On probability measures arising from lattice points on circles

A circle, centered at the origin and with radius chosen so that it has non-empty intersection with the integer lattice \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb Z}^{2}$$\end{document}Z2, gives rise to a probability measure on the unit circle in a natural way. Such measures, and their weak limits, are said to be attainable from lattice points on circles. We investigate the set of attainable measures and show that it contains all extreme points, in the sense of convex geometry, of the set of all probability measures that are invariant under some natural symmetries. Further, the set of attainable measures is closed under convolution, yet there exist symmetric probability measures that are not attainable. To show this, we study the geometry of projections onto a finite number of Fourier coefficients and find that the set of attainable measures has many singularities with a “fractal” structure. This complicated structure in some sense arises from prime powers—singularities do not occur for circles of radius \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\sqrt{n}$$\end{document}n if n is square free.


Introduction
Let S be the set of nonzero integers expressible as a sum of two integer squares. For n ∈ S, let n := { λ = a + bi ∈ Z[i] : a 2 + b 2 = n} denote the intersection of the lattice Z[i] ⊂ C with a circle centered at the origin and of radius √ n. For n ∈ S, let r 2 (n) := | n | denote the cardinality of n ; for n / ∈ S it is convenient to define r 2 (n) = 0. We define a probability measure μ n on the unit circle S 1 := {z ∈ C : |z| = 1} by letting where δ z denotes the Dirac delta function with support at z. The measures μ n are clearly invariant under multiplication by i and under complex conjugation. We say that a measure on S 1 is symmetric if it is invariant under these symmetries. Definition 1.1 A probability measure ν is said to be attainable from lattice points on circles, or simply just attainable, if ν is a weak limit point of the set {μ n } n∈S .
We note that any attainable measure is automatically symmetric. Now, if two integers m, n ∈ S are co-prime, where denotes convolution of measures on S 1 . Thus measures μ n for n a prime power are of particular interest. It turns out that the closure of the set of measures given by μ p e for p ranging over all primes p ≡ 1 mod 4 and exponents e ranging over integers e ≥ 1 contains μ 2 k , as well as μ q 2k for any prime q ≡ 3 mod 4, and any exponent k ≥ 0. (Note that q l ∈ S forces l to be even.) Motivated by the above, we say that a measure μ is prime power attainable of μ is a weak limit point of the set {μ p e } p≡1 mod 4, e≥1 . Similarly, we say that a measure μ is prime attainable if μ is a weak limit point of the set {μ p } p≡1 mod 4 .

Proposition 1.2 The set of attainable measures is closed under convolution. Further,
it is the closure (in the weak topology) of the collection of all convolutions of finitely many prime power attainable measures, i.e., it is topologically generated by the prime power attainable measures.
Hence the set of attainable measures is the smallest closed (in the weak topology) set containing all the prime power attainable measures and closed w.r.t. convolution of probability measures. The set of all symmetric probability measures is clearly a convex set, hence equals the convex hull of its extreme points. Quite interestingly, the set of prime attainable measures is exactly the set of extreme points. Now, since the set of attainable measures contains the extreme points, and is closed under convolution one might wonder if all symmetric probability measures are attainable? By studying Fourier coefficients of attainable measures we shall show that not all symmetric measures are attainable.
Given a measure μ on S 1 and k ∈ Z, define the k-th Fourier coefficient of μ bŷ If μ is symmetric it is straightforward to see thatμ(k) = 0 unless 4|k. Since μ is a probability measure,μ(0) = 1, hence the first two informative Fourier coefficients areμ(4) andμ(8); note thatμ(−k) =μ(k) for all k since μ is both real and even (i.e. it is invariant under complex conjugation).
Note that points lying clearly above the red curve, but below the green one, are quite rare. However, "spikes" in the region |μ(n)| ≤ 1/3 are clearly present.

Square free attainable measures
As we shall see, the spikes in the region |μ(4)| ≤ 1/3 are limits of measures μ n where n is divisible by p e for e ≥ 2, but for measures arising from square free n ∈ S, the structure is much simpler.
We say that a measure μ is square free attainable if μ is a limit point of the set {μ n : n ∈ S and n is square free}. The set of square free attainable measures is also closed under convolution, and it is easy to see that it is generated by the set {μ p } p≡1 mod 4 , whose closure is the set of prime attainable measures.

