# Convergence of self-adjoint elements in $\sigma$-weak topology

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?

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 $$1/2$$.