The Symmetric Subset Problem in Continuous Ramsey Theory
Greg Martin and Kevin O’Bryant
CONTENTS A symmetric subset of the reals is one that remains invariant 1. Introduction under some reflection x → c − x. We consider, for any 0 < 2. Lower Bounds for ∆(ε) ε ≤ 1, the largest real number ∆(ε) such that every subset of 3. Upper Bounds for ∆(ε) [0, 1] with measure greater than ε contains a symmetric subset 4. Some Remaining Questions with measure ∆(ε). In this paper we establish upper and lower Acknowledgments bounds for ∆(ε) of the same order of magnitude: For example, − 11 ≤ ≤ 2 References we prove that ∆(ε)=2ε 1 for 16 ε 1 and that 0.59ε < 2 ≤ 11 ∆(ε) < 0.8ε for 0 <ε 16 . This continuous problem is intimately connected with a cor- responding discrete problem. A set S of integers is called a ∗ B [g] setifforanygivenm there are at most g ordered pairs (s1,s2) ∈ S × S with s1 + s2 = m; in the case g =2,these are better known as Sidon sets. Our lower bound on ∆(ε) im- B∗ g { , ,...,n} plies that every [ ] set√ contained in 1 2 has cardi- nality less than 1.30036 gn. This improves a result of Green for ∗ g ≥ 30. Conversely, we use a probabilistic construction of B [g] sets to establish an upper bound on ∆(ε) for small ε.
1. INTRODUCTION AsetC ⊆ R is symmetric if there exists a number c (the center of C) such that c + x ∈ C if and only if c − x ∈ C. Given a set A ⊆ [0, 1) of positive measure, is there necessarily a symmetric subset C ⊆ A of positive measure? The main topic of this paper is to determine how large, in terms of the Lebesgue measure of A,one may take the symmetric set C. In other words, for each ε>0 we are interested in ⎧ ⎫ ⎨ every measurable subset of [0, 1) of ⎬ ∆(ε):=sup⎩ δ : measure ε contains a symmetric ⎭ . subset of measure δ (1–1) It is not immediately obvious, although it turns out to be true, that ∆(ε) > 0. 2000 AMS Subject Classification: Primary 05D99; We have dubbed this sort of question “continuous Secondary 42A16, 11B83 Ramsey theory,” and we direct the reader to later in this Keywords: Ramsey theory, continuous combinatorics, Sidon sets section for problems with a similar flavor.
c A K Peters, Ltd. 1058-6458/2007 $ 0.50 per page Experimental Mathematics 16:2, page 145 146 Experimental Mathematics, Vol. 16 (2007), No. 2
We determine a lower bound for ∆(ε) using tools and ∆(ε)/ε2 methods from harmonic analysis, some of which were 1 spurred by ideas from nonstandard analysis and the the- ory of wavelets. We also construct sets without large 0.9 symmetric subsets using results from probabilistic num- ber theory. These two lines of attack complement each 0.8 other, and our bounds on ∆(ε) yield new results in ad- ditive number theory as well. 0.7 The following theorem, proved in Sections 2.1 and 3.4, states some fundamental properties of the function ∆(ε). 0.6
Theorem 1.1. The function ∆(ε) is continuous and in ε 0.2 0.4 0.6 0.8 1 fact satisfies the Lipschitz condition 2 FIGURE 1. Upper and lower bounds for ∆(ε)/ε . |∆(x) − ∆(y)|≤2|x − y| for all x, y ∈ (0, 1]. Furthermore, the function ∆(ε)/ε2 is Perhaps surprisingly, the upper bounds given in The- 2 increasing on (0, 1], and hence limε→0+ ∆(ε)/ε exists. orem 1.2(iv)–(v) are derived from number-theoretic con- siderations. A set S of integers is called a B∗[g]set We turn now to stating our quantitative bounds for if for any given m there are at most g ordered pairs ≥ 1 2 ∈ × ∆(ε). The lower bound ∆(ε) 2 ε , which we call the (s1,s2) S S with s1 + s2 = m. We shall use con- trivial lower bound on ∆(ε) (see Lemma 2.2 below), is structions of large B∗[g] sets to derive upper bounds on not so far from the best we can derive. In fact, the bulk ∆(ε) in Section 3.4. Conversely, our bounds on ∆(ε) im- of this paper is devoted to improving the constant in prove the best known upper bounds on the size of B∗[g] 1 this lower bound from 2 to 0.591389. Moreover, we are sets for large g, as we show in [Martin and O’Bryant 07]. able to establish a complementary upper bound for ∆(ε) See the article [O’Bryant 04] of the second author for a using results on an analogous problem in combinatorial survey of B∗[g]sets. number theory. We note that the difficulty of determining ∆(ε)isin Figure 1 shows the precise upper and lower bounds we stark contrast to the analogous problem in which we con- obtain for ∆(ε)/ε2 as functions of ε, which we present as sider subsets of the circle T := R/Z instead of subsets of Theorem 1.2. the interval [0, 1). In this analogous setting, we com- pletely determine the corresponding function ∆T(ε); in 2 Theorem 1.2. We have: fact, we show (Corollary 3.7) that ∆T(ε)=ε for all ≤ 11 0 <ε 1. As it turns out, the methods that allow the i. ∆(ε)=2ε − 1 for ≤ ε ≤ 1,and∆(ε) ≥ 2ε − 1 for 2 16 T ≤ 1 11 proof of the upper bound ∆ (ε) ε , namely construc- ≤ ε ≤ ; ∗ 2 16 tions of large B [g] sets in Z/N Z, are also helpful to us ∗ ii. ∆(ε) ≥ 0.591389ε2 for all 0 <ε≤ 1; in constructing the large B [g] sets themselves. Schinzel and Schmidt [Schinzel and Schmidt 02] con- iii. ∆(ε) ≥ 0.5546ε2 +0.088079ε3 for all 0 <ε≤ 1; sider the problem of bounding ∗ ≤ 96 2 2 ≤ 11 f f 1 iv. ∆(ε) 121 ε < 0.79339ε for 0 <ε 16 ; B := sup , f f ∗ f∞ 2 v. ∆(ε) ≤ √πε = π ε2 + O(ε3) for all 0 <ε≤ 1. (1+ 1−ε)2 4 where the supremum is taken over all nonnegative func- tions supported on [0, 1]; they showed that 4/π ≤ B< 1.7373. The proof of Theorem 1.2(ii) improves the value Note that π < 0.7854. The upper bound in part (v) 4 1.7373 to 1.691. of the theorem is superior to the one in part (iv) in the √ . We remark briefly on the phrase “continuous Ramsey range 0 <ε< 11 (8 6π − 11π) =0.0201. The five parts 96 theory.” A “coloring theorem” has the following form: of the theorem are proved separately in Proposition 3.10, Proposition 2.15, Proposition 2.18, Corollary 3.14, and Given some fundamental set R colored with Proposition 3.15. a finite number of colors, there exists a Martin and O’Bryant: The Symmetric Subset Problem in Continuous Ramsey Theory 147
highly structured monochromatic subset, pro- established in Section 2.5, we investigate in Section 2.6 vided that R is sufficiently large. the connection between f ∗ f∞ and the Fourier coeffi- cients of f, which, when combined with the results of Sec- The prototypical example is Ramsey’s theorem itself: tion 2.3, allows us to show that ∆(ε) ≥ 0.591389ε2.The However one colors the edges of the complete graph Kn bound given in Section 2.3 and improved in Section 2.6 with r colors, there is a monochromatic complete sub- depends on a kernel function with certain properties; in graph on t vertices, provided that n is sufficiently large in Section 2.7 we discuss how we chose our kernel. In Sec- terms of r and t. Another example is van der Waerden’s tion 2.8, we use a different approach to derive a lower { } theorem: However one colors the integers 1, 2,...,n bound on ∆(ε) that is superior for 3 <ε< 5 . with r colors, there is a monochromatic arithmetic pro- 8 8 gression with t terms, provided that n is sufficiently large 2.1 Easy Bounds for ∆(ε) in terms of r and t. We now turn our attention to the investigation of the In many cases, the coloring aspect of a Ramsey-type function ∆(ε) defined in (2–2). In this section we es- theorem is a ruse, and one may prove a stronger state- tablish several simple lemmas describing basic properties ment in the following form: of ∆. Given some fundamental set R, any large sub- Let λ denote Lebesgue measure on R. We find the set of R contains a highly structured subset, following equivalent definition of ∆(ε) easier to work with provided that R itself is sufficiently large. than the definition given in (1–1): If we define
