Confusion on weak operator continuity

Question

Let $M$, $N$ be von Neumann algebras and $\eta: M \mapsto N$ be linear and weak operator continuous on the unit ball. Is this enough to say that $\eta$ is weak operator continuos at every point? I think it is true, still, I am not sure.

Answer 1

No, because wot-convergence may happen via an unbounded net.

Let $M=B(H)$, with $H$ infinite-dimensional and separable, and $N=\mathbb C$. Fix an orthonormal basis $\{e_n\}$ and let $$ \eta(S)=\sum_n\frac1{n^2}\langle Se_n,e_n\rangle. $$ Then $\eta$ is $\sigma$-weak continuous, since $\eta(S)=\operatorname{Tr}(AS)$, where $A=\sum_n\frac1{n^2}\langle\cdot,e_n\rangle$ is trace-class. In particular, it is weak-operator continuous on balls.

If you now look at this answer, you get a net $\{T_F\}$ such that $T_F\to0$ in the weak operator topology, while $$ \eta(T_F)=\operatorname{Tr}(AT_F)=\frac{\operatorname{Tr}(AP_{F^\perp})}{\operatorname{Tr}(AP_{F^\perp})}=1 $$ for all $F$. So $\eta$ is not weak-operator-continuous.