ANALYTIC ERDÖS-TURÁN CONJECTURES AND ERDÖS-FUCHS THEOREM

We consider and study formal power series, that we call supported series, with real coefficients which are either zero or bounded below by some positive constant. The sequences of such coefficients have a lot of similarity with sequences of natural numbers considered in additive number theory. It is this analogy that we pursue, thus establishing many properties and giving equivalent statements to the well-known Erdos-Turan conjectures in terms of supported series and extending to them a version of Erdos-Fuchs theorem.


Introduction
In a seminal paper of 1941, Erdös and Turán [4] made two conjectures in additive number theory, which have had an important impact on the field.They concern the number r(A,n) of representations of a natural number n as a sum of two elements of a subset A of the set N of natural numbers.One of them, the so-called Erdös-Turán conjecture, still a notorious open question, can be formulated as follows.
(ET) If A is a basis of N, if every natural number is the sum of two elements of A, then the number r(A,n) of such representations is unbounded for n ∈ N. The other one predicted that r(A,n) cannot be asymptotically too well approximated by its average value; more precisely, it is impossible to have n m=0 r(A,m) = cn + O(1) for any positive real number c. Fifteen years later, in another very influential paper, Erdös and Fuchs [3] proved even more than that, namely that n m=0 r(A,m) = cn + o(n 1/4 log −1/2 n) is impossible.This surprising result stirred a lot of interest since it was almost as good as a classical estimate of its kind for the number of lattice points in a circle, specific to the set A of the squares in N and obtained via difficult analytic techniques, while this one was valid for any subset A of N and with a simpler proof.Consequently, several authors presented various versions of the Erdös-Fuchs theorem [1,7,9,10,11].Moreover, the Erdös-Fuchs paper contained the statement of a more general conjecture than (ET), namely the following. (GET for some constant d > 0 and all n in N * = N \ {0}, then the number r(A,n) of representations of n as a sum of two elements of A is unbounded for n ∈ N.
Indeed, (GET) implies (ET) because of the well-known fact [7] that if A is a basis of N, then its elements a n , taken in strictly increasing order, verify the condition a n ≤ dn 2 for some constant d > 0 and all n ∈ N * (it also follows from [5,Lemma 3.15] that if A is a basis of N, then we may take d = 9/16).In addition, Erdös and Fuchs remarked in their paper that their "[3, Theorem 1] remains true for sequences of nonnegative numbers {a k }, not necessarily integers."Similarly, Halberstam and Roth [7, page 98] noted about their statement of the Erdös-Fuchs theorem: "Here we do not assume that the integers in A are distinct.In fact even the assumption that the elements of A are integers is superfluous (cf.Erdös-Fuchs, loc.cit., page 68)."In the light of such facts and remarks, it is natural to explore the extent to which some concepts and questions in additive number theory are independent of the specific nature of the subsets of N under consideration.The quest for a broader context, in which the above conjectures and results still make sense or remain valid, leads to the consideration and study of what we call "supported series."These are formal power series whose coefficients form sequences of nonnegative real numbers resembling the subsets of N in that their nonzero terms are bounded below by some positive constant.We thus establish various equivalent statements to the conjectures (ET) and (GET) in the realm of the supported series, and we extend to them a version of the Erdös-Fuchs theorem, due to Newman [12].More precisely, we prove that for any supported series ) for some real numbers c > 0 and t ≥ 0, then t ≥ 1/4.Our point is that most of the concepts, questions, and techniques pertaining to the additive representation of integers by subsets of N are not just about integers, but have a more general scope and can be naturally extended to the context of the supported series.The definitions, notions, and results presented in the sequel are all aimed at determining the essential features of the underlying ideas and problems, in a general setting, thus shedding more light and allowing for a more direct approach.

Definitions and simple properties
The sets of natural numbers, rational integers, rational numbers, real numbers, and complex numbers are, respectively, denoted by N, Z, Q, R, and C. If E is anyone of these sets, then ] be a power series with real coefficients.The mass function of f is the function F of a real variable x defined by F(x) = n≤x a n .In particular, with mass functions F and G, respectively. (1) where the summations can also be extended to all j,k ∈ N. (4) For any a ∈ R, the mass function of a f is aF.Also, a f = |a| • f and s(a f ) = a 2 s( f ).Moreover, if a = 0, then Supp(a f ) = Supp( f ).( 5) For any m ∈ N, the mass function of Proof.The proofs are mostly straightforward.Note that (1) amounts to a 0 = F(0) and As to (3), the mass function of f g is by definition , with a similar relation exchanging F and G; moreover, the summations can be extended to all j,k ∈ N, since F(x) = G(x) = 0 for x < 0.
