On Fejes Tóth's conjectured maximizer for the sum of angles between lines
aa r X i v : . [ m a t h . M G ] J u l ON FEJES T ´OTH’S CONJECTURED MAXIMIZER FORTHE SUM OF ANGLES BETWEEN LINES
TONGSEOK LIM AND ROBERT J. MCCANN
Abstract.
Choose N unoriented lines through the origin of R d +1 .The sum of the angles between these lines is conjectured to be max-imized if the lines are distributed as evenly as possible amongst thecoordinate axes of some orthonormal basis for R d +1 . For d ≥ α -th power of the renor-malized angles between the lines. We show the conjecture is equiv-alent to this same configuration becoming the unique optimizer (upto rotations) for all α >
1. We establish both the asserted optimal-ity and uniqueness in the limiting case α = ∞ of mildest repulsion.The same conclusions extend to N = ∞ , provided we assume onlyfinitely many of the lines are distinct. Keywords: potential energy minimization, spherical designs, projectivespace, extremal problems of distance geometry, great circle distance,attractive-repulsive potentials, mild repulsion limit, Riesz energyMSC2010 Classification: 90C20, 90C26, 52A40, 58C35, 70G75 Introduction
Choose N unoriented lines through the origin of R d +1 . The sum ofthe angles between these lines is conjectured to be maximized if thelines are distributed as evenly as possible amongst the coordinate axesof some orthonormal basis for R d +1 . When d = 2 this conjecture datesback to Fejes T´oth [8]. For d ≥ α -th power of the renor-malized angles between the lines. Here renormalized means the angles Date : July 20, 2020.
TL is grateful for the support of ShanghaiTech University, and in addition, tothe University of Toronto and its Fields Institute for the Mathematical Sciences,where parts of this work were performed. RM acknowledges partial support of hisresearch by Natural Sciences and Engineering Research Council of Canada GrantsRGPIN-2015-04383 and 2020-04162. c (cid:13) are rescaled to achieve a maximum value of 1 when the lines are or-thogonal, and α parameterizes the effective strength of the repulsionbetween each pair of lines. We show the conjecture is equivalent to thissame configuration becoming the unique optimizer (up to rotations) forall α >
1. We establish both the asserted optimality and uniqueness inthe limiting case α = ∞ of mildest repulsion. The optimality extendsalso to N = ∞ , under the assumption that only finitely many of thelines remain distinct.Let S d = { x ∈ R d +1 | | x | = 1 } be the d -dimensional unit sphere, andlet ρ be the geodesic distance on S d , i.e. ρ ( x, y ) = arccos( x · y ).Define functions Λ : [0 , π ] → R and Λ : S d × S d → R asΛ ( t ) = 2 π min { t, π − t } (1.1) Λ( x, y ) = Λ ( ρ ( x, y )) , (1.2)so that Λ( x, y ) ∈ [0 ,
1] is proportional to the (non-obtuse) angle be-tween the lines x R and y R in R d +1 determined by x and y in S d ,attaining its maximum value Λ( x, y ) = 1 if and only if x · y = 0. Next,we define some measure spaces. Let N ∈ N := { , , . . . } , M ( S d ) = { µ | µ is a nonnegative finite measure on S d } , P ( S d ) = { µ ∈ M ( S d ) | µ ( S d ) = 1 , i.e. µ is a probability } , P N ( S d ) = { µ ∈ P ( S d ) | µ )] ≤ N } , P = N ( S d ) = { µ ∈ P N ( S d ) | µ = 1 N N X i =1 δ x i , x i ∈ S d ∀ i } , P fin ( S d ) = ∞ [ N =1 P N ( S d ) , and P =fin ( S d ) = ∞ [ N =1 P = N ( S d ) . Notice that P =fin ( S d ) denotes the set of probability measures with finitesupport and rational weights on S d . Then we consider the interactionenergy given by µ ∈ M ( S d ),(1.3) E ( µ ) = 12 Z Z Λ( x, y ) dµ ( x ) dµ ( y ) . In 1959, Fejes T´oth [8] considered the d = 2 instance of the followingproblem and several variants(1.4) maximize E ( µ ) over P = N ( S d ) for each fixed N ∈ N . In other words, given N , he was interested in the location of N not nec-essarily distinct points on the sphere which maximizes the sum of theirmutual non-obtuse angles. He conjectured that this energy (discrete N A CONJECTURE OF FEJES T ´OTH (1959) 3 sum) is maximized by the periodically repeated copies of the orthonor-mal basis. More precisely:
Conjecture 1. (Fejes T´oth [8]) Given N ∈ N , µ maximizes (1.4) ifthere is an orthonormal basis v , ..., v d +1 of R d +1 and x , ..., x N ∈ S d such that µ = N P Ni =1 δ x i and x i ∈ { v j , − v j } if i ≡ j mod d + 1.Here δ x is the probability concentrated at x , i.e. δ x ( A ) = 1 if x ∈ A and zero otherwise. We shall also consider a version of the conjecturein which N is unconstrained: Conjecture 1’.
