Drift

The drift of a sequence [math]\displaystyle{ f: {\Bbb Q}^+ \to \{ -1, +1 \} }[/math] is the maximal value of [math]\displaystyle{ |f(mx)+\ldots+f(nx)| }[/math] for all rational x and all natural numbers m<n. The Erdos discrepancy problem is equivalent to showing that the drift is unbounded.

It seems likely that one can show that the drift is at least 3. Here are some preliminary results in this direction:

Proposition 1 If f has drift at most 2, and f(x) = f(2x), then f(3x/2) = f(3x) = -f(x).

Proof If f has drift at most 2, then |f(x)+f(2x)+f(3x)| and |f(x)+f(3x/2)+f(2x)| <= 2, and the claim follows.

Corollary 2 If f has drift at most 2, and the function [math]\displaystyle{ f_2(x) := f(x)f(2x) }[/math] equals 1 for some x, then it equals 1 for [math]\displaystyle{ (3/2)^j x }[/math] for any [math]\displaystyle{ j=0,1,2,\ldots }[/math].

Corollary 3 If there exists a function of drift at most 2, then there exists a function f of drift at most 2 such that [math]\displaystyle{ f_2(x) = f_2(3x/2) }[/math] for all x.

Proof For any fixed [math]\displaystyle{ x_0 }[/math], we look at the functions [math]\displaystyle{ f_j(x) := x \mapsto f((3/2)^{-j} x) }[/math] as [math]\displaystyle{ j \to \infty }[/math]. By Corollary 2, the quantity [math]\displaystyle{ f_j(x_0) f_j(2x_0) }[/math] is eventually constant in j. Taking a subsequence limit as j goes to infinity, we can ensure that [math]\displaystyle{ f_2((3/2)^j x_0) }[/math] is independent of j. We then vary [math]\displaystyle{ x_0 }[/math] over the rationals and use a diagonalisation argument to obtain the result.

The next step seems to be to reduce to a function for which f(3x) = -f(2x) for all x. I thought I could do this, but realised my argument did not work.