Gruber P. Convex and Discrete Geometry

482 Geometry of Numbers
where Hn is a regular hexagon in E 2 of area A(Hn) = A(H)/n and centre at the
origin o.
The following analytic proof is due to the author [437]. It uses the moment lemma
of Fejes Tóth [329], p. 198.
Proof. It is sufficient to prove the theorem for functions f with positive continuous
derivative on (0, +∞). LetS⊆ E2 , #S = n. Then
�
�
�
(2) min{
f (�x − s�)}dx = f (�x − si�)dx,
s∈S
H
i
Di
where the sets Di, i = 1,...,n, are the Dirichlet–Voronoĭ cells in H corresponding
to S. Di is a convex polygon of area ai with vi vertices, say. The moment lemma of
Fejes Tóth says that
�
�
(3) f (�x − si�)dx ≥ f (�x�)dx = M(ai,vi),
Di
Ri
say, where Ri is a regular vi-gon with area ai and centre o. Letg be defined by
g(r 2 ) = f (r) for r ≥ 0. Then g(0) = 0 and g has positive continuous derivative
on (0, +∞). LetG be such that G(0) = 0 and G ′ = g. Finally, let h(a,v) =
a/(v tan(π/v)) for a > 0,v ≥ 3. Clearly, the following hold:
If R is a regular polygon with centre o, area a, and v vertices,
then h 1 2 is its inradius, and
�
(4) M(a,v)=
R
�
f (�x�)dx = 2v
0
π v
h 1/2
cos ψ
�
0
g(r 2 �
)rdrdψ = v
0
π v
�
G
h
cos 2 ψ
�
dψ.
Define M(a,v)for a > 0,v ≥ 3 by the latter integral.
After these preparations the main step of the proof of the theorem is to show that
the moment
(5) M(a,v)is convex for a > 0,v ≥ 3.
Let
I =
π
�v
0
�
g
h
cos 2 ψ
�
dψ
cos2 , J =
ψ
π
�v
0
g ′� h
cos2 ψ
�
dψ
cos4 �
, K = g
ψ
Elementary calculus yields for the second order partial derivatives of M,
Maa = vh 2 a J (> 0), Mav = (ha + vhav)I + vhahv J − πha
Mvv = (2hv + vhvv)I + vh 2 2πa
v J +
v cos2 π
v
�
π
�
− hav K.
v3 v cos 2 π v
h
cos 2 π v
K,
�
.