According to Bohr, the definition of the almost periodic function is: A function $f:\mathbb{R}\rightarrow \mathbb{C}$ is called almost periodic if it is continuous and if for every positive $\epsilon$, there exists a positive number $l$ such that every closed interval of length $l$ contains an $\epsilon$-almost period.

Intuitively, when $\epsilon$ is getting smaller, the corresponding $l$ is getting larger. But is this always correct? If it is correct, may I ask how to prove it? Also, is there any relation between $\epsilon$ and the corresponding $\epsilon$-almost period?

  • $\begingroup$ To complete @Nik Weaver answer, I think that it is easy to show that there exists some $\epsilon_0$ such that $l$ is the same for all $0 < \epsilon <\epsilon_0$ (i.e. $l$ does not increase as $\epsilon$ decreases) if and only if $f$ is periodic. $\endgroup$
    – Nick S
    Jun 24 '20 at 18:01

It seems to me that if an interval is an $\epsilon$-almost period then it is trivially an $\epsilon'$-almost period for any $\epsilon' >\epsilon$. So as $\epsilon$ gets smaller the set of intervals which contain $\epsilon$-almost periods shrinks (i.e., fewer intervals have this property).

As to the second question, consider the function $\sum_1^\infty 2^{-n} e^{ib_nx}$ where $(b_n)$ is any sequence going to infinity. In order to contain a $2^{-n}$-almost period you'll need length on the order of $b_n$. So there's no upper bound on the length of the intervals as a function of $\epsilon$.

  • 1
    $\begingroup$ (The series should be $\sum 2^{-n}e^{ix/b_n}$.) $\endgroup$
    – Nik Weaver
    Aug 30 '18 at 15:56

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