A measure µ maximizes (1.3) over P fin ( S d ) if there isan orthonormal basis v , ..., v d +1 of R d +1 such that µ = P d +1 i =1 ( a i δ v i + b i δ − v i ) with a i , b i ≥ a i + b i = d +1 for all i = 1 , ..., d + 1.Essentially, conjectures 1 (and 1 ′ ) assert that there exists a max-imizer of E over P = N ( S d ) (and over P fin ( S d ) respectively), which issupported on an orthonormal basis. These conjectures have been set-tled for the case d = 1, but for d ≥ α with α >
0, the associated quadratic form(1.5) B α ( µ, ν ) = Z Z Λ α ( x, y ) dµ ( x ) dν ( y ) , µ, ν ∈ M ( S d ) , and the corresponding energy(1.6) E α ( µ ) = 12 B α ( µ, µ ) , α ∈ [1 , ∞ ] . Moreover, let us consider also the limiting case of mildest repulsion:Λ ∞ ( x, y ) = 1 if x · y = 0, and zero otherwise. Now we reformulate theconjectures. Notice the following are if and only if statements. Conjecture 2.
For each α > N ∈ N , µ maximizes (1.6) over P = N ( S d ) if and only if there is an orthonormal basis v , ..., v d +1 of R d +1 and x , ..., x N ∈ S d such that µ = N P Ni =1 δ x i and x i ∈ { v j , − v j } if i ≡ j mod d + 1. [6] notes the original conjecture was stated for S and later stated for all S d in [10]. TONGSEOK LIM AND ROBERT J. MCCANN
Conjecture 2’.
For each α > µ maximizes (1.6) over P fin ( S d ) ifand only if there is an orthonormal basis v , ..., v d +1 of R d +1 such that µ = P d +1 i =1 ( a i δ v i + b i δ − v i ) with a i , b i ≥ a i + b i = d +1 for all i = 1 , ..., d + 1. Remark 1.1.
Due to the symmetry of Λ around t = π/ we have Λ α ( x, y ) = Λ α ( − x, y ) , which indicates that the energy (1.6) is invariantunder the mass transfer between x and − x on S d . Thus the conjecture ismore naturally formulated on projective space P d ( R ) = S d / {±} , wherethe kernel Λ α becomes a monotone function of the distance, i.e. purelyrepulsive. Also, the energy is obviously invariant under the rotationof µ . In view of this, the conjectures 2 and 2’ state that for α > ,the maximizer is unique up to rotation and mass exchange between theantipodes. On the other hand, uniqueness is not expected for α = 1 ; e.g.the uniform probability on S gives the same E -energy as ( δ x + δ y ) with x · y = 0 . Definition 1.2.
For x ∈ S d we denote ˜ x = { x, − x } . For x, y ∈ S d , wedenote x ∼ y if ˜ x = ˜ y . In addition, for µ, ν ∈ M ( S d ) we write µ ≡ ν ifthey are equal up to rotation and identification of the antipodes. Thatis, µ ≡ ν if there exists an orthogonal matrix M ∈ O ( d + 1) such that µ ( A ∪ − A ) = ν ( M ( A ∪ − A )) for every measurable subset A of S d . In this paper we say the optimizer is essentially unique if µ ≡ ν for any optimizers µ, ν for a given problem. In the sequel, we denoteConjecture 1 as C1 , and C2 , C1’ , C2’ similarly. We begin with asimple observation.