Prime power attainable measures
As mentioned before, the spikes in the region |μ(4)| ≤ 1/3 are due to measures μ n for which n is divisible by a prime power p e , for e large. Recall that a measure μ is prime power attainable if μ is a weak limit point of the set {μ p e } p≡1 mod 4,e≥1 . If μ is a prime power attainable measure, then the point (μ(4),μ(8)) can indeed lie above the curve max(x 4 , (2|x| − 1) 2 ) in the region |μ(4)| ≤ 1/3, though this phenomenon only occurs for even exponents (see Fig. 4). In fact, we will show that for every k ∈ Z + there exists prime power attainable μ such that (μ(4),μ(8)) = 1 2k + 1 , 1 .
To be able to give a complete description of A 2 we need a definition. (1) We say that a pair of continuous functions defines a cornered domain between a and x 0 if for all x ∈ (a, x 0 ] one has (2) For a pair of functions f 1 , f 2 as above the corresponding cornered domain between a and x 0 is The functions f 1 and f 2 will be referred to as the "lower and upper" bounds for D a,x 0 ( f 1 , f 2 ) respectively. Theorem 1. 6 The intersection of the set A 2 with the line y = 1 equals Further, for k ≥ 1, let x k = 1 2k+1 be the x-coordinate of a point of the intersection described above. Then, for every k ≥ 1 there exists a pair of continuous piecewise analytic functions f 1;k , f 2;k defining a cornered domain between 0 and x k , so that A 2 admits the following global description: Theorem 1.6 is a rigorous explanation of the thin strips or "spikes" connecting all the reciprocals of odd numbers on y = 1, and the curve y = (2|x| − 1) 2 , as in Fig. 5. We remark that the functions f 1;k and f 2;k can with some effort be computed explicitly. The lower bound f 1;k is given as the (component-wise) product of (x k , 1) by the parabola y = 2x 2 − 1 mapping (1, 1) → (x k , 1); we re-parameterize the resulting curve (x · x k , 2x 2 − 1) so that it corresponds to the function The upper bound f 2;k (x) is of a somewhat more complicated nature, see Definition 6.3; it is analytic around the corner with the slope f 2;k (x k ) = 4 3 (2k + 1) (see the proof of Theorem 1.6 in Sect. 6), and it is plausible that it is (everywhere) analytic. It then follows that the set A 2 has a discontinuity, or a jump, at x = x k (this is a by-product of the fact that the slopes of both f 1;k and f 2;k at x k are positive.)

Discussion
Our interest in attainable measures originates in the study [5] of zero sets ("nodal lines") of random Laplace eigenfunctions on the standard torus T := R 2 /Z 2 . More precisely, for each n ∈ S there is an associated Laplace eigenvalue given by 4π 2 n, with eigenspace dimension equal to r 2 (n). On each such eigenspace there is a natural notion of a "random eigenfunction", and the variance (appropriately normalized) of the nodal line lengths of these random eigenfunctions equals (1 + μ n (4) 2 )/512 + o(1) as r 2 (n) → ∞. It was thus of particular interest to show that the accumulation points of μ n (4) 2 , as n ∈ S tends to infinity in such a way that also the eigenspace dimension r 2 (n) → ∞, is maximal-namely the full interval [0, 1]. This is indeed the case (cf. [5, Section 1.4]), but a very natural question is: which measures are attainable?
In order to obtain asymptotics for the above variance it is essential to assume that the eigenspace dimension grows, and one might wonder if "fewer" measures are attainable under this additional assumption. However, as the following shows, this is not the case (the proof can be found in Sect. 4.4.) Proposition 1.7 A measure μ ∈ P is attainable (i.e. μ ∈ A), if and only if there exists a sequence {n j } such that μ n j ⇒ μ with the additional property that r 2 (n j ) → ∞.

Outline
For the convenience of the reader we briefly outline the contents of the paper. In Sect. 2 we give some explicit examples of attainable and non-attainable measures, and describe our motivation for studying the set of attainable measures. In Sect. 3 we give a brief background on Fourier coefficients of probability measures, and in Sect. 4 we recall some needed facts from number theory along with proving the more basic results above. Section 5 contains the proof of Theorem 1.3 (a complete classification of attainable measures in the region |μ(4)| > 1/3), and Sect. 6 contains the proof of Theorem 1.6 (the complete classification of attainable measures in the region |μ(4)| ≤ 1/3), postponing some required results of technical nature to the appendix. Finally, in Sect. 7, we classify the set of square-free attainable measures. δ e i(π k/2+θ) + δ e i(π k/2−θ) ; recall that denotes convolution on S 1 . For θ = 0, π/4 the measureδ θ is supported at 4 points whereas for all other values of θ the support consists of 8 points. Given an integer m ≥ 1 and θ ∈ [0, π/4], let We note thatδ θ =δ θ,1 , and that a measure μ, a priori invariant under complex conjugation, is symmetric if and only if μ is invariant under convolution withδ 0 ; in this case convolving withδ 0 is a convenient way to ensure that a measure is symmetric.

