Gruber P. Convex and Discrete Geometry

(ii) �b1� ≤2 1 2 (d−1) min � �l� :l ∈ L \{o} �
(iii) �b1�···�bd� ≤2 1 4 d(d−1) d(L)
28 Basis Reduction and Polynomial Algorithms 413
Proof. (i) Since the vectors ˆb j are pairwise orthogonal, condition (3) shows that
Since µ 2 j+1 j ≤ 1 4 by (2),
3
4 � ˆb j� 2 ≤�ˆb j+1� 2 + µ 2 j+1 j � ˆb j � 2 for j = 1,...,d − 1.
follows. Then induction implies that
� ˆb j+1� 2 ≥ 1
2 � ˆb j� 2 for j = 1,...,d − 1
(4) � ˆb j� 2 ≥ 2 i− j � ˆbi� 2 for i, j = 1,...,d, i < j.
In particular,
(5) � ˆb j� 2 ≥ 2 1− j � ˆb1� 2 = 2 1− j �b1� 2 for j = 1,...,d.
Multiplying the latter inequalities for j = 1,...,d, and taking into account the
orthogonality of the system { ˆb1,..., ˆbd} and (1), Statement (i) is obtained as follows:
2 − 1 2 d(d−1) �b1� 2d ≤�ˆb1� 2 ···�ˆbd� 2 = det( ˆb1,..., ˆbd) 2
(ii) Proposition (5) yields
Now apply Lemma 28.1.
(iii) Using (1) and (2), we see that:
So, by (4),
= det(b1,...,bd) 2 = d(L) 2 .
min{� ˆb1�,...,� ˆbd�} ≥ 2 − 1 2 (d−1) �b1�.
j−1
i=1
j−1
�b j � 2 =�ˆb j � 2 �
+ µ 2 ji� ˆbi� 2 ≤�ˆb j � 2 � 1
+
4 � ˆbi� 2 .
i=1
i=1
�b j � 2 � �j−1
1
≤ 1 + 2
j−i�
� ˆb j �
2 2 ≤ 2 j−1 � ˆb j� 2 .
Multiplying this for j = 1,...,d, yields (iii)
�b1� 2 ···�bd� 2 ≤ 2 1 2 d(d−1) � ˆb1� 2 ···�ˆbd� 2 = 2 1 2 d(d−1) d(L). ⊓⊔