*I'm aware that the following question is at best a refined version of at least 2 questions which are already on this site. I think it is justified however in that it is more precise and has some new content in it. If, however, anyone decides after reading this question that it is a duplicate I apologize in advance*.

Fix a positive integer $n$. For every real number $m$ define the symbol class $S^m \subset C^{\infty}(\mathbb{R}^n_x \times \mathbb{R}^n_{\xi})$ consisting of functions whose derivatives are all bounded in the $x$-direction and which grows at most like $|\xi|^m$ in the $\xi$ direction (this is not a precise definition but hopefully for those who are familiar its clear what class of symbols i'm using).

Any such symbol (element of $S^m$ for $m \in \mathbb{R}$), defines a pseudo-differential operator which acts continuously on the space of Schwartz functions $\mathcal{S} := \mathcal{S}(\mathbb{R}^n$) and extends to a continuous endomorphism of tempered distributions $\mathcal{S}'$. Here are some facts about this construction:

The map from the symbols to endomorphisms of $\mathcal{S}$ is one to one. And so it endows the symbols with a non-commutative multiplication coming from composition (this multiplication can also be phrased without reference to $\mathcal{S}$ and is given by a certain combination of Fourier transform and convolution). Call this algebra $\Psi DO$.

The composition defined above respects the increasing filtration (by order) defined by setting $\Psi DO^{\le m}$ to be all operators that come from symbols in $S^{l}$ for $l \le m$.

For every real number $1 \lt p \lt \infty$ there's a decreasing filtration (the Sobolev filtration) indexed by the real numbers (say $s \in \mathbb{R}$) on $\mathcal{S}'$ where the $s$-filtered piece is the Sobolev space $W^{s,p} \subset \mathcal{S}'$ (it also has the nice property that $\bigcap_{s\in \mathbb{R}} W^{s,p} = \mathcal{S}$ and $\bigcup_{s\in \mathbb{R}} W^{s,p} = \mathcal{S}'$).

The order filtration on $\Psi DO$ respects the Sobolev filtrations on $\mathcal{S}'$ (for every $1 \lt p \lt \infty$). This is just the (rather non-trivial statement) that every $P \in \Psi DO^{\le m}$ gives a bounded linear operator $P : W^{s,p} \to W^{s-m,p}$ for all $p \in (0,\infty), s \in \mathbb{R}$.

All operators in $\Psi DO$ are pseudo-local, that is they do not increase the microsupport (or wave front sets) of distributions.

The subalgebra $\Psi DO^{- \infty} := \bigcap_m \Psi DO^{m}$ is a (filtered) two sided ideal and the quotient $\Psi DO / \Psi DO^{-\infty}$ is complete for the induced filtration.

My question is whether these properties characterize $\Psi DO$'s, more precisely:

Question:Let $\mathcal{A} \subset End(\mathcal{S}')$ be a subalgebra of continuous endomorphisms of the space of tempered distributions. For every $p \in (0,\infty)$ the Sobolev filtration on $\mathcal{S}'$ induces an increasing filtration on $\mathcal{A}$ by setting $\mathcal{A}^{\le m, p} := \{ P \in \mathcal{A} | P: W^{s,p} \to W^{s-m,p} , \forall s \in \mathbb{R}\}$. Suppose $\mathcal{A}$ satisfies the following 4 properties inspired from the above discussion:

- (
Pseudo-locality) All operators in $\mathcal{A}$ are pseudolocal (microsupport non-increasing).- (
Exhaustion & Strictness) For every $p \in (0,\infty)$ the induced filtration is strictly increasing, i.e. $\mathcal{A}^{\le m,p} \subsetneq \mathcal{A}^{\le l,p}$ whenever $m \lt l$, and exaustive, i.e. $\mathcal{A}^{\infty}:= \bigcup_m \mathcal{A}^{\le m ,p} = \mathcal{A}$- (
Constancy in $p$) For all $1 \lt p \lt q \lt \infty$ the induced filtrations agree. In other words $\mathcal{A}^{\le m,p} = \mathcal{A}^{\le m, q}$ for all $m \in \mathbb{R}$.- (
Completeness) The quotient $\mathcal{A}/\mathcal{A}^{- \infty}$ is complete for the induced filtration.Is it true that $\mathcal{A} = \Psi DO$ ? If not perhaps its true if we require that $\mathcal{A}$ be the smallest subalgebra satisfying the above conditions?