Is the set of translations of a function compact?

Question

Let $X=BUC(\mathbb{R})$ be the Banach space of real bounded uniformly continuous functions on $\mathbb{R}$ equipped with the supremum norm. Let $f\in X$, then the subset $$\{f_a:t\mapsto f(t+a), \ \ a\in\mathbb{R} \}\subset X$$ is a bounded subset of $X$. Is it compact ?

Answer 1

Translations are usually the simplest way to construct bounded non-compact sets in function spaces. This example is no exception. Take $f=e^{-x^2}$. If the family $f_a$ had a limit point that should be the null function, as you can see by fixing a point $x$ and letting $a\to \infty$. But something is wrong, as $f_a$ cannot converge uniformly to the null function, because $\|f_a\|=1$. So the set $\{f_a\}$ is not compact.

Answer 2

Denote by $\mathcal T$ the set of translates. Define $f_a\star f_b:=f_{a+b}$: then $(\mathcal T,\star)$ is a compact topological group. The map $\Psi\colon a\mapsto f_a$ is a continuous bijective morphism and we deduce that $f$ is periodic.

If we merely assume that $\mathcal T$ has a compact closure, then $f$ is almost periodic: it can be approximated for the uniform norm by linear combinations of functions of the form $t\mapsto e^{i\lambda t}$, $\lambda\in\mathbb R$.