Does a function exists to make this set dense? Let $f \in L^2(\mathbb{R})$, then we can show that $\overline{\text{span}}\{f(t-a)\}_{a \in \mathbb{R}} = L^2(\mathbb{R})$ if and only if $\hat{f}(\xi) \neq 0$ almost surely, by using the uniqueness of the Fourier transform. I am wondering if we can relax this assumption? That is, does there exists $f \in L^2(\mathbb{R})$ such that $\overline{\text{span}}\{f(t-a)\}_{a \in \mathbb{Q}} = L^2(\mathbb{R})$ (now running over only rational numbers)?
Initially my (wrong) idea was to choose $f$ so that $\hat{f} = \mathbb{1}_{[0,1]}$. This would then say that if $g \in L^2(\mathbb{R})$ with $0 = \langle g, f(t-a)\rangle_{\mathbb{R}} = \langle \hat{g}, e^{-2 \pi i a \xi} \hat{f}\rangle_{\mathbb{R}} = \langle \hat{g}, e^{-2 \pi i a \xi}\rangle_{[0,1]}$. Since this holds for every $a \in \mathbb{Q}$ it in particular holds for all integers. Thus $\hat{g} = 0$ and hence $g = 0$. But this only guarantees being $0$ on $[0,1]$!
The second thing that comes to mind is since we know that if we had $0 = \langle g, f(t-a)\rangle$ for all $a \in \mathbb{R}$, then this would imply that $g = 0$ and hence we have $\overline{\text{span}}\{f(t-a)\}_{a \in \mathbb{R}} = L^2(\mathbb{R})$. If we only have the assumption that $0 = \langle g, f(t-a)\rangle$ for all $a \in \mathbb{Q}$ could we use the density of $\mathbb{Q}$ to extend conclude that $0 = \langle g, f(t-a)\rangle$ for all $a \in \mathbb{R}$?
Perhaps there is a simple function $f$ where this clearly holds, but nothing comes to mind. Thanks.
 A: The conclusion is valid for any dense subset $A$ of the real line.
Equivalently it remains to show that the functions
$$e^{-2\pi i a\xi}\widehat{f}(\xi),\qquad a\in A,$$ span a dense subset of $L^2(\mathbb{R}).$  Call the closure of this span by $V.$
Let $a\in \mathbb{R}.$ There exists a sequence $a_n\in A$ such that $a_n\to a.$ Then
\begin{equation}\|e^{-2\pi i a_n\xi} \widehat{f}(\xi)-e^{-2\pi i a\xi} \widehat{f}(\xi)\|_2\to 0 \ \ \ \  \ \ \ \ \ \ (*)
\end{equation}
Therefore the functions $e^{-2\pi ia\xi}\widehat{f}(\xi)$ belong to $V$ for any $a\in \mathbb{R}.$ Now we can use the fact that the conclusion is true for $A=\mathbb{R}.$
Proof of $(*)$:
$$\|e^{-2\pi i a_n\xi} \widehat{f}(\xi)-e^{-2\pi i a\xi} \widehat{f}(\xi)\|_2^2=\int\limits_{-\infty}^\infty |e^{-2\pi i(a_n-a)\xi}-1|^2\,|\widehat{f}(\xi)|^2\,d\xi \\ =
\int\limits_{-N}^N |e^{-2\pi i(a_n-a)\xi}-1|^2\,|\widehat{f}(\xi)|^2\,d\xi
+\int\limits_{|\xi|>N} |e^{-2\pi i(a_n-a)\xi}-1|^2\,|\widehat{f}(\xi)|^2\,d\xi \\ \le \int\limits_{-N}^N |e^{-2\pi i(a_n-a)\xi}-1|^2\,|\widehat{f}(\xi)|^2\,d\xi
+4\int\limits_{|\xi|>N} |\widehat{f}(\xi)|^2\,d\xi $$
For given $\varepsilon>0,$ choose $N$ such that the last integral is less than $\varepsilon/2.$ Once $N$ is fixed, the first integral tends to $0$ when $n\to \infty,$ as the functions $e^{-2\pi i(a_n-a)\xi}$ tend to $1$ uniformly in the interval $[-N,N].$
