3
$\begingroup$

[Question previously asked on Math.SE]

Let $N$ be a von Neumann algebra, and $A$ be a dense $∗$-subalgebra of $N$ (in the ultraweak topology) with $A''=N$. Is it true that:

For any $x∈N^+$, there exists a increasing net $(x_j)$ in $A^+$ such that $x_j \to x$ in the ultraweak topology ?

The case of $A$ being an ideal of $N$ seems known (it is right?) but my question is about the general case.

(Precision: $A$ is not supposed to be normed closed.)

Improve this question
$\endgroup$
7
  • 2
    $\begingroup$ Have you considered the case when $N$ is abelian? E.g., $N = L^\infty[0,1]$, and $A = C[0,1]$. $\endgroup$
    – Jesse Peterson
    Commented Jul 26, 2012 at 23:53
  • $\begingroup$ 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? $\endgroup$
    – Jon Bannon
    Commented Jul 27, 2012 at 1:47
  • 6
    $\begingroup$ 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. $\endgroup$
    – Steven Deprez
    Commented Jul 27, 2012 at 9:36
  • 1
    $\begingroup$ @Jon: $f_t$ need not be continuous if $f$ isn't. $\endgroup$
    – Jesse Peterson
    Commented Jul 27, 2012 at 16:03
  • 1
    $\begingroup$ 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 $\ell^\infty([0,1]\cap\mathbb{Q})$. $\endgroup$
    – Tristan Bice
    Commented Jan 3, 2015 at 13:10

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .