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. (Note that the property that [math]\displaystyle{ f_2((3/2)^j x_0) }[/math] is independent of j is preserved under these sorts of scaling limits in the 3/2 direction.)

Define a 23-slice to be any set of the form [math]\displaystyle{ \{ 2^a 3^b x: a,b \in {\Bbb Z} \} }[/math] for some rational x.

Corollary 4 If there exists a function of drift at most 2, then there exists a function f of drift at most 2 such that on each 23-slice S, we either have [math]\displaystyle{ f(2x)=-f(x) }[/math] for all x in S, or [math]\displaystyle{ f(3x)=-2f(x) }[/math] for all x in S.

Proof We begin with a function f as given by Corollary 3. Pick a slice, and suppose that the first claim fails on this slice, thus f(2x)=f(x) for some x in S. By Proposition 1, we see that f(3x/2)=-f(x). By Corollary 3, we conclude that [math]\displaystyle{ f(2^j (3/2) x) = - f(2^j x) }[/math] for all integers j. In particular, f(2(3/2) x) = f(3/2 x). Iterating this, we see that f( 2^j (3/2)^{i+1} x) = - f(2^j (3/2)^i x) for all integers j and natural numbers i. Taking a scaling limit along 3/2 as in Corollary 3, we may then assume that f(3y)=-2f(y) for all y in this slice S. Repeating this argument once for each slice and diagonalising we obtain the claim.

NB: This is the maximum one can say just by looking at the 23-smooth numbers 1,2,3,4; if f has the structure of Corollary 4, then the drift inside 1,2,3,4 is automatically at most 2.