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?

  • 1
    $\begingroup$ 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)? $\endgroup$
    – mcd
    Jul 6, 2018 at 1:17
  • $\begingroup$ @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. $\endgroup$ Jul 6, 2018 at 6:16
  • $\begingroup$ @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 $\endgroup$ Jul 6, 2018 at 6:42
  • 2
    $\begingroup$ 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. $\endgroup$
    – mcd
    Jul 6, 2018 at 7:19

2 Answers 2


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.

  • $\begingroup$ 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). $\endgroup$ 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.

  • $\begingroup$ Could you elaborate on why the second algebra you give satisfies 3? it doesn't seem obvious but maybe im missing something. $\endgroup$ Jul 9, 2018 at 12:16
  • $\begingroup$ @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 $S^0$. $\endgroup$
    – Bazin
    Jul 9, 2018 at 17:26
  • $\begingroup$ 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. $\endgroup$ Jul 9, 2018 at 17:49
  • $\begingroup$ @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. $\endgroup$
    – Bazin
    Jul 9, 2018 at 19:16
  • $\begingroup$ @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 $\Sigma^1$. $\endgroup$
    – mcd
    Jul 10, 2018 at 8:56

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