Yannis Dimopoulos
email: yannis@cs.ucy.ac.cy

Pavlos Moraitis
email: pavlos@mi.parisdescartes.fr

Leila Amgoud
email: amgoud@irit.fr

Extending Argumentation to make good Decisions

Argumentation has been acknowledged as a powerful mechanism for automated decision making. In this context several recent works have studied the problem of accommodating preference information in argumentation. The majority of these studies rely on Dung's abstract argumentation framework and its underlying acceptability semantics. In this paper we show that Dung's acceptability semantics, when applied to a preference-based argumentation framework for decision making purposes, may lead to counter intuitive results, as it does not take appropriately into account the preference information. To remedy this we propose a new acceptability semantics, called super-stable extension semantics, and present some of its properties. Moreover, we show that argumentation can be understood as a multiple criteria decision problem, making in this way results from decision theory applicable to argumentation.

Introduction

In many decision making situations we are confronted with a set of alternatives or options each of which has its own advantages that can be expressed as different arguments supporting that alternative. For instance, in a car purchase scenario, one argument that supports small cars is that they have low running cost, while another argument that favors big cars is that they have better safety features. The final decision is usually based upon the preferences one has over the arguments, or more generally how arguments relate to each other.

It is therefore not surprising that during the last years, argumentation has been acknowledged as a powerful mechanism for automating the decision making process of autonomous agents. Several recent works (see e.g. [START_REF] Kakas | Argumentation based decision making for autonomous agents[END_REF][START_REF] Amgoud | A general argumentation framework for inference and decision making[END_REF][START_REF] Amgoud | Making decisions through preferencebased argumentation[END_REF][START_REF] Bonet | Arguing for decisions: A qualitative model of decision making[END_REF][START_REF] Amgoud | Explaining qualitative decision under uncertainty by argumentation[END_REF]) have emphasized the role of agents' preferences in the evaluation of their arguments within a particular class of argumentation frameworks called preference-based argumentation frameworks. The majority of these frameworks are using the acceptability semantics of the Dung's abstract argumentation framework [START_REF] Dung | On the acceptability of arguments and its fundamental role in nonmonotonic reasoning, logic programming and n-person games[END_REF].

In [START_REF] Dimopoulos | Theoretical and Computational Properties of Preference-based Argumentation[END_REF], it has been shown that preference-based argumentation under the stable extension semantics is essentially a method for making decisions that are supported by "good" or "strong" arguments. Roughly speaking, a set of arguments E is a stable extension if every argument of E is strictly preferred to any other argument that is not included in E.

In this work we show that the stable extensions semantics of Dung's framework when applied to decision making may lead to counterintuitive results and therefore fail to deliver the correct conclusions.

More precisely, we show that stable extensions consider as equally good two sets of arguments (and therefore the options they support), although for every argument of the second set, the first set contains a more preferred argument. One may understand that in this case the agent could randomly select an option that is supported either by an argument from the first or the second set and this could be a wrong decision if these arguments support incorrect conclusions. This problem relates to a similar problem identified independently by Horty in [START_REF] Horty | Argument construction and reinstatement in logics for defeasible reasoning[END_REF] in the context of the use of argumentation for defeasible reasoning. For this reason we propose a new semantics called super-stable extension which allows to fix this problem.

Finally, in this paper we show the correspondence between argumentation and multi-criteria decision making. Then we emphasize that an aggregation method like regime [START_REF] Hinloopen | The regime method: A new multicriteria technique[END_REF] can be an alternative approach for defining a ranking on the set of arguments supporting the options and consequently on the options themselves.

Basics of argumentation

Argumentation is a reasoning model based on the following main steps: i) constructing arguments and counter-arguments, ii) defining the strengths of those arguments, and iii) defining the justified conclusions. Argumentation systems are built around an underlying logical language and an associated notion of logical consequence, defining the notion of argument. The argument construction is a monotonic process: new knowledge cannot rule out an argument but only gives rise to new arguments which may interact with the first argument. Arguments may be conflicting for different reasons.

Definition 1 (Argumentation system [START_REF] Dung | On the acceptability of arguments and its fundamental role in nonmonotonic reasoning, logic programming and n-person games[END_REF]) An argumentation system is a pair T = (A, R). A is a set of arguments and R ⊆ A × A is an attack relation. We say that an argument a attacks an argument b iff (a, b) ∈ R. Among all the arguments, it is important to know which arguments to keep for inferring conclusions. In [START_REF] Dung | On the acceptability of arguments and its fundamental role in nonmonotonic reasoning, logic programming and n-person games[END_REF], different acceptability semantics have been proposed. The basic idea behind these semantics is the following: for a rational agent, an argument a i is acceptable if he can defend a i against all attacks. All the arguments acceptable for a rational agent will be gathered in a so-called extension. An extension must satisfy a consistency requirement and must defend all its elements. Definition 2 (Conflict-free, Defence [START_REF] Dung | On the acceptability of arguments and its fundamental role in nonmonotonic reasoning, logic programming and n-person games[END_REF]) Let B ⊆ A, and a i ∈ A.

-B is conflict-free iff ∄ a i , a j ∈ B s.t. (a i , a j ) ∈ R. -B defends a i iff ∀ a j ∈ A, if (a j , a i ) ∈ R, then ∃ a k ∈ B s.t. (a k , a j ) ∈ R.
The main semantics introduced by Dung are summarized in the following definition.

Definition 3 (Acceptability semantics [START_REF] Dung | On the acceptability of arguments and its fundamental role in nonmonotonic reasoning, logic programming and n-person games[END_REF]) Let B be a conflict-free set of arguments.

-B is admissible iff it defends any argument in B.

-B is a preferred extension iff it is a maximal (w.r.t ⊆) admissible extension.

-B is a stable extension iff it is a preferred extension that attacks any argument in A\B.

Example 1 Let T = (A, R) be an argumentation theory where A = {α 1 , α 2 , α 3 , α 4 } is the set of the arguments and R = {(a 1 , a 2 ), (a 2 , a 1 ), (a 1 , a 4 ), (a 2 , a 3 )} is the set of attacks. This argumentation theory has two stable extensions

E 1 = {α 1 , α 3 } and E 2 = {α 2 , α 4 }
3 Preference-based Argumentation Framework: Properties and Limitations

In [START_REF] Amgoud | On the acceptability of arguments in preference-based argumentation framework[END_REF] the basic argumentation framework of Dung has been extended into preference-based argumentation theory (PBAT). The framework has been further developed and studied in [START_REF] Dimopoulos | Theoretical and Computational Properties of Preference-based Argumentation[END_REF]. The basic idea of a PBAT is to consider two binary relations between arguments:

1. A conflict relation, denoted by C, that is based on the logical links between arguments. 2. A preference relation, denoted by , that captures the idea that some arguments are stronger than others. Indeed, for two arguments a, b ∈ A, a b means that a is at least as good as b. The relation is assumed to be a partial pre-order (that is reflexive and transitive). The relation ≻ denotes the corresponding strict relation. That is, a ≻ b iff a b and b a.

The two relations are combined into a unique attack relation, denoted by R, and the Dung's semantics are applied on the resulting framework. In what follows, we focus on a particular class of PBATs, presented in [START_REF] Dimopoulos | Theoretical and Computational Properties of Preference-based Argumentation[END_REF], where the conflict relation C is irreflexive and symmetric. In [START_REF] Amgoud | Making decisions through preferencebased argumentation[END_REF] the impact of the preference relation on an argumentation system has been studied. After defining a relation £ on the powerset 2 A of the arguments of a PBAT T = (A, R), it has been shown that the stable extensions of T correspond to the most preferred elements of 2 A wrt this relation.

Definition 5 ([3]) Let T = (A, R) be a PBAT built on an underlying pre-order . If A 1 , A 2 ∈ 2 A , with A 1 = A 2 , then A 1 £ A 2 iff

one of following holds:

-

A 1 ⊃ A 2 -for all a, b such that a ∈ A 1 \ A 2 and b ∈ A 2 \ A 1 , it holds that a ≻ b
The exact correspondence between the relation £ and stable extensions is as follows. One feature of the £ relation is that it may not be transitive. Consider for instance the theory of the previous example, and observe that {a, b}£{b, c}£{c}. However, it is not the case that {a, b} £ {c}.

The second important observation, which is the main focus of this work, relates to the conclusions sanctioned by preference-based argumentation under the stable model semantics. The following example, borrowed from [START_REF] Horty | Argument construction and reinstatement in logics for defeasible reasoning[END_REF], shows clearly that these results can be counterintuitive even in simple cases. Example 4 The story of the example is about conclusions that can been drawn regarding the financial situation of a person based on arguments built on information about her occupation and residence. Let's suppose that layers are, in general, considered to be wealthy, but a certain subclass, the pubic defenders, are considered not to be. Consider now an area in Paris -say, Passy -containing a large number of expensive private homes along with a much smaller number of middle-income rental properties. Thus the residents of Passy can be generally considered to be wealthy although the renters, are considered to not to be. Assume that Ann is a public defender (P Da), and therefore a layer (La), who rents in Passy (Ra), and is therefore a resident of Passy (P a).

If we assume that W a represents the proposition that Ann is wealthy, the arguments that can be generated in an underlying propositional language from the above story are a 1 = {P Da, P Da -→ La, La -→ W a}, a 2 = {P Da, P Da -→ ¬W a} a 3 = {Ra, Ra -→ P a, P a -→ W a}, a 4 = {Ra, Ra -→ ¬W a}.

From the above arguments we generate the PBAT T =(A, R), where A = {a 1 , a 2 , a 3 , a 4 }. The attack relation R is composed from the conflict relation C = {(a 1 , a 2 ), (a 2 , a 1 ), (a 1 , a 4 ), (a 4 , a 1 ), (a 3 , a 2 ), (a 2 , a 3 ), (a 3 , a 4 ), (a 4 , a 3 )}, and the preference relation that is defined as a 2 ≻ a 1 and a 4 ≻ a 3 , whereas all other pairs of arguments are incomparable.

Theory T has two extensions, namely E 1 = {a 1 , a 3 } and E 2 = {a 2 , a 4 }. The first extension supports the conclusion that Ann is wealthy, whereas the the second that she is not. Intuitively however one would conclude that Ann is not wealthy. In other words we could argue that the second extension is more preferred than the first, as for every argument of the first it contains a more preferred argument.

The relation £ on the subset of A with two elements is depicted in figure 3. Note again that £ is not transitive. Indeed, it holds that {a 2 , a 4 } £ {a 2 , a 3 } and {a 2 , a 3 } £ {a 1 , a 3 }, but {a 2 , a 4 } £{a 1 , a 3 }. Although, a 2 ≻ a 1 and a 4 ≻ a 3 , the stable model semantics does not render {a 2 , a 4 } better than {a 1 , a 3 }, because a 2 ≻ a 3 and a 4 ≻ a 1 .

The main purpose of this work is to provide a preliminary study of the problem of the conclusions sanctioned by the state-of-the-art argumentation, and identify possible solutions by borrowing ideas from decision theory.

As we noted in section 3, the stable model semantics can lead to counter-intuitive results. To remedy the situation we present a new semantics for preferencebased argumentation called super-stable extensions semantics. The main idea is to only accept conclusions drawn under the stable model semantics from a PBAT T that correspond to the conclusions that are sanctioned by some other PBAT T ′ which is obtained from T by removing incomparability. Therefore, the new semantics may differ from the standard stable model semantics only on theories with incomparability. As we will discuss in the next section theories without incomparability always sanction the correct conclusion under the stable extension semantics.

Before we proceed to the definition of the new semantics, we recall some useful concepts. A relation on a set S is total if for all a, b ∈ S with a = b, a b or b a. A strict total order on a set S is an asymmetric (hence irreflexive), transitive and total relation on S. The notion of an extension of a relation is used in decision theory and economics eg. ( [START_REF] Hansson | Choice structures and preference relations[END_REF], [START_REF] Donaldson | A quasiordering is the intersection of orderings[END_REF]) Definition 6 A binary relation E on S is an extension of a pre-order on S if E is a pre-order on S such that E ⊇ and for all a, b ∈ S if a ≻ b then a ≻ E b. An extension of a pre-order that is complete (ie., for all a, b ∈ S, a b or b a) is called ordering extension of .

Hansson [START_REF] Hansson | Choice structures and preference relations[END_REF] has shown that every pre-order has an ordering extension. Moreover, Donaldson and Weymark [START_REF] Donaldson | A quasiordering is the intersection of orderings[END_REF] proved that a pre-order is the intersection of its ordering extensions. Definition 7 A strict total order ≻ s on a set S is a strict ordering of a total pre-order if for all a, b ∈ S if a ≻ b then a ≻ s b. A strict total order is a strict ordering of a pre-order if it is a strict ordering of one of its ordering extensions.

The following definition extends the notions of ordering extension and strict ordering to the case of PBATs. The following result that relates the stable extensions of a PBAT and the stable extensions of its ordering completions and strict projections is easily provable.

Proposition 1 Let T be a PBAT, T o one of its ordering completions, and T s one of its strict projections. If E o is a stable extension of T o , then it is also a stable extension of T . Moreover, if E s is a stable extension of T s , then it is also a stable extension of T . However, it is not the case that a stable extension of T is a stable extension of some T o built on a pre-order ≻ o that is an ordering extension of . Consider for instance again the theory of example 4, and its stable extension E 1 = {a 1 , a 3 }. To see that there is no PBAT T o which is an ordering completion of T and has E 1 as a stable extension, observe that for E 1 to be a stable extension it must be the case that a 1 a 4 and a 3 a 2 . However, together with a 2 ≻ a 1 these would mean that a 3 a 4 , which is impossible given that a 4 ≻ a 3 .

We now proceed with the definition of the new semantics for preferencebased argumentation. As noted earlier, the basic idea is to only accept a set of arguments as an extension of a PBAT T if this set is a stable extension of an ordering completion of T . More formally the concept is defined as follows.

Definition 9 Let T = (A, R) be a PBAT built on an underlying pre-order and conflict relation C. A stable extension E of T is a super-stable extension of T if it is the stable extension of an ordering extension of T .

By the results of [START_REF] Hansson | Choice structures and preference relations[END_REF] we know that every pre-order has an ordering extension. Therefore, every PBAT has an ordering completion which is itself a PBAT. From [START_REF] Dimopoulos | Theoretical and Computational Properties of Preference-based Argumentation[END_REF], we know that every PBAT has a stable extension. By combining these two results we obtain the following property for super-stable extensions.

Proposition 2 Every PBAT has a super-stable extension.

