Polarization Optimality of Equally Spaced Points on the Circle for Discrete Potentials

Discrete & Computational Geometry, Apr 2013

We prove a conjecture of Ambrus, Ball and Erdélyi that equally spaced points maximize the minimum of discrete potentials on the unit circle whenever the potential is of the form $$\begin{aligned} \sum _{k=1}^n f(d(z,z_k)), \end{aligned}$$ where \(f:[0,\pi ]\rightarrow [0,\infty ]\) is non-increasing and convex and \(d(z,w)\) denotes the geodesic distance between z and w on the circle.

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Polarization Optimality of Equally Spaced Points on the Circle for Discrete Potentials

Discrete Comput Geom Polarization Optimality of Equally Spaced Points on the Circle for Discrete Potentials Edward B. Saff 0 Douglas P. Hardin 0 Amos P. Kendall 0 0 D. P. Hardin ( We prove a conjecture of Ambrus, Ball and Erdélyi that equally spaced points maximize the minimum of discrete potentials on the unit circle whenever the potential is of the form Polarization · Chebyshev constants · Roots of unity · Potentials · - where f : [0, π ] → [0, ∞] is non-increasing and convex and d(z, w) denotes the geodesic distance between z and w on the circle. Primary 52A40 · 30C15 1 Introduction and Main Results Let S1 := {z = x + iy | x , y ∈ R, x 2 + y2 = 1} denote the unit circle in the complex plane C. For z, w ∈ S1, we denote by d(z, w) the geodesic (shortest arclength) distance between z and w. Let f : [0, π ] → [0, ∞] be non-increasing and convex on (0, π ] Mnf (S1) := sup ωn∈(S1)n M f (ωn; S1), and the f -polarization of ωn, In this note, we are chiefly concerned with the n-point f -polarization of S1 (also called the nth f -Chebyshev constant of S1), with f (0) = limθ→0+ f (θ ). It then follows that f is a continuous extended real-valued function on [0, π ]. For a list of n points (not necessarily distinct) ωn = (z1, . . . , zn) ∈ (S1)n, we consider the f -potential of ωn, (1) (2) (3) (4) (5) which has been the subject of several recent papers (e.g., [ 1,2,5,6 ]). In the case (relating to Euclidean distance) when f (θ ) = fs (θ ) := |eiθ − 1|−s = (2 sin |θ /2|)−s , s > 0, we abbreviate the notation for the above quantities by writing U s (ωn; z) := n k=1 M s (ωn; S1) := min z∈S1 k=1 |z − zk |s , Mns (S1) := sup ωn∈(S1)n M s (ωn; S1). fs (d(z, zk )) = n 1 n 1 k=1 |z − zk |s , The main result of this note is the following theorem conjectured by Ambrus et al. [ 2 ]. Its proof is given in the next section. Theorem 1 Let f : [0, π ] → [0, ∞] be non-increasing and convex on (0, π ] with f (0) = limθ→0+ f (θ ). If ωn is any configuration of n distinct equally spaced points on S1, then M f (ωn; S1) = Mnf (S1). Moreover, if the convexity condition is replaced by strict convexity, then such configurations are the only ones that achieve this equality. Applying this theorem to the case of fs given in (4) we immediately obtain the following. Corollary 2 Let s > 0 and ωn∗ := {ei2πk/n : k = 1, 2, . . . , n}. If (z1, . . . , zn ) ∈ (S1)n, then 1 z∈S1 k=1 |z − zk |s ≤ M s (ωn∗; S1) = Mns (S1), min (6) with equality if and only if (z1, . . . , zn ) consists of distinct equally spaced points. The following representation of M s (ωn∗; S1) in terms of Riesz s-energy