Let $f: M\to N$ be a bounded linear map between von Neumann algebras.
Assume that for every net $0 \le x_i\nearrow x$, we have that $f(x_i)\to f(x)$ $\sigma$-strongly. Is it true that $f$ is normal, i.e. $\sigma$-weakly continuous?
Let $\omega \in N_*$. Then we need to check that $\omega \circ f\in M_*$.
Now, if $f$ would be $*$-homomorphism, I think I can prove this: we can use that $\eta \in M_*$ iff $$\eta(\sum_{i\in I} e_i) = \sum_{i\in I} \eta(e_i)$$ for all families of orthogonal projections $\{e_i\}_i\subseteq M$.
So, let $\{e_i\}_i$ be such a family. Then $$(\omega \circ f)(\sum_{i\in I}e_i)= \omega(\sum_{i\in I} f(e_i))= \sum_{i\in I} \omega(f(e_i))$$ where the first equality uses the assumption and the second equality uses that $\{f(e_i)\}_i$ is again a family of orthogonal projections.
My questions: (1) Do I need the assumption that $f$ is a $*$-homomorphism, or is there another argument that avoids this?
(2) Is the converse of the above true? I.e. if $f$ is normal, do we have that $f(x_i)\to f(x)$ $\sigma$-strongly if $0 \le x_i\nearrow x$?
Thanks in advance for your help!
As requested, here are my comments promoted to an answer:
You should go through your argument again - you never need to use $f$ is a $\ast$-homomorphism in the first place if, for the second inequality, instead of using $f(e_i)$ is a family of orthogonal projections, you just use $\omega$ is $\sigma$-weakly and therefore also $\sigma$-strongly continuous.
As for the converse, it is definitely true when $f$ is positive, since an increasing net of positive element converges $\sigma$-weakly to $x$ iff it converges $\sigma$-strongly to $x$. Not sure whether this is generally true for bounded linear maps though.