Let be a probability space and a measure-preserving transformation. Let be Banach spaces of measurable functions on , and the corresponding norms. Given and , the corresponding th correlation is
One sometimes sees the statement of exponential decay given in the following form, which is formally stronger than (2): there are constants and , independent of , such that
In fact, using the Baire category theorem, one can prove that (2) implies (3) under a mild condition on the Banach spaces ; this is the goal of this post, to show that can be chosen uniformly over all , and that can be chosen to have the form . This seems like the sort of thing which is likely known to experts, but I am not aware of the reference in the literature. (I would be happy to learn a reference!)
Proposition 1 Let be a probability measure-preserving transformation and Banach spaces of measurable functions on with the following properties:
- given any , we have and ;
- the inclusions are continuous;
- for every , the map given by is continuous, and similarly for the map when is fixed;
- the map given by is bounded w.r.t. .
To prove the proposition, start by fixing and , and consider the function given by
It follows from (2) that
Moreover, the sets are nested (smaller gives a bigger set, larger gives a bigger set) and so it suffices to take the union over rational values of , meaning that we can treat (5) as a countable union. In particular, by the Baire category theorem there are such that the closure of has non-empty interior. The next step is to show that
- is closed, so it itself has non-empty interior;
- in fact, contains a neighbourhood of the origin.
For the first of these, observe that by the assumptions we placed on the Banach spaces , there is a constant such that
for every . In particular,
Given a sequence such that w.r.t. as , it follows that
and we conclude that , so this set is closed. In particular, there is and such that if , then . By the same token , and now the sublinearity property (4) gives
and so . This shows that contains a neighbourhood of 0, and writing , we see that for every we have , and so
Thus we conclude that
To complete the proof of the proposition, it suffices to apply the same argument once more. Writing
we see from (2) that , and so once again there are such that the closure of has non-empty interior. Given a sequence with , we have for all and all , and so , demonstrating that this set is closed.
Thus there are and such that implies . In particular this gives , and so for every and every we have
But then for every we can consider , which has , and so
which proves (3) and the proposition.