Theories without incomparability

If a PBAT T contains no incomparability, T is an ordering completion of itself. Therefore, any stable extension of T is by definition a super-stable extension of T . In this section we prove that for this class of theories a correspondence holds between their stable extensions and the stable extensions of their strict projections.

For a PBAT without incomparability, T = (A, R), we define the level of argument a ∈ A, denoted by l(a), recursively as follows -l(a) = 1 for all a such that there is no b ∈ A s.t. b ≻ a -l(a) = k for all a such that for all a ′ ∈ A s.t. a ′ ≻ a it holds that l(a ′ ) < k, and ∃a ′′ ∈ A s.t. a ′ ≻ a and l(a ′ ) = k -1

The following lemma relates the level of arguments with relation ≻ and will be used in the proof of the main result of this section (proposition 3 below). A direct consequence of the previous lemma is that l(a) ≤ l(b) iff a b. Also note that since a super-stable extension of a PBAT T = (A, R) is also a stable extension of T , it holds that for all a ∈ E there exist b ∈ E such that (b, a) ∈ R.

From the above we conclude that for all a ∈ E there exist b ∈ E such that (a, b) ∈ C and l(b) ≤ l(a). Note the above property does not hold in general for theories that contain incomparability.

Proposition 3 Let T be a PBAT without incomparability. Every stable extension of T is a stable extension of some strict projection of T .

Proof. We prove the claim by defining a strict projection of T , T s = (A, R s ), for which E is a stable extension. Let ≻ E + s be a strict projection on the arguments of E. Similarly, let ≻ E - s be a strict projection on the arguments of A\E. Finally, let ≻ E +,- s be the binary relation on

(E × (A \ E)) ∪ ((A \ E) × E) such that for any pair of arguments a ∈ E and b ∈ E -if l(a) > l(b) then b ≻ E +,- s a -if l(a) ≤ l(b) then a ≻ E +,- s b where l(a) is the level of argument a in theory T . Define ≻ s =≻ E + s ∪ ≻ E - s ∪ ≻ E +,- s .
We first show that ≻ s is strict total order. It is easy to verify that ≻ s is asymmetric and total by construction. We show that ≻ s is transitive. Let a, b, c ∈ A, such that a ≻ s b and b ≻ s c. We need to show that a ≻ s c. We proceed by case analysis. For the case where a, b, c ∈ E or a, b, c ∈ A \ E, transitivity follows by construction. Assume now that a ∈ E and b, c ∈ E. Then, by construction, l(a) ≤ l(b). Moreover, since b ≻ s c, by lemma 1, it must be the case l(b) < l(c). Hence, l(a) < l(c), which, again by lemma 1, means that a ≻ s c. The case where a ∈ E, b ∈ E, c ∈ E is similar. Now suppose that a, b ∈ E, and c ∈ E. Since a ≻ s b, by lemma 1, we obtain that l(a) < l(b). Moreover, by construction, it must hold that l(b) ≤ l(c). Hence, l(a) < l(c), and therefore a ≻ s c. The remaining cases where a ∈ E can be proved analogously.

Finally, we prove that E is a stable extension of T s = (A, R s ). First note that E is conflict free. Now assume that a ∈ E and define D(a) = {b|b ∈ E and (b, a) ∈ R}. Since E is a stable extension, it must be D(a) = ∅. Let b ∈ D(a), and assume that (b, a) ∈ R s . By construction, it must hold that l(a) < l(b), which by lemma 1 implies a ≻ b. This however contradicts b ∈ D(a). Therefore, (b, a) ∈ R s , which means that E is a stable extension of T s .

⊓ ⊔

By combining the definition of a super-stable extension with proposition 3 we obtain the following strong property regarding super-stable extensions.

Proposition 4 Every super-stable extension of a PBAT T is a stable extension of some strict projection of T .

A multi-criteria view of PBAT

In the previous sections we investigated how standard argumentation semantics can be extended to accommodate preference information on the arguments. In this section we change our perspective and explore a more direct link between argumentation and decision theory. More specifically, we interpret arguments as criteria and regard preferences as information on the relative importance of these criteria. Under this perspective argumentation can be understood as Multiple Criteria Decision Problem (MCDP). We start our analysis with a definition of the problem that leaves out some of its aspects that are not directly relevant to our purposes.

Definition 10 A Multiple Criteria Decision Problem (MCDP) is a triple P = (A, K, ) where -A = {a 1 , . . . , a n } is the set of attributes.

A set of values is associated with each attribute, denoted by v(a 1 ), . . . , v(a n ) -K = {≫ a1 , . . . , ≫ an } is the set of criteria. A criterion ≫ ai is a pre-order associated with the values of an attribute a i is pre-order on the criteria An alternative l wrt to a MCDP P = (A, K, ) is any l ∈ v(a 1 ) × . . . × v(a n ).

We denote the set of alternatives by L P .

In certain situations, a solution to a MCDP is a ranking relation ¤ on the set of alternatives L, ie. ¤ ⊂ L × L. Usually a solution to a MCDP has to satisfy certain properties [START_REF] Bouyssou | Multicriteria Evaluation and Decision Models: stepping stones for the analyst[END_REF].

The following definition shows that argumentation can be transformed in a meaningful way into a MCDP.

Definition 11 Given a PBAT T = (A, R), where A = {a 1 , . . . , a n }, we define its corresponding MCDP M T = (A T , K T , T ) as follows:

-A T = A, with v(a i ) = {a + i , a - i }, for each a i ∈ A T . -K T = {≫ 1 , . . . , ≫ n }, where ≫ i , for 1 ≤ i ≤ n, is defined as the preference a + i ≫ i a - i . -T =
The following is an example of a translation of a specific PBAT into a MCDP.

Example 5 Consider the PBAT T = (A, R), where A = {a 1 , a 2 , a 3 }, and the underlying preference relation defined as:

a 1 ≻ a 2 , a 1 ≻ a 3 , a 2 a 3 , a 3 a 2 .
The corresponding MCDP is defined as M T = (A T , K T , T ), where:

-A T = {a 1 , a 2 , a 3 }, with v(a 1 ) = {a + 1 , a - 1 }, v(a 2 ) = {a + 2 , a - 2 }, v(a 3 ) = {a + 3 , a - 3 }. -K T = {≫ 1 , ≫ 2 , ≫ 3 }, with a + 1 ≫ 1 a - 1 , a + 2 ≫ 2 a - 2 , a + 3 ≫ 3 a - 3 . -T =
Several methods have been proposed in the literature, the applicability of which in many cases depends on the features of the MCDP at hand. Among the methods for tackling MCDPs that appear in the literature and are applicable to the case we consider, it seems that the Regime method [START_REF] Hinloopen | The regime method: A new multicriteria technique[END_REF] is the closest to the spirit of the stable extensions semantics. Here we discuss a simplified version of the method as it appears in [START_REF] Moffett | Incorporating multiple criteria into the design of conservation area networks: A minireview with recommendations[END_REF].

The Regime method works as follows. For any two alternatives A i , A j , let K + be the set of criteria according to which A i is better than A j , and K -be the set of criteria according to which A j is better than A i . Regime ranks A i better than A j , denoted by A i £ t R A j , if K + = ∅ and there is an injective map from K -to K + by which each criterion in K -is mapped to a more important criterion in K + . The set of optimal alternatives is then {A o : ∀i¬(A i £ t R A o )}. We can easily define a preference order on the sets of arguments of a PBAT that captures the Regime method. To do this we associate to any set of arguments A, an alternative

A t = {a + |a ∈ A} ∪ {a -|a ∈ A}. Definition 12 Let T = (A, R) be a PBAT and M T = (A T , K T , T ) its corre- sponding MCDP. For any A 1 , A 2 ⊆ 2 A , it holds that A 1 £ R A 2 if A t 1 £ t R A t 2 .
We can now define the notion of a regime extension of a PBAT by characterizing it in a way similar to the stable extensions.

′ ∈ 2 A such that E ′ £ R E, there are a 1 , a 2 ∈ E ′ such that (a 1 , a 2 ) ∈ C.
We now apply the previous definition to the story of 4, and observe that it yields the correct result.

Example 6 Consider again the theory T of example 4. The corresponding MCDP M T = (A T , K T , T ) can be defined as outlined above. Consider the two sets of arguments E 1 = {a 1 , a 3 } and E 2 = {a 2 , a 4 } of T which correspond to the alternatives E t 1 = {a + 1 , a - 2 , a + 3 , a - 4 } and E t 2 = {a - 1 , a + 2 , a - 3 , a + 4 }. For the comparison E t 2 £ t R E t 1 we have that K + = {a 2 , a 4 }, K -= {a 1 , a 3 }, and the mapping a 1 → a 2 , a 3 → a 4 . For the comparison E t 1 £ t R E t 2 we have that K + = {a 1 , a 3 } K -= {a 2 , a 4 }, but there is no suitable mapping. Therefore we conclude E 2 £ R E 1 . Moreover, E 2 is a regime extension of T .

Conclusion and Future Work

In this paper we pointed out that Dung's stable extensions semantics when applied in preference-based argumentation frameworks for decision making purposes lead to counter-intuitive conclusions. A similar problem has been identified also by Horty in [START_REF] Horty | Argument construction and reinstatement in logics for defeasible reasoning[END_REF]. This mainly holds for argumentation theories where the preference relation used for defining the relative strength of individual arguments contains incomparability. To resolve this problem we proposed a new acceptability semantics called super-stable extensions which allows to capture the conclusions corresponding to the good decisions and to avoid the counter intuitive ones which could correspond to bad decisions. Moreover, we showed that preferencebased argumentation can be understood as a multiple-criteria decision problem allowing to that way the exploration of the application of theoretical results of the decision theory in argumentation. Therefore, this work can been seen as an attempt to bring new ideas from decision theory to argumentation.

Our future work concerns the definition of a binary relation £ SS on the sets of arguments of a PBAT that will be proved exactly the preference relation that is induced by the super-stable extensions semantics and to prove the correspondence between both. This will be the equivalent result of the one we proved in [START_REF] Dimopoulos | Theoretical and Computational Properties of Preference-based Argumentation[END_REF] between the preference £ and the stable extensions.

Definition 4 (Example 2

 42 Preference-based Argumentation Theory (PBAT)) ([3]) Given an irreflexive and symmetric conflict relation C and a preference relation on a set of arguments A, a preference-based argumentation theory (PBAT) on A is an argumentation system T = (A, R), where (a, b) ∈ R iff (a, b) ∈ C and b ≻ a. It follows directly from the definition that if (a, b) ∈ C and a b and b a, then (a, b) ∈ R. Moreover, if (a, b) ∈ C and a, b are either indifferent or incompatible in , then (a, b) ∈ R and (b, a) ∈ R. Also note that if (a, b) ∈ C, then either (a, b) ∈ R or (b, a) ∈ R. Finally, if (a, b) ∈ R and (b, a) / ∈ R, then a ≻ b. The following example illustrates some features of PBATs. Let A = {a, b, c, d} be a set of arguments, and C the conflict relation on A defined as C = {(a, b), (b, a), (b, c), (c, b), (c, d), (d, c)}. Moreover, let the preference relation contain transitive closure of the set of pairs a b, b c, c d, and d c. The corresponding PBAT is T = (A, R), where R = {(a, b), (b, c), (c, d), (d, c)}. Theory T has two stable extensions, E 1 = {a, c} and E 2 = {a, d}.

Theorem 1 (

 1 [START_REF] Amgoud | Making decisions through preferencebased argumentation[END_REF]) Let T = (A, R) be a PBAT built on an underlying pre-order and a conflict relation C. E is a stable extension of T iff there are no arguments a, b ∈ E s.t. (a, b) ∈ C, and for all A ∈ 2 A such that A £ E, there are a 1 , a 2 ∈ A such that (a 1 , a 2 ) ∈ C.The example below explains the link between £ and stable extensions. Example 3 Let T = (A, R) be a PBAT with A = {a, b, c} and R composed from the conflict relation C = {(a, b), (b, a)(a, c), (c, a)} and preference relation that contains the pairs a ≻ b and a ≻ c, and marks all other pairs of arguments as indifferent. The relation £ on 2 A induced by contains the pairs {a}£{b, c}, {a} £ {b}, {a} £ {c}. Since the sets {a, b, c}, {a, b}, {a, c} are ruled out by C, the set E = {a} is the stable extension of T .

Fig. 1 .

 1 Fig. 1. The preference relation induced on sets of arguments.

Definition 8

 8 Let T = (A, R) be a PBAT on an underlying pre-order and a conflict relation C. The PBAT T o = (A, R o ), on an underlying relation o and the conflict relation C, is a ordering completion of T if o is an ordering extension of . The PBAT T s = (A, R s ), on an underlying relation ≻ s and the conflict relation C, is a strict projection of T if ≻ s is a strict ordering of .

Lemma 1 .

 1 Let T = (A, R) be a PBAT without incomparability on an underlying pre-order . For every a, b ∈ A, l(a) < l(b) iff a ≻ b. Proof. The property that for every a, b ∈ A if a ≻ b then l(a) < l(b), follows directly from the definition of the level of an argument. We prove now that if l(a) < l(b) then a ≻ b. Assume a, b ∈ A with l(a) < l(b). Clearly it can not be the case b ≻ a, because then l(b) < l(a). Assume that a b and b a, and l(b) = m (hence, l(a) < m). Then, there must be c ∈ A s.t l(c) = m -1 and c ≻ b. Therefore, it must be the case that c ≻ a, and hence l(a) ≥ m. But this contradicts l(a) < m. Therefore it holds that a ≻ b. ⊓ ⊔

Definition 13 A

 13 set of arguments E is a regime extension of a PBAT T = (A, R) if there are no arguments a, b ∈ E s.t. (a, b) ∈ C, and for all E


John M Campbell
email: jmaxwellcampbell@gmail.com

On a problem due to Glasser on analytically tractable moments

Keywords: Complete elliptic integral, multiple elliptic integral, Feynman diagram PACS numbers: 02.30.-f, 02.30.Gp

Glasser, in 2011, introduced a remarkable integral identity of physical interest and conjectured, along with Wan, that the evaluation

provides the unique analytically tractable moment of K 2 on a sub-unit interval, where K denotes the complete elliptic integral of the first kind, and where

We show how a case of Clausen's product identity related to Ramanujan's series for 1 π may be applied, via an integration argument derived from our past work in fractional analysis and Fourier-Legendre theory, to show how higher moments of K 2 on the same sub-unit interval may be evaluated analytically in terms of the Γ-function, thereby disproving Glasser and Wan's conjecture. This and Glasser's moment formula are motivated by how closely related moment formulas for powers of K arise in the study of Feynman diagrams.

