Now we can start to prove the multiple recurrence theorem in the weak mixing case. Again the material is from Furstenberg’s book ‘Recurrence in Ergodic Theory and Combinatorial Number Theory’ which, very unfortunately, is hard to find a copy of.
Definition: A sequence converges in density to if there exists of density s.t. for all neighborhood of , and .
We denote .
Theorem: For weakly mixing,
then , we have
In particular, this implies for any with , by taking we have
Hence we may pick for which
Establishes the multiple recurrence theorem.
To prove the theorem we need the following:
Lemma 1: Let be a bounded sequence in Hilbert space , if as , as . Then converges weakly in density to
In order to prove the lemma 1, we need:
Lemma 2: Given a family of density subsets of , indexed by density set . Then for all of positive upper density, for all . There exists , such that whenever and .
Proof of lemma 2: We’ll show this by induction. For , since there is no such , the statement is vacant.
We’ll proceed by induction: Suppose for , there exists of positive upper density and integers such that and for all , and .
For , we shall find with positive upper density and where and for all , and .
Let has positive upper density .
Claim: has positive upper density.
Since , let .
Hence there is at most translates of that pairwise intersects in sets of density .
Let be the largest number of such sets, let be a set of numbers where has density .
i.e. has density .
Therefore for any , has positive upper density for some . Hence . is syntactic with bounded gap hence has positive upper density.
(Hence for each )
has positive upper density, has density , has positive upper density.
satisfied the desired property by construction. Hence we have finished the induction.
Proof of lemma 1:
Suppose not. We have some ,
has positive upper density.
Let , let has density .
, let has density .
Apply lemma 2 to , we get:
For all . There exists , such that whenever , and .
Set . Hence
On the other hand, since is bounded in is also bounded (independent of ). Suppose for all ,
then we have
For large , contradiction.
Hence must have density .
Proof of the theorem:
By corollary 2 of the theorem in part 1, since is weak mixing, is weak mixing for all .
We proceed by induction on . For , the statement is implied by our lemma 2 in part 1.
Suppose the theorem holds for , let ,
By induction hypothesis,
Hence it suffice to show
For all , set
For each , let
Since is measure preserving, we substitute ,
Apply induction hypothesis, to the weak mixing transformation and re-enumerate
By lemma 2 in part 1, we have as .
We are now able to apply lemma 2 to , which gives under the weak topology.
i.e. for any , we have .
Establishes the induction step.
Remark: This works for any group of commutative weakly mixing transformations. i.e. if is a group of measure preserving transformations, all non-identity elements of are weakly mixing. are distinct elements in , then as .