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.

  • $\begingroup$ 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? $\endgroup$ – bartgol Jul 29 '14 at 18:44
  • $\begingroup$ 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. $\endgroup$ – bartgol Jul 29 '14 at 18:51
  • 4
    $\begingroup$ 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. $\endgroup$ – J. E. Pascoe Jul 29 '14 at 18:56
  • $\begingroup$ Oh, of course. I read the question a little too fast. Silly me. =P $\endgroup$ – bartgol Jul 29 '14 at 19:05
  • $\begingroup$ 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. $\endgroup$ – 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.

| cite | improve this answer | |
  • 2
    $\begingroup$ Sorry, maybe I'm just being dense, but why does linear dependence imply the space is finite dimensional? Certainly if you have one dependence relation, you will have infinitely many, but it doesn't seem obvious at first glance that I should get a finite dimensional space. $\endgroup$ – Peter Luthy Jul 29 '14 at 18:48
  • 3
    $\begingroup$ 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. $\endgroup$ – J. E. Pascoe Jul 29 '14 at 18:54
  • 1
    $\begingroup$ @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... $\endgroup$ – bartgol Jul 29 '14 at 19:10
  • 1
    $\begingroup$ 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. $\endgroup$ – J. E. Pascoe Jul 29 '14 at 19:17
  • 1
    $\begingroup$ $\cos x$ isn't in $L^2.$ $\endgroup$ – J. E. Pascoe Jul 30 '14 at 0:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.