Singular support: equivalent definition

Question

Let $U\subset\mathbb{R}^{d}$ be an open set. The singular support of a distribution $u\in\mathcal{D}^{\prime}(U)$ is defined to be the compliment of the set of points, which have a neighbourhood in which $u$ can be identified with a smooth function. In a paper Fourier integral operators. I of Hörmander, it is claimed that

$$\DeclareMathOperator\singsupp{sing supp}\singsupp(u)=\bigcap_{\varphi\in C^{\infty}_{c}(U),\varphi u\in C^{\infty}(U)}\{x\in U\mid \varphi(x)=0\}.$$

However, I don't see why it is the case. Does anyone know how to see that this is the case?

My attempt:

The direction $\Leftarrow$ is quite clear I think: Let $x\notin\singsupp(u)$. Then there is an open neighbourhood $N$ of $x$ on which $u\vert_{N}\in C^{\infty}$. Let $\chi\in C_{c}^{\infty}(U)$ be supported in $N$ such that $\chi(x)\neq 0$. Then $\chi u\vert_{N}\in C^{\infty}(U)$ and hence $x\notin\bigcap_{\varphi\in C^{\infty}_{c}(U),\varphi u\in C^{\infty}(U)}\{x\in U\mid \varphi(x)=0\}$.

What is left is the other direction. I essentially have to show that if there exists a test function $\varphi\in C^{\infty}_{c}(U)$ such that $\varphi u\in C^{\infty}(U)$ and $\varphi(x)\neq 0$, then $x\notin\singsupp(u)$.

Answer 1

The basic idea is the one put forward by Pierre PC's comment above. More precisely, let $u,\varphi,x$ be as in the last paragraph of the OP. There is no loss of generality in assuming that $\varphi(x)=\lambda>0$. Then by continuity of $\varphi$ at $x$ there is an open neighborhood $V\ni x$, $V\subset U$ such that $\varphi(x')>\frac{\lambda}{2}$ for all $x'\in V$, so that by the chain rule $\psi=\frac{1}{\varphi|_V}\in C^\infty(V)$. One then clearly has that $u|_V=\psi(\varphi u)|_V\in C^\infty(V)$ (exercise: check this!), thus proving the direction $\Rightarrow$ of the claim.

By the way, this (together with the reasoning used in the OP to get the $\Leftarrow$ direction of the claim) is essentially the same argument used to prove a similar characterization of the support of a distribution: $$\mathrm{supp}\,u = \bigcap_{\varphi\in C^\infty_c(U),\phi u\equiv 0}\{x\in U\ |\ \phi(x)=0\}\ .$$

Answer 2

Let $U$ be an open subset of $\mathbb R^d$ and let $u\in \mathscr D'(U)$. Then we have $$ (\text{supp } u)^c=\{x\in U, \exists V \text{open neighborhood of $x$ such that}\ u_{\vert V}=0\}, \tag{1}$$ and $$ (\text{singsupp } u)^c=\{x\in U, \exists V \text{open neighborhood of $x$ such that}\ u_{\vert V}\in C^\infty(V)\}. \tag{2}$$ Note that restrict a distribution to an open set always makes sense. Also we get from (2) that \begin{multline} (\text{singsupp } u)^c=\{ x\in U, \exists V \text{open neighborhood of $x$ such that} \\ \exists \phi\in C^\infty_c(V), x\in \text{supp}\phi, \quad \phi u\in C^\infty_c(V)\}. \tag{3} \end{multline} To get the equivalence between (2) and (3), you have to notice that for $\phi\in C^\infty_c(V)$, $V$ open neighborhood of $x_0$ $$ \phi u\in C^\infty_c(V), \phi(x_0)\not=0\Longrightarrow \text{$u$ is smooth near $x_0$}. $$ To prove the latter implication, you just have to notice that if $\chi \in C^\infty_c(\{x, \vert\phi(x)\vert>\frac12\vert\phi(x_0)\vert\})$, we have $$ \chi u=\frac{\chi }{\phi} \phi u. $$