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Let $M$ be a closed manifold, and let $P:C^{\infty}(M)\rightarrow C^{\infty}(M)$ be a continuous linear map in the smooth Fréchet topologies. In what is to come, if it helps, one can assume further that $P:H^{s}(M)\rightarrow H^{s+m}(M)$ continously for all $s\in\mathbb{R}$ for some $m>0$.

Suppose that there are pseudodifferential operators $E,A: C^{\infty}(M)\rightarrow C^{\infty}(M)$, with $E$ elliptic, such that, $$ EP=A. $$ Can one determine that $P$ must also be a pseudodifferential operator? It seems that this cannot be observed directly, because if one composes both sides by a parametrix $E^{-1}$ for $E$, it holds that $$ P+KP=E^{-1}A $$ where $E^{-1}E=I+K$, $K$ being a smoothing integral operator. $KP$ seems to be smoothing but I find myself unable to explain why it is within the calculus.

Another approach I was thinking about is to use the Schwartz kernel theorem to apriori construct the symbol of $P$, and show by the composition above that it must be in the right class, but I got a bit lost in the details.

Any help or reference on this will be most appreciated.

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  • $\begingroup$ Isn’t the composition of a smoothing operator with any operator of finite order smoothing because the output is always smooth? $\endgroup$
    – Deane Yang
    Commented Feb 6, 2023 at 17:28
  • $\begingroup$ By finite order you mean its range is finite dimensional? But $P$ has no finite dimensional range $\endgroup$
    – MyShepherd
    Commented Feb 6, 2023 at 17:53
  • $\begingroup$ No. Order of differentiation. The $m$ in your question. $\endgroup$
    – Deane Yang
    Commented Feb 6, 2023 at 19:41
  • $\begingroup$ I don't know that $P:\mathcal{D}'(M)\rightarrow \mathcal{D}'(M)$. Even if I did, does the fact that $KP:\mathcal{D}'(M)\rightarrow C^{\infty}(M)$ alone would imply that $KP$ is in the calculus? $\endgroup$
    – MyShepherd
    Commented Feb 6, 2023 at 22:30
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    $\begingroup$ I believe that the Schwartz kernel theorem tells you the operator can be written as an integral operator where the kernel is a distribution. If the operator is smoothing the kernel is smooth. Such an operator fits the definition of a pseudodifferential operator. But you shouldn’t take my word for it. You should work out the details yourself. $\endgroup$
    – Deane Yang
    Commented Feb 7, 2023 at 17:39

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