-k_{n}\lambda_{n}c_{2n}(\beta))\in\mathbb{Z}^{3}\cap\mathcal{D}(\varepsilon).$

In other words, we have found $u_{n}\in\mathbb{Z}^{3}$ with nonzero first
component which satisfies

$0<|f(u_{n})|\leq\varepsilon.$

In particular,

$\inf_{u\in\mathbb{Z}^{3},u_{1}\neq 0}|f(u)|=0.$

Hence $(\alpha,\beta)$ satisfies the Littlewood conjecture. The theorem 1.2 is
proved.

$\square$

##  5\. Comments on Theorem 1.2

Our main theorem gives sufficent conditions which ensure the validity of the
Littlewood conjecture up to some restrictions involving the denominators of
the convergents . The natural question is whether or not, there exists a pair
$(\alpha,\beta)$ of badly approximable numbers which satisfies both conditions
and such that $1$, $\alpha$ and $\beta$ are $\mathbb{Q}$-algebraically
independent.

Even if it is not the aim of our method, finding an explicit pair
$(\alpha,\beta)$ of quadratic irrationals seems extremely difficult. In our
case the main difficulty is to get an explicit of the denominators of the
convergents of an irrational quadratic number. However, for a special case of
quadratic real numbers with continued fraction expansion of the form
$[\overline{b,a}]$, with $a$ dividing $b$, there is an explicit formula for
the denominators of the convergents (e.g. [9] §10.13 Theorem 178). We will
restrict to the case when $a=b$. Those numbers form a family of irrational
quadratic numbers which generalizes the golden ratio
$\varphi=(1+\sqrt{5})/2=[1;\overline{1}]$.

###  5.1. Denominators of the convergents of quadratic irrational numbers

Let $b\geq 1$ be an integer, we define the real number
$\beta=[b;\overline{b}]$, for $b=1$ we recover the golden ratio. That is not
difficult to see that $\beta$ is a irrational quadratic number satisfying the
equation $\beta^{2}-b\beta-1=0$, thus

$\beta=[b;\overline{b}]=\dfrac{b+\sqrt{b^{2}+4}}{2}.$

The denominators of the convergents are given by the reccurence relation

$q_{n+1}(\beta)=bq_{n}(\beta)+q_{n-1}(\beta).$

The general expression of $q_{n}(\beta)$ is

$q_{n}(\beta)=\dfrac{\beta^{n+1}-{\beta^{\prime}}^{n+1}}{\beta-\beta^{\prime}}.$

where $\beta^{\prime}=\dfrac{b-\sqrt{b^{2}+4}}{2}$ , that is,

$q_{n}(\beta)=\dfrac{1}{2^{n+2}}\dfrac{(b+\sqrt{b^{2}+4})^{n+1}-(b-\sqrt{b^{2}+4})^{n+1}}{\sqrt{b^{2}+4}}.$

Using binomial expansion, one can prove that

$q_{n}(\beta)=\sum_{k=0}^{\lfloor n/2\rfloor}\left(\begin{array}[]{c}n-k\\\
k\par\end{array}\right)b^{n-2k}.$

For even indices we get,

$q_{2n}(\beta)=\sum_{k=0}^{n}\left(\begin{array}[]{c}2n-k\\\
k\par\end{array}\right)b^{2(n-k)}.$

An important fact is that $q_{2n}(\beta)$ can be seen as the value of a monic
polynomial of degree $2n$ with integral coefficients at $b$. Analogously, if
we consider $\alpha=[a;\overline{a}]$ with $b/a\notin\mathbb{Z}$ then
$\alpha\in\mathbb{Q}[\sqrt{a^{2}+4}]$ and $\beta\in\mathbb{Q}[\sqrt{b^{2}+4}]$
with the property that $\\{1,\alpha,\beta\\}$ are linearly independent over
$\mathbb{Q}$.

###  5.2. The condition on the size of $q_{2n}(\alpha)$ relatively to
$q_{2n}(\alpha)$ in Theorem 1.2

######  Lemma 5.1.

Given a real number $0\leqslant\eta<1/3$, there exists a positive real number
$b_{c}(\eta)$, such that the interval $[b^{11/12+\eta/4},b]$ contains at least
one positive integer as soon as $b>b_{c}(\eta)$.