Introduction

In 2011, Glasser [1] introduced the integral identity such that

F (xt) dt (1)
for a given function F , and this was inspired by physics-related applications of double integrals of the above form. Glasser [1] applied the closely related identity to show that

1/ √ 2 0 k K 2 (k) dk = πG 4 , (3)
where the complete elliptic integrals of the first and second kinds are respectively such that

K(k) = π/2 0 dθ 1 -k 2 sin 2 θ , E(k) = π/2 0 1 -k 2 sin 2 θ dθ,
and where G = 1 1 2 -1 3 2 + 1 5 2 -• • • denotes Catalan's constant. Both Glasser and Wan [1] conjectured that Glasser's formula in (3) provides the only analytically tractable moment of K 2 on a sub-unit interval; see also [2,3,4,5,6] for closely related work due to Wan et al. on multiple elliptic integrals. So, this problem due to Glasser and Wan [1] is given by whether or not explicit evaluations for moments of the form

x 0 k m K 2 (k) dk (4)
may be obtained for x < 1, apart from Glasser's evaluation highlighted in (3). We succeed in formulating a technique that we apply to obtain explicit, analytic evaluations of ( 4) for odd natural numbers m and for the same sub-unit interval, with x = 1 √ 2 . In contrast to the work of Glasser [1], our method relies on a fractional analysis-based approach explored in a our previous work [START_REF] Campbell | Integral Transforms Spec[END_REF], together with a case of Clausen's product identity related to Ramanujan's series for 1 π [8]. We introduce and prove the following analytic evaluations, each of which disproves the Glasser-Wan conjecture, and our techniques apply to the infinite family of odd moments suggested below:

1/ √ 2 0 k 3 K 2 (k) dk = Gπ 8 + π 3 4Γ 4 1 4 - Γ 4 1 4 256π (5)
and

1/ √ 2 0 k 5 K 2 (k) dk = (22G -1)π 256 + π 3 4Γ 4 1 4 - Γ 4 1 4 256π (6)
and

1/ √ 2 0 k 7 K 2 (k) dk = (34G -3)π 512 + 11π 3 48Γ 4 1 4 - 95Γ 4 1 4 27648π . (7)
It seems that the Glasser-Wan conjecture indicated above has not previously been considered, subsequent to [1], with regard to past research [9,10] based on Glasser's work in [1].

Preliminaries and background

Glasser's derivation of the main result in [1], i.e., the integral transform in (1), relied on the use of the Bessel transform and manipulations based on the modified Struve function, to prove (2) in the derivation of the main result from [1]. Glasser's proof of the moment formula in (3) was then obtained via (2) by setting F = K and by utilizing the relation

π/2 0 K(a sin θ) dθ = K 2   1 - √ 1 -a 2 2  
given in an equivalent form in Glasser's previous work on definite integrals involving K [11]. As in this past work by Glasser [11], we express that triple integrals that naturally emerge within physics in the study of three-dimensional lattices may often be reduced to β α F (u)K(u) du for an elementary function F , and this has served as a main source of motivation behind the author's past work on integrals involving K [12,13]. In contrast to Glasser's uses of the Bessel transform and the modified Struve function, the crux of our disproof of the Glasser-Wan conjecture is given by a special case of the Clausen hypergeometric product identity.

Clausen's product identity is such that

2 F 1 a, b a + b + 1 2 z 2 = 3 F 2 2a, a + b, 2b a + b + 1 2 , 2a + 2b z ,
and this may be thought of as a key toward the following modular identity that required in the classical derivation of Ramanujan's series for 1 π [15, §5.5]:

2K(k) π 2 = 3 F 2 1 2 , 1 2 , 1 2 1, 1 (2k √ 1 -k 2 ) 2 ,
Equivalently,

π 2 4 ∞ n=0 1 64 n 2n n 3 y n = K 2 1 - √ 1 -y 2 , (8)
and we are to apply the identity in (8) in a similar way relative to our past work in Fourier-Legendre theory and fractional calculus [START_REF] Campbell | Integral Transforms Spec[END_REF], as described below.

The method of semi-integration by parts (SIBP) was introduced in [START_REF] Campbell | Integral Transforms Spec[END_REF] and may, informally, be thought of as a combination of fractional calculus and past techniques from [START_REF] Campbell | [END_REF] on the use of Fourier-Legendre expansions to evaluate integrals involving K. By applying the fractional derivative operators D 1/2 x α = Γ(α+1) Γ(α+1/2) x α-1/2 and D -1/2 x α = Γ(α+1) Γ(α+3/2) x α+1/2 to an analytic function with a Maclaurin series involving a power of central binomial coefficients, these operators may be thought of as having the effect of reducing the power of this binomial factor. This approach led to the evaluation of the equivalent integral expressions

1/2 0 K 2 ( √ x) dx = 1 4 1 0 1 √ 1 -y K 2 1 - √ 1 -y 2 dy, (9)
where this equivalence is given by a change of variables. By then expanding the latter integrand according to the consequence in (8) of Clausen's product identity, and by then integrating term-by-term as in [START_REF] Campbell | Integral Transforms Spec[END_REF], the beta integral valuation

1 0 y n √ 1 -y dy = 2 2n+1
(2n + 1) 2n n can be thought of as reducing the power of the central binomial coefficient in (8), yielding the known evaluation

∞ n=0 1 16 n 2n n 2 1 2n + 1 = 4G π , (10)
which was proved via SIBP in [START_REF] Campbell | Integral Transforms Spec[END_REF] and which is equivalent to Glasser's moment formula (3), proved in [1].

Main results

As indicated above, the result highlighted as Theorem 1 below gives a disproof of the Glasser-Wan conjecture. Glasser expressed how the research interest in the integral in ( 3) is motivated by how expressions of a simliar nature arise in the study of Feynman diagrams and related areas in quantum field theory [14]; this further motivates our new results as in Theorem 1 below.

Theorem 1. The moment formula in (5) holds.

Proof. We begin by rewriting the integral in (5) so that

1/ √ 2 0 k 3 K 2 (k) dk = 1 2 1/2 0 x K 2 ( √ x) dx. (11)
By the same change of variables as in (9) (cf. [START_REF] Campbell | Integral Transforms Spec[END_REF]), this yields

1/ √ 2 0 k 3 K 2 (k) dk = 1 16 1 0 1 - √ 1 -y √ 1 -y K 2 1 - √ 1 -y 2 dy. (12)
By then applying the Clausen-derived identity in (8), the Dominated Convergence Theorem allows us to interchange the operations of definite integration and infinite summation, so that the equivalent form in (10) of Glasser's moment formula [1] gives us that

1/ √ 2 0 k 3 K 2 (k) dk = Gπ 8 - π 2 64 ∞ n=0 1 64 n 2n n 3 1 n + 1 (13)
The ordinary generating function corresponding to the summand of the series in ( 13) may be evaluated as below, with convergence for |z| < 1. This may be proved via term-by-term differentiation for convergent power series, so as to obtain an equivalence to the case of Clausen's product identity in (8):

-8 π 2 z 2E 2 1- √ 1-z 2 -2(1 + √ 1 -z)E 1- √ 1-z 2 K 1- √ 1-z 2 + 1 + √ 1 -z -z K 2 1- √ 1-z 2 .
. So, according to the elliptic integral singular values [15, §1]

K 1 √ 2 = Γ 2 1 4 4 √ π , E 1 √ 2 = 4Γ 2 3 4 + Γ 2 1 4 8 √ π , (14)

Conclusion

We conclude by describing two further topics to explore that concern the new results given above. A further reason as to the research interest in Theorems 1-3 has to do with how equivalent formulations of these results can be applied to obtain new, fast-converging hypergeometric series for fundamental constants. By rewriting the integrals in these theorems over the latter interval in (11) and according to the variable change in (11), we then proceed to rewrite the integrand factor K 2 ( √ x) as a double hypergeometric series according a term-by-term expansion of the Maclaurin series of K( √ x) times itself. By then integrating over the latter interval in (11) this results in double series of convergence rate 1 2 with respect to both index sets. By then taking linear combinations to isolate a given constant among evaluations as in ( 5)-( 7), we may obtain evaluations as in for x ∈ [0, 1/2] given by Cantarini in 2022 [17] can be used to obtain alternative proofs of Theorems 1-3, letting P n (x) denote the Legendre polynomial of order n. We encourage the exploration as to how Cantarini's quasi-FL identity could be used to generalize or otherwise strengthen Theorems 1-3.

2 π = m,n≥0

0 1 0

 1 dθ sin ϕF (x sin ϕ sin θ) = π 2

0π 2x x 0 F

 0 dθ sin ϕF (x sin ϕ sin θ) -(t) dt = 0 (2)

Acknowledgements

The author is very grateful to acknowledge support from a Killam Postdoctoral Fellowship from the Killam Trusts.

by setting z = 1, we obtain an analytic form for the infinite series in (13).

Theorem 2. The moment formula in (6) holds.

Proof. Applying the same change of variables that led to (12), we obtain that

We again apply the consequence of the Clausen identity in (8), and the Dominated Convergence Theorem may be applied to justify the reversal of the order of integration and infinite summation, yielding

and we may proceed as before, with an index shift applied to (10).

Theorem 3. The moment formula in (7) holds.

Proof. Mimicking the above proofs, we write

We again expand the latter integrand according to (8), so as to apply the Dominated Convergence Theorem, to apply term-by-term integration. In this case, we would require the evaluation of the power series

, and we again apply the elliptic integral singular values in (14).

We may continue as above to evaluate higher moments, as in


Philippe Jaming

Saba Chadi

FROM INGHAM TO NAZAROV'S INEQUALITY: A SURVEY ON SOME TRIGONOMETRIC INEQUALITIES

Keywords: 2020 Mathematics Subject Classification. 42A75 Ingham's Inequality, Littlewood problem, non-harmonic Fourier series, lacunary series

The aim of this paper is to give an overview of some inequalities about -norms ( = 1 or = 2) of harmonic (periodic) and non-harmonic trigonometric polynomials. Among the material covered, we mention Ingham's Inequality about 2 norms of non-harmonic trigonometric polynomials, the proof of the Littlewood conjecture by Mc Gehee, Pigno and Smith on the lower bound of the 1 norm of harmonic trigonometric polynomials as well as its counterpart in the non-harmonic case due to Nazarov. For the latter one, we give a quantitative estimate that completes our recent result with an estimate of 1 -norms over small intervals.

We also give some stronger lower bounds when the frequencies satisfy some more restrictive conditions (lacunary Fourier series, "multi-step arithmetic sequences").

Most proofs are close to existing ones and some open questions are mentioned at the end.

Introduction 2.

2 estimates 2.1. Ingham's inequalities 2.2. The condition > 1 2.3. Sequences with large gaps 3.

1 estimates with integer frequencies 3.1. The 1 norm of the Dirichlet kernel 3.2. Lacunary trigonometric polynomials 3.3. The proof of Littlewood's conjecture by Mc Gehee, Pigno and Smith 3.4. Strongly multidimensional sets 3.5. Newman's extremal trigonometric polynomial 4.

1 estimates with real frequencies 4.1. Some results for the Besicovitch norm 4.2. Nazarov's theorem [START_REF] Cohen | On a conjecture of Littlewood and idempotent measures[END_REF]. Some open problems References

INTRODUCTION

The aim of this paper is to give an overview of the main estimates of -norms of harmonic and nonharmonic trigonometric polynomials, when = 2 and = 1. In many fields of mathematics, ranging from number theory (see e.g. the previous survey [START_REF] Belov | On the conjecture of Littlewood and minima of even trigonometric polynomials[END_REF]) to signal processing and PDEs, one is lead to investigate such norms. Our main motivation comes from the use of Ingham's inequality (lower and upper estimates of 2 -norms of non-harmonic trigonometric polynomials) in control theory of PDEs. We refer the interested reader to the book by Komornik and Loreti [START_REF] Komornik | LORETI Fourier Series in Control Theory[END_REF] (iii) Assume further that for = 0, … , -1, +1 -≥ 1, then, for every > 1, there exists a constant ( ) such that, for every 0 , … , ∈ C, Moreover, (a) for ≥ 72 we can take ( ) = 1 122 ;

(b) for 1 < ≤ 2, ( ) = ( -1) 15∕2 .

Let us comment on this theorem. To start with, the theorem is essentially a quantitative version of a result of F. Nazarov [START_REF] Nazarov | On a proof of the Littlewood conjecture by McGehee, Pigno and Smith[END_REF] and is due to the authors together with K. Kellay [START_REF] Ph | SABA The Littlewood problem and non-harmonic Fourier series[END_REF]. Only the case 1 < ≤ 2 has not been presented before. To be more precise, when the 's are integers, this is of course the result of Mc Gehee, Pigno and Smith while Point 2 comes from a slight modification of their argument by Stegeman and Yabuta. Note that the constants are the same as those in Theorem B (and are actually a bit better). While it is obvious that this inequality implies Theorem B (with the same constant), there is an elegant argument by Hudson and Leckband [11] that allows to show that the converse is also true. We will present this argument below. Point 3 is due to Nazarov with non-explicit constant. A direct proof of this theorem is given in [START_REF] Ph | SABA The Littlewood problem and non-harmonic Fourier series[END_REF] with the explicit constants mentioned above. Only the case of small has not been presented so far and is thus the main novelty of the present paper. It is obtained by a very mild modification of Nazarov's proof. Note also that once we have established Point 3 for some 0 , it is valid for all ≥ 0 so that we actually recover Nazarov's original result (with much worse constants than the ones stated above).

The remaining of the paper is organised as follows: Section 2 is devoted to the 2 case and we prove the inequalities in Ingham's Theorem in Section 2.1 and the necessity of the condition > 1 in Section 2.2. Further, Haraux's argument is presented in Section 2. [START_REF] Brevig | VUKOTIĆ The multiplicative Hilbert matrix[END_REF].

In Section 3, we investigate the 1 norms when the frequencies are integers. We start with the classical asymptotic estimate of the Dirichlet kernel and devote Section 2.2 to lacunary trigonometric polynomials. Section 3.3 is devoted to Mc Gehee, Pigno, Smith's proof of Littlewood's conjecture. In Section 3.4 we present Hanson's result on strongly multidimensional sequences and we conclude in Section 3.5 with the result of Newman. Section 4 is devoted to the case of non-harmonic trigonometric polynomials. We start in Section 4.1 with the argument of Hudson and Leckband and then extend the case of lacunary trigonometric polynomials to the non-harmonic setting and the Besicovitch 1 -norms. We conclude this section with the proof of the quantitative version of Nazarov's theorem for small .