Some examples of attainable and unattainable measures
Given θ ∈ [0, π/4] let τ θ denote the symmetric probability measure with uniform distribution on the four arcs given by Using some well known number theory given below (cf. Sect. 4) it is straightforward to show that τ θ is attainable for all θ ∈ [0, π/4]. In particular, dμ Haar = dτ π/4 , the Haar measure on S 1 normalized to be a probability measure, is attainable. In fact, it is well known (see e.g. [2]) that there exists a density one subsequence {n j } ⊆ S, for which the corresponding lattice points n j become equidistributed on the circle; this gives another construction of dμ Haar as an attainable measure. It is also possible to construct other singular measures. In Sect. 4 we will outline a construction of attainable measures, uniformly supported on Cantor sets. Moreover, if q is a prime congruent to 3 modulo 4 it is well known that the solutions to a 2 +b 2 = q 2 are given by (a, b) = (0, ±q), or (±q, 0), thusδ 0 is attainable. A subtler fact, due to Cilleruelo, is that there exists sequences {n j } j≥1 for which n j has very singular angular distribution even though the number of points r 2 (n j ) tends to infinity. Namely, it is possible to force all angles to be arbitrarily close to integer multiples of π/2, hence 1 4 3 k=0 δ i k is an accumulation point of dμ n j as n j → ∞ in such a way that r 2 (n j ) → ∞.
We may also construct some explicit unattainable probability measures on S 1 satisfying all the symmetries; in fact the following corollary of Theorem 1.6 constructs explicit unattainable measures, remarkably supported on 8 points only-the minimum possible for symmetric unattainable measures. for some k ≥ 1.

Some notation and de-symmetrization of probability measures
It is convenient to work with two models: either with the unit circle embedded in C, or Rather than working with {μ n } and its weak partial limits, for notational convenience we work with their de-symmetrized variants, i.e.
Further, let A ⊆ P be the set of all weak partial limits of {ν n } i.e. all probability measures μ ∈ P such that there exists a sequence {n j } with The set A defined above is the de-symmetrization of the collection of attainable measures via (10); by abuse of notation we will refer to the elements of A as attainable measures. One may restate Proposition 1.2 as stating that A is closed w.r.t. convolutions; thus A is an abelian monoid with identity δ 0 ∈ A. The effect of the de-symmetrization (10) is that for all m ∈ Z ν n (m) = μ n (4m); since by the π/2-rotation invariance of μ n , μ(k) = 0 unless k is divisible by 4, this transformation preserves all the information.

Measure classification on the Fourier side
We would like to study the image of A under Fourier transform, or, rather, its projections into finite dimensional spaces. Since A ⊆ P we first study the Fourier image of the latter; a proper inclusion of the image of A inside the image of P would automatically imply the existence of unattainable measures μ ∈ P\A.
For θ ∈ (0, π) let υ θ be the probability measure and for the limiting values θ = 0, π we denote υ 0 = δ 0 and υ π = δ π . As for θ ∈ [0, π], δ θ are the de-symmetrizations ofδ θ/4 in (9), attainable by Proposition 1.2 (see also Lemma 4.1 below), and it then follows that υ θ ∈ A. Clearly (see e.g. [6, Chapter 1]) the set P is the convex hull of Let P k ⊆ R k be the image of P under the projection F k : P → R k given by i.e. P k = F k (P) are the first k Fourier coefficients of the measure μ as μ varies in P.
Recalling the invariance (11) for μ ∈ P we may write where γ k is the curve for θ ∈ [0, 2π ]. Thus P k = F k (P) could be regarded as a convex combination of points lying on γ k (corresponding to υ θ ); it would be then reasonable to expect P k to be equal to the convex hull of γ k . This intuition was made rigorous in a more general scenario by F. Riesz [7] in a classical theorem on the generalized moments problem (cf. [6], Chapter 1, Theorem 3.5 on p. 16). The sets P k are the convex hulls of the curves γ k in R k indeed. Interestingly, since cos(mθ) is a polynomial in cos(θ ), the curve γ k is algebraic. As a concrete example, for k = 2 the image P 2 of P under is the convex hull of the parabola y = 2x 2 − 1, x ∈ [−1, 1], i.e. the set as shown in Fig. 1, to the left. Analogously to the above, define (cf. (5), and bear in mind the de-symmetrization (10)). Since, by the definition, A is closed in P (i.e. the weak limit set of A satisfies A ⊆ A), if follows that for every k ≥ 2, A k is closed in P k in the usual sense. The shell y = 2x 2 − 1 of the convex hull P 2 is (uniquely) attained by the family {υ θ : θ ∈ [0, π]} of measures as in (12) with the Fourier coefficients Finally, it is worth mentioning that the set A is not convex, as A 2 contains the parabola whose points correspond to the measures (12), though not its convex hull. (In other words, had A been convex, that would force all symmetric measures to be attainable.)