Such a result is called a “density theorem.” For example, D(A):=sup{λ(C): C ⊆ A, C is symmetric}, (2–1) van der Waerden’s theorem is a special case of the den- sity theorem of Szemer´edi: Every subset of {1, 2,...,n} then with cardinality at least δn contains a t-term arithmetic progression, provided that n is sufficiently large in terms ∆(ε):=inf{D(A): A ⊆ [0, 1),λ(A)=ε}. (2–2) of δ and t. Ramsey theory is the study of such theorems on differ- Lemma 2.1. ∆(ε) ≥ 2ε − 1 for all 1/2 ≤ ε ≤ 1. ent types of structures. By “continuous Ramsey theory” we refer to Ramsey-type problems on continuous mea- Proof: For A ⊆ [0, 1), the centrally symmetric set A ∩ sure spaces. In particular, this paper is concerned with a (1 − A) has measure equal to density-Ramsey problem on the structure [0, 1) ⊆ R with − − ∪ − Lebesgue measure. The type of substructure we focus on λ(A)+λ(1 A) λ(A (1 A)) is a symmetric subset. =2λ(A) − λ(A ∪ (1 − A)) ≥ 2λ(A) − 1. Other appearances of continuous Ramsey theory in the ≥ − literature are in the work of [Swierczkowski´ 58] (see also Therefore D(A) 2λ(A) 1 from the definition (2–1) of [Guy 94, problem C17], [Banakh et al. 00], [Schinzel and the function D. Taking the infimum over all subsets A of ≥ − Schmidt 02], and [Chung et al. 02]). In all cases, there is [0, 1) with measure ε yields ∆(ε) 2ε 1 as claimed. an analogous discrete-Ramsey-theory problem. However, While this bound may seem obvious, it is in many situ- see [Chung et al. 02] for an interesting example in which ations the state of the art. As we show in Proposition 3.10 the quantities involved in the discrete setting do not tend below, ∆(ε) actually equals 2ε − 1for 11 ≤ ε ≤ 1; and in the limit to the analogous quantity in the continuous 16 ∆(ε) ≥ 2ε − 1 is the best lower bound of which we are setting. ≤ 11 aware in the range 0.61522 ε< 16 =0.6875. One is tempted to try to sharpen the bound ∆(ε) ≥ ∆( ) − 1 2. LOWER BOUNDS FOR ε 2ε 1 by considering the symmetric subsets with center 3 , 1 ,or 2 , for example, instead of merely 1 . Unfortunately, We give easy lower bounds for ∆(ε) and prove that ∆(ε) 2 3 2 it can be shown that given any ε ≥ 1 and any finite set is continuous in Section 2.1. Section 2.2 makes explicit 2 {c ,c ,...,c }, one can construct a sequence S of sets, the connection between ∆(ε) and harmonic analysis. Sec- 1 2 n k each with measure ε, that satisfies tion 2.3 gives a simple, but quite good, lower bound on ∆(ε). In Section 2.4, we give a more general form of the lim max {λ (Sk ∩ (2ci − Sk))} =2ε − 1. argument of Section 2.3. Using an analytic inequality k→∞ 1≤i≤n 148 Experimental Mathematics, Vol. 16 (2007), No. 2
Thus, no improvement is possible with this sort of argu- rather than with ⊕, we would rather avoid any potential ment. confusion with the function ∆ featured prominently in this paper. ≥ 1 2 Lemma 2.2. (Trivial lower bound.) ∆(ε) 2 ε for all 0 ≤ ε ≤ 1. Lemma 2.4. If S and T are two sets of real numbers, then |D(S) − D(T )|≤2λ(S ⊕ T ). Proof: Given a subset A of [0, 1) of measure ε,letA(x) denote the indicator function of A, so that the integral Proof: Let E be any symmetric subset of S, and let c of A(x) over the interval [0, 1) equals ε. If we define be the center of E, so that E =2c − E. Define F = 1 − ∩ ∩ − f(c):= 0 A(x)A(2c x) dx, then f(c) is the measure E T (2c T ), which is a symmetric subset of T with of the largest symmetric subset of A with center c,and center c. We can write λ(F ) using the inclusion–exclusion we seek to maximize f(c). But f is clearly supported on formula [0, 1), and so − − ∪ λ(F )=λ(E)+λ(T )+λ(2c T ) λ(E T ) 1 − ∪ − − ∪ − D(A) = max f(c) ≥ f(c) dc λ(E (2c T )) λ(T (2c T )) 0≤c≤1 0 + λ(E ∪ T ∪ (2c − T )). 1 1 − = A(x)A(2c x) dc dx Rearranging terms, and noting that T ∪ (2c − T ) ⊆ E ∪ 0 0 − ∪ − 1 2 x dw T (2c T ), we see that = A(x) A(w) dx 0 −x 2 − ≤− − − ∪ λ(E) λ(F ) λ(T ) λ(2c T )+λ(E T ) 1 1 1 1 2 + λ(E ∪ (2c − T )). = A(x) dx A(w) dw = 2 ε , 2 0 0 Because reflecting a set in the point c does not change ⊆ − − since A(w) is supported on [0, 1] [ x, 2 x]. Since A its measure, this is the same as was an arbitrary subset of [0, 1) of measure ε,wehave ≥ 1 2 − shown that ∆(ε) 2 ε . λ(E) λ(F ) ≤−λ(T ) − λ(T )+λ(E ∪ T )+λ(E ∪ (2c − T )) It is obvious from the definition of ∆ that ∆(ε)isan = −λ(T ) − λ(T )+λ(E ∪ T )+λ((2c − E) ∪ T ) increasing function; the next lemma shows that ∆(ε)/ε is also an increasing function. Later in this paper (see =2 λ(E ∪ T ) − λ(T ) 2 Proposition 3.12), we will show that ∆(ε)/ε is an in- ≤ 2 λ(S ∪ T ) − λ(T ) =2λ(S \ T ) ≤ 2λ(S ⊕ T ). creasing function. Therefore, since F is a symmetric subset of T , Lemma 2.3. ∆(ε) ≤ ∆(x) ε for all 0 ≤ ε ≤ x ≤ 1.In x λ(E) ≤ λ(F )+2λ(S ⊕ T ) ≤ D(T )+2λ(S ⊕ T ). particular, ∆(ε) ≤ ε. Taking the supremum over all symmetric subsets E of S, { ∈ } Proof: If tA := ta: a A is a scaled copy of a set we conclude that D(S) ≤ D(T )+2λ(S ⊕ T ). If we now A, then clearly D(tA)=tD(A). Applying this with any exchange the roles of S and T , we see that the proof is ⊆ ε ≤ set A [0, 1) of measure x and with t = x 1, we see complete. ε that x A is a subset of [0, 1) with measure ε,andsoby ≤ ε ε the definition of ∆ we have ∆(ε) D( x A)= x D(A). Lemma 2.5. The function ∆ satisfies the Lipschitz con- ⊆ Taking the infimum over all sets A [0, 1) of measure x, dition |∆(x) − ∆(y)|≤2|x − y| for all x and y in [0, 1]. ≤ ε we conclude that ∆(ε) x ∆(x). The second assertion of In particular, ∆ is continuous. the lemma follows from the first assertion with the trivial value ∆(1) = 1. Proof: Without loss of generality assume that y D(S) ≥ D(T )−2(x−y) ≥ ∆(x)−2(x−y) by the definition We note that for any fixed sequence a = {aj}j∈Z,the p of ∆. Taking the infimum over all sets S ⊆ [0, 1) of -norm ap is a decreasing function of p. To see this, p measure y yields ∆(y) ≥ ∆(x) − 2(x − y) as desired. suppose that 1 ≤ p ≤ q<∞ and a ∈ . Then |aj|≤ ∈ Z | |q−p ≤ q−p − ≥ a p for all j , whence aj a p (since q p | |q ≤ q−p| |p 2.2 Notation 0) and so aj a p aj . Summing both sides over ∈ Z q ≤ q−p p q There are many ways to define the basic objects of all j yields a q a p a p = a p, and taking Fourier analysis; we follow [Folland 84]. Unless specif- qth roots gives the desired inequality aq ≤ap. ically noted otherwise, all integrals are over the circle Finally, we define a “pdf,” short for “probability den- 1 2 group T := R/Z; for example, L denotes the class of sity function,” to be a nonnegative function in L whose | | L1-norm (which is necessarily finite, since T is a finite functions f for which T f( x) dx is finite. For each in- ˆ −2πijx measure space) equals 1. Also, we single out a special teger j, we define f(j):= f(x)e dx,sothatfor ∈ 1 ∞ ˆ 2πijx type of pdf called an “nif,” short for “normalized indica- any function f L ,wehavef(x)= j=−∞ f(j)e ∗ tor function,” which is a pdf that takes only one nonzero almost everywhere. We define the convolution f g(c):= f(x)g(c − x) dx, and we note that f ∗ g(j)=fˆ(j)ˆg(j) value, that value necessarily being the reciprocal of the measure of the support of the function. (We exclude the for every integer j; in particular, f ∗ f(j)=fˆ(j)2. possibility that an nif takes the value 0 almost every- We define the usual Lp-norms ⊆ T where.) Specifically, we define for each E the nif 1/p f := |f(x)|p dx p λ(E)−1,x∈ E, fE(x):= and 0,x ∈ E. ˆ ˆ 2 2 f∞ := lim fp =sup y : λ({x: |f(x)| >y}) > 0 . Note that if f is a pdf, then 1 = f(0) = f(0) = f 1 = p→∞ f ∗ f1. With these definitions, H¨older’s inequality is valid: If We are now ready to reformulate the function ∆(ε)in 1 1 terms of this notation. p and q are conjugate exponents, that is, p + q =1, then fg1 ≤fpgq. We also note that f ∗ g1 = f1g1 when f and g are nonnegative functions; in Lemma 2.6. We have 2 ˆ 2 particular, f ∗ f1 = f = f(0) . We shall also 1 2 1 2 1 ε inf g ∗ g∞ ≤ ε inf f ∗ f∞ =∆(ε), p 2 g 2 f employ the -norms for bi-infinite sequences: If a = { } | |p 1/p aj j∈Z, then a p := j∈Z aj and a ∞ := the first infimum being taken over all pdfs g that are sup- lim →∞ a =sup∈Z |a |. Although we use the same − 1 1 p p j j ported on 4 , 4 , and the second infimum being taken notation for the Lp-andp-norms, no confusion should − 1 1 over all nifs f whose support is a subset of 4 , 4 of arise, since the object inside the norm symbol will be ei- ε measure 2 . ther a function on T or its sequence of Fourier coefficients. With this notation, we recall Parseval’s identity Proof: The inequality is trivial, since every nif is a pdf; it remains to prove the equality. f(x)g(x) dx = fˆ(j)ˆg(−j) ⊆ For each measurable A [0, 1), define EA := 1 − 1 ∈ ⊆ − 1 1 2 (a 2 ): a A 4 , 4 .ThesetsA and EA differ (assuming that the integral and sum both converge); in only by translation and scaling, so that λ(A)=2λ(EA) ˆ − particular, if f = g is real-valued (so that f( j)isthe and D(A)=2D(EA). Thus ˆ ˆ conjugate of f(j) for all j), this becomes f2 = f2. ∆(ε):=inf{D(A): A ⊆ [0, 1),λ(A)=ε} The Hausdorff–Young inequality, fˆq ≤fp whenever p and q are conjugate exponents with 1 ≤ p ≤ 2 ≤ q ≤∞, 2 D(A) ⊆ = ε inf 2 : A [0, 1),λ(A)=ε can be thought of as a generalization of this latter version λ(A) of Parseval’s identity. We also require the definition 2D(EA) = ε2 inf : A ⊆ [0, 1),λ(A)=ε (2λ(E ))2 1/p A p map = |a(j)| (2–3) 1 D(E) = ε2 inf : E ⊆ − 1 , 1 ,λ(E)= ε . |j|≥m 2 λ(E)2 4 4 2 { } ⊆ − 1 1 ε for any sequence a = aj j∈Z, so that 0 a p = a p,for For each E 4 , 4 with λ(E)= 2 , the function fE(x) − 1 1 ε example. is an nif supported on a subset of 4 , 4 with measure 2 , 150 Experimental Mathematics, Vol. 16 (2007), No. 2 ∗ 2 ≤ ∗ ∗ ∗ 2 ≤ and it is clear that every such nif arises from some set E. (f f) f f ∞(f f), integration yields f f 2 D(E) ∗ ∗ ∗ ∗ Thus, it remains only to show that λ(E)2 = fE fE ∞, f f ∞ f f 1 = f f ∞. Combining the last three 2 ∗ ˆ −4 ≤ˆ4 ∗ 2 ≤ i.e., that D(E)=λ(E) fE fE ∞. sentences, we see that K 4/3 f 4 = f f 2 ⊆ − 1 1 f ∗ f∞ as claimed. Fix E 4 , 4 , and let E(x) be the indicator func- −1 tion of E. Note that fE(x)=λ(E) E(x). The maximal This reasonably simple theorem already allows us to symmetric subset of E with center c is E ∩ (2c − E), and this has measure E(x)E(2c − x) dx.Thus give a nontrivial lower bound for ∆(ε). The step function 1, 0 ≤|x|≤ 1 , D(E):=sup{λ(C): C ⊆ E, C is symmetric} 4 K1(x):= − 2π4 1 | |≤ 1 1 4 4 3 4/3− 8/3 , 4 < x 2 , π +24ζ( 3 ) (5+2 2 ) =sup E(x)E(2c − x) dx c −4 π4 has Kˆ =1+ 3 > 1.074 (the elab- 1 4/3 4/3− 4 3 8 (2 1) ζ( 3 ) =sup λ(E)fE(x)λ(E)fE(2c − x) dx c orate constant used in the definition of K1 was chosen ˆ to minimize K14/3). A careful reader may complain 2 − = λ(E) sup fE(x)fE(2c x) dx that K is not continuous. The continuity condition is c 1 2 not essential, however, since we may approximate K1 by = λ(E) sup fE ∗ fE(2c) c a continuous function L with L4/3 arbitrarily close to 2 = λ(E) fE ∗ fE∞ , K1 4/3. Green [Green 01] used a discretization of the kernel as desired. function The convolution in Lemma 2.6 may be taken over R 1, 0 ≤|x|≤ 1 , K (x):= 4 or over T, the two settings being equivalent since f ∗ f is 2 − − 4 − 3 1 | |≤ 1 1 α + α 40(2x 1) 2 , 4 < x 2 , supported on an interval of length 1. In fact, the reason 1 ˆ −4 8 we scale f to be supported on an interval of length 2 with a suitably chosen α to get K2 4/3 > 7 > 1.142. is so that we may replace convolution over R,whichis ˆ −4 We get a slightly larger value of K 4/3 in the follow- the natural place to study ∆(ε), with convolution over ing corollary with a much more complicated kernel. See T , which is the natural place to do harmonic analysis. Section 2.7 for a discussion of how we came to find our kernel. 2.3 The Basic Argument We begin the process of improving upon the trivial lower − 1 1 Corollary 2.8. If f is a pdf supported on 4 , 4 , then bound for ∆(ε) by stating a simple version of our method ∗ 2 ≥ that illustrates the ideas and techniques involved. f f 2 1.14915. Consequently, ∆(ε) ≥ 0.574575ε2 for all 0 ≤ ε ≤ 1. Proposition 2.7. Let K be any continuous function on T satisfying K(x) ≥ 1 when x ∈ − 1 , 1 ,andletf be a pdf 4 4 Proof: Set supported on − 1 , 1 .Then 4 4 ⎧ 1 − ⎪1, if 0 ≤|x|≤ , ∗ ≥ ∗ 2 ≥ˆ 4 ⎨⎪ 4 f f ∞ f f 2 K 4/3. 1.2015 K (x)= 2 −1 √1−2x 3 ⎪0.6644 + 0.3356 π tan − , ⎩⎪ 4x 1 if 1 ≤|x|≤ 1 . Proof: We have 4 2 (2–4) The function K (x) is pictured in Figure 2. ˆ ˆ 3 1= f(x) dx ≤ f(x)K(x) dx = f(j)K(−j) ˆ We do not know how to rigorously bound K34/3, j ˆ but we can rigorously bound K44/3, where K4 is a by Parseval’s identity. H¨older’s inequality now gives 1 ≤ piecewise linear function “close” to K3. Specifically, let ˆ ˆ ˆ −4 ≤ f 4 K 4/3, which we restate as the inequality K 4/3 K4(x) be the even piecewise linear function with corners fˆ4. at 4 ˆ4 | ˆ |4 | ∗ |2 ∗ 2 Now f 4 = f(j) = f f(j) = f f 2 1 1 t 1 t j j (0, 1), , 1 , + ,K + , by another application of Parseval’s identity. Since 4 4 4 × 104 3 4 4 × 104 Martin and O’Bryant: The Symmetric Subset Problem in Continuous Ramsey Theory 151 K (x) 3 2. We have searched for more advantageous kernel 1 ˆ functions K(x) for which we can compute K4/3 0.75 in an accurate way. A detailed discussion of our search for the best kernel functions is in Section 2.7. 0.5 0.25 3. The application of Parseval’s identity can be re- placed with the Hausdorff–Young inequality, which ‒0.5 ‒0.25 0.25 0.5 ∗ ≥ˆ −q leads to the conclusion f f ∞ K p , where ≤ 4 ≥ p 3 and q 4 are conjugate exponents. Nu- FIGURE 2. The function K3(x). 