Proposition 1.3. C1 and C2 are equivalent, and so are C1’ and
C2’ . Proof.
Firstly it is clear that C2 implies C1 by taking α ց
1. Forthe converse, assume C1 holds but C2 fails for some α >
1, so thatthere is a maximizer ν of E α which is not of the type described in C1 . Then there exist x, y ∈ spt( ν ) such that x · y / ∈ { , , − } . Let µ be a maximizer of E of the type described in C1 . Then we have thecontradiction E ( µ ) = E α ( µ ) ≤ E α ( ν ) < E ( ν ) ≤ E ( µ )where the strict inequality follows from x · y / ∈ { , , − } . The equiva-lence between C1’ and
C2’ can similarly be understood. QEDIt is clear that if the conclusion of C2 (resp. C2’ ) holds for some α >
1, then the same conclusion also holds for all α ′ satisfying α ′ > α . In thispaper we resolve C2 and C2’ in the mildest case α = ∞ , meaning two N A CONJECTURE OF FEJES T ´OTH (1959) 5 particles on the sphere interact precisely when they are perpendicular.We hope that this may yield an interesting new approach to the FejesT´oth conjecture in general dimensions.
Theorem 1.4.
Fix α = ∞ . Then the conclusion of Conjectures 2 and2’ hold in every dimension d ∈ N . In section 2 of this paper the special case α = ∞ of C2 is establishedusing an induction on both the dimension d and number of particles N (see Theorem 2.1 for more detailed statement); we then deduce from itthe more general statement of Theorem 1.4 in section 3.2. Maximization of E ∞ over P = N ( S d )Let V be a D -dimensional subspace of R D +1 , so that V ∩ S D is a( D − M = N ( S D ) = N P = N ( S D ) = { µ | µ = N X i =1 δ x i , x i ∈ S D } , M VN,k ( S D ) = { µ ∈ M = N ( S D ) | µ ( V ) = k } for k = 0 , , ..., N,E D,N = max µ ∈M = N ( S D ) E ∞ ( µ ) , E D,N,k = max µ ∈M VN,k ( S D ) E ∞ ( µ ) . Theorem 2.1.
Let
D, N ∈ N and α = ∞ . Then:(a) The conclusion of C2 holds. That is, µ maximizes E ∞ over M = N ( S D ) if and only if there is an orthonormal basis v , ..., v D +1 of R D +1 and x , ..., x N ∈ S D so that µ = P Ni =1 δ x i and x i ∼ v j if i ≡ j mod D + 1 .(b) Assume k ≥ DND +1 . Then µ = P Ni =1 δ x i maximizes E ∞ over M VN,k ( S D ) if and only if x i ∼ p for all x i / ∈ V where ˜ p = S D ∩ V ⊥ , and there existsan orthonormal basis v , ..., v D of V such that either (cid:12)(cid:12) { i | x i ∼ v j } (cid:12)(cid:12) = (cid:6) kD (cid:7) or (cid:12)(cid:12) { i | x i ∼ v j } (cid:12)(cid:12) = (cid:4) kD (cid:5) for every j = 1 , ..., D . Remark 2.2.
The conclusion of Theorem 2.1(b) holds for k = ⌊ DND +1 ⌋ as well, because in this case (along with k = ⌈ DND +1 ⌉ ) the structure ofmaximizers over M VN,k ( S D ) given in (b) coincides with the structure ofmaximizers over M = N ( S D ) given in (a), as one can easily check. To prove the theorem, we shall first prove the following lemma.
Lemma 2.3.
