Let $\mathcal S'=\mathcal S'(\mathbb R^n)$ be the Schwartz distribution space.
Suppose $A\colon\mathcal S'\to\mathcal S'$ is linear, continuous and microlocal.
By being microlocal I mean that the wave front sets satisfy $WF(Af)\subset WF(f)$ for all $f$.
(For another version, one could consider the singular supports instead of wave fronts, but I assume the answer wouldn't be different.)
Does it follow that $A$ is a pseudo-differential operator?


This is a variation on [this question about continuous endomorphisms on the Schwartz space](https://mathoverflow.net/q/285164/55893).
There one only assumed linearity and continuity, and the answer was negative.
The other question was about $\mathcal S$ instead of $\mathcal S'$, but I need distributions to allow singularities.


I realize that there are many classes of pseudo-differential operators.
The question is whether such a microlocal operator is always a ΨDO of some kind.
Any details on what kind would be very interesting, of course.



Is every continuous microlocal operator a pseudo-differential operator?
This is more like a longish series of comments somewhat complementing Ilya Zakharevich's answers rather than an answer by itself. First of all, notice that since $\mathscr{S}(\mathbb{R}^n)$ embeds continuously (and densely) into $\mathscr{S}'(\mathbb{R}^n)$, any continuous linear map $A:\mathscr{S}'(\mathbb{R}^n)\rightarrow\mathscr{S}'(\mathbb{R}^n)$ induces by restriction a unique continuous linear map $A|\_{\mathscr{S}(\mathbb{R}^n)}:\mathscr{S}(\mathbb{R}^n)\rightarrow\mathscr{S}'(\mathbb{R}^n)$, which we also denote by $A$ not to overburden our notation. By Schwartz's kernel theorem, we must have $$(A\phi\_1)(\phi\_2)=K\_A(\phi\_1\otimes\phi\_2)\ ,\quad\forall\phi\_1\ ,\,\phi\_2\in\mathscr{S}(\mathbb{R}^n)$$ for a unique $K\_A\in\mathscr{S}'(\mathbb{R}^{2n})$. 


One sees from the calculus of wave front sets that microlocality of $A$ $$\mathrm{WF}(Au)\subset\mathrm{WF}(u)\ ,\quad\forall u\in\mathscr{S}'(\mathbb{R}^n)$$ is entailed by the condition $$\mathrm{WF}(K\_A)\subset\{(x,x;\xi,-\xi)\ |\ (x;\xi)\in\mathbb{R}^{2n}\smallsetminus 0\}\ ,$$ which all pseudodifferential operators with (say) smooth symbols must satisfy by the stationary phase formula, but the converse (as far as I know) is not necessarily true. 


To have a better idea of what the would-be symbol of $A$ should look like, we have that $$\sigma\_A(x,\xi)=A(e^{-i\langle x-\cdot,\xi\rangle})(x)$$ defines for each $\xi\in\mathbb{R}^n$ a tempered distribution $\sigma\_A$ in $x$ which is smooth and polynomially bounded in $\xi$ together with all its $x$- and $\xi$-derivatives. By microlocality, $\sigma\_A$ should be smooth in $x$ as well. However, this alone does not guarantee that $\sigma\_A$ fits into any decent symbolic calculus since the polynomial bounds on its derivatives in $x$ and $\xi$ may vary too wildly.


One can say more if we know that $A$ is translation invariant in the sense that $$A(u(\cdot+x))=(Au)(\cdot+x)$$ for all $x\in\mathbb{R}^n$, $u\in\mathscr{S}(\mathbb{R}^n)$. Due to microlocality, this can only be the case if $A$ is a convolution operator, that is, there is a unique $F\_A\in\mathscr{S}'(\mathbb{R}^n)$ such that $$Au=F\_A\*u\ ,\quad\forall u\in\mathscr{S}'(\mathbb{R}^n)\ .$$ Equivalently, $K\_A$ is the convolution kernel associated to $F\_A$: $$K\_A(x,y)=F\_A(x-y)\ .$$ The Fourier transform $\sigma\_A$ of $F\_A$ is in this case the obvious candidate for the symbol of $A$ (in this case, $\sigma\_A$ no longer depends on $x$, of course). The image of $\mathscr{S}(\mathbb{R}^n)$ under $A$ consists of smooth functions with at most polynomial growth together with all their derivatives, but that is pretty much it - the convolution with *any* $F\in\mathscr{S}'(\mathbb{R}^n)$ will have the same property. Microlocality of $A$ follows if in addition $\mathrm{WF}(F\_A)\subset\{(0;\xi)\ |\ \xi\in\mathbb{R}^n\smallsetminus 0\}$, which is equivalent to the above condition on $\mathrm{WF}(K\_A)$ in the translation invariant case, but again this (as far as I know) is not a necessary condition.


If we could show that: 


1. $(1+\|\xi\|^2)^{-k}\sigma\_A\in L^\infty(\mathbb{R}^n)$ for some $k\in\mathbb{N}$, which amounts to saying that $(1-\Delta)^{-k}A$ is a bounded operator in $L^2(\mathbb{R}^n)$ by the Plancherel formula ($\Delta$ is the Laplacian in $\mathbb{R}^n$), and
2. The commutators of $(1-\Delta)^{-k}A$ with (multiplication by) polynomials are also bounded operators in $L^2(\mathbb{R}^n)$,


one could appeal to the Beals-Cordes commutator characterization of pseudodifferential operators of order zero since $A$ commutes with derivatives in the translation invariant case (notice that (2.) and the latter form of (1.) still make sense in the non-translation-invariant case). In the general (non-translation-invariant) case, we need as well to show that 


3. The commutators of $(1-\Delta)^{-k}A$ with derivatives are also bounded operators in $L^2(\mathbb{R}^n)$


in order to use the Beals-Cordes criterion, which precisely guarantees the kind of polynomial bounds on the derivatives of $\sigma\_A$ needed for it to be a symbol (of order $2k$). I have no idea if microlocality is enough to yield (1.)-(3.) (or even (1.)-(2.) in the translation-invariant case), but it seems a rather tall order to me.