Definition 2.3.The set of all subsets of N will be written as (N).Let P ∈ (N).The characteristic function of P is the function χ P defined on N by χ P (n) = 1 or 0 according as n ∈ P or n ∈ P. The companion series of P is f P = p∈P X p = ∞ n=0 χ P (n)X n .The counting function of P is defined for x ∈ R + by P(x) =| P ∩ [0,x]|, where |E| denotes the cardinality of the set E.
For P,Q ∈ (N) and n ∈ N, we set r(P,Q;n) = |{(p, q) ∈ P × Q : p + q = n}|; we further set s(P,Q) = sup{r(P,Q;n) : n ∈ N} in N. The sumset of P and Q is P + Q = {p + q : p ∈ P, q ∈ Q}.In particular, if P = Q, we write r(P,n) = r(P,P;n) and s(P) = s(P,P).We say that P is a basis of N if P + P = N.
For two subsets and p n ≤ q n for all positive integers n not exceeding |Q| (here, |P| and |Q| may be finite or infinite).
Lemma 2.4.Let P, Q be subsets of N.
(1) The mass function of the companion series f P coincides with the counting function P(x) = n≤x χ P (n) for x ∈ R + .Moreover, Supp( f P ) = P. ( Remark 2.5.To every subset P of N corresponds its companion series f P .In a certain sense, this embeds There are already two partial orders defined in (N), namely the set inclusion ⊂ and the relation .Both relations can be extended to with mass functions F and G, respectively, we say that f is contained by g and write f g if a n ≤ b n for all n ∈ N.
We also say that f is subordinated to g and write f f .The converse is false, as shown by the example where g) and 0 h, then f h gh (resp., f h gh) and, in particular, a f ag (resp., a f ag) for all a ∈ R + * .
Proof.Property (1) follows from the relation , and h = ∞ n=0 c n X n with respective mass functions F, G, and H, then, by Lemma 2.2, the mass functions of f h and Example 2.8.The set of squares in N * will be written as S = {n 2 : n ∈ N * }.We have S(x) = [ √ x ] for all x ∈ R + , where [y] is the integer part of the real number y.For an infinite subset P of N, we have P S, that is, ] and c ∈ R + * , we say that f is c-supported if, for any n ∈ N, the condition a n = 0 implies that a n ≥ c.We denote by Ᏺ c the set of all csupported power series in R[[X]].We say that the series f is supported if it is c-supported for some c ∈ R + * .We denote by Ᏺ the set of all supported series in R The set Ᏺ is closed under addition and multiplication.
(3) The series f lies in Ᏺ if and only if inf{a n : n ∈ P} > 0 or f = 0.In particular, a constant a lies in Ᏺ if and only if Remark 2.11.The following construction affords a better grasp of some features of the order relation .Let with mass functions F and G, respectively.For a real number p n ≤ x < p n+1 , we clearly have F(x) = F(p n ) = n j=1 c j and G(x) = n, while for x < p 1 , we have F(x) = G(x) = 0. Note also that Supp( f ) ⊂ P. Then the condition f P f is equivalent to the following one: ( * ) n j=1 c j ≤ n for all n ∈ N * .Different choices for the sequence (c n ) provide various examples and counterexamples.Here, we just give two illustrations.
(1) For any infinite subset P of N, there exists f ∈ Ᏺ such that Supp( f ) = P and its mass function F satisfies F(x) ≥ x 2 for all large enough real numbers x.Indeed, just take (2) For any infinite subset P of N, there exists f ∈ Ᏺ such that f = ∞ and f P f .Indeed, just take f (X) = ∞ n=1 c n X pn with c n = k if n = k 2 for some k ∈ N * , and c n = 0 otherwise.Clearly, f ∈ Ᏺ and f = ∞.Moreover, for any n ∈ N * , there is a unique m ∈ N * such that m 2 ≤ n < (m + 1) 2 , and we have

The extended class of Erdös-Turán sets
Definition 3.1.We say that an element f of Ᏺ belongs to the extended class Ꮿ(EET) of Erdös-Turán sets if any supported power series which is subordinated to f has infinite size; that is, for any g ∈ Ᏺ such that g f , we have s(g) = ∞.We say that an infinite subset P of N belongs to the class Ꮿ(ET) of Erdös-Turán sets if for any infinite subset Q of N such that Q P, we have s(Q) = ∞.Remark 3.2.Note first that if f ∈ Ꮿ(EET) and P = Supp( f ), then, since f f , we have The class Ꮿ(ET) was defined in [6].Now let P be an infinite subset of N. If f P ∈ Ꮿ(EET), then the relation Q P, which by Lemma 2.7 means that f Q f P , implies that Lemma 2.4).Thus, if f P ∈ Ꮿ(EET), then P ∈ Ꮿ(ET).However, as indicated below, it is not known if the converse holds.
