Lemma 3.1: Suppose a (0, , 0)-leveraged-splitter is applied
to a sequence of 2N elements, and let (1−)N k
(1+)N and l = 2N −k, where 0 < < 1 and 0 < 1.
Then at most ( + )N of the k largest elements end up in
the left half of the sequence and at most ( + )N of the l
smallest elements end up in the right half of the sequence.