Find a pair of functions $f,g:\mathbb{R}\to\mathbb{R}$ such that:

  • $f$ is smooth and compactly supported (say, on $[0,1]$ but this isn't crucial),
  • $g(x)>0$ for all $x\in\mathbb{R}$, $\int g(x)\,dx=1$ (i.e. $g$ is a strictly positive density), and
  • $f*g=0$.

If we remove the condition that $g$ is a strictly positive density, then this is possible by choosing $g$ such that its Fourier transform is a sum of point masses (e.g. something like $1+\sin^2 x$) and choosing $f$ so that its Fourier transform has prescribed zeroes at these point masses. But I am not sure if the condition that $g$ is a density changes anything.

A stronger reformulation (due to Paley-Wiener) of this problem is: Is there a strictly positive density whose Fourier transform has a finite number of prescribed zeroes?

  • $\begingroup$ If you want $g$ be in $L^1$, its Fourier transform will never be just a sum of point masses. So I think your construction won't work. You can still try to find $f$ and $g$ whose Fourier transforms have non-intersecting supports, but as you want $f$ to be compactly supported, my intuition tells it wont be possible "at infinity" $\endgroup$ – Iiro Ullin May 15 at 22:39
  • $\begingroup$ @IiroUllin Interesting! Is there a reference/proof for your claim about $L^1$ functions? Also, since $g>0$ everywhere, its support is the entire real line (apologies if this wasn't clear). $\endgroup$ – Jeff S May 15 at 22:55
  • $\begingroup$ Well, functions whose Fourier transforms are point masses are the so-called almost periodic (or plain periodic) functions. They are not integrable over $\mathbb R$. Can't immediately recall a reference or a simple argument why that is true. I'll return to this when/if I think of something concrete and nobody else answers this before that :) (...Also, I was talking about the support of the FT of $g$, not $g$ itself.) $\endgroup$ – Iiro Ullin May 15 at 23:08
  • 1
    $\begingroup$ K, so it's easier to think this way: $g\in L^1$ implies that $\hat g\in L^\infty$, so can't have any point masses... $\endgroup$ – Iiro Ullin May 15 at 23:32
  • 2
    $\begingroup$ The Riemann-Lebesgue lemma says the Fourier transform of an $L^1$ function is continuous (immediate from dominated convergence) and vanishes at infinity. $\endgroup$ – Nate Eldredge May 15 at 23:38

With your hypotheses above, $\widehat{g}:\mathbb{R}\to\mathbb{R}$ is a uniformly continuous function such that $\displaystyle \lim_{|\gamma|\to\infty} \widehat{g}(\gamma) = 0$, and $$\widehat{f}(z) := \int f(t)e^{-2\pi itz} dt \hspace{28mm} (z\in\mathbb{C})$$ is an analytic function, for which the set $\{z\in\mathbb{C}: \widehat{f}(z)=0\}$ cannot have accumulation points unless $\widehat{f}=0$ on $\mathbb{C}$. Thus, $f*g=0$ iff $\widehat{f} \widehat{g}=0$ which implies either $f=0$ or $g=0$.


This is not possible, even if we only assume $f \in C_c(\mathbb{R})$ and $g \in L^1(\mathbb{R})$.

Of course it is equivalent to ask for $\hat{f} \hat{g} \equiv 0$. By the Riemann-Lebesgue lemma (or just dominated convergence), $\hat{f}$ and $\hat{g}$ are both continuous.

Now I claim that the set $\{\hat{f} = 0\}$ is closed (obviously) and nowhere dense, unless $f \equiv 0$. Say $f$ is supported in $[-a,a]$, and suppose that $\hat{f} = 0$ on some interval $(\omega_0 - \epsilon, \omega_0 + \epsilon)$. By replacing $f(x)$ with $e^{i \omega_0 x} f(x)$ we can assume without loss of generality that $\omega_0 = 0$, so $\hat{f} = 0$ on $(-\epsilon, \epsilon)$. In particular this means $\hat{f}^{(n)}(0)=0$ for every $n$, which by differentiating under the integral sign shows that $\int_{-a}^a x^n f(x)\,dx = 0$ for all $n$. By the Weierstrass approximation theorem it follows that $f \equiv 0$.

So the set $\{\hat{f} \ne 0\}$ is dense in $\mathbb{R}$. If we had $\hat{f} \hat{g} \equiv 0$, then the set $\{\hat{g}=0\}$ must also be dense. But $\hat{g}$ is continuous, so this can only happen if $\hat{g} \equiv 0$ and thus $g \equiv 0$.

  • $\begingroup$ Thanks! The generalization you are suggesting is intriguing, so I want to make sure I understand. 1) Should the last line read "So the set $\{\hat{f} = 0\}$ is nowhere dense in $\mathbb{R}$. If we had $\hat{f} \hat{g} \equiv 0$, then the set $\{\hat{g}=0\}$ must be dense."? 2) Why do you assume $\hat{f}$ vanishes on an interval to prove the zero set is nowhere dense? What prevents $\hat{f}$ from having isolated zeroes? $\endgroup$ – Jeff S May 16 at 1:26
  • $\begingroup$ (1) Thanks, that's a typo, I meant that $\{\hat{f} \ne 0\}$ is dense. Fixed now. (2) Since $\{\hat{f}=0\}$ is closed, if it isn't nowhere dense then it has nonempty interior, which is to say that it contains an interval. $\endgroup$ – Nate Eldredge May 16 at 1:30
  • $\begingroup$ @JeffS: In fact we don't even need $f$ to be continuous, merely $L^1$ and compactly supported; we just need an extra approximation step after the Weierstrass approximation theorem. Btw I think Onur's answer is better since it ties this in with standard facts. $\endgroup$ – Nate Eldredge May 16 at 1:33
  • $\begingroup$ @ChristianRemling: Yeah, that's why I think Onur's answer is better. I forgot that fact and rederived a weaker version. $\endgroup$ – Nate Eldredge May 16 at 18:00
  • $\begingroup$ I can't seem to see @ChristianRemling's comment. Since Onur's answer seems to have more support, I swapped it out. I still think Nate's answer is quite nice. $\endgroup$ – Jeff S May 17 at 15:34

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.