Vanishing convolution between density and compactly supported function

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?

• 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" – Iiro Ullin May 15 at 22:39
• @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). – Jeff S May 15 at 22:55
• 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.) – Iiro Ullin May 15 at 23:08
• 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... – Iiro Ullin May 15 at 23:32
• The Riemann-Lebesgue lemma says the Fourier transform of an $L^1$ function is continuous (immediate from dominated convergence) and vanishes at infinity. – 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 $$g \equiv 0$$.

• 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? – Jeff S May 16 at 1:26
• (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. – Nate Eldredge May 16 at 1:30
• @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. – Nate Eldredge May 16 at 1:33
• @ChristianRemling: Yeah, that's why I think Onur's answer is better. I forgot that fact and rederived a weaker version. – Nate Eldredge May 16 at 18:00
• 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. – Jeff S May 17 at 15:34