As proved in ERT1, the existence of ergodic averages
implies recurrence results for measure-preserving system (from now on, denoted by mps). A natural question is to ask about some kind of generalized ergodic averages and its implications in recurrence. By generalized ergodic averages we mean expressions like

where
is a sequence of positive integers. Von Neumann’s Theorem shows that convergence holds if
, where
are positive integers: just apply the result to the transformation
and the function
. In this post, we prove that the same result holds if
, where
is a polynomial such that
, for every
. We can assume, without lost of generality, that
. In fact,
and
satisfies the required condition.
Theorem 1 (H. Furstenberg) If
is a mps and
is a polynomial such that
, for every
, and
, then the limit

converges in
for every
.
Again, this theorem is Hilbertian in nature and follows from a more general version for unitary operators.
Theorem 2 If
is a unitary operator on a Hilbert space
and
is a polynomial such that
, for every
, and
, then the sequence of operators

converges pointwise in norm.
Proof: The idea is the same as in Von Neumann’s Theorem: we look for an orthogonal decomposition
such that the behaviour of
is understood in each component.
will represent the structured component of
and
the randomic one in the following sense:
- the long-time behaviour of elements
is (almost-)periodic.
- the long-time behaviour of elements of
self-amortizes and converges to zero.
Unfortunately, the decomposition of Von Neumann’s Theorem does not work here. In fact, let
be periodic with respect to
, say
,
. If we write
,
,

converges to
, because

This means that every periodic point of
has a structured behaviour with respect to
. For this reason, let

Exercise 1 Prove that the set of
for which the sequence
converges is a closed subspace of
.
By Exercise 1, the sequence
converges whenever
. By linearity, it remains to prove convergence for
. Such subspace is characterized by

This follows from Von Neumann’s Theorem: if
is the decomposition with respect to
,
, then
is equal to
and its orthogonal complement is given by

which proves (1). This means that
for every
and for every degree-one polynomial
,
. The proof will be complete if we show that the same happens for larger-degree polynomials. By induction, suppose that

for every polynomial
such that
,
and
. Consider
such that
,
and
. We wish to reduce the convergence
to one of the form
, with
(which we know to be true, by the induction hypothesis). This is done with the use of Van der Corput’s Trick (see this lecture of Terry Tao for a broader discussion on this trick).
Theorem 3 (Van der Corput Trick) If
is a bounded sequence such that
for every
, then
Exercise 2 Prove the above theorem. (Hint: this is Theorem 2.2 of this survey of Vitaly Bergelson.)
We’re done if the sequence
,
, satisfies the conditions of Theorem 3. In fact, as
is unitary,

where
is a polynomial of smaller degree, and so (2) is satisfied. This concludes the proof of Theorem 2. 
The method used above is one of the main principles Ergodic Ramsey Theory: the dichotomy between structure and randomness, decomposing the object of study into these two components. Usually, we first define the structured one, in terms of the desired ergodic averages, so that convergence follows almost directly from the definition. Its orthogonal complement is the randomic component and convergence along it is proved using Van der Corput like theorems. For a further discussion on this dichotomy, the reader is referred to this paper of Terence Tao. Observe that the same method applies to prove the following
Theorem 4 If
is a unitary operator on a Hilbert space
and
is a polynomial such that
, for every
, and
, then the sequence of operators

converges pointwise in norm.
Now it’s time to obtain the recurrence consequences (which, as expected, will be stronger than those in ERT1). Let
be the orthogonal projection. We’ll proceed exactly as in Proposition 6 of ERT1, except that the notation will be heavier.
Proposition 5 Let
be such that
. Then
and
.
Proof: Consider the subspaces
,
. By approximation, if each projection
of
into
satisfies
and
, the same happens to
. Fix
and consider the function
. Then
(Exercise 3) and
. Because
minimizes the distance of
to
, we have
. In addition, if we had
, then

implying that
for some
. Integrating, we conclude

a contradiction. 
Theorem 6 If
is a mps,
is a polynomial such that
, for every
,
, and
such that
, then the set

is syndetic.
Proof: If
, then from (2) the expression
converges to
as
. Since

Theorem
guarantees the conclusion. 
Previous posts: ERT0, ERT1.