Number theoretic background
We start by giving a brief summary on the structure of n (equivalently, μ n or their desymmetrized by (10) versions ν n ) given the prime decomposition of n. These results follow from the (unique) prime factorization of Gaussian integers, see e.g. [1]. First, for every "split" prime p ≡ 1 mod 4, there exists an angle θ p ∈ [0, π], such that the measure ν p arising from p is given by More generally, if a split prime p occurs to a power p e , we find that the resulting measure is given by and hence, in particular, (recall the de-symmetrization (10)). Both the {ν n } and 1 4 r 2 (n) are multiplicative in the sense that for n 1 , n 2 co-prime numbers (n 1 , n 2 ) = 1, and r 2 (n 1 )r 2 (n 2 ) = 4r 2 (n 1 n 2 ).
In particular, r 2 (n) = 0 unless n is of the form n = 2 a p e 1 1 · . . . · p e k k q 2r 1 1 · . . . · q 2r l l , for p i ≡ 1 mod 4, q j ≡ 3 mod 4 primes (in particular, all the exponents of primes ≡ 3 mod 4 are even); in this case and By Hecke's celebrated result [3,4] the angles θ p are equidistributed in [0, π/4]: In particular, the following lemma is an immediate consequence.

Proof of Proposition 1.2
Proof We will prove the equivalent de-symmetrized version of the statement, i.e. that if γ 1 , γ 2 ∈ A then Let {m k }, {n k } ⊆ S be two sequences so that ν m k ⇒ γ 1 , ν n k ⇒ γ 2 . We would like to invoke the multiplicativity (16) of {ν n }; we cannot apply it directly, as n k and m k may fail to be co-prime. To this end rather than using ν m k we are going to substitute 1 it with ν m k chosen to approximate ν m k , so that m k is co-prime to n k , via Lemma 4.1.
In the remaining part of the proof we shall argue that provided we care to choose m k so that ν m k approximates ν m k sufficiently well.
To this end it is more convenient to work with the space of Fourier coefficients; the weak convergence of probability measures corresponds to point-wise convergence of the Fourier coefficients. By Lemma 4.1 we may replace m k with m k co-prime to n k that satisfies for every j ≤ k It then readily follows that ν m k ⇒ γ 1 , and hence we establish (17), which in turn implies that γ 1 γ 2 ∈ A, finally yielding the closedness of A w.r.t. convolutions. As for the second assertion, if μ ∈ A, then μ is a weak limit of a sequence ν n j for some {n j } ⊆ S.

Cantor sets are attainable
By Proposition 1.2, A is closed under convolution, it contains [5] uniform measures supported on symmetric intervals [−θ, θ], as well as symmetric sums (δ θ + δ −θ )/2 for all θ > 0. Thus, by using an "additive" construction of Cantor sets, we easily see that uniform measures supported on Cantor sets are attainable.
Namely, given θ > 0, let C n,θ be the n-th level Cantor set obtained by starting with the interval [−θ, θ] and deleting the middle third part of the interval: C 0,θ consists of one closed interval [−θ, θ], and C n+1,θ ⊂ C n,θ is the union of the 2 n+1 intervals obtained by removing the middle third in each of the 2 n intervals that C n,θ consists of. Now, where denotes disjoint union, and C n+1,θ/3 + α denotes the translation of the set C n+1,θ/3 by α.
Since C 0,θ is a symmetric interval, the measure corresponding to its characteristic function is attainable, as mentioned above. Further, since convolving (δ θ + δ −θ )/2 with a uniform measure having support on some set D yields a measure with support on (D +θ)∪(D −θ), uniform measures supported on C n,θ are attainable by induction, via (18). Letting n → ∞ we find that measures with uniform support on Cantor sets are attainable.

