Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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".

share|improve this question
    
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 $\|f\|_{L^\infty}\leq 1$". –  Matthew Daws Dec 13 '10 at 15:49
    
The question wants the definition of uniformly almost periodic functions in the paper and not in the infinitary ergodic theory setting. –  user4949 Dec 13 '10 at 16:38
1  
I think uniform almost periodicity here mean being an element of $UAP^d$. –  Dee Dec 14 '10 at 11:44
add comment

2 Answers 2

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.

share|improve this answer
    
+1, as I agree (and I said this in my comment above as well!) –  Matthew Daws Dec 28 '10 at 14:50
add comment

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

share|improve this answer
1  
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. –  user4949 Dec 13 '10 at 14:54
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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