In the last section, we present a few open questions.

2 ESTIMATES

2.1. Ingham's inequalities. The aim of this section is to show the following: let

Λ = { } ∈Z ⊂ R be a 1-separated sequence, | -| ≥ 1 if ≠ . Let (Λ)
be the set of (non-harmonic) trigonometric polynomials

(Λ) = ( ) ∶= ∑ ∈Z 2 ∶ ( ) ∈Z ⊂ C with finite support .
Note that, if ∈ (Λ) is given, then the 's are determined by

= lim →+∞ 1 ∫ ∕2 -∕2 ( ) -2 d .
We can then define two natural norms on (Λ), namely

‖ ‖ 2 ([-∕2, ∕2]) ∶= 1 ∫ ∕2 -∕2 | ( )| 2 d 1 2
and

‖ ‖ 2 ∶= ∑ ∈Z | | 2 1 2
.

Our aim in this section is to show that, when > 1, these two norms are equivalent. This is done by proving two inequalities. The first one is the direct inequality:

Proposition 2.1 (Ingham's direct inequality). Let ( ) ∈Z be a finitely supported sequence of complex numbers and ( ) ∈Z be a sequence of real numbers with +1 -≥ 1. For every > 0,

(2.1) 1 ∫ ∕2 -∕2 | | | | | ∑ ∈Z 2 | | | | | 2 d ≤ 2 + 1 ∑ ∈Z | | 2 .
Proof. We consider the function on R defined by

ℎ( ) = cos when | | ≤ 1 2 0 otherwise .
As ℎ is real and even, its Fourier transform is given by ĥ

( ) = 2 ∫ 1∕2 0 cos cos 2 d = 2 cos 1 -4 2
with the understanding that ĥ(1∕2) = 1 2

. From this, one shows that ĥ(

) ≥ 1 2 for | | ≤ 1 2 . Finally, let ( ) = ℎ * ℎ( ) = ⎧ ⎪ ⎨ ⎪ ⎩ sin | | -(| | -1) cos 2 when | | ≤ 1 0 otherwise .
One easily shows that is even, non-negative, supported in [-1, 1] and that (0) = 1 2 . Further its Fourier transform is ̂ ( ) = ĥ( ) 2 . In particular, ̂ ( ) ≥ 0 and ̂ ( )

≤ 1 4 for | | ≤ 1 2 .
But then, if ( ) ∈Z is finitely supported and

( ) = ∑ ∈Z 2 , ∫ 1∕2 -1∕2 | ( )| 2 d ≤ 4 ∫ 1∕2 -1∕2 ̂ ( )| ( )| 2 d ≤ 4 ∫ R ̂ ( ) | | | | | ∑ ∈Z 2 | | | | | 2 d = 4 ∑ , ∈Z ∫ R ̂ ( ) 2 ( -) d = 4 ∑ , ∈Z ( -).
Note that the sums are actually finite. Further, if

≠ then | -| ≥ 1 and, as is supported in [-1, 1],
we then have ( -) = 0. This implies that

∫ 1∕2 -1∕2 | ( )| 2 d ≤ 4 (0) ∑ ∈Z | | 2
so that the inequality is proven for = 1 since 4 (0) = 2.

For < 1 we simply write

1 ∫ ∕2 -∕2 | ( )| 2 d ≤ 1 ∫ 1∕2 -1∕2 | ( )| 2 d ≤ 2 ∑ ∈Z | | 2 . for 1 < ≤ 2 3 2 64 for ≥ 2 .
Proof. We will do so in three steps. We first establish this inequality for 1 < ≤ 2.

Let be defined by ( ) = 1 [-1∕2,1∕2] cos and notice that

̂ ( ) = 2 cos 1 -4 2 .
Note that being non-negative, ̂ is maximal at 0 (which may be checked directly).

Next consider * and notice that, as is non-negative, even, continuous with support [-1∕2, 1∕2], then * is non-negative, even, continuous with support [-1, 1] and its Fourier transform is

̂ * = ̂ 2 . Next ∈ 1 (R) with ′ = -1 [-1∕2,1∕2] sin and ̂ ′ ( ) = 4 cos 1 -4 2 thus ̂ ′ * ′ ( ) = -(2 ) 2 ̂ 2 ( )
We now consider = 2 2 * + ′ * ′ so that is continuous, real valued, even and supported in

[-1, 1]. ̂ ( ) = 2 2 -4 2 ̂ 2 ( )
is even (so is the Fourier transform of ̂ ) and in 1 . Further ̂ is non-negative on [-∕2, ∕2] and negative on R ⧵ [-∕2, ∕2].

∑ =1 | | 2 ≥ 2 32 ∑ =1 | | 2 , establishing (2.2) with = 2 32
.

Finally, we notice that, with a change of variable, and a simple limiting argument to remove the condition on the support of ( ), we have just proved the following; Theorem 2.3 (Ingham). Let > 0 and > 1 . There exist two constants = ( ), = ( ) such that -for every sequence of real numbers Λ = { } ∈Z such that +1 -≥ ; -for every sequence ( ) ∈Z ∈ 2 (Z, C),

In other words, for (2.2) to hold for = 1 for this sequence ( ) and every sequence ( ) then (1) = 0 so that the condition > 1 is necessary for this inequality to be interesting.

Proof. Let > 0 and define, for

| | < 1, ( ) = (1 + ) -= exp -arctan ℑ( ) 1+ℜ( ) |1 + | , | | < 1.
Of course, we may also write as a power series

( ) = +∞ ∑ =0 (-) !
where ( ) 0 = 1, (-) = -(--1) ⋯ (--+ 1). Next define

( , ) = 2ℜ ( +1) ( 2 ) = ( +1) ( 2 ) + -( +1) ( -2 ) = +∞ ∑ =0 (-) ! 2 + +1 2 + +∞ ∑ =0 (-) ! -2 + +1 2 .
Now set Λ = { } ∈Z with = + + 1 2 when ≥ 0 and = + 1 -+ 1 2 for ≤ -1, then +1 -≥ 1 (and even = 1 excepted for 0 --1 = 1 + ). In particular, if we set

, ( ) = ∑ =0 (-) ! 2 + +1 2 + ∑ =0 (-) ! | | | | 2 = 2 ∫ 1∕2 -1∕2 | ( 2 )| 2 d .
It follows that, if we had

(2.6) ∫ 1∕2 -1∕2 | , ( )| 2 d ≥ ∑ ∈Z | , ( )| 2
then, letting → +∞, we would also have

(2.7) ∫ 1∕2 -1∕2 | ( , )| 2 d ≥ 2 ∫ 1∕2 -1∕2 | ( 2 )| 2 d
for every 0 < < 1 and every 0 < < 1.

But, if we fix ∈] -1∕2, 1∕2[ then, when → 1,

( ±2 ) = 1 1 + ±2 → 1 1 + ±2 = ∓ 2 cos
(this is where we use that ≤ 1) while

| ( ±2 )| 2 = 1 (1 -) 2 + 4 cos 2 ) ≤ 1 4 cos 2 for 1 2 < < 1.
Similar bounds follow for ( , ):

( , ) = ( +1) ( 2 ) + -( +1) ( -2 ) → + - 2 cos = 1 2 -1 cos -1 while | ( , )| 2 ≤ 1 cos 2 .
When 2 < 1 the majorants are integrable so that we can let → 1 in (2.7). This leads to

(2.8) 2 2-2 ∫ 1∕2 -1∕2 d cos 2 -2 ≥ 2 1-2 ∫ 1∕2 -1∕2 d cos 2 .
Letting → 1 2 the right hand side goes to +∞ while the left hand side stays bounded. We obtain the desired contradiction. Assume that there exists 0 < ≤ 1 ≤ such that, for every ( ) ∈Λ finitely supported sequence of complex numbers,

∑ ∈Λ | | 2 ≤ 1 ∫ ∕2 -∕2 | | | | | ∑ ∈Λ 2 | | | | | d ≤ ∑ ∈Λ | | 2 .
Let ∈ R ⧵ Λ and, for sake of simplicity, assume that

(Λ ∪ { }) ≥ 1. Let 0 < < min , 1 4
, then there is a with 0 < < such that, for every

( ) ∈Λ∪{ } ∑ ∈Λ | | 2 ≤ 1 + ∫ + 2 -+ 2 | | | | | | ∑ ∈Λ∪{ } 2 | | | | | | d Moreover, can be taken of the form = 4
for some absolute constant > 0 ( = 8 -4 would do).

Note that, as is arbitrarily small, as > (Λ) -1 is arbitrarily near to (Λ ⧵ { 0 }) -1 so is + .

Proof. Set ( ) = ∑ ∈Λ∪{ } 2 . For 0 < < 1 4
, we define

( ) = ( ) - 1 ∫ ∕2 -∕2 -2 ( + ) d = ∑ ∈Λ 1 - sin ( -) ( -) 1 (Λ) , (2.9) 1 ∫ ∕2 -∕2 | ( )| 2 d ≥ ∑ ∈Λ | ( )| 2 ≥ ∑ ∈Λ | | 2 ≥ 4 100 ∑ ∈Λ | | 2
On the other hand, from Cauchy-Schwarz,

| ( )| 2 ≤ 2| ( )| 2 + 2 | | | | | 1 ∫ ∕2 -∕2 -2 0 ( + ) d | | | | | 2 ≤ 2| ( )| 2 + 2 1 ∫ ∕2 -∕2 | ( + )| 2 d = 2| ( )| 2 + 1 ∫ + - | ( )| 2 d . It follows that 1 ∫ ∕2 -∕2 | ( )| 2 d ≤ 2 ∫ ∕2 -∕2 | ( )| 2 d + 2 1 ∫ ∕2 -∕2 ∫ + - | ( )| 2 , d d = 2 ∫ ∕2 -∕2 | ( )| 2 d + 2 ∫ ∕2+ -∕2- | ( )| 2 1 ∫ min( ∕2, + ∕2) max(-∕2, -∕2) d d ≤ 2 ∫ ∕2 -∕2 | ( )| 2 d + 2 ∫ ∕2+ ∕2 -∕2-∕2 | ( )| 2 d ≤ 4 1 ∫ + 2 -+ 2 | ( )| 2 d ≤ 8 1 + ∫ + 2 -+ 2 | ( )| 2 d
since we assumed that + ≤ 2 . With (2.9), we have thus shown that (2.10)

∑ ∈Λ | | 2 ≤ 800 4 1 + ∫ + 2 -+ 2 | ( )| 2 d . It remains to add | | 2 on the left hand side. As = - ∑ ∈Λ 2 , | | 2 = 1 ∫ 2 -2 | | 2 d ≤ 2 1 ∫ 2 -2
where we used again that + ≤ 2 and the hypothesis of the lemma. But then, from (2.10), we get

| | 2 ≤ 4 + 1600 4 1 + ∫ + 2 -+ 2 | ( )| 2 d .
Adding this to (2.10), we finally get that

∑ ∈Λ∪{ } | | 2 ≤ 4 + 800 2 + 1 4 1 + ∫ + 2 -+ 2 | ( )| 2 d ≤ 1 4 1 + ∫ + 2 -+ 2 | ( )| 2 d with -1 = 800 × 3 + 4 since ( 0 ) ≤ ( 0 ), ≤ 1 and ( 0 ) ≥ 1.
We thus obtain the inequality with = 4 as claimed.

We can now use this lemma to improve Ingham's Inequality.

Corollary 2.6. Let Λ be a sequence such that

(Λ) = min ≠ ′ ∈Λ | -′ | ≥ 1 and let = ( 1 , … , ) ⊂ Λ. Let > 1 (Λ ⧵ )
. Then there exists two constants 0 < < 1 < such that, for every finitely supported sequence ( ) ∈Λ ,

(2.11) ∑ d ≤ ∑ ∈Λ | | 2 with = 6( + 1) = 6 1 + 1 .
In particular, the upper bound in (2.11) is already established. 4 . We can therefore apply Haraux's lemma again, till we reach Λ 0 . At each step, the constant -is replace by --1 = 4 - -while the integral ranges over -+( +1) 2 , +( +1) 2 . This finally leads to

Now let 1 (Λ ⧵ ) < < and = -. One might take = 1 2 + 1 (Λ ⧵ ) so that = 1 2 - 1 (Λ ⧵ ) From Hauraux's lemma, we get 1 + ∫ with -1 =
1 + ∫ + 2 d ≥ 0 ∑ ∈Λ | | 2 with 0 = 4 1 2 ⋯ .
As + = , this establishes (2.11).

Note that, taking = 8 -4 we obtain

= 2 2 6 3 4 2 13
-1

∏ =0 1 + 1 .
3.

( ) = ∑ =- 2 = sin(2 + 1) sin .
The following is a classical estimate of the 1 -norm of this kernel, which is called the Lebesgue constant.

Lemma 3.1. When → +∞, ‖ ‖ ‖ ‖1 ∶= ∫ 1∕2 -1∕2 | ( )| d = 4 2 ln + (1).
Proof. The proof is based on the following inequality which is easy to establish and whose proof is left to

the reader: for - 2 ≤ ≤ 2 , | | 1 - 2 3! ≤ | sin | ≤ | |. In particular, for | | ≤ 1 2 , 1 | | ≤ 1 | sin | ≤ 1 | | 1 1 -2 2 ∕6 ≤ 1 | | + | | 3 since 1 1-≤ 1 + 2 for | | ≤ 1 2 . It follows that ∫ 1∕2 -1∕2 | sin(2 + 1) | | | d ≤ ∫ 1∕2 -1∕2 | sin(2 + 1) | sin d ≤ ∫ 1∕2 -1∕2 | sin(2 + 1) | | | d + ∫ 1∕2 -1∕2 | sin(2 + 1) | | | 3 d . As ∫ 1∕2 -1∕2 | sin(2 + 1) | | | 3 d ≤ ∫ 1∕2 -1∕2 | | 3 d = 12 we get ‖ ‖ ‖ ‖1 = ∫ 1∕2 -1∕2 | sin(2 + 1) | | | d + (1).
Using parity and the change of variable = (2 + 1) we obtain

∫ 1∕2 -1∕2 | sin(2 + 1) | | | d = 2 ∫ 1∕2 0 | sin(2 + 1) | | | d = 2 ∫ + ∕2 0 | sin | d . Note that ∫ + ∕2 | sin | d ≤ ∫ + ∕2 d = ln + ∕2 = (1∕ ) while ∫ 0 | sin | d ≤ ∫ 0 1 d = .
It remains to estimate

2 ∫ | sin | d = 2 -1 ∑ =1 ∫ ( +1) | sin | d = 2 -1 ∑ =1 ∫ 0 | sin | + d = 2 -1 ∑ =1 ∫ 0 sin + d as sin ≥ 0 on [0, ]. But then 2 ( + 1) = ∫ 0 sin + d ≤ ∫ 0 sin + d ≤ ∫ 0 sin d = 2 .
Writing

-1 ∑ =1 1 + 1 = ∑ =1 1 -1 = -1 ∑ =1 1 -1 + 1 , it follows that 2 ∫ | sin | d = 4 2 ∑ =1 1 + (1).
Recalling that ∑ =1 1 = ln + (1) the result follows.

