proofs

146 On a lemma of Littlewood and Offord
Erdős was right, but it took twenty years until Gyula Katona and Daniel
Kleitman independently came up with a proof for the complex numbers
(or, what is the same, for the plane R 2 ). Their proofs used explicitly the
2-dimensionality of the plane, and it was not at all clear how they could be
extended to cover finite dimensional real vector spaces.
But then in 1970 Kleitman proved the full conjecture on Hilbert spaces
with an argument of stunning simplicity. In fact, he proved even more. His
argument is a prime example of what you can do when you find the right
induction hypothesis.
A word of comfort for all readers who are not familiar with the notion of
a Hilbert space: We do not really need general Hilbert spaces. Since we
only deal with finitely many vectors a i , it is enough to consider the real
space R d with the usual scalar product. Here is Kleitman’s result.
Theorem. Let a 1 , . . . ,a n be vectors in R d , each of length
at least 1, and let R 1 , . . . , R k be k open regions of R d , where
|x − y| < 2 for any x, y that lie in the same region R i .
Then the number of linear combinations ∑ n
i=1 ε ia i , ε i ∈ {1, −1},
that can lie in the union ⋃ i R i of the regions is at most the sum of
the k largest binomial coefficients ( n
j)
.
In particular, we get the bound ( n
⌊n/2⌋)
for k = 1.
Before turning to the proof note that the bound is exact for
a 1 = . . . = a n = a = (1, 0, . . .,0) T .
Indeed, for even n we obtain ( ) (
n
n/2 sums equal to 0, n
n/2−1)
sums equal to
(−2)a, ( n
n/2+1)
sums equal to 2a, and so on. Choosing balls of radius 1
around
−2⌈ k−1
k−1
2
⌉a, . . . (−2)a, 0, 2a, . . . 2⌊
2 ⌋a,
we obtain
( )
n
⌊ n−k+1 + . . . +
2
⌋
( ) ( ) n n
n−2
+
n
+
2 2
( ) ( n
n+2
+ . . . +
2
n
⌊ n+k−1
2
⌋
sums lying in these k balls, and this is our promised expression, since the
largest binomial coefficients are centered around the middle (see page 12).
A similar reasoning works when n is odd.
Proof. We may assume, without loss of generality, that the regions R i
are disjoint, and will do so from now on. The key to the proof is the recursion
of the binomial coefficients, which tells us how the largest binomial
coefficients of n and n − 1 are related. Set r = ⌊ n−k+1
2
⌋, s = ⌊ n+k−1
2
⌋,
then ( ) (
n
r , n
(
r+1)
, . . .,
n
s)
are the k largest binomial coefficients for n. The
recursion ( ) (
n
i = n−1
) (
i + n−1
i−1)
implies
)