Let $f\in L^1(\mathbb{R})$ and let $V_f$ be the closed linear subspace of $L^1(\mathbb{R})$ generated by the translates $f(\cdot - y)$ of $f$. If $V_f=L^1(\mathbb{R})$, I want to show that $\hat{f}$ never vanishes.

We have $\hat{f}(\xi_0)=0$ iff $\hat{h}(\xi_0)=0$ for $h\in V_f$, but I'm not sure how to proceed beyond this. The Riemann-Lebesgue Lemma gives us that $L^1(\mathbb{R})$ is sent to $C_0(\mathbb{R})$ under the Fourier transform, $C_0(\mathbb{R})$ seems to contain functions that vanish...

I got this problem from 3.3 here: http://www.math.ucdavis.edu/~jlirion/course_notes/Prelim_Solutions.pdf.

EDITED: I originally neglected to say that Vf is the closed subspace generated by the translates, not the translates themselves. I don't know if it is reasonable to suppose that this subspace is L1 itself or just a subset...

  • 5
    $\begingroup$ @NateEldredge: Indeed, in the linked paper $V_f$ is the closed linear subspace generated by the translates of $f$ and not the set of translates alone. $\endgroup$ – Giuseppe Negro May 9 '12 at 16:51
  • $\begingroup$ Yikes, thanks for spotting this. $\endgroup$ – user21725 May 9 '12 at 16:52
  • $\begingroup$ @Eric: I think you need to re-edit. It is probably correct with equality. But you need to say "the closed subspace generated by" and not "the space of" $\endgroup$ – Martin Argerami May 9 '12 at 16:53
  • $\begingroup$ I seems to be a corollary of the first part, for example take $f(t)=e^{-\alpha t}\chi_{t\geq 0}$ with $\alpha>0$. We should have that $\frac 1{\alpha-i\xi_0}=0$ which is not possible. $\endgroup$ – Davide Giraudo May 9 '12 at 16:56
  • 3
    $\begingroup$ The point is that $L^1$ contains functions $g$ such that $\hat{g}(\xi_0) \ne 0$, and so those can't be in $V_f$. $\endgroup$ – Robert Israel May 9 '12 at 16:59

Assume by contradiction that

$$\widehat{f}(\psi) =0$$

Then, if $h$ is a translate of $f$ it is easy to show that


Now, let $g \in V_f$. Then $g= \lim h_n$ where $h_n$ are linear combinations of translates of $f$. Thus

$$\widehat{h_n}(\psi) =0$$

and since $h_n \to g$ in $L^1$ we get $\widehat{h_n}(\psi) \to \widehat{g}(\psi)$.

Thus, we indeed get $$ \widehat{g}(\psi) =0 \forall g \in V_f$$

Now to get the contradiction, we need to use the fact that for each $\psi \in \widehat{G}$ there exists some $g \in L^1$ so that $\widehat{g}(\psi) \neq 0$.

This is easy, let $u$ be any compactly supported continuous function on $\hat{G}$ so that $u(\psi) \neq 0$. Then $g= \check{u*\tilde{u}}$ works, where $\check{}$ is the inverse Fourier Transform. This last part can probably be proven much easier, for example if $g$ is non-zero, then $\hat{g}$ is not vanishing at some point, and then by multiplying $g$ by the right character $\widehat{e^{...}g}$ is not vanishing at $\psi$.

  • $\begingroup$ Thanks! I understand the "much easier" way to prove the last, but I don't entirely understand why you take the convolution of $u$ with its conjugate, would you explain this? $\endgroup$ – user21725 May 9 '12 at 17:06
  • 1
    $\begingroup$ Because $\check{u}$ is not necesarily an $L^1$ function, but it is an $L^2$. Then $|\check{u}|^2$ is $L^1$, and $|\check{u}|^2=\check{u*\tilde{u}}$. $\endgroup$ – N. S. May 9 '12 at 17:13
  • $\begingroup$ Appreciate the elaboration, thanks. $\endgroup$ – user21725 May 9 '12 at 17:15

Given $f\in L^1(\mathbb{R})$ let $$I(f)=\operatorname{clos}(\operatorname{span}(f_\lambda:\lambda\in\mathbb{R}))$$ be the $L^1$-closure of the linear span of the translates $f_\lambda$ ($f_\lambda(x)=f(x+\lambda)$). Then the Wiener tauberian (~1932) theorem says that $I(f)=L^1(\mathbb{R})$ if and only if $\hat{f}$ is non-zero.

In fact this can be lifted into the subject of commutative Banach algebras and Gelfand theory.


Your Answer

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