Lacunary trigonometric polynomials.

In this section, we consider a sequence of integers ( ) such that +1 ≥ > 1. Such sequences are called -lacunary in the sense of Hadamard (or simply lacunary).

First note that a -lacunary sequence is a finite union of ′ -lacunary sequences with ′ ≥ 3. Indeed, if ≥ 3 there is nothing to prove and for 1 < < 3, take an integer such that ≥ 3 and write ( ) = + then ( ) is -Lacunary and { } = ⋃ -1 =0 { ( ) }. Next, for ≥ 3, -lacunary sequences have a particular arithmetic property: Lemma 3.2. Let ≥ 3 and ( ) ≥0 a sequence such that 0 ≥ 1 and +1 ≥ . Consider two finite sequences , ∈ {-1, 0, 1} for = 0, … , and assume that

(3.1) ∑ =0 = ∑ =0 then = for every .
In other words, an integer can be represented in at most one way as

∑ ± . Such a sequence is called quasi-independent. Proof. First observe that ≤ 1 - = for = 0, … , .
Assume that (3.1) holds and define = -so that

∑ =0 = 0.
Assume towards a contradiction that there is an such that ≠ 0. Without loss of generality, we may assume that the largest such is and, up to exchanging and , that ≥ 1.

Observe that ∈ {-2, -1, 0, 1, 2} so that we obtain the desired contradiction writing

0 = ∑ =0 = + -1 ∑ =0 ≥ -2 -1 ∑ =0 ≥ -2 -1 ∑ =0 = 1 - 2 -1 -1 = +1 -3 + 2 ( -1) > 0 since ≥ 3 and > 0.
Note that this result is valid when the 's are real, not only for integers. The aim of this section is to prove that trigonometric polynomials with lacunary frequencies have large 1 -norms of which the following estimate is a particular case: there exists a constant > 0 such that, for every ,

∫ 1∕2 -1∕2 | | | | | | ∑ =1 2 | | | | | | d ≥ √ .
This follows from a more general theorem which estimates -norms of lacunary Fourier series. The aim of this section is to present this result. To do so, we follow closely [25, Chapter V.8] which goes through Rademacher series. First let us introduce those series.

To start, let us recall that  = {[ 2 --1 , ( + 1)2 --1 [, = 0, … , 2 +1 -1} the dyadic intervals of generation and  = ⋃ ≥0  the set of all dyadic intervals. Also, if , ∈  then either ∩ = ∅ or ⊂ or ⊂ . The Rademacher functions of generation are then functions that take alternative values +1 and -1 on successive intervals in  , that is

( ) = 2 +1 -1 ∑ =0 (-1) 1 ] 2 --1 ,( +1)2 --1 [ = sign sin(2 2 ) .
The first observation is that, if ∈  and > then takes the value +1 on half of and -1 on the other half so that ∫ = 0. A first consequence is that is orthogonal to in 2 ([0, 1]) since is constant on each ∈  so that ∫ = 0 and  is a covering of [0, 1]. Moreover, as | | = 1, the family ( ) ≥0 is an orthonormal sequence in 2 ([0, 1]).

In particular, we now fix a sequence ( ) ≥0 such that

+∞ ∑ =0 | | 2 converges, we can define = +∞ ∑ =0
and this series converges in 2 ([0, 1]) thus ∈ 2 ([0, 1]). We actually have a bit better:

Theorem 3.3. If +∞ ∑ =0 | | 2 < +∞ then there exists ∈ 2 ([0, 1]) defined by = +∞ ∑ =0
and this series converges both in 2 ([0, 1]) and almost everywhere.

Proof. The 2 convergence has already been established. Further, let = ∫ be the indefinite integral of and let ⊂ [0, 1] be the set of Lebesgue points of so that | | = 1 and on , ′ exists and is finite. Now let, [ ] be the -th partial sum of this series

 [ ]( ) = ∑ =0 ( ). As  [ ] → in 2 ([0, 1]), for every 0 ≤ < ≤ 1, | | | | | ∫ ( ) - [ ]( ) d | | | | | ≤ ∫ 1 0 | ( ) - [ ]( )| d ≤ ∫ 1 0 | ( ) -[ ]( )| 2 d 1∕2 → 0.
We have just shown that, if is an interval, then

∫  [ ] → ∫ thus also, if we fix ≥ 1, ∫ ( [ ] -  -1 [ ]) → ∫ ( - -1 [ ]). On the other hand, if ∈  -1 and ≥ , then ∫ = 0 so that ∫  [ ] = ∫  -1 [ ]. Letting → +∞ we obtain that ∫ ( ) d = ∫  -1 [ ]( ) d for every ∈  -1 .
Next, let 0 ∈ not a dyadic rational ( 0 ≠ 2 , , ∈ N) and let =] 2 -, ( + 1)2 -[ be such that

0 ∈ ∩ . Then, as  -1 [ ] is constant over  -1 [ ]( 0 ) = 1 | | ∫  [ ]( ) d = 1 | | ∫ ( ) d → ′ ( 0 )
when → +∞.

The second result is that is actually in every space:

Theorem 3.4. Let ( ) ∈ 2 and = +∞ ∑ =0
. Then, for 1 ≤ < +∞, ∈ ([0, 1]). Moreover, there exists , , depending on only, such that

+∞ ∑ =0 | | 2 1 2 ≤ ∫ 1 0 | ( )| d 1 ≤ +∞ ∑ =0 | | 2 1 2
.

Proof. Let us first notice that the theorem holds for = 2 since

∶= ∫ 1 0 | ( )| 2 d 1 2 = +∞ ∑ =0 | | 2 1 2
i.e. the inequalities are equalities with 2 = 2 = 1.

Next, let us notice that this implies the lower bound when > 2 with = 1 since then, with Hölder

∫ 1 0 | ( )| d 1 ≥ ∫ 1 0 | ( )| 2 d 1 2
= and also implies the upper bound with = 1 for < 2 since now Hölder implies that

∫ 1 0 | ( )| d 1 ≤ ∫ 1 0 | ( )| 2 d 1 2 =
Further, as Hölder also implies that if 2( -1) < ≤ 2 for some integer ≥ 2, then

∫ 1 0 | ( )| d 1 ≤ ∫ 1 0 | ( )| 2 d 1 2
so that the upper bound

(3.2) ∫ 1 0 | ( )| 2 d 1 2 ≤ 2
would also implies the upper bound for 2( -1) < ≤ 2 with ≤ 2 .

Next, let us show that the upper bound for = 4 implies the lower bound for < 2. Assume for the moment that we are able to prove that

∫ 1 0 | ( )| 4 d 1 4 ≤ 4 .
Let 1 ≤ < 2 and write 2 = + 4(1 -), that is, take = 2 4 -. Then, from Hölder

2 = ∫ 1 0 | ( )| 2 d = ∫ 1 0 | ( )| | ( )| 4(1-) d ≤ ∫ 1 0 | ( )| d ∫ 1 0 | ( )| 4 d 1- ≤ ( 4 ) 4(1-) ∫ 1 0 | ( )| d = ( 4 ) 2- ∫ 1 0 | ( )| d thus ∫ 1 0 | ( )| d 1 ≥ 1-4-

4

.

So it remains to prove (3.2). Notice also that it is enough to prove this inequality with real 's. The constant in the complex case is then multiplied by 2: write

= + where = ∑ ℜ( ) and = ∑ ℑ( ) . Then ‖ ‖ 2 ≤ ‖ ‖ 2 + ‖ ‖ 2 ≤ R 2 ⎡ ⎢ ⎢ ⎣ +∞ ∑ =0 |ℜ( )| 2 1 2 + +∞ ∑ =0 |ℑ( )| 2 1 2 ⎤ ⎥ ⎥ ⎦ ≤ 2 R 2 since |ℜ( )|, |ℑ( )| ≤ | |.
To conclude, we write

∫ 1 0 [ ]( ) 2 d = ∑ 0 +⋯+ =2 0 ,…, 0 0 ⋯ ∫ 1 0 0 0 ( ) ⋯ ( ) d
where ≥ 0 for every and

0 ,…, = ( 0 + ⋯ + )! 0 ! ⋯ ! . Now observe that ∫ 1 0 0 0 ( ) ⋯ ( ) d
1 all the 's are even 0 otherwise and that

∑ =0 2 = ∑ 0 +⋯+ = 0 ,…, ( 2 0 ) 0 ⋯ ( 2 ) .
Further, when 0 + ⋯ + = ,

2 0 ,…,2 0 ,…, = ( + 1)( + 2) ⋯ 2 ∏ =0 ( + 1)( + 2) ⋯ 2 ≤ ( + 1)( + 2) ⋯ 2 2 ≤
(with the convention that on the denominator is ( + 1)( + 2) ⋯ 2 = 1 when = 0). It follows that

∫ 1 0 [ ]( ) 2 d ≤ ∑ =0 | | 2 .
As [ ] → a.e., we conclude that

∫ 1 0 | ( )| 2 d 1 2 ≤ 1∕2 +∞ ∑ =0 | | 2 1 2 that is 2 = 2 1∕2 .
The estimate 2 = 2 1∕2 allows to improve a bit the result: Proof. Let us fix > 0. We first show that if ∶= ‖ ‖ 2 is small enough, then exp(

| | 2 ) ∈ 1 ([0, 1]). Indeed (3.3) ∫ 1 0 exp( | ( )| 2 ) d = +∞ ∑ =0 ! ∫ 1 0 | ( )| 2 d ≤ +∞ ∑ =0 ! (4 2 ) . But ! ≤ +∞ ∑ =0 ! = so that ∫ 1 0 exp( | ( )| 2 ) d ≤ +∞ ∑ =0 (4 2 ) = 1 1 -4 2 < +∞ provided 2 < 1 4
.

Next, take any ∈ 1 (0, 1), and apply the first part to

- [ ] = +∞ ∑ = +1 . As 2 ∶= +∞ ∑ = +1 | | 2 → 0, for large enough 2 < 1 8 thus exp(2 | - [ ]| 2 ) ∈ 1 ([0, 1]). Finally, as | | 2 ≤ 2| - [ ]| + 2| [ ]| 2 , we have exp( | | 2 ) ≤ exp(2 | - [ ]| 2 ) exp(2 | [ ]| 2 ) ∈ 1 since | [ ]| ∈ ∞ thus also exp(2 | [ ]| 2 ) ∈ ∞ .
Next, we consider series of the form

+∞ ∑ =0 2 ( ).
The idea is that such series are of the form ∑ ± 2 , that is, choosing ∈ (0, 1) at random, we randomly change the sign of . Our first result is the following: Theorem 3.6. Let ( ) ∈ 2 and ( ) = +∞ ∑ =0 ( ) 2 . Then, for almost every ∈ (0, 1), the series converges almost everywhere in ∈ (0, 1) and ∈ ([0, 1]) for every 1 ≤ < +∞.

Proof. Let be the set of ( , ) ∈ [0, 1] 2 where the series defining converges.

According to Theorem 3.3, for every

∈ [0, 1], the set 2 = {( , ) ∈ } has measure | | = 1. It follows that | | = 1 but then, for almost every ∈ [0, 1], 1 = {( , ) ∈ } has also measure | 1 | = 1. Next, set = ‖ ‖ 2 and fix ≥ 1. As in (3.3), (3.4) ! ∫ 1 0 | ( )| 2 d = +∞ ∑ =0 ! ∫ 1 0 | ( )| 2 d = ∫ 1 0 exp( | ( )| 2 ) d ≤ 1 1 -4 2 provided < 1 4 2
. It follows that

∫ 1 0 ∫ 1 0 | ( )| 2 d d = ∫ 1 0 ∫ 1 0 | ( )| 2 d d ≤ ! (1 -4 2 ) < +∞.
But then, for every , there is a set

⊂ [0, 1] with | | = 0 such that, if ∈ [0, 1] ⧵ , ∫ 1 0 | ( )| 2 d < +∞. Setting = ⋃ , | | = 0
and, for every ∈ [0, 1] ⧵ , for every , ∈ 2 . Using the inclusion of 2 ([0, 1]) ⊂ ([0, 1]) when ≤ 2 , we obtain that, for almost every , ∈ ([0, 1]) for every ≥ 1, as claimed.

We can now prove the main result of this section: Theorem 3.7. Let > 1 and ( ) ≥0 be a -lacunary sequence of integers, 0 ≥ 1 and +1 ≥

. Let 1 ≤ < +∞. There are two constants , , , such that, if

( ) ≥0 ∈ 2 , then ( ) = ∑ ≥0 2 is in ([0, 1]) with (3.5) , +∞ ∑ =0 | | 2 1 2 ≤ ∫ 1 0 | | | | | | ∑ ≥0 2 | | | | | | d 1 ≤ , +∞ ∑ =0 | | 2 1 2
.

Remark 3.8. Note that a simple change of variable also shows that, for every integer ,

(3.6) , +∞ ∑ =0 | | 2 1 2 ≤ 1 ∫ ∕2 -∕2 | | | | | | ∑ ≥0 2 | | | | | | d 1 ≤ , +∞ ∑ =0 | | 2 1 2
.

Also, we may assume that → , , , are continuous.

Proof. The beginning of the proof is the same as for Theorem 3.4. Parseval's identify shows that (3.5) is satisfied when = 2 with 2, = 2, = 1. The lower bound is then automatically satisfied for ≥ 2 with , = 1 while the upper bound is satisfied for ≤ 2 with 2, = 1. Finally, if we establish the upper bound for > 2, using Hölder's inequality in the same way as in the proof of Theorem 3.4, the lower bound follows for < 2 with 2, =

. Also, it is enough to prove the upper bound when = 2 , ≥ 2 and then, if 2( -1) < ≤ 2 , , ≤ 2 , . Another reduction is that, by homogeneity, it is enough to prove the theorem when

+∞ ∑ =0 | | 2 = 1.
A further restriction is that it is enough to prove the theorem for ≥ 3. Indeed, for 1 < < 3, we introduce an integer such that ≥ 3 and write ( ) = + for = 0, … , -1. Then ( ) +1 ≥ ( ) . If the theorem is established when ≥ 3 then, for each , the upper bound in (3.5) reads

∫ 1 0 | | | | | | ∑ ≥0 + 2 ( ) | | | | | | d 1 ≤ , +∞ ∑ =0 | + | 2 1 2
.