Let F d,n = E ∞ ( µ d,n ) and F d,n,k = E ∞ ( ν d,n,k ) where µ d,n , ν d,n,k are the (conjectured) maximizers over M = n ( S d ) and M Vn,k ( S d ) respectively as in Theorem 2.1. If n − ≥ d ≥ and dnd +1 ≤ k ≤ n − then (2.1) F d,n − k + F d − ,n < F d,n,k TONGSEOK LIM AND ROBERT J. MCCANN and there is a monotonicity F d,n,k > F d,n,k +1 , (2.2) F d,n, ⌊ dnd +1 ⌋ = F d,n, ⌈ dnd +1 ⌉ . (2.3)Of course, E d,n = F d,n and E d,n,k = F d,n,k once the theorem is proved. Proof of Lemma 2.3.
By the structure of maximizers µ described inTheorem 2.1 (a)–(b) respectively, it follows F d,m +1 − F d,m = m − j md + 1 k = l dd + 1 m m for every m , and(2.4) F d,n,k = F d − ,k + k ( n − k ) . (2.5)Firstly (2.3) follows from Remark 2.2. To see (2.2), by using (2.4), (2.5)we find (2.2) is equivalent to 2 k − n + 1 > ⌈ d − d k ⌉ . Notice this is alsoequivalent to 2 k − n ≥ d − d k , that is, k ≥ dnd +1 which is what we assumed.From now on we will prove (2.1). Note that (2.1) is equivalent to(2.6) F d,n − k + F d − ,n − F d − ,k < k ( n − k ) . Firstly, notice F d,n − k is clearly dominated by E ∞ ( ν ) where ν assignsthe total mass n − k uniformly onto an orthonormal basis of R d +1 ,hence F d,n − k ≤ (cid:16) ( d + 1) d (cid:17)(cid:16) n − kd + 1 (cid:17) = d ( n − k ) d + 1) . Secondly, observe (note that the sum is not void as n ≥ d + 1) F d − ,n − F d − ,k = n − X m = k l d − d m m = e + n − X m = k d − d m where e = e ( n, k ) ≤ n − k = ( d − n − k )( n + k − d + e. N A CONJECTURE OF FEJES T ´OTH (1959) 7
It follows F d,n − k + F d − ,n − F d − ,k ≤ d ( n − k ) d + 1) + ( d − n − k )( n + k − d + e ≤ d ( n − k ) d + 1) + ( d )( n − k )( n + k − d + 1) + e = d ( n − k )( d + 1) (cid:16) n − (cid:17) + e. In view of this, (2.6) will follow if we show(2.7) e avg := en − k < d d + 1) + k − dnd + 1 . Since the “average error” e avg ≤
1, (2.7) is immediate if k ≥ ⌈ dnd +1 ⌉ + 1.So we only need to show (2.7) for k = ⌈ dnd +1 ⌉ , which we henceforthassume. Now let l := n − k = ⌊ nd +1 ⌋ so that ( n, k ) = (cid:0) ( d + 1) l + r, dl + r (cid:1) for some r ∈ { , , ..., d } . Then since n = ( d + 1) l + r , d d + 1) + l dnd + 1 m − dnd + 1 = d d + 1) + rd + 1 = d + 2 r d + 1) . On the other hand, since k = dl + r , by definition of e , e = rd + r + 1 d + · · · + d − d + 0 + 1 d + · · · + d − d + 0 + 1 d + . . . where the sum consists of l terms. By the periodically increasing prop-erty of this sequence and the property of arithmetic average, we seethat e avg = e/l is maximized when l = d − r , that is e avg = el ≤ d − r (cid:16) rd + r + 1 d + · · · + d − d (cid:17) = d + r − d . Now it is easy to check d +2 r d +1) > d + r − d , which proves (2.7). QED Proof of Theorem 2.1.
