# 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 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.

• That set is not linearly independent (what if $f$ is odd?). Perhaps you meant that there is an infinite subset of $V$ that is linearly independent? – bartgol Jul 29 '14 at 18:44
• Actually, I'm pretty sure I can build a function $f\in L^2$ that is equal to 1 on every integer. You just need to make the spikes narrower and narrower as you get to infinity. That function would not be in $H^1$, of course, but that's another story. – bartgol Jul 29 '14 at 18:51
• Bartgol, he's talking about the functions $f(x-k) \in L^2,$ not the values of $f$ at the integers which wouldn't be well defined anyway. – J. E. Pascoe Jul 29 '14 at 18:56
• Oh, of course. I read the question a little too fast. Silly me. =P – bartgol Jul 29 '14 at 19:05
• I wonder if anything could be done with the bi-infinite sequence $\langle a \rangle$ defined by $a_i=\sqrt{\int_i^{i+1}f^2(x)dx}$ so $|| f ||=\sum_{-\infty}^{\infty}a_i^2.$ Dependence of shifts of $f$ do not correspond to dependence of shifts of $\langle a \rangle$ (i.e. a linear recurrence) but maybe some inequality can be brough to bear. – Aaron Meyerowitz Jul 29 '14 at 21:34

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.
• Well suppose you had a linear dependence, i.e. there $\sum^n_{k=-l} a_kf(x-k) =0,$ where $a_{-l}$ and $a_n$ are non-zero. By applying the shift, we get that $\sum^n_{k=-l} a_kf(x-(k+t)) =0.$ That is, for any $m > n$ or $m< -l$ we can write $f(m)$ in terms of a recurrence relation. – J. E. Pascoe Jul 29 '14 at 18:54
• @Pascoe: I agree that that inner product goes to 0, but I don't see why that would imply the norm of $||f(x-k_0)||$ goes to 0 too. It's like trying to get strong convergence from weak convergence... – bartgol Jul 29 '14 at 19:10
• I guess the simplest answer would be that the function $\|v\|_S = (\sum_{k\in S} \langle v,f(x-k) \rangle^{2})^{1/2}$ defines a norm on $V$ which is finite dimensional, and all norms on a finite dimensional space are equivalent in the sense that they induce the same topology. – J. E. Pascoe Jul 29 '14 at 19:17
• $\cos x$ isn't in $L^2.$ – J. E. Pascoe Jul 30 '14 at 0:06