L. Thomas Ramsey. University of Hawaii. October 28, 1994

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1 PROPORTIONS OF SIDON SETS ARE I 0 SUBSETS L. Thomas Ramsey University of Hawaii October 28, 1994 Abstract. It is proved that proportions of Sidon sets are I 0 subsets of controlled degree. That is, a set E is Sidon if only if, there are r > 0 positive integer n such that, for every finite subset F E, there is H F with the cardinality of H at least r times the cardinality of F N(H) n (N(H) is a measure of the degree of being I 0 ). This paper leaves open David Grow s question of whether Sidon sets are finite unions of I 0 sets. Introduction An I 0 degree, N(E), will be defined below; it is finite if only if E is an I 0 set allows a quantification of being I 0. The purpose of this paper is to prove the following theorem, which offers weak affirmative evidence to David Grow s question: must Sidon sets be finite unions of I 0 sets [G]? Theorem 1. Let Γ be a discrete abelian group. Then E Γ is Sidon if only if, there are some real r > 0 positive integer n such that, for all finite F E, there is some H F for which H r F N(E) n. 1 In what follows, Γ is a discrete, abelian group G its compact dual. M(G) is the Banach algebra of bounded Borel measures on G; M d (G) is the subalgebra of M(G) consisting of discrete measures. For E Γ, B(E) is the Banach algebra of the restrictions to E of Fourier transforms of measures µ M(G); B d (E) consists of the restrictions to E of Fourier transforms of measures µ M d (G). The closure of B d (E) in l (E) is called AP (E) (the almost periodic functions restricted to E). E Γ is said to be Sidon if only if B(E) = l (E) [LR]; E is called an I 0 set if only if AP (E) = l (E) [HR]. The following definition offers a measure of being an I 0 set Mathematics Subject Classification. 43A56. Key words phrases. Sidon, I 0 -set, almost periodic functions. 1 H denotes the cardinality of H. Unless otherwise specified, variables such as n denote positive integers. 1 Typeset by AMS-TEX

2 2 L. THOMAS RAMSEY Definition. Let D(N) denote the set of discrete measures µ on G for which µ = N c j δ tj, j=1 where c j 1 t j G for each j. For E Γ δ R +, let AP (E, N, δ) be the set of f l (E) for which there exists µ D(N) such that f ˆµ E δ. E is said to be I(N, δ) if the unit ball in l (E) is a subset of AP (E, N, δ). The I 0 degree of E, N(E), is defined to be the first N such that E is I(N, 1/2); if no such N exists, N(E) is set equal to. By the following theorem, I 0 sets are exactly those for which N(E) <. The following theorem was known to Kahane, Mèla, Ramsey Wells much earlier, but the authors like Kalton s more recent formulation proof ([Kl],[Kh], [M], [RW]). Theorem 2. For any discrete abelian group Γ E Γ, the following are equivalent: (1) E is an I 0 -set. (2) There is some real δ (0, 1) some N for which E is I(N, δ). (3) There is some real δ (0, 1) some M R + such that, for all f in the unit ball of l (E), there are points g j G complex numbers c j with c j Mδ j for which f = ˆµ E where µ = c j δ gj. (4) For all real δ (0, 1) there is some N for which E is I(N, δ). (5) B d (E) = l (E). Corollary 3. For any discrete abelian group Γ E Γ, if E is I(N, δ) for some real δ (0, 1), then condition (3) holds with M = 1/δ δ 1/N in the role of δ. j=1 Proof. This is implicit in Kalton s proof, made explicit in [R]. One can weaken the conditions of interpolation still attain an equivalent degree for I 0 sets [R]. Definition. Let C 1 C 2 be closed subsets of the complex plane. For E Γ, E is said to be J(N, C 1, C 2 ) if only if, for all F E, there is some µ D(N) such that ˆµ(F ) C 1 ˆµ(E\F ) C 2. When C 1 = {z R(z) δ}, C 2 = {z R(z) δ}, J(N, C 1, C 2 ) is abbreviated as J(N, δ). J(E) is defined to be the first N such that E is J(N, 1/2); if no such N exists, J(E) is set equal to. The next theorem is proved in [R], shows that E is I 0 if only if J(E) <.