] be such that 0 f and 0 g, with mass functions F and G, respectively.Let P = Supp( f ) and Q = Supp(g).Then the following hold.
(1) follows directly from the definitions. ( (3) This follows from (1), since f + g ∈ Ᏺ and f + g f .(4) By Lemma 2.2, the mass function of Then h lies in Ᏺ, and its mass function H is given, for where u is a polynomial with coefficients in R + .Moreover, by Lemma 2.2, s(X m h) = s(h) = ∞.Hence s(g + c 0 X m ) = s(X m h + u) = ∞ and therefore s(g) = ∞ by Proposition 3.3 (11).Thus X m f ∈ Ꮿ(EET).
(5) Let f = ∞ n=0 a n X n and g = ∞ n=0 b n X n in Ᏺ, and h = f g (also in Ᏺ by Lemma 2.10), with mass functions F, G, and H respectively.Since g = 0, there exists m ∈ N such that b m > 0. By Lemma 2.2, an inequality which also holds trivially for x < m.Thus, if u ∈ Ᏺ is such that u h, then its mass function is a polynomial with coefficients in R + , and therefore s(g) = ∞ by Proposition 3.3 (11).(7) Assume that f + g ∈ Ꮿ(EET).Then, for any h ∈ Ᏺ such that h f , we have h + g f + g, by Lemma 2.7, so that s(h + g) = ∞, and therefore, by Proposition 3.3 (11), since 0 g and g is a polynomial, s(h) = ∞.Thus f ∈ Ꮿ(EET).The converse follows from (3) above.(8) Assume that f g ∈ Ꮿ(EET).Then, for any h ∈ Ᏺ such that h f , we have hg f g, by Lemma 2.7, so that s(hg) = ∞, and therefore, by Proposition 3.3 (11), since 0 g, g = 0 and g is a polynomial, s(h) = ∞.Thus f ∈ Ꮿ(EET).The converse follows from (5) above.
with mass functions F and G, respectively, we say that g is asymptotically subordinated to f if G(x) ≥ F(x) for all large enough real numbers x.
(1) For any m ∈ N, the series f lies in Ꮿ(EET) if and only if f m lies in Ꮿ(EET). ( The series f lies in Ꮿ(EET) if and only if, for any g asymptotically subordinated to f in Ᏺ, we have s(g Proof.The first property follows from Proposition 3.6 (7) since f m differs from f by the polynomial consisting of the first m terms of f .The second property follows from Proposition 3.6(6).
be a subset of N, identified to the sequence (p n ) of its elements indexed in strictly increasing order.For every k ∈ N * , the kth ray of P is the set Remark 3.10.Simple properties for an infinite subset (1) For x ∈ R + and k,n ∈ N * , we have P k (x) = n if and only if p kn ≤ x < p k(n+1) .Thus if P k (x) = n, then kn ≤ P(x) < k(n + 1).( 2) For x ∈ R + , we have kP k (x) ≤ P(x) < kP k (x) + k.
(3) For x ∈ R + and k,n ∈ N * , we have P k (x) ≥ n if and only if p kn ≤ x.Lemma 3.11.
(3) If kQ(x) ≤ P(x) < kQ(x) + k for all x ∈ R + , then p kn ≤ q n < p kn+k for all n ∈ N * .(4) Let d ∈ R + * .The inequality p n ≤ dn 2 holds for large enough n in N * if and only if that is, q n ≤ p kn , for n ∈ N * .Moreover, if we have q n = p kn for some n, then taking x = p kn = q n , we get P(x) = kn = kQ(x).Thus, if the inequality in the assumption is strict, then so it is in the conclusion.Similarly, if we have q n < p kn for some n, then taking q n ≤ x < p kn , we get P(x) < kn ≤ kQ(x).Thus if there is equality in the assumption, then so it is in the conclusion.Hence the results.