Given D , the theorem is trivial if N ≤ D . Sowe assume N ≥ D + 1. We will proceed by an induction on N and D .Given d, n ∈ N , assume:Induction Hypothesis (IH): Theorem 2.1 (a)–(b) holds for every N ∈ N if D ≤ d −
1, and for every N ≤ n − D = d .First we will deduce (a) for ( D, N ) = ( d, n ). Let q, r ∈ Z be integerssuch that n − q ( d +1)+ r with r ∈ { , . . . , d } . By IH(a), we can finda maximizer µ over M = n − ( S d ) such that µ ( V ) = dq + r = (cid:6) d ( n − d +1 (cid:7) = TONGSEOK LIM AND ROBERT J. MCCANN (cid:4) dnd +1 (cid:5) and µ ( { p } ) = q = n − − (cid:4) dnd +1 (cid:5) . Define µ = µ + δ p ∈ M = n ( S d ).We will show that µ is the unique maximizer over M = n ( S d ). Observe E ∞ ( µ ) = E ∞ ( µ ) + B ∞ ( µ , δ p ) = E d,n − + j dnd + 1 k . Let ν be any element in M = n − ( S d ) and let ν = ν + δ p . Then E ∞ ( ν ) = E ∞ ( ν ) + B ∞ ( ν , δ p ) = E ∞ ( ν ) + ν ( V ) . If E ∞ ( ν ) = E d,n − , then ν ≡ µ by IH(a) and hence ν ( V ) ≤ (cid:4) dnd +1 (cid:5) .Moreover ν ( V ) = (cid:4) dnd +1 (cid:5) yields ν ≡ µ . We proved: E ∞ ( ν ) = E d,n − and E ∞ ( ν ) ≥ E ∞ ( µ ) implies ν ≡ µ. Now suppose E ∞ ( ν ) < E d,n − but still E ∞ ( ν ) ≥ E ∞ ( µ ). This implies ν ( V ) > (cid:4) dnd +1 (cid:5) . Set k = ν ( V ) and k ′ = (cid:4) dnd +1 (cid:5) , and observe that IH(2)and (2.5) imply E ∞ ( µ ) = E ∞ ( µ ) + k ′ = E d,n − ,k ′ + k ′ = E d − ,k ′ + k ′ ( n − k ′ ) = F d,n,k ′ ,E ∞ ( ν ) = E ∞ ( ν ) + k ≤ E d,n − ,k + k = E d − ,k + k ( n − k ) = F d,n,k . (2.2),(2.3) then gives E ∞ ( ν ) ≤ E ∞ ( µ ), yielding E ∞ ( ν ) = E ∞ ( µ ) and k = ⌈ dnd +1 ⌉ . Now the implied equality E ∞ ( ν ) = E d,n − ,k tells us thestructure of ν via IH(b), which clearly implies µ ≡ ν , as in the proofof (2.3). This completes the proof of IH(a) for N = n .Now we will deduce (b) for ( D, N ) = ( d, n ) to complete the induc-tion. Fix k ≥ (cid:6) dnd +1 (cid:7) and assume µ ∈ M Vn,k ( S d ) achieves the maximum E ∞ ( µ ) = E d,n,k . We will firstly show that µ (˜ p ) > µ = µ + δ p , so E ∞ ( µ ) = E ∞ ( µ ) + k .Hence E ∞ ( µ ) = E d,n,k implies E ∞ ( µ ) = E d,n − ,k . Then IH(b) appliedto µ yields µ must be of the type described in (b), as desired.Hence from now on we will assume µ (˜ p ) = 0 and show this yields thecontradiction E ∞ ( µ ) < E d,n,k . Let H = S d \ ( V ∪ ˜ p ), and denote µ H , µ V as the restriction of µ onto H and V respectively, so that µ = µ H + µ V .Let W be the set of all ( d − V . Then there N A CONJECTURE OF FEJES T ´OTH (1959) 9 exists W ∈ W such that µ ( W ) = max X ∈W µ ( X ). Now observe E ∞ ( µ ) = E ∞ ( µ H ) + E ∞ ( µ V ) + B ∞ ( µ H , µ V )= E ∞ ( µ H ) + E ∞ ( µ V ) + X x ∈ spt( µ H ) µ ( x ) B ∞ ( δ x , µ V )= E ∞ ( µ H ) + E ∞ ( µ V ) + X x ∈ spt( µ H ) µ ( x ) µ ( V ∩ { x } ⊥ ) ≤ E ∞ ( µ H ) + E ∞ ( µ V ) + ( n − k ) µ ( W ) . Note that E ∞ ( µ H ) ≤ E d,n − k . For E ∞ ( µ V ) + ( n − k ) µ ( W ), observe E ∞ ( µ V ) + ( n − k ) µ ( W ) = E ∞ (cid:0) µ V + ( n − k ) δ z (cid:1) for some z ∈ V ∩ W ⊥ , which yields E ∞ ( µ V ) + ( n − k ) µ ( W ) ≤ E d − ,n in particular. Then byIH and Lemma 2.3, we indeed have E ∞ ( µ ) ≤ E d,n − k + E d − ,n = F d,n − k + F d − ,n < F d,n,k ≤ E d,n,k . This completes the induction and proves Theorem 2.1. QED3.