Proof of Proposition 1.7
Proof We are going to make use of a (de-symmetrized) Cilleruelo sequence n j , i.e. ν n j ⇒ δ 0 and r 2 (n j ) → ∞. Let μ ∈ A be an attainable measure and assume that ν m j ⇒ μ. Using the same idea as in the course of proof of Proposition 1.2 above we may assume with no loss of generality that (n j , m j ) = 1 are co-prime (recall that {n j } is a Cilleruelo sequence of our choice). Then so that the sequence {n j · m j } is as required.

Some conventions related to Fourier analysis
We adapt the following conventions. The k-th Fourier coefficient of a measure μ ∈ P is given by clearly | μ(k)| ≤ 1. The convolution of two probability measures μ, μ ∈ P is the probability measure μ μ defined as With the above conventions we have It is easy to compute the Fourier coefficients of υ θ;M as in (15) to be for M = 1, G 2 (θ ) = cos(θ ) is consistent with (14).
By the definition of A and A k = F k (A) and in light of Lemma 4.1, we can describe A k geometrically as the smallest multiplicative set, closed in P k , containing all the curves 2 i.e. A k is the closed multiplicative subset of P k generated by the above curves. Similarly, the set corresponding to the square-free attainable measures A 0 k is the smallest closed multiplicative set containing the single curve From this point on we will fix k = 2 and suppress the k-dependence in the various notation, e.g. γ A will stand for γ 2;A . The curves for 2 ≤ A ≤ 20 are displayed in Fig. 4, separately for odd and even M = A − 1.

Proof of Theorem 1.3
The two statements of Theorem where M(x) is given by (3).

Proof of Proposition 5.1: attainable measures lie under the max curve for x > 1/3
In what follows, by componentwise product we will mean Definition 5.3 (Totally positive and mixed sign points) Let A + 2 ⊆ A 2 be the set of totally positive attainable points admitting a representation as finite componentwise products is the set of mixed sign attainable points admitting representation (23) with at least one y i < 0.
Note that a point in A 2 may be both totally positive and of mixed sign, i.e. A + 2 may intersect A − 2 . Furthermore, a priori it may be in neither of these. However, by the definition of A 2 , it is the closure of the union of the sets defined: Therefore to prove the inequality (21) on A 2 it is sufficient to prove the same for points in A + 2 and A − 2 separately. These are established in Lemma 5.4 and Proposition 5.5, proved in Sects. 5.4 and 5.5 respectively.
We are now in a position to prove Proposition 5.1.

Proof of Lemma 5.4: the mixed sign points A − 2 lie under the max curve
To pursue the proof of Lemma 5.4 we will need some further notation.
Recall the Definition 5.3 of totally positive attainable points A + 2 , and componentwise product of points (22). It is obvious that the points of either B 1 and B 2 are all lying under the max curve, i.e. if Therefore the following lemma implies Lemma 5.4.

Lemma 5.7
If (x, y) ∈ A − 2 is a mixed sign attainable point then To prove Lemma 5.7 we establish the following two auxiliary lemmas whose proof is postponed until immediately after the proof of Lemma 5.7.
Proof of Lemma 5.7 assuming the auxiliary lemmas. Let 2 with y ≤ 0, then (x, y) ∈ B 2 by Lemma 5.8; hence we may assume y > 0. Let (x i , y i ) be as in (23), which according to the Definition 5.3 have mixed signs. Since y ≥ 0 we can in fact find i = j for which y i , y j < 0, and without loss of generality we may assume that (i, j) = (1, 2). Letting We further note that both (x 1 , y 1 ) and (x 2 , y 2 ) lie in B 2 . Thus by Lemma 5.9, Since |x|,ỹ ≤ 1, the result follows on noting that B 1 is mapped into itself by any map of the form provided that 0 ≤ |α|, β ≤ 1.

Proofs of the auxiliary Lemmas 5.8 and 5.9
Proof of Lemma 5.8 The assumptions are equivalent to (x, y) ∈ P 2 with y ≤ 0. The statement follows immediately upon using the explicit description (13) of P 2 : Proof of Lemma 5.9 The case of either point having zero y-coordinate is trivial, so we may assume that both p 1 , p 2 have negative y-coordinates, and it suffices to prove the statement for points p 1 , p 2 having minimal y-coordinates, i.e., and we may further assume ab = 0 as otherwise the statement is trivial. By symmetry it suffices to consider the case a, b ∈ (0, 1/ √ 2). Thus, if we fix c ∈ (0, 1/2) it suffices to determine the maximum of (2a 2 − 1)(2b 2 − 1) subject to the constraint ab = c. Taking logs we find that the constraint is given by log a + log b = log c and we wish to maximize log(1 − 2a 2 ) + log(1 − 2b 2 ).

