Timeline for Dense subalgebras of von Neumann algebras and increasing nets
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 3, 2015 at 13:10 | comment | added | Tristan Bice | Even if you suppose $A$ to be norm closed you still have counterexamples. E.g. consider $C[0,1]$ as a weakly dense C*-subalgebra of $\ell^\infty[0,1]$ rather than $L^\infty[0,1]$. Then the characteristic function of any point in $[0,1]$ certainly can not be approximated from below by positive continuous functions. If you also want the von Neumann algebra to be countably generated like $L^\infty[0,1]$, then replace $\ell^\infty[0,1]$ with $\ell^\infty([0,1]\cap\mathbb{Q})$. | |
| Jul 28, 2012 at 1:44 | comment | added | Jon Bannon | @Jesse: I wasn't clear. I wanted "$A$ strictly smaller" to indicate that there would be trouble looking at proper *-subalgebras of $C[0,1]$, not that $A$ should be an arbitrary $*$-subalgebra of $L^{\infty}$. So, in that case, $f$ is cts. (Steven's counterexample did the trick already so my comment is irrelevant!) | |
| Jul 27, 2012 at 16:03 | comment | added | Jesse Peterson | @Jon: $f_t$ need not be continuous if $f$ isn't. | |
| Jul 27, 2012 at 10:15 | comment | added | Jon Bannon | Steven's got it. Whatever you're trying to do, Michael, you're going to need some extra conditions on your dense *-subalgebra! | |
| Jul 27, 2012 at 9:36 | comment | added | Steven Deprez | Michael specifies explicitly that the subalgebra $A$ does not have to be norm-closed. In that case, an easy counter-example is the following: Consider the algebra of polynomials $A$ in $N=L^{\infty}([-1,1])$. Consider $f=1_{[0,1]}$ the characteristic function of half this interval. The constant function $0$ is the only positive function in $A$ that is also below $f$. Hence you can not approximate $f$ from below by positive polynomials. | |
| Jul 27, 2012 at 1:47 | comment | added | Jon Bannon | If $A=C[0,1]$ as in Jesse's example above, for positive $t$ the functions $f_{t}(x)=min\{f(x),t\}$ will give us an increasing net converging to $f$. The trouble is that when $A$ is strictly smaller than $C[0,1]$ we can't know if such functions are to be found in $A$. Isn't this your problem, Michael? | |
| Jul 26, 2012 at 23:53 | comment | added | Jesse Peterson | Have you considered the case when $N$ is abelian? E.g., $N = L^\infty[0,1]$, and $A = C[0,1]$. | |
| Jul 26, 2012 at 21:08 | history | asked | Michael | CC BY-SA 3.0 |