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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?

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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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 $g=0$.

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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 $g \equiv 0$.

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  • $\begingroup$ 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

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