Proof. Let us consider the fonction $\psi(x):=\dfrac{\log(x-1)}{\log x}$ for
$x>1$. We claim that $[b^{11/12+\eta/4},b]\cap\mathbb{N}\neq\emptyset$ if and
only if $\psi(b)>\dfrac{11}{12}+\dfrac{\eta}{4}$. Indeed, saying that
$[b^{11/12+\eta/4},b]\cap\mathbb{N}\neq\emptyset$ is equivalent to

$b-b^{11/12+\eta/4}>1.$

The latter holds if and only if

$\log(b-1)>\left(\dfrac{11}{12}+\dfrac{\eta}{4}\right)\log b.$

Thus the claim. Moreover, for $x\geq 2$

$\psi^{\prime}(x):=\dfrac{x\log x-(x-1)\log(x)}{x(x-1)(\log x)^{2}}>0.$

the function $\psi$ is increasing, tends to 1 as $x\rightarrow\infty$ and thus
$\psi$ realizes a bijection $[2,\infty)$ onto $[0,1)$ In particular, there
exists a unique value $b_{c}(\eta)>1$ such that
$\psi(b_{c}(\eta))=\dfrac{1}{12}+\dfrac{\eta}{4}$. Hence, the claim tells us
that $[b^{11/12+\eta/4},b]\cap\mathbb{N}\neq\emptyset$ as soon as
$b>b_{c}(\eta)$. The lemma is proved.

$\square$

Remark. It is possible to compute numerically the values of $b_{c}(\eta)$. For
instance one has,

  * •

For $\eta=0$, one has $b_{c}(0)=6.78199...$

  * •

For $\eta=1/100$, one has $b_{c}(10^{-2})=6.912...$

  * •

For $\eta=1/10$, one has $b_{c}(10^{-1})=8.514...$

  * •

For $\eta=1/4$, one has $b_{c}(1/4)=17.332...$

  * $\vdots$

  * •

For $\eta=1/3$, one has $b_{c}(1/3)=\infty.$

The smallest possible integer $b$ for which
$[b^{11/12+\eta/4},b]\cap\mathbb{N}\neq\emptyset$ is $b=7$. Let us assess the
values of $\eta$ which makes
$[7^{11/12+\eta/4},7]\cap\mathbb{N}\neq\emptyset$.

For $\eta=0$, one can check that

$7^{11/12}=5.95...<6<7.$

Thus the interval $[7^{11/12},7]$ contains $6$. The solution $b=7$ works not
only for $\eta=0$, but also for $\eta=\dfrac{4}{250}=\dfrac{2}{125}$, indeed
for this value one has

$7^{11/12+1/250}=5.998...<6<7$

thus the interval $[7^{11/12+1/250},7]$ contains $6$. However, for $\eta=1/50$
we have that $7^{11/12+1/200}=6.01...$ and then the interval
$[7^{11/12+1/50},7]$ contains no integer. Thus for $\eta\geq 1/50$ there are
no chance that $[7^{11/12+\eta/4},7]$ contains $6$. The critical value
$\eta_{b}$ for $b=7$ is somewhere between $\dfrac{2}{125}=0.016$ and
$\dfrac{1}{50}=0.020$.

######  Lemma 5.2.

For a fixed real $0<\eta<1/3$, one can find a couple of positive integers
$(a,b)$ such that for any integer $n$ large enough we have

$q_{2n}(\beta)^{\frac{11}{12}+\frac{\eta}{4}}\leq q_{2n}(\alpha)\leq
q_{2n}(\beta)$

where $\alpha=[a;\overline{a}]$ and $\beta=[b;\overline{b}]$.

Proof. The previous lemma ensures that there exists a couple of integers
$(a,b)$ such that

$b^{\frac{11}{12}+\frac{\eta_{n}}{4}}<a<b$

as soon as $b>b_{c}(\eta)$. We define the corresponding numbers
$\alpha=[a;\overline{a}]$ and $\beta=[b;\overline{b}]$. We have seen earlier
that $q_{2n}(\beta)$ (resp. $q_{2n}(\alpha)$) is a monic polynomial of degree
$2n$ in $b$ (resp. $a$). Therefore we have the following asymptotic
comparisons,

$\dfrac{q_{2n}(\beta)^{\frac{11}{12}+\frac{\eta}{4}}}{q_{2n}(\alpha)}=\dfrac{\left(b^{\frac{11}{12}+\frac{\eta}{4}}\right)^{2n}(1+o(1))}{a^{2n}(1+o(1))}\sim_{n\rightarrow\infty}\left(\dfrac{b^{\frac{11}{12}+\frac{\eta}{4}}}{a}\right)^{2n}\leq
1$

since $b^{\frac{11}{12}+\frac{\eta}{4}}<a$.