was observed in [ 2 ]: where M s (ωn∗; S1) = Es (S1; 2n) 2n − Es (S1; n) n Es (S1; n) := n n inf ωn∈(S1)n j=1 kk==1j |z j − zk |s 1 , . ⎧ 2ζ (s) Mns (S1) ∼ ⎪⎪⎪⎨⎪⎪ ((12/ππ))s n(2losg−n1,)ns , ss >= 11 ,, ⎪⎪⎩⎪⎪⎪ √2−πs 1−s 2 s 1 − 2 n, s ∈ [0, 1), Thus, applying the asymptotic formulas for Es (S1; n) given in [ 3 ], we obtain the dominant term of Mns (S1) as n → ∞: where ζ (s) denotes the classical Riemann zeta function and an ∼ bn means that limn→∞ an /bn = 1. These asymptotics, but for M s (ωn∗; S1), were stated in [ 2 ]1. For s an even integer, say s = 2m, the precise value of Mn2m (S1) = M 2m (ωn∗; S1) can be expressed in finite terms, as can be seen from formula (1.20) in [ 3 ]. Corollary 3 We have 2 Mn2m (S1) = (2π )2m m k=1 n2k ζ (2k)αm−k (2m)(22k − 1), m ∈ N, where α j (s) is defined via the power series for sinc z = (sin π z)/(π z): ∞ j=0 (sinc z)−s = α j (s)z2 j , α0(s) = 1. 1 We remark that there is a factor of 2/(2π)p missing in the asymptotics given in [ 2 ] for the case p := s > 1. In particular, 2 n2 Mn2(S1) = (2π )2 n2ζ (2) = 4 , The case s = 2 of the above corollary was first proved in [ 1,2 ] and the case s = 4 was first proved in [5]. We remark that an alternative formula for α j (s) is α j (s) = (−1) j B2(sj)(s/2) (2 j )! (2π )2 j , j = 0, 1, 2, . . . , where B(jα)(x ) denotes the generalized Bernoulli polynomial. Asymptotic formulas for Mnf (S1) for certain other functions f can be obtained from the asymptotic formulas given in [ 4 ]. As other consequences of Theorem 1, we immediately deduce that equally spaced points are optimal for the following problems: and n min max ωn∈(S1)n z∈S1 k=1 |z − zk |α (0 < α ≤ 1), n max min ωn∈(S1)n z∈S1 k=1 log 1 |z − zk | , with the solution to (8) being well-known. Furthermore, various generalizations of the polarization problem for Riesz potentials for configurations on S1 are worthy of consideration, such as minimizing the potential on circles concentric with S1. 2 Proof of Theorem 1 For distinct points z1, z2 ∈ S1, we let z1z2 denote the closed subarc of S1 from z1 to z2 traversed in the counterclockwise direction. We further let γ (z1z2) denote the length of z1z2 (thus, γ (z1z2) equals either d(z1, z2) or 2π − d(z1, z2)). Observe that the points z1 and z2 partition S1 into two subarcs: z1z2 and z2z1. The following lemma (see proof of Lemma 1 in [ 2 ]) is a simple consequence of the convexity and monotonicity of the function f and is used to show that any n-point configuration ωn ⊂ S1 such that M f (ωn; S1) = Mnf (S1) must have the property that any local minimum of U f (ωn; ·) is a global minimum of this function (Fig. 1). For φ ∈ R and z ∈ S1, we let ρφ (z) := eiφ z denote the counterclockwise rotation of z by the angle φ. Lemma 4 ([ 2 ]) Let z1, z2 ∈ S1 and 0 < ε < γ (z2z1)/2. Then with f as in Theorem 1, U f ((z1, z2); z) ≤ U f ((ρ−ε(z1), ρε(z2)); z) (9) for z in the subarc ρε(z2)ρ−ε(z1), while the reverse inequality holds for z in the subarc z1z2. If f is strictly convex on (0, π ], then these inequalities are strict. If z1 = z2, then we set z1z2 = {z1} and z2z1 = S1. We now assume that ωn = (z1, . . . , zn) is ordered in a counterclockwise manner and also that the indexing is extended periodically so that zk+n = zk for k ∈ Z. For 1 ≤ k ≤ n and ∈ R, we define τk, : (S1)n → (S1)n by τk, (z1, . . . , zk , zk+1, . . . , zn) := (z1, . . . , ρ− (zk ), ρ (zk+1), . . . , zn). If zk−1 = zk and zk+1 = zk+2, then τk, (ωn) retains the ordering of ωn for positive and sufficiently small. Given := ( 1, . . . , n)T ∈ Rn, let τ := τn, n ◦ · · · ◦ τ2, 2 ◦ τ1, 1 and ωn := τ (ωn). Letting αk := γ (zk zk+1) and αk := γ (zk zk+1) for k = 1, . . . , n, we obtain the system of n linear equations: αk = αk − k−1 + 2 k − k+1 (1 ≤ k ≤ n), (10) which is satisfied as long as clockwise. Let n k=1 αk = 2π or, equivalently, if ωn is ordered countersep(ωn) := 1≤ ≤n min α . Then (10) holds if in which case, the configurations max | k | ≤ (1/4)sep(ωn), 1≤k≤n ωn(,) := τn, ◦ · · · ◦ τ2, 2 ◦ τ1, 1 (ωn) ( = 1, . . . , n) are all ordered counterclockwise. If the components of may replace the ‘(1/4)’ in (11) with ‘(1/2)’. are nonnegative, then we Lemma 5 Suppose ωn = (z1, . . . , zn) and ωn = (z1, . . . , zn) are n-point configurations on S1 ordered in a counterclockwise manner. Then there is a unique ∗ = ( 1∗, . . . , n∗) ∈ Rn so that (a) k∗ ≥ 0, k = 1, . . . , n, (b) ∗j = 0 for some j ∈ {1, . . . , n}, and (c) τ ∗ (ωn) is a rotation of ωn. Proof The system (10) can be expressed in the form A = β, (11) (12) (13) where ⎛ 2 ⎜ −1 ⎜ . A := ⎜⎜ .. ⎝⎜ 0 −1 ... ⎟⎟⎟⎟ ; ⎜⎜ := ⎜⎜ ⎜ ⎝ ⎛ 1 ⎞ ...2 ⎟⎟⎟⎟ , and β := ⎜⎜⎜⎜ ⎟⎠ ⎜⎝ n ⎛ α1 − α1 ⎞ α2 −... α2 ⎟⎟⎟⎟⎟ . ⎠ αn − αn I(t1,i1s, e.l.e. m,1e)nTta.rSyintoce vβeTri1fy=that knk=e1r(Aαk −= α(kr)an=ge0A,)t⊥he =linesaprasny(s1te)m, w(1h3e)real1way=s has a solution . Let j ∈ {1, . . . , n} satisfy j = min1≤k≤n k . Then subtracting j 1 from , we obtain the desired ∗. Since ker A = span 1, there is at most one solution of (13) satisfying properties (a) and (b), showing that ∗ is unique. Part (c) holds as a direct result of the fact that both ωn and ωn are ordered counterclockwise. Lemma 6 Let n = (z1, . . . , zn) be a configuration of n distinct points on S1 ordered counterclockwise, and with f as in Theorem 1, suppose = ( 1, . . . , n) ∈ Rn is such that (a) 0 ≤ k ≤ (1/2)sep( n) for k = 1, . . . , n, and (b) there is some j ∈ {1, . . . , n} for which j = 0. Let n := τ ( n) = (z1, . . . , zn). Then z j z j+1 ⊂ z j z j+1 and U f ( n; z) ≤ U f ( n; z) (z ∈ z j z j+1). (14) If f is strictly convex on (0, π ] and is strict. k > 0 for at least one k, then the inequality (14) We remark that k = 0 for all k = 1, . . . , n is equivalent to saying that the points are equally spaced. Proof Recalling (12), it follows from condition (a) that (z1( ), . . . , zn( )) := ωn(,) are counterclockwise ordered. Since j = 0 and k ≥ 0 for k = 1, . . . , n, the points z(j ) and z(j+)1 are moved at most once as varies from 1 to n and move toward each other, while remaining in the complement of all other subarcs zk( )zk(+)1, i.e., z j z j+1 = z(jn)z(jn+)1 ⊆ z(j )z(j+)1 ⊆ zk(+)1zk( ), for k ∈ {1, . . . , n} \ { j } and ∈ {1, . . . , n}. Lemma 4 implies that, for = 1, . . . , n, k > 0 for some k = 1, . . . , n. we have U f (ωn( −1); z) ≤>U0f.