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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.

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.

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 $m$)

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.

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: I understand now the initial phrasing of the question was naive. Let me correct: What if one asks aditionally that the operator $T$ be pseudo-local (i.e. s.t. its extension to an endomorphism of the dual fixes the wave front set of all Schwartz distributions). To summarize: Is there a pseudo-local continuous endomorphism of $\mathcal{S}$ which is not a pseudodifferential 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 $m$)