Remember one of the characterizations we got in ERT8 of weak mixing: a mps is weak mixing if and only if, for any ,
that is, if and only if the sequence of functions converges in the -norm to .
For our interests, the notion of weak mixing will be useful only if the above property extends to multiple functions, because, as we saw in ERT4, the convergence of these sequence corresponds to multiple recurrence. This actually holds, as we will see below, and is called weak mixing implies weak mixing of all orders. Such property will follow from two main ingredients:
- The product characterization of weak mixing can be extended to multiple products.
- The van der Corput trick allows an inductive argument.
It is worth mentioning that this is the second time van der Corput trick appears in these lectures (the first one was in ERT2). The reader not used to it might think it is just a technical step in the proof, but it is actually an ingredient present in many situations of Ergodic Ramsey Theory when dealing with random components of mps. It has many versions, each of them to the purpose of particular notions of multiple recurrence. We first discuss the one we need.
1. The van der Corput trick
Theorem 1 (van der Corput trick) If is a bounded sequence in a Hilbert space and if
Proof: Take such that . Notice that, for a fixed ,
goes to zero as and so the assertion is equivalent to
By the triangle inequality,
which goes to zero as .
The significance of this inequality is that it replaces the task of bounding a sum of coefficients by that of bounding a sum of “differentiated” coefficients . This trick is thus useful in “polynomial” type situations when the differentiated coefficients are often simpler (have smaller order) than the original coefficients.
The above theorem is written in a modern fashion. The original van der Corput trick is actually known as van der Corput difference theorem and comes from the theory of uniform distributions.
Definition 2 We say that a sequence is uniformly distributed if, for every interval ,
Exercise 1 Given a sequence , prove that the following assertions are equivalent.
- is uniformly distributed.
- For every continuous , where is the Lebesgue measure.
- (Weyl criterion) For any ,
Proof: By Weyl criterion and the assumption, . Fix , and define . Then
converges to zero as . By van der Corput trick,
converges to zero which, again by Weyl criterion, proves that is uniformly distributed.
The previous result implies Weyl equidistribution theorem, which is a generalization of Kronecker theorem on the uniform distribution of , for irrational .
Theorem 4 (Weyl) If is a polynomial with at least one of its coefficients irrational, then , , is uniformly distributed.
Proof: Follows from Kronecker theorem by sucessive applications of Theorem 3.
2. Weak mixing implies weak mixing of all orders
Remember that is weak mixing if and only if is ergodic whenever is ergodic. For each , let .
Proposition 5 If is weak mixing, then is ergodic, for every integers . In fact, it is weak mixing.
Proof: Note that is weak mixing, for any . To prove this, consider and of zero density such that
Let . This set has zero density, as
and the last fraction converges to zero.
We proceed by induction on . The case was proved in the last paragraph. Suppose that, for every integers , is ergodic. If ,
is the product of an ergodic and a weak mixing system, which proves our assertion.
The main result of this section is
Theorem 6 Let be weak mixing. Then, for any ,
in the -norm.
The proof will consist in an inductive argument, with the use of van der Corput trick to reduce the case to and so on. By linearity of the expression, we assume that . In this setting, we want to prove that
where represents the norm of .
Proof: The case follows from the ergodicity of :
Suppose the result is true for and take . Define
Given , let . Then
Rewrite the above expression as
where and is given by
As is ergodic,
By the van der Corput trick,
which concludes the proof.
Corollary 7 (Multiple Poincaré Recurrence for weak mixing mps) If is weak mixing and , then
for every .
Proof: Just consider in (1).