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.](https://projecteuclid.org/download/pdf_1/euclid.cmp/1103920456)


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?
> 
> 
> 



Can we characterize a periodic function by the compactness of the set of its translates?
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.


