Linear independence of translates of a function $\{ \phi(\cdot - x) : x \in \mathcal F \}$ Let $0 \neq \phi \in L^2(\mathbb R^n)$ be a square-integrable function and $\mathcal F \subset \mathbb R^n$ a finite set. If we are in the one-dimensional setting $n=1$ then the set of translates of $\phi$ by $\mathcal F$, i.e.
$$
\{ \phi(\cdot - x) : x \in \mathcal F \} \subset L^2(\mathbb R)
$$
is linear independent. I found this result in the book Christensen, In introduction to frames and Riesz bases, p. 228.
I was wondering if this statement holds in arbitrary dimensions, i.e. without restriction of $n$ to $n=1$. If yes, does somebody knows a reference for this?
Thanks in advance!
 A: I do not know what Christensen's proof is but here is a simple proof that works in $R^n$. Suppose that translates are linearly dependent:
$$\sum c_j\phi(x-t_j)\equiv 0,$$
where $t_j$ are all distinct.
Take Fourier transform; shift correspnds to multilication on an exponential:
$$\sum c_j e^{-it_j\cdot s}\hat{\phi}(s)\equiv 0.$$
But the multiplier
$$m(s):=\sum c_j e^{-it_js}$$ is a non-zero entire function, since all $t_j$ are distinct, so its zeros make a proper analytic subset of $R^n$, and such a set must be of zero measure. Therefore $\hat{\phi}(s)=0$
almost everywhere, so $\phi=0.$
A: A standard proof of the independence of the exponentials $f_j(x)=e^{ix\cdot \xi_j}$ is the following. Take a vector $\omega \in \mathbb R^n$ such that $\omega\cdot (\xi_k-\xi_j)\neq 0$ for $k \neq j$ (this follows by induction on the numbers of vectors)  and let $D=\omega \cdot \nabla$. Then $D(f_j)=i\omega \cdot \xi_j f_j$ so that the $f_j$ are eigenvectors associated to distinct eigenavalues of a linear operator.
