INGHAM TYPE INEQUALITIES IN LATTICES VILMOS KOMORNIK, ANNA CHIARA LAI, AND PAOLA LORETI Abstract. A classical theorem of Ingham extended Parseval's formula of the trigonometrical system to arbitrary families of exponentials satis- fying a uniform gap condition. Later his result was extended to several dimensions, but the optimal integration domains have only been deter- mined in very few cases. The purpose of this paper is to determine the optimal connected integration domains for all regular two-dimensional lattices. 1. Introduction A classical theorem of Ingham [7] extended the Parseval's formula of the trigonometrical system to arbitrary families of exponentials satisfying a uni- form gap condition. Later Beurling [3] determined the critical length of the intervals on which these estimates hold. Kahane [8] extended these results to several dimensions. His theorem was improved and generalized in [1] (see also [10]), but the optimal integration domains have only been determined in very particular cases. The purpose of this paper is to determine the optimal connected integra- tion domains for all regular two-dimensional lattices. 2. A general framework N N Consider M disjoint translates of Z by vectors u1; : : : ; uM 2 R , and consider the functions of the form M M X X i(uj +k;x) X i(uj ;x) arXiv:1512.04212v3 [math.CA] 29 Jan 2016 f(x) = ajke =: e fj(x) j=1 k2ZN j=1 with square summable complex coefficients ajk. Let us observe that the functions X i(k;x) (2.1) fj(x) = ajke k2ZN 2010 Mathematics Subject Classification. 42B05, 52C20. Key words and phrases. Fourier series, combinatorics, non-harmonic analysis, lattices, tilings. 1 2 V. KOMORNIK, A. C. LAI, AND P. LORETI are 2π-periodical in each variable, so that Z 1 2 X 2 (2.2) jfj(x)j dx = jajkj jΩ0j Ω0 k2ZN N on Ω0 := (0; 2π) by Parseval's equality for multiple Fourier series, where N and jΩ0j = (2π) denotes the volume of the cube Ω0. N Next we consider M vectors v1; : : : ; vM 2 R satisfying the following two conditions: (A1) the coordinates of each vk are multiples of 2π; M i(uj ;vk) (A2) the matrix E := e j;k=1 is invertible. It follows from (A1) that the translated sets Ωk := vk + Ω0; k = 1;:::;M are non-overlapping, i.e., their interiors are pairwise disjoint. N Finally we fix an invertible linear transformation L of R , we introduce the lattice M [ ∗ N N Λ := L uj + Z ⊂ R j=1 (here L∗ denotes the adjoint of L) and the set −1 N Ω := L (Ω1 [···[ ΩM ) ⊂ R : Remark 2.1. Let us emphasize that the volume of jΩj of Ω does not depend N on the particular choice of M and the vectors v1; : : : ; vM 2 R satisfying (A1). We prove the following Ingham type generalization of Parseval's formula: Theorem 2.2. Assume (A1) and (A2). (i) There exist two positive constants c1; c2 such that 2 Z X 2 X i(λ,x) X 2 (2.3) c1 jaλj ≤ aλe dx ≤ c2 jaλj λ2Λ Ω λ2Λ λ2Λ for all square summable families (aλ)λ2Λ of complex coefficients. (ii) The estimates fail if we remove any non-empty open subset from Ω. Proof. Let us first consider the case where L is the identity map. INGHAM TYPE INEQUALITIES IN LATTICES 3 Using (A1) we have Z M Z 2 X 2 jf(x)j dx = jf(vk + x)j dx Ω k=1 Ω0 2 M Z M X X i(uj ;vk+x) = e fj(vk + x) dx k=1 Ω0 j=1 2 M Z M X X i(uj ;vk) i(uj ;x) = e · e fj(x) dx: k=1 Ω0 j=1 0 0 Furthermore, by (A2) there exist two positive constants c1; c2 such that 2 M M M 2 0 X i(uj ;x) X X i(uj ;vk) i(uj ;x) c e fj(x) ≤ e · e fj(x) 1 j=1 k=1 j=1 M X 2 0 i(uj ;x) ≤ c2 e fj(x) ; j=1 or equivalently 2 M M M M 0 X 2 X X i(uj ;vk) i(uj ;x) 0 X 2 c jfj(x)j ≤ e · e fj(x) ≤ c jfj(x)j 1 2 j=1 k=1 j=1 j=1 for all x. Integrating over Ω0 and using the last equality we obtain the estimates M Z Z M Z 0 X 2 2 0 X 2 c1 jfj(x)j dx ≤ jf(x)j dx ≤ c2 jfj(x)j dx j=1 Ω0 Ω j=1 Ω0 Since jΩj = M jΩ0j by (A2), using (2.2) the estimates (2.3) follow with 0 0 