3 PROPORTIONS OF SIDON SETS ARE I 0 SUBSETS 3 Theorem 4. The following are equivalent: (1) E is an I 0 set. (2) E is J(N, C 1, C 2 ) for some N disjoint subsets C 1 C 2. (3) For all real δ (0, 1), there is some N such that E is J(N, δ). The next lemma relates J(N, δ) to J(E). Lemma 5. If E is J(N, δ) for some δ (0, 1), then J(E) KN where K = 1/(2δ). Proof. Assume that E is J(N, δ). Then, for any F E, there is some µ D N such that ( γ F ) (R(ˆµ(γ)) δ), ( γ E\F ) (R(ˆµ(γ)) δ). Because K 1/(2δ), Kδ 1/2 thus ) R ( Kµ(γ) 1/2, for γ E, while ) R ( Kµ(γ) 1/2, for γ (E\F ). One can write Kµ as a sum of KN point masses with complex coefficients bounded by 1 in absolute value. Thus E is J(KN, 1/2) J(E) KN. It is readily evident that J(E) N(E). bounded relation between J(E) N(E): In [R], it is proved that there is a Theorem 6. There is a function φ with φ(z + ) Z + such that, for all discrete abelian groups Γ all E Γ, J(E) N(E) φ(j(e)). A key ingredient of the proof of Theorem 1 is this theorem [P]: Theorem 7. E is a Sidon set if only if, there is some δ > 0 with the following property: for every finite A E, there are points g j G, 1 j N with N 2 δ A, such that sup γ(g i ) γ(g j ) δ, for all i j. γ A The last ingredients of the proof are Elton s theorem about sign-embeddings of l n 1 into real Banach spaces [E] Pajor s generalization of Elton s theorem to complex Banach spaces [Pa]. The proof given in this paper does not quote their theorems verbatim; rather, parts of the their proofs are adapted to this situation.

4 4 L. THOMAS RAMSEY Proof of Theorem 1 Sufficiency. Suppose that E Γ has some real r > 0 positive integer N such that, for every finite subset F E, ( H F ) ( H r F N(H) n). Then H is I(n, 1/2). By Corollary 3, condition (3) of Theorem 2 holds with M = 2 δ = (1/2) 1/n. It follows that, for every f in the unit ball of l (E), there is some µ M d (G) such that ˆµ H = f µ Md (G) L = 2 2 j/n <. For all f l (H), there is a constant L which depends only on n such that Since f B(H) f Bd (H), one has j=1 f Bd (H) L f l (H). f B(H) L f l (H). Thus H is a Sidon set with Sidon constant at most L, with L independent of F E. That suffices to make E be Sidon, by Corollary 2.3 of [P]. Necessity. Suppose that E is Sidon. Apply Theorem 7. There is some δ > 0 such that, for all finite F E, there are at least 2 δ F points g j of G such that, for i j, (1) sup γ(g j ) γ(g i ) δ. γ F Necessarily, δ 2. Let F E of cardinality n. Enumerate F as γ 1,..., γ n. Choose p so that τ = 2π/p < δ/2. To be specific, let p = 1 + 4π/δ. Let T denote the unit circle in the complex plane. Partition T into disjoint arcs, T k, 1 k p, of the form T k = { e iθ (k 1)τ θ < kτ }. Let Q = (1 2 δ/2 ) 1. set τ = τ/q. Partition each T k into Q arcs U k,m of the form U k,m = { e iθ (k 1)τ + (m 1)τ θ < (k 1)τ + mτ }, for 1 m Q. Finally, let S 0 denote a set of at least 2 δ F points of G which satisfy inequality (1). Define S i inductively. Let S i k = { g S i 1 γ i (g) T k } S i k,m = { g S i 1 γ i (g) U k,m }.

5 PROPORTIONS OF SIDON SETS ARE I 0 SUBSETS 5 Then S i 1 = p k=1 Si k S i k = Q m=1 Si k,m. There is some m(i, k) such that S i k,m(i,k) Q 1 S i k. So, Let Then By induction one has p k=1 Si k,m(i,k) Q 1 S i 1. S i = S i 1 \ p k=1 Si k,m(i,k). S i (1 Q 1 ) S i 1. S n (1 Q 1 ) n S 0. Note that Q (1 2 δ/2 ) 1 ; consequently, Therefore, (1 Q 1 ) 2 δ/2. S n (1 Q 1 ) n S 0 (2 δ/2 ) n 2 δn = 2 nδ/2. For 1 i n 1 k < p, let I i,k be the arc between U k,m(i,k) U k+1,m(i,k+1). For k = p, let I i,k be the arc between U k,m(i,k) U 1,m(i,1). Necessarily, (2) I i,k { e iθ (k 1)τ + τ θ < (k + 1)τ τ }. The length ( hence the diameter) of each of these arcs is at most (2 2/Q)τ < 2 (δ/2) = δ. It is possible for I i,k =, which will happen when k < p, m(i, k) = Q m(i, k + 1) = 1; it will also happen when k = p, m(i, k) = Q m(i, 1) = 1. Otherwise, e ikτ is in the closure of I i,k : it is in I i,k when m(i, k + 1) > 1 when k = p m(i, 1) > 1. When m(i, k + 1) = 1 I i,k, {e iθ (k 1)τ + (Q 1)τ θ < kτ} I i,k. Likewise, when m(i, 1) = 1 I i,p, {e iθ (p 1)τ + (Q 1)τ θ < pτ} I i,p.

