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If the symbol $p(x,\xi)$ of a pseudodifferential operator $P$ has compact $x$-support, then for any Schwartz function $f$, $Pf$ has compact $x$-support.

Is the reverse true? Namely that if some PDO $P$ with a symbol $p(x,\xi)$ from some reasonable symbol class has a property that $Pf$ has compact $x$-support for any Schwartz $f$, then does it imply that $p(x,\xi)$ has compact $x$-support?

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Yes. Denote by $S_K$, $K$ compact, the closed subspace of Schwartz space $S$ consisting of all $f$ such that the support of $Pf$ is contained in $K$. The assumption and the Baire category theorem imply that $S=S_K$ for some $K$. It follows that the $x$-support of the Schwartz kernel $k(x,y)$ of $P$ is contained in $K$, hence also the $x$-support of the symbol $p(x,\xi)=\int e^{i\xi y} k(x,x-y)\,\mathrm dy$.

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  • $\begingroup$ Hi, can you indicate in a few words how Baire category is used? Thanks! $\endgroup$
    – Willie Wong
    Commented Oct 1, 2013 at 8:45
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    $\begingroup$ Let $(K_j)$ be an exhausting sequence of compact subsets of euclidean space. Using the assumption every $f\in S$ belongs to some $S_{K_j}$. By the Baire category theorem, for some $j$ the closure of $S_{K_j}$ has an interior point, so $S_{K_j}$ itself has an interior point, because it is closed. A linear subspace containing an interior point also contains a neighbourhood (wrt the surrounding space) of the origin. Therefore, $S_{K_j}=S$. $\endgroup$
    – user80744
    Commented Oct 1, 2013 at 10:20

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