In a similar way, using the fact that $a<b$ one gets

$\dfrac{q_{2n}(\beta)}{q_{2n}(\alpha)}=\dfrac{b^{2n}(1+o(1))}{a^{2n}(1+o(1))}\sim_{n\rightarrow\infty}\left(\dfrac{b}{a}\right)^{2n}\geq
1.$

To sum up, we obtain the following for $n\geq n_{0}$

$q_{2n}(\beta)^{\frac{11}{12}+\frac{\eta}{4}}\leq q_{2n}(\alpha)\leq
q_{2n}(\beta).$

$\square$

Now given $0<\eta<1/3$, we have found a couple of badly approximable numbers
$(\alpha,\beta)$ of the form $\alpha=[a;\overline{a}]$ and
$\beta=[b;\overline{b}]$ which satisfies the first assumption of Theorem 1.2
as proved in Lemma 5.2.

$\square$

Remark. The couple $(\alpha,\beta)$ obtained in the Lemma satisfies the linear
condition $\\{1,\alpha,\beta\\}$ is a linearly independent family over
$\mathbb{Z}$ if $a$ does not divide $b$.

###  5.3. The condition on the least common multiple of $q_{2n}(\alpha)$ and
$q_{2n}(\beta)$

This condition is unfortunately very difficult to check in pratice. As we have
noticed earlier it is almost impossible to find the general expression of the
denominators of the convergents of an irrational real number. For the
simpliest case, when $\beta=[b;\overline{b}]$, one has

$q_{2n}(\beta)=\sum_{k=0}^{n}\left(\begin{array}[]{c}2n-k\\\
k\par\end{array}\right)b^{2(n-k)}.$

Already in this case, it seems difficult to find the prime decomposition of
$q_{2n}(\beta)$ even if $b$ is known. Let us consider the pair
$(\alpha,\beta)$ obtained in Lemma 5.2 and let us write their respective prime
decomposition for $n\geq n_{0}$

$q_{2n}(\alpha)=p_{1}^{\alpha_{1,n}}\ldots p_{r_{n}}^{\alpha_{r_{n},n}}$ and
$q_{2n}(\beta)=p_{1}^{\beta_{1,n}}\ldots p_{r_{n}}^{\beta_{r_{n},n}}$

where $\alpha_{i,n}$, $\beta_{i,n}$ are nonnegative integers not both zero for
every $i=1,.,r_{n}$ which we denote simply $\alpha_{i},\beta_{i}$. The prime
numbers $p_{1}<\ldots<p_{r_{n}}$ are given in increasing order. In the one
hand, after applying the logarithm in Lemma 5.2 we infer that

$\mu_{n}\sum_{i=1}^{r_{n}}\beta_{i}\log
p_{i}\leq\sum_{i=1}^{r_{n}}\alpha_{i}\log
p_{i}\leq\sum_{i=1}^{r_{n}}\beta_{i}\log p_{i}$ (47)

where $\mu_{n}=11/12+\eta_{n}/4$. In the other hand, the lcm condition for
$q_{2n}(\alpha)$ and $q_{2n}(\beta)$ reads

$\sum_{i=1}^{r_{n}}\beta_{i}\log
p_{i}\leq\sum_{i=1}^{r_{n}}\max\left(\alpha_{i},\beta_{i}\right)\log
p_{i}\leq(1+\eta_{n})\sum_{i=1}^{r_{n}}\beta_{i}\log p_{i}.$ (48)

If the latter condition is satisfied therefore the pair $(\alpha,\beta)$ of
Lemma 5.2 fullfills the hypothesis of Theorem 1.2. Let us precise the
condition 48. Set $I_{n}=\\{1\leqslant i\leqslant r_{n},\ \ \
\alpha_{i}>\beta_{i}\\}$ which is non empty otherwise $q_{2n}(\alpha)$ divides
$q_{2n}(\beta)$. Therefore,

