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Assume $u, v \in \mathcal{D}'(\mathbb{R}^n)$ are distributions with compact support. Denote by $\operatorname{WF}(\bullet) \subset T^*\mathbb{R}^n \setminus 0$ the wavefront set of a distribution $\bullet$. If $\operatorname{WF}(u) \cap \operatorname{WF}(v) = \emptyset$, then their product $uv$ is well defined. If $uv = 0$, does this imply that at an open and dense set of points one of the two distributions vanish, that is, $\operatorname{supp}(u)^c \cup \operatorname{supp}(v)^c \subset \mathbb{R}^n$ is open and dense?

For the background on the wavefront set and distribution theory, see Chapter 8 of Hörmander's book The Analysis of Linear Partial Differential Operators I. For products under the wavefront set condition, see Theorem 8.2.10 of the same book. I've also asked this question on Mathematics StackExchange.

Remarks:

  • As noted below by Vinicius in the comments, if $u, v \in C^\infty_c(\mathbb{R})$ with $\operatorname{supp}(u) = [-1, 0]$ and $\operatorname{supp}(v) = [0, 1]$, then $uv = 0$ but $\operatorname{supp}(u)^c \cup \operatorname{supp}(v)^c = \mathbb{R}\setminus 0$, so open and dense is the most we can hope for.

  • If $u \in C_c^\infty(\mathbb{R}^n)$ and $v \in \mathcal{D}'(\mathbb{R}^n)$, then $uv = 0$ implies that $v = 0$ on the open set $\{u \neq 0\}$. But for every $x \in \mathbb{R}^n$ for which $u(x) = 0$, either $u = 0$ in a neighbourhood of $x$ or there is a sequence $x_n \to x$ such that $u(x_n) \neq 0$. In either case the density of $\operatorname{supp}(u)^c \cup \operatorname{supp}(v)^c$ clearly follows. The same argument works if both distributions are in $C^0(\mathbb{R}^n)$ or in suitable Lebesgue spaces.

  • The same question makes sense also if $u$ and $v$ don't have compact support. In that case, if for simplicity we set $n = 2$ and there are transversal smooth vector fields $X$ and $Y$ such that $X u = 0$ and $Y v = 0$, then the wavefront set condition is automatically satisfied. In fact, in a suitable local coordinate system adapted to $X$ and $Y$, one can show that $uv$ is locally a tensor product and so at each point either $u = 0$ or $v = 0$. One should keep in mind the trivial case when $X = \partial_{x_1}$ and $Y = \partial_{x_2}$.

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    $\begingroup$ If you take $u,v \in C^{\infty}_c(\mathbb{R})$ with $\text{supp}\,u=[-1,0]$ and $\text{supp}\,v=[0,1]$, then $uv\equiv 0$ but $(\text{supp}\,u)^{c}\cup (\text{supp}\,v)^{c}=\mathbb{R}\setminus \{0\}$. Maybe one can hope that $(\text{supp}\,u)^{c}\cup (\text{supp}\,v)^{c}$ is a dense open subset of $\mathbb{R}^n$ in general. $\endgroup$
    – Vinícius Novelli
    Commented Jul 11, 2022 at 22:54
  • $\begingroup$ Thank you! That is quite obvious now... as the statement is not even correct for smooth functions! I will modify the question. Your suggestion sounds good, as in that case the claim is true for smooth functions: indeed if $u$ and $v$ are smooth, then $uv = 0$ implies $u = 0$ on $\{v \neq 0\}$ and vice versa. So $\operatorname{supp}(u)^c \cup \operatorname{supp}(v)^c$ contains an open and dense set. $\endgroup$
    – Ceka
    Commented Jul 11, 2022 at 23:06

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