Proof of Proposition 5.5: totally positive points A + 2 corresponding to prime powers Lemma 5.10 The function sin t t is decreasing and is
Proof Taking derivatives, this amounts to the fact that tan t > t on (0, π/2).
Proof of Proposition 5.5. If A = 2, the points lying on the curve γ 2 are of the form (x, y) = γ 2 (t) = (t, 2t 2 − 1), and it is straightforward to check that 2t 2 − 1 ≤ t 4 . For A ≥ 3, since we assume that x > 1/3 and are attainable (recalling the notation (22) for componentwise multiplication). On the other hand it is clear that the union of the family of the parabolas as x ranges over [0, 1], is exactly the set Concerning points under the other curve y = (2x − 1) 2 we may employ the multiplicativity of A 2 again to yield that the curve 1] is attainable; this curve in turn can be re-parameterized as {(t, (2t − 1) 2 )} t∈ [0,1] . A similar argument to the above shows that function i.e. as the parameter x varies along [0, 1] the parabolas 1]. Hence all the points under the latter curve are attainable, as claimed.
Upon using Taylor series, we find that, as n → ∞, Since this holds for any fixed α > 0, bearing in mind that A is closed in P (and hence the set A 2 ⊆ [−1, 1] 2 is closed in the usual sense), we indeed find that the curve (x, x 4 ) lies in the attainable set for every x ∈ (0, 1). It is easy to see that also (0, 0) and (1, 1) are attainable. By reflecting the curve (x, x 4 ) (for x ≥ 0) in the x-axis (using that (−1, 1) is attainable and multiplying) we find that (x, 6 Proof of Theorem 1.6: fractal structure for x < 1 3 It is obvious that the second assertion of Theorem 1.6 implies the first part, so we only need to prove the second one. However, since the proof of the second assertion is fairly complicated we give a brief outline of how the first assertion can be deduced, and then indicate how to augment the argument to give the second assertion. We are to understand the closure of all the points (x, y) of the form with A i ≥ 2 arbitrary integers. Using that G A (π/2+t) is either even or odd (depending on the parity of A) and that G A (2(π/2+t)) is even, together with signs of x-coordinates being irrelevant (since (x, y) is attainable if and only if (−x, y) is attainable) we may assume that t i ∈ 0, π 2 for all i. A curve (x 0 , y 0 ) = (G A 0 (t 0 ), G A 0 (2t 0 )) turns out to intersect the line y = 1 with |x| ≤ 1 3 only for A 0 odd, and further forces t 0 = π 2 , and x = ± 1 A . Hence the point (x, y) as in (28) satisfies y = 1 only for A i odd and t i = π 2 for all i ≤ K , whence (x, y) = (± 1 A , 1) with A = K i=1 A i . To prove the second assertion we investigate a (fairly large) neighborhood of the point ( 1 A , 1); given an odd A we consider all finite products (28) with A = K i=1 A i and t i ≈ π 2 (and A i ≥ 3.) We will prove that all products (x, y) of this form will stay between two curves defined below; after taking logarithms this will amount to the fortunate log-convexity of the curves (G A 0 (t), G A 0 (2t)), A 0 ≥ 3 odd, in the suitable range (see Lemma 6.8 below). We argue that this property is invariant with respect to multiplying by curves (G A 1 (t), G A 1 (2t)) for A 1 ≥ 2 even, and also for odd A 1 ≥ 3 for t near π/2.

Proof of the second assertion of Theorem 1.6
To prove the main result of the present section we will need the following results. (The proofs of Propositions 6.1 and 6.2 are postponed to Appendices 1 and 2, respectively.)

Proposition 6.1 Let {A i } i be a finite collection of integers A i ≥ 2, and consider a point (x, y) of the form
where all t i ∈ [0, π/2]. Assume that one of the following is satisfied: • There exists i such that A i ≥ 3 is odd and • There exists i such that A i is even and t i ≥ π/(2 A i ).

Proposition 6.2 Let A ≥ 3 be an odd number, and
the supremum taken w.r.t. all (t i ) i≤K lying in Then for every and moreover the map x → i 0 (x) is piecewise constant. In particular, the function g {A i } (x) is continuous, analytic in some (left) neighbourhood of x = 1 A , and piecewise analytic on (0, 1 A ].
We may finally define the function f 2;k introduced in Theorem 1.6. 3 The reason for appearing is that the supremum is attained by having t i = π 2 for i = i 0 and the maximum taken w.r.t. all non-trivial factorizations of 2k + 1, i.e., all sets of (odd) integers , whose product is 2k + 1.
(2) By the definition of g {A i } and f 2;k , if (x, y) is of the form with all A i ≥ 3 odd, x > 0, and if in addition for all i we have (whence x ≤ 1 2k+1 via (32)), then necessarily where k is defined as in (3) Proposition 6.2 implies that for k ≥ 1 and x < 1 2k+1 , a maximum w.r.t. all (odd) divisors A > 1 of 2k + 1; the latter yields an algorithm for computing f 2;k (x), reducing the original problem into maximizing a finite set of numbers.
The following 3 results will be proven in Appendix 2.

