Vanishing convolution between density and compactly supported function

Question

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?

Answer 1

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 $g=0$.

Answer 2

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 $g \equiv 0$.