But then, with the triangular inequality in ,

∫ 1 0 | | | | | | ∑ ≥0 2 | | | | | | d 1 = ⎛ ⎜ ⎜ ⎝ ∫ 1 0 | | | | | | -1 ∑ =0 ∑ ≥0 + 2 ( ) | | | | | | d ⎞ ⎟ ⎟ ⎠ 1 ≤ -1 ∑ =0 ∫ 1 0 | | | | | | ∑ ≥0 + 2 ( ) | | | | | | d 1 ≤ , -1 ∑ =0 +∞ ∑ =0 | + | 2 1 2 ≤ 1∕2 , ⎛ ⎜ ⎜ ⎝ -1 ∑ =0 +∞ ∑ =0 | + | 2 ⎞ ⎟ ⎟ ⎠ 1 2 = 1∕2 , ∑ ≥0 | | 2 1 2
, where we have used Cauchy-Schwarz in R in the next to last line.

A last reduction comes from the observation that, for every

∫ 1 0 exp( | ( )| 2 ) d = +∞ ∑ =0 ! ∫ 1 0 | ( )| 2 d ≥ ! ∫ 1 0 | ( )| 2 d .
It is therefore enough to prove that there is a ( ) and a > 0 such that, if < ( )

(3.7) ∫ 1 0 exp( | ( )| 2 ) d ≤
which would then imply that

∫ 1 0 | ( )| 2 d ≤ ! as desired.
In order to prove (3.7), let us introduce

( ) = ∑ ≥0 ( ) 2 .
Integrating (3.4) with respect to and using Fubini, we deduce that

∫ 1 0 ∫ 1 0 exp( | ( )| 2 ) d d ≤ ∶= 1 1 -4 2 .
But then, there is an 0 (that we can assume not to be a dyadic rational 0 ≠ 2 ∕ ) such that

∫ 1 0 exp( | 0 ( )| 2 ) d ≤ .
Next, we consider the Riesz product

( ) = ∏ =0 1 + ( 0 ) cos 2 = ∏ =0 1 + ( 0 ) 2 + -2 2 = ∑ ∈Z 2
where the Fourier coefficients have the following property:

-0 = 1 -= 0 if is an integer that is not of the form ∑ ± , in particular when | | > ∑ =0 -if = ∑ with ∈ {-1, 0, 1}. As > 3, this 's are unique. Then = ∏ ≠0 ( 0 ) 2 . In particular, = ( 0 )
2 for = 0, … , and = 0 for > . As a consequence, the partial sums of the Fourier series of are given by

[ ]( ) ∶= ∑ =0 2 = ∑ =0 ( 0 ) 2 2 = 2 ∫ 1 0 0 ( ) ( -) d .
Note that ≥ 0 and ∫ 1 0 ( ) d = 0 = 1 so that = ( ) d is a probability measure. As ( ) = exp( 2) is increasing and convex, we apply Jensen's inequality (with the measure ) to obtain

1 2 [ ]( ) ≤ ∫ 1 0 | 0 ( )| ( -) d ≤ ∫ 1 0 | 0 ( )| | ( -) d .
Integrating over [0, 1] and using Fubini, we get

∫ 1 0 1 2 [ ]( ) d ≤ ∫ 1 0 | 0 ( )| ∫ 1 0 ( -) d d = ∫ 1 0 | 0 ( )| d ≤ .
Letting → +∞, we obtain such that, if ( ) ∈N is an increasing sequence of integers and ( ) is a complex sequence with finite support, then

+∞ ∑ =0 | | + 1 ⩽ ∫ 1 0 | | | | | | +∞ ∑ =0 2 | | | | | | d .
The proof given below will give = 1 96 which is not the best possible. We avoid minor technicalities which would lead to = 1 30 (see e.g. [START_REF] Devore | LORENTZ Constructive Approximation[END_REF][START_REF] Trigub | BELINSKY Fourier Analysis and Approximation of Functions[END_REF]). The proof in [START_REF] Ph | SABA The Littlewood problem and non-harmonic Fourier series[END_REF] requires the introduction of various parameters and a cumbersome optimization of those parameters in the final step and leads to = 1 26 .

Note also that the sum starts at 0, we have so far been unable to obtain a lower bound when the sum runs over Z.

Strategy of the proof.

Notation 3.10. For a function ∈ 2 ([0, 1]) and ∈ Z, we write

̂ ( ) = ( ) = ∫ 1 0 ( ) -2 d
for the Fourier coefficients of . Its Fourier series is then

( ) = ∑ ∈Z ( ) 2
and Parseval's relation reads

∫ 1 0 | ( )| 2 d = ∑ ∈Z | ( )| 2
and can also be written in the form

∫ 1 0 ( ) ( ) d = ∑ ∈Z ( ) -( ).
We fix a trigonometric polynomial

(3.8) ( ) = ∑ =0 2 and = ∑ =0 | | + 1 .
We then write | | = with complex numbers of modulus 1 and we introduce

(3.9) 0 ( ) = ∑ =0 + 1 -2
.

Then by Parseval identity, written in the second form we have

(3.10) ∶= ∑ =0 | | + 1 = ∑ =0 ̂ ( ) ̂ 0 (-) = ∫ 1 0 ( ) 0 ( ) d . so that ⩽ ‖ ‖ 1 ‖ 0 ‖ ∞ ([0,1]) .
The issue is that, typically we have no control over the ∞ norm of 0 other than the trivial and explosive control by ∑ 1 +1 , so we will correct 0 into another test function 1 as follows;

(1) The ∞ norm of the corrected function 1 is controlled by a constant , ‖ 1 ‖ ∞ ≤ .

(2) ̂ 1 (-) only differs a little from ̂ 0 (-) for 0 ⩽ ⩽ , say

| ̂ 1 (-) -̂ 0 (-)| ≤ ̂ 0 (-) 2 = 1 2( + 1)
, while we impose no condition on the behavior of ̂ 1 ( ) for ≠ -.

1 0 | ( )| d ≤ ∫ 1 0 | ( )| d
which is the expected result with = 1 2 . The proof of the Theorem B thus amounts to proving the following lemma:

Lemma 3.11. There exists a universal constant and a

1 ∈ ∞ such that (1) ‖ 1 ‖ ∞ ⩽ (2) | ̂ 1 (-) -̂ 0 (-)| ⩽ 1 2 | ̂ 0 (-)| = 1 2( + 1)
for 0 ⩽ ⩽ where 0 is the function given by (3.9).

The remaining of this section is devoted to the proof of this lemma. Lemma 3.11. First note that, up to eventually adding extra zeros to the sequence ( ) and adding + = + to the sequence ( ), we may assume that = 2 +1 -2.

Proof of

We start by decomposing 0 into a sum of dyadic blocs on which the amplitude | ̂ 0 (-)| = 1 + 1 is more or less constant. More precisely, for = 0, … we set = [2 , 2 +1 [ and

= ∑ +1∈ + 1 -2
.

The function 0 now appears as the partial sum of order of the series ∑ , in other words

0 = ∑ =0 .
Let us start with a simple lemma: Lemma 3.12. With the above notations, we have

(1) ‖ ‖ ∞ ([0,1]) ⩽ 1. (2) ‖ ‖ 2 ([0,1]) ⩽ 2 -∕2 . Proof. By construction, | | = 2 hence ‖ ‖ ∞ ⩽ ∑ +1∈ 1 + 1 ⩽ | | 2 = 1. As | +1 ( )| ≤ 1 and | ( )| ≤ , we get | +1 ( )| ≤ . It remains to prove that ⩽ 2 .
To do so, it suffices to see that

-⩽ 1 - -1 for 0 ⩽ ⩽ 1,
yielding the result immediately.

Lemma 3.15. For 0 ≤ ≤ and = 0, … , , let , = -, with

, = ℎ +1 + … + ℎ when < 0 when = . Then = ∑ =0 , .
Proof. By induction on , when = 0, 0,0 = 0 thus 0,0 = 1 and, indeed, we have

0 = 0 = 0 0,0 .
Assume now that the formula has been established at rank -1 and let us show that = ∑ =0 , . By construction, we have

= -1 -ℎ + = -1 ∑ =0 , -1 -ℎ + .
with the induction hypothesis. It remains to notice that , = -, = 1 and that, for = 0, … , -1,

, = , -1 + ℎ thus , = , -1 -ℎ so that, indeed, we have = ∑ =0 , as claimed.
Recall that

0 = ∑ =0
and we set

1 = = ∑ =0 ,
where the dependence on comes from the definition of the , 's, in particular

‖ 1 ‖ ∞ ⩽ .
The first part of Lemma 3.11 is thus established and it remains to prove the second part. To do so, we start by some intermediary results; Lemma 3.16. If ∈ ∞ (Hardy space) and ( ) ⩾ 0, then -∈ ∞ and

‖ --1‖ 2 ⩽ ‖ ‖ 2 .
Proof. Since ∞ is a Banach algebra, the partial sums

∑ =0 (-1) ! of -are elements of ∞ . Moreover, since is bounded, these sommes converge uniformly toward -, with -∈ ∞ . Finally, if ∈ C and ℜ( ) ⩾ 0, | --1| =
Next let us introduce the following notations:

(1) Let ∈ 1 ([0, 1]), the spectrum of , denoted by spec( ) is the set of non zero Fourier coefficients, more precisely spec( ) = { ∈ Z ∶ ̂ ( ) ≠ 0}. It is easy to show that, if , ∈ 2 ([0, 1]), then spec( ) ⊂ spec( ) + spec( ) = { + ∶ ∈ spec( ), ∈ spec( )}.

(2) Let A be a subset of [1, [, we denote by Λ the set of , ∈ i.e

Λ = { , ∈ }. Lemma 3.17. Let + 1 ∈ = [2 , 2 +1 [ then ̂ , (-) = 0 if < .
Proof. We must show that -∉ spec( , ). By contradiction, we suppose that -∈ ( ). But since spec( ) ⊂ -Λ and, from Lemma 3.16, spec( , ) ⊂ N, thus spec( , ) ⊂ -Λ + N. However, since < , is to the left of , then -Λ is completely to the left of -Λ + N (since

+ 1 ∈ , then -∈ -Λ by definition of Λ ). -Λ - -Λ Proof of (2) in Lemma 3.11. Let ∈ [0, [ and ⩽ such that + 1 ∈ . Hence ̂ 1 (-) = ∑ =0 ̂ , (-) = ∑ = ̂ , (-)
with Lemma 3.17.

On the other hand, ̂ 0 (-) = + 1 while ̂ (-) = + 1 if + 1 ∈ i.e. if = and ̂ (-) = 0

otherwise. So we can write

̂ 0 (-) = ∑ = ̂ (-), hence | | | ̂ 1 (-) -̂ 0 (-) | | | = ∑ = [ ( , -1)](-) = ∑ = -[ ( , -1)]. But | -[ ( , -1)]| ≤ ‖ ( , -1)‖ 1 ([0,1)] ≤ ‖ ‖ 2 ([0,1)] ‖ , -1‖ 2 ([0,1)] = ‖ ‖ 2 ([0,1)] ‖ -, -1‖ 2 ([0,1)]
by definition of , . Using Lemma 3.16 and then Lemma 3.15, we obtain

| -[ ( , -1)]| ⩽ ‖ ‖ 2 ‖ , ‖ 2 ⩽ ‖ ‖ 2 ∑ = +1 ‖ℎ ‖ 2 ⩽ ‖ ‖ 2 √ 2 ∑ = +1
‖ ‖ 2 with Lemma 3.13. Then, from Lemma 3.12, we obtain

| -[ ( , -1)]| ≤ 2 -∕2 √ 2 +∞ ∑ = +1 2 -∕2 = 2 -∕2 √ 2 2 -( +1)∕2 1 -2 -1∕2 .
and, for sufficiently large we have | (4 ; )| = 1 ⩽ 2 . Therefore, there exists a minimal 0 such that | (4 0 ; 0 )| ⩽ 2 0 . Let = 4 0 , and = 0 then using (3.12) and the definition of 0

| | ⩽ | ( ; )| = | (4 0 ; 0 )| ⩽ 2 0 = 1 2 .
In particular

(3.13) | | 1 3 ⩽ 1 2
.

By minimality of 0 -1

1 2 2 = 2 0 -1 ⩽ | | | | 4 0 -1 ; 0 -1 | | | | ≤ 3 ∑ =0 | | | | 4 0 ; 0 -1 + 4 0 -1 | | | | ≤ 4| (4 0 ; 0 )| = 4| ( ; )|
by definition of 0 . We thus get

| ( ; )| ⩾ 1 2 8 ⩾ | | 1 3
8 , with (3.13).

Lemma 3.20. Let > 0 and let and be positive integers with (2 + ) < . Suppose is a finite set of integers, and let

( ) = ∑ ∈ ( ) 2
where

( ) = ∑ | |⩽ , 2
Let and with > 4 and suppose ( ; ) = { 1 , … , } then we have

‖ ‖ 1 ([0,1] ⩾ 1 32 (2 + ln(1 + 2 )) ∑ =1 ‖ ‖ 1 ([0,1]) 2 - 2
We split the remaining of this section into two parts. In the first one, we show that Lemmas 3.19 and 3.20 imply Theorem 3.11. In the second one, we prove the Lemma 3.20. Theorem 3.18. Let > 0 , be two integers satisfying the conditions of the theorem and let be strongly ( , , )-regular. Thus, there are two integers , with > (2 + ) , such that we can write

Proof of

= ⋃ ∈ ( + ), with | | ⩾ and ⊂ {-, … , } with | | ⩾ . We can then write ( ) ∶= ∑ ∈ 2 = ∑ ∈ ( ) 2 with ( ) = ∑ ∈ 2 = ∑ =- , 2
with , = 1 if ∈ and , = 0 otherwise.

Assume first that there exists 1 ∈ such that

(3.14) ‖ 1 ‖ 1 ([0,1]) ⩾ 2 9
ln( ) ln( ).

We then choose ⩾ 16 7

in such a way that there is an such that ( , ) = { 1 }. Hence by Lemma 3.20,

‖ ‖ 1 ([0,1]) ≥ ‖ 1 ‖ 1 ([0,1]) 32 2 + ln 1 + 2 2 - 2 ≥ ‖ 1 ‖ 1 ([0,1]) 2 6 2 + ln 1 + 2 1 - 7 8 .
As > 2 , using (3.14), we conclude that, in this case

‖ ‖ 1 ([0,1]) ≥ 2 9 2 ln ln 2 + ln 1 + 2
which establishes the theorem.