Maximization of E ∞ over P fin ( S d )In this section we show that the same measures maximize the qua-dratic form E ∞ ( µ ) among all measures P fin ( S d ) having finite support,without imposing any upper bound N on the size of the support orrationality on the values of µ . We start with a lemma showing thatrational quadratic forms on the standard simplex admit rational ex-trema. Lemma 3.1.
Let A be a ( d +1) × ( d +1) symmetric matrix with rationalentries. Let ∆ d be the d -dimensional standard simplex in R d +1 , i.e. ∆ d = { x = ( x , ..., x d +1 ) ∈ R d +1 | d +1 X i =1 x i = 1 , x i ≥ ∀ i } , ∆ od = { x = ( x , ..., x d +1 ) ∈ ∆ d | x i > for all i } its relative interior, and ∂ ∆ d = ∆ d \ ∆ od its relative boundary. Define f : ∆ d → R by f ( x ) = x T Ax .(a) Then f has maxima and mimima on ∆ d whose entries are rational.(b) If f has interior extrema on ∆ d , then it has interior extrema whoseentries are rational. That is, suppose there is x ∈ ∆ od such that f ( x ) = M := max ∆ d f . Then there is x ′ ∈ ∆ od whose entries are rational and f ( x ′ ) = M . The same holds for minima. Proof.
We first prove (b), and we will only prove for the maximumand proceed by an induction on d . Let x ∈ ∆ od attain f ( x ) = M . Thennote that the gradient ∇ f ( x ) = Ax must be perpendicular to ∆ d at theoptimum x , that is, Ax = ce for some c ∈ R and e = (1 , , ..., ∈ R d +1 .We firstly consider the case x is the only maximum of f over ∆ d .If A is invertible, then x = cA − e . Now since A has rational entriesso does A − . Hence x has rational entries if and only if c is rational.But as P i x i = 1, c must be rational and we are done. But if A is notinvertible, there is a nonzero y ∈ R d +1 with Ay = 0. Now if y · e = 0,then x + ty ∈ ∆ d for all small t ∈ R , and we have g ( t ) := f ( x + ty ) = 12 ( x + ty ) T A ( x + ty ) = f ( x ) , violating the unique optimality assumed of x , thus y · e = 0. Let H be the hyperplane containing ∆ d . By multiplying a constant we canassume y ∈ H . Now if y = x , observe the unique optimality of x impliesthe system Az = 0 and P i z i = 1 has a unique solution (that is, x ).Then x will be found via elementary row operations on this system,and the rationality of A implies rationality of x , yielding the lemma.On the other hand if y = x , let v = y − x and z = x + tv = (1 − t ) x + ty .Then z ∈ ∆ d for all small t ∈ R , and g ( t ) = f ( z ) = (1 − t ) f ( x ), againviolating the unique optimality of x .It remains to consider the case x ∈ ∆ od is not the unique maximum,so there is x ′ ∈ ∆ d , x ′ = x and f ( x ′ ) = M . Consider the line L = { z | z = (1 − t ) x + tx ′ , t ∈ R } , and the two distinct points { y, y ′ } = L ∩ ∂ ∆ d . Observe then there exist two unique subsimplexes ∆ k , ∆ k ′ of dimensions k and k ′ respectively, such that y, y ′ lie in the relativeinteriors of ∆ k , ∆ k ′ , denoted by ∆ ok , ∆ ok ′ . Here a single point is its ownrelative interior, by convention. Notice ∆ ok ∩ ∆ ok ′ = ∅ . Now observe that f is constant on L , since with v = x ′ − x we have g ( t ) := f ( x + tv ) = 12 ( x + tv ) T A ( x + tv ) = f ( x ) + t f ( v )where we used the fact