# Extension of a compactly supported pseudo-differential operator

Let $$\Omega$$ be a open subset of $$\mathbb{R}^{d}$$ and $$P \in \Psi^{m}(\Omega)$$ a compactly supported pseudo-differential operator, that is, the kernel of $$P$$, is compact. Is it true that $$P$$ extends continuously from $$H_{loc}^{s}(\Omega)$$ into $$H^{s-m}_{comp}(\Omega).$$

I can show that $$P:H_{loc}^{s}(\Omega) \longrightarrow H_{loc}^{s-m}(\Omega)$$ and $$P=\varphi P$$, where $$\varphi \in C_0^{\infty}(\Omega)$$ and $$\varphi=1$$ in a neighborhood of $$K'=\pi_x(\operatorname{supp}k_P)$$, where $$k_P$$ denotes the (Schwartz) kernel of the operator $$P$$.

This is a strong indication that the statement is true.

• I am not entirely following. I thought a compactly supported PsiDO is the one whose action on functions that support outside a compact set vanishes. The kernel of P may not be finite dimensional and may not be compact. – Bombyx mori Nov 11 '19 at 22:44
• I am following the definition of Petersen's book. Introduction To The Fourier Transform And Pseudo-differential Operators. He gives this definition of a compactly supported operator on page 250 and states the existence of the function $\phi$ verifying $\varphi P=P$. Perhaps the result of the Folland book may help: For every $\phi \in C_0^\infty(\Omega)$ there is a $\psi \in C_0^\infty(\Omega)$ such that $||\phi Pu||_{s-m}\leq C_{\phi,s}||\psi u ||_s$. – Victor Hugo Nov 11 '19 at 22:52
• Taking $\phi=\varphi$ we obtain $||Pu||_{s-m}=||\phi Pu||_{s-m}\leq C_{\phi,s}||\psi u ||_s$, for all $u \in C_0^\infty(\Omega)$ and some $\psi \in C_0^\infty(\Omega)$. Is this enough to guarantee extension? – Victor Hugo Nov 11 '19 at 22:56
• So using a density argument and the fact that the spaces involved are Fréchet, the result seems to follow. – Victor Hugo Nov 11 '19 at 23:25