Suppose that $f$ is in $L^2(\mathbb{R})$ and consider the set of integer translates of this function, $V=\{f(x-k):k\in\mathbb{Z}\}$. This set is linearly independent: taking the Fourier transform of the finite sum $\sum a\_k f(x-k)$ one gets $p(e^{2\pi i\xi})\widehat{f}(\xi)$ for some polynomial $p$. If the sum is zero and $f$ is nonzero, then $p$ must be zero on some set of positive measure; this is an infinite set, implying that $p$ must be the zero polynomial and so each $a\_k$ must be zero. I find this to be an especially nice application of the Fourier transform.


My question is this: does there exist a proof of this fact which does not use the Fourier transform? The $L^2$ condition could be modified, but obviously one needs some kind of integrability condition to disallow the constant functions. One can prove this using a variant of the Fourier transform, so I should say that I'm really looking for a proof where you don't integrate against complex exponentials.


As for why $V$ would be an interesting thing for mathematicians to look at: the closure of the span of $V$ (in $L^2$) is one of the fundamental objects in wavelet theory --- a principal shift-invariant space.



Prove that ..., f(x-2), f(x-1), f(x), f(x+1), f(x+2),... is algebraically linearly independent without the Fourier transform
Suppose there is some linear dependence. If the set is linearly dependent, space $V$ should be finite dimensional.
Fix a finite subset $S$ of $\mathbb{Z}$ so that supposedly the set of $f(x-k)$ where $k\in S$ would span.


Note that as $k\_0 \rightarrow \infty,$ $\langle f(x-k\_0), f(x-k) \rangle \rightarrow 0,$ which would imply $\|f(x-k\_0)\|$ went to $0,$ but $\|f(x-k\_0)\|$ is constant, which is a contradiction.