that v · Ax = v · ce = 0. Moreover g (0) = g (1)implies f ( v ) = 0, thus g is constant, hence f ( y ) = f ( y ′ ) = M . Thensince k, k ′ < d , the induction hypothesis is applicable and there exist z ∈ ∆ ok and z ′ ∈ ∆ ok ′ such that f ( z ) = f ( z ′ ) = M and the entries of z, z ′ are rational. Now since the quadratic form f is constant on thelines xz and xz ′ as shown above, f = M on the line zz ′ . Finally, notice zz ′ ∩ ∆ od = ∅ ; otherwise the segment zz ′ is in ∂ ∆ d , thus in a subsimplex∆ d − of dimension d −
1. Then { z, z ′ } ⊆ ∆ d − , yielding { y, y ′ } ⊆ ∆ d − and hence yy ′ ⊆ ∆ d − . But this contradicts to the fact x ∈ yy ′ . Hence N A CONJECTURE OF FEJES T ´OTH (1959) 11 zz ′ ∩ ∆ od = ∅ , yielding { (1 − t ) z + tz ′ | t ∈ (0 , } ⊆ ∆ od . We concludeany rational t ∈ (0 ,
1) yields the rational maximum (1 − t ) z + tz ′ ∈ ∆ od .Now part (a) is immediate from part (b) since any (extremal) point x ∈ ∆ d belongs to the relative interior of a subsimplex ∆ ok . QED Theorem 3.2.
A measure µ maximizes E ∞ over P fin ( S d ) if and onlyif there is an orthonormal basis v , ..., v d +1 of R d +1 such that µ ≡ d +1 P d +1 i =1 δ v i . Proof.
Let µ ∗ be a conjectured (essentially unique) maximizer, so that E ∞ ( µ ∗ ) = ( d +1) d d +1) = d d +1) =: M ∗ . Choose any µ ∈ P fin ( S d ). Wewill show E ∞ ( µ ) ≤ E ∞ ( µ ∗ ), and the inequality is strict unless µ ≡ µ ∗ .To this end, set A = spt( µ ) and P ( A ) = { µ ∈ P fin ( S d ) | spt( µ ) ⊆ A } and M A := max P ( A ) E ∞ . We separate two possible cases, according towhether or not equality holds in E ∞ ( µ ) ≤ M A .In the first case where E ∞ ( µ ) = M A , notice that Lemma 3.1(b)implies there exists ν ∈ P ( A ) such that spt( ν ) = A , E ∞ ( ν ) = M A and ν ( x ) is rational for every x ∈ A . Then Theorem 2.1(a) implies M A ≤ M ∗ thanks to the rationality, and furthermore equality holdsif and only if ν ≡ µ ∗ (here we use the standard fact that the inter-action energy P i = j x i x j is uniquely maximized over the hyperplane { x ∈ R d | P i x i = 1 } by the uniform probability vector; this can, forexample, be seen as a consequence of the Perron-Frobenius theorem).Hence in particular M A = M ∗ yields spt( µ ) = spt( ν ) ⊆ S d +1 i =1 { u i , − u i } for some orthonormal basis u , ..., u d +1 , and this clearly implies µ ≡ µ ∗ again by the same standard fact.In the second case E ∞ ( µ ) < M A , choose any ν ∈ P ( A ) attaining E ∞ ( ν ) = M A . Let B = spt( ν ) and define P ( B ) and M B as above.Then by the above argument we have M B ≤ M ∗ . But notice M B = E ∞ ( ν ) = M A , which yields E ∞ ( µ ) < M A = M B ≤ M ∗ .We conclude µ ∗ is the essentially unique maximizer on P fin ( S d ). QED References [1] Ralph Alexander and Kenneth B. Stolarsky. Extremal problems of distancegeometry related to energy integrals.
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Tongseok Lim: Krannert School of ManagementPurdue University, West Lafayette, Indiana 47907
E-mail address : [email protected] / [email protected] Robert J. McCann: Department of MathematicsUniversity of Toronto, Toronto ON Canada
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