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?
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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.



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


