Limits with better properties

Click here to return to the main Polymath5 page

From an integer sequence to a function defined on the rationals

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 n!a 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

Improved properties in the limit

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.)

Dichotomies

Suppose, as we may, that we have a rational example [math]\displaystyle{ (y_a) }[/math] with discrepancy at most C. That instantly gives us a whole host of other examples, since for every rational b the function [math]\displaystyle{ y_{ba} }[/math] is an example, and any pointwise limit of a subsequence of these further examples (which is the same as an element of the closure of the set of those examples in the product topology) is again an example. We have already seen that we can get improved properties in the limit. What happens if we try to force the limit to take certain values in certain places by choosing an appropriate sequence of dilations?

To get some handle on this question, let us look at a couple of simple examples. Suppose, for instance, that I want to find an example [math]\displaystyle{ (z_a) }[/math] such that [math]\displaystyle{ z_1=z_2 }[/math]. If I can find any r such that [math]\displaystyle{ y_r=y_{2r} }[/math] then I am done, since I can simply set [math]\displaystyle{ z_a=y_{ra} }[/math]. So if I cannot find an example with [math]\displaystyle{ z_1=z_2 }[/math] then I know that [math]\displaystyle{ y_a=-y_{2a} }[/math] for every a, which is a very strong multiplicativity property.

This is not particularly interesting, because the property [math]\displaystyle{ z_1=z_2 }[/math] will rapidly be lost if one dilates the function [math]\displaystyle{ z_a }[/math] and passes to limits. So let us try to do something more infinitary.

Suppose that I want to pass to an example that is multiplicative. I can do this if for every finite multiset A of rationals there exists some rational r such that [math]\displaystyle{ y_ry_{rab}=y_{ra}y_{rb} }[/math] for every a,b in A. The reason is that if I have such an r for a given set A and I set [math]\displaystyle{ z_a=y_{ar}/y_r }[/math], then [math]\displaystyle{ z_{ab}=z_az_b }[/math] for every a,b in A. Taking a collection of sets A whose union is all of [math]\displaystyle{ \mathbb{Q} }[/math] and passing to the limit of the corresponding z-sequences, we obtain an example that is multiplicative everywhere.

This is basically just a simple compactness argument that says either there is a multiplicative example or there is some finite set A such that for every r we find a "failure of A-multiplicativity at r". For example, if A is the multiset {2,2,3}, then it tells us that there is no r such that both the equalities [math]\displaystyle{ y_ry_{4r}=y_{2r}^2 }[/math] and [math]\displaystyle{ y_ry_{6r}=y_{2r}y_{3r} }[/math] hold.

This is quite a strong non-multiplicativity property. It would be interesting to investigate experimentally whether insisting on non-multiplicativity of this kind forces the discrepancy to become large.