# Is every continuous endomorphism of the Schwartz space a pseudo-differential operator?

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

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.

• Sorry, I realize now how stupid that was... Thanks. – Saal Hardali Nov 3 '17 at 17:29
• OK but does pseudo-locality suffice? – მამუკა ჯიბლაძე Nov 3 '17 at 18:02
• @მამუკაჯიბლაძე Yeah, that's what i wanted to ask too! – Saal Hardali Nov 3 '17 at 19:09
• @SaalHardali I asked a follow-up question. – Joonas Ilmavirta Nov 3 '17 at 22:38