Lemma 6.7
For every x 1 , x 2 ∈ [0, 1] the following inequality holds: We are finally in a position to prove Theorem 1.6 (with the first assertion following from the second.) Proof of the second assertion of Theorem 1.6 assuming the results above We first prove that any point (x, y) ∈ A 2 with 0 < x ≤ 1 3 either satisfies y ≤ (2x − 1) 2 or (x, y) ∈ D 0,x k ( f 1;k , f 2;k ) for some k ≥ 1, i.e. establish the inclusion ⊆ of (7). Since A 2 is the closure (in R 2 ) of the set of finite products with some A i ≥ 2, t i ∈ [0, π], and the set on the r.h.s. of (7) is closed in {x > 0}, it is sufficient to prove it for the finite products (36). Thus let (x, y) be given by a finite product (36); by the invariance of A 2 w.r.t. x → −x we may assume that all t i , i ≤ K satisfy t i ∈ [0, π/2]. If there exists either an odd A i such that t i ∈ [ π 2 A i , π 2 − π 2 A i ], or an even A i such that t i ∈ [ π 2 A i , π 2 ], then one of the sufficient conditions of Proposition 6.1 is satisfied, implying that y ≤ (2x − 1) 2 , so that our present statement holds.
We may then assume that for all odd A i we have either t i ∈ [0, π 2 A i ) or t i ∈ ( π 2 − π 2 A i , π 2 ], and for all even A i we have t i ∈ [0, π 2 A i ). Up to reordering the indexes, we may assume that K = K 1 + K 2 with K 1 > 0, and where all the A i with i ≤ K 1 are odd and t i ∈ [ π 2 − π 2 A i , π 2 ], and for all K 1 + 1 ≤ i ≤ K 2 we have whether the corresponding A i is odd or even. Let be the product of the first K 1 odd A i . We claim that, with k as defined in (38), necessarily Define and so that (x, y) = (x 0 , y 0 ) · (x 1 , y 1 ).
Since by the above, (x 0 , y 0 ) satisfies the assumptions of (33), we have y 0 ≤ g {A i } i≤K 1 (x 0 ), and by Proposition 6.2 there exists i 0 ≤ K 1 and t 0 ∈ ( π 2 − π 2 A i 0 , π 2 ], so that and For the sake of brevity of notation we assume with no loss of generality that i 0 = 1, and consider the curve η A 1 in R 2 >0 as in Lemma 6.5; by the virtue of the latter lemma we may re-parameterize η A 1 as (z, h A 1 (z)) in the range z ∈ (−∞, 0], and 0 < h A 1 (x) ≤ 4 3 everywhere. Hence, on noting that all the logarithms involved are negative, the mean value theorem gives that Note that by (42) and the definition of h A 1 as a re-parametrization of (34), we have (recall that we assumed that i 0 = 1). Substituting the latter into (44) implies that there exist a number Equivalently, and by (43). Note that for the choice t 1 = θ 1 and t i = π 2 for 2 ≤ i ≤ K 1 , we have by (45) On the other hand, (37) implies that for every is decreasing for t ∈ [0, π/A] and it is enough to show that G A (π/(2 A)) = (A sin(π/(2 A))) −1 > 1/3; this in turn follows from sin(x)/x being decreasing on [0, π].) Hence Proposition 5.5 is applicable for each of the terms on the r.h.s. of (40), and therefore their product (x 1 , y 1 ) satisfies The inequality (49) together with (48) and the fact that x 4/3 > x 4 for x < 1 yield that as in (41), which is the second inequality of (39).
To prove the first inequality of (39) we use Corollary 6.6 to yield y 0 ≥ (Ax 0 ) 4/3 with A as in (38). These combined imply where we used the obvious inequality x 4 ≥ 2x 2 − 1, valid on [−1, 1]. Finally, an application of the inequality (35) of Lemma 6.7 yields by the definition (8) of f 1;k , and recalling that x k = 1 2k+1 . Conversely, we need to prove that any point (x, y) satisfying necessarily lies in A 2 . To this end fix a number k ≥ 1 and consider all the points (x, y) with s ∈ (0, 1 2k+1 ], t ∈ (0, 1] (recalling the notation (22) for componentwise multiplication). Note that by the multiplicativity of A 2 (Proposition 1.2) all the points of the form (50) are attainable, i.e., (x, y) ∈ A 2 . Since

