Gruber P. Convex and Discrete Geometry

29 Packing of Balls and Positive Quadratic Forms 425
(2) Let {B d + t : t ∈ T } be a packing of the unit ball B d . Then
�
f (�s − t�) ≤ 1 for each s ∈ E d .
t∈T
Since {B d + t : t ∈ T } is a packing, any two distinct vectors t ∈ T have distance
at least 2. Hence
�
�t j − tk� 2 ≥ 4n(n − 1) for any distinct t1,...,tn ∈ T.
j,k
An application of Blichfeldt’s inequality (1) implies that:
Thus
�
�s − t j� 2 ≥ 2(n − 1) for any distinct t1,...,tn ∈ T and s ∈ E d .
j
�
f (�s − t�) = �
f (�s − t j�) = � � 1
1 −
2 �s − t j� 2�
t∈T
j
≤ n − (n − 1) = 1fors ∈ E d ,
where, for given s, the points t1,...,tn are precisely the points t of T with �s −t� <
√ 2, i.e. those points t of T with f (�s − t�) >0. The proof of (2) is complete.
In the third step we show the following:
(3) Let {B d + t : t ∈ T } be a packing of Bd . Then its upper density is at most
d + 2
2
2
− d 2 .
Let K be the cube {x :|xi| ≤1}. Proposition (2) and the definition of f then yield
the following:
V � (τ + 2 √ 2)K � �
� �
≥
f (�s − t�) � ds
≥
=
�
(τ+2 √ 2)K
�
t∈T ∩(τ+ √ 2)K
√
2B d
√
+t
2
�
t∈T ∩(τ+ √ 2)K 0
= 2
d + 2 2 d 2
�
t∈T
f (�s − t�) ds =
S(rB d ) f (r) dr =
�
t∈T ∩(τ+ √ 2)K
V (B d ) ≥ 2 d 2 +1
d + 2
j
�
�
t∈T ∩(τ+ √ 2)K
√
2B d
�
t∈T ∩(τ+ √ 2)K
dV(B d �
)
f (�s�) ds
√ 2
0
r d−1�
1 −
�
V � (B d + t) ∩ τ K � ,
t∈T
r 2�
dr
2