Limits with better properties

Let [math]\displaystyle{ (x_n) }[/math] be an infinite [math]\displaystyle{ \pm 1 }[/math] sequence of discrepancy at most C. Using this sequence, we can define a "rational sequence" [math]\displaystyle{ (y_a)_{a\in\mathbb{Q}} }[/math] in one or other of the following two ways (which look different but are essentially the same).

Method 1

For each positive rational number a and each positive integer n, let [math]\displaystyle{ f_n(a)=x_{n!a} }[/math] if n!a is an integer and let it be undefined (or arbitrarily defined) otherwise. Pick a subsequence of the functions [math]\displaystyle{ f_n }[/math] that converges pointwise and define [math]\displaystyle{ y_a }[/math] to be the limit of the values of [math]\displaystyle{ f_n(a) }[/math] along this subsequence.

Method 2

Define the functions [math]\displaystyle{ f_n }[/math] as above, but this time define [math]\displaystyle{ y_a }[/math] to be the limit of [math]\displaystyle{ f_n(a) }[/math] along a non-principal ultrafilter U. It is simple to check that the result will again be a pointwise limit of a subsequence of the functions [math]\displaystyle{ f_n }[/math].

Properties of the resulting function

Let us begin by proving that every sequence of the form [math]\displaystyle{ y_a,y_{2a},y_{3a},\dots }[/math] (which we shall call an HAP-subsequence) has discrepancy at most C.

This is trivial. Let [math]\displaystyle{ g_1,g_2,g_3,\dots }[/math] be a subsequence of the [math]\displaystyle{ f_n }[/math] such that [math]\displaystyle{ f_n(a) }[/math] converges to [math]\displaystyle{ y_a }[/math] for every rational number a. Let m be any positive integer. Then there exists some n such that [math]\displaystyle{ g_n(ka)=y_{ka} }[/math] for every k between 1 and m. We may also pick n large enough so that [math]\displaystyle{ n!a }[/math] is an integer. If we do so, then the numbers n!a, 2n!a,...,mn!a form an HAP, which implies that the partial sum [math]\displaystyle{ g_n(a)+\dots+g_n(ma) }[/math] has absolute value at most C, since it equals the sum [math]\displaystyle{ x_d+\dots+x_{md} }[/math], where d=n!a.

Thus, this construction gives us a way of passing from an example that works over the integers to an example that works over the rationals. However, it does more than that, as the following argument shows. Let us suppose that the r-sequence and the s-sequence of [math]\displaystyle{ (x_n) }[/math] are equal: that is, [math]\displaystyle{ x_{rn}=x_{sn} }[/math] for every positive integer n. What does this tell us about the function [math]\displaystyle{ y_a }[/math]? Obviously, it implies immediately that [math]\displaystyle{ y_{ra}=y_{sa} }[/math] for every rational number a. From that we deduce that [math]\displaystyle{ y_a=y{sa/r} }[/math] for every rational number a (applying the previous result to a/r) and also that [math]\displaystyle{ y_a=y_{ar/s} }[/math]. In other words, once we have passed to the rational limit, we find that the function is invariant under dilation by r/s, which implies that it is also invariant under dilation by s/r.

Further investigation

If we know only that two HAP-subsequences of the sequence [math]\displaystyle{ (x_n) }[/math] are approximately equal, then it is still possible that we might be able to choose the pointwise limit carefully so as to ensure that in the limit we have exact invariance. It would be very interesting to reach a clear understanding of when this is possible and when it isn't. (If the two sequences differ everywhere on some further HAP-subsequence, for example, then it probably isn't possible. But then we might not be inclined to say that the two sequences are approximately equal.)