Proof of Proposition by convexity
The convexity of the component-wise logarithm of a curve implies that finite products of points lying on that curve would stay below it. We aim at eventually proving that all the curves γ A = (G A (t), G A (2t)), A ≥ 3 odd, t ∈ π 2 − π after taking logarithm, is equivalent to the statement of Proposition 6.2 (see the proof of Proposition 6.2 below); the latter follow from finite products of points on a curve, with the property above, staying below that curve.

Lemma 6.8 Let η A be the curve
Then for every z ∈ (−∞, 0] there exists an Before giving a proof for Lemma 6.9 we may finally prove Proposition 6.2. Proof of Proposition 6.2 assuming Lemmas 6.8 and 6.9 Let A = 2k + 1 ≥ 3 be odd, and (38) be an arbitrary factorization of A into integers A i ≥ 3. Consider the curves } i≤K as defined in (34). By Lemma 6.8 all of the η A i can be re-parametrized as (z i , h A i (z i )) on (−∞, 0], with h A i convex analytic, and h(0) = 0.
Hence, by Lemma 6.9 for every z ∈ (0, 1 A ] there exists i 0 = i 0 (x), so that Note that, after taking logarithms, More formally, recalling the definition (34) of η A i and (z i , h A i (z i )) being a re-parametrization of η A i , the function h(z) defined as in (51), on noting that z = log Ax, satisfies h(log(Ax)) = log sup where (31)), and hence (52) is The latter equality together with Lemma 6.9 then imply that we have for some i 0 ≤ K ; since h i 0 is a re-parametrization of η A i 0 , this is equivalent to which is the first statement of the present proposition, at least for x > 0. For x = 0 it is sufficient to notice that for all i ≤ K , To see that the map x → i 0 (x) is in fact piecewise constant on [0, 1 A ] (with finitely many pieces), we note that it is readily shown that on (0, 1 A ], g {A i } i≤K is a maximum of finitely many analytic curves (namely, , and vanishes at 0, which happens to lye on all of them. Since such a collection of analytic curves may only intersect in finitely many points for x ∈ [0, 1 A ], it follows that i 0 (x) is uniquely determined as the maximum of these outside of finitely many points (that include (0, 0)), and i 0 is constant between any two such consecutive points.
Proof of Lemma 6.9 It is easy to check that with the assumptions of the present lemma, the function H : (−∞, 0] K → R defined by is a convex function. Now fix t < 0 and consider the set (t) is a compact convex domain, and it is evident that . . . , t k ). Now, a convex function cannot attain a maximum in the interior of a convex domain (all the local extrema of a convex function are necessarily minima). Hence there exists an index i 1 ≤ K so that for some (t i ) ∈ (t) with t i 1 = 0, i.e. one of the elements of (t i ) must vanish. By induction, we find that all but one element of (t i ) vanish, say t i = 0 for i = i 0 , whence t i 0 = t, and h(t) = h i 0 (t), as h i (0) = 0 for i = i 0 by the assumptions of the present lemma.

Proof of Theorem 1.4: square-free attainable measures
Proof Recall that we de-symmetrized all the probability measures by an analogue of (10). First we show that (4) holds for any square-free attainable measure; as the first inequality in (4) holds for every probability measure (13) it only remains to show that every point (x, y) = (μ(1),μ(2)) corresponding to a square-free attainable μ satisfies (21).
Ifỹ > 0 and y i ≥ 0 for all i, then y i = 2x 2 i − 1 ≤ x 4 i for all i as it is easy to check the latter inequality explicitly, consequentlyỹ ≤x 4 . Since (21) holds on the collection of all products (53), it also holds on its closure, namely for square-free attainable measures. This concludes the proof of the necessity of the inequality (4).
As we may assume with no loss of generality that x > 0 (note that y > 0 by the assumption of t i being near π/2) the latter is equivalent to log y ≥ 4 3 log(Ax).
Note that, with η A defined as in Lemma 6.5, η A (t) = (z, h A (z)) = (log(Ax), log(y)), with h A analytic convex, h A (0) = 0, and a straightforward computation shows that h A (0) = 4 3 . By the convexity of η A then the curve lies above its tangent line at the origin, i.e. (65) follows.