TITLE: What is the smallest $C^*$-algebra containing the "standard" pseudodifferential operators?
QUESTION [8 upvotes]: 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 $\Psi^0$?

REPLY [4 votes]: 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.