6 6 L. THOMAS RAMSEY For all other j k, there is an arc of length τ between I i,k e ijτ (e.g., U k,m(i,k) or U k+1,m(i,k+1) when 1 k < p). Each sequence {k i } n i=1, with 1 k i p, defines a cylinder in l (F ) of the following form: W [{k i } n i=1] = { f l (F ) f(γ i ) I i,ki }. For g G, let f g (γ) = γ(g) for γ F. Because these cylinders are disjoint, each f g is in at most one of them. S n was chosen to guarantee the f g would be in at least one cylinder for g S n. For g S n, let h(g) = {k i } n i=1 define the cylinder which contains f g. Because each cylinder has diameter less than δ, inequality (1) implies that each cylinder contains at most one f g for g S n. Hence S n = h(s n ). For any subset H F, let Π H be this projection: for f l (F ), Π H (f) = f H. By Corollary 2 of [Pa, p. 742], there is a constant c > 0 which depends only on δ/2 p (which itself depends only on δ) such that there are some H F integers a < b from [1, p] such that H c F {a, b} H Π H (h(s n )). Case 1: (a b) mod p 2. On the circle, there are two arcs between e iaτ e ibτ. Choose c so that e icτ is the center of the shorter of these two arcs, a c a + p. Necessarily c a c b. c is either a half-integer or an integer. If c is an integer, then e icτ is separated by arcs of length τ from I i,a I i,b. If c is a half-integer, c 1/2 c + 1/2 are both integers which are distinct from a b. Since there are arcs of length τ between each of e i(c 1/2)τ e i(c+1/2)τ each of I i,a I i,b, there are arcs of length τ between e icτ each of I i,a I i,b. Case 1A. Assume that a < c < b. Let z 2 I i,b z 1 I i,a. Then where x y can be chosen to satisfy I i,b = { e iθ x θ y }, x bτ τ + τ y bτ + τ τ. [See equation (2).] Moreover, since e i cτ is separated from I i,b by an arc of length τ, both cτ < bτ x < bτ, we have cτ + τ x. Because e icτ is the center of the shorter of the two arcs between e iaτ e ibτ, bτ cτ π/2.

7 PROPORTIONS OF SIDON SETS ARE I 0 SUBSETS 7 Since δ 2 τ < δ/2 ( τ > 0), we have z 2 = e iθ with Hence cτ + τ θ < cτ + π/ e icτ z 2 = e i(θ cτ), with τ θ cτ π/ Thus e icτ z 2 is in the top half-plane, with R(e icτ z 2 ) τ = min{sin(τ ), sin(π/2 + 1)} > 0. Likewise, e icτ z 1 is in the lower half-plane, with R(e icτ z 1 ) τ < 0. Because {a, b} H Π H (h(s n )), for any A H there is some g S n such that h(g)(γ) = b for γ A h(g)(γ) = a for γ H\A. Let µ = e icτ δ g ; µ D(1). Because h(g)(γ i ) = b if only if γ i (g) I i,b, for γ A we have R( e icτ δ g (γ)) = R(e icτ γ(g)) τ Likewise, for γ i H\A, γ i (g) I i,a hence R( e icτ δ g (γ)) = R(e icτ γ(g)) τ This proves that H is J(1, τ ). Case 1B. Assume that b < c < a + p. Let z 2 I i,a z 1 I i,b. Then z 2 = e iθ with cτ + τ θ < cτ + π/2 + 1, e icτ z 2 = e i(θ cτ), with τ θ cτ < π/ Thus e icτ z 2 is in the top half-plane, with R(e icτ z 2 ) τ > 0. Likewise, e icτ z 1 is in the lower half-plane, with R(e icτ z 1 ) τ < 0. Because {a, b} H Π B (h(s n )), for any A B there is some g S n such that h(g)(γ) = a for γ A h(g)(γ) = b for γ H\A. Let µ = e icτ δ g ; again, µ D(1). Because h(g)(γ i ) = a if only if γ i (g) I i,a, for γ A we have R( e icτ δ g (γ)) = R(e icτ γ(g)) τ. Likewise, for γ i H\A, γ i (g) I i,b hence R( e icτ δ g (γ)) = R(e icτ γ(g)) τ.

