Is $\Psi^0(\mathbb{R})$ (pseudodifferential operators with symbols obeying $ |\partial^\alpha_x \partial^\beta_\xi a(x,\xi)| \leq C_{\alpha,\beta} (1+|\xi|)^{-|\beta|} $ ) a $C^*$-algebra?

In other words, is $\Psi^0(\mathbb{R})$ is closed in $\mathcal{L}(L^2(\mathbb{R}))$ in the operator norm topology?

If not, then is there any nice characterization by the $C^*$-algebra generated by $\Psi^0$? Alternatively, what is the strongest (or just a reasonable) topology on $\mathcal{L}(L^2(\mathbb{R}))$ such that $\Psi^0$ is a closed subspace?

Edit: Per Yemon Choi's comments below, the above question seems somewhat hopeless. As described here, $\Psi^0(\mathbb{R})$ is a Fréchet $*$-algebra with a topology stronger than the operator topology. I assume that this is the topology given by the seminorms on symbols: $$ \Vert a \Vert_{\alpha,\beta} = \sup_{x,\xi \in \mathbb{R}} (1+|\xi|)^{|\beta|} |\partial^\alpha_x \partial^\beta_\xi a(x,\xi)|. $$

So, in addition to the above question, I am adding the following question, to make it so that there might be an answer:

Is there a reasonable description of the smallest $C^*$-algebra containing $\Psi^0$?

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    $\begingroup$ My extremely hazy recollection is that $\Psi^0$ is not norm-closed in ${\mathcal L}(L^2)$, and that there is no reasonable topology to put on the latter to make $\Psi^0$ closed therein. Very roughly, think of Schwarz functions inside $L^2(R)$ - the former are naturally a Frechet space, the latter is naturally a Banach space, and as far as I know no good has come of trying to change the topology on either to make it fit better with the other. $\endgroup$
    – Yemon Choi
    Aug 30, 2010 at 3:23
  • $\begingroup$ I'm aware that my comment isn't helpful in terms of answering the original question; I'll try to come back later with either references or an actual calculation $\endgroup$
    – Yemon Choi
    Aug 30, 2010 at 3:25
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    $\begingroup$ From MR90d:47016 (Gramsch, Kaballo, IEOT 1989) "Some of the main tools used by the authors are the so-called $\Psi$- and $\Psi^*$-algebras, for which an appropriate analytic perturbation theory holds (they are Fréchet algebras with an open group of invertible elements and continuous inversion). We mention that many algebras of pseudodifferential operators are $\Psi$-algebras." $\endgroup$
    – Yemon Choi
    Aug 30, 2010 at 3:29
  • $\begingroup$ Thanks! I've edited the question to take your remarks into account. $\endgroup$ Aug 30, 2010 at 4:04

2 Answers 2


I have to confess to being more confused by the theory of pseudodifferential operators than I should be, but I think an answer to a question at least related to yours is briefly discussed in chapter 2 of Higson and Roe's Analytic K-homology.

Begin with an open subset $U$ of $\mathbb{R}^n$ and consider a smooth complex valued function $\sigma$ on $T^*(U)$ with the following properties:

  • $\sigma(x, t \xi) = \sigma(x, \xi)$ for $t \geq 1$, $|\xi| \geq 1$
  • $\sigma(x, \xi)$ vanishes for $x$ outside of some compact subset of $U$

Define the pseudodifferential operator associated to $\sigma$ to be the operator

$D_\sigma f(x) = \frac{1}{(2 \pi)^n} \int \sigma(x, \xi) \hat{f}(\xi) e^{i(x,\xi)} d\xi$

The first condition on $\sigma$ above implies that $\sigma$ gives rise to a function on the cosphere bundle $S^*M$ (the symbol of $D_\sigma$). $D_\sigma$ extends to a bounded operator on $L^2(U)$, so consider the C*-algebra $\mathcal{B}(U)$ generated by the $D_\sigma$. Higson and Roe assert that the map sending $D_\sigma$ to its symbol extends to a surjective $*$-homomorphism from $\mathcal{B}(U)$ to $C_0(S^* U)$ whose kernel is precisely the C* algebra of compact operators on $L^2(U)$.

Thus a certain class of pseudodifferential operators generates a C* algebra extension of $C_0(U)$ by the compact operators. I make no claim about the relationship between $\mathcal{B}(U)$ and your $\Psi^0$, but maybe this statement is still useful to you.

  • $\begingroup$ Thanks! This is very neat! Btw, I think that the $*$-algebra of the $D_\sigma$ is (up to compact operators) called "polyhomogeneous pseudodifferential operators" (at least by Melrose in one of his notes somewhere). $\endgroup$ Aug 31, 2010 at 1:52

A remark to add to Paul's answer: yes $\mathcal{B}(U)$ is precisely the $C^*$-algebra generated by $\Psi^0$, as follows.

Let $\mathcal{A}$ be the $C^\ast$-closure of $\Psi^0$. The image of the principal symbol map $\mathrm{Symb}\colon \Psi^0 \to C_0(S^\ast U)$ is dense, and so the image of $\mathcal{A}$ is all of $C_0(S^\ast U)$. Thus, for any $T\in\mathcal{B}(U)$, we can find $T^\prime \in \mathcal{A}$ with the same symbol. Following Paul, $T-T^\prime$ is a compact operator. The compacts are in $\mathcal{A}$ (since the smoothing operators are dense therein). Thus, $T$ is in $\mathcal{A}$.

edit (AlexE): To phrase this result in another way, we have the so-called pseudodifferential operator extension (a short exact sequence) $$0 \to \mathcal{K} \to \mathcal{A} \stackrel{\mathrm{Symb}}\longrightarrow C_0(S^\ast U) \to 0,$$ where $\mathcal{K}$ are the compact operators.


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