Gruber P. Convex and Discrete Geometry

382 Geometry of Numbers
For this, it is sufficient to prove the following implication:
(2) Let x ∈ E d . Then x ∈ 1
2 (λ1 +···+λd)C + l for suitable l ∈ L.
To see this, choose d linearly independent points l1,...,ld ∈ L such that li ∈ λiC.
Represent x in the form x = x1l1 +···+xdld with xi ∈ R. Choose u1,...,ud ∈ Z
such that |xi − ui| ≤ 1 2 and let l = u1l1 +···+udld. Then
x − l = (x1 − u1)l1 +···+(xd − ud)ld ∈ λ1 λd
C +···+
2 2 C
= 1
2 (λ1 +···+λd)C.
The proof of (2), and thus of (1), is complete, concluding the proof of the right-hand
inequality. ⊓⊔
A Result of Perron and Khintchine on Diophantine Approximation
We first state a result due to Mahler [1939].
Lemma 23.1. Let
P = � x :|ai1x1 +···+aidxd| ≤1, i = 1,...,d � ,
Q = � x :|bi1x1 +···+bidxd| ≤1, i = 1,...,d �
be two parallelotopes in E d such that A ∗ = (aik) −T = (bik) = B. Let λP =
λ1(P, Z d ) and λQ = λ1(Q, Z d ). Then
Proof. To see that
note that
λP ≤ � � 1
d λQ| det A| d−1 ,
λQ ≤ � d λP| det B|
� 1
d−1 .
P ∗ = � y :|b11y1 +···+b1d yd|+···+|bd1y1 +···+bddyd| ≤1 � ,
{x :|xi| ≤1} ∗ ={y : x · y ≤ 1 for all x with |xi| ≤1} ={y :|y1|+···+|yd| ≤1}
and, for an o-symmetric convex body C,
(A −1 C) ∗ ={z : z · A −1 x ≤ for all x ∈ C}
={A T A −T z : z T A −1 x = (A −T z) T x = A −T z · x ≤ 1 for all x ∈ C}
={A T y : y · x ≤ 1 for all x ∈ C} =A T C ∗ .
Thus,
P ∗ = � A −1 {x :|xi| ≤1} � ∗ = A T {x :|xi| ≤1} ∗
= B −1 {y :|y1|+···+|yd| ≤1}
={z :|b11z1 +···+b1d zd|+···+|bd1z1 +···+bddzd|},