# Characterizing pseudo-differential operators as a subalgebra of continuous endomorphisms of tempered distributions

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:

1. (Pseudo-locality) All operators in $$\mathcal{A}$$ are pseudolocal (microsupport non-increasing).
2. (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. $$\mathcal{A}^{\infty}:= \bigcup_m \mathcal{A}^{\le m ,p} = \mathcal{A}$$
3. (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 $$m \in \mathbb{R}$$.
4. (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?

• How to you take $\rho,\delta$-classes into account? And something like SG or Shubin (there are some results about $L^p$ as far as I remember)?
– mcd
Jul 6, 2018 at 1:17
• @mcd In my symbol classes $\delta$ is always $0$, I got the impression that for arbitrary $\delta$ the completeness condition In my question may fail. In any case if i'm wrong about that I was hoping that perhaps the extra minimality comdition will take care of this. Jul 6, 2018 at 6:16
• @mcd The constancy in $p$ condition (3) might be a problem with nonzero $\delta$ as well. As for "SG" or "Shubin" i'm not familiar with these terms Jul 6, 2018 at 6:42
• There is a book by Nicola and Rodino about global pseudodifferential operators (they use other names for the calculi: SG=G and Shubin=$\Gamma$), My point was that different $\rho,\delta$ give you different calculi which all might fulfill your conditions.
– mcd
Jul 6, 2018 at 7:19

Let me try: Consider $S^m_{1,0}$ and $SG^{m,0}$, where the second class of SG-symbols $SG^{m_\psi,m_e}$ is defined by the estimates $$|\partial_x^\alpha \partial_\xi^\beta a(x,\xi)| \lesssim_{\alpha,\beta} \langle x\rangle^{m_e-|\alpha|} \langle \xi\rangle^{m_\psi-|\beta|}.$$ Clearly, $SG^{m,0}$ is a subset of $S^m_{1,0}$, therefore $L^p$-boundedness follows from the $L^p$-boundedness of Kohn-Nirenberg pseudos.

I haven't thought about the completeness, but I don't see a big difference between Kohn-Nirenberg and SG there.

• It seems like this could work. Although i'm not so familiar with SG symbols. I should brush up on that first (regarding the completness as you say). Jul 11, 2018 at 8:07

I do not believe that the listed properties characterize the operators with symbols in $\cup_m S^m_{1,0}$. Let me first make precise the definition of $S^m_{1,0}$ for a given $m\in \mathbb R$: this is the Fréchet space of $C^\infty$ complex-valued functions on $\mathbb R^n_x \times\mathbb R^n_\xi$ such that for all $\alpha, \beta\in \mathbb N^n$, $$\sup_{ (x,\xi)\in \mathbb R^{2n}} \bigl\vert(\partial_x^\alpha\partial_\xi^\beta a)(x,\xi) \bigr\vert(1+\vert \xi\vert)^{-m+\vert \beta\vert}<+\infty.$$ Of course the four listed properties hold true for the algebra $\mathcal A$ of operators on $\mathscr S'(\mathbb R^n)$ with symbol $\in \cup_m S^m_{1,0}$.

Consider now the Fréchet space $\Sigma^m$ of $C^\infty$ complex-valued functions on $\mathbb R^n_x \times\mathbb R^n_\xi$ such that for all $\alpha, \beta\in \mathbb N^n$, $$\sup_{ (x,\xi)\in \mathbb R^{2n}} \bigl\vert(\partial_x^\alpha\partial_\xi^\beta a)(x,\xi) \bigr\vert(1+\vert \xi\vert+\vert x\vert)^{-2m+\vert \alpha\vert+\vert \beta\vert}<+\infty.$$ Then the algebra $\mathcal B$ operators on $\mathscr S'(\mathbb R^n)$ with symbol $\in \cup_m \Sigma^m$ satisfies the four required properties.

• Could you elaborate on why the second algebra you give satisfies 3? it doesn't seem obvious but maybe im missing something. Jul 9, 2018 at 12:16
• @Saal Hardali Considering the harmonic Oscillator $\mathcal H=-\Delta_x+\vert x\vert^2$, you can define the scale of Hilbert spaces $\mathscr H^m$ as the temperate distributions $u$ such that $\mathcal H^m u\in L^2$ and you can prove that $\mathscr H^m$ is also the set of $u$ such that $(\text{Op}a) u$ belongs to $L^2$ for any $a\in \Sigma^m.$ Another point is to prove that $\text{Op}\Sigma^0$ is included in the bounded operators on $L^p$ for $1<p<+\infty$, but it is a consequence of the same result for $S^0$. Jul 9, 2018 at 17:26
• hmmm, i'm not sure I understand your point about the boundness, could you say a few words about the boundness? This is the part i'm least sure about. Also the completeness isn't at all obvious to me for the second class of symbols you propose. Jul 9, 2018 at 17:49
• @Saal Hardali You mean certainly the $L^p$ boundedness. It is a delicate matter since some classes of pseudo-differential operators are bounded on $L^2$ and not on $L^p$. Here it is simpler since $\Sigma^m\subset S^{2m}_{1,0}$ and an operator with symbol in the larger set is indeed bounded from $W^{s,p}$ into $W^{s-2m,p}$. Maybe I was too quick for the completeness, but I do not see why it would be different, only the Sobolev filtration is different with the definition of the spaces $\mathscr H^m$ in my answer. Jul 9, 2018 at 19:16
• @Bazin for $m > 0$ the inclusion is not true, since you have "global" $S_{1,0}^m$ estimates; but the Shubin symbol are allowed to grow in $x$, for example the harmonic oscillator is not in $S^2_{1,0}$, but obviously in $\Sigma^1$.
– mcd
Jul 10, 2018 at 8:56