4 merically, the values (p, q)=(3 , 4) appear to be optimal. However, Beckner’s sharpening [Beckner t =0, 1,...,104. We calculate (using Proposition 2.16 75] of the Hausdorff–Young inequality leads to the −q ˆ stronger conclusion f ∗ f∞ ≥ C(q)Kˆ , where below) that K44/3 < 0.9658413. Therefore, by Propo- p q − 2 q/2−1 q sition 2.7 we have C(q)= 2 (1 q ) = 2e + O(1). We have not ex- perimented to see whether a larger lower bound can ∗ 2 ≥ −4 f f 2 (0.9658413) > 1.14915. be obtained from this stronger inequality by tak- ing q>4. 1 2 Using Lemma 2.6, we now have ∆(ε) > 2 ε (1.14915) > 2 2 ≤ 0.574575ε . 4. Notice that we used the inequality g 2 g ∞ g 1 with the function g = f ∗ f. This inequality is The constants in the definition (2–4) of K3(x) were nu- sharp exactly when the function g takes only one ˆ merically optimized to minimize K44/3 and otherwise nonzero value (i.e., when g is an nif), but the con- have no special significance. The definition of K3(x)is volution f ∗ f never behaves that way. Perhaps certainly not obvious, and there are much simpler kernels for these autoconvolutions, an analogous inequality that do give nontrivial bounds. In Section 2.7 below, we with a stronger constant than 1 could be established. indicate the experiments that led to our choice. Unfortunately, we have not been able to realize any We note that the function success with this idea, although we believe Conjec- ture 2.9 below. If true, the conjecture implies the √ 4/π − 1 1 2 , ˆ ˆ − volution in d dimensions. 1. Instead of considering the sum j f(j)K( j)asa whole, we separate the central terms from the tails Proposition 2.10. Let K be any continuous function on and establish inequalities that depend on the two d Td satisfying K(¯x) ≥ 1 when x¯ ∈ − 1 , 1 ,andletf in distinct ways. This generalized form of the above 2h 2h − 1 1 d argument is expounded in the next section. The suc- be a pdf supported on 2h , 2h .Then cess of this generalization relies on certain inequal- ∗h ≥ ∗h2 ≥ˆ −2h ities restricting the possible values of these central f ∞ f 2 K 2h/(2h−1). coefficients; establishing these restrictions is the goal d of Sections 2.5 and 2.6. The final lower bound de- Every subset of [0, 1] with measure ε contains a symmet- d 2 rived from these methods is given in Section 2.6. ric subset with measure (0.574575) ε . 152 Experimental Mathematics, Vol. 16 (2007), No. 2 Proof: The proof proceeds as above, with the conjugate have exponents 2h , 2h in place of 4 , 4 , and the ker- 2h−1 3 |M|≤ |fˆ(j)Kˆ (j)| nel function K(x ,x ,...,x )=K(x )K(x ) ···K(x ) 1 2 d 1 2 d |j|≥m ⎛ ⎞ ⎛ ⎞ in place of the kernel function K(x) defined in the proof 1/4 3/4 of Corollary 2.8. The second assertion of the proposition ≤ ⎝ | ˆ |4⎠ ⎝ | ˆ |4/3⎠ follows on taking h =2. f(j) K(j) |j|≥m |j|≥m ⎛ ⎞ 1/4 2.4 The Main Bound ⎝ ˆ 4⎠ ˆ = |f(j)| mK4/3, We now present a more subtle version of Proposition 2.7. |j|≥m Recall that the notation nap was defined in (2–3). We which we recast in the form also use z to denote the real part of the complex num- ber z. M 4 |fˆ(j)|4 ≥ . ˆ |j|≥m m K 4/3 Proposition 2.11. Let m ≥ 1. Suppose that f is a pdf supported on − 1 , 1 and that K is even, continuous, | ˆ |4 4 4 We add |j| With x := fˆ(j), the bound of Proposition 2.11 be- ∗ 2 | ˆ |4 j f f 2 = f(j) (2–5) comes j∈Z m−1 4 m−1 m−1 − Kˆ − Kˆ j x 4 1 (0) 2 j=1 ( ) j 4 M 4 1+ +2 x . ≥ 1+ +2 | fˆ(j)| . ˆ j ˆ m K 4/3 j=1 m K 4/3 j=1 This is a quartic polynomial in the xj, and consequently it is not difficult to minimize, giving an absolute lower ∗ 2 Proof: The equality follows from Parseval’s formula bound on f f 2. This minimum occurs at 1/3 j−1 (Kˆ (j)) 1 − Kˆ (0) − 2 Kˆ (i)xi ∗ 2 | ∗ |2 | ˆ |4 i=1 f f 2 = f f(j) = f(j) . x = , j ˆ 4/3 j j j K 4/3 where (Kˆ (j))1/3 is the real cube root of Kˆ (j). A substi- As in the proof of Proposition 2.7, we have tution and simplification of the resulting expression then yields ⎧ ⎫ 1= f(x)K(x) dx = fˆ(j)Kˆ (−j) ⎨ m−1 − ⎬ 1 − Kˆ (0) − 2 Kˆ (j)x 4 m1 j j=1 j 4 min 1+ +2 xj xj ∈R ⎩ ˆ ⎭ = fˆ(j)Kˆ (−j)+ fˆ(j)Kˆ (−j). m K 4/3 j=1 |j| (the details of this calculation are given in Section 2.7), ˆ ˆ which we can also write as M = |j|≥m f(j)K(j). Tak- so that Proposition 2.11, by itself, does not give a bound ∗ 2 ing absolute values and applying H¨older’s inequality, we on f f 2 different from that given by Proposition 2.7. Martin and O’Bryant: The Symmetric Subset Problem in Continuous Ramsey Theory 153 ¯ However, we shall obtain additional information on We say that f is more focused than f (and f is less ˆ ∗ ¯ ∈ 1 ∈ T f(j) in terms of f f ∞ in Section 2.6 below, and this focused than f) if for all z 0, 2 and all r we have information can be combined with Proposition 2.11 to r+z z provide a stronger lower bound on f ∗ f∞ than that f ≤ f.¯ given by Proposition 2.7. r−z −z For example, f sdr is more focused than f. In fact, we − 1 1 Corollary 2.12. Let f be a pdf supported on 4 , 4 ,and introduce this terminology because it refines the notion ˆ set x1 := f(1).Then of symmetric decreasing rearrangement in a way that is useful for us. To give another example, if f is a non- ∗ 2 ≥ | ˆ |4 f f 2 f(j) negative function, set f¯ to be f∞ times the indicator ∈Z j function of the interval − 1 , 1 ; then f¯ is more 2 f ∞ 2 f ∞ ≥ 1+2x4 +(1.53890149 − 2.26425375x )4. 1 1 focused than f. Proof: Set Lemma 2.13. Let u(x) be a symmetric decreasing func- ¯ ¯ ⎧ tion, and let h, h be pdfs with h more focused than h. ⎪1if|x|≤ 1 , Then for all r ∈ T, ⎨⎪ 4 0.546335 K (x)= − − 1 − 1.61707 5 ⎪1 1 4( 2 x) h(x − r)u(x) dx ≤ h¯(x)u(x) dx. ⎩⎪ 1 | |≤ 1 if 4 < x 2 . Denote by K6(x) the even piecewise linear function with Proof: Without loss of generality we may assume that corners at r = 0, since if h¯(x) is more focused than h(x), then it is − 1 1 t 1 t also more focused than h(x r). Also, without loss of (0, 1), , 1 , + ,K + , ¯ 4 4 4 × 104 5 4 4 × 104 generality we may assume that h, h are continuous and strictly positive on T, since any nonnegative function in 4 1 1 t =1,...,10 . We find (using Proposition 2.16) L can be L -approximated by such functions. ˆ . ˆ . z ¯ z ¯ that K6(0) =0.631932628, K6(1) =0.270776892, and Define H(z)= −z h(t) dt and H(z)= −z h(t) dt,so ˆ . 