1
$\begingroup$

Do we say that a function $f$ is uniformly almost periodic in the aforementioned proof if $f$ is bounded (in the sense that $||f||_{L^\infty}\leq 1$) and that there exists a natural number $d>0$ such that $f\in UAP^d$?

Edit: the proof in question appears in Tao's paper "A quantitative ergodic theory proof of Szemerédi's theorem".

$\endgroup$
3
  • $\begingroup$ I'm not really sure what this question wants. From a quick read of the paper (available here: combinatorics.org/Volume_13/PDF/v13i1r99.pdf) the answer is "yes, but I don't see why we need $\|f\|_{L^\infty}\leq 1$". $\endgroup$ Dec 13, 2010 at 15:49
  • $\begingroup$ The question wants the definition of uniformly almost periodic functions in the paper and not in the infinitary ergodic theory setting. $\endgroup$
    – user4949
    Dec 13, 2010 at 16:38
  • 1
    $\begingroup$ I think uniform almost periodicity here mean being an element of $UAP^d$. $\endgroup$
    – user4324
    Dec 14, 2010 at 11:44

2 Answers 2

1
$\begingroup$

As I mentioned in my comment above, a uniform almost periodic function (of order $d$) is an element of $UAP^d$ for some $d$. The bound $||f||_{L^\infty}\leq1$ is unnecessary.

$\endgroup$
1
  • $\begingroup$ +1, as I agree (and I said this in my comment above as well!) $\endgroup$ Dec 28, 2010 at 14:50
-1
$\begingroup$

I think it means the continuous functions on the Bohr compactification of the reals.

http://en.wikipedia.org/wiki/Almost_periodic_function#Uniform_or_Bohr_or_Bochner_almost_periodic_functions

$\endgroup$
1
  • 1
    $\begingroup$ Well, the setting in Tao's proof is finitary, in the sense that the function $f$ above is defined on a finite cyclic group $Z_N$. Thus, continuous functions on the Bohr compactification of the reals might not be the answer. $\endgroup$
    – user4949
    Dec 13, 2010 at 14:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.