$\sum_{i=1}^{r_{n}}\max\left(\alpha_{i},\beta_{i}\right)\log p_{i}=\sum_{i\in
I_{n}}\alpha_{i}\log p_{i}+\sum_{i\notin I_{n}}\beta_{i}\log p_{i}.$

Thus,

$\sum_{i\in I_{n}}\beta_{i}\log p_{i}\leq\sum_{i\in I_{n}}\alpha_{i}\log
p_{i}\leq\sum_{i\in I_{n}}\beta_{i}\log
p_{i}+\eta_{n}\sum_{i=1}^{r_{n}}\beta_{i}\log p_{i}.$ (49)

Or also,

$0\leq\sum_{i\in I_{n}}(\alpha_{i}-\beta_{i})\log
p_{i}\leq\eta_{n}\left(\sum_{i=1}^{r_{n}}\beta_{i}\log p_{i}\right).$ (50)

A sufficent condition which implies 50,

$|I_{n}|\max_{i\in I_{n}}[(\alpha_{i}-\beta_{i})\log
p_{i}]\leq\eta_{n}r_{n}\min_{1\leqslant i\leqslant r_{n}}\beta_{i}\log p_{i}.$
(51)

also,

$|I_{n}|\max_{i\in I_{n}}(\alpha_{i}-\beta_{i})\log
p_{r_{n}}\leq\eta_{n}r_{n}\log p_{1}\min_{1\leqslant i\leqslant
r_{n}}\beta_{i}.$ (52)

Therefore a sufficient condition which implies 49 is given by

$|I_{n}|\max_{i\in I_{n}}(\alpha_{i}-\beta_{i})\leq\eta_{n}\dfrac{r_{n}}{\log
p_{r_{n}}}\log p_{1}\min_{1\leqslant i\leqslant r_{n}}\beta_{i}.$ (53)

A very important feature of irrational quadratic numbers (also true for
irrational cubics) is that the greatest prime factor of the denominator of its
convergents tends to infinity, this result is due to K. Mahler [13]. Thus, the
sequence $(p_{r_{n}})_{n\geq 1}$ tends to infinity as $n$ gets large, and in
particular $(r_{n})_{n\geq 1}$ too. To sum up, the pair $(\alpha,\beta)$
satisfies the assumptions of the theorem if it fullfills Lemma 5.2 and the
inequality 53 for the prime decomposition. The fact that $\alpha$ and $\beta$
lies in two distinct quadratic fields which are not equivalent implies that
$\\{1,\alpha,\beta\\}$ are independent over $\mathbb{Q}$, thus providing a
nontrivial example.

##  6\. Final comments

The main theorem gives a class of pairs of reals which satisfies the
Littlewood conjecture. We ignore if this class is empty, in the latter case it
means that an eventual counterexample of the conjecture has to be found in
this class. The first interesting question concerns the size of the class of
pairs satisfying the main theorem. More precisely, let $\Omega_{\eta}$ be the
set of pairs satisfying the hypothesis of the previous theorem and $\Omega$ be
the union of all $\Omega_{\eta}$ over all the sequences $\eta$ valued in
$[0,1/3]$. A natural question is whether the Hausdorff dimension
$\dim_{H}(\Omega)$ is positive or not ? In the positive case, eventually
finding bounds for $\dim_{H}(\Omega)$.
The method of the paper can be refined in order to go beyond the restricted
class $\Omega$. Indeed, requiring that the measure of $I_{j}^{n}$ tends to
infinity faster than $\mathrm{lcm}(q_{2n}(\alpha),q_{2n}(\beta))$ is a strong
assumption. By a finer analysis of the coefficients using for instance,
algebraic formula for the roots instead of the trigonometric method, might
lead to a better localization of the roots and hopefully allowing us to relax
the two assumptions of the theorem. From a far reaching point of view we
expect that any improvement of this method for instance using analytic ou
combinatorial number theory could help to solve the problem of the existence
of a multiple of $l_{n}$ in one of the $I_{j}$’s without requiring that their
lengths have to diverge rapidily. The latter point is the heart of the
problem. Presumably, this method can also be applied to some variations around
the conjecture like the $p$-adic analog for instance.

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