1 ¯ 1 2 K6 4/3 =0.239175395. Apply Proposition 2.11 with that H( 2 )=H( 2 ) = 1, and note that the more-focused ≤ ¯ ∈ 1 m = 2 to finish the proof. hypothesis implies that H(z) H(z) for all z 0, 2 . Now, h is continuous and strictly positive, which implies 1 2.5 Some Useful Inequalities that H is differentiable and strictly increasing on 0, 2 −1 Hardy, Littlewood, and P´olya [Hardy et al. 88] call a since H (z)=h(z)+h(− z). Therefore H exists as 1 function u(x) symmetric decreasing if u(x)=u(−x)and a function from [0, 1] to 0, 2 . Similar comments hold −1 u(x) ≥ u(y) for all 0 ≤ x ≤ y, and they call for H¯ . Since H ≤ H¯ , we see that H¯ −1(s) ≤ H−1(s) for all f sdr(x):=inf{y : λ ({t: f(t) ≥ y}) ≤ 2|x|} s ∈ [0, 1]. Then, since H−1(s)andH−1(s) are positive and u is decreasing for positive arguments, we conclude the symmetric decreasing rearrangement of f. For exam- that u(H−1(s)) ≤ u(H¯ −1(s)), and so ple, if f is the indicator function of a set with measure µ, sdr 1 1 then f is simply the indicator function of the interval u(H−1(s)) ds ≤ u(H¯ −1(s)) ds. (2–6) − µ µ ( 2 , 2 ). Another example is any function f defined on 0 0 − 2a an interval [ a, a] that is periodic with period n , where On the other hand, making the change of variables s = n is a positive integer, and that is symmetric decreasing H(t), we see that − a a sdr x on the subinterval n , n ; then f (x)=f( n ) for all −1 1 1 1 H (1) x ∈ [−a, a]. In particular, on the interval − , ,we −1 4 4 u(H (s)) ds = u(t)H (t) dt sdr have cos (2πjx) = cos(2πx) for any nonzero integer j. 0 0 We shall need the following result [Hardy et al. 88, The- 1/2 = u(t)(h(t)+h(−t)) dt orem 378]: 0 ≤ sdr sdr = u(t)h(t) dt, f(x)u(x) dx f (x)u (x) dx. T 154 Experimental Mathematics, Vol. 16 (2007), No. 2 1 ¯ −1 ≥ 2 ≤ ≤ since u is symmetric. Similarly, 0 u(H (s)) ds = Proposition 2.15. ∆(ε) 0.591389ε for all 0 ε 1. ¯ T u(t)h(t) dt, and so inequality (2–6) becomes u(t)h(t) dt ≤ u(t)h¯(t) dt as desired. The gist of the proof of Proposition 2.15 is that if f ∗ f∞ is small, then fˆ(1) is small by Lemma 2.14, ∗ 2 2.6 The Full Bound and so f f 2 is not very small by Corollary 2.12, whence To use Proposition 2.11 to bound ∆(ε), we need to de- f ∗ f∞ is not small. If f ∗ f∞ < 1.182778, then we velop a better understanding of the central Fourier coef- get a contradiction. ficients fˆ(j) for small j. In particular, we wish to apply Proof: Let f be a pdf supported on − 1 , 1 , and assume Proposition 2.11 with m = 2, i.e., we need to develop the 4 4 that connections between f ∗ f∞ and the real part of the ˆ f ∗ f∞ < 1.182778. (2–7) Fourier coefficient f(1). We turn now to bounding |fˆ(j)| in terms of f ∗ f∞. ˆ − 1 1 Set x1 := f(1). Since f is supported on 4 , 4 ,wesee ∗ The guiding principle is that if f f is very concentrated that x1 > 0. By Lemma 2.14, then f ∗ f∞ will be large, and if f ∗ f is not very con- centrated then |fˆ(j)| will be small. Green [Green 01, 0 K(x):=αK1(x)+(1− α)K2(x). Proposition 2.16. Let T be a positive integer, n a nonneg- We have ative integer, and p ≥ 1 a real number. For each integer ≤ ≤ 1 t 0 t T , define xt := 4 + 4T ,andletyt be an arbitrary Kˆ p =(α +(1− α)Kˆ (0))p +(1− α)p Kˆ p p 2 1 p real number, except that y0 =1.Let K(x) be the even − − p − p T 1 =(1 (1 α)M) +(1 α) N, (2–10) function on that is linear on 0, 4 and on each of the intervals [xt−1,xt](1≤ t ≤ T ), satisfying K(0) = 1 and where we have set M := 1 − Kˆ (0) and N := Kˆ p. 2 1 p K(xt)=yt (0 ≤ t ≤ T ).Then Taking the derivative with respect to α, we obtain 1/p nKˆ p = (2Λp(dA) · z) , p−1 − p−1 1 − − p(1 α) M − M N , 1 α where d is the T -dimensional vector d =(y1 − y0,y2 − − × q/p y1,...,yT yT −1), A is the T 4T matrix whose (t, k)th − M 1 the only root of which is α =1 M q +N q/p (where p + component is 1 q = 1). It is straightforward (albeit tedious) to check by substituting α into the second derivative of the expression Atk = cos(2π(n + k − 1)xt) − cos(2π(n + k − 1)xt−1), (2–10) that this value of α yields a local maximum for and z is the 4T -dimensional vector Kˆ p. The maximum value attained is then calculated p 1−p to equal N M q + N q/p , which is easily computed z 2 −p j j+1 j+4T −1 =(8Tπ ) ζ(2p, 4T ),ζ(2p, 4T ),...,ζ(2p, 4T ) . from the known function K2. 4 Notice that when p = 3 (so q = 4), applying Proposi- tion 2.7 with our optimal function K yields Proof: Note that −3 ∗ 2 ≥ˆ −4 4 3 −1/3 f f 2 K 4/3 = N M + N 1/2 Kˆ (−j)=Kˆ (j)= K(u) cos(2πju) du 4 3 ˆ 4 M + N (1 − K2(0)) −1/2 = =1+ , 3 4 1/4 N 1Kˆ 4/3 =2 cos(2πju) du 0 whereupon we recover the conclusion of Proposition 2.11 T xt with m =1. +2 (mtu + bt) cos(2πju) du, Haar wavelets provide a convenient basis for t=1 xt−1 2 − 1 1 L 2 , 2 . We have numerically optimized the co- efficients in various spaces of potential kernel functions K spanned by short sums of Haar wavelets to minimize where mt and bt are the slope and y-intercept of the K4/3 within those spaces. The resulting functions are line going through (xt−1,yt−1) and (xt,yt). If we define 156 Experimental Mathematics, Vol. 16 (2007), No. 2 K(x) K(x) 1 1 N = 1 N = 3 0.5 0.5 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 ‒0.5 ‒0.5 ‒1 ‒1 ‒1.5 ‒1.5 K(x) K(x) 1 1 0.5 N = 7 0.5 N = 15 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 ‒0.5 ‒0.5 ‒1 ‒1 ‒1.5 ‒1.5 K(x) K(x) 1 1 N = 31 N = 63 0.5 0.5 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 ‒0.5 ‒0.5 ‒1 ‒1 ‒1.5 ‒1.5 K(x) N = 127 FIGURE 3. Optimal kernels generated by Haar wavelets. 2 2 π j ˆ The first term of this expression is C(j):= 2T K(j), then integrating by parts we have πj T π j m x b πjx sin 2 + ( t t + t) sin(2 t) 1/4 2T π2j2 1 t=1 C(j)= sin(2πju) T 2πj − (mtxt−1 + bt) sin(2πjxt−1) 0 xt 2 2 T π j mtu + bt T + sin(2πju) πj j T 2πj = sin π 2 + (mtxt + bt) sin(2πjxt) t=1 2T xt−1 t=1 xt π2j2 T m T−1 + t cos(2πju) . (2–11) − T (2πj)2 (mt+1xt + bt+1) sin(2πjxt) . t=1 xt−1 t=0 Martin and O’Bryant: The Symmetric Subset Problem in Continuous Ramsey Theory 157 1 1 Since mt+1xt+bt+1 = yt = mtxt+bt and x0 = 4 , xT = 2 , In performing these numerical optimizations, we have this entire expression is a telescoping sum whose value is found that “good” kernels K(x) ∈ K have a very negative 1 + zero. Equation (2–11) thus becomes slope at x = 4 . See Figure 3, for example, where the “N =2j − 1” picture denotes the step function K for 2 2 T xt π j mt which Kˆ is minimal among all step functions whose C(j)= cos(2πju) 4/3 T (2πj)2 1 Z t=1 xt−1 discontinuities all lie within the set 2j+2 . T Viewing graphs of these numerically optimized kernels = (y − y − ) (cos(2πjx ) − cos(2πjx − )) suggests that functions of the form t t 1 t t 1 t=1 | |≤ 1 (2–12) 1, x 4 , Kd1,d2 (x)= − − 1 − d1 d2 1 | |≤ 1 1 (1 (4( 2 x)) ) , 4 < x 2 , yt−yt−1 using mt = − =4T (yt − yt−1). Each xt is rational xt xt−1 −∞ 1 + and can be written with denominator 4T , so we see that which have slope at x = 4 , may be very good. 1 3 the sequence of normalized Fourier coefficients C(j)is (Note that the graph of K2,1/2(x) between 4 and 4 is periodic with period 4T . the lower half of an ellipse.) More good candidates are ˆ functions of the form We proceed to compute n K p with n positive and p ≥ 1: | |≤ 1 1, x 4 , K (x)= e1 e3 ∞ e1,e2,e3 2 −1 (1−2x) 1 | |≤ 1 p tan − e2 , < x , p p π (4x 1) 4 2 nKˆ p = |Kˆ (j)| =2 |Kˆ (j)| |j|≥n j=n where e1,e2,ande3 are positive. We have used a func- ∞ p tion of the form Kd1,d2 in the proof of Corollary 2.12 and 2T =2 C(j) a function of the form Ke1,e2,e3 in the proof of Corol- π2j2 j=n lary 2.8. ∞ 2T p |C(j)|p =2 . ∆(ε) ε = 1 π2 j2p 2.8 A Lower Bound for around 2 j=n We begin with a fundamental relationship between fˆ(1) Because of the periodicity of C(j), we may write this as and fˆ(2). p 1 1 nKˆ p − Lemma 2.17. Let f be a pdf supported on 4 , 4 .Then ⎛ ⎞ − ∞ 2 2T p n+4T 1 2 fˆ(1) − 1 ≤ fˆ(2) ≤ 2( fˆ(1)) − 1. =2 ⎝ |C(j)|p (4Tr+ j)−2p⎠ π2 j=n r=0 ⎛ ⎞ Proof: To prove the first inequality, set L (x)= p n+4T −1 b 2T − ≥ =2 ⎝ |C(j)|pζ 2p, j ⎠ , b cos(2πx) cos(4πx) (with b 0) and observe that for (4T )2π2 4T − 1 ≤ ≤ 1 ≤ b2 j=n 4 x 4 ,wehaveLb(x) 1+ 8 .Thus (2–13) 2 b2 ≥ ˆ ˆ − 1+ 8 f(x)Lb(x) dx = f(j)Lb( j) which concludes the proof. j=−2 Proposition 2.16 is useful in two ways. The first is = b fˆ(1) − fˆ(2). that only d depends on the chosen values y . That is, the 2 t Rearranging, we arrive at fˆ(2) ≥ b( fˆ(1)) − 1 − b . vector z and the matrix A may be precomputed (assum- 8 Setting b =4 fˆ(1), we find that fˆ(2) ≥ 2( fˆ(1))2 − 1. ing that T is reasonably small), enabling us to compute As for the second inequality, since nKˆ p quickly enough as a function of d to numerically optimize the yt. The second use is through (2–13). For L2(x) := 2 cos(2πx) − cos(4πx) a given K,wesety = K(x ), whereupon C(j) is com- t t is at least 1 for − 1 ≤ x ≤ 1 ,wehave puted for each j using the formula in (2–12). Thus we 4 4 2 can use (2–13) to compute nKˆ1p with arbitrary accu- 1 ≤ f(x)L(x) dx = fˆ(j)Lˆ(−j) racy, where K1 is almost equal to K. We have found j=−2 that with T = 10000 one can generally compute Kˆ n 1 p ˆ − ˆ quickly. =2( f(1)) f(2). 