We will thus assume that, for each ∈ ,

(3.15) ‖ ‖ 1 ([0,1]) ⩽ 2 9
ln( ) ln( ).

Note that, from Theorem B,

‖ ‖ 1 ([0,1]) ⩾ ln( )
so that 2 9 ⩽ ln( ), in particular, ≥ 8. We then take and given by Lemma 3.19 applied to the set so that = | ( ; )| satisfies (3.16)

1 3 8 ⩽ ⩽ 1 2 .
We write ( ; ) = { 1 < … < } From Lemma 3.20, we get that 13 .

‖ ‖ 1 ([0,1]) ⩾ 1 2 5 2 + ln 1 + 2 ∑ =1 ‖ ‖ 1 2 - 2 = 1 2 5 2 + ln 1 + 2 ) ( 1 -2 ) with 1 = 2 ∑ =1 ‖ ‖ 1 and 2 = 2 ∑ =1 ‖ ‖ 1 . Next, as ‖ ‖ 1 ⩾ ln( ), 1 ≥ 2 ln( ) 2 ∑ =1 1 ≥ 2 ln( ) 2
We have established that The first one is a simple lemma about numerical integration of trigonometric polynomials:

‖ ‖ 1 ([0,1]) ⩾ 1
d ≤ -1 ∑ =0 ∫ +1 ∫ | ′ ( )| d d = -1 ∑ =0 ∫ +1 | ′ ( )| ∫ +1 d d ≤ 1 -1 ∑ =0 ∫ +1 | ′ ( )| d = ∫ 1 0 | ′ ( )| d
where we have used Fubini and the bound

∫ +1 d ≤ 1 when ≤ ≤ + 1 . We conclude with Bernstein's inequality, ‖ ′ ‖ 1 ([0,1]) ⩽ 2 ‖ ‖ 1 ([0, 1
f ( ) = 1 -| | +1 1 -,…, ( ) so that ( ) =  f ( ) = ∑ | |⩽ 1 - | | + 1 2 = 1 + 1
(sin( ( + 1) )) 2 (sin( )) 2 .

Lemma 3.22. Let , , , be integers with 2 ⩽ < . Then there exists a function , with the following properties:

(1)

, ( ) = 1 for | | ⩽ , (2) , ( ) = 0 for | | ⩾ + 2 (3) when + ⩾ 2 + 4 , 1 + -1 ∑ =- | | | | |  [ , ] + | | | | | ⩽ 16 (2 + ln(1 + ∕ )) Proof. Define , ( ) = 1 d + * f -1 ( ) = 1 ∑ ∈Z 1 {--,…, + } ( -)1 {-+1,…, +1} ( ) = 1 ∑ | |⩽ -1 | -|⩽ + 1 - | | . First, for | | ⩽ , if | | ⩽ -1, then | -| ⩽ | | + | | ≤ + -1, so that , ( ) = 1 ∑ | |⩽ -1 1 - | | = 1, since ∑ | |⩽ -1 1 - | | = 0 ∑ =1- 1 + + -1 ∑ =1 1 - = + 1 2 + -1 2 = .
On the other hand, if

| | ⩾ + 2 and | | ≤ -1 then | -| ≥ | | -| | ≥ + + 1 so that the sum defining
, is empty and , = 0. To prove the last item, the Convolution Theorem shows that

 [ , ]( ) = 1 + ( ) -1 ( ).
As + and -1 are both even, so is , thus

∫ 1 0 | [ , ]( )| d = 2 ∫ 1 2 0 | [ , ]( )| d = 2( 1 + 2 + 3 )
where

1 = 1 ∫ 1 + 0 | + ( ) -1 ( )| d 2 = 1 ∫ 1 1 + | + ( ) -1 ( )| d , 3 = 1 ∫ 1 2 1 | + ( ) -1 ( )| d .
∑ ∈ℤ ( ) ( ) 2 = -1 ∑ =- ( ) ∑ ∈ℤ ( + )+ 2 ( ( + )+ ) = ⟨ , ⟩ = -1 ∑ =- ⟨ , ⟩⟨ , ⟩ = 1 + -1 ∑ =- -1 ∑ =- ( ) -2 + -1 ∑ =- ∑ ∈ℤ ( + )+ 2 ( ( + )+ ) 2 + = 1 + -1 ∑ =-  [ 0 ] + -1 ∑ =- ∑ ∈ℤ ( + )+ 2 [( ( + )+ ) + + ] ).
Noticing that 2 + = 2 [( ( + )+ ) + ] , we may write

∑ ∈ℤ ( ) ( ) 2 = 1 + -1 ∑ =-  [ (0) ] + -1 ∑ =- ∑ ∈ℤ ( + )+ 2 ( ( + )+ )( + + ) = 1 + -1 ∑ =-  [ (0) ] + It follows that ∑ ∈ℤ , ( ) 2 = ∑ ∈ℤ ∑ -⩽ ⩽ + 2 [( + ) ] = ∑ =0 mod ∈ ∑ -⩽ ⩽ + 2 2 = ∑ ∈ ( ;0) 2 ∑ -⩽ ⩽ + 2 = ∑ ∈ ( ;0) ( ) 2 . = ∑ =0 2 ∑ =0 -2 ∶= ∑ =- 2 . when | | ≥ 1.
The first identity allows to conclude.

We can now state the first crucial result in this proof.

Lemma 4.4.

There exist 1 > 0, 1 > 0 such that, if 0 < < 1 , 0 < 1 2 < 1 -√ . Moreover, let

= 1 -1 2 + 1 -1 2
and let be such that = 2 ≥ -7∕2 . Then, for 0 ⩽ ⩽ ,

∑ 0⩽ ⩽ ≠ |̂ ( -)| + ⩽ 1 - + .
Proof. The Taylor expansion when → 0 of 1 -√ is of the form

1 - √ = 2 + ( 4 ),
with > 0. Thus, if 1 < , for small enough, 0 < 1 2 < 1 -√ . Next, notice that 0 < 1 -+ 1 -< 1 if 0 < < 1 -√ which shows that 0 < < 1. For future use, note that there is a > 0 such that (4.8) = 2 + ( 4 ).

We will further assume that is small enough for -7∕2 ≥ max since we chose = 8 and = 10. Note that, for every > 0, the power 7∕2 could be reduced to 3 + by taking large enough, but could not be reduced below 3 with this construction.

We can now turn to the estimate itself. Set = 1 2 and split the sum in the left hand side of the main inequality into two sums The result is obtained if we prove the two estimates , we obtain the expected bound 2 ⩽ ∕(1 -) + .

The following lemma is a first step towards proving Theorem 4.2 and is a consequence of Lemma 4.4.

Proof. First, as | | is real valued, its zero Fourier coefficient 0, is also real, and for the remaining ones , = -, for every ≥ 1. A direct computation then shows that Re(ℎ ) = | | which is less than 1 by lemma 4.7 while Parseval shows that ‖ℎ ‖ 2 ⩽ √ 2‖ ‖ 2 .

We now define a sequence ( ) ⩾ inductively through = and +1 = -ℎ +1 + +1 where 0 < < 1 is a small parameter that we will adjust later. Further set

∶= sup 0< ≤1 1 --= 1 sup 0< ≤ 1 --= 1 1 --.
From the next to last identity, it is easy to obtain the following simple bound:

1 ≤ ≤ 2 .
Lemma 4.9. For all ⩽ ⩽ ,

‖ ‖ ∞ ⩽ 2 .
Proof. By definition of , if ≤ and 0 ≤ ≤ 1, then -+ ≤ -+ ≤ . We can now prove by induction over that | | ≤ from which the lemma follows. First, when = 0, from Lemma 4.7 we get Proof. The first part is obtained by a simple induction on . Moreover, the lemma 4.8 implies

‖ 0 ‖ ∞ = ‖ 0 ‖ ∞ ≤ 1 ≤ . Assume now that ‖ ‖ ∞ ≤ , then
‖ , ‖ 2 = ‖ ‖ ‖ ‖ ‖ ‖ ∑ = +1 ℎ ‖ ‖ ‖ ‖ ‖ ‖2 ⩽ ∑ = +1 ‖ℎ ‖ 2 ⩽ √ 2 ∑ = +1 ‖ ‖ 2 ⩽ √ 2(| | + 1) ∞ ∑ = +1 2 -2 = √ 2(| | + 1) √ 2 -1 2 -2 ,
as stated.

Next, we will need the following simple well-known lemma:

Lemma 4.11. If ∈ ∞ and ( ) ⩾ 0, then -∈ ∞ and

‖ --1‖ 2 ⩽ ‖ ‖ 2 .

2. 3 .

 3 Sequences with large gaps. Lemma 2.5 (Haraux). Let Λ ⊂ R be a sequence such that (Λ) = inf ≠ ∈Z | -| > 1 and let > (Λ) -1 .

Corollary 3 . 5 .=0.

 35 Let ( ) ∈ 2 and = +∞ ∑ Then, for every > 0, exp( | | 2 ) ∈ 1 ([0, 1]).

3 . 3 .Theorem 3 . 9 (

 3339 as claimed (up to ∕2 replacing . The proof of Littlewood's conjecture by Mc Gehee, Pigno and Smith. We will now give the proof of the Littlewood conjecture, following closely[START_REF]QUEFFÉLEC Twelve landmarks of twentieth-century analysis[END_REF]. Mc Gehee, Pigno, Smith). There exists a constant ⩽ 4 2

2 .

 2 ]) . For a finitely supported sequence ( ) ∈Z we define its discrete Fourier transform (or -Fourier trans-If , are two finitely supported sequences, their convolution is the sequence * defined by * ( )) * ( ) = ∑ ∈Z ( -) ( ). The Convolution Theorem is also valid here:  [ * ]( ) =  [ ]( ) [ ]( ). Two classical examples are -the Dirichlet kernel: set d = 1 -,…, so that ( ) =  [d ]

Proof.

  Write elements of C + as ( -, … , -1 ) and the scalar product ⟨ , ( + )+ ) for = -, … , -1 and = ( ) ∈ C + . Then, by periodicity of( )

2 .

 2 , if is an index corresponding to 1 , then | -| ⩾ | -| = ( We shall now bound 2 . In this sum, + ⩾ (1 -)( + ) and | -| ⩾ 1

|

  +1 ( )| = | ( ) -ℎ +1 ( ) + +1 ( )| ≤ | ( )| -ℜ ℎ +1 ( ) + | +1 ( )| = | ( )| -| +1 ( )| + | +1 ( )|.As | +1 ( )| ≤ 1 and | ( )| ≤ , we get | +1 ( )| ≤ as claimed. Lemma 4.10. Let ⩽ ⩽ . For = , … , we define , = -, with , = ℎ +1 + … + ℎ if < 0 if = .

  Proof ofLemma 3.20. The rest of this chapter consist in proving Lemma 3.20. the proof is divided into several lemmas.

				2 5 2 + ln 1 + 2	2	ln( ) ln( ) 2 3	-	1 2 13
			=	2 (2 9 ) 2 2 + ln 1 + 2	2 10 ln( ) ln( ) -	1 2
					2
			⩾	(2 9 ) 2 2 + ln(1 + 2 )	ln( ) ln( )
	since ln( ) ln( ) 2	⩾	1 2 10 -1	.
	3.4.2.

If is a function, we write * 1 = * for the convolution of with itself and then define inductively * +1 = ( * ) * .

we then conclude that

On the other hand, Parseval implies that

as claimed.

Now, write the Fourier series of each | | ∈ 2 ([0, 1])

To each | |, we associate ℎ ∈ 2 ([-, ]) defined via its Fourier series as

Lemma 3.13. For 0 ≤ ≤ , the following properties hold [START_REF] Belov | On the conjecture of Littlewood and minima of even trigonometric polynomials[END_REF]

Proof. We now define a sequence ( ) =0,…, inductively through 0 = 0 and +1 = -ℎ +1 + +1

where 0 < ≤ 1 is a real number to be adjusted later on. Further set

Proof. By definition of , if ≤ and 0 ≤ ≤ 1, then -+ ≤ -+ ≤ . We can now prove by induction over that | | ≤ . First, when = 0, from Lemma 3.12 we get

We conclude that

On the other hand, as

Choosing = 1 24 gives the result.

Note that, when = 1 24 , ≤ 48 so that we can take = 48 in Lemma 3.11, leading to = 1 96 in Theorem 3.9.

3.4. Strongly multidimensional sets. Let > 0 and ( , ) ∈ ℕ 2 . A subset of ℤ is ( ; , )-strongly 2-dimensional if there exists numbers and with > (2 + ) such that

for some set containing at least integers and subsets ⊆ {-, … , } verifying | | ⩾ .

Theorem 3.18 (Hanson [10]). Let > 0 and , be two positive integers satisfying (3.11) ⩾ 3 2 21 3 ln( ) 3 and ⩾ 3 2 21 3 ln( ) 3 ln( ) 3 , where is the constant in theorem B. Suppose is ( ; , ) strongly 2-dimensional subset of ℤ. Then

ln( ) ln( )

Combining this result with Theorem 3.3 in [START_REF] Shao | On character sums and exponential sums over generalized arithmetic progressions[END_REF], we see that this theorem is also best possible up to the constant.

Given a set ⊆ ℤ, a positive integer , and an arbitrary integer , we define

The proof of Theorem 3.18 is a direct consequence of two lemmas. The first one is the following: Lemma 3.19. Let be a set of integers with | | ⩾ 8. Then there are positive integers q and s such that

Proof. For each ⩾ 1, we choose any such that | (4 ; )| is maximal. But, on one hand,

and on the other hand, for fixed, the sets (4 ; ) are disjoints, so at least one of them has cardinality larger than 4 - | |. In particular, (3.12)

For = 1, we thus have

On the other hand, if = mod then = mod so that, for any (4 ; ) ⊂ (4 ; ) for < ,

We have

It follows that

since , are positive integers.

Using the explicit expressions of + and -1 , we have

using that sin ≤ for ≥ 0 and that sin ≥ 2 for 0 ≤ ≤ 1 2 . Finally, for 3 , we do the same computation to bound

Grouping all terms and slightly upper bounding the numerical constants, we obtain

By Lemma 3.21 (and the 1-periodicity of  [ , ]), we obtain

But = + 2 -1 and + ⩾ 2 + 4 so

Hence we get

⩽ 16 2 + ln 1 + , concluding the proof.

