Let $\mathcal{S}:= \mathcal{S}(\mathbb{R}^n)$ be the Schwartz space of smooth functions with rapid decay. The question is pretty simply stated in the title. Pseudo-differential act continuously on the space $\mathcal{S}$, it is therefore natural to wonder whether they are the only ones:

Is there a continuous operator $T:\mathcal{S} \to \mathcal{S}$ which is not a pseudo-differential operator? i.e. not of the form: $$T(f)(x) = \int_{\mathbb{R}^n}a(x,\xi)\hat f(\xi)e^{ix\xi}d\xi$$

For some appropriate symbol $a(x,\xi)$ (of class $S^m$ for some $m$)

EDIT: For a more interesting follow up version of this question (which is not trivially false) see here.


1 Answer 1


No, the most obvious example is the reflection operator: $Rf(x) = f(-x)$ this is not pseudolocal (in fact the $\xi$-compotent of the wavefront set gets a sign flip). Also the Fourier transform.

More generally, every compactly supported Fourier integral operator with non-trivial Lagrangian (not the co-normal to the diagonal) is not a pseudodifferential operator, but preserves the Schwartz-space.

  • 1
    $\begingroup$ Sorry, I realize now how stupid that was... Thanks. $\endgroup$ Nov 3, 2017 at 17:29
  • $\begingroup$ OK but does pseudo-locality suffice? $\endgroup$ Nov 3, 2017 at 18:02
  • $\begingroup$ @მამუკაჯიბლაძე Yeah, that's what i wanted to ask too! $\endgroup$ Nov 3, 2017 at 19:09
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    $\begingroup$ @SaalHardali I asked a follow-up question. $\endgroup$ Nov 3, 2017 at 22:38

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