158 Experimental Mathematics, Vol. 16 (2007), No. 2 Rearranging, we arrive at fˆ(2) ≤ 2( fˆ(1)) − 1. interval 2 If we define F := max{ fˆ(1), − fˆ(2)}, then 3. UPPER BOUNDS FOR ∆(ε) 3.1 Inequalities Relating ∆(ε) and R(g, n) f(x)Lb(x) dx = fˆ(2) − b fˆ(1) ≥−(b +1)F A symmetric set consists of pairs (x, y) all with a fixed x+y × on the one hand, and midpoint c = 2 . If there are few pairs in E E with a given sum 2c, then there will be no large symmetric sdr sdr subset of E with center c. We take advantage of the f(x)Lb(x) dx ≤ f (x)Lb (x) dx constructions of large integer sets whose pairwise sums ε/4 2 sdr repeat at most g times to construct large real subsets of = ε Lb (x) dx −ε/4 [0, 1) with no large symmetric subsets. More precisely, a set S of integers is called a B∗[g] set if for any given m sdr on the other, where Lb (x) is the symmetric decreasing there are at most g ordered pairs (s1,s2) ∈ S × S with − 1 1 rearrangement of Lb(x) on the interval 4 , 4 .Thus s1 + s2 = m. (In the case g = 2, these are better known as Sidon sets.) Define − ε/4 ≥ 1 2 sdr F Lb (x) dx. | | ⊆{ } ∗ b +1ε −ε/4 R(g,n) := max S : S 1, 2,...,n ,Sis a B [g]set . (3–1) The right-hand side may be computed explicitly as a function of ε and b and then the value of b chosen in Proposition 3.1. For any integers n ≥ g ≥ 1, we have terms of ε to maximize the resulting expression. One R(g,n) ≤ g 5 ∆ n n . finds that for ε< 8 , the optimal choice of b lies in the Martin and O’Bryant: The Symmetric Subset Problem in Continuous Ramsey Theory 159 Proof: Let S ⊆{1, 2,...,n} be a B∗[g] set with |S| = Proof: Combining the hypothesized lower bound ∆(ε) ≥ R(g,n). Define δε2 with Proposition 3.1, we find that s − 1 s 2 A(S):= , , (3–2) R(g,n) ≤ R(g,n) ≤ g n n δ ∆ , s∈S n n n √ and note that A(S) ⊆ [0, 1] and that the measure of A(S) which is equivalent to R(g,n) ≤ δ−1/2 gn. is exactly R(g,n)/n. Thus it suffices to show that the g We have been unable to prove or disprove that largest symmetric subset of A(S) has measure at most n . Notice that the set A(S) is a finite union of intervals, − R(g,n) ∆(ε) 1/2 and so the function λ A(S) ∩ (2c − A(S)) , which gives √ lim lim = inf 2 , the measure of the largest symmetric subset of A(S) with g→∞ n→∞ gn 0<ε<1 ε center c, is piecewise linear. (Figure 4 contains a typical i.e., that Corollary 3.2 is best possible as g →∞.At example of the set A(S) portrayed in dark gray below the any rate, for small g it is possible to do better by taking c-axis, together with the function λ A(S) ∩ (2c − A(S)) advantage of the shape of the set A(S) used in the proof shown as the upper boundary of the light-gray region of Proposition 3.1. This is the subject of the companion above the c-axis, for S = {1, 2, 3, 5, 8, 13}.) Without loss paper [Martin and O’Bryant 07] of the authors. of generality, therefore, we may restrict our attention to Proposition 3.1 provides a one-sided inequality linking those symmetric subsets of A(S) whose center c is the ∆(ε)andR(g,n). It will also be useful for us to prove a midpoint of endpoints of any two intervals s−1 , s .In n n theoretical result showing that the problems of determin- other words, we may assume that 2nc ∈ Z. ing the asymptotics of the two functions are, in a weak Suppose u and v are elements of A(S) such that u+v = 2 sense, equivalent. In particular, the following proposition c. Write u = s1 − 1 + x and v = s2 − 1 + y for integers n 2n n 2n implies that the trivial lower bound ∆(ε) ≥ 1 ε2 and the s ,s ∈ S and real numbers x, y satisfying |x|, |y| < 1 . √ 2 1 2 2n trivial upper bound R(g,n) ≤ 2gn are actually equiva- (We may ignore the possibility that nu or nv is an integer, lent. Further, any nontrivial lower bound on ∆(ε) gives since this is a measure-zero event for any fixed c.) Then a nontrivial upper bound on R(g,n), and vice versa. 2nc = n(u + v)=s + s − 1+n(x + y), and since 1 2 2nc, s1,ands2 are all integers, we see that n(x + y)is Proposition 3.3. ∆(ε)=inf g : n ≥ g ≥ 1, R(g,n) ≥ ε also an integer. But |n(x + y)| < 1, so x + y =0and n n for all 0 ≤ ε ≤ 1. s1 + s2 =2nc +1. Since S is a B∗[g] set, there are at most g solutions Proof: That ∆(ε) is bounded above by the right-hand (s1,s2) to the equation s1 + s2 =2nc + 1. If it happens s1−1 s1 side follows immediately from Proposition 3.1 and the that s1 = s2, the interval n , n (a set of measure 1 ) is contributed to the symmetric subset with center fact that ∆ is an increasing function. For the comple- n ⊆ s1−1 s1 ∪ s2−1 s2 mentary inequality, let S [0, 1) with λ(S)=ε. Basic c. Otherwise, the set n , n n , n (a set of measure 2 ) is contributed to the symmetric subset with Lebesgue measure theory tells us that given any η>0, n there exists a finite union T of open intervals such that center c, but this counts for the two solutions (s1,s2) λ(S⊕T ) <η, and it is easily seen that T can be chosen to and (s2,s1). In total, then, the largest symmetric subset g meet the following criteria: T ⊆ [0, 1), the endpoints of having center c has measure at most n . This establishes the proposition. the finitely many intervals that T comprises are rational, and λ(T ) >ε. Choosing a common denominator n for Using Proposition 3.1, we can translate lower bounds the endpoints of the intervals that T comprises, we may m−1 m on ∆(ε) into upper bounds on R(g,n), as in Corollary 3.2. write T = m∈M n , n (up to a finite set of points) for some set of integers M ⊆{1,...,n}; most likely we ≤ 2 ≤ Corollary√ 3.2. If δ inf0<ε<1 ∆(ε)/ε , then R(g,n) have greatly increased the number of intervals of which δ−1/2 gn for all n ≥ g ≥ 1. T consists by writing it in this manner, and M contains many consecutive integers. Let g be the maximal num- We remark that we may take δ =0.591389 by Theo- ber of solutions (m1,m2) ∈ M × M to m1 + m2 = k as ≤ ∗ rem 1.2(ii),√ and so this corollary implies that R(g,n) k varies over all integers, so that M is a B [g] set and 1.30036 gn. This improves the previous best bound on thus |M|≤R(g,n) by the definition of R. It follows that R(g,n) (given in [Green 01]) for g ≥ 30 and n large. ε<λ(T )=|M|/n ≤ R(g,n)/n.Now,T is exactly the 160 Experimental Mathematics, Vol. 16 (2007), No. 2 1 2 3 5 8 13 FIGURE 4. A(S), and the function λ(A(S) ∩ (2c − A(S))), with S = {1, 2, 3, 5, 8, 13}. g set A(M) as defined in (3–2); hence D(T )= n as we saw Lemma 3.4. Let p1,...,pn be real numbers in the range in the proof of Proposition 3.1. Therefore by Lemma 2.4, [0, 1],andsetp =(p1 + ··· + pn)/n. Define mutually g independent random variables X1,...,Xn such that Xi D(S) ≥ D(T ) − 2η = − 2η − n takes the value 1 pi with probability pi and the value − − g R(g,n) pi with probability 1 pi (so that the expectation of ≥ inf : n ≥ g ≥ 1, ≥ ε − 2η. ··· n n each Xi is zero), and define X = X1 + + Xn.Then for any positive number a, Taking the infimum over appropriate sets S and noting −a2 a3 that η>0 was arbitrary, we derive the desired inequality Pr[X>a] < exp + ≥ g ≥ ≥ R(g,n) ≥ 2pn 2p2n2 ∆(ε) inf n : n g 1, n ε . and 3.2 Probabilistic Constructions of −a2 ∗[ ](mod ) Pr[X<−a] < exp . B g n Sets 2pn We begin by considering a modular version of B∗[g]sets. AsetS is a B∗[g] (mod n) set if for any given m there ∈ × ≡ are at most g ordered pairs (s1,s2) S S with s1 +s2 Proof: These assertions are Theorems A.11 and A.13 of m (mod n) (equivalently, if the coefficients of the least- [Alon and Spencer 00]. s 2 n − degree representative of s∈S z (mod z 1) are { } ∗ bounded by g). For example, the set 0, 1, 2, 4 is a B [3] Lemma 3.5. Let p1,...,pn be real numbers in the range { } ∗ (mod 7) set, and 0, 1, 3, 7 is a B [2] (mod 12) set. Note [0, 1],andsetE = p1 +···+pn. Define mutually indepen- ≡ { } that 7 + 7 1 + 1 (mod 12), so that 0, 1, 3, 7 is not dent random variables Y1,...,Yn such that Yi takes the a “modular Sidon set” as defined by some authors, for value 1 with probability pi and the value 0 with probability example [Graham and Sloane 80] or [Guy 94, Problem 1 − pi, and define Y = Y1 + ···+ Yn (so that the expec- 2 C10]. −a tation of Y equals E). Then Pr[Y>E+ a] < exp 3E Just as we defined R(g,n) to be the largest possi- for any real number 0 Proposition 3.6. For every 0 <ε≤ 1, there is a se- Define ∆T(ε) to be the supremum of those real num- quence of ordered pairs (nj ,gj) of positive integers such bers δ such that every subset of T with measure ε has that C(gj ,nj ) ε and gj ε2. a subset with measure δ that is fixed by a reflection nj nj t → c − t. The function ∆T(ε) stands in relation to Proof: Let n be an odd integer. We define a random C(g,n)as∆(ε) stands to R(g,n). However, it turns out subset S of {1,...,n} as follows: for every 1 ≤ i ≤ n,let that ∆T is much easier to understand. Yi be 1 with probability ε and 0 with probability 1 − ε { T with the Yi mutually independent, and let S := i: Yi = Corollary 3.7. Every subset of with measure ε contains } | | n asymmetricsubsetwithmeasureε2, and this is best pos- 1 . We see that√ S = i=1 Yi has expectation E = εn. Setting a = εn log 4, Lemma 3.5 gives sible for every ε: 2 1 ∆T(ε)=ε Pr |S| <εn− εn log 4 < . 2 for all 0 ≤ ε ≤ 1. Now for any integer k, define the random variable Proof: In the proof of the trivial lower bound for ∆(ε) { ≤ ≤ ≡ Rk := # 1 c, d n: c + d k (mod n), (Lemma 2.2), we saw that every subset of [0, 1] with mea- Y = Y =1} c d sure ε contains a symmetric subset with measure at least 1 ε2. The proof is easily modified to show that every sub- = YcYd , 2 T c+d≡k (mod n) set of with measure ε contains a symmetric subset with 2 2 measure ε . This shows that ∆T(ε) ≥ ε for all ε.On so that R is the number of representations of k (mod n) k the other hand, the proof of Proposition 3.1 is also easily C(g,n) g as the sum of two elements of S. Observe that Rk is the modified to show that ∆T ≤ , as is the proof − n n sum of n 1 random variables taking the value 1 with of Lemma 2.5 to show that ∆T is continuous. Then, by 2 probability ε and the value 0 otherwise, plus one random virtue of Proposition 3.6 and the monotonicity of ∆T,we −1 ≡ ≡ 2 variable (corresponding to c d 2 k (mod n)) tak- have ∆T(ε) ≤ ε . ing the value 1 with probability ε and the value 0 other- − 2 wise. Therefore the expectation of Rk is E =(n 1)ε +ε. 3.3 Probabilistic Constructions of B∗[g] Sets Setting a = 3((n − 1)ε2 + ε) log 2n, and noting that We can use the probabilistic methods employed in Sec- a ∗ probability, S is a large B [g] set with g not much bigger Schinzel and Schmidt [Schinzel and Schmidt 02] con- 1 than γ. jectured that among all pdfs supported on 0, 2 ,the The expected size of S is function √1 ,x∈ [0, 1 ], γ f(x)= 2x 2 E := p = 1+ (3–4) 0, otherwise, 0 j πj 1≤j≤n 1≤j<γ/π γ/π≤j≤n has the property that f ∗ f∞ is minimal. We have γ n γ γn γ = + dt + O(1) = 2 − + O(1). ⎧ π πt π π ⎪ π ,x∈ [0, 1 ], γ/π ⎨ 2 √ 2 √ f ∗ f(x)= π − − ∈ 1 ⎪ 2 2 arctan 2x 1,x [ 2 , 1], If we set a0 := 2E0 log 3, then Lemma 3.5 tells us that ⎩ 0, otherwise, −a2 1 | | − 0 π Pr[ S 11 ≤ ≤ 11 ≤ for 16 ε 1, it suffices to prove simply that ∆ 16 We can immediately deduce two nice consequences of 3 8 . this proposition. For any positive integer g, it was shown by the authors ∆(ε) + [Martin and O’Bryant 06, Theorem 2(vi)] that Corollary 3.13. limε→0 ε2 exists. − ≥ R(g,3g g/3 +1) g +2 g/3 + g/6 . Proof: This follows from the fact that the function ∆(ε)/ε2 is increasing and bounded below by 1 on (0, 1] We combine this with Proposition 3.1 and the monotonic- 2 by the trivial lower bound (Lemma 2.2). ityof∆toseethat g R(g,3g −g/3 +1) ∆(ε) ≤ 96 ε2 for 0 ≤ ε ≤ 11 . ≥ Corollary 3.14. 121 16 − ∆ − 3g g/3 +1 3g g/3 +1 g +2g/3 + g/6 Proof: This follows from the value ∆ 11 = 3 calculated ≥ ∆ . 16 8 3g −g/3 +1 in Proposition 3.10 and the fact that the function ∆(ε)/ε2 is increasing. Since ∆ is continuous by Lemma 2.5, we may take the →∞ 11 ≤ 3 The corollary above proves part (iv) of Theorem 1.2, limit of both sides as g to obtain ∆ 16 8 as desired. leaving only part (v) yet to be established. The following proposition finishes the proof of Theorem 1.2. Remark 3.11. In light of the Lipschitz condition |∆(x) − ∆(ε) √π 2 ≤ for all 0 <ε≤ 1. ∆(y)|≤2|x − y|, the lower bound ∆(ε) ≥ 2ε − 1for Proposition 3.15. ε (1+ 1−ε)2 ≤ all 0 <ε 1 also follows easily from the trivial value √ ∆(1) = 1. Proof: Define α := 1− 1 − ε,sothat2α−α2 = ε.Ifwe 2 set γ = πα n in the proof of Proposition 3.8, then√ the sets Proposition 3.12. The function ∆(ε)/ε2 is increasing on constructed are B∗[g] sets with g = πα2n + O( n log n) (0, 1]. and have size at least πα2n2 πα2n Proof: The starting point of our proof is the inequality E0 − a0 =2 − + O(1 + a0) π π [Martin and O’Bryant 06, Theorem 2(v)] √ =(2α − α2)n + O((γn)1/4)=εn + O( n) ≤ R(g,x)C(h, y) R(gh,xy). from (3–4). With the monotonicity of ∆(ε) and Proposition 3.1, this Therefore, for these values of g and n, √ gives R(g,n) εn + O( n) ∆ ≥ ∆ → ∆(ε) n n R(g,x) C(h, y) R(gh,xy) gh ∆ ≤ ∆ ≤ . as n goes to infinity, by the continuity of ∆. On the other x y xy xy hand, we see by Proposition 3.1 that R(gi,xi) √ Choose 0 <ε<ε0.Letgi,xi be such that → ε0 2 xi −2 R(g,n) g πα n + O( n log n) gi ε ∆ ≤ = and → ∆(ε ), which is possible by Proposition 3.3. 