There are two equivalent definitions of mixing for a measure-preserving dynamical system . One is in terms of sets:

for all measurable . The other is in terms of functions:

for all . In both cases one may refer to the quantity inside the absolute value as the *correlation function*, and write it as or , so that (1) and (2) become and , respectively.

Then it is natural to ask how quickly mixing occurs; that is, how quickly the correlation functions decay to 0. Is the convergence exponential in ? Polynomial? Something else?

In a previous post, we discussed one method for establishing exponential decay of correlations using a spectral gap for a certain operator associated to the system. The result there stated that as long as and are chosen from a suitable class of test functions (often Hölder continuous functions), then decays exponentially.

A comment was made in that post that it is important to work with sufficiently regular test functions, because for arbitrary measurable functions, or for the setwise correlations , the decay may happen arbitrarily slowly even when the system is “as mixing as possible”. But no examples were given there — so I’d like to explain this a little more completely here.

Let be the space of infinite sequences of 0s and 1s, and let be the shift map. Let be -Bernoulli measure, so that writing for the *-cylinder* containing all sequences in that begin with the symbols , we have . This is isomorphic to the doubling map on the circle with Lebesgue measure, which was discussed in the previous post and which has exponential decay of correlations for Hölder observables . Nevertheless, we will show that there are sets such that decays quite slowly.

We can think of as modeling sequences of coin flips, with tails corresponding to the symbol 0, and heads to 1.

Let be the set of all sequences in that begin with 1. Then corresponds to the event that the first flip is heads, and corresponds to the event that the st flip is heads.

The description of is a little more involved. Given , let , so that if and only if (at least) one of the symbols is 1. This corresponds to the event that heads appears at least once on or after the th flip and before the th flip. Note that , and so .

Fix an increasing sequence of integers with , and let be the set of all sequences such that every subword contains at least one occurrence of the symbol 1.

Using the interpretation as coin flips, we see that the events are independent, and so

We recall the following elementary lemma.

Lemma 1Let be a sequence of real numbers such that for all . Then if and only if .

It follows from Lemma 1 that if and only if

Thus from now on we assume that the sequence is chosen such that (4) holds. Note that the correlation function can be computed via conditional probabilities: using

(notice that this uses shift-invariance of ), we have

Recalling that is the intersection of the independent events , we let be such that , and observe that . Thus

where the last equality uses the expression for from (3). Together with the expression (5) for in terms of the conditional probability , this gives

where we emphasise that is a function of .

Example 1Take , so that , and . Since , we see that (4) is satisfied, so that .Moreover, satisfies , and so in particular . This implies that , and so we can estimate the correlation function using (6) by

The example shows that the correlation function of sets may only decay polynomially, even if the system has exponential decay of correlations for regular observable functions. In fact, we can use the construction above to produce sets with correlations that decay more or less as slowly as we like along some subsequence of times .

Theorem 2Let be any sequence of positive numbers converging to 0. Then there is a sequence such that the sets from the construction above have .

The result says that no matter how slowly converges to 0, we can choose such that is not bounded above by for any constant .

To prove Theorem 2, we will choose such that is large whenever with even. Let be any increasing sequences of integers (conditions on these will be imposed soon) and define recursively by

Because are increasing, we have

so (4) holds and . The idea is that we will take , so that when , we have small relative to , and hence to .

Let’s flesh this out a little. Given , it follows from (6) that

On the other hand, . Now we may take to be any increasing sequence ( will do just fine) and define recursively in terms of and . We do this by observing that since , given any and , we can take large enough so that for every , we have

In particular, (7) gives for every . Sending completes the proof of Theorem 2.