Approximating a positive element in a von Neumann algebra by positive elements in a weakly dense C*-subalgebra

Question

The noncommutative Lusin Theorem (as stated in "Non-commutative extension of Lusin's theorem", reiterated in Theory of Operator Algebras I) provides a number of ways to approximate elements of a von Neumann algebra with elements of a particular weakly dense C-subalgebra. The general idea is that we have some normal positive functional $\psi$ on a von Neumann algebra $M$, and $A$ is a weakly dense concrete C-subalgebra of $M$. We have some element $a \in M$, and some projection $e \in M$. We want to find some projection $e_0 \leq e$ with $\psi(e - e_0)$ small and some $a_0 \in A$ such that $a_0 e_0 = a e_0$ and $\| a_0 \|$ has some estimate on it. Moreover, the theorem posits that if $a$ meets certain other assumptions (e.g. self-adjoint or unitary), then we can also demand the same thing of $a_0$, though this might alter our estimate on $\| a_0 \|$.

My question is this: If $a$ is positive in $M$, then can we choose this approximation $a_0$ to also be positive?

The answer in the "commutative" setting is yes, and a pretty prompt one at that. If I have some $f \in L^\infty(X, \mu)$, where $(X, \mu)$ is a finite Radon space enjoying perhaps some additional regularity conditions, then Lusin's Theorem tells me that I can find some measurable set $A \subseteq X$ such that $\mu(X \setminus A)$ is arbitrarily small, and such that there exists $g \in C(X)$ for which $f \vert_A = g \vert_A$. But if I know beforehand that $f$ is nonnegative (i.e., it takes nonnegative values almost everywhere), then I can just replace $g$ with $g^+$ (i.e. the positive part of $g$) and my $g^+$ is no worse an approximation of $f$ than $g$, since $|f(x) - g^+(x)| \leq |f(x) - g(x)|$, due to $f$ being nonnegative.

I want to able to say something here in the noncommutative case: If $a \in M$ is positive, then once I have my $a_0$, I can replace it with $a_0^+$ and get no worse of an approximation to $a$. Can I do this? Perhaps more importantly, how would I go about showing this? I assume it's a pretty elementary computation, but I've been banging my head against it and just don't see how it's supposed to work.

Thanks!

Answer 1

You can always use the commutative setting. If $a\geq0$, then $W^*(a)$ is a commutative von Neumann algebra with $a\in W^*(a)\subset M$. Now you can find $a_0\in W^*(a)$ as in your argument.

Answer 2

Here is an another answer, which I believe answers an even more general question.

Theorem: If $M\subseteq B(H)$ is a von Neumann algebra and let $A \subseteq B(H)$ be a $\sigma$-weakly dense $C^*$-subalgebra (or equivalently, weakly dense $C^*$-subalgebra) of $M$. Then if $x\in M_+$, there exists a bounded net of positive elements $\{a_\lambda\}_{\lambda \in \Lambda}\subseteq A$ such that $a_\lambda \to x$ in the strong topology.

Proof: Let $x \in M_+$. By Kaplansky's density theorem, there exists a bounded net $\{b_\lambda\}_{\lambda \in \Lambda}\subseteq A$ such that $b_\lambda \to x$ in the $\sigma$-strong topology. By replacing $b_\lambda$ by $2^{-1}(b_\lambda + b_\lambda^*)$, we may assume that the net consists of self-adjoint elements. But then, consider the functional calculus $$f^+: \mathscr{L}_{\mathbb{R}} = \{x \in B(H): x=x^*\} \to B(H): x \mapsto f^+(x)$$ where we write $f^+ = \max\{f,0\}$. It is strongly continuous (see for instance theorem II.4.7 in Takesaki's book), and thus we deduce that $$b_\lambda^+ = f^+(b_\lambda) \to f^+(x) = x^+ = x$$ strongly, so $\{a_\lambda:= b_\lambda^+\}_{\lambda \in \Lambda}$ is the desired net.