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 nonincreasing 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 nonincreasing 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 npoint 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 realvalued
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 nonincreasing 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 senergy 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 wellknown. 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 npoint 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 npoint
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 nondecreasing 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 DMS0808093 and
DMS1109266.
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