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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.

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  • $\begingroup$ 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. $\endgroup$
    – Bombyx mori
    Commented Nov 11, 2019 at 22:44
  • $\begingroup$ 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 $||\phi Pu||_{s-m}\leq C_{\phi,s}||\psi u ||_s$. $\endgroup$
    – Math
    Commented Nov 11, 2019 at 22:52
  • $\begingroup$ 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? $\endgroup$
    – Math
    Commented Nov 11, 2019 at 22:56
  • $\begingroup$ So using a density argument and the fact that the spaces involved are Fréchet, the result seems to follow. $\endgroup$
    – Math
    Commented Nov 11, 2019 at 23:25

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