c1 = c1 jΩ0j and c2 = c2 jΩ0j. Now we show that the above estimates fail if we remove from Ω a non- empty open subset !. We may assume that ! ⊂ Ωk for some k 2 f1;:::;Mg. Let f 2 L2(Ω) be the characteristic function of !. Thanks to Assumption (A2) the linear system M X i(uj ;vk) i(uj ;x) f(vk + x) = e e fj(x); k = 1;:::;M j=1 has a unique solution i(uj ;x) e fj(x); j = 1;:::;M 2 for each x 2 Ω0, and f1; : : : ; fM 2 L (Ω0). Extending the functions fj by 2π-periodicity in each variable, we get (2.1) for each j with square summable 4 V. KOMORNIK, A. C. LAI, AND P. LORETI coefficients ajk. Since, furthermore, M X i(uj ;x) f(x) = e fj(x) j=1 by Assumption (A1), we conclude that M X X i(uj +k;x) f(x) = ajke j=1 k2ZN in Ω. Since ! has a positive measure, the coefficients ajk do not vanish identi- cally. On the other hand, Z jf(x)j2 dx = 0; Ωn! so that the first estimate of (2.2) fails. In order to complete the proof of the first part of the theorem it suffices to show that if the estimates (2.3) hold for some Λ and Ω, and L is an N invertible linear transformation of R , then the estimates 2 1 Z ∗ X 2 X i(L λ,x) X 2 c1 jaλj ≤ −1 aλe dx ≤ c2 jaλj jL (Ω)j −1 λ2Λ L (Ω) λ2Λ λ2Λ also hold. This follows from the change of variable formula: if x = Lx0, then 2 2 Z Z X i(k;x) X i(L∗k;x0) 0 ake dx = jdet Lj ake dx ; Ω L−1(Ω) k2ZN k2ZN where det L denotes the determinant of L, and −1 jΩj L (Ω) = : jdet Lj Since L transforms non-empty open sets into non-empty open sets, the second part of the theorem also holds in the general case. Remarks 2.3. (i) A standard application of the triangle inequality implies that the as- sumptions (A1) and (A2) are not necessary for the second inequality in (2.3). (ii) The assumption (A2) is not necessary for Part (ii) of the theorem. This may be shown by taking a maximal subset of the vectors for which the corresponding columns of the matrix E are linearly independent, and by completing this subset to a new set of vectors satisfying (A1) and (A2). INGHAM TYPE INEQUALITIES IN LATTICES 5 Given a lattice M [ ∗ N N (2.4) Λ := L uj + Z ⊂ R j=1 we may wonder whether we there exists another representation M0 [ ∗ N N (2.5) Λ = L0 u~j + Z ⊂ R j=1 with another invertible matrix L0 and a smaller integer M0. As we will see in the rest of the paper choosing the minimal M may substantially simplify the study of optimal integration domains. The following simple condition will allow us to determine the smallest M N in all but one of the examples in this work. Given two points a; b 2 R , let us introduce the lattice Λ(a; b) := fa + k(b − a): k 2 Zg generated by a and b. N Lemma 2.4. If there exist M points a1; : : : ; aM 2 R such that Λ(ai; ak) 6⊂ Λ for all i 6= k, then the number M in the representation (2.4) of Λ is the smallest possible. ∗ N Proof. If two points ai and ak belong to the same set L0 u~j + Z in another representation (2.5), then ∗ N Λ(ai; ak) ⊂ L0 u~j + Z ⊂ Λ; contradicting our hypothesis. Therefore each point ai corresponds to a dif- ferent j, and thus M ≤ M0. 3. Triangular and hexagonal lattices We illustrate Theorem 2.2 by two examples. 3.1. Regular triangular lattice. Choosing1 N = 2;M = 1; u1 = v1 = (0; 0) and 1 0 p L = 1 3 2 2 (as usual, we identify the linear transformations with their matrices in the 2 canonical basis of R ), ( p ! ) k 3k Λ = k + 2 ; 2 : k 2 2 1 2 2 Z 1We write the vectors as row vectors but we consider them as column vectors in the computations with matrices. 6 V. KOMORNIK, A. C. LAI, AND P. LORETI 6 4 2 1 2 3 4 5 6 -2 −1 (a) Λ (b) L (Ω0) Figure 1. Hexagonal lattice is a triangular lattice formed by equilateral triangles of unit side. Further- more, since −1 1 0 L = p−1 p2 ; 3 3 −1 Ω = L (Ω0) is a parallelogram of vertices 2π 4π 2π (0; 0); 2π; −p ; 0; p ; 2π; p : 3 3 3 2 Its area is equal to 8pπ ; see Figure 1.