Timeline for Product of distributions under wavefront set condition is zero

8 events

when toggle format	what		by	license	comment
Jul 15, 2022 at 10:05	history	edited	Ceka	CC BY-SA 4.0	added another example in which the claim holds
Jul 13, 2022 at 15:15	history	edited	Ceka	CC BY-SA 4.0	Added an example in which the claim holds.
Jul 12, 2022 at 10:38	history	edited	Ceka	CC BY-SA 4.0	added an example for which the result holds
Jul 11, 2022 at 23:08	history	edited	Ceka	CC BY-SA 4.0	re-formulated the question, the way it was stated it was false
Jul 11, 2022 at 23:06	comment	added	Ceka		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.
Jul 11, 2022 at 22:54	comment	added	Vinícius Novelli		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.
Jul 11, 2022 at 18:19	history	edited	LSpice	CC BY-SA 4.0	Title of Hörmander's book
Jul 11, 2022 at 18:13	history	asked	Ceka	CC BY-SA 4.0