Lemma 3.23. Let , be positive integers and = (-), … , ( -1) ∈ C + . We extend into -an + -periodic sequence ( ) ( + ) + = ( ), for = -, … , -1 and ∈ Z; -a finitely supported sequence (0) by setting (0) ( ) = ( ) for = -, … , -1 and (0) ( ) = 0 for ≥ and for ≤ --1.

Then

From this, we deduce that

with the change of variable = + + and periodicity of ⟶ ∑ 2 .

Lemma 3.24. Let , and be positive integers with (2 + 2 ) + 4 ⩽ for some > 0 and ⩾ 4 . Suppose is a finite set of integers and, for each ∈ let be a trigonometric polynomial of degree at most . Then for any integer , we have:

Proof. Assume we can prove the lemma for = 0, that is, for any sequence ( ) ∈Z of trigonometric polynomials of degree at most and any finite set ,

Then, replacing withand replacing ( ) with ( + ) we get

So there is no loss of generality is assuming = 0. Next, we write = ∑

⌉ the smallest integer larger than 2 and let , the sequence from Lemma 3.22. Since ( ) , to be the -periodic sequence defined by ( ) , ( + ) = , ( ) for = --2 , … , + 2 and ( ) , ( ) = 0 for all other 's. We take , > + 2 such that + = and then ( ) , ( ) = 0 for = -, … , --2 -1 and for = + 2 + 1, … , -1. From Lemma 3.23, we get

Combining (3.17 , ( ) = .

Finally we get

We can now prove the lemma.

Proof of lemma 3.20. Write ( ; ) = { 1 , … , } and write each in the form = + . Applying lemma 3.24 yields

and

It remains to show that

Let us start with 1 : using the 1-periodicity in ,

with Theorem B. Applying Lemma 3.21 to and using that 2 < 1 2 with our hypothesis on and , we get

and the desired estimate of 1 follows immediately.

Let us now estimate 2 . For ∈ [ , + 1], we have

From the 1-periodicity in , the integral of this quantity over [ , + 1] is the same as the integral over

with Bernstein's inequality.

Newman's extremal trigonometric polynomial.

Finally, let us insist on the fact that all results in this section are worse case scenarios i.e. the lowest possible 1 norm of a trigonometric polynomial with eventually a constraint on the spectrum or on the coefficients.

If one looks at the maximal possible value of the 1 -norm under the constraint that all coefficients have modulus 1, then, from Cauchy-Schwarz

for all 's. However, if one wants to reach the bound √ + 1, one would need that equality holds in the Cauchy-Schwarz Inequality so

should be constant. But, it is easy to show that the only trigonometric polynomials with constant modulus are monomials (thus in our case, equality can only occur when = 0). Indeed, consider the polynomial

which is now an equality between meromorphic functions over C ⧵ {0}. As this is true on {| | = 1} it is true over all of C ⧵ {0}, in particular, has only 0 as zero thus ( ) = . So the natural question is, how near to √ + 1 one can get. A result by E. Beller and D. J. Newman [START_REF] Beller | An 1 extremal problem for polynomials[END_REF] shows that one can obtain √ - [START_REF] Belov | On the conjecture of Littlewood and minima of even trigonometric polynomials[END_REF]. For sake of simplicity, we will only present a previous result by D. J. Newman [START_REF] Newman | An 1 Extremal Problem for Polynomials[END_REF] which contains similar ingredients: Theorem 3.25 (Newman,[START_REF] Newman | An 1 Extremal Problem for Polynomials[END_REF]). There is an absolute constant > 0 and an integer 0 such that, for every ≥ 0 there is a sequence 0 , … , with | | = 1 for all and such that

Proof. The example is related to Gauss sums. We fix and set = exp + 1 , = 2 and define

The theorem is based on the following lemma: We postpone the proof of the lemma and show how we can conclude with it. Parseval's identity gives ‖ ‖ 2 = √ + 1 and then Hölder's inequality implies

.

From the lemma, we deduce that there is a constant > 0 such that ∫

for large enough.

It remains to prove the lemma:

Proof of Lemma 3.26. We write

On the other hand, from Parseval,

To compute the last sum, we split it into 4 parts

As | +1-| = | | we have 4 ≤ 1 and 3 ≤ 2 so that we only need to show that 1 and 3 are both ( 3∕2 ).

For 1 , we use the estimate

Further, as for

as claimed.

4. Hudson and Leckband that shows that the estimate for the Besicovitch norm follows from Theorem B. The converse is of course true as well and the result was obtained directly in [START_REF] Ph | SABA The Littlewood problem and non-harmonic Fourier series[END_REF] and even allowed to slightly improve the constant. Theorem 4.1 (Hudson & Leckband [11]). For ( ) ≥0 be real numbers with +1 -> 0 and let ( ) ≥0 be a sequence of complex numbers with finite support. Then

where is the same constant as in Theorem B.

Moreover, assume that ( ) is -lacunary i.e. 0 ≥ 1 and +1 ≥ . Then, for 1 ≤ < +∞,

with , , , the constants in Theorem 3.7.

Actually, only the first part is in [START_REF] Hudson | LECKBAND Hardy's inequality and fractal measures[END_REF]. The second part is here obtained with almost the same proof and seems to be new.

Proof. Let 0 , … , be complex numbers, 0 < 1 < ⋯ < be real numbers and

Let > 0. By a lemma of Dirichlet ([25, p 235], [START_REF]GRAFAKOS Classical Fourier analysis[END_REF]), there is an increasing sequence of integers ( ) ≥1 and, for each ≥ 1 a finite family of integers ( , ) =0,…, such that

Define the -periodic function

and note that, for ∈ [-∕2, ∕2],

But then

We can now conclude as follows. First, in the general case, As Ψ is -periodic, we may apply Theorem B to obtain

Letting → +∞ and then → 0 we obtain Theorem 4.1:

Let us now assume further that +1 ≥ . Let ̃ be such that 1 < ̃ < and, assume that has been chosen such that -(1 + ) > ̃ . Observe that

, and = we obtain

.

From (4.1) we conclude that

The result follows by letting → 0 and then ̃ → .

Nazarov's theorem. Theorem 4.2 (Quantitative version of Nazarov theorem for small ).

There exists positive constants * , * such that, for 0 < < * the following holds: for every 0 < ... < of real numbers satisfying +1 -⩾ 1 and 0 , ..., complex numbers,

First, we start by some preliminary notations and results. Let = ∶= [-1+ 2 , 1+ 2 ]. Let us fix ( ) =0,…, ⊂ R with +1 -≥ 1 for every , ( ) =0,…, a sequence of complex numbers and write

, so that we must find such that:

We define

where is a large integer that we will adjust through the proof. This integer will be of the form = 2 . We will prove that (4.6)

‖ ‖ 1 ( ) ≥ and, as + ≤ ( + 1) , ≥ so that we obtain the desired inequality (4.5) with a constant = .

The meaning of is the following. Consider: -a new sequence of frequencies ( ̃ ) ∈Z such that ̃ = -for = , … , + . and then ̃ = 0 + -for < and ̃ = + -for > . In particular, we still have | ̃ -̃ | ≥ 1. In other words, the sequence ( ) =0,…, is completed into a sequence ( ) ∈Z that is still 1-separated and then shifting it by . -A new sequence of complex numbers ( ̃ ) ∈Z with ̃ = -for = , … , + and ̃ = 0 for other 's. In other words, the sequence ( ) =0,…, is completed into a sequence ( ) ∈Z by 0-padding it and then shifting it by . Then (4.6) reads

with the convention that 0∕0 = 0. Note that, up to adding 0 terms at the end of the sequence , we may assume that + is of the form 2 -1 for some integer . This will allow us to write

Next, as in Ingham's proof, we will introduce an auxiliary function. Again, we consider

whose Fourier transform is given by ĥ( ) = 2 cos( ) 1 -4 2 .

We will need to smooth a bit this function to obtain a better decay of the Fourier transform and thereby slightly enlarge its support. More precisely, let = 10, = 8 and let

and define its + -fold self convolution 1 = * + .

Clearly is non-negative, even and with support [-2 , 2 ]. Finally, we define as

In the following lemma, we list the properties needed on . They are all established via easy calculus and straight forward Fourier analysis.

Lemma 4.3.

There is a 0 > 0 and a 0 > 0 such that, if 0 < < 0 then, [START_REF] Belov | On the conjecture of Littlewood and minima of even trigonometric polynomials[END_REF]

From now on, we will assume that 0 < < 7 3 < 1 so that ≥ 1. In particular, when

Proof of Lemma 4.3. The first one is simple Fourier analysis.

For the second one, we write

For the third one, we notice that 4 2 -1 ≥ 3 2 when ≥ 1 thus

. Thus this fact is established with 0 = 1

For the last one, we take 0 small enough to have sinc

Proof. By definition of ,

By applying the lemma 4.4, that

Using the fact that = | | and the triangular inequality, we obtain

from which the lemma follows immediately.

Construction of ̃

Recall that, from our assumption on and , we can write

where we set  = { ∈ N ∶ 2 ≤ < 2 +1 } and

.

Before we estimate the norms of the 's, let us recall Hilbert's inequality (see e.g. [4, Chapter 10]).

Lemma 4.6 (Hilbert's inequality). Let 1 , … , be real numbers with | -| ≥ 1 when ≠ , and let 1 , … , be complex numbers. We have

We can now prove the following:

The first estimate is obtained by straightforward computations: for all ,

For the second one, set = + . Then

-

-

We then apply Hilbert's inequality 4.6 to the last two sums to obtain

We then write the Fourier series expansion of | | ∈ 2 ( ) as

and define ℎ ∈ 2 ( ) via its Fourier series expansion

Lemma 4.8. For ⩽ ⩽ , the following properties hold [START_REF] Belov | On the conjecture of Littlewood and minima of even trigonometric polynomials[END_REF]

Proof. Since ∞ is a Banach algebra, the partial sums ∑ =0 (-1) ! of -are elements of ∞ . Moreover, since is bounded, these sums converge uniformly toward -, with -∈ ∞ . Finally, if ∈ C and ℜ( ) ⩾ 0,

In our case = ( ), and we have | and by integration we have the desired inequality. Lemma 4.12. Assume that 0 < ⩽

. Then, for

(2) The Fourier series of , ( ) -1 writes , ( )

Proof. By lemma 4.11, ‖ -1‖ 2 ⩽ ‖ ‖ 2 . Then, since , is analytic, its Fourier series writes

But then, with Parseval

which implies the claimed bound.

Now recall that

and define ̃ = .

In particular, from Lemma 4.9, we have

The key estimates here is the following; Lemma 4.13. Once again, we use the notations of the lemma 4.4. There exists 2 > 0 such that, if 0 < < 2 and ≥ -7∕2 then (4.12)

where is the function defined in (4.7).

Proof. It is enough to prove that, under the conditions of the lemma, for 0 ⩽ ⩽ we have

Once (4.13) is established, it will then be enough to multiply the left hand side by and to use the triangular inequality

We fix ∈ [0, ] and let the index such that + ∈  . We define , 1 and 2 as follows

).

From there, we obtain

).

So finally we get

).

Let ( ) = , if + ∈ . Since ̂ is an even function, we can write

.

From Lemma 4.3, recall that, with = 0

+2 ),

we have

Further, from (4.8) and

12 . Thus, choosing sufficiently large for = 2 -7∕2 and ≤ 2 for some 2 > 0 small enough, we are able to apply (4.14) and obtain, with (4.15) + ) .

We can now bound . Since

since last sum has at most 2 -1 ⩽ + . We will now bound 2 .

According to the lemmas 4.7 (1) and 4.12 (1) we get

and + ∈ then

Finally, we deduce that

and we obtain 2 ⩽ 3

.

Note that, from (4.8) and | | ≤ 2, we can take = 4 2 for some 4 > 0. It is now easy to deduce the theorem 4.2 using the 2 inequalities (4.10) and (4.12).

Proof of theorem 4.2. Let

as previously defined. Recall that in (4.10) , we have shown that

with (4.12). It follows that

But, if is small enough and ≥ -7∕2 , -from Lemma 4.3, ‖ ‖ ∞ ≤ 2 ;

-from (4.8), = 2 + ( 4)

.

Therefore, there are two absolute constants * and * such that, if ≤ * ,

As noticed at the start of the proof, this implies that

for every , every sequence of real numbers ( ) ≥0 with +1 -≥ 1 and every complex sequence ( ) ≥0 .

We have not fully optimised the proof, by taking sufficiently large and ∕ sufficiently large, one can replace 15∕2 by 7+ for any fixed .

SOME OPEN PROBLEMS

The 2 theory of exponential sums is rather well understood. This is not the case for 1 theory for which there are still many open questions. Let us mention a few of them.

(1) There is a major difference between the sums that appear in Ingham's Theorem and those that appear in Mc Geehe, Pigno, Smith and Nazarov's Theorems. In the 2 case, the sums can be two sided and not in the 1 case. The proof given here does not work in this case (this is due to the construction of ̃ ) and we are tempted to conjecture the following Question 1. Let > 1 and ( ) ∈Z a real sequence such that +1 -≥ 1 for every and has same sign as . Does there exist a constant such that, for every and every sequence ( ) =-,…, of complex numbers,

(2) The second question is the optimality of the condition = 1 in Nazarov's Theorem. In view of Ingham's counterexample, it is tempting to conjecture that > 1 is requested (as observed by Nazarov, his proof requires this). We think the estimate shown here is not optimal.

(3) Another question related to the previous one comes from Haraux's Theorem. When +1 -→ +∞, can be chosen arbitrarily small. A key element of Haraux's strategy is that the lower and upper bounds in Ingham's inequality are both multiples of the 2 -norm of the coefficients. This is no longer the case in the 1 setting.

In forthcomming work, we have been able to use a compactness argument to show that one may take arbitrarily small in Nazarov's Theorem when +1 -→ +∞. However, in doing so, we loose control of constants. This leads to the following: Question 3. Prove a quantitative version Nazarov's Theorem with arbitrarily small when +1 -→ +∞.

(4) Are 1 ([-, ])-norms of lacunary non-harmonic Fourier series comparable to the 2 -norms of the coefficients? This is the case for Besikovich norms. Note that

The question is then Question 4. Find a (gap) condition on (Λ ) and on that implies that there is a constant > 0 such that, for every ( ) with finite support

.

One may also go the other way and ask whether one can still obtain a result like Theorem B when +1 -→ 0 but with a smaller power on the denominator. For instance, when = ln then the Fourier series become Dirichlet series The following question was asked in [START_REF] Brevig | VUKOTIĆ The multiplicative Hilbert matrix[END_REF]:

Question 5. Is it true that, for every sequence ( ) ≥1 with finite support,