** — 1. Introduction — **

When studying measure preserving systems (defined below) there are many important classes that are worth studying separately. One way to distinguish between different classes is the level of “mixing” or “randomness” of the system. In this post, a measure preserving system is a quadruple where is a set, is -algebra, is a probability measure on and is an invertible bi-measurable map that preserves the measure, i.e., for all (where, as usual, ). In some cases where they are clear, we may omit the reference to or .

For instance, if there exists a nontrivial set (nontrivial means that neither nor ) which is preserved under , then the system is as far from mixing as possible – the set doesn’t mix with the rest of the system. A system where this does **not** happen is called ergodic. Hence ergodic systems are in some sense more “chaotic” than non-ergodic systems because for any set with positive measure, we have that is the whole space (up to measure).

One of the best ways to express the mixing behavior of an ergodic system is the Ergodic Theorem:

Proposition 1 (von Neumann’s Ergodic Theorem)Let be an ergodic probability preserving system and let . ThenIn particular, for we have

This shows that requiring ergodicity, which is not hard to see is equivalent to the property that for any two sets with there exists some such that , implies that the measure of the intersections actually behaves rather regularly. This also suggests some weak form of asymptotic independence (in a probabilistic sense) between any set and the orbit of a set . In fact, a much stronger level of mixing is characterized precisely by this asymptotic independence:

Definition 2 (Strong mixing)A measure preserving system is called strongly mixing if for any two sets we have

It should be immediate to see that a strongly mixing system is always ergodic, however, the converse does not hold. An example to have in mind here are the circle rotations. Let be the circle group and let . Consider the map defined by (the sum is defined). Then the system is ergodic if and only if . This can be proved directly, analyzing the behavior of intervals, or using Fourier analysis. However, this system is not mixing! To see this, consider and small intervals, it should be clear that there are arbitrarily large for which does not intersect . Intuitively speaking this happens because the image of an interval is still an interval, it does not mix on .

In this post we will analyze a notion called weak mixing. A weakly mixing system is always ergodic, but not always strongly mixing. I will show the equivalence of several properties about a measure preserving system to weak mixing, the fact that so many distinct properties are equivalent gives some evidence that weak mixing is in some ways a better notion than strong mixing. Moreover, the notion of weak mixing plays a very important role on Furstenberg’s proof of the Szemeredi’s Theorem. Instead of giving a definition of weak mixing I will instead present a Theorem stating that several properties of a measure preserving system are equivalent. A system satisfying one (and hence all) of those properties is called weakly mixing.

Before I state the Theorem, I need to defined product system: given two mps and we can form the product where is the -algebra in the cartesian product generated by the rectangles (sets of the form where and ) and is the product measure, so that in particular . Also, for a subset we define its upper density by . Here, as usual, for a finite set we use the notation to denote its cardinality.

Theorem 3Let be a mps. Then the following are equivalent

- For any two sets we have
- For any we have
- is ergodic.
- For every ergodic m.p.s. , the product is ergodic.
- For any there exists a subset with upper density such that for . Moreover if is separable we can choose independent of .
- For any with there is such that .
- If and the orbit closure is compact (both the closure and compactness are in the strong topology) then is a constant.
- If a.e. for some function and some then is a constant function.

In the next section I prove this Theorem, for now I will just present some remarks about it. The first condition is a strengthening of the ergodic Theorem and is a clear consequence of the strongly mixing property. This explains why we call such systems weakly mixing. The second property is just a rewriting of the first, with the characteristic functions being replaced with general functions.

The third condition is more surprising, this may be the easiest way to check that the circle rotations , defined above, are not weak mixing. This is also the easiest way to define weak mixing, and when extending the notion of weak mixing to other settings (relative weak mixing, weakly mixing unitary operator on Hilbert spaces, weakly mixing action of groups different than , etc) this seems to be the best property to use. The fourth condition is even more surprising as the implication (3)(4) seems quite unlikely a priori. This condition implies also that if is weak mixing, then is also ergodic, and so is . Therefore we get the amusing property that is ergodic if and only if it is weak mixing!

The fifth condition is another immediate weakening of the strongly mixing condition and should be compared with (1). To be completely clear, to say that for means that for all there is some such that for all , we have .

The condition (6) reflects the mixing philosophy as a strengthening of the ergodicity assumption. Moreover, it hints at why the condition (3) is equivalent to the other properties. Finally the conditions (7) and (8) seems unrelated with the others a priori, but provide a very useful spectral characterization of weakly mixing systems.

** — 2. Proof of the Theorem 3 — **

- (1)(5)
Fix and set . Observe that

Taking the limit as we conclude that for all For each let be the smallest positive integer such that for all we have and make

Now observe that for all , hence for each , choosing such that we have and hence . Taking (note that also because all have density) we conclude that .

Finally, for each , let , then if we also have and so concluding the proof.

In the case when is separable, let be a countable dense family. For each let be such that and for . As above we construct a set of density such that for all there exists such that . It is not hard to check that this set satisfies the conditions, we omit the details.

- (5)(6)
Let be such that , let be such that , and

Clearly , hence we can find such that both and .

- (6)(8)
We proceed by contradiction. Assume that satisfies (6) but not (8). Then is ergodic and there is some non-constant eigenfunction with for some . Since preserves the measure we conclude that , and because the system is ergodic this implies that is a constant and hence we can assume .

Let be such that , where . Since is not constant, we can find two disjoint intervals , both of length smaller than some and that are more than apart (in the circle metric, so that and are identified as the same point) and such that both and have positive measure. Also make . We now have that

But because the length of is smaller than and is more that apart from (and hence also is more that apart from ) we conclude that can not intersect both and at the same time. In other words either or , contradicting (6) as desired.