(2) If x ≥ p k , there is a unique n ∈ N * such that p kn ≤ x < p k(n+1) , and we have P(x) < k(n + 1) ≤ k(Q(x) + 1), since q n ≤ p kn ≤ x, by the assumption.If 0 ≤ x < p k , then P(x) < k ≤ kQ(x) + k.Hence the result in all cases.
(4) The first condition means that P(dn 2 ) ≥ n for large enough n ∈ N * , while the second one means that P(dx 2 ) ≥ [x] for large enough x ∈ R + .The first one implies the second upon taking n = [x]; and the converse implication is trivial.
(5) The first condition means that the sequence (p n /n 2 ) is bounded, while the second one means that the function x/P(x) 2 is bounded for x ≥ p 1 .Now, for p n ≤ x < p n+1 , one has P(x) = n, so that p n /n 2 ≤ x/P(x) 2 < p n+1 /n 2 .Hence the equivalence of the two conditions.Theorem 3.12.Let f ∈ Ᏺ with Supp( f ) = P. Also, let A be a nonempty subset of N. For every k ∈ N * , let P k (resp.,A k ) denote the kth ray of P (resp., A). ( and let F and G be the mass functions of f and g, respectively, so that F(x) ≤ G(x) for x ∈ R + .There exists c ∈ R + * such that f ∈ Ᏺ c , and therefore cP(x) ≤ F(x) ≤ G(x) ≤ g Q(x) for x ∈ R + by Proposition 3.3.Now, if g = ∞, then s(g) = ∞ by Corollary 3.4.Otherwise, let k be a positive integer ≥ g /c.Then P(x) ≤ kQ(x) for x ∈ R + , and therefore q n ≤ p kn for n ∈ N * , that is, Q P k by Lemma 3.11.Since P k ∈ Ꮿ(ET), it follows that s(Q) = ∞ and therefore s(g) = ∞ by Corollary 3.4.Thus f ∈ Ꮿ(EET).
(2) Assume that f < ∞ and f ∈ Ꮿ(EET).Let k ∈ N * and let Q = {q 1 < q 2 < •••} be an infinite subset of N such that Q P k .Then P(x) < kQ(x) + k for x ∈ R + by Lemma 3.11.Thus, by Proposition 3.3, the mass function Then h ∈ Ᏺ and the mass function of h is given by H(x) = tQ(x) + t (by Lemmas 2.2 and 2.4), so it satisfies F(x) ≤ H(x) for x ∈ R + .Hence h f and since f ∈ Ꮿ(EET), therefore s(h ) by Lemmas 2.2 and 2.4.It follows that s(Q) = ∞.Thus P k ∈ Ꮿ(ET).This shows that if f ∈ Ꮿ(EET) then P k ∈ Ꮿ(ET) for all k.The converse follows from (1).
Remark 3.13.The results in Theorem 3.12 raise the question of determining the infinite subsets P of N all of whose rays P k lie in Ꮿ(ET).In particular, one may ask whether if P ∈ Ꮿ(ET), then P k ∈ Ꮿ(ET) for all k ∈ N * .A partial answer is provided in what follows.
The caliber of P is cal(P) = liminf n→∞ (p n /n 2 ) in R + .We say that P belongs to the restricted class Ꮿ(RET) of Erdös-Turán sets if cal(P) = 0.
Remark 3.15.In [6], we showed that Ꮿ(RET) is a subset of Ꮿ(ET) and that the conjecture (GET) is equivalent to the assertion that Ꮿ(RET) Ꮿ(ET).
(ET) If P is a basis of N, then s(P) = ∞.(GET) For any infinite subset 2 for some d ∈ R + * and all n ∈ N * , then s(P) = ∞.Moreover, since in the condition p n ≤ dn 2 we can assume that d ∈ N * (upon replacing d by any integer ≥ d), then, in view of Definition 3.1, (GET) can be restated as follows.
(GET) For any d ∈ N * , the set dS lies in Ꮿ(ET).
Definition 4.2.Let f = ∞ n=0 a n X n .We call f a supported basis of N if f is a supported series such that f 2 = ∞ n=0 r( f ,n)X n has all its coefficients r( f ,n) > 0; that is, if f ∈ Ᏺ and Supp( f ) is a basis of N by Proposition 3.3.
We consider the following analytic versions of the conjectures (ET) and (GET).
