In my previous post I presented an ergodic theoretical proof of Roth’s Theorem, assuming the Koopman-von Neumann Decomposition (and some other minor facts). In this post I present a proof of this Decomposition and moreover prove that the compact vectors form a factor.

** — 1. Introduction — **

Recall that a measure preserving system is a quadruple where is a Borel space, is a probability measure and is an invertible measure preserving map. We denote by the Hilbert space of all functions on and define the unitary operator by . In this post we are interested in writing as the direct sum of two orthogonal subspaces, one contains the compact vectors and the other contains the weak-mixing vectors. We defined these objects now:

Definition 1Let . We say that is acompactoralmost periodicfunction if the orbit closure is compact as a subset of with the strong topology. The set of all compact functions is denoted by .

Definition 2Let . We say that is aweak-mixingfunction ifThe set of all weak-mixing functions is denoted by .

We can now state the Koopman-von Neumann Decomposition:

Theorem 3 (Koopman-von Neumann)The set is a closed subspace of and the it’s orthogonal complement is .

** — 2. Proof of the Koopman-von Neumann Decomposition — **

We separate the proof into some lemmas:

Lemma 4Let be a function that commutes with (so that ) and is uniformly continuous.Then that for also .

We remark that, for instance, the function satisfies the hypothesis.

*Proof:*

Let and fix . Let be such that if are such that and then .

Also let and be finite covers of the orbit closure of and (respectively) by balls with diameter less than . Then for any we have and for some depending on . Therefore and by construction has diameter at most . This implies that the orbit of is contained in the union of finitely many sets with diameter (namely ) and therefore is totaly bounded. Hence is a compact function.

Now I will state a fact about weak-mixing functions. In the proof I will need the van der Corput Trick that I proved in my previous post.

*Proof:*

Let . We have

By the van der Corput Trick we conclude that

and thus

We can now prove that and are orthogonal sets:

*Proof:*

Fix . Let be such that the ball cover the orbit of . Thus for a fixed let be such that . We have

Thus we have

Since , using lema 5 with for large enough we have that the second term can be made smaller than . Thus we have and since was arbitrary we conclude that actually .

We now need a converse of the previous lemma:

Lemma 7Let be not weak mixing. Then there exist some such that .

*Proof:* Consider the operator . This is rank one and thus a Hilbert-Schmidt operator. In the Hilbert space of all Hilbert-Schmidt operators on (with the Hilbert-Schmidt norm) define the unitary operator by . Thus .

By the von Neumann’s Mean Ergodic Theorem, we have that

exists, is in and is invariant under . In other words commutes with .

We will prove that satisfies the claims. Note that so

The last inequality is from the definition of weak mixing function (which is not) and we can pass the limit outside the inner product because convergence in the Hilbert-Schmidt norm implies convergence in the weak operator topology.

Now all that remains to prove is that . But since is an Hilbert-Schmidt operator (and hence compact) and commutes with , the orbit of (under ) is the image under of the orbit of . But since is unitary, the orbit of is bounded and therefore the orbit of is pre-compact. In other words, is a compact vector.

We are now ready to prove the Koopman-von Neumann Decomposition

*Proof:* of Theorem 3

First note that is a -invariant set, because and have the same orbit, and it’s closed under scalar multiplication, because the scalar multiple of a compact set is still compact.

We claim that is a closed set because the orbit of a limit point of functions in has a totaly bounded (and hence pre-compact) orbit. Indeed let and fix . Choose such that . Since is compact there exists a finite set such that the balls cover the orbit of . For each let be such that . Then

Now by Lemma 4 we have that is closed under addition and hence is a closed invariant subspace of . By Lemma’s 6 and 7 we conclude that if and only if it is orthogonal to , hence and this concludes the proof.

** — 3. The Kronecker Factor — **

Given a measure preserving system we call a factor to a sub--algebra of which is invariant under , i.e., such that for any we have that as well.

Proposition 8 (Kronecker factor)The family consisting of all sets such that is a factor. We call the Kronecker factor of .

*Proof:*

Since is an invariant subspace of we have that is an invariant family of subsets. Also since the constant function is in (and it is closed for sums) we get that is invariant under complements.

Since the function given by is uniformly continuous, by Lemma 4 and the Monotone Convergence Theorem we get that is also closed under countable unions, so it’s indeed a -algebra.

The important fact about this factor is the following

Proposition 9Let . Then if and only if it is measurable with respect to the Kronecker factor .

*Proof:*

If is measurable with respect to then can be approximated by simple functions in . Since is a closed subspace also .

Reciprocally if , then for each we have, by Lemma 4, that , so also for each and so . Taking the limit when and using the Dominated Convergence Theorem we conclude that the characteristic function of is in and so is measurable with respect to as desired.

An immediate Corollary of this is this Proposition left unproved in my last post.

As another application of Lemma 4 I now finish the proof of Roth’s Theorem (which in my last post I showed follows from the following theorem), using two key Propositions from my last post:

Theorem 10 (Roth’s Theorem, ergodic version)Let be an ergodic measure preserving system and let have positive measure. Then there exists such that .

*Proof:* Let be the characteristic function of and using Theorem 3 split it as where and . Note that because otherwise the function is also in (by Lemma 4) and would be closer to then . Moreover, representing the constant function equal to (which is compact and thus orthogonal to ) simply by we have

We showed that is in the conditions of Lemma 13 in my previous post. Moreover, by Lemma 12 in that post, we conclude that

Therefore there is at least on such that .

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