2 2 xi 0 n ε n ε n By Proposition 3.6, we may choose sequences of integers πα2 C(hj ,yj ) ε hj ε 2 log n hj and yj such that and as = + O 2 yj ε0 yj ε0 (2α − α2)2 ε n j →∞. This implies π → √π = + o(1) 2 (2 − α)2 − 2 R(gi,xi) C(hj,yj ) gi hj ε (1 + 1 ε) ε and ∆(ε0) , xi yj xi yj ε0 as n goes to infinity. Combining these two inequalities ∆(ε) √π yields 2 ≤ as desired. so that, again using the monotonicity and continuity ε (1+ 1−ε)2 of ∆, 4. SOME REMAINING QUESTIONS ε2 g h R(g ,x ) C(h ,y ) ∆(ε ) i j ≥ ∆ i i j j ∆(ε) 0 2 We group the problems in this section into three cate- ε0 xiyj xi yj gories, although some problems do not fit clearly into ∆(ε) ∆(ε0) as j →∞. This shows that 2 ≤ 2 as desired. any of the categories and others fit into more than one. ε ε0 164 Experimental Mathematics, Vol. 16 (2007), No. 2 4.1 Properties of the Function ∆(ε) 4.2 Artifacts of Our Proof 2 The first open problem on the list must of course be the Let K be the class of functions K ∈ L (T) satisfying ≤ ≤ ≥ − 1 1 ˆ exact determination of ∆(ε) for all values 0 ε 1. In K(x) 1on 4 , 4 . How small can we make K p for ≤ ≤ 4 the course of our investigations, we have come to believe 1 p 2? We are especially interested in p = 3 , but a the following assertion. solution for any p may be enlightening. To give some perspective to this problem, note that a − π 2 ≤ ˆ Conjecture 4.1. ∆(ε) = max 2ε 1, 4 ε for all 0 trivial upper bound for infK∈K K p can be found by tak- ε ≤ 1. ing K to be identically equal to 1, which yields Kˆ p =1. One can find functions that improve upon this trivial Notice that the upper bounds given in Theorem 1.2 choice; for example, the function K defined in (2–4) is an ˆ . are not too far from this conjecture, the difference be- example in which K4/3 =0.96585. On the other hand, 96 . π . p tween the constants 121 =0.79339 and 4 =0.78540 in since the -norm of a sequence is a decreasing function the middle range for ε being the only discrepancy. In of p, Parseval’s identity immediately gives us the lower fact, we believe it might be possible to prove that the ex- bound pression in Conjecture 4.1 is indeed an upper bound for 1/4 1/2 ∆(ε) by a more refined application of the probabilistic 2 Kˆ p ≥Kˆ 2 = K2 ≥ 1 dt method employed in Section 3.3. The key would be to −1/4 show that the various events R >γ+1+a are more or 1 . k = √ =0.70711, less independent of one another (as it stands, we have to 2 assume the worst—that they are all mutually exclusive— and of course √1 is the exact minimum for p =2. in obtaining our bound for the probability of obtaining a 2 “bad” set). We remark that Proposition 2.7 and the function b(x) There are some intermediate qualitative results about defined after the proof of Corollary 2.8 provide a stronger ≤ ≤ 4 the function ∆(ε) that might be easier to resolve. It lower bound for 1 p 3 . By direct computation we ∗ 2 seems likely that ∆(ε) is convex, for example, but we have 1.14939 > b b 2, and by Proposition 2.7 we have ∗ 2 ≥ˆ −4 ∈ K have not been able to prove this. A first step toward b b 2 K 4/3 for any K . Together these imply clarifying the nature of ∆(ε) might be to prove that ˆ ≥ˆ 4 that K p K 4/3 > 0.96579. In particular, for p = 3 we know the value of inf ∈K Kˆ to within one part in | − | K 4/3 ∆(x) ∆(y) { } ten thousand. The problem of determining the actual in- | − | max x, y . x y fimum for 1 4.3 The Analogous Problem for Other Sets [Beckner 75] W. Beckner. “Inequalities in Fourier Analysis.” More generally, for any subset E of an abelian group Ann. of Math. 102 (1975), 159–182. endowed with a measure, we can define [Banakh et al. 00] T. Banakh, O. Verbitsky, and Ya. Voro- bets. “A Ramsey Treatment of Symmetry.” Electron. J. Combin. 7:1 (2000), Research Paper 52, 25 pp. ∆E(ε):=inf{D(A): A ⊆ E, λ(A)=ε}, [Chung et al. 02] Fan Chung, Paul Erd˝os, and Ronald Gra- where D(A) is defined in the same way as in (2–1). For ham. “On Sparse Sets Hitting Linear Forms.” In Number example, ∆[0,1](ε) is the function ∆(ε) we have been con- Theory for the Millennium, I, pp. 257–272. Natick, MA: A K Peters, 2002. sidering throughout this paper, and ∆T(ε) was considered in Section 3.2. [Cilleruelo et al. 02] J. Cilleruelo, I. Ruzsa, and C. Trujillo. Most of the work in this paper generalizes easily from “Upper and Lower Bounds for Finite Bh[g] Sequences, g> E =[0, 1] to E =[0, 1]d. We have had difficulties, how- 1.” J. Number Theory 97:1 (2002), 26–34. ever, in finding good kernel functions in higher dimen- [Folland 84] Gerald B. Folland. Real Analysis.NewYork: sions. That is, we need functions K(¯x) such that John Wiley, 1984. [Green 01] Ben Green. “The Number of Squares and Bh[g] 4/3 Kˆ ¯j Sets.” Acta Arithmetica 100:4 (2001), 365–390. ¯j∈Zd [Graham and Sloane 80] R. L. Graham and N. J. A. Sloane. “On Additive Bases and Harmonious Graphs.” SIAM J. is as small as possible, while K(¯x) ≥ 1 if all components Algebraic Discrete Methods 1:4 (1980), 382–404. ofx ¯ are less than 1 in absolute value. This restricts K on 4 [Guy 94] Richard K. Guy. Unsolved Problems in Number one-half of the space in one dimension, one-quarter of the Theory, second edition. New York: Springer-Verlag, 1994. −d space in two dimensions, and only 2 of the space in d [Hardy et al. 88] G. H. Hardy, J. E. Littlewood, and dimensions. For this reason one might expect that better G. P´olya. Inequalities. Cambridge: Cambridge University kernels exist in higher dimensions, but the computational Press, 1988. difficulties have prevented us from finding them. [Martin and O’Bryant 06] Greg Martin and Kevin O’Bryant. “Constructions of Generalized Sidon Sets.” J. Comb. Thy. ACKNOWLEDGMENTS Ser. A 113:4 (2006), 591–607. The authors thank Heini Halberstam for thoughtful readings [Martin and O’Bryant 07] Greg Martin and Kevin O’Bryant. of this manuscript and helpful suggestions. The first author “Upper Bounds for Generalized Sidon Sets.” In prepara- was supported in part by grants from the Natural Sciences tion, 2007. and Engineering Research Council. The second author was [O’Bryant 04] Kevin O’Bryant. “A Complete Annotated Bib- supported in part by the NSF under grant DMS 0202460, by liography of Work Related to Sidon Sequences.” Elec. J. an NSF-VIGRE Fellowship, and by the Research Foundation Combin. Article DS11, 2004. of City University of New York under grant PSCOOC-37-110. [Schinzel and Schmidt 02] A. Schinzel and W. M. Schmidt. “Comparison of L1-andL∞-Norms of Squares of Polyno- REFERENCES mials.” Acta Arithmetica 104:3 (2002), 283–296. [Alon and Spencer 00] Noga Alon and Joel H. Spencer. The [Swierczkowski´ 58] S. Swierczkowski.´ “On the Intersection Probabilistic Method, second edition. New York: Wiley- of a Linear Set with the Translation of Its Complement.” Interscience, 2000. Colloq. Math. 5 (1958), 185–197. Greg Martin, Department of Mathematics, University of British Columbia, Vancouver, BC, V6T 1Z2, Canada ([email protected]) Kevin O’Bryant, The City University of New York, College of Staten Island, 2800 Victory Blvd, Staten Island, NY 10314 ([email protected]), Received November 8, 2005; accepted May 24, 2006.