Proof.(1) Assume first that (AET) holds.Let P be a basis of N. Then f P = ∞ n=0 χ P (n)X n lies in Ᏺ 1 and Supp( f P ) = P, so that f P is a supported basis.Therefore, by the assumption, s( f P ) = ∞, that is, s(P) = ∞ by Lemma 2.4.Thus (ET) holds.
Conversely, assume that (ET) holds.Let f be a supported basis of N. Then P = Supp( f ) is a basis of N, so that, by the assumption, s(P) = ∞.Moreover, f ∈ Ᏺ c , for some c ∈ R + * .Therefore s( f ) ≥ c 2 s(P) by Proposition 3.3.Hence s( f ) = ∞.Thus (AET) holds.
(2) By Theorem 3.12, (GAET) holds if and only if S k , which is equal to k 2 S, lies in ∈ Ꮿ(ET) for all k ∈ N * .On the other hand, by Remark 4.1, (GET) holds if and only if dS lies in Ꮿ(ET) for all d ∈ N * .Thus (GET) trivially implies (GAET).But the latter also implies the former in view of the fact that if P ∈ Ꮿ(ET) and if Q is an infinite subset of N such that Q P, then Q ∈ Ꮿ(ET), as can be easily seen from Definition 3.1.Therefore for any d ∈ N * , taking k ∈ N * such that k 2 ≥ d, if k 2 S ∈ Ꮿ(ET), in virtue of (GAET), then dS ∈ Ꮿ(ET) since dS k 2 S. Thus (GAET) is equivalent to (GET).

Remark 4.4.
There is an asymptotic version (ETa) of (ET), which is equivalent to (ET) (cf.[7]).First, we note that a subset P of N is called an asymptotic basis of N if r(P,n) > 0 for all large enough n in N.Then, we state the following.
(ETa) If P is an asymptotic basis of N, then s(P) = ∞.We can similarly state an asymptotic form of (AET), namely, (AETa).
(AETa) If f is an asymptotic supported basis, that is, if f ∈ Ᏺ and r( f ,n) > 0 for all large enough n, then s( f ) = ∞.By the same argument as in the proof of Theorem 4.3(1), we see that (AETa) is equivalent to (ETa), and since (ETa) is equivalent to (ET), then all four statements (ET), (ETa), (AET), and (AETa) are equivalent.

A version of the Erdös-Fuchs theorem
In this section, we present a version of the Erdös-Fuchs theorem for supported series, along the lines in Newman [12], trying to be as explicit and complete as possible in the proofs.Namely, we establish the following result.
Theorem 5.1.Let f = ∞ n=0 a n X n be any supported series in R[[X]]; and let where r( f ,n) = i+ j=n a i a j (n ∈ N).For any c ∈ R + * , if n k=0 (r( f ,k) − c) = O(n t ) for some t ∈ R + , then t ≥ 1/4.Remark 5.2.The version of the Erdös-Fuchs theorem given by Newman [12] reads: if A is a subset of N, and if c ∈ R + * is such that n k=0 (r(A,k) − c) = O(n t ) for some t ∈ R + , then t ≥ 1/4.Theorem 5.1 generalizes this result by extending it to all sets (or sequences), A = {a n : n ∈ N}, of nonnegative real numbers whose nonzero elements are bounded below by a positive constant.This is done by introducing and studying the corresponding formal power series f = ∞ n=0 a n X n , having such sequences as coefficients, that we here call the supported series.The point of this generalization is that such properties, as the Erdös-Fuchs theorem, are not exclusively characteristic of sequences of natural numbers, but belong to a much broader class of sequences of real numbers.It is to be further noted that the version of Newman that we here extend is slightly weaker than the original one by Erdös and Fuchs [3], which asserts that the relation n k=0 (r(A,k) − c) = o(n 1/4 log −1/2 n) is impossible.However, the truly far-reaching generalization of the latter result is the one presented by Montgomery and Vaughan [11], credited by them to an unpublished manuscript of Jurkat and described as first appearing in the Ph.D. thesis of Hayashi [8], namely: for any subset A of N and any c ∈ R + * , the relation n k=0 (r(A,k) − c) = o(n 1/4 ) is impossible.Its extension to our context states that for any supported series f = ∞ n=0 a n X n in R[[X]], with f 2 = ∞ n=0 r( f ,n)X n , and any c ∈ R + * , the relation n k=0 (r( f ,k) − c) = o(n 1/4 ) is impossible.This is a natural and more difficult generalization, for another occasion.