0
$\begingroup$

In a von Neumann algebra, if $A_{\alpha}$ converges to $0$ in the $\sigma$-weak topology, do the positive parts $(A_{\alpha})_{+}$ necessarily converge to $0$ in the $\sigma$-weak topology?

$\endgroup$

1 Answer 1

3
$\begingroup$

Nope, not even in the abelian case. Work in $L^\infty[0,1]$. Let $f_n$ be the function which is alternately plus and minus $1$ on the subintervals $[\frac{i}{2^n}, \frac{i+1}{2^n}]$. Then $f_n \to 0$ weak* but the positive parts converge to $1/2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.