- (7)(8)
This is trivial because the orbit of an eigenfunction is one dimensional, hence its closure is compact.

- (2) (7)
We proceed by contradiction. Assume that (2) holds but there exists some non-constant function such that the orbit closure is compact. Observe that replacing with if needed, we can assume that . We want to prove that , so for the sake of a contradiction let’s assume otherwise.

Let be positive and let be such that the balls cover the orbit of . Then for each there is some such that . We have

Thus, averaging over we obtain

Now using (2) we can find such that the right hand side of the above equation is smaller than which is a contradiction.

- (8)(2)
This is essentially the hard part of the Koopman-von Neumann Decomposition. I posted before on this blog about this theorem, and so I will not present a proof here. Using the notation from that post, the Proposition 5 there says that every function in satisfies our condition (2). Moreover, if there are no non-constant eigenfunctions for , then is just the subspace of constant functions.

- (2)(4)
Let and . Applying Cauchy-Schwartz with and using (2) we get

By the triangular inequality we get the same result replacing and with finite linear combinations of functions of the form , in other words, we can choose . Since is dense in we conclude that the same result holds for any .

Hence satisfies (2) and hence it is clearly ergodic.

- (4)(3)
It suffices to show that if (4) holds, then is ergodic. To see this assume that is not ergodic and let be an invariant set such that . Let be the (ergodic) one point system. Then is invariant for and so wouldn’t also be ergodic.

- (3)(1)}
Applying the von Neumann’s Ergodic Theorem (Proposition 1) to the characteristic functions and we obtain

The left hand side of the above equation is

and the right hand side is . Replacing with we can assume that and thus we can conclude (applying the Cauchy-Schwartz inequality and the ergodic theorem):

** — 3. Some more equivalent notions — **

In this section we give three more properties of a measure preserving system that happen to be equivalent to weak mixing. The first proposition shows that weak mixing and “totally” weak mixing (in analogy with total ergodicity) turn out to be the same concept for actions.

Proposition 4Let be a measure preserving system and let be a positive integer. Then is weak mixing if and only if the system is.

*Proof:* We will use the condition (5) from the Theorem 3. Assume first that is weak mixing and let be measurable sets. Let be the set with density such that for . But the the set also has density and clearly for .

Now assume that the system is weak mixing and let be measurable sets. For each , let be the set with density such that for . But then the set still has density and clearly for .

For the next two theorems we will need to use the following generalization of the classical van der Corput Lemma due to Bergelson:

Proposition 5[vdC generalized] Let be a bounded sequence in a Hilbert space . Ifthen as .

I have used this version before on this blog, its proof can be found in this post.

The next theorem states the fact that a system is weak mixing if and only if the sets are asymptotically independent, in the same spirit of the ergodic theorem, that states that a system is ergodic if and only if the sets are asymptotically independent.

Theorem 6Let be an invertible measure preserving system. Then the system is weak-mixing if and only if for any set we have

*Proof:* We only prove one implication (that weak mixing implies this property). The derivation of the other implication is more complicated.

Let and for each let . Also let (so that ) and let . Note that is the constant function for any .

Note that . Since is weak mixing, is still ergodic, so we have

so converge in Cesàro to . Therefore to conclude the proof of the first implication it suffices to show that

This will be achieved if we prove that converges to in the Cesàro sense and the weak topology. We will use the van der Corput trick. Let . Thus from invariance of under

Applying again the ergodic theorem we thus conclude that

Since for any , we have

and so, since is weak mixing, we conclude that

Now we are in the conditions of van der Corput trick and can conclude that indeed converges to in the Cesaro sense which, as we saw above implies the desired convergence.

The next theorem states that the sets become asymptotically independent as .

Theorem 7Let be an invertible measure preserving system. Then the system is weak-mixing if and only if for any set we have

*Proof:* Again we only prove the easy implication: we assume that the system is weak mixing.

Fix and fix . For each let . Let be such that, for we have

Note that for such we get

For each let be such that for we have, for each :

Therefore, if and we get

Our final criterion for weak mixing is related with limits along ultrafilters. We recall that given an ultrafilter and a sequence taking values in some Hausdorff space, we define the limit of along to be some point such that for every neighborhood of the set is in . We recall that if the space is compact, then every sequence converges along (to a unique limit). I posted about this before and I gave a proof of that fact there.

We will be interested in a special class of ultrafilters, called minimal idempotent ultrafilters. A nice introduction of this topic is given in this survey by Bergelson. I explored some properties of general idempotent ultrafilters in a previous post. Minimal idempotents are a subclass of idempotent ultrafilters, for us it will suffice to know that if is such an ultrafilter and then is piecewise syndetic, and in particular, the set is syndetic.

Theorem 8Let be a measure preserving system and let be a minimal idempotent. Then the system is weak mixing if and only if for every we have

*Proof:* Assume first that the system is not weak mixing. By the property (8) of the Theorem 3 there exists some non-constant such that a.e. for some . We can assume that . But then

We recall that because preserves the measure. Since is idempotent we get

Therefore and hence .

Now we assume that the system is weak mixing. Let and let (note that the limit always exists in the weak topology because the ball with radius is compact). Then because is idempotent we get that

We observe that

so in this case in the strong topology. Now fix and note that let . But then, for any we have and thus for any we have that . Because is syndetic, there is some such that any can be decomposed as where and . Therefore the balls with cover the orbit of and hence the orbit closure of is compact. Since the system is weak mixing we conclude by the condition (7) of the Theorem 3 that must be constant.

To conclude that is indeed just note that the inner product (where is the constant function equal to ) is always .

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