Pointwise convergence of almost periodic function

This question is inspired by this post and definitions I use can be found there.

Fix $f$ a real-valued almost periodic (but not periodic) function. For each $t\in\mathbb{R}$, define $f_t(x) = f(x-t)$ for each $x\in\mathbb{R}$. By the definition of an almost periodic function, $\overline{\{f_t\}_{t\in \mathbb{R}}}^{\|\cdot\|_{\infty}}$the closure of the set $\{f_t\}_{t\in \mathbb{R}}$ is compact with respect to $\|\cdot\|_{\infty}$ the supremum norm. As a result, for any $g\in C_0(\mathbb{R})$, $\overline{\{g_t\}_{t\in \mathbb{R}}}^{\|\cdot\|_{\infty}}$ is not compact.

My question is the following: could there be a $g\in C_0(\mathbb{R})$ such that $g$ is a clustered point of the set $\{f_t\}_{t\in \mathbb{R}}$ with respect to the pointwise convergence topology or the weak topology? Since the set of translation of a function from $\partial \overline{\{f_t\}_{t\in \mathbb{R}}}^{\|\cdot\|_{\infty}}$ the boundary points of $\{f_t\}_{t\in \mathbb{R}}$ with respect to $\|\cdot\|_{\infty}$, is not necessarily included in $\{f_t\}_{t\in \mathbb{R}}$, then I wonder if I choose a topology that is properly weaker than the $\|\cdot\|_{\infty}$-topology, can I find a function whose set of translation is "large" enough to be non-compact.