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Given a function $f$, let us define the translates $f_t(x)=f(x-t)$. A (Bochner) almost-periodic function is a bounded continuous function on $\mathbb R^\nu$ such that the set of functions $\{f_t\vert t\in\mathbb R^\nu\}$ form a precompact set with respect to the supremum norm (a precompact set is a set whose closure is compact).

This definition is taken from Appendix 1 of this paper.

I was wondering if we insisted that the set is compact rather than just precompact, is this equivalent to $f$ being periodic? That is,

Is it true that a bounded continuous function $f$ on $\mathbb R^\nu$ is periodic if and only if the set of functions $\{f_t\vert t\in\mathbb R^\nu\}$ form a compact set with respect to the supremum norm?

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  • $\begingroup$ And if you pick a continuous function vanishing at infinity ? $\endgroup$
    – Ayman Moussa
    Commented Oct 22, 2019 at 9:27
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    $\begingroup$ @Ayman Moussa if the function vanishes at infinity, the closure of the set of translates is certainly not compact, so that function is not almost periodic $\endgroup$
    – Darren Ong
    Commented Oct 22, 2019 at 10:10
  • $\begingroup$ Oh sorry. I read too quickly. With the answer below (which is nice by the way), I guess that simply observing that $t\mapsto f_t$ is a continuous bijection from $\mathbb{R}$ to a compact set $K$ is not enough : can someone give a simple example of such a continuous map which is not a homeomorphism ? Of course the other way around ($f:K\rightarrow \mathbb{R}$) is not possible. $\endgroup$
    – Ayman Moussa
    Commented Oct 24, 2019 at 9:29
  • $\begingroup$ Note that in higher dimensions compactness of the set of translates is equivalent to the function being fully periodic. $\endgroup$
    – Nick S
    Commented Aug 13, 2020 at 0:52

1 Answer 1

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I would also like to focus on the case $\nu =1$ (to avoid some slightly annoying but probably trivial bookkeeping issues).

Suppose that $\{f_t\}$ is compact. This shows first of all that $f$ is uniformly continuous, or else we could shift problematic points to zero, say, to obtain a sequence with no convergent subsequence.

Now if $f$ wasn't periodic, then $d(s,t)=\|f_s-f_t\|$ defines a translation invariant metric on $\mathbb R$, and $(\mathbb R,d)$ is compact, by assumption. The identity map $(\mathbb R, |\cdot |)\to (\mathbb R, d)$ is continuous (since $f$ is uniformly continuous). This implies that $(\mathbb R,d)$ is still pathwise connected. Moreover, $\mathbb R$ is a topological group also with the metric $d$: for example, if $d(s_n,s), d(t_n,t)\to 0$, then \begin{align*} \|f_{s_n+t_n}-f_{s+t}\| &= \|f_{s_n} - f_{s+t-t_n}\| \le \| f_{s_n}-f_s\| + \|f_s-f_{s+t-t_n}\| \\ & = \|f_s-f_{s_n}\| + \|f_{t_n}-f_t\| \to 0 . \end{align*}

A pathwise connected compact abelian metric group is a torus (see Theorem 8.46(iii) of the tome of Hoffmann and Morris for this step), but clearly this is absurd here since a torus has torsion and $\mathbb R$ doesn't. So $f$ is periodic.

The other direction is of course trivial.

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    $\begingroup$ Great argument! Instead of using the characterization of pathwise connected compact abelian metric groups one could also use that every continuous group isomorphism between two separable locally compact groups is automatically a homeomorphism (this follows from a Baire category argument). $\endgroup$
    – Jochen Glueck
    Commented Oct 22, 2019 at 20:22
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    $\begingroup$ @JochenGlueck: Thanks! I was actually just wondering if there shouldn't be a way to wrap it up with less intimidating (to me) machinery. Also, I don't really understand anything in HM, I just remembered the statement because I had quoted it before, in a completely different context (not on MO). $\endgroup$
    – Christian Remling
    Commented Oct 22, 2019 at 20:26
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    $\begingroup$ By the way, your argument also shows more generally that, if $(T_t)_{t \in \mathbb{R}}$ is a $C_0$-group on a Banach space $X$ and $x\in X$ has compact orbit, then the orbit is periodic. I think that's quite nice. $\endgroup$
    – Jochen Glueck
    Commented Oct 23, 2019 at 10:07

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