(ωHn(en);cze), (f1o4r)z h∈olzd(sj )azn(jd+)1th(ewihneerqeuωaln(i0t)y is strict if f is := ωn) and the inequality is strict if strictly convex and We now proceed with the proof of Theorem 1. Let ωn = (z1, . . . , zn) be a nonequally spaced configuration of n (not necessarily distinct) points on S1 ordered counterclockwise. By Lemma 5, there is some equally spaced configuration ωn (i.e., αk = 2π/n for k = 1, . . . , n) and some ∗ = ( 1∗, . . . , n∗) such that (a) ωn = τ ∗ (ωn), (b) k∗ ≥ 0 for k = 1, . . . , n, and (c) ∗j = 0 for some j ∈ {1, . . . , n}. Then (10) holds with αk := γ (zk , zk+1) and αk := 2π/n. Since ωn is not equally spaced, we have k∗ > 0 for at least one value of k. For 0 ≤ t ≤ 1, let ωnt := τ(t ∗)(ωn) = (z1t, . . . , znt) and, for k = 1, . . . , n, let αkt := γ (zkt zkt+1). Recalling (10), observe that αkt = αk − t ( k−1 + 2 k − k+1) = αk + t (2π/n − αk ) = (1 − t )αk + t (2π/n), for 0 ≤ t ≤ 1 and k = 1, . . . , n, and so sep(ωnt) ≥ t (2π/n). Now let 0 < t < s < min(1, t (1 + π/(n D))), where D := max{ k : 1 ≤ k ≤ n}. Then Lemma 6 (with n = ωnt, = (s − t ) ∗, and n = τ ( n) = ωns) implies that zsj zsj+1 ⊆ ztj ztj+1 and that U f (ωnt; z) ≤ U f (ωns; z) (z ∈ zsj zsj+1), (15) where the inequality is sharp if f is strictly convex. Consider the function Observe that h(t ) := min{U f (ωnt ; z) : z ∈ ztj ztj+1}, (0 ≤ t ≤ 1). h(t ) ≤ min{U f (ωnt ; z) : z ∈ z j z j+1} ≤ min{U f (ωns ; z) : z ∈ z j z j+1} = h(s), s s s s for 0 < t < s < min(1, t (1 + π/(n D))). It is then easy to verify that h is nondecreasing on (0, 1). Since ωnt depends continuously on t, the function h is continuous on [ 0, 1 ] and thus h is non-decreasing on [ 0, 1 ]. We then obtain the desired inequality M f (ωn ; S1) ≤ h(0) ≤ h(1) = M f (ωn ; S1), where the last equality is a consequence of the fact that ωn is an equally spaced configuration and so the minimum of U f (ωn ; z) over S1 is the same as the minimum over z j z j+1. If f is strictly convex, then h(0) < h(1) showing that any optimal f -polarization configuration must be equally spaced. This completes the proof of Theorem 1. Acknowledgements We thank the referees for their helpful suggestions to improve the manuscript. This research was supported, in part, by the U.S. National Science Foundation under Grants DMS-0808093 and DMS-1109266. 1. Ambrus , G.: Analytic and probabilistic problems in discrete geometry . Ph.D. Thesis , University College London ( 2009 ) 2. Ambrus , G. , Ball , K. , Erdélyi T. : Chebyshev constants for the unit circle . Bull. Lond. Math. Soc . 45 ( 2 ), 236 - 248 ( 2013 ) 3. Brauchart , J.S. , Hardin , D.P. , Saff , E.B. : The Riesz energy of the N th roots of unity: an asymptotic expansion for large N . Bull. Lond. Math. Soc . 41 ( 4 ), 621 - 633 ( 2009 ) 4. Brauchart , J.S. , Hardin , D.P. , Saff , E.B. : Discrete energy asymptotics on a Riemannian circle . Unif. Distrib. Theory 6 , 77 - 108 ( 2011 ) 5. Erdélyi , T. , Saff , E.B. : Riesz polarization inequalities in higher dimensions , J. Approx. Theory 171 , 128 - 147 ( 2013 ) 6. Nikolov , N. , Rafailov , R.: On the sum of powered distances to certain sets of points on the circle . Pac. J. Math. 253 ( 1 ), 157 - 168 ( 2011 )


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Douglas P. Hardin, Amos P. Kendall, Edward B. Saff. Polarization Optimality of Equally Spaced Points on the Circle for Discrete Potentials, Discrete & Computational Geometry, 2013, 236-243, DOI: 10.1007/s00454-013-9502-4