Closure of all operator of order $0$ in $B(L^2(R^n))$

Question

The following question arose as I was playing around a little bit with pseudo-differential operators and K-theory and so on.

Let $H^s$ be the Sobolev space of s-times weakly differentiable functions $f \in L^2(R^n)$ with the usual inner product $\langle \cdot, \cdot \rangle_s$. Note that $H^0 = L^2(R^n)$.

Let $B_0 \subset B(L^2(R^n))$ be the subset of all bounded operators $H^0 \to H^0$ which restrict to a bounded operator $H^s \to H^s$ for every s. Let them be called operator of order 0.

How does the closure of $B_0$ in $B(L^2(R^n))$ look like?

I couldn't find any reference in the internet and as I tried myself I got stuck, since I don't know how to approach this question.

$B_0$ contains surely all multiplication operator corresponding to smooth functions $f \in C_c^\infty(R^n)$ which are compactly supported. So the closure of $B_0$ contains the whole $C_0(R^n)$. Furthermore, all compact operator are in $\bar B_0$.

And it of course contains $\Psi_0$ - the pseudo-differential operator of order 0. But I remember somewhere reading, that the closure of $\Psi_0$ w.r.t. the L^2-L^2-operator norm varies, depending on the concrete definition of pseudo-differential operator you use.

May it be that we have $\bar B_0 = B(L^2(R^n))$?

Thanks in advance.

Answer 1

There is a paper on related subjects:

F. Mantlik, Norm closure of operator algebras with symbolic structure. Math. Nachr. 201, 91-116 (1999).

I could read only its Zentralblatt review: http://www.zentralblatt-math.org/zmath/en/advanced/?q=an:0947.47054&format=complete

creating an impression that the answer is more complicated.

Answer 2

I think I figured out the answer for the last question on myself. The closure of $B_0$ can't be the whole set of all bounded operator.

This is due to the fact that the set $\Psi_{-\infty}$ of all smoothing operator is a two-sided ideal in $B_0$. After passing to the closures, $\bar \Psi_{-\infty} \subset \bar B_0$ is still a two-sided ideal.

The closure of all smoothing operator is strictly larger than the set of all compact operator, so $\bar B_0$ can't be the algebra of all bounded operator, since this has only one two-sided ideal (the compact operator).