This is the theorem 8.9 of the book Introduction to PDE, Folland, G. B. I'm trying to complete the proof details. I would like to know how to justify the "Afirmation", with more rigor.

Every pseudo-differential operator is pseudo-local, that is, if $P\in \Psi^m(\Omega)$, then, for all $u\in \mathcal{E}'(\Omega)$, $\hbox{ sing supp } Pu \subset \hbox{ sing supp } u $.

Proof: Let $P\in \Psi^m(\Omega)$ and $u\in \mathcal{E}'(\Omega)$. Given $y_0 \notin \hbox{ sing supp } u$, choose a neighborhood $V$ of $\hbox{sing supp} u$ in $\Omega$ which do not meets the set $\{y_0\}$. Choose $\varphi \in C_0^\infty(\Omega)$ verifying $\hbox{supp } \varphi \subset V$, such that $\varphi =1$ in a neighborhood $U \subset V$ of $\hbox{ sing supp } u$ and define $u_1=\varphi u$ and $u_2 = (1-\varphi)u$. Then $u = u_1+u_2$, $\hbox{ supp } u_1 \subset V$ and $u_2= (1-\varphi)u\in C_0^\infty(\Omega)$. Indeed, $\hbox{ supp } u_1 = \hbox{ supp }(\varphi u) \subset \hbox{ supp } (\varphi)\cap \hbox{ supp } u\subset V$ and, in addition, $\hbox{ supp }(u_2) \subset \hbox{ supp }((1-\varphi))\cap \hbox{ supp } u\subset \hbox{ supp } u$ and $\hbox{ supp } u$ is a compact set because $u\in \mathcal{E}'(\Omega)$. To show that $u_2 \in C^\infty(\Omega)$, observe that if $x \in \Omega \setminus \hbox{ sing supp } u$, then $u_2$ is $C^\infty$ in $x$. Thus, it is sufficient to prove that $u_2$ is $C^\infty$ in $U$. Let $\psi \in \mathcal{D}(U)$, then $\langle u_2|_{U}, \psi \rangle = \langle u, (1- \varphi)|_{U} \widetilde{\psi}|_{U} \rangle =0$, since $(1-\varphi)|_{U}=0$, where $\widetilde{\psi}$ denotes the extension of $\psi$ zero outside of $U$. Thus, $u_2|_{U}=0$ and consequently $u_2 \in C^\infty(\Omega)$.

Hence, $Pu=Pu_1 + Pu_2$, and $P u_2$ is a $C^\infty(\Omega)$ function since $u_2\in C_0^\infty(\Omega)$. If $k_p$ is the distributional kernel of $P$, the Theorem 8.8 gives that $k_p(x, y)$ is a $C^\infty$ function in $(\Omega\times \Omega) \backslash \Delta_{\Omega \times \Omega}$. In particular, $k_p(x,y)$ is a $C^\infty$ function for $x \notin V$ and $y \in V$, that is, if $Z$ is a neighborhood of $x$ which do not meets the set $V$, then $k_p \in C^\infty(Z \times V)$. Thus, for $x \notin V$ we have \begin{eqnarray*} P u_1(x) = \left<k_p(x,\cdot),u_1 \right>_{\mathcal{E}(V),\mathcal{E}'(V)}. \end{eqnarray*}

In fact, let $u \in \mathcal{E}'(V)$. Since $C_0^\infty(V)$ is dense in $ \mathcal{E}'(V)$, there exists $(u_n) \subset C_0^\infty(V)$ such that $u_n \rightarrow u$ in $ \mathcal{E}'(V)$. Then, \begin{eqnarray*} Pu_{n}(x)& = & \int_{\mathbb{R}^{n}} e^{2 \pi i x \cdot \xi} p(x, \xi) \hat{u}_{n}(\xi) d \xi\\ & = & \int_{\mathbb{R}^{n}} \int_{\mathbb{R}^{n}} e^{2 \pi i (x - y) \cdot\xi} p(x, \xi) u_{n}(y) dy d \xi \\ & = & \int_{\mathbb{R}^{n}} \left[\int_{\mathbb{R}^{n}} e^{2 \pi i (x - y)\cdot \xi}p(x, \xi) d \xi \right] u_{n}(y) dy \\ & = & \int_{\mathbb{R}^{n}}{k_p (x, y)u_{n}(y)}dy \\ & = & \left< k_p (x,\cdot),u_n \right>_{\mathcal{E}(V),\mathcal{E}'(V)}. \end{eqnarray*} Passing to the limit when $n$ goes to infinity, we obtain the desired equality.

Afirmation: This implies that $D^\alpha P u_1(x) = \left<D_x^\alpha k_p(x,\cdot), u_1\right>_{\mathcal{E}(V), \mathcal{E}'(V)}$, and hence we deduce that the aplication $u_1 \mapsto Pu_1|_Z$ is linear and continous from $\mathcal{E}'(V)$ into $C^\infty(Z)$.

I do not know if it is possible to identify a distribution with compact support defined in $\mathcal{E}'(\Omega)$ with a distribution defined in $\mathcal{E}'(V)$, even if $\hbox{ supp } u_1 \subset V \subset \Omega$. Justify this it's important to the next step "$Pu_1$ is $C^\infty$ in $W$ ".

Next, we shall prove that $\hbox{ sing supp } Pu_1 \subset V$. To this end, let $U_{Pu_1}$ be the largest open set on which $Pu_1\in C^\infty(U_{P u_1})$. So, we have to prove that $\Omega \backslash U_{Pu_1} \subset V$, i.e., $\Omega\backslash V \subset U_{Pu_1}$. We will argue by absurd. Let's assume that $x_0\in \Omega\backslash V $ and that $x_0 \notin U_{Pu_1}$. If $x_0 \in \Omega$ and $x_0\notin V$, then there exists a neighbourhood $W$ of $x_0$ such that $W\cap V=\emptyset$. Then, $P u_1$ is also a $C^\infty$ function in $W$. Note that $W \subset \Omega\setminus V$ and by the above comments we have the desired. Thus, $x_0 \in U_{Pu_1}$, which is a absurd. Then, $\hbox{ sing supp } Pu_1 \subset V$ as we desired to prove. Since $y_0 \notin V$, then $y_0 \notin \hbox{ sing supp } Pu_1$. Hence, $y_0 \in U_{Pu_1}$. From this, we deduce that there exists a open set $\omega \subset \Omega$ such that $Pu_1|_{\omega} \in C^\infty(\Omega)$. So, $Pu|_\omega=Pu_1|_{\omega}+Pu_2|_{\omega} \in C^\infty(\omega)$, since $Pu_2 \in C^\infty(\omega)$. Hence, $y_0 \notin\hbox{ sing supp } Pu$. $\blacksquare$

  • $\begingroup$ but $u$ is function or distribution, or both $\endgroup$
    – user89940
    May 22 at 17:30


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