8 8 L. THOMAS RAMSEY This proves that H is J(1, τ ). Case 2A: b = a + 1. Because {a, b} H Π H (h(s n )), for every A H there are g 1 g 2 such that while ( γ A) (h(g 1 )(γ) = b h(g 2 )(γ) = a), ( γ H\A) (h(g 2 )(γ) = a h(g 2 )(γ) = b). The arc U i,m(i,b) is between I i,b I i,a. Let U i,m(i,b) = {e iθ a θ < b }, for some a b such that aτ a < b bτ. If z I i,b, then z = e iθ for some θ such that b θ < bτ + τ τ. [See inclusion (2).] Likewise, if z I i,a, then z = e iθ for some θ such that aτ τ + τ θ < a. Thus, when γ i (g 1 ) I i,b γ i (g 2 ) I i,a, with γ i (g 1 g 2 ) = γ i (g 1 )/γ i (g 2 ) = e iθ τ b a < θ < (b a)τ + 2τ 2τ = (3 2/Q)τ < 3. Thus, when γ A, γ(g 1 g 2 ) is in the upper half-plane R(γ(g 1 g 2 )) τ = min{sin(τ ), sin(3)}. When γ i (g 1 ) I i,a γ i (g 2 ) I i,b, then with γ i (g 1 g 2 ) = γ i (g 1 )/γ i (g 2 ) = e iθ 3 < ( 3 + 2/Q)τ < θ < a b = τ. Thus, when γ H\A, This puts γ i (g 1 g 2 ) in the lower half plane with R(γ(g 1 g 2 )) τ. This makes H a J(1, τ ) set. Case 2B: a = 1 b = p. This is just like Case 2A, if one treats a as p + 1 switch the roles of a b. Set τ = min{τ, τ }. Every finite F E has H F such that H c F H is J(1, τ ). By Lemma 5, J(H) 2/τ. By Theorem 5, N(H) φ(j(h)) for a function φ which is independent of E. Note that J(H) depends only on τ which in turn depends only on δ ( is independent of F E). Also, c depends only on δ, is independent of F E.

9 PROPORTIONS OF SIDON SETS ARE I 0 SUBSETS 9 References [E] John Elton, Sign-Embeddings of l n (September, 1983), TAMS, [G] David Grow, Sidon Sets I 0 -Sets 53 (1987), Colloquium Mathematicum, [HR] S. Hartman C. Ryll-Nardzewski, Almost Periodic Extensions of Functions 12 (1964), Colloquium Mathematicum, [Kh] J.-P. Kahane, Ensembles de Ryll-Nardzewski et ensembles de Helson 15 (1966), Colloquium Mathematicum, [Kl] J. N. Kalton, On Vector-Valued Inequalities For Sidon Sets Sets of Interpolation 54 (1993), Colloquium Mathematicum, [LR] Jorge M. López Kenneth A. Ross, Sidon Sets, Marcel Dekker, Inc., New York, 1975, p. 4. [M] J.-F. Méla, Sur les ensembles d interpolation de C. Ryll-Nardzewski et de S. Hartman 29 (1968), Colloquium Mathematicum, [Pa] Alain Pajor, Plongement de l n 1 dans les espaces de Banach complexes 296 (May, 1983), CRAS, [P] Gilles Pisier, Conditions D Entropie Et Caracterisations Arithmetique Des Ensembles de Sidon. [RW] L. Thomas Ramsey Benjamin B. Wells, Interpolation Sets in Bounded Groups 10(1) (1984), Houston Journal of Mathematics, [R] L. Thomas Ramsey, I 0 -Sets Are Finitely Describable, preprint. [P] Gilles Pisier, Arithmetic Characterization of Sidon Sets 8 (1983), Bull. AMS, Mathematics, Keller Hall, 2565 The Mall, Honolulu, Hawaii ramsey@math.hawaii.edu or ramsey@uhunix.uhcc.hawaii.edu