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?



Convergence of self-adjoint elements in $\sigma$-weak topology
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 $1/2$.


